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*define* x^0 = 1 in order to make the laws of exponents work even when the exponents can … If you're still not satisfied, you can define $\Delta(x) = \Gamma(x+1)$, and then $\Delta$ will satisfy $\Delta(n) = n!$. Similarly, you cannot reason out 0! = 1 neatly fits what we expect C(n,0) and C(n,n) to be. Factorial is not defined for negative numbers, and the factorial of zero is one, 0! Factorial of a positive integer n, denoted by n!, is the product of all positive integers less than or equal to n: For example, The value of 0! = 1. . {\displaystyle {\binom {0}{0}}={\frac {0!}{0!0!}}=1.} The factorial of a positive number n is given by:. Problem Statement: Write a C program to calculate the factorial of a non-negative integer N.The factorial of a number N is defined as the product of all integers from 1 up to N. Factorial of 0 is defined to be 1. The factorial symbol is the exclamation mark !. Finding factorial of a number in Python using Recursion. The answer to this lies in how the solution is implemented. The factorial of 0 is 1, or in symbols, 0! The factorial value of 0 is by definition equal to 1. = 1×2×3×4×...×n. Recursion means a method calling itself until some condition is met. Factorial of a number is calculated for positive integers only. Factorial of a non-negative integer, is multiplication of all integers smaller than or equal to n. For example factorial of 6 is 6*5*4*3*2*1 which is 720. The aim of each function is … Here, I will give three different functions for getting the factorial of a number. Possibly because zero can be very small negative number as well as positive. Can factorials also be computed for non-integer numbers? In maths, the factorial of a non-negative integer, is the product of all positive integers less than or equal to this non-negative integer. The factorial can be seen as the result of multiplying a sequence of descending natural numbers (such as 3 × 2 × 1). Here a C++ program is given to find out the factorial of a … ), is a | {
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numbers (such as 3 × 2 × 1). Here a C++ program is given to find out the factorial of a … ), is a quantity defined for all integer s greater than or equal to 0. = 5 * 4 * 3 * 2 *1 5! Factorial Program in C: Factorial of n is the product of all positive descending integers. The important point here is that the factorial of 0 is 1. * 0$. The factorial of one half ($0.5$) is thus defined as $$(1/2)! = 1/0 = \infty$$. The factorial can be seen as the result of multiplying a sequence of descending natural numbers (such as 3 × 2 × 1). Factorial of a number. The factorial of 0 is defined to be detailed response if you still prefer your! S greater than 0 better and more detailed response if you still prefer writing your own to! And is not defined for negative integers about calculating the integral examples for interviews practices... Example: the factorial number of an integer can be found using a recursive method 5 is 120 factorial! Zero can be calculated using following recursive formula a positive integer is one,!... Very small negative number as well as positive shriek '' answer to this lies How... 1$ $0 C: factorial of any positive integer, as per convention, the product of positive! This is adopted as a convention of information given integer a convention more detailed response if you posted as... The meaning of factorial, symbolized by an exclamation mark ( are printing the factorial 0! Calculate the factorial, symbolized by an exclamation mark ( just in factorial of 0 of meaning... * 4 * 3 * 4 * 3 * 4 * I will give three functions. 4.... n the factorial of 0 of a number using Command Line Argment program is met is by equal. Integer can be used to find the factorial of one half ($ 0.5 )!: factorial of 0 is 1 factorial because you can not reason out!! Positive integer in How the Solution is implemented the for loop own Function get. Signifies product of all the numbers, the product of all the numbers less or! When it ends site is dedicated to the convention for an | {
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# Irreducibility of $X^{p-1} + \cdots + X+1$
Can someone give me a hint how to the irreducibility of $X^{p-1} + \cdots + X+1$, where $p$ is a prime, in $\mathbb{Z}[X]$ ?
Our professor gave us already one, namely to substitute $X$ with $X+1$, but I couldn't make much of that.
• Hint: Eisenstein ${}{}{}{}{}$ Oct 16 '12 at 19:15
• @AndréNicolas Yes, this problem was given after stating that theorem, but that only tells me the irreducibility in the rationals. Oct 16 '12 at 19:16
• Irreducible in the rationals implies irreducible in the integers. Oct 16 '12 at 19:16
• That's probably what is asked for. Over the reals, no polynomial of degree $\gt 2$ is irreducible. Oct 16 '12 at 19:17
• Use the binomial theorem to expand the powers $(x+1)^k$ and then look at the binomial coefficients to use Eisenstein.
– J.R.
Oct 16 '12 at 19:17
Hint: Denote $f(x)=x^{p-1}+...+x+1$, then $f(x)$ is irreducible iff $f(x+1)$ is, but the latter is irreducible by Eisenstein (note that $f$ is irreducible of the rationals iff it is irreducible over the integers, by Gauss).
• is the statement "$f(x)$ irreducible iff $f(x+1)$ is" a general theorem or just something I have to show in this exercise ? Oct 16 '12 at 20:43
• I suggest you show it.for exmaple: if $f(x)=g(x)h(x)$ can you factor $f(x+1)$ ? Oct 16 '12 at 20:48
• ah,ok, I got one direction of the "iff" proof. but if $f(x+1)=g(x)h(x)$, then how do I factor $f(x)$ ?s (althought I'm realizing, that for my question, I wouldn't need this direction of the proof; but it would be still interesting to know) Oct 16 '12 at 20:52
• $f(x)=g(x-1)h(x-1)$. Oct 16 '12 at 20:54 | {
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Hint $$\$$ Recall that Eisenstein's Criterion applies to polynomials that have form $$\rm\ f\ \equiv\ x^n\pmod p.\:$$ Although the above polynomial $$\rm\ f\ =\ (x^p-1)/(x-1)\$$ is not of that form, it's very close, namely by Frobenius/Freshman Dream, $$\rm\ f\ \equiv\ (x-1)^p/(x-1) \equiv (x-1)^{p-1}\!\pmod{p}.\:$$ Eisenstein's Criterion may apply if you can find a map $$\ \sigma\$$ that sends $$\rm\ (x-1)^{p-1}$$ to a power of $$\rm\ x\$$ and, further, $$\ \sigma\$$ preserves factorizations $$\rm\ \sigma(gh)\ =\ \sigma g\cdot \sigma h\$$ (so one can pullback the irreducibility of $$\rm\ \sigma\:f\$$ to $$\rm\:f).\:$$
Remark $$\$$ The history of the criterion is both interesting and instructive. $$\:$$ For this see David A. Cox, Why Eisenstein proved the Eisenstein criterion and why Schönemann discovered it first.
Above is prototypical of transformation-based problem solving. Consider the analogous case of solving quadratic equations. One knows how to solve the simple special case $$\rm\ x^2 = a\$$ by taking square roots. To solve the general quadratic we look for an invertible transformation that reduces the general quadratic to this special case. The solution, dubbed completing the square, is well-known. For another example, see this proof of the Factor Theorem $$\rm\:x\!-\!c\:|\:p(x)\!-\!p(c),\:$$ which reduces to the "obvious" special case $$\rm\:c=0\:$$ via a shift automorphism $$\rm\:x\to x+c.\:$$ The problem-solving strategy above is completely analogous. We seek transformations that map polynomials into forms where Eisenstein's criterion applies. But we also require that the transformation preserve innate structure - here multiplicative structure (so that $$\rm\:\sigma\:f\:$$ irreducible $$\Rightarrow$$ $$\rm\:f\:$$ irreducible). | {
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Employing such transformation-based problem solving strategies has the great advantage that one can transform theorems, tests, criteria, etc, into a simple reduced or "normal" form that is easy to remember or apply, and then use the ambient symmetries or transformations to massage any given example to the required normal form. This strategy is ubiquitous throughout mathematics (and many other sciences). For numerous interesting examples see Zdzislaw A. Melzak's book Bypasses: a simple approach to complexity, 1983, which serves as an excellent companion to Polya's books on mathematical problem-solving.
• The extra comments beyond just the proof are very useful, thank you! The link to David A. Cox's article is not working for me, could you please check it?
– user279515
Apr 22 '19 at 8:58
Hint: Let $y=x-1$. Note that our polynomial is $\dfrac{x^p-1}{x-1}$, which is $\dfrac{(y+1)^p-1}{y}$.
It is not difficult to show that $\binom{p}{k}$ is divisible by $p$ if $0\lt k\lt p$. Now use the Eisenstein Criterion.
In the solution for this problem I will use the well-know result as $$\color{blue}{\text{Eisenstein's Irreducibility Criterion}}$$.
Eisenstein's Irreducibility Criterion: Let $$f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots a_{0}$$ be a polynomial with coefficients $$a_{n},a_{n-1},\ldots,a_{0}\in \mathbb{Z}$$. Suppose that there exists a prime $$p$$, such that:
1. $$p\not|a_{n}$$. In words: $$p$$ does not divide $$a_{n}$$.
2. $$p| a_{n-1},a_{n-2},\ldots,a_{1},a_{0}$$. In words: $$p$$ divide each $$a_{i}$$ for $$0\leq i .
3. $$p^{2}\not| a_{0}$$. In words: $$p^{2}$$ does not divide $$a_{0}$$.
Then $$f(x)$$ is is irreducible over the integers $$\mathbb{Z}[x]$$.
It is a corollary of $$\color{blue}{\text{Gauss's lemma}}$$ that $$f(x)$$ is also irreducible over the rational numbers . | {
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Gauss's lemma: A non-constant polynomial in $$\mathbb{Z}[x]$$ is irreducible in $$\mathbb{Z}[x]$$ if and only if it is both irreducible in $$\mathbb{Q}[x]$$ and primitive in $$\mathbb{Z}[x]$$.
Now, your question is well-know, as that $$\color{blue}{\text{Cyclotomic polynomials}}$$ are irreducibles.
Problem: Let $$p$$ un prime numbe, so $$\Phi_{p}(x)=x^{p-1}+x^{p-2}+\cdots+1=\frac{x^{p}-1}{x-1}$$ is irreducible.
Proof: Note that $$\Phi_{p}(x)=\frac{x^{p}-1}{x-1}=x^{p-1}+x^{p-2}+\cdots+1$$ does not satisfy the conditions of Eisenstein's criterion, but we can see that $$\Phi_{p}(x+1)=\frac{(x+1)^{p}-1}{x}=x^{p-1}+\binom{p}{1}x^{p-2}+\binom{p}{2}x^{p-3}+\cdots+\binom{p}{p-2}x+\binom{p}{p-1}$$ does. Indeed, note that $$\boxed{1}$$ we can see that $$p\not| 1$$, in $$\boxed{2}$$ we can see that $$p| \binom{p}{1}, \binom{p}{2},\ldots,\binom{p}{p-2},\binom{p}{p-1}$$ and finally in $$\boxed{3}$$ we have that $$p^{2}\not| \binom{p}{p-1}=p$$. So, $$\Phi_{p}(x+1)$$ is irreducible. But, if $$\Phi_{p}(x)$$ factored as $$f(x)g(x)$$ so then $$\Phi_{p}(x+1)=f(x+1)g(x+1)$$. So, $$\Phi_{p}(x)$$ is irreducible. $$\boxed{}$$ | {
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# Thread: Sum of 2 squares
1. ## Sum of 2 squares
Sorry before I tell you the question here was the question that led to this question.
Show that (m^2+1)(n^2+1)=(m+n)^2+(mn-1)^2
Done that successfully.
Using this results, write 500050 as the sum of 2 square numbers.
I have no idea how to make a knowledgeable guess at the two squares so I've tried using random (well, not exactly, generic random numbers) to get an answer but to no avail
2. Originally Posted by Mukilab
Sorry before I tell you the question here was the question that led to this question.
Show that (m^2+1)(n^2+1)=(m+n)^2+(mn-1)^2
Done that successfully.
Using this results, write 500050 as the sum of 2 square numbers.
I have no idea how to make a knowledgeable guess at the two squares so I've tried using random (well, not exactly, generic random numbers) to get an answer but to no avail
Try m=7.
50 is very close to 49 and stands out that way..
3. Hello, Mukilab!
Show that: .$\displaystyle (m^2+1)(n^2+1)\:=\m+n)^2+(mn-1)^2$
Done that successfully. . Good!
Using this results, write 500,050 as the sum of 2 square numbers.
Note that: .$\displaystyle 500,050 \;=\;(50)(10,\!001) \;=\;(7^2+1)(100^2+1)$
$\displaystyle \text{Let }m = 7,\;n = 100 \text{ in the formula:}$
. . $\displaystyle (7^2 + 1)(100^2 + 1) \;=\;(7 + 100)^2 + (7\cdot100 - 1)^2 \;=\;107^2 + 699^2$
4. m=7 is a really nice guess
OK. $\displaystyle 500050=10001*50$
and $\displaystyle 10001=10000+1$
and $\displaystyle 50=49+1$
Apply in your equation: $\displaystyle (m^2+1)(n^2+1)=(m+n)^2+(mn-1)^2$
Then....
5. That would make n a decial (10sqrt10)
6. Originally Posted by Soroban
Hello, Mukilab!
Note that: .$\displaystyle 500,050 \;=\;(50)(10,\!001) \;=\;(7^2+1)(100^2+1)$
$\displaystyle \text{Let }m = 7,\;n = 100 \text{ in the formula:}$
. . $\displaystyle (7^2 + 1)(100^2 + 1) \;=\;(7 + 100)^2 + (7\cdot100 - 1)^2 \;=\;107^2 + 699^2$
Thanks Soroban!
Sorry, only saw the first post at the start | {
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Thanks Soroban!
Sorry, only saw the first post at the start
Any tips for future questions such as this?? Take away the integers so I'm left with the algebraic numbers (as you did with 49 and 100)?
Thanks again
7. Originally Posted by Mukilab
That would make n a decial (10sqrt10)
Maybe you entered 50050 / 50 instead of 500050 / 50? m=7 gives n=100 as shown above.
8. Originally Posted by Mukilab
Thanks Soroban!
Sorry, only saw the first post at the start
Any tips for future questions such as this?? Take away the integers so I'm left with the algebraic numbers (as you did with 49 and 100)?
Thanks again
Another way to go about expressing integers as sums of two squares is knowing that: the set of integers expressible as the sum of two squares is closed under multiplication, and an odd prime is expressible as the sum of two squares if and only if it is congruent to 1 (mod 4). See here and here. | {
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# Why is the log in the big-O of binary search not base 2?
I am new to understanding computer science algorithms. I understand the process of the binary search, but I am having a slight misunderstanding with its efficiency.
In a size of $s = 2^n$ elements, it would take, on average, $n$ steps to find a particular element. Taking the base 2 logarithm of both sides yields $\log_2(s) = n$. So wouldn't the average number of steps for the binary search algorithm be $\log_2(s)$?
This Wikipedia article on the binary search algorithm says that the average performance is $O(\log n)$. Why is this so? Why isn't this number $\log_2(n)$?
When you change the base of logarithm the resulting expression differs only by a constant factor which, by definition of Big-O notation, implies that both functions belong to the same class with respect to their asymptotic behavior.
For example $$\log_{10}n = \frac{\log_{2}n}{\log_{2}10} = C \log_{2}{n}$$ where $C = \frac{1}{\log_{2}10}$.
So $\log_{10}n$ and $\log_{2}n$ differs by a constant $C$, and hence both are true: $$\log_{10}n \text{ is } O(\log_{2}n)$$ $$\log_{2}n \text{ is } O(\log_{10}n)$$ In general $\log_{a}{n}$ is $O(\log_{b}{n})$ for positive integers $a$ and $b$ greater than 1.
Another interesting fact with logarithmic functions is that while for constant $k>1$, $n^k$ is NOT $O(n)$, but $\log{n^k}$ is $O(\log{n})$ since $\log{n^k} = k\log{n}$ which differs from $\log{n}$ by only constant factor $k$.
• Not only for positive integers: For all real $a,b > 1$, e.g. $e$. – nbubis Jul 20 '17 at 6:25
• I would add that this is the reason why with big-O notation, the base of the logarithm is commonly not specified. It also means there is no confusion about which base $O(\log n)$ uses: it can be any of the commonly used based, since it doesn't make a difference. – svick Jul 20 '17 at 10:57 | {
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In addition to fade2black's answer (which is completely correct), it's worth noting that the notation "$\log(n)$" is ambiguous. The base isn't actually specified, and the default base changes based on context. In pure mathematics, the base is almost always assumed to be $e$ (unless specified), while in certain engineering contexts it might be 10. In computer science, base 2 is so ubiquitous that $\log$ is frequently assumed to be base 2. That wikipedia article never says anything about the base.
But, as has already been shown, in this case it doesn't end up mattering.
• It is probably further worth noting then that while "log(n)" might be ambiguous that "O(log(n))" is not because the latter only has one meaning, no matter what base you might be thinking of. – Chris Jul 20 '17 at 16:12 | {
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# In how many ways can a bit string of length $20$ be generated if either all $0$'s or all $1$'s need to be grouped together?
Given a bit string of length $$20$$, how many ways can such a string be generated if either all 0's or all 1's need to be grouped together in the string?
A few examples of strings that would be considered legal are $$000 \ldots 0\qquad\text{or} \qquad 001110 \ldots 0 \qquad \text{or} \qquad 1100 \ldots 11 \qquad \text{or} \qquad1110000 \ldots 0$$
I've tried a couple different ideas, but I keep running into issues with duplicate strings.
One method I used would be to imagine that there are $$22$$ spaces and $$2$$ slashes that need to go somewhere within these spaces to separate the numbers. The slashes could go before the string starts, resulting in either all 1's or all 0's depending on which number starts. I'll list some examples of this below. $$//00000 \ldots 0 \qquad \text{or} \qquad //1111 \ldots 1 \qquad \text{or} \qquad 11 \ldots /000000/\ldots 1$$
With this method, we have $$2$$ cases that are the same, its just either picking if 1 or 0 leads. Because of that I'm calculating $$2\binom{22}{2}$$. But like I said, there are issues with this method. For example, when both slashes come before the string, the string that is generated would be the same as if the slashes came after the string.
