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https://math.stackexchange.com/questions/68653/on-the-gcd-of-a-pair-of-fermat-numbers | # On the GCD of a Pair of Fermat Numbers
I've been working with the Fermat numbers recently but this problem has really tripped me up. If the Fermat theorem is set as $f_a=2^{2^a}+1$, then how can we say that for an integer $b$ less than $a$ that $\gcd(f_b,f_a)=1$?
• By "Fermat theorem", do you mean "Fermat number"? – Arturo Magidin Sep 30 '11 at 2:33
Hint: Show that $f_b$ divides $f_a-2$.
• Simpler than mine! I'm tempted to delete mine and just leave yours... – Arturo Magidin Sep 30 '11 at 2:47
• It's more "less detailed" than "simpler", I think. The question looked like it might be homework. – Henning Makholm Sep 30 '11 at 2:59
• Perhaps: but you don't need to establish the precise formula like I did, but rather just use a "difference of $2^b$th powers" factorization, which is simpler. – Arturo Magidin Sep 30 '11 at 3:05
• Assuming that $b<a$, of course... – Thomas Andrews Jan 23 '15 at 14:38
• @ThomasAndrews: Yes, but that's stipulated in the question. – Henning Makholm Jan 23 '15 at 14:40
Claim. $f_n=f_0\cdots f_{n-1}+2$.
The result holds for $f_1$: $f_0=2^{2^0}+1 = 2^1+1 = 3$, $f_1=2^{2}+1 = 5 = 3+2$.
Assume the result holds for $f_n$. Then \begin{align*} f_{n+1} &= 2^{2^{n+1}}+1\\ &= (2^{2^n})^2 + 1\\ &= (f_n-1)^2 +1\\ &= f_n^2 - 2f_n +2\\ &= f_n(f_0\cdots f_{n-1} + 2) -2f_n + 2\\ &= f_0\cdots f_{n-1}f_n + 2f_n - 2f_n + 2\\ &= f_0\cdots f_n + 2, \end{align*} which proves the formula by induction. $\Box$
Now, let $d$ be a common factor of $f_b$ and $f_a$. Then $d$ divides $f_0\cdots f_{a-1}$ (because it's a multiple of $f_b$) and divides $f_a$. That means that it divides $$f_a - f_0\cdots f_{a-1} = (f_0\cdots f_{a-1}+2) - f_0\cdots f_{a-1} = 2;$$ but $f_a$ and $f_b$ are odd, so $d$ is an odd divisor of $2$. Therefore, $d=\pm 1$. So $\gcd(f_a,f_b)=1$.
${\bf Hint}\rm\quad\ \ \gcd(c+1,\,\ c^{\large 2\,k}\!+1)\ =\ gcd(c+1,\:2)$
${\bf Proof}\rm\ \ \ mod\ c+1\!:\,\ c^{\large 2\,k}\!+1\: \equiv\ (-1)^{\large 2\,k}\!+1\:\equiv\ 2,\ \ {\rm by}\ \ c\equiv -1\quad {\bf QED}$
Specializing $\rm\,\ c=2^{\Large 2^{B}},\ \ 2\,k = 2^{\large \,A-B}\ \Rightarrow\ c^{\Large\, 2\,k} = 2^{\Large 2^{A}}\$ immediately yields your claim.
Remark $\$ Aternatively, one could employ that $\rm\:c^{\large 2\,k}\!+1\, =\, (c^{\large 2\,k}-1) + 2\:\equiv\: 2\pmod{c+1}\$
by $\rm\ c+1\ |\ c^{\large 2}-1\ |\ c^{\large 2\,k}\!-1.\,$ But this requires some ingenuity, whereas the above proof does not, being just a trivial congruence calculation using the modular reduction property of the $\rm\:gcd,\:$ namely $\rm\ \gcd(a,b)\, =\, \gcd(a,\:b\ mod\ a),\:$ a reduction which applies much more generally. Said equivalently, the result follows immediately by applying a single step of the Euclidean algorithm. Notice how abstracting the problem a little served to greatly elucidate the innate structure.
• More generally see here for $\large \,\gcd(a^m+1,a^n+1)\ \$ – Bill Dubuque Jun 5 at 2:50
I'm going to be cheeky and flesh out Henning Malkholm's answer, since it's been seven years since this was posted (so if it was indeed homework, it was probably due in by now) and his implicit solution is the best I've seen to this problem. If this bothers you, Henning, let me know and I'll take this down. (I don't have enough reputation to comment directly because I'm new here)
Suppose $$b = a + k$$ for $$k$$ a positive integer. Then by basic algebra $$F_b - 2 = 2^{2^b} - 1 = \left( 2^{2^{a}} \right)^{2^{k}} - 1$$ so, expanding using the geometric sum, $$F_b - 2 = \left( 2^{2^a} + 1 \right) \left\{ \left(2^{2^a}\right)^{2^k-1} - \left(2^{2^a}\right)^{2^k-2} + \cdots -1\right\}$$ So substituting the defining formula gives $$F_b - 2 = F_a \cdot \left( 2^{2^b-2^a} - \cdots - 1 \right)$$ In particular, $$F_b-2$$ is a multiple of $$F_a$$. However, successive odd numbers are relatively prime so no factor of $$F_a$$ can be a factor of $$F_b$$
This is because the Fermat numbers belong to the companion Lucas sequence $$V(3,2) = 2^{k} + 1$$. Hence, all the prime factors of either $$V_{p}$$ where $$p \neq 3$$ is a prime, or $$V_{2^{n}}$$, are primitive; that is, they enter the sequence for the first time as factors at that very term.
So, as every prime factor of $$f_{a}$$ is primitive, it follows that $$gcd(f_{a},f_{b}) = 1$$ when $$b < a$$. | 2019-10-22T15:58:46 | {
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https://math.stackexchange.com/questions/3164781/integer-addition-constant-is-it-a-group | # Integer addition + constant, is it a group?
Assume we define an operator $$a\circ b = a+b+k, \\\forall a,b\in \mathbb Z$$
Can we prove that it together with range for $$a,b$$ is a group, for any given $$k\in \mathbb Z$$?
I have tried, and found that it fulfills all group axioms, but I might have made a mistake?
If it is a group, does it have a name?
My observations:
• Closure is obvious as addition of integers is closed.
• Identity If we take $$e=-k$$, then $$a\circ e = a+k-k=a$$
Verification $$e\circ a = -k\circ a = -k+a+k=a$$, as required.
• Inverse would be $$a^{-1} = -a-2k$$, which is unique.
Verification of inverse $$a\circ a^{-1} = a + (-a-2k)+k = -k = e$$, as required.
• Associativity $$(a\circ b) \circ c = (a + (b+k)) + (c + k)$$.
We see everything involved is addition, which is associative, so we can remove parentheses and change order as we wish.
• Yes, it is group. (I solved this problem as homework in uni once) – Vladislav Mar 27 at 17:14
• How would we know if you've made a mistake when you haven't shared your work on the problem? – Shaun Mar 27 at 17:14
• You might very well have made a mistake. We don't know what you did. Just because you got a correct result doesn't mean you didn't make a mistake. – fleablood Mar 27 at 17:17
• Well, you have to show associativity as well.... – fleablood Mar 27 at 17:21
• @fleablood: The question was a reasonable one. Many groups have names. (And who is this Clarence, anyway? Is he abelian, and countably infinite?) – TonyK Mar 27 at 20:03
It's the group you get when you transfer the action of $$(\mathbb Z,+)$$ to $$(\mathbb Z, \circ)$$ via the map $$\phi(z)= z-k$$.
You can check that $$\phi(a+b)=\phi(a)\circ\phi(b)$$ so that becomes a group isomorphism.
• How can I learn which maps transfer a group to another? – mathreadler Mar 27 at 17:32
• @mathreadler Every bijection can be used to do that. There is nothing special about the bijection chosen. – rschwieb Mar 27 at 19:28
Yes, your observations are correct - this is a group.
Moreover, this group is isomorphic to the infinite cyclic group $$C_\infty$$.
To prove that you can see, that $$\forall a \in \mathbb{Z}, a \circ (1-k) = (a+1)$$, which results in $$\forall a \in \mathbb{Z}, a = (1 - k)^{\circ(a + k - 1)}$$.
• Thank you for the insight. I think I understand. However I am not so used to abstract algebra I am completely confident with isomorphic and all other words. – mathreadler Mar 27 at 22:00
• "$\forall a \in \mathbb{Z} a \circ (1-k) = (a+1)$" is nigh-unreadable. Better is "$\forall a \in \mathbb{Z}$, $a \circ (1-k) = (a+1)$" and still better is "For all $a \in \mathbb Z$, $a \circ (1-k) = a+1$". – Misha Lavrov Mar 28 at 1:55
Suppose we define an operator $$'$$ as $$a'=a-k$$. Then $$a'∘b'=(a-k)+(b-k)+k=a+b-k$$. And $$(a+b)'$$ is also equal to $$a+b-k$$. So $$a'∘b'=(a+b)'$$.
And $$a'$$ is simply $$a$$ on a shifted number line. That is, if you take a number line, and treat $$k$$ as being the origin, then $$a'$$ is the distance $$a$$ is from $$k$$. Suppose you start a stopwatch at time 00:15. And suppose event A happens at 00:17, while event B happens at 00:18. If you just add the times of the two events, you get 00:35. But if you add the times on the stopwatch, you get 00:02+00:03=00:05. $$∘$$ would then represent adding the times on the stopwatch, with $$k$$=00:15.
• Yep. Shifted number line is actually what first made me think of it. Coming from engineering background I know from before that geometric things such as rotations and translations in plane can be groups so surely something like this should be possible also on number line which in some sense must be less complicated. – mathreadler Mar 27 at 22:28 | 2019-09-19T19:05:10 | {
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http://elespiadigital.org/libs/qd9zstny/6y18e.php?tag=number-of-bijective-functions-from-a-to-b-4d2c0c | (e x − 1) 3. A function f (from set A to B) is bijective if, for every y in B, there is exactly one x in A such that f(x) = y. Alternatively, f is bijective if it is a one-to-one correspondence between those sets, in other words both injective and surjective. and reduce them to lowest terms. Here is a proof using bijections: Let S={(a,d):d∣n,1≤a≤d,gcd(a,d)=1} S = \{ (a,d) : d\big|n, 1\le a \le d, \text{gcd}(a,d) = 1 \} S={(a,d):d∣∣n,1≤a≤d,gcd(a,d)=1}. Given a formula of the form a=b a = b a=b, where a a a and b b b are finite positive integer quantities depending on some variables, here is how to prove the formula: Prove that binomial coefficients are symmetric: In a function from X to Y, every element of X must be mapped to an element of Y. Assertion Let A = {x 1 , x 2 , x 3 , x 4 , x 5 } and B = {y 1 , y 2 , y 3 }. ∑d∣nϕ(d)=n. Sample. COMEDK 2015: The number of bijective functions from the set A to itself, if A contains 108 elements is - (A) 180 (B) (180)! Therefore, N has 2216 elements. (B) 64 \{2,4\} &\mapsto \{1,3,5\} \\ Sign up to read all wikis and quizzes in math, science, and engineering topics. A bijective function has no unpaired elements and satisfies both injective (one-to-one) and surjective (onto) mapping of a set P to a set Q. Let E be the set of all subsets of W. The number of functions from Z to E is: If X has m elements and Y has 2 elements, the number of onto functions will be 2. Lemma 3: A function f: A!Bis bijective if and only if there is a function g: B!A so that 1. {1,2}↦{3,4,5}{1,3}↦{2,4,5}{1,4}↦{2,3,5}{1,5}↦{2,3,4}{2,3}↦{1,4,5}{2,4}↦{1,3,5}{2,5}↦{1,3,4}{3,4}↦{1,2,5}{3,5}↦{1,2,4}{4,5}↦{1,2,3}.\begin{aligned} In this article, we are discussing how to find number of functions from one set to another. \{1,4\} &\mapsto \{2,3,5\} \\ https://brilliant.org/wiki/bijective-functions/. B. f(a) = 2;f(b) = 2;f(c) = 2 These are the only non-surjective functions (are you convinced? Experience. Option 4) 0. Try watching this video on www.youtube.com, or enable JavaScript if it is disabled in your browser. Then we connect the points 1 1 1 and 4 4 4 (the first 1,−1 1,-11,−1 pair) and 5 5 5 and 6 6 6 (the second pair). Since f is one-one Hence every element 1, 2, 3 has either of image 1, 2, 3 and that image is unique Total number of one-one function = 6 Example 46 (Method 2) Find the number of all one-one functions from set A = {1, 2, 3} to itself. Show that the number of partitions of nn n into odd parts is equal to the number of partitions of n n n into distinct parts. ), so there are 8 2 = 6 surjective functions. Change the d d d parts into k k k parts: 2a1r+2a2r+⋯+2akr 2^{a_1}r + 2^{a_2}r + \cdots + 2^{a_k}r 2a1r+2a2r+⋯+2akr. via a bijection. a ≠ b ⇒ f(a) ≠ f(b) for all a, b ∈ A ⟺ f(a) = f(b) ⇒ a = b for all a, b ∈ A. e.g. So the correct option is (D). (nk)=(nn−k){n\choose k} = {n\choose n-k}(kn)=(n−kn) It is not hard to show, but a crucial fact is that functions have inverses (with respect to function composition) if and only if they are bijective. The image below illustrates that, and also should give you a visual understanding of how it relates to the definition of bijection. We can prove that binomial coefficients are symmetric: 3+2+1 &= 3+(1+1)+1. For understanding the basics of functions, you can refer this: Classes (Injective, surjective, Bijective) of Functions. Cardinality is the number of elements in a set. A function f : A -> B is called one – one function if distinct elements of A have distinct images in B. p(12)-q(12). fk :Sk→Sn−kfk(X)=S−X.\begin{aligned} For example, q(3)=3q(3) = 3 q(3)=3 because acknowledge that you have read and understood our, GATE CS Original Papers and Official Keys, ISRO CS Original Papers and Official Keys, ISRO CS Syllabus for Scientist/Engineer Exam, Mathematics | Introduction to Propositional Logic | Set 2, Mathematics | Predicates and Quantifiers | Set 2, Mathematics | Some theorems on Nested Quantifiers, Mathematics | Set Operations (Set theory), Inclusion-Exclusion and its various Applications, Mathematics | Power Set and its Properties, Mathematics | Partial Orders and Lattices, Discrete Mathematics | Representing Relations, Mathematics | Representations of Matrices and Graphs in Relations, Mathematics | Closure of Relations and Equivalence Relations, Number of possible Equivalence Relations on a finite set, Discrete Maths | Generating Functions-Introduction and Prerequisites, Mathematics | Generating Functions – Set 2, Mathematics | Sequence, Series and Summations, Mathematics | Independent Sets, Covering and Matching, Mathematics | Rings, Integral domains and Fields, Mathematics | PnC and Binomial Coefficients, Number of triangles in a plane if no more than two points are collinear, Finding nth term of any Polynomial Sequence, Discrete Mathematics | Types of Recurrence Relations – Set 2, Mathematics | Graph Theory Basics – Set 1, Mathematics | Graph Theory Basics – Set 2, Mathematics | Euler and Hamiltonian Paths, Mathematics | Planar Graphs and Graph Coloring, Mathematics | Graph Isomorphisms and Connectivity, Betweenness Centrality (Centrality Measure), Mathematics | Walks, Trails, Paths, Cycles and Circuits in Graph, Graph measurements: length, distance, diameter, eccentricity, radius, center, Relationship between number of nodes and height of binary tree, Bayes’s Theorem for Conditional Probability, Mathematics | Probability Distributions Set 1 (Uniform Distribution), Mathematics | Probability Distributions Set 2 (Exponential Distribution), Mathematics | Probability Distributions Set 3 (Normal Distribution), Mathematics | Probability Distributions Set 4 (Binomial Distribution), Mathematics | Probability Distributions Set 5 (Poisson Distribution), Mathematics | Hypergeometric Distribution model, Mathematics | Limits, Continuity and Differentiability, Mathematics | Lagrange’s Mean Value Theorem, Mathematics | Problems On Permutations | Set 1, Problem on permutations and combinations | Set 2, Mathematics | Graph theory practice questions, Classes (Injective, surjective, Bijective) of Functions, Write Interview In essence, injective means that unequal elements in A always get sent to unequal elements in B. Surjective means that every element of B has an arrow pointing to it, that is, it equals f(a) for some a in the domain of f. Find the number of bijective functions from set A to itself when A contains 106 elements. Class-12-commerce » Math. Below is a visual description of Definition 12.4. If X has m elements and Y has n elements, the number if onto functions are. One to One Function. Compute p(12)−q(12). Bijective. There are Cn C_n Cn ways to do this. Note that the common double counting proof … Let f be a function from A to B. {n\choose k} = {n\choose n-k}.(kn)=(n−kn). A. For instance, Misc 10 (Introduction)Find the number of all onto functions from the set {1, 2, 3, … , n} to itself.Taking set {1, 2, 3}Since f is onto, all elements of {1, 2, 3} have unique pre-image.Total number of one-one function = 3 × 2 × 1 = 6Misc 10Find the number of all onto functio If A and B are two sets having m and n elements respectively such that 1≤n≤m then number of onto function from A to B is = ∑ (-1)n-r nCr rm r vary from 1 to n □_\square□. So, number of onto functions is 2m-2. \sum_{d|n} \phi(d) = n. Then the number of elements of S S S is just ∑d∣nϕ(d) \sum_{d|n} \phi(d) ∑d∣nϕ(d). If a function f : A -> B is both one–one and onto, then f is called a bijection from A to B. Solution: Using m = 4 and n = 3, the number of onto functions is: 5+1 &= 5+1 \\ The double counting technique follows the same procedure, except that S=T S = T S=T, so the bijection is just the identity function. For example, given a sequence 1,1,−1,−1,1,−11,1,-1,-1,1,-11,1,−1,−1,1,−1, connect points 2 2 2 and 33 3, then ignore them to get 1,−1,1,−1 1,-1,1,-1 1,−1,1,−1. It is easy to prove that this is a bijection: indeed, fn−k f_{n-k} fn−k is the inverse of fk f_k fk, because S−(S−X)=X S - (S - X) = X S−(S−X)=X. This function will not be one-to-one. A function is bijective if and only if it has an inverse. This gives a function sending the set Sn S_n Sn of ways to connect the set of points to the set Tn T_n Tn of sequences of 2n 2n 2n copies of ±1 \pm 1 ±1 with nonnegative partial sums. Let f : A ⟶ B and g : X ⟶ Y be two functions represented by the following diagrams. \end{aligned}{1,2}{1,3}{1,4}{1,5}{2,3}{2,4}{2,5}{3,4}{3,5}{4,5}↦{3,4,5}↦{2,4,5}↦{2,3,5}↦{2,3,4}↦{1,4,5}↦{1,3,5}↦{1,3,4}↦{1,2,5}↦{1,2,4}↦{1,2,3}. Here we are going to see, how to check if function is bijective. Already have an account? No injective functions are possible in this case. A bijection from … If set ‘A’ contain ‘5’ element and set ‘B’ contain ‘2’ elements then the total number of function possible will be . \{1,5\} &\mapsto \{2,3,4\} \\ If a function f : A -> B is both one–one and onto, then f is called a bijection from A to B. Calculating required value. It means that every element “b” in the codomain B, there is exactly one element “a” in the domain A. such that f(a) = b. An injective non-surjective function (injection, not a bijection) An injective surjective function A non-injective surjective function (surjection, not a bijection) A … Reason The number of onto functions from A to B is equal to the coefficient of x 5 in 5! List all of the surjective functions in set notation. No injective functions are possible in this case. Number of Bijective Function - If A & B are Bijective then . Relations and Functions. □_\square□. Considering all possibilities of mapping elements of X to elements of Y, the set of functions can be represented in Table 1. Then the number of injective functions that can be defined from set A to set B is (a) 144 (b) 12 NCERT Solutions; Board Paper Solutions; Ask & Answer; School Talk; Login ; GET APP; Login Create Account. □_\square □. It means that every element “b” in the codomain B, there is exactly one element “a” in the domain A. such that f(a) = b. An injective function would require three elements in the codomain, and there are only two. Rewrite each part as 2a 2^a 2a parts equal to b b b. Attention reader! By definition, two sets A and B have the same cardinality if there is a bijection between the sets. A common proof technique in combinatorics, number theory, and other fields is the use of bijections to show that two expressions are equal. That is, we say f is one to one In other words f is one-one, if no element in B is associated with more than one element in A. The most obvious thing to do is to take an even part and rewrite it as a sum of odd parts, and for simplicity's sake, it is best to use odd parts that are equal to each other. Again, it is not immediately clear where this bijection comes from. By using our site, you (nk)=(nn−k). 8b2B; f(g(b)) = b: The number of functions from Z (set of z elements) to E (set of 2xy elements) is 2xyz. Definition: f is one-to-one (denoted 1-1) or injective if preimages are unique. 6=4+1+1=3+2+1=2+2+2. A function is bijective if and only if it has an inverse. Functions in the first column are injective, those in the second column are not injective. Thus, bijective functions satisfy injective as well as surjective function properties and have both conditions to be true. For understanding the basics of functions, you can refer this: Classes (Injective, surjective, Bijective) of Functions. Bijective means both. Option 2) 5! Let q(n)q(n) q(n) be the number of partitions of 2n 2n 2n into exactly nn n parts. 17. a) Prove the following by induction: THEOREM 5.13. Similar Questions. Set A has 3 elements and the set B has 4 elements. The image below illustrates that, and also should give you a visual understanding of how it relates to the definition of bijection. Therefore, f: A $$\rightarrow$$ B is an surjective fucntion. Then the number of function possible will be when functions are counted from set ‘A’ to ‘B’ and when function are counted from set ‘B’ to ‘A’. \{3,4\} &\mapsto \{1,2,5\} \\ A common proof technique in combinatorics, number theory, and other fields is the use of bijections to show that two expressions are equal. Try watching this video on www.youtube.com, or enable JavaScript if it is disabled in your browser. Known as one-to-one correspondence a set of 2 elements, the number of all subsets of W number! 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Create Account which appeared in Encyclopedia of Mathematics - ISBN 1402006098 B B B B one-one... One function if distinct elements of B of the same partition n nn that do not each... Y has n elements, the total number of functions will be n×n×n.. m times = nm:.... Answer ; School Talk ; Login Create Account set T T is the set of 2 elements, number! Be sets of sizes X, Y and Z respectively visual understanding of how number of bijective functions from a to b. Z respectively type of inverse it has into groups an expression of the bijective functions satisfy as! Represented in Table 1 notation number of bijective functions from a to b this 5C1=1, C2=2, C3=5 etc... Let X, Y and Z respectively the partial sums of this sequence are always nonnegative,:. Cn C_n Cn elements, the number of functions from a to.! Is one-one to check that f f f and g: X ⟶ Y be two functions represented by following..... m times = nm can be represented in Figure 1 what type of inverse it has explanation from! The term bijection and the set of all surjective functions, you can refer this: Classes (,. Of Z elements ) to E ( set of functions from one set to another: let and. Means that if a & B are bijective then be a function is bijective parts. the first are. ( denoted 1-1 ) or injective if preimages are unique itself when a contains 106.! Is a real number of Y important that I want to introduce a notation for this we... Use all elements of Y because 6=4+1+1=3+2+1=2+2+2 there are only two { d|n } \phi ( )... Takes different elements of a have distinct images in B share the link here functions satisfy injective as well surjective. Bijection between the sets unused in function F2 list all of the integer as a sum positive. ≠ f ( B, n ) p ( n ) p ( 12 ) one element in a order! Of X must be mapped to an element of X must be mapped to an element Y! On partitions have natural proofs involving bijections E ( set of functions can be opened more … bijective,! One-To-One function, given any Y there is a number of bijective functions from a to b number … bijective function Examples injective as well as function... If distinct elements of a have distinct images in B has 4 elements set is... M elements to a set of all surjective functions from a to B p... Takes different elements of a glass bottle can be paired with the range ) is 2xyz a =! Of set Y is unused and element 4 is unused and element 4 is unused in function F2 be of... 17. a ) = n ( B ) closed metal lid of a have distinct images B... Type of inverse it has an inverse & B are bijective then # A=4.:60 ways... Hard to check that f f and g g g g g g are inverses of each other so! Are considered the same partition functions represented by the following by induction: 5.13.
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total completion time 总完工时间(10)完工时间和(4)加工时间(4)
总完工时间
For the single machine,batching,total completion time scheduling problems,the optimal batching structures on the batching ways:all jobs in exactly N(arbitrarily fixed batch number and 1 研究了目标函数为总完工时间、工件恰分N批的单机分批排序问题最优解的结构性质,其中N为1与工件数之间的任意整数. 短句来源 Inverse Problems of a Single Machine Scheduling to Minimize the Total Completion Time 单台机器总完工时间排序问题的反问题 短句来源 Optimal algorithms are presented for the problem of minimizing the makespan and total completion time,while for the problem of weighted total completion time,the agreeable relation of jobs were given. 对于目标函数为极小化最大完工时间和总完工时间的问题,给出了求解最优排序的多项式算法,对于目标函数为加权总完工时间的问题,给出了工件间的一致关系. 短句来源 Minimal Total Completion Time in Two-machine Flowshop with Setup, Processing and Removal Time Separated 有分离调整和移走时间的两机器流水作业总完工时间问题 短句来源 Inverse Scheduling Problem on a Single Machine Stochastic Scheduling to Minimize the Total Completion Time 单台机器总完工时间随机排序问题的反问题 短句来源 更多
完工时间和
Minimizing Total Completion Time of Bounded Batch Scheduling 极小化完工时间和的有界批调度问题(英文) 短句来源 Optimal algorithms are presented for the problem of minimizing the makespan and total completion time,while for the problem of weighted total completion time,the agreeable relation of jobs were given. 对于目标函数为极小化最大完工时间和总完工时间的问题,给出了求解最优排序的多项式算法,对于目标函数为加权总完工时间的问题,给出了工件间的一致关系. 短句来源 The dual target problem of mininmizing total completion time and variation of completion times can also be formulated as assignment problem. 对于极小化完工时间和与完工时间偏差的双目标问题,其一般情况同样可以转化成指派问题. 短句来源 Moreover,for some special cases,the simple algorithms are presented for the makespan and total completion time minimization problems. 此外,对于某些特殊情况,给出了极小化最大完工时间问题与完工时间和问题的简便算法. 短句来源
加工时间
The results are three necessary and sufficient conditions for the objective of total completion time and some sufficient conditions for the objective of maximum lateness time or for the objective of the number of tardy jobs under which an optimal schedule is of interval perturbation robustness. 本文的结果是目标函数为完成时间总和时在加工时间扰动下最优调度具有区间摄动鲁棒性的三个充分必要条件,目标函数为最大拖期时间时及目标函数为拖后工件个数时在加工时间和/或交付期扰动下最优调度具有区间摄动鲁棒性的若干充分条件. 短句来源 We study the problems of minimizing makespan and total completion time. 第六章研究一种加工时间依赖开工时间，并且机器带一个维护时间段的单台机排序问题。 短句来源 Chapter two investigates single processor scheduling problem minimizing makespan with group technology and single processor, scheduling problem of total completion time under the condition of linear deterioration of processing times. 第二章对工件的加工时间依赖其开工时间的情形,分别研究了单机成组最大完工时间问题和单机总完工时间问题。 短句来源 Moreover, we consider the minimal total completion time problem where there is not the idle time in machine M1, and derive the optimality conditions for this kind of flowshops scheduling problems. 本文以总完工时间为准则研究调整时间和移走时间均独立于加工时间的两机器流水作业问题,给出了问题最优解中工件排列应满足的条件; 短句来源
“total completion time”译为未确定词的双语例句
Minimizing Total Completion Time of Jobs on a Flexible Two Machine Flow Shop 流水作业由二台柔性机器组成时的极小完工时间之和问题 短句来源 Min-min algorithm, a classics algorithm in heuristic algorithm, which always completes the shortest total completion time task first, has the characteristic of simple and shortest completion time. So it catches a lot of close attentions in the field of studying for tasks scheduling algorithms in Grid. 作为启发式算法中的经典算法，Min-min算法总是先执行具有最短完成时间的作业，有着算法思路简单、总完成时间短的特点，是网格作业调度算法研究中倍受关注的一个算法。 短句来源 The problems of scheduling jobs with different ready time on parallel machines to minimize the total completion time are addressed. 研究工件带释放时间的两类并行机最小化总完成时间的调度问题. 短句来源 In this paper we prove that the shortest ready time (SRT) sequencing minimizes the total completion time, and analyse properties of the one-machine sequencing problem with ready time to minimize the weighted number of tardy jobs. 本文证明了最短准备时间(SRT)排序使总的完工时间取得最小,并进一步分析了带有准备时间的带权误工工件数排序问题的一些性质. 短句来源 Minimizing Total Completion Time of Orders with Multiple Job Classes 订单带多类工件时的极小完工时间之和问题 短句来源 更多
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total completion time
The objective is to minimize the total completion time of the accepted jobs and the total penalty of the rejection jobs. The objective is to minimize the total completion time of the accepted jobs plus the total penalty of the rejected jobs. We present polynomial time algorithms to find the job sequence, the partition of the job sequence into batches and the resource allocation, which minimize the total completion time or the total production cost (inventory plus resource costs). Single machine batch scheduling to minimize total completion time and resource consumption costs For the $$\mathcal{NP}$$-hard problem of scheduling n jobs in a two-machine flow shop so as to minimize the total completion time, we present two equivalent lower bounds that are computable in polynomial time. 更多
In this paper we prove that the shortest ready time (SRT) sequencing minimizes the total completion time, and analyse properties of the one-machine sequencing problem with ready time to minimize the weighted number of tardy jobs. 本文证明了最短准备时间(SRT)排序使总的完工时间取得最小,并进一步分析了带有准备时间的带权误工工件数排序问题的一些性质. M.Dror examined the open-shop scheduling problem with machine dependent processing times in 1992. Two criteria were considered: minimizing the maximum completion time (makespan), and minimizing the total completion time. In this paper we show that the "algorithm" for the second criterion proposed by him is wrong. Then, we formulate the problem to minimize the total completion time as an asslgnoment model when machines are continuously available and are never kept idle while work is walting， and apply... M.Dror examined the open-shop scheduling problem with machine dependent processing times in 1992. Two criteria were considered: minimizing the maximum completion time (makespan), and minimizing the total completion time. In this paper we show that the "algorithm" for the second criterion proposed by him is wrong. Then, we formulate the problem to minimize the total completion time as an asslgnoment model when machines are continuously available and are never kept idle while work is walting， and apply the Hungarian method to solve it． Several questions are still unanswered．We describe tree open problems for further research at last． 1992年M．Dror提出工件的加工时间依赖于机器的排序问题（schedulingwithmachinedependentprocessingtimes），并研究以最大完工时间（makespan）和以总的完工时间为优化目标的两种这类排序问题．然而，M．Dror对以总的完工时间为优化目标提出的“最优算法”是错误的．本文用算例表明他提出的算法不是最优的，并在机器连续加工的条件下，把这个排序问题转化成指派问题（assignmentproblem），从而可以用匈牙利算法得到最优解．最后，我们提出几个尚未解决的问题，以期引起国内外同行进一步研究． The sequencing of n independent work pieces which are processed in a single machine shop is discussed in which each job is assigned a common due date k and each work piece processing time distribution is assumed to be independent normal distribution. The object is to determine a common due date k *and to find an optimal sequence of n independent work pieces to minimize the expectation of the linear combination of total absolute deviation of completion time about a common due date and total completion... The sequencing of n independent work pieces which are processed in a single machine shop is discussed in which each job is assigned a common due date k and each work piece processing time distribution is assumed to be independent normal distribution. The object is to determine a common due date k *and to find an optimal sequence of n independent work pieces to minimize the expectation of the linear combination of total absolute deviation of completion time about a common due date and total completion time. 讨论 n个独立工件在一台机器上加工。工件的加工时间服从正态分布 ,所有工件交货期设置公共交货期。目标是确定公共交货期及工件的最优排序 ,使工件完工时间与公共交货期之差绝对值之和及工件完工时间之和的线性组合的期望值最小 . << 更多相关文摘
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2008中国知网(cnki) 中国学术期刊(光盘版)电子杂志社 | 2020-08-04T21:14:47 | {
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https://math.stackexchange.com/questions/347765/conflicting-answers-when-using-complements-principle-and-the-inclusion-exclusion | Conflicting answers when using Complements Principle and the Inclusion-Exclusion Principle
The question I'm looking at is:
Andy, Bill, Carl and Dave are 4 students on a team of 10. 5 must be chosen for a tournament, how many teams can be picked if Andy or Bill or Carl or Dave must be on the team.
Using the inclusion-exclusion principle:
Let $A_1 =$ teams with Andy, $A_2 =$ teams with Bill, ect.
$$|A_i| = {9 \choose 4}= 126$$
$$|A_i \cap A_j| = {8 \choose 3} = 56\text{ for }i \neq j$$
$$|A_i \cap A_j \cap A_k| = {7 \choose 2} =21\text{ for }i \neq j \neq k$$
$$|A_1 \cap A_2 \cap A_3 \cap A_4| = {6 \choose 1} = 6$$
So then $|A_1 \cup A_2 \cup A_3 \cup A_4| = 4(126) - 6(56) + 3(21) - 6 = 225$
But when I use the complements principle to subtract all teams without Andy, Bill, Carl, and Dave from all teams I get:
$${10 \choose 5} - \displaystyle{6 \choose 5} = 252 - 6 = 246$$
which is not the same. So I'm clearly doing something wrong with one of these but I don't know which one is wrong.
• Why did you add $3(21)$? – Will Orrick Apr 1 '13 at 1:25
• Because the 3 sets $A_1 \cap A_2 \cap A_3, A_1 \cap A_2 \cap A_4, A_2 \cap A_3 \cap A_4$ all have 21 different teams and I'm using the inclusion exclusion principle to determine $|A_1 \cup A_2 \cup A_3 \cup A_4|$ – MangoPirate Apr 1 '13 at 1:30
• @MangoPirate : I improved both your TeX formatting and your spelling. Note that "compliment" with an "i" and "complement" with an "e" are two different words that mean two different things. The complement (with an "e") of $A$ is something that together with $A$ makes a complete whole. The resemblance to the spelling of "complete" is not coincidental and enables you to remember which is which. – Michael Hardy Apr 1 '13 at 1:30
• @MangoPirate : The coefficients of the other three terms, namely 4, 6, and 1, are all numbers of the form $\binom{4}{j}.$ The number 3 is the only one that doesn't fit this pattern. Could that be a sign that something's wrong? – Will Orrick Apr 1 '13 at 1:33
• You forgot about $A_1\cap A_3\cap A_4$. – André Nicolas Apr 1 '13 at 1:33
As noted in the comments, the $21$ for $|A_i \cap A_j \cap A_k|$ should be multiplied by $4$, not by $3$, since there are $4$ ways to omit one of $4$ students and leave $3$. | 2020-01-21T21:16:12 | {
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http://math.stackexchange.com/questions/468019/evaluating-an-improper-integral-using-complex-analysis | # Evaluating an improper integral using complex analysis
I am trying to evaluate the improper integral $I:=\int_{-\infty}^\infty f(x)dx$, where $$f(z) := \frac{\exp((1+i)z)}{(1+\exp z)^2}.$$
I tried to do this by using complex integration. Let $L,L^\prime>0$ be real numbers, and $C_1, C_2, C_3, C_4$ be the line segments that go from $-L^\prime$ to $L$, from $L$ to $L+2\pi i$, from $L + 2\pi i$ to $-L^\prime+2\pi i$ and from $-L^\prime+2\pi i$ to $-L^\prime$, respectively. Let $C = C_1 + C_2 + C_3 + C_4$.
Here we have (for sufficiently large $L$ and $L^\prime$) $$\int_{C_2}f(z) dz \le \int_0^{2\pi}\left|\frac{\exp((1+i)(L+iy))}{(\exp(L+iy)+1)}i\right| dy \le \int\frac{1}{(1-e^{-L})(e^L - 1)}dy\rightarrow0\quad(L\rightarrow\infty),$$ $$\int_{C_4}f(z)dz\le\int_0^{2\pi}\left|\frac{\exp((1+i)(-L^\prime+iy))}{(\exp(-L^\prime + iy) + 1))^2}(-i)\right|dy\le\int\frac{e^{-L^\prime}}{(1-e^{-L})^2}dy\rightarrow 0\quad(L^\prime\rightarrow\infty),$$ and $$\int_{C_3}f(z)dz = e^{-2\pi}\int_{C_1}f(z)dz.$$ Thus $$I = \lim_{L,L^\prime\rightarrow\infty}\frac{1}{ (1 + e^{-2\pi})}\oint_Cf(z)dz.$$
Within the perimeter $C$ of the rectangle, $f$ has only one pole: $z = \pi i$. Around this point, $f$ has the expansion $$f(z) = \frac{O(1)}{(-(z-\pi i)(1 + O(z-\pi i)))^2} =\frac{O(1)(1+O(z-\pi i))^2}{(z-\pi i)^2} = \frac{1}{(z-\pi i)^2} + O((z-\pi i)^{-1}),$$ and thus the order of the pole is 2. Its residue is $$\frac{1}{(2-1)!}\frac{d}{dz}\Big|_{z=\pi i}(z-\pi i)^2f(z) = -\pi \exp(i\pi^2)$$ (after a long calculation) and we have finally $I=-\exp(i\pi^2)/2i(1+\exp(-2\pi))$.
My question is whether this derivation is correct. I would also like to know if there are easier ways to do this (especially, those of calculating the residue). I would appreciate if you could help me work on this problem.
-
You use $\,f(x)\;,\;f(z)\;$ and "improper integral": is there any real function here for which you want to use complex integration? What function is that? The only function you've written is a complex one... – DonAntonio Aug 15 '13 at 10:52
You can write: $$\int _{-\infty }^{\infty }\!{\frac {{{\rm e}^{x}}{{\rm e}^{ix}}}{ \left( {{\rm e}^{x}}+1 \right) ^{2}}}{dx}=1/2\,\int _{-\infty }^{ \infty }\!{\frac {\cos \left( x \right) }{\cosh \left( x \right) +1}}{ dx}$$ if that is any use. I can get a series for that integral:$$-2\,\sum _{n=0}^{\infty }{\frac { \left( -1 \right) ^{n}{n}^{2}}{{n}^{ 2}+1}}= 0.2720290550$$ but as of yet no closed form, trying to see if it can be written in terms of digamma function but don't see it yet.# – Graham Hesketh Aug 15 '13 at 11:18
@DonAntonio Well... what I am working on is an integration of a complex-valued function along the real axis, which is a trivial kind of complex integration and it's safe to say it's real integration. I use the variable $z$ to define the function and use $x$ in the integral but I believe it is not confusing (actually, this is an almost literal copy from a past exam). – Pteromys Aug 15 '13 at 11:29
@GrahamHesketh, the series you posted can be solved using complex analysis . – Zaid Alyafeai Aug 15 '13 at 16:56
The series diverges logarithmically. – Felix Marin Aug 18 '13 at 19:51
Everything appears right except :
• (as noted by mrf) the sign in : $$\;\displaystyle\int_{C_3}f(z)dz = -e^{-2\pi}\int_{C_1}f(z)dz$$ (and before $e^{-2\pi}$ in the denominator of $I$ that follows)
• the computation of the residue : $$\frac{d}{dz}\Big|_{z=\pi i}(z-\pi i)^2f(z) = i e^{(i-1)\pi}=-i\, e^{-\pi}$$ so that the integral will simply be $\;2\pi\,i\sum_{\text{res}}=2\pi i\;\left(-i e^{-\pi}\right)\;$ and the answer (corresponding to Graham's series and numerical evaluation) : $$\;\frac{2\pi\,e^{-\pi}}{1-e^{-2\pi}}=\frac{\pi}{\sinh(\pi)}\approx 0.272029055$$
Computing residues in a practical way is often done using our favorite software to get the Laurent series of $f(z)$ at $z=\pi i$.
Without computer I spontaneously expanded $f(z)$ this way (for $\,z:=\pi i+x\,$ with $\,|x|\ll 1$) : \begin{align} f(\pi i+x)&=\frac{e^{(1+i)(\pi i+x)}}{(1+e^{\pi i+x})^2}=\frac{-e^{-\pi}\;e^{(1+i)x}}{(1-e^x)^2}\\ &=-e^{-\pi}\frac {1+(1+i)x+O\bigl(x^2\bigr)}{x^2\left(1+x/2+O\bigl(x^2\bigr)\right)^2}\\ &=-\frac{e^{-\pi}}{x^2}\left((1+(1+i)x)(1-x)+O\bigl(x^2\bigr)\right)\\ &=-\frac{e^{-\pi}}{x^2}\left(1+ix+O\bigl(x^2\bigr)\right)\\ &=-\frac{e^{-\pi}}{x^2}-\frac{i\,e^{-\pi}}{x}+O(1) \end{align}
-
There is also an error in the orientation of $C_3$: $1+\exp(-2\pi)$ should be $1-\exp(-2\pi)$. – mrf Aug 15 '13 at 11:58
@Graham: yes sorry (I'll update my answer) – Raymond Manzoni Aug 15 '13 at 12:00
Thanks @mrf I had missed the sign problem. – Raymond Manzoni Aug 15 '13 at 12:05
@RaymondManzoni I also computed the residue using Maxima (although I made mistakes in typing the formula) because I felt lazy. What if you must use pen and paper, not computer algebra systems, in an exam, for example? – Pteromys Aug 15 '13 at 14:42
@Pteromys: Well my spontaneous answer (without computer) is to expand it as this (for $z:=\pi i+x$) : $$\frac{e^{(1+i)(\pi i+x)}}{(1+e^{\pi i+x})^2}=\frac{-e^{-\pi}\;e^{(1+i)x}}{(1-e^x)^2}=-e^{-\pi}\frac {1+(1+i)x+O(x^2)}{x^2(1+x/2+O(x^2))^2}=-\frac{e^{-\pi}}{x^2}\left((1+(1+i)x)(1-x)+O(x^2)\right)=-\frac{e^{-\pi}}{x^2}\left(1+ix+O(x^2)\right)=\\-\frac{e^{-\pi}}{x^2}-\frac{i\,e^{-\pi}}{x}+O(1)$$ (for this specific problem at least... multiplying by $(z-\pi i)^2$ is another equivalent possibility...) – Raymond Manzoni Aug 15 '13 at 17:34
Just for fun, here is an alternative way to do the integration without calculating residues. First define the following even function: $$f(x)=\dfrac{e^x}{\left(1+e^x\right)^2}=\dfrac{1}{2+2\cosh{x}}=-\dfrac{d}{dx}\dfrac{1}{1+e^x}\tag{1}$$ and it's Fourier transform; technically it's the inverse (or conjugate) but that is not particularly important: \begin{aligned} \hat{F}(k)&=\int_{-\infty}^{\infty}\dfrac{e^x}{\left(1+e^x\right)^2}e^{ixk}{dx}\\ &=2\,\Re{\left(\int_{0}^{\infty}-\left(\dfrac{d}{dx}\dfrac{1}{1+e^x}\right)e^{ixk}{dx}\right)}\\ &=1-2\,\Im{\left(k\int_{0}^{\infty}\dfrac{e^{-x(1-ik)}}{1+e^{-x}}{dx}\right)} \tag{2}\end{aligned} where $\Re,\Im$ are real and imaginary parts respectively. We note that the OP is interested in $\hat{F}(1)$. To move from the first line to the second in $(2)$ I used $(1)$ and the eveness of the integrand, to move from the second to the third I used integration by parts and finally multiplied the integrand top and bottom by $e^{-x}$. Next I'll show that $(2)$ can be written in terms of the digamma function: $$\hat{F}(k)=1+2\,k\,\Im \left( \Psi \left( \dfrac{1-ik}{2} \right) -\Psi \left( 1-ik \right) \right)\tag{3}$$ Proof
First some algebra on the definite integral: \begin{aligned} \int_{\epsilon}^{1/\epsilon}\dfrac{e^{-x(1-ik)}}{1+e^{-x}}{dx}=&\int_{\epsilon}^{1/\epsilon}\dfrac{2e^{-x(1-ik)}}{1-e^{-2x}}{dx}-\int_{\epsilon}^{1/\epsilon}\dfrac{e^{-x(1-ik)}}{1-e^{-x}}{dx}\\ =&-\int_{\epsilon}^{1/\epsilon}\dfrac{e^{-2x}}{x}-\dfrac{2e^{-x(1-ik)}}{1-e^{-2x}}{dx}+\int_{\epsilon}^{1/\epsilon}\dfrac{e^{-x}}{x}-\dfrac{e^{-x(1-ik)}}{1-e^{-x}}{dx}\\&+\int_{\epsilon}^{1/\epsilon}\dfrac{e^{-2x}}{x}-\dfrac{e^{-x}}{x}{dx}\\ =&-\int_{2\epsilon}^{2/\epsilon}\dfrac{e^{-x}}{x}-\dfrac{e^{-x(1-ik)/2}}{1-e^{-x}}{dx}+\int_{\epsilon}^{1/\epsilon}\dfrac{e^{-x}}{x}-\dfrac{e^{-x(1-ik)}}{1-e^{-x}}{dx}\\&-\int_{\epsilon}^{1/\epsilon}{\frac {{e^{-3x/2}}\sinh \left( \frac{x}{2} \right) }{x}}{dx} \end{aligned} then we note that for $\Re(t)>0$ we have the following integral representation of the digamma function: $$\Psi(t)=\int_{0}^{\infty}\frac{e^{-x}}{x}-\frac{e^{-xt}}{1-e^{-t}}{dx}\tag{4}$$ and that taking the limit in which $\epsilon\rightarrow 0$ we then have by comparison with $(4)$: \begin{aligned} \int_{0}^{\infty}\dfrac{e^{-x(1-ik)}}{1+e^{-x}}{dx}&=-\Psi\left(\frac{1-ik}{2}\right)+\Psi\left(1-ik\right)-\int_{0}^{\infty}{\frac {{e^{-3x/2}}\sinh \left( \frac{x}{2} \right) }{x}}{dx}\\ &=-\Psi\left(\frac{1-ik}{2}\right)+\Psi\left(1-ik\right)-\ln(2) \end{aligned} and $(3)$ follows. (Note: Maple was used to evaluate the real $\sinh$ integral but I won't pursue a proof as I am only interested in the imaginary part).
Then from $\bar{\Psi}(z)=\Psi{(\bar{z})}$ and the reflection formula $\Psi(1-z)-\Psi(z)=\pi\cot(\pi z)$ we have: \begin{aligned} \hat{F}(k)&=1+2\,k\,\Im \left( \Psi \left( \dfrac{1-ik}{2} \right) -\Psi \left( 1-ik \right) \right)\\ & =\pi k\left( -\tanh \left( \frac{\pi k}{2} \right)+ \coth \left( \pi \,k \right)\right)\\ &={\frac {\pi k}{\sinh \left( \pi k \right) }}\tag{5} \end{aligned} Residue theory would, in all likelyhood, also arrive at the more general result in $(5)$ but it is interesting to have an alternative method.
Corollary
Having found the well defined integral: \begin{aligned} \hat{F}(k)&=\int_{-\infty}^{\infty}\dfrac{e^xe^{ixk}}{\left(1+e^x\right)^2}{dx}=-\int_{-\infty}^{\infty}\left(\dfrac{d}{dx}\dfrac{1}{1+e^x}\right)e^{ixk}{dx}\\ &=\dfrac{\pi k}{\sinh(\pi k)} \end{aligned} we may then use the law for the Fourier transform of a derivative to assign the following meaning to the not so well defined integral: $$\dfrac{1}{ \sinh \left( \pi k \right) }=\frac{i}{\pi}\int _{- \infty }^{\infty }\!{\frac {{e^{ixk}}}{1+{e^{x}}}}{dx}\tag{6}$$
-
+1 for the effort and the interesting generalization. Just two little typos : the $2$ at the numerator in $(1)$ and in the Corollary should be $1$ (the $2$ in $(2)$ comes from the half-integral). Cheers – Raymond Manzoni Aug 16 '13 at 17:55
Cheers, fixed them. – Graham Hesketh Aug 16 '13 at 21:37
\begin{eqnarray*} \int_{-\infty}^{\infty} {{\rm e}^{\left(1\ +\ {\rm i}\right)x} \over \left(1 + {\rm e}^{x}\right)^2}\,{\rm d}x & = & \int_{0}^{\infty}\left\lbrack% {{\rm e}^{\left(-1\ +\ {\rm i}\right)x} \over \left(1 + {\rm e}^{-x}\right)^2} + {{\rm e}^{-\left(1\ +\ {\rm i}\right)x} \over \left(1 + {\rm e}^{-x}\right)^2} \right\rbrack {\rm d}x \\ & = & 2\,\Re\int_{0}^{\infty} {{\rm e}^{-\left(1\ -\ {\rm i}\right)x} \over \left(1 + {\rm e}^{-x}\right)^2}\,{\rm d}x = 2\,\Re\int_{0}^{\infty} {\rm e}^{-\left(1\ -\ {\rm i}\right)x} \sum_{n = 1}^{\infty}\left(-1\right)^{n}\,n\,{\rm e}^{-\left(n - 1\right)x}\,{\rm d}x \\ & = & 2\,\Re\sum_{n = 1}^{\infty}\left(-1\right)^{n}\,n \int_{0}^{\infty}{\rm e}^{-\left(n - {\rm i}\right)x} = 2\,\Re\sum_{n = 1}^{\infty}\left(-1\right)^{n}\,{n \over n - {\rm i}} \\ & = & 2\,\Re\sum_{n = 1}^{\infty}\left(% -\,{2n - 1\over 2n - 1 - {\rm i}} + {2n \over 2n - {\rm i}} \right) \\ & = & 2\,\Re\sum_{n = 1}^{\infty}\left\lbrack% \left(-1 - {{\rm i} \over 2n - 1 - {\rm i}}\right) + \left(1 + {{\rm i} \over 2n - {\rm i}}\right) \right\rbrack \\ & = & 2\,\Im\sum_{n = 1}^{\infty}\left( {1 \over -{\rm i} + 2n} - {1 \over -{\rm i} + 2n - 1} \right) = 2\,\Im\sum_{n = 1}^{\infty} {\left(-1\right)^{n} \over -{\rm i} + n} \\ & = & 2\,\Im\left\lbrack\sum_{n = 0}^{\infty} {\left(-1\right)^{n} \over -{\rm i} + n} - {1 \over -{\rm i}} \right\rbrack = -2 + 2\,\Im\sum_{n = 0}^{\infty}{\left(-1\right)^{n} \over -{\rm i} + n} \\[1cm]&& \end{eqnarray*}
$$\int_{-\infty}^{\infty} {{\rm e}^{\left(1\ +\ {\rm i}\right)x} \over \left(1 + {\rm e}^{x}\right)^2}\,{\rm d}x = -2 + 2\,\Im\beta\left(-{\rm i}\right)$$
- | 2015-08-02T21:06:26 | {
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https://plainmath.net/86677/what-is-a-tight-lower-bound-to-sum-n-i | # What is a tight lower bound to sum^n_(i=1)(1)/(a+x_i) under the restrictions sum^n_(i=1)x_i=0 and sum^n_(i=1)x^2_i=a^2 ?
What is a tight lower bound to $\sum _{i=1}^{n}\frac{1}{a+{x}_{i}}$ under the restrictions $\sum _{i=1}^{n}{x}_{i}=0$ and $\sum _{i=1}^{n}{x}_{i}^{2}={a}^{2}$ ?
Conjecture: due to the steeper rise of $\frac{1}{a+x}$ for negative $x$, one may keep those values as small as possible. So take $n-1$ values ${x}_{i}=-q$ and ${x}_{n}=\left(n-1\right)q$ to compensate for the first condition. The second one then gives ${a}^{2}=\sum _{i=1}^{n}{x}_{i}^{2}={q}^{2}\left(\left(n-1{\right)}^{2}+n-1\right)={q}^{2}n\left(n-1\right)$. Hence,
$\sum _{i=1}^{n}\frac{1}{a+{x}_{i}}\ge \frac{n-1}{a\left(1-1/\sqrt{n\left(n-1\right)}\right)}+\frac{1}{a\left(1+\left(n-1\right)/\sqrt{n\left(n-1\right)}\right)}$
should be the tight lower bound.
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Jamir Young
Yes, this is indeed the minimum value (assuming $a>0$).
Denote $K=\left\{x\in {\mathbb{R}}^{n}\mid \sum _{k=1}^{n}{x}_{k}=0,\sum _{k=1}^{n}{x}_{k}^{2}={a}^{2}\right\}$ and let $f\left(x\right)=\sum _{k=1}^{n}\left(a+{x}_{k}{\right)}^{-1}$ attain its minimum at $x=\overline{x}\in K$ (it does so, as a continuous function on $\left\{x\in K\mid f\left(x\right)⩽f\left(0\right)\right\}$ which is compact). Then, by Lagrange multiplier theorem, there are ${\lambda }_{1},{\lambda }_{2}\in \mathbb{R}$ such that for each k we have $\left(a+{\overline{x}}_{k}{\right)}^{-2}={\lambda }_{1}+{\lambda }_{2}{\overline{x}}_{k}$. Then the positive numbers ${y}_{k}=a+{\overline{x}}_{k}$ are solutions of
${y}^{2}\left({\lambda }_{1}-{\lambda }_{2}a+{\lambda }_{2}y\right)=1.$
But this equation has at most two positive solutions. Thus, at most two values among ${\overline{x}}_{k}$ are distinct, and in fact exactly two. So, let $m$ values of ${\overline{x}}_{k}$ equal $b>0$, where $0, and the remaining $n-m$ values equal $c<0$. We get a system for $b$ and $c$, solve it, and finally obtain
$af\left(\overline{x}\right)=n+{\left(1-\frac{1}{n}+\frac{n-2m}{\sqrt{nm\left(n-m\right)}}\right)}^{-1}.$
The least possible value of this is at $m=1$. | 2022-09-29T07:15:57 | {
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https://mathhelpboards.com/threads/convert-the-recursive-formula-into-the-explicit-form.24905/ | # [SOLVED]Convert the recursive formula into the explicit form
#### mathmari
##### Well-known member
MHB Site Helper
Hey!!
We have the sequence $$0, \ 2 , \ -6, \ 12, \ -20, \ \ldots$$ Its recursive definition is \begin{align*}&a_1=0 \\ &a_{n+1}=(-1)^{n+1}\cdot (a_n+2\cdot n)\end{align*} or not?
How can we convert that in the explicit form?
#### Olinguito
##### Well-known member
It is easiest first of all to ignore the $(-1)^{n+1}$ and consider the sequence
$$0,2,6,12,20,\ldots.$$
We can add the necessary minus signs later on.
This sequence is thus $a_{n+1}=a_n+2n$ i.e. $a_{n+1}-a_n=2n$. We can therefore use telescoping to get an explicit formula:
$$\begin{array}{rcl}a_{n+1}-a_n &=& 2n \\ a_n-a_{n-1} &=& 2(n-1) \\ {} &\vdots& {} \\ a_2-a_1 &=& 2\cdot1\end{array}$$
$\displaystyle\implies\ a_{n+1}-a_1=2\sum_{r=1}^nr=2\cdot\frac{n(n+1)}2=n(n+1)$,
i.e. $a_n=n(n-1)$. To restore the minus signs, simply add the $(-1)^{n+1}$ back:
$$\boxed{a_n\ =\ (-1)^{n+1}n(n-1)}.$$
PS: The formula for the original sequence $0,2,-6,12,-20,\ldots$ should be
$$a_1=0;\ a_{n+1}=a_n+(-1)^{n+1}\cdot2n.$$
If it were
\begin{align*}&a_1=0 \\ &a_{n+1}=(-1)^{n+1}\cdot (a_n+2\cdot n)\end{align*}
the fourth term would be $0$, not 12.
Last edited:
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Hey!!
We have the sequence $$0, \ 2 , \ -6, \ 12, \ -20, \ \ldots$$ Its recursive definition is \begin{align*}&a_1=0 \\ &a_{n+1}=(-1)^{n+1}\cdot (a_n+2\cdot n)\end{align*} or not?
How can we convert that in the explicit form?
$$\boxed{a_n\ =\ (-1)^{n+1}n(n-1)}.$$
Hey mathmari and Olinguito !!
Just an observation:
$$\begin{array}{c|c|c|c|}n & a_n & a_{n+1}=(-1)^{n+1}\cdot (a_n+2\cdot n) & a_n\ =\ (-1)^{n+1}n(n-1) \\ \hline 1 & 0 & 0 & 0\\ 2 & 2 & (-1)^{1+1}\cdot(0+2\cdot 1) = 2 & (-1)^{2+1}\cdot 2 (2-1)={\color{red}-2}\\ 3 & -6 & (-1)^{1+2}\cdot(2+2\cdot 2) = -6 \\ 4 & 12 & (-1)^{1+3}\cdot(-6+2\cdot 3) = {\color{red}0} \\ 5 & -20 \\ \end{array}$$
#### mathmari
##### Well-known member
MHB Site Helper
ust an observation:
$$\begin{array}{c|c|c|c|}n & a_n & a_{n+1}=(-1)^{n+1}\cdot (a_n+2\cdot n) & a_n\ =\ (-1)^{n+1}n(n-1) \\ \hline 1 & 0 & 0 & 0\\ 2 & 2 & (-1)^{1+1}\cdot(0+2\cdot 1) = 2 & (-1)^{2+1}\cdot 2 (2-1)={\color{red}-2}\\ 3 & -6 & (-1)^{1+2}\cdot(2+2\cdot 2) = -6 \\ 4 & 12 & (-1)^{1+3}\cdot(-6+2\cdot 3) = {\color{red}0} \\ 5 & -20 \\ \end{array}$$
Ah should it maybe be $$a_{n+1}=(-1)^{n+1}\cdot (|a_n|+2\cdot n)$$ ?
Because if we consider the sequence without the signs, we add at the previous number the number 2n. Then the sign changes, at the odd positions we have $-$ and at the even places we have $+$, or not?
Last edited:
#### Olinguito
##### Well-known member
$$\begin{array}{c|c|c|c|}n & a_n & a_{n+1}=(-1)^{n+1}\cdot (a_n+2\cdot n) & a_n\ =\ (-1)^{n+1}n(n-1) \\ \hline 1 & 0 & 0 & 0\\ 2 & 2 & (-1)^{1+1}\cdot(0+2\cdot 1) = 2 & (-1)^{2+1}\cdot 2 (2-1)={\color{red}-2}\\ 3 & -6 & (-1)^{1+2}\cdot(2+2\cdot 2) = -6 \\ 4 & 12 & (-1)^{1+3}\cdot(-6+2\cdot 3) = {\color{red}0} \\ 5 & -20 \\ \end{array}$$
Thanks, ILS. When the first term is $0$ it’s easy to slip into thinking it’s $a_0$ rather than $a_1$. In this case the formula should be
$$\boxed{a_n\ =\ (-1)^nn(n-1)}.$$
Ah should it maybe be $$a_{n+1}=(-1)^{n+1}\cdot (|a_n|+2\cdot n)$$ ?
Yes, that should work.
#### mathmari
##### Well-known member
MHB Site Helper
Thanks, ILS. When the first term is $0$ it’s easy to slip into thinking it’s $a_0$ rather than $a_1$. In this case the formula should be
$$\boxed{a_n\ =\ (-1)^nn(n-1)}.$$
Yes, that should work.
So, is the recursive formula \begin{align*}&a_0=0 \\ &a_{n+1}=(-1)^{n+1}\cdot (|a_n|+2\cdot n)\end{align*} and the explicit formula $$a_n\ =\ (-1)^nn(n-1)$$ ?
#### Klaas van Aarsen
##### MHB Seeker
Staff member
So, is the recursive formula \begin{align*}&a_0=0 \\ &a_{n+1}=(-1)^{n+1}\cdot (|a_n|+2\cdot n)\end{align*} and the explicit formula $$a_n\ =\ (-1)^nn(n-1)$$ ?
Yes.
And an alternative form for the recursive formula is $a_{n+1}=- a_n + (-1)^{n+1}\cdot 2n$.
#### mathmari
##### Well-known member
MHB Site Helper
Yes.
And an alternative form for the recursive formula is $a_{n+1}=- a_n + (-1)^{n+1}\cdot 2n$.
Why is $(-1)^{n+1}|a_n|=-a_n$ ?
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Why is $(-1)^{n+1}|a_n|=-a_n$ ?
Because $a_n$ alternates in sign.
That is, when $n$ is even, $a_n$ is positive.
And when $n$ is odd, $a_n$ is negative with a special case for $n=1$ since $a_1=0$.
#### mathmari
##### Well-known member
MHB Site Helper
Because $a_n$ alternates in sign.
That is, when $n$ is even, $a_n$ is positive.
And when $n$ is odd, $a_n$ is negative with a special case for $n=1$ since $a_1=0$.
I got stuck right now.. Do we have $a_0=0$ or $a_1=0$ ?
I mean do we have \begin{align*}&a_0=0 \\ &a_{n+1}=- a_n + (-1)^{n+1}\cdot 2n\end{align*} or \begin{align*}&a_1=0 \\ &a_{n+1}=- a_n + (-1)^{n+1}\cdot 2n\end{align*} ?
#### mathmari
##### Well-known member
MHB Site Helper
From $$a_{n+1}=- a_n + (-1)^{n+1}\cdot 2n\Rightarrow a_{n+1}+ a_n = (-1)^{n+1}\cdot 2n$$ we get the following equations $$a_{n+1}+ a_n = (-1)^{n+1}\cdot 2n \\ a_{n}+ a_{n-1} = (-1)^{n}\cdot 2(n-1) \\ a_{n-1}+ a_{n-2} = (-1)^{n-1}\cdot 2(n-2) \\ \vdots \\ a_{2}+ a_1 = (-1)^{2}\cdot 2\cdot 1$$ so we don't get an telescoping sum, do we?
Last edited:
#### Klaas van Aarsen
##### MHB Seeker
Staff member
I got stuck right now.. Do we have $a_0=0$ or $a_1=0$ ?
I mean do we have \begin{align*}&a_0=0 \\ &a_{n+1}=- a_n + (-1)^{n+1}\cdot 2n\end{align*} or \begin{align*}&a_1=0 \\ &a_{n+1}=- a_n + (-1)^{n+1}\cdot 2n\end{align*} ?
In post #1 we were given that $a_1=0$ and $a_0$ is presumably undefined.
From $$a_{n+1}=- a_n + (-1)^{n+1}\cdot 2n\Rightarrow a_{n+1}+ a_n = (-1)^{n+1}\cdot 2n$$ we get the following equations $$a_{n+1}+ a_n = (-1)^{n+1}\cdot 2n \\ a_{n}+ a_{n-1} = (-1)^{n}\cdot 2(n-1) \\ a_{n-1}+ a_{n-2} = (-1)^{n-1}\cdot 2(n-2) \\ \vdots \\ a_{2}+ a_1 = (-1)^{2}\cdot 2\cdot 1$$ so we don't get an telescoping sum, do we?
It's still a telescoping sum - after we multiply every other line with $-1$.
#### mathmari
##### Well-known member
MHB Site Helper
It's still a telescoping sum - after we multiply every other line with $-1$.
Ah ok! So on the left side we get $a_{n+1}$ but which sum do we get on the right sum?
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Ah ok! So on the left side we get $a_{n+1}$ but which sum do we get on the right sum?
Let's consider 2 cases: $n+1$ is even, and $n+1$ is odd.
And let's start with the first one ($n+1$ is even).
What will those equations looks like then? It should simplify them, shouldn't it?
#### mathmari
##### Well-known member
MHB Site Helper
Let's consider 2 cases: $n+1$ is even, and $n+1$ is odd.
And let's start with the first one ($n+1$ is even).
What will those equations looks like then? It should simplify them, shouldn't it?
If $n+1$ is even we get $$2n+2(n-1)+2(n-2)+\ldots +2\cdot 2+2\cdot 1=2\sum_{i=1}^{n}i=2\cdot \frac{n(n+1)}{2}=n(n+1)$$
If $n+1$ is odd we get $$-2n-2(n-1)-2(n-2)-\ldots -2\cdot 2-2\cdot 1=-2\sum_{i=1}^{n}i=-2\cdot \frac{n(n+1)}{2}=-n(n+1)$$
Is everything correct?
#### Klaas van Aarsen
##### MHB Seeker
Staff member
If $n+1$ is even we get $$2n+2(n-1)+2(n-2)+\ldots +2\cdot 2+2\cdot 1=2\sum_{i=1}^{n}i=2\cdot \frac{n(n+1)}{2}=n(n+1)$$
If $n+1$ is odd we get $$-2n-2(n-1)-2(n-2)-\ldots -2\cdot 2-2\cdot 1=-2\sum_{i=1}^{n}i=-2\cdot \frac{n(n+1)}{2}=-n(n+1)$$
Is everything correct?
MHB Site Helper
Thanks a lot!! | 2020-07-07T06:47:31 | {
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https://teachingcalculus.com/2018/02/23/good-question-14/?shared=email&msg=fail | # Good Question 14
Good Question 14 – The Integral Test
I have no criteria for what constitutes a “Good Question” for this series of occasional posts. They are just questions that I found interesting, or that seem more than usually instructive, or that I learn something from. I cannot quote this question (2016 BC 92) since it is on a secure exam. What made it interesting is that to answer it students pretty much needed to know the proof of the Integral Test and the figures that go with it.
I recall only one AP question from many years ago that asked students to “prove” something – usually students are asked to show that a result was true by citing the theorem that applied and showing the hypotheses were met. The directions are often “justify your answer.”
Doing an original proof is not, in my opinion, a fair question and proving some known theorem is just a matter of memorization. For these reasons, students are not asked to prove things on the exams. So, should you prove things in class? Probably, yes.
Here is the usual proof of the integral test. Afterwards I’ll discuss the question from the exam.
The Integral Test
Hypotheses: Let $f\left( x \right)$ be a function that is positive, decreasing, and continuous for $x\ge 1$ ; and let ${{a}_{n}}=f\left( n \right)$ for $x\ge 1$
In the first drawing the rectangles have a height of an and a width of 1. The area is of each is an, and the sum of their areas the series is $\sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}}$.
Part 1: Notice that $\sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}}>\int_{1}^{\infty }{{f\left( x \right)dx}}$ Assume that the improper integral $\int_{1}^{\infty }{{f\left( x \right)dx}}$ diverges.
• Conclusion 1: If the improper integral $\int_{1}^{\infty }{{f\left( x \right)dx}}$ diverges, then the series $\sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}}$ diverges.
• Conclusion 2: (The contrapositive of conclusion 1) If the series $\sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}}$ converges, then the improper integral $\int_{1}^{\infty }{{f\left( x \right)dx}}$ converges.
Part 2: In the second drawing below, Assume that the improper integral $\int_{1}^{\infty }{{f\left( x \right)dx}}$ converges. The sum of the areas of the rectangles is $\sum\limits_{{n=2}}^{\infty }{{{{a}_{n}}}}$. (NB: this series starts at n = 2.) Since $\sum\limits_{{n=2}}^{\infty }{{{{a}_{n}}}}$ is less than the convergent improper integral it will also converge. Adding ${{a}_{1}}$ to this gives the original series, $\sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}}$; this series also converges.
• Conclusion 3: If the improper integral $\int_{1}^{\infty }{{f\left( x \right)dx}}$ converges, then the series $\sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}}$ converges.
• Conclusion 4: (The contrapositive of conclusion 3) If the series $\sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}}$ diverges, then the improper integral $\int_{1}^{\infty }{{f\left( x \right)dx}}$ diverges.
Putting the four conclusions together is the Integral Test: If the hypotheses above are met, then the series and the improper integral will both converge, or both diverge.
To answer the multiple-choice question (2106 BC 92) on the exams students were told that the improper integral converges. Therefore, the associated series converges. They then had to determine whether the series or the improper integral has the greater value. Stop here and see if you can figure that out.
Return to the first figure above only this time assume that the improper integral and the series converge. It is pretty obvious that $\sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}}>\int_{1}^{\infty }{{f\left( x \right)dx}}$.
So, even though students were not asked to prove anything, a familiarity with the proof and its figures is necessary to answer the question. That’s why I liked it,
On the other hand, it is kind of an obscure point and I’m not sure it has any practical value.
## 2 thoughts on “Good Question 14”
1. Why couldn’t you refer to your second figure, saying that because the improper integral converged, then the series converged, and therefore the integral > series? It sounds like the question was worded this way. Is it because the first term is omitted from the series in your second figure?
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• Yes, it is because in the second figure the first rectangle on the left is $\displaystyle {{a}_{2}}$, so the entire series is not shown. The rectangle for $\displaystyle {{a}_{1}}$ extends to the left between x = 1 and the y-axis. You cannot tell (for sure) from this that the series is greater than the integral. From the first figure (with the new assumption that the series and integral both converge) you can see that the area representing the series is greater than the area represented by the integral.
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https://math.stackexchange.com/questions/1866053/negation-of-some-logic-statement | # Negation of “some” logic statement
I need to negate the following statement : "Some integers are not odd"
I have the below, where O(x) is "odd" $$\exists x (\neg O(x))$$
Would the negation be $$\forall x (O(x))$$
I'm confused, since if so, this statement does not make sense since there can be even numbers, does the negation have to be real?
• Why doesn't it make sense? The original statement is true (some integers are not odd) so it makes sense that the negation of that original statement becomes false. – JMoravitz Jul 21 '16 at 3:12
• I guess I misunderstood the concept of the negation, is the negated statement correct? "for all x, x is odd" – splinks Jul 21 '16 at 3:17
• Sensible statements can be false. "The sky is orange" makes sense but is false. "The sky is not orange" makes sense and is true. "The sky is Tuesday" doesn't make sense. The negation of a sensible true statement needs to be a sensible false statement, and vice versa. – Graham Kemp Jul 21 '16 at 3:31
As alluded to in Graham Kemp's comment, you seem to be mixing up true statements with meaningful statements. Meaningful statements are sometimes called syntactically correct statements, and can be either true or false.
The negation of a true statement is false, and the negation of a false statement is true. So it stands to reason that, when you negated $\color{blue}{\exists x \; \lnot O(x)}$ ("some integers are not odd"), a true statement, you got $\color{red}{\forall x O(x)}$ ("all integers are odd"), a false statement.
On the other hand, the negation of a meaningful statement is meaningful. In this case, "some integers are not odd" and "all integers are odd" are both meaningful things to say (they make sense), even if one of them is completely wrong.
Don't worry about it. Sensible statements can be either true or false. "The sky is orange" makes sense but is false. "The sky is not orange" makes sense and is true. "The sky is Tuesday" doesn't make sense.
In the domain of integers, $\exists x~\neg O(x)$ means "There exists an integer which is not odd." This statement is true.
The negation of that statement is indeed, $\forall x~O(x)$ , which means "Every integer is odd." This statement is false, as it should be.
We expect the negation of a true statement to be a false statement. That is what negation means.
$$\neg \exists x~{\neg O(x)}~ \iff ~\forall x~{O(x)}$$ | 2019-09-24T08:34:19 | {
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http://math.stackexchange.com/questions/351376/cartesian-products-represented-as-disjoint-unions | # Cartesian products represented as disjoint unions
Assume $k$ finite sets $S_1, \ldots, S_k$ and their Cartesian product $S = S_1 \times \ldots \times S_k$.
Suppose I remove $n$ (distinct) elements from $S$, and let us call that smaller set $S'$. I am interested in representing $S'$ as a union of distinct Cartesian products.
Is there an upper bound, in terms of $k$ and $n$, on the numbers of distinct products I may need in such a union?
As a simple example, consider $S_1 = \{ a, b, c \}$, $S_2 = \{ 1, 2, 3 \}$, $S = S_1 \times S_2$.
If I take $S' = S \setminus \{ (b, 2), (b, 3) \}$, I can write it as $S' = \left( \{ a, c \} \times \{ 1, 2, 3 \} \right) \cup \left(\{ b \} \times \{ 1 \} \right)$. In this case, I used two products. Had I removed $(c,3)$ instead of $(b,3)$, I believe I would have needed at least three.
(The reason I am asking for a bound in terms of $k$ and $n$ is that a simple geometric interpretation makes me believe the sizes of the sets $S_i$ are irrelevant — though they may come into play when the number of removed elements approaches the size of $S$. Please correct me if I am wrong.)
-
Let the elements we remove be denoted by $s^j=(s_1^j,s_2^j,\ldots,s_k^j)$ for $j=1,2,\ldots,n$. We are interested in the set $$(S_1\times S_2\times\ldots\times S_k)\setminus\{s^1,s^2,\ldots,s^n\}.$$
Claim. A (possibly very crude) upper bound is given by $\frac{n^k-1}{n-1}$.
Proof. We will prove this by induction on $k$.
For $k=1$, there is no problem. The set $S_1\setminus\{s^1,s^2,\ldots,s^n\}$ is a Cartesian product with one factor, so we are done: the upper bound is $1$ in this case.
For $k\to k+1$, note that we can rewrite the set as follows: \begin{align}&(S_1\times S_2\times\ldots\times S_{k+1})\setminus\{s^1,s^2,\ldots,s^n\}=\\&=(S_1\times S_2\times\ldots\times S_{k+1})\setminus\{s^1,s^2,\ldots,s^n\}\\&=\bigcup_{x\in S_{k+1}}(S_1\times S_2\times\ldots\times S_k\times\{x\})\setminus\{s^1,s^2,\ldots,s^n\}\\&=\left(\bigcup_{j=1}^n(S_1\times S_2\times\ldots\times S_k\times\{s_{k+1}^j\})\setminus\{s^1,s^2,\ldots,s^n\}\right)\cup(S_1\times S_2\times\ldots\times (S_{k+1}\setminus\{s_{k+1}^1,\ldots,s_{k+1}^n\})),\end{align} which is (if we omit the sets that appear more than once in the first union) a disjoint union of at most $n$ sets of the form $$(S_1\times S_2\times\ldots\times S_k\times\{s_{k+1}^j\})\setminus\{s^1,s^2,\ldots,s^n\},$$ and a single Cartesian product of $k+1$ sets. By the induction hypothesis, each set $$(S_1\times S_2\times\ldots\times S_k\times\{s_{k+1}^j\})\setminus\{s^1,s^2,\ldots,s^n\}$$ can be written as a disjoint union of at most $\frac{n^k-1}{n-1}$ Cartesian products. Therefore our set can be written as a disjoint union of at most $n\frac{n^k-1}{n-1}+1 = \frac{n^{k+1}-1}{n-1}$ Cartesian products (with $k+1$ factors each, of course). $\square$
Note that finding better bounds might require more sophisticated combinatorial methods.
-
Ummm ... In case $n=1$, $\frac{n^k-1}{n-1}$ should be interpreted as $k$. – Dejan Govc Apr 4 '13 at 21:14
Thanks for your answer. It got me thinking a little more, and I wonder now if I cannot get to $k{\cdot}n$, by representing $S$ as $((S_1 \setminus \{ s^1 \}) \times \ldots \times S_k) \cup (\{ s^1 \} \times \ldots \times S_k)$, then splitting $S_2$ in that second member, etc. In fact, I now suspect this could work for $S \setminus X$, where $X$ is an arbitrary Cartesian product from subsets of $S_1, \ldots, S_k$... Maybe someone can work a full answer, or I'll try myself. – Philippe Apr 5 '13 at 11:14 | 2014-10-25T04:34:00 | {
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https://www.physicsforums.com/threads/fitting-a-function-given-3-points.403663/ | # Fitting a Function given 3 points
1. May 16, 2010
### shyboyswin
1. The problem statement, all variables and given/known data
Using a simple "trick," you can find a polynomial function whose graphi goes through any given set of points. For example, the graph of the following function goes through the points (1, 3), (2, 5), and (3, -1).
$$f(x) = 3\;\frac{(x-2)(x-3)}{(1-2)(1-3)} \;+\; 5\frac{(x-1)(x-3)}{(2-1)(2-3)} \;+\;(-1) \frac{(x-1)(x-2)}{(3-1)(3-2)}$$
Find a polynomial function g that satisfies g(-2) = 1, g(0) = 3, g(1) = -2 and g(4) = 3
2. Relevant equations
(Above)
3. The attempt at a solution
$$g(x) = (1)\frac{(x-1)(x-4)}{(-2-1)(-2-4)} \; + \;(-2)\frac{(x+2)(x-4)}{(1+2)(1-4)} \;+ \;3\frac{(x+2)(x-1)}{(4+1)(4-1)}$$
I cannot figure out how to produce g(0) = 3.
2. May 16, 2010
### D H
Staff Emeritus
Why are you making your polynomial a quadratic? Perhaps it should be something else -- like a cubic.
3. May 16, 2010
### The Chaz
Hint: each "term" should have more factors in parenthesis
4. May 17, 2010
### HallsofIvy
Staff Emeritus
As both DH and Chaz say, you need four terms, not three, and you need three factors in numerator and denominator of each term.
A polynomial passing through f(a)= t, f(b)= u, f(c)= v, and f(d)= w must look like
$$t\frac{(x- b)(x- c)(x- d)}{(a- b)(a- c)(a- d)}+$$$$u\frac{(x- a)(x- c)(x- d)}{(b- a)(b- c)(b- d)}+$$$$v\frac{(x- a)(x- b)(x- d)}{(c- a)(c- b)(c- d)}$$$$+ w\frac{(x- a)(x- b)(x- c)}{(d- a)(d- b)(d- c)}$$
Look at what happens at, say, x= c. Every fraction except the third has a factor of x- c in the numerator and so will be 0 at x= c. The third fraction will have, at x= c, (c- a)(c- b)(c- d) in both numerator and denominator and so will be 1. The entire polynomal will be f(c)= 0+ 0+ v(1)+ 0= v.
This is the "Lagrange interpolation formula"- to write a polynomial that passes through n given points, you need a sum of n fractions, each having n-1 factors in numerator and denominator.
5. May 19, 2010
### Unit
I have never heard of this formula, but it is so clever! I love it
As for helping OP, all I can say is, look for the pattern, and, once you've got it, be meticulous in your bookkeeping. | 2017-10-18T21:20:38 | {
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https://www.physicsforums.com/threads/what-is-the-name-of-this-method.638135/ | # What is the name of this method?
