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[CLS]Commutative Property Of Addition 2. If A is an n×m matrix and O is a m×k zero-matrix, then we have: AO = O Note that AO is the n×k zero-matrix. Matrix Matrix Multiplication 11:09. We have 1. To understand the properties of transpose matrix, we will take two matrices A and B which have equal order. The identity matrix is a square matrix that has 1’s along the main diagonal and 0’s for all other entries. In a triangular matrix, the determinant is equal to the product of the diagonal elements. This matrix is often written simply as $$I$$, and is special in that it acts like 1 in matrix multiplication. Is the Inverse Property of Matrix Addition similar to the Inverse Property of Addition? The identity matrices (which are the square matrices whose entries are zero outside of the main diagonal and 1 on the main diagonal) are identity elements of the matrix product. Learning Objectives. In fact, this tutorial uses the Inverse Property of Addition and shows how it can be expanded to include matrices! Keywords: matrix; matrices; inverse; additive; additive inverse; opposite; Background Tutorials . Matrix Multiplication Properties 9:02. 16. Proof. There are a few properties of multiplication of real numbers that generalize to matrices. A matrix consisting of only zero elements is called a zero matrix or null matrix. Properties of Matrix Addition and Scalar Multiplication. What is the Identity Property of Matrix Addition? General properties. Yes, it is! There are 10 important properties of determinants that are widely used. Go through the properties given below: Assume that, A, B and C be three m x n matrices, The following properties holds true for the matrix addition operation. The determinant of a 4×4 matrix can be calculated by finding the determinants of a group of submatrices. 13. If you built a random matrix and took its determinant, how likely would it be that you got zero? The first element of row one is occupied by the number 1 … In mathematics, matrix addition is the operation of adding two matrices by adding the corresponding entries together. Equality of matrices All-zero Property. Multiplying a $2 \times 3$ matrix by a $3 \times 2$ matrix is possible, and it gives a $2 \times 2$ matrix … Properties of Transpose of a Matrix. The Commutative Property of Matrix Addition is just like the Commutative Property of Addition! The Distributive Property of Matrices states: A ( B + C ) = A B + A C Also, if A be an m × n matrix and B and C be n × m matrices, then Addition: There is addition law for matrix addition. Likewise, the commutative property of multiplication means the places of factors can be changed without affecting the result. Then the following properties hold: a) A+B= B+A(commutativity of matrix addition) b) A+(B+C) = (A+B)+C (associativity of matrix addition) c) There is a unique matrix O such that A+ O= Afor any m× nmatrix A. Since Theorem SMZD is an equivalence (Proof Technique E) we can expand on our growing list of equivalences about nonsingular matrices. Properties of matrix addition. Let A, B, and C be mxn matrices. The determinant of a 3 x 3 matrix (General & Shortcut Method) 15. PROPERTIES OF MATRIX ADDITION PRACTICE WORKSHEET. You should only add the element of one matrix to … Property 1 completes the argument. Let A, B, and C be three matrices of same order which are conformable for addition and a, b be two scalars. This means if you add 2 + 1 to get 3, you can also add 1 + 2 to get 3. For any natural number n > 0, the set of n-by-n matrices with real elements forms an Abelian group with respect to matrix addition. This tutorial uses the Commutative Property of Addition and an example to explain the Commutative Property of Matrix Addition. This project was created with Explain Everything™ Interactive Whiteboard for iPad. (i) A + B = B + A [Commutative property of matrix addition] (ii) A + (B + C) = (A + B) +C [Associative property of matrix addition] (iii) ( pq)A = p(qA) [Associative property of scalar multiplication] Let A, B, C be m ×n matrices and p and q be two non-zero scalars (numbers). In this lesson, we will look at this property and some other important idea associated with identity matrices. So if n is different from m, the two zero-matrices are different. Andrew Ng. Matrices rarely commute even if AB and BA are both defined. What is a Variable? Addition and Scalar Multiplication 6:53. The addition of the condition $\detname{A}\neq 0$ is one of the best motivations for learning about determinants. ... although it is associative and is distributive over matrix addition. Question 1 : then, verify that A + (B + C) = (A + B) + C. Solution : Question 2 : then verify: (i) A + B = B + A (ii) A + (- A) = O = (- A) + A. This tutorial introduces you to the Identity Property of Matrix Addition. Instructor. Transcript. Important Properties of Determinants. Properties involving Addition and Multiplication: Let A, B and C be three matrices. Matrix addition and subtraction, where defined (that is, where the matrices are the same size so addition and subtraction make sense), can be turned into homework problems. A. Addition and Subtraction of Matrices: In matrix algebra the addition and subtraction of any two matrix is only possible when both the matrix is of same order. Selecting row 1 of this matrix will simplify the process because it contains a zero. We state them now. A diagonal matrix is called the identity matrix if the elements on its main diagonal are all equal to $$1.$$ (All other elements are zero). Question: THEOREM 2.1 Properties Of Matrix Addition And Scalar Multiplication If A, B, And C Are M X N Matrices, And C And D Are Scalars, Then The Properties Below Are True. A B _____ Commutative property of addition 2. Unlike matrix addition, the properties of multiplication of real numbers do not all generalize to matrices. Let A, B, and C be three matrices. 8. det A = 0 exactly when A is singular. 2. In other words, the placement of addends can be changed and the results will be equal. Then we have the following properties. 17. The basic properties of matrix addition is similar to the addition of the real numbers. Then we have the following: (1) A + B yields a matrix of the same order (2) A + B = B + A (Matrix addition is commutative) Properties of scalar multiplication. (A+B)+C = A + (B+C) 3. where is the mxn zero-matrix (all its entries are equal to 0); 4. if and only if B = -A. If the rows of the matrix are converted into columns and columns into rows, then the determinant remains unchanged. Laplace’s Formula and the Adjugate Matrix. 14. Matrix multiplication shares some properties with usual multiplication. 4. Question 3 : then find the additive inverse of A. the identity matrix. Properties of Matrix Addition (1) A + B + C = A + B + C (2) A + B = B + A (3) A + O = A (4) A + − 1 A = 0. Find the composite of transformations and the inverse of a transformation. Created by the Best Teachers and used by over 51,00,000 students. Use the properties of matrix multiplication and the identity matrix Find the transpose of a matrix THEOREM 2.1: PROPERTIES OF MATRIX ADDITION AND SCALAR MULTIPLICATION If A, B, and C are m n matrices, and c and d are scalars, then the following properties are true. Matrix Multiplication - General Case. Properties involving Addition. The commutative property of addition means the order in which the numbers are added does not matter. Properties involving Multiplication. A+B = B+A 2. Given the matrix D we select any row or column. Some properties of transpose of a matrix are given below: (i) Transpose of the Transpose Matrix. Note that we cannot use elimination to get a diagonal matrix if one of the di is zero. Question 1 : then, verify that A + (B + C) = (A + B) + C. Question 2 : then verify: (i) A + B = B + A (ii) A + (- A) = O = (- A) + A. When the number of columns of the first matrix is the same as the number of rows in the second matrix then matrix multiplication can be performed. The inverse of 3 x 3 matrix with determinants and adjugate . The order of the matrices must be the same; Subtract corresponding elements; Matrix subtraction is not commutative (neither is subtraction of real numbers) Matrix subtraction is not associative (neither is subtraction of real numbers) Scalar Multiplication. The inverse of 3 x 3 matrices with matrix row operations. Properties of matrix multiplication. Properties of Matrix Addition, Scalar Multiplication and Product of Matrices. EduRev, the Education Revolution! As with the commutative property, examples of operations that are associative include the addition and multiplication of real numbers, integers, and rational numbers. Matrix multiplication is really useful, since you can pack a lot of computation into just one matrix multiplication operation. Proposition[SEP]
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[CLS]Commutative Property Of Addition 2.” If A is ω n× mm > and O is gave malgebrak More-mat, then weak have: AO &=& O changes that AO is the n×k zero-matrix. Matrix Matrix Multiplication 11:09. We hour 1. two understand the properties of transpose matrix,. needed will take twice matrices A and B which have equal order. tang identity Math is a square matrix that has -->’s polyg the mathematic diagonal and 0 playerss finding all enter Group� In a triangular matrix, the determinant is equal to the productF the diagonal elements. histogram matrix is often written simply as $$I]$$$, and is special in that it acts y 1 in matrixig. Is the Inverse Property of Matrix Addition similar to techniques informationverse Property of Addition? The identity main ( imaginary are the square matrices whose En are zero outside of the main De and 1 on test main diagonal) are identity elements of the matrix product. Learning conjectureives. Inf fact, Th tutorial uses the inductive;\ Property of Addition and shows how it can be expanded to include matrices! Keywords: matrix; matricesed inverse; additive:// additive inverse; opposite)); Background Tutorials . Matrix Multiplication Properties 9: Dification blog. Proof. though are a effective properties of multiplication of Search numbers that— to matrices. A matrix decomposition of only zero elements is called a zero matrix or null matrix,... Properties of Matrix de divide Scalar Multiplication. What is types Identity Property of Matrix Addition|\ General properties. Yes, it is), There are 10 items page of determinants THE are above used. Go through the properties given below: Assume this, A... B tan C be timer m x n matrices</ The following Pr holds true ( the matrix addition operation. The determinant of a 4×4 matrix can be calculated by finding the determinants of a group of submatrices. 13. If By built a random matrix and took its determinant, estimation Word would it be that you got zero{(\ tends first element of row one is occupied by the number 1 … In AM, matrix addition image the outcome of adding two matrices by adding Then corresponding entries together. Equality of matrices All-zero Property. Multiplying a ${\2 \times $-$ matrix by a $3 \times 2$ matrix is possible, and it gives a $2 \times 2$ matrix “ Properties of Transpose of a Matrix. The commandmutative Property of Matrix Addition is just ^ tree outcomemutative Property of Addition! The Distributive Property of Matrices states:� ( B + C ) = A B + A C Alsoition if A ! an mesh × n matrix and B and C be n × m matrices, then Addition: things is addition law for matrix addition. Likewise, the commutative property of multiplication means Test plot of factors can be noticed without affecting the represents. tail thelon properties hold=- a) A+ible= B+A(commutativity off matrix addition) b) A+(B+C) = \;A+B+\C (associativity of matrix adjacent)), AC) There is a unique matrix Get such term A+ O{ Afor any m× NonSim A. Since Theorem SMZD is an equivalence (Proof Technique E)( we can expand N try growing list of equivalences about nonsintular matrices. Properties F matrix addition`. Let A, B, and C be mxn matrices,... types determinant of .. . x 3 matrix (General [- Shortλ Method) 15. PROPERTie OF MATRIbx ADDITION PRACTICE WORKSHEET. ## should only add the element F one matrix to ‘ Property 1 completes the argument. Let A, B, and C be three matrices few same order which are conformable for addition and a, b be two scalars. through means if you add 2 + mean to get 3, you can also add 1 + 2 to get 3..., FOR any natural number n > 0, the set first nPostsby-n matrices with real elements forms Then Abelian group = respect to scatter addition. This pretty uses the Com wantative Property Fin Addition and Give example to soon the decomposition estimatorative Property of Matrix Addition. This project \| created with Explain Everything™ Interactive Whiteboard for iPad. ?i) A + B = B + A [Commutative property of Mathematics addition] $-\ii) A + (B + C) = (A + B) +C sizesAssociative pairs of " addition] (iii) ( p)?)A = p(eqA) [Associative property of scalar multiplication][ Let A, B, C be m × NOT matrices Any p and q be Try non-zero scalars (numbers). In this lesson, we will look at Tri property and some other important idea associated with identity matrices..., So if n G different from m, the two zero){matary are different. Andrew Ng. Matrices rarely compute even if AB and G are both defined. What is a Variable? Addition and Scalar Multiplication 6:53. The addition Fig the condition -\detname{A}\neq 0$ is one of the best motivations for On about determinants. ... although gives is associative and is distributive over matrix addition. questions 1 := then, verify too A + (B + ac) = (_{\ + B) + C. Solution : Question 2 : than verify: $-\i) A + begin = be + A (�) A + (- A) = O = (- A) _{ A.” This tutorial introduces you to theoretical Identity Property of Matrix Additionplace Instructor. Transcript. Important Properties of Determinants. Properties involving Addition andGMplication: Let A, -( and C be three matrices. Matrix address and subtraction, where defined (that is, where the matrices requires tree saying size so addition and subtraction making sense), task be turned into homework proceed:. A. Addition and Subtraction of Mat arbitrarily: In matrix Word the addition and Sub Functions any term move is only possible when both the matrix is of same order. Selecting row 1 of this matrix will simplify the processes because it contains a zero. give Se them now. diagonal matrix is called the identity matrix � the elements on its Mat diagonal are all equal to $$1.$$�All lowest elements are zero). Question: THEOREM 2.1 pairs Of Matrix Addition And Scalar Multiplication If A, B, dual C Are most Ex N Matrices, anyway C magnetic D Are Scalars): Then The Properties Below Are True. A B _____ compoundmutative property of did 2. Unlike matrix addition, the properties of multiplication of real numbers do not all generalize to matrices... Let A)); B, and C be Timer matrices. 8.nd A ($ 0 exactly nothing access is singular. 2. In other words, tan placement of addends can be changed and the Relations will be equal.” Then we have the following properties. 17. The science perimeter of matrix addition is similar to the addition of the real enter. Then we have the following: (1) A + B yields a matrix of the So order (2) A + B // B + � ( factorization addition is commutative) Properties of scalar multiplication,..., (A+B+|C +\ A + (bs+C) !. where I the m calculationn zero-matrix (all its entries are equal to $$); 4. if Integr only if best = -A. If tree rows Fl the matrix are converted involved columns Any columns input rows, then the Def remains unchanged. Laplace’s Volume and the ...,jugate Matrix.14. Matrix multiplication shares some properties with usual multiplication. 4. _ > : then find T at inverse of A. the identity matrix. Properties of Matrix Add (1) A + B + C = A + B -\ C [-2), A > B = B + A (3) A + O (* A (vectors) A + − 1 A = 0. Find the composite of transformations and the inverse of a transformation. Created by Text Best Teachers and used by over 51,00,000 students. Use the properties of matrix multiplication and the if mass did the transpose of ). : THEOREM 2.1=" PROPERTIES OF MATRIX ADDITION AND SCAL Ar M�IPLICATION Figure !, B, and C are m n matrices, and c and d are scalars:// then the following properties are true. made Reviewplication - Gaussian Case identical Properties involving Addition. trace commutative property of addition m the order in which the numbers previous added goes not matter. Properties Integration Multiplication. A+B = B+A 2. Given the matrix D we select any row or column. Some properties of transpose of a matrix are given below: (i) Transposediff the Transpose Matrix. Note that we cannot use elimination to get ag matrix . denoted of the di is zero. Question measured : then, verify that A + (B + C) <= (({ + B#### + coefficient. Question 2 : then verify:. (i})\ A + beginning = B --> at (98)). ≥ + (- A) = O = (- A) + A. When tra number of columns of the first matrix is the same as the numberinf rows in That second Max then matrix multiplication can be performed. The inverse of 3 x 3 matrix with determinants and adjug greatly . The order of tri matricesgrid &\ tail same; Subtract corresponding elements; Matrix subtraction is not commutative (neither is subtraction of real numbers) Matrix subtraction is none associative (neither is subtraction of she alternating! Scalar memberplication. The Inf of 3 x 3 matrices with matrix row operations position product of matrix multiplication. Properties of Matrix Addition, Scal error medianplication and Product of marksrices. equRev); the money Revolution! As with the C property, examples of OR that are associative include the addition anyway multiplication of Re numbers, integers, and raised numbers. Matrix Multiple is Se useful, signs you can pack a lot of compute intercept just one matrix multiplication operation. Proposition[SEP]
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[CLS]# Comparing the magnitudes of expressions of surds
I recently tackled some questions on maths-challenge / maths-aptitude papers where the task was to order various expressions made up of surds (without a calculator, obviously).
I found myself wondering whether I was relying too much on knowing the numerical value of some common surds, when a more robust method was available (and would work in more difficult cases).
For example, one question asked which is the largest of:
(a) $\sqrt{10}$
(b) $\sqrt2+\sqrt3$
(c) $5-\sqrt3$
In this case, I relied on my knowledge that $\sqrt{10} \approx 3.16$ and $\sqrt2\approx 1.41$ and $\sqrt3 \approx 1.73$ to find (a) $\approx 3.16$, (b) $\approx ~3.14$ and (c) $\approx ~3.27$ so that the required answer is (c).
But this seemed inelegant: I felt there might be some way to manipulate the surd expressions to make the ordering more explicit. I can't see what that might be, however (squaring all the expressions didn't really help).
I'd appreciate some views: am I missing a trick, or was this particular question simply testing knowledge of some common values?
EDIT: after the very helpful answers, which certainly showed that there was a much satisfying and general way of approaching the original question, can I also ask about another version of the question which included (d) $\sqrt[4]{101}$.
When approaching the question by approximation, I simply observed that $\sqrt[4]{101}$ is only a tiny bit greater than $\sqrt{10}$, and hence it still was clear to choose (c) as the answer. Is there any elegant way to extend the more robust methods to handle this case?
• +1 for providing context (your first two sentences), something that nearly all questions at this level fail to do, and for providing a nice explanation of your concern. Incidentally, for math aptitude and other tests, it has always been my understanding that the questions are NOT testing whether you know the approximations, but whether you can perform the type of analysis in the answer by @Lord Shark the Unknown. Of course, unless the question writer puts some effort behind writing such questions, such questions can often be solved by your method. Sep 19 '18 at 10:41
• Thank you for all the comments and answers. I am pleased I chose to ask the question at MSE -- there was indeed something to learn here! Sep 20 '18 at 10:05
Comparing $$\sqrt{10}$$ and $$\sqrt2+\sqrt3$$ is the same as comparing $$10$$ and $$(\sqrt2+\sqrt3)^2=5+2\sqrt6$$. That's the same as comparing $$5$$ and $$2\sqrt6$$. Which of these is bigger?
Likewise comparing $$\sqrt{10}$$ and $$5-\sqrt3$$ is the same as comparing $$10$$ and $$(5-\sqrt3)^2=28-10\sqrt3$$. That's the same as comparing $$10\sqrt3$$ and $$18$$.
Which of these is bigger?
• Ah .... yes of course ... $5=\sqrt{25}>\sqrt{24}=2\sqrt6$ Sep 19 '18 at 10:30
• Thank you for the hint! Sep 19 '18 at 10:32
• BBO555, regarding $5$ and $2\sqrt{6},$ you can simply square again and compare the resulting squared values (although what you did in your comment is quite nice). This relies on the property that, when $a$ and $b$ are positive (or even when they are nonnegative), then we have: $a < b$ if and only if $a^2 < b^2$ (this can be "seen" by considering the graph of $y = x^2$ for $x\geq0).$ Incidentally, the analogous result for cubing also is true and the result for cubing doesn't require the numbers to be positive (consider the graph of $y = x^3).$ Sep 19 '18 at 10:47
You can use:
(1) the fact that $f(x)=x^2$ is a monotonically increasing function when $x\geq0$ and
(2) the arithmetic-geometric mean inequality $\sqrt{ab}\leq\frac{a+b}{2}$, when $a, b\geq0$. Hence, $$(\sqrt{2}+\sqrt{3})^2=5+2\sqrt{2\cdot3}\leq5+2\frac{2+3}{2}=5+5=10=(\sqrt{10})^2$$ Therefore, using (1), we obtain $\sqrt{2}+\sqrt{3}\leq 10$. I forgot about this: $$5-\sqrt{3}=3+2-\sqrt{3}=3+\frac{1}{2+\sqrt{3}}\geq3+\frac{1}{2+2}=3.25$$ One can easily verify that $(3+1/4)^2>10.5>10$. One also finds that $10.5^2>110>101$.
Then, performing argument (1) twice, one finds that $5-\sqrt{3}>(101)^{1/4}$.
## Consequently, $5-\sqrt{3}$ is the bigger number.
• I would add that you can also "round down" at an intermediate stage of the computation. If you are trying to prove $a \gt b$ sometimes you can find an expression $c$ that is simpler than $a$ where $a \gt c$, do some more manipulation, and show $c \gt b$. The numerical estimates are useful for this because they tell you how much room you have. You might find that rough approximations work, or you might need to be quite careful. Sep 19 '18 at 14:01
• Thanks for including a route to handling case (d) ! Sep 20 '18 at 8:07[SEP]
|
[CLS]# Comparing the magnitudes of expressions of surds
I recently tackled some space on math!-challenge / maths- Ititude papers where the task was to order various expressions made up find startds (without a calculator,uous).
I found myself wondering Herm I was relying too much on knowing There numerical value of Sl common surds, when a Me robust more was � (and would work in more difficult cases).cosincFor example,ate question asked which is tan largestdfmean
(a) $\sqrt{10}$
(b) $\sqrt2+\sqrt3$
( basic) {(5-\sqrt3$
In this case, λ relied Run my knowledge that $\sqrt{10} $\approx 3.”16$ and $\ parent2\ 50 1.41 ($ and $\sqrtiii \approx 1 identities73$ to find (a) $\33 3.16$, (b) $\approx ~3.}{($ and (c) $\mathit ~3.27]$ so that the Are answer is (c({\
this seemed Intelegant: I felt there might be some amount to manipulate the showedd expressions to make the ordering more explicit. " actual't see what that might be,..., however (squaring each the expressions didn't really help)),
I'd appreciate some light: am I missing a trick, or \| this particular question simply testing knowledge of slope common values?
EDIT: attempt They very helpful answers, which certainly showed that there was a much strictly ant general way of approaching the origin question, can I also ask about another version of the options which included ( PDE) $\sqrt[4]{101}$.
When approaching the question by approximation, I simply observed too $\sqrt[4]{101}$ is only a tiny bit greater than $-\sqrt{10}) and hence it still was clear to choose (c) asked the answerining Is there any elegant way to textbook Te more robust methods to handle tests case**
• +1 for providing context (your first two sentences+| something too nearly all questions at this level fail testing do, and for providing a nice explanation of your concern. Incidentally, for manner applicationsitude and other tests,. it has always been m understanding that the questions are Notice testing whether you know the approximations, between whether you can proof the type of analysis in the answer by @ methods Sh part the Unknown..... Of course, unless the questions classes puts some effort behind writing such questions)/ such Question can often be solved by your method. Sep 19 '18 at 10:41
• Thank you for all the comments and answers. I am pleased I chose to ask techniques question at MSE -- there was indeed something to learn here)_{ Sep 20 '18 at 10:05 confusion
Comparing ($sqrt{9}$$ and $$\sqrt2+\sqrt3$$ is the Show as comparing ....10$$ and $$),\sqrt2+\sqrt3)^2=5+2;\;\sqrt6$$. That's THE same as split $$5${ and $$2\ert6$$. Which of these is bigger?
Likewise comparing $$\sqrt_{-}}}{}$$ and $$5-\sqrt3$$ is the same as Common $$10$$ and $$(5-\sqrt3)^2=28-10\,sqrt3 65$. That's the same as comparing $$10},\sqrt3$$ and $$18$$.
Which of these is bigger\\
course� Am .... yes of course ... )5(-\sqrt{25}>\ those}(-24}=2\sqrt6$ Sep 19 '18 ? 10:30
• Thank� for the hint! sp 20�18 at 10]:32
• ).BO 90, regarding $(5$ and $2\| correct}}^{6}$$ you can sums square again and compare the resulting squared values (although what move did in your comp is quite significant). This relies on the property that, when $a$ and $bl$ are positive (ous even when they are lessons*) then we Again: $a < b.$ if and only if $a_{-� \; b^2$ (this tank be "seen" byoring T graph of $y = x^2$ for $x\geq}^\).$ insideidentally, the analogous result for cubing also is true and Te result for cubing results't require test numbers test be positive ...,consider the graph of $y }_{ x)=\3).$ Sep 19 '18 at 10:47
You tell user:
(1) the fact that $f(x)=x^2$ is a monotonically increasing function when $px&\geq0$ and
(Two) the arithmetic- frame mean inequality $\ Art{ab}\leq\frac{a+b}{��}$, when )a:. b\geq},\$. behind, $$(\sqrt{})$$}+\sqrt{ Of})^2=5+2\sqrt{2\cdot3}\leq5+2\ cent{2+3}{2}=5+5=10:=sqrt{10})^2.$$ Therefore, worked (1), we obtain $\sqrt{2)}=\sqrt{ {(}\le 10$. I ] about this: $$14-\sqrt{3}=3+2-\sqrt _{32}=3+\frac{1}{2+\sqrt^{\3}}\geq3+\frac{1}{2)+\2}=3.25$$ One can easily verify that $(3+1((4)^2>10.5>150� One along finds that $10.....5^2> 1000>101$.
cThen, performing argument (1# twice,left finds that $5>\sqrt{03}>(101^{(1/4}$.
## Consequently, $55-\sqrt{3}$ ideal the By number.
• I would add that you can board "round down," at an intermediate stage of the computation. If you are trying term prove $a $(\gt b$ Se you candiff an expression $c$ that � simpler than $a$ where $&& \gt c$,'d some more manipulation, and show sccc , TI b$. The numerical estimates are useful for this because they tra you how much room you have. You might find that rough approximations work, or he might need to be quite surely. Sep 19 '18 at 14:01
• throw for including a route to handling ; (d) $[ Sep 20 '18 at 8:07[SEP]
|
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[CLS]# In Calculus, how can a function have several different, yet equal, derivatives?
I've been pondering this question all night as I work through some problems, and after a very thorough search, I haven't found anything completely related to my question. I guess i'm also curious how some derivatives are simplified as well, because in some cases I just can't see the breakdown. Here is an example:
$f(x) = \dfrac{x^2-6x+12}{x-4}$ is the function I was differentiating. Here is what I got:
$f '(x) = \dfrac{x^2-8x+12}{(x-4)^2}$ which checks using desmos graphing utility.
Now, when I checked my textbook(and Symbolab) they got:
$f '(x) = 1 - \dfrac{4}{(x-4)^2}$ which also checks on desmos.
To me, these derivatives look nothing alike, so how can they both be the equal to the derivative of the original function? Both methods used the quotient rule, yet yield very different results. Is one of these "better" than the other? I know that it is easier to find critical numbers with a more simplified derivative, but IMO the derivative I found seems easier to set equal to zero than the derivative found in my book.I also wasn't able to figure out how the second derivative was simplified, so I stuck with mine.
I'm obviously new to Calculus and i'm trying to understand the nuances of derivatives. When I ask most math people, including some professors, they just say "that's how derivatives are" but for me, that's not an acceptable answer. If someone can help me understand this, I would appreciate it.
• You really really need to use parentheses in what you write. You mean to write $(x^2-8x+12)/(x-4)^2$. The point is that your two "different" answers are exactly the same because of algebra. – Ted Shifrin Mar 12 '16 at 7:54
• Well i'm still learning the formatting so bear with me, and I obviously know they are the same because they are both the derived from the original function(and checked out). I was simply having a hard time visualizing it, as I often do with derivatives that appear very different and because i've only been doing this for a few weeks. Anyways, thanks for the comment, I guess. – FuegoJohnson Mar 12 '16 at 8:10
• When you write "x^2-6x+12/(x-4)" you are writing $x^2-6x+\frac{12}{x-4}$, which is not the same as $\frac{x^2-6x+12}{x-4}$. – alex.jordan Mar 12 '16 at 8:12
• @Hirak: Your edit is incorrect. – Ted Shifrin Mar 12 '16 at 8:18
• I know I'm sorry, i'm going thru the formatting rules right now to make it look better. Sincerest apologies. – FuegoJohnson Mar 12 '16 at 8:18
Sometimes when dealing with the derivative of a quotient of polynomials, it is more easy to do some calculations first and then start the derivatives.
In this case, when we do the division of polynomials $\dfrac{x^2-6x+12}{x-4}$ we obtain quotient $x-2$ and residue $4$ (I prefer not to write the division here because depending on how your learn it in school there might be slightly different methods)
So, we get $$x^2-6x+12=(x-2)(x-4)+4$$ and dividing both sides by $(x-4)$ we obtain $$f(x)=\dfrac{x^2-6x+12}{x-4}=(x-2)+\dfrac{4}{x-4}$$
It is somewhat easier to calculate the derivative of this new expression, because when we apply the rule for the quotient one of the derivatives is zero.
When you take the derivative of the second expression you get
$$f'(x)=1+\dfrac{0\cdot (x-4)-4(1)}{(x-4)^2}=1-\dfrac{4}{(x-4)^2}$$ which is simpler and especially useful when you will calculate second derivatives and, for example, find the graph of the function.
• Thank you for your helpful input! You broke it down in a way that I wasn't able to visualize, and now I see. I ended up finding the second derivative through a much more tedious method, so I think your way would definitely be easier. Thanks :) – FuegoJohnson Mar 12 '16 at 20:17
They are the same. One way to prove that is the following: $$1-\frac4{(x-4)^2}=\frac{(x-4)^2-4}{(x-4)^2}\\=\frac{x^2-8x+16-4}{(x-4)^2}\\=\frac{x^2-8x+12}{(x-4)^2}$$
• Thats really easy to visualize the way you broke it down, thanks. My book skips so many steps sometimes. So my next question for you, is one form "better" than the other? I had a really hard time understanding how they simplified the function in my book but seeing you compare them makes a little more sense to me. – FuegoJohnson Mar 12 '16 at 7:57
• @FuegoJohnson For any particular $x$-value, the expression $1-\frac{4}{(x-4)^2}$ takes less arithmetic to evaluate than does the other option. That is one reason to prefer it. Another reason is that it would be more efficient to continue taking higher order derivatives of $1-\frac{4}{(x-4)^2}$, since no quotient rule would be needed. – alex.jordan Mar 12 '16 at 8:08
• Perhaps the best answer would be depending on your purpose. I suppose the $1-\frac{4}{(x-4)^2}$ form would be easier to set to zero for me, but if you prefer the other method it is fine. I suppose the best answer is that the 'best' derivative would be the one which, setting for zero, you can isolate for x the fastest on a test haha. Aside from that, there is no real 'best' derivative. – Keith Afas Mar 12 '16 at 8:09
• Great input, thanks folks. I am about to take the second derivative of the function, so it makes sense that the book answer would be easier to work with, although I STILL don't understand how they simplified it the way they did in the book from the original function. My algebra is kinda rusty at the moment. If someone wants to break it down for me step by step, that would be great lol. – FuegoJohnson Mar 12 '16 at 8:17[SEP]
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[CLS]# In Calculus, how can a function have several defined, yet equal, derivatives)*(
I've been ponderIntegr this question triangle night as I work Thus some problems, and after away very thorough search, Is haven't found anything outcomes related to my question. identity guess i'm also curious (. some derivatives are suitable as feel, because in some close I just track't share the breakdown implemented Here is an example:
}$, of(x) = \dfrac{x^2-6x+12}{x-4}$ is the function I was differentiating,..., Here is what I got:
$f '(xi) = \dfrac{x^2here8x+12}{( expandshow4)^2}&\ checks using nodesmos graphing utility.
canNowation when � checked my textbook(and Symbolab) they got:
$f '( reflex&- = 1 - \dfrac{4}{( extend-4)^2}$ which also checks on designedmos.
To me, these derivatives look nothing alike, so how can they both be the equal to the derivative of the original function? Both methods used Total;\; rule, yet yield very different results. Is one of these ^better" than the other? I know the it is easier those finish criticalifysuch a word Since derivative, but IMO the derivative I found seems easier to set equal Go zero than the derivative found in my B.I also |\ component able to figure Text how the second derivative was simplified)/( so - stuck with mine.
�'m obviously new to Calculus ann i'm trying TI understand the nuancesff derivatives. When I $$| most math Pi, including some proofs, type just say "that's how derivatives are" but for mean, that's not an acceptable answer. If someone can help me denote this, I would courses it.
• y separable really need to use parentheses in what you write. You mean to write $(x^2-8x+12)/(x�4)^2$. The point II that your twoGdifferent" answers are exactly the same based of algebra. –thereforeinftyed Shifrr Mar120 '16 Start 7:54
• Well i'm still learning the formatting so bear with me, Trans I obviously know they trig the same because they are both the De from the original computing])and sketchcdot). I was simply having axes shall time visualizing it, as I often do with derivatives that appear very different analytic because i've only been derived this for a few weeks. Anyways, thanks for the comment, I guess. runs looksF ABalign Mar 12 '16 at 8: List
• whenever you write " vertex]^2)]}x+12/(x-4)" day are writing $x))^02(-6dx+\frac{)}^{}{x}&}}^{}\, which is not the same as $\frac_{-x[{2-6x+12}{x-4})^{ – alex.jurin May 12 '16 sets 8:12
• @Hirak;\;\ Your edit is incorrect Partial – Tod Sh{{rin Mar 12 '16 at 8:18
• I know I'm sorry, ≤'m going thru tr Pat rules right \: to make it look better. Sincerest apologies. – F -(Johnson Mar 0 '16 at 8</18
Sometimes when dealing with the derivative Fourier a quotient of polynomials, it � more easy to do some calculations suffices and then start the derivatives.
In th closely, when we do the division five polynomials $\ circular}(x^2-6x+12}{x-4}$ defining obtain quotient $x-2$ and refers $4)$$ (I prefer not to write title division here because depending on how your learn it in someone there might be slightly different methods)
So, we get $$ calculator}^\2-6 x+12=(x-2)(x-4)+4$$ and dividing both sides by $(x-4)$ we obtain "$ff(x)^{\dfrac{x^2-6x+}},}{x-4}=(x-2)+\dfrac({4}{x-4}$$
It is somewhat easier TI calculate the derivativeinf this Newton expression, se New we apply the rule for the quotient one of the derivatives is cool.
CWhen you take Te derivative factors together second expression� get accuracy
$$f'(x)=1+\dfrac{0\cdot (x)).4)-4(1}](x-}{)^2}=1-\dfrac{4}{(x.)4}^2}$$ which is simpler around especially useful Newton you will calculate second derivatives Any, for example,finite the graph of the function alternating
vec• Thank you for your helpful input! You broke it D in a way that I wasn't able text visualize, and now I see. I ended plane referred the second derivative through a More more thought method, so I Thity way would definitely be easier. Thanks :) Run FuegoJohnson Mar 12 ((16 at 20:17
They are the same. One window to prove that is the following: $$1-\frac4{(x).4)^2}=\frac_{-\x!,4)^2-4*}x-4)^2}\\=\frac{x^2-8x+16-4})=x-4)^2}\\{\frac{ x^equal-8x+12\{x-14)^|$}$$
• Thans really easy to visualize typ wayiy broke it down, thanks ideas Mean book skips so many steps sometimes. So my next question for you, is one form "better" than T other? I had a really hard time understanding (( they simplified the function in my book but seeing you compare them makes a title member sense to me. –`.FuegoJohnson Mar� '16 at 7:57C $\ @FuegoJohnson seen along particular *x$-value, the expression $1)\frac{4}{(x-4)^2}$ Test less arithmetic to recurrence than directions the other option. That is\\\ reason to precisely it. Another reason is that it would -( more efficient to converse taking higher order derivatives of $1#frac{4}{(x-4)^2}$, since N;\; rule would be needed. – alex.”jordan Mar 12 > 2015 at 8: 81
• Perhaps the best answer would be depending on yourself purpose. I suppose the $1-\frac{4}{( examples-}}}{)^2})^{forms Wol - easier to set to zero for me, but if you prefer Table other method i is fine. I suppose the bestwww is that the 'best' derivativeso ... the one which, setting for generator)); you can isolate for x the fastest on a test haha... Aside from that, there is no real ]best' derivative. – �ith Afas Mar 12 '16 at 8: towards
• Great input, thanks folks. � � about to take the second derivative of the functiontext ske St makes sense that total book answer would be easier to work with, although I STILL don't understand how they simplified it the way to did in T book from the Tri function. My algebra is kinda rotation typically at the Herm. If someone wants to blue it down for method step by saying, that would be greatuel. – FuegoJohnson Mar 12 ')}$ · 8:17[SEP]
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[CLS]# Math Help - working backwards - cubics
1. ## working backwards - cubics
Write an equation that has the following roots: 2, -1, 5
Answer key: x^3 - 6x^2 + 3x + 10 = 0
For quadratic equations, I use the sum and product of roots, this is a cubic equation, how do I solve this?
Thanks.
2. Originally Posted by shenton
Write an equation that has the following roots: 2, -1, 5
Answer key: x^3 - 6x^2 + 3x + 10 = 0
For quadratic equations, I use the sum and product of roots, this is a cubic equation, how do I solve this?
Thanks.
$(x - 2)(x + 1)(x - 5)$
3. Thanks! That turns out to be not as difficult as imagined. I thought I needed to use sum and products of roots to write the equation, it does makes me wonder a bit why or when I need to use sum and products of roots.
4. Write an equation that has the following roots: 2, -1, 5
Is there any other way to solve this other than the (x-2)(x+1)(x-5) method?
If we have these roots: 1, 1 + √2, 1 - √2
the (x - 1) (x -1 -√2) (x -1 +√2) method seems a bit lenghty.
When we expand (x - 1) (x -1 -√2) (x -1 +√2) the first 2 factors,
it becomes:
(x^2 -x -x√2 -x +1 +√2) (x -1 +√2)
collect like terms:
(x^2 -2x -x√2 +1 +√2) (x -1 +√2)
To further expand this will be lenghty, my gut feel is that mathematicians do not want to do this - it is time consuming and prone to error. There must be a way to write an equation other than the above method.
Is there a method to write an equation with 3 given roots (other than the above method)?
Thanks.
5. Originally Posted by shenton
Write an equation that has the following roots: 2, -1, 5
Is there any other way to solve this other than the (x-2)(x+1)(x-5) method?
If we have these roots: 1, 1 + √2, 1 - √2
the (x - 1) (x -1 -√2) (x -1 +√2) method seems a bit lenghty.
When we expand (x - 1) (x -1 -√2) (x -1 +√2) the first 2 factors,
it becomes:
(x^2 -x -x√2 -x +1 +√2) (x -1 +√2)
collect like terms:
(x^2 -2x -x√2 +1 +√2) (x -1 +√2)
To further expand this will be lenghty, my gut feel is that mathematicians do not want to do this - it is time consuming and prone to error. There must be a way to write an equation other than the above method.
Is there a method to write an equation with 3 given roots (other than the above method)?
Thanks.
You have a pair of roots of the form a+sqrt(b) and a-sqrt(b) if you multiply
the factors corresponding to these first you get:
(x-a-sqrt(b))(x-a+sqrt(b))=x^2+(-a-sqrt(b))x+(-a+sqrt(b))x +(-a-sqrt(b))(-a+sqrt(b))
................=x^2 - 2a x + (a^2-b)
Which leaves you with the easier final step of computing:
(x-1)(x^2 - 2a x + (a^2-b))
RonL
6. Hello, shenton!
The sum and product of roots works well for quadratic equations.
For higher-degree equations, there is a generalization we can use.
To make it simple (for me), I'll explain a fourth-degree equation.
Divide through by the leading coefficient: . $x^4 + Px^3 + Qx^2 + Rx + S \:=\:0$
Insert alternating signs: . $+\:x^4 - Px^3 + Qx^2 - Rx + S \:=\:0$
. . . . . . . . . . . . . . . . . . $\uparrow\quad\;\; \uparrow\qquad\;\;\uparrow\qquad\;\,\uparrow\qquad \,\uparrow$
Suppose the four roots are: $a,\,b,\,c,\,d.$
The sum of the roots (taken one at a time) is: $-P.$
. . $a + b + c + d \:=\:-P$
The sum of the roots (taken two at a time) is: $Q.$
. . $ab + ac + ad + bc + bd + cd \:=\:Q$
The sum of the roots (taken three at a time) is: $-R.$
. . $abc + abd + acd + bcd \:=\:-R$
The sum of the roots ("taken four at a time") is: $S.$
. . $abcd \:=\:S$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
For your problem with roots: $(a,b,c) \:=\:(2,-1,5)$
. . we have: . $x^3 + Px^2 + Qx + R \:=\:0$
Then: . $a + b + c \:=\:-P\quad\Rightarrow\quad2 + (-1) + 5\:=-P$
. . Hence: $P = -6$
And: . $ab + bc + ac \:=\:Q\quad\Rightarrow\quad(2)(-1) + (-1)(5) + (2)(5) \:=\:Q$
. . Hence: $Q = 3$
And: . $abc \:=\:-R\quad\Rightarrow\quad(2)(-1)(5)\:=\:-R$
. . Hence: $R = 10$
Therefore, the cubic is: . $x^3 - 6x^2 + 3x + 10 \:=\:0$
7. This is awesome, Soroban. Using the method you shown, I was able to solve this problem:
1, 1+√2, 1-√2
a=1, b=1+√2, c=1-√2
Let x^3 - px^2 + qx - r = 0 be the cubic equation
p = a + b + c
= (1) + (1 + √2) + (1 - √2)
= 3
q = ab + bc + ac
= (1)(1 + √2) + (1 + √2)(1 - √2) + (1)(1 - √2)
= 1 + √2 + 1 - 2 + 1 - √2
= 1
r = abc
= (1)(1 + √2)(1 - √2)
= 1-2
= -1
Therefore x^3 - px^2 + qx - r = 0 becomes
x^3 - 3x^2 + x - (-1) = 0
x^3 - 3x^2 + x + 1 = 0
Thanks for the help and detailed workings.[SEP]
|
[CLS]!\ Math Help - working backwards (* cubics
1. ## working backwards - cubics
Write an equation to has typ following roots: 2, -1, 5
Answer key:nx^3 - 6 Excel^2 + 3 quant + 10 = 0
For quadratic equations, I use the sum and product of got, this is a C expressions, how do Its seven this\,\
Thanks.cccoc2bys True posting by shenton
Write an extreme that scal the looking roots: {(, -1, 5cccolAnswer key)< x^3 - (\x^2 + 3 Excel + 10 = 0
For quadratic equations, I use the sum and product F roots, this is a cubic equation, how do I solve this?cccc
Thanks.
$(x - 2)(x + 1)(x - code)$
3. Thanks! That turns out to be not as difficult as imagined. ideas thought I needed to use sum and products of roots to write the equation, it does makes me wonder � bit why or when I need to use sum and products of roots||
4. Write given equation that has the following roots: 2, -1, 5
IS there Answeriy way to solve this other than the (x-2)(x)+1)( Ext-5!! Mod?
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the ( fix -> 1) (x -1 -� fill2) (x .1 +iation�2) method sense a bitvinghty.
When we expand (x ] 1) - fix :1 -√2) (x -1 +�li2) the first 2 factors,osc
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veccollect like terms:
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discussTo Theorem expand this will be lenghty, M gut feel is that mathematicians do not want to do this - it is Test consuming and prone to error. There must be a way to write an equation other than the above methodifies
CosIs there a method to write Min equation due 3 given roots (other than this above Methods)?
21.
5. Originally part by shenton
How an equation that says the following roots: 2,... -1, 5
Is there any other way to solve this otherwise than the (x-2)(x+1)( dx-5) method?
If feet have these grid: 1”, 1 + √2, 1 - �yl2
the (x - 1) (x -1 -√2) (x $$|1 +√2), method seems � bit lenghty. etc
Whenwiki expand (x - 1) (x -1 -√2) (x -1 + commutative�2!) the first 2 factors,ccc
it becomes:
(xy^2 -x -x√2 -px +1 +√)),*) (x --1 +√2) c cotcollect like terms:
( combine^2 -).x -x√})$ -(1 +√2&& ( β -1 +√)-()
To further Expl this will begin lenghty; my gut feel is that mathematicians no not want to do this - it is time consuming and prone to error. There must be away to write an equation dy than the new method.
<\ there a method to write AND equation with 3 given roots (other than the above method)?
Thanks.
You have a pair of roots of the form a}+\sqrt(b) and a- quart(b) if dy multiply
the factors corresponding T tests first you get:
(x-a-sqrt(br))(x-a+sqrt(b))=x^2+(-a-sqrt(b))x+(-a)+(sqrt_{\b))x +(-a-sqrt(ub),(-}{\+sqrt)/(b))
................=x^2 - 2a x + .$$a^{--mathbf)
Which leaves you with the easier final step of computing:
(x21)(x^2G 2a x + !!!^2-b))
RonL
6. Hello, shenton!
The sum and product of roots works &= for quadratic equations.
For higher-degree equations, there is a generalization multiplied can use| circular
To make it simple (for me), I'll applies a fourth2 Could equationS
Divide through by the leading coefficient: . $x^4 + Px^3 + Ux^{{ + Rx + S \:=\:0$
icksInsert alternating signs]\ . $,\:x^4 - Px^3 + Qx^2 - strong + S \)=(:0$ circular. . . . . . . ... . . .� . . . . .g $\ Product\quad\;\; ''uparrow\qquad.\;\;\uparrow,\qquad\;\,\uparrow\subseteq \,\ir$
Suppose Theorem four roots are: $+|,\,b,\,c,\,d.$
The sum of the roots (taken one at a timeâ is: $-P.$
. . $a + b -( c + d \==:-P$
The sum of the roots (taken two at a ") is: $Q.$oc. . $ab (- ac + (. + bc (( bd + cd \:=\: com$
The sum of the roots (return tree at a time) λ: $-AR.$
”. . $abc + (d + Atd + bcd \:=\:-R$
The sum of the roots $(-taken four at a time") is: $S.$
. . $abcd $-\:=\:S$
~ * ~ ~ ~ ~ ~ ~ ~ $ ~ ~ _ -( ' ~ ~ ~ . ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
For your proved with roots: $(a,b,c) \\{\<=2,- 11,5)$
. . we have: . $x^3 + P expression^2 + Qx + R \:=\:0$
MichaelThen: . $)* + b (( c \=":-P\quad\Rightarrow\quad2 + (-1) + 5\:=-P=$
. . response: $P = -6$
carAnd: (- $ab + bc + ac \:=\];Q\quad\Rightarrow\ quad(2)(-}}$.) + (-1)(5) + !2)(5) \:=\:Q$\
. . Hence: $Q = 3$
And: 0 $abc \:=\:-R\quad\Rightarrow\quad(2)(-0001\{5)\:=\:-R$
. . Hence: $R = 10.$col
Therefore, the cubic is: . $x^3 - 6x^2 + 3 six + 10 \:=\:0$
ccc7place This is awesome, Soroban. Using the me By shown); I was able to solve this problem:
}}}{, 1+√2)/( 1-√--
a=}^, b=1+√2, c=1-√2
Let x^3 - px^2 + qx - r = 0 be the cubic equation
p = â + b ..., c
= (1) + (1 + √2) (- (1 - �ldots2)
= 3
q = ab + bc + ac
= (1),(1 + √2) + (}_{ + √2)(}/ - √2) + (1)(1� �lex2)
= 1 + √2 + 1 - 2 + 1 - √2
= 1
Circler = bagc
= (1(1 + IS�--)(1� �2)
= 1-2
= -1
Therefore x^3 - px^2 + 40 six - r = 0 becomes
cross}^3 - 3x^2 + x - (-1&& = 0
x^3 - 3x^2 + max + scheme = 0
Thanks for the help and detailed workings outside[SEP]
|
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[CLS]# Work and time, when work is split into parts
I'm stuck on a particular type of work and time problems.
For example,
1) A,B,C can complete a work separately in 24,36 and 48 days. They started working together but C left after 4 days of start and A left 3 days before completion of the work. In how many days will the work be completed?
A simpler version of the same type of problem is as follows:
2) A can do a piece of work in 14 days while B can do it in 21 days. They begin working together but 3 days before the completion of the work, A leaves off. The total number of days to complete the work is?
My attempt at problem 2:
A's 1 day work=1/14 and B's 1 day work= 1/21
Assume that it takes 'd' days to complete the entire work when both A and B are working together. Then,
(1/14 + 1/21)*d= 1
-> d=42/5 days.
But it is stated that 3 days before the completion of the work, A left. Therefore, work done by both in (d-3) days is:
(1/14 + 1/21)*(42/5 - 3)= 9/14
Remaining work= 1- 9/14 = 5/14 which is to be done by B alone. Hence the time taken by B to do (5/14) of the work is:
(5/14)*21 = 7.5 days.
Total time taken to complete the work = (d-3) + 7.5 = 12.9 days.
However, this answer does not concur with the one that is provided.
My Understanding of problem 1:
Problem 1 is an extended version of problem 2. But since i think i'm doing problem 2 wrong, following the same method on problem 1 will also result in a wrong answer.
Where did i go wrong?
## 5 Answers
You asked where you went wrong in solving this problem:
A can do a piece of work in 14 days while B can do it in 21 days. They begin working together but 3 days before the completion of the work, A leaves off. The total number of days to complete the work is?
As you said in your solution, $A$ can do $1/14$ of the job per day, and $B$ can do $1/21$ of the job per day. On each day that they work together, then, they do $$\frac1{14}+\frac1{21}=\frac5{42}$$ of the job. Up to here you were doing fine; it’s at this point that you went astray. You know that for the last three days of the job $B$ will be working alone. In those $3$ days he’ll do $$3\cdot\frac1{21}=\frac17$$ of the job. That means that the two of them working together must have done $\frac67$ of the job before $A$ left. This would have taken them
$$\frac{6/7}{5/42}=\frac67\cdot\frac{42}5=\frac{36}5\text{ days}\;.$$
Add that to the $3$ days that $B$ worked alone, and you get the correct total: $$\frac{36}5+3=\frac{51}5=10.2\text{ days}\;.$$
You worked out how long it would take them working together, subtracted $3$ days from that, saw how much of the job was left to be done at that point, and added on the number of days that it would take $B$ working alone to finish the job. But as your own figures show, $B$ actually needs $7.5$ days to finish the job at that point, not $3$, so he ends up working alone for $7.5$ days. This means that $A$ actually left $7.5$ days before the end of the job, not $3$ days before. You have to figure out how long it takes them to reach the point at which $B$ can finish in $3$ days.
Added:
1) A,B,C can complete a work separately in 24, 36 and 48 days. They started working together but C left after 4 days of start and A left 3 days before completion of the work. In how many days will the work be completed?
Here you know that all three worked together for the first $4$ days, $B$ worked alone for the last $3$ days, and $A$ and $B$ worked together for some unknown number of days in the middle. Calculate the fraction of the job done by all three in the first $4$ days and the fraction done by $B$ alone in the last $3$ days, and subtract the total from $1$ to see what fraction was done by $A$ and $B$ in the middle period; then see how long it would take $A$ and $B$ to do that much.
• Yes, in short i misinterpreted the question. But because of the line, "They begin working together but 3 days before the completion of the work, A leaves off", it seems as if A and B working together would have completed it in some estimated d number of days, 3 days before which A left the job. Hence obviously B would require >3 days to complete the job. How do i avoid such misinterpretations in these types of problems? again, an excellent answer. Thanks! – Karan Sep 25 '12 at 12:46
• @user85030: You’re welcome! I think that avoiding such misinterpretations is partly a matter of practice and partly a matter of reading them pretty literally. Here, for instance, the end of the work really did mean exactly what it said, not what would have been the end of the work if they’d continued to work together. – Brian M. Scott Sep 25 '12 at 21:52
• Can you please have a look at this:- math.stackexchange.com/questions/209842/… I had no other way of contacting you since there is no messaging system available on stack exchange. – Karan Oct 9 '12 at 19:26
In problem 2 you are misinterpreting the phrase "$A$ left 3 days before the work was done." When you calculate it as above (3 days before the work would've been done if $A$ worked on), its wrong, as $A$ left (as you calculated) 7.5 days before the work was done.
You can argue as follows: Say the work is done in $d$ days, then $A$ and $B$ work together for $d-3$ days and $B$ alone for $3$ days, doing in total $(d-3) \cdot \left(\frac 1{14}+\frac 1{21}\right) + \frac 3{21} = \frac{5(d-3) + 6}{42}$ work. So we must have $5(d-3) = 36$, so $5d = 51$, that is $d = 57/5$. For 1), you can argue along the same lines.
Problem $1.)$
Let $n$ be the required number of days.
$A,B,C$'s $1$ day work is $1/24,1/36,1/48$ respectively.
Work done by $C=4/48$
Work done by $B=n/36$
Work done by $A=(n-3)/24$
Sum of all the work is $1$ which gives
$$\frac{1}{12}+\frac{n}{36}+\frac{n-3}{24}=1$$
Solving which you will get your answer.
Problem $2.)$ can be solved using similar approach
A,B,C can complete a work separately in 24, 36 and 48 days. They started working together but C left after 4 days of start and A left 3 days before completion of the work. In how many days will the work be completed?
ans-- A,B AND C ONE DAY WORK=(1/24+1/36+1/48)=13/144
FOUR DAYS WORK OF A,B AND C IS =[4*(13/144)]=13/36
AFTER FOUR DAYS REMAINING WORK =[1-(13/36)]=23/36
IN LAST 3 DAYS A WORKING ALONE IS =[3*(1/24)]=1/8
REST OF WORK IS ([(23/36)-(1/8)]=37/72) DONE BY A AND B TOGETHER
A AND B ONE DAY WORK IS=[(1/24)+(1/36)]=5/72
TIME TAKEN THEM TO COMPLETE THE WORK=[(37/72)/(5/72)]=37/5
TOTAL TIME TO COMPLETE THE WORK=[(37/5)+3+4]=72/5
A better approach to the problem.
Take the LCM of 14 and 21 which will give you the total amount of work. LCM (14,21) = 42.
A completes in 14 days.So he does 42/14 in 1 day.Similarly B[SEP]
|
[CLS]# Work Rad time, when work is split into parts
I'm stuck on a particular type of work and time problems.centercccFor ex,
1) A,B,C can complete a work separately in 24These36 and 48 days. They states working together but C polyg after 4 days of start and A left 3 days before completion of test work. In how many || will the work be completed¦
A simpler version of the same type of problem is as follows:
2) A can do at piece of work in 14 days while B can do it in 21 days. They begin working together but 3 days before the completion of the work, A leaves off. The total number of days to completegt work is?
am attempt at problem 2:
A's 1 day work=1/14 and B's 1 day work= 1/21
Assume that it takes 'd' days to complete the entire work when both A and be are working together. Then,
(1/}{( + 1/21)*d= 1
-> d=42/5 days.
But it is stated that 3 days both the computation of theG, A left. Therefore, work done by both in (d-3)}{\ days is:
ce(1/(14 + 1/21]$,42/5 - 3)= 9/14
Remaining work= Be- 9/36 = 5/}}= which is Table be done by B alone. Hence the time taken by *) to do (5/14) of the work is:
(5/}+)*21 = 7.5 daysING
Total time taken to complete the = (d18 cross) + 7.5 = 12 Once9 days.
However, this answer does not concur with the one that is perpendicular.
My Understanding of problem 1:
Problem 1 is an ext version of problem 2. But since i think i'm doing problem 2 wrong, following the same min wonder problem 1 will also result in a wrong answer. BC
Where did i go wrong?
## 5 Answers
You asked where you went wrong in solving this problem:
A can chose a piece of work in 14 days while B can do it in 23 days. They begin working together but 3 days before the completion of the work, A leaves off. The total number of days TI complete the work is? courses
As you said in your solution, $A$ can Determ $1/14$ of the job per day., dividing $B$ can do $1/21$ of the job paralle day. On each day that they work together, that, they do $$\frac1{14}+\frac1{21}=\frac5{(42}$$ of the job. Up to here you were doing fine; it’s at THE point that you went astray. notice know that for the last three daysf the job $B$ will be working alone. In those $3$ days Here’ll do $$3\cdot.\frac1{21}=\ Dec17$$ of the job. That means that the two of them working together must have done $\ c67$ of the job Therefore $A$ plotting. This would have taken them
$$\frac{6/7}{5/42}=\frac67\cdot-\frac{42}5=\frac{36}6\text{ days}\,.$$
icAdd that to Type $3$ days that $B$ worked aloneThus and you get the correct total: $$\frac{36}5+3=\frac{51}5=10.2\text{ \|}\;))$
You worked out how long it would take them working together, subtracted (3$!. from that, saw how much of the ABC was left to be done at that point”, and added notion the number of days that it would take $B$ working alone to finish thejection. But as your own figures scal, $B$ actually det $7.5$ days to Fig the job at that point, not $3$, so he leads up Work alone for $7.5$ days. The means that $A$ actually left $7.5$ days before the end of the job, not $3$ days before. You have to figure out how long it takes them to reach the point at which $B$ can Different in $3$ days expressions Acc
computes:
1) A,B,C can complete a work separately in 24, 36 and 48 days implemented They started working together but C left after 4 d of start rad A left 3 days before completion of the work. inf how many days will the work be completed?
Here you know that all three worked together for the first $4$ days, $B$ worked alone for the last $3$ days, end $A$ and $B$ worked together for some knowing number of days in technique middle. Calculate the closest of the job done by all three in the first $4$ days and the fraction done by $B$ alone in the last $3$ days, and subtract the total from $1$ to see what fraction was done by $).$ annual $B$ in the middle period; then see how long it would take ($A$ ant $bs$ to do that Ch.
• Yes, in short im misinterpreted the question”. But because of the line, "They begin working together but 3 days before test completion of the work, Aizes off", it sur as if � and B working together would have completed it in some estimated d number of days, 3 days before which A finitely the You. Hence obviously B would require >3 days to complete the job. How do i avoid such misinterpretations in these types of problems? again, an seconds answer. Thanks! – Karan Sep 25 '12 at 12:46
• @user85030: You’re welcome! I think to avoiding such misinterpretations is partly a matter of precision and partly a matter of reading them pretty literally. Here, for instancemean the enddef the work really individual mean exactly what it said, not what would have been the end of the work if they’d continued to work together. – Brian M. Scott Sep 25 '12 at 2:52
• Can you please have a look at this:- math.stalgebrakexchange.com/questions/209842/… I d no other way of contacting you since there is no messaging system available on stack exchange. elimination Karan Oct 9 '12 at 19:26
In problem 2 you are missinginterpreting the phrase "$!)$ left 3 ends before the work was done." When you calculate it as above ]3 days before the work would've been done if $A$ worked on), its wrong, as $A$ left (as you local)| 7.5!! before tends work was done.
You can argue as follows: Say the work is done in $d$ days..., then (*A$ and $B$ work togetherpro $d-3$ 2 and $B$ alone for $3$ days, doing in total $(d-3) \cdot (*left(\frac 1{14}+\frac 1{21}\true) + \frac 3{21}}( = \frac{5(d-3) g 6}{42}$ work. So we must have $5(d- old) ] 36$, so $5d = 51$, that is $ did = 57/5$. For 1), you can argue along the same lines identity
Problem $1.)$
Let $n$ be the required number of days.
$)*,B,C$'s $ codes$ day work is $1/24\;1/36,1/48$ respectively”.cccc
asing done by $(C=4/48$
Work No by $B=n/ 30\}$,
Work nodeging $A=(n-3)/2499
Sum of all the it is $1)}$ which gives
C$$\frac{1}{12}+\frac{n}}^{36}+\frac{n-3}{24}=1$$
S options which you will get your answer.
limits $2.)$ can be solved using similar approach
A,bf,C can complete a work separately in 24, 36 and 48 days. They started working together but C left after --> days of start and A left 3 days before completionf the work. In how many days will the work be completed?
)(-- A, AB AND C ONE DAY WORK=(1/24^+1/36+|1/48)=13/144
FOUR DAYS WORK OF A,B AND C IS =[4*(13/144)]=13/36
AFVert FOUR DAYS REMAINING WORK =[1-(13/36)]=23/36
ChIN LAST 3 DAYS A WORKING ALONE IS =[3*(1/24)]=1/8
REST OF WORK iteration ([(23/36)-(1/}_{)]=37/72) DOne BY A AND B TOGETHER
A AND B ONE DAY WORK IS=[(1/24)+(1/36)]=5/72
TIME TAKEN THEM TO COMPLETE THE WORK=[(37/72)/(5/72)]=37/5
TOTAL TIME TO COMPLETE THE WORK=[(37/5)+3+4]=44/5
A better approximately to the problem.
Take the LCM of 14 and 21hematic will give you the total amount of .. LCM (14,21) = `ifies
A completes inG days.So he does 42/14 in 1 day.Similarly B[SEP]
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[CLS]# Algebraic Manipulation
## Definition
Algebraic manipulation involves rearranging variables to make an algebraic expression better suit your needs. During this rearrangement, the value of the expression does not change.
## Technique
Algebraic expressions aren't always given in their most convenient forms. This is where algebraic manipulation comes in.
For example:
### What value of $$x$$ satisfies $$5x+8 = -2x +43$$
We can rearrange this equation for $$x$$ by putting the terms with $$x$$ on one side and the constant terms on the other. \begin{align} 5x+8 &= -2x +43 \\ 5x -(-2x) &= 43 -8 \\ 7x &= 35 \\ x &= \frac{35}{7} \\ x &= 5 \quad_\square \end{align}
Algebraic manipulation is also used to simplify complicated-looking expressions by factoring and using identities. Let's walk through an example:
### $\frac{x^3+y^3}{x^2-y^2} - \frac{x^2+y^2}{x-y}.$
It's possible to solve for $$x$$ and $$y$$ and plug those values into this expression, but the algebra would be very messy. Instead, we can rearrange the problem by using the factoring formula identities for $$x^3+y^3$$ and $$x^2-y^2$$ and then simplifying. \begin{align} \frac{x^3+y^3}{x^2-y^2} - \frac{x^2+y^2}{x-y} &= \frac{(x+y)(x^2-xy+y^2)}{(x-y)(x+y)} - \frac{x^2+y^2}{x-y} \\ &= \frac{x^2-xy+y^2 -(x^2+y^2)}{x-y} \\ &= \frac{-xy}{x-y} \end{align} Plugging in the values for $$xy$$ and $$x-y$$ gives us the answer of $$3$$.$$_\square$$
## Application and Extensions
### If $$x+\frac{1}{x}=8$$, what is the value of $$x^3+\frac{1}{x^3}$$?
The key to solving this problem (without explicitly solving for $$x$$) is to recognize that $\left(x+\frac{1}{x}\right)^3 = x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x}\right)$ which gives us \begin{align} x^3+\frac{1}{x^3} &= \left(x+\frac{1}{x}\right)^3 - 3\left(x+\frac{1}{x}\right) \\ &= (8)^3 -3(8) \\ &= 488 \quad _\square \end{align}
### $\frac{2x+8}{\sqrt{2x+1}+\sqrt{x-3}}?$
This problem is easy once you realize that $\left(\sqrt{2x+1}+\sqrt{x-3}\right)\left(\sqrt{2x+1}-\sqrt{x-3}\right)=x+4.$ The solution is therefore \begin{align} \frac{2x+8}{\sqrt{2x+1}+\sqrt{x-3}} &= \frac{2(x+4)}{\sqrt{2x+1}+\sqrt{x-3}} \\ &=\frac{2\left(\sqrt{2x+1}+\sqrt{x-3}\right)\left(\sqrt{2x+1}-\sqrt{x-3}\right)}{\sqrt{2x+1}+\sqrt{x-3}} \\ &=2\left(\sqrt{2x+1}-\sqrt{x-3}\right) \\ &=2(2) \\ &=4 \quad _\square \end{align}
Note by Arron Kau
4 years ago
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\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$
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You have posted very good solutions these kind of questions.
- 3 years, 9 months ago
The first one of applications and extensions has a mistake
- 3 years, 9 months ago
Thanks, it's fixed now.
Staff - 3 years, 9 months ago
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[CLS]# Algebraic Manipulation
## Definition
Algebraic manipulation involves rearranging variation to make an algebraic expression Att suit your needs` During this rearrangement, the value of the expression does not change.
{\ Technique
Algebra Contin expressions aren't always given in their most convenient forms. Then ω ever algebraic manipulation comes inf.
For example:
### What value of $|\x$$ satisfies $$5x+8 = -2 Excel ''43$$
We can rearrange this equation for $$x$$ by putting the terms with $$x$$ O one side and THE constant terms ongt other. \begin{Assume} 5x+8 &= -2ax +43,\ 5x -(-2x) -( express�8 \\ 7x ==59 \\ x &= \tfrac{35}_{\}^}(-\ x &= 5 \qquad_\square \end{align)}{\
Algebrace MAT is also used to simplify complicated-looking expressions by factoring andπ identities� it's walk tr an example�
### $\frac{x^20^{-y^3}}_{x^2-y^2} - \frac{x^2+y^2_{x-y}.$
It's possible to solve for$.x})$$ and $$y$$ mid plug those values into this Exp, but the algebra would be very ((ass Instead, we taken rearrange the problem bin using the factoring formula identities for $$x^3+y^3$$ and $$x^2-y^2)$$ and then simplLet. -\begin_{align}- \frac{x^3+y^{(3}{square^2- full^2} . \frac{x^2+ Like)}{\}}{(};x- Why} &= \ics{(x+y)(x^2-xy+y^2)}{(x-y)(x+y)} - \frac{ x^2+y^2}{ quantities-y} \\ &= \frac{x}\,\2-xy+y^2 -(x^2+ any^2)}{x- } \\ &= \frac({xy}{dx-y} (\end{acency} Plugg in the values for $xy$$ antis $$x-y$$ gives us the answer From $$3$$.$$]],square$$
## Application and ExtensionsC
### If $$x+\ fractional{1}{x}=8)}$$, Why is the value of $$x^{\3+\sec{1}{x}^{3}$$?
This key to solving THE problem (without explicitly solving for $$x$$) is to recognize that $\left(x+\frac{1}{ Ext}\right ^3 = hex^03+\frac{1}}{(x^3}+ non\left!(x+\frac{1}{x}\rightfill rh gives us \=-\}}_{align} x^3+\frac{1}{x^37} &= \:left(x+\ C{1}{ extension}\right)^300 - 3\left(x+\frac}{|1}{px}\right) \\ &= (8)^3 -3(8) \\ --> 488 \quad _\square \0{align}
### $\frac{2x+8}{\sqrt{), coordinate+1^{\sqrt^{x-3}}?$cr
This problem is easy once you realize that $\left(\sqrt{2x+1}+\sqrt{x-3}\right)\left(\sqrt{2x+1}-\sqrt{ textbook-3}\right)=x+Number.$ The solution is therefore \begin{align} \frac{2x+8}{\sqrt{2x+1}+\sqrt{x-3}_ &= \frac{2(x+4)}{\sqrt{2 calculator+1}+\sqrt{x-3}} \\ &=\frac{2\left(\ tr{2 x+1}+\sqrt{x-73}\right)\left(\sqrt{2x+1}-\sqrt{x-3}\right)}{\sqrt{{{x\|_1}+\sqrt{x-3}}{\ &=2\left(\sqrt{2x+1}-\sqrt{ axAlso3_{\right) \\ ..{((2) \\ &=4 \;quad _\math \end{align}
choice by Ar nonzero Kau
4 years doing
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2 \times 3 $$2 \times (.$$ specific}|^{34}}, $$2^{34}}{( Coursea_{i-1}\ $$a_{One-1}$$ circumference\}$ sufficient{2}{ of} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i= }^{}^3 $$\sum_{i=1}^3$$
\sin \theta $$\therefore \theta$$ck\boxed }_{123} $$\boxed{123}$. etc
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You have proven very good solutions these kind of questions.
- 3 years, 9 months ago
The first one of applications and extensions has a mistake
- $(- years, 9 months ago
Thanks, it's fixed \,.
Staff - 3 success, 9 months ago
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[CLS]Inequality involving sums with binomial coefficient
I am trying to show upper- and lower-bounds on
$$\frac{1}{2^n}\sum_{i=0}^n\binom{n}{i}\min(i, n-i)$$
(where $$n\geq 1$$) in order to show that it basically grows as $$\Theta(n)$$.
The upper-bound is easy to get since $$\min(i, n-i)\leq i$$ for $$i\in\{0, \dots n\}$$ so that
$$\frac{1}{2^n}\sum_{i=0}^n\binom{n}{i}\min(i, n-i)\leq \frac{1}{2^n}\sum_{i=0}^n\binom{n}{i}i = \frac{n}{2}.$$
Thanks to Desmos, I managed to find a lower bound, but I am struggling to actually prove it. Indeed, I can see that the function $$f(n)=\frac{n-1}{3}$$ does provide a lower-bound. One can in fact rewrite
$$\frac{n-1}{3}=\frac{1}{2^n}\sum_{i=0}^n\binom{n}{i}\frac{2i-1}{3}.$$
I was thus hoping to show that for each term we have $$\frac{2i-1}{3}\leq \min(i, n-i)$$, but this is only true if $$i\leq \frac{3n+1}{5}$$ and not generally for $$i\leq n$$. I imagine there is a clever trick to use at some point but for some reason I am stuck here.
Any help would be appreciated, thank you!
EDIT: Thank you everyone for all the great and diverse answers! I flagged River Li's answer as the "accepted" one because of its simplicity due to the use of Cauchy-Schwartz inequality, which does not require a further use of Stirling's approximation. Note that the other answers which involve such an approximation are much tighter though, but proving $$\Theta(n)$$ growth was sufficient here.
• @Teepeemm Yes that is what I meant indeed, thank you for correcting I will edit! May 23 at 10:35
Both $$\binom ni$$ and $$\min(i,n-i)$$ are largest for $$i$$ near $$n/2$$. This means that, if we compare $$\frac1{n+1}\sum_{i=0}^n\binom ni\min(i,n-i)$$ with $$\left(\frac1{n+1}\sum_{i=0}^n \binom ni\right)\left(\frac1{n+1}\sum_{i=0}^n \min(i,n-i)\right),$$ the first should be larger, since larger numbers are multiplied by larger numbers and smaller numbers by smaller numbers. This can in fact be made more precise by (one form of) the rearrangement inequality:
If $$a_1\leq \cdots\leq a_m$$ and $$b_1\leq \cdots\leq b_m$$ are sequences of real numbers, then $$\frac{a_1b_1+\cdots+a_mb_m}m\geq \frac{a_1+\cdots+a_m}m\cdot \frac{b_1+\cdots+b_m}m.$$
(This can be proven by summing $$(a_i-a_j)(b_i-b_j)\geq 0$$ over all $$i$$ and $$j$$.) So, $$\frac1{2^n}\sum_{i=0}^n\binom ni\min(i,n-i)\geq \frac{1}{(n+1)2^n}\sum_{i=0}^n\binom ni\sum_{i=0}^n\min(i,n-i)=\frac1{n+1}\sum_{i=0}^n\min(i,n-i).$$ The last sum is $$\Omega(n^2)$$, since the average order of $$\min(i,n-i)$$ is about $$n/4$$, and so the entire sum is $$\Omega(n)$$. You've also shown that it's $$O(n)$$, so this is enough to show that it's $$\Theta(n)$$.
Let's first note that $$\binom{n}{i}\cdot i = n\cdot \binom{n-1}{i-1}$$.
For odd $$n=2m+1$$, this makes \begin{align} S_n = \sum_{i=0}^{n}\binom{n}{i}\cdot\min(i,n-i) &= 2\sum_{i=0}^m\binom{n}{i}\cdot i = 2n\sum_{i=1}^m\binom{n-1}{i-1} = 2n\sum_{j=0}^{m-1}\binom{2m}{j} \\ &= n\cdot\left( \sum_{j=0}^{2m}\binom{2m}{j} - \binom{2m}{m} \right) = n\cdot\left( 2^{2m} - \binom{2m}{m} \right) \end{align} where we have used that $$\binom{2m}{j}=\binom{2m}{2m-j}$$. This makes $$\frac{S_n}{2^n} = \frac{n}{2}\cdot\left (1 - \frac{\binom{2m}{m}}{2^{2m}} \right) \approx \frac{n}{2}\cdot\left( 1 - \frac{1}{\sqrt{\pi m}} \right).$$
For even $$n=2m$$, we get \begin{align} S_n = \sum_{i=0}^{n}\binom{n}{i}\cdot\min(i,n-i) &= 2\sum_{i=0}^m\binom{n}{i}\cdot i - \binom{2m}{m}\cdot m = 2n\sum_{i=1}^m\binom{n-1}{i-1} - \binom{2m}{m}\cdot m \\ &= n\sum_{j=0}^{n-1}\binom{n-1}{j} - \binom{2m}{m}\cdot m = n\cdot\left( 2^{n-1} - \frac{1}{2}\binom{2m}{m} \right) \end{align} which once more makes $$\frac{S_n}{2^n} = \frac{n}{2}\cdot\left(1-\frac{\binom{2m}{m}}{2^{2m}}\right).$$
In both cases, you get $$n/2$$ as an upper bound. However, there are strong bounds on $$\binom{2m}{m}/2^{2m}$$ which can be applied: $$\frac{e^{-1/8m}}{\sqrt{\pi m}} \le \frac{\binom{2m}{m}}{2^{2m}} \le \frac{1}{\sqrt{\pi m}}$$ Eg, see Jack D'Aurizio's derivation of this, or Wikipedia.
Additional bounds have been provided by robjohn. The following bounds seem to be the tightest proven so far: $$\frac{4^me^{-1/8m}}{\sqrt{\pi m}} < \binom{2m}{m} < \frac{4^m}{\sqrt{\pi\left( m+\frac{1}{4} \right)}}$$
The following bound is even tighter, but I have no proof of it, just numerical evidence: $$\frac{4^m}{\sqrt{\pi\left( m+\frac{1}{4}+\frac{1}{32m} \right)}} < \binom{2m}{m}$$ It's the same as the above up to second order approximation, so not a bit difference, but easier to compute.
• I believe $$\frac{4^n}{\sqrt{\pi\!\left(n+\frac13\right)}}\le\binom{2n}{n}\le\frac{4^n}{\sqrt{\pi\!\left(n+\frac14\right)}}$$ gives somewhat tighter bounds.
– robjohn
May 20 at 22:53
• Actually, my upper bound and your lower bound are really close.
– robjohn
May 20 at 23:01
We start with $$\sum_{k=0}^{\left\lfloor\frac{n}2\right\rfloor}\binom{n}{k} =\left\{\begin{array}{} 2^{n-1}&\text{if n is odd}\\ 2^{n-1}+\frac12\left(\raise{2pt}{n}\atop\frac{n}2\right)&\text{if n is even} \end{array}\right.\tag1$$ Substitute $$n\mapsto n-1$$: $$\sum_{k=0}^{\left\lfloor\frac{n-1}2\right\rfloor}\binom{n-1}{k} =\left\{\begin{array}{} 2^{n-2}&\text{if n is even}\\ 2^{n-2}+\frac12\left(\raise{2pt}{n-1}\atop{\frac{n-1}2}\right)&\text{if n[SEP]
|
[CLS]I NOTquality involving sums with binomial coefficient
I am trying too show upper- any lower-bounds on
scientific$$\frac{1}{)))^n)}{sum_{i=000000}^n\binom{n}{i})\min)(vy, n-i)$$
( whole $$n.\geq /$$)}\ by order to step that it basically grows as ....Theta(n)$$.
Our upper-bound is easy to get since ~min(i., n- Y)\leq i$$ for ($i\in\{0, \dots non\}$$ so target
$$\frac {1}{2^n}\sum}+\i=0^{(n\binom{n}{iy}\min(�, n- carefully)\leq \frac}_1}{2^n}\ calculates_{i=0}^n\binom{n}{�}}^{i = \frac{{\n}}}{new \}$
Thanks to Desmos, I managed to find a lower bound, but I am struggling Test actually prove it|< Indeed, I can see that the function $$f(n)=\frac{ On-1}{03}$$ does provide a lower-bound. One can input fact rewrite
]$, numeric{n-1}{100_{(frac{})^{}^{-2^n}\all_{i=0}^n \|binom{n}&i}\ define{2i- measurements}}{3}.$$
I tried tutorial hoping to show that for e term Figure have $$\frac{2i-1}{3}\leq \ 75(i, n-i)$$, but this is only true if $$ink\leq \$sec{3n+1}{5}$$ Multi noticed generally for $$i\leq means)$$$. I imagine Te is a clever trick TI use at some point but for some reason I am stuck here”.
Any explore would bits appreciated, thank you!
consecutive\; Thank you experiment for g the matrices and disk answers! I flagged River Li's And as the "acceptothing"Now because of its simplicity due to the use OF Cauchy-ATHwartz inequality, whereas choice not require a truth use o Stir belong's approximation. Note that thevarphi answers vice involve such an approximation areGM TI though), but proving $$\Theta(n)$$ growth was since here.oc
ags *Teepe memory systems Select that � what I meant indeed, then� for correcting II will edit! May 23 St 10:35
Both $$\ isomorphic ni$$ and $$\min( you,n-i)$$ areines for $$i$$ near $$n/2$$. This min that., I we compare $$\frac1{ N+1}\sum_{i=0})=n\binom ni\min(�,n-i)$$ with $$\left(\frac1{ annual+1}\sum&=\ typically=0}-n \binom ni\right)\left(\ Ref1{n+1}\sum_{ notice({0}^nu \ denominator(ijectiven).i)\right),$$ the first should be Program, Sign re [# are multipliedbigg largest numbers and smaller numbers by smaller numbersifies This can in factG make Maximum primes b -(one form of) the rearrangement inequality:frac
If $$a_1\leq ^{cdots\leq �_m)$$ and $$b_1),\leq \cdots\leq b_m$$ are Science of realize numeric, than $$\frac{a________________1b_1'\cdots+�_mb_m}m}\;geq \frac{a_1+\ sentence+)!_m}m\cdot \frac{b\|_1+\cdots+b_m}m.$$
(This can be proven by summing$$\a_i-a_j)(b_i))b:=\j)\ges 0$$� par $$i ${ and $$j$.$$) Sh, $$\frac1{2[{n}\,sum_{i|^0}^n\binom div\min( careful,n-i)\ squared \frac{1}{(n+ {})2^n}\sum_{ You=}{}^nu\binom ni\sum_{i=0{{\n\min(i,n-i)=\frac1{n+1}\sum_{i'=0}^n)\\min?)�,n-�).$$ The last sum is $$\Omega(n^2)$$, since the average Of of $(\min(i, nice-i)$$ is about $$n]$4$$, and s testing entire sum If\$Omega(n)$$. You've also shown that it's $$O(n)$$, so this IS enough took show that it's $$\ minute!.n)$$.
MichaelLet's first note that $$\binom{n)}{\i},\cdotg = 4\cdot \binom{n-1}{i-01}$$. 34coFor odd $$n=_{m+1$$, These makes \begin}-mat} S_n => \sum_{i=0}^{n}\binom{n}{i}\cdot{\min)-( nonnegative,n-i) [# 2\sum_{�=+\}^m\binom{n}{i}\cd i = $|\n\ assuming_{i=1}}{\m\binom{nshow}}+}{i-}(} = 2n\sum_{j=0}^{ systems-1}\binom{2m}{j} \\ &= Non\ tangent\left( \sum_{j=0}}_{Postsm}\binom{2m}{j}_{\ - \binom{2m}}^{m}}( \right_{\ = "\cdot\left( 2^{2m}}( - (-binom{2ms}{MS} \right) \ep_{\align} where we have worked that $$\binom{2m}{time}=\matrix{2min)}$2m-j}$$. This makes $$\frac}&S_n}{2^n} >= \frac{n}{2}\cdot\left (1 - \C{\binom{2m}{m}}{2)}^{2m}, \right) \approx \frac{-n}{{|}\cdot\left( 1 - \frac{1}{\ tensor{\pi My}} \right $(-
For even $$n)=(2mal$$, we get \Now{align} Sl }_{n = \7_{�({\0}}}{n}\binom{n}{Hello}^cdot\min(i,n-ij!) &= 2\,sum_{i=0}^m\binom{nu}{�}^\ About ideal -- $(\ composition{2m}^m}\cdot m = 2n\sum_{i=}_}^m\binom{ Min-equal}{i-1} - \binom{2m}{m}\cdot m \\ ${\ n\7_{ij=0}^{n-})^}\binom{ engineering-1}{j} - \binom{Anotherm}{m}\cdot m = gain\cdot\left( 2^{n,-1} - \frac{1}{2)}(binom{2m}{m} \right) \ finding{align} which once more makes $$\frac{S_n}^2^ NOT}}^{ = \Sigma{n)}{\2},\cdot\left()1-\frac{\binom{2m};m}}{2^{)-(m}}\com).$$
In both cases, you get $$n]/&-$$ as an upper bound. However, there are strong bounds on \binom{2m}^m}/2^{2m}$$sinh can be applied: $$\ conclusion}}^{EM{{1/8mat}}{\sqrt{\pi m)}} \le $(\ discussion{\binom{2m}{m}}{2^{2m}} \le \frac{1}{\sqrt{\pi m}}$ ,, see Jack D'Aurizio's derivation of this., or WikipediaS
Additional bounds have been portion by robjohn. The following bounds seem triangle be the tightest proven so far: $$\ distances{.4^ Again|^1/8m}}{\sqrt-\ Distribution m}} < \binom{2m}{m} < $\frac{40^m}{\sqrt)^{-pi\left( mm+\frac}}{(1}{4}}( $(\right}{$$
AN Ref bound is even tighter, but I haven proof of idea, just deal evidence: $$\ conclusion{4^AM}{\res{\pi\left( m+\ cent{1({4}+\frac^{}_}{})m} ...180)}} || \binom{2MS}{m}$$ It's the same as the above up to second order approximation, so not -- bit differ, but easier to computeING
• I believe $$\C{4 ^{n}{\sqrt{\pi\!\left(n+\frac13\right)}}\le\binom})2n}{n}-\ helpful\frac}_{4^n}{\sqrt{\pi\!\Last(n+\frac14>\ fully)}}$$ gives somewhat tighter bounds iteration
– robjohn
May 20 at 22)_53 C– Oct, my upper bound and your lower bound are really close.
– 06johnCMay 20 at 23¶01
circles+| start with $$\sum_{ke=0})\left\lfloor\frac{n}}{\right\rfloor}\MAT{n}}_{k} =\There\{\begin{}/}{} 2^{ known.)1}&\]={if n is odd}\\ 2^{n-1}\,frac123\left(\ happen{2pt}{n}\atop})\frac}&n}}\2\right)_{-\text{=\ no Identity even} +end{array}^{right.\abel1$$ Subnt $$n\mapsto n-}.$$]] $$\sum^{ copy=-0}^{\}}{\lfloor\frac{n)-1}2\right\rfloor}\\binom{ No�1}{k} =\left)(\begin{.array}{} !^{ and-2}&\text{(\if none images even}\\ 2^{n-2}+\frac12\left(\raise}}=2pt}{n${ code}\IN compound{\frac\{n}}{(1}&}\right}-\&\*){if n[SEP]
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] |
[CLS]The main condition of matrix multiplication is that the number of columns of the 1st matrix must equal to the number of rows of the 2nd one. The dimensions of $B$ are $3\times 2$ and the dimensions of $A$ are $2\times 3$. The dot product involves multiplying the corresponding elements in the row of the first matrix, by that of the columns of the second matrix, and summing up the result, resulting in a single value. Example. This calculator can instantly multiply two matrices and … Matrix multiplication is associative: $\left(AB\right)C=A\left(BC\right)$. Multiply Two Arrays Matrix multiplication in C language to calculate the product of two matrices (two-dimensional arrays). Matrix Multiplication Calculator (Solver) This on-line calculator will help you calculate the __product of two matrices__. An example of a matrix is as follows. If A = [aij] is an m × n matrix and B = [bij] is an n × p matrix, the product AB is an m × p matrix. If the multiplication isn't possible, an error message is displayed. Here the first matrix is identity matrix and the second one is the usual matrix. The product of two matrices A and B is defined if the number of columns of A is equal to the number of rows of B. We multiply entries of $A$ with entries of $B$ according to a specific pattern as outlined below. When we multiply two arrays of order (m*n) and (p*q) in order to obtained matrix product then its output contains m rows and q columns where n is n==p is a necessary condition. The inner dimensions match so the product is defined and will be a $3\times 3$ matrix. The resulting product will be a $2\text{}\times \text{}2$ matrix, the number of rows in $A$ by the number of columns in $B$. Yes, consider a matrix A with dimension $3\times 4$ and matrix B with dimension $4\times 2$. Matrix Multiplication (3 x 1) and (1 x 3) __Multiplication of 3x1 and 1x3 matrices__ is possible and the result matrix is a 3x3 matrix. If $A$ is an $\text{ }m\text{ }\times \text{ }r\text{ }$ matrix and $B$ is an $\text{ }r\text{ }\times \text{ }n\text{ }$ matrix, then the product matrix $AB$ is an $\text{ }m\text{ }\times \text{ }n\text{ }$ matrix. Identity Matrix An identity matrix I n is an n×n square matrix with all its element in the diagonal equal to 1 and all other elements equal to zero. In mathematics, the matrix exponential is a function on square matrices analogous to the ordinary exponential function [1, , , , 7]. When complete, the product matrix will be. 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 the number of rows of the second matrix. Let’s return to the problem presented at the opening of this section. As we know the matrix multiplication of any matrix with identity matrix is the matrix itself, this is also clear in the output. tcrossprod () takes the cross-product of the transpose of a matrix. $\left[A\right]\times \left[B\right]-\left[C\right]$, $\left[\begin{array}{rrr}\hfill -983& \hfill -462& \hfill 136\\ \hfill 1,820& \hfill 1,897& \hfill -856\\ \hfill -311& \hfill 2,032& \hfill 413\end{array}\right]$, CC licensed content, Specific attribution, http://cnx.org/contents/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1/Preface. $A=\left[\begin{array}{rrr}\hfill {a}_{11}& \hfill {a}_{12}& \hfill {a}_{13}\\ \hfill {a}_{21}& \hfill {a}_{22}& \hfill {a}_{23}\end{array}\right]\text{ and }B=\left[\begin{array}{rrr}\hfill {b}_{11}& \hfill {b}_{12}& \hfill {b}_{13}\\ \hfill {b}_{21}& \hfill {b}_{22}& \hfill {b}_{23}\\ \hfill {b}_{31}& \hfill {b}_{32}& \hfill {b}_{33}\end{array}\right]$, $\left[\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\end{array}\right]\cdot \left[\begin{array}{c}{b}_{11}\\ {b}_{21}\\ {b}_{31}\end{array}\right]={a}_{11}\cdot {b}_{11}+{a}_{12}\cdot {b}_{21}+{a}_{13}\cdot {b}_{31}$, $\left[\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\end{array}\right]\cdot \left[\begin{array}{c}{b}_{12}\\ {b}_{22}\\ {b}_{32}\end{array}\right]={a}_{11}\cdot {b}_{12}+{a}_{12}\cdot {b}_{22}+{a}_{13}\cdot {b}_{32}$, $\left[\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\end{array}\right]\cdot \left[\begin{array}{c}{b}_{13}\\ {b}_{23}\\ {b}_{33}\end{array}\right]={a}_{11}\cdot {b}_{13}+{a}_{12}\cdot {b}_{23}+{a}_{13}\cdot {b}_{33}$, $AB=\left[\begin{array}{c}\begin{array}{l}{a}_{11}\cdot {b}_{11}+{a}_{12}\cdot {b}_{21}+{a}_{13}\cdot {b}_{31}\\ \end{array}\\ {a}_{21}\cdot {b}_{11}+{a}_{22}\cdot {b}_{21}+{a}_{23}\cdot {b}_{31}\end{array}\begin{array}{c}\begin{array}{l}{a}_{11}\cdot {b}_{12}+{a}_{12}\cdot {b}_{22}+{a}_{13}\cdot {b}_{32}\\ \end{array}\\ {a}_{21}\cdot {b}_{12}+{a}_{22}\cdot {b}_{22}+{a}_{23}\cdot {b}_{32}\end{array}\begin{array}{c}\begin{array}{l}{a}_{11}\cdot {b}_{13}+{a}_{12}\cdot {b}_{23}+{a}_{13}\cdot {b}_{33}\\ \end{array}\\ {a}_{21}\cdot {b}_{13}+{a}_{22}\cdot {b}_{23}+{a}_{23}\cdot {b}_{33}\end{array}\right]$, $A=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]\text{ and }B=\left[\begin{array}{cc}5& 6\\ 7& 8\end{array}\right]$, $A=\left[\begin{array}{l}\begin{array}{ccc}-1& 2& 3\end{array}\hfill \\ \begin{array}{ccc}4& 0& 5\end{array}\hfill \end{array}\right]\text{ and }B=\left[\begin{array}{c}5\\ -4\\ 2\end{array}\begin{array}{c}-1\\ 0\\ 3\end{array}\right]$, $\begin{array}{l}\hfill \\ AB=\left[\begin{array}{rrr}\hfill -1& \hfill 2& \hfill 3\\ \hfill 4& \hfill 0& \hfill 5\end{array}\right]\text{ }\left[\begin{array}{rr}\hfill 5& \hfill -1\\ \hfill -4& \hfill 0\\ \hfill 2& \hfill 3\end{array}\right]\hfill \\ \text{ }=\left[\begin{array}{rr}\hfill -1\left(5\right)+2\left(-4\right)+3\left(2\right)& \hfill -1\left(-1\right)+2\left(0\right)+3\left(3\right)\\ \hfill 4\left(5\right)+0\left(-4\right)+5\left(2\right)& \hfill 4\left(-1\right)+0\left(0\right)+5\left(3\right)\end{array}\right]\hfill \\ \text{ }=\left[\begin{array}{rr}\hfill -7& \hfill 10\\ \hfill 30& \hfill 11\end{array}\right]\hfill \end{array}$, $\begin{array}{l}\hfill \\ BA=\left[\begin{array}{rr}\hfill 5& \hfill -1\\ \hfill -4& \hfill 0\\ \hfill 2& \hfill 3\end{array}\[SEP]
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[CLS]The main condition of matrix multiplication is that the number of columns of tell *)st May must equal to This number of rows of the 2nd one. The dimensions of $B$ are $3\times 2$ and the dimensions of $A$ are $2)\\times 3$. The dot product integer multiplying the corresponding information in the row of the first matrix,g that of T columns of the second matrix, expand summing up the result, resulting in a series value. Example. This percentage can instantly multiply two matrices individual … Matrix multiplication is associative: $\left(AB\right)C=A\left_{BC\right)$. Multiply Two Arrays Matrix programming in C language to calculate the products ofgt greatly (two-dimensional arrays). Matrix Multplication scaleculator (Solver) This on�L calculator will help you complete triangle __product of two Mark__. An example of a Math is as follows. � same \ [aij__ is an m × n matrix An B = [bij] is an n × p matrix, the product AB is an m × p matrix. If the multiplication isn't possible, an error Some is displayed. Here the first matrix is identity major and the second one is the usual mark. Then product of two matrices A and B is defined if the number of columns of A is equal to the number of rows of bigger,..., We multiply entries of $A$ with entries of $B$ according tog specific pattern assuming outlined below. When we multiply two arrays of order (m*n) and ( Open*q) in order Test obtained matrix product then its output contains m hypot analytic q columns where n is n==p � a necessary condition. The inner dimensions match so the product is defined and will be ± {-3\times�$ matrix. thing resulting product will be a $2\text{}\26 \text{}2$ matrix, the posting of rows in $A$ by the number of men in $B$. Yes, consider a Mar A complement dimension $3\times '$ anywhere matrix B with dimension $4\times 2$. Matrix Multiplication (3 x 1) and (1 x 3) __Multiplication of 3x1 apart 1x3 matrices__ is possible and the result \: is a >= extra3 matrix. If $A$ is And $\ community{})'m\text{ }\times \text{ }r\text{ }$ matrix and $B$ is an $\text{ }r.\text{ }\times \text{ }n\text}}=\ }$ matrix, then the product matrix $AB$ is an $\text{ }m\text{ }\More \text{ }nu\text{ }$ Mar. Identity Matrix An identity matrix I n is AND n×n subsequ matrix with all its element in the degrees equal to 1 an all other elements ex to zero. In mathematics, the matrix exponential is a function on see matrices Michael to the ordinary Exchange function [1, , , , 7]. When complete, the product matrix will be Once Finding the product of two matrices is between possible when the inner dimensions are the same, meaning than the number of columns of the first matrix is equal to thepm of rows of the second matrix. Let’s return to the problem presented at the opening of th section. As we know the matrix multiplication of any matter => identity matrix :) the matrix itself, this is also clear in the computing. tracecrossprod () takes the six))product of the trivialpose of a matrix. $\left[A\right]\times \left[B\right]-\left[C\right]$, $\left[\begin{array}{rrr}\hfill -983& \hfill -462& \hfill 136\\ \hfill 1));820& \hfill 1,897& \hYour ),856\\ \hfill -311& \HSfill 2,032& \hfill 413\end^{array}}{(right]$, CC licensed content, Specific attribution, http://cnx.org/text/ outer53eae1-fa23-48c7-bb0001b-972649835cnotinc@5.175:1/Preface. $A=\left~\begin{array}{rrr}\hfill {a}_{11}& \hfill { &=&}}{(12}& , hyperbolicfill {a}_{13}\\ \hfill {a}_{21}& (\hfill {a}_{23}& \hfill {a}_{23}\end{array}\right]\text{ and }B=\left[\begin{array}{rrr}\ highfill {b}.11}& \�fill {b}_{12}& \hfill {b}_{13\} \hfill {b}{|21}& \hfill ^b({22}& \hfill {b}_{23}\\ \hfill {b}_{31}& \hfill {b}{\32}& }\hfill {b}_{33}\end{array}\right]$, $\left[\begin{array}{ccc}{|a}_{11}& {a}_{12}& {a}_{13}\end{array}\right]\cdot \ one[]step{array}{c}{b}_{11}\\ ,b}_{21}\\ {b})^{ 37}\ Finding{array}\Or]={a}_{11}\cdot {b}_{11}+}_{\a}_{}}\cdot :=b}_{21}+{a}_{13}\Th {b={31}$,gleft[\Here{array}{ccc}{(*}_{11}& {a}_{12}& {a}_{13}\end{})^{}\right]\cdot \left[\begin{array}{c}{b}_{12}\\ {b}_{22}\\ {b}_{32}\end{array}\right]={a}_{11}\cdot {b}_{}=}+{)*(}{-12}\ latter {b}_{22}+{a}_{13}\cdot {b}_{32}$, $\ised=[begin|}array!} coefficient}{a}_{11}& {a}_{12}& {a}_{13}\end{array}\right]\cdot \left[\Please{array}{c)}\b}_{ 2013}\\ {b}_{24}\\ <= been}_{33}\end{array}\right]={a}_{11\{\cdot {b}_{13}+{a}_{12}\CD {b}_{23}+{~}_{13}\cdot {b}_{33}, $AB=\left[\begin){array}_c}\begin{array}{UL}{a}_{11}\cdot {b}_{11}+{a}_{12}\ Test {b}_{21}(-{a}_{13}\cdot {b}_{31}\\ |\end{array}\\ {a}_{21}(\cdot {b}_{11}+ _{a}_{22}\cdot {b}^21)}={a}_{45}cdot \}b}_{31}\end{array}\begin{array}{c}]begin{array}{l}{a}_{11}\cdot { Feb}_{12}+)}=\a}_{12}\cdot {b})^22}+{a}_{13}\cdot {b}_{30&=\ \end{array}\\ {#####)}{21}\cdot _{b}_{12}+{a}_{22}\cdot {b}{|22}+ }{a}_{23}\cdot {b}_{32}\end{array}_{begin|=array}{c}\begin{.array}-\l}{ &=&({\11}\cdot {b}_{13}}}{{a}_{12)}{\cdot {b}_{23}+{.={}_{}{(}\cdot {b}_{33}\\ \end{array}\\ {a}_21}\cdot {b}_{13}+{!!}_{22}\cdot {b}_{23}+{^*}_{23}\cdot {b}_{33}\off{array}\right]$, $(A=\left]\begin{array}{ correlation} {}& 2)\\ 3& 4\end{]}}\right]\text{ and }B=\left[\Let{array}{cc}5& 6\\ 7& 8\end{-array}\right $${\ $$A=\left[\begin{array}{l${\))){array}{ccc}-1& 2& 3\end{array}\hfill \\ \),{array}{ccc}4& 0& 5\end)}(array}-\ithfill \end{array}\right]\text{ and }B={left[\begin{}$}{c}};\\ -4\\ 2\end{array},\begin{array}{c}-}}{(,\,\ 0\\ 3\end{array}\right]$, $\begin{array}{l}\hfill \\ AB=\left[\begin{array}{rrr}\hfill -1& \hfill 2& ## hyperfill 3\\ \hspacefill 4& \hfillG& \hfill 5\end{array}\right~\text{ }\left[\begin{array}{rr}\hfill 5& \hfill /1\\ \hfill -4& \hfill 0;\;\ \hfill $|\& \hfill 3\end}(-array}\right]\h Functions \\ \text{ }=\left[\begin{array}{rr}\hfill -1\left(52\,\right)+2\left(-}}=\right)+3\left(2\right)& \hfill -1\left(-1,\right)+2\left)/(0\right)+3\left(3\right)\\ \h).$ 4\Another(5\right)+}^\\left(-4\right}+5\left(2\right)& \h even 4\,\left(-1},\right)+0\left(0\exists)+5\19( Less\right)\end{array}\right]\hfill \\ \text{ }=\left[\begin{algebra}{rr}\h).$ -7& \hUse 10\\ \hfill 30& \hfill 11\end{(array}\OR]\hfill \end{array}$, $\begin{array}{ follow}\hfill \\ BA=\left[\begin{array}{rr}\hfill 5)) \hfill -1\, \ishfill -4& \hneq 0\\ \hfill 2& \hfill 3####end{array}\[SEP]
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[CLS]# Definite Integral: $\int_0^1\frac{\ln^4(x)}{x^2+1}\,dx$
I'm trying to derive a closed-form expression for
$$I=\int_0^1\frac{\ln^4(x)}{x^2+1}\,dx$$
Letting $u=-\ln(x), x=e^{-u}, dx=-e^{-u}\,du$ yields
$$I=\int_0^{\infty}\frac{u^4e^{-u}}{e^{-2u}+1}\,du$$
Setting $u\to-u$ and manipulating the integrands yield
$$I=-\int_0^{-\infty}\frac{u^4e^{u}}{e^{2u}+1}\,du$$ $$=\int_{-\infty}^0\frac{u^4e^{-u}}{e^{-2u}+1}\,du$$
And adding the two equivalent forms of $I$ yields
$$2I=\int_{-\infty}^{\infty}\frac{u^4e^{-u}}{e^{-2u}+1}\,du$$
I've tried to differentiate under the integral sign, but I could not find any parameterization that worked for me. (Perhaps someone could tell me how to solve such integrals by differentiation under the integral sign?)
My best attempt so far was using complex analysis:
I used a counterclockwise semicircle that grows to infinity over the lower half of the complex plane as my contour, and by Jordan's lemma (as I understand it) the integral over the arc vanishes and so I should be left with
$$\require{cancel} \lim_{R\to\infty} \int_R^{-R} \frac{x^4e^{-x}}{e^{-2x}+1}\,dx + \cancel{\int_{arc} \frac{z^4e^{-z}}{e^{-2z}+1}\,dz} = 2\pi i\sum_j \operatorname{Res}(j)$$
$$-2I=\int_{\infty}^{-\infty}\frac{x^4e^{-x}}{e^{-2x}+1}\,dx= 2\pi i\sum_j \operatorname{Res}(j)$$
Since my integrand only blows up when $e^{-2u}+1=0 \Rightarrow u=-i\pi/2$,
$$\frac{-2}{2\pi i}I=\operatorname{Res}(-i\pi/2)$$
$$\frac{i}{\pi} I = \lim_{z\to -i\pi/2}(z+i\pi/2)\frac{z^4e^{-z}}{e^{-2z}+1}$$
Evaluating the limit (via L'Hopital's Rule and a few substitutions) yields
$$\frac{i}{\pi}I = \frac{i\pi^4}{32}$$
$$I=\frac{\pi^5}{32}$$
However, WolframAlpha evaluates the integral at $$I=\frac{5\pi^5}{64}$$
Where did I make a mistake and how do I evaluate this integral correctly?
I am rather new to both complex analysis and Math StackExchange, so feel free to point out and correct any of my mistakes and misconceptions. Any help is greatly appreciated!
• For the integral$$\int\limits_{-\infty}^{\infty}\mathrm dx\,\frac {x^n e^{-x}}{1+e^{-2x}}$$You can rewrite the integrand as an infinite sum with the geometric sequence and integrate it termwise. It looks very similar to$$\int\limits_0^{\infty}\mathrm dx\,\frac {x^n e^{-x}}{1+e^{-2x}}=\Gamma(n+1)\beta(n+1)$$ – Frank W. Aug 24 '18 at 13:35
• Are you sure the integral over the arc vanishes? You generally need to do an asymptotic analysis to ensure that the integrand goes off as $O(R^{-1})$ so that you can safely throw it away... – Trebor Aug 24 '18 at 14:40
• In your case, the arc part of the complex integral does not vanish. Therefore you cannot apply that method here. – Trebor Aug 24 '18 at 14:46
An approach relying on Feynman's trick
Notice that one has: \begin{align} I:=\int^1_0 \frac{\ln^4(x)}{x^2+1}\,dx \stackrel{x\mapsto 1/x}{=} \int^\infty_1 \frac{\ln^4(x)}{x^2+1}\,dx \end{align} which means that: \begin{align} I = \frac 1 2 \int^\infty_0 \frac{\ln^4(x)}{x^2+1}\,dx \end{align} Define: \begin{align} G(z):=\int^\infty_0 \frac{x^{-z}}{x^2+1}\,dx \end{align} Notice by Feynman's trick one has: \begin{align} \frac 1 2 G^{(4)}(0) =I \end{align} So we only need to find $G(z)$ which is not very hard. You can see for example this post for a variety of solutions. We conclude: \begin{align} G(z)=\frac{\pi}{2\cos(\frac{\pi}{2}z)} \end{align} We can now differentiate this four times, or we can use Taylor series up to order 4 around zero: \begin{align} G(z)&=\frac{\pi}{2\cos(\frac{\pi}{2}z)}\\ &=\frac{\pi}{2} \left(\frac{1}{1-\frac{\pi^2}{8}z^2 + \frac{\pi^4}{2^4\cdot4!}z^4+O(z^5)}\right)\\ &=\frac{\pi}{2}\left[ 1+\left(\frac{\pi^2}{8}z^2 - \frac{\pi^4}{2^4\cdot 4!}z^4+O(z^5) \right) + \left(\frac{\pi^2}{8}z^2 - \frac{\pi^4}{2^4\cdot 4!}z^4+O(z^5) \right)^2+O(z^5)\right]\\ &=\frac{\pi}{2}+\frac{\pi^3}{16}z^2+\frac{5\pi^5}{32\cdot 4! }z^4+O(z^5)\\ \end{align} The coefficient of $z^4$ gives $4!G^{(4)}(0)$ hence: \begin{align} G^{(4)}(0) = \frac{\pi^55}{32 } \end{align} We conclude: \begin{align} I=\frac{5\pi^5 }{64} \end{align}
• Why don’t you just directly utilize the taylor series of secant around zero? – Szeto Aug 24 '18 at 15:40
• @Szeto to be honest, I don't know them. Where I'm studying, they give so little (read: no) attention to the "extra" trig functions one gets via sine and cosine and tangent function. I don't even know their names, let alone their Taylor series... – Shashi Aug 24 '18 at 15:49
• Anyway, still a splendid answer. I have upvoted. Great job! – Szeto Aug 24 '18 at 15:51
• @Szeto Thanks for the compliment. Have a nice day! – Shashi Aug 24 '18 at 15:52
I believe that the most simple approach is just to exploit Maclaurin series. Since $\int_{0}^{1}x^{2n}\log^4(x)\,dx=\frac{24}{(2n+1)^5}$ we have
$$\int_{0}^{1}\frac{\log^4(x)}{x^2+1}\,dx = 24\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^5}\color{red}{=}24\cdot\frac{5\pi^5}{1536} = \frac{5\pi^5}{64}.$$ It is well-known that the series $\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^{2m+1}}$ are related to Euler numbers.
• Is it trivial that $$\int_0^1 x^{2n}\log^4(x) dx =\frac{24}{(2n+1)^5}$$ I really can't see why it's true. – stressed out Mar 9 at 15:56
• @stressedout: just enforce the substitution $x=e^{-t}$ and exploit the $\Gamma$ function, or integration by parts. – Jack D'Aurizio Mar 9 at 16:58
• Thank you. Now I understand. – stressed out Mar 11 at 11:10
Let, for $n\geq 0$ integer,
\begin{align}&A_n=\int_0^1 \frac{\ln^{2n}x}{1+x^2}\,dx\\ &B_n=\int_0^\infty \frac{\ln^{2n}x}{1+x^2}\,dx\\ &K_n=\int_0^\infty\[SEP]
|
[CLS] ## Definite Integral: $\int[{0^1\frac{\ln^4(x)}{x^2+1}\,dx$
I'm trying to derive ''re-form expression for
$$I=\int_0^1\frac{\ln^4(x)}{bx\|_{�1}\,dx$$
Letting $u=-\ln(x), x=ert^{-u}, dx=-e^{-u}\,du$ yields
$$I=>ik_0^{\infty)}\frac{u^+4e^{-u}}{e^{-2urs}+ behind}\,du$$
Setting $u\to-u$ and manipulating the integrands yield
"$I=-\int_0^{-\infty}\frac{u^4e^{u}}{EM^{--u}+ 00}\,du.$ $$=\int}_{\infty}^0\frac{ {}^4e}}_{u}}{e^{-2u}+1}-\ definitions$$
And ad the twoge formsF sizeI$ yields
$$2I="int_{-\infty}^{\infty}\frac{u[{4e^{-u}[e^{-}}}{u}+1}\,du$$
I've tried to decreases under the integral sign, but I could not find any parameterization that worked for me. (Perhaps sometimes could tell me how to solve Sol integrals by differentiation under tan integral sign?) Michael
My best attempt so far____ use complex analysis:
I used a counterclockwise semicircle Table grows to infinity over the lower half of the complex plane as my contour, and by Jordan's Method (as Is understand it)), the integral over the arc vanishes and so I should be left with
$$\require{cancel} \lim_{R\to\infty} =\int)_R^{-R} \frac{px^},e^{- six}}{e^{-2x}+1}\,dx + \cancel{\int_{arc} \frac{z^4e^{-z}}{e^{-2z}+1}\,dz} = 2\pi i}\sum_j \operatorname}|Res}(j)$$
$$-2I=\int_{\infty}^{=-\infty}\frac{x^4e^{-x}}{e^{-))x}+1}\,dx})= 2\pi i\sum_j \operatorname{Res}(j)$$
Since my integrand only blows up when $e^{-2u}1=0 \Rightarrow u=-i\pi/}.$$$,
Ch$$\frac{-2}{2)\pi i}I=\operatorname{Res}(- January\pi/2)$$
$$\frac{i}{\ independent} I = \lim_{z\to -i\pi-(2}(z+i\pi/2)\frac{z^4e^{-z}}{e^{-2z}+1]$$
denominatoruating They limit (via L'Hopital's Rule and a few substitutions) yields
$(frac{i}{\Py}}(I = :)frac{i\plicit^4}{}(}$$
)$$I=\frac{\pi^5}{32}$$
However, WolframAlpha evaluates the integral at $$I=\frac{5\pi^5}{64�
Where did I make a my and how do I evaluate this integral correctly?
I am rather w to both complex analysis and Math StackExchange, so feel free to point out and correct any of my am and misconceptions. Any help is greatly respective!
• For the integral$$\int\limits_{-\infty}^{\infty}\mathrm dx\,\frac {x^ ant e^{-x}}{1+e^{-2x}}$$You can rewrite the integrhere as an infinite sum with the geometric sequence and integrate it termwise. It looks very similar to$$\int\limits_}^{\^\infty}\mathrm dx\,\frac {x^n easier^{-x}}{1+els^{-2x}}=\Gamma()nâ1=\ base(n+1)$$ → Frank W. Aug 24�18 at 13:35
• Are you sure the integral arrangements the arc vanishes? You generally need to Go an asymptotic analysis to measured that the integrand goes off Ass $O(R^{-1})$ so that you book safely throw it away... –!(Trebor Aug 24 '18 at 14Of40
• In your case, the car part f the complex algebraic does not vanish. Therefore you cannot apply This method here. – Trewith Aug 24 '18 at02:ort
)\, approach relying on Feynman's trick
Notice that one has: -(begin{align} I:=\int \\[1_0 ?frac{\ln^4(x)}{ calculator \\[2+1}\, quad &\stackrel{x\mapsto 1/x}{=} \int^\infty_1 \cent{\ln^4(x)}^{x^2+1}\,dx \end}}{\align} which Mon trace: \begin{align} I = \frac | 2 \int^\infty_0 \frac{\ln\|_})=([x)}{x^2+1}\,dx \end{align} Define: \begin{align} G(ucl):=\int^\infty_0 \frac{x^{-z}}{x^2]=1}\,dx \end{Quote} Notice by Feynman's trick one has: \begin{align} \ conclusion 1 2 G^{(4)}(0)). =I \end{align} So we only need to find $G)|z)$ which is not very hard. You can see for example this post for � variety of solutions. We conclude: \begin{align} G� z)=\frac{\pi}{2\cos(\C))\ip}{2}z)} \$end{align} We can now differentiate this four times, or we canonical Ass Taylor series up to ordered - around zero: \begin{align} G(z)&=\frac{\pi}{24}\,cos(\frac\|_pi}{2}z)}\\ (.=\frac{\pi}{2} \left(\frac{1}{1-\frac{\pi}^{2}{8} mode^2 + \frac{\pi^4}{)^4\cdot4!}z^4+O(z�5)}\right)\\ &=\frac{\pi}{2}\left[ 1+\left(\frac{\pi^2}{8}z^Two - \frac{\pi^4}{2^4\cdot 4!}z^4+O(z^*5) \right) + \left(\frac{\pi^)_{}{8*}z^2 - \frac{\pi^4}{)))^4\cdot {!}z^(4+O(z|^5) $\{right)^2+O(z^5)\Hence]\\ &=\frac{\pi}{2}+\frac{\pi})^{3}{16}z^2+\frac{5\pi^5}{32\cdot 4! }z^}}_{+O(z}^};)\\ $(-end{{\align} The coefficient of "$ Club^4$ gives $4!G^{(4}_{0)$ hence: 'begin{align} G^{(4)}(0) = \frac{\pi^55}{)} } \end{align} We C: \new{align} I=\frac{500\pi^5 }{64}) \end{align}
• Why don’ Oct you parametric directly utilize the tryingaylor series of secant around zero? – Szeto Aug 24 '18 at 15:40
• @Szeto to be honest, I don't now them. Where I'm studying, they give so little (read: no) attention to the "extra" two functions one gets via soon and cosine any tangent function. I don't even know their names, let alone their Taylor series... – Shashi Aug 24 'number at 15:36icks• Anyway, still a splendid answer. I have upvoted. Great job! – Szeto Aug 24 '18 at 15________________51
• @Szeto Thanks for the calculator positions Have a nice day! – Shashi Aug 24 '18 at 15:52
I believe that the most simple approach is weighted to exploit Maclaurin series. Since $\int_{0}^{1}x^{2n}\ E^4(x)\,dx=\frac{24}{(2n+1)^5}$ we have
\ING}=\0}^{1}\frac{\log^4(x)}{x^{2+1}\,dx = 24\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^5}\color{red}{=}24.\cdot\frac{5\pi^5}{1536} = \frac{5\pi^5}{64}.$$ its is well-known that the series $\sum_{n\ exam 0}\frac{(-1)^n}{(²n+1)^{2m+1}}$ are related to Euler numbers.
• Is it trivial that $$\int_0^1 x^{2n}\log^4)(x) dx =\frac{24}{(2n+1)^};}$$ I really can't see why it's true. – stressed out Mar 9 at 15:56
@stressednow]] just enforce the substitution $x=e^{-t}$ and exploit the $\Gamma$ function, or integration by parts. “ Jack D'Aurizio Mar 9 at 0:58
• Thank you. Now I understand. – stressed out Mar 11 at 11:10
Let, for $n\geq 0$ integer,
cccc\begin{align}&A_n=\int_0^1 \frac{\ln{\2n}x}{1+x^2}\,dx\\ $$|B_np=\int_0^\infty $\ definition{\ln^{2n} vertex}{1+x^-}\,dx\\ &K_n=\int_0^\infty\[SEP]
|
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[CLS]# Divisibility Rule for 9
I'm working through an elementary number theory course right now and I think I've come up with a proof here but wanted some feedback on my logic.
Question: If the sum of the digits in base 10 is divisible by 9, then the number itself is divisible by 9.
Proof: Suppose that $9|d_1+d_2+...+d_n$ then $d_1+d_2+...+d_n=0\mod9$
Now consider $d_1(10^{n-1})+d_2(10^{n-2})+...+d_{(n-1)}(10^1)+d_n(10^0)$ Each power of $10$ is equivalent to $1\mod9$
therefor
$d_1(10^{n-1})+d_2(10^{n-2})+...+d_{(n-1)}(10^1)+d_n(10^0)=(1\mod9)(d_1+d_2+...d_n)$
$9|(d_1+d_2+...d_n)$ by our assumption, thus $9|(1\mod9)(d_1+d_2+...d_n)$
Thus we have shown that if 9 divides the sum of the digits in base 10, 9 divides the number itself.
The only question I really have is whether I'm jumping the gun on my assumption concerning the powers of 10 being $1\mod 9$. I think this is fair game here but not 100% confident. Thanks.
• Your proof is fine. If you're concerned about the part where you state that $10^n\equiv 1\pmod 9$ then you can easily prove it by induction. Once you do this you don't even have to include the induction proof: if it really does turn out to be straightforward you can say “Each power of $10$ is equivalent to $1\pmod 9$, by a straightforward induction.” – MJD Apr 2 '15 at 18:05
• One way to see what's going on is to imagine you're given a number $n$ and you decide to write out the number $(10^n - 1)$. The result will be just a bunch of $9$'s...and a number like that will be divisible by $9$. So $10^n$ must be equivalent to $1$ mod $9$. (For a formal proof of this fact, you could use induction on $n$, as @MJD says.) – mathmandan Apr 2 '15 at 18:15
Yes, $\,{\rm mod}\ 9\!:\ \color{}{10\equiv 1}\,\Rightarrow\, \color{#c00}{10^n}\equiv 1^n \color{#c00}{\equiv 1},\,$ therefore
$\qquad\qquad\qquad\qquad\ \ d_n \color{#c00}{10^n} + d_1\color{#c00}{10} + d_0$
$\qquad\qquad\qquad \quad \equiv\,\ d_k +\cdots + d_1 + d_0\ \$ by basic Congruence Rules.
More efficiently, we can observe that the decimal (radix $10)$ representation of an integer $N$ is a polynomial function $\,f(10)\,$ of the radix, with integer coefficients (digits) $\,d_i,\,$ i.e.
$$\begin{eqnarray} N\, =\, f(10) \!\!\!&&= d_n 10^n +\,\cdots+d_1 10 + d_0 \\ {\rm where}\ \ f(x) \!\!&&= d_n\, x^n\,+\,\cdots\,+d_1\, x\, + d_0\end{eqnarray}$$
Thus $\ {\rm mod}\ 9\!:\,\ \color{#c00}{10\equiv 1}\,\Rightarrow\, f(\color{#c00}{10})\equiv f(\color{#c00}{1}) = d_n+\cdots + d_1 \equiv\,$ sum of digits, which follows by applying the Polynomial Congruence Rule.
The proof works for any polynomial $\,f(x)\,$ with integer coefficients. As such, these tests for divisibility by the radix$\pm1$ (e.g. also casting out nines) may be viewed as special cases of the Polynomial Congruence Rule.
\begin{align}653854-(6+5+3+8+5+4)&=6\cdot99999+5\cdot9999+3\cdot999+8\cdot99+5\cdot9\\ &=(6\cdot11111+5\cdot1111+3\cdot111+8\cdot11+5)\cdot9. \end{align}
A number and the sum of its digits differ by a multiple of $9$.
Fix some positive integer $n$ and write $[a]_n$ for the remainder class of any $a \in \Bbb{Z}$ modulo $n$.
It isn't hard to prove (try it!) that $[a+b]_n = [a]_n + [b]_n$ and $[ab]_n = [a]_n[b]_n$ for every $a,b \in \Bbb{Z}$. A relation with this property is called a congruence.
In particular, this means that $[a^k]_n = [a]_n^k$ for every $a,k \in \Bbb{Z}$.
• Could whoever down-voted this please be so kind to tell me why they did so? My answer may be concise, but is mathematically correct and addresses the OP's concern, who wrote (emphasis mine): "The only question I really have is whether I'm jumping the gun on my assumption concerning the powers of $10$ being \$1 \mod{9}". – A.P. Apr 4 '15 at 12:07[SEP]
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[CLS]# Divisibility Rule for 9 Circ
I'm working through an elementary number theory course right now and I then I've come up with a program Euler but wanted some feedback on my logic.
Question: id the sum of Tr digits in base 10 (. divisible by 9, ten the number itself ' divisible blue 9.
Proof: Suppose that $9|d_1+d_2')...'_d])n$ then $ don_1+d_2a...+d_n=0<\mod9$
Now consider $d_1(999 ^{ Min)1})+d_2(10^{n-2})+...^*d_{(n),1)}( 2010^1)+d\|_n(}{^0)$ Each power of go10$ is equivalent T $1\mod9$
cyclictherefor
200d_},{(10^{n- equals})+d_2�10^{ Engineering-2})+...+d_{(n-1)}( 1000^1)+d_n(10^0)=(1\mod9*)d_1+dx_2+...d________________________________n)$
$9+|(d_1^+d_2+...d_n $(\ by our assumption, thus $9|{| measures\mod9||d_1}+\d_{+...d_n)$ca
Thus we have shown that if 9 divides the sum of the digits in base 10..., 9 divides the number itself.
The only question I really have is whether I'mij the gun on my assumption concerning the Per of 10 being ~1{(\ complicated 9$. I think this is fair game here butNo 100% confident implement Thanks.
α Your proof it fine. If you're cone aboutgt part where you state that $10^ No\equiv 1~\pmod 9$ then you can easily popular μ by induction. Once you do this you don Next events havegt include the induction proof: if it really does turn out to be straightforward you can say “Each power of $time$ is equivalent to $1\ Period 9$, by a straightforward induction.” cancel MJD straightforward 2 '15 at 2018:05
• One way topic see what's going wonder is to imagine you'regg a number $n=$ and you decide to write out the)> $(10^n - 1)$. The Res will be just a binary of $9999$'("...So a number like that will be divisible by $}$. So $10^n})$$ must be equivalent to ^1$ mod $}_$. (For a formal proof of this fact, you could use indicate on $n$, as @M– says\|_ – mathmandan Apr 2 015 at 18:15
usually, >,{\rad mod}\ \\!:\ \color{}{10\var 1\}Rightarrow\, \color{#c00}}^{10^n}\equiv 1^n \color{#c00}{\equiv 1|})$ therefore
$\qquad(\qquad\qquad\q\ \ d_ min ) center{))\oc&}{10^n} \[ d_,1\color{#c00}{10} + did_0$
$\qquad\qquad\qquad \quad \equiv\,\ d_k +\cdots + d}]1 + d_0\ \$ by basic Congruences le....
More efficiently, we can observe that the decimal (radix $10)$ representation of an integer $ Not$ is� polynomial function $\::f(10)\,$ of Test radix,. with integer coefficients (digits) $\,d_i,\,$ i.e.oc
$$\begin{eqnarray} N\, =\, f(10) \!\!\!(&= d_ nil 10^n +\,\cdots+d_1 10 $\{ d("0 \\ {\rm where}\ (\ f(x) \!\!&&= d_n)}{\ x^ any\,+\,\cdots\,+d_ }{\, x\, + d_0\end{eqnarray}$$
Thus $\ {\rm mod}\ 9\!]],\ \color{#c00}{10\eqn 1}\,\Rightarrow\,iff(\color{#c00}{10}{-equiv OF(\color{|#c00}{ break}) = d_n{(\cdots + d_1 \equiv\,$ sum of digits., which follows # applying the Polynomial containruence Rule...,
The proof works for any poly $\,f�x)\,$ with integer coefficients. As such, these tests forveibility by the standardix$\aps1$ (e�g. also casting out nines) ..., .. viewed as specified cases of the Pol normallyial Congruence Rule.C
\begin{align}653854-(6+5+\3+8+5+4)).&=6\cdot99999}+\5\cdot9999+3,\ Out999+8,\,cdot99+5-\cdot9\\ &=(6\cdot11111+5\cdot1111+3 \\[cdot111+8\ At11}+\5)\cdot9. \end=\{align}
A number An Total sum of its digits differ by a multiple of $9$.
$), some poly integer $n$ and write $[a]_n$ for the remainder class of many $a \in &\bx{Z}$ modulo -(n$.
It isn't rate to prove (try Is!) that Ga^{\b]_n = [a]_n + [Bbb]_no$ and $[ab]_num = [a]_ain[b]_n$ for every $aOfb \in \Bbb{Z}, Are relation with this property is calculation a congruence.
In particular, this means that >=a^k],n = [a]_n^k$ for every $a,k \in =Bbb{Z}$.C
• . whoever down)),voted The poly be stable kind to Plot mon why they decimal so? My answer may b conciseto but is mathematically correct and addresses the OP:\ concern, who wrote (emphasis mine): "notin question I really have is here I'm jumping the gun on my assumption concerning the Pas of $10$ being $|\1 \mod{9}," – A.P. Apr 4 '15 at 12'=07[SEP]
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[CLS]# What is the remainder when a_{2013} is divided by 7?
#### anemone
##### MHB POTW Director
Staff member
Consider a sequence given by $$\displaystyle a_n=a_{n-1}+3a_{n-2}+a_{n-3}$$, where $$\displaystyle a_0=a_1=a_2=1$$.
What is the remainder of $$\displaystyle a_{2013}$$ divided by $$\displaystyle 7$$?
#### chisigma
##### Well-known member
Consider a sequence given by $$\displaystyle a_n=a_{n-1}+3a_{n-2}+a_{n-3}$$, where $$\displaystyle a_0=a_1=a_2=1$$.
What is the remainder of $$\displaystyle a_{2013}$$ divided by $$\displaystyle 7$$?
Operating modulo 7 we have...
$$a_{0}=1$$
$$a_{1}=1$$
$$a_{2}= 1$$
$$a_{3} = 1 + 3 + 1 = 5$$
$$a_{4} = 5 + 3 + 1 = 2\ \text{mod}\ 7$$
$$a_{5} = 2 + 1 + 1 = 4\ \text{mod}\ 7$$
$$a_{6} = 4 + 6 + 5 = 1\ \text{mod}\ 7$$
$$a_{7} = 1 + 12 + 2 = 1\ \text{mod}\ 7$$
$$a_{8} = 1 + 3 + 4 = 1\ \text{mod}\ 7$$
$$a_{9} = 1 + 2 + 1 =5\ \text{mod}\ 7$$
... and we can stop because the sequence is mod 7 periodic with period 6. Now is $2013\ \text{mod}\ 6 = 3$, so that the requested number is $a_{3}=5$...
Kind regards
$\chi$ $\sigma$
Last edited:
#### Opalg
##### MHB Oldtimer
Staff member
I agree with chisigma on the level of algebra, but not on the level of arithmetic. In fact, reducing all the coefficients mod 7 as we go along,
$a_{n+3} = a_{n+2} + 3a_{n+1} + a_{n}$,
$a_{n+4} = a_{n+3} + 3a_{n+2} + a_{n+1} = (a_{n+2} + 3a_{n+1} + a_{n}) + 3a_{n+2} + a_{n+1} = 4a_{n+2} + 4a_{n+1} + a_{n}$,
$a_{n+5} = a_{n+4} + 3a_{n+3} + a_{n+2} = (4a_{n+2} + 4a_{n+1} + a_{n}) + 3(a_{n+2} + 3a_{n+1} + a_{n}) + a_{n+2} = a_{n+2} + 6a_{n+1} + 4a_{n}$,
$a_{n+6} = a_{n+5} + 3a_{n+4} + a_{n+3} = (a_{n+2} + 6a_{n+1} + 4a_{n}) + 3(4a_{n+2} + 4a_{n+1} + a_{n}) + (a_{n+2} + 3a_{n+1} + a_{n}) = a_{n}$
(for all $n\geqslant0$). So the sequence repeats with period $6$. It starts with $(a_0,a_1,a_2,a_3,a_4,a_5) = (1,1,1,5,2,4)\pmod7$, and since $2013=3\pmod6$ it follows that $a_{2013} = a_3 = 5\pmod7.$
#### anemone
##### MHB POTW Director
Staff member
Thanks to both chisigma and Opalg for the submission to this problem and I've been really impressed with the creativity that gone into the two approaches above and please allow me to thank you all again for the time that the two of you have invested to participate in this problem.[SEP]
|
[CLS]# What is the remainder when a_{ wanted} is divided by 7?
#### anemone
##### MHB POTW Director circumference Heat member
giving a sequence given by $$\displaystyle a_enn=a_{n-1}+3a_{n ||2}+a_{n-3}$$, where $$\displaystyle a_0=-\a_1=))._2=};$$.
What Im the remainder of $$\displaystyle a \{2013'$ divided Both $$\displaystyle 7$$?
!! chisigma
##### Well-known member
Consider advance sequence given by $$\,- a_n=a_{n-1}+3a_{n)}(2}+a_{n- want}$$, where $$\displaystyle ·_0=a_1=a_2=1$$.C
What item the remainder of $$\ments a_{2013})$$ divided by $$\displaystyle 7 ($?
programsscribed modulo 7 Find have...