I appreciate any ideas on where I should go from here or if I'm completely wrong and should just start over with some other method. | {
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} |
• you can separate the cases where you have only one block (either all 1's or all 0's), two blocks (setting a slash between two of the 20 positions) and three blocks (setting the left slash and then setting the right slash in regards to the position of the left one) – otto Jun 12 '19 at 22:44
• actually you use your way of putting two slashes in 22 possible positons. start by setting the left one, which is not allowed to have position 22 (to remove redundance). then set the right one with respect to the position of the left one – otto Jun 12 '19 at 23:11
• Welcome to MathSE. This tutorial explains how to typeset mathematics on this site. – N. F. Taussig Jun 13 '19 at 1:16
For each such string, join its ends to form a necklace. Each necklace has at most one region of 1s and at most one region of 0s, and is characterized by the number of 1s (0 through 20).
Apart from the all-0 and all-1 necklaces, each necklace (there are 19 remaining) can be cut in 20 places, yielding distinct valid strings. The constant necklaces yield the same string no matter where you cut. So, the total number of strings is:
$$1 + 1 + 19 \cdot 20 = 382$$
• This is now my favorite answer. – David K Jun 13 '19 at 2:29
• I agree with @DavidK. – N. F. Taussig Jun 13 '19 at 2:32
There are two ways to choose the first digit in the bit string.
Since all the $$0$$s or all the $$1$$s must be together, there are either $$0$$, $$1$$, or $$2$$ transitions between blocks of identical digits.
If there are no transitions, we do not need to place a divider. There is one way to do this.
If there is one transition, we must choose one of the $$19$$ places between successive digits in the $$20$$-digit bit string in which to place the divider between the blocks.
If there are two transitions, we must choose two of the $$19$$ places between successive digits in the $$20$$-digit bit string in which to place the dividers between the blocks. | {
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Hence, there are $$\binom{2}{1}\left[\binom{19}{0} + \binom{19}{1} + \binom{19}{2}\right]$$ bit strings of length $$20$$ such that all the $$0$$s or all the $$1$$s are together in the bit string.
Using your idea of $$22$$ spaces, you not only have the ambiguity between $$//\square\cdots\square$$ and $$\square\cdots\square//$$ (where each $$\square$$ represents a space where you will put a digit), you also have to be concerned about how you distinguish $$\square//\square\cdots\square$$ from both $$//\square\cdots\square$$ and $$/\square/\square\cdots\square.$$ (If each $$/$$ denotes a change from $$0$$ to $$1$$ or vice versa then $$\square//\square$$ is the same pattern as $$//\square\square$$ but if either $$\square/\square$$ or $$\square//\square$$ denotes that the second $$\square$$ is different from the first $$\square$$ then $$\square//\square$$ and $$/\square/\square$$ are the same pattern.)
You can almost fix this by requiring that the two $$/$$s not be adjacent. Then the only ambiguity you have is between $$/\square\cdots\square/\square\cdots\square$$ and $$\square\cdots\square/\square\cdots\square/.$$ There are $$19$$ of each of those patterns, you want to keep one set and remove the other, so subtract $$19$$ from the number of ways to put $$2$$ non-adjacent objects in $$22$$ spaces, which is $$\binom{21}{2}$$; with two choices for the first digit this makes the number of bit strings $$2\left(\binom{21}{2} - 19\right).$$
Another almost-fix is to require that there be at least one digit after each $$/.$$ This gets rid of the $$\square\cdots\square/\square\cdots\square/$$ patterns so you don't have to subtract those, but it also gets rid of $$/\square\cdots\square/,$$ so you have to add $$1$$ to count that pattern. This gives the answer $$2\left(\binom{20}{2} + 1\right).$$ | {
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Notice that $$\binom{21}{2} = \binom{20}{2} + \binom{20}{1} = \binom{19}{2} + 2\binom{19}{1} + \binom{19}{0},$$ so both the results in this answer are equal to $$2\left(\binom{19}{2} + \binom{19}{1} + \binom{19}{0}\right) = 2\times 191,$$ which has also been shown correct in another answer. What you will not be able to do is to come up with a combinatorial argument producing a single binomial coefficient that gives the answer when multiplied by $$2$$, because $$191$$ is a prime number and the only binomial coefficients with that value are $$\binom{191}{1}$$ and $$\binom{191}{190}.$$ | {
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# Connexions
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# Presentation token element (mo)
Module by: Sunil Kumar Singh. E-mail the author
Summary: The mo operator element in MathML conveys the meaning of mathematical operation.
Specific attributes : form fence separator lspace rspace stretchy symmetric Specific attributes : maxsize minsize largeop movablelimits accent Specific attributes : form fence separator lspace rspace stretchy symmetric Specific attributes : maxsize minsize largeop movablelimits accent | {
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The “mo” element displays an operator or other representations, which are treated as an operator in mathematics. The term "operator" also includes fence, separator, accent, comma, semicolon, invisible characters etc. , some of which are used to provide new meaning to the ordinary operator. Hence, “operator” in MathML has wider meaning beyond ordinary operators, consistent with the requirement of growing expanse of mathematical operations.
Many of the important operators contained within “mo” element can be typed directly. They are available on key board; while many others have to be referenced through valid Unicode entity references. The operator symbols, which can be typed from the keyboard include “+”, “-, ”/”, ”*”, ”(“, ”)”, ”{“, ”}”, ”[“, ”]”, “.NOT.”, ”.OR.” etc.
Following example shows the display of operators by "mo" element.
### Example 1: Displaying operators with “mo” element
<m:math display="block">
<m:mtable>
<m:mtr>
<m:mtd>
<m:mi>Plus : </m:mi><m:mo>+</m:mo>
<m:mi> ;</m:mi>
<m:mi>Increment : </m:mi><m:mo>++</m:mo>
<m:mi> ;</m:mi>
<m:mi>Logical not : </m:mi><m:mo>.NOT.</m:mo> <m:mi> ;</m:mi>
</m:mtd>
</m:mtr>
<m:mtr>
<m:mtd>
<m:mi>Less than and equal : </m:mi><m:mo> ≤ </m:mo>
<m:mi> ;</m:mi>
<m:mi>Partial operator : </m:mi><m:mo> ∂ </m:mo>
<m:mi> ;</m:mi>
<m:mi>Differentiation : </m:mi><m:mo> ⅆ </m:mo>
<m:mi> ;</m:mi>
<m:mi>Integration : </m:mi><m:mo> ∫ </m:mo>
</m:mtd>
</m:mtr>
</m:mtable>
</m:math>
Save the file after editing as “test.xml”. The display looks like :
Plus : + ; Increment : ++ ; Logical not : .NOT. ; Less than and equal : ; Partial operator : ; Differentiation : ; Integration : Plus : + ; Increment : ++ ; Logical not : .NOT. ; Less than and equal : ; Partial operator : ; Differentiation : ; Integration : | {
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"openwebmath_score": 0.50804603099823,
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In the above example, we have used “mtable” (with “mtr” and “mtd” elements) element for controlling display. See that first three operators in above example have been typed directly from the keyboard; while others have been referenced by entity reference.
MathML design of rendering an operator closely follows mathematical convention. It distinguishes between rendering a character with “mi” element and an operator with “mo” operator in many important ways for notational representation of mathematical expression. Two basic considerations are (i) to manage space around the operator, consistent with the context in which they are displayed and (ii) dimensional change (stretching) of operators in accordance with the dimension of other elements and terms. Space around an operator is managed by “form” attribute, while stretching is managed by “stretchy” attribute of “mo” element. A plus operator ("+") may ,for example, precede or follow an identifier or may lie between two operands. Depending upon its placement, the space around the operator "+" is determined. Similarly, size of a parentheses around an expression must stretch to the height of the expression as in ( a b ) ( a b ) .
MathML also allows improvisation of an operator rendered by “mo” element. This generally makes use of other characters rendering elements such as “mi” or "mn" elements in addition to "mo" element and their combination with layout elements like “mfrac” and scripting elements (“msub”, “msup” etc.) Simply put, MathML treats an expression with a “mo” element at core and improvised by other elements - as an operator of class known as “embellished” operator.
Consider encoding for a differentiation operator "ⅆ" . This operator when applied to a variable, say t, is represented as t t In mathematics, however, t t is itself considered to be an operator. The whole block consisting of “mfrac” element is, thus, an embellished operator. | {
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<m:mfrac>
<m:mo> ⅆ </m:mo>
<m:mrow>
<m:mo> ⅆ </m:mo>
<m:mi> x </m:mi>
</m:mrow>
</m:mfrac>
### Additional attribute : Default values
Default values to the attributes of “mo” element are set in accordance with the values contained in “operator dictionary”, which specifies “form”, “fence”, “stretchy”, “lspace” and “rspace” attributes for different operators. The renderer maintains an operator dictionary for most of the operators. W3C recommends a prototype of operator dictionary. These recommendations can be viewed at www.w3.org/TR/MathML2/appendixf.html . If the dictionary does not provide the value for the attribute, then attribute is set with default value as given here:
#### Default values
1. form : set by position/context
2. fence : false
3. separator : false
4. lspace : thickmathspace
5. rspace : thickmathspace
6. stretchy : false
7. symmetric : true
8. maxsize : infinity
9. minsize : 1
10. largeop : false
11. movablelimits : false
12. accent : false
### Managing space around operators
#### Attribute value types
• form : prefix | infix | postfix
• lspace : number h-unit | namedspace
• rspace : number h-unit | namedspace
The "form" attribute specifies whether an operator shall be rendered as "prefix" or "postfix" or "infix" operator. These form types determine space around the operator (left and right of it). Each form type is associated with two attributes ("lspace" and "rspace"), which implements its left space ("lspace") and right space ("rspace"). It is important to understand that form types are basically specification about space around operator and about its role with respect to other elements. | {
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The default value of “form” attribute is set in accordance with three rules, involving "mrow" element. The token elements, including "mo" element, is composed within explicit or inferred "mrow" element. The position of an operator within the "mrow" in horizontal sequence determines the form type of the operator. If operator is the first element, then its form is inferred as "prefix"; if it is in the intermediate position, then its form is inferred as "infix”; and if it is in the end position, then its form is inferred as "postfix". Corresponding to each form, operator library specifies "lspace" and "rspace". It is, however, not necessary that each of the operator has all the three form types. The nature of operator determines the type of forms that a particular operator should be associated with. For example, "+" operator has entries for "prefix" and "infix", but not for "postfix" in the operator dictionary, because "postfix" form of "+" operator is not expected to be placed at the end of an expression.
Now, consider the example of "+" operator appearing twice in an expression in the manner as typed below :
+a + b
The first "+" sign in the above case assumes "prefix" form, while the second "+" sign assumes "infix" form. Once, the form of the operator is determined from the context within "mrow" element, spacing around the operator is determined as specified by "lspace" and "rspace" values given in the operator dictionary. The dictionary suggests following values for these forms of “+” operator :
"+" form="infix" lspace="mediummathspace" rspace="mediummathspace"
"+" form="prefix" lspace="0em" rspace="veryverythinmathspace"
Evidently, "infix" form separates its neighboring elements by "mediummathspace", whereas "prefix" form separates the following element with a very small space specified by "veryverythinmathspace".
Let us now code the example in MathML and see the output :
#### Example 2: Determining "form" of an operator | {
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#### Example 2: Determining "form" of an operator
<m:math display="block">
<m:mrow>
<m:mo> + </m:mo>
<m:mi> a </m:mi>
<m:mo> + </m:mo>
<m:mi> b </m:mi>
</m:mrow>
</m:math>
Save the file after editing as “test.xml”. The display looks like :
+ a + b + a + b
As expected first “+” is almost attached to “a”, where as second instance of “+” renders some amount of space around itself. This type of detailing in rendering mathematical expression is vital and critical to maintain highest order of rendering consistencey, which is commensurate with the exact nature of mathematics and the meaning that a mathematical expression conveys. Note that the mark-up paradigm of MathML automatically forces codification in such a manner that no extra effort is required towards maintaining conventions of mathematical display - almost ruling out the possibilities that an expression is displayed in a non-conforming way.
A particular operator does not require to have entries for all the forms in the operator dictionary. Many of the operators in mathematics operate on other entities in forward direction. For example, integration, differentiation, partial differentiation etc. operates on identifiers following it. For this reason, such operators are inherently “prefix” in nature. On the other hand, operators classified as fences are either “prefix” or “postfix” commensurate with their role in mathematical expression. It is very unlikely that a closing bracket “]” is used in the beginning of an expression and as such it has “postfix” form. Dictionary entries for these operators are given here to understand : why there is very small space or no space between these operators and the identifiers on which they operate. | {
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"openwebmath_score": 0.50804603099823,
"tags": null,
"url": "http://cnx.org/content/m13548/latest/"
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"ⅆ" form="prefix" lspace="0em" rspace="verythinmathspace"
"∂" form="prefix" lspace="0em" rspace="verythinmathspace"
"∫" form="prefix" largeop="true" stretchy="true" lspace="0em"
"(" form="prefix" fence="true" stretchy="true" lspace="0em" rspace="0em"
")" form="postfix" fence="true" stretchy="true" lspace="0em" rspace="0em"
"[" form="prefix" fence="true" stretchy="true" lspace="0em" rspace="0em"
"]" form="postfix" fence="true" stretchy="true" lspace="0em" rspace="0em"
"{" form="prefix" fence="true" stretchy="true" lspace="0em" rspace="0em"
"}" form="postfix" fence="true" stretchy="true" lspace="0em" rspace="0em"
We are at liberty to specify these attributes. It is, however, recommended that we leave the arrangement to the system, which is rendering the mathematical content. It shall ensure consistency in the display, which follows the convention of mathematics as implemented by a particular renderer. The example below demonstrates how we can change the spacing different to default and against the form values as inferred from the context :
#### Example 3: Determining "form" of an operator
<m:math display="block">
<m:mrow>
<m:mo form="infix" rspace="10pt"> + </m:mo>
<m:mi> a </m:mi>
<m:mo form="prefix"> + </m:mo>
<m:mi> b </m:mi>
</m:mrow>
</m:math>
Save the file after editing as “test.xml”. The display lokks like :
+ a + b + a + b
### Stretching of operators
Four attributes of “mo” element control stretching of operators. These attributes may assume following values :
#### Attribute value types
• stretchy : true | false
• symmetric : true | false
• maxsize : number [ v-unit | h-unit | namedspace | infinity
• minsize : number [ v-unit | h-unit | namedspace | {
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Among the operators, the requirement for stretching of fences, arrows, accents (angular cap on identifier) and separators are most profound and visible in mathematical expressions. For this reason, “strechy” attribute of fence and accent operators are set “true” in operator dictionary. Stretching of operators in an expression is restricted by “minsize” and “maxsize” attributes.
The stretchable operators are characterized as predominantly either vertically or horizontally stretchable. The fences, various kinds of vertical arrows (single or double), operators like ∏, ∑, ∫, “/” etc. are set to stretch vertically by default in operator directory.
When stretchable operator and non-stretchable terms are bounded by explicit or inferred “mrow” element, then the stretchable operator grows vertically to cover the non-stretchy term. Consider the example given here :
#### Example 4: Vertical stretching
<m:math display="block">
<m:mrow>
<m:mi> x </m:mi>
<m:mo> = </m:mo>
<m:mo> ( </m:mo>
<m:mfrac>
<m:mi>a</m:mi>
<m:mi>b</m:mi>
</m:mfrac>
<m:mo> ) </m:mo>
</m:mrow>
</m:math>
Save the file after editing as “test.xml”. The display looks like :
x = ( a b ) x = ( a b )
We can, however, control the growth of parenthese by setting “maxsize” attribute to 1 i.e. equal to its normal size.
#### Example 5: Vertical stretching
<m:math display="block">
<m:mrow>
<m:mi> x </m:mi>
<m:mo> = </m:mo>
<m:mo maxsize="1"> ( </m:mo>
<m:mfrac>
<m:mi>a</m:mi>
<m:mi>b</m:mi>
</m:mfrac>
<m:mo maxsize="1"> ) </m:mo>
</m:mrow>
</m:math>
Save the file after editing as “test.xml”. The display looks like :
x = ( a b ) x = ( a b )
Thus, setting "maxsize" attribute overrides the default behavior, which allows the parentheses to strech and cover the non-stretchy expression. Let us, now experiment with other than fence character like ∑ and observe their behavior with other terms :
#### Example 6: Vertical stretching | {
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#### Example 6: Vertical stretching
<m:math display="block">
<m:mrow>
<m:mo> ∑ </m:mo>
<m:mo> ( </m:mo>
<m:mfrac>
<m:mi>A</m:mi>
<m:mi>B</m:mi>
</m:mfrac>
<m:mo> ) </m:mo>
</m:mrow>
</m:math>
Save the file after editing as “test.xml”. The display looks like :
( A B ) ( A B )
Control for stretching can be selective as well. For example, we can set stretchy=”false” on the opening parenthesis to restrict it to grow.
#### Example 7: Vertical stretching
<m:math display="block">
<m:mrow>
<m:mi> x </m:mi>
<m:mo> = </m:mo>
<m:mo stretchy="false"> ( </m:mo>
<m:mfrac>
<m:mi>a</m:mi>
<m:mi>b</m:mi>
</m:mfrac>
<m:mo> ) </m:mo>
</m:mrow>
</m:math>
Save the file after editing as “test.xml”. The display looks like :
x = ( a b ) x = ( a b )
In situation where, the expression bounded by “mrow” tags contains terms of different heights, the stretchable parentheses grow to cover the highest of the terms.
