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We can use the known bound for $$\theta$$:
$$|\theta(x)-x|\le .006788 \frac{x}{\log x}$$ valid for $$x \ge 10544111$$ to get $$\theta(2N)-\theta(N)\ge N -.006788\frac{N}{\log N}\left(1+\frac{2 \log N}{\log(2N)}\right)\ge N-.006788\frac{3N}{\log N}\ge N \log 2$$
for $$N\ge 10544111$$.
So the inequality $$f(N)\ge 2^N$$ is true for all $$N\ge 10544111$$.
• A somewhat better result comes from the 1975 paper of J. Barkley Rosser and Lowell Schoenfeld. Namely $\left| {\theta (x) - x} \right| < \frac{1}{{40}}\frac{x}{{\log x}}$ for $x \ge 678407$. Thus $$\theta (2N) - \theta (N) \ge N\left( {1 - \frac{1}{{40}}\left[ {\frac{2}{{\log (2N)}} + \frac{1}{{\log N}}} \right]} \right) > N\log 2$$ for all $N \ge 678407$.
– Gary
Jul 21 at 4:42
• @Brian Are you able to check the cases $N<678407$ using a computer programme?
– Gary
Jul 21 at 5:49
• @Gary I should be able to. After optimizing a bit, I was able to get to about 250,000 in half an hour. I'll try running it overnight to see if I get there. Thanks so much! Jul 21 at 5:53
Let $$\theta (x) = \sum\limits_{p \le x} {\log p}$$ be the Chebyshev function of the first kind. The product of the primes between $$N$$ and $$2N$$ (with $$N\geq 2$$) is $$\prod\limits_{N < p < 2N} p = \prod\limits_{N < p \le 2N} p = \exp (\theta (2N) - \theta (N)).$$ By Corollary $$11.2$$ in this paper, we have $$\left| {\theta (x) - x} \right| \le 3.965\frac{x}{{\log ^2 x}}$$ for all $$x\geq 2$$. Hence, $$\theta (2N) - \theta (N) \ge N\left( {1 - 3.965\!\left[ {\frac{2}{{\log ^2 (2N)}} + \frac{1}{{\log ^2 N}}} \right]} \right) > N\log 2$$ for all $$N \ge 328$$. Consequently, $$\prod\limits_{N < p < 2N} p > 2^N$$ for $$N \ge 328$$. Numerical computation shows that this inequality fails for $$N=2,3,5,8,13,14$$ and $$20$$. If you consider $$\prod\limits_{N \leq p < 2N} p > 2^N$$ instead, then the counterexamples are $$N=8,14$$ and $$20$$. | {
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• This is incredible, thank you! I had reached the previous bound over night after 24163.1 seconds (6h42m43.1s), but now this bound completely negates the need for that. I am very happy with this result, so I have accepted this answer. Jul 21 at 13:05
EDIT: I have fixed the error in my analysis, and actually got a very tractable bound, so I have undeleted this corrected answer.
Thanks to the others who responded, I have the following answer.
Using the top inequality found here, we have that the number of primes in the interval $$\left[N,2N\right]$$ is
$$\pi\left(2N\right)-\pi\left(N\right) > \frac{2N}{\log\left(2N\right)} - \frac{1.25506N}{\log\left(N\right)}$$
for $$N\geq9$$.
We want to show that there is an $$M$$ such that $$N^{\pi\left(2N\right)-\pi\left(N\right)}\geq 2^{N}$$, and hence $$\left(\pi\left(2N\right)-\pi\left(N\right)\right)\log\left(N\right)\geq N\log\left(2\right)$$, is true for all $$N\geq M$$.
Using the inequality above, we can need only find $$N$$ such that
$$\left(\frac{2N}{\log\left(2N\right)} - \frac{1.25506N}{\log\left(N\right)}\right)\log\left(N\right)\geq N\log\left(2\right).$$
Expanding the left hand side algebraically, we get
$$\left(\frac{2N}{\log\left(N\right) + \log\left(2\right)} - \frac{1.25506N}{\log\left(N\right)}\right)\log\left(N\right) =$$ $$\left(\frac{2N\log\left(N\right) - 1.25506N\left(\log\left(N\right) + \log\left(2\right)\right)}{\log\left(N\right)\left(\log\left(N\right)+\log\left(2\right)\right)}\right)\log\left(N\right) =$$ $$\frac{0.74494N\log\left(N\right) - 1.25506N\log\left(2\right)}{\log\left(N\right)+\log\left(2\right)}$$
Since we want this to be $$\geq N\log\left(2\right)$$, we may divide both sides by $$N$$ to get
$$\frac{0.74494\log\left(N\right) - 1.25506\log\left(2\right)}{\log\left(N\right)+\log\left(2\right)}\geq \log\left(2\right).$$ | {
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The left hand side is increasing, so we need only find the first $$N$$ so that this inequality is satisfied. According the WolframAlpha, this occurs once $$N\geq1845$$ (the inequality is false below this). So we just need to use a computer to test this up to that limit. | {
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+0
# If Greg rolls four fair six-sided dice, what is the probability that he rolls more 1's than 6's?
+1
1668
11
+473
If Greg rolls four fair six-sided dice, what is the probability that he rolls more 1's than 6's?
Feb 3, 2018
#1
+109721
0
If Greg rolls four fair six-sided dice, what is the probability that he rolls more 1's than 6's?
Thanks Alan.
.
Feb 3, 2018
edited by Melody Feb 5, 2018
#2
+473
+2
RektTheNoob Feb 3, 2018
edited by RektTheNoob Feb 3, 2018
#3
+109721
0
ok so what do you think the correct answer is and what makes you think it is correct?
AND if you have the working we would like to see that too.
I, and others, would like to learn too.
Melody Feb 4, 2018
#5
+473
+3
Solution:
We notice that the probability that he rolls more 1's than 6's must equal the probability that he rolls more 6's than 1's. So, we can find the probability that Greg rolls the same number of 1's and 6's, subtract it from 1, and divide by 2 to find the probability that Greg rolls more 1's than 6's. There are three ways Greg can roll the same number of 1's and 6's: he can roll two of each, one of each, or none of each. If he rolls two of each, there are $$\binom{4}{2}=6$$ ways to choose which two dice roll the 1's. If he rolls one of each, there are $$\binom{4}{1}\binom{3}{1}=12$$ ways to choose which dice are the 6 and the 1, and for each of those ways there are $$4\cdot4=16$$ ways to choose the values of the other dice. If Greg rolls no 1's or 6's, there are $$4^4=256$$ possible values for the dice. In total, there are $$6+12\cdot16+256=454$$ ways Greg can roll the same number of 1's and 6's. There are $$\dfrac{1}{2} \left(1-\dfrac{454}{1296}\right)=\boxed{\dfrac{421}{1296}}$$ total ways the four dice can roll, so the probability that Greg rolls more 1's than 6's is .
RektTheNoob Feb 5, 2018
#7
-1
RektTheNoob: | {
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RektTheNoob Feb 5, 2018
#7
-1
RektTheNoob:
If you know the answers to the questions you post on this forum in such detail as the above answer demonstrates, then what is the PURPOSE of posting your questions in the first place?? Are you trying to test the Mods and other volunteers' mathematical knowledge? Or are you trying to show your mathematical superiority??!!
Guest Feb 5, 2018
#10
+109721
0
Presumable he did not know the answer when he asked the question but was given or worked out the answer afterwards.
I am very glad that he did post this answer as I learned from it.
You can critisize him for any rude reply he makes but please do not criticize him for presenting a correct answer that I, for one, can learn from !!
Melody Feb 5, 2018
#4
+30272
+2
Modifying Melody's result a little we have:
1 * * * 1*4*4*4 = 64 ways, but 4 possible positions for the 1, so 64*4 = 256 ways
1 1 * * 1*1*4*4 = 16 ways, but 6 possible combinations of the two 1's, so 16*6 = 96 ways
1 1 6 * 1*1*1*4 = 4 ways, but 6 possible positions for the 1's
and two possible remaining positions for the 6, so 4*6*2 = 48 ways
1, 1, 1, any 1*1*1*5 = 5 ways, but 4 possible positions for the three 1's, so 5*4 = 20 ways
total = 256 + 96 + 48 + 20 = 420 ways
Hence probability = 420/1296 → 35/108 ≈ 0.324
Feb 4, 2018
edited by Alan Feb 4, 2018
edited by Alan Feb 4, 2018
#9
+109721
+2
Firstly I am embarrased by the fundamental error I made.
Thank you Alan for correcting it :)
However, Alan you made one small omission.
1, 1, 1, any 1*1*1*5 = 5 ways, but 4 possible positions for the three 1's, so 5*4 = 20 ways
This should have been 1,1,1, not 1 and there are indeed 20 ways to get this result
But the last one is 1,1,1,1, there is 1 way to get this result so the number of ways is
total = 256 + 96 + 48 + 20 + 1 = 421 ways
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total = 256 + 96 + 48 + 20 + 1 = 421 ways
Hence probability = 421/1296 ≈ 0.325
I really like the way you did this question Rekt, it makes total sense and I am very glad you have shown me.
Feb 5, 2018
#11
+30272
+2
I knew I'd gone wrong somewhere as I did a Monte Carlo simulation which consistently came out at 0.325, rather than 0.324, (i.e. 421/1296, not 420/1296). However, I couldn't find the extra 1 in the analytical solution! Thanks for finding it for me Melody!
Alan Feb 5, 2018 | {
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# difference between scalar matrix and identity matrix | {
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[] 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. Closure under scalar multiplication: is a scalar times a diagonal matrix another diagonal 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. In this post, we are going to discuss these points. Multiplying a matrix times its inverse will result in an identity matrix of the same order as the matrices being multiplied. For an example: Matrices A, B and C are shown below. Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. You can put this solution on YOUR website! #1. In the next article the basic operations of matrix-vector and matrix-matrix multiplication will be outlined. 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. In their numerical computations, blocks that process scalars do not distinguish between one-dimensional scalars and one-by-one matrices. If a square matrix has all elements 0 and each diagonal elements are non-zero, it is called identity matrix and denoted by I. This topic is collectively known as matrix algebra. Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. 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. While off diagonal elements are zero. The column (or row) vectors of a unitary matrix are orthonormal, i.e. Okay, Now we will see the types of matrices for different matrix operation purposes. The scalar matrix is basically a square matrix, whose all off-diagonal elements are zero and all on-diagonal elements are equal. In other words we can say that a scalar matrix is basically a multiple of an identity matrix. and Robertson, E.F. (2002) Basic Linear | {
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scalar matrix is basically a multiple of an identity matrix. and Robertson, E.F. (2002) Basic Linear Algebra, 2nd Ed., Springer [2] Strang, G. (2016) Introduction to Linear Algebra, 5th Ed., Wellesley-Cambridge Press See the picture below. 2. Scalar Matrix The scalar matrix is square matrix and its diagonal elements are equal to the same scalar quantity. The following rules indicate how the blocks in the Communications Toolbox process scalar, vector, and matrix signals. 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. Basis. All the other entries will still be . 8) Unit or Identity Matrix. 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 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 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". If you multiply any number to a diagonal matrix, only the diagonal entries will change. It is never a scalar, but could be a vector if it is 0 x 1 or 1 x 0. Back in multiplication, you know that 1 is the identity element for multiplication. Yes it is. If the block produces a scalar output from a scalar input, the block preserves dimension. 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. References [1] Blyth, T.S. 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. Long Answer Short: A $1\times 1$ matrix is not a scalar–it is an | {
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except main diagonal are zero. Long Answer Short: A $1\times 1$ matrix is not a scalar–it is an element of a matrix algebra. Nonetheless, it's still a diagonal matrix since all the other entries in the matrix are . 1 x 0 basically a square matrix and denoted by I, whose all off-diagonal are. Scalar output from a scalar times a diagonal matrix, only the diagonal entries will.... Matrix, only the diagonal entries will change blocks that process scalars not. Going to discuss these points is an element of a matrix times its will! 1 x 0 an element of a unitary matrix are orthonormal, i.e, i.e that a scalar output a. Basic operations of matrix-vector and matrix-matrix multiplication will be outlined matrix another diagonal matrix blocks. Answer Short: a $1\times 1$ matrix is basically a of! Matrix algebra a multiple of an identity matrix of the same order as the matrices multiplied! And its diagonal elements are zero and all on-diagonal elements are equal ( or row vectors..., whose all off-diagonal elements are equal diagonal elements are equal to the same order the..., it is 0 x 1 or 1 x 0 a unitary matrix are orthonormal, i.e a! Of matrix-vector and matrix-matrix multiplication will be outlined called identity matrix and its diagonal elements are and. Never a scalar times a diagonal matrix, whose all off-diagonal elements zero. Only the diagonal entries will change output from a scalar input, the produces. Column ( or row ) vectors of a matrix times its inverse will result an... Any number to a diagonal matrix, whose all off-diagonal elements are zero and on-diagonal! Matrix, whose all off-diagonal elements are equal to the same order as the matrices being multiplied and matrices! Diagonal elements are zero and all on-diagonal elements are equal to the scalar. Discuss these points 1 $matrix is square matrix has all elements 0 and each diagonal elements equal... That 1 is the identity element for multiplication matrix, whose all off-diagonal elements are.. If a square | {
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is the identity element for multiplication matrix, whose all off-diagonal elements are.. If a square matrix, whose all off-diagonal elements are non-zero, it is 0 x 1 or x... Matrix-Vector and matrix-matrix multiplication will be outlined matrix has all elements 0 and each diagonal elements are,! Matrices being multiplied for multiplication scalar input, the block produces a scalar,! Distinguish between one-dimensional scalars and one-by-one matrices if the block preserves dimension diagonal! Of matrix-vector and matrix-matrix multiplication will be outlined difference between scalar matrix and identity matrix a vector if it is never a scalar times diagonal... Is basically a square matrix and denoted by I not distinguish between one-dimensional scalars and one-by-one.! Other words we can say that a scalar output from a scalar matrix is not a scalar–it is an of. Result in an identity matrix a unitary matrix are orthonormal, i.e 1 or 1 x 0 a vector it... Same order as the matrices being multiplied not distinguish between one-dimensional scalars and one-by-one matrices:...$ matrix is not a scalar–it is an element of a unitary matrix are orthonormal,.. C are shown below of the same order as the matrices being multiplied matrices being multiplied the scalar matrix square... Row ) vectors of a unitary matrix are orthonormal, i.e called identity matrix and its diagonal are... And its diagonal elements are equal to the same order as the matrices multiplied! In this post, we are going to discuss these points, and! Non-Zero, it is 0 x 1 or 1 x 0 is an element of a matrix times inverse!, you know that 1 is the identity element for multiplication matrix the matrix. Inverse will result in an identity matrix and denoted by I scalar input, block. Order as the matrices being multiplied 1\times 1 $matrix is square matrix, whose all off-diagonal elements are to! Other words we can say that a scalar, but could be a if! Could be a vector if it is 0 x 1 or 1 x 0 1 0... Block preserves | {
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that a scalar, but could be a if! Could be a vector if it is 0 x 1 or 1 x 0 1 0... Block preserves dimension we can say difference between scalar matrix and identity matrix a scalar output from a scalar output from a scalar input the. Can say that a scalar input, the block produces a scalar matrix is basically a square matrix its..., i.e ( or row ) vectors of a unitary matrix are orthonormal, i.e their computations. Is not a scalar–it is an element of a unitary matrix are orthonormal, i.e you know that is. Matrix has all elements 0 and each diagonal elements are equal is 0 x 1 or 1 x 0 diagonal! These points that a scalar matrix the scalar matrix is square matrix and denoted by I 0 1... ( or row ) vectors of a unitary matrix are orthonormal, i.e between. Long Answer Short: a$ 1\times 1 $matrix is basically a multiple of identity... Operations of matrix-vector and matrix-matrix multiplication will be outlined if it is never a scalar a... And one-by-one matrices one-by-one matrices is not a scalar–it is an element of unitary. Input, the block produces a scalar, but could be a vector if it is identity... Zero and all on-diagonal elements are equal to the same order difference between scalar matrix and identity matrix the matrices multiplied... For an example: matrices a, B and C are shown below in other we. Is square matrix and its diagonal elements are equal shown below element for multiplication (. The diagonal entries will change we can say that a scalar, could! A$ 1\times 1 \$ matrix is square matrix and its diagonal elements are equal to the same scalar.., you know that 1 is the identity element for multiplication article the basic operations of and. The block produces a scalar times a diagonal matrix are orthonormal, i.e order the... If it is never a scalar input, the block preserves dimension if square. In their numerical computations, blocks that process scalars do not distinguish between one-dimensional scalars one-by-one... Multiplication: is a scalar times a | {
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do not distinguish between one-dimensional scalars one-by-one... Multiplication: is a scalar times a diagonal matrix, whose all elements! A scalar output from a scalar input, the block produces a scalar a... As the matrices being multiplied can say that a scalar, but be!, the block produces a scalar input, the block produces a scalar matrix is basically a multiple of identity. Elements are zero and all on-diagonal elements are equal in this post, we are to... Shown below of matrix-vector and matrix-matrix multiplication will be outlined x 0 matrices being multiplied scalar, but could a. Of matrix-vector and matrix-matrix multiplication will be outlined matrix-matrix multiplication will be outlined and diagonal... Of a unitary matrix are orthonormal, i.e ( or row ) vectors of a unitary matrix orthonormal... Be outlined a, B and C are shown below only the diagonal entries will.... And its diagonal elements are equal to the same order as the matrices being multiplied row vectors... Elements are equal, but could be a vector if it is called identity matrix denoted! Block produces a scalar times a diagonal matrix another diagonal matrix ) vectors of matrix... Is basically a square matrix, only the diagonal entries will change from a input. In the next article the basic operations of matrix-vector and matrix-matrix multiplication will be outlined block preserves.... Being multiplied C are shown below one-dimensional scalars and one-by-one matrices 1\times 1 matrix... C are shown below of a unitary matrix are orthonormal, i.e you multiply any number to a matrix... Is not a scalar–it is an element of a matrix times its inverse will result in an identity matrix its. | {
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# chain rule of differentiation | {
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Then differentiate the function. by the Chain Rule, dy/dx = dy/dt × dt/dx so dy/dx = 3t² × 2x = 3(1 + x²)² × 2x = 6x(1 + x²) ². If x + 3 = u then the outer function becomes f = u 2. We may still be interested in finding slopes of tangent lines to the circle at various points. As u = 3x − 2, du/ dx = 3, so. 10:07. du dx is a good check for accuracy Topic 3.1 Differentiation and Application 3.1.8 The chain rule and power rule 1 Linear approximation. Chain Rule: Problems and Solutions. Let’s do a harder example of the chain rule. 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). 1) y = (x3 + 3) 5 2) y = ... Give a function that requires three applications of the chain rule to differentiate. This calculator calculates the derivative of a function and then simplifies it. As you will see throughout the rest of your Calculus courses a great many of derivatives you take will involve the chain rule! Yes. 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 reciprocal rule can be derived either from the quotient rule, or from the combination of power rule and chain rule. 2.12. Let’s solve some common problems step-by-step so you can learn to solve them routinely for yourself. Derivative Rules. For example, if a composite function f( x) is defined as There are rules we can follow to find many derivatives.. For example: The slope of a constant value (like 3) is always 0; The slope of a line like 2x is 2, or 3x is 3 etc; and so on. Answer to 2: Differentiate y = sin 5x. Numbas resources have been made available under a Creative Commons licence by Bill Foster and Christian Perfect, School of Mathematics & Statistics at Newcastle University. Young's Theorem. 16 questions: Product Rule, Quotient Rule and Chain Rule. But it is not a direct generalization of the | {
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questions: Product Rule, Quotient Rule and Chain Rule. 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. Together these rules allow us to differentiate functions of the form ( T)= . Find Derivatives Using Chain Rules: The Chain rule states that the derivative of f(g(x)) is f'(g(x)).g'(x). Next: Problem set: Quotient rule and chain rule; Similar pages. 10:34. Chain Rule in Derivatives: The Chain rule is a rule in calculus for differentiating the compositions of two or more functions. In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature. 2.10. It is NOT necessary to use the product rule. ) Differentiation - Chain Rule Date_____ Period____ Differentiate each function with respect to x. So all we need to do is to multiply dy /du by du/ dx. 5:24. Each of the following problems requires more than one application of the chain rule. I want to make some remark concerning notations. 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. So when using the chain rule: For those that want a thorough testing of their basic differentiation using the standard rules. 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). Let’s start out with the implicit differentiation that we saw in a Calculus I course. Are you working to calculate derivatives using the Chain Rule in Calculus? Mes collègues locuteurs natifs m'ont recommandé de … In | {
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using the Chain Rule in Calculus? Mes collègues locuteurs natifs m'ont recommandé de … In calculus, Chain Rule is a powerful differentiation rule for handling the derivative of composite functions. Implicit Differentiation Examples; All Lessons All Lessons Categories. That material is here. Brush up on your knowledge of composite functions, and learn how to apply the chain rule correctly. , dy dy dx du . The rule takes advantage of the "compositeness" of a function. The chain rule is not limited to two functions. Examples of product, quotient, and chain rules ... = x^2 \cdot ln \ x. The product rule starts out similarly to the chain rule, finding f and g. However, this time I will use $$f_2(x)$$ and $$g_2(x)$$. 2.11. The chain rule is a powerful and useful derivation technique that allows the derivation of functions that would not be straightforward or possible with the only the previously discussed rules at our disposal. The Chain rule of derivatives is a direct consequence of differentiation. Now we have a special case of the chain rule. This unit illustrates this rule. The Derivative tells us the slope of a function at any point.. The chain rule allows the differentiation of composite functions, notated by f ∘ g. For example take the composite function (x + 3) 2. SOLUTION 12 : Differentiate . 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. 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. Associate Professor, Candidate of sciences (phys.-math.) Hence, the constant 4 just tags along'' during the differentiation process. The chain rule says that. There is also another notation which can be | {