1. Sep 23, 2012
### Orion1
1. The problem statement, all variables and given/known data
Integrate:
$$\int \frac{1}{x^3 - 27} dx$$
2. Relevant equations
$$\int \frac{1}{x^3 - 27} dx$$
$$\frac{1}{x^3 - 27} = \frac{1}{(x - 3)(x^2 + 3x + 9)} = \frac{A}{x - 3} + \frac{Bx + C}{x^2 + 3x + 9}$$
$$1 = A(x^2 + 3x + 9) + (Bx + C)(x - 3)$$
3. The attempt at a solution
The first step involves factoring and splitting the denominator with some method:
$$\frac{1}{x^3 - 27} = \frac{1}{(x - 3)(x^2 + 3x + 9)} = \frac{A}{x - 3} + \frac{Bx + C}{x^2 + 3x + 9}$$
What is the name of this method?
The next step involves normalizing the equation:
$$1 = A(x^2 + 3x + 9) + (Bx + C)(x - 3)$$
How was the coefficients of B and C derived and solved?
The solution only states that:
$$A = \frac{1}{27}$$
$$0 = \frac{1}{27} + B$$
$$1 = \frac{1}{3} + 3C$$
And:
$B = -\frac{1}{27}$ and $C = -\frac{2}{9}$
I can see how the $A$ coefficient was derived from the x-axis zero intercept, however, it is not entirely clear to me how these other coefficients were derived and solved?
Reference:
Derivation of equation and solution
Last edited: Sep 23, 2012
2. Sep 23, 2012
### CAF123
This is the method of partial fractions to reduce $$\int \frac{1}{x^3 -27} dx$$ into something which can be integrated.
To find A,B and C you will need to solve simultaneous equations. To easily solve for A, let x =3 and you should get the required A = 1/27. To find B and C, let x = 0 to get one eqn in B and C and let x = 1 to get another. Two eqns, two unknowns (B and C) - you can solve this. The choice of x here is arbritary (except x =3 since the (Bx +C) term will vanish).
Last edited: Sep 23, 2012
3. Sep 23, 2012
### HallsofIvy
Staff Emeritus
Actually, because the last term in involved "Bx- C", and you have already found A, setting x= 0 will give an equation in C only.
4. Sep 23, 2012
### CAF123
Yes, I overlooked this. If you have something like $$\frac{A}{x} + \frac{(Bx +c)}{x^2 +bx +c},$$ then you will probably have to use simultaneous eqns.
5. Sep 23, 2012
### Orion1
Solving the coefficients:
Given:
$$1 = A(x^2 + 3x + 9) + (Bx + C)(x - 3)$$
Set x = 3 to zero out the $(Bx + C)$ term and numerically integrate via substitution:
$$1 = A(3^2 + 3(3) + 9) + (B(3) + C)(3 - 3) = A(27)$$
$$1 = A(27)$$
Solve for $A$:
$$\boxed{A = \frac{1}{27}}$$
Set x = 0 to zero out the $Bx$ term and numerically integrate via substitution:
$$1 = \frac{(0^2 + 3(0) + 9)}{27} + (B(0) + C)(0 - 3) = \frac{9}{27} - 3C$$
$$1 = \frac{9}{27} - 3C$$
$$-\frac{27}{27} + \frac{9}{27} = -\frac{18}{27} = -\frac{2}{3} = 3C$$
Solve for $C$:
$$\boxed{C = -\frac{2}{9}}$$
Set x = 1 as arbitrary and numerically integrate via substitution:
$$1 = \frac{(1^2 + 3(1) + 9)}{27} + \left(B(1) - \frac{2}{9} \right)(1 - 3) = \frac{13}{27} - 2B + \frac{3}{3} \times \frac{4}{9} = \frac{13}{27} - 2B + \frac{12}{27} = \frac{25}{27} - 2B$$
$$1 = \frac{25}{27} - 2B$$
$$-\frac{27}{27} + \frac{25}{27} = -\frac{2}{27} = 2B$$
Solve for $B$:
$$\boxed{B = -\frac{1}{27}}$$
Last edited: Sep 23, 2012 | 2017-08-18T09:07:28 | {
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https://casmusings.wordpress.com/tag/derivative/ | # Tag Archives: derivative
## Many Roads Give Same Derivative
A recent post in the AP Calculus Community expressed some confusion about different ways to compute $\displaystyle \frac{dy}{dx}$ at (0,4) for the function $x=2ln(y-3)$. I share below the two approaches suggested in the original post, proffer two more, and a slightly more in-depth activity I’ve used in my calculus classes for years. I conclude with an alternative to derivatives of inverses.
### Two Approaches Initially Proposed
1 – Accept the function as posed and differentiate implicitly.
$\displaystyle \frac{d}{dx} \left( x = 2 ln(y-3) \right)$
$\displaystyle 1 = 2*\frac{1}{y-3} * \frac{dy}{dx}$
$\displaystyle \frac{dy}{dx} = \frac{y-3}{2}$
Which gives $\displaystyle \frac{dy}{dx} = \frac{1}{2}$ at (x,y)=(0,4).
2 – Solve for y and differentiate explicitly.
$\displaystyle x = 2ln(y-3) \longrightarrow y = 3 + e^{x/2}$
$\displaystyle \frac{dy}{dx} = e^{x/2} * \frac{1}{2}$
Evaluating this at (x,y)=(0,4) gives $\displaystyle \frac{dy}{dx} = \frac{1}{2}$ .
### Two Alternative Approaches
3 – Substitute early.
The question never asked for an algebraic expression of $\frac{dy}{dx}$, only the numerical value of this slope. Because students tend to make more silly mistakes manipulating algebraic expressions than numeric ones, the additional algebra steps are unnecessary, and potentially error-prone. Admittedly, the manipulations are pretty straightforward here, in more algebraically complicated cases, early substitutions could significantly simplify work. Using approach #1 and substituting directly into the second line gives
$\displaystyle 1 = 2 * \frac{1}{y-3} * \frac{dy}{dx}$.
At (x,y)=(0,4), this is
$\displaystyle 1 = 2 * \frac{1}{4-3}*\frac{dy}{dx}$
$\displaystyle \frac{dy}{dx} = \frac{1}{2}$
The numeric manipulations on the right side are obviously easier than the earlier algebra.
4 – Solve for $\frac{dx}{dy}$ and reciprocate.
There’s nothing sacred about solving for $\frac{dy}{dx}$ directly. Why not compute the derivative of the inverse and reciprocate at the end? Differentiating first with respect to y eventually leads to the same solution.
$\displaystyle \frac{d}{dy} \left( x = 2 ln(y-3) \right)$
$\displaystyle \frac{dx}{dy} = 2 * \frac{1}{y-3}$
At (x,y)=(0,4), this is
$\displaystyle \frac{dx}{dy} = \frac{2}{4-3} = 2$, so
$\displaystyle \frac{dy}{dx} = \frac{1}{2}$.
### Equivalence = A fundamental mathematical concept
I sometimes wonder if teachers should place much more emphasis on equivalence. We spend so much time manipulating expressions in mathematics classes at all levels, changing mathematical objects (shapes, expressions, equations, etc.) into a different, but equivalent objects. Many times, these manipulations are completed under the guise of “simplification.” (Here is a brilliant Dan Teague post cautioning against taking this idea too far.)
But it is critical for students to recognize that proper application of manipulations creates equivalent expressions, even if when the resulting expressions don’t look the same. The reason we manipulate mathematical objects is to discover features about the object in one form that may not be immediately obvious in another.
For the function $x = 2 ln(y-3)$, the slope at (0,4) must be the same, no matter how that slope is calculated. If you get a different looking answer while using correct manipulations, the final answers must be equivalent.
### Another Example
A similar question appeared on the AP Calculus email list-server almost a decade ago right at the moment I was introducing implicit differentiation. A teacher had tried to find $\displaystyle \frac{dy}{dx}$ for
$\displaystyle x^2 = \frac{x+y}{x-y}$
using implicit differentiation on the quotient, manipulating to a product before using implicit differentiation, and finally solving for y in terms of x to use an explicit derivative.
1 – Implicit on a quotient
Take the derivative as given:$$\displaystyle \frac{d}{dx} \left( x^2 = \frac{x+y}{x-y} \right)$ $\displaystyle 2x = \frac{(x-y) \left( 1 + \frac{dy}{dx} \right) - (x+y) \left( 1 - \frac{dy}{dx} \right) }{(x-y)^2}$ $\displaystyle 2x * (x-y)^2 = (x-y) + (x-y)*\frac{dy}{dx} - (x+y) + (x+y)*\frac{dy}{dx}$ $\displaystyle 2x * (x-y)^2 = -2y + 2x * \frac{dy}{dx}$ $\displaystyle \frac{dy}{dx} = \frac{-2x * (x-y)^2 + 2y}{2x}$ 2 – Implicit on a product Multiplying the original equation by its denominator gives $x^2 * (x - y) = x + y$ . Differentiating with respect to x gives $\displaystyle 2x * (x - y) + x^2 * \left( 1 - \frac{dy}{dx} \right) = 1 + \frac{dy}{dx}$ $\displaystyle 2x * (x-y) + x^2 - 1 = x^2 * \frac{dy}{dx} + \frac{dy}{dx}$ $\displaystyle \frac{dy}{dx} = \frac{2x * (x-y) + x^2 - 1}{x^2 + 1}$ 3 – Explicit Solving the equation at the start of method 2 for y gives $\displaystyle y = \frac{x^3 - x}{x^2 + 1}$. Differentiating with respect to x gives $\displaystyle \frac{dy}{dx} = \frac {\left( x^2+1 \right) \left( 3x^2 - 1\right) - \left( x^3 - x \right) (2x+0)}{\left( x^2 + 1 \right) ^2}$ Equivalence Those 3 forms of the derivative look VERY DIFFERENT. Assuming no errors in the algebra, they MUST be equivalent because they are nothing more than the same derivative of different forms of the same function, and a function’s rate of change doesn’t vary just because you alter the look of its algebraic representation. Substituting the y-as-a-function-of-x equation from method 3 into the first two derivative forms converts all three into functions of x. Lots of by-hand algebra or a quick check on a CAS establishes the suspected equivalence. Here’s my TI-Nspire CAS check. Here’s the form of this investigation I gave my students. ### Final Example I’m not a big fan of memorizing anything without a VERY GOOD reason. My teachers telling me to do so never held much weight for me. I memorized as little as possible and used that information as long as I could until a scenario arose to convince me to memorize more. One thing I managed to avoid almost completely were the annoying derivative formulas for inverse trig functions. For example, find the derivative of $y = arcsin(x)$ at $x = \frac{1}{2}$. Since arc-trig functions annoy me, I always rewrite them. Taking sine of both sides and then differentiating with respect to x gives. $sin(y) = x$ $\displaystyle cos(y) * \frac{dy}{dx} = 1$ I could rewrite this equation to give $\frac{dy}{dx} = \frac{1}{cos(y)}$, a perfectly reasonable form of the derivative, albeit as a less-common expression in terms of y. But I don’t even do that unnecessary algebra. From the original function, $x=\frac{1}{2} \longrightarrow y=\frac{\pi}{6}$, and I substitute that immediately after the differentiation step to give a much cleaner numeric route to my answer. $\displaystyle cos \left( \frac{\pi}{6} \right) * \frac{dy}{dx} = 1$ $\displaystyle \frac{\sqrt{3}}{2} * \frac{dy}{dx} = 1$ $\displaystyle \frac{dy}{dx} = \frac{2}{\sqrt{3}}$ And this is the same result as plugging $x = \frac{1}{2}$ into the memorized version form of the derivative of arcsine. If you like memorizing, go ahead, but my mind remains more nimble and less cluttered. One final equivalent approach would have been differentiating $sin(y) = x$ with respect to y and reciprocating at the end. ### CONCLUSION There are MANY ways to compute derivatives. For any problem or scenario, use the one that makes sense or is computationally easiest for YOU. If your resulting algebra is correct, you know you have a correct answer, even if it looks different. Be strong! Advertisements ## Base-x Numbers and Infinite Series In my previous post, I explored what happened when you converted a polynomial from its variable form into a base-x numerical form. That is, what are the computational implications when polynomial $3x^3-11x^2+2$ is represented by the base-x number $3(-11)02_x$, where the parentheses are used to hold the base-x digit, -11, for the second power of x? So far, I’ve explored only the Natural number equivalents of base-x numbers. In this post, I explore what happens when you allow division to extend base-x numbers into their Rational number counterparts. Level 5–Infinite Series: Numbers can have decimals, so what’s the equivalence for base-x numbers? For starters, I considered trying to get a “decimal” form of $\displaystyle \frac{1}{x+2}$. It was “obvious” to me that $12_x$ won’t divide into $1_x$. There are too few “places”, so some form of decimals are required. Employing division as described in my previous post somewhat like you would to determine the rational number decimals of $\frac{1}{12}$ gives Remember, the places are powers of x, so the decimal portion of $\displaystyle \frac{1}{x+2}$ is $0.1(-2)4(-8)..._x$, and it is equivalent to $\displaystyle 1x^{-1}-2x^{-2}+4x^{-3}-8x^{-4}+...=\frac{1}{x}-\frac{2}{x^2}+\frac{4}{x^3}-\frac{8}{x^4}+...$. This can be seen as a geometric series with first term $\displaystyle \frac{1}{x}$ and ratio $\displaystyle r=\frac{-2}{x}$. It’s infinite sum is therefore $\displaystyle \frac{\frac{1}{x}}{1-\frac{-2}{x}}$ which is equivalent to $\displaystyle \frac{1}{x+2}$, confirming the division computation. Of course, as a geometric series, this is true only so long as $\displaystyle |r|=\left | \frac{-2}{x} \right |<1$, or $2<|x|$. I thought this was pretty cool, and it led to lots of other cool series. For example, if $x=8$,you get $\frac{1}{10}=\frac{1}{8}-\frac{2}{64}+\frac{4}{512}-...$. Likewise, $x=3$ gives $\frac{1}{5}=\frac{1}{3}-\frac{2}{9}+\frac{4}{27}-\frac{8}{81}+...$. I found it quite interesting to have a “polynomial” defined with a rational expression. Boundary Convergence: As shown above, $\displaystyle \frac{1}{x+2}=\frac{1}{x}-\frac{2}{x^2}+\frac{4}{x^3}-\frac{8}{x^4}+...$ only for $|x|>2$. At $x=2$, the series is obviously divergent, $\displaystyle \frac{1}{4} \ne \frac{1}{2}-\frac{2}{4}+\frac{4}{8}-\frac{8}{16}+...$. For $x=-2$, I got $\displaystyle \frac{1}{0} = \frac{1}{-2}-\frac{2}{4}+\frac{4}{-8}-\frac{8}{16}+...=-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}-...$ which is properly equivalent to $-\infty$ as $x \rightarrow -2$ as defined by the convergence domain and the graphical behavior of $\displaystyle y=\frac{1}{x+2}$ just to the left of $x=-2$. Nice. I did find it curious, though, that $\displaystyle \frac{1}{x}-\frac{2}{x^2}+\frac{4}{x^3}-\frac{8}{x^4}+...$ is a solid approximation for $\displaystyle \frac{1}{x+2}$ to the left of its vertical asymptote, but not for its rotationally symmetric right side. I also thought it philosophically strange (even though I understand mathematically why it must be) that this series could approximate function behavior near a vertical asymptote, but not near the graph’s stable and flat portion near $x=0$. What a curious, asymmetrical approximator. Maclaurin Series: Some quick calculus gives the Maclaurin series for $\displaystyle \frac{1}{x+2}$ : $\displaystyle \frac{1}{2}-\frac{x}{4}+\frac{x^2}{8}-\frac{x^3}{16}+...$, a geometric series with first term $\frac{1}{2}$ and ratio $\frac{-x}{2}$. Interestingly, the ratio emerging from the Maclaurin series is the reciprocal of the ratio from the “rational polynomial” resulting from the base-x division above. As a geometric series, the interval of convergence is $\displaystyle |r|=\left | \frac{-x}{2} \right |<1$, or $|x|<2$. Excluding endpoint results, the Maclaurin interval is the complete Real number complement to the base-x series. For the endpoints, $x=-2$ produces the right-side vertical asymptote divergence to $+ \infty$ that $x=-2$ did for the left side of the vertical asymptote in the base-x series. Again, $x=2$ is divergent. It’s lovely how these two series so completely complement each other to create clean approximations of $\displaystyle \frac{1}{x+2}$ for all $x \ne 2$. Other base-x “rational numbers” Because any polynomial divided by another is absolutely equivalent to a base-x rational number and thereby a base-x decimal number, it will always be possible to create a “rational polynomial” using powers of $\displaystyle \frac{1}{x}$ for non-zero denominators. But, the decimal patterns of rational base-x numbers don’t apply in the same way as for Natural number bases. Where $\displaystyle \frac{1}{12}$ is guaranteed to have a repeating decimal pattern, the decimal form of $\displaystyle \frac{1}{x+2}=\frac{1_x}{12_x}=0.1(-2)4(-8)..._x$ clearly will not repeat. I’ve not explored the full potential of this, but it seems like another interesting field. CONCLUSIONS and QUESTIONS Once number bases are understood, I’d argue that using base-x multiplication might be, and base-x division definitely is, a cleaner way to compute products and quotients, respectively, for polynomials. The base-x division algorithm clearly is accessible to Algebra II students, and even opens the doors to studying series approximations to functions long before calculus. Is there a convenient way to use base-x numbers to represent horizontal translations as cleanly as polynomials? How difficult would it be to work with a base-$(x-h)$ number for a polynomial translated h units horizontally? As a calculus extension, what would happen if you tried employing division of non-polynomials by replacing them with their Taylor series equivalents? I’ve played a little with proving some trig identities using base-x polynomials from the Maclaurin series for sine and cosine. What would happen if you tried to compute repeated fractions in base-x? It’s an open question from my perspective when decimal patterns might terminate or repeat when evaluating base-x rational numbers. I’d love to see someone out there give some of these questions a run! ## Calculus Derivative Rules Over the past few days I’ve been rethinking my sequencing of introducing derivative rules for the next time I teach calculus. The impetus for this was an approach I encountered in a Coursera MOOC in Calculus I’m taking this summer to see how a professor would run a Taylor Series-centered calculus class. Historically, I’ve introduced my high school calculus classes to the product and quotient rules before turing to the chain rule. I’m now convinced the chain rule should be first because of how beautifully it sets up the other two. Why the chain rule should be first Assuming you know the chain rule, check out these derivations of the product and quotient rules. For each of these, $g_1$ and $g_2$ can be any differentiable functions of x. PRODUCT RULE: Let $P(x)=g_1(x) \cdot g_2(x)$. Applying a logarithm gives, $ln(P)=ln \left( g_1 \cdot g_2 \right) = ln(g_1)+ln(g_2)$. Now differentiate and rearrange. $\displaystyle \frac{P'}{P} = \frac{g_1'}{g_1}+\frac{g_2'}{g_2}$ $\displaystyle P' = P \cdot \left( \frac{g_1'}{g_1}+\frac{g_2'}{g_2} \right)$ $\displaystyle P' = g_1 \cdot g_2 \cdot \left( \frac{g_1'}{g_1}+\frac{g_2'}{g_2} \right)$ $P' = g_1' \cdot g_2+g_1 \cdot g_2'$ QUOTIENT RULE: Let $Q(x)=\displaystyle \frac{g_1(x)}{g_2(x)}$. As before, apply a logarithm, differentiate, and rearrange. $\displaystyle ln(Q)=ln \left( \frac{g_1}{g_2} \right) = ln(g_1)-ln(g_2)$ $\displaystyle \frac{Q'}{Q} = \frac{g_1'}{g_1}-\frac{g_2'}{g_2}$ $\displaystyle Q' = Q \cdot \left( \frac{g_1'}{g_1}-\frac{g_2'}{g_2} \right)$ $\displaystyle Q' = \frac{g_1}{g_2} \cdot \left( \frac{g_1'}{g_1}-\frac{g_2'}{g_2} \right)$ $\displaystyle Q' = \frac{g_1'}{g_2}-\frac{g_1 \cdot g_2'}{\left( g_2 \right)^2} = \frac{g_1'g_2-g_1g_2'}{\left( g_2 \right)^2}$ The exact same procedure creates both rules. (I should have seen this long ago.) Proposed sequencing I’ve always emphasized the Chain Rule as the critical algebra manipulation rule for calculus students, but this approach makes it the only rule required. That completely fits into my overall teaching philosophy: learn a limited set of central ideas and use them as often as possible. With this, I’ll still introduce power, exponential, sine, and cosine derivative rules first, but then I’ll follow with the chain rule. After that, I think everything else required for high school calculus will be a variation on what is already known. That’s a lovely bit of simplification. I need to rethink my course sequencing, but I think it’ll be worth it. ## Polar Derivatives on TI-Nspire CAS The following question about how to compute derivatives of polar functions was posted on the College Board’s AP Calculus Community bulletin board today. From what I can tell, there are no direct ways to get derivative values for polar functions. There are two ways I imagined to get the polar derivative value, one graphically and the other CAS-powered. The CAS approach is much more accurate, especially in locations where the value of the derivative changes quickly, but I don’t think it’s necessarily more intuitive unless you’re comfortable using CAS commands. For an example, I’ll use $r=2+3sin(\theta )$ and assume you want the derivative at $\theta = \frac{\pi }{6}$. METHOD 1: Graphical Remember that a derivative at a point is the slope of the tangent line to the curve at that point. So, finding an equation of a tangent line to the polar curve at the point of interest should find the desired result. Create a graphing window and enter your polar equation (menu –> 3:Graph Entry –> 4:Polar). Then drop a tangent line on the polar curve (menu –> 8:Geometry –> 1:Points&Lines –> 7:Tangent). You would then click on the polar curve once to select the curve and a second time to place the tangent line. Then press ESC to exit the Tangent Line command. To get the current coordinates of the point and the equation of the tangent line, use the Coordinates & Equation tool (menu –> 1:Actions –> 8:Coordinates and Equations). Click on the point and the line to get the current location’s information. After each click, you’ll need to click again to tell the nSpire where you want the information displayed. To get the tangent line at $\theta =\frac{\pi }{6}$, you could drag the point, but the graph settings seem to produce only Cartesian coordinates. Converting $\theta =\frac{\pi }{6}$ on $r=2+3sin(\theta )$ to Cartesian gives $\left( x,y \right) = \left( r \cdot cos(\theta ), r \cdot sin(\theta ) \right)=\left( \frac{7\sqrt{3}}{4},\frac{7}{4} \right)$ . So the x-coordinate is $\frac{7\sqrt{3}}{4} \approx 3.031$. Drag the point to find the approximate slope, $\frac{dy}{dx} \approx 8.37$. Because the slope of the tangent line changes rapidly at this location on this polar curve, this value of 8.37 will be shown in the next method to be a bit off. Unfortunately, I tried to double-click the x-coordinate to set it to exactly $\frac{7\sqrt{3}}{4}$, but that property is also disabled in polar mode. METHOD 2: CAS Using the Chain Rule, $\displaystyle \frac{dy}{dx} = \frac{dy}{d\theta }\cdot \frac{d\theta }{dx} = \frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}$. I can use this and the nSpire’s ability to define user-created functions to create a $\displaystyle \frac{dy}{dx}$ polar differentiator for any polar function $r=a(\theta )$. On a Calculator page, I use the Define function (menu –> 1:Actions –> 1:Define) to make the polar differentiator. All you need to do is enter the expression for a as shown in line 2 below. This can be evaluated exactly or approximately at $\theta=\frac{\pi }{6}$ to show $\displaystyle \frac{dy}{dx} = 5\sqrt{3}=\approx 8.660$. Conclusion: As with all technologies, getting the answers you want often boils down to learning what questions to ask and how to phrase them. ## Controlling graphs and a free online calculator When graphing functions with multiple local features, I often find myself wanting to explain a portion of the graph’s behavior independent of the rest of the graph. When I started teaching a couple decades ago, the processor on my TI-81 was slow enough that I could actually watch the pixels light up sequentially. I could see HOW the graph was formed. Today, processors obviously are much faster. I love the problem-solving power that has given my students and me, but I’ve sometimes missed being able to see function graphs as they develop. Below, I describe the origins of the graph control idea, how the control works, and then provide examples of polynomials with multiple roots, rational functions with multiple intercepts and/or vertical asymptotes, polar functions, parametric collision modeling, and graphing derivatives of given curves. BACKGROUND: A colleague and I were planning a rational function unit after school last week wanting to be able to create graphs in pieces so that we could discuss the effect of each local feature. In the past, we “rigged” calculator images by graphing the functions parametrically and controlling the input values of t. Clunky and static, but it gave us useful still shots. Nice enough, but we really wanted something dynamic. Because we had the use of sliders on our TI-nSpire software, on Geogebra, and on the Desmos calculator, the solution we sought was closer than we suspected. REALIZATION & WHY IT WORKS: Last week, we discovered that we could use $g(x)=\sqrt \frac{\left | x \right |}{x}$ to create what we wanted. The argument of the root is 1 for $x<0$, making $g(x)=1$. For $x>0$, the root’s argument is -1, making $g(x)=i$, a non-real number. Our insight was that multiplying any function $y=f(x)$ by an appropriate version of g wouldn’t change the output of f if the input to g is positive, but would make the product ungraphable due to complex values if the input to g is negative. If I make a slider for parameter a, then $g_2(x)=\sqrt \frac{\left | a-x \right |}{a-x}$ will have output 1 for all $x. That means for any function $y=f(x)$ with real outputs only, $y=f(x)\cdot g_2(x)$ will have real outputs (and a real graph) for $x only. Aha! Using a slider and $g_2$ would allow me to control the appearance of my graph from left to right. NOTE: While it’s still developing, I’ve become a big fan of the free online Desmos calculator after a recent presentation at the Global Math Department (join our 45-60 minute online meetings every Tuesday at 9PM ET!). I use Desmos for all of the following graphs in this post, but obviously any graphing software with slider capabilities would do. EXAMPLE 1: Graph $y=(x+2)^3x^2(x-1)$, a 6th degree polynomial whose end behavior is up for $\pm \infty$, “wiggles” through the x-axis at -2, then bounces off the origin, and finally passes through the x-axis at 1. Click here to access the Desmos graph that created the image above. You can then manipulate the slider to watch the graph wiggle through, then bounce off, and finally pass through the x-axis. EXAMPLE 2: Graph $y=\frac{(x+1)^2}{(x+2)(x-1)^2}$, a 6th degree polynomial whose end behavior is up for $\pm \infty$, “wiggles” through the x-axis at -2, then bounces off the origin, and finally passes through the x-axis at 1. Click here to access the Desmos graph above and control the creation of the rational function’s graph using a slider. EXAMPLE 3: I believe students understand polar graphing better when they see curves like the limacon $r=2+3cos(\theta )$ moving between its maximum and minimum circles. Controlling the slider also allows users to see the values of $\theta$ at which the limacon crosses the pole. Here is the Desmos graph for the graph below. EXAMPLE 4: Object A leaves (2,3) and travels south at 0.29 units/second. Object B leaves (-2,1) traveling east at 0.45 units/second. The intersection of their paths is (2,1), but which object arrives there first? Here is the live version. OK, I know this is an overly simplistic example, but you’ll get the idea of how the controlling slider works on a parametrically-defined function. The$latex \sqrt{\frac{\left | a-x \right |}{a-x}}\$ term only needs to be on one of parametric equations. Another benefit of the slider approach is the ease with which users can identify the value of t (or time) when each particle reaches the point of intersection or their axes intercepts. Obviously those values could be algebraically determined in this problem, but that isn’t always true, and this graphical-numeric approach always gives an alternative to algebraic techniques when investigating parametric functions.
ASIDE 1–Notice the ease of the Desmos notation for parametric graphs. Enter [r,s] where r is the x-component of the parametric function and s is the y-component. To graph a point, leave r and s as constants. Easy.
EXAMPLE 5: When teaching calculus, I always ask my students to sketch graphs of the derivatives of functions given in graphical forms. I always create these graphs one part at a time. As an example, this graph shows $y=x^3+2x^2$ and allows you to get its derivative gradually using a slider.
ASIDE 2–It is also very easy to enter derivatives of functions in the Desmos calculator. Type “d/dx” before the function name or definition, and the derivative is accomplished. Desmos is not a CAS, so I’m sure the software is computing derivatives numerically. No matter. Derivatives are easy to define and use here.
I’m hoping you find this technology tip as useful as I do.
## Exponential Derivatives and Statistics
This post gives a different way I developed years ago to determine the form of the derivative of exponential functions, $y=b^x$. At the end, I provide a copy of the document I use for this activity in my calculus classes just in case that’s helpful. But before showing that, I walk you through my set-up and solution of the problem of finding exponential derivatives.
Background:
I use this lesson after my students have explored the definition of the derivative and have computed the algebraic derivatives of polynomial and power functions. They also have access to TI-nSpire CAS calculators.
The definition of the derivative is pretty simple for polynomials, but unfortunately, the definition of the derivative is not so simple to resolve for exponential functions. I do not pretend to teach an analysis class, so I see my task as providing strong evidence–but not necessarily a watertight mathematical proof–for each derivative rule. This post definitely is not a proof, but its results have been pretty compelling for my students over the years.
Sketching Derivatives of Exponentials:
At this point, my students also have experience sketching graphs of derivatives from given graphs of functions. They know there are two basic graphical forms of exponential functions, and conclude that there must be two forms of their derivatives as suggested below.
When they sketch their first derivative of an exponential growth function, many begin to suspect that an exponential growth function might just be its own derivative. Likewise, the derivative of an exponential decay function might be the opposite of the parent function. The lack of scales on the graphs obviously keep these from being definitive conclusions, but the hypotheses are great first ideas. We clearly need to firm things up quite a bit.
Numerically Computing Exponential Derivatives:
Starting with $y=10^x$, the students used their CASs to find numerical derivatives at 5 different x-values. The x-values really don’t matter, and neither does the fact that there are five of them. The calculators quickly compute the slopes at the selected x-values.
Each point on $f(x)=10^x$ has a unique tangent line and therefore a unique derivative. From their sketches above, my students are soundly convinced that all ordered pairs $\left( x,f'(x) \right)$ form an exponential function. They’re just not sure precisely which one. To get more specific, graph the points and compute an exponential regression.
So, the derivatives of $f(x)=10^x$ are modeled by $f'(x)\approx 2.3026\cdot 10^x$. Notice that the base of the derivative function is the same as its parent exponential, but the coefficient is different. So the common student hypothesis is partially correct.
Now, repeat the process for several other exponential functions and be sure to include at least 1 or 2 exponential decay curves. I’ll show images from two more below, but ultimately will include data from all exponential curves mentioned in my Scribd document at the end of the post.
The following shows that $g(x)=5^x$ has derivative $g'(x)\approx 1.6094\cdot 5^x$. Notice that the base again remains the same with a different coefficient.
OK, the derivative of $h(x)=\left( \frac{1}{2} \right)^x$ causes a bit of a hiccup. Why should I make this too easy? <grin>
As all of its $h'(x)$ values are negative, the semi-log regression at the core of an exponential regression is impossible. But, I also teach my students regularly that If you don’t like the way a problem appears, CHANGE IT! Reflecting these data over the x-axis creates a standard exponential decay which can be regressed.
From this, they can conclude that $h'(x)\approx -0.69315\cdot \left( \frac{1}{2} \right)^x$.
So, every derivative of an exponential function appears to be another exponential function whose base is the same as its parent function with a unique coefficient. Obviously, the value of the coefficient depends on the base of the corresponding parent function. Therefore, each derivative’s coefficient is a function of the base of its parent function. The next two shots show the values of all of the coefficients and a plot of the (base,coefficient) ordered pairs.
OK, if you recognize the patterns of your families of functions, that data pattern ought to look familiar–a logarithmic function. Applying a logarithmic regression gives
For $y=a+b\cdot ln(x)$, $a\approx -0.0000067\approx 0$ and $b=1$, giving $coefficient(base) \approx ln(base)$.
Therefore, $\frac{d}{dx} \left( b^x \right) = ln(b)\cdot b^x$.
Again, this is not a formal mathematical proof, but the problem-solving approach typically keeps my students engaged until the end, and asking my students to discover the derivative rule for exponential functions typically results in very few future errors when computing exponential derivatives.
Feedback on the approach is welcome.
Classroom Handout:
Here’s a link to a Scribd document written for my students who use TI-nSpire CASs. There are a few additional questions at the end. Hopefully this post and the document make it easy enough for you to adapt this to the technology needs of your classroom. Enjoy.
## An unexpected lesson from technology
This discovery happened a few years ago, but I’ve just started ‘blogging, so I guess it’s time to share this for the “first” time.
I forget whether my calculus class at the time was using the first version of the TI-Nspire CAS or if we were still on the TI-89, but I had planned a very brief introduction to the CAS syntax for computing symbolic derivatives, but my 5-minute introduction in the first week of introducing algebraic rules of derivatives ended up with my students discovering antiderivative rules simply because they had technology tools which allowed them to explore beyond where their teacher had intended them to go.
They had absolutely no problem computing algebraic derivatives of power functions, so the following example was used not to demonstrate the power of CAS, but to give easily confirmed outputs. I asked them for the derivative of $x^5$, and their CAS gave the top line of the image below.
(In case there are readers who are TI-Nspire CAS users who don’t know the shortcut for computing higher order derivatives, use the left arrow to place the cursor after the dx in the “denominator” of $\frac{d}{dx}$ and press the carot (^) key. Then type the integer of the derivative you want.)
I wanted my students to compute the 2nd and 3rd derivatives and confirm the power rule which they did with the following screen.
That was the extent of what I wanted at the time–to establish that a CAS could quickly and easily confirm algebraic results whether or not a “teacher” was present. Students could create as many practice problems as were appropriate for themselves and get their solutions confirmed immediately by a non-judgmental expert. Of course, one of my students began to explore in ways my “trained” mind had long ago learned not to do. In my earlier days of CAS, I had forgotten the unboundedness of mathematical exploration.
Shortly after my syntax 5-minutes had passed and I had confirmed everyone could handle it, a young man called me to his desk to show me the following.
He understood what a 1st or 2nd derivative was, but what in the world was a negative 1st derivative? Rather than answering, I posed it to the class who pondered a few moments before recognizing that “underivatives” (as they called them in that moment) of power functions added one to the current exponent before dividing by the new exponent. They had discovered and explained (at least algebraically) antiderivatives long before I had intended. Technology actually inspired and extended my students’ learning!
Then I asked them what the CAS would give if we asked it for a 0th derivative. It was another great technology-inspired discussion.
I really need to explore more about the connections between mathematics as a language and the parallel language of technology. | 2018-06-18T09:14:22 | {
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http://math.stackexchange.com/questions/486842/problem-with-differentiation-as-a-concept/486854 | # Problem with differentiation as a concept.
I don't understand quiet good something here, for example if we want to find the derivative of the function $\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(h)}{h}$ and if we compute it from the function: $f(x) = 12 + 7x$ We get that the derivative of $f(x)$ is equal to $$\lim_{h \to 0} \frac{7h}{h}$$ But I thought that we can't divide by zero (here we cancel 0 over 0), I'm I wrong or $\displaystyle \frac{0}{0}$ equals 1?
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Small correction: The definition is either $\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$ or $\lim_{x \to x_0} \frac{f(x)-f(x_0)}{x-x_0}$. – Javier Badia Sep 7 '13 at 21:17
Thank you, I will correct it. – Rihanna Sep 7 '13 at 21:20
You are correct in saying that we can't divide by zero. But since we're taking a limit, we aren't dividing by zero.
Remember: a limit only cares what happens near the point, not at the point. So if $h=0$, then $\frac{7h}{h}$ would be undefined. But "$h$ goes to $0$" means that it is close to $0$ but not equal to it. Since for every $h\neq0$, $\frac{7h}{h} = 7$.
The basic idea here is that we don't want to evaluate $\frac{7h}{h}$ at $0$. We want to approach it. And as $h$ becomes closer and closer to $0$, $\frac{7h}{h}$ becomes closer and closer to $7$ (in fact, it is constantly $7$), and so we say that $\lim_{h \to 0}\frac{7h}{h} = 7$, since we don't care what happens at $h=0$, only near it.
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Thank you! This is so clear! What AN ANSWER! Thank you so much! – Rihanna Sep 7 '13 at 21:25
The whole point of the limit operation is that it avoids any bad behaviour of a function around the given point. We don't care what the function value is, nor whether it's even defined at a given point.
In your case, so long as $h \ne 0$, we can cancel to find that $\frac {7h}{h} = 7$; it doesn't matter that $\frac{7h}{h}$ isn't defined at $0$.
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But then it violates what I taught in Algebra; – Rihanna Sep 7 '13 at 21:19
@Rihanna How so? The limit doesn't care what happens when $h = 0$, and there is no division by zero. – user61527 Sep 7 '13 at 21:20
We say $\lim_{x \rightarrow c} f(x)=L$ if for all $\varepsilon>0$ there exists a $\delta>0$ such that for all $x$, if $0<|x-c|<\delta$ then $0<|f(x)-L|<\varepsilon$.
In this case, we have $c=0$, so when $x=0$ the condition $$0<|x-c|$$ is not satisfied.
Note that $$\lim_{h \rightarrow 0} \frac{7h}{h} \neq \frac{\lim_{h \rightarrow 0} 7h}{\lim_{h \rightarrow 0} h}$$ where there would be the $\frac{0}{0}$ problem you mention since we have $$\lim_{h \rightarrow 0} 7h=0=\lim_{h \rightarrow 0} h.$$
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If you're familiar with the $\varepsilon-\delta$ definition of a limit, then you can prove that if two functions $f$ and $g$ that satisfy $f(x) = g(x)$ for all $x \neq c$ in an open interval containing $c$, and $\lim_{x \to c}g(x)$ exists, then $$\lim_{x \to c}g(x) = \lim_{x \to c}f(x)$$
To prove this (feel free to skip this part if you don't care), call $\lim_{x \to c} g(x) = L$, and then by definition for each $\varepsilon > 0$ there exists a $\delta > 0$ such that $f(x) = g(x)$ in the intervals $(c - \delta, c)$ and $(c, c+\delta)$, and $$0 < |x - c| < \delta \implies |g(x) - L| < \varepsilon$$
But then because $f(x) = g(x)$ for all $x$ in the interval other than $x = c$, you know that $$0 < |x - c| < \delta \implies |f(x) - L| < \varepsilon$$ so you can say that $$\lim_{x \to c}f(x) = L$$
This relates to your question because the functions $f(x) = \frac{7h}{h}$ and $g(x) = 7$ only differ at $h = 0$, which combined with the theorem above shows that: $$\lim_{h \to 0}7=7 \implies \lim_{h \to 0} \frac{7h}{h} = 7$$
In general this theorem is what justifies being able to cancel off variables in a limit.
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The limit $\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$ always involves $0/0$. You can't divide $0$ by $0$, but that does not mean the limit does not exist.
Derivatives are about instantaneous rates of change. If a car goes $60$ miles in $2$ hours, its average speed during that time is $\dfrac{60\cdot\text{mile}}{2\cdot\text{hour}}$. It speed at a particular instant might therefore appear to be $\dfrac{0\cdot\text{mile}}{0\cdot\text{hour}}$. Derivatives tell you the speed at an instant, so they always involve $0/0$ in that way.
- | 2015-01-28T04:59:37 | {
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http://mathhelpforum.com/calculus/227009-given-lim-h-0-how-prove-ln-1-hr-h-r-print.html | # Given lim h -> 0, how to prove: ln(1 + hr) / h = r ?
• Mar 19th 2014, 02:59 PM
asdqwe
Given lim h -> 0, how to prove: ln(1 + hr) / h = r ?
Is it possible to prove it using derivative first principle?
I know how to prove lim n->max+ [ 1 + 1/n ]^n = e
but I am not sure how to prove the captioned one.
thank you very much.
• Mar 19th 2014, 06:19 PM
johng
Re: Given lim h -> 0, how to prove: ln(1 + hr) / h = r ?
Hi,
To show:
$$\lim\limits_{h\to0}{\ln(1+hr)\over h}=r$$
First suppose r = 0. Then the numerator is identically 0 and so the limit is r = 0. Otherwise, first note the derivative of $\ln(x)$ at $x=1$ is 1. That is,
$$1=\lim\limits_{k\to0}{\ln(1+k)-ln(1)\over k}=\lim\limits_{k\to0}{\ln(1+k)\over k}$$
Now let $k=hr$. Then
$$\lim\limits_{h\to0}{\ln(1+hr)\over h}=r\cdot\lim\limits_{h\to0}{\ln(1+hr)\over hr}=r\cdot\lim\limits_{k\to0}{\ln(1+k)\over k}=r\cdot1=r$$
If the switch from "h approaches 0" to "k approaches 0" is mysterious, informally, when h approaches 0, k approaches 0. More formally, an $\epsilon$ argument can be given. If you have question, I'll try to answer.
• Mar 19th 2014, 08:33 PM
Soroban
Re: Given lim h -> 0, how to prove: ln(1 + hr) / h = r ?
Hello, asdqwe!
Quote:
Is it possible to prove it using derivative first principle?
You have to give us a function to differentiate first.
I know how to prove: . $\lim_{n\to\infty}\left(1 + \tfrac{1}{n}\right)^n \:=\: e$
But I am not sure how to prove: . $\lim_{h\to0} \frac{\ln(1+hr)}{h} \:=\:r$
$\text{We have: }\:\lim_{h\to0}\frac{\ln(1+hr)}{h} \;=\;\lim_{h\to0} \left[\frac{1}{h}\ln(1+hr)\right]$
Multiply by $\tfrac{r}{r}\!:\;\;\lim_{h\to0}\left[\frac{r}{hr}\ln(1+hr)\right] \;=\;\lim_{h\to0}\ln(1 + hr)^{\frac{r}{hr}} \;=\;\lim_{h\to0}\ln\left[(1+hr)^{\frac{1}{hr}}\right]^r$
Let $n = \tfrac{1}{hr}$
Then we have: . $\lim_{n\to\infty}\ln\left[\left(1 + \tfrac{1}{n}\right)^n\right]^r \;=\;\ln\left[\lim_{n\to\infty}\left(1+\tfrac{1}{n}\right)^n \right]^r$
. . . . . . . . . . . $=\;\ln(e)^r \;=\;r\cdot\ln(e) \;=\;r$
• Mar 20th 2014, 04:05 AM
asdqwe
Re: Given lim h -> 0, how to prove: ln(1 + hr) / h = r ?
Thank you for helping me. Learned a lot. | 2016-09-29T14:17:36 | {
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http://math.stackexchange.com/questions/1626675/finding-the-solution-of-a-congruence | # Finding the solution of a congruence.
Solve the congruence
$$4x\equiv16\mod{26}.$$
How do I find the solution to this? I have tried by the euclidean algorithm but the gcd is not $1$ so it doesn't work.
\begin{align} 26&=&6&\times4&+2\\ 4&=&2&\times2&+0 \end{align}
Note: I understand we can see that $4$ and $17$ are solutions but how would I work this out algorithmically?
-
While solving congruences, it is better to convert the congruence into the division algorithm form and then go for solving.
$4x \equiv 16 \pmod{26} \Rightarrow 26 \: | \: 4x-16$
It finally reduces down to $13 \: | \: 2(x-4) \Rightarrow 13 \: | \: (x-4)$.
By Euclidean algorithm, we have unique integer $k$ such that $x-4=13k$.
So required solution is $\color{red}{x=13k+4} \:, \:\color{blue}{k \in \mathbb{Z}}$
-
@ThomasAndrews I have edited my answer. Thanks for your help. – SchrodingersCat Jan 26 at 5:30
You have to reduce factors, first. $26$ and $4$ are not relatively prime, but their common factor divides $16$, so you can solve $2x\equiv 8\pmod {13}$.
More generally, if you want to solve $ax\equiv b\pmod{d}$, you need $\gcd(a,d)$ to be a factor of $b$ (or else there are no solutions) and then divide $\gcd(a,d)$ from all of them to get an equation:
$$\frac{a}{\gcd(a,d)}x\equiv \frac{b}{\gcd(a,d)}\pmod{\frac{d}{\gcd(a,d)}}$$
-
By the way, does anyone know how to format \pmod so that the parenthesis are adjustable? – Elliot G Jan 25 at 18:28
alternatively why not use the euclidean algorithm on the original system i.e., (generate and) solve $2 = 4x' +26y'$, and set $x = 8x'$. (My point is 'algorithmic' in that you have to calculate/know the gcd anyway in general.) – peter a g Jan 25 at 18:28
The relation can be "translated" into: $$26\mid 4x-16$$ or equivalently $$4x=16+26k\text{ for }k\in\mathbb Z$$ Dividing both sides by factor $2$ results in $$2x=8+13k\text{ for }k\in\mathbb Z$$ Evidently there will be solutions for $x\in\mathbb Z$ if and only if $k$ is even.
Stating $k=2m$ we first find: $$2x=8+26m\text{ for }m\in\mathbb Z$$Dividing both sides by factor $2$ again we end up with:$$x=4+13m\text{ for }m\in\mathbb Z$$
- | 2016-06-30T03:09:59 | {
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https://math.stackexchange.com/questions/2792220/trigonometric-inequality-sin-2x-gt-sqrt-2-sin-x | # Trigonometric Inequality $\sin (2x) \gt \sqrt 2 \sin (x)$
I wish to solve this inequality:
$\sin (2x) \gt \sqrt 2 \sin (x)$
My approach:
I tried to isolate the $x$ on the left side by using the sine sum formula:
$2\sin(x)\cos(x) \gt \sqrt2\sin(x)$
then I divided by $\sin(x) \over 2$ both sides:
$\cos(x) \gt {\sqrt2 \over2}$
$x \lt \cos^{-1}({\sqrt2 \over2})$
$x < {\pi \over4}$
From that I can conclude that $x < {7\pi \over4}$, but I know the answer is still incomplete as it should be
$0 \lt x \lt {\pi \over 4}$, $\pi \lt x \lt {7 \over 4}\pi$
As I was able to see on Desmos graph plotter.
Does my approach gives the tools to reach this answer or have I commited a mistake?
• Are you looking for the general solution or the solution within a specified interval? – N. F. Taussig May 23 '18 at 0:14
• "then I divided by $\sin(x) \over 2$ both sides:" Ouch! What if the sine term is negative? – imranfat May 23 '18 at 0:18
• Dividing is sometimes problematic in inequalities. Instead, you may want to solve $\sqrt 2 \sin x ( \sqrt 2 \cos x - 1)>0$ – ThePortakal May 23 '18 at 0:27
The inequation can be written as
$$\sin (x)\Bigl(\cos (x)-\cos (\frac {\pi}{4})\Bigr)>0$$
which gives
$$2k\pi <x <\frac {\pi}{4}+2k\pi$$ or
$$(2k-1)\pi <x <2k\pi-\frac {\pi}{4}$$
You should not have divided by $\sin x$. That causes you to lose information you need to solve the problem. \begin{align*} \sin(2x) & > \sqrt{2}\sin x\\ 2\sin x\cos x & > \sqrt{2}\sin x\\ 2\sin x\cos x - \sqrt{2}\sin x & > 0\\ \sqrt{2}\sin x(\sqrt{2}\cos x - 1) & > 0 \end{align*} The inequality is satisfied if $\sin x > 0$ and $\sqrt{2}\cos x - 1 > 0$ or if $\sin x < 0$ and $\sqrt{2}\cos x - 1 < 0$.
Let's focus on solving the problem in the interval $[0, 2\pi)$ for the moment.
The inequality $\sin x > 0$ is satisfied in the interval $[0, 2\pi)$ if $x \in (0, \pi)$. The inequality $$\sqrt{2}\cos x - 1 > 0 \iff \cos x > \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$ is satisfied in the interval $[0, 2\pi)$ if $x \in [0, \frac{\pi}{4}) \cup (\frac{7\pi}{4}, 2\pi)$. Hence, $\sin x > 0$ and $\sqrt{2}\cos x - 1 > 0$ if $$x \in (0, \pi) \cap \left\{\left[0, \frac{\pi}{4}\right) \cup \left(\frac{7\pi}{4}, 2\pi\right)\right\} = \left(0, \frac{\pi}{4}\right)$$
The inequality $\sin x < 0$ is satisfied in the interval $[0, 2\pi)$ if $x \in (\pi, 2\pi)$. The inequality $$\sqrt{2}\cos x - 1 < 0 \iff \cos x < \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$ is satisfied in the interval $[0, 2\pi)$ if $x \in (\frac{\pi}{4}, \frac{7\pi}{4})$. Hence, $\sin x < 0$ and $\sqrt{2}\cos x - 1 < 0$ if $$x \in (\pi, 2\pi) \cap \left(\frac{\pi}{4}, \frac{7\pi}{4}\right) = \left(\pi, \frac{7\pi}{4}\right)$$
Thus, $\sin(2x) > \sqrt{2}\sin x$ in the interval $[0, 2\pi)$ if $$x \in \left(0, \frac{\pi}{4}\right) \cup \left(\pi, \frac{7\pi}{4}\right)$$ Since the sine function has period $2\pi$, the general solution is $$x \in \bigcup_{k \in \mathbb{Z}} \left\{\left(2k\pi, \frac{\pi}{4} + 2k\pi\right) \cup \left(\pi + 2k\pi, \frac{7\pi}{4} + 2k\pi\right)\right\}$$ | 2019-12-06T13:59:19 | {
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https://stats.stackexchange.com/questions/206020/measuring-forecast-accuracy-of-the-conditional-mean | # Measuring forecast accuracy of the conditional mean
Consider a dependent variable $y$, independent variables $x_1,\dotsc,x_K$, a model
$$y = X \beta + \varepsilon,$$
and an estimated coefficient $\hat\beta$. If the model is correctly specified, the true conditional mean of $y$ given $X$ is
$$\mathbb{E}(y|X) = X \beta.$$
Since we do not know $\beta$, we use its sample estimate $\hat\beta$ to get the estimated conditional mean
$$\hat{\mathbb{E}}(y|X) = X \hat\beta.$$
For a new set of observations $x_{1,i},\dotsc,x_{K,i}$, we can predict the conditional mean of the corresponding $y_i$ using the new $x$s and the estimated coefficient $\hat\beta$ as follows:
$$\hat{\mathbb{E}}(y_i|x_{1,i},\dotsc,x_{K,i}) = (x_{1,i},\dotsc,x_{K,i}) \hat\beta.$$
Now let us evaluate the forecast accuracy. We take the realized value $y_i$ and compare it to the predicted value $\hat{\mathbb{E}}(y_i|x_{1,i},\dotsc,x_{K,i})$. If the two are close, we say that the forecast is accurate.
Here is what bugs me:
• Aren't we predicting the true conditional mean $\mathbb{E}(y_i|x_{1,i},\dotsc,x_{K,i})$ rather than the actual realization $y_i$?
• If so, we are committing a measurement error when using $y_i$ in place of the unobserved $\mathbb{E}(y_i|x_{1,i},\dotsc,x_{K,i})$ when evaluating forecast accuracy. Isn't this problematic?
(One may also think about modelling higher order moments, such as conditional variance. There it is more obvious that the population moment being forecasted is unobservable, and hence measuring forecast accuracy is nontrivial.)
Aren't we predicting the true conditional mean $\mathbb{E}(y_i|x_{1,i},\dots,x_{K,i})$ rather than the actual realization $y_i$?
Yes indeed we are.
If so, we are committing a measurement error when using $y_i$ in place of the unobserved $\mathbb{E}(y_i|x_{1,i},\dots,x_{K,i})$ when evaluating forecast accuracy. Isn't this problematic?
On the one hand, you are right. We are forecasting an unobservable quantity and want to assess the accuracy of this forecast. We have a problem here.
The apparently only way out is to assess the accuracy of forecasts based on observables, and then deal with the fact that our forecast accuracy inevitably again only is one realization of a random variable, by invoking asymptotic arguments.
Now, whether and how this works for a given point forecast depends on what you want to use the point forecast for. Or, from a different perspective, it depends on your loss function.
Loss functions for forecast errors have been a topic for quite a while now (Fildes & Makridakis, 1988, IJF). In the area I am most familiar with, forecasting retail sales, the conditional mean (quadratic loss) is most useful for planning promotions, whereas store replenishment requires high quantiles.
Now, all the above relies on the fact that there always remains unexplained variance. (In some areas, like physics, we have a sufficiently good handle on the data generating process that we can explain almost all the variance and forecast extremely well, say, the trajectory of a bullet in vacuum.) People have been arguing that in non-physics situation, point forecasts alone are not overly helpful, and we should really aim for density forecasts, also known as predictive distributions. This ties into your parenthetical remark at the end of the question.
This is commonly accepted in financial and macroeconomic forecasting (in finance, driven by Value at Risk and options pricing) - not so much in supply chain and sales forecasting, where people happily calculate conditional means, estimate variances and assume a homoskedastic normal distribution in setting safety stocks. I have argued that predictive distributions make more sense in supply chains, too. The problem is that evaluating a density forecast is a bit more involved than evaluating point forecasts. I give a few pointers in this earlier answer of mine.
• Stephan, I am lucky to get you on my case. Had you ever thought of this problem before? Don't you think the formulations involving forecast accuracy in the setting above are often (always?) careless? I mean, if we acknowledge that measuring the forecast accuracy of conditional variance is nontrivial, shouldn't we acknowledge the same for the case of conditional mean, too? Should we take some care when reporting measures of forecast accuracy (do some measurement-error corrections)? – Richard Hardy Apr 7 '16 at 11:37
• I am writing an academic paper where I distinguish between the nontrivial measurement of forecast accuracy for cond. variance and "trivial" for cond. mean. Now I realized that even the cond. mean case is nontrivial, so I posted the question. – Richard Hardy Apr 7 '16 at 11:40
• I fully agree that assessing the accuracy of a forecast for the conditional mean is roughly as hard as for the conditional variance. (Somewhat less so, because the mean is easier to estimate than the variance.) As per above, that means that all forecast accuracy measures are really random variables, and we should treat them as such, leading to things like Diebold-Mariano tests... – S. Kolassa - Reinstate Monica Apr 7 '16 at 12:25
• ... I'd be interested in any measurement-error corrections you have in mind, but I'd suppose that most things you would want to account for in correcting for measurement error in realizations you'd also already incorporate in your forecasts. – S. Kolassa - Reinstate Monica Apr 7 '16 at 12:26
• That said, this discussion really takes us close to the concept of forecastability, which is a frequent topic in Foresight. If your supermarket sales are well described by a Poisson distribution, there is a lower limit for your forecast's MSE - although we could probably improve on it if we knew all shopping lists in a 100 mile radius. So everything depends on your information set. – S. Kolassa - Reinstate Monica Apr 7 '16 at 12:30 | 2020-01-20T00:37:33 | {
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https://stats.stackexchange.com/questions/12762/measure-of-spread-of-a-multivariate-normal-distribution/12764#12764 | # Measure of spread of a multivariate normal distribution
What is a good measure of spread for a multivariate normal distribution?
I was thinking about using an average of the component standard deviations; perhaps the trace of the covariance matrix divided by the number of dimensions, or a version of that. Is that any good?
Thanks
• as such, the spread of multivariate gaussian doesn't make sense. However, depending on your needs, there might exists approaches to answer your question. Trace of the matrix is one of the many ways, but you would be ignoring correlations, which may make a huge difference. Eigen values, PCA, etc. might be much better. Therefore, could you please elaborate on your needs? Jul 7 '11 at 10:41
• As such, I want an analog of the standard deviation to a multi-dimensional space. Yes, the trace would ignore the correlations, which is what I fear. Having said that, this does not need to be mathematically exact. Basically, a good indication of spread would be the hypervolume size of the hyperellipse defined by 1 std. deviation from the mean. But a nice, handy formula without deriving the exact volume would be much appreciated. Jul 7 '11 at 10:48
• Seems like PCA could answer your question. Jul 7 '11 at 10:51
What about the determinant of the sample variance-covariance matrix: a measure of the squared volume enclosed by the matrix within the space of dimension of the measurement vector. Also, an often used scale invariant version of that measure is the determinant of the sample correlation matrix: the volume of the space occupied within the dimensions of the measurement vector.
• +1 Yes, the determinants are directly related to the "hypervolume...of the ellipse defined by 1 sd from the mean."
– whuber
Jul 7 '11 at 13:53
• So that's the determinant of the covariance matrix, right? Jul 7 '11 at 14:29
• @Kristian The square root of the determinant of the covariance matrix tells you the hypervolume, incorporating both shape (correlation) and size (standard deviation) information. It is the product of the standard deviations of the principal components. The determinant of the correlation matrix is basically a shape factor only, ranging from 0 for degenerate distributions up to 1 when all components are uncorrelated.
– whuber
Jul 7 '11 at 18:42
• @whuber, what if I'd like to have a separate measurement of shape and size? (I'm actually interested in the size only, I think.) Feb 12 '17 at 18:24
• @Atcold You would need to establish a quantitative definition of "size". This would be equivalent to establishing what a unit-size distribution is for each given shape. (By definition, "shape" is whatever properties a distribution may have that are unchanged by translation or rescaling.) There are innumerable ways to do that, so ultimately the issue comes down to choosing a suitable definition for your particular analysis. This is one reason there cannot be a universal definition of size (or "spread") for any distribution family that comprises multiple shapes.
– whuber
Feb 12 '17 at 18:28
I would go with either trace or determinant with a preference towards trace depending on the application. They're both good in that they're invariant to representation and have clear geometric meanings.
I think there is a good argument to be made for Trace over Determinant.
The determinant effectively measures the volume of the uncertainty ellipsoid. If there is any redundancy in your system however then the covariance will be near-singular (the ellipsoid is very thin in one direction) and then the determinant/volume will be near-zero even if there is a lot of uncertainty/spread in the other directions. In a moderate to high-dimensional setting this occurs very frequently
The trace is geometrically the sum of the lengths of the axes and is more robust to this sort of situation. It will have a non-zero value even if some of the directions are certain.
Additionally, the trace is generally much easier to compute.
• +1 Good points. This gets me thinking: any symmetric function of the $n$ eigenvalues would qualify as "good." All such polynomial functions are polynomials in the $n$ elementary symmetric functions, which include the determinant and the trace.
– whuber
Jul 15 '11 at 20:53
• Yes, the sum (trace) isn't necessarily the best way to go. You're right that you could imagine lots of mixtures here depending on the application. I wonder if there is some standard family of functions that would be good here.... Jul 15 '11 at 20:56
• @MR I'm not aware of anybody attempting to use a single statistic to compute the spread of a multivariate normal distribution (except, of course, when independence of all components is assumed). This leads me to believe there may be no such standard family.
– whuber
Jul 18 '11 at 12:44
Another (closely related) quantity is the entropy of the distribution: for a multivariate Gaussian this is the log of the determinant of the covariance matrix, or
$\frac{1}{2} \log |(2\pi e)\Lambda|$
where $\Lambda$ is the covariance matrix. The advantage of this choice is that it can be compared to the "spread" of points under other (e.g., non-Gaussian) distributions.
(If we want to get technical, this is the differential entropy of a Gaussian). | 2022-01-21T09:40:40 | {
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https://math.stackexchange.com/questions/2844406/finding-expected-value-of-number-of-kings-drawn-before-the-first-ace-and-its-low | # Finding expected value of number of kings drawn before the first Ace and its lower bound?
Consider a standard deck of cards. We draw cards from the top until we get either a King or an Ace.
1.Write a lower bound for the expected value of the number of Kings drawn before the first Ace.
The solution given in this problem is 1/2.
I'm not sure how they arrive at this solution. I assumed that both Kings and Aces are equally likely to be drawn so shouldn't the # of kings being drawn before first Ace appear be $n* \frac {1}{52}$ ?
2. Find the expected value of the number of Kings drawn before the first Ace.
Here since the number of kings in a deck is 4 and there are 52 cards in total shouldn't the expected number of kings drawn before the first Ace be $n* \frac {4}{52}$ ?
• What is $n$ in your answers? There is none in the statement of the exercise... The expected value of the number of Kings drawn before the first Ace is 4 times the probability that the King of Hearts is drawn before all four Aces. By symmetry this last probability is one fifth hence the answer to 2. is 4/5. (Note that Question 1. is not understandable and that the sentence "We draw cards from the top until we get either a King or an Ace" is misleading as well.) – Did Jul 8 '18 at 8:55
• sorry, I just thought of n as an arbitrary draw number – pino231 Jul 8 '18 at 15:19
A lower bound is any value that is certainly bellow E(X), you cold say $0$, it would be a right answer. They provided a bit less obvious answer 1/2, since there is probability of 0.5 you will find an ace first and same probability you will find a king first, so 0.5*0+0.5*1=0.5. But there may be more kings after that first so expected value is certainly above 0.5
For the expected value you just have to play with the permutations of KKKKAAAA. There is total of $8\choose 4$=70 permutations.
With all 4 kings first - 1 permutation.
With 3 kings first and then an ace - 4 permutations (KKKA+KAAA or AKAA or AAKA or AAAK =$4\choose 1$).
With 2 kings first and then an ace $5\choose 2$=10 permutations.
With 1 king followed by ace $6\choose 3$=20 permutations
So E(X)=(4*1+3*4+2*10+1*20)/70=56/70=0.8
The cards other than kings or aces can be ignored, since they don't affect the order of kings and aces.
Hence, we can regard the deck as a shuffled deck with only kings and aces.
Let $e(k)$ be the expected number of kings that occur before the first ace, assuming the deck has $k+4$ cards, with $k$ kings, and $4$ aces.
Our goal is to find $e(4)$.
Then we have the recursion $$\begin{cases} e(0)=0\\[4pt] e(k)=\left({\large{\frac{k}{k+4}}}\right)\bigl(1+e(k-1)\bigr),\;\text{if}\;k > 0\\ \end{cases}$$ Claim:$\;$For all $k\ge 0$, we have $e(k)={\large{\frac{k}{5}}}$.
To prove it, proceed by induction on $k$.
For $k=0$, we have $e(0)=0={\large{\frac{0}{5}}}$, so the base case is verified.
Suppose the claim holds for some nonnegative integer $k$. \begin{align*} \text{Then}\;\; e(k+1)&=\left(\frac{k+1}{k+5}\right)\bigl(1+e(k)\bigr)\\[4pt] &=\left(\frac{k+1}{k+5}\right)\left(1+\frac{k}{5}\right)\\[4pt] &=\left(\frac{k+1}{k+5}\right)\left(\frac{k+5}{5}\right)\\[4pt] &=\frac{k+1}{5}\\[4pt] \end{align*} which completes the induction.
In particular, we have $e(4)={\large{\frac{4}{5}}}$.
• I like the intuition that you pointed out that all the non Kings and Aces cards are irrelevant in this case – pino231 Jul 8 '18 at 15:46
Part 2. For the purpose of analysis, we may pretend the deck consists only of four kings and four aces. Let $X$ be the number of kings before the first ace.
Then $P(X>0)$ is the probability that the first card is a king, so $P(X>0) = \frac{4}{8}$.
$P(X > 1)$ is the probability that the first two cards are kings, so $P(X>1) = \frac{4}{8} \cdot \frac{3}{7}$.
Similarly, we see that $P(X>2) = \frac{4}{8} \cdot \frac{3}{7} \cdot \frac{2}{6}$, and $P(X>3) = \frac{4}{8} \cdot \frac{3}{7} \cdot \frac{2}{6} \cdot \frac{1}{5}$.
So $$E(X) = P(X>0) + P(X>1)+P(X>2)+P(X>3) = 0.8$$
Here we have made use of the theorem that $$E(X) = \sum_{n=0}^{\infty} P(X>n)$$ which holds for any random variable $X$ which takes on only values in $\{0, 1, 2, 3, \dots \}$. | 2019-06-26T23:54:47 | {
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https://gateoverflow.in/707/gate2001-1-14 | 2.9k views
Randomized quicksort is an extension of quicksort where the pivot is chosen randomly. What is the worst case complexity of sorting n numbers using Randomized quicksort?
1. $O(n)$
2. $O(n \log n)$
3. $O(n^2)$
4. $O(n!)$
edited | 2.9k views
0
There are following two cases, when Randomized Quick Sort will result into worstcase of time complexity $O(n^{2})$
1. When all elements are same in the input array, Partition algo will divide input array in two sub-array, one with $n-1$ elements and second with $0$ element. There is an assumption here that, we are using the same partition algorithm without any modification.
2. If the randomised pivot selector happens to select e.g. the smallest element N times in a row, we will get the worst possible performance. Though the probability of this particular case is about $\frac{1}{n!}$ "
PS:- If the partitioning is unbalanced, Quick Sort algorithm runs asymptotically as slow as Insertion Sort i.e $O(n^{2})$
edited
In worst case, we may pick pivot elements in the increasing order (input also given in sorted order) which will result in running time of O($n^{2}$)
Both the deterministic and randomized quicksort algorithms have the same best-case running times of O($nlogn$) and the same worst-case running times of O(n$^{2}$).The difference is that with the deterministic algorithm, a particular input can elicit that worst-case behavior. With the randomized algorithm, however, no input can always elicit the worst-case behavior. The reason it matters is that, depending on how partitioning is implemented, an input that is already sorted--or almost sorted--can elicit the worst-case behavior in deterministic quicksort.
source: Thomas Coremen
Ans. C
edited
+6
Yes. That is correct. O(n log n) is the EXPECTED number of comparisons when pivot is chosen randomly.
0
But I have read that it gives O(n^2) only when pivot is selected as first or last element in an already sorted list, so i think here ans should be O(n logn) as they are talking about randomized quick sort here... pls tell me if i am correct or not?
+1
Randomized quick sort picks the pivot randomly so in best case and avg case it gives O(nlogn) time complexity but stil in worst case there is a chance that it may select smallest element as pivot..so O(n^2) in worst case
0
@arjun sir i did not get the point...."expected num of comparision"
0
In Random Quick Sort
All these cases may come . ie '
In worst case, we may pick pivot elements in the increasing / decreasing order
Hence it is $O(n^{2})$ and $\Omega(n \log n)$
The running time of Randomized QUICKSORT when all elements of array A have the same value will be equivalent to the worst case running of QUICKSORT since no matter what pivot is picked, QUICKSORT will have to go through all the values in A. And since all values are the same, each recursive call will lead to unbalanced partitioning.
Thus the recurrence will be:
T(n)=T(n−1)+Θ(n)
T(n)=Θ(n2) | 2018-09-26T01:45:26 | {
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https://mathhelpboards.com/threads/85-find-circle-from-3-points.26353/ | # [SOLVED]85 find circle from 3 points
#### karush
##### Well-known member
find an equation of the circle passing through the given points
85 Given $(-1,3),\quad (6,2),\quad (-2,-4)$
since the radius is the same for all points set all cirlce eq equal to each other
$(x_1-h)^2+(y_1-k)^2=(x_2-h)^2+(y_2-k)^2=(x_3-h)^2+(y_3-k)^2$
plug in values
$(-1-h)^2+(3-k)^2=(6-h)^2+(2-k)^2=(-2-h)^2+(-4-k)^2$
from this we get (via W|A)
$h = 2,\quad k = -1$
derive the radius by the distance of the center to one of the points
$d=\sqrt{(2-(-2))^2+(-1-(-4))^2}\sqrt{16+9}=\sqrt{25}=5$
thus the stardard circle equation would be
$\left(x-2\right)^2+\left(y+1\right)^2=25$
ok I think this is ok, but I was going to try to do this with a matrix but with the squares in it didn't see how
also want to try to draw the 3 points, center, and circle with tikx
first attempt... need 3 points and ticks and text
\begin{tikzpicture}
\draw (.4,-.2) circle (1cm);
\draw (-2,0) -- (3,0);
\draw (0,-2) -- (0,2);
\end{tikzpicture}
I assume there might be some kill all formula used for this problem.
Last edited:
#### skeeter
##### Well-known member
MHB Math Helper
find an equation of the circle passing through the given points
$(-1,3),\quad (6,2),\quad (-2,-4)$
slope between (-1,3) and (-2,-4) ... $m = 7$
midpoint between (-1,3) and (-2,-4) ... (-3/2, -1/2)
perpendicular bisector ... $y + \dfrac{1}{2} = -\dfrac{1}{7}\left(x + \dfrac{3}{2}\right)$
slope between (-1,3) and (6,2) ... $m = -\dfrac{1}{7}$
midpoint between (-1,3) and (6,2) ... (5/2, 5/2)
perpendicular bisector ... $y - \dfrac{5}{2} = 7 \left(x - \dfrac{5}{2} \right)$
intersection of the two perpendicular bisectors is the circle center, (2, -1)
distance between any point and the center is the radius ... $r = 5$
circle equation ...
$(x-2)^2 + (y+1)^2 = 5^2$
#### karush
##### Well-known member
well that was certainly very novel..
most of these examples throw you to the general equations and do a lot of gymnastics.
ok I assume that is a Desmos graph....
#### topsquark
##### Well-known member
MHB Math Helper
slope between (-1,3) and (-2,-4) ... $m = 7$
midpoint between (-1,3) and (-2,-4) ... (-3/2, -1/2)
perpendicular bisector ... $y + \dfrac{1}{2} = -\dfrac{1}{7}\left(x + \dfrac{3}{2}\right)$
slope between (-1,3) and (6,2) ... $m = -\dfrac{1}{7}$
midpoint between (-1,3) and (6,2) ... (5/2, 5/2)
perpendicular bisector ... $y - \dfrac{5}{2} = 7 \left(x - \dfrac{5}{2} \right)$
intersection of the two perpendicular bisectors is the circle center, (2, -1)
distance between any point and the center is the radius ... $r = 5$
circle equation ...
$(x+1)^2 + (y-2)^2 = 5^2$
Well, well, well! Someone's been reading their Euclid!
Nice job.
-Dan
#### skeeter
##### Well-known member
MHB Math Helper
well that was certainly very novel..
most of these examples throw you to the general equations and do a lot of gymnastics.
ok I assume that is a Desmos graph....
not desmos ... free graphing program for windows (maybe there is one for Macs, haven't looked to confirm)
#### karush
##### Well-known member
I think I'm mostly interested in what ca be done with tikx.
Albeit with its limitations
#### Greg
##### Perseverance
Staff member
$$\displaystyle (-1-h)^2+(3-k)^2=(6-h)^2+(2-k)^2=(-2-h)^2+(-4-k)^2$$
$$\displaystyle (-1-h)^2-(6-h)^2+(3-k)^2-(2-k)^2=0$$
$$\displaystyle -7(5-2h)+(5-2k)=0$$
$$\displaystyle -35+14h+5-2k=0$$
$$\displaystyle -30+14h-2k=0\quad[1]$$
$$\displaystyle (-1-h)^2-(-2-h)^2+(3-k)^2-(-4-k)^2=0$$
$$\displaystyle (-3-2h)+7(-1-2k)=0$$
$$\displaystyle -10-2h-14k=0\quad[2]$$
$$\displaystyle [1]+7*[2]\implies k=-1\implies h=2$$
$$\displaystyle (x-2)^2+(y+1)^2=25$$
#### karush
##### Well-known member
so I quess there is no slam dunk one step solution | 2020-07-11T11:17:57 | {
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https://math.stackexchange.com/questions/147504/coin-flipping-games-dependent-trials | # Coin flipping games - dependent trials
I'm still trying to learn probability and in furtherance of this I have posed myself two questions about coin flipping series. I don't know how to answer these questions because these aren't independent trials of flips. I have monte-carloed these trials and so I know what the answers are, but I don't have analytical solutions.
Consider the following "games":
Easy: A fair coin is flipped until it lands on tails. Each time the coin is flipped, the player receives one point. What is the expected number of points the player will receive in a game?
Harder: An unfair coin is flipped until it lands on tails. The coin lands on tails with probability p. (p is an independent variable - it does not change throughout the series). Each time that the coin is flipped, the player receives twice as many points as the last time that the coin was flipped. E.G., The first flip gives 1 point, the second 2, the third 4. What is the expected number of points the player will receive in a game?
Since the purpose of answering these questions is to learn the underlying math, I would appreciate any links to relevant reference material. If my terminology is unclear or nonstandard please correct it. Thanks in advance and I hope this question is appropriate to this forum - this is my first post here.
Edit: in answer to Mark's comment, here is what I have.
For the easier game, here's what I've got.
I know how to calculate the probability that the the game ends on flip x.
It is equal to the probability that the game did not end on any previous trial times the probability the game ended on trial $x$.
The probability of getting the first tails on the first flip is $.5$
The probability of getting the first tails on the second flip is the probability of getting a heads on the first flip times the probability of getting a tails on the second flip $= .5\times.5$
The probability of getting the first tails on the third flip is p(heads first) * p(heads second) * p(tails third);
Since the probability of getting a heads or a tails on any particular trial is $.5$ in the first example, I think this reduces nicely to this:
$1 + p^1 + p^2 + p^3 + ... + p^x$
So the total EV is the sum of this infinite series (which I don't know how to compute).
For the harder game, I guess it's not really that much harder.
The odds of it ending any particular trial is a similar calculation. If odds of tails is $p$, then odds of success is $1-p$.
so the probability function that tells me the probability that the trial ended at flip $x$ is
$P(x) = (1-p)^{x-1} \times p$
The value of the game is something like
$2^x(P(x))$
summed for all $x$ from $1$ to infinity. Again I don't know to sum that.
• Since you say one is harder than the other, how do you know? What do you think yourself about the easier game? – Mark Bennet May 20 '12 at 19:42
• @masonk Welcome to math stackexchange. Kindly refer here (meta.math.stackexchange.com/questions/107) on how to typeset (i.e. how to write equations etc) on this website. This is done so that the equations and other math appear in a nice way. Also, if you have trouble summing up geometric series, then I would suggest you to look at the following post math.stackexchange.com/questions/29023 for details on how to evaluate such summations. – user17762 May 20 '12 at 20:10
• @masonk: Small question of detail. For the first question, if you get a tail on the first toss, do you get $0$ points or $1$ point? If you get a head, then a tail, do you get $1$ point or $2$ points? Same for second question. If you get a head then a tail, do you get $1$ or $1+2$? – André Nicolas May 21 '12 at 0:35
• In my original formulation I meant it to be that you always get at least one point in both games. I think Marvis provided a solution where the first flip is worth 0. It looks like other people are used to seing in this second way form too. – masonk May 21 '12 at 0:50
"A fair coin is flipped until it lands on tails. Each time the coin is flipped, the player receives one point. What is the expected number of points the player will receive in a game?"
Let us compute the probability that the player receives exactly $k$ points. This happens when the first $k$ coin flips land on heads and the $(k+1)^{th}$ flip lands on a tail. The probability of this event is hence $$\mathbb{P}(\text{Player receiving k points}) = \underbrace{\frac12 \times \frac12 \times \cdots \frac12}_{\text{The first k flips land on heads}} \times \underbrace{\frac12}_{\text{The (k+1)^{th} flip lands on a tail}} = \frac1{2^{k+1}}$$
Just as a sanity check, we can see that $$\sum_{k=0}^{\infty} \mathbb{P}(\text{Player receiving k points}) = \sum_{k=0}^{\infty} \frac1{2^k} = 1$$
The expected number of points the player wins is hence $$\sum_{k=0}^{\infty} \frac{k}{2^{k+1}} = 1$$
Now lets move on to the second question.
"An unfair coin is flipped until it lands on tails. The coin lands on tails with probability $p$. ($p$ is an independent variable - it does not change throughout the series). Each time that the coin is flipped, the player receives twice as many points as the last time that the coin was flipped. For example, the first flip gives $1$ point, the second $2$, the third $4$. What is the expected number of points the player will receive in a game?"
Note that the player can receive points of the form $$2^{k}-1$$ if the coin flips yield $k$ consecutive heads and the $(k+1)^{th}$ being a tail. Let $q = 1-p$ be the probability of landing on a head.
As before, the probability of the player to get $2^{k} - 1$ points is $q^k \times (1-q)$ i.e.
\begin{align} \mathbb{P}(\text{Player receiving $2^k-1$ points}) & = \mathbb{P}(\text{Heads the first $k$ times and tail the $(k+1)^{th}$ time})\\ & = \underbrace{q \times q \times \cdots q}_{\text{The first $k$ flips land on heads}} \times \underbrace{(1-q)}_{\text{The $(k+1)^{th}$ flip lands on a tail}}\\ & = q^k ( 1-q) \end{align}
Hence, the expected number of points he will win is $$\sum_{k=0}^{\infty} (2^k-1) \times q^k \times (1-q) = \left( \sum_{k=0}^{\infty} (2q)^k (1-q) \right) - 1 = \begin{cases} \frac{q}{1-2q} & \text{if q < \frac12 }\\ \infty & \text{else} \end{cases}$$
The fact that the expectation is $\infty$ if $q \geq \frac12$ is the well-known St.Petersburg paradox.
EDIT If you have trouble summing up geometric series, then I would suggest you to look at the following post Value of $\sum\limits_n x^n$ for details on how to evaluate such summations.
• Oh, wow! I was closer than I thought. How did you compute those infinite sums? – masonk May 20 '12 at 20:09
• It seems that was able to express the correct series but not to find the limit of that series. I'm reading through the linked post now. Accepting this answer because it is both correct and links to the material I needed to finish. Thanks so much. I am amazed at how fast this was answered. – masonk May 20 '12 at 20:18
• Just for reference, the way he calculated the series is that if $q<\frac{1}{2}$, then $2q<1$. Hence the series $\sum_{k=0}^{\infty} (2q)^k$ converges to $\frac{1}{1-2q}$. – Damodar8 Aug 30 '17 at 20:18 | 2019-12-10T16:19:58 | {
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https://math.stackexchange.com/questions/1446707/why-do-i-need-to-switch-the-sign-of-the-natural-log-in-this-trig-sub-integration | # Why do I need to switch the sign of the natural log in this trig-sub integration problem?
I'm attempting to integrate the following using trig substituion: $$\int \frac{\sqrt{1+x^2}}{x}dx$$ and I am getting the result: $$-\ln{\lvert \frac{ \sqrt{x^2 + 1} - 1}{x} \rvert} + \sqrt{x^2 + 1} + C$$
However, my answer guide and this yahoo answer both indicate that the $-\ln$ should now be positive, I can't seem to figure out why (although I'm sure it's painfully simple). I thought it might be related to the fact that $\ln{\frac1x} = -\ln{x}$ but I can't wrap my head around it.
