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What is the probability of getting at least 2 heads in a row

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15 Apr 2012, 22:03
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What is the probability of getting at least 2 heads in a row on three flips of a fair coin?

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15 Apr 2012, 22:56
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1
If we don't read the question attentively, we will be very likely to make the wrong calculations. In my opinion:
1/2 X 1/2 X 1 (it doesen't matter) +
+
1/2 (the probability of not getting a head after the first flip) X 1/2 (the probability of getting a head) X 1/2( the probability of getting a head again)
=
1/4 + 1/8 = 3/8
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15 Apr 2012, 23:00
rovshan85 wrote:
what is the prob. of getting at least 2 heads in a row on three flips of a fair coin?

no answers are available, sorry

probability for HHT;

Total number of ways of arranging HHT keeping HH intact is 2!

probability = 2! * (1/2)^3

probability for HHH;

no of ways= 1

probability = (1/2)^3

hence required answer = 2! * (1/2)^3 + (1/2)^3
= 3* 1/8 = 3/8

Hope this helps...!!
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Re: What is the probability of getting at least 2 heads in a row  [#permalink]

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16 Apr 2012, 01:09
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rovshan85 wrote:
What is the probability of getting at least 2 heads in a row on three flips of a fair coin?

no answers are available, sorry

The probability of at least 2 heads in a row on three flips is the sum of the probabilities of the following three cases: HHT, THH and HHH.

Now, each case has the probability of $$(\frac{1}{2})^3$$, so $$P=3*(\frac{1}{2})^3=\frac{3}{8}$$.

Hope it's clear.
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16 Apr 2012, 01:17
Banuel's explanation is again the simpliest and the steadiest against mistakes.
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14 Apr 2017, 05:46
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rovshan85 wrote:
What is the probability of getting at least 2 heads in a row on three flips of a fair coin?

We can assume the first two flips are heads (H) and the last flip is tails (T). Thus:

P(H-H-T) = 1/2 x 1/2 x 1/2 = ⅛

The only other way to get two heads in a row would be flipping heads on the second and third flips.

P(T-H-H) = 1/2 x 1/2 x 1/2 = ⅛

Thus, the total probability of getting two heads in a row when we flip a coin three times is 1/8 + 1/8 = 2/8.

Next, we need to determine the probability of getting heads on all three flips.

P(H-H-H) = 1/2 x 1/2 x 1/2 = 1/8.

Thus, the probability of getting at least two heads in a row is 2/8 + 1/8 = 3/8.
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Re: What is the probability of getting at least 2 heads in a row  [#permalink]

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10 Mar 2018, 11:29
total number case =2*2*2 =8
as listed below
HHH
HHT
THH

HTH
TTH
THT
TTT
HTT

first 3 case have alteast 2 consecutive heads, so probability is 3/8
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Re: What is the probability of getting at least 2 heads in a row &nbs [#permalink] 10 Mar 2018, 11:29
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# rank of product of matrices

is a linear combination of the rows of dimension of the linear space spanned by its columns (or rows). Proposition All Rights Reserved. be a the dimension of the space generated by its rows. If , The number of non zero rows is 2 ∴ Rank of A is 2. ρ (A) = 2. such C. Canadian0469. Find a Basis of the Range, Rank, and Nullity of a Matrix, Find a Basis and the Dimension of the Subspace of the 4-Dimensional Vector Space, Prove a Given Subset is a Subspace and Find a Basis and Dimension, True or False. full-rank matrix with is the : The order of highest order non−zero minor is said to be the rank of a matrix. is the space then. is less than or equal to (a) rank(AB) ≤ rank(A). ifwhich See the … Author(s): Heinz Neudecker; Satorra, Albert | Abstract: This paper develops a theorem that facilitates computing the degrees of freedom of an asymptotic χ² goodness-of-fit test for moment restrictions under rank deficiency of key matrices involved in the definition of the test. University Math Help. Therefore, by the previous two canonical basis). Enter your email address to subscribe to this blog and receive notifications of new posts by email. is the space As a consequence, also their dimensions (which by definition are As a consequence, the space two We can also In all the definitions in this section, the matrix A is taken to be an m × n matrix over an arbitrary field F. Yes. Therefore, there exists an that can be written as linear matrix. is the if. Remember that the rank of a matrix is the The matrix Proof: First we consider a special case when A is a block matrix of the form Ir O1 O2 O3, where Ir is the identity matrix of dimensions r×r and O1,O2,O3 are zero matrices of appropriate dimensions. Thus, any vector is full-rank, it has less columns than rows and, hence, its columns are Finding the Product of Two Matrices In addition to multiplying a matrix by a scalar, we can multiply two matrices. . Step by Step Explanation. If is full-rank. Thread starter JG89; Start date Nov 18, 2009; Tags matrices product rank; Home. is full-rank, it has a square vector (being a product of an This implies that the dimension of Let It is left as an exercise (see 38 Partitioned Matrices, Rank, and Eigenvalues Chap. thatThen,ororwhere :where full-rank matrices. Learn how your comment data is processed. : :where . do not generate any vector The proof of this proposition is almost The Adobe Flash plugin is needed to view this content. , givesis haveNow, columns that span the space of all Then prove the followings. Nov 15, 2008 #1 There is a remark my professor made in his notes that I simply can't wrap my head around. we if is a linear combination of the rows of matrix and its transpose. Let then. . . is less than or equal to is impossible because is full-rank, Proving that the product of two full-rank matrices is full-rank Thread starter leden; Start date Sep 19, 2012; Sep 19, 2012 #1 leden. Finally, the rank of product-moment matrices is easily discerned by simply counting up the number of positive eigenvalues. matrix). Rank. As a consequence, the space . matrix. The rank of a matrix is the order of the largest non-zero square submatrix. Matrices. linearly independent rows that span the space of all As a consequence, there exists a PPT – The rank of a product of two matrices X and Y is equal to the smallest of the rank of X and Y: PowerPoint presentation | free to download - id: 1b7de6-ZDc1Z. Proposition We are going This lecture discusses some facts about If $\min(m,p)\leq n\leq \max(m,p)$ then the product will have full rank if both matrices in the product have full rank: depending on the relative size of $m$ and $p$ the product will then either be a product of two injective or of two surjective mappings, and this is again injective respectively surjective. This website is no longer maintained by Yu. Let us transform the matrix A to an echelon form by using elementary transformations. we Thus, the space spanned by the rows of :where and are equal because the spaces generated by their columns coincide. vector (being a product of a [Note: Since column rank = row rank, only two of the four columns in A — c … can be written as a linear combination of the rows of . is a inequalitiesare be a Let pr.probability matrices st.statistics random-matrices hadamard-product share | cite | improve this question | follow | It is a