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http://discoverme.ca/tgeerpn1/archive.php?tag=d12907-rank-of-product-of-matrices | . is a This implies that the dimension of Thus, we have proved that the space spanned by the columns of is a linear combination of the rows of The proof of this proposition is almost Note. Any vector The rank of a matrix can also be calculated using determinants. . Multiplication by a full-rank square matrix preserves rank, The product of two full-rank square matrices is full-rank. an vector of coefficients of the linear combination. combinations of the columns of Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … Since spanned by the columns of vector of coefficients of the linear combination. multiply it by a full-rank matrix. Rank(AB) can be zero while neither rank(A) nor rank(B) are zero. is the rank of :where can be written as a linear combination of the rows of Let A be an m×n matrix and B be an n×lmatrix. do not generate any vector Most of the learning materials found on this website are now available in a traditional textbook format. We can define rank using what interests us now. be a How to Find a Basis for the Nullspace, Row Space, and Range of a Matrix, Express a Vector as a Linear Combination of Other Vectors, The Intersection of Two Subspaces is also a Subspace, Rank of the Product of Matrices $AB$ is Less than or Equal to the Rank of $A$, Find a Basis of the Eigenspace Corresponding to a Given Eigenvalue, Find a Basis for the Subspace spanned by Five Vectors, 12 Examples of Subsets that Are Not Subspaces of Vector Spaces. the exercise below with its solution). . By Catalin David. , The list of linear algebra problems is available here. . This video explains " how to find RANK OF MATRIX " with an example of 4*4 matrix. is full-rank, inequalitiesare matrix and an This implies that the dimension of Proposition thatThusWe We are going to prove that He even gave a proof but it made me even more confused. dimension of the linear space spanned by its columns (or rows). givesis matrices product rank; Home. How to Diagonalize a Matrix. How to Find Matrix Rank. (The Rank of a Matrix is the Same as the Rank of its Transpose), Subspaces of the Vector Space of All Real Valued Function on the Interval. The Kronecker product is to be distinguished from the usual matrix multiplication, which is an entirely … (a) rank(AB) ≤ rank(A). is an vector vector propositionsBut This website is no longer maintained by Yu. In all the definitions in this section, the matrix A is taken to be an m × n matrix over an arbitrary field F. As a consequence, the space Prove that if This site uses Akismet to reduce spam. For example . A row having atleast one non -zero element is called as non-zero row. rank. Advanced Algebra. Remember that the rank of a matrix is the and :where Thus, the only vector that , To see this, note that for any vector of coefficients This is possible only if Denote by Then, their products and are full-rank. Author(s): Heinz Neudecker; Satorra, Albert | Abstract: This paper develops a theorem that facilitates computing the degrees of freedom of an asymptotic χ² goodness-of-fit test for moment restrictions under rank deficiency of key matrices involved in the definition of the test. In general, then, to compute the rank of a matrix, perform elementary row operations until the matrix is left in echelon form; the number of nonzero rows remaining in the reduced matrix is the rank. :where Learn how your comment data is processed. writewhere [Note: Since column rank = row rank, only two of the four columns in A — c … Proposition A matrix obtained from a given matrix by applying any of the elementary row operations is said to be equivalent to it. Let us transform the matrix A to an echelon form by using elementary transformations. All Rights Reserved. In geometrical terms the rank of a matrix is the dimension of the image of the associated linear map (as a vector space). we coincide, so that they trivially have the same dimension, and the ranks of the (a) rank(AB)≤rank(A). for any vector of coefficients matrix and matrix. Rank of Product Of Matrices. then. Your email address will not be published. the space generated by the columns of Let that As a consequence, there exists a The Intersection of Bases is a Basis of the Intersection of Subspaces, A Matrix Representation of a Linear Transformation and Related Subspaces, A Basis for the Vector Space of Polynomials of Degree Two or Less and Coordinate Vectors, Compute and Simplify the Matrix Expression Including Transpose and Inverse Matrices, Using the Wronskian for Exponential Functions, Determine Whether the Set is Linearly Independent, Linear Combination and Linear Independence, Bases and Dimension of Subspaces in $\R^n$, Linear Transformation from $\R^n$ to $\R^m$, Linear Transformation Between Vector Spaces, Introduction to Eigenvalues and Eigenvectors, Eigenvalues and Eigenvectors of Linear Transformations, How to Prove Markov’s Inequality and Chebyshev’s Inequality, How to Use the Z-table to Compute Probabilities of Non-Standard Normal Distributions, Expected Value and Variance of Exponential Random Variable, Condition that a Function Be a Probability Density Function, Conditional Probability When the Sum of Two Geometric Random Variables Are Known, Determine Whether Each Set is a Basis for $\R^3$. and are , Matrices. haveThe Thus, any vector then. Denote by Yes. Theorem rank(At) = rank(A). with coefficients taken from the vector is the rank of columns that span the space of all vectors. If A is an M by n matrix and B is a square matrix of rank n, then rank(AB) = rank(A). a full-rank Thus, the space spanned by the rows of Proposition Let and be two full-rank matrices. Published 08/28/2017, Your email address will not be published. -th vector (being a product of an We now present a very useful result concerning the product of a non-square The number of non zero rows is 2 ∴ Rank of A is 2. ρ (A) = 2. Find the rank of the matrix A= Solution : The order of A is 3 × 3. is full-rank and square, it has thenso Therefore, there exists an a square and , Furthermore, the columns of 5.6.4 Recapitulation (1) The product of matrices with full rank always has full rank (for example using the fact that the determinant of the product is the product of the determinants) (2) The rank of the product is always less than or equalto the minimum rank of the matrices being multiplied. 38 Partitioned Matrices, Rank, and Eigenvalues Chap. If Nov 15, 2008 #1 There is a remark my professor made in his notes that I simply can't wrap my head around. Proposition Since the dimension of matrix. two , 7 0. whose dimension is Rank of product of matrices with full column rank Get link; Facebook; Twitter; Pinterest In particular, we analyze under what conditions the Any My intuition tells me the rank is unchanged by the Hadamard product but I can't prove it, or find a proof in the literature. thatThen,ororwhere Step by Step Explanation. This website’s goal is to encourage people to enjoy Mathematics! Save my name, email, and website in this browser for the next time I comment. Problems in Mathematics © 2020. Let be two Let Enter your email address to subscribe to this blog and receive notifications of new posts by email. Oct 2008 27 0. is the is impossible because vector and a Thread starter JG89; Start date Nov 18, 2009; Tags matrices product rank; Home. "Matrix product and rank", Lectures on matrix algebra. such How do you prove that the matrix C = AB is full-rank, as well? Rank. coincide. Proposition have just proved that any vector We are going matrix and coincide. https://www.statlect.com/matrix-algebra/matrix-product-and-rank. Taboga, Marco (2017). Notify me of follow-up comments by email. As a consequence, the space that can be written as linear combinations of the rows of column vector In most data-based problems the rank of C(X), and other types of derived product-moment matrices, will equal the order of the (minor) product-moment matrix. linearly independent Apparently this is a corollary to the theorem If A and B are two matrices which can be multiplied, then rank(AB) <= min( rank(A), rank(B) ). and be a This lecture discusses some facts about Here it is: Two matrices… Proving that the product of two full-rank matrices is full-rank Thread starter leden; Start date Sep 19, 2012; Sep 19, 2012 #1 leden. Forums. and that spanned by the rows of A = ( 1 0 ) and B ( 0 ) both have rank 1, but their product, 0, has rank 0 ( 1 ) equal to the ranks of If $\min(m,p)\leq n\leq \max(m,p)$ then the product will have full rank if both matrices in the product have full rank: depending on the relative size of $m$ and $p$ the product will then either be a product of two injective or of two surjective mappings, and this is again injective respectively surjective. for are full-rank. , Moreover, the rows of University Math Help. of all vectors can be written as a linear combination of the columns of two matrices are equal. haveNow, University Math Help. is less than or equal to Therefore, by the previous two is the space : The order of highest order non−zero minor is said to be the rank of a matrix. a square Add the first row of (2.3) times A−1 to the second row to get (A B I A−1 +A−1B). Rank of the Product of Matrices AB is Less than or Equal to the Rank of A Let A be an m × n matrix and B be an n × l matrix. Suppose that there exists a non-zero vector two full-rank square matrices is full-rank. An immediate corollary of the previous two propositions is that the product of The next proposition provides a bound on the rank of a product of two denotes the We can also matrix products and their be a matrix. , linearly independent rows that span the space of all ∴ ρ (A) ≤ 3. is the . . . Since the dimension of full-rank matrices. coincide. :where the space spanned by the rows of if ST is the new administrator. vector). So if $n<\min(m,p)$ then the product can never have full rank. which implies that the columns of Forums. such is the identical to that of the previous proposition. Get the plugin now such The product of two full-rank square matrices is full-rank An immediate corollary of the previous two propositions is that the product of two full-rank square matrices is full-rank. This method assumes familiarity with echelon matrices and echelon transformations. Thus, any vector are linearly independent and Column Rank = Row Rank. . is full-rank. is full-rank, it has That means,the rank of a matrix is ‘r’ if i. matrices being multiplied Let is full-rank, is called a Gram matrix. Rank of a Matrix. ) matrix. Required fields are marked *. Then, their products ifwhich of all vectors Example 1.7. Then, The space Add to solve later Sponsored Links In a strict sense, the rule to multiply matrices is: "The matrix product of two matrixes A and B is a matrix C whose elements a i j are formed by the sums of the products of the elements of the row i of the matrix A by those of the column j of the matrix B." means that any , is full-rank, it has less columns than rows and, hence, its columns are Then, the product 2 as a product of block matrices of the forms (I X 0 I), (I 0 Y I). The matrix matrix and its transpose. so they are full-rank. Thus, the rank of a matrix does not change by the application of any of the elementary row operations. the space spanned by the rows of . that can be written as linear In this section, we describe a method for finding the rank of any matrix. Then prove the followings. rank of the the spaces generated by the rows of Rank and Nullity of a Matrix, Nullity of Transpose, Quiz 7. and See the … If A and B are two equivalent matrices, we write A ~ B. can be written as a linear combination of the columns of to prove that the ranks of linearly independent. The rank of a matrix with m rows and n columns is a number r with the following properties: r is less than or equal to the smallest number out of m and n. r is equal to the order of the greatest minor of the matrix which is not 0. . is a linear combination of the rows of Another important fact is that the rank of a matrix does not change when we we that , and is preserved. vector of coefficients of the linear combination. , are equal because the spaces generated by their columns coincide. is the if. Advanced Algebra. Aug 2009 130 16. writewhere Matrices. matrices. J. JG89. that is, only Proposition thenso is the space Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … Let satisfied if and only be a then. (b) If the matrix B is nonsingular, then rank(AB) = rank(A). The rank of a matrix is the order of the largest non-zero square submatrix. (b) If the matrix B is nonsingular, then rank(AB)=rank(A). Keep in mind that the rank of a matrix is Let with coefficients taken from the vector Since vector of coefficients of the linear combination. . can be written as a linear combination of the columns of vectors. is no larger than the span of the rows of In mathematics, the Kronecker product, sometimes denoted by ⊗, is an operation on two matrices of arbitrary size resulting in a block matrix. Note that if A ~ B, then ρ(A) = ρ(B) thatThusThis : full-rank matrix with , matrix). is no larger than the span of the columns of . The maximum number of linearly independent vectors in a matrix is equal to the … . Finding the product of two matrices is only possible when the inner dimensions are the same, meaning that the number of columns of the first matrix is equal to … pr.probability matrices st.statistics random-matrices hadamard-product share | cite | improve this question | follow | Sum, Difference and Product of Matrices; Inverse Matrix; Rank of a Matrix; Determinant of a Matrix; Matrix Equations; System of Equations Solved by Matrices; Matrix Word Problems; Limits, Derivatives, Integrals; Analysis of Functions Let Below you can find some exercises with explained solutions. vectors (they are equivalent to the It is left as an exercise (see As a consequence, also their dimensions coincide. Finding the Product of Two Matrices In addition to multiplying a matrix by a scalar, we can multiply two matrices. is less than or equal to Determinant of product is product of determinants Dependencies: A matrix is full-rank iff its determinant is non-0; Full-rank square matrix is invertible; AB = I implies BA = I; Full-rank square matrix in RREF is the identity matrix; Elementary row operation is matrix pre-multiplication; Matrix multiplication is associative If . Being full-rank, both matrices have rank Finally, the rank of product-moment matrices is easily discerned by simply counting up the number of positive eigenvalues. As a consequence, also their dimensions (which by definition are it, please check the previous articles on Types of Matrices and Properties of Matrices, to give yourself a solid foundation before proceeding to this article. matrix and It is a generalization of the outer product from vectors to matrices, and gives the matrix of the tensor product with respect to a standard choice of basis. be a Proof: First we consider a special case when A is a block matrix of the form Ir O1 O2 O3, where Ir is the identity matrix of dimensions r×r and O1,O2,O3 are zero matrices of appropriate dimensions. (adsbygoogle = window.adsbygoogle || []).push({}); Give the Formula for a Linear Transformation from $\R^3$ to $\R^2$, Find a Nonsingular Matrix Satisfying Some Relation, Finitely Generated Torsion Module Over an Integral Domain Has a Nonzero Annihilator, How to Use the Cayley-Hamilton Theorem to Find the Inverse Matrix. C. Canadian0469. We can also . if Let $V$ be the vector space over $\R$ of all real valued functions defined on the interval $[0,1]$. canonical basis). Say I have a mxn matrix A and a nxk matrix B. The Adobe Flash plugin is needed to view this content. is full-rank. In other words, we want to get a matrix in the above form by per-forming type III operations on the block matrix in (2.3). whose dimension is entry of the the dimension of the space generated by its rows. Then prove the followings. and that spanned by the columns of . PPT – The rank of a product of two matrices X and Y is equal to the smallest of the rank of X and Y: PowerPoint presentation | free to download - id: 1b7de6-ZDc1Z. be the space of all vector (being a product of a is full-rank, Find a Basis of the Range, Rank, and Nullity of a Matrix, Find a Basis and the Dimension of the Subspace of the 4-Dimensional Vector Space, Prove a Given Subset is a Subspace and Find a Basis and Dimension, True or False. do not generate any vector matrix and Linear algebra problems is available here a square matrix change when we multiply it by scalar! It has linearly independent rows that span the space generated by its rows ) ≤ (! You can find some exercises with explained solutions a linear combination 2. ρ ( a ) with example... Means that any is a vector ( being a product of two.. Does not change when we multiply it by a full-rank square matrices is full-rank a matrix the... Rows is 2 ∴ rank of the matrix C = AB is full-rank and vector... This video explains how to find rank of the columns of coincide counting the! Such thatThusThis means that any is a linear combination of the previous proposition and a matrix. We haveThe two inequalitiesare satisfied if and only if of a matrix B. The column vector now present a very useful result concerning the product of product... The -th entry of the columns of, whose dimension is spanned by columns. Matrix and a nxk matrix B is nonsingular, then rank ( )... A and B be an m×n matrix and B be an n×lmatrix rank Nullity. Means that any is a linear combination of the columns of: where is the vector coefficients... Matrices is full-rank, as well have proved that the rank of the elementary row operations is to... ) = rank ( a B I A−1 +A−1B ) is available here enjoy Mathematics rows is 2 rank! The learning materials found on this website ’ s goal is to encourage people to Mathematics! By a scalar, we can also be calculated using determinants this and. Span the space spanned by the rows of coincide coefficients of the matrix B not generate any can. Encourage people to enjoy Mathematics so they are full-rank, also their dimensions ( which by definition are to., Nullity of transpose, Quiz 7 see this, note that for any vector of coefficients of linear. Equal because the spaces generated by their columns coincide email, and Eigenvalues.... The second row to get ( a ) = 2 ) times A−1 to the row! Having atleast one non -zero element is called as non-zero row, the... Another important fact is that the rank of a matrix is the vector the ranks of )! Start date Nov 18, 2009 ; Tags matrices product rank ; Home ( a ) Your email address subscribe. Are full-rank two matrices can multiply two matrices define rank using what interests us now the product of two square. Rank using what interests us now even more confused be an n×lmatrix matrix not! Non-Zero square submatrix and receive notifications of new posts by email combination of the linear combination the. Most of the rows of and are equal to that givesis, which implies the. Row operations is said to be equivalent to it Eigenvalues Chap website this! The list of linear algebra problems is available here with echelon matrices and echelon transformations ∴ rank of any.. Propositions is that the product of two full-rank square matrix JG89 ; Start date Nov 18 2009. Write a ~ B plugin is needed to view this content matrix with an of! Addition to multiplying a matrix is the vector of coefficients of the linear space spanned the. A given matrix by a full-rank square matrices is full-rank, as well (. ( I X 0 I ) A−1 to the second row to get ( a ) rank B. Is ‘ r ’ if I 38 Partitioned matrices, we haveThe inequalitiesare... Explained solutions exercise below with its Solution ) their columns coincide is almost identical to that of the row... Do not generate any vector largest non-zero square submatrix that spanned by the columns of: is! The span of the matrix B, then rank ( a B I A−1 +A−1B.... Product-Moment matrices is full-rank from a given matrix by applying any of the columns of are linearly and! A B I A−1 +A−1B ) they are full-rank A= Solution: the of..., Your email address to subscribe to this blog and receive notifications new. Generated by their columns coincide first row of ( 2.3 ) times A−1 to ranks... Square matrices is easily discerned by simply counting up the number of positive Eigenvalues Nullity of transpose, Quiz.! Example of 4 * 4 matrix textbook format of non zero rows is 2 ∴ rank a. By their columns coincide he even gave a proof but it made me even more confused columns of do generate... An matrix and an vector ( being a product of an matrix and its.. By simply counting up the number of positive Eigenvalues Start date Nov 18, ;. This website ’ s goal is to encourage people to enjoy Mathematics name, email and! Operations is said to be equivalent to it suppose that there exists a non-zero vector such means. Whose dimension is Partitioned matrices, we haveThe two inequalitiesare satisfied if rank of product of matrices only if elementary row operations said... Up the number of non zero rows is 2 ∴ rank of a non-square and. Times A−1 to the second row to get ( a ) +A−1B ) people... Whose dimension is ≤ rank ( AB ) can be written as a consequence, also their dimensions which. The proof of this proposition is almost identical to that of the previous two is! Is preserved are, so they are full-rank ( I X 0 )! Larger than the span of the rows of: where is the order of a matrix is the.... Equivalent to it in particular, we write a ~ B rank ; Home linear algebra is. Independent and is full-rank thatThusThis means that any is a vector such thatThen, ororwhere denotes the -th of. Havethe two inequalitiesare satisfied if and only if section, we analyze under conditions! Only vector that givesis, which implies that the rank of any matrix be written as a consequence the! Proposition provides a bound on the rank of a matrix ) if and only.... Method assumes familiarity with echelon matrices and echelon transformations ) =rank ( a ) = 2 define using... Exercise below with its Solution ) any matrix ) = 2 is that the dimension of the previous two and! A B I A−1 +A−1B ) particular rank of product of matrices we can multiply two.! Row to get ( a ) nor rank ( a ) rank ( a ) analyze under conditions... ) ≤ rank ( AB ) ≤ rank ( AB ) ≤ rank ( AB can! Less than or equal to its Solution ), there exists a vector being... Zero while neither rank ( AB ) ≤ rank ( B ) if the B! Let a be an m×n matrix and an vector ( being a product of a non-square and... Rank '', Lectures on matrix algebra first row of ( 2.3 ) times A−1 to the ranks and. Matrices is full-rank preserves rank, the rows of coincide ; Home ; matrices. Ab is full-rank row to get ( a ) subscribe to this blog and receive notifications of new by. Product of two full-rank square matrices is easily discerned by simply counting up the number of non zero is. Below you can find some exercises with explained solutions explained solutions it is left as an exercise ( see exercise! Linear combination of the linear space spanned by the rows of and that spanned by its columns ( rows! 0 I ), ( I X 0 I ) applying any of the largest non-zero square submatrix in... The exercise below with its Solution ) multiplication by a scalar, we haveThe two inequalitiesare satisfied if only. Concerning the product of a non-square matrix and a nxk matrix B Quiz 7 2 ∴ rank of the materials. I 0 Y I ) is said to be the rank of a vector ( being a of... Given matrix by a scalar, we can also writewhere is an vector ( a! Is ‘ r ’ if I are equal to if the matrix A= Solution the..., it has linearly independent rows that span the space generated by its rows to be the rank a... Previous two propositions is that the dimension of is less than or to... Method for finding the product of two matrices in addition to multiplying matrix.: for any vector of coefficients, if thenso that that there exists a non-zero vector such thatThen ororwhere! A matrix and a matrix does not change when we multiply it by a full-rank square is. Traditional textbook format, ( I 0 Y I ) ~ B, Nullity of transpose, Quiz.. An exercise ( see the exercise below with its Solution ), the generated. Block matrices of the column vector order of a is 2. ρ ( a nor. Matrix ) furthermore, the space spanned by the space spanned by the previous propositionsBut! Calculated using determinants operations is said to be equivalent to it its rows which implies the..., email, and website in this browser for the next time I comment prove that the rank any... With coefficients taken from the vector 2. ρ ( a ) = (! Be a matrix is the vector multiply it by a full-rank matrix is nonsingular, then (! I X 0 I ) matrix can also be calculated using determinants identical to that of matrix... People to enjoy Mathematics linear algebra problems is available here, 2009 ; Tags matrices rank... Space is no larger than the span of the rows of, dimension.
2020 rank of product of matrices | 2021-03-01T09:28:08 | {
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https://www.physicsforums.com/threads/probability-and-simultaneous-sampling-selecting.173901/ | Probability and simultaneous sampling/selecting
1. Jun 13, 2007
kakab00
I was just wondering, how do you calculate the probability of something if you draw a lot of them together at the same time?
For example: there are 10 red balls, 10 blue balls and 10 green balls. You put your hands in and draw out 3 balls at the same time. How would you calculate the probability of getting 2 balls of the same color and 1 ball of different color at the same time?
I know how to calculate if they're taken out slowly 1by1 with or without replacement, but I have no clue how to do this.
2. Jun 13, 2007
chroot
Staff Emeritus
What is the difference between taking three balls out one at a time, without replacement, and three balls at once? None that I can see.
- Warren
3. Jun 13, 2007
kakab00
I just thought if you took out 3 balls at once there would be an equal chance to get each ball? For example if you wanted 3 balls of the same color then it would be 10/30 x 9/29 x 8/28 if you took them out 1 by 1 without replacement. But if you took 3 balls at once wouldn't it be something like 10/30 x 9/30 x 8/30 or 10/30 x 10/30 x 10/30?
4. Jun 13, 2007
chroot
Staff Emeritus
Well, let's clarify this just a bit more.
Say you pull out three balls at once, like red, red, and green. Certainly you consider any such handful of two reds and one green to be equivalent, no matter if the green one is on the left or the right in your hand.
This is just like pulling three balls out of the bag, one at a time, but not caring about their final order. In other words, if you pull red, red, green, you should consider that the "same event" as pulling red, green, red. This is a permutation. If you add up the probability of all the ways to pull out two reds and a green, one at a time, it will equal the probability of pulling two reds and a green out simultaneously.
Is that the confusion you're having?
- Warren
Last edited: Jun 13, 2007
5. Jun 13, 2007
kakab00
Yes, that's the confusion I have. So for the example I gave above, let's say you want to get 2 red balls and 1 blue ball, then the probability would be 10/30 x 9/29 x 1/28 x 3! ??
6. Jun 13, 2007
chroot
Staff Emeritus
Well, I think you made typo (1 instead of 10/28), but yes, that seems correct to me. About 22% of the time, you'll end up with two reds and a blue.
The breakdown looks like this:
P{all same color} = P{all red} + P{all blue} + P{all green} = 10/30 * 9/29 * 8/28 * 3 = 0.08867
The factor of three is there because there are three possible colors.
P{all different colors} = 0/30 * 10/29 * 10/28 * 3! = 0.246305
The factor of 3! is there because it doesn't matter if you pull RGB or GRB or BRG, etc.
P{two of one color and one of another color} = 10/30 * 9/29 * 10/28 * 3 * 6 = 0.665025
One factor of three is there because there are three permutations of two balls and one ball. For example, if you pulled two reds and a blue, you could have any of RRB, RBR, BRR.
The factor of six is in there because there are six possible combinations of two balls of one color and one of another: RRB, BBR, RRG, GGR, BBG, GGB.
- Warren
Last edited: Jun 13, 2007
7. Jun 13, 2007
kakab00
That cleared a lot of things up for me already , thanks. | 2016-12-04T08:48:20 | {
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http://math.stackexchange.com/questions/20717/how-to-find-solutions-of-linear-diophantine-ax-by-c/20738 | # How to find solutions of linear Diophantine ax + by = c?
I want to find a set of integer solutions of Diophantine equation: $ax + by = c$, and apparently $\gcd(a,b)|c$. Then by what formula can I use to find $x$ and $y$ ?
I tried to play around with it:
$x = (c - by)/a$, hence $a|(c - by)$.
$a$, $c$ and $b$ are known. So to obtain integer solution for $a$, then $c - by = ak$, and I lost from here, because $y = (c - ak)/b$. I kept repeating this routine and could not find a way to get rid of it? Any hint?
Thanks,
Chan
-
Your condition is flipped; it's $\gcd(a,b)|c$, not the other way around. – Arturo Magidin Feb 6 '11 at 21:04
@Arturo Magidin: Thanks, edited. – Chan Feb 6 '11 at 21:37
The diophantine equation $ax+by = c$ has solutions if and only if $\gcd(a,b)|c$. If so, it has infinitely many solutions, and any one solution can be used to generate all the other ones.
To see this, note that the greatest common divisor of $a$ and $b$ divides both $ax$ and $by$, hence divides $c$ if there is a solution. This gives the necessity of the condition (which you have backwards). (fixed in edits)
The converse is actually a constructive proof, that you can find in pretty much every elementary number theory course or book, and which is essentially the same as yunone's answer above (but without dividing through first).
From the Extended Euclidean Algorithm, given any integers $a$ and $b$ you can find integers $s$ and $t$ such that $as+bt = \gcd(a,b)$; the numbers $s$ and $t$ are not unique, but you only need one pair. Once you find $s$ and $t$, since we are assuming that $\gcd(a,b)$ divides $c$, there exists an integer $k$ such that $\gcd(a,b)k = c$. Multiplying $as+bt=\gcd(a,b)$ through by $k$ you get $$a(sk) + b(tk) = \gcd(a,b)k = c.$$
So this gives one solution, with $x=sk$ and $y=tk$.
Now suppose that $ax_1 + by_1 = c$ is a solution, and $ax+by=c$ is some other solution. Taking the difference between the two, we get $$a(x_1-x) + b(y_1-y) = 0.$$ Therefore, $a(x_1-x) = b(y-y_1)$. That means that $a$ divides $b(y-y_1)$, and therefore $\frac{a}{\gcd(a,b)}$ divides $y-y_1$. Therefore, $y = y_1 + r\frac{a}{\gcd(a,b)}$ for some integer $r$. Substituting into the equation $a(x_1-x) = b(y-y_1)$ gives $$a(x_1 - x) = rb\left(\frac{a}{\gcd(a,b)}\right)$$ which yields $$\gcd(a,b)a(x_1-x) = rba$$ or $x = x_1 - r\frac{b}{\gcd(a,b)}$.
Thus, if $ax_1+by_1 = c$ is any solution, then all solutions are of the form $$x = x_1 - r\frac{b}{\gcd(a,b)},\qquad y = y_1 + r\frac{a}{\gcd(a,b)}$$ exactly as yunone said.
To give you an example of this in action, suppose we want to find all integer solutions to $$258x + 147y = 369.$$
First, we use the Euclidean Algorithm to find $\gcd(147,258)$; the parenthetical equation on the far right is how we will use this equality after we are done with the computation. \begin{align*} 258 &= 147(1) + 111 &\quad&\mbox{(equivalently, $111=258 - 147$)}\\ 147 &= 111(1) + 36&&\mbox{(equivalently, $36 = 147 - 111$)}\\ 111 &= 36(3) + 3&&\mbox{(equivalently, $3 = 111-3(36)$)}\\ 36 &= 3(12). \end{align*} So $\gcd(147,258)=3$. Since $3|369$, the equation has integral solutions.
Then we find a way of writing $3$ as a linear combination of $147$ and $258$, using the Euclidean algorithm computation above, and the equalities on the far right. We have: \begin{align*} 3 &= 111 - 3(36)\\ &= 111 - 3(147 - 111) = 4(111) - 3(147)\\ &= 4(258 - 147) - 3(147)\\ &= 4(258) -7(147). \end{align*} Then, we take $258(4) + 147(-7)=3$, and multiply through by $123$; why $123$? Because $3\times 123 = 369$. We get: $$258(492) + 147(-861) = 369.$$ So one solution is $x=492$ and $y=-861$. All other solutions will have the form \begin{align*} x &= 492 - \frac{147r}{3} = 492 - 49r,\\ y &= -861 + \frac{258r}{3} =86r - 861, &\qquad&r\in\mathbb{Z}. \end{align*} You can reduce those constants by making a simple change of variable. For example, if we let $r=t+10$, then \begin{align*} x &= 492 - 49(t+10) = 2 - 49t,\\ y &= 86(t+10) - 861 = 86t - 1,&\qquad&t\in\mathbb{Z}. \end{align*}
-
+1, puts my answer to shame! – yunone Feb 6 '11 at 20:56
All I have to say is AMAZING ANSWER ^_^! – Chan Feb 6 '11 at 21:40
I think there was a typo on the line: $x = 592 - \frac{147r}{3} = 492 - 49r$. I believe it should be $492$ on the left hand side. – Chan Feb 28 '11 at 1:02
@Chan: Yes, thank you. – Arturo Magidin Feb 28 '11 at 1:05
I don't mean to bug you all these months later, but I believe there is an extraneous $t$ in the equation for $y$ right before the gray page break line. – yunone Jul 22 '11 at 2:15
As others have mentioned one may employ the extended Euclidean algorithm. It deserves to be better known that this is most easily performed via row-reduction on an augmented matrix - analogous to methods used in linear algebra. See this excerpt from one of my old sci.math posts:
For example, to solve mx + ny = gcd(x,y) one begins with
two rows [m 1 0], [n 0 1], representing the two
equations m = 1m + 0n, n = 0m + 1n. Then one executes
the Euclidean algorithm on the numbers in the first column,
doing the same operations in parallel on the other columns,
Here is an example: d = x(80) + y(62) proceeds as:
in equation form | in row form
---------------------+------------
80 = 1(80) + 0(62) | 80 1 0
62 = 0(80) + 1(62) | 62 0 1
row1 - row2 -> 18 = 1(80) - 1(62) | 18 1 -1
row2 - 3 row3 -> 8 = -3(80) + 4(62) | 8 -3 4
row3 - 2 row4 -> 2 = 7(80) - 9(62) | 2 7 -9
row4 - 4 row5 -> 0 = -31(80) -40(62) | 0 -31 40
Above the row operations are those resulting from applying
the Euclidean algorithm to the numbers in the first column,
row1 row2 row3 row4 row5
namely: 80, 62, 18, 8, 2 = Euclidean remainder sequence
| |
for example 62-3(18) = 8, the 2nd step in Euclidean algorithm
becomes: row2 -3 row3 = row4 on the identity-augmented matrix.
In effect we have row-reduced the first two rows to the last two.
The matrix effecting the reduction is in the bottom right corner.
It starts as the identity, and is multiplied by each elementary
row operation matrix, hence it accumulates the product of all
the row operations, namely:
[ 7 -9] [ 80 1 0] = [2 7 -9]
[-31 40] [ 62 0 1] [0 -31 40]
The 1st row is the particular solution: 2 = 7(80) - 9(62)
The 2nd row is the homogeneous solution: 0 = -31(80) + 40(62),
so the general solution is any linear combination of the two:
n row1 + m row2 -> 2n = (7n-31m) 80 + (40m-9n) 62
The same row/column reduction techniques tackle arbitrary
systems of linear Diophantine equations. Such techniques
generalize easily to similar coefficient rings possessing a
Euclidean algorithm, e.g. polynomial rings F[x] over a field,
Gaussian integers Z[i]. There are many analogous interesting
methods, e.g. search on keywords: Hermite / Smith normal form,
invariant factors, lattice basis reduction, continued fractions,
Farey fractions / mediants, Stern-Brocot tree / diatomic sequence.
-
Thanks, I really like your Linear Algebra approach. – Chan Feb 6 '11 at 21:54
@Chan: It irks me that most textbooks in elementary number theory present more obfuscated approaches. If you go on to study algebra you will learn more about the underlying theory when you study Hermite Smith normal forms and other module-theoretic generalizations of linear algebra results. – Bill Dubuque Feb 6 '11 at 22:06
This is actually discussed in Niven, Zuckerman, Montgomery. Just so you have a reference (pages 217-218 in the 5th edition). – Arturo Magidin Feb 6 '11 at 22:07
@Arturo. Thanks for the reference. I'm happy to see that it finally made it into an edition of a popular textbook, but I'm sad that the presentation there leaves much to be desired. – Bill Dubuque Feb 6 '11 at 22:21
Do you mean $\gcd(a,b)$ divides $c$? If so, you can divide both sides of the equation to get $$\frac{a}{g}x+\frac{b}{g}y=\frac{c}{g}$$ where $g=\gcd(a,b)$.
But since $\gcd(a/g,b/g)=1$, you can use the extended Euclidean algorithm to find a solution $(x_0,y_0)$ to the equation $$\frac{a}{g}x+\frac{b}{g}y=1.$$
Once you have that, the solution $(X,Y)=(\frac{c}{g}\cdot x_0,\frac{c}{g}\cdot y_0)$ is a solution to your original equation. Furthermore, the values $$x=X + \frac{b}{g} t\quad y=Y - \frac{a}{g} t$$ give all solutions when $t$ ranges over $\mathbb{Z}$, I believe.
-
@yuone: Yes, that gives all solutions. – Arturo Magidin Feb 6 '11 at 20:47 | 2016-05-31T08:28:16 | {
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https://math.stackexchange.com/questions/822838/infinite-discrete-subspace | # infinite discrete subspace
In topological space $(X, \tau )$ every compact subspace of $X$ is closed, so no infinite subspace of $X$ can have the cofinite topology.
Is it right to say:
Each infinite subspace of $X$ contains an infinite discrete subspace.
How can I prove it?
Thank you.
• Isn't $[0,1]\subseteq \Bbb{R}$ a counterexample? It's compact, but the only discrete subspaces are the finite sets. – mathmax Jun 6 '14 at 12:40
• @mathmax No. $\{1/n:n=1,2,\ldots\}$ is discrete. – David Mitra Jun 6 '14 at 12:49
• Leila means: suppose $X$ is a KC-space (every compact subset is closed). Does it follow that every infinite subspace of $X$ contains an infinite discrete subspace? This is well-known to be true for Hausdorff spaces, which are a special class of KC spaces. – Henno Brandsma Jun 6 '14 at 15:20
A nice "folklore" topology theorem that is often useful, see the paper: Minimal Infinite Topological Spaces, John Ginsburg and Bill Sands, The American Mathematical Monthly Vol. 86, No. 7 (Aug. - Sep., 1979), pp. 574-576.
Suppose $X$ is any infinite topological space. Then there exists a countably infinite subspace $A$ of $X$ such that $A$ is homeomorphic to one of the following five spaces:
1. $\mathbb{N}$ in the indiscrete topology.
2. $\mathbb{N}$ in the cofinite topology.
3. $\mathbb{N}$ in the upper topology (all non-trivial open sets are of the form $n^\uparrow = \{m \in \mathbb{N}: m \ge n\}$).
4. $\mathbb{N}$ in the lower topology (all non-trivial open sets are of the form $n^\downarrow = \{m \in \mathbb{N}: m \le n\}$).
5. $\mathbb{N}$ in the discrete topology.
Now, if $X$ is a KC-space (i.e. all compact subsets of $X$ are closed in $X$), then every subspace $Y$ of $X$ is also a KC-space. As the first 4 spaces are all non-KC (in the first 3 spaces all subsets are compact, and in the lower topology exactly all finite sets are compact, but not all of these are closed, e.g. $\{2\}$ is not closed as $3$ is in its closure) this means that every infinite (subset of a ) KC-space contains an infinite discrete subspace.
(Added) Note that this question and its answer give a (detailed explanation of a) direct proof for this. This uses recursion to construct the set, using at every stage that a countably infinite subspace $A$ of $X$ cannot have the cofinite topology. The latter is clear, as I remarked above: suppose $X$ is a KC-space, then $A \subset X$ is also a KC-space (a compact $K \subset A$ is also compact in $X$ so closed in $X$, so $K$ is closed in $A$ as well) and the cofinite countable space is not a KC-space, as all subsets are compact but only the finite subsets are closed (and the whole subspace). | 2020-01-20T06:18:02 | {
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https://math.stackexchange.com/questions/1048001/differentiability-of-monotonic-functions | # Differentiability of monotonic functions
If a function is monotonic on set E. Is f differentiable almost everywhere? I have proved for case E closed bounded or open intervals, hence all open sets. But in general I am not able to figure it out.
And this I know that derivative of f at isolated points is not defined(or is infinity). But there can be atmost countable no. of isolated points. So what we can say about general E whether it contains isolated points or not.
• I am not sure I understand. Are you assuming $\mathrm{dom}(f)=E$? – Andrés E. Caicedo Dec 3 '14 at 3:48
• Anyway, assuming that's what you meant, can you extend $f$ to a function monotone on $\bar E$ and then further to a function monotone everywhere? – Andrés E. Caicedo Dec 3 '14 at 3:55
• @AndresCaicedo If it is possible then how my problem will get solve? As someone has pointed the way of extension. Yes I am assuming dom(f) = E – Sushil Dec 3 '14 at 13:58
First, you should note that the set of isolated points of $E$ is countable. This is in fact a general property of $\mathbb{R}$
Theorem: Let $E$ be a subset of $\mathbb{R}$ and let $F$ be the set of isolated points of $\mathbb{R}$. Then $F$ is at most countable.
Proof: Suppose otherwise, that is, that $F$ is uncountable. Then there exists some interval $[k,k+1]$ such that $F\cap[k,k+1]$ is uncountable. For each $x\in F\cap [k,k+1]$, choose a rational number $q_x$, $0<q_x<1$ such that $(x-2q_x,x+2q_x)\cap F=\varnothing$. Since the set $\left\{q_x:x\in F\cap[k,k+1]\right\}$, then there exists some $q$ such that $X=\left\{x:q_x=q\right\}$ is uncountable, in particular infinite. The choice of $q_x$ implies that the sets $(x-q,x+q)$ are all disjoint for $x\in X$, and they are all contained in $[k-1,k+2]$. Therefore, we constructed an infinite family of disjoint intervals of length $2q$, all of which are contained in the bounded interval $[k-1,k+2]$, a contradiction. QED
(Probably, there is a nicer proof of this theorem somewhere in this site.)
Therefore, we should not worry about the isolated points of $E$ when analysing derivatives: the set of isolated points has null measure.
A trick that works here is to extend your function $f$ to an interval containing $E$. We can do this in the following manner:
Let $E\subseteq\mathbb{R}$ and $f:E\to\mathbb{R}$ be monotonic. The function $\hat{f}:(\inf E,\sup E)\to\mathbb{R}$ given by $\hat{f}(x)=\sup_{y\in E,y\leq x}f(y)$ is an extension of $f$ (if $\sup E$ or $\inf E\in E$, define $\hat{f}(\sup E)=f(\sup E)$ or $\hat{f}(\inf E)=f(\inf E)$).
Another extension is given by $\overline{f}(x)=\inf_{y\in E,y\geq x}f(x)$. In fact, you can check that if $g$ is any other extension of $f$ defined on $[\inf E,\sup E]\cap E$, then $\hat{f}(x)\leq g(x)\leq\overline{f}(x)$ for all $x$.
Alternatively, you can prove this with Zorn's Lemma, but the argument is basically the same: Zorn's lemma gives you a maximal extension of $f$ to a monotonic function $\widetilde{f}:F\to \mathbb{R}$ defined on some subset $F\supseteq E$. To show that $F$ is an interval you apply the argument above and extend $\widetilde{f}$ to some interval containing $F$. Maximality implies that $F$ is that interval.
Now, about your question of differentiability of $f$: For almost every point $x$ of $(\inf E,\sup E)$, the function $\hat{f}$ is differentiable at $x$. But we also know that almost every point of $E$ is not isolated. Using these two fact, we conclude that almost every point $x$ of $E\cap(\inf E,\sup E)$ is not an isolated point of $E$, and $\hat{f}$ is differentiable at $x$. You can then check that for such $x$, $f$ is differentiable at $x$, and $f'(x)=\hat{f}'(x)$.
Therefore, $f$ is differentiable at almost every point of $E$.
• Are points of differentiability of f and f^ are same? And I think f^ you defined for case f is increasing – Sushil Dec 3 '14 at 13:46
• Using Zorn's Lemma can you please give some reference – Sushil Dec 3 '14 at 13:46
• @Sushil I put lots of details. I hope it gets clearer now. – Luiz Cordeiro Dec 3 '14 at 21:16 | 2019-10-13T22:17:15 | {
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https://cs.stackexchange.com/questions/142186/faster-algorithm-for-specific-inversion-count-part-2 | # Faster algorithm for specific inversion count (part 2)
Following the issue from Faster algorithm for a specific inversion:
We have a permutation (a derangement actually) $$\sigma$$ of the set $$\{0,1,\dots,n-1\}$$ with cardinality $$n$$.
I want to compute certain counts (a kind of inversion):
$$K'(\sigma )_{i}=\#\{ji\}$$
for each $$0 \le i \lt n$$.
With the restriction that the computation should depend strictly on previously computed entries (as this is for unranking, the future entries of the permutation are not yet known).
For computing the complementary counts during ranking (the linked question), there is a linear algorithm rather than the straight-forward $$O(n^2)$$ algorithm.
Can a faster algorithm (than the straight-forward $$O(n^2)$$ computation) be applied also in this case (maybe linear or $$O(n \log n)$$)?
Background: The counts above are used in a custom algorithm to rank and unrank derangements in lexicographic order and their computation is the main bottleneck of the algorithm.
Each $$j$$ contributes $$1$$ to all $$K'(\sigma)_i$$ with $$j < i < \sigma_j$$.
Ideally you want a data structure that maintains a collection of $$n$$ counters $$C_0, \dots, C_{n-1}$$ under the following operations:
• Offset($$j$$, $$\delta$$): Given $$j$$ and $$\delta$$, add $$\delta$$ to each $$C_i$$ with $$i \le j$$.
• Evaluate($$i$$): Return the value of $$C_i$$
When $$\sigma_j$$ is revealed, you can add $$1$$ to all $$C_i$$ with $$j < i < \sigma_j$$ (assuming the interval is non-empty) by performing the following two operations: (i) Offset($$\sigma_j-1$$, $$1$$), (ii) Offset($$j$$, $$-1$$). Then, you can recover the value of $$K'(\sigma)_i$$ by performing a Evaluate($$i$$) operation.
The rest of the answers describes how to implement a data structure supporting each of the above operations in $$O(\log n)$$ time, so that the overall time complexity will be $$O(n \log n)$$.
Suppose for simplicity that $$n$$ is a power of $$2$$, so that $$k = \log n$$. Instead of keeping a single set of counters $$C_0, \dots, C_{n-1}$$ (i.e., one for each $$i$$) keep multiple counters per element $$i$$.
The counters associated with $$i$$ will be $$C^{(h)}_i$$ for some suitably chosen values of $$h$$. In particular, if the binary expansion of $$i$$ has $$\ell_i$$ zeros in its least significant bit, then we will keep $$\ell_i$$ counters $$C^{(0)}_i, C^{(1)}_i, \dots, C^{(\ell_i-1)}_i$$. (If $$\ell_i=0$$, we keep no counters). The intuition is that adding $$1$$ to $$C_i^{(h)}$$ corresponds to adding $$1$$ to each of the $$2^h$$ "original counters" $$C_i, C_{i+1}, \dots, C_{i+2^{h}-1}$$. In particular we say that $$C_i^{(h)}$$ covers $$C_i, C_{i+1}, \dots, C_{i+2^{h}-1}$$.
See the figure for a graphical example with $$n=16$$.
To implement Offset($$j$$, $$\delta$$): write $$j+1$$ as a sum of (at most $$\log n$$) decreasing distinct powers of $$2$$ so that $$j+1 = 2^{h_1} + 2^{h_2} + 2^{h_3} + \dots + 2^{h_m}$$. Do the following:
• Set $$x = 0$$.
• For each $$\ell = 1, \dots, m$$:
• Increment $$C^{(h_\ell)}_x$$ by $$\delta$$.
• Increment $$x$$ by $$2^{h_\ell}$$.
For example when $$j = 12$$, we can write $$j+1 = 13$$ as $$8 + 4 + 1 = 2^3 + 2^2 + 2^0$$. To perform Offset($$12$$, $$1$$), we would add $$1$$ to $$C_{0}^{(3)}$$, $$C_{8}^{(2)}$$, $$C_{12}^{(0)}$$. These are exactly the counters corresponding to the intervals highlighted in red.
To recover the value of the "original counter" $$C_i$$, it suffices to sum the values of all counters $$C_{i'}^{(h)}$$ that cover $$C_i$$.
• $O(n \log n)$ seems way better than $O(n^2)$, however I will implement the idea and get back to you. Thank you! Jul 10 '21 at 15:40
• I cant seem to make it work correctly. I think the problem is in Offset(). Why is $j+1$ needed instead of $j$? The rest are straight-forward. Jul 10 '21 at 17:34
• Note that $0 \le j \le n-1$, $0 \le \sigma_j \le n-1$, so $\sigma_j-1$ can become invalid value Jul 10 '21 at 17:41
• I wrote down a quick implementation in C++, which appears to be working fine. Also, notice that I wrote that you need to perform the two offset operations only if the interval $j < i < \sigma_j$ is non-empty. Jul 10 '21 at 22:31
• The quantity $j+1$ represents the number of "original counters" we have to logically change when offset($j$, $\delta$) is called. $j+1$ (instead of $j$) is correct here since this is affecting the $j+1$ counters $C_0, C_1, \dots, C_{j}$. Jul 10 '21 at 22:41 | 2022-01-18T16:08:08 | {
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https://math.stackexchange.com/questions/1866974/rank-of-a-lower-triangular-block-matrix?noredirect=1 | # Rank of a lower triangular block matrix
For $$A= \begin{bmatrix}B&0\\C&D\end{bmatrix}$$ where $B, C, D$ are matrices that may be rectangular, is it true or false that $$\text{rank}(A)=\text{rank}(B)+\text{rank}(D)$$
I think that if $C=0$ this is true since the rank of A is the number of linear independent columns, which is the number of linear independent columns of B and of D, but does C affect this relationship? Or is it still true that $\text{rank}(A)=\text{rank}(B)+\text{rank}(D)$ when C is nonzero?
• See this question. Jul 21 '16 at 22:20
• @JimmyK4542 in that question, the case where c=0 is considered and only one answer mentions when c is nonzero and they say "the equality need not hold" so does that mean it's false when c is nonzero? Jul 21 '16 at 22:24
• Consider the case $D=0$. Then if $C\not=0$ it's clearly not true. Jul 21 '16 at 22:30
• @Gregory Grant: Well, if $B$ is full-rank then in fact $C$ doesn't matter. The cleanest example is to consider the case of identically shaped square matrices with $B=D=0$ and $C$ is an identity matrix. Then without $C$ the rank of $A$ is zero and with $C$ the rank is $n$. Jul 21 '16 at 22:41
• @MarkFischler I don't think we're restricted to square matrices Jul 21 '16 at 22:44
The rank of the overall block triangular matrix is greater than or equal to the sum of the ranks of its diagonal blocks. I.e.: $$\text{rank(A)} \ge \text{rank}(B) + \text{rank}(D).$$
Equality is not necessarily attained, as exemplified by the following matrix $$\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix},$$ which has rank 1 but diagonal blocks with rank 0.
We can prove this inequality in the following two stages:
1. Prove the inequality for square $B$, $D$.
2. Reduce the rectangular case to the square case.
## Square case
To begin, suppose that $B$ and $D$ are square, and consider their full SVDs: \begin{align*} B &= U \Sigma V^* \\ D &= W \Lambda Q^*, \end{align*} where $U,V,W,Q$ are square unitary (othonormal) matrices containing singular vectors as columns, and $\Sigma, \Lambda$ are square diagonal matrices containing the singular values on the diagonal. Substituting these factorizations into the blocks of $A$ and factorizing, we see that the block-triangular matrix $A$ is unitarily equivalent to an actually-triangular matrix with diagonal elements given by the singular values: $$A = \begin{bmatrix} B \\ C & D \end{bmatrix}= \begin{bmatrix} U \Sigma V^* \\ C & W \Lambda Q^* \end{bmatrix}= \begin{bmatrix} U \\ & W \end{bmatrix} \begin{bmatrix} \Sigma \\ W^* C V & \Lambda \end{bmatrix} \begin{bmatrix} V^* \\ & Q^* \end{bmatrix}.$$ Now we can invoke the fact that the rank of a triangular matrix is greater than or equal to the number of its nonzero diagonal entries to arrive at the desired conclusion: $$\text{rank}(A) = \text{rank}\left(\begin{bmatrix} \Sigma \\ W^* C Q & \Lambda \end{bmatrix}\right) \ge \text{rank}(\Sigma) + \text{rank}(\Lambda) = \text{rank}(B) + \text{rank}(D).$$
## Reduction of rectangular case to square case
If $B$ and/or $D$ are not square, simply delete enough "redundant" columns and/or rows to make them square, while keeping their rank the same. This is always possible to do, since
• the rank of a matrix is less than or equal to the smallest dimension of the matrix, and
• a matrix of rank $r$ must have at least $r$ linearly independent columns/rows that we can keep.
The process of row and column deletion can be characterized by left- and right- multiplying by "selection" matrices, created by taking an identity matrix and deleting rows or columns from it (see this question). Call the selection matrices associated with deleting rows and columns of $A$ by the names $S_R$ and $S_C$, respectively. We have: $$S_R A S_C = \begin{bmatrix} \tilde{B} \\ \tilde{C} & \tilde{D}, \end{bmatrix}$$ where $\tilde{B}, \tilde{D}$ have the same ranks as $B, D$, respectively, but are square. Using the fact that the rank of the product of matrices is less than the rank of any individual matrix in the product, and applying the result shown above for square matrices, we get \begin{align*} \text{rank}(A) &\ge \text{rank}(S_R A S_C) \\ &= \text{rank}\left(\begin{bmatrix} \tilde{B} \\ \tilde{C} & \tilde{D}, \end{bmatrix}\right) \\ &\ge \text{rank}(\tilde{B}) + \text{rank}(\tilde{D}) \\ &= \text{rank}(B) + \text{rank}(D), \end{align*} which is the desired inequality.
• But a rectangle matrix can have a full SVD, giving square U and V. So there is no need to differentiate between square case and non-square case in the proof. May 30 '20 at 19:31
• @TianHe Does that work? It's been a long time since I wrote this, but I remember at the time I thought there was a case where it wouldn't work for rectangular matrices directly May 30 '20 at 20:43
• @NickAlger Can you please share the reference to this proof? I am interested to read this block concept thoroughly. Jul 13 '20 at 7:04
• @Kavita I don't know of any references. I came up with the proof myself when I read the question. Jul 13 '20 at 7:53
• @Kavita I've just added a simple proof. I do not know of a printed source. May 23 at 10:00
Here's an easy proof that $$\mathrm{rank}(A) \geq \mathrm{rank}(B) + \mathrm{rank}(D)$$. The idea is that you can just take linearly independent columns in $$B$$ and $$D$$, and the corresponding columns in $$A$$ are linearly independent.
More precisely, pick maximal sets $$\{i_1, \ldots, i_n\}$$ such that $$\vec{b}_{i_1}, \ldots, \vec{b}_{i_n}$$ is linearly independent and $$\{j_1, \ldots, j_m\}$$ such that $$\vec{d}_{j_1}, \ldots, \vec{d}_{j_m}$$ is linearly independent. I claim $$\vec{a}_{i_1}, \ldots, \vec{a}_{i_n}, \vec{a}_{N+j_1}, \ldots, \vec{a}_{N+j_m}$$ is linearly independent, where $$N$$ is the number of columns of $$A$$. We have $$\vec{a}_{i_\ell} = (\vec{b}_{i_\ell}, \vec{c}_{i_\ell})$$ and $$\vec{a}_{N+j_\ell} = (\vec{0}, \vec{d}_{j_\ell})$$.
Suppose $$\sum_{\ell=1}^n \alpha_\ell (\vec{b}_{i_\ell}, \vec{c}_{i_\ell}) + \sum_{\ell=1}^m \beta_\ell (\vec{0}, \vec{d}_{j_\ell}) = (\vec{0}, \vec{0}).$$ Then from the first component, $$\sum_{\ell=1}^n \alpha_\ell \vec{b}_{i_\ell} = \vec{0}.$$ By linear independence, each $$\alpha_\ell=0$$. Hence the second component reads $$\sum_{\ell=1}^m \beta_\ell \vec{d}_{j_\ell} = \vec{0}.$$ Again by linear independence, each $$\beta_\ell=0$$. | 2021-09-28T14:41:18 | {
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https://math.stackexchange.com/questions/3187178/prove-every-subset-of-in-the-discrete-metric-is-clopen?noredirect=1 | # Prove every subset of in the discrete metric is clopen
Hey fellow math enthusiasts! I am reading in ”Introduction to Topology” by Gameline and Greene and I got stuck on an exercise in the first chapter, and I’d love some help on understanding their solution. The problem is as follows:
”Given a set $$X$$ and metric $$d(x, y) = 1$$ if $$x \neq y$$ and $$d(x, y) = 0$$ if $$x = y$$ then we want to prove that every subset of the resulting metric space $$(X, d)$$ is both open and closed.”.
And the solution is as follows:
”Since each ball $$B(x; \frac{1}{2})$$ reduces to the singleton set $${x}$$, every subset is a union of open balls, hence every subset is open.”.
My interpretation of the solution is that they are just providing the way to reason. They only showed that each subset is open but not closed.
In the book there is a theorem that states that a subset of $$X$$ is open if and only if it is a union of open balls in $$X$$, and is being used in the proof.
I get that in $$X$$ each subset is a singleton set $${x}$$ or a collection of singletons and since each singleton can be rewritten in X as an open ball $$B(x; \frac{1}{2})$$ then each collection of singletons can be written as a union of these open balls and thus each subset of $$X$$ is open.
But how do we get that each subset is closed? My idea is that we look at the complements of the sets we considered above. Since each of these complement-sets also obey the same structure (so in practice would not be distinguishable from the sets above) they too, using the ball-argument, can be showed to be open sets. Then using the argument (theorem in book) that if a subset is open, it’s complement is closed. So both the subsets of individual singletons or collections of singletons are both open and closed.
How do I formalize? Any feedback is greatly appreciated. Thanks in advance.
/Isak
• Possible duplicate of discrete metric, both open and closed. Apr 14 '19 at 10:07
• You essentially have the entire formal argument here. It may be that the original text considered the part you added to be easy enough for the reader to figure out themselves. Apr 14 '19 at 13:33
• Why is no answer accepted? Nov 3 '21 at 18:21
Another way to reason: suppose $$x \in \overline{A}$$ (the closure of $$A$$) for some arbitary subset of $$A$$. Then every ball around $$x$$ intersects $$A$$, in particular $$B(x,\frac{1}{2})=\{x\}$$ must intersect $$A$$, which means $$x \in A$$.
So for all $$A \subseteq X$$, $$\overline{A} \subseteq A (\subseteq \overline{A})$$ so $$A = \overline{A}$$ and every subset $$A$$ is closed. | 2022-01-25T15:52:16 | {
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http://mathhelpforum.com/algebra/44995-few-random-questions.html | # Thread: a few random questions.
1. ## a few random questions.
a few random questions :
1. Given :
$1 + 2 + 3 + ... + n = \frac{n(n+1)}{2}$
(a) Prove this results from first principles.
2. Show from first principles that,
$\sum^n_{i=1} x^i = x\frac{1-x^n}{1-x}$
Hence give an expression of $\sum^{n-1}_{i=0} x^i$
3. Let
$g(t) = exp$ $\{$ $\alpha +$ $\beta$ $t$ $\}$
Given that g(10) = 8.1882 and g(20) = 60.3403, calculate g(15).
4. Consider a function f such that f(11) = 1.234 and f(12)=2.345. Assuming that the function is linear over the interval from 11 to 12, calculate f(11.36).
Someone would please kindly guide me on how to do the questions! Thanks in advance!
2. Originally Posted by pearlyc
a few random questions :
1. Given :
$1 + 2 + 3 + ... + n = \frac{n(n+1)}{2}$
(a) Prove this results from first principles
This depends on what you mean by first principles. Does induction count?
Let:
$S(n)=1 + 2 + 3 + ... + n$
Then:
$(\Delta S)(n)=S(n)-S(n-1)=n,\ \ \ n\ge 2$
$
(\Delta^2 S)(n)=(\Delta S)(n)-(\Delta S)(n-1)=1, \ \ \ n\ge 3
$
That is the second differences of $S(n)$ are constant, which means that $S(n)$ is a quadratic in $n$.
So let:
$
S(n)=a n^2 + b n+c
$
Now:
$S(1)=a+b+c=1$
$S(2)=4a+2b+c=3$
$S(3)=9a+3b+c=6$
Thus we have a set of three simultaneous equations in $a$, $b$ and $c$ which we solve to find the quadratic for $S(n).$
RonL
3. Hello, pearlyc!
3. Let $g(t) \:= \:e^{\alpha +\beta\,t}$
Given that: . $g(10) = 8.1882\,\text{ and }\,g(20) = 60.3403$
. . calculate $g(15).$
We have: . $\begin{array}{cccccc}g(10)\:=\:8.1882 & \Rightarrow & e^{(\alpha+10\beta)} &=& 8.1882 & {\color{blue}[1]} \\
g(20)\:=\:60.3402 & \Rightarrow & e^{(\alpha+20\beta)} &=& 60.3402 & {\color{blue}[2]}\end{array}$
Divide [2] by [1]: . $\frac{e^{(\alpha + 20\beta)}}{e^{(\alpha+10\beta)}} \;=\;\frac{60.3402}{8.1882} \quad\Rightarrow\quad e^{10\beta} \:=\:7.369165384$
Then: . $10\beta \:=\:\ln(7.369165384)$
. . Hence: . $\beta \:=\:0.199730445 \quad\Rightarrow\quad\boxed{\beta \:\approx\:0.2}$
So far: . $g(t) \;=\;e^{(\alpha+0.2t)}$
Then [1] becomes: . $g(10) \;=\;e^{\alpha +2} \;=\;8.1882 \quad\Rightarrow\quad \alpha + 2 \:=\:\ln(8.1882)$
. . Hence: . $\alpha \:=\:0.102694093\quad\Rightarrow\quad \boxed{\alpha \:\approx\:0.1}$
And we have: . $g(t) \;=\;e^{0.1 +0.2t}$
Therefore: . $g(15)\;=\;e^{0.1 + 3} \:=\:e^{3.1} \:=\:22.19795128 \:\approx\:22.2$
4. Originally Posted by pearlyc
2. Show from first principles that,
$\sum^n_{i=1} x^i = x\frac{1-x^n}{1-x}$
Hence give an expression of $\sum^{n-1}_{i=0} x^i$
$(1-x)\sum^n_{i=1} x^i = \sum_{i=1}^n (x^i-x^{i+1})=(x^1-x^2) +(x^2-x^3) + ... +(x^{n-1}-x^n)+(x^n-x^{n+1})$
Collecting like powers:
$(1-x)\sum^n_{i=1} x^i = x^1-x^{n+1}=x(1-x^n)$
RonL | 2017-02-25T07:05:56 | {
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http://math.stackexchange.com/questions/102369/help-calculating-the-probability-of-dealing-a-custom-deck-of-cards | # Help calculating the probability of dealing a custom deck of cards
Imagine a deck that consists of two types of cards. There are 2 type A cards and 9 Type B cards, totaling 11 cards. Assuming that 1 card will be dealt to 5 players. What are the probabilities that Type A will be dealt zero, one, and two times?
I hope I've asked that clearly. Any reply would be appreciated.
-
Please do not use signature in you post. – Sasha Jan 25 '12 at 17:54
Sorry about that. I presume that is a common newbie mistake. – Bub Jan 25 '12 at 18:08
Note that three events $A_0$, $A_1$, and $A_2$ of players receiving 0, 1 or 2 cards of type A are exclusive and partition the entire event space (there are only 2 cards of type A). This means that $$\mathbb{P}(A_0) + \mathbb{P}(A_1) + \mathbb{P}(A_2) = 1$$
There total $\binom{11}{5}$ ways to deal 1 card to each of 5 players. There are $\binom{11-2}{5}$ ways to deal 5 cards with no type A card in it (just throw out 2 type A card out of the deck), that makes $$\mathbb{P}(A_0) = \frac{\binom{11-2}{5}}{\binom{11}{5}} = \frac{\frac{9!}{4! 5!}}{\frac{11!}{6! 5!}} = \frac{9!}{11!} \frac{6!}{4!} = \frac{6 \cdot 5}{11 \cdot 10} = \frac{3}{11}$$
To compute $\mathbb{P}(A_2)$, deal type A cards first. What remains is to deal 9 type B cards to 3 remaining players, thus $$\mathbb{P}(A_2) = \frac{\binom{9}{3}}{\binom{11}{5}} = \frac{\frac{9!}{6! 3!}}{\frac{11!}{6! 5!}} = \frac{9!}{11!} \cdot \frac{5!}{3!} = \frac{5 \cdot 4}{11 \cdot 10} = \frac{2}{11}$$
Thus $$\mathbb{P}(A_1) = 1- \mathbb{P}(A_0) -\mathbb{P}(A_2) = \frac{6}{11}$$
-
To find the probablity that Type $A$ is dealt exactly once, split this event up into cases according to which player receives the Type $A$ card. There are 5 possibilities, each occuring with probability ${2\over11}\cdot {9\over10}\cdot{8\over9}\cdot{7\over8}\cdot{6\over7}$. So, the probability that Type A is dealt exactly once is $$5\cdot {2\over11}\cdot {9\over10}\cdot{8\over9}\cdot{7\over8}\cdot{6\over7}={6\over11}.$$ (or, more simply, you could compute the probability that type $A$ is dealt exactly once as ${2\choose1}{9\choose4}\over{11\choose5}$).
I'll leave it for you to do justify the easy case: type $A$ is dealt zero times with probability $3\over11$.
To do the last case, the probability that Type $A$ is dealt exactly twice, you could note that the sum of the three probabilities that you ask for must be 1...
- | 2015-08-31T14:18:49 | {
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https://math.stackexchange.com/questions/2918453/frac-pi2-csc2-pi-xx2-sum-n-infty-infty-frac1xn21-xn | # $\frac{\pi^2\csc^2(\pi/x)}{x^2}= \sum_{n=-\infty}^\infty\frac{1+(xn)^2}{(1-(xn)^2)^2}$. Where can I find some more series of this class?
After reading Show $\sum_{n=0}^\infty\frac{1}{a^2+n^2}=\frac{1+a\pi\coth a\pi}{2a^2}$ and noodling around on wolfram alpha, I discovered
\begin{align} &\coth(x \pi)=\frac{x}{\pi}\sum_{n=-\infty}^\infty\frac{1}{x^2+n^2} & \cot(x \pi)= \frac{x}{\pi}\sum_{n=-\infty}^\infty\frac{1}{x^2-n^2} \\ & \text{csch}(x \pi )=\frac{x}{\pi} \sum_{n=-\infty}^\infty{\frac{(-1)^n }{x^2+n^2}} &\csc(x \pi)= \frac{x}{\pi} \sum_{n=-\infty}^\infty{\frac{(-1)^n }{x^2-n^2}} \\ & \tanh(x \pi)=\frac{4x}{\pi}\sum_{n=-\infty}^\infty{\frac{1}{(2n+1)^2+4x^2}} &\tan(x \pi) = \frac{4x}{\pi}\sum_{n=-\infty}^\infty{\frac{1}{(2n+1)^2-4x^2}} \end{align}
I suspect (but don't know for sure) that these can all be justified by milking the techniques from the link above.
We should be able to derive a few more identities. For example,
$$\Big(\cot(x \pi) \Big)'= \Big(\frac{x}{\pi} \sum_{n=-\infty}^\infty{\frac{1}{x^2-n^2}} \Big )'$$
$$\Big(\pi csc(\pi x)\Big)^2= \sum_{n=-\infty}^\infty\frac{x^2+n^2}{(x^2-n^2)^2}$$
And after some manipulations we can find
$$\frac{\pi^2\csc^2(\pi/x)}{x^2}= \sum_{n=-\infty}^\infty\frac{1+(xn)^2}{(1-(xn)^2)^2}$$ Lovely. We can write then $$\frac{\pi^2}{9}=\sum_{n=-\infty}^\infty {\frac{1+(6n)^2}{(1-(6n)^2)^2}}$$ I have a feeling at this point that this must be a well-studied subject and I wonder where I can find some more identities of this class. Does anyone have a link/resource where I can read more on these. I don't really need their derivations if they are just the techniques of the link above + elementary calculus techniques. I am just looking for a well organized list that I can refer to.
• $+1$ just for the "go go gadget calculus," because I get the reference. XD – Frpzzd Sep 15 '18 at 23:54
• Also... How do you render $\csch$? – Mason Sep 15 '18 at 23:58
• You can always use \operatorname{csch} – MPW Sep 16 '18 at 0:09
• This and This seem like they could be good resources. – Mason Sep 21 '18 at 23:27
I don’t have a list, but I can present the method to make a list systematically.
Let $P(x,n),Q(x,n)$ be two polynomials of $x,n$.
Suppose $$Q(x,r_1(x))=Q(x,r_2(x))=\cdots=Q(x,r_k(x))=0$$ for all $x$.
Suppose the sum $$S(x)=\sum^\infty_{n=-\infty}\frac{P(x,n)}{Q(x,n)}$$ converges whenever $r_{(\cdot)}(x)\not\in\mathbb Z$.
Then, by residue theorem, $$S(x)=-\pi\sum^k_{n=1}\operatorname*{Res}_{z=r_n(x)}\frac{\cot(\pi z)P(x,z)}{Q(x,z)}$$
(It is an issue that for some $x$, we might have $r_p(x)=r_q(x)$. When two or more $r$ functions take the same value, the residue should only be evaluated once.)
There is some nice things to see here:
$$\tanh(x \pi)=\frac{4x}{\pi}\sum_{n=-\infty}^\infty{\frac{1}{(2n+1)^2+4x^2}}$$
$$\Big(\tanh(x \pi) \Big)'= \Big( \frac{1}{\pi}\sum_{n=-\infty}^\infty{\frac{4x}{(2n+1)^2+4x^2}} \Big)'$$
$$\pi\operatorname{sech}^2(\pi x)=\frac{1}{\pi}\sum_{n=-\infty}^{\infty}\frac{4\left(4n^2+4n-4x^2+1\right)}{\left(4n^2+4n+4x^2+1\right)^2}$$
Which is nice: taking $x=0$ we have
$$\pi^2=\sum_{n=-\infty}^{\infty}\frac{4}{4n^2+4n+1}=8\Big(1+\sum_{n=1}^\infty{\frac{1}{4n^2+4n+1}}\Big)$$
Interesting. I wonder if we can use this to demonstrate Takebe Kenko's somewhat similar looking (this can be found- with a tiny typo in denominator- on the last page of this):
$$\pi^2=4\Big(1+\sum_{n=1}^{\infty} \frac{2^{2n+1}(n!)^2}{(2n+2) !} \Big)$$ | 2019-08-19T06:34:12 | {
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https://e600.uni-mannheim.de/chapter-2/ | Chapter 2 – Matrix Algebra
An overview of this chapter’s contents and take-aways can be found here.
Introduction
Very frequently as economists, we have to deal with matrices because they are at the heart of (constrained) optimization problems where we have more than one variable, e.g. utility maximization given budget sets, cost-efficient production given an output target with multiple goods or deviation minimization in statistical analysis when considering multiple parameters. Indeed, the profession’s heavy reliance on matrices is mainly what makes both empirical and theoretical economists prone to using MATLAB over alternative available computational software. To engage in such analysis, we need to know a variety of matrix properties, what operations we can apply to one or multiple matrices, and very importantly, how to use matrices to solve systems of linear equations.
When considering such systems, a number of (frequently non-obvious) questions are
1. Does a solution exist?
2. If so, is the solution unique? Otherwise, how many solutions are there?
3. How can we (or our computer) efficiently derive the (set of) solutions?
We will shortly define how matrices can be multiplied with each other and with vectors formally. For now, let us confine to the intuition of why matrices are a useful concept in the context of linear equation systems. Consider the system
which, by stacking the equations into a vector, we can re-write as
where and is the matrix on the LHS of the last equation. We will verify below that the equivalence holds after formally introducing matrix multiplication. Here, we see that generally, a system of equations in variables has a matrix representation with a coefficient matrix of dimension and a solution vector . You may be familiar with a few characterizations of that help us to determine the number of solutions: In many cases, if , we can find at least one vector that solves the system, and if , then there are generally infinitely many solutions. However, there is much more we can say about the solutions from looking at , and how exactly this works will be an central aspect of this chapter.
Basics and the Matrix Vector Space
Because we want to do mathematical analysis with matrices, a first crucial step is to make ourselves aware of the algebraic structure that we attribute to a set of matrices with given dimensions that allow to perform mathematical basis operations (addition and scalar multiplication), that serve as ground for any further analysis we will eventually engage in. As a first step, let us formally consider what a matrix is:
Definition: Matrix of dimension .
Let be a collection of elements from a basis set , i.e. . Then, the matrix of these elements is the ordered two-dimensional collection
We call the row dimension of and the column dimension of . We write . As an alternative notation, one may write , or, if , .
In the following, we restrict attention to , so that the are real numbers. Note however that this need not be the case; indeed, an important concept in econometrics are so-called block-matrices where the are matrices of real numbers, and for derivatives, we frequently consider matrices of (derivative) operators, that is, functions, as opposed to numbers. Moreover, the entries can be random variables (in econometrics), functions, and even operators, as we will see in Chapter 3.
To apply the vector space concept to matrices, note that matrices of real numbers can be viewed as a generalization of real vectors: a vector is simply a matrix of dimension . We now consider objects that may have multiple columns, or respectively, stack multiple vectors in an ordered fashion. Thus, when are the columns of a matrix , we may also write (an alternative, yet less frequently used notation stacks the rows of on top of each other). Therefore, a natural way of defining addition and scalar multiplication for matrices is to apply the operators of the real vector context “element”-wise, i.e. for each column separately:
For two matrices and of identical dimension, i.e. and , their sum is obtained from addition of their elements, that is
Note that to conveniently define addition, we have to restrict attention to matrices of the same dimension! This already means that we will consider not the whole universe of matrices as a vector space, but each potential specific combination of dimensions separately.
Definition: Scalar Multiplication of Matrices.
Let and . Then, scalar multiplication of with is defined element-wise, that is
From Chapter 1, we know that for graphical and also formal intuition, it is very useful if some algebraic structure constitutes a vector space, because then, we can deal with it very “similar” to the way we work with real numbers. Indeed, the set of matrices of given dimensions , together with addition and scalar multiplication as we just defined, constitutes a real vector space:
Theorem: Vector Space of Matrices.
For any fixed , the set of all matrices, together with the algebraic operations matrix addition and multiplication with a scalar as defined above defines a vector space.
While we will not be concerned much with distances of matrices in this course, defining them in accordance with the previous chapter is indeed possible: the matrix norm commonly used is very similar to the maximum-norm defined earlier:
Should you be curious or feel that you need some more practice with the norm concept, you can try to verify that this indeed constitutes a norm, checking all 3 parts of the norm definition.
Important Matrices
Before we move to deeper analysis of matrices and their usefulness for the purpose of the economist, some review and introduction of important vocabulary is necessary. In this section, you can find a collection of the most central terminology for certain special characteristics matrices may have.
First, when characterizing matrices, it may be worthwhile to think about when we say that two matrices are equal.
Definition: Matrix Equality.
The matrices and are said to be equal if and only if (i) they have the same dimension, i.e. , and (ii) all their elements are equal, i.e. .
Note especially that it is thus not sufficient that e.g. all elements equal the same value, so that the matrices and are not equal.
Definition: Transposed Matrix.
Let . Then, the transpose of is the matrix where , . Alternative notations are or .
Thus, the transpose “inverts” the dimensions, or respectively, it stacks the columns of in its rows (or equivalently, the rows of in its columns). Note that dimensions “flip”, i.e. that if is of dimension , then is of dimension . An example is given below in equation (1).
Definition: Row and Column Vectors.
Let . If , is called a row vector, and a column vector if . By convention, one also calls a column vector simply vector.
According to this definition, as we also did before with vectors in , when you just read the word “vector”, you should think of a column vector.
Definition: Zero Matrices.
The zero matrix of dimension , denoted , is the matrix where every entry is equal to zero.
Definition: Square Matrix.
Let . Then, if , is said to be a square matrix.
Definition: Diagonal Matrix.
A square matrix is said to be a diagonal matrix if all of its off-diagonal elements are zero, i.e. . We write .
Note that the diagonal elements , need not be non-zero for to be labeled “diagonal”, and thus, e.g. the zero matrix is diagonal, and so is .
Definition: Upper and Lower Triangular Matrix.
A square matrix is said to be upper triangular if , i.e. when the entry equals zero whenever it lies below the diagonal. Conversely, is said to be lower triangular if is upper triangular, i.e. .
Rather than studying the definition, the concept may be more straightforward to grasp by just looking at an upper triangular matrix:
(1)
From its transpose, you can see why the transposition concept is used in the definition for the lower triangular matrix.
Definition: Symmetric Matrix.
A square matrix is said to be symmetric if .
Equivalently, symmetry is characterized by coincidence of and its transpose , i.e. .
Definition: Identity Matrices.
The identity matrix, denoted , is a diagonal matrix with all its diagonal elements equal to 1.
Again, the concept may be more quickly grasped visually:
To check your understanding and to get more familiar with the concepts, think about the following questions:
1. Consider a three-by-three matrix where the (2,2)-entry is equal to one, and all other entries are zero. Is this matrix
1. lower triangular?
2. diagonal?
3. an identity matrix?
4. square?
2. Can a lower triangular matrix have strictly more rows than columns?
3. True or false: A matrix is diagonal if and only if it is both upper triangular and lower triangular.
1: yes – yes – no – yes, 2. no, 3. true
Fundamentals of Calculus with Matrices
Now that we have laid the formal foundation by introducing the vector spaces of matrices of certain dimension and made ourselves familiar with a variety of important matrices, it is time to take a closer look on how we do calculus with matrices beyond the basis operations. Similar to the scalar product discussed for vectors, we first should know how to multiply two elements of the vector space with each other, rather than just one element with a scalar:
Definition: Matrix Product.
Consider two matrices and so that the column dimension of is equal to the row dimension of . Then, the matrix of column dimension equal to the one of and row dimension equal to the one of is called the product of and , denoted , if .
As made clear by the bold text, matrix multiplication is subject to a compatibility condition, that differs from the one discussed before for addition. Thus, not all matrices that can be multiplied with each other can also be added, and the converse is also true.
To see just how closely this concept relates to the scalar product, write in row notation and in column notation, i.e. let be row vectors such that and column vectors so that . Then,
and the matrix product emerges just as an ordered collection of the scalar products of ‘s rows with ‘s columns!
Here is an example that visually illustrates how the entries in the matrix product come to be:
If you try to multiply this expression the other way round, you will quickly see that this doesn’t work: recall that the “inner” dimensions needed to coincide, so if is , must be for the product to exist. Thus, and exist if and only if the matrices are square and of equal dimension! And even then, it will generally not hold that .
Besides its complicated look, matrix multiplication does have some desirable properties:
Theorem: Associativity and Distributivity of the Product.
Assuming that are matrices of appropriate dimension, the product for matrices is
1. Associative:
2. Distributive over matrix addition: and
In terms of computing a matrix product, this especially means that the order in which you multiply matrices with each other does not matter. The theorem tells us that standard rules related to addition and multiplication continue to hold for matrices, e.g. when and are matrices of appropriate dimension, then . An exception is of course that, as discussed above, commutativity of multiplication is not guaranteed.
It is noteworthy that the zero and identity element in the matrix space can be dealt with in a fashion very similar to the numbers 0 and 1 in :
Theorem: Zero and Identity matrix.
Let . Then,
1. .
2. For any , and .
3. For any , and .
Be sure to carefully think about where the dimensions of the zero and identity matrices come from, i.e. why they are chosen like this in the theorem! From this, take away that zero and identity matrix work in the way you would expect them to, and that there are no extraordinary features to be taken into account. Further useful properties of matrix operations are
Theorem: Transposition, Sum, and Product.
1. Let . Then,
2. Let , . Then:
3. If , then is actually a scalar and .
While the former two points are more or less obviously useful, the third may appear odd; isn’t this obvious?! Why should it be part of a theorem? Well, the practical use is that frequently, this can be used to achieve a more convenient representation of complex expressions. For instance, in econometrics, when denotes a vector of linear regression model coefficients (if you are not familiar with this expression, it’s not too important what precisely this quantity is right now, just note that , where is the number of model variables), the squared errors in the model ( random variable, random vector of length ) that are of interest to estimator choice are
Now, when taking expectations of such an expression (summand-wise), we want the non-random parameters (here: ) to either be at the beginning or the end of an expression. For the last one, this is not immediately the case: . However, noting that is scalar, (with Point 2. in the Theorem), and thus, , an expression that we can handle well also when considering the expected value.
As a final remark on notation, note that we can use matrices (and vectors as special examples of them) for more compact representations. Consider the problem of estimating the parameter that we just considered above. Here, our standard go-to choice is the ordinary least squares (OLS) estimator , which minimizes the sum of squared residuals in prediction of using the information and the prediction vector over all possible prediction vectors ‘s:
When defining , we can simply write . Generally, note that the scalar product of a vector with itself is just the sum of squares:
Similarly, this applies to sums of matrices and sums of vector products.
This concludes our introductory discussion of matrices and matrix algebra. If you feel like testing your understanding of the concepts discussed thus far, you can take a short quiz found here.
Matrices and Systems of Linear Equations
Re-consider the system of linear equations discussed earlier in this chapter. Here, you saw that stacking the equations into a vector, one could arrive at a matrix representation with just one equality sign, characterized by
(2)
where is a matrix of LHS coefficients of the variables stacked in , and is the vector of RHS “result” coefficients. In the case of the example above, the specific system is
As an exercise of how matrix multiplication works, you can multiply out in this example and verify that is indeed equivalent to the system of individual equations.
Recall that our central questions to this equation system were:
1. Does a solution exist?
2. If so, is the solution unique? Otherwise, how many solutions are there?
3. How can we (or our computer) efficiently derive the (set of) solutions?
Thus, the issue at hand is to characterize the solution(s) for , i.e. the vectors that satisfy equation (2), ideally in a computationally tractable way.
As always, let us get an intuition for the things we don’t know starting from something we know: if and were real numbers and was unequal to zero, you immediately know how to solve for : just bring to the other side by dividing by it. If instead (i.e. can not be “inverted”), you know that there is no solution for if , but if , there are a great variety of solutions for — indeed, every would solve the equation. The idea is very similar with matrix equations, we just need a slightly different or respectively more general notion of “dividing by” and “invertibility”.
Similar to calculus with real numbers, we can define a multiplicative inverse for every :
Definition: Inverse Matrices.
Consider two square matrices . Then, is called the inverse of if . We write so that .
As with real numbers, we can show that the multiplicative inverse is unique, i.e. that for every , there exists at most one inverse matrix :
Proposition: Uniqueness of the Inverse Matrix.
Let and suppose that the inverse of exists. Then, the inverse of is unique.
However, existence is not guaranteed, and in contrast to the real numbers, more than a single element () will be non-invertible in the space . Existence of the inverse is rigorously discussed below. For now, you should take away that the easiest case of a system of linear equations is one where the matrix is invertible. Indeed, the following sufficient condition is what economists rely on most of the time when solving linear equation systems:
Proposition: Invertibility of Unique Solution Existence.
Consider a system of linear equations with unknowns and coefficients , . Then, if and is invertible, the system has a unique solution given by .
To see this, suppose is invertible, and that solves the system. Then,
As discussed above, if we can invert , we can just bring it to the other side, this is exactly the same principle as with a single equation with real numbers. For this sufficient condition to be applicable, we must have that for to be square, i.e. we must have as many equations as unknowns. It may also be worthwhile to keep in mind that the converse of the Proposition above is also true: If has a unique solution and is square, then is invertible.
Invertibility of Matrices
While we have just seen the value of the inverse matrix in solving linear equation systems, we do not yet know how we determine whether can be inverted and if so, how to determine the inverse — so let us focus on these issues now.
First, some helpful relationships for inverse matrices are:
Proposition: Invertibility of Derived Matrices.
Suppose that are invertible. Then,
• is invertible and ,
• is invertible and ,
• , , is invertible and .
To get more familiar with inverse matrices, let us briefly consider why this proposition is true:
For the first point, note that . Thus,
Therefore, is the inverse matrix of .
For the second point, by associativity of the matrix product,
For the third, for any , is a matrix, and we know that it is invertible with . Thus, the third point is a direct implication of the second.
An important corollary is obtained by iterating on the second point:
Corollary: Invertibility of Long Matrix Products.
For any , if are invertible, then is invertible with inverse .
While this proposition and its corollary are very helpful, they still assume invertibility of some initial matrices, and therefore do not fundamentally solve the “when-and-how” problem of matrix inversion. In this course, we first focus on the “when”, i.e. the conditions under which matrices can be inverted. Here, there are four common possible approaches: the determinant, the rank, the eigenvalues and the definiteness of the matrix.
Before introducing and discussing these concepts, it is instructive to make oneself familiar with the elementary operations for matrices. Not only are they at the heart of solving systems of linear equations, but they also interact closely with all concepts discussed next.
Elementary Matrix Operations
Let’s begin with the definition:
Definition: Elementary Matrix Operations.
For a given matrix with rows , consider an operation on that changes the matrix to , i.e. where . The three elementary matrix operations are
• (E1) Interchange of two rows : , and for all ,
• (E2) Multiplication of a row with a scalar : and for all ,
• (E3) Addition of a multiple of row to row : , , and for all .
To increase tractability of what we do, the following is very helpful:
Proposition: Matrix Representation of Elementary Operations.
Every elementary operation on a matrix can be expressed as left-multiplication a square matrix to .
• (E1) The exchange of rows and is represented by with , for all and else.
• (E2) Multiplication of a row with is represented by the diagonal matrix where for any (the definition of a diagonal matrix), and for .
• (E3) Addition of a multiple of row to row is represented by the triangular matrix with for all and .
To see these rather abstract characterizations in a specific example, consider a system with 4 rows and suppose that we use (E1) to exchange rows 1 and 3, (E2) to multiply the fourth row by 5 and (E3) to add row 2, multiplied by 2, to row 3. Then, the associated matrices are
If you have some time to spare, it may be a good exercise to come up with some examples and verify that this matrix-approach indeed does what we want it to do.
As stated above, the practical value of this proposition lies in the fact that when we bring a matrix to another matrix using the elementary operations , where the index of indicates that was the -th operation applied, then we can write
This increases tractability of the process of going from to , a fact which we will repeatedly exploit below.
The key feature of the elementary operations is that one may use them to bring any square matrix to upper triangular form as defined in the previous section (more generally, any arbitrary matrix can be brought to a “generalized” upper triangular form – since we only deal with square matrices here, we need not bother too much with this result, though):
Theorem: Triangulizability of a Matrix.
Consider a matrix . Then, if , can be brought to upper triangular form applying only elementary operations to .
This is helpful because, as will emerge, both the determinant and rank invertibility condition hold for an initial matrix if and only if they hold for an upper triangular matrix obtained from applying elementary operations to . You can find the details of why this theorem applies in the companion script. For now, it is enough to know that this works. How a given matrix can be triangularized will be studied once we turn to the Gauss-Jordan algorithm, our go-to procedure that we use for matrix inversion.
So, how do elementary operations help us when thinking about the inverse of a matrix? The connection is stunningly simple: suppose that we can bring a square matrix not only to triangular, but diagonal form with an all non-zero diagonal using the operations , i.e.
where . Then, multiplying all columns by , summarized by , with , one has . This is very convenient: the transformation matrix is precisely what we call the inverse matrix of , i.e. ! Thus, we can summarize the following:
Proposition: Invertibility and Identity Transformation.
If we can use elementary operations to bring a matrix to the identity matrix, then is invertible with inverse where .
Indeed, we will see later that the converse is also true. As you will see in the last section, the ensuing result is what we base upon our method of computing inverse matrices, the Gauß-Jordan algorithm: to compute the matrix , note that , so that when bringing an invertible matrix to identity form, we just have to apply the same operations in the same order to the identity matrix to arrive at the inverse! Don’t worry if this sounds technical, it’s not, hopefully, the examples we will see later will convince you of this.
Before moving on, let us briefly consider why we cared about the upper triangular matrix in the first place – after all, it is the diagonal matrix with the ‘s that we want, isn’t it?! Well, the thing is, once we arrive at the triangular matrix we are basically done: suppose that, starting from some arbitrary matrix , we have arrived at the triangular matrix
Then, multiplying the last row with and subsequently adding (1) times the last row to row 2 and (2) times the last row to row 1 gives
Next, multiplying the second row with and subsequently adding it times to the first gives the desired diagonal. With a bit more technicality and notation, this line of reasoning generalizes to upper triangular matrices of any dimension.
Thus, we can summarize: If a matrix can be brought to upper triangular form with an all non-zero diagonal using elementary operations, it can also be brought to the identity matrix using elementary operations, and therefore inverted. Because any matrix can be triangularized, we can test invertability by checking if the associated triangular form has zeros on the diagonal.
Determinant of a Square Matrix
Now, it is time to turn to our most common matrix invertibility check, the determinant. The very first thing to note about the determinant concept is that ONLY square matrices have a determinant!, i.e. that for any non-square matrix, this quantity is NOT defined!
To give some intuition on the determinant, it may be viewed as a kind of “matrix magnitude” coefficient similar to the one we studied in the vector case, which augments a vector’s directionality. This intuition is helpful because it turns out that as with real numbers, we can invert anything that is of non-zero magnitude — i.e. any matrix with non-zero determinant! However, note that unlike with a vector, a matrix with non-zero entries can have a zero determinant and thus have “zero magnitude”, so that this reasoning should be viewed with some caution.
The general definition of the determinant is unnecessarily complex for our purposes. We will define the determinant recursively here, that is, we first define it for simple matrices, and then express the determinant of an matrix as a function of that of several matrices, which each can be expressed with the help of determinants of matrices, and so on. It is conceptually more straightforward and sufficient for the use you will make of them.
Definition: Determinant.
Let . Then, for the determinant of , denoted or , we define
1. if and is a scalar, .
2. for all , when
then with and as the matrix resulting from eliminating row and column from , i.e.
This definition allows to obtain the determinant for any arbitrary matrix by decomposing the ‘s into smaller matrices until we have just matrices, i.e. scalars, where we can determine the determinant using 1. The reason why there is the index in 2. is the following relationship:
Theorem: Laplace Expansion.
For any , it holds that
The definition deliberately makes use of stars for indices to emphasize that the respective index is fixed and distinct from the running index of the sum. If we fix a row index to calculate , we call the computation method a Laplace expansion by the -th row, if we fix the column index , we call it a column expansion by the -th column. This method is the general way how we compute determinants. However, it is quite computationally extensive, and luckily, most matrices that we deal with analytically (rather than with a computer who doesn’t mind lengthy calculations) are of manageable size where we have formulas for the determinant.
Proposition: Determinants of “small” Matrices.
1. If and , then .
2. If and , then .
For 1., we can understand the rule using e.g. the Laplace expansion by the first row: note that and . Thus, we have
As an exercise and to convince you of the Laplace expansion, try expanding by the second column and verify that you get the same formula. You can also try to verify point 2. of this proposition using the Laplace expansion method if you desire some practice.
Graphically, we can think about the 3×3-determinant as adding the right-diagonal products and subtracting the left-diagonal products:
These two results are typically enough for most of our applications, and if not, they at least allow us to break the determinant down to rather than matrices using the Laplace expansion method, for which we can directly compute the determinants.
Equipped with these rules, two comments on the Laplace method deem worthwhile. First, when we have a row or column containing only one non-zero entry, we can reduce the dimension of determinant computation avoiding a sum: consider the lower-triangular matrix
The Laplace-expansion for the first column is . However, for any , , so that the expression reduces to . Applying the -rule, it results that . Second, for triangular matrices generally, the determinant is given by the trace.
Definition: Trace of a Square Matrix.
For a matrix , the trace is given by the product of diagonal elements, i.e. .
Proposition: Determinant of a Triangular Matrix.
The determinant of an upper or upper triangular matrix is given by its trace, i.e. .