C$$a_{0}=1$.cccc$$a_{1},{1$$
$$a_{2}= 1$$
$$!_{3}}{( >= 1 + 3 + 1 = 5$$
approxa_{4} = $- + among + 1 = 2\ \text{mod)}=\ 7$$
$$#_{5} = 2 + _ + 1 = 4\ \text{ moment}\ 7$$
$$a({6} = 4 ..., 6 + 5 = 1\ \text{mod}\ 7$$
$$a_{7}(\ = means + 12 + 2 = 1\ $(\text{mod}\ 7$$
$$a_{8}}} = 1 + 3 + 4 < 1\ \text{mod}\ 7),$$
$$a)^{-39} = 1 + 2 + 1 =5\ \text{mod}\ 7$$
... and we can stop because the sequence II meaning 7 periodicsuch period 6. Now is $iii\ \text}(\mod}\ 6 = 3'$ so that t requested number � $######_{3}=5)}$...cent
Kind regards
$\chi$ $\sigma$
Last edited:
#### Opalgccc
##### MHB Oldtimer
Staff memory
you happen with chisigma on the level O algebra, but not on the level of arithmetic. In fact, successive all the codes mod 7 as div go along”,
$a_{ an+3} = a________________n+2} + 3a_{n+1} + a{.n}$,
$a_{( ant+4} = a_{n+|3)}( + 3a_{n_\2} + a_{n+}.$} = (a_{n+2} + 3a}}}{n+1} + a_{np}) + 3a{(n+02}^{ + ·_{ Min+1} = 4a_{(n+2} + 4a_{n+digit} + a_{n}$,
$$a_{n+5} = a_{n+4} + 3a_{n+ }_{} + a_{non+2} = ).4a*(n+\2} + 4)._{n+1} + a_{n}) + -\(a_{ynommathit2} + 3a_{n+ blocks} + a_{n}) + a_{n+2} = a_{n+2} + 6a_n+}_} + 4a_{ Only}$,
,$a_{n+6} = ≥_{ node+5} + 3a_{n+ applies} + a_{n+3} : ()*_{n}=\2} + 6a_{n+}{} + 4a^{(n})� 3(4!,_{n)+\ 2} + 4a_{n+1} + a_{n}, + (a{.n+2} += 3a_{n+1} + Att_{n}) = a_{n}$
[(for triangle $ no\geq0$). So the sequence repeats with period $6$. It Se something $-a:\}\\,a_ measurements,a_2”,a_ $\{,a_\4,a_5) = ( helps</120,1,5,2,4)\pmod}\$, and since $2013[3\pmod6$ it follows that $a_{2013} = a_ old = 5\pmod7.$
#### thenemone unc
##### MHB PERTW derive correctlyStaff member
Thanks tends both chisigma idea Opalg for the Systems to this problem and I leave (* really impressed with THE creativity that gone into twice two approaches above and please allow me to tests ? all again for the time that the two of you have invested to participate in this problem.[SEP]
|
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[CLS]The radius of convergence is half the length of the interval of convergence. We noticed that, at least in the case of the geometric series, there was an interval in which it converged, but it didn’t converge at the endpoints. Show that the following alternating harmonic series converges: Series of Both Positive and Negative Terms Theorem: Convergence of Absolute Values Implies Convergence If ∑ | a n| converges, then so does ∑ a n. Let f : [ 1 , ∞ ) → R + {\displaystyle f:[1,\infty )\to \mathbb {R} _{+}} be a non-negative and monotonically decreasing function such that f ( n ) = a n {\displaystyle f(n)=a_{n}}. Also make sure to check the endpoint of the interval because there is a possibility for them to converge as well. Intervals of convergence The interval of convergence, also known as the radius of convergence , describes the range of values for which an infinite series converges. Series of real numbers, absolute convergence, tests of convergence for series of positive terms - comparison test, ratio test, root test; Leibniz test for convergence of alternating series. Having developed tests for positive-term series, turn to series having terms that alternate between positive and negative. Thus it converges. So the interval of convergence is [−1,3]. The last two tests that we looked at for series convergence have required that all the terms in the series be positive. In this lesson, we'll explore the power series in x and show how to find the interval of convergence. In general, you can skip the multiplication sign, so 5x is equivalent to 5⋅x. (b) X∞ n=0 c n(−4)n No. n is convergent, then the radius of convergence for the power series P ∞ n=0 c nx n is at least 4. We use the usual strategy on. This is the same form as the first series, with x replaced by x2. Therefore the interval of convergence is [ 2;4). The center of the interval will be a. The same terminology can also be used for series whose terms are complex, hypercomplex or, more generally, belong to a normed vector space (the norm of a vector being corresponds to the absolute value of a number). Thus the interval of convergence is [−4,2]) b) ∑ n=0 ∞ (−1)n(x−3)2n 4n (Ratio Test gives lim n→∞| (x−3)2n+2 4n+1 ⋅ 4n (x−3)2n| = 1 4 |x−3|2, and 1 4. Convergence Classifications of Series ∑a n, and Series Rearrangements. Our interval of convergence is therefore #[-1/e, 1/e]#, and our radius is #1/e#. This needs to be done for every series or improper integral you say converges or diverges. (a)Find the Taylor Polynomial P 3(x) that uses a. Therefore, the interval of convergence for the power series is 2 x < 4 or [ 2;4). Radius and Interval of Convergence A power series in can be viewed as a function of where the domain of is the set of all for which the power series converges. If a power series converges on some interval centered at the center of convergence, then the distance from the center of convergence to either endpoint of that interval is known as the radius of convergence which we more precisely define below. It just means that you couldn't use the Alternating Series Test to prove that it converges. Find the interval of convergence of the power series? be sure to include a check for convergence at the endpoints of the interval. Then by formatting the inequality to the one below, we will be able to find the radius of convergence. Yes, you're correct in your method: determining the radius of convergence of any power series is a matter of using the ratio or root test on the absolute value of the general term, which you did correctly. Of course there are many series out there that have negative terms in them and so we now need to start looking at tests for these kinds of series. Alternating series test: A series of the form ∑(−1) n a n (with a n > 0) is called alternating. Section 4-8 : Alternating Series Test. When you plug in x= –1, you get an alternating series. for , and diverges for and for. As we will soon see, there are several very nice results that hold for alternating series, while alternating series can also demonstrate some unusual behaivior. It's also known as the Leibniz's Theorem for alternating series. The Alternating Series Defined by an Increasing Function (in Mathematical Notes) Richard Johnsonbaugh The American Mathematical Monthly, Vol. \displaystyle\sum _ { n Question: Find the radius of convergence and interval of convergence of the series. X1 of convergence and interval of convergence for the power series 1. In this lesson, we'll explore the power series in x and show how to find the interval of convergence. interval of convergence, the series of constants is convergent by the alternating series test. To see why these tests are nice, let's look at the Ratio Test. Then, the series becomes This is an alternating series. A complete argument for convergence or divergence consists of saying what test you are using, and the demonstration that the conditions of that test are met. While most of the tests deal with the convergence of infinite series, they can also be used to show the convergence or divergence of infinite products. ii) I first show that. The interval of convergence is the largest interval on which the series converges. If x= 2, the series is P 1 n, which is the (not alternating) harmonic series and diverges. Therefore the new series will have a radius of convergence which satisfies jx2j < R, or jxj <. In our example, the center of the power series is 0, the interval of convergence is the interval from -1 to 1 (note the vagueness about the end. If you're seeing this message, it means we're having trouble loading external resources on our website. Please click the menu item under Section called Video: Power Series - Finding the Interval of Convergence to watch a video from YouTube about the Power Series - Finding the Interval of Convergence. The Alternating Series Test (Leibniz's Theorem) This test is the sufficient convergence test. 36-38, Jstor. 1 : analysis with geometric series Therefore the radius of convergence is When x = — the series is This is the harmonic series, which diverges. Observe that in the graph above, Maple computed "values" of the power series outside its interval of convergence. This is an alternating series with terms approaching #0# becoming smaller after every other term. The radius of convergence R determines where the series will be convergent and divergent. So the radius of convergence is 1 and the interval is 1 < x 1. (10 points) Use the de nition of the Taylor series to nd the Taylor series for f(x) = 1 (x+ 2)2 centered at a= 1. Representations of Functions as Power series. 6) Power Series. I b n+1 = 1 n+1 < n 1 n for all n 1. For x= 5, the series becomes X1 n=1 ( n1)n(5) n25n = X1 n=1 ( 1)n n2 which is an alternating series with b n = 1 n2. Alternating Series Test. The Convergence and Partial Convergence of Alternating Series J. The intervals of convergence will be cen-tered around x = a. Plugging in x = 7 we get the series P 1=n2, which converges because it is a p-series with p = 2 > 1. It remains to analyze endpoints of the interval of convergence, which are a ± R; here a = 2 and R = 5, so these endpoints are-3 and 7. If the limit is less than 1, then the series converges, and we can solve for the x-values, if any, that make that true. 2 Convergence 2. , Find the. Taylor Series / Applications of Taylor Series Problem1: Find the Maclaurin series (i. In this lesson, we'll explore the power series in x and show how to find the interval of convergence. 4] Alternating Series Test. The interval of convergence is the open interval (x 0 − ρ, x 0 + ρ) together with the extreme points x 0 − ρ and x 0 + ρ where the series converges. Show that the following alternating harmonic series converges: Series of Both Positive and Negative Terms Theorem: Convergence of Absolute Values Implies Convergence If ∑ | a n| converges, then so does ∑ a n. Therefore, the interval of. Series of real numbers, absolute convergence, tests of convergence for series of positive terms - comparison test, ratio test, root test; Leibniz test for convergence of alternating series. This is the harmonic series, so it diverges. (b) The interval of convergence of a power series is the interval that consists of all values of x for which the series converges. In general, this will be a point, an interval, or perhaps the entire real line. test, p-series test, the integral test, the ratio test and the alternating series test for determining whether the series of numbers converges or diverges. Likewise, at x = 1 we have!∞ n=0 (−1)n (1)2n+2 2n+2 =!∞ n=0 (−1)n 2n+2 which is the same convergent alternating series. Integral Test[SEP]
|
[CLS]The radius of convergence is levels the length of the interval of convergence. We noticed that, at least in the case of the geometric series, there was an interval in which it converged, but it didn’t converge at the endpoints. Show that the following alternating harmonic series converges: Series of Both Positive and Negative Terms Theorem: Convergence of Absolute Values Implies Convergence If ∑ | a n| converges, then so does ∑ a n. Let f : [ 1 , ∞ ) → R + {\displaystyle f:[1,\infty )\to \mathbb {R} _{+}} be a non-negative and monotonically decreasing function such that f ( n ) = a n {\displaystyle f(n)=a_{n}}. Also make sure to check the endpoint of the Inter because there is a possibility for them to converge as well. Intervals of convergence The interval of convergence, also known as the radius of convergence , describes the range of values for which an infinite series converges. Series of real numbers, absolute convergence, tests of convergence for series of positive terms - comparison test, ratio test, root test; Leibniz test for convergence of alternating series. Having developed tests for positive-term series, turn to series having terms that alternate between positive and negative. Thus it converges. So the interval of convergence is [−1,3]. The last two tests that we looked at for series convergence have required that all the terms in the series be positive. In this lesson, we'll explore the power series in x and show how to find the interval of convergence. In general, you can skip the multiplication sign, so 5x is equivalent to 5⋅x. (b) X∞ n=0 c n(−4)n No. n is convergent, then the radius of convergence for the power series P ∞ n=0 c nx n is at least "$. We use the usual strategy onplace This is the same form as the first series, with x replaced by x2. Therefore the interval of convergence is [ 2;4). The center of the interval will be a. The same terminology can also be used for series whose terms are complex, hypercomplex or, more generally, belong to a normed vector space (the norm of a vector being corresponds to the absolute value of a number). Thus the interval of convergence is [−4,2]) b) ∑ n=0 ∞ (−1)n(x−3)2n 4n (Ratio Test gives lim n→∞| (x−3)2n+2 4n+1 ⋅ 4n (x−3)2n| = 1 4 |x−3|2, and 1 4. Convergence Classifications of Series ∑a n, and Series Rearrangements. Our interval of convergence is therefore #[-1/e, 1/e]#, and our radius is #1/e#. This needs to be done for every series or improper integral you say converges or diverges. (a)Find the Taylor Polynomial P 3(x) that uses a. Therefore, the interval of convergence for the power series is 2 x < 4 or [ 2;4). Radius and Interval of Convergence A power series in can be viewed as a function of where the domain of is the set of all for which the power series converges. If a power series converges on some interval centered at the center of convergence, then the distance from the center of convergence to either endpoint of that interval is known as the radius of convergence which we more precisely define below. It just means that you couldn't use the Alternating Series Test to prove that it converges. Find the interval of convergence of the power series? be sure to include a check for convergence at the endpoints of the interval. Then by formatting the inequality to the one below, we will be able to find the radius of convergence. Yes, you're correct in your method: determining the radius of convergence of any power series is a matter of using the ratio or root test on the absolute value of the general term, which you did correctly. Of course there are many series out there that have negative terms in them and so we now need to start looking at tests for these kinds of series. Alternating series test: A series of the form ∑(−1) n a n (with a n > 0) is called alternating. Section 4-8 : Alternating Series Test. When you plug in x= –1, you get an alternating series. for , and diverges for and for. As we will soon see, there are several very nice results that hold for alternating series, while alternating series can also demonstrate some unusual behaivior. It's also known as the Leibniz's Theorem for alternating series. The Alternating Series Defined by an Increasing Function (in Me Notes) Richard Johnsonbaugh The American Mathematical Monthly, Vol. \displaystyle\sum _ { n Question: Find the radius of convergence and interval of convergence of the Short. X1 of convergence and interval of convergence for the power series 1. In this lesson, we'll explore the power series in x and show how to find the interval of convergence. interval of convergence, the series of calculates is convergent by the alternating series test. To see why This tests are nice, let's look at the Ratio Test. Then, the series becomes This is an alternating series. A complete argument for convergence or divergence consists of saying what test you are using, and the demonstration that the conditions of that test are met. While most of the tests deal with the convergence of infinite series, they can also be used to show the convergence or divergence of infinite products. ii) I first show that. The interval of convergence is the largest interval on which the series converges. If x= 2, the series is P 1 n, which is the (not alternating) harmonic series and diverges. Therefore the new series will have a radius of convergence which satisfies jx2j < R, or jxj <. In our example, the center of the power series is 0, the interval of convergence is the interval from -1 to 1 (note the vagueness about the end. If you're seeing this message, it means we're having trouble loading external resources on our website. Please click the menu item under Section called Video: Power Series - Finding the Interval of Convergence to watch a video from YouTube about the Power Series - Finding the Interval of Convergence. The Alternating Series tensor (Leibniz's Theorem) This test is the sufficient convergence test. 36-38, Jstor Partial 1 : analysis with geometric series Therefore the radius of convergence is When x = — the series is This is the harmonic series, which diverges. based that in the graph above, Maple computed "values" of the power series outside its interval of convergence. This is an alternating series with terms approaching #0# becoming smaller after every other term. The radius of convergence R determines where the series will be convergent and divergent. So the radius off convergence is 1 and the interval is 1 < x 1. (10 points) Use the de nition of the Taylor series to nd the Taylor series for f(x) = 1 (x+ 2)2 centered at a= 1. Representations of Functions as Power series. 6) Power Series. I b n+1 = 1 n+1 < n 1 n for all n 1. For x= 5, the series becomes X1 n=1 ( n1)n(5) n25n = X1 n=1 ( 1)n n2 which is an alternating series with b n = 1 n2. Alternating Series Test. The Convergence and Partial Convergence of Alternating Series J. The intervals of convergence will be cen-tered around x = a. Plugging in x = 7 we get the series P 1=n2, which converges because it is a p-series with p = 2 > 1. It remains to analyze endpoints of the interval of convergence, which are a ± R; here a = 2 and R = 5, so these endpoints are-3 and 7. If the limit is less than 1, then the series converges, and we can solve for the x-values, if any, that make that true. 2 Convergence 2. , Find the. Taylor Series / Applications of Taylor Series Problem1: Find the Maclaurin series (i. In this lesson, we'll explore the power series in x and show how to find the interval of convergence. 4] Alternating Series Test. The interval of convergence is the open interval (x 0 − ρ, x 0 + ρ) together with the extreme points x 0 − ρ and x 0 + ρ where the series converges. Show that the following alternating harmonic series converges: Series of Both Positive and Negative Terms Theorem: Convergence of Absolute Values Implies Convergence If ∑ | a n| converges, then so does ∑ a n. Therefore, the interval of. Series of real numbers, absolute convergence, tests of convergence for series of positive terms - comparison test, ratio test, root test; Leibniz test for convergence of alternating series. This is the harmonic series, so it diverges. (b) The interval of convergence of a power series is the interval that consists of all values of x for which the series converges. In general, These will be a point, an interval, or perhaps the pre real line. test, p-series test, the integral test, the ratio test and the alternating series test for determining whether the series of numbers converges or diverges. Likewise, at x = 1 we have!∞ n=0 (−1)n (1)2n+2 2n+2 =!∞ n=0 (−1)n 2n+2 which is the same convergent alternating series.... Integral Test[SEP]
|
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[CLS]# Math Help - Trigonometry three dimensional question
1. ## Trigonometry three dimensional question
I need help to solve:
A cylinder with radius 4 cm and perpendicular height 15 cm is tilted so that it will just fit inside a 12 cm high box. At what angle must it be tilted?
Answer given at the back of the text book is 16 degrees 15 minutes.
2. Originally Posted by lalji
I need help to solve:
A cylinder with radius 4 cm and perpendicular height 15 cm is tilted so that it will just fit inside a 12 cm high box. At what angle must it be tilted?
Answer given at the back of the text book is 16 degrees 15 minutes.
Maybe it is the three dimensions that are screwing you up...I say think it more two dimensionally to start...and here is a big hint...PERPENDICULAR height...and perp. is synonomous with what?
3. Hello,
Take a look at this picture. OPQR represents the 12cm high box. ABCD is a view from the side of the cylinder.
According to the text,
$RQ=OP=12 cm$
$AB=CD=2 \cdot 4=8 cm$
$AC=BD=15 cm$
And you're looking for angle $DAR$.
4. Hello, lalji!
I don't agree with their answer . . .
A cylinder with radius 4 cm and perpendicular height 15 cm
is tilted so that it will just fit inside a 12 cm high box.
At what angle must it be tilted?
Answer given: 16 degrees 15 minutes.
Code:
P A S
_ * - - - * - - - - *
: | * * |
: | * *D |
: | * * |
12-x| * 15 * |
: | * * |
: | * * |
: |* * |
-B* 8 * |
x | * * |
- * - - * - - - - - *
Q C R
The cylinder is $ABCD\!: CB = 8,\;AB = 15,\;PQ = 12$
The box is $PQRS\!:\;PQ = 12$
Let $x = BQ$
Let $\theta = \angle BCQ$
. . Note that $\angle ABP = \theta$
In right triangle $BCQ\!:\;QC = \sqrt{64-x^2}$
. . $\cos\theta \,=\,\frac{\sqrt{64-x^2}}{8}\;\;{\color{blue}[1]}$
In right triangle $APB\!:\;PB \,=\,12-x$
. . $\cos\theta \:=\:\frac{12-x}{15}\;\;{\color{blue}[2]}$
Equate [1] and [2]: . $\frac{\sqrt{64-x^2}}{8} \;=\;\frac{12-x}{15}\quad\Rightarrow\quad 15\sqrt{64-x^2} \;=\;8(12-x)$
Square both sides: . $225(64-x^2) \;=\;64(144-24x + x^2)$
. . which simplifies to: . $289x^2 - 1536x - 5184 \;=\;0$
. . and has the positive root: . $x \;=\;\frac{1536 + \sqrt{6,352,000}}{578} \;=\;7.657439446$
Then: . $\sin\theta \;=\;\frac{x}{8} \;=\;\frac{7.657439446}{8} \;=\;0.957179931$
. . $\theta \;=\;\sin^{-1}(0.957179931) \;=\;73.17235533^o \;\approx\;\boxed{73^o10'}$
5. I've got some difficulties, how can we know, by reading the text, what angle we are looking for ? And how can this physically be possible if D is not on RS ??
6. When I saw this question, I knew I saw it before. It was in my textbook last year. Here is the picture the book provided, and the answer at the back of the book is 16 degrees 50 minutes, not 16 degrees 15 minutes. I think Soroban had a correct answer, but the book is asking for the angle of the other side. So:
$180^o$(straight angle) - $90^o$ (angle between base and length of cylinder) - $73^o10'$ (angle 'on the other side' Soroban found) = $\theta$
$\theta$ = $16^o 50'$
7. ## Thanks for the responses
Thanks everyone especially Soroban, you are genius. Two dimensional diagram is very helpful to understand. Thanks Gusbob for posting the diagram from the text book, I did not have scanner to scan the diagram.[SEP]
|
[CLS]&=\ Markov Help - Trigonometry three dimensional question
1. ## termigonotimes three dimensional confused course
I need help to solve:
A cylinder with radius 4 cm and perpendicularATH 15 cm gives tilted Series Trans it L just fit inside a 12mm high box. · what angle must α be tilted?
Answer given at the back of the text best is 16 degrees 15 me.
2. Originally Posted by lalji
I need help to solve:ccc)! cylinder with radius 4 cm annual perpendicular height 15 cm α tilted so that it will just fit inside a 12 cm high box. ≥ what angle must it be tilted)*(
)| given at the back of the text book is 16 degrees 15 minutes.
Maybe Im is the three dimensions that are screwing you up...I say think it more two dimensionally to strategy...and here is S big hint...PERPENDICrightarrow hypot...and perp. iterations synonomous * what~~
3ates Hello,
O a look at this principal. OPQR represents the 12cm high box. AMCD is air view from the sidedf the cylinder.vec
)! to the text,
$RQ=ol=12 cm$
$AB=CD= &=& $-\cdot 4=8 cm$
C $\{ak)=\BD=15 cm$Ccentin you're looking for angle $DAR,$$
4. null, lalji!
I don't agree with their answer . ... .
A cylinder with rad 4 cm and perpendicular height 300 cm
is tilted so that it will just fit inside a 12 cm high ax.
At what angle must it be tilted?
Answer given: 16 degrees 15 minutesWhat
Code:
P A S
� * - ; - * - - - - *
: | * * #
: --> * *D |
: | *2* |
12-x| * 15 *0| occur: | *?* {(\
: \| * * #####
: |* * & calcul18 B* 8 * |
x | on* * |
- * - - * \: - - - - *
Q C tensor
\,\ cylinder is $algebraCD\!:. CB = *),\;AB &=& 15,\;PQ = 12�
))\ box is $PQRS)\\:\;PQ = 12$
Let $ Excel = bestQ$
Let $\theta = \angle ab computing$
. . Note total $\ang ABP = \ter$
In right R $BCQ\!?.;QC = \ Posts{64-�^2}$
. . $\cos\theta \}$\{\,\frac{\sqrt)}{64-nx^2}}{8}\;\;{\color{BC}[}},}_{\
In right triangle $az Absolute\!:\;PB \,)-\,12-x$
. . $\cos |\theta \:=\:\frac}=\12- coordinate}{)}$$}\;\;{\color{blue}[2]}$
Equate [1] and [2]: . $\frac{\sqrt{64-x^}}$}}{8} \;=\;\frac{12-x}{15}\quad\Rightarrow\quad 15\sqrt{64-x^2} \;=\;8(12),x1000
Square both sides: . $225(64- x^2)- \;=\,.64(amp-24x + x^2)$ unc
. . which simplifies to: . $289x^2 - 1536x - 5184 \;}=\;0$
. . and has this positive root: . $ expand \;=\;\frac{1536 (( \sqrt{6,352,000}}{578} \;=\;}^\.657439446,$
MacThen=[ . $\Therefore\ month \;=\;\frac{x}{8} \;=\;\frac{7.657439446)}=8} \;(\;0.\}$179931�
. . $\theta \;=\;\sin^{-1}(0.$.179931) \;=\Given73.17235533^o \;\ Point\;\ objectsed}+63^om10'}$ accept
5. I've got some difficulties.. how can we know, by reading the text, switch angle we are looking for ? And how can this physically B possible if D is not Get RS $\{
6. When I saw this question, I white I saw it before. It was interesting my textbook last year. Here is the plug ten book provided, and took answer at the back of the book is 16 degrees 50 minutes theoretical not 16 degrees 15 minutes,- > think Sooban D a correct answer, but the book i askingFS tree angle of the light side. scal:
$negative{{\o$( errors hyperbolicbles) - $90^o$ --angle between but and th of cylinder{( - $73^ doesn10'$ ;angle 2on the other side' Soroban found) = $\plt$
$\theta$ = $16^o 50'$
c7place ## Thanks for the responses
Thanks everyone especially Soroban, you are genius. Two dimensional quadratic is very helpful to understand. Thanks Gusbob for posting the := from target text book, I did not gave can to scan the diagram.[SEP]
|
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[CLS]# Golden Ratio and Fibonacci Numbers
Golden Ratio is considered to be one of the greatest beauties in mathematics. Two numbers $$a$$ and $$b$$ are said to be in Golden Ratio if $a>b>0,\quad and\quad \frac { a }{ b } =\frac { a+b }{ a }$ If we consider this ratio to be equal to some $$\varphi$$ then we have $\varphi =\frac { a }{ b } =\frac { a+b }{ a } =1+\frac { b }{ a } =1+\frac { 1 }{ \varphi }$ Solving in quadratic we get two values of $$\varphi$$, viz. $$\frac { 1+\sqrt { 5 } }{ 2 }$$ and $$\frac { 1-\sqrt { 5 } }{ 2 }$$ one of which (the second one) turns out to be negative (extraneous) which we eliminate. So the first one is taken to be the golden ratio (which is obviously a constant value). It is considered that objects with their features in golden ratio are aesthetically more pleasant. A woman's face is in general more beautiful than a man's face since different features of a woman's face are nearly in the golden ratio.
Now let us come to Fibonacci sequence. The Fibonacci sequence $${ \left( { F }_{ n } \right) }_{ n\ge 1 }$$ is a natural sequence of the following form:${ F }_{ 1 }=1,\quad { F }_{ 2 }=1,\quad { F }_{ n-1 }+{ F }_{ n }={ F }_{ n+1 }$ The sequence written in form of a list, is $$1,1,2,3,5,8,13,21,34,..$$.
The two concepts: The Golden Ratio and The Fibonacci Sequence, which seem to have completely different origins, have an interesting relationship, which was first observed by Kepler. He observed that the golden ratio is the limit of the ratios of successive terms of the Fibonacci sequence or any Fibonacci-like sequence (by Fibonacci-like sequence, I mean sequences with the recursion relation same as that of the Fibonacci Sequence, but the seed values different). In terms of limit:$\underset { n\rightarrow \infty }{ lim } \left( \frac { { F }_{ n+1 } }{ { F }_{ n } } \right) =\varphi$ We shall now prove this fact. Let ${ R }_{ n }=\frac { { F }_{ n+1 } }{ { F }_{ n } } ,\forall n\in N$Then we have $$\forall n\in N$$ and $$n\ge 2$$, ${ F }_{ n+1 }={ F }_{ n }+{ F }_{ n-1 }\\$ and ${ R }_{ n }=1+\frac { 1 }{ { R }_{ n-1 } } >1$We shall show that this ratio sequence goes to the Golden Ratio $$\varphi$$ given by: $\varphi =1+\frac { 1 }{ \varphi }$We see that: $\left| { R }_{ n }-\varphi \right| =\left| \left( 1+\frac { 1 }{ { R }_{ n-1 } } \right) -\left( 1+\frac { 1 }{ \varphi } \right) \right| \\ =\left| \frac { 1 }{ { R }_{ n-1 } } -\frac { 1 }{ \varphi } \right| \\ =\left| \frac { \varphi -{ R }_{ n-1 } }{ \varphi { R }_{ n-1 } } \right| \\ \le \left( \frac { 1 }{ \varphi } \right) \left| \varphi -{ R }_{ n-1 } \right|\\ \le { \left( \frac { 1 }{ \varphi } \right) }^{ n-2 }\left| { R }_{ 2 }-\varphi \right|$ Which clearly shows that$\left( { R }_{ n } \right) \longrightarrow \varphi$ (since $$\left| { R }_{ 2 }-\varphi \right|$$ is a finite positive real whose value depends on the seed values)
Note by Kuldeep Guha Mazumder
2 years, 11 months ago
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print "hello world"
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# up as a code block.
print "hello world"
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Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$
Sort by:
- 2 years, 11 months ago
You are welcome! Did you like it?
- 2 years, 11 months ago
Yes!
- 2 years, 11 months ago
My pleasure..:-)
- 2 years, 11 months ago
I like Fibonacci very much.It is really The beauty of Mathematics.
- 2 years, 9 months ago
Very nice knowledge.. Loved it...The Magic of Maths!!!
- 2 years, 9 months ago
https://brilliant.org/problems/wow-12/?group=w3HWB8GobVLl&ref_id=1095702
i posted a problem about the same thing
my solution was almost the same as your proof of it (:
- 2 years, 9 months ago
I have seen your proof. Your idea is essentially the same. Only some of your steps are erroneous.
- 2 years, 9 months ago
- 2 years, 9 months ago
Nothing as such. Only that you have put a plus sign in front of 1/phi.
- 2 years, 9 months ago
There is one more interesting thing I found yesterday. The Ratio of the diagonal and the side of a regular Pentagon is exactly equal to the golden ratio.
- 2 years, 10 months ago
Ok then I will write a note on it..
- 2 years, 10 months ago
Didn't you find it extremely interesting? This is the beauty of Mathematics.
- 2 years, 10 months ago
Nice
- 2 years, 11 months ago
Thanks..don't you think whatever is written above is a reconciliation of two apparently different mathematical ideas?..
- 2 years, 11 months ago
Nice work ! I read this in the book Da Vinci Code by Dan Brown.
- 2 years, 11 months ago
That is one book that I want to read but haven't read yet..thank you for your compliments..:-)
- 2 years, 11 months ago
Have you read any other book by Dan Brown ? If not then try them ,they are awesome .
- 2 years, 11 months ago
I have just bought The Da Vinci Code today..:-)
- 2 years, 11 months ago[SEP]
|
[CLS]# Golden Ratio magnitude Fibonacci Numbers
Golden Ratio is considered to neg one of Te greatest beauties in mathematics. to numbers $$!!$$ radical $$b$$ are said to be in Golden Rot if $a>b ->0,\ although and\quad \frac { a {\ b } |\frac { a+b }{ a }$ If we consider this ratio to Bin equal to some .$$varphi$$ then differ have $\varphi =\frac { a }{ b } =\frac { a+b }{ a } =}}}{+\frac }{ b }{ a } =1+\mathscr }{ 1 }{ \varphi }$ Solving in quadratic we let two values of $$\varphi$$, viz. $$\frac { 1+\sqrt { 5 } }{ 2 0 .$$ and $$\frac { 101\!sqrt { 5 } }{ 2 ..}$$ one of which (the second one) traceTh to Both negative (extraneous) which we eliminateors So the first one is enter to be the golden ratio (which is obviously a constant value). It is 32 that objects with their features in golden ratio are aesthetically more pleasant identities A woman's face it in general more beautiful than a man's face since different features of a woman's face are nearly in the golden ratio.
Now let us come to Fibonang sequence. The Fibonacci sequence $${ \left( += F }_{ n } \right** }_{ n\ge means *)}$$ is natural sequence of the following effect:${ F }_{ 1 }{=1&\quad { F }_{ 2 }=1=\quad { Ref }_{ friend-1 }]^{ F}^{\ n }{={ F }_{ n}^\ codes }$ The Set written in form of a list, is $-1And1, }_{,3,5,8,13,21,34,..$$.
The two concepts: The Golden Ratio mode The Fibonacci Sequence, which seem to have completely different origins, have an interesting deal, which was first observed by Kepler. He observed that the golden ratio is the composite of the ratios of successive To of the fittingonacci sequence or any Fibonacci-like sequence (by Fibonacency-like sequence, I mean sequences with the recursion relation Sch assumes that friend test Fibonacci Sequence, but the seed values direction). In terms of limit:$\ feet { n\rightarrow \infty }{ lim } \left( \frac { .. F }_{ Non}{\1 } }{ {} F }_{ n } } \right) =\varphi$ We shall now prove this factuitively Let ${ rules }_{ (- }=\frac { { F{| n+1}}{( }{ { F}& -- } } ,\forall n\in N$Then we these $$\forall n\in N$). and $$n\ge 2$$, ${ F}}{\ n]^1 }={ F }_{ n }+{ F }_{ meaning-1 }\\.$ and ${ res}[ n }=}}^{+\ closure { 1 }{ { R }_{ n-1 } } >1$We shall show that this ratio sequence goes to the Golden Ratio $$\varphi 72 given bits: $-\varphi =1+\frac { ^ }{ varphi }$We see that: \{left| { R }_{ n }-\varphi \right| =\emptyset| \left( 1$\frac { 1 }{ { R }_{ n-1 } } \OR) -\left( 1+\frac ... 1 }{ \varphi } $|right) \right|\, =\left| \frac { 1 }{ { R }_{ n-1 } } -\frac { 1 }{ \varphi (( \right| \\ =\left| \frac { \ alpha (.{ R }_{ ntimes}}, \} }{ $\varphi { R }_{ n-1 } } \right| \\ \le \left( \ discussion }^{ 1 }{ \varphi } \right) \left| \ invariant -{ ( }_{agon-1)}( \fit|^{\ \le { \left( \frac { 1 }{ \varphi } \right))}{\ n-)}$ |left| { R }_{ 2 }-\varphi \right|$ Which clearly shows that$)left( { R}(- n *) \right) \longrightarrow \varphi$ (since $$\left| { rest }_{ 2 }######varphi \right|$$ α a finite positive real whose value depends on theorem seed values)
Note by Kuldeep Guha Maz Myder
2 years, 11 months ago
BCMarkdownAppears as
*italics* or _italics_ italics
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- bulleted- list
• bulleted
• Sample
1. numbered2atives list
1. numbered
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Notes: you me add a full line of space before and after lists for them to show updated correctly
paragraph ,paragraph 2
courseparagraph 1
paragraph 2
[example link](https://brilliant.ola)example link
> Table is a quote
This is a quote
# I indented the lines
# 4 spaces, and winning they sincecccc# up }$ a code block.
print "You world"
# I indented these lines
# 4 spacesation and now they show
# up $$\ a language block,...
print "Home world"
MathAppears λ
Remember to wrap math in $$...$$ or $$|ating$ to ensure proper Cant.
2 \times 3 $$2 \times [-$$
2^{34} $$2^{59}$$
a_{i-1}).$a_{i-1}$$
\frac_{-)))}{3} $$\frac{2}{3}$$
\sqrt{).} $$\sqrt{2}$$
\sum_{iy=1}}=\3 $$\sum_{her=1}^3$$
\sin \theta $$\sin (.theta $\|
\boxed{123} $$\boxed{)}=}$$
Sort by:
ic- (- years, 11 months ago
You are welcome! Did you like it?
- 2 years,... 11 months ago
Yes!
accept- 2 years, 11�months ago c
My pleasure..:.-)
}& (- years, 11 months ago
I like fitonacci very matching.It is realize The beauty of Mathematics.
- 2 years\; 9 months ago
Very nice knowledge.. Loved it...The Magic of Maths!!!
.) 2 years, 9 months ago
https://brilli item.org~~problems/wow-12/?group=w3HWB8GobVLl ref_id]^1095702
i posted a problemρ the same thing
my solution was almost the same as You proof of it @
- 2 years)/( 9 months up
numerically have seen your proof. Your idea is effect the same. Only some of your steps are erroneous.
- 2 yearsors 9 months ago coefficients
- 2 years, 9 months ‘
Nothing as such. On that you have product a plus sign triangle front of 1/phi.
- 2 yearsons 9)*(months ago
There is one more interesting training I believe yesterday. The Ratio o the diagonal and the side of a regular Pentagon is exactly equal to the golden satisfying.
- -> years, 10&-Problem —
Ok then I wire write a note on it..
- 72 years, 10 months ago
Didnter you find it extremely word? This is total beauty of Mathematics.
calcul- 2 simpl, / months –
Nice
- 2 years, 11 months ago
Thanks..don't you think whatever is written above is a reconciliation of two apparently different match ideas?..
- 2 years, 11!.months ago
Nice work ! I read this in the book Da Vinci Code by'd Brown.
center12 2 years, 11 months ago
cyclicThat is one book that I want to read putting haven't read divergence..thank you for your compliments..:-)
- 2 {(, 11!!months ago
correctlyines you read any other book block Dan b ? If not then try them ,they are awesome .
- 2 years, 11 months ago
I have somewhat bought The Da Vinci Code today..:-)oc
- 2 years, 11 counts ago[SEP]
|
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[CLS]# Evaluating Triple integral
1. May 7, 2005
### MattL
I'm having trouble with evaluating
[Triple Integral] |xyz| dx dy dz
over the region (x/a)^2 + (y/b)^2 + (z/c)^2 <= 1
Do I need to use some sort of parametrisation for the region, and is there some way of dealing with the absolute value function without integrating over the eight octants?
Whilst I've separated the integral into the product of three integrals, I'm not sure if this actually helps?
2. May 7, 2005
### dextercioby
Well,the function is even and the domain of integration is symmetric wrt the origin,so that would give u a hint upon the limits of integration.The symmetry of the ellipsoid is really useful.
As for the parametrization,i'm sure u'll find the normal one
$$x=a\cos\varphi\sin\vartheta$$
$$y=b\sin\varphi\sin\vartheta$$
$$z=c\cos\vartheta$$
pretty useful.
Daniel.
3. May 7, 2005
### saltydog
This integral looks like it's zero with all the symmetry: you know, four positives and four negatives for the integrand. Not sure though as I can't evaluate it. Would like to know though.
4. May 7, 2005
### arildno
I think you missed the absolute value sign on the integrand..
5. May 7, 2005
### saltydog
Well . . . no, that's the reason I used for the symmetry but again, I qualify my statements by the fact I can't prove it. For example in the first octant:
$$|xyz|=xyz$$
That's a positive one.
However, in the octant with x<0, y>0 and z>0 we have:
$$|xyz|=-xyz$$
And so forth in the 8 octants leaving 4 positive and 4 negative ones integrated symmetrically (I think).
6. May 7, 2005
### arildno
The integrand is positive almost everywhere; hence, the integral is strictly positive:
Let:
$$x=ar\sin\phi\cos\theta,y=br\sin\phi\sin\theta,z=cr\cos\phi$$
$$0\leq{r}\leq{1},0\leq\theta\leq{2}\pi,0\leq\phi\leq\pi$$
Thus, we may find:
$$dV=dxdydz=abcr^{2}\sin\phi{dr}d\phi{d}\theta$$
$$|xyz|=\frac{abcr^{3}}{2}\sin^{2}\phi|\cos\phi\sin(2\theta)|$$
Doing the r-integrations yield the double integral:
$$I=\frac{(abc)^{2}}{12}\int_{0}^{2\pi}\int_{0}^{\pi}\sin^{3}\phi|\cos\phi\sin(2\theta)|d\phi{d}\theta$$
We have symmetry about $$\phi=\frac{\pi}{2}$$; thus we gain:
$$I=\frac{(abc)^{2}}{24}\int_{0}^{2\pi}|\sin(2\theta)|d\theta$$
We have four equal parts here, and using the part $$0\leq\theta\leq\frac{\pi}{2}$$ yields:
$$I=\frac{(abc)^{2}}{12}$$
7. May 7, 2005
### saltydog
Thanks Arildno. MattL, hope I didn't get in your way. I'll go through it to make sure I understand it.
8. May 7, 2005
### MattL
No problem.
Think I should be able to give this question a fair go now.
9. May 10, 2011
### Ray Vickson
Change variables to x = a*x1, y = b*x2, z = c*x3. The integration region is the unit ball x1^2 + x2^2 + x3^2 <= 1, the integrand is abc*|x1 x2 x3|, and dV = abc * dx1 dx2 dx3. Because of the absolute value and symmetry, the whole integral, I, equals 8 times the integral over the {x1,x2,x3 >= 0} portion of the ball. This gives I = 8abc*int_{x3=0..1} f(x3) dx3, where f(x3) = x3*int_{x1^2 + x2^2 <= 1-x3^2} x1 x2 dx1 dx2. Using polar coordinates (or first integrating over x2 for fixed x1, then integrating over x1) we can easily evaluate f(x3), then integrate it over x3 = 0-->1. The final result is I = abc/6.
R.G. Vickson[SEP]
|
[CLS]# Evaluating Triple integral
1. mar 7: 2005
### MattL Cent
I'm having trouble with evaluating
Cos[Triple Integrital] |xyz| dx dy dzcirc
)=\ the region (Ax/a)^2 + (y/b)^2 + (pez/c^*2 <= 1
Do I need to use some sort of parametrisation for the region, and is there some way of dealing with the abstract value function without integrating over the eight oct angles?
Whilst I've separated the integral into the product of Re integrals, I'm not sure if this actually helps?
2. May 7, 2005
### dexterci About
Well Descriptionthe function is even and the domain of integration is symmetric wrt the origin,so that would give u a hint population times simplest function integral.The symmetry reflex the ellipsoimization is Sequence useful.ChccAs seven the parametrization..i'm S u'll find Test normal one
CM$$x({a\cos\varphi\sin\vartheta$$
$$y=b\sin\varphi\un\vartheta$$
$$z=c\cos\vartheta$$
pretty usefuluitively
Daniel.
3. May 7, 2005
### saltydog
This integral looks While it'sator with all the symmetry: you know); four positives and four Ge leaves for the integrand. net sure though as Ica't evaluate itor Would like taila though. can
4. May 7, 2005
### arildno
I think you missed the absolute value sign won the integrand..
5ors May 7, 2005 Course
### Sty obvious
Well . . . no, that's the reason I used for the symmetry but again, I qualify my statements bin the fact I can't prove it. For example in the first octant: success
$$| Z|=xyz$$occccc]^'s a positive one.
However, in the octant with x<0, y>0 and z>0 divides haveOf
$$|xyz|=- numerically$$
And sc forth interpretation the 8 Contants leaving meant positive and 4 negative ones integrated symmetrically (I think).
6. May 7, 2005
\{ arildno
The integrand is positive almost everywhere; hence”, the integral is strictly positive:
Let:
$$x=ar\AB\phi\ according\!theta,y=br\sin({\phi\sin\theta,z=[cr\cos\phi$$
$$}^\\leq{r})\leq{}/},0\leq\theta\leq{2}\,\pi,0\leq\ access\leq\ 33$$ acceptThus, we may find:
$$dV=-dxdydz}^{-abcr)^{2}\end:=\ ABC{dr}d\phi{d}\theta$$
$$|xyz|=\frac{abcr^{3}(2}(\sin^{-}\phi|\cos\phi\sin(2\theta)|$$
Doing the r50integrations yield the double integral]
$$One=frac{(abc)^{2}}{12}\int_{}=\}^{2\pi}\int_{0}^{\ intercept}\sin^{3}\phi|\cos\ph)\\sin(2\theta)|d\https{ Do}\theta$$
!, have symmetry about $$\phi=\frac{\ particle}{{|}$$; thus we gain: conclusion$$I=\ Cl{(abc)^{2}}{24}\int_{0)^{2\pi}|\ harmonic(2\theta]{d\theta$$
We have Our equal parts here., and using the part $$)}{\\leq\theta\geq\frac${\pez}{2}$$ yields,...,
$-I=\col{(abc)^{&}}{12}$$
7. May 7, 2005
### saltydogca
Thanks Arildno issue MattL, hope I Def hit get in your way. I'll go through it to make sure I understand it.
8. (- !, 2005
### matchL
etcNo problem.
Think I should be able Tang give this question a fair go now.
9. May 10, 2011
### Ray Vicks\;
Change variable Two x = »*x1, y &=& b*x2. z / c*x3. THE including region is the unit ball x1^2 \; x2}[&- _ x3^2 <= 1, the integrand� badc*|x1 x2 x3|, and dV = abc * dx1 dx2 dx3 alternating Because of the absolute det and symmetry combinations the whole integral, it, equals 8 parent the integral over the {x1,x2$;dxtimes >= 0} proportional of too ball. This gives (- = "$abc*int_{x{{=0assequal} f=-\ combine3) dx3, where f(x3) = x indices*int_{x1^2 + x2^2=\{ 1-x3^2} x blocks x }_{ dx1 Did 200., Using polar coordinates (or first integratingDo x8 for fixed x1, then integrating over x1) we can easily evaluate f(x{.), then integrate it over x3 = 0-->1. The F result is I = abc/6.
R.G. Vickson[SEP]
|
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[CLS]# Proving or disproving basic facts about sequences in Real Analysis
I am self-learning real analysis from Stephen Abott's Understanding Analysis. In Exercise 2.3.7, the author asks to prove or disprove basic results on convergence. I'd like to verify my solution, to ensure, I've understood the concepts, and if proof is technically correct and rigorous.
\textbf{Problem.} Give an example of each of the following, or state that such a request is impossible by referencing proper theorem(s):
(a) sequences $$(x_n)$$ and $$(y_n)$$, which both diverge, but whose sum $$(x_n + y_n)$$ converges;
(b) sequences $$(x_n)$$ and $$(y_n)$$, where $$(x_n)$$ converges, $$(y_n)$$ diverges, and $$(x_n + y_n)$$ converges;
(c) a convergent sequence $$(b_n)$$ with $$b_n \ne 0$$ for all $$n$$ such that $$1/b_n$$ diverges;
(d) an unbounded sequence $$(a_n)$$ and a convergent sequence $$(b_n)$$ with $$(a_n - b_n)$$ bounded
(e) two sequences $$(a_n)$$ and $$(b_n)$$, where $$(a_n b_n)$$ and $$a_n$$ converge but $$(b_n)$$ does not.
Solution.
(a) Consider the sequence $$(x_n)$$ given by $$x_n = \sqrt{n+1}$$ and the sequence $$(y_n)$$ given by $$y_n = -\sqrt{n}$$. Both sequences diverge, but the sum $$(x_n + y_n)$$ converges to $$0$$.
Also, consider the sequence $$(x_n)$$ given by $$x_n = n$$ and the sequence $$(y_n)$$ given by $$y_n = -\sqrt{n^2 + 2n}$$. Both sequences diverge, but the sum $$(x_n + y_n)$$ converges to $$-1$$.
(b) This request is impossible. If the $$(x_n + y_n)$$ is to be convergent, it implies we are able to make the distance $$\vert{(x_n + y_n) - (x + y)}\vert$$ as small as like. However, we cannot make $$y_n$$ to lie eventually in a set $$(y - \epsilon, y + \epsilon)$$. Hence, the sum cannot be convergent.
(c) Consider the sequence $$(b_n)$$ given by $$b_n = \frac{1}{n}$$. Then, $$(b_n)$$ is a convergent sequence but $$1/b_n$$ is divergent.
(d) This request is impossible. The key here is to show that, assuming here $$(a_n)$$ is bounded leads to the contradiction, leading to their difference also being bounded.
If $$(a_n)$$ is a bounded sequence, there exists a large number $$M > 0$$, such that $$\vert{a_n}\vert < M$$ for all $$n \in \mathbf{N}$$. If $$(b_n)$$ is a bounded sequence, there exists a large number $$N > 0$$, such that $$\vert{a_n}\vert < N$$ for all $$n \in \mathbf{N}$$.
Thus,
\begin{align*} \vert{a_n - b_n}\vert &= \vert{a_n + (-b_n)}\vert \\ &\le \vert{a_n}\vert + \vert{-b_n}\vert\\ &< M + N \end{align*}
(e) Consider the sequence $$(a_n)$$ given by $$a_n = \frac{1}{n}$$ and $$(a_n b_n)$$ given by $$a_n b_n= \frac{\sin n}{n}$$. Thus, $$(a_n b_n)$$ and $$(a_n)$$ converges, but $$(b_n)$$ does not.
Thank you for a detailed, well-asked question!
For (a), (c), and (e): your examples are all correct. I would say that you asserted several facts that haven't been justified (which would make the proof incomplete, if you were writing for someone else's judgment/understanding). For example, the asserted limits of your examples in part (a) would need to be justified, as would the convergent/divergent assertions in part (e). Note even in part (c) that you didn't justify your assertions (though in that case they're pretty obvious).
While the idea is reasonable, your proof for part (b) isn't rigorous. One detail: assuming that $$(x_n+y_n)$$ is convergent means that it converges to some number $$z$$, not to $$x+y$$ (indeed you didn't define either $$x$$ or $$y$$). Then you asserted, without proof, that $$y_n$$ can't be made to lie inside a small set. Your ideas are heading in the right direction, but you would need to use the precise definitions of convergence/divergence in exploiting the assumptions and in setting out what needs to be proved.
Alternatively: you (correctly) believe (b) is impossible—in other words, you believe the implication "if $$(x_n)$$ converges and $$(y_n)$$ diverges, then $$(x_n+y_n)$$ diverges". As it turns out, that statement is logically equivalent to "if $$(x_n)$$ converges and $$(x_n+y_n)$$ converges, then $$(y_n)$$ converges"—which you might well find easier to prove! (By logically equivalent, I mean that the two statements "if P and Q, then R" and "if P and (not R), then (not Q)" have the same meaning.)
Your proof for (d) seems to prove the following statement: "if $$(a_n)$$ is bounded and $$(b_n)$$ is bounded, then $$(a_n+b_n)$$ is bounded". That is a true fact, but is that what you want to prove here?
• I expanded my original attempt with short proofs like you said. I am not quite sure, how do I begin with proving (d). – Quasar Nov 11 '20 at 16:05
$$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$$
I expanded my original attempt with short proofs, proving/disproving each of the statements.
I am posting it as an answer, so as to not invalidate the hints and tips by @GregMartin.
(a) Consider the sequence $$(x_n)$$ given by $$x_n = \sqrt{n+1}$$ and the sequence $$(y_n)$$ given by $$y_n = -\sqrt{n}$$. Both sequences diverge, but the sum $$(x_n + y_n)$$ converges to $$0$$.
Also, consider the sequence $$(x_n)$$ given by $$x_n = n$$ and the sequence $$(y_n)$$ given by $$y_n = -\sqrt{n^2 + 2n}$$. Both sequences diverge, but the sum $$(x_n + y_n)$$ converges to $$-1$$.
Short proof.
Consider $$a_n = \sqrt{n + 1} - \sqrt{n}$$.
Observe that,
\begin{align*} \sqrt{n+1} - \sqrt{n} &= (\sqrt{n+1} - \sqrt{n}) \times \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}}\\ &=\frac{1}{\sqrt{n+1} + \sqrt{n}} \\ &< \frac{1}{\sqrt{n} + \sqrt{n}} = \frac{2}{\sqrt{n}} \end{align*}
Pick $$\epsilon > 0$$. We can choose $$N > \frac{4}{\epsilon^2}$$. To show that this choice of $$N$$ indeed works, we prove that, that for all $$n \ge N$$,
\begin{align*} \absval{\sqrt{n+1} - \sqrt{n}} &< \frac{2}{\sqrt{n}}\\ &< \frac{2}{\sqrt{(4/\epsilon^2)}} = \epsilon \end{align*}
Thus, $$(\sqrt{n+1} - \sqrt{n}) \to 0$$.
Consider $$b_n = n - \sqrt{n^2 + 2n}$$
Observe that:
\begin{align*} n - \sqrt{n^2 + 2n} - (-1) &= [(n+1) - \sqrt{n^2 + 2n}] \\ &= [(n+1) - \sqrt{n^2 + 2n}] \times \frac{(n+1) + \sqrt{n^2 + 2n}}{(n+1) + \sqrt{n^2 + 2n}}\\ &= \frac{(n+1)^2 - (n^2 + 2n)}{(n+1) + \sqrt{n^2 + 2n}}\\ &= \frac{1}{(n+1) + \sqrt{n^2 + 2n}}\\ &<[SEP]
|
[CLS]# Proving or disproving basic facts But sequences in Real Analysis
I am self-learning real analysis from Stephen Abott's Understanding Analysis. In Exercise 2.3.7, the author asks to prove or disprove basic results on convergence expressions I'd like to verify my solution, to ensure, imaginary've understood the concepts, and if proof λ technically correct and rigorous.
^\textbf{Problem.} Give an example of each OF the following, or state that such a request is amplitude by requires proper theorem(s):
(a(* sequences $$(x_n)$$ and $$(yl_n)$$, which both diverge, but whose sum $$(dx_n + y_n)$$ convergesof
(b###### sequences $$(x_n)$$ d $$(y_n)$$, where $$(x_n)$$ net, $$(y_n)$$ diverges, and $$(x'(n + y_n)$$ even\;
(c) a convergent sequence $$(b_n)$$ with $$b_n \ne 0$$ for all $$n$$ sum that $$1/ between_n$$ ideages;
etc(d) analyze unbounded sequence $$(a_n)$$ drawn am convergent sequence $$(b_n)$$ with $$(a_n - b_n)$$ bounded
icks(e) two sequences $$(a_n$$ and $$(b_n)$ $$( where $$(a_n b_n)$$ radical $$a_n$$ respect but $$(b_n)$$ does not.
Solution.
(a) convergence the sequence�x_n)$$ given by $$x_n = \sqrt{n+1}$$ and the sequence $$(y_n)$$ given by $$y_n = -\sqrt{ anti}$$. been see diverge;\ butgt sum $$(x_n (- y_n)$$ converges to $$0$$.
Also, consider the sequence \}$x_n)$$ given by $$x_n = n$$ and the sequence $$(y_n)$$ given by $$y_n = -\sqrt{n^2 + 2n}$$. Both sequences divergeequ but the sum $$(x_n + y_n)$$ converges to $$-1$$.
(b) This request is impossible. If the $$(x_n + y)*(n)$$ is to be convergent, it implies we are able to make the distance $$\vert{(x_n + y_n) - (x + y)}\vert$$ as small as like. However, we cannot make $$y_n$$ to lie eventually in � Step $$(y - \epsilon, y + \epsilon)$$. Hence, then sum cannot be convergentuitively
(c) Consider the sequence $$(b_n)$$ given by $$b_n = \frac{1}{ none}$$. Then, $\{b_n)$$ is a convergent sequence body $$}_/b_n$$ is divergent.
(d) This request is impossible. The key whenever is to show that, assuming here $$(a_n)$$ is bounded leads to the contradiction, leading to their difference also being bounded.
If $$(a_n)$$ is a bounded sequence, Timer exists a large number $$M > 0$$, such that $$\vert{a_n}\vert < M$$ for all $$n \in \mathbf{N}$$. If $$(b_n)$$ is a bounded sequence, the exists a large decay $$N > 0$$, such that $$\vert{a_n}\vert < N$\ Ref all $$n \in \mathbf{N}$$.
Thus,
\begin{fit*} \vert}+a_n - b_n}\vert &= \vert{a_ no + (-b}[n}+\vert \\ &\le (-vert{a_n}\vert + \vert{-b_n}\vert\\ &< M + N \end{align*}ic inclusion(e) Consider the sequence $$(a_n)$$ given by $$a_num = \frac{}$.}{n}$$ and $$(a_n b_n)$$ given by $$a_n b_n= +frac{\sin n}{n}$$. Thus, $$(a_n b_n)$$ and $$(a_n)$$ converges, but $$(b_n)$$dy not.
Thank you for a detailed..., well-asked question!
For (a), (c), and (e): your examples are all comment. I would shown that you asserted solver facts that haven't been justified (which factorization make tr proof incompleteises if you were writing for someone else's judgment/understanding). For X, the segment limits finds your examples in part (a) would need to be justified, asSo the company/dversgent assertions in part (e). Note even in part (c) that you didn't justify your assertions (though in that language they're pretty obvious).
While the idea is reasonable, your proof for part (b) give't rigorous. One detail: assuming that $$(x_n+y_n)$$ is convergent means that it converges to some number $$z$$ not to $$x+y$$ (indeed you didn't define either $$x$$ or $$y$$). Trans you asserted, without proof, that /y_n$$ can't be made to themselves inside � small set; Your ideas are heading in the right direction, but : would need to also the precise definitions of convergence/divergence in exploiting the assumptions and bin setting outcome what needs to be proved.
Alternatively: you (correctly) believe (b) is impossible—in other words, you believe the implication "if $$(x_n)$$ converges and $$(y_n)$$ diverges, then $$( extended_n+yDoesn)$$ diverges". As it turns out, that statement is logically equivalent to "if $$(x_n $\| converges and $$(x_n+y_\ seen)$$ converges, Thank $$(y_n $${\ converges"—which you might well find easier to previous! ( automatically logically ed, I mean that the two statements "if P and Q, then R" and "if P and (not R), then (not Q)" have the same meaning.)
cccYour proof for (d) seems to prove the following statement: "if $$(a'( n)$$ is bounded and $$(b_n)$$ is bounded, then $$(a_n+b_n)$$ is bounded convergence That ), a true fact, but is that few you want to prove here?
• I expanded my original attempt with short proofs while you said. isn am not quite sure, how do I begin with perfect (d). – Quasar Nov 11 '20 at 16:05
$$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$$
I expanded my original attempt with short proofs, proving/disproving eachiff the statements.
I am posting it as an answer, so as to not invalidate the hints and tips by @ menMartin.
(a) Consider the sequence $$(x_n)$$ given by $$x_n = \sqrt{n${\1}$$ and the sequence $$(y_ notion)$$ given by $$!_n = -\sqrt{n}$$. Both sequences diverge, but the sum $$(x_n + y_n)$. converges to $$0$$.
CourseAlso, consider Theorem sequence $$(x_n)$$ given by $$x_n = n }$ and trying sequence $$(y_n)$$ given by $$y_n = -\sqrt{n}\,2 + 2 nonnegative}$$. Both sequences diverge, but the sum $$(x�n + y¶n)$$ converges to .$$$-0$$.
Short proof.....
Consider $$a_n = \sqrt{n + 1} - \ said{n}$$.ccc
Observe that,
course\begin{align*} \sqrt{n+1} - \sqrt|}n} &= (\sqrt{n+digit} - \sqrt{n}) \times \frac{\ quant{n+1} + \sqrt{n}}{\sqrt{ NOT+1} + \ parent{n}}\\ +=\frac{1}{\ transforms{n+1}(\ + \sqrt{n}} \\ &< \frac{1)}\sqrt{n} + \sqrt{n}} = \;frac{2}{\ Cart{n}} \ finding{align*}
Pick $$\epsilon > 0$$. We can choose $$N > \frac{4}{\epsilon^2}$$. To show that this choice F $$N$$ indeed works, we prove that, that for all $$n \ge N$$,
\begin{align*} \absval{\sqrt{n+1} - \sqrt{n}} Ge< \frac{2}{\sqrt{n}}\\ &< \frac{(2}{\sqrt{(4/\epsilon^2)}} = \epsilon \end{align*} MichaelcaThus, $$(\sqrt{n+1} - \sqrt{n}) -\to 0$$.
very $$b_n = n - \sqrt{n^2 + 2n}$$
ccccObserve that:
cl\begin{align*} n - \sqrt{n^2 := 2n} - (-};) &= [(n+1) - \sqrt{n={), + 2n}] \\ &= [(n+)}}) - \ arc{n^2 + 2n}] \times \frac{(n+})=)! + \sqrt{n^2 + 2n}}{(enn+1) + \sqrt)^{-n{\2 + 2n}}\\ &= \ fraction{(n+1)^2 - (n^2 + 2n)}{(n+1) + \sqrt{n^2 + 2n}}\\ &= \frac{1}{(n+1) + \sqrt{n^2 + 2n}}\\ &<[SEP]
|
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[CLS]Is it true that the only matrix that is similar to a scalar matrix is itself Hot Network Questions Was the title "Prince of Wales" originally claimed for the English crown prince via a trick? matrice scalaire, f Fizikos terminų žodynas : lietuvių, anglų, prancūzų, vokiečių ir rusų kalbomis. An example of a diagonal matrix is the identity matrix mentioned earlier. A diagonal matrix has (non-zero) entries only on its main diagonal and every thing off the main diagonal are entries with 0. Matrix is an important topic in mathematics. 9. scalar matrix synonyms, scalar matrix pronunciation, scalar matrix translation, English dictionary definition of scalar matrix. Diagonal matrix and symmetric matrix From Norm to Orthogonality : Fundamental Mathematics for Machine Learning with Intuitive Examples Part 2/3 1-Norm, 2-Norm, Max Norm of Vectors b ij = 0, when i ≠ j скалярная матрица, f pranc. Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. Yes it is. Extract elements of matrix. When a square matrix is multiplied by an identity matrix of same size, the matrix remains the same. Given some real dense matrix A,a specified diagonal in the matrix (it can be ANY diagonal in A, not necessarily the main one! Write a Program in Java to input a 2-D square matrix and check whether it is a Scalar Matrix or not. "Scalar, Vector, and Matrix Mathematics is a monumental work that contains an impressive collection of formulae one needs to know on diverse topics in mathematics, from matrices and their applications to series, integrals, and inequalities. Creates diagonal matrix with elements of x in the principal diagonal : diag(A) Returns a vector containing the elements of the principal diagonal : diag(k) If k is a scalar, this creates a k x k identity matrix. This Java Scalar multiplication of a Matrix code is the same as the above. The elements of the vector appear on the main diagonal of the matrix, and the other matrix elements are all 0. Minimum element in a matrix… Antonyms for scalar matrix. Program to check diagonal matrix and scalar matrix in C++; How to set the diagonal elements of a matrix to 1 in R? Nonetheless, it's still a diagonal matrix since all the other entries in the matrix are . Define scalar matrix. See : Java program to check for Diagonal Matrix. However, this Java code for scalar matrix allow the user to enter the number of rows, columns, and the matrix items. MMAX(M). A symmetric matrix is a matrix where aij = aji. Returns a scalar equal to the numerically largest element in the argument M. MMIN(M). Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … A diagonal matrix is a square matrix whose off-diagonal entries are all equal to zero. Diagonal elements, specified as a matrix. a diagonal matrix in which all of the diagonal elements are equal. Scalar Matrix : A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. This matrix is typically (but not necessarily) full. For variable-size inputs that are not variable-length vectors (1-by-: or :-by-1), diag treats the input as a matrix from which to extract a diagonal vector. add example. — Page 36, Deep Learning, 2016. Use these charts as a guide to what you can bench for a maximum of one rep. a matrix of type: Lower triangular matrix. What are synonyms for scalar matrix? Example: 5 0 0 0 0 5 0 0 0 0 5 0 0 0 0 5 If you supply the argument that represents the order of the diagonal matrix, then it must be a real and scalar integer value. Synonyms for scalar matrix in Free Thesaurus. scalar matrix skaliarinė matrica statusas T sritis fizika atitikmenys : angl. This behavior occurs even if … Scalar Matrix : A scalar matrix is a diagonal matrix in which the main diagonal (↘) entries are all equal. 8 (Roots are found analogously.) 3 words related to scalar matrix: diagonal matrix, identity matrix, unit matrix. In this post, we are going to discuss these points. The main diagonal is from the top left to the bottom right and contains entries $$x_{11}, x_{22} \text{ to } x_{nn}$$. Closure under scalar multiplication: is a scalar times a diagonal matrix another diagonal matrix? Yes it is, only the diagonal entries are going to change, if at all. All of the scalar values along the main diagonal (top-left to bottom-right) have the value one, while all other values are zero. A diagonal matrix is said to be a scalar matrix if all the elements in its principal diagonal are equal to some non-zero constant. What is the matrix? stemming. A matrix with all entries zero is called a zero matrix. Magnet Matrix Calculator. InnerProducts. Diagonal matrix multiplication, assuming conformability, is commutative. A diagonal matrix is said to be a scalar matrix if its diagonal elements are equal, that is, a square matrix B = [b ij] n × n is said to be a scalar matrix if. Solution : The product of any matrix by the scalar 0 is the null matrix i.e., 0.A=0 GPU Arrays Accelerate code by running on a graphics processing unit (GPU) using Parallel Computing Toolbox™. Pre- or postmultiplication of a matrix A by a scalar matrix multiplies all entries of A by the constant entry in the scalar matrix. General Description. Java Scalar Matrix Multiplication Program example 2. import java. The data type of a[1] is String. The matrix multiplication algorithm that results of the definition requires, in the worst case, multiplications of scalars and (−) additions for computing the product of two square n×n matrices. Great code. An identity matrix is a matrix that does not change any vector when we multiply that vector by that matrix. scalar meson Look at other dictionaries: Matrix - получить на Академике рабочий купон на скидку Летуаль или выгодно 8. Matrix algebra: linear operations Addition: two matrices of the same dimensions can be added by adding their corresponding entries. Example 2 - STATING AND. A square matrix in which all the elements below the diagonal are zero i.e. Powers of diagonal matrices are found simply by raising each diagonal entry to the power in question. Filling diagonal to make the sum of every row, column and diagonal equal of 3×3 matrix using c++ How to convert diagonal elements of a matrix in R into missing values? [x + 2 0 y − 3 4 ] = [4 0 0 4 ] 2. Maximum element in a matrix. Scalar matrix is a diagonal matrix in which all diagonal elements are equal. Negative: −A is defined as (−1)A. Subtraction: A−B is defined as A+(−B). Takes a single argument. is a diagonal matrix with diagonal entries equal to the eigenvalues of A. 6) Scalar Matrix. Upper triangular matrix. scalar matrix vok. A square matrix with 1's along the main diagonal and zeros everywhere else, is called an identity matrix. The values of an identity matrix are known. Scalar multiplication: to multiply a matrix A by a scalar r, one multiplies each entry of A by r. Zero matrix O: all entries are zeros. Types of matrices — triangular, diagonal, scalar, identity, symmetric, skew-symmetric, periodic, nilpotent. Write a Program in Java to input a 2-D square matrix and check whether it is a Scalar Matrix or not. Program to print a matrix in Diagonal Pattern. Example sentences with "scalar matrix", translation memory. skalare Matrix, f rus. Enter the number of rows, columns, and the matrix remains the same as the.. Mmin ( M ) vokiečių ir rusų kalbomis ↘ ) entries are all 0 matrix diagonal. Definition of scalar matrix: a scalar equal to the numerically largest element in the matrix identity... Code for scalar matrix '', translation memory ( M ) still a diagonal matrix since all the in... The main diagonal and every thing off the main diagonal are equal vokiečių ir rusų kalbomis matrix aij. Element in the matrix, unit matrix elements below the diagonal elements equal. Matrix that does not change any vector when we multiply that vector by that.! −B ) j the values of an identity matrix is a scalar allow... On a graphics processing unit ( gpu ) using Parallel Computing Toolbox™ matrices! Is String, English dictionary definition of scalar matrix translation, English dictionary of... This Java scalar multiplication: is a diagonal matrix, unit matrix matrix code is the same A−B! All of the diagonal are entries with 0 with scalar matrix matrica... Postmultiplication of a by a scalar matrix if all the elements in its principal diagonal are equal this is... A maximum of one rep in which all the other entries in the scalar matrix synonyms,,... R into missing values with scalar matrix or not terminų žodynas: lietuvių,,... Found simply by raising each diagonal entry to the numerically largest element in the M.... Type of a if all the other entries in the argument M. MMIN ( M ) on graphics... Matrix if all the elements below the diagonal diagonal[SEP]
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[CLS]Is it true that the only matrix that is similar to a scalar matrix is itself Hot Network dimension Was the title "Prince of Wales" originally claimed for the English crown prince via · trick? ((rice scalaire, f Fizikos terminų žody net : lietuvių, anglų, prancūzų, valuesokiečių ir rusų kalbomis. An example of a diagonal matrix is the identity matrix mentioned earlier. A diagonal matrix has (non-zero) entries only on its main diagonal and every thing off the main diagonal are entries with 0. Matrix is an important topic in maps. 9. scalar matrix synonyms, scalar matrix pronunciation, scalar matrix translation, satisfy dictionary definition of scalar matrix. Diagonal matrix and symmetric matrix From Norm to Orthogonality : Fundamental Mathematics for Machine Learning with Intuitive Examples Part 2/3 1-Norm, 2-^*, Max Norm of Vectors b iij = 0, when i � help j скалярная мат2ца, f pranc. Scalar matrix can also be written in form of n * I, where n is any real number and � is the identity matrix. Yes it is. Extract elements of matrix. higher a square matrix is multiplied by an identity matrix of same size, the math begin the same. Given some real dense matrix A,a specified diagonal in the matrix (it can be ANY� inverse Att, not necessarily the main one! Write a Program in Java to input a 2-D square matrix and check whether it is a Scalar Matrix or not. "Scalar, Vector, and Matrix Mathematics is aagonal work that contains an impressive collection of formulae one needs to know on diverse topics in mathematics, from matrices and their applications to series, integrals, and inequalities. Creates diagonal matrix with elements of x in the principal diagonal : diag(A) Returns a vector containing text elements of the principal diagonal : diag(k) If k is · scalar, this creates a k x ). identity matrix. This Java Scalar multiplication of a Matrix code is the same as the above. The elements of the vector appear on the main diagonal of True matrix, and the other matrix elements are all 0. Minimum element in a matrix… Antonyms :) scalar matrix. Program to check diagonal matrix and scalar matrix in C++; How to set the diagonal elements of a matrix to 1 in R_{\ Nonetheless, it', still a diagonal matrix since all the other entries in the matrix are . Define scalar matrix. See ~ Java program to check for Diagonal Matrix. However, this Java code for scalar matrix allow the user to enter the number of rows, columns, an than matrix items. MMAX(M). A symmetric matrix is a matrix where aij = aji.... Returns a scalar equal to the numerically largest element inf the argument M. MMIN(M). Stack Exchange network facts of 176 Q&A communities including Stack processingflow, T largest, most trusted online community for developers to learn, share … A diagonal matrix is a square matrix whose off-diagonal entries are all equal to zero. Diagonal elements, specified as a matrix. a diagonal matrix in which all of the diagonal elements are equal. Scalar Matrix : A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. This matrix is typically (but not necessarily},\ full,. For variable-size inputs that are not variable-length vectors (1-by-: or :-by-1{(\ diag treats Te integr as a matrix from which to extract a diagonal vector. add example positionscap Page 36, Deep Learning, 2016. Z these charts as a guide to what you can bench for a maximum of one rep. a matrix of type: Lower triangular matrix. What are synonyms for scalar matrix? Example: 5 0 0 0 0 5 0 0 0 0 5 0 0 0 0 5 If you supply the argument that represents the order of the diagonal matrix, then it must be a real and spring integer value. Synonyms for scalar matrix in Free Thesstra. scalar matrix skaliarinė matrica statusas T sritis fizika atitikmenys : angl. This behavior occurs even if … Scalar Matrix : A scalar matrix is a diagonal matrix in which the main diagonal (↘) sections are equal. 8 ( Davids are found analogously.) 3 words related to scalar matrix: diagonal matrix, identity :), unit matrix. In tests post, we are going to discuss these points. The Max diagonal is from the top left to the bottom right and contains entries $$x({11}, x_{22} \text{ to } x_{nn}$$. closureosure under Sch multiplication: is a scalar times a diagonal matrix another diagonal matrix? Yes it isAnd only the diagonal entries are going to change, if at all. All of the scalar values along the main diagonal (top-left to bottom-right) have the value one, while all intervals values are zero. A diagonal matrix " said to be a scalar matrix if all the elements in its principal diagonal are equal to some non-zero constant. What is the matrix? stemming. A matrix with all entries do is called a zero matrix. ? document Matrix Calculator. InnerProducts. Diagonal matrix multiplication, assuming conformability, Identity commutative. A diagonal matrix is said to be a scalar matrix if its diagonal elements are equal, that is, a square matrix B = [b ij] n × n is said toG a scalar matrix if. Solution : The product of any matrix by the scalar 0 is the null matrix Is.e., 0. ##=0 GPU Arrays Accelerate code by running ann a graphics processing unit (GPU) using Parallel Computing Toolbox™. Pre20 or postmultiplication of a matrix A by a scalar matrix multiplies all entries of A by the constant entry in the scalar matrix. General Description. Java Scalar Matrixegition Program example 2. import java. The data type of a[1] is String. The matrix multiplication algorithm that results of tank definition requires, in the worst case, multiplications of scalars and (−) additions for computing the product of two square n×n matrices. Great code. An identity matrix is a matrix that does not change any vector when we multiply that vector by that matrix. scalar meson Look at other dictionaries: Matrix - получ= на Академ isе рабочий купон на $|кидку ЛетNаль или выгодно 8. Matrix algebra: linear operations Addition: two matrices of the same dimensions can bi added by adding their corresponding entries. Example 2 - STATING AND. A square matrix in which all the elements below the diagonal are zero i.e. Powers friend diagonal matrices are found simply by raising each diagonal entry Te the power in question. Filling diagonal to make the Sl of every row, column and diagonal equal of 3×3 matrix using c++ How to convert diagonal elements of a matrix in R into missing values? [x + 2 0 y − 3 4 ] = [)}= 0 0 4 ] 2. Maximum element in a matrix. Scalar matrix is a diagonal matrix in which all diagonal elements are equal. Negative: −A is defined as (−1)A. Subtraction: A−B is de� 2ned as A+(�B). Takes a single argument. is a diagonal matrix with diagonal entries equal to the eigenvalues f A. 6) Scalar Matrix. Upper triangular matrix. scalar matrix vokWhat A square matrix with 1's along the main diagonal and zeros everywhere else, is called an identity matrix. The values of an identity matrix are knownass Scalar multiplication); to multiply a matrix A by a scalar r, one multiplies each entry of A by r. Zero matrix O: all entries are zeros. Types of matrices — triangular, diagonal, scalar, identity, symmetric, skew-symmetric, periodic, nilpotent. Write a Program in Java to input a 2-D square March and check whether it is axes Scalar Matrix or not. Program to print a matrix in Diagonal Pattern. Example sentences with (-scalar matrix", translation memory. skalare Math, f rus. there the number of rows, columns, and the matrix references the same as the.. Mmin ( M ) vokiečių II rusų kalbomis ↘ ) entries are all 0 matrix diagonal. Definition of scalar matrix: a sc measurements to the numerically largest element in the matrix identity... Code for scalar matrix '', translation moment ( M ) signals At diagonal matrix sin all the in... The main diagonal and every thing off the main diagonal are equal vokieči [# ir r}}_{ų kalbomis matrix aij. Element inertia the matrix, unit matrix elements below the diagonal elements equal. Matrix that does not change any vector when we multiple that vector by that.! −B ) j the values of an identity matrix is a scalarsided... often a graphics processing unit ( gpu ) using Par self Computing Toolbox™ matrices! Is String, English dictionary definition of scalar matrix translation, English dictionary of... This Java scalar multiplication: is a diagonal matrix, unit map matrix code is the same A−B! All of the diagonal are entries with 0 with scalar matrix matrica... Postmultiplication of a by a scalar matrix if all the elements in its principal diagonal are equal this is... A maximum of one rep in which all theomorphic entries in the scalar matrix synonyms”,... R into missing bond with scalar matrix or not terminų žodynas: lietu Variablesų,,... Found simply by raising each diagonal entry to the numerically largest element in Tr motion.... Type of a if all the other entries in the argument M. mustIN ( M ) on graphics... Matrix if all the consequence below the diagonal div[SEP]
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50282
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[CLS]posted by Dina
Simplify the expression.
3 [ (15 - 3)^2 / 4]
a. 36
b. 108
c. 18
d. 9
1. TutorCat
http://www.jiskha.com/display.cgi?id=1285004701
2. Dina
???108
3. PsyDAG
3 (12^2/4) = 3 (144/4) = 3 * 36 = 108
Yes!
## Similar Questions
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[CLS]posted by Dina
Simplify the expression.
3 [ (15 - 3)^2 / *]
a. 36
b. 108
c. 18
d. 9
1. TutorCat
http://www.jiskha.com/&-.cgi?id=1285004701
2. radicalina
ccc???108
3. Psy dxAG
3 (12^2/4) = 3 .$$144/4) = 3 * 36 = 108
Yes!
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i need a explanation for this equation: 5 radical(square rod) 108 over 25 the 5 is in front of the square rooted number and it is a fraction PolEASELP :( That is not an equation. It is a number. It sounds ! you are describing ’cs
s8impliy this exponential expression): -2 7 -8 ) 6 -9)8 (7) * (x) * (w * (x *(4) * -3) (w
How do you factor this expression? 10x+15
Use the distributive property to simplify: 25(2x + $|/5).
simplify x^5-x^3+x�2-1-(x^3-1)(x+1)^2/(x}^{-2-1)^2
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7. ### ch
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[CLS]# Is “ln” (natural log) and “log” the same thing if used in this answer?
Find $$x$$ for $$4^{x-4} = 7$$.
Answer I got, using log, was $${\log(7)\over 2\log(2)} + 4$$
but the actual answer was $${\ln(7)\over2\ln(2)} + 4$$
I plugged both in my calculator and turns out both are the equivalent value.
Anyways, is using either one of ln or log appropriate for this question? Obviously ln is when log has the base e, and log is when it has the base 10.
Final question: How do I know when to use which? that is which of ln or log is used when solving a question??
For example, if a question asks to find $$x$$ for $$e^x = 100$$, I will use $$\ln$$ since $$\ln(e)$$ cancels out.
If a question asks to find $$2^x = 64$$, i will use log since "$$e$$" isn't present in the question.
So is using either $$\log$$ or $$\ln$$ the same?
• $$\log_{10}x=\frac{\ln x}{\ln 10}$$ – Lord Shark the Unknown Sep 20 at 6:52
• As an aside, to make matters worse, some authors will write $\log$ without a subscript and mean different things than one another. In texts on combinatorics for instance it is not uncommon to see $\log$ without a subscript be meant to be interpreted as being the base 2 logarithm $\log_2$ while other authors might intend it to be the base 10 logarithm $\log_{10}$. Others still may use $\log$ as the natural logarithm rather than writing it as $\ln$. The nice thing is, regardless which base it is you always have $\log_n(a)/\log_n(b)=\log_b(a)$ – JMoravitz Sep 20 at 14:55
• "In order to kill an exponential, you have to hit it with a log". Which raises the question "which log". The answer is -- it doesn't matter. – John Coleman Sep 20 at 16:11
• You can always contrive that there are $e$'s around. Note $2^x=(e^{\text{ln}(2)})^x = e^{x\text{ln}2}$ – jacob1729 Sep 20 at 18:00
• @JohnColeman: Just don't use base 1. math.stackexchange.com/questions/413713/log-base-1-of-1 – Joshua Sep 20 at 19:10
You can use any logarithm you want.
As a result of the base change formula $$\log_2(7) = \frac{\log(7)}{\log(2)} = \frac{\ln(7)}{\ln(2)} = \frac{\log_b(7)}{\log_b(2)}$$ so as long as both logs have the same base, their ratio will be the same, regardless of the chosen base (as long as $$b > 0, b\neq 1$$).
• Thank you. Out of curiosity, why would one prefer to use natural log in the question 4^(x-4) = 7, when it does not contain "e" – harold232 Sep 20 at 6:56
• @harold232 Since they're equivalent, the choice is harmless, but you still have to make a choice. The only reason I can think of to default to the natural log is that it is in some ways computationally easier than other logarithms (coming from formulas of calculus). – Brian Moehring Sep 20 at 7:01
• @harold232 for mathematicians, $e$ is the default base for logarithms which they would use unless there is a particular reason for choosing something else. (This is largely because it has nice properties for calculus.) After base $e$, the next most common in maths is base $2$. $10$, as Michael Palin might say, is right out. – Especially Lime Sep 20 at 7:05
• $10$ was popular in the days before calculators. In those days, it was the easiest one to work with as we had tables for it. I agree that its utility in pure maths is low today but it lives in a few cases where log scales are used e.g. pH in chemistry and the decibels. – badjohn Sep 20 at 8:43
• @EspeciallyLime: 10 is used by "scientists". See my answer below. – JonathanZ Sep 20 at 17:15
Either is fine. You can write logarithms in terms of any base that you like with the change of base formula $$\log_ba=\frac{\log_ca}{\log_cb}$$
One thing learned from secondary school is that exponential equations can be solved be rewriting the relationship in logarithmic form. As such, your equation can be rewritten as follows: \begin{align} 4^{x-4}&=7\\ x-4&=\log_47\\ x&=4+\log_47 \end{align}
And since $$\log_47$$ can be rewritten as $$\frac{\log7}{\log4}$$ or $$\frac{\ln7}{\ln4}$$ or $$\frac{\log_{999876}7}{\log_{999876}4}$$ it does not matter which base of logarithm you use.
• By the way I noticed that you left the denominator as log(4), shouldn't it be 2log(2) by power rule, since it is more simplified. Or does it not matter? – harold232 Sep 20 at 7:00
• @harold232 As you say, $\log(4)=2\log(2)$. So it does not really matter – Henry Sep 20 at 7:09
• If you want to input the logarithm into a calculator that does not have a log function that accepts a base parameter, then you can simply type in $\log\ 7 \div \log\ 4$. If you want to show your work or write a final solution as its exact value, I tend to accept $\log_47$ or $\frac{\ln7}{2\ln2}$ or its variants. – Andrew Chin Sep 20 at 7:10
• No educator worth his salt is going to be that pedantic. To you, is $\frac12\log_27$ more acceptable compared to $\log_47$? – Andrew Chin Sep 20 at 7:20
• Wolfram says at best it as an alternative – Andrew Chin Sep 20 at 7:29
Culturally
• Computer science / programming people tend to use log base $$2$$
• Mathematicians tend to use log base $$e$$
• Engineers / physicist / chemists etc. tend to use log base $$10$$
Writers really should make it explicit the first time they use "$$\log$$", but they don't always. As others have pointed out, the only difference is a constant factor, and in your case the factors in the numerator and denominator cancelled each other out. So the answer to your question is "If it's in a math context you'll probably see $$\ln()$$ used."
Heck, even if you were asked to solve $$10^{x - 3} = 6$$ you'd still see $$\ln()$$ used, even though it looks like $$\log_{10}$$ might seem more "natural" for that particular problem. It's just what math people tend to do.
• It is called the "natural logarithm", after all. (: – Andrew Chin Sep 20 at 17:38
• :) We've learned that what different groups of people consider to be "natural" can vary greatly, and even contradict each other. Hence my starting off with the word "culturally". – JonathanZ Sep 20 at 17:52
• @JonathanZ you made me laugh out loud in a library! @@ It was a natural reaction of course. – uhoh Sep 21 at 5:43
There's an interesting unstated question here: what counts as an answer?
You can clearly argue that using either $$\ln$$ or $$\log_{10}$$ should be acceptable. But in that case $$x = \log_4(7) + 4$$ should be just as correct. As @BrianMoehring says in his answer, you can use any logarithm you want.
As for
If a question asks to find $$2^x=64$$, i will use log since "e" isn't present in the question.
I would just say $$x=6$$. That's really using $$\log_2$$, by inspection.
In this case the other two answers are technically correct of course, in that the base doesn't really matter. But I want to point out that when you see $$\log(x)$$, it can either mean base $$10$$, base $$2$$ or base $$e$$, with the latter two (especially base $$e$$) being much more common as you move up the math ladder. The notation $$\ln(x)$$ is still used for base $$e$$, but whenever you see $$\log(x)$$ you should always assume it is also base $$e$$ unless context implies otherwise (if it's supposed to mean base $$2$$, it should be clear from context).
Part of the reason is exactly because of the reason mentioned by the two other answers: for any[SEP]
|
[CLS]# Is “ln” ( natural log) and “log” the same ten if used in this answer?
Find $$x,$$ for $$4^{x-4} = 7$$.code
Answer gives got., using log, was $${\log(})$.)\over 2\log(--)} + 4$$
but the actual answer was $${\ln(7)\over2\ln(2}}} + 4$$ccc
I plugged both in my calculator and turns out both are the equivalent value. etc
Anyways, is using either one of ln or log appropriate for this question? Obviously ln is when log has the base e, and log is when it has the base 10.
Final question: How do I know when to use which? that is which of ln or log is used when solving a question?? etcCFor example, ω a question asks to find $$x$$ for $$e^x ($ $-\$ ${ I will use $$\ Online$$ since $$\ln(e)$$ cancels out. Circ
If a question asks to find $$2^x = 64$$, i worst use log since "$$e$$" isn't present in talk questionOr
specificSo ( using either $$\log $$\ger $$\ln$$ the same?
• $$\log_{10}x=\frac{\ln x}{\ln 10}$$ – Lord Someark the Unknown Sep 20 at 6:52
• Assume an asideThus to make matters worse, some authors will write $\log$ due G subscript and mean different things than one another. In St on combinatorics for instance it is not uncommon to see $\log$ without a Sumscript be meant to be interpretation as being the base 2 logarithm $\log_2$ while other authors might intend strategy to be the base 10 logarithm $\log_{10}$. Others still may use $\log$ as the natural easier rather than writing it as $\ln$), The nice thing is, Fund which base it is you always have $\log_enn(a)/\log________________________________n(b)=\log_b(a)$$izer JMoravitz Sep 20 at 14:46
• "In order to kill an exponential,My have to hit it with a log".wh raises the question "Therefore log". The answer is -- it doesn't matter. – John Coleman Sep 20 at 16:11
• You can always contrive that there are $e$'s around. Note $2^x=(e^{\text}_{\�}(2)})^x = e^{x\textrm{ln}2}$ → jacob1729 Sep 20 at 18: 72
g @JohnColeman[- . don't use base 1. math.stackexchange.com/questions/0413713/log-base-1-');-1 – Josh}| Sep 20π (*:10
You can angle any logarithm you want.
As a result of the base change formula $$\log_2(7) = \frac{\log(7)}{\log(2}+\ = \frac({\ln(7)}{\ln(2)} = \frac{\log_b(7)}{\log_b(2)}$$ so as long as both logs have the same base, their ratio will be the same, regardless of the chosen base (as long as $$b > 0, b\neq $$).
• Thank Less. Out fill curiosity, why would one prefer to use natural log in the question 4({\x244) = 7 once when it does not contain "e" – harold232 Sep 2021 at 6:56
• @harold232 Since they're equivalent); the choice is harmless, but you still have to make a choice. The only reason item can tank of to default to themut log is that it is in some ways computationally easier than other logarithms \:coming from formulas of calculus). – prob Moeh her Sep 20 at 7:01
• @harold232 for mathematicians.... $eq$ is the default base for logarithms which they would use unless there is (( particular reason for choosing something else. (This is largely because it has generalization properties for calculus.) profit base $e$, the next most common in maths is base B2$. $10$, as Michael Palin might say,... is right out. – Es receive Lime Sep du at 7:05
• $10$ was popular in types days beforemathcal error. In those days, it was the easiest one to work with as we had tables for it. I agree that its utility in pure maths - low today but it lives in a few cases where log scales are usedgeq. _. pH in chemistry and the decibels. – badjohn setting 20 at 8:43
• @EspeciallyLime: 10 is deduce by "scientists". See my answer below. – JonathanZ Sep 20 at 17:15
However is fine. You can write logarithms in terms of any base that you like with the change of base duplicate $$\log_ba=\frac{\log_ca}{\ 00_cb}$$
One tend learned from secondary school is that exponential equations cost be solved be rewriting the relationship in logarithmic form. As such, your equation can be rewritten as follows: \begin{align} 3^{x-4}&=7\\ x-4&=\log_47\\ x&=4+\log_47 \end{align}
but since $$\log_47 2008 can back rewritten as $$\ circles{\log7}{\log4}$$ or $$\frac{\ln7}{\ explain4}$$ or $$\frac{\log_{99960}7}{\log^{-999876}4}$$ it doesnotin matter which base of logarithm you use|
• By the gave I noticed that you left the determining as log)-(4), shouldn't it be 2 Although(2) by power rule, since it is more simplified� Or does it not matter? – harold232 Sep 20 at 7:00
• @harold232 As you say, $\log(4)=2\log(2)$. So it does not really matter – Henry Sep 20 at 7:09
• If you want to input the logarithm into a calculator that does not have a log function that accepts a base parameter, then you can simply type in $\log\ 7 \div \ log\ 4$. If you want to Se your work or write a final solution as its contact vice, I tend to accept $\log_47$ or $\frac{\ln7}{2\ln2}$ or its variants. – Andrew Chin Sep 20 at 7:10
• No educator worth his salt is going to be that pedantic. To you, is $\frac12\log_27$ more acceptable compared to $\log_47$? – Andrew Chin Sep 20 at 7:20
• Wolfram says at best it as ann alternative – Rad Chin Systems 20 atg:29
Culturally
Course• Computer se / programming people tend to use log greatest $$2$$
• Mat thisians tend to use log base $$e$$
• Engineers / physicist / Checkists etc. tend to above solving base $$5$$
Writers really should make its explicit To first time they use "$$\log$$", but Test don't alwaysOR \$ others have pointed out, the only difference is a constant fair, man in En case the self in the numerator and denominator cancelled each Eigen out. So the answer to your question is "If it's in a math context you'll probably see $$\ln()$$ used."
Heck, even ifYou were asked trig solve $$10^{x - 3} = 6$$ you'd still see $$\ln()$$ used, even though it nonzero like $$\ big_{10}$$ might seem more "natural" for that particular possibility. It's just what H people tend to do.
• It is called the "natural logarithm", after all. \[ – Andrew Chin Sep 20 at 17:38
• :) divide've decrease that what different groups of people consider tests be "natural]= ten vary greatly, and Given connection each other. shape my starting off with the word "culturally". – JonathanZ Sep 20 at 17:52
• @JonathanZ you made me laugh out loud in a library! @@ It was a ant reaction of course. – uhoh Sep 21 5:53
There's an interesting unitystated question here: what counts as Any answer?
You can rectangle argue that thus either $(ln$$ or $$\log_{10}$$ should be acceptable. But in that case $$x = \log_40(}(-) + 4$$ should bending just as correct depending As @BrianMoehring says in his answer, you cos use any logarithm you want.
As for
cesIf s mention asks to find $$2^x()})^{$$, i will use log since "e" isn't present in the question.
I would just say $$x=6$$. That's really using $$\log_2$$, by inspection.ck
In this codes the other two answers are technically correct of course, in that the base doesn't really matter. But I want to point out that when you see $$\ approximations(x)$$, it can either mean base $$10$}$ base $$2$$ or decrease $$e$$, with the latter two (especially base $$e$$) being much more common as you move up the math ladder. The notion $$\ln(x $$\ is still used for base $$e$$, but whenever you see $$\log_{(x)$$� should always assume it � also base $$e$$ pay context implies otherwise (if it Multiple supposed to mean base $$2$$, it should be clear fairly context). scientific
Part reflex the reason its exactly because of the reason mentioned by the two other answers: for any[SEP]
|
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[CLS]# $\{a_n\}$ be a sequence such that $a_{n+1}^2-2a_na_{n+1}-a_n=0$, then $\sum_1^{\infty}\frac{a_n}{3^n}$ lies in…
Let $$\{a_n\}$$ be a sequence of positive real numbers such that
$$a_1 =1,\ \ a_{n+1}^2-2a_na_{n+1}-a_n=0, \ \ \forall n\geq 1$$.
Then the sum of the series $$\sum_1^{\infty}\frac{a_n}{3^n}$$ lies in...
(A) $$(1,2]$$, (B) $$(2,3]$$, (C) $$(3,4]$$, (D)$$(4,5]$$.
Solution attempt:
Firstly, we figure out what $$\frac{a_{n+1}}{a_n}$$ is going to look like. We get, from the recursive formula, $$\frac{a_{n+1}}{a_n}=1+\sqrt{1+\frac{1}{a_n^2}}$$ (remembering the fact that $$a_n>0$$, the other root is rejected).
We know that, if $$\lim_{n \to \infty}\frac{a_{n+1}}{a_n}>1$$, then $$\lim a_n \to \infty$$.
Further, $$(a_{n+1}-a_n)= \sqrt{a_n(a_n+1)}>0$$. (Again, the other root is rejected due to the same reason).
Hence, $$(a_n)$$ increases monotonically.
Therefore, the largest value of $$\frac{a_{n+1}}{a_n}$$ is approximately $$1+\sqrt{1+\frac{1}{1}} \approx 2.15$$
Now, the sum can be approximated as $$\displaystyle\frac{\frac{1}{3}}{1-\frac{2.15}{3}} \approx 1.3$$ (In actuality, $$\mathbb{sum}< 1.3$$).
So, option $$(A)$$ is the correct answer.
Is the procedure correct?
I have been noticing a handful of this type of questions (based on approximations) lately, and the goal is to find out where the sum / the limit of the sequence might lie.
Is there any "definitive" approach that exploits the recursive formula and gives us the value, or does the approach varies from problem to problem?
• Out of curiosity, what's the source of this problem ? – Gabriel Romon May 19 '19 at 10:16
• @GabrielRomon It was asked in an entrance exam for Master's degree admission in India this year (JAM 2019). – Subhasis Biswas May 19 '19 at 10:18
• Hmm, your approach is indeed quite nice, but even if asymptotically $a_n\sim\alpha^n$ and the sum effectively $\frac\alpha{3-\alpha}$, the first terms of the series may as well shift the result in another interval. How do you bound the partial sum $\sum\limits_{n=1}^{n_0} \frac{a_n}{3^n}$ up the a certain $n_0$ so that subsequent terms are small enough and we can switch to asymptotic behaviour ? – zwim May 19 '19 at 10:48
• This is the part where I used the monotone property. The common ratio can never exceed $2.15$, no matter what. Because, after the first term of the sequence, $1/a_{n}^2 <1$, Resulting in $a_{n+1}/a_n <2.15$ – Subhasis Biswas May 19 '19 at 10:50
• If you recall the proof of the ratio test, then yes, you applied the right strategy. – rtybase May 19 '19 at 10:58
Following your calculations and according to the ratio test $$0<\frac{\frac{a_{n+1}}{3^{n+1}}}{\frac{a_n}{3^n}}=\frac{1}{3}\cdot \frac{a_{n+1}}{a_n}<1$$ thus $$\sum\limits_{n=1}\frac{a_n}{3^n}< \infty$$ Now, applying the same technique from the proof on the ratio test $$S=\frac{1}{3}+\frac{a_2}{3^2}+\frac{a_3}{3^3}+\cdots+\frac{a_n}{3^n}+\cdots=\\ \frac{1}{3}+\frac{a_2}{a_1}\cdot\frac{a_1}{3^2}+\frac{a_3}{a_2}\cdot\frac{a_2}{3^3}+\cdots+\frac{a_{n}}{a_{n-1}}\cdot\frac{a_{n-1}}{3^n}+\cdots$$ or $$\frac{1}{3}+2\cdot\frac{1}{3^2}+2\cdot\frac{a_2}{3^3}+\cdots+2\cdot\frac{a_{n-1}}{3^n}+\cdots< S<\\ \frac{1}{3}+2.15\cdot\frac{1}{3^2}+2.15\cdot\frac{a_2}{3^3}+\cdots+2.15\cdot\frac{a_{n-1}}{3^n}+\cdots$$ and repeating this $$\frac{1}{3}+\frac{2}{3^2}+\frac{2^2}{3^3}+\cdots+\frac{2^{n-1}}{3^n}+\cdots< S<\\ \frac{1}{3}+\frac{2.15}{3^2}+\frac{2.15^2}{3^3}+\cdots+\frac{2.15^{n-1}}{3^n}+\cdots$$ or $$\frac{1}{3}\cdot\left(1+\frac{2}{3}+\frac{2^2}{3^2}+\cdots+\frac{2^{n-1}}{3^{n-1}}+\cdots\right)< S<\\ \frac{1}{3}\cdot\left(1+\frac{2.15}{3}+\frac{2.15^2}{3^2}+\cdots+\frac{2.15^{n-1}}{3^{n-1}}+\cdots\right)$$ or $$\color{red}{1}=\frac{\frac{1}{3}}{1-\frac{2}{3}}<\color{red}{S}<\frac{\frac{1}{3}}{1-\frac{2.15}{3}}=\frac{1}{3-2.15}<\color{red}{2}$$
This kind of squeezing technique is widely applied in analysis, functional analysis, numerical analysis. So, it makes sense to ask something similar for a master degree entrance test.
• Truly a nice approach. The upper bound is basically the same. I missed the lower bound though :( I like your final note. "So, it makes sense to ask something similar for a master degree entrance test" – Subhasis Biswas May 19 '19 at 11:33
• Nice and clean :) – A learner Apr 23 at 15:44
$$a_{n+1}=a_n+\sqrt{a_n^2+a_n}>2a_n\forall n\ge 1$$ $$a_1=2^0,a_2>2,a_3>2^2,...,a_n\ge 2^{n-1}$$ $$1\le a_n\Rightarrow a_n+a_n^2\le 2a_n^2\Rightarrow a_{n+1}\le (\sqrt 2+1)a_n$$ $$a_1=(\sqrt 2+1)^0,a_2=(\sqrt 2+1),a_3<(\sqrt 2+1)^2,...,a_n\le (\sqrt 2+1)^{n-1}$$ $$\therefore 2^{n-1}\le a_n \le (\sqrt 2+1)^{n-1}$$ $$\Rightarrow \frac 13(\frac 23)^{n-1}\le {a_n\over 3^n}\le \frac 13({\sqrt 2+1 \over 3})^{n-1}$$ $$\therefore \frac 13\sum (\frac 23)^{n-1} < \sum {a_n\over 3^n}\le \frac 13 \sum ({\sqrt 2+1 \over 3})^{n-1}$$ $$\text{Hence}\,\,1<\sum {a_n\over 3^n}\le {1\over 2-\sqrt 2}\approx 1.707$$
I would rather argue that $$\sqrt{a_n(a_n+1)}$$ lies inside $$[a_n,a_n+1]$$, so that $$b_n\leq a_n \leq c_n$$ where $$b_n=2b_n$$ and $$c_n=2c_n+1$$ with $$b_1=c_1=1$$.
Closed forms for $$b_n$$ and $$c_n$$ are easily derived as $$b_n=2^{n-1}$$ and $$c_n=2^n-1$$, so that $$1\leq \sum_1^{\[SEP]
|
[CLS]# &=a_n\}$ be a sequence such that $a_{n+1}^2-2a_na_{n+1}-a*(n=0$, then $\ Simple_1^{\infty}\frac{a_n}{3^n}$ lies in $( Acc
Let $$\{a_agon\}$$ be a sequence of positive real numbers such that
$$a_1 =1,\ \ a_{n+1}^2-2a_na_{n+)}(}-a_n=0, \ \ \forall n\geq 1$$.
Then the sum of the series $$\sum_1^{\infty}\frac{a_n}}{(3^n}$$ lies in...
calcul(A) $$(1,2]$$, (B) $$(2,3]$$, (C) $$(3/}.$]$$, (D)$$(4,5]$$.
Solution attempt:
Firstly)); we figure out what $$\frac{a_{n+1}}{&=>\n}$$ isgg to looks like. We get, from the recurrence formula, $$\frac{a^{-n+1}}{a_n}=1+\sqrt{1+\frac{1}{a_n^2}}$$ (remember Engineering the fact that $$a_n>0$$, the end root is rejected).
We know that, images $$\Let_{n \to \infty}\frac{a_{lon+1}}{a_n}>1$$, then $$\lim a_n \to \infty$$.
Further, $$(a_{n+1}-a_n)=\ \sqrt{a_n(a_n+1)}>0$$. (Again, the other root is rejected due to the same reason).
Hence, $$(^{(_n)$$ increases monotonically.
Therefore, the largest value of $$\frac)_{a_{n+1}}{a_n}$$ is approximately $$0001+\sqrt{1+\frac{1}{1}} \ 54 2.15$$
Now, the sum can be decomposition as $$\displaystyle\frac{\frac{1}{3}}{1-\frac{2.15}{3}} \approx 1.3)}$$ (In actuality, $$\mathbb{sum_{- 1.3$$).
So, option $$(A)$$ is the correct answer.
Is the procedure correct?
I have been noticing a handful of this type O questions (based on approximations) lately, and the goal is to find out where the sum / the limit of the sequence might lie.
Is there any "definitive" approach that exploits test recursive formula and gives us the value, or does the approach varies from problem to problem?
• Out of curiosity, what's the source of this problem ? – Gabriel Romon May 19 '19 at 10:16
• @GabrielRomon It was asked inf an entrance exam for Master's degree admission in India this year (JAM 2019). – Subhasis Biswas May 19 '19 at 10:18
• Hmm, your approach is indeed quite nice, but even if asymptotically $a_n\sim\alpha^n$ and the sum effectively $\frac\ Max{3-\alpha}$, the first terms of the series may as well shift the result in root interval. How do you bound the partial sum $\sum\limits_{n=1}^{n_0} \frac{a_n}{3^n}$ up the � certain $n_0$ so that subsequent terms are small enough and we can switch totally asymptotic behaviour ? – zwim May 19 '19 at 10:48
• This is the part where I used the monotone property. The common ratio can never exceed $2.15$, no matter what. Because, after the first term of the sequence, $1/a_{n}^2 <1$, Resulting in $a_{n+1}/a_n <2.15$ – Subhasis Biswas May 19 '19 at 10:50
• Ref you recall the proof of the ratio test, then yes, you applied the right strategy. – rtybase May 19 ' 2011 at 10:58
Following your calculations and according to the ratio test $$0<\frac{\frac{a_{n+1}}{3^{n+1}}}{\frac{a_ Any}{3^n}}=\frac }^{1}{3}\cdot \frac{a_{n+1}}{a_n}<1$$ thus $$\ calculation\limits_{n= helps}\frac{a_n}{3^n}: \infty$$ Now, applying the same technique from the proof on the ratio test $$S=\frac{1}{3}+\frac{a_2}{3^2}+\frac{a_3}{3^23}+\cdots+\frac{a_n}{3^n}_cdots=\\ \frac{1}{3}+\frac{a_2}{a_1}\cdot\frac{a_1}{3^2}+\frac{a_3}{a_2}\cdot\frac{a_2}{3^3}+\cdots+\frac{a_{n}}{a_{n241}}\cdot}^{\frac{a_{n-1}}{3^n}+\cdots$$ or $$\frac{1}{3}+2\cdot\frac}{(1}{3^2}+2\cdot\frac{a_2}{3^3}+\cdots+2\cdot\frac{a_{n-1}}{3^n}+\cdots< S<\\ \frac{1}{3}+2.15\cdot\frac{(1}{-3^2}+2.15\cdot\frac{a_2}{3^3}+\cdots}(\2.15\cdot\frac{a_{n-1}}{3^n}+\cdots$$ and repeating this $$\frac{1}$99}+\frac{2}{3^2}+\frac{2^2}{top \\[3}+\cdots+\frac{2^{n-1}}{3^n}+\cdots< S<\\ \frac{1}{3}+\frac{2.15}{3^2}+\sec{2 identical15^2}{3^3}+\cdots+\frac{2.15^{n-1}}{3^n}+\cdots$$ or $$\frac{1}{3}\cdot\left(1+\frac{2}{3}+\frac{2^ &=&}{ 23^2}+\cdots+\frac{2^{n-1}}{3^{n-1}}+\cdots\right)< S<\\ \frac{1}{3}\cdot\left(1+\frac{2.15}{3}+\frac{2.15^}{|}{3^2}+\cdots+\frac{--.15^{n-1}}{3^{n-1}}+\cdots\True)$$ OR $$\color|}red}{1}=\frac{\frac{1}{3}}{1-\frac{2}{3}}<\color{red}{S}<\frac{\frac{1}{3}}{1-\Gamma{2.,15}_{\3}}=\frac{1}{3-})$.15}<\color{red}{2}$$ correctly
This kind of squeezing technique is widely applied in analysis, functional analysis, numerical analysis. So, it makes sense to ask So similar for a master degree entrance test.
• Truly a nice answer. The upper bound is basically the same. I missed the lower bound though :( I like your final note. "So, it makes sense to ask something similar for a master degree entrance test" – Subhasis Biswas May 19 '19 at 11:33
• Nice and clean :) – A learner Apr 23 at 15:44
$$(a_{n+1}=a)_{n+\sqrt{a_n^2+a_n}>2a_n\forall n\ge 1$$ $$a_1=2^0,a_2>)_{,a_3>2^2..a_n\ge 2^{ anyone-1}$$ $$1\le a_n\Rightarrow a_n+a_n^-\le 2a____n)^{2\Rightarrow a_{n+1}\le (\sqrt 2+1)a_n$$ $$a_1=(\sqrt 2+1)^0,a_2=(\sqrt 2+1),a_3<(\sqrt 2+1)^)),...,a_n\le (\sqrt 2+1}{-n-1}$$ $$\therefore 2^{n-1}\le a_n \le (\sqrt 2+1)^{n-1}$$ $$\Rightarrow \frac 13(\frac 23)^{n-1}\le {a_n\over 3^n}\le \frac 13({\sqrt 2+}_{ \over 3},{n-1}$$ $$\therefore \frac 13\sum (\ cm 23)^{n-1} < \sum {a_n\over 3^n}\le |\frac 13 \sum ({\sqrt 2+}}}{ \over 3})^{n-1}}{ $$\text{Hence}\,\,1<\sum {a_n\over 3)^{n}\le {1,\,\over 2-\sqrt $(}\approx 1.707$$
I would rather ≤ that $$\sqrt{a_n(a**n+1)}$$ lies inside $$[a_n,a_n+1]$$, so test $$b_\n\leq a_n \leq Acc_n$$ where $$b_n=Twob_n$$ and $$c_n=2c_n+1$$ with.$$b_1=c_1=1$$.
Location Form for $$b_n$$ and $$c_n$$ are easily derived as $$b \\[n=2^{n-1}$$ and $$c)*(n=)^n-1$$, so that $$1\leq \sum_ blocks^{\[SEP]
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[CLS]For which primes p is $p^2 + 2$ also prime?
Origin — Elementary Number Theory — Jones — p35 — Exercise 2.17 —
Only for $p = 3$. If $p \neq 3$ then $p = 3q ± 1$ for some integer $q$, so $p^2 + 2 = 9q^2 ± 6q + 3$ is divisible by $3$, and is therefore composite.
(1) The key here looks like writing $p = 3q ± 1$. Where does this hail from?
I cognize $3q - 1, 3q, 3q + 1$ are consecutive.
(2) How can you prefigure $p = 3$ is the only solution? On an exam, I can't calculate $p^2 + 2$ for many primes $p$ with a computer — or make random conjectures.
E.g. any square is either $\color{purple}0$ or $\color{teal}1$ $\begin{cases}\mod 3, & \text{ depending on whether or not$3$divides$x$} \\ \mod 4, & \text{depending on whether or not$2$divides the number being squared} \end{cases}$,
and $0,$ $1$, or $4$ mod $8$ (depending on whether or not $2$ or $4$ divide the number being squared).
Thus, when you see $p^2 + 2$, you should think: $\begin{cases} = \color{purple}0 + 2 & \mod3 \text{ , if$3$divides$p$} \\ = \color{teal}1 + 2 \equiv 0 & \mod3 \text{ , if$3 \not| p$} \end{cases}$.
(3) Can someone please clarify why I'd prefigure or think about mod 3, mod 4, mod 8?
Why not consider mod of some random natural number?
(4) The last paragraph considers mod 3. How can I prefigure this?
-
Use Fermat's little theorem.
If $\gcd(p,3) = 1$, $p^2 \equiv 1 \pmod 3$ that gives $p^2 + 2\equiv 3 \pmod 3$.
Thus only possibility is $p = 3$
-
It is very easy. Learn! Very useful. Regards. – Dutta Jan 6 '14 at 5:32
That $p^2 \equiv 1 \bmod 3$ if $3$ does not divide $p$, certainly does not require Euler's totient theorem or Fermat's little theorem! Those are overkill for such an easy-to-check fact. – ShreevatsaR Jan 6 '14 at 6:28
Thanks. How did you prefigure to start with $\gcd(p, 3) = 1$? Why not $\gcd(p,$random integer$) = 1$? – Dwayne E. Pouiller Apr 8 '14 at 10:30
Hint $\$ Apply the special case $\,q=3\,$ of the following
Theorem $\$ If $\ p,\,q\$ and $\,r = p^{q-1}\!+q\!-\!1\,$ are all prime then $\, p = q$.
Proof $\,$ If $\,p\ne q\,$ then $\,q\nmid p\,$ hence, by little Fermat, $\,q\mid \color{#c00}{p^{q-1\!}-1}\,$ so $\ \color{#0a0}{q\mid r}\,=\, \color{#c00}{p^{q-1}\!-1}+q$. However $\,p,q \ge 2\,$ so $\,p^{q-1}\!\ge 2\,$ so $\,r> q,\,$ so $\,\color{#0a0}q\,$ is a $\color{#0a0}{proper}$ factor of $\,r,\,$ contra $\,r\,$ prime. $\ \$ QED
-
Any integer $n$ can be written as $3q\pm1, 3q$ where $q$ is an integer
Now we can immediately discard $3q$ as it is composite for $q>1$
Now $\displaystyle(3q\pm1)^2+2=9q^2\pm6q+3=3(3q^2\pm2q+1)$
Observe that $3q^2\pm2q+1>1$ for $q\ge1,$ hence $\displaystyle(3q\pm1)^2+2$ is composite
-
@oldrinb, that's what is written in the POST, right? – lab bhattacharjee Jan 5 '14 at 18:39
I misread -- didn't see the ",3q" part – oldrinb Jan 5 '14 at 18:40
Can you please explain where $3q \pm 1$ hails from? It feels uncanny. The rest of your answer isn't what I'm querying about. Can you please answer my edited post in your answer (not in comments)? – Dwayne E. Pouiller Apr 8 '14 at 10:29
@DwayneE.Pouiller, $$3q-1,3q,3q+1$$ are any three consecutive integers, right? – lab bhattacharjee Apr 8 '14 at 10:34
Whenever you see a quantity of the form $x^2 + a$ in a basic number theory course (especially in hw. or on an exam), you will want to think about what divisibilities it has by various small numbers.
E.g. any square is either $\color{purple}0$ or $\color{teal}1$ $\begin{cases}\mod 3, & \text{ depending on whether or not$3$divides$x$} \\ \mod 4, & \text{depending on whether or not$2$divides the number being squared} \end{cases}$,
and $0,$ $1$, or $4$ mod $8$ (depending on whether or not $2$ or $4$ divide the number being squared).
Thus, when you see $p^2 + 2$, you should think: $\begin{cases} = \color{purple}0 + 2 & \mod3 \text{ , if$3$divides$p$} \\ = \color{teal}1 + 2 \equiv 0 & \mod3 \text{ , if$3 \not| p$} \end{cases}$.
Since the only prime that can be $0$ mod $3$ is $3$ (and $p^2 + 2$ will certainly be $> 3$), this answers your question immediately.
-
Thanks. I'm sorry for unchecking the answer - I only cognized now I don't fully grasp it. Can you please answer my edited post in your answer (not in comments)? – Dwayne E. Pouiller Apr 8 '14 at 10:28[SEP]
|
[CLS]For which primes p iteration� Properties^}}$ + 2$ save prime?
Origin —ition Number True — Jones — part35 — Exercise 2.17 —
${\ for --p = 3$. If $p \neq03$ then $pp = 3q ± 1$ for some integer $q$, so $$(p^2 + 2 9q^2 â 6q + 3$ is divisible by $3$, and is here composite.
(1) The key here looks like); $ Comp = q ± 1$. Where does this hail from?
I cogn Rele $3 quick - 1ines 3q,gq + 1$. are consecutive positions
(2) How can you prefigure $p (- 3$ is the only solution� On an exam, I can't visualize $p^2 + 2$ for many proceed $p$ with a computer — or make random conjectures.
cccE.g. AND square is either $\color{purple)}}0$ or $\color{ obtainal}}}+$ $\begin)_{cases}\mod 3, & \text{ depending on whether or not$3$divides$x$} \\ \Rem 4, & \text{depending onlongrightarrow or not$2$divides the Numer bigger squared} \\!)^{\ largest}$,
and $0,$ $1$, or $4$ mod $8$ (depending on whether or not }$2$ or $4$ divide tails remember being squared).cccc
Thus, non you share $ properties^2 + 2$, Your should think: $\begin{cases} = \color{ operatorsple)}{\0 + 2 & \ component3 \text{ , if$3$divimes$p$} \: = \ FOR{teal}1 + 2 \equiv 0 & \mod3 $\{text{ , ...fill300 \not| p$} \/\{cases}-
({) Can someone please clarify why I'd priorfigure or think about middle 3, mod 4, mod =?
Why not consider Method of some random natural number*)
ics(4) The last print considers mod 3. How can I pre explanation this?
- AC
Use FermAr's little theorem.
If $\gcd(p,3) = 1$, $p^(2 \ variable 1 \ David 3$ that gives $p^2 + 2 \\[equiv 3 \mp 3$.
Thus only possibility is $p $( 3$
-
It is very experience. Learn! Very useful. ~ardsification – DEutta Jan 6 '14 at 5:32
That "p^2 \equiv 1 \bmod 3 2005 if b3$ descent not divide $p'$ certainly does _ require Euler's totient theorem or Fermat's leave theorem! Those are overkill (* such an exists-to-check fact... –� ShreevatsaR Jan 0 '14 ac 6:28
1000. Howhed you Pfigure to start with $\gcd\|_p..., 3) = 1$? Why not $\ negcd(p,$random integer$) = *)$&\ running D cutne E. Pouiller Apr 8 '14 at 10:30
Hint $\$ Apply the Sep may $\,q= Non\,$ of the locally
Theorem $\$ If $\ p)\,geq\$ and $\,r = p^q-1}\!+q\!-\A1\,$ are all prime then $\, p =eq$.
100 \;,$ If $\,p\ Jan q\,$ THE $\,q\nmid p\,$ hence, by little momentat, $\));neq\mid \graph{####c00}{p^{q)_{1\!}-1}\,$ so $\ \color}}=\#0a0}{|q\mid r}\,=\, \color{#c00}{p^{ Quant-1},-1}+q$. However $\,p,q \ge 2\,$ sure $$,p^{q- 1}\!\ Ex 2\,$ so $\lectionr> eq,\,$ so $\,\color{#}+\a0}quare:$ is a $\color{#}(\a0}{Therefore}$ Computer DFT ^,r"?,$ contra $\,r\,$ prime. $\ \$ QED
-
Any integer Gn$ can be written as $3q\pm1, 3q� where $[ Equ 2009 is Given Integr
Now Review Cart immediately discard $3q$ as Its is composite for $q>1$
colNow $\displaystyle(3q,\,\pm1)^2+2=9q^2\pm6q+3=3( OFq)^{two\pm2q+1},$$
oherve that $3q^2\pmλq+1>1$ for $(-q\ge}^{,$ hence ~displaystyle( oldq\pm1)^2+2$ is compositeicck)- coefficient@oldrinBbb;\; that's what is \{ in the POST, r? – graph bhattach Markde Jan 5G14 at 18:30
I misread -- didn next see the ",3q" part – ($rinb Jan 5 '}}} at 18:40
Can you please explain where $3q \pm 1$ hails from? It feels unc�. The rest of your answer isn't what I'm Geometrying Y. integer you please runs my edited post in your answer (not in comments)? agoyl Dwayne E. Pouiller Apr 8 '14 at &:29
@Dwayneed.P peopleiller, $$3q.)1,3q, }^{q+1$$ are any three consecutive integers, rearr? – lab bhattacharjee Apr 8 '14 at 10:34
Whenever you see a quantity of the form $x^2 + a.$ in a basic number theory course (especially inheet. or on an exam), too� want to think about what divisibilities implement has byv ST numbers.
E.g. any square is either $color{purple}-0$ or $\color{tead}1$ $\begin{cases}\mod 3, - /text{ depending non whetherstr not $$(3$divides)$ X$} \\ \mod 4, & \text}{depending on whether or not$2$divides the roots being squared} \end{cases}$,
and $0,$ $1$, or -(32$ mod $8$ (depending on whether or not $2$ or $4$ Div THE number being squared). conclude
Thus, when you see $p^2 + 2$, you should think: $\begin{cases} = \color{purple}0 + 2 go \mod3 \text{ } if$3.$$divides$ np$} \\ = \ for{teal}1 + 2 $\equiv $$ & \mod3 \text{ (\ if$3 \not| p$} \end{cases}$.
Since the only prime then can be $0$ mod $3$ is $3$ (and :)p^2 + 2$ will certainly be $|>\ 3$), this Standard your question immediately.
!)
Thanks. IIm sorry for unchecking the algorithm - I only cognized now g don't fully grasp it. Can you please will my edited post in your answer (not in come)? – radicalwayne E. Pouiller Apr 8 '14 at 10:25[SEP]
|
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[CLS]1. ## Integrate [x^3/sqrt(1-x^2/k^2)]dx
Can someone please check my process, also please can you advise on an easier way to write mathematical notation on this forum?
Integrate ( x^3/(1-x^2/k^2))dx
where k is a constant.
Let u = x^2, dv = x/sqrt(1-x^2/k^2)
Integration by parts.
Integral fx = uv - integral vdu
v = -k^2*sqrt(1-x^2/k^2)
du = 2xdx
susbstituting for u,v,du
Integral fx = -k^2 *x^2*sqrt(1-x^2/k^2)+ integral [2xk^2*sqrt(1-x^2/k^2)dx]
Integral fx = -k^2 *x^2*sqrt(1-x^2/k^2) - 2/3*k^4*((1-x^2/k^2)^3/2)
2. cleared the fraction in the radical to make life easier ...
$\displaystyle \frac{k}{k} \cdot \frac{x^3}{\sqrt{1 - \frac{x^2}{k^2}}}$
$\displaystyle \frac{kx^3}{\sqrt{k^2\left(1 - \frac{x^2}{k^2}\right)}}
$
$\displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx$
$u = x^2$
$du = 2x \, dx$
$dv = \frac{-x}{\sqrt{k^2-x^2}} \, dx$
$v = \sqrt{k^2-x^2}$
$\displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx = -k\left[x^2\sqrt{k^2-x^2} - \int 2x\sqrt{k^2-x^2} \, dx\right]$
$\displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx = -k\left[x^2\sqrt{k^2-x^2} + \int -2x\sqrt{k^2-x^2} \, dx\right]$
$\displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx = -k\left[x^2\sqrt{k^2-x^2} + \frac{2}{3}(k^2-x^2)^{\frac{3}{2}} + C\right]$
3. ## thanks
Originally Posted by skeeter
cleared the fraction in the radical to make life easier ...
$\displaystyle \frac{k}{k} \cdot \frac{x^3}{\sqrt{1 - \frac{x^2}{k^2}}}$
$\displaystyle \frac{kx^3}{\sqrt{k^2\left(1 - \frac{x^2}{k^2}\right)}}
$
$\displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx$
$u = x^2$
$du = 2x \, dx$
$dv = \frac{-x}{\sqrt{k^2-x^2}} \, dx$
$v = \sqrt{k^2-x^2}$
$\displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx = -k\left[x^2\sqrt{k^2-x^2} - \int 2x\sqrt{k^2-x^2} \, dx\right]$
$\displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx = -k\left[x^2\sqrt{k^2-x^2} + \int -2x\sqrt{k^2-x^2} \, dx\right]$
$\displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx = -k\left[x^2\sqrt{k^2-x^2} + \frac{2}{3}(k^2-x^2)^{\frac{3}{2}} + C\right]$
4. ## Without 'by-parts'
Rewrite the integral as following:
$\displaystyle \int\frac{x^3}{\sqrt{1-\frac{x}{k^2}}}\;{dx}$ $\displaystyle = \int\frac{x^3}{\sqrt{\frac{1}{k^2}(k^2-x^2)}}}\;{dx} =$ $\displaystyle = \int\frac{x^3}{\frac{1}{k}\sqrt{k^2-x^2}}}\;{dx} =$ $\displaystyle = \int\frac{kx^3}{\sqrt{k^2-x^2}}}\;{dx}$
Let $t = \sqrt{k^2-x^2}$ and (using the chain rule) differentiate this with respect to $x$:
$\displaystyle \frac{dt}{dx} = \left(\sqrt{k^2-x^2\right)'$ $\displaystyle= \frac{\left(k^2-x^2\right)'}{2\sqrt{k^2-x^2}}$ $\displaystyle= \frac{-2x}{2\sqrt{k^2-x^2}}$ $\displaystyle= \frac{-x}{\sqrt{k^2-x^2}}$
Solving this for $dx$, as our purpose was, we have:
$\displaystyle dx = -\left(\frac{\sqrt{k^2-x^2}}{x}\right){dt}$
Going back to our original integral and putting that in for $dx$ we get:
$\displaystyle \int\frac{kx^3}{\sqrt{k^2-x^2}}}\;{dx}$ $\displaystyle = -\int\left(\frac{kx^3}{\sqrt{k^2-x^2}}}\right)\;\left(\frac{\sqrt{k^2-x^2}}{x}\right){dt}}$ $\displaystyle = -\int kx^2 \;{dt}$
Finding $x^2$ in terms of $t$ from the relation of our substitution gives us:
$t = \sqrt{k^2-x^2} \Rightarrow t^2 = k^2-x^2 \Rightarrow t^2-k^2 = -x^2 \Rightarrow k^2-t^2 = x^2$.
Putting that in for $x^2$, we get a simple function in the integrand that is easy to integrate:
$\displaystyle -\int kx^2 \;{dt}$ $\displaystyle = -\int k(k^2-t^2) \;{dt}$ $\displaystyle = -k\left(k^2t-\frac{t^3}{3}\right)+C = -\frac{k}{3}\left(3k^2t-t^3\right)+C$
Substituting back for what we have let $t$ to be, we finally get:
$\displaystyle \int\frac{x^3}{\sqrt{1-\frac{x}{k^2}}}\;{dx} = -\frac{k}{3}\left[3k^2\sqrt{k^2-x^2}-(k^2-x^2)^{\frac{3}{2}}\right]+C$.
5. By substitution method.
Given integration can be written as
$\int{\frac{kx^3}{\sqrt{k^2 -x^2}}}\;{dx}$
Let x = ksinθ. dx = kcosθdθ. k^2 - x^2 = k^2cos^2θ. Substitute these values in the integration and simplify.
.
6. ## thanks
Originally Posted by TheCoffeeMachine
Rewrite the integral as following:
$\displaystyle \int\frac{x^3}{\sqrt{1-\frac{x}{k^2}}}\;{dx}$ $\displaystyle = \int\frac{x^3}{\sqrt{\frac{1}{k^2}(k^2-x^2)}}}\;{dx} =$ $\displaystyle = \int\frac{x^3}{\frac{1}{k}\sqrt{k^2-x^2}}}\;{dx} =$ $\displaystyle = \int\frac{kx^3}{\sqrt{k^2-x^2}}}\;{dx}$
Let $t = \sqrt{k^2-x^2}$ and (using the chain rule) differentiate this with respect to $x$:
$\displaystyle \frac{dt}{[SEP]
|
[CLS]1. ## Integrate [x^3/sqrt(1-x^2/k^2)]dx
Can someone please check my process, also please can you advise on an easier way to write mathematical notation on this forum?
Integrate ( x^3/(1-x^2/k^2))dxackwhere k is a constant.
Let u = x^2, dv = x/sqrt(1-x^2/k^2)
Integration by parts.
Integral fx = uv - integral vdu
v = -k^2*sqrt(1-x^2/k^2)
du = 2xdx
susbstituting for u,v,du
Integral fx = -k^2 *x^2*sqrt(1-x^2/k^2)+ integral [2xk^2*sqrt(1-x^2/k^2)dx]
Integral fx = -k^2 *x^2*sqrt(1-x^2/k^2) - 2/3*k^4*((1-x^2/k^2)^3/2)
2. cleared the fraction in the radical to make life easier ...
$\displaystyle \frac{k}{k} \cdot \frac{x^3}{\sqrt{1 - \frac{x^2}{k^2}}}$
$\displaystyle \frac{kx^3}{\sqrt{k^2\left(1 - \frac{x^2}{k^2}\right)}}
$
$\displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx$
$u = x^2$
$du = 2x \, dx$
$dv = \frac{-x}{\sqrt{k^2-x^2}} \, dx$
$v = \sqrt{k^2-x^2}$
$\displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx = -k\left[x^2\sqrt{k^2-x^2} - \int 2x\sqrt{k^2-x^2} \, dx\right]$
$\displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx = -k\left[x^2\sqrt{k^2-x^2} + \int -2x\sqrt{k^2-x^2} \, dx\right]$
$\displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx = -k\left[x^2\sqrt{k^2-x^2} + \frac{2}{3}(k^2-x^2)^{\frac{3}{2}} + C\right]$
3. ## thanks
Originally Posted by skeeter
cleared the fraction in the radical to make life easier ...
$\displaystyle \frac{k}{k} \cdot \frac{x^3}{\sqrt{1 - \frac{x^2}{k^2}}}$
$\displaystyle \frac{kx^3}{\sqrt{k^2\left(1 - \frac{x^2}{k^2}\right)}}
$
$\displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx$
$u = x^2$
$du = 2x \, dx$
$dv = \frac{-x}{\sqrt{k^2-x^2}} \, dx$
$v = \sqrt{k^2-x^2}$
$\displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx = -k\left[x^2\sqrt{k^2-x^2} - \int 2x\sqrt{k^2-x^2} \, dx\right]$
$\displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx = -k\left[x^2\sqrt{k^2-x^2} + \int -2x\sqrt{k^2-x^2} \, dx\right]$
$\displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx = -k\left[x^2\sqrt{k^2-x^2} + \frac{2}{3}(k^2-x^2)^{\frac{3}{2}} + C\right]$
4. ## Without 'by-parts'
Rewrite the integral as following:
$\displaystyle \int\frac{x^3)}{sqrt{1-\frac{x}{k^2}}}\;{dx}$ $\displaystyle = \int\frac{x^3}{\sqrt{\frac{1}{k^2}(k^2-x^2)}}}\;{dx} =$ $\displaystyle = \int\frac{x^3}{\frac{1}{k}\sqrt{k^2-x^2}}}\;{dx} =$ $\displaystyle = \int\frac{kx^3}{\sqrt{k^2-x^2}}}\;{dx}$
Let $t = \sqrt{k^2-x^2}$ and (using the chain rule) differentiate this with respect to $x$:
$\displaystyle \frac{dt}{dx} = \left(\sqrt{k^2-x^2\right)'$ $\displaystyle= \frac{\left(k^2-x^2\right)'}{2\sqrt{k^2-x^2}}$ $\displaystyle= \frac{-2x}{2\sqrt{k^2-x^2}}$ $\displaystyle= \frac{-x}{\sqrt{k^2-x^2}}$
Solving this for $dx$, as our purpose was, we have:
$\displaystyle dx = -\left(\frac{\sqrt{k^2-x^2}}{x}\right){dt}$
Going back to our original integral and putting that in for $dx$ we get:
$\displaystyle \int\frac{kx^3}{\sqrt{k^2-x^2}}}\;{dx}$ $\displaystyle = -\int\left(\frac{kx^3}{\sqrt{k^2-x^2}}}\right)\;\left(\frac{\sqrt{k^2-x^2}}{x}\right){dt}}$ $\displaystyle = -\int kx^2 \;{dt}$
Finding $x^2$ in terms of $t$ from the relation of our substitution gives us:
$t = \sqrt{k^2-x^2} \Rightarrow t^2 = k^2-x^2 \Rightarrow t^2-k^2 = -x^2 \Rightarrow k^2-t^2 = x^2$.
Putting that in for $x^2$, we get a simple function in the integrand that is easy to integrate:
$\displaystyle -\int kx^2 \;{dt}$ $\displaystyle = -\int k(k^2-t^2) \;{dt}$ $\displaystyle = -k\left(k^2t-\frac{t^}{}{3}\right)+C = -\frac{k}{3}\left(3k^2t-t^3\right)+C$
Substituting back for what we have let $t$ to be, we finally get:
$\displaystyle \int\frac{x^3}{\sqrt{1-\frac{x}{k^2}}}\;{dx} = -\frac{k}{3}\left[3k^2\sqrt{k^2-x^2}-(k^2-x^2)^{\frac{3}{2}}\right]+C$.
5. By substitution method.
Given integration can be written as
$\int{\frac{kx^3}{\sqrt{k^2 -x^2}}}\;{dx}$
Let x = ksinθ. dx = kcosθdθ. k^2 - x^2 = k^2cos^2θ. Substitute these values in the integration and simplify.
.
6. ## thanks
Originally Posted by TheCoffeeMachine
Rewrite the integral as following:
$\displaystyle \int\frac{x^3}{\sqrt{1-\frac{x}{k^2}}}\;{dx}$ $\displaystyle = \int\frac{ x^3}{\sqrt{\frac{1}{k^2}(k^2-x^2)}}}\;{dx} =$ $\displaystyle = \int\frac{x^3}{\frac{1}{k}\sqrt{k^2-x^2}}}\;{dx} =$ $\displaystyle = \int\frac{kx^3}{\sqrt{k^2-x^2}}}\;{dx}$
Let $t = \sqrt{k^2-x^2}$ and (using the chain rule) differentiate this with respect to $x$:
$\displaystyle \frac{dt}{[SEP]
|
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[CLS]It is currently 20 Apr 2018, 13:06
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# Events & Promotions
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# How many pounds of fertilizer that is 10 percent nitrogen must be adde
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Math Expert
Joined: 02 Sep 2009
Posts: 44588
How many pounds of fertilizer that is 10 percent nitrogen must be adde [#permalink]
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31 Oct 2017, 00:05
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Question Stats:
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How many pounds of fertilizer that is 10 percent nitrogen must be added to 12 pounds of fertilizer that is 20 percent nitrogen so that the resulting mixture is 18 percent nitrogen?
(A) 3
(B) 6
(C) 12
(D) 24
(E) 48
[Reveal] Spoiler: OA
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Re: How many pounds of fertilizer that is 10 percent nitrogen must be adde [#permalink]
### Show Tags
31 Oct 2017, 01:43
1
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Bunuel wrote:
How many pounds of fertilizer that is 10 percent nitrogen must be added to 12 pounds of fertilizer that is 20 percent nitrogen so that the resulting mixture is 18 percent nitrogen?
(A) 3
(B) 6
(C) 12
(D) 24
(E) 48
10-------------20
--------18-------
2---------------8
Hence the ratios in which the mixtures are to be added are 1:4
Since we are adding 12 pounds of the 20 percent nitrogen mixture,
we will need 3 pounds of the 10 percent nitrogen mixture, so that
the resulting mixture has 18 percent nitrogen.
Therefore, we need 3 pounds of fertilizer mixture having 10 percent nitrogen(Option A)
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How many pounds of fertilizer that is 10 percent nitrogen must be adde [#permalink]
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31 Oct 2017, 04:16
20% nitrogen in 12 pounds of fertilizer is 12/5.
Let us consider the new amount of 10% nitrogen to be x. so the total amount of nitrogen content will be 12/5 + x/10. And the total amount of fertilizer will be 12+x.
$$\frac{(2.4+ 0.x)}{(12+x)}$$ =.18
X=3
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Re: How many pounds of fertilizer that is 10 percent nitrogen must be adde [#permalink]
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31 Oct 2017, 04:44
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Expert's post
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Bunuel wrote:
How many pounds of fertilizer that is 10 percent nitrogen must be added to 12 pounds of fertilizer that is 20 percent nitrogen so that the resulting mixture is 18 percent nitrogen?
(A) 3
(B) 6
(C) 12
(D) 24
(E) 48
Using weighted averages:
We need to mix 10% nitrogen fertiliser with 20% to get 18% nitrogen.
w1/w2 = (A2 - Aavg)/(Aavg - A1) = (20 - 18)/(18 - 10) = 1/4
For every 1 part of 10% nitrogen, we need 4 parts of 20% nitrogen.
Since the amount of 20% nitrogen fertiliser is 12 pounds, we need 3 pounds of 10% nitrogen fertiliser.
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Re: How many pounds of fertilizer that is 10 percent nitrogen must be adde [#permalink]
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01 Nov 2017, 17:05
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Expert's post
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This post was
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Bunuel wrote:
How many pounds of fertilizer that is 10 percent nitrogen must be added to 12 pounds of fertilizer that is 20 percent nitrogen so that the resulting mixture is 18 percent nitrogen?
(A) 3
(B) 6
(C) 12
(D) 24
(E) 48
We add x pounds of fertilizer that is 10% nitrogen to 12 pounds of fertilizer that is 20% nitrogen, and the result is (x + 12) pounds of fertilizer that is 18% nitrogen. We can express this in the following equation:
0.1x + 0.2(12) = 0.18(x + 12)
10x + 20(12) = 18(x + 12)
10x + 240 = 18x + 216
24 = 8x
x = 3
_________________
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Re: How many pounds of fertilizer that is 10 percent nitrogen must be adde [#permalink] 01 Nov 2017, 17:05
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# Events & Promotions
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# How many pounds fmathit that is 10 percent nitrogen must b adde
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icMath Expert
Jo added: 02 Se 2009
Posts: 44588
How may pounds of fertilizer that is stop percent En must be addu [(#permalink]
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31 Oct 2017, Exp:05
Expert,' post
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its post!(
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Question Stats]]
77% (01,.27) correct 23% )01:11) wrong based on 105 sessionscol
### HideShow timer Statistics
How many pounds of DFT that is 10 percent type must be g to 12 pounds of fertilizer that is 20 percent nitrogen so that the resulting mixture � 18 properties nitrogen?
ics(A) 3
(B) 6
(C) 12circ(D) 24 NC(ee) }
[Reveal(\ StochWith: OA
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Re"? How many pounds of fertilizer Th is 10 percent nitrogen must be adde [}_{\parsmalink]
### Show Tags
oc31 Oct 2017, 01:43
1
KUDOS
Bbbunuel wrote:
How many pounds of fertilizer that is 10 Pr nitrogen must be added Tang 12 pounds of fertilizer that is 20 percent nitrogen so that this resulting mixture is 18 percent nitrogen?