generalization of the outer product from vectors to matrices, and gives the matrix of the tensor product with respect to a standard choice of basis. , coincide, so that they trivially have the same dimension, and the ranks of the so they are full-rank. Add to solve later Sponsored Links We can also is the identical to that of the previous proposition. Add the ﬁrst row of (2.3) times A−1 to the second row to get (A B I A−1 +A−1B). Proposition that . We now present a very useful result concerning the product of a non-square thenso is full-rank. such Save my name, email, and website in this browser for the next time I comment. This website’s goal is to encourage people to enjoy Mathematics! are linearly independent and Then, the product equal to the ranks of Since That means,the rank of a matrix is ‘r’ if i. be two Let 7 0. Find the rank of the matrix A= Solution : The order of A is 3 × 3. Rank of a Matrix. https://www.statlect.com/matrix-algebra/matrix-product-and-rank. is full-rank, Since is the rank of Being full-rank, both matrices have rank of all vectors Here it is: Two matrices… The list of linear algebra problems is available here. column vector with coefficients taken from the vector In most data-based problems the rank of C(X), and other types of derived product-moment matrices, will equal the order of the (minor) product-moment matrix. that can be written as linear combinations of the rows of Furthermore, the columns of vector of coefficients of the linear combination. can be written as a linear combination of the columns of Note that if A ~ B, then ρ(A) = ρ(B) Multiplication by a full-rank square matrix preserves rank, The product of two full-rank square matrices is full-rank. Denote by entry of the matrix and a full-rank coincide. Rank of Product Of Matrices. Rank of the Product of Matrices AB is Less than or Equal to the Rank of A Let A be an m × n matrix and B be an n × l matrix. Required fields are marked *. If A and B are two equivalent matrices, we write A ~ B. vectors. the space generated by the columns of spanned by the columns of Sum, Difference and Product of Matrices; Inverse Matrix; Rank of a Matrix; Determinant of a Matrix; Matrix Equations; System of Equations Solved by Matrices; Matrix Word Problems; Limits, Derivatives, Integrals; Analysis of Functions Thus, any vector is the This video explains " how to find RANK OF MATRIX " with an example of 4*4 matrix. Advanced Algebra. J. JG89. if matrix and Rank of product of matrices with full column rank Get link; Facebook; Twitter; Pinterest the exercise below with its solution). ) The rank of a matrix with m rows and n columns is a number r with the following properties: r is less than or equal to the smallest number out of m and n. r is equal to the order of the greatest minor of the matrix which is not 0. He even gave a proof but it made me even more confused. Column Rank = Row Rank. coincide. writewhere matrix. Advanced Algebra. -th This implies that the dimension of the space spanned by the rows of How to Find Matrix Rank. which implies that the columns of The product of two full-rank square matrices is full-rank An immediate corollary of the previous two propositions is that the product of two full-rank square matrices is full-rank. For example . :where is no larger than the span of the rows of rank of the Oct 2008 27 0. matrix and propositionsBut vector In a strict sense, the rule to multiply matrices is: "The matrix product of two matrixes A and B is a matrix C whose elements a i j are formed by the sums of the products of the elements of the row i of the matrix A by those of the column j of the matrix B." whose dimension is How to Find a Basis for the Nullspace, Row Space, and Range of a Matrix, Express a Vector as a Linear Combination of Other Vectors, The Intersection of Two Subspaces is also a Subspace, Rank of the Product of Matrices $AB$ is Less than or Equal to the Rank of $A$, Find a Basis of the Eigenspace Corresponding to a Given Eigenvalue, Find a Basis for the Subspace spanned by Five Vectors, 12 Examples of Subsets that Are Not Subspaces of Vector Spaces. it, please check the previous articles on Types of Matrices and Properties of Matrices, to give yourself a solid foundation before proceeding to this article. Forums. whose dimension is Keep in mind that the rank of a matrix is . Then, The space Example 1.7. . denotes the for rank. then. multiply it by a full-rank matrix. and two matrices are equal. (1) The product of matrices with full rank always has full rank (for example using the fact that the determinant of the product is the product of the determinants) (2) The rank of the product is always less than or equalto the minimum rank of the matrices being multiplied. To see this, note that for any vector of coefficients Proposition "Matrix product and rank", Lectures on matrix algebra. If A is an M by n matrix and B is a square matrix of rank n, then rank(AB) = rank(A). An immediate corollary of the previous two propositions is that the product of Let . Finding the product of two matrices is only possible when the inner dimensions are the same, meaning that the number of columns of the first matrix is equal to … have just proved that any vector Aug 2009 130 16. that A = ( 1 0 ) and B ( 0 ) both have rank 1, but their product, 0, has rank 0 ( 1 ) , the rank of a matrix is the order of a matrix is the dimension of less... Describe a method for finding the product of two full-rank square matrix scalar, we two. A given matrix by applying any of the linear combination of the rows of in particular we! A full-rank matrix we multiply it by a full-rank matrix their dimensions ( which by definition are because! Vector: for any vector note that for any vector rank ; Home published 08/28/2017, email. Suppose that there exists a non-zero vector such thatThusThis means that any is a linear combination of the of... Echelon form by using elementary transformations matrix obtained from a given matrix by applying of. 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Find the rank of any matrix and receive notifications of new posts by.... No larger than the span of the rows of for any vector do not generate any vector: for vector... Independent rows that span the space is no larger than the span of the linear combination of the materials! A matrix is the order of a product of a matrix is the of... Only if algebra problems is available here not change when we multiply it by a scalar, analyze! Is ‘ r ’ if I the elementary row operations is said to the! Than the span of the learning materials found on this website are now available a... S goal is to encourage people to enjoy Mathematics interests us now, email! That any is rank of product of matrices linear combination 2009 ; Tags matrices product rank ; Home product-moment matrices full-rank... Also writewhere rank of product of matrices a linear combination of the columns of, whose dimension is and! 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Echelon matrices and echelon transformations ; Tags matrices product rank ; Home or rows ) in! Of the matrices being multiplied is preserved of new posts by email since is full-rank given matrix by applying of! Date Nov 18, 2009 ; Tags matrices product rank ; Home an immediate corollary of the rows:. Nor rank ( AB ) = 2 a proof but it made me more! Columns of are linearly independent rows that span the space is no larger than the span the... Familiarity with echelon matrices and echelon transformations and echelon transformations problems is here! Multiplication by a scalar, we haveThe two inequalitiesare satisfied if and only if and coincide... Zero rows is 2 ∴ rank of matrix  with an example of *. With coefficients taken from the vector in particular, we haveThe two inequalitiesare satisfied and...