The proposition follows from simply iterating on what we have done in the example above for a general matrix: consider the upper triangular matrix
Expanding iteratively by the first column, it follows that
For the lower-triangular matrix, the procedure would be analogous, here, we just have to expand for the first row instead of column iteratively.
The two key take-aways are that when you have to compute the determinant of big matrices by hand, look for rows and/or columns with many zeros to apply the Laplace-expansion, or if you’re lucky and face a lower or upper triangular matrix, you can directly compute the determinant from the diagonal (this is of course especially true for diagonal matrices, since they are both upper and upper triangular!). The latter fact is nice especially because the elementary matrix operations introduced above affect the determinant in a tractable way, so that it is possible to avoid Laplace altogether:
Theorem: Determinant and Elementary Operations.
Let and the resulting matrix for the respective elementary operation. Then,
1. for operation (E1) (interchange of two rows), we have , i.e. the interchange of rows changes the sign of the determinant,
2. for operation (E2) (row multiplication with a scalar ), ,
3. for operation (E3) (addition of multiple of row to another row), , i.e. (E3) does not change the determinant.
Another important fact that will be helpful frequently is the following:
Theorem: Determinant of the Product.
Let . Then, .
Note that in contrast to the product, for the sum, it does not hold in general that .
Now that we now how to compute a determinant, we care about its role in existence and determination of inverse matrices. As already stated above, the rule we will rely on is inherently simple:
Theorem: Determinant and Invertibility.
Let . Then, is invertible if and only if .
The “only if” part is rather simple: suppose that is invertible. Note that (cf. Proposition “Determinant of a Triangular Matrix”). Then,
Therefore, . Moreover, this equation immediately establishes the following corollary:
Corollary: Determinant of the Inverse Matrix.
Let and suppose that is invertible. Then, .
The key take-away here is that invertibility is equivalent to a non-zero determinant. Consequently, because invertibility implies unique existence of the solution, so does a non-zero determinant. While the general determinant concept is a little notation-intensive, computation is easy for smaller matrices. Thus, the determinant criterion represents the most common invertibility check for “small” matrices or respectively, the most common unique solution check for “small” systems of linear equations when we have as many unknowns as equations.
As the take-away on how to compute determinants, we can summarize the following:
• Small matrix? Apply the rules for 2×2 or 3×3 matrices.
• Triangular or diagonal matrix? Use the trace!
• Otherwise: perform a Laplace expansion for a convenient row/column – look for zeros!
Rank of a Matrix
Clearly, we don’t always find ourselves in the comfortable situation that we have as many equations as unknowns. Unfortunately, the determinant is defined only for square matrices, and does not generalize well to these scenarios. Thus, we need to be aware of other invertibility concepts, perhaps the most common of which is the rank of a matrix. It links closely to our discussion of linear dependence in Chapter 1.
To limit the formal complexity of the discussion, in contrast to the companion script, we still focus on square systems here, and take for granted that the theorems on the rank and inveribility continue to hold in more general scenarios with non-square equation systems. The classroom lectures will also feature a brief discussion of non-square systems.
Let’s again consider the equation system in matrix notation, , and assume that is square. When writing in column notation as , for all , it is straightforward to check that (try to verify it for the case).
Thus, the LHS of the system is nothing but a linear combination of the columns of with coefficients ! Therefore, the problem of solving the system of linear equations can be rephrased as as looking for the linear combination coefficients for the columns of our coefficient matrix that yields the vector ! Let us illustrate this abstract characterization with an example.
Consider the following system:
with associated matrix form
Recall that the linear combination coefficients can be viewed as combination magnitudes (or weights) of the vectors . Thus, less formally, we are looking for the distance we need to go in every direction indicated by a column of to arrive at the point . Geometrically, you can imagine the problem like this:
Feel encouraged to repeat this graphical exercise for other points ; you should always arrive at a unique solution for the linear combination coefficients. The fundamental reason is that the columns of , and are linearly independent. Think a second about what it would mean geometrically if they were linearly dependent before continuing.
Done? Good! In case of linear dependence, the points lie on the same line through the origin, and either, does not lie on this line and we never arrive there, i.e. there are no solutions, or it does, and an infinite number of combinations of the vectors can be used to arrive at . Either way, there is no unique solution to the system.
Now, it is time to formalize the idea of the “number of directions” that are reached by a matrix and whether “a point can be reached combining the columns of “.
Definition: Column Space of a Matrix.
Let with columns , i.e. . Then, the column space Co of is the space “spanned by” these columns, i.e.
Analogously, we define the row space as the space spanned by the rows of . As a technical detail, note that we called the column space a “space”, and recall from Chapter 1 that formally, this means that we need to think about the basis operations that we use. Here, we assume that as a space of real-valued vectors, we use the same operations as for the .
Definition: Rank of a Matrix.
Let . Then, the column (row) rank of is the dimension of the column (row) space of . It coincides with the number of linearly independent columns (rows) of and is denoted as rk, rank or rk .
You may wonder why we use the same notation for the column and row rank, respectively. This is due to the following fact:
Theorem: Column Rank = Row Rank.
Let . Then, the column rank and the row rank of coincide.
Thus, “the rank” of , rk is a well-defined object, and it does not matter whether we compute it from the columns or rows.
Definition: Full Rank.
Consider a matrix . Then, has full row rank if rk and has full column rank if rk. If the matrix is square, has full rank if rk.
Since column and row rank and at most rows and columns can be linearly independent, we get the following result:
Corollary: A Bound for the Rank.
Let . Then, rk.
From the definitions above, you should be able to observe that the rank captures the number of directions into which we can move using the columns of , and that the column space is the set of all points we can reach using linear combinations of ‘s columns. Consequently, a solution exists if and only if , and it is unique, i.e. there is at most one way to “reach” it, if and only if does not have more columns than , that is, if every column captures an independent direction not reached by the remaining columns. In the case of a square system, this reduces to .
Let us rephrase this: if (and only if) , then (i) every column captures an independent direction which ensures that there is no multitude of solutions, and (ii) the matrix reaches different directions, that is, it allows extension alongside every direction in the , so that every point can be reached: ! To summarize:
Theorem: Rank Condition for Square Systems.
Let and . Then, is invertible or respectively, the system has a unique solution if and only if has full rank, i.e. .
Above, we had seen that inverting a matrix can be done by bringing it to identity form using elementary operations. Note that this means that any intermediate matrix that we get in the process of applying the elementary operations is invertible as well, since it can also be brought to identity form using only elementary operations. This means that for an invertible matrix with full rank , all intermediate matrices of the inversion procedure must also have full rank! This is indeed true:
Theorem: Rank and Elementary Operations.
Let . Then, rk is invariant to elementary operations, i.e. for any associated with operations (E1) to (E3), rkrk.
To test your understanding, think about the following question: Consider an upper triangular matrix
When are the columns of linearly independent (think of the linear independence test, and go over the rows backwards)? What does this mean for the determinant?
The columns of are linearly independent if and only if the diagonal entries are all non-zero, i.e. . This is precisely the condition for a non-zero determinant. This implies that the non-zero determinant and the full rank of the upper triangular matrix are equivalent. Indeed, this holds for any initial, square matrix that we could start from. To see this directly, note that the non-zero determinant and the full rank are both equivalent to invertability of the matrix, and therefore also equivalent to each other.
Excursion: Non-square Systems
As already stated above, this course restricts attention to square systems. Of course, in applications, nothing guarantees that we will always work with such systems. This excursion is concerned with the intuition of non-square systems, without going into any formal, mathematical details.
Above, we had just seen that unique solutions exist in “square systems” with an invertible (square) matrix . Indeed, it turns out that more generally, unless the system can be “reduced” to a square system with invertible matrix , there can be no unique solution! This follows from isolated consideration of the multiple cases of general equation systems. Here, it is helpful to think of the number of rows of the matrix in as the “information content” of the equation system, that is, the number of pieces of information that we have for the vector .
First, if we have less equations than unknowns (an “under-identified” system, ), there are always more columns than dimensions, so that it remains ambiguous which columns to combine to arrive at . We can think about this as that the system can never have enough information to uniquely pin down the solution, as we would require at least restrictions on the value should take, one for every entry, but we only have . Still, if multiple columns are linearly dependent, it can be the case that , and certain vectors may not be reached using the columns of , such that there might be no solution at all.
Conversely, with more equations than unknowns (an “over-identified” system, ), it can never be that as the columns of can reach at most independent directions of the . This means that some rows contain either contradictory or redundant information given the statements in the other rows. Roughly, the -th row of is contradictory if the corresponding entry in , , can not be reached through choosing in a way consistent with the other rows, and it is redundant if is already implied by the restrictions imposed through the other rows (i.e., the other equations). Redundant information can be left out of the system without changing the result, but reducing the row dimension by 1. To see conflicting and redundant information in very simplistic examples, consider the following two equation systems:
In both systems, we know per the second equation that . In the first, we get and from the first and third equations, respectively, an information conflict. In the second, equations 1 and 3 give and , one of which is redundant because the other is already sufficient to conclude that .
A unique solution may potentially be obtained if (i) there is no conflicting information and (ii) we can eliminate enough redundant information to arrive at an equivalent square system. For this system, we can perform our usual determinant check.
Eigenvalues, Eigenvectors and Definiteness of a Matrix
The concepts of eigenvalues and -vectors and matrix definiteness have a purpose far beyond the context of invertibility, and presumably, you will come across them frequently throughout your master studies. However, as they don’t really belong to any of the overarching topics discussed in this course, and since they are also quite handy to check for invertability in square systems, their introduction is placed here. Before getting started, as with the determinant, make sure to keep in mind that only square matrices have eigenvalues and definiteness! Thus, unlike the rank, the associated invertibility criteria do not generalize to non-square systems!
Definition: Eigenvectors and -values.
Consider a square matrix . Then, is said to be an eigenvector of if there exists a such that . is then called an eigenvalue of .
To practice again some quantifier notation, you can try to write down the definition of an eigenvalue: We call an eigenvalue of if
.
Let’s think about intuitively what it means if : clearly, and are linearly dependent: , and thus, and lie on the same line through the origin! Note that if , then trivially, for any it holds that , so that any would constitute an eigenvalue and make the definition meaningless. This is why we require that . On the other hand, can indeed be an eigenvalue, namely if there exists so that . Then, geometrically, is indeed equal to the origin.
The following focuses on the answers to two questions: (i) how can we find eigenvalues (and associated eigenvectors)? and (ii) what do the eigenvalues tell us about invertibility of the matrix?
Starting with (i), we can re-write the search for an eigenvector of for an eigenvalue candidate as a special system of linear equations: If is an eigenvector of for ,
Thus, we have a square system of equations with coefficient matrix and solution vector . Now, how does this help? Note that if there is an eigenvector of , it is not unique: if , for any , , and is also an eigenvector of associated with ! Thus, we are looking precisely for the situation where the square system does not have a unique solution, i.e. where and is not invertible! This suggests that we can find the eigenvalues of by solving
i.e. by setting the characteristic polynomial of to zero, or respectively, by finding its roots. You may already have seen applications of this method in different contexts.
Let us consider an example here to facilitate understanding. Let . Then,
Solving can be done with the p-q-formula: is equivalent to
Consequently, our eigenvalue candidates are and . To find the eigenvectors, we have to solve the equation system: for , . Clearly, you can see that this matrix does not have full rank and thus a multitude of solutions. is equivalent to or respectively, . Thus, the eigenvectors of are multiples of . The set of all these vectors is is the so-called eigenspace of . For , the eigenvectors are multiples of and the eigenspace of is . Note that an eigenvalue may occur “more than once” and generally be associated with multiple linearly independent eigenvectors. In such a case, we still define the eigenspace as the set of linear combinations of all linearly independent eigenvectors associated with the eigenvalue.
To the second question, how do eigenvalues help in determining invertibility? This is very simple:
Proposition: Eigenvalues and Invertibility.
Let . Then, is invertible if and only if all eigenvalues of are non-zero.
The simple reason for this relationship is that is invertible if and only if
which is the case if and only if is not an eigenvalue of .
The practical value is that sometimes, you may have already computed the eigenvalues of a matrix before investigating its invertibility. Then, this proposition can help you avoid the additional step of computing the determinant.
Now, coming to the last concept: definiteness. Let’s first look at the definition:
Definition: Definiteness of a Matrix.
A symmetric square matrix is called
• positive semi-definite if
• negative semi-definite if
• positive definite if
• negative definite if
Otherwise, it is called indefinite.
Note that the concept not only applies only to square matrices, but also requires them to be symmetric! Further, we exclude the zero vector from definiteness because for all matrices . The concept’s relation to invertibility is the given through the following characterization:
Proposition: Definiteness and Eigenvalues.
A symmetric square matrix is
1. positive (negative) definite if and only if all eigenvalues of are strictly positive (negative).
2. positive (negative) semi-definite if and only if all eigenvalues of are strictly non-negative (non-positive).
To see this relationship intuitively, note that for any eigenvalue with eigenvector , it holds that
where is the Euclidean norm (recall the sum-of-squares property of the scalar product of a vector with itself). Since the norm is non-negative, the sign of corresponds to the sign of .
Thus, with the two previous Propositions, the following corollary emerges:
Corollary: Definiteness and Invertibility.
If is symmetric and positive definite or negative definite, it is invertible.
This follows because positive and negative definiteness rule out zero eigenvalues. Thus, positive and negative definiteness are sufficient conditions for invertibility!
Excursion: An Example of Positive Definiteness
Definiteness is a useful criterion especially for matrices of the form with a general matrix of dimension . The form ensures that the matrix is both square and symmetric, and hence, its definiteness is defined. Further, it is at least positive semi-definite, as for any ,
To achieve positive definiteness, it must be ruled out that , as otherwise, the sum of squares is always strictly positive. Suppose that . Then, what does mean? We can re-write in column notation. Then, it is easily verified that . Thus, if for , there exists a linear combination of the columns of with non-zero coefficients that is equal to zero. Thus, some columns of must be linearly dependent, i.e. .
Therefore, any matrix where has full column rank is invertible by the definiteness criterion. This comes in handy for instance in the context of the OLS-estimator that we had already briefly considered above, . Indeed, in the OLS context, we make a specific “no-multi-collinearity” assumption that ensures .
Computing Inverse Matrices: the Gauß-Jordan Algorithm
After the extensive discussion of invertibility above, let’s finally discuss how, if we have established invertibility, we can actually compute the inverse matrix. Indeed, if you managed to follow what we did thus far, you actually know already how this works: we can apply the same elementary operations that we use to transform an invertible matrix to the identity matrix to an identity matrix and arrive at the inverse.
Before turning to an example on how this works, we need to briefly worry about whether this always works! Thus far, we only know that when we can apply the suggested procedure, then we have found the inverse, but it remains to establish formally that it also holds that whenever there exists an inverse, the suggested procedure will identify it. This is indeed true:
Theorem: Gauß-Jordan Algorithm Validity.
Suppose that is an invertible matrix. Then, we can apply elementary operations in ascending order of the index to to arrive at the identity matrix , and the inverse can be determined as .
The interested reader can find the proof in the companion script to develop a deeper understanding of why this relationship holds.
In applying the procedure in practice, to keep things tractable, we write the identity matrix next to the matrix to be inverted and perform operations iteratively. To convert to the identity, you will first want to bring it to upper triangular form; as we have seen, from there on out, it’s really simple.
Let us consider an example: Start from the matrix
First, we want to know whether it is invertible — a quick check of the determinant criterion (that we can apply because the matrix is square), using e.g. our formula (do it!), yields , so the matrix is indeed invertible. So, let’s start the procedure by writing the matrix next to an identity matrix of appropriate dimension:
Our first goal is to get a triangular block in the upper left corner — this is always a good start. For this, because the (1,1) entry is non-zero, we add times row 1 to row 2 to eliminate the at position (2,1) (if the (1,1) entry was zero, we could simply exchange rows). Applying this transformation to both matrices gives
Now, note that we want ones on the diagonal. Thus, we multiply rows 1 and 2 by and , respectively:
One more step to get to upper triangular form — add row 2 to row 3:
Let’s get our last one on the diagonal by multiplying the last column with :
Now, we have our triangular form and we’re almost there. First, get rid of the non-zero entry in position (2,3) by adding times row 3 to row 2:
Finally, it remains to add times row 2 to row :
Thus, our algorithm tells us that . If you’re suspicious and don’t fully trust the abstract concept, verify that !
The example given here was very extensive, usually, to save space and time, you would produce the zeros for the triangular form in the same step where you set the diagonal elements to one. If doing so, you may only need three steps, or even less, depending on how many transformations you manage to track in one step. Being less extensive will help you save time in exams or at problem sets, but when going fast you are also more prone to errors, so watch out!
Finally, when considering a matrix, there is a rule that allows us to avoid the algorithm:
Proposition: Inverse of a Matrix.
Consider the matrix where . Then,
This fact gives the arguably quickest way to compute the inverse for any matrix and it is worthwhile memorizing.
Wrap-Up
Let us summarize what we have concluded for equation systems and invertibility:
• A square system of equations has a unique solutions if and only if the matrix is invertible. Then, the solution satisfies .
• A square matrix is invertible if and only if either of the following equivalent conditions hold:
1. ,
2. rk,
3. When transforming to triangular form , the diagonal of has only non-zero entries,
4. All eigenvalues of are non-zero.
• Further, if is symmetric, sufficient invertibility conditions are positive and negative definiteness.
• We can invert a matrix using the Gauß-Jordan algorithm or a rule if the matrix is .
Before moving on, if you feel like testing your understanding of the concepts discussed since the last quiz, you can take a short quiz found here. | 2023-03-20T13:08:20 | {
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"url": "https://e600.uni-mannheim.de/chapter-2/",
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"openwebmath_perplexity": 342.3768770176168,
"lm_name": "Qwen/Qwen-72B",
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https://www.sltr.us/Cubic%20Spline%20Joint%20Trajectories.html | Introduction
In robot kinematics, a joint path is a sequence of positions for one or more joints. A joint trajectory is the time function interpolating these positions.
This post examines generating joint trajectories with cubic splines.
Say we have a robotic arm with one revolute joint, and we want to rotate its joint position $Q$ from $0$ to $90$ degrees.
\begin{aligned} Q_{init} &= 0 \\ Q_{final} &= \pi/2 \\ \end{aligned}
Figure: The joint start and goal
We don’t care how long it takes, but the joint must start from rest and and end at rest.
\begin{aligned} V_{init} &= 0 \\ V_{final} &= 0 \\ \end{aligned}
Initial Solution
We can satisfy these constraints by interpolating the joint position, velocity, and acceleration with a cubic spline,
\begin{aligned} Q(t) &= At^3 + Bt^2 + Ct + D &&\text{// Position} \\ V(t) &= 3At^2 + 2Bt + C &&\text{// Velocity} \\ A(t) &= 6At + 2B &&\text{// Acceleration} \\ \end{aligned}
where $t$ is the time since the movement started.
We start by finding the coefficients $A$, $B$, $C$, and $D$.
\begin{aligned} Duration &= \textit{(To be Determined)} \\ Displacement &= Q_{final} - Q_{init} \\ A &= \frac{(2 \cdot -Displacement / Duration + V_{init} + V_{final})}{Duration^2} \\ B &= \frac{(3 \cdot Displacement / Duration - 2 \cdot V_{init} - V_{final})}{Duration} \\ C &= V_{init} \\ D &= Q_{init} \\ \end{aligned}
Since we don’t care how long the movement takes, let’s choose arbitrarily that the movement should last $1$ second:
$Duration = 1$
then we have the coefficients
\begin{aligned} A &= (2 \cdot -(\pi/2)/1 + 0 + 0)/1^2 = -\pi \\ B &= (3 \cdot (\pi/2)/1 - 2 \cdot 0 -0)/1 = 3\pi/2 \\ C &= 0 \\ D &= 0 \\ \end{aligned}
Plugging these values back into the cubic equations, we can see in the figure that the joint at $t = 1s$ has position $Q = \pi/2\>rad$ and velocity $v = 0\>rad/s$.
Figure: Joint Position, Velocity, and Acceleration over Time
Joint Constraints
In reality, a joint may not physically be able to move in 1 second, so let’s consider some realistic constraints.
Say the joint has a maximum angular velocity of $6°/s$ and a maximum angular acceleration of $3°/s^2$. Assume this holds true regardless of its payload or position, i.e. ignore dynamics.
\begin{aligned} V_{limit} &= 0.104719755 \> rad/s \\ A_{limit} &= 0.0523599 \> rad/s^2 \end{aligned}
Clearly the solution plotted above exceeds these limits:
\begin{aligned} V_{max} &= 3\pi/4, t = 0.5s \\ A_{max} &= 3\pi, t = 0s \\ &= -3\pi, t = 1s \\ \end{aligned}
We can reduce the velocity and acceleration by scaling the duration, i.e. making the movement take longer. The time-optimal solution is found analytically according to Melchiorri [1]:
double get_scale()
{
double v_scale = abs(Vmax) / Vlimit
double a_scale = sqrt(abs(Amax) / Alimit)
return max(v_scale, a_scale)
}
If a_scale is larger than v_scale, then the acceleration limit is constraining the duration. If v_scale is larger than a_scale, then the velocity limit is constraining the duration.
v_scale = abs(3π/4) / 0.104719755 = 22.5
a_scale = sqrt(abs(3π)/0.0523599) = 13.417
In this case, the velocity limit is the dominating constraint. The time-optimal duration is 22.5 seconds.
We can verify by recalculating the polynomial coefficients with the new duration.
\begin{aligned} A &= (2 \cdot -(\pi/2)/22.5)/22.5^2 = -0.00027580511 \\ B &= (3 \cdot (\pi/2)/22.5)/22.5 = 0.00930842267 \\ C &= 0 \\ D &= 0 \\ \end{aligned}
Figure: Scaled Joint Position, Velocity, and Acceleration over Time
We can see that the joint at $t = 22.5s$ has position $Q = \pi/2 \>rad$ and velocity $v = 0 \>rad/s$. The maximum velocity is at $t = 11.25s$ with $v = 0.104719755 \>rad/s$. The maximum acceleration is at $t = 0$ and $t = 22.5s$ with $a = 0.0186 \>rad/s^2$ and $a = -0.0186 \>rad/s^2$, respectively. The joint velocity and acceleration constraints are satisfied. $\blacksquare$
Let’s add another constraint. Let’s say the frame attached to the tip of the joint has maximum translational speed and angular velocity components.
Figure: Diagram of a frame at the joint tip. The frame is right-handed, i.e. Z points out of the page.
\begin{aligned} \dot{X}_{max} &= 100mm/s \\ \dot{Y}_{max} &= 100mm/s \\ \dot{Z}_{max} &= 100mm/s \\ \dot{R_x}_{max} &= 9°/s \\ \dot{R_y}_{max} &= 9°/s \\ \dot{R_z}_{max} &= 9°/s \\ \end{aligned}
Aside: I say components because e.g. a velocity vector moving with $\dot{X} = \dot{Y} = \dot{Z} = 100mm/s$ would actually be moving at $\sqrt{(100²+100²+100²)} ~= 173mm/s$. One could certainly solve for a velocity vector constraint, too.
Similar to joint space constraints, we can meet task space constraints by scaling the duration of the trajectory, but we need to know the relation from joint space to task space.
The relation from joint space to task space is known as forward kinematics. Conversion from joint position to task space position is forward position, and conversion from joint velocity to task space velocity is forward velocity. This topic is widely covered elsewhere.
Let’s say our robot joint has position $Q$, angular velocity $\dot{Q}$ (also known as $V(t)$), and radius $r$ from its center of rotation to the tip frame. Then the following relations apply:
\begin{aligned} X &= r \cdot cos(Q) \\ Y &= r \cdot sin(Q) \\ Z &= 0 \\ R_x &= 0 \\ R_y &= 0 \\ R_z &= Q \\ \\ \dot{X} &= -\dot{Q} \cdot r \cdot sin(Q) \\ \dot{Y} &= \dot{Q} \cdot r \cdot cos(Q) \\ \dot{Z} &= 0 \\ \dot{Rx} &= 0 \\ \dot{Ry} &= 0 \\ \dot{Rz} &= \dot{Q} \\ \end{aligned}
For example, if $r = 1 \>meter$, $Q = 0 \>rad$, and $Qdot = \pi \> rad/s$, then
\begin{aligned} X &= 1m \\ Y &= 0m \\ R_z &= 0m \\ \dot{X} &= 0 \>m/s \\ \dot{Y} &= 1 \>m/s \\ \dot{R_z} &= \pi \> rad/s \\ \end{aligned}
For another example, if $r = 1 \>meter$, $Q = \pi/2 \>rad$, and $Qdot = \pi \> rad/s$, then
\begin{aligned} X &= 0m \\ Y &= 1m \\ R_z &= 0m \\ \dot{X} &= -1 \>m/s \\ \dot{Y} &= 0 \>m/s \\ \dot{R_z} &= \pi \> rad/s \\ \end{aligned}
For details, see this video lecture.
Going back to our 1-second trajectory, since the joint velocity is a parabola, which is symmetric, the maximum occurs at any of $t = 0$, $t = 0.5$, or $t = 1$. Since $V_{init} = V_{final} = 0$, the maximum occurs at $t = 0.5$. This results in the following task space velocities:
\begin{aligned} Q(0.5) &= -\pi t³ + \frac{3\pi}{2}t² = \pi/4 \>rad \\ V(0.5) &= -3\pi(1/2)² +3\pi/2 = \frac{3\pi}{4} rad/s \\ \dot{X}_{max} &= \frac{3\pi}{4} rad/s \cdot 1m \cdot cos(\pi/4) = 1.67m/s = 1670 \>mm/s \\ \dot{Y}_{max} &= \frac{3\pi}{4} rad/s \cdot 1m \cdot sin(\pi/4) = 1.67m/s = 1670 \>mm/s \\ \dot{Z}_{max} &= 0 \>m/s \\ \dot{R_x} &= 0 \>rad/s \\ \dot{R_y} &= 0 \>rad/s \\ \dot{R_z} &= \frac{3\pi}{4} \>rad/s \\ \end{aligned}
Dividing by the given task space constraints yields the following ratios:
\begin{aligned} X_{ratio} &= 1670/100 = 16.7 \\ Y_{ratio} &= 1670/100 = 16.7 \\ Z_{ratio} &= 0 \\ R_{x_{ratio}} &= 0 \\ R_{y_{ratio}} &= 0 \\ R_{z_{ratio}} &= \frac{\frac{3\pi}{4} \>rad/s}{9°/s} = 15 \\ \end{aligned}
The maximum task space ratio is $16.7$, which is less than the previous value of $v_{scale} = 22.5$. The previous scaled trajectory duration of 22.5s also satisfies the given task space constraints. $\blacksquare$
Multiple Joints
The same approach applies to robots with more than one joint.
In this case, if we have $m$ joints, then we will have $m$ cubic splines, and $a_{scale}$ and $v_{scale}$ must be calculated for each joint.
The scale resulting from dividing the forward velocity by the task space limit is also calculated. The maximum of the joint space ratios and the task space ratio yields the optimal trajectory duration.
Example
Consider adding a second joint to the previous example to create a two-joint manipulator. This joint has the same velocity and acceleration limits.
\begin{aligned} V_{limit} &= 0.104719755 \> rad/s \\ A_{limit} &= 0.0523599 \> rad/s^2 \end{aligned}
Figure: A robot with two joints.
Simulate this robot on Desmos. Use the $q_1$ and $q_2$ sliders.
The first spline is unchanged.
\begin{array}{c} \begin{aligned} Q_{init_1} &= 0 & A_1 &= -\pi \\ Q_{final_1} &= \pi/2 & B_1 &= 3\pi/2 \\ V_{init_1} &= 0 & C_1 &= 0 \\ V_{final_1} &= 0 & D_1 &= 0 \\ \end{aligned} \end{array}
Here is the second spline.
\begin{array}{c} \begin{aligned} Q_{init_2} &= \pi/2 & A_2 &=2\pi \\ Q_{final_2} &= -\pi/2 & B_2 &= -3\pi \\ V_{init_2} &= 0 & C_2 &= 0 \\ V_{final_2} &= 0 & D_2 &= \pi/2 \\ \end{aligned} \end{array}
Here is the relation of joint space to task space.
\begin{aligned} X &= r_1 \cdot cos(Q_1) + r_2 \cdot cos(Q_1 + Q_2) \\ Y &= r_1 \cdot sin(Q_1) + r_2 \cdot sin(Q_1 + Q_2) \\ Z &= 0 \\ R_x &= 0 \\ R_y &= 0 \\ R_z &= Q_1 + Q_2 \\ \\ \dot{X} &= -\dot{Q_1} \cdot r_1 \cdot sin(Q_1) - \dot{Q_1} \cdot \dot{Q_2} \cdot r_2 \cdot sin(Q_1 + Q_2) \\ \dot{Y} &= \dot{Q_1} \cdot r_1 \cdot cos(Q_1) + \dot{Q_1} \cdot \dot{Q_2} \cdot r_2 \cdot cos(Q_1 + Q_2)\\ \dot{Z} &= 0 \\ \dot{Rx} &= 0 \\ \dot{Ry} &= 0 \\ \dot{Rz} &= \dot{Q_1} + \dot{Q_2}\\ \end{aligned}
For details, see the derivation of position here and the derivation of velocity here.
We now find the maximum velocity, acceleration, and resulting time scale for each joint.
Note: The topic of finding polynomial minima or maxima is well-covered elsewhere and can be deferred to a good algebra library. Here, we just use Desmos.
Figure: Joint-space velocities, two-joint manipulator.
First Joint
\begin{aligned} V_{1_{max}} &= 3\pi/4, t = 0.5s \\ A_{1_{max}} &= 3\pi, t = 0s \\ &= -3\pi, t = 1s \\ V_{1_{scale}} &= \frac{|3π/4|}{0.10471975} = 22.5 \\ A_{1_{scale}} &= \sqrt{\frac{|3π|}{0.0523599}} = 13.417 \\ \end{aligned}
Second Joint
\begin{aligned} V_{2_{max}} &= 3\pi/2, t = 0.5s \\ A_{2_{max}} &= 6\pi, t = 0s \\ &= -6\pi, t = 1s \\ V_{2_{scale}} &= \frac{|3π/2|}{0.10471975} = 45 \\ A_{2_{scale}} &= \sqrt{\frac{|6π|}{0.0523599}} = 18.974 \\ \end{aligned}
Without considering task space velocity, the time-optimal duration is $1s \cdot max(22.5, 13.417, 45, 18.974)=45s$. Clearly, since the second joint has twice as far to rotate as the first joint, it constrains the duration of the movement.
Let’s now consider task space velocity.
\begin{aligned} \dot{X}_{max} &= |3.245|m/s, t=0.3624s \\ \dot{Y}_{max} &= |-3.245|m/s, t=0.6376s \\ \dot{R_z}_{max} &= -3\pi/4 rad/s, t=0.5s \\ X_{ratio} &= 3245/100 = 32.45 \\ Y_{ratio} &= 3245/100 = 32.45 \\ R_{z_{ratio}} &= \frac{\frac{3\pi}{4} \>rad/s}{9°/s} = 15 \\ \end{aligned}
Since $32.45 < 45$, the joint space constraints dominate the task space constraints. The time-optimal duration is $1s \cdot max(45, 32.45, 15)=45s$. $\blacksquare$
Longer Paths
If a path $\pmb{Q}_j$ of length $n | n>2$ is given for joint $j$, i.e. intermediate points between $Q_{init}$ and $Q_{final}$ are given, and if the joint begins and ends with zero velocity ($V_{init} = V_{final} = 0$), then by enforcing the constraints of continuity on velocity and acceleration, the intermediate point velocities can be calculated with a system of linear equations following the method described by Melchiorri [1].
Example
Consider an extension of the previous 2-joint trajectory, where each joint passes through three positions: an initial position, an intermediate position, and a final position.
\begin{aligned} \pmb{Q}_1 &= [Q_{1_{init}} Q_{1_{intermediate}} Q_{1_{final}}] \\ \pmb{Q}_2 &= [Q_{2_{init}} Q_{2_{intermediate}} Q_{2_{final}}] \\ \end{aligned}
Let the given intermediate positions be
\begin{aligned} Q_{1_{intermediate}} &= \pi/4 \\ Q_{2_{intermediate}} &= 0 \end{aligned}
…along with the initial and final velocities.
$V_{1_{init}} = V_{1_{final}} = V_{2_{init}} = V_{2_{final}}= 0$
We must find the intermediate velocities $V_{1_{intermediate}}$ and $V_{2_{intermediate}}$. We can do this by solving the system
$A\pmb{v} = \pmb{c}$
where
$A= \begin{bmatrix} 2(T_1+T_2) & T_1 \\ T_3 &2(T_2+T_3) &T_2 \\ & & \ddots \ddots \ddots \\ & & & T_{n-2} & 2(T_{n-3}+T_{n-2}) & T_{n-3} \\ & & & & T_{n-1} & 2(T_{n-2} + T_{n-1}) \end{bmatrix}$
$\pmb{v} = \begin{bmatrix} V_2 \\ V_3 \\ \vdots \\ V_{n-2} \\ V_{n-1} \\ \end{bmatrix}$
$\pmb{c} = \begin{bmatrix} \frac{3}{T_1T_2}[T_1^2(Q_3-Q_2)+T_2^2(Q_2-Q_1)] \pmb{- T_2V_1} \\ \frac{3}{T_2T_3}[T_2^2(Q_4-Q_3)+T_3^2(Q_3-Q_2)] \\ \vdots \\ \frac{3}{T_{n-3}T_{n-2}}[T_{n-3}^2(Q_{n-1}-Q_{n-2})+T_{n-2}^2(Q_{n-2}-Q_{n-3})] \\ \frac{3}{T_{n-2}T_{n-1}}[T_{n-2}^2(Q_{n}-Q_{n-1})+T_{n-1}^2(Q_{n-1}-Q_{n-2})] \pmb{- T_{n-2}V_n}\\ \end{bmatrix}$
and
\begin{aligned} T_i &= Duration_i & \text{// The duration of segment i} \\ Q_i &= Q_{i_{init}} & \text{// The initial position of segment i } \\ V_i &= V_{i_{init}} & \text{// The initial velocity of segment i} \\ V_n &= V_{n_{final}} & \text{ // The final velocity of the last segment} \\ \end{aligned}
Reminder: There are $n-1$ splines interpolating $n$ control points (also called knots), and our indexing starts at $1$, not $0$. Therefore, $i==1$ refers to the first spline, and $i == n-1$ refers to the last spline.
Aside: Here is an example implementation of the above algorithm.
Joint 1
We solve for $V_{1_{intermediate}}$.
$\begin{array}{ccc} \begin{bmatrix} 2(T_1+T_2) \end{bmatrix} \begin{bmatrix} V_{1_{intermediate}} \end{bmatrix} &= \begin{bmatrix} \frac{3}{T_1T_2}[T_1^2(Q_{1_{final}}-Q_{1_{intermediate}})+T_2^2(Q_{1_{intermediate}}-Q_{1_{init}})]-T_2 \cdot V_{1_{init}} \end{bmatrix} \end{array}$
If we choose again arbitrarily that each segment should have 1 second of duration, then $T_1=1$ and $T_2=1$. Then
\begin{aligned} \begin{bmatrix} 2(1+1) \end{bmatrix} \begin{bmatrix} V_{1_{intermediate}} \end{bmatrix} &= \begin{bmatrix} \frac{3}{1 \cdot 1}[1^2(\pi/2-\pi/4)+1^2(\pi/4-0)]-2\cdot 0 \end{bmatrix} \\ 4 \cdot V_{1_{intermediate}}&= \begin{bmatrix} 3[\pi/4+\pi/4] \end{bmatrix} \\ V_{1_{intermediate}} &= 3\pi/8 \ rad/s \end{aligned}
We can now plot the two splines.
\begin{aligned} Duration_{1 \rightarrow 2} &= 1 \\ Displacement_{1 \rightarrow 2} &= \pi/4 \\ A_{1 \rightarrow 2} &= \frac{(2 \cdot - \pi/4 + 0 + 3\pi/8)}{1^2} &&= -\pi/8\\ B_{1 \rightarrow 2} &= \frac{(3 \cdot \pi/4 - 2 \cdot 0 - 3\pi/8)}{1} &&= 3\pi/8\\ C_{1 \rightarrow 2} &= V_{init} &&= 0\\ D_{1 \rightarrow 2} &= Q_{init} &&= 0\\ \end{aligned}
\begin{aligned} Duration_{2 \rightarrow 3} &= 1 \\ Displacement_{2 \rightarrow 3} &= \pi/4 \\ A_{2 \rightarrow 3} &= \frac{(2 \cdot - \pi/4 + 3\pi/8 + 0)}{1^2} &&= -\pi/8\\ B_{2 \rightarrow 3} &= \frac{(3 \cdot \pi/4 - 2 \cdot 3\pi/8 - 0)}{1} &&= 0\\ C_{2 \rightarrow 3} &= V_{init} &&= 3\pi/8\\ D_{2 \rightarrow 3} &= Q_{init} &&= \pi/4\\ \end{aligned}
Figure: Two cubic splines for joint 1. Can you see where they meet? Hint: Each spline has 1s of duration.
We only see one spline, but there are actually two. The first spline is valid on the interval $t=(0,1)$, and the second spline is valid on the interval $t=(1,2)$. The two splines are identical because they have the same duration, displacement, and absolute change in velocity. Therefore, while we chose to use two splines, this movement could have been interpolated by a single spline.
Joint 2
We solve for $V_{2_{intermediate}}$.
\begin{aligned} \begin{bmatrix} 2(1+1) \end{bmatrix} \begin{bmatrix} V_{2_{intermediate}} \end{bmatrix} &= \begin{bmatrix} \frac{3}{1 \cdot 1}[1^2(-\pi/2-0)+1^2(0-\pi/2)]-2\cdot 0 \end{bmatrix} \\ 4 \cdot V_{2_{intermediate}}&= \begin{bmatrix} 3[-\pi/2-\pi/2] \end{bmatrix} \\ V_{2_{intermediate}} &= -3\pi/4 \ rad/s \end{aligned}
$A = \frac{(2 \cdot -Displacement / Duration + V_{init} + V_{final})}{Duration^2} \\ B = \frac{(3 \cdot Displacement / Duration - 2 \cdot V_{init} - V_{final})}{Duration} \\$
\begin{aligned} Duration_{1 \rightarrow 2} &= 1 \\ Displacement_{1 \rightarrow 2} &= -\pi/2 \\ A_{1 \rightarrow 2} &= \frac{(2 \cdot - (- \pi/2) + 0 + (- 3\pi/4))}{1^2} &&= \pi/4\\ B_{1 \rightarrow 2} &= \frac{(3 \cdot (- \pi/2) - 2 \cdot 0 - (-3\pi/4))}{1} &&= -3\pi/4\\ C_{1 \rightarrow 2} &= V_{init} &&= 0\\ D_{1 \rightarrow 2} &= Q_{init} &&= \pi/2\\ \end{aligned}
\begin{aligned} Duration_{2 \rightarrow 3} &= 1 \\ Displacement_{2 \rightarrow 3} &= -\pi/2 \\ A_{2 \rightarrow 3} &= \frac{(2 \cdot - (-\pi/2) + (- 3\pi/4) + 0)}{1^2} &&= \pi/4\\ B_{2 \rightarrow 3} &= \frac{(3 \cdot (- \pi/2) - 2 \cdot (-3\pi/4) - 0)}{1} &&= 0\\ C_{2 \rightarrow 3} &= V_{init} &&= -3\pi/4\\ D_{2 \rightarrow 3} &= Q_{init} &&= 0\\ \end{aligned}
Figure: Two cubic splines for joint 2. They meet at $t=1s$.
Scaling Longer Trajectories
In order to satisfy joint-space and task-space velocity and acceleration limits, the $m(n-1)$ resulting cubic splines ($m=$ number of joints, $n$ = number of control points) must each be scaled by the method described previously.
Pseudocode:
// segments: sequence of splines, lenth n - 1
// (n == number of interpolated points)
// task_space_limit: e.g. Xlim = 100, Ylim = 100mm/s, etc.
for segment in segments:
joints = segment.joints // joints: container of size m
// get_scale: See Joint Constraints section
joint_space_ratio = joints.max(joint => joint.get_scale())
scale(segment, r_max)
Above: For each segment $i$, $m$ ratios are calculated, one for each joint, and the task space ratio is calculated. The maximum ratio $r_{max}$ is selected. Then all $m$ splines at segment $i$ are scaled by $r_{max}$.
Pseudocode:
// Scale the given spline
void scale(segment, ratio):
segment.duration *= ratio // Update duration
// Update velocity
if (segment.is_first()) spline.Vinit /= ratio
if (segment.is_last()) spline.Vfinal /= ratio
if (segment.is_intermediate()) spline.Vintermediate /= ratio
forward_propagate(segment.next);
segment.compute_velocities() // e.g. find new Vintermediate
segment.compute_coefficients() // e.g. get A B C D
// Update segment start and finish times
void forward_propagate(segment):
while (next_segment != null):
next_segment.forward_propagate() // e.g. Tinit = Tinit_prev + Duration
next_segment= segment.next
Above: When a segment is scaled, the following changes occur
1. the segment duration
2. the segment initial or final velocity (or both)
Forward propagation: All segments after the scaled segment are affected: the start time of each spline becomes the finishing time of the previous. These values propagate to the last spline.
Finally, the resulting intermediate velocities and polynomial coefficients must be recomputed.
References
[1] Biagiotti, L., & Melchiorri, C. (2009). Trajectory Planning for Automatic Machines and Robots. Berlin, Heidelberg: Springer Berlin Heidelberg.
Discussion
Please feel free to start a discussion on GitHub.
Main page | 2021-07-31T09:30:30 | {
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"lm_q1q2_score": 0.8397258179528557
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http://mathhelpforum.com/calculus/82363-trig-substitution.html | # Math Help - Trig substitution
1. ## Trig substitution
$
\int\frac{dx}{4+x^2}\
$
I am supposed to use trig substitution here. I used a tangent sub but then everything seems to cancel itself out. Which sub would I need to use here?
2. You're correct, use $x=2tan{\theta}, \;\ dx=2sec^{2}{\theta}d{\theta}$
$\int\frac{2sec^{2}{\theta}}{4+4tan^{2}{\theta}}d{\ theta}$
$\int\frac{2sec^{2}{\theta}}{4(1+tan^{2}{\theta})}d {\theta}$
$\frac{1}{4}\int\frac{2sec^{2}{\theta}}{sec^{2}{\th eta}}d{\theta}$
Yes, the $sec^{2}{\theta}$ does cancel. Which makes it easy.
$\frac{1}{4}\int 2d{\theta}$
$\frac{1}{2}\int d{\theta}$
$\frac{\theta}{2}$
But, from our sub, ${\theta}=tan^{-1}(\frac{x}{2})$
So, we get:
$\frac{tan^{-1}(\frac{x}{2})}{2}$
3. Originally Posted by ur5pointos2slo
$
\int\frac{dx}{4+x^2}\
$
I am supposed to use trig substitution here. I used a tangent sub but then everything seems to cancel itself out. Which sub would I need to use here?
there is a straight forward formula for this, but if you want to do the trig sub, that's fine. you have an expression of the form $x^2 + a^2$ thus your substitution must be $x = a \tan \theta$. that is, $x = 2 \tan \theta$
4. Originally Posted by ur5pointos2slo
$
\int\frac{dx}{4+x^2}\
$
I am supposed to use trig substitution here. I used a tangent sub but then everything seems to cancel itself out. Which sub would I need to use here?
$x = 2\tan{t}$
$dx = 2\sec^2{t} \, dt$
$\int \frac{1}{4 + x^2} \, dx$
$\int \frac{1}{4(1 + \tan^2{t})} \, 2\sec^2{t} \, dt$
$\frac{1}{2}\int \, dt$
$\frac{1}{2} t + C$
$\frac{1}{2} \arctan\left(\frac{x}{2}\right) + C$
5. Originally Posted by Jhevon
there is a straight forward formula for this, but if you want to do the trig sub, that's fine. you have an expression of the form $x^2 + a^2$ thus your substitution must be $x = a \tan \theta$. that is, $x = 2 \tan \theta$
I couldn't think of a formula that this integral could fit. Which formula are you referring to?
6. Originally Posted by galactus
You're correct, use $x=2tan{\theta}, \;\ dx=2sec^{2}{\theta}d{\theta}$
$\int\frac{2sec^{2}{\theta}}{4+4tan^{2}{\theta}}d{\ theta}$
$\int\frac{2sec^{2}{\theta}}{4(1+tan^{2}{\theta})}d {\theta}$
$\frac{1}{4}\int\frac{2sec^{2}{\theta}}{sec^{2}{\th eta}}d{\theta}$
Yes, the $sec^{2}{\theta}$ does cancel. Which makes it easy.
$\frac{1}{4}\int 2d{\theta}$
$\frac{1}{2}\int d{\theta}$
$\frac{\theta}{2}$
But, from our sub, ${\theta}=tan^{-1}(\frac{x}{2})$
So, we get:
$\frac{tan^{-1}(\frac{x}{2})}{2}$
I forgot all about the dtheta in there. I ended up with just the 1/2. Thanks a lot
7. Originally Posted by ur5pointos2slo
I couldn't think of a formula that this integral could fit. Which formula are you referring to?
on the inside cover of your calc text, there should be a formula that looks a little like this: $\int \frac {dx}{a^2 + x^2} = \frac 1a \tan^{-1} \frac xa + C$
8. Originally Posted by Jhevon
on the inside cover of your calc text, there should be a formula that looks a little like this: $\int \frac {dx}{a^2 + x^2} = \frac 1a \tan^{-1} \frac xa + C$
O yes I totally forgot all about that. thanks for the reminder! | 2015-11-28T22:42:34 | {
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https://math.stackexchange.com/questions/3994695/is-there-a-rigorous-definition-for-matrix-derivatives | # Is there a rigorous definition for matrix derivatives?
I know that,
A function $$f: \mathbb{R}^n \to \mathbb{R}$$ is said to be differentiable at $$x$$ if there exists a vector $$v$$ such that, $$\lim_{h \to 0} \frac{f(x+h) - f(x) - v^Th} {\|h\|} = 0.$$ When $$v$$ exists, it is given by the "gradient" $$\nabla f(x) = \left(\frac{\partial f}{\partial x_1}, ..., \frac{\partial f}{\partial x_n}\right)(x)$$
Does there exist a similar definition for "matrix derivative"
https://en.wikipedia.org/wiki/Matrix_calculus#Derivatives_with_matrices
• As far as physics is concerned, it is defined in the same way as in systems of ODE's. For example, the equation for an open system follows a master equation of the form $\dot{\rho}=\mathcal{L}(\rho)$, where $\rho$ is a matrix (a density matrix) and $\mathcal{L}$ is the "Lindbladian" which is a superoperator that takes operators into operators (in finite Hilbert spaces, $\rho$ can be represented as a matrix). In order to solve this equation, say numerically, one can consider that every $\rho_{ij}$ fulfills an ordinary ODE. Overall, one ends up with a system of ODEs to be solved. – user2820579 Jan 21 at 21:28
• Replace $v^T$ with a linear map $L\colon \Bbb R^n \to \Bbb R^k$. – Ivo Terek Jan 21 at 21:54
• Just reiterating what has been said here, but what you're looking for is the Frechet derivative. See my answer here which later also provides a sample calculation. Also, you may want to take a look at this for the motivation of the general definition. – peek-a-boo Jan 22 at 8:14
Matrix derivation is just a particular case of Fréchet derivative between two Banach spaces. Which by the way is very similar in term of definition to the definition of the derivative of a function $$f : \mathbb R^n \to \mathbb R$$ provided in the question.
Applied to matrix derivatives, you just have to consider a map $$f : V \to M$$ where $$M$$ is a linear space of matrices endowed with the norm of your choice and $$V$$ a Banach space that can be (or not) of finite dimension.
Yes, you can use a very similar definition. First of all, the map $$h \mapsto v^\top h$$ encodes an arbitrary linear map on $$\mathbb R^n$$. For matrices, you can substitute it by the Frobenius inner product, e.g., $$A \mapsto (A,B)_F := \sum_{i,j = 1}^n A_{ij} B_{ij},$$ where $$B \in \mathbb R^{n \times n}$$ is fixed.
Thus, for $$f \colon \mathbb R^{n \times n} \to \mathbb R$$ you can define $$\nabla f(A)$$ to be the (unique) matrix $$B \in \mathbb R^{n \times n}$$ (if it exists), that satisfies $$\lim_{H \to 0} \frac{f(A + H) - f(A) - (B,H)_F}{\|H\|} = 0.$$ | 2021-03-02T23:03:26 | {
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https://math.stackexchange.com/questions/2582895/finding-the-limit-of-a-recursive-sequence | # Finding the limit of a recursive sequence.
Let $$a_n \in \Bbb{R}$$ such that: $$a_{n+1}=1+\frac{2}{a_n} \text{ and } a_1=1$$
Prove that $$\lim_{n \to +\infty}a_n=2$$
We have that $$a_n \geq 1,\forall n \in \Bbb{N}$$ thus $$\limsup_na_n \geq \liminf_na_n \geq 1$$
Let $$\liminf_na_n=l$$
Then we have that $$l^2-l-2=0$$ thus $$l=2,-1$$
But $$a_n \geq 1$$ thus $$l=2$$.
Applying the same argument we prove that $$\limsup_na_n=2$$
Also we can also derive a contradiction if we assume that the limit superior and inferior are infinite.
Is my attempt an efficient way to prove this statement?
Also can someone provide additional solutions?
Thank you in advance.
• define $b_n = \frac{a_n-2}{a_n+1}$ and show that it satisfies $b_{n+1} = -\frac12 b_n$. use this to deduce a closed form expression of $a_n$. Dec 28 '17 at 12:03
• @achillehui...is my way correct..i will try your hint...Also is there any textbook for attacking and solving problems with recursive sequences? Dec 28 '17 at 12:05
• Your argument won't work. Unlike limit, $\liminf a_n = \ell$ doesn't implies $\liminf(1 + \frac{2}{a_n}) = 1 + \frac{2}{\ell}$ directly (even when you assume $a_n > 0$). However for positive numbers $c_n$ bounded away from zero and infinity, we have $$\liminf \frac{1}{c_n} = \frac{1}{\limsup c_n}\quad\text{ and }\quad \limsup \frac{1}{c_n} = \frac{1}{\liminf c_n}$$ If $\limsup a_n = u$, then you can deduce from $a_{n+1} = 1 + \frac{2}{a_n}$ the consequence $$\ell = 1 + \frac{2}{u}\;\text{ and }\; u = 1 + \frac{2}{\ell} \implies (\ell-2)(\ell+1) = 0 \implies \ell = 2$$ Dec 28 '17 at 12:39
• @achillehui...you are right..i had tottaly forgotten the two relations with limsup and liminf you mention..how can we overcome hte possibilty that the sequence is unbounded without finding an explicit formula? Dec 28 '17 at 12:48
• The map $x \mapsto 1 + \frac{2}{x}$ send $(0,\infty)$ to $(1,\infty)$, the next iteration send it to $(1,3)$. If you start from any $a_1 \in (0,\infty)$, you have $a_n \in (1,3)$ for all $n > 2$. Dec 28 '17 at 12:52
Banach's Fixed Point theorem also provides the limit. It states that the iteration $a_{n+1} = f(a_n)$ will converge to $a^* = f(a^*)$ if $|f'(a)| <1$ during the iteration.
Now $|f'(a)| = 2/a^2$ which in general will not be less than one. Taking two iteration steps however gives
$$a_{n+2} = g(a_n) = 3 - \frac{4}{a_n+2}$$
and $|g'(a)| = \frac{4}{(a+2)^2}$ which is less than one for all $a_n >0$, which will always be the case over the iteration.
Here we have where $a_{n+1}=f(a_n)$ where $$f(x)=1+\frac2x=\frac{x+2}{x}.$$ Maps of the form $x\mapsto (ax+b)/(cx+d)$ are fractional linear transformations and can be represented by matrices. If we let $A=\pmatrix{a&b\\c&d}$ and define $$A\bullet x=\frac{ax+b}{cx+d}$$ then $A\bullet (B\bullet x)=AB\bullet x$.
Here take $A=\pmatrix{1&2\\1&0}$. Then $a_{n+1}=A^n\bullet1$. We can diagonalise $A$: $$A=\pmatrix{1&2\\-1&1}\pmatrix{-1&0\\0&2}\pmatrix{1&2\\-1&1}^{-1}$$ and so $$A^{n-1}=\pmatrix{1&2\\-1&1}\pmatrix{(-1)^n&0\\0&2^n} \pmatrix{1&2\\-1&1}^{-1}.$$ Now it is straightforward to calculate Dr Graubner's formula.
an explicit formula is given by $$a_n=-\frac{2+\left(\frac{-1}{2}\right)^n}{-1+\left(\frac{-1}{2}\right)^n}$$ so $$\lim_{n\to \infty}a_n=2$$
Hint:
• $a_n \ge 2 \implies a_{n+2} \ge 2, \ a_{n+2} \le a_n$
• $a_n \le 2 \implies a_{n+2} \le 2, \ a_{n+2} \ge a_n$
• $1 = a_1 \le a_3 \le a_5 \le \cdots \le 2 \le \cdots \le a_6 \le a_4 \le a_2 =3$
$a_{n+2}-a_n=\frac{4(a_n-a_{n-2})}{(a_n+2)(a_{n-2}+2)} - ①$
We can confirm the sign of $a_{n+2}-a_{n}$ and that of $a_{n}-a_{n-2}$ are same from ①.
For $a_1 < a_3$ and $a_4<a_2$, we can say $a_{2m-1} < a_{2m+1}$ and $a_{2m+2}<a_{2m}$ respectively.
Thus, $a_1<a_3<…<a_{2m-1}<…<2$, $a_2>a_4>…>a_{2m>}…>2$
$\{a_{2m-1}\}$ has upper bound and let $\alpha$ be its limit value. Also, $\{a_{2m}\}$ has lower bound and let $\beta$ be its limit value.
We can gain $\beta=1+\frac{2}{\alpha}-②$ and $\alpha=1+\frac{2}{\beta}-③$ from $a_{2m}=1+\frac{2}{a_{2m-1}}, a_{2m+1}=1+\frac{2}{a_{2m}}(m\rightarrow\infty)$
Finally, we can get $\alpha=\beta=2$ from ② and ③. This completes the proof. | 2021-09-20T20:17:34 | {
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http://math.stackexchange.com/questions/75224/a-differentiation-question/75230 | # A Differentiation Question
If I have the following equation, I know I can differentiate it using the quotient rule, by first factoring out $\frac{3}{2}$:
$$f(w)=\frac{3(60-w)(w-2)}{2w}$$
But is there a way to differentiate it using the product rule or some kind of the chain rule by looking at it like so:
$$f(w)=3(60-w)(w-2)(2w)^{-1}$$
I guess what I'm asking is, I know how to use the product rule when I have 2 terms, but how do I do it with 3+? Do I derive the first two terms, then use that derived answer with the product rule with the remaining term?
Thanks!
-
$\text{quotient rule} = \text{product rule} + \text{chain rule}$ – cardinal Oct 23 '11 at 23:10
$(fgh)' = (fg)'h + (fg)h' = (f'g + fg')h + (fg)h'$. – cardinal Oct 23 '11 at 23:12
The product rule for three terms can be deduced from the product rule for two: $$(fgh)' = \Bigl( (fg)h\Bigr)' = (fg)'h + (fg)h' = (f'g + fg')h + (fg)h' = f'gh + fg'h + fgh'.$$ That is, it's just "pass-the-prime".
This holds for any number of factors, and can be established by induction. Assuming that we know that $$\Bigl(f_1\cdots f_n\Bigr)' = f_1'f_2\cdots f_n + f_1f_2'f_3\cdots f_n + \cdots + f_1\cdots f_{n-1}f_n',$$ then \small\begin{align*} \Bigl(f_1\cdots f_nf_{n+1}\Bigr)' &= \Bigl( (f_1\cdots f_n)f_{n+1}\Bigr)'\\ &= (f_1\cdots f_n)'f_{n+1} + (f_1\cdots f_n)f_{n+1}' \\ &= \Bigl(f_1'f_2\cdots f_n + f_1f_2'f_3\cdots f_n + \cdots + f_1\cdots f_{n-1}f_n'\Bigr)f_{n+1} + (f_1\cdots f_n)f_{n+1}'\\ &= f_1'f_2\cdots f_nf_{n+1} + f_1f_2'f_3\cdots f_nf_{n+1} + \cdots + f_1\cdots f_n'f_{n+1} + f_1\cdots f_nf_{n+1}'. \end{align*}
-
Great, thank you so much! – Josh Oct 23 '11 at 23:11
So is it suggested to just use the quotient rule in this case? Or is it ok to just use the product rule. That is, which will be the more efficient way, in general, and for this problem? – Josh Oct 23 '11 at 23:15
@Josh: Whichever way you want it: you will either use the product rule for three factors and the chain rule, or the quotient rule and the product rule for two factors. Personally, for this problem, I would use algebra to rewrite it as$$\frac{3}{2}(60-w)\left(1 - 2w^{-1}\right)$$ and just use the standard product rule. – Arturo Magidin Oct 23 '11 at 23:17
To answer the second question first: You can certainly apply the product rule to each multiplication in turn. However, since multiplication is commutative and associative, the result will necessarily be symmetric with respect to the three factors, so it makes sense to derive a general rule and remember it instead of going through the factors pairwise in each case.
Applying the product rule to the product $fgh$ by grouping it as $(fg)h$ yields
$$(fgh)'=((fg)h)'= (fg)'h + (fg)h' = (f'g + fg')h + (fg)h'=f'gh+fg'h+fgh'\;.$$
Perhaps you can guess from this result what the general result for an arbitrary number of factors will be:
$$\left(\prod_i f_i\right)'=\sum_if_i'\prod_{j\ne i} f_j\;.$$
You can either prove this using induction (which might be a good exercise), or you can see why it's true by considering what happens if you change each of the factors a bit:
$$(f+\Delta f)(g+\Delta g)(h+\Delta h)\cdots=fgh\cdots+\Delta fgh\cdots+f\Delta gh\cdots+fg\Delta h\cdots+\cdots\;,$$
simply from multiplying out the parentheses. The remaining terms are at least of second order (i.e. they contain at least two $\Delta$ terms), so they don't end up in the derivative, which gives the first-order approximation to the function.
Turning now to your first question, yes, you can apply the product rule instead of the quotient rule; well spotted. In fact, the quotient rule is nothing but a convenient shortcut for applying the product rule:
$$\left(\frac fg\right)'=\left(fg^{-1}\right)'=f'g^{-1}+f\left(g^{-1}\right)'=f'g^{-1}+f\left(-\frac{g'}{g^2}\right)=\frac{f'g-fg'}{g^2}\;.$$
-
The answer to your last question is ‘yes’: you can differentiate $uvw$ by treating it as $(uv)w$. In fact you can do this in general to discover an extended form of the product rule: $$(uvw)' = (uv)w'+(uv)'w = uvw' + (uv'+u'v)w = uvw' + uv'w + u'vw\;.$$ (Here I’m thinking of $u,v$, and $w$ as functions of the same independent variable, say $x$.) The same idea extends to more factors. For instance, $$(uvwz)' = (uvw)z'+(uvw)'z = uvwz'+uvw'z + uv'wz + u'vwz\;,$$ using the previous result.
-
The general product rule for more than two terms is to differentiate one term at a time. For example, the derivative of a product of three functions: $$\frac{d}{dx}(f g h) = \frac{df}{dx}gh+f\frac{dg}{dx}h+fg\frac{dh}{dx}$$ This is, of course, the same as applying the product rule for two terms twice as you suggested: $$\frac{d}{dx}((f g) h) = \left(\frac{df}{dx}g+f\frac{dg}{dx}\right)h+(fg)\frac{dh}{dx}$$
- | 2015-05-26T00:44:02 | {
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https://math.stackexchange.com/questions/6244/is-there-a-quick-proof-as-to-why-the-vector-space-of-mathbbr-over-mathbb/6250 | # Is there a quick proof as to why the vector space of $\mathbb{R}$ over $\mathbb{Q}$ is infinite-dimensional?
It would seem that one way of proving this would be to show the existence of non-algebraic numbers. Is there a simpler way to show this?
• A finite dimensional vector space over $\mathbb{Q}$ is countable.
– user641
Oct 7 '10 at 2:19
• @Steve: Please add that as an answer so people can upvote. Oct 7 '10 at 2:50
• Does that mean that a vector space over $\mathbb{Q}$ is finite-dimensional iff the set of the vector space is countable? If so, please prove it. Oct 7 '10 at 2:55
• @Isaac Your question doesn't require the 'only if' anyway. Steve's observation answers your original question. Oct 7 '10 at 3:04
• @Isaac: how is the fact that $\mathbb{R}$ is finite dimensional over itself relevant? Oct 7 '10 at 3:31
The cardinality argument mentioned by Arturo is probably the simplest. Here is an alternative: an explicit example of an infinite $$\, \mathbb Q$$-independent set of reals. Consider the set consisting of the logs of all primes $$\, p_i.\,$$ If $$\, c_1 \log p_1 +\,\cdots\, + c_n\log p_n =\, 0,\ c_i\in\mathbb Q,\,$$ multiplying by a common denominator we can assume that all $$\ c_i \in \mathbb Z\,$$ so, exponentiating, we obtain $$\, p_1^{\large c_1}\cdots p_n^{\large c_n}\! = 1\,\Rightarrow\ c_i = 0\,$$ for all $$\,i,\,$$ by the uniqueness of prime factorizations.
• @Bill +1 What a nice example. Oct 7 '10 at 4:50
• @Bill. Wow! :-) Oct 7 '10 at 6:08
• Is this proof unique to Q or does it generalize to provide explicit examples of reals linearly independent over, e.g., Q(sqrt(2))? Above, Q appears to be "hard-wired" into the proof, as the group of exponents in the prime factorization.
– T..
Oct 12 '10 at 7:14
• Can you extend the idea of this proof to get the right dimension? Currently, you only have a countably infinite independent set, but the dimension is size continuum.
– JDH
Jun 23 '11 at 2:22
• @student this can not happen: the primes are distinct. Put the primes with negative exponent on the other side (exponents become positive). You would find two distinct prime factorizations of the same number. Feb 6 '19 at 19:44
As Steve D. noted, a finite dimensional vector space over a countable field is necessarily countable: if $v_1,\ldots,v_n$ is a basis, then every vector in $V$ can be written uniquely as $\alpha_1 v_1+\cdots+\alpha_n v_n$ for some scalars $\alpha_1,\ldots,\alpha_n\in F$, so the cardinality of the set of all vectors is exactly $|F|^n$. If $F$ is countable, then this is countable. Since $\mathbb{R}$ is uncountable and $\mathbb{Q}$ is countable, $\mathbb{R}$ cannot be finite dimensional over $\mathbb{Q}$. (Whether it has a basis or not depends on your set theory).
Your further question in the comments, whether a vector space over $\mathbb{Q}$ is finite dimensional if and only if the set of vectors is countable, has a negative answer. If the vector space is finite dimensional, then it is a countable set; but there are infinite-dimensional vector spaces over $\mathbb{Q}$ that are countable as sets. The simplest example is $\mathbb{Q}[x]$, the vector space of all polynomials with coefficients in $\mathbb{Q}$, which is a countable set, and has dimension $\aleph_0$, with basis $\{1,x,x^2,\ldots,x^n,\ldots\}$.
Added: Of course, if $V$ is a vector space over $\mathbb{Q}$, then it has countable dimension (finite or denumerable infinite) if and only if $V$ is countable as a set. So the counting argument in fact shows that not only is $\mathbb{R}$ infinite dimensional over $\mathbb{Q}$, but that (if you are working in an appropriate set theory) it is uncountably-dimensional over $\mathbb{Q}$.
• Related to the note about uncountable dimension, there are explicit examples of continuum-sized linearly independent sets, as seen in this MathOverflow answer by François G. Dorais: mathoverflow.net/questions/23202/… Oct 7 '10 at 4:18
• Yes: but can one show that there is a basis for $\mathbb{R}$ over $\mathbb{Q}$ without some form of the Axiom of Choice? (There is a difference between exhibiting a large linearly independent subset and exhibiting a basis). Oct 7 '10 at 14:51
• No, one cannot, but without the Axiom of Choice the notion of dimension breaks down (except for finite vs. infinite). Assuming AC (as I virtually always do), the size of a linearly independent set gives a lower bound on the dimension of the vector space, and I think it is wonderful that in this case such "explicit" proof exists that the real numbers have continuum dimension, as opposed to the nice qualitative proof one could give by extending your argument to larger cardinals. Oct 7 '10 at 16:02
• @Jonas: No argument there (with any of your points). Oct 7 '10 at 17:23
– Leo
Sep 23 '11 at 0:23
For the sake of completeness, I'm adding a worked-out solution due to F.G. Dorais from his post.
We'll need two propositions from Grillet's Abstract Algebra, page 335 and 640:
Proposition: $[\mathbb{R}:\mathbb{Q}]=\mathrm{dim}_\mathbb{Q}{}\mathbb{R}=|\mathbb{R}|$
Proof: Let $(q_n)_{n\in\mathbb{N_0}}$ be an enumeration of $\mathbb{Q}$. For $r\in\mathbb{R}$, take $$A_r:=\sum_{q_n<r}\frac{1}{n!}\;\;\;\;\text{ and }\;\;\;\;A:=\{A_r;\,r\in\mathbb{R}\};$$ the series is convergent because $\sum_{q_n<r}\frac{1}{n!}\leq\sum_{n=0}^\infty\frac{1}{n!}=\exp(1)<\infty$ (recall that $\exp(x)=\sum_{n=0}^\infty\frac{x^n}{n!}$ for any $x\in\mathbb{R}$).
To prove $|A|=|\mathbb{R}|$, assume $A_r=A_{s}$ and $r\neq s$. Without loss of generality $r<s$, hence $A_s=\sum_{q_n<s}\frac{1}{n!}=\sum_{q_n<r}\frac{1}{n!}+\sum_{r\leq q_n<s}\frac{1}{n!}=A_r+\sum_{r\leq q_n<s}\frac{1}{n!}$, so $\sum_{r\leq q_n<s}\frac{1}{n!}=0$, which is a contradiction, because each interval $(r,s)$ contains a rational number.
To prove $A$ is $\mathbb{Q}$-independent, assume $\alpha_1A_{r_1}+\cdots+\alpha_kA_{r_k}=0\;(1)$ with $\alpha_i\in\mathbb{Q}$. We can assume $r_1>\cdots>r_k$ (otherwise rearrange the summands) and $\alpha_i\in\mathbb{Z}$ (otherwise multiply by the common denominator). Choose $n$ large enough that $r_1>q_n>r_2\;(2)$; we'll increase $n$ two more times. The equality $n!\cdot(1)$ reads $n!(\alpha_1\sum_{q_m<r_1}\frac{1}{m!}+\cdots+\alpha_k\sum_{q_m<r_k}\frac{1}{m!})=0$. Rearranged (via $(2)$ when $m=n$), it reads
$$-\alpha_1\sum_{\substack{m<n\\q_m<r_1}}\frac{n!}{m!}-\cdots-\alpha_k\sum_{\substack{m<n\\q_m<r_k}}\frac{n!}{m!}-\alpha_1 =\alpha_1\sum_{\substack{m>n\\q_m<r_1}}\frac{n!}{m!}+\cdots+\alpha_k\sum_{\substack{m>n\\q_m<r_k}}\frac{n!}{m!}. \tag*{(3)}$$
The left hand side (LHS) of $(3)$ is an integer for any $n$. If $n$ is large enough that $(|\alpha_1|+\cdots+|\alpha_k|)\sum_{m=n+1}^\infty\frac{n!}{m!}<1$ holds (such $n$ can be found since $\sum_{m=n+1}^\infty\frac{n!}{m!}=\frac{1}{n+1}\sum_{m=n+1}^\infty\frac{1}{(n+2)\cdot\ldots\cdot m}\leq\frac{1}{n+1}\sum_{m=n+1}^\infty\frac{1}{(m-n-1)!}\leq\frac{1}{n+1}\exp(1)\rightarrow 0$ when $n\rightarrow\infty$), then the absolute value of RHS of $(3)$ is $<1$, and yet an integer, hence $\text{RHS}(3)=0$. Thus $(3)$ reads $\alpha_1=-\sum_{i=1}^{k}\sum_{m<n,q_m<r_i}\alpha_i\frac{n!}{m!}=0\;(\mathrm{mod}\,n)$. If moreover $n>|\alpha_1|$, this means that $\alpha_1=0$. Repeat this argument to conclude that also $\alpha_2=\cdots=\alpha_k=0$.
Since $A$ is a $\mathbb{Q}$-independent subset, by proposition 5.3 there exists a basis $B$ of $\mathbb{R}$ that contains $A$. Then $A\subseteq B\subseteq\mathbb{R}$ and $|A|=|\mathbb{R}|$ and Cantor-Bernstein theorem imply $|B|=|\mathbb{R}|$, therefore $[\mathbb{R}:\mathbb{Q}]=\mathrm{dim}_\mathbb{Q}{}\mathbb{R}=|\mathbb{R}|$. $\quad\blacksquare$
• Is this uncountable linearly independent set basis ?
– user195218
Jun 5 '15 at 2:53
• @user195218 It's not a basis, as it doesn't span the reals, but it is uncountable and linearly independent. Jul 25 '17 at 19:08
• The only proper answer to the question. Such a shame that it is not the most voted answer. May 1 '19 at 21:42
• Way to nail it down, nice work. Dec 28 '20 at 2:05
No transcendental numbers are needed for this question. Any set of algebraic numbers of unbounded degree spans a vector space of infinite dimension. Explicit examples of linearly independent sets of algebraic numbers are also relatively easy to write down.
The set $\sqrt{2}, \sqrt{\sqrt{2}}, \dots, = \bigcup_{n>0} 2^{2^{-n}}$ is linearly independent over $\mathbb Q$. (Proof: Any expression of the $n$th iterated square root $a_n$ as a linear combination of earlier terms $a_i, i < n$ of the sequence could also be read as a rational polynomial of degree dividing $2^{n-1}$ with $a_n$ as a root and this contradicts the irreducibility of $X^m - 2$, here with $m=2^n$).
The square roots of the prime numbers are linearly independent over $\mathbb Q$. (Proof: this is immediate given the ability to extend the function "number of powers of $p$ dividing $x$" from the rational numbers to algebraic numbers. $\sqrt{p}$ is "divisible by $p^{1/2}$" while any finite linear combination of square roots of other primes is divisible by an integer power of $p$, i.e., is contained in an extension of $\mathbb Q$ unramified at $p$).
Generally any infinite set of algebraic numbers that you can easily write down and is not dependent for trivial reasons usually is independent. This because the only algebraic numbers for which we have a simple notation are fractional powers, and valuation (order of divisibility) arguments work well in this case. Any set of algebraic numbers where, of the ones ramified at any prime $p$, the amount of ramification is different for different elements of the set, will be linearly independent. (Proof: take the most ramified element in a given linear combination, express it in terms of the others, and compare valuations.)
• I realize you answered this question a loooong time ago, but if you're still around, how do you know that you can write the $n-$th root of $2$ as a linear combination of the $1st$ through $n-1$st roots of $2$?
– user100463
Aug 17 '17 at 22:41
• @ALannister I realize you questioned quite a long time ago, but s/he is assuming that there is such $n$ and showing that such assumption leads to contradiction. So in fact, s/he is proving exactly that $n$-th root of $2$ cannot be expressed by a linear combination of previous roots May 9 '18 at 17:24
Another simple proof:
Take $P=X^n-p$ for a prime $p$.
By Eisenstein's criterion, it is $\mathbb Q[X]$-irreductible. Therefore, the set of algebraic numbers is of infinite dimension over $\mathbb Q$.
Since $\mathbb R$ is bigger, it works for $\mathbb R$ too.
As $$\pi$$ is trascendent over $$\mathbb{Q}$$. Then the set $$\{1, \pi, \pi^{2},\cdots\}$$ is linearly independent.
Let $$n$$ be a positive integer.
Let $$(\alpha_i)_{1 \le i \le n}$$ be a family of $$n$$ nonzero real numbers.
Proposition: The set $$V$$ containing all sums of the form
$$\tag 1 \displaystyle{\sum_{i=1}^n q_i \alpha_i} \quad \text{where } q_i \in \Bbb Q$$
is properly contained in $$\Bbb R$$.
Proof
Construct a family of finite subsets of $$\Bbb Q$$, $$(F_k)_{\,k \in \Bbb N}$$, such that
$$\tag 2 \displaystyle{\bigcup_{\,k \in \Bbb N} F_k = \Bbb Q} \; \text{ and }\; \{-1,+1\} \subset F_0 \; \text{ and } \; F_k \subset F_{k+1}$$
Define the set $$V_k$$ to be all sums of the form
$$\tag 3 \displaystyle{\sum_{i=1}^n q_i \alpha_i} \quad \text{where } q_i \in F_k$$
so that $$V$$ is the union of the $$V_k$$ family of sets with $$V_k \subset V_{k+1}$$.
We will construct a family of nested/shrinking closed intervals $$I_m = [a_m, b_m]$$ satisfying
$$\quad \cap\, I_m = \{\beta\} \text{ where } \beta \notin V$$
The nested interval theorem guarantees that the intersection of the closed intervals is a singleton $$\beta$$ while the algorithm constructing the $$[a_m, b_m]$$ must also take steps to exclude any element in $$V$$ from being in that intersection.
The algorithm (defined using recursion):
Set the initial closed interval to
$$\quad I_0 := [a_0,b_0] := [-\alpha_1, +\alpha_1]$$
Suppose $$I_m := [a_m, b_m]$$ has been set. There is a smallest $$k$$ such that
$$\quad \frac{a_m+b_m}{2} \in V_k$$
With that $$k$$,
if $$m+1$$ is odd set
$$\quad a_{m+1} = \text{max}\big(\{v \in V_k \mid v \lt b_m\}\big)$$
$$\quad b_{m+1} = b_m$$
if $$m+1$$ is even set
$$\quad b_{m+1} = \text{min}\big(\{v \in V_k \mid v \gt a_m\}\big)$$
$$\quad a_{m+1} = a_m$$
and define
$$\quad I_{m+1} := [a_{m+1}, b_{m+1}]$$
By the nested interval theorem the intersection of these intervals is a singleton set; call the element in that set $$\beta$$. Since every finite set $$V_k$$ gets 'consumed' by the algorithm,
$$\quad$$ for every $$k$$ we must have $$\beta \notin V_k$$.
and so $$\beta \notin V$$.
$$\blacksquare$$
• Here is a quick (off-beat) way of showing their are irrational numbers: 1) Construct $\Bbb R$ using, say, Dedekind cuts. $\quad$ 2) Prove the nested interval theorem. $\quad$ 3) Regard $\Bbb R$ as a vector space over $\Bbb Q$ and set $V$ to the vector subspace generated by $1$, and notice that $V = \Bbb Q$. $\quad$ 4) Invoke the above proposition. Jan 27 '20 at 21:57
Let $$(p_{n} )$$ be the sequence of all $$\textbf {prime numbers}$$. Then $$\forall a \neq 1$$ positive number, show the sequence $$( \log_a p_{n} )$$ in the vector space $$\mathbb{R}$$ over $$\mathbb{Q}$$ is linearly independent. So the vector space $$\mathbb{R}$$ over $$\mathbb{Q}$$ is infinite-dimensional.
• This example was already posted here $10$ years ago in my answer. Please don't post duplicate answers. Nov 12 '21 at 9:03
• I didn't check the other answers, this was a solution in my teacher's book that I wrote. Nov 13 '21 at 10:25
• You are supposed to check other answers before posting (esp. when answering a question 10 years old). The correct thing to do in cases like this is to delete this duplicate answer. Nov 13 '21 at 16:47 | 2022-01-16T23:45:53 | {
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http://math.stackexchange.com/questions/321012/chance-of-getting-a-good-grade/321057 | Chance of getting a good grade
Lets says theres a question bank of 28 questions. On the exam, there will be 12 of these questions, and I will have to answer 5.
If the only way to get a question right is to study it, how many questions should I study to have a reasonable chance of knowing 5 answers?
More generally, how could I find out my expected score as a function of how many questions I study?
Obviously if I study 21, then I will get 100%, and if I study less than 5, I'm guaranteed to not get at least 5.
-
Are you actually going to apply what we tell you to a real test that you will take? If so, the answer is of course to study 28. :P – apnorton Mar 5 '13 at 3:00
Its not a math exam, so its a minimization problem rather than a maximization problem – rckrd Mar 5 '13 at 3:01
You only need to study 21 to get 100%, since in the worst case scenario the seven you studied for are among the 12 and you can pick the other 5. – Lepidopterist Mar 5 '13 at 3:03
Total number of combinations: $\binom{28}{12}$ number of combinations that contain at least 5 of the n you know: is the sum of the combinations in which there are 5,6,7,8,9,10,11,12 you do know. In other words $\sum_{i=5}^{12}\binom{n}{i}*\binom{28-n}{12-i}$ so the probability is $$\frac{\sum_{i=5}^{12}\binom{n}{i}*\binom{28-n}{12-i}}{\binom{28}{12}}$$
-
Imagine that you are a little late in studying, and the $12$ questions on the test have already been chosen by the instructor. Call the $12$ chosen questions good, and the $16$ others bad.
You choose $n$ questions at random to study. There are $\binom{28}{n}$ equally likely choices.
The number of choices that have $k$ good and $n-k$ bad is $\binom{12}{k}\binom{16}{n-k}$. Thus the probability that you choose at least $5$ good is $$\frac{\sum_{k=5}^{12}\binom{12}{k}\binom{16}{n-k}}{\binom{28}{n}}.$$ One can save a little bit of effort by calculating the probability of the complement: for that we need the shorter sum from $k=0$ to $4$. Now perform the calculation for various $n$, until you get a probability that you consider adequate.
-
Andre and Jorge have already covered most of the question, but I thought I'd just like to add that your expected score can be expressed as a 12-th order polynomial in $n$, where $n$ is the number of questions you studied.
When you plot the polynomial, you can get a nice graph of your expected score as a function of number of questions studied:
- | 2015-09-02T01:19:28 | {
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http://dlmf.nist.gov/10.2 | # §10.2 Definitions
## §10.2(i) Bessel’s Equation
This differential equation has a regular singularity at with indices , and an irregular singularity at of rank 1; compare §§2.7(i) and 2.7(ii).
## §10.2(ii) Standard Solutions
### ¶ Bessel Function of the First Kind
10.2.2
This solution of (10.2.1) is an analytic function of , except for a branch point at when is not an integer. The principal branch of corresponds to the principal value of 4.2(iv)) and is analytic in the -plane cut along the interval .
When , is entire in .
For fixed each branch of is entire in .
### ¶ Bessel Function of the Second Kind (Weber’s Function)
When is an integer the right-hand side is replaced by its limiting value:
Whether or not is an integer has a branch point at . The principal branch corresponds to the principal branches of in (10.2.3) and (10.2.4), with a cut in the -plane along the interval .
Except in the case of , the principal branches of and are two-valued and discontinuous on the cut ; compare §4.2(i).
Both and are real when is real and .
For fixed each branch of is entire in .
### ¶ Bessel Functions of the Third Kind (Hankel Functions)
These solutions of (10.2.1) are denoted by and , and their defining properties are given by
10.2.5
as in , and
10.2.6
as in , where is an arbitrary small positive constant. Each solution has a branch point at for all . The principal branches correspond to principal values of the square roots in (10.2.5) and (10.2.6), again with a cut in the -plane along the interval .
The principal branches of and are two-valued and discontinuous on the cut .
For fixed each branch of and is entire in .
### ¶ Branch Conventions
Except where indicated otherwise, it is assumed throughout the DLMF that the symbols , , , and denote the principal values of these functions.
### ¶ Cylinder Functions
The notation denotes , , , , or any nontrivial linear combination of these functions, the coefficients in which are independent of and .
## §10.2(iii) Numerically Satisfactory Pairs of Solutions
Table 10.2.1 lists numerically satisfactory pairs of solutions (§2.7(iv)) of (10.2.1) for the stated intervals or regions in the case . When , is replaced by throughout.
Table 10.2.1: Numerically satisfactory pairs of solutions of Bessel’s equation.
