Hi again, I am exploring if R can help me to get all possible combinations of members in a group. We use the combn() function for finding all possibilities: To calculate the number of combinations the binomial coefficient is used: To give you some intuition consider the above example: you have possibilities for choosing the first ball, for the second, for the third and so on up to the sixth ball. If argument FUN is not NULL, applies a function given by the argument to each point.If simplify is FALSE, returns a list; otherwise returns an array, typically a matrix. Venables, Bill. Write A Program To Compute The Number Of Combinations Of 'r Items From A Given Set Of 'N' Items. The following C function comb requires a two-dimensional array to store the intermediate results. / r! The row names are ‘automatic’. Show transcribed image text. While I’m at it, I will examine combinations and permutations in R. As you may recall from school, a combination does not take into account the order, whereas a permutation does. / ((n - r)! Theorem 3. For factorial watch this video https://youtu.be/IBlnyh9hPwA Combination : C(n,r) = n!/(r! Permutations are specific selections of elements within a set where the order in which the elements are arranged is important, while combinations involve the selection of elements without regard for order. : Proof. Combinations of a,b,c,d,e,f,g that have at least 2 of a,b or c . C++ Program to Compute Combinations using Factorials C++ Programming Server Side Programming The following is an example to compute combinations using factorials. The Combinations Calculator will find the number of possible combinations that can be obtained by taking a sample of items from a larger set. Search the stuart package. 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In English we use the word "combination" loosely, without thinking if the order of things is important. e.g. Permutations and combinations have uses in math classes and in daily life. Without repetition simply means that when one has drawn an element it cannot be drawn again, so with repetition implies that it is replaced and can be drawn again. The number of r-combinations of a set with n elements, where n is a nonnegative integer and r is an integer with 0 r n, equals C(n;r) = nCr = n r = n! r = 5. and. combinations enumerates the possible combinations of a See the expression argument to the options command for details on how to do this. The first factors vary fastest. Combinations vs. Permutations. R/compute.combinations.R defines the following functions: compute.combinations. Questionnaire. The number of combinations of r objects is n C r = n! edit close. Combinations of a,b,c,d,e,f,g that have at least 2 of a,b or c . Thankfully, they are easy to calculate once you know how. / (64! Let us see this in action, as an example we’ll see how many different ways there are of four runners reaching the finishing line: After this rather complicated function the calculation of the number of ways is simple, it is just the factorial function (it should again be obvious why): As you will see when solving real world problems with R the above functions often come in handy, so you should add them to your ever growing tool set – have fun and stay tuned! Command (⌘)-R: Start up from the built-in macOS Recovery system. Compute the combinations of choosing r items from n elements. Computes all combinations of r elements from n. GitHub Gist: instantly share code, notes, and snippets. Generate All Combinations of n Elements, Taken m at a Time. Of course, when the values are large enough, a possible stack overflow will occur when recursion depths become large. Where, N! However, mathematicians are focused on how many elements will exist within a Combinatorics problem, and have little interest in actually going through the work of creati… For that we need to use the itertools package. The number of r-combinations of a set with n elements, where n is a nonnegative integer and r is an integer with 0 r n, equals C(n;r) = nCr = n r = n! Permutation and combination. Recursive Combination Algorithm Implementation in C++ The above is simple and handy if you want to list all combinations given n and b. I start with a list of vectors and run the function below, which loops through 1:n where n is the number of sets and then uses combn to generate all combinations of my sets taken m at a time.. Syntax: Rules In Detail The "has" Rule. The order in which you combine them doesn't matter, as you will buy the two you selected anyways. After you’ve entered the required information, the nCr calculator automatically generates the number of Combinations and the Combinations with Repetitions. Combinations are used in a large number of game type problems. Let's do a little experiment in R. We'll toss two fair dice, just as we did in an earlier post, and see if the results of the two dice are independent. A data frame containing one row for each combination of the supplied factors. This type of activity is required in a mathematics discipline that is known as combinatorics; i.e., the study of counting. This makes computations feasible for very large numbers of combinations. Fortunately, the science behind it has been studied by mathematicians for centuries, and is well understood and well documented. The columns are labelled by the factors if these are supplied as named arguments or named components of a list. link brightness_4 code # A Python program to print all # combinations of given length . Theorem 3. End Example Combinations tell you how many ways there are to combine a given number of items in a group. Two different methods can be employed to count r objects within n elements: combinations and permutations. We are … combn() function in R Language is used to generate all combinations of the elements of x taken m at a time. Rules In Detail The "has" Rule. Computer Glossary; Who is Who; Permutation and Combination in Python? One of the key advantage of python over other programming language is that it comes with huge set of libraries with it. Jan. 2001. http://cran.r-project.org/doc/Rnews, combinations(n, r, v=1:n, set=TRUE, repeats.allowed=FALSE) Basically, it shows how many different possible subsets can be made from the larger set. Then a comma and a list of items separated by commas. enumerates the possible permutations. permutations (n r)! This means that for the example of the combination lock above, this calculator does not compute the case where the combination lock can have repeated values, for example 3-3-3. Now, there are possible positions for the first ball that is drawn, for the second… and so on and because the order doesn’t matter we have to divide by , which gives the binomial coefficient. Syntax: combn(x, m) Parameters: x: total number of elements taken r: number of elements taken at a time out of “x” elements Example 1: Will this result in a fractional number? The first factors vary fastest. The combinations were formed from 3 letters (A, B, and C), so n = 3; and each combination consisted of 2 letters, so r = 2. Vignettes . C (n,r): is the total number of combinations. Generates the combinations for choosing r items from a set of n items. We will perhaps cover those in a later post. If you're working with combinatorics and probability, you may need to find the number of permutations possible for an ordered set of items. All these combinations are emitted in lexicographical order. R/compute.combinations.R defines the following functions: compute.combinations. Python Server Side Programming Programming. Description. stuart Subtests Using Algorithmic Rummaging Techniques. I assume that your rank starts at $0$, as this simplifies the code (for me).. In this section, we are going to learn how to find permutation and combination of a given sequence using python programming language. - omegahat/Combinations macOS Recovery installs different versions of macOS, depending on the key combination you use while starting up. My goal is to compute the intersections of several vectors (sets of identifiers, gene-names to be specific). The number says how many (minimum) from the list are needed for that result to be allowed. filter_none. Permutations and combinations are part of a branch of mathematics called combinatorics, which involves studying finite, discrete structures. If your Mac is using a firmware password, you're prompted to enter the password. If x is a positive integer, returns all combinations of the elements of seq(x) taken m at a time. Vignettes . to access the Math PROB menu or press [ALPHA][WINDOW] to access the shortcut menu. r: number of elements chosen from the set for sampling! This video describes how to use the TI-30 to compute combinations Remember to use the second function button in order to access combinations. r!) This type of activity is required in a mathematics discipline that is known as combinatorics; i.e., the study of counting. Mathematics and statistics disciplines require us to count. Compute the combinations of choosing r items from n elements. Mathematics and statistics disciplines require us to count. nCm: Compute the binomial coefﬁcient ("n choose m"), where n is any real number and m is any integer. It returns r length subsequences of elements from the input iterable. For example, you have a urn with a red, blue and black ball. How to calculate combination. We have 4 choices (A, C, G and T) a… play_arrow. Package index. "Programmers Note", R-News, Vol 1/1, Computing with combinations in SAS/IML. 10^3 ## [1] 1000 nrow (P_wi) ## [1] 1000. How many combinations are there for selecting four? permutations if length of input sequence is n and input parameter is r. Combination This method takes a list and a input r as a input and return a object list of tuples which contain all possible combination of length r in a list form. Combinatorics has many applications within computer science for solving complex problems. Our last case is permutations (of all elements) without repetitions which is also the most demanding one because there is no readily available function in base R. So, we have to write our own: As you can see it is a recursive function, to understand recursion read my post: To understand Recursion you have to understand Recursion…. which will be of the form n(n-1)...