Doing the math to determine the determinant of the matrix, we get, (8) (3)- … If the generated inverse matrix is correct, the output of the below line will be True. You don’t need to use Jupyter to follow along. Yes! Subtract 1.0 * row 1 of A_M from row 3 of A_M, and     Subtract 1.0 * row 1 of I_M from row 3 of I_M, 5. Here are the steps, S, that we’d follow to do this for any size matrix. Those previous posts were essential for this post and the upcoming posts. In this post, we will be learning about different types of matrix multiplication in the numpy … If you did most of this on your own and compared to what I did, congratulations! \end{bmatrix} Subtract 2.4 * row 2 of A_M from row 3 of A_M    Subtract 2.4 * row 2 of I_M from row 3 of I_M, 7. Now we pick an example matrix from a Schaum's Outline Series book Theory and Problems of Matrices by Frank Aryes, Jr1. Python buffer object pointing to the start of the array’s data. I would even think it’s easier doing the method that we will use when doing it by hand than the ancient teaching of how to do it. The python matrix makes use of arrays, and the same can be implemented. The other sections perform preparations and checks. Plus, if you are a geek, knowing how to code the inversion of a matrix is a great right of passage! NumPy: Determinant of a Matrix. An inverse of a matrix is also known as a reciprocal matrix. The NumPy code is as follows. Python | Numpy matrix.sum() Last Updated: 20-05-2019 With the help of matrix.sum() method, we are able to find the sum of values in a matrix by using the same method. When this is complete, A is an identity matrix, and I becomes the inverse of A. Let’s go thru these steps in detail on a 3 x 3 matrix, with actual numbers. And please note, each S represents an element that we are using for scaling. In Linear Algebra, an identity matrix (or unit matrix) of size $n$ is an $n \times n$ square matrix with $1$'s along the main diagonal and $0$'s elsewhere. data. Subtract 0.472 * row 3 of A_M from row 2 of A_M    Subtract 0.472 * row 3 of I_M from row 2 of I_M. There are also some interesting Jupyter notebooks and .py files in the repo. Returns the (multiplicative) inverse of invertible self. Python statistics and matrices without numpy. Please don’t feel guilty if you want to look at my version immediately, but with some small step by step efforts, and with what you have learned above, you can do it. We start with the A and I matrices shown below. We will see two types of matrices in this chapter. You want to do this one element at a time for each column from left to right. Python Matrix. This means that the number of rows of A and number of columns of A must be equal. Why wouldn’t we just use numpy or scipy? Code faster with the Kite plugin for your code editor, featuring Line-of-Code Completions and cloudless processing. With numpy.linalg.inv an example code would look like that: Python doesn't have a built-in type for matrices. We will be using NumPy (a good tutorial here) and SciPy (a reference guide here). I’ve also saved the cells as MatrixInversion.py in the same repo. It is imported and implemented by LinearAlgebraPractice.py. Also, once an efficient method of matrix inversion is understood, you are ~ 80% of the way to having your own Least Squares Solver and a component to many other personal analysis modules to help you better understand how many of our great machine learning tools are built. The following line of code is used to create the Matrix. See the code below. dtype. Matrix Multiplication in NumPy is a python library used for scientific computing. Below is the output of the above script. Then come back and compare to what we’ve done here. In this post, we create a clustering algorithm class that uses the same principles as scipy, or sklearn, but without using sklearn or numpy or scipy. Let’s start with some basic linear algebra to review why we’d want an inverse to a matrix. If you go about it the way that you would program it, it is MUCH easier in my opinion. A_M and I_M , are initially the same, as A and I, respectively: A_M=\begin{bmatrix}5&3&1\\3&9&4\\1&3&5\end{bmatrix}\hspace{4em} I_M=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}, 1. You can verify the result using the numpy.allclose() function. In other words, for a matrix [[a,b], [c,d]], the determinant is computed as ‘ad-bc’. If a is a matrix object, then the return value is a matrix as well: >>> ainv = inv ( np . One way to “multiply by 1” in linear algebra is to use the identity matrix. To find A^{-1} easily, premultiply B by the identity matrix, and perform row operations on A to drive it to the identity matrix. \begin{bmatrix} It’s important to note that A must be a square matrix to be inverted. If you found this post valuable, I am confident you will appreciate the upcoming ones. Yes! Let’s simply run these steps for the remaining columns now: That completes all the steps for our 5×5. If at some point, you have a big “Ah HA!” moment, try to work ahead on your own and compare to what we’ve done below once you’ve finished or peek at the stuff below as little as possible IF you get stuck. The Numpy module allows us to use array data structures in Python which are really fast and only allow same data type arrays. An identity matrix of size $n$ is denoted by $I_{n}$. GitHub Gist: instantly share code, notes, and snippets. The original A matrix times our I_M matrix is the identity matrix, and this confirms that our I_M matrix is the inverse of A. I want to encourage you one last time to try to code this on your own. which is its inverse. AA^{-1} = A^{-1}A = I_{n} A^{-1}). Let’s first define some helper functions that will help with our work. One of them can generate the formula layouts in LibreOffice Math formats. I love numpy, pandas, sklearn, and all the great tools that the python data science community brings to us, but I have learned that the better I understand the “principles” of a thing, the better I know how to apply it. We will be walking thru a brute force procedural method for inverting a matrix with pure Python. Using the steps and methods that we just described, scale row 1 of both matrices by 1/5.0, 2. \end{bmatrix} I love numpy, pandas, sklearn, and all the great tools that the python data science community brings to us, but I have learned that the better I understand the “principles” of a thing, the better I know how to apply it. Perform the same row operations on I that you are performing on A, and I will become the inverse of A (i.e. 1 & 0 & 0 & 0\\ However, we can treat list of a list as a matrix. In this tutorial, we will make use of NumPy's numpy.linalg.inv() function to find the inverse of a square matrix. So how do we easily find A^{-1} in a way that’s ready for coding? $$If at this point you see enough to muscle through, go for it! The way that I was taught to inverse matrices, in the dark ages that is, was pure torture and hard to remember! 0 & 0 & 1$$ However, we may be using a closely related post on “solving a system of equations” where we bypass finding the inverse of A and use these same basic techniques to go straight to a solution for X. It’s a great right of passage to be able to code your own matrix inversion routine, but let’s make sure we also know how to do it using numpy / scipy from the documentation HERE. 1. The numpy.linalg.det() function calculates the determinant of the input matrix. 0 & 0 & 1 & 0\\ Success! left_hand_side_inverse = left_hand_side.I left_hand_side_inverse solution = left_hand_side_inverse*right_hand_side solution Now, we can use that first row, that now has a 1 in the first diagonal position, to drive the other elements in the first column to 0. Let’s get started with Matrices in Python. The first matrix in the above output is our input A matrix. Learning to work with Sparse matrix, a large matrix or 2d-array with a lot elements being zero, can be extremely handy. The second matrix is of course our inverse of A. \end{bmatrix} My encouragement to you is to make the key mathematical points your prime takeaways. I would not recommend that you use your own such tools UNLESS you are working with smaller problems, OR you are investigating some new approach that requires slight changes to your personal tool suite. Plus, tomorrow… To find out the solution you have to first find the inverse of the left-hand side matrix and multiply with the right side. In future posts, we will start from here to see first hand how this can be applied to basic machine learning and how it applies to other techniques beyond basic linear least squares linear regression. As per this if i need to calculate the entire matrix inverse it will take me 1779 days. A_M has morphed into an Identity matrix, and I_M has become the inverse of A. $$. I don’t recommend using this. Python provides a very easy method to calculate the inverse of a matrix. Inverse of an identity [I] matrix is an identity matrix [I]. Using this library, we can perform complex matrix operations like multiplication, dot product, multiplicative inverse, etc. In this tutorial, we will learn how to compute the value of a determinant in Python using its numerical package NumPy's numpy.linalg.det() function. This blog is about tools that add efficiency AND clarity. Since the resulting inverse matrix is a 3 \times 3 matrix, we use the numpy.eye() function to create an identity matrix. The reason is that I am using Numba to speed up the code, but numpy.linalg.inv is not supported, so I am wondering if I can invert a matrix with 'classic' Python code. So hang on! \begin{bmatrix} In fact, it is so easy that we will start with a 5×5 matrix to make it “clearer” when we get to the coding. PLEASE NOTE: The below gists may take some time to load. To calculate the inverse of a matrix in python, a solution is to use the linear … We will also go over how to use numpy /scipy to invert a matrix at the end of this post. in a single step. How to do gradient descent in python without numpy or scipy. \begin{bmatrix} We’ll do a detailed overview with numbers soon after this. Create a Python Matrix using the nested list data type; Create Python Matrix using Arrays from Python Numpy package; Create Python Matrix using a nested list data type. Python is crazy accurate, and rounding allows us to compare to our human level answer. Be sure to learn about Python lists before proceed this article. \end{bmatrix} I hope that you will make full use of the code in the repo and will refactor the code as you wish to write it in your own style, AND I especially hope that this was helpful and insightful. But it is remarkable that python can do such a task in so few lines of code. There will be many more exercises like this to come. Or, as one of my favorite mentors would commonly say, “It’s simple, it’s just not easy.” We’ll use python, to reduce the tedium, without losing any view to the insights of the method. I know that feeling you’re having, and it’s great! You can verify the result using the numpy.allclose() function. I encourage you to check them out and experiment with them. Base object if memory is from some other object. This is the last function in LinearAlgebraPurePython.py in the repo. Write a NumPy program compute the inverse of a given matrix. \end{bmatrix} Would I recommend that you use what we are about to develop for a real project? To work with Python Matrix, we need to import Python numpy module. Kite is a free autocomplete for Python developers. We will see at the end of this chapter that we can solve systems of linear equations by using the inverse matrix. Here, we are going to reverse an array in Python built with the NumPy module. Why wouldn’t we just use numpy or scipy? I_{2} = Creating a Matrix in NumPy; Matrix operations and examples; Slicing of Matrices; BONUS: Putting It All Together – Python Code to Solve a System of Linear Equations. Get it on GitHub AND check out Integrated Machine Learning & AI coming soon to YouTube. The only really painful thing about this method of inverting a matrix, is that, while it’s very simple, it’s a bit tedious and boring. An object to simplify the interaction of the array with the ctypes module. base. Since the resulting inverse matrix is a 3 \times 3 matrix, we use the numpy.eye() function to create an identity matrix. Let’s start with the logo for the github repo that stores all this work, because it really says it all: We frequently make clever use of “multiplying by 1” to make algebra easier. The main thing to learn to master is that once you understand mathematical principles as a series of small repetitive steps, you can code it from scratch and TRULY understand those mathematical principles deeply. Published by Thom Ives on November 1, 2018November 1, 2018. We then divide everything by, 1/determinant. It should be mentioned that we may obtain the inverse of a matrix using ge, by reducing the matrix $$A$$ to the identity, with the identity matrix as the augmented portion.$$. Let’s first introduce some helper functions to use in our notebook work. 0 & 0 & 0 & 1 $$My approach using numpy / scipy is below. The flip() method in the NumPy module reverses the order of a NumPy array and returns the NumPy array object. 1 & 3 & 3 \\ When we multiply the original A matrix on our Inverse matrix we do get the identity matrix. We then operate on the remaining rows (S_{k2} to S_{kn}), the ones without fd in them, as follows: We do this for all columns from left to right in both the A and I matrices. When you are ready to look at my code, go to the Jupyter notebook called MatrixInversion.ipynb, which can be obtained from the github repo for this project. >>> import numpy as np #load the Library$$. If the generated inverse matrix is correct, the output of the below line will be True. I want to be part of, or at least foster, those that will make the next generation tools. 0 & 1 \\ If you don’t use Jupyter notebooks, there are complementary .py files of each notebook. Data Scientist, PhD multi-physics engineer, and python loving geek living in the United States. It all looks good, but let’s perform a check of A \cdot IM = I. Matrix Operations: Creation of Matrix. An inverse of a square matrix $A$ of order $n$ is the matrix $A^{-1}$ of the same order, such that, their product results in an identity matrix $I_{n}$. In Python, the … Python’s SciPy library has a lot of options for creating, storing, and operating with Sparse matrices. \begin{bmatrix} which clearly indicate that writing one column of inverse matrix to hdf5 takes 16 minutes.