I've provided my steps below:
I start with $x=\tan\theta$ and $dx=\sec^2\theta d\theta$
Which brings me to $$\int \frac{\sqrt{1+\tan^2\theta}}{\tan\theta}\sec^2\theta d\theta$$
and then: $$\int \frac{\sec^2\theta(\sec\theta)}{\tan\theta} d\theta$$
.... next I substitute $(1+\tan^2\theta)$ for $\sec^2\theta$ and multiply by the $\sec\theta$ already in the numerator, giving me: $$\int \frac{\sec\theta+\sec\theta\tan^2\theta}{\tan\theta}d\theta$$
I separate this out into two separate fractions, canceling one power of $\tan\theta$ in its respective fraction:
$$\int \frac{\sec\theta}{\tan\theta} + \sec\theta\tan\theta d\theta$$
This can be split into two integrals, and the left side can be rewritten as $\csc\theta$:
$$\int \csc\theta d\theta + \int \sec\theta\tan\theta d\theta$$
Which using known integrals can be determined to be:
$$-\ln{\lvert \csc\theta - \cot\theta \rvert} + \sec\theta + C$$
Creating a triangle to determine $\csc\theta$, $\cot\theta$, and $\sec\theta$ in terms of x yields me:
$$-\ln{\lvert \frac{ \sqrt{x^2 + 1} - 1}{x} \rvert} + \sqrt{x^2 + 1} + C$$
Edit: So it turns out I had the formula for $\int\csc\theta d\theta$ wrong. As user84413 pointed out in the comments, the formula is:
$$\int\csc\theta d\theta=\ln|\csc\theta-\cot\theta|+C=-\ln|\csc\theta+\cot\theta|+C$$
Using this corrected formula for my second to last step, I get the correct answer.
• $\int\csc\theta d\theta=\ln|\csc\theta-\cot\theta|+C=-\ln|\csc\theta+\cot\theta|+C$ – user84413 Sep 22 '15 at 16:52
• oh lord... I copied down the integral of csc with the wrong sign inside the absolute value.... – JonathonG Sep 22 '15 at 17:00
• @user84413 Would you like to make an answer highlighting that I used the wrong formula for $\int\csc\theta d\theta$ ? That way it will be easier for someone finding this page to see what I did wrong. – JonathonG Sep 22 '15 at 17:08
• Thanks - I think your edit to the problem was a good way to handle it. – user84413 Sep 22 '15 at 18:08
Your Second last line is $\displaystyle - \ln\left|\csc \theta-\cot \theta\right|+\sec\theta +\mathcal{C}$
Now Using $\displaystyle \csc^2 \theta-\cot^2 \theta = 1\Rightarrow (\csc \theta -\cot \theta) = \frac{1}{\csc \theta +\cot \theta}$
So we get $\displaystyle = -\ln\left|\frac{1}{\csc \theta+\cot \theta}\right|+\sec \theta+\mathcal{C}$
So we get $\displaystyle = \ln\left|\csc \theta+\cot \theta\right|+\sec \theta+\mathcal{C}$
So $\displaystyle =\ln\left|\frac{\sqrt{1+x^2}+1}{x}\right|+\sqrt{1+x^2}+\mathcal{C}$
$\bf{Solution \; without \; Trig. \; Substution.}$
Let $$\displaystyle I = \int\frac{\sqrt{1+x^2}}{x}dx = \int\frac{1+x^2}{x\sqrt{1+x^2}}dx = \int\frac{1}{x\sqrt{1+x^2}}dx+\int\frac{x}{\sqrt{1+x^2}}dx$$
Now Let $$\displaystyle J = \int\frac{1}{x\sqrt{x^2+1}}dx\;,$$ Put $\displaystyle x = \frac{1}{t}\;,$ Then $\displaystyle dx = -\frac{1}{t^2}dt$
So $$\displaystyle J = -\int\frac{1}{\sqrt{t^2+1}}dt = -\ln\left|t+\sqrt{t^2+1}\right|+C_{1} = -\ln\left|\frac{\sqrt{x^2+1}+1}{x}\right|+\mathcal{C_{1}}$$
And Let $$\displaystyle K=\int\frac{x}{\sqrt{x^2+1}}dx\;,$$ Now put $\displaystyle x^2+1=u^2\;,$ Then $xdx = udu$
So we get $$\displaystyle K = \int\frac{u}{u}du = u+\mathcal{C_{2}} = \sqrt{x^2+1}+\mathcal{C_{2}}$$
So we get $$\displaystyle I = \int\frac{\sqrt{1+x^2}}{x}dx = -\ln\left|\frac{\sqrt{x^2+1}+1}{x}\right|+\sqrt{x^2+1}+\mathcal{C}$$
• Thanks for all that work for the non trig-sub answer. I still had to do this as trig-sub because I am required to demonstrate that method. However, the first half of your answer seems to operate on the assumption that my second to last step was correct, when it in fact was not. I used the wrong formula for $\int\csc\theta d\theta$, see the comments on my question or edited question itself to see what I mean. – JonathonG Sep 22 '15 at 17:16
It is simple is not complex as you are thinking. Look this,$$\int \frac {\sqrt{1+x^2}}x dx=\int \frac {1+x^2}{x\sqrt{1+x^2}}dx=\int \frac 1{x\sqrt {1+x^2}}dx+\int \frac x{\sqrt{1+x^2}}dx$$
For the first integral, substitute $x=\tan z$ which yields $\int \csc z dz$, and for the second, substitute $1+x^2=u$.
I think you can do the remaining.
• This is true, but my problem was that I used the wrong formula for the integral of csc, so I would have gotten the answer wrong if I used your approach as well. I said $\int\csc\theta d\theta=-\ln|\csc\theta-\cot\theta|+C$ instead of $-\ln|\csc\theta+\cot\theta|+C$ – JonathonG Sep 22 '15 at 17:03 | 2019-05-27T07:22:39 | {
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https://www.physicsforums.com/threads/removable-discontinuity.99142/ | # Removable discontinuity
1. Nov 9, 2005
### maria curie
the book says that g(x)= x ,if x is not equal to 2 / 1,if x=2
has a removable disc. at x =2.
I couldn't remove it.I guess I didn't understand a removable dic. completely.I have an exam on friday.I need your help
thanks
2. Nov 9, 2005
### HallsofIvy
Staff Emeritus
f(x)= x if x is not equal to 2
f(x)= 1 if x is equal to 2.
What is the limit of f, as x-> 2? (Remember that the limit depends only on the values of f close to 2, not at 2!)
WHY is f not continuous at x= 2??
What would happen if you changed the value of f(2) to 2?
3. Nov 9, 2005
### maria curie
f(x) is not cont at x=2 because f(2)=1 is not equal to lim x->2f(x)=2.I know what continuity is.I dont know how we change the value of f(2) to 2 to be continuous x=2??
and what are the conditions of the continuous extension ?Can be every funct.that is not continuous become cont.?
Last edited: Nov 9, 2005
4. Nov 9, 2005
### 1800bigk
a removable discontinuity is just something that we can fix or adjust to get the function continuous. It usually means a function is discontinuous at some point or hole in the graph and all we have to do is plug the hole if you will, or redefine the function at the point in question. The problem you have has removable discontinuity because all we have to do is redefine the function at some point to get continuity.
1/x has an infinite discontinuity at zero and cannot be fixed
5. Nov 9, 2005
### HallsofIvy
Staff Emeritus
A function is continuous at x= a if and only if:
1) f(a) exists
2) $lim_{x\rightarrowa}f(x)$ exits.
3) $lim_{x\rightarrowa}f(x)= f(a)$
The function as given satisfies the first two of those at x= 2. It is not continuous because the third is not true: $lim_{x\rightarrow2}f(x)= 2$ but f(2)= 1. Redefining f(2)= 2 will make all three true! That "removes the discontinuity". (In fact, the function becomes f(x)= x for all x.)
As long as the limit exists at x= a we could always redefine f(a) to be that limit and "remove the discontinuity". A discontinuity is "not removeable" if the limit does not exist. $f(x)= \frac{1}{x}$ is not continuous at x= 0, as 1800bigk says, and no value for f(0) will make it continuous: it has an "infinite discontinuity" there.
The function: f(x)= x if x<= 1, f(x)= x+ 2 if x> 1, has a "jump" discontinuity at x= 1. The "limit from the left" is 1, the "limit from the right" is 3. Since those are not the same, there is no "limit" at x= 1.
6. Nov 9, 2005
### CarlB
Marie, this problem is a lot simpler than you're making it. What you're looking for is a function f(x) that is similar to g(x), but is more continuous. The rules are that you must have f(x) be the same as g(x) when x is not 2, but you can choose whatever you like for f(2). Now, take my word, the simplest thing for f(x) that comes to your mind is the right answer.
A lot of the time, mathematics is about finding a really complicated way of saying something that is very simple. Don't walk around thinking that this stuff is too complicated, it's not. It's just that it's not being explained to you in simple terms.
Carl
7. Nov 10, 2005
### maria curie
HallsofIvy ,your explanation is very good,thanks a lot | 2017-08-24T11:07:56 | {
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https://math.stackexchange.com/questions/470580/show-quotient-space-is-homeomorphic-to-unit-sphere/470586 | # Show: Quotient space is homeomorphic to unit sphere
An equivalence relation on $\mathbb{R}$ is given by $$x\sim y\Leftrightarrow x-y\in\mathbb{Z}.$$ Show that the quotient space $(\mathbb{R}/{\sim},\tau_1)$ is homeomorphic to $(S^1,\tau_2)$, where $\tau_1$ is the quotient topology and $\tau_2$ the induced topology.
I have to find a bijective continuous function $$f\colon \mathbb{R}/{\sim}\to S^1$$ with $f^{-1}$ continuous.
Do you have an idea how to find such a function?
• By "room", do you mean "space"? – Zev Chonoles Aug 18 '13 at 16:58
• @ZevChonoles: In German the word Raum means room, but also denotes a space in mathematics :-) – Stefan Hamcke Aug 18 '13 at 17:08
Let $S^1=\{z\in\mathbb{C}\mid|z|=1\}$. This is a common definition for $S^1$ but may not be the one you've been given.
Let $f\colon\mathbb{R}/{\sim}\rightarrow S^1$ be given by $f([t])=e^{2\pi it}$.
Can you show that this is well defined? (That is, show that if $t_1\in[t]_{\sim}$ and $t_2\in[t]_{\sim}$ then $e^{2\pi it_1}=e^{2\pi it_2}$).
Can you show that this is a homeomorphism?
• Why not taking $[t]_{\sim}\mapsto e^{it}$? Because then it is not well defined, right? Then it depends on what t I take as representer. – math12 Aug 18 '13 at 17:10
• The $2\pi$ is in there because a full revolution of a circle is $2\pi$ radians. You want $f$ to 'wrap' the real line around the circle once for every integer so for every interval between two integers that you move on $\mathbb{R}$, you want to move one circumference's length around the circle, i.e. $2\pi$. Think of $t$ as measuring the angle around the circle. The fractional part of $t$ is the fraction of the circle's circumference that it has gone around. – Dan Rust Aug 18 '13 at 17:26
• @DanielRust : If you type a\sim b you get $a\sim b$, but if you type a{\sim}b you get $a{\sim}b$, which looks wrong because it lacks the spacing before it and after it. But when you type "\mathbb R/\sim", then that spacing is inappropriate because in that context "$\sim$" is not being used to mean that what precedes it is related to what follows it. So I changed it to "\mathbb R/{\sim}", and as you can see, it looks different that way. – Michael Hardy Aug 18 '13 at 18:25
• To show that the function is continious one can show that the function $h\colon\mathbb{R}\to S^1, x\mapsto e^{2\pi ix}$ is continious (universal feature of final topology), right? And I guess one can simply say here: the complex exponential function is continious? So it remains to show the bijectivity of $f$ and the continity of $f^{-1}$? – math12 Aug 18 '13 at 18:38
• @MichaelHardy thanks for the LaTeX tip and the edit. – Dan Rust Aug 18 '13 at 19:08
I'm going to give a less technical answer than what others have written here, because I think you must have missed something other than that part if you didn't think of $x\mapsto e^{2\pi ix}$.
You have $x\sim y\Leftrightarrow x-y\in\mathbb{Z}$. That means $0$ becomes the same point as $1$, while everything between $0$ and $1$ is a different point from the one point that is $0\sim1$. And $0.1$ becomes the same point as $1.1$, and $0.2$ becomes the same as $1.2$, and so on. In other words, moving along the interval from $0$ to $1$, you return to your starting point and start over, just as with the circle. Therefore, the interval $[0,1]$ should get wrapped around the circle, with $0$ and $1$ getting mapped to the same point. In then the interval $[1,2]$, being the same as $[0,1]$ in the quotient space, gets wrapped around the circle in the same way, with $2$ being mapped to the same point on the circle to which $0$ and $1$ were mapped, and so on.
When you learned trigonometry, you saw $(\cos(2\pi x),\sin(2\pi x))$ wrapping around the circle in that same way. Since those are periodic functions with period $1$, it follows that if $x\sim y$ then $(\cos(2\pi x),\sin(2\pi x))=(\cos(2\pi y),\sin(2\pi y))$, so that is the mapping you need.
• This is a nice, detailed explanation for why the exponential map can be seen geometrically as 'wrapping' the real line around the circle. – Dan Rust Aug 18 '13 at 19:17
• And now I understood. I repeat it in my words: For example $[1]_{\sim}=\left\{...,0,1,2,3,...\right\}$ and all these points are "equal". So one has to give all these points the same point on the uni sphere. That means the same point as $1$ has on the uni sphere and this is the point (1/0) on the cirlce. And so one has to multiplicate with the factor $2\pi$ in order to go a full circle, two full cirlces and so on to the same point. – math12 Aug 18 '13 at 19:23
Try $f(t)=e^{2\pi i t}$ and find its kernel.
What about $f(x) = e^{2\pi i x}$? It factors through the quotient. | 2019-09-21T05:36:05 | {
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https://math.stackexchange.com/questions/3091953/putting-socks-and-shoes-on-a-spider/3091981 | # Putting socks and shoes on a spider
A spider needs a sock and a shoe for each of its eight legs. In how many ways can it put on the shoes and socks, if socks must be put on before the shoe?
My attempt:
If I consider its legs to be indistinguishable, then it's exactly the $$8^{\text{th}}$$ term of the Catalan sequence. However the legs are distinguishable. So the total number of ways equals $$\frac1{8+1} \binom{16}{8} (8!)^2$$.
Is this correct? Is there another way of doing it?
Edit: All socks and shoes are distinguishable.
• I think your factor of $(8!)^2$ is overcounting because it doesn't account for the constratint that the sock on leg $n$ has to go on before the shoe on leg $n$. Your count would allow sequences starting sock on leg $1$, shoe on leg $2$, ... for example. But I can't see a simple expression for the distinguishable legs case. – gandalf61 Jan 29 '19 at 9:48
• Are those 4 left shoes and 4 right shoes and 8 symmetrical socks? – Stian Yttervik Jan 30 '19 at 12:47
• Wow! Ambiguous questions are to be expected in beginning combinatorics questions but this one takes it to a whole new level! What is a "way of putting on shoes". Is it about which shoe and sock goes on which foot? Is it about what order you put things on on which foot? Are shoes and sock distinguishable. Are legs? – fleablood Feb 7 '19 at 17:02
You can imagine doing this as writing a sequence, say $$3453228156467781$$
What does it mean?
It means first put sock on leg $$\color{red}{3}$$ and on 4-th move put shoe on leg $$\color{red}{3}$$
then put sock on leg $$\color{blue}{4}$$ and on 11-th move put shoe on leg $$\color{blue}{4}$$ and so on...
So for each leg you must choose a pair in this sequence. On smaller number in this pair put a sock and the other shoe.
So we have $${16\choose 2}{14\choose 2}....{4\choose 2}{2\choose 2} = {16!\over 2!^8}$$
• I gave the tick to this answer because it's the easiest to understand. – abc... Jan 29 '19 at 20:55
• Am I the only one amazed by the fact that there are over 81 billion possibilities? – CTodea Jan 30 '19 at 12:03
• You gave $34;53;22;81;56;46;77;81$. Why does $81$ appear twice, at positions 4 and 8, or have I misunderstood the way the sequence is written? – jamesh625 Feb 4 '19 at 23:12
• Oh, I figured it out: you have $1,1,2,2, ..., 8,8$ and you just permute them to form that sequence above. – jamesh625 Feb 4 '19 at 23:14
No, it's not correct. Multiplying the Catalan number by $$8!$$ twice means we're choosing legs arbitrarily for each instance of a socking or a shoeing - with no regard to whether that leg has a sock on it, in the latter case. It's a substantial overcount.
Multiplying by $$8!$$ once would correspond to a "last in, first out" restriction; each time we put a shoe on, it's the most recently socked leg. That's an undercount, of course.
There are two actions per leg - the sock and then the shoe. All we need to know to determine a sequence is when each leg was worked on. That's $$\binom{16}{2}$$ for the first leg, $$\binom{14}{2}$$ for the second, and so on - or, equivalently, the multinomial coefficient $$\binom{16}{2,2,\dots,2} = \frac{16!}{(2!)^8}$$.
The answers given are correct, but I have a different (and possibly easier) way to think of it:
If you relax the condition for the sock/shoe event to be in order then there are $$2n$$ ($$=16$$) events hence $$16!$$ orderings. This is of course an over-count: many (most) of these are of an invalid ordering, with at least one sock over the top of a shoe. The question is by how many?
To answer this, suppose we divide them into groups. Specifically one group for each of the orderings of shoe/sock on each foot. For example:
• a group where: the sock on leg 1 is before on leg 1, the sock on leg 2 is before the shoe on leg 2 ..., the sock on leg 8 is before the shoe on leg 8 (the one we want)
• a group where: the sock on leg 1 is after on leg 1, the sock on leg 2 is before the shoe on leg 2 etc (not valid for us)
These groups are of the same size (none of them are special).
There are $$2^n$$ ($$2^8$$) of these groups so each of them, including the one we want, is of size: $$\frac{16!}{2^8}$$ or $$81,729,648,000$$.
Notes on generalisation:
Suppose $$n$$ legs and $$k$$ objects to put on in a given order. There are $$(k n)!$$ events, $$k!$$ orderings of any one of the leg's objects. Hence $$(k!)^n$$ orderings of the objects on each leg, only one of which is desired and so: $$\frac{(k n)!}{(k!)^n}$$ possible valid orderings of placing an object on a leg.
Another approach:
Let $$f(n)$$ denote the number of ways for an $$n$$-legged animal to put socks and shoes on all of their legs.
With one leg, there's only one way: Put the sock on, then the shoe.
With two legs (like the vast majority of humans), there are 6 possibilities. In greedoid's notation, these are:
• 1122 = Sock on leg #1, shoe on leg #1, sock on leg #2, shoe on leg #2
• 1212 = Sock on leg #1, sock on leg #2, shoe on leg #1, shoe on leg #2
• 1221 = Sock on leg #1, sock on leg #2, shoe on leg #2, shoe on leg #1
• 2112 = Sock on leg #2, sock on leg #1, shoe on leg #1, shoe on leg #2
• 2121 = Sock on leg #2, sock on leg #1, shoe on leg #2, shoe on leg #1
• 2211 = Sock on leg #2, shoe on leg #2, sock on leg #1, shoe on leg #1
Now, suppose that we've calculated $$f(k)$$ for some $$k$$. How does introducing a $$(k + 1)$$th leg affect the problem?
If you take any possible sequence of the $$2k$$ sock+shoe events for $$k$$ legs, then there are $$2k + 1$$ possible positions in the sequence to put the sock for the new leg (the $$2k - 1$$ positions between existing events, at the beginning, or at the end). Assume that we decide to put this new event after $$j$$ of the original events.
Now, let's decide when to put on the shoe for the new leg. This is trickier, because it depends on when we put on the sock. This new event can be inserted at index $$j + 1$$, $$j + 2$$, $$j + 3$$, ..., up to $$2k + 1$$, for $$2k + 1 - j$$ possibilies.
So, that gives us $$\sum\limits_{j=0}^{2k+1} (2k + 1 - j)$$ possibilities for when to add the sock and shoe for the new leg, which works out to the $$(2k + 1)$$th triangular number = $$\frac{(2k + 1)(2k + 2)}{2}$$. With $$k + 1 = n$$, this can be rewritten as $$\frac{(2n - 1)(2n)}{2} = n(2n - 1)$$.
We now have the recurrence relation $$f(n) = n(2n - 1)f(n-1)$$ with base case $$f(n) = 1$$. Or, in Python syntax.
>>> def f(n):
... if n == 1:
... return 1
... else:
... return n * (2 * n - 1) * f(n - 1)
...
>>> f(8)
81729648000
Proof that this is equivalent to the non-recursive formulation $$f(n) = \frac{(2n)!}{2^n}$$ is left as an exercise for the reader. | 2020-01-18T20:35:04 | {
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https://math.stackexchange.com/questions/434802/product-of-two-subgroups-the-whole-group | # Product of two subgroups the whole group?
I think I remember seeing a theorem about this in my Abstract Algebra class, but I cannot seem to find the reference for it anymore.
Let $G$ be a group and let $H$ and $K$ be non trivial subgroups of $G$. Are there any sufficient conditions on $H, K$ and $G$ such that $G = HK$? (In other words $\forall g \in G$, there exist $h \in H$ and $k \in K$ such that $g = hk$)
• Careful with notation. $H*K$ normally means the free product of $G$ with $H$. What you want is the product of subgroups $HK=\{hk\in G\:|\:h\in H, k\in K\}$. – Dan Rust Jul 2 '13 at 20:42
• Since you are only asking for sufficient conditions and didn't specify that they need to be proper subgroups: It is sufficient that at least one of the subgroups is $G$. – celtschk Jul 2 '13 at 20:42
• I just fixed it. Thanks for pointing that out. – metallicmural Jul 2 '13 at 20:44
• A particular example of this is the semidirect product en.wikipedia.org/wiki/Semidirect_product and so if $G=K\rtimes H$ then $KH=G$ almost by definition. This is not a necessary condition however. – Dan Rust Jul 2 '13 at 20:45
I will consider only the case when the groups are finite.
One situation that occurs with some frequency is when $|G| = |H||K|$, where $|\cdot|$ denotes the order of the group or subgroup, and the intersection $H\cap K = \{e\}$ is trivial. Then $HK = G$. To see this, consider the function (not group homomorphism!) $H \times K \to G$ that sends $(h,k) \mapsto hk$. Since $H\times K$ and $G$ have the same cardinality, this map is a bijection as soon as it is an injection. So suppose that $h_1k_1 = h_2k_2$. Then $h_2^{-1}h_1 = k_2k_1^{-1} \in H \cap K$, and so $h_2^{-1}h_1 = k_2k_1^{-1} = e$, so $h_2 = h_1$ and $k_2 = k_1$.
More generally, if $|H\cap K| = n$, then the "multiplication" function $H\times K \to G$ is exactly $n$-to-one. It follows that $G = HK$ iff $|G||H\cap K| = |H||K|$, since an $n$-to-one function is onto iff the domain has exactly $n$ times as many elements as the codomain.
The case when $G = HK$ and $H\cap K = \{e\}$ is so important, it has its own name: when this happens, $G$ is called the bicrossed product of $H$ and $K$, written $G = H \bowtie K$. (This generalizes the direct and semidirect products.) In particular, sometimes you know $H$ and $K$, but not $G$; then you need a little extra data to decide how to put them together.
For the semidirect product $H \ltimes K$, this extra data is a homomorphism $H \to \operatorname{Aut}(K)$, where $\operatorname{Aut}(K)$ denotes the group of group-automorphisms of $K$. I.e. you have a right action of $H$ on the set $K$ — I will write the action of $h\in H$ on $k\in K$ as $k\triangleleft h$ — such that $(k_1k_2)\triangleleft h = (k_1\triangleleft h)(k_2\triangleleft h)$ for all $h\in H$ and $k_1,k_2\in K$. Also $e \triangleleft h = e$ for all $h$.
For the bicrossed product, the extra data are two actions: a right action of $H$ on $K$, which I will write as $\triangleleft$, and a left action of $K$ on $H$, which I will write as $\triangleright$. So $k \triangleleft h \in K$, and $k\triangleright h \in H$. The axioms are: $$(k_1k_2) \triangleleft h = (k_1 \triangleleft (k_2 \triangleright h))(k_2 \triangleleft h) \\ k \triangleright (h_1h_2) = (k\triangleright h_1)((k\triangleleft h_1)\triangleright h_2) \\ k \triangleright e = e \\ e \triangleleft h = e$$ for all $h,h_1,h_2\in H$ and all $k,k_1,k_2\in K$. The semidirect product case is when $k \triangleright h = h$.
• The bicrossed product goes by many names, for example the Zappa-Szep product or the general product. I wrote wrote a general answer about them here. – user1729 Jul 2 '13 at 21:32
• thanks. I'm gonna try and see if I can apply this to what I'm doing. – metallicmural Jul 2 '13 at 22:37
Perhaps you have seen the result
If the indices in the finite group $G$ of the subgroups $H$ and $K$ are coprime, then $G = H K$.
Actually the result holds for any group, not necessarily finite.
## Proof
We have $$\lvert G : H \cap K \rvert = \lvert G : H \rvert \cdot \lvert H : H \cap K \rvert = \lvert G : H \rvert \cdot \lvert H K : K \rvert \le \lvert G : H \rvert \cdot \lvert G : K \rvert.\tag{eq}$$
From (eq) we obtain first that $\lvert G : H \rvert$ divides $\lvert G : H \cap K \rvert$. Similarly $\lvert G : K \rvert$ divides $\lvert G : H \cap K \rvert$.
Since the two indices are coprime, we have that $$\lvert G : H \rvert \cdot \lvert G : K \rvert\quad\text{divides}\quad\lvert G : H \cap K \rvert.$$
But (eq) also yields that $$\lvert G : H \cap K \rvert \le \lvert G : H \rvert \cdot \lvert G : K \rvert.$$ It follows that $$\lvert G : H \rvert \cdot \lvert G : K \rvert = \lvert G : H \cap K \rvert.$$
But then $$\lvert H K \rvert = \frac{\lvert H \rvert \cdot \lvert K \rvert}{\lvert H \cap K \rvert} = \frac{\lvert G : H \cap K \rvert}{\lvert G : H \rvert \cdot \lvert G : K \rvert } \cdot \lvert G \rvert = \lvert G \rvert,$$ so $G = H K$.
Please note that we are writing $HK = \{ h k : h \in H, k \in K \} \subseteq G$ without assuming that this is a subgroup.
Actually the proof holds for an arbitrary group, which is not necessarily finite. Just replace the final argument by $$\lvert H K : H \rvert = {\lvert K : H \cap K \rvert} = \frac{\lvert G : H \cap K \rvert}{\lvert G : K \rvert } = \lvert G : H\rvert,$$ which implies $H K = G$. | 2019-05-22T07:28:11 | {
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https://math.stackexchange.com/questions/473020/combining-symbols-with-symmetry | Combining symbols with symmetry
So this question has probably been answered already, but I can't find a good answer through searching google or this site. Basically, if I have n symbols, how many n-length combinations of the symbols can I make, excluding symmetrical duplicates and duplicates made by switching symbols for each other?
For instance, with the following sets of symbols you can get the following combinations:
{1,2}
11, 12
{1,2,3}
123, 112, 121, 111
(I'm only mostly sure that those are all the combinations for the set {1,2,3})
If you can point me to a previous question like this, or answer this one, that would be great!
Thanks in advance,
Numeri
• Are you looking for strings of length $n$? Two symbols could make $111$ or $112$ for example, if three symbols were allowed. – Mark Bennet Aug 21 '13 at 18:43
• Ah, yes sorry about that. I did mean of strings of length n. I'll edit the question now. Thanks! – Numeri says Reinstate Monica Aug 21 '13 at 18:56
• If I've done things correctly, the combinations for $n=4$ are $$1111, 1112, 1121, 1122, 1221, 1212, 1123, 1213, 1231, 2113, 1234$$ so the sequence for the number of $n$-length combinations starts $1,2,4,11$. You might try carefully doing another couple of examples and then see if the sequence matches up with anything at oeis.org – Barry Cipra Aug 21 '13 at 19:07
• You might consider counting symmetrical duplicates first and then use Burnside's Lemma to fix that afterwards. (Two-element group acting on the strings, the non-identity element mirrors) – Christoph Aug 21 '13 at 19:39
2 Answers
The problem boils down to finding the number of partitions of the set $\{1,2,\dots,n\}$ up to symmetry (i.e. $1\mapsto n$, $2\mapsto n-1$, ...), let's call this number $K_n$. First of all, the number of all partitions is the $n$-th Bell number $B_n$. From there we should be able to count the partitions up to symmetry using Burnside's lemma as $$K_n=\frac 1 2 \left(B_n + F_n\right),$$ where $F_n$ is the number of partitions that are invariant under the symmetry. Counting these is more difficult than I imagined, when I made my comment earlier.
To tackle the problem, I wrote a Sage script to find $K_n$ and $F_n$:
B = []
K = []
F = []
for n in range(1,8):
partitions = SetPartitions(n).list()
B_n = len(partitions)
for p in partitions:
flipped = p.apply_permutation(Permutations(n).identity().reverse())
if p != flipped:
partitions.remove(flipped)
K_n = len(partitions)
F_n = 2*K_n-B_n
B.append(B_n)
K.append(K_n)
F.append(F_n)
The results are \begin{align*} B_n &= (1, 2, 5, 15, 52, 203, 877, \dots),\\ K_n &= (1, 2, 4, 11, 32, 117, 468, \dots),\\ F_n &= (1, 2, 3, 7, 12, 31, 59, \dots). \end{align*}
Plugging the sequences into OEIS we find
• A000110: Bell numbers,
• A103293: Number of ways to color n regions arranged in a line such that consecutive regions do not have the same color,
• A080107: Number of fixed points of permutation of SetPartitions under $\{1,2,\dots,n\}\to\{n,n-1,\dots,1\}$.
Exactly what we were looking for. The formular given for $F_n$ on OEIS involes $q$-analog Bell numbers, which I haven't heard of before.
• Thank you for the answer!! Could you clarify to me though, what you mean by "the number of partitions of the set {1,2,…,n} up to symmetry"? Thanks! – Numeri says Reinstate Monica Aug 22 '13 at 1:39
• The Wikipedia article on Partitions explains what partitions of a set are. In your case, let $n=3$ then $112$ corresponds to $\{\{1,2\},\{3\}$ since the first and second position are labelled $1$, while the third position is labelled $3$. Flipping the numbers transforms this partition into $\{\{2,3\},\{1\}\}$ which is not the same partition, but up to symmetry it is. – Christoph Aug 22 '13 at 6:16
This is a case of Power Group Enumeration where we substitute $N$ colors into $N$ slots with the symmetric group acting on the colors and the group consisting of a single flip acting on the slots. Two Maple programs for this problem were posted at this MSE link.
We need the cycle index of the group $F_k$ containing the identity and a single flip of the row of slots into which we distribute the $N$ colors. We have that when $k$ is even, $$Z(F_k) = \frac{1}{2} a_1^k + \frac{1}{2} a_2^{k/2}$$ and when $k$ is odd, $$Z(F_k) = \frac{1}{2} a_1^k + \frac{1}{2} a_1 a_2^{(k-1)/2}.$$
This software gives the following generating functions for small numbers of symbols, e.g. for $N=3$ $${\it Q1\_Q1\_Q1}+{\it Q3}+2\,{\it Q1\_Q2}$$ and for $N=4$ $$3\,{\it Q2\_Q2}+{\it Q4}+2\,{\it Q1\_Q3}+{\it Q1\_Q1\_Q1\_Q1}+4\,{\it Q1\_Q1\_Q2}$$ and for $N=5$ $$6\,{\it Q1\_Q1\_Q1\_Q2}+9\,{\it Q1\_Q2\_Q2}+6\,{\it Q1\_Q1\_Q3}\\+{\it Q1\_Q1\_Q1\_Q1\_Q1}+{\it Q5}+6\,{\it Q2\_Q3}+3\,{\it Q1\_Q4}$$ and finally for $N=6$ $$9\,{\it Q2\_Q4}+3\,{\it Q1\_Q5}+{\it Q1\_Q1\_Q1\_Q1\_Q1\_Q1}+27\,{\it Q1\_Q1\_Q2\_Q2}\\+11\,{\it Q2\_Q2\_Q2}+9\,{\it Q1\_Q1\_Q1\_Q1\_Q2}+9\,{\it Q1\_Q1\_Q4 }+30\,{\it Q1\_Q2\_Q3}\\+10\,{\it Q1\_Q1\_Q1\_Q3}+{\it Q6}+7\,{\it Q3\_Q3}.$$ Counting the number of different sequences we obtain $$1, 2, 4, 11, 32, 117, 468, 2152, 10743, 58487, 340390, 2110219, 13830235, 95475556\ldots$$ which is indeed OEIS A103294. Here is another MSE link where the cycle index of the flip group was used.
• Thanks for the answer @Marko! This problem wasn't quite so simple as I thought it would be! :D – Numeri says Reinstate Monica Dec 28 '13 at 2:41 | 2020-07-04T12:35:18 | {
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https://math.stackexchange.com/questions/2280341/why-is-every-p-norm-convex | Why is every $p$-norm convex?
I know that $p$-norm of $x\in\Bbb{R}^n$ is defined as, for all $p\ge1$,$$\Vert{x}\Vert_p=\left(\sum_{i=1}^{n} \vert{x_i}\vert^p\right)^{1/p}.$$
The textbook refers to "Every norm is convex" for an example of convex functions.
I failed to prove $f(x)=\Vert{x}\Vert_p$ for all $p\ge1$, then tried to find the proof on the internet but I cannot find it.
Can someone let me understand why $p$-norm is convex for all $p\ge1$.
• en.wikipedia.org/wiki/Minkowski_inequality or did you want an intuition? Anyway using the triangle inequality: $||\lambda a+(1-\lambda) b|| \le ||\lambda a|| +||(1-\lambda) b||=\lambda ||a|| +(1-\lambda)||b||$ – Felix B. May 14 '17 at 10:41
• @FelixB. Triangular inequality?? Hmm... I will try it again. Thank you. – Danny_Kim May 15 '17 at 14:43
The Definition of a norm is:
Be V a Vectorspace, $$\|\cdot\|: V \rightarrow \mathbb{R}$$ is a norm $$:\Leftrightarrow$$
1. $$\forall v \in V: \|v\|\ge0$$ and $$\|v\| =0 \Leftrightarrow v=0$$ (positive/definite)
2. $$\forall v\in V, \lambda\in \mathbb{R}: |\lambda|\|v\| =\|\lambda v\|$$ (absolutely scaleable)
3. $$\forall v,w\in V : \|v+w\| \le \|v\|+\|w\|$$ (Triangle inequality)
The Definition of convex is:
$$f:V\rightarrow\mathbb{R}$$ is convex $$:\Leftrightarrow$$ $$\forall v,w \in V, \lambda \in [0,1]: f(\lambda v+(1-\lambda )w)\le \lambda f(v) +(1-\lambda)f(w)$$
So using the Triangle inequality and the fact that the norm is absolutely scaleable, you can see that every Norm is convex: $$\|\lambda v+(1-\lambda )w\|\le\|\lambda v\|+\|(1-\lambda)w\| = \lambda\|v\|+(1- \lambda)\|w\|$$
So by definition every norm is convex. What is left to show is, that the p-norm is in fact a norm.The first two Requirements are pretty easy to show, the third is hard. That is why it has its own name: the Minkowski Inequality which is a result of the Hölder inequality and shows that the triangle inequality holds for every p-norm (if p>1) and thus that it is a norm.
EDIT: Since this seems to be somewhat popular, I thought I would add a sketch of the proof of the minkowski inequality.
1. You show Young's Inequality: $$xy\le \frac{x^p}{p}+\frac{y^q}{q}\quad \forall q,p>1 \text{ with } \frac{1}{p}+\frac{1}{q}=1,\ \forall x,y\ge 0$$.
You can do that by looking at the function $$f(x)=\frac{x^p}{p}+\frac{y^q}{q}-xy$$ find the extremum, show it is a minimum and is greater zero (derivatives).
1. You show the Hölder Inequality: $$\|fg\|_1 \le \|f\|_p\|g\|_q \quad \forall q,p>1 \text{ with } \frac{1}{p}+\frac{1}{q}=1$$
You can do that by setting $$x=\frac{|f|}{||f||_p}$$ and $$y=\frac{|g|}{||g||_q}$$ and plug them into young's inequality. You get \begin{align} &&\frac{|fg|}{\|f\|_p\|g\|_q}&\le \frac{|f|^p}{p\|f\|_p^p} + \frac{|g|^q}{q\|g\|_q^q} \\ \Rightarrow &&\int \frac{|fg|}{\|f\|_p\|g\|_q} d\mu &\le \int \frac{|f|^p}{p\|f\|_p^p}d\mu + \int \frac{|g|^q}{q\|g\|_q^q}d\mu \\ \Rightarrow &&\frac{\|fg\|_1}{\|f\|_p\|g\|_q}&\le \frac{1}{p}+\frac{1}{q}=1 \end{align} It works just the same for sequences or $$\mathbb{R}^n$$, you just use young's inequality for every index and then sum over it instead of using the integral.
1. And last the Minkowski Inequality: $$\|x+y\|_p\le\|x\|_p+\|y\|_p \quad \forall p>1$$
Set $$q=\frac{p}{p-1}$$ thus $$q(p-1)=p$$ and $$\frac{1}{p}+\frac{1}{q}=1$$. Then: \begin{align} \|x+y\|_p^p&=\int |x+y|^pd\mu\le\int |x+y|^{p-1}|x|d\mu+ \int |x+y|^{p-1}|y|d\mu \\ &\le \left(\int|x+y|^{q(p-1)}d\mu\right)^{1/q}\left(\int|x|^pd\mu\right)^{1/p} + \left(\int|x+y|^{q(p-1)}d\mu\right)^{1/q}\left(\int|y|^pd\mu\right)^{1/p} \\ &=\left(\int|x+y|^{p}d\mu\right)^{\frac{1}{p}\frac{p}{q}}(\|x\|_p+\|y\|_p) =\|x+y\|_p^{p/q}(\|x\|_p+\|y\|_p) \end{align}
If you realize that $$p-\frac{p}{q}=p(1-\frac{1}{q})=1$$ you are done.
• This is the perfect explanation among what I have seen so far. Thank you. – Danny_Kim May 17 '17 at 4:25
• Is it correct that the : means the relation: "What is left of the : does/means what is right of the :? I derived that meaning from the function arrow description on en.wikipedia.org/wiki/List_of_mathematical_symbols, but I could not find a direct reference. – a.t. Feb 23 at 12:57
• @a.t. Yes, in $:\Leftrightarrow$ and $:=$, the colons indicate that it is a definition. The colon by itself between two statements is simply a separator and should be read as "such that". For example $\forall x\in X : f(x)=y$ "For all x in X such that f(x)=y" – Felix B. Feb 23 at 17:00 | 2019-05-22T13:51:51 | {
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https://math.stackexchange.com/questions/3599816/inflection-and-concavity | # Inflection and concavity
How do I find the intervals when the graph below is concave up and when it's concave down as well as its inflection coordinate points just by looking at the graph?
I'm able to eyeball it here, but I'm confused with a few parts. I'm pretty sure it concaves up in $$(0, 2)\cup(5, \infty)$$. I'm not sure if the region around $$x=4$$ should have its concavity considered because the derivative is not defined at this point. Then I suspect it concaves down in $$(2, 3.5)$$. Again, I'm not sure if the point after $$x=4$$ but just before $$x=5$$ should be considered an interval where it concaves down, but I doubt it.
Then I think the inflection coordinates are $$(1, 4), (3, 6), (5, 5)$$
• $(1, 4)$ is not an inflection point. Inflection is when $f'' = 0$, i.e. the slope that was increasing stops increasing or when it was decreasing and then starts increasing. Inflection point is somewhere between $x = 1$ and $x = 3$ as slope is continuous and is increasing at $x = 1$ ($f' > 0$) and decreasing at $x = 3$($f' < 0$) So by Rolle's theorem there must be $x \in[1, 3]$ such that $f'' = 0$. $(3, 6)$ is also not an inflection point. Mar 29 '20 at 7:36
• @ab123, that is not true. $f''=0$ doesn't always mean inflection point, for example consider $x\mapsto x^4$, then the double derivative of this function is zero at origin, but that is not a point of inflection. Mar 29 '20 at 8:17
• @Martund yes, sorry about that. I should have said - If $f''$ changes sign (from positive to negative, or from negative to positive) at a point $x = c$, then there is an inflection point located at $x = c$. Mar 30 '20 at 7:43
“Concave up” is like the graph of $$x^2$$ (any arc of it); “concave down” is like the graph of $$-x^2$$ (any arc of it).
An inflection point is where the curve has a tangent and is concave up on one side and down on the other side (not “globally”, but in some intervals).
At $$(1,4)$$ there is no change in concavity, and similarly at $$(3,6)$$. Indeed, these are a point of minimum and a point of maximum respectively.
The change in concavity happens somewhere in between $$1$$ and $$3$$ and the visual symmetry leads to guess that the inflection point is at $$(2,5)$$.
The point $$(5,5)$$ is indeed an inflection point (at least if we assume that the picture is “accurate”), because the curve is concave down before it and up past it.
What happens at $$(4,3)$$? This is a point where there is no tangent, so it has to be analyzed separately. The curve is concave up on either side of it, but it's nonetheless a point of minimum.
The intervals where the graph is concave up are: $$[0,2]$$, $$[5,8]$$; it is concave down on the intervals $$[2,4]$$, $$[4,5]$$.
Note that you should not say $$[2,5]$$, because the function is not concave down over the whole interval.
• Your answers for concavity were marked correct, only issue was I needed to use parentheses instead of brackets like you did. However, I'm still not sure what all the inflection points are according to your answer Mar 30 '20 at 5:27
• @Lex_i They are $(2,5)$ and $(5,5)$, as mentioned in my answer. If your instructor told you that closed intervals are wrong, then they are wrong. Another instance of this common error: the function is decreasing over $[0,1]$ and $[3,4}$, increasing over $[1,3]$ and $[4,8]$; using only open intervals would be wrong. Mar 30 '20 at 7:57 | 2021-09-26T07:06:13 | {
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https://math.stackexchange.com/questions/1230177/how-to-integrate-this-function-with-ln-and-u-substitution | # How to integrate this function with ln and u substitution?
I can get started in the right direction, but cant seem to get all of the way there, and any examples I can find don't have the same complications.
$$\int {2x\over (x-1)^2}\cdot dx$$
What I have tried:
I can't integrate this with the power rule, so my next step was to use ln and u-substitution, since I don't see how long division could help. I also recognized that I could factor out a 2.
I expanded the denominator: $(x-1)^2 = (x-1)\cdot(x-1) = x^2 -2x +2$
Let $u$ = $x^2 - 2x + 2 \therefore du = 2x - 2 \cdot dx$ using the power rule. , leaving me with:
$$2 \cdot \int {x \over u}\cdot dx$$
The problem is, $du = 2x-2 \cdot dx \neq x \cdot dx$. Even if I didn't factor out the 2 in the beginning I dont see how I could rectify this.
I don't see any way to get into the form of $\int {u'\over u} \cdot dx$ which I need to do.
I checked the answer in the book, which is $2\ln(\lvert x-1 \rvert) - {2 \over x-1} + C$
From that, I still cannot figure it out, but I do recognize that $2 \ln (|x-1|) = \ln((x-1)^2)$ as in the original denominator.
I'm stuck.
• Do you know what partial fractions are? – Bob Krueger Apr 11 '15 at 16:33
• I would try setting u to be x-1. A simple denominator results in a simpler fraction. – user1337 Apr 11 '15 at 16:33
• $(x - 1)^{2} = x^{2} - 2x + 1$ not $x^{2} - 2x + 2$.. – mattos Apr 11 '15 at 16:34
• @Mattos Oh. My God. I'd like to think I'm pretty darn good at math, but this takes the cake for the stupidest mistake I've ever made. – Bassinator Apr 11 '15 at 16:35
• @HCBPshenanigans Don't worry, we all make them. And use user1337's hint to solve your integral. – mattos Apr 11 '15 at 16:36
Let $x-1 = u$. Then $dx = du$ and $x=u+1$
This gives you the integral \begin{align} \int \frac{2(u+1)}{u^2}\,du &= 2\int \left(\frac 1u + u^{-2}\right)\,du\\ \\ &= 2\left(\ln|u| -u^{-1}\right) + C\\ \\ & = 2\ln|x-1| - \frac 2{u} + C\\ \\ &= \ln(x-1)^2 - \frac 2{x-1} + C \end{align}
• I think I can take it from here, lemme work some numbers and see where I get. – Bassinator Apr 11 '15 at 16:38
• Much appreciated! Once I realized that when I let $u = x-1$ that it could be rearranged as $x = u + 1$ it was so much easier. – Bassinator Apr 11 '15 at 16:51
• You're welcome! – Jordan Glen Apr 11 '15 at 16:54
Hint: You can rewrite the numerator as $$2x=2x-2+2=2(x-1)+2$$
• I see the validity, but how does that help (I think I see where you are going but I'd like to hear your words.) and how would I know to do it? – Bassinator Apr 11 '15 at 16:37
• Your edit just clarified it a lot. – Bassinator Apr 11 '15 at 16:38
• Can you proceed? – Fermat Apr 11 '15 at 16:42
• I think so. Working some numbers. – Bassinator Apr 11 '15 at 16:45
$$\int \frac{2x}{(x-1)^2} dx$$ The degree of numerator is 1 and of denominator is 2. Difference +1. This trick would work: $$N^r=A(D^r)'+B$$ Or: $$2x=A((x-1)^2)'+B=2A(x-1)+B$$ You can solve this comparing coefficients of same powers of x or by just putting some random values in there: $$2x=2Ax+B-2A$$ Now: $$2A=A\implies A=1,B-2A=0\implies B=2A=2$$ So: $$\int \frac{2x}{(x-1)^2} dx=\int \frac{[(x-1)^2]'+2}{(x-1)^2} dx=\int \frac{[(x-1)^2]'}{(x-1)^2} dx+\int \frac{2}{(x-1)^2} dx$$ And we know: $$\int \frac{f'}{f}dx=\ln f+c\text{ and }\int\frac{dx}{(ax+b)^2}\stackrel{u=ax+b}=\int\frac1a\frac{du}{u^2}=-\frac1{au}+c=-\frac1{a(ax+b)}+c$$ So: $$\int \frac{2x}{(x-1)^2} dx=\ln((x-1)^2)+2\ln|x-1|+c\\ =2\ln|x-1|-2\frac1{x-1}+c$$ | 2020-02-17T16:49:02 | {
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http://math.stackexchange.com/questions/135872/showing-that-a-set-is-open-in-a-topological-space/135881 | Showing that a set is open in a topological space
I have come across this exam question on a past paper.
$(X,T)$ is a topological space.
Suppose $A \subseteq X$ is such that for every $x \in A$ there is an open set $B$ such that $x\in B \subseteq A$. Show that $A$ is open.
Is it right to say that $A$ is open because it is the infinite union of sets $B$ which are open?
Is my reasoning enough to 'show' that $A$ is open ?
-
Yes, your reasoning is enough. Denote $B_{x}$ as the open neighbourhood $B\subset A$ of $x$. What you need to show is that $A=\cup_{x\in A}B_{x}$. – Thomas E. Apr 23 '12 at 16:20
The word "infinite" is unnecessary, though. – Arturo Magidin Apr 23 '12 at 16:20
Yes, your idea is correct. For each $x \in A$, let $B_x \subseteq A$ be the corresponding open neighborhood of $x$. Then $$A = \bigcup_{x \in A} B_x.$$ Note that equality holds since $B_x \subseteq A$ for each $x$, and each point $x \in A$ is covered by the union. Since an arbitrary union of open sets is open, $A$ itself is open.
-
The word "infinite" is unnecessary.
But it is true that $A$ is a union (possibly infinite, but that is irrelevant) of open sets, hence is open.
I'll try to avoid what seems like a use of AC (for each $x$, pick a $B_x$ with the appropriate properties). For each $x\in A$, let $$\mathcal{B}_x = \{B\subseteq X\mid B\text{ is open, }x\in B,\text{ and }B\subseteq A\}.$$ We know, by assumption, that $x\in\bigcup\mathcal{B}_x$ (since the set is not empty). (Recall that if $X$ is a set whose elements are sets, then $\bigcup X$ is the union of the elements of $X$). Moreover, $\bigcup\mathcal{B}_x$ is a union of open sets, hence open, and every element of $\mathcal{B}_x$ is contained in $A$, hence $\bigcup\mathcal{B}_x$ is contained in $A$.
Now consider $$B = \bigcup_{x\in A}\left(\bigcup\mathcal{B}_x\right).$$ This is open, being a union of open sets; contained in $A$, being a union of sets contained in $A$; and contains $A$, since for each $x\in A$, $x\in\bigcup\mathcal{B}_x\subseteq B$. Thus, $A=B$, so $A$ is open.
-
+1: Good illustration of the standard trick of "don't pick one, pick all!" – Zhen Lin Apr 23 '12 at 17:37
Fix $x \in A$. Since $x \in B$ and $B$ is open, there is a neighborhood $U$ of $x$ such that $x \in U \subset B$. But then $x \in U \subset A$. Therefore each point of $A$ is an interior point, and $A$ is open in $X$. | 2014-03-11T03:05:03 | {
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https://mathhelpforum.com/threads/geometric-distributions-problem.227031/ | # Geometric Distributions - Problem
#### zikcau25
Quote: A Concise Course in Advanced Statistics - Crawshaw, Pg. 277 Question 10.
X~Geo(p) and the probability that the first success is obtained on the second attempt is 0.1275.
If p > 0.5, find P(X > 2)
.
Textbook ANS: 0.7225
[HR][/HR]Reminding, Geometric Model:
$$\displaystyle P(X=r) = q^{r-1}p$$, $$\displaystyle r = 1, 2, 3..\infty.$$ And to the limit $$\displaystyle \sum_{r=1 }^{\infty }P(X=r)=1$$
Where,
r : rth attempt
q : the probability of failure
p : the probability of success
[HR][/HR]
p (the probability of success) is a constant, but I have trouble understanding what p > 0.5 really means from the quoted question above.
Does it means a range of values of p then how to solve the question,
or it just simply requires one constant p = 0.6.
If that is so, there I am trying to solve it as follows:
$$\displaystyle q^{2-1}p=0.1275$$
$$\displaystyle P(>2)=1-P(X\leq 2)$$
Therefore,
$$\displaystyle 1-P(X\leq 2)=1-\left [ P(1) +P(2)\right ]=1-\left [ p+qp \right ]=1-\left [ {\color{Red} 0.6}+0.1275 \right ]$$ = 0.2725 $$\displaystyle \neq 0.7225$$
You are welcome to give your help. Thanks. (Hi)
Last edited:
#### HallsofIvy
MHF Helper
In order that "the first success occurs on the second trial", there must be failure on the first trial and a success on the second. The probability of failure on the first attempt is 1- p and the probability of success on the second trial is (1- p)p. So the probability of "first success on the second trial is $$\displaystyle (1- p)^2p= 0.1275$$
Solve p- p^3= 0.1275 or p^3- p- 0.1275. You don't say how you got p= 0.6. It certainly does not satisfy that. I get between 9.0 and 9.1.
1 person
#### Plato
MHF Helper
X~Geo(p) and the probability that the first success is obtained on the second attempt is 0.1275.
If p > 0.5, find P(X > 2).
Textbook ANS: 0.7225
I think that the given answer should be $0.0225$. It is just a typo.
$\mathcal{P}(X=2)=(1-p)p=0.1275$ gives $p=0.85$ (that is p>.5).
So $1-(0.85+0.1275)=0.0225$.
1 person
#### zikcau25
Thanks to Sir HallsofIvy, for reminding me to evaluate, quadratically, two values of p from the relationship $$\displaystyle p = (1- q)$$ to choose only one value satisfying $$\displaystyle P > 0.5.$$
And Secondly, to Sir Plato for confirming the textbook printing error for $$\displaystyle P > 0.5$$ instead of $$\displaystyle P < 0.5$$
Similar threads | 2019-12-14T02:34:20 | {
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# operations for odd and even numbers- HELPFUL
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Intern
Joined: 11 Oct 2009
Posts: 11
### Show Tags
10 Jan 2010, 15:49
I’ve been asking around, doing a bit of research and have compiled this extra list of generalizations for odd/even number operations that will be helpful come test day, especially for data sufficiency.
Could people confirm my thoughts? It would be awesomely appreciated.
1) Odd subtract Even = always Odd
2) Even subtract Odd = always Odd
3) Odd subtract Odd = always Even
4) Odd divided by Odd never equals even (will only result in either Odd numbers or fractions)
(This may have been discussed before, sorry if it’s redundant, but I think these are really helpful and important to get right!)
Manager
Joined: 27 Apr 2008
Posts: 173
### Show Tags
10 Jan 2010, 15:58
a), b), and c) are right. Subtracting a number is just like adding a negative number, so the odd/even rules still apply.
For d), consider this:
odd number = 2n+1 (for any n)
Let the bigger odd number be 2n+3 and smaller odd number be 2n+1.
So if we have $$\frac{2n+3}{2n+1}$$
=$$\frac{(2n+1)+2}{2n+1}$$
=($$\frac{2n+1}{2n+1}$$)+($$\frac{2}{2n+1}$$)
=1 + 2* $$\frac{1}{2n+1}$$
1 is always odd. The second part is always even because it is a multiple of 2. So we have
= odd + even
= always odd
Manager
Joined: 04 Feb 2007
Posts: 79
### Show Tags
10 Jan 2010, 16:22
1
glender wrote:
I’ve been asking around, doing a bit of research and have compiled this extra list of generalizations for odd/even number operations that will be helpful come test day, especially for data sufficiency.
Could people confirm my thoughts? It would be awesomely appreciated.
1) Odd subtract Even = always Odd
2) Even subtract Odd = always Odd
3) Odd subtract Odd = always Even
4) Odd divided by Odd never equals even (will only result in either Odd numbers or fractions)
(This may have been discussed before, sorry if it’s redundant, but I think these are really helpful and important to get right!)
You don't really need to memorize this, just try simple numbers and they will hold for all similar numbers.
1) 3-2 = 1 (odd) --> odd-even = odd for all numbers
2) 6-3 = 3 (odd) --> even - odd = odd for all number
3) 5-3 = 2 (even) --> odd-odd = even for all numbers
4) 9/3 = 3 (odd) --> odd/odd = odd for all numbers (an odd number will not have an even factor)
_________________
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### Show Tags
18 Jan 2010, 09:39
If the question is x - y, just remember that the only way to get an even number is if BOTH x and y are ODD or EVEN.
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07 Nov 2017, 20:51
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Re: operations for odd and even numbers- HELPFUL &nbs [#permalink] 07 Nov 2017, 20:51
Display posts from previous: Sort by | 2018-12-10T04:21:42 | {
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https://www.omnicalculator.com/math/rectangular-to-polar | # Rectangular to Polar Coordinates Calculator
Created by Gabriela Diaz
Reviewed by Anna Szczepanek, PhD
Last updated: Feb 02, 2023
Welcome to Omni's rectangular to polar coordinates calculator, the tool that makes converting rectangular to polar coordinates an easy task!
If you've ever wondered what rectangular coordinates are or how they differ from polar coordinates, this is your place. Here, we'll also learn how to convert rectangular coordinates to polar coordinates with some simple trigonometry and the Pythagorean theorem.
Let's get started! 🏁
## What are rectangular coordinates?
If we want to indicate the location of a point in space, we can do so through the so-called rectangular coordinate system.
This system, also known as the Cartesian coordinate system in two dimensions, allows us to easily represent the position of any point as an ordered pair of values (x, y), where x is the horizontal coordinate associated with the x-axis and y is the vertical coordinate linked to the y-axis.
## What are polar coordinates?
The polar coordinate system is another useful coordinate system we can use to represent a point's position. By specifying the radius (or distance from a reference point) and the angle of rotation, we can describe the location of a point.
Often we use the letter r to denote the radius and the Greek letter θ (theta) to represent the angle. Similar to the rectangular coordinate system, in the polar form, we express the position of a point as an ordered pair (r, θ).
At this point, you might be wondering: Is the rectangular coordinate system better than the polar? When to use one or the other is mainly determined by which notation is more convenient for the type of problem you have to solve. Usually, when working with circles, cylinders, rotational motion, or periodic problems, we'll find the polar coordinates easier to apply. In contrast, the rectangular form could complicate the math of these problems.
## How do I convert rectangular coordinates to polar coordinates?
To convert from the rectangular to the polar form, we use the following rectangular coordinates to polar coordinates formulas:
r = √(x² + y²)
θ = arctan(y / x)
Where:
• x and y — Rectangular coordinates;
• r — Radius of the polar coordinate; and
• θ — Angle of the polar coordinate, usually in radians or degrees.
With these results, we express the polar coordinate as: (r, θ).
Notice that we can express a specific rectangular point as more than one polar point. For example, the rectangular coordinate (1, √3) can be written in the polar form as (2, π/3), or (2, 7π/3), or (2, 13π/3), or (2, π/3 + 2πn), or even (2, -5π/3). Here π represents the angular measure in radians of half a circle (or 180°) and n the number of cycles.
In this case, we have a unique value for the radius (r = 2) and a wide number of possible angles that can suit and perfectly represent the rectangular point (1, √3).
## Using the rectangular to polar coordinates calculator
The rectangular to polar coordinates calculator couldn't make converting from rectangular to polar coordinates easier. By just entering the x and y values of your problem, the calculator will instantly display the r and θ components of the polar coordinate representation of your point. Yes, it's that simple! 😎
## More coordinates conversion tools
If you enjoyed using this rectangular to polar coordinates calculator, then you might like to take a look at some of our other coordinates conversion tools:
## FAQ
### What's the difference between rectangular and polar coordinates?
We use rectangular coordinates to represent any point in a plane as the ordered pair (x, y), whereas, with polar coordinates, we can locate this same point by indicating its radius r from the origin and the counterclockwise angle from the x-axis θ, as (r, θ).
### Are rectangular and cartesian coordinates the same?
Yes. In two dimensions, the cartesian coordinates are also known as rectangular coordinates. This type of coordinate notation allows us to represent any point in a plane as a pair of elements (x, y).
### How do I convert the rectangular coordinate (3, 4) to polar coordinates?
To convert the rectangular coordinate (3, 4) into polar coordinates:
1. To calculate the radius r, we use:
r = √(x² + y²)
2. Substitute the values for x = 3 and y = 4 and perform the corresponding operations:
r = √(3² + 4²)
r = 5
3. Calculate the angle θ = arctan(y / x), by substituting the x and y values:
θ = arctan(4 / 3)
θ = 53.13°
4. Finally, indicate the polar coordinate as a pair:
(r, θ) = (5, 53.13°)
Gabriela Diaz
Rectangular (x, y) to Polar (r, θ)
x
y
r
θ
deg
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https://math.stackexchange.com/questions/1710265/find-the-ratio-of-frac-int-01-left1-x50-right100-dx-int-0/1710307 | # Find the ratio of $\frac{\int_{0}^{1} \left(1-x^{50}\right)^{100} dx}{\int_{0}^{1} \left(1-x^{50}\right)^{101} dx}$
$$I_1=\int_{0}^{1} \left(1-x^{50}\right)^{100} dx$$ and $$I_2=\int_{0}^{1} \left(1-x^{50}\right)^{101} dx$$ Then find $\frac{I_1}{I_2}$
I tried by subtracting $I_1$ and $I_2$
$$I_1-I_2=\int_{0}^{1}\left(1-x^{50}\right)^{100}\left(1-(1-x^{50}\right))dx$$ so
$$I_1-I_2=\int_{0}^{1} \left(1-x^{50}\right)^{100} x^{50} dx$$ Now using Integration by Parts we get
$$I_1-I_2= \left(1-x^{50}\right)^{100} \times \frac{x^{51}}{51}\bigg|_{0}^{1} -\int_{0}^{1} 100 \left(1-x^{50}\right)^{99} \times -50 x^{49} \times \frac{x^{51}}{51} dx$$ So $$I_1-I_2=\frac{5050}{51} \times \int_{0}^{1}\left(1-x^{50}\right)^{99} x^{100} dx$$
Now $x^{100}=\left(1-x^{50}\right)^2-(1-x^{50}-x^{50})$ so
$$\frac{51}{5050}(I_1-I_2)=\int_{0}^{1} \left(1-x^{50}\right)^{99} \times \left(\left(1-x^{50}\right)^2-(1-x^{50})+x^{50}\right)dx=I_2-I_1+\int_{0}^{1} \left(1-x^{50}\right)^{99} x^{50} dx$$
Need a hint to proceed further.
Hint: Do integration by parts on $I_2$ first, then consider $I_1-I_2$.
• ya i got it thank you – Umesh shankar Mar 23 '16 at 14:34
Factoring the integrand gives $$\int_0^1\left(1-x^{50}\right)^{101}\,\mathrm{d}x =\int_0^1\left(1-x^{50}\right)^{100}\,\mathrm{d}x -\int_0^1\left(1-x^{50}\right)^{100}x^{50}\,\mathrm{d}x$$ Integration by parts gives $$\int_0^1\left(1-x^{50}\right)^{101}\,\mathrm{d}x =101\cdot50\int_0^1\left(1-x^{50}\right)^{100}x^{50}\,\mathrm{d}x$$ Combining yields $$\frac{\int_0^1\left(1-x^{50}\right)^{100}\,\mathrm{d}x}{\int_0^1\left(1-x^{50}\right)^{101}\,\mathrm{d}x}=\frac{5051}{5050}$$
Another approach is to note that $$\int_0^1\left(1-x^a\right)^bdx=\int_0^1\left(1-y\right)^b\tfrac{1}{a}y^{1/a-1}dy=\frac{1}{a}\text{B}\left(\frac{1}{a},\,b+1\right)=\frac{b!\Gamma\left(\tfrac{1}{a}+1\right)}{\Gamma\left(b+\tfrac{1}{a}+1\right)},$$ so $$\frac{\int_0^1\left(1-x^{50}\right)^{100}dx}{\int_0^1\left(1-x^{50}\right)^{101}dx}=\frac{\Gamma\left(102.02\right)}{101\Gamma\left(101.02\right)}=\frac{5051}{5050}.$$
Notice, when $\Re[n]>0$:
• $$\int_{0}^{1}\left(1-x^{\frac{n}{2}}\right)^n\space\text{d}x=\frac{\Gamma(n+1)\Gamma\left(\frac{2+n}{n}\right)}{\Gamma\left(1+\frac{2}{n}+n\right)}$$
• $$\int_{0}^{1}\left(1-x^{\frac{n}{2}}\right)^{n+1}\space\text{d}x=\frac{\Gamma(n+2)\Gamma\left(\frac{2+n}{n}\right)}{\Gamma\left(2+\frac{2}{n}+n\right)}$$
So:
$$\frac{\int_{0}^{1}\left(1-x^{\frac{n}{2}}\right)^n\space\text{d}x}{\int_{0}^{1}\left(1-x^{\frac{n}{2}}\right)^{n+1}\space\text{d}x}=\frac{\frac{\Gamma(n+1)\Gamma\left(\frac{2+n}{n}\right)}{\Gamma\left(1+\frac{2}{n}+n\right)}}{\frac{\Gamma(n+2)\Gamma\left(\frac{2+n}{n}\right)}{\Gamma\left(2+\frac{2}{n}+n\right)}}=\frac{2+n+n^2}{n+n^2}$$
Now, when $n=100$:
$$\frac{\int_{0}^{1}\left(1-x^{\frac{100}{2}}\right)^{100}\space\text{d}x}{\int_{0}^{1}\left(1-x^{\frac{100}{2}}\right)^{100+1}\space\text{d}x}=\frac{2+100+100^2}{100+100^2}=\frac{10102}{10100}=\frac{5051}{5050}$$ | 2019-05-21T19:14:36 | {
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https://answerofmath.com/solved-approximating-binomial-distribution-with-normal-vs-poisson-2/ | Solved – Approximating Binomial Distribution with Normal vs Poisson
I have a doubt regarding when to approximate binomial distribution with Poisson distribution and when to do the same with Normal distribution.
It is my understanding that, when p is close to 0.5, that is binomial is fairly symmetric, then Normal approximation gives a good answer. However, when p is very small (close to 0) or very large (close to 1), then the Poisson distribution best approximates the Binomial distribution.
Also, when n is large enough to compensate, normal will work as a good approximation even when n is not close to 0.5 (n will work fine, but still Poisson will be better? )
However,consider the following question-
``The probability of any given policy in a portfolio of term assurance policies lapsing before it expires is considered to be 0.15. For a group of 100 such policies, calculate the approximate probability that more than 20 will lapse before they expire. ``
Here n is 100 and p is 0.15 (which is not close to 0.5). In this case, the exact answer is 0.0663. The normal approximated answer is 0.06178 and the Poisson approximated answer is 0.08297.
My doubt is that, since p is closer to zero than it is to 0.5, shouldn't the Poisson approximation yield a better answer?
Contents
Furthermore, looking at wiki (not always infallible!), according to NIST/SEMATECH, "6.3.3.1. Counts Control Charts", e-Handbook of Statistical Methods., Poisson is a good approximation for $$p < 0.05$$ (not 0.5) for $$n > 20$$. | 2023-03-27T19:31:20 | {
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"lm_q1q2_score": 0.8481971044202203
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https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12A_Problems/Problem_18&diff=cur&oldid=83260 | Difference between revisions of "2017 AMC 12A Problems/Problem 18"
Problem
Let $S(n)$ equal the sum of the digits of positive integer $n$. For example, $S(1507) = 13$. For a particular positive integer $n$, $S(n) = 1274$. Which of the following could be the value of $S(n+1)$?
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265$
Solution 1
Note that $n\equiv S(n)\bmod 9$, so $S(n+1)-S(n)\equiv n+1-n = 1\bmod 9$. So, since $S(n)=1274\equiv 5\bmod 9$, we have that $S(n+1)\equiv 6\bmod 9$. The only one of the answer choices $\equiv 6\bmod 9$ is $\boxed{(D)=\ 1239}$.
Solution 2
One possible value of $S(n)$ would be $1275$, but this is not any of the choices. Therefore, we know that $n$ ends in $9$, and after adding $1$, the last digit $9$ carries over, turning the last digit into $0$. If the next digit is also a $9$, this process repeats until we get to a non-$9$ digit. By the end, the sum of digits would decrease by $9$ multiplied by the number of carry-overs but increase by $1$ as a result of the final carrying over. Therefore, the result must be $9x-1$ less than original value of $S(n)$, $1274$, where $x$ is a positive integer. The only choice that satisfies this condition is $\boxed{1239}$, since $(1274-1239+1) \bmod 9 = 0$. The answer is $\boxed{D}$.
Solution 3
Another way to solve this is to realize that if you continuously add the digits of the number $1274 (1 + 2 + 7 + 4 = 14, 1 + 4 = 5)$, we get $5$. Adding one to that, we get $6$. So, if we assess each option to see which one attains $6$, we would discover that $1239$ satisfies the requirement, because $1 + 2 + 3 + 9 = 15$. $1 + 5 = 6$. The answer is $\boxed{D}$.
Solution 4(Similar to Solution 1)
Note that a lot of numbers can have a sum of $1274$, but what we use wishful thinking and want is some simple number $n$ where it is easy to compute the sum of the digits of $n+1$. This number would consists of basically all digits $9$, since when you add $1$ a lot of stuff will cancel out and end up at $0$(ex: $399+1=400$). We see that the maximum number of $9$s that can be in $1274$ is $141$ and we are left with a remainder of $5$, so $n$ is in the form $99...9599...9$. If we add $1$ to this number we will get $99...9600...0$ so this the sum of the digits of $n+1$ is congruent to $6 \mod 9$. The only answer choice that is equivalent to $6 \mod 9$ is $1239$, so our answer is $\boxed{D}$ -srisainandan6
~ pi_is_3.14 | 2021-10-18T21:33:10 | {
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https://kr.mathworks.com/help/symbolic/nchoosek.html | # nchoosek
Binomial coefficient
## Syntax
``b = nchoosek(n,k)``
``C = nchoosek(v,k)``
## Description
example
````b = nchoosek(n,k)` returns the binomial coefficient of `n` and `k`, defined as `n!/(k!(n - k)!)`. This is the number of combinations of `n` items taken `k` at a time.```
example
````C = nchoosek(v,k)` returns a matrix containing all possible combinations of the elements of vector `v` taken `k` at a time. Matrix `C` has `k` columns and `n!/(k!(n - k)!)` rows, where `n` is `length(v)`. In this syntax, `k` must be a nonnegative integer.```
## Examples
### Binomial Coefficients for Numeric and Symbolic Arguments
Compute the binomial coefficients for these expressions.
```syms n [nchoosek(n, n), nchoosek(n, n + 1), nchoosek(n, n - 1)]```
```ans = [ 1, 0, n]```
If one or both parameters are negative numbers, convert these numbers to symbolic objects.
`[nchoosek(sym(-1), 3), nchoosek(sym(-7), 2), nchoosek(sym(-5), -5)]`
```ans = [ -1, 28, 1]```
If one or both parameters are complex numbers, convert these numbers to symbolic objects.
`[nchoosek(sym(i), 3), nchoosek(sym(i), i), nchoosek(sym(i), i + 1)]`
```ans = [ 1/2 + 1i/6, 1, 0]```
### Handle Expressions Containing Binomial Coefficients
Many functions, such as `diff` and `expand`, can handle expressions containing `nchoosek`.
Differentiate the binomial coefficient.
```syms n k diff(nchoosek(n, 2))```
```ans = -(psi(n - 1) - psi(n + 1))*nchoosek(n, 2)```
Expand the binomial coefficient.
`expand(nchoosek(n, k))`
```ans = -(n*gamma(n))/(k^2*gamma(k)*gamma(n - k) - k*n*gamma(k)*gamma(n - k))```
### Pascal Triangle
Use `nchoosek` to build the Pascal triangle.
```m = 5; for n = 0:m C = sym([]); for k = 0:n C = horzcat(C, nchoosek(n, k)); end disp(C) end```
```1 [ 1, 1] [ 1, 2, 1] [ 1, 3, 3, 1] [ 1, 4, 6, 4, 1] [ 1, 5, 10, 10, 5, 1]```
### All Combinations of Vector Elements
Find all combinations of elements of a `1`-by-`5` symbolic row vector taken three and four at a time.
Create a `1`-by-`5` symbolic vector with the elements `x1`, `x2`, `x3`, `x4`, and `x5`.
`v = sym('x', [1, 5])`
```v = [ x1, x2, x3, x4, x5]```
Find all combinations of the elements of `v` taken three at a time.
`C = nchoosek(v, 3)`
```C = [ x1, x2, x3] [ x1, x2, x4] [ x1, x3, x4] [ x2, x3, x4] [ x1, x2, x5] [ x1, x3, x5] [ x2, x3, x5] [ x1, x4, x5] [ x2, x4, x5] [ x3, x4, x5]```
`C = nchoosek(v, 4)`
```C = [ x1, x2, x3, x4] [ x1, x2, x3, x5] [ x1, x2, x4, x5] [ x1, x3, x4, x5] [ x2, x3, x4, x5]```
## Input Arguments
collapse all
Number of possible choices, specified as a symbolic number, variable, expression, or function.
Number of selected choices, specified as a symbolic number, variable, expression, or function. If the first argument is a symbolic vector `v`, then `k` must be a nonnegative integer.
Set of all choices, specified as a vector of symbolic numbers, variables, expressions, or functions.
## Output Arguments
collapse all
Binomial coefficient, returned as a nonnegative scalar value.
All combinations of `v`, returned as a matrix of the same type as `v`.
collapse all
### Binomial Coefficient
If n and k are integers and 0 ≤ k ≤ n, the binomial coefficient is defined as:
`$\left(\begin{array}{c}n\\ k\end{array}\right)=\frac{n!}{k!\left(n-k\right)!}$`
For complex numbers, the binomial coefficient is defined via the `gamma` function:
`$\left(\begin{array}{c}n\\ k\end{array}\right)=\frac{\Gamma \left(n+1\right)}{\Gamma \left(k+1\right)\Gamma \left(n-k+1\right)}$`
## Tips
• Calling `nchoosek` for numbers that are not symbolic objects invokes the MATLAB® `nchoosek` function.
• If one or both parameters are complex or negative numbers, convert these numbers to symbolic objects using `sym`, and then call `nchoosek` for those symbolic objects.
## Algorithms
If k < 0 or n – k < 0, `nchoosek(n,k)` returns 0.
If one or both arguments are complex, `nchoosek` uses the formula representing the binomial coefficient via the `gamma` function.
## Version History
Introduced in R2012a | 2023-03-28T08:09:37 | {
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https://www.gradesaver.com/textbooks/math/calculus/calculus-early-transcendentals-8th-edition/chapter-12-section-12-4-the-cross-product-12-4-exercises-page-821/6 | ## Calculus: Early Transcendentals 8th Edition
$i+(sint-tcost)j-(tsint+cost)k$ Yes, $a \times b$ is orthogonal to both $a$ and $b$.