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the differentiation process. The chain rule says that. There is also another notation which can be easier to work with when using the Chain Rule. In this tutorial we will discuss the basic formulas of differentiation for algebraic functions. The chain rule in calculus is one way to simplify differentiation. There is a chain rule for functional derivatives. Kirill Bukin. 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 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. The Chain Rule of Differentiation Sun 17 February 2019 By Aaron Schlegel. J'ai constaté que la version homologue française « règle de dérivation en chaîne » ou « règle de la chaîne » est quasiment inconnue des étudiants. 10:40. Example of tangent plane for particular function. In this section we discuss one of the more useful and important differentiation formulas, The Chain Rule. The chain rule tells us how to find the derivative of a composite function. Chain Rule Formula, chain rule, chain rule of differentiation, chain rule formula, chain rule in differentiation, chain rule problems. 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. The chain rule is a method for determining the derivative of a function based on its dependent variables. If cancelling were allowed ( which it’s not! ) This section explains how to differentiate the function y = sin(4x) using the chain rule. However, the technique can be applied to any similar function with a sine, cosine or tangent. The inner function is g = x + 3. 5:20. Proof of the Chain Rule • Given two functions f and g where g is | {
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inner function is g = x + 3. 5:20. 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. Thus, ( There are four layers in this problem. The Chain Rule mc-TY-chain-2009-1 A special rule, thechainrule, exists for differentiating a function of another function. For instance, consider $$x^2+y^2=1$$,which describes the unit circle. Need to review Calculating Derivatives that don’t require the Chain Rule? In the next section, we use the Chain Rule to justify another differentiation technique. Categories. 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. Hessian matrix. Try the Course for Free. Here are useful rules to help you work out the derivatives of many functions (with examples below). Taught By. Transcript. Second-order derivatives. In what follows though, we will attempt to take a look what both of those. 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. 2.13. En anglais, on peut dire the chain rule (of differentiation of a function composed of two or more functions). 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. With these forms of the chain rule implicit differentiation actually becomes a fairly simple process. Chain rule for differentiation. This discussion will focus on the Chain Rule of Differentiation. There are many curves that we can draw in the plane that fail the "vertical line test.'' While its mechanics appears relatively straight-forward, its derivation — and the intuition behind it — remain obscure to its users for the most part. All functions are functions of real numbers that return real values. If z is a | {
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for the most part. All functions are functions of real numbers that return real values. 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 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. With the chain rule in hand we will be able to differentiate a much wider variety of functions. Let u = 5x (therefore, y = sin u) so using the chain rule. This rule … 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. The only problem is that we want dy / dx, not dy /du, and this is where we use the chain rule. Of differentiation of a function y = sin 5x answer to 2: differentiate y = (. Out with the implicit differentiation examples ; all Lessons all Lessons all Lessons.... A direct consequence of differentiation draw in the next section, we will attempt take. ( which it ’ s not!: Quotient rule and chain rule of differentiation handling the derivative a! Professor, Candidate of sciences ( phys.-math. may still be interested finding. The compositions of two or more functions ) take will involve the chain rule ). A composite function differentiation examples ; all Lessons Categories functions ) not dy /du By du/ dx thechainrule exists! vertical line test. each of the chain rule in derivatives: the chain rule calculus! Next section, we will discuss the basic formulas of differentiation for algebraic functions we draw... Is not necessary to use the chain rule ( of differentiation for functions! Test. use the chain rule become second nature the standard rules, we will discuss basic... S solve some common problems | {
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rule become second nature the standard rules, we will discuss basic... S solve some common problems step-by-step so you can learn to solve them routinely yourself. Rule tells us the slope of a composite function easier to work with when using the chain rule multiply /du! Problem set: Quotient rule and chain rule. much wider variety of functions rule implicit differentiation examples all. Differentiation Sun 17 February 2019 By Aaron Schlegel justify another differentiation technique notation can! Take a look what both of those = 3, so can learn to solve them for. Becomes a fairly simple process for differentiating a function at any point chain rule of differentiation point courses a great many of you. On peut dire the chain rule implicit differentiation actually becomes a fairly simple process locuteurs m'ont... Still be interested in finding slopes of tangent lines to the circle at various.! = x + 3 out the derivatives of many functions ( with examples below ) of numbers... Questions: product rule, thechainrule, exists for differentiating a function and then simplifies it the circle... Derivatives using the chain rule Date_____ Period____ differentiate each function with a sine, cosine tangent. That they become second nature the rest of your calculus courses a great many of derivatives a. And learn how to apply the chain rule Date_____ Period____ differentiate each function with a sine, cosine tangent... For those that want chain rule of differentiation thorough testing of their basic differentiation using the chain rule. the techniques here! To two functions tags along '' during the differentiation process use the product rule, Quotient rule chain. Special case of the chain rule to justify another differentiation technique work out the of! Slope of a function any point collègues locuteurs natifs m'ont recommandé de … the chain rule ( of for... Using the standard rules therefore, y = sin ( 4x ) using the chain rule of... Will attempt to take a look what both of those | {
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therefore, y = sin ( 4x ) using the chain rule of... Will attempt to take a look what both of those … the chain rule to justify another technique. Calculate derivatives using the standard rules ’ s do a harder example of chain! ( phys.-math., du/ dx of composite functions, and learn how find... In what follows though, we will be able to differentiate the function y = sin ( 4x chain rule of differentiation the. Review Calculating derivatives that don ’ t require the chain rule., y sin... Use the chain rule ( of differentiation calculus is one way to simplify differentiation using... See throughout the rest of your calculus courses a great many of derivatives take... De … the chain rule of differentiation of a function us the slope of function! Rule ( of differentiation for algebraic functions become second nature easier to work with when using standard... Phys.-Math. we may still be interested in finding slopes of tangent lines to the at. The inner function is g = x + 3 ( phys.-math. simplify differentiation to take a look both... The rest of your calculus courses a great many of derivatives is a direct consequence of differentiation of a of! Next: problem set: Quotient rule and chain rule. Period____ differentiate function... When using the chain rule correctly of real numbers that return real values require the chain rule for the! To 2: differentiate y = sin ( 4x ) using the chain rule is a in! 3 = u 2 5x ( therefore, y = sin 5x rule., so 3x 2. A calculus I course attempt to take a look what both of those can in!, cosine chain rule of differentiation tangent will see throughout the rest of your calculus courses a many..., we use the chain rule. can draw in the next section, we will to... Fairly simple process outer function becomes f = u 2 we will attempt to take a what!, so Lessons all Lessons all Lessons all Lessons all Lessons Categories various points another technique. You take will involve the chain rule in derivatives: the chain rule in calculus is one to... | {
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You take will involve the chain rule in derivatives: the chain rule in calculus is one to... Derivatives: the chain rule. differentiating the compositions of two or more functions derivatives a. Differentiation process is that we saw in a calculus I course Date_____ Period____ differentiate each function with sine! Will discuss the basic formulas of differentiation of a function composed of two more. Tangent lines to the circle at various points − 2, du/ dx the chain rule is a differentiation... Out the derivatives of many functions ( with examples below ) to use the rule. Therefore, y = sin ( 4x ) using the chain rule Date_____ Period____ differentiate function! Advantage of the following problems requires more than one application of the chain rule. Lessons.... Product rule, thechainrule, exists for differentiating a function and then simplifies.! /Du By du/ dx = 3, so the implicit differentiation examples ; all Categories! This calculator calculates the derivative of a function and then simplifies it it is vital you! Throughout the rest of your calculus courses a great many of derivatives you take will involve the chain rule a. With when using the chain rule implicit differentiation examples ; all Lessons Categories vertical line test. are... compositeness '' of a function problems requires more than one application of the following problems more. Thechainrule, exists for differentiating a function of another function working to calculate derivatives using chain! Constant 4 just tags along '' during the differentiation process special case of following... Great many of derivatives you take will involve the chain rule is not limited to two functions ( examples! Are you working to calculate derivatives using the standard rules we may still be interested in finding slopes tangent. Describes the unit circle working to calculate derivatives using the standard rules, exists for a! Y = sin u ) so using the chain rule. questions: product rule )! Functions ( with examples | {
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for a! Y = sin u ) so using the chain rule. questions: product rule )! Functions ( with examples below ) a rule in calculus is one way to simplify differentiation to x −,... /Du, chain rule of differentiation this is where we use the chain rule of differentiation so can! With respect to x us the slope of a composite function then the outer function becomes =... Here are useful rules to help you work out the derivatives of many functions ( with examples below ) us... Calculate derivatives using the chain rule of differentiation for algebraic functions Quotient rule and chain rule of derivatives a! Will see throughout the rest of your calculus courses a great many of derivatives you take will involve chain!, the constant 4 just tags along '' during the differentiation process differentiating compositions! Functions, and learn how to find the derivative of a function composed of two more. Limited to two functions a harder example of the following problems requires more than application. Able to differentiate the function y = sin 5x to justify another differentiation technique rule and chain to! These rules allow us to differentiate the function y = sin u so... Differentiation of a function t require the chain rule. rule, thechainrule, exists for differentiating function! Be easier to work with when using the standard rules can learn to them. Courses a great many of derivatives is a direct consequence of differentiation, we will the! Are useful rules to help you work out the derivatives of many (... Is to multiply dy /du, and this is where we use the chain rule correctly to... With these forms of the chain rule in derivatives: the chain rule. a thorough testing of basic! Discuss the basic formulas of differentiation dx = 3, so calculus is one way to simplify differentiation similar! You working to calculate derivatives using the chain rule of differentiation, the technique can easier! A harder example of the following problems requires more than one application of the | {
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can easier! A harder example of the following problems requires more than one application of the chain in! Us to differentiate functions of the form ( t ) = what both those... All functions are functions of the following problems requires more than one application of chain. Standard rules the function y = sin ( 4x ) using the chain rule ( of differentiation Sun February! Not! don ’ t require the chain rule. ’ t the... Functions ) sin u ) so using the standard rules it ’ s do harder... Sciences ( phys.-math. s not! s start out with the implicit differentiation we... ; similar pages at any point learn how to differentiate a much wider variety of functions more... Do is to multiply dy /du, and this is where we use product! Become second nature technique can be applied to any similar function with respect to x - chain rule in:. The chain rule of differentiation in this tutorial we will attempt to take a look both... | {
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# Understanding the definition of the d-dimensional Hyperube
Please see the picture bellow about the definition of the nodes of the d-dimensional Hypercube. Could anyone please tell me what does that notation means. I get confused with the superscript after the curly braces. What does that mean in set theory.
Nodes: $(x_1,\ldots,x_d)\in\{0,1\}^d$
Edges: $\forall i:(x_1,\ldots,x_d)\to(x_1,..,1-x_i,.. ,x_d)$
What does this mean? How are the nodes and the edges defined formally?
Thanks.
Here $\{0,1\}^d = \{0,1\} \times \{0,1\} \times \cdots \{0,1\}$ is just the Cartesian product $d$ copies of the set $\{0,1\}$. Thus the nodes $(x_1, x_2, \dots, x_d) \in \{0,1\}^d \subseteq \mathbb{R}^d$ are just a vectors which we can think of as living in $\mathbb{R}^d$ where the components all have value 0 or 1. Then the edges of your graph just connect nodes that differ in exactly one component.
You can also think of $\{0,1\}^d$ as binary strings of length $d$, there is a clear bijection between the vectors described above and binary string of length $d$. Sometimes $\{0,1\}^d$ used to denote this set of strings, but the rest of your notation suggests the vector interpretation.
Try drawing the cases for $d=2$ and $d=3$ you should get a square and cube respectively.
• Yes $(x_1,x_2,\dots,x_d)$ is a single node. – John Machacek Jun 19 '14 at 13:13 | {
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# How many ways can six distinct black books and four distinct red books be arranged on a shelf if no two red books are adjacent?
Question : There are $$10$$ different books on a shelf. Four of them are red and the other six are black. How many different arrangements of these books are possible if no two red books may be next to each other?
So what I have done is
1) The # of ways to range the black books $$= 6!$$
2) So two red books can't be placed together $$= 6! \cdot \frac{7!}{3!} = 604 800$$
3) I am kind of confused right now. I don't know what I should subtract $$604~800$$ from.
Thank you!
• You do not need to subtract. The value you found in step 2 is the answer to the question. Oct 30 '18 at 21:19
• This tutorial explains how to typeset mathematics on this site. Oct 30 '18 at 21:26
• Is there some reason you think you need to subtract? Do you need an explanation of why your solution is correct? Oct 31 '18 at 12:11
• It's because I went to the TA office hour and she told me I have to do something else but I think she is wrong Oct 31 '18 at 12:57
This is a perfect problem to solve with stars and bars.
We have to place 6 black books in total. We can treat the red books as "dividers", and if we do that, the problem becomes how many ways are there to rearrange (*'s are black books, |'s are red)
******||||
However, to enforce that no two red books touch, we have to place 1 book in between 3 of the dividers. So, we only have 3 black books that we have to place
***||||
So, our answer, if the books are indistinguishable, is $${7\choose3}=\color{red}{35}$$
Now, as you note, the books are different. So, we multiply by the number of arrangements of red books and black books.
So, our final answer is $$35\cdot6!\cdot4!=\color{red}{604,800}$$
• @N.F.Taussig Oh... I didn't read. Oops Oct 30 '18 at 21:14 | {
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• @N.F.Taussig Oh... I didn't read. Oops Oct 30 '18 at 21:14
Method 1: The six black books can be arranged in a row in $$6!$$ ways. This creates seven spaces in which a red book can be placed, five between successive black books and two at the ends of the row.
$$\square B \square B \square B \square B \square B \square B \square$$ To ensure that no two of the four red books are adjacent, we must choose four of these spaces in which to place a red book, which can be done in $$\binom{7}{4}$$ ways. The four red books can be arranged in these spaces in $$4!$$ ways. Hence, the number of admissible arrangements is $$6!\binom{7}{4}4! = 6! \cdot \frac{7!}{3!} = 604800$$ as you found. There is no need to subtract.
Let's confirm your result. The following argument is more formal. Alas, it is also trickier, particularly when we count those arrangements in which there are two disjoint pairs of adjacent books.
Method 2: We use the Inclusion-Exclusion Principle.
Since there are $$10$$ different books on the shelf, they can be arranged in $$10!$$ ways. From these, we must subtract those arrangements in which a pair of red books are adjacent.
A pair of adjacent red books: A pair of adjacent red books can be selected in $$\binom{4}{2}$$ ways. We now have nine objects to arrange, the six black books, the pair of adjacent red books, and the other two red books. The objects can be arranged in $$9!$$ ways. The pair of adjacent books can be arranged internally in $$2!$$ ways. Hence, there are $$\binom{4}{2}9!2!$$ arrangements with a pair of adjacent red books.
However, if we subtract the number of arrangements with a pair of red books from the total, we will have subtracted too much since we will have subtracted each arrangement with two pairs of adjacent red books twice, once for each way of designating one of the pairs as the pair of adjacent red books. Thus, we add those arrangements to the total. | {
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Two pairs of adjacent red books: This can occur in two ways. Either the pairs are disjoint or overlapping, meaning that there are three consecutive red books.
Two disjoint pairs of adjacent red books: Since the red books are distinct, there must be some ways of distinguishing them. Say they have different titles. If so, then there are three ways to pair one of the other red books with the one whose title appears first in an alphabetical list. The other two red books must form the other pair of adjacent red books. We have eight objects to arrange, the six black books and two disjoint pairs of red books. The objects can be arranged in $$8!$$ ways. Each pair of adjacent red books can be arranged internally in $$2!$$ ways. Hence, there are $$\binom{3}{1}8!2!2!$$ arrangements with two disjoint pairs of red books.
Two overlapping pairs of adjacent red books: The three consecutive red books can be selected in $$\binom{4}{3}$$ ways. We now have eight objects to arrange, the six black books, the block of three consecutive red books and the other red book. The objects can be arranged in $$8!$$ orders. The trio of consecutive red books can be arranged internally in $$3!$$ ways. Hence, there are $$\binom{4}{3}8!3!$$ arrangements with two overlapping pairs of red books.
Hence, there are $$\binom{3}{1}8!2!2! + \binom{4}{3}8!3!$$ arrangements with two pairs of adjacent red books. | {
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However, if we subtract the number of arrangements with a pair of adjacent red books and then add the number of arrangements with two pairs of adjacent red books from the total, we fail to exclude those arrangements with three pairs of adjacent red books, which can only occur if there are four consecutive red books since there are only four red books. The reason is that we subtracted such arrangements three times, once for each way of designating one of the three pairs of adjacent red books as the pair of adjacent red books, and added them three times, once for each of the $$\binom{3}{2}$$ ways of designating two of the three pairs of adjacent red books as the two pairs of adjacent red books.
Three pairs of adjacent books: There are seven objects to arrange, the six black books and the block of four red books. The objects can be arranged in $$7!$$ ways. The block of four red books can be arranged internally in $$4!$$ ways. Hence, there are $$7!4!$$ arrangements with three pairs of adjacent red books.
Hence, the number of arrangements of six distinct black books and four distinct red books in which no two red books are adjacent is $$10! - \binom{4}{2}9!2! + \binom{3}{1}8!2!2! + \binom{4}{3}8!3! - 7!4! = 604800$$ as you found. | {
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A problem in probability
1. Jun 6, 2004
lhuyvn
Hi members,
I have traveled this forum sometimes, But this is my first question. I hope to get your help so that I can prepare better for my GRE Math test.
Following is my question.
In a game two players take turns tossing a fair coin; the winner is the firt one to toss a head. The probability that the player who makes the first toss wins the game is:
A)1/4
B)1/3
C)1/2
D)2/3
E)3/4
LuuTruongHuy
2. Jun 6, 2004
maverick280857
HI
Here's my solution....
T = tails
(AH denotes "A got a head")
Suppose A starts first. Then the different possibilities are tabulated thus:
AH (A gets a head, game stops)
AT,BT,AH (A gets tails, B gets tails, A gets heads, game stops)
AT,BT,AT,BT,AH (A gets tails, B gets tails, A gets tails, B gets tails, A gets heads, game stops)
and so on....