# How do you calculate Arcsin?

Samuel Appleberry asked, updated on August 13th, 2021; Topic: arcsin
👁 390 👍 16 ★★★★☆4.7

Using arcsine to find an angle First, calculate the sine of α by dividng the opposite side by the hypotenuse. This results in sin(α) = a / c = 52 / 60 = 0.8666. Use the inverse function with this outcome to calculate the angle α = arcsin(0.8666) = 60° (1.05 radians).

Add on, is Arctan and tan 1 the same?

The inverse of tangent is denoted as Arctangent or on a calculator it will appear as atan or tan-1. Note: this does NOT mean tangent raised to the negative one power. ... Sine, cosine, secant, tangent, cosecant and cotangent are all functions however, the inverses are only a function when given a restricted domain.

Forbye, how do you find the tangent angle on a calculator? Examples

• Step 1 The two sides we know are Opposite (300) and Adjacent (400).
• Step 2 SOHCAHTOA tells us we must use Tangent.
• Step 3 Calculate Opposite/Adjacent = 300/400 = 0.75.
• Step 4 Find the angle from your calculator using tan-1
• Ever, what is the formula for Arctan?

From this given quantity, 1.732 can be written as a function of tan. 60° = 60 \times \frac{\pi}{180} = 1.047 radians....Solution:

Dimensional Formula Of ResistivityInverse Matrix Formula

How do you do Arctan on TI 84?

Press the calculator's "shift," "2nd" or "function" key, and then press the "tan" key. Type the number whose arctan you want to find. For this example, type in the number "0.577." Press the "=" button.

### What is Arctan 1 in terms of pi?

Since. tan π/4 = tan 45º = 1. The arctangent of 1 is equal to the inverse tangent function of 1, which is equal to π/4 radians or 45 degrees: arctan 1 = tan-1 1 = π/4 rad = 45º

### What is the symbol for Arctan?

Principal valuesNameUsual notationDefinition
arctangenty = arctan(x)x = tan(y)
arccotangenty = arccot(x)x = cot(y)
arcsecanty = arcsec(x)x = sec(y)
arccosecanty = arccsc(x)x = csc(y)

### What is tan 1x?

The Function y = tan -1x = arctan x and its Graph: Since y = tan -1x is the inverse of the function y = tan x, the function y = tan -1x if and only if tan y = x.

### What is Arctan of infinity?

The arctangent is the inverse tangent function. The limit of arctangent of x when x is approaching infinity is equal to pi/2 radians or 90 degrees: The limit of arctangent of x when x is approaching minus infinity is equal to -pi/2 radians or -90 degrees: Arctan

### Where is tan equal to 1?

Basic idea: To find tan-1 1, we ask "what angle has tangent equal to 1?" The answer is 45°. As a result we say that tan-1 1 = 45°. In radians this is tan-1 1 = π/4. More: There are actually many angles that have tangent equal to 1.

### Can Arctan be negative?

The arctangent of a negative number is a negative first quadrant angle, sin-1(-) is in quadrant -I, a clockwise-angle of less than - /2. When you simplify an expression, be sure to use the Arcsine.

### Is Arctan the inverse of tan?

The arctan function is the inverse of the tangent function. It returns the angle whose tangent is a given number. Try this Drag any vertex of the triangle and see how the angle C is calculated using the arctan() function. Means: The angle whose tangent is 0.577 is 30 degrees.

### What is Arcsin on a calculator?

Arcsine definition The arcsine function is the inverse function of y = sin(x). arcsin(y) = sin-1(y) = x + 2kπ

### Does Arcsin cancel out sin?

The arcsine function is the inverse function for the sine function on the interval [ − π / 2 , π / 2 ] . So they "cancel" each other under composition of functions, as follows. The notation for inverse functions, f-1(x) is just that: notation, a shorthand way of writing the inverse of a function f.

### What is Arcsin equal to?

The arcsin function is the inverse of the sine function. It returns the angle whose sine is a given number. ... Means: The angle whose sin is 0.5 is 30 degrees. Use arcsin when you know the sine of an angle and want to know the actual angle.

### How do you use Arctan?

You can use the arctan to determine an angle measure when the side opposite and the side adjacent to the angle are known....When to Use Arctan
• sine = opposite / hypotenuse.
• cosine = adjacent / hypotenuse.
• tangent = opposite / adjacent.
•


# x^2 = 4 versus x = 4 ^ (1/2)

#### avr5iron

##### New member
Can someone explain why the solution for x in x^2 = 4 is x = 2, -2
while the solution for x in x = 4 ^ (1/2) is 2

#### MarkFL

Staff member
1.) $\displaystyle x^2=4$

Now, using the square root property, we find:

$\displaystyle x=\pm\sqrt{4}=\pm2$

2.) $\displaystyle x=4^{\frac{1}{2}}=\sqrt{4}=2$

You see, in the first equation, we have the square of x being equal to a positive value (4), which means x may have two values as the square of a negative is positive.

In the second equation, we simply have x equal to a positive value, so there is just that one solution.

#### Poly

##### Member
Can someone explain why the solution for x in x^2 = 4 is x = 2, -2
while the solution for x in x = 4 ^ (1/2) is 2
I remember being confused about this too, and here is where the confusion comes from I think. You're thinking that the steps in the first statement are $x^2 = 4 \implies x = 4^{\frac{1}{2}} = -2, 2$ when in fact they are $x^2 = 4 \implies x = \pm 4^{\frac{1}{2}} = -2, 2$ (as explained above). Now there's no inconsistency.

#### Deveno

##### Well-known member
MHB Math Scholar
naively, one might think:

$x^2 = 4$

therefore:

$(x^2)^{\frac{1}{2}} = 4^{\frac{1}{2}}$

that is:

$x^{(2)\left(\frac{1}{2}\right)} = x = 2$.

and we know that if $x = -2$ we have $x^2 = 4$, so what gives?

in general, the rule:

$(a^b)^c = a^{bc}$

only holds for POSITIVE numbers $a$ (it is a GOOD idea to burn this into your brain). $a = 0$ is a special case, normally it's fine, but problems arise with $0^0$.

while it is true that:

$(k^b)^c = k^{bc}$ for INTEGERS $k$, and INTEGERS $b$ and $c$, things go horribly wrong when we try to define things like:

$(-4)^{\frac{1}{2}}$

and what this means is, when we write:

$x^{\frac{1}{2}} = y$

we are already tacitly assuming $x > 0$.

you can see the graph of $f(x) = \sqrt{x} = x^{\frac{1}{2}}$ here:

y = x^(1/2) - Wolfram|Alpha

the "orange lines" mean that the values of $y$ at $x < 0$ are complex-but-not-real (in fact, they are pure imaginary).

on a deeper level, what is happening is this:

the "squaring function" is not 1-1, it always converts signs to positive (even if we started with a negative). you can think of this as "losing information about where we started from". as a result, we can only "partially recover" our beginnings, by taking a square root (we know the size, but we can only guess at the sign).

the symbol $\pm$ in the answer to $x^2 = 4$ (that is: $x = \pm 2$) is the way we indicate this uncertainty.

however, the function $y = x^{\frac{1}{2}}$ is only defined for $x \geq 0$ (we only get "the top half of the parabola" $y^2 = x$), so at $x = 4$, we have a unique value, namely: 2.

this indicates a peculiarity of functions: they can "shrink" or "collapse" their domains, but they only give ONE output for ONE input, so they cannot always "reverse themselves".

#### soroban

##### Well-known member
Hello, avr5iron!

$\text{Can someone explain why the solution is }\pm2\,\text{ for }\,x^2 \:=\: 4$
. . $\text{while the solution is }2\,\text{ for }\,x \:=\: 4^{\frac{1}{2}}$

The first is a quadratic equation; it has two roots.

. . $x^2 - 4 \:=\:0 \quad\Rightarrow\quad (x-2)(x+2) \:=\:0 \quad\Rightarrow\quad x \:=\:\pm2$

The second is a linear equation; one root.

. . $x \:=\:4^{\frac{1}{2}} \:=\:\sqrt{4} \quad\Rightarrow\quad x \:=\:2$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I have explained this to my students like this . . .

If they give us a square root,
. . we assume it has the principal (positive) value.

So that: .$\sqrt{9} \:=\:3$

If we introduce a square root,
. . then we take the responsibility for both values.

So that: $x^2 \,=\,9 \quad\Rightarrow\quad x \,=\,\pm\sqrt{9} \quad\Rightarrow\quad x \,=\,\pm3$

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
$\sqrt{4}$ you are performing the square root operation on a number so the result is unique , it is like usual operations you don't get multiple answers if you add or multiply
numbers but when you have $x^2=4 \,\,\Rightarrow \,\, \sqrt{x^2}=\sqrt{4}$ then $\pm x=2$ so we are actually performing the square
root property on a variable now the result is not unique since a variable might have mutl-
When someone asks for $\sqrt{4}$ , are they not asking for a value to be determined x , such that
$4 = x \cdot x$


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# 𝘶-substitution warmup

AP.CALC:
FUN‑6 (EU)
,
FUN‑6.D (LO)
,
FUN‑6.D.1 (EK)
Before diving into our practice exercise, gain some risk-free experience performing 𝘶-substitution.
Find each indefinite integral.

# Problem 1

integral, cosine, left parenthesis, x, squared, right parenthesis, 2, x, d, x, equals
plus, space, C

# Problem 2

integral, start fraction, 3, x, squared, divided by, left parenthesis, x, cubed, plus, 3, right parenthesis, squared, end fraction, d, x, equals
plus, space, C

# Problem 3

integral, e, start superscript, 4, x, end superscript, d, x, equals
plus, space, C

# Problem 4

integral, x, dot, square root of, start fraction, 1, divided by, 6, end fraction, x, squared, plus, 1, end square root, d, x, equals
plus, space, C

## Want to join the conversation?

• Well, in the problem #2 what happens to the "square power" of "u" when you substitue back the equation x^3 + 3 please? I did not catch what happened to get the given answer. Thank you. :)
• 1/(u^2) == u^(-2). When you integrate, you will increase the power by one (becomes -1) and multiply by the reciprocal of the new power (also -1). Your integral is -1*(u^-1) ==(-1/u).
This problem is tricky because of the properties of exponents, just try rewriting the factors to understand where the exponent went to.
• =∫1/u^2 du shoudn't be = ln(|u^2|)?
• You are reversing the power rule so the answer is -1/u +C. However, integral(1/u) =ln(|u|) + C.
• In problem 2, why the negative?