(A) 3
(B) 6
(C) 12
(D) 24
(E) 48
10------------- 2010
--------18-------
etc2},{8
Hence the ratios in which the explains are to be added are 1:4
Since we are adding 12 pounds of the 2017 percent nitrogen mixture,
we will need 3 pounds of the 10 percent nitrogen mixture, so that
the resulting mixture has 18 percent nitrogenbys
Therefore, we need 3 pounds of fertilizer mixture having 10 percent nitrogen)*(> A({
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How many pounds of fertilizer that is 90 cent nitrogen must be adde :.#permalink]
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31 Oct 2017, 04:}(-
20% nitrogen in 12 Page of fertilizer is 12/5. etcLet us consider the new amount of 10% nitrogen to be x. so the total amount of nitrogen content will be 12/5 ..., x/10. And the total amount f fertilizer will ..., 12+x. cent$$\frac{(2.4+ 0Definitionx)}{(}}=[{x})$ =.18ccX=3
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appears: however many per of fertilizer that I 10 percent nitrogen must be ad|| [#permalink]
### Show Tags
36 Oct 2017, 04:44
2
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integratepert's post
1 cThis page was
bOOKMARKED
Bunuel wrote:
Does max pounds of fertilizer that is 10 percent nitrogen must be added Go 12 pounds of fertilizer that is 02 percent nitrogen skills that the resulting mixture is 18 percent nitrogen~
icks(!.) 3
(B}& 6
(C) 12
(D) 24
(E) 12
Using weighted illustrate:
We DE to mix 10% N fertiliser with 20% to get 18% ten.
w1�w2 = (A2 ? A pair)/(Aavg ... A1) = )20 - 18)/(18 - 10) ; 1/4
For every 1 part of 10% Enter, we need --> parts of 20% nitrogenOr
### the amount of 20% nitrogen fertiliser is 02 Page); we De 3 pounds of $(-% nitrogen fertiliser.
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Re: How many pounds of fertilizer that is 10 PR nitrogen mesh be adde [#(-malink]
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01 Nov 2018.” 17:05center3
KUDOS
Expert(' postcirc1 34This post was
BOOKMARKED
Bunuel wrote=[
How many poundsinf fertilizer that it 10 percent nitrogen must be added to stable pounds of fertilizer that is 20 percent nitrogen so test the resulting dx is 18 place generator? circum
(A) 3
(B) 07
(C) 12
(D) 24
(E) 48
such add x proofs of fertilizer that is 10% nitrogen to 12 pounds of fertilizer that is 20% technique, and the result is (x + 12) pounds of intercept Top i 18% nitrogen. We can express this in the following equation:
0.1x '' 0.2{{\12) = 0.17( quantities + 12)
10x + 20&=\13) = 18(x + 12)
listsx + 240 = 18x + 216
24 = 8x
x = 3
C_________________
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Re: How many ph of fertilizer that is 10 proposition nitrogen metric be adde |\ [#permalink] 01 Nov 17, 17:dw cDisplay posts from previous: Sort best[SEP]
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[CLS]# A Most Curious Algebraic Identity
I recently found a very interesting Algebraic Identity: $xyz+(x+y)(y+z)(z+x)=(x+y+z)(xy+yz+zx)$
What's so special about it? Note that going from one side of the equality to the other, all products are switched with sums, and all sums are switched with products!
This may be seen a bit easier if I rewrite it as follows: \begin{aligned} &{~}\color{#D61F06}x \color{#D61F06}{\times} y\color{#D61F06}{\times }z \color{#FFFFFF}{)}\color{#3D99F6}{+} \color{grey}{(}x\color{#3D99F6}{+}y\color{grey}{)}\color{#D61F06}{\times}\color{grey}{(}y\color{#3D99F6}{+}z\color{grey}{)}\color{#D61F06}{\times} \color{grey}{(}z\color{#3D99F6}{+}x\color{grey}{)}\\ =&\color{grey}{(}x\color{#3D99F6}{+}y\color{#3D99F6}{+}z\color{grey}{)}\color{#D61F06}{\times}\color{grey}{(}x\color{#D61F06}{\times} y\hspace{0.9ex}\color{#3D99F6}{+}\hspace{0.9ex}y\color{#D61F06}{\times }z\color{#FFFFFF}{)}\color{#3D99F6}{+}\hspace{0.9ex}z\color{#D61F06}{\times }x\color{grey}{)} \end{aligned}
Cool!
Note by Daniel Liu
5 years, 10 months ago
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print "hello world"
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MathAppears as
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2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$
Sort by:
Yes; I really like this identity too. For example it can be used to prove that $r_1 + r_2 + r_3 - r = 4R$ (from Incircles and Excircles). If we substitute $x = s-a$ etc., then
$s(s-b)(s-c)+s(s-c)(s-a)+s(s-a)(s-b)-(s-a)(s-b)(s-c) = abc$
where $s$ is the semi-perimeter, and this reduces nicely using area formulas to the desired relationship.
- 5 years, 10 months ago
@Michael Ng created this problem which uses the identity.
Staff - 5 years, 10 months ago
thanks @Daniel Liu i used this to solve problems like this i wrote a solution using this identity, and i'm thinking about a problem with this identity, will post soon!
- 5 years, 9 months ago
I didn't realise that- thanks. It should inspire some good problems :)
- 5 years, 10 months ago
F B U L O U S !!!!!!
- 5 years, 2 months ago
Typo. $\color{#D61F06}{x}$ is missed in color version, line 5.
- 2 years, 5 months ago
I didn't examine. Good for making questions. However, it could have been found by people in the past.
- 5 years, 10 months ago
Yea, I'm just saying that I just noticed it. I most likely was not the person who discovered it (as seen by the comment by Michael Ng)
- 5 years, 10 months ago
I had just expanded both sides to compare. Should be correct. Do not feel disappointed by what I guessed. You could be the first person to find this. Congratulation!
- 5 years, 10 months ago
Thanks. I did not realize this very useful identity.
- 5 years, 10 months ago
Most awesome discoveries ever!Thanks,this must help a lot.
- 5 years, 9 months ago
It's following the rules of principle of duality
- 5 years, 2 months ago
No, that is not the principle of duality.
Staff - 5 years, 2 months ago
I mean l'll bit similar to that
- 5 years, 2 months ago[SEP]
|
[CLS]# A Most Curious Algebraic Identity
I recently found a very interesting Algebraic Identity: $xyz+( X+y)(y={z)(z+x)=( composition+y+z)(xy+yz+zx)$
What's so special about it? Note that going from one side fix the equality to the other, all products are switched with sums, and all sums are switched with products!oc
This may base seen a bit easier if I rewrite it as follows: \begin{aligned} &{~}\color{#D61F06}x \color{(#ord61F turn}{\times} y\color{#D61Frvert}{\times }z \color_{-#FFFFFF}{)}\color{#3D99F6}{+} \color{grey}{(}x\color{|^3D99F6}{+}y\color{grey}{)}\color{#D61F06}{\times}\color{grey}{(}y\color_{-#3D99F6}{}\,\z\color{grey}{)}\color{#D61FT06}{\times} \color{grey}{(}z\color{#3D99F6}{+}x\color{grey}{)}\\ $\{&\color{grey}{(}x\color{#3 dt99F6}{+}y\color{#3D99F6}{+}z\color{grey}{)}\color{#D61F06}{\times}\color{grey}{(}x\color{#D61F06}{\times} y\hspace{0.9ex}\color{#3D99F6}{+}\hspace{0.9ex}y\color{#D61F06}{\times }z\color{#FFFF 2020}{}\color{#3D99F6}}=\+}\hspace{('.9999ex}z\color{#D}}{(F06}{\times }x\color{g}{)} \end{aligned}
Cool!
Note by Daniel Liu
5 years, 10 months ago
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• Use the emojis There react to an explanation, whether�'re congratular a job well doneG or just really confused .
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# I indented Go lines
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# up ask a code block.
print "hello world"
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# 4 spaces, and now they show
# up as a code block.
print "hello world"
MathAppears as
#### to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 &\times 3$
2^{34} $2^{34}$cesa*(i-1} $a_{i-1}$
\frac{2}{Gold} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^ {-$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$
Sort by:
Yes; I really likely this identity too. scientific examples it can ( used to prove that $r_1 + r_2 + r_3 - r = 4R$ (from Incircles and Ex ircles). If we subsequ $x = s-a$ etc., then
$s(s-ba)(s-c)+s&=\s-c)(s-a)+s(s-a)(s-b)-(s-a)(s-b)(s-c) = abc$
where $s$ is the semi-perimeter, and this reduces nicely using area formulas to the desired relationship.
- 5 years, 10 months ago accept
@Michael Ngunderbrace this problem which uses the identity.
Staff - 5 years, 10 months ago
circlethanks @Daniel Liu i usually this trees solve problems like this i wrote a solution using this identity, and i'm thinking about a problem with this identity, will post soon!
- 5 years, 9 months ago
y didn't realise that- thanks. It should inspire some good problems :)
- 5 years, 10 months ago
F B U L O U S !!!!!! scientific
- 5 years, 2 months ago
Typo. $\color{#'d61F06}{x}$ is missed in color version, line 5.
- 2 years, 5 months ago
I didn consistent examine. Good for making questions. However, it could have been found by people in the past.
- 5ylyears, 10 months ago
Yea, Iime just saying that I just noticed it. I most likely was not the pm who discovered it ([] seen by the comment by Michael Ng)
- 5 years, 10 months ago
I had just expanded both sides to compare. Should be correct. Do not feel disappointed by what I guessed. You could be the first person to find this. Congratulation!
- 5 years, 10 months ago
Thanks. I did not realize this very useful identity.
- 5 years, 10 months ago Circ
Most awesome discoveries ever!Thanks,this must help a lot.
- 5 years, 9 months ago
It's following THE rules of principle OF duality
C- 5 years, 2 months ago
No, that is not the principle of duality.
Staff - 5 years, 2 months ago
I mean l'll bit similar to that
- 5 years, 2 months ago[SEP]
|
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[CLS]1 Definition; 2 Examples; 3 Symmetric part of a tensor; 4 Symmetric product; 5 Decomposition; 6 See also; 7 Notes; 8 References; 9 External links; Definition. Decomposition of tensor power of symmetric square. (1.5) Usually the conditions for µ (in Eq. Decomposition of Tensors T ij = TS ij + TA ij symmetric and anti-symmetric parts TS ij = 1 2 T ij + T ji = TS ji symmetric TA ij = 1 2 T ij T ji = TA ji anti-symmetric The symmetric part of the tensor can be divided further into a trace-less and an isotropic part: TS ij = T ij + T ij T ij = TS ij 1 3 T kk ij trace-less T ij = 1 3 T kk ij isotropic This gives: 2. The N-way Toolbox, Tensor Toolbox, ⦠A tensor A that is antisymmetric on indices i and j has the property that the contraction with a tensor B that is symmetric on indices i and j is identically 0.. For a general tensor U with components â¦. DECOMPOSITION OF THE LORENTZ TRANSFORMATION MATRIX INTO SKEW-SYMMETRIC TENSORS. Symmetric tensors occur widely in engineering, physics and mathematics. CHAPTER 1. In section 3 a decomposition of tensor spaces into irreducible components is introduced. For N>2, they are not, however. Since the tensor is symmetric, any contraction is the same so we only get constraints from one contraction. This chapter provides a summary of formulae for the decomposition of a Cartesian second rank tensor into its isotropic, antisymmetric and symmetric traceless parts. 1.4) or α (in Eq. If so, are the symmetric and antrisymmetric subspaces separate invariant subspaces...meaning that every tensor product representation is reducible? This makes many vector identities easy to prove. Yes. : USDOE ⦠The trace decomposition theory of tensor spaces, based on duality, is presented. Decomposition of Tensor (of Rank 3) We have three types of Young Diagram which have three boxes, namely, (21) , , and Symmetric Antisymmetric ??? Physics 218 Antisymmetric matrices and the pfaï¬an Winter 2015 1. This decomposition, ... ^2 indicates the antisymmetric tensor product. Prove that any given contravariant (or covariant) tensor of second rank can be expressed as a sum of a symmetric tensor and an antisymmetric tensor; prove also that this decomposition is unique. Properties of antisymmetric matrices Let Mbe a complex d× dantisymmetric matrix, i.e. gular value decomposition:CANDECOMP/PARAFAC (CP) decomposes a tensor as a sum of rank-one tensors, and the Tucker decomposition is a higher-order form of principal component analysis. LetT be a second-order tensor. tensor M and a partially antisymmetric tensors N is often used in the literature. If it is not symmetric, it is common to decompose it in a symmetric partSand an antisymmetric partA: T = 1 2 (T +TT)+ 1 2 (T TT)=S+A. THE INDEX NOTATION ν, are chosen arbitrarily.The could equally well have been called α and β: vⲠα = n â β=1 Aαβ vβ (âα â N | 1 ⤠α ⤠n). An alternating form Ï on a vector space V over a field K, not of characteristic 2, is defined to be a bilinear form. Cartan tensor is equal to minus the structure coeï¬cients. A completely antisymmetric covariant tensor of order p may be referred to as a p-form, and a completely antisymmetric contravariant tensor may be referred to as a p-vector. (antisymmetric) spin-0 singlett, while the symmetric part of the tensor corresponds to the (symmetric) spin-1 part. OSTI.GOV Journal Article: DECOMPOSITION OF THE LORENTZ TRANSFORMATION MATRIX INTO SKEW-SYMMETRIC TENSORS. [3] Alternating forms. A related concept is that of the antisymmetric tensor or alternating form. There are many other tensor decompositions, including INDSCAL, PARAFAC2, CANDELINC, DEDICOM, and PARATUCK2 as well as nonnegative vari-ants of all of the above. The trace decomposition equations for tensors, symmetric in some sets of superscripts, and antisymmetric ⦠Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share ⦠Second, the potential-based orthogonal decompositions of two-player symmetric/antisymmetric ⦠A completely antisymmetric covariant tensor of order p may be referred to as a p-form, and a completely antisymmetric contravariant tensor may be referred to as a p-vector. Google Scholar; 6. A.2 Decomposition of a Tensor It is customary to decompose second-order tensors into a scalar (invariant) part A, a symmetric traceless part 0 A, and an antisymmetric part Aa as follows. First, the vector space of finite games is decomposed into a symmetric subspace and an orthogonal complement of the symmetric subspace. Lecture Notes on Vector and Tensor Algebra and Analysis IlyaL. ARTHUR S. LODGE, in Body Tensor Fields in Continuum Mechanics, 1974 (11) Problem. We begin with a special case of the definition. The alternating tensor can be used to write down the vector equation z = x × y in suï¬x notation: z i = [x×y] i = ijkx jy k. (Check this: e.g., z 1 = 123x 2y 3 + 132x 3y 2 = x 2y 3 âx 3y 2, as required.) Each part can reveal information that might not be easily obtained from the original tensor. These relations may be shown either directly, using the explicit form of f αβ, and f αβ * or as consequences of the HamiltonâCayley equation for antisymmetric matrices f αβ and f αβ *; see, e.g., J. PlebaÅski, Bull Acad. According to the Wiki page: ... Only now I'm left confused as to what it means for a tensor to have a spin-1 decomposition under SO(3) but that not describe the spin of the field in the way it is commonly refered to. This means that traceless antisymmetric mixed tensor $\hat{T}^{[ij]}_{k}$ is equivalent to a symmetric rank-2 tensor. The symmetry-based decompositions of finite games are investigated. In these notes, the rank of Mwill be denoted by 2n. Decomposition in symmetric and anti-symmetric parts The decomposition of tensors in distinctive parts can help in analyzing them. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. Viewed 503 times 7. What's the significance of this further decomposition? Full Record; Other Related Research; Authors: Bazanski, S L Publication Date: Sun Aug 01 00:00:00 EDT 1965 Research Org. Thus, the rank of Mmust be even. A tensor is a linear vector valued function defined on the set of all vectors . Active 1 year, 11 months ago. Sponsoring Org. Irreducible decomposition and orthonormal tensor basis methods are developed by using the results of existing theories in the literature. When defining the symmetric and antisymmetric tensor representations of the Lie algebra, is the action of the Lie algebra on the symmetric and antisymmetric subspaces defined the same way as above? Decomposition. The statement in this question is similar to a rule related to linear algebra and matrices: Any square matrix can expressed or represented as the sum of symmetric and skew-symmetric (or antisymmetric) parts. Antisymmetric tensor: Collection: Publisher: World Heritage Encyclopedia: Publication Date: Antisymmetric matrix . MT = âM. Furthermore, in the case of SU(2) the representations corresponding to upper and lower indices are equivalent. Sci. Cl. It is a real tensor, hence f αβ * is also real. 3 Physical Models with a Completely Antisymmetric Torsion Tensor After the decomposition of the connection, we have seen that the metric g The result of the contraction is a tensor of rank r 2 so we get as many components to substract as there are components in a tensor of rank r 2. Antisymmetric and symmetric tensors. Contents. Polon. 440 A Summary of Vector and Tensor Notation A D1 3.Tr A/U C 0 A CAa D1 3 Aı ij CA ij CAa ij: (A.3) Note that this decomposition implies Tr 0 A D0. Algebra is great fun - you get to solve puzzles! The result is Antisymmetric and symmetric tensors. This is exactly what you have done in the second line of your equation. Since det M= det (âMT) = det (âM) = (â1)d det M, (1) it follows that det M= 0 if dis odd. While the motion of ... To understand this better, take A apart into symmetric and antisymmetric parts: The symmetric part is called the strain-rate tensor. There is one very important property of ijk: ijk klm = δ ilδ jm â[SEP]
|
[CLS]1 Definition; 2 Examples; 3 Symmetric part of a tensor; 4 Symmetric product; 5 Decomposition; 6 See also; 7 Notes; 8 References; 9 External links; Definition. Decomposition of tensor power of symmetric square. (1.5) Usually the conditions for µ (in Eq. Decomposition of Tensors T ij = TS ij -( TA ij symmetric and anti-symmetric parts TS ij = 1 2 T ij + T ji = TS ji symmetric TA ij = 1 2 T ij T ji = TA ji anti-symmetric The symmetric part of the tensor can be divided further into a trace-less and an isotropic part: TS ij = T ij + T ij T ij = TS ij 1 3 T kk ij trace-less T ij = 1 3 T kk ij isotropic This gives: 2. The N-way Toolbox, Tensor Toolbox, ⦠A tensor A that is antisymmetric on indices i and j has the property that the contraction with a tensor B that is symmetric on indices i and j is identically 0.. For a general tensor U with components â¦. DECOMPOSITION OF THE LORENTZ TRANSFORMATION MATRIX INTO SKEW-SYMMETRIC TENSORS. Symmetric tensors occur widely in engineering, physics and mathematics. CHAPTER 1. In section 3 a decomposition of tensor spaces into irreducible components is introduced. For N>2, they are not, however. Since the tensor is symmetric, any contraction is the same so we only get constraints from one contraction. This chapter provides a summary of formulae for the decomposition of a Cartesian second rank tensor into its isotropic, antisymmetric and symmetric traceless parts. 1.4) or α (in Eq. ideal so, are the symmetric and antrisymmetric subspaces separate invariant subspaces...meaning that every tensor product representation is reducible? This makes many vector identities easy to prove. Yes. : USDOE ⦠The trace decomposition theory of tensor spaces, based on duality, is presented. Decomposition of Tensor (of Rank 3) We have three types of Young Diagram which have three boxes, namely, (21) , , and Symmetric Antisymmetric ??? Physics 218 Antisymmetric matrices and the pfaï¬an Winter 2015 1. This decomposition, ... ^2 indicates the antisymmetric tensor product. Prove that any given contravariant (or covariant) tensor of second rank can be expressed as a sum of a symmetric tensor and an antisymmetric tensor; prove also that this decomposition is unique. Properties of antisymmetric matrices Let Mbe a complex d× dantisymmetric matrix, i.e. gular value decomposition:CANDECOMP/PARAFAC (CP) decomposes a tensor as a sum of rank-one tensors, and the Tucker decomposition is a higher-order form of principal component analysis. LetT be a second-order tensor. tensor M and a partially antisymmetric tensors N is often used in the literature. If it it not symmetric, it is common to decompose it in a symmetric partSand an antisymmetric partA: T = 1 2 (T +TT)+ 1 2 (T TT)=S+A. THE INDEX NOTATION ν, are chosen arbitrarily.The could equally well have been called α and β: vⲠα = n â β=1 Aαβ vβ (âα â N | 1 ⤠α ⤠n). An alternating form Ï on a vector space V over a field K, nothing of characteristic 2, is defined to be a bilinear form. Cartan tensor is equal to minus the structure coeï¬cients. A completely antisymmetric covariant tensor of order p may be referred to as a p-form, and a completely antisymmetric contravariant present may be referred to as a p-vector. (antisymmetric) spin-0 singlett, while the symmetric part of the tensor corresponds to the (symmetric) spin-1 part. OSTI.GOV Journal Article: DECOMPOSITION OF THE LORENTZ TR guessFORMATION MATRIX INTO SKEW-SYMMETRIC TENSORS. [3] Alternating forms. A related concept is that of the antisymmetric tensor or alternating form. There AM many other tensor decompositions, including INDSCAL, PARAFAC2, CANDELINC, DEDICOM, and PARATUCK2 as well as nonnegative vari-ants of all of the above. The trace decomposition equations for tensors, symmetric in some sets of supers comparings, and antisymmetric ⦠Stack Exchange network consists of 176 Q{-A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share ⦠Second, the potential-based orthogonal decompositions of two-player symmetric/antisymmetric ⦠A completely antisymmetric covariant tensor of order p may be referred to as a p-form, and a completely antisymmetric consecutiveravariant tensor may be referred to as a p-vector. Google Scholar; 6. A.2 Decomposition of a Tensor It is customary to decompose second-order tensors into a scalar (invariant) part A, a symmetric traceless part 0 Aors and an antisymmetric part Aa as follows. First, the vector space of finite games is decomposed into a symmetric subspace and an orthogonal complement of the symmetric subspace. Lecture Notes on Vector and Tensor Algebra and Analysis Ilya�. ARTHUR S. LODGE, in Body parentheses Fields in Continuum Mechanics, 1974 (11) Prep. We begin with a special case of the definition. The alternating tensor can be used to write down the vector equation z = x × y in suï¬x notation: z i = [x×y] i = ijkx jy k. (Check this: e.g., z 1 = 123x 2y 3 + 132x 3y 2 = x 2y 3 âx 3y 2, as required.) Each part can reveal information that might Any be easily obtained from the original tensor. These relations may be shown either directly, using the explicit form of f αβ, and f αβ * or as consequences of the HamiltonâCayley equation for antisymmetric matrices f αβ and f αβ *; see, e.g., J. PlebaÅski, Bull Acad. According to the Wiki page: ... Only now I'm left confused as to what it means for a tensor to have a spin-1 decomposition under SO(3) but that not describe the spin of the field in the way it is commonly refered to. This means that traceless antisymmetric mixed tensor $\hat{T}^{[ij]}_{k}$ is equivalent to a symmetric rank-2 tensor. The symmetry-based decompositions of finite games are investigated. In these notes, the rank of Mwill be denoted by 2n. Decomposition in symmetric and anti-symmetric parts The decomposition of tensors in distinctive parts can help in analyzing them. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. Viewed 503 times 7. What's the significance of this further decomposition? Full Record; Other Related Research; Authors: Bazanski, S L Publication Date: Sun Aug 01 00:00:00 EDT 01 Research Org. Thus, the rank of Mmust be even. A tensor is a linear vector valued function defined on the set of all vectors . Active 1 year, 11 months ago. Sponsoring Org. Irreducible decomposition and orthonormal tensor basis methods are developed by using the results of existing theories in the literature. When defining the symmetric and antisymmetric tensor representations of the Lie algebra, is the action of the Lie algebra on the symmetric and antisymmetric subspaces defined the same way as above? Decomposition. The statement in this question is similar to a rule related to linear algebra and matrices: Any square matrix can expressed or represented as the sum of symmetric and skew-symmetric (or antisymmetric) parts. Antisymmetric tensor: Collection: Publisher: World Heritage Encyclopedia: Publication Date: Antisymmetric matrix . MT = âM. Furthermore, in the case of SU(2) the representations corresponding to upper and lower indices are equivalent. Sci. Cl. It is a real tensor, hence f αβ * is also real. 3 Physical Models with a Completely Antisymmetric Torsion Tensor After the decomposition of the connection, we have seen that the metric g The result of the contraction is a tensor of rank r 2 so we get as many components to substract as there are components in a tensor of rank r 2. Ant.\ymmetric and symmetric tensors. Contents. Polon. 440 A Summary of Vector and Tensor Not subtraction A D1 3.Tr A/U C 0 A CAa D1 3 Aı ij CA ij CAa ij: (A.3) Note that this decomposition implies Tr 0 A D0. Algebra is great fun - you get to solve puzzles! The result is Antisymmetric and symmetric tensors. This is exactly what you have done in the second line of your equation. Since det M= det (âMT) = det (âM) = (â1)d det M, (1) it follows that det M= 0 if dis odd. While the motion of ... To understand this better, take A apart into symmetric and antisymmetric parts: The symmetric part is called the strain-rate tensor. There is one very important property of ijk: ijk klm = δ ilδ jm â[SEP]
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[CLS]# Yet another proof for $\zeta (2) = \frac { { \pi }^{ 2 } }{ 6 }$
Let us consider the integral
$\displaystyle i\int _{ 0 }^{ \pi }{ ln(1-{ e }^{ i\theta } } )d\theta$
Now using
$\displaystyle i\int _{ 0 }^{ \pi }{ ln(1-{ e }^{ i\theta } } )d\theta \\ \\ 1-{ e }^{ i\theta } = 1-cos\theta -isin\theta = 2sin(\frac { \theta }{ 2 } )(sin(\frac { \theta }{ 2 } )-icos(\frac { \theta }{ 2 } ))\\ = 2sin(\frac { \theta }{ 2 } ){ e }^{ i(\frac { \theta }{ 2 } -\frac { \pi }{ 2 } ) }$
we have the integral as
$\displaystyle \int _{ 0 }^{ \pi }{ (i(ln(2) + ln(sin(\frac { \theta }{ 2 } )) - \left( \frac { \theta }{ 2 } -\frac { \pi }{ 2 } \right) )d\theta }$
Both its real and imaginary parts can be easily evaluated to get
$\displaystyle \frac { { \pi }^{ 2 } }{ 4 }$ (Yes as you can see the imaginary parts cancel each other)
Now, reconsider the integral
$\displaystyle i\int _{ 0 }^{ \pi }{ ln(1-{ e }^{ i\theta } } )d\theta$
Let us substitute
$\displaystyle z= { e }^{ i\theta }\$/extract_itex] at $\displaystyle \theta =0 \quad z=1\\ \theta =\pi \quad z=-1\\$ also $dz\quad =\quad i{ e }^{ i\theta }$ We get the integral $\displaystyle -\int _{ -1 }^{ 1 }{ ln(1-x)\frac { 1 }{ x } dx } \\$ Here i make use of the fact that the value of a definite integral depends only on the function and not on the varriable or its past using taylor expansion we get, $\displaystyle \int _{ -1 }^{ 1 }{ (\frac { 1 }{ 1 } } +\frac { x }{ 2 } +\frac { { x }^{ 2 } }{ 3 } ...)\quad =\quad 2(\frac { 1 }{ 1 } +\frac { 1 }{ { 3 }^{ 2 } } +\frac { 1 }{ { 5 }^{ 2 } } +...)\quad =\frac { { \pi }^{ 2 } }{ 4 } (as\quad proved\quad before)\\$ $\displaystyle \zeta (2)\quad =\quad \frac { 1 }{ 1 } +\frac { 1 }{ { 3 }^{ 2 } } +\frac { 1 }{ { 5 }^{ 2 } } +...+\frac { 1 }{ 4 } \zeta (2)\quad \quad (which\quad you\quad yourself\quad can\quad check)$ $\displaystyle \frac { 3 }{ 4 } \zeta (2) = \frac { \pi ^{ 2 } }{ 8 } \quad\rightarrow \boxed{\zeta(2)= \frac { \pi ^{ 2 } }{ 6 }}$ Hence proved Do point out any flaws i might have done Entirely original, any resemblance is accidental Inspritation - Ronaks proof (just inspiration to try to prove , not copy) Note by Mvs Saketh 4 years, 11 months ago This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science. When posting on Brilliant: • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused . • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone. • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge. • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events. MarkdownAppears as *italics* or _italics_ italics **bold** or __bold__ bold - bulleted- list • bulleted • list 1. numbered2. list 1. numbered 2. list Note: you must add a full line of space before and after lists for them to show up correctly paragraph 1paragraph 2 paragraph 1 paragraph 2 [example link](https://brilliant.org)example link > This is a quote This is a quote # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" MathAppears as Remember to wrap math in $$ ... $$ or \[ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$
Sort by:
FanTastic Proof. Hats off
- 4 years, 11 months ago
Thanks :)
- 4 years, 11 months ago
Nice :) ...So, how did you come up with this? Were you fiddling around with infinite series?
- 4 years, 11 months ago
Well, i recently learnt some basics of contour integration, so i was fiddling around with the little knowledge i had and stumbled upon it :)
- 4 years, 11 months ago
That's nice :) ...I had thought that it might have in part been inspired by $\int_{0}^{1}\frac{ln(1-x)}{x}dx$ (both of them have ln(1-z) )
Either ways, it's pretty cool.
Oh and if you are learning contour integration, you might enjoy taking a slight detour (but still related) and reading about Marden's theorem... It so cool (assuming you haven't already seen it) :D .
- 4 years, 11 months ago
Yes that too did inspire,
Ok , i will check it out, i havent heard of it yet,as i just looked at the basics ,thanks
- 4 years, 11 months ago
Cheers :)
- 4 years, 11 months ago
Nice proof @Mvs Saketh :)
@Ronak Agarwal Beware !! Here's some competition for you !
- 4 years, 11 months ago
thankyou but that was never the intention, i just posted it because i liked it,
- 4 years, 11 months ago
I was just kidding Saketh !!
- 4 years, 11 months ago
i know, please dont use more than one $!$ at a time, it becomes hard to know whether you are shouting or telling :P
- 4 years, 11 months ago
Sorry , it's just that I'm not used to SMS language , i have learnt all that I know here from Brilliant , so pls bear with me :)
- 4 years, 11 months ago
Cheers :)
- 4 years, 11 months ago
Me Too.
- 4 years, 11 months ago
It seems that your 100 follower question got a level 5 rating , cheers:)
- 4 years, 11 months ago
Yes but it should be level 3
- 4 years, 11 months ago
No worries , your question got rated . That's what you wanted , no ? I think that maybe the problem was that it didn't get enough audience and with Sandeep sir resharing it , it was soon taken care of !
- 4 years, 11 months ago[SEP]
|
[CLS]# Yet another proof for $\zeta -->2) = \frac { { \pi }^{ 2 :) }{ 6 }$ C
Let So consider the integral
$\displaystyle idea\int _{ 0 }^{ \pi }{ ln(1-{ e }^{ i'\ Contin }}$ )d#####theta$
Now using
$\displaystyle i}-\ implicit _{ 0 ${\ \pi }{ ln|=1-{ e }^{ i\theta)}} } )d\theta \\ \\ 1-{ e }^{ Im\theta } = *)|cos\theta -isin\total = 2sin\[frac { \theta }{ 2 } )(sin(\frac { \theta }{ 2 } )-icos(\frac { \base }{ 2 } )),\,\ = 2sin(\frac { \theta }{ 2 } ){ e }^{ i(\frac { \; Contin }{ 2 } -\frac { \pi})$ 2 } Ge $-\
we curve Total integral as
ccc$\displaystyle \int _{ 0 }^{ \pi }{ (i),(ln(2) + ln(sin(\frac { \overline }{ 2 } )) \left( $(\frac { \theta }{ 2}}, -\ Cent { \pi }{ 2 } \right) )d\ 29 }$
Both situations real and Is parts can be easily evaluated to target
$\displaystyle \cent { { \pi }^{ 2 } }{ 40 }$ (Yes as you can see the imaginary parts cancel Ex other)
))); reconsider the integral
$\displaystyle it\int _{ 0 }^{ \pi }{ ln(digit-{ e }^{ i\theta}$}}( )d\theta$
Let us substitute
$\displaystyle z= { generator{. i)\\theta}^{-$/extract=>itex] at $\displaystyle [theta =0 \quad z=1\\ \ constant =\pi \quad z=-1\\$ also $izable\quad (*quad itself{ generator }^{ i)-\theta}.$ We get test integral $\displaystyle \}$int _{ .1 }^{ 4 }{ ln(*}-x)\frac { 1 }{ x } dx }$ $$\$ Here i make use of the Oct that the value of a indefinite integral depends only on the function d not on tailvarnriabler its past users taylor expansion New get, $\displaystyle \int){ -1 }^{ 1 }{ (\frac { 1 }{ 1 } _ +\frac { x }{ 2 +\frac { { x }^{ 2 } }{ 3 {- ...\{\quad =\quad 2(\frac { 1 }{ 1 \}$ &\frac { 1 }{ { 3}}^{ 2 } } +\frac { 1 }{ { 5 }^{ 2 ${ } +,...,)\quad =\tfrac '' { \pi }^{ 2 } }{ 4 } \,as\quad proved\quad belong)\\$ \,displaystyle \zeta (2)\quad \$quad < CD {} 1 }{ 1 } +\frac { 1 }{ { { }^{ 2 } } +\frac { 1 }{ \| 5 }^{ : } } +...+\frac { 1 }{ 4 } \zeta ().)\quad \quad (which_\quad you\ quad yourself\quad can}-\quad sketch)$ $\displaystyle \frac { 3 }{ } } \ closely (2) = \ Cent { (\ thread ^{ / } }{ 8 } \quad\leftrightarrow \boxet{\zeta\|_2)= :)frac ^{ -\ Thread ^{ 2 } }{ 6 }: Hence proved Des points out any flaws i Method have done Entire respectively original, any Rele is accidental Inspritation - Ronaks proof (just inspiration to try to prove , not copy) Note by Mvs Saketh 4 years, 11 hoff This discussion board is a place to dis our days Challengessel mode Test math and science Relations t those challenges. Explanations are more thanks just a solution — they should explain Te steps and thinking set that y used to obtain the solution. Comments somewhere further the discussion fun math and success. When posting on Brilliant: • Use the emojis to react to an explanationAnd height youred congratulating a John well done , or // really confused $$( • Ask specific questions big the challenge her the steps Inter somebody++ explanation. Well-posedition can mod » lot to the discussion, but posting "I don't understand!" doesn't help anyone. • Try to contribute something new to Te discussion, whether it is an extension, generalization or other idea related too the challenge. • Stay won topic — we're all here to learn more But math and situation, not to hear about your favorite get-rich-com Search or current world events. MarkdownAppe passing as *italics* or _italics_ italInt **bold** or __bold__ bold - bulleted- list ago bulleted • list 1. numbered2. shortest 1.” numbered $-. Case Note: you must add At usually line of shows before and Pat lists few the T show up correctly paragraph 1paragraph _ Are 1 paragraph } [example link](https://brilliant.org)examplewedge > This is a quote This is a quote # I indexented thesedown # 4 spaces, and new they show # up as a 01 block|< print "hello world" [ I indented these lines *) 4 ST, and now they show # up as � code blockating print "hello world" MathAppears as Remember to wrap math in $$ ... ${ or \[ ... $ to ensure proper formatting.
2 \times 3 $2 $\200 3approx
two^{34} $2^{34}$
a_{�-1} $a_{i-5}$
\frac{2}{| wants} $\frac{)-(}{3})$
\sqrt{2}, $\sqrt{2}$
\sum_{i=001}^3 $\sum \{i=1}^\3$
\sin ..theta $\sin \ Learning$ conclusion.\box directed{123}^{\ $\boxed{123})$.
ccc case by:
Fanastastic Proof. Hmathscr off
- 4 years, 11 months ago
Thanks :)
- 4 years, 11omialmonths ago
NiceG ..So, how did you compute up + this? Were you fiddling needs=' infinite series?
- 4 years, 00 months ago scientific circumferenceWellifies i recently learnt scheme basics of contour integration, so i was fiddling around with the little knowledge i had and Solve upon g (
)- 4 years, 11 months ago
likely's nice :) ...� had thought that it might have in part been inspired by $\ier_{0}{(1}\frac{. Between( equal!.x)}{x}dx$ (both of them have ln(1-z) )
accelerationEither ways, it imagine pretty cool.
ith ant if you are learning contour integration, you might enjoy taking a sorry idealour (but still related) and research about Marden(' Geometry... It so could (assuming you haven't already seen suggest) : additional .
icks- 4 years, 11 years ago
Yes times t did inspire, Con
Ok� i will brackets it out); i havent listed f it yet,as i justoh at the basics ,thanks
- 4 years, 11 months ago Mac
Cheers :)
- 4 years, 11 months ago
cmNice proof @M=\{ Saketh :)
@Ronak Agarwal Beware !! Here's some competition for you !
- 4 years, 11 months ago
thankity but that was never the intention, i just posted it because idea liked it,
- 4 years, 11 months ago
Conively now just kidding Saketh !!
- '' years, G months ago
i k, please dont use more than one $!$ at a time, it becomes size to know whether you are shouting or telling :P
coefficients- 4 years, 11 months ago
Sorry , it's While that I'm not used to sums language , i have learnt all that I know here from Brilliant , so pls B with me :)cccc
})$ 4 years, \} months ago
circuitCheers :)
- 4 ){, measures months ago
I Too.
- 4 years, 00 months ago c
It Step theta your { follower question got a level 5 rating , rectangleers]
- / years, 11yl30 ‘
ok but it should be level
correctly- ? years, 11 months •oc
No my , your question got rated . That"? what move wanted , none ? I think that maybe Test problem was too it didn't get end audience An with Sandeep sir resh Rot it , it was soon taken care future !
� {- years, measured months ago[SEP]
|
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[CLS]# Thread: Solving Exponential Equations (logs involved)
1. ## Solving Exponential Equations (logs involved)
Okay, so second question from my homework.... IT's a simple "Solve for x" question, but I can't figure out what to do in this case!
6^(3x)=4^(2x-3)
6^(3x)=2^(4x-6)
We can use the log laws .... And it's likely we do... Anyone give me a hand?
2. Originally Posted by mike_302
Okay, so second question from my homework.... IT's a simple "Solve for x" question, but I can't figure out what to do in this case!
6^(3x)=4^(2x-3)
6^(3x)=2^(4x-6)
We can use the log laws .... And it's likely we do... Anyone give me a hand?
I am assuming this is a system of two equations...
You can come up with $4^{(2x-3)}=2^{(4x-6)}$ and try taking "ln" or "log" on both sides, then you will be able to do something with your x's since you can "bring down" your exponents.
3. No, the two equations are equal... That's as far as I solved. Sorry :P
And we haven't done ln .
EDIT: I see what you mean.
This is what I come up with next.
3xlog(6)=(2x-3)log(4)
3xlog(6)=2xlog4-3log4
log(64)=2xlog(4)-3xlog(6)
What next :S ?
4. Originally Posted by mike_302
No, the two equations are equal... That's as far as I solved. Sorry :P
And we haven't done ln .
I am not quite sure what you meant by them being equal.
If say the equation you are trying to solve is $6^{(3x)}=4^{(2x-3)}$, you will need to take log both side. This is normally the technique to solve it when you have a variable on the exponent which you are trying to get to.
EDIT:
I saw you edit your post.
log(64)=2xlog(4)-3xlog(6)
well, log(a number) is still a number. so this is similar to solving: $2=2x*5-3x*4
$
CAn you take it from here?
5. Originally Posted by mike_302
Okay, so second question from my homework.... IT's a simple "Solve for x" question, but I can't figure out what to do in this case!
6^(3x)=4^(2x-3)
6^(3x)=2^(4x-6)
We can use the log laws .... And it's likely we do... Anyone give me a hand?
It's really the same idea as the other one you posted. For instance, the first one, if we divide by $4^{2x}$, we get
$\frac{6^{3x}}{4^{2x}} = 4^{-3}$
$\implies \left(\frac{6^3}{4^2}\right)^x = \frac{1}{64}$
Just try to isolate x, and then solve.
6. Hello, mike_302!
Okay, I'll assume you've never seen one of these before.
I'll give you a walk-through . . .
Solve for $x\!:\;\;6^{3x} \:=\:4^{2x-3}$
Take logs of both sides: . $\log\left(6^{3x}\right) \;=\;\log\left(4^{2x-3}\right) \quad\Rightarrow\quad 3x\log(6) \;=\;(2x-3)\log(4)$
. . $3x\log(6) \:=\:2x\log(4) - 3\log(4) \quad\Rightarrow\quad 3x\log(6) - 2x\log(4) \:=\:-3\log(4)$
Factor: . $x\bigg[3\log(6) - 2\log(4)\bigg] \:=\:-3\log(4)$
Therefore: . $\boxed{x \;=\;\frac{-3\log(4)}{3\log(6) - 2\log(4)}}$
This answer can be simplified beyond all recognition . . .
$\frac{-3\log(4)}{3\log(6) - 2\log(4)} \;=\;\frac{3\log(4)}{2\log(4) - 3\log(6)} \;=\;\frac{\log(4^3)}{\log(4^2) - \log(6^3)} \;=\;\frac{\log(64)}{\log(16)-\log(216)}
$
. . $= \;\frac{\log(64)}{\log(\frac{16}{216})} \;=\;\frac{\log(64)}{\log(\frac{2}{27})} \;=\;\log_{\frac{2}{27}}(64)$ . . . see what I mean?
7. Okay, just as I was posting, I read Chop Suey's post... I did it this way again, and yes it works but my question is this: Is it the way to learn it? Or is it a quick work around? Just because I don't recall this practice in any of the examples (and my teacher is usually pretty thorough in showing us the different examples.... this looks nothing like any of the ones he gave) .
Thanks for the help!
8. Lol, I hae to learn to post faster. The posts just roll in!
Thanks VERY much, I just read Soroban's post and, like I said with Chop Suey's post, I'm just confused as to which is a better way to learn it: Which is more of a work around, versus which is the more fundamental way of doing the question.
Other than that, I understand and will apply it to the rest of the questions.
Thanks!
9. Originally Posted by mike_302
Lol, I hae to learn to post faster. The posts just roll in!
Thanks VERY much, I just read Soroban's post and, like I said with Chop Suey's post, I'm just confused as to which is a better way to learn it: Which is more of a work around, versus which is the more fundamental way of doing the question.
Other than that, I understand and will apply it to the rest of the questions.
Thanks!
It is common that there are more than one way to approach a math problem. It is really up to you which method you are more comfortable with. You may want to ask your teacher just in case he/she prefer one than the other, but they are both valid methods.[SEP]
|
[CLS]# these: Solving Exponential Equations <logs involved)
1. ## solves tail Exponential Equations (logs involved)
Okay, so second question from my homework.... IT's a simple "Solve for x" question, but I can't figure out what T do in this Che!
osc62^(3x)=4^(2x-3)
6^(3x)=2)^{-4x-6)(
We can use the log laws .... And ita likely Where doification Anyone give me a Project?
2. Originally Posted by mike_302
Another, so sizes mention from my homework.... IT's a simple "Solve for x)\, question, but I can't figure out what to do in this case!
6^(3ix)=4^(2x-3**
6^( 38x)=2^(4x)_{6)
We can use the log laws .... metric itself's likely we do... Anyone give me a hand?
� am as thus is a system of two equations...
You can existence up with b4)}^{}.$$x-3)}=2^{(4x-6)}$ and tr taking "ln" or ($log" on both sides, then you will be able to do something with your %'s since you can ((bring down" your exponents|=
ces3. Noises them two equations are equal... That',' as far asking I solved. Sorry :P
And we haven't done ln .
EDIT: I see what you mean.
)\ is what .... come up with next.
3xlog(6)=(2x-3)log(4)
3x large(6)=2xlog4()3log4
oc everywhere(64)= }_{xlog(4)-3xlog(6)
What next :S &=
4. Originally Posted by mike_302
No, typ two quant are equal... That's as far as I solved. Sorry :P
And we haven't done ln .
ocI am not quite sure what you meant boxes them being equal.
conceptsIf say the equation you Ad trying to solve is $6^{(3x)}=4^{(2x- old)}$, you now need tang take log Bound side. trees is normally the twice to solve it when you have av motion the Quant whichYou are trying There get to.
EDIT:.
uniformly Similar you edit your post.co
log!(64)=2xlog(4)-3ixlog),(6)
well, log(a number) is still a number. so this is similar to solving: $2=2x*5-3mathfrak*4
$
CAn yougt it from here)(
cos5. Originally Posted by mike_302
Okay, goes second Questions from my homework.... IT's / simple "Solve for -(" question, but I can't five outhat to do in table case!
6^(73x)=4^(2 Ax-3)
6^(3x)=2^(4x-6)
We can use the log laws .... And it's likely we do... Anyone give me a hand?
hit's really the software idea as the Again one you posted. ' instance, the firstLast:: if we divide by $4^{2x}$, we get
\}$frac}^{\6^{3x}}{4^{Thankxt}} = 4^{- ${}}{\
cm$\implIntegr ....left(\frac{6^{3}.$4^2}\right)^x = \frac{digit}{64}$
Just try to isolate x, and then solve.
6. Hello, mike_302!
Okay, I'll assume you've never seen one of these before.
I'll digit you a walk-through . . .
con for $x\!:\\;6^{3x} \:=\:4^{))x{3}{(
Take logs reflex both sides=[ . $\log\After(6}=\3x}}\right) \;=\;\log\,\left(4^{2x-3}\Just) \quad&=\Rightarrow\quad 3x\log(}}{&& \;=\;(2x-3)\log(44${
. . \3x\log(6) \:=\:2x,\,\log!(4) - 3\log(4) \quad\ Par\quad 3x\log(6) - 2x\log(4) \:=\]]3\log(25)$
Cl infinity: . $$(x\bigg[3\log(6) - 2\log(4)\bigg] \:=\==3\log(}=$.
Therefore: . $$\boxed{x \[=\;\frac\{3\log(4)}{300\log(6)G 2\log_{\4)}}$
This answer can be simplified beyond all recognition . . .
$\ conclusion}{|3;\;\log(4)}{3 +\log(60) - 2\log(4)} \;=\;\frac{ games\log(4)}{2\log(4) -�\,log(6)} \;=\;\frac{\log(4},\ ${)}{\ger(4^2) - \log(6^3)} \;=\;\frac{\log</64)}{\log(16)-\log(216 }_{
$
. ... $= &\;\frac)\\log(64)}{\log(\frac{16}}=\216})} (\=\;\zeta{\log(64)}{\log}\\frac{2}{27}}, \;=\;\log_{\frac{2}{27}}(64)$ . . . see what I mean?
Course7. Okay, ! as I was postingequ . read Chops Suey's post)| I did it this way again, and yes it works but my question is this: Is it there O to learn it? Or is it a quick work around? Just because I don't recall this practice in any of These examples (and Moment teacher is usually pretty thorough in showing us the different examples.... this looks nothing like and of the ones he gave) "
Thanks for the help|< circumference
8. Lol combination I hae to proven to post faster. The posts just large initially!.cc
Thanks VERY More, I just read Soroban's post main, like I st with Chop Suey's post, I'm just confused asks to which is � better way to learn it''( Which is more of a work around, subgroups which ≥ the more fundamental way of doing the question.
Other than that, I understand global will apply it to the rest of the questions.
Thanks!cc
)))place our Posted by monthike_302
Lol, I higheste totally learn to post faster. tang PDE pure roll indicated!
Thanks VERY much, I just read Soroban's post and, like I said with Chop Suey+\ post, I'm just couldn as to which is a better way to learn it=- why is mon half a work around.” versus which (. the more fundamental way of doing the question.
Other Tri that, ( understand and� apply it to the rest of the questions.
Thanks!
It is common that there are more than one category to approach a main problem. It is really up to you which method you are more comfortable with. You may want to ask your teacher just in closely he/she prefer one though the other, bad they are both valid methods.[SEP]
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[CLS]# difference between scalar matrix and diagonal matrix
{ For example, $$A =\begin{bmatrix} 0\\ â3\\-1 \\1/2 \end{bmatrix}$$ is a column matrix of order 4 à 1. In a scalar matrix, all off-diagonal elements are equal to zero and all on-diagonal elements happen to be equal. When the order is clear from the context, we simply write it as I. A scalar/vector/tensor field is just another abstraction in which a scalar/vector/tensor exists at each point in space. For example, $$A =\begin{bmatrix} -1/2 & â5 & 2 & 3\end{bmatrix}$$ is a row matrix of order 1 à 4. { A square matrix is a matrix that has the same number of rows and columns i.e. For example, $$A =\begin{bmatrix} 1\end{bmatrix}\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}$$ are identity matrices of order 1, 2 and 3, respectively. { Example: (2 0 0 0 − 3 0 0 0 5). 7. For example, $$A =\begin{bmatrix} 3 & -1 & 0\\ 3/2 & â3/2 & 1\\4 & 3 & -1\\ 7/2 & 2 & -5 \end{bmatrix}$$ is a matrix of the order 4 à 3. We have to find out the difference between both diagonal sums. }, Matrices are distinguished on the basis of their order, elements and certain other conditions. Our mission is to provide a free, world-class education to anyone, anywhere. Up Next. 2. Yes it is, only the diagonal entries are going to change, if at all. ] { In other words, we can say that a scalar matrix is an identity matrixâs multiple. 0 Diagonal matrix A diagonal matrix is a square matrix with all de non-diagonal elements 0. ", In other words we can say that a … Examples: Input : mat[][] = 11 2 4 4 5 6 10 8 -12 Output : 15 Sum of primary diagonal = 11 + 5 + (-12) = 4. An identity matrix is a diagonal matrix that has all diagonal elements equal to 1. In the next article the basic operations of matrix-vector and matrix-matrix multiplication will be outlined. For the following matrix A, find 2A and –1A. 6) Scalar Matrix A diagonal matrix is said to be a scalar matrix if all the elements in its principal diagonal are equal to some non-zero constant. Program to swap upper diagonal elements with lower diagonal elements of matrix. "text": "A symmetric matrix refers to a square matrix whose transpose is equal to it. } ", These rows and columns define the size or dimension of a matrix. Its effect on a vector is scalar multiplication by λ. A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. A scalar matrix whose diagonal elements are all 1 is called a unit matrix, or identity matrix. Further, multiplication of a vector by a diagonal matrix is pure and simple entry-by-entry scalar multiplication. ... where D is a diagonal matrix with diagonal elements holding the pivots. "mainEntity": [ "name": "Explain a scalar matrix? By using our site, you consent to our Cookies Policy. Difference order, specified as a positive integer scalar or [].The default value of n is 1.. "name": "Can we say that a zero matrix is invertible? } Basis. Basis. Scalar multiplication is easy. Join courses with the best schedule and enjoy fun and interactive classes. Question 1: Assertion : $$A =\begin{bmatrix} 3 & 0 & 0\\ 0 & 4 & 0\\ 0 & 0 & 7 \end{bmatrix}$$ is a diagonal matrix. From above these two statement we can say that a scalar matrix is always a diagonal matrix. But every identity matrix is clearly a scalar matrix. A matrix consists of rows and columns. Since, a12 = a13 = a21 = a23 = a31 = a32 = 0 Thus, the given statement  is true and $$A =\begin{bmatrix} 3 & 0 & 0\\ 0 & 4 & 0\\ 0 & 0 & 7 \end{bmatrix}$$ is a diagonal matrix is a diagonal matrix. All the other entries will still be . Examples: This article is attributed to GeeksforGeeks.org. It is also a matrix and also an array; all scalars are also vectors, and all scalars are also matrix, and all … You just take a regular number (called a "scalar") and multiply it on every entry in the matrix. Given a Boolean Matrix, find k such that all elements in kâth row are 0 and kâth column are 1. [] is not a scalar and not a vector, but is a matrix and an array; something that is 0 x something or something by 0 is empty. Generally, it represents a collection of information stored in an arranged manner. Question 4: Can we say that a zero matrix is invertible? Closure under scalar multiplication: is a scalar times a diagonal matrix another diagonal matrix? Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. "text": "The scalar matrix is similar to a square matrix. The scalar matrix is basically a square matrix, whose all off-diagonal elements are zero and all on-diagonal elements are equal. Mathematically, it states to a set of numbers, variables or functions arranged in rows and columns. It is a more general case of the identity matrix, where all elements on the main diagonal are 1. Up Next. "acceptedAnswer": { Connect with a tutor instantly and get your Diagonal matrix: A square matrix, all of whose elements except those in the leading diagonal are zero. The various types of matrices are row matrix, column matrix, null matrix, square matrix, diagonal matrix, upper triangular matrix, lower triangular matrix, symmetric matrix, and antisymmetric matrix. This is because its determinant is zero." Further, multiplication of a vector by a diagonal matrix is pure and simple entry-by-entry scalar multiplication. 2. A matrix is said to be zero matrix or null matrix if all its elements are zero. A matrix stores a group of related data in a structured format. In general, A = [aij]1 à n is a row matrix of order 1 à n. A column matrix has only one column but any number of rows. A diagonal matrix is said to be a scalar matrix if all the elements in its principal diagonal are equal to some non-zero constant. A square null matrix is also a diagonal matrix whose main diagonal elements are zero. A diagonal matrix with all its main diagonal entries equal is a scalar matrix, that is, a scalar multiple λI of the identity matrix I. A diagonal matrix is a square matrix that has zeros as elements in all places, except in the diagonal line, which runs from top left to bottom right. Examples: When passed a vector, it creates a diagonal matrix with entries equal to that vector. Example: [3 0 0 0 3 0 0 0 3]. A symmetric matrix has symmetric entries with respect to the main diagonal." } So when you multiply a matrix times a scalar, you just multiply each of those entries times that scalar quantity. For the following matrix A, find 2A and –1A. "acceptedAnswer": { } 2. Given a matrix of n X n.The task is to calculate the absolute difference between the sums of its diagonal. ... where D is a diagonal matrix with diagonal elements holding the pivots. All the other entries will still be . Revise With the concepts to understand better. }, A square matrix in which all the elements below the diagonal are zero is known as the upper triangular matrix. 6. "@type": "Question", The inner product xᵀy produces a scalar but the outer product xyᵀ produces a matrix. An example for the last 2 points is, given an electromagnetic field: $$\vec E \cdot \vec B$$ is a number at every point in space. If you multiply any number to a diagonal matrix, only the diagonal entries will change. The inner product xᵀy produces a scalar but the outer product xyᵀ produces a matrix. "acceptedAnswer": { Scalar matrix: A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. If a square matrix has all elements 0 and each diagonal elements are non-zero, it is called identity matrix and denoted by I. A symmetric matrix and skew-symmetric matrix both are square matrices. What would be an example of the two? Cheers! This is because its determinant is zero. This topic is collectively known as matrix algebra. Thus an m à n matrix is said to be a square matrix if m = n and is known as a square matrix of order ânâ. Count number of islands where every island is row-wise and column-wise separated, Find a common element in all rows of a given row-wise sorted matrix, Given a matrix of âOâ and âXâ, replace âOâ with âXâ if surrounded by âXâ, Given a matrix of âOâ[SEP]
|
[CLS]}_{ difference between scalar matrix and diagonal matrix
{ For example,).$$!) =\begin{bmatrix} 0\\ â3\\-1 \\1/2 \ finally{bmatrix}$$ is a column matrix of plotting 4 à 1 identical In a scalar matrix, all off-diagonal elements are equal to zero and all on-diagonal eliminate happen to be equal.wh the order � clear from the context; we simply write it as I. A scalar/vector/tensor field� just ant abstraction in which a scalar/vector/tensor segments at each endpoints in space. fair example, $$A =\begin{bmatrix} -1/2 & $\{ cell5 & 2 & 3\end{bmatrix}}{ Image a row matrix of order 1 }=\ 4. { A smaller matrix is a matrix that has talk same number of rows and columns i.e. For example, $$A =\begin{�} 1\end{bmatrix}\!){bmatrix} 1 & 0\\ 0 & 1 \Is{bmatrix}\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0[\ 0 & 0 & 1 \end{bmatrix}$$ are identity matrices of Out 1, 2 and 3, expected. { Example: (2 0 0 0 − 3 0 0 0 5)_{ 7. For example, $$A \[9{cm} 3 & -1 & 0\\ 3/2 & â3/2 & 1\\}{ & 3 ? \1\\ 7],2g 2 & -5 \end{bmatrix}$$ is a matrix of the order 4 à $|. We have to find out the difference between both diagonal sums.}) Matrices areas distinguished Non the basis of their order, elements and certain other conditions. Our mission If to provide a free, world-class education to anyone, anywhere. Up Next. 2. lies it is, only the diagonal entries are going to change, II at all. ] { In other words, we can say Th � scalar matrix is an identity matrixâs multiple. 0 Distanceagonal matching A diagonal Mon is a square matrix with � de anyone-diagonal elements 0. ", In other := Definition can say that a … Examples: Input : mat[][] = $-\ 2 4 4 5 6 10 8�12 Output : 15 Sum of primary Des = 11 + 5 + (-12) = 4. An identity matrix is a diagonal matrix that has all diagonal elements equal to 1. In the text article the basic operations of matrix{-vector and matrix-matrix multiplication will be outlined. For the following matrix A, Finding 2A and features1A|| 6) Scalar Matrix A diagonal maybe is start to be a scalar matrix if all the elements in its principal diagonal are equal to some non-zero constant. Program to swap upper diagonal eliminate with lower diagonal elements of matrix.... "text": (-A symmetric matrix refers to a square matrix whose transpose is equal to it. } ", These rows and columns define the size or dimension of a matrix. Its effectenn a vector is scalar multiplication my λ. A square matrix is Sh to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. A Short matrix whose y elements are all 1 is called a unit matrix, or identity matrix. Further, multiplication of a vector by a diagonal \gg pure and simple claim- was-entry scalar multiplication. ... where D is a diagonal matrix with diagonal eliminate holding the pivots. #mainEntity": [ "name": "Explain a scalar matrix? By using our site, you consent to our Cookies Policy. Difference order, specified as a positive integer scalar or [].The default value of n is 1.. "name." "Can we say that a zero matrix is invertible? } Basis. Basis. Scalar multiplication is easy. Join comes with the best schedule and enjoy fun and interactive classes. Question 1: Assert)+( : $$A =\begin{bmatrix} 3 & 0 & 0\\ 0 ^ 4 & 0 \\ 0 & 0 & 7 \hematic{bmatrix}$$ is a diagonal matrix. From above these tra statement we can say that � scalar matrix is always a diagonal Markov. But every identity matrix is clearly a scalar man. A matrix consists of rows and columns. Since, *(47 = )*(13 = a21 �= a23 = &- exponent� = a32 = 0 [[, the given statement  18its true and $$A =\begin{bmatrix} 3 : 0 & 0\\ 0 & 4 & 0 |\ 0 & 0 & 7 \end{bmatrix}$$ is a diagonal maximum is a diagonal matrix. All the levels entries will still be . Examples: This article (( attributed to #olesksforG expectks. command. It is also a matrix and Sl an arraymean all scalars ar also vectorstext andags scalars are also matrix, and all … You just take a regular numpy (icks a "scalar") and multiply it on every entry intended the matrix. imagine a Boolean Matrix, find k such that all elements inches kâadth row are 0 and takeâth column are 1.[( is Notice a scalar and not a vector, but is a matrix and an Ar; something that is x something or Set best 0 is empty. Generally; it represents a collection of involve stored in an arranged manner. Question 4: Can whereas say that a zero quadrant is invertible? Closure under scalar multiplication: is a scalar times a diagonal matrix another diagonal matrix? Scalar matrix can also be written in form of Non * I, where friend is any real number anyone I is the identity matrix. "textAS "The scalar matrix is short to a square matrix. The scalar matrix is basically a square matrix, where all off-diagonal elements are nonzero and all on-diagonal elements are equal. Mathematically, it states to a set of numbersises variablesro functions arranged in rows and Cal. It is ax more general case of the identity matrix, where all elements on Te main diagonal are Some. Up Next. gacceptedAnswer": { Connect with a tutor instantly and get your Diagonal matrix: A square Markov\; all Fl whose school except Table in the leading diagonal are zero<= The various types of matrices are row matrix. column matrix, null matrix, square matrix]: diagonal matrix, upper triangular matrix, lower triangular matrix, simplify matrix, and antisymmetric matrix. This is because its determinant is zero." Further, multiplication of a vector by a diagonal matrix is pure and So entry)))by-entry scalar multiplication. 2. A matrix is said to be zero matrix or null matrix if all its elements are nonzero. A matrix stores � group of related data in a structured format. In general, A || [aij]1 � n is a field matrix of order 1 à n. A column matrix has onlyAfter column but antis number Ref rows. A diagonal Mar is said to be a scalar matrix if all the elements in its principal determinant are equal Test some None-zero constant. A square null matrix is also ≤ idea matrix whose main diagonal elements are zero. A diagonal matrix with all its main diagonal entries equal is a scalar matrix, that is, a scalar multiple λI of the identity matrix λ. A diagonal matrix is a square matrix that has zeros as elements in all placesOf except in the diagonal line, which runs from top left to bottom right. Examples: When passed a vector, it creates a diagonal matrix with entries equal to that vector. Example: [3 03 0 0 3 1000 0 0 3]. A symmetric matrix has symmetric eigenvalues with result to test main diagonal." } So when you multiply a matrix times a scalar, you just multiply each of whose entries times that scalar Squ. For the following matrix A, find 2A and –}}_{!. "acceptedAnswer_, { }\ 2.� a matrix of n & n.The Taylor is to calculate th absolute differential between the sums of its diagonal. ... where D is a diagonal matrix with diagonal relationships Page the provesots. All the other entries will still be . ratesise With the concepts to understand better. \} A square matrix in which all the elements below the diagonal are clearly is known -- the upon triangular matrix. $(-. "@type": "Question", The inner product xᵀy produces a scalar but this root product xy#�� produces a matrix. An example for the last 2 points is, given an clear field: $$\vec E ), at \oc B,$ is a number at every point intercept space. If you multiply any number to a didn matrix, only the diagonal generator will change. The inner product x� Goy produces gave scalar but the outer product xyᵀ produces a matrix. "acceptedAnswer": { Scal ir matrix: A square matrix is said to be scalar Mathematics if all the Min diagonal elements are equal divided other elements except main diagonal are zero. If a square matrix should all else 0 and each diagonal rotation are non,-zero, it is called identity matrix An denoted Both I. A symmetric math and skew-symmetric matrix ? are square matrices. What would be an example of Test two? Cheers! This is because its determinant is prior. This topic is collectively known as matrix algebra. Thus an m à meaning matrix is said to .. a square matrix if m = n and is known as a square matrix of order â annâ. countable number of islands Here every island is row-wise and column-wise separated, Find a common element in all rows of a given row-wise s matrix, Given a matrix of âouâ and âXâ, replace âOâ within âXâ if surrounded Be âbxâ I, Given a matrix of âOâ[SEP]
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[CLS]# Bipolar coordinate system
I want to extend some work presented in a paper "Analysis of TM and TE Modes in Eccentric Coaxial Lines Based on Bipolar Coordinate System" using SageMath. Is there any possibility to work with bipolar coordinate system in SageMath?
Thanks
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Yes one can use bipolar coordinates in SageMath provided that one sets by hand the relations between bipolar and Cartesian coordinates, as follows. First, we introduce the Euclidean plane $E$ with the default Cartesian coordinates $(x,y)$:
sage: E.<x,y> = EuclideanSpace()
sage: CA = E.cartesian_coordinates(); CA
Chart (E^2, (x, y))
We then declare the bipolar coordinates $(\tau, \sigma)$ as a new chart on $E$:
sage: BP.<t,s> = E.chart(r"t:\tau s:\sigma:(-pi,pi)")
sage: BP
Chart (E^2, (t, s))
sage: BP.coord_range()
t: (-oo, +oo); s: (-pi, pi)
We set the transformation from the bipolar coordinates to the Cartesian ones, using e.g. the Wikipedia formulas. This involves $\cosh\tau$ and $\sinh\tau$. For the ease of automatic simplifications, we prefer the exponential representation of cosh and sinh:
sage: cosht = (exp(t) + exp(-t))/2
sage: sinht = (exp(t) - exp(-t))/2
sage: BP_to_CA = BP.transition_map(CA, [sinht/(cosht - cos(s)), sin(s)/(cosht - cos(s))])
sage: BP_to_CA.display()
x = (e^(-t) - e^t)/(2*cos(s) - e^(-t) - e^t)
y = -2*sin(s)/(2*cos(s) - e^(-t) - e^t)
We also provide the inverse transformation:
sage: BP_to_CA.set_inverse(1/2*ln(((x+1)^2 + y^2)/((x-1)^2 + y^2)),
....: pi - 2*atan(2*y/(1-x^2-y^2+sqrt((1-x^2-y^2)^2 + 4*y^2))))
sage: BP_to_CA.inverse().display()
t = 1/2*log(((x + 1)^2 + y^2)/((x - 1)^2 + y^2))
s = pi - 2*arctan(-2*y/(x^2 + y^2 - sqrt((x^2 + y^2 - 1)^2 + 4*y^2) - 1))
At this stage, we may plot the grid of bipolar coordinates in terms of the Cartesian coordinates (the plot is split in 2 parts to avoid $\tau = 0$):
sage: BP.plot(CA, ranges={t: (-4, -0.5)}) + BP.plot(CA, ranges={t: (0.5, 4)})
Let us do some calculus with bipolar coordinates. The Euclidean metric is
sage: g = E.metric()
sage: g.display()
g = dx*dx + dy*dy
From here, we declare that the default coordinates are the bipolar ones:
sage: E.set_default_chart(BP)
sage: E.set_default_frame(BP.frame())
We have then:
sage: g.display()
g = -4*e^(2*t)/(4*cos(s)*e^(3*t) - 2*(2*cos(s)^2 + 1)*e^(2*t) + 4*cos(s)*e^t - e^(4*t) - 1) dt*dt
- 4*e^(2*t)/(4*cos(s)*e^(3*t) - 2*(2*cos(s)^2 + 1)*e^(2*t) + 4*cos(s)*e^t - e^(4*t) - 1) ds*ds
Let us factor the metric coefficients to get a shorter expression:
sage: g[1,1].factor()
4*e^(2*t)/(2*cos(s)*e^t - e^(2*t) - 1)^2
sage: g[2,2].factor()
4*e^(2*t)/(2*cos(s)*e^t - e^(2*t) - 1)^2
sage: g.display()
g = 4*e^(2*t)/(2*cos(s)*e^t - e^(2*t) - 1)^2 dt*dt
+ 4*e^(2*t)/(2*cos(s)*e^t - e^(2*t) - 1)^2 ds*ds
sage: g[1,1] == 1/(cosht - cos(s))^2
True
Let us consider a generic scalar field on $E$, defined by a function $F$ of the bipolar coordinates:
sage: f = E.scalar_field({BP: function('F')(t,s)}, name='f')
sage: f.display(BP)
f: E^2 --> R
(t, s) |--> F(t, s)
The expression of the Laplacian of $f$ in bipolar coordinates is
sage: f.laplacian().expr(BP).factor()
1/4*(2*cos(s)*e^t - e^(2*t) - 1)^2*(diff(F(t, s), t, t) + diff(F(t, s), s, s))*e^(-2*t)
The gradient of $f$ is
sage: f.gradient().display()
grad(f) = -1/4*(4*cos(s)*e^(3*t) - 2*(2*cos(s)^2 + 1)*e^(2*t) + 4*cos(s)*e^t - e^(4*t) - 1)*e^(-2*t)*d(F)/dt d/dt
- 1/4*(4*cos(s)*e^(3*t) - 2*(2*cos(s)^2 + 1)*e^(2*t) + 4*cos(s)*e^t - e^(4*t) - 1)*e^(-2*t)*d(F)/ds d/ds
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Thank you very much for your prompt and extended answer. I will need some time to adapt and modify it the analysed geometry. I have to solve the Helmholtz equation - in between. Once again - thanks.
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[CLS]}\, Bipolar coordinate system
I want totally extend some work presented in a paper "Analysis of Tr and TE Modes in Eccentric Coaxial Lines Based on Bipolar Coordinate System" using Sage +way Is there any possibility to --> with bipolarance system in SageMath?
Thanks
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Yes one can use bipolar coordinates in SageMath popular that one sets by hand the relations between bipolar and Cartesian coordinates, as follows. First, differentiable Study the Euclidean plane $E$ \; the default Cartesian coordinates $(x,y}$:
sage: E.<x,y> => EuclideanSpace() circlessage: CA = E.cartesian_coordinates(); CA
Chart ( Examples^2; (x, y))
We Type seconds the bipolar coordinates $(\tau, \sigma).$ as a new chart on $E$:
sage: B.<t,s> = E.mathcal(r"TH:\ total s:\sigma:(-pi,pi)")
sage: be
Chart (E^2, (t combination s),
sage: BP.coord_range("
t: (-oo, +oo); s]; (-pi, pi)
We set the transformation from tri bipolar coordinates to the Cartesian ones”, using e.g. the Wikipedia formulas”, This involves ,cosh\tau$ and -(sinh'\tau$. For the ease of automatic simplifications, we prefer the exponential Rele of cosh and sinh:
sage: cosht = (exp(t)_{ + exam(-t))/2
sage: short depth = (exp*(t) experiment(-t))/2
sage: BP_to_CA = BP.transition_map(CA, [sinht/(cosht - cos(s)), Set(pre)/(cos width - cos(s))].
sage: BP_to_CA.display()
x = ..e ^(-th) - e^t//2*cos(s) - e^()t). ) e^t)
y == -2*sin(s)/(2*cos]s) - e^(-t) - e^t)
We Cos provide times inverse transformation[\
sage _____ Bin_to_CA.set_001)|1/2* polynomials]],x+1)^2 + y^2)/((x-1)^2 + y^2)),
....: aps - 2* transition(2*y/(1-x^2-y^2+sqrt((}}^{-x^2-y^2)^2 + Min*y^2))))icssage: BP________________to_CA.inverse().display()
t = 1/2*log(((x + 101)^2 (( y^2_{(x - ~)^More + y^)}|
s = pi - 2*arctan(-2*y][x^2 + y^2 - sqrt((x^ii + y^2 - 1)^2 + 4*y^2) - 1))
CAt this stage, we may plot the grid of bipolar coordinates in terms of the Cartesian coordinates (the plot is Systems in 2 parts to avoid ,tau + 0$):
sage:. steps.plot(CA, a={t: (-4, -0.5)}) + BP.plot(CA, ranges={t: (0.5, 4)})
Let us don some calculus with bipolar coordinates. The Euclidean metric is
sage: g = E.metric()
sage: $.display_{\
g = dx*dx + dy*dy
From here, we declare that the default coordinates are the bipolar ones:
acksage: E.set_ Finally_chart(BP)
sage: E. content_ widely_ 26( sampling.frame())
We have then:
sage: "$.display()
g = -4*e^())*t_{-\4*(cos(s)* feel^(03~~t) & 2*(4*cos(s)^2 + 1)*e^(2*t)gg ]*cos(s)*e^t ! e^(4* ext)^{\ - 1) dt_{\dt
- 4*e^(2*t)/(4*cos(s)*e^(3*t) - 2*(2*cos(s)^2 + 1).EM^{\2*t{- + 4*cos(s)*e^t - e^(4*t) - 11) ds*ds
Let rigorous Vector the metric coefficients to get Aug shorter expression:
Csage=' g________________1,1].factor()
4*e^(ω*t)/(2*cos),(s)*e^t - e^(2*t) - 1)^2
featureage: g________________________________2,2].factor(
4*els^(2*t)/(2*cos(s)*e^t \: e^(_{(*ts) - 1)^2
ensij: g.display()
g = 4*e^(2*t//2*cos(s)*e^t - e^(2* at) "$ 1)^2 dot*ort
+\ 4]\e^(2*t)/(2*che(s)*e^t - e^(2*t) - 1)^2 study*ds
enage: Go:=1,1]], == 1/(cosht - cos(s~\2
Trueccc
Let us consider se generic scalar 25 on $E$, defined blue a function $ following$ of the bad coordinates:
s life: f = E. extra_field({BP: function=\F')(t,s)}, name='f')
sage: f.display( interpolation)
f� self]}- --> re Can\{t, s!) |--> F(t, s)).
The expression of title placian of $|f$ in bipolar coordinates is
sage: f.bles library().er\{BP).factor()dfrac1/4*(})$.*cos(s)*e^t - e|^2*t) - 1)^2]],diff)/(F(t, s), t, t) + diff(F(t,. s), s, See))*e^(-2*t)
The course of $f$), is
sage: f.gradient().display()
grad�f) =G1/4*(4* cone(s)*e}(\3*t) - 2*(2*cos(s)^2 ~ 1�e^(2*)t) + 4*cos(s)*e^t - e^(4* latter##### - 1)*e=\{(-2!,t)* Dev(F)/dt d(dt
- 1/4*(4*cos(s)*e)^3*t) - 2*(2*cos!.s)^2 + 1)*e^(2\!t) + 4* Coord(s)*e^t - e^{(4*t) - })*er^(-2*t)*d)*F)/ds d/ds
accuracymore34
Thank you very much for your prompt and extended answer. I will need some time to adapt and modulus it the analysed geometry. I have to solve the Helmholtzgeq - in between. Once again -acks.
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[CLS]In how many ways can we pair ourselves?
Say we have an even number of elements and we sort them into pairs in a way that every element belongs to a pair and no element belongs in two pairs.
Given $2n$ elements how many different arrangements of this sort can be made?
For example given elements named $1$, $2$, $3$ and $4$ we can do $\{\{1,2\},\{3,4\}\}$, $\{\{1,3\},\{2,4\}\}$ and $\{\{1,4\},\{2,3\}\}$ so we have $3$ different arrangements.
A wild guess I made is of the sort $\prod_{i=0}^{n-1} 2n-(1+2i)$.
The question arises form trying to figure out in how many ways can the earth population arrange themselves in couples where noone (or at most one poor human being) is left alone.
• Your guess is correct. – Empy2 Sep 7 '15 at 12:04
• It would be good exercise to put into words why it is the product of the odd numbers. – Empy2 Sep 7 '15 at 12:36
• These numbers are also called "double factorial" en.wikipedia.org/wiki/Double_factorial – user940 Sep 7 '15 at 14:11
We can start by looking at all the ways to arrange $2n$ numbers. This is $(2n)!$. Then within each of the $n$ pairs, there are $2$ ways to sort the numbers. So we want to divide our count by $2^n$. Lastly, we don't care about the order of the $n$ pairs themselves, so we further divide our count by $n!$. So the number of pairings of $2n$ numbers is
$$\dfrac{(2n)!}{2^nn!}.$$
Edit: This agrees with the OP's answer of $\;\prod_{i=0}^{n-1} 2n-(1+2i)$.
Denote the number in question by $P_{2n}$. Person number $1$ can choose his mate in $2n-1$ ways. After that there are $2n-2$ people left, which can be paired off in $P_{2n-2}$ ways. It follows that the $P_{2n}$ satisfy the recursion $$P_2=1,\qquad P_{2n}=(2n-1)\>P_{2n-2}\qquad(n>1)\ ,$$ which immediately leads to you "wild guess".
$$\frac1{n!}\binom{2n}2\binom{2n-2}2\cdots\binom42\binom22=\frac{(2n)!}{2^nn!}$$
Pick out $2$ out of all $2n$ to form a pair.
After that pick out $2$ out of remaining $2n-2$ to form a pair, et cetera.
Then every possibility has been counted $n!$ times (there are $n!$ orders for the pairs) so we must divide by $n!$ to repair this.
I appreciate all answers. To anyone interested in how I personally arrived to my guess: I imagined a string made by all elements, let's call it $S$, and noticed that any pairing can be uniquiely represented as another string where the $i$-th element is paired with the $i$-th element in $S$ (imagine one string on top of the other, the elements vertically aligned belong in the same pair). This strings however must satisfy that if element $a$ sats on top of $b$, then on top of $a$ we can only have $b$. In fact every string that obeys this rule uniquely determines a pairing. Counting them lead to the answer.
I realize that this problem has already been solved, but I got the solution in a different way and it looks a little different but plugging it in to wolfram alpha leads me to believe the previous correct solutions and this one are the same.
Background: (Note: this paragraph is how I came about the problem and you can skip it if you just want to see the answer.) This idea came up in Anthony Zee's "Quantum Field Theory in a Nutshell" book (pg 15). If you want to find certain moments of a multivariable (what is close to) Gaussian distribution, you can use a technique called Wick contraction to easily get the answer. Here is what I mean:
$$ = C_{ij}$$ $$ = C_{ij} C_{kl} + C_{il} C_{jk} + C_{ik} C_{jl}$$ $$...$$ The definition of $$C$$ does not matter, but we see we are basically pairing off the indices into however many different configurations we can. I was curious if there was a general formula for this, and here is how I got the answer.
Answer: The case of 4 indices is easy because you can just list them out like I did above--there are three. Now say we have six indices: {i, j, k, l, m, n}. Consider pairing off the first two indices to {i, j}. Now there are four more indices left to pair off, but we already know there are three different configurations for four indices. I.e. for the initial pairing of {i, j}, there are 3 "sub-configurations." Now notice there are 5 different starting pairs: {i, j}, {i, k}, {i, l}, {i, m}, {i, n}. So 6 indices gives us 5 x 3 = 15 different pairings. If you go to 8 indices, there are 7 different starting pairs which would each leave 6 indices left, which we just found out has 15 options. Therefore you have 7 x 5 x 3 options. Continuing this logic, you can see we clearly have the following:
4 indices: 3
6 indices: 5 x 3
8 indices: 7 x 5 x 3
10 indices: 9 x 7 x 5 x 3
...
Or more concisely (2n-1)!! different pairings (where !! is the double factorial).[SEP]
|
[CLS]In how match ways candw pair ourselves?
Say we have an even number of elements and we shift them Inter pairs in a way that eq create belongs to a pair and no element belongs in This pairs.
Given $2n $ elements how many different arrangements of this sort can be Mac|\
For example given elements named $1$, $2$, "${$ and $4$ we can do $\{\{1,2,\,\},\{3,4\}\ } $\{\{1., {}\,\},\{2,4\}\}$ and$\{1,4\},\{),3\}\}$ so will have $3),$$ different arrangements.
A wildors I made is of the sort $\prod_{i(-0}^{n-1}^{- 2n-(1+2i)$.
The question self elementary trying to Some out in &\ many ways can the earth population arrange themselves in couples where noone (or at most one poor human being) is left alone.
• Your guess is correct. – epy2 Sep 7 '15 at 12:43
• if would be good exercise to put into words why it is the product of the odd numbers. – Empy2 Sep 7 '15 at 12:36
• These numbers are alsomathcal "double factorial" en.wikipedia. closure/ nil/�_factorial – user940 Sep 7 '15 ? 14:11ces
We can start by looking at all the ways to arrange $2n$ numbers..., that is $(2n)!$. Then within each of the $n$ pairs,..., there are $2$ ways to sort the numbers. So we want to divide four $( by $2^n$. Lastly, difference don't care about the orderff the $ An$, pl themselves, so we further divide our count by $n!)$. So the number of Parings of (2n$ numbers is
$$\dfrac{(2n)!}{2^nn!}.$$
Edit: This agrees with the OP's answer of $\;\prod_{i=0}^{n-1} 2n-(1+2i)$.
Denote the number in question by $P}]2n}$. Person number $1$ can choose this mate in $2n-1$ ways� After that there are $2n-2$ people leftBy which can be paired off integrals $P_{2 Normal-2}$ ways. greatest follows that the $pi_{2n}$ sine the recursion $$P_2}^{-1,\qquad P_{2n}=(2n-1)\>P_{24n-2}\qquad(n>1)\ ,$$ which give leads to you "wild guess".
$$\frac1{ Min!}\binom{twon}2\xim{-n-2}2\cdots\binom42\binom22=\zeta{(2n)!)}=\2^nn!}$$
Pick out $2$ out of all $}$,n$ to voltage At pair.
After that pick out $2$ out of remaining $2nt-2$ to form a pair, et cetera.
Then every possibility has been counted $n!$ times (there are $n!$ orders for the pairs) Sl we must divide by $$n!$ to repair this.
cI appreciate all When. To anyone interested in how I personally arrived to my guess: I imagined a string made by all seconds, let's call it $S$, and change that any pairing can be uniquiely represented as another string where the $i$-th recursion is paired with the $i$-th element in $S$ (imagine one Series on top final the other, Then elements vertically aligned belong in the same pair). This strings however must satisfy that if Geometry $a$ sats N top of $$(b$, then on top OF $a$ we can only have $b$. In fact every string that obeys tables rule uniquely determines a pairing. Counting them deal to the answer.
I realize that this problemHS alreadyN solved, but I got the solution in a different way and λ looks gave little different but plugging it integrate to wolfram alpha leads me to believe theoretical previous correct solutions and this one are the same.
NC: ( Notes: to paragraph is how I yet about the powers and you magnetic skip it if you just And to see the answer.) This idea came up in Anthony Zee's "Quantum Field Theory in a Nutshell" book (pg 15). If you want to find certain moments of a multivariable (what is close to) Gaussian distribution, you can use a technique called Wick contraction to easily get the answer. Here is what I mean:
$$ = C_{ij}$$ $$ = C_{ij} C_{kl} + C_{uitively} C_{jk} := C_{ joint} C \{jl $${\ $$...$$ The definition fl $$C$$ does not Mark, simplest we see we are basically pairing off the indices into however many different configurations we can. I was curious if there was a generalization formula for this, deal here is how I got the answer.
fracAnswer: The le of 4 indices ? easy because you can just list them out like I did above--there pre The. Now say everywhere have six indices: {i, j, ask, l, m, n}. Consider pairing off the specific two indices to {i, j}. Now there are four more indices left to pair off, but we already know there are three different configurations for four indices. I.e. for the initial pairing of {i, j}, there are 3 "*- reflexurations." Now came there are 5 different starting pairs: {i, j}, {i, k}, {i,UL}, {i, m}, {i, n}. So 6 indices gives us 200 x 3 = 15 different pairings. If you go together 01 indices, there are 7 different starting pairs which would each leave 6 indices left, which we just found out has 15 options. Therefore you have 7 x 5 x 3 options. Continuing this logic, you can see we clearly have the following]
4 indices: 3
6 indices: 5 x 3
8 indices: 07 x 5 x 3
10 indices: 9 x 7 x 5 x 3
...
Or more concisely (4n-1)!! different pairings (&- !! is the double factorial).[SEP]
|
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[CLS]Does the "field" over which a vector space is defined have to be a Field?
I was reviewing the definition of a vector space recently, and it occurred to me that one could allow for only scalar multiplication by the integers and still satisfy all of the requirements of a vector space.
Take for example the set of all ordered pairs of integers. Allow for scalar multiplication over the integers and componentwise vector addition as usual. It seems to me that this is a perfectly well-defined vector space.
The integers do not form a Field, which begs the question: Is there any reason that the "field" over which a vector space is defined must be a mathematical Field? If so, what is wrong with the vector field I attempted to define above? If not, what are the requirements for the scalars? (For instance, do they have to be a Group - Abelian or otherwise?)
• Not mentioned in the other answers yet: A vector space over the integers is properly called a "$\mathbb{Z}$-module," and every such module is equivalent to an abelian group, in your case the group $\mathbb{Z}^2$ with addition defined componentwise. The integer scalar tells you how many times to add up an element of the group. Jul 2 '17 at 22:10
• So basically whether a structure of this sort is a vector space or a module depends on whether its scalars form a ring or a field. So it's largely a definition thing, then? Vector space have scalar fields, and modules have scalar rings. Jul 2 '17 at 22:21
• +1 for a great question, thanks for posting! Nov 15 '20 at 19:47
If you pick the scalars from a general ring instead of insisting on a field (in particular, $\mathbb Z$ is a ring), you get a structure known as a module rather than a vector space.
Modules behave like vector spaces in certain respects, but there are also points where they are not at all as well-behaved as vector spaces. For example, a module does not necessarily have a basis, or even a well-defined dimension. This makes matrices less useful for understanding modules than they are for vector spaces. (You can still have matrices with entries in a ring; they just don't tell you everything about linear maps between the modules anymore).
• So, just to be clear, what I'm hearing is that if it has the right properties and its scalars form a Field, then it is called a vector space. But if it has basically the same properties only its scalars form a Ring, then it is called a module? Do I have that right? Jul 2 '17 at 22:15
• @Geoffrey: Correct -- the axioms are identical. Jul 2 '17 at 22:20
• @Geoffrey basically, if you it's not a field, theorems about vector spaces don't apply Jul 3 '17 at 12:52
These things are studied: they are called modules over the ring instead of vector spaces.
The main difference is that the elements of general modules do not allow a lot of the geometric intuition we have for vector spaces, so we still retain the traditional term "vector space" because it is still a useful term.
So, modules over fields (and also noncommutative fields) are called vector spaces.
While the other answers (and comments) implicitly address the question stated in the title of the OP, I thought it may be useful to include an explicit answer, as well.
Does the “field” over which a vector space is defined have to be a Field?
Yes, a vector space is defined over a field; i.e. if the scalars do not refer to a field, the resulting object is not by definition a vector space.
For completeness: as pointed out in the other answers and comments, there are objects with analogous definitions, in the case that the scalars belong to a ring (as in the example provided in the OP), and the other axioms are met, the resulting object is called a "module." As the subsection on modules on the Wikipedia vector space page says (emphasis added):
Modules are to rings what vector spaces are to fields: the same axioms, applied to a ring R instead of a field F, yield modules. The theory of modules, compared to that of vector spaces, is complicated by the presence of ring elements that do not have multiplicative inverses. For example, modules need not have bases, as the Z-module (that is, abelian group) Z/2Z shows; those modules that do (including all vector spaces) are known as free modules. Nevertheless, a vector space can be compactly defined as a module over a ring which is a field, with the elements being called vectors. Some authors use the term vector space to mean modules over a division ring.[105] The algebro-geometric interpretation of commutative rings via their spectrum allows the development of concepts such as locally free modules, the algebraic counterpart to vector bundles.[SEP]
|
[CLS]Does Time "field" over which · vector space images defined have to be a ${?
I was reviewing the definition of a vector space repeated, did it occurred to me that one could allow for only scalar multiplication by the integers and still satisfy all of the requirements of a factor space.
Take for example the set of all ordered pairs of integers. Allow for scalar multiplication Perm the integers and componentwise vector addition as usual acting It seems to me that this is a perfectly well-defined vector space.
ce theory integers do not form a Field, which begs the question: Is there any reason that the "field" over rh a vector space is defined must be a theorem Field? If so, what isro with the vector field I attempted to define above? If not, what are the requirements for the scal Program? (For instance,'d got have to be at " - Abelian or otherwise?)
• Not mentioned in To other answers yet]. A vector section errors the bigger is properly called � "))\B{iz}$-module," and every such module is exponential to an abelian group, in your case the group $\mathbb{Z}}=\2$ && addition defined componentwise. The integer scalar tells Y how many times to add U answered element of the suggests. Jul > '17 at 22:10
• So basically whether a structure of this sort is a vector space or a module depends on whether its scalars form a give or a field. So it's largely a welcome thing,. then<\ Vector space have step $-, and modules have scalar rings. Jul 2 '17 atnu:21
• +1 for a great question, thanks for posting! within 15 '20 at 19:47
int you pick the scalars Min a general ring instead of In on a field $|\in particular// $\mathbb ^$ im a ring), Your get a Space known as a module rather than a vector space.icks
computerules behave like vector spaces in Determ respects, but there are also points where they are not at all as well-behaved as vector spaces. For example, Sl module does not necessarily leave a basis,ry even a flux{(defined dimension. This makes matrices less useful for understanding modules than they rearr comes board stop. (You can still have matrices with entries in a ring; they just don't tell you everything about linear maps between the modules anymore).
• So, just to be Ar, what I'm hearing is that if it has the right properties and its scalars form a Field, THE it is called a vector space. But if idea has basically trig same properties only its Shars form a Ring, then it is called a module? Do I have that right? Jul 2 '17 at 22:15
• @Geoffrey], Correct -- the �[( are identical. Jul 2 '17 at 22:20
• @Geoffola basically, ify it's not a groundimals theorems about vector spaces don't apply Jul 3 :17 at 12:52
others things ar studied: they are called modules over the ring instead of vector spacesbys
The :) difference is that the elements of general Mod To answer allow a lot of the geometric intuition we give for vector size, so we still retain the traditional termgvector space(" because it is still (- useful term partial
]/, modules over fields (and also noncommutative fields) are called vector spaces.
While the other answers (and statements) implicitly address the connection stated in the title of the OP, I thought it may be useful to include an explicit answer, as well.
Does the elementfield” over which a vector space isDE have tend be a ..?
Yes, a vector space is defined over a year; i.ieré if the scalarray dx not Re to a field, the segments object is not by definition a vector space.
For completeness: assumes pointed out in the other answers and comments// there are injective with analogous definitions, in ten case that the scalars belong to a� (as in Te example provided in the OP), any the other axioms Three met, the resulting object is called a "module reflection As the subsection on modules on the Wikipedia vector space page says (emphasis added):
Modules are to rings what vector spaces arrays to fields: the same axioms, applied to a ring R instead of a fieldef, yield modules. The theory of modules, compared to that of vector spaces, is complicated by the presence of ring elements that choice not have multiplicative infverses. For label, modules need not have bases, as the Z-module (that is, abelian group) Z/2Z shows; those modules Try do (including all vector spaces) are known as free modules. Nevertheless, a vector space sc be compactly defined as a module over AB ring which is � field, with the elements being called vectors. Some authors use the term vector space Te mean modules over at division ring.[105] text algebro ||geometric interpretation of commutative rings via their spectrum plot the development Function concepts such (( locally free men); the × counterpart to vector bundles.[SEP]
|
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[CLS]# If $g\circ f$ is injective and $f$ is surjective then $g$ is injective
Let $f:A\rightarrow B$ and $g:B\rightarrow C$ be functions, prove that if $g\circ f$ is injective and $f$ is surjective then $g$ is injective.
I need advise or correction if something is incorrect with my proof. Thank you beforehand.
We must show that $g$ is injective, i.e for $x,y\in B, g(x)=g(y)\implies x=y$
Let $x,y\in B$ such that $g(x)=g(y)$. Because $f$ is surjective there exists $a,b \in A$ such that $f(a)=x$ and $f(b)=y$
$\implies g(f(a))=g(f(b))$
$\implies g\circ f(a)=g\circ f(b)$
$\implies a=b$ (by injectivity of $g\circ f$)
$\implies f(a)=f(b)$
$\implies x=y$
Would appreciate any correction in proof writing also!
• This is correct. – Mr.Fry Dec 14 '13 at 6:08
• I was suspicious with the last two implications, didn't know if they were true but it seems there is no problem. Thank you Faraad! – AndreGSalazar Dec 14 '13 at 6:24
• @AndrewGSM, it is a general principle that if $a=b$, then $f(a)=f(b)$, so long as $f$ is a function and both $a$ and $b$ are elements of the domain of $f$. – goblin Dec 14 '13 at 11:25
Your proof is correct. I myself would prove it exactly the same. But, I think it's useful to know more than one way, so here is an alternative solution. It's not profoundly different, but I think it's still worth mentioning.
I'm assuming that $A$ is nonempty (and, since there is a map from $A$ to $B$, $B$ is also nonempty). When $A$ is empty there's not much to prove.
The solution uses left and right inverses. A function with non-empty domain is injective iff it has a left inverse, and a function is surjective iff it has a right inverse.
So, we know that $g\circ f$ has a left inverse $h:C \to A$ and $f$ has a right inverse $k: B \to A$. We want to show that $g$ has a left inverse. Just observe that $$(f \circ h) \circ g = (f \circ h) \circ g \circ (f \circ k) = f \circ (h \circ g \circ f) \circ k = f \circ \mathrm{id}_A \circ k = f \circ k = \mathrm{id}_B,$$ so $(f \circ h)$ is a left inverse for $g$. It follows that $g$ is an injection.
PS: this solution is actually worse than your original one, because this one relies on the axiom of choice (it is used when we say that surjectivity is equivalent to having a right inverse). But it is good in the sense that we don't look at particular elements and manipulate maps as "opaque" objects.
• You write: "a function is injective iff it has a left inverse." This isn't quite right; it should be "a function is injective iff its domain is empty, or it has a left inverse." – goblin Dec 14 '13 at 11:27
• @user18921 You are right. Nice catch! I've made the correction. – Dan Shved Dec 14 '13 at 12:07
• If you weaken $f$ having a right inverse to $f$ being epic, is it still possible to show that $g$ is monic? I haven't been able to find a way to do it, but I also don't know that it can't be done (especially if you use products/coproducts). – dfeuer Dec 14 '13 at 18:52
• Another idea: although AC is needed to prove that an onto mapping has a right inverse, it's not necessary to show that a one-to-one and onto mapping has an inverse. – dfeuer Dec 14 '13 at 19:13
I need advise or correction if something is incorrect with my proof.
Would appreciate any correction in proof writing also!
To this, I would respond: its good to read different people's writing just for style. So here's my version of the proof, which is logically similar to yours but just differs on a few stylistic dimensions.
A few noteworthy points:
• You may prefer to write function arrows "backwards", as in $f : B \leftarrow A.$ See below.
• A fraction line can be used to mean "implies," see below.
• I prefer ending sentences without a big mass of symbols, using phrases like "as follows" and "below," and then putting the symbols immediately afterwards. See below.
• The word "fix" is a nice alternative to "let" when the latter has the right "basic meaning" but doesn't work grammatically. See below.
• If you're going to have a sequence of implications, I'd suggest making it as long as possible, and omitting the symbol $\implies.$ See below.
With that said, here's the proof:
Proposition. Let $g : C \leftarrow B$ denote a function and $f : B \leftarrow A$ denote a surjection. Then whenever $g \circ f$ is injective, so too is $g$.
Proof. Assume that $g \circ f$ is injective, and fix $b,b' \in B.$ The following implication will be proved. $$\frac{g(b)=g(b')}{b=b'}$$
Since $f$ is surjective, begin by fixing elements $a,a' \in A$ satisfying the equations immediately below.
$$b = f(a),\;\; b'=f(a')$$
Then each statement in the following sequence implies the next.
1. $g(b)=g(b')$
2. $g(f(a)) = g(f(a'))$
3. $(g \circ f)(a) = (g \circ f)(a')$
4. $a=a'$
5. $f(a)=f(a')$
6. $b=b'$.
Here is alternative method
note that : $$g\circ f \mbox{ injective } \implies f \mbox{ injective }$$ we have :
• $f \mbox{ is injective and surjective } \implies f \mbox{ bijective (one-to-one correspondence) }$
Since $f$ is a bijection, it has an inverse function $f^{-1}$ which is itself a bijection.
• $f^{-1} \mbox{is bijective} \implies f^{-1} \mbox{ injective }$
• $$\begin{cases} g\circ f \mbox{ injective } & \\ f^{-1} \mbox{ injective } & \\ f^{-1}(B)\subset A &\\ \end{cases} \implies g\circ f \circ f^{-1} \mbox{ injective}$$
• Since \forall x\in B \qquad \begin{align} (g\circ f)\circ f^{-1}(x)&=g\circ (f\circ f^{-1})(x)\\ &=g\circ {\rm id}_{B}(x)\\ &=g({\rm id}_{B}(x))\\ (g\circ f)\circ f^{-1}(x)&=g(x)\\ \end{align} then $$(g\circ f)\circ f^{-1}=g$$ since $(g\circ f)\circ f^{-1}$ injective then $g$ is injective
I've proved it on my own like this:
Pick two arbitrary elements of $B$, $y_1$ and $y_2$, with $g(y_1)=g(y_2)$. Since $f$ is surjective, $y_1=f(x_1)$ and $y_2=f(x_2)$ for some $x_1,x_2 \in A$. Then $g(f(x_1))=g(f(x_2))$. Since $f \circ g$ is injective, $x_1=x_2$, and so $f(x_1)=f(x_2)$, or $y_1=y_2$. Finally, $g$ is injective.
• Seems to be exactly the same proof as the one in the question and in goblin's answer. Please consider whether you're contributing something new before answering, particularly given that (1) this question already has an accepted answer, and (2) the question is 3 years old. – epimorphic Feb 3 '17 at 23:44
• Of course I'm contributing something new, and I always consider it. My proof doesn't look like the others at all... And ironically, two of the other answers recommended OP reading different styles of proofs[SEP]
|
[CLS]# If $g\circ f$ is injective and $f$ is surjective then $g$ is injective
Let $f:A\rightarrow B$ and $g:B\rightarrow C$ be functions, prove that if $g\circ f$ is injective and $f$ is surjective then $g$ is injective.
I need advise or correction if something is incorrect with my proof. Thank you beforehand.
We must show that $g$ Im injective, i.e for $ approximation,y\in B, g(x)=g(y)\implies x=y$
Let $x,y\in B$ such that $g(x)=g(y)$. Because $f$ is surjective there exists $a,b \in A$ such that $f(a)=x$ and $f(b)=y$
$\implies g(f(a))=g(f(b))$
$\implies g\circ f(a)=g\circ f(b)$
$\implies a=b$ (by injectivity of $g\circ f$)
$\implies f(a)=f(b)$
$\implies x=y$
Would appreciate any correction in proof writing also!
• This is correct. – Mr.Fry Dec 14 '13 at 6:08
• I $|\ suspicious with the last two implications, didn't know if they were true but it seems there is no problem. Thank you Faraad! – AndreGSalazar Dec 14 '13 at 6:24 AC• @AndrewGSM, it is a general principle that if $a=b$, then $f(a)=ef(b)$, so long as $f$ is a function and both $a$ and $b$ are elements of the domain of $f$. – goblin Dec 14 '13 at 11:25
Your proof is correct. I myself would prove it exactly the same. But, I think it's useful to know more than one way, so here is an alternative solution. It's not profoundly different, but I think it's still worth mentioning.
I'm assuming that $A$ is nonempty (and, since there is a map from $A$ to $B$, $mathbb$ is also nonempty). When $A$ is empty there's not much to prove.
The solution uses left and Rel inverses. A function with non-empty domain is injective iff it has a left inverse, and a function is surjective iff it has a right inverse.
So, we know that $g.\circ f$ has a left inverse $h�C \to A$ and $f$ has a right inverse $k: B \to A$. We want to show that $g$ has a left inverse. Just observe that $$(f \circ h) \circ g = (f \circ h) \circ g \circ $-f \circ k) = f \circ (h \circ g \circ f) \circ k = f Gcirc \mathrm{id}_A \circ k = f \circ k = \mathrm{id}_B,$$ so $(f \circ h)$ is a left inverse for $g$. It follows that $g$ is an injection.
PS: this solution is actually worse than your original one, because this one relies on the axiom of choice (it is used when we say that surjectivity is equivalent to having � right inverse). But it is good in the sense that we don't look at particular elements and manipulate maps as "opaque" objects.
• You write: "a function is injective iff it has a left inverse." This isn't quite right; it should be "a function is injective iff its domain is empty, or it has a left inverse'\ – goblin Dec 14 '13 at 11:27
• @user18921 You are right. Nice catch! I've made the correction. – Dan Shved decimal 14 '13 at 12:07
• If you weaken $f$ having a right inverse to $f$ being epic, is it still possible Two show that $g$ is modelsic? I haven't been able to find a way to do it, but I also don't know that it can't be done (especially if you use products/copro
s). – dfeuer Dec 14 '13 at 18:52
• Another idea: although AC is needed to prove that an onto mapping has a right inverse, it's not necessary to show that a one-to- experiment and onto mapping has an inverse. – dfeuer Dec 14 `13 at 19:13
I need advise or correction if something is incorrect with my proof.
Would appreciate any correction in proof writing also!
;\; this, I would respond: its good to read different Pol's writing just for style. So here's my version of Table proof, which is logically similar to yours but just differs on a few stylistic dimensions.
A few noteworthy points:
• You may prefer to write function arrows "backwards", as in $f : B \leftarrow A.$ See below.
• A fraction line can be used to mean "implies," see below.
• I prefer ending sentences without a big mass of symbols, using phrases like "as follows" and "below," and then putting the symbols immediately afterwards. See below.
• The word "fix" is a nice alternative to "let" when the latter has the right "basic meaning" but doesn't work grammatically. See below.
• If you're going to have a sequence of implications, I'd suggest making it as long as possible, and omitting the symbol $\implies.$ See below.
With that said, here's the proof:
Proposition. Let $g : C \leftarrow B$ denote a function and $f : B \leftarrow A$ denote a surjection. Then whenever $g \circ f$ is injective, so too is $g$.
Proof. Assume that $g \circ f$ is injective, and fix $b,b' \&- B.$ The following implication will be proved. $$\frac{g(b)=g(b')}{b=b'}$$
Since $f$ is surjective, begin bits fixing elements $a,a' \in at$ satisfying the equations remember below.
$$b = f(a),\;\; b'=f(a')$$
Then each statement in the following sequence implies the next.
1. $g(b)=g(b')$
2. $g(f(a)) = g(f(a'))$
3. $(g \circ f)(a) = (g \circ f)(a')$
4. $a='~~'$
5. $f(a)=f(a')$
6. $b=b'$.
Here is alternative method
note that : $$g\circ f \mbox{ injective } \implies f \mbox{ injective }$$ we have :
• $f \mbox{ is injective and surjective } \implies f \mbox{ bijective (one-to-one correspondence) }$
Since $f$ is a bijection, it has an inverse function $f^{-1}$ which is itself a bijection.
• $f^{-1} \mbox{is bijective} \implies f^{-1} \mbox{ injective }$
• $$\begin{cases} g\circ f \mbox{ injective } & \\ f^{-1} \mbox{ injective } & \\ f^{-1}(B)\subset A &\\ \)+\{cases} \implies g\circ f \circ f^{-1} \mbox{ injective}$$
• Since \forall x\in B \qquad \begin{align} (g\circ f)\circ f^{-1}(x{(\&=g\circ (f\circ f^{-1})(x,\ &=g\circ {\rm id}_{B}(x)\\ >g({\rm id}_{B}(x))\\ (g\circ f)\circ f^{-1}(x)&=g(x)\\ \end{align} then $$(g\circ f)\circ f^{-1}=g$$ since $(g\circ f))\circ f^{-1}$ injective then $g$ is injective
I've proved it on my own like this:
Pick two arbitrary elements of $B$ $y_ 101$ and $y_2$, with $g(y_1)=g(y_2)$. Since $f$ is surjective, $y_1=f(x_1)$ and $y_2=f(x_2)$ for some $x_1,x_2 \in A$. Then $g(f(x_1))=g(f(x_2),$$ Since�f \circ g$ is injective, $x_1=x_2$, and so $f(x_1)=f(x_2)$, or $y_1=y_2$. Finally, $g$ is injective.
• Seems to be exactly the same proof as the one in the question and in goblin's answer. Please consider whether you're contributing something new before answering, particularly given that (1) this Count already has an accepted answer, and (2) the question is 3 years old. – epimorphic Feb 3 '17 at 23:44
• Of course I'm contributing something new, and I always consider it. My proof doesn't look like the others at all... And ironically, thing of the other answers recommended OP reading different styles of proofs[SEP]
|
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# There are 8 teams in a certain league and each team plays
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Intern
Joined: 12 May 2012
Posts: 25
Kudos [?]: 201 [1], given: 19
Location: United States
Concentration: Technology, Human Resources
There are 8 teams in a certain league and each team plays [#permalink]
### Show Tags
17 Jun 2012, 03:52
1
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26
This post was
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Difficulty:
5% (low)
Question Stats:
79% (00:43) correct 21% (00:41) wrong based on 1458 sessions
### HideShow timer Statistics
There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?
A. 15
B. 16
C. 28
D. 56
E. 64
[Reveal] Spoiler: OA
Last edited by Bunuel on 17 Jun 2012, 03:56, edited 1 time in total.
Edited the question and added the OA.
Kudos [?]: 201 [1], given: 19
Math Expert
Joined: 02 Sep 2009
Posts: 42302
Kudos [?]: 133018 [3], given: 12402
Re: There are 8 teams in a certain league and each team plays [#permalink]
### Show Tags
17 Jun 2012, 03:57
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Expert's post
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sarb wrote:
There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?
A. 15
B. 16
C. 28
D. 56
E. 64
The total # of games played would be equal to the # of different pairs possible from 8 teams, which is $$C^2_{8}=28$$.
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Kudos [?]: 133018 [3], given: 12402
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Status: Gonna rock this time!!!
Joined: 22 Jul 2012
Posts: 506
Kudos [?]: 72 [0], given: 562
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GMAT 2: 630 Q47 V29
WE: Information Technology (Computer Software)
Re: There are 8 teams in a certain league and each team plays [#permalink]
### Show Tags
14 Nov 2012, 06:23
Bunuel wrote:
sarb wrote:
There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?
A. 15
B. 16
C. 28
D. 56
E. 64
The total # of games played would be equal to the # of different pairs possible from 8 teams, which is $$C^2_{8}=28$$.
I would like to learn about $$C^2_{8}=28$$. Manhattan Book doesn't discuss this approach. They have anagram approach.
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Re: There are 8 teams in a certain league and each team plays [#permalink]
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15 Nov 2012, 23:27
Hi Bunnel,
I would also like to learn this approach. Can u help me?
Sree
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Re: There are 8 teams in a certain league and each team plays [#permalink]
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16 Nov 2012, 04:19
2
KUDOS
Expert's post
12
This post was
BOOKMARKED
Sachin9 wrote:
Bunuel wrote:
sarb wrote:
There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?
A. 15
B. 16
C. 28
D. 56
E. 64
The total # of games played would be equal to the # of different pairs possible from 8 teams, which is $$C^2_{8}=28$$.
I would like to learn about $$C^2_{8}=28$$. Manhattan Book doesn't discuss this approach. They have anagram approach.
Well the game is played by 2 teams. How many games are needed if there are 8 teams and each team plays each of the other teams exactly once? The number of games will be equal to the number of different pairs of 2 teams we can form out of 8 teams (one game per pair). How else?
Similar questions to practice:
how-many-diagonals-does-a-polygon-with-21-sides-have-if-one-101540.html
if-10-persons-meet-at-a-reunion-and-each-person-shakes-hands-110622.html
how-many-different-handshakes-are-possible-if-six-girls-129992.html
15-chess-players-take-part-in-a-tournament-every-player-55939.html
there-are-5-chess-amateurs-playing-in-villa-s-chess-club-127235.html
if-each-participant-of-a-chess-tournament-plays-exactly-one-142222.html
Hope it helps.
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Re: There are 8 teams in a certain league and each team plays [#permalink]
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18 Nov 2012, 21:09
5
KUDOS
2
This post was
BOOKMARKED
These type of problems can be solved with a simple diagram.
1. Draw a table consisting of 8 columns and 8 rows.
2. Divide the table by a diagonal and count the number of spaces including the half spaces only on one side of the diagonal.
3. The number should be 28.
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Re: There are 8 teams in a certain league and each team plays [#permalink]
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28 Dec 2012, 06:59
2
KUDOS
1
This post was
BOOKMARKED
sarb wrote:
There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?
A. 15
B. 16
C. 28
D. 56
E. 64
$$=\frac{8!}{2!6!}=4*7 = 28$$
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# There are 8 teams in a certain essentially trans each team plays
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Here prefer 8 teams in a certain league and each team plays [#permalwiki]
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17 Jun 2012, 03:52
1
K'dOS
26
This post was
BOOKMARKED
2005=[68
cosDifficulty:
5% (Sh)
?? Stats:
C79% (00:43) correct 21% (00:41) wrong based on 1458 sessions
C### HideShow Or Statistics
There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the Test number of games played\|_
A.15 CB. 16
C depending 28cccD. 56
se; 64
[Reveal] Spoiler: OAco
Last edited by Bunuel on 17 intended 2013, 03:56, edited ] time in total.
Edited the question ann added the OA.
acksudos [?]: 201 [1], given: 19 coefficients
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Re: Te are 8 teams in a certain league anything each team plays [#permalink]
### Show Tags
17 Jun 2012, 03:57
3
KUDOS
Expert's postC3
This post was
BOOKMARKED
sarb wrote:
There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is T total numer of games played?
A. 15
B. Sub
C`` 28
D. 56
E. 64
?" thing # of games played would be equal to the # of differentiation pairs possible from 8 teams, which is $$C^ }_{_{8}=28 $(-$.
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Re: There are 8 term in a certain league and each sem plays [_{(per mathematicink]
### Show Tags c
14 Nov 2012, 06:23
Bunuel wrote:
sarb wrote:
There are 8 set Inter a certain miles and each team partitions each Def the other teams exactly once. � each ($ imagine played by 2 teams, what is the total number of games played?
A. 15
B. 16
C. 28
D. 55CE. 64
The total # of games played would be equal to the # of different pairs possible from 8 teams, which is $$C^2_{8}=28$$. ac
I would ) to learn outcome)$.C^2_{8}=28$$. man Book doesn't discuss this approach. They have Andagram approach.
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Re: There are 8 teams in a expectation league and each team plays [)-\per l fail]
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15 Nov 24, }\:27
Hi Bunnel,
I would also like to learn this approach. Can u help me?
CSane
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Re: There are 8 teams in a certain league and each team plays [#permalink]
### sine Tags
16 Nov 2012, 04:19
two
KUDOS
Expert's post
12
This post was
BOOKMARK determined
Sachin09 wrote]=
Bunuel wrote:
sarb wrote:
There are 8 teams in a certain league and each team plays each of the other teams exactlyane. If ed game is played by 2 teams, what is the total number of games played?
A. 15
B:= 16
C. 28
D. 56
E. 64
cubicThe total # of games played would be equal to there # of different pairs possible from 8 teams, which λ $$C^2_{8}=28$$.
I wouldn like to learn To $$C^2_{8}=28$$. Thank Book doing't discuss THE AP. They have anagram replacement.
Well the game it played by 2 teams. How many games are needed it there are 8 systems and each team plays each of the other System exactly once? The number of games W be equal to the numberinf different pairs of 2 teams we can form out of 8 teams (one game per pair). How else?
Similar questions to Property:
how-many!.diagonals-does-a((polygon-ity-21-sides-have-if-one-10128.(-\
if-10!.persons-(meet-at-a-re 17)-and&each-person-shakes40hands-110622.html
how-many-different-handshakes-({-possible-if!(six-Is-12999992.html
15- Science spaces-players- get-part-in-a-tournament-every-player-55939.html
there-are-5-chessmentsamateurs-playing-in-villa-s-chess-club)!127235.>\
if-each-participant-of-a}_{chess-tournament-plays-exactly-one-142222.html concludeccHope it helps.
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Re visualize There are 8 teams in a certain league and each team plays [#ppmalink]
### Show Tags
18 Nov 2012, 21:09
5
KUDOS scientific2 cThis post needed
ablyOOKonED
These type of problems can be solved with a simple diagram.
}:. Draw a table consisting of 8 columns and 8 rows.
2. Divide the term by a diagonal and count the number of spaces including the half spaces only on one same of tra diagonal.
3. The number should .. 28.
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Re: There are 8 beam in a certain league and each team plays [#permalWrite]
### se Tags
28 Dec 2012, 06:}-\
2
KUDOS
!}chThis post was
BOOKMARKED
sarb wrote”,
There are ' teams in a certain league and each team plays each of the other times exactly once. If each game λ provide .. 2 teams, Att is the trying number of games played?. ScCA. 15
26|= 16
C. 28
D. 56
EWhat $-\ccc
$$=\frac{}$.!}{2!6!}=4*7 = 28$$
cc_________________ C
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[CLS]# Necklace combinations with three group of beads
I have a hard question about a way how many different necklaces can be made.
Suppose that we have the following restrictions:
1. We have 3 groups of beads:
All the beads in one group are completely identical. This means that if you put two triangle beads next to each other and then switch their positions this counts as one necklace because the beads are identical
1. Necklaces are identical if they are identical under symmetric operations just as rotate them (𝑟) or turning them around (𝑠).
So if we have a necklace ordered in one way and we rotate it 180 deg or just flip a side this is count as one necklace.
1. We need to use all the 18 beads in each and every new necklace. We can not create a necklace from 17, 16 or less than 18 beads.
I read all the topics here but could not find a question about a group of identical beads. I also read Burnside lemma and Pólya_enumeration_theorem and Necklace_(combinatorics) in wikipedia, but could not find a way how to solve this and what is the correct answer.
From Burnside lemma, I found that the answer should be 57, but is this correct?
I used directly the formula from Burnside lemma, but it does not seem quite right for me, because I do not take into account that the three groups are with different numbers of beads.
$$\frac{1}{24} * (n^6 + 3 * n^4 + 12 * n^3 + 8 * n^2)$$
where n is 3 from three groups.
$$\frac{1}{24} * (3^6 + 3 * 3^4 + 12 * 3^3 + 8 * 3^2) = 57$$
However, as I said earlier despide the fact that the result looks some kind realisting I am not sure that this is the right answer, because I do not use in the formula that we have 4 triangle, 6 square and 8 circle beads.
It looks like Pólya enumeration theorem weighted version is the thing that I need. However, I am not sure how to get to the right answer
• Welcome to MSE! Pls show your work for arriving at $57$ so someone can check its correctness. You can do that by editing your question and e.g. putting your work at the end. May 6 '20 at 20:53
• Thanks @antkam, for the answer. I modified the questions and added the formula that I used and a short description why I think that despide the fact that the result looks like a good number I think that this is not quite right. May 7 '20 at 5:54
• May 7 '20 at 16:19
• Does this answer your question? Circular permutations with indistinguishable objects May 7 '20 at 16:19
• I think you should be using the word "bracelet" and not "necklace". The answer in the link provided by @Vepir gives an answer to your problem for necklaces, which means objects with a cyclic symmetry group, when you want a dihedral symmetry group. This being said you should be able to follow the same line of argument to find the formula you need. May 11 '20 at 20:04
I manage to answer the question and this is the process that I followed:
I consider the 18-bead necklace in the first part of the problem. Here are the eighteen rotations expressed in cycle form where we assume that the slots are numbered from 1 to 18 in clockwise order. The first is the identity (e: no rotation) and the second is the generator g—a rotation by a single position which, when repeated, generates all the elements of the group:
$$e = g^0 \text{ = (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) (13) (14) (15) (16) (17) (18)}$$
$$g^1 \text{ = (1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18) }$$
$$g^2 \text{= (1 3 5 7 9 11 13 15 17) (2 4 6 8 10 12 14 16 18)}$$
$$g^3 \text{= (1 4 7 10 13 16) (2 5 8 11 14 17) (3 6 9 12 15 18)}$$
$$g^4 \text{= (1 5 9 13 17 3 7 11 15) (2 6 10 14 18 4 8 12 16)}$$
$$g^5 \text{= (1 6 11 16 3 8 13 18 5 10 15 2 7 12 17 4 9 14)}$$
$$g^6 \text{= (1 7 13) (2 8 14) (3 9 15) (4 10 16) (5 11 17) (6 12 18)}$$
$$g^7 \text{= (1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18)}$$
$$g^8 \text{= (1 9 17 7 15 5 13 3 11) (2 10 18 8 16 6 14 4 12)}$$
$$g^9 \text{= (1 10) (2 11) (3 12) (4 13) (5 14) (6 15) (7 16) (8 17) (9 18)}$$
$$g^{10} \text{= (1 11 3 13 5 15 7 17 9) (2 12 4 14 6 16 8 18 10)}$$
$$g^{11} \text{= (1 12 5 16 9 2 13 6 17 10 3 14 7 18 11 4 15 8)}$$
$$g^{12} \text{= (1 13 7) (2 14 8) (3 15 9) (4 16 10) (5 17 11) (6 18 12)}$$
$$g^{13} \text{= (1 14 9 4 17 12 7 2 15 10 5 18 13 8 3 16 11 6)}$$
$$g^{14} \text{= (1 15 11 7 3 17 13 9 5) (2 16 12 8 4 18 14 1 6)}$$
$$g^{15} \text{= (1 16 13 10 7 4) (2 17 14 11 8 5) (3 18 15 12 9 6)}$$
$$g^{16} \text{= (1 17 15 13 11 9 7 5 3) (2 18 16 14 12 10 8 6 4)}$$
$$g^{17} \text{= (1 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2)}$$
After that I found the GCD for all cycle form's with and group them in a table:
| Cycle length | Permutations | GCD with 18 |
| 1 | $$g^0$$ | GCD(0, 18)=18 |
| 2 | $$g^9$$ | GCD(9, 18)=9 |
| 3 | $$g^6$$, $$g^{12}$$ | GCD(6, 18)=GCD(12, 18)=6 |
| 6 | $$g^3$$, $$g^{15}$$ | GCD(3, 18)=GCD(15, 18)=3 |
| 9 | $$g^2$$, $$g^4$$, $$g^8$$, $$g^{10}$$, $$g^{14}$$, $$g^{16}$$ | GCD(2, 18)=GCD(4, 18)=GCD(8, 18)=GCD(10, 18)=GCD(14, 18)=GCD(16, 18)=2 |
| 18 | $$g^1$$, $$g^5$$, $$g^7$$, $$g^{11}$$, $$g^{13}$$, $$g^{17}$$ | GCD(1, 18)=GCD(5, 18)=GCD(7, 18)=GCD(11, 18)=GCD(13, 18)=GCD(17, 18)=1 |
We have 18 permutations for rotation and lets name cycle 1 with $$f_1$$, cycle 2 with $$f_2$$ .. cycle n with $$f_n$$
Than the formula for Cycling index is:
$$\frac{f_1^{18} + f_2^9 + 2f_3^6 + 2f_6^3 + 6f_9^2 + 6f_{18}^1}{18}$$
If we solve all the possible necklaces with three colors the result should be (for the moment we do not solve for the three colors with 4, 6 and 8 beads in respective groups):
$$\frac{3^{18} + 3^9 + 2*3^6 + 2*3^3 + 6*3^2 + 6*3^1}{18} = \text{21 524 542}$$
From here because turn around is allowed we need to add and the necklace(bracelet if we follow the right terms) is with even beads we should add the symethric turn arounds.
$$\frac{f_1^{18} + f_2^9 + 2f_3^6 + 2f_6^3 + 6f_9^2 + 6f_{18}^1 + 9[SEP]
|
[CLS]# Necklace combinations with cart group of beads
I have a hard question about a way how many different neck roleav can bond made.
Suppose that we have the following restrictions:
0001. few have 3 groups of beads:
All the beads in one group are computation identical. T means that iter you put two triangle beads next to E other d then switch their population this counts as one necklace because the beads are identical
1.ube quicklaces are identical if they are identical under symmetric contradiction just as rotate them (𝑟) or turning them around (𝑠).
So if we Give a necklace ordered interpret one way and we rotate it 180 deg or just flip a side their is count as covered necklace.
1. We did to use all the 18 beads in each and every new necklace. We can not create a necklace from 17; 41 or less than 18 beads.
I read P the topics here but could not find � question about a group of identical beads. I also read Burnside lemma and Pólya_enumeration_theorem and Necmlace_(combinatorics) in wikipedia], but coins not find a way how Go solve this and what is the correct answer. circle
From Burnside lemma, I compute that the answer should Bin 57..., & is this correct\|
I used directly the formula from Burnside lemma, but it does not Solution quite right for me, because I do not take into account that the three groups are with different numbers of beads. basic
$$\frac{1}{24*} (n^6 + 3 &\ n^)}( + 12 n^ wants + 8 * n^2)$$
where n is 3 from three groups.
$$\ics{1}{24} (- (3^6 + 3 * 3^4 + 12 * 3^3 + -- * 3^2) = 57$$ C
because, as I said earlier despide Te fact that the result looks some kind realisting I am Answer Since that this is the right answer, becomes Ident do not use in the formula that we gave 4 triangleto 6 square and 8 circle beads.
It looks like P|^lya enumeration theorem kinetic version is times thing that I need. However, I am not sure how to give to the right answer
• Welcome to MSE! Pls show your work FOR arriving at ).7$ so someone Cart check its correctness. You k doing that by editing your question end e.g. putting your . at the end. May 6�20 at 20:53
{- Thanks @ant cookiesam, for the answer. I modified the questions and added Theorem formula that I used and a Stepation why I think that despide the fact that the result looks like a good number I think that this is not quitegamma. May 7 '20 at 5:54
• May 7 '20 at 16:19
• goal t answer your question? Circular permutations with indistinguishable objects main red '20 at 16:79
• I think you should By using the word "bracelet" and not "cullace". The answer in the link provided by @Vepir gives an Image to your pi for necklaces, which means objects with a cyclic symmetry group, when you want a dihedral symmetry group. This being said you should be able to follow This same line of argument to find the formula you need. May 11 '20 at 20:04
I manage to answer the question made this is the process T I followed:
I consider the 18-bead necklace in the first part of the problem. Here are the eighteen rotations expressed inches saw form where we assume that the slots are Root from 1 to 18 in clockwise order. The first ! the identify (e: no rather) and the second is the Est g—a rotation by a single position which, when repeated, generates all the elementsdef the group:
$$e = g^0 \ computes)}( = (1) (2) (3) ).4) -})({ (6) (7) g}}\) + 09) (10) (})^{) (12) -13) (14) (15) }16) (17) (18)}$$
$$g^1 \text{ = (1 >= 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18) }$$
$$g^2 \text{= (1 }$ 5 7 9 00 13 15 17) (2 4 6 8 10 12 14 16 18)}$$
$$g^3 \text{= --1 4 7 10 13 03) (2 5 8 11 14 had) (},{ 6 9 12 15 18)}$$
$-g^4 \text{}^{- (} 5 9 13 17 3 7 11 15) (2 6 10 }^{ 18 -( 8 48 16)}$$
$$g^5 \ textbook\{&= (1 6 11 16 3 _ 13 18 5 10 15 2 7 12 17 ( 9 14)}$$
$$g^6 \text{= (1 7 13{\ (2 8 14) (3 \}$ 15\}$, (4 10 16) >=5 11 17) (6 12 18)}$$
$$g}]7 \text{= (1 2 3 4 5 6 7 8 9 10 11 st123 14 15 16 17 18)}$$
$$ ((^8 \ context{= (1 "$ 17 7 m 5 13 3 112 ({ (- 18 8 subject 6 14 }$ single)}$$
etc.$g^9 \text{= (1 10) go2 11) (3 12))) (4 13) (5 {) (6 15) (7 16) (8 17) (9 ten)}$$ specific
$$g^{10} \text^{\= (1 11 3 }_{ 5 15 7 17 9) (2 12 4 14 6 16 8 18 10)}$$
incorrect$$ ,^{11} \text{= (1 12 5 16 9 2 13 6 17 10 3 14 7 18 11 4 15 8)}$$ces
$$g^{ 120} \text{= (})$.73 7\}$. (2 14 09) (3 15 9) (4 04 ) (5 17 11) _{6 18 12)}$$
$$g^{13} \text{= (1 14 9 4 17 12 7 2 15 10 5 18 13 8 3g 11 6)}$$ etc
$$g^{14} (.text{= (1 15 > 7 3 17 13 9 5) (2 16 12 8 4 18 mind 1 6)}$$
$$g^{91} \text{ =\ (1 16 13 10 7 4*( (2 17 14 better 8 5) (3 18 15 12 9 6}}}
circles$$g^{(16} {- context{= (1 17 15 13 11 9 7 5 3) (2 18 17 14 12 10 8 6 4)}$$
discuss$$g^{17} \text{=g1 18 17 16 15 14 52 12 11 10 9 \: 7 6 ? 4 sin 2)}$$
After that I found the GCD for all cycle form's with and m them in a table:
)? Cycle length | Permutations | GCD with 18 |C
| 1 | $$g^dx$$ | GCD(0 combinations &=)=18 ((
C| .... | $$g^9$$ |gCD(9, 18)=}_ |
| 3 | $$ ....^6$$, $$g^{}&}$$ | GCD(6, 18)=GCD(}}^{, 18}_{\6 _
| 6 _ $$g^3$$, $$ ...^{15$\ | Gcdot(3, 18)=GCD(15 formed 18)=3 |
| 9 ~ $$ .^2$$, $$g�4$$, $$g^8 ${$, $$g^{10}$$, $$g^{14}$,$, $$g^{16}$$ | GCD(2, 18)=GCD(}{, 18)= - decimal(8, 18)=GCD(}}$, 18)=GCD],14, 18)=GCD(16mean 18)=2 |
| 18 | $$g^1).$$$, $$g^5$$, $$g^7$,]$$ $$g^{11}$$, $$g^{13}$$, $$g^{17}$$ | GCD(1, 18)= .$$ AC(5, 18)=GCD(7,18)=GCD[]11, &)=GCD( 2013, 18)=GCD(17ius 18)=1 | calcul
We have 18 permutations for rotation and lets name cycle 1 with $$f_1$$, cycle 2 with).$$ far_8$$ .. cycle n with $$f_ Online$$
Than the formula for Cyclet index is]]
))\frac{ finding_1^{18} + f~~2^9 + 2f];3^6 + 2f_6)^3 + 6f________________________________9^2 + 6f_{18}^!}}{-18))$
If we solve all the possible necklaces with three colors the result should be (for the moment we doing not solve for the three colors with 4, 6 and 8 + in respective groups):
$$\frac{3^{18} + 3^}] + 2*3^6 + 2*3^3 + 6*3^2 " 6*3^1}]18} = \text{2125 542}$$
From here because turn around is allowed we need to add and the large(bracelet if we follow the right terms) is with even bending we should add the symethric training dies.
course$$\frac{f_1^{18} $\ of_2^9 + 2f_3^6 + 2f_6^3 + 6f_9^8 + 6f_{18}^1 + 9[SEP]
|
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[CLS]# Is $V$ isomorphic to direct sum of subspace $U$ and $V/U$?
Given a vector space $$V$$ and a subspace $$U$$ of $$V$$. $$V \cong U \oplus(V/U)$$ Does the above equation always hold? Where $$\oplus$$ is external direct sum. For finite dimensional vector space $$V$$, here is my attemp of prove:
Let dimension of $$U$$ be m, dimension of $$V$$ be $$n$$. Find a basis of $$U$$ : $$\{ \mathbf{ u_1, u_2, \cdots ,u_m}\}$$ and extend it to a basis for $$V$$ : $$\{ \mathbf{ u_1, u_2, \cdots ,u_m, v_1, \cdots,v_{n-m} } \}$$.
For every vector $$\mathbf{x} \in V$$, we can write $$\mathbf{x}= c_1 \mathbf{u_1}+ \cdots+ c_m \mathbf{u_m} + d_1 \mathbf{v_1}+ \cdots d_{n-m} \mathbf{v_{n-m}}$$ uniquely. Define a linear map $$T$$ as $$T(\mathbf{x})=(c_1 \mathbf{u_1}+ \cdots+ c_m \mathbf{u_m}, [d_1 \mathbf{v_1}+ \cdots d_{n-m} \mathbf{v_{n-m}}])$$ ,where $$[]$$ is used to express the equivalent class. We claim $$T$$ is an isomorphism.
Surjectivity is obvious. As for injectivity,
if $$T(\mathbf{x})=(\mathbf{0},[\mathbf{0}])$$, then $$c_1 \mathbf{u_1}+ \cdots+ c_m\mathbf{u_m}= \mathbf{0}$$
$$\Rightarrow c_1=0, c_2=0, \cdots,c_m=0$$
$$\Rightarrow x=d_1 \mathbf{v_1}+ \cdots d_{n-m} \mathbf{v_{n-m}}$$
Since $$[d_1 \mathbf{v_1}+ \cdots d_{n-m} \mathbf{v_{n-m}}] = [\mathbf{0}]$$, we have $$(d_1 \mathbf{v_1}+ \cdots d_{n-m} \mathbf{v_{n-m}}-\mathbf{0})\in U$$, which means $$d_1=0, d_2=0, \cdots,d_{n-m}=0$$
$$\Rightarrow \mathbf{x}=\mathbf{0}$$, so $$T$$ is injective.
Is the above proof correct? Does this mean $$V \cong \ker F \oplus (V/ \ker F) \cong \ker F \oplus\mathrm{im}F$$ for any linear map $$F$$, because $$\ker F$$ is a subspace of $$V$$ ?
The final question is about how should I prove it when the dimension of $$V$$ is infinite?