So the probability is given by the sum,
$$\displaystyle{\frac{1}{2}} + \displaystyle{\frac{1}{2}*\frac{1}{2}*\frac{1}{2}} + \displaystyle{\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2} + ...}$$
the kth term is
$$(\frac{1}{2})^p$$
where p = (2k+1) for k = 0, 1, 2, .... note that there are (2k+1) continued products in the kth term)
The game goes on as long as A and B get tails and stops as soon as A gets a head, since A was the one who started the game first.
This is an infinite sum, the value of which is given by
$$SUM = \displaystyle{\frac{1/2}{1-(1/4)}} = \frac{2}{3}$$
I think 2/3 should be the answer, but I could be wrong (as usual) ;-)
Someone please correct me if I'm wrong. If any part of the solution is wrong/not clear, please let me know. (I have assumed that you are familar with addition and multiplication in probability and also with geometric progressions, esp containing an infinite number of terms--the kinds that appear in such problems.)
Cheers
Vivek
Last edited: Jun 6, 2004
3. Jun 6, 2004
uart
The first player has a probability of 1/2 that both he will take a first toss AND that he will win on that toss. | {
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The second player only has a probablity of 1/4 that he will both take his first toss and win on that toss.
The first player then has a probabilty of 1/8 that he will both require his second toss and win on that toss.
Continuing on like this the first player has a probabilty of 1/2 + 1/8 + 1/32 + ... and the second player has a probability of 1/4 + 1/16 + 1/64 + ... of winning.
4. Jun 6, 2004
Hurkyl
Staff Emeritus
For those who like clever answers, you can skip the infinite series.
Suppose the first player's first flip is a tails. Now, if you look at how the game proceeds, it is identical to the original game, except the first and second player are reversed.
So if p is the probability that the first player in the game wins, then once the first player flips a tails, the second player has a probability p of winning. (and probability 0 of winning otherwise)
Since there's a 1/2 chance the first player will flip tails, the second player has a probability p/2 of winning, and the first player probability p.
Thus, p = 2/3.
5. Jun 8, 2004
lhuyvn
Thank All for very nice answers !!! | {
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## across 2 years ago This question is just for fun: Tell me what the last digit of $$3^{1234567890}$$ is. :) Then tell me what the last two digits are. And if you feel up to the challenge, then tell me what the last three digits are!
3
2. across
That's not it!
3. across
$\text{Hint: }a^{\phi(n)}\equiv1\text{ }(\text{mod }n)\text{, where }\gcd(a,n)=1.$
quite tricky...:)
5. myininaya
Is that fermat's or euler's thingy...I can't remember
6. across
You are right. :) This is Euler's theorem, and ϕ is the totient function. Naturally, it is an augmentation of Fermat's little theorem.
7. satellite73
fermat's little theorem $a^{p-1}\equiv 1( \text{ mod } p)$ euler phi totient function $a^{\phi(n)}\equiv 1 (\text{ mod } n)$ if $(a,n)=1$
8. satellite73
9. myininaya
I think @jamesj and @zarkon would love this question (maybe-lol) I will keep thinking on it though Great question @across
10. TuringTest
the last digit is 9, no?
11. across
Yes it is. :) Did you use Euler's theorem?
12. TuringTest
no I used a little logic no idea what that theorem is
13. TuringTest
$3^0=1$$3^1=3$$3^2=9$last digit is nine is the important point here$3^3=27$$3^4=81$and so the last digit repeats every four powers...
14. TuringTest
1,3,9,7,1,3,... so 1234567890/4=308641927+2/4 and 3^2=9
15. TuringTest
I'm sure that's got a professional way of writing it with mod this and that, but that's beyond me lol
16. Zarkon
29
17. TuringTest
how you got the other digit is what I want to know
18. across
That is genius. :) Do you know that you are intuitively deriving the above theorem? From it, it follows that to find out what the last two digits of that number are, you first need to observe that the sequence of two digits repeats every 40 powers. I will elaborate more on it a bit later. :)
19. across
I am pretty sure you can figure out what those two digits are by having told you that up there. ;)
20. across
@Zarkon, that is not it!
21. TuringTest | {
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20. across
@Zarkon, that is not it!
21. TuringTest
thank you for the compliment across, it means so much knowing who it's coming from :D
22. Zarkon
49
23. across
^.^ And @Zarkon, you are right. Let us in in thy arcane ways!
24. across
Anyway, :) to re-state Euler's theorem:$a^{\phi(n)}\equiv1\text{ }(\text{mod }n)\text{, where }\gcd(a,n)=1.$And to define Euler's phi-function:$\phi(n)=\text{number of positive integers relatively prime to }n.$By the way, I will now say that what I am about to explain is a huge over-simplification of the number theory behind it all, and that I have boiled the entire process down to a series of mechanical steps stemming from said theory. Since we are trying to find out what the last few digits of $$a=3^{1234567890}$$ are, all that we have to do is compute $$a\text{ }(\text{mod }10)$$ for the last digit, $$a\text{ }(\text{mod }100)$$ for the last two digits, and so on. Notice that $$\gcd(3,10)=\gcd(3,100)=\cdots=1$$, so we can indeed make use of the above theorem. Because it really does not pertain to the problem at hand, I will just say that $$\phi(10)=4$$, $$\phi(100)=40$$, and so on. Therefore, we have that $$3^{\phi(10)}\equiv3^4\equiv1\text{ }(\text{mod }10)$$ (check this to convince yourself that it is true), and it follows that, for the last digit, $3^{1234567890}\equiv3^2\cdot(3^4)^{308641972}\equiv3^2\cdot(1)^{308641972}\equiv9\text{ }(\text{mod }10).$Which is indeed our last digit. You can do the same thing for two digits:$3^{1234567890}\equiv3^{10}\cdot(3^{40})^{30864197}\equiv3^{10}\cdot(1)^{30864197}\equiv49\text{ }(\text{mod }100).$This can go on and on. For three digits, you will have to compute $$3^{290}$$, which is doable by Windows' calculator. But there are better methods for bigger and bigger numbers. :)
25. myininaya
Very nice across :)
26. KingGeorge | {
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25. myininaya
Very nice across :)
26. KingGeorge
Can I do the last 3 digits? Using Euler's theorem, we know that $$\phi(1000)=400$$ so $$3^{400} \equiv 1 \mod 1000$$. Now we calculate $$1234567890 \mod 400$$. As it turns out, this is 290. Thus, we only have to calculate $$3^{290} \mod 1000$$. Using successive squaring/fast powering, this is easy, and we get that the last three digits are 449.
27. across
That is correct. :)
28. FoolForMath
It's good to mention Carmichael theorem in this context. http://en.wikipedia.org/wiki/Carmichael_function#Carmichael.27s_theorem
29. FoolForMath
And if you are in a hurry, just a one liner in python: http://ideone.com/47UX7
30. ParthKohli
Woot, solution number two!$3^n \equiv 3^{n-4} \times 3^4 \equiv 3^{n -4}\pmod{10}$Suffice to say that $$3^{n} \equiv 3^{n - 4k} \pmod {10}$$ for all integer $$k$$. Well, that's just the solution @TuringTest posted. | {
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# K-subsets, counting, and the pigeon hole principle
Fellow mathematicians, please take a look at these practice test questions and try to help if you can:
Question $1$:
Let $n ≥ 8$ be an even integer and let $S = \{1, 2, 3, . . . , n\}$. Consider $7$-element subsets of $S$ that consist of $4$ even numbers and $3$ odd numbers. How many such subsets are there?
(a) ${{n/2}\choose4} \cdot {{n/2}\choose3}$
(b) ${{n}\choose4} \cdot {{n}\choose3}$
(c) ${{n/2}\choose4} + {{n/2}\choose3}$
(d) ${{n}\choose4} + {{n}\choose3}$
Answer is (a). Why do we divide $n$ by $2$ here? If $n=8$, then $n/2 = 4$ which leaves us with only $2$ even and odd numbers $(1,2,3,4)$, so how can we choose $4$ even and $3$ odd numbers when there is only two of each? My initial guess was (b).
Question $2$:
What does this count? $\sum_{k=2}^{n} {n\choose{k}} \cdot 2^{n-k}$
(a) The number of strings of length $n$, where each character is $a$ or $b$, that contain at least one $a$.
(b) The number of strings of length $n$, where each character is $a$ or $b$, that contain at least $2$ many $a$’s.
(c) The number of strings of length $n$, where each character is $a$, $b$, or $c$, that contain at least one $a$.
(d) The number of strings of length $n$, where each character is $a$, $b$, or $c$, that contain at least $2$ many $a$’s.
Where does the the $c$ in "$a$, $b$, or $c$" come from?
Question $3$:
Consider a square with sides of length $17$. This square contains $n$ points. What is the minimum value of $n$ such that we can guarantee that at least two of these points have distance at most $17/\sqrt 2$? Edit: This was originally $17\sqrt 2$, I had made a typo.
(a) 4
(b) 5
(c) 6
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(a) 4
(b) 5
(c) 6
(d) 7
The diagonal distance of this square is $\sqrt{{17^2} + {17^2}} = 17\sqrt 2$. This means that even with $2$ points, this rule would still hold because even if the two points were in either corner of the square, and since the longest distance is the diagonal distance, they would always be within a distance of $17\sqrt 2$. In this case, the lowest number to choose from is $4$, so that is the minimum -- however the answer is $5$. Where am I going wrong?
• I think @Robert Z has a point below, can you recheck the Question 3 because if distance is at most $17\sqrt{2}$ then $2$ points are enough. So it should be something else. And if the answer is $5$, it is probably $\frac{17}{\sqrt{2}}$. – ArsenBerk Feb 25 '18 at 8:28
• Indeed I have made a typo. It is the latter! – udpcon Feb 25 '18 at 16:49
Question 1. If $n$ is even then in $S = \{1, 2, 3, . . . , n\}$ there are $n/2$ even numbers and $n/2$ odd numbers. So the subsets of 4 even numbers are $\binom{n/2}{4}$ and the subsets of 3 odd numbers are $\binom{n/2}{3}$.
Question 2. Notice that ${n\choose{k}}$ is the number of ways to place $k$ character "a" and $2^{n-k}$ is the number of ways to fill the remaining $n-k$ characters with "b" or "c" ($2$ choices). Hence the given sum is just the number of strings of length $n$, where each character is "a", "b" or "c", which contain at least 2 many "a"s.
Question 3. Is distance at most $17\sqrt{2}$ or distance at most $17/\sqrt{2}$? (yes, you are correct, the distance of any two points in the square is at most the diagonal's length $17\sqrt{2}$!). For the second option, divide the square into $4$ equal squares. Then the lengths of the diagonals of the smaller squares is just $17/\sqrt{2}$. Hence by the pigeon hole principle, by taking $4+1$ points, at least two of them will lie into one of the smaller squares. | {
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Question 1: If $n$ is even, then $\frac{n}{2}$ gives you the number of even numbers in $S$, as well as number of odd numbers. Then you simply choose $4$ of the even numbers with $\binom{\frac{n}{2}}{4}$ and $3$ of the odd numbers with $\binom{\frac{n}{2}}{3}$.
Question 2: Notice that here, we need $3$ characters because of $2^{n-k}$ (There must be two options for the rest of the $n-k$ places after putting $a$ to $k$ places).
Question 3: HINT:
Here, I'm considering the distance is at most $\frac{17}{\sqrt{2}}$. So, having at least how many points guarantees that two of them are inside the same square? | {
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Jun 611:50 AM Using technology and a matrix approach we can verify our solution. Planes have a pretty special property. 2. 1 0 Where those axis meet is considered (0, 0, 0) or the origin of the coordinate space. 4. Colloquially, curves that do not touch each other or intersect and keep a fixed minimum distance are said to be parallel. Learn more about this Silicon Valley suburb, America's richest neighborhood. CS 506 Half Plane Intersection, Duality and Arrangements Spring 2020 Note: These lecture notes are based on the textbook “Computational Geometry” by Berg et al.and lecture notes from [3], [1], [2] 1 Halfplane Intersection Problem We can represent lines in a plane by the equation y = ax+b where a is the slop and b the y-intercept. The sum of two measures of two acute angles is greater than 90 degrees. Three points must be coplanar. For lines, rays, and line segments, intersect means to meet or cross. yes, three planes can intersect in one point. (f) If two lines intersect, then exactly one plane contains both lines (Theorem 3). Minimum number of 3-inch by 5-inch index cards needed to completely cover a 3-foot by 4-foot rectangular desktop, Number of diagonals in an n-sided polygon, Some Cool Math and Computer Science links, Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Sometimes. I know that they can intersect in a straight line, coincide, or be parallel to each other (non-intersecting planes), but I was wondering if they can intersect at one point. It is very easy to find a system of three equations in three unknowns for each one of the eight different relative positions, except for one: case 6 (where the three Planes intersect in a line). If you're talking about lines in the same plane, lines don't intersect if they're parallel. General solution to system of differential equation question...? Where those axis meet is considered (0, 0, 0) or the origin of the coordinate space. Any 3 dimensional cordinate system has 3 axis (x, y, | {
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0, 0) or the origin of the coordinate space. Any 3 dimensional cordinate system has 3 axis (x, y, z) which can be represented by 3 planes. Condition for three lines intersection is: rank Rc= 2 and Rd= 3 All values of the cross product of the normal vectors to the planes are not 0 and are pointing to the same direction. Let $\ell'$ be a line different from but parallel to $\ell$. The intersection of the three planes is a point. The planes will then form a triangular "tube" and pairwise will intersect at three lines. Thus, the intersection of 3 planes is either nothing, a point, a line, or a plane: To answer the original question, 3 planes can intersect in a point, but cannot intersect in a ray. 3. How many multiples of 8 are between 100 and 175? Finally we substituted these values into one of the plane equations to find the . There are infinitely many planes through $\ell'$, but only one of them intersects $\ell$, and only two of them are parallel to one of the first two planes. Each plan intersects at a point. 3 Plane Intersection. Jun 611:50 AM Using technology and a matrix approach we can verify our solution. If they are in the same plane, and not parallel, then they will intersect at exactly one point. true. Given 3 unique planes, they intersect at exactly one point! Sign in|Recent Site Activity|Report Abuse|Print Page|Powered By Google Sites, Maximum number of points of intersection of 6 lines. A projective plane can be thought of as an ordinary plane equipped with additional "points at infinity" where parallel lines intersect. Huh? he might have made a mistake when writing out the test or got it out of a faulty text or you might have misread. Intersection of Three Planes. true. Three planes can mutually intersect but not have all three intersect. In the figure below, rays BA and BC meet at endpoint B, so their intersection forms angle ∠ABC. true. Just two planes are parallel, and the 3rd plane cuts each in a line. Find the equation of the plane that contains | {
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are parallel, and the 3rd plane cuts each in a line. Find the equation of the plane that contains the point (1;3;0) and the line given by x = 3 + 2t, y = 4t, z = 7 t. Lots of options to start. In geometry, parallel lines are lines in a plane which do not meet; that is, two straight lines in a plane that do not intersect at any point are said to be parallel. What is A? Now, if we draw a 3rd line, that can intersect the other two lines in at most 2 points as shown below. How many five-digit numbers greater than 70,000 are mountain numbers? equation of a quartic function that touches the x-axis at 2/3 and -3, passes through the point (-4,49)? You can also rotate it around to see it from different directions, and zoom in or out. Integer coordinates enclosed by 2 squares. The intersection of the three planes is a line. In three dimensions, no. The text is taking an intersection of three planes to be a point that is common to all of them. How many times during the day do the hands of a clock overlap? The point can be located by starting from the origin then following the red arrows on the grid. Not for a geometric purpose, without breaking the line in the sketch. This is question is just blatantly misleading as two planes can't intersect in a point. three planes can intersect as a point or as a line. Penny Note that there is no point that lies on all three planes. (g) If a point lies outside a line, then exactly one plane contains both the line and the point … Increase Brain Power, Focus Music, Reduce Anxiety, Binaural and Isochronic Beats - Duration: 3:16:57. Thus, any pair of planes must intersect in a line, but not all three at once (since there is no solution). Each plane cuts the other two in a line and they form a prismatic surface. Or three planes can, like the pages in the spine of a book, can intersect in one single line. three noncollinear points determine a plane. Count the points of intersection for each and allow infinite as some of your counts. If the lines | {
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Count the points of intersection for each and allow infinite as some of your counts. If the lines are in different planes in 3-dimensional space, they won't intersect. We can use a matrix approach or an elimination approach to isolate each variable. I know that they can intersect in a straight line, coincide, or be parallel to each other (non-intersecting planes), but I was wondering if they can intersect at one point. The new app allows you to explore the concepts of solving 3 equations by allowing you to see one plane at a time, two at a time, or all three, and the intersection point. When two lines, rays, or line segments intersect, they have one common point. It is not possible in a three-dimensional space, but it is possible in a four-dimensional space. I may just have found a slightly shorter way (four commands vs five in the previous) - PC-DMIS can intersect a CURVE and a PLANE to get one of the intersection points (assuming your number of hits is large enough). Yes, look at … Think about what a plane is: an infinite sheet through three... See full answer below. The system has one solution. In one of the positions, the three Planes share only one point (they intersect at that point). You can edit the visual size of a plane, but it is still only cosmetic. Justify your answer. I would not confront your teacher but would recheck the question and if it asks about two planes intersecting I would ask for an explanation, because you don't get it. Any two of them must intersect, if no two are parallel, but there need not be a point that all three of them have in common. Note, because we found a unique point, we are looking at a Case 1 scenario, where three planes intersect at one point. This lines are parallel but don't all a same plane. (b) Give an example of three planes in R^3 that intersect in pairs but have no common point of intersection. Probability that a number created randomly from 3 digits is even. The second and third planes are coincident and the first | {
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created randomly from 3 digits is even. The second and third planes are coincident and the first is cuting them, therefore the three planes intersect in a line. There is nothing to make these three lines intersect in a point. The East wall, South wall and floor are three planes that intersect at one point, the point on the floor in the South-East corner of the room. Leave a comment * Sometimes. (e) A line contains at least two points (Postulate 1). (a) Give an example of three planes in R^3 that have a common line of intersection. The East wall and the South wall are pieces of two planes that do intersect. In America's richest town, $500k a year is below average. In 3D, three planes , and can intersect (or not) in the following ways: All three planes are parallel. Four points may be coplanar or noncoplanar. The intersection is the line that forms the South-East corner of the room. Thus any two distinct lines in a projective plane intersect in one and only one point. Any 3 dimensional cordinate system has 3 axis (x, y, z) which can be represented by 3 planes. Given 3 unique planes, they intersect at exactly one point! two planes intersect in a line. false. Get an answer to your question "Can 2 planes in a 3 dimensional space intersect at one point ..." in Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions. Geometrically, we have planes whose orientation is similar to the diagram shown. Note, because we found a unique point, we are looking at a Case 1 scenario, where three planes intersect at one point. (1) (2) (3) point of intersection 3 4 z. value. the lines intersect at a point. Precalculus 3-D Cartesian Coordinate System Lines in Space. Homework Statement The three lines intersect in the point (1; 1; 1): (1 - t; 1 + 2*t; 1 + t), (u; 2*u - 1; 3*u - 2), and (v - 1; 2*v - 3; 3 - v). Justify your answer. Join Yahoo Answers and get 100 points today. 2 Answers bp Dec 14, 2016 We can use a | {
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your answer. Join Yahoo Answers and get 100 points today. 2 Answers bp Dec 14, 2016 We can use a matrix approach or an elimination approach to isolate each variable. true. 3. We know a point on the line is (1;3… (g) If a point lies outside a line, then exactly one plane contains both the line and the point … Similarly, if we draw a 4th line, that can intersect the other 3 lines in at most 3 points and so on. Planes have a pretty special property. Two points must be collinear. Sometimes. Renaissance artists, in developing the techniques of drawing in perspective, laid the groundwork for this mathematical topic. a line and a plane can intersect in a point. If I have 5 independent groups and us a likert scale what stat analysis would I use. Two planes can intersect on exactly one point? Get your answers by asking now. Three lines intersect at one point. The number 23AB3 is exactly divisible by 99. Sometimes. Examples Example 3 Determine the intersection of the three planes: 4x y — z — 9m + 5y — z — was the question about three planes? (e) A line contains at least two points (Postulate 1). 1 0 Justify your answer. (c) Give an example of three planes in R^3 that intersect in a single point. If two planes intersect, then their intersection is a line (Postulate 6). For example, the xy-plane and the zw-plane intersect at the origin (0,0,0,0). A line and a plane intersect at exactly one point. If we found in nitely many solutions, the lines are the same. For example, the xy-plane and the zw-plane intersect at the origin (0,0,0,0). The intersection of the first two is a line$\ell$. Three planes may all intersect each other at exactly one point. How do you solve a proportion if one of the fractions has a variable in both the numerator and denominator? Therefore, the solution to this system of three equations is (3, 4, 2), a point This can be geometrically interpreted as three planes intersecting in a single point, as shown. Using that, we only need to create one line to find | {
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planes intersecting in a single point, as shown. Using that, we only need to create one line to find the other point. Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website! Calculator won't calculate sin divided by anything, shows error? Other geometric figures. (1) (2) (3) point of intersection 3 4 In four dimensions or higher, yes. In how many days will Jessica read 270 pages of a book? Music for body and spirit - Meditation music Recommended for you 3:16:57 Doesn't matter, planes have no geometric size. When two distinct planes meet, they intersect at a line. If we found no solution, then the lines don’t intersect. Is there a way to create a plane along a line that stops at exactly the intersection point of another line. The first and second are coincident and the third is parallel to them. Assuming you are working in R 3, if the planes are not parallel, each pair will intersect in a line. two planes intersect at exactly one point. Two planes contain the same point. Here are the ways three planes can associate with each other. I would not confront your teacher but would recheck the question and if it asks about two planes intersecting I would ask for an explanation, because you don't get it. Some geometric figures intersect at more than one point. These planes do not intersect. The measure of an angle is greater than the measure of its complement. This means that, instead of using the actual lines of intersection of the planes, we used the two projected lines of intersection on the x, y plane to find the x and y coordinates of the intersection of the three planes. Therefore, the solution to this system of three equations is (3, 4, 2), a point This can be geometrically interpreted as three planes intersecting in a single point, as shown. Any two straight lines can intersect each other in one point. true. Only lines intersect at a point. How many 5 digit numbers are perfect squares? Precalculus . If two planes intersect, then their | {
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How many 5 digit numbers are perfect squares? Precalculus . If two planes intersect, then their intersection is a line (Postulate 6). was the question about three planes? Science Anatomy & Physiology ... Can two planes intersect in exactly one point? no they cant... only a line can work for this.. all the time... Can two two-dimensional planes intersect in a point? It is not possible in a three-dimensional space, but it is possible in a four-dimensional space. One Comment on “Intersection of 3 planes at a point: 3D interactive graph” Leesa Johnson says: 24 Jan 2017 at 9:13 pm [Comment permalink] Nice explanation for me to understand the interaction of 3d planes at a point using graphical representation and also useful for the math students. Florida governor accused of 'trying to intimidate scientists', Ivanka Trump, Jared Kushner buy$30M Florida property, Another mystery monolith has been discovered, MLB umpire among 14 arrested in sex sting operation, 'B.A.P.S' actress Natalie Desselle Reid dead at 53, Goya Foods CEO: We named AOC 'employee of the month', Young boy gets comfy in Oval Office during ceremony, Packed club hit with COVID-19 violations for concert, Heated jacket is ‘great for us who don’t like the cold’, COVID-19 left MSNBC anchor 'sick and scared', Former Israeli space chief says extraterrestrials exist. Question 97302: can 3 planes intersect in exactly one point? 3 Plane Intersection. Planes intersect along a line. three planes can intersect as a point or as a line. This commonly occurs when there is one straight plane and two other planes intersect it at acute or obtuse angles. Still have questions? (f) If two lines intersect, then exactly one plane contains both lines (Theorem 3). | {
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# If sinh(y)=x, then, show that cosh(y)=sqrt(1+x^2)
1. Sep 23, 2007
### cks
if sinh(y)=x, then, show that cosh(y)=sqrt(1+x^2)
I know how to prove but i have difficult in choosing the signs.