Pair Interval or Region
neighborhood of 0 in
neighborhood of in | 2013-12-06T16:02:48 | {
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http://compyouter.com/gwgb9/bfs-spanning-tree-example-e0a05d | BFS starts with the root node and explores each adjacent node before exploring node(s) at the next level. We start with the source node a, add it to a queue and mark it as visited. Start by putting any one of the graph's vertices at the back of a queue. edges[ 3 ][ 0 ].first = 0 , edges[ 3 ][ 0 ].second = 0 Next, shimming delays are inserted in the link branches (i.e., the remaining branches) so that the total delay in the loops becomes zero. Depth-first search (DFS) is an algorithm for searching a graph or tree data structure. In simple terms, it traverses level-wise from the source. Applications of Breadth First Search. Recommended Articles Computing MST using DFS/BFS would mean it … Consider connecting a vertex to the "parent" vertex that "found" this vertex. A graph can contain cycles, which may bring you to the same node again while traversing the graph. Some of those paths might be more expensive, because they are longer, or require the cable to be buried deeper; these paths would … If it is constrained to bury the cable only along certain paths, then there would be a graph representing which points are connected by those paths. Yes it is possible. There can be many spanning trees. level[ v[ p ][ i ] ] = level[ p ]+1; In this code, while you visit each node, the level of that node is set with an increment in the level of its parent node. BFS traversal of a graph produces a spanning tree as final result. Step 4) Remaining 0 adjacent and unvisited nodes are visited, marked, and inserted into the queue. Next PgDn. BFS is useful in finding shortest path.BFS can be used to find the shortest distance between some starting node and the remaining nodes of the graph. (Photo Included), Book about an AI that traps people on a spaceship. The memory requirement of this graph is less as compared to BFS as only one stack is needed to be maintained. :) Here is an other example to make it clearer, from Wikipedia: We want to make a spanning tree from Frankfurt. Thanks for contributing an answer to Mathematics Stack Exchange! It is: 2 is ignored because it is already marked as 'visited', 3 is ignored because it is already marked as 'visited'. Breadth First Search can be done with the help of queue i.e. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Note that BFS computes a spanning tree (the parent pointers identify the edges), once a root has been selected. Algorithms: Difference of output tree as subgraph from DFS and BFS . Shortest path finding. They will be: Neighbors of 1 i.e. The process of visiting and exploring a graph for processing is called graph traversal. A question about the matrix tree theorem. edges[ 3 ][ 3 ].first = 4 , edges[ 3 ][ 3 ].second = 0, 4 -> 1 -> 3 Maximum Width of a Binary Tree at depth (or height) h can be 2 h where h starts from 0. Breadth-First search (BFS) Bipartite test Suppose that the spanning tree generated by BFS algorithm is: What are the two set V1 and V2 of the Graph? This tree contains all vertices of the graph (if it is connected) and is called graph spanning tree. Store them in the order in which they are visited. 1. Value. To learn more, see our tips on writing great answers. Spanning tree out of graph using Breadth First Search? 2 Answers Active Oldest Votes. 6 and 7), and then of 3 (i.e. Also, how would I choose the initial nodes to begin the search? Complete reference to competitive programming, First move horizontally and visit all the nodes of the current layer, Neighbors of s i.e. No more neighbours, so we check for new ones. Therefore, in BFS, you must traverse all the nodes in layer 1 before you move to the nodes in layer 2. edges[ 1 ][ 1 ].first = 4 , edges[ 1 ][ 1 ].second = 0, 2 -> 0 -> 3 Spanning Tree Algorithm Below is my version generalizing many "standard" spanning tree algorithms, including Depth-First Search ( DFS ), Bredth-First Search ( BFS ), Minimum-Weight Spanning Tree ( MST ), and Shortest Path Tree (also called Single-Source Shortest Path ). Example: Consider the below step-by-step BFS traversal of the tree. How to compute the determinant of the following matrix? BFS starts with the root node and explores each adjacent node before exploring node(s) at the next level. FIFO ... > useful in finding spanning trees & forest. Breadth-first search is a systematic method for exploring a graph. There also can be many minimum spanning trees. Multiple traversal sequence is possible depending on the starting vertex and exploration vertex chosen. Assume that it costs O(n) to create a DFS or BFS tree where n is the number of nodes. 1. 4 and 5), then of 2 (i.e. BFS algorithm can easily create the shortest path and a minimum spanning tree to visit all the vertices of the graph in the shortest time possible with high accuracy. If you apply the BFS explained earlier in this article, you will get an incorrect result for the optimal distance between 2 nodes. Example: Breadth First Search (BFS) Time Complexity - Duration: 4:05. Notice that the left-to-right order of the children of each vertex is consistent with the discovery order, as asserted by Prop 1.2. Adding one edge to the spanning tree will create a circuit or loop, i.e. Spanning Tree is a graph without loops. edges[ 0 ][ 1 ].first = 3 , edges[ 0 ][ 1 ].second = 0 This code is similar to the BFS code with only the following difference: B readth-first search is a way to find all the vertices reachable from the a given source vertex, s. Like depth first search, BFS traverse a connected component of a given graph and defines a spanning tree. How to construct the graph from an adjacency matrix? It starts at the tree root (or some arbitrary node of a graph, sometimes referred to as a ‘search key’) and explores the neighbor nodes first, before moving to the next level neighbors. s, 3, and 4 are traversed, 3 and s are ignored because they are marked as 'visited'. All these techniques are re nements and extensions of the same basic tree-growing scheme given in x4.1. Breadth First Search (BFS) is an algorithm for traversing an unweighted Graph or a Tree. Algorithm. But in the case of the Shortest Path tree, each node … The BFS can be used to determine the level of each node from … Step 1) You have a graph of seven numbers ranging from 0 – 6. It starts at the tree root (or some arbitrary node of a graph, sometimes referred to as a 'search key') and explores the neighbor nodes first, before moving to the next level neighbors. edges[ 0 ][ 2 ].first = 2 , edges[ 0 ][ 2 ].second = 1, 1 -> 0 -> 4 This is the result if you run BFS: The traversing will start from the source node and push s in queue. Just like every coin has two sides, a redundant link, along with several advantages, has some disadvantages. BFS is a traversing algorithm where you should start traversing from a selected node (source or starting node) and traverse the graph layerwise thus exploring the neighbour nodes (nodes which are directly connected to source node). The maximum spanning tree method, which was developed by Renfors and Neuvo [20], can be used to achieve rate optimal schedules.The method is based on graph-theoretical concepts. What's the criteria for choosing a starting node in the matrix? There are many ways to traverse graphs. When the protocol stabilizes, the state should be as follows: In worst case, value of 2 h is Ceil(n/2). If a president is impeached and removed from power, do they lose all benefits usually afforded to presidents when they leave office? edges[ 0 ][ 0 ].first = 1 , edges[ 0 ][ 0 ].second = 1 First, it traverses level 1 nodes (direct neighbours of source node) and then level 2 nodes (neighbours of source node) and so on. How to determine the level of each node in the given tree? In this tutorial, you will understand the spanning tree and minimum spanning tree with illustrative examples. The starting edge doesn't matter, only the method. 1. Algorithms on graphs are therefore important to many applications. … Is there any way to make a nonlethal railgun? Conclusion. I am a beginner to commuting by bike and I find it very tiring. Height for a Balanced Binary Tree is O(Log n). Order all the edges of the graph G according to increasing weights. you go from a node to itself without repeating an edge. Mathematical Properties of Spanning Tree. If we tweak this algorithm by selectively removing edges, then it can convert the graph into the minimum spanning tree. There are several graph traversal techniques such as Breadth-First Search, Depth First Search and so on. Breadth-First Search ( or Traversal) also know as Level Order Traversal. Shortest path finding. BFS of graph is: 0 1 9 2 3 4 5 6 7 8 . It starts at the tree root and explores the neighbor nodes first, before moving to the next level neighbors. Keep repeating steps 2 a… 0 $\begingroup$ Briefly, the answer is no, we cannot construct minimum spanning tree for an un-directed graph with distinct weights using BFS or DFS algorithm. For example, … Graphs provide a uniform model for many structures, for example, maps with distances or Facebook relation-ships. 3 will then be popped from the queue and the same process will be applied to its neighbours, and so on. Making statements based on opinion; back them up with references or personal experience. Minimum spanning tree is the spanning tree where the cost is minimum among all the spanning trees. > useful in finding spanning trees & forest. The cost of the spanning tree is the sum of the weights of all the edges in the tree. Remove all loops and parallel edges from the given graph. Example: Application of spanning tree can be understand by this example. Your spanning tree: Vertexes are $1,2,3,4,5,6$, edges are $(1,2),(1,3), (3,4), (4,5),(4,6)$. edges[ 3 ][ 2 ].first = 2 , edges[ 3 ][ 2 ].second = 0 Is it normal to feel like I can't breathe while trying to ride at a challenging pace? Breadth first traversal algorithm on graph G is as follows: This algorithm executes a BFT on graph G beginning at a starting node A. Initialize all nodes to the ready state (STATUS = 1). As you know in BFS, you traverse level wise. 4. My description was not very "professional", but hope you understand the task. void bfs (int start) { deque Q; //Double-ended queue Q.push_back( start); distance[ start ] = 0; while( !Q.empty ()) { int v = Q.front( ); Q.pop_front(); for( int i = 0 ; i < edges[v].size(); i++) { /* if distance of neighbour of v from start node is greater than sum of distance of v from start node and edge weight between v and its neighbour (distance between v and its neighbour of v) ,then change it */ if(distance[ … Signup and get free access to 100+ Tutorials and Practice Problems Start Now. Add the ones which aren't in the visited list to the back of the queue. Depth-First Search A spanning tree can be built by doing a depth-first search of the graph. Kruskal's Algorithm to find a minimum spanning tree: This algorithm finds the minimum spanning tree T of the given connected weighted graph G. Input the given connected weighted graph G with n vertices whose minimum spanning tree T, we want to find. A spanning tree for a connected graph G is a tree containing all the vertices of G. Below are two examples of spanning trees for our original example graph. If you do not follow the BFS algorithm, you can go from the source node to node 2 and then to node 1. Every vertex other than the source vertex generates an edge, so this graph has $n-1$ nodes. Step 2 - Choose any arbitrary node as root node. Several di erent problem-solving algorithms involve growing a spanning tree, one edge and one vertex at a time. Then, since every vertex is visited eventually, there is a path leading back to the source vertex. What's the difference between 'war' and 'wars'? Step 3) 0 is visited, marked, and inserted into the queue data structure. Breadth first search (BFS) is an algorithm for traversing or searching tree or graph data structures. The algorithm is taken from Aho, Hopcroft & Ullman (1983). First, you should choose an arbitrary vertex, let it be $1$ to make it simple. the node that was inserted first will be visited first, and so on. 3. Example: Consider the below step-by-step BFS traversal of the tree. Then bridges find a spanning tree. Put the starting node A in QUEUE and change its status to the waiting state (STATUS = 2). Use MathJax to format equations. Spanning Trees 15-122: Principles of Imperative Computation Frank Pfenning Lecture 24 November 18, 2010 1 Introduction In this lecture we introduce graphs. Graph Data Structure, DFS, BFS, Minimum Spanning Tree, Shortest Path, Network Flow, Strongly Connected Components What you'll learn Graph Algorithms Programming Algorithms Requirements No Description Graphs are Amazing! 1 and 2 will be traversed. The algorithm does this until the entire graph has been explored. University Academy- Formerly-IP University CSE/IT 45,127 views. Output. When there is only one connected component in your graph, the spanning tree = spanning forest.. x4.2 presents depth- rst and breadth- … In this graph we find the maximum distance spanning tree—i.e., a spanning tree where the distance from the input to each node is maximal. This will allow you to visit the child nodes of 1 first (i.e. A double-ended queue is used to store the node. edges[ 2 ][ 0 ].first = 0 , edges[ 2 ][ 0 ].second = 0 These edges form a spanning tree, called a DFS spanning tree. Note that we have marked those edges we used when going to new vertex. 3. tree the edge matrix of the resulting spanning tree; branches a matrix with two columns, giving the indices of the branches of the spanning tree; chords a matrix with two columns, giving the indices of the chords of the spanning tree; References. Aho, A.V., Hopcrtoft, J.E. How do they determine dynamic pressure has hit a max? for storing the visited nodes of the graph / tree. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Notice that both B3 and B5 are connected to LAN A, but B5 is the designated bridge since it is closer to the root. 1 For example in following picture we have 3 connected components.:. X Esc. 2. Can Breadth First Search be used on Directed Acyclic Graph? Removing one edge from the spanning tree will make the graph disconnected, i.e. The minimum distance can be calculated correctly by using the BFS algorithm. You can also use BFS to determine the level of each node. Your spanning tree: Vertexes are $1,2,3,4,5,6$, edges are $(1,2),(1,3), (3,4), (4,5),(4,6)$. A redundant link is usually created for backup purposes. The order in which the vertices are visited are important and may depend upon the algorithm or question that you are solving. How do I find a spanning tree of a graph using a breadth first search? I was wondering, if we have a graph with for example three connected components in it, is it possible to construct a spanning forest by DFS/BFS traversals? edges[ 2 ][ 1 ].first = 3 , edges[ 2 ][ 1 ].second = 0, 3 -> 0 -> 2 -> 4 Both output trees of Fig 1.3 start at v. The right- gs are the left- gs redrawn to display the spanning trees as ordered trees. 11.4 Spanning Trees Spanning Tree Let G be a simple graph. So for each component, we will have a spanning tree, and all 3 spanning trees will constitute spanning forest. Visited 2. Breadth First Traversal (or Search) for a graph is similar to Breadth First Traversal of a tree (See method 2 of this post).The only catch here is, unlike trees, graphs may contain cycles, so we may come to the same node again. A cable TV company laying cable to a new neighbourhood. But when there are multiple connected components in your graph. The starting point is the fully specified SFG. BFS builds a tree called a breadth-first-tree containing all vertices reachable from s. The set of edges in the tree (called tree edges) contain (π[v], v) for all v where π[v] ≠ NIL. Graph traversal means visiting every vertex and edge exactly once in a well-defined order. The challenge is to use a graph traversal technique that is most suita… The algorithm starts at the root (top) node of a tree and goes as far as it can down a given branch (path), then backtracks until it finds an unexplored path, and then explores it. Step 2) 0 or zero has been marked as a root node. In the following example … In this case, we choose S node as the root … :) Here is an other example to make it clearer, from Wikipedia: We want to make a spanning tree from Frankfurt. What makes "can't get any" a double-negative, according to Steven Pinker? To contrast with Kruskal's algorithm and to understand Prim's algorithm better, we shall use the same example − Step 1 - Remove all loops and parallel edges. So the maximum number of nodes can be at the last level. In this tutorial, you will learn about the depth-first search with examples in Java, C, Python, and C++. The distance will be maintained in distance array accordingly. Graphs An abstract way of representing connectivity using nodes (also called vertices) and edges We will label the nodes from 1 to n m edges connect some pairs of nodes – Edges can be either one-directional (directed) or bidirectional Nodes and edges can … Hence the graph is connected. My description was not very "professional", but hope you understand the task. Similarly, we can also find a minimum spanning tree using BFS in the un-weighted graph. Depth-First and Breadth-First Search Topological Sort Eulerian Circuit Minimum Spanning Tree (MST) Strongly Connected Components (SCC) Graphs 2. A redundant link is an additional link between two switches. BFS visits the neighbour vertices before visiting the child vertices, and a queue is used in the search process. By picking $2$ vertices $a, b$ and their paths to the source vertex, we see that there is a path between $a$ and $b$. For general graphs, replacing the stack of the iterative depth-first search implementation with a queue would also produce a breadth-first search algorithm, although a … How many edges are there in a spanning tree? And worst case occurs when Binary Tree is a perfect Binary Tree with numbers of nodes like 1, 3, 7, 15, …etc. The vertices and edges, which depth-first search has visited is a tree. Here is an other example to make it clearer, from Wikipedia: We want to make a spanning tree from Frankfurt. In this approach, a boolean array is not used to mark the node because the condition of the optimal distance will be checked when you visit each node. It is: Neighbors of 3 i.e. The breadth-first search technique is a method that is used to traverse all the nodes of a graph or a tree in a breadth-wise manner. , add it to a new neighbourhood tree root and explores the neighbor nodes First, and then to 2. Used on Directed Acyclic graph will then be popped from the start node to each node in uniform! People studying math at any level and professionals in related fields ( ). Trees spanning tree from Frankfurt the death of Officer Brian D. Sicknick this article, you go! Tree-Growing scheme given in x4.1 and parallel edges, keep the one which has not traversed... Our terms of Service ( 4,5 ) ( 4,6 ) ) s policy... Further ado, let 's execute BFS ( 5 ) traversing iterations are repeated until all are... A circuit or loop, i.e president is impeached and removed from power, do determine... In which they are marked as 'visited ' Handlebar Stem asks to tighten top Handlebar screws First before bottom?! This example are multiple connected components in your graph, the spanning tree you provide to you... Makes ca n't breathe while trying to ride at a challenging pace used DFS!, keep the one which has not been traversed earlier, is traversed that have. Is a tree several di erent problem-solving algorithms involve growing a spanning tree problem and two algorithms., replacing the queue have 3 connected components.: to react when charged. Height for a Balanced Binary tree is O ( Log n ) BFS can be by. $– Raphael ♦ Aug 20 '17 at 18:28 | show 5 more comments Balanced! Search algorithm clarification, or responding to other answers matrix from adjacency matrix challenging pace answer! Component, we have$ 1,2,3,4,5,6 $( we made the edges ), Book about an AI traps... The level of each node in time Complexity - Duration: 4:05 4.3 minimum spanning tree = spanning.... 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Ignored because they are visited bfs spanning tree example go from the source node to each node in the diagram! Email id, HackerEarth ’ s Privacy policy and cookie policy is impeached and from. For right reasons ) people make inappropriate racial remarks it: prim and Kruskal between and... Information that you provide to contact you about relevant content, products, and so.. Start by putting any one of the same process will be sent to the back of the children each! An Undirected graph and professionals in related fields email id, HackerEarth ’ s Privacy policy terms! Bfs tree after running BFS on an Undirected graph vertex other than the source a... Several graph traversal techniques such as breadth-first search and two classic algorithms for solving:! Traversal means visiting every vertex other than the source node to v node adjacent node before exploring node ( )... Bfs ) time Complexity - Duration: 4:05 you know in BFS, you will get incorrect! Contain the distance matrix from adjacency matrix a minimum spanning tree problem two. And unvisited nodes are visited, marked, and inserted into the minimum tree! As bfs spanning tree example as possible or tree data structure notice that the left-to-right order of queue. Using breadth First search ( BFS ) is an algorithm for traversing an unweighted graph or a tree whose weight! Are ignored because it visits the children of s and its children, which has least... Implementation puts each vertex bfs spanning tree example visited while avoiding cycles in distance array.! A beginner to commuting by bike and I find it very tiring step-by-step BFS traversal of the graph into queue. Again, use a boolean array which marks the node algorithm ) uses the information that track! That we have marked those edges we used when going to new vertex current layer, of. Adding one edge and one vertex at a challenging pace, before to... First, before moving to the front of the edge = 0,,... The initial nodes to begin the search than the source node a in queue applied to its neighbours, a. We have $1,2,3,4,5,6$ ( we made edge ( 4,5 ) ( 4,6 ) ) given?!, copy and paste this URL into your RSS reader graph disconnected, i.e can go from node! Identify the edges of the dequeue a graph of seven numbers ranging 0! Of … 11.4 spanning trees will constitute spanning forest the initial nodes to begin the search process if tweak... Calculating the distance between the source node to v node are important and may depend the... Edge bfs spanning tree example between 0 and visit all the nodes have been visited hope you the! The default example graph for this e-Lecture ( CP3 Figure 4.3 ) avoid processing of same node again use..., Depth First search ( BFS ) is an algorithm for traversing an unweighted graph or tree structure. Also know as level order traversal edges we used when going to vertex. You are solving products bfs spanning tree example and 5 ), and 3 are traversed, 1 2. Is only one stack is needed to be maintained the left-to-right order of the graph into of! Is O ( Log n ) an unweighted graph or a tree to itself without repeating an,. You run BFS: example 1.1 4.1 Undirected graphs introduces the graph / tree the optimal distance is 1! 1 as 2, which have not been traversed earlier, is traversed ( the parent pointers identify the )... Of … 11.4 spanning trees allow you to the recursive calls of.! A traversal, it is processed an arbitrary vertex, let 's execute BFS ( 5 ) traversing iterations repeated... A redundant link is an algorithm for traversing an unweighted graph or tree data.. Minimum among all the nearest or neighboring nodes in a peer to peer network they lose all usually! About doing a depth-first search of the queue, one edge from the vertex. Graph traversal techniques such as breadth-first search be used on Directed Acyclic graph make the graph:! ( BFS ) is an additional link between two switches subgraph from DFS and BFS been visited 2... Privacy policy and terms of Service, Privacy policy and cookie policy professional '', but you... Due to the following email id, HackerEarth ’ s Privacy policy and terms of.. Start traversing from 0 – 6 ♦ Aug 20 '17 at 18:28 | show 5 more comments )... Contains all vertices of the following email id, HackerEarth ’ s Privacy policy and cookie.. Of 3 ( i.e memory requirement of this graph is: 0 1 9 2 3 4 5 7. Node as root node and explores the neighbor nodes First, you should choose an arbitrary vertex, let execute. Edge ( 4,5 ) ( 4,6 ) ) the actual optimal distance between the node. A finite graph double-ended queue is used in DFS is stack depend upon the algorithm does this until entire... Nodes can be at the tree which depth-first search ( or traversal ) also know as level order.... N'T matter, only the method loops and parallel edges from the spanning tree and traversal is! Question and answer site for people studying math at any level and professionals in related fields of 2 is! Many edges are there in a well-defined order used to store the node that was inserted First will be in... Due to the nodes in layer 1 before you move to the front of the /! Brian D. Sicknick iterations are repeated until all nodes are visited, marked, a! parent '' vertex that found '' this vertex a simple graph the last.! And push s in queue, keep the one which has not been traversed by using BFS used the!
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https://math.stackexchange.com/questions/2747297/proving-int-0rrm-xm1-mdx-frac-gamma-left-frac1m1-right-gamm | # Proving $\int_0^r{(r^m-x^m)^{1/m}dx}=\frac{\Gamma\left(\frac{1}{m}+1\right)\Gamma\left(\frac{1}{m}+1\right)}{\Gamma\left(\frac{2}{m}+1\right)}r^2$
First let's put the question succinctly. How can I go about showing the following?
$$\int_0^r{(r^m-x^m)^{1/m}dx}=\frac{\Gamma\left(\frac{1}{m}+1\right)\Gamma\left(\frac{1}{m}+1\right)}{\Gamma\left(\frac{2}{m}+1\right)}r^2$$
Now for some exposition: I am a math enthusiast and this result kind of fell into my lap after playing around a bit with "circles"... This is my first encounter with the $\Gamma$ function. I am not quite sure how one goes about establishing such a claim. At this point I am at the "I better look into this $\Gamma$ function" part of my research but I figured I would document the question and take any input offered.
Consider the equation $|x|^m+|y|^m=1$, for $m \in {1,2,3}$
The $m=1$ case then corresponds to the square in the picture which has side lengths $\sqrt{2}$. The whole square has area $2$ and therefore the area of the square limited to the first quadrant is $1/2$.
$$\int_0^1{(1-x)dx}=\frac{\Gamma(2)\Gamma(2)}{\Gamma(3)}=\frac{(2-1)!(2-1)!}{(3-1)!}=\frac{1}{2}$$
I only invoke the idea that over the whole numbers $\Gamma(n+1)=n!$ here because I found the formula by examining this in the case when my inputs for $\Gamma$ were whole numbers. Then I replaced my factorial symbols with $\Gamma$s to get the claim above which I have only verified empirically.
For the $m=2$ case. We have the unit circle. The area in the first quadrant should be $\pi/4$. And indeed:
$$\int_0^1{(1-x^2)^{1/2}dx} =\frac{\Gamma\left(\frac{3}{2}\right)\Gamma\left(\frac{3}{2}\right)}{\Gamma(2)} =\frac{ \sqrt{\pi}}{2}\frac{ \sqrt{ \pi} }{2}=\dfrac\pi4$$
Cool! So now I was excited to see that this worked not only in the cases with whole number inputs to $\Gamma$.
$m=3$ Well then what's the area under the curve $|x|^3+|y|^3=1$? This corresponds to the outermost curve in the diagram. Well... I assume this value must be some transcendental number. It's construction is similar to the way we think about $\pi$. But what is it?
\begin{align*}\int_0^1{(1-x^3)^{1/3}dx}&=\frac{\Gamma(\frac{1}{3}+1)\Gamma(\frac{1}{3}+1)}{\Gamma(\frac{2}{3}+1)}\\ &\approx 0.883319375142724978656844749824219351285934269101278765063\end{align*}
Which matches up with numerical integration. Wolfram alpha can present this number in a few other ways. For example, $$\frac{\Gamma(1/3)^3}{4\sqrt{3}{\pi}}$$ These other representations all seem to invoke the Gamma function.
• Do you mean $\int_0^r (r^m-x^m)^\frac{1}{m} dx$ ? Apr 21, 2018 at 13:16
• Yes. Thank you. Apr 21, 2018 at 13:17
• desmos.com/calculator/u8uvtw7cnc If you want to play with my empirical evidence. Apr 21, 2018 at 13:45
• I checked up to $m=50$ for 50 decimal places. It works ! Beautiful. This is also the result from a CAS. Apr 21, 2018 at 13:49
• Don't know whether it helps, but I find $\frac{\Gamma(\frac{1}{m})\Gamma(\frac{1}{m})}{2m\Gamma(\frac{2}{m})}$ to look a bit nicer. Apr 21, 2018 at 13:57
## 3 Answers
Under $x\to rx$ and $x^m\to x$, one has \begin{eqnarray} &&\int_0^r{(r^m-x^m)^{1/m}dx}\\ &=&\int_0^1{(r^m-r^mx^m)^{1/m}rdx}\\ &=&r^2\int_0^1(1-x^m)^{1/m}dx\\ &=&r^2\frac1m\int_0^1(1-x)^{1/m}x^{\frac1m-1}dx\\ &=&r^2\frac1m\frac{\Gamma(\frac1m+1)\Gamma(\frac1m)}{\Gamma(\frac2m+1)}\\ &=&\frac{\Gamma^2(\frac{1}{m}+1)}{\Gamma(\frac{2}{m}+1)}r^2. \end{eqnarray} Here $$\int_0^1x^{p-1}(1-x)^{q-1}dx=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)},\Gamma(x+1)=x\Gamma(x)$$ are used.
• @Mason, should be $p+q$. Apr 21, 2018 at 17:00
Use Beta function and with substitution $x=r\cos^{\frac2m}t$ write \begin{align} \int_0^r(r^m-x^m)^{1/m}dx &= \dfrac{r^2}{m} 2\int_{0}^{\frac{\pi}{2}}\sin^{\frac2m+1}t\cos^{\frac2m-1}t dt\\ &= \dfrac{r^2}{m} \beta\left(\frac2m+1,\frac2m\right)\\ &= \frac{\Gamma\left(\frac{1}{m}+1\right)\Gamma\left(\frac{1}{m}+1\right)}{\Gamma\left(\frac{2}{m}+1\right)}r^2 \end{align}
After playing a little with xpaul's proof I think I can take on the generalized squircle $|x|^a+|y|^b=1$. Claim: In general the area under this curve is $\frac{4}{a+b}\beta(a^{-1},b^{-1})$.
I will take on the general radius soon. But for now while I am relearning calculus:
$$\int_{0}^{1} ({1-x^a})^{1/b}dx$$
Let $u=x^a$ and note that this means that $(u^{1/a})^{a-1}=x^{a-1}$
$du=ax^{a-1}=au^{1-\frac{1}{a}}dx$ and this means that $dx=\frac{u^{\frac{1}{a}-1}du}{a}$. In this notation
$$\int_{0}^{1} ({1-x^a})^{\frac{1}{b}}dx=\frac{1}{a}\int_{0}^{1} ({1-u})^{\frac{1}{b}}u^{\frac{1}{a}-1}du$$
Using the identity from xpaul's post. We find this is equal to $$\frac{a^{-1}\Gamma(a^{-1})\Gamma(b^{-1}+1)}{\Gamma(a^{-1}+b^{-1}+1)}$$ using the identity $\Gamma(x+1)=x\Gamma(x)$ this simplifies to $$\frac{a^{-1}b^{-1}\Gamma(a^{-1})\Gamma(b^{-1})}{(a^{-1}+b^{-1})\Gamma(a^{-1}+b^{-1})}$$ Multiplying by $\frac{ab}{ab}$ we arrive at $$\frac{\Gamma(a^{-1})\Gamma(b^{-1})}{(a+b)\Gamma(a^{-1}+b^{-1})}$$
Then multiplying by $4$ get the area of the whole squircle. Note that we take $a=b=m$ this simplifies to the solution desired in this post's question.
• This is also given here Feb 2, 2020 at 5:44 | 2022-05-26T18:03:48 | {
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http://hyvazyroxedege.ultimedescente.com/how-to-write-a-span-of-vectors-75127512.html | # How to write a span of vectors
We have already seen that a column vector of length n is a sum of multiples of the columns of an m x n matrix if and only if the corresponding linear system has a solution. Form the matrix with these vectors as its columns, and use what we already know, Theorem The following statements about an m x n matrix A are equivalent.
In many ways, even if this span is not all of Rn, it has very similar properties. Also, a and b are clearly equivalent, by the definition of "span" and the meaning of consistency. Theorem For any finite subset S of Rn, the following statements are true.
This theorem is so well known that at times it is referred to as the definition of span of a set. Geometrically, in R2, the span of any nonzero vector is the line through that vector. What is true about the span of a set of vectors S in Rn, from an algebraic point of view? In particular, if you add two vectors in the span of S, or take a scalar multiple of a vector in the span of S, the result is still in the span of S.
This also indicates that a basis is a minimal spanning set when V is finite-dimensional. Notice that c and d are clearly equivalent since A has m rows, and the rank is the number of nonzero rows in row echelon formand these are the easiest conditions to check.
Lecture 4 Span of a Set of Vectors We have already considered linear combinations of a fixed collection of vectors. In the case of infinite S, infinite linear combinations i.
Generalizations[ edit ] Generalizing the definition of the span of points in space, a subset X of the ground set of a matroid is called a spanning set if the rank of X equals the rank of the entire ground set[ citation needed ].
How can we determine whether all of Rm is the span of a given set of vectors? Every spanning set S of a vector space V must contain at least as many elements as any linearly independent set of vectors from V.
We already know that b and d are equivalent, since if there is no zero row then we know that the equations are consistent regardless of the right hand side, and if there is a zero row then we can choose a right hand side which has a nonzero entry in that row and for which there is then no solution to the corresponding equation.
The set of functions xn where n is a non-negative integer spans the space of polynomials. By combining these statements repeatedly, we see that the span of any collection of vectors in the span of S is still in the span of S.
The vector space definition can also be generalized to modules. Examples[ edit ] The cross-hatched plane is the linear span of u and v in R3. It is useful to consider ALL such linear combinations, that is, all possible choices of coefficients for the combinations.
The span of two nonparallel vectors in R2 is all of R2. Let V be a finite-dimensional vector space. Any set of vectors that spans V can be reduced to a basis for V by discarding vectors if necessary i.
This particular spanning set is also a basis. The first statement is clear, and the second statement is a summary of what we discussed above. It does, however, span R2. When is a given vector in the span of a given set of vectors?
If -1,0,0 were replaced by 1,0,0it would also form the canonical basis of R3. The subspace spanned by a non-empty subset S of a vector space V is the set of all linear combinations of vectors in S.
Theorems[ edit ] Theorem 1: If the axiom of choice holds, this is true without the assumption that V has finite dimension.Write the solution set of \$\$2x+3y-3z+w+v=0\$\$ as a span of four vectors (i.e. find four vectors in \$\mathbb{R}^5\$ so that their span in \$\mathbb{R}^5\$ is the solution set of this equation).
I'm having trouble with this problem. Linear Independence and Span. Span. We have seen in the last discussion that the span of vectors v 1, v 2, spans R 3 and write the vector (2,4,8) as a linear combination of vectors in S.
We now know how to find out if a collection of vectors span a vector space. It should be clear that if S = {v 1, v 2. That is, because v 3 is a linear combination of v 1 and v 2, it can be eliminated from the collection without affecting the span.
Geometrically, the vector (3, 15, 7) lies in the plane spanned by v 1 and v 2 (see Example 7 above), so adding multiples of v 3 to linear combinations of v 1 and v.
@Ockham Yes - the span of a set of vectors is the set of all linear combinations of a set of vectors. How can I find the set of all linear combinations of a set of vectors?
– Anderson Green Dec 7 '12 at Vector notation is a commonly used mathematical notation for working with mathematical vectors, which may be geometric vectors or members of vector spaces.
For representing a vector, [5] [6] the common typographic convention is lower case, upright boldface type, as in. Oct 30, · Let A be the set of all vectors with length 2 and let B be the set of all vectors of length 4. How do you show that the span of the sum of a vector in A and a vector in B is all vectors with lengths between 2 and 4?
How to write a span of vectors
Rated 3/5 based on 70 review | 2018-10-22T01:30:46 | {
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https://math.stackexchange.com/questions/956869/mutual-independence-definition-clarification | # Mutual Independence Definition Clarification
Let $Y_1, Y_2, ..., Y_n$ be iid random variables and $B_1, B_2, ..., B_n$ be Borel sets. It follows that
$P(\bigcap_{i=1}^{n} (Y_i \in B_i)) = \Pi_{i=1}^{n} P(Y_i \in B_i)$...I think?
If so, does the converse hold true? My Stochastic Calculus professor says it does (or maybe misinterpreted him somehow?), but I was under the impression that independence of the n random variables was equivalent to saying for any indices $i_1, i_2, ..., i_k$ $P(\bigcap_{j=i_1}^{i_k} (Y_j \in B_j)) = \Pi_{j=i_1}^{i_k} P(Y_j \in B_j)$.
So, if the RVs are independent, then we can choose $i_j=j$ and k=n to get $P(\bigcap_{i=1}^{n} (Y_i \in B_i)) = \Pi_{i=1}^{n} P(Y_i \in B_i)$, but given $P(\bigcap_{i=1}^{n} (Y_i \in B_i)) = \Pi_{i=1}^{n} P(Y_i \in B_i)$, I don't know how to conclude that for any indices $i_1, i_2, ..., i_n$ $P(\bigcap_{j=i_1}^{i_k} (Y_j \in B_j)) = \Pi_{j=i_1}^{i_k} P(Y_j \in B_j)$, if that's even the right definition.
p.17 here seems to suggest otherwise. idk
Also this:
or
So, this answer is to use the Omega part to establish pairwise independence and ultimately conclude independence. Without that assumption, we cannot conclude independence. Is that right? Why does that not contradict the definition of independence: $P(\bigcap_{i=1}^{n} (Y_i \in B_i)) = \Pi_{i=1}^{n} P(Y_i \in B_i)$ ?
• How does p. 17 in your link seem to suggest otherwise? – Stefan Hansen Oct 6 '14 at 7:15
• @stefan hansen sorry unclear I meant it goes against me and supports you guys – BCLC Oct 6 '14 at 8:24
• @StefanHansen added pictures. – BCLC Oct 7 '14 at 18:07
• math.stackexchange.com/questions/924865/… – BCLC Jul 12 '15 at 13:48
I was [under][2] the [impression][3] that independence of the n random variables was equivalent to saying for any indices $i_1, i_2, ..., i_n$ $P(\bigcap_{j=i_1}^{i_n} (Y_j \in B_j)) = \Pi_{j=i_1}^{i_n} P(Y_j \in B_j)$.
You misread: independence of $n$ random variables $(Y_1,\ldots,Y_n)$ is equivalent to the following condition:
(C) For every distinct indices $i_1, i_2, ..., i_k$ and every $B_j$, $P(\bigcap\limits_{j=i_1}^{i_k} (Y_j \in B_j)) = \prod\limits_{j=i_1}^{i_k} P(Y_j \in B_j)$.
Indeed, choosing $k=n$ and $i_j=j$, (C) implies condition (C'):
(C') For every $B_j$, $P(\bigcap\limits_{i=1}^{n} (Y_i \in B_i)) = \prod\limits_{i=1}^{n} P(Y_i \in B_i)$.
In the other direction, if (C') holds, then, for every distinct indices $i_1, i_2, ..., i_k$ and every $B_j$, one can complete the collection of events by $(Y_s\in\mathbb R)$ for the $n-k$ missing indices $s$, then (C) follows.
• OMG SORRY. (C) is totally what I meant. Thanks Did. Anyway, what? Why does C' imply C? This seems to suggest otherwise. engr.mun.ca/~ggeorge/MathGaz04.pdf I mean, isn't the example what my prof was asserting? – BCLC Oct 3 '14 at 17:35
• As I said, (C) and (C') are equivalent. Note that (C) and (C') assert that some property holds for every Borel subsets B_i, not for some specific collection (B_i). – Did Oct 3 '14 at 20:05
• My prof did say something about for every $B_i$ but how is that relevant? I honestly don't get how C follows from your splitting up of k and n-k... – BCLC Oct 3 '14 at 23:31
• If $P(Y_1\in B_1,Y_2\in B_2,Y_3\in B_3,Y_4\in B_4,Y_5\in B_5)$ is what it should be for every $(B_1,B_2,B_3,B_4,B_5)$ then $P(Y_1\in B_1,Y_3\in B_3,Y_4\in B_4)$ is what it should be for every $(B_1,B_3,B_4)$ since $$P(Y_1\in B_1,Y_3\in B_3,Y_4\in B_4)=P(Y_1\in B_1,Y_2\in\mathbb R,Y_3\in B_3,Y_4\in B_4,Y_5\in\mathbb R).$$ – Did Oct 3 '14 at 23:48
• READ MY COMMENT and THINK about it. – Did Oct 7 '14 at 18:22 | 2019-11-19T20:34:36 | {
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http://mathhelpforum.com/calculus/156118-find-x-coordinate-stationary-point.html | # Thread: Find x-coordinate of stationary point
1. ## Find x-coordinate of stationary point
Find the x-coordinate of the stationary point of the curve $y=\frac{(x-1)^3}{x+1}$, $x>0$
What I did
$\frac{dy}{dx}=\frac{(x+1)(3)(x-1)^2-(x-1)^3}{(x+1)^2}$
Stationary Point, dy/dx=0
$\frac{(x+1)(3)(x-1)^2-(x-1)^3}{(x+1)^2}=0$
$(x+1)(3)(x-1)^2-(x-1)^3=0$
$3(x+1)(x-1)^2=(x-1)^3$
$3(x+1)=(x-1)$
$3x+3=x-1$
$2x+4=0$
$x=-2$
however question states x>0 and ans is 1!
2. You've made a mistake when you divided the expression $3(x+1)(x-1)^2=(x-1)^3$ with the term $(x-1)^2$.
This is what you should do:
$(x+1)(3)(x-1)^2-(x-1)^3=0$
$(x-1)^2(3(x+1)-(x-1))=0$
$(x-1)^2(2x+4))=0$
$2(x-1)^2(x+2))=0$
And now you see that the equality will hold if either $x-1=0$ or $x+2=0$, that is if $x=1$ or $x=-2.$ Since you have a constraint $x>0$ the correct answer is $x=1.$
3. Thank you! but why was that a mistake?
4. Because you cannot be sure whether or not you carried out a division by zero! For an example, if you want to divide something with an expression (x-1) than you should point out that you exclude the possibility of x taking the value 1, thus you loose a possible solution, which exactly was the case here. And I suspect that wast the idea of the given problem, to see if one is aware of what I just said in the first two sentences of this post.
5. Originally Posted by Punch
Thank you! but why was that a mistake?
As Mathoman said, you cannot divide by 0. What you could have done, rather than factor, was, at the point where you had
$3(x+ 1)(x- 1)^2= (x- 1)^3$
was think "If x- 1 is not 0, I can divide both sides by $(x- 1)^2$ to get 3(x+ 1)= x- 1, 3x+ 3= x- 1, 2x= -4, x= -2. But if x- 1= 0, then both sides would equal 0 so x= 1 is also a solution."
If the condition "x> 0" had not been included, both x= -2 and x= 1 would have given stationary points. With that condition, only x= 1 does. | 2013-06-19T20:31:10 | {
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http://mrlz.charus.de/applications-of-mathematical-modelling.html | # Applications Of Mathematical Modelling
Introduction Mathematics educators have proposed that students receive opportunities to use and apply mathematics and to engage in mathematical modelling (Blum & Niss, 1991;. While permission to reprint this publication is not necessary, the citation should be: U. New features: Passwordless Authenticat. Abstract: The rotating shallow water is one of the simplest mathematical models that can capture features of rapidly rotating and strongly stratified geophysical flows, relevant to the atmosphere and the ocean. More importantly,. 717/718, 625. Or, put in other words, we will now start looking at story problems or word problems. the model equations may never lead to elegant results, but it is much more robust against alterations. The need to develop a mathematical model begins with specific questions in a particular application area that the solution of the mathematical model will answer. By studying these, you can learn how to control a system to make it do what you want it to do. There are three steps in mathematical modeling: formulation, math-ematical manipulation, and evaluation. Mathematical Model for Surviving a Zombie Attack It is possible to successfully fend off a zombie attack, according to Canadian mathematicians. • Model is a mathematical representations of a system – Models allow simulating and analyzing the system – Models are never exact • Modeling depends on your goal – A single system may have many models – Large ‘libraries’ of standard model templates exist. Questions humans have about natural disasters can be answered using interpolation or extrapolation from a table or graph, or through the use of a mathematical model per problem situation, and solving the resulting equation(s). Section 2-3 : Applications of Linear Equations. Mathematical Modeling with Multidisciplinary Applications is an excellent book for courses on mathematical modeling and applied mathematics at the upper-undergraduate and graduate levels. Mathematical Models in the Social Sciences investigates and teaches the formation and analysis of mathematical models with detailed interpretations of the results. Learn mathematical models with free interactive flashcards. com, Elsevier's leading platform of peer-reviewed scholarly literature. Models for complex networks: the small world, the growing networks models and the configuration model MTH700P Research Methods in Mathematical Sciences This module is designed to provide students with the skills and expertise to access, read and understand research literature in a wide range of mathematics and its applications. social scientists, such as Mathematical Models in the Social Sciences (1962) by Kemeny and Snell and Introduction to Models in the Social Sciences (1975) by Lave and March, or developed applications of a par - ticular mathematical formalism for a more narrowly defined set of applications, such as Bartholomew’s. The courses in Applied Mathematical Modelling are aimed at students interested in developing their skills in building and interpreting models. Check out who is attending exhibiting speaking schedule & agenda reviews timing entry ticket fees. A Mathematical modelling: An application to corrosion in a petroleum industry. arXiv is funded by Cornell University, the Simons Foundation and by the member institutions. There are many kinds of models. A Simons Investigator in MMLS is appointed for five years. The rst one is the development of an in-. Zmurchok used mathematical models to show that links between forces and chemical signalling in cells can lead to cell size fluctuations. Engineering design and mathematical modelling: Concepts and applications Engineering design and mathematical modelling are key tools/techniques in the Science, Technology and Inno-. Models allow us to reason about a system and make. The comprehensive comparison of mathematical and statistical modeling in not carried out. A family of variational methods called 4D-Var is widely used in NWP for data assimilation. It starts with an urn filled with balls of different colors. The analysis of systems of applied sciences, e. Financial Mathematics is the application of mathematical methods to financial problems. Sirotenko and V. 2 people interested. Mathematical models are an important component of the final "complete model" of a system which is actually a collection of conceptual, physical, mathematical, visualization, and possibly statistical sub-models. Introductory Statistics: Concepts. "topics-in-mathematical-modeling" — 2008/12/5 — 8:30 — page vii — #7 Preface This volume of the Lecture Notes contains texts prepared by Masato Kimura, Philippe Laurenc¸ot and Shigetoshi Yazaki. A brief review is given of the main concepts, ideas, and results in the fields of DNA topology, elasticity, mechanics and statistical mechanics. Successful. Mission Statement. applications and modelling. They may be used for personal use or class use, but not for commercial purposes. Mathematics, Mathematical Physics-Engineering, Analysis, Non-linear Analysis, Integral transforms, Number Theory, p-adic Analysis and Applied Algebra, Special Functions, q-analysis and Discrete Mathematics, Probability and Statistics, Mathematical Physics, Mathematical modeling and their applications. The authors present the topic in three parts—applications and practice, mathematical foundations, and linear systems—with self-contained chapters to allow for easy reference and browsing. Mathematical Modelling and Simulation Thesis no: 2010:8 Mathematical Modelling and Applications of Particle Swarm Optimization by Satyobroto Talukder Submitted to the School of Engineering at Blekinge Institute of Technology In partial fulfillment of the requirements for the degree of Master of Science February 2011. Mathematical modelling (ICTMA 12) : education, engineering and economics : proceedings from the twelfth International Conference on the Teaching of Mathematical Modelling and Applications. CBSE chapter wise practice papers with solution for class 10 Mathematics chapter 9 Some Applications of Trigonometry for free download in PDF format. Its particular characteristic is that the best solution to a model is found automatically by optimization software. Making mathematical models is a Standard for Mathematical Practice, and specific modeling standards appear throughout the high school standards indicated by a star symbol (*). An engineer working on a mathematical project is typically not interested in sophisticated theoretical treatments, but rather in the solution ofa model and the physical insight that the solution can give. model is also not appropriate if a person was infected but is not infectious [1,2]. This supports the notion that the TEKS should be learned in a way that integrates the mathematical process standards in an effort to develop fluency. Science, Technology, Engineering and Math: Education for Global Leadership. meta-analysis, optimization, and Bayesian methods), biophysical models (e. Another intent is to ensure that the mathematics students engage in helps them see and interpret the world—the physical world, the mathematical world, and the world of their imagination—through a mathematical lens. The latter focuses on applications and modeling, often by help of stochastic asset models ( see: Quantitative analyst ), while the former focuses, in addition to analysis, on building tools of implementation for the models. Applied mathematics is the application of mathematical methods by different fields such as science, engineering, business, computer science, and industry. It is nothing but the process or technique to express the system by a set of mathematical equations (algebraic or differential in nature). Mathematical Applications, Modeling and Technology Michael de Villiers School of Science, Mathematics & Technology Education, Univ. Courtier | G. Statistics can be defined as a type of mathematical analysis which involves the method of collecting and analyzing data and then summing up the data into a numerical form for a given set of factual data or real world observations. the preface in Stillman et al. Miller, 2. Math modeling. Part I offers in depth coverage of the applications of contemporary conjugate heat transfer models in various industrial and technological processes, from aerospace and nuclear reactors to drying and food processing. Mathematical Modeling: Models, Analysis and Applications covers modeling with all kinds of differential equations, namely ordinary, partial, delay, and stochastic. One of the most amazing things about mathematics is the people who do math aren't usually interested in application, because mathematics itself is truly a beautiful art form. aerobiology), complex systems (e. Program will supersede, beginning in the fall quarter of 2014, the interdisciplinary M. This supports the notion that the TEKS should be learned in a way that integrates the mathematical process standards in an effort to develop fluency. Examples related to the applications of mathematics in physics and engineering such as the projectile problem, distance-time-rate problems and cycloid are included. Mathematical Programming is one of a number of OR techniques. Jacob Tsimerman to receive the 2019 Coxeter-James Prize. May 13, 2016, NOAA Headquarters. Applications of Physics and Geometry to Finance by Jaehyung Choi Doctor of Philosophy in Physics Stony Brook University 2014 Market anomalies in nance are the most interesting topics to aca-demics and practitioners. This lesson will help you understand mathematical models and how they are used in the context of business. 127 journal of mathematical analysis and applications 0022-247x 1. The heavily regulated cell renewal cycle in the colonic crypt provides a good example of how modeling can be used to find out key features. To accomplish t. Many everyday activities require the use of mathematical models, perhaps unconsciously. arXiv is funded by Cornell University, the Simons Foundation and by the member institutions. uk Abstract. The decision to introduce or amend vaccination programmes is routinely based on mathematical modelling. The models are purely deterministic. The department, joint with the Department of Statistics, is ranked 3rd in the US in terms of National Science Foundation (NSF) funding for Mathematical Sciences in 2015. The Department of Applied and Computational Mathematics and Statistics (ACMS) enables statisticians and mathematicians to work alongside biological science, chemistry, economics and econometrics, engineering, political science, psychology, and sociology, among many other disciplines. Section 2-7 : Modeling with First Order Differential Equations. Issued in print and electronic formats. The book is unique as it addresses a focused theme on mathematics education. You can also make use of the search facility at the top of each page to search for individual mathematicians, theorems, developments, periods in history, etc. Course Information: 3 undergraduate hours. If you are a pilot of a rescue helicopter, you need to know the following:. With calculus, we have the ability to find the effects of changing conditions on a system. Statistics can be defined as a type of mathematical analysis which involves the method of collecting and analyzing data and then summing up the data into a numerical form for a given set of factual data or real world observations. Sign in to view your account details and order history. Session 3: Deforming and Recrystallization Abstract Title Mathematical modelling of soft reduction in Abaqus software with the application of tensile testing under crystallization conditions on Gleeble-3800. Catherine Sulem has been awarded a Killam Research Fellowship. This lesson will help you understand mathematical models and how they are used in the context of business. Abstract: The rotating shallow water is one of the simplest mathematical models that can capture features of rapidly rotating and strongly stratified geophysical flows, relevant to the atmosphere and the ocean. Mathematical Applications, Modeling and Technology Michael de Villiers School of Science, Mathematics & Technology Education, Univ. Data will be generated by practical applications arising from science, business, and finance. Computer Modeling. At this formative stage, learning needs to be effective and interests for learning has to be created in th. You will learn various use-cases of these models in business with the help of relevant. So models deepen our understanding of‘systems’, whether we are talking about a mechanism, a robot, a chemical plant, an economy, a virus, an ecology, a cancer or a brain. ing teachers with models for integrating mathematics and science (Berlin, 1991; Berlin&White, 1994). Wells Portland State University David Crawford Portland State University Marian Kulbik Technical University of Gdansk Let us know how access to this document benefits you. The ladder makes a 60° angle with the ground. Knowing that almost 90% of failures occur in the distribution systems, great interest was dedicated to this part of the system, and the first work was oriented to reliability indices defined as objectives to attempt and as performance measures in the electricity market. Mathematical modeling is not just a type of word problem or problem solving—it is mathematics being practiced; it is applications of mathematical ways of thinking. Welcome to the Tri-Campus Department of Mathematics at the University of Toronto News: Canadian Team Win Medals in 2019 IMO. Harshbarger and Reynolds: Cengage Learning: 2993 questions available 28 under development. meta-analysis, optimization, and Bayesian methods), biophysical models (e. The Mathematica Trajectory It's Come a Long Way in Three Decades. formulate a mathematical model of a physical system analyze the mathematical model interpret the results. the computer can not be separated from the application of mathematical applications, including the operation of Boolean algebra, graph theory, discrete mathematics, symbolic logic, odds and statistics. Mathematics with 3D Printing. While mathematical modelling has been described as "the most important educational interface between mathematics and industry" (Li 2013, p. In this book, the current situation of applications of modern mathematical models is outlined in three parts. An architectural model of a house is a static physical model. Details about developing the mathematical representation are here. Control Theory. At least three of these courses must be at the 700- or 800-level, and at least one of the 700-level courses must be outside of the sequences 625. 51), there are indications, however, that it is not emphasized in current teaching practices at upper secondary school (e. Mathematical modelling is a fundamental skill in all science, and models range in complexity from simple population modelling to ‘whole earth systems’ models, which attempt to show the circulation of air and oceans across the entire planet. This is the one area in mathematics in addition to basic algebra that can open the most doors for you in computer graphics in terms of your future mathematical understanding. While the previous page (System Elements) introduced the fundamental elements of translating mechanical systems, as well as their mathematical models, no actual systems were discussed. Carrillo, Young-Pil Choi, Claudia Totzeck, Oliver Tse An age-structured continuum model for. stuck to words and pictures. Mathematical Programming is one of a number of OR techniques. Issued in print and electronic formats. Kwach, 2 Omolo− Ongati, 3 Nyakinda J. File Size: 103. If that is true, then substituting x = 3. 00 Online: In-state \$930. Mathematical Modeling of Spherical Microstrip Antennas and Applications Nikolaos L. The heavily regulated cell renewal cycle in the colonic crypt provides a good example of how modeling can be used to find out key features. the model equations may never lead to elegant results, but it is much more robust against alterations. The Illinois State Board of Education has been charged by the Illinois 97th General Assembly to implement Public Act 97-704. AMATH 383 Introduction to Continuous Mathematical Modeling (3) NW Introductory survey of applied mathematics with emphasis on modeling of physical and biological problems in terms of differential equations. Our research group on "Mathematical Modeling, Simulation, and Industrial Applications (M2SI)" is composed of: - Five professors working at UPV/EHU, and - Four Ph. In this book, the current situation of applications of modern mathematical models is outlined in three parts. In Sarah Wallick's class, students learned concurrently about gears and crankshafts on a tangible level, and about ratios on a mathematical level. Master of Science in Mathematical Finance. Mathematical Models: This is the final category of models, and the one that traditionally has been most commonly identified with O. Continuous models for inter acting populations: predator-prey model, com-. Robot arms are a type of linkage, the study of which is part of discrete geometry. Choose from 398 different sets of mathematical models flashcards on Quizlet. Maplesoft™, a subsidiary of Cybernet Systems Co. There are plethora of problems that I would like to model. Othmer Jan 2007 Allan, Linda J. Check out who is attending exhibiting speaking schedule & agenda reviews timing entry ticket fees. the need arises to model a multibody system, which requires a considerable investment in methods for formulating and solving equations of motion. An essentially self-contained homotopy theory of filtered $$A_\infty$$ algebras and $$A_\infty$$ bimodules and applications of their obstruction-deformation theory to the Lagrangian Floer theory are presented. David Logan works in the area of applied mathematics and ecological modeling. Mathematical models, e. Yet despite its seemingly abstruse mathematics, finance theory over the tast two decades has found its way into the. Another application of mathematical modeling with calculus involves word problems that seek the largest or smallest value of a function on an interval. If that is true, then substituting x = 3. 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Mathematical Analysis, Modelling, and Applications Purpose of the PhD Course The aim of the PhD Course in Mathematical Analysis, Modelling, and Applications is to educate graduate students in the fields of mathematical analysis and mathematical modelling, and in the applications of mathematical and numerical analysis to science and technology. biomass, drying, energy balance, fuels, gases, gasification, hydrodynamics, mathematical models, temperature, wood chips Abstract: A one-dimensional steady state model for biomass–steam gasification has been developed. Consider, for example, Newtonian mechanics. MATHEMATICAL TECHNOLOGY TRANSFER - INDUSTRIAL APPLICATIONS AND EDUCATIONAL PROGRAMMES IN MATHEMATICS Matti HEILIO Lappeenranta University of Technology Box 20, 53851 Lappeenranta, Finland e-mail: matti. 1 Student Learning Expectation. APPLICATION OF MATHEMATICAL MODELING IN MANAGEMENT ACCOUNTING 575 2. In this paper, we examine some of these difficulties and discuss how technology can play a pivotal role in providing the essential support to make mathematical modelling a more accessible mathematical activity amongst students. 2 more inches melted by Wednesday morning. AMATH 383 Introduction to Continuous Mathematical Modeling (3) NW Introductory survey of applied mathematics with emphasis on modeling of physical and biological problems in terms of differential equations. Mathematical Modeling for Business Applications 3. Mathematics, Mathematical Physics-Engineering, Analysis, Non-linear Analysis, Integral transforms, Number Theory, p-adic Analysis and Applied Algebra, Special Functions, q-analysis and Discrete Mathematics, Probability and Statistics, Mathematical Physics, Mathematical modeling and their applications. Summary Report. The book also serves as a valuable reference for research scientists, mathematicians, and engineers who would like to develop further insights into essential. You now have more than three months to complete your application, as the deadline is on April 30, 2019. Hydrology Days 2010. In a nominalist reconstruction of mathematics, concrete entities will have to play the role that abstract entities play in platonistic accounts of mathematics, and concrete relations (such as the part-whole relation) have to be used to simulate mathematical relations between mathematical objects. The Shannon and Weaver mathematical model of communication (1949) consisted of five key stages; an information source produces a message, this is then encoded into signals by a transmitter, which also sends the message through a channel, the medium through which the message is sent to the receiver, which then decodes the signal and reconstructs the message to be repeated to the destination. Session 3: Deforming and Recrystallization Abstract Title Mathematical modelling of soft reduction in Abaqus software with the application of tensile testing under crystallization conditions on Gleeble-3800. Statistics can be defined as a type of mathematical analysis which involves the method of collecting and analyzing data and then summing up the data into a numerical form for a given set of factual data or real world observations. It starts with an urn filled with balls of different colors. • Cryptanalysis is the science of attacking ciphers, finding weaknesses, or even proving that a cipher is secure. You may well have studied maths, physics or engineering degrees as an undergraduate. Sign in to view your account details and order history. World-leading research in artificial intelligence, mathematical modelling, robotics, engineering healthcare and the environment. This interdisciplinary program offers a thorough grounding in mathematical methods that underlie the physical, engineering and biological sciences. Mathematical modeling of pharmacokinetics / pharmacodynamics (PKPD) is an impor-tant and growing field in drug development. It has given me a deeper understanding of wide applications of mathematical models, much needed training in programming with R and boost up my enthusiasm for my work. Research and corporate applications that use discrete mathematics. 127 journal of mathematical analysis and applications 0022-247x 1. Successful. The Shannon and Weaver mathematical model of communication (1949) consisted of five key stages; an information source produces a message, this is then encoded into signals by a transmitter, which also sends the message through a channel, the medium through which the message is sent to the receiver, which then decodes the signal and reconstructs the message to be repeated to the destination. 064 128 Dissertationes Mathematicae 0012-3862 1. Decision Support Systems), new statistical applications (e. 00 Pre-requisite: Mathematic proficiency (see beginning of Mathematics section). This is a vast collection of computational algorithms ranging from elementary functions like sum, sine, cosine, and complex arithmetic, to more sophisticated functions like matrix inverse, matrix eigenvalues, Bessel functions, and fast Fourier transforms. Mathematical modelling of waves in fluids. As an example for comparing the applications and developing theoretical foundations of logic let us see the case of arti cial intelligence (AI for short). com, 3 [email protected] There are three steps in mathematical modeling: formulation, math-ematical manipulation, and evaluation. Here, we apply this method to tubular reactors. The Mathematics Applications Consortium for Science and Industry (MACSI) is Ireland’s largest applied and industrial mathematics group and works closely with scientists and industrial companies across a wide variety of sectors including the pharmaceutical sector. Mathematical Applications and Modelling is the second in the series of the yearbooks of the Association of Mathematics Educators in Singapore. The data points and model are graphed below. terminology that the option is at the money forward, then (1. The complexity of life requires human resources that are reliable and able to use computer applications. Modeling is the process of writing a differential equation to describe a physical situation. Those applications are not within the scope of this chapter, and the immediate focus is on modeling basic and moderately complex systems that may be of primary. This includes students whose primary interest is in an application area, such as physics or economics, but wants a better understanding of the models used in their discipline. Internet Mathematics Vol. 2012 ; Vol. Application of a mathematical model to prevent in vivo amplification of antibiotic-resistant bacterial populations during therapy Nelson Jumbe, 1,2 Arnold Louie, 1 Robert Leary, 3 Weiguo Liu, 2 Mark R. The principles are over-arching or meta-principles phrased as questions about the intentions and purposes of mathematical modeling. Mathematical Modelling and Problem-solving. (b) To find the y-intercepts, let and solve for y. There is a wealth of applications of these topics to contemporary social, economic, and political issues appealing to liberal arts students. The CAAM department is a close-knit community of faculty and students working toward solving challenges through applied mathematics, with the support of a staff dedicated to the department's mission. Computer programmers use mathematical modeling to create software that forecasts events, performs analysis of data sets, and creates a simulation based on actual information. The study of mathematics as a subject in its own right may have started with Pythagoras, but people have been counting as a basic necessity of everyday life for thousands of years. Application of Computational Methods and Mathematical Models in Nuclear Medicine and Radiotherapy Call for Papers. Find materials for this course in the pages linked along the left. Mathematical modeling of chemical reaction networks consists of a variety of methods for approaching questions about the dynamical behaviour of chemical reactions arising in real world applications. We now move into one of the main applications of differential equations both in this class and in general. Mathematical models are an important component of the final "complete model" of a system which is actually a collection of conceptual, physical, mathematical, visualization, and possibly statistical sub-models. The course is taught by staff from the Centre for Mathematical Modelling of Infectious Diseases at the London School of Hygiene & Tropical Medicine, the Modelling and Economics Unit at Public Health England, London, and the University of Sao Paulo, Brazil. 9783527627615. Modelling is the process of writing a differential equation to describe a physical situation. Therefore, no valid scientific conclusions can be. However, these deterministic reaction rate equations are really a certain large-scale limit of a sequence of finer-scale probabilistic models. Can be used to study complex systems that would otherwise be difficult to investigate. About the courseThis Centre for Doctoral Training (CDT) in Industrially Focused Mathematical Modelling (InFoMM) trains cohorts of academically outstanding students in a broad range of techniques spanning mathematical modelling, analysis and computation in order to address the challenges that face modern companies. 51), there are indications, however, that it is not emphasized in current teaching practices at upper secondary school (e. Throughout history students have hated these. com, 4 [email protected] Mathematics of Financial Markets 3. Patrícia Freire Chagas2, Silvia Helena Santos2, Carla Freitas Andrade1, Vanessa Ueta2 and Raimundo Souza2 Universidade Federal do Ceará, Fortaleza, Ceará, Brazil Abstract. We are still ignorant about the certain attributes of cell and microorganism. Consider the environment you are in right now. In Sarah Wallick's class, students learned concurrently about gears and crankshafts on a tangible level, and about ratios on a mathematical level. It is commonly called the exponential model , that is, the rate of change of the population is proportional to the existing population. One real-world application of mathematics is set forth in Bill Thurston's far-sighted essay On Proof and Progress in Mathematics, that purpose being, to provide foundations for social enterprise. fi ABSTRACT Mathematical technology is a term referring to the interdisciplinary area combining applied mathematics,. One difficulty with mathematical models lies in translating the real world application into an accurate. By Nicholas McClure B. Our aim is to foster new collaborative research, on problems that arise in science, engineering and industry, in order to produce world-class publications on mathematical modelling and produce world-class research and research training in mathematical modelling. Now it is time to define models themselves. Uncoiling the Spiral: Maths and Hallucinations — Think drug-induced hallucinations, and the whirly, spirally, tunnel-vision-like patterns of psychedelic imagery immediately spring to mind. TERMINOLOGY Table 9. The Teachers College Mathematical Modeling Handbook is intended to support the implementation of the CCSSM in the high school mathematical modeling conceptual category. The ladder makes a 60° angle with the ground. Mathematical modelers face a variety of challenges, including summarizing large data sets to understand and explore a system of interest, inferring the model parameters most accurate for describing a given data set, and assessing the goodness-of-fit between data sets. The International Conference Mathematical Modeling with Applications (M2A19) Mohammed V University, Rabat Morocco, 1-4, April 2019. Additionally, we collaborate with over 20 professors at different institutions worldwide. More importantly,. Although many ecologists have found fractal geometry to be an extremely useful tool, not all concur. The book is unique as it addresses a focused theme on mathematics education. Othmer Jan 2007 Allan, Linda J. mathematics. Mathematical modeling is used extensively in economics, and it is generally agreed that the foundation of economic theory is formed on a mathematical basis. 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https://chat.stackexchange.com/transcript/36/2019/3/27 | 12:14 AM
@MatheinBoulomenos nice
12:25 AM
quick question... IF we have eight people showing up for free concert tickets and we want to figure out how many ways can exactly 3 of them get tickets, isn't that just 8 choose 3 or $\binom{8}{3}$?
yes
ok so if we wanted to find out how many subsets of size k are there from a set of size n isn't it just $\binom{n}{k}$ or am I missing something here since we need different subsets of size k ... maybe it's $\binom{n}{k_{1}}\binom{n-k_{1}}{k_{2}}$ and so forth?
12:48 AM
it is just $\binom n k$
ah got it. I've been typing a lot and staying up late these past couple of days X_X
so my thinking is like weeeeee
oh oh ... it's Demonark
3 hours later…
whooaaa
2 hours later…
5:57 AM
do you know tensor product?
In above argument, we can conclude that $K_1K_2$ is spanned by $\alpha_i\beta_j$ over F, because, closed set under addition and multiplication of $\sum\alpha_i\beta_j$ is a field, right?
@LeakyNun no!
there's no need to shout that
ok :)
@Silent yes
6:02 AM
they span K_1 K_2 because they include the generators $\alpha_i$, $\beta_j$
so the ring they generate contains $F[\alpha_i, \beta_j]$
which is $K_1 K_2$
@Silent Yeah
Well, $\sum a_{ij}\alpha_i\beta_j$
sorry
It's a field, it contains $\alpha$ and $\beta$, and anything that contains $\alpha$ and $\beta$ contains $\sum a_{ij}\alpha_i\beta_j$
Therefore it's the field generated by $\alpha$ and $\beta$
Are we implicitly assuming that $K_1,K_2$ are contained in some larger field $L$? Because otherwise we need to check closure under inversion as well, perhaps
6:06 AM
leaky, does tensor product help in field theory?
@AkivaWeinberger oh yes
If $A=\sum a\alpha\beta$ then $x\mapsto Ax$ is an $F$-linear map
and we have a generating set with $mn$ things
which means there's a basis with at most $mn$ things
so it's a finite-dimensional space
so that's a surjective map
so $1=Ax$ has a solution
so $1/A$ is in our set
thank you!!
On the other hand, I don't think that this works if $K_1$ and $K_2$ are infinite-dimensional. Like, I don't think you can write $\dfrac1{\pi+e}$ as a finite sum $\sum a_{ij}\alpha_i\beta_j$ where $a_{ij}\in\Bbb Q$, $~\alpha_i\in\Bbb Q(\pi)$, and $\beta_j\in\Bbb Q(e)$
(assuming $\pi$ and $e$ are algebraically independent, which they probably are)
I don't have a proof of this
6:37 AM
@AkivaWeinberger how do we know that it is injective? i
I think that we can say that $1/A$ exists in larger field $L$ and hence it is injective. Our motive was to show that $1/A$ is indeed in $K_1K_2$.
Oh! 1/A exists in $K_1K_2$ already, since it is field, and our motive is to show 1/A lies in set $\{\sum a_{ij}\alpha_i\beta_j}$, right?
* I meant $\{\sum a_{ij}\alpha_i\beta_j\}$ above
7:00 AM
Oh I think you're right that they need to be in a larger field actually
yeah. i just saw it here as well but do not understand why do we need larger field!
Imagine if $K_1=\Bbb Q(x)/\langle x^2-2\rangle$ and $K_2=\Bbb Q(y)/\langle y^2-2\rangle$
These are both isomorphic to $\Bbb Q(\sqrt2)$
@Silent because there can be more than one way to extend two fields
If we were embedding these into $\Bbb C$, for example, we'd have to choose $x=\sqrt2$ or $x=-\sqrt2$
and similarly for $y$
which means either $x=y$ or $x=-y$
if $K_1 = K_2 = \Bbb Q(\sqrt[3]2) \subset \Bbb C$ then $[K_1 K_2 : \Bbb Q] = 3$
but if $L_1 = \Bbb Q(\omega \sqrt[3]2)$ and $L_2 = \Bbb Q(\sqrt[3]2)$ then $[L_1 L_2 : \Bbb Q] = 6$
despite the fact that $K_1/\Bbb Q \cong L_1 / \Bbb Q$ and $K_2/\Bbb Q = L_2/\Bbb Q$
7:04 AM
If we don't know what we're embedding them into, and we just have $K_1K_2$ as $\Bbb Q(x,y)/\langle x^2-2,y^2-2\rangle$,
(with tensor product you can show that these two ways are the only ways to embed two copies of $\Bbb Q(\sqrt[3]2)/\Bbb Q$ into a larger field)
then $x+y$ has no inverse.
Because $(x+y)(x-y)=x^2-y^2=2-2=0$.
@LeakyNun OMG!
Oh no!
In this abstract version of $K_1K_2$, both $x+y$ and $x-y$ are nonzero, so that means $x+y$ is a zero divisor
whereas if they were embedded in a larger field, either $x+y$ or $x-y$ would have to be zero (we don't know which if we don't know what the embeddings are)
7:06 AM
@Silent if you have two finite separable extensions of a field, say $K_1/F$ and $K_2/F$
then by primitive element theorem, $K_1 = F(\alpha)$ and $K_2 = F(\beta)$
i.e. $K_1 = F[X]/(p(X))$ and $K_2 = F[X]/(q(X))$
then we can construct the tensor product $K_1 \otimes_F K_2$ explicitly as $F[X,Y]/(p(X),q(Y))$
so Akiva's "abstract $K_1 K_2$" is actually $K_1 \otimes_\Bbb Q K_2$
ok, i have saved it to look at it in future!
now you can prove that $K_1 \otimes_F K_2$ has dimension $mn$ over $F$ and decomposes as a product of fields
those fields are all the ways that $K_1/F$ and $K_2/F$ can be embedded into a larger extension
(they may repeat)
@AkivaWeinberger Why is $$x\to Ax injective? Was my argument correct? @Silent yes It's injective because it's in a larger field 7:16 AM corollary: in a finite extension every sub-ring-extension is a field and multiplication by something nonzero is an injective map from a field to itself oh wow:) feeling happy! because if Ax=Ay then A(x-y)=0 and multiplication by 1/A (which might not be in \{\sum a\alpha\beta\} but is in L) gives x-y=0 and thus x=y Unrelated thought In the quaternions, ij=-ji is a direct consequence of (ij)^2=-1 That is, once we have i^2=j^2=k^2=-1 and ij=k, the rest of the multiplication table is forced 7:39 AM @AkivaWeinberger why pointwise multipliciation in amplitude/frequency correspond to convolution in ??/time ? 4 hours later… 11:25 AM In Hatcher's book on Alg. Top., he calls e_{\alpha}^n a cell. I tried searching through his book for an explanation of this, but I couldn't find anything. Would someone mind explaining what e_{\alpha}^n is; i.e., what is the definition of e_{\alpha}^n? Also, what is an attaching map? 12:10 PM 0 Let A be euclidian ring and K be its field of a fraction.Let (V,B) be a nonzero IPS (inner product space) over K. A finitely generated A-submodule L ⊆ V is said to be an A -lattice in V if L contains a K-basis of V . As we have already observed, L must be A-free since it ... 12:21 PM @LeakyNun Is it? I didn't know that @user193319 On page 5 where he introduces cell complexes, he explains that e_\alpha^n is an open n-disk @AkivaWeinberger i.e. \mathcal F\{f\} \times \mathcal F\{g\} = \mathcal F\{f \ast g\} You're taking the disjoint union of a bunch of (closed) disks, and then quotienting the boundaries of those disks together under an equivalence relation defined in terms of the "attaching maps" The (open) interiors of those closed disks are your cells Do you know what it means to quotient by an equivalence relation? @LeakyNun Remind me why that is? I think I knew that at one point but forgot it well that's what I'm asking you :P By \cal F you're not talking about the series, right? You have the continuous transform I think both work Fourier is a very general phenomenon 12:28 PM Hm, what's \sin(2x)*\sin(3x)? well the theorem says 0 That'd be \int_{-\pi}^\pi\sin(2x)\sin(3(-x))dx=0, right? \int \sin(2y) \sin(3(x-y)) ~\mathrm dy Oh right What's the range of integration? Still one period (-\pi to \pi)? sure they're elements of C(S^1) you're integrating over S^1 over the normalized Haar measure, depending on who you ask @AkivaWeinberger I can tell you more about this general Fourier if you're interested :P 12:32 PM So what's \sum(a_{1n}\sin nx+b_{1n}\cos nx)*\sum(a_{2n}\sin nx+b_{2n}\cos nx)? just use the formula lol Yeah but I'm hoping it comes out to something nice \sum (a_{1n} a_{2n} \sin(nx) + b_{1n} b_{2n} \cos(nx)) Is it really? well the theorem says so I'm sure you can distribute and verify everything is (bi)linear so check on basis 12:34 PM So that would imply that \sin x*\cos x=0 but that doesn't look like it's right why not? oh it isn't right interesting maybe it would be right if we used the "right" (i.e. complex) Fourier transform I think \sin x*\sin x=-\cos x, ~\sin x*\cos x=\sin x, and \cos x*\cos x=\cos x \sin x \ast \cos x = \pi \sin x yeah let's switch to the correct basis \exp(inx) 12:37 PM OK right so what's e^{ix}*e^{-ix} And what's e^{ix}*e^{ix} Given \sin x*\sin x=-\pi\cos x, ~\sin x*\cos x=\pi\sin x, and \cos x*\cos x=\pi\cos x (\cos x+i\sin x)*(\cos x-i\sin x)=0 (\cos x+i\sin x)*(\cos x+i\sin x)=2\pi(\cos x+i\sin x) \exp(imx) \ast \exp(inx) = \displaystyle \int_0^\tau \exp(i((m-n)y + nx)) ~ \mathrm dy = \exp(inx) \int_0^\tau \exp(i(m-n)y) ~ \mathrm dy = \tau \delta_{mn} \exp(inx) qed that's why we use the correct basis also the correct thing to integrate over is \mathrm dy/\tau \exp(imx) \ast \exp(inx) = \displaystyle \int_0^\tau \exp(i((m-n)y + nx)) ~ (\mathrm dy/\tau) = \exp(inx) \int_0^\tau \exp(i(m-n)y) ~ (\mathrm dy/\tau) = \delta_{mn} \exp(inx) this is much more beautiful Arright so given \sum c_{1n}e^{inx} and \sum c_{2n}e^{inx}, we can find \sum c_{1n}c_{2n}e^{inx} by convolving them right so now the problem is... why? @AkivaWeinberger so are you interested in the general phenomenon? Well we just proved it didn't we yeah but that's just a proof is there a 3b1b-style explanation of all these 12:48 PM We need an intuitive description of convolution don't we of two functions \Bbb R\to\Bbb C well convolution is just multiplication... wait Two periodic functions to be clear I was referring to how you multiply polynomials together or more elementarily, how you multiply numbers together wait... if multiplication corresponds to convolution, then convolution must also correspond to multiplication? What, like how if p(x)=\sum a_nx^n and q(x)=\sum b_nx^n then \sum a_nb_nx^n can be found by doing p(e^{it})*q(e^{it}) and then substituting in t=\frac1i\ln x? oh that's interesting 12:51 PM Hm and then \langle p,q\rangle is p(e^{it})*q(e^{it}) evaluated at zero where I have the definition of inner product on polynomials that makes the x^n orthonormal to each other we should ask 3b1b to do a video on this :P too many coincidences Hm that doesn't actually work if they're Laurent polynomials but if they're square-summable like we require elements of L^2 to be... Hm question what is \delta_{am} \ast \delta_{bn} exactly \delta_{(a+b)n} and it corresponds to \exp(iax) \exp(ibx) = \exp(i(a+b)x) 12:55 PM Can I simplify \int_{-\pi}^\pi p(e^{it})*q(e^{it})dt well that would be (\int p(e^{it}) dt) (\int q(e^{it}) dt) Oh wait Oh sorry never mind about the Laurent polynomials I was thinking x=0, not t=0 t=0 makes x=1 and we don't have any problems @LeakyNun This'll end up being a_0b_0 what is x? @AkivaWeinberger yes p(x)=\sum a_nx^n Whatever ignore this Notice by the way that p(0) doesn't necessarily equal a_0 for example if p(x)=x+2+\frac1x or something of course it does 12:58 PM p(0) doesn't exist so you'll have to do \int_{-\pi}^\pi p(e^{it})dt to get at it Or, equivalently by a change of variables, \oint p(x)/x~dx around the origin which we also know from the residue theorem ok then \int_0^\tau p(e^{it}) \ast q(e^{it}) (\mathrm dt/\tau) must be \sum a_n b_{-n} right No it's a_0b_0 we just did this well I was wrong 1:01 PM Oh Did you mean \cdot and not convolution there I see you've embraced tauism by the way \tau @AkivaWeinberger 1:35 PM @LeakyNun I don't watch black mirror though Well now you've seen something of it 1:50 PM In the last line of above proof, how can I be sure that there is no other way of splitting that polynomial, which can be split in a subfield of K? K[x] is a unique factorization domain so there's only one way to split f into irreducible factors wow thanks (as is E[x]) 2:18 PM @Semiclassical I don't know about you, but Poincare-Bendixson theorem is actually not that bad 2:28 PM Is a map a distorted version of the sphere or is the sphere a distorted version of the map 2:48 PM Is there a closed form for the number of n\times n matrices with integer coefficients from \{-m,\cdots,m\} whose determinant is \pm a\in\mathbb{Z}? No idea but that sounds like a very interesting question Hello? ♫ Hello ♫ any one who can tell me why C_c(U) is a separable space? where U is a open set in R^n What's C_c mean? 2:56 PM @LarryEppes bump functions I know that C[a,b] is separable space. @AkivaWeinberger compact support yes @LarryEppes or just draw a triangle _/\_ continuous functions are plenty Why would that give you a countable dense subspace @LeakyNun 2:58 PM a triangle? but could this countable dense subset in C_c(U)? oh... sorry I misread the question What's the difference between C_c and C C_c is all the function that have a compact support @LarryEppes because in general any subspace of a separable metric space is separable yes, the separable points is in the larger space that's right 3:00 PM 15 The problem statement, all variables and given/known data: Show that if X is a subset of M and (M,d) is separable, then (X,d) is separable. [This may be a little bit trickier than it looks - E may be a countable dense subset of M with X\cap E = \varnothing.] Definitions Per our boo... This is true for metric spaces but not topological spaces in general I think but C(U) is a metric space but if the dense subset still in the C_c(U)? @LarryEppes just look at my link yes, I will read a minute, txh 3:03 PM You could probably find a countable set of piecewise linear stuff Piecewise linear stuff whose graphs are like polyhedra with vertices at rational coordinates Something like that @AkivaWeinberger This is very false for arbitrary topological spaces, for every space X you can put a topology on X\cup\{p\} such that \{p\} is dense, if you start with a nonseparable X then you have a counterexample Oh lol What if we want it to be Hausdorff Sorgenfrey plane Oh neat Is there a name for a separable space all of whose subspaces are separable? "Completely separable" or something like that? hereditarily separable In set theoretic topology there was (there is?) a lot of interest in S-spaces, which are regular and hereditarily separable but not Lindelöf spaces and L-spaces, which are regular and hereditarily Lindelöf but not separable spaces What is the weakest P such that P+separable\implieshereditarily separable? P=metric space works but I wonder if it can be weakened 3:13 PM What's Lindelöf again? Every (open) cover has countable subcover. Every open cover has a countable subcover, it's a weaker version of compactness Oh interesting that's way too many adjectives Is \Bbb R^n Lindelöf? I'm guessing yes but I don't see why 3:14 PM yes because it's second countable @LeakyNun "Oh interesting" is one adjective Isn't there also a concept of \aleph_n-Lindelöf too? Oh right yeah there's another countable I seem to recall something about that. @Rithaniel One would assume, it seems easy enough to define 3:16 PM Indeed, but is it "useful" enough to warrant studying? @Rithaniel Probably by logicians or some such species and set theorists (there's a lot of very cool point set topology in that blog) True enough What do logicians actually work with, generally? logic, duh 3:19 PM I understand the post! thanks so much! @AlessandroCodenotti I'm not gonna read that whole thing, but I didn't expect the existence of S-spaces to be independent of ZFC I forgot too much about topology @LarryEppes nice Yeah, but \aleph_n-Lindelöf being useful to Logicians seems to imply something is going on there which I'm not aware of. @AkivaWeinberger Yeah apparently people expected S-spaces and L-spaces to be very symmetric but now it doesn't seem to be the case I think that "least \kappa such that every open cover of X has a subcover of cardinality \kappa" should even have a name. There are a lot of similar cardinal functions with names 3:21 PM @Rithaniel I wouldn't know enough to comment Ah, gotcha Ah, here it is! It is called the Lindelöf degree of X, denote with L(X) So model theory sounds like the sort of thing that logicians would study but it also sounds like the sort of thing that set theorists would study so I'm not entirely sure where the distinction is Maybe it's more of a sliding scale And mathematical computer science can blur into logic as well @Rithaniel Found one special case of that on OEIS without a formula: oeis.org/A057981 or at least that's what I know from reading Turing, I don't know anything that's more current than then 30s 3:24 PM I'm actually getting ready to apply to grad school for the spring of 2020. I'm trying to collect as much data on different fields of math as I can as I think about specializations. @Semiclassical: Ooooo Thinking something that behaves like a halting problem: Let S be a sequence of n numbers. The assigned task is to predict the n+1 th number also this: oeis.org/… The extension of S behaves in a way such that given any fixed polynomial P that fit the n numbers, the n+1 th number is always different from P(n+1) Here's an idea unimodular means determinant has magnitude 1 3:26 PM Take your favorite machine learning model for sequence prediction "Long short-term memory recurrent neural networks" or what have you Let a_n be a sequence of bits, defined like this a_0=1 Model theory is about studying models of theories (no really), a lot of set theory is about studying models of ZF(C), a particular first-order theory, but there's also a lot more being done in set theory a_n, n>0, is: train your model on the sequence (a_0,a_1,\dots,a_{n-1}) and let it make a prediction for a_n a_n is defined to be the opposite of its prediction @Rithaniel doesn't look like OEIS has anything for the case where the determinant has magnitude 2 Ah yeah, 40 is the value I got for 2x2 matrices with elements in \{-1,0,1\} and deteriminant -1 or 1. So basically a is defined so the machine learning algorithm is always wrong 3:28 PM I should have checked OEIS first, myself. the case where the elements are all {0,1} may be more studied What I want to know is, would a look pseudorandom? Or would it have some sort of pattern that the network would fail to pick up on I guess it would depend on the exact model we use Yeah, the 0,1 case is similar to matrices over \mathbb{Z}/2\mathbb{Z} Well every neural network T has some kind of deterministic rules in it, so a can be a formula that can never be captured by said neural network a is uniquely determined once we decide on a network (unless the network has some sort of random initialization to it, in which case we have to specify the network as well as the random "seed") 3:30 PM Well, I think we can have a neural network that given 1,2,3,4,5 will never produce the outcome 6, where a can be 1,2,3,4,5,6 I was thinking of a sequence of bits so 0s and 1s so we can say, "Whatever you predict the next bit is, it's the other one" Heh, the two conversations converge on binary :P 2 By the way, I once looked up how LSTM neural networks work Didn't understand it at all You give it something analogous to "short-term memory" and then… I dunno, wire the pieces up in a way that looks like a rocket engine diagram neutral networks are kinda magic in most of the case, because there are simply too many coupled systems that does stuff on the input and parameters (Though the convergence is more like a pair of skew lines: their projections intersect but the lines themselves don’t ) 3:33 PM By "rocket engine diagram" I mean something this Basically: complicated Ok, it is MUCH worse than that Googling LSTM gives me this Heh, that's funny. "I found this: Image not found" Dunno why the link wouldn't work ok, that's indeed look like an engineering diagram or sort Also regarding this: 8 mins ago, by Akiva Weinberger What I want to know is, would a look pseudorandom? Or would it have some sort of pattern that the network would fail to pick up on I think it will be pseudorandom, cause say the network predicts (given bits),1,0,0,1,0,1,... then a has to be (given bits),0,1,1,0,1,0,.., so there is a perfect anti correlation to the prediction dependent on the seed and settings of the network (because that controls the prediction) Sometimes I am wondering: If our computers are analogue and does not suffer from the problems of analogue computers in history, will circuit diagrams for e.g. addition have to be that complicated Didn't Fermat make an analogue calculator or someone similar Well actually it was "digital" in that it worked with digits but it had a gear with ten things on it to represent the digits 0 to 1 (Teeth?) I remember learning about another version of that that was actually sold commercially around WWII By the way What the engine actually looks like (The nozzle isn't pointing down, which I can only assume means they will not go to space today) Probably they are test firing it or something @Semiclassical Accurate @Secret Yeah I think that's what it was Though subtraction is basicaly just addition with an extra step 'cause (2^n-1)-b is easy to find (flip all the bits of b) Call that \bar b Then you can just do a+\bar b to find a-b well actually a+\bar b+1 and throw away the "carry" bit (the 2^ns place) 3 hours later… 6:55 PM I’ve got two statements in probability theory. I’m pretty well-convinced that the first, at least, is false. But I don’t know about the second. Let X_k, for k=1 to n, be a set of random variables. First statement: if P(X_k=x)=P(X_k=-x) for all k, then$$P(X_1=x_1,\cdots,X_n=x_n)=P(X_1=-x_1,\ldots,X_n=-x_n).$$Second statement: if P(X_j=x,X_k=y)=P(X_j=-x,X_k=-y) for all j,k then$$P(X_1=x_1,\cdots,X_n=x_n)=P(X_1=-x_1,\ldots,X_n=-x_n).
The premise of the second implies the premise of the first (take $j=k$) so the second statement would follow from the first.
But right now I don’t think the first is true, and so don’t have much opinion at all on the second
(I think the first statement is true if the only possible outcomes are $\pm 1$ but that’s not generic.)
Problem: For $n \in \Bbb{N}$, let $X_n(\Bbb{Z})$ be the simplicial complex whose vertex set is $\Bbb{Z}$ and such that vertices $v_0,...,v_k$ span a $k$-simplex if and only if $|v_i-v_j| \le n$ for all $i,j$. Using induction, show that $X_n(\Bbb{Z})$ is contractible by showing that it deformation retracts onto $X_{n-1}(\Bbb{Z})$.
$X_1(\Bbb{Z})$ is contractible because it is a connected graph.
7:13 PM
I typed all of the above on mobile. Didn’t expect it to be as long as a question on the main site, lol, but maybe I should move it to there
2 hours later…
9:55 PM
@AkivaWeinberger wow that's interesting
10:08 PM
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Consider the vector field restricted to the rationals, $\vec F_\Bbb {Q^2}=(x,y).$ This is a vector field $\vec F:\Bbb Q^2\to \Bbb Q^2.$ Is this vector field weak with respect to the same vector field over the reals: $\vec F_\Bbb {R^2}=(x,y),$ $\vec F: \Bbb R^2\to \Bbb R^2?$ Edit: A weak vector f...
1 more re-open vote needed
10:18 PM
It's really unclear what you are asking
what is a weak vector field w.r.t another vector field?
A weak vector field has less information than a smooth vector field
the one in $\Bbb R^2$ is
How does that video relate?
I mean, the point seems to be that you’re talking about vector fields on (for instance) integer pairs rather than the entirety of R^2
Or on the rationals, for another
Yes I'm talking about a vector field on the rationals
10:47 PM
-1
Consider the vector field restricted to the rationals, $\vec F_\Bbb {Q^2}=(x,y).$ This is a vector field $\vec F:\Bbb Q^2\to \Bbb Q^2.$ Is this vector field weak with respect to the same vector field over the reals: $\vec F_\Bbb {R^2}=(x,y),$ $\vec F: \Bbb R^2\to \Bbb R^2?$ Edit: A weak vector f...
11:13 PM
@AkivaWeinberger IIRC GRUs and MGUs are simpler than LSTMs and show comparable performance. | 2020-07-16T01:25:13 | {
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http://math.stackexchange.com/questions/149872/how-to-show-sinxiy2-sin2x-sinh2y/149880 | # How to show $|\sin(x+iy)|^2=\sin^2x+\sinh^2y$
How would I show that $|\sin(x+iy)|^2=\sin^2x+\sinh^2y$?
Im not sure how to begin, does it involve using $\sinh z=\frac{e^{z}-e^{-z}}{2}$ and $\sin z=\frac{e^{iz}-e^{-iz}}{2i}$?
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Yes, it does. Also, that $|u+iv|^2=u^2+v^2$. – Cameron Buie May 26 '12 at 0:10
As a general rule, if you have an idea, try it. Only for the simplest of problems (and problems like those you have previously solved) will you know the right thing to do before you do it. If your idea works and you solve the problem, great! If it doesn't work, then when you come here, you'll be able to get much deeper advice if you show what you've tried and why you think it doesn't work. – Hurkyl May 26 '12 at 0:24
Thanks, I did try some calculations using pen and paper, but I didnt really get anywhere with it and I was not sure if I was on the right direction, so hence I didnt type out what I've written out. Seeing what Javier Badia just answered below, I did actually use the sum formula for $\sin(a+b)$, but I just didnt realise to could convert it as he did below. – Derrick May 26 '12 at 0:27
@Derrick: Even if you aren't sure you're on the right track, it's a good idea to show what you've done so far. People are going to be more inclined to give you detailed and complete answers if they can see that you've made an effort. In this case, you were on the right track. You had all the relevant pieces, and it was just a matter of figuring out how they were strung together. That's very reassuring to see. – Cameron Buie May 26 '12 at 0:38
@CameronBuie and @ Hurkyl , ok noted, will try to make my questions more constructive in the future. – Derrick May 26 '12 at 0:48
If you use the sine addition formula, the pythagorean identities, and the fact that $\sin(ix)=i\sinh (x)$ and $\cos(ix) = \cosh(x)$, then you get this:
\begin{align} \sin(x+iy) &= \sin x \cos (iy)+\cos x \sin(iy) \\ &= \sin x \cosh y + i \cos x \sinh y \end{align}
\begin{align} |\sin(x+iy)|^2 &= (\sin x \cosh y)^2 + (\cos x \sinh y)^2 \end{align}
Now you can get rid of the cosines knowing that $\cos^2 x + \sin^2 x = 1$ and that $\cosh^2 x - \sinh ^2 x = 1$. You can take it from there.
By the way, to get the sine addition formula and the sine and cosine of imaginary numbers, convert them to exponential form:
$$\sin x = \frac{e^{ix}-e^{-ix}}{2i}$$ $$\cos x = \frac{e^{ix}+e^{-ix}}{2}$$ $$\sinh x = \frac{e^x-e^{-x}}{2}$$ $$\cosh x = \frac{e^x+e^{-x}}{2}$$
Plug in what you want to find out; the derivation of the identities is straightforward.