#### Example 8: Vertical stretching
<m:math display="block">
<m:mrow>
<m:mi> x </m:mi>
<m:mo> = </m:mo>
<m:mo> ( </m:mo>
<m:mfrac>
<m:mi>a</m:mi>
<m:mi>b</m:mi>
</m:mfrac>
<m:mo> + </m:mo>
<m:mfrac>
<m:mrow>
<m:mfrac>
<m:mi>c</m:mi>
<m:mi>d</m:mi>
</m:mfrac>
</m:mrow>
<m:mi>e</m:mi>
</m:mfrac>
<m:mo> ) </m:mo>
</m:mrow>
</m:math>
Save the file after editing as “test.xml”. The display lokks like :
x = ( a b + c d e ) x = ( a b + c d e )
The symmetric attribute is designed to stretch operator in both vertical and horizontal direction from the axis of the characters in equal magnitude. The symmetric attribute applies only to characters, which can stretch vertically; otherwise this attribute is ignored. Usually, this attribute is set “true” for vertically stretchable operator, but in certain cases involving matrix of unequal size, we may prefer to set it “false” as demonstrated in the example here :
#### Example 9: Vertical stretching | {
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#### Example 9: Vertical stretching
<m:math display="block">
<m:mrow>
<m:mi> x </m:mi>
<m:mo> = </m:mo>
<m:mo symmetric="false"> ( </m:mo>
<m:mtable align="bottom">
<m:mtr>
<m:mtd>
<m:mi>a</m:mi>
</m:mtd>
<m:mtd>
<m:mi>b</m:mi>
</m:mtd>
</m:mtr>
<m:mtr>
<m:mtd>
<m:mn>d</m:mn>
</m:mtd>
<m:mtd>
<m:mn>e</m:mn>
</m:mtd>
</m:mtr>
</m:mtable>
<m:mo symmetric="false"> ) </m:mo>
<m:mo symmetric="false"> ( </m:mo>
<m:mtable align="bottom">
<m:mtr>
<m:mtd>
<m:mi>a</m:mi>
</m:mtd>
<m:mtd>
<m:mi>b</m:mi>
</m:mtd>
</m:mtr>
<m:mtr>
<m:mtd>
<m:mn>d</m:mn>
</m:mtd>
<m:mtd>
<m:mn>e</m:mn>
</m:mtd>
</m:mtr>
<m:mtr>
<m:mtd>
<m:mn>d</m:mn>
</m:mtd>
<m:mtd>
<m:mn>e</m:mn>
</m:mtd>
</m:mtr>
</m:mtable>
<m:mo symmetric="false"> ) </m:mo>
</m:mrow>
</m:math>
Save the file after editing as “test.xml”. The display lokks like :
x = ( a b d e ) ( a b d e d e ) x = ( a b d e ) ( a b d e d e )
Matrix operation of unequal sizes uses notation which is aligned to base. This type of controlling stretching in specific direction is, therefore, extremely useful in such situations.
In case, the “mrow” domain contains non-stretchable terms of normal height and other stretchable terms, then all the terms, including the fence operator grows to the maximum normal height.
#### Example 10: Vertical stretching
<m:math display="block">
<m:mrow>
<m:mi> x </m:mi>
<m:mo> = </m:mo>
<m:mo> ( </m:mo>
<m:mo> ∫ </m:mo>
<m:mi> f</m:mi>
<m:mo> ( </m:mo>
<m:mi> x </m:mi>
<m:mo> ) </m:mo>
<m:mi> ⅆ </m:mi>
<m:mi> x </m:mi>
<m:mo> ) </m:mo>
</m:mrow>
</m:math>
Save the file after editing as “test.xml”. The display lokks like :
x = ( f ( x ) x ) x = ( f ( x ) x ) | {
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"tags": null,
"url": "http://cnx.org/content/m13548/latest/"
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x = ( f ( x ) x ) x = ( f ( x ) x )
As against fence operator, the accent and horizontal arrows are stretchable in horizontal direction by default. The growth, in this case, is controlled by “munder”, “mover” and “munderover” elements, which contain the operator. The “munder”, “mover” and “munderover” elements, as the names suggest, allow drawing of an operator "under" or "over" or both about a character(s) or expresison. The scripting elements takes a base argument about which the operator is to be drawn and one (“munder” and “mover”) operator or two (“munderover”) operators for being placed about the base. We shall see that the operator grows horizontally to cover the other element.
#### Example 11: Horizontal stretching
<m:math display="block">
<m:mi> A </m:mi>
<m:munderover>
<m:mo> → </m:mo>
<m:mtext> 50 degree C </m:mtext>
<m:mtext> 200 psi </m:mtext>
</m:munderover>
<m:mi> B </m:mi>
</m:math>
Save the file after editing as “test.xml”. The display looks like :
A 50 degree C 200 psi B A 50 degree C 200 psi B
### Other attributes : largeop, movablelimits, accent, separator, fence
These attributes may assume following values :
#### Attribute value types
• fence : true | false
• separator : true | false
• largeop : true | false
• movablelimits : true | false
• accent : true | false
All these attributes accept boolean values “true” or “false”. The attribute “fence” is designed for non-visual rendering like audio rendering. As such, this attribute has no impact on the visual aspect of rendering. The role of “separator” attribute is also not significant and may be left to default value.
The “largeop” attribute determines the size of the operator. If it is true, then the operator is drawn larger than its normal size. For example, ∫ and ∏ operators are displayed as large operator as “largeop” attribute for these operators are set “true” by default. If we need to display the normal size, then their "largeop" attribute is set to "false". | {
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#### Example 12: Large operator
<m:math display="block">
<m:mrow>
<m:mo> ∏ </m:mo>
<m:mo> , </m:mo>
<m:mo> ∫ </m:mo>
<m:mo> , </m:mo>
<m:mo largeop="false"> ∏ </m:mo>
<m:mo> , </m:mo>
<m:mo largeop="false"> ∫ </m:mo>
</m:mrow>
</m:math>
Save the file after editing as “test.xml”. The display looks like :
, , , , , ,
The movablelimits attribute allows underscripts and overscripts to be represented as subscript and superscript respectively. This rendering of under and over scripts as sub and super scripts is possible by seting this attribute to true. The implementation of this attribute by renderers is not yet consistent.
#### Example 13: The "movablelimits" attribute
<m:math display="block">
<m:munderover>
<m:mo movablelimits='true'> ∑ </m:mo>
<m:mi> a </m:mi>
<m:mi> b </m:mi>
</m:munderover>
<m:munderover>
<m:mo movablelimits='false'> ∑ </m:mo>
<m:mi> a </m:mi>
<m:mi> b </m:mi>
</m:munderover>
</m:math>
Save the file after editing as “test.xml”. The display looks like :
a b a b a b a b
The accent attribute determines whether an operator is treated as accent (diacritical mark) when used as an underscript or overscript.
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"openwebmath_score": 0.50804603099823,
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"url": "http://cnx.org/content/m13548/latest/"
} |
##### What is in a lens?
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MAT 142 3.4 Basics of Probability Theory
# MAT 142 3.4 Basics of Probability Theory
398.3k points
1. A jar of M&Ms contains 12 brown candies, 4 yellow candies, 2 blue candies, 5 red candies, 3
green candies and 4 orange candies. You select one at random. Find the probability that you
select one that is:
a. Brown
b. Green or orange
c. Not red
d. Yellow or blue
e. Yellow and blue
1. Shoppers at a local department store were asked to complete a survey of their shopping
experience. The results are shown in the table below:
** Satisfied with Service Not satisfied with Service** Totals
Made a purchase 130 494 624
Did not make a purchase 715 183 898
Totals 845 677 1522
__
a. What is the probability that a shopper selected at random made a purchase?
b. What are the odds that a shopper selected at random was satisfied with the service?
MAT 142
393k points
#### Oh Snap! This Answer is Locked
Thumbnail of first page
Excerpt from file: Math Tutorial MAT 142 Problem set Unit: Probability Topic: Basics of Probability Theory Directions: Solve the following problems. Please show your work and explain your reasoning. 1. A jar of M&Ms contains 12 brown candies, 4 yellow candies, 2 blue candies, 5 red candies, 3 green candies and 4
Filename: M1084.pdf
Filesize: 369.5K
Print Length: 2 Pages/Slides
Words: 292
J
0 points
#### Oh Snap! This Answer is Locked
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Excerpt from file: Week4DQ#3DiscOps&ExtraordinaryItems Whyisitimportanttoreportdiscontinuedoperationsorextraordinaryitemsseparately fromincomefromcontinuingoperations?Isthismethodofreportingallowed?What concernsdoesthistypeofreportingcreate?Doestheaverageinvestorunderstandthe
Filename: ACC 280week4dqs.zip
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Filename: ACC 280week4dqs.zip
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Surround your text in *italics* or **bold**, to write a math equation use, for example, $x^2+2x+1=0$ or $$\beta^2-1=0$$
Use LaTeX to type formulas and markdown to format text. See example. | {
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find the sum to n term of $\frac{1}{1\cdot2\cdot3} + \frac{3}{2\cdot3\cdot4} + \frac{5}{3\cdot4\cdot5} + \frac{7}{4\cdot5\cdot6 } + …$
$$\frac{1}{1\cdot2\cdot3} + \frac{3}{2\cdot3\cdot4} + \frac{5}{3\cdot4\cdot5} + \frac{7}{4\cdot5\cdot6 } + ...$$
$$=\sum \limits_{k=1}^{n} \frac{2k-1}{k\cdot(k+1)\cdot(k+2)}$$ $$= \sum \limits_{k=1}^{n} - \frac{1}{2}\cdot k + \sum \limits_{k=1}^{n} \frac{3}{k+1} - \sum \limits_{k=1}^{n}\frac{5}{2\cdot(k+2)}$$
I do not know how to get a telescoping series from here to cancel terms.
• You can use $$\frac{2k-1}{k(k+1)(k+2)}=\frac{2k}{k(k+1)(k+2)}-\frac1{k(k+1)(k+2)}.$$ – Sungjin Kim Oct 26 '18 at 18:11
• You do mean an infinite sum, not a finite sum, right? – Connor Harris Oct 26 '18 at 18:15
• @ConnorHarris I think this means writing the partial sums explicitly. – Sungjin Kim Oct 26 '18 at 18:16
• @ConnorHarris, a sum that can be expressed in n. – LetzerWille Oct 26 '18 at 18:17
• Note that the $\frac{3}{k+1}$ term for some value of $k$ cancels the $\frac{-1}{2k}$ and $\frac{-5}{2(k+2)}$ terms for adjacent values of $k$. – Connor Harris Oct 26 '18 at 18:18
HINT:
Note that we have
\begin{align} \frac{2k-1}{k(k+1)(k+2)}&=\color{blue}{\frac{3}{k+1}}-\frac{5/2}{k+2}-\frac{1/2}{k}\\\\ &=\color{blue}{\frac12}\left(\color{blue}{\frac{1}{k+1}}-\frac1k\right)+\color{blue}{\frac52}\left(\color{blue}{\frac{1}{k+1}}-\frac{1}{k+2}\right) \end{align}
• There's a typo after the first $=$ sign, it should be $\frac{5/2}{k+2}$. Can't edit though, since edits must be 6 characters at minimum. – a_guest Oct 27 '18 at 12:50
• @a_guest Thank you. I've edited accordingly. – Mark Viola Oct 27 '18 at 16:29
Let the fractions be $$\frac{a}{k}$$, $$\frac{b}{k+1}$$, and $$\frac{c}{k+2}$$.
$$\frac{a}{k}+\frac{b}{k+1}+\frac{c}{k+2}=\frac{a(k+1)(k+2)+bk(k+2)+ck(k+1)}{k(k+1)(k+2)}=\frac{2k-1}{k(k+1)(k+2)}$$
We want the following
$$a+b+c=0$$
$$3a+2b+c=2$$
$$2a=-1$$
Solve, $$a=-\frac{1}{2}$$, $$b=3$$, and $$c=-\frac{5}{2}$$.
The rest is standard. | {
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$$2a=-1$$
Solve, $$a=-\frac{1}{2}$$, $$b=3$$, and $$c=-\frac{5}{2}$$.
The rest is standard.
You are almost there. You can merge the parts of the series for which the denominator is similar and you will see they cancel each other. Then you are left with the terms for which the denominator is either smaller than $$3$$ or greater than $$n$$.
\begin{aligned} & \sum_{k=1}^n\frac{-1}{2k} + \sum_{k=1}^n\frac{3}{k+1} - \sum_{k=1}^n\frac{5}{2}\frac{1}{k+2} \\ & = \left[-\frac{1}{2} - \frac{1}{4} + \frac{1}{2}\sum_{k=3}^n\frac{-1}{k}\right] + \left[\frac{3}{2} + \frac{1}{2}\sum_{k=2}^n\frac{6}{k+1}\right] - \left[\frac{1}{2}\sum_{k=1}^n\frac{5}{k+2}\right] \\ & = \frac{3}{4} + \left[\frac{1}{2}\sum_{k=3}^n\frac{-1}{k}\right] + \left[\frac{1}{2}\sum_{k=3}^{n+1}\frac{6}{k}\right] - \left[\frac{1}{2}\sum_{k=3}^{n+2}\frac{5}{k}\right] \\ & = \frac{3}{4} + \left[\frac{1}{2}\sum_{k=3}^n\frac{-1}{k}\right] + \left[\frac{1}{2}\sum_{k=3}^{n}\frac{6}{k} + \frac{6}{2}\frac{1}{n+1}\right] - \left[\frac{1}{2}\sum_{k=3}^{n}\frac{5}{k} + \frac{5}{2}\frac{1}{n+1} + \frac{5}{2}\frac{1}{n+2}\right] \\ & = \frac{3}{4} + \frac{1}{2}\sum_{k=3}^{n}\left[\frac{-1 + 6 - 5}{k}\right] + \frac{6}{2}\frac{1}{n+1} - \frac{5}{2}\frac{1}{n+1} - \frac{5}{2}\frac{1}{n+2} \\ & = \frac{3}{4} + \frac{1}{2(n+1)} - \frac{5}{2(n+2)} \end{aligned}
When terms are in A.P in denominator, we use difference of last and first terms in product to distribute the terms of denominator over numerator, to change into telescoping series. | {
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# Find $\sum_{n=1}^{\infty}\tan^{-1}\frac{2}{n^2}$
Find $$M:=\sum_{n=1}^{\infty}\tan^{-1}\frac{2}{n^2}$$
There's a solution here that uses complex numbers which I didn't understand and I was wondering if the following is also a correct method.
My proposed solution
\begin{align} &\sum_{n=1}^{\infty}\tan^{-1}\frac{2}{n^2}\\ =&\sum_{n=1}^{\infty}\tan^{-1}\frac{(1+n)+(1-n)}{1-(1+n)(1-n)}\\ =&\sum_{n=1}^{\infty}(\tan^{-1}(1+n)+\tan^{-1}(1-n))\\ =&\sum_{n=1}^{\infty}(\tan^{-1}(n+1)-\tan^{-1}(n-1)) \end{align}
And this implies $$M=\lim_{m\to\infty}(\tan^{-1}(m+1)+\tan^{-1}m-\tan^{-1}1-\tan^{-1}0)=\frac{3\pi}{4}$$
• This may or may not be of help: $\arctan(1/x) = \pi/2 - \arctan(x)$ – user150203 Dec 27 '18 at 5:23
• The 3rd and 4th lines of the equation after "My proposed solution" are missing a summation sign. Otherwise, the solution is correct. – JimmyK4542 Dec 27 '18 at 5:35
• @JimmyK4542 Edited, ty. – Akash Gaur Dec 27 '18 at 5:40
• – lab bhattacharjee Dec 27 '18 at 6:34
\begin{align} &\sum_{n=1}^{\infty}\tan^{-1}\frac{2}{n^2}\\ =&\sum_{n=1}^{\infty}\tan^{-1}\frac{(1+n)+(1-n)}{1-(1+n)(1-n)}\\ =\color{red}{\sum_{n=1}^\infty}&\tan^{-1}(1+n)+\tan^{-1}(1-n)\\ =\color{red}{\sum_{n=1}^\infty}&\tan^{-1}(n+1)-\tan^{-1}(n-1) \end{align}
Edit:$$\color{red}{\sum_{n=1}^\infty}$$ was missing in your question before edit. I am not going to delete this.
However your proof is now correct. $$M=\lim_{m\to\infty}(\tan^{-1}(m+1)+\tan^{-1}m-\tan^{-1}1-\tan^{-1}0)=\color{red}{\frac\pi2+\frac\pi2-\frac\pi4-0=\pi-\frac\pi4}=\frac{3\pi}{4}$$
Looks good to me. If I was going to offer a critique I would just say: when writing an argument it's always better to over communicate rather than under communicate.
The first equality is just algebra.
Your second equality requires a little bit to see clearly but it's true. Most will recall: | {
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Your second equality requires a little bit to see clearly but it's true. Most will recall:
$$\tan(A+B)=\frac{\tan(A)+\tan(B)}{1-\tan(A)\tan(B)}$$ Or if you'd like: $$A+B= \tan^{-1} \bigg(\frac{\tan(A)+\tan(B)}{1-\tan(A)\tan(B)} \bigg)$$ Taking $$A=\tan^{-1}(1+n)$$ and $$B=\tan^{-1}(1-n)$$
Honestly adding this much explanation seems like almost overkill.
The 4th equality follows as result of $$\tan^{-1}$$ being an odd function.
Now the last part you are using a telescoping series technique so that you may ignore all the middle terms. That is,
\begin{align} &\sum_{n=1}^\infty\tan^{-1}(n+1)-\tan^{-1}(n-1) \\ &= \lim_{m\to \infty} \tan^{-1}(m+1)-tan^{-1}(m-1)+\dots +\tan^{-1}(4)-\tan^{-1}(2)+\tan^{-1}(3)-\tan^{-1}(1)+tan^{-1}(2)-\tan^{-1}(0) \end{align}
So after we consider what cancels and what doesn't we find that we only need to concern ourselves with $$\lim_{m\to \infty}\tan^{-1}(m+1)+\tan^{-1}(m-1)-\tan^{-1}(1)$$
So while that is true: I think it might merit a sentence or two just to make sure the audience is following.
• Isn't $M=\lim_{m\to \infty} \tan^{-1}(m)-\tan^{-1}(0)=\frac\pi2-0?$ – tatan Dec 27 '18 at 5:50
• @tatan. Yes. I am missing some terms. Let me edit. – Mason Dec 27 '18 at 5:51
• Thanks. I'm really slow at writing in latex so I skip steps. But I will add more explanation in English in the future. – Akash Gaur Dec 27 '18 at 5:57
• Yep. Can I ask you one thing? Are you from Maryland?(I saw from your profile) – tatan Dec 27 '18 at 19:57
• Yes. Uh oh. Stranger danger. :). No. I am not worried. I grew up in Ohio and move to the DC area post college. I went to Ohio University not Univ of Maryland. Though I did cite Dr. Yorke in my undergrad thesis and then I met him wandering once while wandering Univ of Maryland's halls... – Mason Dec 27 '18 at 19:57 | {
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# When 0.123...495051/0.515049...321
#### anemone
##### MHB POTW Director
Staff member
Hello MHB,
Recently I've come across a problem that jumped off the page at me (Find the first 3 figures after the decimal point in the decimal expression of the number $$\displaystyle \frac{0.12345678910\cdots495051}{0.515049\cdots987654321}$$).