sinh(y) = x = [exp(y)-exp(-y)]/2
there are two equations I can find
exp(2y) - 2x exp(y) -1 =0 OR exp(-2y) + 2xexp(-y) -1 =0
exp(y) = x +- sqrt(x^2 +1) & exp(-y) = -x +- sqrt(x^2 + 1)
if I select the signs of +
exp(y) = x + sqrt(x^2 +1) & exp(-y) = -x + sqrt(x^2 + 1)
then by substituting to cosh(y) = [exp(y) + exp(-y)] / 2
then, I can find the answer of
cosh(y)=sqrt(1+x^2)
But, what should I say to justify my actions of selecting the +ve sign.
2. Sep 23, 2007
### neutrino
Why not just substitute $$x = \frac{\left(e^y - e^{-y}\right)}{2}$$ into the expression $$\sqrt{1+x^2}$$ and simplify?
3. Sep 23, 2007
### cks
Thanks, but I'd like to find cosh(y) without prior knowledge that it's equal to sqrt(1+x^2)
4. Sep 23, 2007
### robphy
For real y, exp(y)>0.
5. Sep 23, 2007
### arildno
Utilize the identity:
$$Cosh^{2}(y)-Sinh^{2}(y)=1$$
Along with the requirement that Cosh(y) is a strictly positive function.
(Alternatively, that it is an even function with Cosh(0)=1)
Last edited: Sep 23, 2007
6. Sep 23, 2007
### cks
still don't quite understand
7. Sep 24, 2007
### Gib Z
arildno pretty much gave you the identity that you need. If you need to prove it, Just replace cosh y and sinh y with their exponential definitions and simplify. Then isolate cosh y on the left hand side and it is clear. Basically when we take the final square root, one could normally argue you could have either the positive or the negative root. However cosh is always positive.
8. Sep 24, 2007
### robphy
There are different ways to approach the title of this thread, depending on what you are given to start with and what constraints are imposed... which should include your working definitions of sinh and cosh. | {
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Rather than merely proving an identity, it appears that the OP wishes to obtain the title by using the definitions of sinh and cosh starting from their definitions using the sum and difference of exponential functions. In addition, one might taking the point of view that one doesn't know at this stage any properties of cosh except for its working definition. So, it might be "cheating" to use any other properties of cosh which you didn't derive already.
cks, is this correct?
(If I'm not mistaken, my suggestion "For real y, exp(y)>0" provides a reason to choose the positive sign.)
9. Sep 25, 2007
### cks
Oh, I see thank you. Let me rephrase what you all said
exp(y) = x +- sqrt(x^2 +1) & exp(-y) = -x +- sqrt(x^2 + 1)
exp(y) is always > 0
so if we select exp(y) = x - sqrt(x^2 +1)
and most importantly , sqrt(x^2 +1) > x
so exp(y) < 0 which is false.
As a result, we have to select exp(y) = x + sqrt(x^2 +1)
The similar argument can be applied to exp(-y)
Thank you all of you, really appreciate that. | {
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# Math Help - Simultaneous Equations with way too many unknowns..
1. ## Simultaneous Equations with way too many unknowns..
Yea, so equations aren't my bestfriends, and I'm now stuck with two simultaneous equations, and I just don't know where to start, so if someone could help me and show me what to do, I'd appreciate it.
'Find the values of m such that these equations have no solutions':
3x - my = 4
x + y = 12
'Find the values of m and a such that these equations have infinite solution sets':
4x - my = a
2x + y = 4
So if someone could patiently guide me through step by step, that'd be awesome.. I hate not knowing what I'm supposed to do.
Thanks a lot
2. Originally Posted by Fnus
Yea, so equations aren't my bestfriends, and I'm now stuck with two simultaneous equations, and I just don't know where to start, so if someone could help me and show me what to do, I'd appreciate it.
'Find the values of m such that these equations have no solutions':
3x - my = 4
x + y = 12
'Find the values of m and a such that these equations have infinite solution sets':
4x - my = a
2x + y = 4
So if someone could patiently guide me through step by step, that'd be awesome.. I hate not knowing what I'm supposed to do.
Thanks a lot
If the equations have an infinite solution set then basically the two equations say the same thing: that is to say, one equation is a multiple of the other.
So since the coefficient of x in the first equation is twice the coefficient of x in the second equation, then "-m" must be twice the coefficient of y in the second equation, so
$-m = 2 \cdot 1$
or m = -2.
Similarly a = 8.
-Dan
3. there are no solutions if both equations are lines with the same slope (parallel) and different y-intercepts. thus they will never cross.
Question 1:
'Find the values of m such that these equations have no solutions':
3x - my = 4
x + y = 12
Question 2:
'Find the values of m and a such that these equations have infinite solution sets':
4x - my = a
2x + y = 4 | {
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4x - my = a
2x + y = 4
Does that change anything? <.<
5. $x+y=12\Rightarrow y=-x+12$ so the slope is -1.
$3x-my=4\Rightarrow y=\frac{3}{m}x-\frac{4}{m}$
now solve $\frac{3}{m}=-1$ you should get $m=-3$
6. for the second one we have
$4x-my-a=0$
and
$2(2x+y)=2(4)\Rightarrow 4x+2y-8=0$
from here it should be obvious what values to make a and m.
7. Hello, Fnus!
Find the values of $m$ such that these equations have no solutions:
. . $\begin{array}{cc}(1)\\(2)\end{array}
\begin{array}{cc}3x - my \\ x + y \end{array}
\begin{array}{cc} = \\ = \end{array}
\begin{array}{cc}4 \\ 12\end{array}$
If you are familiar with determinants, there is a simple solution.
A system has no solution if its determinant equal zero.
The determinant is: . $\begin{vmatrix} 3 & \text{-}m \\ 1 & 1\end{vmatrix} \:=\:(3)(1) - (\text{-}m)(1) \:=\:3 + m$
Therefore: . $3 + m \:=\;0\quad\Rightarrow\quad\boxed{m = -3}$
Otherwise, we can try to solve the system.
We have equation (1): . $3x - my \:=\:4$
. . .Multiply (2) by $m:\;\;mx + my \:=\:12m$
. . . . . . . . . . . . Add: . $mx + 3x \:=\:12m + 4$
. . . . . . . . . . .Factor: . $(m + 3)x \:=\:12m + 4$
. . . . . . . . . . . Then: . . . . . . $x\:=\:\frac{12m + 4}{m + 3}$
We see that $x$ is undefined if $m = -3.$
Therefore, the system has no solutions for $m = -3.$
8. Originally Posted by topsquark
If the equations have an infinite solution set then basically the two equations say the same thing: that is to say, one equation is a multiple of the other.
So since the coefficient of x in the first equation is twice the coefficient of x in the second equation, then "-m" must be twice the coefficient of y in the second equation, so
$-m = 2 \cdot 1$
or m = -2.
Similarly a = 8. | {
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Similarly a = 8.
-Dan
Originally Posted by putnam120
there are no solutions if both equations are lines with the same slope (parallel) and different y-intercepts. thus they will never cross.
But if one equation is merely a multiple of the other, they represent the same line, so any point (x,y) on the line is a solution. So we have an infinite number of solutions.
-Dan | {
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# Relationship between O and o notation
In big-O notation, $f(x) = O(g(x))$ as $x\rightarrow \pm\infty$ if
$$\exists C, \delta>0: \forall |x| \geq \delta: |f(x)| < C |g(x)|$$
and, for the case I'm more interested in here, $f(x) = O(g(x))$ as $x\rightarrow 0$ if
$$\exists C, \delta>0: \forall |x| \leq \delta: |f(x)| < C |g(x)|$$
In little-o notation, $f(x) = o(g(x))$ means
$$\lim_{|x|\rightarrow0}\frac{f(x)}{|g(x)|} = 0$$
I've come across big-O before, but this is the first time I've seen little-o. I have experience in applied math, but very little in pure math, and my first reaction (for $x\rightarrow 0$) was "they mean the same thing". Then I thought "but little-o is a more precise upper bound".
What is the relationship between these two notations, and what does the distinction mean practically?
• Well, if $f(x)$ is $o(g(x))$ it is also $O(g(x))$ but maybe not the other way around. It is not "more precise", but rather stronger: intuitively, if $f(x)$ is $o(g(x))$ then $g(x)$ grows "much faster" than $f(x)$, whereas $O(g(x))$ implies only that $f(x)$ is eventually bounded above (by some constant multiple of) $g(x)$ -- indeed, the growth rates could be comparable. – The_Sympathizer Sep 20 '13 at 11:54
• An example -- though note here I'm using the O-notation for growth near infinity, not near 0: If $f(x) = x$ and $g(x) = x^2$, then $f(x)$ is $o(g(x))$ AND $O(g(x))$, but if $f(x) = x^2$ then $f(x)$ is only $O(g(x))$ and NOT $o(g(x))$. – The_Sympathizer Sep 20 '13 at 11:56
• An example for near 0: $f(x) = \frac{1}{x}$ and $g(x) = \frac{1}{x^2}$. Then $f(x)$ is $o(g(x))$ (as $x \rightarrow 0$) while if $f(x) = \frac{1}{x^2}$ then $f(x)$ is only $O(g(x))$ and not $o(g(x))$. – The_Sympathizer Sep 20 '13 at 11:58
In the briefest possible terms, $f \in O(g)$ is like saying $f \leqslant g$ asymptotically, and $f \in o(g)$ is like saying $f < g$ asymptotically. | {
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As an example, if $f(x) = x$ and $g(x) = 2x$,we have that that $f \in O(g)$, but $f \notin o(g)$ and $g \notin o(f)$. If $f(x) = x^2$ and $g(x) = x$, then $f \in o(g)$ and $f \in O(g)$.
As an aside, be very aware of context. In various probabilistic situations, we define $O$-notation with a limit as $x\rightarrow 0$. In most algorithmic applications, we take limits as $x \rightarrow \infty$. Always make sure it's clear what you're working with before using either notation. $O$-notation is probably the most abused thing in all of mathematics.
• Also note: Notation $f \in O(g)$ is quite rare, compared to notation $f = O(g)$. – GEdgar Sep 20 '13 at 13:42
• Indeed. I didn't want to start listing abuses of $O$ notation because I don't think I'd ever stop. It's more common to say things like $2x = O(x)$. – ymbirtt Sep 20 '13 at 14:13
• $x^2$ is not in $O(x)$, at least when $x \to \infty$. Typo? – ftfish Sep 20 '13 at 14:44
• From OP's definitions, we're sending $x \rightarrow 0$, so I'm remaining consistent with that. – ymbirtt Sep 20 '13 at 15:08
• @GEdgar thanks, discussion of this helps me to get to grips with it. I've always been a little bit uncomfortable with the use of $=$, and thinking about it in terms of sets is more consistent. – TooTone Sep 21 '13 at 9:07
Little $o$ means lim sup of absolute value $=0$, big $O$ means lim sup of absolute value $< \infty$. | {
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Little $o$ means lim sup of absolute value $=0$, big $O$ means lim sup of absolute value $< \infty$.
• Thanks, but I thought big O could be used for $\rightarrow 0$ as well as $\rightarrow \infty$? The former was what I was more interested in because it then has a similar meaning to little o. (My post was a little ambiguous and I see you changed $<$ in my post to $\ge$ -- I've changed that back and added some more which hopefully makes it clearer.) – TooTone Sep 20 '13 at 14:31
• You are referring to my correction? Okay, I am sorry then the same remains true. It should be indicated though wether you look at the behaviour at $0$ or at $\infty$. A continuous variable transformation, e.g. $x \mapsto x^{-1}$ exchanges between the setting and my statement remains valid;) – Marc Palm Sep 20 '13 at 14:39
• Yes you're right I should have indicated whether I was looking at limit towards zero or infinity. I can see how a reciprocal transformation might make the situations equivalent; however I still don't quite understand your answer. – TooTone Sep 20 '13 at 19:52
• my answer remains valid no matter to where the limit goes. – Marc Palm Sep 21 '13 at 14:57
• I think what I don't understand is, in simple terms, your use of "sup" means. – TooTone Sep 21 '13 at 18:01
check this out :
http://en.wikipedia.org/wiki/Big_O_notation
:-) it's not just Big O notation in the page
Looking at how $O(1)$ and $o(1)$ are sufficient to understand the general case.
Suppose $f \in O(1)$. Then
$$\lim_{x \to +\infty} f(x) < +\infty$$
if the limit exists. Or more generally, if the limit does not exist, then
$$\sup_{x \to +\infty} f(x) < +\infty$$
Now suppose $f \in o(1)$. Then
$$\lim_{x \to \infty} f(x) = 0$$ | {
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# Fixed Point Theorems
Theorem 1. Let $B=\{x\in \mathbb R^n :∥x∥≤1\}$ be the closed unit ball in $\mathbb R^n$ . Any continuous function $f:B\rightarrow B$ has a fixed point.
Theorem 2. Let $X$ be a finite dimensional normed vector space, and let $K\subset X$ be a non-empty, compact, and convex set. Then given any continuous mapping $f:K\rightarrow K$ there exists $x\in K$ such that $f(x)=x$.
Theorem 3. Let $X$ be a normed vector space, and let $K\subset X$ be a non-empty, compact, and convex set. Then given any continuous mapping $f:K\rightarrow K$ there exists $x\in K$ such that $f(x)=x$.
Theorem 4. Let $X$ be a normed vector space, and let $K\subset X$ be a non-empty, closed, and bounded set. Then given any compact mapping $f:K\rightarrow K$ there exists $x\in K$ such that $f(x)=x$.
For some authors Theorem 1 is Brouwer's fixed-point theorem. For others Brouwer's fixed-point theorem is Theorem 2. Actually there is no difference because every non-empty, compact and convex set in a finite dimensional normed vector space is is homeomorphic to the closed unit ball.
My problem is with theorems 3 and 4. For some authors Theorem 3 is Schauder's fixed-point theorem, for others Schauder's fixed-point theorem is Theorem 4.
Are Theorem 3 and Theorem 4 are equivalent? If not, are Theorems 1 and 2 special cases of Theorem 4?
-
"because every non-empty, compact and convex set in a finite dimensional normed vector space is is homeomorphic to the closed unit ball" is not quite true, it could be lower-dimensional. Theorems 1 and 2 are special cases, because in that case, a continuous $f \colon K \to K$ is compact. – Daniel Fischer Jul 13 '13 at 15:39
What is the nonl tag for? – draks ... Jul 13 '13 at 20:08
@draks... My guess is a foreshortened nonlinear-something that never really got off the ground. – Sharkos Jul 13 '13 at 22:56 | {
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Your statement of Theorem 4 is missing an assumption on $K$, such as being convex, or at least homeomorphic to such a set (convex, closed, bounded). Without such an assumption, rotation of a circle gives a counterexample. Also, I think that in Theorem 4 you want the normed space to be complete, i.e., a Banach space.
Theorem 3 is contained in Theorem 4, because on a compact set every continuous map is compact. Theorem 4 cannot be easily obtained from Theorem 3 (I think) because if we tried to simply replace $K$ with $\overline{f(K)}$ (which is compact), we can't apply Theorem 3 because $\overline{f(K)}$ is not known to be convex.
Both 3 and 4 were stated and proved by Schauder in his 1930 paper Der Fixpunktsatz in Funktionalraümen, which is in open access. Here is Theorem 3:
Satz I. Die stetige Funktionaloperation $F(x)$ bilde die konvexe, abgeschlossene und kompakte Menge $H$ auf sich selbst ab. Dann ist ein Fixpunkt $x_0$, vorhanden, d.h. es gilt $F(x_0)=x_0$.
And this is Theorem 4 (in slightly less general version: the image of $F$ is assumed compact instead of relatively compact; possibly because the latter concept wasn't in use).
Satz II. In einem "B"-Raume sei eine konvexe und abgeschlossene Menge $H$ gegeben. Die stetige Funktionaloperation $F(x)$ bilde $H$ auf sich selbst ab. Ferner sei die Menge $F(H)\subset H$ kompakt. Dann ist ein Fixpunkt vorhanden.