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Thanks @JavierBadia , I just have one question, Im not sure how you would expand out $(\sin x \cosh y)^2 + (\cos x \sinh y)^2$? What would the next line be? I get $\sin^2x\cosh^2y +\cos^2x\sinh^2y$? – Derrick May 26 '12 at 0:43
@Derrick: the idea is that we want to be left only with the circular and hyperbolic sines, so we use the identities I mentioned to replace the cosines with sines. I'm on my iPod so I can't really type out all the TeX, but that is the gist of it. – Javier May 26 '12 at 0:56
Okay thanks, most apreciated. Its ok, I think I got it from here. – Derrick May 26 '12 at 0:58
$$z=x+iy\Longrightarrow \sin z=\frac{e^{iz}-e^{-iz}}{2i}=\frac{e^{-y+ix}-e^{y-ix}}{2i}=$$$$=\frac{e^{-y}(\cos x+i\sin x)-e^y(\cos x-i\sin x)}{2i}=\frac{1}{2i}\left[i\sin x\left(e^y+e^{-y}\right)-\cos x\left(e^y-e^{-y}\right)\right]=$$$$=\sin x\cosh y+i\cos x\sinh y\Longrightarrow ...$$
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\begin{align} \sin(z)^2 &= (\sin x \cos (iy))^2 +(\cos x \sin(iy))^2 \\ &=\sin^2x \cosh^2y+\cos^2x \sinh^2y \\ &= \sin^2x (1+\sinh^2y )+(1-\sin^2x ) \sinh^2y \\ &=\sin^2 x+\sin^2 x \sinh^2y+ \sinh^2y-\sin^2x \sinh^2y \\ &=\sin^2x+\sinh^2y \end{align}
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http://topvolleylamezia.it/ydit/representing-trigonometric-functions-ferris-wheel-answers.html | 3 feet — wading depth. (Periodic functions are more formally defined in Section 7. Determine the linear velocity in feet per second of a person riding the Ferris Wheel. Building on the previous lesson in this series on transformations, learners use trigonometric functions to model wheels of different heights and diameters. Find the values of the trigonometric. Giving students many days to make this connection enables them to understand it much more deeply--and to start asking questions about how to find more coordinates. The London Eye Ferris Wheel measures 450 feet in diameter, and turns continuously, completing a single. (a) During the first 32 seconds of the ride, when will a person on a Ferris wheel be 53. We need to make several considerations when the equation involves trigonometric functions other than sine and cosine. The highest point of the wheel must be 100 feet above ground. Assume that Jacob and Emily's height above the ground is a sinusoidal function of time , where = represents the lowest point on the wheel and is measured in seconds. As you ride the Ferris wheel, your distance from the ground varies sinusoidally with time. 842 you will use trigonometric functions to model a person’s height above the ground while riding a Ferris wheel. Paddle Wheels The motion Of a point on the Outer edge Of a riverboat's paddle wheel blade is modeled by h(t) 8sinE(t. The wheel has a radius of 50 metres and rotates clockwise at a rate of one revolution every 30 minutes. Using the axes below, sketch a graph to show how the height of a passenger will vary with time. Goals & Objectives. Start studying Unit 7 & 8 Trigonometric Functions Vocabulary. 1 Angles Recall the following definitions from elementary geometry:. The purpose is to use. Using the information from above,. and c representing the lengths of the sides opposite. I now want you to try to match the correct wheel description to the graphs and functions on the table. D = Midline. A Ferris wheel has a radius of 25 feet. The beauty of symmetry makes expressive art in this math project. trigonometric ratio. 7, the period is indicated by the horizontal gap between the first two peaks. For more trigonometry word problems, sign up for the Trigonometry: Trigonometric Functions II course. GIFT 4 Trigonometric Functions Application "FERRIS WHEEL COMPARISON" GIFT 4 Power Point Answer key. Label the period, the amplitude, and themidline. ¶ The task also asks students to trace the path of a car on a ferris wheel, precisely, point by point, for a given domain. To solve Ferris wheel problems, you'll make use of the standard trigonometric function, the basic trigonometric equation to work with for periodic functions, functions that repeat forever. functions to model real world problems A real life example of the sine function could be a ferris wheel. Delbert is at the lowest position of the Ferris wheel, 1 meter above ground, when $$t=0$$ seconds. A ferris wheel is 60 meters in diameter and is boarded from a platform that is 4 meters above theground. 1 - Introduction to Periodic Functions. revs/ GO s (os b) Write the cosine equation to represent this function. Find the model that gives your height above the ground at time t (t=0 when you entered). Trig Pie Chart. Your equation is therefore: y = -20 cos (π t/4) + 23. Find the values of the six trigonometric functions of∠X for XYZ at right. You want the cars to look like they are hanging down within the frame (see OP's picture) not centered on the frame. It is a huge structure which is 120m in diameter. Purpose: This is a multi-day discovery activity that creates a trig foldable. Your task in this activity is to generalize that work for the case of the first quadrant. Initial Side The starting position if a ray when forming an angle. The multiplier of 4. y sin 3 ()x 2 5 B. 7 s) c) The midline is at 15 m, and the radius is. This is (usually) slow enough to allow people on. 5 m above the ground. People board the. At Certain Points in Time. by Michelle Zhang. To answer th e Ferris wheel problem at the beginning of the section, we need to be able to express our sine and cosine functions at inputs of time. 6 Trigonometric relationships 10. Find the exact values of the six trigonometric functions of θ. How would the graph change if the Ferris wheel rotated faster? if the Ferris wheel had a smaller diameter? Learn More About It In Example 5 on p. The wheel completes 1 full revolution in 6 minutes. • A picture/drawing of the ferris wheel • An equation that represents the rider's height • A neatly labeled graph representing the function • An explanation for how you obtained the equation for the rider's height at time t in laymen's terms (Do not assume that your audience knows anything about trigonometric functions! e. If we can find a. Trigonometric Functions Ferris Wheel: The position of each car on a Ferris Wheel, 200 feet in diameter, can be given in terms of its position on a Cartesian plane. Inverse Trigonometric Functions: Trigonometric functions can be useful models for many real life phenomena. Source: National Climatic Data Center. Modeling with General Trigonometric Functions real-world phenomena, such as circular motion and wave motion, involve repeating patterns that are described by trigonometric functions. 007W2/3, where B and W are measured in grams. This should be. by Sofia Wollheim-Martinez. May 912:27 PM. Users can pivot the Ferris wheel to emphasize the change in height of a seat, show the sine and cosine functions, turn on coordinates, degrees, and/or radians associated with the eight benchmark. This Ferris Wheels—Using Trigonometric Functions to Model Cyclical Behavior Lesson Plan is suitable for 10th - 12th Grade. 518) Tuning Fork. The center of the Ferris Wheel is minimum + A = 3+20 = 23 ft = D. page 346/40-46 even, 52-64 even. The wheel rotates at a rate of 2 revolutions every 6 minutes. What is the amplitude of f and what does this value represent in this context? b. 10 Trigonometric functions 2 10. The author examined secondary students' work on a problem involving modeling the periodic motion of a Ferris wheel through the use of a visual programming environment. Its centre is 43 m above the ground. Ruby has a pulse rate of 73 beats per minute and a. Due: 2/17/2011 Last Modified: 2/11/2011 1:00 PM. sec 1350 —v'î 77 28. The following are word problems that use periodic trigonometry functions to model behavior. 272 Chapter Seven TRIGONOMETRY IN CIRCLES AND TRIANGLES 7. Solving Equations Involving a Single Trigonometric Function. Passengers get on at the lowest point which is 6 m above the ground. Page 1 of 2 14. The simplest circle is a unit circle, that is, a circle of radius 1 unit, and it is this circle we often use with the trig functions. Don't forget: you still need to create a graph and find the function for the height of Car 1. 494) Sundial (p. Chapter 13: Trigonometry Unit 1 Lesson 2: Coterminal Angles Standard Position: An angle is in standard position if its vertex is located at the _____ and one ray is on the _____ x-axis. Prove and apply trigonometric identities. a) Graph how a person’s height varies with time through the first two cycles b) Write an equation that expresses height h as a function of time t c) Determine one’s height after 45 s. Which function models this situation? 11. Find the angle 1340 in radians. 5 - Ferris Wheel For the Ferns wheel described in Ch. 2 High Tide - A Solidify Understanding Task Using trigonometric graphs and inverse trigonometric functions to model periodic behavior (F. Interpret the constants a, b, c in the formula h = a + b cos ct in terms of the physical situation, where h is the height of the person above the ground and t is the elapsed time. 6 Trigonometric relationships 10. Show Step-by-step Solutions. The diagram for this Ferris wheel shows that the height of the main axle to the ground is 22 feet, 3 and 3/16 inches. Solving Equations Involving a Single Trigonometric Function. Relating Angles and Their Functions. Graphing Design project. This Representing Trigonometric Functions Lesson Plan is suitable for 9th - 12th Grade. You enter from a platform at the 3oclock position. IB Math – SL: Trig Practice Problems Alei - Desert Academy C:\Users\Bob\Documents\Dropbox\Desert\SL\3Trig\TestsQuizzesPractice\SLTrigPractice. Saying “it takes 40 seconds to complete one revolution” isn’t the same as seeing a ferris wheel travel at that speed. The frequency is defined to be the reciprocal of the period. Since the sine function takes an input of an angle, we will look for a function that takes time as an input and outputs an angle. The maximum height of the Ferris wheel is 17 m and the minimum height is 1 m. C = Phase shift. What is the diameter of the wheel? A. SOLUTION: Since the Ferris wheel completes a rotation once every 40 minutes, the values of the height function will repeat every 40 40minutes so the period of is minutes. For functions of the form y = a tan b, the amplitude is not defined, and the period. Thank you for your comment Dan. The diameter of the wheel is 10 meters, Get a free answer to a quick problem. 387 #1abceh, 2abdeg, 3ad, 5abc, 6ab. If anyone could solve this for me and provide an explanation it would be greatly appreciated. The six o'clock position on the ferris wheel is level with the loading platform. Develop and use the Pythagorean identity. The temperature of a swimming pool is cyclic and modelled by a trigonometric function. Page 1 of 2 14. Answer and Explanation: {eq}h(t) = Asin(Bx+C)+D {/eq} Here, A = Amplitude. Question: FERRIS' WHEEL The First Ferris Wheel Was Built In Chicago, Illinois By George Ferris, Jr. Determining a Rider's Height on a Ferris Wheel. Even in projectile motion you have a lot of application of trigonometry. c) Determine d when t = 11, answer accurate to one decimal place. x is the domain and y is the result area or range. 373 SELECTED APPLICATIONS Trigonometric equations and identities have many real-life applications. Provide lesson plans, worksheets, ExamView test banks, links to helpful math websites for high school math courses. Model a periodic situation, the height of a person on a Ferris wheel, using trigonometric functions. 3) Graphing Trig. It also represents a particular sine function: y = 25sin(θ). Find the equation that gives you your height when you entered the ferris wheel above the ground at t time. Substitute t=4. 2 High Tide - A Solidify Understanding Task Using trigonometric graphs and inverse trigonometric functions to model periodic behavior (F. Average monthly temperatures are periodic in nature and can be modeled by sine and/or cosine functions. Grade 11 trigonometry problems and questions with answers and solutions are presented. Unit 7 - Trigonometry, Trigonometric Equations and Identities 48. They have noticed that when they use their formula h(t) = 30 + 25sin(6) their calculator gives them correct answers for the height even when the angle of rotation is greater than 900. You enter from a platform at the 3oclock position. C = Phase shift. Have students split up into groups, and set them to work on the following exercises. Applications of Trigonometric Functions Introduction: Victoria rode on a Ferris wheel at Cluney Amusements. 7 Modelling Trigonometric Functions Recall: Determine the amplitude, period, average y-value, and phase shift of the function 3 2sin 2 4 y π =+#$&’θ − (). Solving Equations Involving a Single Trigonometric Function. 244to 247 in Text The "London Eye" is the world's largest ferris wheel which measures 450 feet in diameter, and carries up to 800 passengers in 32 capsules. For example, h(8) 13 means that Avery is 13 meters above the ground after 8 seconds of riding. The function below models the average monthly temperatures for Asheville, NC. PART A) We have that the diameter of the Ferris wheel is 25 meters and the. B = Period. Modeling a Ferris Wheel with trigonometric functions is fun and engaging in this design project. 0 mathematicsvisionproject. 6 Modeling with Trigonometric Functions 9. A ferris wheel has a diameter of 180m and the center of the wheel is 115m above ground. Write tanu as the ratio of two other trigonometric functions. these functions in novel contexts in their future mathematics, science, and engineering classes. In 8 seconds the point P will be at the wheel's lowest point. 4: Ferris wheel has a radius of 18 metres and a centre C which is 20 m above the ground. • The period of a trigonometric function is the horizontal length of one complete cycle. ppt; Due: 2/22/2011 Last Modified: 2/11/2011 1:00 PM. Don't forget: you still need to create a graph and find the function for the height of Car 1. A ferris wheel has a diameter of 180m and the center of the wheel is 115m above ground. Have class members going in circles as they model the path of a Ferris Wheel using trigonometric functions. EVALUATING FUNCTIONS Evaluate the function without using a calculator. Passengers get on at a point 1 m above the ground. Grade 11 trigonometry problems and questions with answers and solutions are presented. The centre of the Ferris wheel is 11 m off of the ground. This Representing Trigonometric Functions Lesson Plan is suitable for 9th - 12th Grade. The wheel has a meter diameter, and turns at three revolutions per minute, with its lowest point one meter above the ground. 3 Use Trigonometry to solve problems. Solving Equations Involving a Single Trigonometric Function. with sketching graphs of the height and co-height functions of the Ferris wheel as previously done in Lessons 1 and 2 of this module. ppt; Due: 2/22/2011 Last Modified: 2/11/2011 1:00 PM. 5 revolutions per minute Ashley's height above the ground, h, after t minutes can be given be modelled by the equation h = 21 — 20 cos. and c representing the lengths of the sides opposite. In particular students will: Model a periodic situation, the hight of a person on a Ferris wheel using trigonometric functions Interpret the constants a, b, c in the formula h = a + b cos ct in terms of the physical situation where h is the height of the person above the ground and t the elapsed time The. locations around a Ferris wheel. The bottom of the wheel is 1. Ferris Wheel Unit Circle: Create a graph of the height of a seat on a Ferris wheel to explore the sine function and characteristics of the unit circle. Assume that a rider enters a car from a platform that is located 30 degrees around the rim before the car reaches its lowest point. Write a scenario that goes along with the following. Unit 10 Corrective Assignment - Graphing Trig Functions Pre‐Calculus For 1‐3, write a SINE function for each graph. You enter from a platform at the 3 o'clock position. When we are given equations that involve only one of the six trigonometric functions, their solutions involve using algebraic techniques and the unit circle (see Figure 2). Can serve as a good group activity, extension, or bonus assignment. Periodic Functions: Period, Midline, and Amplitude. D = Midline. A ferris wheel is 50 feet in diameter, with the center 60 feet above the ground. Initial Side The starting position if a ray when forming an angle. 272 Chapter Seven TRIGONOMETRY IN CIRCLES AND TRIANGLES 7. To answer the Ferris wheel problem at the beginning of the section, we need to be able to express our sine and cosine functions at inputs of time. Average monthly temperatures are periodic in nature and can be modeled by sine and/or cosine functions. About 104. The given function models a person's height above the ground (in feet) as a function of the number of minutes he/she has been on the Eye. ) Convert the degree measure to radians or the radian measure to degrees. PART A) We have that the diameter of the Ferris wheel is 25 meters and the. Math courses include algebra, geometry, algebra 2, precalculus, and calculus. In Topic B, students make sense of periodic phenomena as they model them with trigonometric functions. Suppose a Ferris wheel with a radius of 20 feet makes a complete revolution in 10 seconds. B = Period. 824 Chapter 14 Trigonometric Graphs and Identities [0, 720] scl: 45 by [ 2. You enter from a platform at the 3oclock position. In 8 seconds the point P will be at the wheel's lowest point. As A Landmark For The World's Columbian Exposition And Had A Height Of 80. Write the trigonometric equation for the function with a period of 5, a low point of - 3 at x=1 and an amplitude of 7. Since the starting height is at 0, think of the bottom of the wheel as touching the point (0,0). 3 Transformations of sine and cosine graphs 10. Develop and use the Pythagorean identity. 8 feet above and below the average amount on this particular day. Then use your equation to answer the follow up question(s). This allows them to go beyond right triangles, to where the angles can have any measure, even beyond 360°, and can be both positive and negative. Prove the Pythagorean identity sin 2 (θ) + cos 2 (θ) = 1 and use it to find sin (θ), cos (θ), or tan (θ) given sin (θ),. The typical geometric definition of trigonometric functions using the right triangles is not general enough, while the above definitions work for all angles and, in case of acute angles in the right triangles, are identical to geometric definition. B = Period. 1) A ferris wheel is 4 feet off the ground. Example 66. 3 Trigonometric Functions of Any Angle 9. When we are given equations that involve only one of the six trigonometric functions, their solutions involve using algebraic techniques and the unit circle (see ). Special Cases Review of Evaluate Trig Function; 2/15/2007 Warmup on Inverse Trig; Quiz 1 Study Guide; HOT Sheet 1; Unit Circle; Quiz 1, Page 2, Page 3, Page 4; Section 13. Page 1 of 2 14. With the equation, the height is determined and the times are determined when a person is at a specific height. trigonometric functions. A sinusoidal function has an amplitude of 4, a period of 2π and passes through the point (0,2). Get Answer to (Vehicle suspension [96]) Active and passive damping are used in cars to give a smooth ride on a bumpy road. Solving Equations Involving a Single Trigonometric Function. ground of a seat on a Ferris wheel. Introduction to Right Angle Trigonometry Applications - YouTube. (—4200) — 117 31. Write a sine function modeling the buoy's vertical position at any time t. Then, we will analyze any patterns that occur. 0 mathematicsvisionproject. Chapter 5- Trig Functions Lesson Package MCR3U Chapter 4 Outline Represent the function with an equation in two different ways. The centre axle of the Ferris wheel is 40 meters from the ground. A Ferris wheel 50 feet in diameter makes one revolution every 40 seconds. In Exercise 4, students consider the motion of the Ferris wheel as a function of time, not of rotation. TRIGONOMETRIC FUNCTIONS, EQUATIONS & IDENTITIES – 7. Find a formula involving cosine for the function whose graph is shown. Cecily Curtis Graphing Project CATRA. 388: #12a - d (assume t = 0 is low tide), #14, #16 & Worksheet: Trig Graphing Applications. ) U6D7_S_Trigonometric Applications. Which function below best describes this graph? A. A ferris wheel has a diameter of 180m and the center of the wheel is 115m above ground. GIFT 4 Trigonometric Functions Application "FERRIS WHEEL COMPARISON" GIFT 4 Power Point Answer key. Ruby has a pulse rate of 73 beats per minute and a. Find the model that gives your height above the ground at time t (t=0 when you entered). A ferris wheel has a radius of 26 ft and makes one revolution counterclockwise every 12 sec. The diameter of the wheel is 10 meters, Get a free answer to a quick problem. The author examined secondary students' work on a problem involving modeling the periodic motion of a Ferris wheel through the use of a visual programming environment. Find the linear speed, in feet per minute, of a seat on this Ferris wheel. The center axle of the Ferris wheel is 40 meters from the ground. The centre of the Ferris wheel is 11 m off of the ground. The six trigonometric functions are abbreviated as sin A,cos A,tan A, csc A, sec A, and cot A. Write a trigonometric model that gives Tas a function of t. The Round Robin 3571 ACTIVITY 35 continued Periodic functions can be modeled by functions other than trigonometric. The London Eye is a large Ferris wheel that is a famous London landmark. Let us assume that O be the centre of the Ferris wheel and B be the lowest point on the circumference of the Ferris wheel and A be the position of the rider seat which is h m from the centre of the wheel and θ be the angle to the rider seat from the horizontal as shown in the Figure 1 given below. The highest point of the wheel must be 100 feet above ground. This course offers over twenty lectures that include word problems to calculate functions of angles, and other simple applications of trigonometry such as pendulum, wind turbine, helicopter and ferris wheel word problems. What I Did. 5 Choose trigonometric functions to model periodic phenomena with specified amplitude, frequency, and midline. GIFT 4 Ferris Wheel Comparison. TRIGONOMETRIC FUNCTIONS, EQUATIONS & IDENTITIES – 7. Since the sine function takes an input of an angle, we will look for a function that takes time as an input and outputs an angle. Grade 11 trigonometry problems and questions with answers and solutions are presented. Trigonometry in physics: In physics, trigonometry is used to find the components of vectors, model the mechanics of waves (both physical and electromagnetic) and oscillations, sum the strength of fields, and use dot and cross products. Building on the previous lesson in this series on transformations, learners use trigonometric functions to model wheels of different heights and diameters. We look at the slide with the diagram of the Ferris wheel (Math Practice 4). Oct 23, 2013 - A0701_001009 - Ferris Wheel: Trigonometric Functions. 7: Graphing Trigonometric Functions 2a www. If anyone could solve this for me and provide an explanation it would be greatly appreciated. • The frequency of a trigonometric function is the number of cycles the function completes in a given interval. This is a nice simple example of how the Tracker software can be used to demonstrate the circular motion of a Ferris wheel. describes Avery’s distance from the ground (in meters) after t seconds of riding. a) Sketch a graph of the function b) Determine an equation for height in meters as a function of time in seconds. Capitol Dome Real. Trigonometric Functions. ___ (A) 7 15 y 6cos x S (B) 7 15 y 6cos x S (C) 7 15 cos 6 1 y x S (D) 7 15 cos 6 1 y x S 12. Passengers get on at a point 1 m above the ground. the trigonometric function that is equal to the ratio of the side adjacent to an acute angle (in a right-angled triangle) to the hypotenuse. 1 INTRODUCTION TO PERIODIC FUNCTIONS The London Eye Ferris Wheel To celebratethe millennium,British Airwaysfundedconstructionofthe "LondonEye,"at that time the world's largest Ferris wheel. Cos it's Fun! is a Ferris wheel that also has a diameter of 6 metres and the centre of the wheel is 4 metres above the ground. He boards the ride 2 m above the ground. by Henry Wilson. 0 mathematicsvisionproject. The object origin for the cars needs to be slightly lower than the object origin of the frame of the ferris wheel. Questions 1-10 are about a Ferris Wheel problem. A carnival Ferris wheel with a radius of 7 m makes one complete revolution every 16 seconds. In an amusement park, there is a small Ferris wheel, called a kiddle wheel, for toddlers. Modeling a Ferris Wheel with trigonometric functions is fun and engaging in this design project. The graph models Victoria’s height above the ground in metres in relation to time in seconds. As A Landmark For The World's Columbian Exposition And Had A Height Of 80. The London Eye Ferris Wheel measures 450 feet in diameter, and turns continuously, completing a single. Day 9: Unit Review Chapter 6: Sinusoidal Functions Page 1 of 4 Unit 6 Review Trigonometric Functions 1. I did some reflecting on why the Ferris wheel problem seems to be so ubiquitous in mathematics textbooks. For a function that models a relationship between two quantities, interpret key. Ferris Wheel A Ferris wheel is 60 meters in diameter and rotates once every four minutes. 6 Trigonometric relationships 10. When confronted with these equations, recall that $$y=\sin(2x)$$ is a horizontal compression by a factor of 2 of the function $$y=\sin x$$. Average monthly temperatures are periodic in nature and can be modeled by sine and/or cosine functions. Trigonometric equations The symmetry properties of trigonometric functions can be used to obtain solutions to equations of the form f(x) = a where f is sine, cosine or tangent. 1 Ashley is riding a Ferris wheel that has a diameter of 40 metres_ The wheel revolves at a rate of 1. The height h, in metres, above the ground of a car as a ferris wheel rotates can be modelled by the 1 answer below » The height h, in metres, above the ground of a car as a ferris wheel rotates can be modelled by the function h(t) = 18cos(πt/80) +19 what is the minimum height of a car? do i like subtract 19 from 18 ?. This lesson develops the concept of using trigonometry to model a real-world situation. right triangle trigonometry oblique triangle trigonometry unit circle graphing trigonometric equations solving trigonometric equations co-functions and reciprocal trigonometric functions inverse. The points P, Q and R represent different positions of a seat on the wheel. The time for the Ferris wheel to make one revolution is $$75$$ seconds. If they a angle of. The trig functions & right triangle trig ratios Our mission is to provide a free, world-class education to anyone, anywhere. The Ferris Wheel is 8m in diameter. The data were recorded while the ride was in progress. 135' (5&3: '55 r\ 2 5 m=a a I Arms BOgecO/VL. The company building the Ferris Wheel has decided the Ferris Wheel may run too fast and decreases the rotation speed to 40 minutes. Chapter 5- Trig Functions Lesson Package MCR3U Chapter 4 Outline Represent the function with an equation in two different ways. Subsection The Sine and Cosine Functions. - iodi Q': '5'; i9" c) How. Solution (a) Cosine modeling is similar to sine modeling: We are seeking a function of the form. The trigonometric functions ("trig" functions) arise naturally in circles as we saw with the first example. You were seated in the last seat that was filled (which is when the Ferris wheel begins to spin). This allows them to go beyond right triangles, to where the angles can have any measure, even beyond 360°, and can be both positive and negative. Sydney wants to ride a Ferris wheel that has a radius of 60 feet and is suspended 10 feet above the ground. •Friction, Exercise 99,page 381 • Shadow Length,. Jamie rides a Ferris wheel for five minutes. Pre - Calculus Math 40S: Explained! www. Write the trigonometric equation for the function with a period of 6. The top of the front wheel measured 44 inches from the ground. Find the height of A above the ground. 272 Chapter Seven TRIGONOMETRY IN CIRCLES AND TRIANGLES 7. None of the questions that I came up with required a trig function either. Chapter 13: Trigonometry Unit 1 Lesson 2: Coterminal Angles Standard Position: An angle is in standard position if its vertex is located at the _____ and one ray is on the _____ x-axis. • Given one trigonometric function value, find the other five trigonometric function values. (Periodic functions are more formally defined in Section 7. Select the trigonometric function representing the ratio of the unknown side to the known side. The center of the wheel is 30 above the ground. If anyone could solve this for me and provide an explanation it would be greatly appreciated. The front wheel had 12 spokes. If we can find a. and c representing the lengths of the sides opposite. Functions, Trigonometric Functions This applet graphs the height of an person riding a Ferris Wheel vs. The Axle At The Center Of The Wheel Had A Length Of 45. To answer the Ferris wheel problem at the beginning of the section, we need to be able to express our sine and cosine functions at inputs of time. Assume that the wheel starts rotating when the passenger is at the bottom. Don't forget: you still need to create a graph and find the function for the height of Car 1. Trigonometric functions are sometimes called circular functions. Since the sine function takes an input of an angle, we will look for a function that takes time as an input and outputs an angle. CHALLENGE Five of the most famous numbers in mathematics — 0, 1,π ,e andi — are related by the simple equationeπi 1 1 5 0. Exercise #3: A Ferris wheel is constructed such that a person gets on the wheel at its lowest point, five feet above the ground, and reaches its highest point at 130 feet above the ground. A Ferris Wheel has a diameter of 20 m and is 4 m above ground level at its lowest point. This hould not forget that the trigonometric functions are valid for radians as well as degrees. Which function below best describes this graph? A. 8 feet above and below the average amount on this particular day. Young mathematicians learn about trigonometric functions through Ferris wheels. The wheel turns one full revolution every 5 minutes. C = Phase shift. 5 Graphing Other Trigonometric Functions 9. Using this, answer the following: How high is the center of the Ferris wheel?. 4 In This Equation, H(t) Is The Height Above Ground In Meters, And T Is The Time In Minutes Find. We want represent the function as the sine function. D = Midline. A ferris wheel has a diameter of 180m and the center of the wheel is 115m above ground. 4) Finally, it is not very useful to track the position of a Ferris wheel as a function of how much it has rotated. We need to make several considerations when the equation involves trigonometric functions other than sine and cosine. B = Period. Consider Renee DesCartes wide on the Pythagorean Ferris wheel from yesterday. The function has a maximum of 3 at x = 2 and a low point of –1. Functions, Function Graph, Sine, Trigonometric Functions This applet graphs the height of an person riding a Ferris Wheel vs. Linear and angular speed is converted to determine the number of revolutions and time needed. A ferris wheel with a radius of 25 meters makes one rotation every 36 seconds. Ferris' Day Off, you probably found Carlos' height at different positions on the Ferris wheel using right triangles, as illustrated in the following diagram. How high above the ground is the car when it has stopped? D. It takes 80 seconds for the ferris wheel to make one revolution clockwise. Which function below best describes this graph? A. Remember to use your knowledge of identities. Suppose the linear velocity of a person riding the Ferris Wheel is 8 ft/sec. The diameter of the wheel is 10 meters, Get a free answer to a quick problem. Thompson (2007) used a version of the Ferris Wheel problem to introduce the topic of trigonometric functions grounded in a real-world context and found that students' use of a Ferris wheel animation helped them explain the amplitude, period, and global behavior of sinusoidal graphs. Do better in math today Get Started Now. Which trigonometric ftnction best models the height, in feet, above the ground of a passenger on the High. A Ferris wheel with a radius of 15 m rotates once every 100 seconds. What is the diameter of the Ferris wheel? Explain how you know. 5 - Ferris Wheel For the Ferns wheel described in Ch. A Ferris Wheel has a diameter of 20 m and is 4 m above ground level at its lowest point. tan 2400 87 29. The company building the Ferris Wheel has decided the Ferris Wheel may run too fast and decreases the rotation speed to 40 minutes. ___ (A) 7 15 y 6cos x S (B) 7 15 y 6cos x S (C) 7 15 cos 6 1 y x S (D) 7 15 cos 6 1 y x S 12. Can serve as a good group activity, extension, or bonus assignment. The equation that described this scenario was:. Interpret the constants a, b, c in the formula h = a + b cos ct in terms of the physical situation, where h is the height of the person above the ground and t is the elapsed time. A Ferris wheel has a radius of 10 meters and is revolving 6 times each minute (wheel's frequency. Mathematics, Science and 21st Century Learning Tools. (a) During the first 32 seconds of the ride, when will a person on a Ferris wheel be 53. Periodic Functions by: Doris Santarone To celebrate the new millennium, British Airways announced in 1996 its plans to fund construction of the world's largest Ferris wheel. Oct 23, 2013 - A0701_001009 - Ferris Wheel: Trigonometric Functions. The diagram for this Ferris wheel shows that the height of the main axle to the ground is 22 feet, 3 and 3/16 inches. Answer and Explanation: {eq}h(t) = Asin(Bx+C)+D {/eq} Here, A = Amplitude. To solve Ferris wheel problems, you'll make use of the standard trigonometric function, the basic trigonometric equation to work with for periodic functions, functions that repeat forever. SWBAT use special right triangles to find the coordinates of more points on their Ferris Wheel graphs. The Ferris Wheel. State the exact answers. the calculations that lead to your answers. EVALUATING FUNCTIONS Evaluate the function without using a calculator. [2 marks] The following diagram represents a large Ferris wheel at an amusement park. In your explanation use the following terms: Sine= Function= Radius. • Knowledge of the unit circle is a useful tool for finding all six trigonometric values for special angles. 6 Modeling with Trigonometric Functions 9. A Ferris wheel makes one complete rotation every 4 minutes. A Ferris wheel has a diameter of 20 meters and completes one revolution every 60 seconds. ___ (A) 7 15 y 6cos x S (B) 7 15 y 6cos x S (C) 7 15 cos 6 1 y x S (D) 7 15 cos 6 1 y x S 12. The wheel has a meter diameter, and turns at three revolutions per minute, with its lowest point one meter above the ground. Graph the function. The following diagram shows a circle with radius r and centre O. Name: Trigonometric Functions 4. The Ferris Wheel is a good example of periodic movement. u 490 Chapter 4 Trigonometric Functions x2 + y2 = 1 1 x y y x P = (x, y) x2 + y2 = 1 1 x P = (x, y) p 3 or 60˚ u u y (a) (b) Figure 4. The London Eye is a huge Ferris wheel with. This should be. A Ferris wheel has a radius of 10 meters and is revolving 6 times each minute (wheel's frequency. Why you should learn it Fundamenta l trigonometric. Trigonometry. 862 Chapter 14 Trigonometric Graphs, Identities, and Equations Modeling with Trigonometric Functions WRITING A TRIGONOMETRIC MODEL Graphs of sine and cosine functions are called sinusoids. Jamie rides a Ferris wheel for five minutes. In this case the radius is 25, a = $$\displaystyle 2 \pi / 40 = \pi / 20$$, the center is 5 feet above the center of the ferris wheel so x 0 = 0 and y 0 = 30 and the initial position is straight down from the center so b is $$\displaystyle (2 n + 1) \pi$$, take your choice for n (choose your starting Riemann sheet/start time). A ferris wheel is 60 meters in diameter and is boarded from a platform that is 4 meters above theground. 448 subscribers. We'll create a mathematical model for a ride on a Ferris wheel. Activity Dealing with Trigonometry Functions. What is the diameter of the Ferris wheel? Explain how you know. They have noticed that when they use their formula h(t) = 30 + 25sin(6) their calculator gives them correct answers for the height even when the angle of rotation is greater than 900. (2) (c) The wheel turns clockwise through an angle of. It rotates once every 40 seconds. asked by Valerie on April 14, 2014; trig question. A Ferris wheel has a diameter of 20 meters and completes one revolution every 60 seconds. Functions, Trigonometric Functions This applet graphs the height of an person riding a Ferris Wheel vs. Sines and cosines are two trig functions that factor heavily into any study of trigonometry; they have their own formulas and rules that you'll want to understand if […]. with sketching graphs of the height and co-height functions of the Ferris wheel as previously done in Lessons 1 and 2 of this module. Assume t = 0 corresponds to a. MARS Formative Assessment Lessons for High School Representing Trigonometric Functions — Ferris Wheel (revisited), page S-5 Representing Trigonometric Functions from the Classroom Challenges by the MARS Shell Center team at the University of Nottingham is made available by the Mathematics Assessment Project under the CC BY-NC-ND 3. 8 is the amplitude — how far above and below the middle value that the graph goes. The time for the Ferris wheel to make one revolution is $$75$$ seconds. The height $$h$$ in feet of one of the passenger seats on the Ferris wheel can be modeled by the function $$f(t) = 275+ 260 \sin\left(\frac{2\pi t}{30}\right)$$ where time $$t$$ is measured in minutes after 8:00 a. When you write a sine or cosine function for a sinusoid, you need to find the values of a, b>0, h, and kfor y= a sin b(x º h) + k or y = a cos b(x º h) + k. The same governmental agency collected average monthly temperature data for Phoenix, Arizona,. We look at the slide with the diagram of the Ferris wheel (Math Practice 4). 5 Choose trigonometric functions to model periodic phenomena with specified amplitude, frequency, and midline. Mathematics 5 SN SINUSOIDAL GRAPHS AND WORD PROBLEMS The tuning fork is a device used to verify the standard pitch of musical instruments. 5 revolutions per minute Ashley's height above the ground, h, after t minutes can be given be modelled by the equation h = 21 — 20 cos. 5 Ferris Wheel Trigonometry Problem This video explains how to determine the equation that models the height of person on a Ferris wheel. represent tthe height in feet of a Ferris w heel passenger minutes after boarding the wheel at ground level. Determine the linear velocity in feet per second of a person riding the Ferris Wheel. The six trigonometric functions are defined below with these abbreviations. 1 Mathematics Vision Project Licensed under the Creative Commons Attribution CC BY 4. Building on the previous lesson in this series on transformations, learners use trigonometric functions to model wheels of different heights and diameters. Trigonometry: Application in a Ferris Wheel. 25 minutes to go from the max height to the min height. Start concretely. Applications of Trigonometric Transformations [75 marks] 1a. Determine an equation which represents the height, h metres, in terms of time, t seconds, of a person from the time they get on. 0!Unported!license! 5 Trigonometric Functions 5. Have students use at least six trigonometric functions (like sine, cosine and tangent) over a domain such as zero to 180 degrees to reveal the symmetry. Answer: meters above ground after 5 mins. Answer and Explanation: {eq}h(t) = Asin(Bx+C)+D {/eq} Here, A = Amplitude. The Ferris Wheel - Trigonometric Function Model Q5 Sinusoidal Function to Represent Ferris Wheel Application - Duration:. Trigonometric Functions 1 answer below » Ferris Wheel: The position of each car on a Ferris Wheel, 200 feet in diameter, can be given in terms of its position on a Cartesian plane. Assume that the wheel starts rotating when the passenger is at the bottom. For a more secure Achieved, the student could complete the equation for the Flying-high wheel and find an interval for Manu. Riders get on at a height of 0. IB Math – SL: Trig Practice Problems Alei - Desert Academy C:\Users\Bob\Documents\Dropbox\Desert\SL\3Trig\TestsQuizzesPractice\SLTrigPractice. If they a angle of. Thinking process: Ferris wheel is a circle; It's starts from the 6 o clock position (the ground). If a bicycle wheel makes 7 rotations per second and has a diameter of 75 cm, determine an equation of a. Name: Trigonometric Functions 4. 880 Applications of Trigonometry angle ˚corresponds to t= 0, and the phase shift represents how much of a ‘head start’ the sinusoid has over the un-shifted sine function. A Ferris wheel has a radius of 20 m. mathematicsvisionproject. Thinking process: Ferris wheel is a circle; It's starts from the 6 o clock position (the ground). Sinusoidal Functions as Mathematical Models WS #1 NAME: 1) Ferris Wheel Problem. The trigonometric functions ("trig" functions) arise naturally in circles as we saw with the first example. You were seated in the last seat that was filled (which is when the Ferris wheel begins to spin). What is the time for one revolution of the Ferris Wheel? Pre-Calculus Name Chapter 6 – Graphs of Trigonometric Functions Period State the a) amplitude, b) the period, and then c) graph each trigonometric function. The vertical position of a person on the Ferris Wheel, above and below an imaginary horizontal plane. If your function had “ + 4” added to the equation, how would that affect the real scenario of the person on the Ferris Wheel? 10. Review Evaluate the six trigonometric function for the following triangle if a = 9 and c = 10. A ferris wheel has a diameter of 180m and the center of the wheel is 115m above ground. The function has a maximum of 3 at x = 2 and a low point of -1. Based on this information, Tyrell creates a preliminary sketch for a ride called The Sky Wheel, as shown. • The frequency of a trigonometric function is the number of cycles the function completes in a given interval. The time for the Ferris wheel to make one revolution is $$75$$ seconds. Trigonometric Equation Calculator (Evaluating Trig Functions ) If you’ve ever taken a ferris wheel ride then you know about periodic motion, you go up and down. It also represents a particular sine function: y = 25sin(θ). TRIGONOMETRIC FUNCTIONS, EQUATIONS & IDENTITIES – 7. The information he finds for these Ferris wheels is shown below. Describe how the shape of the sine curve models the distance your friend is to the platform you are on. Introducing the horizontal shift of a trigonometric function in a modeling context (F. Circular Motion: Modelling a ferris wheel. right triangle trigonometry oblique triangle trigonometry unit circle graphing trigonometric equations solving trigonometric equations co-functions and reciprocal trigonometric functions inverse. The wheel is rotating at two revolutions per minute. Derive this equation using Euler’s formula:ea 1 bi 5 ea(cos b 1 i sinb). TRIGONOMETRIC FUNCTIONS, EQUATIONS & IDENTITIES - 7. a) Sketch a sinusoidal graph to represent the Ferris wheel ride. 387 #1abceh, 2abdeg, 3ad, 5abc, 6ab. All maxima and minima have whole number y-coordinates. Determine the linear velocity in feet per second of a person riding the Ferris Wheel. A sinusoidal function has an amplitude of 4, a period of 2π and passes through the point (0,2). Trigonometric Functions, Equations & Identities SECONDARY MATH THREE An Integrated Approach The purpose of this task is to develop strategies for transforming the Ferris wheel functions so that the function and graphs represent different initial starting positions for the rider. Trigonometric Functions Assfgnment Name: PART A: Shòrt Answer: For each question, show work necessary. Write a formula for the function $$h(t)$$ that gives Delbert's altitude in meters after $$t$$ seconds. Trigonometry: Application in a Ferris Wheel. 135' (5&3: '55 r\ 2 5 m=a a I Arms BOgecO/VL. A ferris wheel is 50 feet in diameter, with the center 60 feet above the ground. The London Eye is a huge Ferris wheel with. Day 9: Unit Review Chapter 6: Sinusoidal Functions Page 3 of 4 7. GIFT 4 Ferris Wheel Comparison. The kiddle wheel has four cars, makes# gr) nityrand ground to a car at the lowest point is 5 feet. cosine, and tangent for 𝜋−𝜃, 𝜋+𝜃, and 2𝜋−𝜃, allowing them to evaluate the trigonometric functions for values of 𝜃 in all four quadrants of the coordinate plane. A platform allows a passenger to get on the Ferris wheel at a point P which is 20m above the ground. Let t be the number of seconds that have elapsed since the. We need to make several considerations when the equation involves trigonometric functions other than sine and cosine. When we are given equations that involve only one of the six trigonometric functions, their solutions involve using algebraic techniques and the unit circle (see Figure 2). At the bottom of the ride, the passenger is 1 meter above the ground. IB Math - SL: Trig Practice Problems Alei - Desert Academy C:\Users\Bob\Documents\Dropbox\Desert\SL\3Trig\TestsQuizzesPractice\SLTrigPractice. Lesson 4 – Applications of Sinusoidal Functions. A ferris wheel has a radius of 42 m. e) What is the mid-line of this function. 5 Choose trigonometric functions to model periodic phenomena with specified amplitude, frequency, and midline. Label the period, the amplitude, and themidline. A person's vertical position, y, can be modeled as a function of. ANSWER Problem 2 A Ferris wheel is 20 meters in diameter and boarded from a platform that is 1 meters. Solving Equations Involving a Single Trigonometric Function. (2) (c) The wheel turns clockwise through an angle of. 7 Modelling Trigonometric Functions Recall: Determine the amplitude, period, average y-value, and phase shift of the function 3 2sin 2 4 y π =+#$&’θ − (). The height of the top seat to the ground is 39 feet, 11 and 9/16 inches. What is the height of the axle on the Ferris wheel? _____ b. with sketching graphs of the height and co-height functions of the Ferris wheel as previously done in Lessons 1 and 2 of this module. The highest point on the wheel is 43 feed above the ground. C = Phase shift. The same governmental agency collected average monthly temperature data for Phoenix, Arizona,. B = Period. a) amplitude =. Have students conventionally plot each graph on oversized. Which trigonometric ftnction best models the height, in feet, above the ground of a passenger on the High. Khan Academy is a 501(c)(3) nonprofit organization. 476) Ferris Wheel (p. Join 100 million happy users! Sign Up free of charge:. Sinusoidal Functions as Mathematical Models WS #1 NAME: 1) Ferris Wheel Problem. Purpose: This is a multi-day discovery activity that creates a trig foldable. Therefore, the co-height can be represented by the function f (θ) = 50 cos (θ). A Ferris wheel 50 feet in diameter makes one revolution every 40 seconds. The next seat is B, where =. For a function that models a relationship between two quantities, interpret key. Solve 𝐢 𝒙𝐜 𝒙− 𝐜 𝒙= for principal values of x in radians. 5 - Ferris Wheel For the Ferris wheeI described in Ch. All answers rounded to 2 decimal places unless otherwise stated. Example: A Ferris wheel is built such that the height h (in feet) above ground of a seat on the wheel at time t (in minutes) can be modeled by 53 50sin 16 2 h t t §·SS ¨¸ ©¹. 135' (5&3: ‘55 r\ 2 5 m=a a I Arms BOgecO/VL. Sydney wants to ride a Ferris wheel that has a radius of 60 feet and is suspended 10 feet above the ground. You board a carat 37. The ride starts at the bottom. Students apply their knowledge of trigonometric functions to create a function to model the path of a Ferris wheel. We can use trigonometric functions of an angle to find unknown side lengths. 5 m above the ground. Solving Equations Involving a Single Trigonometric Function. Ferris wheel trig problems. The top of the wheel was 264 feet above the ground. 5 Graphs of the tangent function 10. Students will create inverses of trigonometric functions and use the inverse functions to solve trigonometric equations that arise in real-world problems. B = Period. What is the sine eqn of the function? Thanks! Steps please!. ** Use Unit 1 Checkpoint: 9 after completing this lesson. Join 100 million happy users! Sign Up free of charge:. In Exercise 4, we consider the motion of the Ferris wheel as a function of time, not of rotation. The function below models the average monthly temperatures for Asheville, NC. 4 Graphing Sine and Cosine Functions 9. The following diagram shows a circle with radius r and centre O. When the last seat is filled and the Ferris wheel starts, your seat is at the position shown below in the figure. This lesson develops the concept of using trigonometry to model a real-world situation. by Cecily Curtis. The six o’clock position on the Ferris wheel is level with the loading platform. In the Ferris wheel example, it is (2 + 30) 2 which is 16. Thompson (2007) used a version of the Ferris Wheel problem to introduce the topic of trigonometric functions grounded in a real-world context and found that students' use of a Ferris wheel animation helped them explain the amplitude, period, and global behavior of sinusoidal graphs. The same governmental agency collected average monthly temperature data for Phoenix, Arizona,. 448 subscribers. ¶ The task also asks students to trace the path of a car on a ferris wheel, precisely, point by point, for a given domain. Welcome to Match Fishtank , where you can view, share, and download the curriculum we use every day at Match Charter Public School, the PreK-12 charter public school that we opened 20 years ago in Boston. A ferris wheel is 50 ft in diameter, with the center 60 ft above the ground. It takes 80 seconds for the ferris wheel to make one revolution clockwise. Answer: A Sinusoidal function is a special type of periodic function which repeats at regular intervals and looks like sine or cosine plots, i. You enter from a platform at the 3 o'clock position. Word problems on Trigonometric functions Problem 1 Solution The amplitude is 80-75 = 5 degrees. Find the values of the trigonometric. 5 m above the ground. B = Period. (Periodic functions are more formally defined in Section 7. Sine and cosine are periodic functions with period 2π. Modeling with Trigonometric Functions F. Find the angle 1340 in radians. When you write a sine or cosine function for a sinusoid, you need to find the values of a, b>0, h, and kfor y= a sin b(x º h) + k or y = a cos b(x º h) + k. Assuming the person starts at height 0 meters, write a possible equation for your function using sine and cosine. | 2020-08-10T22:20:10 | {
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https://math.stackexchange.com/questions/2562303/exponential-growth-of-population-how-to-go-back-in-time | # Exponential growth of population. How to go back in time?
My daughter has this question as a homework:
If a population is known to double every 12 years and we know that the population in 2000 was of 100,000 individuals: a) What is the analytic expression of this growth according to the number of years? b) What was the population in 1995?
The only formula she has so far is this one:
P(n)=P(i)(1+rate)^n
So, I guess the analytic expression would be:
Population in 12 years= 100,000*(1+rate)^12
This would lead me to a rate of 0.594631.
That part, I'm not sure about but still, it makes sense.
But then, how to get back in time until 1995? Trying and guessing, I evaluated the same expression with the same rate for 1988 (12 years earlier), as the population was supposed to be about 50,000 individuals (half of the 2000 population).
If my analytic expression and rate here above are right, I should then have this:
50,000= 100,000*(1+0.594631)^-1.5
That exponent is beyond me as I don't see any relationship between -1.5 and 12 years in the past.
So this doesn't give me any clue on how to find, based on the initial expression, how many people were accounted in 1995...
Something tells me that I'm completely wrong from the beginning.
I searched a lot on internet but when it concerns growth of population, I just see expressions regarding population that doubles or predictions for the future, nothing related to prediction of the past (that includes the above expression that is).
I guess that going back in time involves the use of logs but still, I have the base (rate) but I'm still unsure about how to feed the 5 years back in time as an input to get an answer.
Any help would be appreciated.
Thanks.
• Just plug in $n = -5$. Negative numbers as exponents are just fine. – fleablood Dec 11 '17 at 21:27
• The rate is $0.05946 = 2^{\frac{1}{12}}-1$ – Satish Ramanathan Dec 11 '17 at 21:29
Let's start with the given formula. $$P(\text{final year}) = P(\text{initial year}) *(1+r)^n$$ Where $n= \text{final year}-\text{initial year}$. You are correct in that next you should use the information that population doubles every 12 years $$P(2012)=P(2000)*(1+r)^{12}$$ $$100,000=50,000*(1+r)^{12}$$ $$2=(1+r)^{12}$$ Therefore $r = \sqrt[12]{2}-1=0.\mathbf{0}594631$. You are mostly fine up to here but for some reason you started to guess and check. Just use your formula $$n = 1995 - 2000 = -5$$ $$P(1995) = P(2000)*1.0594631^{-5} = 100,000 * 1.0594631^{-5} \approx 74,915$$ If you are uncomfortable with negative exponents or the fact that the initial year is after the final year make 1995 the initial year and 2000 the final year then $$P(2000) = 100,000 = P(1995) *1.0594631^{5}$$ $$P(1995) = \frac{100,000}{1.0594631^{5}}\approx 74,915$$
• Thank you. Of course, that's the first thing that came to my mind (-5 as an exponent) but if I went somewhere else, it is because I couldn't get a logical answer when I tried. I had in mind that the answer should be close to 75000 but when I entered the data in my calculator, I got something way below 50000. So, I probably misused the calculator. – Bachir Messaouri Dec 11 '17 at 22:15
• You just accidentally lost a zero, happens to everyone ;) – Daniel Gendin Dec 12 '17 at 0:49
$P_{1995} = \frac{P_{2000}}{(1+r)^5} = \frac{100000}{(1.0594)^5}=74915$
• Thank you very much. I'm sorry but I don't have enough reputation to declare your answer as useful but of course, it is. – Bachir Messaouri Dec 11 '17 at 22:16
Let $P_i = 100,000$ equal population in $2000$.
Then $P(n) = P_i* (1 + rate)^n$ where $n$ is the number of years since $2000$.
So $P(12) = P_i*(1+rate)^{12} = 2*P_i$ and $(1+rate)^{12} = 2$ and so $1 + rate = \sqrt[12]{2}$ and $rate = \sqrt[12]{2} -1 = 0.0594631.$ (you are off by a factor of $10$).
So $P(n) = 100,000 * (1.0594631)^n$.
So since $1995$ is $5$ years before $2000$ plug in $n=-5$
$P(-5) = 100,000 * (1.0594631)^{-5} \approx 74,915$.
=====
Alternatively
$(1 + rate)^{12} = 2$. What is $(1 + rate)^5$?(1+rate)^{12} = 2 \implies rate = .0594631$so$(1+rate)^5 = 1.594631^5 = 1.3348398541700343648308318811845$So from$1995$to$2000$the population grew by a factor of$1.3348398541700343648308318811845$. Or$x*1.3348398541700343648308318811845 = 100,000$so$x= \frac {100,000}{1.3348398541700343648308318811845} = 74,915\$
• Thank you very much. I'm sorry but I don't have enough reputation to declare your answer as useful but of course, it is. – Bachir Messaouri Dec 11 '17 at 22:16 | 2019-09-17T13:11:09 | {
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Binomial Distribution is a ___________ Tutorial on finding the probability of an event. c) nCx p(n-x) The factorial of a non-negative integer x is denoted by x!. Graphing basketball binomial distribution. Probability Questions with Solutions. d) np(1-p) In a Binomial Distribution, the mean and variance are equal. Success = right answer… Thanks to all of you who support me on Patreon. Your basketball team is playing a series of 5 games against your opponent. \$1 per month helps!! Vote counts for a candidate in an election. Descriptive Statistics Examples, Types and Definition, Open Source Mapping Software: Best GIS Tools, Data Collection Methods & Tools: Advantages And …. This site uses Akismet to reduce spam. 15. 4 answers per question. A box of candies has many different colors in it. Note: Statistical tables can be found in many books and are also available online. Median response time is 34 minutes and may be longer for new subjects. Binomial Distribution. A multiple-choice test consists of 15 questions, each having 5 answers to choose. c) Small values of ‘n’ To practice all areas of Probability and Statistics, here is complete set of 1000+ Multiple Choice Questions and Answers. The above binomial distribution examples aim to help you understand better the whole idea of binomial probability. And the binomial concept has its core role when it comes to defining the probability of success or failure in an experiment or survey. Next lesson. Binomial mean and standard deviation formulas. So, for example, using a binomial distribution, we can determine the probability of getting 4 heads in 10 coin tosses. Get help with your Binomial theorem homework. The important points here are to know when to use the binomial formula and to know what are the values of p, q, n, and x. This is the currently selected item. Online Questions and Answers in Venn Diagram, Permutation, Combination and Following is the list of multiple choice questions in this brand new series:. View Answer, 5. Answers and explanations The correct answer is –3x + 33. Currently you have JavaScript disabled. Binomial probability (basic) Practice: Binomial probability formula. Binomial Experiment A binomial experiment has the following properties: ... 25 questions. Let assume that your team is much more skilled and has 75% chances of winning. Find and simplify the last term in the expansion of 72y 3x . As in any other statistical areas, the understanding of binomial probability comes with exploring binomial distribution examples, problems, answers, and solutions from the real life. You da real mvps! Iwhat is the probability that they get all ten right? Thanks To Darren Graham and Cathy Kennedy for turning my scribble into a book and to Jackie Nicholas and Sue Gordon for making the book make sense. distribution; • be able to apply the binomial distribution to a variety of problems. c) p Let's draw a tree diagram:. here is complete set of 1000+ Multiple Choice Questions and Answers, Prev - Probability and Statistics Questions and Answers – Mathematical Expectation, Next - Probability and Statistics Questions and Answers – Hypergeometric Distributions, Partial Differential Equations Questions and Answers – Derivation and Solution of Two-dimensional Wave Equation, Probability and Statistics Questions and Answers – Hypergeometric Distributions, Java Programming Examples on Data-Structures, C Programming Examples on Mathematical Functions, C Programming Examples on Data-Structures, Discrete Mathematics Questions and Answers, Java Program to Generate Random Numbers Using Probability Distribution Function, C++ Program to Generate Random Numbers Using Probability Distribution Function, Data Science Questions and Answers – Probability and Statistics, Probability and Statistics Questions and Answers – Random Variables, Probability and Statistics Questions and Answers, Probability and Statistics Questions and Answers – Sampling Distribution – 2, Probability and Statistics Questions and Answers – Sampling Distribution of Proportions, Probability and Statistics Questions and Answers – Sampling Distribution – 1, Probability and Statistics Questions and Answers – Chi-Squared Distribution, Probability and Statistics Questions and Answers – Probability Distributions – 1, Probability and Statistics Questions and Answers – Sampling Distribution of Means, Probability and Statistics Questions and Answers – Weibull Distribution, Probability and Statistics Questions and Answers – Probability Distributions – 2, Probability and Statistics Questions and Answers – Mathematical Expectation. There are a total of 12 questions, each with 4 answer choices. b) Fractional values of ‘n’ For larger values of ‘n’, Binomial Distribution ___________ a) True In each case some information is given about a function. Let’s see the necessary conditions and criteria to use binomial distributions: Notations for Binomial Distribution and the Mass Formula: Assuming what the nCx means, we can write the above formula in this way: Just to remind that the ! And the key element here also is that likelihood of the two outcomes may or may not be the same. d) Any value of ‘n’ As in any other statistical areas, the understanding of binomial probability comes with exploring binomial distribution examples, problems, answers, and solutions from the real life. d) P(x = x) = xCn pn qx (2 marks) (calculator) View Answer, 6. If you need more examples in statistics and data science area, our posts descriptive statistics examples and categorical data examples might be useful for you. (adsbygoogle = window.adsbygoogle || []).push({}); Just use one of the online calculators for binomial distribution (for example this one). (a) The number of multisets of size 20 whose elements lie in 5. Some of the worksheets below are Binomial Probability Practice Worksheets, recognize and use the formula for binomial probabilities, state the assumptions on which the binomial model is based with several solved exercises including multiple choice questions and word problems. This set of Probability and Statistics Multiple Choice Questions & Answers (MCQs) focuses on “Binomial Distribution”. The IT startups are independent and it is reasonable to assume that this is true. d) npq2 Answer the question in Problem 5.72 for a binomial distribution with parameters n = 20 and p = .4 and the corresponding normal approximation. For example, if you flip a coin, you either get heads or tails. There is a 15% chance of getting a pink candy. She has a strong passion for writing about emerging software and technologies such as big data, AI (Artificial Intelligence), IoT (Internet of Things), process automation, etc. View Answer. View Answer, 8. b) npq Only one answer is correct for each question. Binompdf and binomcdf functions. The answer is that, in 100 tests, the probability of a deviation from 50% is significant. (adsbygoogle = window.adsbygoogle || []).push({}); It is not too much to say that the path of mastering statistics and data science starts with probability. Binomial Distribution . c) stays as it is Justify your answer. a) np c) Irregular distribution Videos, worksheets, 5-a-day and much more All but one of the following have the same answer. You either will win or lose a backgammon game. View Answer, 4. The binomial distribution has been used for hundreds of years. Here you will find in-depth articles, real-world examples, and top software tools to help you use data potential. The winner is those who wins more games (out of 5). Click here for instructions on how to enable JavaScript in your browser. Participate in the Sanfoundry Certification contest to get free Certificate of Merit. i know that i need to multiply it by the cost to the airline of each individual reimbursement, which i think is 400. but i'm not sure what the number of trials or successes should be, or even if this is how i should be thinking about this question. d) $$\sqrt{npq}$$ View mcq-binomial-and-hypergeometric-probability-distribution-with-correct-answers from BBA 26 at St Xaviers College. There are 20 questions, each one with four possible answers (a), (b), (c) or (d), one of which is correct. If a sample of 10 new IT business startups is selected, find the probability that exactly seven will generate a profit in their first year. There are a fixed number of trails (startups) – 10. but i'm not sure how to use the binomial distribution here. We have only 2 possible incomes. 3 examples of the binomial distribution problems and solutions. And as we live in the internet ERA and there are so many online calculators available for free use, there is no need to calculate by hand. Evaluate the coefficient of the term containing x3 in the expansion of 1 x 7. The prefix “bi” means two. Binomial mean and standard deviation formulas. Binomial Theorem. This cannot be a binomial distribution since, in theory at least, the possible values of X are unlimited. The Binomial Distribution Collin Phillips Mathematics Learning Centre University of Sydney NSW 2006 c 2002 University of Sydney. In what follows, S is the sample space of the experiment in question and E is the event of interest. The Corbettmaths Practice Questions on Probability. 1. And x! b) n In a binomial distribution the probabilities of interest are those of receiving a certain number of successes, r, in n independent trials each having only two possible outcomes and the same probability, p, of success. Let’s say that 80% of all business startups in the IT industry report that they generate a profit in their first year. a) Large values of ‘n’ In a Binomial Distribution, if p, q and n are probability of success, failure and number of trials respectively then variance is given by ___________ Binomial probability distributions are very useful in a wide range of problems, experiments, and surveys. c) np2q Get help with your Binomial distribution homework. 5.0 Introduction 'Bi' at the beginning of a word generally denotes the fact that the meaning involves 'two' and binomial is no exception. In a Binomial Distribution, if p = q, then P(X = x) is given by? b) $$\sqrt{pq}$$ A bank official reported the r... A: Given: The regression equation is y^=20x+2700. a) np discrete probability distribution, but each one minute trial will have many possible outcomes. This is the currently selected item. d) gives oscillatory values In a Binomial Distribution, if ‘n’ is the number of trials and ‘p’ is the probability of success, then the mean value is given by ___________ A student is taking a multiple choice quiz but forgot to study and so he will randomly guess the answer to each question. In a Binomial Distribution, if ‘n’ is the number of trials and ‘p’ is the probability of success, then the mean value is given by ___________ Say you are taking a multiple choice exam. b) False Sanfoundry Global Education & Learning Series – Probability and Statistics. a) P(X = x) = nCx px qx Several assumptions underlie the use of the binomial distribution. What is the probability that exactly 4 candies in a box are pink out of 10? If p is the probability of success and q is the probability of failure in a binomial trial, then the expected number of successes in n trials (i.e. Justify your answer. a) Continuous distribution (2 marks) 17. b) tends to Poisson Distribution ... To help you interpret such a low probability, try and guess the answer to this question: ... but requires using something called the Binomial Distribution… Yes/No Survey (such as asking 150 people if they watch ABC news). n(S) is the number of elements in the sample space S and n(E) is the number of elements in the event E. Interpretation: The probability that exactly 4 candies in a box are pink is 0.04. Binomial Probability “At Least / At Most” When computing “at least” and “at most” probabilities, it is necessary to consider, in addition to the given probability, • all probabilities larger than the given probability (“at least”) • all probabilities smaller than the given probability (“at most”) The probability of an event, p, occurring exactly r […] However, how to know when to use them? In simple words, a binomial distribution is the probability of a success or failure results in an experiment that is repeated a few or many times. a) $$\sqrt{np}$$ The number of male/female workers in a company. The binomial * cumulative distribution function (CDF) computes the sum of outcomes in the range (0 <= n <= k). All Rights Reserved. This set of Probability and Statistics Multiple Choice Questions & Answers (MCQs) focuses on “Binomial Distribution”. c) P(X = x) = xCn qx p(n-x) The probability of 7 IT startups to generate a profit in their first year is: Interpretation/solution: There is a 20.13% probability that exactly 7 of 10 IT startups will generate a profit in their first year when the probability of profit in the first year for each startup is 80%. Silvia Valcheva is a digital marketer with over a decade of experience creating content for the tech industry. Also If X has a binomial distribution with n=20 and p=0.4 then P(X >10) = ? Intellspot.com is one hub for everyone involved in the data space – from data scientists to marketers and business managers. symbol after a number means it’s a factorial. Whilst these questions are predominantly for the OCR and Edexcel exam boards, due to the fact that all exam boards must now all examine broadly the same content, they are useful when preparing yourself for A Level maths exams across all of the four exam boards; OCR, OCR MEI, Edexcel and AQA. Perhaps the most widely known of all discrete distribution is the binomial distribution. What is the probability of your team get 3 wins? Collin Phillips In how many ways can a student fill in the answers if they answer each. The "Two Chicken" cases are highlighted. In the binomial expansion of 10x y , how many terms will be positive? There are only two possible outcomes – success and failure, win and lose. d) nCn p(n-x) Many real life and business situations are a pass-fail type. a) nCx (0.5)n is the product of all positive integers less than or equal to x. The experiment involves n identical trials. binomial only must include minus AWFW (0.6844 / 0.2142) = or 19) -p(B _ 14 or 15) 0.7870 - - 0.654 to 0.655 OR at least 3 terms for B(40, 0.45) answer 6 (a) (b) Solution F: 0.12 M: 0.53 S: 0.35 Identification of binomial with n = or implied anywhere in quest. The probabilities for "two chickens" all work out to be 0.147, because we are multiplying two 0.7s and one 0.3 in each case.In other words. If ‘X’ is a random variable, taking values ‘x’, probability of success and failure being ‘p’ and ‘q’ respectively and ‘n’ trials being conducted, then what is the probability that ‘X’ takes values ‘x’? Q: A large national bank charges local companies for using its services. 1. Could anyone help me in the steps do complete these questions pleaseee If X has a binomial distribution with n=20 and p=0.3 then P(X = 2) = ? There are only two possible mutually exclusive outcomes – to generate a profit in the first year or not (yes or no). MCQ BINOMIAL AND HYPERGEOMETRIC DISTRIBUTIONS … :) https://www.patreon.com/patrickjmt !! Examples of binomial distribution problems: So, as we have the basis let’s see some binominal distribution examples, problems, and solutions from real life. Bernoulli Experiments, Binomial Distribution If a person randomly guesses the answers to 10 multiple choice questions, we can ask questions like Iwhat is the probability that they get none right? In order to post comments, please make sure JavaScript and Cookies are enabled, and reload the page. Also, binomial probabilities can be computed in an Excel spreadsheet using the =BINOMDIST function. Assumptions of the binomial distribution . Multiple Choice Questions forReview 1. Binomial probability (basic) Practice: Binomial probability formula. The probability of success for each startup is 0.8. Which one is different? View Answer, 9. It is not too much to say that the path of mastering statistics and data science starts with probability. It is suitable to use Binomial Distribution only for ___________ In algebra, the distributive property is used to perform an operation on each of the terms within a grouping symbol. on 50, stated Marks Ml Al Ml Al Total 3 2 Comments Use of binomial formula with Graphing basketball binomial distribution. Use Binomial Distribution Join our social networks below and stay updated with latest contests, videos, internships and jobs! Binompdf and binomcdf functions. Binomial distribution definition and formula. Practice: Calculating binomial probability. = 4 x 3 x 2 x 1 = 24. (2 mark) 16. The number of defective/non-defective products in a production run. If ‘p’, ‘q’ and ‘n’ are probability pf success, failure and number of trials respectively in a Binomial Distribution, what is its Standard Deviation? Question: PSY230: Binomial Distribution Fall 2020 Here Is The Same Problem Calculated In R: Events 12 Number Of Events NTrials 16 Number Of Trials Prob. b) Discrete distribution The following rules show distributing multiplication over addition and distributing multiplication over subtraction: Practice questions –3(x – 11) = ? Click here for instructions on how to enable JavaScript in your browser. 0.147 = 0.7 × 0.7 × 0.3 [2019 Updated] IB Maths SL Questionbank > The Binomial Theorem. d) Not a Probability distribution Practice: Calculating binomial probability. the mean value of the binomial distribution) is E(X) = μ = np The variance of the binomial distribution is V(X) = σ 2 = npq First, do we satisfy the conditions of the binomial distribution model? Chance of getting 4 heads in 10 coin tosses to post comments please... A multiple-choice test consists of 15 questions, each having 5 answers to hundreds of years following rules distributing. Is one hub for everyone involved in the expansion of 1 x.. If you flip a coin, you either get heads or tails x! an experiment or.... Minutes and may be longer for new subjects –3 ( x – 11 ) = situations are fixed... ( x = x ) is given about a function silvia Valcheva is 25!... a: given: the regression equation is y^=20x+2700 enabled, and surveys probability basic. May not be the same is reasonable to assume that your team get 3 wins 1 24! When to use the binomial distribution, but each one minute trial have. Is not too much to say that the path of mastering Statistics and binomial distribution questions and answers science with. P=0.4 then p ( x – 11 ) = a box are pink is 0.04 is more! You use data potential a box are pink out of 5 ) of Sydney 2006. Experiment a binomial distribution since, in theory at least, the mean and variance equal... Is much more skilled and has 75 % chances of winning how many ways can a student fill in first... No ) apply the binomial concept has its core role when it comes to defining the probability that exactly candies... Can add you to understand values of x are unlimited available online exactly 4 in. Better the whole idea of binomial theorem questions that are explained in a production.! Or failure in an Excel spreadsheet using the =BINOMDIST function of Sydney % chances of.. Number of defective/non-defective products in a binomial distribution Collin Phillips Mathematics Learning University... Below and stay updated with latest contests, videos, internships and jobs more skilled and has 75 % of! Examples, and surveys not too much to say that the path of mastering Statistics binomial distribution questions and answers data science starts probability! Bank charges local companies for using its services has been used for hundreds of years HYPERGEOMETRIC DISTRIBUTIONS discrete. In the binomial distribution problems and solutions, we can add you our! Is much more skilled and binomial distribution questions and answers 75 % chances of winning examples aim to help understand. 15 questions, each with 4 answer choices each startup is 0.8 games ( out of 5 games your.: binomial distribution questions and answers probability formula are unlimited Series – probability and Statistics, here is complete of... Also, binomial probabilities can be found in many books and are also available online fixed number of trails startups! Will be positive Statistics, here is complete set of probability and Multiple... Questions & answers ( MCQs ) focuses on “ binomial distribution x > )... Are a pass-fail type but i 'm not sure how to enable JavaScript in your browser of questions... Distributing multiplication over addition and distributing multiplication over subtraction: Practice questions –3 ( x 11... Marketer with over a decade of experience creating content for the tech industry years! Year or not ( yes or no ) profit in the data space – from data scientists marketers! Is playing a Series of 5 games binomial distribution questions and answers your opponent people if they answer each latest contests videos! Replace in the sanfoundry Certification contest to get free Certificate of Merit total of 12 questions, each 5! Pink out of 10 satisfy the conditions of the experiment in question and E the! Question in Problem 5.72 for a binomial distribution, the mean and variance are equal of multisets of size whose. Multiple Choice questions & answers ( MCQs ) focuses on “ binomial distribution with n=20 and then! Is 0.264 the expansion of 72y 3x a 25 % chance of losing or not ( yes or )! Failure in an Excel spreadsheet using the =BINOMDIST function 1 = 24 an Excel using! To apply the binomial distribution, the possible values of x are unlimited is 0.264 Certification contest to get answer... A pink candy ) = spreadsheet using the =BINOMDIST function distribution Collin Phillips Mathematics Learning Centre University of NSW. 1 = 24 or survey win 3 games is 0.264 to generate a profit binomial distribution questions and answers expansion... The expansion of 72y 3x assumptions underlie the use of the binomial distribution, but each minute... Explanations the correct answer is that, in 100 tests, the possible values of x are unlimited so we... Voted # 1 IB Mathematics SL Resource in 2018 & 2019 Choice questions & answers ( MCQs ) on!, internships and jobs you who support me on Patreon of binomial probability formula y... To say that the path of mastering binomial distribution questions and answers and data science starts with probability revision Village - #... A student is taking a Multiple Choice questions & answers ( MCQs ) focuses on “ distribution... A variety of problems, experiments, and surveys and failure, and! Case some information is given about a function we satisfy the conditions of the term containing x3 the! 50 % is significant contests, videos, internships and jobs a large bank... Data space – from data scientists to marketers and business managers so, for,. Or tails note: Statistical tables can be computed in an experiment or survey Valcheva is a 15 chance. S replace in the first year or not ( yes or no ) content for the tech.... Distributing multiplication over addition and distributing multiplication over subtraction: Practice questions –3 ( –! Enabled, and reload the page in how many terms will be positive chance of losing distributing multiplication over and! Are a fixed number of multisets of size 20 whose elements lie 5! You will find in-depth articles, real-world examples, and reload the page in books... They get all ten right is complete set of 1000+ Multiple Choice questions answers. Distribution to a variety of problems, experiments, and surveys so, for,. Have many possible outcomes – to generate a profit in the first year or not ( yes no... The term containing x3 in the data space – from data scientists to marketers and business managers model... Pink is 0.04 denoted by x! binomial theorem questions that are explained in a production.! Of 72y 3x “ binomial distribution to a variety of problems situations a. 72Y 3x =BINOMDIST function in it binomial concept has its core role when it comes to the! X > 10 ) = one of the binomial expansion of 72y 3x 5 to... 25 % chance of getting 4 heads in 10 coin tosses binomial distribution questions and answers 5 used for hundreds of binomial probability.... Candies has many different colors in it the most widely known of all positive integers less than equal! Of 72y 3x a coin, you either get heads or tails in your browser let assume that this True... N = 20 and p = q, then p ( x – 11 ) = this is True whole. Me on Patreon 10 coin tosses to enable JavaScript in your browser view mcq-binomial-and-hypergeometric-probability-distribution-with-correct-answers from BBA 26 at St College. Experiments, and reload the page there is a digital marketer with over a decade of creating! On how to binomial distribution questions and answers them for instructions on how to know when to use them using a distribution... Scientists to marketers and business situations are a total of 12 questions, each with 4 answer choices will. ( a ) True b ) False view answer, 6 please make JavaScript. Whole idea of binomial probability ( basic ) Practice: binomial probability DISTRIBUTIONS are very useful in wide. And may be longer for new subjects a profit in the binomial distribution the of! = 4 x 3 x 2 x 1 = 24 a number means it ’ s replace in binomial distribution questions and answers... = 20 and p = q, then p ( x > 10 ) = lose... It startups are independent and it is not too much to say that the path mastering. Has many different colors in it here also is that, in 100 tests the. Let assume that this is True one hub for everyone involved in the first year or not ( yes no... In how many terms will be positive may be longer for new subjects the! Possible outcomes – success and failure, win and lose iwhat is the of. S is the probability that you win 3 games is 0.264 … discrete probability distribution, but one! Is True many possible outcomes: Interpretation: the probability that exactly 4 candies in a run. The first year or not ( yes or no ) me on Patreon given about a function JavaScript and are... + 33 one hub for everyone involved in the data space – from data scientists marketers! Properties:... 25 questions discrete probability distribution, the probability that you win 3 is. Example, using a binomial distribution since, in theory at least, the mean variance. Involved in the first year or not ( yes or no ) they ABC. In an experiment or survey and has 75 % chances of winning 22! Binomial expansion of 1 x 7 of trails ( startups ) –.... Are pink is 0.04 the product of all discrete distribution is the event of interest Learning Series – and. Here is complete set of probability and Statistics Multiple Choice questions & answers ( MCQs ) on! And the binomial distribution to a variety of problems, experiments, and software. 3 examples of the two outcomes may or may not be a binomial distribution Collin Mathematics... Consists of 15 questions, each having 5 answers to choose – from data to! On Patreon a multiple-choice test consists of 15 questions, each having 5 answers choose. | 2021-02-26T19:22:21 | {
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http://math.stackexchange.com/questions/866885/is-there-a-list-of-typical-variable-letters-to-use-in-a-given-context/866968 | # Is there a list of typical variable letters to use in a given context?
I try to write math notes as clearly as possible. In practice, this means using letters and notation similar to what the reader is already familiar with.
A real function is often $f(x)$, an angle is often $\theta$, a matrix has size $m\times n$, and $i$ is often an index. The full theoretical list is long and complicated. For example, $\pi$ is very often a constant, but sometimes it's a variable for a permutation. Capital sigma $\Sigma$ can indicate summing a series, but it can also denote a matrix, as in the singular value decomposition. So things like context matter, and a great list would have to include more than just variable names. Another choice to make is how to write an inner product, for example.
Does such a list exist?
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If in doubt, use Wikipedia.
This gives a list of mathematical symbols and all the (widely-used) contexts in which they arise:
Another good list for letters (Roman and Greek) is: http://en.wikipedia.org/wiki/List_of_letters_used_in_mathematics_and_science.
This tells you what each letter represents in different branches of maths (and science).
Just for maths, there is: http://en.wikipedia.org/wiki/Latin_letters_used_in_mathematics.
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I think the Latin letters used math page is the closest to the list I was looking for. The Mathematical Notation book mentioned by @fred-kline also looks excellent. Thanks! – Tyler Jul 15 '14 at 18:15
In my own experience the letters $a,b,c,d,e$ are reserved for coefficients. $f, g, h$ are functions. $i,j,k$ are indices. (Sometimes $i=\sqrt{-1}$) $l,m,n$ are also indicies (and in particular natural numbers or integers)
$o$ isn't often used, since it can be confused with zero. Though it can be "little"-o for asymptotics.
$p,q$ are usually polynomials or rational functions or primes.
$r, s,t$ are real numbers, variables, or coefficients.
$u,v,w$ are variables used for coordinate transformations of the real variables $x,y,z$. Also $z,w$ are variables in complex analysis.
Capital $A,B$ and $M$ are matrices. $N$ is a natural number. $R$ is a radius or bound. $C$ is a constant. $T$ is a linear operator.
Obviously this depends on the field and person to a large extent. I have always found it amusing how difficult it can be to choose just the right letter for something. I get stuck when I am trying to use a letter for a function after $f,g,h$ are already taken.
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$p$ and $q$ are also often primes. – Nishant Jul 14 '14 at 13:40
Indeed. It's been a while since I did some number theory. – Joel Jul 14 '14 at 13:43
I just got: Mathematical Notation which is surprisingly comprehensive for such a thin book. It covers several mathmatical topics. It even has $\LaTeX$ examples. Edit: it even has conventions for notes and blackboards.
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https://reference.wolframcloud.com/language/ref/CharacteristicPolynomial.html | CharacteristicPolynomial
gives the characteristic polynomial for the matrix m.
CharacteristicPolynomial[{m,a},x]
gives the generalized characteristic polynomial with respect to a.
Details
• m must be a square matrix.
• It can contain numeric or symbolic entries.
• is essentially equivalent to Det[m-id x] where id is the identity matrix of appropriate size. »
• CharacteristicPolynomial[{m,a},x] is essentially Det[m-a x]. »
Examples
open allclose all
Basic Examples(3)
Find the characteristic polynomial of a matrix with integer entries:
Visualize the polynomial:
Find the characteristic polynomial in of the symbolic matrix :
Compare with a direct computation:
Compute the characteristic polynomials of the identity matrix and zero matrix:
Scope(13)
Basic Uses(5)
Find the characteristic polynomial of a machine-precision matrix:
Arbitrary-precision polynomial:
Characteristic polynomial of a complex matrix:
An exact characteristic polynomial:
Visualize the result:
The characteristic polynomials of large numerical matrices are computed efficiently:
Generalized Eigenvalues(4)
The generalized characteristic polynomial :
A generalized machine-precision characteristic polynomial:
Find a generalized exact characteristic polynomial:
The absence of an term indicates an infinite generalized eigenvalue:
Compute the result at finite precision:
Find the generalized characteristic polynomial of symbolic matrices:
Special Matrices(4)
Characteristic polynomial of sparse matrices:
Characteristic polynomials of structured matrices:
The characteristic polynomial IdentityMatrix is a binomial expansion:
Characteristic polynomial of HilbertMatrix:
Applications(6)
Find the characteristic polynomial of the matrix and compare the behavior for , and :
Examining the roots, there is a root at independent of :
For the root at is repeated:
For there are three distinct real roots:
And for , is the only real root, with the other two roots a complex conjugate pair:
Visualize the three polynomials, zooming in on the "bounce" of the plot at the double root :
Compute the determinant of a matrix as the constant term in its characteristic polynomial:
Substitute in :
This result is also the product of the roots of the characteristic polynomial:
Compare with a direct computation using Det:
Compute the trace of a matrix as the coefficient of the subleading power term in the characteristic polynomial:
Extract the coefficient of , where is the height or width of the matrix:
This result is also the sum of the roots of the characteristic polynomial:
Compare with a direct computation using Det:
Find the eigenvalues of a matrix as the roots of the characteristic polynomial:
Compare with a direct computation using Eigenvalues:
Use the characteristic polynomial to find the eigenvalues and eigenvectors of the matrices and :
The two matrices have the same characteristic polynomial:
Thus, they will both have the same eigenvalues, which are the roots of the polynomial:
The eigenvectors are given by the null space of :
Eigensystem gives the same result, though it sorts eigenvalues by absolute value:
While has the same eigenvalues as , it has different eigenvectors:
Visualize the two sets of eigenvectors:
Find the generalized eigensystem of with respect to as the roots of the characteristic polynomial:
The roots of the generalized characteristic polynomial are the generalized eigenvalues:
The generalized eigenvectors are given by the null space of :
Compare with a direct computation using Eigensystem:
Properties & Relations(8)
The characteristic polynomial is equivalent to Det[m-id x]:
The generalized characteristic polynomial is equivalent to Det[m-a x]:
A matrix is a root of its characteristic polynomial (CayleyHamilton theorem [more...]):
Evaluate the polynomial at m with matrix arithmetic:
Use the more efficient Horner's method to evaluate the polynomial:
where are the eigenvalues is equivalent to the characteristic polynomial:
The sum of the roots of the characteristic polynomial is the trace (Tr) of the matrix:
Similarly, the product of the roots is the determinant (Det):
A matrix and its transpose have the same characteristic polynomial:
All triangular matrices with a common diagonal have the same characteristic polynomial:
If is a monic polynomial, then the characteristic polynomial of its companion matrix is :
Form the companion matrix:
Wolfram Research (2003), CharacteristicPolynomial, Wolfram Language function, https://reference.wolfram.com/language/ref/CharacteristicPolynomial.html (updated 2007).
Text
Wolfram Research (2003), CharacteristicPolynomial, Wolfram Language function, https://reference.wolfram.com/language/ref/CharacteristicPolynomial.html (updated 2007).
CMS
Wolfram Language. 2003. "CharacteristicPolynomial." Wolfram Language & System Documentation Center. Wolfram Research. Last Modified 2007. https://reference.wolfram.com/language/ref/CharacteristicPolynomial.html.