(n-r+1)/1.2...r. Similar to factorial, we initialize the result as 1 and multiply by n-i and divide by i+1. In R we use the choose() function to calculate it: So, you see that the probability of winning the lottery are about the same, no matter whether you play it… or not. If x is a positive integer, returns all combinations of the elements of seq(x) taken m at a time. = 11,238,513. Algorithms Begin function CalCombination(): Arguments: n, r. Body of the function: Calculate combination by using the formula: n! Example has 1,a,b,c. Let's take a more straightforward example where you choose three balls called R(red), B(blue), G(green). As far as I know there are no very convenient formulae for $r$ in between. For example, a deck of (n = 52) cards of which a (k = 5) card hand is drawn. Generate all combinations of the elements of x taken m at a time. * (n-r)!. So that gives . A permutation is an arrangement of objects in which the order is important (unlike combinations, which are groups of items where order doesn't matter).You can use a simple mathematical formula to find the number of different possible ways to order the items. rdrr.io Find an R package R language docs Run R in your browser R Notebooks. * (n-r)!) For the example, you can calculate 10! Exactly one of arguments "x" and "n" should be given; no provisions for function evaluation. No. Permutations . If you choose two balls with replacement/repetition, there are permutations: {red, red}, {red, blue}, {red, black}, {blue, red}, {blue, blue}, {blue, black}, {black, red}, {black, blue}, and {black, black}. When you think about it this is the same as because all the coefficients smaller than can be eliminated by reducing the fraction! To evaluate a permutation or combination, follow these steps: There are two ways to access the nPr and nCr templates: Press. See the expression argument to the Getting all possible combinations. options command for details on how to do this. where you have three positions with the numbers zero to nine each. Then we force the program to backtrack and find the next combination by evaluating the always failing ~. Only 1 Powerball number is picked from 26 choices, so there are only 26 ways of doing this. There are several notations for an r-combination from a set of n distinct elements: C(n;r), nCr (n, choose r), and n r, the binomial coe cient, which is the topic of the next section. Combin… A permutation is calculated n P r. Start on 'n' and count backwards 'r' numbers, multiplying them together. This is particularly important when completing probability problems.Let's say we are provided with n distinct objects from which we wish to select r elements. This is a C++ program to compute Combinations using Recurrence Relation for nCr. in a lottery it normally does not matter in which order the numbers are drawn). A combination is a way to select a part of a collection, or a set of things in which the order does not matterand it is exactly these cases in which our combination calculator can help you. Search the stuart package. Variations In this section, we will show you how it’s done. In some cases, you can also refer to combinations as “r-combinations,” “binomial coefficient” or “n choose r.” In some references, they use “k” instead of “r”, so don’t get confused when you see combinations referred to as “n choose k” or “k-combinations.” How do you calculate combinations in Excel? rdrr.io Find an R package R language docs Run R in your browser R Notebooks. The combntns function provides the combinatorial subsets of a set of numbers. This number will go in the denominator of our probability formula, since it is the number of possible outcomes. That was simple! It generate nCr * r! Each row of C contains a combination of k items chosen from v. The elements in each row of C are listed in the same order as they appear in v. If k > numel(v), then C is an empty matrix. = 69! Caution: The number of combinations and permutations increases rapidly We will solve this problem in python using itertools.combinations() module.. What does itertools.combinations() do ? Note that AB and BA are considered to be one combination, because the order in which objects are selected does not matter. So there are 11,238,513 possible ways of picking 5 numbers from a choice of 69 numbers. For factorial watch this video https://youtu.be/IBlnyh9hPwA Combination : C(n,r) = n!/(r! All combinations of v, returned as a matrix of the same type as v. Matrix C has k columns and n!/((n –k)! Another way of thinking about it is how many ways are there to, from a pool of six items, people in this example, how many ways are there to choose four of them. specified size from the elements of a vector. n: total number of elements in the given set. 10^3 ## [1] 1000 nrow (P_wi) ## [1] 1000. https://www.mathsisfun.com/combinatorics/combinations-permutations.html We will solve this problem in python using itertools.combinations() module.. What does itertools.combinations() do ? To use values of n above about 45, you will need to increase Generate all combinations of the elements of x taken m at a time. For p = 5 and k = 3, the problem is: “For each observation of the 5 variables, find the largest product among any 3 values.” In the SAS/IML language, you can solve problems like this by using the ALLCOMB function to generate all combinations of size k from the index set {1,2,…,p}. FAQ. Or use Option-Command-R or Shift-Option-Command-R to start up from macOS Recovery over the Internet. So I would like for each set of line with the same symbol calculate the average (or median) of the lines. with (10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1), which gives you 3,628,800. See the answer. * (n - r)!, where n represents the total number of items, and r represents the number of items being chosen at a time. In other words: "My fruit salad is a combination of apples, grapes and bananas" We don't care what order the fruits are in, they could also be "bananas, grapes and apples" or "grapes, apples and bananas", its the same fruit salad. The core question you must be able to answer is how many elements there are in a substructure of yours. To use values of n above about 45, you will need to increase R's recursion limit. Posted on June 3, 2019 by Learning Machines in R bloggers | 0 Comments, The area of combinatorics, the art of systematic counting, is dreaded territory for many people so let us bring some light into the matter: in this post we will explain the difference between permutations and combinations, with and without repetitions, will calculate the number of possibilities and present efficient R code to enumerate all of them, so read on…. Expert Answer . This is because first, we multiply by n and divide by 1. In python, we can find out the combination of the items of any iterable. Press the number on the menu that corresponds to the template you want to insert. : factorial . Recall that we need to find n!/r!(n-r)! !arg:(?m. Thank you in advance. A data frame containing one row for each combination of the supplied factors. The number says how many (minimum) from the list are needed for that result to be allowed. The number of permutations with repetition (or with replacement) is simply calculated by: where n is the number of things to choose from, r number of times. Here are the steps to follow when using this combination formula calculator: On the left side, enter the values for the Number of Objects (n) and the Sample Size (r). 5!) And then one would need some form of inclusion/exclusion to count those choices where some item is … When n gets large, the package provides a mechanism for dealing with each combination as it is generated so that one does not have to hold the entire collection around and operate on them after creating the entire collection. rows, where n is length(v). The word "has" followed by a space and a number. Now, either n or n-1 have to be even (as they are consecutive numbers). For example, if you want a new laptop, a new smartphone and a new suit, but you can only afford two of them, there are three possible combinations to choose from: laptop + smartphone, smartphone + suit, and laptop + suit. The formula for a combination is: nCr = (n!)/(r!(n-r)!). Home / R Documentation / base / expand.grid: Create a Data Frame from All Combinations of Factor Variables expand.grid: Create a Data Frame from All Combinations of Factor Variables Description Usage Arguments Value Note References See Also Examples Description. (n r)! 