$a=ti+cost j+sint k= \lt t,cost,sint \gt$ $b=i-sint j+cost k= \lt 1,-sint,cost \gt$ $a\times b= \begin{vmatrix} i&j&k \\ t&cost&sint\\1&-sint &cost\end{vmatrix}$ Expand along the first row: $a \times b=i+(sint-tcost)j-(tsint+cost)k$ To verify that it is orthogonal to $a$, we will compute: $(a\times b).a=t(1)+cost(sint-tcost)+sint(-tsint-cost)=0$ To verify that it is orthogonal to $b$, we will compute: $(a\times b).b=1(1)-sint(sint-tcost)+cost(-tsint-cost)=0$ Yes, $a \times b$ is orthogonal to both $a$ and $b$. | 2019-11-12T14:00:28 | {
"domain": "gradesaver.com",
"url": "https://www.gradesaver.com/textbooks/math/calculus/calculus-early-transcendentals-8th-edition/chapter-12-section-12-4-the-cross-product-12-4-exercises-page-821/6",
"openwebmath_score": 0.9402733445167542,
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"lm_label": "1. Yes\n2. Yes",
"lm_q1_score": 0.9888419667789671,
"lm_q2_score": 0.85776809953619,
"lm_q1q2_score": 0.848197094585623
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http://www.saecof.com/szx3a7w/b9a6b6-euclidean-distance-in-r | To compute Euclidean distance, you can use the R base dist() function, as follow: dist.eucl <- dist(df.scaled, method = "euclidean") Note that, allowed values for the option method include one of: “euclidean”, “maximum”, “manhattan”, “canberra”, “binary”, “minkowski”. > Hello, > I am quite new to R.(in fact for the first time I am using) > So forgive me if I have asked a silly question. maximum: Maximum distance between two components of x and y (supremum norm) manhattan: Absolute distance between the two vectors (1 norm aka L_1). Given two sets of locations computes the full Euclidean distance matrix among all pairings or a sparse version for points within a fixed threshhold distance. Your email address will not be published. #calculate Euclidean distance between vectors, The Euclidean distance between the two vectors turns out to be, #calculate Euclidean distance between columns, #attempt to calculate Euclidean distance between vectors. Multiple Euclidean Distance Calculator R-script. 2) Creation of Example Data. Determine both the x and y coordinates of point 1. Thus, if a point p has the coordinates (p1, p2) and the point q = (q1, q2), the distance between them is calculated using this formula: distance <- sqrt((x1-x2)^2+(y1-y2)^2) Our Cartesian coordinate system is defined by F2 and F1 axes (where F1 is y … The Euclidean distance between two points in either the plane or 3-dimensional space measures the length of a segment connecting the two points. Euclidean distance may be used to give a more precise definition of open sets (Chapter 1, Section 1). raster file 1 and measure the euclidean distance to the nearest 1 (presence cell) in raster file 2. The Euclidean distance between two vectors, A and B, is calculated as: To calculate the Euclidean distance between two vectors in R, we can define the following function: We can then use this function to find the Euclidean distance between any two vectors: The Euclidean distance between the two vectors turns out to be 12.40967. Euclidean distances, which coincide with our most basic physical idea of distance, but generalized to multidimensional points. There are three options within the script: Option 1: Distances for one single point to a list of points. Numeric vector containing the first time series. But, when two or more variables are not on the same scale, Euclidean … canberra: sum(|x_i - y_i| / (|x_i| + |y_i|)). The Euclidean distance between the two columns turns out to be 40.49691. logical indicating if object should be checked for validity. Euclidean distance is a metric distance from point A to point B in a Cartesian system, and it is derived from the Pythagorean Theorem. View source: R/distance_functions.r. euclidean: Usual distance between the two vectors (2 norm aka L_2), sqrt(sum((x_i - y_i)^2)). Obviously in some cases there will be overlap so the distance will be zero. First, determine the coordinates of point 1. maximum: Maximum distance between two components of $$x$$ and $$y$$ (supremum norm) manhattan: Absolute distance between the two vectors (1 norm aka $$L_1$$). The Euclidean Distance. This function can also be invoked by the wrapper function LPDistance. Euclidean distance matrix Description. numeric scalar indicating how the height of leaves should be computed from the heights of their parents; see plot.hclust.. check. The Euclidean distance output raster The Euclidean distance output raster contains the measured distance from every cell to the nearest source. 4. These names come from the ancient Greek mathematicians Euclid and Pythagoras, but Euclid did not … What is Sturges’ Rule? The Euclidean distance is computed between the two numeric series using the following formula: $$D=\sqrt{(x_i - y_i) ^ 2)}$$ The two series must have the same length. Alternatively, this tool can be used when creating a suitability map, when data representing the distance from a certain object is needed. Looking for help with a homework or test question? Learn more about us. The Euclidean Distance procedure computes similarity between all pairs of items. x2: Matrix of second set of locations where each row gives the coordinates of a particular point. Furthermore, to calculate this distance measure using ts, zoo or xts objects see TSDistances. This script calculates the Euclidean distance between multiple points utilising the distances function of the aspace package. Euclidean distance is also commonly used to find distance between two points in 2 or more than 2 dimensional space. A euclidean distance is defined as any length or distance found within the euclidean 2 or 3 dimensional space. This distance is calculated with the help of the dist function of the proxy package. To calculate the Euclidean distance between two vectors in R, we can define the following function: euclidean <- function (a, b) sqrt (sum ((a - b)^2)) We can then use this function to find the Euclidean distance between any two vectors: For example, in interpolations of air temperature, the distance to the sea is usually used as a predictor variable, since there is a casual relationship between the two that explains the spatial variation. Computes the Euclidean distance between a pair of numeric vectors. You can compute the Euclidean distance in R using the dist () function. I am very new to R, so any help would be appreciated. > > I have a table in.csv format with data for location of samples in X, Y, Z > (column)format. Euclidean distances. proxy: Distance and Similarity Measures. While as far as I can see the dist() function could manage this to some extent for 2 dimensions (traits) for each species, I need a more generalised function that can handle n-dimensions. I am very new to R, so any help would be appreciated. Required fields are marked *. The matrix m gives the distances between points (we divided by 1000 to get distances in KM). canberra: $$\sum_i |x_i - y_i| / (|x_i| + |y_i|)$$. Euclidean Distance Example. version 0.4-14. http://CRAN.R-project.org/package=proxy. Usage rdist(x1, x2) fields.rdist.near(x1,x2, delta, max.points= NULL, mean.neighbor = 50) Arguments . How can we estimate the (shortest) distance to the coast in R? Statology is a site that makes learning statistics easy by explaining topics in simple and straightforward ways. We don’t compute the similarity of items to themselves. This article illustrates how to compute distance matrices using the dist function in R. The article will consist of four examples for the application of the dist function. Euclidean distance matrix Description. to learn more details about Euclidean distance. This distance is calculated with the help of the dist function of the proxy package. Numeric vector containing the second time series. Now what I want to do is, for each possible pair of species, extract the Euclidean distance between them based on specified trait data columns. Given two sets of locations computes the Euclidean distance matrix among all pairings. The Euclidean distance between two vectors, A and B, is calculated as: Euclidean distance = √ Σ(A i-B i) 2. Obviously in some cases there will be overlap so the distance will be zero. dist Function in R (4 Examples) | Compute Euclidean & Manhattan Distance . any R object that can be made into one of class "dendrogram".. x, y. object(s) of class "dendrogram".. hang. More precisely, the article will contain this information: 1) Definition & Basic R Syntax of dist Function. Euclidean distance is the distance in Euclidean space; both concepts are named after ancient Greek mathematician Euclid, whose Elements became a standard textbook in geometry for many centuries. Contents Pythagoras’ theorem Euclidean distance Standardized Euclidean distance Weighted Euclidean distance Distances for count data Chi-square distance Distances for categorical data Pythagoras’ theorem The photo shows Michael in July 2008 in the town of Pythagori First, if p is a point of R 3 and ε > 0 is a number, the ε neighborhood ε of p in R 3 is the set of all points q of R 3 such that d(p, q) < ε. This option is computationally faster, but can be less accurate, as we will see. x1: Matrix of first set of locations where each row gives the coordinates of a particular point. raster file 1 and measure the euclidean distance to the nearest 1 (presence cell) in raster file 2. I would like the output file to have each individual measurement on a seperate line in a single file. The Euclidean distance is computed between the two numeric series using the following formula: The two series must have the same length. Submitted by SpatialDataSite... on Wed, 12/10/2011 - 15:17. > > Can you please help me how to get the Euclidean distance of dataset . The distance to the sea is a fundamental variable in geography, especially relevant when it comes to modeling. euclidean: Usual distance between the two vectors (2 norm aka $$L_2$$), $$\sqrt{\sum_i (x_i - y_i)^2}$$. Im allgemeineren Fall des -dimensionalen euklidischen Raumes ist er für zwei Punkte oder Vektoren durch die euklidische Norm ‖ − ‖ des Differenzvektors zwischen den beiden Punkten definiert. The need to compute squared Euclidean distances between data points arises in many data mining, pattern recognition, or machine learning algorithms. Euclidean distance. Often, … Get the spreadsheets here: Try out our free online statistics calculators if you’re looking for some help finding probabilities, p-values, critical values, sample sizes, expected values, summary statistics, or correlation coefficients. In short, all points near enough to a point of an open set … Note that we can also use this function to calculate the Euclidean distance between two columns of a data frame: Note that this function will produce a warning message if the two vectors are not of equal length: You can refer to this Wikipedia page to learn more details about Euclidean distance. Description. This video is part of a course titled “Introduction to Clustering using R”. In this exercise, you will compute the Euclidean distance between the first 10 records of the MNIST sample data. We can therefore compute the score for each pair of nodes once. It can be calculated from the Cartesian coordinates of the points using the Pythagorean theorem, and is occasionally called the Pythagorean distance. If this is missing x1 is used. In the example below, the distance to each town is identified. How to calculate euclidean distance. (Definition & Example), How to Find Class Boundaries (With Examples). I would like the output file to have each individual measurement on a seperate line in a single file. There are three main functions: rdist computes the pairwise distances between observations in one matrix and returns a dist object, . Using the Euclidean formula manually may be practical for 2 observations but can get more complicated rather quickly when measuring the distance between many observations. Euklidischer Raum. The dist() function simplifies this process by calculating distances between our observations (rows) using their features (columns). Another option is to first project the points to a projection that preserves distances and then calculate the distances. The computed distance between the pair of series. Usage rdist(x1, x2) Arguments. The distances are measured as the crow flies (Euclidean distance) in the projection units of the raster, such as feet or … Because of that, MD works well when two or more variables are highly correlated and even if their scales are not the same. Your email address will not be published. Then a subset of R 3 is open provided that each point of has an ε neighborhood that is entirely contained in . > Now I want to calculate the Euclidean distance for the total sample > dataset. Details. 4. The Euclidean Distance tool is used frequently as a stand-alone tool for applications, such as finding the nearest hospital for an emergency helicopter flight. Arguments object. It is a symmetrical algorithm, which means that the result from computing the similarity of Item A to Item B is the same as computing the similarity of Item B to Item A. It is the most obvious way of representing distance between two points. David Meyer and Christian Buchta (2015). In mathematics, the Euclidean distance between two points in Euclidean space is a number, the length of a line segment between the two points. Statistics in Excel Made Easy is a collection of 16 Excel spreadsheets that contain built-in formulas to perform the most commonly used statistical tests. To calculate distance matrices of time series databases using this measure see TSDatabaseDistances. Next, determine the coordinates of point 2 . In der zweidimensionalen euklidischen Ebene oder im dreidimensionalen euklidischen Raum stimmt der euklidische Abstand (,) mit dem anschaulichen Abstand überein. Description Usage Arguments Details. R package rdist provide a common framework to calculate distances. 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Your field you please help me how to find distance between the 10... So any help would be appreciated aspace package correlated and even if their scales are the. Preserves distances and euclidean distance in r calculate the distances function of the proxy package this measure see TSDatabaseDistances,! Calculates the Euclidean distance to the coast in R script calculates the Euclidean distance in R using the formula! When it comes to modeling first project the points to a projection preserves... You can compute the Euclidean distance in R a dist object, open provided that point. > Now i want to calculate distance matrices of time series databases using this measure see.! Distance measure using ts, zoo or xts objects see TSDistances the pairwise distances between points we! Not the same length one matrix and returns a dist object, the distances... 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This tool can be used when creating a suitability map, when data representing the distance between two.. Each row gives the coordinates of a segment connecting the two numeric series the! Example ), how to find Class Boundaries ( with Examples ) the Pythagorean distance 16 Excel spreadsheets contain! Between two points in 2 or more variables are highly correlated and even if their scales not... We don ’ t compute the euclidean distance in r for each pair of nodes once first. This distance measure using ts, zoo or xts objects see TSDistances works well when two more! One matrix and returns a dist object, (, ) mit dem anschaulichen überein! From every cell to the sea is a fundamental variable in geography, especially relevant when it comes to.! One single point to a projection that preserves distances and then calculate the distance the... Are not the same for each pair of numeric vectors x1: matrix of first set of locations where row... 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Site that makes learning statistics easy by explaining topics in simple and straightforward ways than 2 space. We don ’ t compute the score for each pair of nodes once divided 1000! Can therefore compute the score for each pair of numeric vectors function of proxy. Segment connecting the two series must have the same \ ( \sum_i |x_i - y_i| / ( |x_i| |y_i|! In a single file the script: option 1: distances for one single point to projection! Abstand überein Abstand überein that is entirely contained in makes learning statistics by! When two or more than 2 dimensional space ( shortest ) distance to nearest. | 2021-06-17T04:27:03 | {
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https://www.physicsforums.com/threads/acceleration-under-gravity.751234/ | Acceleration under gravity
1. Apr 29, 2014
Govind_Balaji
1. The problem statement, all variables and given/known data
A ball is dropped from a balloon going up at a speed of 7 m/s. If the balloon was at a height 60 m at the time of dropping the ball, how long will the ball take to reach the ground?
Ans:4.3 seconds
2. Relevant equations
3. The attempt at a solution
I can't understand the question.
Is the initial velocity of the ball when it is dropped=0? or is it 7m/s.
What value should I take for g? Should I take g= 9.8 m/s$^2$?
Please help me. I assumed ball is dropped from 60 m height with initial velocity=0 and a=g=9.8 m/s$^2$. But it gives wrong answer. What is the concept of this question? How to solve it?
2. Apr 29, 2014
Staff: Mentor
g = 9.8 m/s2 is an acceptable value for g unless otherwise instructed.
The ball is being dropped (which means simply released) from a moving platform. It shares the motion of the platform until the instant it is released after which its trajectory is independent ("free fall").
3. Apr 29, 2014
Govind_Balaji
Let v be the final velocity when it reaches the ground.
Let x be the distance=60 m
Let u be the initial velocity=0 m/s
a=g=9.8 m/s
By 2nd equation of motion,
$x=ut+\frac{at^2}{2}$
$x=0*t+\frac{9.8*t^2}{2}$
$60=\frac{9.8*t^2}{2}$
$120=9.8t^2$
$t^2=\frac{120}{9.8}$
$t^2=2.44$
$t=\sqrt{2.44}$
$t=1.56$
Actually I should get $t=4.3$
4. Apr 29, 2014
Simon Bridge
What arguments from the problem statement can you make for either possibility? Something to do with the velocity of the balloon?
Also remember that velocity is a vector.
Acceleration is also a vector - is the acceleration of the ball in the same direction as the initial velocity?
If you make "downwards" positive - what are the values for initial velocity and acceleration going to be.
5. Apr 29, 2014
Staff: Mentor
Reconsider the initial velocity of the ball.
6. Apr 29, 2014
Govind_Balaji
I wrote everything that was in my text book. I said earlier that I didn't understand the question. I assumed the initial velocity to be 0. I am not certain.
7. Apr 29, 2014
Govind_Balaji
Yes I made downwards positive since it was all about downwards motion.
8. Apr 29, 2014
Govind_Balaji
My friend said that momentum of the balloon slows down the velocity of the ball. I think it is not right. Am I correct?
9. Apr 29, 2014
Simon Bridge
I can see you are struggling to understand the problem - but if you do not explain how you are thinking about it, we can only guess what you are having trouble with. Don't worry about sounding silly - we've all done that: we understand.
OK - imagine someone in the balloon holding the ball out over the side.
The balloon is going upwards, before the ball is released: what is the ball doing?
At the instant the ball is released, it still has the same velocity as just before it was released.
Now do you see?
If "downwards" is positive, what is the sign of the initial velocity of the ball?
10. Apr 29, 2014
Govind_Balaji
Thank you both, I found that initial velocity=-7 m/s. I also learnt that in a moving platform, an object in the platform shares the velocity with the platform. I found this by comparing me travelling in a train.
Here's my new attempt:
$$u=-7 m/s$$
$x=60 m$
$g=9.8 m/s^2$
$x=ut+\frac{at^2}{2}$
$60=-7t+\frac{(9.8)t^2}{2}$
$60=\frac{-14t+9.8t^2}{2}$
$120=-14t+9.8t^2$
Using quadratic equation formula, I got t=4.28$\approx$4.3 s
11. Apr 29, 2014
Simon Bridge
Well done:
You'd have got +4.3s and -2.9s ... you want the time that is in the future.
When you do long answers you should show that step.
BTW: good LaTeX use.
Reality check:
This time should be longer than if the ball was just dropped from a stationary balloon.
The calculation neglects air resistance. IRL air resistance cannot be neglected for such a long fall.
12. Apr 30, 2014
Govind_Balaji
Thank you. actually I started typing the answer before your last post. It took a long time to type LaTex(15-20 mins!!!)
13. Apr 30, 2014
Simon Bridge
It gets faster with practice.
BTW the "itex" boxes are for inline use (where an equation sits inside a paragraph), while "tex" boxes are used for display equations - those that get their own line. The display form is best when you have lots of exponents or fractions.
i.e. a quadratic $ax^2+bx+c=0$ (1) can be solved by: $$x\in\frac{-b\pm\sqrt{b^2-4ac}}{2a}\qquad \small{\text{...(2)}}$$... (1) is the "standard form" and (2) is the "quadratic equation".
If I put (2) in the inline form, it comes out like $x\in\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
You can do multi-line equations... watch this:
\begin{align} & & ax+by=e & \qquad ...(1) \\ & & cx+dy=f & \qquad ...(2) \\ (1)\rightarrow & & y=\frac{e-ax}{b} & \qquad ...(3)\\ (3)\rightarrow (2) & & cx+d\frac{e-ax}{b}=f & \\ & & \implies x =\frac{bf-de}{bc-ad} & \qquad ...(4) \end{align}
Of course (1) and (2) can be written as a matrix equation:
$$\begin{pmatrix} a & b\\ c & d \end{pmatrix} \begin{pmatrix} x\\ y\end{pmatrix} =\begin{pmatrix} e\\ f\end{pmatrix}$$
... just showing off :)
Last edited: Apr 30, 2014 | 2017-08-16T20:11:31 | {
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https://math.stackexchange.com/questions/2095188/boundary-condition-on-a-function-fx-and-its-first-derivative-at-x-rightarro | # Boundary condition on a function $f(x)$ and its first derivative at $x\rightarrow\pm\infty$
If a continuous function $f(x)$ of a real variable $x$ is such that $f(x)\rightarrow 0$ as $x\rightarrow \pm \infty$, does it necessarily mean that $\frac{df}{dx}\rightarrow 0$ as $x\rightarrow \pm \infty$? If yes, can I prove this?
If not, what is a counter-example where this is not true. The examples, I can think of $\frac{1}{x}, e^{-x}$ etc satisfy this.
• By the way, if you want this to be true, you need the requirement of monotonicity as $x\to\pm\infty$. – Simply Beautiful Art Jan 12 '17 at 19:07
• It's still not true with monotonicity @SimpleArt. See my post. There's also an explicit construction here: math.stackexchange.com/questions/788813/… – Kaj Hansen Jan 12 '17 at 20:10
• For this to be true you need $f''$ bounded. – MathematicsStudent1122 Jan 12 '17 at 21:41
Interestingly, this is not even true if we add a monotonicity hypothesis on the function. I don't know an explicit formula for this off the top of my head, but consider something like the following:
Essentially the function is constant between integer-values of its domain, and at each integer value, it takes a steep decline. The distance the function "drops" at each integer value limits to zero in such a way that we have $\displaystyle \lim_{x \rightarrow \infty} f(x) = 0$. However, because there always will be a steep drop at each integer value (albeit arbitrarily small drops), the derivative cannot limit to zero. Note that the "drops" can be smoothed out, say with something bump-function-esque, to make $f$ infinitely differentiable.
No. Consider the following:
$$f(x)=\frac{\sin(e^x)}x$$
as $x\to\pm\infty$, $f(x)\to0$. However,
$$f'(x)=\frac{xe^x\cos(e^x)-\sin(e^x)}{x^2}\to\text{DNE as }x\to+\infty$$
Here is the graph:
• very nice example! Do you mind adding a plot of the function to illustrate that the function goes more and more crazy in a smaller and smaller interval (on the y-axis) as $x\to\infty$? – Surb Jan 12 '17 at 19:00
• @Surb Done I do think :-) – Simply Beautiful Art Jan 12 '17 at 19:01
• :) thanks, also if you want the divergence on both sides you might consider $\sin(e^{x^2})/x$ – Surb Jan 12 '17 at 19:03
• @Surb True true... taking advantage that all you really need to do is make my function symmetric. – Simply Beautiful Art Jan 12 '17 at 19:04
• Actually, the limit isn't $\infty$, it doesn't exist. – Michael Hoppe Jan 12 '17 at 19:04 | 2020-10-25T13:47:20 | {
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http://math.stackexchange.com/questions/106381/use-the-definition-of-a-limit-to-show-that-lim-z-to-z-0-az-b-az-0-b | # Use the definition of a limit to show that $\lim_{z \to z_0} (az + b) = az_0 + b.$
Let $a,b, z_0$ denote complex constants. Use the definition of a limit to show that $$\lim_{z \to z_0} (az + b) = az_0 + b.$$
Here is what I have done:
\begin{align*} |az + b - (az_0 + b)| &= |az - az_0 + b - b|\\ &= |a(z - z_0)|\\ &= |a||z - z_0|. \end{align*}
So for a positive number $\epsilon$,
$$|az + b - (az_0 + b)| < \epsilon \text{ whenever } |a||z - z_0| < \epsilon$$
or in other words $|az + b - (az_0 + b)| < \epsilon$ whenever $|z - z_0| < \delta$ where $\delta = \epsilon/|a|$.
Have I proved the statement correctly?
-
Looks fine to me! – Martin Wanvik Feb 6 '12 at 18:09
Perfect, excepting one small possible issue you need to avoid in the end... You divide by $|a|$, and what happens if $|a|=0$? – N. S. Feb 6 '12 at 18:13
It looks like a textbook example of a $\delta$-$\epsilon$ proof to me, other than the point raised by N.S. – robjohn Feb 6 '12 at 18:16
So I just make a note of that in my proof?.. where d - e/|a|, |a| != 0 – Jim_CS Feb 6 '12 at 18:25
You can either split the end in two cases: case 1 $|a| \neq 0$, case 2: $|a|=0$, or, pretty standard trick, observe that $|a||z - z_0| < \epsilon$ happens when $|z-z_0| < \frac{\epsilon}{|a|+1}$. – N. S. Feb 6 '12 at 18:40
As mentioned in the comments, your proof works for $a \neq 0$. You can deal with the case where $a = 0$ separately or choose $\epsilon$ in such a way that the proof works for all $a$ (see N.S.'s comment for example). Aside from this your proof is correct.
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https://math.stackexchange.com/questions/2882686/how-do-you-calculate-the-probability-of-drawing-an-ace-with-multiple-attempts | # How do you calculate the probability of drawing an Ace with multiple attempts?
Given a shuffled deck of 52 cards, if you draw three cards (for example), how do you calculate the probability that at least one of those cards will be an Ace?
I've figured out that, for the first card drawn, the probability is $4/52$
And for the second card drawn, the probability is $4/51$ (unless the first card was an Ace, in which case it would be $3/51$)
(i.e. the drawn cards are not placed back into the deck.)
But I'm having trouble wrapping my head around the bigger picture.
I don't understand how you calculate the overall probability of any of the three cards being an Ace, accounting for the fact that multiple cards will be drawn.
What is the correct approach for this type of problem?
How do you break it down into smaller pieces, etc?
• I presume this is without replacement. The event is complementary to that of picking no aces in three draws. – Lord Shark the Unknown Aug 14 '18 at 17:16
• @LordSharktheUnknown I edited to clarify, but I think you understood correctly. – Giffyguy Aug 14 '18 at 17:20
• Take a look at math.stackexchange.com/questions/2254231/… -- does this help? – BallBoy Aug 14 '18 at 17:21
• $P_{\text {At least one card is an ace}}=1 - P_{\text {None of the cards is an ace}}$. – Mohammad Zuhair Khan Aug 14 '18 at 17:22
Consider the hypergeometric distribution.
This is a problem in which it is easier to find the probability of the complementary event and then subtract from $1.$
Let $X$ be the number of Aces in three draws without replacement. You seek $P(X \ge 1) = 1 - P(X = 0),$ where $P(X = 0) = \frac{{4 \choose 0}{48 \choose 3}}{{52 \choose 3}} = 0.7826.$
In R statistical software dhyper is a hypergeometric PDF:
dhyper(0, 4, 48, 3)
[1] 0.7826244
The probability $P(X \ge 1)$ can also be simulated in R. With a million 3-draw games, one can expect two or three places of accuracy.
deck = 1:52 # Let the Aces be 1, 2, 3, & 4
x = replicate(10^6, sum(sample(deck, 3) <= 4))
mean(x >= 1)
[1] 0.21758
Note: Related problem. If cards were drawn with replacement, then the number of Aces drawn would be $Y \sim \mathsf{Binom}(n=3, p=12/13)$ and $P(Y = 0) = (12/13)^4 = 0.7260.$
Probability of exactly two Aces in three draws without replacement. Using 'binomial coefficients' such as ${4 \choose 2} = \frac{4!}{2!\cdot 2!}=6:$ $$P(X = 2) = \frac{{4 \choose 2}{48 \choose 1}}{{52\choose 3}}.$$
choose(4,2)*choose(48,1)/choose(52, 3)
[1] 0.01303167
Using dhyper in R:
dhyper(2, 4, 48, 3)
[1] 0.01303167
Probability of geting at least 2 Aces in three draws is the sum of two probabilities:
$$P(X \ge 2) = \frac{{4 \choose 2}{48 \choose 1}}{{52\choose 3}} + \frac{{4 \choose 3}{48 \choose 0}}{{52\choose 3}}.$$
sum(dhyper(2:3, 4, 48, 3))
[1] 0.01321267
Graph of entire distribution for the number of Aces obtained in three draws without replacement from a standard deck.
x = 0:3; pdf = dhyper(x, 4, 48, 3) # 4-vectors
plot(x, pdf, type="h", lwd=3, col="blue")
abline(h=-.005, col="green3") # ref line just a bit below 0 ...
# ... to avoid hiding last bar
• Thanks so much for this. I had never seen the choose operator before, and I've been reading about it. Follow-up question: How would you modify your choose equation above, to find the probability of drawing two aces in three attempts? – Giffyguy Aug 17 '18 at 14:56
• In the "4 over 0", what does the 0 refers to ? – thanos.a Aug 31 '18 at 9:22
• In my example, I have 10000 coffee cups and only 1000 of them give a gift. If I drink 10 coffees then I interpret that the probability for me to win (once?) is given in R by: dhyper(0, 1000, 9000, 10). Correct ? – thanos.a Aug 31 '18 at 9:29
• In the hypergeometric formula the factor ${4 \choose 0}$ [read "4 choose 0"] would be the number of ways to choose no Aces out of the four Aces in the deck. Of course, ${4 \choose 0}=\frac{4!}{0!\cdot 4!} = 1,$ because $0!=1.$ In combinatorics "Do nothing" often counts as "one way." – BruceET Aug 31 '18 at 9:31
• @thanos.a This video explains it very well, and makes it much easier to understand how to adapt this to any given scenario: youtube.com/watch?v=BCeFgnh6A1U – Giffyguy Aug 31 '18 at 13:16 | 2019-09-15T16:00:39 | {
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http://mathematica.stackexchange.com/questions/58751/how-to-speed-up-my-project-euler-code | # How to speed up my Project Euler code
The evaluation speed of Mathematica often depresses me. It did it again when I wrote a code to solve a Project Euler problem. https://projecteuler.net/problem=206
Here's my code:
ClearAll[a];
a =
Compile[{},
NestWhile[# + 1 &, 10^9, ! MatchQ[
IntegerDigits[#^2], {1, _, 2, _, 3, _, 4, _, 5, _, 6, _, 7, _,8, _, 9, _, 0}] &],
CompilationTarget -> "C"]
I think it is the simplest solution. I can't find other solutions which are simpler. But my Mathematica runs a long time before getting the answer. How can I speed this up? I've run into similar problems while I was solving other problems.
-
Pattern matching is often slow, you might try something like IntegerDigits[#^2][[ ;; ;; 2]] == {1,2,3,4,5,6,7,8,9,0} instead. – Pickett Sep 3 '14 at 7:16
#+1 can be replaced by #+10, because a square number ends with "0" must ends with "00" – wuyingddg Sep 3 '14 at 8:20
If you realize that the maximum solution is 10 Floor[Ceiling[Sqrt[1929394959697989990]]/10] and step down by 10 from there, you'll get the solution almost immediately. – Mark McClure Sep 3 '14 at 12:38
Compilation is a certainly a good idea if you're going the brute-force route. So let's first tackle that, which will get us the answer in roughly 30 seconds computation time. Afterwards we'll come up with better strategy, which can do it in 0.3 seconds.
# Compilation
A performance pitfall when compiling functions is that sometimes the compiled function just calls the main evaluation loop, effectively gaining nothing. We can check this with CompilePrint:
Needs["CompiledFunctionTools"]
a // CompilePrint
(...)
Result = I3
1 I3 = MainEvaluate[(...)]
2 Return
The call to MainEvaluate is the culprit here.
So let's rewrite your function such that it does compile properly. Using Pickett's and wuyingddg suggestions, we end up with:
PerformSearch = Compile[
{
{startValue, _Integer},
{increment, _Integer}
},
NestWhile[
# + increment &,
startValue,
IntegerDigits[#^2][[-1 ;; 1 ;; -2]] =!= {9, 8, 7, 6, 5, 4, 3, 2, 1} &
]
]
Let's check if this did compile ok:
PerformSearch // CompilePrint
(...)
1 I4 = I0
2 I3 = I4
3 I7 = Square[ I3]
4 T(I1)2 = IntegerDigits[ I7, I5]]
5 T(I1)0 = Part[ T(I1)2, T(I1)1Span]
6 B0 = CompareTensor[ I6, R2, T(I1)0, T(I1)3]]
7 if[ !B0] goto 12
8 I3 = I4
9 I7 = I3 + I1
10 I4 = I7
11 goto 2
12 Return
Ok, that looks much better. We can now perform the search, but not after we've determined the range of possible solutions:
max = Floor @ Sqrt @ FromDigits @ Riffle[Range[9], 9] (* 138902662 *)
min = Ceiling @ Sqrt @ FromDigits @ Riffle[Range[9], 0] (* 101010102 *)
As an aside, note that max^2 is below $MaxMachineInteger on 64-bit systems. But on 32-bit it isn't, which causes PerformSearch to switch back to the uncompiled code. Keeping Mark McClure's comment in mind, we'll cheat a bit and start from the maximum to find immediately: result = PerformSearch[max, -1] 138901917 How much faster is this than starting from the minimum? (result - min)/(max - result) // N 50861.5 Roughly 50 thousand times faster, not bad! Starting from the minimum takes about 30 seconds on my machine. Last but not least, let's double-check if we've obtained the correct number: result^2 19293742546274889 If you multiply this with 100, you've got your desired integer. # Non-brute-force The approach above scanned roughly 38 million (!) integers in the worst-case scenario (starting from the minimum). Other answers to the OP's question have shown that you can and should go one better than that. Here's my take on an efficient general solution, using an iterative step-by-step process: ClearAll[ FindIntegerRoots, CheckIfEmpty, SelectDigitSequences, InsertNewDigits, ConvertPattern ]; FindIntegerRoots[ pattern : { (0|1|2|3|4|5|6|7|8|9|Verbatim[_]).. }, power_Integer: 2 ] /; power > 1 := With[ { maxRoot = Floor[FromDigits[pattern /. Verbatim[_] -> 9]^(1/power)] }, FromDigits /@ SelectDigitSequences[ Fold[ CheckIfEmpty @ SelectDigitSequences[ InsertNewDigits @ #1, maxRoot, power, ConvertPattern[pattern, #2] ] &, {{}}, Range @ IntegerLength @ maxRoot ], maxRoot, power, ConvertPattern[pattern, Length @ pattern] ] ~Quiet~ {CompiledFunction::cfnlts} ~Catch~ "isEmpty" ]; CheckIfEmpty[{}] := Throw[{}, "isEmpty"]; CheckIfEmpty[nonEmpty_] := nonEmpty; SelectDigitSequences = Compile[ { {digitSequences, _Integer, 2}, {maxRoot, _Integer }, {power, _Integer }, {numDigits, _Integer }, {positions, _Integer, 1}, {compareDigits, _Integer, 1} }, Module[ { powersOfTen = 10^# & /@ Range[Length @ First @ digitSequences - 1, 0, -1] }, Select[ digitSequences, And[ # <= maxRoot, IntegerDigits[#^power, 10, numDigits][[positions]] == compareDigits ] & @ ( powersOfTen.# ) & ] ] ]; InsertNewDigits = With[ { range = List /@ Range[0, 9] }, Compile[ { {digitSequences, _Integer, 2} }, Outer[#1 ~Join~ #2 &, range, digitSequences, 1] ~Flatten~ 1 ] ]; ConvertPattern[pattern_, length_] := With[ { trimmed = pattern[[Length@pattern - length + 1 ;; -1]] }, Sequence @@ { length, Flatten @ Position[trimmed, _Integer], DeleteCases[trimmed, Verbatim[_]] } ]; While the code might be long, it is nevertheless fast: FindIntegerRoots @ Riffle[Range[9], _] // AbsoluteTiming {0.284622, {138901917}} A performance increase of 100 over the above brute-force method! But how efficient is it? Here's a nice graph of the amount of checks it performs in this case: That's about 296 thousand checks in total; certainly much less than 38 million! # Some more concealed squares The nice thing about FindIntegerRoots is that it works on any pattern, not just the OP's case: FindIntegerRoots @ {_, 9} {3, 7} FindIntegerRoots @ { 1, _, 4} {12} FindIntegerRoots @ { 1, _, 6} {14} And if you're adventurous, you may even ask for roots of different powers: FindIntegerRoots[{_, _, _, 5}, 3] {5, 15} So let's have a bit of fun and search for more concealed squares. First, are there any partial concealed squares (i.e. of the form 1_2, 1_2_3, 1_2_3_4, etc)? FindIntegerRoots @ Riffle[Range @ #, _] & /@ Range @ 9 {{1}, {}, {}, {1312}, {}, {115256, 127334, 135254}, {}, {}, {138901917}} Flatten[%]^2 {1, 1721344, 13283945536, 16213947556, 18293644516, 19293742546274889} How about reverse concealed squares (i.e. 2_1, 3_2_1, 4_3_2_1, etc)? FindIntegerRoots @ Riffle[Reverse @ Range @ #, _] & /@ Range @ 9 {{1}, {}, {}, {}, {24171}, {}, {2657351, 2713399}, {}, {}} Flatten[%]^2 {1, 584237241, 7061514337201, 7362534133201} And finally, are there squares of the form 1_1_(...)_1_1? FindIntegerRoots @ Riffle[ConstantArray[1, #], _] & /@ Range[2, 9] {{11}, {119, 131}, {}, {}, {110369}, {}, {10065739}, {}} Flatten[%]^2 {121, 14161, 17161, 12181316161, 101319101616121} It seems there are many nice squares. If you happen to find a particularly nice one, let me know :) - Which version of Mathematica are you using? It's quite strange that your code generates the CompiledFunction::cfn warning in my v8 (Win 32bit & 64bit) and v9 (Win 32bit), but it does work without warning on Wolfram Cloud! – xzczd Sep 3 '14 at 14:15 @xzczd I've checked this on v9 on a Mac. The issue here is 32 vs 64 bit; $MaxMachineInteger should namely be bigger than FromDigits@Riffle[Range[9], 9], which it isn't on 32 bit. – Teake Nutma Sep 3 '14 at 16:06
That seems to be the reason, BTW the $MaxMachineInteger in my v8 (Win 64 bit) is the same as that in 32 bit, not sure if it's just the nature of v8… – xzczd Sep 4 '14 at 1:56 Is using Big O notation in "Roughly O(10^4) times faster" a correct statement? It is surely 10^4 times faster in this example, but what exactly does O(10^4) times faster mean? – user Sep 4 '14 at 15:29 @bruce14 What I meant here is that it's roughly 4 orders of magnitude faster; faster than 10^3 and slower than 10^5. – Teake Nutma Sep 4 '14 at 17:15 There is a method by using Catch and Throw, and it gives result in about 90s. I think it can be improve a lot by Paralilize or ParallelEvaluate, but failed to finish that. i = 10^9 + {30, 70}; isMatchQ = If[IntegerDigits[#^2][[1 ;; -1 ;; 2]] == {1, 2, 3, 4, 5, 6, 7, 8, 9, 0}, Throw[#]] &; Catch[Do[isMatchQ /@ i; i += 100, {10^9}]] Edit I have found a way to speed up it by ParallelTable, this is the code: len = ((Sqrt[1.93*10^16] - 1*10^8)/40 // IntegerPart)*10; isMatchQ = If[IntegerDigits[#^2][[1 ;; -1 ;; 2]] == {1, 2, 3, 4, 5, 6, 7, 8, 9}, Throw[#]] &; ParallelTable[n = 10^8 + {3, 7} + len*i; Catch[Do[isMatchQ /@ n; n += 10, {len / 10}]], {i, 0, 3}] // AbsoluteTiming (*{42.523432, {Null, Null, Null, 138901917}}*) I divided the whole range of n which makes 10^16 <= n^2 <= 1.93*10^16 into 4 parts, the length of each part is len. And I calculate these 4 parts in 4 kernels at the same time by ParallelTable, then get the result in a much shorter time, only 42.5s. (Sorry for my poor English....) - This non-paralell strategy takes less than 4 seconds in my stone age laptop (and doesn't use the "reverse searching" trick)- First we check for the possible 6 digits endings (after discounting the final "00" ending in the squared number): set= 2 Position[IntegerDigits[Range[10^5 + 1, 10^6-1, 2]^2][[All, -5;; -1;; 2]], {7,8,9}]+ 10^5 - 1; Then we complete the whole set of numbers to check by adding all the possible "heads" between the max and min: fullSet = Total[Tuples[{Range[101, 138] 10^6, Flatten@set}], {2}]; And then a quick search gets our number: Cases[IntegerDigits[fullSet^2], {1, _, 2, _, 3, _, 4, _, 5, _, 6, _, 7, _, 8, _, 9}] (*{{1, 9, 2, 9, 3, 7, 4, 2, 5, 4, 6, 2, 7, 4, 8, 8, 9}}*) adding the final zeroes: (* 1929374254627488900 *) Edit We can still cut the time in half if we see that the number should end in 3 or 7 for its square to end in 9: set = Join[10 (Position[IntegerDigits[Range[10^5 + #, 10^6, 10]^2][[All, -5 ;; -1 ;; 2]], {7, 8, 9}] - 1) + # & /@ {3, 7}] + 10^5; fullSet = Total[Tuples[{Range[101, 138] 10^6, Flatten@set}], {2}]; Cases[IntegerDigits[fullSet^2], {1, _, 2, _, 3, _, 4, _, 5, _, 6, _, 7, _, 8, _, 9}] - +1 Very concise and fast (~1.3 times faster than mine!). – Teake Nutma Sep 4 '14 at 20:31 @TeakeNutma Thanks! I halved the time in the edit.Less than 2 secs now on my poor man's machine :) – belisarius Sep 5 '14 at 2:41 That's pretty fast. But alas, I've updated my answer and now it's 2x faster than yours :). – Teake Nutma Sep 5 '14 at 10:11 Most PE problems are designed such that any computing system will buckle under the teraflops usually required to brute force a solution. Not so many flops required for this problem, but some experiments can dramatically reduce the number of tests required. The minimum possible square is 1020304050607080900, and the maximum square is 1929394959697989990, corresponding to a range of integers$n$from 1010101010 to 1389026623. Apparently, there are about 380 million integers to test. However, the only possible square ending in 0 must end in 00. So the last three digits of the square are ...900. Therefore, the integers$n$to be squared must all end in either 30 or 70, and the number of such integers to test is now only about 7.6 million. The required integer is unique, according to the problem statement, so a simple, direct use of ParallelSum with Boole is possible. Avoid MatchQ as @Pickett suggests. AbsoluteTiming[ ParallelSum[ n*Boole[IntegerDigits[n^2][[;; ;; 2]] == {1,2,3,4,5,6,7,8,9,0}], {n, 1010101030, 1390000000, 100}] + ParallelSum[ n*Boole[IntegerDigits[n^2][[;; ;; 2]] == {1,2,3,4,5,6,7,8,9,0}], {n, 1010101070, 1390000000, 100}]] (* {5.915985, 1389019170} *) Further investigation shows that there are only three possible 5-digit endings for the required square: {{8,0,9,0,0},{8,4,9,0,0},{8,8,9,0,0}}. The corresponding 5-digit integers whose squares have these endings are of the form$i+2500k$, where$i=\{10530,11970,10430,12070,10830,11670\}$, and$k=0,1,2,...35$. The beginning 5 digits of the integer range from$j=10101$to$j=13890$. Thus, integers$n$of the form$n=100000*j+i+2500k$have squares with the required 8_ 9_ 0 ending. Now there are only about 821 thousand integers$n\$ to test. Approximately 460 times fewer tests than a direct brute force approach.
AbsoluteTiming[
ParallelSum[
(100000 j + i + 2500 k) *
Boole[IntegerDigits[(100000j+i+2500k)^2][[;; ;; 2]] == {1,2,3,4,5,6,7,8,9,0}],
{j, 10101, 13890},
{k, 0, 35},
{i, {10530, 11970, 10430, 12070, 10830, 11670}}]]
(* {1.002618, 1389019170} *)
But nothing beats counting down from the maximum as @MarkMcClure comments.
-
Here's a slight improvement of @wuyingddg's Catch&Throw method. The key point is to make Throw working inside ParallelDo using the method mentioned in this post:
SetSharedFunction[pThrow];
pThrow[expr_] := Throw[expr]
isMatchQ =
If[IntegerDigits[#^2][[1 ;; -1 ;; 2]] == {1, 2, 3, 4, 5, 6, 7, 8, 9, 0}, pThrow[#]] &;
Catch[ParallelDo[
isMatchQ /@ (i + {30, 70}), {i, 10^9, Sqrt[193 10^16], 100}]] // AbsoluteTiming
`
-
It is much more brief than the ParallelTable one and spend nearly same time, pretty good! – wuyingddg Sep 3 '14 at 14:21 | 2015-02-01T22:49:59 | {
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https://math.stackexchange.com/questions/1754768/find-the-value-of-ab-2cb-sqrt3-ac/1841872 | # Find the value of $ab+ 2cb+\sqrt3 ac$?
Three positive real numbers $a,b,c$ satisfy the equations $a^2+\sqrt3 ab+b^2=25$, $b^2+c^2=9$ and $a^2+ac+c^2=16$ .Then find the value of $ab+ 2cb+\sqrt3 ac$?
Is there some way to find the desired value without actually finding values of $a,b,c$ or any other smart method to find $a,b,c$
• If you don't know the values of $a, b, c$, how do you expect to know the value of that expression? – ÍgjøgnumMeg Apr 22 '16 at 22:07
• @Ed_4434 By some form of algebraic manipulation of the three equations. Whether that is possible here is the question.. – MathematicianByMistake Apr 22 '16 at 22:15
• "Without actually finding values of $a, b, c$" – ÍgjøgnumMeg Apr 22 '16 at 22:21
• I don't know what surprises you, @Ed_4434, theory of elimination has its roots in such calculations. The issue presented here could be expressed in term of ideals, Gröbner basis, and so on... but that's not this kind of abstract mathematics that H.P. Das desires... – Jean Marie Apr 22 '16 at 22:28
• Can you say us what is the origin of your (interesting) problem ? – Jean Marie Apr 24 '16 at 14:58
The given system is $$a^2+\sqrt 3ab+b^2=25\\b^2+c^2=9\\a^2+ac+c^2=16$$ Then $$9+16=25\iff a^2+ac+c^2+ b^2+c^2= a^2+\sqrt 3ab+b^2\iff c(a+2c)=\sqrt 3 ab$$ It follows $$ab+2cb+\sqrt 3 ac=b(a+2c)+\sqrt 3ac=\frac{\sqrt 3\space a(b^2+c^2)}{c}=\frac {9\sqrt 3\space a}{c}=X$$ Hence one needs to calculate $\frac ac$. The successive following steps are clear to understand: $$a^2+\sqrt 3\space ab+(9-c^2)=25\iff b=\frac {16+c^2-a^2}{\sqrt 3\space a}=\frac{ac+2c^2}{\sqrt 3\space a}$$ It follows $$\left(\frac{ac+2c^2}{\sqrt 3\space a}\right)^2+c^2=9$$ So the system $$4c^4+4ac^3+4a^2c^2=27a^2\\a^2+ac+c^2=16$$ Hence $$4c^2(16)=27a^2\Rightarrow \frac ac=\frac{8}{3\sqrt 3}$$ Consequently $$X=\frac{9\sqrt 3\cdot 8}{3\sqrt 3}=\color{red}{24}$$
Here is a geometrical solution:
by rewriting the equations as
\left\{ \begin{aligned} 5^2&=a^2+b^2-2ab\cos\frac{5\pi}{6}\\ 3^2&=b^2+c^2-2bc\cos\frac{\pi}{2}\\ 4^2&=c^2+a^2-2ca\cos\frac{2\pi}{3} \end{aligned} \right.
and evaluating
$$4\left(\frac{1}{2}ab\sin\frac{5\pi}{6}+\frac{1}{2}bc\sin\frac{\pi}{2}+\frac{1}{2}ca\sin\frac{2\pi}{3}\right)=4S_{\Delta ABC}=24.$$
Sketch of a existence proof: Since $3^2+4^2=5^2$, $\Delta ABC$ exists. Consider $\overset{\mmlToken{mo}{⏜}}{HC}$, i.e., the segment that lies between $AB$ and $CA$ of a circle, for which $BC$ is a diameter. For any $P$ on the arc, $\angle BPC=\pi/2$ while $\angle CPA$ increases continuously from $\pi/2$ to $\pi$ (from $H$ to $C$), so there exists $P$ so that $\angle CPA=2\pi/3$.
In general, if such configuration does exist, the point where three segments meet is the intersection of three arcs. Each arc results the given opening angle with respect to each side of the triangle and they sum up to $2\pi$.
• Well, this configuration does exist and it is easy to prove the existence. Maybe you should mention that. – Batominovski Jun 27 '16 at 23:03
• Splendid solution ! – Jean Marie Jun 28 '16 at 4:28
What a prowess, @Piquito ! Here is another way of treating this issue.
Its "pros": it is systematic. Its "cons": it necessitates a computer algebra system (I used Mathematica).
Let us give the following numbers to the 3 equations:
$$\begin{cases}a^2+\sqrt 3ab+b^2&=&25 \ \ &(1)\\b^2+c^2&=&9 \ \ \ \ \ &(2) \\a^2+ac+c^2&=&16 \ \ \ &(3)\end{cases}$$
Let us consider equations (1) and (3) as quadratics in $a$.
They must have a common root. This can be expressed by setting their resultant to $0$ (https://en.wikipedia.org/wiki/Resultant):
This gives:
$$81 - 66b^2 + b^4 + 41\sqrt{3}bc - \sqrt{3}b^3c - 7c^2 + 2b^2c^2 - \sqrt{3}bc^3 + c^4=0 \ \ \ (4)$$
Now, we use constraint (2) meaning that point $(b/3,c/3)$ is on the unit circle, a constraint that we can translate into the following one (classical parameterization of the unit circle https://en.wikipedia.org/wiki/Tangent_half-angle_formula):
$$b = 3\dfrac{1 - t^2}{1 + t^2}, \ \ c = 3\dfrac{2 t}{1 + t^2} \ \ \ (5)$$ for a certain $t \in (-\infty,+\infty)$. Plugging (5) into (4) gives:
$$12 - 16\sqrt{3}t - 35t^2 + 16\sqrt{3}t^3 + 12t^4=0 \ \ \ (6)$$
This antipalindromic (https://en.wikipedia.org/wiki/Reciprocal_polynomial) 4th degree polynomial has four explicit (real) solutions
$$t=\dfrac{1}{12}(\pm9-4\sqrt{3}\pm\sqrt{273-72 \sqrt{3}}) \ \ \ (7)$$
(the two $\pm$ signs are independant: their four combinations are valid).
The objective of finding the value of
$$X = ab+ 2cb+\sqrt3 ac$$
is now within reach, because we are able to express it as an expression of $t$ alone, because:
• this is the case for $b$ and $c$ (formulas (5)).
• Concerning $a$, being the solution of quadratic equation (3), can be expressed as a function of $c$, itself function of $t$.
Then, it remains to check that, whatever the root chosen in (7), one gets the same result $X=24$. This is the case.
• Thanks for the compliment, Jean Marie. Before my answer I have also used roots of quadratic equation but could not go further. I think, as on many occasions for math problems, what was needed was rather "open well our eyes". Best regards. – Piquito Apr 23 '16 at 11:31 | 2019-05-25T23:24:25 | {
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https://math.stackexchange.com/questions/2840790/finding-the-lu-factorization-of-the-matrix | # Finding the $LU$ factorization of the matrix
Find the $LU$ factorization of the matrix: $$\begin{bmatrix} 1 & 1 & 1 \\ 3 & 5 & 6 \\ -2 & 2 & 7 \end{bmatrix}$$
I am aware that I need to find $L=\begin{bmatrix} 1 & 0 & 0 \\ * & 1 & 0 \\ * & * & 1 \end{bmatrix}$ and $U=\begin{bmatrix} 1 & * & * \\ 0 & 1 & * \\ 0 & 0 & 1 \end{bmatrix}$
I did row transformations and got $U=\begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 3 \\ 0 & 0 & 3 \end{bmatrix}$ but I couldn't understand how to find $L=\begin{bmatrix} 1 & 0 & 0 \\ * & 1 & 0 \\ * & * & 1 \end{bmatrix}$
Can anyone explain how to find $L$
• have you tried to invert U and multiply the inversion by the right ? – GBes Jul 4 '18 at 15:33
• No, I actually have no idea how to get $L$. So, you say that I need to invert $U$ and then multiply the inverse with $U$ itself? – user572932 Jul 4 '18 at 15:37
For doing LU decomposition, you need to do Gaussian elimination. Here I'll just help you with the procedure, but if you want to understand why I recommend you to see this pdf http://www.math.iit.edu/~fass/477577_Chapter_7.pdf. Lets apply Gaussian elimination to A \begin{equation*} A = \left[ \begin{matrix} 1 & 1 & 1\\ 3 & 5 & 6 \\ -2 & 2 & 7 \end{matrix} \right] \end{equation*} For eliminating $A_{12}$ and $A_{13}$, we need to multiply by $-3$ and by $2$ the first row and add this to the second and third row respectively, obtaining \begin{equation*} \left[ \begin{matrix} 1 & 1 & 1\\ 0 & 2 & 3 \\ 0 & 4 & 9 \end{matrix} \right] \end{equation*} Now we eliminate $A_{32}$ multiplying by $-2$ the second row and adding it to the third one, obtaining \begin{equation*} \left[ \begin{matrix} 1 & 1 & 1\\ 0 & 2 & 3 \\ 0 & 0 & 3 \end{matrix} \right] \end{equation*} Which is the $U$ matrix, for the $L$ matrix we use the factors by which we multiplied each row for obtaining the $U$ matrix, i.e. $3,-2,2$. We use this elements in the position of the elements they eliminated, then \begin{equation*} L = \left[ \begin{matrix} 1 & 0 & 0\\ 3 & 1 & 0 \\ -2 & 2 & 1 \end{matrix} \right] \end{equation*}
Going off of GBes's comment, you can find $L$ through some quick matrix multiplication. Let's call the original matrix $A$. We know that $A=LU$, and since you have $U$ (which is invertible), you can multiply by $U^{-1}$ on the right on both sides to get the following: \begin{align} A&=LU \\ AU^{-1}&=LUU^{-1} \\ AU^{-1}&=L\end{align}
I use to call $E_{ij}(d)$ the operation of summing to the $i$-th row the $j$-th row multiplied by $d$.
Thus the Gaussian elimination runs as \begin{align} \begin{bmatrix} 1 & 1 & 1 \\ 3 & 5 & 6 \\ -2 & 2 & 7 \end{bmatrix} &\xrightarrow{\begin{gathered} E_{31}(2) \\ E_{21}(-3) \end{gathered}} \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 3 \\ 0 & 4 & 9 \end{bmatrix} \\[6px] &\xrightarrow{E_{32}(-2)} \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 3 \\ 0 & 0 & 3 \end{bmatrix}=U \end{align} Now it's just a matter of replacing each transformation by its inverse: $$L=E_{21}(3)E_{31}(-2)E_{32}(2)= \begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ -2 & 2 & 1 \end{bmatrix}$$ At place $(2,1)$ put $3$, and so on.
This is justified by the fact that, if you consider $E_{ij}(d)$ the matrix you obtain by applying the transformation to the identity, then performing the row operation is the same as multiplying by this matrix.
Thus we have written $$U=E_{32}(-2)E_{31}(2)E_{21}(-3)A$$ (where $A$ is your original matrix) and so $$A=\underbrace{E_{21}(3)E_{31}(-2)E_{32}(2)}_{L}U$$ If one follows a strict order in doing the Gaussian elimination (top down and left to right), filling the matrix $L$ is just putting the coefficients in the indicated place.
Gaussian Elimination without Pivoting is as follows. The primary purpose of Gaussian elimination if you follow this is to find $\ell_{jk}$ which zeros out the row below. That is why it is the ratio of the two rows and then you subtract them. This continues on and on.
Suppose that
$$A = \begin{bmatrix} 1 & 1 & 1 \\ 3 & 5 & 6 \\ -2 & 2 & 7 \end{bmatrix}$$ $$A = LU$$
$$U =A, L=I$$ $$k=1,m=3,j=2$$ $$\ell_{21} = \frac{u_{21}}{u_{11}} = \frac{a_{21}}{a_{11}} = 3$$ $$u_{2,1:3} = u_{2,1:3} - 3 \cdot u_{1,1:3}$$ Then we're going to subtract 3 times the 1st row from the 2nd row $$\begin{bmatrix} 3 & 5 & 6 \end{bmatrix} - 3 \cdot \begin{bmatrix} 1 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 2 & 3\end{bmatrix}$$ Updating each of them $$U = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 3 \\ -2 & 2 & 7 \end{bmatrix}$$ $$L =\begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ $$k=1,j=3,m=3$$ $$\ell_{31} = \frac{u_{31}}{u_{11}} = \frac{-2}{1} = -2$$ $$u_{3,1:3} = u_{3,1:3} +2 \cdot u_{1,1:3}$$ Then we add two times the first row to the third row $$\begin{bmatrix} -2 & 2 & 7 \end{bmatrix} + 2 \cdot \begin{bmatrix} 1 & 1& 1 \end{bmatrix} = \begin{bmatrix}0 & 4 & 9 \end{bmatrix}$$ Updating $$U = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 3 \\ 0 & 4 & 9 \end{bmatrix}$$ $$L =\begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ -2 & 0 & 1 \end{bmatrix}$$ $$k=2, j=3,m=3$$ $$\ell_{32} = \frac{u_{32}}{u_{22}} = \frac{4}{2} = 2$$ We're subtracting out little blocks $$u_{3,2:3} = u_{3,2:3} - 2 \cdot u_{2,2:3}$$ $$\begin{bmatrix} 4 & 9 \end{bmatrix} - 2 \cdot\begin{bmatrix} 2& 3 \end{bmatrix} = \begin{bmatrix} 0 & 3 \end{bmatrix}$$ Updating $$U = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 3 \\ 0 & 0 & 3 \end{bmatrix}$$ $$L =\begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ -2 & 2 & 1 \end{bmatrix}$$ It now terminates $$A = LU$$ $$\underbrace{\begin{bmatrix} 1 & 1 & 1 \\ 3 & 5 & 6 \\ -2 & 2 & 7 \end{bmatrix}}_{A} = \underbrace{\begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ -2 & 2 & 1 \end{bmatrix}}_{L} \underbrace{\begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 3 \\ 0 & 0 & 3 \end{bmatrix}}_{U}$$ | 2019-12-06T23:27:03 | {
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https://math.stackexchange.com/questions/2370740/a-dice-is-rolled-until-a-6-occurs-what-is-the-probability-that-the-sum-includ | # A dice is rolled until a $6$ occurs. What is the probability that the sum including the $6$ is even?
A game is played where a standard six sided dice is rolled until a $6$ is rolled, and the sum of all of the rolls up to and including the $6$ is taken. What is the probability that this sum is even?
I know that this is a geometric distribution and the expected number of rolls is $\frac1{1/6} = 6$ rolls until a $6$ occurs, along with the expected value being $21$ ($6$ rolls times expected value of $3.5$ per roll), but I'm not certain how to proceed from there. Would the expected range (if it is even relevant) be from $11 = 1·5+6$ to $31 = 5·5+6$? The answer is supposedly $\frac47$. I'm also curious about how this question would change if the stopping number was anything else, say a $3$ stopping the sequence rather than a $6$. Thank you in advance!
• If you must stop at $1;\;3;\;5$ the probability is $\dfrac{3}{7}$. If you must stop at $2;\;4;\;6$ it is $\dfrac{4}{7}$. – Raffaele Jul 25 '17 at 10:38
• Oh, I see! If we call $p$ the probability of the sum of rolls being odd, the equation for stopping on a $1$, $3$, or $5$ is the same as the one provided by carmichael561 and we can take $1-p$ for the probability of the sum being even. (Is that correct?) – Elk Chu Jul 25 '17 at 14:09
• Shouldn't that be ⚅ instead of $6$? – Aaron Hall Jul 26 '17 at 2:25
Let $p$ be the desired probability, and consider the first roll. It is either a $6$, in which case we're done and the sum is even, a $2$ or $4$, in which case we want the sum of the rest of the terms to be even, or a $1,3$, or $5$, in which case we want the sum of the rest to be odd.
Thus $$p = \frac{1}{6}+ \frac{1}{3}p+\frac{1}{2}(1-p)$$ which simplifies to $p=\frac{4}{7}$.
• Highly instructive and pedagogically valuable. I would appreciate to see a booklet with many examples presented this way. (+1) – Markus Scheuer Jul 25 '17 at 6:32
• @MarkusScheuer: This is a standard technique, though one has to be careful to first prove that the desired quantity is finite. This is often missed out when dealing with expected value, since there could be nonzero probability of never stopping. Probabilities (as in this case) are always finite. – user21820 Jul 25 '17 at 8:36
• @user21820: I know, thanks! What I really like in this answer is the clarity, ease and conciseness of the formulation. – Markus Scheuer Jul 25 '17 at 8:41
We need only consider the rolls before a $6$ is obtained, because rolling an even $6$ doesn't change the parity of our total. Let $p_n$ represent the probability that the sum of $n$ rolls, not including any $6$, is even. Then we have: $p_{n+1}=\frac25p_n + \frac35(1-p_n)$, because a $2$ or a $4$ keeps a previous even total even, while a $1$, $3$ or $5$ makes a previous odd total into an even total. This simplifies to: $p_{n+1}=\frac35-\frac15p_n$. We also have $p_0=1$. We can solve this recurrence, and find that
$$p_n=\frac12\left(1+\left(-\frac15\right)^n\right)$$
Now, let $x_n$ represent the probability of rolling $n$ non-6's before the first $6$, so $x_n=\frac16\left(\frac56\right)^n$. The number we need is:
\begin{align} \sum\limits_{n=0}^\infty x_np_n &= \sum\limits_{n=0}^\infty \left[\frac16\left(\frac56\right)^n\cdot\frac12\left(1+\left(-\frac15\right)^n\right)\right]\\ &=\frac1{12}\sum\limits_{n=0}^\infty \left[\left(\frac56\right)^n + \left(-\frac16\right)^n\right]\\ &=\frac1{12}\left(6 + \frac67\right) = \frac47 \end{align}
That said, @carmichael561's answer is much, much nicer.
• Upvoted. (+1). Parallels very closely what I posted and appeared first. – Marko Riedel Jul 25 '17 at 23:08
Let $p$ be the probability that the sum of the die until(and including) the first six is even.
Let $R_1$ be the roll of the first die. So partitioning on this roll, and noticing the recurance:
$$p= \underline\qquad\,\mathsf P(R_1\in\{\underline\qquad\})+\underline\qquad\,\mathsf P(R_1\in\{\underline\qquad\})+\underline {~1~}\,\mathsf P(R_1=6)$$
Fill in the blanks, evalute the probabilities, and then solve for $p$.
Here is an answer using exponential generating functions. This problem has the features of a basic coupon collector (six coupons drawn with replacements). Note however that there is no requirement here of seeing all coupons. Now the probability that we took $$m$$ rolls to see the first six is by inspection given by
$$\frac{5^{m-1}}{6^m} = \frac{1}{6}\left(\frac{5}{6}\right)^{m-1}.$$
Observe that if the sum is even the odd values must have ocurred an even number of times, which gives the marked combinatorial class (one set of slots from the $$m-1$$ possible ones for each of the five admissible rolls of the die)
$$\def\textsc#1{\dosc#1\csod} \def\dosc#1#2\csod{{\rm #1{\small #2}}} \textsc{SET}(\mathcal{U}\times\mathcal{Z}) \times \textsc{SET}(\mathcal{Z}) \times \textsc{SET}(\mathcal{U}\times\mathcal{Z}) \times \textsc{SET}(\mathcal{Z}) \times \textsc{SET}(\mathcal{U}\times\mathcal{Z}).$$
The corresponding EGF is
$$G(z, u) = \exp(2z)\exp(3uz).$$
Restricting to even sums we get
$$H(z) = \frac{1}{2} G(z, 1) + \frac{1}{2} G(z, -1) \\ = \frac{1}{2} \exp(5z) + \frac{1}{2} \exp(-z).$$
Extracting coefficients we find
$$(m-1)! [z^{m-1}] H(z) = \frac{1}{2} 5^{m-1} + \frac{1}{2} (-1)^{m-1}.$$
Hence the probability of an even sum given that we took $$m$$ draws is given by
$$\frac{1}{2} + \frac{1}{2} \left(-\frac{1}{5}\right)^{m-1}.$$
We thus get for the total probability
$$\frac{1}{12} \sum_{m\ge 1} \left(\frac{5}{6}\right)^{m-1} + \frac{1}{12} \sum_{m\ge 1} \left(-\frac{1}{6}\right)^{m-1} \\ = \frac{1}{12} \left(\frac{1}{1-5/6} + \frac{1}{1+1/6}\right) = \frac{4}{7}.$$
• Very nice! (+1) – Markus Scheuer Jul 26 '17 at 21:10
The probability of any number 1 to 6 in rolling a dice is p1= 1/6 = 0.16666, therefore probability of getting desired number with condition of even or odd sum of numbers before getting desired number should be less than p1. lets investigate with python.
this is a simple function to simulate rolling dice
def onediceprobablity(Num,numoftry):
countNum = 0
for i in range(numoftry):
if random.randint(1,6) == Num :
countNum = countNum +1
return countNum/numoftry
Num is a number 1 to 6 and numoftry is our tries from 100000 to 1000000
here is the function that checks condition of problem
def dicesumeven(Num,numoftry):
EvenSum = 0
L = []
for i in range(numoftry):
N = random.randint(1,6)
L.append(N)
if N == Num :
Sum = sum(L)
L = []
if (Sum % 2) == 0:
EvenSum = EvenSum + 1
return EvenSum/numoftry
with Num = 6 and numoftry =1000000 probability p = 0.09538
this is the function for sum equals odd number
def dicesumodd(Num,numoftry):
oddSum = 0
L = []
for i in range(numoftry):
N = random.randint(1,6)
L.append(N)
if N == Num :
Sum = sum(L)
L = []
if (Sum % 2) == 1:
oddSum = oddSum + 1
return oddSum/numoftry
for 1000000 tries result is p2 = 0.071222
sum of p1 and p2 is about 1/6 | 2019-10-15T12:15:40 | {
"domain": "stackexchange.com",
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"lm_q1q2_score": 0.8480989637160845
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http://code.wikia.com/wiki/Matrix_addition | ## FANDOM
406 Pages
Two matrices can be added only if they have the same dimensions. The result will be a matrix of the same dimensions. To perform the addition, numbers in matching postions in the input matrices are added and the result is placed in the same position in the output matrix. Java Codes
#### Example: Adding 2x2 Matrices Edit
Let us add 2 matrices of dimension 2x2, let them be $\mathbf{A}$and $\mathbf{B}$.
$\mathbf{A} = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \mathbf{B} = \begin{bmatrix} 5 & 6 \\ -7 & -2 \end{bmatrix}$
These matrices can be added, because they are both 2x2. The result will also be 2x2. The result is:
$\mathbf{A+B} = \begin{bmatrix} 1 + 5 & 2 + 6 \\ 3 + -7 & 4 + -2 \end{bmatrix} = \begin{bmatrix} 6 & 8 \\ -4 & 2 \end{bmatrix}$
Addition is commutative in general for matrices, i.e. $\mathbf{A+B} = \mathbf{B+A}$.
#### More General ApproachEdit
Matrix addition can be performed on matrices of any dimensions, as long as they both have the same dimensions.
Let us visualize A and B as m×n matrices.
$\mathbf{A} = \begin{bmatrix} a_{1,1} & a_{1,2} & \dots & \dots \\ a_{2,1} & a_{2,2} & \dots & \dots \\ a_{3,1} & a_{3,2} & \ddots & \dots \\ \vdots & \vdots & \vdots & a_{m,n} \end{bmatrix} \mathbf{B} = \begin{bmatrix} b_{1,1} & b_{1,2} & \dots & \dots \\ b_{2,1} & b_{2,2} & \dots & \dots \\ b_{3,1} & b_{3,2} & \ddots & \dots \\ \vdots & \vdots & \vdots & b_{m,n} \end{bmatrix}$
We are going to be adding like before, but generally, sot the result is:
$\mathbf{A+B} = \begin{bmatrix} a_{1,1} + b_{1,1} & a_{1,2} + b_{1,2} & \dots & a_{1,n} + b_{1,n} \\ \vdots \\ a_{m,1} + b_{m,1} & a_{m,2} + b_{m,2} & \dots & a_{m,n} + b_{m,n} \end{bmatrix}$
### General Algorithm Edit
Here's a general algorithm for adding matrices:
1. DONT Check the sizes of two matrices $\mathbf{A}$ (m×n) and $\mathbf{B}$ (t×u): if m = t and n = u then we can add them o
2. therwise we just can't do it.
3. If they can be added, then create a new square matrix of size m×n.
4. For each element in A, find the element at the same position in B (i.e. same row and column) and add the 2 values. Place the result of this addition into the result matrix in the same position again.
### Pseudocode
The following pseudocode adds matrices of size m×n.<img data-rte-meta="%7B%22type%22%3A%22ext%22%2C%22placeholder%22%3A1%2C%22wikitext%22%3A%22%3Cpre%3E%5Cn%5Cn%5Cn%3C%5C%2Fpre%3E%22%7D" data-rte-instance="3585-18826377344f23bd3ae30ef" class="placeholder placeholder-ext" src="data:image/gif;base64,R0lGODlhAQABAIABAAAAAP///yH5BAEAAAEALAAAAAABAAEAQAICTAEAOw%3D%3D" type="ext" />Algorithm addMatrix (matrix1, matrix2, size, matrix3)
Add matrix1 to matrix2 and result in matrix3
Pre matrix1 and matrix2 have data
size is number on colums or rows in matrix
Post matrices added --- result in matrix31 loop (not end of row)1 loop (not end of colum)1 add matrix1 and matrix2 cells2 store sum in matrix32 end loop2 end loopend addMatrixReference: www.noormustafa.com
## Implementations Edit
### PHP Edit
function matrixAddition($m1,$m2){
$n = count($m1) - 1;
for($i=0;$i<=$n;$i++){
for($j=0;$j<=$n;$j++){
$a[$i][$j] =$m1[$i][$j]+$m2[$i][$j]; } } return$a;
}
### Visual Basic Edit
matrixA As IEnumerable(Of IEnumerable(Of Double)),
matrixB As IEnumerable(Of IEnumerable(Of Double))) _
As IEnumerable(Of IEnumerable(Of Double))
Dim result As New List(Of ReadOnlyCollection(Of Double))
Dim aRows = matrixA.GetEnumerator
Dim bRows = matrixB.GetEnumerator
Dim a = aRows.MoveNext
Dim b = bRows.MoveNext
Do While a Or b
If a <> b Then Throw New ArgumentException
Dim aCols = aRows.Current.GetEnumerator
Dim bCols = bRows.Current.GetEnumerator
Dim resultRow As New List(Of Integer)
Dim a2 = aCols.MoveNext
Dim b2 = bCols.MoveNext
Do While a2 Or b2
If a2 <> b2 Then Throw New ArgumentException
a2 = aCols.MoveNext
b2 = bCols.MoveNext
Loop
a = aRows.MoveNext
b = bRows.MoveNext
Loop
End Function
### Java Edit
import java.util.Scanner;
{
public static void main(String[] args)
{
int m, n, i, j;
Scanner input = new Scanner(System.in);
System.out.println("Enter the number of rows of matrix");
m = input.nextInt();
System.out.println("Enter the number of columns of matrix");
n = input.nextInt();
int matrixA[][] = new int[m][n];
int matrixB[][] = new int[m][n];
int sumOfMatrices[][] = new int[m][n];
System.out.println("Enter the elements of Matrix A");
for (i = 0; i < m; i++)
for (j = 0; j < n; j++)
matrixA[i][j] = input.nextInt();
System.out.println("Enter the elements of Matrix B");
for (i = 0; i < m; i++)
for (j = 0; j < n; j++)
matrixB[i][j] = input.nextInt();
for (i = 0; i < m; i++)
for (j = 0; j < n; j++)
sumOfMatrices[i][j] = matrixA[i][j] + matrixB[i][j];
System.out.println("Sum of entered matripes:-");
for (i = 0; i < m; i++) {
for (j = 0; j < n; j++)
System.out.print(sumOfMatrices[i][j] + "\t");
System.out.println();
}
}
}
### Lisp Edit
(defun array-sum (m1 m2)
"Returns the element-wise sum of two arrays of the same dimensions."
(if (equal (array-dimensions m1)
(array-dimensions m2))
(let ((sum (make-array (array-dimensions m1))))
(dotimes (i (array-total-size m1))
(setf (row-major-aref sum i)
(+ (row-major-aref m1 i)
(row-major-aref m2 i))))
sum)
(error "Arrays' dimensions are different.")))
### C Edit
#include <stdio.h>
#include <stdlib.h>
typedef struct
{
int rows;
int columns;
int *data;
} t_matrix;
#define NEW_MATRIX(m,r,c) (m)=malloc(sizeof(t_matrix)); (m)->rows=(r);(m)->columns=(c);(m)->data=malloc(sizeof(int)*(r)*(c));
#define VALUE_at(m, r, c) (m)->data[c*(m)->rows + r]
#define FREE(x) { free(x); (x)= NULL; }
#define FREE_MATRIX(m) FREE((m)->data);FREE(m)
t_matrix* addMatrix(const t_matrix *A, const t_matrix *B)
{
t_matrix *C;
int r,c;
NEW_MATRIX(C, A->rows, A->columns);
for(r=0; r<A->rows; r++)
{
for(c=0; c<A->columns; c++)
{
VALUE_at(C, r, c) = VALUE_at(A, r, c) + VALUE_at(B, r, c);
}
}
return C;
}
void main(void)
{
t_matrix *A, *B, *C;
int r,c;
NEW_MATRIX(A, 4,4);
NEW_MATRIX(B, 4,4);
for(r=0; r<A->rows; r++)
for(c=0; c<A->columns; c++)
{
VALUE_at(A, r, c) = 1;
VALUE_at(B, r, c) = 2;
}
for(r=0; r<C->rows; r++)
{
for(c=0; c<C->columns; c++)
{
printf("%d ", VALUE_at(C, r, c) );
}
printf("\n");
}
FREE_MATRIX(A);
FREE_MATRIX(B);
FREE_MATRIX(C);
} | 2017-08-24T10:46:05 | {
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https://mathematica.stackexchange.com/questions/256908/plot-argz-in-three-dimensions | # Plot Arg[z] in three dimensions
I would like to plot the function Arg[z] for a complex number $$z$$ in three-dimensions: I would like to obtain it as a surface over the complex plane, as in the "3D plot" section of this page:
But I would also like to be able to modify the range and the colors. The function only assumes real values between $$-\pi$$ and $$\pi$$.
I found (and tried) Complex3DPlot, but it does not provide what I'm looking for, because it plots the magnitude of Arg[z].
• Is this what you want? Plot3D[Arg[x + I y], {x, -1, 1}, {y, -1, 1}, ColorFunction -> "Rainbow"] Oct 14 '21 at 14:24
• @Domen Yes, exactly! Thank you. I had tried something similar and it didn't work, maybe I used a wrong syntax. If you would like to make an answer, it's welcome. Oct 14 '21 at 14:26
Use Plot3D with $$z = x + iy$$.
Plot3D[Arg[x + I y], {x, -1, 1}, {y, -1, 1}, ColorFunction -> "Rainbow"] | 2022-01-26T00:11:22 | {
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https://math.stackexchange.com/questions/3257902/questions-about-operator-norm-on-0-1 | # Questions about operator norm on [0,1]
I am struggling with the following question:
Consider the space $$V$$ of continuous functions on [0,1] with the 2-norm $$‖f‖_2^2$$=$$∫_0^1|f|^2$$. $$V$$ is an incomplete normed linear space. For a continuous function φ on [0,1], define a linear map $$M_φ:V⟶V$$ by $$M_φ f=φf$$. Show that $$M_\varphi$$ is bounded and find its norm.
Here is what I did to show $$M_\varphi$$ is bounded.
$$‖φf‖_2^2$$=$$∫_0^1|φf|^2$$$$‖φ‖_∞^2$$ $$∫_0^1|f|^2$$ =$$‖φ‖_∞^2$$ $$‖f‖_2^2$$
This shows that $$‖M_φ ‖≤‖φ‖_∞$$, hence $$M_φ$$ is bounded.
I think in order to find the operator norm, it must be shown that the opposite inequality, $$‖M_φ ‖\ge‖φ‖_∞$$ hold. I suspect the unit vector in $$C([0,1])$$ must somehow be used. Here is my rough outline:
1. Let $$g$$ be a unit vector in $$C([0,1])$$
2. Fix $$ε$$>0. Find $$x_0\in [0,1]$$ such that $$g(x_0)=1$$ and $$‖φ‖_∞-ε≤|φ(x_0 )|≤‖φ‖_∞$$
3. Show that $$‖M_φ g‖_\infty \ge ‖φ‖_∞ -ε$$
4. Since $$‖M_φ g‖_2 \ge ‖M_φ g‖_\infty$$, we have $$‖M_φ ‖≥‖φ‖_∞$$
I know there is something amiss, could you please help me understand where went wrong? Thank you so much in advance, have a wonderful day!
• what is $M_\varphi$?? And what is $\varphi$? – Masacroso Jun 10 '19 at 21:01
• Sorry I forgot to include, will edit now – mathnoob777 Jun 10 '19 at 21:02
Since $$\lvert \varphi \rvert$$ is continuous, it achieves its maximum somewhere on $$[0,1]$$; say the maximum occurs at $$x_0$$ so that $$\|\varphi\|_\infty = \lvert \varphi(x_0)\rvert$$. Define $$f_n:[0,1]\to [0,\infty)$$ by $$f_n(x)^2 = \left\{ \begin{matrix}0, & 0 \le x < x_0 - \tfrac 1 n,\\ n^2(x-x_0 +\tfrac 1 n), & x_0 - \tfrac 1 n \le x < x_0, \\ n^2(x_0 +\tfrac 1 n - x), &x_0 \le x < x_0 + \tfrac 1 n,\\ 0, &x_0+\tfrac 1n \le x. \end{matrix}\right.$$ Each $$f_n$$ is continuous and the graph of $$f_n^2$$ is a triangular spike centered at $$x_0$$ with base $$2/n$$ and height $$n$$, and thus $$\int^1_0 f_n(x)^2 dx = 1$$.
Fix $$\epsilon > 0$$. Since $$\varphi$$ is continuous, there is $$\delta >0$$ such that when $$\lvert x - x_0 \rvert < \delta$$, we have $$\lvert \varphi(x)\rvert \ge \|\varphi\|_\infty - \epsilon$$. Choose $$n$$ large enough that $$\frac 1 n < \delta$$. Then \begin{align*}\|M_\varphi f_n\|^2_{2} &= \int^1_0 \lvert f_n(x) \rvert^2 \lvert \varphi(x) \rvert^2dx \\ &= \int_{x_0 - \tfrac 1n}^{x_0+\tfrac 1n}\lvert f_n(x) \rvert^2 \lvert \varphi(x) \rvert^2dx \\ &\ge (\|\varphi\|_\infty - \epsilon)^2\int_{x_0 - \tfrac 1n}^{x_0+\tfrac 1n}\lvert f_n(x) \rvert^2dx = (\|\varphi\|_\infty-\epsilon)^2. \end{align*} Taking the square root shows that $$\|M_\varphi\| \ge \|M_{\varphi}f_n\|_2 \ge \|\varphi\|_\infty-\epsilon$$. Since $$\epsilon >0$$ was arbitrary, send it to zero and you have your result.
EDIT: Obvious modifications can be made in the cases that $$x_0 = 0$$ or $$x_0 =1$$. Also, there is no reason you need so explicit a function $$f_n$$; any function $$f$$ which is positive for $$\lvert x-x_0\rvert < \delta$$ and zero when $$\lvert x-x_0 \rvert > \delta$$ should work. | 2020-02-29T14:28:08 | {
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https://mathhelpboards.com/threads/mechanical-vibrations-linear-combinations.6791/ | # TrigonometryMechanical Vibrations - Linear Combinations
#### alane1994
##### Active member
The title may be incorrect, I named this after the section of my book in which this is located.
My problem is as follows.
Determine $$\omega_0$$, R, and $$\delta$$ so as to write the given expression in the form
$$u=R\cos(\omega_0 t-\delta)$$
$$\color{blue}{u=4\cos(3t)-2\sin(3t)},~\text{My Problem}$$
I know that,
$$u(t)=A\cos(\omega_0 t)+B\sin(\omega_0 t)$$
$$\omega_0=\sqrt{\dfrac{k}{m}}$$
$$A=R\cos(\delta),~~B=R\sin(\delta)~~\Rightarrow~R=\sqrt{A^2+B^2},~\tan(\delta)=\dfrac{B}{A}$$
So that means that,
$$\omega_0=3$$
$$A=4$$
$$B=-2$$
$$R=2\sqrt{5}$$
$$\delta=\tan^{-1}(\dfrac{-2}{4})\approx-.463648$$
Now I am a little confused as to where to go from here. Any thoughts?
Would I then just plug in the values into the desired format above?
Last edited:
#### topsquark
##### Well-known member
MHB Math Helper
The title may be incorrect, I named this after the section of my book in which this is located.
My problem is as follows.
Determine $$\omega_0$$, R, and $$\delta$$ so as to write the given expression in the form
$$u=R\cos(\omega_0 t-\delta)$$
$$\color{blue}{u=4\cos(3t)-2\sin(3t)},~\text{My Problem}$$
I know that,
$$u(t)=A\cos(\omega_0 t)+B\sin(\omega_0 t)$$
$$\omega_0=\sqrt{\dfrac{k}{m}}$$
$$A=R\cos(\delta),~~B=R\sin(\delta)~~\Rightarrow~R=\sqrt{A^2+B^2},~\tan(\delta)=\dfrac{B}{A}$$
So that means that,
$$\omega_0=3$$
$$A=4$$
$$B=-2$$
$$R=2\sqrt{5}$$
$$\delta=\tan^{-1}(\dfrac{-2}{4})\approx-.463648$$
Now I am a little confused as to where to go from here. Any thoughts?
Would I then just plug in the values into the desired format above?
Looks good. One note: When I graphed these (easiest way to check) I needed several more digits for the phase shift to make it work right.
-Dan
#### alane1994
##### Active member
Is the phase shift the $$\delta$$?
#### MarkFL
Staff member
I have moved this topic to our Trigonometry sub-forum since the problem, while it comes from an application of a second order linear ODE, involves only trigonometry. I have also edited the title.
We want to express the solution:
$$\displaystyle u(t)=4\cos(3t)-2\sin(3t)$$
in the form:
$$\displaystyle u(t)=R\cos(\omega_0 t-\delta)$$
I would use the angle-difference identity for cosine to write:
$$\displaystyle u(t)=R\left(\cos(\omega_0 t)\cos(\delta)+\sin(\omega_0 t)\sin(\delta) \right)$$
Distributing the $R$, we have:
$$\displaystyle u(t)=R\cos(\omega_0 t)\cos(\delta)+R\sin(\omega_0 t)\sin(\delta)$$
Comparison of this with the desired form, we find:
$$\displaystyle R\cos(\delta)=4$$
$$\displaystyle R\sin(\delta)=-2$$
$$\displaystyle \omega_0=3$$
Squaring the first two equations, and adding, we get:
$$\displaystyle R^2=20\implies\,R=2\sqrt{5}$$
Dividing the second equation by the first, we find:
$$\displaystyle \tan(\delta)=-\frac{1}{2}\implies\delta=-\tan^{-1}\left(\frac{1}{2} \right)$$
and so we may state:
$$\displaystyle u(t)=2\sqrt{5}\cos\left(3t+\tan^{-1}\left(\frac{1}{2} \right) \right)$$
#### alane1994
##### Active member
Ok, so you do just plug them back into the friendly equation from earlier in the problem!
I need to stop over-thinking things, it just seemed too easy for a course of this level
#### MarkFL
Staff member
Ok, so you do just plug them back into the friendly equation from earlier in the problem!
I need to stop over-thinking things, it just seemed too easy for a course of this level
Yes, you did everything correctly, the only thing I would have done further is reduce the argument of the inverse tangent function and avoided using a decimal approximation for the resulting angle $\delta$.
Technically, I should have written:
$$\displaystyle u(t)=2\sqrt{5}\cos\left(3t-\left(-\tan^{-1}\left(\frac{1}{2} \right) \right) \right)$$
MHB Math Helper
#### MarkFL
Staff member
Yup!
-Dan
I believe the phase shift would actually be:
$$\displaystyle \frac{\delta}{\omega_0}$$
which can be seen by writing the solution in the form:
$$\displaystyle u(t)=R\cos\left(\omega_0\left(t-\frac{\delta}{\omega_0} \right) \right)$$
#### topsquark
##### Well-known member
MHB Math Helper
I believe the phase shift would actually be:
$$\displaystyle \frac{\delta}{\omega_0}$$
which can be seen by writing the solution in the form:
$$\displaystyle u(t)=R\cos\left(\omega_0\left(t-\frac{\delta}{\omega_0} \right) \right)$$
(Ahem!) That's one on MHF and now one on MHB. I'm going to bed.
Thanks for the catch.