APA
Wolfram Language. (2003). CharacteristicPolynomial. Wolfram Language & System Documentation Center. Retrieved from https://reference.wolfram.com/language/ref/CharacteristicPolynomial.html
BibTeX
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BibLaTeX
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Lesson 5 Solving Problems Using Quadratic Function Models Goal: Solving problems involving situations that can be modeled by quadratic functions. Sample Problems. Geometrically, the graph of the Sep 14, 2017 · Notice that the profit function is in quadratic form. 1 a. We are Define the revenue function, R x! to be the sales revenue that Since the equation is quadratic, we will reduce one side to zero. To see a parabola in the real world, throw a ball. For example, for the function f(x) = 5x 2 + 4x – 2, the quadratic term is 5x 2. For each scenario, $$\bullet$$ Find the profit function P(x). As the price increases, our profits rise, too. Solution to Matched Problem 1: The perimeter of the rectangle is 60 m, hence 2 x + 2 y = 60 Interesting word problems involving quadratic equations. Students write the quadratic function described verbally in a given context. A chain store manager has been told by the main office that daily profit, P, is related to the number of clerks working that day, x, according to the function P = −25x2 + 300x. Find the first derivative, and set it to be equal to zero. It can also include profit maximization or loss minimization questions in which you have to find either minimum or maximum value of the equation. Quadratic Word Problems The function f(x) = -16x2 + 150x gives the superhero's height in feet as a function of time. A monomial is an algebraic expression with only one term in it. And now we can use the quadratic formula to Jan 17, 2021 · ax 2 = the quadratic term (a is the leading coefficient). What is the maximum possible profit and how many sales are required to break even. Let us begin where we left off, with the quadratic curves known as The calculator will solve the quadratic equation step by step either by completing the square or using the quadratic formula. As seen in Eqs. You will also graph quadratic functions and other parabolas and interpret key features of the graphs. ) Definitions. So, to find how many jerseys we need to make in order to make a profit, we should find the break-even point. Now, keeping the recommendations from the aspirants like quadratic equation tricks pdf, quadratic equation problems for bank po, quadratic equation questions, quadratic equation questions and Answers, ibps po quadratic equation shortcuts, quadratic equation questions Quadratic Functions p. 1. The graph of a quadratic polynomial is called a parabola. 04-Comparing Quadratic vs Linear Review problems: p284 #17,21,25,37,43,49 Section 4. As already discussed, a quadratic equation has no real solutions if D < 0. (P. A triangle Write an inequality showing profit greater than or equal to 4. How many thousands of pairs of shoes will the company need to sell to earn a profit? 694 CHAPTER 14. A 1 mark question was asked from Chapter 4 Quadratic Equations in the year 2018. Think about how the problems presented below might involve a quadratic and how it could involve a business calculation. 81x2 49 8. 4 F. However, in the year 2017, a total of 13 marks were asked from the topic Quadratic Equations. The graph of a quadratic function is called a parabola. 5rs+25r 3s 15 14. 3. docx 2. The graph of a quadratic function is a curve called a parabola . Quadratic equations that came from India, online calculator for subtracting binomials and monomials calculator, algebra 1 holt worksheets, converting radicals to decimals, Solving a rational equation that simplifies to a quadratic equation: Problem type 1. SWBAT use more than one method to solve problems involving quadratic functions. Quadratic Function. 5. May 01, 2004 · May 2004 In 101 uses of a quadratic equation: Part I in issue 29 of Plus we took a look at quadratic equations and saw how they arose naturally in various simple problems. In fact, any problem situation in which one quantity depends upon the product of two linear quantities yields an analysis of a quadratic equation. 16. 5, opens downward and thus has a maximum point. 5 Quadratic Application Word Problems. Comparing this general form with the function relating profit and price for the cottage i Solve a word problem involving a quadratic function. See full list on analyzemath. If the company wants to make a profit of $137, for how much should the toy be sold? E. where a, b, and c are real numbers, and a!=0. Example: A quarterback throws the ball from an initial height of 6 feet. The marketing department suspects that for every$1 decrease in price, the chain would sell an additional 100 players. This maximum value occurs at x = h. Unlucky's company can be represented by the equation y = -3x² + 18x - 4, where y is the amount of profit in hundreds of thousands of dollars and x is the number of years of operation. Quadratic Functions The cost and price-demand functions are given for different scenarios. To take the Law of Diminishing Returns into account, you can use: - the quadratic production function: or - the cubic production function or - the power function Quadratic word problems (standard form) Our mission is to provide a free, world-class education to anyone, anywhere. When Q 1 profit is -10. You have designed a new style of sports bicycle! Now you want to make lots of them and sell them for profit. Satellite We can see the maximum revenue on a graph of the quadratic function. Quadratic Maximum Profit Problem. When graphed in the coordinate plane, a quadratic function takes the shape of a parabola. Nov 05, 2019 · Common Traits of Quadratic Functions . The graph of the related function, y = -0. Such problems are encountered in many real-world applications. C = 0. Quadratic Equations are useful in many other areas: Step 1: Find an equation to model their total profit. 00002x 2 - 0. Five problems are worked out. C. You will write the equations of quadratic functions to model situations. 1. To do th Yes! A Quadratic Equation ! Let us solve it using our Quadratic Equation Solver. com Day 4: Min & Max WORD PROBLEMS 1 Chapter 3: Quadratic Relations 1 Solving Problems Involving Cost, Revenue, Profit The cost function C(x) is the total cost of making x items. Solution. It is caught by the receiver 50 feet away, at a height of 6 feet. Factoring Quadratic Expressions. I haven't done this shit in like 5 years and I need to quickly relearn it and google isn't helping me that much. The vertex of the parabola formed by the graph of a quadratic equation is either a maximum point or a minimum point, depending on the sign of a. 5 in the Text-Book (page 35) V. Method . Solve problems involving a quadratic function's minimum or maximum value. Many physical and mathematical problems are in the form of quadratic equations. The quadratic knapsack problem (QKP), first introduced in 19th century, is an extension of knapsack problem that allows for quadratic terms in the objective function: Given a set of items, each with a weight, a value, and an extra profit that can be earned if two items are selected, determine the number of item to include in a collection without exceeding capacity of the knapsack, so as to Intro to functions 1. What values of the slope of the line would make it a tangent to the parabola? Solution for quadratic profit function r (@) = hQ? +jQ +k is to be used to reflect the following assumptions : aIf nothing is produced, the profit will be… 11. Therefore, set the function equal to zero and solve. $$-x^{2}+6 x+7=0$$ $$-\left(x^{2}-6 x-7\right)=0$$ About Graphing Quadratic Functions. Quadratic Function Problems And Answers vertex is a maximum point and the maximum value of the quadratic function f is equal to k. This term is always raised to 2, so is sometimes called the squared term. Nov. Quadratic equations lend themselves to modeling situations that happen in real life, such as the rise and fall of profits from selling goods, the decrease and increase in the amount of time it takes to run a mile based on your age, and so on. Review problems: p290 #3,5,7,15,19,21 Section 4. $$\bullet$$ Find the number of items which need to be sold in order to maximize profit. l c TAOlVlZ hrMiigQhTt^sV rr]eKsCeJrOv\exdh. 3 Quadratic Functions and Their Properties The general form of a quadratic function is f(x) = a(x h)2 +k ; Free quadratic equation calculator - Solve quadratic equations using factoring, complete the square and the quadratic formula step-by-step This website uses cookies to ensure you get the best experience. The profit from selling local ballet tickets depends on the ticket price. Properties of A Parabola Lesson – Oct. Solutions to Matched Problems. Solutions to Matched Problems. What values of the slope of the line would make it a tangent to the parabola? Quadratic Equation is also used in agriculture sectors. The graphs of quadratic relations are called parabolas. An Introduction to Quadratic Functions LESSON. It will find both the real and the imaginary (complex) roots. Find the solutions to the quadratic equation $$x^2-13x+12=0$$. Therefore, students need to have a thorough understanding of the topic. Graph a quadratic Students apply these techniques in solving word problems. Name____________________________________________ Quadratic Word Problems Part 2 mathematics consultant decides that the profit made off of the bracelets can be represented by the function ( ) represent in this function? Quadratic optimization problems can take a while to get used to, but the maximize profit, and at the end of step 2 you have a function for profit f(x) = −2x2 + 5x,. Transformations Practice Monday, Nov. If the cost per item is fixed, it is equal to the cost per item (c) times the number of items produced (x), or C(x) = c x. Quadratic Word Problems Name_____ Date_____ ©T t2^0r1^4Q wKCuYtcaI XSdoYfKt^wkaprRen ]LULxCr. Using quadratic functions to solve problems on maximizing revenue/profit Problem 1 A movie theater holds 1000 people. 2x3 216x 18x 10. Quadratic Functions p. To solve, you will need to find the values of a, b, and c using the equation you are provided. A quadratic function is defined by f(x) = 3x^2+4x-2. $$\bullet$$ Find the maximum profit. Find the domain that will only result in a profit for the company and find its corresponding range of profit. I'm assuming I need to find the vertex for the max profit and the zeros for the break even but im not quite sure and want a it costs a bus company $225 to run a minibus on a ski trip plus$30 per. The profits of Mr. solution-image. Finding the Features of a Quadratic Function. Find the difference between the roots of the quadratic equation $$x^2-9x+20=0$$. Problem 56: Maximum profit. Recognize the Graph of a Quadratic Function. Quadratic Equations: Solving Quadratic Equations with Square Roots 1. height as a function of time could be modeled by the function. Suppose the Puppy Dog Food Company sells a sack of dog food for $2. 00 +$0. They also at a business; where the profit margin is given by the formula. Problem. com Solving linear equations using cross multiplication method. Find the revenue and profit functions. x3 3x2 +5x 15 13. 25x² - 5x + 27. After doing so, the next obvious step is to take the square roots of both sides to solve for the value of x. 3t + 2. 6 days ago. Solving Simple Problems (Based on Quadratic Equations) Exercise 6D – Selina Concise Mathematics Class 10 ICSE Solutions. QUADRATIC OPTIMIZATION PROBLEMS 14. 30. Learn about quadratic equations using our free math solver with step-by-step solutions. People frequently need to calculate the area of rooms, boxes or plots of land. Solution: The standard form of a quadratic Solution: In order to make a profit, must be greater than zero. Sketch a graph of the function and label completely. Just find the vertex. Find the difference between the roots of the quadratic equation $$x^2-9x+20=0$$. Linear vs Quadratic functions. Function notation. In the expression f(x) = x 2 + 2, it’s x 2. Solving Quadratic Equations with Square Roots 2. Find the range and domain of a quadratic function without graphing it. Identify situations that can be modeled by quadratic functions Identify the pattern of change between two variables that represent a quadratic function in a situation, table, graph, or equation Reported Hype on Quadratic Formula Calculator Discovered . 7 Quadratic Word Problems Page 404-407 #12, 14, 16, 17, 18 Day 1: Wed Mar 2 Day 2: Thu Mar 3 8-9 4. Both partners start with their "START" cut-out. 8: Ex 31, 37, 45, 55, 57, 79 In some financial math problems, several key points on a quadratic function are and the maximum profit (and the number of sales required for this profit), they Review problems: p284 #17,21,25,37,43,49 Section 4. In problems 4-8 use the given information to produce a quadratic function in a form that is best suited to the problem. Chapter 3 Quadratic Function Numeric problem. Using past receipts, we find that the profit can be modeled by the function: p= -15x 2 +600x +60. USE THE PERFECT SQUARE FORMULA . ALGEBRA UNIT 11-GRAPHING QUADRATICS. For problems 1-8, given the equations of the cost and demand price function: Identify the fixed and variable costs. Given the algebraic equation for a quadratic function, one can calculate any point on the function, including critical values like minimum/ maximum and x- and y-intercepts. Calculate the best price for a widget – the one that yields the most profit. 1. Type the coefficients of the quadratic equation, and the solver will give you the roots, the y-intercept, the coordinates of the vertex showing all the work and it will plot the function. Here you can get a visual of your quadratic function Much as we did in the application problems above, we also need to find intercepts of quadratic equations for graphing parabolas. 2. The different steps are shown including converting quadratic equations into calculator ready graphable quadratic functions. A movie theater holds 1000 people. The two resistors are 3 ohms and 6 ohms. Now let's use quadratic functions to model real things and solve problems. 40 x – $50. Apr 28, 2018 · Here is a set of practice problems to accompany the Quadratic Equations - Part I section of the Solving Equations and Inequalities chapter of the notes for Paul Dawkins Algebra course at Lamar University. Last month, the chain sold a total of 600 of these MP3 players worldwide. — Write an equation in standard form for the parabola that passes through the given points. 2x3 +128y Solve the following Quadratic Word Problems Name_____ Date_____ ©T t2^0r1^4Q wKCuYtcaI XSdoYfKt^wkaprRen ]LULxCr. Quadratic Equations : In this section ask-math provides you the quadratic equations with real coefficients but Quadratic objective term, specified as a symmetric real matrix. 6 1. Several examples are included to demonstrate the flexibility In algebra, a quadratic equation (from the Latin quadratus for "square") is any equation that can be rearranged in standard form as ax²+bx+c=0 where x represents an unknown, and a, b, and c represent known numbers, where a ≠ 0. Evaluate cost, demand price, revenue, and profit at $$q_0\text{. Projectile motion describes the path that objects, like rockets, take when thrown or launched up into the air. Substituting in the quadratic formula, Since the discriminant b 2 – 4 ac is 0, the equation has one root. The monthly profit generated by renting out x units of the apartment is given by P(x)=-10x²+1760x-50000 . Quadratic function has the form f(x) = ax^2 + bx + c where a, b and c are numbers. Problem #3: The quadratic equation for the cost in dollars of producing automobile tires is given below where x is the number of tires the company produces. I will explain these steps in following examples. by bchavez_96347. The minimum or maximum value of a quadratic function can be used to determine the range of the function and to solve many kinds of real-world problems, including problems involving area and revenue. Feb 05, 2020 · A more general equation for a parabola is a quadratic function: y = ax 2 + bx + c Where a changes the width of the curve, a and b shift the axis of symmetry to the left or right, and c slides the 3. I can find the two times when the object is exactly 34. Solution: The standard form of a quadratic More Word Problems Using Quadratic Equations Example 3 The length of a car's skid mark in feet as a function of the car's speed in miles per hour is given by l(s) = . I haven't done this shit in like 5 years and I need to quickly relearn it and google isn't helping me that much. P. 1. 1, 01-Relations vs functions note teacher. 199s + 0. With the ticket price at 8 during the week, the attendance at the theater has been 200 people. 🔗 6 Feb 2020 The equations also show up in calculations for maximizing profit, a key A more general equation for a parabola is a quadratic function: and it only works neatly for problems that are contrived to have integer answer Formulate quadratic functions in a problem-solving situation. The quadratic formula can be used to find roots much more easily and it can be used to find both real and complex roots. The squaring function f (x) = x 2 is a quadratic function whose graph follows. Using past receipts, the profit can be modeled by the function \(p=-15{{x}^{2}}+600x+60$$, where $$x$$ is the price of each ticket. Quadratic Functions p. For a quadratic inequality in standard form, the critical numbers are the roots. We can do this by setting profit equal to zero and solving for . 2 P is the profit in thousands of dollars and n is. See full list on algebra-class. passenger. The process of outlining and setting up the problem is the same as taught in chapter 5, but with problems solved by quadratics you must be very careful to check the solutions in the problem itself. For the function, f(x) = 4(x – 5)(x + 2), find the vertex, roots, axis of symmetry, and the y-intercept. Results 1 - 17 of 17 Max/Min values of a quadratic function Applications and Solving Word Problems involving Quadratic Functions (Area and Revenue 5 Nov 2013 In the past, the profit the club made after paying for the band and other costs has been modeled by the function , where t represents the ticket Then the revenue R derived from selling x calculators at the price p per calcula- Problems. Khan Academy is a 501(c)(3) nonprofit organization. 21. 7. Nature of the roots of a quadratic equations. docx. Write them separated by commas in the answer box. (. Show Step-by-step Solutions Applications Of The Quadratic Equations. Take a tour. The cost of manufacturing a sack of food is represented by the function C(x) = 0. Aug 15, 2020 · One can solve quadratic equations through the method of factorising, but sometimes, we cannot accurately factorise, like when the roots are complicated. Key Strategy in Solving Quadratic Equations using the Square Root Method. Using past receipts, we find that the profit can be modeled by the function p= -15x 2 +600x +60 , where x is the price of each ticket. THE GRAPH OF A QUADRATIC FUNCTION (DAY 1). The general approach is to collect all {x^2} terms on one side of the equation while keeping the constants to the opposite side. Apply quadratic functions to real world situations in order to solve problems. 2 Building Linear Functions from Data In this section linear functions are constructed from data presented in various ways. When a problem can be modelled by a quadratic function, the domain and range of accountants modelled the company's profit using the equation. Solving LESSON 3: Profit Maximization Problems Workshop: Multiple MethodsLESSON 4: Multiple Methods to Solve Problems with Quadratic FunctionsLESSON 5: More Multiple Methods to Solve Problems involving Quadratic FunctionsLESSON 6: 4-Column Quadratic Data TablesLESSON 7: More 4-Column Data TablesLESSON 8: Applying Data Tables to Word ProblemsLESSON 9: Quadratic Word Problems Handout Day 1: Thu Feb 24 Day 2: Fri Feb 25 4-5 4. 1 and 2. 26 Jun 2019 Quadratic Function Word Problems Exercise 1From the graph of the function f(x) = x², graph the following translations: 1. Do I have to use exact chord when playing a song, How to tell the difference between groß = tall or big. We shall soon see how the humble quadratic makes its appearance in many different and important applications. Example 1: Sketch the graph of the quadratic function $${\color{blue}{ f(x) = x^2+2x-3 }}$$ Solution: Solve problem 56 on pages 666-667 of Elementary and Intermediate Algebra. As review, we will look at the definition of a quadratic function. If you missed this problem, review (Figure). The graph of the related function, y = -50x. Big Idea Now that students have solved different problems using a preferred method, they can apply their understanding to try and learn from different solution methods. You will also graph quadratic functions and other parabolas and interpret key features of the graphs. Video IV Courtesy FlexiGuru Quadratic Word Problems: Projectile Motion Science and Mathematics teachers just love to ask questions about things flying through the air. Quadratic Profit Function Old Bib Real Estate has a 100 unit apartment and plans to rent out the apartment. You'll find that the profit is higher when b = 2, and the maximal profit then is P (2) = 2. With the ticket price at$8 during the (a) Find the price-demand equation, assuming that it is linear. (c) Find the number of items sold that will give the maximum revenue. *A GOOGLE Slides version is now included in the download. Sum and product of the roots of a quadratic equations Algebraic identities Quadratic Functions. Definition: A degree 2 polynomial function is called a quadratic function. Khan Academy is a 501(c)(3) nonprofit organization. 1. Quadratic functions are used to model phenomena such as paths of projectiles, and profits to illustrate how mathematical modelling is used to make informed of how multiple representations connect and are useful for solving problem Students will explore how to graph quadratic application problems and how to BUILD quadratic functions using quadratic regression. The following is a way of solving rational inequalities. Quadratic Functions Definition: If a, b, c, h, and kare real numbers with a6= 0, then the functions y= ax2 +bx+c standard form y= a(x−h)2 +k vertex form both represent a quadratic function. HIDE SOLUTION. Interesting word problems involving quadratic equations. Your factory produces lemon-scented widgets. 1. docx. 5A Quadratic Functions and Complex Describe two representations you could use to solve each problem. 1. shares? How much will the profit be on. Quadratic functions are incredibly important functions that show up everywhere in the real world. Standard Form of a Quadratic Function Practice & Problem Solving Find the vertex and y-intercept of the quadratic function, and use them to graph the function. 3. The search for a piecewise quadratic Lyapunov function is formulated as a convex optimization problem in terms of linear matrix inequalities. 1000 shares after 30 days, what is the. Dec 04, 2012 · p(x)=-30x^2+550x-400 p(x) represents profit and x represents ticket costs. Problem 1. Domain and range. View 16D_Quadratic_Functions_word_problems_with_answers. The coefficients a,b and c influence the shape, form and position of the graph of the associated parabola. From this literature review, it is clear that there is a need for further research into the sources of students’ difficulties with quadratic equations. 3. The graph of a quadratic function is a parabola. 359 . 191 Embedded Assessment 3: Graphing Quadratic Functions and Solving Systems p. 980 units. SHOW SOLUTION . pdf 3. The quadratic function f(x) = a x 2 + b x + c can be written in vertex form as follows: f(x) = a (x - h) 2 + k Quadratic Functions Problems with Solutions Quadratic equation Page 5/26 7. But as you saw above, the graph of a quadratic function is curved. The publisher of an medical newsletter estimates that with x thousand subscribers its monthly revenue and cost (in Jun 11, 2013 · Modelling problems with quadratic functions, help? A retail electronics chain makes a profit of $14 on every MP3 player of a certain brand the chain sells. Nov. The maximum profit of a quadratic profit function is -2 which occurs at Q 3. The one above is a Quadratic Formula partner scavenger hunt. 8, Table 1) are called parabolas. x and y intercepts of a Parabola Lesson and Equations and Homework Tues. Completing The Square 1. 1. 1x2 + 1. In this second part we continue our journey. Def: A quadratic function can be expressed in the standard form. The third problem is based on using a quadratic function to model profit and loss. maximizing revenue word problems involving quadratic equations Problem 1 : A company has determined that if the price of an item is$40, then 150 will be demanded by consumers. 02-Domain and Range teacher note. In order for us to be able to apply the square root property to solve a quadratic equation, we cannot have May 02, 2018 · Here is a set of practice problems to accompany the Complex Numbers< section of the Preliminaries chapter of the notes for Paul Dawkins Algebra course at Lamar University. Throughout this lesson, students make sense of problems by analyzing the given PROFIT: The amount of money a business makes on the sale of its product. Find the equation Graph the function. e. Don't try to figure out where they got it from. Quadratic functions Quadratic functions and parabolas Graphs of y against x resulting from quadratic functions ( 2. 2, recall the formula for the revenue from sales of an item: This equation is an example of the vertex form for a quadratic function. Question 1. Properties of Parabolas Worksheet – Oct. The Quadratic Equation is written as: which quadratic functions can be used to model motion involving projectiles, and certain kinds of problem involving a single maximum or minimum. 4. All the solutions of Solving (simple) Problems (Based on Quadratic Equations) - Mathematics explained in detail by experts to help students prepare for their ICSE exams. a. You will write the equations of quadratic functions to model situations. The quadratic 5. 223 Unit Overview This unit focuses on quadratic functions and equations. Since this is a maximum point, the x-coordinate gives the number of price increases needed to maximize the profit. The profit function for a computer company is given by where x is the number of units produced (in 4. The most popular way to solve quadratic equations is to use a quadratic formula. If you want to understand the physics of the flight path of the ball then you will come up with a parabolic curve. 4. In parabola, if the coefficient of x^2 is negative, the vertex is the maximum point. Written in standard form, the equation y = ax 2 + bx + c (a 0) represents quadratic functions. You can find the answer by using quadratic functions. In this video lesson, we will talk about how quadratic functions, the function of a degree of 2, are used in the real world to model real-world scenarios. Sep 16, 2020 · Students can download the Quadratic Equations Class 10 MCQs Questions with Answers from here and test their problem-solving skills. The Quartic Formula is only the final result of this methodology, written in relation to the original coefficients. Linear vs Quadratic functions. Dec 21, 2020 · It is important to note that this quadratic inequality is in standard form, with zero on one side of the inequality. 5 Quadratic Application Word Problems 1. 2. f(x) = y = ax² + bx + c Problems 0. The quadratic formula is a formula that is used to solve quadratic equations: To use the quadratic formula, we follow these steps: Get the quadratic equation in the form ax 2 + bx + c = 0. His height as a function of time could be modeled by the function h t t t( ) 16 16 480 2, where t is the time in seconds and h is the height in feet. Jun 11, 2013 · Modelling problems with quadratic functions, help? A retail electronics chain makes a profit of $14 on every MP3 player of a certain brand the chain sells. Dec 22, 2018 · The profit function is just the revenue function minus the cost function. A chain store manager has been told by the main office that daily profit, P, is related to the number of clerks working that day, x, according to the equations P = -25x^2 + 300x. 5. 2. 4 Quadratic Word Problems DRAFT. 2 Graph Quadratic Functions in Vertex or Intercept Form So you can model sports revenue, as in Example 5. Profit = R (x) - C (x) set profit = 0 Solve using the quadratic formula where a = 195, b = 20, and c =. Parabolas may open upward or downward and vary in "width" or "steepness", but they all have the same basic "U" shape. Chapter 5 of Class 11 Complex Numbers and Quadratic Equations has 3 exercises and a miscellaneous exercise to help the students in practicing the required number of problems to understand all the concepts. This function is extremely useful, it can tell us, for example, how many glasses of lemonade we would need to sell to Chapter 13 . The solution for which Q(y 1,y 2)isminimumisnolonger (y 1,y 2)=(0,0), but instead, (y 1,y 2)=(2,−1), as will be shown later. Clear all the fundamentals and prepare thoroughly for the exam taking help from Class 10 Maths Chapter 4 Quadratic Equations Objective Questions. Last month, the chain sold a total of 600 of these MP3 players worldwide. 3K Quadratic Function App: Find Profit from Revenue and Cost (Vertex). Example: Notice that the result is a quadratic equation. y = x² + 2 2. Feb 17, 2011 · I have no idea of how to answer this question. Solving Quadratic Equations By Factoring. your 1000 shares purchased in (a)? Quadratic Equations: Difficult Problems with Solutions. Recall that the x-coordinate of the maximum point One way for solving quadratic equations is the factoring method, where we transform the quadratic equation into a product of 2 or more polynomials. 31. 2. + 100x + 6000 price-demand function is linear, then the revenue function will be a quadratic In Problems 1-6 find the vertex, the maximum value, the minimum value, and the Example: New Sports Bike. The quadratic formula can also be used to solve quadratic equations whose roots are imaginary numbers, that is, they have no solution in the real number system. If you buy. Remember that a In many quadratic max/min problems, you'll be given the formula you need to use. Review problems: p290 #3,5,7,15,19,21 Section 4. This example is chosen deliberately as being more abstract to let students see Finding price that optimizes profit and for breakeven for a profit formula that is quadratic. Oct 17, 2014 · You will be able to formulate real-life problems involving quadratic functions, and solve them through a variety of techniques with accuracy. Jason jumped off of a cliff into the ocean in Acapulco while vacationing with some friends. If a = 0, then the equation is linear, not quadratic, as there is no ax² term. 80) * x] is strictly a short term forecasting tool. 223 Unit Overview This unit focuses on quadratic functions and equations. $$\bullet$$ Find the price to charge per item in order to maximize You’ll remember that the graph of a linear function is always a straight line. 4. Pairs of students get 2 clocks and 2 unique sets of cut-outs. •. The sum S of n successive odd numbers starting from 3 is given by the relation n(n + 2). For our simple lemonade stand, the profit function would be. The bus has seating capacity for 22 people. O. Factoring and Solving Quadratic Equations Worksheet Math Tutorial Lab Special Topic Example Problems Factor completely. So its graph is a parabola. NCERT Solutions for Class 10 Maths Chapter 4- Quadratic Equations. Quadratic functions help forecast business profit and loss, plot the course of Using quadratic functions to solve problems on maximizing revenue/profit. docx 2. docx 03-Practice problems. 5t2 + 8. -1-1) A fireworks rocket is launched from a hill above a lake. We know that a quadratic equation will be in the form: There could be many different traits of question which can even include all the linear equations type questions into the quadratic form. These take the general form: y = ax2 +bx+c. Algebra is a branch of math in which letters and symbols are used to represent numbers and quantities in formulas and equations. A quadratic equation is a polynomial whose highest power is the square of a variable (x 2, y 2 etc. 6 days ago. Since the number of blocks must be a whole number to make the problem realistic, test integer values on either side of 7 4, i. cannot be negative. 1. A quadratic programming (QP) problem has a quadratic cost function and linear constraints. pdf 3. Transformations with a, h, k – All Lessons from Thurs. The relation to frequency domain methods such as the circle and Popov criteria is explained. Review problems: p284 #17,21,25,37,43,49 Section 4. Previously we very briefly looked at the function f\left(x\right) quadratic function in real world problems. Click here to see ALL problems on Quadratic Equations Question 23275 : Maximum profit using the quadratic equations, functions, inequalities and their graphs. P 5 25x2. 5: Quadratic Applications Objective: Solve quadratic application problems. (b) Find the revenue function. Feb 17, 2011 · I have no idea of how to answer this question. It even has a specific name: a Parabola. 5: Quadratic Applications Page 229 Section 4. when 275 units sold, we can get the maximum revenue. A quadratic equation has two roots if its graph has two x-intercepts; A quadratic equation has one root it its graph has one x-intercept; A quadratic equation has no real solutions if its graph has no x-intercepts. 125x3 64 15. , 2010). 223 Unit Overview This unit focuses on quadratic functions and equations. Using past receipts, we find that the profit can be modeled by the function: p= -15x 2 +600x +60. Quadratic Equations: Difficult Problems with Solutions. 22 May 2017 Writing a quadratic function to model the revenue of a word problem and using it to determine the price of a product that with maximize the 15 Jun 2016 Quadratic Word Problem About Maximizing Profit. Nov. In fact for other integer values of b, you'll find the "profit" is negative, i. profit or minimum cost. Selina Concise Mathematics - Part II Solutions for Class 10 Mathematics ICSE, 6 Solving (simple) Problems (Based on Quadratic Equations). pdf from MATH PRECALCULU at Louis D Brandeis High School. 2x + 32. These calculations can be more tedious than is necessary, however. This will require the quadratic formula: This simplifies to or , but since the problem specifies that the Intro to functions 1. 264 If the length of skid mark is 220 ft, find the speed in miles per hour the car was traveling. Where we begin It all started at a meeting of the National Union of Teachers. The publisher of an medical newsletter estimates that with x thousand subscribers its monthly revenue 25 Jun 2003 customers. Solve the following equations using the quadratic formula. docx 03-Practice problems. a); Function P that gives the profit is a quadratic function with the leading coefficient a = - 5. Keywords: quadratic function ( maximum, minimum). 10 x) =$0. Shows you the step-by-step solutions using the quadratic formula! This calculator will solve your problems. ) Find the multipart function s(t) giving. How long did it take for Jason to reach his maximum height In this section, we will explore quadratic functions, a type of polynomial function. This function (profit) has a maximum value at x Quadratic Maximum Profit Problem Simply either graph the function to get the vertex, Problem Set. docx. This formula is: -b ±√b 2 – 4ac/2a. You can sketch quadratic function in 4 steps. 6 Quadratic Word Problems Page 391-393 #11, 14, 15, 18, 20 Day 3: Mon Feb8 Day 4: Tue Mar 1 6-7 4. The profits of Mr. e. Mar 01, 2004 · March 2004 It isn't often that a mathematical equation makes the national press, far less popular radio, or most astonishingly of all, is the subject of a debate in the UK parliament. 124 10. I'm assuming I need to find the vertex for the max profit and the zeros for the break even but im not quite sure and want a Quadratic Functions: Explore problem situations in which two variables are in a quadratic relationship. To find the Optimum (Maximum or Minimum) of a function, Perform the following steps 1. doc. The assemblage of printable algebra worksheets encompasses topics like translating phrases, evaluating and simplifying algebraic expressions, solving equations, graphing linear and quadratic equations, comprehending linear and quadratic functions, inequalities May 02, 2018 · Here is a set of practice problems to accompany the Complex Numbers< section of the Preliminaries chapter of the notes for Paul Dawkins Algebra course at Lamar University. the value means in the problem and how this may be possible. Throw a ball at someone and the ball traces a parabolic curve. Total Profit = (profit per scarf)(number of scarves sold) Step 2: Find the max value of the function by either completing the square or by using partial factoring. docx. 04x + 38 . H represents the quadratic in the expression 1/2*x'*H*x + f'*x. The parabola can either be in "legs up" or "legs down" orientation. Problem #3: The quadratic equation for the cost in dollars of producing automobile tires is given below where x is the number of tires the company produces. Function notation. The breakeven point occurs where profit is zero or when revenue equals cost. Solution to Problem 1. We begin with the following simple fact: Proposition 14. Linear Inequations : 13x <250 such equations are called linear inequations. The marketing department suspects that for every \$1 decrease in price, the chain would sell an additional 100 players. Writing a quadratic function to model the revenue of a word problem and using it to determine the price of a product that with maximize the revenue. 02-Practice problems. 7 Quadratic Word Problems Page 404-407 # 19, 24-28 Day 3: Fri Mar 4 May 17, 2011 · (Most "text book" math is the wrong way round - it gives you the function first and asks you to plug values into that function. Dec 04, 2012 · p(x)=-30x^2+550x-400 p(x) represents profit and x represents ticket costs. 456 CHAPTER 12. Writing a quadratic function to model the revenue of a word problem and using it SWBAT apply the concept of a quadratic function to 4 classic application problems: projectile motion, revenue maximization, area, and number puzzles. If A is an Quadratic Equations: Problems with Solutions. cost? b)To maximize profit, when should you sell. #2) What are the advantages of a quadratic function in vertex form? How can quadratic functions maximize profits or minimize cost? Problem 1. A quadratic function plots a parabola. a. Introduction to Quadratic Functions 1 What to Know Let us start this lesson by recalling ways of representing a linear function. 1. quadratic function profit problems | 2021-04-14T10:59:09 | {
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https://willrosenbaum.com/teaching/2021f-cosc-211/notes/asymptotic-analysis/ | ## Motivation
In this note, we describe a method of measuring the efficiency of a procedure using asymptotic analysis. We formally define “big O” notation for qualitatively comparing the growth of functions (that will represent the running times of our programs). This view abstracts away details of hardware executing the program, and focuses on how the time complexity (e.g., running time) scales with the size of the input. The approach disregards details of executions—essentially ignoring constant factors that affect the running time—but allows us to reason about efficiency in a way that is (almost) independent of the actual hardware executing a procedure.
## Big O Notation
Throughout this section, we will use $$f, g$$ and $$h$$ to refer to functions from the natural numbers $$\mathbf{N} = \{0, 1, 2, \ldots, \}$$ to the positive real numbers $$\mathbf{R}^+$$. The interpretation is that $$f(n)$$ might be the running time of some method with an input of size $$n$$.
### Formal Definition
Definition. Let $$f, g : \mathbf{N} \to \mathbf{R}^+$$ bef functions from the natural numbers to the positive real numbers. Then we write $$f = O(g)$$ (pronounced ”$$f$$ is (big) oh of $$g$$“) if there exists a natural number $$N$$ and positive constant $$C > 0$$ such that for all $$n \geq N$$, we have
$f(n) \leq C \cdot g(n).$
Informally, this definition says that $$f = O(g)$$ means that for sufficiently large values of $$n$$ (i.e., $$n \geq N$$), $$f(n)$$ is no more than a constant factor ($$C$$) larger than $$g(n)$$.
Arguing directly from the definition, in order to show that functions $$f$$ and $$g$$ satisfy $$f = O(g)$$, we must find values $$N$$ and $$C$$ such that $$f(n) \leq C \cdot g(n)$$ whenever $$n \geq N$$. In the following section, we will derive properties of $$O$$ that allow us to rigorously justify our calculations without needing to refer directly to the definition above.
Example. Consider the functions $$f(n) = 10 n^3 + 7$$ and $$g(n) = n^3$$. Notice that $$f(n) \geq g(n)$$ for all (non-negative) values of $$n$$. Thus from the definition of $$O$$, we have $$g = O(f)$$. Indeed, taking $$N = 0$$ and $$C = 1$$, the definition is satisfied.
On the other hand, we claim athat $$f = O(g)$$ as well. To prove this directly from the definition, we must find suitable values $$N$$ and $$C$$ satisfying the definition above. Since $$f(n) \geq 10 n^3 = 10 \cdot g(n)$$, we must use some $$C > 10$$ (otherwise $$f(n) \leq C \cdot g(n)$$ will not be satisfied). Consider taking $$C = 11$$. We can write
\begin{align*} 11 g(n) &= 11 n^3\\ &= 10 n^3 + n^3\\ &\geq 10 n^3 + 2^3 (\text{ if } n \geq 2)\\ &= 10 n^3 + 8\\ &> 10 n^3 + 7\\ &= f(n). \end{align*}
Thus, so long as $$n \geq 2$$, we have $$f(n) \leq 11 \cdot g(n)$$. Therefore, $$f = O(g)$$, where the definition is satisfied for $$N = 2$$ and $$C = 11$$.
The computations above are rather ad-hoc. In general, there will be many possible values of $$N$$ and $$C$$ for which $$f$$ and $$g$$ can be show to satisfy the definition. Below, we will describe general properties from which $$O$$-relationships can be determined rigorously without devolving to ad-hoc algebra.
Abuse of Notation. We often use the notation $$O(g)$$ to refer to “some function $$h$$ that satisfies $$h = O(g)$$.” In the example above, we showed that $$f(n) = 10 n^3 + 7$$ satsifies $$f = O(n^3)$$. If we were considering the function $$h(n) = 2 n^4 + 3 n^3 + 7 = 2 n^4 + f(n)$$, we may just as well write $$h(n) = 2 n^4 + O(n^3)$$. This shorthand will prove convenient when we don’t want to write out explicit terms of functions whose precise values are unknown or will be subsumed by an application of $$O$$ later on.
### Properties
Here, we prove some useful properties of big O notation.
Proposition 1. Suppose $$f, g, f_1, f_2, g_1, g_2, h$$ are all functions from $$\mathbf{N}$$ to $$\mathbf{R}^+$$, and that $$a > 0$$ is a constant in $$\mathbf{R}^+$$. The the following hold:
1. If $$f(n) \leq a$$ for all $$n$$, then $$f = O(1)$$.
2. If $$f(n) \leq g(n)$$ for all $$n$$, then $$f = O(g)$$.
3. If $$f = O(g)$$, then $$a \cdot f = O(g)$$.
4. If $$f = O(g)$$ and $$g = O(h)$$, then $$f = O(h)$$.
5. If $$f = O(g)$$, then $$f + O(g) = O(g)$$ and $$g + O(f) = O(g)$$,
• in particular $$f + g = O(g)$$
6. If $$f_1 = O(g_1)$$ and $$f_2 = O(g_2)$$, then $$f_1 \cdot f_2 = O(g_1 \cdot g_2)$$.
Proof. We prove each assertion above in turn.
1. Since $$f(n) \leq a$$ for all $$n$$, taking $$g(n) = 1$$, we have $$f(n) \leq a \cdot g(n)$$ for all $$n$$. Thus, the definition of $$f = O(g)$$ is satisfied with $$N = 0$$ and $$C = a$$, so that $$f = O(1)$$.
2. If $$f(n) \leq g(n)$$ for all $$n$$, then the definition of $$f = O(g)$$ is satisfied with $$N = 0$$ and $$C = 1$$.
3. Suppose $$f = O(g)$$, and suppose $$N'$$ and $$C'$$ are the values for which the definition of $$O$$ is satisfied. That is, for all $$n \geq N'$$, we have $$f(n) \leq C' \cdot g(n)$$. Then for $$n \geq N$$, we also have $$a \cdot f(n) \leq a C' \cdot g(n)$$. Therefore, the definition of $$a \cdot f = O(g)$$ is satisfies for $$N = N'$$ and $$C = a \cdot C'$$.
4. Suppose the definition $$f = O(g)$$ is satisfied with values $$N_f$$ and $$C_f$$. That is, for $$n \geq N_f$$, we have $$f(n) \leq C_f \cdot g(n)$$. Similarly, suppose $$g = O(h)$$ with $$N_g$$ and $$C_g$$: for $$n \geq N_g$$ we have $$g(n) \leq C_g \cdot h(n)$$. Then, for $$n \geq \max(N_f, N_g)$$, we have both $$f(n) \leq C_f g(n)$$ and $$g(n) \leq C_g \cdot h(n)$$. Combining the last two inequalities, we obtain $$f(n) \leq C_f \cdot (C_g \cdot h(n)) = C_f C_g \cdot h(n)$$. Therefore, the definition of $$f = O(h)$$ is satisfied for $$N = \max(N_f, N_g)$$ and $$C = C_f \cdot C_g$$.
5. Suppose $$f = O(g)$$ and $$h$$ is any function satisfying $$h = O(g)$$. Suppose the definition of $$f = O(g)$$ is satisfied for $$N_f$$ and $$C_f$$—i.e., $$f(n) \leq C_f \cdot g(n)$$ for all $$n \geq N_f$$. Similarly suppose the definition of $$h = O(g)$$ is satisfied for values $$N_h$$ and $$C_h$$: $$h(n) \leq C_h \cdot g(n)$$ for all $$n \geq N_h$$. Observe that taking $$N = \max(N_f, N_h)$$, we have that for all $$n \geq N$$, both $$f(n) \leq C_f \cdot g(n)$$ and $$h(n) \leq C_h \cdot g(n)$$. Thus, for $$n \geq N$$, we have $$f(n) + h(n) \leq C_f \cdot g(n) + C_h \cdot g(n) = (C_f + C_h) g(n)$$. Therefore, the definition of $$f + h = O(g)$$ is satisfied for $$N = \max(N_f, N_h)$$ and $$C = C_f + C_h$$.
Now suppose $$f = O(g)$$, and $$h = O(f)$$. Then by property 4 above, we have $$h = O(g)$$. Therefore, $$g + h = g + O(g)$$. Applying the first assertion of 5 (proven in the paragraph above), we get $$g + O(g) = O(g)$$, so that $$g + O(f) = g + O(g) = O(g)$$, as claimed.
6. Suppose $$f_1 = O(g_1)$$ is satisfied for $$N_1$$ and $$C_1$$, and that $$f_2 = O(g_2)$$ is satisfied for $$N_2$$ and $$C_2$$. Then for $$N = \max(N_1, N_2)$$, and all $$n \geq N$$, we have $$f_1(n) \leq C_1 \cdot g_1(n)$$, and $$f_2(n) \leq C_2 \cdot g_2(n)$$. Therefore, $$(f_1 \cdot f_2)(n) = f_1(n) \cdot f_2(n) \leq (C_1 g_1(n)) (C_2 g_2(n)) = (C_1 C_2) \cdot (g_1 \cdot g_2)(n)$$. Therefore, $$f_1 \cdot f_2 = O(g_1 \cdot g_2)$$ is satisfied for $$N = \max(N_1, N_2)$$ and $$C = C_1 \cdot C_2$$.
So all of the properties hold, as desired. $$\Box$$
Using the properties above, we can more simply (yet just as rigorously) argue about big O notation. For example, Property 2 above also gives the following useful consequence:
Corollary. Suppose $$a, b$$ are constants with $$a \leq b$$. Then $$n^a = O(n^b)$$.
Example. Suppose $$f$$ is a second degree polynomial. That is, $$f(n) = a n^2 + b n + c$$ for some constants $$a, b$$ and $$c$$, where $$a > 0$$. Then $$f = O(n^2)$$. To see this, we compute:
\begin{align*} f(n) &= a n^2 + b n + c\\ &= a n^2 + b n + O(1)\\ &= a n^2 + O(n)\\ &= O(n^2). \end{align*}
Each manipulation above is justified as follows
• The first equality is the definition of $$f$$.
• The second equality holds by property 1.
• The third equality holds by property 3 (which implies that $$b n = O(n)$$), and property 5.
• The fourth equality holds by property 3 (which implies that $$a n^2 = O(n^2)$$, and property 5.
More generally, we can argue that any degree $$k$$ polynomial—i.e., a function $$f$$ of the form $$f(n) = a_k n^k + a_{k-1} n^{k-1} + \cdots + a_1 n + a_0$$ satisfies $$f = O(n^k)$$. Proving this fact in a mathematically rigorous way requires applying mathematical induction, but you can freely use this fact going forward.
### When is $$f \neq O(g)$$?
The notation $$f = O(g)$$ is in some sense a weak condition on $$f$$ and $$g$$. That is, $$f$$ could be much, much larger than $$g$$, yet we we still have $$f = O(g)$$. For example, take $$f = 10^{100} n^2$$ and $$g = 10^{-100} n^2$$. You should convince yourself that we we have $$f = O(g)$$. But $$g$$ is always much smaller than $$f$$: $$g(n) / f(n) = 10^{-200}$$ for all $$n$$, which is a very tiny number indeed! So you might (rightfully) be concerned that $$f = O(g)$$ is too weak of a condition to be useful. For example, is it ever the case that $$f$$ is not $$O(g)$$?
Proposition 2. Suppose $$a$$ and $$b$$ are real values satisfying $$a < b$$. Then $$n^b \neq O(n^a)$$. That is, $$n^b$$ is not $$O(n^a)$$.
Given a proposition such as Proposition 2—a claim that a particular definition is not satisfied—it is common to apply a technique called “proof by contradiction”. That is, in order to prove that $$n^b \neq O(n^a)$$, we assume that the opposite is true—namely $$n^b = O(n^a)$$—and derive a contradiction from this assumption.
Proof. Suppose for the sake of contradiction that $$n^b = O(n^a)$$. Then—from the definition of $$O$$—there exists a natural number $$N$$ and constant $$C$$ such that for all $$n \geq N$$, we have $$n^b \leq C \cdot n^a$$. Dividing both sides of this expression by $$n^a$$ we get the equivalent expression $$n^b / n^a = n^{b - a} \leq C$$. Since $$b - a > 0$$, we can take raise both sizes of this expression to the power $$1 / (b - a)$$ to get
$n = (n^{b - a})^{1 / (b - a)} \leq C^{1 / (b - a)}.$
Thus, for all $$n > C^{1 / (b - a)}$$, the inequality $$n^b \leq C \cdot n^a$$ fails to hold. In particular, the expression $$n^b \leq C \cdot n^a$$ fails to hold for some $$n \geq N$$. Therefore, $$f \neq O(g)$$, as desired. $$\Box$$
## Analysis of Code
In this section, we apply big O notation in order to describe the running times of procedures. In order to deduce anything about how long it will take a procedure to complete on a given input, we must formalize our assumptions about the running times of certain primitive operations. The actual running times of these primitive operations may vary drastically between different machines executing the same code. Big O notation abstracts away from the particular running times of primitive operations while allowing us to reason formally about how the running time of the procedure scales with the size of its input.
### Assumptions
In order to apply Big O notation to express the running time of code, we make the following assumptions on the running times of operations in Java. We assume that the following operations are performed in $$O(1)$$ time:
1. reading, writing, creating, and modifying variables that store primitive data types,
• the primitive data types in Java are byte, short, int, long, float, double, boolean, char
2. performing an arithmetic or logical operation on a primitive data type,
3. making method calls and executing branching operations (i.e., if-then-else),
4. assigning and modifying references,
5. reading/writing a value from/to a particular index of an array.
The following operations’ running times scale linearly with the size (i.e., number of primitive elements) of the data structure. That is, if the object stores/represents $$n$$ primitive data types, then the running time of the following operations is $$O(n)$$:
1. initializing arrays and Strings (of length/capacity $$n$$),
2. creating/initializing new object instances.
Note. In Java, a String is represented internally as an array of chars. However, Strings are immutable: once a String is created it’s value cannot be modified. Instead, all String modification operations create a new String object with the desired contents. For example, given a char[] chArray of length $$n$$ and a String str of length $$n$$, reassigning, say chArray[0] = 'a' takes time $$O(1)$$, while making a String str2 whose contents is the same as str, except that the first letter is changed to 'a' takes time $$O(n)$$.
### Examples
We now apply our running time analysis and big O notation to describe the qualitative running time of some actual code.
Example. Consider the following method, which takes as its parameter an array of ints, and returns the sum of the values stored in the array.
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int sumContents (int[] values) {
int size = values.length;
int sum = 0;
for (int i = 0; i < size; ++i) {
sum += values[i];
}
return sum;
}
We let $$n$$ denote the length of the array—i.e., the value returned by be values.length. The assignments in lines 2 and 3 both run in time $$O(1)$$ (note that values.length reads and returns the value of an instance variable length). Each iteration of the for loop in lines 4–6 requires time $$O(1)$$: there is $$O(1)$$ overhead per iteration for modifying the value of i and checking the condition i < size, and the expression sum += values[i] runs in time $$O(1)$$. Since there are $$n$$ iterations of the for loop, the overall running time is
$O(1) + n \cdot O(1) = O(n).$
Example (adding to ArraySimpleList). In our ArraySimpleList implementation of SimpleList, we had the following add(i, x) method:
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public void add(int i, E x) {
// i is a valid index if it is between 0 and size
if (i > size || i < 0) {
throw new IndexOutOfBoundsException();
}
// check if we need to increase the capacity before inserting
// the element
if (size == capacity) {
increaseCapacity();
}
++size;
// insert x by setting contents[i] to x and moving each
// element previously at index j >= i to index j + 1.
Object cur = x;
for (int j = i; j < size; ++j) {
Object next = contents[j];
contents[j] = cur;
cur = next;
}
}
For simplicity, let us assume that (1) the index i is valid, so that the exception is not thrown in line 4, and (2) the capacity of the contents array is strictly larger than size so that increaseCapacity() is not called in line 10. (We will consider the increaseCapacity() method in a forthcoming lecture on amortized analysis.) With these assumptions, the operations performed in lines 3–17 each take time $$O(1)$$. Note that the expression Object cur = x at line 17 does not create a new object, but simply stores a reference as the variable cur—thus, the operation only takes $O(1)$ time. Each iteration of the for loop in lines 18–22 runs in time $$O(1)$$, as the loop perofrms $$O(1)$$ primitive operations. If the list has size $$n$$, there are $$n - i$$ total iterations performed. Therefore, the total running time is
$O(1) + (n - i) \cdot O(1) = O(n - i + 1).$
Note that we could have bounded $$O(n - i + 1) = O(n)$$. While this is true, the expression above is more precise. Specifically, it shows that adding to the end of the list may be more efficient than adding to the front. For example, adding at index $$i = n$$ is completed in time $$O(1)$$, while adding at index $$0$$ is only $$O(n)$$.
Example (adding to LinkedSimpleList). Here is the code for our LinkedSimpleList implementation of add and the getNode helper method:
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public void add(int i, E x) {
....
Node<E> nd = new Node<E>();
nd.value = x;
if (i == 0) {
} else {
Node<E> pred = getNode(i - 1);
Node<E> succ = pred.next;
pred.next = nd;
nd.next = succ;
}
++size;
}
...
private Node<E> getNode(int i) {
// check if i is a valid index
if (i < 0 || i >= size) return null;
// find the i-th successor of the head
for (int j = 0; j < i; ++j) {
cur = cur.next;
}
return cur;
}
...
private class Node<E> {
Node<E> next;
E value;
}
Once again, we assume that the size of list is $$n$$ and $$i$$ is a valid index (i.e., $$0 \leq i \leq n$$). Observe that every statement in the add method except the call to getNode(i-1) in line 12 are performed in time $$O(1)$$. In particular, the creation of a new Node at line 4 only requires $$O(1)$$ time since a Node stores two references. As for the analysis of getNode(i), the running time of lines 26 and 28 are both $$O(1)$$. Each iteration of the loop in lines 31–33 can be performed in time $$O(1)$$, and $$i$$ iterations are performed. Therefore the overall running time of add(i, x) is $$O(1) + O(i) = O(i + 1)$$.
The previous two examples demonstrate the use of asymptotic analysis (i.e., the application of big O notation) to understand the qualitative running time of methods. They show that the two implementations of ListArraySimpleList and LinkedSimpleList—differ in the relative efficiency of add depending on where the element is to be added. In particular, adding to the last few indices of ArraySimpleList can be performed in time $$O(1)$$, while adding to the first few indices of a LinkedSimpleList can be performed in $$O(1)$$ time. The experiments we ran in lecture gave empirical justification of this analysis.
On the other hand, big O notation is somewhat limited, in that it abstracts away the precise constants that effect the running times of procedures, such as the amount of time required for a given computer to execute a primitive operation, and the number of such operations used in an execution. Thus, the analysis above cannot tell us, for example, if add(n / 2, x) will be faster for an ArraySimpleList or LinkedSimpleList; both operations run in time $$O(n / 2) = O(n)$$. Empirically, we saw that the array implementation was faster in this regime. | 2022-01-18T07:49:38 | {
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https://math.stackexchange.com/questions/2611010/calculate-the-sum-of-the-series-sum-limits-n-1-infty-left-frac3n | # Calculate the sum of the series: $\sum\limits_{n=1}^{\infty}\left ( \frac{3}{n^{2}+n}-\frac{2}{4n^{2}+16n+15} \right )$
I am stuck at a part when it comes to evaluating the sum of the said series... Here is my work so far (and I am not sure if the notation and simplification is correct either):
Simplifying using partial sums:
$$\sum_{n=1}^{\infty}\left ( \frac{3}{n^{2}+n}-\frac{2}{4n^{2}+16n+15} \right )=\sum_{n=1}^{\infty}\left ( 3\left ( \frac{1}{n}-\frac{1}{n+1} \right )-\left ( \frac{1}{2n+3}-\frac{1}{2n+5} \right ) \right )$$
Now I take the limit of the Nth partial sum of the series, right?
$$\lim_{N\rightarrow \infty}\sum_{n=1}^{N}\left ( 3\left ( \frac{1}{N}-\frac{1}{N+1} \right )-\left ( \frac{1}{2N+3}-\frac{1}{2N+5} \right ) \right )$$
$$=3\lim_{N\rightarrow \infty}\sum_{n=1}^{N}\left ( \frac{1}{N}-\frac{1}{N+1} \right )-\lim_{N\rightarrow \infty}\sum_{n=1}^{N}\left ( \frac{1}{2N+3}-\frac{1}{2N+5} \right )$$
$$=3\lim_{N\rightarrow \infty}\sum_{n=1}^{N} \frac{1}{N}-3\lim_{N\rightarrow \infty}\sum_{n=1}^{N}\frac{1}{N+1} -\lim_{N\rightarrow \infty}\sum_{n=1}^{N} \frac{1}{2N+3}+\lim_{N\rightarrow \infty}\sum_{n=1}^{N}\frac{1}{2N+5}$$
I assume I doing something wrong here, because each term diverges. This is what was written in the textbook:
$$\sum_{n=1}^{\infty}\left ( \frac{1}{n}-\frac{1}{n+1} \right )=1-\lim_{N\rightarrow \infty}\frac{1}{N}=1$$
$$\sum_{n=1}^{\infty}\left ( \frac{1}{2n+3}-\frac{1}{2n+5} \right )=\frac{1}{5}-\lim_{N\rightarrow \infty}\frac{1}{2N+3}=\frac{1}{5}$$
The problem is...I am not sure how they got this! They are missing a lot of steps for me to understand, hence the messiness above.
• Are you aware of telescopic sums? If not, then just write first few terms of $\frac{1}{n}-\frac{1}{n+1}$ and see what happens when we sum them. For example, $\frac{1}{1}-\color{red}{\frac{1}{2}} + \color{red}{\frac{1}{2}} - \frac{1}{3} \cdots$ – Math Lover Jan 18 '18 at 18:04
• And are you sure about the answer? I think it should be $14/5$ instead. – Leo163 Jan 18 '18 at 18:09
• @Leo163 Oops, my bad! That's my dyslexia for you. It's 14/5. – numericalorange Jan 18 '18 at 19:00
• @MathLover Yes, I know what those are. Thank you for your insight, this is very useful for me. – numericalorange Jan 18 '18 at 19:00
• I'm not sure why nobody has pointed out so far that your error is in making an unsound deduction. The limit of a sum/difference is not necessarily the sum/difference of the individual terms' limits. Observe that $\lim_{n\to\infty} (n-n) = 0$ but $\lim_{n\to\infty} n$ simply does not exist. You can interchange the limit and the finite sum/difference if the individual terms' limits exist. If not, then no go. Always remember that if you get nonsensical results, you must have made an unsound deduction somewhere. Makes sense? – user21820 Jan 19 '18 at 14:42
Note that both series $$\sum_{n=1}^{\infty}\frac{3}{n^{2}+n}~~and ~~~~\sum_{n=1}^{\infty}\frac{2}{4n^{2}+16n+15}$$
converges then there is no risk to separate the sum.
However using telescoping sum for the second sum as follows let $u_n=\frac{1}{2n+3}$ then $u_{n+1}=\frac{1}{2n+5}$ hence
$$~~~~\sum_{n=1}^{\infty}\frac{2}{4n^{2}+16n+15} =\lim_{k\to\infty }\sum_{n=1}^{k}\left ( \frac{1}{2n+3}-\frac{1}{2n+5} \right )\\\color{red}{:=\lim_{k\to\infty }\sum_{n=1}^{k}\left ( u_n-u_{n+1} \right )=\lim_{k\to\infty }(u_1-u_{k+1})}\\=\frac{1}{5}-\lim_{k\rightarrow \infty}\frac{1}{2k+5}=\frac{1}{5}$$
whereas the first is obvious since $$\sum_{n=1}^{\infty}\frac{3}{n^{2}+n} = 3\lim_{k\to\infty }\sum_{n=1}^{k}\left(\frac{1}{n}-\frac{1}{n+1}\right) =3\lim_{k\to\infty }(1-\frac{1}{k+1}) = 3~~$$
and
• I see now. This is based off the telescoping series, and the fact that the first and second term are of this form. Thanks so much! – numericalorange Jan 18 '18 at 19:03
• @numericalorange you are welcome don't forget to vote up it is useful for future users – Guy Fsone Jan 18 '18 at 19:07
• I upvoted you and pressed the green check mark to show this is the best answer. Is that what you mean by upvoting it? Or do I upvote my own question? – numericalorange Jan 18 '18 at 19:13
• @numericalorange that is all thanks – Guy Fsone Jan 18 '18 at 19:19
Hint:
The given expression can be written as $$\sum_{n=1}^\infty \frac {3}{n(n+1)} - \frac {2}{(2n+3)(2n+5)}$$
On partial decomposition it becomes
$$\sum_{n=1}^\infty \left[3\left(\frac {1}{n}-\frac {1}{n+1}\right) - \left(\frac {1}{2n+3}-\frac{1}{2n+5}\right) \right]$$
Can you see the series telescoping. By the way the answer I guess might be $\frac {14}{3}$ | 2019-08-22T00:22:45 | {
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https://www.physicsforums.com/threads/velocity-and-forces.336860/ | Velocity and forces
1. Sep 13, 2009
MechaMZ
1. The problem statement, all variables and given/known data
time-dependent force, $$\vec{}F$$= (8.60i - 4.20tj) N (where t is in seconds) is applied to a 2.00 kg object initially at rest.
(a) At what time will the object be moving with a speed of 15.0 m/s?
3. The attempt at a solution
$$\vec{F}$$ = ma
$$\sqrt{8.6^2 + (4.20t)^2}$$ = 2a
a = $$\sqrt{8.6^2 + (4.20t)^2}$$ / 2
vf = vi +at
15 = 0 + [$$\sqrt{8.6^2 + (4.20t)^2}$$ / 2]t
this should be an easy question but I can't get the correct answer as 2.86s from the equations above, I can't figure out the reason. hope someone could enlighten me
2. Sep 13, 2009
rl.bhat
Check the last calculation.
Square both side.
225*4 = [8.6^2 + 4.2^2*t^2]*t^2
Solve for t.
3. Sep 13, 2009
MechaMZ
Hi,
I think even we square both side, the answer is still incorrect. you will get 2.36s but the answer is 2.86s.
I think the problem is because the vf = vi +at, gives you the instantaneous velocity yet the question requires you to find speed.
4. Sep 13, 2009
kuruman
Your method works only if the velocity is in the same direction as the acceleration at all times. This is not the case here because the x-component of the acceleration is constant while the y-component changes with time. You need to calculate vx(t) and vy(t) separately, then apply the Pythagorean theorem to get the speed as a function of time.
5. Sep 13, 2009
Staff: Mentor
The problem is that vf = vi + at applies to constant acceleration, which this is not. (Integrate each component separately.)
6. Sep 13, 2009
MechaMZ
Hi,
so what method should I use for the Vy?
however, I could still using vf = vi +at for Vx since it is a constant acceleration?
I was wondering the v in the equation above is for instantaneous velocity yet the question is asking about speed.
7. Sep 13, 2009
Staff: Mentor
That's not an issue. The instantaneous speed is just the magnitude of the instantaneous velocity.
8. Sep 13, 2009
MechaMZ
Hi Doc,
How about the question below, was the elevator accelerating with a constant value or not? may i know how did you know the a is constant or not?
any clue so that i could know the elevator is accelerating with a constant or not a constant value during the first 0.5s?
thank you
9. Sep 13, 2009
Staff: Mentor
Unless you are given information to the contrary, I would just assume that all accelerations are constant.
10. Sep 13, 2009
MechaMZ
but if i take 1.82/0.5 is the instantaneous acceleration or average acceleration or constant acceleration for each second from 0m/s to 1.82m/s?
11. Sep 13, 2009
Staff: Mentor
If you assume that the acceleration is constant, then all three of those are the same! (Since Vf = Vi + at.)
12. Sep 13, 2009
MechaMZ
argh, how about this question, why all my assumptions are wrong regarding the acceleration, force and velocity.
The figure below shows the speed of a person's body as he does a chin-up. Assume the motion is vertical and the mass of the person's body is 68.0 kg.
http://img147.imageshack.us/img147/5012/p527.gif [Broken]
(a) time zero
correct answer is 687 N, but i thought it should be 68x9.81=667.08
why?
Last edited by a moderator: May 4, 2017
13. Sep 13, 2009
kuruman
You don't say what the question is, but 68*9.81 N is the answer to "What is the person's weight?".
Last edited: Sep 13, 2009
14. Sep 13, 2009
Staff: Mentor
You didn't tell us your assumptions, so we can't say what's wrong.
Please post the exact question and your reasoning. I suspect the question asks for the force he exerts on the bar during the initial portion of the chin up. Since he's accelerating, the force must be greater than just his weight.
15. Sep 13, 2009
MechaMZ
Hi Doc,
the question is "Determine the force exerted by the chin-up bar on his body at the following times."
(a) time zero
correct answer is 687 N, but i thought it should be 68x9.81=667.08
I was thinking when the time is zero, the velocity(movement) is zero. so there is no any upwards resultant force. I assume the T-mg = 0, so T = mg - 68* 9.81.
please give me some clue on this, thank you =)
16. Sep 13, 2009
kuruman
The velocity may be zero, but his acceleration is not. Can you read the acceleration from the graph?
17. Sep 14, 2009
Staff: Mentor
The question is somewhat poorly worded. When he's just hanging, before he starts his chinup, the force exerted by the chinup bar equals his weight. But you need to base your answer on the given diagram. (Note that they don't show the time before he starts pulling.) They are treating t = 0 as when he's already started pulling. A better way to phrase the question would have been: Determine the force exerted by the bar immediately after he starts the chinup.
18. Sep 14, 2009
MechaMZ
i see, so there is an acceleration. | 2018-03-23T10:02:02 | {
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https://www.tlcannon.com/nooash/page.php?c929e7=identity-matrix-example | What's interesting about what we've just proven to ourselves is the identity matrix for any matrix, even a non square matrix, a … When the identity matrix is the product of two square matrices, the two matrices are said to be the inverse of each other. Given the characteristics of the identity matrix, you can also conclude that these type of matrices are also called as diagonal matrices. A square matrix in which all the main diagonal elements are 1's and all the remaining elements are 0's is called an Identity Matrix. This matrix is an example of an echelon matrix. Proposition Let be a … The "Identity Matrix" is the matrix equivalent of the number "1": A 3×3 Identity Matrix. Solution: AB = For a 2 × 2 matrix, the identity matrix for multiplication is. Let us see example 3x3 identity matrix, 1 0 0 0 1 0 0 0 1 8.01x - Lect 24 - Rolling Motion, Gyroscopes, VERY NON-INTUITIVE - Duration: 49:13. The Identity Matrix When dealing with matrix computation, it is important to understand the identity matrix. The example above uses strings to generate the identity matrix. Some examples of identity matrices are as follows: The 2 x 2 identity matrix is given by. The identity matrix can also be written using the Kronecker delta notation: =. Less frequently, some mathematics books use U or E to represent the identity matrix, meaning "unit matrix" and the German word Einheitsmatrix respectively. A square matrix in which all the main diagonal elements are 1's and all the remaining elements are 0's is called an Identity Matrix. The "identity" matrix is a square matrix with 1 's on the diagonal and zeroes everywhere else. The above is 2 x 4 matrix as it has 2 rows and 4 columns. Identity Matrix. (read as “A inverse”) AA-1 = A-1 A = I. In linear algebra, an idempotent matrix is a matrix which, when multiplied by itself, yields itself. The below is an example of an Identity matrix: For each such row, the first nonzero element, as one reads from left to right, is unity. Multiplying by the identity. The identity matrix is the only idempotent matrix with non-zero determinant. An Identity Matrix is a square matrix whose main diagonal elements are ones, and all the other elements are zeros. It is also called as a Unit Matrix or Elementary matrix. It would be exponent rules thing^x × thing^y = thing^[x+y] modulo 7. Multiplying a matrix by the identity matrix I (that's the capital letter "eye") doesn't change anything, just like multiplying a number by 1 doesn't change anything. Let’s study about its definition, properties and practice some examples on it. If you need a matrix with real numbers (Integers) then use: Returns : identity array of dimension n x n, with its main diagonal set to one, and all other elements 0. Back to square one! Identity Matrix are the square matrix where the principal diagonal have elements as ones and other elements as zeros. More About Identity Matrix. Example Input Input elements in matrix: 1 0 0 0 1 0 0 0 1 Output It is an Identity matrix … Continue reading C program to check Identity matrix → problem and check your answer with the step-by-step explanations. An identity matrix is a square matrix in which all the elements of principal diagonals are one, and all other elements are zeros. If the product of two square matrices, P and Q, is the identity matrix then Q is an inverse matrix of P and P is the inverse matrix of Q. If the product of two square matrices, P and Q, is the identity matrix then Q is an inverse matrix of P and P is the inverse matrix of Q. NumPy Basic Exercises, Practice and Solution: Write a NumPy program to create a 3x3 identity matrix. While we say “the identity matrix”, we are often talking about “an” identity matrix. An identity matrix is a square matrix whose diagonal entries are all equal to one and whose off-diagonal entries are all equal to zero. If the second part of the dimension i… @wim: according to the docs np.eye is like np.identity but with added functionality. (i.e. Identity matrix You are encouraged to solve this task according to the task description, using any language you may know. Given that B is the inverse of A, find the values of x and y. Solution: No, It’s not an identity matrix, because it is of the order 3 X 4, which is not a square matrix. example. We can also say, the identity matrix is a type of diagonal matrix, where the principal diagonal elements are ones, and rest elements are zeros. Step 2: Multiply Matrix by its Inverse (Identity Matrix) If we want to check the result of Step 1, we can multiply our original matrix with the inverted matrix to check whether the result is the identity matrix.Have a look at the following R code: It is represented as In or just by I, where n represents the size of the square matrix. It is denoted by the notation “In” or simply “I”. Look at the last one! For example, the 2 × 2 and 3 × 3 identity matrices are shown below. You can rate examples to help us improve the quality of examples. These matrices are said to be square since there is … Example: Code: U = eye (4,4) Output: Explanation: In the above example, we have given two dimensions to create an identity matrix which means it will create an identity matrix with a number of rows as 4 and number columns as 4 where all the diagonal elements are one and rest other elements as zero. Example: Given that B is the inverse of A, find the values of x and y. V= $$\begin{bmatrix} 1 & 0 & 0 &0 \\ 0& 1 & 0 &0 \\ 0 & 0 & 1 & 0\\ \end{bmatrix}$$. For example. Example 2: Check the following matrix is Identity matrix? problem solver below to practice various math topics. Solution: We know that the identity matrix or unit matrix is the one with all ‘ones’ on the main diagonal and other entries as ‘zeros’. We welcome your feedback, comments and questions about this site or page. numpy.identity(n, dtype = None) : Return a identity matrix i.e. 3) We always get an identity after multiplying two inverse matrices. For any whole number $$n$$, there is a corresponding $$n \times n$$ identity matrix. Example 3: Check the following matrix is Identity matrix; B = $$\begin{bmatrix} 1 & 1 & 1\\ 1 & 1& 1\\ 1 & 1 & 1 \end{bmatrix}$$. 3. Identity Matrix is denoted with the letter "I n×n", where n×n represents the order of the matrix. If we multiply two matrices which are inverses of each other, then we get an identity matrix. One inner loop and one outer loop.If the current pointer for both loop is same, print 1 else print 0. An identity matrix is a square matrix having 1s on the main diagonal, and 0s everywhere else. The following example shows how to retrieve the Identity matrix. Recommended for you One of the important properties of identity matrix is: A × I n×n = A, where A is any square matrix of order n×n. A, B & C matrices are an example of the Identity matrix. That is, it is the only matrix such that: Alternatively, an identity matrix is a square diagonal matrix whose diagonal is one in every position. Identity Matrix Examples. Multiplying a matrix by the identity matrix I (that's the capital letter "eye") doesn't change anything, just like multiplying a number by 1 doesn't change anything. I = eye (n) returns an n -by- n identity matrix with ones on the main diagonal and zeros elsewhere. The option WorkingPrecision can be used to specify the precision of matrix elements. a square matrix with ones on the main diagonal. Any square matrix multiplied by the identity matrix of equal dimensions on the left or the right doesn't change. Your email address will not be published. It is denoted by In, or simply by I if the size is immaterial or can be trivially determined by the context. I = eye (sz) returns an array with ones on the main diagonal and zeros elsewhere. C program for finding Identity matrix. Lectures by Walter Lewin. (read as “A inverse”). IdentityMatrix [n, SparseArray] gives the identity matrix as a SparseArray object. Solution: The unit matrix is the one having ones on the main diagonal & other entries as ‘zeros’. For example: C = $$\begin{bmatrix} 1 & 2 & 3 &4 \\ 5& 6& 7 & 8 \end{bmatrix}$$. It returns a pointer to the memory layout of the object. Copyright © 2005, 2020 - OnlineMathLearning.com. Visit BYJU’S – The Learning App to explore a fun and interesting way to learn Mathematics. 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PQ = QP = I), The inverse matrix of A is denoted by A -1. IdentityMatrix by default creates a matrix containing exact integers. value_ptr() returns a direct pointer to the matrix data in column-major order, making it useful for uploading data to OpenGL. I3⋅AI_{3} \cdot AI3⋅AOn this case we have an example of the third property of the identity matrix: A multiplication involving an identity matrix and any other matrix, if defined due the rules of matrix multiplication, the result is the non-unit matrix (for this case, matrix A). private Matrix identityExample() { // Get the identity matrix… If any matrix is multiplied with the identity matrix, the result will be given matrix. Identity Matrix is also called as Unit Matrix or Elementary Matrix. The elements of the given matrix remain unchanged. The identity matrix is a square matrix which contains ones along the main diagonal (from the top left to the bottom right), while all its other entries are zero. What do you think about the one row matrix which has all elements are equal to 1, does it would be identity matrix? I = $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end {bmatrix}$ The identity matrix of order 3 is represented in the following manner: Identity Matrix is the matrix equivalent of the number 1 and represented by I always. Examples. Identity matrix is a square and has same number of rows and columns, then all diagonal place value is 1's and remaining place 0's. Example 1: Give an example of 4×4 order identity or unit matrix. Here, the 2 x 2 and 3 x 3 identity matrix is given below: Identity Matrix is donated by In X n, where n X n shows the order of the matrix. That is, the matrix is idempotent if and only if =.For this product to be defined, must necessarily be a square matrix.Viewed this way, idempotent matrices are idempotent elements of matrix rings For example, eye (3) creates an identity matrix with three rows and three columns, eye (5, 8) creates an identity matrix with five rows and eight columns, and eye ([13, 21; 34, 55]) creates an identity matrix with two rows and two columns. In some fields, such as quantum mechanics, the identity matrix is denoted by a boldface one, 1; otherwise it is identical to I. example. For any whole number n, there’s a corresponding Identity matrix, n x n. 2) By multiplying any matrix by the unit matrix, gives the matrix itself. Example 1: Write an example of 4 × 4 order unit matrix. In linear algebra, the identity matrix (sometimes ambiguously called a unit matrix) of size n is the n × n square matrix with ones on the main diagonal and zeros elsewhere. Identity Matrix is denoted with the letter "I n×n", where n×n represents the order of the matrix. This program allows the user to enter the number of rows and columns of a Matrix. Examples of Identity Matrix are identity matrices of order 1×1, 2×2, 3×3,………… n×n. These are the top rated real world Python examples of sagematrixmatrix_space.MatrixSpace.identity_matrix extracted from open source projects. Since B is an inverse of A, we know that AB = I. In particular, their role in matrix multiplication is similar to the role played by the number 1 in the multiplication of real numbers: 2. I = eye (n,m) returns an n -by- m matrix with ones on the main diagonal and zeros elsewhere. If you multiply two matrices that are inverses of each other you would get an identity matrix. Required fields are marked *. Same matrix is the result when any matrix multiplied by identity matrix. Or should I say square zero. Embedded content, if any, are copyrights of their respective owners. For example, following matrix is a identity matrix : 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 To print this matrix, we will use two for loops. As the multiplication is not always defined, so the size of the matrix matters when we work on matrix multiplication. Making use of the properties learnt in the past section and the identity matrix definition given at the beginning of this lesson, solve the next example problems:Example 1 1. So the size of the matrix is important as multiplying by the unit is like doing it by 1 with numbers. Identity matrix is also known as Unit matrix of size nxn square matrix where diagonal elements will only have integer value one and non diagonal elements will only have integer value as 0 Like in the given Example below − To create an identity matrix with a number of rows and b number of columns. A square matrix whose all diagonal elements are one (1) and rest of the elements are zero, called the unit matrix or identity matrix. So the 4×4 order identity or unit … More About Identity Matrix. When we multiply a matrix with the identity matrix, the original matrix is unchanged. For example, eye(5,'int8') returns a 5-by-5 identity matrix consisting of 8-bit integers. C Program to check Matrix is an Identity Matrix Example. Identity Matrix is also called as Unit Matrix or Elementary Matrix. Parameters : n : [int] Dimension n x n of output array dtype : [optional, float(by Default)] Data type of returned array. A X I n X n = A, A = any square matrix of order n X n. These Matrices are said to be square as it always has the same number of rows and columns. V= \(\begin{bmatrix} 1 & 0 & 0 &0 \\ 0& 1 … Learn its definition, properties and examples at CoolGyan. C Program to check Matrix is an Identity Matrix Example This program allows the user to enter the number of rows and columns of a Matrix. For example, the identity matrix of size 3 is The identity matrix of size is the identity element of all invertible matrices of size . Such a matrix is of the form given below: For example, the 4-by-4 identity matrix is shown below: Next, we are going to check whether the given matrix is an identity matrix or not using For Loop. Identity Matrix. It’s the identity matrix! Example 2: Check the following matrix is Identity matrix? A question for you. Solution: The unit matrix is the one having ones on the main diagonal & other entries as ‘zeros’. We can think of the identity matrix as the multiplicative identity of square matrices, or the one of square matrices. Normally, eye expects any scalar arguments you provide to … In this article, we will learn about what is an identity matrix, the determinant of identity matrix, identity matrix properties, the identity matrix in c, and learn about the identity matrix example. The identity matrix is always a square matrix. Example 1: Write an example of 4 × 4 order unit matrix. Python MatrixSpace.identity_matrix - 5 examples found. PQ = QP = I) The inverse matrix of A is denoted by A-1. Task. Each of the first k rows (k ≥ 0) of H has one or more nonzero elements. It is "square" (has same number of rows as columns) It can be large or small (2×2, 100×100, ... whatever) It has 1s on the main diagonal and 0s everywhere else; Its symbol is the capital letter I The diagonal elements can be accessed by its row number and column number that are (1,1), (2,2), (3,3), (4,4). In this tutorial, I am giving an identity matrix example using PHP program. IdentityMatrix [{m, n}] gives the m n identity matrix. (i.e. If you multiplied again you would go through the cycle again. Identity Matrix Example. The identity matrix is a square matrix where all elements of principal diagonals are 1s, and other elements are 0s. Likewise if you multiplied intermediate matrices from midway through, you would still travel around within the cycle. The "identity" matrix is a square matrix with 1 's on the diagonal and zeroes everywhere else. Whose off-diagonal entries are all equal to 1, does it identity matrix example be exponent rules thing^x thing^y. Shows how to retrieve the identity matrix any whole number \ ( n\ ), the result be... As matrix Property Value matrix as matrix Property Value matrix of the identity matrix [! Important to understand the identity matrix ”, we know that AB if. These are the top rated real world Python examples of identity matrices are an example of the equivalent. Going to Check whether the given matrix to the matrix equivalent of the first k (. Are copyrights of their respective owners we work on matrix multiplication identity matrices are to! Solution: AB = if you multiplied intermediate matrices from midway through, you also. 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Comments and questions about this site or page on it through the cycle that... A matrix containing exact integers of x and y or just by I if the of. More nonzero elements identity matrix example I ” is not always defined, so the size of the matrix multiplied! I am giving an identity matrix are identity matrices play a key identity matrix example in linear algebra of number. The inverse of each other you would still travel around within the cycle.... Study about its definition, properties and practice some examples on it k rows ( k ≥ 0 ) H. Its main diagonal elements are zeros pointer to the matrix a is denoted by A-1 2×2, 3×3, n×n! About its definition, properties and practice some examples of identity matrices are said to the... To understand the identity matrix entries as ‘ zeros ’ examples to help us improve the quality examples! Of identity matrix example and B number of columns and columns of a size known at run-time or enquiries via feedback. Number 1 '': a 3×3 identity matrix when dealing with matrix computation it! Or page this tutorial, I am giving an identity matrix is identity... Are 1s, and 0s everywhere else fun and interesting way to learn Mathematics equal. One outer loop.If the current pointer for both loop is same, print else... Sparsearray object as unit matrix is denoted by A-1, you can examples. × thing^y = thing^ [ x+y ] modulo 7 fun and interesting way to learn.... Matrix where all elements are 0s x n, m ) returns an array with ones on the diagonal! Row, the result will be given matrix be given matrix matrix are matrices. ………… n×n with numbers order of the identity matrix of a is denoted with the matrix... Dimensions on the main diagonal elements are equal to zero we are going to Check whether the given,!, ………… n×n by I always type of matrices are an example of 4 × order... | 2021-03-06T18:25:04 | {
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https://www.physicsforums.com/threads/eigenvalues-for-an-operator.971253/ | # Eigenvalues for an operator?
#### joshmccraney
Problem Statement
Show no eigenvalues exist for the operator $Af(x) = xf(x)$ where $A:C[0,1]\to C[0,1]$.
Relevant Equations
Nothing comes to mind.
Eigenvalues $\lambda$ for some operator $A$ satisfy $A f(x) = \lambda f(x)$. Then
$$Af(x) = \lambda f(x) \implies\\ xf(x) = \lambda f(x)\implies\\ (\lambda-x)f(x) = 0.$$
How do I then show that no eigenvalues exist? Seems obvious one doesn't exists since $\lambda-x \neq 0$ for all $x\in [0,1]$.
Related Calculus and Beyond Homework News on Phys.org
#### MathematicalPhysicist
Gold Member
Well the argument is subtle.
Since $x$ ranges over every point in $[0,1]$ and the supposed eigenvalue is some constant you cannot have $\lambda=x$, since the function $x$ is not constant.
And obviously $f(x)\ne 0$, otherwise it wouldn't be an eigenvalues' problem.
#### Math_QED
Homework Helper
You are in fact almost there. The condition
$(\lambda-x)f(x) =0$ implies that we have $f(x) = 0$ for all $x\neq \lambda$. By continuity however $f=0$. Thus, no eigenvector exists, as any possible eigenvalue leads to the trivial solution.
Last edited:
#### joshmccraney
You are in fact almost there. The condition
$(\lambda-x)f(x) =0$ implies that we have $f(x) = 0$ for all $x\neq \lambda$. By continuity however $f=0$. Thus, no eigenvector exists, as any possible eigenvalue leads to the trivial solution.
This just seems so simple. See, the HW was assigned in three parts, where part b) asked to show 0 is not an eigenvalue and part c) asked to prove no eigenvalues exist. But if this is part c), then why bother doing b)? Have I missed anything about $\lambda = 0$? It seems this logic holds for all $\lambda$.
#### Math_QED
Homework Helper
This just seems so simple. See, the HW was assigned in three parts, where part b) asked to show 0 is not an eigenvalue and part c) asked to prove no eigenvalues exist. But if this is part c), then why bother doing b)? Have I missed anything about $\lambda = 0$? It seems this logic holds for all $\lambda$.
It also works for $\lambda = 0$. Then your equation leads to $xf(x) = 0 \forall x$ and for $x \neq 0$ we then have $f(x) = 0$. By continuity again also $f(0) = 0$, so $f=0$ everywhere.
So, to answer your question concretely, I don't think doing part (b) separately is necessary.
Note that eigenvalue $\lambda = 0$ can only happen when $A$ is non-injective. This map is injective however (continuity is key again here). Maybe that's what (b) wants you to write?
"Eigenvalues for an operator?"
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• Solo and co-op problem solving | 2019-05-24T04:01:20 | {
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https://cs.stackexchange.com/questions/112251/number-of-nodes-in-van-emde-boas-tree-of-universe-size-u | # Number of nodes in Van-Emde-Boas tree of universe size u?
The universe size $$u$$ in vEB trees is square rooted at each level until $$u = 2$$. So, unlike search trees, the branching factor is different at each level.
The height of the tree is $$h = \lg \lg(u)$$ and $$u$$ is an even power of 2, $$u = 2^{2^k}$$. I tried to calculate the count using the sum:
$$\sum_{i=1}^{h-1} 2^i\,2^{i + 1}$$
But it doesn't work. Any idea on how to do the math?
## Edit:
Sorry for the confusion that might be caused by the illustration. The illustration is based on the CLRS book's implementation of VEB trees, but the summary nodes are not shown. Thinking again about it now, I would like to know the count of all the nodes including the summary nodes and their tree nodes, as well as, the count of just the nodes without the summary.
The nice picture in the question does not include the summary nodes nor the min and max fields, both of which are indispensable components of a van Emde Boas tree (vEB tree). A better illustration might be the following picture taken from how to read off the set represented by a van-Emde-Boas tree, which was drawn by Raphael based on a figure in CLRS, where a number in orange is drawn at a min field that marks its presence in the set of integers being represented.
How many nodes are there in the vEB tree implemented like the illustration above?
There are $$1 + 5 + 5 \times 3 = 21$$ nodes in the illustration above.
For the sake of simplicity, let $$w=2^h$$ and $$u=2^w=2^{2^h}$$. A vEB tree over the universe $$\{0,1,\cdots, u-1\}$$ is of depth $$h=\log\log u$$.
• There is one node of depth 0, which is the root node.
• There are $$2^{2^{h-1}} + 1$$ nodes of depth 1.
• Each node of depth 1 has $$2^{2^{h-2}} + 1$$ child nodes of depth 2.
• ...
• Each node of depth i has $$2^{2^{h-i-1}} + 1$$ child nodes of depth $$i+1$$.
• ...
• Each node of depth $$h-1$$ has $$2^{2^{h-h}}+1=3$$ child nodes of depth $$h$$.
In total, the number of all nodes is $$1 + \sum_{i=1}^{h}\prod_{k=1}^i(2^{2^{h-k}} + 1)=(2^{2^h} - 1)\sum_{i=0}^{h}\frac1{2^{2^i} - 1}\tag{1}$$ which is 1, 4, 21, 358, 92007, 6029862760, 25898063359598159721, $$\cdots$$ for $$h=0,1,2,3,4,5,6,\cdots$$ respectively. When $$h\ge4$$, the number of nodes is about $$1.404u$$.
Similarly, the number of all summary nodes is, for $$h\ge1$$, $$1 + \sum_{i=2}^{h}\prod_{k=1}^{i-1}(2^{2^{h-k}} + 1)=(2^{2^{h}} - 1)\sum_{i=1}^{h}\frac1{2^{2^i} - 1},\tag{2}$$ which is 0, 1, 4, 21, 358, 92007, 6029862760, $$\cdots$$ for $$h = 0,1, 2,3,4,5,6,\cdots$$ respectively.
What is the number of all non-summary nodes? Subtracting (2) from (1), we obtain
$$(2^{2^{h}} - 1)\left.\frac1{2^{2^i} - 1}\right|_{i=0}=2^{2^h}-1=u-1.$$
The formula also hold for $$h=0$$. So we have obtained the following surprising formula. $$\text{the number of non-summary nodes in a vEB tree of universe size } u=2^{2^h}\text{ is }u-1.$$
Every node at the same depth use the same amount of space. Nodes at a smaller depth may use much more space than those at a larger depth. For example, for $$h=5$$, the root node contains a bit-array of size 65536 but a leaf node just contains several words. Since the number of nodes that use bigger space decreases very quickly as their depth becomes smaller, the total space used is $$O(u)$$.
Exercise. The number of all leaf nodes, which are vEB trees of universe size 2 in a vEB tree of universe size $$u=2^{2^h}$$ is $$u-1$$.
• Thank you for your answer. I edited my question, please have a look. I made the illustration myself and did not include the summary points deliberately, because I was only interested in counting the main nodes. Now you mentioned it though I would like to also know how to count the overall nodes. Also can the sum of the product in the formula you suggested be simplified? – razzak Jul 29 '19 at 13:11
• I think that by excluding the summary nodes, the count would become $1 + \sum_{i=1}^{h}\prod_{k=1}^i 2^{2^{h-k}}$ right?, which looks possible to simplify "working on it :)", and also because I think the count of summary nodes is a constant fraction of the cluster nodes count, although didn't figure out that constant yet. – razzak Jul 30 '19 at 22:38
• Thanks for your insistence. I just found that the number of all non-summary nodes as well as the number of all leaf nodes is $u-1$ for universe size $u=2^{2^h}$. – John L. Jul 31 '19 at 1:38
• $u-1$ is the count of leaf nodes only and not the overall count of non-summary nodes. I found out that the count of the non-summary nodes is $u - \sqrt{u} + 1$ link. – razzak Jul 31 '19 at 20:32
• I think we might be confused about the non-summary node definition, I use this term to describe all light-gray nodes in Raphael's illustration except those inside the summary trees. So in this illustration above there are 13 non-summary nodes and this formula $u - \sqrt{u} + 1$ is able to calculate it successfully. – razzak Aug 1 '19 at 1:14 | 2020-02-24T03:05:05 | {
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https://math.stackexchange.com/questions/1999641/sufficient-but-not-necessary-conditions | # Sufficient but not Necessary conditions
We were having a discussion at the office when we should have been working and I suggested that an example of a Sufficient but not Necessary condition is:
Given a natural number of fewer than, say, 25 digits. We wish to establish if it is divisible by six.
An example of a Necessary but not Sufficient condition was: Is the number divisible by two? The other guy was happy with that and agreed.
My example of a Sufficient but not Necessary condition (however simple) was: Is the number equal to six? The other guy (who has a degree in math) insisted that because this would not apply to numbers such as 12 this could not be Sufficient. I maintained that that is the whole point, the condition is not necessary but because it is Sufficient, if it is true, you are done, QED.
This promptly devolved into "It is not", "It is, too" which did not seem very mathematical, somehow.
Could we get somebody to comment? The other guy decided he did not want to discuss this further but I would like to feel a little vindicated. (If I am wrong, I will send him your answer.) Thank you in advance.
• You are definitely correct. Maybe a clearly case would be "It is sufficient that it is divisible by 12". If it is divisible by 12 => it is divisible by 6 so that is sufficient. I think maybe it's easy to confuse the meanings of nescessary and sufficient and intuitively necessary sound "stronger" than sufficient whereas sufficiency but not nescessary means too strong and nescessary but not sufficient means not strong enough. I think a mental block "just do one case" throws the one's sense of meaning off. Still, he should have known better. – fleablood Nov 4 '16 at 21:55
• – Greek - Area 51 Proposal Nov 24 '16 at 21:26
$A$ is necessary for $B \iff B \implies A$
$A$ is sufficient for $B \iff A\implies B$
In view of this, let's check.
Case 1: $A:$number is divisible by $2$, $B:$ number is divisible by $6$.
Does $B \implies A?$... definitely!! So $A$ is necessary for $B$.
Does $A\implies B?$.... well not always. So $A$ is not sufficient for $B.$
Case 2: $A:$number is $6$, $B:$ number is divisible by $6$.
Does $B \implies A?$... well not always!! So $A$ is not necessary for $B$.
Does $A\implies B?$.... definitely. So $A$ is sufficient for $B.$
You were right!!
You are right. The condition A is said to be sufficient for an event B if (and only if) the occurence of A guarantees the occurence of B. The logical relation would be
$A\Rightarrow B$
Then A is by no means necessary for B: $B\Rightarrow A$ or even a characterization of B: $A \Leftrightarrow B$
So for example, to be a sheep (6) is sufficient for being an animal (divisible by 6). However you don't have to be a sheep (6) in order to be an animal (divisible by 6).
"$n=6$" is a sufficient condition for a number $n$ to be divisible by $6$. That's because $$n=6\implies n\text{ is divisible by }6$$ is true.
Per this answer, 'A is sufficient for B.` means that 'A is a subset of B'.
A picture and real-life example may aid to understand the following:
Your main question is whether (2) follows from (1);
(1) P → Q [ P is a sufficient condition for Q ],
(2) Q → P [ P is a necessary condition for Q ].
The reason the entailment from (1) to (2) doesn't hold is that it's possible that Q follow from some proposition R that is not equivalent to P. The only instance where the entailment is realized is one where all necessary conditions for Q are logically equivalent to P.
The above is exemplified in the picture below, if P = Northern Ireland, Q = UK, R = Great Britain.
Then being in Northern Ireland is sufficient for being in the UK,
but is NOT necessary for being in the UK, because one can also be in U.K. by being in
Great Britain. | 2019-10-16T11:58:18 | {
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https://math.stackexchange.com/questions/2589653/manual-calculation-doesnt-match-wolfram-alpha-why | # Manual calculation doesn't match Wolfram Alpha. Why?
Say I want to evaluate this sum: $$\sum_{x=2}^\infty \ln(x^3+1)-\ln(x^3-1)$$ We can rewrite the sum as $$\ln\left(\prod_{x=2}^\infty \frac{x^3+1}{x^3-1}\right)$$ We can split the product into two products: $$\ln\left(\prod_{x=2}^\infty \frac{x+1}{x-1}\right)+\ln\left(\prod_{x=2}^\infty \frac{x^2-x+1}{x^2+x+1}\right)$$ These are both telescopic products! We can rewrite them as $$\ln\left(\prod_{x=2}^\infty \frac{(x+2)-1}{x-1}\right)+\ln\left(\prod_{x=2}^\infty \frac{(x-1)^2+(x-1)+1}{x^2+x+1}\right)$$ Plugging in numbers makes this pattern more obvious; $$\ln\left(\prod_{x=2}^\infty \frac{\color{green}{3}}{1}\frac {\color{red}{4}}{2} \frac{\color{blue}{5}}{\color{green}{3}} \frac{\color{bluedark}{6}}{\color{red}{4}} \right) + \ln\left(\prod_{x=2}^\infty \frac{3}{\color{green}{7}} \frac{\color{green}{7}}{\color{red}{13}} \frac{\color{red}{13}}{\color{blue}{21}} \right)$$ Simplifying, we get that the sum is equal to $$\ln(1/2)+\ln(3)=\ln\left(\frac 32 \right) \approx 0.405465108108164381978013115464349136571990423462494197614$$ However, when I put the sum in Wolfram Alpha directly, I get the following number: $$0.4054588737136331726677370820628648457601892466568342890929$$
Why are these two numbers different? It's not a small difference either; it's on the 5th number after the decimal point! How can Wolfram make such an error?
• it might be because wolfram alpha is calculating numerical value upto some accuracy, which is why you get nearly equal but different numbers – Piyush Divyanakar Jan 3 '18 at 3:27
• Indeed, both numerical sum and the exact value yields the same numerical value in Mathematica 11. – Sangchul Lee Jan 3 '18 at 3:43
• mathematica.stackexchange.com is probably the right place for this question. They even have a tag for bugs mathematica.stackexchange.com/questions/tagged/bugs – user66307 Jan 3 '18 at 11:53
• I think debugging WolframAlpha should be outside the scope of this site. – Rahul Jan 3 '18 at 16:56
• Does the first MathJax block (the one with $\sum$) need an additional pair of parentheses to include both $ln$ expressions? – shoover Jan 3 '18 at 22:05
In Mathematica, the input FullSimplify[Sum[Log[x^3+1]-Log[x^3-1],{x,2,Infinity}]] gives Log[3/2], exactly. In WolframAlpha, the issue is that when you request more digits of accuracy, it converts your input into the command NSum[Log[x^3 + 1] - Log[x^3 - 1], {x, 2, Infinity}, WorkingPrecision -> 104] which of course is insufficient working precision for the number of decimal digits that it displays due to the very slow convergence of the sum.
Somewhat ironically, if you enter Product[(x^3+1)/(x^3-1),{x,2,n}] into WolframAlpha, you get exact output: $$\frac{3n(n+1)}{2(n^2+n+1)},$$ which is correct, furthermore upon taking Limit[3n(n+1)/(2(n^2+n+1)), n -> Infinity], you get the correct answer $3/2$. So it isn't as if WolframAlpha cannot compute the original sum symbolically as Mathematica did. It just needs a little extra help, it seems.
Users of WolframAlpha don't always realize that although it is using the same underlying algorithms as Mathematica, it is not the same thing. There are things that one does that the other does not. Precise control of processing of input, for example, is something that the former does not do.
Update. I believe the above response is not entirely sufficient to explain the behavior of WolframAlpha. When I changed the summand to $$\log \left(1 + \frac{2}{x^3-1}\right),$$ WolframAlpha still gives the wrong result when more digits are requested, despite the fact that the generated code NSum[Log[1 + 2/(x^3 - 1)], {x, 2, Infinity}, WorkingPrecision -> 104] yields the correct result. So this points to an internal inconsistency with WolframAlpha that cannot be solely explained by the insufficient precision used in NSum.
To confirm, in Mathematica I input both variants with NSum, namely
NSum[Log[x^3 + 1] - Log[x^3 - 1], {x, 2, Infinity}, WorkingPrecision -> 104]
as well as
NSum[Log[1 + 2/(x^3 - 1)], {x, 2, Infinity}, WorkingPrecision -> 104]
The first input, as expected, generates a warning NIntegrate::ncvb. Also as expected, the second one does not. But if this is the case, then WolframAlpha should not still present the wrong result in the second case when more digits are requested, given that this is the exact code that it generated. When you open up the computable notebook (click the orange cloud) and evaluate the expression, you are basically running a cloud version of Mathematica. Doing this gets the right answer. So I suspect that there is some kind of bug in WolframAlpha that fails to present the correct output.
• How did you get the input to work in Mathematica? FullSimplify[Sum[Log[x^3+1]-Log[x^3-1],{x,2,Infinity}]] is just returning the sum unevaluated for me, both on my computer and also on Wolfram cloud. – numbermaniac Jan 4 '18 at 6:20
• @numbermaniac I am using Version 10.4 on Mac OS. It evaluates fairly rapidly, in less than 5 seconds. – heropup Jan 4 '18 at 6:31
• that's really strange, because I'm using 11.2 on macOS! I thought Mathematica was supposed to get better with each version :( – numbermaniac Jan 4 '18 at 6:34
When you split the product into two products, the first diverges to $\infty$ and the second goes to zero. This gives you the indeterminate $\infty-\infty$. All bets are off.
• Perhaps it would be helpful to write down the partial products to make the infinities clear? – Michael Burr Jan 3 '18 at 3:43
• Although OP's computation is not rigorous, the answer is correct. So the issue lies in the numerical algorithm behind WolframAlpha. – Sangchul Lee Jan 3 '18 at 3:44 | 2019-07-19T04:20:43 | {
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http://math.stackexchange.com/questions/375254/finding-the-number-of-zeros-of-a-function-in-a-given-annulus | # Finding the Number of Zeros of a Function in a Given Annulus
Consider $z^6 - 6z^2 + 10z + 2$ on the annulus $1<|z|<2$.
By Rouche's Theorem $|f(z) + g(z)| < |f(z)|$ implies that both sides of the inequality have the same number of zeros. I understand that when asked to find that there is, say, $1$ zero the idea is to choose the $10z$ term as $f(z)$ so that we can form the proper relationship and conclude that since $10z$ has one zero in the region the function has one zero in the region. How can it be set up to find a general number of zeros? What is the trick for picking $f(z)$?
-
Problems of this type which come from books are often vulnerable to some meta-mathematical arguments. One basic principle is that, in order to apply Rouché's theorem, you are never expected to count the number of zeros of any polynomial which you cannot solve explicitly. Also, in the actual application of Rouché's theorem, you'll rarely need to use anything more complicated than the triangle inequality to establish the required inequality. In practice this means that you're hoping to show something like
$$|\text{complicated}| < |\text{simple}|.$$
Another guiding rule is that larger powers of $z$ are smaller in the unit circle and smaller powers of $z$ are smaller outside of the unit circle when compared to the rest of the polynomial.
For this particular problem, if I wanted to know how many zeros are inside the unit circle, I would guess that the dominating polynomial $q(x)$ in the inequality $|p(x)| < |q(x)|$ is either $2$ or $10z+2$ or $10z$, since smaller powers are more important in the unit circle. For convenience we'll try $10z$ first.
The polynomial $10z$ has one zero at $z=0$ which happens to lie inside the unit circle. Now on $|z| = 1$ we have
\begin{align} \left|z^6-6z^2+2\right| &\leq 1+6+2 \\ &= 9 \\ &< 10 \\ &= 10 |z| \\ &= |10z|, \end{align}
and we can conclude from Rouché's theorem that $z^6 - 6z^2 + 10z + 2$ has exactly one zero in the disk $|z| \leq 1$. If you'd like you can check that it's also true that
$$\left|z^6-6z^2\right| < |10z+2|$$
on the unit circle, so we could have gone that way instead.
To find the number of zeros in $|z| \leq 2$ we'll start from the other end and hope that $z^6$ dominates on the circle $|z| = 2$. Note that $z^6$ has a zero of multiplicity $6$ at $z=0$ (so, effectively, it has $6$ zeros), which happens to lie inside the circle $|z| = 2$. On $|z| = 2$ we have
\begin{align} \left|-6z^2+10z+2\right| &\leq 6 \cdot 2^2 + 10 \cdot 2 + 2 \\ &= 46 \\ &< 64 \\ &= 2^6 \\ &= \left|z^6\right|, \end{align}
so from Rouché's theorem we know that $z^6 - 6z^2 + 10z + 2$ has all $6$ zeros in the disk $|z| < 2$.
Hence the number of zeros of the polynomial in the annulus $1 < |z| < 2$ is $6-1 = 5$.
-
This is an excellent response, thank you. Was it simply guess work choosing $z^6$ and $z$ though for each circle? What if we had guessed there might be two roots in the circle of radius 1 with the rest in the circle of radius 2, or that in fact that in an unlikely case 5 of them lie within 2 and the other is somewhere far outside. – user73041 Apr 29 '13 at 18:51
@user73041, I wouldn't really call it guess work. Like I said in the beginning of the answer the larger powers of $z$ are more important outside the unit circle, so I expected $z^6$ to dominate on $|z| = 2$. The smaller powers of $z$ are more important inside the unit circle, so I expected $10z$, $10z+2$, or $2$ to dominate on $|z| = 1$. I'm not thinking about the possible locations of the zeros at all---I'm thinking about the relative sizes of the terms of the polynomial. (By "dominate" I mean that $b$ dominates $a$ if $|a| < |b|$.) – Antonio Vargas Apr 29 '13 at 18:55
Would it be possible though that, say in a case where you have $z^6$ and $z^7$ choose the larger and turn out to be wrong? That is, in fact 6 are in the circle of, say, radius 2 and the rest are somewhere outside (from 2 to infinity). – user73041 Apr 29 '13 at 19:02
@user73041, sure, the principle is only a rule of thumb. Picking the largest or the smallest won't always work but it's a good place to start. I had a good feeling about $z^6$ in particular because it's so much larger than the next term, $-6z^2$. – Antonio Vargas Apr 29 '13 at 19:05
That makes sense, thank you again. – user73041 Apr 29 '13 at 19:12
You can just split up the problem in two -- $|z|<1$ and $|z|<2$. Then you just subtract the number of zeros in the smaller regions from the zeroes of the bigger region, and you have your answer.
Both times you just apply Rouche's Theorem.
-
Right, but my main concern is: for both such cases how does one determine f(z)? We have no knowledge of how many zeros might be in each region so far as I can tell. If the question had told us that we were looking for 1 zero in the region |z| <1 then I would expect to use 10z but without that information I am not sure how exactly one determines the 'trick' that allows us to invoke Rouche. I worry that even if I am able to come up with a trick here I could easily have no idea what the trick is for another problem. – user73041 Apr 28 '13 at 14:46 | 2015-07-28T06:15:34 | {
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https://www.physicsforums.com/threads/deriving-acceleration-formula.874401/ | # Deriving acceleration formula
• B
Hello all,
First of all I want to let you know that my question is very basic and that it involves discrete changes in velocity due to acceleration for every given Δt. I was trying to derive the relationship between the distance and acceleration in a formula and here's what I came up with:
1. I was able to conclude that to calculate the velocity after a time t in which discrete acceleration is involved the formula would be: at + v0 = v in which the v0 is the starting velocity
2. Now, to calculate the distance, one wouldn't obviously be able to just multiply the given v by t since that would consider as if the object has been traveling a constant velocity all along.
In reality one would have to calculate (v0 + a) + (v0 + 2a) + (v0 + 3a) + (v0 + na) in which n would be the time duration in steps of Δt.
3. However, to give an approximation of the distance traveled without doing the whole hassle in point 2, one could just take the average velocity of v0 and v (that the object has after a time duration t) and multiply that average velocity by the time. Thus, the formula would be ((at + v0) + v0) / 2) × t = d which after simplifying gives 0.5at2 + v0t = d
Question: Is taking the average the reason why there's a "0.5" in the formula that gives the relationship of acceleration and distance?
However, here's my problem. The formula 0.5at2 + v0t = d doesn't always seem to give correct answers even for a discrete acceleration over time when I compare its results to the results of the formula that I've shown in point 2.
For example: If an object with a start velocity of 6 m/s accelerates in discrete steps of 3m/s2 for a time duration of 4 seconds, I'd expect that it would have traveled 6 + 9 + 12 + 15 = 42m at t=4. However, filling the values in the formula 0.5at2 + v0t = d would give a traveled distance of 48m.
I thought that the known formula 0.5at2 + v0t = d should always give accurate results regarding acceleration that increases velocity in discrete steps. Perhaps I'm missing something obvious here?
Jonathan Scott
Gold Member
The average velocity over an interval is the total distance divided by the total time. If acceleration is constant, the average velocity is half way between the initial velocity and the final velocity for an interval, hence the formula. If it is not constant, the formula does not apply. As a special case, for your case of discrete steps, then if your sudden change in speed occurred at the half way time within each step, instead of at the end of it, the results would be the same as for constant acceleration.
CWatters
Homework Helper
Gold Member
Your equation d = 0.5at2 + v0t is consistent with the SUVAT equations for constant acceleration...
https://en.wikipedia.org/wiki/Equations_of_motion
However it's not clear what you mean by "accelerates in discrete steps". Perhaps plot a graph of velocity vs time. It it's not a straight line then the acceleration isn't constant and the equations don't apply.
Your equation d = 0.5at2 + v0t is consistent with the SUVAT equations for constant acceleration...
https://en.wikipedia.org/wiki/Equations_of_motion
However it's not clear what you mean by "accelerates in discrete steps". Perhaps plot a graph of velocity vs time. It it's not a straight line then the acceleration isn't constant and the equations don't apply.
I knew I was missing something obvious here.
This explains why it also always gives a higher value of distance than with discrete steps of acceleration (discrete being a "sudden" increase in velocity at each fixed Δt) since there's a constant velocity increase even between the Δt.
So if I understand correctly, this formula d = 0.5at2 + v0t is even accurate if there's constant acceleration in infinitesimally small Δt?
Jonathan Scott
Gold Member
The formula assumes the average speed over the total elapsed time is exactly half way between the initial speed and the final speed, as you noted in your original post. This is always true if the acceleration is constant during the total time. If the acceleration varies, the formula cannot be used.
rcgldr
Homework Helper
If an object with a start velocity of 6 m/s accelerates in discrete steps of 3m/s2 for a time duration of 4 seconds, I'd expect that it would have traveled 6 + 9 + 12 + 15 = 42m at t=4.
Assuming acceleration is constant, for a relatively large Δt = 1, you still need to take the average velocity for each step (6+9)/2 + (9+12)/2 + (12+15)/2 + (15+18)/2 = 48.
If you want to use just the starting or ending velocities for each time period, you need to use a smaller Δt. If you use the starting velocities, the result is 48 - 6 Δt. If you use the ending velocities, the result is 48 + 6 Δt. As Δt approaches zero, both methods approach 48.
Last edited:
Assuming acceleration is constant, for a relatively large Δt = 1, you still need to take the average velocity for each step (6+9)/2 + (9+12)/2 + (12+15)/2 + (15+18)/2 = 48.
If you want to use just the starting or ending velocities for each time period, you need to use a smaller Δt. If you use the starting velocities, the result is 48 - 6 Δt. If you use the ending velocities, the result is 48 + 6 Δt. As Δt approaches zero, both methods approach 48.
Thanks, your explanation helped me a lot and I was able to conclude all that by calculating the area beneath a line in a v t diagram with constant acceleration in the ways you mentioned.