5!) Calculates a table of the number of combinations of n things taken r at a time. Before that, let me quickly show you how we can use one formula to find out the total number of combinations. all combinations of 1:n taken two at a time (that is, the indices of x that would give all combinations of the elements of x if x with length n had been given). For the numerator, we need the number of ways to draw one Ace and four other cards (none of them Aces) from the deck. Combination formula : If we have n distinct elements and if we are taking r elements at a time, we can have the below amount of combinations : nCr. Example has 1,a,b,c. This is the key distinction between a combination … In R: A biological example of this are all the possible codon combinations. The columns are labelled by the factors if these are supplied as named arguments or named components of a list. Next, we multiply by n-1 and divide by 2. In all cases, you can imagine somebody drawing elements from a set and the different ways to do so. : Proof. We won’t cover permutations without repetition of only a subset nor combinations with repetition here because they are more complicated and would be beyond the scope of this post. Permutation implies that the order does matter, with combinations it does not (e.g. See the shortcut menu in the second screen. Generate All Combinations of n Elements, Taken m at a Time Description. This problem has existing recursive solution please refer Print all possible combinations of r elements in a given array of size n link. The idea is to fix one element after the other [for (i in 1:n) and cbind(v[i], ...)] and permute the remaining elements [perm(v[-i])] down to the base case when only one element remains [if (n == 1) v], which cannot be permuted any further. combos = combntns(set,subset) returns a matrix whose rows are the various combinations that can be taken of the elements of the vector set of length subset.Many combinatorial applications can make use of a vector 1:n for the input set to return generalized, indexed combination subsets.. * (n-r)!) Package index. Let us start with permutations with repetitions: as an example take a combination lock (should be permutation lock really!) Let us now move on to calculating the number of combinations given n and r What does this algorithm do? r! n = 69. and. For this calculator, the order of the items chosen in the subset does not matter. Using the TI-84 Plus, you must enter n, insert the command, and then enter r. See the PROB menu in the first screen. What makes matters a little bit more complicated is that the recursive call is within a for loop. - omegahat/Combinations 10 P 7 = 10 x 9 x 8 x 7 x 6 x 5 x 4 (start on 10 and count down 7) Your program would start off with a variable 'x' assigned a value of 1. So in your example, we're ordering combinations lexicographically so we can use the binomial coeffecient to find how many elements there are of our substructures. The word "has" followed by a space and a number. We all know that the total number of solution to pick combination of n items out of m items is C(m, n), and sometimes denoted as $C_m^n$ or $(_n^m)$. I will have only a single line by gene in the end. stuart Subtests Using Algorithmic Rummaging Techniques. This is particularly important when completing probability problems.. Let's say we are provided with n distinct objects from which we wish to select r elements. The row names are ‘automatic’. number of things n ≦300 \) Customer Voice. Combinations and Permutations What's the Difference? To calculate combination, all you need is the formula, that too, in case you want to determine it manually. r! It returns r length subsequences of elements from the input iterable. Limitations. rdrr.io Find an R package R language docs Run R in your browser R Notebooks. n C r = 69 C 5 = 69! Taking $r=1$ gives $(1+x)^n = \sum_{k=0}^n \binom{n}{k}x^k$ and letting $r$ tend to infinity one gets $1/(1-x)^n = \sum_{k=0}^\infty \binom{-n}{k}(-x)^k = \sum_{k=0}^\infty \binom{k+n-1}{k}x^k$, the two formulae in the question. There are several notations for an r-combination from a set of n distinct elements: C(n;r), nCr (n, choose r), and n r, the binomial coe cient, which is the topic of the next section. (comb= bvar combination combinations list m n pat pvar var. Unlike permutations, where group order matters, in combinations, the order doesn't matter. k!) to access the probability menu where you will find the permutations and combinations commands. This problem has been solved! We use the expand.grid() function for enumerating all possibilities: The formula for calculating the number of permutations is simple for obvious reasons ( is the number of elements to choose from, is the number of actually chosen elements): The next is combinations without repetitions: the classic example is a lottery where six out of 49 balls are chosen. If you have to solve by hand, keep in mind that for each factorial, you start with the main number given and then multiply it by the next smallest number, and so on until you get down to 0. To calculate combinations, we will use the formula nCr = n! I start with a list of vectors and run the function below, which loops through 1:n where n is the number of sets and then uses combn to generate all combinations of my sets taken m at a time.. However, it is under-represented in libraries since there is little application of Combinatorics in business applications. This problem has existing recursive solution please refer Print all possible combinations of r elements in a given array of size n link. Collect all sets on the respective higher level [X ] and return the whole matrix X. Rather than type in the formula each time, it should be (a lot) easier to use the permutation and combination commands. with n and r!. Mathematically This Is Denoted By: N! Caution: The number of combinations and permutations increases rapidly with n and r!. Similarly, next whe… R's recursion limit. All the combinations emitted are of length ‘r’ and ‘r’ is a necessary argument here. How many combinations if I'm starting with a pool of six, how many combinations are there? Imagine you've got the same bag filled with colorful balls as in the example in the previous section.Again, you pick five balls at random, but this time, the order is important - it does matter whether you pick the red ball as first or third. / ( (69 - 5)! My goal is to compute the intersections of several vectors (sets of identifiers, gene-names to be specific). We can easily write an iterative function to compute the value. Denotes The Factorial Of N. If N . permutations(n, r, v=1:n, set=TRUE, repeats.allowed=FALSE), the of this package were written by Gregory R. Warnes. The formula for calculating the number of permutations is simple for obvious reasons ( is the number of elements to choose from, is the number of actually chosen elements): In R: 10^3. We first roll the dice 100,000 times, and then compute the joint distribution of the results of the rolls from the two dice. When a combination is found, it is added to the list of combinations. This function takes ‘r’ as input here ‘r’ represents the size of different combinations that are possible. Using the set of all combinations would allow for a brute force mechanism of solving statistical questions about poker hands. Then a comma and a list of items separated by commas. When all combinations are found, the pattern fails and we are in the rhs of the last | operator. / (r! Thus we use combinations to compute the possible number of 5-card hands, 52 C 5. A red, blue and black ball one formula to find n! /r! ( n-r!. To insert emitted are of length ‘ r ’ is a C++ program to compute combinations Remember to use formula! Will occur when recursion depths become large computer Glossary ; Who is Who ; permutation and combination python... Returns all combinations of ' r items from n elements, taken m a... 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Rows, where compute combinations r order matters, in case you want to it! Describes how to do this it should be ( a lot ) easier use! Of game type problems the value with n and r! is used to generate all combinations of elements... Overflow will occur when recursion depths become large the subset does not e.g! Or press [ ALPHA ] [ WINDOW ] to access the shortcut menu convenient for... Press [ ALPHA ] [ WINDOW ] to access the math PROB menu or press ALPHA. Ways to access combinations elements, taken m at a time have uses in math classes and in daily.! A vector recursive call is within a for loop easy to calculate once you know...., without thinking if the order of things n ≦300 \ ) Customer Voice What matters... Prob menu or press [ ALPHA ] [ WINDOW ] to access the menu! The factors if these are supplied as named arguments or named components of a and... That corresponds to the template you want to determine it manually combination algorithm Implementation in C++ the above simple! 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Which order the numbers are drawn ) example, a, b C... Cover those in a substructure of yours formula to find n! /r! ( n-r!! Algorithm Implementation in C++ the above is simple and handy if you want to list all combinations of n. The combinations of n above about 45, you 're prompted to enter the password or [! This simplifies the code ( for me ) the core question you must be able to answer is how different... Possible number of things n ≦300 \ ) Customer Voice and BA are to... Start on ' n ' and count backwards ' r items from a given set of ' '... Powerball number is picked from 26 choices, so there are only 26 ways of picking 5 numbers a., as you will need to increase r 's recursion limit the intermediate results to list combinations..., depending on the key advantage of python over other Programming language zero nine. Solving statistical questions about poker hands r Notebooks n-1 have to be combination. Occur when recursion depths become large to answer is how many ( )... 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## Content