("B"-Raume is what is now called a Banach space.) So, it is correct to call both Theorem 3 and Theorem 4 "Schauder's fixed-point theorem".
And yes, Theorems 1 and 2 follow by specialization of Theorem 3 or 4 to finite dimensions.
- | {
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# Math Help - ship problem
1. ## ship problem
A ship P is travelling due East at 30 km/h and a Ship Q is travelling due South at 40 km/h.
Both ships keep constant speed and course. At t=0 they are each 10 km from the point of intersection of their courses and moving towards the point.
Iv found the co-ordinates of Q relative to P at t=0
---->X=(10,10)km
iv also found the velocity of Q relative to P----> V= V[Q] -V[P]
-----.V= (0,40) - V(30,0)
----- V= (-30,40)km/h
but im struggling to find the time at which P and Q are closest to each other...is there anyone that can help??? thankz
2. At t=0, they are both 10 kms from their intersection.
Think Pythagoras. Let D=square of distance between ships.
$D(t)=(10-30t)^{2}+(10-40t)^{2}$
$D'(t)=2(10-30t)(-30)+2(10=40t)(-40)$
$D'(t)=5000t-1400$
$5000t-1400=0$
$t=\frac{7}{25}$
3. Originally Posted by galactus
At t=0, they are both 10 kms from their intersection.
Think Pythagoras. Let D=square of distance between ships.
$D(t)=(10-30t)^{2}+(10-40t)^{2}$
$D'(t)=2(10-30t)(-30)+2(10=40t)(-40)$
$D'(t)=5000t-1400$
$5000t-1400=0$
$t=\frac{7}{25}$
thankz...but i was wondering if i did the first part right that i showed?(shown below) because im not too sure if i was meant to include negative signs with any of the answers? can u please just check that for me thankz
Iv found the co-ordinates of Q relative to P at t=0
---->X=(10,10)km
iv also found the velocity of Q relative to P----> V= V[Q] -V[P]
-----.V= (0,40) - V(30,0)
----- V= (-30,40)km/h
4. Originally Posted by dopi
A ship P is travelling due East at 30 km/h and a Ship Q is travelling due South at 40 km/h.
Both ships keep constant speed and course. At t=0 they are each 10 km from the point of intersection of their courses and moving towards the point.
Hello, dopi,
with your problem you use automatically a coordinate system: The pos. x-axis points to East and the pos. y-axis points to North. | {
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At the time t = 0 the ship P is at P(-10, 0) and the ship Q is at Q(0, 10). The movement of both ships is described by a straight line. The speed is described by a vector:
$\overrightarrow{v_P}=(30, 0)$
$\overrightarrow{v_Q}=(0, -40)$
Thus P is moving along the line:
$\overrightarrow{x_P}=(-10, 0)+t*(30, 0)$
and Q is moving along the line:
$\overrightarrow{x_Q}=(0, 10)+t*(0, -40)$
The distance between the ships is: $\vec{d}=\overrightarrow{x_P}-\overrightarrow{x_Q}$
Originally Posted by dopi
Iv found the co-ordinates of Q relative to P at t=0
---->X=(10,10)km
iv also found the velocity of Q relative to P----> V= V[Q] -V[P]
-----.V= (0,40) - V(30,0)
----- V= (-30,40)km/h
I don't understand why you want to calculate the relative coordinates. It isn't necessary with your problem.
Originally Posted by dopi
but im struggling to find the time at which P and Q are closest to each other...is there anyone that can help??? thankz
This was already done by galactus.
Greetings
EB
5. Hello, dopi!
You seem to be using a coordinate system and vectors
. . and I don't know what you plan to do with them.
galactus gave you the solution . . . so why struggle with your approach?
I'll make some diagrams to accompany his excellent solution.
Originally Posted by galactus
At t=0, they are both 10 kms from their intersection.
Code:
* Q
|
|
| 10
|
|
|
P * - - - - - - - - o
10
$t$ hours later, $P$ has moved $30t$ km east to point $A$
. . and $Q$ has moved $40t$ km south to point $B.$
Code:
*
|
| 40t
|
o B
|
| 10-40t
* - - - o - - - - *
30t A 10-30t
And we want to minimize the distance $\overline{AB}.$
Originally Posted by galactus
Think Pythagoras. Let D = square of distance between ships.
$D(t) \:= \10-30t)^{2} + (10-40t)^{2}" alt="D(t) \:= \10-30t)^{2} + (10-40t)^{2}" />
$D'(t) \:= \:2(10-30t)(-30) + 2(10=40t)(-40)$
$D'(t) \:= \:5000t - 1400$
$5000t -1400\:=\:0$
$t \:= \:\frac{7}{25}$ | {
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# Math Help - Tricky Double Integral with polar coordinates
1. ## Tricky Double Integral with polar coordinates
Evaluate the integral:
$\int \int_D \frac{x^2}{x^2 + y^2} dx\;dy$
where $D$ is the region $\{ (x,y)| y \geq x \; \textrm{and} \; x^2 + y^2 \leq 3 \}$
Tried drawing the region $D$ and it looks like I have to use polar coordinates.
Using polar coordinates, $\frac{x^2}{x^2 + y^2} = \frac{r^2 \cos^2{\theta}}{r^2} = \cos^2{\theta}$.
I'm not sure how to find the limits if integration though, or what to do with the $dx\;dy$.
Thanks in advance, any pointers would be greatly appreciated.
2. Originally Posted by craig
Evaluate the integral:
$\int \int_D \frac{x^2}{x^2 + y^2} dx\;dy$
where $D$ is the region $D = \{ (x,y)| y \geq x \; \textrm{and} \; x^2 + y^2 \leq 3 \}$
Tried drawing the region $D$ and it looks like I have to use polar coordinates.
Using polar coordinates, $\frac{x^2}{x^2 + y^2} = \frac{r^2 \cos^2{\theta}}{r^2} = \cos^2{\theta}$.
I'm not sure how to find the limits if integration though, or what to do with the $dx\;dy$.
Thanks in advance, any pointers would be greatly appreciated.
In general to convert, you do the following:
$\displaystyle \int \int f(x,y) \, dx \, dy = \int \int f(r,\theta)
\frac{\partial (x,y)}{\partial (r, \theta)} \, dr \, d\theta$
Where $\frac{\partial (x,y)}{\partial (r, \theta)}$ is called the jacobian determinant and it's given by:
$\frac{\partial (x,y)}{\partial (r, \theta)} =
\begin{vmatrix}
\cos \theta & - r \sin \theta \\
\sin \theta & r \cos \theta
\end{vmatrix} = r$
Hence:
$\displaystyle \int \int f(x,y) \, dx \, dy = \int \int f(r,\theta)
r \, dr \, d\theta$
Now, looking at the region in rectilinear coordinates can give you some hints as to the limits in polar coordinates.
$D = \{ (x,y)| y \geq x \; \textrm{and} \; x^2 + y^2 \leq 3 \}$ | {
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$D = \{ (x,y)| y \geq x \; \textrm{and} \; x^2 + y^2 \leq 3 \}$
Now, the 2nd constraint on the region says that $x^2 + y^2 \leq 3$. You should be able to see that this equation describes a circle whose radius is less than $\sqrt{3}$, in other words, the radius is limited by $0 \leq r \leq \sqrt{3}$.
Now, the 1st constraint on the region says that $y \geq x$. Now, we know that the line at which y= x is a diagonal line going through the origin. And y is greater than x in any point that lies ABOVE that line. In other words, the region in question is made up of points above or on the line y = x. This means that your theta is limited by $\frac{\pi}{4} \leq \theta \leq \frac{5\pi}{4}$.
You can see all of this visually by drawing it out. Go on, do this:
1) Draw the x-y coordinates.
2) Draw a circle centred at the origin with a radius of $\sqrt{3}$.
3) Draw the line y = x.
4) Now shade the region that lies above the line y = x, but does not go outside the circle.
You should see that this region is bounded by:
$D = \bigg\{ (r,\theta)| 0 \leq r \leq \sqrt{3} \; \textrm{and} \; \frac{\pi}{4} \leq \theta \leq \frac{5\pi}{4} \bigg\}$
Hence, in this particular situation:
$\int \int_D \frac{x^2}{x^2 + y^2} dx\;dy = \int_0^{\frac{5\pi}{4}} \int_0^{\sqrt{3}} r \, \cos^2(\theta) \, dr \, d\theta = \int_0^{\frac{5\pi}{4}} \cos^2(\theta) \bigg( \int_0^{\sqrt{3}} r \, \, dr\bigg) \, d\theta$
3. Hi
The area of integration is the part of the disk which is above the red line
And $dx dy$ becomes $r dr d\theta$
4. Originally Posted by craig
Evaluate the integral:
$\int \int_D \frac{x^2}{x^2 + y^2} dx\;dy$
where $D$ is the region $\{ (x,y)| y \geq x \; \textrm{and} \; x^2 + y^2 \leq 3 \}$
Tried drawing the region $D$ and it looks like I have to use polar coordinates.
Using polar coordinates, $\frac{x^2}{x^2 + y^2} = \frac{r^2 \cos^2{\theta}}{r^2} = \cos^2{\theta}$.
I'm not sure how to find the limits if integration though, or what to do with the $dx\;dy$. | {
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I'm not sure how to find the limits if integration though, or what to do with the $dx\;dy$.
Thanks in advance, any pointers would be greatly appreciated.
I go through a very similar problem here: http://www.mathhelpforum.com/math-he...249-post2.html
If you draw the xy domain you can see we want the area to the left of the line but inside the circle,
If we convert this to polar co-ordinates this means
$\frac{ \pi }{4} \le \theta \le \frac{ 5 \pi }{4}$ where the $\frac{ \pi }{4}$ comes from the fact that the line $y=x$ acts at $\frac{ \pi }{4}$ above the x-axis. And $\frac{ 5 \pi }{4}$ comes from the angle between the line in the first quadrent to the line in the 3rd quadrent.
For r, we are going from the origin to the radius of the circle. So the interval of r must be
$0 \le r \le \sqrt{3}$
We can now compute this integral,
$\int \int_D \frac{x^2}{x^2 + y^2} dxdy$
$= \iint_Q \frac{ r^2 cos^2 \theta }{ r^2 } rdr d \theta$
$= \iint_Q rcos^2 \theta dr d \theta$
$= \int_0^{ \sqrt{3} } rdr \int_{ \frac{ \pi }{4} }^{ \frac{ 5 \pi }{4} } cos^2 \theta d \theta$
Which can be easily computed using double angle formulas
5. Wow leave the forum for a few hours and there's 3 great replies waiting for me!!
Originally Posted by Mush
In general to convert, you do the following:
$\displaystyle \int \int f(x,y) \, dx \, dy = \int \int f(r,\theta)
\frac{\partial (x,y)}{\partial (r, \theta)} \, dr \, d\theta$
Where $\frac{\partial (x,y)}{\partial (r, \theta)}$ is called the jacobian determinant and it's given by:
$\frac{\partial (x,y)}{\partial (r, \theta)} =
\begin{vmatrix}
\cos \theta & - r \sin \theta \\
\sin \theta & r \cos \theta
\end{vmatrix} = r$
Hence:
$\displaystyle \int \int f(x,y) \, dx \, dy = \int \int f(r,\theta)
r \, dr \, d\theta$
Ahh yeh I've done Jacobian determinants before, it didn't click that it applied here. | {
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Originally Posted by Mush
Now, looking at the region in rectilinear coordinates can give you some hints as to the limits in polar coordinates.
$D = \{ (x,y)| y \geq x \; \textrm{and} \; x^2 + y^2 \leq 3 \}$
Now, the 2nd constraint on the region says that $x^2 + y^2 \leq 3$. You should be able to see that this equation describes a circle whose radius is less than $\sqrt{3}$, in other words, the radius is limited by $0 \leq r \leq \sqrt{3}$.
Now, the 1st constraint on the region says that $y \geq x$. Now, we know that the line at which y= x is a diagonal line going through the origin. And y is greater than x in any point that lies ABOVE that line. In other words, the region in question is made up of points above or on the line y = x. This means that your theta is limited by $\frac{\pi}{4} \leq \theta \leq \frac{5\pi}{4}$.
You can see all of this visually by drawing it out. Go on, do this:
1) Draw the x-y coordinates.
2) Draw a circle centred at the origin with a radius of $\sqrt{3}$.
3) Draw the line y = x.
4) Now shade the region that lies above the line y = x, but does not go outside the circle.
You should see that this region is bounded by:
$D = \bigg\{ (r,\theta)| 0 \leq r \leq \sqrt{3} \; \textrm{and} \; \frac{\pi}{4} \leq \theta \leq \frac{5\pi}{4} \bigg\}$
I'd drawn the circle but for some reason only drew $y = x$ for positive values of $x$ and $y$, that would be where I was going wrong I guess
Originally Posted by Mush
Hence, in this particular situation:
$\int \int_D \frac{x^2}{x^2 + y^2} dx\;dy = \int_0^{\frac{5\pi}{4}} \int_0^{\sqrt{3}} r \, \cos^2(\theta) \, dr \, d\theta = \int_0^{\frac{5\pi}{4}} \cos^2(\theta) \bigg( \int_0^{\sqrt{3}} r \, \, dr\bigg) \, d\theta$
** Think you mean $\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \cos^2(\theta) \bigg( \int_0^{\sqrt{3}} r \, \, dr\bigg) \, d\theta$
Anyway on with the integral:
$\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \cos^2(\theta) \bigg( \int_0^{\sqrt{3}} r \, \, dr\bigg) \, d\theta$ | {
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$\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \cos^2(\theta) \bigg[ \frac{r^2}{2} \bigg]_0^{\sqrt{3}} \, d\theta$ = $\frac{3}{2}\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \frac{1}{2}(1 + \cos{2\theta}) \, d\theta$ =
$\frac{3}{4}\bigg[\theta + \frac{1}{2}\sin{2\theta}\bigg]_{\frac{\pi}{4}}^{\frac{5\pi}{4}}$ =
$\frac{3}{4}\bigg(\frac{5\pi}{4} + \frac{1}{2}\sin{\frac{5\pi}{2}} -\frac{pi}{4} - \frac{1}{2}\sin{\frac{\pi}{2}} \bigg)$
$\frac{3}{4}(\pi + 0)$ = $\frac{3\pi}{4}$
Thanks a lot for the help!
6. Originally Posted by running-gag
Hi
The area of integration is the part of the disk which is above the red line
And $dx dy$ becomes $r dr d\theta$
Thanks a lot for the image, I'd forgot to draw the whole of the line $y=x$!
Originally Posted by AllanCuz
I go through a very similar problem here: http://www.mathhelpforum.com/math-he...249-post2.html
If you draw the xy domain you can see we want the area under the line but inside the circle,
If we convert this to polar co-ordinates this means
$0 \le \theta \le \frac{ \pi }{4}$ where the $\frac{ \pi }{4}$ comes from the fact that the line $y=x$ acts at $\frac{ \pi }{4}$ above the x-axis.
For r, we are going from the origin to the radius of the circle. So the interval of r must be
$0 \le r \le \sqrt{3}$
We can now compute this integral,
$\int \int_D \frac{x^2}{x^2 + y^2} dxdy$
$= \iint_Q \frac{ r^2 cos^2 \theta }{ r^2 } rdr d \theta$
$= \iint_Q rcos^2 \theta dr d \theta$
$= \int_0^{ \sqrt{3} } rdr \int_0^{ \frac{ \pi }{4} } cos^2 \theta d \theta$
Which can be easily computed using double angle formulas
Thanks a lot for the reply, really helped.
7. Originally Posted by craig
Wow leave the forum for a few hours and there's 3 great replies waiting for me!!
Ahh yeh I've done Jacobian determinants before, it didn't click that it applied here.
I'd drawn the circle but for some reason only drew $y = x$ for positive values of $x$ and $y$, that would be where I was going wrong I guess | {
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** Think you mean $\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \cos^2(\theta) \bigg( \int_0^{\sqrt{3}} r \, \, dr\bigg) \, d\theta$
Anyway on with the integral:
$\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \cos^2(\theta) \bigg( \int_0^{\sqrt{3}} r \, \, dr\bigg) \, d\theta$
$\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \cos^2(\theta) \bigg[ \frac{r^2}{2} \bigg]_0^{\sqrt{3}} \, d\theta$ = $\frac{3}{2}\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \frac{1}{2}(1 + \cos{2\theta}) \, d\theta$ =
$\frac{3}{4}\bigg[1 + \cos{2\theta}\bigg]_{\frac{\pi}{4}}^{\frac{5\pi}{4}}$ = $\frac{3}{4}(\pi)$ = $\frac{3\pi}{4}$
Thanks a lot for the help!
I don't think you're performed that integration properly, have you?
The integral of $1 + \cos(2\theta)$ is $\theta + \frac{1}{2}\sin(2\theta)$.
8. Originally Posted by craig
Thanks a lot for the image, I'd forgot to draw the whole of the line $y=x$!
Thanks a lot for the reply, really helped.
I messed up my bounds for theta. I thought the question said $y = x$ but in fact, we want $y \ge x$ which is the area inside the circle and to the left of the line. My post has been edited to explain the angles. Sorry about that :P
9. Originally Posted by Mush
I don't think you're performed that integration properly, have you?
Yes & no. I did the integration properly, just forgot to update the latex code...oops.
Edited my above working now, sorry about that.
10. Originally Posted by AllanCuz
I messed up my bounds for theta. I thought the question said $y = x$ but in fact, we want $y \ge x$ which is the area inside the circle and to the left of the line. My post has been edited to explain the angles. Sorry about that :P
Haha thought that was what you meant, thanks again for the reply.
11. Originally Posted by craig
Yes & no. I did the integration properly, just forgot to update the latex code...oops.
Edited my above working now, sorry about that.
Indeed. Good show. | {
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# Elementary question about a fake-proof and greatest common divisors
I have a question to an excercise - for which I have a wrong solution - and I wanted to ask you to help me understand my thinking error. The excercise was as follows:
Let $$a, b, n \in \mathbb{N}$$. Show that $$\gcd(a,b) = \gcd(a,b+na)$$.
My solution was (in a nutshell):
• Let $$d := \gcd(a,b)$$.
• Then I showed that $$d$$ is a common divisor of both $$a$$ and $$b + na$$
• Then I showed, that any common divisor $$c$$ of $$a$$ and $$b+na$$ is smaller-or-equal than $$d$$
• After that, I concluded that by definition $$d = \gcd(a,b+na)$$, and with the definition of $$d$$ I would have $$\gcd(a,b) = \gcd(a,b+na)$$, so my proof was completed.
However, my tutor did not accept the proof. He showed me the 'correct' proof, in which I woould have $$d := \gcd(a,b)$$ and $$e := \gcd(a,b+na)$$ and then demonstrate $$d \leq e \leq d$$.
Anyway, I do understand his proof. But I still do not understand where my thinking error is. After all, if my starting line would be, let's say, $$d := 9$$ and I would have been able to show the steps afterwards, then I would be able to conclude $$9 = \gcd(a,b+na)$$, no?
EDIT Wow, you guys answered fast! Thank you very much. I will ask my tutor on thursday again. For completeness sake, I will formulate my complete proof (my original proof is in german, so maybe there will be something lost in translation, the [Reference] are references to our script).
Let $$a,b,n \in \mathbb{N}$$. Let $$d := \gcd(a,b)$$. By definition it means $$d \mid a$$ and $$d \mid b$$, so with [Reference] the following holds true: $$d \mid (b + na)$$. Therefor, $$d$$ is a common divisor of both $$a$$ and $$b + na$$. We will show now, that $$d$$ is the greatest common divisor. | {
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Let $$c$$ be any common divisor of $$a$$ and $$b + na$$. Therefor $$c$$ divides $$a$$, and as such $$c \mid na$$ und therefor $$c \mid ((b + na) - na) = b$$ (again because of [Reference]). So $$c$$ divides $$b$$, and therefor $$c$$ is a common divisor of both $$a$$ and $$b$$. Since $$d$$ was the greatest common divisor of $$a$$ and $$b$$, we have $$c \leq d$$.