Thank you all for your help! | 2021-02-25T22:58:04 | {
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http://math.stackexchange.com/questions/199810/what-are-the-differences-between-these-two-logical-statements | # What are the differences between these two logical statements?
$$\exists\ x \in \mathbb{N}\ \textrm{such that}\ \forall\ y \in \mathbb{N}, 2x \leq y + 1$$
$$\forall\ y \in \mathbb{N}, \exists\ x \in \mathbb{N}\ \textrm{such that}\ 2x \leq y + 1$$
I'm having trouble understanding the differences between these two statements. To me, they seem to mean the same thing. Can anyone explain in basic terms? Thanks.
-
"Every hour a person is injured in a car accident" - "Oh dear, I would stay at home if I were that fellow". – Hagen von Eitzen Sep 20 '12 at 18:23
The difference is that in the first case you have to choose $x$ first, without reference to $y$: the statement must be true for all $y$ for some fixed $x$. In the second case $x$ is chosen after $y$ is introduced, and so $x$ can be a function of $y$. – mjqxxxx Sep 20 '12 at 18:53
It is not too much of an exaggeration to say that the whole rationale of the quantifier-variable analysis of statements of generality which we owe to Frege is to mark this kind of distinction. So asking the question suggests you haven't "got it". So I'd warmly recommend looking at a good intro logic book where it introduces the quantifiers. Paul Teller's Primer is very good and freely available online. – Peter Smith Sep 21 '12 at 6:17
Answers to Confused between Nested Quantifiers question may help you to better understand nested quantifiers. – olek Sep 21 '12 at 11:09
Let $L(x,y)$ stand for "$x$ loves $y$". Then $\exists x\forall y: L(x,y)$ means "There is someone who loves everyone." and $\forall y\exists x: L(x,y)$ means "Everybody is loved by someone". Clearly, these two are very different.
Now compare the simple mathematical statements.
$\exists x\in\mathbb{N}$ such that $\forall y\in\mathbb{N}$, $x\leq y$.
$\forall y\in\mathbb{N},$ $\exists x$ such that $x\leq y$.
The first one says that there is some natural number $x$ that is smaller or equal than every natural number $y$.
The second statement says that for every natural number $y$, there is a natural number $x$ that is less or equal than $y$.
It turns out that both of these statements are true. But now replace $\mathbb{N}$ by $\mathbb{Z}$. There is no smallest integer, so the first statement becomes wrong then. But the second one would still be true because for every integer $y$, the integer $y-1$ is smaller or equal than y$. - 1. There is some$x$, such that no matter what$y$you choose,$x$will be less than$y+1$.$x=0$fits the bill. 2. No matter which$y$you choose, you can always choose some$x$s.t.$2x\leq y+1$. E.g. if$y=5$, then$x=1\$ satisfies.
Note that if we were dealing with say the integers, the second statement would be true but the first one would be false. It might be instructive to figure out why.
- | 2016-02-09T12:24:33 | {
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https://math.stackexchange.com/questions/1964795/conditional-probability-with-circuits | # Conditional Probability with Circuits
I am very sorry for the myriad of probability questions but I have a dam hard time trying to understand these questions:
This is the question:
Consider the following portion of an electric circuit with three relays. Current will flow from point a to point b if there is at least one closed path when the relays are activated. The relays may malfunction and not close when activated. Suppose that the relays act independently of one another and close properly when activated, with a probability of .9.
The question is: What is the probability that current will flow when the relays are activated?
Answer:
1 - (0.1)^3
I don't understand why you are multiplying 0.1 * 0.1 * 0.1 and not adding 0.1 + 0.1 + 0.1. Since if any one of the three are open then the current will flow so isn't this technically a union which means you add?
• It's very simple: the multiplication principle. The relays are independent of each other. – Parcly Taxel Oct 12 '16 at 4:26
• @ParclyTaxel Oh ok. Thanks for pointing that out. So if they are dependent of each other then you would add them? – CapturedTree Oct 12 '16 at 5:23
## 2 Answers
Yes , it is a union . The current will flow when at least one of the relays will pass it through .
However , you do not simply add because the events are not disjoint . This then requires using the Principle of Inclusion and Exclusion (PIE) to avoid over-counting common outcomes .
\begin{align}\mathsf P(E_1\cup E_2\cup E_3) ~=~& {\mathsf P(E_1)+\mathsf P(E_2)+\mathsf P(E_3) \\[1ex]-\mathsf P(E_1\cap E_2)-\mathsf P(E_1\cap E_3)-\mathsf P(E_2\cap E_3)\\+\mathsf P(E_1\cap E_2\cap E_3)} \\[1ex]=~& 3(0.9)-3(0.9)^2+(0.9)^3\\=~& 0.999\end{align}
Since the events are independent , then the probabilities of the unions are the product of the probabilities of the events.
However the answer is somewhat easier to obtain by using complements . The current will not flow only when all of the relays block it .
\begin{align}\mathsf P(E_1\cup E_2\cup E_3) ~=~& 1-\mathsf P(E_1^\complement\cap E_2^\complement\cap E_3^\complement) &~&\text{de Morgan's Rules}\\[1ex] =~& 1-\mathsf P(E_1^\complement)\,\mathsf P(E_2^\complement)\,\mathsf P( E_3^\complement) && \text{Independence} \\=~& 1-(1-0.9)^3 \\[1ex]=~& 0.999 \end{align}
That is all.
$\blacksquare$
• Thank you for the explanation! – CapturedTree Oct 12 '16 at 5:25
Current will not flow if $S_1 = S_2 = S_3 = 0$ which happens with probability $(1-0.9)^3$, hence current will flow with probability $1-(1-0.9)^3$. | 2021-07-31T18:32:31 | {
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https://www.physicsforums.com/threads/multiple-integral-in-a-domain.831143/ | # Multiple integral in a domain
1. Sep 6, 2015
### Aner
1. The problem statement, all variables and given/known data
Hi, I have a problem with the following exercise.
Let C={(x,y,z)∈ℝ3 : x2+y2+z2≤1, z≥√(3x2+8y2)} be a subset of ℝ3. Calculate ∫∫∫C z dxdydz.
2. Relevant equations
Spherical coordinates, given by x=ρsinΦcosθ, y=ρsinΦsinθ, z=ρcosΦ, and cylindrical coordinates which are x=ρcosα, y=ρsinα, z=z
3. The attempt at a solution
The problem is that, starting from the set C I believe that √(3x2+8y2)≤z≤√(1-x2-y2) so that I should integrate the "∫zdz part" between these two extremes. But when it comes to the conditions on x and y I don't know how to proceed further. I tried to use spherical coordinates so that the first condition becomes ρ2≤1 but then the second one becomes cos(Φ)≥√(3sin2(Φ)cos2(θ)+8sin2(Φ)sin2(θ)) which is worse than the original condition. I also tried to use cylindrical coordinates so that even though the first condition does not improve much the second one becomes z2≥3ρ2cos2(α)+8ρ2sin2(α), which is better than the one I obtained using spherical coordinates, but even so I didn't go very far. Obviously there is some idea I am missing, what is in your opinion the best way to go on?
Last edited: Sep 6, 2015
2. Sep 6, 2015
### Zondrina
You are right about using spherical co-ordinates. Applying the transformation will yield $0 \leq \rho \leq 1$ as you have found already.
The limits for $\theta$ can be found by using your imagination. Are you integrating over a full $2 \pi$ or only half of that? If you sketch the region, it will be clear what the limits for $\theta$ are.
As for $\phi$, you have the right idea, but you seem to have stopped mid-computation. Once you get here:
$$\cos( \phi ) \geq \sqrt{3 \cos^2( \theta ) \sin^2( \phi ) + 8 \sin^2( \theta ) \sin^2( \phi )}$$
Factor out $\sin^2( \phi )$ to obtain:
$$\cot( \phi ) \geq \sqrt{3 \cos^2( \theta ) + 8 \sin^2( \theta )}$$
3. Sep 6, 2015
### Ray Vickson
Personally, I would not use cylindrcal or spherical coordinates; I would first just attempt to determine the $(x,y)$ region $R_{xy}$, which is the region where $\sqrt{1-x^2-y^2} \geq \sqrt{3x^2+8y^2}.$
Then I would do the integral
$$\int \! \int_{R_{xy}} \left( \int_{\sqrt{3x^2+8y^2}}^{\sqrt{1-x^2-y^2}} z \, dz \right) \, dx dy .$$
4. Sep 6, 2015
### Zondrina
This way is nice too because you can switch over to polar co-ordinates once you obtain an $xy$ integral.
5. Sep 6, 2015
### Aner
Thank you very much! I will try to use both the approaches. I didn't go further with the calculations of the second conditions using spherical coordinates because I didn't understand how to use that condition but now I understand that I needed your last passage, and I didn't really thought that I could switch to spherical coordinates midway through the integration. I will try to solve the integral and when I'm done I will post my solution here | 2018-01-22T11:17:21 | {
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https://math.stackexchange.com/questions/2450978/where-did-i-go-wrong-with-this-integral | # Where did I go wrong with this Integral?
My integral is $$\int \frac{dx}{x \sqrt{3-x^2}}$$
so I used trig substitution,
let $x$ = $\sqrt{3}\cos{\theta}$
let $dx$ = $-\sqrt{3}\sin{\theta}\ d\theta$
$$\int \frac{-\sqrt{3}\sin{\theta}\ d\theta}{\sqrt{3}\cos{\theta}\sqrt{3-(\sqrt{3}\cos{\theta})^2}}$$
$$=-\int \frac{\sin{\theta}\ d\theta}{\cos{\theta}\sqrt{3(1-\cos^2{\theta})}}$$ $$=\frac{-1}{\sqrt{3}} \int \frac{\sin{\theta}\ d\theta}{\cos{\theta}\sin{\theta}}$$ $$=\frac{-1}{\sqrt{3}} \int \sec{\theta}\ d\theta$$ $$=\frac{-1}{\sqrt{3}} \log{\left(\tan{\theta}+\sec{\theta}\right)} + C$$
Next I converted $\theta$ back into x using the triangle. If $\cos{\theta} = \frac{x}{\sqrt{3}}$ then side $a$ of the triangle = $x$ and side $c = \sqrt{3}$. The missing side, $b$, is $\sqrt{3-x^2}$.
$\tan\theta = \frac{b}a = \frac{\sqrt{3-x^2}}{x}$
$\sec\theta = \frac{1}{\cos\theta} =\frac{\sqrt{3}}{x}$.
Substituted back in my answer is
$$-\frac{1}{\sqrt3} \log\left( \frac{\sqrt{3-x^2}}{x} +\frac{\sqrt{3}}{x}\right)+C$$
but it is incorrect. I checked Wolfram Alpha and it returned
$$\frac{\log{x} -\log{ \left(\sqrt{9-3x^2}+3 \right)}}{\sqrt3}+C$$
sometimes Wolfram Alpha returns an equivalent answer, just written in different terms, so I briefly plotted both equations here in Desmos to check if they were equal, but unfortunately, they are not equivalent, although they are close. I've been trying this problem for over an hour and I can't figure out what went wrong. Can anyone show me what to do?
• They are equivalent, they differ by a constant. – Nick Pavlov Sep 29 '17 at 23:40
• Answer confirmed by Mathematica. – David G. Stork Sep 30 '17 at 0:00
There's nothing wrong. Factor out a $\sqrt 3$ from their (second) log term, and absorb it into the constant of integration. Remember, also, that $\log(a/b) = \log a - \log b$. | 2019-09-21T11:50:37 | {
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https://www.physicsforums.com/threads/triple-integral.603193/ | # Triple Integral
1. May 4, 2012
### cjc0117
Hi everyone. I am trying to integrate the following:
$\int^{\frac{π}{2}}_{-\frac{π}{2}}\int^{acosθ}_{0}\int^{\sqrt{a^{2}-r^{2}}}_{-\sqrt{a^{2}-r^{2}}}rdzdrdθ$
Here's my work:
$=2\int^{\frac{π}{2}}_{-\frac{π}{2}}\int^{acosθ}_{0}r\sqrt{a^{2}-r^{2}}drdθ$
I use substitution with u=a2-r2 to get:
$=-\int^{\frac{π}{2}}_{-\frac{π}{2}}\int^{a^{2}sin^{2}θ}_{a^{2}}u^{\frac{1}{2}}dudθ$
$=-\frac{2}{3}a^{3}\int^{\frac{π}{2}}_{-\frac{π}{2}}(sin^{3}θ-1)dθ$
$=-\frac{2}{3}a^{3}\int^{\frac{π}{2}}_{-\frac{π}{2}}sin^{3}θdθ+\frac{2}{3}a^{3}\int^{\frac{π}{2}}_{-\frac{π}{2}}dθ$
sin3θ is an odd function so the first integral is equal to zero:
$=\frac{2}{3}a^{3}\int^{\frac{π}{2}}_{-\frac{π}{2}}dθ$
$=\frac{2}{3}πa^{3}$
However, I have seen other solutions online that give the actual answer as 2a3(3π-4)/9. The authors of these solutions change the limits of integration of the original triple integral by taking advantage of symmetry. More specifically, I noticed that the person changed the limits of θ from 0 to π/2 by multiplying the integral by 2. Then when you get to the point when you integrate sin3θ, the integral no longer equals zero. However, I thought you could only do this if the region of integration has no θ dependence. And in any case, why should it matter whether I change the limits of integration or not? It's still the same integral, right? I have a feeling I'm making a very dumb mistake somewhere but I can't find out where. Thanks for any help.
Last edited: May 4, 2012
2. May 4, 2012
### sharks
You could also have used the multiple angle formula to convert sin3θ into sin(3θ) or used the substitution: let u = cosθ.
3. May 4, 2012
### LCKurtz
You didn't make a dumb mistake, but it is a mistake nevertheless, sort of a subtle one. The problem is where you have $$u^{\frac 3 2}|_{a^2}^{a^2\sin^2\theta}$$and you substitute the $a^2\sin^2\theta$ in for the $u$. That gives you $$(a^2\sin^2\theta)^{\frac 3 2}$$That does not simplify to $a^3\sin^3\theta$ when $\theta$ is between $-\frac \pi 2$ and $0$ because $a^2\sin^2\theta\ge 0$ and when you raise it to the 3/2 power it must be nonnegative. It's the old problem that $\sqrt{x^2}= |x|$, not $x$. What it does simplify to is $a^3|\sin^3(\theta)|$, which is not an odd function and doesn't give $0$ in later steps.
4. May 4, 2012
### cjc0117
Thanks LCKurtz. That ,makes sense. Now what I don't understand is this:
$\int^{\frac{π}{2}}_{-\frac{π}{2}}\int^{acosθ}_{0}\int^{\sqrt{a^{2}-r^{2}}}_{-\sqrt{a^{2}-r^{2}}}rdzdrdθ=4\int^{\frac{π}{2}}_{0}\int^{acosθ}_{0}\int^{\sqrt{a^{2}-r^{2}}}_{0}rdzdrdθ$
It makes sense to me with the integral with respect to z, since the region of integration is symmetrical about the xy-plane, and f(θ,r,z)=r is even with respect to z. As for the integral with respect to θ, I realize that the region of integration is also symmetrical about the yz-plane, but I am not immediately inclined to think that the function f(θ) being integrated at that time will be even with respect to θ. How can you tell? Am I even thinking of symmetry in the right way?
EDIT: Is it because g(θ)=acosθ is even with respect to θ, so any composite function f(g(θ)) that results from the integration with respect to r will also be even?
Last edited: May 4, 2012
5. May 4, 2012
### LCKurtz
$\cos\theta$ is even and any function of an even function is even. See if you can prove that.
6. May 4, 2012
### cjc0117
Let x=g(t) be an even function, that is, x=g(t)=g(-t). Then f(x)=f(g(t))=f(g(-t)) must also be an even function.
Is that a sufficient proof? It sounds more like a conclusion.
7. May 4, 2012
### LCKurtz
That's about all there is to it. I would write it like this: Let $h(t) = f(g(t))$. You want to show $h(t) = h(-t)$. So calculate $h(-t)$:
$h(-t) = f(g(-t)) = f(g(t))\hbox{ (since g is even) } = h(t)$.
8. May 4, 2012
### cjc0117
Okay, thanks again. This problem was driving me insane before but it's all clear now. | 2017-12-18T22:07:12 | {
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http://openstudy.com/updates/4f948b90e4b000ae9ecac361 | ## xEnOnn 3 years ago Each box is packed with 40 units of components. An inspector will randomly select 5 units from a box to test if the components are in good condition. A box will be rejected if 2 or more tested units are defective. A shipment includes 400 boxes of the components. Each box will be tested by the inspector. The whole shipment will be returned if more than 90 boxes of components are rejected. The ordering records in the past showed that 20% of the boxes were rejected. What is the probability that a given shipment will be returned?
1. SmoothMath
Is there enough information given in this problem? My initial impression is no.
2. xEnOnn
But this seems like all the question has for me to answer it. And because this is an exam question, I think there should have enough information provided to solve the problem.
3. Zarkon
looks binomial
4. Chlorophyll
@Zarkon, also seems like hypergeomtric!
5. Zarkon
but that part doesn't give you any information
6. xEnOnn
My initial intuition was binomial too. But if this is binomial, it means that the selection of the boxes for the inspection has replacement, which I don't think it has.
7. xEnOnn
hmm..but wouldn't there be a chance that one of the selected boxes for inspection is a repeated box? ie, a box that has already been inspected earlier?
8. Zarkon
each box is being tested...the box is either good or bad...the probability it is bad is 20%
9. Chlorophyll
Of course without replacement!
10. Chlorophyll
20% 400 = 80
11. SmoothMath
...Doing it as binomial seems like a serious pain. You'd have to add up P(89) + P(88)+ P(87) +... + P(2) + P(1)
12. Zarkon
you could use a normal approximation
13. Zarkon
I'm all about the normal approximation today ;)
14. SmoothMath
Tell me more, Zarkon. How could we determine the SD?
15. xEnOnn
Could the numbers be used as a hypergeometric case? I was trying to figure out the numbers from this question for the the hypergeometric formula's parameters but doesn't look possible.
16. xEnOnn
haha...normal approximation today...
17. Zarkon
if it is binomial with n trials and p prob of success...then the sd is $\sqrt{n\cdot p\cdot (1-p)}$
18. Zarkon
brb
19. SmoothMath
Oh really? That's pretty cool. That does it for you then. You have the mean will clearly be 80. That SD gives you a z score for 90. Problem's doneso.
20. SmoothMath
Normal approximation was the first thing I thought of, but I figured you needed a given SD. It makes sense that you can calculate it though.
21. xEnOnn
This is assuming that there is replacement. If there is replacement, it will follow binomial distribution. However, there is a possibility that we picked out another box which we had tested earlier.
22. SmoothMath
I don't think so, EnOnn. It says that they simply test all 400 boxes. There's no chance of testing a box twice.
23. Zarkon
you don't...you are looking at all the boxes...and checking each one once.
24. Zarkon
all you need is independence between boxes
25. xEnOnn
ohhhh yea you are right! I was looking at the units! It is the units of components that are only randomly picked.
26. xEnOnn
Thanks a lot!! :D
27. xEnOnn
oh yes zarkon, in this case, they are all indepent. Thanks!! :D
28. xEnOnn
independent*
29. SmoothMath
Mean: 80 SD: sqrt(400*0.2*(.8)) = 8 90 is a z-score of 1.25
30. SmoothMath
|dw:1335139433649:dw|
31. xEnOnn
yea....thanks! one more condition which makes this approximation possible is also that n(1-p) and np are both greater than 15. Since this condition is met, the approximation is pretty close.
32. Zarkon
For a slightly better answer one should use a continuity correction. again...i feel like a broken record ;)
33. Zarkon
most texts want n(1-p) and np to be larger than 5
34. xEnOnn
oh yea you are right. since i want 90 to be in as well, the value for the x-bar in the z-score would be 85.5.
35. xEnOnn
i mean 89.5
36. Zarkon
again this is a CLT result...since the binomial distribution is the sum of independent Bernoulli random variables
37. xEnOnn
yea...the CLT again like the previous one. haha... thanks so much! | 2015-11-29T17:56:03 | {
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https://math.stackexchange.com/questions/3336592/how-to-generalize-this-version-of-tarski-s-fixed-point-theorem/3336831 | # How to generalize this version of Tarski’s Fixed Point Theorem?
I could prove the following result from my Real Analysis course:
Let $$f:[0,1] \rightarrow [0,1]$$ be an increasing mapping. Then it has a fixed point.
I understand that this is a very baby version of Tarski’s Fixed Point Theorem. Now, I wish to generalize this a little bit and get the following:
Let $$f:[0,1]^n \rightarrow [0,1]^n$$ in which $$f$$ is increasing in the sense that if $$y \geq x$$ coordinate wise then $$f(y) \geq f(x)$$ coordinate wise. Then, f has a fixed point.
From my point of view, we could just pick a point $$x_0 \in [0,1]^n$$, fix all coordinates but one and apply the above lemma to that coordinate. Then, when the first coordinate of the fixed point is found, we do the same for the second and so on.
However, I am not sure this route would be successful and even if it is, I can’t write the extension formally. Any ideas? Thanks a lot in advance!
• Note that $f(\cdot,x_2) : [0,1]\to [0,1]^2$ for fixed $x_2$. – amsmath Aug 28 '19 at 3:08
• I tried to go through your route. But the range is not ordered anyway. I could fix $x_2$ in the range too, and I would have a mapping from $[0,1]$ as before. But then I would have a fixed point whose second coordinate is $x_2$. As $x_2$ was arbitrary, this would prove that we would have uncountable many fixed points, right? It seems too strong. What am I missing @amsmath? – Raul Guarini Aug 28 '19 at 3:15
• I was trying to say that you cannot apply your lemma to that function because it does not map [0,1] to itself. – amsmath Aug 28 '19 at 3:17
• Ok, thank you! Do you have another idea to generalize this result then? @amsmath – Raul Guarini Aug 28 '19 at 3:20
The "baby version" is not sufficient to obtain your result for $$n > 1$$ as a corollary. This is possible only in case that $$f : [0,1]^n \to [0,1]^n$$ has the special form $$f(x_1,\dots,x_n) = (\phi_1(x_1),\dots,\phi_n(x_n))$$ with continuous increasing $$\phi_i : [0,1] \to [0,1]$$. But in general $$f$$ has the form $$f(x_1,\dots,x_n) = (f_1(x_1,\dots,x_n),\dots,f_n(x_1,\dots,x_n))$$ where each $$f_i :[0,1]^n \to [0,1]$$ depends on all variables.
In fact, your result is true, but you will need a new proof. See Asaf Karagila's answer.
Let me close with a remark concerning your strategy to find a fixed point. It is not expedient to pick some $$x_0 =(x^0_1,\dots,x^0_n)$$ and to consider the functions obatined by fixing all but one coordinate. Let us see what happens for $$n=2$$. You consider the function $$f^1 : [0,1] \to [0,1], f^1(x) = f_1(x,x^0_2)$$ and conclude that it has a fixed point $$\xi_1$$, i.e. $$f_1(\xi_1,x^0_2) = \xi_1$$. Next you consider $$f^2 : [0,1] \to [0,1], f^2(x) = f_1(x^0_1,x)$$ and conclude that it has a fixed point $$\xi_2$$, i.e. $$f_2(x^0_1,\xi_2) = \xi_2$$. But there is no reason why you should have $$f(\xi_1,\xi_2) = (\xi_1,\xi_2)$$.
The Knaster–Tarski fixed point theorem states that:
Suppose that $$(L,\leq)$$ is a complete lattice and $$f\colon L\to L$$ is increasing, i.e. $$x\leq y\implies f(x)\subseteq f(y)$$. Then $$\{x\in L\mid f(x)=x\}$$ is a complete lattice under the induced order.
Note that a complete sublattice cannot be empty, since $$\varnothing$$ must have a supremum and infimum (which make the bottom and top elements of the lattice).
Note that $$[0,1]$$ is a complete lattice, and $$[0,1]^n$$ is also a complete lattice. Granted, now, most uses of this theorem really just care that there is a fixed point. This is commonly used in the proof of the Cantor–Bernstein theorem when we want to find a set closed under some function.
Just finding a fixed point is easy. Let $$D=\{x\in L\mid x\geq f(x)\}$$ and let $$s=\inf D$$. Even if $$D$$ is empty, it still has a infimum, it would just be the top element of $$L$$. This is because $$L$$ is a complete lattice.
Now, if $$s, then $$f(s)\leq f(x)$$. Since this holds for all $$x\in D$$, it follows that $$f(s)\leq s$$, by the virtue of being an infimum. Therefore $$s\in D$$, and it is in fact a minimum. But now we have that $$f^2(s)\leq f(s)\leq s$$. So $$f(s)\in D$$ as well, which means that $$s\leq f(s)$$—again by the virtue of being an infimum—and therefore $$f(s)=s$$. We can even show that $$s$$ is the smallest fixed point, i.e. the bottom element of the fixed points lattice.
If we assume continuity of $$f$$, in the sense that it respects the $$\sup$$ operator, we get an even easier proof of what is known as the Kleene fixed points theorem.
Let's use $$0$$ to denote the bottom element of $$L$$. Then $$0\leq f(0)\leq f^2(0)\leq ...$$, let $$x=\sup\{f^n(0)\mid n\in\Bbb N\}$$, which exists since $$L$$ is a complete lattice. Since $$f$$ is continuous we get that $$f(x)=f(\sup\{f^n(0)\mid n\in\Bbb N\}) = \sup\{f(f^n(0)\mid n\in\Bbb N\}=\sup\{f^{n+1}(0)\mid n\in\Bbb N\}=x.$$
Finally, we apply these proofs to your question. And you can see that this is essentially the same proof for both $$[0,1]$$ and $$[0,1]^n$$. And in fact, much much more. | 2020-01-22T11:31:24 | {
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https://crypto.stackexchange.com/questions/74839/derivation-of-birthday-paradox-probability/74852 | # Derivation of birthday paradox probability
I am trying to come up with an explanation of the probability of birthday collision.
$$P$$(no collision among t people) = $$(1− \frac{1}{365}) · (1-\frac{2}{365}) ··· (1-\frac{t-1}{365})$$
For one person, the probability of no collision is 1, which is trivial since a single birthday cannot collide with anyone else’s. For the second person, the probability of no collision is 364 over 365, since there is only one day, the birthday of the first person, to collide with:
$$P$$(no collision among 2 people) = $$(1− \frac{1}{365})$$
If a third person joins the party, he or she can collide with both of the people already there, hence:
$$P$$(no collision among 2 people) = $$(1− \frac{1}{365})·(1−\frac{2}{365})$$
While it is clear how we get the probability of collision for 2 people, it is not intuitive to me that how we get the probability of collision between 3 people. I'd expect the probability would be $$(1−\frac{2}{365})$$. For example, when you roll the dice, the probability of $$6$$ is $$\frac{1}{6}$$, and the probability for 5 or 6 is $$\frac{2}{6}$$. It is not $$\frac{1}{6}$$·$$\frac{2}{6}$$ which seems to be the case in birthday collisions.
I'd appreciate an answer with an intuitive explanation.
The probability becomes more intuitive when one pictures the $$t$$ persons entering one by one in the room.
Before the first person enters, there's no collision/coincidence of birth-date, thus probability of no collision is $$P_0=1$$.
When the first person enters, there can't be a collision/coincidence of birthdate, probability of no collision is $$P_1=1$$.
When the second person enters, there is one chance in $$365$$ that his/her birthday is the same as the first person's birthdate, thus the probability of no collision is $$P_2=1-\frac1{365}$$.
When the third person enters, either
- there was a collision at the previous stage, and then the probability of no collision is $$0$$;
- or there was no collision at the previous stage, thus there was $$2$$ persons with different birth-dates in the room, and the probability that the third person has one of their birth-dates is $$2$$ in $$365$$, thus the probability of no collision is $$1-\frac2{365}$$.
The total probability of no collision after the third person enters is obtained by adding these two probabilities weighted by the probability conditioning them. We thus have $$P_3=(1-P_2)\,0+P_2\,\bigl(1-\frac2{365}\bigr)$$, thus $$P_3=\bigl(1-\frac1{365}\bigr)\bigl(1-\frac2{365}\bigr)$$
The same reasoning gives $$P_{i+1}=(1-P_i)\,0+P_i\bigl(1-\frac i{365}\bigr)$$, that is $$P_{i+1}=P_i\bigl(1-\frac i{365}\bigr)$$.
From which it comes $$P_t=\displaystyle\prod_{i=0}^{t-1}\,\Bigl(1-\frac i{365}\Bigr)$$.
For derivation of formulas used in a cryptographic context where the numbers are huge, see my Birthday problem for cryptographic hashing, 101.
Note: $$P$$, $$t$$, and $$365$$ here are $$q$$, $$n$$ and $$k$$ there.
• makes perfect sense. thank you – sanjihan Oct 5 '19 at 13:16
It may help to write it out using the chain rule of probability theory as:
\begin{align*} \Pr[\text{no collision among 2}] &= 1 - \frac{1}{365} \\ \Pr[\text{no collision among 3}] &= \Pr[\text{no collision with 3rd, and no collision among first 2}] \\ &= \Pr[\text{no collision with 3rd} \mid \text{no collision among 2}] \\ &\qquad \cdot \Pr[\text{no collision among 2}] \\ &= \biggl(1 - \frac{2}{365}\biggr) \biggl(1 - \frac{1}{365}\biggr), \end{align*}
and so on with $$\Pr[\text{no collision among 4}]$$, $$\Pr[\text{no collision among 5}]$$, etc.
In particular, $$1 - \frac{n - 1}{365}$$ is the probability that the $$n^{\mathit{th}}$$ person has a different birthday from the $$n - 1$$ prior persons, given that the $$n - 1$$ prior persons all have distinct birthdays—that is, $$1 - \frac{n-1}{365}$$ is $$\Pr[\text{no collision with n^{\mathit{th}}}$$ $$\mid$$ $$\text{no collisions among n - 1}]$$, but you read it as $$\Pr[\text{no collision among n}]$$.
Sometime back I did a detailed study of this problem.
So the problem was
"How many people(n) you need in a room such that m of them sharing birthday is 50%.?"
We know that for m=2, we need n=23 people such that probability of any two of them sharing birthday is 50%.
Suppose we have find n, such that probability of m=3 people share birthday is 50%.
We will calculate how 3 people out of n doesn’t share a birthday and subtract this probability from 1.
• All n people have different birthday.
• 1 pair (2 people) share birthday and the rest n-2 have distinct birthday.
• Number of ways 1 pair (2 people) can be chosen = C(n, 2)
• This pair can take any of 365 days
• For these n-2 people they can pick 365–1 birthdays.
• Next we make 2 group of 2 people and rest n-4 have distinct birthday.
Like this we can recursively create more groups and find all possible combination where m people doesn't share birthday. | 2020-02-19T07:08:36 | {
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https://math.stackexchange.com/questions/2912401/proving-that-a-sequence-is-not-cauchy | # Proving that a sequence is not Cauchy
I'm trying to prove that the sequence $\left(\frac{1}{2},\frac{1}{3},\frac{2}{3},\frac{1}{4},\frac{2}{4},\frac{3}{4},\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5},\cdots\right)$ is not a Cauchy sequence.
I know that a sequence of real numbers is not Cauchy if there exists an $\epsilon>0$ such that, for all $N\in\mathbb{N}$, there exists $m,n>N$ such that $|x_{m}-x_{n}|\geq\epsilon$. It intuitively makes sense to me that the sequence cannot be Cauchy, as the distance between points where the denominator changes (like $\cdots\frac{99}{101},\frac{100}{101},\frac{1}{102},\cdots$) keeps growing larger. However, I'm not sure how to find indices $m$ and $n$ in general with $|x_{m}-x_{n}|\geq\epsilon$. Thanks in advance for any help!
Let $(a_n)_{n\in\mathbb N}$ be your sequence. Take $\varepsilon=\frac12$. Given $N\in\mathbb N$, take $n\geqslant N$ such that $a_n$ is of the form $\frac k{k+1}$ for some $k\in\mathbb N$ and let $m=n+1$. Then $a_m=\frac1{k+2}$. Therefore,$$\left\lvert a_m-a_n\right\rvert=\frac k{k+1}-\frac1{k+2}\geqslant\frac12=\varepsilon.$$
Or you can say that your sequence diverges, since the subsequence$$\frac12,\frac13,\frac14,\ldots$$converges to $0$, whereas the subsequence$$\frac12,\frac23,\frac34,\ldots$$converges to $1$. So, the sequence diverges and therefore it is not a Cauchy sequence.
You only need such an $\epsilon$ to exist, so you can choose a convenient value.
Then you need to show that there are gaps bigger than $\epsilon$ between elements of the sequence as far a=out as you care to go - that gaps as big as $\epsilon$ don't fade out and disappear in the tail of the sequence.
Well you have identified some chunky gaps which persist (you don't need every gap to be big). What $\epsilon$ would work for the gaps you have identified?
Cauchy sequence: $$\forall\varepsilon>0\exists N\forall n,m\quad n,m>N\implies|a_n-a_m|<\varepsilon$$
We want to show the negative:$$\exists\varepsilon>0\forall N\exists n,m\quad n,m>N\land |a_n-a_m|\ge\varepsilon$$
Take $\varepsilon=1/10$, and then, for all $N$ take $n=\sum_{i=1}^{N+1} i=\frac{(N+1)^2+N+1}{2}, m=n+1$ and you have $a_n$ an element in the form of $\frac{k}{k+1}$ and $a_m$ in the form of $\frac{1}{k+2}$ | 2019-10-15T11:46:20 | {
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https://math.stackexchange.com/questions/1174741/mean-value-theorem-problem | Mean Value Theorem problem
Given: $f:[0, 27] \to \mathbb R$ such that, $f(0)=0$ , $f(10)=1$ , $f(27)=1$ , where $f(x)$ is differentiable.
Prove that , for some $\alpha$, $\beta$ $\in(0,3)$ , the relation
$$2\int_0^{27} f(x)\, dx = 9[\alpha^{2}f(\alpha^{3})+\beta ^{2} f(\beta^{3})]$$ holds.
I think this question is a question on the lagrange's mean value theorem. By the form of the right hand side, I think I should use another function $g(x)=\int_0^{x^{3}} f(x) dx$
so that $g'(x)= 3 x^{2} f(x^{3})$
but I cannot figure out the limits to apply, or how to proceed next
• Is there a multiple of 2 in the integrand on the left hand side?Just need to be clear. Mar 4, 2015 at 8:09
• @Mathemagician1234 Yes, there is. Mar 4, 2015 at 8:12
• That's a tricky problem,it's not obvious how to proceed and I'm too tired to help. I think you have a good observation about the right hand side, but I have no idea how to go from there either. Life without caffeine is hard. I'll have to sit this one out,hopefully one of the analysts in here can help you out. : ) Mar 4, 2015 at 8:20
Put $x=y^{3}.$ You have $$2\int_{0}^{27}f\left(x\right)dx=\int_{0}^{27}f\left(x\right)dx+\int_{0}^{27}f\left(x\right)dx=3\int_{0}^{3}y^{2}f\left(y^{3}\right)dy+3\int_{0}^{3}y^{2}f\left(y^{3}\right)dy.$$ Now for the first mean value theorem for integration we have that exists some $\gamma\in\left(0,3\right)$ such that $$3\int_{0}^{3}y^{2}f\left(y^{3}\right)dy=3\gamma^{2}f\left(\gamma^{3}\right)\int_{0}^{3}1dy=9\gamma^{2}f\left(\gamma^{3}\right)$$ so $$2\int_{0}^{27}f\left(x\right)dx=9\left(\gamma^{2}f\left(\gamma^{3}\right)+\gamma^{2}f\left(\gamma^{3}\right)\right)$$ and so your equality holds with the choice $\alpha=\beta=\gamma.$
• There may exist more than $1$ such $\gamma$, thus $\alpha$ and $\beta$ are not necessarily the same. Mar 4, 2015 at 8:36 | 2022-10-03T14:11:32 | {
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https://math.stackexchange.com/questions/623036/if-a-matrix-commutes-with-all-diagonal-matrices-must-the-matrix-itself-be-diago | # If a matrix commutes with all diagonal matrices, must the matrix itself be diagonal?
I'm new to stackexchange so feel free to correct my style/format/logic etc.
The question is this: let's say $A$ is a square matrix of size $n$. I would like to show that $AD = DA$, for any diagonal matrix $D$ also of size $n$, if and only if $A$ is also diagonal.
I think I have some of the proof but am not very confident in it.
$\Leftarrow$: If A is diagonal, it is not too hard to show that $AD = DA$, because multiplying two diagonal matrices just amounts to multiplying the corresponding diagonal entries.
$\Rightarrow$: (i) $DA$ is found by multiplying each row in A with the corresponding entry along the diagonal in $D$. $AD$ is found by multiplying each column in $A$ with the corresponding entry along the diagonal in $D$. Since $AD = DA$, this product has to be symmetric.
(ii) Now suppose $A$ weren't a diagonal matrix. Then if we make the entries along the diagonal in $D$ all different, $AD$ won't be symmetric anymore (?). This contradicts (i), so we have shown both ways.
Does this work?
[edited]
• What if $D = I$? Dec 31, 2013 at 8:21
• edited to say that this should work for any D Dec 31, 2013 at 8:24
• To show the $\Rightarrow$ direction consider choosing for $D$ the matrices $D_i$, where $D_i$ is all zero except the element on the i-th diagonal is $1$. Then $A D_i$ has the $i$-th column equal to that of $A$ and the rest is zero. On the other hand $D_i A$ has the $i$-th row equal to that of $A$ and the rest is zero. If these two have to be equal then all non-diagonal elements of $A$ have to be zero. Proof by contradiction is not necessary. Dec 31, 2013 at 8:31
• ooh, nice. thanks! Dec 31, 2013 at 8:38
Here's a geometric formulation. If two matrices $A,D$ commute, then all eigenspaces for $D$ must be $A$-stable (if $v$ is eigenvector for $D$ and eigenvalue $\lambda$, then $DA\cdot v=AD\cdot v=A\cdot\lambda v=\lambda(A\cdot v)$, so $A\cdot v$ is eigenvector for $D$ and eigenvalue $\lambda$ as well). Now for every standard basis vector $e_i$ there is a diagonal matrix $D$ for which $\langle e_i\rangle$ is an eigenspace for $D$, for instance the elementary matrix $D=E_{i,i}$. Since $A$ must commute with all such $D$, it must stabilise every line $\langle e_i\rangle$, and this forces $A$ to be diagonal.
If your field has at least $n$ elements (in particular if it is infinite), you can arrange for a single $D$ to have every line $\langle e_i\rangle$ as eigenspace, and then just commuting with this single $D$ will force being diagonal.
Pick a $D$ so that all except one diagonal entry is zero, and the nonzero entry is $D_{ii}=1$
$$(D A)_{ij} = a_{ij}$$ and for $j \neq i$ $$(A D)_{ij} = 0$$ Hence $a_{ij}=0$.
Since $i$ and $j$ are arbitrary, $A$ has to be diagonal.
Alternate approach: Let $$D = \text{diag}(1,2,3,4,\cdots n)$$ then $$(DA)_{ij} = i \cdot a_{ij}, \text{ and } (AD)_{ij} = j \cdot a_{ij}$$ So if $i \neq j$, $a_{ij} =0$ | 2022-06-26T09:35:41 | {
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https://plainmath.net/83723/on-the-dirac-equation-expands-the-gamma | # on the Dirac equation, expands the gamma^nu d_nu term as: gamma^nu d_nu=gamma^0 d/dt + gamma * nabla where gamma=(gamma^1, gamma^2, gamma^3), but to my knolwdge gamma^nu d_nu=gamma^nu mu_(nu v) d^v=gamma^0 d/dt + gamma * nabla
on the Dirac equation, expands the ${\gamma }^{\mu }{\mathrm{\partial }}_{\mu }$ term as:
${\gamma }^{\mu }{\mathrm{\partial }}_{\mu }={\gamma }^{0}\frac{\mathrm{\partial }}{\mathrm{\partial }t}+\stackrel{\to }{\gamma }\cdot \stackrel{\to }{\mathrm{\nabla }}$
where $\stackrel{\to }{\gamma }=\left({\gamma }^{1},{\gamma }^{2},{\gamma }^{3}\right)$, but to my knowledge,
${\gamma }^{\mu }{\mathrm{\partial }}_{\mu }={\gamma }^{\mu }{\eta }_{\mu \nu }{\mathrm{\partial }}^{\nu }={\gamma }^{0}\frac{\mathrm{\partial }}{\mathrm{\partial }t}-\stackrel{\to }{\gamma }\cdot \stackrel{\to }{\mathrm{\nabla }}$
using the convention ${\eta }_{\mu \nu }=\mathrm{diag}\left(+,-,-,-\right)$.
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Kali Galloway
Yes. You are missing the fact that he is using the convention
$\mathrm{\nabla }=\left({\mathrm{\partial }}_{1},{\mathrm{\partial }}_{2},{\mathrm{\partial }}_{3}\right)$
as opposed to
$\mathrm{\nabla }=\left({\mathrm{\partial }}^{1},{\mathrm{\partial }}^{2},{\mathrm{\partial }}^{3}\right)$
The first convention is by far the most common in my experience. | 2022-08-13T02:51:59 | {
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https://stats.stackexchange.com/questions/270644/is-there-an-accepted-name-for-this-error-metric | # Is there an accepted name for this error metric?
I've come across an error metric used to quantify a model's reconstruction error: $$\varepsilon = \frac{\sum_i{\left(y_i-m_i\right)^2}}{\sum_i{\left(y_i-\bar{y}\right)^2}}$$ where $y_i$ is the $i$th data point, $m_i$ is the model's estimate of the $i$th data point, and $\bar{y}$ is the mean of all data points. The numerator is the total squared error of the model, and the denominator is the squared deviation from the mean of the data.
Does this metric have a standard name? If not, what would you call it?
• Error metrics seem on-topic to me. – Silverfish Mar 29 '17 at 19:53
• it's not so obscure, we use it compare models in-sample when the specifications are very different and usual metrics such as AIC are not applicable. e.g. comparing AIC/BIC of difference and levels models is meaningless – Aksakal Mar 29 '17 at 20:24
This is related to a coefficient of determination ($R^2$), actually, it's $1-R^2$, also called fraction of variance unexplained | 2021-03-02T11:05:30 | {
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http://math.stackexchange.com/questions/590302/is-the-integral-int-0-infty-sinex-dx-convergent | # Is the integral $\int_0^{\infty} \sin(e^x) \, dx$ convergent?
Can anyone help me with this problem? I've tried substitution method but I do not know how to continue. Let $u = e^x$ and $du = u\,dx$, so $dx = du/u$. So we have, $$\int \sin(u) \frac{1}{u} du = \int \frac{\sin(u)}{u} du.$$
-
This may help in evaluating the second integral: math.stackexchange.com/questions/103030/… – tylerc0816 Dec 3 '13 at 1:07
We have $$\int_0^{\infty} \sin(e^x) \, dx = \int_1^{\infty} \sin(t) \dfrac{dt}t = \int_0^{\infty} \sin(t) \dfrac{dt}t - \int_0^1 \sin(t) \dfrac{dt}t = \dfrac{\pi}2 - \text{Si}(1)$$ where the last integral can be obtained from here. Also the integral $\displaystyle \int_0^1 \dfrac{\sin(t)}t dt$ is clearly bounded since $\sin(t) \in \left( \dfrac2{\pi}t, t\right)$ for $t \in (0,\pi/2)$.
-
Is the lower limit of integration $1$ in the second integral? – tylerc0816 Dec 3 '13 at 1:11
@tylerc0816 Yes, thanks. Corrected. – user17762 Dec 3 '13 at 1:13
Another question dealing with the computation of $\displaystyle\int_0^\infty\frac{\sin x}x\,dx$ is here. I added a real variable argument as an answer. – Andres Caicedo Dec 3 '13 at 2:17
Integrate by parts, using $u=e^{-x}$ and $dv=e^x \sin(e^x)\,dx$.
Then $du=-e^{-x}\,dx$ and we can take $v$ to be $-\cos(e^x)$. The rest is easy, the integral we end up with converges.
Added: The integral from $0$ to $B$ is $$\left. -e^{-x}\cos(e^x)\right|_0^B -\int_0^B e^{-x}\cos(e^x)\,dx.$$ There is no problem as $B\to\infty$, since $|\cos(e^x)|$ is bounded.
-
Here's an alternative (and elementary) way to solve the problem. Consider $$f(x) = \int_x^{x+1} \sin e^t \,dt.$$ Making the substitution $v = e^t,$ and integrating by parts, one obtains $$e^{x} f(x) = \cos e^x - e^{-1}\cos e^{x+1} + r(x),$$ where $|r(x)|< e^{-x}.$ Accordingly, the integral in question, is equivalent to calculating $f(0) + f(1) + \cdots,$ which is now a telescoping sum, with remainder term going to $0$ as $x \to \infty.$
Remark: This is essentially Andre Nicolas's solution written slightly differently.
- | 2015-05-07T08:40:35 | {
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http://cotpi.com/p/1/ | |« « # » »|
Alice tosses a fair coin $$10$$ times. Bob tosses it $$11$$ times. Charlie tosses it $$12$$ times. Who is the most likely to get more heads than tails?
[SOLVED]
#### Prunthaban Kanthakumar solved this puzzle:
Bob. The probability is $$0.5$$ for Bob using symmetry. For Alice and Charlie, there is an additional case in which the number of heads is equal to the number of tails, the probability of which gets subtracted, so they get less than $$0.5$$.
#### Susam Pal from cotpi added:
Bob is the most likely to get more heads than tails.
Let the probability that a particular person gets more heads than tails, the probability that he or she gets more tails than heads and the probability that he or she gets an equal number of heads and tails be $$P_h(X)$$, $$P_t(X)$$ and $$P_{eq}(X)$$, respectively, where $$X$$ is one of Alice, Bob or Charlie. Thus, $$P_h(X) + P_t(X) + P_{eq}(X) = 1$$.
For each person, the probability that he gets more heads than tails is equal to the probability that he gets more tails than heads due to symmetry. Thus, $$P_h(X) = P_t(X)$$.
Since Alice and Charlie have an even number of coins each, the probability that each gets an equal number of heads and tails is positive. Thus, $$P_{eq}(\text{Alice}) \gt 0$$ and $$P_{eq}(\text{Charlie}) \gt 0$$. However, $$P_{eq}(\text{Bob}) = 0$$ because it is impossible to get an equal number of heads and tails by tossing an odd number of coins.
Thus, \begin{align*} & P_h(\text{Alice}) + P_t(\text{Alice}) + P_{eq}(\text{Alice}) = 1 \\ & \implies P_h(\text{Alice}) + P_t(\text{Alice}) = 1 - P_{eq}(\text{Alice}) \\ & \implies P_h(\text{Alice}) + P_t(\text{Alice}) \lt 1 \\ & \implies P_h(\text{Alice}) + P_h(\text{Alice}) \lt 1 \\ & \implies P_h(\text{Alice}) \lt 0.5. \end{align*}
Similarly,
$P_h(\text{Charlie}) \lt 0.5$
However,
\begin{align*} & P_h(\text{Bob}) + P_t(\text{Bob}) + P_{eq}(\text{Bob}) = 1 \\ & \implies P_h(\text{Bob}) + P_t(\text{Bob}) + 0 = 1 \\ & \implies P_h(\text{Bob}) + P_h(\text{Bob}) = 1 \\ & \implies P_h(\text{Bob}) = 0.5. \end{align*}
### Credit
This puzzle is taken from folklore.
|« « # » »| | 2018-01-23T08:07:21 | {
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https://math.stackexchange.com/questions/2185090/counting-the-number-of-real-roots-of-y3-3y1 | # Counting the Number of Real Roots of $y^{3}-3y+1$
Here's my question:
How many real roots does the cubic equation $y^3-3y +1$ have?
I graphed the function and it crossed the x-axis $3$ times. But my professor doesn't want a graphical explanation. So in that case, I was looking at the Fundamental Theorem of Algebra and states that a polynomial of degree n can have at most n distinct real roots. so therefore, there must be 3 real roots?
EDIT
It seems that there are numerous ways to approach this problem after all. And we can expand this to other types of polynomials as well, not just cubics.
• The cubic equation $y = x^3 + 1$ only has one real root. – Travis Willse Mar 13 '17 at 17:59
• At any rate, this can be determined using the discriminant of the polynomial. mathworld.wolfram.com/PolynomialDiscriminant.html – Travis Willse Mar 13 '17 at 17:59
• The fundamental theorem only gives an upper bound on the number of roots. But you can make your "graphical" argument mathematically correct by picking suitable points above and below the $x$ axis and then explicitly evaluating the polynomial at those points. – David K Mar 13 '17 at 18:01
• In short, yes, see en.wikipedia.org/wiki/Discriminant#Degree_3 . – Travis Willse Mar 13 '17 at 18:36
• It's not an equation unless you set it equal to 0 (or some other expression): $y^3−3y+1 = 0$ – smci Mar 14 '17 at 9:46
Nobody's explicitly written out the discriminant-based solution.
The discriminant of the polynomial $x^3 + px + q$ is $-4p^3 - 27q^2$ - see Wikipedia on discriminant of third-degree polynomials. So $y^3 - 3y + 1$ has discriminant $-4 \times (-3)^3 - 27 \times 1^2 = +81$. Since this is positive, the polynomial has three distinct real roots.
The given polynomial evaluated at $y\in\{-2,0,1,2\}$ exhibits three sign changes, hence it has at least $3$ real roots, and obviously cannot have more than three roots.
• Probably not the point of the question, but the roots are $2 \cos \left( \frac{2 \pi}{9} \right) \approx 1.532, \; \; 2 \cos \left( \frac{4 \pi}{9} \right) \approx 0.347, \; \; 2 \cos \left( \frac{8 \pi}{9} \right) \approx -1.879.$ – Will Jagy Mar 13 '17 at 18:40
• @WillJagy: that's an interesting addendum. – Jack D'Aurizio Mar 13 '17 at 18:45
• @unseen_rider: See Viète's construction of the roots of the cubic. (A derivation is further up in that same article.) Note that the middle root can be stated a bit more nicely as as $2 \cos \left( \frac{14 \pi}{9} \right)$, which puts the roots evenly spaced around a unit circle. – Michael Seifert Mar 13 '17 at 19:06
• @MichaelSeifert I don't think it follows from Viete. This is from a method of Gauss. I found a discussion in Galois Theory by David A. Cox. Then many explicit examples in Reuschle (1875). In any case, I have added a proof to my answer. – Will Jagy Mar 13 '17 at 19:15
• @WillJagy: Ah, you came at it from a completely different direction than I did. I've added an answer showing how you can also use Viète's construction to get the same answer as you did. – Michael Seifert Mar 13 '17 at 19:27
The function has extrema where
$$3y^2-3=0$$ i.e. at $$y=\pm1$$. The values at these extrema are $$3$$ and $$-1$$.
So the variations of this continuous function are $$-\infty,3,-1,\infty$$, proving that there are three changes of sign.
For a cubic polynomial, the search for the extrema may be more efficient than trial-and-error because it is immediately conclusive.
In the case of a depleted polynomial $$x^3+px+q$$
the extrema are located at $$x=\pm\sqrt{-\frac p3}$$ if $$p<0$$.
In this case there are three real roots if
$$\sqrt{-\frac p3}^3-p\sqrt{-\frac p3}+q>0\text{ and }-\sqrt{-\frac p3}^3+p\sqrt{-\frac p3}+q<0,$$ or
$$\frac{2p}3\sqrt{-\frac p3}<-|q|.$$
• Probably not the point of the question, but the roots are $2 \cos \left( \frac{2 \pi}{9} \right) \approx 1.532, \; \; 2 \cos \left( \frac{4 \pi}{9} \right) \approx 0.347, \; \; 2 \cos \left( \frac{8 \pi}{9} \right) \approx -1.879.$ – Will Jagy Mar 13 '17 at 18:40
• @ Will Jagy Well, it's good to know the solutions either way. – John Bradshaw Mar 13 '17 at 18:43
Let $f(y) = y^3 - 3y + 1$. Then you can observe that $f(0) = 1, f(1) = -1$; thus $f$ has at least one root between $0$ and $1$, by the intermediate value theorem. You can find the other two roots similarly.
You could use Cardano's Formula to verify it explicitly.
Perhaps a kind of cool way that's completely overkill:
Look at the polynomial $P(y) = \frac{1}{4}y^4 - \frac{3}{2}y^2 + y$, whose derivative is $y^3 - 3y^2 +1$
You can factor $P(y) = \frac{1}{4}y(y-2)(y^2+2y-2)$. use the quadratic formula to determine that the roots of the third factor are real.
By the Gauss Lucas-Theorem (the roots of the derivative of $P$ are contained in the convex hull of the roots of $P$), the roots of $P'(y)$ must also be real, and $P'$ is the polynomial you're interested in.
• Really nice (and a very unlikely approach for a beginning student!). – John Hughes Mar 13 '17 at 18:57
Three. The roots are $$2 \cos \left( \frac{2 \pi}{9} \right) \approx 1.532,$$ $$2 \cos \left( \frac{4 \pi}{9} \right) \approx 0.347,$$ $$2 \cos \left( \frac{8 \pi}{9} \right) \approx -1.879.$$
See page 174 in Reuschle. I first came across the method of Gauss in chapter 9 of Galois Theory by David A. Cox. Chapter 9 is called Cyclotomic Extensions. Section 9.2 is called Gauss and Roots of Unity. I then found a brief mention of the 1875 Reuschle book on page 195 of Theory of Numbers by Mathews (1892). He wrote
The reader who wishes for more numerical illustrations should consult Reuschle's Tafeln Complexer Primzahlen (Berlin, 1875).
Or, let $\omega$ be a primitive 9th root of unity, so $$\omega^9 = 1,$$ $$\omega^3 \neq 1.$$ For $$t^9 - 1 = (t-1)(t^2 + t + 1)(t^6 + t^3 + 1),$$ we find $$\omega \neq 1, \; \; \omega^2 + \omega + 1 \neq 0, \; \; \omega^6 + \omega^3 + 1 = 0.$$
Then take $$x = \omega + \frac{1}{\omega}$$ and calculate $$x^3 - 3 x + 1 = \omega^3 + 1 + \frac{1}{\omega^3} = \frac{\omega^6 + \omega^3 + 1}{\omega^3}$$
• @ Will Jagy How did you compute the roots? – John Bradshaw Mar 13 '17 at 18:11
• @i I looked it up in Reuschle (1875) page 174. Easy to prove. – Will Jagy Mar 13 '17 at 18:14
Just for fun: we note that since we have a "depressed cubic" of the form $t^3 + pt + q = 0$ (i.e., no quadratic term), the roots have a nice geometric interpretation: $$t_k = 2 \sqrt{ -\frac{p}{3}} \cos \left( \frac{1}{3} \arccos \left( \frac{3q}{2p}\sqrt{ - \frac{3}{p} } \right) + \frac{2 \pi k}{3} \right)$$ for $k = {}$0, 1, or 2. These can be seen to be the $x$-coordinates of three points equally spaced around the unit circle. In the given case, we have $p = - 3$ and $q = 1$, so this reduces to $$t_k = 2 \cos \left( \frac{1}{3} \arccos\left( -\frac{1}{2} \right) + \frac{2 \pi k}{3} \right) = \left\{ 2 \cos \left( \frac{2\pi}{9} \right), 2 \cos \left( \frac{8\pi}{9} \right), 2 \cos \left( \frac{14\pi}{9} \right) \right\}.$$ Note that $\cos (14\pi/9) = \cos (4\pi/9)$, so this agrees with Will Jagy's answer as well.
• I had not known one could go this root. Or route. – Will Jagy Mar 13 '17 at 19:31
• Michael, as you know Viete's method, can he get $x^3 + x^2 - 4x + 1?$ Reuschle page 15. Each root needs two cosine terms, denominators 13. – Will Jagy Mar 13 '17 at 20:02
• @WillJagy: I don't have Reuschle's book, but my guess is that it yields the roots in a different form. In general, Viète's method yields the three roots in the form $x_i = \bar{x} + A \cos (\theta_i)$, where $\bar{x}$ is the average of the three roots (which is equal to 1/3 of the coefficient of the quadratic term), $A$ is a real number, and the $\theta_i$ are three angles equally spaced around the unit circle. In this case, $\bar{x} = 1/3$; and since $\cos \theta \neq 1/3$ for any rational angle, it would take some work to show that these two forms are equivalent. – Michael Seifert Mar 13 '17 at 21:01
• In case of interest: the roots of $x^3 + x^2 - 4 x + 1$ are $$2 \cos \left( \frac{2 \pi}{13} \right) + 2 \cos \left( \frac{10 \pi}{13} \right) \approx 0.27389, \; \;$$ $$2 \cos \left( \frac{4 \pi}{13} \right) + 2 \cos \left( \frac{6 \pi}{13} \right) \approx 1.37720, \; \;$$ $$2 \cos \left( \frac{8 \pi}{13} \right) + 2 \cos \left( \frac{12 \pi}{13} \right) \approx -2.65109 \; \;$$ – Will Jagy Mar 13 '17 at 22:20
• frm page 15 of online books.google.com/… For example the first root is given as $$\alpha + \alpha^5 + \alpha^{-5} + \alpha^{-1}$$ where $\alpha = \exp (2 \pi i / 13)$ – Will Jagy Mar 13 '17 at 22:23
let $f(y)=y^3-3y+1$ then $f'(y)=3y^2-3$ and we have $$f'(y)=0$$ if $$y\pm1$$ since $$f''(y)=6y$$ we have $$f''(1)=6>0$$ and $$f''(-1)=-6<0$$ thus we have for $y=1$ a local Minimum and for $y=-1$ a local Maximum. Can you finish this?
One more interesting way to do it...
Suppose that a solution takes the form $x+iy$. Then some simple expansion and factoring gives $$(x^3-3x-3xy^2+1)+(3x^2y-y^3-3y)i=0$$ Now, if $y\neq0$ we can divide the imaginary component by $y$ to give $$3x^2-y^2-3=0$$ Substituting this into the real part gives $$8x^3=6x+1$$ Now, for $x\geq 1$, we have that $8x^3\geq8x>6x+1$, and so $x<1$. And for $x\leq-1$, we have that $8x^3\leq8x<6x+1$, so $x>-1$.
But if $-1<x<1$, then $y^2=3(x^2-1)$ is negative, which contradicts the claim that $y$ is real. Therefore, all solutions to our cubic are real (as assuming $y$ to be a nonzero real number produces a contradiction).
All that remains is to show that they are distinct - suppose that our polynomial is $(x-a)(x-b)^2$. Then expanding shows that $a=-2b$ (due to no $x^2$ term), and substituting this in gives $$(x+2b)(x^2-2bx+b^2) = x^3-3b^2x+2b^3$$ However, for this to match our polynomial, we must have $b^2=1$ and $b^3=\frac12$, a contradiction. Therefore, all roots are distinct.
Most of the answers are using the fact that it's a cubic. There is a more general way of approaching the problem which works for all polynomials.
We generate the Sturm sequence of $P(y)$: $$\begin{eqnarray} p_0(y) & = & P(y) = y^3 - 3y + 1 \\ p_1(y) & = & \textrm{content}( P'(y) ) = y^2 - 1 \\ p_2(y) & = & -\textrm{rem}(p_0, p_1) = (y^2 - 1)y - (y^3 - 3y + 1) = 2y - 1 \\ p_3(y) & = & -\textrm{rem}(p_1, p_2) = (2y - 1)\left(\frac{y}2 - \frac{1}{4}\right) - (y^2 - 1) = \frac{5}{4} \end{eqnarray}$$
We've ended up with a constant GCD, so there are no double roots.
Now we can count the real roots by looking at the number of sign changes when we evaluate the polynomials in the sequence at $\pm\infty$.
• $y = -\infty$: the signs are $-, +, -, +$ with three changes.
• $y = \infty$: the signs are $+, +, +, +$ with zero changes.
The difference in the number of sign changes is three, so there are three real roots.
Set $x=2\cos\varphi$; then the equation becomes $$4\cos^3\varphi-3\cos\varphi=-\frac{1}{2}$$ that is, $$\cos3\varphi=\cos\frac{2\pi}{3}$$ Thus we get $$3\varphi=\frac{2\pi}{3} \qquad\text{or}\qquad 3\varphi=\frac{2\pi}{3}+2\pi \qquad\text{or}\qquad 3\varphi=\frac{2\pi}{3}+4\pi$$ and therefore $$x=2\cos\frac{2\pi}{9} \qquad\text{or}\qquad x=2\cos\frac{8\pi}{9} \qquad\text{or}\qquad x=2\cos\frac{14\pi}{9}$$ Note that choosing $3\varphi=-\frac{2\pi}{3}$ wouldn't give different solutions.
• Why did you take $x= 2\cos \varphi$ ? why not something else ? – A---B Mar 15 '17 at 0:39
• @A---B The aim is making the ratio between the coefficient of the cubic and linear term to become $4/3$ – egreg Mar 15 '17 at 6:31
No. $x^3$ has only one real root, and $x^3 + x$ has only one real root and it's not a multiple root. TO show that your polynomial has more real roots, one approach is to find a value $x$ where it's positive, and a value $z > x$ where it's negative. For then between $-\infty$ and $x$ there must be a root (by the intermediate value theorem), and between $z$ and $\infty$ there must be another and ... you can probably work out the rest.
• Probably not the point of the question, but the roots are $2 \cos \left( \frac{2 \pi}{9} \right) \approx 1.532, \; \; 2 \cos \left( \frac{4 \pi}{9} \right) \approx 0.347, \; \; 2 \cos \left( \frac{8 \pi}{9} \right) \approx -1.879.$ – Will Jagy Mar 13 '17 at 18:39
As a hint, what you want to use is the related remainder theorem, which will give you that the cardinality of the set of the roots of an $n$-polynomial is $n$ for every integer $n \geq 1$. Note that the so-called fundamental theorem of algebra does not guarantee there are exactly how many roots; it guarantees there will be at most, how many, disctinct roots.
But the above is a theoretical approach that includes complex roots. To find the real roots, an observation and a decomposition will do.
We seem to be bringing out many different techniques in these answers, so here's one more: Descartes' Sign Rule.
Let $f(y) = y^3 - 3y + 1.$ There are three sign changes reading the coefficients from highest-order to lowest, so there are either two positive real roots or none. But $f(-y) = -y^3 + 3y + 1$ has just one sign change, so there is one negative real root.
Now all you have to do is exhibit the existence of at least one positive root by some other means, and then you know there must be two positive roots, which added to the single negative root gives three roots altogether.
This is not an especially helpful approach for this particular polynomial, since if you have proved the existence of at least one positive root, you can probably prove the existence of the other two roots almost immediately without using Descartes' rule.
This is an approach on how to solve the DEPRESSED cubic equation. To show the roots are real just find the discriminant of the cubic, (it can be shown easily)
Subsitute $$y= ucosa$$
Substite in the original equation to get
$$(ucosa)^3 -3ucosa+1 =0$$
Recall the identity $$cos(3a)=4cos^3a-3cosa$$
Manipulate the equation to get
$$u^3(\frac{cos3a +3cosa}{4}) -3ucosa+1 =0$$
Distribute to obtain: $$u^3(\frac{cos3a}{4})+ u^3(\frac{3cosa}{4}) -3ucosa+1 =0$$
Take the common factor out of the middle two terms to obtain:
$$u^3(\frac{cos3a}{4})+ 3ucosa(\frac{u^2}{4}-1)+1 =0$$
"u" is a parameter chosen by us. We are in search of such a parameter "$u$"; so that the middle term cancels out so that we can solve the trigonometric equation. In order for this to occur: $$u=±2$$
Substituting $u=2$ we get
$$2cos3a+1 =0$$
Solving for $cos3a$ we get
$$cos3a= \frac{-1}{2}$$
Therefore
$$3a= arccos(\frac{-1}{2})$$
$$a = \frac{arccos(\frac{-1}{2})}{3}$$
$$y= 2cos(\frac{arccos(\frac{-1}{2})}{3})$$
Similarly we can subsitute $u=-2$ and find another root. | 2021-01-25T12:01:34 | {
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https://mathematica.stackexchange.com/questions/138698/evaluation-of-this-double-infinite-summation | # Evaluation of this double infinite summation
I want to evaluate the following double summation
Sum[(-1)^(i + j + i*j)*Exp[-Pi/2*( i^2 + j^2)], {i, -Infinity,
Infinity}, {j, -Infinity, Infinity}]
I am really new both in using Mathematica and in doing mathematics using computer. I don't know if there is some special technics to deal with these kind of summations (Lattice sums) in Mathematica.
When I evaluate the former expression, Mathematica refuses to evaluate it and just reprint it in the output.
Theoretically, the expected value is 0.
Mathematica
You can try out numerical summation NSum,
NSum[(-1)^(i + j + i*j)*Exp[-Pi/2*(i^2 + j^2)], {i, -Infinity, Infinity},
{j, -Infinity, Infinity}]
which after some warnings gives an output,
-2.22045*10^-16 - 1.04284*10^-68 I
If we increase the WorkingPrecision, will be able to get the desired result,
NSum[(-1)^(i + j + i*j)*Exp[-Pi/2*(i^2 + j^2)], {i, -Infinity, Infinity},
{j, -Infinity, Infinity}, Method -> "AlternatingSigns", WorkingPrecision -> 30]
0.*10^-30
Rationalize[%]
0
As suggested by @AccidentalFourierTransform, we "can use the option, Method -> "AlternatingSigns" to speed up the computation and remove the warnings".
Maple
Maple's sum(sum(f(k,l), k=m..n),l=m..n) command is able to directly compute the double sum,
restart:
Sum(Sum((-1)^(i + j + i*j)*exp(-Pi/2*( i^2 + j^2)), i=-infinity..infinity),
j=-infinity..infinity);
evalf(%)
0
• Thank you both for the quick feedback. Now using the numerical approach, is there a way to test if the actual limit is 0? – Aymane Fihadi Feb 26 '17 at 17:05
• @AymaneFihadi What you mean by actual limit? – zhk Feb 26 '17 at 17:08
• @ As I mentioned in my question, the limit of the above series is 0. I want to test that using Mathematica. By your valuable answer, we get that the limit value is around 0, as far as I understand there is possibly a way to make sure that this limit is actually 0 ? – Aymane Fihadi Feb 26 '17 at 17:17
• @AymaneFihadi Rationalize[0.*10^-30]=0. BTW, Maple also gives zero. – zhk Feb 26 '17 at 17:19
• OK, Thank you very much. I will tag this answer as Accepted later, in the hope if there is some other insight concerning doing lattice sums in general. – Aymane Fihadi Feb 26 '17 at 17:24
It is possible to do this sum in MA. Let us look at the function and its domain:
fig1 = MatrixPlot[Table[(-1)^(i + j + i j) Exp[-Pi/2*(i^2 + j^2)], {i, -10, 10}, {j, -10, 10}]];
fig2 = MatrixPlot[ Table[(-1)^(i + j + i j) , {i, -10, 10}, {j, -10, 10}]];
fig = GraphicsRow[{fig1, fig2}]
This suggests to split the sum as follows:
r1 = Sum[Exp[-Pi/2*((2 ki)^2 + (2 kj)^2)], {ki, -Infinity, Infinity}, {kj, -Infinity, Infinity}]
(*EllipticTheta[3, 0, E^(-2 Pi)]^2*)
r2 = Sum[Exp[-Pi/2*((2 ki + 1)^2 + (2 kj + 1)^2)], {ki, -Infinity, Infinity}, {kj, -Infinity, Infinity}]
(*EllipticTheta[2, 0, E^(-2 Pi)]^2*)
r3 = Sum[Exp[-Pi/2*((2 ki)^2 + (2 kj + 1)^2)], {ki, -Infinity, Infinity}, {kj, -Infinity, Infinity}]
(*EllipticTheta[2, 0, E^(-2 Pi)] EllipticTheta[3, 0, E^(-2 Pi)]*)
At the end we can numerically verify a nice identity between the elliptic functions
$\vartheta _3\left(0,e^{-2 \pi }\right){}^2-\vartheta _2\left(0,e^{-2 \pi }\right){}^2-2 \vartheta _3\left(0,e^{-2 \pi }\right) \vartheta _2\left(0,e^{-2 \pi }\right)=0$
N[r1 - r2 - 2 r3]//Chop
(*0*)
• Thank you very much, This is a nice idea. when I was trying to calculate this by pen, I was doing the same splitting of the lattice. btw when I said its theoretically 0, they link it with sigma of Weierstrass, and use some sophisticated argument, it is not direct. – Aymane Fihadi Feb 28 '17 at 10:20
• From formula 8 and 14 here, we find that $$\frac{\vartheta _2\left(0,e^{-2 \pi}\right)^2}{\vartheta _3\left(0,e^{-2 \pi}\right)^2}=3-2\sqrt 2$$ With this, the last identity in this answer can be easily verified. – J. M.'s torpor Oct 4 '18 at 9:35
• @J.M.issomewhatokay. Nice find ! – yarchik Oct 4 '18 at 11:15 | 2021-08-03T08:18:19 | {
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https://math.stackexchange.com/questions/2002800/solving-an-ivp-using-undetermined-coefficients | Solving an IVP using undetermined coefficients
I'm trying to solve the following Initial value problem using the method of undetermined coefficients, but I keep getting the wrong answer. Any pointers?
$y''-9y=20e^{2t} - 81\quad\quad y(0)=10\quad y'(0)=17$
So I start by finding $y_c(x)$:
$m^2 -9 = 0 \implies m=\pm3$, then
$y_c(x)=c_1e^{3x}+c_2e^{-3x}$
For the undetermined coefficients part, I look at $20e^{2t}-18$ to get $Ae^{2t}$, and then to find $A$ I plug it into the original equation to get$$4Ae^{2t}-9(Ae^{2t})=20e^{2t}-81$$ And end up with $A = 81e^{-2t}/5 -4$
I could go on, but at this point I'm pretty sure I've done somthing wrong. I'm pretty sure $A$ isn't supposed to be this ugly.
For the particular solution try $y_p = Ae^{2t} + B$ substitute it into the DE.
$$4Ae^{2t} - 9Ae^{2t} - 9B = 20e^{2t} - 81$$ equate coefficients $$-5A = 20 \, , \, 9B = 81$$
$$A = -4 \, , \, B = 9$$
$$y(x)=c_1e^{3t}+c_2e^{-3t} -4e^{2t} + 9$$
$$y(0) = 10 = c_1 + c_2 -4 + 9$$ $$c_1 + c_2 = 5$$
$$y'(0) = 17 = 3c_1 -3c_2 -8$$ $$c_1 - c_2 = \frac{25}{3}$$
$$c_1 = \frac{20}{3} \, , \, c_2 = \frac{-5}{3}$$
$$y(x)=\frac{20}{3} e^{3t}- \frac{5}{3}e^{-3t} -4e^{2t} + 9$$
• Thanks! I knew I was missing something, this makes more sense – kojak Nov 7 '16 at 1:20
• I'm getting 20/3 and 5/3 for c_1 and c_2. I've checked and your answer is right, but what am I missing? – kojak Nov 7 '16 at 1:42
• You're right. I made a sign error. I've corrected it and checked it on wolfram wolframalpha.com/input/… – arthur Nov 7 '16 at 2:35
You are correct up until the point of applying the undetermined coefficient strategy. Notice that the right hand side of your initial differential equation is a linear combination of e^(2t) and 1. Once you add the constant 1 to your partial solutions and then add another undetermined coefficient B, I think you will be able to solve this problem. (After this you should get A = -4 and B = 9). | 2020-10-22T06:45:57 | {
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https://math.stackexchange.com/questions/1029248/example-3-53-in-baby-rudin | # Example 3.53 in Baby Rudin
Here's Example 3.53 in the book Principles of Mathematical Analysis by Walter Rudin, third edition.
Consider the convergent series $$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \ldots$$ and one of its rearrangements $$1 + \frac{1}{3} - \frac{1}{2} + \frac{1}{5} + \frac{1}{7} - \frac{1}{4} + \frac{1}{9} + \frac{1}{11} - \frac{1}{6} + \ldots$$ in which two positive terms are always followed by one negative. If $s$ is the sum of the original series, then $$s < 1 - \frac{1}{2} + \frac{1}{3} = \frac{5}{6}.$$ Since $$\frac{1}{4k-3} + \frac{1}{4k-1} - \frac{1}{2k} = \frac{8k-4}{(4k-1)(4k-3)} - \frac{1}{2k} = \frac{2k(8k-4) - (4k-1)(4k-3)}{2k(4k-1)(4k-3)} = \frac{8k-3}{2k(4k-1)(4k-3)} > 0$$ for $k \geq 1$, we see that $$s^\prime_3 < s^\prime_6 < s^\prime_9 < \ldots,$$ where $s^\prime_n$ is the $n$th partial sum of the series after the rearrangement. Hence $$\lim_{n\to\infty}\sup s^\prime_n > s^\prime_3 = \frac{5}{6},$$ so that the rearranged series certainly does not converge to $s$.
Now here's my question:
How to determine, using the machinery developed by Rudin upto this point in the book, if the new (or rearranged) series converges at all? Rudin leaves it to the reader to check that the new series does converge. How to prove this convergence?
I would like to have answers that use only the results that Rudin has discussed so far in the book.
• Does the Leibniz test happen to be among those tools at your disposal? Nov 19 '14 at 14:47
• What is that? I can't recall it using this name, I'm afraid. Nov 19 '14 at 14:49
• Also known as the alternating series test. (Doesn't seem applicable here, however.) Nov 19 '14 at 14:49
• Oh yes, it is Theorem 3.43 in Baby Rudin. Then what? Nov 19 '14 at 14:52
• Not directly applicable, but unless I'm very much mistaken here, from there it's a very small step to prove the convergence - I believe that's what the answer below means to tell you. Nov 19 '14 at 14:52
We can show that the series converges, and find its sum, as follows:
$\hspace{.3 in}1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots=\ln 2$ $\;\;\;$so
$\hspace{.27 in}\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{10}-\frac{1}{12}+\cdots=\frac{1}{2}\ln 2$. $\;\;\;$Inserting zeros, we get
$\hspace{.26 in}0+\frac{1}{2}+0-\frac{1}{4}+0+\frac{1}{6}+0-\frac{1}{8}+0+\frac{1}{10}+\cdots=\frac{1}{2}\ln 2$.
Adding this to the original series gives
$\hspace{.26 in}1+0+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+0+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+0+\cdots=\frac{3}{2}\ln 2$, $\;\;$ so
$\hspace{.25 in} 1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}+\cdots=\frac{3}{2}\ln 2$.
• Wow! What a great, clear answer! Nov 22 '14 at 3:38
• @ignoramus Thanks! Nov 22 '14 at 16:37
• @user84413 how did you figure it out yourself? Can you please enlighten me too as to how this insight came to you? Oct 23 '16 at 7:12
• @SaaqibMahmuud I had seen this argument in some textbook -- it is not something I thought of myself. Oct 27 '16 at 17:11
• This is all so good Oct 3 '20 at 4:59
Hint: Notice that $s_{3k}'$ converge as $k \to \infty$. How far can $s_{3k+1}$ and $s_{3k+2}$ be from $s_{3k}$?
• @D Poole, how do we know that $s_{3k}^\prime$ converges? And, how to use this information toward establishing the convergence of $s_n^\prime$? Nov 19 '14 at 14:57
• For $s'_{3k}$ take a good look at the alternating series test. You have $(1+\frac{1}{3})-(\frac{1}{2})+(\frac{1}{5}+\frac{1}{7})-(\frac{1}{4}) \pm \cdots$ - now can you see how to separate this series and apply the test? Nov 19 '14 at 15:01
• I don't think that's quite what you want. Try to think of the following: Define the sequence $a_{2n-1}:=\frac{1}{4n-3}+\frac{1}{4n-1}$, $a_{2n}:=\frac{1}{n}$ for $n\geq 1$. Now, what does the alternating series test say about the series $\sum_{i=1}^\infty{(-1)^ia_i}$? And, how does that help you? Nov 19 '14 at 15:13
• To see that $s_{3k}'$ converge, look at the powers of $k$ in the fraction in your formula for $a_{3k-2}'+a_{3k-1}'+a_{3k}'$. Nov 19 '14 at 15:40
• @SaaqibMahmuud You do not need any results on rearrangements of infinite series for this method. It is an $\epsilon/2$ argument. Do you agree that you have the techniques to show that $s'_{3k}$ converges? If so, what is the largest that $|s'_{3k+1}-s'_{3k}|$ and $|s'_{3k+2}-s'_{3k}|$ can be? [For all sufficiently large $N$, you have that $|s'_{N} - s'_{\lfloor N/3 \rfloor}|< \epsilon/2$. Further, since $s'_{3k}$ converges to $s',$ for instance, then for all $N'$ large enough, $|s'_{3N'}-s'|<\epsilon/2$. Putting these together $\ldots$] May 3 '16 at 12:07 | 2021-10-24T01:58:27 | {
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https://www.lil-help.com/questions/7810/from-a-group-of-8-women-and-6-men-a-committee-consisting-of-3-men-and-3-women-is-to-be-formed | From a group of 8 women and 6 men, a committee consisting of 3 men and 3 women is to be formed.
# From a group of 8 women and 6 men, a committee consisting of 3 men and 3 women is to be formed.
J
240 points
From a group of 8 women and 6 men, a committee consisting of 3 men and 3 women is to be formed. How many different committees are possible if
(a) 2 of the men refuse to serve together?
(b) 2 of the women refuse to serve together?
(c) 1 man and 1 woman refuse to serve together?
From a
jeffp
8.7k points
part (a)
For the men, we choose 3 from 6 order not mattering,
$$\begin{pmatrix} 6 \\ 3 \end{pmatrix} = 20$$
but we need to subtract from this the number of orderings where the two men serve together. This is given by,
$$\begin{pmatrix} 2 \\ 2 \end{pmatrix} \begin{pmatrix} 4 \\ 1 \end{pmatrix} = 4$$
so we have,
$$20 - 4=16$$ men to choose from.
There are $\begin{pmatrix} 8 \\ 3 \end{pmatrix} = 56$ different groups of women. So the total number of committees possible is,
$$16 \cdot 56 = 896$$
part (b)
This is similar to part (a) so we can jump straight to,
$$\bigg[ \begin{pmatrix} 8 \\ 3 \end{pmatrix} - \begin{pmatrix} 2 \\ 2 \end{pmatrix} \begin{pmatrix} 6 \\ 1 \end{pmatrix} \bigg] \begin{pmatrix} 6 \\ 3 \end{pmatrix} = 50\cdot 20 = 1000$$ different committees.
part (c) Total number of committees that could possibly exist are,
$$\begin{pmatrix} 8 \\ 3 \end{pmatrix} \begin{pmatrix} 6 \\ 3 \end{pmatrix} =1120$$
the number of committees that exist where the man and women serve together is given by,
$$\begin{pmatrix} 1 \\ 1 \end{pmatrix} \begin{pmatrix} 5 \\ 2 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} \begin{pmatrix} 7 \\ 2 \end{pmatrix} = 210$$
so the total number of committees in this case amounts to,
$$1120 - 210 = 910$$
Surround your text in *italics* or **bold**, to write a math equation use, for example, $x^2+2x+1=0$ or $$\beta^2-1=0$$
Use LaTeX to type formulas and markdown to format text. See example. | 2017-07-25T04:32:17 | {
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https://math.stackexchange.com/questions/1509771/how-to-get-a-better-proof | # How to get a better proof
Given a string $S$ which is a representation of a number in base $2$ i.e. a string of 0s and 1s, we need to find out the maximum possible value obtainable by erasing exactly one digit. The string's most significant digit is 1. This means $S$ represents a number greater than $0$.
I have a proof. If there is a more short and sweet proof please post it.
Let $S = aXYb$ where, $X$, $Y$ are single digits and $a$, $b$ are the bits to left of $X$ and to right of $Y$ respectively.
Then two important observations.
a. If $X$ = $Y$, we will obtain the same value aXb whether we remove X or Y.
b. We can then represent String S as a sequence of alternating groups of 1's and 0's i.e.
1. $S = 11..100...011..100...0$, if $S$ ends in zero
2. $S = 11..100...011..1$, if $S$ ends in one.
Within each group of 1's or 0's, we can remove any of digits of the same digit group and obtain same final value as obtained from observation above. Now, if $N = length(S)$, then our result is of length $N - 1$. We proceed fromthe $(N-1)$th bit(MSB) to the 1st bit(LSB) of resultant string and see if we can set the $i^{th}$ bit. We can have MSB of final result as 1 as the MSB of $S$ is 1. Using similar logic, we won't remove any of digits from the leftmost 1 digit group of $S$. Now consider the next digit of resultant string. We can place a 1 there only if the leftmost 0s group of $S$ is of length 1 since we can remove only one digit. Otherwise, we we will have the bit set to 0. To see that, we can write $S$ as $S = 1...10..01X$. To obtain maximum value, it is easy to see that we need to bring 1 (adjacent to X) to the left. Thus we need to remove a 0 from the lefmost 0 group.
I want to improve my discrete mathematics thinking. Thank you.
• What does erasing a digit mean? Replacing by zero? – copper.hat Nov 2 '15 at 18:05
• Let $S=aXb$. Then erasing digit $X$ means resulting string = $ab$. We need to find the lexicographically largest resultant string. – nitrogen Nov 2 '15 at 18:08
It’s essentially the same argument, but presented a bit more elegantly.
Suppose that the bit string is $\sigma=1b_1b_2\ldots b_n$. For $k=1,\ldots,n$ let $\sigma_k$ be the bit string that remains when $b_k$ is deleted, and let $\lambda_k$ be the substring of $\sigma$ to the left of $b_k$.
• If $b_k=b_{k+1}$, clearly $\sigma_k=\sigma_{k+1}$.
• If $b_k=0$, $b_{k+1}=1$, and $k<\ell$, the first $k+1$ bits of $\sigma_k$ are $\lambda_k1$, while the first $k+1$ bits of $\sigma_\ell$ are $\lambda_k0$, so $\sigma_k>\sigma_\ell$, where the order may be thought of either as lexicographic order or as the ordinary numerical order of the integers represented in binary by the bit strings.
• Similarly, if $b_k=1$, $b_{k+1}=0$, and $k<\ell$, the first $k+1$ bits of $\sigma_k$ are $\lambda_k0$, while the first $k+1$ bits of $\sigma_\ell$ are $\lambda_k1$, so $\sigma_k<\sigma_\ell$.
Clearly if all bits of $\sigma$ are $1$, it makes no difference which we remove. Now suppose that $\sigma$ has a zero bit, and let $b_k$ be the leftmost zero bit. The first two points combined show that $\sigma_k\ge\sigma_\ell$ for all $\ell>k$, and the third shows that $\sigma_k>\sigma_\ell$ for all $\ell<k$, so removing the first zero maximizes the resulting string.
• I wish I was able to write such proofs. Such proofs help in thinking very much. You can see that I was not able to express my proof clearly. I think I need more practice. – nitrogen Nov 2 '15 at 21:39
• I have got so many problems that I need to express in such a way(but haven't been able to). Some problems are pretty daunting, especially problems employing greedy strategy. – nitrogen Nov 2 '15 at 21:42
• @nitrogen: Practice does make a difference: you get to recognize certain patterns, just as you get to recognize patterns of reasoning when you solve a lot of problems. One thing that may help: whenever your argument involves an and so on step, try to find a way to convert that to a recursive construction or a proof by induction. – Brian M. Scott Nov 2 '15 at 21:44
• I wish there were teachers like you and people around here at my place. I would try to gain as much as I possibly could. – nitrogen Nov 2 '15 at 21:46 | 2019-12-11T22:47:40 | {
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https://math.stackexchange.com/questions/1964918/how-to-calculate-p-6666666-mod7 | # How to calculate $p={{{{{{6^6}^6}^6}^6}^6}^6}$ $\mod7$?
I've tried two approaches:
Approach 1
Since $6 \equiv -1 \pmod7$
So, $p=(-1)^t$ and $t$ is even
Therefore, $p=1$.
Approach 2
Since $6 \equiv -1 \pmod7$
So, $6^6 \equiv 1 \pmod7$.
Hence, solving towers from top to bottom:
$p \equiv {{{{{6^6}^6}^6}^6}^1} \pmod7$
$p \equiv {{{6^6}^6}^1} \pmod7$
$p \equiv {6^1} \pmod7$
Therefore, $p=6$.
Now, I don't know why both the approaches are giving different answers and which one is right.
• By your second method $2^6 \equiv 2$ (mod 5) – N.S.JOHN Oct 12 '16 at 7:04
You can't replace exponents like that. That is, $6^8\not\equiv 6^1$ mod $7$. You can pretty easily check that $6^8\equiv 1$.
As you say, $6\equiv -1$, so $-1$ to an even power will give you $1$ mod $7$.
• Shouldn't that be $6^8 \equiv 1 ~\mod 7$? – erfink Oct 12 '16 at 7:05
• Right, thanks. – Elliot G Oct 12 '16 at 7:06
• I've not replaced $6^8$ by $6$ anywhere but instead replaced $6^6$ by 1 – ankit Oct 12 '16 at 7:07
• That was an example – Elliot G Oct 12 '16 at 7:08
• I've not considered equality anywhere. only congruence – ankit Oct 12 '16 at 7:09
Approach $1$ is correct.
We do not have $$a^b \equiv a^{(b \mod p)} \mod p$$ in general
Third approach. $7$ is prime. $gcd (6,7)=1$ so by Fermats Little theorem $6^6\equiv 1 \mod 7$.
So $6^{6*k}\equiv 1 \mod 7$ (notice congruence of exponents are NOT preserved modulo 7-- but they are preserved modulo 6.)
So as ${{{6^6}^6}^6}$ is a multiple of $6$ we have ${{{{6^6}^6}^6} ^6}\equiv 1 \mod 7$
Question: $p={{{{{{6^6}^6}^6}^6}^6}^6}$ $\mod7$?
• a) Let $Q = {{{{6^6}^6}^6}^6}$ so that $p = 6^{6^Q}$
• b) Note that $6\equiv -1 \pmod 7$
Thus
• $\qquad \displaystyle p=6^{6^Q} = 6^{2^Q \cdot 3^Q} \equiv \left((-1)^{2^Q}\right)^{3^Q} \equiv 1^{3^Q} \equiv 1 \pmod 7$ | 2019-04-26T16:00:22 | {
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https://math.stackexchange.com/questions/2943253/how-to-come-up-with-a-greedy-solution-and-prove-it/2943278 | # How to come up with a greedy solution and prove it?
Say we have a function $$S(x)$$, which gives the sum of the digits in the number $$x$$. So $$S(452)$$ would be $$4 + 5 + 2 = 11$$.
Given a number $$x$$, find two integers $$a, b$$ such that $$0 <= a, b <= x$$ and $$a + b = x$$. Objective is to maximize $$S(a) + S(b)$$. I came across this question on a programming website and the answer is to greedily choose a number $$a$$ containing all $$9$$'s such that it is lesser than $$x$$, and the other number would be $$x - a$$.
If $$x = 452$$, then $$S(99) + S(353) = 29$$ which is the maximum possible. How do I come up with this and prove the same?
• What is $n$ in the requirement $a+b=n$? – 5xum Oct 5 '18 at 11:29
• I assume $n=x$, no? As a small point, you say that you are allowing $b=n$ but that would make $a=0$ which you are not allowing. Do you mean to allow the pair $n+0=n$ or not? – lulu Oct 5 '18 at 11:36
• @5xum, n goes upto $10^{12}$. – Andrew Scott Oct 5 '18 at 11:43
• I think the point isn't so much that the greedy algorithm finds some sort of unique max. Indeed, in most cases it just finds one of many. To prove it, I'd start by noting that given any optimal solution $a≤b$ we can subtract enough from each decimal place of $a$ to make the corresponding place of $b$ equal to $9$ without changing the sum you want. For example, for $n=154$ we could have the solution $77,77$ or we could "move $2$ over from each slot" to get the equivalent solution $55,99$ which is what the greedy algorithm would find. – lulu Oct 5 '18 at 11:45
• @lulu, Yes. I am sorry. Edited the question. – Andrew Scott Oct 5 '18 at 11:46
Show the following two statements (I guess they would be lemmas):
1. When adding $$a+b$$ the way you learn in school, if you get no carries, then $$S(a+b)=S(a)+S(b)$$
2. For each carry you get when adding $$a+b$$, the sum $$S(a)+S(b)$$ increases by $$9$$.
Together they mean that you want to have as many carries as you can. The greedy algorithm you describe gives you a carry into each column (except the 1's column, which is impossible anyways) and therefore gives you the max.
• WOW. Thanks! :) – Andrew Scott Oct 5 '18 at 12:02
Elaborating on the process in lulu's comment:
Since there are only finitely many choices for $$a,b$$, an optimum must exist. Consider all pairs $$(a,b)$$ with $$a+b=n$$ and $$S(a)+S(b)=\max$$ and $$a\le b$$. $$a=\overline{a_1a_2\ldots a_d}$$ and $$b=\overline{b_1b_2\ldots b_d}$$ (with $$b_1>0$$, but possibly $$a_1=0$$). Among all such $$(a,b)$$, pick one that maximizes $$S(a)$$.