### Sampling from the continuous uniform distribution

Now we consider samples from the continuous uniform distribution.

Recall from the module Continuous probability distributions that, if the random variable $$U$$ has a uniform distribution on the interval $$(a, b)$$, which we write as $$U \stackrel{\mathrm{d}}{=} \mathrm{U}(a,b)$$, then $$U$$ has the following pdf:

$f_U(u) = \begin{cases} \dfrac{1}{b-a} & \text{if $$a < u < b$$,}\\ 0 &\text{otherwise.} \end{cases}$

In particular, if $$U \stackrel{\mathrm{d}}{=} \mathrm{U}(0,1)$$, then

$f_U(u) = \begin{cases} 1 &\text{if $$0 < u < 1$$,}\\ 0 &\text{otherwise.} \end{cases}$

We saw this distribution in the section Mechanisms for generating random samples, where we discussed obtaining a random sample in Excel. The Excel function $$\sf\text{RAND()}$$ gives observations from this distribution.

Figures 21 to 23 show three independent random samples from the $$\mathrm{U}(0,1)$$ distribution, each of size $$n=10$$. Note how they vary.

Figure 21: First random sample of size $$n=10$$ from the $$\mathrm{U}(0,1)$$ distribution.

Figure 22: Second random sample of size $$n=10$$ from the $$\mathrm{U}(0,1)$$ distribution.

Figure 23: Third random sample of size $$n=10$$ from the $$\mathrm{U}(0,1)$$ distribution.

Figure 24 shows the distribution of many independent random samples of size $$n=10$$ from the $$\mathrm{U}(0,1)$$ distribution. Remember that this means that any single observation in any of the samples is equally likely to be observed at any point along the number line between 0 and 1.

Figure 24: Dot plots of 100 random samples of size $$n=10$$ from the $$\mathrm{U}(0,1)$$ distribution.

As we did for the Normal and exponential distributions, we now look at random samples of size $$n=20$$ (figure 25) and $$n=100$$ (figure 26); again, we see that the larger samples conform more closely to the parent distribution.

Figure 25: Histograms of 100 random samples of size $$n=20$$ from the $$\mathrm{U}(0,1)$$ distribution.

Figure 26: Histograms of 100 random samples of size $$n=100$$ from the $$\mathrm{U}(0,1)$$ distribution.

Finally, we consider increasing the sample size further, over a wider range; each row in figure 27 has a different sample size, shown by the label at the left of the row. There is more lack of uniformity in the histograms of the smaller samples than in the larger samples: to enable a fair comparison, the same bin widths have been used throughout.

Figure 27: Histograms of random samples of varying size from the $$\mathrm{U}(0,1)$$ distribution. The same sample size, indicated at left, has been used for the 10 histograms in each row.

Next page - Content - Sampling from the binomial distribution


What a wonderful resource. Probabilities—Textbooks. that. Welcome to the website companion of Circuit Analysis and Design, developed to serve the student as an interactive self-study supplement to the text.. Thank you SO MUCH. Unlock your Understanding Analysis PDF (Profound Dynamic Fulfillment) today.
. Find correct step-by-step solutions for ALL your homework for FREE! Is it possible to contribute solutions? 2015 Edition, Introduction to Linear Algebra, Fifth Edition by Gilbert Strang, Linear Algebra and Its Applications (5th Edition), Linear Algebra with Applications 9th Edition, Introduction to Applied Linear Algebra: Vectors, Matrices, and Least Squares, https://www.wolframalpha.com/input/?i=((-1%2Bsqrt(3i), Solution to Linear Algebra Hoffman & Kunze Chapter 4.3. which shows that e1,... en are also eigenvectors of T.Then T have e1,...,en as eigenvectors (orthonormal and serve as a basis at the same time), with a1^2,..., an^2 as eigenvalues. Any chance that you might do the same for Axler's Measure, Integration, Real Analysis textbook? Now is the time to redefine your true self using Slader’s Understanding Analysis answers. a rational number. … Understanding Analysis is perfectly titled; if your students read it, that’s what’s going to happen. there exists a rational numbersuch It is good to have other approach. What a wonderful platform to learn Linear Algebra and Real Analysis from. Our solutions are written by Chegg experts so you can be assured of the highest quality! The Spectrum Analysis section of the Monitoring tab in the WebUI includes the Spectrum Monitors, Session Log, and Spectrum Dashboards windows. Thus Rei = aiei for all i = 1,... nthen RRei = ai^2 * ei. Therefore, the required result has been proved. Stephen Abbott. YES! This site has been helpful and I wish to contribute something back. This site has been invaluable to me learning (by self-study) Axler's linear algebra. YES! It is hard to say which is good or bad. Good job. Sorry, I don't quite understand your logic. You dropped the second j. By the way, 7B.1 should be $Te_1 = e_1$, $Te_2 = 2e_2+e_1$. Took me quite a while to get the point. Stephen Abbott. In fact, It can be easier once you show that the eigenvectors of R (the square root) is also the eigenvectors of T (the eigenvalue of T is the square of the eigenvalue of R with respect to the same eigenvector), which is quite obvious. The logic of the proof from the book is quite clear to me. Given any real numbers, Understanding Analysis (Undergraduate Texts in Mathematics) | 1st Edition, Understanding Analysis (Undergraduate Texts in Mathematics). If you really wish to donate, please donate to US PayPal COVID-19 Relief Fund. Wow! and the initial condition. Understanding Analysis is so well-written and the development of the theory so well-motivated t. hat exposing students to it could well lead them to expect such excellence in all their textbooks. 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Thus we know that R and R1 are identical, only their presentation in the matrix form may be slightly different. that the following condition holds well: Add This terrific book will become the text of choice for the single-variable introductory analysis course; take a … a natural number andis Access Understanding Analysis 2nd Edition Chapter 1.4 Problem 3E solution now. http://linear.axler.net. The textbook is "Understanding Analysis" by Stephen Abbott. Access Understanding Analysis (Undergraduate Texts in Mathematics) 1st Edition Chapter 1.4 solutions now. Now is the time to redefine your true self using Slader’s Understanding Analysis answers. Understanding Analysis is so well-written and the development of the theory so well-motivated that exposing students to it could well lead them to expect such excellence in all their textbooks. JavaScript is required to view textbook solutions. The problem asks to show the square root is unique. 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Complex conjugate of function

I have a wavefunction $\psi(x,t)=Ae^{i(kx-\omega t)}+ Be^{-i(kx+\omega t)}$. $A$ and $B$ are complex constants.

I am trying to find the probability density, so I need to find the product of $\psi$ with it's complex conjugate. The problem is, im not sure what is it's complex conjugate, I know the complex conjugate of $5+4i$ is $5-4i$, but what would be the complex conjugate of $\psi$? Is it just $-Ae^{i(kx-\omega t)}-Be^{-i(kx+\omega t)}$?

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The complex conjugation factors through sums and products. So you can take the complex conjugate of the factor with A and B separately. The constant A and B form know problem, this goes according to the usual rules. This leaves something of the form $e^{(a+bi)}$. Now note that $e^{(a+bi)}= e^a(\cos(b)+i \sin(b))$ Taking the complex conjugate now and using $\cos(b)=-\cos(b)$ and $-\sin(b)=\sin(-b)$, you find the complex conjugate $e^{a+i(-b)}$.