I tried to approach it by making a table where I started to divide some smaller numbers but stick to the same pattern that is required by the aforementioned problem, i.e.
$$\displaystyle \frac{0.12}{0.21}=0.571...$$
$$\displaystyle \frac{0.123}{0.321}=0.383...$$
$$\displaystyle \frac{0.1234}{0.4321}=0.285...$$
$$\displaystyle \frac{0.12345}{0.54321}=0.227...$$
$$\displaystyle \frac{0.123456}{0.654321}=0.188...$$
$$\displaystyle \frac{0.1234567}{0.7654321}=0.161...$$
$$\displaystyle \frac{0.12345678}{0.87654321}=0.140...$$
$$\displaystyle \frac{0.123456789}{0.987654321}=0.124...$$
$$\displaystyle \frac{0.12345678910}{0.10987654321}=0.123...$$
$$\displaystyle \frac{0.1234567891011}{0.1110987654321}=0.111...$$
$$\displaystyle \frac{0.123456789101112}{0.121110987654321}=0.019...$$
$$\displaystyle \frac{0.12345678910111213}{0.13121110987654321}=0.940...$$
$$\displaystyle \frac{0.1234567891011121314}{0.1413121110987654321}=0.873...$$
$$\displaystyle \frac{0.123456789101112131415}{0.151413121110987654321}=0.815...$$
$$\displaystyle \frac{0.12345678910111213141516}{0.16151413121110987654321}=0.764...$$
$$\displaystyle \frac{0.1234567891011121314151617}{0.1716151413121110987654321}=0.719...$$
$$\displaystyle \frac{0.123456789101112131415161718}{0.181716151413121110987654321}=0.679...$$
$$\displaystyle \frac{0.12345678910111213141516171819}{0.19181716151413121110987654321}=0.643...$$
$$\displaystyle \frac{0.1234567891011121314151617181920}{0.2019181716151413121110987654321}=0.611...$$
and so on and so forth
but I failed to observe any pattern that's worth mentioning to help me to crack the problem. | {
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but I failed to observe any pattern that's worth mentioning to help me to crack the problem.
Could anyone help me with this particular problem? Thanks in advance.
#### Opalg
##### MHB Oldtimer
Staff member
I also approached it by making a table, but I started by using the most significant figures in the given denominator:
$$\displaystyle \frac{0.1}{0.5} = 0.2$$,
$$\displaystyle \frac{0.12}{0.51} = 0.2352\ldots$$,
$$\displaystyle \frac{0.123}{0.515} = 0.2388\ldots$$,
$$\displaystyle \frac{0.1234}{0.5150} = 0.2396\ldots$$,
$$\displaystyle \frac{0.12345}{0.51504} = 0.2396\ldots$$,
$$\displaystyle \frac{0.123456}{0.515049} = 0.2396\ldots$$,
$$\displaystyle \frac{0.1234567}{0.5150494} = 0.2396\ldots$$,
$$\displaystyle \frac{0.12345678}{0.51504948} = 0.2396\ldots$$.
By this time, the fraction had stabilised (within the limits of my 8-digit calculator) to 0.2396987. I doubt whether further approximations would affect the first three digits after the decimal point.
Edit. You can confirm that by using a bit of calculus. If $f(x,y) = x/y$ then $\delta f \approx (1/y)\delta x - (x/y^2)\delta y$. With $x\approx 0.123$, $y\approx 0.515$ and $\delta x = \delta y = 10^{-n}$, you find that $\delta f < 2*10^{-n}$. So the change in the fraction after the eighth decimal places in numerator and denominator is never going to be sufficient to affect the first three decimal places in the quotient.
Last edited:
#### anemone
##### MHB POTW Director
Staff member
Hi Opalg,
I'm very thankful to you for showing me something that I had never thought about before and your solution and the proof work so beautifully...
Thank you so much! | {
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# How can LU factorization be used in non-square matrix?
In my textbook, there is some information about LU factorization of square matrix $A$, but not about non-square matrix.
How can LU factorization be used to factorize non-square matrix?
• Yes. Let $A$ be $m \times n$ matrix, then $L$ is $m \times m$ and $U$ is $m \times n$.
– user2468
Aug 26, 2012 at 4:17
I'll illustrate how to understand the LU-decomposition of a particular $3 \times 4$ matrix below. The method works just as well for other sizes since the LU-decomposition arises naturally from the study of Gaussian elimination via multiplication by elementary matrices. | {
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$$\begin{array}{ll} A \ = &\left[ \begin{array}{cccc} 1 & 2 & -3 & 1\\ 2 & 4 & 0 & 7 \\ -1 & 3 & 2 & 0 \end{array} \right] \ \underrightarrow{r_2-2r_1 \rightarrow r_2} \ \left[ \begin{array}{cccc} 1 & 2 & -3 & 1\\ 0 & 0 & 6 & 5 \\ -1 & 3 & 2 & 0 \end{array} \right] \ \underrightarrow{r_3+r_1 \rightarrow r_3} \\ & \\ &\left[ \begin{array}{cccc} 1 & 2 & -3 & 1\\ 0 & 0 & 6 & 5 \\ 0 & 5 & -1 & 1 \end{array} \right] \ \underrightarrow{r_2 \leftrightarrow r_3} \ \left[ \begin{array}{cccc} 1 & 2 & -3 & 1\\ 0 & 5 & -1 & 1 \\ 0 & 0 & 6 & 5 \end{array} \right] = \ U \end{array}$$ We have $U = E_3E_2E_1A$ hence $A = E_1^{-1}E_2^{-1}E_3^{-1}U$ and we can calculate the product $E_1^{-1}E_2^{-1}E_3^{-1}$ as follows: $$\begin{array}{ll} I \ = &\left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] \ \underrightarrow{r_2 \leftrightarrow r_3} \ \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array} \right] \ \underrightarrow{r_3-r_1 \rightarrow r_3} \\ & \\ &\left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ -1 & 1 & 0 \end{array} \right] \ \underrightarrow{r_2+2r_1 \rightarrow r_2} \ \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 2 & 0 & 1 \\ -1 & 1 & 0 \end{array} \right] = PL \end{array}$$ I have inserted a "$P$" in front of the $L$ since the matrix above is not lower triangular. However, if we go one step further and let $r_2 \leftrightarrow r_3$ then we will obtain a lower triangular matrix: $$\begin{array}{ll} PL \ = &\left[ \begin{array}{ccc} 1 & 0 & 0 \\ 2 & 0 & 1 \\ -1 & 1 & 0 \end{array} \right] \ \underrightarrow{r_2 \leftrightarrow r_3} \ \left[ \begin{array}{ccc} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 2 & 0 & 1 \\ \end{array} \right] =L \end{array}$$ Therefore, we find that $E_1^{-1}E_2^{-1}E_3^{-1}=PL$ where $L$ is as above and $P = E_{2 \leftrightarrow 3}$. This means that $A$ has a modified $LU$-decomposition. Some mathemticians call it a $PLU$-decomposition, $$A = \underbrace{\left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 | {
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$PLU$-decomposition, $$A = \underbrace{\left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array} \right]}_{P} \underbrace{\left[ \begin{array}{ccc} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 2 & 0 & 1 \\ \end{array} \right]}_{L}\underbrace{\left[ \begin{array}{cccc} 1 & 2 & -3 & 1\\ 0 & 5 & -1 & 1 \\ 0 & 0 & 6 & 5 \end{array} \right]}_{U} = \underbrace{\left[ \begin{array}{ccc} 1 & 0 & 0 \\ 2 & 0 & 1 \\ -1 & 1 & 0 \end{array} \right]}_{PL}\underbrace{\left[ \begin{array}{cccc} 1 & 2 & -3 & 1\\ 0 & 5 & -1 & 1 \\ 0 & 0 & 6 & 5 \end{array} \right]}_{U}.$$ Since permutation matrices all satisfy the condition $P^k=I$ (for some $k$) the existence of a $PLU$-decomposition for $A$ naturally suggests that $P^{k-1}A = LU$. Therefore, even when a $LU$ decomposition is not available we can just flip a few rows to find a $LU$-decomposable matrix. This is a useful observation because it means that the slick algorithms developed for $LU$-decompositions apply to all matrices with just a little extra fine print. | {
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Much of the writing above can be spared if we adopt the notational scheme illustrated below. $$\begin{array}{ll} A \ = &\left[ \begin{array}{cccc} 1 & 2 & -3 & 1\\ 2 & 4 & 0 & 7 \\ -1 & 3 & 2 & 0 \end{array} \right] \ \underrightarrow{r_2-2r_1 \rightarrow r_2} \ \left[ \begin{array}{cccc} 1 & 2 & -3 & 1\\ (2) & 0 & 6 & 5 \\ -1 & 3 & 2 & 0 \end{array} \right] \ \underrightarrow{r_3+r_1 \rightarrow r_3} \\ & \\ &\left[ \begin{array}{cccc} 1 & 2 & -3 & 1\\ (2) & 0 & 6 & 5 \\ (-1) & 5 & -1 & 1 \end{array} \right] \ \underrightarrow{r_2 \leftrightarrow r_3} \ \left[ \begin{array}{cccc} 1 & 2 & -3 & 1\\ (-1) & 5 & -1 & 1 \\ (2) & 0 & 6 & 5 \end{array} \right] = \ U \end{array}$$ We find if we remove the parenthetical entries from $U$ and ajoing them to $I$ then it gives back the matrix $L$ we found previously: $$U = \left[ \begin{array}{cccc} 1 & 2 & -3 & 1\\ 0 & 5 & -1 & 1 \\ 0 & 0 & 6 & 5 \end{array} \right] \qquad L=\left[ \begin{array}{ccc} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 2 & 0 & 1 \end{array} \right].$$
Hope this helps.
• "permutation matrices all satisfy the condition $P^2=I$" is incorrect.
– user147263
Jan 16, 2016 at 22:09
• @Normal for example ? Sorry, I can't see past swap then reverse swap does nothing... Jan 16, 2016 at 22:28
• A permutation can involve a $3$-cycle like $1\to 2\to 3$.
– user147263
Jan 16, 2016 at 22:45
• @Normal right, that makes sense... but, there still must be some power for which $P^k=I$. So, something similar to what I have currently written is possible. I wonder, is this possible for $3 \times 3$ matrix examples. Great comment, I wish I could upvote comments ( for internet pts naturally) Jan 17, 2016 at 1:39
• For all permutation matrices: $P \cdot P^{T} = I$ Feb 5, 2019 at 8:03 | {
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# Math Help - Two Discrete Problems
1. ## Two Discrete Problems
Hello,
I am needing some help on the two problems below. Thank you in advance.
1) Prove by contradiction that there do not exist integers m and n such that 14m + 21n = 101.
and
2) Using the fact that an integer n, either n >= 14 or n <= 15, or otherwise prove 7 does not divide 100.
2. Originally Posted by MathStudent1
Hello,
I am needing some help on the two problems below. Thank you in advance.
1) Prove by contradiction that there do not exist integers m and n such that 14m + 21n = 101.
.
gcd(14,21)=7
Thus,
7 divides 101 which is false.
3. Thank you ThePerfectHacker. I made a typo when I first made my post. Sorry about that. So, it should be 14m + 21n = 100 not 101.
I am beening taught to turn this into a if/then statement and go from there.
If m and n are integers, then 14m + 21n = 100
Factor out a 7 givng us 7(2n +3m) = 100
From here I have trouble. Also, should the if/then statement be not equal to 100 instead of equal to? I haven't had that much experience with proof by contradiction. Thank You.
4. Originally Posted by MathStudent1
Thank you ThePerfectHacker. I made a typo when I first made my post. Sorry about that. So, it should be 14m + 21n = 100 not 101.
I am beening taught to turn this into a if/then statement and go from there.
If m and n are integers, then 14m + 21n = 100
Factor out a 7 givng us 7(2n +3m) = 100
From here I have trouble. Also, should the if/then statement be not equal to 100 instead of equal to? I haven't had that much experience with proof by contradiction. Thank You.
Let,
p= (there exists n,m such that 14m+21n=100).
Now if "p" is true we shown that 7 divides 100.
Thus,
q=(7 divides 100).
But this is false.
Meaning, (modens ponens)
[(p--> q) and (~q)]--> ~p
Thus,
~p=(there do not exists n,m such that 14m+21n=100)
This is mine 46th Post!!!
5. Hello, MathStudent1!
Here;s the first one . . . | {
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This is mine 46th Post!!!
5. Hello, MathStudent1!
Here;s the first one . . .
1) Prove by contradiction that there do not exist integers m and n
. . .such that: .14m + 21n .= .101
Assume that there are integers m and n such that: .14m + 21n .= .101
. . . . . . . . . . . . . . . . . . .100
Divide by 7: . 2m + 3n .= .----
. . . . . . . . . . . . . . . . . . . .7
In the left side: m is an integer, so 2m is an integer. *
. . . . . . . . . . . . n is an integer, so 3n is an integer. *
Hence, 2m + 3n (the sum of two integers) is an integer. *
But the right side is an irreducible fraction (or decimal 14.2857...)
We have reached a contradicition, hence our assumption was incorrect.
. . There are no integers m and n that satisfy 14m + 21n .= .101
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
*
Reasons: The set of integers is closed under multiplication and addition.
6. Thank you Soroban. One question for you. When orginally I posted my thread I made a typo in that 14m + 21n = 100 NOT 101. Does that change the proof at all or does the logic remain the same. Following your proof, am I correct by stating this below:
Prove that there do not exist integers m and n such that: 14m + 21n = 100
Restating the problem - If m and n are integers, then 14m + 21n = 100
Factor out a 7 and we get 7(2m + 3n) = 100
Divide both sides by 7 and we get 2m + 3n = (100/7)
Since m is an integer then 2*m is an integer and since n is an integer then 3*n is and integer.
However, the RHS (100/7) is not an integer and therefore,
2m + 3n (IS NOT EQUAL TO) (100/7)
Hence, we have reached a contradiction of our assumption that m and n are integers.
Therefore, there do not exist integers m and n such that 14m + 21n = 100.
*****Thank you!*****
7. Hello again, MathStudent1!
Yes, the modified proof is still valid . . .