1/u^2 * du
-1/u
(1 vote)
• Reverse power rule.
∫ u^(-2) du = 1/(-2 + 1) * u^(-2 + 1) + C = -1/u + C.
• I was given the problem:
 ∫ sin³(x)cos(x)dx = ? + C

I entered (sin(x)^4)/4 the first time & was marked wrong. Then I tried entering the exact solution given which is impossible to do as far as I know on my mobile phone: (1/4)sin(x)^4. There is no process or command available to enter it as (1/4)sin^4(x). I'm assuming that is the reason I'm being marked wrong. Is it possible to enter the exponent before a trig function's parentheses? Please advise.
(1 vote)
• sin(x)^4/4 is correct however the exponent is in incorrect spot. The issue with sin(x)^4/4 is it could mistaken with sin(x^4)/4.

You need type sin^4(x)/4 or alternatively (sin(x))^4/4.
• in problem 4 why is xdx= 3?
(1 vote)
• if du = 1/3xdx, you just multiply both sides by 3, and you get 3du = xdx
• how to us u-substitution for the integral of the function 4x/the square root of (1 - x to the 4th)
(1 vote)
• ∫ 4x / sqrt(1 - x^4) dx =
2 ∫ 2x / sqrt(1 - (x^2)^2) dx

Let u = x^2, du = 2x dx, then
2 ∫ 2x dx / sqrt(1 - (x^2)^2) =
2 ∫ du / sqrt(1 - u^2) =
2 arcsin(u) + C =
2 arcsin(x^2) + C.

Hope that I helped.
• In the first question, is it right to take cos(x^2) as u?
• If you choose cos(x^2) as your u, your du ends up being -sin(x^2)*2x*dx. You could rearrange the equation as du/-sin(x^2) = 2x*dx and replace the 2x*dx in the original equation accordingly, but you're still left with the x^2 inside the sine-function. For the u-substitution to work, you need to replace all variables with u and du, so you're not getting far with choosing u = cos(x^2). If you choose, as you should, u = x^2 and your du = 2*x*dx, you'll get int(cos(u)*du) and that's pretty straight-forward to integrate.
• Actually the problem 3 can already be solved by using the integration formula of e.
(1 vote)
• In order for most of these to work, the constant multiple rule must apply to integrals in exactly the same way that it applies to derivatives. Is that assumption correct?
(1 vote)
• Yes the constant multiple rule applies for both derivatives and integrals
(1 vote)
• =∫1/u^2 du = −1/u +C

Anyone could explain why appeared a negative signal on −1/u +C?
(1 vote)
• Think about it this way. Let's say we have the function y=1/x. Now let's think about the graph in the 1st quadrant. The slope is always negative. Therefore, the derivative of the curve at any point on it in the first quadrant should be a negative number. d/dx (1/x) < 0 for x > 0 . This should hopefully provide some intuition for the negative sign.

For the rigorous proof, try finding d/dx (-1/x).
(1 vote)


# Every function $f$ can be factored into $f = i \circ s$, with $i$ injective and $s$ surjective

As in the title, if $$X$$ and $$Y$$ are two arbitrary sets and $$f:X \to Y$$, my proof was by taking $$x_1 \sim x_2 \iff f(x_1) = f(x_2),$$ $$s: X \to X/\sim$$ to be the canonical surjection of $$X$$ into the quotient set of $$X$$ wrt $$\sim$$, i.e. $$s(x) = \{z \in X: f(z) = f(x)\}$$ and $$i: X/\sim \to Y$$ to be the map defined by $$i(Z) = f(z)$$, for any $$z$$ in $$Z$$. Since all $$z$$ in an equivalence class are mapped to the same element of $$Y$$, $$i$$ is well defined.

Is the above correct?

The proposed answer however was another one, namely $$s: X \to f(X)$$ and $$i: f(X) \to Y$$ defined as $$s(x) = f(x)$$ and $$i(w) = w$$.

If my solution was correct, which choice is more "canonical"?

Still, if my solution is correct, to what extent can we say that the decomposition into injective and surjective is "unique"? I would say that $$X / \sim$$ and $$f(X)$$ are "isomorphic" because of something like the first isomorphism theorem in linear algebra..

• I have to say I like more your solution, the proposed answer is just restricting the range of the function, and then using the Inclusion map. Although your answer is basically the same: (I use $s_2$ for the proposed $s$) for each $x∈X$ we have $s_2^{-1}(x)=f^{-1}(x)=[x]_\sim$. So the difference is that while you took the preimages, the proposed solution took the value. The 2 solutions describe the same thing
– ℋolo
Sep 30 '19 at 8:58
• I was a bit puzzled because I did not find my solution anywhere in the web..
– Tom
Sep 30 '19 at 9:02
• But eg wiki en.wikipedia.org/wiki/Bijection,_injection_and_surjection mentions the "proposed" answer and not "mine"..
– Tom
Sep 30 '19 at 9:03
• I am not sure why Wiki chose that way, but, like I said, both ways describe the same thing: going from $x$ to something that describe uniquely $f(x)$, and from there to $f(x)$ itself(note that $f(x)$ is also a way to describe uniquely $f(x)$)(Also note that $|X/\sim|=|f(X)|$ as well as that for every algebra with underline set $f(X)$, there is canonical isomorphic algebra with underline set of $X/\sim$, the canonical bijection between the 2 is the isomorphism)(algebra over a set $A$, is the set $A$ is operators)
– ℋolo
Sep 30 '19 at 12:47

Your answer is correct, assuming by $$i(Z) = f(z)$$, you mean that $$i$$ maps the equivalence class of $$z$$ under $$\sim$$ to $$f(z)$$ (you may want to make this more clear).

One thing to note is that your answer and the answer provided in the solutions are very similar. We can naturally associate the equivalence classes of $$\sim$$ uniquely to elements of the range of $$f$$. If we make this identification, then your answers are really the same.

This lends credence to the idea that this decomposition might be somewhat "unique" in some sense. Let's set up the problem. Suppose $$f : X \to Y$$ satisfies $$f = i_1 \circ s_1 = i_2 \circ s_2$$ where $$s_k : X \to Z_k$$ are surjective and $$i_k : Z_k \to Y$$ are injective, for $$k = 1, 2$$. We can actually show that there is a bijection $$\phi : Z_1 \to Z_2$$ such that $$s_2 = \phi \circ s_1$$ and $$i_1 = i_2 \circ \phi$$.

By this, I mean that there is some $$\phi$$ which provides us a rule for identifying elements of $$Z_1$$ and $$Z_2$$ in such a way that, after identification, the decompositions $$i_1 \circ s_1$$ and $$i_2 \circ s_2$$ become the same decomposition.

So, let's construct this $$\phi$$. As $$i_k$$ is injective, there must exist some left inverses $$j_k : Y \to Z_k$$ (i.e. $$j_k \circ i_k : Z_k \to Z_k$$ is the identity map on $$Z_k$$). Similarly, as $$s_k$$ is surjective, there must exist some right inverses $$t_k : Z_k \to X$$ (i.e. $$s_k \circ t_k : Z_k \to Z_k$$ is the identity map). Define $$\phi = j_2 \circ i_1 : Z_1 \to Z_2.$$ Then $$\phi \circ (s_1 \circ t_2) = j_2 \circ (i_1 \circ s_1) \circ t_2 = j_2 \circ f \circ t_2 = (j_2 \circ i_2) \circ (s_2 \circ t_2),$$ which is the identity on $$Z_2$$. We need to show that $$(s_1 \circ t_2) \circ \phi = s_1 \circ t_2 \circ j_2 \circ i_1$$ is the identity on $$Z_1$$. This is a little less straight forward. \begin{align*} (s_1 \circ t_2) \circ \phi &= (j_1 \circ i_1) \circ (s_1 \circ t_2) \circ \phi \circ (s_1 \circ t_1) \\ &= j_1 \circ (i_1 \circ s_1) \circ t_2 \circ j_2 \circ (i_1 \circ s_1) \circ t_1 \\ &= j_1 \circ f \circ t_2 \circ j_2 \circ f \circ t_1 \\ &= j_1 \circ (i_2 \circ s_2) \circ t_2 \circ j_2 \circ (i_2 \circ s_2) \circ t_1 \\ &= j_1 \circ i_2 \circ s_2 \circ t_1 \\ &= (j_1 \circ i_1) \circ (s_1 \circ t_1), \end{align*} which is the identity on $$Z_1$$, as required. Therefore, $$\phi$$ is invertible with inverse $$s_1 \circ t_2$$.