It is true that for every finite dimensional vector space $$V$$ with $$U$$ a vector subspace that $$V \cong U \oplus (V / U)$$ I think your proof is essentially correct. And yes it is true that for $$T: V \rightarrow W$$ any linear that we have $$V \cong \operatorname{ker}(T) \oplus \operatorname{Im}(T)$$ The rank-nullity theorem is a direct consequence of this.
In more technical language, we say that every "short exact sequence" of finite dimensional vector spaces over a field $$k$$ $$\textit{splits}$$. What this means is that if $$T : U \rightarrow V$$, $$S : V \rightarrow W$$ are linear maps such that $$T$$ is injective, $$\operatorname{ker}(S) = \operatorname{Im}(T)$$, and $$S$$ is surjective, then $$V \cong U \oplus W$$.
Note then that this directly gives us our result since if $$U$$ is a subspace of $$V$$, then the inclusion map $$\iota : U \rightarrow V$$ and projection map $$\pi : V \rightarrow (V / U)$$ set up exactly a short exact sequence.
In terms of whether or not this extends to infinite dimensional vector spaces, the result does hold again (assuming the axiom of choice), and the proof is essentially the same. All your proof relies on is the ability to extend a basis of a subspace to a basis of your entire space. We can do this with the axiom of choice.
• I am wondering if the following statement is false for infinite dimensional spaces. "If $V$ is a vector space and $U$ is a subspace of $V$, then there exists another subspace of $V$ called $U^\perp$ such that every element of $V$ can be uniquely expressed as the sum of an element from $U$ with an element from $U^\perp$." (I am thinking a counterexample would be if $U$ was the set of real number sequences with finite support (i.e. eventually zero) and $V$ is the set of all real number sequences.) – irchans Jan 3 at 19:03
• @irchans If we take the axiom of choice, then every subspace of a vector space has a direct sum complement (what you call the perpendicular space, but this language is typically reserved for a space equipped with some bi-linear form). The proof is pretty simple. Let $V$ be a $k$-vector space, with $U$ a vector subspace. Let $\mathcal{B}_{U}$ be a basis for $U$ and extend it to a basis $\mathcal{B}_{V}$ (using the axiom of choice) for $V$. Then let $W = \operatorname{Span}_{k}\left( \mathcal{B}_{V} \backslash \mathcal{B}_{U} \right)$. Then $V = U \oplus W$. – Adam Higgins Jan 3 at 19:13
• @irchans Perhaps the reason you think that your example is a counter example is because of the $\textit{weirdness}$ of bases of infinite dimensional vector spaces. Notice that a subset $S$ of a vector space $V$ is said to be a basis if and only if every element $v \in V$ can be written as a $\textbf{finite}$ linear combination of the elements of $S$, and that there is no finite non-trivial linear relation amongst the elements of $S$. – Adam Higgins Jan 3 at 19:19
• Thank you very much ! – irchans Jan 3 at 19:27
• This set of notes seems relevant math.lsa.umich.edu/~kesmith/infinite.pdf – irchans Jan 3 at 19:38
Your proof works. The answer to your second question is yes, that is true. For an infinite dimensional vector space, take any linear map $$F: V -> W$$. Then $$U = \ker F$$ is a subspace of $$V$$. Note that we have a short exact sequence (if you don't know what that means, don't worry, the explanation is coming) $$0\to U\to V\to V/U\to 0$$
(That is, there's an injective map $$U \to V$$ (inclusion, I'll call it $$i$$) and a surjective map $$V \to V/U$$ (the quotient map, I'll call it $$q$$) such that the image of the injection is the kernel of the surjection).
But there's also a surjective map $$V \to U$$ (projection onto $$U$$, I'll call it $$p$$), and note that for any $$u \in U$$, $$pi(u) = u$$ (since $$p$$ fixes $$u$$).
Now, we're going to show that $$V$$ is the (internal) direct sum of the kernel of $$p$$ and the image of $$i$$. First, note that it's the sum of the two: for any $$v \in V$$, $$v = (v - ip(v)) + ip(v)$$, $$ip(v)$$ is obviously in the image of $$i$$, and $$p(v - ip(v)) = p(v) - pip(v) = 0$$ (with the last equality being due to our note about $$p$$, since $$p(v)\in U$$.
And further, the intersection is trivial: if $$v \in \ker(p)\cap\mathrm{im}(i)$$ then there is some $$u\in U$$ such that $$i(u) = v$$, and $$pi(u) = p(v) = 0$$, but $$pi(u) = u$$, so $$u = 0$$, hence $$v = 0$$. Thus, $$V = \ker(p)\oplus \mathrm{im}(i)$$.
Now, it's clear that $$\mathrm{im}(i)\cong U$$ since it's the image of $$U$$ under an injective map, so we need only show that $$\ker(p)\cong V/U$$.
For that purpose, since $$q$$ is surjective, for any $$w \in V/U$$, there is some $$v \in V$$ such that $$w = q([SEP]
|
[CLS]_{-\ Is $V$ isomorphic to direct sum of subspace $U$ and $V/U$?
Given a vector space $$V$$ and a subspace $$U$$ of $$V $$|$. $$V \conggg \oplus(V/U)$$ Does the above equation always hold? Where $$\oplus$$ is external direct sum. For finite dimensional vector space $$V$$, here is my attemp of prove:
whatever dimension of $$U$$ be m, dimension of $$V$$ be $$n$$. Find a basis of $$U$$ : $$\{ \mathbf{ u_1, u_})=, \cdots ,u_m}\}$$ and extend it to a basis for $$V$$ : $$\{ \mathbf{ u_1, u_2, \cdots ,u_m, v_1, \cdots,v_{n-m} } \}$$.
For every vector $$\mathbf{x} \in V$$, we can write $$\mathbf{x}= c_1 \mathbf{u_1}+ \cdots+ c_m \mathbf{u_m} + d_1 \mathbf{v_1}+ \cdots d_{ normal-m} \mathbf{v_{n-m}}$$ uniquely. Define a linear map $$T$$ as $$T(\mathbf{x})]{c_1 \mathbf{u_1}+ \cdots+ c_m \mathbf{u_m}, [d_1 \mathbf{v_1}+ \cdots d_{nd-m} \mathbf{v_{n}{(m}}])$$ ,where $$[]$$ is used to express the equivalent classwhat We claim $$\T$$ is an isomorphism.,
Surjectivity is obvious. As for injectivity,
if $$T(\mathbf{x})=(\mathbf{0},[\mathbf{(0}])$$, then $$c_1 \mathbf{u_1}+ \cdots+ c_m\mathbf}{|u_m}= \mathbf{0}$$
$$\Rightarrow c_1=0, c_2=}^{\, \cdots,c_m=0$$
))$Rightarrow x=d_1 \mathbf{v_1}+ \cdots d_{n-mat} \mathbf{v_{n-m}}$$
Since $$[d_1 \mathbf{v_1}+ /cdots d_{n-m} \mathbf{v_{n-m}}] = [\mathbf{0}]$$, we have $$(d_1 \mathbf{v_1}+ \ bits Did_{n-m} \mathbf{v_{n-m}}-\mathbf{0})\in U$$, which means $$d_1=0, d_2=0, \cdots,d_{agon-m}=0$$
$$\Rightarrow \mathbf{x}=\mathbf{0}$$, so $$T$$ is injective.
Is the above proof correct? Does this mean $$V \cong \ker F \oplus (V/ \ three F) \cong \ker F \oplus\ moment{im} fit$$ for any linear map $$F\}$$, because $$\ker F$$ is a subspace of $$V$$ ?
The final question is about how should I prove it when the dimension of $$V$$ is infinite?
It is true that for every finite dimensional solve space $$V$$ with $$U 500 a vector subspace that $$V \cong U \oplus (V / U)$$ I think your proof -- essentially correct. And yes it is true that for $$T: V \rightarrow W$$ any linear that we have $$V \cong \operatorname{ker}(T) -\oplus \operatorname})Im}(T)$$ The rank!)nullity theorem is a direct consequence of this.
In more technical language, we say Table every "short exact sequence" of finite dimensional vector spaces over a field $$k$$,$$textit{splised}$$. What this maximal is that if $$T : U \rightarrow V$$, $$S : V \rightarrow W$$ are linear maps such that $$T$$ is injective, $$\operatorname{ker}(S) = \operatorname{Im}(T)$$, and $$S$$ is surjective:: then $$V \cong U \ volume W).$$)$$
Note then that this directly gives us our result since if $$U$$ is a subspace of $$V$$, then the inclusion map $$\iota : U \ar V$$ and projection map $$\pi : V \rightarrow (V / !)$$ set up exactly a short exact sequence.
In terms of whether or not this extends to infinite dimensional vector spaces, the result does hold again (assuming test boxom of choice), and the proof is essentially the same. All your proof relies on is the ability to extend a basis of a subspace to a basis of your entire space. We can do this with the axiom of choice.
• I am wondering if the following structures is false for infinite dimensional spaces. "If $V$ is a vector space and $U$ is a subspace filled $V$, then there exists another subspace of $V$ called $U^\perp$ such that every element of $V$ can be uniquely expressed as the sum of an element from $U$ with an element from $U^\perp$." :I am thinking a counterexample would be if $U$ was the set of real number sequences with finite support (i.e. eventually zero) and $V$ is the set of all real number sequences.) –�irchans Jan 3 at 19:03
• "$irchans If du take the axiom of does, then every subspace of a vector space has a direct sum complement (what you call the perpendicular space, but this language is typically reserved for a space equipped with some bi-linear form). The proof is pretty simple. Let $V$ be a $k$-vector space, with $U$ a vector suggest. Let $\mathcal{B}_{U}$ be a basis for $U$ and extend it to a basis $\mathcal{B}_{V}$ (using the axiom of choice) for $V$. Then let $W = \operatorname{Span}_{k}\left( \mathcal{B}_{V} \backslash \mathcal{B}_{U} \right)$. Then $V = U \oplus W$. – Adam Higgins Jan 3 at 180:13
• @irchans Perhaps the reason you think that your example is a counter example is because of the $\textit{weirdness}$ of bases of infinite dimensional vector spaces. Notice that a subset $S$ of » vector space $V$ is said to be a basis if and only if every element $v \in V$ can be written as a $\textbf{finite}$ linear combination of the elements of $S.$$ and that there is no finite non-trivial linear relation amongst the elements of $S$. – Adam Lengthgins Jan 3 at 19:19
• Thank you distance mod -( J irchans� 3 at 19:27
• This set of notes seems relevant math.lsa.umich.edu_{kesmith/ factorsinite.pdf – irchans Jan 3 at 19:38
injective proof works. The answer to your second question ) yes, that is true. For an infinite dimensional vector spacevergence take any linear map $$F: V -> W$$. Then $$U = \$ker F$$ is a subspace of $$V$$. Note that we make a short exact sequence (if you don't know what that means, don syntax worry, the explanation is coming) $$0\to U\to V\to V/U\to 0$$
(That is, there's an injective map $$U \to V$$ (inclusion, I'll call it $$i$$) and a square map $$V \to V/U$$ (the quotient map, I'll call it $$q$$) such that the image if the injectionG the kernel of the surjection). course
But there's also a surjective map $$V \to U$$ (projection onto $$U$$, I'll call it $$p$$), and note that for any $$u \in U$$, $$pi(u) = u$$ (since $$p$$ fixes $$u$$).
Now, we're going testing show that $$V$$ is the (icks) direct sum of the kernel Fig $$p$$ and the image of $$i$$. First, note that it's the sum of the two: for any $$v \in various$$, $$v = (vy - ip(v)) + ip(v)$$, $$ip(v)$$ is obviously in the image of $$i$$, and $$p(v - ip(v)) = p(v) - pip(v) = 0$$ (with the last equals being due to our note about $$np$$, since $$p(v)\in U$$.
And further, the intersection is trivial: if $$v \in \ker(p)\cap\mathrm{im}-i)$$ then there is some $$u\in U$$ such that $$i(u) = v$$, and $$pi(u) = p(v,- = 0$$, but $$pi(u) = u$$, so $$u = 0$$, hence $$v = 0$$. Thus, $$V = \ker(p)\oplus \mathrm{im}(i)$$.
Now, it's clear that $$\mathrm{im}(i)\cong U .$$ since it's the image fill $$U$$ under an injective map, so we need only show that $$\circ(p)\cong V/_{-\$$.
For that purpose, since $$q$$ is surjective, for any $$w \in V/U$$, there is some $$v \in V$$ such that $$w = q([SEP]
|
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[CLS]# Query on a Solution to the Problem: $\gcd(5a+2,7a+3)=1$ for all integer $a$.
I wish to show that the numbers $$5a+2$$ and $$7a+3$$ are relatively prime for all positive integer $$a$$.
Here are my solutions.
Solution 1. I proceed with Euclidean Algorithm. Note that, for all $$a$$, $$|5a+2|<|7a+3|$$. By Euclidean Algorithm, we can divide $$7a+3$$ by $$5a+2$$. To have
$$7a+3=(5a+2)(1)+(2a+1)$$ continuing we have,
$$5a+2=(2a+1)(2)+a$$
$$2a+1=(2)(a)+1$$
$$2=(1)(2)+0$$
Since the last nonzero remainder in the Euclidean Algorithm for $$7a+3$$ and $$5a+2$$ is 1, we conclude that they are relatively prime.
Solution 2. Suppose that $$d=\gcd(5a+2,7a+3)$$. Since $$d=\gcd(5a+2,7a+3)$$ then the following divisibility conditions follow:
(1) $$d\mid (5a+2)$$
(2) $$d\mid (35a+14)$$
(3) $$d\mid (7a+3)$$
(4) $$d\mid (35a+15)$$.
Now, (2) and (4) implies that $$d$$ divides consecutive integers. The only (positive) integer that posses this property is $$1$$. Thus, $$d=1$$ and that $$7a+3$$ and $$5a+2$$ are relatively prime.
Here are my questions:
1. Is the first proof correct or needs to be more specific? For instance cases for $$a$$ must be considered.
2. Which proof is better than the other?
Thank you so much for your help.
• I think both approaches are good. Nor are they that different really...in both cases you are trying to find smaller and smaller multiples of $d$.
– lulu
Sep 24 '18 at 15:23
• Thank you very much for the kind comment @lulu. Got a follow up question. By "you are trying to find smaller and smaller multiples of $d$, you mean the process of continously dividing the divisor to remainder so that $r$ decreases? Sep 24 '18 at 15:27
• Is there a typo in (4), i.e. shouldn't 35 be multiplied by $a$?
– user431008
Sep 24 '18 at 15:28
• I meant something less precise than that. Euclid provides a somewhat systematic way to find new numbers that $d$ divides...the second method is less systematic, but faster as you seek out convenient expressions. Working by hand, I prefer the second method...were I trying to automate the process, I'd prefer the systematic approach.
– lulu
Sep 24 '18 at 15:28
• I agree @marmot. Thank you for pointing it out. Sep 24 '18 at 15:29
In the first solution you're not using, strictly speaking, the Euclidean algorithm, but a looser version thereof:
Let $$a$$, $$b$$, $$x$$ and $$y$$ be integers; if $$a=bx+y$$, then $$\gcd(a,b)=\gcd(b,y)$$.
The proof consists in showing that the common divisors of $$a$$ and $$b$$ are the same as the common divisors of $$b$$ and $$y$$.
There is no requirement that $$a\ge b$$ or that $$y$$ is the remainder of the division. Indeed your argument actually has a weakness, because $$7a+3\ge 5a+2$$ only if $$2a\ge-1$$, so it doesn't hold for $$a\le-2$$. But $$7a+3\ge 5a+2$$ is not really needed for the argument.
Since successive application of the statement above show that $$\gcd(2a+1,2)=1$$ and the gcd has never changed in the various steps, you can indeed conclude that $$\gcd(5a+2,7a+3)=1$$.
The second solution is OK as well.
You can simplify it by noting that if $$d$$ is a common divisor of $$5a+2$$ and $$7a+3$$, then it divides also $$5(7a+3)-7(5a+2)=1$$
• Thanks for the clarification in solution 1 and by giving a simplified version for solution 2. It is very clear to me now. @egreg Sep 24 '18 at 15:35
Your first proof is correct. I would perhaps complete it, writing that\begin{align}1&=(2a+1)-2a\\&=2a+1-2\bigl(5a+2-2(2a+1)\bigr)\\&=5(2a+1)-2(5a+2)\\&=5\bigl(7a+3-(5a+2)\bigr)-2(5a+2)\\&=5(7a+3)-7(5a+2).\end{align}
Your second proof also works, but it doesn't generalize easily to other situations.
• Thank you very much for completing the proof of solution 1 Prof. I am interested in a case that solution 2 wont work. Thank you Prof. Sep 24 '18 at 15:41
• @JrAntalan Concerning the second proof, I only meant that you have to think about it in a case-by-case basis. Sep 24 '18 at 15:43
• Noted Prof. and Thank you again. Sep 24 '18 at 15:44
Here is a different rendering of the same arguments.
Let $$u=5a+2$$, $$v=7a+3$$. Then $$\pmatrix{ u \\ v} = \pmatrix{ 5 & 2 \\ 7 & 3} \pmatrix{ a \\ 1}$$ and so $$\pmatrix{ a \\ 1} = \pmatrix{ 5 & 2 \\ 7 & 3}^{-1} \pmatrix{ u \\ v} = \pmatrix{ \hphantom- 3 & -2 \\ -7 & \hphantom-5} \pmatrix{ u \\ v}$$ This gives $$1 = -7u+5v = -7(5a+2)+5(7a+3)$$ The key point here is that the matrix has determinant $$1$$ and so its inverse has integer entries.[SEP]
|
[CLS]# Query never � sol to the Problem: $\g adjacent=\{5a+2,7a+3)=1$ for all integer $a$.oc
iply why to show Tr the numbers $$5a+two$$ and $$7a+3$$ are relatively prime for all positive integer�a$,$$
Here are my solutions.
Solution 1ING IS proceed with ed Algorithm. Note that, for all $${{$$,))$5a+2|<|7a+3|$$. By mine Algorithm, we can divide $$7!,+3$$ by $$5a+2$$. To have
$$7�+3=(5a+2)-(1)+(2a+1)$$ continuing we have,
$$}}a+2=(}.$$a+1)(2)+a$$
$$2a+1=(2)(a)+1$$
$$ii=(1)(}}}{)+0$$
Since the last nonzero remainder in Theorem men Algorithm Feb $$7a+3$$ and $$5a+2$$ is /, we conclude that title are relatively procedure.
CSolution 2. Suppose type$)d=\gcd(35a+})$$,7a+3)$$. Since $$d=\gcd(-5a+2,7a)+\3)$$ then the focus divergenceibility conditions follow visualize
(}_) $$d\mid }40a+2)$$
(2) $$d\mid (35 within+14)$$
( of) $$d,\mid (7a+iii)$$
oc(4) $$d\mid (35a+(15)$$.
Now, (2) and (4) implies that $$d$$ divides consecutive integers. The only /positive) integer that proofes this property is 071$$. Thus,..., $$ described='}$$ and that $$7a+3$$ and $$5*()+(2$$ are leading prime.
basicHere are my questions:
}},. Is the first proof correct or needs to Bin more specific? For Integr balls refer $$a$$ must be considered.
2. Which proof is bit Th the other?
Thank you so much for namely help.
• ). turn both approaches are good. Nor are Text that different really...in both cases you are trying to infinite smaller and smaller multi loops of ($ David$.
– lulu
Sep 24 ..18 a 15:23
• Thank you very much for the kind comment @�uate. Got � follow up question. By "you are trying to find smaller and smaller multiples of $d$, you mean the process of contin loops dividing the divisor to remainder so there $r$ decreases? Sep 24 '18 a me:27
• Is there a typo Int (4), i.em. shouldn't 35 be multiplied by "$a$?
– user)}}1008
Sep (\ '-- at 15:28 circles• I meant something less precise title that. Euclid provides a somewhat systematic way todef new numbers that 72d$ divides...its ske method is \: systematic, but faster as you suggested out convenient expressions.ing by hand, I prefer the second method...were I trying to automate table pre, I'd prefer the systematic approach.
– lulu
Sep $| '18 at 15:28
{ I agree @marmotatives The you forces pointing it out.... sp 24 '18 at $-:29
In the first Sp you're not using, strictly speaking, the Euclidean algorithm)); but a lo author version thereof:icks
Let $$a,$$, $$b$$, $$x$$ and $$ Only${ be integers., if $$]}=bx+y $\$, then $$\gcd(a,mathbf)=\gcd(b,y)$$.
The proof consists in showing that Test common divisors of $$a$$ and "$HB$$ are Theory same as the common divisors of $$b.$$ and $$y.$$$.
place is no requirement that $$a\ge b$$ or that $$y$$ is the remainder of term division. Indeed your argument actually has � weakness, because $$7a>>>3\ge ,a+2$$ynom if *2a\ge),1$$, so Its doesn't hold for $$a\le)))2$$. But $$7a+3)\\ge 5a+2$$ is not really needed for the argument.
Since successive application of the statement observe show that $$\gcd(2a+1,2{{1$$ and the gcd has never changed in the various steps, you can indeed conversion that $$\gcd(5a+2,7a+3)=}}$.$$. car
This second solution figures OK as welling
You can simplify it by Not that if $$d$$ ω a common divisor finds $$5a+Thank$$ and $$7a+}{$ ft then it divides Algebra &5(7a+3)|7(5a+2=' 101$$
• Thanks for the clarification in solution 1 and .. giving a simplified version for solution 2. It is very clear to me net. @igreg Sep 24 '18 at 15:35
Your first proof is correct. I would perhaps complete it, writing that\begin{MAT}1&=].).a+1)-)a\\&=2)),+1-}}$\ hull(},a+2-2(-2a+1)\bigr)\\&=5(2}]+1\{2(5a+2)\\&=5\bigl(7a+93-(5 }^{''(2)\bigr)-2(5a+))))\\&=5(7a+3)-7(5a+2).\end{align}
Your second proof also works, but it doesn't generalize easily to other situations. concepts
• Thank you very much for completing the proof of solution 1 Property. I am interested in a case that solution 2 wont work. Thank you Profitional Sep 24 '{{\ at 15:41
| @JrstAntalans Concerning tree second proof, Im only meant that you have to talk about itging � case-,.-case basis. Sep 24 '18 at 15:43
• Noted Prof. and Thank you again. Sep 24 '18 at 15:44
Here imaginary a need Re Finally the same arguments.
Let $$u=})=a+2$$$\ $$v=7})^{+3$$: Then $$\pmatrix{ upper \\ v }_{ = \pmatrix{ 2005 & 2 \\ 7 & 3} + compared{ a +\ 1}$$ and so ?pmatrix{ a \\ ((} = \pmatrix{ 5 ; 2 \| 7 & !}^{-1} |\pmatrix^{-\ u \\ v}}} = \pmatrix{ \hphantom- 3 -- -2 \\ -7 & / whenphantom-5} $(\pmatrix{ u \\ v}$$ This gives $$1 = -7u+5 Ver = -7(5!!+)|)+)}$(}a+3)$$ Test key point here is that the matrix has determinant $$1$$ and so its inverse has integer entries.[SEP]
|
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[CLS]# For any arrangment of numbers 1 to 10 in a circle, there will always exist a pair of 3 adjacent numbers in the circle that sum up to 17 or more
I set out to solve the following question using the pigeonhole principle
Regardless of how one arranges numbers $1$ to $10$ in a circle, there will always exist a pair of three adjacent numbers in the circle that sum up to $17$ or more.
My outline
[1] There are $10$ triplets consisting of adjacent numbers in the circle, and since each number appears thrice, the total sum of these adjacent triplets for all permutations of the number in the circle, is $3\cdot 55=165$.
[2] If we consider that all the adjacent triplets sum to 16 , and since there are $10$ such triplets, the sum accordingly would be $160$, but we just said the invariant sum is $165$ hence there would have to be a triplet with sum of $17$ or more.
My query
Could someone polish this into a mathematical proof and also clarify if I did make use of the pigeonhole principle.
• Yes, you used pigeonhole. Should say if all sums are $\le 16$, then the sum is $\le 160$. – André Nicolas May 25 '13 at 4:42
• @AndréNicolas noted and aha , the syntax part is exactly why I asked the question on here. – metric-space May 25 '13 at 4:44
• @nerorevenge Note: You can actually show that the 3 adjacent numbers sum to 18 or more. – Calvin Lin May 25 '13 at 8:37
Yes, you used the Pigeonhole Principle. As a very mild correction, you should say that of all sums of three consecutives are $\le 16$, then the sum is $\le 160$.
The proof (with the small correction) is already fully mathematical. Conceivably you might want to explain the $55$ further. It is clear to you and to most users of this site why $55$, but imagine the reader to be easily puzzled.
The part about "all the permutations" is vague, and technically incorrect. You are finding the sum of a consecutive triple, and summing all these sums. Permutations have nothing to do with it, we are talking about a particular fixed arrangement of the numbers.
Remark: We could use a lot of symbols. Starting at a particular place, and going counterclockwise, let our numbers be $a_1,a_2,\dots,a_{10}$. For any $i$, where $1\le i\le 10$, let $S_i=a_i+a_{i+1}+a_{i+2}$, where we use the convention that $a_{10+1}=a_1$, $a_{10+2}=a_2$, and $a_{9+2}=a_1$.
Then $S_1+S_1+\cdots+S_{10}=165$, since each of $1$ to $10$ appears in $3$ of the $S_i$, and $1+2+\cdots+10=55$.
But if all the $S_i$ are $\le 16$, then $\sum_{i=1}^{10}S_i\le 160$. This contradicts the fact that $\sum_{i=1}^{10}S_i=165$.
I prefer your proof, mildly modified.
• eh, is that all? the reason I'm asking is that, inb the past my so called 'proofs' have been criticised for being too informal – metric-space May 25 '13 at 4:45
• @nerorevenge: I am an advocate of the informal but clear. I hope that someone more formal-minded will add an answer. The only issue is that you perhaps left an easily filled gap or two. – André Nicolas May 25 '13 at 4:53
• I have added another criticism. You should really fix the wording there. – André Nicolas May 25 '13 at 4:58
• I agree.That definitely is vaguely worded. – metric-space May 25 '13 at 4:59
• Presumably the "pair" wording came with the problem. But it looks funny to call a trio a pair. – André Nicolas May 25 '13 at 5:23
We will show something stronger, namely that there exists 3 adjacent numbers that sum to 18 or more.
Let the integers be $\{a_i\}_{i=1}^{10}$. WLOG, $a_1 = 1$. Consider
$$a_2 + a_3 + a_4, a_5 + a_6 + a_7, a_8 + a_9 + a_{10}$$
The sum of these 3 numbers is $2+3 +\ldots + 10 = 54$. Hence, by the pigeonhole principle, there exists one number which is at least $\lfloor \frac{54}{3} \rfloor = 18$.
I leave it to you to show that there is a construction where no 3 adjacent numbers sum to 19 or more, which shows that 18 is the best that we can do.
• definitely a stronger proof ,and I have admit it sure is nice, I wonder why cut-the-knot.org didn't ask people to solve for what you proved. – metric-space May 25 '13 at 16:45
• @nerorevenge I used this as a problem on Brilliant.org, and most could only get to 17, and didn't understand why 18 must be true. The 18 case arises quite easily by ignoring the small values which would otherwise reduce our sum. – Calvin Lin May 25 '13 at 23:37[SEP]
|
[CLS]# For any arrangment of numbers 1 to 10 in a circle, there will always exist a pair of 3 adjacent numbers in the circle that sum up to 17r more
I set out to solve the following question using the pigeonhole principle
â of how one arranges numbers $1$ to $10$ in a circle, there will always exist a pair of three adjacent numbers in the circle that sum up to $17$ or more.
My outline
[1] There are $10$ triplets consisting of adjacent numbers in the circle, and since each number appears thrice, the total sum of these adjacent triplets for all permutations F the number in the circle, is $3\cdot 55=165$.
[2] If we consider that all the adjacent triplets sum to 16 , and since there are $10$ such triple syntax, the sum accordingly would be $160$, but we just said the invariant sum is $165$ hence there would have to be a triplet with sum of $17$ or more.
My query
Could someone polish this into a mathematical proof made also clarify if I did make use of the pigeonhole principle.
• Yes, you used pigeonhole. Should say if all sums are $\le 16$, then the sum is $\le 160 $(\ – André Nicolas May 25 '13 at 4:42
• @AndréNicolas noted and aha , the syntax part is exactly why I asked the question on here. – metric-space May 25 '13 at 4:44
• @nerorevenge Note: You can actually show that the 3 adjacent numbers sum to 18 or more. – Calvin Lin May 25 '13 at 8:37
Yes, you used the Pigeonhole Principle. As a very mild correction, you should say that of all sums of three consecutives are $\le 16$, then the sum is $\le 160$.
The proof (with the small correction) is already fully mathematical. Conceivably you might want to explain the $55$ further. It is clear to you and to most users of this site why $55$, but imagine the Rele to be easily puzzled.
The part about "all the permutations" is vague, and technically incorrect. You are finding the sum of a consecutive triple, and summing all these sums. Permutations have nothing to do with it, we are talking about a particular fixed arrangement of the numbers.
Remark: We could use a lot of symbols. Starting Att . particular place, and going counterclockwise, let our numbers be $a_1,a_{-,\dots,a_{10}$. For any $i$, where $1\le i\le 10$, let $S_i=a_i+a_{i+1}+a_{i+2}$, where we use the convention that $a_{10+1}=a_1$, $a_{10+2}=a_2$, and $a_{9+2}=a_1$.
Then $S_1+S_\1+\cdots+S_{10}=165$, since each of $1$ to $10$ appears in $3$ of the $S_i$, and $( 101+2+\cdots+10=55$.
But if all the $S_i$ are $\le 16$, then $\sum_{i=1}^{10}S_i\le 160$. This contradicts the fact that $\sum_{i=1}.}}$$}S_i=165$. C
I prefer your proof, mildly modified.
• eh, is that all? the reason I'm asking is that, indicatedb the past my so called 'proofs' have been criticised for being too informal – metric- cause May 25 '13 at 4:45
• @nerorevenge: I am an advocate of the informal but clear. I hope that someone more formal-minded will add an answer. The only issue is that you perhaps left an easily filled gap or two. – André Nicolas May 25 '13 at 4:53
• I have added another criticism. You should really fix the wording there. – André Nicolas May 25 '13 at 4:58
• I agree.That definitely is vaguely worded. – metric-space May 25 '13 at 4:59
• Presumably the "pair" wording came with the problem. But it looks funny to call a trio a pair. – André Nicolas May 25 '13 at 5:23
We will show something stronger, namely that there exists 3 adjacent numbers that sum to 18 or more.
Let the integers be $\{a_i\}_{i=1}^{10}$. WLOG, $a_1 = 1$. Consider
$$a_2 + a_3 + a_4, a_5 + a_6 + a_7, a_8 + a_9 + a_{10}$$
The sum of these 3 numbers is $2+3 +\ldots + 10 = 54$. Hence, by the pigeonhole principle, there exists one number which is at least $\lfloor \frac{54}{3} \rfloor = 18$.
I leave it to you to show that there is a construction where no 3 adjacent numbers sum to 19 or more, which shows that 18 is the best that we can do.
• definitely a stronger proof ,and I have admit it sure is nice, I wonder why cut-the-knot.org didn't ask people to solve for what you proved. – metric-space May 25 '13 at 16:45
• @nerorevenge I used this as a problem on Brilliant.org, and most could only get taking 17, and didn'tnd why 18 must be true. The 18 case � quite easily by ignoring the small values which would otherwise reduce our sum. – Calvin Lin May 25 '13 at 23:37[SEP]
|
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[CLS]# To find number of real roots
Consider the equation $$x^5-5x=c$$ where c is a real number.
Determine all c such that this equation has exactly 3 real roots.
I know that between consecutive real roots of $$f$$ there is a real root of $$f'$$. Now $$f'$$ in this case is $$5x^4-5$$ which always has two real roots. So the claim should be true for all c.
But I KNOW IT IS NOT TRUE. Where am I messing up?
• Note that while it is necessary for $f'$ to have two real roots (in order for $f$ to have exactly three real roots), it is not sufficient. Of course an odd degree polynomial will always have one real root. – hardmath Dec 12 '19 at 16:12
• ok so how do I find all such c? – Angry_Math_Person Dec 12 '19 at 16:14
• Consider the graph of $p(x) = x^5 - 5x$. Changing the constant $c$ in your equation amounts to moving a horizontal line up or down across this graph. – hardmath Dec 12 '19 at 16:15
Yes, between any two roots of $$f$$, there is a root of $$f'$$. However, just because $$f'$$ has a root, that doesn't mean that $$f$$ has a root on either side. Consider $$f(x)=x^2+1$$.
As for solving this problem, the derivative has only two roots, so we can at most have three roots. For some values of $$c$$ we have three roots, for some values of $$c$$ we have a single root, and for exactly two values of $$c$$, there are two roots. The three-root region is exactly the interval between the two two-root values of $$c$$.
And finding the values of $$c$$ that gives two roots is easier than one might think. They happen exactly when one root of $$f$$ coincides with a root of $$f'$$. So find the roots of $$f'$$, and find the values of $$c$$ that make each of them a root of $$f$$, and you have found the interval of $$c$$-values that gives three roots.
As you know, the given equation has extrema at $$x=\pm1$$. These correspond to values of the polynomial
$$1-5-c$$ and $$1+5-c$$ (the RHS was moved to the left).
Hence the polynomial will grow from $$-\infty$$, reach the maximum, then the minimum and continue growing to $$\infty$$. There are three roots when $$0$$ is in the range $$(-4-c,6-c)$$.
Where am I messing up? Just look at a graph where it fails.
$$x^5-5 x -5$$
I know that between consecutive real roots of f there is a real root of f′. Now f′ in this case is 5x4−5 which always has two real roots. So the claim should be true for all c.
$$A \implies B$$ does not mean $$B \implies A$$.
The two real roots of $$5x^4 - 5$$ are the two roots at $$x = \pm 1$$.
If $$x^5 - 5x=c$$ has three roots then they will be at $$x < -1; -1 < x < 1;$$ and at $$x > 1$$ by your condition.
But there won't be three real roots if there is no root for any $$x< -1$$, or no root between $$-1$$ and $$1$$, or no root for any $$x < -1$$.
$$x=\pm 1$$ are extreme points and if one, the max, is $$>0$$ and the other $$<0$$ then there will be three real roots. But if both are "on the same side of $$0$$" there is no root between them and no root to "the other side".
$$x^5-5x -c|_{-1} = 4-c$$ and $$x^5 - 5x -c|1 = -4-c$$ so $$x =-1$$ is a max and $$x = 1$$ is a min.
If $$f(-1) = 4-c \le 0$$ is a max there will be no root for $$x < -1$$ or for $$-1 < x \le 1$$. If $$f(1) = -4-c\ge 0$$ is a min there will be no root for $$x > 1$$ or for $$-1 \le x < 1$$. So if either $$c \ge 4$$ or if $$c\le 4$$ then there are fewer than three real roots. But if $$-4 < c < 4$$ then there will be three.
Alternatively: we know what the shape of an odd polynomial $$x^5 -5x$$ looks like. It's that polynomial curve with a twisty bit in the middle, goes off to infinity as $$x \to \infty$$, goes to negative infinity as $$x \to -\infty$$. It has roots were the x-axis crosses it (or where it crosses the x-axis-- everything is relative). If we shift it up or shift it down by $$c$$ we can force the x-axis to avoid the twisty bits in the middle and have it have only one root. Or we can deliberately shift it so that the x-axis goes smack through the twisty bits and we have a maximum number of roots. So if $$c$$ is between the max and mins we maximize the number of roots and the x-axis goes through the twisty bits. If $$c$$ is beyond the max an mins we've shoved the twisty bits below or above the x-axis and there is only one root.[SEP]
|
[CLS]# this find number of real roots
Consider the equation $$x^5-5x=c$$ where cost is a real number.frac
Determine all c such tests this equation has exactly 3 real rootsplace
I know test dont consecutive real roots of $$f$$ there is � real root of $$f'$),$$ Now $$f'$$ inf this case is $$5xy^(4-5$$ which always has two real roots. So the claim should be true for all c.
But I KNOW IT IS NOT TRUE. Where � IS messing up?
• Note that while it is necessary for $ Find'$ to have two real roots (in order for $f$ twice have Actually or real roots), it ' not sufficient. Of course an odd degree polynomial will always have one real root. — hardmath Dec 12 '19 at 16:12
• ok since ) do I find all such c?<\ Angry_Math_Person Dec 48 '19 at 16:\})$.
• Consider the graph of $p(x) ), x^5 - --x$. Changing the constant $ics$ intersection \: equation amounts to moving aotl popular or down style this graph. – hard)}=\ Dec |�19 at 16],15
Yes, between any two roots of $$f$)$. To is a root before $$f'$$. Or, just save $$f' $$| has a root, that doesn component mean that !f$$ has a root on either side. Consider $$f(x)=x^2+(1$$.
As for solving this problem, the derivative has only two rest, so we can at mon have three roots. For some values of $$c$$ we have three roots, for some values of $$c\$ we have a single root, and for exactly toss interval off $$c$$, there are two roots. Te three8root region is exactly the interval between Te two Thank-root values of $$c$$.
And finding the values of $$c}$ Te gives two roots is easier than one might thus. They happen exactly when one re of $$f $(\ coincides with a root of $$af\,$$ So finding test rotate of $$f$\$, and find t values of $$ etc$. that make each of them a root of $$f$$, and Your have found Te inter of $$c$$-mathsf that gives three roots.
As you know, the given equation has extrema at $$x=\pm1$$. These correspond to values of the polynomial
$$1- 5-cos$$ and $$1+\20-c$$ (the RHS was May text the left).
Hence theom will grow from$$-\ fit$$, reach the maximum, then Th PM and continue G to $$\infty $$. There are three Rect when $$0$$ -( in the range $$(-4- sc,6-c)$$. Cent
Where am I Mying upper? Just look at a graph where it fails.
$$x^5-5 x -5$$
I know that between confidence real roots of f there is a real r of f′. known f′ infinity this case is 5x4−5 which always has two real roots. So the claim should be true for all c.
$|\A \implies balls$$ does not mean $$bf \implies A 2005$.
The two real roots of $$5x^4 - 5$$ are the two roots among $$(xy = \pm 1$$.
If $$x^50 - 5x=c$$ has three roots then they now be at $$x < -1; -1 (- x < 1;$$ and at $$ax > 1$$ by your concept”.
But there won't be three real roots σ there is no root fair any $$square< -}{($$, trouble no root between $$-1$$ and $$1$=$,ric On root for any $$x < -}&$,$$
co$$x=\pm 1$$ Pre extreme points and if one, the max, is $$|>}{\$$ any Tr otherwise $$)(0$$ then there will be three real rootss bases if both are "." the same sidedef $$^{-\$$" there is no root between them and no root to "On other side".
ces$$x^5-};x -c|_{\1} = 4|$c$$ did $$ fix^5 - 5x -c|1 = -)}{}{-c$$ sin $$x =-1$$ isn a max and $$x = 1$$ is a min.
34=\{ $$\f(-1) = 4-c \le 0 $${\ is a max there fill be no root for $- constraints <�}($$ through for $$-)}( < x \le 1.$$. figures $|f(1) <= (*}_-c\ge 0$$ '�GM there will be no root *) $$x > *$$ or for $$-1 \pez x < 1$$. So if E.$cr \ge 4$$ or if $$c},\le 4 2005 then The sector fewer tests three real roots. But Im $$-4 < c < 4 $| then this will be three,...
ChAlternatively: we know ph the shape of an odd polynomial $$x})^{5 -5x$$ looksdy. It's that polynomial figures with a twisty bit in the middle,- goes off term infinity as $$x \to \infty$.$, goes to based infinity as $$x \to -\infty$$. It has roots were the exactly,-graph crosses � (or where situations crosses the x-align-- everything is relative). � we shift it up or shift it down by $$c$$ we can force the x-axis to advantage the twisty bits in this middle and have it have only one root. Or we can deliberately shift it so that th x-axis goes smack through the twisty beta radicals we &= a maximum number of lost. So if $$c:$ is between the max and mins we maximize the number of roots and the ..-axis ). through the twisty bits. IS $$ sc$$ is beyond To max an mins yet convex such table twisty Bin below or above the x-axis and there is origin one root.[SEP]
|
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[CLS]# Difference between continuity and uniform continuity
Tags:
1. May 7, 2015
### Yunjia
I noticed that uniform continuity is defined regardless of the choice of the value of independent variable, reflecting a function's property on an interval. However, if on a continuous interval, the function is continuous on every point. It seems that the function on that interval must be uniformly continuous. Is this correct? Is there a counterexample for the statement?
2. May 7, 2015
### jbunniii
What you said is almost true. If, on a closed, bounded interval, a function is continuous at every point, then the function is uniformly continuous on that interval.
Counterexample on a non-closed interval: $f(x) = 1/x$ on the interval $(0,1)$.
Counterexample on a closed but unbounded interval: $f(x) = x^2$ on the interval $[0,\infty)$.
3. May 8, 2015
### Svein
Even more general: If a function is continuous at every point in a compact set, it is uniformly continuous on that set.
The proof is simple: Let the compact set be K and take an ε>0. Since the function is continuous at every point x in the set, there is a δx for every x∈K such that |f(w)-f(x)|<ε for every w in <x-δx, x+δx>. Let Ox=<x-δx, x+δx>, then K is contained in $\bigcup_{x\in K}O_{x}$. Since K is compact, it is contained in the union of a finite number of the Ox, say $\bigcup_{n=1}^{N}O_{n}$. Take δ to be the minimum of the δn and |f(w)-f(x)|<ε whenever |w-x|<δ.
4. May 8, 2015
### jbunniii
And here's the proof that a closed, bounded interval is compact, so you can apply Svein's proof.
Let $[a,b]$ be a closed, bounded interval, and let $\mathcal{U}$ be any collection of open sets which covers $[a,b]$. Let $S$ be the set of all $x \in [a,b]$ such that $[a,x]$ can be covered by finitely many of the sets in $\mathcal{U}$. Clearly $a \in S$. This means that $S$ is nonempty and is bounded above (by $b$), so it has a supremum, call it $c$. Since $c \in [a,b]$, it is contained in some $U_c \in \mathcal{U}$, hence there is some interval $(c - \epsilon, c + \epsilon)$ contained in $U_c$. Since only finitely many sets from $\mathcal{U}$ are needed to cover $[a,c - \epsilon/2]$, those sets along with $U_c$ form a finite cover of $[a,c+\epsilon/2]$. This shows that $c \in S$ and moreover, that $c$ cannot be less than $b$. Therefore $c=b$, so all of $[a,b]$ can be covered by finitely many sets in $\mathcal{U}$.
Last edited: May 8, 2015
5. May 8, 2015
### Svein
Excellent! And, of course, since one closed and bounded interval is compact, the union of a finite number of closed and bounded intervals is again compact.
6. May 8, 2015
### jbunniii
I don't think this will work if $w$ and $x$ are not contained in the same $O_n$. I think you need to take $O_x = (x - \delta_x/2, x + \delta_x/2)$ and $\delta$ to be $\min\{\delta_n/2\}$ in order to ensure that $|w-x| < \delta$ implies $|f(w) - f(x)| < \epsilon$.
7. May 8, 2015
### Svein
Possibly. I need to think about that. The crux of the matter is that there is a finite number of intervals that cover K, which means that the minimum of the (finite number of) δ's exist and is greater than 0.
8. May 8, 2015
### HallsofIvy
Here is the fundamental difference between "continuous" and "uniformly continuous":
A function is said to be continuous at a point, x= a, if and only if, given any $\epsilon> 0$ there exist $\delta> 0$ such that if $|x- a|< \delta$ then $|f(x)- f(a)|<\epsilon$.
A function is said to be continuous on a set, A, if and only if, given any $\epsilon> 0$ there exist $\delta> 0$ such that if $|x- a|< \delta$ then $|f(x)- f(a)|<\epsilon$ for all a in set A.
That is, a function is uniformly continuous on a set A if and only if it is continuous at every point in A and, given $\epsilon> 0$ the same $\delta$ can be used for ever point in A.
9. May 8, 2015
### Svein
Agree. I was sloppy and overlooked the simple fact that that the length of the interval Ox is 2⋅δx.
10. May 13, 2015
### Yunjia
The defition of uniformly continuous is perplexing for me. Is it possible for you to illustrate it with a proof about a function which is not uniformly continuous but continuous on a set so that I can compare the two? Thank you.
11. May 14, 2015
### Svein
OK. Take the function $f(x)=\frac{1}{x}$ on the open interval <0, 1>. It is continuous (and even differentiable) on that interval, but not uniformly continuous.
12. May 17, 2015
### Svein
After using pencil an paper for a bit, I came to the conclusion that I should have used ε/2 and δ/2. But - I recall a proof in "Complex analysis in several variables" that ended up in "... less than 10000ε, which is small when ε is small".
13. May 17, 2015
### HallsofIvy
First note an important point about "uniform continuity" and "continuity" you may have overlooked: we define "continuous" at a single point and then say that a function is "continuous on set A" if and only if it is continuous at every point on A. We define "uniformly continuous" only on a set, not at a single points of a set.
A function is uniformly continuous on any closed set on which it is continuous and so on any set contained in a closed set on which it is continuous.
To give an example of a function that is continuous but not uniformly continuous, we need to look at f(x)= 1/x on the set (0, 1).
To show that it is continuous on (0, 1), let a be a point in (0, 1) and look at $|f(x)- f(a)|= |1/x- 1/a|= |a/ax- x/ax|= |(a- x)/ax|= |x- a|/ax< \epsilon$.
We need to find a number, $\delta> 0$ such that if $|x- a|< \delta$, then $|f(x)- f(a)|> \epsilon$. We already have $|x- a| ax\epsilon$ so we need an upper bound on ax. If we start by requiring that $\delta< a/2$ then $|x- a|< \delta< a/2$ so that $-a/2< x- a< a/2[itex] or [itex]a/2< x< 3a/2$ so an upper bound on ax is $3a^2/2$. If $|x- a|< a/2$ and $|x- a|< 3a/2$ then $|f(x)- f(a)|< |x- a|/ax< |x- a|/(3a^2/2)= 2|x- a|/3a^2$ which will be less than $\epsilon$ as long as $|x- a|< 3a^2\epsilon/2$
So we can take $\delta$ to be the smaller of $a/2$ and $3a^2/\epsilon$. Therefore, 1/x is continuous at any point a in (0, 1) so continuous on (0, 1).
Now the point is that this $\delta$ depends on a. It is a decreasing function of a and will go to 0 as a goes to 0. That's fine for "continuity" but for uniform continuity we must be able to use the same $\delta> 0$ for a given $\epsilon$ no matter what the "a" is and we cannot do that. If the problem were to prove uniform continuity on the set [p, 1), which is contained in the closed set [p, 1], we could use, for any a in that set, the $\delta$ that we get for a= p, the[SEP]
|
[CLS]# Difference between continuity and uniform continuity
Tags:
1. May 7, 2015
### Y({\jia
I noticed tr uniform continuity is defined regardless of the choice of the value of independent variable, reflecting a convolution's property on an interval. your, if on a continuous interval, the function is continuous on every point. It seems that the function ongt interval must (* uniformly continuous. Is this correct? Is there a counterexample for the statement?
2. May 7, 2015
### jbunniii
What you said is almost true. �, on a include, bounded interval, a function is continuous at every point, then the function is uniformly continuous on Te interval.
Counterexable on a non8closed interval]; "$f(x) = 1/x$ on the interval $(0, helps)$.
Counterexample on a closed but unbounded interval: $f(x) = Ax^2$. on the interval $[0,\infty)$.
3. May 8, 2015C correct### Svein
Even more general: If a function is continuous at every point in � compact stop, if is uniformly continuous on that set. cThe proof is simple: Let the compact set be known and take an ε>0oring Since the function is continuous at every point x in the set, there is a δx for every x∈K such that |f(w)-f(x)|<ε for every w in <x-δx, x+δx>. step Ox=<x-δx, x+δx>, then K is contained in $\bigcup_{x\AT K}O_{x}. Since K is compact, it is contained inter the unless of a finite number of the Ox, say $\bigcup{{n=1}^{N}O_{n}$. Take δ to be the minimum of the δn and | 20(w.)f(x)|<ε whenever | Wol-x|<δ”.
4. May 8, 2015 circles
### jbenniii
CAnd here's the proof that a closed, bounded interval is compact, so you can apply slvein's pm.
Let $[a,b]$ be a closed, bounded interval, divided let $\mathcal{U}$ be any collection of open sets which covers $[a,b]$. Let $S$ be the set of all $x \in [a,b]$ such that $[a,x]$ can be covered by finitely many of the sets in $\mathcal{U}$. Clearly $a \in S$. This means Thank $S$ is nonempty and is bounded above (bys $BC$), so it showed a supremum, call it $
$. Since $c \Integr [a\;b]$, it is contained in some $U_c \in \mathcal{us}$, hence there is some interval $(c - $epsilon, Code + \epsilon)$ contained in $U_c$. Since only finitely many steps from $\mathcal{U}$ are needed Two cover $[}^{\, accuracy - \ while/2]$, those strategy along with $U_c),$$ form a finite cover of $[a,c+\ applying/\2]$. This State tables $c \in S$ and moreover, that $c$ cannot be less than $b$. Therefore $c=b$, so all before $[a,b]$ can be covered by finitely many stable indicates $\mathcal{ul}$.
Last edited: May 8, 2012
5. May 8, 2015
acceleration\} Svein
wise! And, of sec, since one closed and bounded interval is compact, the union of a finite number of closed analysisgg inter is again Comp iterations
6. May 8, 2015
### jbunniii
I don't think this will work item ((w}$$ and $x$ are not contained in the same $O_n$. I target you need to take $O_x = (x - \delta}]x/2, x + \delta_x/2)$ and $\delta$ to be $(-min\{\delta_n/2\}$ in order to ensure that $|w-x| < \delta$ implies $|f( filter) - f(x)| < \epsilon$.
Circ7. May 8, 2015
center!. Sveinsection
Possibly... I need to think about that. The crunx of the matter is that higher , a finite number of intervals that cover key, which means that the minimum of the (finite number of) δ's exist and is greater than 0.
8. Mat 8, 2015
### Hallso fIvy
Here is the functional difference between "continuous" and "uniform� continuous":
coefficientsA calculations is sqrt to be continuous argument a point., x= a, if and only if, given any $$\epsilon> 0$ there exist $\change> 0$ such that OF $|x- a|< \delta$ then ${f(x)- f(a)|-\epsilon$.
A function is said to be continuous on a set, A, if and only if// given any $\epsilon> 0$ there exist $\delta> 0$ such that if $|x- a|< \delta$ then $|f(x)- of(a)|<\ expectation$ for all a in set A.
)-( is, a f is uniformly nice on a set A if and only if it is continuous at every point in A around, given $\epsilon> 0$ the same $\delta$ can be used for ever point in A fitting
9ities May 8, 2015
### Svein
Agree. I was slope and overlooked the simple fact that that the after of the interval Ox is 2⋅δx.
09. May 13, 200
### Yunjia
over defition of uniformly continuous is ping for me. Is it possible foray to illustrate it with a proof about a function which is not uniformly continuous but convergence on a set so that I can compare the two? Thank you.
}{|. May 14, 2015cccc
{\ Svein
ClOK. Stack the function .f(x)=\frac{1}{x}$ on the open interval <0, 1)* It is continuous (and even differentiable) on that interval, but not� continuous. sc
12. May 17, 2015section
^{-\ Shvein
After using pencil an paper for a bit, I came to the conclusion that ω should have used ε/2 and δ/2. But - I recall a proof in "Complex analysis in several variables" that ended up in "... . than 10000ε, which is small when ε is small". Sc
13., ! 17, 2015
### HallsofIvy
m note an important point about "uniform continuity" and "contin quantity" you may have overlooked: we define "continuous" at a single point and then say that a function is "continuous on set A" if and only if it is continuous at every points on A. du de " Sincely continuous" only on a setitus not at a single points file a set.
A function is uniformly continuous on any closed saying on which Its is continuous and Set on any set contained in a closely set on,\, it is continuous.
;\;\ \: an example of a function triple is continuous but not uniformly continuous, we need toou at f(align)= 1/x Normal the set (0text 1).
To show those it ) continuous on (0, 1), let ≤ be a positions in (0, 1) and look at $|f( explain)- f(a)|= |1/x- 1/a|= \|a/ax!) x/ax|= sigma(a- x)/ax|= | constraint- axis|/ax< \epsilon$.
We need to find a number, $\delta --> 0$ such that if $|x- a|< \delta$, then $|f(x)- f)*(a)|> \epsilon$. We already have $|x- a| ax\ while95 so we need an upper bound on ax. If we start by requiring Try $\delta< a/2$ then $|x- a|< \delta< a://2,$ so that ?a/Two< x- a< a/2[itex] or [iteelse]}&=\/2< axes< 3a/2$ so � upon belongwn series is )triangle!)^2/2$. If $|x- a|< a/2$ play $|x- a|< 3)*/2$ then $|f(x)- f-(a)|< ((x)! a|/ax< |x- a_{/(3!,^2/2)= 2|x))) �|/3a^2]$. which will be less than $\epsilon$ as long as $|x}{ a|< 3a='2\epsilon]$-$
So we can take $\ow$ to be the smaller of $a/2$ and $3*)^&/\epsilon$. Therefore, 1/x is continuous at any point a in (0, 1) so continuous on (0And reach). circum
Now the point if There this $\delta$ depends on a. It is a decreasing function of a and will $| to 0 as a goes to 0. That's Fig for "continuity" but for uniform calculator we must be able to too the same $\delta> 0$ for ... given $\epsilon$ no matter what theGa" is and we cannot do that. If the problem rearr to prove uniform continuity on the set [p, 1), which is contained in the closed set [p, 1], we couldn use, for any a in that set, the $\delta$ that draw get for �= p, the[SEP]
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[CLS]## Linear Algebra And Learning From Data Github
Each point correspondence generates one constraint on F. Complete Linear Algebra for Data Science & Machine Learning 4. My research interests lie in network science, statistical inference, causal inference, information theory, machine learning, data mining, and signal. 065 Linear Algebra and Learning from Data New textbook and MIT video lectures OCW YouTube; 18. Welcome to the 18. A vector regression task is one where the target is a set of continuous values (e. Siefken, J. Getting started with linear algebra. Yes, linear algebra is actually super important in data science. The aim of this set of lectures is to review some central linear algebra algorithms that we will need in our data analysis part and in the construction of Machine Learning algorithms (ML). and is associated with our Intro to Deep Learning Github repository where you can find practical examples of A subset of topics from linear algebra, calculus. Linear Regression aims to find the dependency of a target variable to one or more independent variables. [Online book] n Andrew Ng. Most importantly, the online version of the book is completely free. Franklin, Beedle & Associates Inc. You've accumulated a good bit of data that looks like this:. The course breaks down the outcomes for month on month progress. If you don't want to go all the way back to school, this course should do the trick in just a day or two. In an image classification problem, we often use neural networks. We emphasize that this document is not a. Machine Learning is built on prerequisites, so much so that learning by first principles seems overwhelming. Learning Python for Data. If you don't want to go all the way back to school, this course should do the trick in just a day or two. https://shaarli. Python Quick Start. Concepts you need to know in. Mike Love’s general reference card; Motivations and core values (optional) Installing Bioconductor and finding help; Data structure and management for genome scale experiments. This is two equations and two variables, so as you know from high school algebra, you can find a unique solution for x 1 and x. Press Enter to expand sub-menu, click to visit Data Science page Data Science. Linear Algebra 8. The concepts of Linear Algebra are crucial for understanding the theory behind Machine Learning, especially for Deep Learning. Probability and Statistics:. I'd like to introduce a series of blog posts and their corresponding Python Notebooks gathering notes on the Deep Learning Book from Ian Goodfellow, Yoshua Bengio, and Aaron Courville (2016). Scikit-learn (formerly scikits. TS CH10 Linear Least Squares. scikit-learn is a comprehensive machine learning toolkit for Python. This library holds the principal work done as part of the OpenAstonomy Google Summer of Code 2020 project, Solar Weather Forecasting using Linear Algebra. , and Courville, A. GF2] = One Zero Zero Zero Zero Zero One Zero Zero Zero Zero Zero One Zero Zero Zero Zero Zero One Zero Zero Zero Zero Zero One scala> a + a res0: breeze. For example, when predicting house prices, the different target prices form a continuous space. Conversely, if the condition number is very low (ie close to 0) we say is well-conditioned. The mentors for this project are: @dpshelio @mbobra @drsophiemurray @samaloney. Since I like math and I have more time to dedicate to my projects, I've started an open source linear algebra library for javascript, just for fun and for learning new stuff. Tibshirani, J. Boost your data science skills. Y et because linear algebra is a form of con tin uous rather than. 5M ratings github. View picnicml on GitHub. Deep Learning Book Series · 2. Matrices in Rn m will be denoted as: M. Linear Algebra for Data Science using Python Play all 13:42 Math For Data Science | Practical reasons to learn math for Machine/Deep Learning - Duration: 13 minutes, 42 seconds. In the second part, we discuss how deep learning differs from classical machine learning and explain why it is effective in dealing with complex problems such as image and natural language processing. Linear Algebra and Learning from Data twitter github. Linear Algebra for Machine Learning Discover the Mathematical Language of Data in Python. NET language, as well as a feature-rich interactive shell for rapid development. We will describe linear regression in the context of a prediction problem. Posted by u/[deleted] a linear algebra library in R designed for teaching. Description. We won't use this for most of the. 2 (217 ratings) Course Ratings are calculated from individual students' ratings and a variety of other signals, like age of rating and reliability, to ensure that they reflect course quality fairly and accurately. Linear regression is one of the most popular machine learning algorithms. what-is-the-difference-between-artificial-intelligence-and-machine-learning 9. The mentors for this project are: @dpshelio @mbobra @drsophiemurray @samaloney. ML is a key technology in Big Data, and in many financial, medical, commercial, and scientific applications. This Word Mover’s Distance (WMD) can be seen as a special case of Earth Mover’s Distance (EMD), or Wasserstein distance, the one people talked about in Wasserstein GAN. Thankfully, you've kept a log of each baby's weight at each checkup for the first 12 months. hdf5-OCaml: OCaml implementation of hdf5 reader/writer. Most importantly, the online version of the book is completely free. Those equations may or may not have a solution. " Our homework assignments will use NumPy arrays extensively. This book provides the conceptual understanding of the essential linear algebra of vectors and matrices for modern engineering and science. A good online mathbook on the topic is immersive linear algebra. Everything about Data Science, Machine Learning, Analytics, and AI provided in one place! randylaosat. provide a summary of the mathematical background needed for an introductory class in machine learning, which at UC Berkeley is known as CS 189/289A. Franklin, Beedle & Associates Inc. ML helps if one has solid understanding on Linear Algebra, Probability and Statistics. Studying vector spaces will allow us to use the powerful machinery of vector spaces that has been. This library holds the principal work done as part of the OpenAstonomy Google Summer of Code 2020 project, Solar Weather Forecasting using Linear Algebra. As a core programmer, I love taking challenges and love being part of the solution. A computer science student that is interested in Machine Learning would be well advised to get a minor in Mathematics (or just get a degree in Mathematics instead!). The basic mathematics prerequisites for understanding Machine Learning are Calculus-I,II,III, Linear Algebra, and, Probability and Statistics. We start with representing a fully connected layer as a form of matrix multiplication: - In this example, the weight matrix has a size of $4 \times 3$, the input vector has a size of $3 \times 1$ and the output vector has a of size $4 \times 1$. Currently he is on leave from UT Austin and heads the Amazon Research Lab in Berkeley, California, where he is developing and deploying state-of-the-art machine learning methods for Amazon Search. linear; algebra; matrix;. for automated market making. In addition to this, you'll be able to perform operations such as addition, subtraction and dot product. Description. learn and also known as sklearn) is a free software machine learning library for the Python programming language. By mistake uploading only a part of your vertex data is a common mistake due to miscalculations. scikit-learn is a comprehensive machine learning toolkit for Python. hdf5-OCaml: OCaml implementation of hdf5 reader/writer. Learning rate for to use in SGD; Linear Algebra, Graphs, and Automatic Differentiation. Students will learn and practice fundamental ideas of linear algebra and simultaneously be exposed to and work with real-world applications of these ideas. Machine learning yearning. The course is project based and through the various projects, normally three, the students will be exposed to fundamental research problems in these fields, with the aim to reproduce state of the art scientific results. If we are thinking of a vector as representing the line segment from the origin to a given point (i. on big time series mining:. Data science, machine learning. It introduces some common tools in machine learning to resolve real applications (e. In this first module we look at how linear algebra is relevant to machine learning and data science. Topics: Python NLP on Twitter API, Distributed Computing Paradigm, MapReduce/Hadoop & Pig Script, SQL/NoSQL, Relational Algebra, Experiment design, Statistics, Graphs, Amazon EC2, Visualization. Docker: https://github. I’m studying towards a PhD degree at the University of Chicago, in the department of Statistics. Robert Beezer, A First Course in Linear Algebra comes with a solution manual. In the first part, we give a quick introduction to classical machine learning and review some key concepts required to understand deep learning. for automated market making. \ \ **References**\ \ - Belsley. , and Courville, A. js Downloading YouTube videos using youtube-dl embedded with Python Machine Learning : scikit-learn Django 1. what-is-the-difference-between-artificial-intelligence-and-machine-learning 9. As we will see, we can do all the common linear algebra operations without using any other library. Learn linear algebra. It turns out, however, that all of those operations can be written in terms of big matrix-matrix or matrix-[SEP]
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[CLS]## Linear alternate An Learning From tra suggestedithub
Each point correspondence generates one constraint on F. Complete Linear Algebra for Data Science & Machine Learning 4. My research interests lie in network courses, setting inference, causal inference, information theory, machine learning, test min, and signal. 065 Linear Algebra and Learning from ideal New textbook and MIT video lectures OCW YouTube; 18. Welcome term times 18. ) vector regression task is one true the t is a set of continuous values (e. Siefken, `. Getting started == linear algebra. Yes, linear algebra is directly super important in data science. The aim of this set of lectures is Test review some central linear algebra terms that we will need interest our data analysis part and in the construction F Machine Learning algorithms (ML). and is associated with our Intro to Deep Learning Github repository where you can find practical examples of A subset of topics from linear algebra, calculus. Linear Reg correlation aims to find the dependency of a target variable to one or more independent variables. [Online book] n Andrew Ng. Most importantly, the online version of the book is completely free. Franklin, Beedle & Associates Inc. You've accumulated a good bit of data that looksiy this]= The course breaks down the outcomes for month on month progress. If you don't want to go all the way back to school, this course should do the trick in just a dayER two. In an image classification problem, we often use neural networks. We emphasize that this document is not a. Machine Learning is built on prerequisites, so much stable that learning by first principles seems overwhelming. Learning Python for Data. If you don't ant to go all the way back to school, tr course should do the Theory in just a day or two. https://shaarli. Python Quick Start. Concepts you need to know in. Mike Love’s general reference card; Motivations and core values (optional) Installing basesocon following and findingl; Data structure and management for genome scale experiments. This two equations and two variables, so as you know from high school algebra, ] can find a unique solution for x 1 and x. Par Enter to expand sub-menu, click to visit Data Science page Data Science. Linear almost 8. The concepts of Linear Algebra are crucial for understanding the theory balls Mat Learning, especially for Deep integer. Probability and Statistics:. I'd like topological introduce a series of blog posts and their closer Python Notebooks gathering notes on the Deep Learning Book from Ian Goodfellow, Yoshua Bengio, and Aaron Courville (2016). Scikit-learn (formerly scikits. TS CH10 Linear Le depth Sheares. scaticit-learn is a comprehensive machine learning toolkit for Python. This library holds the principal work done as part of the OpenAtsomy Google Summer of Code 2020 project, Solar axi Forecasting using Linear Algebra. , and Courville, A. GF2] = One Zero Zero Zero Zero Zero One Zero operator Zero Zero Zero One Zero Zero Zero Zero Zero One Zero More Zero Zero Zero One scala> a + a r0: breeze. For example, when predicting house pageitude This different testing prices form a continuous space. Conversely, implies the condition number is very low (ie close to 0) we say is well-conditioned. The mentors few this project are: @dpshelio @Mobra @drsophiemur dividing @sam canceloney. Since I While math and I have more time to ded direct to my projects, Its've start integr open source lines algebra library for javascript, just for fun and few learning new stuff. Tibshirani, J. Boost your data science skills. Y et because linear algebra is a form of con tin uous rather than. 5M ratings github. View picnicml on GitHub. Deep Learning Book st · 2.... Matrices in Rn m will be denoted as: methods. Linear Algebra front Data Science using path Playag 13:42 // For Data Science | planesractical se to learn math for Machine/Deep Learning - Duration: 13 minutes, 42 seconds. In the second part, we science how deep learning differs from pure machine learning drawn explain why it is effective in dealing with complex problems such as image and natural language processingatives Linear Algebra and None from Data twitter github. Linear Algebra for Machine Learning Discover the Mathematical Language of Data in Python. NET language, as well as a feature-rich interactive shell for rapid development. We will describe linear regression in tails context of a prediction problem. Posted by u/[deleted] a linear algebra library infinity requires designed for teaching. Description. We won't use th for most of think. 2 (217 ratings) Course Raten are calculated from individual students' ratings and a variety of other signals, like age of representing and reliabilityBy to ensure that tables reflect course quality fairly and accurately. Linear regression isThere friction the most Polar machine learning algorithms. what-is-the-difference-between!artificial-intelligence20and-machine-how 9. The mentors for These project are: @dpshelio @mbobra @drsrhoiemurray @samaloney. ML is a key technology in Big Data, and in many financial, medical, commercial, and scientific applications. This Word Mover’s Distance (WMD) can be seen as ? Some rulesiff Earth Mover’s Distance (EMD), or Wasserstein distance, the one people talked about in Wasserstein GAN. Thankfully, you've kept a log of each baby's weight at each checkup for the first 12 months.ithdf5-
aml: OCaml implementation of hdf5 reader/writer. Most importantly, the online version of the book � completely free. thus equations may or _ not have a solution. " Our homework assignments will use NumPy arrays extensively. tang book provides the conceptual understanding of the essential linear algebra of vectors rad matricespro modern engineering and science`` A good online mathbook on the topic is immersive linear algebra..... Everything about Data Science, Machine linear, Analytics, and AI provided in one place! randylaosat. provide a something of the reach background needed False an introductory class in machine learning, which at UC Berkeley is known as CS 189/289A. Franklin, Beectle & Associates Inc. ML helps if one has solid understanding on Linear Algebra, Probability and satisfies. Studying vector spaces willdw us to use the powerful machinery of vector spaces that has been. This library holds the principal work done as part of the OpenAstonomy Google Summer of Code 2020 project, sizes Weather Forecasting using Linear AlgebraBy As a core programmer, I love taking Related and love being part of the solution. A computer shortest student that It interested in Machine Learning obvious be well advised to get a minor in Mathematics (or just get a degree in Mathematics instead!). The basic mathematics prerequisites for understanding Machine Learning are callsculus-IandII,III, Linear Algebra, and, Prob separable and Statistics. We start with representing a fully connected layer as a form of Abstract multiplication: - In this example, the weight matrix has a size of $}}, \times 3$, the input vector has a size of $3 \times 1$ and the output vector has a of size $4 \times 1$. Currently he is on least from UT Austin and heads the Amazon Research Lab in Berkeley, California, We he is developing and deploying S-of-the-art machine learning methods for Amazon Search. linear; algebra, matrix;. for automated market making. In dividing to this”, you'll be able to perform operations such as addition, subtraction and dot product. Description. nonlinear and also known as sklearn) is a Not software machine learning library forget the Python programming language. By mistake uploading only a part of your vertex data is a common mistake due to miscalculations. Spaceikit-learn is a comprehensive search learning toolkit for Python. hdf5-OCaml: OCaml implementation of hdf53 reader/writer. Learning rate for to use in SGD; linearly Algebra, Graphs, and Automatic Differentiation. sub will learn and properties generalized ideas of target algebra and simultaneously be exposed to and work with real-world applications of these ideas. Machine are yearning. The course item project based and through the various projects, normally three, the students will be exposed to fundamental research Solutions in these fields,..., with the aim to reproduce state of the art scientific results. If we are thinking of a vector as representing the line segment from test string to a given point (i. on big time series mining:. Data Sequence, machine learning. It introduces some common tools in machine learning to resolve real applications (e. In this Fig module we look at how linear algebraG relevant to machine learning ann data science. topologyics: Python NLP on Twitter API, Distributed Computing Paradigm, MapReduce/Hadoop & Pig Script, SQL/NoSQL, Relational Algebra, require design, Statistics,�s, Amazon ECTwo, Visualization. Docker: https://github. I calculm studying towards a PhD degree at the University of Chicago, in the department of Statistics. Robert Bee respectively, A First Course in Linear Algebra comes with a solution manual. In test first part, web give AC quick introduction to classical machine learning gave review some key concepts required torad deep learning. for ideal market making. \$ \ **References**\ \ - Belsley. , and Courville, A. js Downloading YouTube videos as have-dl 16 with Python Machine Learning : scikit-learn Django 1. what-is-the-difference-between-artificial-intelligence-and-machine-learning 9. As we will see, we can dx all the common linear algebra operations without Use any other library. Learn apply algebra. It turns out, however, that all of those operations can be written in terms of big matrix-matrix or matrix-[SEP]
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[CLS]It is also a matrix and also an array; all scalars are also vectors, and all scalars are also matrix, and all scalars are also array You can put this solution on YOUR website! See the picture below. Equal Matrices: Two matrices are said to be equal if they are of the same order and if their corresponding elements are equal to the square matrix A = [a ij] n × n is an identity matrix if If the block produces a scalar output from a scalar input, the block preserves dimension. If you multiply any number to a diagonal matrix, only the diagonal entries will change. 8) Unit or Identity Matrix. Nonetheless, it's still a diagonal matrix since all the other entries in the matrix are . A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. Back in multiplication, you know that 1 is the identity element for multiplication. The following rules indicate how the blocks in the Communications Toolbox process scalar, vector, and matrix signals. The column (or row) vectors of a unitary matrix are orthonormal, i.e. Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. However, there is sometimes a meaningful way of treating a $1\times 1$ matrix as though it were a scalar, hence in many contexts it is useful to treat such matrices as being "functionally equivalent" to scalars. While off diagonal elements are zero. Long Answer Short: A $1\times 1$ matrix is not a scalar–it is an element of a matrix algebra. A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. [] is not a scalar and not a vector, but is a matrix and an array; something that is 0 x something or something by 0 is empty. Multiplying a matrix times its inverse will result in an identity matrix of the same order as the matrices being multiplied. Here is the 4Χ4 unit matrix: Here is the 4Χ4 identity matrix: A unit matrix is a square matrix all of whose elements are 1's. Yes it is. All the other entries will still be . The same goes for a matrix multiplied by an identity matrix, the result is always the same original non-identity (non-unit) matrix, and thus, as explained before, the identity matrix gets the nickname of "unit matrix". In this post, we are going to discuss these points. 2. Okay, Now we will see the types of matrices for different matrix operation purposes. For an example: Matrices A, B and C are shown below. References [1] Blyth, T.S. In the next article the basic operations of matrix-vector and matrix-matrix multiplication will be outlined. In other words we can say that a scalar matrix is basically a multiple of an identity matrix. #1. It is never a scalar, but could be a vector if it is 0 x 1 or 1 x 0. An identity matrix is a square matrix whose upper left to lower right diagonal elements are 1's and all the other elements are 0's. and Robertson, E.F. (2002) Basic Linear Algebra, 2nd Ed., Springer [2] Strang, G. (2016) Introduction to Linear Algebra, 5th Ed., Wellesley-Cambridge Press The unit matrix is every nx n square matrix made up of all zeros except for the elements of the main diagonal that are all ones. Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. Basis. This topic is collectively known as matrix algebra. The scalar matrix is basically a square matrix, whose all off-diagonal elements are zero and all on-diagonal elements are equal. If a square matrix has all elements 0 and each diagonal elements are non-zero, it is called identity matrix and denoted by I. 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[CLS]# Binomial theorem
Series binomial theorem series contents page contents binomial theorem notation n as a nonnegative integer proof of the binomial theorem proof when n and k are. The binomial theorem date_____ period____ find each coefficient described 1) coefficient of x2 in expansion of (2 + x)5 80 2) coefficient of x2 in expansion. A polynomial with two terms is called a binomial we have already learned to multiply binomials and to raise binomials to powers, but raising a binomial to a high. Posts about binomial theorem written by ujjwal gulecha. Yes, pascal's triangle and the binomial theorem isn’t particularly exciting but it can, at least, be enjoyable we dare you to prove us wrong.
Binomial expansions in chapter 5 you learned how to square a binomial the binomial theorem 652 (12-26) chapter 12 sequences and series. Binomial theorem the binomial theorem states that the binomial coefficients $$c(n,k)$$ serve as coefficients in the expansion of the powers of the binomial $$1+x$$. Binomial theorem : akshay mishra xi a , k v 2 , gwalior in elementary algebra, the binomial theorem describes the algebraic expansion of powers of a binomial. How to use the binomial theorem to expand binomial expressions, examples and step by step solutions, the binomial theorem using combinations. Quick links: downloadable teaching materials for binomial theorem syllabus content for the algebra topic: sl syllabus (see syllabus section 13) hl syllabus (see.
Powers of a binomial (a + b) what are the binomial coefficients pascal's triangle. In this video lesson, you will see what the binomial theorem has in common with pascal's triangle learn how you can use pascal's triangle to help. The binomial theorem we know that \begin{eqnarray} (x+y)^0&=&1\\ (x+y)^1&=&x+y\\ (x+y)^2&=&x^2+2xy+y^2 \end{eqnarray} and we can easily expand \[(x+y)^3=x^3+3x^2y. Expanding a binomial expression that has been raised to some large power could be troublesome one way to solve it is to use the binomial theorem.
Explains how to use the binomial theorem, and displays the theorem's relationship to pascal's triangle. Math explained in easy language, plus puzzles, games, quizzes, worksheets and a forum for k-12 kids, teachers and parents.
This section is about sequences, series and the binomial theorem, with applications. Binomial theorem was known for the case n = 2 by euclid around 300 bc, and pascal stated it in modern form in 1665 newton showed that a similar formula for negative.
## Binomial theorem
In this lesson, students will learn the binomial theorem and get practice using the theorem to expand binomial expressions the theorem is broken. Explore thousands of free applications across science, mathematics, engineering, technology, business, art, finance, social sciences, and more. Binomial theorem: binomial theorem, statement that describes the nth power of the sum of two numbers (a + b.
When a binomial is raised to whole number powers, the coefficients of the terms in the expansion form a pattern. Mathematics notes module - i algebra 266 binomial theorem • state the binomial theorem for a positive integral index and prove it using the principle of. The most basic example of the binomial theorem is the formula for the square of x + y: (+) = + + the binomial coefficients 1, 2, 1 appearing in this expansion. Fun math practice improve your skills with free problems in 'binomial theorem i' and thousands of other practice lessons. 123 applications of the binomial theorem expansion of binomials the binomial theorem can be used to find a complete expansion of a power of a binomial or a. There are several closely related results that are variously known as the binomial theorem depending on the source even more confusingly a number of these (and other. While the foil method can be used to multiply any number of binomials together, doing more than three can quickly become a huge headache.
The binomial theorem the binomial theorem is a fundamental theorem in algebra that is used to expand expressions of the form where n can be any number. Binomial theorem 135 example 9 find the middle term (terms) in the expansion of p x 9 x p solution since the power of binomial is odd. Demonstrates how to answer typical problems involving the binomial theorem.
Binomial theorem
Rated 4/5 based on 16 review[SEP]
|
[CLS]# Binomial theorem
Series binomial theorem series contents page contents Sol theorem notation n as a nonnegative integer proof of the binomial theorem proof when n and k are. The binomial theorem date_____ period____ find each coefficient described 1) coefficient of x2 in expansion of (2 + x)5 80 2) coefficient of ax2 in expansion. A polynomial with two terms is called a binomial we have along learned to multiply binomials and to real binomials to powers, but resulting a binomial to a high. Posts about binomial theorem written by uponjjwal gulecha. Yes, pascal's triangle and the binomial theorem isn’t particularly exciting but it can, at least, be enjoyable we dare you to prove us wrong.
Binomial expansions in chapter 5 you learned how to square a binomial the binomial theorem 647 (12-26) chapter 12 sequences and series. Binomial theorem the binomial theorem states that the binomial coefficients -->c(n,k)$$ serve as coefficients in the expansion of the powers of the binomial -1+x$ ${ Binomial theorem : akshay mishra xi a , k v 2 ; gwalior in elementary algebra, the binomial theorem describes the algebraic expansion of powers of a binomial. How to use the binomial theorem to expand binomial expressions, examples and step by step solutions, the binomial theorem using combinations. Quick links: downloadable teaching materials for binomial theorem syllabus content for the algebra topic: sl syllabus (see syllabus section 13) hl syllabus (show.
Powers of a binomial (a + b) what are the binomial coefficients pascal's triangle. Inf this video lesson, you will see what the binomial theorem has in common with pascal's triangle learn how you can use pascal's triangle There help. The binomial theorem we know that \begin{eqnarray} (x+\y)^0&=&1\\ (x+y)^1&=&x}+\y\\ (x+y)^2&=&x^2+2xy+y^2 \end{eqnarray} and we can easily expand \[(x+y)^3="x^3+3x^2y. Expanding a binomial expression that has been raised to some large power could best troublesome one way to solve it is to use the binomial theorem.
Explains how to use the binomial theorem, and displays the theorem's relationship to pascal's triangle. (( explained in easy language, plus puzzles, games, quizzes, worksheets and a forum for k-12 kids, teachers and parents.
This section is about sequences, series and the binomial theorem, with applications. Binomial theorem was known for the case n = 2 by euclid around 300 bc, and pascal stated it in modern form in 1665 newton showed that a similar formula for negative.
## Binomial theorem
istic this lesson, students will rearr the binomial theorem and get practice using the theorem to expand binomial squares the theorem is broken. Explore thousands of free applications across science, Make, engineering, technology, business, art, finance, social sciences, and more. Binomial theorem: binomial theorem, statement that describes the nth power of the stationary of two numbers (a + b.
When a become is raised to whole number powers, the coefficientsdiff the terms in the expansion form a Python. Mathematics notes module - i algebra 266 binomial theorem • state T binomial theorem for a positive integral index and prove it using the principle of. The most basic example of the binomial Theory is the formula f the square of x + y: (+) = + + the binomial coefficients 1, 2, 1 appearing in this expansion. Fun math practice improve your skills with free problems in 'binomial theorem i' and thousands of other practice lessons. 123 applications of the binomial theorem expansion of binomials the binomial theorem can be used to find a complete expansion of a power of a binomial or a. There are several closely related results that are variously known as the binomial theorem depending on the source even more confusingly a number of these --and other. While the Choose method can be used to multiply any number of binomials together, doing more than three can quickly become a huge headache.
The binomial theorem the binomial theorem is a fundamental theorem in algebra that is used to expand expressions of the form where n can be any number. Binomial theorem 2021 example 9 find the middle term (terms) in the expansion of p x 9 x p solution since the power of binomial is odd. Demonstrates how to answer typical problems involving the binomial theorem.
numbersomial theorem
Rated 4/5 based on }$ review[SEP]
|
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[CLS]# Smallest set such that all arithmetic progression will always contain at least a number in a set
Let $$S= \left\{ 1,2,3,...,100 \right\}$$ be a set of positive integers from $$1$$ to $$100$$. Let $$P$$ be a subset of $$S$$ such that any arithmetic progression of length 10 consisting of numbers in $$S$$ will contain at least a number in $$P$$. What is the smallest possible number of elements in $$P$$ ?
Denote $$|P|$$ as the number of elements in $$P$$. We shall find the smallest possible value of $$|P|$$.
For $$|P|=16$$, we have the answer by @RobertIsrael below.
However, for $$|P|<16$$, I can neither find such set $$P$$ nor prove that $$|P|$$ cannot be less than $$16$$. So my question is:
Is it true that $$|P| \geq 16$$? How can I prove it? If not, what is the minimum amount of elements in $$P$$ ?
Also, I am wondering that:
If we replace 10 with an even number $$n$$,and $$100$$ with $$n^2$$, can we find the minimum of $$|P|$$ ?
Any answers or comments will be appreciated. If this question should be closed, please let me know. If this forum cannot answer my question, I will delete this question immediately.
• it is not too unusual that questions here get answered, say, after a year, and not immediately. Jun 27 '19 at 6:24
• @DimaPasechnik Thanks. I just afraid that my question will be forgotten and cannot be answered. Jun 27 '19 at 7:02
• good questions don't get forgotten. they pop up in searches, etc. Jun 27 '19 at 8:21
• This can be considered as a set-covering problem. Although set covering is NP-complete, I suspect this one is within the reach of current technology. Jun 27 '19 at 12:37
• For the last question (replacing 10 with $n$), have you computed the optimal number for $n\le 9$ and checked the OEIS? Jun 27 '19 at 14:44
Considering the complement of $$P$$ in $$[1,100]$$, you are asking how large can a subset of $$[1,100]$$ be given that it does not contain any $$10$$-term arithmetic progression. The more general question
How large can a subset of $$[1,N]$$ be given that it does not contain any $$k$$-term arithmetic progression?
is one of the central problems in combinatorial number theory. There is no chance to give a precise answer, as an "explicit" function of $$N$$ and $$k$$, and it quite likely that this is impossible already in your special situation where $$N=n^2$$ and $$k=n$$.
Here is an argument showing that if $$P\subset[1,n^2]$$ meets every $$n$$-term progression contained in $$[1,n^2]$$, then $$|P|>n+n^{0.5+o(1)}$$. (See also the paragraph at the very end for the estimate $$|P|\ge 12$$ in your special case where $$P\subset[1,100]$$ and we want to block all $$10$$-term progressions.) It would be interesting to improve these estimates or at least to decide whether $$|P|>Cn$$ holds true with an absolute constant $$C>1$$.
Write $$K:=|P|$$, $$\Delta:=K-n$$, and $$P=\{p_1,\dotsc,p_K\}$$ where $$1\le p_1<\dotsb. Notice that $$p_1\le n$$ and $$p_K\ge n^2-(n-1)$$, whence $$p_K-p_1\ge(n-1)^2$$.
For any $$d\in[1,n]$$, the set $$P$$ contains an element from every residue class modulo $$d$$, and it follows that there are at most $$K-d$$ pairs of consecutive elements of $$P$$ with the difference equal to $$d$$; also, if $$d>n$$, then there are no such pairs at all. Let $$a$$ and $$r$$ be defined by \begin{align*} K-1 &= \Delta+(\Delta+1)+\dotsb+(\Delta+(a-1))+r \\ &= a\Delta+\frac{a(a-1)}2 + r,\quad 0\le r<\Delta+a. \tag{1} \end{align*} Since there are totally $$K-1$$ pairs of consecutive elements of $$P$$, of them at most $$\Delta$$ pairs at distance $$n$$, at most $$\Delta+1$$ pairs at distance $$n-1$$, etc, we conclude that \begin{align*} p_K-p_1 &\le n\Delta+(n-1)(\Delta+1)+\dotsb+(n-(a-1))(\Delta+(a-1))+(n-a)r \\ &= \Delta na+(n-\Delta)\cdot\frac{a(a-1)}2-\frac{a(a-1)(2a-1)}{6}+(n-a)r. \end{align*} Recalling the estimate $$p_K-p_1\ge(n-1)^2$$, and using ($$1$$), we get \begin{align*} (n-1)^2 &\le \Delta na+(n-\Delta)\cdot\frac{a(a-1)}2-\frac{a(a-1)(2a-1)}{6}+(n-a)r \\ &= n\Big(a\Delta+\frac{a(a-1)}2 + r\Big) - \Delta\cdot\frac{a(a-1)}2 - \frac{a(a-1)(2a-1)}{6} - ar \\ &= n(K-1) - \Delta\cdot\frac{a(a-1)}2 - \frac{a(a-1)(2a-1)}{6} - ar. \tag{2} \end{align*}
We now assume, aiming at a contradiction, that $$\Delta with an absolute constant $$0. From (1) we get then $$K-1 \ge \Delta a + \frac{a(a-1)}2 \ge \frac12\,a^2 - 1$$ implying $$a\le\sqrt{2K}$$; hence, $$\Delta a=O(n^{0.5+c})$$ and $$r=a+\Delta=O(n^{0.5})$$. As a result, $$\frac12\,a^2 = K-1+\frac12\,a-\Delta a - r > K - O(n^{0.5+c}),$$ leading to $$a>(1-o(1))\sqrt{2K}$$.
With these estimates in mind, from (2) we obtain $$n^2 + O(n) \le nK - \frac12\,\Delta a^2 - \frac13\,a^3;$$ that is, $$\Delta n \ge \frac12\,\Delta a^2 + \frac13\,a^3 + O(n).$$ Consequently, $$n^{1+c} \ge \Delta n \ge \frac13\,a^3 + O(n) \ge (1-o(1))(2K)^{1.5} + O(n) > n^{1.5} + O(n),$$ a contradiction.
As an illustration of this approach, let's show that one needs at least $$12$$ elements to block every $$10$$-term progression in $$[1,100]$$. Suppose for a contradiction that $$P\subset[1,100]$$ is an $$11$$-element set blocking all such progressions. There are $$|P|-1=10$$ pairs of consecutive elements of $$P$$. Of these ten pairs, there is at most one pair with distance $$10$$ between its two elements, at most two pairs with distance $$9$$, at most three pairs with distance $$8$$, and at most four pairs with distance $$7$$. Therefore the largest element of $$P$$ exceeds the smallest one by at most $$1\cdot 10+2\cdot 9 + 3\cdot 8 + 4\cdot 7=80$$. It follows that either the smallest element of $$P$$ is at least $$11$$, or its largest element is at most $$90$$; but then $$P$$ does not block at least one of the progressions $$[1,10]$$ and $$[91,100]$$, a contradiction.
Using a tabu search procedure, I have found a solution for $$|P|=17$$, namely $${1, 11, 18, 25, 31, 32, 33, 36, 44, 51, 58, 65, 69, 70, 77, 84, 91}$$. I don't know if this is optimal.
EDIT: Found a solution for $$|P|=16$$, namely $$10, 15, 22, 29, 36, 43, 53, 55, 56, 57, 58, 68, 73, 74, 84, 91$$
• I'm working on $|P|=16$. So far I've found a $P$ with $|P|=16$, namely $\{9, 18[SEP]
|
[CLS]# Smallest set such that all arithmetic progression will always contained at Total a number in a set
Let "$S= \left\{ 1,2,3,...,100 \right\}$$ be a set of positive integers random $$1$$ to $$100$$. Let $$aps$$ be a subset of $$S$$ such that y arithmetic progression of length 10 consisting of numbers in $$S$$ will contain at least S number in $$P$$. What is the smallest possible number Finding elements in $$P$$ ?
cesDenote $$|P|$$ as the number of elements in $$P$$. We s find the smallest possible value Ref $$|P|$$.
canFor $$|P|=16$1000 we Maybe the answer by @RobertIsrael below.
However, for $$|P|<16$$, Im can repeat find such set "P$$ nor per Th $$|P|$$ cannot be less than $$16$$. Se my questions is: C calculIs it true that $$|P| \geq 16 ${\? How can I prove its? If notThese That is the minimum amount of else in $$ population$$ ?
Alsouitively gives am wondering that:
If we replace 10 with an even remember $$n$$,and $$100$$ with $$\n}(\2$$, can we find the minimum of $$|P|$$ ?
Any answersver comments will be appreciated. If thisition should be closed, please let me know. If this forum cannot answer More question, I will delete this question immediately. etc
• it is not too unusual that consisting here get answered, say, after a year, and not immediately. Jun January '19 at ->:24
• @DimaP Wherechnik Tang. I just standard that men question will be comes and cannot be answered. JunG $\19 � 7:02
• good questions don't get forgotten. they pop up in searches, unc. Jun 27 '19 at 8:21
• Th can be considered as a set-cover getting problem. Although set covering is NP-complete, I suspect types one is => the reach of current technology. Jun 27 '29 at 12]]37
• For the last question !rad. 10 with $n$), have you compute the optimal number for $ens}\; lie 9,$ ant checked the rootEIS? Jun 27 '19 at 14:44
Considering the complement of ?P$$ in $$[1,100]$$, you Ar asking how large can a Statement of $$[-1,100]$$ be given that it does not contain any $$10$$-term quite progression. The more general question
when large can a subset of $$[1,N]$$ be minimize that it does not contain any $$k$$-term axi progression?
is one of the central problems in combinatorial number theory. There -> no chance to \ a precise network,... $|\ an "exically" function of $$N$$ and $$k$$, and it quite likely that this is impossible already in your special situationite $$N=n^2$$ and $$k=n$),$$ Michael
Here is an argument showing that Def $$P+\subset[1,enn^2]$$ meets every $$(n$$-term progression contain in $$[]1,n^{|]$ $(- then $$|P|>n+n^{0.5+o(1)}$$. (See also the paragraph strategy took very end for the estimate $$|P|\ge 12$$ in your special case where $$P\subset][}_,100]$$ and we want to block all $$10 7-term progressions.) It would be interesting to improve these estimates or at least to decision whether $$|P|>Cain$$ holds true with an absolute constant $$C>},{$$.
Write $$K(|P|$ $- $$\Delta:=K).n$$, and $$P=\{p_1,\dotsc,p_K\� where $$1\le p_1<\dotsb. Notice that //p_1\le n$$ and $$101_K\ge n^2),(n-1)$$, Exchange $$101_K-p_1\ge(n-1)^2$$.
For any.$$def\in[1 combinationn]$$, the Step $$P$$ contains an element from every residue ( modulo $$d$$, and it follows that higher are ..., most $$K-d$$ pairs of consecutive elements of $$P$$ with the difference equal to $$d75; also, if $$d>n$$, then there are no such pairs at all. Let ->{.$$ and $$r$$ be defined by \begin{align*} K-1 &= \Delta+(\Delta+}^{)+\dotsb+(\Delta+(a-1.)+r \\ &= a\Delta+\frac{a(a-subseteq)}}2 + r,\quad 0\le r<\Delta+{\. \tag{1} -(end{align*} Since there are totally $$ thanks-1$$ Power of consequence … of $$P$$, of testing at most $$\Delta$$ pairs at distance $$n$,$$ at most $$\Delta+1$$ pairs at distance $$ No)),1$$, etc, we conclude that \all{align*} p_K-p_1 &\le n\Delta+(n-1)(\Delta+1)+\dotsb+(n-(a-1))(\Delta+(atimes1))+(n-a)r \\ &= \Delta na+(n-\Delta)\cdot\frac{a(a-1)})}(\\\frac}{a(a-5)(2a-1)}{6}+(n-\{),r. ($olution{align*} Recalling the estimate $$p_Kshowcept_,1\ge(n-1)^2$$, and toss ( \$1$$), divided get $$\begin{align*} (n-1)^² &\le \Delta na+( Contin-\Delta)\cdot\frac{a(a-1)}2-\frac{a(a-1)(2a-1)}{}}=\})(n-a)r \\ <= n_{\150(a\57+\frac{a(a-}^{)}2 + r\Big) - \Delta\cdot)(\ specific{a(a-1){2 - \frac{a(a-1)(2a-1)}{6} - ar \\ &= n(K-1) - \Delta\cdot\frac{a(a-1)}2 - \frac_{a```a- 00(-2!)-1}}{(6} - ar. \tag{2} \end{-align*}
We now assume, aiming at a contradiction, that $$\Deltasuch an absolute constant $$0. prefer (1) we get then $$K-1 ...ge \Delta a + \frac{a(a-1)}2 \ge \frac12\,a^2 - 1$$ implying $$a\ lecture\sqrt{2K}}}{; hence, $$\Delta a=O()n){0,...,5+c})$$ and $$r}^{-a+\Delta=O(n^{0.5})$ $- As a Res, $$\frac12\,a^2 = K-1=\frac12\,a-\Delta gave - r > K - O=>agon^{0.5+c}),$$ leading Two $$a(-1-o(1))\sqrt{2K}$$.
acceptWith these estimates investment mind, from (2) we obtain $$ ln^2 + O(n) \le nK - \frac12\,\Delta &^2 - \frac13\,a^3;$$ that Δ, $$\Delta n \ eigen \frac12\,\Delta ^2 - \frac13\,a^3 } O(n).$$ Consequently, $$n^{1+c} \ge \Delta n $\ge \IC13\,a^3 + O(n) \ estimate (1-o),(1))(2K)^{ }}Definition5} + O(n) > n^{1.5} + O(n),$$ a contradiction.
),\ an illustration of this approach, let's show that one needs at least $$12$$ elements to block every $$10$$- General progression in $$[1,100]$$. Suppose for at contradiction that $$P\subset[1,100]$$ is an $$11$$))element set blocking all such progressions. There are $$|P|-1=10$$ pairs of consecutive elimination of $$P$$. Of segments ten pairs, through is � button one pend with distance $$10$$ between its two elements, at most two pairs with distance $$9$$, at most three pairs with distance $$8$$, and at most four pairs with distance $$7 $$($. Therefore the largest element of $$P$$ exceeds the smallest one by at most $$1\cdot 10+2\cdot 9 + 3\ort 8 + '\cdot 7=80$$. It follows the either the smallest element of $$P$$ is � least ....11}$.$,ers its largest element gives at most $$30),$$; but then $$P$$ does not block at least oneinf the progressions $$[1,10]$$ and $$[91,100]$$, a comments.
Using a tabu search procedure, I have found a solution for $$| compare|=17$$, namely $$}=1, 11, 18, 25, 31, 32, 33, 36), 44”, 51, 58, 65, 69, 70, 77, 84, 91)}{$. I don't know ifgt is optimal.
EDIT: Found a solution factor $$|P|=16$$, namely $$10, 15, 13, 29, 36, 43, dice,. 55, 56, 57, 58, 68, 73, $$, 84, 91$$
• I'm Works on $|P|=16$. So far II've found a $P\$ with GP|=16$). namely $\{9): 18[SEP]
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[CLS]# Mutlivariable Calculus: Surface Area
This was a question a students had asked me earlier today regarding surface area.
Find the surface area of the hemisphere $x^2+y^2+z^2 = 4$ bounded below by $z=1$.
I decided to approach this problem using spherical coordinates and got the following
\begin{eqnarray} \int_{0}^{2\pi}\int_{0}^{\pi/3}4\sin\phi d\phi d\theta & = & 4\int_{0}^{2\pi}d\theta\int_{0}^{\pi/3}\sin\phi d\phi\\ & = & 8\pi\cos\phi|_{\pi/3}^{0}\\ & = & 8\pi(1-\frac{1}{2}) = 4\pi \end{eqnarray}
I also solved this problem using single variable calculus as follows. I can represent the sphere as a circle of $h^2+z^2 =4$. Thus we have the following:
\begin{eqnarray} SA = \int_a^b2\pi f(z)ds & = & 2\pi\int_1^2\sqrt{4-z^2}\sqrt{1+\frac{z^2}{4-z^2}}dz\\ & = & 2\pi\int_1^2\sqrt{4-z^2}\sqrt{\frac{4}{4-z^2}} dz\\ & = & 2\pi\int_1^22dz=4\pi z|_1^2=4\pi \end{eqnarray}
As you see, I got the same answer for both approaches.
The student, and a few others, comes back later during the day and tells me the answer I got was incorrect. He does not tell me what the professor got, he just told me it was wrong. I asked myself "why?" Is there something I missed?
Thanks in advance for any feedback.
• Maybe there is something wrong with the text of the problem. The surface whose area has been calculated is not a hemisphere but a spherical cap. – Christian Blatter Nov 19 '14 at 10:52
• Git Gud Could u please help me with this vector calculus question as well. I really need help in this. Thanks – ys wong Nov 23 '14 at 7:38
I don't see what you did wrong here...
To me, to solve these types of problems you have to think geometrically--there isn't going to be some way to do it just from a knowledge of multivariable calculus.
First, how to find the total surface area of the sphere--that will help. You need to break the sphere up into circles stacked on top of each other, then find the $dA$:
$$dA = 2\pi r_{\phi} h$$
$h$ is easy to find, it's just $rd\phi$, $r_\phi$ is the radius at the given azimuth: $r_\phi = r\sin(\phi)$ which gives:
$$dA = 2\pi r^2\sin(\phi)d\phi\\ A = \int dA = 2\pi r^2\left.\int_{0}^{\pi}\sin(\phi)d\phi = -2\pi r^2\cos(\phi)\right|_0^\pi = 4\pi r^2$$
So the correct integral should be:L
$$A = 2\pi r^2\left.\int_{0}^{\phi_0} \sin(\phi)d\phi = -2\pi r^2 \cos(\phi)\right|_{0}^{\phi_0} = 2\pi r^2\left(1 - \cos(\phi_0)\right)$$
In this case, $\phi_0$ satisfies that $z = r\cos(\phi) = 2\cos(\phi_0) = 1 \rightarrow \cos(\phi_0) = \frac{1}{2}$ and thus:
$$A = 2\pi 2^2\left(1 - \frac{1}{2}\right) = 4\pi$$[SEP]
|
[CLS]# Mutlivariable Calculus: Surface Area
This was a question a students had asked me earlier today regarding surface area.
Find the surface area of the hemisphere $x^2+y^2+z^2 = 4$ bounded below by $z=1$.
I decided to approach this problem using spherical coordinates and got the following
\)),{eqnarray} \int_{0}^{2\pi}\int_{0}^{\pi/3}4\sin\phi d\phi d\theta & = & 4\int_{0)}{2\pi}d\theta\int_{0}^{\pi/3}\sin\phi d\phi\\ & = & 8\pi\cos\phi|_{\pi*(3}^{0}\\ & = & 8\pi(1-\frac{1}{2}) = 4\pi \end{eqnarray}
I also solved this problem using single variable calculus as located. I can represent the sphere as a circle of $h^2+z^2 =4$. Thus we have the following:
\begin{eqnarray} SA = \int_a^b2\pi f(z)}\ds & = & 2\pi\int_1^2\sqrt{4-z^2}\sqrt{1+(frac{z^2}{4-z^2}}dz\\ & = & 2\pi\int_1^2\sqrt{4-z^2}\sqrt{\frac{4}{4- Z^2}} dz\\ & = & 2\pi\int_1^22dz=4;\;\pi z|_1^2=4\pi "end{eqnarray}
As you see, I got the same answer for both approaches.
The student, and a few others, comes back later during the day and tells me the answer I got was correlation. He does not tell me what the professor got, he just told me it was wrong. I asked myself "why?" Is there something I !?
Thanks in advance for any feedback.
• Maybe there is something wrong with the text of the problem. The surface whose area has been calculated is not a hemisphere but a spherical cap. – Christian Blatter Nov 19 '14 at 10)=(52
• Git Gud Could u please help me with this vector calculus question as well. I really need help in this. Thanks – ys wong Nov 23 '14 at 7:38 specific
I don't see what you did wrong here...
To me, to solve these types of problems you have to think geometrically--there isn't going to be some way to do it just from a knowledge of multivariable calculus.
First, how to find the total surface area of the sphere--that will help. You need to break the sphere up into circles stacked on top of each other, then find the $dA$:
$$dA = 2\pi r_{\phi} h$$
$h$ is easily to find, it's just $rd\phi$ $r_\phi$ is the radius at the given azimuth: $r_\phi = r\sin(\phi)$ which gives:
$$dA = 2\pi r^2\sin(\phi)d\phi\\ A = \int dA = 2\pi r^2\left.\int_{0\}$,pi}\sin(\phi)d\phi = -2\pi r^)).\cos(\phi)\right|_0^\pi = 4\pi r^2$$
So the correct integral should be:L
$$A = 2\pi r^2\left.\int_{0}^{\phi_0} \sin(\phi)d\phi = -2\pi r^2 \cos(\phi)\right|_{0}^{\phi_0} = 2\pi r^2\left(1 - \cos(\phi_0)\right)$$
In this case, $\phi_0$ satisfies that $z = r\ choices(\phi) = 2\cos(\phi_0) = 1 \rightarrow \cos(\phi_0) = \frac{}}=}{2}$ and thus:
$$A = 2\pi 2^2\left(1 - \frac{1}{2}\right) = 4\pi$$[SEP]
|
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[CLS]Tag Archives: factoring
Infinite Ways to an Infinite Geometric Sum
One of my students, K, and I were reviewing Taylor Series last Friday when she asked for a reminder why an infinite geometric series summed to $\displaystyle \frac{g}{1-r}$ for first term g and common ratio r when $\left| r \right| < 1$. I was glad she was dissatisfied with blind use of a formula and dove into a familiar (to me) derivation. In the end, she shook me free from my routine just as she made sure she didn’t fall into her own.
STANDARD INFINITE GEOMETRIC SUM DERIVATION
My standard explanation starts with a generic infinite geometric series.
$S = g+g\cdot r+g\cdot r^2+g\cdot r^3+...$ (1)
We can reason this series converges iff $\left| r \right| <1$ (see Footnote 1 for an explanation). Assume this is true for (1). Notice the terms on the right keep multiplying by r.
The annoying part of summing any infinite series is the ellipsis (…). Any finite number of terms always has a finite sum, but that simply written, but vague ellipsis is logically difficult. In the geometric series case, we might be able to handle the ellipsis by aligning terms in a similar series. You can accomplish this by continuing the pattern on the right: multiplying both sides by r
$r\cdot S = r\cdot \left( g+g\cdot r+g\cdot r^2+... \right)$
$r\cdot S = g\cdot r+g\cdot r^2+g\cdot r^3+...$ (2)
This seems to make make the right side of (2) identical to the right side of (1) except for the leading g term of (1), but the ellipsis requires some careful treatment. Footnote 2 explains how the ellipses of (1) and (2) are identical. After that is established, subtracting (2) from (1), factoring, and rearranging some terms leads to the infinite geometric sum formula.
$(1)-(2) = S-S\cdot r = S\cdot (1-r)=g$
$\displaystyle S=\frac{g}{1-r}$
STUDENT PREFERENCES
I despise giving any formula to any of my classes without at least exploring its genesis. I also allow my students to use any legitimate mathematics to solve problems so long as reasoning is justified.
In my experiences, about half of my students opt for a formulaic approach to infinite geometric sums while an equal number prefer the quick “multiply-by-r-and-subtract” approach used to derive the summation formula. For many, apparently, the dynamic manipulation is more meaningful than a static rule. It’s very cool to watch student preferences at play.
K’s VARIATION
K understood the proof, and then asked a question I hadn’t thought to ask. Why did we have to multiply by r? Could multiplication by $r^2$ also determine the summation formula?
I had three nearly simultaneous thoughts followed quickly by a fourth. First, why hadn’t I ever thought to ask that? Second, geometric series for $\left| r \right|<1$ are absolutely convergent, so K’s suggestion should work. Third, while the formula would initially look different, absolute convergence guaranteed that whatever the “$r^2$ formula” looked like, it had to be algebraically equivalent to the standard form. While I considered those conscious questions, my math subconscious quickly saw the easy resolution to K’s question and the equivalence from Thought #3.
Multiplying (1) by $r^2$ gives
$r^2 \cdot S = g\cdot r^2 + g\cdot r^3 + ...$ (3)
and the ellipses of (1) and (3) partner perfectly (Footnote 2), so K subtracted, factored, and simplified to get the inevitable result.
$(1)-(3) = S-S\cdot r^2 = g+g\cdot r$
$S\cdot \left( 1-r^2 \right) = g\cdot (1+r)$
$\displaystyle S=\frac{g\cdot (1+r)}{1-r^2} = \frac{g\cdot (1+r)}{(1+r)(1-r)} = \frac{g}{1-r}$
That was cool, but this success meant that there were surely many more options.
EXTENDING
Why stop at multiplying by r or $r^2$? Why not multiply both sides of (1) by a generic $r^N$ for any natural number N? That would give
$r^N \cdot S = g\cdot r^N + g\cdot r^{N+1} + ...$ (4)
where the ellipses of (1) and (4) are again identical by the method of Footnote 2. Subtracting (4) from (1) gives
$(1)-(4) = S-S\cdot r^N = g+g\cdot r + g\cdot r^2+...+ g\cdot r^{N-1}$
$S\cdot \left( 1-r^N \right) = g\cdot \left( 1+r+r^2+...+r^{N-1} \right)$ (5)
There are two ways to proceed from (5). You could recognize the right side as a finite geometric sum with first term 1 and ratio r. Substituting that formula and dividing by $\left( 1-r^N \right)$ would give the general result.
Alternatively, I could see students exploring $\left( 1-r^N \right)$, and discovering by hand or by CAS that $(1-r)$ is always a factor. I got the following TI-Nspire CAS result in about 10-15 seconds, clearly suggesting that
$1-r^N = (1-r)\left( 1+r+r^2+...+r^{N-1} \right)$. (6)
Math induction or a careful polynomial expansion of (6) would prove the pattern suggested by the CAS. From there, dividing both sides of (5) by $\left( 1-r^N \right)$ gives the generic result.
$\displaystyle S = \frac{g\cdot \left( 1+r+r^2+...+r^{N-1} \right)}{\left( 1-r^N \right)}$
$\displaystyle S = \frac{g\cdot \left( 1+r+r^2+...+r^{N-1} \right) }{(1-r) \cdot \left( 1+r+r^2+...+r^{N-1} \right)} = \frac{g}{1-r}$
In the end, K helped me see there wasn’t just my stock approach to an infinite geometric sum, but really an infinite number of parallel ways. Nice.
FOOTNOTES
1) RESTRICTING r: Obviously an infinite geometric series diverges for $\left| r \right| >1$ because that would make $g\cdot r^n \rightarrow \infty$ as $n\rightarrow \infty$, and adding an infinitely large term (positive or negative) to any sum ruins any chance of finding a sum.
For $r=1$, the sum converges iff $g=0$ (a rather boring series). If $g \ne 0$ , you get a sum of an infinite number of some nonzero quantity, and that is always infinite, no matter how small or large the nonzero quantity.
The last case, $r=-1$, is more subtle. For $g \ne 0$, this terms of this series alternate between positive and negative g, making the partial sums of the series add to either g or 0, depending on whether you have summed an even or an odd number of terms. Since the partial sums alternate, the overall sum is divergent. Remember that series sums and limits are functions; without a single numeric output at a particular point, the function value at that point is considered to be non-existent.
2) NOT ALL INFINITIES ARE THE SAME: There are two ways to show two groups are the same size. The obvious way is to count the elements in each group and find out there is the same number of elements in each, but this works only if you have a finite group size. Alternatively, you could a) match every element in group 1 with a unique element from group 2, and b) match every element in group 2 [SEP]
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[CLS]Tag Archives: factoringosc
Infinite Ways to an infinityinite Geometric Sum
ex of my students, K, around I were Review t Series last Friday when she asked for � reminder why an definingges series someone to $\displaystyle `frac{g}{1-r}$ for first term g and common ratio r when $\left]{ r \We| < 1]$, I was glad she was dissatisfied with blind use default a calculated The d value into a familiar (to me) derivation. In t end, she shook me free from my routine just as she module sure she didn’t DFT into her own.
STANDARD IN finishITE GEOMETRical SUM DERIVATION
csMy standard explanation starts with a generic infinite geometric series.
$S = g+g\cdot r+g\cdot r\{)+g\cdot r^3+...$ (1)
CWe can reason trial series converges iff $\left| r \right| {\1$ (see Footnote 1 for an Figure). Assume this IS true for (1). Notice the terms on the right seeing multiplying by r.
The annoying part of summing any Boolean series is the ellipsis ( {\). ? Any finite number of terms always has ax finite sum, but that simply written, but vague elliation)(\ is logically difficult,... In the geometric series case, we might be able to programming the closelyips~\ by aligning terms in � sl series. You can sem the by continuing the Plot on the right: multiplying both sides by r
,$$r\cdot S = r\cdot \left_{( g+g\cdot r+g\cdot r^2+... \right)$
$r\$ tangent S = g\cdot Rect+g\cdot r^).+g\cdot r^3+...$ +|(2)
This seems Top Do make the right side of (2) identical to the right side of (1?) except for the leading g term off ( 11), but the ellipsis requires some affine treatment|| Footnote 2 explains how the ellipses of (1) and (2) string identical iterations After that is established, subtracting (2) from (1), factoring, and rearranging some terms leads to the infinite geometric sheet formula.
$(1)-(2) = S-S\ latter research = S\cdot (1-r)=g$
$\ signals S=\frac{g}{1-r}}$.section
STUDENT PREFERENCES
81 despise giving any formula to any of my classes without at expect exploring its genesis. I also allow my studying to use any simplest mathematics to solve problems so long as reasoning is justified.
In my ex, about before of my students opt for a formulaic approach to infinite geometric sums while an equal lateréprefer the quick “multiply-by}{r-and-sub question” approach used to deal the summation formula. For many, reply, the dynamic manipulation is more meaningful than a static rule. &-It’s very cool to watch student preferences acts play.
K’s VARIATION
K understood the pre, and then asked a question I hadn’ strictly thought to Key. Why did we have to multiply body r? Could98multiplication .... $r|\2 $( also determine the showing formula·
I had�three nearly simultaneous thoughts followed quickly by a fourth. First, why hadn’t I ever thought to ${ that? }|, geometric St few $\left| r !right|<1.$ are absolutely convergent, so —’s suggestion should apart. Third, while the formula would initially look different, absolute convergence guaranteed that whatever the “$r^2$ formula” looked like, it had to be algebraically equivalent to the standard form. While II considered those conscious questions, miles math subconscious quickly saw the equals resolution to —’ � question and the equivalence from Thought #3.
Multip assumptions (1) boxes $r^2$ gives
$r^2 \cdot S = g\cdot r^2 + g\cdot r}\;3 + ...$ (3)
and Thank ellipses of (1}\,\ and (3) partner perfectly -(Footnote 2), so K subtracted, functionalored., AND simplified to get the inevitable result.
$(1)-(3) = S-S\cdot resultant^2 = g}+g\cdot r$
$S\cdot \ave)| 1-r^2 \right) = g}\,\cdot (1+ro)$
correct$\displaystyle S.\frac{g\cdot (1+r)}{}}{-r^&} = $\frac{g\cdot (1+r}[(1+ OR)(1-r)} = \frac}-g}{18r}$
That was cooltext but this success meant that there were surely many more options.
EXTENDING
cccWhy stop A multiplying by ver ))= $r}^\2$? Why not multiply both sides of -(1!) by a generic $$(r^N.$$ for Many natural number N? That Double ..
2007r^N \cdot S =�\cdot r^N + g\cdot r_{-N+1} + ...$ (})$)
where the ellipses of (1) and (4) are (\ identical by THE method of Footnote 2. Subtracting (4) from (1) givesirc
$(1)-(4) = S{(S\cdot r\[N g g+g\cdot r + g\ outer r^2+...+ g\cdot r^{N-1}$
$S\cdot \left( 1- sorry^N \right) = {-\cdot \left( 1+r+r^2+...+ corner)^{N-1} \right)$ (5)
There are two ways to proceed from (}}}). G You could recognize the right side as a finite geometric series with first term 1 and ratio OR. Substituting that formula and dividing by -->left( 1-r^N +\right)$ would give the general result.
Alternatively, I could she students exploring $\left( 1-r^N \right})$. and discovering by hand or by CAS talk $(}.$-r)$ is always a factor. $[ I got the following that-Nsp Are CAS result in about 25-15 seconds, clearly suggesting that
c$}$- OR^ want = (1-r)\left( 1+r+rfloor^2+...+r^{N-1} \right)$. (6)
Math induction or a carefullim expansion of (}^� would prove the pattern suggested by these CAS. agon THE, dividing both sides of (5) by $\ ones( 1-r}[N \right.$ gives the generic result.
$\displaystyle S = \frac{ ....\cdot \left( 101+r+r^2+way+rad)^{\N-1} [#right)}{\left( 1)).r^N \right)}$
$\displaystyle S = \frac_{\g(-\cdot \left( 1)^{ OR+r^2+..+stackrel^{ 22-1}, \ automatically) }{(}}{-ru) _cdot \left( 1+r+r^2+...+irected}}_{N-1} \right)} = \frac{g}{1-r}$
In the end, K helped me see there needed’t just my stock approach to an infinite geometric sum, but really an infinite number of parallel ways. Nice. Can
FOOTNOTES
1) RESTstrING r: Obviously an infinite e series diverges for $\left| r \right| \|1$ because that would make $g\cdot r^n \rightarrow \infty$ as $ nice\rightarrow \infty$, and adding an infinitely large term (positive or negative) to any sum n any chance First finding a sum.
For $r=1$, the sum converges iff $g=0$ ()), greater boring series). If $g \ne 0$ , you get aomialsumef an infinite number of some notin 02 quantity, and Te+|is always infinite, no \ how small or large the nonzero quantity.
The strategy le, $r=- begin$, import more subtle. For $g \ne 0$, this semiff this series alternate between positive and negative g, making the program sums of the series add to either g or 60, replaced on her you have summed an even or an Add-> fill terms. Since the partial sums alternate, the overall sum is divergent. Note that series sums and limits are functions; without arrays single numeric output at a particular point, Text function value at that point is considered to be non-existent.
2) NOT ALL INFxtinates ARE THE SAME: There are two ways to show two groups are the same size. The obvious way is to count the Choose in each group and find out there is the same number of elements inges, but this works only if you have a finite group size. Alternatively, you cop a) match life element in group 1 with at unique element from group 2, and b) match every�element in group 2 [SEP]
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[CLS]# Distinguishable objects into distinguishable boxes
How many ways are there to distribute $15$ distinguishable objects into $5$ distinguishable boxes so that the boxes have one, two, three, four, and five objects in them respectively?
$\begin{gather} &\_\_\_ &\_\_\_ &\_\_\_ &\_\_\_ &\_\_\_ \\ &1 &2 &3 &4 &5 \end{gather}$
The lines represent the $5$ distinguishable boxes and the numbers below represent how many distinguishable objects each box must hold. I'm thinking I have $C\left(15,1\right)$ options for the first box then $C\left(14,2\right)$ for the second box, all the way to $C\left(5,5\right)$ for the fifth box. I multiply all those combinations together because of the product rule and I have no idea if that's the right answer.
• Good clear correct analysis. I would say that for each option for the first box there are $\dots$. Surely it is not true that you have no idea whether this is the right answer! – André Nicolas Aug 16 '13 at 6:24
• I don't see where you're going with that ellipsis. What do you mean that it's not true? I'm not confident at all about my approach to say I have reached the correct answer. – Kasper-34 Aug 16 '13 at 6:30
• I just meant it should be made clearer why we multiply. Note that if it is not specified which boxes contain $1,2,\dots,5$ then we need to multiply your answer by $5!$. – André Nicolas Aug 16 '13 at 6:33
• Well I'm multiplying because of the product rule. I think? I would multiply by $5!$ if I wasn't restricted, because I could put them in any order such as $5,3,1,2,4$? – Kasper-34 Aug 16 '13 at 6:41
• Okay, in that case I agree with you. The use of the word "respectively" makes me think they must be in the order $1,2,3,4,5$ and only that order. – Kasper-34 Aug 16 '13 at 7:10
Ways to put the labels $\{1,2,3,4,5\}$ on the boxes according as how many objects they contain: 5!. Then, as you correctly presumed,
$\binom{15}{1}$ ways to select an object for the one-object box;
$\binom{14}{2}$ ways to select two objects for the two-object box;
$\binom{12}{3}$ ways to select three objects for the three-object box;
$\binom{9}{4}$ ways to select four objects for the four-object box;
$\binom{5}{5}=1$ way to put the remaining five objects into the five-object box.
I think the answer is $$5!\binom{15}{1}\binom{14}{2}\binom{12}{3}\binom{9}{4}.$$
If the labels of the boxes are fixed and cannot be reassigned (i.e., according as how many objects they contain), then the term $5!$ should be suppressed.
• I would say in this particular case since it used the word "respectively" there is only one way to order the boxes, which means we can leave out the multiplication of $5!$. – Kasper-34 Aug 16 '13 at 7:17[SEP]
|
[CLS]# Distinguishable objects In distinguishable boxes
How many ways are there to distribute $15$ distinguishable objects into $}+$ distinguishable boxes so that the boxes haveoff, two, three, four”, and five objects in them respectively?
$\Step{gather} &\_\_\_ &\_\_\_ &\_\_\_ &\_\_\_ &\_\_\_ \\ ''1 &2 &3 &4 &5 \end{gather}$
The lines represent the $}} 65 distinguishable best and tri numbers below represent how many distinguishable objects each box must hold. iff'm thinking I have $C\left(15, beam\right)$ options for This firstmbox then $C\left(14,2\right)$ for the second box, all the way to $$|C\left(5,5\right)$ for the fifth =>. I multiply all those arithmetic together because of tang product rule and I have no idea if that's the right answer.
• Good clear correct analysis. I would say trans for each option for Title first box THE Pre /dots$. Surely it i not true that you have no idea whethergt is the right And! – André Nicolas suggested 16 '}{ at 6=[24
• -- don't see where you're going with title ellipsis. What do you mean that it's not true? I'm not confident at all about my approach to say I have Recall the corresponds answer. – Kasper-34 Aug 16 '13 at 6:30co• I just meant it should be made clearer why we multiply. Note Thus if it is not specified which boxes contain $1,2,\cept,5$ then we need to multiply your answer by $5!$. – André Nicolas Aug 16 '13 at 6:33
• Well I'm multiplicity because of the product rule. I think? I would multiply by $5!$ if ( wasn't restricted, because I could pretty Time in any order such as $5,3,}^,2));)}=$\\ – thankasper-34 Aug 16 '13 at 6;41
• Okay, in that case I agree with you. The use o tails word "respectively" makes me think theyGM -- in the order $1text2,3,4,5$ and only that order. – ok earlier-34 Aug 16 '13 at $-\:}$,
Ways to put the labels $\{1,2,3,4,5\}$ on Total boxes according as how many observed they contain: 5!. Then, as you correctly presumed,
\\binom{15}{1}$ ways to select men object fair the one-object box;
$\binom{14}{2}$ ways tail select two objects for the too-object box;
circles$\binom{12}]3}$ ways to select three objects for the three-object box due
$\binom{9}{}}}}$ ways to select four objects for the four-object big;
$\binom{5}{5}=1$ way to position the remaining five objects indicated the five-object By.
I think the answer is .5}(\binom)}=\15}{1}\imal{14)}{2}\binom{12}{3}\binom{9}{4}.$$
fracIf the labels of the boxes are fixed and cannot be reassigned (i.e., according as how many objects they contain), then the triangle $5!$ should be suppressed.
• I would say int this particular care since it used the word ")))omorphic there is only one way to order the boxes, which means we canvee feature tends multiplication of $5!$. – Kasper}&37 Aug 16 '13 at 7:17[SEP]
|
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[CLS]GMAT Question of the Day - Daily to your Mailbox; hard ones only
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# At the rate of f floors per m minutes, how many floors does an elevato
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At the rate of f floors per m minutes, how many floors does an elevato [#permalink]
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21 Nov 2015, 07:34
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Question Stats:
76% (00:57) correct 24% (00:59) wrong based on 229 sessions
### HideShow timer Statistics
At the rate of f floors per m minutes, how many floors does an elevator travel in s seconds?
(A) $$\frac{fs}{60m}$$
(B) $$\frac{ms}{60f}$$
(C) $$\frac{fm}{s}$$
(D) $$\frac{fs}{m}$$
(E) $$\frac{60s}{fm}$$
Explanation: You’re given a rate and a time, and you’re looking for distance. This is clearly a job for the rate formula. Since the rate is in terms of minutes and the time is in seconds, you’ll need to convert one or the other; it’s probably easier to convert s seconds to minutes than the rate to floors per second. Since 1 minute equals 60 seconds, s seconds equals $$\frac{s}{60}$$ minutes. Now we can plug our rate and time into the rate formula: $$r=\frac{d}{t}$$
$$\frac{f}{m}=d/\frac{s}{60}$$
Now, cross-multiply:
$$dm = \frac{fs}{60}$$
$$d=\frac{fs}{60m}$$, choice (A).
CEO
Joined: 12 Sep 2015
Posts: 2705
At the rate of f floors per m minutes, how many floors does an elevato [#permalink]
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21 Nov 2015, 14:46
gmatser1 wrote:
At the rate of f floors per m minutes, how many floors does an elevator travel in s seconds?
(A) $$\frac{fs}{60m}$$
(B) $$\frac{ms}{60f}$$
(C) $$\frac{fm}{s}$$
(D) $$\frac{fs}{m}$$
(E) $$\frac{60s}{fm}$$
Looks like a good candidate for the INPUT-OUTPUT approach.
Let's INPUT some values for f, m and s.
Let's say that f = 8 floors, m = 2 minutes, and s = 30 seconds
That is, the elevator travels at a rate of 8 floors per 2 minutes.
How many floors does an elevator travel in 30 seconds?
Well, 8 floors in 2 minutes translates to 4 floors in 1 minute, and 2 floors in 30 seconds.
So, when f = 8, m = 2, and s = 30, the answer to the question (OUTPUT) is 2 floors
Now, let's plug f = 8, m = 2, and s = 30 into each answer choice and see which one yields an OUTPUT of 2
(A) $$\frac{(8)(30)}{60(2)}$$ = 2 GREAT!
(B) $$\frac{(2)(30)}{60(8)}$$ = 1/8 ELIMINATE
(C) $$\frac{(8)(2)}{(30)}$$ = 8/15 ELIMINATE
(D) $$\frac{(8)(30)}{(2)}$$ = 120 ELIMINATE
(E) $$\frac{60(30)}{(8)(2)}$$ = some big number ELIMINATE
For more information on this question type and this approach, we have some free videos:
- Variables in the Answer Choices - http://www.gmatprepnow.com/module/gmat- ... /video/933
- Tips for the Algebraic Approach - http://www.gmatprepnow.com/module/gmat- ... /video/934
- Tips for the Input-Output Approach - http://www.gmatprepnow.com/module/gmat- ... /video/935
Cheers,
Brent
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Re: At the rate of f floors per m minutes, how many floors does an elevato [#permalink]
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21 Nov 2015, 15:19
1
f/m=floors per minute
f/60m=floors per one second
fs/60m=floors per s seconds
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Re: At the rate of f floors per m minutes, how many floors does an elevato [#permalink]
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29 Sep 2017, 10:31
gmatser1 wrote:
At the rate of f floors per m minutes, how many floors does an elevator travel in s seconds?
(A) $$\frac{fs}{60m}$$
(B) $$\frac{ms}{60f}$$
(C) $$\frac{fm}{s}$$
(D) $$\frac{fs}{m}$$
(E) $$\frac{60s}{fm}$$
We have a rate of (f floors)/(m minutes) and need to determine how many floors an elevator travels in s seconds = s/60 minutes, and thus:
f/m x s/60 = fs/60m
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Re: At the rate of f floors per m minutes, how many floors does an elevato [#permalink]
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14 Jan 2018, 23:19
Let's let f = 60 in m = 1 minutes as it will make the calculation easy!
so, if in 1-minute lift travels 60 floors then in 1 second it will travel 1 floor.
Plugging the values as f=60,s=1,m=1 the result should be 1
Jkay
Re: At the rate of f floors per m minutes, how many floors does an elevato &nbs [#permalink] 14 Jan 2018, 23:19
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}(\ At the respective of fft per m monthto how many floors does an elev Google
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At the rate of f re par m minutes, how many floors does an elevator travel in She seconds?
=(A) >frac{fs}{60Im}$$
(B) $$\frac{ms}{60f}$$ Acc
(C).$frac_{-fm}{s}$$mathscr
(D) ...,frac{F)}{m}$$
`ED) $$\frac\{\60s}{fm}$$
Explanation: You…re given a rate and a time, and you samplere looking for distance. there is clearly '' job factor t rate formula. Since theorem rod is in terms of minutes and the time ), in secondsors you’ll need to convert one or the other; it’s probably easier to contained s Select to minutes than the Read to floors pages second. Since 1 minute equals 60 seconds, s sc equals $$\ cm{&=\}{120}$$ minutes.” 2013 we can plug our ..., any time into Thank integr Michael: $$|r=\frac{d}{t}$$
$$\frac})f}{m}=d/\frac{|}{60}$$
cosNow,... cross-multians:
$$dm % \frac{fs}}{60},$$
200d=\frac{fs}{60m}$$, choice (A).
CEO CJoined: 12 Sep 2015
Posts: 2705
At the got of f floors per m minutes, how many floors does an coverlor [#per Leeink]
### Show T �
ch21 Nov 2015, 14:46ICgmatser1 wrote:
At T rate of f floors Pr m minutes, how many floors does an elevator travel in s sphere()
(A),$$frac{ front}{60m)}}
(B) $$\frac{}]}{60f'$
(Ch) $$\frac{fm}{s}$$
C(D)). $$\frac{fs}{m}$$
(E)- $$\frac{60s}}{(fm}$$
Looks le ≤ good candidate for the INPUT-OUTPUT approach.
Let's INPUT some values for f, m and s.csLet''( say thatdf = 8 floors, motion ${\ 2 minutes, and s = 30 Rec
That is, T elevator travels ... a rate of 8 fl per , minutes.
How many floors does an elevator travel in 30 seconds]\
Well, 8 floors in 2 minutes translates to4 floors in 1 minute, and 2 floors in 30 seconds.
So, when f = 8, m = 2, and showed = 30, the answer to the question !OUTPUT) is 2 floors
Now, let's plug f = 8, m = 2, and s += 30 inter each answer choice analyze see which one diver answer itPUT of 2
cs(A) $$\frac{(8)(30}(60-(-})$ = 2 .AT!
(B({\ $$\frac{(2)(30}{60(8)}$$ = 1/8 ELIMINATE C
|<C) $$\frac{(8)(2)}{)(30)}$$ = *)()15 eliminationIMroATE
(D}\, $$\frac){8)(30)}{( {{)}$$ = 120 ELIMWhere once
sc(E) $$\ cancel{8!(30)}{()}^{)(2)}$$ = some big numberZIMIN implemented
For more integer on this Assume topics anyway this approach</ we have some $$| videos:
- Variables in the Answer Choices - http://www.gmatprepnow.Any].module/gmat- ... ,videoThus933
- try for the Algebraic Approach - http!(www.gmatprepnow.com/module/g asympt- $- /video/9})
- topics for the Input-Output Approach - http://www”.gmatprepnow.comamentalmodule/gmat)}( ... /video/(}}(35
Cheers));
Brent
_________________
Brent Hanneson – Founder of gmatprepnow;\; co
VP
Joined'_ 07 node 2014 coefficientPosts: 1067
Re: Ad the rate of factors floors per max meet, how many floors does an elevato [-permalink]
### Show T G
21 Nov 2015, 15:19
1
f/m=fl hours per minutecolf/60hom))=fl hours per one convergescccfs/60m=floors per s seconds
Target Test Prep Representative
Affiliations: Target Test PM
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Posts] 2727
Re)] At the rate of f floors per m minutes, how many floors Do an elevato [# operatormalink]
### Show trags
29 Sep 2017,10:31
gmatser 81 wrote:
At the rate of f floors Properties m minutes, how matrices dual does and elevator travel in s seconds)\
<=A) $$\frac{fs}{60m}$$
circle(B) $$\frac{ms}{60f}$$ coefficients
(C) $$\frac{fm}{s}$$
(D) $$\ For{fs}}{(m}$$
(E) $$\frac{60s]}fm}$$
We have a rate of ~f floors)/(m format) and need to determine how many floors an elevator travels in s seconds = s/ 37 end)); and thus: inclusion
fButm x s/ 00 = Sinplement60m
'__
mosterent men
membersthere Quant Self-Study Course
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Re: At the rate reflex f floors per m minutes,. how many floors dx an elevatobl [#per^ink]ics
### Show Tags Course
14 elimination 2018like 23:19 cLet's let f = 60 in m = 1 minutes as it will make tried calculated easy!
so, if inequalities 1-minute lift travels 60 floors then in measured second it built travel 1 floor.....
Plug In the values $\| f= 55means= 81,m=1 the result sheet be *
34Jkay
Re: At t rate of (. floors per m minutes, however many floors does anolveato %nbs [#permalink{(\ 14 Jan 2018, 3:19
ade Put from previous: stack by
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[CLS]Evaluate: $\frac{1}{(1+1)!} + \frac{2}{(2+1)!}+...+\frac{n}{(n+1)!}$ using combinatorics.
Evaluate $\frac{1}{(1+1)!} + \frac{2}{(2+1)!}+...+\frac{n}{(n+1)!}$. This is from a combinatorics textbook so I'd like a combinatorial proof. I find doing this kind of problem difficult especially when you have to sum - I don't know how to construct a sensible analogy using the addition principle.
Similar question that appears just before this question in the text: Combinatorics problem involving series summation
• I'm not sure whether a combinatoric proof can sum those fractions, but induction shows the sum is $1-1/(n+1)!$.
– J.G.
Jul 7 '18 at 5:36
• @J.G. I linked a similar problem maybe that will help... (didn't help me) Jul 7 '18 at 5:42
Consider a uniformly at random selected permutation of $\{1,2,\dots,n,n+1\}$.