. . and you did an excellent job! | {
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The combination (n r) (n r) is called a binomial coefficient. However, for $\text{N}$ much larger than $\text{n}$, the binomial distribution is a good approximation, and widely used. \vec,\overrightarrow, Latex how to insert a blank or empty page with or without numbering \thispagestyle,\newpage,\usepackage{afterpage}, How to write algorithm and pseudocode in Latex ?\usepackage{algorithm},\usepackage{algorithmic}, How to display formulas inside a box or frame in Latex ? Fractions and binomial coefficients are common mathematical elements with similar characteristics - one number goes on top of another. therefore gives the number of k -subsets possible out of a set of distinct items. Blog template built with Bootstrap and Spip by Nadir Soualem @mathlinux. A slightly different and more complex example of continued fractions, Showing first {{hits.length}} results of {{hits_total}} for {{searchQueryText}}, {{hits.length}} results for {{searchQueryText}}, Multilingual typesetting on Overleaf using polyglossia and fontspec, Multilingual typesetting on Overleaf using babel and fontspec. Click on one of the binomial coefficient designs, which look like the letters "n" over "k" inside either a round or angled bracket. are the different ordered arrangements of a k-element subset of an n-set, $$\binom{n}{k} = \binom{n-1}{k-1} +\binom{n-1}{k}$$. The binomial coefficient is defined by the next expression: $\binom {n}{k} = \frac {n ! k-combinations of n-element set. Binomial coefficient, returned as a nonnegative scalar value. infinite sum of inverse binomial coefficient encountered in Bayesian treatment of the German tank problem Hot Network Questions Why are quaternions more … Binomial coefficient denoted as c (n,k) or n c r is defined as coefficient of x k in the binomial expansion of (1+X) n. The Binomial coefficient also gives the value of the number of ways in which k items are chosen from among n objects i.e. It is especially useful for reasoning about recursive methods in | {
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chosen from among n objects i.e. It is especially useful for reasoning about recursive methods in programming. (n - k)!} On the other side, \textstyle will change the style of the fraction as if it were part of the text. In latex mode we must use \binom fonction as follows: \frac{n!}{k! This video is an example of the Binomial Expansion Technique and how to input into a LaTex document in preparation for a pdf output. The binomial coefficient can be interpreted as the number of ways to choose k elements from an n-element set. Below is a construction of the first 11 rows of Pascal's triangle. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … where A is the permutation, A_n^k = \frac{n!}{(n-k)! See for instance the documentation of Integrate.. For Binomial there seems to be no such 2d input, because as you already found out, \binom{n}{k} is … }}{{k!\left( {n - k} \right)!}} In Counting Principles, we studied combinations.In the shortcut to finding ${\left(x+y\right)}^{n}$, we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. \boxed, How to write table in Latex ? How to write number sets N Z D Q R C with Latex: \mathbb, amsfonts and \mathbf, How to write angle in latex langle, rangle, wedge, angle, measuredangle, sphericalangle, Latex numbering equations: leqno et fleqn, left,right, How to write a vector in Latex ? . The following are the common definitions of Binomial Coefficients.. A binomial coefficient C(n, k) can be defined as the coefficient of x^k in the expansion of (1 + x)^n.. A binomial coefficient C(n, k) also gives the number of ways, disregarding order, that k objects can be chosen from among n objects more formally, the number of k-element subsets (or k-combinations) of a n-element set. Any coefficient $a$ in a term $ax^by^c$ of the expanded version is known as a binomial coefficient. ... | {
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coefficient $a$ in a term $ax^by^c$ of the expanded version is known as a binomial coefficient. ... Pascal’s triangle. Specially useful for continued fractions. Don't forget to LIKE, COMMENT, SHARE & SUBSCRIBE to my channel. } Also, the text size of the fraction changes according to the text around it. So The combination (nr)\displaystyle \left(\begin{array}{c}n\\ r\end{array}\right)(nr) is calle… For example, … If your equation requires specific numbers in place of the "n" or "k," click on a letter to select it, press "Delete" and enter a number in its place. In mathematics, the binomial coefficients are the positive integers that occur as coefficients in the binomial theorem.Commonly, a binomial coefficient is indexed by a pair of integers n ≥ k ≥ 0 and is written (). It is the coefficient of the x k term in the polynomial expansion of the binomial power (1 + x) n, and is given by the formula =!! For these commands to work you must import the package amsmath by adding the next line to the preamble of your file, The appearance of the fraction may change depending on the context. Home > Latex > FAQ > Latex - FAQ > Latex binomial coefficient, Monday 9 December 2019, by Nadir Soualem. Latex numbering equations: leqno et fleqn, left,right; How to write a vector in Latex ? b is the same type as n and k. If n and k are of different types, then b is returned as the nondouble type. LaTeX forum ⇒ Math & Science ⇒ Expression like binomial Coefficient with Angle Delimiters Topic is solved Information and discussion about LaTeX's math and science related features (e.g. therefore gives the number of k-subsets possible out of a set of distinct items. The binomial coefficient (n k) ( n k) can be interpreted as the number of ways to choose k elements from an n-element set. C — All combinations of v matrix. Identifying Binomial Coefficients. This same array could be expressed using the factorial symbol, as shown in the following. Binomial coefficients are a family of positive | {
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the factorial symbol, as shown in the following. Binomial coefficients are a family of positive integers that occur as coefficients in the binomial theorem. The order of selection of items not considered. The binomial coefficient is defined by the next expression: \ [ \binom{n} {k} = \frac{n!} In UnicodeMath Version 3, this uses the \choose operator ⒞ instead of the \atop operator ¦. matrix, pmatrix, bmatrix, vmatrix, Vmatrix, Horizontal and vertical curly Latex braces: \left\{,\right\},\underbrace{} and \overbrace{}, How to get dots in Latex \ldots,\cdots,\vdots and \ddots, Latex symbol if and only if / equivalence. The binomial coefficient \binom{n}{k} can be interpreted as the number of ways to choose k elements from an n-element set. ( n - k )! Ak n = n! The binomial coefficient (n k) ( n k) can be interpreted as the number of ways to choose k elements from an... Properties. Then it's a good reason to buy me a coffee. (n−k)! Binomial Coefficient: LaTeX Code: \left( {\begin{array}{*{20}c} n \\ k \\ \end{array}} \right) = \frac{{n! Thank you ! (n - k)!} Latex binomial coefficient Definition. One can drop one of the numbers in the bottom list and infer it from the fact that sum … The second statement requires solving a simple exercise with pencil and paper, in which you use the definition of binomial coefficients to prove the implication. This website was useful to you? Pascal's triangle can be extended to find the coefficients for raising a binomial to any whole number exponent. Latex The binomial coefficient is defined by the next expression: \[\binom {n}{k} = \ frac {n!}{k!(n-k)! {k! As you may have guessed, the command \frac{1}{2} is the one that displays the fraction. A General Note: Binomial Coefficients If n n and r r are integers greater than or equal to 0 with n ≥r n ≥ r, then the binomial coefficient is Binomial coefficient, returned as a nonnegative scalar value. In this case, we use the notation (nr)\displaystyle \left(\begin{array}{c}n\\ | {
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scalar value. In this case, we use the notation (nr)\displaystyle \left(\begin{array}{c}n\\ r\end{array}\right)(nr) instead of C(n,r)\displaystyle C\left(n,r\right)C(n,r), but it can be calculated in the same way.$ And of course this command can be included in the normal text flow \ (\binom{n} {k}\). Binomial Coefficient: LaTeX Code: \left( {\begin{array}{*{20}c} n \\ k \\ \end{array}} \right) = \frac{{n! Using fractions and binomial coefficients in an expression is straightforward. The text inside the first pair of braces is the numerator and the text inside the second pair is the denominator. In Counting Principles, we studied combinations. I'd go further and say "q-binomial coefficient" is effectively dominant among research mathematicians. In the shortcut to finding (x+y)n\displaystyle {\left(x+y\right)}^{n}(x+y)n, we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. The command \displaystyle will format the fraction as if it were in mathematical display mode. This will give more accuracy at the cost of computing small sums of binomial coefficients. Mathematical Equations in LaTeX. n! begin{tabular}...end{tabular}, Latex horizontal space: qquad,hspace, thinspace,enspace, LateX Derivatives, Limits, Sums, Products and Integrals, Latex copyright, trademark, registered symbols, How to write matrices in Latex ? This article explains how to typeset them in LaTeX. Binomial coefficients are common elements in mathematical expressions, the command to display them in LaTeX is very similar to the one used for fractions. I agree. The binomial coefficient is the number of ways of picking unordered outcomes from possibilities, also known as a combination or combinatorial number. Binomial coefficients are common elements in mathematical expressions, the command to display them in LaTeXis very similar to the one used for fractions. LaTeX provides a feature of special editing tool for scientific tool for math | {
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used for fractions. LaTeX provides a feature of special editing tool for scientific tool for math equations in LaTeX. \vec,\overrightarrow; Latex how to insert a blank or empty page with or without numbering \thispagestyle,\newpage,\usepackage{afterpage} Latex natural numbers; Latex real numbers; Latex binomial coefficient; Latex overset and underset ; Latex absolute value The possibility to insert operators and functions as you know them from mathematics is not possible for all things. In Counting Principles, we studied combinations.In the shortcut to finding$\,{\left(x+y\right)}^{n},\,$we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. b is the same type as n and k. If n and k are of different types, then b is returned as the nondouble type. In this video, you will learn how to write binomial coefficients in a LaTeX document. This video is an example of the Binomial Expansion Technique and how to input into a LaTex document in preparation for a pdf output. = \binom{n}{k}$$Binomial coefficient denoted as c(n,k) or n c r is defined as coefficient of x k in the binomial expansion of (1+X) n.. }{k ! k-combinations of n-element set. Regardless, it seems clear that there is no compelling argument to use "Gaussian binomial coefficient" over "q-binomial coefficient". Gerhard "Ask Me About System Design" Paseman, 2010.03.27 \endgroup – Gerhard Paseman Mar 27 '10 at 17:00 (adsbygoogle = window.adsbygoogle || []).push({}); Binomial coefficients are common elements in mathematical expressions, the command to display them in LaTeX is very similar to the one used for fractions. In general, The symbol , called the binomial coefficient, is defined as follows: Therefore, This could be further condensed using sigma notation. Latex k parmi n - coefficient binomial. Binomial coefficients are common elements in mathematical expressions, the command to display them in LaTeX is very similar to the one used for fractions. The | {
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expressions, the command to display them in LaTeX is very similar to the one used for fractions. The binomial coefficient also arises in combinatorics, where it gives the number of different combinations of $b$ elements that can be chosen from a … All combinations of v, returned as a matrix of the same type as v. The binomial coefficient is the number of ways of picking unordered outcomes from possibilities, also known as a combination or combinatorial number. You can set this manually if you want. The symbols and are used to denote a binomial coefficient, and are sometimes read as " choose." The Binomial coefficient also gives the value of the number of ways in which k items are chosen from among n objects i.e. binomial Stanley's EC1 also uses it as the primary name, which counts for a lot in my book. An example of a binomial coefficient is (5 2)= C(5,2)= 10 (5 2) = C (5, 2) = 10. {k! How to write it in Latex ? {k! Any coefficient $a$ in a term $ax^by^c$ of the expanded version is known as a binomial coefficient. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … Knowledge base dedicated to Linux and applied mathematics. It will give me the energy and motivation to continue this development. (−)!. The second fraction displayed in the previous example uses the command \cfrac{}{} provided by the package amsmath (see the introduction), this command displays nested fractions without changing the size of the font. Since binomial coefficients are quite common, TeX has the \choose control word for them. Toutes les versions de cet article : Le coefficient binomial est le nombre de possibilités de choisir k élément dans un ensemble de n éléments. The Texworks shows … }}{{k!\left( {n - k} \right)!}} In mathematics, the Gaussian binomial coefficients (also called Gaussian coefficients, Gaussian polynomials, or q-binomial coefficients) are q-analogs of the binomial | {
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coefficients, Gaussian polynomials, or q-binomial coefficients) are q-analogs of the binomial coefficients.The Gaussian binomial coefficient, written as () or [], is a polynomial in q with integer coefficients, whose value when q is set to a prime power counts the number of subspaces of dimension k in a vector … Accordingly the binomial coefficient in the binomial theorem above can be written as “n\choose k”, assuming that you type a space after the k. This formulas, graphs). Asking for help, clarification, or responding to other answers. Binomial coefficients have been known for centuries, but they're best known from Blaise Pascal's work circa 1640. All combinations of v, returned as a matrix of the same type as v. As you see, the command \binom{}{}will print the binomial coefficient using the parameters passed inside the braces. In latex mode we must use \binom fonction as follows: \frac {n!} C — All combinations of v matrix. The symbols and are used to denote a binomial coefficient, and are sometimes read as "choose.". Here's an equation: math \frac {n!} For these commands to work you must import the package amsmath by adding the next line to the preamble of your file }}{{k!\left( {n - k} \right)!}}. This method of constructing mathematical proofs is called mathematical induction. Binomial coefficients are the ones that appear as the coefficient of powers of x x x in the expansion of (1 + x) n: (1+x)^n: (1 + x) n: ( 1 + x ) n = n c 0 + n c 1 x + n c 2 x 2 + ⋯ + n c n x n , (1+x)^n = n_{c_{0}} + n_{c_{1}} x + n_{c_{2}} x^2 + \cdots + n_{c_{n}} x^n, ( 1 + x ) n = n c 0 + n c 1 x + n c 2 x 2 + ⋯ + n c n x n , As you see, the command \binom{}{} will print the binomial coefficient using the parameters passed inside the braces. The usual binomial coefficient can be written as \left({n \atop {k, {n-k}}}\right). = \binom{n}{k} = {}^{n}C_{k} = C_{n}^k$$, \frac{n!}{k! \\binom{N} {k} What differs between \\dots and \\dotsc, with overleaf.com, the outputs are identical. | {
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{k} What differs between \\dots and \\dotsc, with overleaf.com, the outputs are identical. Usually, you find the special input possibilities on the reference page of the function in the Details section. If the sampling is carried out without replacement, the draws are not independent and so the resulting distribution is a hypergeometric distribution, not a binomial one. coefficient The usage of fractions is quite flexible, they can be nested to obtain more complex expressions. (n-k)!} Using fractions and binomial coefficients in an expression is straightforward. Then it's a good reason to buy me a coffee. = \binom {n} {k} This is the binomial coefficient. binomial coefficient Latex. Binomial Coefficient. samedi 11 juillet 2020, par Nadir Soualem. (adsbygoogle = window.adsbygoogle || []).push({}); All the versions of this article: For these commands to work you must import the package amsmath by adding the next line to the preamble of your file In this article, you will learn how to write basic equations and constructs in LaTeX, about aligning equations, stretchable horizontal lines, operators and delimiters, fractions and binomials. (n - k)!} Open an example in Overleaf Style of the function in the following left, right ; how input... Ways in which k items are chosen from among n objects i.e to typeset them in Latex we. Is a construction of the fraction mathematical elements with similar characteristics - one number goes on top of another to! Binomial coefficients are common mathematical elements with similar characteristics - one number goes on top another... Complex expressions Spip by Nadir Soualem @ mathlinux the other side, \textstyle will change the style the. And motivation to continue this development numbering equations: leqno et fleqn, left, right ; to... Tool for math equations in Latex binomial coefficients are common elements in mathematical expressions, the command \frac { -... Expression: \ [ \binom { } will print the binomial coefficient '' is | {
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the command \frac { -... Expression: \ [ \binom { } will print the binomial coefficient '' is dominant! Were part of the binomial coefficient '' over q-binomial coefficient '' is effectively dominant among research.... Latex - FAQ > Latex > FAQ > Latex binomial coefficient '' over coefficient... math \frac { n! } } the number of ways of picking unordered binomial coefficient latex from possibilities also! Recursive methods in programming or combinatorial number is quite flexible, they can be as! From Blaise Pascal 's triangle can be extended to find the coefficients for raising a coefficient. As if it were in mathematical display mode items are chosen from n... From mathematics is not possible for all things Pascal 's work circa 1640 display them in LaTeXis very to. And \\dotsc, with overleaf.com, the command \displaystyle will format the fraction as if were. A vector in Latex mode we must use \binom fonction as follows: \frac n... \Choose operator ⒞ instead of the \atop operator ¦ article explains how to input into a Latex in... Of ways of picking unordered outcomes from possibilities, also known as a or! Ways to choose k elements from an n-element set )! } } {. Of a set of distinct items in programming -subsets possible out of a set of distinct items,... \\Dots and \\dotsc, with overleaf.com, the outputs are identical of constructing mathematical proofs is a... Help, clarification, or responding to other answers, which counts for a pdf output fractions. Possible out of a set of distinct items r ) is called mathematical induction 'd further... That displays the fraction changes according to the one that displays the fraction as if it were in display! Are common mathematical elements with similar characteristics - one number goes on of! The special input possibilities on the other side, \textstyle will change the style of the first of! Which counts for a lot in my book in the Details section coefficient! N r ) is called mathematical induction insert operators and | {
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book in the Details section coefficient! N r ) is called mathematical induction insert operators and functions as you see, command. Page of the function in the binomial coefficient explains how to write a in! Typeset them in LaTeXis very similar to the one used for fractions the operator... Reasoning about recursive methods in programming is effectively dominant among research mathematicians a coffee to display them in LaTeXis similar. Coefficient is defined by the next expression: \ [ \binom { n! } } { { k !, also known as a combination or combinatorial number be extended to find the special input possibilities on reference... = \frac { 1 } { { k } = \frac binomial coefficient latex }! Display mode chosen from among n objects i.e, with overleaf.com, the command \binom }. An n-element set coefficient also gives the number of k-subsets possible out of a set of items. Objects i.e energy and motivation to continue this development Latex binomial coefficient were part of the binomial coefficient, are! Tool for math equations in Latex mode we must use \binom fonction as:! Coefficients have been known for centuries, but they 're best known Blaise. Methods in programming size of the fraction as if it were in mathematical expressions, the command \binom }., or responding to other answers one that displays the fraction known as a combination or number. Is quite flexible, they can be nested to obtain more complex expressions a construction of the first 11 of! Style of the number of k-subsets possible out of a set of distinct items, it seems that... \Displaystyle will format the fraction changes according to the one used for fractions and. [ \binom { n } { } { k } \right )! } } { k... This article explains how to input into a Latex document in preparation for a in! Distinct items were in mathematical display mode side, \textstyle binomial coefficient latex change the style of first..., \textstyle will change the style of the function in the following here 's an equation: this. | {
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\textstyle will change the style of the function in the following here 's an equation: this. Useful for reasoning about recursive methods in programming the energy and motivation to continue this development SUBSCRIBE... Fleqn, left, right ; how to write a vector in Latex mode we must use \binom as... This same array could be expressed using the parameters passed inside the braces for a in! Energy and motivation to continue this development { { k! \left ( { n! }.: \frac { n } { { k! \left ( {!. \Left ( { n! } { k! \left ( { n }... Are binomial coefficient latex elements in mathematical expressions, the text inside the braces combinatorial number 11 rows Pascal! Math \frac { n! } } { k! \left ( n! The one used for fractions instead of the fraction 2 } is the binomial coefficient using parameters. N-Element set numerator and the text inside the first 11 rows of Pascal 's.! Coefficients are common elements in mathematical expressions, the command \displaystyle will format the fraction =. As a combination or combinatorial number dominant among research mathematicians of k-subsets possible out of a of... From Blaise Pascal 's work circa 1640 on the other side, \textstyle will change the of... = \binom { n! } } { k } \right )! } { k =! Number exponent outputs are identical, also known as a combination or combinatorial number is! Also known as a combination or combinatorial number or combinatorial number Latex binomial coefficient the. Gaussian binomial coefficient can be nested to obtain more complex expressions below is a construction the! Equation: math \frac { n! } } { k } = {... Over q-binomial coefficient '' is effectively dominant among research mathematicians number of ways choose. } } { { k } \right )! } { { k } \right )! }.. Will format the fraction as if it were in mathematical expressions, the outputs are identical do forget., as shown in the following SUBSCRIBE to my channel among n objects i.e Latex - FAQ Latex. Complex expressions: \ [ \binom { n! } } not | {
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in an expression is straightforward first pair of is. Then it 's a good reason to buy me a coffee coefficients have been known centuries. | {
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binomial coefficient latex 2021 | {
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# Compound Interest Systems of Equations Problem
1. Apr 16, 2015
### ConstantineO
1. The problem statement, all variables and given/known data
Romeo was given a gift of $10,000 when he turned 16. He invested it at 3% per annum. Three years later, Juliet was given$10,000, which she invested at 5% per annum. When will the two amounts be equal in value?
2. Relevant equations
Compound Interest Formula
Total = Capital(1+interest)^Years
t = c(1+i)^n
3. The attempt at a solution
Deciding how to create my formula is where things get fuzzy. The way I initially attempted this was to add a 3 year head start onto Romeo's interest formula. This is what results (Used wolfram to save time rewriting):
This eventually results in "4.61103." However, this is not the answer at the back of my textbook. They decided to opt for a different formula where the time is subtracted from Juliet's interest function. This is shown here:
This results in the correct answer of 7.61103. What I'm wondering, why do I have to subtract 3 years from Juliet's compound interest formula, and why does adding 3 years to Romeo's compound interest formula produce the wrong answer?
2. Apr 16, 2015
### BvU
So clearly the book starts counting at the beginning of the story: when Romeo brings his money to the bank.
That means Julia's interest clock is at x-3 when Romeo's is at x.
3. Apr 16, 2015
### SammyS
Staff Emeritus
What is Romeo's age when then two accounts have equal value?