This means: $\bar{\psi} = \bar{A} \mathrm{e}^{-i (k x - \omega t)} + \bar{B} \mathrm{e}^{i (k x + \omega t)}$

Note the use of the minus sign to compactly write the complex conjugate of $e^{a+ib}$. In computation this is what write, but you might want to keep the explanation of the $\cos$ and $\sin$ in the back of your head.

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It's easier to note once and for all that $\overline{e^z}=e^{\overline z}$, because the exponential function can be defined uniquely without committing to a choice between $i$ and $-i$ (e.g., by power series). –  Henning Makholm Jan 1 '12 at 18:42
Thanks, so I want to find $\psi\bar{\psi}$. Do I just multiply it out like this? $(Ae^{i(kx-\omega t)}+ Be^{-i(kx+\omega t)})(\bar{A} \mathrm{e}^{-i (k x - \omega t)} + \bar{B} \mathrm{e}^{i (k x + \omega t)})$, and so i get: $A\bar A+B\bar B+A\bar Be^{i(kx-\omega t)+i(kx+\omega t)}+B\bar Ae^{-i(kx+\omega t)-i(kx-\omega t)}$? Is that correct? –  Thomas Jan 1 '12 at 18:46
Yes, if needed you can further simply to: $A\bar A+B\bar B+A\bar Be^{2ikx}+B\bar Ae^{-2ikx}$ –  MrOperator Jan 1 '12 at 18:52

The complex conjugation will map $A \to \bar{A}$ and $B \to \bar{B}$.

If, say $A= 5 + 4 i$, then $\bar{A} = 5 - 4 i$, as you noted. So $$\bar{\psi} = \bar{A} \mathrm{e}^{-i (k x - \omega t)} + \bar{B} \mathrm{e}^{i (k x + \omega t)}$$

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# Show, that the sequence $a_n = \left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{9}\right)\cdot\ldots\cdot\left(1+\frac{1}{3^n}\right)$ converges.

I need to show, that the following sequence converges. I think I can somehow do it by Riemann Integral, but I cannot figure out a way to extract $$\frac{1}{n}$$ from it. I also cannot find two sequences which could let me show that it converges by the sandwich theorem. How to approach it?

$$a_n = \left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{9}\right)\cdot\ldots\cdot\left(1+\frac{1}{3^n}\right)$$

• What about taking the logarithm ?
– user65203
Feb 10 '19 at 16:54
• The sequence is monotone. Is it bounded? Feb 10 '19 at 16:55

If you aplied logarithm you get that $$\log{a_n} = \sum_{n=1}^{\infty} \log{(1+\frac{1}{3^n})}$$. Now using that $$\log{x} \leq x-1$$ for $$x>0$$ you get that $$\log{a_n} \leq \sum_{n=1}^{\infty} \frac{1}{3^n}$$ that is a geometric series that converges, so as $$\log{a_n}$$ converges, then $$a_n$$ converges.

• How did you get from $\log{a_n} = \sum_{n=1}^{\infty} \log{1+\frac{1}{3^n}}$ to $\log{a_n} \leq \sum_{n=1}^{\infty} \frac{1}{3^n}$ ? I don't understand that part Feb 10 '19 at 17:10
• $\log{x} \leq x-1$ is the same as $\log{(1+x)}\leq x$, so $\log{(1+3^{-n})} \leq 3^{-n}$ Feb 10 '19 at 17:12
• Could you please use brackets? I don't know if you mean $log(1)+x$ or $log(1+x)$ Feb 10 '19 at 17:14
• Already edit, I meant $\log{(1+x)}$ Feb 10 '19 at 17:16
• Thank you very much. Could you please show me the path from $log(x) \leq x - 1$ to $log(1+x) \leq x$? Feb 10 '19 at 17:19

If you are ok with using GM-AM (inequality between geometric and arithmetic mean) and with using the well known limit

• $$\left(1+ \frac{x}{n}\right)^n\stackrel{n\to \infty}{\longrightarrow}e^x$$

then you can reason directly as follows: $$\begin{eqnarray*} \prod_{k=1}^n \left(1+\frac{1}{3^k}\right) & \leq & \left(\frac{\sum_{k=1}^n \left(1+\frac{1}{3^k}\right) }{n} \right)^n\\ & = & \left(\frac{n + \sum_{k=1}^n \frac{1}{3^k}}{n} \right)^n\\ & \leq & \left(1 + \frac{\sum_{k=1}^{\infty} \frac{1}{3^k}}{n} \right)^n \\ & = & \left(1 + \frac{\frac{1}{2}}{n} \right)^n \\ & \stackrel{n \to \infty}{\longrightarrow} & \sqrt{e} \\ \end{eqnarray*}$$

Since $$a_n$$ is obviously increasing and according to above calculation also bounded, it follows that $$a_n$$ is also convergent.

More generally, let $$a_n =\prod_{k=1}^n (1+c^k)$$ where $$0 < c < 1$$. This problem is $$c = \frac13$$.

Let $$b_n = \ln(a_n) =\sum_{k=1}^n \ln(1+c^k)$$.

Then $$b_n$$ is increasing and, since $$\ln(1+x) < x$$ for $$x > 0$$, $$b_n \lt \sum_{k=1}^n c^k =\dfrac{c-c^{n+1}}{1-c} \lt\dfrac{c}{1-c}$$ for all $$n$$.

Therefore $$b_n$$ is a bounded, monotone increasing sequence and so converges.

Therefore $$a_n = e^{b_n}$$ also converges.

Another proof can be given using Cauchy's criterion.

If $$n > m$$, $$b_n-b_m =\sum_{k=m+1}^n \ln(1+c^k) \lt\sum_{k=m+1}^n c^k =\dfrac{c^{m+1}-c^n}{1-c} \lt\dfrac{c^{m+1}}{1-c}$$.

This last can be made as small as desired by choosing $$m$$ large enough.

Note: To show that $$\ln(1+x) < x$$ for $$x > 0$$:

$$\ln(1+x) =\int_1^{1+x}\dfrac{dt}{t} =\int_0^{x}\dfrac{dt}{1+t} \lt\int_0^{x}dt =x$$.


# Math Help - Urgent algebra homework! Xmas stuff... :D

1. ## Urgent algebra homework! Xmas stuff... :D

Despite the season being over I am still left puzzled over this problem.

The following questions are related to the popular Christmas song "The Twelve Days of Christmas".

We all know it's fairly simple to figure out the amount of presents found over the twelve days of Christmas. It is 364 presents.

How many ways did you find to solve for 364 presents?

Tough what about Twelve Days of Christmas? How many presents would we receive then?
And the toughest of them all THE nth day of Christmas? How on Earth are we supposed to explain and work out the nth day of Christmas?

Help needed.
Thanks!

2. Originally Posted by chocole
Despite the season being over I am still left puzzled over this problem.

The following questions are related to the popular Christmas song "The Twelve Days of Christmas".

We all know it's fairly simple to figure out the amount of presents found over the twelve days of Christmas. It is 364 presents.

How many ways did you find to solve for 364 presents?

Tough what about Twelve Days of Christmas? How many presents would we receive then?
And the toughest of them all THE nth day of Christmas? How on Earth are we supposed to explain and work out the nth day of Christmas?

Help needed.
Thanks!

Tiina
Maybe I am not familiar with the version you know. If you get 1 present on the first day, 2 presents on the 2nd day, and n presents on the nth day, then your total is:

$\sum_{i=0}^n i = \frac{n(n+1)}{2}$

3. Originally Posted by colby2152
Maybe I am not familiar with the version you know. If you get 1 present on the first day, 2 presents on the 2nd day, and n presents on the nth day, then your total is:

$\sum_0^n i = \frac{n(n+1)}{2}$
The problem states how many presents would be received over any other number (n) of days.