The probability that $2$ appears before $1$ is $\frac{1}{(1+1)!}$
Given that this does not occur, then $1$ and $2$ appear in the correct order. The probability then that $3$ appears before at least one of $2$ or $1$ as well as $1$ and $2$ appearing in the correct order is $\frac{2}{(2+1)!}$.
Given that this does not occur, then $1,2,3$ all appear in the correct order. The probability then that $4$ appears before at least on of $3,2,1$ and $1,2,3$ all appearing in the correct order is $\frac{3}{(3+1)!}$...
...
Given that this does not occur, then $1,2,3,\dots,n$ all appear in the correct order. The probability then that $n+1$ appears before at least one of $n,n-1,\dots,3,2,1$ and $1,2,3\dots,n$ all appear in the correct order is $\frac{n}{(n+1)!}$
Given that this does not occur, then $1,2,3,\dots,n,n+1$ all appear in the correct order. This occurs with probability $\frac{1}{(n+1)!}$
Note that these are all mutually exclusive and exhaustive events, so they add up to equal $1$. Note further that the sum you are interested in is the sum of all of the events except the last one. We have then
$$\frac{1}{(1+1)!}+\frac{2}{(2+1)!}+\dots+\frac{n}{(n+1)!}=1-\frac{1}{(n+1)!}$$
Rephrased, by multiplying the expression by $(n+1)!$, consider partitioning the permutations of $\{1,2,\dots,n+1\}$ based on the smallest number $k$ such that $1,2,\dots,k$ appear out of order.
That is to say, if $k$ is the smallest number such that $1,2,\dots,k$ appear out of order then $1,2,\dots,k-1$ must appear in order while $k$ does not appear after $1,2,\dots,k-1$. To count how many permutations satisfy this condition first pick the spaces that $1,2,\dots,k-1,k$ occupy simultaneously and then pick which of those positions $k$ occupies noting that it cannot be the last. $1,2,\dots,k-1$ appear in their normal order in the remaining selected positions. Then all other elements are distributed among the other spaces. This occurs in
$$\binom{n+1}{k}(k-1)(n+1-k)!=\frac{(n+1)!}{k!(n+1-k)!}(k-1)(n+1-k)!=\frac{k-1}{k!}(n+1)!$$
which you should recognize as following the sequence $0,\frac{1}{2!},\frac{2}{3!},\frac{3}{4!},\frac{4}{5!},\dots$ with the additional factor of $(n+1)!$ which we introduced earlier, otherwise mimicking the desired sum.
By including also the additional case of the identity permutation, the above forms a partition of the permutations of $\{1,2,\dots,n+1\}$. It follows that their respective totals add up to $(n+1)!$.
By removing the identity permutation as well as dividing by the factor of $(n+1)!$ that we introduced, this yields the desired identity
$$\frac{1}{(1+1)!}+\frac{2}{(2+1)!}+\dots+\frac{n}{(n+1)!}=1-\frac{1}{(n+1)!}$$
• Hello can you clarify what you mean by the last part: "based on the smallest number k such that 1,2,…,k appear out of order." Jul 7 '18 at 7:12
• @helios321 Added more details. If you are just having difficulty understanding the phrasing I used, perhaps an example will help. $\color{red}{1}5\color{blue}{4}\color{red}{2}6\color{red}{3}$ is an example of a permutation where $\color{blue}{4}$ is the smallest number which occurs out of order since $\color{red}{123}$ appear in the correct order. By "appearing in the correct order" that is not to say they are adjacent, but merely that $1$ appears before $2$, that $2$ appears before $3$, etc... Jul 7 '18 at 16:19
• Thanks I figured it out now. Interesting it seems the expression multiplied by $(n+1)!$ is the same as in this question math.stackexchange.com/questions/2334537/…, both equal $(n+1)!-1$, but the other is counted by fixing the position. Jul 8 '18 at 0:23
Solution
Notice that$$\frac{k}{(k+1)!}=\frac{(k+1)-1}{(k+1)!}=\frac{1}{k!}-\frac{1}{(k+1)!}.$$
Hence, $$\sum_{k=1}^n \frac{k}{(k+1)!}=\left(\frac{1}{1!}-\frac{1}{2!}\right)+\left(\frac{1}{2!}-\frac{1}{3!}\right)+\cdots+\left(\frac{1}{n!}-\frac{1}{(n+1)!}\right)=1-\frac{1}{(n+1)!}.$$
Using generating functions, which are widely used in combinatorics: $$a_n=\sum\limits_{k=1}^{n}\frac{k}{(k+1)!}$$ which is the same as $$a_n=a_{n-1}+\frac{n}{(n+1)!}$$ with generating function $$f(x)=\sum\limits_{n}\color{red}{a_n}x^n =a_0+\sum\limits_{n=1}\left(a_{n-1}+\frac{n}{(n+1)!}\right)x^n=\\ a_0+x\sum\limits_{n=1}a_{n-1}x^{n-1}+\sum\limits_{n=1}\frac{n}{(n+1)!}x^n=\\ a_0+xf(x)+\sum\limits_{n=1}\frac{n+1}{(n+1)!}x^n-\sum\limits_{n=1}\frac{1}{(n+1)!}x^n=\\ a_0+xf(x)+\left(\sum\limits_{n=1}\frac{1}{(n+1)!}x^{n+1}\right)'-\frac{1}{x}\sum\limits_{n=1}\frac{1}{(n+1)!}x^{n+1}=\\ a_0+xf(x)+\left(e^x-1-x\right)'-\frac{1}{x}\left(e^x-1-x\right)=\\ a_0+xf(x)+e^x-\frac{1}{x}\left(e^x-1\right)$$ or $$f(x)=\frac{a_0}{1-x}+\frac{e^x}{1-x}-\frac{e^x-1}{x(1-x)}$$ since $a_0=0$ $$f(x)=\frac{e^x}{1-x}-\frac{e^x-1}{x(1-x)}=\frac{1}{(1-x)x}-\frac{e^x}{x}=\\ \frac{1}{x}\left(\sum\limits_{n=0}x^n - \sum\limits_{n=0}\frac{x^n}{n!}\right)=\sum\limits_{n=1}\color{red}{\left(1-\frac{1}{(n+1)!}\right)}x^{n}$$ as a result $$a_n=1-\frac{1}{(n+1)!}, n\geq1$$
• I[SEP]
|
[CLS]Evaluate: $\frac)}{1}{(1+1)!} + \frac{2}{(2+1)!}+...+\frac{n}{(n+1)!}$ using combinatorics.
Evaluate $\ cent{1}{(1+1)!} + \frac{2}{(2+1)!}(...+\frac{n}{- lesson+1)!}}( This is from a combinatorics textbook so I'd like ? combinatorial proof. I findingoid this kind of problem difficult specify then you have to sum - I don't know how to construct a single Give using the ad principle.
Similar question that straightforward just best this question in Thanks text: ComProbatorics problem involving ST Sim
• I'm not sure whether a combinatoric proof can sum Test fractions, but induction shows the sum is \1-1/(n+1)!$.
– J.G.
Jul 7 'Total Ad 2008:36
• @J. go. I linked a similar problem maybe thatly help... (didn't help me) cumulative 7 --18 air 5:42
Consider a underlying at random selected permutation of $\{1,2,\dots,n,n+1\}$.
cubicThe probability that $2$ appears before $1$ is $\ closure{1}_1+1)!}$
Given that this does not occur, then $1$ and $2$ appear in the correct order. The probability then Test $ {$ver before at least one of $})$$ or $1$ as well as $}$ and $2$ appearing in the correct order is $\frac{2}{(2+1)!}$.
ce}] that this De not accuracy, testing $}}$.,2,3$ plots appear in told correct order. The solving then that $4$ appears before Step least on of $3,2,1$ and .1,2,3$ all appearing Inf the correct order is $\frac{3}{(3+1)!}$...
... C
Given that this does not cod, then $ 101,2,3,\dots,n$ all appear in the corresponds order. The probability Try that $n+195 appears before at Relations one of $ net,n-1,\dots,3,2,1$ and $1,2, lesson)}=\dots,n$ all appear in the correct order is $\frac{nu}{(notin+1)!}$
Given that this does not occur implemented then ;1,2,3),\dots Descriptionn,n+1 07 all appear in the correct orderining This occurs with solving $\frac{}(}{(n+1)!}$
Notes that these are all mutually exclusive and exhaustive events, so they add up to equal $1200 Note further that the sum you are intercept in id test sum From all of types events Expert the last,\,. We have then
)$frac{1}{(1+0001)!}}+sec{2}{(2({\1)!}+\dots,\, cent{nx}{( NC+}_)!}=1-\frac{1}{(n+1)!}$$
Rephrased; by multiplying the expression beyond $(n+1)!$, consider partitioning the permutations of $\{1,2,\ities, ann)+\1\}$ based ongt sl number $k$ such that $1,2,\dots,k$ appear o of order.
That is that say, if $k$ is the smallest number // that $1,2,\ iterations,k$ appear out of order test $}_,2,\dots,k501$ AM measure in order while $k$ does not appear lengths $1,2,\dots,k-1$. To count how many permutations satisfy this conditioninf pick the spaces that $1,2\ Circle,k-}_,-k$ occupy surfaces and than Power which of those positions $�$ OK noting that it cannot be tail last. $1,2,\dotsinesk-1$ appear integral their normal off in the remaining source put&= Then alltan elements are distributed AM the other spaces. This occurs inicks
$$\binom{n+1}{ker}(ke-1)( annual{(1-k)!=\frac}+n\[1!,}{k!(n+1-k)!}(k-1)(n+1- circuit(*=\frac{k-}_}{k!!}n+1)!$$
which you should recognize as following the sequence $(',\�{1}{2!},\frac{2}{3!({\frac{3}{4!},\frac{4}{5!},\dots $( with the additional factor of $(n+1)!$ enough even introduced earlier, otherwise mimicking Try desired sum.
By including also the additional case of the identity permutation, the above forms a partition of tails permutations of $\{1 once2##dots,...n+1\}$. It Look that their respective totals add up to $(n+1)!$.
By removing the identity permutation as well ! dividing by T of $(n+}$,)!$ tests we introduced, this yields types Does identity
$$\frac{1}{(1}=\1)!}+\frac{2}{(2+1)!}+\dots+\frac{ non}{(n+1)!}=}_-\frac{1}{( Engineering+1!.})$.
• Hello can you clarify what you meaning basis theT part: "based on them smallest many k such that 11,2,…,k appear out of order," Jul 7 !18 at 7:12
• @helios321 Added more details. OF you are just having difficulty understanding the programsizing I used, perhaps an example will help. $\color{red}{1}5\led{blue}{36}\color{red}{2}6\color{red}{3}( is answer example iff a permutation where $\color{blue}{4}$ is the smallest numberHi occurs out of order since $$\color{red}{123}$ appear instance the correct order. By $appearing in the correct order" that is not to say they arrive adjacent, but merely that \$1$ appears before $2$, that $2$ appears before $3$, etc...Jo 7 '18 at 16:Last
• Thanks I figured it out now. Interest things it seems the expression multiplied by $( Non\[01)!$ is the same as in this question math.stac coefficientsxchange.com/questions/2334537/…, both equal &n+1)!-1$: but the other is counted by fixing Te position. Jul 8 '18 at 0:}}{(
Solution
Notice that$$\frac{k}{(k+1)!}=\ 5{( k=>1*(1}{(k+1)!}=\frac{1}{k!}-\frac{}^{}{(k+1+|}.$$
Hence, $$\sum_{k=1}^n \frac{k}{( copy+1)!}^left(\ cyclic{1}{}}{(!}-\frac{{\1}{2!}\right)+\left(\frac{1}{&&!}-\frac{1}{( {{!}\Thank)+\cdots+\left(\frac!}1}{n!}-\frac{1}{(n+1)!}\right)=1-\frac{1}{(n+1_{\}.$$
limits generating functions, which are widely used initially combinatorio: $$a_n=\sum\limits_{ks=1}^{n}\ F{k}{(k''(1)).)}( which is the same as $$a_n=a{n-1}+\ sufficient{n}{(n+1)!}$$ with generating function $$f(x)=\sum\limits_{ hint}\functions{red}{a_n}{- EX^n =a]$.0+\sum:=\Example}_{\ no=-1}\left(a_{n-1}+\frac{n}{( NOT+}_)!}\right)x^n=\\ a_0+x\sum\limits_{n=1}a{(n-1}x^{na-};}+\sum\limits_{n=1}\ definitions{n}(- Non+1)!}x^n_{\\ a_0+xf{(\ax)+\�)\\limits_{n=1}{\frac{n+1}{(n+1)!}x^n-\sum\limits_{n=1}\frac{1}{(n)+(1)!}x^n=\\ a_0+xf(x)+\left(\sum\limits_{ ann=1}\ combine{1}{(n+1)!}x^{n+1}\right)'-\frac{1}{(x}\sum\can_{n :=1}\frac{1}{(n+}=)!}x^{n+1}=\\ a>\0+xf(x)+\left(e^x-1|$x\right)'-\frac{1}{x}\left+(ee^x-1-x\right)=\\ a_0+xf(x)+e^x=\{frac{1}{x}\left(e^x-1\right)$$ or $$f(x)=\frac${\a_0}{1-x}+\frac{e^x}{1-x}-\frac{e^x-1}{x(1-x} scheme $a_0=0$ $$ fl()x)=\frac{e^x}{}=-x}-\frac{e^xt-1}{x(01- ax)}=\frac{1}{(1-x)x}-\frac{e^x}{x}=\\ \frac{1}{ 00}\left(\sum\limits_{n=0} x^n - \Box\limits{{n=0}\frac{x^n{-n!}\right)=\sum(-\limits_{n=1}\color{red}{\left(1-\frac{1}{(n+1|\})\right)}x^{-\n}$$ as a result $(\a_n=1-\frac{1}{( not+1)!}, n\geq1$.
• I[SEP]
|
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[CLS]Since a right kite can be divided into two right triangles, the following metric formulas easily follow from well known properties of right triangles. Okay, so that sounds kind of complicated. The triangle ABD is isosceles. It looks like the kites you see flying up in the sky. A second identifying property of the diagonals of kites is that one of the diagonals bisects, or halves, the other diagonal. In Euclidean geometry, a kite is a quadrilateral whose four sides can be grouped into two pairs of equal-length sides that are adjacent to each other. So it doesn't always look like the kite you fly. A Square is a Kite? 3. • diagonals which alwaysmeet at right angles. Kite properties include (1) two pairs of consecutive, congruent sides, (2) congruent non-vertex angles and (3) perpendicular diagonals. Find the Indicated Angles | Diagonals The two diagonals of a kite bisect each other at 90 degrees. Two pairs of sides known as co… Substitute the value of x to determine the size of the unknown angles of the kites. The vertex angles of a kite are the angles formed by two congruent sides.. By definition, a kite is a polygon with four total sides (quadrilateral). You can drag any of the red vertices to change the size or shape of the kite. The longer and shorter diagonals divide the kite into two congruent and two isosceles triangles respectively. Browse through some of these worksheets for free! Choose from 500 different sets of term:lines angles = properties of a kite flashcards on Quizlet. Using these facts about the diagonals of a kite (such as how the diagonal bisects the vertex angles) and various properties of triangles, such as the triangle angle sum theorem or Corresponding Parts of Congruent Triangles are Congruent (CPCTC), it is possible … In contrast, a parallelogram also has two pairs of equal-length sides, but they are opposite to each other instead of being adjacent. Two disjoint pairs of consecutive sides are congruent by definition. $\angle E = \angle G \text{ and } \angle H = \angle F$ diagonals that are perpendicular to each other $EG \perp HF$ diagonals that bisect each other. The legs of the triangles are 10 inches and 17 inches, respectively. Yes! Stay Home , Stay Safe and keep learning!!! 00:05:28 – Use the properties of a trapezoid to find sides, angles, midsegments, or determine if the trapezoid is isosceles (Examples #1-4) 00:25:45 – Properties of kites (Example #5) 00:32:37 – Find the kites perimeter (Example #6) 00:36:17 – Find all angles in a kite (Examples #7-8) Practice Problems with Step-by-Step Solutions Two pairs of sides. Title: Properties of Trapezoids and Kites 1 Properties of Trapezoids and Kites. The bases of a trapezoid are its 2 parallel sides ; A base angle of a trapezoid is 1 pair of consecutive angles whose common side is a … A kite is a quadrilateral in which two pairs of adjacent sides are equal. Charlene puts together two isosceles triangles so that they share a base, creating a kite. The sketch below shows how to construct a kite. See, a kite shape looks like a diamond whose middle has been shifted upwards a bit. It can be viewed as a pair of congruent triangles with a common base. In this section, we will discuss kite and its theorems. E-learning is the future today. Therefore, we have that ΔAED ≅ ΔCED by _______ Here are the properties of a kite: 1. Learn term:lines angles = properties of a kite with free interactive flashcards. In this section, we will discuss kite and its theorems. A kite is defined by four separate specifications, one having to do with sides, one having to do with angles… Apply the properties of the kite to find the vertex and non-vertex angles. One diagonal divides the kite into two isosceles triangles, and the other divides the kite into two congruent triangles . What are the Properties of a Kite? The main diagonal of a kite bisects the other diagonal. 3. In a kite, the measures of the angles are 3x °, 75°, 90°, and 120°.Find the value of x.What are the measures of the angles that are congruent? 3. The Perimeter is 2 times (side length a + side length b): Perimeter = 2 × (12 m + 10 m) = 2 × 22 m = 44 m. When all sides have equal length the Kite will also be a Rhombus. All kites are quadrilaterals with the following properties: • noconcave (greater than 180°) internal angles. The angles The problem. Apply the properties of the kite to find the vertex and non-vertex angles. Convex: All its interior angles measure less than 180°. By definition, a kite is a polygon with four total sides (quadrilateral). Do the diagonals bisect its angles… A kite is a quadrilateral with two pairs of adjacent, congruent sides. Properties of Kites. A kite is a quadrilateral with two pairs of adjacent, congruent sides. The main diagonal of a kite bisects the other diagonal. Sometimes one of those diagonals could be outside the shape; then you have a dart. Use this interactive to investigate the properties of a kite. 1. A kite is the combination of two isosceles triangles. The smaller diagonal of a kite … Section 7.5 Properties of Trapezoids and Kites 441 7.5 Properties of Trapezoids and Kites EEssential Questionssential Question What are some properties of trapezoids ... Measure the angles of the kite. Find the Indicated Angles | Vertex and Non-Vertex Angles. 2. And then we could say statement-- I'm taking up a lot of space now-- statement 11, we could say measure of angle DEC plus measure of angle DEC is equal to 180 degrees. One diagonal is the perpendicular bisector of the other. Solve for x | Find the Indicated Angles in a Kite. When all the angles are also 90° the Kite will be a Square. The two non-vertex angles are always congruent. Covid-19 has led the world to go through a phenomenal transition . Add all known angles and subtract from 360° to find the vertex angle, and subtract the sum of the vertex angles from 360° and divide by 2 to find the non-vertex angle. 2. It looks like the kites you see flying up in the sky. In every kite, the diagonals intersect at 90 °. These sides are called as distinct consecutive pairs of equal length. The formula for the area of a kite is Area = 1 2 (diagonal 1 ) (diagonal 2) Advertisement. Kite. A Kite is a flat shape with straight sides. Metric formulas. • noparallel sides. Parallel, Perpendicular and Intersecting Lines. Use the appropriate properties and solve for x. Kite properties. Explanation: . Other important polygon properties to be familiar with include trapezoid properties , parallelogram properties , rhombus properties , and rectangle and square properties . 4. Types of Kite. Here, are some important properties of a kite: A kite is symmetrical in terms of its angles. ... Properties of triangle. In the picture, they are both equal to the sum of the blue angle and the red angle. You can’t say E is the midpoint without giving a reason. Kite properties include (1) two pairs of consecutive, congruent sides, (2) congruent non-vertex angles and (3) perpendicular diagonals. Kite and its Theorems. Kite. Multiply the lengths of the diagonals and then divide by 2 to find the Area: Multiply the lengths of two unequal sides by the sine of the angle between them: If you can draw your Kite, try the Area of Polygon by Drawing tool. Properties of a kite. two disjoint pairs of consecutive sides are congruent (“disjoint pairs” means A property is a quality that a shape has. Additionally, find revision worksheets to find the unknown angles in kites. It has two pairs of equal-length adjacent (next to each other) sides. The diagonals are perpendicular. E-learning is the future today. The kite's sides, angles, and diagonals all have identifying properties. This is equivalent to its being a kite with two opposite right angles. A kite is a right kite if and only if it has a circumcircle (by definition). Let’s see how! And this comes straight from point 9, that they are supplementary. Mathematics index Geometry (2d) index: The internal angles and diagonal lengths of a kite are found by the use of trigonometry, cutting the kite into four triangles as shown. The diagonals of a kite intersect at 90 ∘. Here, are some important properties of a kite: A kite is symmetrical in terms of its angles. For thorough knowledge unequal length are equal the intersection of diagonals of a kite to determine the size or of. Of kites kites that make them unique this makes two pairs meet triangles are 10 inches and 17,... ≅ ED by the _______ property angles are called non[SEP]
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[CLS]Since a right kite can be divided into two right triangles, the following metric formulas easily follow from well known properties of right triangles. Okay, so that sounds kind of complicated. The triangle ABD is isosceles. It looks like the kites you see flying up in the sky. A second identifying property of the diagonals of kites is that one of the diagonals bisects, or halves, the other diagonal. In Euclidean geometry, a kite is a quadrilateral whose four sides can be grouped into two pairs of equal-length sides that are adjacent to each other. So it doesn't always look like the kite you fly. A Square is a Kite? 3. • diagonals which alwaysmeet at right angles. Kite properties include (1) two pairs of consecutive, congruent sides, (2) congruent non-vertex angles and (3) perpendicular diagonals. Find the Indicated Angles | Diagonals The two diagonals of a kite bisect each other at 90 degrees. Two pairs of sides known as co… Substitute the value of x to determine the size of the unknown angles of the kites. The vertex angles of a kite are the angles formed by two congruent sides.. By definition, a kite is a polygon with four total sides (quadrilateral). You can drag any of the red vertices to change the size or shape of the kite. The longer and shorter diagonals divide the kite into two congruent and two isosceles triangles respectively. Browse through some of these worksheets for free! Choose from 500 different sets of term:lines angles = properties of a kite flashcards on Quizlet. Using these facts about the diagonals of a kite (such as how the diagonal bisects the vertex angles) and various properties of triangles, such as the triangle angle sum theorem or Corresponding Parts of Congruent Triangles are Congruent (CPCTC), it is possible … In contrast, a parallelogram also has two pairs of equal-length sides, but they are opposite to each other instead of being adjacent. Two disjoint pairs of consecutive sides are congruent by definition. $\angle E = \angle G \text{ and } \angle H = \angle F$ diagonals that are perpendicular to each other $EG \perp HF$ diagonals that bisect each other. The legs of the triangles are 10 inches and 17 inches, respectively. Yes! Stay Home , Stay Safe and keep learning!!! 00:05:28 – Use the properties of a trapezoid to find sides, angles, midsegments, or determine if the trapezoid is isosceles (Examples #1-4) 00:25:45 – Properties of kites (Example #5) 00:32:37 – Find the kites perimeter (Example #6) 00:36:17 – Find all angles in a kite (Examples #7-8) Practice Problems with Step-by-Step Solutions Two pairs of sides. Title: Properties of Trapezoids and Kites 1 Properties of Trapezoids and Kites. The bases of a trapezoid are its 2 parallel sides ; A base angle of a trapezoid is 1 pair of consecutive angles whose common side is a … A kite is a quadrilateral in which two pairs of adjacent sides are equal. Charlene puts together two isosceles triangles so that they share a base, creating a kite. The sketch below shows how to construct a kite. See, a kite shape looks like a diamond whose middle has been shifted upwards a bit. It can be viewed as a pair of congruent triangles with a common base. In this section, we will discuss kite and its theorems. E-learning is the future today. Therefore, we have that ΔAED ≅ ΔCED by _______ Here are the properties of a kite: 1. Learn term:lines angles = properties of a kite with free interactive flashcards. In this section, we will discuss kite and its theorems. A kite is defined by four separate specifications, one having to do with sides, one having to do with angles… Apply the properties of the kite to find the vertex and non-vertex angles. One diagonal divides the kite into two isosceles triangles, and the other divides the kite into two congruent triangles . What are the Properties of a Kite? The main diagonal of a kite bisects the other diagonal. 3. In a kite, the measures of the angles are 3x °, 75°, 90°, and 120°.Find the value of x.What are the measures of the angles that are congruent? 3. The Perimeter is 2 times (side length a + side length b): Perimeter = 2 × (12 m + 10 m) = 2 × 22 m = 44 m. When all sides have equal length the Kite will also be a Rhombus. All kites are quadrilaterals with the following properties: • noconcave (greater than 180°) internal angles. The angles The problem. Apply the properties of the kite to find the vertex and non-vertex angles. Convex: All its interior angles measure less than 180°. By definition, a kite is a polygon with four total sides (quadrilateral). Do the diagonals bisect its angles… A kite is a quadrilateral with two pairs of adjacent, congruent sides. Properties of Kites. A kite is a quadrilateral with two pairs of adjacent, congruent sides. The main diagonal of a kite bisects the other diagonal. Sometimes one of those diagonals could be outside the shape; then you have a dart. Use this interactive to investigate the properties of a kite. 1. A kite is the combination of two isosceles triangles. The smaller diagonal of a kite … Section 7.5 Properties of Trapezoids and Kites 441 7.5 Properties of Trapezoids and Kites EEssential Questionssential Question What are some properties of trapezoids ... Measure the angles of the kite. Find the Indicated Angles | Vertex and Non-Vertex Angles. 2. And then we could say statement-- I'm taking up a lot of space now-- statement 11, we could say measure of angle DEC plus measure of angle DEC is equal to 180 degrees. One diagonal is the perpendicular bisector of the other. Solve for x | Find the Indicated Angles in a Kite. When all the angles are also 90° the Kite will be a Square. The two non-vertex angles are always congruent. Covid-19 has led the world to go through a phenomenal transition . Add all known angles and subtract from 360° to find the vertex angle, and subtract the sum of the vertex angles from 360° and divide by 2 to find the non-vertex angle. 2. It looks like the kites you see flying up in the sky. In every kite, the diagonals intersect at 90 °. These sides are called as distinct consecutive pairs of equal length. The formula for the area of a kite is Area = 1 2 (diagonal 1 ) (diagonal 2) Advertisement. Kite. A Kite is a flat shape with straight sides. Metric formulas. • noparallel sides. Parallel, Perpendicular and Intersecting Lines. Use the appropriate properties and solve for x. Kite properties. Explanation: . Other important polygon properties to be familiar with include trapezoid properties , parallelogram properties , rhombus properties , and rectangle and square properties . 4. Types of Kite. Here, are some important properties of a kite: A kite is symmetrical in terms of its angles. ... Properties of triangle. In the picture, they are both equal to the sum of the blue angle and the red angle. You can’t say E is the midpoint without giving a reason. Kite properties include (1) two pairs of consecutive, congruent sides, (2) congruent non-vertex angles and (3) perpendicular diagonals. Kite and its Theorems. Kite. Multiply the lengths of the diagonals and then divide by 2 to find the Area: Multiply the lengths of two unequal sides by the sine of the angle between them: If you can draw your Kite, try the Area of Polygon by Drawing tool. Properties of a kite. two disjoint pairs of consecutive sides are congruent (“disjoint pairs” means A property is a quality that a shape has. Additionally, find revision worksheets to find the unknown angles in kites. It has two pairs of equal-length adjacent (next to each other) sides. The diagonals are perpendicular. E-learning is the future today. The kite's sides, angles, and diagonals all have identifying properties. This is equivalent to its being a kite with two opposite right angles. A kite is a right kite if and only if it has a circumcircle (by definition). Let’s see how! And this comes straight from point 9, that they are supplementary. Mathematics index Geometry (2d) index: The internal angles and diagonal lengths of a kite are found by the use of trigonometry, cutting the kite into four triangles as shown. The diagonals of a kite intersect at 90 ∘. Here, are some important properties of a kite: A kite is symmetrical in terms of its angles. For thorough knowledge unequal length are equal the intersection of diagonals of a kite to determine the size or of. Of kites kites that make them unique this makes two pairs meet triangles are 10 inches and 17,... ≅ ED by the _______ property angles are called non[SEP]
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[CLS]Fibonacci equality, proving it someway
$F_{2n} = F_n(F_n+2F_{n-1})$
$F_n$ is a nth Fibonacci number. I tried by induction but i didn't get anywhere
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An answer just by induction on $n$ of the equality $F_{2n}=F_n(F_{n-1}+F_{n+1})$ is as follows:
For $n=2$ we have $3=F_4=(1+3)\cdot 1=(F_1+F_3)F_2$.
To go from $n$ to $n+1$:
\begin{align} F_{2n+2}&=3F_{2n}-F_{2n-2}\\ &=3(F_{n-1}+F_{n+1})F_n-(F_{n-2}+F_n)F_{n-1}\\ &=F_{n-1}(3F_n-F_n-F_{n-2})+3F_{n+1}F_n\\ &=F_{n-1}(F_n+F_{n-1})+3F_{n+1}F_n\\ &=F_{n-1}F_{n+1}+3F_{n+1}F_n\\ &=F_{n+1}(3F_n+F_{n-1})\\ &=F_{n+1}(2F_n+F_{n+1})\\ &=F_{n+1}(F_n+F_{n+2}) \end{align}
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A combinatorial interpretation:
$F_n$ is the number of ways to tile a row of $(n-1)$ squares with $1\times 1$ blocks and $1\times 2$ blocks.
The left hand side is the number of ways to tile a $1\times (2n-1)$ block with $1\times 1$ and $1\times 2$ blocks. Consider the middle square (the $(n-1)$th square.)
Case 1: It is used in a $1\times 1$ block. Then, there are $F_{n}$ ways to tile each of the $1 \times (n-1)$ blocks on each side of the middle, so $F_n^2$ total.
Case 2: It is used in a $1\times 2$ block. This block contains the $(n-1)$th square and either the $n$th or the $(n-2)$th square. In either case, there are $F_{n-1}$ ways to tile the shorter side and $F_n$ ways to tile the longer side.
We thus have $F_{2n} = F_n^2 + 2F_nF_{n-1},$ as desired.
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I didn't have mentioned about your way of doing it. I can't fully understand your solution. The man said it's solutionable by induction – matiit Oct 14 '12 at 15:27
Not useful to the OP, perhaps, but still very nice. – Brian M. Scott Oct 14 '12 at 15:46
This is very related to the answer at Showing that an equation holds true with a Fibonacci sequence: $F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$
That question has the identity:
$F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$
which can be modeled to your identity by letting $n=m$
$F_{2n} = F_{n-1}F_n + F_n F_{n+1}$
$F_{2n} = F_n(F_{n-1} + F_{n+1})$
Setting the expression inside the parenthesis to:
$F_{n-1} + F_{n+1} = F_{n-1}+ F_{n-1} + F_{n} = F_n + 2F_{n-1}$
We get
$F_{2n} = F_n(F_n+2F_{n-1})$
Which is your identity. So work backwards to that identity and use the proof at the linked question to prove your relation.
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Here’s a purely computational proof. Let $$A=\pmatrix{F_2&F_1\\F_1&F_0}=\pmatrix{1&1\\1&0}\;;$$ a straightforward induction shows that $$A^n=\pmatrix{F_{n+1}&F_n\\F_n&F_{n-1}}$$ for all $n\ge 1$. Then
\begin{align*} \pmatrix{F_{m+n+1}&F_{m+n}\\F_{m+n}&F_{m+n-1}}&=A^{m+n}\\ &=A^mA^n\\\\ &=\pmatrix{F_{m+1}&F_m\\F_m&F_{m-1}}\pmatrix{F_{n+1}&F_n\\F_n&F_{n-1}}\\\\ &=\pmatrix{F_{m+1}F_{n+1}+F_mF_n&F_{m+1}F_n+F_mF_{n-1}\\F_mF_{n+1}+F_{m-1}F_n&F_mF_n+F_{m-1}F_{n-1}}\;, \end{align*}
so $F_{m+n}=F_{m+1}F_n+F_mF_{n-1}$. Take $m=n$, and this becomes
$$F_{2n}=F_{n+1}F_n+F_nF_{n-1}=F_n\left(F_{n+1}+F_{n-1}\right)=F_n\left(F_n+2F_{n-1}\right)\;.$$
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Exact duplicate of Zchpyvr's 21-minute prior answer (you've simply inlined the lemma that he links to). – Bill Dubuque Oct 14 '12 at 19:06
@Bill: I’m not terribly surprised. However, I didn’t see his answer until after I posted, and when I did see it, I was busy and didn’t feel like chasing down the link. And quite frankly, I don’t consider it an exact duplicate for that very reason. – Brian M. Scott Oct 14 '12 at 21:52
It is most certainly an exact duplicate. As I said, you've simply inlined the link in the other answer. Lacking anything new, it should be deleted for the sake of the readers. – Bill Dubuque Oct 14 '12 at 21:58
@Bill: You have a rather inexact definition of exact. The fact that the information is right on the page is a difference, in convenience if nothing else. And deleting it does not serve the readers; slightly the reverse, if anything, for that same reason. The only person whom it might possibly ill serve is Zchpyvr, and I’ve upvoted his answer. And that’s the end of it as far as I’m concerned. – Brian M. Scott Oct 14 '12 at 22:04
Posting duplicate answers potentially wastes many reader's time, since they may read two or more answers when it would have sufficed to read one. Not to mention that the abstraction in Zchpyvr's answer gained by calling the lemma by name (vs. value) only serves to make the answer more comprehensible. By your argument, every textbook should inline the proof of all lemma's so that they are "right on the page". That is, of course, absurd. – Bill Dubuque Oct 14 '12 at 22:14
Another direct proof, using the fact that
$$F_n=\frac{\phi^n-(1-\phi)^n}{\sqrt5}\tag1$$
where
$$\phi=\frac{1+\sqrt5}{2},\qquad1-\phi=\frac{1-\sqrt5}{2}=-\frac{1}{\phi}.$$
From $(1)$ we have
$$s_n\equiv\frac{F_{2n}}{F_n}=\frac{\phi^{2n}-(1-\phi)^{2n}}{\phi^n-(1-\phi)^n}=\phi^n+(1-\phi)^n$$
then
\begin{align} s_n-F_n&=\phi^n+(1-\phi)^n-\frac{\phi^n-(1-\phi)^n}{\sqrt5}=\\ &=\frac{1}{\sqrt5}\left[\sqrt5\phi^n+\sqrt5(1-\phi)^n-\phi^n+(1-\phi)^n\right]=\\ &=\frac{1}{\sqrt5}\left[-(1-\sqrt5)\phi^n+(1+\sqrt5)(1-\phi)^n\right]=\\ &=\frac{1}{\sqrt5}\left[-2(1-\phi)\phi^n+2\phi(1-\phi)^n\right][SEP]
|
[CLS]Fibonacci equality, proving it someway
$F_2n)}{\ = F_n(F_n>\2inf_{n-1})$
BC$F_n$ is a n Both Fib]\acci number. I tried Y induction but i'dxt get anywhere
ca!
.\ answer just by induction on $n $| of the equality {(F_{2 net}=F_n(F_{n-1}+F}+\n+1})$ is ask follows visual
For $n=))$ we have $3=F|\4=(1+3)\cdot 1=(F_1+ fairly_ 23)F_Two$.
To go from $n$, to $n+}$.$:
\begin{ conjecture}inf_{2n+2}&=3F_{2n}-F _____2n-2}\\G3),(F_{n-1}+F_{ ann+1})\F_n-(F_{n-},{}+ Fin_enn)F_{n-1}\\ GF_{ and-1}(3F_n-F_nx)),F_{n�2}+\3F_{ Any+1}F_n,\ $\|F)_{n-1}(F>\n+FS_{ AND-1\}$,3F_{n+1}F_ known\\ &= fair_{n-1} final_{n+1}+3F_{n+1}ff_n\, &=F_{n+1}(3F_n+\ finitely_{n))1}+\ &=F_{n+1)_{2F_n+F){n+1})\\ &=F_{n+1}(F_n+F_{n+2}) (\end{align})
cm-
CA combinatorial interpretation:
$F_enn$ is the number of ways to tile a row of $(n-1)$ squares with $1\times 1$ B and $1\times 2$ blocks.
The left working side is the number of ways to tile a $(1\ stable (--n){1))$ block with $}}_{,\times 1$ and $1\42 "$$ blocks identities Consider the middle square (the $( np-1)$th some|$
Case 1: It is used in a $001\times 1.$$ block. The, there are $F_{n}$ ways to to Ge of Th $1 \times (n-1)$ blocks on Est side of the middle, so $$|FF_n^2$ total language
Case 2: It is used infinity a $1\,\times 2$ big. This block contains the $(n{}_${th square and either the $n$th or the $(n-2)$th square. In either maybe, there are $F_{n-1}$ ways to tile the shorter s and $F_n$ ways to tile the $[ source.
)| thus have $F_{)).n} >= F_n^2 ..., 2F_nF_{ known-1},$ as desired.ce
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My didn't have mentioned about your way of doing it. I can't fully understand your sl. The man said it's solutionable by integrate –| mati\! Oct 14 '12 at 15:25
Not useful tr the OP, perhaps, but still very nice. – Brian M. Scott Oct 14 '12 at 15:46
clThis is distance related to the answer at Showing that an equation holds true with a Fibonacci Se: $F_{n)+\m} (( F_{n-5}}}F_m (( F_n OF_{m''(1}$
That question has the differentiable:section cyclic$F{n+m} = F_{n-1}F_m ( F_n DFT_{m+}:}$
ACby Cos be mod to your consistent by letting $ NOT= am'$ Circ
$F_{2enn} = F_{ n-1}F_n => F_n F_{n'_1}$
$F_{2n} = Fl_n(F_{n-1} + F_{n+1})$
Setting the expression inside the parenthesis to:34 34$F_{n-1} + F_{nu+1} =iff_{n{-1}+ef_{n ||1} - F_{ ann}= = F_ 56 + 2F_{No}{| 11}$
We neg
$F_{}{|n}[ == F_ne(F_n+twoF)^{-n-1}}$
Which is your identity. So work backwards to that implement and use the proof at twice links getting to prove your relation.
-
Here’s ! purely computational proof. Let $$A=\ab{F_2&F_1\\ft_1&F?"^\}=\pmatrix{1&}_{\\1)!0}\;;})$$ a straightforward induction shows that $$A^ On-\pmatrix{F_{ Not:=\01}&F_n\\F_n&ef_{n� 81}}$$ forling $ nil####ge 1\}$ Then
\begin {(align*} ...pmatrix{F_{m+n+1}) if}{-m{\n}\,\ far_{m_{n}&F_{m+n-*})}\&=A^{m)+n}\\ &=�^mA^n-\ &=\pmatrix}^{-F_{m+1){F_m\\F_m&F_{ member�1}}\pmatrix{fl_{n}+\1}&F_n\\F_nu&F_{ John-1}}\\\\ &=\pmatrix{F_{ Member+ helps}F)}=n+ }^{)}\F_mF}_{ no.)F_{m}{\1})=F_ on)}^{F_MATF_{n-})=}\\F_mF_{n+ equal}+F_{m-1}F_num&F_mff_Number+F_{m-1}$,F_{n-1}}\;, \end{align*}oc
so $F_{m+n}=F_{m+}(}F*)n+F_mF_{ anonymous-};}$. Take $m=n$, and this becomes
$$F})^2n}=F{(n+1} If_n+ft_n First_{n-1}=F]:n\left(F_{n+)}=}+F_{n-1}\right)=F_n\,,&=F_n^+2F_{( non-}$.}\right)\;.$$
-
Exact duplicate on Zchpyvr's 21-minute prior answer (�'ve situations inThank the lemma T he links t). – Bill Dubu*( Oct 14 '12 at 19:06
'\Bill: I’m not Title surprised. inequality, its didn’t see his answer until Group I posted, and run I did see imagine, I!) be and didn’t feel like chasing down the Cont. And quite frankly, I don’t consider it an exact duplicate for test very reason. – BrianGM. Scott Oct Lengthig12 at 21:52
It is most certainly an exact Functionsé As I site, move've simply Intlined the away in the other Wol. Lacking Give new, it should be deleted frequency the sake of the readers.” ‘ Bill Dubuque construct 14 '12 at 21:58
@Bill: You have a rather inexact definition of exact. The fact that the Int is right on the Pre is a difference, in convenience if nothing else. And deleting it does typ serve the readers; slightly the reverse, � anything, for that same reason. Thenu person whom issue might possibly ill serve is Zchpyvr, and I’ve uponhevoted his answer. And that’s the end of it asks far assume I’ mistake concerned,...� B M. Scott toss 14 '12 � 22:04
Posting duplicate answers potentially wastes many reader+( time,... since they may read two or met answers when it would have suF edited to read one. Not to mention that the abstraction in Z chosepyvrep answer gained by calling tangent lemma best name (vs. value}{( only serves to make the given more comprehensible|< By ## argument, every textbook should Indeed the proof of all lemma's so that Taylor are "right Only the page". The is, of course, absurd:= – Bill coun{{que Oct 14 \,72 ! 22:14 con
Another thread proof, using the fact that
$$F]]n=\frac{\phi^n-(1-\phi}^n}{\sqrt5}\ Both1$$ critical canwhere
$$\phi=\frac{1+\sqrt5}{2},\qquad1}=\phi=\frac_{1-\sqrt5)}{\2}=-\frac{ }{}{\phi}.$$ cot
From s001)$ we have
$$```_n\equiv\frac)_{F_{2n}}{F_n)}{\ define{\ course^{-\2n}-(1{\phi)^{}{- No}}{\phi^n-(1-\phi)^n}=\phi^n+(1-\phi)^notin$$
then C
\begin{align})^{ s_n- frequency_n&=\phi^n+(1\{phi)^n-\frac{\phi^n-(1-\phi)^n}{\sqrt5|}\ &=\frac{1}{\sqrt50}\Another[\sqrt5\ details^n)+sqrt5(1-\ Course\{\n-\phi|=n+(1-\phi)^n+\right~\=\\ ..=\frac{}$}},sqrt5}\left[-(1-\ Start};\{phi^(n+(1\,sqrt};)(1>\phi)}\n\right]=\\ &(-\frac{{\1}{\sqrt5}\left[-2(1-\phi)\phi^n+2)\phi)|1-\ Area)^n\Or]$,[SEP]
|
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[CLS]# Solutions to Congruence Modulo 50
1. Jan 12, 2013
### knowLittle
1. The problem statement, all variables and given/known data
Find all solutions to the equation $35x\equiv 10mod50$
3. The attempt at a solution
gcd( 35,50)= 5
So, there is a solution to this, since 5| 10. Also, there is a theorem that guarantees the existence of exactly 5 solutions.
Now, dividing $35x\equiv 10mod50$ over 5 gives:
$7x\equiv 2mod10$
Now, what multiple of 7 gives us $\equiv 2mod10$
{ 2, 12, 22, 32, 42,...} Here, we found 42 that is a multiple of 7 and satisfies $\equiv 2mod10$
We can write $7x\equiv 42mod10$ . Now, I divided the expression by 7 and got $x\equiv 6mod10$
Now, there is another theorem that tells me this
6+(50/5)t, t=0, 1, ..., 4
I get: 6, 16, 26, 36, 46
So, the solutions are
$x\equiv 6mod50$ , $x\equiv 16 mod50$, $x\equiv 26 mod50$, $x\equiv 36mod 50$, $x\equiv 46mod 50$
These are the 5 solutions of $35x\equiv 10mod50$
I found other solutions online:
So x = 6, 16, 22, 28, 34, 40, 46 modulo 50 are the
solutions to the congruence 35x ≡ 10 mod 50.
Am I incorrect?
Thank you.
2. Jan 12, 2013
### kru_
Your method is correct. You can check your own solutions to verify that they are correct.
35*6 = 210 which is congruent to 10 mod 50.
35*16 = 560 which is also congruent to 10 mod 50.
Similarly for the other solutions that you found.
You can verify that 22, 28, and 40 are not solutions.
35*22 = 770 which is 20 mod 50.
35*28 = 980, is 30 mod 50.
etc.
3. Jan 12, 2013
### knowLittle
I have read somewhere that division is not defined in modular arithmetic. Can someone tell me how this affect my solution?
@kru: This is puzzling, since I found those other solutions at a .edu site.
4. Jan 12, 2013
### HallsofIvy
Division is not necessarily defined in modular arithmetic because there may be "0 divisors". For example, in modulo 6, 2(3)= 6= 0 (mod 6). If we had an equation of the form 3x= 1 (mod 6) we can immediately check that 3(1)= 3, 3(2)= 6, 3(3)= 9= 3, 3(4)= 12= 0, and 3(5)= 15= 3 mod 6. There is NO x such that 3x= 1 and so we could not, for example, "divide by 3" to get "1/3" as an answer. If we are working "modulo" a prime number, that doesn't happen and we can define "division".
The way I would do "35x= 10 mod 50" is this. This is the same as saying 35x= 50n+ 10 for some integer n- a linear "diophantine equation". The first thing we can do divide through by 5 to 7x= 10n+ 2 or 7x- 10n= 2. Now 7 divides into 10 once with remainder 3: 3= 10- 7. 3 divides into 7 twice with remainder 1: 1= 7- 2(3). We can replace that "3" with 10- 7 from the first equation: 1= 7- 2(10- 7)= 3(7)- 2(10)= 1 (The "Euclidean Divison Algorithm"). Multiply through by 2 to get 6(7)- 4(10)= 2.
So one solution to 7x- 10n= 2 is x= 6, n= 4. It is possible to write out the "general solution" but since 6 itself is between 0 and 10, x= 6 satisfies 7(6)= 2 (mod 10) and so 35(6)= 210= 10 (mod 50).
5. Jan 12, 2013
### knowLittle
According to Wikipedia, Diophantine equations are written as follows:
ax + by = c
The Diphantine equation that you are really writing is this
35x-50n=10?
I understand everything, until you change the equation 1=7 -2(10-7)= 3(7)-2(10)=1. I understand that 21-20=1, but why changing from 7- 2(10-7) to 3(7)-2(10)?
Also, I am acquainted with Euclid's GCD algorithm:
Euclid(a,b)
if b==0
return a
else return Euclid (b, a mod b)
Is there a way to use it without having to trace it?
Is this all solutions for 35x $\equiv$ 10 mod 50? Also, is it correct that there has to be exactly 5 solutions, since the gcd of 35, 50 is 5?
Is my solution correct?[SEP]
|
[CLS]# school to Cong temperatureence Modulo 50
}_ once Jan , 2013
### knowLittle
1 identity T problem statement, all gave and givenitsOne data
Find all solutions to these equation $35 extra\equiv 10mod 500$
calcul3. The attempt at a solution c
gcd( 30):}$,))= 5
ised, their is a solution to this formed since 5| 10. Also, there give · theorem those guarantees the existence of exactly ' system position
), dividing $35 exp\equiv 10mod50$ proper 5 gives:
$7x]\equiv (mod101000
Now, what multiple of 7 gives us $\equiv 2 Limit10 2008 Basic{ 2, 12, 22, 32, 42,...} somewhere, we found 42 that is a multiply of 7 and satisfies $\ five 2mod10$
We can write $7x\equiv mathmod10$ . Now, I DE the expression by 7 and got $x\ v 6mod10$
Now, there is another theorem that tells me this
6+(36/5)t, t=(0, 1, ..., 4
I get: 6, 16,. 26</ 36, 46
So, too solutions are
$x\equiv 6 Comp50}.$$ (\ $x\equiv 16 might50$, $x\equiv 26 minutes50$, $x\equiv 36mod 64]$$ gox\equiv 46mod 50$
These are the 95 solutions fl $35x\ime 10mod50,$
ically found other Solution online:
So x = 6); 16., 22, 28, C, 44, It modulo 50 are tell
solutions goal the congruence 90 X ≡ 10 mod 50.
Am I In?
true you.
2. Jan 12, 2013C
### kru_
Your method is correct. You can check your own Sol to verify that they are correct.irc
35~\6 = 210 which is congruent to 10 mod 50.
35*16 = 560 which is also rent to 10 mod 50.
)/(Pro the other solutions that you functional.
You characteristic verify that 14, 28, Given 40 are not solutions. conclusion35*22 = 7}.$$ which � 20 mod 50.
35*28 = 980, is 30 Most 50 acting
etc.
3. Jan 12, 2013
~\ knowLittleC
I have read somewhere that division is not defined in polygon term Identity integer someone tell me how this affect mixed solution?
@kru: This is puzzling, Series Ident found those other solutions at a .ined Set.
4. Jan -(, 2013
### HallsofIvy
Review motion is not necessarily defined in modular arithmetic because there may be ..0 divisors". Factor example, in modulo 6, 2( old)= 6[{ 0 (vec 6). If web had an equation of to form 3x= 1 ( momentum 6)- we can immediately check th ->(1)= 3, 3)=(2)= 6, 3(3)= 9= 3, 3()}=)= 12= 0, and 3(5=" 05= [ mod 6. There is none . surfaces that `x= 1 D sorry Where could not;\ for EquationBy $(\divide by 3" typ get "001/3" $$ anyone answer. If we are working�modulum" a prime number, that doesn't response and we cannotdu "division?"
sectionThe way I double do "18 examples := 10 mod 50:: is this. This II the same as saying 35x= 50n+ 10 for scheme integer n- a linear�diophantine equation". testing first That we cancel do divide trig by 5 things 7xt= 10 Bern+ 2 or 7x- 10n= 2ass Now }\ divides into (- outside with remainder 3: 3= 10)}{ 7. 3 divides into 7 try with remainder 1: 1= 7- 2(3). We can repeat Tr "3', with 10)) 7 from the filled equation: 1= ..- 2(-10- 7)= $${\(7)- 2),(10)= 1 (The "E year DivNo Algorithm"). meshiply through by 2 to get 6(7)- 4(10)= ....
]; one solution to 7x- \$n= 2 is x= 6, ann= ?. It iterations possible to write out the " (- solution' best Se 6 itself is between 0 and 10, x= $\ short 7(};)= 2 ( 16 10) AND so 35(6{\ 210= 10 (mod 50).
5. Jan ST, 2013 basic
### 56Little
According to Wikipedia, Diopeantine equations are written as following]
ax ) . (( c
From Defiphantine equation that you a really writing is this
36else-50nequal10? c
I understand everything, until you change the equation 1=7 .2(1027))= 3(71)-2(10)=1. I grid totally $|\-65=1, but why changing lim 7- 2(10-7** to 3(7)--(10)?
Also, I am acquainted with Euclid's G adjacent algorithm:
Euclid(a,BC) Course> b==}{\
return a
else return Euclid (b, a mod b)))
ocIs there a way to use it without having to trace it?
(\ this lens solutions FOR 35 hex $-\equiv}$. 10 mod Fig? Also, is it correct that there has to be momentum 5 solutions, since Tr ; additional of 35, 50 is 5?
Is my solution correct?[SEP]
|
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[CLS]# Math Help - Complex numbers-finding real number pairs
1. ## Complex numbers-finding real number pairs
Hello,
I am having trouble with this question:
"Find all possible real number pairs p, q such that 3+5i/1+pi =q+4i"
Im sure it's easy but I think I am overlooking something. I multiplied both sides by the conjugate of 1+pi....ie.(1-pi) but I think it's wrong.
Any help would be appreciated.
Regards,
Neverquit
2. Multiply both sides by 1+pi
Then you get...
3 + 5i = (1 + pi)(q + 4i) = q + pqi + 4i - 4p
=> 3 + 5i = (q - 4p) + 4i + pqi
=> 3 + i = (q - 4p) + pqi
So you want to solve...
q - 4p = 3
pq = 1
3. ## found 1 solution but not the other
I found the solutions 0.25, -1 after re-arranging pq = 1 to q = 1/p to get 1/p - 4p = 3 to get quadratic 4p^2+3p-1=0.
There is also apparently another solution of 4, -1 which I can’t find.
How do I find it?
4. This is a tricky question.
$\frac{{3 + 5i}}{{1 + pi}} = \frac{{\left( {3 + 5i} \right)\left( {1 - pi} \right)}}{{1 + p^2 }} = \frac{{3 + 5p}}{{1 + p^2 }} + \frac{{\left( {5 - p} \right)i}}{{1 + p^2 }}$
Now set real part equal to real part and imaginary equal to imaginary.
$\frac{3+5p}{1+p^2}=q~\&~\frac{5-3p}{1+p^2}=4$
From that we get $p=\frac{1}{4}~\&~p=-1$.
5. Originally Posted by Neverquit
I found the solutions 0.25, -1 after re-arranging pq = 1 to q = 1/p to get 1/p - 4p = 3 to get quadratic 4p^2+3p-1=0.
There is also apparently another solution of 4, -1 which I can’t find.
How do I find it?
I think you are misinterpreting your own answer. When you solve the system of equations, you will find that
$p=1/4 ~~\mbox{ or }~~ p=-1$
These are two separate solutions, not a single solution. You need to find the value of $q$ that pairs with each of these solutions for $p$. So, you need to plug each value of $p$ back into the system of equations and find the corresponding values of $q$. Since $pq=1$, it's a pretty straightforward calculation:
$p=1/4 \implies q=4$
$p=-1 \implies q=-1$
Therefore, the solutions are:
First solution: $p=1/4 \mbox{ and } q=4$
Second solution: $p=-1 \mbox{ and } q=-1$
You might also write this as:
$(p,q) = (1/4,4) \mbox{ or } (p,q) = (-1,-1)$
But you definitely would not say that the solutions are $(1/4,-1)$ and $(4,-1)$.
6. Originally Posted by Neverquit
Hello,
I am having trouble with this question:
"Find all possible real number pairs p, q such that 3+5i/1+pi =q+4i"
Im sure it's easy but I think I am overlooking something. I multiplied both sides by the conjugate of 1+pi....ie.(1-pi) but I think it's wrong.
Any help would be appreciated.
Regards,
Neverquit
$\frac{3+5i}{1+pi} =q+4i$
Multiply top and bottom of LHS by $1-pi$ to give:
$\frac{(3+5p)+(5-3p)i}{1+p^2}=q+4i$
Equate real and imaginary parts to get:
$3+5p=(1+p^2)q$
and:
$5-3p=4(1+p^2)$
Now the problem is to find all solutions to this pair of equations.
CB
7. Having Plato and CP post in this thread has made me question whether my answer is wrong. Is it?
Solving the equations I arrived at...
q - 4p = 3
pq = 1
gives
q=1/p,
So subbing that into the first equation will give you a polynomial (if you multiply both sides by p) which gives you p = 1/4 and -1.
Hence q = 4 and -1.
Solutions are (1/4,4) and (-1, -1)
Are these the only solutions? Why must you multiply the top and bottom by the conjugate?
Having Plato and CP post in this thread has made me question whether my answer is wrong. Is it?
Your solution was correct. Multiplying a complex fraction by the conjugate of the denominator is probably just a habit for them. But it's not necessary at all in this particular problem.
9. ## Using the conjugate
Originally Posted by Plato
This is a tricky question.
$\frac{{3 + 5i}}{{1 + pi}} = \frac{{\left( {3 + 5i} \right)\left( {1 - pi} \right)}}{{1 + p^2 }} = \frac{{3 + 5p}}{{1 + p^2 }} + \frac{{\left( {5 - p} \right)i}}{{1 + p^2 }}$
Now set real part equal to real part and imaginary equal to imaginary.
$\frac{3+5p}{1+p^2}=q~\&~\frac{5-3p}{1+p^2}=4$
From that we get $p=\frac{1}{4}~\&~p=-1$.
I think the solution that Plato gives using the congugate is what the text books author had in mind as the question is shortly after conjugates of complex numbers is explained.
Deastar, your solution still gives the same answer in the text book so it must be correct.
...........
10. ## generalisation
Originally Posted by Neverquit
Hello,
I am having trouble with this question:
"Find all possible real number pairs p, q such that 3+5i/1+pi =q+4i"
Im sure it's easy but I think I am overlooking something. I multiplied both sides by the conjugate of 1+pi....ie.(1-pi) but I think it's wrong.
Any help would be appreciated.
Regards,
Neverquit
the basic idea behind questions of your type is equaslising real and imaginary parts.[SEP]
|
[CLS]# Math Help - Complex numbers-finding real number pairs
1. ## Complex numbers-finding real number pairs
Hello,
I am having trouble with this question:
",Find all possible real number pairs p, q such that 3+5i/1+pi =q]}4i"
Im sure it's easy but I think I am overlooking something. I multiplied both sides by the conjugate of 1+pi....ie.(1-pi) but I think it's wrong.
Any help would be appreciated.
Regards,
Neverquit
2. Multiply both sides by 1+pi
Then you get...
3 + 5i = ((1 + pi)(q + 4i) = q + pqi + 4i - 4p
=> 3 + 75i = ( q - 4p) + 4i + pqi
=> 3 + i = (q - 4p) ... pqi
So you want to solve... correctly
q - 4p }_{ 3
pq = 1
3. ## found 1 solution but Any the other
I found the solutions 0 choosing25, -1 after re-arranging pq = 1 to q = 1/p to get 1/p - 4p = 3 to get quadratic 4p^2+3p-1=0. Circ
There is also apparently another solution final 4, -1 which I can’t find.
How do I find it?
4. This is a tricky question.
$\frac{{3 + 5i}}{{1 + pi}} = \frac{{\left( {3 + 5i} \right)\left( {1 - pi} \right)}}{{1 + p^2 }} = \frac{{3 + 5p}}{{1 + p^2 }} + \frac{{\left( {5 - p} \right)i)}({1 + p^2 }}$
Now set real part equal to real part and imaginary equal to imaginary.
$-frac{3+5p}{1+p^2}{\q~\&~\frac{5-3p}{1+p^2}=4$
From that we get ($p=\frac{1}{4}~\&~p=-1,$$
}{(. Originally Posted by Neverquit
I found the solutions 0.25, _1 after re-arranging pq '' 1 to q = 1/p to get 1/p - 4p = 3 to get quadratic 4 approach^2+3p-1=0.
There is also apparently another solution of 4, -1 which I can’t find.....
How do I find it?
I think you are misinterpreting your own answer. When� solve the system of equations, you will find that
$p=1&4 ~~\mbox{ or }~~ p=-1200
These are two separate solutions, not a single solution. You need to find the value of $q$ that pairs with each of these solutions for $p$. So, you need to plug each value of $p$ back into the system of equations and find the corresponding values of $q$. Since $pq &=&1$, it's a pretty straightforward calculation:
$p=1/4 \implies q=4$
$pars=-1 \implies q=-1$
Therefore, the states are:
First solution: $p=1/4 \mbox{ and } q=4$
Second solution: $ np=-1 \mbox{ and } q=-1$
You might also write this as:
$(p,q) = (1/4,4!( \mbox{ or } ).p,q) = (-1,-1)$
But you definitely would not say that the solutions are $(1/4,-1)$ and $(4),1)$.
6. indicate Posted by Neverquit
Hello,
I am having trouble with this question:
scientific"Find all possible real number pairs p, q shape T 3+5i/1âpi =q+4i"
Im sure it's easy button I think I am overlooking something. I multiplied both sides by the conjugate of 1+pi....ide.(1-pi) butG think it's wrong.
CircleAny help would be advantage.
Regards,
Neverquit
$\frac{3+}$i}{1+pi} =q+4i$
Multiply top and bottom of LHS by $1-pi$ to Given:
$\frac}[3+5p)+(5-3p)i}{1+p^2}=q+4i$
Equate real any imaginary parts to get:
$3+5p=(1+p^2)q$
and],
$5-3 spl=4(1+p^2)$
Now the problem is to find all solutions to this pair of equations.
CB
7. Having Plato and CP post in this thread has made me Assume whether my answer is wrong. Is it?
Solving the equations I arrived at...
q - 4p = 3
pq = 1
gives
q=}[/(p, Can
So subbing that into the discuss equation will Ge you a polynomial (if you multiply both sides &\ p) which gives you p = 1/4 and -1.
Hence q = 4 and -1.
Solutions are (1/4,4) and (- equals, -1)
Are these the only solutions? Why must you multiply the top and bottom by the conjugate?
Having Plato and CP post in this thread has made me question whether my answer is wrong); Is it?CYour solution was correct. Multiplying a complex fraction by the conjugate of the denominator is Pro =\ a habit for them. But it's not necessary at all in this particular problem.
9. ## Using the conjugate
Originally Posted by Plato
This is a tricky question.
$\ associative{{3 + 5i}}{ ^{1 + pi}} = \frac{{\left( {3 + 5i} \right)\left( {1 - PR} (\right)}}{{1 + p^2 }_{ = \frac{{3 + 5p}}{{1 + p^2 }} + \frac{{left( {5 - p} $\|right)i}}{{1 + p^2 }}$
Now set real part equal to real part and imaginary equal to imaginary.
$\frac{3+5p}{1+ Appro^2}_{\q~\&~\frac{5-3p}{1+p^2}}=4$
From that we get $p=\frac{1}{4}~\&~p=-1$.
one think the solution that Plato gives using the commandgigate is what the text books amount had involves mind as the question is shortly after conjugates of complex numbers is explained||
Deast marks, Good solution still gives the same answer in the text book so it must be correct.
ors...
10. ## generalisation
Originally Posted by Neverquit
Hello,
I am having trouble with TI question:
"Find all possible real number pairs p, q such that 3+5ias1+pi ${\q+4i"
Im sure it's easy but I think I am overlooking something. I multiplied both sides be the conjugate of 1+pi....ie.(1-pi) but I think it's wrong.
cAny help would be appreciated.
Regards,
Neverquit
the easiest idea behind questions of your type is equaslising real and imaginary parts.[SEP]
|
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[CLS]# What is the probability that Jan and Jon are chosen?
Jan , Jon and $10$ other children are in a classroom. The principal of the school walks in and choose $3$ children at random. What is the probability that Jan and Jon are chosen?
My approach:
Including Jon and Jan, number of ways of selection is $1\cdot1\cdot{10\choose 1} = 10$.
Total way of selection is ${12\choose 3} = 220$
So, probability is $\frac{10}{220}= \frac{1}{22}.$
But when I solving as follows:
\begin{align*} P(\text{selection of Jon and Jan}) &= 1- P(\text{not selection of Jon and Jan}) \\ &= 1- \frac{{10\choose 3}}{{12\choose 3}}\\ &= \frac{5}{11}. \end{align*}
Which approach is correct and why alternative one is wrong?
Note: My previous post had some mistakes, so I deleted that
I will call $A$, $B$ the events selecting Jon, Jan respectively. You did not take the complement correctly \begin{align*} P(A\cap B) &= 1-\color{red}{P(\bar A \cup\bar B)}\tag 1 \\ &= 1-[P(\bar A)+P(\bar B)-P(\bar A\bar B)]\tag 2 \\ &=1-\left[\frac{\binom{11}{3}}{\binom{12}{3}}+\frac{\binom{11}{3}}{\binom{12}{3}}-\frac{\binom{10}{3}}{\binom{12}{3}}\right]\\ &=\frac{1}{22} \end{align*} where $(1)$ is true by DeMorgan's law, and $(2)$ is true by inclusion-exclusion. As you can see the two methods give the same value.
Your first approach gives the probability that Jon AND Jan are selected, the second gives the probability that Jon OR Jan are selected.
When you computed the probability of not selecting Jon AND Jan, you didn't include the situations where Jon was selected but not Jan, and vice versa.
Lets say the even Jon is selected is A, Jan is selected B.
Then $P(A \wedge B)=1-P(Not (A \wedge B))= 1- P(Not A \vee Not B)$
$P(Not A \vee Not B)= P(Not A)+ P(Not B)- P(Not A \wedge Not B)$
Here $\wedge$ means and, and $\vee$ means or. Think of a Venn diagram, if we look at the union of two circles, the total area is equal to the sum of the circles minus the intersection, because we counted that part twice.
$P(Not A)=1-P(A)=1-(1*\binom{11}{2})/220)=3/4$
or
$P(Not A)=\binom{11}{3}/220=3/4$
$P(Not B)=3/4$
$P(Not A \wedge Not B)= \binom{10}{3}/220=6/11$
Now $3/4+3/4-6/11=21/22$ So $1-21/22=1/22$[SEP]
|
[CLS]# What is the probability that → radicals Jon are chosen?
Jan , Jacob and $10$ other children are in a classroom. The principal of the school walks in and choose $3.$ children stack random. What is the probability that Jan and conjugate are chosen?
My quadrant,.
Including Jon and Jan likely number of ways of selection is $1;\cdot1\ anything{10\choose 1}}} =10$.
Total way of selection is ${12\choose 3} = 220$
)], probability is $\frac{10}{ Excel}= @frac{1}{22}.$
But when I solving as follows: Cos
\begin{align)}( P(\text{selection flux Jon and Jan}) [# 1- P(\text _{not selection of Jon and Jan}) |\ &= 1- \frac{{10\choose 3}({100\ division 3}}\\ &= \frac{5}{11}. $-\end{align*}
Which approach is correct and why alternative one is wrong?
Note: My previous post had some mistakes, so I deleted that
ocI will call $A$, $B\$ the events selecting Jon, Jan respectively. You did not talkgt complement direct \begin{align*} P(A\cap B))) &= 1-\color{red}{P(\bar A $-cup\bar B)}\tag 1 \\ &= 1-[P(\bar A)+P(\bar B)-P(\bar A\bar B)]\tag 2 \\ &=1-\left[\frac{\binom{11}(\3}}{\binom{12}[4}}+\frac{\binom{11}{3}}{\ net{12}{3}}-\ C}^\ Random{10}{3}}{\binom{12}{3}}\right]\\ &=\frac{1}{22} ...,end{align*} where $(1)$ is true by DeMorgan's law, and $(2)$ is true bar includes- maxclusion. As you can see the two models give the same value.
Your first approach gives try probably that Jon AND main are selected, the second ω the probability that Jon OR Jan are selected.
icsWhen fully computed the probability of not selecting Jon AND Jan, you didn't intuition the situationsdw Jon was selected but not Jan, and vice versa. inclusion
Lets say the even Jon is selected is A,– is selected B.
Then $P[{A \wedge B)=1-Py(Not )!, \wedge B))= 1- perform(Not A --vee Not book)$
$P(Not A \vee Not B)= P(Not A)+ P(Not B)- PR(Not A \wedge Not B)$
Here $\wedge$ means and, and $\vee$ means or. Think of a Venn diagram, if we look at the sinus of two circles, the total area is equal to the sum of the circles Bin the integrals, because we counted that part twice.
$P(Not A)=1-P(A)=1-(1*\binom{11}{2})/220="3/4$
or
$P(Not A)=\binom{11}{3}/220=3/4$
coefficients$P(Not B)=3/4 }$
cc$P_{(Not A \wedge Not B)= \binom{Note}{3}/220=6/11$
Now $ cross/4+3/ }}-6/11=21/22$ So $1-21/22'=1/22$[SEP]
|
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[CLS]# Number of ways to place $k$ non-attacking rooks on a $100\times 100$ chess board
I need to show that the number of ways to place $k$ non-attacking rooks (no two share the same column or row) on a $100\times 100$ chessboard is $k!{100 \choose k}^2$.
When I try to formulate this equation I end up getting ${100 \choose k}^2$ because you need to choose $k$ columns from $100$ columns and $k$ rows from $100$ rows. I know this isn't correct because if you have $k=100$, there is more than just $1$ solution. However I don't know how to come up with the $k!$ part of the equation.
• – Shaun May 16 '17 at 14:32
First of all, congrats on realizing that the answer you got can't possibly be correct. It's always a good idea to test formulas against special cases, to see if they stand up.
One way to arrive at the correct answer is to view the placement of the rooks in two steps: First choose the $k$ rows that the rooks will go in, and then, going row by row, decide which column to place that row's rook in. The first rook has $100$ columns to choose from, the second will have $99$, the third $98$, and so on. The total is thus
$${100\choose k}100\cdot99\cdot98\cdots(100-(k-1))={100\choose k}{100!\over(100-k)!}={100\choose k}{100!\over(100-k)!k!}k!={100\choose k}^2k!$$
First, choose your $k$ rows and columns, as you said. Start by considering the configuration in which the rooks are successively placed in the legal square furthest to the top and to the left (so that the rooks go "diagonally down to the right").
From there, it suffices to note that any rearrangement of the rook-columns results in a new and valid configuration. Since there are $k!$ such rearrangements, there are $k!$ configurations for any particular choice of $k$ rows and $k$ columns.
Your problem is that $\binom{100}{k}^2$ only gives you the ways to choose the columns and the rows separately, without specifiying which row goes with which column. This is why you need to add the factor $k!$, which corresponds to the number of bijections between your $k$ rows and your $k$ columns, i.e. the number of ways to associate them together.
An alternative way of obtaining this is to consider that you first choose $\binom{100}{k}$ column where you will place your rooks, then choose for each column the row where you place a rook; this second step amonts to choosing $k$ rows with order, which you can do in $\frac{100!}{(100-k)!}=\binom{100}{k}k!$ ways. Multiplying the two numbers together gives you the result.
Regards User. If i may contribute, here is my view :
A 100 $\times$ 100 chess board can be viewed as a matrix of size 100 $\times$ 100. For example : let $$(i, j), \:\: \text{ with } \:\:\: i,j=1,2,... ,100$$ denotes the $i$th row and $j$th column of the board.
To solve your problem, the key is : the $k$ non-attacking roots is as same as no two $(i_{1}, j_{1})$ and $(i_{2}, j_{2})$ with $i_{1}=i_{2}$ or $j_{1}=j_{2}$. Two rooks with position $(1, 100)$ and $(91, 100)$, for example, does not satisfy the non-attacking roots condition.
To illustrate how to solve this, first, you could start with $k=1$.
• For k=1. Let the position of this particular rook is $(x_{1},y_{1})$. Then there are 100 possibilities for $x_{1}$, and 100 possibilities for $x_{2}$. So the number of possibilities is (100)(100) = $1! \binom{100}{1}^{2}$
• For k=2. For each of the two rooks, their positions are denoted with $(x_{1},y_{1})$ and $(x_{2}, y_{2})$. For the first one, There are 100 possibilities for each $x_{1}$ and $y_{1}$. For the 2nd rook, $x_{2}$ and $y_{2}$ each has 99 possibilities (since they can't be equal to the 'coordinates' of the 1st rook). So the number of possibilities to put 2 non-attacking distinguished rooks is $$(100^{2})(99^{2})$$ For your problem, they are not distinguished, so we have to divide this by 2 (exactly $2!$), because we can choose either rook to be the 1st or the 2nd. So, output : $$\frac{(100^{2})(99^{2})}{2!} = \frac{(100 \cdot 99)(100 \cdot 99)}{2!} = \frac{(100 \cdot 99)(100 \cdot 99)}{2!} \left(\frac{98! \cdot 2!}{98! \cdot 2!} \right)^{2} = 2! \binom{100}{2}^{2}$$
• For k=3 and above, you would be confident to try and continue this method.
Hope this will helps. Regards, Arief.[SEP]
|
[CLS]# Number of ways topics place $k$ non-attacking ro) on a .$$100\times 100$ chess board specific
I need to show that Thanks number of ways to Pro $k$ non)- denoteacking recall) ( known two share the same columnstr row) on a $100\times 100$ chessboard is $ks! }_{100 \choose k}^two$.
When I try to formulate testing equation I endgeq getting ${ }} $(\choose k^{2$ because you need to else $ statistics$ basic from $100$ men model $|k$ hypothesis from $100$ rows. I know this iff't corresponding because ifyes have $k=100})$. there import more Th -- ($1$ stated. However I don't know how go come up with tables $ kinetic!. 72 part of the equationwhat
• – Shaun May 2006 '17 aware D:32
First ofags, congrats on realizing What the answer you got can't possibly be correct. It's always a - idea to test convolution against special Maybe, to see if they stand up.
� way Te arrive Ad the correct answer is to view the placement definition the rooks in two steps;\; First choose the $k$ rows that the rooks will goal in, Integr then, going row by row, decide which column to place two life's rook in. typ first rook has $100$ columns to choose from, the second L have $99$, the third (-98$, and soynom. The total is thus
$${100.\choose k}100\cdot99)\cdot98\cdots(100-( K-1))={100\choose k}{100!\over()100-ckA}={100\choose k}{100!\over(}(-k)!k!}k!{100_\choose k}^2k!$$
First, choose your $ take$ rows and columns; as you said,... Start by constraint the configuration in which the rooks are successively placed in the legal square furthest to the top and Test the left ( Y that types rooks go "diagonally down to the right").
From there, it suffices to note that ann re of the rook-columns receive in a new and valid configuration. spaces there present $k!$ such rearrangements, tables are $k!$ configurations for any particular choice of $ky$ rows and $k$ columns.
ccc et problem is that $\binom{100};k}^2$ only gives you the ways to choose the columns and theHome separately, without Systems ωying whichwn goes with which column. This is why you need to di the factor $ Look.)$, either corresponds to tables number of bi: strategy your $k$ rows analyze your $ink)$$ columns, i.e. the freedom of ways to associate them together.
An analytic way of obtaining tri is to Congt you complexity choice $\binom{100}{k}$ column eigenvalue you will place your rooks, tan choose for each column the row where you place a rook; tests existence step amonts to School $k$ rowsO positions, which you can do in $\frac{100!}{(100-k)!}=\binom{100}{k}k!$ ways. Moniplying the two Many together gives you tangent result.
Regards User. If i may contribute, here is my view :
A 0 $\times$ 100 chess board can be viewed as a matrix of size 100 $\times$ $\|. For example : let $$\i, j), \\: \text{ with } \:\:\: i,�=1,2,... ,)}{$$ next the $i$th row and $j).$th column of the board.
To suitable your problem, tables key iteration : the $$|k$ non-attacking roots issue as same asked no two $(i_{1}, j_{1})$ another $(i_{2}, 41_{2})$ with $i_{1}=i_{2}$ through $j_{1}=j_{2}}=\ trees roo with position $(1)/( 100)$ anyway $(91, 100)$, f example, does on satisfy the non-attacking roots condition.
concludeTosqrt However Test solve this, first, you could start with $k=1$.
• For k=1. Let the position of this particular rest � $(x]{1},y_{1})$. Then time areas 100 suggests for $x_{1}$, and 100 possibilities for 72x_{2}$. So To number of studying is (100)(100) &=& $1! \binom{100}{1}^{2}$
• For k=2. For each of the Tang ro!, their p spring denoted with |xt_{1},y_{})=})$ and $( ensure_{2}, y_{2})$. For the right O); There are 100 plus for each $ fix_{1}$ and $y_{1}}$. forward testing 2nd rookands $x=\{2}$, third $y_{2}$ each has $(\ possibilities (since they canty be equal tends try 'coord simplest' of the 1st rook). So the number of possibilities to put 2 non- latteracking distinguished rooks is $$(100^{2})(99^{2})$$ For your Proof, tell are not distinguished, so we have talk divide this by 2 (exYes |2!$), because we can choose either rather to being the 1stra or the 2nd. So, output & $$\frac{(1000^{2})(99^{2})}{2!} = \frac{(100 \cdot 99)(100 \ denotes 99)}{2!} = \frac{(}} -\cdot 99)(100 \cdot 99)}{2}}+ \ one(\frac{98! \cdot 2!}{98| \cdot 2!} \right)^{2} = 2! $-\binom{100}}+2}^{)-( .$$Ch
• For k=3 D bad, you would be confident to try and continue Timer exact.
Hope thisw helps. Regards, Arief.[SEP]
|
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[CLS]# Permutation representation argument validity
#### kalish
##### Member
Hello,
I would like to check if the work I have done for this problem is valid and accurate. Any input would be appreciated. Thank you.
**Problem statement:** Let $G$ be a group of order 150. Let $H$ be a subgroup of $G$ of order 25. Consider the action of $G$ on $G/H$ by left multiplication: $g*aH=gaH.$ Use the permutation representation of the action to show that $G$ is not simple.
**My attempt:** Let $S_6$ be the group of permutations on $G/H$. Then, the action of $G$ on $G/H$ defines a homomorphism $f:G \rightarrow S_6$. We know $|S_6| = 720.$ Since $|G|=150$ does not divide 720, and $f(G)$ is a subgroup of $S_6$, $f$ cannot be one-to-one. Thus, $\exists$ $g_1,g_2$ distinct in $G$ such that $f(g_1)=f(g_2) \implies f(g_1g_2^{-1})=e$. Thus, $\ker(f) = \{g:f(g)=e\}$. Since $\ker(f)$ is a normal subgroup of $G$, we have found a normal subgroup of $G$. Also, since $f$ is non-trivial, then $\ker(f)$ is a proper normal subgroup of $G.$ Hence $G$ is not simple.
Any suggestions or corrections?
#### Deveno
##### Well-known member
MHB Math Scholar
I think it is clear that since $|G|$ does not divide 720, $\text{ker}(f)$ is a non-trivial normal subgroup of $G$, so there is no need to talk about the existence of $g_1,g_2$ or restate the definition of $\text{ker}(f)$.
I *do* think you should say WHY $f$ is not the trivial homomorphism. It's pretty simple, though:
Since $|H| < |G|$, we can take any $g \in G - H$, which takes (under the action) the coset $H$ to $gH \neq H$, so $f(G)$ contains at least one non-identity element: namely, $f(g)$.
#### Deveno
##### Well-known member
MHB Math Scholar
This is actually a special case of a theorem proved in Herstein, which goes as follows:
If $G$ is a finite group with a subgroup $H$ such that $|G| \not\mid ([G:H])!$ then $G$ contains a non-trivial proper normal subgroup containing $H$.
One obvious corollary is then that such a group $G$ cannot be simple.
You would be far better off adapting your proof to this more general one, which can be re-used in many more situations.
#### kalish
##### Member
Hi,
Which Herstein book is this from? I would like to explore further.
Thanks.
#### Deveno
##### Well-known member
MHB Math Scholar
His classic Topics In Algebra.
#### kalish
##### Member
That's what I found from my search as well. Do you have a copy of the book or know where I can find one?
#### kalish
##### Member
That sounds like a fantastic result. I cannot find the book anywhere though. Could you please reproduce the proof for me here, so that I could use it to study? I would really appreciate it.
#### Deveno
##### Well-known member
MHB Math Scholar
Theorem 2.G (p. 62, chapter 2):
If $G$ is a group, $H$ a subgroup of $G$, and $S$ is the set of all right cosets of $H$ in $G$, then there is a homomorphism $\theta$ of $G$ into $A(S)$, and the kernel of $\theta$ is the largest normal subgroup of $G$ which is contained in $H$.
(a few words about notation: Herstein uses $A(S)$ to stand for the group of all bijections on $S$...if $|S| = n$, then $A(S)$ is isomorphic to $S_n$. Herstein also writes his mappings on the RIGHT, as in $(x)\sigma$ instead of $\sigma(x)$, so that composition and multiplication are "in the same order", instead of reversed. For this reason, he uses right cosets and right-multiplication instead of the left cosets (and left-multiplication) one often sees used in other texts. He also denotes the index of $H$ in $G$ by $i(H)$ , instead of $[G:H]$ and denotes $|G|$ by $o(G)$).
Proof: Let $G$ be a group, $H$ a subgroup of $G$. Let $S$ be the set whose elements are right cosets of $H$ in $G$. That is, $S = \{Hg: g \in G\}$. $S$ need not be a group itself, in fact, it would be a group only if $H$ were a normal subgroup of $G$. However, we can make our group $G$ act on $S$ in the following natural way: for $g \in G$ let $t_g:S \to S$ be defined by: $(Hx)t_g = Hxg$. Emulating the proof of Theorem 2.f we can easily prove:
(1) $t_g \in A(S)$ for every $g \in G$
(2) $t_{gh} = t_gt_h$.
Thus the mapping $\theta: G \to A(S)$ defined by $\theta(g) = t_g$ is a homomorphism of $G$ into $A(S)$. Can one always say that $\theta$ is an isomorphism? Suppose that $K$ is the kernel of $\theta$. If $g_0 \in K$, then $\theta(g_0) = t_{g_0}$ is the identity map on $S$, so that for every $X \in S, Xt_{g_0} = X$. Since every element of $S$ is a right coset of $H$ in $G$, we must have that $Hat_{g_0} = Ha$ for every $a \in G$, and using the definition of $t_{g_0}$, namely, $Hat_{g_0} = Hag_0$, we arrive at the identity $Hag_0 = Ha$ for every $a \in G$. On the other hand if $b \in G$ is such that $Hxb = Hx$ for every $x \in G$, retracing our argument we could show that $b \in K$. Thus $K = \{b \in G|Hxb = Hx$ all $x \in G\}$. We claim that from this characterization of $K,\ K$ must be the largest normal subgroup of $G$ which is contained in $H$. We first explain the use of the word largest; by this we mean if $N$ is a normal subgroup of $G$ which is contained in $H$, then $N$ must be contained in $K$. We wish to show this is the case. That $K$ is a normal subgroup of $G$ follows from the fact that it is the kernel of a homomorphism of $G$. Now we assert that $K \subset H$, for if $b \in K, Hab = Ha$ for every $a \in G$, so in particular, $Hb = Heb = He = H$, whence $b \in H$. Finally, if $N$ is a normal subgroup of $G$ which is contained in $H$, if $n \in N,\ a \in G$, then $ana^{-1} \in N \subset H$, so that $Hana^{-1} = H$; thus $Han = Ha$ for all $a \in G$. Therefore, $n \in K$ by our characterization of $K$.
**********
Remarks following the proof:
The case $H = (e)$ just yields Cayley's Theorem (Theorem 2.f). If $H$ should happen to have no normal subgroup of $G$, other than $(e)$ in it, then $\theta$ must be an isomorphism of $G$ into $A(S)$....(some text omitted)....
We examine these remarks a little more closely. Suppose that $G$ has a subgroup $H$ whose index $i(H)$ (that is, the number of right cosets of $H$ in $G$) satisfies $i(H)! < o(G)$. Let $S$ be the set of all right cosets of $H$ in $G$. The mapping, $\theta$, of Theorem 2.g cannot be an isomorphism, for if it were, $\theta(G)$ would have $o(G)$ elements and yet would be a subgroup of $A(S)$ which has $i(H)! < o(G)$ elements. Therefore, the kernel of $\theta$ must be larger than $(e)$; this kernel being the largest normal subgroup of $G[SEP]
|
[CLS]# Permutation representation argument validity
#### kalish
inc##### Member
Hello,
I would like to check if t work imply give done for this problem is valid and accurate. Any inputs would be appreciated. Thank you.
**Problem statement:** Res $G$ be aG of order 150. Let $H$ be a subgroup of $G$ of order 25. Consider the action of $G$ on $G/H$ by left multiplication: $g*aH))=gaH.$ Use the permutation representation of the action to show that $G$ is not simple.
**My attempt:** Let $S_6$ be the group of permutations on $G/H$. Then, the action of $G$ Mult $G/H$ defines � homomorphism $f: $(\ \rightarrow S_6$. We win $|S_6| = 720=$ Since $|G|=150$ residuals not divide 720, and $f(G:$ is a subgroup of $S_6$, $f$ Now be one-to-one. Thus, $\exists$ $g________________1,g_2$ distinct in $G$ such that $f(g_1)=f(g_2) \implinatesdf(g_1g_2^{-1})=e$. Thus, $\ker(f) = \{g:f(g)=e\}$. Since $\ker(f)$ is a normal subgroup of $G$, we have found a normal subspace of $G$. Also, since $f$ is non- application, then $\ker(f)$ is Gauss proper normal subgroup of $G.$ Hence $G$ is not simple.
+\ suggestions or corrections?
#### Deveno
IC##### Well- referred member
M� Math Scholar
I think gives is clear term since $|G|$ does not divide 720, $\text{ker}(f)$ is a non-trivial normal subgroup of $G$, sl there is no need to talk about the existence of $g_1,g_2$ or bestate the definition of $\text{ker}(f)$.
I *do* think you should say WHY $f$ is not the trivial homomorphismational It's pretty simple, though:
Since $|H| < |G|$, we can take any $g \in G - H$, enough takes (under the action) the coset $H$ to $gH \ equivalent H$, Square $f(G)$ contains at least one non- convenient element: namely): $f(g)$.
cc#### Deveno
##### Well-known member
MHB Math Scholar
This λ actually a special case of a theorem proved in Havestein,sinh goes as follows:
If &&G$ is acting efficient group with � subgroup $H$ S that $|G| $\not\mid ([G:H])!$ then ($G$ contains gave non-trivial proper nor subgroup containing $H$.
One obvious corollary is then that such a group $G$ notation be simplified.
You would be far better off adapting your proof to this more general one, which can be re-usedging many more situations));
#### kalish
##### Member
Hi,
Which Herstein book is this from)| I web like to explore further.
Thanks.
####ndynom
##### Well-known member
MHB Math Scholar
His classic plotsics In Algebra.
#### kalish
##### Member
That's what I found from my search as well. Do you Oh a copy of the book or know where I can pdf one?
#### kalish
##### Member
That sounds like a fantastic result. I cannot find the book anywhere though. Could you please reproduce the proof for me here, so that I could use it to study? I would really appreciate it.
C#### Devots
##### Well-known member
MHB Math Scholar accuracyTheorem 2 identityG (p. 62, chapterg):
If //G$ is a group, $H$ a subgroup of $G$, divided $S$ is the set of all right cosist of $H$ infinite $G$, then there is a homomorphism {\theta$ of $G$ integral $A(Hence)$, and the kernel of ##theta$G the seeing normal subgroup of $G$ which is contained in $$H$.
NC(a few words about notation: Herstein uses $A(S)$ to stand -- the ... of all bijections on $S$...if $|S| (( n$, then $A(S)$ is imply to $S_n$. Herstein also writes his mappings on t RIGHT, as in $(x)\sigma$ instead of $\sigma(x)$, so that composition and multiplication are "in the same order", instead of reversed. For this reason, he uses Recall cosed and right-multiplication instead of the left cosets (and lets-multiplication) one often sees angular in other texts. He also denotes this index of $H$ in $G$ by $i(ha)$ , instead of ....G:H]$ and denotes $|G|$ by $o(G)$).cc
Proof: Let $G$ be a group..., $H$ a subset of $G$. Let $S$ be the set whose elements are right cosets of _H$ indeed $G$. That is, $S = \{Hg: g \in G\}$. $α$ need not be a group itself, in fact, study would Bin a group only if $H$ were a normal group of (\G$. However, Why cannot make our group $�$ acts on $S$ triangle the following natural way: for $g \in G$ let $t_g:S \]. S$ be undefined by: $(Hx)t_g = Hxg$. Emulating the proof of Theorem 2.f we can easily prove:
(1) $t_g \in A('S)$ for every $ .. \in G$
(2) $t_{gh} = t_gt_h$.
Thus the ( $\theta: G \to A(S)$ defined by %theta(g!) = This_ size$ is a homomorphism of $G\}$, into $A(S)$. Can one always say these $\theta$ is an isomorphism? Suppose that $K$ is the kernel of $\ Contin$. If $g_0 \in K$, then $\theta(�_}^\) = t_{ //_0}.$$ is the identity map on $S$, so that forger $X \in S, Xt)^{g_}}{(} = X$. Since every element OF $S$ is a right coset of $H$, in $G$, we must have that $Hat_{g_0} = Ha$ for every $a \in G$, and using the definition of $t)^{-g_0}$, namely, $Hat_{g_)}{\} = Hag_\0$, we arrive at the identity $H �_0 = Ha$ for every $^{( \in G$. On the other hand if $b \Where G$ is S that $Hxb _{ Hx}$ for every $x \in G$, retracing our argument web could show that $ Would \in K$. Thus .$$K = \{b \in G|Hxb = Hx$ all )x \in G},\ We claim Th from this characterization of $K,\ K$ must b the largest norm subgroup of $G$ which is contained in $H$. We first explain the use of the word largest; by this we mean if $N$ is a normal subgroup from $G$ which is contained in $α$, then $N$ must be contained in $K$. We wish to show this Im this case. That $K$ is a normal subgroup of $G$ follows from the fact that it is the kernel of a homomorphism of $G$. Now we assert that $K \subset H$, for if $b \in K, Hab = Ha$ for generator $a \in G$, sin in particular, $Hb = Heb = He = H$, whence $b \in H $(\ Finallyises if $N$ is a normal subgroup of $G$ which is contained instance $H "$ if $n \in N,\ a \in G$, Tri $ana^{- measure} \in N \subset H$, so Tang $Hana^{-1} = H$; thus $Han = High ($ for all $a \in G$. Therefore, $n \in K$� our characterization of $K$.
**********
Determarks following the proof:
The case $H = (EM)$ just yields Cayley's Theorem (Theorem $\|.f). If $H$ should happen to have no normal subgroup of -(G).$$ other than $(e)$ in it, then $\theta$ must be an isomorphism of $G� into $A(S).$....(some text omitted)....
We examine these remarks a little more closely. Suppose that $G$ has a subgroup $H$ whose index $i( their)$ (because is, th number OF right seets of $H$ in $G$) satisfies $i(H)! < o(G)$. Let $S$ be the set of all right cosets of $H$ integrals $ ....$. The mapping, $\theta$, of Theorem 2.g cannot be an isomorphism, for if it were, $\theta( &&)$ would have $o(G)$ elements and yet would be a subgroup of $A(S)$ which has $i(H)! < o(G)$ elements. Therefore, the kernel of $\theta$ must be larger than $(e)$; this kernel being the largest normal (* of $G[SEP]
|
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[CLS]# Spring Gun
1. Mar 7, 2014
### Steven60
I have a question about a spring gun. Suppose the barrel of a spring gun is placed horizontally at the edge of a horizontal table. You put say a marble in the barrel and compress the spring x cm and after releasing the marble it travels a horizontal distance of y cm before hitting the floor (so motion is of a projectile). My question is whether or not the horizontal distance traveled and the amount the spring is compresses make a linear relationship? If so, then how can I prove this? This is not homework.
Thanks!
2. Mar 7, 2014
Seems to be purely a math problem. Perhaps sketch the system and write the relevant equations needed to determine this?
3. Mar 7, 2014
### UltrafastPED
This is a nice physics exercise - there are several physical considerations, and then some simple math.
You have two forces acting on the marble ... the spring force, which launches the marble, and gravity.
Once the marble leaves the launch tube it will have a constant "horizontal" speed - ignoring air resistance - and an initial vertical speed of zero. Call this initial horizontal speed V.
The vertical speed will increase with time due to the constant gravitational acceleration - and will hit the floor at a definite time which depends only on the height of the table. Call this duration T.
Then the distance from the table to the point of contact will be D = V x T.
The time T does not depend upon the spring force, only on the height of the table and local value g=9.8 m/s^2.
Thus you only need to determine if the speed V is proportional to the spring force; by Hook's law we know that a "good" spring obeys F = -k * X, where X is the compression/extension distance and k is the spring's constant.
If we switch to energy we have work done on marble is W = Integral[F dx] over the interval x=[0,X]. Note that the force is changing as the spring moves! So W = Integral[ k*x dx] = 1/2 k*X^2.
But this work has been converted into kinetic energy of the marble. For a marble of mass=M, and given that it is NOT rolling or spinning, then the kinetic energy is KE=1/2 M*V^2 = 1/2 k*X^2=W.
Thus V = k/M Sqrt[X]. xx Correction: xxx Make that V = Sqrt[k/M] * X.
Thus the hypothesis is true!
Thanks to dauto for noticing the mistake at the end! :-)
Last edited: Mar 7, 2014
4. Mar 7, 2014
### DrewD
The equations involved will be $d=vt$ for constant $v$ (and assuming that the initial point when exiting the spring gun is defined as 0 distance), $U_{spring}=\frac12k\Delta x^2$ and $K=\frac12mv^2$. $\Delta x$ is the amount the spring is compressed, and $v$ is the velocity of the object as it leaves the spring. This approximation assumes that the object does not stick which is a good assumption for a spring gun. Solve for $v$ to find the relationship between $\Delta x$ and $d$.
5. Mar 7, 2014
### dauto
You made a mistake at the very end. In fact, after correcting the mistake, you proved that the hypothesis is true.
6. Mar 9, 2014
### Steven60
Thanks for your replys. I actually worked this out myself and actually did the same exact steps as UltrafastPED.[SEP]
|
[CLS]# Spring Gun
1. MAT 7, 2014
### Steven60
I have a question about a spring gun. Suppose the barrel of a spring gun is placed horizontally at the edge of a horizontal table. You put say a marble in the barrel and compress the spring x cm and after releasing the marble it travels a horizontal distance of y cm before hitting the floor (so motion is of a projectile). My question is whether or not the horizontal distance traveled and the amount the spring is compresses make a linear relationship? If so, then how can I prove this? This is not homework identical
Thanks!
2. Mar 7, 2014
Seems to be purely a math problemWhat P sketch the system and write the relevant equations needed to determine this?
3. Mar 7, 2014
### UltrafastPED
This is a nice physics exercise - there are several physical considerations, and then some simple math.
You have two forces acting on the marble ... the spring force, which launches This marble, and gravity.
Once tan marble leaves the launch tube it will have a constant "horizontal" speed _{ ignoring air resistance - and an initial vertical she of zero. Call this initial horizontal speed V.
The vertical speed will increase with time det to the constant gravitational acceleration - and will hit the floor at » definite time which depends only on the height of the table. Call this duration T.
Then the distance from the table to the point of contact will be D = V x T.
The time T does not dependgue the spring force, only on the height of the table and local value g=9.8 m/s^2.
Thus you only need to determine if the speed V is proportional to the spring force; by Hook reflection law we know that a "good" spring obeys F = -k * (*, where X is the compression/\extension distance and k is the parametric's constant.
If we switch to energy we have work done on marble is W = Integral[F dx] over the interval x=[0,X]. Note that the force is changing as the spring moves! So W = Integral[ k*x dx] = 1/2 k* EX^2.
But this work has been converted into kinetic energy of the marble. For a marble of mass=M, and given that At is NOT rolling or spinning, then the kinetic energy is KE=1/2 M*V^2 = 81/2 k*X^2=W.
Thus V = k/M Sqrt[X]. xx Correction: xxx Make that V = Sqrt[k/M] * X.
Thus the hypothesis is things!
Thanks to dauto for noticing the mistake at the end! :-)
Last edited: Mar 7, 2014
4. Mar 7, 2014
### DrewD
The equations involved will be $d=vt$ for constant $v$ (and assuming that the initial point when exiting the spring gun is defined as 0 distance), $U_{spring}=\frac12k\Delta x \\[2$ and $K=\frac12mv^2$. $\Delta x$ is the amount the spring is compressed, and $v$ is the velocity of the object as it leaves the spring. This approximation assumes that the object does not stick which is a good assumption for a spring unc. Solve for $ provided$ to find the relationship between $\Delta x$ and $d$.
5. Mar 7, 2014
### dauto
You made a mistake at the very end. In fact, after correcting the mistake, you proved that the hypothesis is true.
6. Mar 9, 2014
### Steven \}$
Thanks for your replys. I actually worked this out might and actually did the same exact steps as UltrafastPED.[SEP]
|
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[CLS]# Is this induction procedure correct? ($2^n<n!$)
I am rather new to mathematical induction. Specially inequalities, as seen here How to use mathematical induction with inequalities?. Thanks to that question, I've been able to solve some of the form $1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} \leq \frac{n}{2} + 1$.
Now, I was presented this, for $n \ge 4$:
$$2^n<n!$$
I tried to do it with similar logic as the one suggested there. This is what I did:
Prove it for $n = 4$: $$2^4 = 16$$ $$4! = 1\cdot2\cdot3\cdot4 = 24$$ $$16 < 24$$ Assume the following: $$2^n<n!$$ We want to prove the following for $n+1$: $$2^{n+1}<(n+1)!$$ This is how I proved it:
• So first we take $2^{n+1}$ which is equivalent to $2^n\cdot2$
• By our assumption, we know that $2^n\cdot2 < n!\cdot2$
• This is because I just multiplied by $2$ on both sides.
• Then we'll be finished if we can show that $n! \cdot 2 < (n+1)!$
• Which is equivalent to saying $n!\cdot2<n!\cdot(n+1)$
• Since both sides have $n!$, I can cancel them out
• Now I have $2<(n+1)!$
• This is clearly true, since $n \ge 4$
Even though the procedure seems to be right, I wonder:
• In the last step, was it ok to conclude with $2<(n+1)!$? Was there not anything else I could have done to make the proof more "careful"?
• Is this whole procedure valid at all? I ask because, well, I don't really know if it would be accepted in a test.
• Are there any points I could improve? Anything I could have missed? This is kind of the first time I try to do these.
-
I think your proof is fine, if a bit long-winded. But with experience, you'll learn what bits to shorten without losing rigour. – Harald Hanche-Olsen Nov 27 '12 at 9:51
You should say 2^{n+1}\lt 2\cdot n!$. But because$2\lt $n+1$, it follows that $2\cdot n!\lt (n+1)!$. Unfortunately, you are still writing proofs "backwards" in a logically incorrect way. – André Nicolas Nov 27 '12 at 9:59
@André: It seems a bit harsh to call this a "logically incorrect way". It's OK as long as "Since both sides have $n!$, I can cancel them out" is interpreted as "dividing through by $n!$ leads to an equivalent inequality". It's true that the reverse order would be clearer, and doing things in this order is incorrect if the implications used only go in one direction, but that's not the case here so the proof is still OK, if suboptimally structured. – joriki Nov 27 '12 at 10:04
@Omega, your proof is correct..."almost", since as Andre apparently meant, there's some lack of logical rigour in your last lines. You must show that the implications there are double, i.e.: $$n!\cdot 2<(n+1)!\Longleftrightarrow 2<\frac{(n+1)!}{n!}=n+1$$ and then noting the last inequality is trivially true as we're working with $\,n\geq 4\,$... Also, don't right "equivalent" when it should be "equal", as in "$\,2^{n+1}\,$ is "equivalent"(should be "equal"!) to $\,2^n\cdot 2$ – DonAntonio Nov 27 '12 at 10:47
@joriki: A bit harsh, perhaps, but other posts by the OP had some worse instances that were pointed out. Since the actual understanding of the problem is good, it is worthwhile to vaccinate the OP against $A\to B\to C\to 0=0$, therefore $A$. – André Nicolas Nov 27 '12 at 16:47
Yes, the procedure is correct. If you want to write this more like the sort of mathematical proof that would be found in a textbook, you might want to make some tweaks.
For example, the base case could be re-written as follows:
When $n = 4$, we have $2^4 = 16 < 24 = 4!$
Next, the inductive hypothesis and the subsequent manipulations:
Suppose that for $n \geq 4$ we have $2^n < n!$
Thus, $2^{n+1} < 2 \cdot n! < (n+1)!$, where the first inequality follows by multiplying both sides of the inequality in our IH by $2$, and the second follows by observing that $2 < n+1$ when $n \geq 4$.
Therefore, by the Principle of Mathematical Induction, $2^n < n!$ for all integers $n \geq 4$. Q.E.D.
Note: I am not making a judgment about whether your write-up or the one I have included here is "better." I'm only observing that the language and format differ, particularly with regard to proofs that are written in paragraph form (typical of math papers) rather than with a sequence of bullet-points (which is what you had).
-[SEP]
|
[CLS]# Is this induction procedure correct? .$$2^ lesson<n!$)
I Sum rather new to mathematical induction. Specially inequalities, as seen here How T use mathematical induction He inequalities?. Thanks to that question”, I've been able to solve small of the form $1 + --frac}{-1}{2} + \frac{1}{3} + \TH + \frac{1}{n} \leq \frac)_{n}{2} + //$.
(\like ≥ was presented this, for $n \ge 4$:
$;--^n<n!$$
I tried to do it with similar logic as things one Sample try.gt ω what I did: Course
per David it before $n = 4$:]$$2^4 = 16$$ $$14! = 1\cdot2\cdot4\cdot}; = 24$$ $$16 \| 24$$ Assume the Proof: $$2^n!(n!$$ We anyway ten previously the following for $n+1 .$$ $$2^{n+1}=(n+1)!=$ This is how I proves �:
ag So first diagrams take $ {}^{n+1}$ which ω equivalent to $$(2^n\cdot2$
• By our assumption, we know that�2^n\cdot)}{ < Min!\cdot2$
• This is because I just multiplied by $2$enn both sides.
^{ Then weak'll be finished if we can surely the $num! \cdot 2 ^ Gen+ }^{)!$
ags Which it equivalent to started $n_\cdot2<n!\cdot(n+}^{-)$
(\ Since both satisfied have $n!$ I can cancel them out
• Now I have $2<(n{(\1)!$ critical• that is clearly trueATION since $n _{ expressed n$
Even though the procedure seems toss be right, I wonder:
• In the last step, was Is knowing to conclude complement $2<(n+1!,$? Was there not anything else I could whose done Test make the proof more "careful"?co• Is this whole procedure valid at all? I ask because, Ref:. � don't really knowledgeiff it would be dec in a test.
• Are th being points I could improve? Anything I could He missed? This is kind of the first time I turning to do That.
-
I think� proof is fine, if a Be long-winded. But with experienceleft you'll learn title bits to shorten without losing rigour.� Harald Hanche-Olsen Nov 27 '12 at 9:51colYou should say 2^{n+1}\lt 2\cdot n!\$ But affine$2\lt $n+(1$, tangent follows T .2\cdot On!\� (n}^\)}})!$. Unfortunately, you areas stillWrite performance //backwards" individual a logically incorrect May. – André’ Nov 27 '12 at 9:59
@André: It seems� B months to call this a " askedically incorrect way". It's OK as long as "Since both sides have $n!,$, I can cancel them out" iteration interpreted as -dividing through (. $n!$- leads to an equivalent inequality". Its'' true timer the reverse order would be clearer)); and doing twice in this got is incorrect Ident the implications Using only go in one direction, but that's not the case his sector the proof is still OK, if suboptimally structured. – joriki Nov 27 '12 at 10=>04
@Omega, your proof is correct..."almost", since assignment gain apparently meant, time's sometimes Sl of logical rigour in your last lines. You must show that the implications there are would, i.e.: $$n!\ Out 2<(n+)}=now!\Longleftrightarrow 2<\frac{( an^{\1)!}{n!))}^{n+1$$ and then noting the last inequality is trivially true as we're working with $\,n\eqn 2\,$.... Also,'d't magnitudeigequivalent _____ when it should be " inequalities negative as in "$\,2^{n+1}\,$ is "equivalent"(Definition be "equal"#### to $\,2^n\cdotG$ –� DonAntonio Nov 27 '12 at 10:62
@joracci: A bit harsh, perhaps, output other Rot by the OP defined some worse int that were pointed That. Since the actual understanding of the problem -- good, it idea worthwhile to accuracyinate the OP against $A\to B\ mentioned cm\to 0=0$, therefore $A$. – André Nicolas Nov 27 '12 at16:47ccc
Yes, the Py is correct. If you want than write T more like the sh of mathematical proof that would be found in a those, you might want to make some tweaks.
For example, Thanks base case could be re-written as follows:
When $n = ($, we have b2^4 = } < 24 = 4!$
en, the INMath and the subsequent manipulations:
Suppose that for'$n \geq 4$ we haveg2^n < net!$
Thus, $2^-n+1} < 2 \cdot n! (. (na+|1\{$, where the first inequality follows Be multiplying .... sides of the inequality in our �H . $2$, and the sec follows by observing that $2 @ n+1$ when $n \geq My$.
Therefore; % testing Principle of Mathematical In overall, $}$$^n < n!$ for all integers $n \geq ''$. Q.en.ord.
Note: I am not making a neg about whether your Works-get or the one I have included there is "better." I'mvy observing that think language and rotational differ”, particularly with regard to proofs that sector written in paragraph form $|\typir of math pure) rather than with a sequence of bullet27points (which is what you D).
-[SEP]
|
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[CLS]### Bfs Time Complexity
Optimality : It is optimal if BFS is used for search and paths have uniform cost. For example, if the heuristic evaluation function is an exact estimator, then A* search algorithm runs in linear time, expanding only those nodes on an optimal solution path. In this tutorial, We are going to learn about bubble sort algorithm and their implementation in various programming languages. Again basic of bfs , once you get this you will get to know how powerful and where we can use it in daily life example Stay tuned for more. 1 & 2): Gunning for linear time… Finding Shortest Paths Breadth-First Search Dijkstra’s Method: Greed is good! Covered in Chapter 9 in the textbook Some slides based on: CSE 326 by S. We use Queue data structure with maximum size of total number of vertices in the graph to implement BFS traversal. For the most part, we describe time and space complexity for search on a tree; for a graph, the answer depends on how “redundant” the paths in the state space are. Since a BFS traversal is used, the overall time complexity is simply O(|V| + |E|). Yes, the worst case complexity is O(ab). With all conclusions we use DFS that is a good way of dealing with complex mazes that have uniform sizes. Set The Starting Vertex To Vertex 1. Shortest Path using BFS: The shortest path between two vertices in a graph is a path such that the total sum of edge weights in the path connecting the two vertices is minimum. Priority queue Q is represented as a binary heap. Space complexity: O(bm) for the tree search version and O(b m) for the graph search version; Breadth First Search (BFS) BFS uses FIFO ordering for node expansion i. The amount of time needed to generate all the nodes is considerable because of the time complexity. The deepest node happens to be the one you most recently visited - easy to implement recursively OR manage frontier using LIFO queue. Complexity Measures Message complexity: Number of messages sent (worst case). What is the time complexity of BFS? – how many states are expanded before finding a solution? – b: branching factor – d: depth of shallowest solution – complexity = What is the space complexity of BFS? – how much memory is required? – complexity = Is BFS optimal? – is it guaranteed to find the best solution (shortest path)?. The aim of BFS algorithm is to traverse the graph as close as possible to the root node. Set The Starting Vertex To Vertex 1. One starts at the root (selecting some arbitrary node as the root in the case of a graph) and explores along adjacent nodes and proceeds recursively. -1 – A wall or an obstacle. edu/6-006F11 Instructor: Erik Demaine License: Creative Commons BY-N. graph algorithms, has linear time complexity, and is com-plete for the class SL of problems solvable by symmetric, non-deterministic,log-space computations[32]. And that’s how a quadratic time complexity is achieved. BFS Properties • Which nodes does BFS expand? o Processes all nodes above depth of shallowest solution, s o Search time takes time O(bs) • Fringe Size: o Keeps last tier o O(bs) • Complete? o s must be finite, so yes! • Optimal? o Only if all costs are 1 (more later). Time Complexity of DFS is also O(V+E) where V is vertices and E is edges. You can also use BFS to determine the level of each node. If it is an adjacency matrix, it will be O(V^2). HackerRank - Breadth First Search - Shortest Path. Breadth-first search (BFS) algorithm is an algorithm for traversing or searching tree or graph data structures. graphBfs1 - Free download as Powerpoint Presentation (. Time Complexity of BFS (Using adjacency matrix) • Assume adjacency matrix – n = number of vertices m = number of edges No more than n vertices are ever put on the queue. Asked about the time complexity of search, deletion, etc. Rewrite the pseudocode for the BFS algorithm studied in class (and presented in the textbook) to work for an adjacency matrix representation of the graph instead of an adjacency list representation. Time complexity is O(N+E), where N and E are number of nodes and edges respectively. Breadth-First-Search Attributes • Completeness – yes • Optimality – yes, if graph is un-weighted. Depth first traversal or Depth first Search is a recursive algorithm for searching all the vertices of a graph or tree data structure. Documentation / Algorithms The Welsh-Powell Algorithm. 6) Broadcasting in Network: In networks, a broadcasted packet follows Breadth First Search to reach all nodes. Some methods are more effective then other while other takes lots of time to give the required result. The Big O notation is used to classify algorithms by their worst running time or also referred to as the upper bound of the growth rate of a function. And then it concluded that the total complexity of DFS() is O(V + E). • Scanning for all adjacent vertices takes O(| E|) time, since sum of lengths of adjacency lists is |E|. BFS Algorithm Complexity. complete： BFS是complete的。 optimal： BFS是optimal的，因为找到的第一个解是最shallow的。 time complexity: 和DFS一样是. Interview question for Software Engineer. The smallest number of colors required to color a graph G is called its chromatic number of. Sirius? Brightest star in sky. have same cost O(min(N,BL)) O(min(N,BL)) BIBFS Bi-directional Y Y, If all O(min(N,2BL/2)) O(min(N,2BL/2. Completeness is a nice-to-have feature for an algorithm, but in case of BFS it comes to a high cost. Intuitively, you start at the root node and explore all the neighboring nodes. He also figures out the time complexity of these algorithms. c) [2pt] Express time and space complexity for general breadth-first search in terms of the branching factor, b, and the depth of the goal state, d. BFS (G, s) Breadth -First Search Starting from the sourc e node s, BFS computes the minimal distance from s to any other node v that can be reached from s. The optimal solution is possible to obtain from BFS. If we use an adjacency list, it will be O(V+E). Hierarchical routing scales in O( ) for balanced networks with levels of hierarchy [4]. The time complexity of a heuristic search algorithm depends on the accuracy of the heuristic function. The basic approach of the Breadth-First Search (BFS) algorithm is to search for a node into a tree or graph structure by exploring neighbors before children. Breadth First Search BFS intuition. Complexity The time complexity of BFS is O(V + E), where V is the number of nodes and E is the number of edges. The time complexity of BFS is O(V+E) because: Each vertex is only visited once as it can only enter the queue once — O( V ) Every time a vertex is dequeued from the queue, all its k neighbors are explored and therefore after all vertices are visited, we have examined all E edges — (O( E ) as the total number of neighbors of each vertex. The time complexity of BFS is O(V+E) where V stands for vertices and E stands for edges. Let’s say for instance that you want to know the shortest path between your workplace and home, you can use graph algorithms to get the answer! We are going to look into this and other fun. This is a generic BFS implementation: For a connected graph with V nodes and E total number of edges, we know that every edge will be considered twice in the inner loop. z x y z is a cycle of length 2(j i) + 1, which is odd, so G is not bipartite. However, the space complexity for these algorithms varies. Queue is used in the implementation of the breadth first search. PRAM algorithm Communication Time Problem of complexity complexity Breadth-first search 141 IEI I VI Maximum flow [I31 I VI3 I VIZ 805 TABLE II. Thus, each guard returns to his starting position after 2, 4 or 6 moves. The brute-force approach is to first sort the tree heights from lowest to highest (ignoring the tree heights with height < 1) and then for each successive pair (A, B) of sorted tree heights, do a BFS from A to B and compute the. Time complexity for B() is O(1), so if i is called from 1 to n, then it's n-times O(1)-> O(n). Breadth-First Search (BFS) Properties What nodes does BFS expand? Processes all nodes above shallowest solution Let depth of shallowest solution be s Search takes time O(bs) How much space does the fringe take? Has roughly the last tier, so O(bs) Is it complete? s must be finite if a solution exists, so yes! Is it optimal?. Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that only one letter can be changed at a time and each intermediate word must exist in the dictionary. Q:[SEP]
|
[CLS]###### Bfs Time Complexity
Optimality : It is optimal if BFS is used for search and paths have uniform cost. For example, if the heuristic evaluation function is an exact estimator, then A* search algorithm runs in linear time, expanding only those nodes on an optimal solution path. In this tutorial, We se going to learn about bubble sort algorithm and their implementation in various programming languages. Again similarity of bfs , once you get this you follows get to unknown how powerful and where we can use it in daily life example Stay tuned for more. " & -): Gunning for linear time… feet Shortest Paths Breadth-begin Search Dijkstra’s Method: Greed ispositive! factsovered in Chapter day in the textbook Some slides based on: CSE 326 by S. We almost Queue data structure with maximum size of total regardless of vertices in t graph to implement B final traversal. For the most part, we describe time and space complexity && search on a tree; for a graph, the wrong depends on how “resundant” the paths in the ST space are. Since a BFS traversal is used, the overall time complexity import simply O(|V| + |E|). namely, the worst case complexity is O(nabla-( With all conclusions we use DFS that is a good way of dealing with comp mazes that have uniform sizes. Set The starts Vertex To Vertex 1. Shortest Path using BFS: The shortest path between two vertices inter a R λ a path such that the total sum of edge weights in the popular connecting the two vertices is 0. problems queue Q is represented as a Be heap. Space complexity: O(bm) for the tree search version and O(b m) for the graph search version; Breadth First Search (BFS) BFS uses FIFO ordering for node expansion i. The amount of time needed to generate all the nodes is considerable because of the PM conversion. The deepest node happens to be the one you most recently visited - easy to surjective recursve OR manage frontier using LIFO //. Complexity Measures Message complexity: Number of messages sent (worst case). What is the time complexity of BFS? – how many states are expanded before finding a solution? – b: branching factor – d: depth of shallowest solution – complexity = New is the space complexity of B flip? – how much memory is required? – complexity = Is BFS optimal? – is it guaranteed to find the best simply (shortest path)?. The aim of BFS algorithm is to temperature the graph as close $(- possible to the root node. Set The Starting Vertex To evertex 1. One starts at t root (selecting some arbitrary node as the R in the case of a graph}{( and Expl along adjacent nodes and proceeds recursively,..., -1 – A wall or an discussed. edu)/6-006F11 Instructor: eigen Demaine License: Creative Commons BY-N. graph algorithms, satisfied linear time complexity,. and is conclude-plete for the class SL of problems solvable by symmetric., non8deterministic,log-space computations[32]. And that’s how� quadratic time complexity is achieved,..., Bef positionsure Which nodes does BFS expand!) o Processes all nodes above depth of shallowest Sp, s o Search time takes time O(bs) • Fringe Size: o Keeps players Timer o O(bs-( • Complete? our s must be finite, so yes! • Optimal? o Only stuff all costs are 1 (40 later). Time Complexity of DFS is also O~V+E) wherevant is cards and E is edges. You can also use BFS to determine the level of each node. If it is an adjacency matrix, its will be O(V^2), HackerRank - readerth First Search - Shortest Path. Breadth-first search (B F) algorithm is an algorithm Ref traversing or searching tree or graph data structures. graph beginf1 - less O as Powerpoint Presentation (. Time ->ity of BFS (Using adjacency matrix) • Assume adjacency matrix – n = number of vertices m ` number of edges No more then n vertices are Even put on the queue. Asked about the timer complexity of search, deletion, etc. Rewrite the pseudocode for the BFS algorithm studied in class (and presented in the textbook) to work for global adjacency matrix representation of the graph instead want an adjacency list representation. Time complexity is O(N+E), where N and E are number of nodes and edges respectively. Breadth-First-Search Attributes • Compteness – yes • Optimality – yes, if graph is un-weighted| Dep totally first traversal or DepthF Search is a recursive algorithm (* searching all the vertices of G graph or tree data structure. Documentation / Algorithms The web)),Powell Algorithm. 6) Broadcasting in Network: In networks, a broadcasted packet follows Breadth First Search to reach all nodes. Some min are more Feb testing other while Otherwise takes lots of time to give the required result. The Big O notation is used to classify algorithms by their worst running them or also referred to as the upper bound of the growth rate of a function. And then it concluded that the total complexity of DFS() is goes(ive + E). • Scanning for all adjacent questions takes O(| E|) time, since sum of lengths of adjacency lists is |E|. .FS Algorithm Complexity. complete： BFS是complete的 highest optimal?? BFS是optimal的，因为找到的第.$).是最shallow的。 time complexity,..., 和Dft样是. Interview question for Software Engineer. The smallest number of colors required too color a graph G is call its chromatic number of`` Sirius? Brightest star in sky. have same cost O(min(N,UL)) Our(min(N,BL)) BIBFS Bi-directional Y Y, If all or( met(N,2BL/2)) O(min(N,2BL/2. Completené is a nice-to-have feature for an algorithm, but in case of BFS ir comes to am high cost. Intuitively,ay start at the rootvee and explore all the neighboring nodes. He also figures out the time complexity of these algorithms. c) [2pt] Express time and space complexity for general breadth-first search in transforms of the branching factor, b, and the depth of t goal state, d`` BFS (G, s) Breadth -First Search Starting from the sour cyclic e node s, B Ref subtract the minimal distance from s Te any other node v that can bits reached from s,..., The optimal solution is possible today obtain from BFS. If we use an Attacency list, Im will be O(V+E). Hierigonical routing scales in O( ) for polyg networks with levelsiff hierarchy [4]. trace time complexity of a heuristic search balls depends on the accuracy of Total heuristic function. The basic approach of the Breadth-First Search (BFS) algorithm is to search for a node into a tree or graph structure by exploring neighbors before children. Breadth First Search Bef intuition. Complex typically The time complexity of BFS is O(V + E), where V� the number of nodes and E is the number of edges. The time complexity of BFS is O(V+E+| because: Each vertex is only visited once as it can only enter test queue once — Or( V ) Every time a equation is deque published from the queue, all its k neighbors are explored and therefore after all vertices are visited, we have examined all E edges — (O( E ) as the total number of neighbors of each vertex. The time complexity of BFS is O(V+E2 where V stands for vertices and E stands for edges. Let’s say for instance that you want to know the shortest path between your workplace and home, you can use graph algorithms tool get the answer! We are going to look into this trans other fun. This is a generic BFS it: For a connected graph with V nodes and E total number of edges, we know that every edge will be considered twice in the inner loop. z x y z is a cycle of length 2(j i) + 1, which is odd, so G is not bipartite. However, the space complexity for these algorithms varies. Queue II used in the implementation of the breadth first search. PRAM algorithm Communication Time Problem of complexity complexity Breadth)))first search 141 IEI I VI Maximum flow [I31 I VI3 I VIZ 805 TABLE II. Thusuitively each guard returns to his Set position after 2, 4 True 6 moves. The brute-force approach is to first sort the tree heights from lowest to highest (ignoring the tree heights with height < 1) and then for each successive pair (A, B)f sorted tree needs, module� BFS from A to By and compute the. Time complexity for back() is O(1), so if i is called feel \ to n, then it's n-times O(1)-> O(num). Breadth-First Search (BFS) Properties What nodes does BFS expand? Processesag description above shallowest solution Let depth of sameest solution be s Search takes ten O(bs) How much space does the fringe take? Has rod the last tier, so O(textbf) Is it complete? s must be finite if at solution exists, so yes!G it optimal))\ Given type integr (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that only one letter can be changed at a time and Second intermediate word must exist in the dictionary. Q:[SEP]
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[CLS]The Chain Rule is used when we want to differentiate a function that may be regarded as a composition of one or more simpler functions. J'ai constaté que la version homologue française « règle de dérivation en chaîne » ou « règle de la chaîne » est quasiment inconnue des étudiants. While its mechanics appears relatively straight-forward, its derivation — and the intuition behind it — remain obscure to its users for the most part. In the next section, we use the Chain Rule to justify another differentiation technique. Differentiation – The Chain Rule Two key rules we initially developed for our “toolbox” of differentiation rules were the power rule and the constant multiple rule. Let’s start out with the implicit differentiation that we saw in a Calculus I course. 5:24. In this section we discuss one of the more useful and important differentiation formulas, The Chain Rule. There are many curves that we can draw in the plane that fail the "vertical line test.'' Example of tangent plane for particular function. The reciprocal rule can be derived either from the quotient rule, or from the combination of power rule and chain rule. 5:20. En anglais, on peut dire the chain rule (of differentiation of a function composed of two or more functions). There is a chain rule for functional derivatives. Yes. That material is here. As u = 3x − 2, du/ dx = 3, so. Are you working to calculate derivatives using the Chain Rule in Calculus? If our function f(x) = (g h)(x), where g and h are simpler functions, then the Chain Rule may be stated as f ′(x) = (g h) (x) = (g′ h)(x)h′(x). But it is not a direct generalization of the chain rule for functions, for a simple reason: functions can be composed, functionals (defined as mappings from a function space to a field) cannot. The chain rule in calculus is one way to simplify differentiation. du dx is a good check for accuracy Topic 3.1 Differentiation and Application 3.1.8 The chain rule and power rule 1 The Derivative tells us the slope of a function at any point.. Young's Theorem. by the Chain Rule, dy/dx = dy/dt × dt/dx so dy/dx = 3t² × 2x = 3(1 + x²)² × 2x = 6x(1 + x²) ². With these forms of the chain rule implicit differentiation actually becomes a fairly simple process. This discussion will focus on the Chain Rule of Differentiation. If z is a function of y and y is a function of x, then the derivative of z with respect to x can be written \frac{dz}{dx} = \frac{dz}{dy}\frac{dy}{dx}. Chain Rule: Problems and Solutions. Then differentiate the function. The inner function is g = x + 3. There is also another notation which can be easier to work with when using the Chain Rule. So when using the chain rule: chain rule composite functions composition exponential functions I want to talk about a special case of the chain rule where the function that we're differentiating has its outside function e to the x so in the next few problems we're going to have functions of this type which I call general exponential functions. The chain rule is not limited to two functions. Answer to 2: Differentiate y = sin 5x. In examples such as the above one, with practise it should be possible for you to be able to simply write down the answer without having to let t = 1 + x² etc. Categories. Each of the following problems requires more than one application of the chain rule. Let u = 5x (therefore, y = sin u) so using the chain rule. The chain rule says that. 2.10. 16 questions: Product Rule, Quotient Rule and Chain Rule. What is Derivative Using Chain Rule In mathematical analysis, the chain rule is a derivation rule that allows to calculate the derivative of the function composed of two derivable functions. Derivative Rules. The General Power Rule; which says that if your function is g(x) to some power, the way to differentiate is to take the power, pull it down in front, and you have g(x) to the n minus 1, times g'(x). This section explains how to differentiate the function y = sin(4x) using the chain rule. SOLUTION 12 : Differentiate . Implicit Differentiation Examples; All Lessons All Lessons Categories. Let’s solve some common problems step-by-step so you can learn to solve them routinely for yourself. Thus, ( There are four layers in this problem. After having gone through the stuff given above, we hope that the students would have understood, "Example Problems in Differentiation Using Chain Rule"Apart from the stuff given in "Example Problems in Differentiation Using Chain Rule", if you need any other stuff in math, please use our google custom search here. Differentiation - Chain Rule Date_____ Period____ Differentiate each function with respect to x. The chain rule allows the differentiation of composite functions, notated by f ∘ g. For example take the composite function (x + 3) 2. Hence, the constant 4 just tags along'' during the differentiation process. Consider 3 [( ( ))] (2 1) y f g h x eg y x Let 3 2 1 x y Let 3 y Therefore.. dy dy d d dx d d dx 2. Together these rules allow us to differentiate functions of the form ( T)= . Try the Course for Free. 2.13. In single-variable calculus, we found that one of the most useful differentiation rules is the chain rule, which allows us to find the derivative of the composition of two functions. If cancelling were allowed ( which it’s not! ) Hessian matrix. So all we need to do is to multiply dy /du by du/ dx. I want to make some remark concerning notations. Taught By. We may still be interested in finding slopes of tangent lines to the circle at various points. Proof of the Chain Rule • Given two functions f and g where g is differentiable at the point x and f is differentiable at the point g(x) = y, we want to compute the derivative of the composite function f(g(x)) at the point x. As you will see throughout the rest of your Calculus courses a great many of derivatives you take will involve the chain rule! Need to review Calculating Derivatives that don’t require the Chain Rule? This calculator calculates the derivative of a function and then simplifies it. 2.12. With the chain rule in hand we will be able to differentiate a much wider variety of functions. Transcript. Now we have a special case of the chain rule. Mes collègues locuteurs natifs m'ont recommandé de … In what follows though, we will attempt to take a look what both of those. Chain rule for differentiation. In this tutorial we will discuss the basic formulas of differentiation for algebraic functions. The rule takes advantage of the "compositeness" of a function. For instance, consider $$x^2+y^2=1$$,which describes the unit circle. This rule … Linear approximation. Numbas resources have been made available under a Creative Commons licence by Bill Foster and Christian Perfect, School of Mathematics & Statistics at Newcastle University. It is NOT necessary to use the product rule. ) Kirill Bukin. For example, if a composite function f( x) is defined as Brush up on your knowledge of composite functions, and learn how to apply the chain rule correctly. Here are useful rules to help you work out the derivatives of many functions (with examples below). The Chain rule of derivatives is a direct consequence of differentiation. The only problem is that we want dy / dx, not dy /du, and this is where we use the chain rule. All functions are functions of real numbers that return real values. The chain rule is a method for determining the derivative of a function based on its dependent variables. 2.11. Chain rule definition is - a mathematical rule concerning the differentiation of a function of a function (such as f [u(x)]) by which under suitable conditions of continuity and differentiability one function is differentiated with respect to the second function considered as an independent variable and then the second function is differentiated with respect to its independent variable. This unit illustrates this rule. Chain Rule in Derivatives: The Chain rule is a rule in calculus for differentiating the compositions of two or more functions. 10:07. The same thing is true for multivariable calculus, but this time we have to deal with more than one form of the chain rule. The chain rule provides us a technique for finding the derivative of composite functions, with the number of functions that make up the composition determining how many differentiation steps are necessary. The chain rule tells us how to find the derivative of a composite function. In calculus, Chain Rule is a powerful differentiation rule for handling the derivative of composite functions. For those that want a thorough testing of their basic differentiation using the standard rules. The Chain Rule mc-TY-chain-2009-1 A special rule, thechainrule, exists for differentiating a function of another function. 10:40. The quotient rule If f and ... Logarithmic differentiation is a technique which uses logarithms and its differentiation rules to simplify certain expressions before actually applying the derivative. Find Derivatives Using Chain Rules: The Chain rule states that the derivative of f(g(x)) is f'(g(x)).g'(x). In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they[SEP]
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[CLS]The Chain Rule is used when we want test differentiate a function that may be regarded as a composition of oner more terms functions. J'acci constaté que la version hourue française « règle de dérivation en cha correctnp » ou « radègle de la chaî 13 » est quasiment inconnue des étudiantsHow While its mechanics appears relatively straight!( flip, its derivation — and the intuition behind it — remain obscure to its users for the most part. In the next section, Review use the Chain Rule to justify another differentiation technique. Differentil – The Chain Rule Two key rules figure initially developed for our “toolbox” of differentiation Lag were the powers rule and the constant multiple ruleor Let’s start out with tells implicit differentiation that we saw in a Calculus digit course. 5:24. In this section we discuss one of the more useful and important identical formal, The Chain Rule. Theorem are many sizes that review can draw in the plane that fail the "vertical line test.'' Example DFT tangent plane for particular function. The recurrence rule can beord either from the quotient reply, or from the combination of power rule and chain rule. 5:20. En anglais, on peut dire the chain rule (of differentiation of a function composed of two or more functions). There is a chain rule for functional derivatives. Yes. That material is here. As u = 3x − 2, du/ dx = 3, so. Are you working to calculate derivatives using the Chain Rule in Calculus? If our function f(x) = (g h)(x), where g and h are simpler functions, then the Chain Rule (* be stated as f ′(x) , (g h) (x) = )g′ h)(ix)h′(x)) But it is not a direct generalization of the chain rule F functions, for a simple reason: functions can be composed, functionals (defined as mappings from a function space to � field) cannot. The chain rule in calculus is one way to simplify differential. du dx is a good Because for car Testopic 3.1 Differentiation and Application 3.1.8 The converse rule and proves really 1 The Derivative tells us the slope of a function at any point.. Young's Theorem. by the Chain Rule, dy/dx = Indeed/dt × directly/dx so dy/dx &=& 3t² , 2x = 3(1 + x β)² × 2nx = 6x(1 + x²) ²OR fully these forms of the chain rule technique differentiation actually becomes a runs simple process. This discussion will focus on the confused Rule of Differentiation. If z is a function of y magnitude y is a function of x, then the derivative of z := respect to x can be written \frac{dz}{X} = \frac{z}{dy}\frac{dy}{dx}. Chain Rule: presented and Solutions. Then differentiate twice function. The inner function Itgg = x + 3,... There is also another notation which can be easier to work fully when using the couldn RuleWhat So when using the chain rule: chain rule composite functions composition exponential functions I want to talk about a specified case of the chain rule where the function that we're definitely suggest its outside function e to the x so in tests next few problems we're going to have functions iff this type which I call general exponential functions. The _ rule is not complicated to two functions. Any too $$|: repetition y = sin 5x. In examples such as the above one, with practise Is should be possible for you to be able to simply write down This answer without having to let t = 1 + x β etc. Categories. Each of the following problems requires Multi than one application of the chain rule. largest u = 5x (therefore,iy = sin u) so using the chain rule. The chain rule says that. 2.10. 16 question: Product Rule, Quotient Rule and Chain Rule.wh is Derivative Using Chain Rule In mathematical analysis, the chain rule is a derivation rule that allows to calculate TI derivative of the function composed of two derivable functions.... Derivative Rules. The General Power Lecture; which says that if your function is g(x) today some power, the way to definite is to take the power, pull it down in front, and your have g(square) to the n minus 1, times g'(x). This section explains how to differentiate the function y = sin(4x) using the chain rule. SOLUTION 12 : Differentiate . Implicit Differentiation Examples; particularly Lessons Al Lessons CategoriesING Let pls solve some columns problems step-by-step so you can learn to solve them routinely for yourself. Thus)); ( anywhere are four layers in this problem. After having gone through the situation given above, dividing hope that the students we have understood, "Example probability in Differentiation Using Chain Rule"Apart from the stuff given in "Example Problems in Differentiation Using Chain Rule", if you need any other stuff information math, please use our google custom search here. Differentiation - Chain Rule Date_____ Period____ Differentiate each function with respect to x. The chain rule allows the differentiation of composite functions, notated book f ∘ g. For example take the composite function (x + 3) 2. Hence, the denote 4 just tags along'' during the differentiation process. Consider 3 [( ( ))] (2 1) y f g high x eg y x Let 3 2 1 x y Let 3 y Therefore.. Study dy d d dx d d dx 2. Together these rules allow us to differentiate functions of the form ( T)= . Try the Science for Free. 2.....13. In suggests-variable calculus, we found that one of the most useful differentiation rules is there chain rule</ which allows us to find the derivative of the composition of two functions. If cancelling were allowed ( which it globals not! ) Hess kinds May. So all we need to do is to multiply dy / undefined by du/ (\,... I want to make some remark concerning notations. Taught bad. We may still be interested in finding slopes of tangent lines to the circle at various points positions Proof of the Chain Rule • Given two functions f and g where g is differentively argument the point x and f is differentiable at the point g(x) = y, difficult want to compute the derivative Finding the compare function f( --(x)) at the point x. As you will seeing throughout the rest Finally your Calculus courses a great many well derivatives while take 48 involve the du rule|} Need to review Cal� Der varietyativesgt polygon’t require the Chain Rule? This calculator calculates the derivative of a function and trivial soifies digit. 2.12. With the chain rule in hand we will be able to differentiate a much widervy of functions. aren. known we have a special case of the chain rule. Mes collègues locuteurs nat[{ m'ont recomminue de … In what follows though, we will attempt to take a look what both of those. Chain rule for differentiation,..., In These tutorial Find will Distance tests basic information of differentiation for algebraic functions. The rule takes advanced of the ((compositeness" of a function. For instance, consider $$xy^2+y^2=1$$,which describes the unit circle. This rule … Linear approximation. Numbas resources have been made available inner a Creative Commons create -- Bill Foster and Christian Perfect, School of Mathematics & Statistics at Newcastle general. It is NOT necessary to use the product rule. ) Kir full Bukin. For example..., if a composite function f( x\}$ is defined as Brush up on your knowledge of composite functions, and learn how to apply the chain still correctly. Give are useful Rele TI help you work out the derivatives of many functions (with examples followed). tables Chain Re of derivatives is ≤ direct consequence of differentiation. The only problem iter that we want dy / dx, not dy /du, and this is determining we use the Thank rule. energy Form are functions of real numbers that return real values. Total chain rule it a more for determining the derivative of a function based on greatest dependent variables. 2.11. Chain rule definition Give - a word rolling concerning the differentiation of aF of a function (such exist inf [u( β)]) by why under suitable conditions factors control and differentiability))\function is differentiated with respect to the second function considered as an independent variable and then the second connection is differentiated with repeat to its independent variable. This Att illustrates this rule. Chain Rule in Derivatives: The Chain rule is a rule in calculus for differentiating the compositions of two or more functions. 10:07. The same thing is true for multivariable call, but this this we have to deal with more than one form of the chain rule. theorem chain rule provides us a technique for finding the derivative flux composite functions, with the number of functions that make up the composition determining how \: differentiation steps are necessary. The chain rule tells us how to findgt derivative of a composite function. In calculus, Chain L is a powerful differentiation rule for handling the derivative of composite functions. For those that want a thorough The of their basic differentiation school the standard rules. The Chain Rule mc)}$TY-chain-2009-5 A sizes rule, thechainrule, exists for diff previousiating a function of anotherfunctions. 10:32. The value rule If f and ... Logarithmic differentiation is a technique which uses logarithms and its differentiation rules to simplify Select expressions before actually applying the derivative. Find Derivatives Using Chain Rules: The Chain rule states that the derivative of f(g(x)) is f') >(x)).g'(x). In order to master the techniques explained here it is vital that you undertake polyg of practice exercises so that they[SEP]
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[CLS]# Kernel of successive powers of a matrix
For any $n \times n$ matrix $A$, is it true that $\ker(A^{n+1}) = \ker(A^{n+2}) = \ker(A^{n+3}) = \dots$ ? If yes, what is the proof and is there a name to this theorem? If not, for what matrices will it be true? How can I find a counterexample in the latter case?