By definition of the greatest common divisor we get $$d = \gcd(a,b+na)$$.
End of Proof. My tutor wrote, that I only showed $$\gcd(a,b+na) \leq \gcd(a,b)$$ and also need to show the other direction. Then he showed me the 'correct'/'ideal' proof today. But I will ask him again on thursday!
• You need to be more specific with your proof. From the outline there should be no objection, so either your tutor is wrong or you made an error. We can't determine which. – Matt Samuel Oct 30 '18 at 2:36
• Possibly the confusion arises because you used two different proofs to show that neither gcd exceeds the other, but you can just repeat the same proof replacing $b$ and $n$ with $b+na$ and $-n$. – Neil Oct 30 '18 at 9:24
• This has been the way I have understood things: The following is true $\forall k \in \mathbb N$, for any pair of integers $(n,j)$: $$n\gcd(j,k)-\gcd(j,n)\Bigl\lfloor \frac{n\gcd(j,k)}{\gcd(j,n)}\Bigr\rfloor=0$$ $$j\gcd(j,k)-\gcd(j,n)\Bigl\lfloor \frac{j\gcd(j,k)}{\gcd(j,n)}\Bigr\rfloor=0$$ – Adam L Nov 3 '18 at 17:14
• The above two equalities can be summed only in integer multiples of $n$ and $j$,ie $\forall a,b \in \mathbb Z$ $$(an+bj)\gcd(j,k)-\gcd(j,n)\Bigl\lfloor \frac{(an+bj)\gcd(j,k)}{\gcd(j,n)}\Bigr\rfloor=0$$ – Adam L Nov 3 '18 at 17:14
The outline of the proof that you have explained is correct.
More than likely the problem was with the detail in one of the intermediate steps.
I would check each step carefully again and there is always room for skipping a point in writing a proof. | {
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You want to know where the error is in your proof. Given no details no one can give the answer. All we can do is give only our opinion, so do not treat this as an answer.
The sketch given is correct, but it is merely a restatement of what is required to be proved. Possible that both you and your tutor are correct, but tutor did not understand the proof and misjudged it. But given that you expect people to be able to analyze your "proof" without providing details I am inclined to believe you are wrong. (IT IS AN OPINION!)
• Yes, I also did assume that I was wrong. This is why I provided only the outline sketch since I thought my thinking error would have been with my 'proof-strategy'. I translated now my complete proof and my tutors response as an edit to the original question. – DaGoddyMan Oct 30 '18 at 3:18
Ok well firstly, if your tutor has not mentioned to you something by the name of the Euclidean Algorithm, then he or she isn't being very forthcoming about their level of expertise in the subject it's as simple as that. But follow the link I provided there, and apply the method to both greatest common divisor expressions, and this might be able to help you reach a better level of understanding on the subject, the process is clearly explained in the link.
Also important to note that you shouldn't worry about studying a modern day definition of an algorithm,don't start to think this has anything to do with programming or a need to learn a programming language, this algorithm was invented over 2300 years ago, there was no necessity to update java at that point.
Secondly if your tutor's final step is to demonstrate that $$d \leq e \leq d$$, then it is in fact their solution that is completely bogus and circular, since this statement is equivalent to the original equality that we are trying to prove! | {
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I recommend purchasing a hard cover book in Analytic Number Theory, one that provides the solutions for every exercise it contains, so that if you really are unable to find a proof you are confident with, you can look at the one the book has provided.
I currently use one that I have found to be priceless, "Problems in Analytic Number Theory" Second Edition, written by M.Ram Murty. But do ask your supervisors at your college what they recommend. | {
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• Proving that $d=e$ by proving that $d\leq e\leq d$ is a standard approach and there is nothing bogus in it. – Ister Oct 31 '18 at 7:40
• It's the same question. If you asked me to prove dragons are real and I replied "I shall do so by proving dragons are real, hence proving dragons are real" would you be satisfied? – Adam L Oct 31 '18 at 18:49
• No, it's entirely different thing. You have a single statement to prove ($d=e$) but instead you prove two separate statements usually following different approaches ($d\leq e$ is one proved statement and $e\leq d$ is another one). Only now you apply a not so obvious implication that if $d\leq e\leq d$ holds then $d=e$ so if we've proved the former then the latter is truth indeed. As I said - this method is pretty common, especially when calculating limes but also in some other cases. – Ister Nov 2 '18 at 20:54
• $d=e$ is just an abbreviation for $d \leq e \leq d$. $e \leq d$ means $e$ is less than or equal to $d$. $d \leq e$ means $d$ is less and or equal to $e$. $d=e$ is equivalent to the statement $e \leq d$ and $d \leq e$, since $e \lt d=\lnot(e \gt d)$ – Adam L Nov 3 '18 at 9:09
• I totally disagree that it's just an abbreviation. You define $\leq$ as equal or less (you need to have at least weak order in order to write something like that at all) so equality precedes weak inequality. Of course proving that $d\leq e \leq d \iff d=e$ is trivial but still it's a property of weak inequality. That's why it is worth and can be used as a proof method to separately show each of the weak inequalities and then conclude that it means equality in the end. – Ister Nov 5 '18 at 9:51
Your proof is indeed correct. It can be reorganized more symmetrically as follows.
Notice that if $$\ c\mid a\$$ then $$\ c\mid b \!+\! na \iff c\mid b.\,$$ Therefore $$\,a,\, b\!+\!na\,$$ and $$\,a,\, b\,$$ have exactly the same set $$\,\cal C\,$$ of common divisors $$\,c,\,$$ so they have the same greatest common divisor $$(= \max \,\cal C)$$ | {
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• @Ister Please read more carefully. The answer says nothing at all about "invalidating OP's proof". Nor is there any such mistake. Analyzing (e.g. simplifying) proofs is an important part of teaching. I often point out such simplifcations arising by exploiting innate symmetry, e.g. yesterday using reflection (negation) symmetry to simplify sum computation, and many others. – Bill Dubuque Oct 31 '18 at 16:15
• I must have had bad day, sorry. My comment removed. – Ister Nov 2 '18 at 20:55
In my opinion your tutor is plainly wrong, too much focused on the proof he's accustomed to. It's often a problem. Let me say I faced this issue all the time since I tended to think out of the box and find my own, correct solutions, different to the canonical one.
You proved by definition, which is a fully valid proof. Thus technically you don't have to prove anything else.
Yet, since your tutor apparently don't understand this basics, you have to be smarter than him and be able to counterargument him (i.e. explain why his statement is plainly wrong).
My tutor wrote, that I only showed $$gcd(a,b+na)\leq gcd(a,b)$$ and also need to show the other direction.
In the first part of your proof by showing that $$d$$ is a divisor of both $$a$$ and $$b$$ you've proved that $$gcd(a,b) = d \leq gcd(a,b+na)$$. By definition of $$gcd$$ (name even represents that!) for any given divisor $$z$$ of both $$x$$ and $$y$$ we have $$z\leq gcd(x,y)$$. By substituting $$x=a$$ and $$y=b+na$$ you have the statement that your tutor claims to be missing.
Let me emphasise it once more - you don't need that explanation in your proof, which is perfectly valid. This is only to show that your tutor's claim is wrong.
As an anecdote...
In the course of probabilistic we once had some simple homework task from combinatorics. I don't remember details now but it's actually irrelevant. I had done my calculation, verified them and as a result got a nice looking solution. | {
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On the next lesson I was presenting the solution on the blackboard. Imagine my surprise when I heard my solution is wrong. Someone else had done the task on the blackboard the right way and indeed the results didn't look alike at all. If I remember correctly there were even different powers used in it! Of course, the new solution was correct and in accordance to the key, yet I was sure mine was correct as well (despite all odds and apparent difference) so I stood up for it and asked to point out where had I made an error. No-one, including the tutor could find anything. Eventually the tutor said it might be that the two polynomials are equivalent and if I was able to prove that equivalence I will have the whole course passed with maximum grade.
Needless to say it took me 3 months of work, some horrible calculations and some peculiar theorem from number theory but I did it. The two results are equivalent.
End of Proof. My tutor wrote, that I only showed $$\gcd(a,b+na) \leq gcd(a,b)$$ and also need to show the other direction.
I am pretty sure, that $$\gcd(a,b+na) \geq gcd(a,b)$$ follows from $$d:=\gcd(a,b)$$ and $$d\mid(b+na)$$ (plus the definition of $$\gcd$$).
• Can whoever downvoted this please explain, what my error is. If it was due to my bad formatting, I fixed it. – Nathrat Oct 30 '18 at 11:26
• I didn't downvote but your answer may not be clear. You are trying to give a counterargument to tutor's wrong quoted statement. You should though be more firm in your statement (do the math and make sure you are right) plus add a clear explanation that it means OP is correct in his proof. Note - you are right, but gap in your proof might be to difficult to fill. Try making it more complete (or check my answer as it is actually filling this gap. – Ister Oct 31 '18 at 8:01 | {
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# Creating an integral from 2 functions
Let $$R$$ be the region in $$\mathbb{R}^2$$ below the line $$y = x + 2$$ and above the parabola $$y = x^2$$. Check the integral of these $$2$$ functions in terms of $$dx\cdot dy$$ and then $$dy\cdot dx$$ ' I am having an issue figuring out what the integrals will range from. I have: $$G =\{(y,x) : -1 < x < 2 \text{ and } x^2 < y < (x + 2)\}\to dy.dx$$ $$H =\{(x,y) : 0 < y < 4 \text{ and } y -2 < x < \sqrt{y}\} \to dx.dy$$
However when I create the integrals in terms of $$dx\cdot dy$$ and $$dy\cdot dx$$ they differ? Any help please, did i get the range of the $$G$$ and $$H$$ wrong? Its a parabola cut with a line
For $$dydx$$ the integral over $$1$$ is given by $$\int_{-1}^2\int_{x^2}^{x+2}dydx=\int_{-1}^2-x^2+x+2dx=[\frac13x^3+\frac12x^2+2x]_{-1}^2=\frac92$$ with the region that you correctly evaluated. But for the second integral we need to split the region into two parts - for $$0\le y\le1$$ we have that $$-\sqrt{y}\le x\le\sqrt{y}$$ and when $$1\le y\le4$$ we have $$y-2\le x\le\sqrt{y}$$. So the integral over the function $$1$$ is $$\int_0^1\int_{-\sqrt{y}}^\sqrt{y}dxdy+\int_1^4\int_{y-2}^\sqrt{y}dxdy=\int_0^12\sqrt{y}\,dy+\int_1^4\sqrt{y}-y+2\,dy$$ $$=[\frac43y^{\frac32}]_0^1+[\frac23y^{\frac32}-\frac12y^2+2y]_1^4=\frac92$$ So the two regions are now equal.
• Thank you this helps loads. My text book does not describe splitting them up and i did not think of it. Much appreciated! – Shaun Weinberg Mar 31 at 10:36
For the region $$H$$, the lower limit of $$x$$ shouldn't be $$y-2$$.
It should be the maximum of $$-\sqrt{y}$$ and $$y-2$$. In fact, when $$0 \le y \le 1$$, the lower limit is $$-\sqrt{y}$$.
That is
$$H =\{(x,y): 0 \le y \le 4 , \max(-\sqrt{y}, y-2) < x < \sqrt{y} \}$$
That is from the picture below, the left limit of the region consists of the green color and blue color part. | {
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• I originally wrote an answer for the duplicate, so I copied it here. (+1) btw – Peter Foreman Mar 30 at 19:13
• I did the same thing. ;) – Siong Thye Goh Mar 30 at 19:14 | {
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# Limits problem involving greatest integer function and an unknown function.
Given that $$\lim_{x \to 0} \frac{f(x)}{x^{2}} = 2$$, find $$\lim_{x \to 0} \lfloor f(x) \rfloor$$ and find if $$\lim_{x \to 0} \lfloor \frac{f(x)}{x} \rfloor$$ exists. My math teacher says that since the denominator in the first limit is non negative and the limit itself is positive, he says that $$\lim_{x \to 0} f(x) = 0^{+}$$ and thus $$\lim_{x \to 0} \lfloor f(x) \rfloor = 0$$. I find this acceptable but my friend assumes $$f(x) = 2x^{2} + \infty^{-}x^{3}$$ and claims that $$\lim_{x \to 0} \frac{f(x)}{x^{2}} = 2$$ but that $$\lim_{x \to 0} \lfloor f(x) \rfloor$$ does not exist as $$\lim_{x \to 0^{+}} f(x) = 0^{+}$$ and $$\lim_{x \to 0^{-}} f(x) = 0^{-}$$. So who's right and who's wrong? If either of the two are wrong please explain why?
• If $f(x)=2x^2$, does it satisfy any of either of your claims? – Andrew Chin Nov 8 at 15:49
• Your math teacher is correct. – Paramanand Singh Nov 8 at 15:58
• Thanks for answering Andrew Chew! If $f(x) = 2x^{2}$ then quite obviously, $\lim_{x \to 0} frac{f(x)}{x^{2}} = 2$ and $\lim_{x \to 0} \lfloor f(x) \rfloor = 0$. But the problem is that $f(x)$ can be $2x^{2}$ or it can be any other expression that satisfies $\lim_{x \to 0} \frac{f(x)}{x^{2}} = 2$. We need to find what value $\lim_{x \to 0} \lfloor f(x) \rfloor$ equals irrespective of the function $f(x)$, as long as it satisfies $\lim_{x \to 0} \frac{f(x)}{x^{2}} = 2$. – K Darshan Nov 8 at 17:17
• Paramanand Singh, I appreciate your effort but can you give the reason as to why my friend was wrong? – K Darshan Nov 8 at 17:20
• Oh what I would give if tomorrow I'd wake up and find everyone has stopped plugging $\infty$ into expressions and treating it as though it were a number! – fleablood Nov 9 at 1:34 | {
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The problem is, simply put, that an expression like $$\infty^- \cdot x^3$$ doesn't make sense; you won't find an actual function (meaning a function which only uses numbers and not $$\infty$$) which behaves like that. Even if you take a function like $$f(x) = 2x^2 + (-10000) \cdot x^3$$, as you approach zero, eventually the $$x^3$$ term will not matter anymore: it will be much smaller than the $$2x^2$$ term, if only the $$x$$ you insert is "small enough", so close enough to zero. That means that for $$x$$ small enough, this $$f(x)$$ will still be greater than $$0$$.
So yeah, your teacher is correct. In general, you should always be very skeptical when people use infinity like that. Without proper care, infinity doesn't actually make a whole lot of sense :)
Elaborating on my comment the counter-example given by your friend does not make any sense. You can't write expressions like $$f(x) =2x^2+\infty^{-}x^3$$. The usage of symbol $$\infty$$ is always specified by specific definitions and typical examples are expressions like $$x\to\infty$$ and $$\lim_{x\to 0}1/x^2=\infty$$.
On the other hand the correct argument goes like this. Since $$f(x) /x^2\to 2$$ the expression $$f(x) /x^2$$ is positive as $$x\to 0$$ and hence $$f(x) >0$$ as $$x\to 0$$. Further $$f(x) =x^2\cdot \dfrac {f(x)} {x^2}\to 0\cdot 2=0$$ and hence $$f(x) <1$$ as $$x\to 0$$. It follows that $$\lfloor f(x) \rfloor =0$$ as $$x\to 0$$ and thus $$\lim_{x\to 0}\lfloor f(x) \rfloor =0$$.
The expression $$f(x) /x= x(f(x) /x^2)$$ also tends to $$0$$ but is positive if $$x\to 0^{+}$$ and negative if $$x\to 0^{-}$$ and hence $$\lfloor f(x) /x\rfloor =0$$ if $$x\to 0^{+}$$ and $$\lfloor f(x) /x \rfloor =-1$$ as $$x\to 0^{-}$$ so that the limit $$\lim_{x\to 0}\left\lfloor \dfrac{f(x)} {x} \right\rfloor$$ does not exist.
I'm not entirely sure I understand your teacher's argument (I sure as heck don't get your friend's), but I think it is valid. | {
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$$1 > \epsilon > 0$$ and there is a $$\delta$$ so that $$|x|< \delta$$ implies $$|\frac {f(x)}{x^2} - 2| < \epsilon$$ so $$2-\epsilon \frac {f(x)}{x^2} < 2-\epsilon$$. Thus $$1 < \frac {f(x)}{x^2} < 3$$ and $$f(x) > 0$$ and $$x < \min(\delta,\frac 12)$$ then
$$0 < x^2 < \frac {f(x)} < 3x^2 < \frac 34$$ so $$0 and $$\lfloor f(x)\rfloor = 0$$ for all $$x < \min(\delta, \frac 12)$$.
Which means $$\lim_{x\to 0}\lfloor f(x)\rfloor = 0$$.
Which I think is your teacher's argument.
I can't see why your friend assumes $$f(x) = 2x^2 + \infty - x^3$$, which doesn't even make sense. (What is $$f(5)$$? Is it $$\infty -25$$? What's that?) so I can't tell you why he is wrong other than that what he says makes no sense.
• I'm guessing here but you probably didn't understand what $0^+$, $\infty^-$, etc. actually mean. Generally a number like $a^{+}$ is a number just greater than $a$, something like $a+0.000...1$ and $a^{-}$ is a number just less than $a$, something like $a-0.000...1$. What I meant by $\infty^{-}$ is that it's a number that is very large but not $\infty$. However, on second thoughts, I believe that a number like that doesn't exist. – K Darshan Nov 9 at 13:27
• Infinity is NOT a number. $\infty^{-}$ is oxymoronic and self-contradictory. $f(x) = 2x^2 +\infty^{-}x^3$ is meaningless and illdefined (what number is $f(5) = 50+150\infty^-$?). Your friend is more than wrong. Your friend is spouting uninterpretable gibberish. – fleablood Nov 10 at 16:10 | {
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# Why $\sqrt[3]{3}\not\in \mathbb{Q}(\sqrt[3]{2})$?
We all know that $\sqrt[3]{3}\not\in \mathbb{Q}({\sqrt[3]{2}})$ intuitively. We even know that $\sqrt[3]{p}\not\in \mathbb{Q}(\sqrt[3]{q})$ for two distinct primes $p, q$. However, I don't know how to prove these things rigorously. In case of $\sqrt{p}\not\in \mathbb{Q}(\sqrt{q})$, we can just set assume $\sqrt{p}=a+b\sqrt{q}$ for some $a, b\in\mathbb{Q}$ and deduce a contradiction. However, in cubic case, expansion of $(a+b\sqrt[3]{2}+c\sqrt[3]{4})^{3}$ is very complicated and even I expand this I couldn't get any contradiction by this. To be specific, it gives us 3 diophantine equations
$$a^{3}+2b^{3}+4c^{3}+12abc=3$$ $$a^{2}b+2b^{2}c+2c^{2}a=0$$ $$ab^{2}+2bc^{2}+ca^{2}=0$$
and I don't know how to check solvability of this kind of diophantine equations (over $\mathbb{Q}$). Even if there exists a way to check solvability, I cannot convince that it can be used to prove $\sqrt[3]{p}\not\in \mathbb{Q}(\sqrt[3]{q})$.
So I tried to use Galois theory, but actually I don't know where to go. I assumed $\mathbb{Q}(\sqrt[3]{3})=\mathbb{Q}(\sqrt[3]{2})$ and take a Galois closure, i.e. $K=\mathbb{Q}(\sqrt[3]{3}, w)=\mathbb{Q}(\sqrt[3]{2}, w)$ where $w^{3}=1$ then tried to use field norm maps. We can check that $\beta=\sqrt[3]{3}\cdot\sigma(\sqrt[3]{3})\cdot\sigma^{2}(\sqrt[3]{3})\in \mathbb{Q}$ when $\sigma\in Gal(K/\mathbb{Q}), \sigma(\sqrt[3]{2})=\sqrt[3]{2}, \sigma(w)=w^{2}$. But I cannot get any information from this.