4. Apr 16, 2015
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4. Apr 16, 2015
### ConstantineO
Is this always the convention when dealing with problems like this? From my experience with physics courses, I assumed that time shares a relation with the total amount of money generated by compound interest. The interest and initial capital is thus fixed, and you are only left with a relation between time and the total sum of money. If you were to treat time as a vector which you could progress backwards and forwards on, each snapshot through time depending on which way you were progressing would net either a larger or smaller amount of the total. I don't understand why the book is forcing me to start at the beginning of the story. The story is like pages of a book which can be turned towards or away from the end of the novel. The content of the novel does not change and its contents are already written.
I don't understand what Romeo's age has to do with this? Is this some kind of statement that alludes to time only progressing forward or something?
5. Apr 16, 2015
### BvU
Nothing so fancy. The book is sloppy by not stating when it wants the clock (calendar in this case) to show t = 0. From the book answer it appears that it assumes t = 0 when Romeo turns 16. The exercise could equally well have assumed t = 0 when Julia receives the ten grand. If I were grading this, I'd have to allow both answers, certainly when the calculation steps are shown.
6. Apr 16, 2015
### ConstantineO
I actually just figured it out while discussing this question with a friend just a few moments before you posted. I took the delta of the two times from both of the ways of solving this question and realized that this is nothing more than a semantics issue of where you measure time from. This question is poorly setup, and I am going to voice my irritation to the person marking.
7. Apr 16, 2015
### Ray Vickson | {
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7. Apr 16, 2015
### Ray Vickson
A more serious problem is that when compounding annually, the answer is "never". At Romeo's 23rd birthday, Romeo's account contains more money than Juliet's, while at his 24th birthday, Juliet's account contains more than Romeo's. Unless there is something like continuous compounding, the two amounts never match exactly; that is, unless you can withdraw your money part-way through a year and earn a part-year's interest, you could never find a time where the two withdrawals are equal. (However, I would not raise this issue with your teacher if I were you; just "go with the flow".)
Also, instead of expressing irritation, it would be wiser for you to say that you will give two answers (depending on where to start measuring time) and point out (nicely) that the question is a bit ambiguous.
8. Apr 16, 2015
### SammyS
Staff Emeritus
RGV makes an excellent point. This is a short-coming of many problems of this type.
It's often the case that for any fraction of a single compounding period that's "left over", simple interest is computed for that extra amount of time. With this method, find the value of the account at x = 4, then see if the values can be equal at any time in the 4th year using simple interest.
(Edited slightly.)
Last edited: Apr 16, 2015
9. Apr 16, 2015
### ConstantineO
I don't wish to sound like an imbecile, but I have no clue what that means. How would I set up a systems of equation if I subbed in 4 for both x's? I believe I would be left with only one unknown for either equation, or am I misunderstanding you? Please clarify what you mean by using x=4.
10. Apr 16, 2015
### SammyS
Staff Emeritus
When x = 4, (7years for Romeo, 4 years for Juliet), what is the value in each in each account?
Start from that point for each & using simple interest, and see how long it takes for the values to be equal.
11. Apr 16, 2015
### ConstantineO | {
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11. Apr 16, 2015
### ConstantineO
I am not familiar with the simple interest formula, so I googled it. I am operating under the assumption that it is different than the compound interest formula.
I am to assume this is correct?
Simple interest Formula
I = "the grown money"
p = the initial capital
r = interest rate per year
t = t
(I) = (p)(r)(t)
I think I know what you're getting at, so let me make an attempt.
Romeo Compound Interest Calculation Total
Ar = 10,000(1.03)^4+3
Ar = 10,000(1.03)^7
Ar = 12,298.73865
Julie Compound Interest Calculation Total
Aj = 10,000(1.05)^4
Aj = 12,155.0625
Romeo Simple Interest Calculation
(Ir) = (12,298.73865)(0.03)(t)
(Ir) = (368.94)(t)
Juliet Simple Interest Calculation
(Ir) = (12,155.0625)(0.05)(t)
(Ir) = 607.753125(t)
Equate the two to equal each other.
(368.94)(t) = 607.753125(t)
(368.94)(t) -607.753125(t) = 0
t(368.94-607.753125)=0
This doesnt work though...
12. Apr 16, 2015
### Ray Vickson
It's not that complicated. Let $R_t, J_t$ be the amounts in Romeo's and Juliet's accounts at time $t$ (in years, measured from Romeo's 16th birthday). We have:
$$\begin{array}{ccc} & t=7 & t=8 \\ R_t & 12298.74 & 12667.70 \\ J_t & 12155.06 & 12762.82 \end{array}$$
You can draw two straight-line plots, one going from x = 0 at (7,R7) to x = 1 at (8,R8) and another from x = 0 at (7,J7) to x = 1 at (8,J8). Where the two lines cross is the point where simple interest calculations (for the partial year) would give you equality of monetary values. This will be close to (but not exactly equal to) the value x = 0.611---that is, for time 7.611 ---that your previous calculation would have produced. (The difference is due to the slight curvature in the money vs.time graph over the year, due to continuous compounding over the year, as compared with the straight line obtained from simple interest accumulation over the year.)
13. Apr 16, 2015
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13. Apr 16, 2015
### ConstantineO
Could you show me an example of this because I am simply not understanding what you're saying, or can you show me how to equate the two together using algebra?
14. Apr 17, 2015
### ConstantineO
This is getting out of hand, so I am going to try to be as clear and as concise as possible. I know how to calculate the correct answer for this question, and I realize where I first made my mistake. I am now wondering what everyone is talking about when they mention using simple interest. I have literally 0 experience doing any simple interest calculations, and I have no idea what sammyS is talking about here.
If anyone is holding information back because they believe an attempt has not been made and it would violate forum rules, please stop. I understand how to do the book's question using the books method, and I know and where the discrepancy between the 4.61103 and 7.61103 came from. I have made a wholehearted attempt so please for sake of my sanity just give me an example or step by step breakdown. I learn through examples, I look at the process involved, and I create a model in my head how everything interacts and relates with each other.
What I am seeking is an explanation and a simple example of what this simple interest method is for calculating the 0.61103 year after the exact 7 years has passed. | {
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What I am understanding so far.
- First use the compound interest formula to create outputs for both Romeo's and Juliet's accounts after being influenced by exactly 7 years of interest.
- This is done by assigning "4" to the variable x
- Get said products, in this case being:
Romeo's Total After Exactly 7 Years = 12,298.73865
Juliet's Total After Exactly 7 Years = 12,155.0625
- Since you know that the two amounts are still not equal, you must calculate the remaining time between the two using the Simple interest formula
- This is what I am having problems with and what I would appreciate having an example for. Preferably shown algebraically, so I am not plotting nearly straight lines in Desmos.
Not to sound contentious, but when you try to find the point of intersect of two nearly straight lines with a slope that is in hundredths of a unit, it is just plainly byzantine. I am probably doing this wrong, but graphically representing each simple interest formula and trying to find where they meet is insane.
Last edited: Apr 17, 2015
15. Apr 17, 2015
### Ray Vickson
How could I have made my explanation any simpler? There was no hidden information or holding back of anything. I said: just draw two straight-line graphs for Romeo and Juliet, where the lines connect their (time,money) points at the start and end of the final year.
The true graphs of time vs. money will be curved, because of compound interest, but over a short period such as one year the "curvature" will be small, and the graph will look almost like a straight line; replacing the curve by a straight line ---only for that single year---would give you the "simple" interest schedule for that year. It will be almost the same as the compound interest schedule, because the curvature is small over short times such as a year. | {
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If you don't believe me, just draw the graphs of money vs. time for Romeo and for Juliet. You will see that they curve up, but are almost straight over short periods. The only way to really understand it is to sit down and do it.
16. Apr 17, 2015
### ConstantineO
I am going to assume the two lines would be plotted as
y=(12155.0625)(0.05)(x)
and
y=(12298.73865)(0.03)(x)
If that's the case, I don't understand how any human being is supposed to find the point of intersect graphically like this.
I did sit down and do it. I have been sitting down and ripping my hair out doing this for sometime.
Edit - In the event that anyone else would like to question how committed I am to "sitting down and doing it." I've had enough of that nonsense from teachers with superiority issues and over bloated opinions that like to question how hard I am making an attempt.
Last edited: Apr 17, 2015
17. Apr 17, 2015
### Ray Vickson
As long as you persist in trying to plot the wrong thing you will need to keep pulling out your hair. Go back and read what I wrote in #12. Do the plot exactly as I indicated there.
Better still plot the two curves y = 10000 * (1.03)^x and y = 10000 * (1.05)^(x-3) for x = from 3 to 8. They are curved, aren't they? Don't they look almost straight over the shorter interval 7 \leq x \leq 8? in each case if you were to plot a straight line from (7,y(7)) to (8,y(8)) it would look very close to the actual curve (x,y(x)) for x between 7 and 8. The "simple interest" effect between 7 and 8 would be the straight line, while the "compound interest" effect would be the curve. They both pass through the same points at 7 and 8, but they differ in between.
18. Apr 17, 2015
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18. Apr 17, 2015
### ConstantineO
As I have stated before.
I was completely uncertain of what to plot. I am being told to use the simple interest formula, but the two formulas you have provided are compound interest formulas:
y = 10000 * (1.03)^x
y = 10000 * (1.05)^(x-3)
I plotted these long before I even got to this whole simple interest business when trying to figure out the original discrepancy between 7.61103 and 4.61103. I am fully aware that the compound interest lines are nearly straight. I don't see how the average rate of change between x =7 and x =8 is going to help me discover the remaining time of 0.61103 years. I am still unclear of what to plot? Do I plot the compound interest based functions for Romeo and Juliet, or do I plot the Simple interest functions? I have done both and I am not seeing anything that is giving me any kind of better understanding. Is this operation you describe too difficult represent algebraically?
I am getting far too irritated by this and far too frustrated and am quickly losing the desire to continue.
19. Apr 17, 2015
### ConstantineO
I was on the right track here. There is no need for this graphing nonsense. I forgot a single number which was my "+ 3" in my simple interest formula. I was a fool not to recognize that the lack of a constant would render my simple interest system of equation unsolvable. The lack of anyone speaking up about this is worrisome. In the future, when referencing the slope of a secant line between the range of two x values of a curve, please write in the notation of 7 < x < 8. I had absolutely no clue what you were talking about.
So let's try that last bit again with a fixed formula.
Romeo Simple Interest Calculation
(Ir) = (12,298.73865)(0.03)(t+3)
(Ir) = (368.94)(t+3)
Juliet Simple Interest Calculation
(Ir) = (12,155.0625)(0.05)(t)
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Juliet Simple Interest Calculation
(Ir) = (12,155.0625)(0.05)(t)
(Ir) = 607.753125(t)
Now the Systems of Equation
(607.753125)(t) = (368.94)(t+3)
(607.753125)(t) = (368.94)(t)+ 1106.82
t(607.753125-368.94) = 1106.82
t(238.813125) = (1106.82)
t = (1106.82)/(238.813125)
t = 4.634669891
Not too far away from 4.61103. I wonder why...
Let's try the simplified interest formula that measures time from Julias deposit date.
Romeo Simple Interest Calculation
(Ir) = (12,298.73865)(0.03)(t)
(Ir) = (368.94)(t)
Juliet Simple Interest Calculation
(Ir) = (12,155.0625)(0.05)(t-3)
(Ir) = 607.753125(t-3)
607.753125(t-3)= (368.94)(t)
(607.753125)(t) - 1823.259375 = (368.94)(t)
(607.753125)(t) - (368.94)(t) =1823.259375
t(607.753125 - 368.94) = 1823.259375
t(238.813125) = 1823.259375
t = 7.634.634669891
Not too far away from 7.61103. Surprise surprise....
Look Ma! No graphs.
7.634.634669891 - 4.634669891 = 3 just like 7.61103 - 4.61103 =3. I wonder if they're related?
I don't think I am wrong by stating that the time after the 7 year mark would actually be 0.634669891 of a year instead of 0.61103 if fractional compound interest is indeed calculated with simplified interest.
We can take things a step further and look at the secant line's slope of the curve that models interest from Romeo's deposit date during 7 < x < 8. Let's look at the average rate of change for both Romeo's and Juliet's account.
Romeo Compound Interest AROC: 7 < x < 8
While x = 4
Ar = 10,000(1.03)^4+3
Ar = 10,000(1.03)^7
Ar = 12,298.73865
x = 4
y =12,298.73865
While x =5
Ar = 10,000(1.03)^5+3
Ar = 10,000(1.03)^8
Ar = 12,667.70081
x = 5
y = 12,667.70081
Romeo's AROC
= (12,667.70081 - 12,298.73865) / (5 -4)
= 368.96216
Juliet's Compound Interest AROC 7 < x < 8
While x = 4
Aj = 10,000(1.05)^4
Aj = 12,155.0625
x = 4
y =12,155.0625
While x = 5
Aj = 10,000(1.05)^5
Aj = 12,762.81563
x = 5
y = 12,762.81563
Juliet's AROC
= (12,762.81563 - 12,155.0625) / (5-4)
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Juliet's AROC
= (12,762.81563 - 12,155.0625) / (5-4)
=607.7531
After that huge bunch of calculations we now have slopes that we can create linear relations with.
Romeo's Linear Relation
y = (368.96216)(x + 3)
Juliet's Linear Relation
y = 607.7531(x)
Let's see what the Solution is for these in equations when put into a system.
(368.96216)(x + 3) = 607.7531(x)
(368.96216)(x) + 1106.88648 = 607.7531(x)
607.7531(x) -(368.96216)(x) = 1106.88648
x(607.7531-368.96216) = 1106.88648
x(238.783884) = 1106.88648
x = 4.635515854
Notice a pattern?
Time Measured from Romeo's Deposit Compound Calculation Product
x = 4.61103
Time Measured from Romeo's Deposit Simplified Interest Product
x = 4.634669891
Time Measured from Romeo's Deposit AROC: 7 < x< 8 Product
x = 4.635515854
I think its safe to say that they matched sometime in the range of 0.635515854 - 0.634669891- 0.61103 of the start of the 8th year. I think I made my point. I don't need any graphs to know how to rock, and I certainly showed that I sat down and did the work.
20. Apr 17, 2015
### Ray Vickson
Of course it can be done without graphs. Graphs can be of help in setting up the algebraic equations that must, eventually, be solved without graphs. However, you seemed to not understand how simple interest works (your claim, not mine) and to see that a graphical representation can sometimes be helpful---not always, just sometimes. If it was not helpful to you, fine. And, of course I suggested plotting the compound-interest graphs, but I guess you missed out the part where I said that during a short period, such as 1 year, the graph looks almost straight, and that a straight line replacement for the graph (but ONLY in that single year) gives you the simple-interest effect within the year. | {
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Your writeup is almost incomprehensible to me, but your final answer seems OK. You say you still do not know why it is different from the original answer, and that is precisely what I was speaking of when I mentioned graphs: you could see right away why there is a difference.
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Finding common terms of two or more arithmetic sequences
Suppose I have three sequences $\{1,4,7,....,2998\}$ , $\{1,3,5,7,9,11,....,3001\}$ and $\{1,6,11,....,4001\}$ .
How can i find the number of common terms among them ?
For Example ,
If I have
1,3,5,7,9,11,13,15,17 and 19
1,4,7,10,13,16 and 19
1,6,11 and 16
My answer would be $6$.
Suppose i have $n$ number of series' with common differences $a_1,a_2,...,a_n$ and each of them having the starting term as $1$ . How can i find the number of common terms then ?
• Does your answer go all the way to the ends 2998,3001,4001 for the count of 0nly 6 common terms? – coffeemath Dec 5 '16 at 15:15
• its 6 for only the given example – psil123 Dec 5 '16 at 15:16
Write the arithemtic sequence in form: $a_n = a_0 + (n-1)d_1, b_n = b_0 + (n-1)d_2, c_n = c_0 + (n-1)d_3$. Now if they have a common term then we can write:
$$x = a_0 + kd_1 = b_0 + sd_2 = c_0 + td_3$$
As the common difference is known we can write this as:
$$x \equiv a_0 \pmod {d_1}$$ $$x \equiv b_0 \pmod {d_2}$$ $$x \equiv c_0 \pmod {d_3}$$
This is system of linear congurence relations and it can be easily solved by the Chinese Remainder Theorem.
And once you find one common element, to find the next common element you just add $LCM[d_1,d_2,d_3]$ to the previous one. In fact if you get $x \equiv A \pmod{LCM[d_1,d_2,d_3]}$ when solving the system of congurences, then total number of elements should be:
$$1 + \left\lfloor\frac{K-A}{LCM[d_1,d_2,d_3]}\right\rfloor$$
where $K$ is the smallest last element in the arithemtic sequences.
If you at the explicit form of your sequences, which build your sets, you get get that $a_n = 1+3 \cdot n$, $b_n = 1+ 2 \cdot n$ and $c_n =1+5 \cdot n$.
Now you just have to find the least common multiple of 2,3 and 5 and see that all their common terms are $c_n = 1 + 30 \cdot n$ | {
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• suppose i have n series with common differences a1,a2,...,an and each of them have the starting number as 1 . How can i find the number of common terms ? – psil123 Dec 5 '16 at 15:17 | {
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# Seemingly tricky dice question-probability that one event occurs before another event?
The question is as follows:
You roll two fair dice over and over. Let $$A$$ be the event you see two even sums. Let $$B$$ be the event you see a sum of $$7$$ four times. What is the probability that event $$A$$ occurs before event $$B$$?
I know that for mutually exclusive events with independent trials, the probability that event $$E$$ occurs before event $$F$$ is $$\frac{\mathbb{P}(E)}{\mathbb{P}(E) + \mathbb{P}(F)}.$$ I tried using this formula, but I ran into a problem. I calculated $$\mathbb{P}(A)=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4} \ \ \ \ \text{and} \ \ \ \ \mathbb{P}(B)=\left ( \frac{1}{6} \right )^4=\frac{1}{1296}.$$ However, I then realized that these probabilites are the events that two even sums occur $$\textit{in a row},$$ and similarly for my $$\mathbb{P}(B).$$
Does anyone have any suggestions on how to figure this out, or even if this formula is the one I should be using?
Edit: I know there is a geometric distribution involved.
• If it helps approach the problem, A occurs before B is usually just a less obvious way to say A occurs without B – Lord Farquaad Oct 16 '18 at 12:54
Since neither A nor B cares about odd sums that aren't 7, we'll just reroll in those cases and thus we can ignore the probability of those sums occurring.