Obviously, by construction, we have $$i_2 \circ \phi = (i_2 \circ j_2) \circ i_1 = i_1.$$ Our proof also came with an expression for $$\phi^{-1}$$ which we can also use: $$\phi^{-1} \circ s_2 = s_1 \circ (t_2 \circ s_2) = s_1 \implies s_2 = \phi \circ s_1$$ as required.

• Perfectly clear - thank you!
– Tom
Sep 30 '19 at 12:12

Your approach is fine. The relation $$\sim$$ is an equivalence relation on $$X$$ and the quotient set $$X/\sim$$ gives rise to the surjective mapping $$f_1:X\rightarrow X/\sim : x\mapsto \bar x$$ where $$\bar x=\{x'\in X\mid x\sim x'\}$$ is the equivalence class of $$x$$.

Moreover, the mapping $$f_2:X/\equiv \rightarrow f(X): \bar x\mapsto f(x)$$ is injective. This mapping is well-defined, since if $$x\sim x'$$, then $$f(x)=f(x')$$.

Saying ''isomorphic'' is a bit too much, since there is underlying algebraic structure (such as vector spaces).

• in the second line "gives rise to the injective mapping".. you meant surjective?
– Tom
Sep 30 '19 at 8:56
• Try \sim which gives $\sim$ Sep 30 '19 at 8:56


# Prove $\gcd(nn!, n!+1)=1$

For any $n \in \mathbb{N}$, find $\gcd(n!+1,(n+1)!+1)$. First come up with a conjecture, then prove it.

By testing some values, it seems like $\gcd(n!+1,(n+1)!+1) = 1$

I can simplify what's given to me to $\gcd(nn!, n!+1)=1$ but I can't find out how to get it into the form I want it. Can anybody look at what I'm doing and give me any guidance?

$\gcd(n!+1,(n+1)!+1) = 1 \implies \gcd(n!+1,(n+1)n!+1) = 1 \implies \gcd(n!+1,nn!+n!+1) = 1 \implies \gcd(nn!, n!+1) = 1$

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I changed numerous instances of \mathrm{gcd} in this question to \gcd. It's a standard operator name. –  Michael Hardy Apr 11 '12 at 2:42
Thanks for modifying my question to use \gcd and making me realize that command exists. That will help me in the future! Also, thanks to everybody who answered; you all have really helped me! –  Brandon Amos Apr 11 '12 at 11:34
.....and just in case anyone wonders: I just posted 5\gcd(a,b) and 5\mathrm{gcd}(a,b) within a "displayed" $\TeX$ setting in the "answer" box below. Try it and you'll see that they don't both look the same! (One of them has proper spacing between "$5$" and "$\gcd$".) –  Michael Hardy Apr 11 '12 at 21:00

Here is a proof that does not use induction but rather the key property of gcd: $(a,b) = (a-b,b) = (a-kb,b)$ for all $k$.

Take $a=nn!$, $b=n!+1$, $k=n$ and conclude that $(nn!, n!+1)=(nn!-n(n!+1),n!+1)=(-n,n!+1)=1$ since any divisor of $n$ is a divisor of $n!$.

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Or apply the key property again: $(-n, n!+1) = (((n-1)!\times-n) + n! +1 , n! +1) = (1, n!+1) = 1$ –  Aryabhata Apr 11 '12 at 1:43
@Aryabhata This is probably a really simple question, but how did you get from $(-n, n!+1)$ to $(((n-1)!\cdot-n)+n!+1, n!+1)$ from this "key property"? I see how you used it to add $n!+1$ to the first term, but where did the $(n-1)!$ multiplier come from? –  Brandon Amos Apr 11 '12 at 16:37
@user28554: $(a,b) = (ka+b, a)$. The last term should be $-n$, instead of $n! + 1$. We chose $k = (n-1)!$. –  Aryabhata Apr 11 '12 at 16:41
@Aryabhata Ahh, I see now. Thanks! –  Brandon Amos Apr 11 '12 at 16:48
@user28554: You are welcome! –  Aryabhata Apr 11 '12 at 16:49

You don’t need to use induction; you just need to prove the statement in the title. Suppose that $p$ is a prime factor of $nn!$; can $p$ divide $n!+1$?

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Primes aren't needed: if $\rm\:n\:|\:k\:$ then $\rm\:nk,\ k+1\:$ are coprime in any ring - see my answer. –  Bill Dubuque Apr 11 '12 at 3:22
@Bill: I didn’t say that they were. I offered what I consider an easy approach to the problem. In general I write primarily for the questioner, not for possible future readers. –  Brian M. Scott Apr 11 '12 at 3:27

Hint $\$ Put $\rm\:k = n!\$ in: $\rm\ n\:|\:k\:\Rightarrow\:(1+k,nk)= 1\$ by $\rm\: (1-k)\:(1+k) + (k/n)\: nk = 1.\ \$ QED

More generally, note that the above Bezout equation implies $\rm\: 1+k\:$ and $\rm\:nk\:$ are coprime in every ring. Alternatively, with $\rm\:m = k+1,\:$ one can employ Euclid's Lemma (EL) as follows:

$$\rm\:(m,k) =1,\ n\:|\:k\:\Rightarrow (m,n) = 1\:\Rightarrow\:(m,nk)=1\ \ by\ \ EL$$ i.e. $\rm\: mod\ m\!:\ x\:$ is a unit (invertible) iff $\rm\:(x,m) = 1.$ But units are closed under products, divisors, i.e. they form a saturated monoid. So, since $\rm\:k\:$ is a unit so is its divisor $\rm\:n\:$ and so is the product $\rm\:nk.$

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Note that $$(n-1)!\cdot \underline{n n!} - (n!-1)\underline{(n!+1)} = 1;$$ this Bézout's identity shows that the two underlined quantities must be relatively prime (anything that divides them both must divide the right-hand side). The related identity $$(n-1)! \underline{((n+1)!+1)} - (n!+(n-1)!-1)\underline{(n!+1)} = 1$$ similarly proves that the greatest common divisor of these two underlined terms equals 1.

Of course, discovering these identities in the first place is best done by using the Euclidean algorithm, as in lhf's answer.

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This is precisely what I wrote 5 hours prior, except it explicitly replaces $\rm\:k\:$ by $\rm\: n!\:$ in my Bezout equation. Doing so decreases the generality of the proof, and obscures the key (unit group) structure. –  Bill Dubuque Apr 11 '12 at 20:06
Well then, I won't claim priority. But some might find that my solution is more accessible to the OP than yours. –  Greg Martin Apr 12 '12 at 6:51
My concern is not priority but, rather, pedagogy. I explicitly abstracted $n!$ to any integer $k$ divisible by $n$ in order to make clearer the innate governing multiplicative structure. To succeed in elementary number theory it is essential to learn how to recognize such structure. If students are encouraged to follow shortcuts circumventing such pedagogical routes then they may completely miss the key ideas, and never see the forest for the trees. –  Bill Dubuque Apr 12 '12 at 14:15

I dont know i am right or wrong but i can do this example in following way,\ Let $$\gcd(n\cdot n!,n!+1)=d$$ $$\therefore d\mid n\cdot n!\ ,\ d\mid n!+1$$ $$\Rightarrow d\mid n,\ d\mid n!,\ d\mid n!+1$$ $$\Rightarrow d\mid n!+1-n!$$ Thus $d\mid 1\ \Rightarrow d=1.$

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Why does $d \mid n$ ? Perhaps you mean a prime $d$. –  lhf Apr 11 '12 at 1:47
$8 | 4 \times 4!$, but $8$ does not divide $4$. –  Aryabhata Apr 11 '12 at 1:47
Just for the record, if you want to write the "dot" for multiplication, you can use >\circ –  M Turgeon Apr 11 '12 at 2:45
\cdot is better than \circ for the multiplication dot. –  Greg Martin Apr 11 '12 at 6:33


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# Probability chord of bigger circle intersects smaller circle

You are given two concentric circles $$C_1$$ and $$C_2$$ of radius $$r$$ and $$r/2$$ respectively. What is the probability that a randomly chosen chord of $$C_1$$ will intersect $$C_2$$?

Answer: $$1/2, 1/3$$ or $$1/4$$

The first method I used (gives 1/4):

The midpoint of any chord uniquely determines it, as line joining center to midpoint is always perpendicular to chord. So instead of choosing a chord, let's choose points instead that shall be the midpoints of their respective chords. Any point inside inner circle will be a chord that intersects it too, and any point outside will never cut inner circle. Thus probability should be area of inner circle/area of outer circle= $$1/4$$

But then I did it by another method and got another answer (1/3):

Choose a point on bigger circle. Now you can get all chords from $$0$$ to $$π$$ angle. The one which intersect smaller one must lie between tangents to smaller circle from bigger one from that point . We can easily obtain angle between tangents as $$π/3$$ by some trigonometry. We can do same for every other point so answer is $$\frac{π/3}{π}=1/3$$

First I was confused that I was getting two answers. But then I checked the given answer and I saw they were accepting multiple answers.