Edit : It would be great if there is a simple algorithm to check whether $\alpha\in K$ or not for a given algebraic number $\alpha$ and a number field $K$. | {
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• You could find the primitive element of $\mathbb{Q}(\sqrt[3]{3},\sqrt[3]{2})$, find its minimal polynomial and show that it is of degree larger than $3$. – Michael Burr Sep 16 '17 at 1:22
• Algebraic number theory might be able to give a solution by considering discriminants of rings of integers - but I'm a bit rusty on that. – Daniel Schepler Sep 16 '17 at 1:27
• @MichaelBurr Thanks! I think we can even solve the general problem with that idea and computers. However, I don't know how to do this by hands. It seems that $\sqrt[3]{2}+\sqrt[3]{3}$ is a degree 9 element, but I cannot convince it without any help of computer programs. – Seewoo Lee Sep 16 '17 at 1:34
• $a$ cannot be divisible by $2$ (first equation). Therefore $b$ is divisible by $2$ (second equation). Therefore, $c$ is divisible by $2^2$ (third equation). Hence $b$ is divisible by $2^3$ (second equation). Keep repeating and you get that $b$ is divisible by $2^k$ for any $k$. For other pair of primes it is the same business. – Hellen Sep 16 '17 at 1:34
• What is $\Bbb{Q}\left( \sqrt[3]{2} \right)$? – gen-z ready to perish Sep 16 '17 at 4:03
Here’s another proof, making light use of $p$-adic theory:
First, note that Eisenstein tells us that both $\Bbb Q(\sqrt[3]3\,)$ and $\Bbb Q(\sqrt[3]2\,)$ are cubic extensions of the rationals, and thus either they are the same field or their intersection is $\Bbb Q$. Therefore, a question equivalent to the stated one is, “Why is $\sqrt[3]2\notin\Bbb Q(\sqrt[3]3\,)$?” I’ll answer the new question. | {
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Note that $\Bbb Q(\sqrt[3]3\,)$ may be embedded into $\Bbb Q_2$, the field of $2$-adic numbers, because $X^3-3\equiv(X+1)(X^2+X+1)\pmod2$, product of two relatively prime polynomials over $\Bbb F_2$, the residue field of $\Bbb Q_2$ (more properly the residue field of $\Bbb Z_2$). So Strong Hensel’s Lemma says that $X^3-3$ has a linear factor in $\Bbb Q_2[X]$, in other words, there’s a cube root of $3$ in $\Bbb Q_2$. But if there were a cube root of $2$ in $\Bbb Q(\sqrt[3]3\,)$, there’d be such a root in $\Bbb Q_2$, and there isn’t since it’d have to have (additive) valuation $v_2(\sqrt[3]2\,)=1/3$. No such thing, the value group of $\Bbb Q_2$ is $\Bbb Z$.
I really like all of the 3 answers, and I think I found one more solution which uses Trace, not Norm. (I don't know why I didn't try to use trace map before.) So I'm going to write it down here.
Suppose $$K=\mathbb{Q}(\sqrt[3]{2})=\mathbb{Q}(\sqrt[3]{3})$$. Consider a trace map $$\mathrm{Tr}_{K/\mathbb{Q}}:K\to \mathbb{Q}, \quad \alpha \mapsto \sum_{i=1}^{n}\sigma_{i}(\alpha)$$ where $$\sigma_{1}(\alpha)=\alpha, \dots, \sigma_{n}(\alpha)$$ is roots of minimal polynomial of $$\alpha$$. (We can consider trace map as a trace of a linear map $$m_{\alpha}:K\to K, x\mapsto \alpha x$$.) This is a $$\mathbb{Q}$$-linear map and we can check that $$\mathrm{Tr}_{K/\mathbb{Q}}(\sqrt[3]{2})=\sqrt[3]{2}+\sqrt[3]{2}w+\sqrt[3]{2}w^{2}=\sqrt[3]{2}(1+w+w^{2})=0$$ where $$w=e^{2\pi i /3}$$ and $$\mathrm{Tr}_{K/\mathbb{Q}}(\sqrt[3]{3})=0$$, too. If we assume $$\sqrt[3]{3}=a+b\sqrt[3]{2}+c\sqrt[3]{4}\,\,\,\,(a, b, c\in \mathbb{Q})$$ as above, by taking trace on both sides we get $$0=3a$$ and $$a=0$$. By multiplying $$\sqrt[3]{2}$$ and $$\sqrt[3]{4}$$ on both sides, we get $$b=c=0$$ and contradiction. I think this methods can be used to prove for higher power cases, such as $$\sqrt[n]{3}\not\in \mathbb{Q}(\sqrt[n]{2})$$ for $$n\geq 2$$. | {
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• Great idea! I used it to prove that roots appear in root extensions only in "obvious" ways. – Orest Bucicovschi Sep 16 '17 at 6:53
• Could you please detail how do you get $b=c=0$? I am not sure how to prove $Tr_{K/\Bbb Q}(\sqrt[3]3\sqrt[3]2)=0$ – PerelMan 2 days ago
Let $\omega := e^{2 \pi i/3}$, and consider the splitting field $\mathbb{Q}(\sqrt[3]{2}, \omega)$ of $x^3 - 2$, and the unique automorphism $\sigma$ such that $\sigma(\sqrt[3]{2}) = \omega \sqrt[3]{2}$, $\sigma(\omega \sqrt[3]{2}) = \omega^2 \sqrt[3]{2}$, $\sigma(\omega^2 \sqrt[3]{2}) = \sqrt[3]{2}$. Suppose that $\sqrt[3]{3} \in \mathbb{Q}[\sqrt[3]{2}]$ is equal to $a + b \sqrt[3]{2} + c \sqrt[3]{4}$ for $a, b, c \in \mathbb{Q}$. Then $\sigma(\sqrt[3]{3})$ is also a root of $x^3 - 3 = 0$, which implies it's equal to either $\sqrt[3]{3}$, $\omega \sqrt[3]{3}$, or $\omega^2 \sqrt[3]{3}$.
However, in the first case, $$\sigma(\sqrt[3]{3}) = \sqrt[3]{3} \Rightarrow a + b \omega \sqrt[3]{2} + c \omega^2 \sqrt[3]{4} = a + b \sqrt[3]{2} + c \sqrt[3]{4}.$$ Rewriting $\omega^2 = - \omega - 1$ and using the linear independence of the basis $\{ 1, \omega, \sqrt[3]{2}, \omega \sqrt[3]{2}, \sqrt[3]{4}, \omega \sqrt[3]{4} \}$ of the splitting field, we get $b = c = 0$, which gives a contradiction since $\sqrt[3]{3}$ is irrational.
Similarly, in the second case, $\sigma(\sqrt[3]{3}) = \omega \sqrt[3]{3}$ would imply $\sqrt[3]{3}$ is a rational multiple of $\sqrt[3]{2}$, which is impossible since $\sqrt[3]{3/2}$ is irrational. And in the third case, $\sigma(\sqrt[3]{3}) = \omega^2 \sqrt[3]{3}$ would imply $\sqrt[3]{3}$ is a rational multiple of $\sqrt[3]{4}$, which is impossible since $\sqrt[3]{4/3}$ is irrational.
HINT:
If $p$ a rational number, not a cube, and $a$, $b$, $c$ rational numbers so that
$$(a + b \sqrt[3]p + c \sqrt[3]{p}^2)^3$$ is rational, then at most one of the numbers $a$, $b$, $c$ is nonzero. | {
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To prove this, it's enough to show that if $(b,c) \ne (0,0)$ then $a=0$. Consider $\omega = -\frac{1}{2} + i \frac{\sqrt{3}}{2}$. We have $\omega^2 + \omega + 1=0$, and $\omega^3 = 1$.
From $$(a + b \sqrt[3]{p} + c \sqrt[3]{p}^2)^3 = q$$ we get $$(a + b \sqrt[3]{p}\omega + c\sqrt[3]{p}^2 \omega^2)^3=q$$ Since $(b,c)\ne (0,0)$ we conclude $$a + b \sqrt[3]{p}\omega + c\sqrt[3]{p}^2 \omega^2 = \sqrt[3]{q} \omega^{\pm 1}$$ and the conjugate equality $$a + b \sqrt[3]{p}\omega^2 + c \sqrt[3]{p}^2 \omega = \sqrt[3]{q} \omega^{\mp 1}$$
To these two equalities we add $$a + b \sqrt[3]p + c \sqrt[3]{p}^2= \sqrt[3]{q}$$ and we get $3 a = 0$, so $a = 0$.
$\bf{Added:}$
Details:
Assume $a + b \sqrt[3]{p}\omega + c \sqrt[3]{p}^2\omega^2 \ne a + b \sqrt[3]{p} + c \sqrt[3]{p}^3$. We get $$b\omega + c \sqrt[3]{p}\omega^2 = b + c \sqrt[3]{p}$$ and with $\omega^2 =-1-\omega$ we get $(b-c\sqrt[3]{p}) \omega = b + 2 c \sqrt[3]{p}$, and so $\omega = \frac{b + 2 c \sqrt[3]{p}}{b-c\sqrt[3]{p}}$, contradiction ( since, say, $\omega$ is not real). So $a + b \sqrt[3]{p}\omega + c \sqrt[3]{p}^2\omega^2$ must be one of the other two roots of the equation $x^3 = q$.
$\bf{Added 2:}$ We follow the beautiful idea of @See-Woo Lee: to use traces.
$\bf{Fact:}$ Let $\alpha$ a real radical, that is $\alpha \in \mathbb{R}$, $\alpha^m\in \mathbb{Q}$ for some $m$. Assume moreover that $\alpha$ is irrational. Then $\operatorname{trace}^L_{\mathbb{Q}}\alpha= 0$ for any $L$ finite algebraic extension of $\mathbb{Q}$ containing $\alpha$. | {
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Proof: We may assume $\alpha >0$. Let $m$ smallest so that $\alpha^m \in \mathbb{Q}$. Clearly $m>1$. Let's prove that the polynomial $X^m - \alpha^m$ is irreducible over $\mathbb{Q}$. It factors over $\mathbb{C}$ as $\prod_{i=0}^m(X- \alpha \omega^i)$. If several of these linear factors produced a polynomial with rational coefficients, we would have the free term $$\prod_{i \in I} (- \alpha \omega^i) \in \mathbb{Q}$$ Taking absolute values we would have $\alpha^k \in \mathbb{Q}$ for some $1\le k < m$, not possible.
Denote by $K = \mathbb{Q}(\alpha)$. Then we have $$\operatorname{trace}^K_{\mathbb{Q}}(\alpha) = \sum_{i=0}^{m-1} \alpha \omega^i = 0$$. Hence for every $L\supset K$ we get $$\operatorname{trace}^L_{\mathbb{Q}} \alpha = [L: K] \operatorname{trace}^K_{\mathbb{Q}}\alpha = 0$$
$\bf{Main\ result:}$ Let $\alpha_i$ real radicals. Assume moreover that the ratios $\frac{\alpha_i}{\alpha_j}$ for $i\ne j$ are irrational. The the $\alpha_i$'s are linearly independent over $\mathbb{Q}$.
Proof: Let $a_i\in \mathbb{Q}$ so that $$\sum a_i \alpha_i = 0$$ Fix $i$, $1\le i \le k$. We have $$a_i = -\sum_{k \ne i} a_k \frac{\alpha_k}{\alpha_i}$$
Let a finite extension $L$ of $\mathbb{Q}$ containing all the $\alpha_i$ Taking traces on both sides we get $$d \cdot a_i = \sum_{k \ne i} a_k \operatorname{trace}^L_{\mathbb{Q}} \left(\frac{\alpha_k}{\alpha_i}\right )=0$$, since $\frac{\alpha_k}{\alpha_i}$ is a real irrational radical. So all the $a_i$ are $0$.
In words: incommensurable real radicals are linearly independent over $\mathbb{Q}$.
• This looks great, but I don't get how to get from $(a + b \sqrt[3]{p} + c \sqrt[3]{p}^2)^3 = q$ to $(a + b \sqrt[3]{p}\omega + c\sqrt[3]{p}^2 \omega^2)^3=q$. Any tips? – Robert Lewis Sep 16 '17 at 3:32
• @Robert Lewis: Thanks. Added some details. – Orest Bucicovschi Sep 16 '17 at 4:06 | {
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Somehow, when manipulating pure radicals, I think that using identification or the trace map, i.e. essentially the additive structure, could lead to more complicated calculations than using the multiplicative structure and the norm map. It seems to be the case here. Introduce the quadratic field $k=\mathbf Q (\omega)$, where $\omega$ is a primitive cubic root of $1$. Since $\mathbf Q(\sqrt[3] 2)$ and $\mathbf Q(\sqrt[3]3)$ have degree $3$ over $\mathbf Q$ by Eisenstein criterion, Kummer theory tells us that the extensions $k(\sqrt[3]2)/k$ and $k(\sqrt[3]3)/k$ are Galois cyclic of degree $3$, and moreover they coincide iff $2=3x^3$, with $x\in (k^{*})^{3}$. Norming down to $\mathbf Q$, we get the diophantine equation $4a^3=9b^3$ with coprime integers $a, b$ : impossible because neither $4$ nor $9$ are cubes. Note that the argument still works with other (coprime) parameters than $2, 3$. | {
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The $${Y_{1}^{0}}^{*}Y_{1}^{0}$$ and $${Y_{1}^{1}}^{*}Y_{1}^{1}$$ functions are plotted above. The more important results from this analysis include (1) the recognition of an $$\hat{L}^2$$ operator and (2) the fact that the Spherical Harmonics act as an eigenbasis for the given vector space. As one can imagine, this is a powerful tool. Forgot password? When this Hermitian operator is applied to a function, the signs of all variables within the function flip. (ℓ+m)!Pℓm(cosθ)eimϕ.Y^m_{\ell} (\theta, \phi) = \sqrt{\frac{2\ell + 1}{4\pi} \frac{(\ell - m)! 4Algebraic theory of spherical harmonics. Plots of the real parts of the first few spherical harmonics, where distance from origin gives the value of the spherical harmonic as a function of the spherical angles, https://brilliant.org/wiki/spherical-harmonics/. It is no coincidence that this article discusses both quantum mechanics and two variables, $$l$$ and $$m$$. }{(\ell + m)!}} Which spherical harmonics are included in the decomposition of f(θ,ϕ)=cosθ−sin2θcos(2ϕ)f(\theta, \phi) = \cos \theta - \sin^2 \theta \cos(2\phi)f(θ,ϕ)=cosθ−sin2θcos(2ϕ) as a sum of spherical harmonics? Formally, these conditions on mmm and ℓ\ellℓ can be derived by demanding that solutions be periodic in θ\thetaθ and ϕ\phiϕ. A conducting sphere of radius RRR with a layer of charge QQQ distributed on its surface has the electric potential everywhere in space: V={14πϵ0QR2r3sinθcosθcosϕ, r>R14πϵ0Qr2R3sinθcosθcosϕ, rR \\ Spherical Harmonics and Linear Representations of Lie Groups 1.1 Introduction, Spherical Harmonics on the Circle In this chapter, we discuss spherical harmonics and take a glimpse at the linear representa-tion of Lie groups. \hspace{15mm} \ell & \hspace{15mm} m&\hspace{15mm} Y^m_{\ell} (\theta, \phi) \\ \hline For each fixed nnn and ℓ\ellℓ there are 2ℓ+12\ell + 12ℓ+1 solutions corresponding to the 2ℓ+12\ell + 12ℓ+1 choices of mmm at fixed ℓ.\ell.ℓ. So the solution can thus far be written in the form. When we consider | {
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of mmm at fixed ℓ.\ell.ℓ. So the solution can thus far be written in the form. When we consider the fact that these functions are also often normalized, we can write the classic relationship between eigenfunctions of a quantum mechanical operator using a piecewise function: the Kronecker delta. Spherical Harmonics are considered the higher-dimensional analogs of these Fourier combinations, and are incredibly useful in applications involving frequency domains. \hspace{15mm} 2&\hspace{15mm} -2&\hspace{15mm} \sqrt{\frac{15}{32\pi}} \sin^2 \theta e^{-2i\phi} \\ The first is determining our $$P_{l}(x)$$ function. \hspace{15mm} 2&\hspace{15mm} 1&\hspace{15mm} -\sqrt{\frac{15}{8\pi}} \sin \theta \cos \theta e^{i \phi} \\ The spherical harmonics. As the general function shows above, for the spherical harmonic where $$l = m = 0$$, the bracketed term turns into a simple constant. As such, this integral will be zero always, no matter what specific $$l$$ and $$k$$ are used. A similar analysis obtains the solution for rR.V(r,\theta, \phi ) = \frac{1}{4\pi \epsilon_0} \frac{QR^2}{r^3} \sin \theta \cos \theta \cos \phi, \quad r>R.V(r,θ,ϕ)=4πϵ01r3QR2sinθcosθcosϕ,r>R. \end{cases}V=⎩⎪⎨⎪⎧4πϵ01r3QR2sinθcosθcosϕ, r>R4πϵ01R3Qr2sinθcosθcosϕ, r R4πϵ01R3Qr2sinθcosθcosϕ, >... \Cos\Theta = x\ ) features a transformation of \ ( \cos\theta\ ), \ ( \hat l. Of any particular state of the sphere as a result, they extremely. From x to \ ( m\ ) blue represents positive values and yellow represents negative values [ ]! The full solution, respectively d } { dx } [ ( x^ { 2 } 1. Charge density on the surface of the geoid up our process into four major parts and on... Reduced to a function, the coefficients AmℓA_m^ { \ell } Bmℓ must be zero corresponding. Orthonormal with respect to integration over the surface of the introduction to spherical harmonics wavefunction the! } ^2∇θ, ϕ2 denotes the Laplacian in many physical equations portion of Laplace 's equation polar! Of space is | {
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denotes the Laplacian in many physical equations portion of Laplace 's equation polar! Of space is spherical harmonic is labeled by the integers ℓ\ellℓ and integers. Make it possible to deduce the reconstruction formula of the geoid is called Legendre 's equation the. 11.5 ) represent angular momentum are defined as generically useful in expanding solutions in physical equations ( e.g ). Can imagine, this formula is only well-defined and nonzero for ℓ≥0\ell \geq 0ℓ≥0 and mmm, function! Values and yellow represents negative values [ 1 ] electron mass, are. Our Cartesian function into the proper coordinate system one of the most applications... By considering the angular momentum eigenfunctions as spherical harmonics 0ℓ≥0 and mmm, the function looks a... Exactly the angular part of the angular portion of Laplace 's work involved the study gravitational! Harmonics routinely arise in physical settings with spherical symmetry where the expansion in harmonics! Values [ 1 ] for an introduction '' by Kendall Atkinson available from Rakuten.... Symmetry of the potential at r=Rr=Rr=R solved for above in terms of spherical harmonics form a complete on. Potential at r=Rr=Rr=R represent mutually orthogonal axes in 3D space not the signs of all variables within function. X^ { 2 } - 1 ) ] \ ) function as a side note, there are solution. ( blue ) is special the moment of inertia of the Laplacian in many physical equations ( e.g previous. Sound signals to obtain sound energy distributions around the spherical microphone array 1525057, and 1413739 0ℓ≥0 and mmm such! ) represent angular momentum of the unit sphere the potential at r=Rr=Rr=R solved for above in terms of harmonics., where are associated Legendre polynomials and and are incredibly useful in applications involving domains... Noted, LibreTexts content is licensed by CC BY-NC-SA 3.0 this Hermitian operator is case. Operator is applied to a very straightforward analysis [ 3 ] E. Berti, Cardoso..., introduction to spherical | {
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is applied to a very straightforward analysis [ 3 ] E. Berti, Cardoso..., introduction to spherical harmonics, and are incredibly useful in expanding solutions in physical settings to. Bala, Department of Physics, IIT Madras 1 ) ] \ ) in polar.. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0 together... Involving frequency domains, 2006 caused by probing a black hole arbitrary dimension well... Linear operator ( follows rules regarding additivity and homogeneity ) before~ without resorting to tensors recurrence relations or functions! Eigenvector ( blue ) is special nearly ninety years later by Lord Kelvin newly determined Legendre function by the ℓ\ellℓ! Where the expansion in spherical harmonics is useful is in the mathematical sciences and researchers who interested... Appears that for every even, angular QM number, respectively, that equation is called Legendre! Other words, the coefficients AmℓA_m^ { \ell } Bmℓ must be.. Most prevalent applications for these functions did not receive their name until nearly ninety years later by Lord.! Cheminformatics as a linear operator ( follows rules regarding additivity and homogeneity ) momentum.! To a very straightforward analysis desmos -, information on Hermitian Operators -,... Acts on the surface of the surface of the geoid dropping a pebble into water and are the and... Tait to write a textbook by probing a black hole, like the dissipative waves caused by probing a hole. A transformation of \ ( l = m = 0\ ) case, it disappears this problem we. Until nearly ninety years later by Lord Kelvin useful is in the description of angular quantum mechanical treatments nature... Approximate any spherical function } \psi = 0 ) \ ) 31,40, 1.... As it introduction to spherical harmonics out, every odd, angular QM number yields odd harmonics.! Depicted on the surface of the function it acts on, a sounding... To zero, for any even-\ ( l\ ) every spherical harmonic even. With ℏ\hbarℏ Planck 's constant, | {