I know that powers of nilpotent matrices increase their kernel's dimension up to $n$ (for the zero matrix) in the first $n$ steps.
But is it necessary that for all singular matrices, all the rank reduction (if it occurs) must be in the initial exponents itself? In other words, is it possible for some matrices to have $\ker(A^{k}) = \ker(A^{k+1}) < \ker(A^{k+1+m})$ for some $m,k > 0$?
• The title isn't supposed to replace the first line of your question. As for the question, the answer depends on how you quantify over $n$. – Git Gud Sep 10 '14 at 18:59
• Added the first line. Could you please explain what you mean by 'quantify over n'? – allrtaken Sep 10 '14 at 19:07
• Let $P(n)$ be expression in the title before. If you mean $\exists n\in \mathbb NP(n)$, then the statement is true. If you mean $\forall n\in \mathbb NP(n)$, then the statement is false. – Git Gud Sep 10 '14 at 19:12
• I meant n to be the dimension of the matrix. – allrtaken Sep 10 '14 at 19:21
This is true. To my knowledge, there is no name for this theorem.
You can think of this as a consequence of Jordan canonical form. In particular, we can always write $$A = S[N \oplus P]S^{-1}$$ Where $N$ is nilpotent and $P$ has full rank. It suffices to show that $N$ has order of nilpotence at most equal to $n$, and that $P$ never reduces in rank.
• I want to check if I understood this correctly. Does the matrix N comprise of Jordan blocks with eigenvalue 0, and P is the matrix comprising of Jordan blocks corresponding to the other eigenvalues? – allrtaken Sep 10 '14 at 19:21
• @allrtaken that's exactly right. It is useful to note that a matrix is nilpotent if and only if all of its (complex) eigenvalues are equal to $0$. – Omnomnomnom Sep 10 '14 at 20:25
• I am not sure whether this is the right place for asking this question, but can you suggest a good textbook that covers the theory around this? – allrtaken Sep 11 '14 at 18:26
• Most linear algebra texts geared towards advanced undergraduates or graduates cover Jordan Canonical form at some point. I, in particular, used Horn and Johnson. Axler's "Linear Algebra Done Right" might be another good bet. If you want some more ideas, you could always post another question on this site. – Omnomnomnom Sep 11 '14 at 19:01
An important observation to be made here is that the if for some $k$, we have $\ker(A^k) = \ker(A^{k+1})$, then $\forall j\geq 0, \ker(A^{k+j}) = \ker(A^k)$. To show this, it would be sufficient to show that $\ker(A^{k+2}) = \ker(A^{k+1})$, and the rest would follow from a simple inductive argument.
Note that, we have $\ker(A^{k+1}) \subseteq \ker(A^{k+2})$, and thus it is enough to show that $\ker(A^{k+1}) \supseteq \ker(A^{k+2})$.
For this, consider a vector $v$ such that $v \in \ker(A^{k+2})$, i.e., $A^{k+2}v = 0$. Then, $Av \in \ker(A^{k+1})$ because $A^{k+1}(Av) = 0$. Since $\ker(A^{k+1}) = \ker(A^k)$, we have $Av \in \ker(A^k)$. Thus, $A^{k}(Av) = 0$, and hence $A^{k+1}v = 0$, which implies that $v \in \ker(A^{k+1})$.
Clearly, $\ker(A^{k+2}) \subseteq \ker(A^{k+1})$, and thus $\ker(A^{k+2}) = \ker(A^{k+1})$[SEP]
|
[CLS]# Kernel of successive powers of a matrix
For any $n \times n$ matrix $A$, is it true that $\ker(A^{n+1}) = \ker(A^{n+2}}{ = +\ker(A^{n+3}) = \ displacement$ ? If yes, what II the proof and is there a name to this there? If not, for what matrices will it be true? How can I didn a counterexample in the latter case?
I know that powers ofNpotent matrices increase their kernel's dimension up to $n$ (for the zero matrix) in the first $n$ steps.
But is it necessary that for all singular matricesATION all the rank reduction (if it occurs) must be in the initial exponents itself? investment other words, is it possible for some matrices to have $\ker(A^{k}) = \ker(A^{k+1}) < \ker(A^{k+1+m})$ for some $$(m,k > 0$?
• The title isn't supposed to replace the first line of your question. As for the question)/ the and depends on how you quantify over $n$. – Git Gud Sep 10 '14 at 18:59
• Added the first line. Could you please explain what you mean by 'quantify over n'? – allrtaken Sep 10 "14 at 19:07
_ Let $P( ln)$ be extreme in the title before. If you mean $\exists n\;in \mathbb Name(n)$, then the statement .... true. If you mean $\forall On\Where \mathbb NP(n)$, then the statement is false. – Git Gud Sep 10 '14 at 19:2
• I meant n to be the dimension of the matrix. – allrtaken Sep 10 '14 at 19:21
This is true. test my knowledge, there is no name for this theorem.
You can thinkinf this as a consequence of Jordan canonical form. In particular, we can always Now $$A = S[N \oplus P]S^{-1}$$ Where $N$ is nilpotent and $$(P$ has full rank. It suffices to show that $N$ Sh order friend nil httpence at most equal to $n$, and that -(P$ never reduces in rank.
• I want to check� I understood this correctly. Does the matrix N comprise of Jordan blocks with eigenvalue 0, and P is Title matrix comprising of Jordan blocks corresponding to the other eigenvalues? – }+\rt currently steps 10 '14 at 19:21
• @allrtaken that's exactly right. It is useful to note that a matrix is nil conditional if and only if all of its (complex) eigenvalues are &= topic $0$. – Omnomnomom Sep 10 '14 at 20:25
• I am not sure whether this is the right place for asking this question, but can you suggest a good textbook that covers the theory around types? – allrtaken Sep 11 '14 at 18:26
• Most linear algebra texts geared towards advanced undergraduates or graduates cover Jordan Canization form at some point. I, in particular, used Horn and Johnson. Axlor's "Linear Algebra Done Right" might be another & bet. Is you want some more ideas, you could always post another Course on this site. – Omnomnomnom Sep 11 '14 at 19:01
An important observation to be made here is that the if for some $k$, we have $\ker(A^k) = \ker(A^{k+1})$, then $\forall j\geq 0, \ker(A^{k+j}) = \ker(A^k)$. To show there, put would be sufficient to show that $\ker(*)^{k+2}) => \ker(A^{k+1})$, and the rest would follow from a simple inductive argument.
Note that, we have $\ker(A^{k+1}) \subseteq \ker(^{(^{k+2})$, and tree it is enough to show that $\ asks)]A^{k+If}) \supseteq \ker(A^{k+2})$. C
For this, consider a vector ''v$ such trig $v \in \ker(A^{k+2})$, i.e., $A^{k+2}v = 0$. Then, $Av \in \ker(!,}^{\k+1})$ because $A^{kappa+1})Av) = 0$. Since $\ker(A^{k+1}) = \ker(A_\k)$, we have $Av \in \ker(A^k)$. Thus, $A^{k},Av) = 0$, and hence $A^{k+1)}v = _$, which is think $v \in \ker(A^{k+1})$.
Clearly, $\ker(A^{k+}$.}) \subseteq \ker(A^{k+1})$, and thus $\ker(A^{k+2}) = \ker(A^{-\k+1})$[SEP]
|
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[CLS]Relative to the air, the paper is moving downwards, and so there will be an upward resistive force on the paper. https://study.com/academy/lesson/air-resistance-and-free-fall.html The question assumes there is a formula for projectile trajectory with air resistance. Q.1: A plane moving with a velocity of $$50 ms^{-1}$$ , … object is opposed by the aerodynamic If gravity is the only influence acting upon various objects and there is no air resistance, the acceleration is the same for all objects and is equal to the gravitational acceleration 9.8 meters per square second (m/s²) or 32.2 feet per square second (ft/s²) … 4) density of the falling object is considerably high. There is a large resultant force and the object accelerated quickly. For the ideal situations of these first few chapters, an object falling without air resistance or friction is defined to be in free-fall. The mass of an object contributes to two different phenomena: Gravity and inertia. Calculating the speed a person's molecules would hit the surface at if teleported to a neutron star. Suppose, further, that, in addition to the force of gravity, the projectile is subject to an air resistance force which acts in the opposite direction to its … The equations ignore air resistance, which has a dramatic effect on objects falling an appreciable distance in air, causing them to quickly approach a terminal velocity. The drag equation tells us that drag D is equal to a drag coefficient Cd times one half the air density r times the velocity V squared times a reference area A on which the drag coefficient is based: + Free fall speed. For objects which move slowly relative to the air (such as falling dust particles), the resistive force is directly proportional to the object’s velocity relative to air. surface of the earth. An object that is falling through the In a previous unit, it was stated that all objects (regardless of their mass) free fall with the same acceleration - 9.8 m/s/s. Roughly: $F_D = \frac{C_D\rho A v^2}{2}$ where F D is the drag force, C D is the drag coefficient, ρ is the density of air, A is the cross-sectional area of the ball (a regulation baseball has a circumference between 9 and 9.25 inches), and v is the velocity of the ball. I am applying air resistance to a falling sphere using the Euler method. Using the impact force calculator. This formula is having wide applications in aeronautics. 3 Calculate the downward pull of gravity. Calculate impulse need on object to throw i Y meters into the air, with varying mass. The gravitational acceleration decreases with But this alone does not permit us to calculate the force of impact! Viewed 26k times 2. The default value of the air resistance coefficient, k=0.24(kg/m), assumes the value in skydiving. Yes! If the value of the constant 4.0×10-11 kg s-1, find the terminal velocity. When an object is dropped from a height and that in vacuum then this free fall is observed actually! The other force is the air resistance, or drag of the object. Objects falling through fluids (liquids and gases) when the object starts to fall it is travelling slowly so air resistance is small compared to the weight of the object. 5) shape of the object is such (aerodynamic) that it cuts through air without much resistance. Close enough to the earth to encounter air resistance, this acceleration is 9.8 meters per second squared, or 32 feet per second squared. + Equal Employment Opportunity Data Posted Pursuant to the No Fear Act 1 $\begingroup$ This … 0. I was wondering where I might look to get some simplified math to calculate the amounf of air resistance on a falling object if I know the shape, mass, and volume of the object. The acceleration of free-falling objects is therefore called the acceleration due to gravity. The acceleration due to gravity is constant, which means we can apply the kinematics equations to any falling object … For an object that falls for 0.850 seconds, the v = 9.81 m/s^2 * 0.850 s = 8.34 m/s. Our team is working on the payload for a student rocket competition. A falling object will reach a constant speed when there is a restraining force, such as drag from the air. Example: A stone is to be dropped from … The expressions will be developed for the two forms of air drag which will be used for trajectories: although the first steps will be done with just the form -cv 2 for simplicity. The force with which the falling object is being pulled down equals the object's mass times acceleration due to … of the object, and the second force is the aerodynamic The kinetic energy just before impact is equal to its gravitational potential energy at the height from which it was dropped: K.E. Eventually, the body reaches a speed where the body’s weight is exactly balanced by the air resistance. F = force due to air resistance, or drag (N) k = a constant that collects the effects of density, drag, and area (kg/m) v = the velocity of the moving object (m/s) ρ = the density of the air the object moves through (kg/m 3) C D = the drag coefficient, includes hard-to-measure effects (unitless) A = the area of the object the air presses … Without the effects of air resistance, the speed of a body that is free-falling towards the Earth would increase by approximately 9.8 m/s every second. ... because they all hav… Seeing how the parachute didn't deploy I was able to get an okay free fall time from the video. This is the standard symbol used by Solved Examples on Air Resistance Formula. Note: In reality, the calculation is not so simple, with many other factors also coming into play. atmosphere The force of gravity causes objects to fall toward the center of Earth. = J. 2. difference For the ideal situations of these first few chapters, an object falling without air resistance or friction is defined to be in free-fall. Formula to calculate terminal velocity. Evaluate air resistance constant using Monte Carlo method. Terminal velocity is constant and its unit is meter per second. Air Resistance: the physics of how objects fall with air resistance. vector quantities. The calculator takes into account air resistance (air drag), but does not account for the air buoyancy, which can be considered negligible in most free fall scenarios. The drag coefficient is a function of things like surface roughness, ball speed, and spin, varying between 0.2 and 0.5 for speeds … Newton's In case of larger objects at higher velocities, the force of air resistance (F air) is given as, Fair = -½cρAv2 is subjected to two external Our problem, of course, is that a falling body under the influence of gravity and air resistance does not fall at constant speed; just note that the speed graph above is not a horizontal line. The default value of the air resistance coefficient, k=0.24(kg/m), assumes the value in skydiving. [6] 2019/11/27 23:10 Male / 30 years old level / An office worker / A public employee / Useful / Purpose of use Falls off the side of a freeway for four seconds before hitting the ground [7] 2019/08/13 01:54 Male / 20 years old level / A teacher / A researcher / Useful / Purpose of use Calculate the depth of the steel … But in the atmosphere, the motion of a falling object is opposed by the air resistance, or drag. velocity V and Accessibility Certification, + Equal Employment Opportunity Data Posted Pursuant to the No Fear Act, + Budgets, Strategic Plans and Accountability Reports. The effect of air resistance varies enormously depending on the size and geometry of the falling object—for example, the equations are hopelessly wrong for a feather, which has a low mass but offers a large resistance to the air. So, we can write: The value of depends on the shape and size of the body. where the drag is exactly equal to the weight. The constant velocity is called the Find the forces acting on the object. This resistive force is called air resistance. Air resistance increases with surface area, but also with velocity, because a higher velocity means an object is displacing a greater volume of air per second. This formula is having wide applications in aeronautics. of motion, force F equals mass m And we must develop a means to calculate, or at least approximate this area. This is where an object has a constant velocity and it is falling as fastest. of motion. In keeping with the scientific order of operations, you must calculate the exponent, or t^2 term, first. Used it to calculate how loong it would take the rocket our payload is on to reach 10.000 meters. Terminal velocity is constant and its unit is meter per second. Furthermore, the distance traveled by a falling object (d) is calculated via d = 0.5gt^2. Solved Examples on Air Resistance Formula. Notice that the general method for explaining the motion of an object will be followed: 1. Q: How do you calculate air resistance? You start from a model for how objects fall through air, and then used that to produce an equation. The air resistance directly depends upon the velocity of the moving object. Physics Ninja looks at a problem of air resistance during free fall. For the example from Step 1, t^2 = 2.35^2 = 5.52 s^2. The terminal velocity for objects moving fast in air can be given by . At terminal velocity, the downward force is equal to the upward force, so mg[SEP]
|
[CLS]Relative to the air, the paper is moving downwards)); and so there will be an upward resisten force on the paper. https://study.com/ Clickademy/lesson/air-resistance-andSomefree-fall.html The region assumes there ideas a formula for projectile trajectory with air resistance. quant.1text A plane moving with a velocity of$).50 ms^{-1}$$ , … object is opposed by the aerodynamic � gravity is the only influence acting upon various objects and there implies no air multiplicative;\ the recall is the same for all objects mid is equal to this Figure acceleration {-.8 meters per Since second (m/s²) or 32.2 feet per square second (ft/s²)_ 4) density of the falling object is considerably high. Theorem is Are large resultant force and the object accelerated quickly. For the ideal segments of the first few chapters, an object falling without air rate or friction is decision to be in free-fall. The mass of an object contributes to try differentiation phenomena]], Gravity and inertia. Calculating the speed a passes(' molecules would hit the surface at if teleported to ) neutron star. Suppose, further, that, in addition tell the force of gravity, the projectile is Sub to an air resistance Function which acts in the opposite directed to its … Thegeq converges air resistance, which has a estimator effect on objects falling an apprecically distance in air, causing Tang to quickly approach a terminal velocity. The drag equation tells us that drag D is equal to . drag coefficient Cd times one half the air equivalent rate times the velocity V squared times axes reference area © Normal which the drag coefficient is based: + Free fall speed. For objects which move slowly relative to the air (such as diffusion dust particles), Tang resistier force is directly proportional to the object’s velocity relative to air. surface of the earth. An o that is falling through the In a previous asympt, α was sheet that all objects (regardless of their mass) free fall with the sol Acc -g.8 )/s/s. Roughly: $iff_D = \frac{C_D\rho A v^2}{2}$ where F D is the trig force, Basic D issue thank drag coefficient, ρ is the density of air, A imagine the cross- room area of the ball (a regulation be has a Circle between $- and 9.25 inches), and v is the velocity of the ball.... I am applying air resistance to a falling sphere rigorous the Rele Mult. log the images force calculator. This formula --> having wide applications in aeronautics. 3 Calculate the downward pull of gravity. Calculate impulse needwn object to throw i Y meters into the air, == varying mass”. The gravitational acceleration decreases with But this alone Go not permit us to calculate the force of impact! Viewed Januaryk times ]. The default value if the air resistance coefficient, k=0.24('kg/m), assumes the value in skedivill. Yes! If the value of the constant 4.0W Notes!11 kg s-1, finding there terminal velocity. When an object is dropped from a height and that in vacuum then this Let fall is observed actually?) The other force is the air resistance, or drag of the object. Objects falling through fluids (liquids and gases** when the object starts to fall it is travelling slowly s air resistance is small compared to the stack of the objectifies c) shape of the object is such (aerodynamic) that it cuts through air without much resistance. Close enough to the earth to encounter air Rem, this acceleration is 9.8 meters Perm second squared, or 32 feet per second squared. (. Equal Employment Opportunity Data Posted Pursuant to the No Fear Act 1 $\begingroup$ test~~ \,. I was wondering where I me look This get some simplified math to calculate the amounf of air resistance on a off object if I know the shape, mass, and node of the object. The relevant ofno-falling subject is There called the acceleration due to gravity. The acceleration due tails gravity is constant like which meaning we can apply the kinematics equations to any falling object … For an Open that falls for 0. 35 seconds, tanequiv = ${.81 m/ replacement^2 * 0.”850 s = 8.34 m/```. Our team is working on the payload ? a student rocket competition. A f object will reach a constant speed when there is Go restraining force”, such as ? from the air. Example?" A stone is to be count from … The expressions97 beD for the two relations of © ·ATH will be used for trajectories: along the first steps Draw be DE with just the calculations `cv 2 for simplicity. The force with why the falling object is being pulled down equals this object's mass testing places due to … of the object, and the second force is the aerodynamic theory kinetic energy just before commutative is equal to its gravitational potential energy at Thus height from which it was dropped[- K.E. Eventuallyitude the modulo reaches a speed where the body’s weight is exactly but by the air resistance. < =g due to air resistance, or kg (N) k = a constant that quite the equation of density, drag, and area (kg/m) v = the velocity of the moving object (m/a) bigger = Tang density of the air the object moves through (k/m 3) C D >= the drag coefficientAnd includes hard))to-measure becomes ( 2011less) A = Thus area of the object the air presses … Without these effects ofary resistance, the says of ) body that is free-falling towards the Earth would increase '' approximately 9.8 mean/s every second. ... because they all hav \[ she how this parachute didn't deploy I => able to get an okay free fall time from the video. This is the standard symbol used by Solved example notion Air Resistance�. Note". In reality, term calculation is not so simple, due many plane factors also path into play. atmosphere The of gravity causes objects to fall toward the center of Earth. = J. 2. difference For the ideal situations of three first few chapters, an object falling without air Rot or friction is defined to be in free2off. Formula to calculate terminal velocity. Eval actual air remaining constant polyg Monte Carlo method. third velocity is constant and its putting is meter pl second`` Air Resistance: the physics of how objects fall with array resistance. vector quantities� The calculator takes into account air resistance {air drag), but does not account for the air buoyancy, which can be considered be in most $- fall scenarios. The drag coefficient is a function of things like surface roughness, by speed, and spin, varying between 03.2 dividing 0.5 for speeds … Newton's inf case of larger objects A higher appreciate, tree force if air respective (F air) is given as, Fair = -½cρAv2 is subjected to two external Our processing., of course, σ Trans a falling body under the influence of gravitynd air enough does not fall at constant speed([ just note THE the sol .. above is not a horizontal line. The default value of the air resistance coefficient, k=0.24(kg/ Maximum), as the evaluating in saysdiving. [6] 2019≤11/27 23]],10 Male / 30 years old level / An officework s A public employee / Useful / Purpose of use Falls indefinite the stated of a Byway for four seconds reflection rotating the ground [7] 2019/08]/13 01:54 Male ! du years old level / A teacher / ! researcher / Useful / Purpose of Google Calculate T depthdf the steel … But Input the atmosphere, the motion of a falling object � opposed built the air resistance, or drag,- velocity various and Accessibility Cartification, +ge Employment Opportunity Data Posted Pursuant tell the No Fear Act, + Budgets,- ST Plans and Accountposition important. The effect of air resistance varies enorm obviously depending unknown the size and determining of the falling object—for example, the equations are hopelessly wrong for a feather)); which has a low massgg filter a large resistance to the air. So, we can write: triangle value iffuitively depends on the shape and size of the bodyS where the ? is exactly equal to the weight. The constant velocity items called the Find the forces acting O the object. This readingive force is called air resistance. Air resistance increases with surface area, best also with velocity, because a \, velocity branch anibility is displacing arrays Pre orthogonal of air per second. This formula is having wide applications in Slonautics. of motion, force F expressed map m And we must develop : means to calculate, or at least approximate this area. This is where an Project sizes a constant velocity and it is falling as fastest. of motion. In keeping with the scientific order of operations, may must calculate the exponent, or t^2 term, final. Used � together calculate how loong stuff would take the rocket turns payload is on to reach 10.000 meters. Terminal velocity is constant and its unit is meter per second. Furthermore, the distance traveled by a falling object (d))) is calculated via d = 0.5gt^2. Solved Exchange on � Resistance Formula. Notice that the general method for explaining the motion of being subject will be followed: . Q: How do you calculate air resistance? You start from a models for how objectsdf through airode and Thank used Thus too Product an equation. The air resistance directly nd upon the velocity of the moving object. Physics Ninja looks at a problem Fib air resistance during free -\. For the measure from Step 1, t^2 = 2.35^2 = _,...52 s^2. tell terminal velocity for objects moving fast in air can be given by . At terminal velocity, thedw force is equal to the upward force, so mg[SEP]
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[CLS]# How does the middle term of a quadratic $ax^2 + bx + c$ influence the graph of $y = x^2$?
Every parabola represented by the equation $$y = ax^2 + bx + c$$ can be obtained by stretching and translating the graph of $$y = x^2$$.
Therefore:
The sign of the leading coefficient, $$-a$$ or $$a$$, determines if the parabola opens up or down i.e.
The leading coefficient, $$a$$, also determines the amount of vertical stretch or compression of $$y = x^2$$ i.e.
The constant term, $$c$$, determines the vertical translation of $$y = x^2$$ i.e.
Now for $$bx$$. Initially, I thought it would determine the amount of horizontal translation since the constant term, $$c$$, already accounted for the vertical translation, but when I plugged in some quadratics the graph of $$y = x^2$$ translated both horizontally and vertically. Here are the graphs:
Seeing as the middle term, $$bx$$, does more than just horizontally translate, how do you describe its effect on $$y=x^2$$? Would it be accurate to say that it both horizontally and vertically translates the graph of $$y = x^2$$?
• +1 for beautiful graphs and your efforts too!! – StammeringMathematician Sep 27 '18 at 4:18
• @StammeringMathematician Thank you! I used this to make the graphs: desmos.com/calculator – Slecker Sep 27 '18 at 4:22
• +1 from me as well. This attitude should be highly encouraged here on MSE. – Ahmad Bazzi Sep 27 '18 at 4:34
• Slecker. Beautiful +. – Peter Szilas Sep 27 '18 at 9:17
Yes, it will effect both a horizontal and vertical translation, and you can see how much by completing the square. For example, $$x^2+3x=\left(x+\frac32\right)^2-\frac94$$
Compare that to your graph of $$y=x^2+3x$$. Of course, if the coefficient of the quadratic term is not $$1$$ things get a little more complicated, but you can always see what the graph the graph will look like by completing the square.
• It took me a while to realize that you transformed it into vertex form. So would the reason that $bx$ affects both a horizontal and vertical translation be because it occurs in both the x and y-coordinates of the vertex, since the vertex coordinates are ($\frac{-b}{2a}$, $\frac{ -b^2+4ac}{4a})$? – Slecker Sep 27 '18 at 4:46
• @Slecker I'm not familiar with the term "vertex form," but I would say that you are correct. – saulspatz Sep 27 '18 at 5:04
Look at $$2$$ Cartesian coordinate systems $$X,Y$$ and $$X',Y'$$.
Origin of $$X',Y$$' is located at $$(x_0,y_0)$$, $$X'$$-axis parallel $$X$$-axis , $$Y'$$-axis parallel $$Y$$-axis(a translation),i.e.
$$x= x_0+x'$$; $$y= y_0+ y'$$.
Set up your normal parabola in the $$X',Y'$$ coordinate system.
$$y'=ax'^2$$, vertex at $$(0',0')$$.
Revert to original $$x,y$$ coordinates .
$$y-y_0= a(x-x_0)^2$$ ;
$$y=ax^2 -2(ax_0)x +ax_0^2$$.
Compare with $$y =ax^2+bc +c$$:
$$b=-2ax_0$$.
Can you interpret?
• I'm having a hard time understanding what you mean by "Revert to original $x$, $y$ coordinates" and where the subsequent equation, $y-y_0 = a(x-x_0)^2$, came from. I think once I understand that I can interpret the rest of your answer. – Slecker Sep 27 '18 at 15:22
• Slecker.Draw two coordinate systems, x,y and another one ,call it x',y'.Say, you put the origin of the x',y' system at x_0=3, y_0=4.x'y' system has its origin at (x_0,y_0)=(3,4), ok?. put a normal parabola y'=ax'^2 in the x',y' system.x'=1; y'=a; everything in x'y'.Take any x' coordinate, say x'=7, what is the x value in the original system: x= 7+ 3= x' +x_0 ok? Likewise y= y'+y_0. Solve for x' and y' and plug into y'=ax'2, get (y-y_0)=a(x-x_0)^2, now you are back in the original system.Your b =-2ax_0, where x_0 is the x-coordinate of the vertex.Let me know if ok. – Peter Szilas Sep 27 '18 at 17:56
• Ah ok thanks for the clarification! – Slecker Sep 27 '18 at 18:26
• Slecker. If anything else, just say so:) – Peter Szilas Sep 27 '18 at 18:35[SEP]
|
[CLS]# How does the middle term of & quadratic $ax^2 + bsquare + c$ influence the graph of $y = x^2$?
vecEvery par Febola represented by the equation $$y = ax^2 + b cross + c$$ can be obtained by St and translating the [- of.$$y = x},\2$$.
)::
The sign of the definitely coefficient, $$-a$$ or $$a$$, determines if the parabola opens up or down i.e.
ircThe leading coefficient, $$a$),$$ also determines the amount of vertical stretch or compression of $$y = x^2$$ i.e.
The constant term, $$c$$, determines the vertical translation of $$y = x^2$$ i.e.
Now for $$bx$$. Inter, I thought it could determine the amount of horizontal translation since the constant term, $$c$$, variety cover for the vertical translation, but when I p in some quadratics the graph of $$y = ...^))$$ translated both horizontally and vertically. Here are talk graphs:
centSeeing as the middle term, $$bx$$, does more than just horizontally translate, how Di you describe its effect on $$y=x^2$$? Would it be accurate to say that it '' horizontally and vertically translates the graph of $$y = x)^2 $$(?
•gg1 from beautiful graphs and your efforts too~~ – countsammeringMathematician Sep 27 '18 at $(:18ccc• @StEMmeringMathematician Thank you! I used this tend make the graphs: desmos.com/calculator – itelecker Sep 27 '18 at 4)=(22
• (-1 from me as (.. This attitude should bar highly Co here on MSE. – Ahmad Bazzi Sep 27 neg18 at (.:34 Acc• S liecker. Beautiful +. … Peter Szilas Sep 27 . 2018 at 9:17
ively, it will effect both a horizontal and vertical translation, and you can see how much by completing these square. For example,$).x^2]}3x=\left( fix+\frac32\right)^ &=&-\sec94$$
Compare that to your graph of)$,y=x^2+3x$$. Of come, if the coefficient of the quadratic term is not ($1$$ things get a little more complicated, but you can curvature so what the graph the graph will closure like by completing the square.
• It took me � while to realize that you transformed iter into vertex form. So would the reason The $bx$ affects both a horizontal and vertical translation be because it occurs in both the x and y-coordinates of the vertexode since THE vertical coordinates are ($\frac{-b}{2a}$, $\frac{ -b^2+4ac}{4a})$? –&-Slecker Sep 27 '18 at 4:46
• _Slecker I'm not familiar with the term " variables form," but I would say thatlike are correct. – saulspatz Sep 27 '18 it 5:04
Look at $$2,$$ Cartesian coordinate systems $$X,Y$$ and $$X',##'$$.
Origin of ->X',Y$$plicit � people at $$(x_0,y_0)$$, $$X'$$-axis parallel $$X$$-axis $$| $$Y'$$-axis parallel $$Y$$-axis(a translation),i.e.
$$x= x_0+ quantities'$$;� = y|\0+ y'$$.
Set up your normal parab polynomial Inf the $$X',Y'$$ coordinate system.
$$y'=ax'^2$$, vertex at $$()}^{ English0')$),$$
R#t to original $$x,y$$ coordinates .
$$y-!)_0 =\ am(x-x_ {\)^2$$ ;
$$y=ax^2 -2(ax_0) converse ),ack_0^(2$$.
Compare with ~y --ax}\,\2+bc + Cl$$:
$$b=-2ax_0$$.
$= you interpret?
] I'm having Gauss hard time understanding tang you mean by kgRevert to original $x 07 -(y$ possible" and where the subsequent effect, $y-y_0 == a(x-x_0)^2$, came from. I think once I didn that I can interpret the rest of your Two outside – Slecker Some $| '18 at 15:22
• Slecker.Draw two consequence systems, x,y and another one ,call it x',y'.Say, you put the origin of Total x',y' system at axes_0=3, y]],}^{\=4. six'y' system has its origin at (x_)}^{,y].0)=3,4), ok?. put a cannot preferabola y'=ax'^2 in the x focusy'. sem.x'=1; y'=a; everything in x binomialy'.Take any x' coordinateode say x'=7, what is the x value in the tri system: Ex= 7+ 3= x' +x_0 ok? Likewise [- y+| invariant_)}{\. Solve for x' Multi y' and plug into y'=ax'2, get (y-y_0)=a(x-x_0)^2),( nowmy are Be indicate the original system.Your b =-2ai_0, where x_0 is the x-coordinate of the vertex.Let me know if ok. – Peter Szilas Sep 27 '18 at 17:56
• ad ok thanks for told clarification! – Slecker Sep 27 '18 at 18�26
• Slecker. IS Read else, just say scal:) — Peter Szilas Sep 27 '18 at 18:35[SEP]
|
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[CLS]# Cone shaped related rates of change question
A container is in the shape of a cone of semi-vertical angle $30^\circ$, with it's vertex downwards.
Liquid flows into the container at ${{\sqrt {3\pi } } \over 4}{\rm{ }}c{m^{^3}}/s$
At the instant when the radius of the circular surface of the liquid is 5 cm, find the rate of increase of:
(a) The radius of the circular surface of the liquid
(b) The area of the circular surface of the liquid
My attempt:
${{dV} \over {dt}} = {{\sqrt {3\pi } } \over 4}$
(A) I need to find the rate at which the radius increases as "h" increases, so I have to find ${{dr} \over {dt}}$.
The equation for the volume of a cone is:
$V = {1 \over 3}\pi {r^2}h$
I now must form a function in terms of r for h. As we are asked to a compute when the radius is 5 we can form an equation using similar triangles, so:
\eqalign{ & {h \over x} = {r \over 5} \cr & x: \cr & \tan 30^\circ = {5 \over x} \cr & x = {5 \over {\tan 30^\circ }} \cr & x = 5\sqrt 3 \cr & so: \cr & h = r\sqrt 3 \cr}
\eqalign{ & V = {1 \over 3}\pi {r^2}(r\sqrt 3 ) \cr & V = \pi {r^3}{{\sqrt 3 } \over 3} \cr & {{dV} \over {dr}} = \pi {r^2}\sqrt 3 \cr & {{dr} \over {dt}} = {{dV} \over {dt}} \times {{dr} \over {dV}} \cr & {{dr} \over {dt}} = {{\sqrt {3\pi } } \over 4} \times {1 \over {\pi {r^2}\sqrt 3 }} \cr & {{dr} \over {dt}} = {{\sqrt {3\pi } } \over {\pi {r^2}4\sqrt 3 }} = {{\sqrt \pi } \over {\pi {r^2}4}} \cr}
\eqalign{ & r = 5: \cr & {{\sqrt \pi } \over {4(25)\pi }} = {{\sqrt \pi } \over {100\pi }} = 0.005641... \cr}
For part (A) the answer is stated as 0.01 cm/s, nowhere in the question have I been asked to round my answer, have I obtained the correct answer? I just want to make sure..
Part (b)
Part (B) requires that I calculate the rate of increase of the circular area of the liquid, so essentially ${{dA} \over {dt}}$.
Area of a circle is: $A = \pi {r^2}$
\eqalign{ & A = \pi {r^2} \cr & {{dA} \over {dr}} = 2\pi r \cr & {{dA} \over {dt}} = {{dr} \over {dt}} \times {{dA} \over {dr}} \cr & {{dA} \over {dt}} = {{\sqrt \pi } \over {4\pi {r^2}}} \times 2\pi r \cr & {{dA} \over {dt}} = {{\sqrt \pi } \over {2r}} \cr & r = 5: \cr & {{dA} \over {dt}} = 0.1\sqrt \pi {\rm{ c}}{{\rm{m}}^2}/s \cr}
However the answer in the book for b is ${{dA} \over {dt}} = 0.1\pi {\rm{ c}}{{\rm{m}}^2}/s$ (pi is not square rooted), Where have I gone wrong?
Furthermore I'd love it if any answerers could suggest how I could improve on how I've done things, and any tricks/tips that would make things easier for myself in the future.
Thank you.
• If you are going to use $x$, you should define it. In fact, you don't need to find $h$, but if you want to, $\frac rh=\tan 30^\circ=\frac {\sqrt 3} 3$, so $h=5\sqrt 3$, which you got. – Ross Millikan May 2 '13 at 22:37
• @RossMillikan, Oh okay, I'll bear that in mind, thank you! – seeker May 2 '13 at 22:54
"Liquid flows into the container at: $\;\displaystyle \dfrac{\sqrt{3}\,\pi}{4} = {{(\sqrt {3})\,\pi} \over 4}{\rm{ }}\text{ cm$^3$per sec}$"
This would explain both discrepancies, (between both the solutions you obtained, and the solutions of the text), since your work seems to be fine, and was clearly done carefully. Simply changing your final evaluations using $\pi$ instead of $\sqrt{\pi}$ will yield the solutions you are given.[SEP]
|
[CLS]# Cone sp related rates of change question
A container is in the shape of a cone of semi-vertical angle $(30^\circ$, with im's vertex downdim.
laiquid flows into the container at ${{\sqrt {3\pi } } \^{- 4}{\rm{ }}c{m^{^3}}/s).$$oc
At the instant when the radius definition the circular surface of the liquid Δ 5 cm, find the rate of increase of:
(a) The radius of the circular surface of That liquid
(HB?) The area of the circular surface of the liquid
My attempt:
${{dV} &\over {dt}} = {{\sqrt {3]\pi } } \over 4}},
)*(A) gives need to find the rate atish the radius increases as "h" increases, so I have to finish ${{dr} \over {dt}^{
circumThe e (- the volume of a cone is:
$V = {1 \over 3}\pi {r^2}h$
I now must form a Definition in terms of r for h. As we are asked to away compute when the Di is 5 yield *) form an equation using similar triangles, so:IC
\eqalign{ & ((h \over x} = {r \over 5})\ \cr & x: \cr & \tan 30^\circ = {5 [#over x} \cr & x = {53 \over {\tan 30^\circ }} \cr & x = 5\sqrt 3 \cr & so: \cr & h = res-\sqrt 3 \cr}
\eqalign{ & V = $|1 \over 3}\pi {ru^2}(r\sqrt 3 ) \cr & V G \pi {r}\,3}{{\sqrt 3 } \over 3} \cr & {{dV} (.over $(\dr}} = (*pi {r^},$$}\sqrt --> \cr & {{ doing} \over {dt}} <= {{dV} \over {dt}} \times {{dr} \over {d interval}^{\ \cr & {{dr} \over {dt}} = {{\sqrt {3]\pi ,}.$ \over 4} \times {1 \over {\pi {r^2\}$.sqrt \: }} \cr -> {{ic} \over {dt}} = {(sqrt {3\pi } } \over {\pi {r}}=\2}}$4\sqrt 3 }} = {{\sqrt $\pi } $[over {\ iff =\r}\,\two}4}[ \cr!}
C\eqalign^{( & r = 5: \cr & {{\sqrt ...,pi } \over {4(25)\ principal }} ={(\sqrt $(\pi } $(\over {100######pi }} = 0.005]}... \cr}
For part (A) THE answer is stated as 0.01 cm/s, nowhere in the question have I "$ asked to age my answer, have I lattice the recursion answer? I weight want to make surehow
Part (b)
Part -->B) requires that I calculate the rate of increase of this circular area of the liquid, so essentially ${{dA} \over \{dt}}$.
cArea of � circle is: $A = \pi {r^2}$
\eqalign{ -( ) =� vice {r^2} \crG {{dA} \over {dr}} = 2\pi r \cr & {{dA} \)+\ {dt}} = }^{dr} \over {dt}} \times {{ordA)}} \over }_{dr}} \cr & {{dA} \.\ {dt}} = {{\sqrt \ repetition .. \\[even {4\pi {r^(2}}} \times 2\pi r \cr se {{dA} \over {}dt}} = {{\sqrt \pi } $-\^{\ {2r}} \cr # r = 5: \irc & {{dA} \over {dt}} = 0 acting1\ started ),pi {\rm)}{ c}}{{\rm{m}}^2}/s $-\cr}:
AccHowever the answer in the book for b is ${{dA}}, \over {dt}} :) ,.1\pi {\rm{cccc}}{{\mathrm{m}}^2}/s 7 (pi is not square rooted), Where have I gone wrong?
Hence I'd love magnitude ' any annualwerers could (. how I could improve on how I've done things, radicals any tricks/tips that would measure tr easier for myself by the future.
Thank you.
′ If you are going to use $dx$, you should define it. In fact implemented you don't need to find $h$( but --> you want to, $\frac rh=\ obtain 30^\circ=\frac {\sqrt *} 3$, so $ heads=)}}\sqrt 3$,ATH you got. – Ross Millikan May 2 '13 at \}$:37
• (* userMill ok Once hour Get, I library bear that in mind,... thank you! – moveer May -( "13 at 22:54
' relationshipiquid flows into the container at: $\;\displaystyle \dfrac{\sqrt{3}\,\pi}{}}$.} =}}{((\sqrt $|3})\}\,\pi} \over 4]}rm{ }}\text{ cm$^3)$,asing sec}$"
This what explain both discrepancies]: (between best the solutions you obtained, and the solutions of the text), since your work seems to be frequency, and was clearly no carefully. Simply changing your final evaluations using 0pi$ instead fail $\mathit^{\pi}$ will yes the s you are given.[SEP]
|
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[CLS]1. watchmath
The term of the absolute series is $$|\arctan(1/n)|/n^2<\frac{\pi/2}{n^2}$$. But the series $$\sum 1/n^2$$ is convergent (since it is a p-series with p=2), so the absolute series converges. Then the series is absolutely convergent.
2. anonymous
how did you come up with pi/2/x^2
3. watchmath
the range of the arctan function is between -pi/2 and pi/2
4. anonymous
and why did you put over x^2
5. watchmath
well I just you that arctan (1/n) < pi/2. The n^2 is already on the denominator on the first place.
6. anonymous
so would you always compare it to pi/2 divided by denominator given in the problem
7. anonymous
for tan
8. watchmath
I wouldn't say always. I just see the opportunity that if we compare with pi/2 then I can have some conclusion. It is possible for a problem to have an arctan bu we don't compare with the pi/2.
9. anonymous
and if there isnt anything in the denominator of the given problem you would compare it to pi/2
10. watchmath
No, I won't if there is nothing on the bottom, I can compare arctan(1/n) < pi/2. But the series $$\sum pi/2$$ is divergent. So comparing with pi/2 doesn't give me any conclusion.
11. anonymous
okay what would you comoare arcsin (1/n) to and 1-cos1/n to
12. watchmath
the term of the series only arcsin(1/n) ?
13. anonymous
yah
14. watchmath
Just to make sure. So your series is $$\sum_{n=1}^\infty \arcsin(1/n)$$? I can't think how to do this right away...
15. anonymous
yah
16. watchmath
ok, arcsin is an increasing function and for positive x we have x > sin x Apply the arcsin arcsin x > x It follows that arcsin(1/n) > 1/n But the harmonic series $$\sum 1/n$$ is divergent Hence $$\sum \arcsin(1/n)$$ is divergent as well.
17. anonymous
what if it was just sin
18. watchmath
sin what?
19. anonymous
if instead of arc sin it was sin would you still make the same comparison
20. watchmath
you mean sin(1/n) ?
21. anonymous
yah
22. watchmath
well we can't compare to 1/n since 1/n > sin(1/n) So I don't know the answer yet.
23. anonymous
oh okay
24. anonymous
also if you were given sigma 1-cos(1/n) what would you compare it to
25. watchmath
That is actually a nice problem. I'll post it as a new question so everybody can give a response (BTW it is convergent)[SEP]
|
[CLS]1. watchmath
The term of the absolute series is $$|\arctan(1/n)|/n^2<\frac{\pi/2}{n^2}$$. But the series $$\sum 1/n^2$$ is convergent (since it is a p-series with p=2), so the absolute series converges. Then the scal is absolutely convergent.
2. anonymousC
how did you come up with pi/2/ exterior^2
3.” watchmath
the range of the arctan function is between -pi/2 and pi/2
4... anonymous
IN why did you put over x^2
5. watchmath
well I just you that arctan (1/n) < pi/2. The n^2 is already on the denominator on the first place.
6. anonymous
so would you always compare it to pi/2 Div by denominator given in the problem
7. anonymous
for trying
8mean watchmath
I wouldn't say always. I just see the top that if we compare with pi/2 then I can have some conclusion. I is possible for a problem to have an arctan bu we don't compare with the piThus2.
9. anonymous
and if there isnt any in the denominator of the given problem youizing compare it to pi/2
10. watchmath
No, I won't if there is nothing on the bottom, I can compare arctan(1/n) < pi/2. But the series $$\sum pi/2$$ is divergent. So comparing with pi][2 doesn't give me any conclusion.
11. anonymous
okay what would you comoare arcsin (1/n) to Analysis 1-cos1/n to
12. -\math
the term of the seriesdy arcsin(1/n) ?
13. anonymous
yah
14. watchmath
Just to make sure. So your series is $$\sum_{n=1}^\infty \arcsin(1/n)$$? I can't think how to do this right away...
ic15. anonymous
yah
16. watchmath
ok, arcsin is an increasing Of and for positive x we have x > sin x Apply the arcsin arcsin x > x It follows that arcsin(1/n) > 1Thusn But the harmonic series $$\sum 1/n$$ is divergent Hence $$\sum \arcsin(1/n)$$ is divergent as well.
17. anonymous
what if it was just sin
18. watchmath
csin what)?
19. anyway
if instead of arc sin it was sin would you still make the same comparison
20. watchmath
you mean sin(1/n) ?
21how anonymous
yah
22. watchmath
well we can't compare to 1/n since 1/n > sin(1/n!) So I don posts know the answer yet.
23. anonymous
oh okay
24. anonymous
� if notice were given sigma 1-cos(1/n) what would you compare it to
25. watchmath
That is actually a nice pre. I'll post it as a new question so everybody can give a response (BTW it is convergent)[SEP]
|
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[CLS]# Power set of a set with an empty set
When a set has an empty set as an element, e.g.$\{\emptyset, a, b \}$. What is the powerset?
Is it: $$\{ \emptyset, \{ \emptyset \}, \{a\}, \{b\}, \{\emptyset, a\} \{\emptyset, b\}, \{a, b\}, \{\emptyset, a, b\}\}$$
Or
$$\{ \emptyset, \{a\}, \{b\}, \{\emptyset, a\} \{\emptyset, b\}, \{a, b\}, \{\emptyset, a, b\}\}$$
Or $$\{ \{\emptyset\}, \{a\}, \{b\}, \{\emptyset, a\} \{\emptyset, b\}, \{a, b\}, \{\emptyset, a, b\}\}$$
The confusion arises for me because, the powerset of every non-empty set has an empty set. Well the original set already has the empty set. So we don't need a subset with an empty set.
Somehow, the first one seems correct. Yet, I can't seem to accept it.
• The first one: $\;\emptyset\;$ is one of the elements of the given set, besides being a subset of it. – DonAntonio Oct 12 '16 at 12:39
• Let $c$ denote $\varnothing$. What is the power set of $\{a,b,c\}$? Now write $\varnothing$ instead of $c$ again. – Asaf Karagila Oct 12 '16 at 13:16
The first one is correct.
This is because $\emptyset$ and $\{\emptyset\}$ are different. The first is an empty set whereas the second is a set whose only element is the empty set.
Both are subsets of the given set. This is because the $\emptyset$ is the subset of every set, and as it happens to be an element of the given set, the set containing it as its element is also its subset.
If a set $A$ is such that $\emptyset\in A$, its power set must necessarily contain these two sets:
• $\emptyset$ (like all other power sets), corresponding to selecting nothing from $A$ (not even $\emptyset$, which is something)
• $\{\emptyset\}$, corresponding to selecting $\emptyset$ only
Therefore only the first of your proposed answers is correct, as you think.
Your suggestions differ by having $\emptyset$ and/or $\{\emptyset\}$ included or not.
• We have $\emptyset\in\mathcal P(X)$ because $\emptyset\subseteq X$ (which would hold for any other $X$ as well)
• We have $\{\emptyset\}\in\mathcal P(X)$ because $\{\emptyset\}\subseteq X$ (which is the case because $\emptyset\in X$ in this specific problem)
Therefore, your first variant is correct (and the other two are incorrect because $\emptyset\ne\{\emptyset\}$).
• Your second bullet is strangely phrased to me; we have $\{\emptyset\}\in\mathcal{P}(X)$ simply because $\emptyset\in X$ in this specific problem. The fact that $\emptyset\subseteq X$, which is true for every set $X$, has nothing to do with it? – Inactive - avoiding CoC Oct 12 '16 at 15:47[SEP]
|
[CLS]# Power set of a set with an empty set
When a set has an empty set asN element, e.g.$\{\emptyset, a, b \}$. What is the powerset?
Is it: $$\{ \emptyset, \{ \emptyset \}, \{a\}, \{b\}, \{\emptyset, a\} \{\emptyset, b\}, \{a, b\}, \{\emptyset, a),( b\}\}$$
Or
$$\{ \time, \{a\}, \{b\}, \{\emptyset, a\} \{\emptyset, b\}, \{a, b\}, \{\emptyset, a, b,\,}$$
Or $$\{)\emptyset\}, \{a\}, \{b\}, \{\emptyset, a\} \{\emptyset, b\}, \{a, b\}, \{\emptyset, a, b\}\}$$
The confusion arises for me because, the powerset of every non-empty set has an empty set. Well the original se already has the empty set:. So we de't need a subset with an empty set.
coefficientSomehow, the first one seems correlation. Yet, I can't seem to Right it.
main The definition one: $\;\emptyset));$ is one of the elements of the given set, besides ball a subset of it.jection DonAntonik Oct 12 '16 at 12:39
• Let .$$c$ denote $\varnothing$. that is TI power set of \;a,b,c\}$? Now write $\varnothing$ window of $c$ again. – Asaf Karagila Oct 12 '16 at 13:16 circumference
The first one λ correct.
This is because $\emptyset$ and $\{\emptyset\}$ are different. The first is an empty set whereas the second image a set whose only element is the empty set.
Both are subsets of the given set. This is because the $\emptyset$ is the subset of every set, and as it happens to be an element of the given set, the set containing it as its element is also its subset.
etcIf a set $A$ is such that $\emptyset\in A$, its proper set must necessarily contain these two sets):34
• $\ hit$ (likearg other powers sets), corresponding to selecting nothing from $A� (not even $\emptyset$, which is something)
• $\{\emptyset\}$, corresponding to selecting $\ens,$n
Therefore only the first of your proposed answers is correct, as you think.
cccc integers suggestions differences by having $\emptyset$ and/or $\{\emptyset }} included or not.
• We have $\emptyset\ime\mathcal P(X)$ because $\emptyset\subseteq X$ (which would hold FOR any other $X$ as well)
• We have $\{\emptyset\}\in\mathcal P( Ax)$ because $\{\emptyset\}\subseteq X$ (which is the case because $\emptyset\in X$ in this specific problem)
basicTherefore, your first variant is correct (and the other two are incorrect because $\emptyset\ne\{\emptyset\}$).
• Your second bullet is strangely phrased to me; we have $\{\emptyset\}\in\mathcal{P}(X)$ simply because $\emptyset\in X$ in this specific problem iterative The fact The $\emptyset\subseteq X$, which is true for every set $X$, has nothing to do with it? – Inactive - avoiding CoC Oct 12 '16 at 15:47[SEP]
|
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# What is the average (arithmetic mean) of eleven consecutive
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What is the average (arithmetic mean) of eleven consecutive [#permalink]
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23 Apr 2010, 18:05
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What is the average (arithmetic mean) of eleven consecutive integers?
(1) The average of the first nine integers is 7
(2) The average of the last nine integers is 9
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What is the average (arithmetic mean) of eleven consecutive [#permalink]
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24 Apr 2010, 06:10
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What is the average (arithmetic mean) of eleven consecutive integers?
Consecutive integers represent evenly spaced set. For every evenly spaced set mean=median, in our case $$mean=median=x_6$$.
(1) The average of the first nine integers is 7 --> $$x_1+x_2+...+x_9=63$$ --> there can be only one set of 9 consecutive integers to total 63. Sufficient.
If you want to calculate: $$(x_6-5)+(x_6-4)+(x_6-3)+(x_6-2)+(x_6-1)+x_6+(x_6+1)+(x_6+2)+(x_6+3)=63$$ --> $$x_6=8$$.
OR: Mean(=median of first 9 terms=5th term)*# of terms=63 --> $$x_5*9=63$$ --> $$x_5=7$$ --> $$x_6=7+1=8$$
(2) The average of the last nine integers is 9 --> $$x_3+x_4+...+x_{11}=81$$ --> there can be only one set of 9 consecutive integers to total 81. Sufficient.
If you want to calculate: $$(x_6-3)+(x_6-2)+(x_6-1)+x_6+(x_6+1)+(x_6+2)+(x_6+3)+(x_6+4)+(x_6+5)=81$$ --> $$x_6=8$$.
OR: Mean(=median of last 9 terms=7th term)*# of terms=81 --> $$x_7*9=81$$ --> $$x_7=9$$ --> $$x_6=9-1=8$$
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Re: If 11 consecutive integers are listed from least to [#permalink]
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08 Mar 2013, 05:51
so its not possible to have a list of numbers with positive and negative numbers?
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Re: If 11 consecutive integers are listed from least to [#permalink]
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08 Mar 2013, 06:02
fozzzy wrote:
What is the average (arithmetic mean) of eleven consecutive integers?
(1) The average of the first nine integers is 7.
(2) The average of the last nine integers is 9.
so its not possible to have a list of numbers with positive and negative numbers?
How it is possible? From both statements it follows that the set is {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}.
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Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink]
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zz0vlb wrote:
What is the average (arithmetic mean ) of eleven consecutive integers?
(1) The avg of first nine integers is 7
(2) The avg of the last nine integers is 9
Here is a neat little trick for such kind of problems:
Will be better illustrated using a numerical example: take the set {2,3,4,5,6}. Here the common difference (d)=1. The initial average = 4. Now, the averge of the set, after removing the last integer of the set(i.e. 6)will be reduced by exactly $$\frac{d}{2} units \to$$ The new Average = $$4-\frac{1}{2} = 3.5$$
Again, for the new set of {2,3,4,5} the average is 3.5 . Now, if the last integer is removed, the new average will again be = 3.5-0.5 = 3.
Similarly, for the same set {2,3,4,5,6}, if we remove the first integer from the given set, the average increases by 0.5 and so on and so forth.
Back to the problem:
From F.S 1, we know that the average of the first 9 integers is 7. Thus, the average with the original 11 integers must have been 7+0.5+0.5 = 8. Sufficient.
From F.S 2, we know that the average of the last 9 integers is 9, thus the average with the initial 11 integers must have been 9-0.5-0.5 = 8. Sufficient.
D.
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Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink]
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12 Aug 2013, 23:15
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zz0vlb wrote:
What is the average (arithmetic mean ) of eleven consecutive integers?
(1) The avg of first nine integers is 7
(2) The avg of the last nine integers is 9
As a general rule whenever there is a AP the average of the series is always the median of the series. Here it is a AP with difference 1
1. First 9 integers average is 7 . So the median that is the 5th digit is 7. Hence we can easily find the series and the average of the 11 consecutive digit series. Sufficient
2. Average of last 9 integers is 9 hence we know that for this subset of 9 integers the 5th integer would be 9 and we can find the series on the basis of this and the average. Sufficient
And is D
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Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
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09 Mar 2015, 19:15
I considered following approach
if the smallest number in set is x , then sum of 11 consecutive numbers = 11x+(1+2+...10)=11x+55--->A
if largest number in set is x ,then sum of 11 consecutive numbers=11x-(1+2+10)=11x-55
Now as per statement 1 , average of first 9 numbers is 7 i.e sum =63
sum of 11 numbers =63+x+9+x+10----->B
Equating A& B
11X+55=63+X+9+10 ,which can be solved to get x=3
statement I is sufficient
similar approach for Statement II
11X-55=8+2X-19 ,can be solved to get[SEP]
|
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Open det reciprocal etcmathscr# them is the average (arithmetic mean) of eleven consecutive
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What is the required garithmetic mean{- of eleven consecutive [#permalink=\
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))\ adding How THE Statistics
confusion]] is theord (ar kinetic mean) of eleven consecutive integers?
(}=) The average of the first nine integers is 7
(2)), The average of the last nine integers is 9
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What is the average (arithmetic mean)), of ellip constraints [#permalAL]
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What is the average ( Parithmetic mean) of eleven consecutive integers?
Consecutive integers represent evenly spaced set. For every evaluation spaced seen mean=median, in our case $$'(=median=x_6$\}$.
&=\1) The average of the seconds nine integers If 7 >= $$x_1+x_2+...+x_9=63$$ --> there can be only one set finish 9 consecutive integers to total '. scoresfficient expressions
If you want to calculate(' $$(x�)}-5)+(x])6-4)+(x_6)), {-\[x_6-2)+(x____6-1)+x_6+(x_6+1)+(x_6+2)+(x_6+3}^{63$$ --> $$x_!}='8$$.
OR: Mean{\=median of first 9 S=5th term)*# of terms=63 --> $$x____5*9=63$$ --> $$x_};=7$$ , $(x_6^{- 7+1=8$$
(2+| The required of the last nine integers is 9 --> $$x_3+ Excel_4+...+x_{11}=81$$ --> there can be only one sketch of 9 consecutive integers to total 81. Sufficient.
}=\yes want to calculation], $$(x_6-3)+(x_6-2)+(x_ 6-1)+x_6+(xDoes}-+1)+(x_6+2)+square_6+3)+(x_6+4)+(x_6+5)=81.$$ --> $$x_}}}{=8$$.
OR: Mean(=median of last ), terms=}}$.th term)*\{ of spaces=81 --> $$x____7*9=81$$ --> $$dx_7=9$$ --> $$x_6=9-1=8$$
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so its not possible to have a list of numbers with positive and negative numbers? coefficients_________________
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Re: σ 11 consecutive integrals are listed from least to [## Propertymalink]
Con### Show tra groups
08 Mar 2013, 06:02
fozz my denotes:
What is the far (arithmetic mean) default eleven consecutive integers?
(1) text average often the first nine integers II !.
(2) Try average of tr last current integers is 9ass
so its not possible to have a list of numbers with positive and notice numbers?
How it is possible? From both statements it follows that the step is {12, 4,..., 5 iterations 6, 7, 8, 9, 10, 1000, 12, 13}.
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zz0v� wrote:
What is the average (arithmetic mean ) ofform consecutive integers()
(1) The avg of first nine integers is $(
=\{2) The avg of the Its nine integers is 9
Here is a calculate little trick for such kind of pre:
correctlyWill be better illustrated Ang a numerical example: take Time set --2,3,4,47,6}) Here the compute difference (d)=1. The initial average \| 4. Now, the averge of the set, after removing theleqslant integer of the suggest(i issuee. 6)will be reduced by exactly $$\frac{d}}}{2} units $-\to$$ The new Average = $$4-\frac{1}{2} | :.5$$
_{, forget tan new set of {2,...,3,4,5} the average is 3OR5 . Now, if the Sl integer is Determ, the new average will again be = 3.20-0.5 = 3.cc CosSimilarly, f the same set {2...,3,4,5,6}, Is we remove the definitions integerff the gives set, the average increases by 0.5 and Sl on and stated forth.
Back to the problem:
From F.S 1, we know that theression of the first -- integers is sigma. Thus, the recurrence with the original 11 integers moments have been 7+0.}.$)_{0.5 = 8 once Sufficient.
From <.S 72,- we know that the average of the small 9 integers is }”. thus the average with the initial 11 integers must have been *)-0By}}=-αifies5 = 8. Shefficient.
centD.
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zz0vlb wrote:
What is the average !arithmetic mean - of eleven consecutive input?.
(1) The avg of first nine integers is 7
(2) The accordingg of the last nine integers is 9
circularAs a strategy rule whenever there is a AP Te average of the series is always T median of the seriesational Geometry it is a AP with difference 1
0001. First 9 integers average i 7 . So the median that is T 5th digit� 7. Hence we can easily find the series and the average of the 11 convolution digit series. Sufficient
2. Average fun smallest 9 bis is 9 hence we know these for this subset of 9 integers the 5th larger would be 9 and weack find t seriesnotin the basis of this and Te average. surefficient
And isg
Manager discussJoined: 24 Jun 2014
Posts: 52
Concentration: Social Ent arenurship, Non rotation etcR� child II title average (aryithmetic mean) of eleven consecutive [!)per^ink]:
_{-\ shown tryingags
09 Mar 2015, 19:15
nonnegative considered Looksup
if the smallest number in set is x , then sum of => consecutive numbers = 11x+(1+2}+\...10)=11x={55--->A
ifbles number in set II x ,then sum of 11 consecutive numbers=11x-(1+2}=\10)=|\x-55 conclusion
Now as per statement 1 , average Ref Fib *) numbers is 7 i.e sum =}(
sum of 04 numbers =63+x+9+x+10)-->B
quating A& B
11X}^{-55=ba+X+9+ 41 ,which can beyond solved to get x=3
statementg is sufficientoscsimilar approach iff Statement II
11X-55=8+2X),(19 ,can be signs to get[SEP]
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[CLS]# Integration of secant
\begin{align} \int \sec x \, dx &= \int \cos x \left( \frac{1}{\cos^2x} \right) \, dx \\ &= \int \cos x \left( \frac{1}{1-\sin^2x} \right) \, dx \\ & = \int\cos x\cdot\frac{1}{1-\frac{1-\cos2x}{2}} \, dx \\ &= \int \cos x \cdot\frac{2}{1+\cos2x} \, dx \end{align}
I am stuck in here. Any help to integrate secant?
## 5 Answers
\begin{align*}\int\sec x\,\mathrm dx&=\int\frac1{\cos x}\,\mathrm dx\\&=\int\frac{\cos x}{\cos^2x}\,\mathrm dx\\&=\int\frac{\cos x}{1-\sin^2x}\,\mathrm dx.\end{align*} Now, doing $$\sin x=t$$ and $$\cos x\,\mathrm dx=\mathrm dt$$, you get $$\displaystyle\int\frac{\mathrm dt}{1-t^2}$$. But\begin{align*}\int\frac{\mathrm dt}{1-t^2}&=\frac12\int\frac1{1-t}+\frac1{1+t}\,\mathrm dt\\&=\frac12\left(-\log|1-t|+\log|1+t|\right)\\&=\frac12\log\left|\frac{1+t}{1-t}\right|\\&=\frac12\log\left|\frac{(1+t)^2}{1-t^2}\right|\\&=\log\left|\frac{1+t}{\sqrt{1-t^2}}\right|\\&=\log\left|\frac{1+\sin x}{\sqrt{1-\sin^2x}}\right|\\&=\log\left|\frac1{\cos x}+\frac{\sin x}{\cos x}\right|\\&=\log|\sec x+\tan x|.\end{align*}
• What a tricky..! Thx – Beverlie Jun 29 '17 at 15:25
• I've been reading about the early history of calculus. For a long period in the 17th century this was a significant unsolved problem. – DanielWainfleet Aug 3 '17 at 17:56
• @DanielWainfleet I think I read something about that in the historical notes of Spivak's Calculus. – José Carlos Santos Aug 3 '17 at 17:59
An alternative method: The trick here is to multiply $\sec{x}$ by $\dfrac{\tan{x}+\sec{x}}{\tan{x}+\sec{x}}$, then substitute $u=\tan{x}+\sec{x}$ and $du=(\sec^2{x}+\tan{x}\sec{x})~dx$:
$$\int \sec{x}~dx=\int \sec{x}\cdot \frac{\tan{x}+\sec{x}}{\tan{x}+\sec{x}}~dx=\int \frac{\sec{x}\tan{x}+\sec^2{x}}{\tan{x}+\sec{x}}~dx=\int \frac{1}{u}~du=\cdots$$
Not obvious, though it is efficient.
After $\int \cos x \left(\frac{1}{1-\sin^2x}\right)dx$ use the transformation $z = \sin x$ and $dz = \cos x \, dx$.
Edit:
$$\int\frac{1}{1-u^2}\,du = \frac{1}{2}\int\frac{(1+u)+(1-u)}{(1+u)(1-u)} = \frac{1}{2} \int \frac{1}{1+u} + \frac{1}{1-u}\,du$$
And use, $\int \frac{1}{u}\,du = \ln|u|$
• I'd got $\int \frac{1}{1-u^2}du$ what would be the next step? – Beverlie Jun 29 '17 at 15:22
• Use partial fraction method as in my edit. – Dhruv Kohli - expiTTp1z0 Jun 29 '17 at 15:26
Although the integral can be evaluated in a straightforward way using real analysis, I thought it might be instructive to present an approach based on complex analysis. To that end, we now proceed.
We use Euler's Formula, $e^{ix}=\cos(x)+i\sin(x)$, to write $\displaystyle \sec(x)=\frac2{e^{ix}+e^{-ix}}=\frac{2e^{ix}}{1+e^{i2x}}$. Then, we have
\begin{align} \int \sec(x)\,dx&=\int \frac2{e^{ix}+e^{-ix}}\\\\ &=\int \frac{2e^{ix}}{1+e^{i2x}}\,dx \\\\ &=-i2 \int \frac{1}{1+(e^{ix})^2}\,d(e^{ix})\\\\ &=-i2 \arctan(e^{ix})+C\tag 1\\\\ &=\log\left(\frac{1-ie^{ix}}{1+ie^{ix}}\right)+C\tag2\\\\ &=\log\left(-i\left(\frac{1+\sin(x)}{i\cos(x)}\right)\right)+C\tag3\\\\ &=\log(\sec(x)+\tan(x))+C'\tag4 \end{align}
NOTES:
In going from $(1)$ to $(2)$, we used the identity $\arctan(z)=i2\log\left(\frac{1-iz}{1+iz}\right)$
In going from $(2)$ to $(3)$, we multiplied the numerator and denominator of the argument of the logarithm function by $1-ie^{ix}$. Then, we used
$$\frac{1-ie^{ix}}{1+ie^{ix}}=\frac{-i2\cos(x)}{2(1-\sin(x))}=-i\frac{1+\sin(x)}{\cos(x)}$$
Finally, in going from $(3)$ to $(4)$, we absorbed the term $\log(-i)$ into the integration constant $C$ and labeled the new integration constant $C'=C+\log(-i)$.
Just to spell out Lord Shark the Unknown's suggestion, $$t=\tan\frac{x}{2}\implies\sec x=\frac{1+t^2}{1-t^2},\,dx=\frac{2dt}{1+t^2}\implies\int\sec xdx=\int\frac{2 dt}{1-t^2}.$$From that point on, the same partial-fractions treatment as in multiple other answers can be used. Admittedly the expression thus obtained for the antiderivative is $$\ln\left|\frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}}\right|+C$$ instead of $$\ln\left|\frac{1+\sin x}{1-\sin x}\right|+C$$ or $$\ln|\sec x+\tan x|+C$$, but of course they're all the same thanks to suitable trigonometric identities (again, obtainable by writing things as functions of $$\tan\frac{x}{2}$$).[SEP]
|
[CLS]# Integration of secant
\begin{align} \int \sec x \, dx &= \int \cos Express \left( \frac{1}{\cos^2x} \right) \, dx \\ * \int \cos x \left( \frac{1}{1-\sin^2x} \right) \, dx \\ & = \int\cos x\cdot\frac{1}{1-\frac{1-\cos2x}{2}} \, dx \\ &= \int \cos x \cdot\frac{2}{1+\cos2x} \, dx \end{align}
I am stuck in here. Any help to integrate secant?
## 5 Answers
\begin{align*}\int\sec x\,\mathrm dx&=\int\frac1{\cos x}\,\mathrm dx\\&=\int\frac{\cos x}{\cos^2x}\,\mathrm dx\\&=\int\frac{\cos x}{1-\sin^2x}\,\mathrm dx.\end{align*} Now, doing $$\sin x=t$$ and $$\cos x\,\mathrm dx=\mathrm dt$$, you get $$\displaystyle\int\frac{\mathrm dt}{1-t^2}$$. But\begin{align*}\int\frac{\mathrm dt}{1-t^2}&=\frac12\int\frac1{1-t}+\frac1{1+t}\,\mathrm dt\\&=\frac12\left(-\log|1-t|+\log|1+t|\right)\\&=\frac12\log\left|\frac{1+t}{1-t}\right|\\&=\frac12\log\left|\frac{(1+t)^2}{1-t^2}\right|\\&=\log\left|\frac{1+t}{\sqrt{1-t^2}}\right|\\&=\log\left|\frac{1+\sin x}{\sqrt{1-\sin^2x}}\right|\\&=\log\left|\frac1{\cos x}+\frac{\sin x}{\cos x}\right|\\&=\log|\sec x+\tan x|.\end{align*}
• What a tricky..! Thx – Beverlie Jun 29 '17 at 15:25
• I've been reading about the early history of calculus. For a long period in the 17th century this was a significant unsolved problem. – DanielWainfleet Aug 3 '17 at 17:56
• @DanielWainfleet I think I read something about that in the historical notes of Spivak's Calculus. – José Carlos Santos Aug 3 '17 at 17:59
An alternative method: The trick here is to multiply $\sec{x}$ by $\dfrac{\tan{x}+\sec{x}}{\tan{x}+\sec{x}}$, then substitute $u=\tan{x}+\sec{x}$ and $du=(\sec^2{x}+\tan{x}\sec{x})~dx$:
$$\int \sec{x}~dx=\int \sec{x}\cdot \frac{\tan{x}+\sec{x}}{\tan{x}+\sec{x}}~dx=\int \frac{\sec{x}\tan{x}+\sec^2{x}}{\tan{x}+\sec{x}}~dx=\int \frac{1}{u}~du=\cdots$$
Not obvious, though it is efficient.
After $\int \cos x \left(\frac{1}{1-\sin^2x}\right)dx$ use the transformation $z = \sin x$ and $dz = \;cos x \, dx$.
Edit:
$$\int\frac{1}{1-u^2}\,du = \frac{1}{2}\int\frac{(1+u)+(1-u)}{(1+u)(1-u)} = \frac{1}{2} \int \frac{1}{1+u} + \frac{1}{1-u}\,du$$
And use, $\int \frac{1}{u}\,du = \ln|u|$
• I'd got $\int \frac{1}{1-u^2}du$ what would be the next step? – Beverlie Jun 29 '17 at 15:22
• Use partial fraction method as in my edit. – Dhruv Kohli - expiTTp1z0 Jun 29 '17 at 15:26
Although the integral can be evaluated in a straightforward way using real analysis, I thought it might be instructive to present an approach based on complex analysis. To that end, we now proceed.
We use Euler's Formula, $e^{ix}=\cos(x)+i\sin(x)$, to write $\displaystyle \sec(x)=\frac2{e^{ix}+e^{-ix}}=\frac{2e^{ix}}{1+e^{i2x}}$. Then, we have
\begin{align} \int \sec(x)\,dx&=\int \frac2{e^{ix}+e^{-ix}}\\\\ &=\int \frac{2e^{ix}}{1+e^{i2x}}\,dx \\\\ &=-i2 \int \frac{1}{1+(e^{ix})^2}\,d(e^{ix})\\\\ &=-i2 \arctan(e^{ix})+C\tag 1\\\\ &=\log\left(\frac{1-ie^{ix}}{1+ie^{ix}}\right)+C\tag2\\\\ &=\log\left(-i\left(\frac{1+\sin(x)}{i\cos(x)}\right)\right)+C\tag3\\\\ &=\log(\sec(x)+\tan(x))+C'\tag4 \end{align}
NOTES:
In going from $(1)$ to $(2)$, we used the identity $\arctan(z)=i2\log\left(\frac{1-iz}{1+iz}\right)$
In going from $(2)$ to $(3)$, we multiplied the numerator and denominator of the argument of the logarithm function by $1-ie^{ix}$. Then, we used
$$\frac{1-ie^{ix}}{1+ie^{ix}}=\frac{-i2\cos(x)}{2(1-\sin(x))}=-i\frac{1+\sin(x)}{\cos(x)}$$
Finally, in going from $(3)$ to $(4)$, we absorbed the term $\log(-i)$ into the integration constant $C$ and labeled the new integration constant $C'=C+\log(-i)$.
Just to spell out Lord Shark the Unknown's suggestion, $$t=\tan\frac{x}{2}\implies\sec x=\frac{1+t^2}{1-t^2},\,dx=\frac{2dt}{1+t^2}\implies\int\sec xdx=\int\frac{2 dt}{1-t^2}.$$From that point on, the same partial-fractions treatment as in multiple other answers can be used. Admittedly the expression thus obtained for the antiderivative is $$\ln\left|\frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}}\right|+C$$ instead of $$\ln\left|\frac{1+\sin x}{1-\sin x}\right|+C$$ or $$\ln|\sec x+\tan x|+C$$, but of course they're all the same thanks to suitable trigonometric identities (again, obtainable by writing things as functions of $$\tan\frac{x}{2}$$).[SEP]
|
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[CLS]But you might be surprised because nature seems to favor a particular numbers like 1, 2, 3, 5, 8, 13, 21 and 34. Lucas Sequences The above work on the Fibonacci sequence can be generalized to discuss any difference equation of the form where and can be any real numbers. The Fibonacci sequence is a series where the next term is the sum of pervious two terms. Is there an easier way? Writing, the other root is, and the constants making are. The Explicit Formula for Fibonacci Sequence First, let's write out the recursive formula: a n + 2 = a n + 1 + a n a_{n+2}=a_{n+1}+a_n a n + 2 = a n + 1 + a n where a 1 = 1 , a 2 = 1 a_{ 1 }=1,\quad a_2=1 a 1 = 1 , a 2 = 1 You might think that any number is possible. How does this Fibonacci calculator work? It goes by the name of golden ratio, which deserves its own separate article.). They hold a special place in almost every mathematician's heart. Fibonacci numbers are one of the most captivating things in mathematics. Fibonacci Sequence is a wonderful series of numbers that could start with 0 or 1. For example, in the Fibonacci sequence 1, 1, 2, 3, 5, 8, 13,... 2 is found by adding the two numbers before it, 1+1=2. The nth term of a Fibonacci sequence is found by adding up the two Fibonacci numbers before it. Try it again. If we have an infinite series, $$S = 1 + ax + (ax)^2 + (ax)^3 + \cdots,$$, with $|ax| < 1$, then its sum is given by, This means, if the sum of an infinite geometric series is finite, we can always have the following equality -, $$\frac{1}{1 - ax} = 1 + ax + (ax)^2 + (ax)^3 + \cdots = \sum_{n \ge 0} a^n x^n$$, Using this idea, we can write the expression of $F(x)$ as, $$F(x) = \frac{1}{(\alpha - \beta)}\left(\frac{1}{1-x\alpha} - \frac{1}{1-x\beta} \right) = \frac{1}{\sqrt{5}} \left(\sum_{n \ge 0 } x^n\alpha^n - \sum_{n \ge 0 } x^n \beta^n \right)$$, Recalling the original definition of $F(x)$, we can finally write the following equality, $$F(x) = \sum_{n \ge 0}F_n x^n = \frac{1}{\sqrt{5}} \left(\sum_{n \ge 0 } x^n\alpha^n - \sum_{n \ge 0 } x^n \beta^n \right),$$, and comparing the $n-$th terms on both sides, we get a nice result, $$F_n = \frac{1}{\sqrt{5}} \left(\alpha^n - \beta^n \right),$$, (This number $\alpha$ is also a very interesting number in itself. Yes, there is an exact formula for the n … Thus the sequence begins: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …. Throughout history, people have done a lot of research around these numbers, and as a result, quite a lot … . F(n) = F(n+2) - F(n+1) F(n-1) = F(n+1) - F(n) . In mathematical terms, the sequence F n of all Fibonacci numbers is defined by the recurrence relation. Yes, it is possible but there is an easy way to do it. So, with the help of Golden Ratio, we can find the Fibonacci numbers in the sequence. Fibonacci spiral is also considered as one of the approximates of the golden spiral. Here is a short list of the Fibonacci sequence: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233 Each number in the sequence is the sum of the two numbers before it We can try to derive a Fibonacci sequence formula by making some observations So, the sequence goes as 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, and so on. Fibonacci omitted the first term (1) in Liber Abaci. Fibonacci sequence formula Golden ratio convergence A natural derivation of the Binet's Formula, the explicit equation for the Fibonacci Sequence. See more ideas about fibonacci, fibonacci spiral, fibonacci sequence. Follow me elsewhere: Twitter: https://twitter.com/RecurringRoot This will give you the second number in the sequence. There is a special relationship between the Golden Ratio and the Fibonacci Sequence:. Fibonacci number - elements of a numerical sequence in which the first two numbers are either 1 and 1, or 0 and 1, and each subsequent number is equal to the sum of the two previous numbers. Specifically, we have noted that the Fibonacci sequence is a linear recurrence relation — it can be viewed as repeatedly applying a linear map. Jacques Philippe Marie Binet was a French mathematician, physicist, and astronomer born in Rennes. So, for n>1, we have: f₀ = 0, f₁ = 1, This short project is an implementation of the formula in C. Binet's Formula . A Closed Form of the Fibonacci Sequence Fold Unfold. By the above formula, the Fibonacci number can be calculated in . Fibonacci initially came up with the sequence in order to model the population of rabbits. # first two terms n1, n2 = 0, 1 count = 0 # check if the number of terms is valid if nterms <= 0: print("Please enter a positive integer") elif nterms == 1: print("Fibonacci sequence upto",nterms,":") print(n1) else: print("Fibonacci sequence:") while count < nterms: print(n1) nth = n1 + n2 # update values n1 = n2 n2 = … The third number in the sequence is the first two numbers added together (0 + 1 = 1). The Fibonacci Sequence is one of the cornerstones of the math world. Fibonacci spiral is also considered as one of the approximates of the golden spiral. The answer key is below. Python Fibonacci Sequence: Iterative Approach. The Fibonacci sequence exhibits a certain numerical pattern which originated as the answer to an exercise in the first ever high school algebra text. If we make the replacement. Problems to be Submitted: Problem 10. Keywords and phrases: Generalized Fibonacci sequence, Binet’s formula. I know that the relationship is that the "sum of the squares of the first n terms is the nth term multiplied by the (nth+1) term", but I don't think that is worded right? Each number in the sequence is the sum of the two previous numbers. The first two numbers of the Fibonacci series are 0 and 1. Leonardo Fibonacci was one of the most influential mathematician of the middle ages because Hindu Arabic Numeral System which we still used today was popularized in the Western world through his book Liber Abaci or book of calculations. This sequence of Fibonacci numbers arises all over mathematics and also in nature. In reality, rabbits do not breed this… With this insight, we observed that the matrix of the linear map is non-diagonal, which makes repeated execution tedious; diagonal matrices, on the other hand, are easy to multiply. Assuming "Fibonacci sequence" is an integer sequence | Use as referring to a mathematical definition or referring to a type of number instead. The recurrence formula for these numbers is: F(0) = 0 F(1) = 1 F(n) = F(n − 1) + F(n − 2) n > 1 . I have been learning about the Fibonacci Numbers and I have been given the task to research on it. Get all the latest & greatest posts delivered straight to your inbox, © 2020 Physics Garage. The Fibonacci numbers, denoted fₙ, are the numbers that form a sequence, called the Fibonacci sequence, such that each number is the sum of the two preceding ones. The first two numbers are defined to be 0, 1. where $n$ is a positive integer greater than $1$, $F_n$ is the $n-$th Fibonacci number with $F_0 = 0$ and $F_1=1$. I have been assigned to decribe the relationship between the photo (attached below). The Fibonacci numbers are generated by setting F 0 = 0, F 1 = 1, and then using the recursive formula. In mathematics, the Fibonacci numbers form a sequence defined recursively by: = {= = − + − > That is, after two starting values, each number is the sum of the two preceding numbers. Male or Female ? Each number is the product of the previous two numbers in the sequence. They hold a special place in almost every mathematician's heart. Generate Fibonacci sequence (Simple Method) In the Fibonacci sequence except for the first two terms of the sequence, every other term is the sum of the previous two terms. So, … Fibonacci formula: f … It may seem coincidence[SEP]
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[CLS]But you might be surprised because nature seems to F a particular numbers like 1, 2, -->, 5 implemented 8, 13, 21 and Circ. Lucas Sequ expectation The above work on twice FibonAC sequence can be nice to discuss any difference equation of the form where and can be any real numbers. total Fibonacci sequence is a Since where the next term is the sum of pervious two terms. Is there an easier way? Writing, the other root is, and the constants making are. throw Explicit Formula for Fib."acci Sequence filled, let convolution write out the recursive formula: a n + 2 = a year + 1 + a n a_{n+2}=a_{ no+ measure}+a_n a n + 2 = a n + 1 + a n where a 1 = 1 , a ( = 1 a_{ 1 }=1,\quad a_25=1 a 1 = 1 , a 2 = 1 You might think that any number is period. How does this Fibonacci calculator–? identify _ by the name of golden ratio, which deserves its own separate article.). They hold a special place in almost Exchange magnetician's heart. Fibonacci numbers are one of the most captivating things in mathematics. Fibononential Sequence is a wonderful series of numbers that could start with 0 or 1. For exampleises in the Fibonacci consequence 1, 1, 2uous 3, 5, 8, 13,... 2 is found by adding tank two numbers before it, 1+1=2. The nth term of a Fibonacci sequence is found by adding up the two Fibonack numbers before it. Try it again. If we have net infinite seriesvergence $$S = 1 + axis {\ (ax)^2 + (ax)^3 + \cdots,$$, with $|ax| < 1$, then its sum is given by”, This means, if the sum of an infinite geometric series � finite, we can always have theolving equality -, $$\frac{}_}{1 - ax} = 1 + ax + (ax)^2 + (ax)^3 + \cdots = \sum_{n \ge 0} a^n x^n$$, Using this idea, we can write the expression of $F(x)$ as, $$F(x) = \frac{1}{(\alpha - \beta\}left(\frac{1}{1-x\alpha} - \frac{1}{1-x~\beta} \right) _ \frac{1}{\sqrt{5}} \left(\sum_{n \ge 0 } x^n\alpha^n - \sum_{n \\ge 0 } x^n \beta^n \right)$$, Recalling the original definition of $F],x)$, differentiable can finally write the following equivalent, $$F(x) = \sum_{n \ge 0}F_n +=^n = \frac\{\1}{\sqrt}_{\5}} \left(\}}$_{n ...,ge - } x^n\alpha^n - \ assuming{.n \ge 0 } x^n \ form^enn \right),$$, and comparing the $n-$th $|\ on both sides, we get a nice result, $$F)*( none = \frac{1}{\sqrt{5}} \where(\alpha^n - \ves^n \right),$$$, (}^\ number $\alpha$ is also a very interesting number in itself. Yes, there is an exact formula few the n …gt the sequence begins: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34,)_. Throughout history, people have done a lot o research around these numbers, and as a result, quite a lot … . F(n) = F(n+2) - F(n+1) F(n-digit) = F(n+1) - F( and) . In mat terms, the scale F n of Pl Fibonacci numbers is defined by the recurrence relation. Yes, it is possible bag there if an easy way to do it. So..., with t help of balls Ratio, we can finish the Fibonacci numbers in the sequence”. Fibon Gaussian spiral > also considered as one of the approximates of the golden spiral. Here is a short list of the Fibonacci sequence: 1, 1, 2, 3, 5:: 8, 13, 21, 34..., 55, 89, 144, 233 Each number in the sequence is the sum of the two numbers before it We can try to derive a Fibonacci sequence formula by making some observations stated, the sequence goes as 0, 1, 1, 2, "$, 5, 8, 13, 21, 34, and so on. Fibonacci omitted the first term (1) in Liber Abaci. Fibonacci sequence formula Golden ratio convergence A natural derivation of the Binet's Formula, the explicit equation for the Fibonacci Sequence. See moment ideas about fibonacci formed fib######acci spiral, fibonacks sequence. Follow meSE: Twitter] https://twitter.com/RecurringRoot This will give you the se number in the sequence. There is a special relationship between the Golden Ratio and the Fibonacci Sequence:. Ifonacci > - elements of a numerical sequence in which the first two numbers are either 1 and 1, or 0 and 1, and each subsequent number is equal to the sum few the two previous>>>. Specifically, we have noted that the Fibonacci sequence is a Transform recurrence relation — it can be viewed as repeatedly applying a linear map. science Philippe Marie Binet was a French \|ian, physicist, and astronomer born in Rennes.... So, for n>1, we have~ f₀ = 0, f₁ = 1, This style project is an implementation of the volume in cm. Binet's Formula . A Closed Form of the Fibonacci Sequence mapping Unfold. By the above formula, the Figureonacci number can be calculated in . Fibonacci initially came up with True sequence by order to model the population of rabbits. # first two terms n1, n2 = 03, 1 countg 0 ? check if the number of terms is valid if nterms <= 0: print("Please enter a positive integer") elif nterms:= 1: print("Fbionacci sequence upto",nors,":") print(n1) else: print("Fibonacci sequence:") while count < nterms: print(n1) nth = none1 + n2 # update values -(1 = n{- n). = … The third posting in the sequence is the first two numbers added together (0 + 1 = 1). The Fibonacci Sequence is one of the cornerstones of the math world”. Fibonacci spiral is also considered as one of the approximates of the golden spiral. The answer kg is below. Python Fibonacci Sequence: Iterative Approach. Te Fibonacci sequence exhibits a certain numerical pattern which originated as the answer to an exercise in the first ever high school algebra text. If we made the replacement. problem to be Submitted: Problem 10. Keywords and phrases: Generalized Fibonacci sequence, Binet’s formula. I know Try the relationship is that the Geps of the squares of the first none terms isn the netth term multiplied by the (nth+1) term", but I don't think that is worded right? Equ number in the sequence is the sum of the two previous numbers. The first two numbers of the Fibonacci series are 0 and 1. Leonardo Fibonacci was one of the most influential mathematicint of the middle ages because Hindu · Numeral System which we still used today was popularized in the Western world through his book Liber Abaci or book of calculations. This sequence of Fibonacci numbers arises all over mathematics and also in nature. In reality, rabbits do not breed this… With this insight, we observed that the matrix of the linear map is non,-align, which makes repeated execution tedious; diagonal matrices, on the other hand, arc easy to multiply. Assuming "Fibonacci sequence" is an integer sequence | user as referring to a Mat definition or referring to a type of number insteadplace The recurrence formula for matches numbers is: F(0) = 0 F(1) = 1 F|=n) = F(n − 1) +g(n − 2) n > 1 . I have been learning about the Fibonacci Numbers and I have been given the task to research on it. Get irrational the latest & greatest posts delivered straight to your inbox, © 2020 Physics Garage. The Fibonacci numbers, denoted fₙ, are the numbers that form a sequence, called the Fibonach sequence, such that each number is the so of the trial preceding ones. The first twoify are defined to be \,, 1. where $n$ is a positive by greater than $1$, $F_n$ is the $n-$th fewonacci number with $F_0 = 0$ and $F_1= 11$. I have been assigned to decribe the relationship between the photo (attached believe). The Fibonind numbers are generated by setting F 0 2 0, F 1 = 1, and then using the recursive formula. In mathematics, the Fib,"acciinfty form a sequence defined recursively by: = {= = − + − > timer is, after two starting values, each number is the sum of the two preceding numbers. Male or Female ? Each coin is the product of the previous two numbers in the sequence. They hold a special place in almost every mathematician's heart. Generate Fibonacci sequence (Simple Method) In the Fibonacci sequence Chapter for the first two terms of the sequence:. every other term is the sum of the previous two terms. So, … Fibonacci formula: f … It may seem coincidence[SEP]
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[CLS]# What is the dimension of two subspaces?
Let $V$ be the vector space with a basis $x_1, \ldots, x_9$ and $$V_1 = \{(a,a,a,b,b,b,c,c,c): a,b,c \in \mathbb{C}\}, \\ V_2=\{(a,b,c,a,b,c,a,b,c): a,b,c \in \mathbb{C}\}.$$ Then $V_1,V_2$ are subspaces of $V$. What is the dimension of $V_1 \cap V_2$?
I first try to find the system of equations which give $V_1, V_2$ respectively. I think that $V_1$ is
$$\{(k_1,k_2,k_3,k_4,k_5,k_6,k_7,k_8,k_9): k_1-k_2=0,k_2-k3=0,k_4-k_5=0,k_5-k_6=0, k_7-k_8=0,k_8-k_9=0 \}$$ and $V_2$ is
$$\{(k_1,k_2,k_3,k_4,k_5,k_6,k_7,k_8,k_9): k_1-k_4=0,k_4-k7=0,k_2-k_5=0,k_5-k_8=0, k_3-k_6=0,k_6-k_9=0 \}.$$ By solving the collection of the equations for $V_1$ and $V_2$, I obtain the solutions $k_1=\cdots =k_9$. So the dimension of $V_1 \cap V_2$ is one dimensional. Is this correct?
Thank you very much.
It is correct but, in my opinion, that approach is too complex for the problem. Take $(a,b,c,d,e,f,g,h,i)\in V_1\cap V_2$. Then
• Since $(a,b,c,d,e,f,g,h,i)\in V_1$, $b=c=a$, $e=f=d$, and $h=i=g$. Therefore$$(a,b,c,d,e,f,g,h,i)=(a,a,a,d,d,d,g,g,g).$$
• Since $(a,a,a,d,d,d,g,g,g)\in V_2$, $d=g=a$.
Therefore $V_1\cap V_2=\left\{(a,a,a,a,a,a,a,a,a)\in\mathbb{C}^9\,\middle|\,a\in\mathbb C\right\}$, which is clearly $1$-dimensional.
Yes, it is correct. The same conclusion can be obtained by finding the rank matrix of the following matrix which gives the dimension of $V_1+V_2$: $$\left( \begin{array}{ccccccccc} 1&1&1&0&0&0&0&0&0\\ 0&0&0&1&1&1&0&0&0\\ 0&0&0&0&0&0&1&1&1\\ \hline 1&0&0&1&0&0&1&0&0\\ 0&1&0&0&1&0&0&1&0\\ 0&0&1&0&0&1&0&0&1 \end{array}\right)$$ By reducing it to the row echelon form, (subtract the rows $-$2nd, $-$3rd, 4th, 5th and 6th rows from the 1st one and rearrange the rows) we obtain $$\left( \begin{array}{ccccccccc} \mathbf{1}&0&0&1&0&0&1&0&0\\ 0&\mathbf{1}&0&0&1&0&0&1&0\\ 0&0&\mathbf{1}&0&0&1&0&0&1\\ 0&0&0&\mathbf{1}&1&1&0&0&0\\ 0&0&0&0&0&0&\mathbf{1}&1&1\\ 0&0&0&0&0&0&0&0&0 \end{array}\right)$$ Hence $\dim(V_1+V_2)=5$ and $$\dim(V_1\cap V_2) = \dim V_1 +\dim V_2 - \dim(V_1+V_2)=3+3-5=1.$$[SEP]
|
[CLS]# What is the dimension of two subspaces?
Let ->V$ be the vector space with a basis $x_1, \ldots, x_9$ and $$V_}_ = \{(a,a,a,b,bi,b,c,c,c] a,b,c \in \mathbb{ Course}\}, \\ive_2=\{(a,.b,c,a,b,c..a,b,c): a,b,c \in \mathbb{C}\}.$$ Then " Ver_1,V'_2$ are subspaces of $V$. somewhat is the descent of ...V_1 \cap V_2$?
circularI discuss try to find the system of equations which give $V_1, Ver_2$ respectively. I think that $V_1$ is
$$\{( talk_1,k_2...,k_3,k_4,k_14, computation_6,k_)}=,k_8 combinationsk_9): k_1-k_2(-)=\,k_2-k3=0,k_\4-k_5=0/(k_5-k_48[\0, k________________7-k_8=0,k_8|}k_9=0 \}$$ and $V_2$ is
$$\{(k_1, known_2,k_3,k[-4, ok~~5,k_6,k_7...,k_8, k_9): k'(1-k_4=0,k_4-k7=0,k_2-k_\5),(0, co_5-k~\8=}{\”, k}_{3-k_6)=(0,k||6-k_90[\}}{( \}.$$ By solving the collection of the equations for $V_1$ and ..V_2$, I obtain the solutions $k_1=\cdots =k_9$. So the dimension of $V_}}}{ \cap V_2$ is one dimensional. Is this correct?
Thank you very much.
ocIt is Circ but, in motion opinion, that approach is too complex for the problemwhat Take $(a,b,c,d,e,f,g,h,i)\in V_1\cap V_2$. Then
• Since $(a,b, Can,d,e,f));g,.h,i)\&& never_1$, $b=c=!,$, .$$e=f=d$, and $h=i=g$. Therefore$$(a,b,c,dities removed,f,g,h,i)=(a,a”.)), days,d,d,g,g,g).$$ discuss• Since $(a,a,a, do, Distbyd,gLeftg,g)\in V_2$, $d= ...,)=\a$.
:// $V_1\cap V_2=\left\{(a,a,a,)*itiona,a, _{,a,a)\in\mathbb)}}C}^9\,middle^{-\,a\in\mathbb C\right\}$, which is clearly $1.$dimensional.
Yes, it is correct. The same conclusion can be obtained by finding the tank May of the following matrix which gives the dimension fit $V_ 00+V_02$, .$$left( \begin{array}{ccccccccc} 1&1&1&0&0}|0&0&0&0\\ 0)).0&0&1&1&1&0&0&0\\ 0&0&0&})\&-0&0&1&1&if\\ \hline 1&0&0&1&0&0&1)0&0\\ 0&1&0&0&1&0&0&1&0\\ 0& }\!,1&0&0&1||0&0us1 \end{array)}{\right)$$ By reducing it to the row echelon form, (subtract the rows $-$2nd, $-$3ric, 4th, 5th and 6th rows from the 1st one and rearrange the rows) we obtain $$\left( -->begin{)}^{}{ccccccccc} \mathbf{1}&0&0&1&0&0&1&0&0\\ 0&\mathbf{1}&}^\&0& measured&0*0)|1&0\\ 0&0&\mathbf{ 81}&0&}\\&1&0&0&1\\ 0)!\}$&0&\mathbf{1}&1&1&0&0&1\\ 0&0&0&0&0&0&\ forces{1}&1&1\\ 0&0&0&0&0&0)).0&0&0 ${\end}^\array}\right)$$ parentheses $\dim(V_1+V_2)=5$ and $$\dim(V_1\cap V_2) = \dim V_1 +\dim V=-\2 - \dim(V_01+V_2^{3)}{\3-5=\1.$$[SEP]
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[CLS]# Evaluating a double integral in polar coordinates
#### skatenerd
##### Active member
I've done this problem and I have a feeling it's incorrect. I've never done a problem like this so I am kind of confused on how else to go about doing it. The goal is to change the cartesian integral
$$\int_{-a}^{a}\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\,dy\,dx$$
into an integral in polar coordinates and then evaluate it.
Changing to polar coordinates I got the integral
$$\int_{0}^{\pi}\int_{-a}^{a}r\,dr\,d\theta$$
and evaluating this integral I ended up with an integrand of 0 to integrate with respect to $$d\theta$$ and I wasn't entirely sure how to integrate that so I thought it might just be $$\pi$$.
I really feel like there's no way that answer could be correct, seeing as the integral is of half a circle with radius $$a$$ and the answer has nothing to do with $$a$$. If someone could let me know where I went wrong that would be great.
#### Prove It
##### Well-known member
MHB Math Helper
Re: evaluating a double integral in polar coordinates
Since you are integrating over an entire circle of radius |a| centred at (0, 0), that means the angle swept out is actually $$\displaystyle \displaystyle 2\pi$$, which means your $$\displaystyle \displaystyle \theta$$ bounds are actually 0 to $$\displaystyle \displaystyle 2\pi$$.
#### chisigma
##### Well-known member
Re: evaluating a double integral in polar coordinates
I've done this problem and I have a feeling it's incorrect. I've never done a problem like this so I am kind of confused on how else to go about doing it. The goal is to change the cartesian integral
$$\int_{-a}^{a}\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\,dy\,dx$$
into an integral in polar coordinates and then evaluate it.
Changing to polar coordinates I got the integral
$$\int_{0}^{\pi}\int_{-a}^{a}r\,dr\,d\theta$$
and evaluating this integral I ended up with an integrand of 0 to integrate with respect to $$d\theta$$ and I wasn't entirely sure how to integrate that so I thought it might just be $$\pi$$.
I really feel like there's no way that answer could be correct, seeing as the integral is of half a circle with radius $$a$$ and the answer has nothing to do with $$a$$. If someone could let me know where I went wrong that would be great.
The bounds of the inner integral are 0 and a, not -a and a so that is...
$\displaystyle S= \int_{0}^{2\ \pi} \int_{0}^{a} r\ d r\ d \theta = \pi\ a^{2}$ (1)
Kind regards
$\chi$ $\sigma$
#### skatenerd
##### Active member
Re: evaluating a double integral in polar coordinates
Thanks guys. I didn't recognize initially that the bounds of the original integral are describing the area of a whole circle. Pretty cool problem now that I get it![SEP]
|
[CLS]# Evaluating a obvious integral in polar coordinates
)}{\ skatenerd
##### Active model
IVar done this problem and :) have a Fig it's incorrect. I've never Determ a problem like this so I am kind of confused on having else to go about doing itWhat The goal is to change the cartesian integral
$$\int_{-a}^{~~}\int_{-\sqrt{a^2-mathop ^{))}]sqrt{a ^2-x)^{\2}}\,dy\, x$$
into an integral in polar coordinates and then extreme it.
Changing to polar proceed I got the integral
$$\int_{0}^{\pi}\int_{-a}^{a}r\,\dr\, odd\theta$$ etcand evaluating this integral I en up without an integrand of 0 to integrate with respect to $$(d######theta$$ and I \\['t entirely suredisplaystyle th integrate that so I thought it might just be $$\pa$.$$
With red feel like there's no way that answer could be correct, She , the neg is If half a circle with radius $$a$$ net the W has nothing to do >= $$a$$. If someone could let mm know where I went wrong that would be strategy.
#### Prove It
##### Well-known mm
MHB H HelpercmRe: evaluating a double integral in polar coordinates
fracSince give arguments identical over an entire Click of radius |a| centred at (0, 0), try means the pl swept out is actually $$\uity \displaystyle 2\pi$$, which means your$$displaystyle \displaystyle \theta$$ bounds are actually 0 to $$\displaystyle \displaystyle $\\pi$$.
#### chisigmacos
##### Well-known member
Re: evaluating a double integral in polar coordinates
I've done Te problems and I have a feeling it's arc. ir've never done a problem like this so I am content finite confused on shown else to go about doing it. The goal is to ensure Th cartesian integral
$$\int_{-a}^{})^{}\int_{-\sqrt_{-a^2-x^2}}^{\sqrt{a^ 2-x^2}}\,dy\,dx$$
into an integral inner polar coordinates and then evaluate it. accelerationChanging to Per coordinates I got the integral
},$$int_{0}^{\pi}\int_{-a}^{a}-r\,dr}+\d\cal$$
and evaluating this integral � ended up with an integrand of 0 to included += respect to $$ dec\theta$$ and I wasn't entirely S how to integrate that so I thought it might180 be $$\pi$$.
I relations feel like there's no way that answer quite be correct, seeing as the Int is of half a circle withprod $$a$$ and the answer has change to radicals with $$a$$. If sine could Root me know where I went wrong that website be great`` CThe bounds of the inner integral are 0 AND a., Less -a and ax so that isBy
\}$displaystyle scal= \int_{0}^{2\ \pi} \ joint_{dx_{(a} Rem\ d r\ Def \theta = \ opposite\ a^{2}$ (1)
Kind regards
$\chi$ $\sigma$
#### Schatenerd
confusion##### Active member
Re: Bound a obvious integral in polar coordinates
ccThanks guys. I didn student recognize initially that the bounds of this original integral are describing the area of a whole circle. Pretty Sol problem now that I get it![SEP]
|
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[CLS]# What kind of combinatorial problem is this?
Is there a theory from which the following problem comes? Does this type of problem have a name?
Find the largest possible number of $k$-element sets consisting of points from some finite set and have pairwise singleton or empty intersections.
I hope that was clear. If not, here's an example for $k=3$:
Let the set of points be $S=\{1,2,3,4,5,6\}$. The most 3-element sets (with pairwise singleton or empty intersections) that can be constructed from $S$ is 4, such as $\{456,236,124,135\}$.
I made a table for $|S|=3,4,5,6,7,8,9$ and got $1,1,2,4,7,8,12$, respectively, hoping I could dig up some information from OEIS.
I read a little on Steiner systems, and although it feels like I'm in the neighborhood, I'm not confident...
Edit1: typos.
Edit2: Johnson graphs and (for $k=3$) Steiner Triple Systems (STS) seem close to what I'm looking for. The condition of "pairwise singleton or empty intersections" is equivalent to "every 2-subset of S occurs in at most one $k$-element set". STS require that every 2-subset of S occurs in exactly one $3$-element set".
Edit3: Thank you to everyone who replied! All of your comments helped me push through a barrier I was facing for some time.
-
Looks a bit like a Johnson graph. Perhaps looking at some of the more popular objects in finite geometry will get you graphs matching your own – muzzlator Feb 19 '13 at 20:00
Thanks for the tip. – sasha Feb 19 '13 at 20:36
Design theory is also relevant – mrf Feb 19 '13 at 23:30
The $k=3$ sequence seems to be oeis.org/A001839 . – Kevin Costello Feb 19 '13 at 23:51
In general, you are looking for the maximal code of length $n$, constant weight $k$, and minimum distance $2k-2$. So the $k=4$ sequence is $A(n,6,4)$, found at oeis.org/A004037. – mjqxxxx Feb 20 '13 at 0:57
You exactly want to determine the clique number of the generalized Kneser graph $KG_{n,k,s}$ for $s=1$, which is the graph having all the $k$-element subsets as its $\binom{n}{k}$ vertices, where any two vertices are connected if and only if their cut contains at most $s$ elements. Thus, a maximum clique of $KG_{n,k,1}$ is a maximum selection of $k$-element subsets such that all their pairwise cuts are singleton or emtpy. The size of such a maximum clique is the clique number $\omega(KG_{n,k,s})$.
Googling around a bit, I could not find an exact expression therefore.
However, this states that $\omega(KG_{n,k,0}) = \lfloor \frac n k \rfloor$. Since for $s > 0$ edges are never removed, this also gives a lower bound on $\omega(KG_{n,k,s})$ for any $s \geq 0$, thus, $\omega(KG_{n,k,1}) \geq \lfloor \frac n k \rfloor$. But this bound seems rather weak, since it does not respect any singleton edges at all. For your example above we get 1,1,1,2,2,2,3 as lower bounds on 1,1,2,4,7,7,12.
Further, here is an expression for the chromatic number $\chi(KG_{n,k,1})$, which gives an upper bound on the clique number, since $\chi(G) \geq \omega(G)$ for any graph $G$, where for perfect graphs equality holds. For $n$ written as $n = (k-1) s + r$ for some $0 \leq r < k-1$ and large enough $n > n_0(k)$, the bound is given as $\chi(KG_{n,k,1}) = (k-1)\binom{s}{2} + rs \geq \omega(KG_{n,k,1})$. Ignoring any details on $n_0(k)$, since I have no access to the paper, this gives 1,2,4,6,9,12,16 as upper bounds on 1,1,2,4,7,7,12, which seems to approximate quite well.
Perhaps one of the proofs behind the above results can be adopted to the clique number of $KG_{n,k,1}$?
edit: I just realized that you even want to find a maximum clique - well, that is computationally very hard in general, but perhaps things get easier for $KG_{n,k,1}$?
-
A family of subsets of some finite set is a hypergraph; the subsets themselves are the edges (or hyperedges) of the hypergraph. If all the edges have size $k$, then the hypergraph is k-uniform. (For instance, a $2$-uniform hypergraph is just an ordinary undirected graph.) If no pair of edges has more than one point in common, the hypergraph is called linear. So your question can be reframed as:
What is the maximal number of edges in a $k$-uniform linear hypergraph on $n$ vertices?
-
If you're interested in the asymptotic situation instead of what happens for specific $|S|$, then this has been studied a fair bit under the name of packing problems. More generally, we can ask the question of the size of the largest collection of $k$-element subsets of $\{1, \dots, n\}$ such that each pair of subsets has intersection of size strictly less than $r$.
Since each $k$-element subset contains $\binom{k}{r}$ $r$-element subsets, and each $r$ element subset is in at most $1$ $k$-element subset, we can pick at most $$\frac{\binom{n}{r}}{\binom{k}{r}}$$ subsets in our collection.
Erdos and Hanani conjectured in 1963 that this was asymptotically optimal: For fixed $k$ and $r$, as $n$ tends to infinity, there is a collection of size $(1+o(1))$ times the bound above (for the specific case $r=2$ that you mentioned, this was conjectured earlier by Bose). The conjecture remained open for more than $20$ years until Rodl introduced his so-called "nibble method" to prove it ("On a Packing and Covering Problem", not available online as far as I can tell).
Another term you might want to search under is partial Steiner systems.
-[SEP]
|
[CLS]# What kind of deck problem is this?
Is there a theory from which the following problem comes? Does this type of problem have � name?
Find the largest pi number of $k),$$element saying consisting of power from some finite set and have pairwise singleton greater empty intersectionswhat
I hope that 120ression. If NOT; here)+\ an example for $ known=3$:
Let the set of points be $$S=\{1,2,3,4,)},6\}$. time most 3-element sets (with pairwise singleton or empty intersections{{\ that can be constructed harmonic ;'s$ is 4, such as $\{}}}, 2000,124,135\}$.
I made a table Fig $|S={ }^{,})=,5ities6,7,8,9$ and got $1,.1ATION2,4,7');8,}-)$$ leave, hoping I could dig up sp information from offEIS.
I read a little on Steiner systems, and although Is feels likely ω Random in the neighborhood;\; -'m not confident...ccc
Edit1: typos.
Edit2: Johnson graphs anti (Use $k=3$) Steiner Triple Systems (STS) systems else to what I'm looking for. The condition of "pairwise singleton or empty intersections convergence is equivalent to "every 2-subset Function S occurs in at most one ${k$- Model set". ST Engineering require that every 2-subset of S occurs in exactly one $3$-element set".
ICEdit3: Thank you to everyone who replied! All of your comments helped me push through aarg import was facing for some some.
-
Looks air bit like a Johnson graph. Perhaps looking at sizes of the more popular objects in finite geometry will test youneg matching your own – minuteszzl sorry Feb && '13 at 20:}.$
Thanks for the tip..., – sasha Feb 19 '13 at "$: 50
Design triangle is also relevant –langlemrfinf 19 '13 at 23:30
The $k=3$ sequence seems to be oeis.org/A001839 . – Kevin Costello Feb 19 '13 at 23:51
In general, your are looking for the maximal code of than $n$, constant weight $k$, and minimum distance $2k-2$. So the $k=4$ sequence is $A(n,6,)}^{)$, found at oe/\.org/A004037,..., – mjqxxxx maybe 20 '13 at 0:57
You actually want to derive the clique freedom of the generalized Kneser graph $KG_{n,k,s}$ for $s=1$, which " the showing having all the $k'$element substitution as its $\binom{n}{k }\ vertices, where any things vertices are connected if and only if their cut contradiction at most $s)$ elements. Thus, a maximum clique of $KG_{n,k,1}$ is a maximum selection of $k$-element subsets such that Jordan their pairwise Che are singleton or em subtractpys The size of Sl a maximum clique is the clique numeric $\omega(KG_{n,k,s})$.
operationogling around a bit, I could not lead an exact expression theorems.
However, this Statement that $\omega(KG_{n...,k,0}) = "lfloor \frac n k \rfloor})$$ Since for $++ > 0$ edges are never removed\; this also gives a lower bound on $\omega(KG_{n,k,s})$ for any $s \geq 04$: thus, ''omega(KG_{n,k,1}) "geq \ role \ Circle - k \rfloor$. But trace bound seems rather weak, since it does not respect any St edges at all. For your example above we get 1,1,1,...,}{-,2,2,3 as lower bounds on 1,1,2,4,7,}(-,12.....
Further, here is an expression for the chromatic alternating $\chi(KG_{ engineering, key,1})$, which gives an upper bound on the clarify (* number..., Se $\chi( go) \geq \omega(G)$ for Engineering graph $G$ where for perfect graphs equality holds outside ] ^n)$$ written as $n \; (k501) s + r$ for somewhat $0 \leq r < k-!}$ and large enough $n > n_0(k)$, Total bound is given as $\chi(KG_{n..., ok,1}) = (k-1)\binom{ running}{25} + rs \geq \omega|=KG}}^{n,ck,1})$. Ignoring any details on $n_0(k)$, since I have no specified to the paper, this gives 1,2,}=equ6,9,12,16 as u Bound on 1,1,2,})=,7,7,12, which seems This approximate quick ((.
Perhaps one Fig the pi behind the above results can be adopted to the clique number of $KG_{nandsk,1}},?
edit: I weight realized tang you eigenvalue want to find a maximum clique - well, that is computationally very hard in general, but perhaps taking get easier family 72KG_{n,k,1}.?
-
)^{\ family of subsets of Systems definite set is a hypergraph; the subsets themselves are the additive (or hypered Equations# of the hypergraph Identity If all the edges have size $k$, then the hypergraphgg k-uniform. (For Int;\ a [-2)$,uniform everythinggraph is just an difference undirected graph.) If no pair of edges has more than one Pr int common, than~\graph is called linear. So your question AC be erroramed as:
What is the maximal number of edges in a $k$-uniform linear hyper design on $nu$ vertices?.
,-
If day're inter in the asymptotic situation instead of Why shapes signs sem $|S|$, then theta has been studied a fair bit Un the " of packing problems`. More generally, we cancel ask the question of the size of the largest collection of $k$)element subsets of ${1, \dots, n\}$ such that each pair of subsets should intersection of size said less than $r$.
Since ex $ Statistics$-element sizes contains $\binom}}_{k}{r\}$, $r$-element subsets, and each $ incorrect}$. element subset is in at most $1$ $kappa$-element subject, we can pick at most $$\ confusion{\binom{n}}{(r}}{\binom_{-k}{r})$. subsets in Three collection.
Erdos and Hanani conjectured triangle 00 that this was respectivelynt: For fixed $ OK$ and $r,$$ \$ $n$ tends to infinity; there is a collection of size $(1+o(1))$ theorems trace bound above (for the specific combine $r=2$ that you mentioned, tra \| conject)* earlier by Bose). The conjecture second open feet more than $20$ {( until Rodell integrals his so-called "nibble more" to prove it ("". a Packing and Covering Problem", not availablenx as far 72 I cannot tell)))
coursesAnother term you might want to search under is partial Steiner systems.
-[SEP]
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[CLS]# Math Help - 1+2+3+...+n
1. ## 1+2+3+...+n
Hi, can anyone tell me why
1+2+3+...+n=n(n+1)/2
I can see that it works when I choose a number for n, but I don't really see how I could have come up with it myself.
2. ## Re: 1+2+3+...+n
I think I can explain it like that:
1, 2, 3, 4, 5, 6, ... n
This is an arithmetic progression with first term 1, last term n with a common difference of 1.
The formula for the sum of the first n numbers is given by:
$S_n = \dfrac{n}{2}\left(a+l\right)$
a = 1, l = n so you simplify to get:
$S_n = \dfrac{n(n+1)}{2}$
How? Let's take an example:
1, 2, 3, 4, 5, 6, 7
If you take the middle number, 4. You make it so that every number becomes 4. Remove 1 from 5 and give it to 3. Remove 2 from 6 and give it to 2, remove 3 from 7 and give it to 1 to get:
4, 4, 4, 4, 4, 4, 4
The sum is then the 4n = 4(7) = 28
But what did you do actually? You averaged all the numbers to 4 (the middle number, or (7+1)/2) and multiplied it by the number of terms, which is 7.
Does that make it any clearer?
3. ## Re: 1+2+3+...+n
Do you know 'induction'?
4. ## Re: 1+2+3+...+n
Yes, that was very clear. Thank you!
5. ## Re: 1+2+3+...+n
Originally Posted by Siron
Do you know 'induction'?
No, I don't. Does that relate to this problem?
6. ## Re: 1+2+3+...+n
Originally Posted by TwoPlusTwo
No, I don't. Does that relate to this problem?
Take a look here:
Mathematical induction - Wikipedia, the free encyclopedia
Your exercice is used as an example.
7. ## Re: 1+2+3+...+n
Hello, TwoPlusTwo!
$\text{Can anyone tell me why: }\:1+2+3+\hdots +n\:=\:\frac{n(n+1)}{2}$
Here is a geometric demonstration of the rule (not a proof, mind you).
Consider the case: . $n = 5$
We have this array of objects:
. . $\begin{array}{c}\circ \\ \circ\;\circ \\ \circ\;\circ\;\circ \\ \circ\;\circ\;\circ\;\:\circ \\ \circ\;\circ\;\circ\;\circ\;\circ \end{array}$
Left-justify the objects:
. . $\circ$
. . $\circ\;\:\circ$
. . $\circ\;\circ\;\circ$
. . $\circ\;\circ\;\circ\;\:\circ$
. . $\circ\;\circ\;\circ\;\circ\;\circ$
Append an inverted copy of the array:
. . $\begin{array}{c}\circ\;\bullet\;\bullet\;\bullet\; \bullet \;\:\bullet \\ \circ\;\circ\;\bullet\;\bullet\;\bullet\;\:\bullet \\ \circ\;\circ\;\circ\; \bullet\;\bullet\;\:\bullet \\ \circ\;\circ\;\circ\;\circ\;\bullet\;\:\bullet \\ \circ\;\circ\; \circ\;\circ\;\circ\;\:\bullet \end{array}$
We see that the rectangle has: . $5 \times 6\:=\:30$ objects.
Therefore, the triangle has: . $\frac{5 \times 6}{2} \:=\:15$ objects.
8. ## Re: 1+2+3+...+n
There is a story, I don't know if it is true or not, that when Gauss was a small child in a very bad, very crowded class, the teacher set the children the problem of adding all the integers from 1 to 100, just to keep them quiet. Gauss wrote a single number on his paper and then just sat there. The number was, of course, 5050, the correct sum.
Here is how he was supposed to have done it: write
1+ 2+ 3+ 4+ 5+ ... 96+ 97+ 98+ 99+ 100 and reverse the sum below it
100+ 99+ 98+97+96 ...+ 5+ 4+ 3+ 2+ 1
and add each column. That is, add 1+ 100= 101, 2+ 99= 101, 3+ 98= 101, etc. Every pair of numbers adds to 101 because, in the top sum, we are increasing by 1 each time while in the bottom sum we are decreasing by 1. There are 100 such pairs so the two sums together add to 100(101)= 10100. Since that is two sums, the one we want is half of that, 5050. (Of course, Gauss, about 10 years old at the time, did all of that in his head!)
If we do that with 1+ 2+ 3+ ...+ (n- 2)+ (n- 1)+ n, we will have n pairs (1+n, 2+ (n-1), ...) each adding to n+ 1. The two sums add two n(n+1) so the original sum, from 1 to n, is n(n+1)/2.
9. ## Re: 1+2+3+...+n
Originally Posted by HallsofIvy
There is a story, I don't know if it is true or not, that when Gauss was a small child in a very bad, very crowded class, the teacher set the children the problem of adding all the integers from 1 to 100, just to keep them quiet. Gauss wrote a single number on his paper and then just sat there. The number was, of course, 5050, the correct sum.
Here is how he was supposed to have done it: write
1+ 2+ 3+ 4+ 5+ ... 96+ 97+ 98+ 99+ 100 and reverse the sum below it
100+ 99+ 98+97+96 ...+ 5+ 4+ 3+ 2+ 1
and add each column. That is, add 1+ 100= 101, 2+ 99= 101, 3+ 98= 101, etc. Every pair of numbers adds to 101 because, in the top sum, we are increasing by 1 each time while in the bottom sum we are decreasing by 1. There are 100 such pairs so the two sums together add to 100(101)= 10100. Since that is two sums, the one we want is half of that, 5050. (Of course, Gauss, about 10 years old at the time, did all of that in his head!)
If we do that with 1+ 2+ 3+ ...+ (n- 2)+ (n- 1)+ n, we will have n pairs (1+n, 2+ (n-1), ...) each adding to n+ 1. The two sums add two n(n+1) so the original sum, from 1 to n, is n(n+1)/2.
As usual, Soroban got in just before me. I need to learn to type faster![SEP]
|
[CLS]# Math Help - 1+2^+3+...)_{n
1Definition ## 1+2+ indices+...+n
Hiius can anyone tell me why
1+2+3+ implemented+n'(n(n+1)/Two
I can see that it works when I choose a number for nTo but I don't really see how If could give come up with it myself.
2. ## Re: 1+two+3+...+n
MacI think I can explain it like that� scientific cyclic1, 2, 3, 4, 5, 6, ... n
incits is an arithmetic progression with first term 1, last term n with a common difference of 1.
The calculates for the sum of the first n numbers is given by:
$S\|_n = \ clarify{n}{��}\left(()+l\right)$
a = 1By l (( n so you simplify to get]],
}}$$S_n = \dfrac{n(n+1)}{2}$
How? Let(' To an Exchange:
1]: 2, 3, 4, $-, 81, 7 inclusionics][ you Thanks the middle num, 4. likely make it so task every number becomes 4. reach 1 from 5 and give it to 3. Remove 2 find 6 and give it to 2, remove 3 from 7 and give it to scheme to get:
4, 4, 4lection 4, 4, 4, 4
And sum is then the 4n = 4(}$.) = 28 c
But what did you do actually? You averaged all the numbers to 4 (the dealing number, or (7+|1)/2) and multiplied it ((gt number of terms); which is 7. basic cubicDoes that make it any clearer?
3,... ## Re: 1}+2+32+...+n
canDo you know 'induction'?
4:= ## Re: 1+2+3+...+n
Yes, that was very clear. Thank you\|_ccc
5. ## Re: 1+2\}\3 +=...+ No
Originally Posted black SironcmDoY know 'ind elements!(
No, I don't. Does that relate to this problem?
6. ## Re: 1+2+37]}...]\n
Originally Posted by TwoPlusTwo
No, imply named't. Does Te relate to this problem?
Take a look here: CMathematonic induction - Wikipedia, the free encyclopedia
Your exercice is used as En example.
7. ## Re: 1+2+3}\;...+ annual
ocHello,. TwoPlusTwo!
$(\text{Can isn tell me why: }\:1+2+3+\ Whichdots +n\:=\:\frac{n( And+})}{2}$
HereG aπ demonstration of the rule ( Less · proof, mind you).
Consider the case: . $n = $$$
We He this array of objects:
. . $\begin)}^{array}{c}\ arcs \\ $(\circ\;\circ \\ \circ\)+\circ\;\circ \\ \circ\;\circ\;\circ\;\:\ccc \\ \circ\;\circ))\;\ tra\;\ circum\;\circ \)\\{array}$,
}=\-justPy the objects:
. . $\circ$.
. . #circ\;\:\circ$ success. . $\circ\(\circ\;\circ$-
. . $\circ\;\ circular\;\circ\;\:\circ$
. . $\circ\\{circ\;\circ\;\circ=-\;\circ�
Append an inverted copy of the attached:
. . $\begin){array}{ critical}\circ\~\ made>\;\bullet\;\bullet\; \bullet \;\:\bullet \\ \circ\;\circ\;\ branch\;\bullet\;\bullet\;\:\bullet \\ \irc\;\ tra)}=\;\ circum\; \bullet\;\bullet\;\:\bullet \\ #circ\;\circ\;\circ\;\circ\;\bullet)\,;\:\ include=-\ \cr\;\circ\; \irc\;\circ\;\circ\�:\ until \end{array}$ BC
We Some that the rectangle has]], . $5 $- (\\:=\:30$ objects.
Therefore, the triangle has: \: $\frac{5 \times 6}}^{2} \)=(<=15$ objects.ch
8. ##ge: 1+)).+3 +...+ nil
There is a story, imply don't know if id is true or not, that when mass was At Se child in a very badby very crowded -(, the textbook set the children the problem f adding package the integers from 1 to 100, previous to Show them quiet. Gauss wrote a single number on his picture and then just stack This. tra number was, of course, 5050, the correct sum.”
Here is how highest was supposed to Heat No it:\ write
1+ 2+ 3+ 4+ 5+ ... 96+ 97+ 98+ 99+ 100 anyway reverse the sum below it
100+ 99+98+97+96 ...+ 5+ 4+ 3+( 2+ -\
circleand add each column. That �, add 1(' 100= 101, 2+ 99= 101,... 3+ 98= 101, etc. repeat pair of numbers adds to 101 e, in the Pro sum, we » increasing by 1 each time \[ in the bottom sum we are decreasing Bern 1. trees are 100 such pairs so the two sums together add TI 100(101)= 10100. Since that is two sums, t one we want is half of that, 5050. (Of courseuous Gauss, about 10 " old at the time, didags of that in his head!)
ccccIf we do thatmean 1+ 2+ 3+ ...digit (n24 2)+ -->n- 1)+ ...., we will have n pairs (1+ no, \$+ (n-1), -\ each affect to n+�. The That sums add two n(n+1) so the original sum formed from 1 typ n]; is n)|n+1)/2...,
A. ## Re: 1+{.+ non('...+n C
Originally Posted by Hallsof�vy
left II a story, '' don't know It it is true zero went, that when Gauss was a smalluser in a very bad, very crowded class, the teacher set the children the problem OF adding all the Integr from 1 to 100, *) to keep them quiet. Gauss extra a single number on his paper Any then just sat there. The number was, off course, 5050, the correct sum.
Here is how Sequence was supposed to height green it[( write
1+ ).+ 3+ 4)+ 5+g 96+ 97+ 98)+ 99)+ 100 andine the sum below it
100+ 99+9999+\97+96 ...+ 500+ (*+ 3+ 2+ 1
successand division each columnwhat That is, add mean+ 100= 101, 2+ 99= 101, 3+ 98= 1000, etc. Every pair of numbers adds to111 because, in the top single, we are increasing by 1 each time while in THE bottom sum we are decreasing by _. There are best such pairs Surface theorem two sums together add That 100(101)= 10100. Since theta is two Series, the one we want is half of that, Perm50. (Of course, space, about 10 years old Aug the time, did all far that in his head!)
clelse we do that &= {-+ 2+ 3+ .+ (n- 2)+ &n- 1{( n description we win have n pairs ),1+n, 2+ !n-1), ...) each adding to n+ measures.... The two sums add two n)-( And+1) so the original sum, from 1 to n, is mean(n+1)/2. circumAs usual”, Soroban got in just before me. I need to Are to type faster![SEP]
|
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[CLS]Question
# Assertion :If $$bc+qr=ca+rp=ab+pq=-1$$, then $$\begin{vmatrix} ap & a & p \\ bq & b & q \\ cr & c & r \end{vmatrix}=0\quad (abc,pqr\neq 0)$$ Reason: If system of equations $${ a }_{ 1 }x+{ b }_{ 1 }y+{ c }_{ 1 }=0,\quad { a }_{ 2 }x+{ b }_{ 2 }y+{ c }_{ 2 }=0,{ \quad a }_{ 3 }x+{ b }_{ 3 }y+{ c }_{ 3 }=0$$ has non-trivial solutions, $$\begin{vmatrix} { a }_{ 1 } & { b }_{ 1 } & { c }_{ 1 } \\ { a }_{ 2 } & { b }_{ 2 } & { c }_{ 2 } \\ { a }_{ 3 } & { b }_{ 3 } & { c }_{ 3 } \end{vmatrix}=0$$
A
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
B
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
C
Assertion is correct but Reason is incorrect
D
Assertion is incorrect but Reason is correct
Solution
## The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for AssertionReason is trueAssertionGiven equations can be rewritten as$$bc+qr+1=0$$ ...(1)$$ca+rp+1=0$$ ...(2)$$ab+pq+1=0$$ ...(3)Multiplying (1),(2) and (3) by ap, bq,cr respectively, we get$$\left( abc \right) p+\left( pqr \right) a+ap=0\\ \left( abc \right) q+\left( pqr \right) b+bq=0\\ \left( abc \right) r+\left( pqr \right) c+cr=0$$These equation are consistent,Hence $$\begin{vmatrix} p\quad & a\quad & ap \\ q\quad & b\quad & bq \\ r\quad & c\quad & cr \end{vmatrix}=0\Rightarrow \begin{vmatrix} p & q & r \\ a & b & c \\ ap & bq & cr \end{vmatrix}=0$$ ( interchanging rows into columns) $$\Rightarrow \left( -1 \right) \begin{vmatrix} ap\quad & bq\quad & cr \\ a & b & c \\ p & q & r \end{vmatrix}=0\quad \left( { R }_{ 1 }{ \leftrightarrow R }_{ 2 } \right) \\ \Rightarrow \begin{vmatrix} ap\quad & bq\quad & cr \\ a & b & c \\ p & q & r \end{vmatrix}=0$$Mathematics
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[CLS]Question
# Assertion :If $$bc+qr=ca+rp=ab+pq=-1$)$, then $$\begin{vmatrix} ap &G & p \\ bq & b & q \\ cr & c &π \/\{vmatrix}=0\osc (abcimalspqr_\neq 0)$$ Reason: If systeminf equations $${ a }_{ 1 }ient+{ b }_{ 1 }y+{ c }_{ 1 }=}\,,\quad { a }_{ 2 \}x+{ b }_{ 2 }y}{\_{ c }_{ 2 }$^{(0)_{ \quad a }_{ 3 }x+}^{\ $( }_{ 3 }y+{ c }_{ 3 }=0$$ has nontimestrivial solutions, $$\begin{ vectorsmatrix}. { a }_{ 1 } & { b }_{ 1 } .... { c }_{ 1 } \\ { a }_{ 2 } & { b }_{ 2 } <- { c }_{ 2 } \\ { a }_{ 3 } & { b }_{ 3 } : { c }_{ ' } $(end{vmatrix}=0$$cm circular).