So I thought about how there could be multiple possible probabilities and only possible reason seemed to be the boundary conditions as I had included diameters in chords in second solution but not in first. However even though there are infinite diameters I still don't think probability should be affected this much as we have infinite points.

Can someone give clarity on this? In particular, what exact conditions are included by which solution, and how will we get the third given answer (1/2)? As far as I can find there are $$3$$ boundary conditions we have to consider carefully-

1. If diameters are included or not
2. Degenerate "chords" that are $$r$$ distance from center, ie they are actually points on the circumference
3. If tangents are included in intersection or not

I believe the discrepancy results from the fact that the problem does not outline how the chords are randomly selected. As you have shown, there are multiple ways to randomly select a chord, and the two ways you have described have different probability distributions.

This can be seen by comparing the PDF of the length of the selected chord in both cases. In your first solution, the probability you select an arbitrary length is uniform. However, in your second solution, shorter chords are more likely to be selected than longer chords. Hence, there is a clear difference in the probability distributions in both selection processes.

As a result, this makes it possible to get different answers. Since the problem failed to elaborate on how chords were chosen, the problem writers were forced to accept multiple answers.

Moreover, the 3 boundary conditions you have outlines will not affect the probability because they have infinitesimal impact on the total probability.

You can get $$\frac{1}{2}$$ by modifying how you select the midpoint of the chord from your first solution. Instead of uniformly selecting a point from the interior of the larger circle, uniformly choose the distance of the midpoint from the center from the range $$[0,r]$$ and then uniformly choose the angle the midpoint makes with some arbitrary point/ray (e.g. define a positive x-axis) from the range $$[0,2\pi)$$. The chord will intersect the smaller circle whenever the chosen distance of the midpoint from the center is $$<\frac{r}{2}$$, which happens exactly $$\boxed{\frac{1}{2}}$$ of the time.

• Thanks for your answer. I also thought the same thing about the boundary conditions but only reason I had my doubts was because we have infinite diameters and infinite tangential chords So they would contribute a lot more than just one chord Jul 27, 2021 at 6:46
• I don't know how to rigorously explain it, but it's kind of like how if you remove a curve of infinite points from a region, the area of the region is still the same even though you removed an infinite number of "things" from the region. Jul 27, 2021 at 6:49
• yeah I think a field of mathematics called "measure theory" is used so quite far beyond high school math Jul 27, 2021 at 6:54


# What is the expected number of coin tosses needed to obtain a head?

Due to my recent misunderstandings regarding the 'expected value' concept I decided to post this question. Although I have easily found the answer on the internet I haven't managed to fully understand it.

I understood that the formula for the expected value is:

$$E(x) = x_1p_1 +x_2*p_2 +...+x_np_n$$

The x's are the possible value that the random variable can take and the p's are the probabilites that this certain value is taken.

So, if I get a head on the first try, then $p_1 = \frac{1}{2} , x_1 = 1$ If I get a head on the second try, then $p_2 = \frac{1}{4} , x_2 = 2$

And then, I woudl have that:

$$E(x) = \frac{1}{2}1+ \frac{1}{4}2 +...$$

So my reasoning led me to an inifnite sum which I don't think I can't evaluate it that easy. In the 'standard' solution of this problem, the expected value is found in a reccurisve manner. So the case in which the head doesn't appear in the first toss is treated reccursively. I haven't understood that step.

My questions are: is my judgement correct? How about that reccursion step? Could somebody explain it to me?

• For fun, I would say 2. =) – Vincent Aug 27 '14 at 15:46
• Yes, I knew that too. :D I just didn't know how we found that answer – Bardo Aug 27 '14 at 15:48

Let $X$ be the number of tosses, and let $e=E(X)$. It is clear that $e$ is finite.

We might get a head on the first toss. This happens with probability $\frac{1}{2}$, and in that case $X=1$.

Or else we might get a tail on the first toss. In that case, we have used up $1$ toss, and we are "starting all over again." So in that case the expected number of additional tosses is $e$. More formally, the conditional expectation of $X$ given that the first toss is a tail is $1+e$.

It follows (Law of Total Expectation) that $$e=(1)\cdot\frac{1}{2}+(1+e)\cdot\frac{1}{2}.$$

This is a linear equation in $e$. Solve.

Remark: The "infinite series" approach gives $$E(X)=1\cdot\frac{1}{2}+2\cdot\frac{1}{2^2}+3\cdot\frac{1}{2^3}+\cdots.$$ This series, and related ones, has been summed repeatedly on MSE.

• Conditioning like I did is a totally standard technique in the calculation of expectation. – André Nicolas Aug 27 '14 at 16:01
• Well, perhaps you can use the series approach for general $p$, and the equation $e=p+(1+e)(1-p)$, and see that they give the same answer. – André Nicolas Aug 27 '14 at 16:09
• Given that the first toss is a tail, $E(X)=1+e$. – André Nicolas Aug 27 '14 at 16:10
• It is not true that with probability $1/2$ you will need $1+e$. What is true is that given the first is tail the total expected number of tosses is $1+e$. – André Nicolas Aug 27 '14 at 16:24
• @Bardo Note that in your previous question in my solution we use essentially the same trick (in a slightly more complex situation). – Aahz Aug 27 '14 at 16:34

Your approach is perfectly fine. The probability of getting the first head in the $n$th trial is $\frac{1}{2^n}$, so we have $$\mathbb{E}(x) = \sum_{ n \geq 1} \frac{n}{2^n}.$$ This infinite sum can be calculated in the following way: first note that $\frac{1}{1-x} = \sum_{n \geq 0} x^n$ for $|x|<1$. Differentiating both sides yields $$\frac{1}{(1-x)^2} = \sum_{n \geq 0} n x^{n-1} = \sum_{n \geq 1} n x^{n-1} = \frac1x \sum_{n \geq 1} n x^n.$$ Pluggin in $x = \frac12$ yields $4 = 2 \sum_{n \geq 1} \frac{n}{2^n} = 2 \mathbb{E}(x)$, or $\mathbb{E}(x) = 2$.

The recursive solution goes, I think, as follows: let $\mathbb{E}(x)$ be the expected number of trials needed. Then the expected number of trials needed after the first trial, given that it was not heads, is also $\mathbb{E}(x)$. In other words, the expected (total) number of trials is $\mathbb{E}(x)+1$ in that case. This gives the equation $$\mathbb{E}(x) = \frac12 + \frac12(\mathbb{E}(x)+1)$$ which gives the same answer $\mathbb{E}(x) = 2$.

• It's good to know that my approach wasn't incorrect, it gives me a little confidence. I am no thinking about that reccursion step. – Bardo Aug 27 '14 at 15:54

To make ends meet... You have been explained by several users that, looking at the toss process itself, one sees that the expectation $E(X)$, that you know is $E(X)=S$, with $$S=\sum_{n\geqslant1}\frac{n}{2^n},$$ solves the relation $$E(X)=1+\frac12E(X).$$ It happens that one can also show directly that $$S=1+\frac12S,$$ this relation following from a shift of indexes. To do so, note that $S=R+T$ with $$R=\sum_{n\geqslant1}\frac{1}{2^n},\qquad T=\sum_{n\geqslant1}\frac{n-1}{2^n},$$ hence, using the change of variable $n=k+1$, $$T=\sum_{k\geqslant0}\frac{k}{2^{k+1}}=\frac12\sum_{k\geqslant0}\frac{k}{2^k}=\frac12\sum_{k\geqslant1}\frac{k}{2^k}=\frac12S,$$ hence the proof would be complete if one knew that $$R=1.$$ To show this, use the same trick once again, that is, note that $$R=\frac12+\sum_{n\geqslant2}\frac{1}{2^n}=\frac12+\sum_{k\geqslant1}\frac{1}{2^{k+1}}=\frac12+\frac12\sum_{k\geqslant1}\frac{1}{2^{k}}==\frac12+\frac12R,$$ hence the proof that $S=2$ is complete.

• Yes, I got it...I knew beforehand the trick of shifting indexes, so I totatly understood your argument..But I am still struggling to undesrtande the simpler argument..Thank you very much! – Bardo Aug 27 '14 at 18:29


# Prove that A Real Root Exists in [-1, 1]

#### anemone

##### MHB POTW Director
Staff member
Given $$\displaystyle f(x)=5tx^4+sx^3+3rx^2+qx+p$$ for $f(x)\in R$. If $r+t=-p$, prove that there is a real root for $f(x)=0$ in $[-1,1]$.