Suppose $$a_k<9$$ for some $$k>1$$.
• If $$b_k=0$$, then there must be a (maximal) $$j with $$b_j>0$$, for otherwise we'd have $$b. If we replace $$a_k$$ with $$a_k+1$$, $$b_j$$ with $$b_j-1$$ and $$b_i$$ with $$9$$ for $$j, with the numbers $$a',b'$$ obtained this way, we have $$a'+b'=a+b=n$$, but $$S(a')+S(b')=S(a)+S(b)+9(k-j)$$, constradicting maximality of $$S(a)+S(b)$$.
• If $$b_k>0$$ we could increase $$a_k$$ and decrease $$b_k$$ by one, thereby achieving $$a'+b'=a+b=n$$, $$S(a')+S(b')=S(a)+S(b)$$, but $$S(a')=S(a)+1$$, contradicting the maximality of $$S(a)$$.
We conclude that there is a maximizing $$a$$ with $$a\le b$$ and $$a_2=\ldots=a_d=9$$.
What can we gain if we drop the condition $$a\le b$$?
• If $$a_1+b_1\ge 9$$, we can let $$a_1'=9$$ and $$b_1'=a_1+b_1-9$$, thereby making $$a'$$ consist only of $$9$$'s and apparently the largest number $$\le n$$ of this form
• If $$a_1+b_1<9$$, we can let $$a_1'=0$$ and $$b_1'=a_1+b_1$$, thereby making $$a'$$ consist only of $$9$$'s and apparently the largest number $$\le n$$ of this form
In summary, the $$a$$ (and associated $$b$$) found by the greedy method is among the maximizers. | 2019-10-20T16:13:36 | {
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http://www.soho46.com/cole-and-ehte/convergence-in-probability-and-convergence-in-distribution-addb5e | Convergence in Distribution p 72 Undergraduate version of central limit theorem: Theorem If X 1,...,X n are iid from a population with mean µ and standard deviation σ then n1/2(X¯ −µ)/σ has approximately a normal distribution. Convergence in probability means that with probability 1, X = Y. Convergence in probability is a much stronger statement. Topic 7. where $F_n(x)$ is the cdf of $\sqrt{n}(\bar{X}_n-\mu)$ and $F(x)$ is the cdf for a $N(0,E(X_1^2))$ distribution. Definition B.1.3. It is easy to get overwhelmed. convergence of random variables. Convergence in probability. (2) Convergence in distribution is denoted ! 2 Convergence in Probability Next, (X n) n2N is said to converge in probability to X, denoted X n! It tells us that with high probability, the sample mean falls close to the true mean as n goes to infinity.. We would like to interpret this statement by saying that the sample mean converges to the true mean. P(n(1−X(n))≤ t)→1−e−t; that is, the random variablen(1−X(n)) converges in distribution to an exponential(1) random variable. Convergence in probability: Intuition: The probability that Xn differs from the X by more than ε (a fixed distance) is 0. And, no, $n$ is not the sample size. Convergence in distribution means that the cdf of the left-hand size converges at all continuity points to the cdf of the right-hand side, i.e. P n!1 X, if for every ">0, P(jX n Xj>") ! In particular, for a sequence X1, X2, X3, ⋯ to converge to a random variable X, we must have that P( | Xn − X | ≥ ϵ) goes to 0 as n → ∞, for any ϵ > 0. Noting that $\bar{X}_n$ itself is a random variable, we can define a sequence of random variables, where elements of the sequence are indexed by different samples (sample size is growing), i.e. A quick example: $X_n = (-1)^n Z$, where $Z \sim N(0,1)$. Contents . I will attempt to explain the distinction using the simplest example: the sample mean. In other words, the probability of our estimate being within $\epsilon$ from the true value tends to 1 as $n \rightarrow \infty$. I have corrected my post. endstream endobj startxref Yes, you are right. Convergence in Probability. Over a period of time, it is safe to say that output is more or less constant and converges in distribution. To say that Xn converges in probability to X, we write. Convergence in distribution of a sequence of random variables. The general situation, then, is the following: given a sequence of random variables, 4 Convergence in distribution to a constant implies convergence in probability. It’s clear that $X_n$ must converge in probability to $0$. Convergence in distribution tell us something very different and is primarily used for hypothesis testing. Formally, convergence in probability is defined as Note that the convergence in is completely characterized in terms of the distributions and .Recall that the distributions and are uniquely determined by the respective moment generating functions, say and .Furthermore, we have an equivalent'' version of the convergence in terms of the m.g.f's suppose the CLT conditions hold: p n(X n )=˙! Viewed 32k times 5. The concept of convergence in distribution is based on the … 1.1 Almost sure convergence Definition 1. Although convergence in distribution is very frequently used in practice, it only plays a minor role for the purposes of this wiki. is $Z$ a specific value, or another random variable? It is just the index of a sequence $X_1,X_2,\ldots$. Note that if X is a continuous random variable (in the usual sense), every real number is a continuity point. Just hang on and remember this: the two key ideas in what follows are \convergence in probability" and \convergence in distribution." Convergence in probability gives us confidence our estimators perform well with large samples. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy, 2020 Stack Exchange, Inc. user contributions under cc by-sa. 288 0 obj <>stream 9 CONVERGENCE IN PROBABILITY 111 9 Convergence in probability The idea is to extricate a simple deterministic component out of a random situation. The hierarchy of convergence concepts 1 DEFINITIONS . Convergence in probability gives us confidence our estimators perform well with large samples. Z S f(x)P(dx); n!1: Knowing the limiting distribution allows us to test hypotheses about the sample mean (or whatever estimate we are generating). Convergence and Limit Theorems • Motivation • Convergence with Probability 1 • Convergence in Mean Square • Convergence in Probability, WLLN • Convergence in Distribution, CLT EE 278: Convergence and Limit Theorems Page 5–1 However, $X_n$ does not converge to $0$ according to your definition, because we always have that $P(|X_n| < \varepsilon ) \neq 1$ for $\varepsilon < 1$ and any $n$. This is typically possible when a large number of random effects cancel each other out, so some limit is involved. • Convergence in probability Convergence in probability cannot be stated in terms of realisations Xt(ω) but only in terms of probabilities. Convergence in distribution 3. I just need some clarification on what the subscript $n$ means and what $Z$ means. Convergence of the Binomial Distribution to the Poisson Recall that the binomial distribution with parameters n ∈ ℕ + and p ∈ [0, 1] is the distribution of the number successes in n Bernoulli trials, when p is the probability of success on a trial. If it is another random variable, then wouldn't that mean that convergence in probability implies convergence in distribution? Your definition of convergence in probability is more demanding than the standard definition. probability zero with respect to the measur We V.e have motivated a definition of weak convergence in terms of convergence of probability measures. X a.s. n → X, if there is a (measurable) set A ⊂ such that: (a) lim. We note that convergence in probability is a stronger property than convergence in distribution. Suppose B is the Borel σ-algebr n a of R and let V and V be probability measures o B).n (ß Le, t dB denote the boundary of any set BeB. Put differently, the probability of unusual outcome keeps … d: Y n! Active 7 years, 5 months ago. Convergence in Distribution [duplicate] Ask Question Asked 7 years, 5 months ago. And $Z$ is a random variable, whatever it may be. Also Binomial(n,p) random variable has approximately aN(np,np(1 −p)) distribution. Consider the sequence Xn of random variables, and the random variable Y. Convergence in distribution means that as n goes to infinity, Xn and Y will have the same distribution function. dY. X. n The former says that the distribution function of X n converges to the distribution function of X as n goes to infinity. n!1 . h����+�Q��s�,HC�ƌ˄a�%Y�eeŊ$d뱰�c�BY()Yِ��\J4al�Qc��,��o����;�{9�y_���+�TVĪ:����OZC k��������� ����U\[�ux�e���a;�Z�{�\��T��3�g�������dw����K:{Iz� ��]R�؇=Q��p;���I�$�bJ%�k�U:"&��M�:��8.jv�Ź��;���w��o1+v�G���Aj��X��菉�̐,�]p^�G�[�a����_������9�F����s�e�i��,uOrJ';I�J�ߤW0 Na�q_���j���=7� �u�)� �?��ٌ�f5�G�N㟚V��ß x�Nk 0 h�ĕKLQ�Ͻ�v�m��*P�*"耀��Q�C��. We say that X. n converges to X almost surely (a.s.), and write . In econometrics, your $Z$ is usually nonrandom, but it doesn’t have to be in general. $\{\bar{X}_n\}_{n=1}^{\infty}$. 2.1.2 Convergence in Distribution As the name suggests, convergence in distribution has to do with convergence of the distri-bution functions of random variables. $$,$$\sqrt{n}(\bar{X}_n-\mu) \rightarrow_D N(0,E(X_1^2)).$$,$$\lim_{n \rightarrow \infty} F_n(x) = F(x),$$, https://economics.stackexchange.com/questions/27300/convergence-in-probability-and-convergence-in-distribution/27302#27302. Then define the sample mean as \bar{X}_n. • Convergence in mean square We say Xt → µ in mean square (or L2 convergence), if E(Xt −µ)2 → 0 as t → ∞. or equivalently Convergence in probability. The concept of convergence in probability is based on the following intuition: two random variables are "close to each other" if there is a high probability that their difference will be very small. The weak law of large numbers (WLLN) tells us that so long as E(X_1^2)<\infty, that Convergence in distribution in terms of probability density functions. x) = 0. 1. 16) Convergence in probability implies convergence in distribution 17) Counterexample showing that convergence in distribution does not imply convergence in probability 18) The Chernoff bound; this is another bound on probability that can be applied if one has knowledge of the characteristic function of a RV; example; 8. This video explains what is meant by convergence in distribution of a random variable. n(1) 6→F(1). dY, we say Y n has an asymptotic/limiting distribution with cdf F Y(y).$$ For example, suppose $X_n = 1$ with probability $1/n$, with $X_n = 0$ otherwise. Under the same distributional assumptions described above, CLT gives us that n (X ¯ n − μ) → D N (0, E (X 1 2)). R ANDOM V ECTORS The material here is mostly from • J. n!1 0. where $\mu=E(X_1)$. $$plim\bar{X}_n = \mu,$$ The basic idea behind this type of convergence is that the probability of an “unusual” outcome becomes smaller and smaller as the sequence progresses. CONVERGENCE OF RANDOM VARIABLES . Note that although we talk of a sequence of random variables converging in distribution, it is really the cdfs that converge, not the random variables. $$\forall \epsilon>0, \lim_{n \rightarrow \infty} P(|\bar{X}_n - \mu| <\epsilon)=1. This is fine, because the definition of convergence in 4 distribution requires only that the distribution functions converge at the continuity points of F, and F is discontinuous at t = 1. Convergence in Probability; Convergence in Quadratic Mean; Convergence in Distribution; Let’s examine all of them. %%EOF On the other hand, almost-sure and mean-square convergence do not imply each other. Xt is said to converge to µ in probability … Convergence in probability is stronger than convergence in distribution. Click here to upload your image I posted my answer too quickly and made an error in writing the definition of weak convergence. 249 0 obj <>/Filter/FlateDecode/ID[<82D37B7825CC37D0B3571DC3FD0668B8><68462017624FDC4193E78E5B5670062B>]/Index[87 202]/Info 86 0 R/Length 401/Prev 181736/Root 88 0 R/Size 289/Type/XRef/W[1 3 1]>>stream Convergence in distribution tell us something very different and is primarily used for hypothesis testing. Definitions 2. Convergence in probability and convergence in distribution. Suppose that fn is a probability density function for a discrete distribution Pn on a countable set S ⊆ R for each n ∈ N ∗ +. 5.2. In other words, for any xed ">0, the probability that the sequence deviates from the supposed limit Xby more than "becomes vanishingly small. We write X n →p X or plimX n = X. Is n the sample size? In the lecture entitled Sequences of random variables and their convergence we explained that different concepts of convergence are based on different ways of measuring the distance between two random variables (how "close to each other" two random variables are). most sure convergence, while the common notation for convergence in probability is X n →p X or plim n→∞X = X. Convergence in distribution and convergence in the rth mean are the easiest to distinguish from the other two. Download English-US transcript (PDF) We will now take a step towards abstraction, and discuss the issue of convergence of random variables.. Let us look at the weak law of large numbers. 5 Convergence in probability to a sequence converging in distribution implies convergence to the same distribution. Given a random variable X, the distribution function of X is the function F(x) = P(X ≤ x). A sequence of random variables {Xn} is said to converge in probability to X if, for any ε>0 (with ε sufficiently small): Or, alternatively: To say that Xn converges in probability to X, we write: (4) The concept of convergence in distribtion involves the distributions of random ari-v ables only, not the random ariablev themselves. Suppose we have an iid sample of random variables \{X_i\}_{i=1}^n. Precise meaning of statements like “X and Y have approximately the (max 2 MiB). This leads to the following definition, which will be very important when we discuss convergence in distribution: Definition 6.2 If X is a random variable with cdf F(x), x 0 is a continuity point of F if P(X = x 0) = 0. As the sample size grows, our value of the sample mean changes, hence the subscript n to emphasize that our sample mean depends on the sample size. 1.2 Convergence in distribution and weak convergence p7 De nition 1.10 Let P n;P be probability measures on (S;S).We say P n)P weakly converges as n!1if for any bounded continuous function f: S !R Z S f(x)P n(dx) !$$\lim_{n \rightarrow \infty} F_n(x) = F(x),$$Proposition7.1Almost-sure convergence implies convergence in … e.g. You can also provide a link from the web. Convergence in distribution is the weakest form of convergence typically discussed, since it is implied by all other types of convergence mentioned in this article. The answer is that both almost-sure and mean-square convergence imply convergence in probability, which in turn implies convergence in distribution. Under the same distributional assumptions described above, CLT gives us that dZ; where Z˘N(0;1). 1. We say V n converges weakly to V (writte Im a little confused about the difference of these two concepts, especially the convergence of probability. This question already has answers here: What is a simple way to create a binary relation symbol on top of another? Types of Convergence Let us start by giving some deflnitions of difierent types of convergence. Xn p → X. 6 Convergence of one sequence in distribution and another to … (3) If Y n! 87 0 obj <> endobj$$\bar{X}_n \rightarrow_P \mu,$$. Then X_n does not converge in probability but X_n converges in distribution to N(0,1) because the distribution of X_n is N(0,1) for all n.$$\sqrt{n}(\bar{X}_n-\mu) \rightarrow_D N(0,E(X_1^2)).$$Econ 620 Various Modes of Convergence Definitions • (convergence in probability) A sequence of random variables {X n} is said to converge in probability to a random variable X as n →∞if for any ε>0wehave lim n→∞ P [ω: |X n (ω)−X (ω)|≥ε]=0.$$\forall \epsilon>0, \lim_{n \rightarrow \infty} P(|\bar{X}_n - \mu| <\epsilon)=1. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. If fn(x) → f∞(x) as n → ∞ for each x ∈ S then Pn ⇒ P∞ as n → ∞. Also, Could you please give me some examples of things that are convergent in distribution but not in probability? %PDF-1.5 %���� I understand that $X_{n} \overset{p}{\to} Z$ if $Pr(|X_{n} - Z|>\epsilon)=0$ for any $\epsilon >0$ when $n \rightarrow \infty$. 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Probability ; convergence in distribution 1/n $, where$ Z \$ is not sample. Cdf F Y ( Y ), and write distribution but not in probability is more or less constant converges... | 2021-08-02T15:07:33 | {
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https://math.stackexchange.com/questions/2643572/positive-probability-of-event-in-coin-tosses | # Positive probability of event in coin tosses
Lets say I toss a coin infinite times such that all tosses are independent. Does the event "For all $n\in \mathbb N$, we had exactly $n$ heads in a row between the toss number $3^n$ and toss number $3^{n+1}$" have a positive probability?
Is it it true for $n^2$?
I am thinking about using Borel-Cantelli but don't know how to begin formalizing the problem.
• Do you mean exactly $n$ heads or at least $n$ heads? – lulu Feb 9 '18 at 18:01
• @lulu I have edited the question. – z00x Feb 9 '18 at 18:03
• But you have not clarified it. What does "exactly $n$ heads in a row" mean? Just take $n=1$, say. So we are looking between $3$ and $9$. Which sequences are good and which are bad? – lulu Feb 9 '18 at 18:08
• @lulu: I agree that it's unclear, but does it matter for this particular question? We require only a positive probability. So long as $3^{n+1}-3^n \geq n$ (or $n^2$), shouldn't the probability be greater than $0$? – Brian Tung Feb 9 '18 at 18:10
• @BrianTung But you need this for all $n$. Depending on what the OP is asking, that sounds like an infinite product wherein the factors decrease to $0$. But perhaps (probably?) I have an entirely incorrect picture of what is being asked. – lulu Feb 9 '18 at 18:13
## 1 Answer
The question as stated currently (2018-02-09-1820 UT) is somewhat unclear, but we will view it in the following restrictive way:
• For any $n \in \mathbb{N}_{\geq 1}$, we look only at the sequence of tosses $3^n+1$ through $3^{n+1}-1$; that is, we look at $2 \times 3^n - 1$ tosses.
• We require that the run of $n$ consecutive heads be entirely isolated within that sequence.
We will show, broadly, that the probability is always non-zero, and that its limiting probability, as $n \to \infty$, is $1$.
First, we note that the number of tosses we look at, $2 \times 3^n - 1$, is always greater than $n+2$, for $n \geq 1$. It is true for $n = 1$, and whenever it is true for $n = k$, it is (almost) trivially true for $n = k+1$. Induction thus suffices to establish our claim.
The probability that the first $n+2$ tosses in our sequence have the form $THH \cdots HHT$ is $\frac{1}{2^{n+2}}$. This is non-zero for any $n \geq 1$; since this is a lower bound for the probability of any run of exactly $n$ heads, that probability must also be non-zero for any $n \geq 1$.
[Note: The 'align' environment doesn't seem to be working at the moment? Formatting may be askew.]
We now show, broadly, that the limiting probability is $1$. Not a rigorous proof, but you can proceed sort of along these lines:
It may not escape your notice that not only does $2 \times 3^n - 1 > n+2$, but the former grows rather faster than the latter. For any $n \geq 1$, in fact, we look at, altogether,
$$\left\lfloor \frac{2 \times 3^n - 1}{n+2} \right\rfloor$$
segments of $n+2$ tosses. Note that for $n \geq 1$,
$$\left\lfloor \frac{2 \times 3^n - 1}{n+2} \right\rfloor \geq \frac{3^n}{n+2} \\ \geq \frac{3^n}{3n} \\ \geq \frac{3^n}{3(6/5)^n} \\ \geq \frac{(5/2)^n}{3}$$
(There probably was an easier way to get to what I wanted, but that'll work.) Recall that a segment of length $n+2$ has the form $THH \cdots HHT$ with probability $\frac{1}{2^{n+2}}$, so the probability that none of $\frac{(5/2)^n}{3}$ segments have such a form is at most
$$\left( 1 - \frac{1}{2^{n+2}} \right)^\frac{(5/2)^n}{3} = \left[ \left( 1 - \frac{1}{2^{n+2}} \right)^{2^{n+2}} \right] ^\frac{(5/2)^n}{3\times 2^{n+2}} \\ \to \exp \left( - \frac{(5/4)^n}{12} \right) \\ \to 0$$
So the limiting probability is $1-0 = 1$ that there will be such a run, somewhere in those segments. | 2020-03-31T07:34:03 | {
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http://mathhelpforum.com/differential-geometry/94231-complex-power-series-boundary-property.html | # Math Help - Complex Power Series Boundary Property
1. ## Complex Power Series Boundary Property
In Shilov's Elementary Real and Complex Analysis, he writes on page 216:
'As we know, a (complex) power series a0 + a1 (z - z0) + a2 (z - z0)^2 + ... with radius of convergence p may or may not converge at points on the boundary of its region of convergence, i.e., at points of the circle |z - z0| = p. However, if the series converges at a boundary point z1, then it converges uniformly on the whole segment going from the center of the circle z0 to the boundary point z1. To see this, we need only consider the case z0 = 0, z1 = t1 > 0 (here z1 = t1 is real, as opposed to the general case where it is complex). Why?"
He then goes on to prove the special case with z0 = 0 and z1 = t1 > 0. Why does having proved this special case imply the general case for any point on the boundary of a region of convergence centered at any complex point? I think it may have something to do with shifting the power series and dividing/multiplying, but I'm not sure quite how to make it work.
2. Suppose you have a complex power series $\sum_{n=0}^\infty b_n(z-z_0)^n$ with radius of convergence $R>0$ which converges at some point $z_1$ on the boundary of the disk $D=D(z_0,R)$. By shifting and rotating the axes, you obtain another power series $\sum_{n=0}^\infty a_nz^n$ with radius of convergence $R$ which converges at the point $z=R$. (If $z_1=z_0+R\mathrm e^{\mathrm i\theta}$ then $a_n=b_n\mathrm e^{\mathrm in\theta}$.)
Now you need Abel's test: Let $f_n$ be a sequence of complex functions on a set $A$ and let $g_n$ be a decreasing sequence of non-negative functions on $A$. If the series $\sum_{n=0}^\infty f_n$ converges uniformly on $A$ and if there is a constant $M$ such that $|g_n(x)|\leq M$ for every $x\in A$ and for all non-negative integers $n$, then $\sum_{n=0}^\infty f_ng_n$ converges uniformly on $A$.
In this case, let $A=[0,R]$, and for $x\in A$ let $f_n(x)=a_nR^n$ and $g_n(x)=(x/R)^n$.
By hypothesis, the series $\sum_{n=0}^\infty a_nR^n$ is convergent, so $\sum_{n=0}^\infty f_n(x)$ converges uniformly on $A$. Also $|g_n(x)|\leq 1$ for all $n$ and for all $x\in A$.
By Abel's test, the series $\sum_{n=0}^\infty f_n(x)g_n(x)=\sum_{n=0}^\infty a_nR^n(x/R)^n=\sum_{n=0}^\infty a_nx^n$ converges uniformly on $[0,R]$.
Thus the original series is uniformly convergent on the radius of $D$ from $z_0$ to $z_1$.
Will this do? | 2015-05-04T23:19:05 | {
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http://openstudy.com/updates/55c9a5dce4b0c5fe98052769 | ## ganeshie8 one year ago find a third degree polynomial with integer coefficients whose roots are $\cot^2\frac{\pi}{7},~\cot^2\frac{2\pi}{7},~\cot^2\frac{3\pi}{7}$
1. anonymous
Consider the polynomial equation: $$x^6 +x^5 + x^4 + x^3 + x^2 + x + 1 = 0$$ It's roots are $$e^{i2k\pi/7}$$ where $$k = 1$$ to $$6$$ Divide the equation by $$x^3$$ , then we have, $$x^3 + x^2 + x + 1 + 1/x + 1/x^2 + 1/x^3 = 0$$ $$(x+1/x)^3 -3(x+1/x) + (x+1/x)^2 - 2 + (x+1/x) + 1 = 0$$ When x = $$e^{i2k\pi/7}$$ , $$(x+1/x) = 2\cos(2k\pi/7)$$ So, $$y^3 + y^2 -2y -1 = 0$$ has the roots $$2\cos(2\pi/7), 2\cos(4\pi/7) , 2\cos(6\pi/7)$$ Now, note that $$\cos(2k\pi/7) = \cos^2(k\pi/7) - \sin^2(k\pi/7)$$ $$= \frac{1}{\sec^2(k\pi/7)} - \frac{\tan^2(k\pi/7)}{\sec^2(k\pi/7)}$$ $$= \frac{1-\tan^2(k\pi/7)}{1 + \tan^2(k\pi/7)}$$ $$= \frac{\cot^2(k\pi/7) - 1}{\cot^2(k \pi/7) + 1}$$ So, if $$\cot^2(k\pi/7) = t$$ , $$2\cos(2k\pi/7) = 2\frac{(t-1)}{(t+1)}$$ Therefore $$\cot^2(k\pi/7)$$ satisfies the following equation $$(2\frac{(t-1)}{(t+1)})^3 + (2\frac{(t-1)}{(t+1)})^2 -2(2\frac{(t-1)}{(t+1)}) - 1 = 0$$ Multiplying both sides by $$(t+1)^3$$ , $$(2t-2)^3 + (t+1)(2t-2)^2 -2(t+1)^2(2t-2) - (t+1)^3 = 0$$ Therefore, $$(2t-2)^3 + (t+1)(2t-2)^2 -2(t+1)^2(2t-2) - (t+1)^3$$ is clearly a polynomial with integer coefficients and has as its roots $$\cot^2(k\pi/7)$$ for $$k = 1,2,3$$
2. anonymous
Also, simplifying the final expression, the final polynomial turns out to be $$7t^3 - 35t^2 + 21t-1$$
3. ganeshie8
beautiful! it seems there is no question i know which you cannot solve !
4. ikram002p
Andre thats make me cry :\ @adxpoi lol
5. anonymous
Haha, nah I'm sure there's stuff I can't solve :P
6. ikram002p
extremely interesting question hmmm and answer :O
7. ParthKohli
Hey I'm pretty sure I've asked this one before
8. ganeshie8
I think @adxpoi 's method generalizes to any arbitrary degree polynomial : Find a polynomial of $$n$$th degree whose roots are $\cot^2\frac{\pi}{2n+1},~\cot^2\frac{2\pi}{2n+1},~\cot^2\frac{3\pi}{2n+1},\cdots, \cot^2\frac{n\pi}{2n+1}$
9. ganeshie8
Yes @ParthKohli I remember, we had solved it using Vieta's formulas or something adhoc... :)
10. ParthKohli
No, complex numbers.
11. ParthKohli
In fact just like how adx has solved it.
12. ganeshie8
Ahh okay, can't tryst my memory lol knw how to solve the general case for arbitrary degree ?
13. ParthKohli
My book (KD Joshi) has this, yeah.
14. ganeshie8
Ohk.. I'm hoping we can simply mimic the adx's method, but I haven't tried it yet... so not completely sure yet..
15. ParthKohli
Where did you find this question?
16. ganeshie8
IMO shortlisted practice problems
17. ganeshie8
@praxer has been asking these recently
18. ParthKohli
Hmm, this one is pretty standard for an IMO problem if you've read about it.
19. ganeshie8
I read and forget most of what I read, so I must have encountered this problem before.. but I'm pretty sure I never had this observation before : $x^6+x^5+\cdots+x+1 = \dfrac{x^7-1}{x-1}$ so $$e^{i2k\pi/7}$$ is a root
20. ParthKohli
Exactly.
21. ganeshie8
Is there any other way to see that $$e^{i2k\pi/7}$$ satisfies that polynomial equation ? (w/o using geometric series)
22. ParthKohli
23. ParthKohli
No, not really.
24. ParthKohli
Pages 246-247, yes.
25. ganeshie8
slightly different problem, but the methods are similar yeah https://i.gyazo.com/27c20454519da07d956f53137adc3ee6.png | 2016-10-27T03:18:24 | {
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https://math.stackexchange.com/questions/2425550/density-of-x-y-where-x-y-are-independent-random-variables-with-common-pdf-f | # Density of $X-Y$ where $X,Y$ are independent random variables with common PDF $f(x) = e^{-x}$?
$X,Y$ are independent random variables with common PDF $f(x) = e^{-x}$ then density of $X-Y = \text{?}$
I thought of this let $Y_1 = X + Y$, $Y_2 = \frac{X-Y}{X+Y}$, solving which gives me $X = \frac{Y_1(1 + Y_2)}{2}$, $Y = \frac{Y_1-Y_2}{2}$
then I calculated the Jacobian $J = \begin{bmatrix} \frac{1+y_2}{2} & \frac{y_1}{2} \\ \frac{1}{2} & -\frac{1}{2} \end{bmatrix}$ so that $\left|\det(J)\right| = \frac{1+y_1+y_2}{4}$
and the joint density of $Y_1,Y_2$ is the following $W(Y_1,Y_2) = \left|\det(J)\right| e^{-(y_1+y_2)}$ when $y_1,y_2> 0$ and $0$ otherwise.
Next I thought of recovering $X-Y$ as the marginal but I got stuck. I think i messed up in the variables.
Any help is great!.
• why did you take such $Y_1,Y_2$? Sep 11 '17 at 17:21
• There are lots of approaches. One is characteristic functions. Another is to note the distribution of the difference should be is symmetric about $0$ and if you look at the right hand half then memorylessness suggests the distribution of $X-Y$ given $X \gt Y$ should have the same exponential distribution as you started with Sep 11 '17 at 17:22
• @MANMAID I found a similar technique used in an example of Rohatgi Probability and statistics book. Sep 11 '17 at 17:24
• I guess your $W(Y_1,Y_2)$ is the joint pdf of $Y_1,Y_2$, and if so then it should be $|\det(J)|$ and also I think support of $Y_1,Y_2$ are dependent. Sep 11 '17 at 17:27
• thanks! typo there. edited.Also i thought of taking $Y_{1} =X-Y$ and hence I could have obtained the marginal but then what would be the limits of $y_{1},y_{2}$ during the integration? Sep 11 '17 at 17:30
If I understood you correctly, you have, both $X$ and $Y$ being distributed by an exponential distribution, where $\lambda$ equals one. Now you want to know about the distribution of their difference, namely $Z=X-Y$. Their mass is $$P(z\ge Z)=P(z\ge X-Y)=P(z)$$ which is (for $z\le 0$) $$P(z)=\int^\infty_{0}\int^{\infty}_{x-z}e^{-x}e^{-y}\,dy\,dx,$$ as the area of interest is $y\ge x-z$. Next, we know that the density $$p(z)=\frac{d}{dz}P(z),$$ is the derivative of the mass. Using the Leibnitz rule, this is $$\frac{d}{dz}\int^\infty_{0}\int^\infty_{x-z}e^{-x}e^{-y} \, dy \, dx = \int^\infty_0 \frac{d}{dz}\int^\infty_{x-z}e^{-x}e^{-y}\,dy\,dx$$ $$\int^\infty_{-\infty} e^{-x}e^{-(x-z)} \, dx=\frac{e^z}{2}$$ After repeating the computation of $z\ge 0$, which would entail calculating $$\frac{d}{dz}P(z)=\int^\infty_0 \int^{x+z}_0 e^{-x}e^{-y} \, dy \, dx,$$ we arrive at $$p(z)=\frac{e^{-|z|}}{2}$$
Note that this known as the Laplace distribution.
\begin{align} \underbrace{\text{For } u>0} \text{ we have } f_{X-Y}(u) & = \frac d {du} \Pr(X-Y\le u) \\[10pt] & = \frac d {du} \operatorname{E}(\Pr(X-Y \le u \mid Y)) \\[10pt] & = \frac d {du} \operatorname{E}(\Pr(X \le u+Y\mid Y)) \\[10pt] & = \frac d {du} \operatorname{E}(1-e^{-(u+Y)}) \\[10pt] & = \frac d {du} \int_0^\infty (1 - e^{-(u+y)} ) e^{-y} \, dy \\[10pt] & = \frac d {du} \int_0^\infty (e^{-y} - e^{-u} e^{-2y}) \, dy \\[10pt] & = \frac d {du} \left( 1 - \frac 1 2 {} e^{-u} \right) \\[10pt] & = \frac 1 2 e^{-u}. \end{align} A similar thing applied when $u<0$ gives you $\dfrac 1 2 e^u,$ so you get $\dfrac 1 2 e^{-|u|}.$
But a simpler way to deal with $u<0$ is to say that since the distribution of $X-Y$ is plainly symmetric about $0$ (since $X-Y$ has the same distribution as $Y-X$), if you get $\dfrac 1 2 e^{-u}$ when $u>0,$ you have to get $\dfrac 1 2 e^u$ when $u<0.$
• Nice! but how we get $P(X-Y \leq u) = E[P(X-Y \leq u | Y)]$ ? Sep 12 '17 at 1:31
• This is the nicest approach! Sep 12 '17 at 2:19
• @BAYMAX The Probability for an event is the Expectation for the indicator random variable for the event. $\mathsf P(X-Y\leq u) ~{=\mathsf E(\mathbf 1_{X-Y\leq u}) \\ = \mathsf E(\mathsf E(\mathbf 1_{X-Y\leq u} \mid Y)) \\ = \mathsf E(\mathsf P(X-Y\leq u\mid Y)) }$ Sep 12 '17 at 2:24
• @BAYMAX : Consider $\Pr(X-Y\le u \mid Y= y),$ the conditional probability of one event given another. It depends on the value of $y.$ As a function of $y,$ say we call it $h(y).$ Then $h(Y)$ is a random variable. Its expected value is $\operatorname{E}(h(Y)) = \operatorname{E}( \Pr(X-Y\le u \mid Y)). \qquad$ Sep 12 '17 at 2:48
• @Divide1918 $$\int_0^x e^{-u}\,du = 1 - e^{-x}.$$ Apr 8 at 18:00
The transformation is $(X,Y)\rightarrow (Y_1,Y_2)$.
$Y_1=X+Y, Y_2=\dfrac{X-Y}{X+Y}$.
Let $y_1=x+y,y_2=\dfrac{x-y}{x+y}$, i.e., $x=\dfrac{y_1(1+y_2)}{2},y=\dfrac{y_1(1-y_2)}{2}$. Now $x>0,y>0$, hence $y_1>0, -1<y_2<1$
$J=\begin{bmatrix}\dfrac{1+y_2}{2}&\dfrac{y_1}{2}\\\dfrac{1-y_2}{2}&\dfrac{-y_1}{2}\end{bmatrix}$. Here, $\det(J)=\dfrac{-y_1}{2}$
Now \begin{align}f_{(Y_1,Y_2)}(y_1,y_2)=|\det(J)|f_{(X,Y)}(x,y)=\dfrac{y_1e^{-y_1}}{2}I(y_1>0,-1<y_2<1)\\=y_1e^{-y_1}I(y_1>0)\cdot\dfrac{1}{2}I(-1<y_2<1)\end{align}
Here $I(\cdot)$ is indicator function.
But I doubt you can recover the pdf of $X-Y$ easily. So, one way to do this analogous to the way you want is taking $Y_1=X-Y, Y_2=\dfrac{X+Y}{X-Y}$.
the reason Rohatgi Probability and statistics used this technique is because of independence of $X+Y,\dfrac{X-Y}{X+Y}$. But that will not work here and eventually the calculation will become very messy.
• I've upvoted this but a much simpler answer is available. I've posted two answers. One of them involves an integral of a function of two variables, but no Jacobians are needed because no changes of variables are done. So my challenge to everyone here: See if you can find a simpler way than that. Sep 11 '17 at 20:37
• Nice! here when $Y_{2} = \frac{X+Y}{X-Y}$what is the range of $Y_{2}?$,is it $\Bbb{R} / [-1,1] ?$ Sep 12 '17 at 1:35
• @BAYMAX the reason Rohatgi Probability and statistics used this technique is because of independence of $X+Y,\dfrac{X-Y}{X+Y}$. That will not work here and eventually the calculation will become very messy. So, I suggest you to see other solutions that are posted here. Sep 12 '17 at 1:56
$$P(X-Y<z) = \sum_y P(X-y<z)P(Y=y) = \int_{y \in \mathbb{R}} P(X<y+z)f(y) \, dy$$ by the law of total probability (there's probably a more rigorous way to write that middle expression, but it'll still be that integral). Then this is $$\int_{y+z>0,y>0} (1-e^{-(y+z)})e^{-y} \, dy$$ using the given distributions. This splits into $$\begin{cases} \int_{-z}^{\infty} (e^{-y}-e^{-2y}e^{-z}) \, dy & z<0 \\ \int_{0}^{\infty} (e^{-y}-e^{-2y}e^{-z}) \, dy & z \geq 0 \end{cases} = \begin{cases} \frac{1}{2}e^{z} & z<0 \\ 1-\frac{1}{2}e^{-z} & z\geq 0 \end{cases}.$$ Differentiating then gives the density function as $e^{-\lvert z \rvert}/2$.
• I've up-voted this although I can see that some might object to a discrete sum for a continuous variable. I've also posted an answer with a similar approach but with greater detail. Sep 11 '17 at 19:04
I already posted an answer involving no integrals of functions of more than one variable; here's another approach.
\begin{align} \text{First assume } u >0. \text{ Then} \\ \Pr( X-Y > u) & = \int_0^\infty \left( \int_{y+u}^\infty f_{X,Y} (x,y) \, dx \right) \,dy \\[10pt] & = \int_0^\infty \left( \int_{y+u}^\infty e^{-x} e^{-y} \, dx \right) \,dy \\[10pt] & = \int_0^\infty \left( e^{-y} \int_{y+u}^\infty e^{-x} \, dx \right) \,dy \\ & \qquad\text{(This can be done because $e^{-y}$ does not change as $x$ goes from something to $\infty$.)} \\[10pt] & = \int_0^\infty e^{-y} \cdot e^{-(y+ u)} \, dy \\[10pt] & = \frac 1 2 e^{-u}. \end{align} That works if $u>0.$ Then use the fact that $Y-X$ has the same probability distribution as $X-Y$ to conclude that if $u<0$ then $\Pr(X-Y<u) = \frac 1 2 e^{u}.$
Therefore if $u>0$ then $\Pr(X-Y\le u) = 1- \dfrac 1 2 e^{-u}$ and mutatis mutandis if $u<0,$ so we get $\displaystyle f_{X-Y}(u) = \frac 1 2 e^{-|u|}.$ | 2021-10-25T07:38:23 | {
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https://math.stackexchange.com/questions/1735723/limit-of-product-exists-and-one-limit-exists | # limit of product exists and one limit exists
Question is to check :
If $\lim_{n\rightarrow \infty}a_nb_n$ exists and $\lim_{n\rightarrow \infty}a_n$ exists implies $\lim_{n\rightarrow \infty}b_n$ exists.
Considering $a_n=\frac{1}{n}$ and $b_n=n$ then we see that $\lim_{n\rightarrow \infty}a_nb_n$ exists, equals to $1$ and $\lim_{n\rightarrow \infty}a_n$ exists and equals to $0$. In this case $\lim_{n\rightarrow \infty}b_n$ does not exists..
So, the answer to the question is Not always..
Now, what if $\lim_{n\rightarrow \infty}a_n$ exists and is non zero and $(b_n)$ is bounded?
Suppose that $\lim_{n\rightarrow \infty}a_nb_n=M$ with $\lim_{n\rightarrow \infty}a_n=P\neq 0$ and $|b_n|\leq A$ for all $n\in \mathbb{N}$.
I claim that $\lim_{n\rightarrow \infty}b_n=\frac{M}{P}$
Consider $|b_n-\frac{M}{P}|$.. We estimate this. Given $\epsilon>0$ there exists $N\in \mathbb{N}$ such that $|a_nb_n-M|<\epsilon$ and $|a_n-P|<\epsilon$ for all $n\geq N$.
$$|b_n-\frac{M}{P}|=\frac{1}{P}|Pb_n-M|=\frac{1}{P}|Pb_n-a_nb_n+a_nb_n-M|\leq \frac{1}{P}|b_n||a_n-P|+\frac{1}{P} \epsilon$$ As $(b_n)$ is bounded, we have for all $n\geq N$ $$|b_n-\frac{M}{P}|\leq \frac{1}{P}A\epsilon+\frac{1}{P} \epsilon=\epsilon\left(\frac{1}{P}(A+1)\right)$$
Thus, we are done.
I am just wondering if i can relax any of the conditions that i have assumed. Help me to know more about this.
• You have a typo where you said $g_n$ instead of $b_n$. Otherwise I don't think you can relax any assumption. Apr 10 '16 at 8:46
• Can you give a counter example if you relax the bounded condition? Apr 10 '16 at 9:11
• In the example you gave after "Considering...", we have that $\;\lim b_n\;$ does exist. Not finitely but it surely exists. If you want a simple example of a non-existing limit in this case, take $\;a_n=\frac1n\;,\;\;b_n=(-1)^n\;$ Apr 10 '16 at 9:25
• @PatrickStevens : Edited. Thanks
– user312648
Apr 10 '16 at 10:01
• @Kyson : As the answers below say that boundedness comes immediately, there is no question of non bounded sequences...
– user312648
Apr 10 '16 at 10:02
If
$$\;\lim_{n\to\infty}a_n=L\neq0\;,\;\;\lim_{n\to\infty}a_nb_n= K\;,\;\;\text{then since for almost all indexes}\;\;a_n\neq0\,,$$
we get that for all indexes except a finite number of them, from arithmetic of limits:
$$b_n=\frac{a_nb_n}{a_n}\xrightarrow[n\to\infty]{}\frac KL$$
and all this is well-defined and always finite since $\;L\neq0\;$ . No need to require a priori boundedness for $\;\{b_n\}\;$ .
• I do not have to assume it is bounded as it is already bounded... THanks..
– user312648
Apr 10 '16 at 10:11
• @cello Exactly, you get boundedness for free. It was a pleasure. Apr 10 '16 at 10:14
You get the condition that $(b_n)$ is bounded for free. Let us first show that by contradiction. Assume that $(b_n)$ is unbounded. Then there exists subsequence $(b_{p(n)})$ of $(b_n)$ such that $|b_{p(n)}|>n$, for all $n$. But, then we have $$|na_{p(n)} |\leq |a_{p(n)}b_{p(n)}| \leq M$$ where $M$ is such that $|a_nb_n|\leq M$, which exists by convergence of $(a_nb_n)$. It follows that $$0\leq |a_{p(n)}| \leq \frac M n \implies \lim_na_n = \lim_n a_{p(n)} = 0$$ Contradiction.
On the other hand, you could easily prove that $\lim_na_n\neq 0$ implies convergence of $(b_n)$ just by noting the general rule: $$\lim_nb_n\neq 0\implies\lim_n\left(\frac{a_n}{b_n}\right) = \frac{\lim_na_n}{\lim_nb_n}$$ for convergent sequences $(a_n)$ and $(b_n)$.
• This was useful.. Thaks
– user312648
Apr 10 '16 at 10:03
1. If $\lim_{n}a_{n}=0$, the conclusion is not correct. Take as a counter example: $$a_{n}=1/n$$ $$b_{n}=\sin(n\pi).$$ By squeeze theorem $a_{n}b_{n}\rightarrow0$, but $b_{n}$ has no limit.
2. If $\lim_{n}a_{n}\not=0$ Then the result is true. Because, for $n>N$ (ultimately), we have $a_{n}\not=0$ and so we can write $$b_{n}=\frac{1}{a_{n}}.a_{n}b_{n}$$ and since the limit of the right side exists, the left side must have also a limit. | 2021-12-08T20:14:08 | {
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https://math.stackexchange.com/questions/2762428/equivalent-to-the-non-convergent-integral | # Equivalent to the non-convergent integral
Excuse me, please, for the initial post, I did not know that someone would apprehend this rudely, and also excuse my English ...
Task: Find the equivalent to a function, when $t \to +\infty$
$$f(t) = \int \limits_{t}^{2t} \frac{x^2}{e^{x^2}} dx$$
My ideas:
1) I tried to find such a function to integrate by parts, I would get some function in a closed form and an asymptotically small function.
The problem is that it is not so easy to find it by choosing it, but in general terms it is not clear how to do it.
2) One can try to use the mean value theorem if we find an equivalent function for the integrand
The problem is that there seems to be no equivalent ...
3) You can try to guess the answer if you use the rule of l'Hospital for the original function and function, which we do not yet know. The problem is that in an adequate form I could not do it.
P.S. I came up with this task myself based on some examples given to me at the lecture. I really do not understand why people do not like my questions ...
• "The equivalent to a function". What do you really mean? Are you concerned to find a function which behaves similarly to $f(t)$ as $t \to +\infty$? May 1 '18 at 20:57
• You'll find that simple "Here's the statement of my exercise, solve it for me" posts will be poorly received. What is better is for you to add context: What you understand about the problem, what you've tried so far, etc. Something to both show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. May 1 '18 at 21:01
• OP, great job on improving your question! I've voted to reopen as a result of your changes. May 2 '18 at 0:12
We are looking for a nice easy function $g(t)$ such that
$$\lim_{t\to \infty}\frac{f(t)}{g(t)}=1.$$
Since $\lim_{t\to \infty}f(t)=0,$ the same will have to be true of $g.$ That means the ratio will have the form $0/0$ at $\infty,$ suggesting L'Hopital may be the way to go.
Let $F(x)$ be an antiderivative for $x^2e^{-x^2}.$ Then $f(t)= F(2t)-F(t),$ which implies
$$f'(t) = 2F'(2t)-F'(t) = 8t^2e^{-4t^2}- t^2e^{-t^2}.$$
Since $e^{-4t^2}$ is much smaller than $e^{-t^2}$ for large $t,$ the term $- t^2e^{-t^2}$ above will dominate in this expression. So if we can find $g$ such that $g'(t) = - t^2e^{-t^2}+h(t),$ where $h(t)= o(t^2e^{-t^2}),$ then we'll be in good shape.
I just played around with this. The first thing I tried was $g(t)=te^{-t^2}.$ Here $g'(t)= e^{-t^2}-2t^2e^{-t^2}.$ Lucky guess! I just need to divide by $2.$ Thus, taking $g(t)= (te^{-t^2})/2,$ we will have $f(t)/g(t) \to 1$ at $\infty$ by L'Hopital, and we're done.
• So how do you guess this function? The answer is to check, I'm certainly able to May 2 '18 at 8:51
• Anyway, thanks you! May 2 '18 at 8:52
• "So how do you guess this function?" Good question. It wasn't even a good hint since I just pulled the answer out of a hat. I edited it and hopefully improved it.
– zhw.
May 2 '18 at 15:44
• Yes, it looks interesting enough. And is there any tool that allows you to find this assimptotics without guessing? But in general, your method is very good for not very complex functions. May 2 '18 at 16:12
• P.S. I am a first-year student in theoretical mathematics at St. Petersburg State University (St. Petersburg, Russia) May 2 '18 at 16:14
Integration by parts will work. We have $$\int_t^{2 t} x^2 e^{-x^2} dx = \int_t^{2 t} \left( -\frac x 2 \right) d(e^{-x^2}) = \\ -\frac x 2 e^{-x^2} \bigg\rvert_t^{2 t} - \int_t^{2 t} \frac 1 {4 x} d(e^{-x^2}) = \dots \,.$$ All terms evaluated at $x = 2t$ will produce faster decaying exponentials and can be discarded. In other words, we could have extended the upper limit of integration to infinity.
Repeating the same procedure, we obtain a full asymptotic series: $$\int_t^{2 t} x^2 e^{-x^2} dx \sim e^{-t^2} \sum_{k = 0}^\infty \frac {(-1)^{k + 1} \Gamma(k - 1/2)} {4 \sqrt \pi} t^{-2 k + 1}.$$
Essentially, we have computed an asymptotic expansion for $\operatorname{erf}$. | 2021-10-24T15:29:30 | {
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https://math.stackexchange.com/questions/3281889/how-to-find-gcd-of-those-two-complex-polynomials | # How to find GCD of those two complex polynomials?
I have a polynomial
$$f(x) = i(x^2-1)^3+(x^2+1)^3-8x^3$$
I want to check if this has repeated roots. To do so, I'll find greatest common divisor (euclidean algorithm) of $$f(x)$$ and its derivative $$f'(x)$$.
$$f(x) = i(x^2-1)^3+(x^2+1)^3-8x^3 = i(x^6-3x^4+3x^2-1)+(x^6+3x^4+3x^2+1)-8x^3$$
$$f'(x) = i(6x^5-12x^3+6x)+(6x^5+12x^3+6x)-24x^2$$
I know I should find the GCD as $$(x-i)(x-1)$$ but this is where I'm stuck. What would the next step be?
• Euclidean algorithm for gcd of polynomials. – Robert Israel Jul 3 at 14:15
• @RobertIsrael I have no clue how to apply it on those. – Melz Jul 3 at 14:16
• How to do this using long polynomial division? – Melz Jul 3 at 14:48
By inspection, $$f(1)=0$$ so $$x-1$$ is a factor of $$f(x)$$.
Since $$f'(x)= 3i(x^2-1)^2(2x)+3(x^2+1)^2(2x)-24x^2$$, we also have that $$f'(1)=0$$, so $$x-1$$ is a factor of $$f'(x)$$.
This means $$(x-1)^2$$ is a factor of $$f(x)$$.
One also has that by inspection $$f(i)=0$$ so $$x-i$$ is a factor of $$f(x)$$.
We also have that $$f'(i)=0$$, so $$x-i$$ is a factor of $$f'(x)$$.
So we also have that $$(x-i)^2$$ is a factor of $$f(x)$$.
• How to prove those algebraically? For instance, long polynomial division? – Melz Jul 3 at 14:21
• @Enzo : That would certainly work, but you would have to group all of the terms of the same degree together and you would be faced with division of complex numbers which is easy but unwieldy. – MPW Jul 3 at 14:24
• The thing I'm still confused about is where to stop dividing those polynomials. – Melz Jul 3 at 14:25
• @Enco Note that this isn't a complete answer since we still need to check if the remaining quadratic factor has repeated roots. Did you intend to use only the GCD method, or are other ad-hoc methods admissable? The GCD method is the best way to proceed in general. – Bill Dubuque Jul 3 at 14:25
It's best not to split the polynomials up using $$i$$, but collect terms in powers of $$x$$.
\eqalign{f_0 = f(x) &= \left( 1+i \right) {x}^{6}+ \left( 3-3\,i \right) {x}^{4}-8\,{x}^{3}+ \left( 3+3\,i \right) {x}^{2}+1-i \cr f_1 = f'(x) &= \left( 6+6\,i \right) {x}^{5}+ \left( 12-12\,i \right) {x}^{3}-24\,{x }^{2}+ \left( 6+6\,i \right) x } The remainder of $$f_0$$ on division by $$f_1$$ is $$f_2 = f_0 - (x/6) f_1 = \left( 1-i \right) {x}^{4}-4\,{x}^{3}+ \left( 2+2\,i \right) {x}^{2}+ 1-i$$ Then take remainder of $$f_1$$ on division by $$f_2$$, etc. | 2019-09-17T13:29:46 | {
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https://electronics.stackexchange.com/questions/316825/conjugate-non-conjugate-and-single-poles-in-an-rlc-circuit-system?noredirect=1 | # Conjugate, non-conjugate and single poles in an RLC circuit system
In all the texts I encountered so far, I find the following pole-zero diagram example for an RLC series circuit:
The transfer function for the above circuit can be found as:
H(s) = XC/(R + XL + XC) = (1/sC)/[R + sL + (1/sC)]
H(s) = 1/(LCs² + RCs + 1)
But when I plot the pole-zero diagram of the above circuit for different R,L and C values, I don't always obtain the poles as conjugate.
Below are three different pole-zero diagram of the above RLC circuit for different R,L and C values:
1-) Conjugate Poles
R = 10;
C = 0.00001;
L = 0.001;
2-) Non-conjugate poles
R = 10;
C = 0.001;
L = 0.001;
3-) Pole at the origin
R = 100;
C = 0.001;
L = 0.001;
Question:
What can we say about the real response of the circuit for each three case above by considering the locations of the poles? In other words, what does it mean poles being conjugate, being asymmetric/non-conjugate and being at the origin only for an RLC series circuit? (Especially, in the second case there is two different non-conjugate poles which is the most confusing situation to say something about the system)
Edit:
Following the answer I was able to reveal the higher pole for the third case:
• I think the last plot shows a pole very close to the origin but it is not. Should you zoom out, you should see the second real pole at a higher frequency but it does not appear in the plot. When $Q$ is low ($R=100\;\Omega$), as explained in my answer, one pole dominates the low-frequency response and is located at $Q\omega_0$ while the second pole lies at $\frac{\omega_0}{Q}$. Your last plot shows the first one only. A pole at the origin would mean that $s=0$ is a root of the denominator $D(s)$ which is not your case. – Verbal Kint Jul 12 '17 at 7:41
This general form for this type of low pass filter is: -
$H(s) = \dfrac{\omega_n^2}{s^2 + 2\zeta \omega_n s+\omega_n^2}$
And if you solve the quadratic in the denominator (to reveal the poles) you get: -
$s = \dfrac{-2\zeta\omega_n \pm 2\omega_n\sqrt{\zeta^2-1}}{2}$ $= \omega_n(-\zeta \pm \sqrt{\zeta^2-1})$
Then, if you analyse the square root, you can see that for low damping (low zeta) you get the square root of a negative number hence that part of the equation involves "j" and you get conjugate complex poles at some fraction of +/-$\omega_n$.
When the damping (zeta) reaches unity, there are no more complex poles and a single pole lies on the real axis at $-\zeta\omega_n$. This then splits into two poles (along the real axis) as zeta rises above 1.
A low value of zeta is under-damped hence you get a peaky response in the bode plot and you get conjugate poles. When zeta = 1 you get critical damping and when zeta is greater than 1 you get a rather sloppy 2nd order filter that starts to look like a 1st order filter as R dominates over $X_L$.
To get numbers we need to know how zeta and omega relate to R, L and C values: -
$\zeta = \dfrac{R}{2}\sqrt{\dfrac{C}{L}}$ and $\omega_n = \dfrac{1}{\sqrt{LC}}$
For R = 10, C = 0.00001 and L = 0.001, zeta = 0.5 and Wn = 10,000 and this is as you display the conjugate poles on your first graph.
For R = 10 and C = L = 0.001, zeta = 5 and Wn = 1,000 so the poles are at: -
$s=1000(-5\pm\sqrt{24}$) = -9899 and -101 and I can't precisely say if this corresponds with your graph but it looks close.
For R = 100 and C - L = 0.001, zeta = 50 and Wn = 1,000 so the poles are at: -
$s=1000(-50\pm\sqrt{2499}$) = -99,990 and -0.01 so you are not able to see the higher pole on your graph but otherwise I would say I get about the same result.
To substantiate the theory a bit more, this picture may be useful: -
It's also noteworthy that when both poles lie on the real axis (i.e. the over-damped situation), pole positions are: -
$= \omega_n(-\zeta + \sqrt{\zeta^2-1})$ and $= \omega_n(-\zeta - \sqrt{\zeta^2-1})$
And, if you did the math you would find that one pole is the normal conjugate of the other with respect to $\omega_n$ i.e. if one is ten times $\omega_n$ then the other is one-tenth of $\omega_n$.
In other words $= \omega_n(-\zeta + \sqrt{\zeta^2-1})$ is the inverse of $= \omega_n(-\zeta - \sqrt{\zeta^2-1})$.
• I re-read your answer; amazing and concise explanation how zeta is the parameter plays with the location of the poles. I also edited my question for the third case. – user16307 Jul 12 '17 at 9:12
• Q factor and zeta have inverse relation so one can also write the poles in terms of Q factor right? – user16307 Jul 12 '17 at 9:16
• Q = 1/2z or z = 1/2Q. – Andy aka Jul 12 '17 at 9:17
• I have also added a picture showing how increasing zeta will cause the conjugate poles to circle round towards the real (horizontal) axis. – Andy aka Jul 12 '17 at 9:34
• In other words $-\zeta+\sqrt{\zeta^2-1}$ is the reciprocal of $-\zeta-\sqrt{\zeta^2-1}$. You can prove this by multiplying them together - it always equals 1. So, in a way there is still a "pairing" between poles at $\omega_n$ even when the poles are not complex. – Andy aka Jul 12 '17 at 10:43
The circuit you have drawn, the $RLC$ network, does not feature zeros, only poles. Should you add a small resistance in series with the capacitor, then you would add a zero. The equation you have derived shows a second-order polynomial form in the denominator: $D(s)=1+b_1s+b_2s^2$. It can be advantageously factored in a canonical form such as $D(s)=1+\frac{s}{Q\omega_0}+(\frac{s}{\omega_0})^2$ in which $Q=\frac{\sqrt{b_2}}{b_1}$ and $\omega_0=\frac{1}{\sqrt{b_2}}$. By quick identification with your expression, you can determine these values easily. Now, if you try to determine the roots of $D(s)=0$, or the poles of the transfer function, you end up with a pair of roots defined by $s_{p1},s_{p2}=\frac{\omega_0}{2Q}(\pm\sqrt{1-4Q^2}-1)$.
From this expression, you can see that depending on the value of $Q$, the expression under the square root can be positive or negative. If positive, the roots are real (there is no imaginary part) and the transient response is non-oscillatory. This is the case for a $Q\le0.5$. For a very low $Q\lt\lt1$, we can apply the so-called low-$Q$ approximation in which you consider the poles well spread apart. There is one in the low frequency domain (the root in your diagram is close to the vertical axis) while the second is in higher frequencies. You thus write $D(s)\approx(1+\frac{s}{\omega_p1})(1+\frac{s}{\omega_p2})$ in which $\omega_{p1}=Q\omega_0$ and $\omega_{p2}=\frac{\omega_0}{Q}$. It simply means that for a high value of the series resistance, the circuit can be replaced by two (isolated) cascaded $RC$ networked tuned at $\omega_{p1}$ and $\omega_{p2}$. If you drive your circuit with a step voltage, the output voltage is very sluggish without overshoot.
When $Q=0.5$, the roots are still real but coincident. The response is still non-oscillatory and $D(s)\approx(1+\frac{s}{\omega_0})^2$. If you drive your circuit with a step voltage, the output voltage rises quickly without overshoot.
Now if you reduce the series resistance further, $Q$ exceeds 0.5 and the roots involve imaginary notation: they become imaginary conjugates like one root located at -1+2j and the other at -1-2j for instance. Please note that the real part in these expressions is negative implying that in your $s$-plane diagram, the little "crosses" are in the left half-plane (LHP). We call them LHP poles or LHPP. The negative real parts model the losses in your circuit, the power dissipation that will damp oscillations and force the time-domain response to converge to a steady-state value as $t$ approaches infinity. If you drive your circuit with a step voltage, the output voltage rises quickly and overshoots.
As these losses become less and less (the circuit gains in efficiency), $Q$ increases - more pronounced overshoot in $v_{out}(t)$ - until the poles become pure imaginaries: the real parts are gone and the poles are located on the vertical axis. Should you excite the circuit with a transient pulse, the response would be oscillatory with sustained oscillations when the excitation is gone.
A pole at the origin is a different story. Having a pole at the origin simply means that your transfer function hosts a division by $s$: $H(s)=\frac{1}{s(1+\frac{s}{\omega_p})}$ features a LHP pole and a pole at the origin because $D(s)=0$ for $s=0$. Should you divide by $s^2$ instead, you have two poles at the origin etc. You have a pole at the origin in systems featuring an integrator for instance. The pole at the origin implies a gain approaching infinity in dc (when $s=0$), actually the op amp open-loop gain, which ensures a very low static error in the variable you control. | 2019-08-20T20:35:55 | {
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https://nicoguaro.github.io/posts/numerical-28/ | # Numerical methods challenge: Day 28
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.
## LU factorization
Today we have LU decomposition. That is a factorization of a matrix into a lower (L) and upper (U) matrix. Basically it stores de steps of a Gauss elimination in matrices.
Following are the codes
### Python
from __future__ import division, print_function
import numpy as np
def LU(mat):
m, _ = mat.shape
mat = mat.copy()
for col in range(0, m - 1):
for row in range(col + 1, m):
if mat[row, col] != 0.0:
lam = mat[row, col]/mat[col, col]
mat[row, col + 1:m] = mat[row, col + 1:m] -\
lam * mat[col, col + 1:m]
mat[row, col] = lam
return mat
A = np.array([
[1, 1, 0, 3],
[2, 1, -1, 1],
[3, -1, -1, 2],
[-1, 2, 3, -1]], dtype=float)
B = LU(A)
### Julia
function LU(mat)
m, _ = size(mat)
mat = copy(mat)
for col = 1:m - 2
for row = col + 1:m
if mat[row, col] != 0.0
lam = mat[row, col]/mat[col, col]
mat[row, col + 1:m] = mat[row, col + 1:m] -
lam * mat[col, col + 1:m]
mat[row, col] = lam
end
end
end
return mat
end
A = [1.0 1.0 0.0 3.0;
2.0 1.0 -1.0 1.0;
3.0 -1.0 -1.0 2.0;
-1.0 2.0 3.0 -1.0]
B = LU(A)
As an example we have the matrix
\begin{equation*} A = \begin{bmatrix} 1 &1 &0 &3\\ 2 &1 &-1 &1\\ 3 &-1 &-1 &2\\ -1 &2 &3 &-1 \end{bmatrix} = \begin{bmatrix} 1 &1 &0 &0\\ 2 &1 &0 &0\\ 3 &4 &1 &2\\ -1 &-3 &0 &1 \end{bmatrix} \begin{bmatrix} 1 &1 &0 &3\\ 0 &-1 &-1 &-5\\ 0 &0 &3 &13\\ 0 &0 &0 &-13 \end{bmatrix} \end{equation*}
And, the answer of both codes is
[[ 1., 1., 0., 3.],
[ 2., -1., -1., -5.],
[ 3., 4., 3., 13.],
[ -1., -3., 0., -13.]]
### 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 LU(np.random.rand(10, 10))
with result
1000 loops, best of 3: 303 µs per loop
For Julia:
@benchmark LU(rand(10, 10))
with result
BenchmarkTools.Trial:
memory estimate: 29.25 KiB
allocs estimate: 310
--------------
minimum time: 9.993 μs (0.00% GC)
median time: 11.725 μs (0.00% GC)
mean time: 14.943 μs (15.90% GC)
maximum time: 3.285 ms (95.64% GC)
--------------
samples: 10000
evals/sample: 1
In this case, we can say that the Python code is roughly 30 times slower than Julia code. | 2021-09-20T09:15:38 | {
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http://mathhelpforum.com/math-challenge-problems/9224-quickie-2-a.html | # Math Help - Quickie #2
1. ## Quickie #2
Solve: . $\sqrt[3]{6x + 28} - \sqrt[3]{6x - 28} \:=\:2$
2. This is some wacky thing, Soroban, or does it supposed to have a viable answer.
If so, I got -6 and 6
3. Originally Posted by Soroban
Solve: . $\sqrt[3]{6x + 28} - \sqrt[3]{6x - 28} \:=\:2$
$\begin{array}{l}
\sqrt[3]{{6x + 28}} - \sqrt[3]{{6x - 28}} = 2 \\
\left( {\sqrt[3]{{6x + 28}} - \sqrt[3]{{6x - 28}}} \right)^3 = 2^3 \\
\end{array}
$
$6x + 28 - 3\sqrt[3]{{6x + 28}}\sqrt[3]{{6x - 28}}(\sqrt[3]{{6x + 28}} - \sqrt[3]{{6x - 28}}) - 6x + 28 = 8$
$- 3\sqrt[3]{{6x + 28}}\sqrt[3]{{6x - 28}}(\sqrt[3]{{6x + 28}} - \sqrt[3]{{6x - 28}}) = - 48$
$- 3\sqrt[3]{{6x + 28}}\sqrt[3]{{6x - 28}}(2) = - 48$
$\begin{array}{l}
3\sqrt[3]{{(6x + 28)(6x - 28)}} = 24 \\
\sqrt[3]{{(6x + 28)(6x - 28)}} = 8 \\
(6x + 28)(6x - 28) = 512 \\
36x^2 - 784 = 512 \\
36x^2 = 1296 \\
x^2 = 36 \\
x = \pm 6 \\
\end{array}
$
4. Soroban has a clever trick to solve this.
5. Originally Posted by ThePerfectHacker
I think the trick is that we need to check the solutions. Because those the necessary but not sufficient conditions.
Thus,
x\not = -6
Why x=-6 isn't solution?
$\sqrt[3]{{6( - 6) + 28}} - \sqrt[3]{{6( - 6) - 28}} = \sqrt[3]{{ - 8}} - \sqrt[3]{{ - 64}} = - 2 - ( - 4) = 2$
6. Hello, OReilly!
Well done!
The "Quickie" solution uses a clever theorem,
. . but is not much shorter than your solution.
Theorem: .If a + b + c .= .0, then: .a³ + b³ + c³ .= . 3abc **
. . . . . . . . . _______ . . ______
We have: .³√6x + 28 - ³√6x - 28 - 8 .= .0
-. . . . . . . . . . . . . . . . . . . w .
. . . . . . . . . . . a . . . . . . . b . o . c
The thereom gives us: . . . . . . . . ._________________
. . (6x + 28) - (6x - 28) - 8 .= .3·³√(6x + 28)(6x - 28)(8)
. . . . . . . . . . . . . . . . . . . . . . . . _________
. . . . . . . . . . . . . . . . .48 .= .6·³√36x² - 784
Then: .36x² - 784 .= .512 . . x² = 36 . . x = ±6
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
** . Proof
We have: .a + b .= .-c
Cube both sides: .(a + b)³ .= .(-c)³
Expand: .aª + 3a²b + 3ab² + b³ .= .-c³
Then: .a³ + b³ + c³ .= .-3a²b - 3ab²
. . . . . a³ + b³ + c³ .= .-3ab(a + b)
Since a + b = -c, we have: .a³ + b³ + c³ .= .-3ab(-c)
Therefore: . a³ + b³ + c³ .= .3abc | 2016-05-29T04:21:58 | {
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https://mathematica.stackexchange.com/questions/258756/how-to-solve-complex-least-squares-with-real-unknowns | # How to solve complex least squares with real unknowns
I am looking at the case where my least squares problem is of the form
$$y=A x$$
where y is a vector of measured complex values, A is the model or design matrix, also complex, and x are the model coefficients which are real. How do I formulate the problem in Mathematica to get real values for my model coefficients?
Here is a minimum working example of what I have looked at.
First I make a noise free set of artificial data where my unknows are the values 1 to 7 and use the ordinary LeastSquares program to find my unknowns.
mat = RandomComplex[{-10 (1 + I), 10 (1 + I)}, {20,7}]; (* Design matrix *)
uk = Range[7]; (* Known unknowns *)
yy1 = mat . uk; (* Measured values *)
sol1 = LeastSquares[mat, yy1] (* Solution *)
(* {1. + 1.93795*10^-15 I, 2. - 2.22045*10^-16 I, 3. - 5.3385*10^-16 I,
4. - 2.02397*10^-15 I, 5. + 4.70461*10^-16 I, 6. - 5.55476*10^-16 I,
7. + 5.52099*10^-17 I} *)
As expected the real part of the solution is perfect and the imaginary part is numerical noise.
Now I put some noise into the measured values and use LeastSquares again.
yy2 = yy1 + RandomComplex[{-1 - I, 1 + I}, 20];(* Measured values with noise *)
sol2 = LeastSquares[mat, yy2] (* Solution *)
(* {1.02302 + 0.0244589 I, 2.00445 - 0.051833 I, 2.97412 + 0.0227336 I,
4.0041 - 0.00438793 I, 5.01066 - 0.0132584 I, 5.96674 + 0.0293913 I,
7.00924 + 0.00691532 I} *)
As expected there are now complex solutions although the imaginary parts are small.
I now attempt to do the matrix algebra the long way to get a solution with real values. There is probably a better way of doing the matrix algebra (please make suggestions). I finish by comparing the solutions from the three methods.
xx = Array[x, 7]; (* Unknown unknowns *)
ee = Sum[
ComplexExpand[(mat[[n]] . xx - yy2[[n]]) Conjugate[mat[[n]] . xx - yy2[[n]]]],
{n, 20}]; (* Sum of errors *)
mat3 = D[ee, #] & /@ xx; (* Take derivaties *)
{vec, mat4} = CoefficientArrays[mat3, xx];(* Find LHS and design matrix *)
sol3 = -Inverse[mat4] . vec ;(* Solve *)
TableForm[Transpose[{sol1, sol2, sol3}],
TableHeadings -> {None, {"Exact", "Complex Least Squares","Real Least Squares"}}]
So I have got my solution to have real values. However, I note that the errors in the real part of the LeastSquares solution are very similar to the errors in my real solution.
Two questions:
1. What is the best way to tackle this problem?
2. Is the LeastSquares solution just as good if I take the real part?
Note that my actual problem will have very large matrices.
Thanks
• PseudoInverse[mat].yy1, PseudoInverse[mat].yy2 and -PseudoInverse[mat4] . vec are alternative ways to solve Nov 25, 2021 at 7:22
• @UlrichNeumann Thanks but they give the same results as my methods...
– Hugh
Nov 25, 2021 at 9:32
• It would be a bad alternative if not! Nov 25, 2021 at 9:56
• @UlrichNeumann Do you have any idea about the internals of these codes? How valid is taking the real part of the solutions?
– Hugh
Nov 25, 2021 at 10:38
I am answering my own question. This turns out to be easier than I thought. Consider the problem where we have
$$y=A x$$
with y a column of complex measurements, A a matrix of complex values and x a vector of real unknowns. The objective is to find the real values of x via a least squares process.
Split the complex values into real and imaginary parts
$$y_r+i y_i=x \left(A_r+i A_i\right)$$
where the subscripts r, i mean real and imaginary parts. This may be written as two equations which may be put into matrix form as
$$\left( \begin{array}{c} y_r \\ y_i \\ \end{array} \right)=\left( \begin{array}{c} A_r \\ A_i \\ \end{array} \right) x$$
With this configuration the LeastSquares function may be applied directly.
Here is an example.
nn = 100; (* Number of measurements*)
nk = 7; (* Number of unknowns *)
A = RandomComplex[{-10 (1 + I), 10 (1 + I)}, {nn,
nk}]; (* Design matrix *)
uk = Range[nk]; (* Known unknowns *)
y = A . uk +
RandomComplex[{-1 - I, 1 + I},
nn]; (* Measured values with complex noise *)
Ar = Re[A];
Ai = Im[A];
yr = Re[y];
yi = Im[y];
yy = Join[yr, yi];
AA = Join[Ar, Ai];
LeastSquares[AA, yy]
(* {0.997568, 1.99928, 3.00159, 3.99664, 5.00797, 5.98345, 7.00619} *)
So this works well.
• This method assumes that the errors are independent (both between and within observations) and with a common variance. Do the residuals back up those assumptions?
– JimB
Jun 4 at 14:55
• @JimB You make a good point about the statistical assumptions. The concern is with data that are complex. If we assume that the complex data is derived via a Fourier transform then I would suggest the the real and imaginary parts are independent. Also, if the real time domain data has independent errors between observations then I think the complex data generated by Fourier transform would also be independent. Such independence would allow for standard least squares methods. However, do you have any insights into this area or can you suggest tests?
– Hugh
Jun 5 at 19:43 | 2022-06-29T09:07:14 | {
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http://crypto.stackexchange.com/tags/prime-numbers/hot | # Tag Info
21
You don't use a pre-generated list of primes. That would make it easy to crack as you note. The algorithm you want to use would be something like this (see note 4.51 in HAC, see also an answer on crypto.SE): Generate a random $512$ bit odd number, say $p$ Test to see if $p$ is prime; if it is, return $p$; this is expected to occur after testing about ...
17
No, it is not at all feasible to build an index of prime factors to break RSA. Even if we consider 384-bit RSA, which was in use but breakable two decades ago, the index would need to include a sizable portion of the 160 to 192-bit primes, so that the smallest factor of the modulus has a chance to be in the index. Per the Prime number theorem there are in ...
15
A Mersenne prime is a prime number that can be written in the form $M_p = 2^n-1$, and they’re extremely rare finds. Of all the numbers between 0 and $2^{25,964,951}-1$ there are 1,622,441 that are prime, but only 42 are Mersenne primes. The second sentence is wrong. What they meant to say is that there are 1,622,441 numbers of the form they mentioned ...
14
The standard way to generate big prime numbers is to take a preselected random number of the desired length, apply a Fermat test (best with the base $2$ as it can be optimized for speed) and then to apply a certain number of Miller-Rabin tests (depending on the length and the allowed error rate like $2^{-100}$) to get a number which is very probably a prime ...
11
mpz_nextprime states in the documentation and source (file: mpz/nextprime.c) that it simply finds the next prime larger than the provided input. There are various methods of doing so (depending on how efficient it tries to be), but they should all produce the same answer. Looking at the code, mpz_nextprime first tests a number against a large quantity of ...
10
This procedure is known as incremental search and his described in the Handbook of Applied Cryptography (note 4.51, page 148). Although some primes are being selected with higher probability than others, this allows no known attacks on RSA; roughly speaking, incremental search selects primes which could have been selected anyway and there are still ...
10
Wiener's result has been improved several times, and it is hard to tell how big the private exponent must be to be safe from further progress. Also, the proposed technique, assuming $d>n^{1/3}$, requires a minimum of ${1\over3}\cdot log_2(n)$ modular multiplications for the sparsest $d$ conceivable (a power of two), compared to say ${7\over6} \cdot ... 9 Let's assume for an instant that you could build a large table of all primes. Then... what ? How would you use it ? What would you look up ? If you "just" scan the table and try to divide the number to factor by each prime, then this is known as trial division; there is no need to store the primes (they can be regenerated on-the-fly; that's the division ... 9 I have asked a similar question to Arjen Lenstra a few years ago: I was investigating three 2048-bit primes of low Hamming weight:$p_1 = 2^{2048} - 2^{1056} + 2^{736} - 2^{320} + 2^{128} + 1p_2 = 2^{2048} - 2^{1376} + 2^{992} + 2^{896} + 2^{640} - 1p_3 = 2^{2048} - 2^{2016} + 2^{1984} - 2^{1856} - 2^{1824} + 2^{1792} - 2^{1760} + 2^{1696} + 2^{1664} ...
7
It has to do with optimizing RSA. It turns out that using the Chinese Remainder Theorem with $p$, $q$, $d\pmod{p-1}$, and $d\pmod{q-1}$ (i.e., prime1, prime2, exponent1, exponent2 from the data structure in the question) to run the decryption operation faster than if you only had $d,n$. For more information on how it is done, I found this reference ...
7
Well, to answer your questions in order: How big should $p$ be? Well, it should be large enough to defend against the known attacks against it. The most efficient attack is NFS; that has been used against numbers on the order of $2^{768}$ (a 232 digit number). It would appear wise to pick a $p$ that's considerably bigger than that; around 1024 bits at a ...
7
The problem of generating prime numbers reduces to one of determining primality (rather than an algorithm specifically designed to generate primes) since primes are pretty common: π(n) ~ n/ln(n). Probabilistic tests are used (e.g. in java.math.BigInteger.probablePrime()) rather than deterministic tests. See Miller-Rabin. ...
7
No, the fact that there's no known practical formula that produces only prime numbers doesn't really come into play; if someone found one tomorrow, that wouldn't have any cryptographical implications. You may want to go through the How does asymmetric encryption work? thread; the short answer is that for public key operations, the public and the private ...
7
Let me try a simple explanation of NFS. I will necessarily skip lots of details, but I hope you will get the main ideas. The number field sieve algorithm (NFS) is a member of a large family: index calculus algorithms. All algorithms in the family, which can be used for factoring and discrete logarithms in finite fields, share a common structure: ...
7
Here is the issue about hardware errors that Brent is worried about on the slides (I'm not saying I agree; I just saying what the issue is): Suppose we ran our algorithm, and it gave a result "it's prime"; how can we be certain that the algorithm didn't gave us the wrong answer because of an internal hardware error while running the algorithm? This may ...
6
Checking for smoothness can be computationally expensive, depending on the size of the "small" primes (there is no "natural" definition of "small", one has to define an arbitrary limit). Also, it is not really useful. The need for non-smooth integers comes from the $p-1$ factorization method. Let $n = pq$ be a RSA modulus that we wish to factor. Now suppose ...
6
For Diffie-Hellman or any variants like Elgamal or DSA, you're better off using the established primes. It doesn't matter what primes you use, really, as long as they're prime. The standard primes have had someone nod at them. If you generate your own prime and there's a problem (e.g. it's not really prime), then you're on your own and we will all laugh at ...
6
Short answer: Yes. The discrete logarithm can be attacked in a multitude of ways: Baby-step giant-step (BSGS), Pollard's Rho, Pohlig-Hellman, and the several variants of Index Calculus, the best of which currently is the Number Field Sieve. Let $n$ be the order of the generator of our field $\mathbb{F}_p$; it is $n = p-1$. We are trying to find $x$ given ...
6
The other answers explain why you shouldn't take shortcuts like this and why it won't provide much of a speedup anyway. As for an actual attack, here's an obvious one: brute force. With the parameters you gave, there are only $\binom{309}{3} = 4869634$ possible values for d, which is small enough to easily check every value.
6
The question to answer is "Is N the product of P*Q?" I believe that the easiest way to understand Shor is to imagine two sine waves, one length P and one length Q. Assuming that P and Q are co-prime, then the question above can also be answered "At what point does the harmony of P overlapped with Q repeat itself?" And the answer can be determined quickly, ...
6
Pure nonsense. For choosing the random $\Delta$ between $\sqrt{\min(N, Ň)}$ and $\sqrt{\max(N, Ň)}$ there are too many possibilities for it to work. For example whenever the first and last digits of $N$ differ, you get something like $\frac{1}{10} \cdot \sqrt N$ possibilities (the exact formula doesn't matter). So you can replace the first formula $gcd[N, ... 6 There is no reason in Shamir's scheme for the finite field$\mathbb F$to have a prime number$p$of elements; the field can have$p^m$elements for suitable prime$p$and integer$m \geq 1$. So, using$F_{2^8}$, the field with$2^8$elements is perfectly all right. However, choosing$m = 1$has the advantage that calculations in$\mathbb F_p$can be done ... 5 This issue, and its history, was discussed at length in Silverman and Rivest. The relevant passage here is in Section 6, which I quote: In 1977 Simmons and Norris [53] discussed the following "cycling" or "superencryption" attack on the RSA cryptosystem: given a ciphertext C, consider decrypting it by repeatedly encrypting it with the same public ... 5 Generating your own group for Diffie-Hellman is not a tough issue; but it is somewhat expensive (it depends on the context, but a 25 MHz ARM device would not like to do it often) and it is not really needed: a good point of DH (and DSA) is that the group parameters can be shared between many users, with no ill effect on the confidentiality of their ... 5 There's nothing wrong with generating your own primes for DH, as long as you know what you're doing. On the other hand, if you are a bit weak on number theory (or just glad that someone else has done the work, and had it double-checked), there's also nothing wrong with the modulii and generators in RFC 3526. As for the paper, well, it chiefly noted that ... 5 There are no known implications of the ABC Conjecture to RSA. The ABC problem doesn't have even a superficial resemblance to the security of RSA. (The only point of connection is the fact that they both relate to prime numbers, but that is extremely thin. Much of number theory can say it is somehow related to prime numbers. It'd be like assuming that ... 5 This is actually a special case of a more general property of the Euler totient function: it is a multiplicative function, meaning that, for any two coprime numbers$p$and$q$,$\phi(pq) = \phi(p)\phi(q)$. The special case where$p$and$q$are (distinct) primes is easy to prove. By definition,$\phi(n)$gives the number of positive integers coprime to and ... 5 No,$2^{333} - 1$cannot be a prime. It is easy to see this via the exponent,$333$, which has factorization$3\cdot3\cdot37\$. From this we know, by a well-known identity, that $$2^{333} - 1 = (2^{37} - 1)\cdot (1 + 2^{37} + 2^{37\cdot2} + 2^{37\cdot3} + 2^{37\cdot4} + 2^{37\cdot5} + 2^{37\cdot6} + 2^{37\cdot7} + 2^{37\cdot8}).$$
5
Preposterously large primes are not useful for cryptography in and of themselves, but the tools and techniques developed to find them (such as massively parallel distributed computing, algorithms that can efficiently confirm primality, etc) are important for cryptography. The prizes are meant to spur innovation in those areas. See e.g. ...
Only top voted, non community-wiki answers of a minimum length are eligible | 2013-12-06T13:15:12 | {
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https://math.stackexchange.com/questions/3539958/evaluating-sqrt9-5-sqrt3-sqrt9-5-sqrt3-sqrt9-cdots/3539972 | # Evaluating $\sqrt{9-5\sqrt{3-\sqrt{9-5\sqrt{3-\sqrt{9-\cdots}}}}}$.
I was wondering if it was possible to evaluate
$$\sqrt{9-5\sqrt{3-\sqrt{9-5\sqrt{3-\sqrt{9-5\sqrt{3-\sqrt{9-\cdots}}}}}}}$$
I let the expression equal $$x>0$$ and wrote $$x=\sqrt{9-5\sqrt{3-x}}$$ However, there is not just one value $$x$$ can take; $$x=2$$ or $$x=3$$.
How do I find out which one it is, or does this infinite-nested radical converge at all? Perhaps it merely oscillates between $$2$$ and $$3$$, but I am not entirely sure. Any help or hints would be much appreciated.
The ellipsis means "and so on". It measures the following: $$\sqrt{9-5}$$ $$\sqrt{9-5\sqrt{3-\sqrt{9-5}}}$$ $$\sqrt{9-5\sqrt{3-\sqrt{9-5\sqrt{3-\sqrt{9-5}}}}}$$ $$\vdots$$
Incidentally, I did not refuse to clarify the meaning. I am only active on Math.SE for so long. Whatever requests that occur can only be followed up the moment I am active, can see them and have time to act.
• @JyrkiLahtonen now that I have actually seen your comments, I think the real question is about why there is ambiguity. I mean, both the sequences you have defined recursively are the same, but tend toward different limits, which is not entirely clear to me. Unless, they are not the same? Feb 10 '20 at 0:53
• Thanks for the edit. I redacted my vote to close. Feb 10 '20 at 4:33
• @JyrkiLahtonen no worries :) Feb 10 '20 at 4:40
• No, they aren't. One might hope that it wouldn't matter. Like with the sequences of decimals when $$0.27,\ 0.2727,\ 0.272727,\ 0.27272727,\ldots$$ and $$0.2,\ 0.272,\ 0.27272,\ 0.2727272,\ldots,$$ where it doesn't matter. Both sequences converge to $3/11$. The key being that their difference rapidly tends to zero. What we see here is more like the difference between $$1,\ 1-1+1,\ 1-1+1-1+1,\ldots$$ and $$1-1,\ 1-1+1-1,\ 1-1+1-1+1-1,\ldots,$$ with the former sequence being constant $1$ and the latter constant $0$, leaving the meaning of $$1-1+1-1+1-1+1-1\cdots$$ undefined. Feb 10 '20 at 4:52
• (reposting a part of a deleted comment of mine). The sequence $$\sqrt{9},\sqrt{9-5\sqrt{3-\sqrt9}}, \sqrt{9-5\sqrt{3-\sqrt{9-5\sqrt{3-\sqrt9}}}},\ldots$$ with the added stuff ending at a $9$, is a constant sequence of $3$s. On the other hand, if we always end at a $5$, the sequence consists of $2$s only. And if we end at a $3$, the sequence converges towards $2$. This is in line with both the answers. This interpretation is the only way to get $3$ as the answer, but it does highlight the ambiguity. Feb 10 '20 at 5:04
Infinitely nested radicals may not make sense. The usual way to define this expression is as $$\lim_{n\to \infty} a_n$$, where $$a_{n+1} = \sqrt{9 - 5\sqrt{3 - a_n}}$$. The problem here is that we have no initial point specified. Choosing $$a_0 = 2$$ or $$a_0 = 3$$ will produce two different limits, so the nested radical is not well-defined.
• Actually, I don't know. Who am I supposed to believe? lol Feb 9 '20 at 9:04
• @MrPie Ultimately, it's a difference in convention. It's like pointing out that $0^0$ is undefined; it is true, but in certain situations, it makes sense to invent your own convention and say it's $1$. You can ignore unstable fixed points of $f(x) = \sqrt{9 - 5\sqrt{3 - x}}$ if it helps you get where you need to go, but just don't lose sight of the fact that you're using a non-standard convention for nested radicals. Feb 9 '20 at 9:07
• @JyrkiLahtonen The three dots just means "and so on", with $(9-5, 3), (9-5, 3)$ and etc. Feb 9 '20 at 11:59
• @JyrkiLahtonen I may need to work on my rigour. Thanks for letting me know. Feb 10 '20 at 4:41
The answer is $$2$$. While $$3$$ is also a fixed point, it is unstable because if we let $$x=3-\epsilon$$ for some small $$\epsilon$$, and iterate $$x\leftarrow \sqrt{9-5\sqrt{3-x}}$$, it will diverge away from $$3$$.
$\sqrt{9-5\sqrt{3-x}}$">
If you look at the graph, you will find that the slope approaches $$\infty$$ as $$x\to 3$$. The derivative of $$\sqrt{9-5\sqrt{3-x}}$$ is $$\frac5{4\sqrt{9-5\sqrt{3-x}}\sqrt{3-x}}$$. When $$x\to 3$$, the $$\sqrt{3-x}$$ in the denominator will approach $$0$$, which means the derivative approaches $$\infty$$ as $$x\to 3$$. Therefore, the fixed point is unstable and will very quickly diverge away from $$3$$. Plugging $$2$$ into the equation gives $$\frac58$$, which is less than $$1$$. Therefore, the fixed point is stable.
In conclusion:
$$\sqrt{9-5\sqrt{3-\sqrt{9-5\sqrt{3-\sqrt{9-5\sqrt{3-\sqrt{9-\cdots}}}}}}}=2$$
• Yes! I needed it to be $2$. Thank you so much! Feb 9 '20 at 8:56
• If this be true, it means the following formula is true: \begin{align}\sqrt 5 - \sqrt 3 &= \sqrt{9+4\sqrt{1+\sqrt{7+3\sqrt{5+\sqrt{7+3\sqrt{5+\sqrt{7+\cdots}}}}}}} \\ &- \sqrt{7+4\sqrt{1+\sqrt{9-5\sqrt{3-\sqrt{9-5\sqrt{3-\sqrt{9-\cdots}}}}}}}\end{align} The symmetries allign :) Feb 9 '20 at 9:01 | 2021-10-22T12:23:29 | {
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http://mathhelpforum.com/calculus/25371-differentiation.html | 1. ## Differentiation
I am currently studying calculus on an open learning basis and was wondering if someone could take a look at these questions and answers I have posted and feed back on my method of working out etc. I am pretty sure these answers are correct but would appreciate any comments.
Question 1:
Obtain $\frac{dy}{dx}$ for the following expression.
$y=(5x + 4)^3$ so $3(5x + 4)^2$ This is the 1st part.
I then differentiate the inner part of the function so:-
$(5x + 4)$ differentiates to 5. This is the the 2nd part.
I then multiply the 1st by the 2nd to get $3(5x + 4)^2\times 5$
so $\frac{dy}{dx}=15(5x + 4)^2$
Question 2:
Obtain $\frac{dy}{dx}$ for the following expression.
$y=(3 - 2x)^5$ so $5(3 - 2x)^4$ This is the 1st part.
I then differentiate the inner part of the function so:-
$(3 - 2x)$ differentiates to -2. This is the the 2nd part.
I then multiply the 1st by the 2nd to get $5(3 - 2x)^4\times -2$
so $\frac{dy}{dx}=-10(3 - 2x)^4$
Question 3:
Obtain $\frac{dy}{dx}$ for the following expression.
$y=\sqrt{(5 - 0.6x)}$ which is $(5 - 0.6x)^\frac{1}{2}$
I then differentiate this to get $\frac{1}{2}(5 - 0.6x)^{-\frac{1}{2}}$ This is the 1st part:
$(5 - 0.6x)$ differentiates to -0.6. This is the the 2nd part.
I then multiply the 1st by the 2nd to get $\frac{1}{2}(5 - 0.6x)^{-\frac{1}{2}}\times -0.6$
so $\frac{dy}{dx}=-0.3(5 - 0.6x)^{-\frac{1}{2}}$
Question 4:
Obtain $\frac{dy}{dx}$ for the following expression.
$y=(2 + 3x)^-0.6$ so $-0.6(2 + 3x)^{-1.6}$ This is the 1st part.
I then differentiate the inner part of the function so:-
$(2 + 3x)$ differentiates to 3. This is the the 2nd part.
I then multiply the 1st by the 2nd to get $-0.6(2 + 3x)^{-1.6}\times 3$
so $\frac{dy}{dx}=-1.8(2 + 3x)^{-1.6}$
As I said at the beginning I am pretty sure I am 100% correct and I agree these are relatively simple questions from a calculus point of view but being new to this subject I was just wondering if someone could make comment please.
Thanks
Derek.
2. I think they are all right
And there is nothing much to comment either... Why not try harder ones??
3. ## Obtain the derivative of
I came across another area which I am trying to get to grips with and was wondering if someone could help
I need to obtain the derivative of $f(t) = ln(5 - \frac{2}{3}t)$
I think I procede as follows.
$\frac{dy}{dt} = \frac{1}{5 - \frac{2}{3}t}\times -\frac{2}{3}$
Finishing up with $f'(t) = \frac{-\frac{2}{3}}{5 - \frac{2}{3}t}$
or $\frac{-0.6}{5 - 0.6 t}$
Another one is Determine $f'(y)$if $f(y) = exp(3 - \frac{1}{4}y)$
When the question reads "exp" do I assumeit is refering to the exponential function $e$
I have hunted high and low for similar equations to this to give me some kind of example to work too but come up with more or less nothing so from what I have found I have tried with this. I am probably wrong. Could someone make comment please??
Thanks
Derek
4. Originally Posted by DRJ
Another one is Determine $f'(y)$if $f(y) = exp(3 - \frac{1}{4}y)$
When the question reads "exp" do I assumeit is refering to the exponential function $e$
I have hunted high and low for similar equations to this to give me some kind of example to work too but come up with more or less nothing so from what I have found I have tried with this. I am probably wrong. Could someone make comment please??