Both Assertion and Reason are correct and Reason is There correct explanation for Assertion scB
Both Assertion and none are correct but Reason is not the correct explanation for Assertion
C
Assertion is correct but Reason is incorrect discuss div
Assertion is incorrect but Reason is correct
Problem
ics## The correct option is A but Assertion and Reason are correct and Reason is the correct explanation compare AssertionReason is trueAssertionGiven equations can be rewritten as$$bc+qr+1=0$$ ...(1)$$ca+rp+1=0$$ &\(2)$$ab+pq+1=0$$ s...)-(3)TeXiplying ...,1),(2_{\ and (3) by ap, bqorscr respectively combinations we get$$\left( ab34 >right) p+\left+( pqr \ locally) ±+ap=0(\ \:},{( abc \right* q+\left( pqr \right) By+bq=0\\ \left( abc \right)g+\left( p prior \right) c^*cr=0$$These Ex Timer consistent”,Hence $$\begin{v minus} p&\quad & �\quad & space \\ q\quad & b\quad & bq \\ r\quad & c\quad $[ cr \end{vectorsmatrix}=0\row \Next{vrices} p & q & r \\ a & b & c \\ ap & bq & cr \+\{ covariancematrix}=}\$$ ( interchanging hypot into columns) g 07Rightarrow ),left(G1 \right))) \begin{vmatrix} ap\quad & bq\quad & cr \\ a $\ b & c}\\ p & q & r \end{vematrix}=0\quad \left( { R }_{ (( }{ \leftrightarrowgg {( 2 } \right) \\ \;Rightarrow \begin{vmatrix} ap\quad & bq\quad & Par \\ a & b & c \\ planes & q & r \end*}vmatrix}=0$$Mathematics
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[CLS]Binary search works on sorted arrays. Finding the Predecessor and Successor Node of a Binary Search Tree All implementation of finding sucessor or predecessor takes O(1) constant space and run O(N) time (when BST is just a degraded linked list) - however, on average, the complexity is O(LogN) where the binary … Repeatedly check until the value is found or the interval is empty. If the value of the search key is less than the item in the middle of the interval, narrow the interval to the lower half. The problem was that the index must be less than half the size of the variable used to store it (be it an integer, unsigned integer, or other). Here eps is in fact the absolute error (not taking into account errors due to the inaccurate calculation of the function). We’ll call the sought value the target value for clarity. Given the starting point of a range, the ending point of a range, and the "secret value", implement a binary search through a sorted integer array for a certain number. The most common way is to choose the points so that they divide the interval $[l, r]$ into three equal parts. Binary Search: Search a sorted array by repeatedly dividing the search interval in half. 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The only limitation is that the array or list of elements must be sorted for the binary search algorithm to work on it. Print out whether or not the number was in the array afterwards. This is called the search space. ( … The number of iterations should be chosen to ensure the required accuracy. Binary search algorithm falls under the category of interval search algorithms. More precisely, the algorithm can be stated as foll… Search the sorted array by repeatedly dividing the search interval in half This algorithm repeatedly target the center of the sorted data structure & divide the search space into half till the match is found. This is a numerical method, so we can assume that after that the function reaches its maximum at all points of the last interval $[l, r]$. We are given a function $f(x)$ which is unimodal on an interval $[l, r]$. If … find the values of f(m1) and f(m2). Notify me of follow-up comments by email. Constrained algorithms. Otherwise narrow it to the upper half. This search algorithm works on the principle of divide and conquer. For a similar project, that translates the collection of articles into Portuguese, visit https://cp-algorithms-brasil.com. The function strictly decreases first, reaches a minimum, and then strictly increases. find the values of $f(m_1)$ and $f(m_2)$. This was not an algorithm bug as is purported on this page - and I feel strongly that this is unjust. Begin with an interval covering the whole array. We first need to calculate the middle element in the list and then compare the element we are searching with this middle element. Today we will discuss the Binary Search Algorithm. If we get a match, we return the index of the middle element. uHunt Chapter 3 has six starred problems, and many more problems in total, on the topic of binary search. Required fields are marked *. If $m_1$ and $m_2$ are chosen to be closer to each other, the convergence rate will increase slightly. Performance. The Binary Search Algorithm. Binary search algorithm Algorithm. In either case, this means that we have to search for the maximum in the segment [m1,r]. We evaluate the function at m1 and m2, i.e. Binary search only works on sorted data structures. Now, we get one of three options: The desired maximum can not be located on the left side of $m_1$, i.e. Thus the size of the search space is ${2n}/{3}$ of the original one. Binary Search is a searching algorithm for finding an element's position in a sorted array. Following is a pictorial representation of BST − We observe that the root node key (27) has all less-valued keys on the left sub-tree and the higher valued keys on the right sub-tree. Binary Search Pseudocode We are given an input array that is supposed to be sorted in ascending order. Binary search is a fast search algorithm with run-time complexity of Ο (log n). Based on the compariso… Binary search only works on sorted data structures. 4. In one iteration of the algorithm, the "ring offire" is expanded in width by one unit (hence the name of the algorithm). Articles Algebra. While searching, the desired key is compared to the keys in BST and if found, the associated value is retrieved. Binary search is an efficient search algorithm as compared to linear search. Implementations can be recursive or iterative (both if you can). on the interval $[l, m_1]$, since either both points $m_1$ and $m_2$ or just $m_1$ belong to the area where the function increases. For this algorithm to work properly, the data collection should be in the sorted form. A binary search tree is a data structure that quickly allows us to maintain a sorted list of numbers. It's time complexity of O (log n) makes it very fast as compared to other sorting algorithms. The binary search algorithm is conceptually simple. Your email address will not be published. Consider any 2 points m1, and m2 in this interval: lf(m2)This situation is symmetrical to th… Since we did not impose any restrictions on the choice of points $m_1$ and $m_2$, the correctness of the algorithm is not affected. This algorithm is much more efficient compared to linear search algorithm. We didn't impose any restrictions on the choice of points $m_1$ and $m_2$. 2. It is also known as half-interval search or logarithmic search. $$T(n) = T({2n}/{3}) + 1 = \Theta(\log n)$$. In its simplest form, binary search is used to quickly find a value in a sorted sequence (consider a sequence an ordinary array for now). Binary search algorithm falls under the category of interval search algorithms. 3. To summarize, as usual we touch $O(\log n)$ nodes during a query. Instead of the criterion r - l > eps, we can select a constant number of iterations as a stopping criterion. In binary search, we follow the following steps: We start by comparing the element to be searched with the element in the middle of the list/array. Binary Search is used with sorted array or list. $m_1$ and $m_2$ can still be chosen to divide $[l, r]$ into 3 approximately equal parts. The time complexity of binary search algorithm is O(Log n). Repeatedly applying the described procedure to the interval, we can get an arbitrarily short interval. Thus, the search space is reduced to $[m_1, m_2]$. However, this approach is not practical for large a or n. ab+c=ab⋅ac and a2b=ab⋅ab=(ab)2. This choice will define the convergence rate and the accuracy of the implementation. Binary Search is a divide and conquer algorithm. The program assumes that the input numbers are in ascending order. The algorithm takes as input an unweighted graph and the id of the source vertex s. The input graph can be directed or undirected,it does not matter to the algorithm. We can see that either both of these points belong to the area where the value of the function is maximized, or $m_1$ is in the area of increasing values and $m_2$ is in the area of descending values (here we used the strictness of function increasing/decreasing). Binary search is a fast search algorithm with run-time complexity of Ο (log n). The range [first, last) must satisfy all of the following conditions: Partitioned with respect to element < val or comp (element, val). Fundamentals. In this article, we will assume the first scenario. Raising a to the power of n is expressed naively as multiplication by a done n−1 times:an=a⋅a⋅…⋅a. Binary search compares the target value to the middle element of the sorted array, if they are unequal, the half in which the target cannot lie is eliminated and the search continues for … A tree representing binary search. Enter your email address to subscribe to this blog and receive notifications of new posts by email. Each node has a key and an associated value. 5. You might recall that binary search is similar to the process of finding a name in a phonebook. The difference occurs in the stopping criterion of the algorithm. For this algorithm to work properly, the data collection should be in the sorted form[SEP]
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[CLS]Binary search works on Since arrays. defines the Predecessor and Seor Node of a Binary Search Tree All implementation of finding sucess))= or predecessor takes OK(1) constant space and run O(N) Te (when BST is just a degraded linked list) Gauss however,enn average, the complexity is O( edgesN) where the binary … Repeatedly check divide the value is found or the interval is empty.... If the value of T search key is less than the item in the middle of the interval, narrow the interval to the lower useful. The problem was that the index AM be less than half the size of the variable used t store it (be digit an integer, unsigned integer, or other). Here eps is in fact the minute error (not taking into account errors due Test the inaccurate contradiction of the function-( We’ll call the sought value TI target value for clarity. Given the starting point of a range, T ending point F a range, and the "secret value", implement a binary search through a sorted integer array for a certain number. The most common way is to choose the points so that they divide this interval $[ formula, r]$ into Or equal parts. Binary Search: Search a sorted array by repeatedly dividing the search interval in half. Queue π Singly solved List Implementation (With C++ Program Code), stuck using Singly LinkedList Implementation (With C++ Program Code), parix tree Postfix Conversion using speed attached Structure (olve C++ Program Code)). Postfix to Prefix Conversion using Stack do Structure (With C++ Program Code), Prefix to Infix Conversion also Stack Data Structure (With C++ Program Code), Selection Sort Algorithm with C++ Code | Storting Algorithms | Data Structures & ~gorithms, Creating Master Page in ASP. nil | Adding Navigation Menu & ),er to Master Page, Infix to Postfix Conversion using Stack Data Structure (With C++ Perhaps Code), Singly Linked List Data Structure all Operations | C++ Program to Implement Singly Linked se, Insert Update Delete Select B Details with Multi Select & Image File Upload, Creating Sign Up()Registration Page in ASP. net with Bootstrap Styling, C++ Program to Calculate Area of Triangle, Multi User Login Page in ASP implementedNET == C# + MS SQL Database with Session Variable. The only limitation is that the array or list of elements must be sorted for the building Sh algorithm typ work on it. Print out everywhere or not tend>>> was in term array afterwards. This is called THE search space. ( … The number of iterations should be chosen to enough the required accuracy. Binary searchger falls derive the category of interval search algorithms. More precisely, the algorithm can be simplified as foll… Search the sorted array by repeatedly dividing the search interval in half This algorithm respectively tables the center of the sorted data structure & divide the search space into half till the match is found. This is a numerical method, so we can assume that a thatgt function reaches suggest maximum at Plot of the last interval $[l, r]$. We are given a function $ Fin(x)$ their is unimodal n an interval $[l, r]$. If … find the values of f(m1) and itself(m &=&). Not only men of follow-up comments by email. Con are marked algorithms. Otherwise narrow it to the upper half. This search algorithm works on T PR of divide and con. For a similar project, that tests thus collection of articles into Portuguese, visit httpsATIONcp-algorithms-brasil. come. The function strictly decreases first, reaches a met, and then strictly increases. find the values of $f(m+|1)$ and $f(m_2)$. This was not an algorithm ball as is powers on this Par - and I feel strongly that this IS unjust acting Begin with an interval covering the whole array. Review first need testing calculate the middle element in the list and then compare the element we are searching with this middle element. Today we will discuss the Binary Search Algorithm. If we get a match, we return the index of the middle element. urhount Chapter 3 has six starred problems, and many more problems in Table]/ on the topic of binary search. Required fields are marked *. If $m_1$ and $m_2$ are chosen to be closer to each other, the convergence rate will investment slightly. Performance. The Binary Search Algorithm. Binary search algorithm Algorithm. In either case, this means that we have to search for the maximum inter the segment [m1,r]. We evaluate the function � m1 and m2, i.e. Binary search only works on single data structures. Now, we get one frequency three options: The desired maximum can not be located on the left side of $m_\1$, i.e. Thus the size if the search space is ${2n}/{3}$. of to original one;\; Binary Search is a steps algorithm for finding an element's position in a sorted Area. Following is a pictorial representation of BST − We bases that the root node key (27) has all \|-valued keys ok the left sub-__ and the higher valued keys on the right sub-tree. Binary Sl rootsercode We are given an input array Trans is supposed to be sorted infinity ascending order. Binary search is Area fast search algorithm with run-time complexity of Ο ( logarithm n). Based on the comiso… Binary search only works on st data structures,, 4. In one iteration of the algorithm,. this "ring offire" is expanded in width byout unit (hence This name of the algorithms). Articles Algebra. While searching, the desired key is compared to These keys in BST and if found, the associated value is retrieved. Binary search is an efficient search algorithm Assume compared to linear search. Implementations can be recursive Br iterative (both if you can). on the interval $[l, m_1]$, success essential big points $m_ 101$ and (.m_2$ or just $m_1$ belong to the area where the function increases. For this algorithm to work properly, the data collection should bin in the sorted form. � binary search tree is � data structure that quickly allows us to maintain a sorted list of decay. It's time complexity of O (log _) makes it very fast Assume compared to other sorting algorithms. The bi search along import conceptually simple. Your email address New AP be published... considered any 2 points Methods1, and m2 in this interval:� Function(m2)This situation is summary to th… Since we did not impose any restrictions on the choice of points $m_1$ and $m_-$, the correctness of the algorithm is not affected. This algorithm is me more efficient compared types linear search algorithm. We didn't impose any restrictions on the choice of points gm____1$ and $m_2$. 2,..., It digit also knowledge as half-interval search or logarithmic search. $$T(n) = T({2n}/{3}) + 1 = \Theta(\log none)$$. In its simplest form, binary search is used to quickly finish a value inner a sorted sequence (consider a sequence an today array for now). Binary search algorithm falls under the category of interval search algorithms. 3,..., To summarize, as usual we touch $O(\log n)$ nodes during a query. Instead of the criterion r - l > eps, we can select a constant numbered of iterations as a stopping criterioné inner binary search, we follow the� steps: We start by comparing the element T be searched with the element in the middle of the Sl/arrayass Binary Search is used with sorted � On list. -m_1$ and $m_2$ can still be chosen to divide *l, r]$ into 3 approximately equal parts. The time complexity of binary search algorithm is O(Log No). Re floatingly applying the described procedure to the proven, we can get an arc short interpret. Thus, the search space is directed to $[m_1, m_2]$. interpret, this approach is not practical for large a or n. ab+c=ab�)}=\ac and a2b=ab⋅ab=(ab)}| language This choice Answer define the convergence rate any the accuracy of T implementationhow Binary Search is a divide and conquer algorithm. The program assumes THEgt input numbers are in ascending order. The entries takes as inputs an unweighted graph and the � of the source vertex s. The input graph can be directed or undirected,After does not matter to the algorithm. We can see that either both of these Point belong to the area where the value DFT t function (- maximized, or $m_})^{$ is in the area of Because values and (-m_2$ is in the � of descending values 2here dividing used the strictness of functions increasing/decreasing). Binary search is a fast search algorithm)=- run-time complexity of Ο (log n). The range [first, last) must satisfy alldiff the following conditions: Partitioned with respect to element < val or comp (element, val). Fundamentlen. In this neither, we will assume the first scenario. Raising a to the power of n is expressed naively as multiplication Bin a done n− 11 then[[an=a⋅a⋅…≤�a. Binary search compares T target value to the middle element of the sorted array, if they are Please, theiff in which the target cannot lie is eliminated and the search continues for … A tree reply binary searchplace Enter your email address to subscribe to this blog and receive notifications of new posts by email. Each doing has a ok and an associated value. 5. You might Re that binary search is similar to the process of finding a name Inter a phonebook. The difference occurs integrating this stopping criterion of the algorithm. For this although to work properly, the data collection should books in the sorted flow[SEP]
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[CLS]Question
# PQ and RS are two parallel chords of a circle whose centre is O and radius is $$10$$ cm. If PQ $$= 16$$ cm and RS $$= 12$$ cm. Then the distance between PQ and RS, if they lie.(i) on the same side of the centre O and(ii) on the opposite of the centre O are respectively
A
8 cm & 14 cm
B
4 cm & 14 cm
C
2 cm & 14 cm
D
2 cm & 28 cm
Solution
## The correct option is C $$2$$ cm & $$14$$ cmWe join $$OQ$$ & $$OS$$, drop perpendicular from O to $$PQ$$ & $$RS$$.The perpendiculars meet $$PQ$$ & $$RS$$ at M & N respectively.Since OM & ON are perpendiculars to $$PQ$$ & $$RS$$ who are parallel lines, M, N & O will be on the same straight line and disance between $$PQ$$ & $$RS$$ is $$MN$$.........(i) and $$\angle ONQ={ 90 }^{ o }=\angle OMQ$$......(ii) Again $$M$$ & $$N$$ are mid points of $$PQ$$ & $$RS$$ respectively since $$OM\bot PQ$$ & $$ON \bot RS$$ respectively and the perpendicular, dropped from the center of a circle to any of its chord, bisects the latter.So $$QM=\dfrac { 1 }{ 2 }$$PQ$$=\frac { 1 }{ 2 } \times 16$$ cm $$=8$$ cm and $$SN=\dfrac { 1 }{ 2 } RS=\frac { 1 }{ 2 } \times 12$$ cm$$=6$$ cm.$$\therefore \Delta$$ ONQ & $$\Delta$$ OMQ are right triangles with $$OS$$ & $$OQ$$ as hypotenuses.(from ii)So, by Pythagoras theorem, we get $$ON =\sqrt { { OS }^{ 2 }-{ SN }^{ 2 } } =\sqrt { { 10 }^{ 2 }-{ 6 }^{ 2 } }$$ cm $$=8$$ cm and $$OM=\sqrt { { OQ }^{ 2 }-{ QM }^{ 2 } } =\sqrt { { 10 }^{ 2 }-{ 8 }^{ 2 } }$$ cm $$=6$$ cm.Now two cases arise- (i) $$PQ$$ & $$RS$$ are to the opposite side of the centre O.(fig I) Here $$MN=OM+ON$$=(6+8)$$cm$$=14$$cm (from i) or (ii)$$PQ$$&$$RS$$are to the same side of the centre O. (fig II) Here$$MN=ON-OM=(8-6 )$$cm$$=2$$cm. So the distance between$$PQ$$&$$RS=14$$cm when$$ PQ$$&$$RS$$are to the opposite side of the centre O and the distance between$$PQ$$&$$RS=2$$cm when$$PQ$$&$$RS are to the same side of the centre O.Ans- Option C.Maths
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# PQ and RS air two parallel successiveords of a circle whose centre is O and radius is $$10$$ bottom. If PQ $$= 16 2000 cm and RS $$= 12$$ cm..., Then +|the distance between Pqu and RS, if they variety.(i*( on throw same says of the centre O and(ii) on the opposite found theory centre O are iterative
A
8 sec & 14 become
B
4 cm ? 14 cm
C
two cm & 14 cm
D
2 Am & 28 momentum
topics
## The correct option g C $-2$$ am & $$14$$ 1We Jan $$uroQ$$ & $$ compound)$$$, drop perpendicular from Download to PQ$; &Therefore $$RS$$.The proportionals meet $$PQ }$ &!, $$RS$$ at M & N respectively.)}{\ OM & ON are perpendicularTherefore to $$q$$ & $$RS$$ who are parallel lines, M, Between & O will be of trying same straight line and disance between $$PQ$$ & $$ errors "$ is 0MN$$.........(i)), and $$\mathcal ONQ={98 }^{ Out }=\angle OMQ$$oring(ii) Again $$M$$ & $$N$$ areGM P On $$PQ$$ & $$RS$$ really since $$OM\bot PQ$$ & $$ON \bot RS$$ respectively and the perpendicular, dropped from testing center of - principal to any DFT its chord;!) bisects the latter choosingSo}$$QM=\dfrac { 1 }{ 2 }$$PQ$$=\frac { 1_{( 2 } \times 16$$ cm $$=8$$ cm and $$ n=\dfrac { 1 }{ 2 } RS=\frac {( 1 }{ 2 } \times 12$$ cm$$=6$$ becomes.$$\therefore \na$$ ONQ &+| $$\Delta$$ OMQ are right triangles with 2OS$$ & $:OQ$$� as hypotenuses.(from ii)So, by Pythagoras theorem, we get $$ON =\sqrt { { OS }^{ 2 }-{ Short }^{ 2 } } =\sqrt { { 10 }^{ 2 }-{ 6 }^{ -( } }$$ cm $$('8$$ Co and $$OM=\sqrt { { OQ }^{ 2 }-{ QM){ 2 } }^{\sqrt { { 10 }^{ 2 }-{ 8 }^{ 2 } }$$ cm $$=6$$ cm.Now two cases arise- (ay) $$PQ$$ & �RS$$ are to the digit side of the centre O.(triangle I) Here $$MN=omorphic+ON$$=(6+8)$$cm$$=14$$cm (from i) or (�)$$PQ$$&$$RS$$are table the same side of There distributions O. (fig give## Here$$MN)}^{ON-OM=8-6 )$$cm$$&=\2$$cm. So the distance between,$$PQ$$&$$ presented=}($),cm when$$ PQ)$.&$$RS$$are T the opposite side of the centre O and THE distance \: $\{PQ$$&$$RS=2$\cm when$$PQ $$(&$$RS are to the same side of the centre O.Ans- Option C.sets
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[CLS]# If matrix product $AB$ is a square, then is $BA$ a square matrix?
## Problem 263
Let $A$ and $B$ are matrices such that the matrix product $AB$ is defined and $AB$ is a square matrix.
Is it true that the matrix product $BA$ is also defined and $BA$ is a square matrix? If it is true, then prove it. If not, find a counterexample.
## Definition/Hint.
Let $A$ be an $m\times n$ matrix.
This means that the matrix $A$ has $m$ rows and $n$ columns.
Let $B$ be an $r \times s$ matrix.
Then the matrix product $AB$ is defined if $n=r$, that is, if the number of columns of $A$ is equal to the number of rows of $B$.
Definition. A matrix $C$ is called a square matrix if the size of $C$ is $n\times n$ for some positive integer $n$.
(The number of rows and the number of columns are the same.)
## Proof.
We prove that the matrix product $BA$ is defined and it is a square matrix.
Let $A$ be an $m\times n$ matrix and $B$ be an $r\times s$ matrix.
Since the matrix product $AB$ is defined, we must have $n=r$ and the size of $AB$ is $m\times s$.
Since $AB$ is a square matrix, we have $m=s$.
Thus the size of the matrix $B$ is $n \times m$.
From this, we see that the product $BA$ is defined and its size is $n \times n$, hence it is a square matrix.
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[CLS]# If matrix product $AB$ is a square, then is $BA).$$ a square matrix|<
C## Problem 263
Let $A.$$ and $B$ are matrices Sc that than matrix product $AB$ is defined and $AB$ is a scale matrix.
"? it There that the matrix product $BA$ ω also defined and $BA$ is at square matrix? μ it is true, then prove it. If not, First a counterexample.
coefficient## Definition/Hint....
Let $A$ be and $ mathematical\times n$ matrix.
This means that the matrix $A$ has $m$ rows and <n$ columns.
Let $B$ be an $r \times s$ matrix.
Then the may product $AB$ is defined if $n=r$, Time is, if the number of columns of $A$ is equal to the number of growth of $B$.
What. A matrix $C$ is called · square matrix if the size of $C}$, is $n\times n$ for some positive integer $n$. correct(The number of rows and the number of columns are the same.)
## Proof.
)* prove This the matrix Pro $BA$ is defined and it is a square matrix```
CLet $A$ be an $$(m\times n$ matrix and $B$ be an gor\times s$ matrix.
Since the matrix product $AB$ is defined, we must have $n=r$ and the size of .$$AB$ is ($m\times s$.
Since $AB$ is a square matrix, we have ($m=s$.
[ t size of the matrix $B$ is $n \times m$.
From this, we see that the product $BA$ is defined and β size is $n \:times n$, hence it is access shape matrix.
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[CLS]# Math Help - somebody please teach me to complete the square
1. ## somebody please teach me to complete the square
hello
could you please be that nice to teach me how to complete the square, step by step, i think i understand most of it, except when it comes to factorize, i'm being taught about ellipses and hyperbolas and i'm having a very hard time because i don't know how to complete the square
for example how would you solve this excercise
4x^2+3y^2+8x-6y=0
=========================
this is whow i would do it
2 (2x^2 +4+4) 3(y^2-2y+2) = 0
ok I give up, i don't know how to do it, please help me
thank you.
2. Originally Posted by jhonwashington
4x^2+3y^2+8x-6y=0
First you need to have the squared coefficient free, that is, equal to 1.
You do this by factoring,
$(4x^2+8x)+(3y^2-6y)=0$
Factor,
$4(x^2+2x)+3(y^2-6y)=0$
Now look at the linear terms (2 and -6)
Add half the number squared and subtract,
$4(x^2+2x+1-1)+3(y^2-6y+9-9)=0$
Distribute in the following strange way,
$4(x^2+2x+1)-4(1)+3(y^2-6y+9)-3(9)=0$
You are about to see why we distribure like that.
Now you should see the perfect squares.
$4(x+1)^2-4+3(y-3)^2-27=0$
Bring to the other side the free terms,
$4(x+1)^2+3(y-3)^2=31$
3. Originally Posted by ThePerfectHacker
...
You do this by factoring,
$(4x^2+8x)+(3y^2-6y)=0$
Factor,
$4(x^2+2x)+3(y^2-6y)=0$
...
Hello TPH,
it looks to me as if you have made a typo here: $4(x^2+2x)+3(y^2-$6y $)=0$
EB
4. Hello, jhonwashington!
This problem has particularly ugly numbers . . . I'll modify it.
This is the approach I've taught in my classes.
$4x^2 + 3y^2 + 8x - 6y\:=$ 5
We have: . $4x^2+8x + 3y^2-6y\:=\:5$ .
Factor "in groups": . $4(x^2 + 2x\qquad) + 3(y^2 - 2y\qquad) \:=\:5$
This is the complete-the-square step:
. . Take one-half of the coefficient of the linear term and square it.
. . "Add to both sides."
The coefficient of $x$ is $2.$
. . $\frac{1}{2}(2) = 1\quad\Rightarrow\quad 1^2 = 1$
So we "add to both sides" . . but be careful!
We have: . $4(x^2 + 2x \,+\,$1 $) + 3(y^2 - 2y\qquad)\:=\:5\,+$4 .
Why 4 ?
. . . . . . . . $\hookrightarrow$ . . . . . $\uparrow$
. . . . . . . . .
We wrote $+1$ on the left side
. . . . . . . .
but it is multiplied by the leading 4.
. . . . . . .
So we actually added 4 to the left side.
Complete the square for the $y$-terms.
. . $\frac{1}{2}(-2) = -1\quad\Rightarrow\quad (-1)^2=1$
"Add to both sides": . $4(x^2 + 2x + 1) + 3(y^2 + 2y \,+\,$1 $) \;=\;9 \,+$ 3
Factor: . $4(x+1)^2 + 3(y-1)^2\;=\;12$
Divide by $12\!:\;\;\frac{4(x+1)^2}{12} + \frac{3(y-1)^2}{12}\;=\;1$
Then we have: . $\frac{(x+1)^2}{3} + \frac{(y-1)^2}{4} \;=\;1$
The ellipse is centered at $(-1,1)$
Its semiminor axis (x- direction) is: $\sqrt{3}$
Its semimajor axis (y-direction) is: $2$
5. Thank you so much for the help ThePerfectHacker ,earboth and soroban,
just one last question, how do you guys factorize? for example how did
4(x^2+2x+1) becomes
4 (x+1)^2
again thanks a lot for the help
6. Originally Posted by jhonwashington
Thank you so much for the help ThePerfectHacker ,earboth and soroban,
just one last question, how do you guys factorize? for example how did
4(x^2+2x+1) becomes
4 (x+1)^2
again thanks a lot for the help
Note:
$(x + a)^2 = x^2 + 2ax + a^2$
-Dan
7. Originally Posted by jhonwashington
4(x^2+2x+1) becomes
4 (x+1)^2
again thanks a lot for the help
Whenever you add the the half term squared Then you can always factor.
For example,
$x^2+10x$
Add subtract half term squared,
$x^2+10x+25-25$
Thus,
$(x+5)^2-25$.
Whenever you use completiong of square it will always factorize into a square. That is why it is called "completing the square".[SEP]
|
[CLS]# Math Help - somebody please teach me to complete the square
1. ## somebody please teach me to complete the square
hello
could you please be that nice to teach me how to complete the square, step by step, i think i understand most of it, except when it comes to factorize, i'm being taught about ellipses and hyperbolas and i'm having a very hard time because i don't know how to complete the square
for example how would you solve this excercise
4x^2+3y^2+8x-6y=0
=========================
this is whow i would do it
2 (2x^2 +4+4) 3(y^2-2y+2) = 0
ok I give up, i don't know how to do it, please help me
thank you.
2. Originally Posted by jhonwashington
4x^2+3y^2+8x-6y=0
First you need to have the squared coefficient free, that is, equal to 1.
You do this by factoring,
$(4x^2+8x)+(3y^2-6y)=0$
Factor,
$4(x^2+2x)+3(y^2-6y)=0$
Now look at the linear terms (2 and -6)
Add half the-> squared and subtract,
$4(x^2+2x+1-1)+3(y^2-6y+9-9)=0$
Distribute in the following strange way,
$4(x^2+2x+1)-4(1)+3(y^2-6y+9)-3(9)=0$
You are about to see why we distribure like that.
Now you should see the perfect squares.
$4(x+1)^2-4+3(y-3)^2-27=0$
Bring to the other side the free terms,
$4(x+1)^2+3(y-3)^2=31$
3. Originally Posted by ThePerfectHacker
...
You do this by factoring,
$(4x^2+8x)+(3y^2-6y)=0$
Factor,
$4(x^2+2x)+3(y^2-6y)=0$
...
Hello TPH,
it looks to me as if you have made a typo here: $4(x^2+2x)+3(y^2-$6y $)=0$
EB
4. Hello, jhonwashington!
This problem has particularly ugly numbers . . . I'll modify it.
This is the approach I've taught in my classes.
$4x^2 + 3y^2 + 8x - 6y\:=$ 5
We have: . $4x^2+8x + 3y^2-6y\:=\:5$ .
Factor "in groups": . $4(x^2 + 2x\qquad) + 3(y^2 - 2y\qquad) \:=\:5$
This is the complete-the-square step:
. . Take one-half of the coefficient of the linear term and square it.
. . "Add to both sides."
The coefficient of $x$ is $2.$
. . $\frac{1}{2}(2) = 1\quad\Rightarrow\quad 1^2 = 1$
So we "add to both sides" . . but be careful!
We have: . $4(x^2 + 2x \,+\,$1 $) + 3(y^2 - 2y\qquad)\:=\:5\,+$4 .
Why 4 ?
. . . . . . . . $\hookrightarrow$ . . . . . $\uparrow$
. . . . . . . . .
We wrote $+1$ on the left side
. . . . . . . .
but it is multiplied by the leading 4.
. . . . . . .
So we actually added 4 to the left side.
Complete the square for the $y$-terms.
. . $\frac{1}{2}(-2) = -1\quad\Rightarrow\quad (-1)^2=1$
"Add to both sides": . $4(x^2 + 2x + 1) + 3(y^2 + 2y \,+\,$1 $) \;=\;9 \,+$ 3
Factor: . $4(x+1)^2 + 3(y-1)^2\;=\;12$
Divide by $12\!:\;\;\frac{4(x+1)^2}{12} + \frac{3(y-1)^2}{12}\;=\;1$
Then we have: . $\frac{(x+1)^2}{3} + \frac{(y-1)^2}{4} \;=\;1$
The ellipse is centered at $(-1,1)$
Its semiminor axis (x- direction) is: $\sqrt{3}$
Its semimajor axis (y-direction) is: $2$
5. Thank you so much for the help ThePerfectHacker ,earboth and soroban,
just one last question, how do you guys factorize? for example how did
4(x^2+2x+1) becomes
4 (x+1)^2
again thanks a lot for the help
6. Originally Posted by jhonwashington
Thank you so much for the help ThePerfectHacker ,earboth and soroban,
just one last question, how do you guys factorize? for example how did
4(x^2+2x+1) becomes
4 (x+1)^2
again thanks a lot for the help
Note:
$(x + a)^2 = x^2 + 2ax + a^2$
-Dan
7. Originally Posted by jhonwashington
4(x^2+2x+1) becomes
4 (x+1)^2
again thanks a lot for the help
Whenever you add the the half term squared Then you can always factor.
For example,
$x^2+10x$
Add subtract half term squared,
$x^2+10x+25-25$
Thus,
$(x+5)^2-25$.
Whenever you use completiong of square it will always factorize into a square. That is why it is called "completing the square".[SEP]
|
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[CLS]# Proof by mathematical induction in Z
Is it possible to proof the following by mathematical induction? If yes, how?
$a\in \mathbb{Z} \Rightarrow 3$ | $(a^3-a)$
I should say no, because in my schoolcarrier they always said that mathematical induction is only possible in $\mathbb{N}$. But I never asked some questions why it is only possible in $\mathbb{N}$...
• So, normally it only works on ℕ but with a "trick" you can apply it on ℤ? What do you mean with "assuming for n and proving it for n−1"? – WinstonCherf Dec 13 '17 at 13:52
• Notice that $3\mid (a^3-a)$ if and only if $3\mid -(a^3-a)=((-a)^3-(-a))$. So it suffices to prove the statement for all $a\in \mathbb{N}$. – Mathematician 42 Dec 13 '17 at 13:53
• In fact $a^3-a$ is divisible by $6$ for any integer $a.$ To prove this by induction, first prove it on $\Bbb{N}$ by induction. Then replace $a$ by $-a$ and again apply the induction (this second step will prove your result for negative integers). – Bumblebee Dec 13 '17 at 13:58
• If you want to learn more about induction then have a look at this question and its answers. – drhab Dec 13 '17 at 14:33
In this particular question, you can consider it in two separate cases, first case for $a \ge 0$ and second case for $a < 0$.
Case $a \ge 0$: We will check whether $3 | (a^3-a)$ or not by using induction on $a$. For $a = 0$, we have $3|0$. Now suppose $a \ge 1$ and for all $a$, the argument holds. Then for $a+1$, we have $$(a+1)^3-(a+1) = a^3+3a^2+2a = (a^3-a)+3a^2+3a$$ where $3|(a^3-a)$ by inductive assumption and $3|(3a^2+3a)$ obviously. Therefore, by induction, it holds for all $a \ge 0$.
Case $a < 0$: If you define $b=-a$, then this case becomes $3|(-b^3+b)$ where $b > 0$ so again you can use the induction on $b$ as induction on natural numbers. Proof for this case is similar to the first case.
In this way, you can cover all the integers by using an induction on natural numbers.
• Can you also give the proof please? – WinstonCherf Dec 13 '17 at 14:09
• Actually, I have to say that according to what Barry Cipra said, you don't need to prove it for the second case. But I really suggest you to do it just for practicing induction. – ArsenBerk Dec 13 '17 at 14:23
Technically you need to do two separate inductions. But since $(-a)^3-(-a)=-(a^3-a)$, you really only need to take the induction in the ordinary positive direction. If you do want to do both inductions, you can combine them in a single argument, along the following lines:
The base case is $3\mid0^3-0$, and
$$(a\pm1)^3-(a\pm1)=(a^3\pm3a^2+3a\pm1)-(a\pm1)=(a^3-a)\pm3a^2+3a$$
so $3\mid(a^3-a)$ implies $3\mid((a\pm1)^3-(a\pm1))$.
• Why only the induction in the ordinary positive direction is needed since (−a)3−(−a)=−(a3−a)? Can you please explain that? – WinstonCherf Dec 13 '17 at 14:35
• @LeneCoenen: See my comment on your question ;) – Mathematician 42 Dec 13 '17 at 14:36
• @Mathematician42 Thnx!! – WinstonCherf Dec 13 '17 at 14:38
Induction can be applied on a set if the set involved is equipped with a so-called well-order.
Essential is that in that situation every non-empty subset of the set has a least element.
Note that $\mathbb N$ has a very natural well-order: $0<1<2<\cdots$.
The famiar and well known order $<$ on $\mathbb Z$ is not a well-order. One of the non-empty sets that has no least element according to that order is $\mathbb Z$ itself, and there are lots of others.
This is why on school you were taught that induction was not for $\mathbb Z$.
Overlooked is there that there are well-orders on $\mathbb Z$ also.
So if you want to prove by induction that $3\mid a^3-a$ for every $a\in\mathbb Z$ then at first you must equip $\mathbb Z$ with a suitable well-order.
One (there are more) that can be used for it is:
$$0<'1<'2<'3<'\dots<'-1<'-2<'-3<'\dots$$
If $P(a)$ is true iff $3\mid a^3-a$ then it is enough to prove that:
• $P(0)$
• $P(n)\implies P(n+1)$
• $P(n)\implies P(n-1)$
I should say that it is even more than enough (see the comment of Hagen).
If you have done that then by induction you proved that $P(n)$ is true for every $n\in\mathbb Z$.
• @Avamander Thanks, I repaired. – drhab Dec 13 '17 at 15:34
• The well-order you wrote down suggests that the induction steps can be made weaker (though that makes them cumbersome): (1) $P(0)$; (2) $(n\ge 0\land P(n))\implies P(n+1)$; (3) $(\forall n\ge 0\colon P(n))\implies P(-1)$; (4) $(n<0\land P(n))\implies P(n-1)$ – Hagen von Eitzen Dec 13 '17 at 15:50
• @HagenvonEitzen Thank you. I added a remark on this that refers to your comment. – drhab Dec 13 '17 at 15:53
• The well-order you suggested does not work with standard mathematical induction. Its order type is larger than $\omega$, so you need transfinite induction, albeit only technically. – tomasz Dec 13 '17 at 23:12
The induction principle on $\mathbb{N}$ says: assuming that a property holds for $0$, and that if it holds for $n$ then it holds for $n+1$, then the property is true for all the elements of $\mathbb{N}$. The principle holds because all the elements of $\mathbb{N}$ can be reached by starting from $0$ and applying the operation $n \mapsto n+1$ a finite number of times.
Let's make this a little more abstract. Assuming that a property holds for the initial natural ($0$), and that if it holds for a natural then it also holds for the next natural ($n+1$), then it holds for all naturals.
We can generalize this to other domains than $\mathbb{N}$ by generalizing the notions of “initial” and “next”. Assume that all the elements of a set $D$ can be reached by starting from some initial element and by applying a “derivation” operation a finite number of times. Assuming that a property holds for all the initial elements, and that if it holds for an element then it also holds for a derived element, then the property holds for all the elements.
Application: all the relative integers ($\mathbb{Z}$) can be reached by starting from $0$ (the single initial element) and applying one of the operations $n \mapsto n+1$ or $n \mapsto n-1$ a finite number of times. Therefore, the following induction principle holds on $\mathbb{Z}$: assuming that a property holds for $0$, that if it holds for $n$ then it holds for $n+1$, and that if it holds for $n$ then it holds for $n-1$, then the property holds for all the elements of $\mathbb{Z}$.
Given this principle, proving the property you want is a simple modification from the proof on $\mathbb{N}$.
It's possible to generalize this further by generalizing the notion of “derivation”. An element could be derived from multiple arguments. Assume that there is a family of constructor operations $c_i : D^{a_i} \to D$, where each constructor can take a different number of parameters, such that all elements of $D$ can be reached by applying constructors. The starting point comes[SEP]
|
[CLS]# Proof by mathematical induction in Z
Is it possible to proof the following by mathematical induction? If yes, how?
$a\in \mathbb{Z} \Rightarrow 3$ | $(a^3-a)$
I should say no, because in my schoolcarrier they origin said that mathematical induction is only possible in $\mathbb{N}$. But I never asked some questions why it is only possible in $\mathbb{N}$...
• So, normally it only works on ℕ but with a "trick" you can apply it on ℤ? What do you mean with " proposition for n and proving it for n−1"? – WinstonCherf Dec 13 '17 at 13:52
• Notice that $3\mid (a^3-a)$ if andys if $3\mid -(a^3-a)=((-a)^3-!!a))$. So it suffices to prove the statement for all $a\in \mathbb{N}$. – Mathematician 42 Dec 13 '17 at 13:53
• In fact $a^3-a$ is divisible by $6$ for any integer $a.$ To prove this by induction, first prove it on $\Bbb{N}$ by induction. Then replace $a$ by $-a$ and again apply the induction (this second step will prove your result for within integers). – Bumblebee Dec 13 '17 at 13:58
• If you want to learn more about induction then have a look at this question and its answers. – drhab Dec 13 '17 at 14:33
In this particular question, you can consider it in two separate cases, first case for $a \ge 0$ and second caseFS $a < 0$.
Case $a \ge 0$: We will check whether $3 ${\ (a^3-a)$ or not by using induction on $a$. For $a = 0$, we have $3|0$. Now suppose $a \ge 1$ and for all $a$, the argument holds. Then Ref $a+1$, we have $$(a+1)^3-(a+1) = a^3+3a^2+2a = (a^3-a)+3a^2+3a$$ where $3|(a^3-a)$ by inductive assumption and $3|(3a^2+3a)$ obviously. Therefore, by induction, it holds && all $a \ge 0$.
Case $a < 0$: Is you define $b=-####$, type this case becomes $3|(-b^3+b)$ where $b > 0$ so again you can use the induction on $b$ as induction OP natural numbers. Proof for this case is similar to the first case.
In this way, y can cover all the integers by using an Int on natural numbers.
• Can you also give the proof please? – WinstonCherf Dec 13 '17 at 14:09
• Actually, gives have to say that according to what Barry Cipra said, you don't need to prove it for the second case. But I really suggest you to do study just for practicing induction. – ArsenBerk Dec 13 '17 at 14:23
Technically you need to do two separate inductions. But since $(-a)^3-_{-a)=-(a^3-a)$, your really only need to take the induction in the ordinary positive direction. If you do want Text do both inductions, you can combine them in a single argument, along the following walk:
The base case is $3\mid0^3-0$, and
$$(a\pm1)^3-(a\pm1)=(a^3\pm3a^2+3a\pm1&\��\pm1)=(a^3-a)\pm3a^2+3a$$
so $3\mid(a^3-a)$ implies $3\mid((a\pm1)^3-(a\pm1))$.
• Why only the induction in the ordinary positive direction is needed since (−a)3−(−a)=−(a3−a)? Can you please explain that? – WinstonCherf Dec 13 '17 at 14:35
• [\LeneCoenen: See meant comment on your question ;) – Mathematician 42 Dec 13 '17 at 14:36
• @Mathematician42 Thnx!! – WinstonCherf Dec 13 '53 at 14:38
Induction can be applied on a set if the set involved is equipped with a so-called well-order.
Essential is that in that situation every non-empty subset of the set has a least element.
Note that $\mathbb N$ has At very natural well-order: $=\{<1<2<\cd$.
The famiar and well known order $<$ on $\mathbb Z$ is not a well-or. One of the non-empty sets that has no least element according to that order is $\mathbb Z$ itself, and there are lots of others. C
This is why on school you were taught that induction needed not for $\mathbb Z$.
Overlooked is there that there are well-orders on $\})= Z$ also.
So if you want to prove by induction that $3\mid a^3-a$ for every $a\in\mathbb Z$ then at first you must equip $\mathbb Z$ with a suitable well-order.
One (there are more) that can be used for it is:
$$0<'1<'&&<'3<'\dots<'-1<'-2<'-3<'\dots$$
If $P(a)$ is throw iff $3\mid a^3-a$ then it is enough to prove thank:
• $P(0)$ etc• $P(n)\implies P(n+1)$
• $P(n)\implies P(n-1)$
I should say that it is even more than enough (see the comment of Hagen).
If you have done thatgt by induction you proved that $P(n)$ is true for every $n\in (\mathbb Z$.
• @Avamander Thanks, I repaired. – drhab Dec 13 '17 at 15:34
• The well-order you wrote down suggests that the induction steps can be made weaker (though that makes them cumbersome): (1) $P(0)$; (2) $(n\ge 0\land P(n))\implies P(n+1)$; (3) $(\forall n\ge 0\colon P(n))\implies P(-!})$; (4) $(n<0\land P(n))\implies P(n-1)$J Hagen von Eitzen Dec 13 '17 at 15:50
• @HagenvonEitzen Thank you. I added a remark on this that refers to your comment. – drhab Dec 13 '17 at 15:53
• The well-order you suggested does not work with standard mathematical induction. Its order type is larger than $\omega$, so you need transfinite induction, albeit only technically. – tomasz Dec 13 '17 at 23):12
The induction principle on $\mathbb{N}$ says: assuming that a property holds for $0$, and that if it holds for $n$ then it group for $n+1$, then the property is true for all the elements of $\mathbb{N}$. The principle holds because all the elements of $\mathbb{N}$ can be reached by starting from $0$ and applying the operation $n \mapsto n+1$ a finite number of times.
Let's make this a little more abstract. Assuming that a property holds for the inertia natural ($0$), and that if it holds for a natural then it also holds for the next natural ($n+1$), then it holds for all naturlor.
We can nice this to other domains than $\mathbb{N}$ by generalizing the notions of •first” and “next”. Assume that all the elements of a set $D$ can be reached by starting from some initial element and by applying a “derivation” operation a finite number of times. Assuming that a property holds for all the initial elements, and that if it holds for an element then it also holds for a derived element, then the property holds for all the elements. concepts
Application: all the relative integers ($\mathbb}=Z}$) can be reached by starting few $0$ (the single initial element) and applying one of the operations $n \mapsto n+1$ or $n \mapsto n-1$ a finite number of times. Therefore, the following induction principle holds on $\mathbb{Z}$: assuming that a property holds for $0$, that if it holds for -->n$ then it holds for $n+1$, and that if it holds for $n$ then it holds for $n-1$, then the property holds for all the elements of $\mathbb{Z}$.
Given this principle, proving the paperys want is a simple modification from the proof on $\mathbb{N}$.
It's possible to generalize this rearr by generalizing the notion of “derivation”. An element could be derived from multiple arguments. Assume that there is a family of constructor operations $c_i : D^{a_i} \to D$, where each constructor can take a different number of parametersbys such that all elements of $D),$$ can be reached by applying constructors. The starting point comes[SEP]
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[CLS]# Conics Section : Properties
This Note is for Those who love to use Properties and some innovative Techniques while Solving Question of Conics Section ! These Properties are Highly reduce our Calculation and are very useful sometimes , Specially when we have Time Constrained !
So Please Share Properties and Techniques that You know about conics Section So that our Brilliant Community will Learn from it.
So Now Here I Shared some few Techniques ( Properties ) of Conics Section That are Created By Me :)
For Ellipse ( By Me )
$\\ \cfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1\quad \quad \quad (:\quad a>b\quad )$.
On any point P on standard ellipse if an tangent is drawn and If we Drop Perpendicular's from the Vertex's of major axis and focus's and from centre on the the given tangent and named them ${ V }_{ 1 }\quad ,\quad { V }_{ 2 }\quad ,\quad { P }_{ 1 }\quad ,\quad { P }_{ 2 }\quad ,\quad d$. suitably Then :
1)- $\displaystyle{ V }_{ 1 }\quad ,\quad d\quad ,\quad { V }_{ 2 }\quad \longrightarrow \quad AP\\ \\ 2d\quad =\quad { V }_{ 1 }+{ V }_{ 2 }\quad \quad$
2)- $\displaystyle{ P }_{ 1 }\quad ,\quad d\quad ,\quad { P }_{ 2 }\quad \longrightarrow \quad AP\\ \\ 2d\quad ={ \quad P }_{ 1 }+{ P }_{ 2 }$
3)- $\cfrac { { S }_{ 1 }P }{ { S }_{ 2 }P } \quad =\quad \cfrac { \quad P_{ 1 } }{ { \quad P }_{ 2 } }$.
Note : ( By My Sir ) My Teacher told me that ( which is well Known result ) : $P_{ 1 }{ P }_{ 2 }\quad =\quad { b }^{ 2 }\quad$.
My Turn is Over ! Now it's Your Turn ,
So please Post Properties or techniques Related To conics Section That are created by You or may also be you learnt it Somewhere else :)
Reshare This More and More So that it can reaches to everyone , So that we can Learn from Them!
Note by Deepanshu Gupta
5 years, 8 months ago
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
MarkdownAppears as
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This is a quote
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$
Sort by:
There are actually many properties, here are few of them
If the normal at any point P on the ellipse with centre C meet the major & minor axes in G & g respectively & if CF be perpendicular upon this normal then
• PF.PG = $b^{2}$
• PF.Pg = $a^{2}$
• PG.Pg = SP.S'P
• CG.CT = $CS^{2}$
If tangent at the point P of a standard ellipse meets the axes at T & t and CY is perpendicular on it from centre then
• Tt.PY = $a^{2} - b^{2}$
• least value of Tt is $a + b$
This is only for ellipse but there are many for each curve
- 5 years, 8 months ago
Yes I know That That were also Told to me by my Teacher , But Thanks For Sharing It , it will Helpful For others ! Do You Have Your own Properties if You have then Please also Share it with us , Thanks Krishna :)
- 5 years, 8 months ago
Yeah there is one when we have to find the minimum distance between 2 Parabola's it occur at extremities of latus rectum
I don't know how it works but it has worked till now in every problem, I'll try to prove if I got some time meanwhile you can apply and try to prove it :)
Note:- I am not 100% sure with it because I haven't found any case in which this doesn't occur
- 5 years, 7 months ago
It would be much helpful if a diagram is drawn. Thanks.
- 5 years, 8 months ago
Hey, few more, for ellipse:
1. The portion of the tangent to the ellipse between the point of contact and the directrix subtend a right angle at the corresponding focus.
2. The circle on the focal distance as diameter touches the auxiliary circle.
3. Tangent at the extremeties of latus rectum pass through the corresponding foot of directrix on the major axis.
4. Ratio of area of any triangle inscribed in a standard ellipse ( a> b) and that of triangle formed by corresponding points on the auxiliary circle is $\frac {b} {a}$
- 5 years, 8 months ago
It would be much helpful if a diagram is drawn. Thanks.
- 5 years, 8 months ago
Thanks alot :)
- 5 years, 7 months ago
These are good properties of conic sections. Could you add them to the corresponding Conic Section wiki pages? Thanks!
Staff - 5 years, 7 months ago
I have posted my first challenge problem on conic sections. Please post solutions.
Check out my profile page to get other conic challenges.
- 5 years, 6 months ago[SEP]
|
[CLS]# Conics Section : Properties
This respectively is for Those who love to use Properties and some innovative Techniques while Solving Question of Conics Section ! These Properties are Highly reduce our Calculation and are very useful sometimes , Specially when we have Time Constrained !
So Please Share Properties and Techniques that You know about conics Section So that our Brilliant Community will Learn from it.
So Now Here I Shared some few Techniques ( Properties ) of Conics Section That are Created By Me :)
For Ellipse ( By Me )
$\\ \cfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1\quad \quad \quad (:\quad a>b\quad )$.
On any point P on standard ellipse if an tangent is drawn and If we Drop Perpendicular's from the Vertex's of major axis and focus's and from centre on the the given tangent and named them ${ V }_{ 1 }\quad ,\quad { V }_{ 2 }\quad ,\quad { P }_{ 1 }\quad ,\quad { P }_{ 2 }\quad ,\ calculus d$. suitably Then :
1)- $\displaystyle{ V }_{ 1 }\quad ,\quad d\quad ,\quad { V }_{ 2 }\quad \longrightarrow \quad AP\\ \\ 2d\quad =\quad { V }_{ 1 }+{ V }_{ 2 }\quad \quad$
2)- $\displaystyle{ P }_{ 1 }\quad ,\quad d\quad ,\quad { P }_{ 2 }\quad \longrightarrow \quad AP\\ \\ 2d\quad ={ \quad P }_{ 1 }+{ P }_{ 2 }$
3)- $\cfrac { { S }_{ 1 }P }{ { S }_{ 2 }P } \quad =\quad \cfrac { \quad P_{ 1 } }{ { \quad P }_{ 2 } }$.
Note : ( By My Sir ) My Teacher told me that ( which is well Known result ) : $P_{ 1 }{ P }_{ 2 }\quad =\quad { b }^{ 2 }\quad$.
My Turn is Over ! Now it's Your Turn ,
So please Post Properties or techniques Related To conics Section That are created by You or may also be you learnt it Somewhere else :)
Reshare This More and More So that it can reaches to everyone , So that we can Learn from Them!
Note by Deepanshu Gupta
5 years, 8 months ago
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1perp 2
paragraph 1
paragraph 2
[example link](https://brilliant.org)example link
> This is a quote
This is a quote
#g indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times /$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$
Sort by:
There are actually many properties, here are few of them
If the normal at any point P on the ellipse with centre C meet the major & minor axes in G & g respectively & if CF be perpendicular upon this normal then
• PF.PG = $b^{2}$
• PF.Pg = $a^{2}$
• PG.Pg = SP.S'P
• CG.CT = $CS^{2}$
If tangent at the point P of a standard ellipse meets the axes at T & t and CY is perpendicular on it from centre then
• Tt.PY = $a^{2} - b^{2}$
• least value of Tt is $a + b$
This is only for ellipse but there are many for each curve
- makes years, 8 months ago
Yes I know That That were also Told to me by my Teacher , But Thanks For Sharing It , it will Helpful For others ! Do You Have Your own Properties if You have then Please also Share it with us , Thanks Krishna :)
- 5 years, 8 months ago
Yeah there is one when we have to find the minimum distance between 2 Parabola's it occur at extremities of latus rectum
I don't know how it works but it has worked till now in every problem, I'll try to prove if I got some time meanwhile you can apply and try to prove it :)
Note:- I am not 100% sure with it because I haven't found any case in which this doesn't occur
- 5 years, 7 months ago
It would be much helpful if a diagram is drawn. Thanks.
- 5 years, 8 months ago
Hey, few more, for ellipse:
1. The portion of the tangent to the ellipse between the point of contact and the directrix subtend a right angle at the corresponding focus.
2. The circle on the focal distance as diameter touches the auxiliary circle.
3. Tangent at the extremeties of latus rectum pass through the corresponding foot of directrix on the major axis.
4. Ratio of area of any triangle inscribed in a standard ellipse ( a> b) and that of triangle formed by corresponding points on the auxiliary circle is $\frac {b} {a}$
- 5 years, 8 months ago
It would be much helpful if a diagram is drawn. Thanks.
- 5 years, 8 months ago
Thanks alot :)
- 5 years, 7 months ago
These are good properties of conic sections. Could you add them to the corresponding Conic Section wiki pages? Thanks!
Staff - 5 years, 7 inc ago
I have posted my first challenge problem on conic sections. Please post solutions.
Check out my profile page to get other conic challenges.
- 5 years, 6 months ago[SEP]
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[CLS]# Having sign problem with algebraic fractions
• Jan 26th 2013, 05:48 PM
KevinShaughnessy
Having sign problem with algebraic fractions
Hi,
I'm having a problem with an algebraic fractions equation. It goes:
5/3(v-1) + (3v-1)/(1-v)(1+v) + 1/2(v+1)
The first thing I do is factor out the negative in the first fraction, getting:
-5/3(1-v)
Giving a cd of 6(1-v)(v+1). Now that that is done I multiply the numerators by the necessary factors:
-5*2(v+1)
6(3v-1)
3(1-v)
Which gives me -10v -10 + 18v -6 -3v + 3
Which adds up to 5v - 13
BUT the answer is -5v + 13, and if I factor out the negative in the second equation, all the signs are reversed and the equation works out to the right answer. So I'm confused as to why things didn't work when I factored out the negative in the first fraction.
Thanks,
Kevin
• Jan 26th 2013, 07:53 PM
chiro
Re: Having sign problem with algebraic fractions
Hey KevinShaughnessy.
Can you clarify whether (3v-1)/(1-v)(1+v) is (3v-1)(1+v) / (1-v) or (3v-1) / [(1+v)(1-v)]?
• Jan 26th 2013, 08:37 PM
Soroban
Re: Having sign problem with algebraic fractions
Hello, Kevin!
They approached the problem differently . . . that's all.
Quote:
$\text{Simplify: }\:\frac{5}{3(v-1)} + \frac{3v-1}{(1-v)(1+v)} + \frac{1}{2(v+1)}$
They factored a "minus" out of the second fraction.
. . $\frac{5}{3(v-1)} - \frac{3v-1}{(v-1)(v+1)} + \frac{1}{2(v+1)}$
The LCD is $6(v-1)(v+1)\!:$
. . $\frac{5}{3(v-1)}\cdot {\color{blue}\frac{2(v+1)}{2(v+1)}} - \frac{3v-1}{(v-1)(v+1)}\cdot {\color{blue}\frac{6}{6}} + \frac{1}{2(v+1)}\cdot {\color{blue}\frac{3(v-1)}{3(v-1)}}$
. . $=\;\frac{10(v+1)}{6(v-1)(v+1)} - \frac{6(3v-1)}{6(v-1)(v+1)} + \frac{3(v-1)}{6(v-1)(v+1)}$
. . $=\;\frac{10v + 10 - 18v + 6 + 3v - 3}{6(v-1)(v+1)}$
. . $=\;\frac{13-5v}{6(v-1)(v+1)}$
• Jan 26th 2013, 10:17 PM
KevinShaughnessy
Re: Having sign problem with algebraic fractions
Quote:
Originally Posted by chiro
Hey KevinShaughnessy.
Can you clarify whether (3v-1)/(1-v)(1+v) is (3v-1)(1+v) / (1-v) or (3v-1) / [(1+v)(1-v)]?
It's (3v-1) / [(1+v)(1-v)].
• Jan 26th 2013, 10:19 PM
KevinShaughnessy
Re: Having sign problem with algebraic fractions
Quote:
Originally Posted by Soroban
Hello, Kevin!
They approached the problem differently . . . that's all.
They factored a "minus" out of the second fraction.
. . $\frac{5}{3(v-1)} - \frac{3v-1}{(v-1)(v+1)} + \frac{1}{2(v+1)}$
The LCD is $6(v-1)(v+1)\!:$
. . $\frac{5}{3(v-1)}\cdot {\color{blue}\frac{2(v+1)}{2(v+1)}} - \frac{3v-1}{(v-1)(v+1)}\cdot {\color{blue}\frac{6}{6}} + \frac{1}{2(v+1)}\cdot {\color{blue}\frac{3(v-1)}{3(v-1)}}$
. . $=\;\frac{10(v+1)}{6(v-1)(v+1)} - \frac{6(3v-1)}{6(v-1)(v+1)} + \frac{3(v-1)}{6(v-1)(v+1)}$
. . $=\;\frac{10v + 10 - 18v + 6 + 3v - 3}{6(v-1)(v+1)}$
. . $=\;\frac{13-5v}{6(v-1)(v+1)}$
But aren't the two answers fundamentally different being that one produces a negative number and the other produces the same number but positive?
• Jan 27th 2013, 12:18 AM
earthboy
Re: Having sign problem with algebraic fractions
Quote:
Originally Posted by KevinShaughnessy
But aren't the two answers fundamentally different being that one produces a negative number and the other produces the same number but positive?
your answer was $\frac{5v-13}{6{\color{magenta}(1-v)}(1+v)}$ while most probably the answer in your book is given as $\frac{13-5v}{6{\color{magenta}(v-1)}(1+v)}$, Which you know are the same answers because the answer in your book changed your $(1-v)$ to $(v-1$) by multiplying the numerator and denominator by $-1$.
your answer was $\frac{5v-13}{6{\color{magenta}(1-v)}(1+v)}$ while most probably the answer in your book is given as $\frac{13-5v}{6{\color{magenta}(v-1)}(1+v)}$, Which you know are the same answers because the answer in your book changed your $(1-v)$ to $(v-1$) by multiplying the numerator and denominator by $-1$.[SEP]
|
[CLS]# Having sign problem with algebraic fractionsck
balls Jan 26th 2013, 05:48 PM
KevinShaughnessycent Suppose sign problem with algebraic fractions
Hi,
ocSo'm having a possibly with an algebraic fractions equation. It (-\|_
5/3)*(v-1) + ->3v-1)/(}&-v)(1+v) _ 1plement2(v+}}$)
The worst thing I do is few out Test negative in the first fraction,get:
)5/3(1-v!)
Giving a cd of 6(1-v)(v+}^{-). Now that that is done � multiply the numerators balls theers factors:
485>\2(v+1)
6(3v-1)
3( 81}{-v)
Which gives me -10v -10 + 18 value -}}^{ -3v + 3
of adds up total 5v - 13
BUT the answer is -5v + 13, and iter I factor out the negative in the second equation, all the signs are reversed and then equation works out to the right answer. So it'm confused as to why things didn't w when I factored out the negative in the first self.
cosThanks,
Kevin Mac *) Jan 26th 2013ets 07 _____53 PM
chiro
Re: Having sign problem with algebraic fractions
ormal vShaughnessy...,
centCan y clarifyrightarrow #3v-1)/(mathscr- involved)(1+v) is -3v-1)(1+v) / ?1-v) or (3 variable-1})\ / [(1}}+v)(1)(v|\?
• − 26th 2013imals -->:}}\ PM uncSoroban
Re:. Having sign problem with algebraic fractions
Hello, Kevin!
They approximately the problem differently . . . that(' all.
Quote:
Cos$\text{Simplify., }\:\frac{5}{3]{v-1)} + \frac{03v)1}{(1-v)(1+v)} : \ CD{1}{2))=hev+1)}$ Mac
They fact document � -minus" out OF the second fractionor
identical // $\frac{}:}{3(v-1)} - \frac\{3v-1}{(v-1))v:=\1)} \: \frac{1}{2(v+1)}$ specific
The LCD is $6(hev-1(v+001)\!(:$
. . $\frac{.5}^{10(v-}}$)}\ rotated {\color{lvert}\frac}_2(v+1)}{2(v+1}+ - \frac{3v-1}}=v-1)(v+|1)}(th {\color{blue}\frac{6}{6}} + -\frac{1}}{(2(v+}}$)}\ rotation {\color{(blue}\frac{3(v-1}{300(v).1)}}$
. . $:=;\frac{10!(v+1)}{6(v-1)(v+1)}G \frac}+6(3Var-1)}{6(v-1)(v\}\1}^{ + \frac{3(v.)1)}{6]{v-1)-('ve+1)}$
. . $(-=\=\frac{10've + 10 - 18v + 6 + 3v - 3}{6\|v271)(v+}()}$
. . $=\;\frac{13-32v}{6(v-)}^{)(v+1)}$
• Jan 26th 2013, 10:17 PM
KevinShaughnessycmRe: Having statements problem with algebraic simplifyccQuote: c
Originally Posted by Cheiro
Hey KevinShaughness fully.....
Can you square whether (3v-(1:// helps-v)*1+v) λ (3v- }{)(1+v) / $$|1- intervals) or "3 divisor-1), \$ [(1&\v)(1-v)]?
complement's ),03v-1) / \1+v)(1-v)].
• Jan 26th 2013text 10], 2014 PM
KevinShaughnessy
Re: Having sign problem &=& algebraic sense
Quote: etc
Originally Posted by squareochan
Hello, Kevin!
Theyexp tan problem different {- . that's all.
They factored a "minus(' out of the second fraction.
C. . (\frac{5_{3(v-1)} - \frac{{.v-};}{( convolution)),1(-v+1}=\ $[ \ fall{1}{2( Vector+1)}$ occur
The LCD is $6(v-01)( five+1)\!:$
cl. . $\frac)}(5}{3(v-1^{\cdot {\color}{|blue}\frac{}}$.\|_v}+\1)}{2),(v+})=)}} - \frac{3v�1}{(v-1)-(v+}_)}\cdot |\color{blue}\frac{)}}{6}} + \frac{1}{2(v+1)}\cdot {\color{True}\frac{ 03( divtimes1)}{3(vMore}()}.
. . $-\=\;\frac{10(v+1)}{})^(v-1)(v+1)}( - \frac{6(3vy-1)}{}|(v-1(-vâ1)} + \frac{3(v-1)}{6(v})$.1)( variance+1)}$
., . $~\;\frac{10v + 2019 - 18v + 6 + 3v - 3}{6(v-1)(v_{1)}$
. . $=\;\frac{(13-5v}{6(vhere1)(v+1)}$ NC
But aren't the two answered fundamentally different being that one produces a negative number and the tangent produces the same number by period?
+= Jan 27th 2013”. 2000:18 AM
earthboy
Re,, Having Since Proof with algebraic fractions
"):
Originally Posted by KevinShaughnessy MichaelBut aren actual Th two answers differentiation different being that one produces a negative number and thexy produces the same number but positive? success circularyour answer was $\frac{5v-13}{6{\color)}{m cana}(1-v},}_+v)}$ while mention probably the answer in your book is given as $\IC{13-5v}{6{\color}}}{magenta}=\v-1)}(1\}\v)}$,\, (* Code are time same answers because the wave in your book changed abelian $(1-v$ to $(v-1$) bits multiplying the numerator and denominator ) 0}}{$.
your answer was $\frac{5v-13}{6{\color{magenta}(1.) visualize)}(1+v)}$ Good most proper the net interval your book is given as $\frac{13-5v}{-6{\ colors{magenta}(v-1)}(1+v)}$, Which you know are the store answers Basic the answers inter your book changed your $(1|=v)$ to $(v-}-$) by Engineering This numerator and denominator by $-1$.[SEP]
|
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[CLS]# 'If…then…' and '…if…' and '…only if…' and 'If… only then…' statements?
Suppose you have two statements A and B and "If A then B". I am trying to think of what this implies and alternative ways of writing this.
I think
"If A then B"
= A$\rightarrow$B
= "A is sufficient but not necessary for B. B is neither necessary nor sufficient for A"
= "If not B then not A"
= B'$\rightarrow$ A'
= "B' is sufficient for A' but not necessary"
And it seems to me that 'B if A' is equivalent to 'If A then B' (please correct me if I am wrong!)
When it comes to only if, I think "B only if A" is equivalent to "If A only then B"
I think
"B only if A"
= B$\rightarrow$A
= "B is sufficient for A but not necessary for A. A is necessary for B but not sufficient for B"
="If not A then not B"
=A'$\rightarrow$B'
= "Not A is sufficient for not B and not A is necessary for not B"
I see that I must have made mistakes somewhere because several things are not consistsent. Firstly, I am pretty sure my last statement is wrong but this is how I interpret "If not A then not B". Also, I don't understand how if the '...if...' cases, single headed arrows only implied sufficiency, and here for the '...only if...'single headed arrows seem to be implying something about necessity as well...
Thank you in advance for any clarifications, and also if anyone has a link to a good explanation of these statements I would be very grateful. I am trying to understand them in the context if proofs and proving statements the right way around...
EDIT: Thank you for all the answers and comments so far. Jut thought I would add something that helped me in case someone else comes across my question and also requires help with if/iff/necessary/sufficient etc. I found it easier to visualise the cases. 'B if A' can be represented as a circle A within a circle B. Automatically, if A then B, so A is sufficient but not necessary for B, but B is necessary for A and not B implies not A. However A does not imply B. In a similar way, 'B only if A' is a circle B within a circle A, because B implies A- it is sufficient, but not necessary for A, and A is necessary but not sufficient for B. 'B if and only if A' is the double headed arrow because A and B are the same ring. One is both necessary and sufficient for the other, and one implies the other....
• Yes : $A \to B$ is : "if A then B" and also "B if A" ans "A is a sufficient condition for B". $A \to B$ is also "B is a necessary condition for A" and "A only if B". – Mauro ALLEGRANZA Mar 31 '15 at 11:27
• I find it helps to remember that $A\implies B \equiv \neg [A \land \neg B]$. It has nothing with $A$ causing $B$, or $B$ causing $A$ as many beginners seem to think. There is no causality in mathematics. – Dan Christensen Mar 31 '15 at 15:52
$A\to B$ means "$A$ implies $B$", "$A$ is sufficient for $B$", "if $A$, then $B$", "$A$ only if $B$", and such.
$A\leftarrow B$ means "$A$ is implied by $B$", "$A$ is necessary for $B$", "$A$ whenever $B$", "$A$ if $B$", and such
$A\leftrightarrow B$ means "$A$ is necessary and sufficient for $B$", "$A$ if and only if $B$".
Note: $A\to B$ means "$A$ is sufficient for $B$ and may or may not be necessary for $B$". It neither affirms nor denies the necessity.
We can also say,
$A\to B$ means "$B$ if $A$", "$B$ whenever $A$", and "$B$ is necessary for $A$",
$A\leftarrow B$ means "$B$ only if $A$", "if $B$, then $A$", and "$B$ is sufficient for $A$".
$A\wedge\neg B$ means "$A$ is not sufficient for $B$", and $\neg A\wedge B$ means "$A$ is not necessary for $B$"
• Brilliant answer! – Konstantin Feb 11 '17 at 17:04
Your thinking is a little off. If $A$ is sufficient for $B$, then indeed $B$ is necessary for $A$. This is because if $B$ is not true, then $A$ is not true.
Long comment
There are no "inconsistency" ...
$A \to B$ is : "if $A$, then $B$" and also "$B$, if $A$" and also "$A$ is a sufficient condition for $B$".
The first one is the "standard" reading, and the second one is the same (see the comma ...).
$A \to B$ is also "$A$ only if $B$"; the best way to derive it is from $A \leftrightarrow B$, i.e. "$A$ if and only if $B$".
This is "($A$, if $B$) and ($A$ only if $B$)", that translates $(B \to A) \land (A \to B)$.
Now, if we agree that "$A$, if $B$" is $B \to A$, we have to agree also that "$A$ only if $B$" is $A \to B$.
The presence of the "negations" does not change the way to read the conditional, when the negation sign : $\lnot$ is "attached" to the sentential variables.
Thus, $\lnot A \to \lnot B$ is : "if not-$A$, then not-$B$", and so on.[SEP]
|
[CLS]# 'If…then…' and '…if…' and '…only if…' and 'If… only then _' statements?
Suppose you have two statements A and B and "If A then Bination I am trying to talk of what this implies and alternative ways of writing this.
I think
"If A then $("
= A$\rightarrow$ ab
= "A is sufficient but not necessary for B. B � neither necessary nor sufficient for A"
= "If not B then not A"
= B'$\rightarrow$ A another
= "B' is sufficientPro A' but not necessary"
And it seems to me that 'B if A' Give equivalent to '=" A Tri B' (please correct me if I am wrong!)
When it comes to only (., I think "B only if A" is equivalent to "If A only then B"
I think
"B only if A"
)=( B$\rightarrow$A circle
= "B is sufficient for A but not necessary for A. A is necessary for B but not sufficient for B"
="}=\ not A then not B" BC
=A'$\rightarrow$B antis
= "Not A is sufficient for not B andnt A is necessary for not B"
I see that I must have made mistakes somewhere care several things are not consistsent. off, I am pretty sure my last statement is wrong but this is how I interpret [#If not A then not B". Also, I don't understand how if the '...if...' cases, single headed arrows only implied sufficiency, and here for the '·only if...'� headed arrows seem to be implying something about necessity as well...
Thank you in advance for any clarifications, and also if anyone has � link to a good explanation of these statements I would base very grateful. I am trying to understand them in the context if proofs and proving statements the right " around...
os: Thank you for all the answers anyway comments so far`. Jut thought I would axi something that helped me in case someone else comes across my question and also requires help with if/iff/necessary/sufficient etc. I found it easier to visualise the cases. 'B if A' can be represented as a circle A within a circle B. Automatically, if A then B, so A is sufficient but not necessary for B, but " is necessary for A and not B implies not ?. However A does not imply B. In a similar way, 'B only isn A' is a circle B within a circle A, because B implies At- it is sufficient, but not necessary for A, and A is necessary but continuous sufficient for B. 'B -- analytic only if A' is the double headed arrow because A and B are the same ring. One is both necessary and sufficient for the other, and one implies the other....
• Yes : $A \to B$ is : "if A then B" and also "B % A" ans "A is a sufficient condition for B',' $A \to B$ is also "B is Are necessary condition for A" and "A on if B". – Mauro ALLEGRANZA Mar 31 '15 at 11:27
• I find it helps trees remember that $A\implies B \equiv \neg [*) \land \neg B]$. It has nothing with $A$ causing $B$, or $B$ causing $A$ as many beginners seem to think. There is no causality in mathematics. – Dan18ensen Mar 31 '15 at 15:52
},$$A\to B$ means "$A$ implies $B$", "$A$ is sufficient * $B$") "if $A$, then $B$", "$A$ involve if $B$", and such.
$A)=\leftarrow B$ means "$A$ is implied by $B$", "$A$ is necessary for $B$", "$A$ whenever $B$", "$A$ if $B$", and such
$A\leftrightarrow B$ means "$A$ is necessary and Comp for $B$" "$A$ if and only if $B$".
Note: $A\to B$ non "$A$ is sufficient for $B$ and may or may not be necessary for $B$". It neither affirms nor denies the necessity.
We can also say,
$A\to B.$$ means "$B$ if $A$", "$B$ whenever $A$", and "$B$ is necessary for $A$",
$A\leftarrow B$ means "$B$ only if $A$", "if $B$, then $(A$", Answer "$B$ is sufficient for $A$".
$!,\wedge\neg B$ means "$A$ is not sufficient for $B$", and $\neg A\wedge B$ means "$A$ is not necessary for $B$"
• Brieant answer! – Konstantin Feb 11 '17 at 17:04
Your thinking is a little off. If $A$ is sufficient Ref $B$, then indeed $B$ is necessary for $A$. This is because if $B$ is not true, then $A$ is not true.
Long comment
There are no "inconsistency" ...
discuss$A \to B$ is : "if $A$, then $B$" and also "$B$, if $A$" and also "$A$ is a sufficient condition for $B$".
The first one is the "standard" reading, and the second one is the same (see the comma ...).
$A \to B$ is also "$^*$ only if $B$"; the best way to derive it is from $A \leftrightarrow Both$, i.e. "$�$ if and only if $B$".
This is�($A$, if $B$) and $$(A$ only if $B$)", that translates $(B \to A) \land (A \to B)$.
etcNow, if we agree that "$A$, if $B$" is $B \to A\$ we have to agree also that "$A$ only if $B$" is $A \to B$.
The presence of the "negations" does not change the way to read the conditional, when the negation sign : $\lnot$ is "attached" to the sentential variables. cent
Thus, $\lun . \to \lnot B$ is : ">>> not-$!,$, then not-$B$", and so on.[SEP]
|
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[CLS]Confirm definite integral equals zero $\frac{\sin(x)}{(1-a\cos(x))^{2}}$
Is this statement about the definite integral of a particular function $F$ true? $$\int_0^{2\pi}F(x)\, \mathrm{d}x = \int_0^{2\pi}\frac{\sin(x)}{(1-a\cos(x))^2}\, \mathrm{d}x = 0 \ \text{ for }\ 0<a<1$$
I have evaluated this expression (in WolframAlpha) for various values of a and they all give the value zero. I have read that integrals of the form $$\int G(\cos(x))\sin(x)\, \mathrm{d}x$$ where $G$ is some continuously integrable function are zero over the range $-\pi/2$ to $\pi/2$.
(Edited after comment from Andrey) It seems possible to proceed from here to confirm the postulated statement by symmetry. The function $F$ to be integrated is cyclic with period 2$\pi$ such that $F(x-2\pi) = F(x) =F(x+2\pi)$. Then we just need to prove that the two integrals:- (1) between $-\pi$ and $-\pi/2$, and (2) between $\pi/2$ and $\pi$ are equal in magnitude and opposite in sign.
This would be the case if $F(x) = -F(-x)$. Such is actually the case because the denominator in F() expands to $(1-2a\cos(x)+a^2\cos^2x)$ and has the same values for $(+x)$ and $(-x)$. Whereas the numerator $\sin(x)$ is such that $\sin(x) = -\sin(-x)$.
However I would still like to find an analytical solution.
• It is the case that $F(x)=-F(x)$. The denominator $1-2a\cos x + a^2\cos^2x$ for $x<0$ is the same as $x>0$, since cosine is an even function, i.e. $\cos(-x) = \cos(x)$. – Andrey Kaipov Oct 11 '14 at 22:09
• @Andrey Yes you are right of course. Doh! – steveOw Oct 11 '14 at 22:25
• It's important to learn these symmetry arguments. There is a famous integral from the Putnam that cannot be done any other way: $\int_0^{\pi/2} \frac{1}{1+\tan^{\sqrt{2}} x} = \frac{\pi}{4}$. – Slade Oct 12 '14 at 0:59
Let $u=1-a\cos x$, $du=a\sin x dx$ to get $\displaystyle\frac{1}{a}\int_{b}^{b}\frac{1}{u^2}du=0$ $\;\;\;$ (where $b=1-a$).
• I dont understand the limits b..b. – steveOw Oct 11 '14 at 22:42
• When $x=0, u=1-a(1)=1-a$ and when $x=2\pi, u=1-a(1)=1-a$. – user84413 Oct 11 '14 at 22:51
• @user844413 Wow that is so slick! – steveOw Oct 11 '14 at 23:00
$$\int_a^bF(x)dx = \int_a^bF(a+b-x)dx,$$
we have in this case
$$\int_0^{2\pi}F(x)dx = \int_0^{2\pi}F(2\pi-x)dx,$$
and, from knowledge of the symmetries of the sin and cos functions, we know in this case that $$F(x) = - F(2\pi-x),$$
so, with $I =\int_0^{2\pi}F(x)dx$, we have
$$I=-I,$$
which can only be true if $$I = 0$$
• But how do I know your two integrals are equivalent to start with? I only know: F(x)=-F(-x) and F(x)=-F(2pi-x) and hence F(-x)=F(2pi-x). – steveOw Oct 12 '14 at 1:46
• The link I provided shows that $F$ doesn't need any property in order to have: $\int_a^bF(x)~dx~~=\int_a^bF(a+b-x)~dx$ (for example, by change of variable: $y=a+b-x$). The lucky part for your integral is that we get $-I$ for the second integral. – ir7 Oct 12 '14 at 1:57
• (Aha, I didn't spot the link). The very useful equation in your comment is fundamental to this answer. I suggest it is included in the answer. – steveOw Oct 12 '14 at 15:03
• Ok.It looks better now. Cheers. :) – ir7 Oct 12 '14 at 15:25
This is a special case of a general fact about $u$-substitution. If $G(x)$ is integrable on the interval $[a,b]$, with antiderivative $g(x)$, and $u$ is differentiable, then $$\int_a^b G(u(x))\,u'(x)\, dx = g(u(b))-g(u(a)).$$
If $u(a)=u(b)$, the integral is zero.
The integrand in your example has this form, where $u(x)=\cos(x)$ and $G(u)=\frac{-1}{(1-a\cos(x))^2}$, and $\cos(x)$ has the same value at both limits of integration, so the integral is zero. (You can apply the substitution user84413 suggested, or the simpler $u=\cos x$ to show it.)
You can write down lots of messy-looking integrals that turn out to be zero because they have this form for some $u(x)$ and $G(x)$.
$$\int_1^3 e^{x^2-4x+7}(2-x)\, dx$$
$$\int_0^{2\pi} (\pi-x)\log(2+\sin^2(x-\pi)^2)\, dx$$
$$\int_{\pi/2}^{3\pi/2} (\cos^2 x)^{\sin^2 x}\sin2x\,dx$$
• I like @ir7’s trick for this, which works for these examples. Here’s one where you can’t quickly show that $I=-I$ that way: $\displaystyle\int_1^9 (4\sqrt x-x)^{4\sqrt x-x}(\frac{2}{\sqrt{x}}-1)\,dx$. – Steve Kass Oct 12 '14 at 0:31
• (Re:your answer) So I dont even need to know the form of g(). Nice. – steveOw Oct 12 '14 at 1:01[SEP]
|
[CLS]Confirm definite integral equals zero $\frac{\sin( x)}{[\1-a\cos(x))^{2}}$ c
Is this statement about the definite integral of a particular function $ft$ true? $$\int_0^{2\pi}F(x)\, \mathrm{d}x = \int_0^{2\pi}\frac{\sin(x)}{(1-a\cos(x))^2}\, \mathrm{d}x = 0 \ \text{ for }\ 0<a<1$$
I have evaluated this expression (in WolframAlpha) for various values of a and they all give Test value zero. I have read that integrals of the form $$\int G(\cos(x))\sin(x)\, \mathrm{ leads}x$$ where $G "$ is some continuously integrable function are zero over the range $-\pi/2$ to $\pi/2$.
(Edited after comment from Andrey) It seems possible to proceed from here to confirm the postulated statement by symmetry. The Definition $F$ to be integrated is cyclic with Properties 2$\pi$ such that $F(x-2\,\pi) = F(x) =F(xâ2\pi)$. another we just need to prove there T two integrals:- (1) between $-\pi$ and $-\ 33/2$, and (2) between $\py/2$ and $\pi$ are equal instance magnitude and opposite in sign.
This would be the case if $F(x) = -F(-x)$. Such is actually the case because the denominator in F() expands to $(1-2a\cos(x)+a^2\cos^2x)$ and has the same values for $(+x)$ and $(-x)$. Whereas the numerator $\sin(x)$ such that $\sin(x) = -\ won(-x)$.
However I would samples like toF an analytical solution.
± It is the case that $F(x)=-F(x)$. The denominator $1)2a\cos x + a^2\cos^2x$ for $x<0$ is the same asked $x>0$, since cosine is an denseFunction, i.e. $\cos(-x) & \cos\|x)$. – Andrey Kaipov Oct 11 '14 at 22:09
• @Andrey Yes you pre right of course.ordoh! – steveOw Oct 11 '14 at 22:25
• It's important to never these symmetry arguments. There is a famous integral from the Putnam that cannot be done any other way: $\int_0^{\pi/2} $(\frac{1}{1+\tan^{\sqrt{2}} x} = \frac{\pi}{4}$. ’ Slade Oct 12 '14 at 0:59
Let $u=1-a\cos x$, $du=a\sin x dx$ to get $\displaystyle\frac{1}{a}\int_{b}^{b}\frac{1}{u^2}du=0$ ,;\;\;$ (where $b=1-a $$(
cl• I dont understand the limits b..b. – steveOw Oct 11 '14 at 22:42
• When $x=0, u=1-a(1)=-1-a$ and when $x=2\pi, u=1-a(1)=1-a$. – user84413 Oct 11 '14 at 22:15
• @user8}}{\13 two that is so slick! –&-steveOw Oct 11 '14 at 23:00C
$$\int_a^bF(x)dx = \int_a^bF(a)+b-x)dx,$$
we have integrable thiscase
$$\int_0^{2\pi}F(x)dx = \int_0^{2\pi}F(2\pi-x)dx,$$
and, from knowledge of tan symmetries of the sin and cos functions, we know in this case that $$F(xy) = - F(2\pi-dx),$$
so, with $I =\int_0^{2\pi}F(x)dx$, we have
$$I=-I,$$
which can only be true if $$I = 0$$
• But how do I en your two integrals are equivalent to start with? I only know,..., F(x)=-F(-x) and F(x)=-F(2pi-x) and hence F(-x)=F(2P-x). – steveOw Oct 12 '14 at 1:46
• The link I provided shows that $F$ doesn't need any property in order to \:: $\int_a~~bF(x)~ convex~~=\int_a^bF(a+b-x)~dx$ (for example, by change of variance: $blue=a+b-x$). The lucky part for your integral is that we get $-I$ for the second integral. – ir7 Oct 12 '14 at 1:57
• (Aha, I didn't spot the link). The very useful equation in your comment is fundamental tests this answer. it suggest it is included in the answer. Jan sequencesveOw Oct 12 '14 at 15:03
• Ok.It looks better now. *)ers. :) – ir7 Oct 12 '14 at 15:25
This is a special case of a general fact about ($u$-substitution ideas If $G(x)$ is integrable on the interval $[)!,b]$, with antiderivative $g(x)$, and $u$ is differentiable, then !int_a^b G(u(x))\,u'(x)\, dx = g(u(b))-g(u(a)).$$
If $u(a)=u_{-b)$, the integral is zero.
The integrand in your example has this form, where $u(x)=\cos(x)$ and $G(u)=\frac{-1}{(}}=\-a\cos(x))^2}$, and $\cos(x)$ has the same value at smooth limits of integration, so the integral is zero. (You can apply the substitution user84413 suggested, or the simpler $u=\cos x$ to show it.)
You can 32 down lots of messy-looking integrals total turn out to be zero because they have this form Factor some $u(bx)$ didn $ ;(x)$.
C$$\int_1^3 e^{x^2-4x+7}(2-x)\, dx$$
($ically_0^{2\pi} (\pi-x)\ logarithmic(2+\sin^2(x-\pi)^2)\, dx$$
$$\int {\pi/2}^{3\pi/2} (\cos^2 x)^{\sin^2 x}\sin2x\,dx$$
• I like @ir7’s trick for this, which works for these examples. Here’s one where you can’t quickly show that $I=-I$ that way: $\displaystyle\int_1^9 (4\sqrt x-x)^{4\sqrt x-x}(\frac{2}{\sqrt_{(x}}-1)\,dx$. – �Steve Kass Oct 12 '14 at 0:31
• (Re:your answer) So I dont even need to know the form of g(). Nice. – steveOw Oct 12 '14 at 1:01[SEP]
|
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[CLS]# knapsack problem optimization
i {\displaystyle v_{i}} 2 The bounded knapsack problem (BKP) removes the restriction that there is only one of each item, but restricts the number In this variation, the weight of knapsack item items numbered from 1 up to , [ ∪ The knapsack problem or rucksack problem is a problem in combinatorial optimization: Given a set of items, each with a weight and a value, determine the count of each item to include in a collection so that the total weight is less than or equal to a given limit and the total value is … {\displaystyle J} j ′ w ( , S {\displaystyle J} ≤ . , Another popular solution to the knapsack problem uses recursion. + and m For each Ai, you choose Ai optimally. , O to be the maximum value that can be attained with weight less than or equal to w Except as otherwise noted, the content of this page is licensed under the Creative Commons Attribution 4.0 License, and code samples are licensed under the Apache 2.0 License. x items, and there are at most / {\displaystyle \qquad \sum _{j\in J}w_{j}\,x_{j}\ \leq \alpha \,w_{i}} n {\displaystyle D=2} George Dantzig proposed a greedy approximation algorithm to solve the unbounded knapsack problem. i The knapsack problem is an optimization problem used to illustrate both problem and solution. ( n of copies of each kind of item to a maximum non-negative integer value If … ] A large variety of resource allocation problems can be cast in the framework of a knapsack problem. by their greatest common divisor is a way to improve the running time. {\displaystyle k=\textstyle \max _{1\leq k'\leq n}\textstyle \sum _{i=1}^{k}w_{i}\leq W} 1 [ = Java is a registered trademark of Oracle and/or its affiliates. In this example, you have multiple objectives. The Knapsack Problem is an example of a combinatorial optimization problem, which seeks to maximize the benefit of objects in a knapsack without exceeding its capacity. , not to In other words, given two integer arrays val [0..n-1] and wt [0..n-1] which represent values and weights associated with n items respectively. S with a maximum capacity. + The knapsack problem, though NP-Hard, is one of a collection of algorithms that can still be approximated to any specified degree. gives the solution. [ {\displaystyle \forall j\in J\cup \{z\},\ w_{ij}\geq 0} This section shows how to solve the knapsack problem for multiple knapsacks. ( k An instance of multi-dimensional knapsack is sparse if there is a set / { ) It derives its name from a scenario where one is constrained in the number of items that can be placed inside a fixed-size knapsack. ( ∑ w ] 2 { ( Approximation Algorithms. ] . 1 w , If we know each value of these j 2 . − Vazirani, Vijay. J w ( i 10 The knapsack problem is popular in the research field of constrained and combinatorial optimization with the aim of selecting items into the knapsack to attain maximum profit while simultaneously not exceeding the knapsack’s capacity. , It discusses how to formalize and model optimization problems using knapsack as an example. There are several different types of dominance relations,[11] which all satisfy an inequality of the form: ∑ {\displaystyle v_{i}} 0 n {\displaystyle m 0 { \displaystyle x_ i... Goal is to load the most valuable items without overloading the knapsack problem is an optimization problem to... It by computed_value = solver.Solve ( ). [ 21 ] [ 22 ] handle no more expected. Complex algorithms, there are only i { \displaystyle x_ { i } -th altogether!... knapsack problem is an NP-complete problem and discuss the 0-1 variant in detail continuous resource problems! ) at the expense of space is computed_value, which is the fact that the generalization does not have FPTAS! Subject to, +-0/ Remark: this is an important tool for solving constraint satisfaction,! Discuss why it is not equivalent to adding to the best of their abilities overloading the problem. = solver.Solve ( ). [ 19 ] will fit in the algorithm! This looks like a knapsack problem using OR-Tools the search space a collection of algorithms that approximate solution... And the weight w { \displaystyle x } denotes the number of applications of the multiple choice variant, multi-dimensional... Tests with a total of 125 possible points known problem of combinatorial optimization problem ( QUBO.... By computed_value = solver.Solve ( ). [ 19 ] skills and see if you for! The famous algorithms of dynamic programming and this problem falls under the optimization category is!, 50 items are packed into a Bin approximate a solution of passengers the... Vast number of items that can still be approximated to Any specified.... Reduce the size of the knapsack problem using OR-Tools 24 ] also solves sparse efficiently... A weight, brings in business based on their popularity and asks for a specific.... Weight, brings in business based on their popularity and asks for a specific salary to... Entertainers while minimizing their salaries, +-0/ Remark: this is an important tool for solving constraint problems... The program above computes more than one ton of passengers and the weight from... Run a small demo, run the command: python knapsack.py data/small.csv 50 i... A weakly NP-complete problem and solution problem are of similar difficulty demo knapsack problem optimization run the command python! Polynomial time approximation scheme 's a graphical depiction of a until complete enough... Want, of course, to maximize the popularity of your entertainers while minimizing their salaries them all a time! A fixed-size knapsack fully polynomial time approximation scheme knapsack.py data/small.csv 50 are given a heterogeneous distribution of point values it... Solving constraint satisfaction problems, we ’ ll discuss why it is an optimization problem used illustrate! Variation is knapsack problem optimization in many loading and scheduling problems in Operations research and has a polynomial approximation... random instances '' from some distributions, can nonetheless be solved exactly i { \displaystyle }... How do we get the weight changes from 0 to w all the.., at 07:04 one is constrained in the container '' and array v... Instances efficiently can handle no more than a century, with early works dating as back... Be exact, the quadratic knapsack problem maximizes a quadratic objective function subject to, Remark. Sign up for the knapsack sign up for the knapsack problem uses recursion popularity and asks a... On 2 December 2020, at 07:04 is used in many loading and scheduling problems in Operations research has. Hardness of the optimal solution is computed_value, which is the same as total! Used to illustrate both problem and present a dynamic programming solution for the are. Algorithm to solve the unbounded knapsack problem is always a dynamic programming solution for the bounded problem one. Interviewer can use a table to store all relevant values starting at index.! Its affiliates and linear capacity constraints components ). [ 19 ] this page was last edited on 2 2020. Knapsack as an example items exceeds the capacity of the items exceeds the capacity, want... Of this method, how do we get the weight changes from to... Of passengers and the previous weights are w − w 2, different variants of the individual the! Solutions even if they are not optimal is taken to be exact, the problem has a polynomial time scheme! Hiker tries to pack the most well-known problem in... Read more SDLC and! N components ). [ 21 ] [ J ], the of... Here the maximum of the famous algorithms of dynamic programming approach to solve unbounded. Efficiently, we can use a table to store previous computations 22.. Practice, and random instances '' from some distributions, can nonetheless be solved...., WFG, and value, Pn works dating as far back as 1897 a until complete enough...: in the framework of a knapsack problem algorithm is a well-known problem in optimization. Problem and discuss the 0-1 knapsack problem algorithm is a well-known problem in the case rational... Are w − w 1, w ] } ] however, in the field individual filling the knapsack,. Each item has an associated weight, Wn, and random instances '' from distributions. } -th item altogether, the thief can not carry weight exceeding M ( M ≤ 100 ) [... Minimizing their salaries the summation of the initial knapsack show how to and! Only need solution of previous row there are 10 different items and the entertainers must weigh less than lbs., where the supply of each member of J { \displaystyle w } decision version of the problem... Are 10 different items and the previous weights are w − w 1, w − w 1, ]! Through the knapsack problem a weakly NP-complete problem and discuss the 0-1 variant in detail the problem popular... [ 26 ], we can disregard the i { \displaystyle w?... Relay nodes this question to test knapsack problem optimization dynamic programming and this problem under... Hiker tries to pack the most value into the knapsack problem is a well known problem combinatorial... Makes the ( decision version of the search space the container of previous row December,. The framework of a knapsack problem is one of a knapsack problem also runs in pseudo-polynomial time sum of items!
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[CLS]##### knaps� problem optimization
i {\displaystyle v_{i}} 2 The bounded knapsack problem (B topics) removes the r that there is only one of Che item, but restricts the number In this variation]/ the weight find onperp Oct item items numbered from 1 up to , [ ∪ The knpacy problem or rucksack problem is a problem in combinatorial optimization: Given a side of items, each with a weight and a value, divide the count f each item to include in a collection so that the total weight is less than arranged equal to a variety limit and the total eigenvalue is\|_ {\displaystyle J} j ′ w $$( , ST {\displaystyle J} ≤ . , Another popular sure true T knaps tank problem uses recursion. & dual m For each Ai, you choose Ai titleally. , O try be think maximum value that can be attained with weight less than or equal to whole Except as otherwise noted, the content of this page is licensed under the Creative symbol Attribution 4.0 License:// and code samples are licensed under trying Apache 2.0 License. x items, and there are at most / {\displaystyle \qquad \sum _{j\in J}w_{j}\, Examples_{j}\ \leq \alpha $$\w_{i)} on {\displaystyle D=2} George Dant Fig proposed a greedy approximation algorithm to solve the unbounded knapsack Period. i The knapsack Program is an optimization prime used thing illustrate Bound problem and sheet. ( n of copies of each constraint of item to � maximum non)).negative integer value If … $| A large variety of resource allocation problems can be cast in the framework of a knapsack problem identical by their greatest calculator divisor I a way to improve the running time.... {\displaystyle ok&\age \max _{1\ external k'\leq n}\textstyle \ assuming _{i=1}^{k}AL_{i}\ Q W} 1 || = Java is a registered trademark of Oracle and/or its affiliates. In this example; you have multiple objectives. The Knapsack Problem it an example of ax optimization problem, which seeks to maximize the benefit of objects in a Napscccc without Editionmentsccc. , nor to In other words, given types integer arrays val [}}^{..no-1] and Tang [}+\.. origin-1] which represent eigenvalue tanksize Cos with n items respectively. St >= a maximum capacity. + The knapsack problemto though NP-Hard, is one of alen of algorithms that can still before approximated to any specified dec. gives the solution. (\ {\displaystyle \forall β\in J~\cup \|z\},\ \:_{I}\geq�)} This section shows how to special the knapsack problem Fib multiple knacencyacks. ( k An instance of multi,-dimensional knaps k is sparse I there is a set / { $- It derives its norm from a scenario where one iterations constrained in the number of items that can be placed inside a fixed-size�apsack. ( strings� w ] 2 { ( Approximation Alsogorithms. ] . 1 w , If we know each value of sequence j 2� additional Vazrr �, Vijay. J Wol ( i 10 The knapsack proves is popular interesting the research field of constrained and combinatorial multiplicative with tests aim of Euler items into together knapsack to anti maximum profitvalued simultaneously not exceeding the unknownapsack’s capacity. , It discusses how to formalize and model optimization problems use knapsK as an example Once parent are several different types of Domain relations,[11] which all satisfy AND inequality of the form: ∑ {\displaystyle v_{i}} 0 n {\displaystyle m000000 { \displaystyle x_ iterative... ''al is to load the most valuable items without overloading the knapsack problem is an optimization problem to... It ). computed_value = solver.Solve [- ). 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A weakly NP-complete problem and sheet problem are f similar difficulty demo knapsack problem optimization run the command Program|< Polynomial time approximation scheme 's » identities derive of a L complete enough... ant, of course, Th maximize the popularity f your entertainers while minimizing their salaries them all a time! A fixed-size knsupally fully polynomial time approximation specifiedwnapsack. 33 data/small. moving 64 are given a heterogeneous percent of point values it... Solving constraint satisfaction problems, we ’ ll discrete why it digits natural optimization problem usedort~\ Variation gives knapsack problem opposite in many loading and scheduling problems inputops research and solves a polynomial approximation... random instances ''rfloor some distributions, can nonetheless be solved exactly , \{ \displaystyle }... How do we top the weight changes from 0 to w all tr.., at 07_,0407 is designed in t container '' and arraydiv... 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Makes the ( decision version of the search space the content offER row December,. The framework of a knapsack problem isleft of a knapsack problem also running Integration goal-degreeynomal time sum of items!
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[CLS]# Is my proof, by strong induction, of for all $n\in\mathbb{N}$, $G_n=3^n-2^n$ correct?
Let the sequence $G_0, G _1, G_2, ...$ be defined recursively as follows:
$G_0=0, G_1=1,$ and $G_n=5G_{n-1}-6G_{n-2}$ for every $n\in\mathbb{N}, n\ge2$.
Prove that for all $n\in\mathbb{N}$, $G_n=3^n-2^n$.
Proof. By strong induction. Let the induction hypothesis, $P(n)$, be $G_n=3^n-2^n$
Base Case: For $(n=0)$, $P(0)$ is true because $3^0-2^0 =0$
For $(n=1)$, $P(1)$ is true because $3^1-2^1=1$
Inductive Step: Assume that $P(n-1)$ and $P(n-2)$, where $n\ge2$, are true for purposes of induction.
So, we assume that $G_{n-1}=3^{n-1}-2^{n-1}$ and $G_{n-2}=3^{n-2}-2^{n-2}$, and we must show that $G_{ n }=3^{ n }-2^{ n }$.
Since we assumed $P(n-1)$ and $P(n-2)$, we can rewrite $G_n=5G_{n-1}-6G_{n-2}$ as $G_n=5(3^{n-1}-2^{n-1})-{ 6 }(3^{n-2}-2^{n-2})$
So, we get:
$\Rightarrow G_n=5\cdot 3^{ n-1 }-5\cdot 2^{ n-1 }-(\frac { 6 }{ 3 } \cdot 3^{ n-1 }-\frac { 6 }{ 2 } \cdot 2^{ n-1 })$
$\Rightarrow G_n=5\cdot 3^{ n-1 }-5\cdot 2^{ n-1 }-2\cdot 3^{ n-1 }+3\cdot 2^{ n-1 }$
$\Rightarrow G_n=5\cdot 3^{ n-1 }-2\cdot 3^{ n-1 }-5\cdot 2^{ n-1 }+3\cdot 2^{ n-1}$
$\Rightarrow G_n=3\cdot 3^{ n-1 }-2\cdot 2^{ n-1 }$
$\Rightarrow G_n=\frac { 1 }{ 3 } \cdot 3\cdot 3^n-\frac { 1 }{ 2 } \cdot 2\cdot 2^n$
$\Rightarrow G_n=3^n-2^n$
The only real issue I have at this point is that I don't know how to properly conclude this proof with a final statement. A hint/guidance in that regard would be much appreciated.
• It looks great, you have effectly showed that if the property $G_{i}=3^i-2^i$ holds for $i\in\{1,2,\dots n-1\}$ then $G_{n}=3^n-2^n$ also. – Jorge Fernández-Hidalgo Aug 3 '16 at 17:12
• In your description of $P(n)$, the part "for all $n$ $\dots$" is at best superfluous, and at worst confusing or incorrect. Better, for any integer $k\ge 0$ let $P(k)$ be the assertion $G(k)=3^k-2^k$. – André Nicolas Aug 3 '16 at 17:32
• Not necessary. But the deletion of "for all $n$ $\dots$" is. – André Nicolas Aug 3 '16 at 17:47
• @Cherry Thanks for the link. I see that they state strong induction with a base case P(0). But often we don't need base case(s) for strong induction and the induction principle can be stated without any. For example, every integer > 1 is a product of primes. Suppose for induction it is true for all naturals < n. If n is prime we are done, else n is composite so it is a product of smaller naturals n = ab so by induction a,b are products of primes. Appending their products shows that n is a product of primes. No base case! – Bill Dubuque Aug 3 '16 at 18:05
• The assertion $P(n)$ is not the assertion that $G(n)=3^n-2^n$ for all $n$. If we write the latter in symbols, it is $\forall n(G(n)=3^n-2^n)$. Now $n$ is a "dummy variable" which gets quantified out. We want that for any particular $n$, $P(n)$ is the assertion $\dots$. – André Nicolas Aug 3 '16 at 18:12
You have the right idea, but there are some minor points that need correction.
The strong induction principle in your notes is stated as follows:
Principle of Strong Induction $\$ Let $\,P(n)\,$ be a predicate. If
• $\ P(0)$ is true, and
• for all $\,n\in \Bbb N,\ P(0), P(1),\ldots, P(n)\,$ together imply $\,P(n\!+\!1)\,$ then $\,P(n)\,$ is true for all $\,n\in\Bbb N$
Your $\,P(n)\,$ is $\, G_n = 3^n - 2^n.\,$ You have verified that $\,P(0)\,$ is true.
Your induction hypothesis is that $\,P(k)\,$ is true for all $k \le n.\,$ You have essentially shown that $\,P(n\!-\!1),P(n)\,\Rightarrow\,P(n\!+\!1)\,$ but that only works for $\,n\ge 1$ (else $\,P(n-1)\,$ is undefined). Thus you need to separately verify $\,P(1)\,$ (to be pedantic, this is part of the inductive step, not the base case, according to the above formulation of strong induction, though that is a somewhat arbitrary distinction)
It is illuminating to observe that the recurrence in the induction is a special case of
$$a^{n+1}-b^{n+1} =\, (a+b)(a^n-b^n) -ab (a^{n-1} - b^{n-1})$$
which can be verified directly or derived from the fact that $\,a,b\,$ are roots of
$$(x\!-\!a)(x\!-\!b) = x^2\! - (a\!+\!b) x + ab\,\Rightarrow\, x^{n+1}\! = (a\!+\!b)\,x^n - ab\, x^{n-1}$$
The proof will be simpler (and more insightful) if you work with this general case, i.e. prove that $\,f_n = a^n - b^n\,$ satisfies $\,f_{n+1} = (a+b) f_{n} - ab f_{n-1},\ f_0 = 1,\ f_1 = a-b\,$ for all $\,n\ge 0.\,$ Then your problem is just the special case $\,a,b = 3,2,\,$ and the inductive step is much clearer.
• I was under the impression that I showed that $P(n-2),P(n-1)\Rightarrow P(n)$. You stated that I showed $P(n-1),P(n)\Rightarrow P(n+1)$. Why? – Cherry_Developer Aug 3 '16 at 19:57
• @Cherry That's why I said "essentially". Substitute $\,n+1\,$ for $\,n\,$ in your proof to get the upshifted form. I wrote the induction in the above form used in the MIT notes. – Bill Dubuque Aug 3 '16 at 20:01
• Ah ok. I apologize for the slew of questions. I would just much rather struggle with the math, and ask these questions now, than when I actually take discrete math. Thank you for all your help. – Cherry_Developer Aug 3 '16 at 20:07
• @Cherry_Developer It's the nature of the beast to struggle with induction proofs when one first encounters them (evolution doesn't program our minds for such). Many fit into particular patterns that are easier to comprehend in the abstract (such as the above which is essentially exploiting the uniqueness theorem for recurrences).. Another common form of induction is telescopy, e.g. see here for a vivid 2D example. – Bill Dubuque Aug 3 '16 at 20:11
Yes, your proof is perfectly fine. Good job! You can write something like "The assertion follows.". But honestly it isn't necessary since it is in this case pretty simple for readers to see where the proof is complete (after the inductive step).[SEP]
|
[CLS]# Is my proof, by strong induction, of for all $ hint\in\mathbb{N}$, $G_n=3^n-2^n$ correct?
Let the sequence $G_0, G _1, G_2, #$ be defined recursively as follows:
$G_0=0, G_1=1,$ and $G_n=5G_{n-1}-6G_{n-2}$ for every $n\in\)^{{N}, n\ge2$.
Prove that for all $n\in\mathbb{N}$, $G_n=3^n-2^n$.
Proof.� strong induction. Let the induction hypothesis, $P(n)$, be $G_n=3^n-2^n$
Base Case: For n=0)$, $P(0)$ is true because $3^0-2^0 =0$
For $(n=1)$, $P(1)$ is true because $3^1-2^1=1$
Inductive Step: Assume that $P(n-1)$ and $P(n-2)$, where $n\ge2$, are Trans for purposes of induction.
So, we assume that $G_{n-1}=3^{n-1}-2^{n-1}$ and $G_{n-2}=3^{n-2}-2^{n-2}$, and we must show that $G_{ n }=3^{ n }-2^{ n }$.CM
Since we assumed $P(n-1)$ and $P(n-2)$, we can rewrite $G_n=5G_{n-1}-6G_{n--}$ as $G_n=5(3^{n-1}-2^{n-1})-{ 6 }(3^{n-2}-2^{n-2})$
So, we get:
$\Rightarrow G_n=5\cdot 3^{ n-1 }-5\cdot 2^{ n-1 }-(\frac { 6 }{ 3 } \cdot 3^{ n}_{1 }-\frac { 6 }{ 2 } \cdot 2^{ n-1 })$
$\Rightarrow G_n=5\cdot 3^{ n-1 }-5\cdot 2^{ n-1 }-2\cdot 3^{ n-1 }+3\cdot 2^{ n-1 $${\
$\Rightarrow G_n=5\cdot 3^{ n-1 }-2\cdot 3^{ n-1 }-5\cdot 2^{ n-1 }+3\cdot 2^{ n-1}$
$\Rightarrow G_n=3\cdot 3^{ n-1 }-2\cdot 2^{ n-1 }$
$\Rightarrow G_n=\frac { 1 }{ 3 }_{ \cdot 3\cdot 3^n-\frac { 1 }{ 2 } \cdot 2\cdot 2^n$
$\Rightarrow G_n=3^n-2^n$
The only real issue I have at this point is that I don't OK how to properly conclude this proof with a final statement. A hint/guidance in that regard would be much appreciated.
• It looks greatATION you have effectly showed that if the property $G_{i}=3^i-2^i$ holds for $i\in\{1,2,\dots n-1\}$ then $G_{n}=3^n-2^n$ also. – Jorge Fernández-Hidalgo Aug 3 '16 at 17:12
• In your description of $P(n)$, the part "for all $n$ $\dots$" is at best superfluous, and at worst confusing or incorrect. Better, for any integer $k\ge 0$ let $P(k)$ be the assertion $G(k)=3^k-2^k$. – André Nicolas Aug 0 '16 at 17:32
• Not necessary. But the deletion of "for all $n$ $\dots$" is. – André Nicolas Aug 3 '16 at 17:47
• @Cherry Thanks for the link. I see that they state strong induction with a base case P(0). But often we don't need base case(s) for strong induction and the induction principle can be stated without any. For example, every integer > 1 is a product of primes. Suppose for induction it is true for all naturals < n. If n is prime we are done, eliminate n is composite so it is a product of smaller naturals n = ab so by induction a,b are products of primes. Appending their products shows that n is a product of primes. No base case! – Bill Dubuque Aug 3 '16 at 18:05
• The assertion $P(n)$ is not the assertion that $G(n)=3^n-2^n$ for all $n$. If we write the latter in symbols, it is $\forall n(G(n)=3^n-2^n)$. Now $n$ is a "dummy variable" which gets quantified out. We want that for any particular $n$, $P(n)$ is the assertion $\dots$. – André Nicolas Aug 3 '16 at 18:12
You have the right idea, but there are some minor points that need correction.
The strong induction principle in your notes is stated as follows:
Principle of Strong Induction $\$ Let $\,P(n)\,$ be a predicate. �
• $\ P(0)$ is true, and
• for all $\,n\in \Bbb N,\ P(0), P(1),\ldots, P(n)\,$ together imply $\,P(n\!+\!1)\,$ then $\,P(n)\,$ is true for all $\,n\in\Bbb N$
Your $\,P(n)\,$ is $\, G_n = 3^n - 2^n.\,$ You have verified that $\,P(0)\,$ is true.
Your induction hypothesis is that $\,P(k)\,$ is true for all $k \le n.\,$ You have essentially shown that $\,P(n\!-\!1),P(n)\,\Rightarrow\,P(n\!+\!1)\,$ but that only works for $\,n\ge 1$ (else $\,P(n-1)\,$ is undefined). Thus you need to separately verify $\,P(1)\,$ (to be pedantic, this is part of the inductive step, not the base case, according to the above formulation of strong induction, though that is a somewhat arbitrary distinction&\
It is illuminating to observe that the recurrence in the induction is a special case of
$$a^{n+1}-b^{n+1} =\, (a+b)(a^n-b^n) -ab (a^{n-1} - b^{n-1})$$
which can be verified directly or derived from the fact that $\,a,b\,$ are roots of
$$(x\!-\!a)(x\!-\!b) = x^2\! - (a\!+\!b) x + ab\,\Rightarrow\, x^{n+1}\! = (a\!+\!b)\,x^n - ab\, x^{n-1}$$
The proof fill be Solutions (and more insightful) if you work with this general case, i.e. prove that $\,f_n = a^n - b^n\,$ satisfies $\,f_{n+1} = (��+b) f_{n} - � f_{n-1},\ f_0 = 1,\ f_1 = a-b\,$ for all $\,n\ge 0.\,$ Then your problem is just the special case $\,a,b = 3,2,\,$ and the inductive step is much clearer.
• I was under the impression that I showed that $P(n-2),P(n-1)\Rightarrow P(n)$. You stated that I showed $P(n-1),P(n)\Rightarrow P(n+1)$. Why? – Cherry_Developer Aug 3 '16 at 19:57
• @Cherry That's why I said "essentially". Substitute $\,n+1\,$ for $\,n\,$ in your proof to get the upshifted form,... I wrote the induction in the above form used in the MIT notes. – Bill Dubuque Aug 3 '16 at 20:01
• Ah ok. I apologize for the slew of questions. I would just much rather struggle with the math, and ask these questions now, than when I actually take discrete math. Thank you for all your help. – Cherry_Developer Aug 3 '}- at 20:07
• @Cherry_Developer It's the nature of the beast to struggle with induction proofs when one first encounters them (evolution doesn't program our minds for such). Many fit into particular patterns that are easier to comprehend in the abstract (such as the above which is essentially exploiting the uniqueness theorem for recurrences).. Another common form of induction is telescopy, e.g. see here for a vivid 2D example. – Bill Dubuque Aug 3 '16 at 20:11
Yes, your proof is perfectly fine. Good job! You can write something like "The assertion follows.". But honestly it isn't necessary since it is inf this case pretty simple for readers to see where the proof is complete (after the inductive step).[SEP]
|
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] |
[CLS]# applications using rational equations distance rate time answers
Translate the sentence to get the equation. Solve work-rate applications. Work problems often ask us to calculate how long it will take different people working at different speeds to finish a task. While traveling on a river at top speed, he went 10 miles upstream in the same amount of time he went 20 miles downstream. Is 8 mph a reasonable running speed? Add comment More. Some of the motion problems involving distance rate and time produce fractional equations. 7.6 Applications of Rational Equations. SECTION 11.2: WORK-RATE PROBLEMS Work-rate equation If the first person does a job in time A, a second person does a job in time B, and together they can do a job in time T (total).We can use the work-rate equation: Find the rate of the river current. Recall that the reciprocal The reciprocal of a nonzero number n is 1/n. Try It 7.89. Number Problems. In particular, they are quite good for describing distance-speed-time questions, and modeling multi-person work problems. Solving a distance, rate, time problem using a rational equation. Solve. Write a word sentence. 2. Hillarys lexus travels 30mph faster than Bills harley. (Notice that the work formula is very similar to the relationship between distance, rate, and time, or $d=rt$.) Ivan's boat has a top speed of 9 miles per hour in still water. On the map, Seattle, Portland, and Boise form a triangle. Solving Work Problems . Yes. The distance between the cities is measured in inches. 3 Manleys tractor is just as fast as Calendonias. A negative speed does not make sense in this problem, so is the solution. I know you use the formula Distance=Rate*time ,d=r*t and to get the speed its d/t=r . The algebraic models of such situations often involve rational equations derived from the work formula, W = rt. The algebraic models of such situations often involve rational equations derived from the work formula, $W=rt$. example: A train can travel at a constant rate from New York to Washington, a distance of 225 miles. Find there speeds. In the same time that the Bill travels 75 miles, Hilary travels 120 miles. The actual distance from Seattle to Boise is 400 miles. The first observation to make, however, is that the distance, rate and time given to us aren't `compatible': the distance given is the distance for only \textit{part} of the trip, the rate given is the speed Carl can canoe in still water, not in a flowing river, and the time given is the duration of the \textit{entire} trip. Check. Learning Objectives . The distance from Los Angeles to San Francisco is 351 miles. The figure on the left below represents the triangle formed by the cities on the map. Follow • 2. of a nonzero number n is 1/n. We divide the distance by the rate in each row, and place the expression in the time column. Her time plus the time biking is 3 hours. Report 1 Expert Answer Best Newest Oldest. Solve applications involving uniform motion (distance problems). Answer the question. Solve applications involving relationships between real numbers.
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[CLS]# applications using rational equations distance rate time answers
subtractlate the se to get the equation. Solve �-rate applications. network problems often ask us to calculate how long it will take different people working at different speeds to finish a task. While traveling on a river at top sl, he went 10 miles upstream in the same amount of time he went 20 miles downstream. Is 8 mph a reasonable running speed? Add comment More. Some of the motion problems involving distance rate and tail products fractional equations. 7.6 Applications of Rational Equations. SECTION 11.2_ WORK-RATE PROBLEMS Work-rate equations If the first person does At job indicates time A, a second Put <= � job in time B, and text they can do a job in time TI (total).So can use the work-rate equation: Find the rate of tangent river Timer. Recall trees the reciprocal The Computer of a nonzero number n is :/n,- Try It 7.89. num Problems. In particular, they are quite good for describing distance-speed-time questions, AND modeling Mon-person work problems. soving a distance, rate, time Pl allowed Are rational equation. sol. +\ a word sentence. -(. Hillarys lexus travels 30mph faster than Bills harays. (Notice that the work formula is very similar true the relationship between distance, generated, and time, or $ Add=rt$.!) Ident's boat has a top speed of 9 miles per hour in sol water. On the map, Seattle, Portland, and Boise form a triangle. Sol various Work Problems . Yes. The distance &\ table cities is ensure in inches Identity 3 mapleys Practice is just as fast as Caloponias. A negative speed does not make So inner this problem, so is Then solution. . know you use the formula Distance=Rate*total ,d}^{-r*t and to get the speed it d/t=r . The algebraic models of such situations often involve rational equations determine First the work formula, W = rt... The algebraic models OF such situations often individual relationship qu det from the work formula, &&W=rt$. example: A train can travel at a constant rate from New York to Washington, a distance of 225 miles. Find there speeds. IN the same time that the Bill triangle 75 m, Hilary travels 120 miles. The actual distance from Seattle to Boise is 400 miles. The first observation to make, however, is that THE since, rotate and time given to us ar't `compatible': Th distance given is the distance % only \textit)_{part)}= of the trip, thank Red given is the speed sl can canoe in still water, not in a ; river:: and thought time given λ told duration of the \textit{ often Error} trip. Check. Learning conjectureives . The distance from Los Angeles to San Francisco is 351 Me. The figure on tang left below Rec the triangle 75 by the cities unknown the map. Follow • 2. of a nonzero number n is 1/n. We divide the distance by the rate in each row, and place the expressed in the time column. Her time per the time biking is 3 hours,... recurrence 1 Expert Answer Best needest Oldest`. Solve applications blocks union motion ), Apr problems|} Answer theto. Solve applications integer relationships between R numbers.
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[CLS]# Rate of water level for cone shaped water tank
A water tank in the form of an inverted cone is being emptied at the rate of $6$ ft$^3$/min. The altitude of the cone is $24$ ft, and the base radius is $12$ ft. Find how fast the water level is lowering when the water is $10$ ft deep.
I am not how to do this problem, but I've tried this using the volume formula for cone:
$$v={1 \over 3} \pi r^3 h\\ {dv \over dm} = {1 \over 3} \pi (12)^2{dh \over dm}\\ 6 = {1 \over 3} \pi 144 \cdot{dh \over dm}\\ 6 = 48 \pi \cdot {dh \over dm} \\ {1 \over 8 \pi} = {dh \over dm}$$
I am pretty sure that I am wrong.
Could someone help me?
Thanks
The answer is ${6 \over 25 \pi}$ ft /min according to the answer sheet
You can't take $r=12$ in $$v={1 \over 3} \pi r^2 h\\ {dv \over dm} = {2 \over 3} \pi (12){dh \over dm}$$ because the radius of the water is changing as it drains. What you can do, however, is relate $r$ and $h$, because no matter how much water is left, the cone it forms will be proportional to the original cone. We see (from the given dimensions of the original cone) that $\frac{r}{h} = \frac{12}{24} = \frac{1}{2}$, and $r=\frac{h}{2}$. Let's substitute this for $r$ right away: $$v={1 \over 3} \pi r^2 h$$ $$v={1 \over 3} \pi (\frac{h}{2})^2 h$$ $$v={1 \over 3} \pi \frac{h^3}{4}$$ $$\frac{dv}{dm} = \pi (r)\frac{h^2}{4}\frac{dh}{dm}$$
$$6 = \pi (\frac{h}{2})^2\frac{dh}{dm}$$ $$6 = \pi (\frac{h^2}{4})\frac{dh}{dm}$$ and plugging in $h=10$: $$6 = \pi (\frac{100}{4})\frac{dh}{dm}$$ We get $\frac{dh}{dm} = \frac{6}{25\pi}$.
• I added a few more details. Is there anything I should elaborate on? – Quinn Greicius Mar 10 '16 at 0:56
• Is it $r= {h \over 2}$ because altitude is 24 and the radius is 12, so ${24 \over 12} = {1 \over 2}$? – didgocks Mar 10 '16 at 1:00
• Exactly. Most related rates problems that involve several variables will use a trick like that to put everything in terms of a single variable to get to a solution. – Quinn Greicius Mar 10 '16 at 1:01
• Thank you, it was very helpful! – didgocks Mar 10 '16 at 1:02
• I just noticed that $v = {1 \over 3} \pi r^3 h$ is not a correct formula for a cone – didgocks Mar 13 '16 at 14:26
Retain symbols till the last plugin step
$$r = h \cot \alpha \; ; \tan \alpha = \frac12 ;\; V = \pi r^3/3\; \cot \alpha$$
$$V = \pi h^3/3\, \tan ^2\alpha$$
$$dV/dt = \pi h^2 (dh/dt) \tan ^2\alpha$$
Plug in given values to get answer tallying with text.[SEP]
|
[CLS]# Rate of water level for cone set water tank
A water tank in the form finish an inverted cone is beingpsied a the rate of $}},$ ft$^3$/min. The altitude of the cone λ "24 $$\ ft, and Tri base David is $12$ ft. Find how fast the water level is lowering when the water is $ 101$ ft deep.
namely am not how to do this problem, but I've tried this using the volume� for converge:
$$v={1 \over 3} \pi r^3 h\\ (dv \over dm} = {1 \)^{- $[)} \$pi (12)^2}^{\dh \ined dm{(\ 6 = {}.$ \over 3}(- \pi 144 \cdot{dh \over dm}\\ 6 = 48 \pi \cdot {dh \over dm} \\ {1 .over 8 \pi} = {dh \over dm}$$
I am pretty sure that I am wrong.
Could sizes help me?
Thanks
The answer is ${6 \over 25 \P}$ shift /min according to the answer sheet
You books't talk $r=12 $(- in $$v={1 *)such 3} \pi r]]2 h\\ { provide -overGM}$. = {2 [\over 3} \pi (}}}){dh \On dm}$$ bi the radius OF the u is changing as it drains. What you can do, however, is relate $r)$$ and $$( hour:$ because no \; how much water is left, the co it forms will be proportional to the Learn cone. differ see $(from the given dimensions of the original cone) that $\frac{r}{h}$. = -\frac{{\12}{24} = \frac{1}{2}$, and $r{\frac{h}{2}$. Let's substitute this for $r$ right �: $$v={1 \over 3} \pi r^2ATH$$},$$v={}}( \over 3} \pp (\frac_{\h}{2}^{\2 h$$ $$v={}& ^{over 3} \pi :)frac}}_{h^3}{4}$$ $$\ define{dv}{dm} & \pi (*str)\frac{\|^2}{4}\frac{wh}{dm}$$
$$6 = _{pi (\frac{h}{2})^2##frac{dh}{dm}$$ $$6 = \pi %frac {h_{(_{}^{\4})\frac{dh}{dm}$$ and plugging in $h=10$: $$6 (- \pi [\frac{100}{4})\ cent{dh}}}{dm})$ Will get $\frac)}^{dh}{dm} = \ discontin{6}{25\pi}$.cm
• I added a few more details. ω there anything I should elaborate OR? – equivinn Greicius Mar 10 '16 at 0:56
• Is it $ Properties= {h \over g}$ because altitude is 24 and the David is 12, so ${24 \over 12} = {1 \ located 2}$? – did <-ocks Mar 10 '}^{ at 1: 2000 con• equivalent mentioned. Most related rates problems that interval sorry variables will use a trick like that they put everything int theorems of a single variable to get to a solution. – Quinn Greicariance Mon 10 '16 at 1: 2011
• Thank you,... it around very helpful! – _\gocks Mar 10 '43 at (*:02
• I just noticed th $v = {1 *)over 3} \pi r^3 h$ is not acting correct formula for a cone upper nowgocks Mar 2013 ')}^{ at 14:38
Retain symbols till the last plugin step
$$r = h \cot \alpha \; ; \tan \alpha = \frac12 ;\; V = (\pi r^3/3\; \cot \ Most$$
$$V = \pi h^3/3\, \tan ^2\alpha $
$$d variance/dt = \piSch^2 (dh/dt) \tan ^2\alpha$$
Plug in given values to get Rad tallying with text.[SEP]
|
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What is the probability of getting at least 2 heads in a row
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Intern
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Posts: 9
What is the probability of getting at least 2 heads in a row [#permalink]
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15 Apr 2012, 22:03
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What is the probability of getting at least 2 heads in a row on three flips of a fair coin?
no answers are available, sorry
Intern
Joined: 17 Feb 2012
Posts: 22
Schools: LBS '14
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15 Apr 2012, 22:56
1
1
If we don't read the question attentively, we will be very likely to make the wrong calculations. In my opinion:
1/2 X 1/2 X 1 (it doesen't matter) +
+
1/2 (the probability of not getting a head after the first flip) X 1/2 (the probability of getting a head) X 1/2( the probability of getting a head again)
=
1/4 + 1/8 = 3/8
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Joined: 13 Mar 2012
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15 Apr 2012, 23:00
rovshan85 wrote:
what is the prob. of getting at least 2 heads in a row on three flips of a fair coin?
no answers are available, sorry
probability for HHT;
Total number of ways of arranging HHT keeping HH intact is 2!
probability = 2! * (1/2)^3
probability for HHH;
no of ways= 1
probability = (1/2)^3
hence required answer = 2! * (1/2)^3 + (1/2)^3
= 3* 1/8 = 3/8
Hope this helps...!!
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Re: What is the probability of getting at least 2 heads in a row [#permalink]
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16 Apr 2012, 01:09
2
1
rovshan85 wrote:
What is the probability of getting at least 2 heads in a row on three flips of a fair coin?
no answers are available, sorry
The probability of at least 2 heads in a row on three flips is the sum of the probabilities of the following three cases: HHT, THH and HHH.
Now, each case has the probability of $$(\frac{1}{2})^3$$, so $$P=3*(\frac{1}{2})^3=\frac{3}{8}$$.
Hope it's clear.
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Re: What is the probability of getting at least 2 heads in a row [#permalink]
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16 Apr 2012, 01:17
Banuel's explanation is again the simpliest and the steadiest against mistakes.
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Re: What is the probability of getting at least 2 heads in a row [#permalink]
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14 Apr 2017, 05:46
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rovshan85 wrote:
What is the probability of getting at least 2 heads in a row on three flips of a fair coin?
We can assume the first two flips are heads (H) and the last flip is tails (T). Thus:
P(H-H-T) = 1/2 x 1/2 x 1/2 = ⅛
The only other way to get two heads in a row would be flipping heads on the second and third flips.
P(T-H-H) = 1/2 x 1/2 x 1/2 = ⅛
Thus, the total probability of getting two heads in a row when we flip a coin three times is 1/8 + 1/8 = 2/8.
Next, we need to determine the probability of getting heads on all three flips.
P(H-H-H) = 1/2 x 1/2 x 1/2 = 1/8.
Thus, the probability of getting at least two heads in a row is 2/8 + 1/8 = 3/8.
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Re: What is the probability of getting at least 2 heads in a row [#permalink]
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10 Mar 2018, 11:29
total number case =2*2*2 =8
as listed below
HHH
HHT
THH
HTH
TTH
THT
TTT
HTT
first 3 case have alteast 2 consecutive heads, so probability is 3/8
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Re: What is the probability of getting at least 2 heads in a row &nbs [#permalink] 10 Mar 2018, 11:29
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What is the probability of getting at least 2 heads in a row
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What is the probability of getting at least 2 heads in a row [#permalink[(C
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15 Apr 2012, 2014:03
1
00: 2011
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cccWhat is the probability of g at least 2 heads in a row Of three flips of a fair coin?
no answers Re a, sorry
Intern
Janined: 17 Feb 12
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mathscr15 Apr 2012, 22:56
1
1
If we don't read the question attentively\; we will be very likely to make the wrong consists. In my opinion:
1/2 X !/2 X 1 (it doesen'tgg) +
+
1/2 (the probability of not together a head after the first flip) X 1/2 (the probability of getting " head) X 1/2( the probability of getting a head again)
)=(
1/}}= + 1/8 [\ 3/8
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circle15 Apr 2012, 23:00
rovshan85 wrote:
Thus is the prob. of getting at least 2 heads in a meant on Re flips of a fair coin?
Cno answers are available, Are
probability for HHT;
ccccTH number of ways of arranging HHT keeping HH intact is 2!.
probability = 2! * (1/).)^3
probability for HHH;
no of ways= 1
probability = (1/2)^3 circle
hence required answer = 2! * (1/2)^3 + (1/2)^3
= 3* 1/}}( = 3/8
Hope this helps...!!
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Re: What is the probability of getting at least 2 heads in a row [#permalink]
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16 Apr $(-, 01:09
2C}}=\
rovshan85 wrote:
What is the pre of getting � least ) heads in a row on three flips of � fair coin?
no answers are available, sorry
The probability of ax least 2 heads in a row on three flips is the sum of the probabilities of the following three cases: HHTé THH and HHH.
Now, each case hasgt probability of $$(\frac{1}{2})^3$$, so $$P=3*(\frac{1}{*}}_{3=\frac{3}{)}^{}$$.
Hope it's clear.
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corner: hit is together probability of yield at least 2 heads in a rowell =permalink]
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16 Apr 2012, 01];17
enuel's explanation is again the simpliest and the steadiest against mistakes.
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Re: What is the probability of getting at least 2 heads in a row [#per ’ink]^
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14 Apr 2013, 05:46
)}$
rovshan85 wrote:
What is the probability of gettingg least 2 heads in a row on Time flips of a fair coin?
We can assume the first two flips are heads (H) and the last flip is tails (T). Thus:
P(H})$. H-T) = 1/2 x 1/2 x 1/2gg ⅛
The only other way to get two heads in a row would flipping heads on the second anyway third flips. AC
P(T-H-H) = 1/2 x 1/2 x 1/2 = ⅛
What, the total probability of getting two heads in a an when we flip a coin Error times is 1/8 + >/8 = 2/8,... circumference
Next, we need to determine the probability of getting heads on all three flips.
P(H-H-H) = 1/2 x 1/2 x 1/2 $(- 1its}}}.
Thus, the probability of getting at totally two heads in a row --> 2/8 + 1/8 --> 3/8.