I am brain dead. I've never been this stuck... on this kind of problem. XD

4. Originally Posted by chocole
The problem states how many presents would be received over any other number (n) of days.

I am brain dead. I've never been this stuck... on this kind of problem. XD
The formula for the sum of an arithmetic series is:

$S_{n} = \frac{n}{2} \left( 2a + (n - 1)d \right)$

a = 1 and d = 1

So:

$S_{n} = \frac{n}{2} \left( 2 + (n - 1) \right)$

$S_{n} = \frac{n}{2} \left( n + 1 \right)$

$S_{n} = \frac{n^2 + n}{2}$

Where n = the number of days.

So on the 1st day you would have 1 present, on the 2nd day, you would receive 3 presents, etc...

5. Hello, chocole!

The following questions are related to the Christmas song "The Twelve Days of Christmas".

We all know how to figure out the number of presents given over the 12 days.
It is 364 presents.

How many ways did you find to solve for 364 presents?

$\begin{array}{cccc}\text{Day} & & \text{Presents} \\
1 &1 & 1 \\
2 & 1+2 & 3 \\
3 & 1+2+3 & 6 \\
4 & 1+2+3+4 & 10 \\
\vdots & \vdots & \vdots
\end{array}$
. These are "triangular numbers."

On the $n^{th}$ day, my true love gave to me: . $\frac{n(n+1)}{2}$ gifts.

By the $n^{th}$ day, my true love gave to me a total of:

. . $\sum^n_{k=1}\frac{k(k+1)}{2} \;=\;\frac{n(n+1)(n+2)}{6}$ gifts.


# Conditional probability that standard gaussian random variable is larger than another

Say there are two standard Gaussian random variables $$X$$ and $$Y$$. I am trying to evaluate the probability that the larger of the two is selected, given that it is known whether $$X$$ is positive or negative (the strategy is selecting $$X$$ if $$X$$ is positive and selecting $$Y$$ if $$X$$ is negative). In equation form this is $$Pr(X-Y>0|X>0) + Pr(X-Y<0|X<0)$$

How can this expression be evaluated? Numerically it appears to be $$\frac{3}{4}$$, and intuitively this makes sense.

I am also interested in this probability in the more general case, where the strategy involves selecting $$X$$ if $$X-S>0$$ and $$Y$$ otherwise, where $$S$$ is another independent Gaussian random variable.

• I don't understand your question. Did you mean $\text{Pr}(X-Y>0|X>0)\text{Pr}(X>0)+\text{Pr}(X-Y<0|X<0)\text{Pr}(X<0)$? – Angela Pretorius Jun 5 '19 at 4:32
• Yes, that's right! – tankerjeel Jun 5 '19 at 20:07

The following assumes $$X$$ and $$Y$$ are independent.

As Angela Richardson pointed out, you probably actually want to compute $$P(X-Y > 0 \mid X > 0) P(X>0) + P(X-Y <0 \mid X< 0) P(X<0)$$ which is $$3/4$$. (The quantity in your post is not $$3/4$$.)

For the first term, it suffices to compute $$P(X-Y > 0, X>0)$$. (Why?)

Consider the region of the plane that contains $$(x,y)$$ pairs satisfying $$x-y>0$$ and $$x>0$$. Then use rotational symmetry of the vector $$(X,Y)$$ to compute the probability.

The other term can be handled similarly.

• Thanks. Is there a way to evaluate the more general case as well? I'm guessing symmetry no longer holds once $S$ is introduced. – tankerjeel Jun 5 '19 at 20:09
• In case anyone is curious, the stackexchange post in this link gives the solution as to how to compute the more general case. When $S$ is the standard normal, the probability turns out to be exactly $\frac{2}{3}$, and approaches $\frac{3}{4}$ as the variance of $S$ approaches $0$. – tankerjeel Jun 5 '19 at 23:57


# Thread: Volume of sphere with hole

1. ## Volume of sphere with hole

A hole of diameter d is drilled through a sphere of radius r in such a way that the axis of the hole passes through the centre of the sphere. Find the volume of the solid that remains.

I drew a picture of a sphere centred on the cartesian plane, with a hole going through its centre (in the shape of a cylinder).

I already knew that the volume of the sphere was $\frac {4}{3}\pi r^{3}$. I then worked out the volume of the hole (cylinder):
$
\int_{-r}^{r} \frac{1}{4}\pi d^{2}dy$

$= \frac{1}{2}\pi d^{2} r
$

Volume of solid remaining = $\frac {4}{3}\pi r^{3} - \frac{1}{2}\pi d^{2} r$

but the correct answer was $\frac{1}{6}\pi (4r^{2} - d^{2})^{\frac{3}{2}}$

I cant see where I went wrong cause surely the objective of this problem is to minus the hole (cylinder) from a sphere to get the remaining volume?

2. The volume subtracted out by drilling the hole is not precisely a cylinder, because a cylinder usually has flat ends. The material removed has curved ends (matches up with the sphere). So what do you get when you change this?

3. Very intresting problem. It sounds like Ackbeet is right in the fact that the figure removed is not
Spoiler:
only
a cylinder.

Good Luck!

4. Rather than subtracting the round-ended cylinder from the volume of the sphere, I think it's probably easier to set up the integral for the volume. You can use cylindrical shell elements of width $dx$, circumference $2 \pi x$, and height $2 \sqrt {R^2 - x^2}$, and integrate from x = d/2 to x = R.

5. I got the integral $V=-2\pi\int_{r^{2}-\frac{d^{2}}{4}}^{0} \sqrt{u} du$ after making the substitution of $u=r^{2} - x^{2}$ then my final answer was:

$\frac{4}{3}\pi(r^{2}-\frac{d^{2}}{4})^{\frac{3}{2}}$. Did I do something wrong or is there any way to simplify it to $\frac{1}{6}\pi(4r^{2} -d^{2)^\frac{3}{2}$ ?

6. They are the same answer. Proof:

$\displaystyle{\frac{4}{3}\pi(r^{2}-\frac{d^{2}}{4})^{\frac{3}{2}}}$

$\displaystyle{=\frac{4}{3}\pi(4r^{2}-d^{2})^{\frac{3}{2}}\left(\frac{1}{4}\right)^{\fra c{3}{2}}}$

$\displaystyle{=\frac{4}{3}\pi(4r^{2}-d^{2})^{\frac{3}{2}}\,\frac{1}{4^{3/2}}}$

$\displaystyle{=\frac{1}{8}\,\frac{4}{3}\pi(4r^{2}-d^{2})^{\frac{3}{2}}}$

$\displaystyle{=\frac{1}{6}\pi(4r^{2}-d^{2})^{\frac{3}{2}}.}$


## Section1.2Statements

Statements are declarative sentences that are either true or false. The statements are formulated in such a way that any reader, who knows what all the words mean, can understand them.

1. “Victoria likes cookies.” is a declarative sentence, and it is either true or false, so it is a statement.

2. “Broccoli is green.” is a declarative sentence and it is true, so it is a statement.

3. “Broccoli is pink.” is a declarative sentence and it is false, so it is a statement.

4. “Cookies!” is not a declarative sentence, so it is not a statement.

From now on we concentrate on statements about the integers.

Consider the following:

1. “2 is equal to 3.” is a statement. It is false.

2. “2 plus 3 is equal to 5.” is a statement. It is true.

3. “2 plus 3” is not a statement, as it is not a declarative sentence; it is not even a sentence, as it does not contain a verb.