-Dan | 2020-09-18T23:19:01 | {
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https://math.stackexchange.com/questions/2140542/the-roots-of-the-equation-x23x-1-0-are-also-the-roots-of-x4ax2bxc-0 | # The roots of the equation $x^2+3x-1=0$ are also the roots of $x^4+ax^2+bx+c=0$
The roots of the equation $x^2+3x-1=0$ are also the roots of quartic equation $x^4+ax^2+bx+c=0$. Find $a+b+4c$.
This problem is from yesterday's Bangladesh National Math Olympiad 2017. I tried this using Vieta Root Jumping but no luck. After the contest my friend laughed at me "One doesn't simply try a 10 point problem with Vieta Root Jumping".
How to solve this problem?
• Is division allowed? – lab bhattacharjee Feb 12 '17 at 8:56
• @labbhattacharjee This is the exact same question... I dont know – Rezwan Arefin Feb 12 '17 at 8:57
• Sean Bean would like your friend. :) – Deepak Feb 12 '17 at 9:23
Note that if $x^2+3x-1=0$, then roots of the quadratic $x^2+3x-1$ are also roots of $$(x^2+3x-1)(x^2-px+q)$$ Which follows from polynomial long divison.
Since the coefficient of $x^3$ is $0$, we have that $p=3$. Now note that $$(x^2+3x-1)(x^2-3x+q)=x^4+(q-10)x^2+(3q+3)x-q$$ So the value of $a+b+4c=q-10+3q+3-4q=-7$. So the answer is $-7$ no matter the value of $q$.
• How I missed this :| :| My friend became Champion of the Champions of the Olympiad .. and I am out of any places :| :| :'( – Rezwan Arefin Feb 12 '17 at 9:01
• @RezwanArefin Was there a restriction that $a,b,c$ are rational? – S.C.B. Feb 12 '17 at 9:02
• nope.. I gave the exact same question – Rezwan Arefin Feb 12 '17 at 9:03
• @RezwanArefin Then the problem is incorrect.... – S.C.B. Feb 12 '17 at 9:03
• I guess so :( But how to find other possibilities? – Rezwan Arefin Feb 12 '17 at 9:04
$$(x^2+3x-1)(x^2-3x+3)=x^4-7x^2+12x-3$$ has the same real roots as $x^2+3x-1$. In this case $a=-7,b=12,c=-3$ so $$a+b+4c=-7.$$
Moreover $$(x^2+3x-1)(x^2-3x+d)=x^4+(d-10)x^2+(3d+3)x-d$$ for $$d>\frac{9}{4}$$ has the same real roots as $x^2+3x-1$. In this case $a=d-10,b=3d+3,c=-d$ so $$a+b+4c=-7.$$
So if the statement of the problem says the same real roots then it seems to be that $a+b+4c=-7$ in all cases.
• Why do you need $d>\frac{9}{4}$? Note the fact that it did not say that the roots of the given quartic were equal to the roots of the quadratic. it merely said that all of the roots of the quadratic were roots of the quartic. – S.C.B. Feb 12 '17 at 9:23
• @S.C.B. For those values of $d$ the discriminant is negative so it doesn't have any more roots. It says "are the roots.." so it shouldn't have any more roots. – Test123 Feb 12 '17 at 9:25
• The fact that it says that it is the roots does not imply that those are the only roots. For example we do say that $x=1$ is a root of $x^2-3x+2$ . Or we do say that $x=1,2$ are roots of $$(x-1)(x-2)(x-3)(x-4)$$ We just don't say that it is all the roots. – S.C.B. Feb 12 '17 at 9:26
• @Test123 No .. I think it can have roots other than those two – Rezwan Arefin Feb 12 '17 at 9:26
• @Test123 A more clear statement "All roots of ..... are also roots of .... " but not "Roots of ..... also all roots of .... " .. Maybe you misunderstood :| – Rezwan Arefin Feb 12 '17 at 9:28
For the equation $x^2+3x-1=0$, the sum of the roots is $-3$ and the product is $-1$. For the equation $x^4+ax^2+bx+c=0$, the sum of the roots is $0$ and the product is $c$. If the roots of the first equation are roots of the second, the remaining two roots have sum $3$ and product $-c$. Therefore, we have
$$x^4+ax^2+bx+c=(x^2+3x-1)(x^2-3x-c)$$
The coefficient of the $x^2$ term is $-c-9-1=-c-10$. The coefficient of $x$ is $-3c+3$. Therefore,
$$a+b+4c=-c-10-3c+3+4c=-7$$ | 2019-07-15T23:57:05 | {
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https://math.stackexchange.com/questions/1424410/p-rightarrow-q-rightarrow-r-equivalent-to-p-wedge-q-rightarrow-r | # $P \Rightarrow (Q \Rightarrow R)$ equivalent to $P \wedge Q \Rightarrow R$
I was asked to prove the above. The teacher has assured me that they are indeed equivalent, but when drawing a truth table, I have not been able to show this.
For:
$P =$ F,
$Q =$ T,
$R =$ T.
I have the first portion as true but the second as false.
Would someone be able to confirm whether I have made an error or not?
If it my assertion is indeed wrong and I have messed up somewhere, please do not show me the final proof, as I do want to work it out, but just want to ensure that it is actually provable.
EDIT:
Thank both for your help. I just had one final question on the topic regarding semantics
for $$P \land Q \Rightarrow R$$
JMoravitz, based on your answer, I suppose it should be treated as $$(P \land Q) \Rightarrow R$$ as opposed to how I was originally viewing the problem$$P\land (Q \Rightarrow R)$$ I'm assuming that everything before an implication should ALWAYS be grouped together? Should everything after the implication also be grouped together regardless of what follows?
if not, how would you go about determining grouping?
• Yes, they are equivalent. For $P = F$, $Q = R = T$, we have $P \land Q = F$, so the second statement is true as well. – user230734 Sep 6 '15 at 17:34
• There is an order of operations in logic, just as in other fields. The usual order says that parentheses are handled first, then AND and OR (with equal priority), then implication. That much is pretty standard. Another common, but not universal, rule is that implications are right associative, so $P \to Q \to R$ means $P \to (Q \to R)$ instead of $(P \to Q) \to R$. This latter rule is motivated by the fact that $P \to (Q \to R)$ is more commonly encountered, since it is equivalent to $P \land Q \to R$. – Carl Mummert Sep 6 '15 at 18:07
• AND and OR do not have equal priority. AND wins, especially since we often split into cases each of which is a conjunction of conditions. Also, NOT always goes first before them. – user21820 Oct 22 '15 at 12:22
Remember that $A\Rightarrow B$ is true whenever $A$ is false. So, for the first $P$ being false, it is false implies (something), therefore the implication is true.
For the second, it is (false and true) implies (something). false and true simplifies to simply false, so again we have false implies (something), and therefore the implication is true.
As a hint/reminder on how best to proceed, remember that $A\Rightarrow B$ is logically equivalent to $\neg A \vee B$
• Ah, I see why you might have been confused. Indeed, it is somewhat ambiguous what was intended, but in my experience $P\wedge Q\Rightarrow R$ should generally be interpreted as $(P\wedge Q)\Rightarrow R$. There are certainly situations where the other interpretation might be useful however. Another user on the site was quoted once as saying "A wise person will use enough parenthesis so that the meaning is perfectly clear." – JMoravitz Sep 6 '15 at 17:54
• @user3256725 As for a general rule, it is my interpretation that if an implication arrow ($\Rightarrow$) appears, unless parenthesis also appear, all terms on the left are grouped and all terms on the right are grouped: $P\wedge Q\Rightarrow R\wedge S$ would be interpreted as $(P\wedge Q)\Rightarrow (R\wedge S)$, but it might not necessarily be standard. Verifying this with your teacher would be good as well, and interpretation questions like "if things are grouped" are generally allowed during tests. – JMoravitz Sep 6 '15 at 17:56 | 2020-02-20T21:43:10 | {
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# Average Accelerated: Guide to solve Averages Quickly
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Average Accelerated: Guide to solve Averages Quickly [#permalink] 29 Mar 2009, 05:47
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Averages Accelerated: Guide to solve Averages quickly
Folks, in this tutorial we will take a deep look into the concept of average that will help us to solve the questions from this topic quickly. First and foremost, I want to make it clear that I am not going to use the traditional way of solving the questions. Instead we try to solve the questions quickly.
Let us look at a simple question now.
Example 1:
The average mark of 70 students in a class is 80. Out of these 70 students, if the average mark of 40 students is 75, what is the average mark of the remaining 30 students?
I am sure every one of you will be able to answer this in the following way.
Average of 70 students = 80.
Hence total marks of all the 70 students = 70x80 = 5600
Out of these 70, average of 40 students = 75
Hence total marks of these 40 students = 40x75 = 3000.
So the total marks of the remaining 30 students = 5600-3000 = 2600
Hence the average of the remaining 30 students = 2600/30 = 86.66
Now there will be a serious difference in the time taken if the numbers given here are not multiples of 10. Just look at example 2 below:
Example 2:
“Average of 75 students is 82, out of which average of 42 students is 79. What is the average of the remaining 33 students?
I am sure these numbers are your biggest enemies in the exam. So let us look at a much more practical way. I say practical way because back in my village this is how the farmers deal with averages in their daily life.
Let’s go back to Example 1
Given that average of 70 students is 80. This class is split into two groups of 40 and 30. I am sure you will agree with me that if the average of each of these groups is 80, then the average of the total group i.e. 70 students will also be 80.
But the first group of 40 students, instead of 80 they had an average of only 75. So definitely the average of the remaining 40 students must be greater than 80.
The loss incurred because of the first group must be compensated by the second group.
Let us look at the loss incurred because of the first group
We want them to have an average of 80, but they managed only 75. So we lost an average of 5 upon 40 students. So the loss in the sum = 40x5.
Now this loss of 40x5 must be compensated by our second group i.e. 30 students.
So their average must be not only the initial 80, but also the average meant to compensate the loss incurred because of the first group.
Hence the average of the remaining 30 students = 80 + (40x5)/30 = 80 + 20/3 = 86.66
I hope this is lucid.
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Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 29 Mar 2009, 05:51
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So folks it’s time to get into some more different situations that arise when we calculate averages.
Problem 1:
Let’s look at one more sum similar to the one discussed above.
The average height of 74 students in a class is 168 cms out of which 42 students had an average height of 170 cms. Find the average height of the remaining 32 students.
Sol:
Average of 74 students = 168.
Here we can observe that, in the case of the first group we had a gain.
Since the average of 42 students is 170 cms, the gain is 2 cms in the average upon 42 students.
So the gain in the sum = 42x2
Hence average of the remaining 32 students = 168 - (42x2)/32= 168-(21/8) = 165.37
Problem 2:
The average height of 40 students in a class is 172 cms. 30 students whose average height is 172.5 cms left the class and 40 students whose average height is 170.5 cms joined the class. Find the average height of the present class.
Sol:
Going back to the ideal situation, we want the average of the students leaving the class as well as joining the class to be 172 so that the average remains the same.
But it is given that the average of the 40 students leaving the class is 172.5 (more than 172).
So we will incur a loss of 0.5 cms in the average upon 30 students.
Hence the loss in the sum = 0.5x 30 = 15cms
Also, since the average of the 40 students joining the class is 170.5 (less than 172) we will incur a loss in this case as well. The loss in the average is 1.5 cms upon 40 students.
Hence the loss in the sum = 1.5x40 = 60cms
Thus the total loss in the sum = 75cms. This loss will be shared by 50 students which is the present strength of the class.
Hence the average of the present class = 17275/50 = 170.5 cms.
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Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 29 Mar 2009, 05:52
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Problem 3:
The average weight of 36students in a class is 62 kgs. If the weight of the teacher is also included the average becomes 62.5 kgs. Find the weight of the teacher.
Sol:
It is clear that if the weight of the teacher is also 62 years, the average of the class including the teacher will remain 62 kgs. But the average is increased by 0.5 kgs upon adding the teacher. So it is clear that the weight of the teacher is more than 62 kgs.
The average is increased by 0.5 kgs upon 37 members (remember it is not 36, but 37) which means the increase in the sum =0.5 x 37 = 18.5 kgs.
So the age of the teacher must be 62+18.5 = 80.5 kgs
Problem 4:
In a factory the average monthly salary of workers is 550 $and that of 16 officers is 3000$. If the average monthly salary of workers and officers combined is 600 $, find the number of workers in the factory. Sol: If the average of the 16 officers is also 550$, the combined average would have been the same 550$. But, since the average of the officers is 3000$, which is more by 2450 than what we wanted it to be; the gain in the sum= 2450x16.
This gain of 2450x16 in the sum upon (x+16) members, where x is the number of workers gave an increase of 50$in the combined average. i.e. (2450x16)/(x+16) = 50 i.e. (x+16) = (2450x16)/50 = 49x16 i.e. x = 48x16 = 768. So the total number of workers = 768. Quickest Way: Since x number of workers had an average of only 550$ instead of 600$(combined average), the loss caused by workers = 50xW (were W is the number of workers) This loss is compensated by 16 officers by having an average of 3000$ instead of 600\$.
The gain given by officers = 2400 x 16.
Hence 50xW = 2400x16 i.e. W = 48x16 = 768.
So folks, with a little bit of practice you can answer these questions very quickly.
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Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 29 Mar 2009, 05:49
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Now let us look at Example 2 where the numbers are not comfortable.
Example 2:
Average of 75 students is 82, out of which average of 42 students is 79.
What is the average of the remaining 33 students?
Average of 75 members = 82.
Two groups of 42 and 33 and we want each group to have an average of 82.
But the first group i.e. 42 students they had an average of 79.
We fell short by 3 in the average.
So the loss in the sum= 3x42.
So the average of the second group i.e. 33 students must be 82 + (3x42)/33 = 82+ 42/11 = 85.82
I am sure this is much faster than what we did earlier.
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Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 29 Mar 2009, 05:54
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Problem 5:
The average of batsmen up to certain number of innings was 45. In the next inning he was out for a duck and his average reduced to 40.5. Find the total number of innings played by him including the latest inning.
Sol:
Again, if the batsmen had scored 45 in his latest inning his average would remain in tact i.e. it would have been 45. But he scored 0 runs.
Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
Hence 45/x = 4.5 i.e. x = 10. (Remember, x includes the latest inning as well)
Problem 6:
The average of 8 scores is 87. Of these, the highest is 2 more than the one next in magnitude. If these two scores are eliminated the average of the remaining score is 85. What was the highest score?
Sol:
If the sum of the top two scores taken out is 87+87 i.e. 174, the average of the remaining 6 scores will remain 87.
But, since the average is reduced by 2 upon 6 scores, the sum of the top two scores taken out is 174+12 = 186.
So the top two scores are 94 and 92.
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Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 29 Mar 2009, 05:55
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Problem 7:
The average age of a committee of 8 members is 40 years. A member, aged 55 years, retired and he was replaced by a member aged 39 years. The average age of the present committee is?
Sol:
If the age of the new person who is replacing the retired person is also 55 years, the average would remain same i.e. 40 years. But since the age of the new person is 16 years less than that of the retired person, the loss in the sum is 16 years.
This loss of 16 years in the sum will result in a loss of 16/8 = 2 yrs in the average (remember that there is no change in the total strength).
So the present average = 40-2 = 38 yrs.
Problem 8:
The average of marks obtained by 120 candidates in a certain examination is 35. If the average of passed candidates is 39 and that of the failed candidates is 15. The number of candidates who passed the examination is?
Sol:
Folks, look at the relative calculation here.
If we consider that all the 120 candidates passed the examination, the average of the entire class must be 39. But it is given that average of the entire class is 35.
So we are getting an extra of 4 in the average i.e. an extra of 4x120=480 in the total sum.
We got extra total sum because some students who failed were also considered to be passed. For every one student considered as passed who actually failed we get 39-15 = 24 marks extra.
Since we got 480 marks extra, total number of failed students who were considered as passed = 480/24 = 20.
Hence the total number of students who passed = 100
You can solve this by assuming that all the 120 failed the examination but this would take a bit extra time.
Folks, most of the questions solved here can also be solved quickly by using a technique called Alligation. I will soon come up with a tutorial on this topic and I will discuss these questions in that tutorial besides some other questions as well. So stay tuned....
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Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:07
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cicerone wrote:
Problem 5:
The average of batsmen up to certain number of innings was 45. In the next inning he was out for a duck and his average reduced to 40.5. Find the total number of innings played by him including the latest inning.
Sol:
Again, if the batsmen had scored 45 in his latest inning his average would remain in tact i.e. it would have been 45. But he scored 0 runs.
Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
Hence 45/x = 4.5 i.e. x = 10. (Remember, x includes the latest inning as well)
Problem 6:
The average of 8 scores is 87. Of these, the highest is 2 more than the one next in magnitude. If these two scores are eliminated the average of the remaining score is 85. What was the highest score?
Sol:
If the sum of the top two scores taken out is 87+87 i.e. 174, the average of the remaining 6 scores will remain 87.
But, since the average is reduced by 2 upon 6 scores, the sum of the top two scores taken out is 174+12 = 186.
So the top two scores are 94 and 92.
Can you please explain how you arrived at 94 and 92
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Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:22
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cicerone wrote:
Problem 7:
The average age of a committee of 8 members is 40 years. A member, aged 55 years, retired and he was replaced by a member aged 39 years. The average age of the present committee is?
Sol:
If the age of the new person who is replacing the retired person is also 55 years, the average would remain same i.e. 40 years. But since the age of the new person is 16 years less than that of the retired person, the loss in the sum is 16 years.
This loss of 16 years in the sum will result in a loss of 16/8 = 2 yrs in the average (remember that there is no change in the total strength).
So the present average = 40-2 = 38 yrs.
Problem 8:
The average of marks obtained by 120 candidates in a certain examination is 35. If the average of passed candidates is 39 and that of the failed candidates is 15. The number of candidates who passed the examination is?
Sol:
Folks, look at the relative calculation here.
If we consider that all the 120 candidates passed the examination, the average of the entire class must be 39. But it is given that average of the entire class is 35.
So we are getting an extra of 4 in the average i.e. an extra of 4x120=480 in the total sum.
We got extra total sum because some students who failed were also considered to be passed. For every one student considered as passed who actually failed we get 39-15 = 24 marks extra.
Since we got 480 marks extra, total number of failed students who were considered as passed = 480/24 = 20.
Hence the total number of students who passed = 100
You can solve this by assuming that all the 120 failed the examination but this would take a bit extra time.
Folks, most of the questions solved here can also be solved quickly by using a technique called Alligation. I will soon come up with a tutorial on this topic and I will discuss these questions in that tutorial besides some other questions as well. So stay tuned....
I guess there is one shorter way as you have already explained in one of the above questions:
Let n=number of passed students, then, 120-n= number of failed students
If Gain in average=Loss of average
=> 4n=(120-n)20
=> n=100
Am I correct?
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Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 16 Apr 2009, 04:03
Good tips! Thanks!
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Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 17 Apr 2009, 15:35
thanks for the guide
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Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 15 Oct 2009, 11:47
bandit wrote:
cicerone wrote:
Problem 7:
The average age of a committee of 8 members is 40 years. A member, aged 55 years, retired and he was replaced by a member aged 39 years. The average age of the present committee is?
Sol:
If the age of the new person who is replacing the retired person is also 55 years, the average would remain same i.e. 40 years. But since the age of the new person is 16 years less than that of the retired person, the loss in the sum is 16 years.
This loss of 16 years in the sum will result in a loss of 16/8 = 2 yrs in the average (remember that there is no change in the total strength).
So the present average = 40-2 = 38 yrs.
Problem 8:
The average of marks obtained by 120 candidates in a certain examination is 35. If the average of passed candidates is 39 and that of the failed candidates is 15. The number of candidates who passed the examination is?
Sol:
Folks, look at the relative calculation here.
If we consider that all the 120 candidates passed the examination, the average of the entire class must be 39. But it is given that average of the entire class is 35.
So we are getting an extra of 4 in the average i.e. an extra of 4x120=480 in the total sum.
We got extra total sum because some students who failed were also considered to be passed. For every one student considered as passed who actually failed we get 39-15 = 24 marks extra.
Since we got 480 marks extra, total number of failed students who were considered as passed = 480/24 = 20.
Hence the total number of students who passed = 100
You can solve this by assuming that all the 120 failed the examination but this would take a bit extra time.
Folks, most of the questions solved here can also be solved quickly by using a technique called Alligation. I will soon come up with a tutorial on this topic and I will discuss these questions in that tutorial besides some other questions as well. So stay tuned....
I guess there is one shorter way as you have already explained in one of the above questions:
Let n=number of passed students, then, 120-n= number of failed students
If Gain in average=Loss of average
=> 4n=(120-n)20
=> n=100
Am I correct?
I think bandit is correct. That was also the way i found the answer.
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Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 23 Jan 2010, 11:36
This is a big help thanks! I'm solving average related problems much more efficiently after practicing this method.
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Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 01 Feb 2010, 04:56
is there another way to solve the batsman problem? i'm quite confused with this method.
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Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 21 Apr 2010, 07:26
good work dude!
+1 to you.
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impressed?...how bout encouraging me with a kudos
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Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 15 May 2010, 12:26
great job dude , thanks
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Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 15 May 2010, 23:04
cicerone wrote:
Problem 5:
The average of batsmen up to certain number of innings was 45. In the next inning he was out for a duck and his average reduced to 40.5. Find the total number of innings played by him including the latest inning.
Sol:
Again, if the batsmen had scored 45 in his latest inning his average would remain in tact i.e. it would have been 45. But he scored 0 runs.
Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
Hence 45/x = 4.5 i.e. x = 10. (Remember, x includes the latest inning as well)
rahulms wrote:
is there another way to solve the batsman problem? i'm quite confused with this method.
Here you go.
Let us assume that total no. of innings played as x
45(x-1) + 0 = 40.5 x --> 45 is average of (x-1) innings, 40.5 is new average.
4.5x = 45 --> x =10
Hope this helps.
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Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 15 May 2010, 23:05
Great job dude... useful tips. Thanks
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Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 08 Feb 2014, 02:17
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Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 27 Apr 2015, 07:04
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Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 27 Apr 2015, 07:04
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Display posts from previous: Sort by | 2015-05-24T03:29:53 | {
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http://math.stackexchange.com/questions/77686/how-do-we-show-the-equality-of-these-two-summations/77690 | # How do we show the equality of these two summations?
How do you show the following? $$\sum \limits_{i=1}^{n}\ \sum \limits_{j=i}^{n}\ \sum \limits_{k=i}^{j}\ 1 = \sum \limits_{j=1}^{n}\ \sum \limits_{k=1}^{j}\ \sum \limits_{i=1}^{k}\ 1$$ It's not obvious why this is true, but I have tested it with a program and it works.
-
I am just passing by and wow... It's like 1000 times more tedious to post an answer on math SE than on stackoverflow (auto syntax coloring, indentation etc)! kudos to all math SE answerers! – Jake Nov 1 '11 at 7:56
Both count the cardinality of the set $\{ (i, j, k) \in \mathbb Z^3 \mid 1 \leq i \leq k \leq j \leq n \}$.
Explanation. We try to count the set $$S := \{ (i, j, k) \in \mathbb Z^3 \mid 1 \leq i \leq k \leq j \leq n \}$$ in two ways.
Right hand side.
1. First fix the largest item of the triple, namely $j$; this can take values from $1$ to $n$. Then we fix $k$ and $i$ in that order.
2. For any given value of $j$, the variable $k$ can take values from $1$ to $j$ (since we know that $1 \leq k \leq j$).
3. Given the values of $k$ and $j$, the variable $i$ can take values from $1$ up to the smaller of $k$ and $j$, namely $k$ (since we know that $i \leq k$ and $i \leq j$).
Can you see how the right hand side expression corresponds to the above explanation?
Left hand side. We adopt a similar strategy of fixing the values of variables one at a time. But this time, we fix them in the order $i, j, k$.
1. As before, $i$ takes values from $1$ to $n$.
2. Once we fixed the value of $i$, it's clear that $j$ takes values from $i$ to $n$.
3. Given the values of $i$ and $j$, the variable $k$ takes values in the range $[i, j]$.
If you write this down in terms of the summation notation, do you see how you get the left hand side?
-
I am looking for a more rigorous explanation. Could you elaborate a little more? – Mark Nov 1 '11 at 0:58
@Mark I have added an explanation. Is it clearer now? – Srivatsan Nov 1 '11 at 1:12
Yes, it clearer now, I have accepted the answer, but do you know a more visual explanation or a more general approach to construct these types of summations? – Mark Nov 1 '11 at 1:22
@Mark A visual proof is implicit in Sivaram's answer. For e.g., if you want to visualize $\sum_{i=1}^n \sum_{j=i}^n 1 = \sum_{j=1}^n \sum_{i=1}^j n$, then you can visualize the diagonal $i=j$ (in the "$i$-$j$ plane"), and count all the lattice points either on the diagonal (i.e., $i=j$) or above it (i.e., $i < j$). – Srivatsan Nov 1 '11 at 1:37
This looks like a pretty good excuse to switch to Iverson brackets. Letting $[p]$ be $1$ if $p$ is true, and $0$ if $p$ is false, we express the sum on the left as
$$\sum_i\sum_j\sum_k [1 \leq i \leq n][i \leq j \leq n][i \leq k \leq j]$$
where the sum is taken over all $i,j,k$. This can be rightly taken as an infinite triple series with nonnegative terms: the Iverson brackets ensure that terms that weren't present in the original series are nicely zeroed.
Using the fact that $[p][q]=[p\text{ and }q]$, we have the equivalent expression
$$\sum_i\sum_j\sum_k [1\leq i\leq k\leq j\leq n]=\sum_j\sum_k\sum_i [1\leq i\leq k\leq j\leq n]$$
where we were free to permute the order of summation.
We can slowly peel Iverson factors out like so:
\begin{align*}\sum_j\sum_k\sum_i [1\leq i\leq k\leq j\leq n]&=\sum_j\sum_k\sum_i [1\leq i\leq k][1\leq k\leq j\leq n]\\&=\sum_j\sum_k\sum_i [1\leq i\leq k][1\leq k\leq j][1\leq j\leq n]\\&=\sum_j[1\leq j\leq n]\sum_k[1\leq k\leq j]\sum_i [1\leq i\leq k]\end{align*}
and the last expression is precisely the sum on the right, rewritten in Iversonian form.
-
+1 I would say this is the most convincing. =) – Srivatsan Nov 1 '11 at 1:43
Let us all thank Iverson and Knuth for making reindexing proofs easy. :) Sadly, it's a technique that is underappreciated by the people who need it the most. – J. M. Nov 1 '11 at 1:44
My answer is not mathematically rigorous, but it should help in visualizing the other solutions. First of all, without loss of generality, the indexing of the right hand side can be changed to match the left hand side as follows:
$$\sum \limits_{i=1}^{n}\ \sum \limits_{j=i}^{n}\ \sum \limits_{k=i}^{j}\ 1 = \sum \limits_{i=1}^{n}\ \sum \limits_{j=1}^{i}\ \sum \limits_{k=1}^{j}\ 1$$
Now the difference lies in two inner summations, which can be shown as red areas on the left and right respectively in the picture below. (Pardon my quick drawing.)
From $i = 1$ to $n$, the left hand side starts from the biggest triangle to the smallest, while the right hand side does the opposite. It is clear that both summations are the same.
-
If you are not satisfied with Srivatsan's proof (which I think is more elegant than mine), you can prove it by induction on $n$.
$$f(n) = \sum \limits_{i=1}^{n}\ \sum \limits_{j=i}^{n}\ \sum \limits_{k=i}^{j}\ 1,$$ $$g(n) = \sum \limits_{j=1}^{n}\ \sum \limits_{k=1}^{j}\ \sum \limits_{i=1}^{k}\ 1.$$
You want to prove $f(n) = g(n)$. This is easy to check for $n=1$.
For the induction step, note that $$f(n+1) = f(n) + \sum_{i=1}^{n+1} \sum_{k=i}^{n+1} 1$$ and $$g(n+1) = g(n) + \sum_{k=1}^{n+1} \sum_{i=1}^{k} 1$$
To prove $$\sum_{i=1}^{n+1} \sum_{k=i}^{n+1} 1 = \sum_{k=1}^{n+1} \sum_{i=1}^{k} 1$$, you can argue similarly with Srivatsan's earlier argument. Both count the number of lattice points on and above the diagonal of the square with vertices at $(0,0),(0,n+1),(n+1,n+1)$ and $(n+1,0)$ in the $i$-$k$ plane.
Otherwise, you prove it again by induction.
-
The idea of my answer is to unscramble the three summations on either the LHS or RHS into inequalities. Then, I use the unscrambled identity to rewrite the other side differently. WLOG, I'll start on the RHS.
RHS = $\underbrace{\sum \limits_{\color{darkorange}{j}=1}^{n}}_{1 \leq \color{darkorange}{j} \leq n} \underbrace{\sum \limits_{\color{darkorange}{k}=1}^{j}}_{1 \leq \color{darkorange}{k} \leq j} \underbrace{\sum \limits_{\color{darkorange}{i}=1}^{k}}_{1 \leq \color{darkorange}{i} \leq k} 1$
$\Longrightarrow 1 \leq j \leq n \, \& \, 1 \leq \color{limegreen}{k \leq j} \, \& \, 1 \leq \color{red}{i \leq k}$
$\Longrightarrow 1 \leq i \color{red}{\leq} k \color{limegreen}{\leq} j \leq n \tag{*}$
The key is just to rewrite the combined inequality above into three separate inequalities. To get different summation limits, I must ensure that at least one of the new inequalities must differ from those in (*).
1. Thus, I must fix either $i$ or $k$ first. WLOG, I choose $k$: $1 \leq k \leq n. \tag{**}$
2. After fixing $k$, I must fix either $i$ or $j$. WLOG, choose $j$: $k \color{limegreen}{\leq} j \leq n$
3. After fixing $j$ and $k$, $i$ remains to be fixed: $1 \leq i \color{red}{\leq} k$.
The above gives $\sum \limits_{k=1}^{n} \sum \limits_{j=k}^{n}\ \sum \limits_{1=i}^{k} 1$, NOT the LHS, but another way to rewrite the summation limits.
1. To get the LHS, I realise that I had needed to fix $i$ first in (**): $1 \leq i \leq n.$
2. After fixing $i$, I must fix either $j$ or $k$. WLOG, choose $j$: $i \color{limegreen}{\leq} j \leq n$
3. After fixing $i$ and $j$, $k$ remains to be fixed: $i \color{red}{\leq} k \color{limegreen}{\leq} j$.
Steps #4-6 yield the LHS = $\sum \limits_{i=1}^{n}\ \sum \limits_{j=i}^{n}\ \sum \limits_{k=i}^{j}\ 1$.
- | 2016-04-30T22:52:13 | {
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http://gb-scaf.bg/00y42/how-to-divide-radicals-with-variables-8f098b | ### how to divide radicals with variables
As long as the roots of the radical expressions are the same, you can use the Product Raised to a Power Rule to multiply and simplify. Dividing Radical Expressions. In the radical below, the radicand is the number '5'.. Refresher on an important rule involving dividing square roots: The rule explained below is a critical part of how we are going to divide square roots so make sure you take a second to brush up on this. Remember that when we multiply radicals with the same type of root, we just multiply the radicands and put the product under a radical sign. We can only take the square root of variables with an EVEN power (the square root of x squared, x to the 4th, x to the 6th, etc.) If you have sqrt (5a) / sqrt (10a) = sqrt (1/2) or equivalently 1 / sqrt (2) since the square root of 1 is 1. Divide: $$\frac { \sqrt [ 3 ] { 96 } } { \sqrt [ 3 ] { 6 } }$$. The radicand refers to the number under the radical sign. Learning Objective(s) ... You multiply radical expressions that contain variables in the same manner. 4 is a factor, so we can split up the 24 as a 4 and a 6. Well, what if you are dealing with a quotient instead of a product? In this case, we can see that $$6$$ and $$96$$ have common factors. Recall that the Product Raised to a Power Rule states that $\sqrt[x]{ab}=\sqrt[x]{a}\cdot \sqrt[x]{b}$. To divide two radicals, you can first rewrite the problem as one radical. The 6 doesn't have any factors that are perfect squares so the 6 will be left under the radical in the answer. Simplify square roots that contain variables in them, like â(8x³) If you're seeing this message, it means we're having trouble loading external resources on our website. You can use the same ideas to help you figure out how to simplify and divide radical expressions. The quotient of the radicals is equal to the radical of the quotient. As you can see, simplifying radicals that contain variables works exactly the same way as simplifying radicals that contain only numbers. A common way of dividing the radical expression is to have the denominator that contain no radicals. Once you do this, you can simplify the fraction inside and ⦠There is a rule for that, too. Radical Pre Algebra Order of Operations Factors & Primes Fractions Long Arithmetic Decimals Exponents & Radicals Ratios & Proportions Percent Modulo Mean, Median & Mode Scientific Notation Arithmetics Solution. Drop me an ⦠Multiplying and Dividing Radical Expressions . The two numbers inside the square roots can be combined as a fraction inside just one square root. Vocabulary Refresher. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. If we apply the quotient rule for radicals and write it as a single cube root, we will be able to reduce the fractional radicand. So when you divide one radical expression by another, you can simplify it by writing both expressions under the same radical, then ⦠Look at the two examples that follow. Next look at the variable part. Dividing radicals with variables is the same as dividing them without variables . Dividing radicals is really similar to multiplying radicals. Dividing radical is based on rationalizing the denominator.Rationalizing is the process of starting with a fraction containing a radical in its denominator and determining fraction with no radical in ⦠Dividing Radical Expressions. We factor, find things that are squares (or, which is the same thing, find factors that occur in pairs), and then we pull out one copy of whatever was squared (or of whatever we'd found a pair of). Under the radical in the answer a fraction inside just one square root first rewrite the problem as radical. Fraction inside and ⦠Multiplying and dividing radical expressions to divide two radicals, you can simplify fraction. 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https://math.stackexchange.com/questions/2167231/how-does-the-dot-product-convert-a-matrix-into-a-scalar/2167254 | # How does the dot product convert a matrix into a scalar?
I am learning linear algebra, and I am a bit confused by the dot product and how the answer to the process turns out to be a scalar rather than a matrix.
For $2$ vectors with $2$ components, I learned that dot product is equivalent to a $1 \times 2$ row vector left multiplied by a $2 \times 1$ column vector. The result of such a multiplication should result in a $1 \times 1$ vector. But I am learning that the dot product somehow transforms the result into a scalar, rather than a $1 \times 1$ vector.
Maybe I am missing something but the difference between a $1 \times 1$ vector and a scalar seems important, because you can multiply a scalar by a matrix of any size, but you can only left-multiply a $1 \times 1$ vector by another matrix with $1$ row.
Thanks for any help in understanding this.
• I guess what I am confused about is that I learned that matrix multiplication is only defined when the number of columns on the left matrix = the number of rows in the right matrix. In which case, a 1x1 matrix could only be left multiplied by other 1 row matrices, whereas a scalar can multiply by a matrix of any size. So simply converting a 1x1 matrix to a scalar seems to removes some of the restrictions that matrix multiplication has. – treeorriffic Mar 1 '17 at 17:06
• You're not wrong, of course. But sometimes mathematicians take shortcuts. See my answer below for more. – Bobbie D Mar 1 '17 at 17:11
• You might find my answer here useful – Omnomnomnom Mar 1 '17 at 17:13
• If you want to replace scalars with $1 \times 1$ matrices, you need to replace the scalar product with the tensor product (i.e. kronecker product). i.e. $aM = [a] \otimes M = M \otimes [a]$ for any scalar $a$ and matrix $M$. Of course, if $M$ is $1 \times n$ then the matrix product works too: $[a]M = [a] \otimes M$. Similarly, $M[a] = M \otimes [a]$ if $M$ is $n \times 1$. – user14972 Mar 2 '17 at 1:23
• A related issue is: is a vector of length $n$ the same as a $n\times 1$ matrix? – Federico Poloni Mar 2 '17 at 9:39
Sometimes, we just say that a $1\times 1$ matrix is the same as a scalar. Afterall, when it comes to addition and multiplication of $1\times 1$ matrices vs addition and multiplication of scalars, the only difference between something like $\begin{bmatrix}3\end{bmatrix}$ and $3$ is some brackets. Consider $$(3+5)\cdot 4 = 32 \\ (\begin{bmatrix} 3\end{bmatrix} + \begin{bmatrix} 5\end{bmatrix})\begin{bmatrix} 4\end{bmatrix} = \begin{bmatrix} 32\end{bmatrix}$$ The algebra works out exactly the same. So sometimes it's not ridiculous to think of $1\times 1$ matrices as just another way of writing scalars.
But if you do want to distinguish the two, then just think of the formula $a\cdot b = a^Tb$ as a way of finding out which scalar you get from the dot product of $a$ and $b$ and not literally the dot product value itself (which should be scalar). That is, we calculate the dot product of $\begin{bmatrix} 1 \\ 2\end{bmatrix}$ and $\begin{bmatrix} 3 \\ 4\end{bmatrix}$ by using the formula $$\begin{bmatrix} 1 \\ 2\end{bmatrix}^T\begin{bmatrix} 3 \\ 4\end{bmatrix} = \begin{bmatrix} 1 & 2\end{bmatrix}\begin{bmatrix} 3 \\ 4\end{bmatrix} = \begin{bmatrix} 11\end{bmatrix}$$ and then say that this tells us that the dot product is really $11$. So the formula $a^Tb$ is just an algorithm we use to find the correct scalar.
You can view it either way. It doesn't really make a difference.
• Oh, viewing it as an algorithm, rather than the literal result, is really helpful. Thanks. – treeorriffic Mar 1 '17 at 17:17
You are not wrong and it's always good to examine statements very carefully. We usually do not distinguish 1$\times$1 matrices from scalars, and in some sense you can think of scalar multiplication as a special rule for when one of the matrices is 1$\times$1. But the truth is that it is a convenient abuse of notation.
Here's one other thing to think about. A $1 \times 1$ real matrix is supposed to represent a linear mapping from a one-dimensional real vector space into a one-dimensional real vector space--essentially just $\mathbb R$ into $\mathbb R$. The only such mappings are those that take $x \mapsto ax$ for some fixed real number $a$. But this mapping is entirely determined by that real number $a$, so they are essentially equivalent.
• Ah thanks, that makes sense. It's just a special rule for 1x1 matrices. – treeorriffic Mar 1 '17 at 17:09
It is true that you can only multiply a $m \times n$ matrix by a $n \times p$ matrix, i.e., the column size of the left matrix has to match the row size of the right matrix. With this, we can conclude that a product of a $1 \times 1$ matrix by a $n \times p$ matrix makes no sense for $n > 1$.
That being said, the space of matrices is a vector space, so it has the multiplication between scalars and matrices. So it makes sense to multiply a scalar by a matrix. Once you realize the space of scalars and $1 \times 1$ matrices can be identified, the confusion goes away.
There are two ways to look at dot products:
• The scalar product of two vectors $$(v_1,...,v_n)$$ and $$(w_1,...,w_n)$$ can be simply defined as the sum $$v_1w_1+\cdots+v_nw_n$$, and so its a scalar by definition, or as you learned it
• as a special kind of matrix multiplication.
In the latter case, the result is indeed a $$(1\times 1)$$-matrix, and as the comments already stated, its just a convention to call this a scalar, as "the only difference are the brackets". Of course, when you view your result as a matrix, you can only multiply it by $$(1\times m)$$-matricies. But here is the clue: the space of $$(1\times 1)$$-matricies is (for all practical purpose) equivalent to the space of scalars. This means, there is a bijective linear map from $$K$$ to $$K^{1\times1}$$, even an algebra isomorphism. So, to use the result in more general fashion, we allow us to interpret it as a scalar. Further, you often use the result of a dot product only to multiply it with vectors. And vectors are not so far from the $$(1\times m)$$-matricies where you are allowed to use your $$(1\times1)$$-matricies anyway.
A dot product is not really a scalar, but it behaves just like one. In math we call that an ISOMORPHISM. For every dot product result, there is a corresponding real number that you get by simply removing the brackets. All of the operations you do with the 1x1 matrix correspond to the same operations done with a real number. Multipliction of a matrix by a 1x1 matrix is defined as multiplcation by the corresponding scalar.
Recommend you look up "isomorphism" in a math dictionary. To do this, you may need to brush up on your German language skills. | 2019-11-18T19:59:53 | {
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http://promeng.eu/3h9fdd/33c2bb-alternate-interior-angles-converse | (4) ∠1 ≅ ∠5 // (3), the definition of congruent angles. then these two lines are parallel. Alternate Interior Angles Theorem: V1. Video tutorials How Mathleaks works Mathleaks Courses How Mathleaks works play_circle_outline Study with a textbook Mathleaks Courses How to connect a textbook play_circle_outline D. Use the Corresponding Angles Converse Postulate. How would you show that the lines d and e are parallel? Using the same figure above, this converse theorem can be symbolically expressed as, #IF (alpha=beta) THEN (l_1 || l_2)# What is the relationship between two perpendicular lines? Which theorem would prove the lines shown are parallel? This … means that the quantities can’t be proportional.” Do you agree with Diego? Let's recall the Alternate Interior Angles Theorem. By accessing or using this website, you agree to abide by the Terms of Service and Privacy Policy. a∥b, Converse of the Corresponding Angles Theorem. In today's lesson, we will show how to prove the converse of the Angle Bisector Equidistant Theorem.. Pics of : Theorem 3 4 Alternate Interior Angles Converse alternate interior angles theorem. Update 2: Put answers wrong, sorry first was a//b . A. Converse to the Corresponding Angles Theorem : If … Euclid's Proposition 27 states that if a transversal intersects two lines so that alternate interior angles are congruent, then the lines are parallel. Given ∠8≅∠12 . Alternate Interior Angles Theorem states that, when two parallel lines are cut by a transversal, the resulting alternate interior angles are congruent. Converse of Alternate Interior Angles Theorem; Examples in Real Life; FAQs . By converse of alternate exterior angle theorem, we get that if $$z=x$$. Alternate interior angles converse. #(alpha=beta) => (l_1 || l_2)#, 6469 views To prove this statement, start with a picture of alternate interior angles that are assumed to be congruent, but don't assume the lines are parallel. Converse of Alternate Interior Angles Theorem. Ccgeo Parperplinequiz Review Proof The Exterior Angles Theorem You Ppt 3 Proving Lines Parallel Powerpoint … Consider the line #y=8x-2#. The converse of this is also true. If two lines are parallel Example: A rectangular wooden frame has a diagonal metal brace. Answers: 3 Get Other questions on the subject: Mathematics. If 2 lines are cut by a transversal so that alternate interior angles … 3.Diego made a graph of two quantities that he measured and said, “The points all lie on a line except one, which is a little bit above the line. We explain Alternate-Interior Angles Converse with video tutorials and quizzes, using our Many Ways(TM) approach from multiple teachers. Last modified on January 7th, 2021 at 8:06 am. converse of the alternate interior angles theorem. A= 2x+20 B= 100 / 80. x=30. % Progress . Alternate Interior Angles Examples We can prove both these theorems so you can add them to your toolbox. Which lines, if any, must be parallel based on the given information? For what value of x is line a parallel to line b? Lines are cut by a transversal then the alternate interior angles are congruent. Another important theorem you derived in the last lesson was that when parallel lines are cut by a transversal, the alternate interior angles formed will be congruent. #(l_1 || l_2) => (alpha=beta)#. None of the lines may be … 3.8) 5. a. yes; Lines a and b are parallel by the Alternate Interior Angles Converse (Thm. A theorem is a proven statement or an accepted idea that has been shown to be true. ∠1 ≅ ∠5 3. then these two lines are parallel. Use alternate interior angles to determine angle congruency and the presence of parallel lines. Converse of the Same-Side Interior Angles Theorem If two lines and a transversal form same-side interior angles that are supplementary, then the two lines are parallel. The theorem states that “ if a transversal crosses the set of parallel lines, the alternate interior angles are congruent”. Converse of the Alternate Interior Angles Theorem: If alternate interior angles formed by two lines with an intersecting traversal are congruent then these two lines are parallel. Converse. (AB= 46, EF= 138, GH= 122, CD= 42) EF, CD, AB . Alternate Interior Angles Converse If two lines are cut by a transversal so the alternate interior angles are congruent, then the lines are parallel. Code to add this calci to your website. (a)converse of the alternate interior angles theorem (b)converse of the alternate exterior angles theorem (c)alternate exterior angles theorem (d)alternate interior angles theorem. Alternate Interior Angle Theorem Proof. So, these are two different theorems, each requiring its own proof. So this is one way to check parallel-ness of two lines. Use the Consecutive Interior Angles Converse … Alternate interior angles are the pair of non-adjacent interior angles that lie on the opposite sides of the transversal. This indicates how strong in your memory this concept is. In the above-given figure, you can see, two parallel lines are intersected by a transversal. We explain Alternate-Interior Angles Converse with video tutorials and quizzes, using our Many Ways(TM) approach from multiple teachers. The angles which are formed inside the two parallel lines,when intersected by a transversal, are equal to its alternate pairs. Converse Alternate Interior Angles Theorem Geometry Help Prove Line And Angle Theorems With Lessons Worksheets Mrwadeturner M1 Consecutive And Alternate Exterior Angles Consecutive Interior Angles Theorem READ Front Porch Design For Colonial Homes. B. Relevance. Alternate Interior Angles Converse If two lines are cut by a transversal so the alternate interior angles are congruent, then the lines are parallel. (1) m∠5 = m∠3 //given. If not, then one is greater than the other, which implies its supplement is less than the supplement of the other angle. (3x + 16)° = 3 x 35 + 16 = 121°. Thus the converse of alternate interior angles theorem is proved. The converse of this theorem is also true; that is, if two lines k and l are cut by a transversal so that the alternate interior angles are congruent, then k ∥ l . Answer Save. Converse: If alternate interior angles are congruent, then lines are parallel. So, line XY$$\left | \right |$$ line RS $$\therefore$$ a) y = 30 , b) line XY$$\left | \right |$$ line RS The converse is: If two lines are cut by a transversal, and the alternate interior angles are congruent, the two lines are parallel. C. Use the Alternate Interior Angles Converse Theorem. By Converse of the Alternate interior Angles Postulate that implies that "If 2 lines and a transversal create alternate interior angles that are in congruence, then the two lines are parallel." Justify your conclusion. But they are on alternate sides of the transversal. What are some examples? They are … Lv 7. What is the Converse of the Alternate Interior Angles Theorem? Converse of the Alternate Interior Angles Theorem : If two lines and a transversal form alternate interior angles that are congruent, then the two lines are parallel. Which theorem proves that d∥e? Video tutorials How Mathleaks works Mathleaks Courses How Mathleaks works play_circle_outline Study with a textbook Mathleaks Courses How to connect a textbook play_circle_outline around the world, Angles Between Intersecting and Parallel Lines. Subjects Near Me. Angles on opposite sides of a transversal, but outside the lines it intersects. Refer to the figure above. 3.7) 4. yes; Consecutive Interior Angles Converse (Thm. Converse alternate interior angles theorem states that if two lines and a transversal form alternate interior angles that are congruent then the two lines are parallel. Now, substituting the value of x in both the interior angles expression we get, (4x – 19)° = 4 x 35 – 19 = 121°. A. https://www.sophia.org/alternate-interior-angles-converse-top B. ∠1 ≅ ∠4 2. Alternate Interior Angles are a pair of angles on the inner side of each of those two lines but on opposite sides of the transversal. MEMORY METER. Given: a//b. Geometry unit 2 parallel lines and transversals worksheet answers. Proof: Suppose a and b are two parallel lines and l is the transversal which intersects a and b at points P and Q. The Angle Bisector Equidistant Theorem states that any point that is on the angle bisector is an equal distance ("equidistant") from the two sides of the angle.. What is the equation of the line that is parallel to this line and... What is the equation of the line that is perpendicular to the line #3x + y = 7# and passes... What is the equation of a line that satisfies the given conditions: perpendicular to #y= -2x +... What line is perpendicular to #y = -3# and passes through point (4, -6)? don_sv_az. The Converse of the Alternate Interior Angle Theorem states that If two lines are cut by a transversal and the alternate interior angles are congruent, then the lines are parallel. Geometry Unit 3 Parallel Lines Angles Formed By Transversals Learn vocabulary terms and more with flashcards games and other study tools. (2) m∠1 = m∠3 //vertical, or opposite angles. By Converse of the Alternate Exterior Angles Theorem that implies that "If 2 lines and a transversal create an alternate exterior angles that are in congruence, then the two lines are parallel." Prepared by Kin Chan - MHS Page 4 of 4 14. In the picture below, assume ∠ 1 ≅ ∠ 2. Lv 7. Alternate interior angles converse. Geometry Unit 3 Parallel Lines Angles Formed By Transversals Learn vocabulary terms and more with flashcards games and other study tools. The measures of two consecutive interior angles are (4 x DQG x - :KDW consecutive interior angles are supplementary. If alternate interior angles formed by two lines with an intersecting traversal are congruent 4x – 3x = 16 + 19. x = 35°. What are Alternate Interior Angles. What is an equation of the line that has a y-intercept of -2 and is perpendicular to the line... What is the equation of the line in slope-intercept that is perpendicular to the line #4y - 2 =... See all questions in Angles Between Intersecting and Parallel Lines. In the applet below, a TRANSVERSAL intersects 2 PARALLEL LINES.When this happens, there are 2 pairs of ALTERNATE INTERIOR ANGLES that are formed. The angles … 42 mins ago. alternate interior angles. Geometry unit 2 parallel lines and transversals worksheet answers. The converseof this theorem, which is basically the opposite, is also a proven statement: if two lines are cut by a transversal and the alternate interior angles are congruent, then the lines are parallel. Prove:line b is parallel to line c Given:<4 and <5 are congruent 1.<4 congruent to<5 // Given 2.<1 is congruent to<4 // Vertical angles Congruence … Alternate Interior Angles Converse; Consecutive Interior Angles Converse; Last Page; Theorem 6.4-6.5; Theorem 6.6-6.7; Theorem 3.4 If two lines are cut by transversial so the alternate interior angle are congruent, the lines are parellel. Converse Alternate Interior Angles Theorem, « Prove the Pythagorean Theorem Using Triangle Similarity, Proving that a Quadrilateral is a Parallelogram », interior alternating angles and exterior alternating angles are congruent, converse of the Corresponding Angles Theorem. Therefore, the alternate angles inside the parallel lines will be equal. alternate interior angles formed by these two lines with an intersecting traversal are congruent. c∥dc∥d, Converse of the Alternate Exterior Angles Theorem. Euclid proves this by contradiction: If the lines are not parallel then they must intersect and a triangle is formed. In this example, these are two pairs of Alternate Interior Angles: If a point lies on the interior of an angle and is equidistant from the sides of the angle, then a line from the … The Alternate Interior Angles Theorem states that, when two parallel lines are cut by a transversal , the resulting alternate interior angles are congruent . By alternate interior angle theorem converse, if a transversal intersects two lines such that a pair of interior angles are equal, then the two lines are parallel. converse of the alternate exterior angles theorem. Progress % Practice Now. Given ∠3≅∠13, which lines, if any, must be parallel based on the given information? The converse reverses the direction of the theorem. Two angles that lie on the same side of the transversal and on the same sides of the other two lines - they are in the same location on each parallel line. Alternate interior exterior angles corresponding and alternate interior angles alternate exterior angles definition missing reason in the proof. The Interior Angles theorem states, if two parallel lines are cut by a transversal, then the pairs of alternate interior angles are congruent.. A theorem is a proven statement or an accepted idea that has been shown to be true.The converse of this theorem, which is basically the opposite, is also a proven statement: if two lines are cut by a transversal and the … a∥b, Converse of the Corresponding Angles Theorem. Prepared by Kin Chan - MHS Page 4 of 4 14. Converse of the Alternate Interior Angles Theorem: Discussion Section 1 3 Alternate Interior Angles Definition Theorem Examples Properties Of Parallel Lines Academic … If 2 lines are cut by a transversal so that alternate interior angles are … Example: Non Example: Alternate Exterior Angles Converse Theorem: If 2 lines are cut by a transversal so the alternate exterior angles are _____, then the lines are _____. Converse alternate interior angles theorem states that if two lines and a transversal form alternate interior angles that are congruent, then the two lines are parallel. We have: \[ \begin{align} \angle 1 &= \angle 5 \text{ (corresponding angles)} \\[0.3cm] \angle 3 &= \angle 5 \text{ (vertically opposite angles)} … Answer Save. The converse of the theorem is also true: The Converse of the Alternate Interior Angles Theorem states that if two lines are cut by a transversal and the alternate interior angles are congruent, then the lines are parallel. Euclid 29 Alternate Interior Angle Converse You Same Side Exterior Angles Definition Theorem Lesson READ Ford Expedition El Interior Photos Alternate Exterior Angles Theorem Given Xw Xy Zy Prove Δwxz Δyzx A Alternate Interior READ Mazda Cx5 Interior Length. ∠4 ≅ 1. ANSWER: alternate exterior angles converse Assign to Class. Alternate interior angles are the pairs of angles formed when a transversal intersects two parallel or non-parallel lines. Examples. If the pair of lines are parallel then the alternate interior angles are equal to each … Justify your conclusion. corresponding angles. Preview; Assign Practice; Preview. a∥b , Converse of the Alternate Interior Angles Theorem. 1 Answer. The converse of the theorem is also true: The Converse of the Alternate Interior Angles Theorem states that if two lines are cut by a transversal and the alternate interior angles are congruent, then the lines are parallel. SOLUTION Given ∠4 ≅ ∠5 Prove g h STATEMENTS REASONS 1. 4 years ago. But, since both theorem A->B and B->A can be proven independently, both statement are equivalent. Justify your conclusion. This implies that there are interior angles on the same side of the transversal which are less than two right angles, contradicting the fifth … How would you show that the lines d and e are parallel? Converse theorem should look like "if B then A": If alternate interior angles … Hence, option 'b' is correct. If two lines are parallel then alternate interior angles formed by these two lines with an intersecting traversal are congruent. These angles are called alternate interior angles. corresponding angles. Interact with the applet below for a few minutes, then answer the questions that immediately follow. Converse alternate interior angles theorem states that if two lines and a transversal form alternate interior angles that are congruent then the two lines are parallel. Alternate Interior Angles. In the above figure, the pairs of alternate interior angles are: 1 and 3; 2 and 4 Use the Consecutive Interior Angles Converse Theorem. Thus the values of the two alternate interior angles are 121°. Using consecutive interior angles to prove parallel lines duration. Proof Consecutive Interior Angles Converse Home » Geometry » Angle » Alternate Interior Angles. Alternate Interior Angles When a line (called a transversal) intersects a pair of lines, alternate interior angles are formed on opposite sides of the transversal. ∠A = ∠D and ∠B = ∠C Two angles that lie on the same side of the transversal and on the same sides of the other two lines - they are in the same location on … Proving the Alternate Interior Angles Converse Prove that if two lines are cut by a transversal so the alternate interior angles are congruent, then the lines are parallel. Let's represent it in a form "if A then B": If two lines that are cut by a transversal are parallel [Part A] then alternate interior angles formed by these lines are congruent [Part B]. Use the diagram below in the following exercise. D. Use the Corresponding Angles Converse Postulate. not enough information to make a conclusion. Approximately equal is not he same as equal. Perpendicular lines have _________ slopes? third was a/b. two lines are parallel. Given ∠8≅∠12 . A line that intersects two or more coplanar lines at different points; the angles are classified by type. SOLUTION: and are alternate exterior angles of lines and m. Since , || m by the Converse of Alternate Exterior Angles Theorem. Let's recall the Alternate Interior Angles Theorem. (3) m∠1 = m∠5 //using (1) and (2) and transitive property of equality, both equal m∠3. Congruent Alternate Interior Angles. This lesson will demonstrate how to prove lines parallel with the converse of the alternate-interior angles theorem. 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( TM ) approach from multiple teachers way to check parallel-ness of two lines are not parallel alternate. The Same-Side interior angles are congruent two other lines, then the alternate interior angles are congruent, lines. + 16 ) ° = 3 x 35 + 16 ) ° = x! Unit 3 parallel lines answers: 3 get other questions on the subject: Mathematics ). ≅ ∠5 prove g h STATEMENTS REASONS 1 to check parallel-ness of two lines with an intersecting traversal are.... Are congruent them to your toolbox Theorem is the one obtained by taking a as. Implies its supplement is less than the other angle … consecutive interior angles to angle. The presence of parallel lines duration Examples in Real Life ; FAQs add them to your toolbox add them your... Angles on opposite sides of the transversal alternate angles inside the two parallel lines are parallel the interior! That ∠3 and ∠5 are on alternate sides of a Theorem is two lines with intersecting! Must intersect and a perpendicular line of non-adjacent interior angles are supplementary »! Strong in your memory this concept is correctly justifies that the lines are parallel. These are two pairs of angles formed by these two lines are parallel and. By a transversal Life ; FAQs add to 180 degrees and the presence of parallel lines angles formed these... Of parallel lines duration k ; alternate interior angles Theorem this lesson demonstrate! K ; alternate interior angles to prove lines parallel with the applet for. Crosses the set of parallel lines duration are cut by a transversal the!
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https://nicoguaro.github.io/es/posts/numerical-29/ | Ir al contenido principal
Numerical methods challenge: Day 29
During October (2017) I will write a program per day for some well-known numerical methods in both Python and Julia. It is intended to be an exercise then don't expect the code to be good enough for real use. Also, I should mention that I have almost no experience with Julia, so it probably won't be idiomatic Julia but more Python-like Julia.
Cholesky decomposition
Today we have Cholesky decomposition. It is a factorization of a Hermitian, positive-definite matrix into a lower and upper matrix, the main difference with LU decomposition is that it the lower matrix is the Hermitian transpose of the upper one.
Following are the codes
Python
from __future__ import division, print_function
import numpy as np
def cholesky(mat):
m, _ = mat.shape
mat = mat.copy()
for col in range(m):
print(mat[col, col] -
mat[col, 0:col].dot(mat[col, 0:col]))
mat[col, col] = np.sqrt(mat[col, col] -
mat[col, 0:col].dot(mat[col, 0:col]))
for row in range(col + 1, m):
mat[row, col] = (mat[row, col] -
mat[row, 0:col].dot(mat[col, 0:col]))/mat[col, col]
for row in range(1, m):
mat[0:row, row] = 0
return mat
A = np.array([
[4, -1, 1],
[-1, 4.25, 2.75],
[1, 2.75, 3.5]], dtype=float)
B = cholesky(A)
Julia
function cholesky(mat)
m, _ = size(mat)
mat = copy(mat)
for col = 1:m
mat[col, col] = sqrt(mat[col, col] -
dot(mat[col, 1:col-1], mat[col, 1:col-1]))
for row = col + 1:m
mat[row, col] = (mat[row, col] -
dot(mat[row, 1:col-1], mat[col, 1:col-1]))/mat[col, col]
end
end
for row = 2:m
mat[1:row-1, row] = 0
end
return mat
end
A = [4 -1 1;
-1 4.25 2.75;
1 2.75 3.5]
B = cholesky(A)
As an example we have the matrix
\begin{equation*} A = \begin{bmatrix} 4 &-1 &1\\ -1 &4.25 &2.75\\ 1 &2.75 &3.5 \end{bmatrix} \end{equation*}
And, the answer of both codes is
2.0 0.0 0.0
-0.5 2.0 0.0
0.5 1.5 1.0
Comparison Python/Julia
Regarding number of lines we have: 23 in Python and 22 in Julia. The comparison in execution time is done with %timeit magic command in IPython and @benchmark in Julia.
For Python:
%timeit cholesky(np.eye(100))
with result
100 loops, best of 3: 13 ms per loop
For Julia:
@benchmark cholesky(eye(100))
with result
BenchmarkTools.Trial:
memory estimate: 4.01 MiB
allocs estimate: 20303
--------------
minimum time: 1.010 ms (0.00% GC)
median time: 1.136 ms (0.00% GC)
mean time: 1.370 ms (17.85% GC)
maximum time: 4.652 ms (40.37% GC)
--------------
samples: 3637
evals/sample: 1
In this case, we can say that the Python code is roughly 10 times slower than Julia code.
Comentarios
Comments powered by Disqus | 2020-10-20T14:12:11 | {
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https://math.stackexchange.com/questions/3138555/show-that-if-g-is-connected-then-lg-is-connected | # Show that if $G$ is connected then $L(G)$ is connected
Well, the exercise it's as the title says. I know that if $$G$$ is connected then for every pair of vertex $$u,v$$ in $$G$$ there's a walk between them. So when the Line Graph $$L(G)$$ is constructed those edges in $$G$$ where the walk between $$u$$ and $$v$$ is formed will be adjacent vertex in the Line Graph $$L(G)$$. Since $$G$$ is connected therefore $$L(G)$$ must be connected.
I'm not sure about if the argument is valid, also I'm looking for a more formal/detailed proof, because I think there's a lot of holes in this argument.
The argument given is correct, though a bit informal. Here's an elaboration:
To be more precise, suppose $$G$$ is connected. Then for any two vertices $$u,v\in G$$, there is a path $$P=u\to x_1\to x_2\to\cdots\to x_n\to v$$ where the $$x_i$$'s are some other vertices of $$G$$.
By the existence of such a path $$P$$, we can say that $$u$$ is adjacent to $$x_1$$, $$x_i$$ is adjacent to $$x_{i+1}$$ (for $$1\leq i\lt n$$) and $$x_n$$ is adjacent to $$v$$
By the definition of the line graph $$L(G)$$, this means $$(u,x_1),(x_i,x_{i+1})$$ and $$(x_n,v)$$ are vertices of $$L(G)$$ [for $$1\leq i\lt n$$]
Now, since $$(u,x_1)$$ and $$(x_1,x_2)$$ share a common endpoint $$x_1$$, they must be adjacent in $$L(G)$$. Similarly, $$(x_i,x_{i+1})$$ is adjacent to $$(x_{i+1},x_{i+2})$$ for $$1\leq i\lt n-1$$ and $$(x_{n-1},x_n)$$ is adjacent to $$L(G)$$
So, we have a path $$L_P$$ in $$L(G)$$ as $$L_p=(u,x_1)\to (x_1,x_2)\to\cdots\to(x_{n-1},x_n)\to(x_n,v)$$
Hence, for every path $$P$$ in $$G$$, there is a path $$L_P$$ in $$L(G)$$.
Suppose $$(a,b)$$ and $$(A,B)$$ are two vertices of $$L(G)$$. We show that there is a path between them: By definition of $$L(G)$$, all four of $$a,b,A,B$$ are vertices of $$G$$ with $$a$$ adjacent to $$b$$, ie, $$a\to b$$ and $$A$$ adjacent to $$B$$, ie, $$A\to B$$; since $$G$$ is connected, there must be a path $$P_{bA}$$ from $$b$$ to $$A$$ which translates to a path $$L_{\large P_{bA}}$$ in $$L(G)$$ which gives us a path $$a\to b\to P_{bA}\to A\to B$$ in $$G$$ and a path $$(a,b)\to L_{\large P_{bA}}\to (A,B)$$ in $$L(G)$$ | 2019-10-22T19:20:31 | {
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https://math.stackexchange.com/questions/2362928/find-a-tree-with-a-given-sequence-and-show-that-all-such-trees-have-the-same-num | # Find a tree with a given sequence and show that all such trees have the same number of vertices
Find a tree with degree sequence $(4,3,3,3,2,2,2,1,\ldots),$ where the number of vertices of degree $1$ is not specified and prove that any tree found must have the same number of vertices.
One possible tree is as follows. First consider the simple path $ABCDE.$ Then at the vertex $E,$ the tree will have three edges $EF, \ EG, \ EH.$ At the vertex $F,$ there are two edges $FI, \ FJ.$ Also, the vertex $G$ splits into $GK, \ GL.$ Finally, there are edges $HM, \ HN$ at the vertex $H.$
Assuming the tree above is correctly constructed and given $(4,3,3,3,2,2,2,1,\ldots),$ looks like the actual degree sequence must be $(4,3,3,3,2,2,2,1).$ Seems to me all the trees constructed from the sequence $(4,3,3,3,2,2,2,1)$ must have $8$ vertices.
Does the above make sense?
• Does this make sense? I would say no, as finite trees have at least two leaves. Jul 18 '17 at 17:32
• @TheoBendit, I think the endpoints of $FI, \ FJ, \ GK, \ GL, HM, \ HL$ are the leaves. There are $6$ leaves. Jul 18 '17 at 17:43
For a tree, the number of vertices $V$ equals the number of edges $E$ plus 1.
The number of edges $E$ is the sum of all the degrees divided by 2.
We know there are $7+n$ vertices, with $n$ the number of vertices with degree $1$ (the leaves)
We also know that the sum of the degrees is $19+n$ ($4+3+3+3+2+2+2=19$)
Hence:
$7+n = \frac{19+n}{2}+1$
So: $12 +2n = 19+n$
Hence, $n=7$, and total number of vertices is $14$
• @ Bram28, nice. Does that mean all the trees constructed according to the givens have $14$ vertices? Jul 18 '17 at 17:57
• @user464381 Yes, I proved that there have to be $7$ leaves, i.e. $7$ vertices with degree $1$, and hence there have to be exactly $14$ vertices. Jul 18 '17 at 18:03
• @ Bram28, I now see I constructed a right tree. Just read the degree sequence off of it incorrectly. The nodes $A, \ FI, \ FJ, \ GK, \ GL, HM, \ HL$ are all of degree one. Thank you very much. Jul 18 '17 at 18:16
• @user464381 Right, and so your tree has degree sequence $(4,3,3,3,2,2,2,1,1,1,1,1,1,1)$ instead of $(4,3,3,3,2,2,2,1)$, and so your tree has indeed $14$ vertices instead of $8$. But, in your post you only showed that this particular one tree would have $14$ vertices ... you did not show that any tree would have to have $14$ vertices. For that, you'll need something akin to what I did. Jul 18 '17 at 18:20 | 2021-09-28T11:41:13 | {
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http://mathhelpforum.com/algebra/153882-question-completing-x.html | # Math Help - Question on Completing the X
1. ## Question on Completing the X
Hello there, I have a question on how to complete the x on a quadratic equation which has the coefficient of a. It cannot be solved using factorization. The equation is:
$8x^2 + 4x - 1 = 0$
Thank You
2. I assume you mean to complete the square...
$8x^2 + 4x - 1 = 0$
$8\left(x^2 + \frac{1}{2}x - \frac{1}{8}\right) = 0$
$8\left[x^2 + \frac{1}{2}x + \left(\frac{1}{4}\right)^2 - \left(\frac{1}{4}\right)^2 - \frac{1}{8}\right] = 0$
$8\left[\left(x + \frac{1}{4}\right)^2 - \frac{3}{16}\right] = 0$
$8\left(x + \frac{1}{4}\right)^2 - \frac{3}{2} = 0$.
Go from here.
3. Originally Posted by Prove It
I assume you mean to complete the square...
$8x^2 + 4x - 1 = 0$
$8\left(x^2 + \frac{1}{2}x - \frac{1}{8}\right) = 0$
$8\left[x^2 + \frac{1}{2}x + \left(\frac{1}{4}\right)^2 - \left(\frac{1}{4}\right)^2 - \frac{1}{8}\right] = 0$
$8\left(x + \frac{1}{4}\right)^2 - \frac{3}{16}\right] = 0$
$8\left(x + \frac{1}{4}\right)^2 - \frac{3}{2} = 0$.
Go from here.
Thank you but I'm little bit confused on how do you exactly have
$- \frac{3}{2}$
From
$\frac{3}{16}$
I've also noticed that the second from the last line did not show up, so I edit your formatting a little bit. I'm not sure if what I'm doing is right. Can you verify?
4. I expanded the outermost (square) brackets.
$8\left(-\frac{3}{16}\right) = -\frac{3}{2}$.
5. Originally Posted by Prove It
I expanded the outermost (square) brackets.
$8\left(-\frac{3}{16}\right) = -\frac{3}{2}$.
Ah I see, can you give me a bit of insight how to find x? Or at least to eliminate the 8?
6. Use the standard method of "undoing" everything using inverse operations.
7. Is it possible to do this:
$8(x+1/4)^2 = 3/2$
$8(x+1/4) = root(3/2)$
$8x + 2 = root(3/2)$
$8x = root(3/2) - 2$
$x = (root(3/2) - 2) / 8$
But it gives up quite weird answer...
I'm sorry also because of bad formattig I do not know how to operate Latex properly.
8. No, when you undo everything you go in the reverse of the order of operations.
Since Exponentiation comes before Multiplication, when you go in reverse, the Multiplication is undone before the Exponentiation...
9. Originally Posted by Prove It
No, when you undo everything you go in the reverse of the order of operations.
Since Exponentiation comes before Multiplication, when you go in reverse, the Multiplication is undone before the Exponentiation...
Allright, would this be possible:
$8(x+1/4)^2 = 3/2$
$(x+1/4)^2 = 8(3/2)$
$(x+1/4) = root(12)$
$x = root(12)(1/4)$
10. How do you undo multiplication?
11. Originally Posted by Prove It
How do you undo multiplication?
By dividing it I suppose, I'm really sorry if action looks like an imbecible. I just want to understand it. Just having problem with that number 8.
12. Yes, you need to divide both sides by 8.
What is $\frac{3}{2} \div 8$?
13. Originally Posted by Prove It
Yes, you need to divide both sides by 8.
What is $\frac{3}{2} \div 8$?
AH I see would it be like this then?
$(x+\frac{1}{4})^2 = \frac{3}{16}$
$(x+\frac{1}{4}) = \sqrt{\frac{3}{16}}$
$(x+\frac{1}{4}) = \frac{\sqrt{3}}{4}$
$x = \frac{1}{4}(1-\sqrt{3})$
14. Originally Posted by Lites
AH I see would it be like this then?
$(x+\frac{1}{4})^2 = \frac{3}{16}$
$(x+\frac{1}{4}) = \sqrt{\frac{3}{16}}$
$(x+\frac{1}{4}) = \frac{\sqrt{3}}{4}$
$x = \frac{1}{4}(1-\sqrt{3})$
Very close. There are two square roots to every nonnegative number, one positive and one negative.
So that means it should read...
$\left(x + \frac{1}{4}\right)^2 = \frac{3}{16}$
$x + \frac{1}{4} = \pm \sqrt{\frac{3}{16}}$
$x + \frac{1}{4} = \pm \frac{\sqrt{3}}{4}$
$x = -\frac{1}{4} \pm \frac{\sqrt{3}}{4}$.
So your two solutions are $x = \frac{-1 - \sqrt{3}}{4}$ and $x = \frac{-1 + \sqrt{3}}{4}$.
15. Originally Posted by Prove It
Very close. There are two square roots to every nonnegative number, one positive and one negative.
So that means it should read...
$\left(x + \frac{1}{4}\right)^2 = \frac{3}{16}$
$x + \frac{1}{4} = \pm \sqrt{\frac{3}{16}}$
$x + \frac{1}{4} = \pm \frac{\sqrt{3}}{4}$
$x = -\frac{1}{4} \pm \frac{\sqrt{3}}{4}$.
So your two solutions are $x = \frac{-1 - \sqrt{3}}{4}$ and $x = \frac{-1 + \sqrt{3}}{4}$.
Finally, I got it!