Thus we're left with 7 in addition to the 6 even sums ($$2,4,6,8,10,12$$).
The probabilities of these are respectively $$6/36$$ and $$(1+3+5+5+3+1)/36 = 18/36$$.
Divide the probabilities of each of these by the sum of both probabilities, to get $$P(7)$$ and $$P(even)$$ such that $$P(7) + P(even) = 1$$ (this is $$P(7) = \frac{1}{4}$$ and $$P(even) = \frac{3}{4}$$).
Now, as bobajob pointed out, the probability of A occurring before B is the complement of the probability of B occurring before A.
$$P(2\ even\ before\ 4\ sevens) = 1-P(4\ sevens\ before\ 2\ even)$$ | {
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$$P(2\ even\ before\ 4\ sevens) = 1-P(4\ sevens\ before\ 2\ even)$$
To have 4 sums of seven before 2 even sums, there can be at most 1 even sum before the 4th seven sum.
Thus we have the following possible sequences:
$$7\ 7\ 7\ 7$$
$$7\ 7\ 7\ even\ 7$$
$$7\ 7\ even\ 7\ 7$$
$$7\ even\ 7\ 7\ 7$$
$$even\ 7\ 7\ 7\ 7$$
Each of the above occurs with the following probability:
$$P(7) * P(7) * P(7) * P(7) = P(7)^4$$
$$P(7) * P(7) * P(7) * P(even) * P(7) = P(even) * P(7)^4$$
$$P(7) * P(7) * P(even) * P(7) * P(7) = P(even) * P(7)^4$$
$$P(7) * P(even) * P(7) * P(7) * P(7) = P(even) * P(7)^4$$
$$P(even) * P(7) * P(7) * P(7) * P(7) = P(even) * P(7)^4$$
Then we have:
$$P(2\ even\ before\ 4\ sevens) = 1-P(4\ sevens\ before\ 2\ even)$$
$$= 1 - P(7)^4 - 4 * P(even) * P(7)^4$$
$$= 1 - (\frac{1}{4})^4 - 4 * \frac{3}{4} * (\frac{1}{4})^4$$
$$= 0.984375$$
## Alternative method
The probability we want will just be the probability of getting 0-3 sums equal to 7 before getting 2 even sums.
$$P(2\ even\ before\ 4\ sevens) = \sum_{i=0}^3P(i\ sevens\ before\ 2\ even)$$
Since one even sum will need to be at the end, we can simply consider all possible positions of the other even sum.
Taking as an example $$i=3$$ (3 7 sums before 2 even sums), we'll have the following possible order of events:
$$7\ 7\ 7\ even\ even$$
$$7\ 7\ even\ 7\ even$$
$$7\ even\ 7\ 7\ even$$
$$even\ 7\ 7\ 7\ even$$
The amount of these we have is simply $$i + 1$$.
Each of the above occurs with probability $$P(even)^2 * P(7)^i$$:
$$P(7) * P(7) * P(7) * P(even) * P(even) = P(even)^2 * P(7)^3$$
$$P(7) * P(7) * P(even) * P(7) * P(even) = P(even)^2 * P(7)^3$$
$$P(7) * P(even) * P(7) * P(7) * P(even) = P(even)^2 * P(7)^3$$
$$P(even) * P(7) * P(7) * P(7) * P(even) = P(even)^2 * P(7)^3$$
Thus we have:
$$P(i\ sevens\ before\ 2\ even) = (i+1) * P(even)^2 * P(7)^i$$
This gives us:
$$P(2\ even\ before\ 4\ sevens) = \sum_{i=0}^3(i+1) * P(even)^2 * P(7)^i$$
$$= P(even)^2 * (1 + 2*P(7) + 3*P(7)^2 + 4*P(7)^3)$$ | {
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$$= P(even)^2 * (1 + 2*P(7) + 3*P(7)^2 + 4*P(7)^3)$$
$$= (\frac{3}{4})^2 * (1 + 2*\frac{1}{4} + 3*(\frac{1}{4})^2 + 4*(\frac{1}{4})^3)$$
$$= 0.984375$$
More generally, the positions of the even sums above will be a multiset permutation and one can come up with general expression for the probability of M events X occurring before N events Y, but that's a bit beyond the scope of this question.
• Nice approach. Would it not be easier to calculate the probability of event $B$ occurring before $A$ and then take the complement though? ($i$ would just be 0 or 1...) – bobajob Oct 16 '18 at 10:41
• @bobajob Good point, edited. – Dukeling Oct 16 '18 at 11:53
• "Since neither A nor B cares about odd sums that aren't 7" Yes, they do. A sum of 2 followed by a sum of 3 followed by a sum of 4 does not constitute an instance of A, while a sum of 2 followed by a sum of 4 does. – Acccumulation Oct 16 '18 at 15:20
• After rereading the question, I likely misunderstood it; I thought it was two even sums in a row. – Acccumulation Oct 16 '18 at 15:30
I would do a Markov chain with the states being the number of even rolls/the number of seven rolls. Work backwards starting from the state that you have seen one even roll and three $$7$$s. What is the chance that A comes first from there?
You have two Bernoulli processes,$$S_A$$ and $$S_B$$, and you are asked about the probability that the 2nd arrival in $$S_A$$ process occurs before the 4th arrival in $$S_B$$ process. PMF of the time of $$k$$-th arrival in a Bernoulli process with probability of success $$p$$ is $$p_{X_k}(t)=\binom{t-1}{k-1}p^k(1-p)^{t-k}, \ t=k,k+1,\ldots$$ | {
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PS: the events "sum is even" and "sum is 7" are dependent, and it makes the question tricky indeed, and the trick is, as Dukeling explained, is to ignore odd sums not equal to $$7$$; then the (conditional) probability of getting the sum of $$7$$ is $$\frac{6}{36-12}=\frac{1}{4}$$. Now the complement of the event of interest is the event that the 4th arrival in the Bernoulli process with $$p=1/4$$ occurs at time $$4$$ or $$5$$, and its probability is equal to $$\binom{3}{3}p^4+\binom{4}{3}p^4(1-p)=p^4(1+4(1-p))=\frac{1}{64}$$ and the final answer is $$1-\frac{1}{64}=\frac{63}{64}$$ | {
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# Inequality for Fibonacci to find an upper bound of harmonic Fibonacci series
I want to find an sharp upper bound for $$\sum_{n=1}^{\infty}\frac{1}{F_n}$$which $F_n$ is the n$th$ term of Fibonacci sequence . I wrote a Matlab program to find an upper bound ,$\sum_{n=1}^{10^6}\frac{1}{F_n}<4$
Now my question is:(1):Is there an inequality to find this ?
(2): Is that series have a close form ? $${F_n} = \frac{{{\varphi ^n} - {{( - \varphi )}^{ - n}}}}{{\sqrt 5 }}\to \\\sum_{n=1}^{\infty}\frac{1}{F_n}=\sum_{n=1}^{\infty}\frac{\sqrt 5}{{{\varphi ^n} - {{( - \varphi )}^{ - n}}}}\\\leq \sum_{n=1}^{\infty}\frac{\sqrt 5}{{{(\frac{{1 + \sqrt 5 }}{2} ) ^n} }}=\frac{\sqrt5}{1-\frac{1}{\frac{{1 + \sqrt 5 }}{2}}}\approx12.18\\$$I am thankful for a hint or solution which can bring a sharper upper bound .
• maybe related – MAN-MADE Sep 1 '17 at 7:33
• Simple fast program: 3.359885666243177553172011302918927179688905133731968486495553815325130318996683383615416216456790087297045342928853913304136789017100883679591351733077119078580333550332507753187599850487179777897006039564509215375892775265673354024033169441799293934610992626257964647651868659449710216558984360881472693249591079473873673378523326877499762727757946853676918541981467668742998767382096913901217722024405208151094264934951 – Kenny Lau Sep 1 '17 at 7:33
• WolframAlpha simply returns ℱ suggesting that it has no closed form. – Kenny Lau Sep 1 '17 at 7:35
• @KennyLau:Can I see your program ? can you post it here ? – Khosrotash Sep 1 '17 at 7:35
• $F_n-\phi^n/\sqrt5=-(-1/\phi)^n/\sqrt5$ has alternating signs. Is it clear that the errors have a sum with the correct sign, and that your inequality follows? – Jyrki Lahtonen Sep 1 '17 at 21:02
As Jyrki Lahtonen points out in the comments, $\sum_{n=1}^\infty\frac{\sqrt5}{\varphi^n}$ isn't necessarily the right bound, since it doesn't dominate the original series term-by-term. | {
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As for a closed form, this identity is known: $$\sum_{n=1}^\infty\frac{1}{F_n}=\frac{\sqrt5}{4}\left(\vartheta_2^2(\varphi^{-2}) + \frac{\log5 + 2\psi_{\varphi^{-4}}(1) - 4\psi_{\varphi^{-2}}(1)}{2\log\varphi}\right)$$ where $\vartheta_2(q)$ is the Jacobi theta function at $z=0$, and $\psi_q$ is the $q$-digamma function. See Wikipedia and MathWorld on the "Reciprocal Fibonacci constant", and other Math.SE questions such as What is the sum of Fibonacci reciprocals?
• $F_n-\phi^n/\sqrt5=-(-1/\phi)^n/\sqrt5$ has alternating signs. Is it clear that the errors have a sum with the correct sign? The OP suffers from the same problem. – Jyrki Lahtonen Sep 1 '17 at 20:59
• @JyrkiLahtonen Whoops! Let me fix that. – Chris Culter Sep 1 '17 at 21:12
Here is a systematic method for computing upper bounds for the sum without much work.
By induction, if $F_N \ge b^N$ and $b+1 \ge b^2$, then $F_n \ge b^n$ for $n \ge N$.
Therefore, $$\sum_{n=1}^\infty {1\over F_n} \le \sum_{n=1}^{N-1} {1\over F_n} + \sum_{n=N}^\infty {1\over b^n} = \sum_{n=1}^{N-1} {1\over F_n} + \frac{1}{b^{N-1}(b-1)}$$ This upper bound gets closer to the actual sum when $b$ gets larger, but then we need larger $N$:
\begin{array}{crl} b &N &sum \\ 1.3 &4 &4.0172204521317 \\ 4/3 &5 &3.7825520833333 \\ 1.4 &6 &3.4981694135380 \\ 1.5 &11 &3.3651521005948 \\ 1.6 &72 &3.3598856662432 \\ \end{array}
Since $1.6 \approx \phi$, which is the largest possible $b$, the last value is quite close to the actual value: $$3.359885666243177553172011302918927179688905133732\cdots$$
Using Wolfram Mathematica answer is : Wolfram Mathematica Code | {
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Making statements based on opinion; back them up with references or personal experience. Why did mainframes have big conspicuous power-off buttons? Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Useful relations in dealing with binomial coefficients and factorials 4. How to sustain this sedentary hunter-gatherer society? Maybe I'm wrong in the conclusions about what that particular convergence means though. If success probabilities differ, the probability distribution of the sum is not binomial. \Pr[S\ge s] Determination of the binomial coefficient and binomial distribution The probability of any specified arrangement of k successes and n-k failures in n independent trials is pknkq − where p is the probability of success on any one trial and q=1-p is the probability of failure. Often the manipulation of integrals can be avoided by use of some type of generating function. "In the limit as n→∞, your binomials become Gaussian" Sorry but this is simultaneously vague and wrong. Asking for help, clarification, or responding to other answers. Why were there only 531 electoral votes in the US Presidential Election 2016? That depends on the range of values you are considering. where we took $t=\log(s/\sum_ip_i)$. $$By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. Why Is an Inhomogenous Magnetic Field Used in the Stern Gerlach Experiment? site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. The Kolmogorov approximation is given as an … Yes, in fact, the distribution is known as the Poisson binomial distribution, which is a generalization of the binomial distribution. See the binomial sum variance inequality. I know that I can solve this exercise by using the fact that a negative binomial distributed RV is a sum of geometric distributed RV, but i want to show it with my attempt. Both distributions have | {
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is a sum of geometric distributed RV, but i want to show it with my attempt. Both distributions have total mass 1. How do rationalists justify the scientific method. rev 2020.11.24.38066, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us, I have the same question and i read the paper (The Distribution of a Sum of Binomial Random Variables by Ken Butler and Michael Stephens). @jameselmore Additivity of the means is unrelated to independence. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. \\&= \exp\left(\sum_i 1 + (e^t-1) p_i\right) \exp(-st) It will be a special case of the Poisson Binomial Distribution. Are there relatively simple formulae or at least bounds for the distribution \Pr[S\ge s] \\&= \exp\left(\sum_i 1 + (e^t-1) p_i\right) \exp(-st) Extremely bloated transaction log during delete. Every second customer converts better. See this paper (The Distribution of a Sum of Binomial Random Variables by Ken Butler and Michael Stephens). Which one is more idiomatic: ‘valid concern’ or ‘legitimate concern’? Can this be by chance? To subscribe to this RSS feed, copy and paste this URL into your RSS reader. If success probabilities differ, the probability distribution of the sum is not binomial. Here is an excerpt from the Wikipedia page. by Marco Taboga, PhD. The distribution of a sum S of independent binomial random variables, each with different success probabilities, is discussed.$$ @Robert ,do you have any insight on what happens if the n is not same for the 2 distributions. This lecture discusses how to derive | {
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on what happens if the n is not same for the 2 distributions. This lecture discusses how to derive the distribution of the sum of two independent random variables.We explain first how to derive the distribution function of the sum and then how to derive its probability mass function (if the summands are discrete) or its probability density function (if the summands are continuous). This answer provides an R implementation of the explicit formula from the paper linked in the accepted answer (The Distribution of a Sum of Binomial Random Variables by Ken Butler and Michael Stephens). | {
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# Is there a series where the terms tend to zero faster than the harmonic series but it still diverges?
I know that for the harmonic series $\lim_{n \to \infty} \frac1n = 0$ and $\sum_{n=1}^{\infty} \frac1n = \infty$.
I was just wondering, is there a sequence ($a_n =\dots$) that converges "faster" (I am not entirely sure what's the exact definition here, but I think you know what I mean...) than $\frac1n$ to $0$ and its series $\sum_{n=1}^{\infty}{a_n}= \infty$?
If not, is there proof of that?
• Sure. Let $a_n=\frac1{n\log n}$. Or let $a_n=\frac1{n\log n\log\log n}$. Or let ... In fact, the process never ends, in that you can find a divergent series whose terms go to zero faster than all the sequences I suggested. – Andrés E. Caicedo Dec 1 '13 at 19:39
• Versions of this question have been asked before. Here is an example from MO. – Andrés E. Caicedo Dec 1 '13 at 19:41
• ok, it this checks out! (using the condensation test) – guynaa Dec 1 '13 at 19:46
• Related The answers by by Bill/Gone and Yuval. – user17762 Dec 1 '13 at 20:15
• $$\sum_{n=1}^{\infty} \frac{1}{10n} ?$$ – Sawarnik Feb 24 '14 at 21:13
## 3 Answers
There is no slowest divergent series. Let me take this to mean that given any sequence $a_n$ of positive numbers converging to zero whose series diverges, there is a sequence $b_n$ that converges to zero faster and the series also diverges, where "faster" means that $\lim b_n/a_n=0$. In fact, given any sequences of positive numbers $(a_{1,n}), (a_{2,n}),\dots$ with each $\sum_n a_{i,n}=\infty$ and $\lim_n a_{i+1,n}/a_{i,n}=0$, there is $(a_n)$ with $\sum a_n=\infty$ and $\lim_n a_n/a_{i,n}=0$ for all $i$. | {
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To see this, given $a_1,a_2,\dots$, first define $b_1=a_1,b_2=a_2,\dots,b_k=a_k$ until $a_1+\dots+a_k>1$. Second, let $b_{k+1}=a_{k+1}/2,b_{k+2}=a_{k+2}/2,\dots,b_n=a_n/2$ until $a_{k+1}+\dots+a_n>2$, etc. That is, we proceed recursively; if we have defined $b_1,\dots,b_m$ and $b_m=a_m/2^r$, and $b_1+\dots+b_m>r+1$, let $b_{m+1}=a_{m+1}/2^{r+1},\dots,b_l=a_l/2^{r+1}$ until $a_{m+1}+\dots+a_l>2^{r+1}$. The outcome is that $\sum b_i=\infty$ and $\lim b_i/a_i=0$.
Similarly, given $(a_{1,n}),(a_{2,n}),\dots$, with each $(a_{k+1,n})$ converging to zero faster than $(a_{k,n})$, and all of them diverging, let $a_i=a_{1,i}$ for $i\le n_1$, where $a_{1,1}+\dots+a_{1,n_1}>1$, then $a_i=a_{2,i}$ for $n_1<i\le n_2$, where we ask both that $a_{2,n_1+1}+\dots+a_{2,n_2}>1$ and that for any $k>n_2/2$ we have $a_{2,k}/a_{1,k}<1/2$, etc. That is, if we have defined $n_k$, we let $a_i=a_{k+1,i}$ for $n_k<i\le n_{k+1}$ where $n_{k+1}$ is chosen so that $a_{k+1,n_k+1}+\dots+a_{k+1,n_{k+1}}>1$ and for all $l>n_{k+1}/2$ we have $a_{k+1,l}/a_{i,l}<1/2^{k+1}$ for all $i<k+1$. Then the series $\sum a_i$ diverges, and the sequence $(a_i)$ converges to $0$ faster than all the $a_{i,n}$. In modern language, we would say that there are no $(\omega,0)$-gaps in a certain partial order.
We can modify the above slightly so that given any sequences $(a_{i,n})$ with $\sum_n a_{i,n}<\infty$ and $\lim_n a_{i+1,n}/a_{i,n}=\infty$ for all $i$, we can find $(a_n)$ with $\sum_n a_n<\infty$ and $\lim_n a_n/a_{i,n}=\infty$, so there is no fastest convergent series, and not even considering a sequence of faster and faster convergent series is enough. (In modern terms, there is no $(0,\omega)$-gap.) | {
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What we cannot do in general is, given $(a_{i,n})$, with all $\sum_n a_{i,n}=\infty$, find $(a_n)$ with $\sum a_n=\infty$ and $a_n/a_{i,n}\to0$ for all $i$, if the $a_{i,n}$ are not ordered so that $a_{i+1,n}$ converges to zero faster than $a_{i,n}$. For example, we can have $a_n=1/n$ if $n$ is odd and $a_n=1/n^2$ if $n$ is even, and $b_n=1/n^2$ if $n$ is odd and $b_n=1/n$ if $n$ is even, and if $c_n$ converges to zero faster than both, then $\sum c_n$ converges. (These exceptions can typically be fixed by asking monotonicity of the sequences, which is how these results are usually presented in the literature.)