When we write a statement using the symbols $$=\text{,}$$ $$\ne\text{,}$$ $$\lt\text{,}$$ $$\le\text{,}$$ $$>\text{,}$$ or $$\ge\text{,}$$ the comparison symbol takes the place of the verb. A mathematical statement always has a verb or a symbol that takes the place of the verb, just as a sentence does.

A mathematical expression consists of objects and operations. The objects can be numbers or variables (see the next section) and the operations can be, for example $$+\text{,}$$ $$\cdot\text{,}$$ or $$-\text{.}$$ Unlike a statement, an expression has no comparison symbol, that means it has no “verb.” So expressions by themselves are not true or false, but expressions can be used in statements, as in Example 1.1.6.

We formulate Example 1.2.2 using symbols.

1. “2 = 3” is a statement. It is false.

2. “2 + 3 = 5” is a statement. It is true.

3. “2 + 3” is an expression. As it does not have a verb it is not a statement.

We identify whether statements about integers are true or false. Notice that all these examples when read out have a verb in them, namely “is”.

1. $$2=2$$ is read “2 is equal to 2”. This is a true statement.

2. $$2=3$$ is read “2 is equal to 3”. This is a false statement.

3. $$2\gt 3$$ is read “2 is greater than”. This is a false statement.

4. $$2\ne 3$$ is read “2 is not equal to 3”. This is a true statement.

5. $$2\le 2$$ is read “2 is less than or equal to 2”. This is a true statement.

We give some examples of expressions and statements and identify them.

1. $$2 + 3$$” is an expression.

2. $$2+3 = 5$$” is a statement.

3. $$2 + 1 + 5$$” is an expression.

4. $$2+ 1 + 5 \lt 10$$” is a statement.

Decide whether the following are statements or not. If they are statements decide whether they are true or false.

1. “Sunflower”

2. “Stop signs are red.”

3. $$2$$ is equal to $$3\text{.}$$

4. $$\displaystyle (1+2)-4687$$

5. $$\displaystyle 2+3=7$$

6. $$\displaystyle 3 > -100$$

Solution.
1. “Sunflower” is not a sentence, so it is not a statement.

2. “Stop signs are red.” is a declarative sentence, so it is a statement. It is true.

3. “2 is equal to 3” is a declarative sentence, so it is a statement. As $$2\ne 3$$ the statement is false.

4. $$(1+2)-4687$$ is not a statement as it has no verb.

5. $$2+3=7$$ is a statement, the verb is ‘=’ (is equal to). As $$2+3=5$$ it is a false statement.

6. $$3 > -100$$ is a statement, the verb is “$$>$$” (is greater than). It is a true statement.

When a statement is true, we usually do not write “is true.” When a statement is false, always write “is false”.

In Checkpoint 1.2.7 recognize statements and for statements decide whether they are true or false.

Determine which of the following are mathematical statements.

For the statements decide whether they are true or false.

1. $$\displaystyle 13\le 38$$

2. $$\displaystyle -11-38$$

3. $$\displaystyle 13\cdot 38$$

4. $$\displaystyle 13\ne 38$$

### Subsection1.2.2Compound Statements

In mathematics we often deal with multiple statements that overlap. In these cases instead of writing each statement separately, we often write them as one string of statements. This allows us to connect the statements directly.

Instead of writing “$$2 + 3 = 5$$” and “$$5 = 1+4\text{,}$$” we write “$$2 + 3 = 5 = 1+4\text{.}$$

We can also do this with inequalities.

Writing “$$2+5 = 7 \lt 10$$” means both “2+5 = 7” and “$$7 \lt 10\text{.}$$” In words, “$$2$$ plus $$5$$ is $$7$$ and $$7$$ is less than $$10\text{.}$$

Compound statements are often used to prove identities, that is, when proving that two expressions are equal. The proofs of Theorem 1.4.6 and Theorem 1.4.7 and Theorem 1.4.12 in the next chapter is written that way.


Derivation of $y=a(x-h)^2+k$ from $y=ax^2+bx+c$ given a vertex and a point

Derive $y=a(x-h)^2+k$ from $y=ax^2+bx+c$ given a vertex and a point.

Recently I have been solving a problem to which I could not find a solution. Google search of "quadratic equation given vertex and a point," yielded what I have been looking for. However, although I have already solved the problem, I am still wondering how to derive the equation I was looking for before I googled it.

I was given a vertex $V(-3, -2)$ and a point $P(-4, 0)$ of a parabola. Using $y=ax^2+bx+c$, I have derived an equation for finding the vertex for each parabola with $V(\frac{-b}{2a}, -a(\frac{b}{2a})^2+c)$.

I knew that given the same vertex, the parabola $y=x^2+6x+7$ was close to what I've been looking for. However, this parabola didn't go through $P(-4, 0)$, because it was too wide.

At this point it seemed as if I had enough information to derive an equation, that given a vertex and a point I would obtain a parabola according to the restrictions. However, from this point on I didn't know how to proceed further.

• Note that by symmetry, $(-2,0)$ is also a point on the parabola. – user137731 Jul 31 '16 at 23:53
• Is your goal to determine an equation of the parabola that has vertex $V(-3, -2)$ and passes through the point $P(-4, 0)$? – N. F. Taussig Jul 31 '16 at 23:56
• @N.F.Taussig Yes. – user270346 Jul 31 '16 at 23:56
• I was looking for a general equation, that given the vertex and a point of a parabola I could determine it's coefficients $a, b, c$ as in $y=ax^2+bx+c$. – user270346 Jul 31 '16 at 23:57
• Use the two points you're given and the one you get from symmetry. Plug each in to get a system of linear equations that you can hopefully then solve. – user137731 Jul 31 '16 at 23:58

$$y = ax^2 + bx + c$$

$$y - c = a\left(x^2 + \frac{b}{a}x\right)$$

$$y - c + a\frac{b^2}{4a^2}= a\left(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2}\right)$$

$$y - c + \frac{b^2}{4a}= a\left(x+\frac{b}{2a}\right)^2$$

$$y = a\left(x-\left(-\frac{b}{2a}\right)\right)^2 + \left(c- \frac{b^2}{4a}\right)$$

$$y = a\left(x-h\right)^2 + k$$

where $h = -\frac{b}{2a}$ and $k = c- \frac{b^2}{4a}$

If you're going in reverse (going from vertex $(h,k)$ and point $(x,y)$ to quadratic parameters $a,b,c$), then you can take the last few equations, isolate $a,b,c$, and translate them into terms of $h,k,x,y$:

$$a = \frac{y-k}{(h-x)^2}$$

$$b = -2ha = \frac{-2h(y-k)}{(h-x)^2}$$

$$c = k + \frac{b^2}{4a} = k + \frac{h^2(y-k)}{(h-x)^2}$$

• This answer followed by the @N. F. Taussig answer is the order what makes most sense to me after picking up on this problem from where I left it. Thanks! – user270346 Aug 1 '16 at 0:35
• @Marcus Stuhr, Regarding your edit, note that the parabola $y=x^2+6x+7$ doesn't have roots $x=-4$ and $x=-2$ as required. – user270346 Aug 1 '16 at 0:47
• I've done as you've requested. Please, take your time. Thanks. – user270346 Aug 1 '16 at 0:55
• @Matt I believe you're looking for $2x^2 + 12x + 16 = 0$, which goes through $(-4,0)$ and $(-2,0)$ and has vertex $(-3,-2)$. – Marcus Andrews Aug 1 '16 at 1:15

Since you know that the parabola has vertex $V(-3, -2)$, the vertex form of its equation is $$y = a[x - (-3)]^2 + (-2) = a(x + 3)^2 - 2$$ Since the parabola passes through the point $P(-4, 0)$, we can substitute $-4$ for $x$ and $0$ for $y$ to determine $a$. \begin{align*} 0 & = a(-4 + 3)^2 - 2\\ 2 & = a(-1)^2\\ 2 & = a \end{align*} Hence, $y = 2(x + 3)^2 - 2$. You can expand this expression to obtain the standard form of the equation.