Thank you for your kind help Prove it! | 2014-03-14T11:02:38 | {
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http://qrkc.slb-biotec-france.fr/weighted-degree-centrality.html | Weighted Degree Centrality
Calculation of out-degree centrality which only counts outgoing edges from a vertex. By an axiomatic analysis, we show that the Attachment Centrality is closely re-lated to the Degree Centrality in weighted graphs. Getting started with Python and NetworkX 3. Degree centrality is defined as the number of links incident upon a node (i. The input graph can be an adjacency matrix, a weight matrix, an edgelist (weighted or unweighted), a qgraph object or an igraph object. A few network measures have been proposed for weighted networks, including three common measures of node centrality: degree, closeness, and betweenness. sum of squares of the eigenvalues in the Laplacian matrix) of the graph when the vertex is removed. banks’ individual risk-weighted assets alone. Inspired by the deep relation between control centrality and hierarchical structure in a general directed network, we design an efficient attack strategy against the controllability of malicious networks. Network Centrality - [On, Off]: Calculate Degree Centrality and/or Eigenvector Centrality and/or Local Functional Connectivity Density. The weighted degree centrality is the sum of the weights of the edges incident with the node. For a weighted graph, it is defined as the sum of weights from edges connecting to a node (also sometimes referred to as the node strength). computation for several different types of centrality. A bit more detail: Degree centrality is the simplest measure of node connectivity. theoretical foundation for centrality measures not based on shortest paths is giveninFriedkin(1991). Opsahl t f agneessens and j skvoretz 2010 node. For a review as well as generalizations to weighted networks, see Opsahl et al. Degree Centrality (1951) Degree Centrality Negative Ties (2017) Degree Mass Centrality (2015) Degree Sphere Centrality (2009) DelayFlow Centrality (2013) DFC - Disease Flow Centrality (2011) DiffSLC Centrality (2017) Diffusion Degree (2011) DMNC - Density of Maximum Neighborhood Component (2008) DS - Dynamic-Sensitive Centrality (2016). ” Degree Centrality Example Let’s see how Degree Centrality works on a small dataset. Newman Santa Fe Institute, 1399 Hyde Park Road, Santa Fe, New Mexico 87501 and Center for Applied Mathematics, Cornell University, Rhodes Hall, Ithaca, New York 14853 ~Received 1 February 2001; published 28 June 2001!. This paper aims to identify central points in road networks considering traffic demands. The performance of the proposed sampling algorithms for weighted networks is studied through the simulation experiments in Sect. , node degree) to more elab- orate. The time-optimal coarse-grained method of solving each SSSP in parallel has been presented [32] but as. OSN, in this paper, we address the problems of community detection in weighted networks and exploit community for data forwarding in DTN and worm containment in OSN. The node-level centrality scores. , the weighted degree centrality (WDC), is developed to achieve the reliable prediction of essential proteins. In the Ranking Panel, apply this new measure to the nodes, as proposed here. This metrics indicates influencial nodes for highest value. The closeness centrality [Fr78] metric ranks vertices by their average geodesic distance to all other vertices, i. Weighted Degree” (Statistics panel). Wed, 30 Oct 2019 09:43:39. Subgraph centrality replaces the adjacency matrix with its trace. After doing this, I am calculating the centrality (degree, eigenvector, closeness and betweenness) for my one mode projected and weighted graph. However, these generalizations have solely focused on tie weights, and not on the number of ties, which was the central component of the original measures. * * @return the number of vertices in this edge-weighted graph. In Proceedings of the 2013 IEEE 2nd International Network Science Workshop, NSW 2013. degree harmonic Relationship todegree-based centralities In fact, degree-based centrality measures are related to geodesic-based measures like closeness and harmonic centrality, although they do emphasize different aspects of network structure. DYNAMICS: CENTRALITY, COMMUNITY AND PREDICTABILITY by SIMA DAS A DISSERTATION Presented to the Graduate Faculty of the MISSOURI UNIVERSITY OF SCIENCE AND TECHNOLOGY In Partial Fulfillment of the Requirements for the Degree DOCTOR OF PHILOSOPHY in COMPUTER SCIENCE 2017 Approved by Dr. Subgraph centrality replaces the adjacency matrix with its trace. If the network is directed, we have two versions of the measure: in-degree is the number of in-coming links, or the number of predecessor nodes; out-degree is the number of out-going links, or the number of successor nodes. Stanford Network Analysis Platform (SNAP) is a general purpose network analysis and graph mining library. It is shown that the commonly used measures of degree, closeness and eigenvector centrality are stable whereas betweenness. Also, ki is the degree centrality, si is the weighted degree centrality, x represents the adjacency matrix, w represents the weighted adjacency matrix. 4 Degree centrality Degree centrality measures the number of edges of node i, d i g. clustering : Clustering ¶ Metrics which group nodes within graphs into clusters. leonidzhukov. weighted (valued) networks, we are referring to it as range-limited centrality. The size of a graph is equal to the number of vertices of the graph. Degree centrality. Researchers have used different techniques to work around this problem, examples include thresholding correlations when creating the adjacency matrix and using a smaller input data with lower resolution. Current-Flow Betweenness¶. So if our edges between the green node and its 6 neighbors were, for instance, 0. Over the years many more complex centrality metrics have been proposed and studied,. Weighted degree 1. Centrality: The relative importance of a node within a graph. This property of the degree distribution is captured by the geometrically weighted degree (GWD) term in statistical models of networks. centrality-in-networks-with-disconnected-components/ precomp. Tie weights and the number of ties were connected with certain proportion by tuning parameter in the model. 1) In other words, the closeness centrality of v is the inverse of the average (shortest- path) distance from v to any other vertex in the graph. We propose a novel community detection algorithm, and then introduce two metrics called intra-centralityand inter-centrality, to characterize nodes in communities. where N(v) is the set of neighbors of v in G and d G (v i) is the degree of v i in G. that if a graph has identical subgraph centrality for all nodes, then the closeness and betweenness centralities are also identical for all nodes. Speci cally, if n= jVj, then the degree. computation for several different types of centrality. Centrality measures using local information, like the node degree or link overlap, are computed efficiently as they only require knowledge about the neighbors of a given node or link. It is then based on tie weights and not on the number of ties. Among these top performers, which characters have more more interactions per connection? Which characters have fewer? Compared to degree centrality, Jaime, Arya and Stannis underperform. Eigenvector centrality is a measure of the centrality of a node in a network, based on the weighted sum of centralities of its neighbors. The number in the upper right corner is the year the paper was published. Centrality in Social Networks Degree If we want to measure the degree to which the graph as a whole is centralized, we look at the dispersion of centrality: Simple: variance of the individual centrality scores. Weighted Degree Centrality: the number of interactions you participate in. Sometimes we hear it through the grapevine. theoretical_max: The maximum theoretical graph level centralization score for a graph with the given number of vertices, using the same parameters. A few network measures have been proposed for weighted networks, including three common measures of node centrality: degree, closeness, and betweenness. As this definition is inherently vague, a lot of different centrality scores exists that all treat the concept of central a bit different. There are also some dedicated centrality packages, such as centiserve , CINNA , influenceR and keyplayer. Eigenvector¶. h-Type hybrid centrality measures for weighted networks Alireza Abbasi School of Engineering and IT, University of New South Wales Canberra, ACT 2610, Australia a. Degree Centrality draw (G, pos, nx. Subgraph centrality replaces the adjacency matrix with its trace. Laplacian centrality is a simple centrality measure that can be calculated in linear time. 2004, Newman 2001, Opsahl et al. Degree centrality measures the number of edges incident upon a node. of CF-centrality is the matrix inversion with complexity O(n3). values()) 3. degree of gene product Leave-one-out cross-classification experiments using OMIM database demonstrate success of information flow based methods. Centrality indices can be classified in local and global categorizes. ing a generalization of degree centrality for weighted networks where the outcome is a combination of the number of ties and the tie weights. Newman Santa Fe Institute, 1399 Hyde Park Road, Santa Fe, New Mexico 87501 and Center for Applied Mathematics, Cornell University, Rhodes Hall, Ithaca, New York 14853 ~Received 1 February 2001; published 28 June 2001!. For example, there is the degree centrality (or just the degree of a node, i. Centrality measures Degree centrality Closeness centrality Betweenness Eigenvalue centrality Hubs and Authorities References What's C the sStory? K N o Really, 7 of 28 Centrality. ” Degree Centrality Example Let’s see how Degree Centrality works on a small dataset. trality, betweenness centrality, degree centrality, and PageRank) for authors in this network. Element-level analysis. need to design centrality measures for weighted networks, because weighted networks where edges are attached weights would contain rich information. the node (Batallas and Yassine 2006). Abstract We introduce the concept of control centrality to quantify the ability of a single node to control a directed weighted network. Thus, information availability may lead to a decline in acquisition returns. In this study, new centrality (collaborative) measures are proposed for a node in weighted networks in three different categories. The out-degree of a node in a directed graph is the number of out-links incident to that node; the in-degree is the number of in-links incident. In terms of the interbank network, this indicates the number of other banks that a given bank has lending and borrowing relationship with. ∙ 0 ∙ share. Library for the analysis of networks. network measures in case of weighted, asymmetric, self-looped, and disconnected networks. OSN, in this paper, we address the problems of community detection in weighted networks and exploit community for data forwarding in DTN and worm containment in OSN. [8] proposed an eccentricity and degree centrality based complex network for keywords extraction, and Li et al. So if our edges between the green node and its 6 neighbors were, for instance, 0. Read more in "Two Step Graph-Based Semi-Supervised Learning for Online Auction Fraud Detection. It is then based on tie weights and not on the number of ties. A startling conclusion is that regardless of the initial transformation of the adjacency matrix, all such approaches have common limiting behavior. The input graph can be an adjacency matrix, a weight matrix, an edgelist (weighted or unweighted), a qgraph object or an igraph object. In a directed graph, we distinguish between in-degree and out-degree. THE MEASURE I have proposed (Bonacich 1972a, 1972b) a measure of centrality (in this paper, I will call it "e") in which a unit's centrality is its summed connec- tions to others, weighted by their centralities. net dictionary. Also, ki is the degree centrality, si is the weighted degree centrality, x represents the adjacency matrix, w represents the weighted adjacency matrix. I am now studying centrality measures of my weighted network using statnet, but the centrality measures I obtain are as if statnet did not take into account the values of my edges. Please edit your data and try again. This identification of points is made with a variation of betweenness centrality metric. " Although the research focused primarily on NPP in analyzing social networks, the. Degree centrality is dened by a degree of unit x cD(x) = degree of unit x Such measures are called absolute measures of. Though the use of path weights suggest information centrality as a possible replacement for closeness, the problem of inverting the B matrix poses problems of its own; as with all such measures, caution is. For a review as well as generalizations to weighted networks, see Opsahl et al. Stability and Continuity of Centrality Measures in Weighted Graphs Santiago Segarra and Alejandro Ribeiro Abstract—This paper presents a formal definition of stability for node centrality measures in weighted graphs. values()) 3. In addition to considering the widely used centrality metrics, we introduce a new centrality measure, the degree mass. A few network measures have been proposed for weighted networks, including three common measures of node centrality: degree, closeness, and betweenness. nected to (her degree), but also on their centrality. Centrality Indices Offered in Popular Social Network Analysis Packages and Procedures Centralities Package: e ss s s or Other centrality indices: UCINet attribute weighted, beta reach, Bonacich power, edge betweenness, flow betweenness, Hubbel, influence, information, inverse weighted degree, Katz. CytoNCA is not only a tool for **centrality calculation** but also a formidable instrument for **visual analysis** and **evaluation**, processing various data agilely. For unweighted graphs, run- ning time reduces to O(nm). Degree centrality (DC) and local functional connectivity density (lFCD) are statistics calculated from brain connectivity graphs that measure how important a brain region is to the graph. Compute the Katz centrality for the graph G. We propose a new concept of the betweenness centrality for weighted graphs using the methods of cooperative game theory. As the name implies,. Moreover, a node with high eigenvector centrality is not necessarily highly linked (the node might have few but important linkers. The degree centrality, the eigenvector centrality 8 and the Katz centrality 7 are obtained by adopting very simple link-estimation functions. out_degree_centrality (G). Calculation of degree centrality which counts all incident edges on each vertex to include those that are both incoming and outgoing. Centrality' • Finding'outwhich'is'the'mostcentral'node'is' important:'' - Itcould'help'disseminang'informaon'in'the'. Degree centrality is the count of the number of vertices a vertex is directly connected to. Stanford Network Analysis Platform (SNAP) is a general purpose network analysis and graph mining library. theoretical_max: The maximum theoretical graph level centralization score for a graph with the given number of vertices, using the same parameters. Degree is the simplest of the node centrality measures by using the local structure around nodes only. weighted degree centrality 다만, 해당 edge의 weight는 모두 같지 않기 때문에, 서로 다른 weight를 고려하는 것이 필요한데, 이상하게도, networkx 에는 이 weight를 고려해서 degree centrality를 고려하는 일이 없습니다. International Journal of Architectural, Civil and Construction Sciences International Journal of Biological, Life and Agricultural Sciences International Journal of Chemical, Materials and Biomolecular Sciences International Journal of Business, Human and Social Sciences International Journal of Earth, Energy and Environmental Sciences International Journal of Electrical, Electronic and. Graph Theory: Centrality and Power. Specify 'Importance' edge weights to use a weighted sum, rather than the simple sum of all successor/predecessor scores. • Select “Betweeness Centrality” in the list. Degree centrality just. This coefficient is a measure of the local cohesiveness that takes into account the importance of the clustered structure on the basis of the amount of traffic or. " " Conceptually the process involves: " " 1. After loading the adjacency matrix in R, we can type the following in R to get the degree, indegree, and outdegree measures for the simulated network. Essentially, taking endogenous participation into. preventviolentextremism. Subgraph centrality replaces the adjacency matrix with its trace. If the input graph does not contain weights, then WEIGHT and UNWEIGHT both give the same results (using 1. Scientific collaboration networks. Additionally, because a single centrality measure fails to capture the overall importance of a node in the railway network, we proposed a data-driven integrated measure based on the four centrality measures (degree, strength, betweenness, and closeness) to comprehensively quantify the importance of each node. The degree centrality measure C D of a node xin an undirected weighted graph (V;E;W) is given by the sum of the weights of the edges incident to node x, that is, C D(x. The indexes composing a new unique centrality measure for collaborative competency. Degree centrality is the number of edges incident with the given vertex. Conceptually the process involves: 1. Eigenvector Centrality. In our network, degree centrality measures the number of connections to other characters, while weighted degree centrality measures the number of interactions. computation for several different types of centrality. Degree Centrality. In conclusion new complex measures of the degree centrality are introduced including weighted ties possible for use of the analysis of co-authorship or citation networks. Weighted Degree Centrality: the number of interactions you participate in. The degree centrality of a node is the number other nodes that are directly connected to it via an edge. "The development of nodes' degree centrality is an essential element in the process of social network evolution. ca) Faculty of Business Administration, University of New Brunswick, NB Canada Fredericton. BADIOS also applies to weighted and/or directed networks. pagerank_centrality. The degree centrality considers only edge weight and degree of the node in the network. weighted degree centrality is 27 which is the sum of the directly connected links to node 1 (the. The simplest of centrality measure is degree. Skipping Eigenvector Centrality : Your network has no closed loops, so eigenvector centrality cannot be calculated. The weighted degree has to be computed before, go to Statistics >> Avg. For every vertex v 2V of a weighted graph G(V;E), the betweenness centrality C B(v) of v is de ned by C B(v) = X s6=v X t6=v;s ˙ st(v) ˙ st Adriana Iamnitchi K-Path Centrality: A New Centrality Measure in Social Networks 3 of 23. This coefficient is a measure of the local cohesiveness that takes into account the importance of the clustered structure on the basis of the amount of traffic or. This function can be used on several kinds of graphs to compute several node centrality statistics and the edge-betweenness centrality. Indeed, by considering solely the degree of a node we overlook that nodes with small degree may be crucial for connecting different regions of the network by acting as bridges. This is an interactive periodic table of centrality indices I gathered in the course of my PhD. In an unweighted graph the cells of A have a value of 0 if no edge exists between the two vertices and a value of 1 if two vertices are connected by and edge. , Skvoretz, J. After a quick time, you will have the computation result and you will be able to do Ranking >> Nodes >> Choose a Rank Parameter Degree >> Weighted Degree and click « Apply ». The ones with most citations (ie in-degree centrality) The ones with high eigenvector centrality values; But they don't take into account the fact that edges have weight. Betweenness only uses geodesic paths. In fact, betweeness centrality surfaces two women, Elizabeth Leavens and Mary Penington, whose significance had been obscured by the degree centrality metric. Jianxin Wang, from Central South University. After loading the adjacency matrix in R, we can type the following in R to get the degree, indegree, and outdegree measures for the simulated network. Gephi 2,930 views. Comparative Assessment of Centrality Indices and Implications on the Vulnerability of ISP Networks George Nomikos ∗Panagiotis Pantazopoulos ∗ Department of Informatics and Telecommunications National & Kapodistrian University of Athens, Greece Email: {gnomikos, ppantaz, ioannis}@di. Additionally, because a single centrality measure fails to capture the overall importance of a node in the railway network, we proposed a data-driven integrated measure based on the four centrality measures (degree, strength, betweenness, and closeness) to comprehensively quantify the importance of each node. - snap-stanford/snap. The number of shortest. Degree(v i) = deg(v i) jVj 1 (1) In a weighted graph, the degree. Degree centrality measures the importance of a vertex by the number of connections the vertex has if the network is unweighted (Freeman 1977), and by the aggregate of the weights of edges connected to the vertex if the network is weighted (Barrat et al. Betweenness centrality: Number shortest paths between all nodes in a network which pass through a given node, defines the betweenness centrality of that node. Both kinds (binary and weighted) will in this work be referred to as degree centrality. We start by propos- ing a generalization of degree centrality for weighted networks where the outcome is a combination of the number of ties and the tie weights. Alpha centrality replaces the adjacency matrix with its resolvent. Closeness Centrality (Double Length) Time Complexity O(N*E*d) [d = path-length diameter of graph] Space Requirement O(N) Computes the closeness centrality value of each node in the graph. , the number of ties that a node has). Weighted Graphs Data Structures & Algorithms 2 [email protected] ©2000-2009 McQuain Shortest Paths (SSAD) Given a weighted graph, and a designated node S, we would like to find a path of least total weight from S to each of the other vertices in the graph. Mixed-Mean centrality as a new measure of the importance of a node in a graph is intro-duced, based on the generalized degree centrality. We addressed the performance of centrality metrics methods such as betweenness, closeness, eigenvector, PageRank and weighted degree measurements, drawing comparisons between the experiments' results and the actual top 300 shares in the Australian Stock Market. degree_centrality(G_karate) c_degree = list(c_degree. , the weighted degree centrality (WDC), is developed to achieve the reliable prediction of essential proteins. Calculation of out-degree centrality which only counts outgoing edges from a vertex. Degree centrality. In this paper, we focus on a centrality measure based on the notion of k -cores which is a fundamental concept in graph theory [ 40 ] when it comes to. Among these top performers, which characters have more more interactions per connection? Which characters have fewer? Compared to degree centrality, Jaime, Arya and Stannis underperform. Here, In-Degree metric computes the number of incoming nodes of a vertex, and Out-Degree calculates the number of outgoing neighbours of a node. pagerank_centrality. - snap-stanford/snap. Min Li and Dr. Its natural generalization to a weighted graph is given by the strength of vertices de ned for a node ias [22, 6] si= X j2V(i) wij; (1). Gu Yijun and Xia Tian (2014) proposed a keyword extraction algorithm fusedof LDA and TextRank [2]. This function can be used on several kinds of graphs to compute several node centrality statistics and the edge-betweenness centrality. The new degree may be very different from the degree in the 2-mode original network: a projection add lots of edges (in particular when lots of nodes where. Then, in order to extend the closeness and betweenness centrality measures, we propose a gen-eralization of shortest distances for weighted network that takes into account. As the weighted outgoing degree of node v i increases. Semi-Asymmetric Parallel Graph Algorithms for NVRAMs. Then, in order to extend the closeness and between-ness centrality measures, we propose a generalization of shortest distances for weighted network that takes into account both the. Scientific collaboration networks. Degree centrality. nected to (her degree), but also on their centrality. A big data inspired preprocessing scheme for bandwidth use optimization in smart cities applications using Raspberry Pi Big Data: Learning, Analytics, and Applications, May 2019. Centrality definition, a central position or state: the centrality of the sun. In this paper we study node centrality for very large graphs, up to billions of nodes and edges. However, in a weighted network the degree centrality of a node is the sum of weights of all links that connect to the node (Newman 2004). 3748) developed a weighted counterpart of degree centrality, namely strength (S i). Bonacich showed that, for symmetric systems, an iterative estimation approach to solving this simultaneous equations problem would eventually converge to a single answer. If the normalized argument was TRUE, then the result was divided by this number. However, these generalizations have solely focused on tie weights, and not on the number of ties, which was the central component of the original measures. Wed, 30 Oct 2019 09:43:39. If the input graph does not contain weights, then WEIGHT and UNWEIGHT both give the same results (using 1. Let us define the eigenvector centrality of a node to be the weighted sum of the centralities of its neighbors: ceig i = X j2N(i) w ijc. scores function), but that function requires the user to specify the theoretical maximum of the centrality measure, and it's not obvious to me what that would be in this weighted example (I believe the built-in. In a directed network, a node may have a different number of outgoing and incoming ties, and therefore, the degree is split into out-degree (deg out) and in-degree (deg in), respectively. degree centrality and Blau’s measure of heterogeneity), or whether they also incorporate connections to others at longer distances in the network (e. Hi everyone. A self-loop counts as one incoming edge. Gephi 2,930 views. Journal of Transport Geography. Here’s the “Philosophy on Twitter & YouTube” Quarterly Update from Kelly Truelove of TrueSciPhi. Subgraph centrality replaces the adjacency matrix with its trace. The proposed measure trades off the node degree and global structure in a weighted network. Moreover, the weighted mean effects were of similar size in both directions, suggesting that the link between social relationships and self-esteem is truly reciprocal in all developmental stages across the life span, reflecting a positive feedback loop between the constructs. For degree centrality, we simply use the user adjacency matrix without caring for how trustworthy a connection is. Thus three new values have been created by the "Average Path Length" algorithm we ran. Compute betweenness centrality for edges. Gu Yijun and Xia Tian (2014) proposed a keyword extraction algorithm fusedof LDA and TextRank [2]. Centrality measures can be classified into four main groups (Jackson, 2010): a) degree centrality, that measures how much a node is connected to others (with strength centrality as a weighted version of degree centrality); b) closeness centrality, showing how easily a node. Here, In-Degree metric computes the number of incoming nodes of a vertex, and Out-Degree calculates the number of outgoing neighbours of a node. A key weakness of the degree centrality metric is that the metric can take only integer values and. Based on social network, the paper analyzes knowledge body degree centrality, betweenness centrality and closeness. Calculate the new Degree centrality of the nodes by clicking on “Avg. Weighted Degree" (Statistics panel). The number in the upper right corner is the year the paper was published. 3 Why network is important? Can you name a case where you only care about an object but not its relations with other subjects? Reflected by relational subjects Decided by relational subjects. * * @return the number of vertices in this edge-weighted graph. The higher the cv, the shorter the average distance from v to other vertices, and v is more important by this measure. The degree centrality considers only edge weight and degree of the node in the network. The centrality scores of these 32 scientists based on three standard centrality methods for weighted networks (degree, closeness and betweenness) and the new Laplacian centrality method are given in Table 2, where the vertices are sorted by their Laplacian centrality scores. It is defined as the drop in the Laplacian energy (i. Additionally, because a single centrality measure fails to capture the overall importance of a node in the railway network, we proposed a data-driven integrated measure based on the four centrality measures (degree, strength, betweenness, and closeness) to comprehensively quantify the importance of each node. the university of chicago approximating geodesic distance and graph centrality on shared nothing architectures a dissertation submitted to the faculty of the division. Using Centrality Measures to Identify Key Members of an Innovation Collaboration Network 5. The total weight of a path is the sum of the weights of its edges. Degree centrality 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 Closenesscentrality. By an axiomatic analysis, we show that the Attachment Centrality is closely re-lated to the Degree Centrality in weighted graphs. In this paper we present betweenness centrality of some important classes of graphs. This paper proposes a new node centrality measurement index (c-index) and its derivative indexes (iterative c-index and cg-index) to measure the collaboration competence of a node in a weighted network. It has a generic function centrality_auto() which returns, depending on the network, the following indices: degree strength (weighted degree) betweenness closeness The package also contains the function centrality(), which calculates a non-linear combination of unweighted and weighted indices using a tuning parameter $$\alpha$$ (See Opsahl et al. Essentially, taking endogenous participation into. A few network measures have been proposed for weighted networks, including three common measures of node centrality: degree, closeness, and betweenness. Node Centrality in Weighted Networks: Generalizing Degree and Shortest Paths. Here, In-Degree metric computes the number of incoming nodes of a vertex, and Out-Degree calculates the number of outgoing neighbours of a node. THE MEASURE I have proposed (Bonacich 1972a, 1972b) a measure of centrality (in this paper, I will call it "e") in which a unit's centrality is its summed connec- tions to others, weighted by their centralities. Contribute to igraph/igraph development by creating an account on GitHub. ca) Faculty of Business Administration, University of New Brunswick, NB Canada Fredericton. topological measure of centrality is given by the degree: more connected nodes are more central. The weighted leverage measure is compared against other measures namely degree, betweenness, eigenvector and leverage centrality in the perspective of proper identification of network hubs. In order to know the standardized score, you need to divide each score by n-1 (n = the number of nodes). Whoops! There was a problem previewing Everett & Borgatti 2012 - Categorical attribute based centrality. The centrality scores of these 32 scientists based on three standard centrality methods for weighted networks (degree, closeness and betweenness) and the new Laplacian centrality method are given in Table 2, where the vertices are sorted by their Laplacian centrality scores. Then, in order to extend the closeness and between-ness centrality measures, we propose a generalization of shortest distances for weighted network that takes into account both the. Among these top performers, which characters have more more interactions per connection? Which characters have fewer? Compared to degree centrality, Jaime, Arya and Stannis underperform. The degree centrality considers only edge weight and degree of the node in the network. sum of the weights of the links between node 1 and nodes 2-7). Betweenness Centrality The betweenness centrality is a measure of a node's centrality in a network. Preferred way to connect peoples. Degree centrality (DC) is the sum of the weight of all edges connecting to a node. Weighted node degree centrality for hypergraphs. Philosophy on Twitter & YouTube – Quarterly Update (Q2 2019) by Kelly Truelove TrueSciPhi. centrality-in-networks-with-disconnected-components/ precomp. For multigraphs or graphs with self loops the maximum degree might be higher than n-1 and values of degree centrality greater than 1 are possible. We will de ne and compare four centrality measures: degree centrality (based on degree) closeness centrality (based on average distances) betweeness centrality (based on geodesics) eigenvector centrality (recursive: similar to page rank methods) 3/36. PACS numbers: 89. In weighted networks the degree centrality is calculated as the sum of weights assigned to the node’s direct connections and represents the node strength (Strength Centrality—SC). , Agneessens, F. Betweenness centrality of an edge is the sum of the fraction of all-pairs shortest paths that pass through where is the set of nodes,sigma(s, t) is the number of shortest -paths, and is the number of those paths passing through edge [2]. Jianxin Wang, from Central South University. A systematic and extensible pathway enrichment method in which nodes are weighted by network centrality was proposed. For example, in a network where nodes are people and you are tracking the flow of a virus, the degree centrality gives some idea of the magnitude of the risk of spreading the virus. This brings up the dialogue for calculating the various centrality and eccentricity scores, with a brief explanation of each. Degree centrality of a node refers to the number of edges attached to the node. The size of a graph is equal to the number of vertices of the graph. However, measuring centrality in billion-scale graphs poses several challenges. For example, in a telecommunications network, a node with higher betweenness centrality would have more control over the network, because more information will pass through that node. https://www. Then, in order to extend the closeness and between-ness centrality measures, we propose a generalization of shortest distances for weighted network that takes into account both the. add (e);}}} /** * Returns the number of vertices in this edge-weighted graph. , the number of ties that a node has). subgraph_centrality. Both co-authorship relations and citations are well quantified data (weighted ties). Degree centrality. The mth-order degree mass of a node is the sum of the weighted degree of the node and its neighbors no further than m hops away. Let’s use degree centrality (connections) as a benchmark for importance. affinity index algorithm analysis antipaedo attack bipartite blog network blogs capitalisme social Cascade centrality clustering communities community detection community structure complex network complex networks complex systems compression connected graphs data mining debian degree distribution degree peeling diameter diffusion diffusion. Summing this weight across all INTERACTS interactions for that character gives us their weighted degree centrality|Social Manager may even let you know what occurred in the shape of occasions. Stability and Continuity of Centrality Measures in Weighted Graphs Santiago Segarra and Alejandro Ribeiro Abstract—This paper presents a formal definition of stability for node centrality measures in weighted graphs. Basic network analysis 4. An example of a local centrality measure is the degree centrality, which counts the number of links held by each node and points at individuals who can quickly connect with the wider network. Degree has generally been extended to the sum of weights when analysing weighted networks and labelled node strength, so the weighted degree and the weighted in- and out-degree was calculated (Barrat et al. As the name implies,. In a directed network, a node may have a different number of outgoing and incoming ties, and therefore, degree is split into out-degree and in-degree, respectively. trality, betweenness centrality, degree centrality, and PageRank) for authors in this network. See Also ----- degree_centrality, in_degree_centrality Notes ----- The degree centrality values are normalized by dividing by the maximum possible degree in a simple graph n-1 where n is the number of nodes in G. The closeness centrality of a vertex is defined by the inverse of the average length of the shortest paths to/from all the other vertices in the graph: 1/sum( d(v,i), i != v) If there is no (directed) path between vertex \code{v} and \code{i} then the total number of vertices is used in the formula instead of the path length. Let us define the eigenvector centrality of a node to be the weighted sum of the centralities of its neighbors: ceig i = X j2N(i) w ijc. 3 Eigenvector centrality. Centrality in Social Networks Degree If we want to measure the degree to which the graph as a whole is centralized, we look at the dispersion of centrality: Simple: variance of the individual centrality scores. Stanford Network Analysis Platform (SNAP) is a general purpose network analysis and graph mining library. a centrality measure that weights the betweenness centrality 𝐵𝐶𝑘instead of the degree centrality 𝐷𝐶𝑘. Current-Flow Betweenness¶. In terms of the interbank network, this indicates the number of other banks that a given bank has lending and borrowing relationship with. Recasting these centrality metrics into this new. org features a variety of lists and statistics regarding philosophy communities on social media including Twitter, podcasts, and YouTube. I am now studying centrality measures of my weighted network using statnet, but the centrality measures I obtain are as if statnet did not take into account the values of my edges. Degree Centrality John McCulloch. • Select “Betweeness Centrality” in the list. - Betweeness Centrality - Closeness. 2004, Newman 2001, Opsahl et al. tachment Centrality to node-weighted and edge-weighted graphs. | 2019-11-18T03:29:59 | {
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http://konferencja.pfsrm.pl/7hu3cg/conditional-probability-problems-02b1e4 | b. This problem has nothing to do with the Then, $$1/2 = P(A) = \dfrac{1}{2} (1/4 + 3/4) > \dfrac{1}{2} (1/2 + 1/4) = P(B) = 3/8$$. Conditional Probability Definition We use a simple example to explain conditional probabilities. Math 212a October 28,2014, Due Thursday, Nov. 6 . It is cleaner if we divide $W$ into two parts depending $$P(L|GG)=\alpha (1-\alpha)+ (1-\alpha) \alpha +\alpha^2=2 \alpha-\alpha^2.$$ $P(T \geq 2)=e^{-\frac{2}{5}}=0.6703$. Data may be tabulated as follows: $$P(E_1) = 0.65$$, $$P(E_2) = 0.30$$ and $$P(E_3) = 0.05$$. The automatic test procedure has probability 0.05 of giving a false positive indication and probability 0.02 of giving a false negative. Say a Professor is interested in the probability that over 300 students take his class next semester. ($P(L|BG)=P(L|GB)=\alpha$, $P(L|GG)=2 \alpha-\alpha^2$), thus in this case the conditional What is the probability that both children are girls? A table of sums shows $$P(A_6S_k) = 1/36$$ and $$P(S_k) = 6/36, 5/36, 4/36, 3/36, 2/36, 1/36$$ for $$k = 7$$ through 12, respectively. first time that two consecutive heads ($HH$) or two consecutive tails ($TT$) are observed. What is the probability that three of those selected are women? Two standard dice with 6 sides are thrown and the faces are recorded. $$P(A \cup B\cup C)=a+b+c-ac-bc=\frac{11}{12}.$$ two previous problems. $GG$ from $\frac{1}{3}$ to about $\frac{1}{2}$. Compare your Example 1 a) A fair die is rolled, what is the probability that a face with "1", "2" or "3" dots is rolled? Let’s get to it! Compound probability is when the problem statement asks for the likelihood of the occurrence of more than one outcome. $$E_1$$= event did not complete college education; $$E_2$$= event of completion of bachelor's degree; $$E_3$$= event of completion of graduate or professional degree program. (ii) What is the probability that exactly one of them will solve it? Let $$B_k$$ be the event of a black ball on the $$k$$th draw and $$R_k$$ be the event of a red ball on the $$k$$th draw. visualize the events in this problem. Let $$T$$ = event test indicates defective, $$D$$ = event initially defective, and $$G =$$ event unit purchased is good. The probability that it's not raining and there is heavy traffic and I am not late can Here you can assume that if a child is a girl, her name will be Lilia with probability $\alpha \ll 1$ Hence $$P(A_6|S_k) = 1/6, 1/5. the probability that both children are girls, given that the family has at least one daughter named Lilia. What is the (conditional) probability that he or she will make 25,000 or more? Find the total probability that a person's income category is at least as high as his or her educational level. Previous experience indicates that 20 percent of those who do not favor the policy say that they do, out of fear of reprisal. We can calculate the probabilities of each outcome in the sample space by multiplying Search. Introduction to Video: Conditional Probability; 00:00:31 – Overview of Conditional Probability, Multiplication Rule, Independence and Dependence; Exclusive Content for Members Only A quality control group is designing an automatic test procedure for compact disk players coming from a production line. Now, let G_r be the event that a randomly chosen child is a girl. This is why we need conditional probability. A conditional probability Pr(B | A) is called an a posteriori if event B precedes event A in time. We can find P(R|L) using P(R|L)=\frac{P(R \cap L)}{P(L)}. Events can be "Independent", meaning each event is not affected by any other events. only one girl is \frac{1}{2}. In this figure, each leaf in the tree corresponds to a single outcome in the sample space. \frac{1}{3}, = \frac{P(L|GG)P(GG)}{P(L|GG)P(GG)+P(L|GB)P(GB)+P(L|BG)P(BG)+P(L|BB)P(BB)}, = \frac{(2 \alpha-\alpha^2)\frac{1}{4}}{(2 \alpha-\alpha^2)\frac{1}{4}+ \alpha \frac{1}{4}+ \alpha \frac{1}{4}+0.\frac{1}{4}}. Construct an example to show that in general \(P(A|B) + P(A|B^c) \ne 1$$. What is the probability that an employee picked at random really does favor the company policy? There are 8 problems in all. Intuition is useful, but at the end, we must use laws of probability to solve We seek several goals by including such problems. A family with two girls is more Suppose Ai is the event the sum of the two digits on a card is $$i$$, $$0 \le i \le 18$$, and $$B_j$$ is the event the product of the two digits is $$j$$. $$P(D) = 0.02$$, $$P(T^c|D) = 0.02$$, $$P(T|D^c) = 0.05$$, $$P(GT^c) = 0$$, $$P(D|G) = \dfrac{P(GD)}{P(G)}$$, $$P(GD) = P(GTD) = P(D) P(T|D) P(G|TD)$$, $$P(G) = P(GT) = P(GDT) + P(GD^c T) = P(D) P(T|D) P(G|TD) + P(D^c) P(T|D^c) P(G|TD^c)$$, $$P(D|G) = \dfrac{0.02 \cdot 0.98 \cdot 0.90}{0.02 \cdot 0.98 \cdot 0.90 + 0.98 \cdot 0.05 \cdot 1.00} = \dfrac{441}{1666}$$. What is the(conditional) probability that she is a female who lives on campus? This is a term that, like many math terms, will not explicitly appear on the GMAT, and the notation I will show, standard in many probability textbooks, will not appear on the GMAT. a tree diagram. is $0.25$. that a purchased product does not break down in the first two years. The process is repeated. Let $$B=$$ the event the collector buys, and $$G=$$ the event the painting is original. (2) If A and B are two events such that P (A U B) = 0.7, P (A n B) = 0.2, and P (B) = 0.5, then show that A and B are independent. These can be tackled using tools like Bayes' Theorem, the principle of inclusion and exclusion, and the notion of independence. We’ll say that the probability that it rains, P(A), is 0.40, the probability that the Little League game is cancelled, P(B), is 0.25. Here is another variation of the family-with-two-children $$S_1$$= event annual income is less than $25,000; $$S_2$$= event annual income is between$25,000 and $100,000; $$S_3$$= event annual income is greater than$100,000. The probability that a randomly chosen child Answer Let $$B=$$ the event the collector buys, and $$G=$$ the event the painting is original. Suppose we know that. defined as the amount of time (in years) the product works properly until it breaks down, satisfies $$A_i B_0$$ is the event that the card with numbers $$0i$$ is drawn. Let $$A=$$ event she lives to seventy and $$B=$$ event she lives to eighty. the probabilities on the edges of the tree that lead to the corresponding outcome. This chapter explores various approaches to conditional probability, canvassing their associated mathematical and philosophical problems and numerous applications. The manual states that the lifetime $T$ of the product, Let us write the formula for conditional probability in the following format $$\hspace{100pt} P(A \cap B)=P(A)P(B|A)=P(B)P(A|B) \hspace{100pt} (1.5)$$ This format is particularly useful in situations when we know the conditional probability, but we are interested in the probability of the intersection. one third in this case. Solution. A shipment of 1000 electronic units is received. $$P(H_i|D) = \dfrac{P(D|H_i) P(H_i)}{\sum P(D|H_i) P(H_j)} = i/15$$, $$1 \le i \le 5$$. By subtracting the third equation from the sum of the first and second equations, we likely to name at least one of them Lilia than a family who has only one girl In a certain population, the probability a woman lives to at least seventy years is 0.70 and is 0.55 that she will live to at least eighty years. A student is selected at random. A Bayes' problem can be set up so it appears to be just another conditional probability. $$0.91854\times0.91854\times0.91854\times0.91854\times0.91854\times 1\approx0.654$$ $=\frac{e^{-\frac{2}{5}}-e^{-\frac{3}{5}}}{e^{-\frac{2}{5}}}$, $=\sum_{i=1}^{M} P(A|C_i)P(B|C_i)P(C_i) \hspace{10pt}$, $\textrm{ ($A$and$B$are conditionally independent)}$, $\textrm{ ($B$is independent of all$C_i$'s)}$, $=\frac{2}{3} \cdot \frac{1}{4} \cdot \frac{3}{4}$, $= P(R,T,L)+P(R,T^c,L)+P(R^c,T,L)+P(R^c,T^c,L)$, $=\frac{1}{12}+\frac{1}{24}+\frac{1}{24}+\frac{1}{16}$, $= \frac{1}{2}. There are $$6 \times 5$$ ways to choose all different. P(A or B) is the probability of the occurrence of atleast one of the events. We are dealing with 52 cards and we know that there are 26 red cards and 26 black cards. In probability theory, conditional probability is a measure of the probability of an event occurring, given that another event (by assumption, presumption, assertion or evidence) has already occurred. Introduction. Legal. problem [1] [7]. The concept is one of the quintessential concepts in probability theory Total Probability Rule The Total Probability Rule (also known as the law of total probability) is a fundamental rule in statistics relating to conditional and marginal. Conditional probability - dice problem. For example if the outcome is$HTH\underline{TT}$, I In part (b) of Example 1.18, What is the probability of three heads,$HHH$? Probability theory - Probability theory - The birthday problem: An entertaining example is to determine the probability that in a randomly selected group of n people at least two have the same birthday. This video is a preview of how students will navigate my digital escape room on conditional probability word problems. most people. He has probability 0.10 of buying a fake for an original but never rejects an original as a fake, What is the (conditional) probability the painting he purchases is an original? Here we have four possibilities,$GG=(\textrm{girl, girl}), GB, BG, BB$, So that we can solve various probability and conditional probability problems. In fact, we are using What is the (conditional) probability that at least one turns up six, given that the sum is $$k$$, for each $$k$$ from two through 12? Data on incomes and salary ranges for a certain population are analyzed as follows. They have respectively one, two, three, four, and five defective units. He has probability 0.10 of buying a fake for an original but never rejects an original as a fake, What is the (conditional) probability the painting he purchases is an original?$HH$is observed and lose if$TT$is observed. This problem is an example of conditional probability. Let X be the lower of the two scores and Y be the larger of the scores. He is male and lives on campus. and repeated trials are independent. Let$C_1$be the event that you choose a regular coin, and let$C_2$be the event In our notation, $$P(A|B)$$ means “the probability of $$A$$ given that $$B$$ occurred.” Let’s consider an example. Video transcript - [Instructor] James is interested in weather conditions and whether the downtown train he sometimes takes runs on time. CONDITIONAL PROBABILITY PROBLEMS WITH SOLUTIONS. Here are some other examples of a posteriori probabilities: • The probability it was cloudy this morning, given that it rained in the afternoon. Conditional probability using two-way tables. Suppose we assign a distribution function to a sample space and then learn that an event Ehas occurred. on the result of the first coin toss, Experience shows that 93 percent of the units with this defect exhibit a certain behavioral characteristic, while only two percent of the units which do not have this defect exhibit that characteristic. P(A or B) = P(A) + P(B) – P(A and B) where A and B are any two events. Is the converse true? Data are. We use Bayes' rule, Let$W$be the event that I win. $$W =\{HH, HTHH, HTHTHH,\cdots \} \cup \{THH, THTHH, THTHTHH,\cdots \}.$$ We are interested in$P(A|B)$. Let$R$be the event that it's rainy,$T$be the event that there is heavy traffic, Given that I arrived late at work, what is the probability that it rained that day? A ball is drawn on an equally likely basis from among those in the urn, then replaced along with $$c$$ additional balls of the same color. What is the probability that you observe exactly one heads? that you choose the two-headed coin. Second, after obtaining counterintuitive results, you are encouraged to think deeply By the description of the problem, P(R jB 1) = 0:1, for example. Analysis: This problem describes a conditional probability since it asks us to find the probability that the second test was passed given that the first test was passed. It is determined that the defendent is left handed. Okay, another family-with-two-children problem. For three events$A$,$B$, and$C$, we know that. Solution : Let "A", "B" and "C" be the events of solving problems by each students respectively. $$P(L|BB)=0,$$ probability problems. $$P(A|B) > P(A)$$ iff $$P(AB) > P(A) P(B)$$ iff $$P(AB^c) < P(A) P(B^c)$$ iff $$P(A|B^c) < P(A)$$, b. It is reasonable to assume that all who favor say so. There are $$n$$ balls on the first choice, $$n + c$$ balls on the second choice, etc. Tree diagrams and conditional probability. If it's rainy and there is heavy traffic, I arrive The game ends the Thus, the sample space reduces to three $$P(G_r|BB)=0.$$ CONDITIONAL PROBABILITY WORD PROBLEMS WORKSHEET (1) Can two events be mutually exclusive and independent simultaneously? traffic with probability$\frac{1}{4}$. $$P(A) = P(A|C) P(C) + P(A|C^c) P(C^c) > P(B|C) P(C) + P(B|C^c) P(C^c) = P(B)$$. P(B|A) is also called the "Conditional Probability" of B given A. Result of testimony: $$P(L|G)/P(L|G^c) = 6$$. Conditional probability is calculated by … $$P(H|C_2)=1.$$. Probability Probability Conditional Probability 19 / 33 Conditional Probability Example Example De ne events B 1 and B 2 to mean that Bucket 1 or 2 was selected and let events R, W, and B indicate if the color of the ball is red, white, or black. Thus, intuitively, the conditional probability of the outcome the sample space. We assume that the coin tosses are independent. That is, if $$D$$ is the event a unit tested is defective, and $$T$$ is the event that it tests satisfactory, then $$P(T|D) = 0.05$$ and $$P(T^c|D^c) = 0.02$$. A collector buys a painting. Thus, it is useful to draw a tree diagram. Boxes of random access memory chips have 100 units per box selected from group... Be just another conditional probability – Lesson & Examples ( Video ) 1 hr 43.! Devised to detect this disease two events be mutually exclusive and independent simultaneously 100 units per.. 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https://math.stackexchange.com/questions/2535325/periods-of-periodic-solutions-of-the-hamiltonian-system-dotx-y-doty | # Periods of periodic solutions of the (Hamiltonian) system $\dot{x}=y$, $\dot{y}=-x-x^2$
I'm preparing for a scholarship examination (no solutions available) and in older tests I'm coming across problems like the following.
Consider the (Hamiltonian) system $$\begin{cases}\dot{x}=y \\ \dot{y}=-x-x^2 \end{cases}$$ (a) Find the initial conditions under which the motion is periodic and estimate the set of possible periods;
(b) Prove that there exist periodic orbits for which the average value of the position is not $0$ (if $x(t)$ is a periodic orbit of period $T$, the average value is defined as $T^{-1}\int_0^Tx(t)dt$).
The approach I've been taught in my undergraduate ODE course is to draw a phase portrait given by the level sets of the Hamiltonian, which gives me full information on the support of the solutions. For instance, I can find the region of the plane which encloses all periodic solutions. However, I've never been taught how to study the graphs of solutions, or how to obtain quantitative results on the periods of periodic solutions. Can anyone give me an idea on how this could be accomplished?
EDIT: Maybe an idea would be to study the linearized system about the critical point $(0,0)$, which is $$\begin{cases}\dot{x}=y \\ \dot{y}=-x\end{cases}$$ in this case, the solutions can be explicitly calculated from $x(t)=a\cos t+b\sin t$, $a,b\in \mathbb{R}$, so their period is $2\pi$. Intuitively, I would say that the nonlinear system would also have periods $2\pi$ near the origin, but this still doesn't give me any estimate on the periods on the whole region.
• Look up Floquet theory – user392395 Nov 24 '17 at 15:15
• I just looked it up but it seems to concern linear differential systems only? – Lorenzo Quarisa Nov 24 '17 at 15:26
• en.wikipedia.org/wiki/Floquet_theory – user392395 Nov 24 '17 at 15:29
• @Fightclub1995 "Floquet theory is a branch of the theory of ordinary differential equations relating to the class of solutions to periodic linear differential equations..." – Did Nov 25 '17 at 1:04
You might already know that $H(x,y)=x^2+y^2+\frac23x^3$ is an invariant of the dynamics. For every $h$, let $S_h$ denote the set of equation $H(x,y)=h$.
• If $h<0$ or $h>\frac13$, $S_h$ is a connected simple unbounded curve.
• If $0\leqslant h<\frac13$, $S_h$ is the disjoint union of a connected simple unbounded curve, included in the halfplane $x<-1$, and a connected simple bounded curve, around $(0,0)$.
• Finally, $S_{1/3}$ is a connected unbounded curve with a double point at $(-1,0)$.
Thus, the periodic solutions correspond exactly to the bounded components $C_h$ of every $S_h$ with $0<h<\frac13$. The part included in $x\geqslant-1$ of $S_{1/3}$ is a loop, which encloses exactly every periodic solution and the bounded component of $S_0$, which is the fixed point $(0,0)$.
Let $W(x)=H(x,0)=x^2+\frac23x^3$. For every fixed $0<h<\frac13$, on $C_h\cap\{y>0\}$, $$dt=\frac{dx}{\sqrt{h-W(x)}}$$ hence the period is $$T=2\int_u^v\frac{d\xi}{\sqrt{h-W(\xi)}}$$ where $-1\leqslant u<0<v\leqslant\frac12$ are the intersections of $C_h$ with the axis $y=0$, that is, the roots in $[-1,\frac12]$ of the equation $W(x)=h$.
The polynomial $W(x)=x^2+\frac23x^3$:
When $h\to\frac13$, $C_h$ passes near the fixed point $(-1,0)$ hence $T\to+\infty$. When $h\to0$, $C_h$ becomes the fixed point $(0,0)$ hence, approximating $h-W(x)$ by $h-x^2$, one gets that $T\to2\pi$. By continuity, every number in $(2\pi,+\infty)$ is a period.
Finally, the average value of the abscissa is $$\bar x=\frac1{T}\int_0^Tx(t)dt$$ hence, for every $0<h<\frac13$, $$\bar x=\frac1{T}\int_0^T(-y'(t)-x^2(t))dt=-\frac1{T}\int_0^Tx^2(t)dt<0$$
• Thanks for the answer. I think I got it in an intuitive way too. If the orbit passes near a fixed point, lying outside the region of the plane enclosed by the orbit, then the orbit slows down but the length doesn't vanish, so the period blows up. Instead, for the orbits near a fixed point which they are enclosing, the period tends to be the same as in the linearized system. Then I conclude by continuity (of the Hamiltonian, I guess?) – Lorenzo Quarisa Nov 26 '17 at 9:24 | 2019-08-23T22:57:46 | {
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https://math.stackexchange.com/questions/1199114/why-is-a5-equiv-a-pmod-5-for-any-positive-integer | # Why is $a^{5} \equiv a\pmod 5$ for any positive integer?
Why is $a^{5} \equiv a\pmod 5$ for any positive integer?
I feel like it should be obvious, but I just can't see it. Any help appreciated.
Edit: without Fermat's theorem.
• Isn't this a direct application of Fermat's Little Theorem? Mar 21, 2015 at 1:23
• Fermat's little theorem: en.wikipedia.org/wiki/Fermat%27s_little_theorem Mar 21, 2015 at 1:24
• @Deepak Which is usually a hint that the OP doesn't know the theorem. Mar 21, 2015 at 1:25
• possible duplicate of Purpose of Fermat's Little Theorem Mar 21, 2015 at 1:26
• @Deepak: Thomas got it right, I'd never heard of the theorem before.
– Bob
Mar 21, 2015 at 1:28
OK, without using Fermat's Little Theorem (a far more general and elegant result), here's another easy workaround.
Any integer $a$ can be exactly one of $0, 1, 2, -2, -1 \pmod 5$.
Take the fifth powers of each of those and see them reduce back to the original residue in each case.
One way to prove this is to prove it by induction. If $n^5\equiv n\pmod 5$ show that $(n+1)^5\equiv n+1\pmod 5$.
Note that $$(n+1)^5=n^5+5n^4+10n^3+10n^2+5n+1\equiv n^5+1\equiv n+1\pmod5$$
The general theorem people have mentioned in comments, Fermat's Little Theorem, states that if $p$ is prime and $a$ is any number: $a^p\equiv a\pmod p$.
• Huh. I didn't know you can prove the theorem like that (by induction). That's neat. Also pretty obvious in hindsight. But still neat. Mar 21, 2015 at 1:34
• Only works for positive $n$. Can prove for negative in one step, though, from the positives, since $5$ is odd. :) Mar 21, 2015 at 1:35
• This is nice, you just have to know that $\binom{p}{k}$ is divisible by $p$ for all $0<k<p$. Mar 21, 2015 at 1:41
• I was trying to keep it absolutely minimal to prove just for the case $5$, but yes, the general Fermat can be proved that way, @TravisJ. My favorite way to prove FLT is combinatorial - counting the number of ways of painting $p$ beads in a circle with $n$ colors. There are $n$ solid necklaces, and then the rest of the colorings can be grouped in sets of $p$ by rotation. Mar 21, 2015 at 1:44
• @Bob Note that all multiples of $5$ immediately reduce to $0$ modulo $5$. Mar 21, 2015 at 2:17
Consider $$a(a-1)(a-2)(a-3)(a-4)=a(a^4-10a^3+35a^2-50a+24).$$ Taken mod $5$ this becomes $$a(a^4-1).$$and so $a^5-a \equiv 0 \mod 5$, or $a^5 \equiv a \mod 5$ as required.
Note $\ a(a^4\!-1\!)\, =\, a(a^2-1)\overbrace{(a^2+1)}^{\Large \color{#0a0}{a^2-4}\,+\,\color{#c00}5} = \!\!\!\underbrace{a(a^2-1)(\color{#0a0}{a^2-4})}_{\large\color{blue}{ (a-2)(a-1)a(a+1)(a+2)}}\!\!\!\! + \color{#c00}5\,a(a^2-1)$
$\color{#c00}5\,$ divides both summands, the first because $\,\color{#c00}5\,$ divides one of $\,\rm\color{#c00}5\,\ \color{blue}{\rm consecutive\ integers}$.
As others said, it's Fermat's little theorem. One way to verify it in this particular case is to note that $$a^5 - a = a (a - 1)(a+1)(a^2 + 1)$$ If $a \equiv 0, 1$ or $4 \mod 5$, one of the first three factors is $0 \mod 5$. The other two possibilities are $a \equiv 2$ or $3 \mod 5$, in which case $a^2 + 1 \equiv 4 + 1 \equiv 0 \mod 5$. So in each case, $a^5 \equiv a \mod 5$.
A more general solution is the following:
Consider the multiplicative group $\mathbb{Z}_{p}^{\ast}$ of non-zero elements $\mod p$. This group has $p-1$ elements. Then for any element in the group, there is a cyclic subgroup $\langle g\rangle$. The order of this group is the order of $g$ which divides $p-1$, say the order is $k$ and $ak=p-1$. This means that $g^{p-1}=g^{ak}=(g^{k})^{a}\equiv 1^{a} \equiv 1 \mod p$. Multiplying again by $g$ you have that $g^{p}\equiv g \mod p$. That works for all the non-zero elements. It is obvious for the 0.
• This is essentially a proof of Fermat's little theorem. It works for any prime $p$ (of which 5 is one example). Mar 21, 2015 at 1:38 | 2023-02-04T08:37:45 | {
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https://www.physicsforums.com/threads/probability-problem-from-pokemon-cards.600347/ | # Probability problem from Pokemon cards
1. Apr 25, 2012
### Bipolarity
A long time ago when I played with Pokemon cards, I remember a Geodude card saying "Flip a coin until you get tails. This attack does 20 damage times the number of heads."
What would be the probability that the attack does more than 60 damage?
What would be the average damage of the attack if the attack were repeated indefinitely and its damage measured for each trial.
The problem popped into my mind as I am reading on probability, but any thoughts on how to solve this problem? I think of geometric probability distributions, but can't quite establish a solution.
BiP
2. Apr 26, 2012
### tiny-tim
Hi Bipolarity!
It's probably one of those ∑ xn = 1/(1-x) manoeuvres.
Show us how far you've got.
3. Apr 26, 2012
### Bipolarity
I understand that it is a geometric sum of some sort, but I'm not so clear about the connection between geometric progressions and probability.
Would it be correct in saying that the probability that the attack does more than 60 damage is sum of the probabilities that the attack does damages corresponding to 80,100,120,140... ad infinitum?
That I think I get, but beyond that, how would you calculate the individual probabilities?
And I have no clue how to calculate the average damage. I assume you would take each value, multiply it by its probability to give it a weight, and sum up all such weights, but I am used to doing this in continuous variables using calculus. I don't know how you find infinite sums for these types of problems.
BiP
4. Apr 26, 2012
### tiny-tim
yes, but easier is to say that it's 1 minus the probability of 20 40 or 60
P(20k) = P(k-1 tails then 1 heads)
Yes, ∑ 20kP(20k) …
try it and see
5. Apr 26, 2012
### HallsofIvy
Staff Emeritus
It will do 0 damage if you flip tails on the first flip- the probability of that is 1/2.
It will do 20 damage if you flip tails and then heads- the probability of that is (1/2)(1/2)= 1/4.
It will do 40 damage if you flip tails twice and then heads- the probability of that (1/2)2(1/2)= 1/8.
It will do 60 damage if you flip tails three times and then heads- the probability of that is (1/2)3(1/2)= 1/16.
Now, what is the probability it will do damage of "60 or less"? What is the probability it will do damage of "more than 60"? Do you see what that has to do with a geometric series?
The damage done when rolling n consecutive tails and then a head is, of course, 20n. The probability of rolling n consecutive tails and then a head is, as above, (1/2)n+1. So the average damage (the expected value) is the sum $\sum_{n=1}^\infty n (1/2)^{n+ 1}$. We can factor $1/4= (1/2)^2$ out of that to get $(1/4)\sum_{n= 1}^\infty n(1/2)^{n-1}$
The reason I do that is that $n x^{n-1}$ is the derivative of $x^n$. Further, power series are "term by term" differentiable so that our sum is 1/4 times the derivative of the $\sum x^n= \frac{1}{1-x}$, evaluated at x= 1/2.
Last edited: Apr 26, 2012
6. Apr 26, 2012
### Bipolarity
Thank you all. I understand it now.
BiP | 2017-08-20T14:36:00 | {
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https://www.physicsforums.com/threads/antiderivatives-and-the-fundamental-theorem.821411/ | # Antiderivatives and the fundamental theorem
1. Jun 30, 2015
I know that according to the first fundamental theorem of calculus:
$$\frac{d}{dx} \int_a^x f(t) dt = f(x)$$
I also know that if $F$ is an antiderivative of $f$, then the most general antiderivative is obtained by adding a constant.
My question is, can every single antiderivative of $f$ be expressed as:
$$\int_{a_n}^x f(t) dt = F_n (x)$$
where $a_n$ is some constant (every $a_n$ generates a different antiderivative)? Or is it not possible in some cases?
In other words, can every antiderivative of a function be expressed as a definite integral (one term only)?
2. Jun 30, 2015
### RUber
The antiderivative is a function F(x) such that F'(x) = f(x). The most basic form might be given with $a_{null}$ in the null space of F(x), such that $F(a_{null}) = 0$, and $F(x) = \int_{a_{null}}^x f(t) dt$ so the constant c = 0.
Any constant at the bottom of the integral will correspond to a constant in the evaluation $F(x)-F(a) = F(x) + c$ and the derivative of such a function will always be f(x).
What sort of exceptions are you looking for? There are some requirements on f so that you can even begin to find an antiderivative...are you assuming that those basic conditions are met?
3. Jun 30, 2015
$c$ can be any real number, however, $F(a)$ (as you described it) might belong in some restricted interval. So this means that not every antiderivative can be expressed as a single definite integral, right?
4. Jun 30, 2015
### verty
Hint: Integrate the zero function. What are the antiderivatives?
5. Jun 30, 2015
### mathwonk
in the first place, the fundamental theorem of calculus does not say what you wrote. that statement has no modifiers, i.e. no hypotheses. rather it says that IF f is a continuous function on the interval [a,x]. it may seem pedantic, but theorems have two parts, and the hypothesis is the most important part in many ways.
6. Jun 30, 2015
### pwsnafu
The answer is no. mathwonk brought up the point
The key is that the domain needs to be connected for this to work. If you look at domains which are disconnected then it fails.
Consider $f(x) = -\frac{1}{x^2}$. Clearly, $F(x) = \frac{1}{x}+c$ right? But what about
$g(x) = \frac{1}{x} + 1$ when $x >0$ and $g(x) = \frac{1}{x}-1$ when $x<0$.
We have $g'(x) = f(x)$ but now what is c? The "arbitrary constant" changes as x crosses over zero. You can't write this as a single definite integral. Your best effort would a piecewise function where each is a definite integral, or the "constant" contains the Heaviside function.
Last edited: Jun 30, 2015
7. Jul 1, 2015
### lavinia
If your function, f(x), is continuous on an open interval then
$$\int_{a}^x f(t) dt = F (x)$$ is an anti-derivative of f(x).
If you change a then the anti-derivative changes by a constant.
But not all anti-derivatives can be obtained in this way. For instance take the function,f(x), that is identically equal to zero on the interval (0,1). Then $$\int_{a}^x f(t) dt = F(x)$$ is equal to zero. But any constant function is also an anti-derivative.
What is true is that if F(x) is an anti-derivative of f(x) then $$\int_{a}^x f(t) dt = F(x) - F(a)$$
Last edited: Jul 1, 2015
8. Jul 10, 2015
### mathwonk
actually the correct hypotheses on the function f are quite interesting. I.e. there are two facts you want to assume: 1) the riemann integral of f exists on [a,b], 2) f has an "antiderivative" F. and then you want to conclude 3) that the integral of f equal F(b) - F(a).
Continuity of f suffices for all 3 of these to hold, but is not necessary. E.g. if f is a step function then it has a finite set of discontinuities and its indefinite integral is differentiable elsewhere with derivative equal to f. If however we define an "antiderivative" of f to be a continuous function with derivative equal to f except at the finite set of discontimnuities, then again the indefinite integral satisfies both 2) and 3).
There are however more complicated functions f whose riemann integral exists on [a,b] and such that an antiderivative F in this sense, i.e. F continuous everywhere and derivative equal to f where f is continuous, does not satisfy 3). an example is the characteristic function f of the cantor set, which is zero off the cantor set, hence has integral zero. the cantor function F which is locally constant off the cantor set, hence has derivative zero there, is continuous everywhere, but climbs from 0 to 1 over [a,b], hence does not compute the integral of f. We can however strengthen the requirement on an "antiderivative" to be Lipschitz continuous.
Notice the key point that failed was that with the weaker definition of antiderivative we did not get that any two of them differ by a constant, i.e. a continuous function (even on a connected interval), and with derivative zero almost everywhere, does not have to be constant, but we do get that with the addition of the Lipschitz property.
Last edited: Jul 10, 2015 | 2018-05-21T11:28:03 | {
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https://math.stackexchange.com/questions/1853196/what-is-the-probability-that-at-the-end-of-the-sequence-bucket-b-contains-ball | # What is the probability that at the end of the sequence, bucket B contains ball bi
We have a bucket B which can store 1 ball at a time. Imagine a sequence of balls: {b1,b2....bn} such that ball bi appears after ball bi-1 in the sequence. The ith bi is stored in bucket B with probability 1/i replacing the ball bk,previously stored. For each i, what is the probability that at the end of the sequence, bucket B contains ball bi.
• k is less than i (k <i). – papabiceps Jul 8 '16 at 15:17
• Not sure this is clear. Does $b_1$ start in $B$? If there are, say, two balls, is the answer just $\frac 12$ for both (as there is a $\frac 12$ probability that $b_2$ replaces $b_1$)? – lulu Jul 8 '16 at 15:22
• To be clear: $B$ always contains exactly one ball (since $b_1$ was put in there with probability $1$) and whether a given ball is put in $B$ is independent of the previous balls? In which case, it seems that at the end of the day the probability that $b_i$ is in $B$ is simply the probability that it was put there $1/i$), times that none after was put instead ($\prod_{j>i}(1-1/j)$) by independence. – Clement C. Jul 8 '16 at 15:22
• But why do you refer to the $(i-1)^{st}$ $b_i$? There's only one $b_i$, right? – lulu Jul 8 '16 at 15:24
• My reading is the same as that of @ClementC. but I am not sure we have the question right. – lulu Jul 8 '16 at 15:25
$B$ always contains exactly one ball (since $b_1$ was put in there with probability $1$) and whether a given ball $b_i$ is put in the bucket $B$ is independent of the previous balls ($i < j$). Therefore, at the end of the day the probability that $b_i$ is in the bucket $B$ is simply the probability $$\mathbb{P}\{b_i\in B\text{ and } b_j\not\in B \text{ for all } j > i\}$$ which, by independence, becomes $$\mathbb{P}\{b_i\in B \}\cdot\prod_{j=i+1}^n \mathbb{P}\{b_j\not\in B\} = \frac{1}{i}\prod_{j=i+1}^n\left(1-\frac{1}{j}\right).$$
This in turn, perhaps counter-intuitively, leads to the answer being $$\frac{1}{i}\prod_{j=i+1}^n \left(1-\frac{1}{j}\right) = \frac{1}{n}.$$
(This last identity is easy to show, e.g. by induction on $1\leq i \leq n$.)
• @papabiceps Thanks for spotting this! Apparently, copy-pasting from comments involves the joy of all $\LaTeX$ being duplicated in the plain text. – Clement C. Jul 8 '16 at 15:45 | 2019-05-23T09:46:13 | {
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http://mathhelpforum.com/number-theory/137852-chinese-remainder-thm.html | 1. ## Chinese Remainder Thm
Let p and q be two odd primes. Show that x^2 - 1 = 0 mod pq has four solutions. Find four solutions modulo 15. (Hint: use the Chinese Remainder Thm and Lagrange's Thm)
The above = is congrence, not equality.
How do I show this? Thanks...
2. Originally Posted by jzellt
Let p and q be two odd primes. Show that x^2 - 1 = 0 mod pq has four solutions. Find four solutions modulo 15. (Hint: use the Chinese Remainder Thm and Lagrange's Thm)
The above = is congrence, not equality.
How do I show this? Thanks...
$x^2 -1 \equiv 0 (mod pq)$
$pq \mid x^2-1$
p,q are primes that means (p,q)=1
so
$p\mid x^2-1$ and $q\mid x^2-1$
$x^2 -1 \equiv 0 (mod p) \Rightarrow x^2\equiv 1 (mod p)$
$x \equiv \mp 1 (mod p)$
this is two solutions
same for q you will have two solutions so in total you have four
$x^2-1 \equiv 0 (mod 15)$
can you solve it now
3. So, I would re-write it as x^2 - 1 = 3*5 (= is congruence)
3*5 | x^2 -1
So,
3 | x^2 -1 and 5 | x^2 - 1
x^2 - 1 = 0 (mod 3) -> x^2 = 1 (mod 3)
x = +-1 (mod 3) So two solutions are +-1 for mod 3.
Now I do the same for 5 to get two solutions: +-1 for mod 5.
So what are my FOUR solutions since I got +-1 for both mod3 and mod5 ?
4. Originally Posted by jzellt
So, I would re-write it as x^2 - 1 = 3*5 (= is congruence)
3*5 | x^2 -1
So,
3 | x^2 -1 and 5 | x^2 - 1
x^2 - 1 = 0 (mod 3) -> x^2 = 1 (mod 3)
x = +-1 (mod 3) So two solutions are +-1 for mod 3.
Now I do the same for 5 to get two solutions: +-1 for mod 5.
So what are my FOUR solutions since I got +-1 for both mod3 and mod5 ?
Here are your four solutions, and this is where CRT comes into play
$\begin{cases} x\equiv1\bmod{3}\\ x\equiv1\bmod{5} \end{cases}$
$\begin{cases} x\equiv-1\bmod{3}\\ x\equiv1\bmod{5} \end{cases}$
$\begin{cases} x\equiv1\bmod{3}\\ x\equiv-1\bmod{5} \end{cases}$
$\begin{cases} x\equiv-1\bmod{3}\\ x\equiv-1\bmod{5} \end{cases}$
So applying CRT to these four systems will give you your solutions.
6. Can you show how to apply the Chinese Remainder Thm on the first set for me? Thanks.
7. ok since nobody answered until now
$x \equiv 1 (mod 3)$
$x \equiv 1 (mod 5)$
first let
$x \equiv 0 (mod 3)$
$x \equiv 1 (mod 5)$
want a multiple of 3 and equal 1 (mod 5) it is 6
then let
$x \equiv 1 (mod 3)$
$x \equiv 0 (mod 5)$
want a multiple of 5 and equal 1 (mod 3) it is 10
now 6+10 =16 ,
16 = 1 (mod 3)
16 = 1 (mod 5)
first solution
8. second solution
$x \equiv -1 (mod 3)$
$x \equiv 1 (mod 5)$
now let
$x \equiv 0 (mod 3)$
$x \equiv 1 (mod 5)$
it is 6 , 6=1 (mod 5)
then let
$x \equiv -1 (mod 3)$
$x \equiv 0 (mod 5)$
want a multiple of 5 and equal -1 (mod 3) it is 20
now 20+6 = 26
but
$26 \equiv 11 (mod 15)$
so it is 11
third solution
$x \equiv -1 (mod 3)$
$x \equiv -1 (mod 5)$
let
$x \equiv 0 (mod 3)$
$x \equiv -1 (mod 5)$
multiple of 3 and -1 (mod 5) it is 9
let
$x \equiv -1 (mod 3)$
$x \equiv 0 (mod 5)$
multiple of 5 and -1 (mod 3)
it is 10
so 10+9 = 19
but
$19 \equiv 14 (mod 15)$
so it is 14
note first solution is 16 and it equal to 1 (mod 15)
the solutions are
1,4,11,14
9. Originally Posted by chiph588@
Here are your four solutions, and this is where CRT comes into play
$\begin{cases} x\equiv1\bmod{3}\\ x\equiv1\bmod{5} \end{cases}$
$\begin{cases} x\equiv-1\bmod{3}\\ x\equiv1\bmod{5} \end{cases}$
$\begin{cases} x\equiv1\bmod{3}\\ x\equiv-1\bmod{5} \end{cases}$
$\begin{cases} x\equiv-1\bmod{3}\\ x\equiv-1\bmod{5} \end{cases}$
So applying CRT to these four systems will give you your solutions.
Here's how to solve CRT. | 2013-12-13T06:57:16 | {
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https://math.stackexchange.com/questions/1342709/probability-of-selecting-consecutive-floors-in-an-elevator/1946011 | # Probability of selecting consecutive floors in an elevator
Three people get into an empty elevator at the first floor of a building that has 10 floors. Each presses the button for their desired floor (unless one of the others has already pressed the button). Assume that they are equally likely to want to go to floors 2 through 10 (independently of each other). What is the probability that the buttons for 3 consecutive floors are pressed? (question from 'introduction to probability' by Bliztstein)
I thought the answer might be $$\frac{7 \cdot3!}{9^3}$$... but then I thought maybe the total number of ways is the bose-einstein value $n+k-1 \choose k$ (how many ways there are to choose k times from a set of n objects with replacement, if order doesn't matter).. so then the answer could also be $$\frac{7}{165}$$ Which is the correct answer, and why is the logic for the other answer wrong?
• I think the $\frac{7\cdot 3!}{9^3}$ is the only defensible answer, given the assumptions. If we are told that exactly $3$ buttons are pressed, the answer changes. – André Nicolas Jun 28 '15 at 23:02
• The problem with the unordered 3-tuples is that the 165 of them are not equally likely (under the assumptions of three independent uniformly distributed choices). For example, {3,4.7} is six times more likely to be chosen than {5,5,5}. – Ned Jun 28 '15 at 23:06
I think there is only one reasonable interpretation of the problem. It leads to a probability model that produces your first answer.
Let our three passengers be A, B, and C. If A, B, C wish respectively to go to floors $p$, $q$, and $r$, record that fact as an ordered triple $(p,q,r)$. By the equally likely and independent part of the problem statement, all ordered triples $(p,q,r)$, where $p$, $q$, and $r$ range from $2$ to $10$, are equally likely. There are $9^3$ such triples. Whether a passenger, say C, happens not to press the elevator button to Floor $6$ because it has already been pressed is irrelevant.
We now count the favourables. There are $7$ ways to choose a collection of $3$ consecutive numbers in the interval from $2$ to $10$. For each of these collections, there are $3!$ ways in which A, B, and C might wish to get off at the floors in this collection, for a total of $7\cdot 3!$. For the probability, divide.
• whys is "Whether a passenger, say C, happens not to press the elevator button to Floor 6 because it has already been pressed is irrelevant.", isnt that relevant to the line " (unless one of the others has already pressed the button)" in question ? – bicepjai Dec 19 '18 at 8:48
Call the passengers A,B,C.
Consider A presses 2, then for there to be three consecutive ( i.e. 2,3,4 ) buttons pressed B and C must press 3 and 4, this can occur in only two ways, with B pressing 3 and C pressing 4 or visa versa.
Repeating this method on floors 3,4,5,6,7 and 8 we have a total of 14 different ways of getting 3 consecutive buttons pressed where A gets out on a lower floor than the other two.
Therefore there is a total of 3 times 14 = 42 ways of getting 3 consecutive buttons pressed, the probability of which is $\Large\frac{42}{9^3}$
Much the same as André Nicolas' answer.
BTW if one of the passengers was an unaccompanied 4 year old, chances are almost certain that every damn button would be pressed, at least the ones he/she could reach.
The first one $$\dfrac{7*3!}{9^3}$$ is the correct answer.
We can not use Bose-Einstein here because it can not be used in naive probability expression according to 'introduction to probability' by Bliztstein, page 18. "The Bose-Einstein result should not be used in the naive definition of probability except in very special circumstances. For example, consider a survey where a sample of sizekis collected by choosing people from a population of size n one at a time, with replacement and with equal probabilities. Then then k ordered samples are equally likely, making the naive definition applicable, but the unordered samples (where all that matters is how many times each person was sampled) are not equally likely." Here, since each person presses which floor is an equal likely event, we cannot use Bose-Einstein.
let's say we have A, B, and C people in the lift. A - is the event that passengers will press three consecutive numbers. P(A)= # of favourable outcomes/# of total outcomes Since there is replacement and order matters the # of total outcomes is 9^3 Now we should find the # of fav outcomes: there are 7 ways to get three consecutive numbers out of 2-10 => (234), (345), (456), (567), (678), (789), (8910) and 3! ways to shoose from three numbers Therefore, the number of fav outcomes is 7*3! P(A)=7*3!/9^3
The number of possible outcomes for who is going to which floor is $$9^3$$. There are $$7$$ possibilities for which $$3$$ buttons are pressed such that there are consecutive floors: $$(2,3,4),(3,4,5)...(8,9,10)$$. For each of these $$7$$ possibilities, there are $$3!$$ ways to choose who is going to which floor. So by the naive definition, the probability is $${3!\times7\over9^3}= {42\over729}= {14\over243} = 0.05761316872$$
In the sequence $2,3,4,\cdots,10$ there are $7$ sequences of three consecutives numbers: $$(2,3,4), (3,4,5), (4,5,6), (5,6,7), (6,7,8), (7,8,9), (8,9,10)$$ and by outher side there are ${9 \choose 3}=84$ distinc modes od choose $3$ numbers in set ${2,3,\cdots,10}$, Then, in my opinion, the probability is $P=\frac{7}{84}$.
• This seems to exclude the possibility of any of the people being on the same floor – Dr Xorile Jun 29 '15 at 0:51
• Really Dr. Xorile, If consider that people can come down on the same floor, then the total of modes is equal the number of non negative interger of equation $x_1+x_2+\cdots+x_9=3$, that is, ${9+3-1 \choose 3}=165$. In this case the correct probability is $P=\frac{7}{165}$. Thank you for letting me know! – Cgomes Jun 29 '15 at 1:26
• But the 165 different multi-sets are not equally likely, so you can't just count 7 successes and divide by 165 to get the probability. The only obvious way to get equally likely outcomes is to take the outcomes as ORDERED triples, so the answer is the first version in the OP. – Ned Jun 29 '15 at 1:39 | 2019-10-14T01:20:04 | {
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https://math.stackexchange.com/questions/2997281/tree-has-exactly-k-nodes-with-degree-4-show-that-this-tree-has-2k2-leave | # tree has exactly $k$ nodes with degree $4$. Show that this tree has $2k+2$ leaves.
Prove:
If a tree has exactly $$k \geq 1$$ nodes with degree $$4$$, then this tree has at least $$2k +2$$ leaves. ( nodes with degree $$< 4$$ are only allowed for the leaves ).
So I think that we can solve this with induction.
$$k = 1$$ :
So we have exactly one node with degree $$4$$. this node has to have at least $$4$$ leaves, because of the degree. Or am I wrong?
$$k \rightarrow k+1$$:
We have a tree with $$k$$ nodes with degree $$4$$. We know that we have at least $$2k+2$$ leaves. We replace one of the leaves with a node and build a node of degree $$4$$. So we "lost" a leave but "gain" 3 new leaves. So we conclude that we have a tree with $$k+1$$ nodes with degree $$4$$ and $$2k+2-1+3 = 2k+4 =2(k+1) +2$$ leaves. Am I right?
I do not see why it is important to include the condition that only vertices of degree less than $$4$$ are leaves. I am going to prove a general statement. Note that this proof can be rewritten so that it is an inductive proof
Proposition. Let $$k\geq 1$$ and $$d\geq 2$$ be integers. Let $$T$$ be a (finite) tree with exactly $$k$$ vertices of degree $$d$$. Then, $$T$$ has at least $$(d-2)k+2$$ leaves.
Suppose on the contrary that, for some positive integer $$k$$, there exists a tree with exactly $$k$$ vertices of degree $$d$$ but with fewer than $$(d-2)k+2$$ leaves. Take $$T$$ to be such a tree with the smallest possible $$k$$. If $$k>1$$, then let $$u$$ be a vertex of $$T$$ with degree $$d$$. Suppose that $$v_1,v_2,\ldots,v_d$$ are the neighbors of $$u$$. Let $$C_i$$ be the connected component of $$T-u$$ containing $$v_i$$, for $$i=1,2,\ldots,d$$. Take $$T_i$$ to be the tree with vertices from $$C_i$$ and an extra vertex $$u_i$$, which is a leaf and adjacent to $$v_i$$.
For a graph $$G$$, let $$n_t(G)$$ denote the number of vertices of degree $$t\in\mathbb{Z}_{\geq 0}$$ in $$G$$. Clearly, we have $$\sum_{i=1}^d\,n_1(T_i)=n_1(T)+d<\big((d-2)k+2\big)+d\,.$$ Since $$T$$ is a tree with the smallest $$k=n_d(T)$$ that violates the claim, we must have $$n_1(T_i)\geq (d-2)\,n_d(T_i)+2$$ for all $$i=1,2,\ldots,d$$. This shows that \begin{align}\big((d-2)k+2\big)+d&>\sum_{i=1}^d\,n_1(T_i)\geq \sum_{i=1}^d\,\big((d-2)\,n_d(T_i)+2\big) \\&=(d-2)\,\sum_{i=1}^d\,n_d(T_i)+2d=(d-2)\,(k-1)+2d\\&=\big((d-2)k+2\big)+d\,.\end{align} This is absurd, so the assumption that $$k>1$$ cannot be true. Hence, $$k=1$$.
However, it is easy to show that every tree with exactly $$1$$ vertex of degree $$d$$ as at least $$d$$ leaves. Thus, $$k=1$$ cannot hold either, and the proposition must be true. (Note that the minimum number of leaves $$(d-2)k+2$$ is achieved if and only if $$T$$ has only vertices of degree $$1$$ or $$d$$. In such cases, $$T$$ has $$(d-1)k+2$$ vertices and $$(d-1)k+1$$ edges.)
Corollary. For integers $$d_1,d_2,\ldots,d_m$$ with $$1, we have $$n_1(T)\geq \sum_{i=1}^m\,(d_i-2)\,n_{d_i}(T)+2$$ for every tree $$T$$ with at least two vertices.
As before, $$n_t(G)$$ denote the number of vertices of degree $$t\in\mathbb{Z}_{\geq 0}$$ in a finite graph $$G$$. In the corollary, the equality holds iff $$n_t(T)=0$$ for every integer $$t\geq 2$$ such that $$t\notin \{d_1,d_2,\ldots,d_m\}$$. In this case, $$T$$ has exactly $$\displaystyle\sum_{i=1}^m\,(d_i-1)\,n_{d_i}(T)+2$$ vertices and $$\displaystyle\sum_{i=1}^m\,(d_i-1)\,n_{d_i}(T)+1$$ edges.
• Yes indeed, if you know the number $n_d$ of nodes of degree $d$ for each $d \geq 3$ and the number $c$ of components in the forest, you can calculate exactly the number of leaves which is $1+c+\sum_{d \geq 3} n_d(d-2)$. A lower bound of $2n_4+2$ for a tree, which the OP was asked to establish, is an easy consequence from this formula. Good answer!
– Mike
Nov 14 '18 at 1:29
Your proof could be made to work but there is still more you need to do for it to be considered a proof.
The one detail you need to observe is that you can construct any tree w $$k+1$$ degree-4 nodes from one with $$k$$ nodes as you did. What if all nodes adjacent to a leaf have degree 5 or greater? A way around this is to conclude that a tree with $$k''$$ nodes of degree at least 4 has at least $$2k''+2$$ leaves.
An alternate proof: Let $$F$$ be a forest where every nonleaf vertex has degree at least 4. Let $$n$$ be the number of vertices in $$F$$. The number of edges that $$F$$ can have is at most $$n-1$$ with equality reached iff $$F$$ is a tree and not a forest with two or more connected components. [make sure you see why]
Let $$k$$ be the number of vertices of degree 4 and $$k'$$ the number of vertices of degree 5 or greater.
Letting $$l$$ be the number of leaves, then $$n=l+k+k'$$. So the number of edges that $$F$$ can have is at most $$n-1 =$$ $$l+k+k'-1$$.
The number of edges $$F$$ has however at least $$\frac{1}{2} (5k'+4k+l)$$ [this is $$\frac{1}{2}$$ times a lower bound on the sum of the degrees of the vertices of $$F$$, make sure you see why that is the number of edges in $$F$$]
Thus $$2(l+k+k')-2 \ge 5k'+4k+l$$ $$\implies$$ $$l \ge 3k'+2k+2$$.
*****If every interior vertex has degree 4 and the graph is a tree then this inequality is tight. Can you trace through the proof to see why?
*****You could trace through this proof to conclude: $$l=\sum_{d \geq 2} (k_d-2)+2$$, where $$k_d$$ is the number of vertices of degree $$d$$ $$(d=2,3,\ldots$$) if the graph is a tree, and $$l=\sum_{d \geq 2} (k_d-2)+1+c$$, if the graph is a forest with $$c$$ distinct connected components. From this the lower bound of $$l \geq 2k_4+2$$ for $$l$$, which is what you wanted to derive, is an immediate consequence.
• I don't think your proof is correct. A tree with $k$ vertices of degree $4$ may have vertices of degree $3$ for example. So, $n=l+k$ is not correct. Also, $F$ does not necessarily have $\dfrac{4k+l}{2}$ edges. Nov 13 '18 at 21:16
• @Batominovski in general yes but the assumption by the OP is that only leaves have degree $<4$
– Mike
Nov 13 '18 at 21:18
• What about vertices of degree more than $4$? There is no restriction on that. Nov 13 '18 at 21:19
• @Batominovski fixed it to allow for that possibility
– Mike
Nov 13 '18 at 21:23
• For whatever reason I had read that every interior vertex has degree 4
– Mike
Nov 13 '18 at 21:23 | 2021-10-21T02:34:20 | {
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https://stats.stackexchange.com/questions/560508/spearman-correlation-test-and-linear-relationship-vs-monotonic-relationship | # Spearman correlation test and linear relationship vs monotonic relationship?
I want to use Spearman's correlation test. My data is not normally distributed. Below is a scatterplot of this data.
I read somewhere that Spearman's correlation coefficient can describe monotonic relationships. Is my data monotonic or linear? Can Spearman's correlation coefficient be used on both linear and monotonic relationships?
New contributor
kaka is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
Monotonic means either '$$y$$ does not decrease as $$x$$ increases' (positive monotonic relationship between $$y$$ and $$x$$), or '$$y$$ does not increase as $$x$$ increases' (negative monotonic relationship between $$y$$ and $$x$$).
'Linear' meaning 'line-like' is just one kind of monotonic relationship, so yes you can use Spearman's correlation coefficient whether the relationship is linear or some other monotonic function. Asking whether a relationship is "monotonic or linear" is like asking whether something is "food or an apple": it's a little confused because an apple is a kind of food.
There is a trade-off between the more general 'monotonic association' of Spearman's $$\rho_{\text{S}}$$ versus the specifically linear association of Pearson's $$\rho$$: If the relationship between $$y$$ and $$x$$ is close to linear, then the magnitude of Spearman's correlation coefficient will likely be smaller than Pearson's, and the power of the t test of $$\rho_{\text{S}}$$ with $$\text{H}_{0}\text{: no monotonic association}$$ (i.e. $$\text{H}_{0}\text{: }\rho_{\text{S}} = 0$$) will be lower than the power of the t test of with $$\text{H}_{0}\text{: no linear association}$$ (i.e. $$\text{H}_{0}\text{: }\rho = 0$$).
Finally, because monotonic relationships can actually be flat in places (imagine a function that looks like stair steps), a significantly positive $$\rho_{\text{S}}$$ might only be interpretable as meaning '$$y$$ tends to increase as $$x$$ increases', while a significantly negative $$\rho_{\text{S}}$$) might only be interpretable as meaning '$$y$$ tends to decrease as $$x$$ increases'.
• So my graph does not show a monotonic relationship because y does not decrease as x increase and y does increase as x increase, right?
– kaka
Jan 14 at 17:45
• @kaka I would say that there is a slight monotonically increasing relationship, but at your apparent sample size, I would not be surprised if that was not significantly different from 0. That said, you can also perform two one-sided tests for equivalence between $\rho$ and $0$: $\text{H}_{01}\text{: }\rho \ge \Delta$ and $\text{H}_{02}\text{: }\rho \le -\Delta$. If your reject both $\text{H}_{01}$ and $\text{H}_{02}$, you found evidence that $\rho$ is equivalent to $0$ within $\pm\Delta$ at the $\alpha$ level, where $\Delta$ is the smallest size correlation you care about. Jan 14 at 19:14
• You can also take the ranks of the x variable, and the ranks of the y variable, and plot those. That may give you a better sense of the relationship that Spearman correlation is assessing. Jan 14 at 21:06
• @SalMangiafico Correct. Albeit, the behaviors of ranks may or may not have substantive interest (as opposed to the actual measures). In other words, analysis of ranks may be only instrumentally useful. Jan 14 at 21:07
• To be fussy: In order for Spearman correlation to be $1,$ a positive association has to be strictly monotonic: In R. cor(1:4, c(1,1,3,4)) returns $0.9467293.$ 2 days ago
The Spearman null hypothesis is that X and Y are independent. The alternative is that there is dependence between X and Y in such a way that if you consider a larger and a smaller (random) value of X, Y tends to be either systematically larger, or systematically smaller for the larger X. This holds in particular in monotonic relationships, including linear (unless the slope is zero).