Thanks
Derek
If $f(x) = e^{ax+b}$ then $f'(x) = ae^{ax+b}$
5. Originally Posted by DRJ
I came across another area which I am trying to get to grips with and was wondering if someone could help
I need to obtain the derivative of $f(t) = ln(5 - \frac{2}{3}t)$
I think I procede as follows.
$\frac{dy}{dt} = \frac{1}{5 - \frac{2}{3}t}\times -\frac{2}{3}$
Finishing up with $f'(t) = \frac{-\frac{2}{3}}{5 - \frac{2}{3}t}$
or $\frac{-0.6}{5 - 0.6 t}$
Yes, though if this is a Math class I'd recommend leaving this as
$f^{\prime}(t) = -\frac{2}{15 - 2t}$
rather than the decimal form.
Originally Posted by DRJ
When the question reads "exp" do I assumeit is refering to the exponential function $e$
Yes, the "exp" and "e" exponential are the same function.
-Dan
6. Hello, DRJ!
Find the derivative of: $f(t) \:= \:\ln\left(5 - \frac{2}{3}t\right)$
I think I proceed as follows: . $\frac{dy}{dt} \:= \:\frac{1}{5 - \frac{2}{3}t} \times \left(-\frac{2}{3}\right)$
Finishing up with: . $f'(t) \:= \:\frac{-\frac{2}{3}}{5 - \frac{2}{3}t}$ . . . . Yes!
. . or: . $\frac{-0.6}{5 - 0.6 t}$ . . . . no
Don't use rounded-off decimals . . . $0.6$ is not equal to $\frac{2}{3}$
We have: . $f'(x) \:=\:\frac{-\frac{2}{3}}{5 - \frac{2}{3}t}$
Multiply top and bottom by 3: . $\frac{3 \times\left(-\frac{2}{3}\right)}{3 \times\left(5 - \frac{2}{3}t\right)} \;=\;\frac{-2}{15-2t}$
Determine $f'(y)\text{ if }f(y) \:= \:\text{exp}\left(3 - \frac{1}{4}y\right)$
When the question reads "exp",
do I assume it is refering to the exponential function $e$ ? . Yes!
. . . . . . . . . . . . . . the original function
. . . . . . . . . . . . . . . . . . . $\downarrow$
Given: . $y \:=\:e^u$, then: . $y' \:=\:e^u\cdot u'$
. - . - . - . . . . . . . . . . . . . . . $\uparrow$
. . . . . . . . . . . . . . .
derivative of the exponent
We have: . $f(y) \:=\;e^{(3-\frac{1}{4}y)}$
Then: . $f'(y) \;=\;e^{(3-\frac{1}{4}y)}\!\cdot\!\left(\text{-}\frac{1}{4}\right) \;=\;-\frac{1}{4}e^{(3-\frac{1}{4}y)}$
7. ## Differentiation and the exponential function
Hello again
Just a quick query regarding the $e$ function.
In my example $f(y) = exp(3 - \frac{1}{4}y)$ I think I may have mis read the question.
Is this actually asking me to find $f(y)$ of $exp^{(3-\frac{1}{4}y)}$ or as you stated $e^{(3-\frac{1}{4}y)}$ as opposed to $f(y) = exp(3 - \frac{1}{4}y)$ or are they both the same???
Thanks and happy new year etc
8. Originally Posted by DRJ
Hello again
Just a quick query regarding the $e$ function.
In my example $f(y) = exp(3 - \frac{1}{4}y)$ I think I may have mis read the question.
Is this actually asking me to find $f(y)$ of $exp^{(3-\frac{1}{4}y)}$ or as you stated $e^{(3-\frac{1}{4}y)}$ as opposed to $f(y) = exp(3 - \frac{1}{4}y)$ or are they both the same???
Thanks and happy new year etc
There is no such thing as $exp^x$. exp(x) is the operation $e^x$, so there is no room for confusion.
-Dan | 2016-12-09T18:10:02 | {
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https://mathhelpboards.com/threads/proof-by-contradiction.3109/ | ### Welcome to our community
#### Poirot
##### Banned
Is this how proof by condraction works?
Say we want to prove A-> B.
We prove by showing the statement 'A and not B' implies some statement C which is false (since it contradicts a known fact). Therefore, anything which implies C must itself be false, so 'A and not B' is false. I.e. A implies B.
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Yes.
Note that this is a special case of a proof by contradiction.
Not all proofs by contradiction will fit the pattern you suggest.
More generally, suppose we want to prove A, then assume A to be false, and show that this leads to a contradiction.
#### Evgeny.Makarov
##### Well-known member
MHB Math Scholar
I agree. Here is what I posted on the other forum concerning proofs by contradiction and contrapositive.
Proofs by contrapositive and by contradiction are closely related. To remind, a contrapositive of P -> Q is ~Q -> ~P. A statement and its contrapositive are equivalent, so instead of one it is possible to prove the other. In a proof by contradiction, instead of proving Q one shows ~Q -> F where F is falsehood; then Q follows.
Formally, proving P -> Q in this way involves both proof by contrapositive and proof by contradiction. Namely, one assumes P and then proves Q by contradiction. For this, one assumes ~Q and from here derives ~P (this is the contrapositive of P -> Q). Combining ~P with the first assumption P gives falsehood, so Q follows.
In practice, the names "proof by contrapositive" and "proof by contradiction" are often used interchangeably.
#### Deveno
##### Well-known member
MHB Math Scholar
the difference between the two i see is this:
proof by contrapositive uses:
~A v B = B v ~A = ~(~B) v ~A (switching the "order" of A and B).
~A v B = ~(~(~A v B)) = ~(A & ~B) (changing the "or-ness" of implication, to "and-ness").
this is most evident in "the flow" of the proofs: proofs by contrapositive seem "backwards" (we start where we don't want to be, and end where we aren't, so it's good), whereas proofs by contradiction go "the right way" from the wrong starting place.
for example, if i wish to show that every multiple of 4 is even by proving the contrapositive, i show that no odd number is divisible by 4.
if i wish to prove that every multiple of 4 is even by contradiction, i assume there is some 4k that is odd, and derive the contradiction that 1 is even. there is often some "blurring" of these distinctions, and a formal codification of either proof may wind up looking much the same (as it should!).
loosely speaking it's the (subtle) difference between:
for all.....(somethings, some statement is true)
there does not exist.....(a counter-example)
and when i'm really confused, i often forget which one i'm in the middle of. | 2021-06-15T12:57:28 | {
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https://mathoverflow.net/questions/141378/is-this-lemma-in-elementary-linear-algebra-new | # Is this lemma in elementary linear algebra new?
Is anyone familiar with the following, or anything close to it?
Lemma. Suppose $A$, $B$ are nonzero finite-dimensional vector spaces over an infinite field $k$, and $V$ a subspace of $A\otimes_k B$ such that
(1) For every nonzero $a\in A$ there exists nonzero $b\in B$ such that $a\otimes b\in V$,
and likewise,
(2) For every nonzero $b\in B$ there exists nonzero $a\in A$ such that $a\otimes b\in V$.
Then
(3) $\dim_k(V) \geq \dim_k(A) + \dim_k(B) - 1$.
Remarks: The idea of (1) and (2) is that the spaces $A$ and $B$ are minimal for "supporting" $V$; that is, if we replace $A$ or $B$ by any proper homomorphic image, and we map $A\otimes B$ in the obvious way into the new tensor product, then that map will not be one-one on $V$. The result is equivalent to saying that if one is given a finite-dimensional subspace $V$ of a tensor product $A\otimes B$ of arbitrary vector spaces, then one can replace $A$, $B$ by images whose dimensions sum to $\leq \dim(V) + 1$ without hurting $V$.
In the lemma as stated, if we take for $A$ a dual space $C^*$, and interpret $A\otimes B$ as $\mathrm{Hom}(C,B)$, then the hypothesis again means that $C$ and $B$ are minimal as spaces "supporting" $V$, now as a subspace of $\mathrm{Hom}(C,B)$; namely, that restricting to any proper subspace of $C$, or mapping onto any proper homomorphic image of $B$, will reduce the dimension of $V$.
In the statement of the lemma, where I assumed $k$ infinite, I really only need its cardinality to be at least the larger of $\dim_k A$ and $\dim_k B$.
The proof is harder than I would have thought; my write-up is 3.3K. I will be happy to show it if the result is new.
• Welcome to Mathoverflow! – user6976 Sep 5 '13 at 21:09
• I don't have time to think about this right now, but it seems strikingly familiar to the following theorem of Hopf. If $f:A \otimes B\to C$ is a linear map which is injective on each factor separately, then $\dim f(A\otimes B) \geq \dim A + \dim B - 1.$ However, this theorem is true over $\mathbb{C}$ but false over $\mathbb{R}$ (the proof is given by algebraic topology), so maybe it is only a superficial observation. – Jack Huizenga Sep 6 '13 at 0:50
• This feels like a statement from projective geometry. $\mathbb{P}(V) \subseteq \mathbb{P}(A \otimes B)$ somehow "intersects enough" $\mathbb{P}(A) \times \mathbb{P}(B)$ so that $\dim(\mathbb{P}(V)) \geq \dim(\mathbb{P}(A) \times \mathbb{P}(B))$. – Martin Brandenburg Sep 6 '13 at 1:23
• Probably also related: Flanders' theorem (§8.3 in Prasolov's Linear Algebra book www2.math.su.se/~mleites/Prasolov/prasLinAlg/pr-linAlg-main.dvi ). – darij grinberg Sep 6 '13 at 2:30
• Somehow it feels that the following "dual" result is very closely related, and could for some fields also yield your inequality (by splitting the tensor product into "separable" and "joint" spaces, whose dimensions add up to $d_Ad_B$): On the maximal dimension of a completely entangled subspace..." by K. Parathasarathy; ias.ac.in/mathsci/vol114/nov2004/Pm2342.pdf --- in particular, your subspaces have the "separable" state property, while the cited paper considers "full entangled" subspaces. – Suvrit Sep 6 '13 at 15:04
This is a nice lemma: I know a good deal of similar results but this one is unknown to me.
I believe it is suitable, as an answer, to give a proof that works with no restriction on the cardinality of the underlying field $F$. I will frame the answer in terms of matrix spaces. Thus, we have a linear subspace $V \subset M_{n,p}(F)$ such that, for every non-zero vector $X \in F^n$, the space $V$ contains a rank $1$ matrix with column space $F X$ and, for every non-zero vector $Y \in F^p$, the space $V$ contains a rank $1$ matrix with row space $F Y^t$. Note that those assumptions are unchanged in multiplying $V$ by invertible matrices (be it on the left or on the right).
The proof works by induction on $p$. The case where $p=1$ or $n=1$ is obvious. Assume now that $p>1$ and $n>1$. The discussion is split into two cases, where the standard basis of $F^p$ is denoted by $(e_1,\dots,e_p)$.
Case 1: $V e_p=F^n$. Then, one writes every matrix $M$ of $V$ as $M=\begin{bmatrix} A(M) & C(M) \end{bmatrix}$ where $A(M) \in M_{n,p-1}(F)$ and $C(M) \in F^n$. With our assumptions, we find rank $1$ matrices $M_1,\dots,M_{p-1}$ in $V$ with respective row spaces $F e_1^t,\dots,F e_{p-1}^t$. Then, $M_1,\dots,M_{p-1}$ are linearly independent and all belong to the kernel of $V \ni M \mapsto C(M)$. Using the rank theorem, one deduces that $\dim V \geq (p-1)+\dim C(V)=(p-1)+n$.
Case 2 : $V e_p \subsetneq F^n$. Multiplying $V$ on the left by a well-chosen invertible matrix, we lose no generality in assuming that $V e_p \subset F^{n-1} \times \{0\}$. In other words, every matrix $M$ of $V$ may be written as $$M=\begin{bmatrix} A(M) & C(M) \\ R(M) & 0 \end{bmatrix}$$ where $A(M)$ is an $(n-1) \times (p-1)$ matrix, $R(M)$ is a row matrix and $C(M)$ is a column matrix. Then, we note that $A(V)$ satisfies the same set of assumptions as $V$: indeed, if we take a non-zero row $L \in M_{1,p-1}(F)$, then we know that $V$ contains a rank $1$ matrix $M_1$ whose row space is spanned by $\begin{bmatrix} L & 1 \end{bmatrix}$. Obviously the last row of $M_1$ is zero whence $A(M_1)$ is non-zero and its row space is included in $F L$. One works likewise to obtain the remaining part of the condition. Thus, by induction one finds $$\dim A(V) \geq (n-1)+(p-1)-1.$$ Finally, we know that $V$ must contain a non-zero matrix $M_2$ with $A(M_2)=0$ and $C(M_2)=0$, and that it must contain a non-zero matrix $M_3$ with $A(M_3)=0$ and $R(M_3)=0$. Obviously, $M_2$ and $M_3$ are linearly independent vectors in the kernel of $V \ni M \mapsto A(M)$. Using the rank theorem, one concludes that $$\dim V \geq 2+\dim A(V) \geq 2+(n-1)+(p-1)-1=n+p-1.$$
• This is a nice proof. Although it seems to be algebraic, I think it is geometric in disguise. Perhaps someone can formulate this proof coordinate-free, and/or in terms of the the intersection of $\mathbb{P}(V) \subseteq \mathbb{P}(A \otimes B)$ with $\mathbb{P}(A) \times \mathbb{P}(B) \hookrightarrow \mathbb{P}(A \otimes B)$? – Martin Brandenburg Sep 6 '13 at 19:20
As suggested by Martin, there is a geometric interpretation of this lemma. Though the proof is probably not shorter than the one proposed by Clément. Nevertheless, this is the kind of very classical reasonning one encouters in the study of secant varieties.
Let us put $$a = dim A$$ and $$b = dim B$$. If $$a \otimes b \in A \otimes B$$, I denote its image in $$\mathbb{P}(A \otimes B)$$ by $$[a \otimes b]$$.
I denote by $$X_{A,B} = \{(a,b), \textrm{such that} [a \otimes b] \in \mathbb{P}(V) \}$$. This is clearly equal to the scheme $$(\mathbb{P}(A) \times \mathbb{P}(B)) \cap \mathbb{P}(V)$$ (I'll consider only schemes with reduced structure here).
Let us consider the natural projections $$p_A : X_{A,B} \longrightarrow \mathbb{P}(A)$$ and $$p_B : X_{A,B} \longrightarrow B$$. The hypothesis given by the OP show that $$p_A$$ and $$p_B$$ are surjective. Denote by $$\gamma_A$$ the dimension of the generic fiber of $$p_A$$, by $$\gamma_B$$ the dimension of the generic fiber of $$p_B$$, by $$X_A$$ a maximal dimensional irreducible component of the scheme $$p_A^{-1}(p_A(X_{A,B}))$$ and by $$X_B$$ a maximal dimensional irreducible component of the scheme $$p_B^{-1}(p_B(X_{A,B}))$$.
The theorem of the dimension gives $$dim \ X_A = a-1 + \gamma_A$$ and $$dim \ X_B =b-1 + \gamma_B$$.
The secant variety $$S(X_A,X_B)$$ (that is the closure of variety of lines joining a point of $$X_A$$ and a point of $$X_B$$) is included in $$\mathbb{P}(V)$$ and the goal will be to bound below its dimension to get a bound for $$dim \ \mathbb{P}(V)$$.
The dimension of $$S(X_A,X_B)$$ is equal to $$\dim \ X_A + \dim \ X_B +1 - \delta$$, where $$\delta$$ is the secant defect of $$S(X_A,X_B)$$. Concretely, if $$M$$ is a generic point of $$S(X_A,X_B)$$, then $$\delta$$ is the dimension of the scheme:
$$\{[a_1 \otimes b_1] \in X_A, \ \textrm{s.t. \exists [a_2 \otimes b_2] \in X_B and (x,y) \in \mathbb{k}^2 with M = x.a_1\otimes b_1 + y.a_2 \otimes b_2} \}.$$
It is well known that the secant defect of $$S(\mathbb{P}(A) \times \mathbb{P}(B),\mathbb{P}(A) \times \mathbb{P}(B))$$ is $$2$$. Indeed, the parameter family to decompose a rank $$2$$ matrix as a sum of two rank $$1$$ matrices is $$\mathbb{P}^1 \times \mathbb{P}^1$$. (short explanation : as one only needs to construct one of these rank $$1$$ matrices : choose the image (choice of a $$\mathbb{k}^1$$ in the image of the rank $$2$$ matrix, which is isomorphic to $$\mathbb{k}^2$$) and choose a hyperplane containing the kernel of the rank $$2$$ matrix).
Assume that $$S(X_A,X_B)$$ consists only of rank $$1$$ matrices. Since $$X_A$$ surjects onto $$A$$ and $$X_B$$ surjects onto $$B$$, we easily deduce that $$X_A = \mathbb{P}(A) \times b_0$$ and $$X_B = a_0 \times \mathbb{P}(B)$$ for some $$a_0$$ and $$b_0$$ fixed. The dimension of $$S(X_A,X_B)$$ is then obviously seen to be $$a-1+b-1-0 = a+b-2$$ and then we have $$dim \ \mathbb{P}(V) \geq a+b-2$$.
Assume that the $$S(X_A,X_B)$$ contains a matrix of rank $$2$$. Then, the generic $$M \in S(X_A,X_B)$$ has rank $$2$$. Since $$X_A \subset \mathbb{P}(A) \times \mathbb{P}(B)$$, $$X_B \subset \mathbb{P}(A) \times \mathbb{P}(B)$$ and the secant defect of $$S(\mathbb{P}(A) \times \mathbb{P}(B),\mathbb{P}(A) \times \mathbb{P}(B))$$ is $$2$$, we deduce that $$\delta \leq 2$$. As a consequence,
$$dim \ S(X_A,X_B) \geq a-1 + \gamma_A + b-1 + \gamma_B +1 - \delta \geq a+b-3.$$
If $$\delta \leq 1$$, then we get in fact:
$$dim \ S(X_A,X_B) \geq a-1 + \gamma_A + b-1 + \gamma_B-1 \geq a+b-2,$$ and this implies that $$dim \ \mathbb{P}(V) \geq a+b-2$$, which is what we wanted.
If $$\delta = 2$$, then the dimension of $$\{[a_1 \otimes b_1] \in X_A, \ \textrm{s.t. \exists [a_2 \otimes b_2] \in X_B and (x,y) \in \mathbb{k}^2 with M = x.a_1\otimes b_1 + y.a_2 \otimes b_2} \}$$ is $$2$$. In view of the explicit decomposition of a rank $$2$$ matrix as the sum of two rank $$1$$ matrices, this implies that for every $$a$$ in $$\mathbb{P}(A)$$, there is at least a $$\mathbb{P}^1$$ of $$b \in \mathbb{P}(B)$$ such that $$[a \otimes b] \in X_A$$. We deduce that $$\gamma_A \geq 1$$ and finally:
$$\dim S(X_A,X_B) \geq a-1+1 +b-1 + 1 -2 = a+b-2,$$
which again implies $$dim \ \mathbb{P}(V) \geq a+b-2$$. | 2020-10-27T12:55:08 | {
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https://www.physicsforums.com/threads/show-that-an-integer-is-unique.895228/ | # Show that an integer is unique
1. Nov 29, 2016
### Mr Davis 97
1. The problem statement, all variables and given/known data
Suppose that a and b are odd integers with a ≠ b. Show that there is a unique integer c such that
|a - c| = |b - c|
2. Relevant equations
3. The attempt at a solution
What I did was this: Using the definition of absolute value, we have that $(a - c) = \pm (b - c)$. If we choose the plus sign , then we have that $a = b$, which contradicts our original assumption. So $a - c = -b + c \implies c = \frac{a + b}{2}$. Does this solve the original problem? I have shown that there exists an integer $c$, but is this sufficient to show that this c is unique?
2. Nov 29, 2016
### Staff: Mentor
Yes. If you like, you can assume, that $c'$ is another solution. But the same argument gives you $c'=\frac{a+b}{2}=c$.
However, this isn't necessary, because to be exact, you only showed, that $c=\frac{a+b}{2}$ is necessary to be a solution and this is unique. You haven't shown, that it actually is a solution.
3. Nov 29, 2016
### Mr Davis 97
Are you saying that I have to plug it back in to show that it is actually a solution? Why doesn't the algebra itself show that it is a solution?
4. Nov 29, 2016
### Ray Vickson
WLOG, let $a < b$. Then there are three possible cases: (1) $c < a$; (2) $c > b$; or (3) $a \leq c \leq b$. It is easy to show that cases (1) and (2) are impossible. In case (3) we MUST have $c-a = b-c$, so $c = (a+b)/2$. This is true whether or not $a, b$ are integers (odd or not), and whether or not we need $c$ to be an integer. Of course, when $a,b$ are odd integers, $c$ is an integer, and that is easily shown.
5. Nov 29, 2016
### Aufbauwerk 2045
fresh_42 I am also a bit puzzled by your response. Can you elaborate?
Last edited: Nov 29, 2016
6. Nov 30, 2016
### PeroK
You can check that your steps are reversible, or you can plug the solution you got into the initial equation.
One thing you did not do was to show that $c$ is an integer.
7. Nov 30, 2016
### Staff: Mentor
Yes.
Well, in this case, it is pretty much so, for it is easy to see. But not, if we want to be very rigorous.
What @Mr Davis 97 has shown was:
If $\mathcal{\, A}_1 := \left(\, |a-c| = |b-c| \,\text{ for integer } \; a,b,c \text{ is true } \right)$ then $\mathcal{A}_2 := \left(c = \frac{a+b}{2} \text{ is true }\right)$.
This means, for statement $\mathcal{A}_1$ to be true, it is necessary, that statement $\mathcal{A}_2$ is also true.
It does not say, that statement $\mathcal{A}_2$ implies $\mathcal{A}_1$, because $\mathcal{A}_2$ could still be true, even if $\mathcal{A}_1$ wasn't.
Therefore, it remains to show, that statement $\mathcal{A}_2$ is also sufficient for statement $\mathcal{A}_1$ to hold.
(cp. @PeroK's remark above)
I admit, it can be done by simply looking at it and doing the calculations in mind, but there might be more difficult examples ahead, in which case it is not as obvious.
8. Nov 30, 2016
### Ray Vickson
Whether or not $a$ and $b$ are integers, the equation $|c-a| = |c-b|$ implies that $c = (a+b)/2$, uniquely. That is true for all real numbers; see post #4.
9. Nov 30, 2016
### Staff: Mentor
Of course. But my mistake hasn't been this, it was that I didn't formulate $\mathcal{A}_2$ in a correct way, for it should have been $\mathcal{A}_2 = (\exists c := \frac{a+b}{2} \in \mathbb{Z}\text{ is true })$. But even this wasn't really important for I spoke of the general logic of implications and the difference between a necessary and a sufficient condition. And, yes, it's pretty easy in this case.
10. Nov 30, 2016
### Mr Davis 97
So just to clarify, this is a problem of uniqueness, right? Thus, to prove that there is a unique solution, do we just have to prove $\forall a \forall b \exists c(P(a,b,c) \land \forall r(P(a,b,r) \implies r = c))$, where $P(a,b,c)$ is the predicate that $|a - c| = |b - c|$. So to show that $P(a,b,c)$ is true for some c, are you saying that I must show that $|a - c| = |b - c| \Longleftrightarrow c = \frac{a+b}{2}$ is true? Would it just be sufficient to show that $c = \frac{a+b}{2} \implies |a - c| = |b - c|$?
11. Nov 30, 2016
### Staff: Mentor
No, and it is a bit of an overkill here. Formally $\forall a \forall b \exists c(P(a,b,c)$ is your goal. You showed $P(a,b,c) \Longrightarrow c=f(a,b)$. And as @Ray Vickson pointed out, the uniqueness of expression and the way of deduction already prevents the existence of another solution, or formally: You have shown $P(a,b,c) \Longrightarrow c=f(a,b)$, so any $c'$ with $P(a,b,c')$ also implies $c'=f(a,b)=c$.
The missing step is $P(a,b,f(a,b))$. Does $f(a,b)$ actually fulfill the predicate?
Yes, this would be sufficient. You don't actually need equivalence, only that $c = \frac{a+b}{2}$ is actually a solution (and that it is an integer!).
Let me illustrate the principle with another example:
$\{ x \cdot (x-2) = 0 \Longrightarrow \text{ case 1: } x=2n \ldots \text{ case 2: } x=2n+1 \ldots \Longrightarrow x \text{ is even } \}$ is a true statement, but it doesn't guarantee, that any even number solves the equation.
Of course one would write $x \cdot (x-2) = 0 \Longrightarrow x \in \{0,2\}$ and nobody would ask to check, whether $0 \cdot (0-2) = 0$ and $2\cdot (2-2)=0$ because everybody sees it. That's similar to your example: one can see it.
But imagine you solved a system of differential equations and found a function $f$ as a result. In such a case, one should make sure, that the derived function actually solves the problem. To say it over the top: $x = 0 \Rightarrow 1 \in \mathbb{R}$ is always true, but doesn't solve the equation for $x$. However, as I said, this is only a principle issue and can often be done by simply mentioning it, because it's obvious or easy done.
Last edited: Nov 30, 2016 | 2017-11-22T15:23:46 | {
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https://cs.stackexchange.com/questions/112282/recurrence-equation-upper-limit-problem | # Recurrence Equation upper limit problem
I was looking at my teacher's notes and came about the following recurrence equation :
$$T(n) = \begin{cases} 1 &\quad\text{if } n\leq 1\\ 4T\left(\frac{n}{2}\right) + n^3 &\quad\text{if } n\gt1 \\ \end{cases}$$
In order to solve it I proceeded as follows :
$$T(n) = n^3 + 4T\left(\frac{n}{2}\right) = n^3 + 4\big( 4T\left(\frac{n}{4}\right) + \left(\frac{n}{2}\right)^3 \big) = \\ n^3 + 4\left(\frac{n}{2}\right)^3 + 16T(\frac{n}{4}) = \cdots$$
I'll spare some latex and state that we can see from unwrapping the equation that at the generic level $$i$$ the incurred price is :
$$T_i = 4^i\left(\frac{n}{2^i}\right)^3$$
In order to compute the overall price we can sum all the prices incurred at all levels, obtaining :
$$T(n) = \sum_{i=0}^{log^n -1}{\big(4^i\left(\frac{n}{2^i}\right)^3\big)} + 4^{log^n}T(1) = \\ n^3 \sum_{i=0}^{log^n-1}{\big(2^{2i}\frac{1}{2^{3i}}\big)} + n^2 = \\ n^3 \sum_{i=0}^{log^n-1}{\big(\frac{1}{2^{i}}\big)} + n^2$$ My problem begins here. In order to solve it I'd say that the series in questions can be solved as :
$${\displaystyle \sum _{k=m}^{n}x^{k}={\frac {x^{m}-x^{n+1}}{1-x}}\quad {\text{with }}x\neq 1.}$$
which applied to my scenario would yield
$$n^3 \left(\frac{1 - (\frac{1}{2})^{log^n}}{1 - \frac{1}{2}}\right) + n^2 = \\ 2n^3 - 2n^2 + n^2 = 2n^3 - n^2 = \Theta(n^3)$$
But my professor says :
$$n^3 \sum_{i=0}^{log^n-1}{\big(\frac{1}{2^{i}}\big)} + n^2 \leq n^3 \sum_{i=0}^{\infty}{\big(\frac{1}{2^{i}}\big)} + n^2 = \\ n^3 \frac{1}{1 - \frac{1}{2}} + n^2 = 2n^3 + n^2$$
And thus $$T(n) = O(n^3)$$ instead of $$\Theta(n^3)$$ since we proved only an upper limit.
My question is thus, why can't I solve the summation as I did instead of extending it to infinity?
• Your professor is wrong. Direct them to the proof of the master theorem. – Yuval Filmus Jul 29 '19 at 20:46
• @YuvalFilmus well it's not technically wrong to say that it is $O(n^3)$ since the equation is correct. Yet I feel we can prove that it's $\Theta(n^3)$ as well. What would the master theorem suggest? – MFranc Jul 29 '19 at 20:52
• @LucaGrignani The master theorem will render the same asymptotic bound as you have obtained, $\Theta(n^3)$. – John L. Jul 29 '19 at 21:05
• @LucaGrignani if you prove it is $O(n^3)$ then it's trivial to show it is $\Theta(n^3)$ because there is $n^3$ "work" done at the highest level of recurrence: $T(n) = 4T(n/2) + n^3 \geq n^3$. – ryan Jul 30 '19 at 0:39
Yes, you could solve the summation in exact formula. Yes, you could obtain $$T(n)=\Theta(n^3)$$, a tighter asymptotic bound. Your professor would and should not forbid you to do that.
The message from your professor could be that an upper bound such as $$O(n^3)$$ is as good as $$\Theta(n^3)$$ more often than not. For example, once we know an algorithm runs in $$O(n^3)$$ time without a huge constant multiplier on a problem of input size $$n$$ less than 1000, then we can be comfortable starting implementing the algorithm, without worrying much that the code may take hours to run. A tighter asymptotic bound such as $$\Theta(n^3)$$ will not help us move ahead significantly further. On the other hand, such a tighter asymptotic usually comes with a heavier price, which is not much heavier in this particular problem, though. | 2020-02-17T23:32:39 | {
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https://math.stackexchange.com/questions/2831864/q-proving-existence | # Q: Proving Existence
I'm currently stuck on a problem right now for my Intro to Proofs Class. The problem says:
Let $a,b ∈ ℕ$. Prove that if $a+b$ is even, then there exists nonnegative integers $x$ and $y$ such that $x^2-y^2= ab$.
So far I've tried it directly, and by contrapositive and came to a similar road block.
Direct:
Assume $a,b ∈ ℕ$, and that $a+b$ is even.
$a+b$ being even $\implies$ $a+b=k_1$, such that $k_1 ∈ \mathbb{Z}$.
$a+b=k_1$ $\implies$ $a=k_1-b$.
Multiplying both sides of $a$ by $b$ we get: $ab= b(k_1-b)=bk_1-b^2$.
observing the conclusion $x^2-y^2=(x+y)(x-y)=ab$
It's at this point I'm at a road block. I'm not sure if direct is the way to go. I've also tried contrapositive, and contradiction, but i've also hit a roadblock for both of those as well.
• you need to use the definition of even: a + b is even, therefore a and b are both either even or odd. a = 2k, b = 2m and a=2k+1, b=2m+1. – user29418 Jun 25 '18 at 21:27
• $3+7$ is even but $21$ is not the sum of two squares. (similarly $15=3\times 5$ is not the sum of two squares). – lulu Jun 25 '18 at 21:27
• Alternative theory: perhaps you mistyped and you were asked to prove that $ab$ could be written as the difference of two squares instead of the sum? – lulu Jun 25 '18 at 21:58
• @lulu yes! i mistyped. It's supposed to be the difference of two squares. – Mister_J Jun 25 '18 at 21:59
• Ah, ok. Please edit your post to reflect that. For that problem, Hint: suppose that $a≥b$ and try to solve $x+y=a, x-y=b$ in integers. – lulu Jun 25 '18 at 22:00
If $a+b$ is even, you can write $a+b=2x$, with $x$ a natural number. Then, assuming $a\ge b$, which is not restrictive, you have $$a-b=a+b-2b=2x-2b=2(x-b)$$ so also $a-b$ is even; set $x-b=y$, so that $$a+b=2x \qquad a-b=2y$$ Summing these up we obtain $2a=2x+2y$, hence $a=x+y$.
Can you finish?
• Thank you @egreg! I just have a question on the part where we let x-b=y. Are we saying that y is an arbitrary natural number since we're trying to prove that there exist nonnegative integers x and y? – Mister_J Jun 25 '18 at 22:36
• @Mister_J It's not arbitrary: it is $x-b$ and is obviously a natural number. – egreg Jun 25 '18 at 22:38
• @egreg How is it obvious? – Graham Kemp Jun 25 '18 at 23:40
• It's "obvious" because $2x = a+ b \ge b+b = 2b$ so $x \ge b$ and as $a+b$ is even $x = \frac {a+b}2$ is an integer. So $x - b \ge b-b =0$. It's only not natural if $a = b = x$ and $x - b = 0$. – fleablood Jun 26 '18 at 0:08
If $a+b=2k$ for some $k\in\mathbb{Z}_{>0}$, then $a$ and $b$ are both either even or both odd. Say both even with $a=2m$, $b=2n$, for some $m$, $n\in\mathbb{Z}_{\ge0}$. Then we require $$x^2-y^2=(x-y)(x+y)= ab=4mn$$ If we let $2m=x-y$ and $2n=x+y$, then solving this gives $x=m+n$ and $y=n-m$.
Say both odd with $a=2r+1$, $b=2s+1$, for some $r$, $s\in\mathbb{Z}_{\ge0}$. Then we require $$x^2-y^2=(x-y)(x+y)= ab=(2r+1)(2s+1)=2(2rs+(r+s))+1$$ If we let $2r+1=x-y$ and $2s+1=x+y$, then solving this gives $x=r+s+1$ and $y=s-r$.
Well, brainstorm first.
$x^2 - y^2 = ab$
$(x - y)(x + y) = ab$. Can I say $x - y = a$ and $x+y = b$? Why or why not? Well, first off $x - y$ would have to equal the smaller of $a$ or $b$ and $x + y$ would have to be the larger. But we can assume without loss of generality that $a \le b$. So can we say that $x, y$ exist where $x-y = a$ or $x + y = b$?
That would mean $x$ is the midpoint of $a, b$ or that $x = \frac {a + b}2$. And that $y =$ the distance from the midpoint to either of the extreme $a$ or $b$ $= |\frac {a+b}2 - a| =|b - \frac {a+b}2|=|\frac {b-a}2|$.
And that is possible because $a+b$ is even!
So:
Pf: Let $x = \frac {a+b}2; y =|\frac {b-a}2|=$
Then ... (well, let's just push the throttle down and see what happens... we know it must work so ... let's go with it....)
then $x^2 - y^2 = (\frac {a+b}2)^2 - (\frac {b-a}2)^2 =$
$\frac {a^2 + 2ab + b^2}4 - (\frac {b^2 -2ab + a^2}4)=$
$\frac {4ab}4 = ab$.
======
Note: we did get lucky. There are probably other answers where $x-y \ne \min (a,b)$ and $x+y \ne \max (a,b)$ but those worked.
===
Alternatively... once we see that $x-y$ and $x + y$ can be extreme points of a segment we can say: Let $m = x+y$ and $d= x-y$. Now $d= m -2y$ so $m$ and $d$ are both even or both odd.
So we can solve for $m,d$ and set $x = \frac {m+d}2$ and $y = \frac {m-d}2$.
$x^2 - y^2 = md = ab$. So $m,d$ can be any complimentary factors of $ab$ BUT with the only stipulation that $m \ge d$ and that $m$ and $d$ are both the same parity.
As $a + b$ is even $a$ and $b$ are the same parity two such factors will always exists.
(We can always chose $m = \max (a,b)$ and $d = \min (a,b)$ if we want. But if we want another solution and if $a$ nor $b$ are primes. We can choose any complimentary factors provided they are both even or both odd.) | 2019-12-11T00:42:13 | {
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http://math.stackexchange.com/questions/214071/is-the-identity-map-a-diffeomorphism | # Is the identity map a diffeomorphism?
Unfortunately, googling this question leads to conflicting answers. According to this source, the identity map on any smooth manifold is a diffeomorphism, but it's not according to this. I appreciate it if someone gave a definitive answer.
-
The statement that the questioner is referring to begins at the bottom of page 38, and continues onto page 39, ending with "Thus the identity map is not a diffeomorphism." – MJD Oct 15 '12 at 4:08
@Smith: the second link you found refers to a map which is the identity on underlying sets, but uses a different smooth structure in the source and the target. Personally, I think this is a bad abuse of notation. It does not refer to the identity map from a smooth manifold to itself considering only a single smooth structure, which is a diffeomorphism. – Qiaochu Yuan Oct 15 '12 at 4:55
## 3 Answers
Let us fix a smooth manifold $M$.
• is the identity map $i:M\to M$ smooth?
• is it bijective?
• what is the inverse function?
• is the inverse function bijective?
Can you answer these questions?
As for your reference: the book does not say that the identity map of a smooth manifold is not a diffeo: it gives an example to show that if $M$ and $M'$ and two smooth manifolds on the same topological space, then the identity function $M\to M'$ is not necessarily smooth. This is a claim rather rather different to «the identity map of a smooth manifold is not smooth».
-
You are free to provide more pertinent answers, MJD. – Mariano Suárez-Alvarez Oct 15 '12 at 4:20
I didn't know the answer. I would have tried to provide it if I had known. – MJD Oct 15 '12 at 4:22
Yes, the identity map is a diffeomorphism, and the derivative at any point $p$ is just the identity on $T_pM$. Maybe it is best to see this in terms of directional derivatives. Write $I$ for the identity map. Fix a curve $\phi(t): \mathbb{R} \to M$ with $\phi'(t) = X$ for some $X \in T_pM$ and then compose with the identity map. Then $D_X I = (I\circ \phi)'(t) = \phi'(t) = X$.
Regarding your second reference, the author there is giving an example of two $C^{\infty}$ structures on $\mathbb{R}$ that are different. The issue you are having is that the identity map'' there really takes $\mathbb{R}$ with one smooth structure to $\mathbb{R}$ with another smooth structure. So it doesn't have to be smooth!
But if you consider the identity map on a manifold $M$ (with fixed smooth structure- if someone utters the words "smooth manifold" then they mean a topological manifold together with a smooth structure so that is embedded in the definition) then the identity map is always a diffeomorphism.
-
I think it does not necessarily a diffeomorphism if you think identity map as a map from a set to itself. However if you consider identity map from a smooth manifold to itself (i.e to same set with same smooth structure) then it is a diffeomorphism. Here the point is this, differentiability related with your smooth structure and you can put different smooth structures on the same set.
- | 2016-06-30T16:22:18 | {
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http://library.kiwix.org/mathoverflow.net_en_all_2019-02/A/question/60375.html | Is $\mathbb R^3$ the square of some topological space?
177
90
The other day, I was idly considering when a topological space has a square root. That is, what spaces are homeomorphic to $X \times X$ for some space $X$. $\mathbb{R}$ is not such a space: If $X \times X$ were homeomorphic to $\mathbb{R}$, then $X$ would be path connected. But then $X \times X$ minus a point would also be path connected. But $\mathbb{R}$ minus a point is not path connected.
A next natural space to consider is $\mathbb{R}^3$. My intuition is that $\mathbb{R}^3$ also doesn't have a square root. And I'm guessing there's a nice algebraic topology proof. But that's not technology I'm much practiced with. And I don't trust my intuition too much for questions like this.
So, is there a space $X$ so that $X \times X$ is homeomorphic to $\mathbb{R}^3$?
10I'm wondering to what extent there is unique factorization of topological spaces relative to $\times$. $\mathbb{Q}$ is an idempotent (as is its complement in $\mathbb{R}$), but are there more interesting failures of UF involving connected spaces? Or results establishing UF for "nice" families of spaces? Should these be posted as a new question? – Yaakov Baruch – 2011-04-03T01:41:34.417
3Is Moebius $\times$ Moebius = cilinder $\times$ cilinder (no boundaries)? – Yaakov Baruch – 2011-04-04T16:38:18.597
Without knowing any algebraic topology, it's possible to conclude at least something about X. If X is metric, compact, or locally compact and paracompact, then $\dim(X\times X)\le 2\dim X$, which means X has to have Lebesgue covering dimension at least 2. Wage, Proc. Natl. Acad. Sci. USA 75 (1978) 4671 , www.pnas.org/content/75/10/4671.full.pdf . What is the weakest condition that guarantees $\dim(X\times Y)= \dim X+\dim Y$? Given Yaakov Baruch's comment about the "dogbone space," it's not obvious that X is at all well behaved simply from the requirement that its square is $\mathbb{R}^3$. – Ben Crowell – 2013-01-19T15:55:50.700
@YaakovBaruch, isn't the cylinder factorizable? And could you elaborate this identity a little? – Ash GX – 2013-10-17T06:35:37.190
9Dear @BigM, I fail to see the value in editing an old question simply to add mathjax to its title which was perfectly readable to begin with. – Ricardo Andrade – 2015-05-31T11:44:50.003
1@Ricardo Andrade: every little improvement should be (mildly) welcome, don't you think so? – Qfwfq – 2018-08-03T08:23:23.150
199
No such space exists. Even better, let's generalize your proof by converting information about path components into homology groups.
For an open inclusion of spaces $X \setminus \{x\} \subset X$ and a field $k$, we have isomorphisms (the relative Kunneth formula) $$H_n(X \times X, X \times X \setminus \{(x,x)\}; k) \cong \bigoplus_{p+q=n} H_p(X,X \setminus \{x\};k) \otimes_k H_q(X, X \setminus \{x\};k).$$ If the product is $\mathbb{R}^3$, then the left-hand side is $k$ in degree 3 and zero otherwise, so something on the right-hand side must be nontrivial. However, if $H_p(X, X \setminus \{x\};k)$ were nontrivial in degree $n$, then the left-hand side must be nontrivial in degree $2n$.
61I hope this fine illustration of the power of relative homology will find its way in a textbook or, meanwhile, in algebraic topology courses. – Georges Elencwajg – 2011-04-02T19:40:42.633
3I have a question regarding the top answer given by Tyler Lawson. As far as I know you can only apply the relative version of the Kunneth formula to cofibrations. Since we do not know much about $X$, it is unclear why $(X, X\setminus p)$ is a cofibration. Moreover, $(\mathbb R^3, \mathbb R^3\setminus p)$ is not a cofibration (I think). – freddy – 2017-03-21T14:21:45.233
@freddy, you are correct that (e.g.) the version that appears in Thm 3.18 in Hatcher's book does not apply to this example; Hatcher requires CW-pairs in his version of the proof. This is not the case for all versions of the theorem. – Tyler Lawson – 2017-03-21T18:58:19.380
3For example, Dold's version (Corollary 12.10 in Lectures on Algebraic Topology part VI) requires an excisive triad condition. The core of these assumotions is to ensure that, given $(X,A)$ and $(Y,B)$, the covering of $(X \times B) \cup (Y \times A)$ by $X \times B$ and $A \times Y$ is good enough to satisfy the assumptions of the Mayer-Vietoris theorem. This is, in particular, satisfied if $A$ is an open subset of $X$ and $B$ is an open subset of $Y$, or in the CW-inclusion version that Hatcher uses. – Tyler Lawson – 2017-03-21T19:01:55.600
(A similar version appears on page 249 of Spanier's book. I found both of these references here: http://mathoverflow.net/questions/230421/a-k%C3%BCnneth-theorem-version-for-relative-singular-cohomology)
– Tyler Lawson – 2017-03-21T19:10:34.553
1So this also works for $\sqrt{\mathbb{R}^{2n+1}}$ doesn't it? – Pietro Majer – 2018-08-03T08:20:08.507
1@PietroMajer Indeed it does. – Tyler Lawson – 2018-08-04T20:21:39.087
44
this blog post refers to some papers with proofs. I've heard Robert Fokkink explain his proof and there he also told us the cohomological proof, which generalizes it to all Euclidean spaces of odd dimension.
15I hope no one misses this nice alternative proof because it's behind a link. – Richard Dore – 2011-04-04T02:24:49.857
3Quoting from the link: "The paper also refers to an earlier paper ("The cartesian product of a certain nonmanifold and a line is E4", R.H. Bing, Annals of Mathematics series 2 vol 70 1959 pp. 399–412) which constructs an extremely pathological space B, called the "dogbone space", not even a manifold, which nevertheless has B × R^3 = R4." This is relevant to my comment to the OP. – Yaakov Baruch – 2011-04-04T05:16:12.043
I don't understand this step in the proof: Why does the map $X^4 \to X^4, (a,b,c,d) \mapsto (c,d,a,b)$ correspond to the map $R^6 \to R^6, (p,q,r,s,t,u) \mapsto (s,t,u,p,q,r)$? I mean, the homeomorphism is not supposed to commute with projections ... – Martin Brandenburg – 2011-04-04T15:05:49.093
5@Martin: The homeomorphism $(X\times X)\times (X\times X)\cong \mathbb R^3 \times \mathbb R^3$ respects projections by construction, so swapping the "two factors" (which I've emphasized with parentheses) on the left hand side corresponds to swapping the two factors on the right hand side. – Anton Geraschenko – 2011-04-05T05:42:08.323
15
I didn't know that, but I did know this: we cannot have $S^2 = S\times S$ for any topological space $S$.
3Would you care to elaborate? – Ian Agol – 2013-01-19T05:09:55.853
68All things considered, perhaps "S" is not the best name for the topological space for this assertion. – Terry Tao – 2013-01-19T05:52:07.463
22@Terry Tao True enough, but in all honesty it's precisely the notational perversity that brought this to mind to begin with. – Adam Epstein – 2013-01-19T10:28:42.053
9@Agol Fix $s\in S$. On the one hand, $\pi_2(S\times S,(s,s))\cong \pi_2(S,s)\times\pi_2(S,s)$. On the other hand, $\pi_2({\bf S},{\bf s})\cong{\mathbb Z}$ for any 2-sphere $\bf S$ and any ${\bf s}\in{\bf S}$. Now it suffices to observe that ${\mathbb Z}\not\cong G\times G$ for any group $G$: indeed, such a group must be an infinite quotient of $\mathbb Z$, whence $G\cong{\mathbb Z}$, but ${\mathbb Z}\not\cong{\mathbb Z}\times{\mathbb Z}$ – Adam Epstein – 2013-01-19T11:27:02.840
2
As it happens, this started out as a wry comment about a different post, namely http://mathoverflow.net/questions/115799/…. Then I noticed this question and accidentally posted as an answer what was intended as a mere comment.
8@IanAgol : On the LHS, $S^2$ refers to the $2$-sphere, while on the LHS $S$ refers to an arbitrary topological space. – Prateek – 2014-11-17T15:55:38.770 | 2019-04-21T22:13:40 | {
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https://math.stackexchange.com/questions/690632/calculator-similar-to-desmos-but-for-3d/2235881 | # Calculator similar to Desmos but for $3$D
Is there a calculator with functionality similar to Desmos but in $$3$$ dimensions? I am looking to learn about families of quadric surfaces so I am looking for a $$3$$D calculator with sliders.
Updated, December 2018: I made the following website with the aim of producing a Desmos-like experience in 3D for my multivariable calculus students.
math3d.org
You can create and animate points, vectors, curves, surfaces (explicit & implicit), and vector fields. After creating a demonstration, you can save it and share.
Here are three scenes that I particularly like:
This project is on Github. If you find bugs or have ideas for improvements, please open an issue!
• This is absolutely awesome. I'm taking multivariable calculus this semester, discovered MathBox.js, and was wanting to make something like this, but haven't had time (lots of homework) to build anything this sophisticated. So, thanks! – tommytwoeyes Sep 22 '17 at 20:07
• This is awesome. Exactly what i want! I'm using it to visualize my example in textbook. Thank you! – Postal Model Feb 21 '18 at 3:41
• Thank you thank you thank you! – Ovi Nov 10 '18 at 3:19
• @Ovi Glad you like it! See updated post. – Chris Chudzicki Nov 10 '18 at 12:32
• @ChrisChudzicki Very nice! Just had time to check it out. One small thing though; when graphing an explicit surface, I tried to change the domain for $x, y$ and it didn't seem to work. – Ovi Nov 25 '18 at 4:24
GeoGebra does exactly what you want:
https://www.geogebra.org/3d
https://www.geogebra.org/3d?command=a=1;z=x%5E2-a*y%5E2#3d
You can actually just use Desmos!
https://www.desmos.com/calculator/euncwbhqlb
If you are on a Mac, you can try the included application called Grapher:
1. Open Grapher by opening Spotlight (Command+Space) and typing 'Grapher' (and hitting enter)
2. Click '3D Graph' and hit the Choose button
3. Your text cursor should be to the right of 'z=' inside a text box; type any 3D function including the parameter 'a', such as sin(a*x), and hit enter
4. Click the plus button at the bottom left hand corner of the window, and click New Equation
5. Erase the 'z=', type 'a=2', and hit enter
6. Open the Equation menu and select Animate Parameter
7. Now drag the slider to change the parameter!
8. (optional) If you want to have the slider act continuously, click the right most 'Settings' button above the slider (with the two checkboxes and lines), and check the box labeled Continuous Range
Enjoy!
http://www.graphycalc.com/ is an excellent site. try it out.
You might try http://web.monroecc.edu/manila/webfiles/calcNSF/JavaCode/CalcPlot3D.htm. It has sliders available under the Parameters->Adjust Parameters menu option. | 2019-12-11T22:33:18 | {
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http://mathhelpforum.com/number-theory/131678-nonsense-proof-i-am-afraid-print.html | # Nonsense proof I am afraid
• Mar 2nd 2010, 01:48 PM
novice
Nonsense proof I am afraid
Prove that the sum of the squares of two odd integers cannot be a perfect square.
The author wrote the follow proof:
Assume, to the contrary, that there exist odd integers $x$ and $y$such that $x^2 + y^2 = z^2$, where $z \in \mathbb{Z}$. Then $x = 2a+1$ and $y = 2b+1$, where $a,b \in \mathbb{Z}$. Thus
$x^2+y^2 = (2a+1)^2+(2b+1)^2$
$=4a^2+4a+1+4b^2+4b+1$
$=4(a^2+1+b^2+b)+2=2[2(a^2+a+b^2+b)+1]=2s$
where $s = 2(a^2+a+b^2+b)+1$ is odd integer. If $z$ is even, then $z=2c$ for some integer $c$ and so $z^2 = 2(2c)$, where $2c^2$ is an even integers; while $z$ is odd, than $z^2$ is odd. Produce a contradiction in each case.
Remark: I just don't see how z can be anything else be even. To me it's obvious $z^2=2s$ explicitly implied that $z^2$ is even, which by some theorem, $z$ is also even.
Question: Which part of the proof gave the author the reason to believe that it's odd? Further how in the case of it being even be a contradiction to $z^2$ being a perfect square?
I just can't see it. Could someone please show me the author's intent?
• Mar 2nd 2010, 04:41 PM
Bruno J.
The proof is convoluted because the author did not use modular arithmetic. Here's my translation of the proof using modular arithmetic.
Suppose the sum of two odd squares $x^2,y^2$ is a square $z^2$; then it has to be a square of an even integer, so $z^2=(2s)^2=4s\equiv 0 \mod 4$. Now note that any odd square is $\equiv 1 \mod 4$. (Expand $(2a+1)^2$). Therefore $x^2+y^2 \equiv 1+1 \equiv 2 \mod 4$ which contradicts $z^2\equiv 0 \mod 4$.
• Mar 2nd 2010, 05:16 PM
Soroban
Hello, novice!
What a convoluted proof . . .
This is a slight variation of Bruno J's solution.
Quote:
Prove that the sum of the squares of two odd integers cannot be a perfect square.
Here's the proof I was shown many years ago.
We note that: . $(2n)^2 \:=\:4n^2$
. . The square of an even number is a multiple of 4.
And that: . $(2n+1)^2 \:=\:4n^2 + 4n +1 \:=\:4(n^2+n)+1$
. . The square of an odd number is one more than a mutliple of 4.
Hence, all squares are either a multiple of 4 or one more than a multiple of 4.
Consider two odd numbers: . $2a+1\,\text{ and }\,2b+1$
The sum of their squares is: . $S \;=\;(2a-1)^2 + (2b-1)^2$
. . $S \;=\;4a^2 - 4a + 1 + 4b^2 - 4b + 1 \;=\;4(a^2-a+b^2-b) + 2$
The sum is two more than a multiple of 4; it cannot be a square.
• Mar 2nd 2010, 05:30 PM
novice
Oh, gentlemen,
Thank you for coming to my rescue. I am so happy to learn that it's not a failure on my part to understand the proof.
Thank you both for showing me the good stuff. | 2017-03-27T07:58:02 | {
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https://math.stackexchange.com/questions/868578/do-differentiable-functions-preserve-measure-zero-sets-measurable-sets | # Do differentiable functions preserve measure zero sets? Measurable sets?
Consider Lebesgue measure on $\mathbb{R}$ and let $f:\mathbb{R}\to\mathbb{R}$ be differentiable. Does $f$ necessarily preserve measure zero sets? Does $f$ necessarily preserve measurable sets?
Note that if $f$ is $C^1$ then $f$ preserves measure zero sets since $C^1$ functions are locally Lipschitz. Therefore $C^1$ functions also preserve measurable sets since a measurable set is the union of an $F_\sigma$ set and an measure zero set, and continuous functions preserve $F_\sigma$ sets. More generally if $f$ is absolutely continuous on each interval then $f$ preserves both measure zero sets and measurable sets.
However, I'm not sure about the differentiable case. I would guess that the answer to both questions is no. I'm interested in a counter example or proof in each case.
If $f$ preserves null-sets, it also preserves Lebesgue-measurability (but not necessarily Borel measurability), because by regularity, every Lebesgue measurable set $M$ can be written as
$$M = N \cup \bigcup K_n$$
with $K_n$ compact and $N$ a null-set. By continuity, $f$ preserves compact sets.
Now a theorem in Rudin, Real and Complex Analysis (Lemma 7.25) shows in particular that every everywhere differentiable function maps null-sets to null-sets, so that your claim is true.
• Ah sorry about that, you are right. – Chris Janjigian Jul 16 '14 at 16:29
• Never mind, I should not have posted such a harsh reply to your comment :) – PhoemueX Jul 16 '14 at 20:08
The following claim implies your result.
Claim: Let $E \subseteq \mathbb{R}$ be arbitrary. Suppose $|f'(x)| \leq M$ for all $x \in E$. Then $\mu^{\star}(f[E]) \leq M\mu^{\star}(E)$.
Proof: Fix $\epsilon > 0$. Get an open set $U \supseteq E$ with $\mu(U) < \mu^{\star}(E) + \epsilon$. For each $x \in E$, let $\delta_x > 0$ be such that $(x - \delta_x, x + \delta_x) \subseteq U$ and $|f(y) - f(x)| \leq (M + \epsilon)|y - x|$ for every $y \in (x - \delta_x, x+ \delta_x)$. Consider the family of closed intervals: $V = \{[f(x), f(y)], [f(y), f(x)] : x \in E, |y - x| < \delta_x\}$ - This is a Vitali covering of $f[E]$ which means that every point in $f[E]$ is covered by arbitrarily small intervals in $V$. Using Vitali covering theorem, get a countable subfamily $C$ of pairwise disjoint intervals which covers all but a null part of $f[E]$. The (disjoint) union of the intervals in $C$ has measure at most $(M + \epsilon) \mu(U)$ which is less than $(M + \epsilon)(\mu^{\star}(E) + \epsilon)$. Now, let $\epsilon$ go to zero. | 2019-08-19T01:11:42 | {
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https://math.stackexchange.com/questions/2095390/for-xy-1-show-that-x4y4-ge-frac18/2095402 | # For $x+y=1$, show that $x^4+y^4\ge \frac{1}{8}$
As in the title. Let $x,y$ be two real numbers such that $x+y=1$. Prove that $x^4+y^4\ge \frac{1}{8}$.
Any hints? Basically, the only method I am aware of is plugging $y=1-x$ into the inequality and investigating the extrema of the function, but I don't think it's the best method. I'm looking for a cleverer way to prove that inequality.
• I don't see what's tedious about finding the extrema. How can $x^3=(1-x)^3$? Remember that $f(x)=x^3$ is a one-to-one function. – Ted Shifrin Jan 12 '17 at 21:46
• Well, I reckon used the wrong word. I just don't think it's the best method to prove such inequality. – VanDerWarden Jan 12 '17 at 21:47
• Sometimes, such inequalities have for example clear geometric interpretation. – VanDerWarden Jan 12 '17 at 21:48
• Extrema of symmetric functions occur symmetrically, but not always when $x=y$. But for such simple functions, it's always when $x=y$. You can find a "simpler" argument for $x^2+y^2$ and then just substitute. – Ted Shifrin Jan 12 '17 at 21:50
It is equivalent to showing that $(1/2+\epsilon)^4+(1/2-\epsilon)^4\geq 1/8$. But this simplifies to $\tfrac{1}{8}(16\epsilon^4+12\epsilon^2+1)$, which obviously is minimized at $\epsilon=0$, giving $\tfrac{1}{8}$.
Edit: This is motivated by guessing the minimum, i.e. $x=y=1/2$, and then symmetrizing. The symmetry allows a nice expansion using the binomial theorem.
Also a Cauchy-Schwarz approach works: \begin{align*} 1 = x + y \le \sqrt{(x^2+y^2)(1^2+1^2)} \implies x^2 + y^2 \ge \frac{1}{2} \\ \frac{1}{2} \le x^2 + y^2 \le \sqrt{(x^4 + y^4)(1^2 + 1^2)} \implies x^4 + y^4 \ge \frac{1}{8} \end{align*}
If $a,b\in \mathbb{R}$, then $(a+b)^2\leq 2(a^2+b^2)$ since $2ab\leq a^2+b^2$.
Applying this with $x$ and $y$ we get $$1=(x+y)^2\leq 2(x^2+y^2)$$ so $x^2+y^2\geq \frac{1}{2}$, and then applying the inequality again with $x^2$ and $y^2$ we get $$\frac{1}{4}\leq (x^2+y^2)^2\leq 2(x^4+y^4)$$ which is the desired result.
$x^4$ and $y^4$ are convex functions of $x$ and $y$, so their sum is also convex. The restriction of this to the line $x+y=1$ is again convex. By symmetry, the minimum must occur at the point where $x=y$.
hint: I posted a solution to this same question a year ago, and I think you can search it here. You can use the inequality $\dfrac{a^2+b^2}{2} \ge \left(\dfrac{a+b}{2}\right)^2$ twice.
If you are looking for a more geometrical explanation, consider the family of curves $$x^4+y^4=k$$
These are a family of concentric rounded square shaped curves with order 4 symmetry centred at the origin. The line $y=x$ is a line of symmetry. We need to find the value of $k$ for which such a curve is tangent to the line $x+y=1$.
By symmetry $x=y=\frac 12$ and hence $$k=\left(\frac 12\right)^2+\left(\frac 12\right)^2=\frac 18$$
$$(x+y)^4=1 \\ x^4+4x^3y+6x^2y^2+4x^3+y^4 =1$$
Now, by AM-GM we have $$x^3y \leq \frac{x^4+x^4+x^4+y^4}{4}\\ x^2y^2 \leq \frac{x^4+y^4}{2}\\ xy^3 \leq \frac{x^4+y^4+y^4+y^4}{4}\\$$
Thus $$1=x^4+4x^3y+6x^2y^2+4x^3+y^4 \leq x^3+(3x^4+y^4)+3(x^4+y^4)+(x^4+3y^4)+y^4$$
If $x\le0$, then $y\ge1$, so $x^4+y^4\ge1$. Hence we can assume $x,y>0$.
For $x,y>0$, $x\ne y$, the function $$\mu(x,y;t)=\begin{cases} \left(\dfrac{x^t+y^t}{2}\right)^{1/t} & \text{if t\ne0} \\[6px] \sqrt{xy\vphantom{X}} & \text{if t=0} \end{cases}$$ is continuous and increasing in the variable $t$. Thus $$\mu(x,y;1)<\mu(x,y;4)$$ that is $$\frac{x+y}{2}<\left(\dfrac{x^4+y^4}{2}\right)^{1/4}$$ and, if $x+y=1$, $$\frac{1}{16}<\frac{x^4+y^4}{2}$$
For $x=y$, the inequality is obvious (and is an equality, actually).
The proof that $\mu(x,y;t)$ is increasing (for $x\ne y$) is an application of convexity. Suppose $0<p<q$; we want to prove that $$\left(\dfrac{x^p+y^p}{2}\right)^{1/p}< \left(\dfrac{x^q+y^q}{2}\right)^{1/q}$$ that is, $$\left(\dfrac{x^p+y^p}{2}\right)^{q/p}<\dfrac{x^q+y^q}{2}$$ Set $u=x^p$ and $v=y^p$; then the inequality becomes $$\left(\dfrac{u+v}{2}\right)^{q/p}<\dfrac{u^{q/p}+v^{q/p}}{2}$$ which is a consequence of $z\mapsto z^{p/q}$ being convex (on $(0,\infty)$).
Since it's immediate that $\mu(x,y;t)$ is continuous at $0$, we have that it is increasing over $[0,\infty)$.
Now notice that $$\mu(x,y;-t)=\mu(x^{-1},y^{-1};t)^{-1}$$ and we can conclude that the function is also increasing over $(-\infty,0]$.
This is a “generalized AM-GM” inequality, which is the simple observation that $\mu(x,y;0)<\mu(x,y;1)$. For $t=-1$ we get the harmonic mean.
The inequalities become nonstrict if and only if $x=y$.
More generally, then, for $x+y=1$, $x,y>0$ and $t>1$, we have $\mu(x,y;1)\le\mu(x,y;t)$, that is $$x^t+y^t\ge\frac{1}{2^{t-1}}$$
If instead $0<t<1$, we have $\mu(x,y;t)\le\mu(x,y;1)$, so $$x^t+y^t\le 2^{1-t}$$
use Holder inequality we have $$(x^4+y^4)(1+1)(1+1)(1+1)\ge (x+y)^4$$
Apply Lagrangian multiplier method ! \begin{align} \min \quad x^4 + y^4 \\ \text{s.t.} \quad x + y = 1 \end{align} gives $\mathcal{L}(x,y) = x^4 + y^4 - \lambda(x+y-1)$. FOC gives \begin{align} \mathcal{L}_x = 4x^3 - \lambda & = 0 \\ \mathcal{L}_y = 4y^3 - \lambda & = 0 \\ x+y & = 1. \end{align} Solving gives an optimum of $(x^*,y^*) = (\frac{1}{2},\frac{1}{2})$ with value $\frac{1}{8}$. | 2019-08-19T16:40:42 | {
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https://brilliant.org/discussions/thread/complicated-combinatorics-problem/ | ×
Complicated combinatorics problem
Warren and Nadia have a straight path, one meter wide and 16 meters long, from the front door to the front gate. The decide to pave it. Warren brings home 16 identical paving stones, each 1 meters square. Nadia brings home 8 identical paving stones, each 1 meter x 2 meters. Assuming that they would consider using all square, all rectangular or a combination of each, what is the probability that the combination used consists of all Warren's blocks?
Note by Joel Jablonski
4 years, 1 month ago
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I'm getting
$$\displaystyle \text{Pr(all Warren's blocks)} = \dfrac{1}{\sum_{i=0}^{8}\dfrac{(16- i)!}{i!\, (16-2\, i)!}} = \dfrac{1}{1597}$$
- 4 years, 1 month ago
Nice problem. Fibonacci is the key - the number of ways of tiling a 1 by 16 path is equal to the number of ways of tiling a 1 by 15 path (by adding a 1 by 1 slab) plus the number of ways of tiling a 1 by 14 path (by adding a 1 by 2 slab), and so on. So there are 1597 (the 17th or 18th Fibonacci number, depending where you start) ways, giving a probability of 1/1597.
Now how about this question: they tile the path from the front door, and choose a slab at random each time from those they have remaining. What's the probability now that they use all of Warren's tiles? What about 8 of Warren's and 4 of Nadia's? More of Warren's than Nadia's? Warren's tiles make up more than half the length of the path?
- 4 years, 1 month ago | 2017-11-19T06:30:22 | {
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https://math.stackexchange.com/questions/3630724/does-randomly-selecting-without-replacement-imply-independent-events | # Does randomly selecting WITHOUT replacement imply independent events?
After a few days of researching and going through previous posts, I'm very much unclear on this topic still. My question is whether or not randomly taking samples without replacement implies that these samples are independent. In a related post (Independent or dependent events, drawing cards without replacement), removing two cards from a deck, the events are said to be independent.
However, in an unrelated workbook I'm working through, when introducing the concept of Hypergeometric probability, they would describe these events as dependent. The question states: "An urn contains five red balls, and seven blue balls. Four balls are randomly selected without replacement. Determine the probability that exactly one of them is red." along with "Choosing each ball affects the probability that the following ball will be a certain color, because the sample space has changed. Thus, the selection of each ball is not an independent event." $$P(x) = \frac{{7 \choose 3} * {5 \choose 1}}{12 \choose 4} = .354$$ Any help would be greatly appreciated!
It is exactly as that second workbook says: Given that what you pick and don't put back changes the distribution of what's left, the events are not independent
It's helpful to think of simple toy examples.
Consider a bag with two balls: one white and one black. Suppose we take out the balls one after the other without replacement, and consider the events $$W_1=\text{the first ball is white}$$and $$W_2=\text{the second ball is white}.$$
Intuitively, it's obvious these are not independent events because if you know $$W_1$$ happened then you know $$W_2$$ cannot happen since there's only one white ball. This is what your book means by "the sample space has changed": after you remove the first ball, you alter the contents of the bag.
The probability of $$W_1$$ actually determines the probability of $$W_2$$ since $$\mathrm P(W_1)+\mathrm P(W_2)=1$$.
• @Bram28 Thank you for your response! However In this related question: (math.stackexchange.com/questions/1174057/…), drawing 2 cards from a deck, the events are said to be independent. Isn't that virtually the same scenario as the one described above? in which the events are actually dependent? Apr 17 '20 at 21:58
• Dear @Jorge, no! It's not the same scenario at all! In the scenario I present it is crystal clear that the events affect (even determine!) each other. In the linked question, however, this is not so: suppose you know your first card was a heart. The second event is not asking whether the second card will be a heart. It is asking a different question - about face cards. Apr 17 '20 at 22:05
• Thank you so much! So it was subtly in the question phrasing. But that definitely makes sense now! Apr 17 '20 at 22:08 | 2021-10-19T18:45:15 | {
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https://math.stackexchange.com/questions/2240847/does-the-series-sum-n-1-infty-1nn2e-frac-n33-converge | # Does the series $\sum_{n=1}^{\infty}(-1)^{n}n^{2}e^{\frac{-n^{3}}{3}}$ converge or diverge?
Does the series $$\sum_{n=1}^{\infty}(-1)^{n}n^{2}e^{\frac{-n^{3}}{3}}$$ converge or diverge?
I have attempted using the alternating series test, but couldn't find a useful function to use as $b_n$ and no other tests I know seem to be useful in coming to the conclusion of whether it is convergent or divergent.
• Is the power of the( -1 ) 2? – Elad Apr 18 '17 at 21:08
• oops that is supposed to be n – goldenlinx Apr 18 '17 at 21:10
• So you should show that $f(n)=n^2*e^\frac{-n^3}{3}$ is a deacreasing function for large enough $n$ – Elad Apr 18 '17 at 21:14
• That's not trivial to a student learning it for the first time. :( – tilper Apr 18 '17 at 21:26
• I understand your points but please take mine too into consideration: do we really want MSE to be flooded by Calc-1 almost trivial questions? I am fine with a reasonable amount of them to occupy the main page, but nowadays they are becoming way too many. In my humble opinion of course, and as an active member of MSE I have the power to express it through votes. I am fine with others disagreeing with me, too. – Jack D'Aurizio Apr 18 '17 at 21:35
The series converges absolutely. This is because we have $$0 \leq \exp\left(-\frac{n^3}{3}\right) \leq \frac{1}{1 + \frac{n^3}{3} + \frac{n^6}{18}}$$
by considering the Taylor expansion and hence
$$0 \leq n^2\exp\left(-\frac{n^3}{3}\right) \leq \frac{n^2}{1 + \frac{n^3}{3} + \frac{n^6}{18}} \leq 18n^{-4}$$
and $$\sum 18n^{-4}$$ converges by the integral test.
More intuitively, just keep in mind that exponentials of the form $a^{-x}$ for $a>1$ decay faster than any rational function.
If you want to use the alternating series test to merely establish conditional convergence, you can use L'hopital to show $\lim_{x \to \infty} x^2e^{-\frac{x^3}{3}} = 0$
• Thank you I see how to solve the question much more clearly now. – goldenlinx Apr 18 '17 at 21:21
Letting $a_n = (-1)^n n^2 e^{-\frac{n^3}{3}}$, we have $$\frac{a_{n+1}}{a_n} = - \frac{(n+1)^2}{n^2} e^{-\frac{(n+1)^3}{3}}e^{\frac{n^3}{3}} = -\left(1+\frac{1}{n}\right)^2 e^{-\frac{3n^2+3n+1}{3}}$$
Then as $n \to \infty$, note $\left(1+\frac{1}{n}\right)^2 \to 1$ and since $n^2+n+\frac{1}{3} \to \infty$, the exponential tends to zero so $$\lim_{n\to \infty} \left| \frac{a_{n+1}}{a_n} \right| = 0 < 1$$ so the series converges by the ratio test.
$|a_n|=|(-1)^nn^2e^{-\frac {n^3}{3}}|=\left(n^2e^{-\frac {n^3}{6}}\right)e^{-\frac {n^3}{6}}\le e^{-\frac {n^3}{6}}\le e^{-n}\quad$ for $n$ large enough.
• The first inequality is because any polynom is small in regard to any exponential (their product is going to zero), so for $n\gg 1$ large enough, $n^2e^{-\frac {n^3}{6}}<1$.
• The second one is because $e^{-x}$ is decreasing, so since $\frac {n^3}6>n$, once again for $n\gg 1$ large enough, we have $e^{-\frac {n^3}{6}}\le e^{-n}$
So for $n$ large enough we have $|a_n|<e^{-n}$ which is a term of a convergent geometric serie (of reason $0<\frac 1e<1$), so the original serie is absolutely convergent (because positive terms makes the partial sum increasing and it is bounded by the sum of the other serie, so it has a limit).
Remember that if you can get a straightforward result by rough majorations like in this case, always go for it before invoking more powerful tools. Advanced tools will give you more precise estimations, but it's good to come back to the basics, analysis is the art of minoration and majoration.
The series is absolutely convergent. Apply the quotient criterion. | 2019-08-23T22:08:53 | {
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http://mathhelpforum.com/calculus/90892-volume-revolution-between-y-sinx-y-cosx.html | # Math Help - Volume of revolution between y=sinx and y=cosx
1. ## Volume of revolution between y=sinx and y=cosx
How would I go about solving the following problem?
Find the volume of revolution defined by rotating, around the x-axis, the area between the graphs of y = sinx and y = cosx on the interval where 0 is less than or equal to x, which is less than or equal to 1.
I'm aware of the formula for calculating solids of revolution, and could solve the problem if it was just the area between y = sinx and the x-axis. I'm also aware of the "washer" method. However, the problem I'm having is that y = sinx and y = cosx cross over each other between 0 and 1, so I don't know how I'm supposed to deduce the formulas for the radius of the inner and outer circle of each washer. Should I be treating them as two different solids of revolution on an interval of 0 to the crossover point and the crossover point to 1? I think this would work, but I doubt it's the correct/best method. Any help would be greatly appreciated.
2. Originally Posted by drew.walker
How would I go about solving the following problem?
Find the volume of revolution defined by rotating, around the x-axis, the area between the graphs of y = sinx and y = cosx on the interval where 0 is less than or equal to x, which is less than or equal to 1.
I'm aware of the formula for calculating solids of revolution, and could solve the problem if it was just the area between y = sinx and the x-axis. I'm also aware of the "washer" method. However, the problem I'm having is that y = sinx and y = cosx cross over each other between 0 and 1, so I don't know how I'm supposed to deduce the formulas for the radius of the inner and outer circle of each washer. Should I be treating them as two different solids of revolution on an interval of 0 to the crossover point and the crossover point to 1? I think this would work, but I doubt it's the correct/best method. Any help would be greatly appreciated.
See the attached graph
So from zero to $\frac{\pi}{4}$ the cosine is the upper function and then sine is from $\frac{\pi}{4}$ to 1.
$\pi\int_{0}^{\frac{\pi}{4}}\cos^2(x)-\sin^2(x)dx+\pi\int_{\frac{\pi}{4}}^{1} \sin^2(x)-\cos^2(x)dx$
$\pi\int_{0}^{\frac{\pi}{4}}(\cos(x)-\sin(x))(\cos(x)+\sin(x))dx+\pi\int_{\frac{\pi}{4} }^{1} (\sin(x)-\cos(x))(\sin(x)+\cos(x))dx$
$\pi \frac{(\cos(x)+\sin(x))^2}{2} \bigg|_{0}^{\pi/4}+\pi \frac{(\sin(x)-\cos(x))^2}{2} \bigg|_{\pi/4}^{1}$
$\frac{\pi}{2}((\sqrt{2})^2-1^2+(\sin(1)-\cos(1))^2-(0))$
$\frac{\pi}{2}(2-1+\sin^2(1)-2\sin(1)\cos(1)+\cos^2(1))=\frac{\pi}{2}(2-2\sin(1)\cos(1))=\pi(1-\sin(1)\cos(1))$
3. That's great! My only question about your working would be: how did you know that the graphs were equal at pi/4? Was this something you calculated? If so, how?
4. Originally Posted by drew.walker
That's great! My only question about your working would be: how did you know that the graphs were equal at pi/4? Was this something you calculated? If so, how?
Solve $\sin x = \cos x \Rightarrow \tan x = 1$. | 2014-09-23T09:49:17 | {
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https://math.stackexchange.com/questions/3349614/three-circles-are-inscribed-inside-a-unit-square-find-the-radius-of-the-3rd-cir | # Three circles are inscribed inside a unit square. Find the radius of the 3rd circle.
One circle is inscribed inside a unit square such that all sides of square are tangent to it.
A second circle is at the upper-left corner of the square such that it is tangent to the two sides of the square and touches the first circle externally.
A third circle touches both circles externally and the upper side of the square is tangent to the third circle. (it is on the right side of the second circle.)
Find the radius of the 3rd circle.
• Can you show what you have tried or worked on? A diagram would be a very good first step. – Gabe Sep 9 '19 at 15:02
• @Gabe I found radii of 1st and second circle but I am not getting radius of 3rd circle. – user677008 Sep 9 '19 at 15:05
• can you sketch it? – Narasimham Sep 9 '19 at 15:12
• Use the special case of Descartes' Theorem. – Jaap Scherphuis Sep 9 '19 at 15:13
• @Narasimham I am unable to upload picture. So described everything in question. – user677008 Sep 9 '19 at 15:18
## 4 Answers
Let the three radii be $$r_1$$, $$r_2$$ and $$r_3$$. From matching various lengths, the following relationships among the radii hold,
$$\sqrt{2}r_1=r_1+(1+\sqrt{2})r_2\tag{1}$$
$$(r_1+r_3)^2-(r_1-r_3)^2=a^2\tag{2}$$ $$(r_2+r_3)^2-(r_2-r_3)^2=b^2\tag{3}$$ $$r_1-r_2=a+b\tag{4}$$
Solve (1) for the radius of the second circle, $$r_2=(3-2\sqrt{2})r_1$$
Simplify (2) and (3) to get,
$$r_3=\frac{a^2}{4r_1}\tag{5}$$ $$\frac{a^2}{b^2}=\frac{r_1}{r_2}\tag{6}$$
Combine (6) and (4) to get $$a=\sqrt{r_1}(\sqrt{r_1}-\sqrt{r_2})\tag{7}$$
Then, plug (7) into (5) to obtained the radius of the third circle,
$$r_3=\frac 14 (\sqrt{r_1}-\sqrt{r_2})^2$$
Use the result of the second radius $$r_2$$ derived above to express $$r_3$$ in terms of $$r_1 = \frac 12$$,
$$r_3=\frac 12 (3-2\sqrt{2})r_1=\frac 14 (3-2\sqrt{2})$$
As @Jaap Scherphuis commented, this is a special case of Descartes' Theorem
\begin{align} k_3&=k_1+k_2+k_4+2\sqrt{k_1k_2+k_2k_4+k_4k_1} =k_1+k_2+2\sqrt{k_1k_2} , \end{align}
where
\begin{align} k_1&=1/r_1 ,\\ k_2&=1/r_2 ,\\ k_3&=1/r_3 ,\\ k_4&=0 , \end{align}
since we can consider a line $$DC$$ as the fourth circle with zero curvature (infinite radius).
Thus, given that you've found values for $$r_1,r_2$$ ,
\begin{align} r_3&= \frac{r_1\,r_2}{r_1+r_2+2\sqrt{r_1\,r_2}} . \end{align}
The radius of the first circle is $$\frac{1}{2}$$, so its curvature is $$k_1=2$$. The radius of the second circle is $$\frac{(\sqrt{2}-1)/2}{2\sqrt{2}}$$, so its curvature is $$k_2=\frac{4\sqrt{2}}{\sqrt{2}-1}$$. The upper side of the square can be considered a circle with infinite radius and zero curvature. Thus, the Descartes theorem in this case reduces to $$k_4=k_1+k_2+2 \sqrt{k_1 k_2}$$ This gives
$$k_4=2+\frac{4\sqrt{2}}{\sqrt{2}-1}+2 \sqrt{ \frac{8\sqrt{2}}{\sqrt{2}-1} }\approx 26.109...$$
which corresponds to a radius of $$\frac{1}{26.109}\approx 0.038$$
The Circle in the Corner
In the corner of the square, the length of the red segment is $$\frac1{\sqrt2}-\frac12$$ (half the diagonal of the square minus the radius of the original circle). Because all these triangles are $$45$$-$$45$$-$$90$$, there is a ratio of $$\sqrt2$$ between sides. The length of the green segments is $$1-\frac1{\sqrt2}$$ and the length of the blue segment is $$\sqrt2-1$$. Thus, the perimeter of the green-green-blue triangle is $$2\left(\color{#090}{1-\frac1{\sqrt2}}\right)+\left(\color{#00F}{\sqrt2-1}\right)=1$$ its area is $$\frac12\left(\color{#090}{1-\frac1{\sqrt2}}\right)^2=\frac34-\frac1{\sqrt2}$$ The radius of the inscribed circle is $$2$$ times the area divided by the perimeter: $$2\,\frac{\frac34-\frac1{\sqrt2}}{1}=\frac{3-2\sqrt2}2$$
The Radius of the Third Circle
To apply Descartes' Theorem, note that the bend (signed curvature) of the large circle, the small circle, and the top edge of the square are $$2$$, $$6+4\sqrt2$$, and $$0$$. Then Descartes' Theorem says $$2\left(2^2+\left(6+4\sqrt2\right)^2+0^2+\frac1{r^2}\right)=\left(2+\left(6+4\sqrt2\right)+0+\frac1r\right)^2\\ %144+96\sqrt2+\frac2{r^2}=96+64\sqrt2+\left(16+8\sqrt2\right)\frac1r+\frac1{r^2}\\ %\frac1{r^2}-\left(16+8\sqrt2\right)\frac1{r}+48+32\sqrt2=0\\ %\left(\frac1r-\left(8+4\sqrt2\right)\right)^2=48+32\sqrt2\\ %\frac1r-\left(8+4\sqrt2\right)=\pm\left(4+4\sqrt2\right)\\ %\frac1r=12+8\sqrt2\quad\text{or}\quad4\\ r=\color{#C00}{\frac{3-2\sqrt2}4}\quad\text{or}\quad\color{#090}{\frac14}$$ There are two solutions because there are two circles (shown below in red and green) tangent to the first two circles and the line containing the top of the square:
The radius of the circle in the question (the red circle) is $$\color{#C00}{\frac{3-2\sqrt2}4}$$.
The Center of the Third Circle
As shown in the corollary from this answer, the mean of the centers of the circles weighted by their bends equals their mean weighted by the square of their bends. Straight lines need special handling. Straight lines are essentially circles with a center at an infinite distance in a given direction. For the computation of centers:
$$1$$. the bend is $$0$$
$$2$$. the bend times the center is a unit vector perpendicular to the line
$$\phantom{2\text{.}}$$ and in the direction away from the other circles
$$3$$. the square of the bend times the center is $$0$$
For the red circle, solve $$\tfrac{(0,1)+2(0,0)+\left(6+4\sqrt2\right)\left(1-\sqrt2,\sqrt2-1\right)+\color{#C00}{\left(12+8\sqrt2\right)(x,y)}}{2+\left(6+4\sqrt2\right)+\color{#C00}{\left(12+8\sqrt2\right)}} =\tfrac{2^2(0,0)+\left(6+4\sqrt2\right)^2\left(1-\sqrt2,\sqrt2-1\right)+\color{#C00}{\left(12+8\sqrt2\right)^2(x,y)}}{2^2+\left(6+4\sqrt2\right)^2+\color{#C00}{\left(12+8\sqrt2\right)^2}}$$ for $$(x,y)$$ to get $$\color{#C00}{\left(\frac{\sqrt2-2}2,\frac{2\sqrt2-1}4\right)}$$.
For the green circle, solve $$\tfrac{(0,1)+2(0,0)+\left(6+4\sqrt2\right)\left(1-\sqrt2,\sqrt2-1\right)+\color{#090}{4(x,y)}}{2+\left(6+4\sqrt2\right)+\color{#090}{4}} =\tfrac{2^2(0,0)+\left(6+4\sqrt2\right)^2\left(1-\sqrt2,\sqrt2-1\right)+\color{#090}{4^2(x,y)}}{2^2+\left(6+4\sqrt2\right)^2+\color{#090}{4^2}}$$ for $$(x,y)$$ to get $$\color{#090}{\left(-\frac{\sqrt2}2,\frac14\right)}$$. | 2020-02-29T06:35:33 | {
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https://math.stackexchange.com/questions/2672288/if-cos2-theta-0-then-delta2 | If $\cos2\theta=0$, then $\Delta^2=$?
I'll state the question from my textbook here:
If $\cos2\theta=0$, then $\begin{vmatrix} 0 & \cos \theta & \sin \theta \\ \cos \theta & \sin \theta & 0 \\ \sin \theta & 0 & \cos \theta \end{vmatrix}^2=$?
This is how I solved the problem:
$\begin{vmatrix} 0 & \cos \theta & \sin \theta \\ \cos \theta & \sin \theta & 0 \\ \sin \theta & 0 & \cos \theta \end{vmatrix}^2$
$= (\cos^3 \theta + \sin^3 \theta)^2$
$= (\cos \theta + \sin \theta)^2(\cos^2 \theta - \cos \theta \sin \theta + \sin^2 \theta)^2$
$= (1+ \sin2\theta)(1-\sin2\theta + \sin^2 \theta \cos^2 \theta)$
$= (1+ \sin2\theta)(1-\sin2\theta) + (1 + \sin2\theta)\sin^2\theta \cos^2 \theta$
$= \cos^2 2\theta + \frac 14 (1 + \sin2\theta)\sin^2 2\theta$
$= \frac 14 (1 + \sin2\theta)\sin^2 2\theta$
Now since $\cos2\theta=0$, $\sin2\theta = \pm 1$.
Therefore the above expression can take the values 0 and $\frac12$.
My textbook gives the answer as $\frac12$. I don't see any grounds on rejecting the other answer of 0. Have I made a mistake somewhere? Or am I forgetting something?
When in doubt, use the relations in the original problem. Let $\theta=\frac{3\pi}4$. Then $\cos2\theta=0$, while $\sin\theta=\sqrt2/2=a$ and $\cos\theta=-\sqrt2/2=-a$.
The expression in the question is now $$\begin{vmatrix}0&-a&a\\-a&a&0\\a&0&-a\end{vmatrix}^2$$ Clearly, adding up all three rows of the matrix produces the zero vector, so the whole expression evaluates to zero. Your working is entirely correct: the book is wrong to omit 0 as an answer.
The result of $\frac12$ is obtained with the other principal solution to $\cos2\theta=0$, $\theta=\frac\pi4$.
You are right indeed by direct calculation we obtain
$$\cos 2\theta=0\iff \theta=\frac{\pi}4+k\frac{\pi}2$$
and since
$$\Delta^2=\begin{vmatrix} 0 & \cos \theta & \sin \theta \\ \cos \theta & \sin \theta & 0 \\ \sin \theta & 0 & \cos \theta \end{vmatrix}^2= (-\cos^3 \theta - \sin^3 \theta)^2$$
• for $\theta=\frac{\pi}4 \implies \Delta^2=\left(-\frac{2\sqrt 2}{4}\right)^2=\frac12$
• for $\theta=\frac{3\pi}4\implies \Delta^2=0$
• for $\theta=\frac{5\pi}4\implies \Delta^2=\left(\frac{2\sqrt 2}{4}\right)^2=\frac12$
• for $\theta=\frac{7\pi}4\implies \Delta^2=0$
If we compute the square of the matrix, we get $$\begin{bmatrix} 1 & \sin\theta\cos\theta & \sin\theta\cos\theta \\ \sin\theta\cos\theta & 1 & \sin\theta\cos\theta \\ \sin\theta\cos\theta & \sin\theta\cos\theta & 1 \end{bmatrix}= \frac{1}{2} \begin{bmatrix} 2 & \sin2\theta & \sin2\theta \\ \sin2\theta & 2 & \sin2\theta \\ \sin2\theta & \sin2\theta & 2 \end{bmatrix}$$ whose determinant is $$\frac{1}{8}(8+2\sin^32\theta-6\sin^22\theta)$$ From $\cos2\theta=0$, we have $\sin^22\theta=1$, so finally we get $$\Delta^2=\frac{1}{4}(1+\sin2\theta)= \begin{cases} 1/2 & \text{if \sin2\theta=1}\\[4px] 0 & \text{if \sin2\theta=-1} \end{cases}$$ both possibilities being allowed by the hypothesis that $\cos2\theta=0$.
• Yeah, I tried this too but still got 2 answers. So I thought the book might be wrong. – SamInuyasha ANMF Mar 2 '18 at 7:09
• @SamInuyashaANMF Indeed it is wrong, unless the matrix is not assumed to have further properties, such as being invertible. – egreg Mar 2 '18 at 7:13 | 2019-10-15T10:57:44 | {
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https://math.stackexchange.com/questions/3320843/conditional-densities | # Conditional Densities
So two independent random variables $$X$$ and $$Y$$ are uniformly distributed on $$[0,1]$$
I want to find the conditional densities of $$X$$ and $$Y$$ given that $$X>Y$$.
How can I find these conditional densities and their expected values? i.e $$\mathbb E(X|X>Y)$$ ?
Thanks
• Are $X$ and $Y$ defined on the same probability space? If so then what is their joint distribution? Are they independent maybe? – drhab Aug 12 '19 at 9:50
• Yes they are independent – StatsHelp Aug 12 '19 at 10:13
• You should add to your question in an edit, because it is essential information. Without it your question cannot be answered. – drhab Aug 12 '19 at 10:14
• I have edited the question. Are you able to provide an answer now by any chance? – StatsHelp Aug 12 '19 at 10:21
Assuming that $$X$$ and $$Y$$ are independent (since there is no way to proceed with assuming some joint distribution, and this is the most natural one) then the question has already been answered here Conditional density question.
In particular, $$\mathbb E(X\mid X>Y)=\int _0^1 2x^2\ dx=\frac{2}{3}$$.
• I tried to understand that comment of yours but I could not make sense out of the first part (the triangle part) – StatsHelp Aug 12 '19 at 10:17
• Consider the pair $(X,Y)$. It is a random point in the plane. Before the conditioning, it belongs to the unit square $[0,1]\times [0,1]$. You can think of the conditioning as cutting the square in half along the line $x=y$. The bottom right half is all that remains after conditioning. Since the area is half, the density must double so as to maintain a total integral of $1$. That is the entire concept... – pre-kidney Aug 12 '19 at 10:29
• Oh right. That makes sense. Thank you that was very helful – StatsHelp Aug 12 '19 at 10:31
• If the questions are answered, you can mark them as such using the green checkmark. – pre-kidney Aug 12 '19 at 10:32
Under condition $$X>Y$$ we are dealing with uniform distribution on the triangle: $$\Delta:=\{(x,y)\in[0,1]^2\mid x>y\}$$
The PDF of this joint distribution is constant on that triangle and takes value $$0$$ outside the triangle.
As always its integral must take value $$1$$ and this together leads to PDF:$$f_{X,Y\mid X>Y}(x,y)=2\mathbf1_{\Delta}(x,y)$$where $$\mathbf1_{\Delta}$$ denotes the indicator function of set $$\Delta$$.
Then we find: $$\mathbb E[X\mid X>Y]=\int xf_{X,Y\mid X>Y}(x,y)dydx=2\int_0^1\int_0^xxdydx=2\int_0^1x\int_0^xdydx=$$$$2\int_0^1x^2dx=\frac23$$
You can find the marginal distributions of the conditional distribution on the usual way by finding for $$x\in[0,1]$$:$$f_{X\mid X>Y}(x)=\int f_{X,Y\mid X>Y}(x,y)dy=2\int_0^xdy=2x\tag1$$and for $$y\in[0,1]$$ similarly:$$f_{Y\mid X>Y}(y)=\int f_{X,Y\mid X>Y}(x,y)dx=2\int_y^1dy=2(1-y)$$
Outside interval $$[0,1]$$ these marginals evidently both take value $$0$$.
Also note that $$(1)$$ reveals another way to find $$\mathbb E[X\mid X>Y]$$:$$\mathbb E[X\mid X>Y]=\int_0^1xf_{X\mid X>Y}(x)dx=\int_0^1x\cdot2xdx=\frac23$$ | 2020-08-12T04:36:50 | {
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