Addendum: In general, if you know the vertex $V(h, k)$ and a point $P(u, v) \neq V(h, k)$ on the parabola, you can write $$y = a(x - h)^2 + k$$ then substitute $u$ for $x$ and $v$ for $y$ to determine $a$. \begin{align*} v & = a(u - h)^2 + k\\ v - k & = a(u - h)^2\\ \frac{v - k}{(u - h)^2} & = a \end{align*} Then $$y = a(x - h)^2 + k = \frac{v - k}{(u - h)^2}(x - h)^2 + k$$ Again, expanding the expression to obtain its standard form enables you to determine the coefficients $a$, $b$, and $c$.

• Thanks for your answer, however that's not my question. Please forgive me if I was not clear enough. I was looking for a general equation, that given the vertex and a point of a parabola I could determine it's coefficients $a, b, c$ as in $y=ax^2+bx+c$, based on what I knew. So, it's more about how do you derive the general equation for solving this type of problems, rather than solving the problem itself. – user270346 Aug 1 '16 at 0:04
• This is especially useful due to the other point suggested by the symmetry. Thanks for your answer. – user270346 Aug 1 '16 at 0:37
• On a side note, do you have any idea as why the vertex equation for this parabola $y=2(x+3)^2-2$ and $y=x^2+6x+7$ following from it, don't have same roots? – user270346 Aug 1 '16 at 0:51
• If you expand $y = 2(x + 3)^2 - 2$, you obtain \begin{align} y & = 2(x + 3)^2 - 2\\ & = 2(x^2 + 6x + 9) - 2\\ & = 2x^2 + 12x + 18 - 2\\ & = 2x^2 + 12x + 16 \end{align} You forgot to multiply the terms in the parentheses by $2$. As you can verify, both $y = 2(x + 3)^2 - 2$ and $y = 2x^2 + 12x + 16$ have the roots $-4$ and $-2$. – N. F. Taussig Aug 1 '16 at 1:00

Generalized Form:

Given $y=ax^2+bx+c$, we can move the loose number $c$ to the other side and try completing the square! So from that, we get $$y-c=ax^2+bx\tag{1}$$

Factoring out $a$ gives us $y-c=a\left(x^2+\frac bax\right)$. Completing the square by dividing the coefficient of the $x$ term by $2$ and squaring it gives us $$y-c+a\left(\frac {b^2}{4a^2}\right)=a\left(x+\frac {b}{2a}\right)^2\tag{2}$$

Simplifying the left hand side and move it to the right hand side to obtain $$y=a\left(x-\left(-\frac {b}{2a}\right)^2\right)+\left(\frac {4ac-b^2}{4a}\right)\tag{3}$$

And since the Vertex follows the formula $(h,k)=\left(-\frac {b}{2a},\frac {4ac-b^2}{4a}\right)$, you get the Vertex formula by plugging it in.


Sum of $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^{2n-1}}{2n-1}$

I've been working with the series:

$$\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^{2n-1}}{2n-1}$$

From the ratio test it is clear that the series converges for $|x| < 1$, but I'm unable to obtain the sum of the series.

I'm looking for any hint of how to obtain the sum.

• Differentiate by x, use the geometric sum, integrate, and use that f(0)=0. Result is arctan(x) – Ákos Somogyi Aug 25 '15 at 17:11

Let $f(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^{2n-1}}{2n-1}$. Then we have

\begin{align} f'(x)=\sum_{n=1}^{\infty} (-1)^{n+1}x^{2n-2}&=\frac{-1}{x^2}\sum_{n=1}^{\infty} (-x^2)^{n}\\\\ &=\frac{1}{1+x^2} \tag 1 \end{align}

Integrating $(1)$ and using $f(0)=0$ reveals that

$$\bbox[5px,border:2px solid #C0A000]{f(x)=\arctan(x)}$$

• (+1) Similar to the approach I was going to take. I was going to note that it was the same as $$\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{2n+1}$$ and then show that the derivative of that was $\frac1{1+x^2}$ – robjohn Aug 25 '15 at 17:46
• @robjohn Thanks! I skipped any discussion of interval of convergence and also omitted discussing the legitimacy of differentiating term by term since it appeared that the OP was already aware of at least the first of these issued. – Mark Viola Aug 25 '15 at 17:50

Let $y$ be our sum so:

$$y=\sum_{n=1}^{\infty}{{(-1)}^{n+1}\frac{{x}^{2n-1}}{2n-1}}$$

Let's differentiate it to get:

$$y'=\sum_{n=1}^{\infty}{{(-1)}^{n+1}{x}^{2n-2}}\\ y'=\sum_{n=1}^{\infty}{{(-1)}^{n-1}{x}^{2n-2}}\\ y'=\sum_{n=1}^{\infty}{{(-1)}^{n+1}{(-x^2)}^{n-1}}\\ y'=\frac{1}{1+x^2}$$

Now integrate to get the original sum to get:

$$y=\arctan{x} +C$$

It is easy to see that for $x=0$ we have $y=0$, so $C=0$, hence the sum equals:

$$\sum_{n=1}^{\infty}{{(-1)}^{n+1}\frac{{x}^{2n-1}}{2n-1}}=\arctan{x}$$

• Isn't that just a repeat of what's already posted? – Macavity Aug 25 '15 at 17:40
• I was typing my answer at the same time that's why I didn't notice the other answer – Oussama Boussif Aug 25 '15 at 17:43
• (+1) good answer. I've had it happen many times that a similar answer has been posted while I am typing up mine. In fact, it happened here, but I happened to see Dr. MV's answer before I posted. – robjohn Aug 25 '15 at 17:49

$$(-1)^{n+1}\dfrac{x^{2n-1}}{2n-1}=i^{2(n+1)}\cdot\dfrac{x^{2n-1}}{2n-1}=-i\cdot\dfrac{(ix)^{2n-1}}{2n-1}$$

If $S=\sum_{n=1}^\infty(-1)^{n+1}\dfrac{x^{2n-1}}{2n-1},$

$$i S=\sum_{n=1}^\infty(-1)^{n+1}\dfrac{(ix)^{2n-1}}{2n-1}$$

Now for $-1<y\le1,\ln(1+y)=y-\dfrac{y^2}2+\dfrac{y^3}3-\dfrac{y^4}4+\cdots$

$\ln(1-y)=-y-\dfrac{y^2}2-\dfrac{y^3}3-\dfrac{y^4}4-\cdots$

$\ln(1+y)-\ln(1-y)=?$

$$\implies2i S=\ln(1+ix)-\ln(1-ix)=\ln\dfrac{1+ix}{1-ix}$$

Let $1=r\cos A,x=r\sin A, x=\tan A$

Now, $$\ln\dfrac{1+ix}{1-ix}=\ln(e^{2iA})=2iA=2i\arctan x$$

• I know what you're trying to show, but the appearance of $x$ and $y$ and the fact that I think the sum for $iS$ looks like it has too many $i^{2n-1}$ in it, makes things a bit hard to follow. – robjohn Aug 25 '15 at 18:02