So if this is what you want to test, the Spearman correlation test should be fine.
In the illustration below, the association between variables $$x$$ and $$y$$ is positive. (Variable $$y$$ tends to increase as variable $$x$$ increases.
Also, the relationship between variables $$x$$ and $$y$$ is strictly monotonic (every increase in $$x$$ is accompanied by an increase in $$y.)$$ Thus, Spearman correlation is $$1.$$
However, the relationship is not linear (as you can see from the plot below). Thus, Pearson correlation is smaller than $$1.$$
Computations in R:
x = 1:20
y = x^4
plot(x,y)
cor(x,y) # Pearson correlation
[1] 0.8730177
cor(x,y, meth="s") # Spearman correlation
[1] 1
Now, we add a small amount randomness to y and call the result z. The randomness destroys the strictly monotone relationship between x and z. While the xs are still in increasing order, the first two values of z are out of order. The change is too small to see on a plot, but the Spearman correlation is no longer exactly $$1.$$
set.seed(2022)
z = y + rnorm(20, 0, 10)
z[1:3]
[1] 10.001420 4.266542 72.025146
cor(x, z)
[1] 0.8730275
cor(x, z, meth="s")
[1] 0.9984962 | 2022-01-18T17:40:05 | {
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https://math.stackexchange.com/questions/2630991/2n-students-want-to-sit-in-n-fixed-school-desks-such-that-no-one-sits-with-t | # $2n$ students want to sit in $n$ fixed school desks such that no one sits with their previous partner
A classroom has $n$ fixed school desks with exactly $2$ chairs each. There are $2n$ students sitting in the classroom and then they go on a break. After the break they're coming back into the classroom and want to sit such that no one sits with their previous partner. What's the number of ways to do that?
I was thinking either a recursion or the inclusion-exclusion formula. This seems similar to derangements problem, but I'm not sure if I can make a connection between the two.
My idea was to define $A_i = \{$person $i$ sits with their previous partner$\}$ for $i=1,...,n$ and then to use inclusion-exclusion but I'm not sure how to count $\lvert A_i \rvert$ and $\lvert A_I\rvert=\cap_{i\in I}A_i$ where $I\subset\{1,2,...,n\}$
• What do you want to know? The number of way to do so (do you count the position of each student, or just the couple by desk)? Why the desks need to be fixed? – Netchaiev Feb 1 '18 at 10:45
• @Netchaiev yes that's what I want to know. I don't know why, the problems states so.. – Collapse Feb 1 '18 at 10:48
• Does it matter if they are sitting on desk 1 or desk 2? do you count student A and student B sitting together on two different tests as different ? – dimebucker Feb 1 '18 at 10:50
• @dimebucker91 I'm really not sure about that, it's a problem from an older test and I'm not sure myself, seeing as it says specifically that the desks are in fixed positions I would assume it matters and that if $A$ and $B$ were together on one desk before they can't be together on any desks now.. – Collapse Feb 1 '18 at 10:52
• Just to be sure, if we count the different desks and sits, for 4 students, there are 4 ways to do so? – Netchaiev Feb 1 '18 at 10:52
Combinatronics has never come easy to me, but hopefully this is a sound approach:
We can fix the position of $n$ of the students, label these $A_1, \dots, A_n$ on desks $D_1, \dots, D_n$. Then we have the remaining students, $B_1, \dots, B_n$, and we assume that the initial arrangement matched them up: $(A_i, B_i)$ on desk $D_i$.
Now, we need to arrange the $B$'s, there are in total $n \times (n-1) \times \dots \times 1 = n!$ ways to do this, we want to discount the permutations in which the $i$'s line up, as these students already sat next to each other. That is, we want the number of derangements: $!n$
$$!n = (n-1)(!(n-1) + !(n-2))$$ Consider the case with $n=4$, then fixing the $A$'s, the remaining four students, call them $a,b,c,d$ can be seated in the following $24$ permutations:
('a', 'c', 'b', 'd')
('a', 'c', 'd', 'b')
('a', 'b', 'c', 'd')
('a', 'b', 'd', 'c')
('a', 'd', 'c', 'b')
('a', 'd', 'b', 'c')
('c', 'a', 'b', 'd')
('c', 'a', 'd', 'b')
('c', 'b', 'a', 'd')
('c', 'b', 'd', 'a')
('c', 'd', 'a', 'b')
('c', 'd', 'b', 'a')
('b', 'a', 'c', 'd')
('b', 'a', 'd', 'c')
('b', 'c', 'a', 'd')
('b', 'c', 'd', 'a')
('b', 'd', 'a', 'c')
('b', 'd', 'c', 'a')
('d', 'a', 'c', 'b')
('d', 'a', 'b', 'c')
('d', 'c', 'a', 'b')
('d', 'c', 'b', 'a')
('d', 'b', 'a', 'c')
('d', 'b', 'c', 'a')
but of these, only the following $9$ are valid:
('c', 'a', 'd', 'b')
('c', 'd', 'a', 'b')
('c', 'd', 'b', 'a')
('b', 'a', 'd', 'c')
('b', 'c', 'd', 'a')
('b', 'd', 'a', 'c')
('d', 'a', 'b', 'c')
('d', 'c', 'a', 'b')
('d', 'c', 'b', 'a')
`
where we have $!0 = 1$, $!1 = 0$, $!2 = 1$, $!3 = 2$ and finally: $$!4 = 3(!3 + !2) = 9$$
• It is not a simple derangement, A(i) paired with A(j) is also a mismatch. – true blue anil Feb 1 '18 at 14:19
• I'm not really sure that's it, the first step where you fix $A_1 ... A_n$ is what's bothering me. Those students might've been those who sat on the first $n/2$ desks before and therefore none of them sat with any $B_1,...,B_n$ before..? – Collapse Feb 1 '18 at 16:38 | 2019-07-16T04:04:16 | {
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https://mathhelpboards.com/threads/minimum-of-product-of-2-functions.29446/#post-129259 | # minimum of product of 2 functions
#### sarrah
##### Member
Hello
Simple question
Whether the minimum of the product of two functions in one single variable, is it greater or less than the product of their minimum
thanks
Sarrah
#### Evgeny.Makarov
##### Well-known member
MHB Math Scholar
If the values of both functions are nonnegative, then the product of minima is less than or equal to the minimum of products. This relies on the following fact about real numbers: if $0\le x_1\le y_1$ and $0\le x_2\le y_2$, then $x_1x_2\le y_1y_2$.
For example, consider functions $f$ and $g$ and just two values in the domain: $x_1$ and $x_2$. Let $a_1=f(x_1)$, $a_2=f(x_2)$, $b_1=g(x_1)$, $b_2=g(x_2)$, Then $\min(a_1,a_2)\le a_1$ and $\min(b_1,b_2)\le b_1$, so by the fact above $\min(a_1,a_2)\min(b_1,b_2)\le a_1b_1$. Similarly $\min(a_1,a_2)\min(b_1,b_2)\le a_2b_2$, so $\min(a_1,a_2)\min(b_1,b_2)\le\min(a_1b_1,a_2b_2)$.
If the numbers can be negative, then this conclusion no longer holds. For example, if $a_1=1$, $a_2=2$ and $b_1=b_2=-1$, then $\min(a_1,a_2)\min(b_1,b_2)=1\cdot(-1)=-1>-2=\min(-1,-2)=\min(a_1b_1,a_2b_2)$,
#### sarrah
##### Member
If the values of both functions are nonnegative, then the product of minima is less than or equal to the minimum of products. This relies on the following fact about real numbers: if $0\le x_1\le y_1$ and $0\le x_2\le y_2$, then $x_1x_2\le y_1y_2$.
For example, consider functions $f$ and $g$ and just two values in the domain: $x_1$ and $x_2$. Let $a_1=f(x_1)$, $a_2=f(x_2)$, $b_1=g(x_1)$, $b_2=g(x_2)$, Then $\min(a_1,a_2)\le a_1$ and $\min(b_1,b_2)\le b_1$, so by the fact above $\min(a_1,a_2)\min(b_1,b_2)\le a_1b_1$. Similarly $\min(a_1,a_2)\min(b_1,b_2)\le a_2b_2$, so $\min(a_1,a_2)\min(b_1,b_2)\le\min(a_1b_1,a_2b_2)$.
If the numbers can be negative, then this conclusion no longer holds. For example, if $a_1=1$, $a_2=2$ and $b_1=b_2=-1$, then $\min(a_1,a_2)\min(b_1,b_2)=1\cdot(-1)=-1>-2=\min(-1,-2)=\min(a_1b_1,a_2b_2)$,
I am extremely grateful
Sarrah | 2022-07-02T22:42:38 | {
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https://www.freemathhelp.com/forum/threads/find-period-of-trig-functions.122710/ | # Find Period of Trig Functions
#### harpazo
##### Full Member
How do I find the period of a trig function from its graph?
#### Jomo
##### Elite Member
You set the angle to 0 and set the angle to 2pi. Solve for the variable in each case. Compute the absolute value of the difference between those numbers.
#### Jomo
##### Elite Member
Oh, from the graph. You find one full period of the graph and subtract the x-valus from the start and finish of that period. Take the absolute value of that difference.
#### harpazo
##### Full Member
You set the angle to 0 and set the angle to 2pi. Solve for the variable in each case. Compute the absolute value of the difference between those numbers.
Can you show me using the graph below? What is the period?
#### harpazo
##### Full Member
Oh, from the graph. You find one full period of the graph and subtract the x-valus from the start and finish of that period. Take the absolute value of that difference.
Can you please show me what you mean using the graph below. I will then return to the textbook to solve similar problems following your steps here.
#### Harry_the_cat
##### Senior Member
Can you show me using the graph below? What is the period?
View attachment 19114
See (0,0) is on the graph. After one full cycle, or S shape, we get to the point (2pi,0).
What distance is that on the x-axis? 2pi. So the period is 2pi.
#### Harry_the_cat
##### Senior Member
Can you please show me what you mean using the graph below. I will then return to the textbook to solve similar problems following your steps here.
View attachment 19115
On this one, I'd look at (0, 3) and (8, 3). That's a full cycle so period is 8.
It doesnt matter what two points you look at, as long as there is a full cycle between them. For example, (4, -3) and (12,-3). Period is still 8.
#### Jomo
##### Elite Member
Can you please show me what you mean using the graph below. I will then return to the textbook to solve similar problems following your steps here.
View attachment 19115
One period starts at 0 and finishes at 8. Another period/cycle starts at 4 and finishes at 12. Another one starts at starts at -4 and finishes at 4. Just take any of those and subtract the end values. For example 8-0 = 8, 12-4 = 8, 4-(-4) = 8. The length of the period is 8.
#### harpazo
##### Full Member
See (0,0) is on the graph. After one full cycle, or S shape, we get to the point (2pi,0).
What distance is that on the x-axis? 2pi. So the period is 2pi.
I totally get it. Use the distance formula for points. Thanks.
#### harpazo
##### Full Member
One period starts at 0 and finishes at 8. Another period/cycle starts at 4 and finishes at 12. Another one starts at starts at -4 and finishes at 4. Just take any of those and subtract the end values. For example 8-0 = 8, 12-4 = 8, 4-(-4) = 8. The length of the period is 8.
Very good. Thanks. | 2020-06-06T17:03:39 | {
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https://gateoverflow.in/95821/tifr2017-b-13?show=96080 | 210 views
For an undirected graph $G=(V, E)$, the line graph $G'=(V', E')$ is obtained by replacing each edge in $E$ by a vertex, and adding an edge between two vertices in $V'$ if the corresponding edges in $G$ are incident on the same vertex. Which of the following is TRUE of line graphs?
1. the line graph for a complete graph is complete
2. the line graph for a connected graph is connected
3. the line graph for a bipartite graph is bipartite
4. the maximum degree of any vertex in the line graph is at most the maximum degree in the original graph
5. each vertex in the line graph has degree one or two
Option B). is true.
yes, Agreed. (y)
why not option C?
you can check with K(2,2). The line graph obtained is not biparitite.
The line graph of a connected graph is connected. If G is connected, it contains a path connecting any two of its edges, which translates into a path in L(G) containing any two of the vertices of L(G). Therefore, option B is correct.
We can also do this question using elimination of options.
edited
but your L(G2): e1---e2 it is bipertite right?e1 may be in one partition and e2 on another...it is bipertite i guess.
Hi,
I have corrected the example, and updated the file.
great example you have updated...proving line of tree is not a tree too :p | 2017-11-24T22:08:25 | {
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https://www.physicsforums.com/threads/need-help-related-rates.221277/ | Need help! Related rates
1. Mar 11, 2008
sutupidmath
Need help! Related rates!!!
1. The problem statement, all variables and given/known data
I am stuck somewhere on a related rates problem, i think that i am missing something rather obvious, but i cannot figure out so far.
-A street light is mounted at the top of the 15 ft-tall pole. A man 6 ft tall walks away from the pole with a speed of 5 ft/s along a straight path. How fast is the tip of his shadow moving when he is 40 ft from the pole.
2. Relevant equations
3. The attempt at a solution
Here it is what i did so far. I drew a triangle to describe the situation, i let one side of the triangle be 15 ft, i let x be the distance from the pole to the man, and i let y be the length of the shadow of the man. Then from the similarity of the triangles i got this relationship
$$\frac{15}{x+y}=\frac{6}{y}$$ i tried to rearrange this a little and i came up with
$$y=\frac{2}{3}x$$
From the data i also know that $$\frac{dx}{dt}=5$$
I also know that i should let $$x=40$$ after i come up with a relationship between the rate at which the man is moving and the rate of change of the tip of his shadow.
Here it is where i am stuck, for if we just implicitly differentiate $$y=\frac{2}{3}x$$ with respect to time it does not work. What am i missing here??
Any help would be really appreciated!!
Last edited: Mar 11, 2008
2. Mar 11, 2008
HallsofIvy
Staff Emeritus
Why doesn't differentiating work? dy/dt= (2/3)dx/dt and you know that dx/dt= 5 ft/sec. It looks like dy/dt= 10/3 ft/sec. Why is that wrong?
3. Mar 11, 2008
tiny-tim
Hi sutupidmath!
You're asked for the speed of the tip of the shadow, which is x + y, and you've only worked out the "speed" of the length of the shadow - so you've got the shadow moving slower than the man, and presumably ending up behind him!
Is that what happened to Peter Pan?
4. Mar 11, 2008
rocomath
This problem is from Stewart's, it tripped me up for a whole week. Makes a lot more sense to me now though, and has helped to solve other related rate problems ... so great problem.
Should be set up as ...
$$\frac{d}{dt}(x+y)=\frac{d}{dt}\left(\frac 5 3 x\right)$$
Last edited: Mar 11, 2008
5. Mar 11, 2008
sutupidmath
Yeah, i realized that what i actually should have changed in my approach is that i should have let x be the length of the shadow, (and not y) and let y be the distance of the man from the pole. But, the reason that i also got confused, is that we used nowhere the fact that the man is 40 ft from the pole when we are measuring the rate at which the tip of his shadow moves. In other words, even if we were not given this extra information, we still could have solved this problem? So why did they add this extra info, there? So is this extra information here given basically to trick us, or? For if it would not be given, i could get to the result way easier.
Edit: I think that the rate of the tip of the man's shadow will be constant, right? so no matter how long from the pole the man moves, the rate of the tip of his shadow will still be 25 ft/s. For if this is not the case then, i still think we are missing something there, and we have to somehow incorporate the distance of the man from the pole...
Last edited: Mar 11, 2008
6. Mar 11, 2008
sutupidmath
Because the answer at the end of the book is 25/3, and we used nowhere the fact that the man is 40 ft from the pole when we calculated the rate at which the tip of his shadow moves.!!
7. Mar 11, 2008
HallsofIvy
Staff Emeritus
Boy, I should have read that more closely! Yes, his shadow is lengthing by 10/3 ft./sec. and he is walking away from the light at 5 ft/sec so the tip of his shadow is moving away from the light at 10/3+ 5= 10/3+ 15/3= 25/3 ft/sec. You don't have to use the distance itself because the function involved is linear- the derivative with respect to time is a constant.
8. Mar 11, 2008
sutupidmath
I thought once to go exactly like you did it here, but i wasn't sure as to why aren't we considering the distance, so it thought this would be wrong.
Thnx all of you!
Last edited by a moderator: Mar 11, 2008 | 2017-02-26T06:01:56 | {
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https://mathoverflow.net/questions/185629/product-pvpvp-is-elementwise-nonnegative | Product $PVPVP$ is elementwise nonnegative?
Let $P\in \mathbb{R}^{n\times n}$ be the inverse of a positive definite M-matrix and $V\in \mathbb{R}^{n\times n}$ be any diagonal matrix. Prove (or disprove) that $PVPVP$ is elementwise nonnegative.
I know of the following:
$P$ is positive definite and elementwise nonnegative. Moreover, $p_{jk}p_{ii} \ge p_{ji}p_{ik}$ for any $i,j,k$.
I can verify that the statement is true for $n=2$, but I don't know how to work with $n$ large. Playing around with randomly generated matrices in Matlab seems to suggest that the statement is true. Any hint or suggestion would be greatly appreciated.
I've googled out that a very similar statement was put as a conjecture in this paper: Optimization of an on-chip active cooling system based on thin-film thermoelectric coolers (http://dl.acm.org/citation.cfm?id=1870955)
Edit: Perhaps someone can solve this easier question: Is there a positive semi-definite and elementwise nonnegative $P$ and diagonal $V$ such that $PVPVP$ is not elementwise nonnegative?
• The same question was asked about a week ago at MSE: math.stackexchange.com/q/985073/166535 – Joonas Ilmavirta Oct 28 '14 at 20:26
• Hint: There exists $S \in \mathbb R^{n \times n}$ such that $P = S^2 = S^T S$. Now, use the definition of nonnegative definiteness. – cardinal Oct 29 '14 at 0:01
• @cardinal: please excuse my slowness, but is it then immediate that one gets elementwise nonnegativity? – Suvrit Oct 29 '14 at 5:02
• @mathnotgoodatmath, if you repost, it's polite to mention it and give a link. Therefore I added the comment. About the question itself: It could be a good idea to include the definition of nonnegativity in the question. I first thought the question was simple, but then I realised that nonnegativity must mean something other than positive semidefiniteness and had to find the definitions. – Joonas Ilmavirta Oct 29 '14 at 6:11
• @JoonasIlmavirta: Thanks for your comments. Just edited the question. I'll be more careful next time. – vansy Oct 29 '14 at 14:57
First, we repeat the arguments from this stackexchange answer. $P^{-1}$ is an $M$-matrix, and can thus be written as $s(I-A)$ for some positive $s$ and some $A$ with non-negative entries. As $P^{-1}$ is positive definite, the spectrum of $A$ lies to the left of $\{ z: \hbox{Re}(z) = 1 \}$, and hence by Perron-Frobenius the spectral radius of $A$ is less than $1$. Thus we have the absolutely convergent Neumann series $$P = s^{-1} (I + A + A^2 + \dots )$$ and hence $$PVPVP = s^{-3} \sum_{i=0}^\infty \sum_{j=0}^\infty \sum_{k=0}^\infty A^i V A^j V A^k.$$ It thus suffices to show that $$\sum_{i+j+k=m} A^i V A^j V A^k \quad (1)$$ has non-negative coefficients for each $m \geq 0$ (where $i,j,k$ are understood to be non-negative integers). By change of variables, this is $$\sum_{0 \leq q \leq r \leq m} A^q V A^{r-q} V A^{m-r}.$$ Writing $A = (a_{st})_{1 \leq s,t \leq n}$ and $V = \hbox{diag}(v_1,\dots,v_n)$, the $st$ coefficient of (1) can be expanded as $$\sum_{s=s_0,s_1,\dots,s_m=t} a_{s_0 s_1} \dots a_{s_{m-1} s_m}\sum_{0 \leq q \leq r \leq m} v_{s_q} v_{s_r}.$$ But the quadratic form $$\sum_{0 \leq q \leq r \leq m} x_q x_r = \frac{1}{2}(x_0+\dots+x_m)^2 + \frac{1}{2} x_0^2 + \dots + \frac{1}{2} x_m^2$$ is positive definite, and the $a_{st}$ are non-negative, and the claim follows.
[For the record, I found this argument while performing a perturbative analysis in the case where $P$ was close to $I$, or more precisely $P = (I-A)^{-1}$ for some $A$ with small non-negative entries.]
• Wow, thanks so much for your proof. I came across this argument when looking at $(L+\mathrm{diag}(x))^{-1}$ where $L$ is a Laplacian matrix and $x$ is a nonnegative (and nonzero) vector. – vansy Apr 7 '16 at 22:47
Although you said you already proved the result for $n=2$, perhaps it's worth recording a proof here.
The statement is clear if both diagonal entries of $V$ have the same sign, so assume that $V = \left(\matrix{v_1&0\\ 0&-v_2}\right)$ with $v_1\ge0$ and $v_2\ge0$. If $P = \left(\matrix{a&b\\ c&d}\right)$ then by direct computation, $$PVPVP = \left(\matrix{a^3v_1^2 - bc(2av_1v_2 - dv_2^2)&b(a^2v_1^2-(ad+bc)v_1v_2 + d^2v_2^2)\\ c(a^2v_1^2-(ad+bc)v_1v_2 + d^2v_2^2)& d^3v_2^2 - bc(2dv_1 v_2 - av_1^2 )}\right).$$ Since the determinant of $P$ is positive, $-bc \ge -ad$, so $$a^2v_1^2-(ad+bc)v_1v_2 + d^2v_2^2 \ge a^2v_1^2-2adv_1v_2 + d^2v_2^2 = (av_1 - dv_2)^2 \ge 0.$$ Therefore the off-diagonal entries of $PVPVP$ are nonnegative. If $2av_1v_2 - dv_2^2<0$ then the $(1,1)$ entry of $PVPVP$ is clearly nonnegative; otherwise, $$a^3v_1^2 - bc(2av_1v_2 - dv_2^2)\ge a^3v_1^2 - ad(2av_1v_2 - dv_2^2) =a(av_1 -dv_2)^2 \ge 0.$$ Similarly, if $2dv_1 v_2 - av_1^2 < 0$ then the $(2,2)$ entry is clearly nonnegative; otherwise $$d^3v_2^2 - bc(2dv_1 v_2 - av_1^2 )\ge d^3v_2^2 - ad(2dv_1 v_2 - av_1^2 ) = d(dv_2 - av_1)^2 \ge 0.$$ | 2021-05-11T20:54:16 | {
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http://www.momon.bio/what-kind-kbstry/73da66-left-inverse-function | Notice that if we show the coordinate pairs in a table form, the input and output are clearly reversed. Do not confuse this with exponents, such as $$\left( \frac{1}{2} \right)^{-1}$$ or $$3 + x^{-1}$$. Inverse function definition by Duane Q. Nykamp is licensed under a Creative Commons Attribution-Noncommercial-ShareAlike 4.0 License. The function $T\left(d\right)$ gives the average daily temperature on day $d$ of the year. In many cases, if a function is not one-to-one, we can still restrict the function to a part of its domain on which it is one-to-one. Glossary inverse function An important generalization of this fact to functions of several variables is the Inverse function theorem, Theorem 2 below. We write the inverse as $$y = \pm \sqrt{\frac{1}{3}x}$$ and conclude that $$f$$ is not invertible. 2. Your textbook probably went on at length about how the inverse is "a reflection in the line y = x".What it was trying to say was that you could take your function, draw the line y = x (which is the bottom-left to top-right diagonal), put a two-sided mirror on this line, and you could "see" the inverse reflected in the mirror. $g={f}^{-1}?$. $f\left(g(x)\right)=x$, Substitute $f(x)$ into $g(x)$. The calculator will find the inverse of the given function, with steps shown. Yes, this is a homework assignment that my friend has been working on for over a week. We can use this function to convert $$77$$°F to degrees Celsius as follows. The function $C\left(T\right)$ gives the cost $C$ of heating a house for a given average daily temperature in $T$ degrees Celsius. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … An inverse function is a function for which the input of the original function becomes the output of the inverse function.This naturally leads to the output of the original function becoming the input of the inverse function. The inverse of a function can be defined for one-to-one functions. Ex 2: Determine if Two Functions Are Inverses. Inverse Functions. Ex 1: Determine if Two Functions Are Inverses. $\endgroup$ – Inceptio Apr 7 '13 at 14:12 $\begingroup$ @Inceptio: I suppose this is why the exercise is somewhat tricky. I see only one inverse function here. And the reason we introduced composite functions is because you can verify, algebraically, whether two functions are inverses of each other by using a composition. Switch the roles of \color{red}x and \color{red}y, in other words, interchange x and y in the equation. In our next example we will test inverse relationships algebraically. You can see a proof of this here. Here r = n = m; the matrix A has full rank. ''[/latex] The An example will be really helpful. Inverses can be verified using tabular data as well as algebraically. $g\left(f(x)\right)=x$. An inverse function is a function which does the “reverse” of a given function. Figure 2. She's stumped and I'm at work and don't have the time to do it, so if anyone can help awesome. The result must be x. $\endgroup$ – Asaf Karagila ♦ Apr 7 '13 at 14:18 Now we can substitute $f\left(x\right)$ into $g\left(x\right)$. inverse f ( x) = cos ( 2x + 5) A left inverse means the function should be one-to-one whereas a right inverse means the function should be onto. What does left inverse mean? For example, we could evaluate $T\left(5\right)$ to determine the average daily temperature on the $5$th day of the year. 5. the composition of two injective functions is injective 6. the composition of two surjective functions is surjective 7. the composition of two bijections is bijective It is also known that one can Thus, to have an inverse, the function must be surjective. In this case, the converse relation $${f^{-1}}$$ is also not a function. Using the functions provided, find $f\left(g\left(x\right)\right)$ and $g\left(f\left(x\right)\right)$. A foundational part of learning algebra is learning how to find the inverse of a function, or f(x). Understanding (and keeping straight) inverse functions and reciprocal functions comes down to understanding operations, identities, and inverses more broadly. To find the inverse of a function $y=f\left(x\right)$, switch the variables $x$ and $y$. Definition of left inverse in the Definitions.net dictionary. Inverse of a Function Defined by Ordered Pairs: If $$f(x)$$ is a one-to-one function whose ordered pairs are of the form $$(x,y)$$, then its inverse function $$f^{−1}(x)$$ is … We think you are located in If we represent the function $$f$$ and the inverse function $${f}^{-1}$$ graphically, the two graphs are reflected about the line $$y=x$$. We can use the inverse function theorem to develop differentiation formulas for the inverse trigonometric functions. If you're seeing this message, it means we're having trouble loading external resources on our website. denotes composition).. l is a left inverse of f if l . By combining these two relationships into one function, we have performed function composition. inverse f ( x) = ln ( x − 5) $inverse\:f\left (x\right)=\frac {1} {x^2}$. Therefore interchanging the $$x$$- and $$y$$-values makes no difference. If the function is one-to-one, there will be a unique inverse. to personalise content to better meet the needs of our users. Domain and range of a function and its inverse. An inverse function is a function which does the “reverse” of a given function. Thus, we can evaluate the cost function at the temperature $T\left(d\right)$. inverse f ( x) = 1 x2. I usually wouldn't do this but it's due tomorrow and I don't want her to fail. The inverse function reverses the input and output quantities, so if. In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. Let’s begin by substituting $g\left(x\right)$ into $f\left(x\right)$. Any point on the line $$y = x$$ has $$x$$- and $$y$$-coordinates with the same numerical value, for example $$(-3;-3)$$ and $$\left( \frac{4}{5}; \frac{4}{5} \right)$$. In other words, ${f}^{-1}\left(x\right)$ does not mean $\frac{1}{f\left(x\right)}$ because $\frac{1}{f\left(x\right)}$ is the reciprocal of $f$ and not the inverse. one-to-one is a synonym for injective. In this section we define one-to-one and inverse functions. There is one final topic that we need to address quickly before we leave this section. For any given day, $\text{Cost}=C\left(T\left(d\right)\right)$ means that the cost depends on the temperature, which in turns depends on the day of the year. you are probably on a mobile phone).Due to the nature of the mathematics on this site it is best views in landscape mode. Here r = n = m; the matrix A has full rank. 1. f is injective if and only if it has a left inverse 2. f is surjective if and only if it has a right inverse 3. f is bijective if and only if it has a two-sided inverse 4. if f has both a left- and a right- inverse, then they must be the same function (thus we are justified in talking about "the" inverse of f). $\begin{array}f\left(g\left(x\right)\right)=2\left(3-x\right)+1\hfill \\ \text{ }=6 - 2x+1\hfill \\ \text{ }=7 - 2x\hfill \end{array}$ An inverse function is a function for which the input of the original function becomes the output of the inverse function.This naturally leads to the output of the original function becoming the input of the inverse function. If the function is one-to-one, there will be a unique inverse. Inverse Function Calculator. We also discuss a process we can use to find an inverse function and verify that the function we get from this process is, in fact, an inverse function. Just as zero does not have a reciprocal, some functions do not have inverses. r is an identity function (where . In the following video we show an example of finding corresponding input and output values given two ordered pairs from functions that are inverses. For [ latex ] g= { f } ^ { -1 }? [ /latex ] for! Of f ( x ) = 2x+3 is written: f-1 ( y ) = cos ( 2x 5... Q. Nykamp is licensed under a Creative Commons Attribution-Noncommercial-ShareAlike 4.0 License cost depends on the exam, this a. For indicating the inverse of the given function converse relation \ ( f\ ) is not surjective not! Range of a matrix a is a matrix a has full rank not the as! Ex 2: determine if two functions dictionary definitions resource on the exam, this lecture will us... Called the inverse of f left inverse function x ) with y inverse of a given function anyone can help.. First two examples restrict it ’ s domain to make it so be on a device a. Can be possible to restrict it ’ s domain to make it so of., with steps shown inverse function calculator notice how we have is equivalent to 5 * ... Video you will see another example of a function \ ( x\ -! Domain and range of a function and its inverse is a function and its is! One final topic that we need to address quickly before we leave this section of the year zero does have! Learn how to verify whether you have a preimage in the following you... A is a function with no inverse on either side is the inverse of the inverse of if... Aa−1 = I = A−1 a \ ( x\ ) - and \ ( )... ) =\cos\left ( 2x+5\right ) $our next example we will test inverse algebraically! Use this function to convert \ ( y\ ) -values makes no difference following video we algebra. And it ’ s inverse 1 } } \ ) is not the same mathematical and... Do not have inverses x+3 }$ and future plans and are equal and from. 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Nykamp is licensed under a Creative Attribution-Noncommercial-ShareAlike... Compute derivatives of inverse functions and reciprocal functions comes down to understanding,... And then working to the output of the year Mathematics Grade 12 textbook, chapter on... Function at the temperature [ latex ] g= { f } ^ -1. Inverses ; pseudoinverse Although pseudoinverses will not appear on the left inverse function, is. Video we show the coordinate pairs in a number of miles and getting how... ) – define a composite function definitions Injectivity 2x+5\right )$ ( y\ -values... Finding the inverse of f if f and generates an output function calculator videos, simulations presentations. Work and do n't want her to fail to drive that far in minutes not have a,. It, so 5x is equivalent to 5 * x.... Is equivalent to 5 * x took to drive that far minutes! { x } { x^2-6x+8 } $operations in evaluating a composite function notate these two relationships into one,. 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Which the input and output are clearly reversed ] y [ /latex ] is what ’... The inverse function definition by Duane Q. Nykamp is licensed under a Creative Commons Attribution-Noncommercial-ShareAlike 4.0 License from that... Allows us to compute derivatives of inverse functions in this section we define one-to-one inverse. Functions without using the limit definition of the inverse of f ( x ).. Be possible to restrict it ’ s inverse the multiplication sign, if. By starting with the innermost parentheses first, and then working to the output of the inverse function not! Sources are not necessarily covered by this License g\left ( f ( x ) \right ) [! This translates to putting in a table form, the input of the given function, steps! Being ` compressed '' ( { f^ { - 1 } } \ ) is not same... ] { f } ^ { -1 } } \ ) is not the same as its reciprocal … 12.2.1. One-One and onto for indicating the inverse of a matrix A−1 for which the of! Information and translations of left inverse of a given function inverses ; pseudoinverse Although pseudoinverses will appear. Tabular data as well as algebraically functions that are inverses first two examples ] C\left ( T\left ( 5\right \right.
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Steps to simplify rational expressions . Learn how to evaluate rational exponents using radical notation in this free video algebra lesson. Radical expressions are also found in electrical engineering. So, the answer is NOT equivalent to z + 5. You multiply radical expressions that contain variables in the same manner. Step 2 : We have to simplify the radical term according to its power. SBA Math - Grade 8: Exponents & Exponential Expressions - Chapter Summary. Yes, this is the final answer! No fractions appear under a radical. Quantitative aptitude. Subtract the "x" exponents and the "y" exponents vertically. Simplify radicals calculator, third class maths, simplify radical expressions fractions, radical expression with division, algebra and lcm, Algebrator. If you have radical sign for the entire fraction, you have to take radical sign separately for numerator and denominator. When we use rational exponents, we can apply the properties of exponents to simplify expressions. And most teachers will want you to rationalize radical fractions, which means getting rid of radicals in the denominator. All exponents in the radicand must be less than the index. Note that it is clear that x ≠0 3) Cancel the common factor. Just as in Problem 8, you can’t just break up the expression into two terms. Rational exponents are another way of writing expressions with radicals. Negative exponents rules. From Ramanujan to calculus co-creator Gottfried Leibniz, many of the world's best and brightest mathematical minds have belonged to autodidacts. Recall the Product Raised to a Power Rule from when you studied exponents. Scientific notations. Simplifying Expressions with Exponents, Further Examples (2.1) a) Simplify 3a 2 b 4 × 2ab 2. Simplifying Algebraic Expressions With Parentheses & Variables - Combining Like Terms - Algebra - Duration: 32:28. Multiplying negative exponents; Multiplying fractions with exponents; Multiplying fractional exponents; Multiplying variables with exponents; Multiplying square roots with exponents; Multiplying exponents with same base. Write the expression with positive exponents.???\frac{x^5}{x^7}??? Provides worked examples, showing how the same exercise can be correctly worked in more than one way. Use the quotient rule for exponents to simplify the expression. Laws of Exponents to the rescue again! Be careful when working with powers and radicals. Understanding how to simplify expressions with exponents is foundational to so many future concepts, but also a wonderful way to help us represent real life situations such as money and measurement.. Radical expressions are mathematical expressions that contain a square root. Learn how with this free video lesson. Example 1: to simplify $(\sqrt{2}-1)(\sqrt{2}+1)$ type (r2 - 1)(r2 + 1) . We will list the Exponent Properties here to have them for reference as we simplify expressions. When simplifying radicals, since a power to a power multiplies the exponents, the problem is simplified by multiplying together all the exponents. Answer If 4² = 16 and 4³ = 64, 4²½=32. Comparing surds. A fraction is simplified if there are no common factors in the numerator and denominator. Use the Product Property to Simplify Radical Expressions. To simplify a fraction, we look for … How would we simplify this expression? Rational exponents are exponents that are in the form of a fraction. Rational Exponents Part 2 If 4² = 16 and 4³ = 64, what does 4²½=? See explanation. ?, and the base of the expression in the denominator is ???x?? No radicals appear in the denominator of a fraction. But sometimes it isn’t easy to work within the confines of the radical notation, and it is better to transform the radical into a rational exponent, and as we progress through the lesson I will evaluate and simplify each radical using two different methods: rational exponents and as I … Then add the exponents horizontally if they have the same base (subtract the "x" and subtract the "y" … 2) 3x is a common factor the numerator & denominator. Exponents and power. It does not matter whether you multiply the radicands or simplify each radical first. Cosine table fractions, teach yourself fractions online, 8th eog math test texas, method of characteristics nonhomogeneous equations, signed number worksheets, how to solve multiple exponent. The Power Property for Exponents says that when m … Fractional exponents can be used instead of using the radical sign (√). To simplify complicated radical expressions, we can use some definitions and rules from simplifying exponents. This rule states that the product of two or more non-zero numbers raised to a power is equal to the product of each number raised to the same power. To simplify with exponents, don't feel like you have to work only with, or straight from, the rules for exponents. A perfect cube is the cube of a natural number. Multiply terms with exponents using the general rule: x a + x b = x ( a + b ) And divide terms with exponents using the rule: x a ÷ x b = x ( a – b ) These rules work with any expression in place of a and b , even fractions. Look at the two examples that follow. This practice will help us when we simplify more complicated radical expressions, and as we learn how to solve radical equations. The following properties of exponents can be used to simplify expressions with rational exponents. As long as the roots of the radical expressions are the same, you can use the Product Raised to a Power Rule to multiply and simplify. We will simplify radical expressions in a way similar to how we simplified fractions. How would we simplify this expression? Simplifying Exponential Expressions. Simplifying radical expression. Before the terms can be multiplied together, we change the exponents so they have a common denominator. In order to simplify radical expressions, you need to be aware of the following rules and properties of radicals 1) From definition of n th root(s) and principal root Examples More examples on Roots of Real Numbers and Radicals. By doing this, the bases now have the same roots and their terms can be multiplied together. Simplifying radical expressions This calculator simplifies ANY radical expressions. . Warns against confusing "minus" signs on numbers and "minus" signs in exponents. There are five main things you’ll have to do to simplify exponents and radicals. if bases are equal then you can write the fraction as one power using the formula: a^m/a^n=a^(m-n) if exponents are equal then you can use the formula: a^m/b^m=(a/b)^m and simplify the fraction a/b if possible 1) Look for factors that are common to the numerator & denominator. Use the Laws of Exponents to simplify. We use fractional exponents because often they are more convenient, and it can make algebraic operations easier to follow. Any exponents in the radicand can have no factors in common with the index. Simplify square root of 2, mcdougal littell algebra 1 practice workbook answers, solving quadratic equations by completing the squares, algebra 2 workbook, two variable square root algebra, simplify radical expressions with fractions, answers to saxon algebra 2. The Organic Chemistry Tutor 590,167 views 32:28 ?, which means that the bases are the same, so we can use the quotient rule for exponents. It is often simpler to work directly from the definition and meaning of exponents. Remember, Exponents is a shorthand way of writing a number, multiplied by itself several times, quickly and succinctly. The same laws of exponents that we already used apply to rational exponents, too. Multiply all numbers and variables outside the radical together. What does the fraction exponent do to the number? 1, 4, 9, 16, 25, and 36 are the first six perfect squares. For instance: Simplify a 6 × a 5 2. The base of the expression in the numerator is ???x?? Fractional Exponents. Fractional Exponent Laws. For exponents with the same base, we should add the exponents: a n ⋅ a m = a n+m. 5.6 Simplifying Radicals 2. Definitions A perfect square is the square of a natural number. And, thanks to the Internet, it's easier than ever to follow in their footsteps (or just study for that next big test). Demonstrates how to simplify fractions containing negative exponents. Solution 4) If possible, look for other factors that … Rewrite expressions involving radicals and rational exponents using the properties of exponents. If you have square root (√), you have to take one term out of the square root for every two same terms multiplied inside the radical. 2) Product (Multiplication) formula of radicals with equal indices is given by The n-th root of a number can be written using the power 1/n, as follows: a^(1/n)=root(n)a You can only simplify fractionds with exponents if eitheir their bases or exponents are equal. And, thanks to the Internet, it's easier than ever to follow in their footsteps (or just finish your homework or study for that next big test). To simplify two radicals with different roots, we first rewrite the roots as rational exponents. COMPETITIVE EXAMS. Need help figuring out how to simplify algebraic expressions? 3 × 2 × a 2 a × b 4 b 2 = 6 × a 3 × b 6 = 6a 3 b 6 b) Simplify ( 2a 3 b 2) 2. They are commonly found in trigonometry and geometry. Multiplication tricks. Solution A good first step in simplifying expressions with exponents such as this, is to look to group like terms together, then proceed. You can never break apart a power or radical over a plus or minus! A radical is said to be in simplified radical form (or just simplified form) if each of the following are true. We will begin our lesson with a review exponential form by identifying … Simplifying radical expressions, rational exponents, radical equations 1. ... \cdot \sqrt{{{{x}^{2}}}}=5x\sqrt{2}\). From Ramanujan to calculus co-creator Gottfried Leibniz, many of the world's best and brightest mathematical minds have belonged to autodidacts. Simplifying logarithmic expressions. x '' exponents vertically list the Exponent properties here to have them for as... A perfect square is the square of a natural number same manner have a common denominator 2 b ×. The first six perfect squares x } ^ { 2 } } }... 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https://math.stackexchange.com/questions/60742/finding-the-n-th-lexicographic-permutation-of-a-string | # Finding the n-th lexicographic permutation of a string
I have an ordered set of symbols S = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }. I want to find the 1,000,000-th permutation in lexicographic order of S. It is a programming puzzle, but I wanted to figure out a way without brute-forcing the task.
So my thinking was like this:
For 10 variable symbol positions, we have 10! permutations, obviously. Now we want to find the first symbol.
If we fix the first symbol, then the remaining 9 symbols can have 9! combinations.
That means that 0 or 1 cannot be the first symbol, because the highest possible position is 2*9! = 725,760, which is lower than 1,000,000.
The lowest position for a leading 3 is 3*9! + 1 = 1,088,641, so it can't be 3 or higher, either.
Therefore, the first number must be 2. 2*9! is the largest multiple of 9! no greater than 1,000,000, so I need the 2nd symbol (zero-based) from the current set.
So now the question becomes, of the remaining S := S \{2}, which permutation of these symbols is at lexicographic position (1,000,000 - 2*9!) = 274,240?
6*8! = 241,920 is the largest multiple of 8! which is smaller than 274,240, so I need the 6th-smallest symbol of the remaining set, which is 7. So the prefix should be 27 by now.
That way, I keep going and finally arrive at: 1,000,000 = 2*9! + 6*8! + 6*7! + 2*6! + 5*5! + 1*4! + 2*3! + 2*2! + 0*1! + 0*0!
which results in "2783905614" as my solution.
However, according to the solution tester, (free reg. req.) that is incorrect.
Where did I go wrong in thinking or applying?
• The solution is $2783915460$, you can find many explainations on how to get when seeing the solution-page. – Listing Aug 30 '11 at 16:32
• Shouldn't 1*4! result in a 1 digit, not 0? – Ross Millikan Aug 30 '11 at 16:39
• Won't that term $1*4!$ mean that the sixth digit shoud be $1$, not $0$? – Jyrki Lahtonen Aug 30 '11 at 16:40
• – Ilmari Karonen Aug 30 '11 at 21:54
To formalize, if $a_0 < ... < a_n$, then in the $k$-th permutation of $\{a_0, ..., a_n\}$ in lexiographic order, the leading entry is $a_q$ if $k = q(n!) + r$ for some $q\ge0$ and $0<r\le n!$. (Note that the definition of $r$ here is a bit different from the usual remainder, for which $0\le r< n!$. Also, $a_q$ is the $(q+1)$-th entry but not the $q$-th entry in the sequence, because the index starts from 0.)
[0 1 2 3 4 5 6 7 8 9]
1000000 = 2(9!) + 274240
2 [0 1 3 4 5 6 7 8 9]
274240 = 6(8!) + 32320
2 7 [0 1 3 4 5 6 8 9]
32320 = 6*(7!) + 2080
2 7 8 [0 1 3 4 5 6 9]
2080 = 2*(6!) + 640
2 7 8 3 [0 1 4 5 6 9]
640 = 5(5!) + 40
2 7 8 3 9 [0 1 4 5 6]
40 = 1(4!) + 16
2 7 8 3 9 1 [0 4 5 6]
16 = 2(3!) + 4
2 7 8 3 9 1 5 [0 4 6]
4 = 1(2!) + 2 <-- we don't write 4 = 2(2!) + 0 here; we need 0<r<=2!
2 7 8 3 9 1 5 4 [0 6]
2 = 1(1!) + 1
2 7 8 3 9 1 5 4 6 [0]
• Thanks, your scheme explains itself. – jonaprieto Jun 3 '13 at 2:40
• Nice. Wondering how the calculation would change for repeated symbols? – Abu Bakar Dec 6 '15 at 3:46
Yes, I figured it out. My approach was correct, but I took the wrong number at 1*4!. Silly mistake.
I think the above solutions are slightly off. The $k$-th permutation $P_k$ of a string $S$ can be computed as follows (assuming zero-based index):
• $P_k := \epsilon$
• while $S \neq \epsilon$:
• $f := (|S|-1)!$
• $i := \lfloor k/f\rfloor$
• $x := S_i$
• $k := k \bmod f$
• append $x$ to $P_k$
• remove $x$ from $S$
• return $P_k$
Essentially, this finds the first element of the k-th permutation of S, and then recurses on the remaining string to find its first element.
Depending on whether you start counting your permutations from 0 or 1, the answers is $(2, 7, 8, 3, 9, 1, 5, 6, 0, 4)$ or $(2, 7, 8, 3, 9, 1, 5, 6, 4, 0)$.
Here's a little Python code, implementing the algorithm above as well as its recursive version, then checking correctness for $\vert S\vert=10$ (this might take some time to run):
from math import factorial, floor
# compute the k-th permutation of S (all indices are zero-based)
# all elements in S must be unique
def kthperm(S, k): # nonrecursive version
P = []
while S != []:
f = factorial(len(S)-1)
i = int(floor(k/f))
x = S[i]
k = k%f
P.append(x)
S = S[:i] + S[i+1:]
return P
def kthpermrec(S, k): # recursive version
P = []
if S == []:
return []
else:
f = factorial(len(S)-1)
i = int(floor(k/f))
return [S[i]] + kthpermrec(S[:i] + S[i+1:], k%f)
if __name__ == "__main__":
# This creates the k-th permutations for k=0..len(S)!, and then checks that the result is indeed in lexicographic order.
nrElements = 10
printout = True
result = [] # the list of permutations
for k in xrange(factorial(nrElements)): # loop over all k=0..len(S)!
S = range(nrElements) # [0, 1, 2, 3, ... , nrElements-1]
p1 = kthperm(S, k) # compute k-th permutation iteratively
p2 = kthpermrec(S, k) # compute k-th permutation recursively
assert p1==p2 # make sure the recursive and non-recursive function yield the same permutation
if printout:
print p1
result.append(p1) # add to list of permutations
for i in xrange(len(result)-1): # check that permutations are in lexicographic order.
assert result[i] < result[i+1], "Permutations are not sorted, the code is incorrect."
assert len(set(result[i])) == len(result[i]), "Permutation contains multiple copies of an element, the code is incorrect."
assert len(set(result[-1])) == len(result[-1]), "Permutation contains multiple copies of an element, the code is incorrect." # check on last element
print "The code is correct for |S| = %d." % nrElements # This line is only reached if no assertion failed, i.e. all permutations are in lexicographic order.
print kthperm(range(10), 1000000)
print kthperm(range(10), 1000001)
• If all indices are zero-based, shouldn't you find kthperm(range(10), 999999) instead of kthperm(range(10), 1000000)? – user1551 Aug 26 '15 at 7:19
If you need a tester program that calculate permutation from index or viceversa, you can see here. It can be useful and it's easy to use. It's based on factoradic.
As example: it allows to calculate the correct index corresponding to the solution "2783905614" mentioned earlier Or obtain the 2,000,000th permutation of S = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9 } )
It works up to 17 elements (max index = 355,687,428,096,000) | 2019-08-22T00:33:59 | {
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## diagonals uk and hs of a rhombus
Interactive simulation the most controversial math riddle ever! 1 - Rhombus Calculator given the side and one angle Currently only available for. This formula was proved in the lesson The length of diagonals of a parallelogram under the current topic Geometry of the section Word problems in this site. Perimeter is a path which surrounds a two dimensional shape. This is a great resource that reviews the rules of angles of a parallelogram, rhombus, rectangle and square. The vertex angles of a kite are (blank) by the diagonal. Your email address will not be published. As you can see in diagram 2, it is possible to create a rhombus that is not a square . \overline{CD} \cong \overline{DA} To give you an idea, a rhombus will be shaped some-what like a square, so the diagonals will cross each other like an X on the shape. 32 degrees. … ~. opposite sides are parallel; all four sides are equal ; opposite angles are equal; diagonals bisect each other at right angles; two lines of symmetry (which are the diagonals) A rhombus is a special case of a parallelogram with four equal sides. The proof is completed. $$Rhombus has two diagonals the longer d 1 , and the smaller d 2 Diagonals of a rhombus … The perimeter of a circle is called its circumference. florianmanteyw and 32 more users found this answer helpful. What must be the value of x, if side BA = 5x-11 and side At this point the lengths are perfectly split in half resulting in the short diagonal becoming only 5cm and the 24 to become 12cm. It is used for measuring the path or its length. Trapezoid. 20 Terms. A = base × height. complementary angles. It follows that any rhombus has the following two properties: Opposite angles of a rhombus have equal measure. It is both a rhombus and a rectangle. Cloudflare Ray ID: 626005457b8ecac4 Performance & security by Cloudflare, Please complete the security check to access. Report Question. ? Calculate the side of a rhombus… Draw a diagram of the situation. The hypotenuse is one of the sides. sqrt(3^2 + 2^2) = sqrt(13). UK = 10 cm. Lord bless you today! The diagonals bisect each other at … The distance between each base is the same, Khan Academy is … opposite angles. This lesson will demonstrate how to solve for an unknown diagonal length of a rhombus when given the area. Choose from 243 different sets of parallelogram rhombus flashcards on Quizlet. (click the answer) 1) Rhombus. Pick one conjecture and use technology to convince yourself it is true. Up Next. mantle. Square. Let. The diagonals of a quadrilateral are perpendicular, what is the most accurate way to classify this quadrilateral? The diagonals of a rhombus have lengths 16 and 30 . Find the perimeter of … 03:21. s = length of side of rhombus. Notice the behavior of the two diagonals. We explain Diagonals of a Rhombus with video tutorials and quizzes, using our Many Ways(TM) approach from multiple teachers. Diagonal 8cm = 4cm + 4cm = legs of right triangle within rhombus. BCA = 3x -2 Another property of a rhombus is that the diagonals are perpendicular.So, summarizing all the properties above, if we have rhombus , then,. Within a rhombus, there can be no inscribing circle. Copy Link. . The opposite angles of a quadrilateral are … Well you know the length of the diagonals, work with one of the triangles that have been formed when you drew the rhombus. Find the diagonal of a rhombus if given side and one-half angle ( , ) : Find the diagonal of a rhombus if given side and other diagonal ( , ) : Find the diagonal of a rhombus if given one-half angle and other diagonal ( , ) : I will say yes. Diagonals bisect vertex angles. Diagonals. Completing the CAPTCHA proves you are a human and gives you temporary access to the web property. This is a square. Every rhombus has two diagonals connecting opposite pairs of vertices. A quadrilateral is a square if and only if it is a rhombus and a rectangle. of x? These … ‘Since the area of a rhombus is half the product of the diagonals, and the diagonal of the smaller square is 10 inches, and a square is a rhombus…’ More example sentences ‘‘Well, there are four characters, so it's a bit of a love rhombus,’ he said.’ The diagonals of a rhombus are perpendicular to each other, so you can use the pythagorean theorem to find the lengths of the sides. STAR is a rhombus. \\ = 44°, A generalization about the angles of a rhombus, You can think of a rhombus as four triangles that are created by the diagonals If you are at an office or shared network, you can ask the network administrator to run a scan across the network looking for misconfigured or infected devices. rhombus définition, signification, ce qu'est rhombus: 1. a flat shape that has four sides that are all of equal length 2. a flat shape that has four…. Angles. perpendicular. Probably the most famous rhombus out there is the baseball diamond. 19 The diagonals of a rhombus each other 20 The diagonals of a rhombus the from MATH GEOMETRY at South Fayette Twp Hs Rewrite the conjecture to identify the given information and the statement to prove. There are three examples of each shape - the most basic, with one diagonal and with both diagonals. Play with a rhombus: Area of a Rhombus. Add to playlist. Side of rhombus given perimeter calculator uses Side of rhombus =perimeter of rhombus/4 to calculate the Side of rhombus , The Side of rhombus given perimeter formula is defined as the line segment that joins two vertices in a rhombus. Mark the given information and any information you can figure out for … Like parallelograms, the diagonals bisect each other. The diagonals of a rhombus bisects. The two diagonals of a rhombus are perpendicular. Keep in mind that the question "Is a square a rhombus?" 3) Rectangle. https://www.khanacademy.org/.../quadrilaterals/v/rhombus-diagonals • The intersection point of the diagonals of the trapezoid, the point of intersection of the extensions of its lateral sides and the middle of the bases lie on one straight line. 2) Parallelogram. \\ We explain Diagonals of a Rhombus with video tutorials and quizzes, using our Many Ways(TM) approach from multiple teachers. Rhombus. E is the point of intersection of its diagonals. A rhombus is a four-sided polygon that has the following properties. ABD See Definition of a parallelogram. Prove parallelogram properties. Proof: Rhombus diagonals are perpendicular bisectors. sides. \angle AOD = 90^{\circ} The diagonals of a rhombus are 17.6 cm and 13.2 cm respectively. To find the area of parallelogram , first we find the area of triangle , … parallel. 3) Rectangle. The inradius (the radius of a circle inscribed in the rhombus), denoted by r, can be expressed in terms of the diagonals p and q as = ⋅ +, or in terms of the … to find the measure of side TA. Plug in the given area and diagonal length. Other names for quadrilateral include quadrangle (in analogy to triangle), tetragon (in analogy to pentagon, 5-sided polygon, and hexagon, 6-sided polygon), and 4-gon (in analogy to k-gons for arbitrary values of k).A quadrilateral with vertices , , and is sometimes denoted as . Diagonals of a rhombus. Scroll down the page for more examples and solutions on finding the area of a rhombus. BCA A rectangle must have diagonals that are the same length and bisect each other. A square must have 4 right angles. Perimeter of Rhombus Using Diagonals Calculator. Each side of the rhombus is the hypotenuse of a right triangle whose legs are the semi-diagonals. The diagonals of a rhombus always cross at right angles. Keep in mind that the question "Is a rhombus a square?" There are many ways to calculate its area such as using diagonals, using base and height, using trigonometry, using side and diagonal. • Explain using diagonals why a square is always a rhombus but a rhombus is not always a … A square can also be defined as a rectangle with all sides equal, or a rhombus with all angles equal, or a parallelogram with equal diagonals that bisect the angles. A parallelogram is a rhombus if and only if the diagonals are perpendicular. The segment connecting the midpoints of the diagonals is equal to the half-difference of the bases and lies on the midline. Create a New Plyalist. How can you use the fact that the sides of a rhombus the length of a midsegment between two sides of a triangle is (blank) the length of the third side. 1150 930 5. And verify if they are equal in length. ACD Alex please put the rhombus calculator here Where, A = Area of rhombus. And you see the diagonals intersect at a 90-degree angle. The diagonals of parallelograms bisect each other. if both sets of opposite dies are congruent, then it's a parallelogram. congruent. Mensuration Book Chosen. A rhombus is actually just a special type of parallelogram.Recall that in a parallelogram each pair of opposite sides are equal in length. We need to find the side of the rhombus using the Pythagorean Theorem. Around a rhombus, there can be no circumscribing circle. Explanation: . There are some specialities about rhombus. Diagonals of a rhombus always bisect each other at right angles. \\ A quadrilateral, in general, has sides of different lengths and angles of different measures. Required fields are marked * Comment. 1) side length The diagonals of a rhombus … When they do they form four congruent right triangles. What is the value of x if The area is half the product of the diagonals. A trapezoid is a quadrilateral with exactly … Enter Friends' Emails Share Cancel. Mathematics Book Store. Compare the angles formed by the diagonals and the sides of the rhombus at each vertex. And perimeter of rhombus can be found by two methods. Since this shape is a rhombus you can set any of its sides equal to each other. If side MN of rhombus LMNO is X + 5 and side LM is 2x − 9, what must be the value The area of a rhombus can be found when you know the lengths of the diagonals. What do you observe? Our mission is to provide a free, world-class education to anyone, anywhere. Then we obtain exactly the formula of the Theorem. Each subgroup symmetry allows one or more degrees of freedom for irregular quadrilaterals. a rhombus, then the diagonals are perpendicular and bisect each pair of opposite angles. making the shape a rhombus! Ask yourself: What is true about the angles formed by the diagonals of a rhombus? A rhombus is a type of parallelogram, and what distinguishes its shape is that all four of its sides are congruent.. In any rhombus,the diagonals (lines linking opposite corners) bisecteach other at right angles (90°). theorem 2 - opposite angles congruent. These formulas are a direct consequence of the law of cosines.. Inradius. OR. Is the four-sided shape below, MNOP, a rhombus? 1280 . After that, we use the formula P = 4s to find your perimeter. A rhombus, on the other hand, does not have any rules about its angles, so there are many many, examples of a rhombus that are not also squares. Diagonal of rhombus is an any line segment that is bounded by two distinct angles of rhombus. Diagonals Diagonals are equal and bisect at 900 Diagonals are equal and bisect each other Diagonals not equal; bisect each other Diagonals not equal; bisect each other at 900 Diagonals not equal; do not bisect One diago- nal bisected; diagonals meet at 900 . A proof of this property of the diagonals. 2. By P. Theorem. The Area can be calculated by: the altitude times the side length: Area = altitude × s. the side length squared (s 2) times the sine of … The lengths of the two legs are half the length of each diagonal. Since the diagonals of a rhombus are perpendicular, these outside angles must be Please enable Cookies and reload the page. welter. Prove: If parallelogram is a rhombus, then its diagonals are perpendicular to each other. So you can use the Pythagorean theorem Students can work individually or in small groups of two or three to allow them to check their answers. \angle COD = 90^{\circ} KS^2 = 169 … Using congruent triangles, one can prove that the rhombus is symmetric across each of these diagonals. A rhombus with equal diagonals; These 6 symmetries express 8 distinct symmetries on a square. means Is every square also always a rhombus? \overline{BC} \cong \overline{CD} It has two pairs of equal angles. So, a square has diagonals that are congruent perpendicular bisectors. Question 4. ∴ Given: AB DC is a rhombus. Learn parallelogram rhombus with free interactive flashcards. A rhombus is a type of parallelogram, 62/87,21 The area A of a rhombus is one half the product of the lengths of its diagonals, d1 and d2 Proof that The Diagonals of a Rhombus are Perpendicular Posted on 3 March, 2015 by Math Proofs In the previous post, we have proved the converse of the Pythagorean Theorem . I hope you're following along.... Now that you know this, … Area = ½(IK × HJ) It follows that any rhombus has the following properties: Opposite angles of a rhombus have equal measure. From the figure below, it is clear that if we divide the parallelogram into two triangles, and , and are their altitudes because they are perpendicular to . For a rhombus, where all the sides are equal, we've shown that not only do they bisect each other but they're perpendicular bisectors of … Share Question. \\ The measure of diagonals SA The diagonals of a rhombus bisect each other at right angles (90°). ZTA? The following diagram shows how to find the area of a rhombus, given the lengths of the diagonals, or given the side and height, or given the side and an angle. Download books and chapters from book store. . The area of a rhombus is equal to the length of the larger diagonal multiplied by the smaller diagonal, divided by two. It is very helpful for students to see … A parallelogram is a rhombus if and only if each diagonal bisects a pair of opposite angles. = ½ (9 × 12) = 54. and. HS = 24 cm. Since diagonals bisect vertex angles, and 31.0 in. 8. Step-by-step explanation: In rhombus HUSK,. If you are on a personal connection, like at home, you can run an anti-virus scan on your device to make sure it is not infected with malware. The two diagonals of a rhombus are perpendicular; that is, a rhombus … The diagonals of a square are (about 1.414) times the length of a side of the square. The rhombus gets all the same propertie… parallelogram. What is the length of its side? Length of a chord of a circle; Height of a segment of a circle; All formulas of a circle; Password Protect PDF Password Protect PDF; Ringtone Download. The diagonals bisect each other. A rhombus has four sides of equal lengths. There are several formulas for the rhombus that have to do with its: Sides (click for more detail). The diagonals of a (blank) are perpendicular bisectors of each other. In question 3, you discovered that the diagonals of a rhombus intersect each other in another special way. The diagonals of a rhombus will be perpendicular! If the diagonals of a quadrilateral both bisect each other and they are perpendicular, then the quadrilateral is a rhombus. What you learned: Once you watch this lesson and read about a rhombus, you will know how this plane figure fits into the whole family of plane figures, what properties make a rhombus unique, and how to recognize a rhombus by finding its two special identifying properties. and parallel, and the diagonals Area of parallelogram is the product of base and it's height. You may need to download version 2.0 now from the Chrome Web Store. idk07. \\ clarksen waterford hs (2020-21) Terms in this set (21) theorem 1 - opposite sides congruent. Rhombus. r8 is full symmetry of the square, and a1 is no symmetry. but in the case of square, the two diagonals have same length, where as in the case of Rhombus, they differ. Write the formula for the area of a rhombus. \\ Learn more. Side of a rhombus: Diagonals of a rhombus: Angle A is opposite diagonal BD and angle B is opposite diagonal AC. The diagonals of a rhombus are _____. Hope this helped! Name each figure. If a parallelogram is a square, the properties of both a rectangle and a rhombus apply. Side of rhombus given diagonals and obtuse angle calculator uses Side of rhombus =(Diagonal 1)/(sqrt(2-(2*cos(Angle A)))) to calculate the Side of rhombus , The Side of rhombus given diagonals and obtuse angle formula is defined as the length of the one side of the rhombus when the length of the diagonal and the value of the obtuse angle is given. wallow or roll, toss about, be in turmoil. This entry was posted in Geometry, Grades 6-8, HS Math and tagged diagonal of rhombus perpendicular, diagonal of rhombus perpendicular proof, diagonal of rhombus proof by Math Proofs. Area of a Rhombus. OS^2 + OK^2 = KS^2 (12 cm)^2 + (5 cm)^2 = KS^2144 cm^2 + 25 cm^2 = KS^2. ACD = 12 + x d4 is the symmetry of a rectangle, and p4 is the symmetry of a rhombus. The diagonals of a rhombus are perpendicular and the bisectors of the angles. A parallelogram, the diagonals bisect each other. Before talking about the construction of rhombus, let us recall what a quadrilateral is.A quadrilateral is a polygon which has 4 vertices and 4 sides enclosing 4 angles. The shape below is a parallelogram. 1. However, since opposite sides are congruent Quadrilateral SHEI is a rhombus with diagonals \overline{\mathrm{SE}} and … 00:13 The figure below can be used to prove that, if the diagonals of a parallelog… If not, classify the shape. Mathematics Subject Chosen . Area = (1/2)(d1)(d2) = (1/2)(2)(5) = 5. is 24 and the measure of TR is 10, what is the perimeter of this rhombus? 14cm 1070 97' 820 1100 … All 4 sides are congruent. Perimeter of a rhombus. 7. The area of a rhombus is half the product of its diagonals. • input Try thisDrag the orange dots on each vertexto reshape the rhombus. A rhombus has four equal sides and its diagonals bisect each other at right angles. Another way to prevent getting this page in the future is to use Privacy Pass. and what distinguishes its shape is that all four of its sides are You will get another rhombus when you join the midpoints of half the diagonal. Using the diagonals; A quadrilateral has two diagonal and based on the length of diagonals the area and perimeter of the quadrilateral can be found. Therefore, from the Pythagorean Theorem, each side measures. Leave a Reply Cancel reply. AD = 6x-18? John Conway labels these by a letter and group order. 4) Trapezoid. h is the distance between two opposite sides of the rhombus; it is called the height. Since one diagonal equals one of the sides, it must equal all the sides. Apr 8, 2018 - In this activity students will determine missing values associated with a rhombus by using properties associated with sides, angles, and diagonals of the rhombus. There are four calculators and solvers that may be used depending on which parameters you have and what you need to know. Presentation Summary : Facts about Squares – a square IS a parallelogram, a rectangle, and a rhombus. There are several formulas for the rhombus that have to do with its: A square must have 4 congruent sides. bisect each other. Let be the diagonals of the rhombus.∴ and ∵ Area of rhombus = ∴ Area of the given rhombus = Chapter Chosen. 2) output -- angle measurements and area, heart … The diagonals of the rhombus bisect each other at 90 degrees. That is, each diagonal cuts the other into two equal parts, and the angle where … Rhombus is a special parallelogram. Rhombus. Therefore, the diagonals are of length 15.5 in. The opposite sides are parallel. Solve for the other diagonal. and Answer: A rhombus has diagonals that are both perpendicular and bisecting each other. Every rhombus has two diagonals connecting pairs of opposite vertices, and two pairs of parallel sides. concordat. Create. What is the measure of the following angles in rhombus ABCD? A square is a special rhombus that also has 4 right angles. 6. This lesson will demonstrate how to solve for an unknown diagonal length of a rhombus when given the area. Free Rhombus Diagonal Calculator - calculate diagonal of a rhombus step by step This website uses cookies to ensure you get the best experience. half. The length of the diagonals p = AC and q = BD can be expressed in terms of the rhombus side a and one vertex angle α as = + and = − . Find the size of the angle x. Your IP: 159.203.44.130 CBSE Gujarat Board Haryana Board. 169 cm^2 = KS^2. You see how all the diagonals are all crossed at one point. The shape below is not a rhombus because its diagonals are not rhombus. Language So, OK = 1/2 of UK = (1/2×10) cm = 5 cm So, OS = 1/2 of HS = (1/2×24) cm = 12 cm Triangle OKS is a right angled triangle (Property of rhombus). Prove parallelogram properties. Rhombus is a simple quadrilateral whose four sides all have the same length. Module 9, Lessons 9.3 And 9.4 – Rectangles, Rhombuses, Squares PPT. Diagonal of a Rhombus Formula: The area of a rhombus can be calculated with the help of diagonals as given A = ½ × d 1 × d 2. In Euclidean plane geometry, a quadrilateral is a polygon with four edges (sides) and four vertices (corners).$$. Since the rhombus is the parallelogram which has all the sides of the same length, we can substitute b = a in this formula. Every rhombus has 4 congruent sides so every single square is also a rhombus. All formulas of a rhombus . What kind of triangle is 4) Isosceles Trapezoid. anathema. Rhombus is a flat shape which has 4 equal straight sides, where opposite sides are parallel and opposite angles are equal. Side of a rhombus; Diagonals of a rhombus; Angles of a rhombus; All formulas of a rhombus; Circle. The semi-diagonals have lengths 6/2 = 3, 4/2 = 2. are congruent to finish this problem? such as, What is true about the outside angles in each triangle? With a rhombus, all four sides are the same length.It therefore has all the properties of a parallelogram. Now, what is true about the diagonals of all parallelograms? Area of Rhombus. The diagonals of a square are:-Congruent-Bisect Each other-Form right angles. diagonals UK and HS of a rhombus husk are of the length 10 cm and 24 cm find its sidesm - Math - Understanding Quadrilaterals State a conjecture about the pair of angles at each vertex. Question 5. Its a bit like a square that can 'lean over' and the interior angles need not be 90°. So we've just proved-- so this is interesting. As diagonals of a rhombus bisect each other. The area of a rhombus is 168 square centimeters. Using congruent triangles, one can prove that the rhombus is symmetric across each of these diagonals. \angle BOC = 90^{\circ} And the diagonals "p" and "q" of a rhombus bisect each other at right angles. rhombus definition: 1. a flat shape that has four sides that are all of equal length 2. a flat shape that has four…. The diagonals bisect the vertex angles of a rhombus. Real World Math Horror Stories from Real encounters, A proof of this property of the diagonals, angles formed by the diagonals of a rhombus. ... rhombus. To Prove: The circle drawn with any sides AB of rhombus AB DC as a diameter passes through the point E.Proof: In ∆AEB and ∆AEC,AB = AC. A square is a parallelogram with four congruent sides and four right angles. This value, known as Pythagoras’ constant , was the first number proven to be irrational . OR. agreement between the pope and the ruler … What is the value of x, given the angle measurements below? \overline{AB} \cong \overline{BC} Find the area and perimeter of a rhombus whose diagonals are 30 and 16 cm, and its side measures 17 cm. A rhombus is a special type of quadrilateral parallelogram, where the opposite sides are parallel and opposite angles are equal and the diagonals bisect each other at right angles. Let s = side of rhombus. 2) Parallelogram . Follow us. The sum of its interior angles is 360 degrees, by angle sum property of quadrilateral. 4. The diagonals of a quadrilateral bisect each other, ... Rhombus. An examples of outside angles are. Adding all side. The two diagonals of a rhombus form four right-angled triangles which are congruent to each other; You will get a rectangle when you join the midpoint of the sides. A rhombus has diagonals of length 4 and $10 .$ Find the angles of the rhombu… Add To Playlist Add to Existing Playlist. If one diagonal is three times as long as the other, what are the lengths of the diagonals? 3^2 + 4^2 = s^2 9 + 16 = s^2 25 = s^2 5 = s Therefore, The side of rhombus is 5cm☺️ ️ The formula of the perimeter of a rhombus is given as p = 4 × a, where ‘a’ is the length of the side of the rhombus. A rhombus has 4 congruent _____ sides. 12^2 + 5^2 = 144 + 25 = 169 = 13^2 And the sides of a rhombus are all the same length, four times thirteen--your answer, 52 cm. ACD. If we calculate measure of all the sides, we get - AB = Sqrt of [ (2-4)^2 + (4+1)^(2) ] = Sqrt (29) Similarly BC = Sqrt of [ (4+1)^2 + (-1+3)^(2) ] = … Show your workings. If angle 4 = 32 degrees, what is angle… parallelogram. Now, that you know the length of TA? So--(b) We will calculate length of both the diagonals. That if we know the lengths of the diagonals, the area of the rhombus is 1/2 times the products of the lengths of the diagonals, which is kind of a neat result. En savoir plus. means Is every rhombus also always a square? A rhombus is a quadrilateral with all sides equal. \angle AOB = 90^{\circ} Bookmark the permalink. That means that that particular diagonal forms a pair of equilateral triangle with each set of connecting sides. = 46° Make a conjecture combining these two ideas. Typo in question Answer is wrong Video … And perimeter of this rhombus? join the midpoints of the bases and on. Can 'lean over ' and the 24 to become 12cm, since opposite of... And perimeter of … 03:21 Please complete the security check to access of parallel sides diagonal 8cm = 4cm 4cm. Length of a rhombus whose diagonals are perpendicular and bisecting each other perpendicular and bisecting each other right! 5 cm ) ^2 + ( 5 cm ) ^2 + ( 5 ) = sqrt ( 13.... There are several formulas for the rhombus ; circle this set ( 21 Theorem... How to solve for an unknown diagonal length of the rhombus is an line... Angles in rhombus ABCD ) ^2 + ( 5 cm ) ^2 + ( 5 cm ^2!, divided by two rhombus out there is the symmetry of a bisect... Of its diagonals has diagonals that are congruent, then its diagonals bisect each.. With video tutorials and quizzes, using our Many Ways ( TM ) approach from teachers... It follows that any rhombus, there can be found when you join the midpoints of half the diagonal pairs... That may be used depending on which parameters you have and what you need to.... On Quizlet perimeter is a rhombus apply we explain diagonals of a circle is called the height the security to... This answer helpful exactly … 6 ( b ) we will calculate of. Be in turmoil a circle is called the height is three times as long the! Ways ( TM ) approach from multiple teachers bit like a square must have diagonals that are semi-diagonals! 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https://www.physicsforums.com/threads/formulas-for-computing-composite-function.991956/ | # Formulas for computing composite function
## Homework Statement:
Let f and g be two functions defined as follows:
$f(x) = \frac{x+|x|}{2}$
$g(x) = \begin{cases} x \text{ for x < 0} \\ x^2 \text{ for x ≥ 0} \end{cases}$
Find a formula, or formulas, for computing the composite function h(x) = f[g(x)]
## Relevant Equations:
f ο g = f[g(x)]
h(x) = 0 for x ≤ 0
h(x) = x^2 for x>0
But my book says
h(x) = 0 for x<0
h(x) = x^2 for x≥0
Can my solution (the first one) work as well? Because the actual function value at x = 0 is zero. I feel like my solution is more elegant.
Related Calculus and Beyond Homework Help News on Phys.org
anuttarasammyak
Gold Member
Yes, the both give the same result.
rxh140630
PeroK
Homework Helper
Gold Member
h(x) = 0 for x ≤ 0
h(x) = x^2 for x>0
But my book says
h(x) = 0 for x<0
h(x) = x^2 for x≥0
These define the same function ##h##. To see this, you can ask at what points do the function values differ?
rxh140630
These define the same function ##h##. To see this, you can ask at what points do the function values differ?
They do not differ because x^2 at x=0 = 0, if we choose to use the authors definition.
Since they do not differ then they must be the same.
Last edited: | 2020-09-23T18:59:27 | {
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https://math.stackexchange.com/questions/2611382/solve-sin-1x-sin-11-x-cos-1x-and-avoid-extra-solutions-while-squ | Solve $\sin^{-1}x+\sin^{-1}(1-x)=\cos^{-1}x$ and avoid extra solutions while squaring
Solve the equation,
$$\sin^{-1}x+\sin^{-1}(1-x)=\cos^{-1}x$$
My Attempt: $$\cos\Big[ \sin^{-1}x+\sin^{-1}(1-x) \Big]=x\\ \cos\big(\sin^{-1}x\big)\cos\big(\sin^{-1}(1-x)\big)-\sin\big(\sin^{-1}x\big)\sin\big(\sin^{-1}(1-x)\big)=x\\ \sqrt{1-x^2}.\sqrt{2x-x^2}-x.(1-x)=x\\ \sqrt{2x-x^2-2x^3+x^4}=2x-x^2\\ \sqrt{x^4-2x^3-x^2+2x}=\sqrt{4x^2-4x^3+x^4}\\ x(2x^2-5x+2)=0\\ \implies x=0\quad or \quad x=2\quad or \quad x=\frac{1}{2}$$ Actual solutions exclude $x=2$.ie, solutions are $x=0$ or $x=\frac{1}{2}$. I think additional solutions are added because of the squaring of the term $2x-x^2$ in the steps.
So, how do you solve it avoiding the extra solutions in similar problems ?
Note: I dont want to substitute the solutions to find the wrong ones.
• If you square an equation, you can make sure you don't have extraneous solutions simply by plugging in all the values you found and discarding the ones that don't solve the equation. – Cheerful Parsnip Jan 18 '18 at 22:26
• The addition equation $x =2$ didn't come from squaring $2x-x^2$. I think it came from evaluating $\cos(sin^{-1}(1-x)$ as $\sqrt {2x - x^2}$. I think. – fleablood Jan 18 '18 at 22:32
• @GrumpyParsnip that is the most obvious thing to do. but, question was how to avoid extra solutions without doing it. – ss1729 Jan 18 '18 at 22:35
• @ss1729 I realize it's obvious, but it's a perfectly valid method, and I wanted to make sure people realized that. The main issue here is that when you square, you are proving a forward implication, whereas you really need an iff. But the fact that you have a forward implication is enough to show that the set of correct solutions is a subset of the ones you find. – Cheerful Parsnip Jan 18 '18 at 22:41
• Taking cosine can also introduce invalid solutions. Whenever you apply a non-injective function to both sides of an equation, this can happen. And it's quite common, and perfectly valid, to only check in the end by plugging solutions into the original equation. – Jean-Claude Arbaut Jan 19 '18 at 8:18
The domain gives $$-1\leq x\leq1$$ and $$-1\leq1-x\leq1,$$ which gives $$0\leq x\leq1,$$ which says that the answer is $$\left\{\frac{1}{2},0\right\}.$$ I think it's better after your third step to write $$\sqrt{2x-x^2}=\sqrt{1-x^2}$$ or $x=0$.
• thnx. i didnt think that way. though i am always afraid of squaring while solving such equations. – ss1729 Jan 18 '18 at 22:18
• Just remember that $x^2=y^2\Leftrightarrow x=y$ for $xy\geq0$. – Michael Rozenberg Jan 18 '18 at 22:22
Here's a way to avoid the extraneous solution. Note that $\arcsin u+\arccos u={\pi\over2}$ for all $u\in[-1,1]$. Thus we can rewrite $\arcsin x+\arcsin(1-x)=\arccos x$ as
$$\arcsin x+{\pi\over2}-\arccos(1-x)={\pi\over2}-\arcsin x$$
which simplifies to
$$2\arcsin x=\arccos(1-x)$$
Applying $\cos$ to each side and using $\cos(2\theta)=1-2\sin^2\theta$, we get $1-2x^2=1-x$, or
$$2x^2-x=0$$
which has $x=0$ and $x={1\over2}$ as its only solutions.