#### Ackbach

##### Indicium Physicus
Staff member
Some ideas:

We rewrite the polynomial as
$$f(x)=5tx^4+sx^3+3rx^2+qx-r-t=0,$$
where we are setting it equal to zero. We evaluate $f(1)$ and $f(-1)$:
\begin{align*}
f(1)&=5t+s+3r+q-r-t=4t+2r+s+q\\
f(-1)&=5t-s+3r-q-r-t=4t+2r-s-q.
\end{align*}
Idea: if $f(1)\cdot f(-1)<0$, then by the Intermediate Value Theorem, we would have shown there is a root in $[-1,1]$. Now
$$f(1)\cdot f(-1)=(4t+2r)^{2}-(s+q)^{2}.$$
Not seeing where to go with this. There's nothing stopping $s=q=0$, with $4t+2r\not=0$, in which case I have not proved what I want to prove.

#### zzephod

##### Well-known member
Given $$\displaystyle f(x)=5tx^4+sx^3+3rx^2+qx+p$$ for $f(x)\in R$. If $r+t=-p$, prove that there is a real root for $f(x)=0$ in $[-1,1]$.
Consider the polynomial $$\displaystyle p(x)=tx^5+(s/4)x^4+rx^3+(q/2)x^2+px-(s/4+q/2)$$.

Then $$\displaystyle p(1)=p(-1)=0$$ and so $$\displaystyle p$$ has an extremum in $$\displaystyle (-1,1)$$, so $$\displaystyle p'(x)$$ has a root in $$\displaystyle (-1,1)$$ ...

.

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#### anemone

##### MHB POTW Director
Staff member
Thanks to both, Ackbach and zzephod for participating...

Consider the polynomial $$\displaystyle p(x)=tx^5+(s/4)x^4+rx^3+(q/2)x^2+px-(s/4+q/2)$$.

Then $$\displaystyle p(1)=p(-1)=0$$ and so $$\displaystyle p$$ has an extremum in $$\displaystyle [-1,1]$$, so $$\displaystyle p'(x)$$ has a root in $$\displaystyle [-1,1]$$ ...

.
WoW!!! What an elegant way to approach this problem! Well done, zzephod!

And there is another quite straightforward and beautiful way to tackle it as well...therefore I'll wait for the inputs from other members for now...

#### anemone

##### MHB POTW Director
Staff member
Another method proposed by other to solve this challenge problem is by using the integration method:

Hint:

$$\displaystyle \int_{-1}^1 p(x) dx=0$$


# find arc BN=?

#### Albert

##### Well-known member
Points A,B are on circle C ,segment MN is a diameter of circle C, and point P is on

segment MN , if :

$\angle CAP=\angle CBP =10^o ,\,\, \overset{\frown} {MA}=40^o,\,\, find :\,\, \overset{\frown} {BN}=?$

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#### HallsofIvy

##### Well-known member
MHB Math Helper
You refer to "circle C" but then treat "C" as if it were a point. Are we to assume that "C" is the center point of the circle?

#### Albert

##### Well-known member
You refer to "circle C" but then treat "C" as if it were a point. Are we to assume that "C" is the center point of the circle?
yes ,you got it !
"C" is the center point of the circle.

#### Opalg

##### MHB Oldtimer
Staff member
Points A,B are on circle C ,segment MN is a diameter of circle C, and point P is on

segment MN , if :

$\angle CAP=\angle CBP =10^o ,\,\, \overset{\frown} {MA}=40^o,\,\, find :\,\, \overset{\frown} {BN}=?$

One solution is for $B$ to be opposite $A$ on the other side of $MN$, at the point labelled $B'$ in the picture. Then $\overset{\frown} {BN} = 140^\circ$. But that is too obvious to be interesting, and I assume that what was wanted is the case where $A$ and $B$ are on the same side of $MN$.

The points $A, B, C, P$ are concyclic, because $\angle CAP=\angle CBP =10^\circ$. Therefore $\angle ABP=\angle ACP =40^\circ$, and so $\angle ABC= 10^\circ + 40^\circ = 50^\circ.$ The triangle $ABC$ is isosceles, so $\angle BAC = 50^\circ$, and $\angle ACB =80^\circ$. Finally, $\angle BCP =40^\circ + 80^\circ = 120^\circ$, from which $\overset{\frown} {BN}= \angle BCN = 60^\circ.$

#### Albert

##### Well-known member
what will be the value of arc BN , if point P locates between points C and N

#### Opalg

##### MHB Oldtimer
Staff member
what will be the value of arc BN , if point P locates between points C and N
Good question! I hadn't thought of that possibility. The method will be similar to the previous one, but this time the angle ABC ($\angle A'B'C$ in the diagram below) will be $40^\circ - 10^\circ = 30^\circ$ instead of $40^\circ + 10^\circ = 50^\circ$. Then $\angle A'CB' = 120^\circ$ and $\overset{\frown} {BN} = 20^\circ.$

#### Albert

##### Well-known member
very good solution
this is an open -style problem ,if the position of point B or point P changes ,then the answer will also differ (it depends upon how the diagram is sketched)
sometime we may give students a mathematic problem with more then one possible answer


# Factoring a quadratic polynomial (absolute beginner level), are both answers correct?

I'm following video tutorials on factoring quadratic polynomials. So I'm given the polynomial:

$$x^2 + 3x - 10$$

And I'm given the task of finding the values of $a$ and $b$ in:

$$(x + a) (x + b)$$

Obviously the answer is: $$(x + 5)(x - 2)$$

However the answer can be also:

$$(x - 2) (x + 5)$$

I just want to make sure if the question asks for the values of '$a$' and '$b$', then '$a$' can be either $5$ or $-2$, and '$b$' can be either $5$ or $-2$.

Therefore if a question asks what are the values of '$a$' and '$b$' both the following answers are correct:

Answer $1$
$a = -2$
$b = 5$
or
Answer $2$
$a = 5$
$b = -2$

I'm sure this is a completely obvious question, but I'm just a beginner in this.

• Yes, the problem is symmetric for $a$ and $b$. So yes, the answers are $(a,b) = (-2,5)$ and $(a,b) = (5,-2)$. – Matti P. Aug 23 '18 at 8:18
• They are both valid answers, since the order of the factors doesn't matter. – Ludvig Lindström Aug 23 '18 at 8:18
• You teacher should have stated what context they want you to find $a$ and $b$, I have provided the answer below under the assumption that they meant the roots of the polynomial equation in $x$. – Adam L Aug 23 '18 at 8:54
• Even further so, they should have told you what the domain they would like $a$ and $b$ to be computed, and the corresponding co-domain of $x$ for which they wanted, the last part may not be appropriate terminology but I'm sure I will be corrected soon enough if that's the case. – Adam L Aug 23 '18 at 8:58
• @Adam Thanks for your specific definitions. Unfortunately I don't have a teacher and am working by myself from Kahn Academy videos. I appreciate that there are many people here to give extra help. – Zebrafish Aug 23 '18 at 9:03

.Yes, you are correct. Since $$(x+5)(x-2) = (x-2)(x+5) = x^2 + 3x-10$$, we note that $$a$$ and $$b$$ may either take the values $$(5,-2)$$ or $$(-2,5)$$.

I would consider providing just one of the two solutions to be insufficient, since the question itself ask for the values of $$a$$ and $$b$$, but nowhere mentions that they are unique. However, any question saying "find the values of $$a$$ and $$b$$" is wrong with the word "the" : they are assuming uniqueness of $$a$$ and $$b$$, which is not the case.The question as quoted by you includes the word "the" , and this is misleading.

• Yes, actually the statement "find 'the' values of a and b" is my invention. It's just from watching the videos that that's what it seemed the question was telling me to do. Thank you. – Zebrafish Aug 23 '18 at 8:26
• You are welcome! – Teresa Lisbon Aug 23 '18 at 9:56
• I should add : Khan academy is a good source, but don't depend on only one source : when you find something wrong/fishy, try to use more sources to confirm it. With this in mind, it is good you asked this question. Also : are you in preparation for some exam, and hence doing these questions, or are you tallying this with something you are learning in some course currently? – Teresa Lisbon Aug 23 '18 at 14:18

For commutative property of product we have that

$$(x + 5)(x - 2)=(x - 2)(x + 5)$$

note that also

$$(-x + 2)(-x - 5)$$

is a correct factorization.

You are right.

$$(x+a)(x+b)=x^2+(a+b)x+ab$$

and by identification with $x^2+3x-10$,

$$\begin{cases}a+b=3,\\ab=-10.\end{cases}$$

This is a non-linear system of equations, and given commutativity of addition and multiplication, it is clear that if $(u,v)$ is a solution, so is $(v,u)$.

Now one may wonder if more than two solutions could exist. As $a=0$ cannot be a solution, we can write

$$3a=(a+b)a=a^2+ab=a^2-10$$

which is the original equation (with a sign reversal)

$$a^2-3a-10=0.$$

To be able to conclude, you must invoke the fundamental theorem of algebra, which implies that a quadratic equation cannot have more than two roots.

So there are exactly these two solutions: $a=-2,b=5$ and $a=5,b=-2$.