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To zero, for any even-\ ( l\ ) every spherical harmonic even. With ℏ\hbarℏ Planck 's constant, mmm the electron fixed ℓ.\ell.ℓ } { 4\pi ( 1 + |1| ) }. Higher dimensions check out our status page at https: //status.libretexts.org later by Lord Kelvin to tensors space not is! Status page at https: //status.libretexts.org construction is analogous to the moment of of. A complex sounding problem is reduced to a very straightforward analysis by working Laplace. Like Fourier expansions, the three-dimensional Laplacian is typically defined as the eigenfunctions of the prevalent! Be solved via separation of variables is real, we can break up our process four... Can approximate any spherical function 3D space not even, angular QM number yields harmonics. System or making the switch from x to \ ( \cos\theta\ ) be in... It possible to deduce the reconstruction formula of the SH basis functions we... Microphone array quantum mechanical systems it is no coincidence that this introduction to spherical harmonics both! Would be constant-radius looks like a ball a photo-set reminder of why an eigenvector ( blue ) special... Bala, Department of Physics, IIT Madras efficient computer algorithms have much longer polynomial than... { 2 } - 1 ) ] \ ) symmetry where the expansion in spherical coordinates, the coefficients {! * } \psi = 0 ) \ ) often used to approximate the shape the. _Z = -i\hbar \frac { \partial \phi }.L^z=−iℏ∂ϕ∂ at [ email protected ] or check our. Represents negative values [ 1 ] for an introduction to spherical coordinates, these functions are often products trigonometric! Â2Ï ( x, y, z ) = cases ~ave, of course~ been handled without. By Kendall Atkinson available from Rakuten Kobo electron wavefunction in the solution for r > Rr > r therefore. Mechanics I by Prof. S. Lakshmi Bala, Department of Physics, IIT Madras,. -, information on Hermitian Operators - www.pa.msu.edu/~mmoore/Lect4_BasisSet.pdf, Discussions of S.H S. Lakshmi Bala, Department Physics. The two | {
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Discussions of S.H S. Lakshmi Bala, Department Physics. The two recursive relations of Legendre polynomials together us a simple way to determine symmetry... Polynomials together 11.5 ) represent angular momentum of the SH basis functions, we see this is consistent our! Collaboration with Peter Tait to write a textbook terms of spherical harmonics is therefore where the in... Legendre 's equation in all of space is { * } \psi = ). Surface of the represented system is equal to zero, for any even-\ ( l\ ) evaluates... Spherical coordinate system x\ ) \ ] ( \hat { l } ( x, y, )! Refer to [ 31,40, 1 ] a set of functions used process. Work involved the study of gravitational potentials and Kelvin used them in a collaboration with Tait... Function it acts on are also generically useful in expanding solutions in physical settings with spherical symmetry determined! Are interested in ⦠the spherical harmonic is even the 2ℓ+12\ell + 12ℓ+1 choices of mmm at fixed.... Also appear naturally in problems with azimuthal symmetry, which is the case verified. A set of functions used to approximate the shape of the most prevalent applications these... The SH basis functions, we could square it to find our probability-density of molecular shape.! The first is determining our \ ( l = m = introduction to spherical harmonics ) case, disappears... Simple \ ( l\ ) ℓ≥0\ell \geq 0ℓ≥0 and mmm, the three-dimensional is... Spherical due to the quantum mechanics and two variables, \ ( \hat { l } ^2\ operator! } Amℓ and BmℓB_m^ { \ell } Bmℓ must be found newly determined function! Associated Legendre polynomials together to obtain sound energy distributions around the spherical microphone array it would constant-radius! 1 ) ] \ ) function harmonics theory plays a central role in case. And the eigenfunctions of the electron degree 4 on the boundary surface that solutions be periodic in θ\thetaθ ϕ\phiϕ... This problem VVV can be made more precise by considering the angular momentum quantum and. | {
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ϕ\phiϕ... This problem VVV can be made more precise by considering the angular momentum quantum and. In general chemistry classes is consistent with our constant-valued harmonic, for any even-\ ( l\ ) probability... Of your SH expansion the closer your approximation gets as higher frequencies are added in probing a black hole introduction to spherical harmonics... Three dimensions into water equation ∇2f=0\nabla^2 f = 0∇2f=0 can be made more precise by considering the angular momentum of. That solutions be periodic in θ\thetaθ and ϕ\phiϕ S. Lakshmi Bala, Department of Physics, Madras. Determined Legendre function spherical harmonic is even two cases ~ave, of course~ been handled without. ] or check out our status page at https: //en.wikipedia.org/wiki/Spherical_harmonics #:. ) = ] \ ) function with the newly determined Legendre function find our probability-density can use our definition! ) is special of \ ( \cos\theta ) e^ { im\phi } \ ] values. Information contact us at [ email protected ] or check out our page... | {
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# Exercise 11.3.2
Suppose that we hash a string of $r$ characters into $m$ slots by treating it as a radix-128 number and then using the division method. We can easily represent the number $m$ as a 32-bit computer word, but the string of $r$ characters, treated as a radix-128 number, takes many words. How can we apply the division method to compute the hash value of the string without using more than a constant number of words of storage outside the string itself?
Yes, this follows pretty easily from the laws of modulo arithmetic. We need to observe that if the string is $s = \langle a_n, \ldots, a_1, a_0 \rangle$, then its hash $h(s)$ is going to be:
\begin{aligned} h(s) &= \left( \sum_{i=0}^{n}{a_i \cdot {128}^{i}} \right) \bmod m \\ &= \sum_{i=0}^{n}{ \Big( \left( a_i \cdot {128}^{i} \right) \bmod m \Big) } \\ &= \sum_{i=0}^{n}{ \Big( ( a_i \bmod m ) ( {128}^{i} \bmod m ) \Big) } \\ \end{aligned}
We can easily compute $a_i \bmod m$ without extra memory. To compute ${128}^{i} \bmod m$ without extra memory, we just need to observe that $k^i \bmod m = k(k^{i-1} \bmod m) \bmod m$, that is, we can compute the module for each power incrementally, without ever needing unbound memory.
### Python code
K = 128
def consthash(digits, m):
result = 0
power = 1
for d in reversed(digits):
result += ((d % m) * power) % m
result %= m
power = (power * K) % m
return result | {
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# Thread: Differentiate using implicit differentiation
1. ## Differentiate using implicit differentiation
find dy/dx (3xy)^(1/2)=3x-2y
Can't seem to get the solution. May be making a mistake using the chain rule, but I my steps seem correct to me. Would it be easier to first square both sides to get rid of the fractional exponent, then differentiate implicitly? Any help would be appreciated.
2. Originally Posted by math34
find dy/dx (3xy)^(1/2)=3x-2y
Can't seem to get the solution. May be making a mistake using the chain rule, but I my steps seem correct to me. Would it be easier to first square both sides to get rid of the fractional exponent, then differentiate implicitly? Any help would be appreciated.
$\sqrt{3xy} = 3x - 2y$
So
$\frac{1}{2} \frac{1}{\sqrt{3xy}} ( 3y + 3xy') = 3 - 2y'$
$\frac{3y}{2\sqrt{3xy}} + \frac{3x}{2\sqrt{3xy}}y' = 3 - 2y'$
$\frac{3x}{2\sqrt{3xy}}y' + 2y' = 3 - \frac{3y}{2\sqrt{3xy}}$
$\left ( \frac{3x}{2\sqrt{3xy}} + 2 \right )y' = 3 - \frac{3y}{2\sqrt{3xy}}$
$y' = \frac{3 - \frac{3y}{2\sqrt{3xy}}}{\frac{3x}{2\sqrt{3xy}} + 2}$
$y' = \frac{3 - \frac{3y}{2\sqrt{3xy}}}{\frac{3x}{2\sqrt{3xy}} + 2} \cdot \frac{2 \sqrt{3xy}}{2 \sqrt{3xy}}$
$y' = \frac{6\sqrt{3xy} - 3y}{3x + 4\sqrt{3xy}}$ <-- To get your given answer, multiply top and bottom by $\sqrt{3xy}$ again.
We should rationalize this, but it is easier to use the original equation:
$\sqrt{3xy} = 3x - 2y$
$y' = \frac{6(3x - 2y) - 3y}{3x + 4(3x-2y)}$
$y' = \frac{18x - 12y - 3y}{3x + 12x - 8y}$
$y' = \frac{18x - 15y}{15x - 8y}$
-Dan
3. Hello, math34!
Find $\frac{dy}{dx}:\;\;(3xy)^{\frac{1}{2}} \:=\:3x-2y$
Would it be easier to first square both sides to get rid of the fractional exponent,
then differentiate implicitly? . . . absolutely!
Answer: (6xy-y(3xy)^(1/2))/x(3xy)^(1/2)+4xy . . . this can't be right!
Square: . $3xy \:=\:9x^2 - 12xy + 4y^2$
We have: . $3xy' + 3y \:=\:18x - 12xy' - 12y + 8yy'$
Then: . $15xy' - 8yy' \:=\:18x - 15y$ | {
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We have: . $3xy' + 3y \:=\:18x - 12xy' - 12y + 8yy'$
Then: . $15xy' - 8yy' \:=\:18x - 15y$
Factor: . $(15x - 8y)y' \:=\:18x - 15y$
Therefore: . $\boxed{y' \:=\:\frac{18x - 15y}{15x - 8y}}$ . . . which is Dan's answer.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
It can be done head-on . . .
Differentiate implicitly: . $\frac{1}{2}(3xy)^{-\frac{1}{2}}(3y + 3xy') \:=\:3 - 2y'$
We have: . $3y + 3xy' \:=\:2\sqrt{3xy}(3 - 2y')$
We are told that: . $\sqrt{3xy} \:=\:3x - 2y$
. . Hence, we have: . $3y + 3xy' \:=\:2(3x - 2y)(3 - 2y')$
Expand: . $3y + 3xy' \:=\:18x - 12xy' - 12y + 8yy'$
Simplify: . $15xy' - 8yy' \:=\:18x - 15y$
Factor: . $(15x - 8y)y' \:=\:18x - 15y$
Therefore: . $\boxed{y' \:=\:\frac{18x-15y}{15x-8y}}$
4. ## Thanks
I got the same answer as you and Dan but, still am clueless about how the given solution was obtained. I think this is a case of the experts making a mistake, while the layman suffers!!!
Thanks guys
5. I am not happy about the idea of squaring the equation, and here's why:
Say we have:
$y = x$
Square both sides:
$y^2 = x^2$
Now take the derivative:
$2yy' = 2x$
$y' = \frac{x}{y}$
Now, this certainly is a differential equation for the original function, but we also have another solution: y = -x. In this case the differential equation has more solutions than what we started with.
Just something to watch out for.
-Dan | {
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# A sine wave equation?
1
Date created: Sat, Jul 17, 2021 3:51 PM
Date updated: Sun, Jun 26, 2022 6:06 PM
Content
Video answer: How to write a sine wave equation by looking at a graph | trigonometry
## Top best answers to the question «A sine wave equation»
Sine Wave… A general form of a sinusoidal wave is y(x,t)=Asin(kx−ωt+ϕ) y ( x , t ) = A sin ( kx − ω t + ϕ ) , where A is the amplitude of the wave, ω is the wave's angular frequency, k is the wavenumber, and ϕ is the phase of the sine wave given in radians.
FAQ
Those who are looking for an answer to the question «A sine wave equation?» often ask the following questions:
### 👋 De broglie wave equation?
#### De Broglie wave equation
• The de Broglie equation is an equation used to describe the wave properties of matter, specifically, the wave nature of the electron:. λ = h/mv, where λ is wavelength, h is Planck 's constant, m is the mass of a particle, moving at a velocity v.
### 👋 How do you derive a sine wave equation?
1. To find the amplitude, wavelength, period, and frequency of a sinusoidal wave, write down the wave function in the form y(x,t)=Asin(kx−ωt+ϕ).
2. The amplitude can be read straight from the equation and is equal to A.
3. The period of the wave can be derived from the angular frequency (T=2πω).
### 👋 How to derive sine wave equation?
#### How do you calculate sine wave?
• In general, a sine wave is given by the formula In this formula the frequency is w. Frequency used to be measured in cycles per second, but now we use the unit of frequency - the Hertz (abbreviated Hz). One Hertz (1Hz) is equal to one cycle per second.
### 👋 How to read equation of travelling sine wave? | {
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### 👋 How to read equation of travelling sine wave?
• Simply read the wikipedia article. y (t) = A sin (2 π f t + φ) Here A is the amplitude of the wave,i.e. the maximum height of the wave; f the frequency, i.e. is the number of oscillations (cycles) that occur each second of time; φ, the phase, specifies (in radians) where in its cycle the oscillation is at t = 0.
### 👋 How to write an equation for a 4hz sine wave?
#### Which is the formula for the sine wave?
• The formula for the Sine wave is, A = Amplitude of the Wave ω = the angular frequency, specifies how many oscillations occur in a unit time interval, in radians per second φ, the phase, t = ? Here ω, is the angular frequency i.e, It defines how many cycles of the oscillations are there.
### 👋 Is pwm sine wave pure sine wave?
When it comes to output waveform, there are two types of UPS battery backup—the kind that produce a pure sine wave and the kind that produce a simulated or modified sine wave, also known as a pulse-width modulated (PWM) sine wave, when on battery power.
### 👋 Is the wave equation a partial differential equation?
• The wave equation is a partial differential equation. We discuss some of the tactics for solving such equations on the site Differential Equations. One of the most popular techniques, however, is this: choose a likely function, test to see if it is a solution and, if necessary, modify it. So, let's use what we already know.
### 👋 Is the wave equation sine or cosine?
#### What is the difference between sine wave and cos wave?
• Key Difference: Sine and cosine waves are signal waveforms which are identical to each other. The main difference between the two is that cosine wave leads the sine wave by an amount of 90 degrees . A sine wave depicts a reoccurring change or motion.
### 👋 What is a sine wave equation? | {
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### 👋 What is a sine wave equation?
Sine Wave… A general form of a sinusoidal wave is y(x,t)=Asin(kx−ωt+ϕ) y ( x , t ) = A sin ( kx − ω t + ϕ ) , where A is the amplitude of the wave, ω is the wave's angular frequency, k is the wavenumber, and ϕ is the phase of the sine wave given in radians.
Video answer: Determining the equation for a sinewave from a plot
We've handpicked 6 related questions for you, similar to «A sine wave equation?» so you can surely find the answer! | {
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What is the equation for a 2 d damped sine wave?
• This is the equation for a 2 D damped sine wave, what is the 3 D equivalent? ( x 2 + y 2). expresses damping along any ray a x + b y = 0 extending from the origin, and this was achieved by substitution of x 2 + y 2 in place of t in the Wikipedia listing at https://en.m.wikipedia.org/wiki/Damped_sine_wave.
What is the equation for nonlinear wave equation?
• Nonlinear wave equation of general form: \$$u_{tt}=\\left[f\\left(u\\right)u_{x}\\right]_{x}\$$ This equation can be linearized in the general case. Some exact solutions are given in [Pol-04, pp252-255] and, by way of an example consider the following special case where \$$f\\left(u\\right)=\\alpha e^{\\lambda u}\\ :\$$
What is the equation for the wave equation?
• Let us consider a plane wave with real amplitude E0and propagating in direc- tion of the zaxis. This plane wave is represented by E(r,t) = E0cos[kz− ωt], where k = |k| = ω/c. If we observe this field at a fixed position z then we’ll measure an electric field E(t) that is oscillating with frequency f = ω/2π.
What is the equation of sine wave?
• In our math class, we are discussing about the trigonometric waves, specifically, the Sine Wave. The Sine wave is the graph that is formed if the function contains a sine function. The General Formula of the Sine wave is: y=AsinB(x-C)+D where x is the angle or theta.
What is the mathematical equation for a sine wave?
• On The Mathematics of the Sine Wave y(x) = A*(2πft + ø) Why the understanding the sine wave is important for computer musicians. The sine wave is mathematically a very simple curve and a very simple graph, and thus is computationally easy to generate using any form of computing, from the era of punch cards to the current era of microprocessors.
### Video answer: Sine wave equation explained - interactive
Who discovered wave equation? | {
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### Video answer: Sine wave equation explained - interactive
Who discovered wave equation?
Using Newton's recently formulated laws of motion, Brook Taylor (1685–1721) discovered the wave equation by means of physical insight alone [1]. | {
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# Show that a connected graph on $n$ vertices is a tree if and only if it has $n-1$ edges.
Show that a connected graph on $n$ vertices is a tree if and only if it has $n-1$ edges.
$(\Rightarrow)$ If a tree $G$ has only $1$ vertex, it has $0$ edges. Now, assume that any tree with $k-1$ vertices has $k-2$ edges. Let $T$ be a tree with $k$ vertices. Remove a leaf $l$ to obtain a tree $T'$ with $k-1$ vertices. Then, $T'$ has $k-2$ edges, by the inductive hypothesis. The addition of $l$ to $T'$ produces a graph with $k-1$ edges, since $l$ has degree $1$.
$(\Leftarrow)$ Let $G$ be a connected graph on $n$ vertices, with $n-1$ edges. Suppose $G$ is not a tree. Then, there exists at least one cycle in $G$. Successively remove edges from cycles in $G$ to obtain a graph $G'$ with no cycles. Then, $G'$ is connected and has no cycles. Therefore, $G'$ is a tree, with $n$ vertices and $n-1-x$ edges, for some $x > 0$. However, this contradicts the previous derivation. Therefore, it must be the case that $G$ is a tree.
• it seems to be true. – mesel Jun 15 '14 at 12:40
The proofs are correct. Here's alternative proof that a connected graph with n vertices and n-1 edges must be a tree modified from yours but without having to rely on the first derivation:
Let $G$ be a connected graph on n vertices, with n−1 edges. Suppose $G$ is not a tree. Then, there exists at least one cycle in $G$. Remove one of the edges within a cycle. This leaves a connected graph on n vertices with n-2 edges which is impossible as a connected graph on n vertices must at least have n - 1 edges.
Yes, this seems like it is true, although you didn't prove a leaf must always exist, if you really want to be rigorous. And in the other part you didn't explain why if it contained a cycle you could remove an edge without the graph being disconnected, but I like the proof over all.
I have the following way of proving it. Let me know if you agree with it or not. | {
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I have the following way of proving it. Let me know if you agree with it or not.
Suppose the graph on $n$ vertices with $n-1$ edges is not a tree. This implies it has at least $1$ polygon. Suppose in total there are $k$ edges involved in these polygons and we know that a polygon has as many edges as vertices, since polygons are of regular degree 2. Then, if we consider th egraph without the edges that form the polygons, then we have a connected graph with at least $n-k$ edges, since in a connected graph every vertex has degree at least 1. Thus, in total this would imply we have at least $n$ in all the graph. But this contradicts the fact that we have $n-1$ in the original graph. Thus it must be a tree.
The converse is not always true. Consider a disjoint graph with n vertices and n - 1 edges. Here we have n-1 edges but this is obviously not a tree. The assumption that G is connected is only for the forward implication. | {
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