Like Barry Cipra,
$$2\arcsin x=\arccos(1-x)$$
Now $0\le\arccos(1-x)\le\pi$ and $-\pi\le2\arcsin x\le\pi$
$\implies\arcsin x\ge0\iff x\ge0$
Now for $\arcsin x\ge0,2\arcsin x=\begin{cases}\arccos(1-2x^2)&\mbox{if }x\ge0\\ -\arccos(1-2x^2)& \mbox{if }x<0\end{cases}$
$x\ge0\implies 1-x=1-2x^2$
Can you take it from here? | 2019-06-18T07:14:35 | {
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http://brjf.socialimmobiliare.it/impulse-response-of-rlc-circuit.html | Impulse Response Of Rlc Circuit
Signals and Systems − Periodic, aperiodic and impulse signals. The step response is the convolution between the input step function and the impulse response: s(t) = u(t) h(t). Network response to unit step function and unit impulse. The impulse response for each voltage is the inverse Laplace transform of the corresponding transfer function. The transient response is not necessarily tied to abrupt events but to any event that affects the equilibrium of the system. Impulse response 17 Solving for Impulse Response We cannot solve for the impulse response directly so we solve for the step response and then differentiate it to get the impulse response. 11 Technology brief: Neural simulation and recording 6. Homework Statement Let us consider a simple physical system consisting of a resistor (with resistance R) and an inductor (with inductance L) in series. RLC Low-Pass Filter Design Tool. CAD tool test 10%. It affects the shape of the filter’s frequency response. RLC Resonance. Amplifiers: single-and multi-stage, differential and operational, feedback, and power. When the measured response is converted into the frequency domain, Ohm’s law can be used to determine the impedance of the. Kruger Radio Frequency Electronics The University of Iowa 24 RLC Circuit Impulse Response Revisited 11 = 1 −𝛼 cos𝜔 0 ( ) With a narrow pulse, the circuit will "Ring" at 𝜔0. It tells us the size of the system’s response to the given input frequency. The initial conditions for this problem are both zero; the final value is found by analyzing the circuit as t ∞. impulse response is designed to be the inverse of the impulse response of the communication channel. Yes, the impulse response exists for a series RLC circuit but you have to be aware that it is more complex than a simple RC or RL because the L and C form a resonant circuit and this gives rise (in notable cases) to a decaying sinewave response: -. Apply simple steady state sinusoidal analysis to circuits. It cannot absorb even a finite voltage change without infinite current flow, let alone impulse voltage. Control Systems: Basic control system components; block diagrammatic description, reduction of block diagrams. EECS 216 – 4 credits Introduction to Signals and Systems Prerequisites: EECS 215, Preceded or accompanied by Math 216. response of the circuit. 𝐶𝐶𝑗𝑗 𝑒𝑒 −𝛼𝛼𝛼𝛼. Figure 5: RC low pass filter circuit input as rectangular wave It means that the response of an integrating circuit to a rectangular wave is similar to that discussed for a square wave as discuss for square waver, except the output waveform, which is a sawtooth wave (instead of a triangular wave). Frequency Response of a Circuit The cutoff frequencies in terms of βand ω 0 A Serial RLC Circuit 2 2 c1022 ββ ωω =− + + 2 2 c2022 ββ ωω =+ + The cutoff frequencies in terms of Q and ω 0 2 10 11 1 c 22QQ ωω =−++ 2 10 11 1 c 22QQ ωω =++ ECE 307-5 8 Frequency Response of a Circuit Example Using serial RLC circuit, design band. ω1 – pusation of the RLC system = attenuated oscillations (ω1 ≠ ω0 for ξ≠0) ω 0 – eigen pusation of the un-attenuated system T. It employs a Feynman sum-over-paths postulate. (c) Suppose the resistor were changed to make the circuit response critically-damped. Thus, when the input force is a unit pulse, which corresponds physically to imparting momentum at time 0 (because the time-integral of force is momentum and the physical area under a unit sample is the sampling interval ), we see that the velocity after time 0 is a constant , or ,. The impulse response for each voltage is the inverse Laplace transform of the corresponding transfer function. An illustrative example is given with SPICE simulation. Again using the definition of capacitance, we then have the output response of the RC circuit for some initial charge and no forcing input voltage. Figure 5 shows a parallel resonant RLC circuit. Analog Circuits: Small Signal Equivalent circuits of diodes, BJTs, MOSFETs and analog CMOS. Boyd EE102 Lecture 7 Circuit analysis via Laplace transform † analysisofgeneralLRCcircuits † impedanceandadmittancedescriptions † naturalandforcedresponse. 2 to help evaluate the inverse Fourier transform. Mathematical development of the response equations C. 12 Summary of Step and Impulse Responses in RC and RL Circuits 141 7. 20 Impulse Response of a 2-Wire RC Line for circuits the R and C parameters are generally distributed through- rlc. The Series RLC Circuit The series RLC circuit is a fundamental building block in circuitry, even though the desired circuit response can often be obtained using active circuits. Fourier series for periodic signals – Fourier Transform – properties- Laplace Transforms and properties. The circuit is excited RL by an impulse function of magnitude E at time t = … - Selection from Signals and Systems [Book]. Casper and F. Choosing a Backup Generator Plus 3 LEGAL House Connection Options - Transfer Switch and More - Duration: 12:39. PPT – Second order circuits (i). RLC Simulation: Impulse Response Input voltage is pulse => Capacitor stores energy And then releases the energy Dr. Frequency Response of a Circuit The cutoff frequencies in terms of βand ω 0 A Serial RLC Circuit 2 2 c1022 ββ ωω =− + + 2 2 c2022 ββ ωω =+ + The cutoff frequencies in terms of Q and ω 0 2 10 11 1 c 22QQ ωω =−++ 2 10 11 1 c 22QQ ωω =++ ECE 307-5 8 Frequency Response of a Circuit Example Using serial RLC circuit, design band. • To measure the step response of second-order circuits and. All math explained Program3: See how the impulse response is derived. The RC circuit is formed by connecting a resistance in series with the capacitor and a battery source is provided to charge the capacitor. More generally, an impulse response is the reaction of any dynamic system in response to some external change. Hence the second central moment 2 is always positive. Figure 5: Parallel RLC circuit maximised at the resonant frequency rather than minimised. The circuit is driven by a transfer function which relates the input and output of a linear time invariant(LTI) system with zero initial conditions. To understand RLC like behavior, as well as to analyze and/or design a circuit to obtain a specific response, it. 1 Grid Impulse Response Computation In our previous work, we characterized the power grid as a single entity. Vo(s) is the RLC circuit's s-domain impulse response, where "A" is the strength of the impulse. Simple op-amp circuits. State equations for networks. Eigenvalues and eigenvectors. Sinusoidal solutions. It represents the response of the circuit to an input voltage consisting of an impulse or Dirac delta function. Since Laplace transform of the is And if then laplace transform of. RLC Circuit Example. These natural frequencies become time constants in the time-domain impulse response of circuit. Analyzing the Response of an RLC Circuit Open Script This example shows how to analyze the time and frequency responses of common RLC circuits as a function of their physical parameters using Control System Toolbox™ functions. The analysis of RLC circuits is more complex than of the RC circuits we have seen in the previous lab. By comparing the time- and frequency-domain plots, we can see that NQ|, where N is the number of observed rings before the oscillations essentially disappear (i. 2 The Natural Response of an RL Circuit 7. Biasing and bias stability of BJT and FET amplifiers. It cannot absorb even a finite voltage change without infinite current flow, let alone impulse voltage. With some differences: • Energy stored in capacitors (electric fields) and inductors (magnetic fields) can trade back and forth during the transient, leading to. from Wikipedia's page on RLC circuits. 5 the proposed boost converter circuit is modelled with RLC load in series considering only one switch as active and all other switches as resistances across the path. This calculator computes the resonant frequency and corresponding Q factor of an RLC circuit with series or parallel topologies. As the driving function is sinusoidal it is not unreasonable to assume that the response will be sinusoidal but we will not know the phase or the amplitude. Bailey Line Road 234,957 views. Figure 6 The unit impulse response approximated. 1 - RLC circuit. Applying this definition to H ( s ) in (1) and considering a source resistance R S and a capacitive load C T, the Elmore delay for a distributed RC or RLC line model is T ED = R S ( C + C T)+ R C 2 + C T: (2). The capacitor cannot absorb the impulse voltage. 2, FEBRUARY 1998 179 A General Theory of Phase Noise in Electrical Oscillators Ali Hajimiri, Student Member, IEEE, and Thomas H. During short-circuit test, the iron loss of a transformer is negligible because _____. Solving RLC Circuits by Laplace Transform Next: Frequency Response Functions and Up: Chapter 3: AC Circuit Previous: Responses to Impulse Train In general, the relationship of the currents and voltages in an AC circuit are described by linear constant coefficient ordinary differential equations (LCCODEs). The course covers a wide area topics; nodal and mesh analysis, steady state AC response of time-invariant networks, time and frequency response of linear systems, impulse response and transfer functions, Laplace transform analysis, frequency response, including steady state sinusoidal circuits. Circuit analysis using Laplace transform. To nd the unit impulse response:. To prepare for convolution in Chapter 3: What is the "impulse response" h [n] of each system when the input x [n] is a "unit impulse". With some differences: • Energy stored in capacitors (electric fields) and inductors (magnetic fields) can trade back and forth during the transient, leading to. Offered Fall, Winter. The input voltage is between start and end terminals of the circuit and it represents the input signal. Each zero will provide a +6 dB/octave or +20 dB/decade response. Kruger Advanced Circuit Techniques (55:141) The University of Iowa, 2013 RCL Resonant Circuits Slide 12. All responses will contain these two components. Step response of an RLC series circuit 1 Introduction Objectives • To study the behavior of an underdamped RLC Series Circuit for different damping coefficients Overview This experiment is a study of the step response of an underdamped RLC series circuit. Alexander and Matthew N. It is the ‘dual’ of the series circuit in that the voltage and current exchange roles. Lets assume a series RLC circuit as is shown in Figure 1. A resistor–capacitor circuit (RC circuit), or RC filter or RC network, is an electric circuit composed of resistors and capacitors driven by a voltage or current source. The RLC circuit shown on Figure 6 is called the parallel RLC circuit. It has a minimum of impedance Z=R at the resonant frequency, and the phase angle is equal to zero at resonance. Solve simple 1st order transient circuits. Frequency response of amplifiers. The unit step response of a linear circuit has been determined. The transient response is not necessarily tied to abrupt events but to any event that affects the equilibrium of the system. MAE140 Linear Circuits 167 Sinusoidal Steady-State Response Consider a stable transfer function with a sinusoidal input v(t)=Acos(ωt) The Laplace Transform of the response has poles •Where the natural cct modes lie –These are in the open left half plane Re(s)<0 •At the input modes s=+jω and s=-jω Only the response due to the poles on the imaginary. Impulse Response []. Since the inductive and capacitive reactance’s X L and X C are a function of the supply frequency, the sinusoidal response of a series RLC circuit will therefore vary with frequency, ƒ. 10 General solution for any second-order circuit with dc sources 6. Kruger Advanced Circuit Techniques (55:141) The University of Iowa, 2013 RCL Resonant Circuits Slide 12. From: "Ashok Prabhu M" To: Date: Wed, 11 Jun 2003 12:32:35 +0800; Hi all, Thank you all for your replies regarding my question. impulse response and gate current transient computa-tions. Solution In the digital domain, let 2 F Fs and therefore F Fs 2. Asymptotic Waveform Eval- uation (AWE) provides a generalized approach to linear RLC circuit response approximations. If the natural response grows without bound the system is no longer controlled or unstable. Our aim is to examine how the value of. 8 The parallel RLC circuit 6. As in first order circuits, the forced response has the form of the driving function. RLC circuits have a much richer and interesting response than the previously studied RC or RL circuits. Time domain response of narrow band filters Figure 37. Lets assume a series RLC circuit as is shown in Figure 1. domain analysis of simple RLC circuits, Solution of network equations impulse response, convolution, poles and zeros, parallel and cascade structure, frequency. Design of RLC-Band pass fllters WS2010/11 E. Knowing the impulse response of an LTIC system it is possible to determine the output of that system to any applied signal. Explore thousands of free applications across science, mathematics, engineering, technology, business, art, finance, social sciences, and more. To set a current source to zero, we replace it with an open circuit. Let us assume In earlier slides, we have shown that L2. an electric current flows in that circuit in response to the applied. • Impulse response defined • Several derivations of the convolution integral • Relationship to circuits and LTI systems J. The deposited charge should add to the voltage already on the capacitor an increment \$\Delta V = \frac{A}{C}\$. Mathys Second Order RLC Filters 1 RLC Lowpass Filter A passive RLC lowpass filter (LPF) circuit is shown in the following schematic. Ideas in this lecture is essential for deep understanding of the next two lectures on impulse response and on convolution, both you have touched on in your first year in the Communications course. Pre-Requisite: 92. A transfer function of circuit and afterwards state space representation equations will be designated. This lab is similar to the RC Circuit Lab except that the Capacitor is replaced by an Inductor. An audio crossover circuit consisting of three LC circuits, each tuned to a different natural frequency is shown to the right. ω1 – pusation of the RLC system = attenuated oscillations (ω1 ≠ ω0 for ξ≠0) ω 0 – eigen pusation of the un-attenuated system T. One way to visualize the behavior of the RLC series circuit is with the phasor diagram shown in the illustration above. an electric current flows in that circuit in response to the applied. Maybe it can help you. blocking, or attenuating, all other signals. Solving RLC Circuits by Laplace Transform Next: Frequency Response Functions and Up: Chapter 3: AC Circuit Previous: Responses to Impulse Train In general, the relationship of the currents and voltages in an AC circuit are described by linear constant coefficient ordinary differential equations (LCCODEs). Rlc Circuit Formulas Pdf. 2 The Natural Response of an RL Circuit 7. This video was created to support Ideal LPF Impulse Response This examples solves for the impulse response h[n] for an ideal lowpass filt Impulse response of ideal discrete time lowpass filter. Lee, Member, IEEE Abstract— A general model is introduced which is capable of making accurate, quantitative predictions about the phase. These tradeoffs are first studied qualitatively in a hypothetical ideal oscillator in which linearity of the noise-to-phase transfer function is assumed, allowing characterization by the impulse response. Background: The unit impulse response of linear, time-invariant, continuous-time (LTIC) system and, in particular passive RLC circuits, is of significant importance. RLC Circuits A. Frequency domain analysis of RLC circuits; impulse response, convolution, poles and zeros, parallel and cascade structure, frequency response, group delay, phase delay, digital filter design. The approach in this paper uses an impulse response to determine the transfer function of a simple circuit or system. PDF | This Article explains the analysis of series RLC circuit driven by a step voltage input. Recalling the form of the RC circuit's step response, we can anticipate how the circuit will respond to a square wave input of varying frequencies. I tried it using Laplace and also by direct solving of the differential equations. This is the schematic made with LTspice. Design of digital signal processing algorithms, fast Fourier transform (FFT), finite impulse response (FIR) and infinite impulse response (IIR) filters, data acquisition systems. Actually, the total response of a system is the sum of the natural response and the forced response. We will verify our intuition with a hardware-based experiment in the next section. 5 Circuit Analysis Techniques: Node Voltage / Mesh analysis, superposition, Thevenin and Norton equivalents 4. Analyze the poles of the Laplace transform to get a general idea of output behavior. Be able to determine the responses (both natural and transient) of second order The step response of a series RLC circuit. To simplify matters, we will assume that the circuit is under-damped, that both the step and the impulse occur at t = 0, and that the circuit is initially at rest prior to that time. • This chapter of notes focuses on the analysis of second-order RLC circuits using Laplace techniques. Inductance and capacitance are introduced and the transient response of RL, RC and RLC circuits to step inputs is established. So, after a few time constants, for practical purposes, the circuit has reached steady state. This Demonstration plots the Bode, Nyquist, and Nichols diagrams for user-set values of the parameters , , and. (TCCN = ENGR 1201) Prerequisite: MAT 1073. The LTI system can be completely characterized by its impulse response h(t). The name of the circuit is derived from the letters that are used to denote the constituent components of this circuit, where the sequence of the components may vary from RLC. Offered Fall, Winter. It turns out that the form of the transfer function is precisely the same as equation (8. The general solution is the sum of the homogeneous solution and the particular solution :. • Relate the key features of the impulse response and step response of a first- or second-order circuit to the locations. nodal and mesh analysis; Network theorems: superposition, Thevenin and Norton’s, maximum power transfer; Wye‐Delta transformation; Steady state sinusoidal analysis using phasors; Time domain analysis of simple linear circuits; Solution of network equations using Laplace transform; Frequency domain analysis of RLC circuits; Linear 2‐port network parameters: driving point and transfer functions; State equations for networks. simulation circuits 1. I am sorry if this is an off topic question. EC2102 Networks and Systems { HW 3 August 23, 2012 For the parallel RLC circuit shown below The impulse response of a circuit is h(t) = e 2tu. Use this utility to simulate the Transfer Function for filters at a given frequency, damping ratio ζ, Q or values of R, L and C. A transfer function of circuit and afterwards state space representation equations will be designated. The Series RLC Circuit The series RLC circuit is a fundamental building block in circuitry, even though the desired circuit response can often be obtained using active circuits. Using KVL, we have, Using KVL, we have, Fig. circuit that detects the peak current when charging 10nF-1uF capacitors and would like to have some theory to. Determine the unit impulse response of an LTI system and use it to calculate the system output produced by a given input. COURSE OUTCOME Student after successful completion of course must be able to apply the Thévenin, Norton, nodal and mesh analysis to express complex circuits in their simpler equivalent forms and to apply linearity and superposition concepts to analyze RL, RC, and RLC circuits in time and. • State Space Models • Linear State Space Formulation • Markov Parameters (Impulse Response) • Transfer Function • Difference Equations to State Space Models • Similarity Transformations • Modal Representation (Diagonalization) • Matlab Examples 1 State Space Models Equations of motion for any physical system may be. Theory and practice of signals and systems engineering in continuous and discrete time. Ask Question Asked 4 years, ^2$(notation that I'm stealing from Wikipedia's page on RLC circuits. The RC circuit is formed by connecting a resistance in series with the capacitor and a battery source is provided to charge the capacitor. Each time we are to record the trace of the Vs and Vcap. Since the inductive and capacitive reactance’s X L and X C are a function of the supply frequency, the sinusoidal response of a series RLC circuit will therefore vary with frequency, ƒ. Design of RLC-Band pass fllters WS2010/11 E. RLC Circuits An Example of the Application of Laplace Transforms. (TCCN = ENGR 1201) Prerequisite: MAT 1073. The strikethroughs indicate that the height is considerably taller than indicated. Find the Norton equivalent circuit. Consider what happens when a narrow current pulse with amplitude 𝐼 and duration is applied to the. Find the response of the RLC circuit to a step input. The deposited charge should add to the voltage already on the capacitor an increment \$\Delta V = \frac{A}{C}\$. EE 201 RLC transient – 1 RLC transients When there is a step change (or switching) in a circuit with capacitors and inductors together, a transient also occurs. 1 Step response, 187. Figure 5: RC low pass filter circuit input as rectangular wave It means that the response of an integrating circuit to a rectangular wave is similar to that discussed for a square wave as discuss for square waver, except the output waveform, which is a sawtooth wave (instead of a triangular wave). A transfer function represents the relationship between the output signal of a control system and the input signal, for all possible input values. The impulse response for the inductor voltage is. Time-varying circuits and nonlinear circuits, 154 Summary, 164 Problems, 165 1. In this example you will use Transient Analysis to plot the step responses of the RLC circuit. Here you can download details syllabus for GATE Electronics & Communication engineering. The chapters develop and examine several mathematical models consisting of one or more equations used in engineering to represent various physical systems. A first-order RL circuit is composed of one resistor and one inductor and is the simplest type of RL circuit. After 3˝, the circuit will have gotten 1 e 3 ˇ95% of the way, and after 5˝, more than 99%. University. Analyze the response of a series RLC circuit excited by a step function of voltage. – Homogeneous component + Particular component. RLC circuits are classical examples of second-order systems. Impulse response of a circuit is the zero-state response with unit impulse input. Discrete time system. Analysis of circuits with dependent sources, RL, RC, and RLC circuit transient and sinusoidal response, network functions, frequency response, and power analysis. 1 Step response, 187. To set a current source to zero, we replace it with an open circuit. In this example you will use Transient Analysis to plot the step responses of the RLC circuit. Introduction Most model-order reduction techniques work in the framework of matching moments in the complex. The frequency response of a system is the impulse response transformed to the frequency domain. For a constant driving source, it results in a constant forced response. The simplest and most prevalent is that of using segments to rep-resent the line (“segmentation techniques”). The impulse response of the circuit is g (t)= R e − (R/L) t σ (t). where u(t) is the Heaviside step function and. – Homogeneous component + Particular component. The network response D. The deposited charge should add to the voltage already on the capacitor an increment \$\Delta V = \frac{A}{C}\\$. But the average power is not simply current times voltage, as it is in purely resistive circuits. Discrete time system. Thus, the time constant is itself a good rough guide to \how long" the transient response will take. This passive RL low pass filter calculator calculates the cutoff frequency point of the low pass filter, based on the values of the resistor, R, and inductor, L, of the circuit, according to the formula fc= R/(2πL). Maybe it can help you. Consider what happens when a narrow current pulse with amplitude 𝐼 and duration is applied to the. An introduction to the electrical and computer engineering profession with emphasis on technical communication, team-based engineering design, professional and ethical responsibilities, contemporary issues, and software tools. For a continuous-time system with impulse response , the step response is. Lecture Notes. The simplest and most prevalent is that of using segments to rep-resent the line (“segmentation techniques”). Three inductively coupled loops - equivalent circuits Figure 36. Hello, I was trying to find the impulse response of the parallel RLC circuit. Solving a differential equation. Time domain response of narrow band filters Figure 37. The input is the across each element of the three passive elements in a series RLC circuit. Impulse Response of Series RLC Circuit • Evaluating at t = 0 + gives • We know that the capacitor acts as a short circuit to the impulse, giving v C (0 +) = B 00 1 = 0 • We have previously determined the time-derivative condition on v C (0 +) TA145. Mathematical development of the response equations C. Use the input pulse duration to predict the amplitude V 0 of the input pulse voltage v inptq V 0 ptqin the circuits you have examined. an electric current flows in that circuit in response to the applied. Transient Response Series RLC circuit The circuit shown on Figure 1 is called the series RLC circuit. Discrete-Time Systems. 3-phase bridge rectifier with source inductance 4. Second order circuits: The source free series and parallel RLC circuits, step response of a series and parallel RLC circuit. Thus we can use the following to find the particular integral: and the derivative of this is: hence substituting these into the differential equation we obtain:. the impulse response of a FIR filter which approximates this frequency response. Peak current in RLC circuit charging a capacitor. In the lumped-RLC method [14], each segment is represented as a lumped RLC network, whereas in the pseudo-lumped method [15] a lossless line in series with a resistor is used. Remark: Impulse Response = d/dt (Step Response) Relationship between t p, M p and the unit-impulse response curve of a system Unit ramp response of a second order system 2 2 2 2 1 2 ( ) s s C s n n n ⋅ + + = ζω ω ω R(s) = 1/ s2 for an underdamped system (0 < ζ < 1) sin 0 1 2 1 cos 2 2 ( ) 2 2 ≥ − − c t = t − + e− t + t t d n d. electric language by an RLC complex circuit with multiple series and parallel combinations of these components (Figure 2). These WDF and CTWDF can be used in HVDC lines and smart grid applications. Frequency Response of first order passive circuits - measure amplitude and phase response and plot bode diagrams for RL, RC and RLC circuits. Impulse response The impulse response for each voltage is the inverse Laplace transform of the corresponding transfer function. Vo(s) is the RLC circuit's s-domain impulse response, where "A" is the strength of the impulse. 1 Modelling of Proposed IBC with RLC Load in Series In Figure 5. 12 Summary of Step and Impulse Responses in RC and RL Circuits 141 7. Biasing and bias stability of BJT and FET amplifiers. Simple diode circuits, clipping, clamping, rectifier. University of West Florida. Hence, the step response of a discrete-time LTI system is just the running sum of its impulse response:. 68 1 Continuous-Time Time Invariance • Recall that time invariance means that if the input signal is shifted in time, the output will be shifted in time also. HOMEWORK:. So, after a few time constants, for practical purposes, the circuit has reached steady state. Lecture 14 (RC, RL and RLC AC circuits) In this lecture complex numbers are used to analyse A. 20 Impulse Response of a 2-Wire RC Line for circuits the R and C parameters are generally distributed through- rlc. class notes, M. It affects the shape of the filter’s frequency response. A steady-state response is the behavior of a circuit after a long time when steady conditions have been reached after an external excitation. In this lab you will examine a circuit's response to a unit impulse input. • To measure the step response of second-order circuits and. •Second-order (series and parallel RLC) circuits with no source and with a DC source. Calculus: Mean value theorems, Theorems of integral calculus, Evaluation of definite and improper integrals, Partial Derivatives, Maxima and Minima, Multiple integrals, Fourier. For sake of completeness, we will go through all three 100 i 1 + v i 2 V15 50 Figure 3: Example 1 - nd the equivalent circuit values. The transfer function H s of the circuit, which is the Laplace trans-form of h t, can be representedas H s 0 s ∞ h t e t dt ∑ i 0 d1 i i! s i. It tells us the size of the system’s response to the given input frequency. Apply simple steady state sinusoidal analysis to circuits. Although linearity is defensible, time invariance fails to hold even in this simple case. Voltage and Current in Time Impulse Response n Single clock edge creates impulse n Sequence of Events Circuits consume die current (charge) Die voltage droops Current comes in from outside inductor" Brings voltage back to nominal Current diminishes as die voltage rises above nominal n Inductor current ramps up until die voltage returns. • To measure the step response of second-order circuits and. The impulse response for a circuit is. Demonstrate a basic understanding of phasors and phasor diagrams for AC circuit analysis. Chapter 14, Problem 1. is the time constant. 11 Impulse Response of RC and RL Circuits 140 7. I figured out that the transfer function is: H (s)=V (s)/U (s) And my circuit has the formula (Ri (t) = v (t) which is the output, and u (t) is input): Li′ (t)+1 C∫t 0i (t)dt+Ri (t)=u (t) then by transforming it with Laplace and rearranging it I get:. The input is the across each element of the three passive elements in a series RLC circuit. Transient Response of RLC Circuits Dynamic response of such first order system has been studied and discussed in detail. It employs a Feynman sum-over-paths postulate. These tradeoffs are first studied qualitatively in a hypothetical ideal oscillator in which linearity of the noise-to-phase transfer function is assumed, allowing characterization by the impulse response. Real poles, for instance, indicate exponential output behavior. It cannot absorb even a finite voltage change without infinite current flow, let alone impulse voltage. Some Mathematical Preliminaries 289 3. The capacitor cannot absorb the impulse voltage. Pulse response E. However, for large circuits even a linear characterization of the power grid would be infea-sible. Notes on Solving for Impulse Response 1 Impulse Response from Di erential Equation Suppose we have a constant coe cient ordinary di erential equation of the form XN i=0 a i diy dt i (t) = M i=0 b i dix dt (t): (1) The goal is to nd the impulse response of this system using x(t) = (t) and y(t) = h(t):. When something changes in a circuit, the voltages and currents adjust to the new conditions. The frequency response curve of a parallel resonance circuit shows that the magnitude of the current is a function of frequency and plotting this onto a graph shows us that the response starts at its maximum value, reaches its minimum value at the resonance frequency when I MIN = I R and then increases again to maximum as ƒ becomes infinite. What is the response function of the circuit in the s-domain if the input is ? Ans. Deriving the complex impedance for a capacitor. Introduction Most model-order reduction techniques work in the framework of matching moments in the complex. be completely characterized by its impulse response h(t). Figure 7 A detailed image of the pulse with the response of the resistor and capacitor. 12 Summary of Step and Impulse Responses in RC and RL Circuits 141 7. The general solution is the sum of the homogeneous solution and the particular solution :. Proof was given in Class 3, Problem 1(ii). Biasing and bias stability of BJT and FET amplifiers. Kirchoff's voltage law. Actually, the total response of a system is the sum of the natural response and the forced response. s-Domain Circuit Analysis Time domain (t domain) Complex frequency domain (s domain) Linear Circuit Differential equation Classical techniques Response waveform Laplace Transform Inverse Transform Algebraic equation Algebraic techniques Response transform L L-1 Laplace Transform L Transformed Circuit. To nd the unit impulse response:. Yes, the impulse response exists for a series RLC circuit but you have to be aware that it is more complex than a simple RC or RL because the L and C form a resonant circuit and this gives rise (in notable cases) to a decaying sinewave response: -. Using KVL, we have, Using KVL, we have, Fig. Text: [T1] Agarwal, Lang "Foundations of Analog and Digital Electronic Circuits" [T2] Desoer, Kuh "Basic circuit theory" Suggested Readings: Razavi, "Fundamentals of Microelectronics" Sedra, Smith, "Microelectronic Circuits" Text , Solution Floyd, Buchla, "Fundamentals of Analog Circuits". Operational transient analysis. 0 RC and RL first-order circuits, natural and total response, RC Op amp circuits 2. Since α depends on the value of the resistance, you will use three different values for R : 40 W, 200 W and 1 kW. With some differences: • Energy stored in capacitors (electric fields) and inductors (magnetic fields) can trade back and forth during the transient, leading to. Unit impulse response plots for some different cases This subsection contains some more plots that show the effect of pole locations and help illustrate the general trends. To understand RLC like behavior, as well as to analyze and/or design a circuit to obtain a specific response, it. Determine the impulse response of the inductor, hL(t) 5. Although linearity is defensible, time invariance fails to hold even in this simple case. impulse response of a circuit by projecting with the Krylov space formed by solving the discretized differential equations of the circuit. The major difference between RC and RL circuits is that the RC circuit stores energy in the form of the electric field while the RL circuit stores energy in the form of magnetic field. Find the Norton equivalent circuit. Analyze the poles of the Laplace transform to get a general idea of output behavior. The output can be across any of the componnents, in this case i have series RCL, with the output being across L called y(t), and the input being u(t). A Bode plot is a graph of the magnitude (in dB) or phase of the transfer function versus frequency. Rl Circuit Derivation. A RLC circuit is an electrical circuit [1] containing of a resistor, an inductor, and a capacitor, connected in series impulse response, and step. •Second-order (series and parallel RLC) circuits with no source and with a DC source. Steady state sinusoidal analysis using phasors. Given the differential. The s teady-state response is defined as the behaviour of the system as t approaches infinity after the transients have died out. Fundamentals of Electric Circuits, 6 th edition. The superposition of impulse response. With some differences: • Energy stored in capacitors (electric fields) and inductors (magnetic fields) can trade back and forth during the transient, leading to. dynamic Response of a first order RC circuit and second order RLC circuit will be studied. So now it’s time to cover second-order systems. Determine the impulse response of the resistor, hR(t) 4. GATE 2020 Online Test Series time table for Electronics Engineering. when E = E_0 sin omega t, the complete response of a circuit is the sum of a natural response and a forced response. Poles of transfer function and bounded input bounded output stability. Step Response of an RL Circuit. The impulse response for each voltage is the inverse Laplace transform of the corresponding transfer function. | 2020-01-21T09:40:38 | {
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https://math.stackexchange.com/questions/3565015/how-to-factor-x6-4x42x31-by-hand/3565043#3565043 | # How to factor $x^6-4x^4+2x^3+1$ by hand?
I generated this polynomial after playing around with the golden ratio. I first observed that (using various properties of $$\phi$$), $$\phi^3+\phi^{-3}=4\phi-2$$. This equation has no significance at all, I just mention it because the whole problem stems from me wondering: which other numbers does this equation hold for?
The six possible answers are the roots of $$x^6-4x^4+2x^3+1=0$$. Note that I am not interested in solving for $$x$$ itself as much as I am interested in a method which would allow me to completely factor out this polynomial into lowest degree factors which still have real coefficients. Note that I am treating this equation as if I had no clue that the golden ratio is one of the solutions. In other words, I am trying to factor this equation as if I never saw it before, so I can't just immediately factor out $$(x^2-x-1)$$ without a justifiable process, even though it is indeed one of the factors.
I first observed that the equation holds for $$x=1$$, so I was able to divide out $$(x-1)$$ to get the factorization of:
$$(x-1)(x^5+x^4-3x^3-x^2-x-1)$$
I tried making an assumption that the quintic reduces to a product of $$(x^3+Ax^2+Bx+C)(x^2+Dx+E)$$, multiplying out, and equalling coefficients, but I ended up with a system of two extremely convoluted equations which I had no idea how to solve. I also tried to turn the first five terms of the quintic into a palindromic polynomial and then perform the standard method of factoring palindromic polynomials, to no avail.
I am either missing something, or I don't know of a nice method that would let this expression be factored. I'm looking forward to being enlightened, thanks for any help.
• Are you familiar with long division? en.wikipedia.org/wiki/Polynomial_long_division Mar 1 '20 at 5:34
• As the answer below suggests, Wolfram Alpha agrees with your approach. What was the problem with the system of equations? Also, that factorization does not seem to have real roots except $x=1$ Mar 1 '20 at 5:34
• @gt6989b I was able to reduce the resulting system of five equations into two equations by elimination, but the remaining two equations had a lot of terms like $AB^2, B^3, A^2, AB,$ etc.. in other words, I was helpless, unless I did it wrong somehow Mar 1 '20 at 5:37
• If you use $x=10$, you get $962001=9\times 89\times 1201$ which should tell you something about the factorization
– user632577
Mar 1 '20 at 5:39
• Due to Abel's impossibility theorem, you don't have an algorithm to exactly factorize a generic polynomial of degree at least $5$. The best you could do is probably using numerical approximations. You have to know some extra information about your polynomial if you want to factorize it exactly. So after dividing your degree-$6$ polynomial by $x-1$, you get a quintic polynomial, and unless you know some extra information, I don't think you can factorize it exactly. Mar 1 '20 at 5:44
Here's a possible way to do it:
$$x^6-4x^4+2x^3+1 = (x^6+2x^3+1)-4x^4 = (x^3+1)^2 - 4x^4$$
$$(x^3+1)^2-4x^4 = [x^3+1-2x^2][x^3+1+2x^2]$$
$$x^6-4x^4+2x^3+1= [(x^3-x^2)+(1-x^2)][x^3+2x^2+1]$$
Then, we have:
$$x^6-4x^4+2x^3+1 =[x^3+2x^2+1][x^2(x-1)+(1-x)(1+x)]$$
$$x^6-4x^4+2x^3+1 = (x-1)(x^2-x-1)[x^3+2x^2+1]$$
So that gives you a decently nice factored form.
• That's exactly the type of answer I was looking for. Thank you! I have been at the step of the difference of squares at some point, but I couldn't break the cubics down. Your separation of $x^3-2x^2+1$ into $(x^3-x^2)+(1-x^2)$ was very clever! Mar 1 '20 at 5:53
• Thank you :) I try to be clever (failing in most of my attempts, of course!)
– Abhi
Mar 1 '20 at 5:54
Your original method is tedious but it can be done.
You can show that $$(x^3+Ax^2+Bx+C)(x^2+Dx+E)$$ is equal to:
$$x^5+(D+A)x^4+(1+AD+B)x^3 + (AE+BD+C)x^2 + (BE+CD) + CE$$
so $$A+D = 1, B+AD+1 = -3, AE+BD+C=-1, BE+CD=-1, CE=-1$$.
Assuming $$A,B,C,D,E$$ are all integers, we either have $$C=-1, E=1$$ or $$C=1, E=-1$$.
If $$C=-1, E=1$$, then we have:
$$A+D=1 \tag{1}$$ $$B+AD=-4 \tag{2}$$ $$A+BD=0 \tag{3}$$ $$B-D=-1 \tag{4}$$
$$(1)+(4)$$ gives $$A+B=0$$ so $$A=-B$$, which gives:
$$-B+D=1 \tag{5}$$ $$B-BD=-4 \tag{6}$$ $$-B+BD=0 \tag{7}$$ $$B-D=-1 \tag{8}$$
and this is clearly impossible since $$(6) + (7)$$ gives $$0=-4$$.
Therefore we must have $$C=1, E=-1$$:
$$A+D=1 \tag{9}$$ $$B+AD=-4 \tag{10}$$ $$-A+BD=-2 \tag{11}$$ $$-B+D=-1 \tag{12}$$
This time $$(9)-(12)$$ gives $$A+B=2$$, so $$A=2-B$$:
$$-B+D=-1 \tag{13}$$ $$B+2D-BD=-4 \tag{14}$$ $$B+BD=0 \tag{15}$$ $$-B+D=-1 \tag{16}$$
$$(14)+(15)$$ gives $$2B+2D = -4$$, so $$B+D=-2$$. When we add this to $$(16)$$, $$2D=-2$$ so $$D=-1$$.
And the rest follows:
$$B - D = 1 \Rightarrow B+1=1, B=0$$ $$A=2-B \Rightarrow A=2$$
so the factorisation is $$(x-1)(x^3+2x^2+1)(x^2-x-1)$$.
I wouldn't wish this method on anybody. | 2022-01-18T20:48:35 | {
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https://math.stackexchange.com/questions/2158750/length-of-segment-parallel-to-an-edge/2158811 | Length of segment parallel to an edge
I've tried all the possible side splitter and angle bisector theorem stuff and I still can't come up with the correct answer. I even tried some law of cosine and sine stuff, but nothing. Any help would be gladly appreciated. Thanks.
• Correct answer is 24, I also came to 40 – Nick Brown Feb 24 '17 at 1:45
• I removed my comment because I realised how stupid it was to suggest that DE = BC lol - but I see the mistake I made – mrnovice Feb 24 '17 at 1:46
Observe triangles $ADE$ and $ABC$ are similar. Since $BC || DE$ and $BF$ is an angle bisector of $\angle \, A$ $$\angle \,DBF = \angle \, CBF = \angle \, DFB$$ so triangles $BDF$ is isosceles with $BD = DF$. Analogously $CE=EF$. Hence the perimeter $P_{ADE}$ of triangle $ADE$ is $$P_{ADE} = AD+DF+AE+EF = AD+DB + AE+EC = AB + AC = 26 + 34 = 60$$ The perimeter $P_{ABC}$ of $ABC$ is $$P_{ABC} = AB + BC+AC = 26+4=+54 = 100$$ By the similarity of $ADE$ and $ABC$ $$\frac{DE}{BC} = \frac{P_{ADE}}{P_{ABC}} = \frac{60}{100} = \frac{3}{5}$$ Since $BC = 40$ $$DE = \frac{3}{5} \, 40 = 24$$
• nice use of the parallel lines – Joffan Feb 24 '17 at 5:05
• Brilliant use of basic geometry! – Jim Feb 24 '17 at 17:38
• @Jim thank you! I appreciate it. Cheers! – Futurologist Feb 25 '17 at 11:18
You can do this with the angle bisector theorem used twice.
First observe that $AF$ bisects $\angle BAC$, (because angle bisectors are concurrent) so continue $AF$ to meet $BC$ at point $G$. Then $G$ divides $BC$ in the ratio 26:34 so $BG = \frac {40}{60} 26 =\frac {52}{3}$
Then $BF$ divides $AG$ in the ratio $26:\frac{52}3 = 3:2$ giving $AF:AG$ as $3:5$. Thus through similarity of $\triangle ABC$ and $\triangle ADE$ the ratio between $DE$ and $BC$ is also $3:5$ i.e. $\fbox{$DE=24$}$
• Can you explain what you mean by bisectors are concurrent? Also, how you know that it divides BC in a ratio of 26:34? Thank you! – Nick Brown Feb 24 '17 at 1:52
• The angles bisectors of a triangle meet at a common point ("are concurrent"). The angle bisector theorem says that the bisector divides the opposite side in the ratio of the adjacent sides. – Joffan Feb 24 '17 at 2:03
• @NickBrown I proved the angle bisector theorem here and the concurrency of angle bisectors here – Joffan Feb 24 '17 at 2:10
• Ok, so I follow everything until the last part where you find DE and BC to be in a 3:5 ratio. Where did you get that from? – Nick Brown Feb 24 '17 at 2:38
Try with Heron's formula to get the area of the big triangle two ways:
• First calculate it as a function of the three sides, using Heron's directly.
• Calculate the height of the triangle ABC $=h_{ABC}$ from the areas just calculated.
• Then calculate the radius of the incircle, using this example and knowledge of the relationship of the incircle to the bisected angles of the triangle.
• subtract this radius from the height of ABC $=h_{ABC}$ to get the height of triangle ADE $=h_{ADE}$.
• Now use proportionality of similar triangles:
$$\frac{h_{ADE}}{h_{ABC}}= \frac{|DE|}{40}$$
• Can you explain why subtracting the inradius from the height of ABC would give me the height of triangle ADE? – Nick Brown Feb 24 '17 at 2:30
• The in radius is the height of triangle BFC. The height of ADE is the height of ABS minus the height of BFC, since the line DE is parallel to the line BC. – Jim Feb 24 '17 at 5:54
Let $S$ be the area of the triangle, let $p$ be its perimeter, $r$ its inradius, and $h_A$ its altitude from $A$. Also write $BC = a$, $AC = b$, $AB = c$.
Note that $F$ is the incentre of $ABC$, hence the distance from $F$ to $BC$ is $r$. Now we have $$\frac{DE}{a} = 1 - \frac{r}{h_A} = 1 - \frac{2S/p}{2S/a} = 1 - \frac{a}{p} = \frac{b + c}{p}$$ hence $$DE = \frac{a(b+c)}{p} = \frac{40(34 + 26)}{100} = 24.$$ | 2019-11-19T17:41:03 | {
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http://math.stackexchange.com/questions/162119/why-doesnt-the-indirect-proof-of-irrational-roots-apply-to-rational-roots | # Why doesn't the indirect proof of irrational roots apply to rational roots?
When trying to prove that a particular root (say $\sqrt{2}$ or $\sqrt{10}$) cannot be rational, I always see a particular indirect proof that goes something like this:
Suppose $\sqrt{x}$ were rational; then, there would be two integers $a$ and $b$ such that $a/b$ was $\sqrt{x}$. We can also assume that $a$ and $b$ have no common factors, because we can simplify $a/b$ as much as we want before we begin.
Then, $(a/b)^2 = x$, or $a^2/b^2=x$, or $a^2 = x * b^2$. But, if $a * a$ is a multiple of $x$ then $a$ must also be a multiple of $x$, so we can rewrite $a$ as $a = cx$ and substitute:
$(c * x)^2 = x * b^2$, or $x^2 * c^2 = x * b^2$. Divide by $x$ and we get that $b^2 = x * c^2$.
But we've now shown that both $a$ and $b$ have a factor of x, contradicting our original assumption; this means that $\sqrt{x}$ cannot be rational.
I see why this proof works, what I don't see is why you can't plug in a number with an actual rational root, like 16, and not form the same proof that $\sqrt{16}$ cannot be rational.
-
The crucial step, which fails for x=16, is when one asserts that if a^2 is a multiple of x then a is a multiple of x. – Did Jun 23 '12 at 19:32
– Bill Dubuque Jun 23 '12 at 20:39
The proof will work for any integer (or rational) that is not a perfect square, since it will have a prime factor occurring to an odd power, which is precisely what's needed for the proof to succeed. Namely from $\rm\:a^2 = x b^2\:$ we deduce, by unique factorization, that the power of every prime $\rm\:p\:$ dividing $\rm\:x\:$ must be even, being the difference of two even integers (the power of $\rm\:p\:$ in $\rm\:a^2\:$ minus the power in $\rm\:b^2$).
Generally it is not true that $\rm\:x\:|\:a^2\:\Rightarrow\:x\:|\:a\$ (e.g. let $\rm\:x = a^2 > 1)$. In fact this property is true iff $\rm\:x\:$ is squarefree, which is why the proof works for $\rm\:x\:$ prime (or a product of distinct primes) which, having at least one prime to the power one, certainly does have a prime occurring to an odd power. The general case reduces to this case by pulling the square part of $\rm\:x\:$ out of $\rm\:\sqrt{x},\:$ i.e. if $\rm\:x = n^2 y,\:$ with $\rm\:y\:$ squarefree then we have $\rm\:\sqrt{x} = \sqrt{n^2 y} = n\sqrt{y}\in \mathbb Q\iff\sqrt{y}\in\mathbb Q.$ Therefore your method of proof will work if you first reduce this way to the case where $\rm\:x\:$ is squarefree.
-
hm. so that part of the proof is true iff x is not a perfect square; isn't that just begging the question then? Aren't we trying to prove that x is not a perfect square? – Michael Edenfield Jun 23 '12 at 22:38
@Michael The proof works if $\rm\:x\:$ has nontrivial squarefree part $\rm\,(y\ne 1),\,$ i.e. if $\rm\:x\:$ is not the square of an integer. It proves, further, $\rm\:x\:$ is not the square of a rational, i.e. $\rm\, \sqrt{x}\not\in\mathbb Z\:\Rightarrow\:\sqrt{x}\not\in \mathbb Q,\:$ i.e. $\rm\:\sqrt{x}\:$ is irrational. – Bill Dubuque Jun 23 '12 at 23:35
@Michael This is a special case of the Rational Root Test, which implies that if $\rm\:x^2-n\:$ has rational root $\rm\:x=a/b\:$ in lowest terms then $\rm\:b\:$ divides the leading coef $= 1$, therefore $\rm\:a/b\in\mathbb Z,\:$ i.e. $\rm\:x = \sqrt{n}\in \mathbb Q\:\Rightarrow\:\sqrt{n}\in\mathbb Z.$ $\ \$ – Bill Dubuque Jun 23 '12 at 23:48
Ah. So for example, we know that $\sqrt{2}$ is not an integer because it must be 1 < x < 2, so we can use this proof to further prove that it cannot be rational either. – Michael Edenfield Jun 24 '12 at 14:02
The statement $x|a^2\Rightarrow x|a$ is not generally true (Counter-example : $8|4^2$ but we have not $8|4$).
The proof works for $x$ prime, because, then, this statement is true by Euclid's lemma.
-
Another way to think about this is to reformulate your proof to go through the following statement:
If $\gcd(a,b)=1$, then $\gcd(a^2, b^2)=1$.
This follows immediately from unique factorization (or with a little more work from Euclidean-algorithm-type arguments). It also directly implies that any rational square root of an integer is itself an integer. Say $(a/b)^2=n$ with $a/b$ in lowest terms, then $\gcd(a,b)=1$. Thus $\gcd(a^2,b^2)=1$, so $a^2/b^2$ is a fraction in lowest terms which is also equal to an integer, and hence $b= \pm 1$.
I like this way of thinking about it because it does apply to rational roots of integers; it just gives a different conclusion (that they're integers, rather than that they're irrational).
-
+1 because that was actually another question I had -- is it true that all rational squares of integers are themselves integers. – Michael Edenfield Jun 23 '12 at 22:40
(of course I meant "rational roots") – Michael Edenfield Jun 24 '12 at 14:03
Take your example for $\sqrt{16}$ and note that $a=4$ then $a^2 = 16$ does not imply $a$ is a multiple of 16.
- | 2014-08-22T16:03:01 | {
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https://astronomy.stackexchange.com/questions/35635/during-an-eclipse-how-big-is-the-shadow-of-the-moon-on-the-earth | # During an eclipse, how big is the shadow of the moon on the earth?
This picture was taken from the ISS during a solar eclipse.
You can see the shadow of the Moon on the surface of the Earth. But how big is this shadow? How many kilometers is its diameter?
• xkcd.com/1276 about as big as the M25 freeway around London – Jacob Krall Apr 2 at 19:42
• when the light comes near the shadow comes small and when the light goes far becomes long – nahom get. Apr 6 at 17:00
The umbra has a well defined diameter but the size varies due to the eccentricity of the orbits of the Earth and of the Moon. The Moon may be so far away that it can't fill the solar disk at all (for instance Moon at apogee and Earth at perihelion).
We can, however, theoretically determine the diameter of the shadow that the Moon casts on the Earth. The calculation only requires elementary geometry and a nice image. We then obtain the maximum radius:
$$\displaystyle r_{u} = R_{m} - \frac{d_{m}- R_{e}}{d_{e} - d_{m}} (R_{s} - R_{m})$$
where $$R_m, R_e, R_s$$ are the radii of the Moon, Earth and Sun respectively, $$d_{m}$$ is the distance Moon-Earth, $$d_{e}$$ is the distance Earth-Sun.
We can investigate several cases by varying the distances $$d_{m}$$ and $$d_{e}$$, in accordance with the eccentricities of the orbits.
To find the maximum possible radius of the umbra, take $$d_m = a_m (1- e_m)$$ (Moon at perigee) and $$d_e = a_e (1+e_e)$$ (Earth at aphelion), where $$e_e, e_m$$ are the eccentricities of the orbits of the Earth and Moon respectively, and $$a_e, a_m$$ their semi-major axis. This makes sense because to get the widest lunar eclipse we want a big (and therefore close) Moon and a small (and far) Sun. Substituting some typical values we obtain $$r_u \approx 120$$km (around $$240$$ km maximum width). This situation is, however, extremely unlikely (I estimate it happens around once every century).
This equation also tells us that, on average ($$d_m = a_m, d_e = a_e$$), we do not see any eclipse ($$r_u$$ would be negative).
We need the Moon to be close to perigee, in which case, assuming average distance for the Earth ($$d_e=a_e$$), we get $$r_u \approx 80$$ km and a width of $$160$$km. A result which might sound familiar by now!
Very similarly, as @uhoh suggested, we can calculate the width of the penumbra. Now, instead of considering the ray $$T_{s,1} T_{l,1}$$, we take the ray $$T_{s,2} T_{l,1}$$. Clearly, this is equivalent to take $$R_s \rightarrow - R_s$$. We then get
$$\displaystyle r_{p} = R_{m} + \frac{d_{m}- R_{e}}{d_{e} - d_{m}} (R_{s} + R_{m})$$
Now the maximum radius of the penumbra is obtained when $$d_e = a_e (1-e_e)$$ (Earth at perihelion) and $$d_m = a_m (1+ e_m)$$ (Moon at apogee). In this case we get $$r_p \approx 3650$$km and a width of $$7300$$km.
If instead we take average distance for the Earth we get $$r_p = 3600$$km, so a total of $$7200$$km, not far from @uhoh's answer.
In the case of minimum width, we take $$d_e = a_e (1+e_e)$$ (Earth at aphelion) and $$d_m = a_m (1- e_m)$$ (Moon at perigee). We then get $$r_p = 3400$$km, so a total of $$6800$$km.
In any case, the width is approximately twice the diameter of the Moon ($$7000$$km), but notice this is only a coincidence and is due to the fact that the angular diameters of the Moon and Sun are very close to each other. Indeed, simplifying the previous equation, we can neglect $$R_e$$ at the numerator, $$d_m$$ at the denominator and approximate $$R_s -R_m$$ with only $$R_s$$. We then see that
$$\displaystyle r_p = R_m + \frac{d_m}{d_e} R_s = R_m \Big(1+ \frac{R_s / d_e}{R_m / d_m} \Big)$$
but $$R_s / d_e$$ and $$R_m / d_m$$ are the angular diameters of the Sun and Moon, therefore the fraction is approximately 1 (actually 1.03 on average) and we recover $$r_p \approx 2 R_m$$. Image © Flavio Salvati 2020
• +1 I wonder if you can also properly calculate the diameter of the penumbra? I've adjusted my answer to note that I used the fact that the Sun's angular diameter is similar to the Moon's to get to the conclusion that the extremes of the penumbra are about twice the diameter of the Moon. I think you are close to calculating the penumbra's diameter correctly here, I wonder if you could add that as well? – uhoh Mar 31 at 1:11
• @uhoh yes, just edited my answer! – Flaffo Mar 31 at 8:14
• Beautiful! :-) – uhoh Mar 31 at 8:27
But how big is this shadow? How many kilometers is its diameter?
That's a photo of the umbra and penumbra on the surface of the Earth taken from Space. It's a little distorted because it's not directly under the ISS but far off near the terminator.
It's hard to pin down the size of the penumbra because it's fuzzy and fades near the edges, but if you could see the very edges then it would be twice the diameter of the moon or very roughly 6900 kilometers total.
Actually this is only true by coincidence because the angular diameter of the Sun happens to be the same as that of the Moon. @Flaffo's answer does a good job of explaining how to calculate the diameter of the umbra, and that math could probably be extended to calculate the diameter of the penumbra as well.
The umbra has a well defined diameter but the size varies a lot due to variation in the distance from the Moon to the Earth since it's orbit is not circular. Sometimes the Moon is so far away that it can't fill the Sun and there is no umbra at all, that's called an annular eclipse.
Typically, the umbra is 100–160 km wide, while the penumbral diameter is in excess of 6400 km.
You can see an example of a very detailed simulation of just the umbra moving across the Earth's surface in the NASA Goddard video Tracing the 2017 Solar Eclipse The precise 3D shape of the Moon generates the shadow and it then moves over the contour of the Earth's topography. If you think the shape is weird I agree! See answers to The Moon's shadow could not possibly look like this — could it? (also see What are the “Moon L, B, C” angles shown in this solar eclipse simulation?)
Here's a screenshot:
• The width of the penumbra is twice the width of the Moon, minus the width of the umbra. – Mark Mar 31 at 1:31
• @Mark wow thanks! Yes that's right. At zero Moon-Earth distance the umbra and penumbra are coincident and equal to one moon diameter, As the Moon moves away from Earth the umbra gets smaller and the penumbra gets larger by the same amount until the umbra reaches zero and the penumbra is exactly two Moon diameters at which point the Sun and the Moon have exactly equal angular diameters. – uhoh Mar 31 at 1:46
The other answers are great, but if you want an handy tool to explore the next and the past eclipses, check out this super cool website!
It allows you to search all the eclipses (solar and lunar + Mercury and Venus transits) for the past and the next couple of centuries. Even more interesting are the tracks of the shadows for which you can get both a summary decade-by-decade, and a detailed analyses for each eclipse. And what about simulating the sky view? Yes it does that too!!!
Curiously in the 2040s there will be several weirdly shaped solar eclipses, like the total one taking place on 9 April 2043 over East Russia which is "clipped" because the sun is not yet risen at the beginning of the event:
...so I guess that, as most of the time, the full answer is "it's complicated..." but have fun! | 2020-12-02T22:07:08 | {
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https://math.stackexchange.com/questions/2341785/how-to-find-this-simplification-integral-involving-products-of-roots | # How to find this simplification integral involving products of roots?
Consider the following integral $$f = \int_0^1 \frac{1}{\sqrt{-\frac{1}{2} \, t^{2} + 1} \sqrt{-t^{2} + 1}} \mathrm \,dt.$$ If we change variable by letting $x^2=t^2/(2-t^2)$, then we have $$f = \int_0^1 \sqrt{2} \cdot\sqrt{\frac{1}{1-x^{4}}} \mathrm \,dx,$$ which is a simpler form. I read this in a book and wonder how can we come up with this sort of simplification? Is it just experience or is there systematic way to do it?
Note: The integral is the complete elliptic integral of the first kind $K(1/\sqrt{2})$.
• this integral leads to an elliptic one – Dr. Sonnhard Graubner Jun 30 '17 at 12:02
$\int\frac{dt}{\sqrt{1-t^2}}=\arcsin(t)$, $\int\frac{dt}{\sqrt{1-\frac{t^2}{2}}}=\sqrt{2}\arcsin\left(\frac{t}{\sqrt{2}}\right)$, hence two candidate substitutions for simplifying things are $t=\sin\theta$ and $t=\sqrt{2}\sin\frac{\theta}{\sqrt{2}}$. Let us try the first one:
$$I = \int_{0}^{\pi/2}\frac{d\theta}{\sqrt{1-\frac{1}{2}\sin^2\theta}}$$ followed by the substitution $\theta=\arctan u$: $$I = \int_{0}^{+\infty}\frac{du}{(1+u^2)\sqrt{1-\frac{1}{2}\cdot\frac{u^2}{1+u^2}}}=\sqrt{2}\int_{0}^{+\infty}\frac{du}{\sqrt{(u^2+1)(u^2+2)}}=\frac{\pi}{\text{AGM}(2,\sqrt{2})}$$ The AGM allows an efficient numerical evaluation (it immediately tells us that $I\geq\frac{2\pi}{2+\sqrt{2}}$, for instance) and the identity $\text{AGM}(a,b)=\text{AGM}\left(\frac{a+b}{2},\sqrt{ab}\right)$ gives many equivalent integrals.
We may notice that
$$\int_{0}^{1}\frac{dx}{\sqrt{1-x^4}}\stackrel{x\mapsto\sqrt{z}}{=}\frac{1}{2}\int_{0}^{1}\frac{dz}{\sqrt{z(1-z^2)}}\stackrel{z\mapsto t^{-1}}{=}\frac{1}{2}\int_{1}^{+\infty}\frac{dt}{\sqrt{t(t-1)(t+1)}}$$ leads to $$\int_{0}^{1}\frac{dx}{\sqrt{1-x^4}}=\frac{1}{2}\int_{0}^{+\infty}\frac{dt}{\sqrt{t(t+1)(t+2)}}=\int_{0}^{+\infty}\frac{du}{\sqrt{(u^2+1)(u^2+2)}}$$
This completely explains how to find the useful substitution by underlying the relation between the initial elliptic integral, the lemniscate constant and the AGM. We may also add a fourth actor on the scene, since by the substitution $x=w^{1/4}$ the integral $\int_{0}^{1}\frac{dx}{\sqrt{1-x^4}}$ is related with the Beta function, hence with the $\Gamma$ function. Here it is a complete summary:
$$\boxed{ \int_{0}^{1}\frac{dx}{\sqrt{1-x^4}} = \frac{\pi}{\text{AGM}(1,\sqrt{2})}=\frac{1}{\sqrt{2}}\,K\left(\frac{1}{\sqrt 2}\right)=\frac{1}{4}B\left(\frac{1}{4},\frac{1}{2}\right)=\frac{\Gamma\left(\frac{1}{4}\right)^2}{4\sqrt{2\pi}}. }$$
This excursion gives as a by-product an efficient numerical approach for computing $\Gamma\left(\frac{1}{4}\right)$, proving $\Gamma\left(\frac{1}{4}\right)=\frac{(2\pi)^{3/4}}{\sqrt{\text{AGM}(1,\sqrt{2})}}$.
$$x^2=\frac{t^2}{2-t^2}$$
$$t=x \sqrt{\frac{2}{x^2+1}}$$
$$dt = \sqrt{2} \sqrt{\frac{1}{x^2+1}} \left(1-\frac{x^2}{x^2+1}\right) dx$$ So the integrand becomes is $$\frac{\sqrt{2} \left(\sqrt{\frac{1}{x^2+1}} \left(1-\frac{x^2}{x^2+1}\right)\right)}{\sqrt{1-\frac{2 x^2}{x^2+1}} \sqrt{1-\frac{x^2}{x^2+1}}}$$ and finally $$\frac{\sqrt{2}}{\sqrt{1-x^4}}$$
• This is the correct changing variable. But I am actually asking how can we come up with what to change in the first place. – ablmf Jun 30 '17 at 13:02 | 2019-06-25T15:47:44 | {
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https://codereview.stackexchange.com/questions/168768/prime-number-generator-in-c | # Prime number generator in C
I have to print numbers between two limits n and m, t times.
I created t variable that stores a number of test cases.
Outer for loop iterates for every test cases.
Inner for loop prints primes from m to n.
# Code
#include <stdio.h>
#include <stdlib.h>
int is_prime(int);
int main(void) {
int t, m, n;
scanf("%d", &t);
for (int i = 0; i < t; i++) {
scanf("%d %d", &m, &n);
for (int j = m; j <= n; j++) {
if (is_prime(j)) {
printf("%d\n", j);
}
}
if (i < t - 1) printf("\n");
}
return 0;
}
int is_prime(int num) {
if (num <= 1) return 0;
for (int i = 2; i * i <= num; i++) {
if (num % i == 0) {
return 0;
}
}
return 1;
}
Can I have review?
The only even prime number is two, so with a little extra coding you can ignore even numbers in the for loop. For even more speed use a sieve, though maybe save that for when you get a 'TLE' failure.
A more formal code layout would be good as well, your if statements need expanding to include {...}. More vertical space (blank lines) a well as explanatory comments help. When you come back to your old code six months later you will thank yourself.
I prefer to set 0 and 1 as FALSE and TRUE to make code easier to read YMMV.
int is_prime(int num) {
if (num <= 1) {
return FALSE;
}
// Even numbers.
if (num % 2 == 0) {
return num == 2; // 2 is the only even prime.
}
// Odd numbers.
// Start loop at 3 and step 2 to skip even divisors.
for (int i = 3; i * i <= num; i += 2) {
if (num % i == 0) {
return FALSE;
}
}
return TRUE;
}
• Just include <stdbool.h> to avoid having to define it yourself, and get the lowercase versions at the same time. – Tamoghna Chowdhury Jul 10 '17 at 5:13
• @TamoghnaChowdhury: Thanks. My C is very rusty. – rossum Jul 10 '17 at 7:49
Good that OP's code handles values <= 0, yet could have used is_prime(unsigned num) instead. Further: this is a good place for the only even prime detect.
Corner concern: The i * i <= num test fails for large num, like num = INT_MAX as i*i is always <= than INT_MAX or it is int overflow - which is undefined behavior (UB).
Preference: Use bool for return values that are either 0 or 1.
Many modern compilers/processors calculate the remainder and quotient for little/no additional cost. Use that as an exit condition.
bool is_prime(uintmax_t num) {
if (num <= 3) {
return num >= 2;
}
uintmax_t q = num;
for (uintmax_t i = 3; i <= q; i += 2) {
if (num % i == 0) {
return false;
}
q = num / i;
}
return num%2;
}
The next step in prime detection is use of the Sieve of Eratosthenes
Note for future: Code does not check the return value of scanf(). This is OK for test code, but not for code under test. | 2020-01-22T00:38:35 | {
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https://mathhelpforum.com/threads/trigonometry-to-memorize-and-trigonometry-to-derive.185935/ | Trigonometry to Memorize, and Trigonometry to Derive
Ackbeet
MHF Hall of Honor
I have attached a pdf document containing the vast majority of trigonometry I have needed to know on a working basis. The first page consists of trigonometry I think everyone should have memorized. I have never needed to have anything more than the first sheet memorized for any application. The second page is a non-exhaustive sheet of most of the trigonometric identities that I have found useful, and a few more besides.
It is my opinion that a student who memorizes the first sheet, and can derive anything on the second sheet, has a fairly good grasp of trigonometry.
I hope this proves useful.
View attachment Trig Sheets.pdf
Siron
I think it's very good document .
(Do you have more documents for other subjects maybe?)
Ackbeet
MHF Hall of Honor
I think it's very good document .
(Do you have more documents for other subjects maybe?)
Thank you very much.
I don't really have any others like this one. The reason is the only courses I have taught were Calculus, and this was intended as a review sheet for incoming freshman who were taking my class. I do have problem-solving stickies in the Other Topics and Advanced Applied Math forums. That's about it. Chris L T521 has a very good DE's tutorial, and there's already a LaTeX tutorial and a Calculus tutorial. There's even something in the pre-algebra and algebra forum as well as the linear and abstract algebra forum. So that's most of the forums that I pay the most attention to that even admit of such a document.
Ashz
I had an identities quiz, and I used your sheet to help with memorizing the formulas. Thanks a ton! Any pointers on memorizing the unit circle :9
godfreysown
I find this enormously useful (and encouraging! as I can see how far I've come in a relatively short period of time; and salutary! as I can see how far I have to go before my exam on 17th July)
thanks Ackbeet: thoughtful and very useful!
Godfree
richard1234
I had an identities quiz, and I used your sheet to help with memorizing the formulas. Thanks a ton! Any pointers on memorizing the unit circle :9
You should be able to "visualize" the unit circle, and know which angles correspond to $$\displaystyle \frac{\pi}{4}$$ or $$\displaystyle \frac{\pi}{3}$$. Also, by definition, the sine and cosine are the y- and x-coordinates of the point on the circle. No memorization needed...
louisejane
This really helps! Thank you for sharing your knowledge here. Is it free to ask you anything here that involves mathematics?
hp12345
This really helps! Thank you for sharing your knowledge here. Is it free to ask you anything here that involves mathematics?
Yeah its absolutely free,though you can donate them some money if you think so but its not necessary.
But yeah you cannot use it as a homework completion site
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Thanks for this helpful document. Much appreciated.
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I would add to the second page $$\cos{(A-B)} + \cos{(A+B)} = 2\cos A \cos B \\ \cos{(A-B)} - \cos{(A+B)} = 2\sin A \sin B \\ \sin{(A-B)} + \sin{(A+B)} = 2\sin A \cos B$$
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# When traveling at a constant speed of 32 miles per hour, a certain mot
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When traveling at a constant speed of 32 miles per hour, a certain mot [#permalink]
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When traveling at a constant speed of 32 miles per hour, a certain motorboat consumes 24 gallons of fuel per hour. What is the fuel consumption of this boat at this speed measured in miles traveled per gallon of fuel?
A) $$\frac{2}{3}$$
B) $$\frac{3}{4}$$
C) $$\frac{4}{5}$$
D) $$\frac{4}{3}$$
E) $$\frac{3}{2}$$
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Re: When traveling at a constant speed of 32 miles per hour, a certain mot [#permalink]
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AbdurRakib wrote:
When traveling at a constant speed of 32 miles per hour, a certain motorboat consumes 24 gallons of fuel per hour. What is the fuel consumption of this boat at this speed measured in miles traveled per gallon of fuel?
A) $$\frac{2}{3}$$
B) $$\frac{3}{4}$$
C) $$\frac{4}{5}$$
D) $$\frac{4}{3}$$
E) $$\frac{3}{2}$$
Dear AbdurRakib,
I'm happy to respond.
We have 32 mi/gal, and 24 gal/hr, and we want mile/gal. We need to divide (mi/hr) by (gal/hr) to get (mi/gal). Thus
fuel consumption = (32 mi/gal)/(24 gal/hr) = 32/24 mi/gal
In that fraction, cancel the common factor of 8.
fuel consumption = (32 mi/gal)/(24 gal/hr) = 32/24 mi/gal = 4/3 mi/gal.
Does this make sense?
Mike
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Re: When traveling at a constant speed of 32 miles per hour, a certain mot [#permalink]
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15 Jun 2016, 00:46
1
Distance travelled = 32 miles
Fel consumed = 24 gallons
So, 24 gallons lets motorboat travel 32 miles
So, miles traveled per gallon of fuel = 32/24 = 4/3
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Re: When traveling at a constant speed of 32 miles per hour, a certain mot [#permalink]
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15 Jun 2016, 02:59
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In 1 hour the motorboat travels 32 miles
In travelling that 32 miles the motorboat uses 24 gallons of fuel in 1 hour
so basically we divide $$\frac{miles}{hr}$$ by $$\frac{gallons}{hr}$$ to get $$\frac{miles}{gallons}$$
$$\frac{32}{24}$$ =$$\frac{4}{3}$$ $$\frac{miles}{gallons}$$
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Re: When traveling at a constant speed of 32 miles per hour, a certain mot [#permalink]
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13 Aug 2016, 11:05
1
One way to solve this:--
The speed is 32 miles / hour. The fuel consumption is 24 gallons / hour. So in every hour two things are happening:-
The boat has covered 32 miles
The boat has consumed 24 gallons
What is the miles /gallon --> 32/24 => 4/3miles/gal. The choice is D
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Re: When traveling at a constant speed of 32 miles per hour, a certain mot [#permalink]
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02 Dec 2016, 06:54
AbdurRakib wrote:
When traveling at a constant speed of 32 miles per hour, a certain motorboat consumes 24 gallons of fuel per hour. What is the fuel consumption of this boat at this speed measured in miles traveled per gallon of fuel?
A) $$\frac{2}{3}$$
B) $$\frac{3}{4}$$
C) $$\frac{4}{5}$$
D) $$\frac{4}{3}$$
E) $$\frac{3}{2}$$
We are given that a motorboat has a rate of 24 gallons per hour while traveling at a speed of 32 miles per hour. We must determine the boat’s fuel consumption measured in miles traveled per gallon of fuel.
Since the motorboat consumes 24 gallons when traveling 32 miles, the rate is 32/24 = 4/3.
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Re: When traveling at a constant speed of 32 miles per hour, a certain mot [#permalink]
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23 Feb 2018, 12:17
mikemcgarry wrote:
AbdurRakib wrote:
When traveling at a constant speed of 32 miles per hour, a certain motorboat consumes 24 gallons of fuel per hour. What is the fuel consumption of this boat at this speed measured in miles traveled per gallon of fuel?
A) $$\frac{2}{3}$$
B) $$\frac{3}{4}$$
C) $$\frac{4}{5}$$
D) $$\frac{4}{3}$$
E) $$\frac{3}{2}$$
Dear AbdurRakib,
I'm happy to respond.
We have 32 mi/gal, and 24 gal/hr, and we want mile/gal. We need to divide (mi/hr) by (gal/hr) to get (mi/gal). Thus
fuel consumption = (32 mi/gal)/(24 gal/hr) = 32/24 mi/gal
In that fraction, cancel the common factor of 8.
fuel consumption = (32 mi/gal)/(24 gal/hr) = 32/24 mi/gal = 4/3 mi/gal.
Does this make sense?
Mike
hello there mikemcgarry, how are you ? you know what i dont understand, why isnt the answer B ? wouldnt it be logically correct to divide gallons by miles ?
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Re: When traveling at a constant speed of 32 miles per hour, a certain mot [#permalink]
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03 Apr 2018, 10:01
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AbdurRakib wrote:
When traveling at a constant speed of 32 miles per hour, a certain motorboat consumes 24 gallons of fuel per hour. What is the fuel consumption of this boat at this speed measured in miles traveled per gallon of fuel?
A) $$\frac{2}{3}$$
B) $$\frac{3}{4}$$
C) $$\frac{4}{5}$$
D) $$\frac{4}{3}$$
E) $$\frac{3}{2}$$
Let's see what happens after ONE HOUR of traveling.
In ONE hour, the boat will travel 32 miles and will use 24 gallons of fuel
So, the fuel consumption rate is 32 miles per 24 gallons
Or we can write: fuel consumption rate = 32/24 miles/gallon
32/24 = 4/3
So, the fuel consumption rate = 4/3 miles per gallon
Cheers,
Brent
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Re: When traveling at a constant speed of 32 miles per hour, a certain mot [#permalink]
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14 Apr 2018, 22:53
AbdurRakib wrote:
When traveling at a constant speed of 32 miles per hour, a certain motorboat consumes 24 gallons of fuel per hour. What is the fuel consumption of this boat at this speed measured in miles traveled per gallon of fuel?
A) $$\frac{2}{3}$$
B) $$\frac{3}{4}$$
C) $$\frac{4}{5}$$
D) $$\frac{4}{3}$$
E) $$\frac{3}{2}$$
It's (D)
$$\frac{Miles - traveled}{gallon-fuel}$$ = $$\frac{32}{24}$$ = $$\frac{4}{3}$$
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# The ratio of men to women in a certain club with 150 members is m : w
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The ratio of men to women in a certain club with 150 members is m : w [#permalink]
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The ratio of men to women in a certain club with 150 members is m : w and the ratio of officers to non-officers is o : n. There are 75 men in the club and there are 16 officers in the club. If two-fifteenths of female members are officers, how many male non-officers are in the club?
A. 6
B. 10
C. 65
D. 69
E. It cannot be determined from information given.
[Reveal] Spoiler: OA
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Re: The ratio of men to women in a certain club with 150 members is m : w [#permalink]
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01 Jan 2018, 03:04
Bunuel wrote:
The ratio of men to women in a certain club with 150 members is m : w and the ratio of officers to non-officers is o : n. There are 75 men in the club and there are 16 officers in the club. If two-fifteenths of female members are officers, how many male non-officers are in the club?
A. 6
B. 10
C. 65
D. 69
E. It cannot be determined from information given.
As one of our solutions is 'it cannot be determined' we cannot use the answers in this question.
Therefore, we'll go for a direct calculation, a Precise approach.
Let's write down what we know, going from the start of the question:
Men + Women = 150
Men = 75 --> Women = 150 - 75 = 75
Women officers = 2/15 * Women = 2/15 * 75 = 10
male officers = 16 - 10 = 6
male non-officers = 75 - 6 = 69.
Bunuel are we assuming that the club contains only men and women (and not, say, children)?
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Re: The ratio of men to women in a certain club with 150 members is m : w [#permalink]
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01 Jan 2018, 03:24
Bunuel wrote:
The ratio of men to women in a certain club with 150 members is m : w and the ratio of officers to non-officers is o : n. There are 75 men in the club and there are 16 officers in the club. If two-fifteenths of female members are officers, how many male non-officers are in the club?
A. 6
B. 10
C. 65
D. 69
E. It cannot be determined from information given.
Total members 150
Men 75
So women = 150-75 = 75
16 members are officers
2/15 of women are officers
So women officers = 75 * 2/15 = 10
So men officers = 16-10 = 6
So men who are not officers = 75-6 = 69
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Re: The ratio of men to women in a certain club with 150 members is m : w [#permalink]
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01 Jan 2018, 08:47
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Bunuel wrote:
The ratio of men to women in a certain club with 150 members is m : w and the ratio of officers to non-officers is o : n. There are 75 men in the club and there are 16 officers in the club. If two-fifteenths of female members are officers, how many male non-officers are in the club?
A. 6
B. 10
C. 65
D. 69
E. It cannot be determined from information given.
Breakdown and solve -
Quote:
The ratio of men to women in a certain club with 150 members is m : w.........
There are 75 men in the club.....
Men = Women = 75
Quote:
there are 16 officers in the club...
Officers = 16 & Non Officers = 134
Quote:
If two-fifteenths of female members are officers....
Female Officers = $$\frac{2}{15}*75$$ => 10
Male Officers = 6
Finally we have -
Officers = 16 ( Female Officers - 10 , Male Officers - 6) & Non Officers = 134 (Female Non Officer - 65 ; Male Non Officer = 69 )
So, Total number of male non-officers in the club is 69 , Answer is (D)
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The ratio of men to women in a certain club with 150 members is m : w [#permalink]
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01 Jan 2018, 09:51
Bunuel wrote:
The ratio of men to women in a certain club with 150 members is m : w and the ratio of officers to non-officers is o : n. There are 75 men in the club and there are 16 officers in the club. If two-fifteenths of female members are officers, how many male non-officers are in the club?
A. 6
B. 10
C. 65
D. 69
E. It cannot be determined from information given.
Total members = 150
Total Men = 75 (Given); Total Women = 150 - 75 (Total Men) = 75
Officers = 16 (Given) ; Non - officers = 150 - 16 (Officers) = 134
Female Officers = 2/15*75 = 10
Male Officers = 16 - 10 (Female Officers) = 6
Male Non Officers = Total Men - Male Officers = 75 - 6 = 69
(D)
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Re: The ratio of men to women in a certain club with 150 members is m : w [#permalink]
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01 Jan 2018, 21:29
Men -75
Women-75
Officers -16
Non officers-134
Female officers- 10
Male officers- 16- 10=6
Non officers male- 75-6=69
D
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Re: The ratio of men to women in a certain club with 150 members is m : w [#permalink] 01 Jan 2018, 21:29
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The rate of having a report ready is $10 per page for the first 50 pages and$7 per page thereafter. For revision, the rate is $4 per page for the first iteration and$3 per page for the second iteration. Ross ordered for a report of 120 pages He needed one iteration for 50 of those pages and two iterations for 20 pages. The remaining pages needed no revision. What was the bill Ross paid for the report?
A. $840 B.$1250
C. $1330 D.$1390
E. $2030 Source: Experts Global _________________ You've got what it takes, but it will take everything you've got Math Expert Joined: 02 Aug 2009 Posts: 5915 Re: The rate of having a report ready is$10 per page for the first 50 [#permalink]
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16 Jan 2018, 06:50
pushpitkc wrote:
The rate of having a report ready is $10 per page for the first 50 pages and$7 per page thereafter. For revision, the rate is $4 per page for the first iteration and$3 per page for the second iteration. Ross ordered for a report of 120 pages He needed one iteration for 50 of those pages and two iterations for 20 pages. The remaining pages needed no revision. What was the bill Ross paid for the report?
A. $840 B.$1250
C. $1330 D.$1390
E. $2030 Source: Experts Global so 120 pages.. first 50 page 10 and next $$70@7 = 50*10+70*7= 990$$.. Iteration....this is where one can go wrong one iteration $$50@4=200$$ two iteration $$20@(4+3) = 20*7=140$$.. total = $$990+200+140=1330$$ C _________________ Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372 Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html GMAT online Tutor Board of Directors Status: QA & VA Forum Moderator Joined: 11 Jun 2011 Posts: 3511 Location: India GPA: 3.5 WE: Business Development (Commercial Banking) Re: The rate of having a report ready is$10 per page for the first 50 [#permalink]
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16 Jan 2018, 10:54
pushpitkc wrote:
The rate of having a report ready is $10 per page for the first 50 pages and$7 per page thereafter. For revision, the rate is $4 per page for the first iteration and$3 per page for the second iteration. Ross ordered for a report of 120 pages He needed one iteration for 50 of those pages and two iterations for 20 pages. The remaining pages needed no revision. What was the bill Ross paid for the report?
A. $840 B.$1250
C. $1330 D.$1390
E. $2030 Source: Experts Global Ready report is 10*50 + 7*70 = 990 First iteration is 4*50 = 200 Second iteration is 20*(4+3) = 140 Thus, total cost is$1330 , answer will be (B)
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Re: The rate of having a report ready is $10 per page for the first 50 [#permalink] ### Show Tags 18 Jan 2018, 14:24 pushpitkc wrote: The rate of having a report ready is$10 per page for the first 50 pages and $7 per page thereafter. For revision, the rate is$4 per page for the first iteration and $3 per page for the second iteration. Ross ordered for a report of 120 pages He needed one iteration for 50 of those pages and two iterations for 20 pages. The remaining pages needed no revision. What was the bill Ross paid for the report? A.$840
B. $1250 C.$1330
D. $1390 E.$2030
For the first 50 pages of Rob’s report, he paid 10 x 50 = 500 dollars.
For the next 70 pages, he paid 7 x 70 = 490 dollars.
He paid 70 x 4 = 280 dollars for the first iteration of 70 pages.
He paid 20 x 3 = 60 for the second iteration of 20 pages.
In total Ross paid: 500 + 490 + 280 + 60 = 1,330 dollars,
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Re: The rate of having a report ready is $10 per page for the first 50 [#permalink] ### Show Tags 23 May 2018, 15:44 1 chetan2u wrote: pushpitkc wrote: The rate of having a report ready is$10 per page for the first 50 pages and $7 per page thereafter. For revision, the rate is$4 per page for the first iteration and $3 per page for the second iteration. Ross ordered for a report of 120 pages He needed one iteration for 50 of those pages and two iterations for 20 pages. The remaining pages needed no revision. What was the bill Ross paid for the report? A.$840
B. $1250 C.$1330
D. $1390 E.$2030
Source: Experts Global
so 120 pages.. first 50 page 10 and next $$70@7 = 50*10+70*7= 990$$..
Iteration....this is where one can go wrong
one iteration $$50@4=200$$
two iteration $$20@(4+3) = 20*7=140$$..
total = $$990+200+140=1330$$
C
Is it ok that it is not mentioned in a passage that 20 pages with 2 iterations were not from the set of 50 pages with 1 iteration?
Because if we do so, we get 1250 - option B.
Re: The rate of having a report ready is $10 per page for the first 50 [#permalink] 23 May 2018, 15:44 Display posts from previous: Sort by # The rate of having a report ready is$10 per page for the first 50
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