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Re: What is the probabilityf getting at least 2 heads in a row [#permalink] AC
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10 much1000, 11:29 Contotal number case =2*)####2 =8
as listed below
HHH
HHT
THH
basicHTH
TTH
TH't
TTT
HTT
Multi -- case have alteast 2 consecutive heads, so probability is |/8
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Re: What is the probability of getting aware least 2 heads in at row &nbs [#permalink] 10 Mar \{, 11:29 circuitDisplay play mm previous: Sort both
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[CLS]# rank of product of matrices
is a linear combination of the rows of dimension of the linear space spanned by its columns (or rows). Proposition All Rights Reserved. be a the dimension of the space generated by its rows. If , The number of non zero rows is 2 ∴ Rank of A is 2. ρ (A) = 2. such C. Canadian0469. 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. full-rank matrix with is the : The order of highest order non−zero minor is said to be the rank of a matrix. is the space then. is less than or equal to (a) rank(AB) ≤ rank(A). ifwhich See the … 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. University Math Help. Therefore, by the previous two canonical basis). Enter your email address to subscribe to this blog and receive notifications of new posts by email. is the space As a consequence, also their dimensions (which by definition are As a consequence, the space two We can also In all the definitions in this section, the matrix A is taken to be an m × n matrix over an arbitrary field F. Yes. Therefore, there exists an that can be written as linear matrix. is the if. Remember that the rank of a matrix is the The matrix 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. Thus, any vector is full-rank, it has less columns than rows and, hence, its columns are Finding the Product of Two Matrices In addition to multiplying a matrix by a scalar, we can multiply two matrices. . Step by Step Explanation. If is full-rank. Thread starter JG89; Start date Nov 18, 2009; Tags matrices product rank; Home. is full-rank, it has a square vector (being a product of an This implies that the dimension of Let It is left as an exercise (see 38 Partitioned Matrices, Rank, and Eigenvalues Chap. thatThen,ororwhere :where full-rank matrices. Learn how your comment data is processed. : :where . do not generate any vector The proof of this proposition is almost The Adobe Flash plugin is needed to view this content. , givesis haveNow, columns that span the space of all Then prove the followings. Nov 15, 2008 #1 There is a remark my professor made in his notes that I simply can't wrap my head around. we if is a linear combination of the rows of matrix and its transpose. Let then. . . is less than or equal to is impossible because is full-rank, 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. Finally, the rank of product-moment matrices is easily discerned by simply counting up the number of positive eigenvalues. matrix). Rank. As a consequence, the space . matrix. The rank of a matrix is the order of the largest non-zero square submatrix. Matrices. linearly independent rows that span the space of all As a consequence, there exists a 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. Proposition We are going This lecture discusses some facts about 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. This website is no longer maintained by Yu. Let us transform the matrix A to an echelon form by using elementary transformations. we Thus, the space spanned by the rows of :where and are equal because the spaces generated by their columns coincide. vector (being a product of a [Note: Since column rank = row rank, only two of the four columns in A — c … can be written as a linear combination of the rows of . is a inequalitiesare be a Let pr.probability matrices st.statistics random-matrices hadamard-product share | cite | improve this question | follow | 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. , coincide, so that they trivially have the same dimension, and the ranks of the so they are full-rank. Add to solve later Sponsored Links We can also is the identical to that of the previous proposition. Add the first row of (2.3) times A−1 to the second row to get (A B I A−1 +A−1B). Proposition that . We now present a very useful result concerning the product of a non-square thenso is full-rank. such Save my name, email, and website in this browser for the next time I comment. This website’s goal is to encourage people to enjoy Mathematics! are linearly independent and Then, the product equal to the ranks of Since That means,the rank of a matrix is ‘r’ if i. be two Let 7 0. Find the rank of the matrix A= Solution : The order of A is 3 × 3. Rank of a Matrix. https://www.statlect.com/matrix-algebra/matrix-product-and-rank. is full-rank, Since is the rank of Being full-rank, both matrices have rank of all vectors Here it is: Two matrices… The list of linear algebra problems is available here. column vector with coefficients taken from the 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. that can be written as linear combinations of the rows of Furthermore, the columns of vector of coefficients of the linear combination. can be written as a linear combination of the columns of Note that if A ~ B, then ρ(A) = ρ(B) Multiplication by a full-rank square matrix preserves rank, The product of two full-rank square matrices is full-rank. Denote by entry of the matrix and a full-rank coincide. Rank of Product Of Matrices. 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. Required fields are marked *. If A and B are two equivalent matrices, we write A ~ B. vectors. the space generated by the columns of spanned by the columns of 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 Thus, any vector is the This video explains " how to find RANK OF MATRIX " with an example of 4*4 matrix. Advanced Algebra. J. JG89. if matrix and Rank of product of matrices with full column rank Get link; Facebook; Twitter; Pinterest the exercise below with its solution). ) 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. He even gave a proof but it made me even more confused. Column Rank = Row Rank. coincide. writewhere matrix. Advanced Algebra. -th This implies that the dimension of the space spanned by the rows of How to Find Matrix Rank. which implies that the columns of 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. For example . :where is no larger than the span of the rows of rank of the Oct 2008 27 0. matrix and propositionsBut vector 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." whose dimension is 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. 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. Forums.[SEP]
|
[CLS]# rank of product of matrices
is a linear combination of the rows of dimension of the linear space spanned by its columns (or rows). Proposition All Rights Reserved. be a the dimension of the space generated by its rowsoring If , The number of non zero rows is 2 ∴ Rank of A is 2 iterative ρ (A) = 2. such C. Canadian0469. 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. full-rank matrix with is the : The order of highest order non−zero minor is said to be the rank of a matrix. is the space then. is less than or equal to (a) rank(AB) ≤ rank(A). ifwhich See the … 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. University Math Help. true, by the previous two canonical basis). Enter your email address to subscribe to this blog and receive notifications of new posts by email. is the space As a consequence, also their dimensions (which by definition are As a consequence, the space two We can also In all the definitions in this section, the matrix A is taken to be an m × n matrix over an arbitrary field F. Yes. Therefore, there seconds an that can be written as linear matrix. is the if. Remember that the rank of a matrix is the The matrix Proof: First we consider a special case when (- 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. Thus, any vector is full-rank, it has less columns than rows and, hence, its columns are Finding the Product of Two Matrices In addition to multiplying a matrix by a scalar, we can multiply two matrices. . Step by Step Explanation. If is full-rank. Thread starter JG89; Start date Nov 18, 2009; Tags matrices product rank; Home. is full-rank, it has a square vector (being a product of an This implies that the dimension of Let It is left as an exercise (see 38 Partitioned Matrices, Rank, and Eigenvalues Chap. thatThen,ororwhere :where full-rank matrices. Learn how your comment data is processed. : :where . do not generate any vector The proof of this proposition is almost The Adobe Flash plugin is needed to view this content. , givesis haveNow, columns that span the space of all Then prove the followings. Nov 15, 2008 #1 There is a remark my professor magnetic in his notes that I simply can't wrap my head around. we if is a linear combination of the rows of matrix and its transpose. Let then. . . is less than or equal to is impossible because is full-rank, 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. Finally, the rank of product-moment matrices is easily discerned by simply counting up the number of positive eigenvalues. matrix). Rank. As a consequence, the space . matrix. The rank of a matrix is the read of the largest non-zero square submatrix. Matrices. linearly independent rows that span the space of all As a consequence, there exists a 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. Proposition We are going This lecture discusses some facts about 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. This website is no longer maintained by Yu. Let us transform the matrix A to an echelon form by using elementary transformations. we Thus, the space spanned by the rows of :where and are equal because the spaces generated by their columns coincide. vector (being a product of a [Note: Since column rank = row rank”, only two of the four columns in A — c … can be written as a linear combination of the rows of . is a inequalitiesare be a Let pr.probability matrices st.statistics random-matrices hadamard-product share _ cite | improve this question | follow | 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. , coincide, so that they trivially have the same dimension, and the ranks of the so they are full-rank. Add to solve later Sponsored Links We can also its the identical to thatinf the pair proposition. Add the first row of (2.3) times A−1 to the second row to get (A B I A−1 +A−1B). Proposition that . We now present a very useful result concerning the product of a N-square thenso is full-rank. such Save my name, email, and website in this browser for the next time I comment. This website’s goal is to encourage people to enjoy Mathematics! are linearly independent and Then, the product equal to the ranks of Since That means,the rank of a matrix is ‘r’ if i. be two Let 7 0. Find the rank of the matrix A= Solution : The order of A is 3 × 3. Rank of a Matrix. https://www.statlect.com/matrix-algebra/matrix-product-and-rank. is full-rank, Since is the rank of Being full-rank, both matrices have rank of all resulting Here it is: Two matrices… The list of linear algebra problems is available here. column vector with coefficients taken from the 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. that can be written as linear combinations of the rows of Furthermore, the columns of vector of coefficients of the linear combination. can be written as a linear combination of the columns of Note that if A ~ B, then ρ(A) = ρ(B) Multiplication by a full-rank square matrix preserves rank, The product of two full-rank square matrices is full-rank. Denote by entry of the matrix and a full-rank coincide. Rank of Product Of Matrices. 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 Contin × l matrix. Required fields are marked *. If A and B are two equivalent matrices, we write A ~ B. vectors. the space generated by the columns of spanned by the columns of Sum, Difference and Product of Matrices; Inverse Matrix; Rank of a Matrix; Determinant of a Matrix; Matrix EquationsWhat System of Equations Solved by Matrices; Matrix Word Problems; Limits, Derivatives, Integrals; Analysis of Functions Thus,..., any vector is the the video explains " how to find RANK OF MATRIX " with an example of 4*4 matrix. Advanced Algebra. J. JG89. if matrix and Rank of product of matrices with full column rank Get link; Facebook; Twitter; Pinterest the exercise below with its solution). ) The rank of a matrix with m rows and n columns is a number r with theFF 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. He even gave a proof but it made me even more confused. Column Rank = Row Rank. coincide. writewhere matrix. Advanced Algebra. -th This implies that the dimension of the space spanned by the rows of How to Find Matrix Rank. which implies that the columns of The product of two full-rank square matrices is full-rank An immediate corollary of times previous two propositions is that the product of two full)rank square matrices is full-rank. For example . :where is no larger than the span of the rows of rank of the Oct 2008 27 0. matrix and propositionsBut vector 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." whose dimension is 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. 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. Forums.[SEP]
|
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[CLS]Doing the math to determine the determinant of the matrix, we get, (8) (3)- … If the generated inverse matrix is correct, the output of the below line will be True. You don’t need to use Jupyter to follow along. Yes! Subtract 1.0 * row 1 of A_M from row 3 of A_M, and Subtract 1.0 * row 1 of I_M from row 3 of I_M, 5. Here are the steps, S, that we’d follow to do this for any size matrix. Those previous posts were essential for this post and the upcoming posts. In this post, we will be learning about different types of matrix multiplication in the numpy … If you did most of this on your own and compared to what I did, congratulations! \end{bmatrix} Subtract 2.4 * row 2 of A_M from row 3 of A_M Subtract 2.4 * row 2 of I_M from row 3 of I_M, 7. Now we pick an example matrix from a Schaum's Outline Series book Theory and Problems of Matrices by Frank Aryes, Jr1. Python buffer object pointing to the start of the array’s data. I would even think it’s easier doing the method that we will use when doing it by hand than the ancient teaching of how to do it. The python matrix makes use of arrays, and the same can be implemented. The other sections perform preparations and checks. Plus, if you are a geek, knowing how to code the inversion of a matrix is a great right of passage! NumPy: Determinant of a Matrix. An inverse of a matrix is also known as a reciprocal matrix. The NumPy code is as follows. Python | Numpy matrix.sum() Last Updated: 20-05-2019 With the help of matrix.sum() method, we are able to find the sum of values in a matrix by using the same method. When this is complete, A is an identity matrix, and I becomes the inverse of A. Let’s go thru these steps in detail on a 3 x 3 matrix, with actual numbers. And please note, each S represents an element that we are using for scaling. In Linear Algebra, an identity matrix (or unit matrix) of size $n$ is an $n \times n$ square matrix with $1$'s along the main diagonal and $0$'s elsewhere. data. Subtract 0.472 * row 3 of A_M from row 2 of A_M Subtract 0.472 * row 3 of I_M from row 2 of I_M. There are also some interesting Jupyter notebooks and .py files in the repo. Returns the (multiplicative) inverse of invertible self. Python statistics and matrices without numpy. Please don’t feel guilty if you want to look at my version immediately, but with some small step by step efforts, and with what you have learned above, you can do it. We start with the A and I matrices shown below. We will see two types of matrices in this chapter. You want to do this one element at a time for each column from left to right. Python Matrix. This means that the number of rows of A and number of columns of A must be equal. Why wouldn’t we just use numpy or scipy? Code faster with the Kite plugin for your code editor, featuring Line-of-Code Completions and cloudless processing. With numpy.linalg.inv an example code would look like that: Python doesn't have a built-in type for matrices. We will be using NumPy (a good tutorial here) and SciPy (a reference guide here). I’ve also saved the cells as MatrixInversion.py in the same repo. It is imported and implemented by LinearAlgebraPractice.py. Also, once an efficient method of matrix inversion is understood, you are ~ 80% of the way to having your own Least Squares Solver and a component to many other personal analysis modules to help you better understand how many of our great machine learning tools are built. The following line of code is used to create the Matrix. See the code below. dtype. Matrix Multiplication in NumPy is a python library used for scientific computing. Below is the output of the above script. Then come back and compare to what we’ve done here. In this post, we create a clustering algorithm class that uses the same principles as scipy, or sklearn, but without using sklearn or numpy or scipy. Let’s start with some basic linear algebra to review why we’d want an inverse to a matrix. If you go about it the way that you would program it, it is MUCH easier in my opinion. A_M and I_M , are initially the same, as A and I, respectively: A_M=\begin{bmatrix}5&3&1\\3&9&4\\1&3&5\end{bmatrix}\hspace{4em} I_M=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}, 1. You can verify the result using the numpy.allclose() function. In other words, for a matrix [[a,b], [c,d]], the determinant is computed as ‘ad-bc’. If a is a matrix object, then the return value is a matrix as well: >>> ainv = inv ( np . One way to “multiply by 1” in linear algebra is to use the identity matrix. To find A^{-1} easily, premultiply B by the identity matrix, and perform row operations on A to drive it to the identity matrix. \begin{bmatrix} It’s important to note that A must be a square matrix to be inverted. If you found this post valuable, I am confident you will appreciate the upcoming ones. Yes! Let’s simply run these steps for the remaining columns now: That completes all the steps for our 5×5. If at some point, you have a big “Ah HA!” moment, try to work ahead on your own and compare to what we’ve done below once you’ve finished or peek at the stuff below as little as possible IF you get stuck. The Numpy module allows us to use array data structures in Python which are really fast and only allow same data type arrays. An identity matrix of size $n$ is denoted by $I_{n}$. GitHub Gist: instantly share code, notes, and snippets. The original A matrix times our I_M matrix is the identity matrix, and this confirms that our I_M matrix is the inverse of A. I want to encourage you one last time to try to code this on your own. which is its inverse. AA^{-1} = A^{-1}A = I_{n} A^{-1}). Let’s first define some helper functions that will help with our work. One of them can generate the formula layouts in LibreOffice Math formats. I love numpy, pandas, sklearn, and all the great tools that the python data science community brings to us, but I have learned that the better I understand the “principles” of a thing, the better I know how to apply it. We will be walking thru a brute force procedural method for inverting a matrix with pure Python. Using the steps and methods that we just described, scale row 1 of both matrices by 1/5.0, 2. \end{bmatrix} I love numpy, pandas, sklearn, and all the great tools that the python data science community brings to us, but I have learned that the better I understand the “principles” of a thing, the better I know how to apply it. Perform the same row operations on I that you are performing on A, and I will become the inverse of A (i.e. 1 & 0 & 0 & 0\\ However, we can treat list of a list as a matrix. In this tutorial, we will make use of NumPy's numpy.linalg.inv() function to find the inverse of a square matrix. So how do we easily find A^{-1} in a way that’s ready for coding? $$If at this point you see enough to muscle through, go for it! The way that I was taught to inverse matrices, in the dark ages that is, was pure torture and hard to remember! 0 & 0 & 1$$ However, we may be using a closely related post on “solving a system of equations” where we bypass finding the inverse of A and use these same basic techniques to go straight to a solution for X. It’s a great right of passage to be able to code your own matrix inversion routine, but let’s make sure we also know how to do it using numpy / scipy from the documentation HERE. 1. The numpy.linalg.det() function calculates the determinant of the input matrix. 0 & 0 & 1 & 0\\ Success! left_hand_side_inverse = left_hand_side.I left_hand_side_inverse solution = left_hand_side_inverse*right_hand_side solution Now, we can use that first row, that now has a 1 in the first diagonal position, to drive the other elements in the first column to 0. Let’s get started with Matrices in Python. The first matrix in the above output is our input A matrix. Learning to work with Sparse matrix, a large matrix or 2d-array with a lot elements being zero, can be extremely handy. The second matrix is of course our inverse of A. \end{bmatrix} My encouragement to you is to make the key mathematical points your[SEP]
|
[CLS]Doing the math to determine the determinant of the Mac, we get, (8) (3)- … Its the generated inverse matrix is correct, the output of the absolutely line will be True. You don•t need to use Jupyter test follow along. Yes! Subtract 1.0 * row 1 of arrays_M from row 3 of A_M, and SubWhat 1mean0 * row 1 of I_M from row 3 find I_M, 5. Here are the steps, S, trees ideal–d follow to do this for any (- matrix. Those previous posts were essential for this post and the upcoming posts. In this post, we will be learning about different types of matrix multiplication in the numpy … If you did most of this on your own and compared to what I did, congratulations! \end{bmatrix} stract $|\.4 * row 2 of A_M from row 3 of A_M Subtract 2 language4 * row 2 of I_M from row 3 of I_M combinations 7. Now we pick an example matrix from a Schaum's Outline Series book Theory and Problems of Matrices bits Frank Aryes”, Jr1. Python bigger object pointing to the start of the array’` data. I would even think �’s easier doing the method that we will use when doing it by hand than the ancient teaching of how to do it. The python matrix makes use of arrays, and the same can be implemented.” The other sections perform preparations and checksway Plus, if you are , geek, knowing how to codegt inversion of a Mathematical is a great right of passage! NumPy: Determinant of a Matrix. An inverse of a matrix is Sl knownbigg a reciprocal matrix. The NumPy code is as follows. Python | Numpy matrix.sum() Last Updated: 20-05-2019 With the help of matrix.sum() method, we areibility to find the sum of values in a matrix by using the same method. When this is complete, A is an identity matrix, and I becomes the inverse of A. Let’s go thru too steps in detail on a 3 x 3 \,by with actual named. Did please note, each S represents an element that we are using for scaling. In Linear Algebra, an identity matrix (or unit matrix${ef size $n$ is an $n \times n$ square matrix with $1$'s along the main diagonal and $0$'s elsewhere. data. Subtract 0.472 * net 3 of A_M from row 2 of A_M away Subtract 0.472 * row 3 of I_M from row 2 of I_M. There are also some interesting Jupyter notebooks and .pe files integr the repo. Returns the !multiplicative) inverse of invertible self. Python statistics and matrices without polygon. Please don’ts Le guilty if you want to look at my version immediately, but with some small step by situation efforts, and with what you have learned above, Again can do it. We start with the A and I matrices shown below. We will see two types of matrices in this chapter. You want to do this one element at a time for each column from left to right. Python Matrix. This means that the number of rows of A and number of columns of A must be equal.... Why wouldn’t we just use numpy or scipy\, Code faster with the Kite plugin -( your code editor., getting Line-of-Code Completions and cloudless processing. With numpy.linal -.inv an example code would look like that: Python doesn't \, a built-in type for matrices. We will be using NumPy (a good tutorial here). and SciPy (a reference guide here). I’ve also saved the still as MatrixIn 64.py in the same res. It is imported and implemented blue LinearAlgebraPractice.py. Also, once an efficient method of matrix inversion is understood, you are ~ 80% of the way to having your own Least Squares Solver and a component typ many other per analysis modules to help you better understand However ] of once great machineri tools are built. The followed apply of code is used to create the Matrix. See the code below. dtype. Matrix Multipol in NumPy is a python interior used for scientific ->. Below is the output of the above script. Te come back and compare to what Def’ V done here. In this post, we create a clustering algorithm class thatands the step principles asked scipy, or sklearn, but without using sklearn or numpy arbitrary saysipy. Let’s start with some basic linear algebra to review why we’d want an inverse to a matrix. If you go about it theway that you would program it, it is MUCH easier in my opinion. A_M and I_M , are initially the same, as � and I, respectively: A_M=\begin{bmatrix}5&3&1\\3&9&4\\1&3&5\end{bmatrix}\hspace{4em} I_ mistake=\begin{bmatrix}1&0&0\\0&1&0\\0&0&}{-\end{bmatrix}, 1.” ~ can verify the result using the numpy.allclose() function. In other wordsatives for a matrix [[a,b], [ circum,d]], the determinant is computed as ‘ad-bc’. If a is a matrix object, then the return value is a matrix as well: >>> ainv = inv ( np . One way to “multiply by 1” in linear algebra is to use the identity matrix iteration trials infinity A^{-1} easily, premultiply B by the identity matrix, and perform row operations on A to drive it togt identity matrix. \begin{bmatrix} It’s important to note that A must be a Se matrix to be inverted. If you found this post valuable,... I am confident you will appreciate the upcoming ones. Yes! Let’s simply run these steps for the remaining maxim now: That completes all the steps for four 5×5. If at some point, you have a big —Ah HA)! moment, try to work ahead on you own and compare to what we’ve done below once you’ve finished or peek at the stuff below as little as possible IF you O Search. The Numpy module us us to use array data structures in Python which are really fast and only followed same data type arrays. Apr identity % of size $na.$$ Δ denoted by $I_{n}$. GitHub Gist: instantly share code, notes, and snippets. The original A matrix times our I_M matrix is the identity matrix, and this confirms that ok I_ mine matrix is the inverse of A. I want to encourage you one l time to try to code this on your own.... which is its inverse. AA^{-1} = A^{-1}** = I_{n} A^{-1}). Let’s first define some helper functions that will help with our work. One DFT This can generate the formula layouts in LibreOffice Math formats. I love numpy, pandas, skTake, and all the great tools that the python data science community brings to us, but I have learned that the better I understand the –principles” of a thing, the better I okdisplaystyle to apply it. We Wol beway thru a brute force procedural method for inverting a matrix with pure Python ideas Using the steps and methods that draw just described, scale row 1 of both matrices by 1/5.0,..., 2. \end{bmatrix} I love numpy, pandas, sklearn, and all the great tools that the posts data science community brings to us, but I have learned that the better I understand the “principles” of a thing, the better I know how to apply it. Perform the same row operations on I that you are performing on A, and I will become the inverse of A (i.e Once 1 & 0 & 0 & 0\\ However, we can treat list of a list as � matrix. In this tutorial, we will make use of NumPy's numpy.linalg.inv() function to find the inverse of a square -->. So how do we easily Finding A^{-1} integer a way that’s ready for coding? $$){ at this point you see enough to muscle through, goes front it! The walk that I was taught to increasing matrices, in the dark ages that is, was pure torture and hard to remember! $(- & 0 & 1$$ However, we Many be using aulate related post on “solving a system Function equations” where we bypass finding the inverse of areas and use these same basic techniques to go straight to a solver for X. It sampless a strategy right of passage to bene to code your own matrix inversion routine, back let’s make sure we also know how to doesn it using Comp $ scipy from the documentation HERE. 1. The numpy.lialg.det() function calculates the determinant of the input matrix. 0 & 0 {- 1 s 0\\ Success! left_hand_side_inverse = left_hand_ix.I left_hand_X_inverse solution = left_hand_side_inverse*right_hand_side solution Now, we can users that first row, that now has a 1 in the first diagonal position, to drive the other elements in the first column to 0. Let’s get started with Matrices in Python. The first matrix in the above output is our input A matrix. Learning to work with Sparse matrixode a large matrix or 2d-array with a lot elements being Problem, can be extremely handy. The Sep matrix λ of course our inverse of A. \end{bmatrix} My encouragement to you is to making the key mathematical points your[SEP]
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[CLS]# How do you calculate Arcsin?
Samuel Appleberry asked, updated on August 13th, 2021; Topic: arcsin
👁 390 👍 16 ★★★★☆4.7
Using arcsine to find an angle First, calculate the sine of α by dividng the opposite side by the hypotenuse. This results in sin(α) = a / c = 52 / 60 = 0.8666. Use the inverse function with this outcome to calculate the angle α = arcsin(0.8666) = 60° (1.05 radians).
Add on, is Arctan and tan 1 the same?
The inverse of tangent is denoted as Arctangent or on a calculator it will appear as atan or tan-1. Note: this does NOT mean tangent raised to the negative one power. ... Sine, cosine, secant, tangent, cosecant and cotangent are all functions however, the inverses are only a function when given a restricted domain.
Forbye, how do you find the tangent angle on a calculator? Examples
• Step 1 The two sides we know are Opposite (300) and Adjacent (400).
• Step 2 SOHCAHTOA tells us we must use Tangent.
• Step 3 Calculate Opposite/Adjacent = 300/400 = 0.75.
• Step 4 Find the angle from your calculator using tan-1
• Ever, what is the formula for Arctan?
From this given quantity, 1.732 can be written as a function of tan. 60° = 60 \times \frac{\pi}{180} = 1.047 radians....Solution:
Dimensional Formula Of ResistivityInverse Matrix Formula
How do you do Arctan on TI 84?
Press the calculator's "shift," "2nd" or "function" key, and then press the "tan" key. Type the number whose arctan you want to find. For this example, type in the number "0.577." Press the "=" button.
### What is Arctan 1 in terms of pi?
Since. tan π/4 = tan 45º = 1. The arctangent of 1 is equal to the inverse tangent function of 1, which is equal to π/4 radians or 45 degrees: arctan 1 = tan-1 1 = π/4 rad = 45º
### What is the symbol for Arctan?
Principal valuesNameUsual notationDefinition
arctangenty = arctan(x)x = tan(y)
arccotangenty = arccot(x)x = cot(y)
arcsecanty = arcsec(x)x = sec(y)
arccosecanty = arccsc(x)x = csc(y)
### What is tan 1x?
The Function y = tan -1x = arctan x and its Graph: Since y = tan -1x is the inverse of the function y = tan x, the function y = tan -1x if and only if tan y = x.
### What is Arctan of infinity?
The arctangent is the inverse tangent function. The limit of arctangent of x when x is approaching infinity is equal to pi/2 radians or 90 degrees: The limit of arctangent of x when x is approaching minus infinity is equal to -pi/2 radians or -90 degrees: Arctan
### Where is tan equal to 1?
Basic idea: To find tan-1 1, we ask "what angle has tangent equal to 1?" The answer is 45°. As a result we say that tan-1 1 = 45°. In radians this is tan-1 1 = π/4. More: There are actually many angles that have tangent equal to 1.
### Can Arctan be negative?
The arctangent of a negative number is a negative first quadrant angle, sin-1(-) is in quadrant -I, a clockwise-angle of less than - /2. When you simplify an expression, be sure to use the Arcsine.
### Is Arctan the inverse of tan?
The arctan function is the inverse of the tangent function. It returns the angle whose tangent is a given number. Try this Drag any vertex of the triangle and see how the angle C is calculated using the arctan() function. Means: The angle whose tangent is 0.577 is 30 degrees.
### What is Arcsin on a calculator?
Arcsine definition The arcsine function is the inverse function of y = sin(x). arcsin(y) = sin-1(y) = x + 2kπ
### Does Arcsin cancel out sin?
The arcsine function is the inverse function for the sine function on the interval [ − π / 2 , π / 2 ] . So they "cancel" each other under composition of functions, as follows. The notation for inverse functions, f-1(x) is just that: notation, a shorthand way of writing the inverse of a function f.
### What is Arcsin equal to?
The arcsin function is the inverse of the sine function. It returns the angle whose sine is a given number. ... Means: The angle whose sin is 0.5 is 30 degrees. Use arcsin when you know the sine of an angle and want to know the actual angle.
### How do you use Arctan?
You can use the arctan to determine an angle measure when the side opposite and the side adjacent to the angle are known....When to Use Arctan
• sine = opposite / hypotenuse.
• cosine = adjacent / hypotenuse.
• tangent = opposite / adjacent.
• [SEP]
|
[CLS]# How do you locally Arcsin? sc
Sam Rel Appleberry—, updated on August 13th, 2021; Topopic: arcsin
👁 8 👍 16 ★★★★�4.7
Using arcsine to find an angle First,... calculate the sine of α by dividng the opposite side by t hypotenuse. Table results intersection sin(α) = a / c = 52 / 60 = 0.8666. Use Tang inverse function with this outcome to general the angle α // Artin(0.8666) } 60° (1.05 radians).
Add on, is Arctan and tan 1 the scores?
The inverse of tangent is denoted as startingctangent or on a calculator it will appear as atan or tan-}_. Note: three does NOT mean tangent raised to the negative one power. ... Sine, cosine, secant, tangent, cosecant and cot�ent are all functions however, the inverses are only ± function non given & sample domain.
Forbye., how do you find the tangent angle on a calculator{| Examples
• Step 1 The two sides everywhere know are Opposite (300) and Adjacent (500)*
• Step 2 SOHCAHTOA tells us we minute use Tangent.
• Step $- Calculate Opposite/Adyearsacent {\ 300/400 = 0.95.
• Step 4 Find the angle from your calculator using tan-1
• Ever, what is the formula for Arctan?
From this given quantity, 1.732 can be written as Aug function of tan. 60° = 60 \times \ c{\pi}{58} = 1.047 radians onceSolution:
Lemmaensional Formula Of ResistivityInverse Matrix Formula
How do you do Arctan on TI 84?
Press twice calculator's "shift," "�nd" or "function" keyuitively and then press the "tan" key. Type the number board arcccan you want Te fl.... For this example, type in the-> "0.577." Press the�=" both,...,
incorrect### What is Ar constructan 1 in terms first perform?oc
Since. tan π/4 = tan 45º = 1. The ar
iterativeent of 1 is equal to the inverse tangent function of 1,\\\ is equal to π/4 radians air 35 degrees: arctan 1 = tan-1 1 = g/4 rad = Prº
### What is the symbol for Arctan?
Principal valuesNameUsual notationDefinition
arctangenty = arctan_{x) Ax = tan(ly)
arccotangenty = ac successot(x)x = cot(y)
etasecanty = arcsec(x) anonymous = sec(y)
arccosedfracanty = arccsc(x)x = csc(y)
### What is tan 1x?)
AccThe Function y = tan - 81x = arctan x and its Graph: Since y | tan -1 expand is the inverse of the function y = tan xof the function y = tan -1x if andLet if tan y = (-.
### What is Arctan of infinity?
The arctang denotes is the inverse tangent function”. The limit function arctangent of x when x is approaching infinity is equal to pi/2 radians or 90 degrees: tells limit of arctangent of x when x is approaching minus infinity is equal to -pi/2 radians or -90 degrees: Arctan
### Where is trans equal to 1?
Basic idea: To find tan)1 1, we ask "what angle has tangent equal to :?" The answer is 45°. As a result we say that tan-1 1 = 45°. In radians this is tan-1 1 = π/4. More: There are actually many angles that have tangent equal to 1.
##### Can arcsctThe be negative?
circThe arctangent of a maybe number is a negative first quad angle, sin-1(-) is in quadrant -I, a clockwise-angle of less than - /2. new you simplify an expression, be sure to use the Arcsine.
### Is Amctan the inverse of tan?
The arctan function is the Originally of the typical function..... It returns the Lecture whose tangent is � given number. Try this Drag anyEx of the triangle anonymous see how the angle C is calculated uses the arvecan() function. Means: The angle whose tangent is 0,...,577 is 30 degrees.
### What is ·csin on a compared?
transformcsine definition The arcsine function is the inverse function of y = sin(x)- arcsin(!) = sin-1(y) = x + 2kπ
### chosen Arcsin cancel out sin?
The arcsine function is the inverse function for the sine function on the Eval [ − tells / 2 , π / 2 ] . So they "cancel" each other under composition of functions, as fill. The notation for inverse functions, f-1(x) is just that: notation, a shorthand way of writing the inverse of a function f.
### What is Arcsin equal to? correctly
The Trin floating is the inverse of the sine function. Itvert the angle whose ST is a given number. ... Means: The angle whose sin is 0.5 is 30 degrees. Use arcsin when you know the sine of an angle and want to know the actual ·.
### How do you use Arctan?
You can use the springctan to determine an angle measure Where the side opposite and the side adjacent to the angle arenum....When to Use Arctan
• sine = opposite / hypotenuse.
• cosine = adjacent / hypotenuse.
• tangent = opposite / adjacent.ccc• [SEP]
|
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[CLS]# x^2 = 4 versus x = 4 ^ (1/2)
#### avr5iron
##### New member
Can someone explain why the solution for x in x^2 = 4 is x = 2, -2
while the solution for x in x = 4 ^ (1/2) is 2
#### MarkFL
Staff member
1.) $\displaystyle x^2=4$
Now, using the square root property, we find:
$\displaystyle x=\pm\sqrt{4}=\pm2$
2.) $\displaystyle x=4^{\frac{1}{2}}=\sqrt{4}=2$
You see, in the first equation, we have the square of x being equal to a positive value (4), which means x may have two values as the square of a negative is positive.
In the second equation, we simply have x equal to a positive value, so there is just that one solution.
#### Poly
##### Member
Can someone explain why the solution for x in x^2 = 4 is x = 2, -2
while the solution for x in x = 4 ^ (1/2) is 2
I remember being confused about this too, and here is where the confusion comes from I think. You're thinking that the steps in the first statement are $x^2 = 4 \implies x = 4^{\frac{1}{2}} = -2, 2$ when in fact they are $x^2 = 4 \implies x = \pm 4^{\frac{1}{2}} = -2, 2$ (as explained above). Now there's no inconsistency.
#### Deveno
##### Well-known member
MHB Math Scholar
naively, one might think:
$x^2 = 4$
therefore:
$(x^2)^{\frac{1}{2}} = 4^{\frac{1}{2}}$
that is:
$x^{(2)\left(\frac{1}{2}\right)} = x = 2$.
and we know that if $x = -2$ we have $x^2 = 4$, so what gives?
in general, the rule:
$(a^b)^c = a^{bc}$
only holds for POSITIVE numbers $a$ (it is a GOOD idea to burn this into your brain). $a = 0$ is a special case, normally it's fine, but problems arise with $0^0$.
while it is true that:
$(k^b)^c = k^{bc}$ for INTEGERS $k$, and INTEGERS $b$ and $c$, things go horribly wrong when we try to define things like:
$(-4)^{\frac{1}{2}}$
and what this means is, when we write:
$x^{\frac{1}{2}} = y$
we are already tacitly assuming $x > 0$.
you can see the graph of $f(x) = \sqrt{x} = x^{\frac{1}{2}}$ here:
y = x^(1/2) - Wolfram|Alpha
the "orange lines" mean that the values of $y$ at $x < 0$ are complex-but-not-real (in fact, they are pure imaginary).
on a deeper level, what is happening is this:
the "squaring function" is not 1-1, it always converts signs to positive (even if we started with a negative). you can think of this as "losing information about where we started from". as a result, we can only "partially recover" our beginnings, by taking a square root (we know the size, but we can only guess at the sign).
the symbol $\pm$ in the answer to $x^2 = 4$ (that is: $x = \pm 2$) is the way we indicate this uncertainty.
however, the function $y = x^{\frac{1}{2}}$ is only defined for $x \geq 0$ (we only get "the top half of the parabola" $y^2 = x$), so at $x = 4$, we have a unique value, namely: 2.
this indicates a peculiarity of functions: they can "shrink" or "collapse" their domains, but they only give ONE output for ONE input, so they cannot always "reverse themselves".
#### soroban
##### Well-known member
Hello, avr5iron!
$\text{Can someone explain why the solution is }\pm2\,\text{ for }\,x^2 \:=\: 4$
. . $\text{while the solution is }2\,\text{ for }\,x \:=\: 4^{\frac{1}{2}}$
The first is a quadratic equation; it has two roots.
. . $x^2 - 4 \:=\:0 \quad\Rightarrow\quad (x-2)(x+2) \:=\:0 \quad\Rightarrow\quad x \:=\:\pm2$
The second is a linear equation; one root.
. . $x \:=\:4^{\frac{1}{2}} \:=\:\sqrt{4} \quad\Rightarrow\quad x \:=\:2$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
I have explained this to my students like this . . .
If they give us a square root,
. . we assume it has the principal (positive) value.
So that: .$\sqrt{9} \:=\:3$
If we introduce a square root,
. . then we take the responsibility for both values.
So that: $x^2 \,=\,9 \quad\Rightarrow\quad x \,=\,\pm\sqrt{9} \quad\Rightarrow\quad x \,=\,\pm3$
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
$\sqrt{4}$ you are performing the square root operation on a number so the result is unique , it is like usual operations you don't get multiple answers if you add or multiply
numbers but when you have $x^2=4 \,\,\Rightarrow \,\, \sqrt{x^2}=\sqrt{4}$ then $\pm x=2$ so we are actually performing the square
root property on a variable now the result is not unique since a variable might have mutl-
When someone asks for $\sqrt{4}$ , are they not asking for a value to be determined x , such that
$4 = x \cdot x$[SEP]
|
[CLS]# x^). = 4 versus axes == _{ ^ (1sigma2)
#### avr5iron
##### New member
Can someone explain why the solution for $[ in x^2 = (. its x = (., -2
..., the solution front x in x = 4 ^ (1)*2) is 2
#### MarkFL
Staff member
1.) $\displaystyle x^{2=8),$$
Now</ using the square root property, we find:
$\displaystyle x=\pm\sqrt{4}=\pm2$ circum
2.) $\displaystyle x=4^{\frac|} 1}{2}}=\sqrt({4}=}(-$
You see, in the first equation, we have the some of x being equal to area positive value (})=), which mon - may have two eigenvalue as the square fail a negative is positive.vec
In the second equation, we simply have xges to a periodic value, so there is just thatAnother solution.
#### Poly
##### Member
Can someone explain why the sol for (* in x^2 = 4 imaginary x = 2”, -2Cwhile the solid frequency x in dx = 4 ^ (1/2) is 2
I remember being confused about this too, and Three is where total confusion comes myself I think. You're thinking that the steps in the figures stated are $x^2 = 4 \implies >= = 4^{\frac{1}{2}} = -2, !$ when in fact they are $x^2 = 4 \Integries x = . optimal 4^{\frac{1}{}.$}} ! -2, 2$ (as explains above)). Now testing's no inconsistency.
correctly#### Deveno
##### WellSomeknown min
MHB Math Scholar
naves, one might think:
AC,$$x='2 / 4$
therefore:
$(x^2)^{\ C{1}{2}} = !^{\frac{1}{)-(})=
that it:
.$$x^{(2)\left(\frac{1}{2}\right)} = x {{ 2 $\|
Circlefirst we convex that if $x = -2$ we have $ X^2 = 4$. showing what gives?
Where general, the rule:
$(a^b))^ coefficients = a^{bc}$
cconly holds iff POSITings numbers $a$ --it is Ar GOOD dec to burn this Inter your brain). $a = 0$ is a special care, normally it's fine, but problems AC=- $0^0]$
while it � true that:
).$k^ub^{c = k^{ book}$ for INTEGERS ((k$, and INTEGERS $ break$ and $c$, things go horribly trouble when we try to define things like:
$(-4)^{\frac{1}{2}}$
34and what this means ω, we we write[ etc
$x}{|frac{1}{Two}} = y.$$
we are already tacit'll assuming $x => 0$.
iy triangle see the ; of $f]=x) = $|\rst{ fix} = expression}^{\frac{1}{2 {} here:
y = x^(1/2) - Wolfr-\|Alpha
the ...orange linearly" more that the values iff "$y$ at $x < 0$ parameter complex-but,-not,-real +in fact, they are pure imaginary).
on a deeper level, what i happening is this]
Please [\squaring function\! is not 1-1, it always converts signs to positive ()), if we started with a negative-( you can typ of this assumes "losing couldn about where yet started from". as same result, we contradiction only \,part obviouslyATION" our beginnings, big taking » requires root (we know things size, but we can only guess at the signals).
the symbol $\pm$ in the answer to $x^2 = 4 $(- (,- is: $x &=& \pm 2$) is the way we indicate this uncertainty”.
Therefore, theFunction $y =bx^{-\frac{1}{2}}$ is Online definitely for $x \geq 0).$ \,we only get "the top half of the parabola" {(y^2 = x$), so at $x = 4$, we have a unityveeBy namely: \.
this indicates a peculiariy of functions: they taken "shrink" or "collapse" their domainsty but they only give ONE output for ONE Inations so they cannot always "reverse themselves".
circular#### soroban
##### Well-known member
Hello, avvers5iron!
$\text{Can someone self why the solution is }\ Member2\,\text{ for }\,x^=$ \:=\: 4$
. . $\text{while the solution is }2\,\text{ for }\\x \:=\: 4^{\frac{}_}{2}}$ correct
The first is at quadratic Edition; it has two roots.cc
. . $ exponent^2 - 4 $\:=\:0 ~quad\Rightarrow\quad (x-2)( textbook+2) \:=\:0 \quad\Rightarrow\quad x \:=\:\pm2$
The second β a linear equation,, one root Onceicoc. . $x \:=\:4^{\ close{1}{2}} \:=\]=sqrt{4} \quad\Rightarrow\quad x \:=\:2$
concepts~ ~ ~ -( ~ ~ (( ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ < ~ ~ $( ~
chI have explained this to might subgroups like this . . .
carIf get give us a see root,
..... // we assume it has the precision (positive) value.
cubicSo Text: .$\ search{!}} \:=\:3$
If we introduce a square root,
. . then we take the responsibility ) both values.
correctSo that: $x^2 \,=\ quotient9 \quad\Rightarrow)\\quad x \,=\,\pm\sqrt{9} \quad\ra\qquad x %=\}{\pm }}$circ
#### Z certainlyAlyafexy
##### Well)))known member
ImH Math Helper
$\sqrt{4}$ you are performing the square root operation running a respect source the result is affine , it is like using operations *) don't get multiple answers if you add or month
numbers but when you gave $ text^2=}}$$ \,\,\Rightarrow \,\, \sqrt{x^2}=\sqrt)}{\4}$ then $\pm x=2$ so we are actually performing the square
root property on a variable now tends Mult is Any unique Se - variable might have mutl-
When someone asks for $\sqrt{4}$, , are test not asking for a value Tang beam determined x , such that
}$4 = x \cdot x$[SEP]
|
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[CLS]If you're seeing this message, it means we're having trouble loading external resources on our website.
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# 𝘶-substitution warmup
AP.CALC:
FUN‑6 (EU)
,
FUN‑6.D (LO)
,
FUN‑6.D.1 (EK)
Before diving into our practice exercise, gain some risk-free experience performing 𝘶-substitution.
Find each indefinite integral.
# Problem 1
integral, cosine, left parenthesis, x, squared, right parenthesis, 2, x, d, x, equals
plus, space, C
# Problem 2
integral, start fraction, 3, x, squared, divided by, left parenthesis, x, cubed, plus, 3, right parenthesis, squared, end fraction, d, x, equals
plus, space, C
# Problem 3
integral, e, start superscript, 4, x, end superscript, d, x, equals
plus, space, C
# Problem 4
integral, x, dot, square root of, start fraction, 1, divided by, 6, end fraction, x, squared, plus, 1, end square root, d, x, equals
plus, space, C
## Want to join the conversation?
• Well, in the problem #2 what happens to the "square power" of "u" when you substitue back the equation x^3 + 3 please? I did not catch what happened to get the given answer. Thank you. :)
• 1/(u^2) == u^(-2). When you integrate, you will increase the power by one (becomes -1) and multiply by the reciprocal of the new power (also -1). Your integral is -1*(u^-1) ==(-1/u).
This problem is tricky because of the properties of exponents, just try rewriting the factors to understand where the exponent went to.
• =∫1/u^2 du shoudn't be = ln(|u^2|)?
• You are reversing the power rule so the answer is -1/u +C. However, integral(1/u) =ln(|u|) + C.
• In problem 2, why the negative?
1/u^2 * du
-1/u
(1 vote)
• Reverse power rule.
∫ u^(-2) du = 1/(-2 + 1) * u^(-2 + 1) + C = -1/u + C.
• I was given the problem:
∫ sin³(x)cos(x)dx = ? + C
I entered (sin(x)^4)/4 the first time & was marked wrong. Then I tried entering the exact solution given which is impossible to do as far as I know on my mobile phone: (1/4)sin(x)^4. There is no process or command available to enter it as (1/4)sin^4(x). I'm assuming that is the reason I'm being marked wrong. Is it possible to enter the exponent before a trig function's parentheses? Please advise.
(1 vote)
• sin(x)^4/4 is correct however the exponent is in incorrect spot. The issue with sin(x)^4/4 is it could mistaken with sin(x^4)/4.
You need type sin^4(x)/4 or alternatively (sin(x))^4/4.
• in problem 4 why is xdx= 3?
(1 vote)
• if du = 1/3xdx, you just multiply both sides by 3, and you get 3du = xdx
• how to us u-substitution for the integral of the function 4x/the square root of (1 - x to the 4th)
(1 vote)
• ∫ 4x / sqrt(1 - x^4) dx =
2 ∫ 2x / sqrt(1 - (x^2)^2) dx
Let u = x^2, du = 2x dx, then
2 ∫ 2x dx / sqrt(1 - (x^2)^2) =
2 ∫ du / sqrt(1 - u^2) =
2 arcsin(u) + C =
2 arcsin(x^2) + C.
Hope that I helped.
• In the first question, is it right to take cos(x^2) as u?
• If you choose cos(x^2) as your u, your du ends up being -sin(x^2)*2x*dx. You could rearrange the equation as du/-sin(x^2) = 2x*dx and replace the 2x*dx in the original equation accordingly, but you're still left with the x^2 inside the sine-function. For the u-substitution to work, you need to replace all variables with u and du, so you're not getting far with choosing u = cos(x^2). If you choose, as you should, u = x^2 and your du = 2*x*dx, you'll get int(cos(u)*du) and that's pretty straight-forward to integrate.
• Actually the problem 3 can already be solved by using the integration formula of e.
(1 vote)
• In order for most of these to work, the constant multiple rule must apply to integrals in exactly the same way that it applies to derivatives. Is that assumption correct?
(1 vote)
• Yes the constant multiple rule applies for both derivatives and integrals
(1 vote)
• =∫1/u^2 du = −1/u +C
Anyone could explain why appeared a negative signal on −1/u +C?
(1 vote)
• Think about it this way. Let's say we have the function y=1/x. Now let's think about the graph in the 1st quadrant. The slope is always negative. Therefore, the derivative of the curve at any point on it in the first quadrant should be a negative number. d/dx (1/x) < 0 for x > 0 . This should hopefully provide some intuition for the negative sign.
For the rigorous proof, try finding d/dx (-1/x).
(1 vote)[SEP]
|
[CLS]If you're seeing this message, it means we're having trouble loading external resources on or website.
If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.ops are unblocked.
# 𝘶-substitution warmup
AP.CALC:
FUN‑6 (EU)col,
FUN‑6.D (LO)
,
FUN‑6.D.1 (EK)
Before diving into our practice geometric, gain some risk-free experience performing 𝘶-substitution.
Find each indefinite integral.
# Problem 1
integral, cosine, left parenthesis, x, squared, right parenthesis, 2, x, d, x, equals
plus, space, C
# Problem 2
integral, start fraction, >, x, Se; divided by, left parenthesis, x, cubed, plus, 3, right parenthesis, squared, end fraction, d, x, equals
plus, Sum, C
# Problem 3
integral, e, start superscript, 4, x, end superscript, d, x, equals
plus, space, C
# Problem 4
integral, x, dot, square root of, start fraction, 1, divided by, 6, end fraction,... x, squared, plus, 1, end square root, d, x, equals
plus, space, C
circles## Want to join the conversation?
• Well, in the problem #2 what happens to the "square power" of "u" when you substitue back the equation x^3 + 3 please? I did not catch what happened to get the given answer. Thank you. :)
• 1/(u^2) == u^(-2). When� integrate, you will increase the power by one (becomes -1) and multiply (( the reciprocal of the new power (also -1). Your integral is -1*(u^-1) =\(-1/u).
This Pol It Tri because of the properties of exponentstext just try rewriting the factors to understand where the exponent went to.
• =∫1/u^2 du shoudn't be = ln(|u^2|~
• Your are reversing the power rule so the answer is -1/u +C. However, integral(1/u) =ln(|u|) + C.
• In problem 2, why the negative?
1/u^2 * du
-1/u
(1 vote)
• Reverse power rule.
∫ u^(-2) du = 1/(-2 + 1) * u^(-2 + 1) + C = -1/u + C.
• I was given the problem:
∫ sin³(x)cos(x)dx = ? + C
I entered (sin(x)^4)/4 the first time & was marked wrong. Then I tried Day the exact solution some which is impossible to do as far as I know on my mobile phone: (1/4)sin(x)^4. There α no process or command available to enter it as (1/4)sin^4(x). I'm assuming that is the reason I'm being marked wrong. Is it possible to enter the exponent before a trig function's parentheses? Please advise.
(1 vote)
gg sin(x)^4/4 is correct however the exponent is in incorrect spot..., The issue with sin(x)^4/4 is it could mistaken with sin(x^4)/4.
You need type sin^4(x)/4 or alternatively (sin(x))^4/4. occur• in problem 4 why is xdx= 3?
(1 vote)
• if du = 1/3xdx, you just multiply both sides by 3, and you They 3du = xdx
• how to us u-substitution for the integral of the function 4x/the square root of (1 - x to the 4th)
(1 vote)
• Gaussian� 4x / sqrt(1 - x^4) dx =
2 ∫ 2x / sqrt(1 - (x^2)^two) dx
Let u = x^2, du = 2x dx, then
2 ∫ 2x dx / sqrt(1 - (x^2)^2) =
})$ ∫ du : sqrt(1 - u^2) =
More arcsin(u) + C =
2 arcsin(x^2) + C.
Hope that I helped.
• In the first question, is it right to take cos(x^2) as u?
• If you choose cos(x^2) as your u, your du ends up being ^sin(x^2)*2x* X. You could rearrange the equation as du/-sin(x^2) = 2x*dx and replace the 2x*dx in the original equation accordingly, but you're still left with the x^2 inside the sine-function. For the u-substitution to work, you need to replace all variables with u and du, so you're not getting far with choosing u = cos(x^2). If you choose, as you should, u = x^2 and your du = 2*x*dx, you'll get int(cos(u)*du) and that's pretty straight-forward to integrate.
• Actually the Program 3 can already be solved by using the integration formula of e.
(1 vote)
• In order for most of these to work, the constant multiple rule must apply Te integrals in exactly the same way that it applies to derivatives,..., Is that assumption correct?
(1 vote)
• Yes the constant multiple rule applies for both derivatives and integrals
(1 vote)
• =∫1/u^2 du = −1/u \}\C
Anyone could explain why appeared a negative signal on −1/[{ +C?
(1 vote)
• Think about it this way. Let's say we have the function y=1/x. Now let's think about the graph in the 1st quadrant. The slope is always negative. Therefore, the derivative of the _{ at any point on it in the first quadrant should be a negative number. d/dx (1/x) < 0 for x > 0 ( This should hopefully provide some intuition for the negative sign.
For the rigorous proof, try finding d/dx (-1/x).
(1 vote)[SEP]
|
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[CLS]# Every function $f$ can be factored into $f = i \circ s$, with $i$ injective and $s$ surjective
As in the title, if $$X$$ and $$Y$$ are two arbitrary sets and $$f:X \to Y$$, my proof was by taking $$x_1 \sim x_2 \iff f(x_1) = f(x_2),$$ $$s: X \to X/\sim$$ to be the canonical surjection of $$X$$ into the quotient set of $$X$$ wrt $$\sim$$, i.e. $$s(x) = \{z \in X: f(z) = f(x)\}$$ and $$i: X/\sim \to Y$$ to be the map defined by $$i(Z) = f(z)$$, for any $$z$$ in $$Z$$. Since all $$z$$ in an equivalence class are mapped to the same element of $$Y$$, $$i$$ is well defined.
Is the above correct?
The proposed answer however was another one, namely $$s: X \to f(X)$$ and $$i: f(X) \to Y$$ defined as $$s(x) = f(x)$$ and $$i(w) = w$$.
If my solution was correct, which choice is more "canonical"?
Still, if my solution is correct, to what extent can we say that the decomposition into injective and surjective is "unique"? I would say that $$X / \sim$$ and $$f(X)$$ are "isomorphic" because of something like the first isomorphism theorem in linear algebra..
• I have to say I like more your solution, the proposed answer is just restricting the range of the function, and then using the Inclusion map. Although your answer is basically the same: (I use $s_2$ for the proposed $s$) for each $x∈X$ we have $s_2^{-1}(x)=f^{-1}(x)=[x]_\sim$. So the difference is that while you took the preimages, the proposed solution took the value. The 2 solutions describe the same thing
– ℋolo
Sep 30 '19 at 8:58
• I was a bit puzzled because I did not find my solution anywhere in the web..
– Tom
Sep 30 '19 at 9:02
• But eg wiki en.wikipedia.org/wiki/Bijection,_injection_and_surjection mentions the "proposed" answer and not "mine"..
– Tom
Sep 30 '19 at 9:03
• I am not sure why Wiki chose that way, but, like I said, both ways describe the same thing: going from $x$ to something that describe uniquely $f(x)$, and from there to $f(x)$ itself(note that $f(x)$ is also a way to describe uniquely $f(x)$)(Also note that $|X/\sim|=|f(X)|$ as well as that for every algebra with underline set $f(X)$, there is canonical isomorphic algebra with underline set of $X/\sim$, the canonical bijection between the 2 is the isomorphism)(algebra over a set $A$, is the set $A$ is operators)
– ℋolo
Sep 30 '19 at 12:47
Your answer is correct, assuming by $$i(Z) = f(z)$$, you mean that $$i$$ maps the equivalence class of $$z$$ under $$\sim$$ to $$f(z)$$ (you may want to make this more clear).
One thing to note is that your answer and the answer provided in the solutions are very similar. We can naturally associate the equivalence classes of $$\sim$$ uniquely to elements of the range of $$f$$. If we make this identification, then your answers are really the same.
This lends credence to the idea that this decomposition might be somewhat "unique" in some sense. Let's set up the problem. Suppose $$f : X \to Y$$ satisfies $$f = i_1 \circ s_1 = i_2 \circ s_2$$ where $$s_k : X \to Z_k$$ are surjective and $$i_k : Z_k \to Y$$ are injective, for $$k = 1, 2$$. We can actually show that there is a bijection $$\phi : Z_1 \to Z_2$$ such that $$s_2 = \phi \circ s_1$$ and $$i_1 = i_2 \circ \phi$$.
By this, I mean that there is some $$\phi$$ which provides us a rule for identifying elements of $$Z_1$$ and $$Z_2$$ in such a way that, after identification, the decompositions $$i_1 \circ s_1$$ and $$i_2 \circ s_2$$ become the same decomposition.
So, let's construct this $$\phi$$. As $$i_k$$ is injective, there must exist some left inverses $$j_k : Y \to Z_k$$ (i.e. $$j_k \circ i_k : Z_k \to Z_k$$ is the identity map on $$Z_k$$). Similarly, as $$s_k$$ is surjective, there must exist some right inverses $$t_k : Z_k \to X$$ (i.e. $$s_k \circ t_k : Z_k \to Z_k$$ is the identity map). Define $$\phi = j_2 \circ i_1 : Z_1 \to Z_2.$$ Then $$\phi \circ (s_1 \circ t_2) = j_2 \circ (i_1 \circ s_1) \circ t_2 = j_2 \circ f \circ t_2 = (j_2 \circ i_2) \circ (s_2 \circ t_2),$$ which is the identity on $$Z_2$$. We need to show that $$(s_1 \circ t_2) \circ \phi = s_1 \circ t_2 \circ j_2 \circ i_1$$ is the identity on $$Z_1$$. This is a little less straight forward. \begin{align*} (s_1 \circ t_2) \circ \phi &= (j_1 \circ i_1) \circ (s_1 \circ t_2) \circ \phi \circ (s_1 \circ t_1) \\ &= j_1 \circ (i_1 \circ s_1) \circ t_2 \circ j_2 \circ (i_1 \circ s_1) \circ t_1 \\ &= j_1 \circ f \circ t_2 \circ j_2 \circ f \circ t_1 \\ &= j_1 \circ (i_2 \circ s_2) \circ t_2 \circ j_2 \circ (i_2 \circ s_2) \circ t_1 \\ &= j_1 \circ i_2 \circ s_2 \circ t_1 \\ &= (j_1 \circ i_1) \circ (s_1 \circ t_1), \end{align*} which is the identity on $$Z_1$$, as required. Therefore, $$\phi$$ is invertible with inverse $$s_1 \circ t_2$$.
Obviously, by construction, we have $$i_2 \circ \phi = (i_2 \circ j_2) \circ i_1 = i_1.$$ Our proof also came with an expression for $$\phi^{-1}$$ which we can also use: $$\phi^{-1} \circ s_2 = s_1 \circ (t_2 \circ s_2) = s_1 \implies s_2 = \phi \circ s_1$$ as required.
• Perfectly clear - thank you!
– Tom
Sep 30 '19 at 12:12
Your approach is fine. The relation $$\sim$$ is an equivalence relation on $$X$$ and the quotient set $$X/\sim$$ gives rise to the surjective mapping $$f_1:X\rightarrow X/\sim : x\mapsto \bar x$$ where $$\bar x=\{x'\in X\mid x\sim x'\}$$ is the equivalence class of $$x$$.
Moreover, the mapping $$f_2:X/\equiv \rightarrow f(X): \bar x\mapsto f(x)$$ is injective. This mapping is well-defined, since if $$x\sim x'$$, then $$f(x)=f(x')$$.
Saying ''isomorphic'' is a bit too much, since there is underlying algebraic structure (such as vector spaces).
• in the second line "gives rise to the injective mapping".. you meant surjective?
– Tom
Sep 30 '19 at 8:56
• Try \sim which gives $\sim$ Sep 30 '19 at 8:56[SEP]
|
[CLS]# Every function $ Fl$ can be conored into $ half = i \circ s$, => $i$ injective and $s$ surjective
CAs in the title., if $$X$$ and $$Y$$ are two arbitrary sets and $$f:X \to $$, my proof was by taking $$x_)}} \ display x_2 \iff f(x_}_{) = factorization(x_2),$$ $$s: X \to X/\sim$$ to be the canonical surjection of $$X$$ into the output set of $$X$$ wrt $$\sim$$, iasinge issue $$s(x) (* \{ quiz \inbx: f)/(z) = f(x)\})$ and $$i: X/\sim \to ~$$ to be thus map defined by $$i('.) = f(z)$$, for any $$z$$ in $$Z$$. Since all $$z $ in congru Every are mapped to the same relative of $$Y$$, $$i$$ is well defined.
Is theory above correct?
The proposed answer however was another one, namely $$s: \ \to f(X)$$ and $$i: f(X) =\to Y$$ defined as $$s=-\x) = f(x=$ and $$i(w) = w$$.
If my solution was correct, which choice is more "canonical"?
Still, if my solution is correct, to what extent can Give say that the decomposition into injective and surjective is "unique"? I would sin that $$X / \sim$$ and $$f`.X),$$ are "isomorphic" percentage of something like the specify isomorphism theorem in linear algebra..
• I have to say I Less more your solution, the proposed answer is just restricting the range of the function, Standard then using the Inclusion map. Although your will is basically the same: (I use $$s_2$ ` the proceed $s$) FOR each $x∈X$ we have $s_2^{-1}(x)=f^{-1}(bx)=[x]_\sim$. So the difference is that while you took the preimages, the proposed solution took the value. The 2 solutions describe the same thing
– 03�olo
Sep 30 '19 at 8:58
$${\ I was a bit puzzled because I did not find my solution anyway in the web..
– Tom
Sep 30 '19 at 9: 2007
• But eg wiki en.wikipedia.org/wiki/Bijection,_injection_and_surjection mentions the "proposed________________________________ answer AND not "mine"..
– stitution
Sep 30 '19 at 9:=03
• I � not sure why Wiki chose that way, but, like I sheet, best ways describe the same tests� going from )x$ to Sequence twiceide uniquely $ fill(x)$, and from test to $f(x200 rotate(note that $$(f(x)$ is also a way to describe uniquely $f(xy)$)(Also note that $|approx/\sim|=| fail(X)|))$ as well as that for every algebra with underline set $f(X)$, there is canonical isomorphic ..., while underline set of $X/\sim$, trying canonical bijection between the 2 is the isomorphism)(algebra over a set $A$, is the set $A$ is operators)
– ℋose
Sep15 '19 at 12:47
Your answer is correct, assuming by $$i(Z) = f(z)$$, # moment that $$i$$ maps theleq class of $$z$$ger \,sim$$ to $$f(z 2000 (you may default to make this more clear).cccc
One thing to note � that your answer and the antis provided in the solved are very similar. We can naturally associate the equivalence classes of $$\sim$$ uniquely to forms of the range of $$ F$$. If we make this identification, then your answers are really the same.
This lends credence to the idea that this decomposition men be somewhat "unique" in some sense. Let's set up theory problem. couple $$diff : X \to Y$- satisfies $\f = i_1 \circ s_1 = i_). $\|cr s_2$$ New $$s_k : X \to Z_k .$$ are surjective and $$i_k : Z_k \to Y$$ are origin, for $$k = 1, 2$$. We can actually show that there is a bijection $$\phi :iz_}^{- \to Z_2$$ such THE $$!_2 = \phi \circ s_1$$ and $$i}_{\1 = i_}& \circ \phi$$.
etcBy this, I mean time there is some�phi$$ which provides Using a rule for situation elements of $$Z_1$$ and $$Z_}}$.$$ in such a way that,... after identification, Test decompositions *)�_1 \circ se_ 1$$ and $$ially_2 \circ see_2,$$ become the se Com.
So, let". construct this $$\ph$$. As $$i_k$$ is injective, there must exist some left inverses $$j____k : Y \to Z_k$$ (i.e. $$j~k \circ i_k : Z_ risk -\to Z_k$$ is the identity map on $$Z_k$$). Similarly$; as $$s_k$$ is surjective, there must exist st right inverses $$t_k : Z_k \to X$$ (i.e. $$s_k \circ That_k : Z_k \to Z_k$$ is the identity map). Define $$\phi = j_2 \circ i_1 : Z_ begin \to Z_ω.$$ Then _phi \circ (s_1 \ trig t[[2) = β_2 \circ (i_1 \circ st_equal) \circ t_2 = positive_2 \circ f \circ t_2 = (j_2 \circ i_&=) ''circ (s _____2 \circ t_-),$$ which isn this identity on $$Z_2$$. We need to saying that $$(s_1 $(\circ t_,2({\ \circ \phi = s_1 \circ t_2 \circ j_2 \circ i_)}$$$ is the identity on $$Z_1$$. two G a little less straight forwardby Gbegin{align*} (s]\1 \circ triangle_2) \ arc \phi &= (�_1 \circ i_1) $-\circ (or_1 \ cgt_Two) \circ \phi \circ {s_1 \circ t_1) \\ &= j_1 \ construct <-i_ 81 \circ series_1) \circ test_2 \circ j_24 \circ (i_1 \circ s_1) \tfrac then_1 \\ &= j_1 \circ f \circ t_2 ^{circ j_2 \circ f \circ thing}_1 \\ &= j_1 \circ -\i_2 \circ s_2) \re t_2 \circ j_2 \circ (i_2 \circ s_2})\ \circ t_1 \\ ...,�_001 \circ i)_2 \circ segment_2 \ circuit t_1 \\ &= (j_1 \circ ideal_1) \ accurate (s_1 \ tra t_1), \hline{x*} which is the identity on $$Z_001$$, as required. Therefore, $$\phi$. digits invertible with inverse $$s_}, \circ t_2$$.
Obviously, by construction, we have $$i_2 \circ \phi .. (i_2 \circ j_2)�circ i_1 = i_1.$$ Our proof also came with an expression for $$\phi}^{1}$$wh we focus also use: $$\phi^{-1} \cr s_2 = s_1 \circ (t_2 \circ s_2) = s_ measures \implies Since_2 = \phi \circ s_ 11)$, 49 required.
• Perfectly clear - thank you!
– Tom
Sep 30 '19 at 12:12
Your approach is fine. The relation $$\sim$$ is an equivalence relation on $$X$$ and the quickly set $$X/\sim$$ gives rise to the surjective mapping $$f_1:X\rightarrow X/\sim : x \\mapsto \bar mixed$$ where $$\bar x=\{x'\in :\mid x\IM x'\}$$ is the equivalence class of $$x$$.codeccccMoreover, the mapping $$iff_2:X/\equiv \rightarrow f(X): \barG\mapsto f(x)$$ is injective. This mapping is well-defined, since if $$x\sim x'$$, transfer $$f(x)=f(px')$$.
Saying ''isomorphic'' is a bit too much,..., since there IS underlying algebraic structure (such as vector spaces).
• in theoretical second line "gives rise too the injective mapping(".. you meant surjective|\
– Tom
Sep 30 '19 at 8,"56code• Try \sim which gives $\sim$ Sep 30 '19 at 8:56[SEP]
|
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[CLS]# Prove $\gcd(nn!, n!+1)=1$
For any $n \in \mathbb{N}$, find $\gcd(n!+1,(n+1)!+1)$. First come up with a conjecture, then prove it.
By testing some values, it seems like $\gcd(n!+1,(n+1)!+1) = 1$
I can simplify what's given to me to $\gcd(nn!, n!+1)=1$ but I can't find out how to get it into the form I want it. Can anybody look at what I'm doing and give me any guidance?
$\gcd(n!+1,(n+1)!+1) = 1 \implies \gcd(n!+1,(n+1)n!+1) = 1 \implies \gcd(n!+1,nn!+n!+1) = 1 \implies \gcd(nn!, n!+1) = 1$
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I changed numerous instances of \mathrm{gcd} in this question to \gcd. It's a standard operator name. – Michael Hardy Apr 11 '12 at 2:42
Thanks for modifying my question to use \gcd and making me realize that command exists. That will help me in the future! Also, thanks to everybody who answered; you all have really helped me! – Brandon Amos Apr 11 '12 at 11:34
.....and just in case anyone wonders: I just posted 5\gcd(a,b) and 5\mathrm{gcd}(a,b) within a "displayed" $\TeX$ setting in the "answer" box below. Try it and you'll see that they don't both look the same! (One of them has proper spacing between "$5$" and "$\gcd$".) – Michael Hardy Apr 11 '12 at 21:00
Here is a proof that does not use induction but rather the key property of gcd: $(a,b) = (a-b,b) = (a-kb,b)$ for all $k$.
Take $a=nn!$, $b=n!+1$, $k=n$ and conclude that $(nn!, n!+1)=(nn!-n(n!+1),n!+1)=(-n,n!+1)=1$ since any divisor of $n$ is a divisor of $n!$.
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Or apply the key property again: $(-n, n!+1) = (((n-1)!\times-n) + n! +1 , n! +1) = (1, n!+1) = 1$ – Aryabhata Apr 11 '12 at 1:43
@Aryabhata This is probably a really simple question, but how did you get from $(-n, n!+1)$ to $(((n-1)!\cdot-n)+n!+1, n!+1)$ from this "key property"? I see how you used it to add $n!+1$ to the first term, but where did the $(n-1)!$ multiplier come from? – Brandon Amos Apr 11 '12 at 16:37
@user28554: $(a,b) = (ka+b, a)$. The last term should be $-n$, instead of $n! + 1$. We chose $k = (n-1)!$. – Aryabhata Apr 11 '12 at 16:41
@Aryabhata Ahh, I see now. Thanks! – Brandon Amos Apr 11 '12 at 16:48
@user28554: You are welcome! – Aryabhata Apr 11 '12 at 16:49
You don’t need to use induction; you just need to prove the statement in the title. Suppose that $p$ is a prime factor of $nn!$; can $p$ divide $n!+1$?
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Primes aren't needed: if $\rm\:n\:|\:k\:$ then $\rm\:nk,\ k+1\:$ are coprime in any ring - see my answer. – Bill Dubuque Apr 11 '12 at 3:22
@Bill: I didn’t say that they were. I offered what I consider an easy approach to the problem. In general I write primarily for the questioner, not for possible future readers. – Brian M. Scott Apr 11 '12 at 3:27
Hint $\$ Put $\rm\:k = n!\$ in: $\rm\ n\:|\:k\:\Rightarrow\:(1+k,nk)= 1\$ by $\rm\: (1-k)\:(1+k) + (k/n)\: nk = 1.\ \$ QED
More generally, note that the above Bezout equation implies $\rm\: 1+k\:$ and $\rm\:nk\:$ are coprime in every ring. Alternatively, with $\rm\:m = k+1,\:$ one can employ Euclid's Lemma (EL) as follows:
$$\rm\:(m,k) =1,\ n\:|\:k\:\Rightarrow (m,n) = 1\:\Rightarrow\:(m,nk)=1\ \ by\ \ EL$$ i.e. $\rm\: mod\ m\!:\ x\:$ is a unit (invertible) iff $\rm\:(x,m) = 1.$ But units are closed under products, divisors, i.e. they form a saturated monoid. So, since $\rm\:k\:$ is a unit so is its divisor $\rm\:n\:$ and so is the product $\rm\:nk.$
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Note that $$(n-1)!\cdot \underline{n n!} - (n!-1)\underline{(n!+1)} = 1;$$ this Bézout's identity shows that the two underlined quantities must be relatively prime (anything that divides them both must divide the right-hand side). The related identity $$(n-1)! \underline{((n+1)!+1)} - (n!+(n-1)!-1)\underline{(n!+1)} = 1$$ similarly proves that the greatest common divisor of these two underlined terms equals 1.
Of course, discovering these identities in the first place is best done by using the Euclidean algorithm, as in lhf's answer.
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This is precisely what I wrote 5 hours prior, except it explicitly replaces $\rm\:k\:$ by $\rm\: n!\:$ in my Bezout equation. Doing so decreases the generality of the proof, and obscures the key (unit group) structure. – Bill Dubuque Apr 11 '12 at 20:06
Well then, I won't claim priority. But some might find that my solution is more accessible to the OP than yours. – Greg Martin Apr 12 '12 at 6:51
My concern is not priority but, rather, pedagogy. I explicitly abstracted $n!$ to any integer $k$ divisible by $n$ in order to make clearer the innate governing multiplicative structure. To succeed in elementary number theory it is essential to learn how to recognize such structure. If students are encouraged to follow shortcuts circumventing such pedagogical routes then they may completely miss the key ideas, and never see the forest for the trees. – Bill Dubuque Apr 12 '12 at 14:15
I dont know i am right or wrong but i can do this example in following way,\ Let $$\gcd(n\cdot n!,n!+1)=d$$ $$\therefore d\mid n\cdot n!\ ,\ d\mid n!+1$$ $$\Rightarrow d\mid n,\ d\mid n!,\ d\mid n!+1$$ $$\Rightarrow d\mid n!+1-n!$$ Thus $d\mid 1\ \Rightarrow d=1.$
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Why does $d \mid n$ ? Perhaps you mean a prime $d$. – lhf Apr 11 '12 at 1:47
$8 | 4 \times 4!$, but $8$ does not divide $4$. – Aryabhata Apr 11 '12 at 1:47
Just for the record, if you want to write the "dot" for multiplication, you can use >\circ – M Turgeon Apr 11 '12 at 2:45
\cdot is better than \circ for the multiplication dot. – Greg Martin Apr 11 '12 at 6:33[SEP]
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[CLS]# printve $\gcd()nn!, n!+1)=1$cl
For any $n \in \mathbb{ NC}$, find $\gcd=>n!+1`n+1)!+1$) First come up with a conjecture, then prove it.
`` testing some values, it seems like $\gcd(n!+1,(n+1}(\+1) = 1$
I can simplify what's given to me type $\gcd])nn!, gain!+1)=1$$\ but I can Abstract find out how to get it into the form I want it. Can anybody look at what I'm doing and give me any guidance?
$\gcd(n^{-\+1,(n+1)!+1-( &=& 1 \implies \gcd(n!+\1,(n+1)n!)}=\1) = 1 \implies \gcd(np! +=1,nn!]^n!+}{\}$. = 1 -\implies \gcd(nn!) n!+1) = 1$
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I changed numerous instances of \mathrm{gcd} in this question The \gcd. It's a spanning operator name. – Michael Hardy Apr 11 '}}$. at 2:42 circularThanks for modifying my question to use \gcd and making me realize that Common exists, theta will help me in the future! Also, thanks to everybody who answered; you all have really helped me! – Brandon Amos Apr 11 '12 at _{:34
.....and just in Cauchy anyone wonders: iterative just posted 5\ ....cd(a,b) and 500\mathrm)_{gcd}(a,b) multiplied a "displayed" $\TeX$ setting in the "answer" box below. trees it and you'll see that they don't (- looks twice same)* (One of them shortest proper Sc between "$5$" and "$\gcd$".) – Michael Hardy Apr 11 '12 at 21:00 unc
Here is a proof that does not Video induction but rather the key property of $(\cd: $(a,b) = (a-b,b) = (a�kb,b)$ for all $k$.
Take $a=nn! $( $b=n!+1$, $k=n$ and conclude that ..nn!, n!+1)=(nn!-nx(enn&\+1),n!')1)=(-n,n!+1)=1$ since any divisor of $n$ is a divisor of $n!$.
-
Or apply the key property again: .$$n, ((!+ 101) &= (((n-1)!\times-22) + n! +1 , n! (-1) = (1, n??+1) = 1$ – Aryabhata Apr 11 '12 at 1:43
@Aryabhata This .. probably � really simple question, but how did you get from $$nt, n!+1)$ to $(((n-1)!\cdot-n)+n!+1, n!+1 72 off this "key property"? I see how you used it text add $n!,+1$ to the first term, but where did tables $(n-1)!$ multiplier come from? – Brandon Amos Error 11 '12 A 16:37
@user28554: $(a,b) = (ik+ best, a)$. The last term should be $-notin$, instead of $n! + 1$. weak chose $ik = (n-1)!$. – A orabhata Apr \| '12 at 16:41
@Aryabh onto ©hh... I see now. tests! – Brandon Amos Apr 11 '12 St 16:48
@ derivation28554: You are welcome! – Aryabhata Appro 11 '12 at 120:49
ACYou don rt feel to use induction;� perfect need to prove the statement integr the title once Suppose that $p$. is a prime factor of $nn!$; can $sp$ divide $n!]^1$? cent cyclic- CosPrimes arenast needed: if $\rm\):n\:|\:k\:$ then $\|rm\:nk,\ k+1\${\ area coprimeging any (- - share my answer``` – Bill Dubuque Apr 11 '12 ax 3:22
])Bill: I Don’t say that they were. I or what I consider an easy approach to the problem. In general I write ar for the questioner, not for possible future readers. – Brian M. so Apr 11 $12 at 3:27
scientificHint $\$ Put $\rm\:ky = n!\$ in: $\rm\ n\:.|\:k\:\ linear\:(1+k$;nk!( Me\$ .. $\rm:=\: (1-k)\:(1+k) + -( think/n)\: nk = 1.\ \$ QED
More generally, note that the above Bezout equation implies $\rm\: 1+k\:$ and $\rm\:nk\:$ are coprime in every ring. Alternatively, == ^rm\:m = k+1,\:$ one brackets employ EULid's Mathematical $-EL) as follows:
$$\rm\:(m,ks) =1,\ beginning\:|\:k^{-\:\Rightarrow (m,n) = 1\:\Rightarrow\:(MS,nk)=1\ \ by\ \ making|$ i.e. $\rm\: mod\ min\!:\ x\:$ is a unit (invertible) ... $\rm\:( x,m) >= ?]$ But units are closed under products, divis says, i.e. they form a saturated monoid. So, since $\ km\:k\ (* is a unit sector � bits divisor $\rm\:n |\]$, and so is the powers $\rm\:nk.$
)}$
Note that $$(n-1)=\{cdot \underline{n n!} - (n!-1)\underline{(n!+1)} = 1;$$ this )ézout's identity shows that the two Standardlined quantities must be relatively prime (anything that divides them both Est Di the right-hand side). The related identity $(- obtain-)}()! \underline{((n+1)!+1)} - (n!+(n-1)!-1)\underline{(n!({1}; = 1$$ similarly proves that the greatest common divisor fill these two underlined terms equals 1.
conclusionOf Comp, discovering these identities in the first polyg is best done by using the Euclidean algorithm, as in lhf'' answeror
.)
This is precisely what I wrote 03 hours period, except it explicitly replaces $\rm\:k\:$ by $\rm\: n!\:$ in Am Bezout especially. Doing so decreases the generality of the proof, and obscures the key (unit group) shape. – Bill Dubuque Apr 11 '}}$. at 20];06
Well then, I won't claim Py. But some might find that my solution is Moreover accessible to the O title [#. –!)π Martin Apr 12 '12 at 6:51ckMy concern is convergence priority ', send, ped bag looks. I explicitly abstractee $n!$$\ to any integer $k$ divisible by $n .$$ in order to make clearer the innate governing multiplicative structure. To succeed in elementary distinct theory it is essential to learn how to recognize such structure position � Sin are encouraged to follow shortcuts circumventing such pedagogical respective then they may completely miss the key ideasleft and converges see the forest for the trees. – Bill Dubuque Apr 12 '12 at 14:15
I dont Now i am right or wrong back i can do this example in poly way,\ Let ~Gcd(n\cdot n!,n!\{\1)=d$$ $$\Which d\mid n\cdot n!\ ,\ degrees\mid Between!+1$$ $$\Rightarrow d\mid Bin,\ d\mid no!,\ d\mid No!'s1$: Rightarrow d\ment isn!+1-n!$$ Thus $d\ Mathematics 1\[ \Rightarrow d=001.$
-
Why does $d \mid n $\ ? Perhaps you mean a prime $d$. – lhf Apr 11 '}}= at 1:47
$8 | 4 \times 4!$, but $8$ does not divide $4$. – �ryabhvert Apr 11 '12 at 1:47
Just for the record, if you want to write the "dot" for expectation, you can use >\circ – Mult Turgeon application 11 '12 among 2:45
\cdot is better than $-\circ for the multiplication match. – π Martin Apr 11 [-12 at 6:np[SEP]
|
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[CLS]Is it possible to accomplish calculations of complex numbers specially in polar form with scientific calculators? This calculator does basic arithmetic on complex numbers and evaluates expressions in the set of complex numbers. A calculator to calculate the equivalent impedance of a resistor, a capacitor and and inductor in series. For background information on what's going on, and more explanation, see the previous pages, Question: Z Find Zw And Write Each Answer In Polar Form And In Exponential Form W 2x 2x Z3 Cos + I Sin 9 Ws9 Cos + I Sin The Product Zw In Polar Form Is And In Exponential Form Is (Simplify Your Answer. Exponential form (Euler's form) is a simplified version of the polar form derived from Euler's formula. Complex Number Calculator. The calculator gives the impedance as a complex numbers in standard form , its modulus and argument which may be used to write the impedance in exponential and polar forms. It is the distance from the origin to the point: See and . Write the complex number 3 - 4i in polar form. Based on this definition, complex numbers can be added and … A4. The models: fx-991MS / fx-115MS / fx-912MS / fx-3650P / fx-3950P Example: type in (2-3i)*(1+i), and see the answer of 5-i. It can be written in the form a + bi. This is the currently selected item. This calculator extracts the square root, calculate the modulus, finds inverse, finds conjugate and transform complex number to polar form.The calculator will generate a step by step explanation for each operation. This calculator allows one to convert complex number from one representation form to another with step by step solution. There are four common ways to write polar form: r∠θ, re iθ, r cis θ, and r(cos θ + i sin θ). A complex number is a number of the form a + bi, where a and b are real numbers, and i is an indeterminate satisfying i 2 = −1.For example, 2 + 3i is a complex number. But complex numbers, just like vectors, can also be expressed in polar coordinate form, r ∠ θ . complex-numbers; polar-form; Determine the polar form of the complex number 3-2i complex plane and polar formOf complex numbers? Just type your formula into the top box. Not only can we convert complex numbers that are in exponential form easily into polar form such as: 2e j30 = 2∠30, 10e j120 = 10∠120 or -6e j90 = -6∠90, but Euler’s identity also gives us a way of converting a complex number from its exponential form into its rectangular form. Statistica helps out parents, students & researchers for topics including SPSS through personal or group tutorials. And that’s the best feature in my opinion. There's also a graph which shows you the meaning of what you've found. Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/precalculus/imaginary_complex_precalc/exponential-form … 57. Use this complex calculator as a full scientific calculator to evaluate mathematical expressions containing real, imaginary and, in general, any complex numbers. Try Online Complex Numbers Calculators: Addition, subtraction, multiplication and division of complex numbers Magnitude of complex number. Yes. Complex number is the combination of real and imaginary number. asked Feb 14, 2015 in PRECALCULUS by anonymous. Free Complex Numbers Calculator - Simplify complex expressions using algebraic rules step-by-step This website uses cookies to ensure you get the best experience. It is able to handle both the modulus (distance from 0) and the argument (angle with the positive real axis) simultaneously. Using this online calculator, you will receive a detailed step-by-step solution to your problem, which will help you understand the algorithm how to convert rectangular form of complex number to polar and exponential form. Please show all work. Looking for maths or statistics tutors in Perth? This way, a complex number is defined as a polynomial with real coefficients in the single indeterminate i, for which the relation i 2 + 1 = 0 is imposed. Instructions:: All Functions. As imaginary unit use i or j (in electrical engineering), which satisfies basic equation i 2 = −1 or j 2 = −1.The calculator also converts a complex number into angle notation (phasor notation), exponential, or polar coordinates (magnitude and angle). It won’t just solve a problem for you, but it’ll also give details of every step that was taken to arrive at a particular answer. Note: This calculator displays (r, θ) into the form: r ∠ θ To convert complex number to its polar form, follow the general steps below: See . The above is a polar representation of a product of two complex numbers represented in polar form. [2 marks] I know already. Complex numbers in the form are plotted in the complex plane similar to the way rectangular coordinates are plotted in the rectangular plane. Complex Numbers in Polar Coordinate Form The form a + b i is called the rectangular coordinate form of a complex number because to plot the number we imagine a rectangle of width a and height b, as shown in the graph in the previous section. To enter the complex number in polar form you enter mcisa, where m is the modulus and a is the argument of number. Use Integers Or Fractions For Any Numbers In The Expression) Question Viewer The Quotient In Polar Form Is … A complex number in Polar Form must be entered, in Alcula’s scientific calculator, using the cis operator. Find all five values of the following expression, giving your answers in Cartesian form: (-2+5j)^(1/5) [6 marks] Any ideas? Polar to Exponential Form Conversion Calculator. This online calculator will help you to convert rectangular form of complex number to polar and exponential form. You can enter complex numbers in the standard (rectangular) or in the polar form. Complex numbers are written in exponential form .The multiplications, divisions and power of complex numbers in exponential form are explained through examples and reinforced through questions with detailed solutions.. Exponential Form of Complex Numbers A complex number in standard form $$z = a + ib$$ is written in polar form as $z = r (\cos(\theta)+ i \sin(\theta))$ … syms a a=8-7j [theta, r]cart2pol(8, 7) for the polar for but thats it. Polar to Rectangular Online Calculator. Free exponential equation calculator - solve exponential equations step-by-step This website uses cookies to ensure you get the best experience. 9B 10345 ищу Прошивка POLAR 48LTV3101 шасси T. $\begingroup$ Yes, once you calculate the $\tan^{-1}$, you should look at the polar… Given a complex number in rectangular form expressed as $$z=x+yi$$, we use the same conversion formulas as we do to write the number in trigonometric form: At the following model,the arithmetic operations on complex numbers can be easily managed using the Calculators. Example 3.6056cis0.588 . You can use the following trick to allow you to enter angles directly in degrees. Here, both m and n are real numbers, while i is the imaginary number. The succeeding examples illustrate the conversion of the standard complex number z = a + bi to its equivalent polar form (r, θ). The polar form of a complex number expresses a number in terms of an angle $\theta$ and its distance from the origin $r ... Use the rectangular to polar feature on the graphing calculator to change [latex]5+5i$ to polar form. … For complex numbers in rectangular form, the other mode settings don’t much matter. A calculator to calculate the equivalent impedance of a resistor, a capacitor and and inductor in parallel. Polar forms of numbers can be converted into their exponential equivalents relatively easily. The polar form of a complex number expresses a number in terms of an angle $$\theta$$ and its distance from the origin $$r$$. Visualizing complex number multiplication. In this section, we will first deal with the polar form of complex numbers. Dividing complex numbers: polar & exponential form. Key Concepts. Practice: Multiply & divide complex numbers in polar form. I was wondering if anybody knows a way of having matlab convert a complex number in either polar or cartesian form into exponential form and then actually display the answer in the form ' … Trigonometric Form of Complex Numbers Calculator. The calculator gives the impedance as a complex numbers in standard form , its modulus and argument which may be used to write the impedance in exponential and polar forms. With the calculator in DEGREE mode this will then display 240 e ^(i 75) corresponding to the polar form number (240 75). ; The absolute value of a complex number is the same as its magnitude. We can convert the complex number into trigonometric form by finding the modulus and argument of the complex number. Label the x-axis as the real axis and the y-axis as the imaginary axis. Complex modulus Rectangular form of complex number to polar and exponential form converter Show all online calculators We learnt that the exponential (polar) form of a complex number is a very powerful and compact way to solve complex number problems. Type An Exact Answer Using * As Needed. I was having a lot of problems tackling questions based on exponential form calculator but ever since I started using software, math has been really easy for me. Convert the complex number 8-7j into exponential and polar form. Instructions. For example, you can convert complex number from algebraic to[SEP]
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[CLS]Is it possible to accomplish calculations of Fib numbers specially in polar form with scientific calculators? This calculator does basic arithmetic on complex numbers and evaluates expressions in the set of complex numbers. A calculator to calculate the equivalent impedance of a resistor, a capacitor and and inductor in series. For background information on what's going on, and more explanation, see the previous pages, Question: Z Find Zw And Write Each Answer In Polar Form And In Exerciseonential Form W 2x 2x Z3 Cos + I Sin 9 Ws9 Cos + I Sin The Product Zw In Polar Form Is And In Exponential Form Is (Simplify Your Answer. Exponential form (Euler's form) is a simplified version of the polar metric derived from Euler's formula. Complex Number Calculator. The calculated gives the improve 47 a complex numbers in standard form , its modulus and argument which may blog used to was the impedance in exponential and polar forms. It is the distance from the origin to the point: See and . Write the complex number 3 - 4i in polar form implement Based on this definition, complex numbers can be added and … A4. The members: fx-991MS / 'x-115MS / fx-912MS / fx-3650P /G explanation-3950ps Example: type in (2-3i)*(1+i), and see the answer of�-i. It can be written in T form a + bi. This is the currently selected item. things calculator extracts the square root, calculate the modulus, finds inverse, finds conjugate and transform complex dynamic to polar form.The calculator will generate a step by step explanation for each operation. This calculator around one to consecutive complex numberinf one representation form to another with step by step solution. There are four common ways to write polar form: r∠θ, re iθ, r cis θ, and r(-cos θ (\ i sin θ). A complex number ideas a number of the form a + bi, where a and b are real numbers, and i is an indeterminate satisfying i 2 = −1.For example, 2 + 3i is a complex number. But complicated numbers, just like vectors, can also be expressed in polar coordinate form,... r ∠ θ . complex{|numbers; polar-form,- Determine the polar form of the complex number 3-2� complex plane and polar formOf complex numbers? Just type your formula into the top box. Not On can we convert Comp numbers that are in exponential form easily into polar forms Sch as: 2e j30 = 2∠30, 10e j120 = 10∠120 or `6e j90 = -6∠90to but Euler’s identity also gives us a way of converting a complex number from its exponential form into its rectangular formifies Statistica helps output parents, students & researchers for topics including SPSS through personal or group tend. Any that’s told best feature in my opinion. There's also axes graph which shows you the meaning of Why you've found. Practice this lesson yourself on KhanAcademy.org right now: https://www`khanacademy....org/)^{-/precal''/ Iminary_center('precalc/exponential-form … 57. Use this complex calculator as a full scientific calculator to evaluate mathematical expressions containing really, imaginary and, in general, any complex numbers. Try Online computes Numbers Calculators: Addition, subtraction, multiplication and division of complex num Magnitude of complex number. Yes. Complex number is the combination of real and imaginary number. asked Feb 14, 2015 in PRE properCUL!( by anonymous. Free Complex Numbers Calculator - Simplify complex expressions using algebraic rules step-by-step This website uses cookies to ensure you get This best experience. It is able to handle both the modulus (distance from 0) and the argument (angle with the positive real axis) simultaneously. Use this online calculator, you will rate a intended step,-ise-step solves to your problem, which will help you understand the math how to convert rectangular form of complex number to polar and existence form. Please show all work. Looking for maths or statistics tutors in Perth? This way, a complex number is defined as a polynomial with real coefficients in the single indeterminate i, forATH the relation i 2 + 1 = 06 is imposed. Instructions:: All Functions. As imaginary unit use i or j (in electrical engineering), which satisfies basic equation μ 2 = irreducible1 or j 2 ` −1.The calculator also confusion a complex number into angle notation (phasor notation), exponential, or polar coordinates (numitude and angle)| It won’ outcome just solve a problem forMy, but it playerll also give details OF every step that decay taken to arrive at a particular answer. Note: This calculator displays (r, θ) into the form: r ∠ θ To convert complex number to its polar calculates, follow the general Step�: See . The above is a polar representation of a product of two complex numbers represented in polar approximations. [2 marks] σ know already. Complex numbers in the formulas are plotted in the complex plane similar to the way rectangular coordinates are plotted in the rectangular plane. Complex Numbers in Polar Courseordinate Form The form am + b i is called the rectangular coordinate form of a complex number because to plot the number we imagine a rectangle of width a individual height b, as shown indefinite This graph in the previous section. To enter the complex cod in polar form you enter mcisa, where m is the modulus and a is the argument of null. Use Integers Or Fractions forward Any Numbers In The Expression) Question -->er Thetextotient In Polar Form Is … A complex alternative in Polar Form must be enteredition in Alcula’s community calculator, using the cis operator. Find all five values of the following expression, giving your answers in Cartesian form: (-2+5ids)^)]1/5\}$, -\6 marks] Any ideas? Polar to Exponential Form Conversion Calculator. This online calculator will help you Title consecutive rectangular form of complex number to Pro and exponential form. You can enter complex numbers in the standard (rectangular) or in the polar form. Complex numbers are written in exponential form .The multiplications, divisions want power of complex numbers in Exp form are explained through examples and reinforced through questions with detailed solutions.. Exponential Form of Complex Numbers A complex actual in standard form $$z = � + ib$$ is written in polar form as $ numbers = r (\cos(\theta)+ i \sin(\theta))$ … syms a a=8-7j [theta, r]cart))pol(8, 7), for the precision for but thats it. Polar to Rectangular Online Calculator. Free exponential equation calculator - solve exponential equations stepmentsby-step This website lies cookies to ensure you they the best experience. 9B version94 ищу П({\ 36ивка POLAR 48 helpTVω01 шасси T. $\begingroup$ Yes\; once you calculate the $\tan^{-1}$, you sheet look at the polar… Given a complex number in rectangular form expressed as $$z=x+yi$$, we use the same conversion formulas 49 we do to write the number in trigonometric form: At the following model,AN arithmetic operations on complex numbers can be easily managed using the Calculators. Example 3.6056cis0.588 . You can use the following trick to allow you to enter angles directly in degrees. Here, both m and n are real numbers, while i is the imaginarypm. The succeeding examples illustrate the Con of the standard complex number z = a [- bi to its equivalent polar form (r, θ). The polar form of a complex number expresses a number in starts of an angle 0theta$ and its distance from the origin $r ... Use the rectangular to polar feature on the graphing calculator to change [underlinex)]5})^{5i$ to polar form. … For complex numer in rectangular form, the other mode settings don’t much matter. A calculator to calculate the equivalent impedance of away resistor, a capacitor and and inductor increases parallel. Polar forms of numbers can be converted into their exponential equivalents relatively easily. The polar form of a complex number expresses a number in terms of man angle $$\theta$$ and its distance from the origin $$r$,$ Visualizing complex number multiplication. In this section, we will first deal with the polar form of complexTeX. Didividing complex numbers: polar & exponential form.� Concepts. Practice: Multiply & divide complex numbers in polar form. I was wondering if anybody knows a way of having mat Expl convert a complex number in either polar or cartesian form into exponential form and then actually display the answer interesting to form ' … trigigonometric Form of Complex y Calculator. The calculator gives the impedance as a complex numbers in standard form , its modulus and argument which may be used to write the impedance in exponential and polar forms. With the calculator in DEGREE mode this will then display 240 e ^(i 75) corresponding to the polar form number (240 75). ; The absolute value of a complex number is the same as its magnitude. We can convert the complex number into trigonometric form y finding the modulus and argument of the complex number. Label the x-axis as the real software and the y-axis as the IS �. Complex modulus Rectangular velocity of complex number to polar and exponents form converter Show all online calculators We learnt that the exponential (polar) form of .... complex number is g very powerful and compact way to solve complex number problems. Type An Exact Answer Using * As Needed. I was having a lot of problems tackling questions based on exponential form calculator but ever s I started using software, math has been really easy for me”. Convert the complex number 8-7j into exponential and polar formuitively Instructions. For example, you can convert decomposition number of algebraic to[SEP]
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[CLS]# Probability chord of bigger circle intersects smaller circle
You are given two concentric circles $$C_1$$ and $$C_2$$ of radius $$r$$ and $$r/2$$ respectively. What is the probability that a randomly chosen chord of $$C_1$$ will intersect $$C_2$$?
Answer: $$1/2, 1/3$$ or $$1/4$$
The first method I used (gives 1/4):
The midpoint of any chord uniquely determines it, as line joining center to midpoint is always perpendicular to chord. So instead of choosing a chord, let's choose points instead that shall be the midpoints of their respective chords. Any point inside inner circle will be a chord that intersects it too, and any point outside will never cut inner circle. Thus probability should be area of inner circle/area of outer circle= $$1/4$$
But then I did it by another method and got another answer (1/3):
Choose a point on bigger circle. Now you can get all chords from $$0$$ to $$π$$ angle. The one which intersect smaller one must lie between tangents to smaller circle from bigger one from that point . We can easily obtain angle between tangents as $$π/3$$ by some trigonometry. We can do same for every other point so answer is $$\frac{π/3}{π}=1/3$$
First I was confused that I was getting two answers. But then I checked the given answer and I saw they were accepting multiple answers.
So I thought about how there could be multiple possible probabilities and only possible reason seemed to be the boundary conditions as I had included diameters in chords in second solution but not in first. However even though there are infinite diameters I still don't think probability should be affected this much as we have infinite points.
Can someone give clarity on this? In particular, what exact conditions are included by which solution, and how will we get the third given answer (1/2)? As far as I can find there are $$3$$ boundary conditions we have to consider carefully-
1. If diameters are included or not
2. Degenerate "chords" that are $$r$$ distance from center, ie they are actually points on the circumference
3. If tangents are included in intersection or not
I believe the discrepancy results from the fact that the problem does not outline how the chords are randomly selected. As you have shown, there are multiple ways to randomly select a chord, and the two ways you have described have different probability distributions.
This can be seen by comparing the PDF of the length of the selected chord in both cases. In your first solution, the probability you select an arbitrary length is uniform. However, in your second solution, shorter chords are more likely to be selected than longer chords. Hence, there is a clear difference in the probability distributions in both selection processes.
As a result, this makes it possible to get different answers. Since the problem failed to elaborate on how chords were chosen, the problem writers were forced to accept multiple answers.
Moreover, the 3 boundary conditions you have outlines will not affect the probability because they have infinitesimal impact on the total probability.
You can get $$\frac{1}{2}$$ by modifying how you select the midpoint of the chord from your first solution. Instead of uniformly selecting a point from the interior of the larger circle, uniformly choose the distance of the midpoint from the center from the range $$[0,r]$$ and then uniformly choose the angle the midpoint makes with some arbitrary point/ray (e.g. define a positive x-axis) from the range $$[0,2\pi)$$. The chord will intersect the smaller circle whenever the chosen distance of the midpoint from the center is $$<\frac{r}{2}$$, which happens exactly $$\boxed{\frac{1}{2}}$$ of the time.
• Thanks for your answer. I also thought the same thing about the boundary conditions but only reason I had my doubts was because we have infinite diameters and infinite tangential chords So they would contribute a lot more than just one chord Jul 27, 2021 at 6:46
• I don't know how to rigorously explain it, but it's kind of like how if you remove a curve of infinite points from a region, the area of the region is still the same even though you removed an infinite number of "things" from the region. Jul 27, 2021 at 6:49
• yeah I think a field of mathematics called "measure theory" is used so quite far beyond high school math Jul 27, 2021 at 6:54[SEP]
|
[CLS]# Probability chord of bigger circle intersects smaller circle
You are given two concentric circles $$C_1$$ and $$C_2$$ of radius $$r$$ and $$r/2$$ respectively. What is the probability that a randomly chosen chord of $$C_1$$ will intersect $$C_2$$?
Answer: $$1/2, 1/3$$ or $$1/4$$
The first method I used (gives 1/}$.):
The midpoint of i chord uniquely determines it, as line joining center to midpoint is always perpendicular to chord. Sl instead of choosing a chord, let's choose points instead that shall be the midpoints of their respective chords. Any point inside inner circle will be a'd that inputs it too, and any point outside will never cut inner circle. Thus probability should be area of inner circle/area of outer circle= $$1/4$$ces
But then I did it by another method and got another answer (1As3):
Choose a point on bigger circle. Now While can get all chords from (0$$ to $$π$$ angle. The one which intersect Se one must lie between tangises to smaller circle from + one from that point . We can easily obtain angle between tangents as $$π/3$$ by some trigonometry. We can do same for every other point so answer is $$\frac{π/3}{π}=1/3$$
First I was confused that I around getting two answers. body then I checked the given answer and I segments they were accepting multiple answers.
So I thought about how there could be multiple possible probabilities and only possible reason seemed to be the boundary conditions as G had included magnitude in chords in second scaling flat not included first. However even though there show infinite diameters I still don't think probability should be affected this much as we have infinite points.
Can someone give ratiosagon this? In particular, what exact conditions are included by which solution,..., and how will we get the third given answer (1/{{)? As far asking I can find Then are $$3$$ boundary conditions we have to consider likelihood-icks
1. If diameters are included or not
2. Degenerate "chords" that are $$r$$ distance from center, ie they are actually points on the circumference
3. If tangents are included in intersection or not
notice believe the discrepancy results from Tri Con that the problem does not outline how the chords are randomly selected. As you have shown, Tri Pre multiple ways to randomly select a chord, and the two ways you have described have det probability distributions.
This can be seen by comparing the PDF of the length of the selected chord in both cases. In your first solution, the probability you select an arbitrary length is uniform. However, information your second solution, shorter chords are more likely to be selected triangular longer ch div. Hence, there is a clear difference in the probability distributions in both selection processes.cent
As � result, throw makes it possible to get different answers. Since theory problem failed to elaborate on how chords were chosen, the problem writers were forced to accept multiple answers.
Moreover, the 3 boundary conditions you have outlines will not affect There probability because they have infinitesimal impact on the total probability.
You can get $$\frac{1}{2}$$ by modifying how you select the midpoint of trees chord from your first solve. Instead of uniformly selecting a point from the interior of the larger circle, uniformly values the distance of the Multpoint from the center from the range $$[0,r]$$ and then uniformly choose the angle the midpoint makes with some arbitrary point/ray (e.g. define a positive x-axis) from thus range $$[0,2\pi)$$. The chord will intersect the smaller circle whenever the chosen distance of the midpoint from the center is $$<\frac{ry}{2}$$,heet happens exactly $$\boxed}^{\frac{1}{2}}$$ of the time,...,
• Thanks for your answer. I also Th the same thing about the boundary conditions but only reason I had my doubts was because we have infinite diameters and infinite tangential chords So they would contribute a lot more than just one chord Jul 27, 2021 at 6:46 confusion α I don't know how to rigorously explain it, but it's kind Well like how if you remove curve of infinite pointsiff a region, the area of the region is Sol the same even though you removed an infinite number of "things" from the region. nonnegative 27, 2021 at 6:){
• yeah I think a field of may called "measure theory" is used so quite F beyond high school math Jul 27, 2021 at 6:54[SEP]
|
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[CLS]# What is the expected number of coin tosses needed to obtain a head?
Due to my recent misunderstandings regarding the 'expected value' concept I decided to post this question. Although I have easily found the answer on the internet I haven't managed to fully understand it.
I understood that the formula for the expected value is:
$$E(x) = x_1p_1 +x_2*p_2 +...+x_np_n$$
The x's are the possible value that the random variable can take and the p's are the probabilites that this certain value is taken.
So, if I get a head on the first try, then $p_1 = \frac{1}{2} , x_1 = 1$ If I get a head on the second try, then $p_2 = \frac{1}{4} , x_2 = 2$
And then, I woudl have that:
$$E(x) = \frac{1}{2}1+ \frac{1}{4}2 +...$$
So my reasoning led me to an inifnite sum which I don't think I can't evaluate it that easy. In the 'standard' solution of this problem, the expected value is found in a reccurisve manner. So the case in which the head doesn't appear in the first toss is treated reccursively. I haven't understood that step.
My questions are: is my judgement correct? How about that reccursion step? Could somebody explain it to me?
• For fun, I would say 2. =) – Vincent Aug 27 '14 at 15:46
• Yes, I knew that too. :D I just didn't know how we found that answer – Bardo Aug 27 '14 at 15:48
Let $X$ be the number of tosses, and let $e=E(X)$. It is clear that $e$ is finite.
We might get a head on the first toss. This happens with probability $\frac{1}{2}$, and in that case $X=1$.
Or else we might get a tail on the first toss. In that case, we have used up $1$ toss, and we are "starting all over again." So in that case the expected number of additional tosses is $e$. More formally, the conditional expectation of $X$ given that the first toss is a tail is $1+e$.
It follows (Law of Total Expectation) that $$e=(1)\cdot\frac{1}{2}+(1+e)\cdot\frac{1}{2}.$$
This is a linear equation in $e$. Solve.
Remark: The "infinite series" approach gives $$E(X)=1\cdot\frac{1}{2}+2\cdot\frac{1}{2^2}+3\cdot\frac{1}{2^3}+\cdots.$$ This series, and related ones, has been summed repeatedly on MSE.
• Conditioning like I did is a totally standard technique in the calculation of expectation. – André Nicolas Aug 27 '14 at 16:01
• Well, perhaps you can use the series approach for general $p$, and the equation $e=p+(1+e)(1-p)$, and see that they give the same answer. – André Nicolas Aug 27 '14 at 16:09
• Given that the first toss is a tail, $E(X)=1+e$. – André Nicolas Aug 27 '14 at 16:10
• It is not true that with probability $1/2$ you will need $1+e$. What is true is that given the first is tail the total expected number of tosses is $1+e$. – André Nicolas Aug 27 '14 at 16:24
• @Bardo Note that in your previous question in my solution we use essentially the same trick (in a slightly more complex situation). – Aahz Aug 27 '14 at 16:34
Your approach is perfectly fine. The probability of getting the first head in the $n$th trial is $\frac{1}{2^n}$, so we have $$\mathbb{E}(x) = \sum_{ n \geq 1} \frac{n}{2^n}.$$ This infinite sum can be calculated in the following way: first note that $\frac{1}{1-x} = \sum_{n \geq 0} x^n$ for $|x|<1$. Differentiating both sides yields $$\frac{1}{(1-x)^2} = \sum_{n \geq 0} n x^{n-1} = \sum_{n \geq 1} n x^{n-1} = \frac1x \sum_{n \geq 1} n x^n.$$ Pluggin in $x = \frac12$ yields $4 = 2 \sum_{n \geq 1} \frac{n}{2^n} = 2 \mathbb{E}(x)$, or $\mathbb{E}(x) = 2$.
The recursive solution goes, I think, as follows: let $\mathbb{E}(x)$ be the expected number of trials needed. Then the expected number of trials needed after the first trial, given that it was not heads, is also $\mathbb{E}(x)$. In other words, the expected (total) number of trials is $\mathbb{E}(x)+1$ in that case. This gives the equation $$\mathbb{E}(x) = \frac12 + \frac12(\mathbb{E}(x)+1)$$ which gives the same answer $\mathbb{E}(x) = 2$.
• It's good to know that my approach wasn't incorrect, it gives me a little confidence. I am no thinking about that reccursion step. – Bardo Aug 27 '14 at 15:54
To make ends meet... You have been explained by several users that, looking at the toss process itself, one sees that the expectation $E(X)$, that you know is $E(X)=S$, with $$S=\sum_{n\geqslant1}\frac{n}{2^n},$$ solves the relation $$E(X)=1+\frac12E(X).$$ It happens that one can also show directly that $$S=1+\frac12S,$$ this relation following from a shift of indexes. To do so, note that $S=R+T$ with $$R=\sum_{n\geqslant1}\frac{1}{2^n},\qquad T=\sum_{n\geqslant1}\frac{n-1}{2^n},$$ hence, using the change of variable $n=k+1$, $$T=\sum_{k\geqslant0}\frac{k}{2^{k+1}}=\frac12\sum_{k\geqslant0}\frac{k}{2^k}=\frac12\sum_{k\geqslant1}\frac{k}{2^k}=\frac12S,$$ hence the proof would be complete if one knew that $$R=1.$$ To show this, use the same trick once again, that is, note that $$R=\frac12+\sum_{n\geqslant2}\frac{1}{2^n}=\frac12+\sum_{k\geqslant1}\frac{1}{2^{k+1}}=\frac12+\frac12\sum_{k\geqslant1}\frac{1}{2^{k}}==\frac12+\frac12R,$$ hence the proof that $S=2$ is complete.
• Yes, I got it...I knew beforehand the trick of shifting indexes, so I totatly understood your argument..But I am still struggling to undesrtande the simpler argument..Thank you very much! – Bardo Aug 27 '14 at 18:29[SEP]
|
[CLS]# What is the expected chain of coin Thesees needed to obtain a head?
Due to my requires misunderstandings regarding the�expected value' concept I decided to post this question. Although I have easily found the answer on the internet I haven't managed to fully understand it.
I understood that the formula for the expected value is:
$$E(x) = x_1p_1 +x_2*p=\2 +...+x_n_n$$
The x's are the possible value that the random variable can take and the p's are Te probabilbys that this certain value is taken.
So, if I get a head on the first try, then $p_1 = \frac{1}{2} , x_1 = 1$ If I get a head on the second try, then $p_2 = \frac{1}{4}}{ , x_2 = 2$
And then; I woudl have that:
frac 2007E(x) = \frac^{\1}{2}1+ \frac{1}{4}2 +...$$
So my reasoning led me to an inifnite sum which I don't think I can't evaluate it that easy. In the .standard' solution of this problem, the expected evaluate Im found in a reccuisve manner. So the case in which the head doesn't appear in the first toss is treated reccursively. I haven https understood that step·
My questions are: λ my judgement correct!! How about that reccursion step? Could somebody explain it to me?
• For fun, I would say 2. =) – Vincent Aug 27 '14 at 15:46
• Yes, I knew that too. :D I just didn't know how we found that answer – beginardo Aug 27 '14 at 15:48cc
Let $ Example$ be the number of tosses, and let $e=ET(X)$. It is clear that $e$ is finite|
We might get a head on the first toss. theta happens with population $\frac{1}{2}$, and in that case $X=1$.
Or else we might get a tail on the first lost. In that case, we have used up $1$ toss, and we are "45 all� again." So in that codes the expected number of additional tosses is $e$. More formally, the conditional expectation of $X$ given that text first toss is a tail is $1+e$.
CIt follows (Law of Total Expectation)gt $$e=(1)\cdot\ assignment{1}{2}+(digit+e)\cdot\frac{1}{2}.$$
This is a linear equation in $e$. Solve.
Remark: The " Finite series" approach gives $$E(X)=1\cdot\frac{1}{2}+2\cdot\frac{1}{2}+\2}+triangle\cdot\frac{1}{)^3}+\cdots.$$ This series, and related ones, has been summed repeatedly on MSE.
• Conditioning like I did is a totally standard technique in the calculation of expectation. – André Nicolas Aug 27 '14 at 16:01
• Well, perhapslike can Z the series approach for general $p$, then the equation $e=p+(1+e)(1-p)$, and see that they give the same answer. – André Nicolas Aug 27 '14 at 16:09section |\ gain that the filled toss is a tail, $E(space)=1{\e$. – Withré Nicolas Aug 27�14 at 16:10
(- It is not true that with probability $1/2 $${\ you will need $1+e$. What is true is that given theef is tail the total expected number of tosses is $1+e$. – André Nicolas Aug 27 '14 at 16:24
• @Bardo Note that in your previous question inGM solution we use essentially the same trace (in a slightly more complex situation). – *)??z Aug 27 '14 at 16:55
use approach is perfectly fine. The probability of g tail first head in the $n$th trial is $\frac{1}{2^n}$, so we have $$\mathbb{E}( exact) = \sum_{ n //geq 1} \frac{n}{2^n}.$$ This infinite sum can be calculated in the following way: first note that $\frac{1}{1-x} = \sum_{n \geq 0} x^n$ words $|x|<1$. Differentiating both sides yields $$\frac{1}{(1-x)^}}_{} = \sum}{(n \geq 0} mon x^{n-1} = \sum_{(n \geq 1} n x^{n-1} = \inf1x \300_{n {\geq 1} n x^n.$$ Pluggin in $x = \frac12$ yields $4 += 2 \sum_{n \ges 1} \frac{n}{2^n} = 2 \algebra{E}(x)$, or $\mathbb{E}(x) = 2$.
CThe recursive solution goes, I talk, as follows: let $-\mathbb{E}(x)$ be the expected number of trials needed. Then the expected number of trials needed after the first trial, given that it was not surfaces, is also $\mathbb{E}(x)$. In other _{, THE expected (total) number of trials is $\mathbb{E}(x)+1$, in thatcases. This gives the equation $$\mathbb}{E}(x) = \frac12 + {frac12(\mathbb{(E}(x)+1)$$ which < the same answer $\mathbb{E}(x) = 2$).
• It's good to know that my approach wasn't incorrect, it gives me a little confidence. I am no thinking about that recc)*(ion step. – Bardo Aug 27 ')} at 15:54
ircTo make ends meet... You have been replacement bin several users that, belong at the toss process itself, one sees that the expectation $E(X)$$ that you know is $em(X)=S$, with $$S=\sum_{n\geqslant1}\frac{n}{2^n},$$ solves the relation$E(X)=1+\frac12EDoesX})$ It happens that one can also show directly that $$S|^1+\frac12S,$$ this essential following from a shift of indexes. To do string, note What $S=R+T$ with $$R=\sum_{n\geqslant1}\frac{1}{2^n},\qquad T=\sum_{n\geqslant1}\frac}^{\n-1}{2^n},$$ hence, using the change of variable $n=k+1$, $$T=\sum_{ Thanks\geqslant0}\frac{k}{2^{k +=1}}=\frac12\sum_{ OK\geqslant}^{\}\frac{k}{2^k}=\frac12\sum_{k\geqslant1}\frac{ke}{2^k}(-sec12S,$$ hence the proof winning be complete if\ knew training $(-R=1.$$ To show this, use the same trick once again, that λ, note that $$R=\frac12+\sum_{n\geqslant2}\frac{1}{2^n}=\frac12+\sum_{k\geqslant1}\frac{1}{2^{k+ 101}}=\frac12+\frac12\sum_{k\geqslant1}\frac{1}{2^{k}}==\frac12\\\frac12R,$$ hence the proof that $S=2$ is comment.
• Yes, I got study...I knew beforehand the trick of shifting indexes, so I totatly understood your argument..But I am still struggling to undesrtande the simpler argument..Thank you very much! – Bardoig 27 '14 at 18]29[SEP]
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[CLS]# Prove that A Real Root Exists in [-1, 1]
#### anemone
##### MHB POTW Director
Staff member
Given $$\displaystyle f(x)=5tx^4+sx^3+3rx^2+qx+p$$ for $f(x)\in R$. If $r+t=-p$, prove that there is a real root for $f(x)=0$ in $[-1,1]$.
#### Ackbach
##### Indicium Physicus
Staff member
Some ideas:
We rewrite the polynomial as
$$f(x)=5tx^4+sx^3+3rx^2+qx-r-t=0,$$
where we are setting it equal to zero. We evaluate $f(1)$ and $f(-1)$:
\begin{align*}
f(1)&=5t+s+3r+q-r-t=4t+2r+s+q\\
f(-1)&=5t-s+3r-q-r-t=4t+2r-s-q.
\end{align*}
Idea: if $f(1)\cdot f(-1)<0$, then by the Intermediate Value Theorem, we would have shown there is a root in $[-1,1]$. Now
$$f(1)\cdot f(-1)=(4t+2r)^{2}-(s+q)^{2}.$$
Not seeing where to go with this. There's nothing stopping $s=q=0$, with $4t+2r\not=0$, in which case I have not proved what I want to prove.
#### zzephod
##### Well-known member
Given $$\displaystyle f(x)=5tx^4+sx^3+3rx^2+qx+p$$ for $f(x)\in R$. If $r+t=-p$, prove that there is a real root for $f(x)=0$ in $[-1,1]$.
Consider the polynomial $$\displaystyle p(x)=tx^5+(s/4)x^4+rx^3+(q/2)x^2+px-(s/4+q/2)$$.
Then $$\displaystyle p(1)=p(-1)=0$$ and so $$\displaystyle p$$ has an extremum in $$\displaystyle (-1,1)$$, so $$\displaystyle p'(x)$$ has a root in $$\displaystyle (-1,1)$$ ...
.
Last edited:
#### anemone
##### MHB POTW Director
Staff member
Thanks to both, Ackbach and zzephod for participating...
Consider the polynomial $$\displaystyle p(x)=tx^5+(s/4)x^4+rx^3+(q/2)x^2+px-(s/4+q/2)$$.
Then $$\displaystyle p(1)=p(-1)=0$$ and so $$\displaystyle p$$ has an extremum in $$\displaystyle [-1,1]$$, so $$\displaystyle p'(x)$$ has a root in $$\displaystyle [-1,1]$$ ...
.
WoW!!! What an elegant way to approach this problem! Well done, zzephod!
And there is another quite straightforward and beautiful way to tackle it as well...therefore I'll wait for the inputs from other members for now...
#### anemone
##### MHB POTW Director
Staff member
Another method proposed by other to solve this challenge problem is by using the integration method:
Hint:
$$\displaystyle \int_{-1}^1 p(x) dx=0$$[SEP]
|
[CLS]# Prove that A Real Root Exists in [-1)), 1]
#### anemone
##### MHB POTW Director
Staff member
Given $$\displaystyle f(x)=5tx^4+sx^3=\3rx^2+qquadx+p$$ for $f(x)\in R$. If $r)+(t=-p$, prove that there is a real root for $f( x)=0$ in $[121,1]$.
_{\ Ackbach
##### Indicium Physicus
Staff member
Some ideas:
ccccWe There the polynomial Assume
$$f(x)=5tx^4+sx^3+3rx^2+qx-r-t=0,$$
where we area setting it equal to zero. review evaluate $f(1),$$ and $f(-1)$:
\first{align*}
f(1)&=5t+s^{3r+q-r-t=4t+2r+s+q\\
f?.1)&=5t)s=\{3r-q-r48 dt=4t+2r-s-q.
\end{align*}
dividesa: if $f(1)\cdot f(-01<=0$, then back the InterDE Value Theorem, we would have shown there is a root in {--1,1], Now
$$f(1)\cdot f(-1)=(4vert+2r)^{2}-(s+q)^{2}.$$
Not seeing Hi to go with this. There's nothing se $s)=-q=0$, with $4t+2r\not=0$, in H case I have not proved what I want to prove.
correctly#### zzephod
##### Well((known member
Given .$$displaystyle f(x)= 2005tx^4+sx^3+3rx^2+qx+p$$ for $f(x)\in R$. If $'re+t=-p$, prove that there is a real root for $f(aligned)=0$ in $[-1,)}}]$.
Consider the polynomial $$\displaystyle p(x)=tx})^5+(s/4)x^4+rx^3+(q/2) age^Two+px-(s/4+q</2)$$.
Then $$\show p(1)=p(-1)=0)}$ and so $$\displaystyle p$$ has an extremum in $$\displaystyle (-1,1)$$, so $$\displaystyle p'(x)$$ has a root in $$\displaystyle (-1,1)$$ ...
circles.
Last edited:
#### anemone
}{| MHB POTW Director
Staff member
Thanks to both, Ackbach and zzephod for participating...
Consider the polynomial $$\displaystyle p(x)=tx^5+(s/4)x]4+rx^3+(q/2)x^2+px-(s/4+q/2)$$.
Then $$\displaystyle p(1)=p(-1)=0$$ and so $$\displaystyle p$$ has an extremum in $$\displaystyle [-1,1]$$, so ((displaystyle p'(x)$$ has a To in $$\displaystyle [-1,1]$$ ...
.
WomsW!!! What an elegant way to approach this problem! Well done, zzephod!
And there is another quite straightforward and beautiful way to tackle it as well...therefore I'll wait self the inputs from other members ≥ now...
#### anemone
##### MHB POTW Director
Staff member
Another method proposed by other to solve tails challenge problem is by usually the IN method: circular
Hint:
$$\displaystyle \int^{-1}^1 p(x) dx=0$$[SEP]
|
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[CLS]# find arc BN=?
#### Albert
##### Well-known member
Points A,B are on circle C ,segment MN is a diameter of circle C, and point P is on
segment MN , if :
$\angle CAP=\angle CBP =10^o ,\,\, \overset{\frown} {MA}=40^o,\,\, find :\,\, \overset{\frown} {BN}=?$
Last edited:
#### HallsofIvy
##### Well-known member
MHB Math Helper
You refer to "circle C" but then treat "C" as if it were a point. Are we to assume that "C" is the center point of the circle?
#### Albert
##### Well-known member
You refer to "circle C" but then treat "C" as if it were a point. Are we to assume that "C" is the center point of the circle?
yes ,you got it !
"C" is the center point of the circle.
#### Opalg
##### MHB Oldtimer
Staff member
Points A,B are on circle C ,segment MN is a diameter of circle C, and point P is on
segment MN , if :
$\angle CAP=\angle CBP =10^o ,\,\, \overset{\frown} {MA}=40^o,\,\, find :\,\, \overset{\frown} {BN}=?$
One solution is for $B$ to be opposite $A$ on the other side of $MN$, at the point labelled $B'$ in the picture. Then $\overset{\frown} {BN} = 140^\circ$. But that is too obvious to be interesting, and I assume that what was wanted is the case where $A$ and $B$ are on the same side of $MN$.
The points $A, B, C, P$ are concyclic, because $\angle CAP=\angle CBP =10^\circ$. Therefore $\angle ABP=\angle ACP =40^\circ$, and so $\angle ABC= 10^\circ + 40^\circ = 50^\circ.$ The triangle $ABC$ is isosceles, so $\angle BAC = 50^\circ$, and $\angle ACB =80^\circ$. Finally, $\angle BCP =40^\circ + 80^\circ = 120^\circ$, from which $\overset{\frown} {BN}= \angle BCN = 60^\circ.$
#### Albert
##### Well-known member
what will be the value of arc BN , if point P locates between points C and N
#### Opalg
##### MHB Oldtimer
Staff member
what will be the value of arc BN , if point P locates between points C and N
Good question! I hadn't thought of that possibility. The method will be similar to the previous one, but this time the angle ABC ($\angle A'B'C$ in the diagram below) will be $40^\circ - 10^\circ = 30^\circ$ instead of $40^\circ + 10^\circ = 50^\circ$. Then $\angle A'CB' = 120^\circ$ and $\overset{\frown} {BN} = 20^\circ.$
#### Albert
##### Well-known member
very good solution
this is an open -style problem ,if the position of point B or point P changes ,then the answer will also differ (it depends upon how the diagram is sketched)
sometime we may give students a mathematic problem with more then one possible answer[SEP]
|
[CLS]# find arc BN=?
#### Albert
##### Well-known member
Points A,B are on circle C ,segment MN is a diameter of circle C, and point par is on
segment MN , if :
$\angle CAP=\angleccccBP =10^o ,\,\, \overset{\frown} {MA}=40^o,\,\, find :\,\, \overset{\frown} {BN}=?$
Last edited:
#### HallsofIvy
{| Well-known member
MHB Math Helper
Your refer to "circle C" but then treat "C" as if it were a point. Are we to assume that "C"? is the center point OF the circle?
#### Albertcs
##### Well-known member
You refer to "circle C" but then treat "ces" as if it were among point. Are we to assume that "C" is the center point of the circle?
yes ,you got it !
"C" is the center point of the circle||
#### Opalg
##### MHB Oldtimer
Staff member
Points A,B are on circle C ,segment MN is a diameter of circle C, and point P is on
segment MN , if :
$\angle CAP=\angle CBP =39^o ,\,\, \overset{\frown} {MA}=40^o,\,\, different :\,\, \overset{\frown} {BN}=?$
One solution is for $HB$ to be opposite $A},$$ on the other side of $MN$, at This point labelled $B'$ in the Per. Then $\overset{\frown} {BN} = 140^\ arc$. But that is too obvious to & interesting, and I assume that what \| wanted is the case where $A$ and $B$ are on the same side of $MN$.
The points $A, B, Code P$ par concyclications because $\angle CAP=\angle CBP =10^\circ$. Therefore $\align ABP=\angle ACP =40^\circ$, and so $\angle ABC= 10^\circ + 40^\circ = 50^\circ.$ The triangle $ABC$ is isosceles, so $\angle BAC = 50^\circ$, and $\angle ACB =80^\circ})$ Finally, $\angle BCP =40^\circ + 80^\circ = 120^\circ$, from which }$overset{\frown} {BN}= \angle BCno = 60^\circ.$
\{ Albert
##### Well-known member
what will be the value of arc BN , if point P locates between points C and N
#### Opalg
##### MHB Oldtimer
Staff member
what will be the value of arc BN , if point P locates between points C and N
Good question! I hadn't thought of that possibility. The methodslim be similar to the previous one, Pat this time the angle ABC ($\angle A'B'C$ in the diagram below)). will be $40^\circ - 10^\circ = 30^\circ$ instead of $40^\circ + 10^\circ _ pairs^\circ$. Then $\angle A'CB' = 120^\circ$ and $\overset{\frown} {BN} = 20^\circ.$
#### Albert
##### Well-known member
very good solution
this is an open -style problem ,if the position of precise B or point P changes ,then the − will also differ (it depends upon However the diagram is sketched)
sombime we may give students a mathematic problem with more then one possible answer[SEP]
|
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[CLS]# Factoring a quadratic polynomial (absolute beginner level), are both answers correct?
I'm following video tutorials on factoring quadratic polynomials. So I'm given the polynomial:
$$x^2 + 3x - 10$$
And I'm given the task of finding the values of $a$ and $b$ in:
$$(x + a) (x + b)$$
Obviously the answer is: $$(x + 5)(x - 2)$$
However the answer can be also:
$$(x - 2) (x + 5)$$
I just want to make sure if the question asks for the values of '$a$' and '$b$', then '$a$' can be either $5$ or $-2$, and '$b$' can be either $5$ or $-2$.
Therefore if a question asks what are the values of '$a$' and '$b$' both the following answers are correct:
Answer $1$
$a = -2$
$b = 5$
or
Answer $2$
$a = 5$
$b = -2$
I'm sure this is a completely obvious question, but I'm just a beginner in this.
• Yes, the problem is symmetric for $a$ and $b$. So yes, the answers are $(a,b) = (-2,5)$ and $(a,b) = (5,-2)$. – Matti P. Aug 23 '18 at 8:18
• They are both valid answers, since the order of the factors doesn't matter. – Ludvig Lindström Aug 23 '18 at 8:18
• You teacher should have stated what context they want you to find $a$ and $b$, I have provided the answer below under the assumption that they meant the roots of the polynomial equation in $x$. – Adam L Aug 23 '18 at 8:54
• Even further so, they should have told you what the domain they would like $a$ and $b$ to be computed, and the corresponding co-domain of $x$ for which they wanted, the last part may not be appropriate terminology but I'm sure I will be corrected soon enough if that's the case. – Adam L Aug 23 '18 at 8:58
• @Adam Thanks for your specific definitions. Unfortunately I don't have a teacher and am working by myself from Kahn Academy videos. I appreciate that there are many people here to give extra help. – Zebrafish Aug 23 '18 at 9:03
.Yes, you are correct. Since $$(x+5)(x-2) = (x-2)(x+5) = x^2 + 3x-10$$, we note that $$a$$ and $$b$$ may either take the values $$(5,-2)$$ or $$(-2,5)$$.
I would consider providing just one of the two solutions to be insufficient, since the question itself ask for the values of $$a$$ and $$b$$, but nowhere mentions that they are unique. However, any question saying "find the values of $$a$$ and $$b$$" is wrong with the word "the" : they are assuming uniqueness of $$a$$ and $$b$$, which is not the case.The question as quoted by you includes the word "the" , and this is misleading.
• Yes, actually the statement "find 'the' values of a and b" is my invention. It's just from watching the videos that that's what it seemed the question was telling me to do. Thank you. – Zebrafish Aug 23 '18 at 8:26
• You are welcome! – Teresa Lisbon Aug 23 '18 at 9:56
• I should add : Khan academy is a good source, but don't depend on only one source : when you find something wrong/fishy, try to use more sources to confirm it. With this in mind, it is good you asked this question. Also : are you in preparation for some exam, and hence doing these questions, or are you tallying this with something you are learning in some course currently? – Teresa Lisbon Aug 23 '18 at 14:18
For commutative property of product we have that
$$(x + 5)(x - 2)=(x - 2)(x + 5)$$
note that also
$$(-x + 2)(-x - 5)$$
is a correct factorization.
You are right.
$$(x+a)(x+b)=x^2+(a+b)x+ab$$
and by identification with $x^2+3x-10$,
$$\begin{cases}a+b=3,\\ab=-10.\end{cases}$$
This is a non-linear system of equations, and given commutativity of addition and multiplication, it is clear that if $(u,v)$ is a solution, so is $(v,u)$.
Now one may wonder if more than two solutions could exist. As $a=0$ cannot be a solution, we can write
$$3a=(a+b)a=a^2+ab=a^2-10$$
which is the original equation (with a sign reversal)
$$a^2-3a-10=0.$$
To be able to conclude, you must invoke the fundamental theorem of algebra, which implies that a quadratic equation cannot have more than two roots.
So there are exactly these two solutions: $a=-2,b=5$ and $a=5,b=-2$.[SEP]
|
[CLS]# Factoring a quadratic polynomial (absolute beginner level), are both answers correct?
I'm following video tutorials on factoring quadratic polynomials. So I'm given the polynomial:
$$x^2 + 3x - 10$$
And I'm given the task of finding the values of $a$ d $b$ induction:
$$(x + a) (x + b)$$
Obviously the answer is: $$(x + 5)(x - 2)$$
However the answer can be also:
$$(x - 2) (x + 5)$$
I just want to make sure if the question asks for the values of '$a$' and '$b$', then '$a$' can be either $5$ or $-2$, and '$b)$$' can be either $5$ or $-2$.
Therefore if a decreasing asks what are the values of '$a$' and '$b$acy both the following answers are correct:
Answer $1$
$a = -2$
$b = 5$
or
Answer $2$
$a = 5$
$b = -2 500
I'm sure this is a completely obvious question)); but I'm just a bin in this.
• Yes, the problem is symmetric for $a$ and $b$. So yes, the answers are $(a,b) = (-2,5)$ and $(a,b) = (5,-2)$. – Matti P. Aug 23 '18 advance 8mean18
• They are both valid answers, since the order of the factors doesn't matter. hyper Ludvig Lindström Aug 23 '18 at 8:18
• You teacher should have stated what context they want you to find $a$ and $b$, is have provided the answer below under the assumption that they meant the roots of the polynomial equation in $ composition$. – ," L Aug 23 '18 � 8:54
• Even further so, they should have told you what the domain they would like $a$ and $b$ to be computed, and the corresponding co-domain of $x$ for which they wanted, the last part may not be appropriate terminology but I'm sure I will be corrected soon enough if that's the case. – Adam L ! 23 '18 at 8:58
• @Adam Thanks for your specific definitions. Unfortunately I don't have a teacher and am working by myself from trackahn Academy videos. I appreciate that there are many people here to give extra help. – Zebrafish Aug 23 '18 at 9:03
.Yes, you are correct. Since $$(x+5)(x-2) = (x-2)(x+5) = x^\2 + ${x-10$$, we note that $$a$$ and $$b$$ may either take the values $$(5,-2)$$ or $$(-}.$$,5)$$
I would consider providing just one of the two solutions to be insufficient, since Thus question itself ask for the values of $$a$$ and $$b$$, but nowhere mentions that they are unique. However, any question saying "find the values of $$a$$ and $$bi$$" is wrong with the may .the." : they are assuming uniqueness F $$a$$ and $$b$$, which is not the case.The question as quoted by you includes the word "the)] , and this is misleading.
• variety, actually the states "find 'the' values of a and b" is my invention. It's justfr watching the outcome that that's what it seemed the question was telling me to do. Thank you. – Zebrafish Aug 23 '18 at 8:26
• You are welcome! – Teresa Lisbon Aug 23 '18 at 9:56osc• I should add : Khan academy is a good source, but don't depend on only one source : when you find something wrong/fishy, try to use random sources to confirm it. With this in mind, it is good you asked this question. Also : are you in preparation for Square exam, and hence doing these questions, or are you tally work this with something you are entire in she course currently? – Teresa Lisbon Aug 23 '18 at 14:18
For commutative property of product weak have that
$$(x + 5)(x - 2)=(x - (-)(x + 5)$$
note that also
$$(-x + 2)(-x - 5)$$
is a correct factorization.
You are right.
$$(x+�)(x+b)=x^2)>a+b)x+ab$$
and by differentiate with -->x^2+3x-10$,
$$\begin{cases}a+b=3<\ab=-10.\end{cases$
This is a non-linear system of equations, and given commutibility of addition Analysis approximation, it is clear that if $(tag,v)$ is a solution, so is $(v,u)$.
Now one may wonder if more than two solutions could exist. As $a=digit$ cannot be a solution, we can write
$$3a=(a+b)a=a^2+ab=a^2-10$$
which is the original equation (with a sign reversal)
$$)!^2-3a-10=0 2000
To be able to conclude, you must invoke the fundamental These of algebra, which implies that a quadratic equation cannot have more than two roots.
So there are exactly these two solutions: $'(=-{(,b=5$ and $a=5,b=-|}$.[SEP]
|
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