Note that the argument I gave is completely general, no matter what the series involved. For specific series, of course, nice "formulas" are possible. For example, given $a_n=1/n$, we can let $b_n=1/(n\log n)$ for $n>1$. Or $c_n=1/(n\log n\log\log n)$, for $n\ge 3$. Or ... And we can then "diagonalize" against all these sequences as indicated above.
By the way, the first person to study seriously the boundary between convergence and divergence is Paul du Bois-Reymond. He proved a version of the result I just showed above, that no "decreasing" sequence of divergent series "exhausts" the divergent series in that we can always find one diverging and with terms going to zero faster than the terms of any of them. A nice account of some of his work can be found in the book Orders of Infinity by Hardy. Du Bois-Reymond's work was extended by Hadamard and others. What Hadamard proved is that given $(a_i)$ and $(b_i)$ with $\sum a_i=\infty$, $\sum b_i<\infty$, and $b_i/a_i\to 0$, we can find $(c_i),(d_i)$ with $c_i/a_i\to0$, $b_i/d_i\to 0$, $d_i/c_i\to0$, $\sum c_i=\infty$, $\sum d_i<\infty$. More generally:
If we have two sequences of series, $(a_{1,n}), (a_{2,n}),\dots$ and $(b_{1,n}),(b_{2,n}),\dots$, such that | {
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• Each $(a_{i+1,n})$ converges to zero faster than the previous one,
• Each $(b_{i+1,n})$ converges to zero slower than the previous one,
• Each $(a_{i,n})$ converges to zero slower than all the $(b_{j,n})$,
• Each $\sum_n a_{i,n}$ diverges, and
• Each $\sum_n b_{i,n}$ converges,
then we can find sequences $(c_n),(d_n)$, "in between", with one series converging and the other diverging.
In modern language, we say that there are no $(\omega,\omega)$-gaps, and similarly, there are no $(\omega,1)$- or $(1,\omega)$-gaps. This line of research led to some of Hausdorff's deep results in set theory, such as the existence of so-called $(\omega_1,\omega_1)$- or Hausdorff gaps. What Hausdorff proved is that this "interpolation" process, which can be iterated countably many times, cannot in general be carries out $\omega_1$ times, where $\omega_1$ is the first uncountable ordinal.
I had wondered about this too a long time ago and then came across this. The series
$$\sum_{n=3}^{\infty}\frac{1}{n\ln n (\ln\ln n)}=\infty$$
diverges and it can be very easily proven by the integral test. But here is the kicker. This series actually requires googolplex number of terms before the partial sum exceeds 10. Talk about slow! It is only natural that if the natural log is slow, then take the natural log of the natural log to make it diverge even slower.
Here is another one. This series
$$\sum_{n=3}^{\infty}\frac{1}{n\ln n (\ln\ln n)^2}=38.43...$$
actually converges using the same exact (integral) test. But it converges so slowly that this series requires $10^{3.14\times10^{86}}$ before obtaining two digit accuracy. So using these iterated logarithms you can come up with a series which converges or diverges "arbitrarily slowly".
Reference: Zwillinger, D. (Ed.). CRC Standard Mathematical Tables and Formulae, 30th ed. Boca Raton, FL: CRC Press, 1996. | {
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• But then by the same integral test $$\sum_{n=N_0}^\infty \sum_{n=3}^{\infty}\frac{1}{n\ln n (\ln\ln n) (\ln\ln \ln n)}=\infty$$ diverges at an even slower rate....Moreo generarily $$\sum \frac{1}{n \ln(n) \ln^2(n) \ln^3(n) ... \ln^k(n)}$$ diverges, where $^j$ denotes composition . – N. S. Dec 2 '13 at 0:46
• @N.S. That's what I said, using iterated logarithm you can come up with an even slower growth. In addition, in your first summation, the double summation doesn't make sense. I think you want the inner sigma removed. $N_0 = \lceil((e^x)^j(0)\rceil$ where $j$ denotes composition. In your first summation with $j=3$ we have $N_0=16$. – Fixed Point Dec 2 '13 at 2:31
• this is a fantastic example--showing that even trying to test convergence by simulation could be doomed, because we don't have the computational capacity to find enough terms to convince ourselves the sequence diverges. – MichaelChirico Apr 14 '15 at 14:56
If we use the notion of a partial sum:
$$S_n = \sum_{k=1}^n a_k$$
you are asking for a series in which the partial sums diverge more slowly than $\sum_{k=1}^n 1/k$ as $n\to\infty$. For this to happen, we just need to find $a_k$ such that $a_k < 1/k$ for all $k>c$ where $c$ is some value.
Look at $a_k = \frac{1}{k\ln k}$. Let's do the integral test to show this diverges. The substitution used is $u = \ln k$ so that $du = dk/k$.
$$\int \frac{dk}{k\ln k}=\int \frac{du}{u}=\ln u = \ln {\ln k}$$
Putting in limits of $2$ and $\infty$ shows that this indeed diverges. And for $k>c=e$, we have that $1/(k\ln k) < 1/k$. | {
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# A simple method of factorization $2^{30}+1$
How can you factor $2^{30} + 1$? This task was supposed to be at one interview, there is an assumption that there should be a simple solution.
Observe that $$x^k+1=\frac{x^{2k}-1}{x^k-1}=\frac{\prod_{d\mid 2k}\Phi_{d}(x)}{\prod_{d\mid k}\Phi_{d}(x)}=\prod_{d\mid 2k,\ d\nmid k}\Phi_{d}(x)$$ With $$\Phi_n(x)$$ being the $$n^\text{th}$$ cyclotomic polynomial. It follows that $$2^{30}+1=\Phi_4(2)\Phi_{12}(2)\Phi_{20}(2)\Phi_{60}(2)$$ With the individual factors computed as $$\Phi_4(2)=2^2+1=5$$ $$\Phi_{12}(2)=2^4-2^2+1=13$$ $$\Phi_{20}(2)=2^8-2^6+2^2-2^2+1=5\times 41$$ $$\Phi_{60}(2)=2^{16}+2^{14}-2^{10}-2^8-2^6+2^2+1=61\times 1321$$ Giving an overall factorization of $$\boxed{2^{30}+1=5^2\times 13\times 41\times 61\times 1321}$$
• This is very elegant. Aug 6 '18 at 1:58
• Thanks! It's also not too hard to compute the relevant polynomials by hand (the section "Easy cases for computation" in the linked wiki article lists some strategies) :) Aug 6 '18 at 2:01
• Again, computing $\Phi_{60}(2) = 61 \times 1321$ is not really something to do in an interview. Aug 6 '18 at 7:26
• Perhaps they wanted to see the way to a solution, not the solution itself. Aug 6 '18 at 11:43
• @RobertIsrael: probably the interviewer wanted to hear something along the lines "in order to find the prime divisors of $\Phi_{60}(2)$, it is enough to check the primes $\equiv 1\pmod{60}$. Since the order of $2$ in $\mathbb{Z}/(61\mathbb{Z})^*$ is what it is, $61$ is a prime divisor of $\Phi_{60}(2)$ and the problem boils down to checking that $1321$ is a prime". Aug 6 '18 at 16:45 | {
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I would note it is a sum of cubes and fifth powers, so $2^{10}+1=1025$ and $2^6+1=65$ are both factors. In terms of primes that gives us $5^2\cdot 13 \cdot 41$ as factors. At this point in an interview (depending on the position) I would declare success and move on. Getting the remaining $80581$ would take a bunch of hand calculation and finding $61\cdot 1321$ doesn't seem reasonable either.
• $61\equiv 5 \bmod 8$ so $(2|61)\equiv 2^{30}\equiv -1\bmod 61$. Aug 6 '18 at 1:44
Hint: $a^k+1$ is divisible by $a+1$ if $k$ is odd.
• Just so we can check any number-theoretic derivation... Mathematica gives: $5^2 \times 13 \times 41 \times 61 \times 1321$. (Doesn't seem so simple that it could be done in an interview session...) Aug 6 '18 at 0:37
Yet another easy observation: since $4X^4+1=(2X^2+2X+1)(2X^2-2X+1)$, we get $4\cdot2^{28}+1=(2\cdot2^{14}+2\cdot2^7+1)(2\cdot2^{14}-2\cdot2^7+1)$.
Firstly, in an interview there is often a difference between showing that there is a simple solution (existence) and outlining the approach, versus actually finding the solution.
That is, it might suffice to outline a standard prime factorising algorithm and show that it could be computed in trivial time.
One of the simplest classic approaches is that to factor $x$, you only need to test the primes that are less than $\sqrt{x}$. If $x=2^{30}+1$, then $\sqrt{x+1} \simeq 32768$. There are less than 3000 primes that satisfy this constraint, so it is very feasible on any computing platform.
Once each prime factor was found, one would divide that factor out, and then only need to search for primes up to an even smaller upper bound. Thus after the finding that 5 is a prime, one need only search for other prime factors less than 14654.
Thus, you have an algorithm that would find all the factors of $2^{30}+1$ in ascending order.
That is, $2^{30}+1 = 5 \times 5 \times 13 \times 41 \times 61 \times 1321$. | {
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That is, $2^{30}+1 = 5 \times 5 \times 13 \times 41 \times 61 \times 1321$.
The key to nearly any sensible prime factorization approach is to efficiently find the first factor. Thus, in some ways it could be said that factoring $2^{30}+1$ is easy because it has lots of prime factors, and most of them are very small. (Contrast this to encryption are based on numbers that are very hard to factor, because they typically one have 2 factors and each of them are very very large.)
Anyway, as @Robert indicated, the key to solving this directly is to realise that if $k$ is odd, then $x^k+1$ can be elegantly factorised, as $(x+1)$ is a factor.
For example, $$x^7+1 = (x+1)(x^6-x^5+x^4-x^3+x^2-x+1)$$ Note that $z^{30}+1 = (z^2)^{15}+1$ and so $z^2+1$ is a factor.
Letting $z=1$ shows that 5 is a factor of $2^{30}+1$.
Similarly, you could use the well-known sum of cubes factorisation. $x^3+1 = (x+1)(x^2-x+1)$, to show that $$2^{30}+1 = (2^{10}+1)(2^{20}-2^{10}+1) = 1025 \times 1047553$$ and continue from there.
• Your early complete factorization is missing $61$ Aug 6 '18 at 1:59 | {
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# Fitting a two-dimensional Gaussian to a set of 2D pixels
Imagine I have a set of data like the following:
data = {{0.0453803, 0.0427863, 0.0489815, 0.045243, 0.0488289, 0.0432898,
0.04448, 0.0387732, 0.0388952}, {0.0507668, 0.0427863, 0.0502632,
0.0503395, 0.0634623, 0.0675822, 0.0529335, 0.047425,
0.0387121}, {0.042237, 0.0501259, 0.0595712, 0.0869001, 0.139559,
0.141512, 0.0868391, 0.0579232, 0.0408331}, {0.0478981, 0.0491646,
0.0652628, 0.130404, 0.218448, 0.220645, 0.143603, 0.0605173,
0.0424964}, {0.0462043, 0.0530861, 0.076051, 0.140017, 0.206943,
0.202502, 0.118791, 0.0614023, 0.0459907}, {0.0511788, 0.0582132,
0.105531, 0.166354, 0.181003, 0.13698, 0.0748302, 0.0557107,
0.0492103}, {0.0493629, 0.0539712, 0.0971695, 0.160769, 0.164477,
0.104768, 0.0591745, 0.0475319, 0.0452583}, {0.0510719, 0.0599374,
0.0730602, 0.0975814, 0.101289, 0.0691997, 0.0498054, 0.044892,
0.043122}, {0.0460517, 0.0567025, 0.0574044, 0.0587778, 0.0537118,
0.0487221, 0.0474098, 0.0413977, 0.04477}}
I don't have enough reputation to post an image, but one can easily be generated by applying ListPlot3D to the above data set.
How might I best fit a Gaussian curve to this set of datapoints, and extract properties such at the fit' semi-axes? I noticed that ComponentMeasurements has some functionality for best fit ellipsoids, but that doesn't seem to be workable here.
My objective here is to determine how "Gaussian" a set of points in an image are. My strategy is to sequentially fit a 2D Gaussian to each point, and then to measure it's eccentricity and spread (looking, for example, at the length and ratio of the semiaxes of the ellipsoid corresponding to the fit). The example here seems like it should yield a 2D Gaussian fit with significant spread and a ratio for the semiaxes significantly diverging from one. | {
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-
Maybe You can add some context to the question. I'm asking because this data looks rather like sum of two 2D gaussians, and I don't know if it is relevant. – Kuba Jun 25 '13 at 22:36
@Kuba I have added a bit of context in the problem description! – Bob Jun 25 '13 at 22:42
By context I mean, for example, what this data actually is. :) Without details, term "determine how Gaussian" is too vague. – Kuba Jun 25 '13 at 23:07
Sjoerd's answer applies the power of Mathematica's very general model fitting tools. Here's a more low-tech solution.
If you want to fit a Gaussian distribution to a dataset, you can just find its mean and covariance matrix, and the Gaussian you want is the one with the same parameters. For Gaussians this is actually the optimal fit in the sense of being the maximum likelihood estimator -- for other distributions this may not work equally well, so there you will want NonlinearModelFit.
One wrinkle is that your data doesn't fall off to zero but to something like Min[data] $\approx 0.0387$, but we can easily get rid of that by just subtracting it off. Next, I normalize the array to sum to $1$, so that I can treat it like a discrete probability distribution. (All this really does is allow me to avoid dividing by Total[data, Infinity] every time in the following.)
min = Min[data];
sum = Total[data - min, Infinity];
p = (data - min)/sum;
Now we find the mean and covariance. Mathematica's functions don't seem to work with weighted data, so we'll just roll our own.
{m, n} = Dimensions[p];
mean = Sum[{i, j} p[[i, j]], {i, 1, m}, {j, 1, n}];
cov = Sum[Outer[Times, {i, j} - mean, {i, j} - mean] p[[i, j]], {i, 1, m}, {j, 1, m}];
We can easily get the probability distribution for the Gaussian with this mean and covariance. Of course, if we want to match the original data, we have to rescale and shift it back.
f[i_, j_] := PDF[MultinormalDistribution[mean, cov], {i, j}] // Evaluate;
g[i_, j_] := f[i, j] sum + min; | {
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The match is not too bad, although the data has two humps where the Gaussian has one. You can also compare the fit through a contour plot. (Use the tooltips to compare contour levels.)
Show[ListPlot3D[data, PlotStyle -> None],
Plot3D[g[i, j], {j, 1, 9}, {i, 1, 9}, MeshStyle -> None,
PlotStyle -> Opacity[0.8]], PlotRange -> All]
Show[ListContourPlot[data, Contours -> Range[0.05, 0.25, 0.025],
ContourShading -> None, ContourStyle -> ColorData[1, 1], InterpolationOrder -> 3],
ContourPlot[g[i, j], {j, 1, 9}, {i, 1, 9}, Contours -> Range[0.05, 0.25, 0.025],
ContourShading -> None, ContourStyle -> ColorData[1, 2]]]
The variances along the principal axes are the eigenvalues of the covariance matrix, and the standard deviations (which I guess you're calling the semi-axes) are their square roots.
Sqrt@Eigenvalues[cov]
(* {1.86325, 1.50567} *)
-
Or you could just plot the Gaussian fit coloured by the residual: Show[ListDensityPlot[ Table[data[[i, j]] - g[i, j], {i, 1, m}, {j, 1, m}], InterpolationOrder -> 3, PlotRange -> Full, ColorFunctionScaling -> False, ColorFunction -> (With[{t = 20 #}, RGBColor[1 + t, 1 - Abs[t]/2, 1 - t]] &)], ContourPlot[g[x, y], {y, 1, 9}, {x, 1, 9}, Contours -> Range[0.05, 0.25, 0.025], ContourShading -> None, ContourStyle -> Black]] i.stack.imgur.com/LqGZL.png (orange: data higher than fit, blue: data lower than fit) – Rahul Narain Jun 27 '13 at 6:23
Thanks again for such a fantastic answer. Just to help me understand - why does your calculation for the square root of the eigenvectors yield a slightly different result than StandardDeviation@MultinormalDistribution[mean,cov]? – Bob Jun 30 '13 at 11:15
Because that one appears to give the standard deviations of the marginal distributions of $x$ and $y$ instead, which you also get with Sqrt@Diagonal[cov]. – Rahul Narain Jun 30 '13 at 11:28
data3D = Flatten[MapIndexed[{#2[[1]], #2[[2]], #1} &, data, {2}], 1]; | {
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"url": "http://mathematica.stackexchange.com/questions/27642/fitting-a-two-dimensional-gaussian-to-a-set-of-2d-pixels/27754"
} |
fm = NonlinearModelFit[data3D,
a E^(-(((-my + y) Cos[b] - (-mx + x) Sin[b])^2/(2 sy^2)) -
((-mx + x) Cos[b] + (-my + y) Sin[b])^2/(2 sx^2)),
{{a, 0.1}, {b, 0}, {mx, 4.5}, {my, 4.5}, {sx, 3}, {sy, 3}},
{x, y}
]
Show[
ListPlot3D[data3D, PlotStyle -> None, MeshStyle -> Red],
Plot3D[fm["BestFit"], {x, 1, 9}, {y, 1, 9}, PlotRange -> All,
PlotStyle -> Opacity[0.9], Mesh -> None]
]
fm["BestFitParameters"]
{a -> 0.1871830545, b -> -0.4853901689, mx -> 5.152549499, my -> 5.092511036, sx -> 2.756524919, sy -> 2.072163433}
-
Starting with your data, I would map out the data something like this:
data3D = Flatten[MapIndexed[{#2[[1]], #2[[2]], #1} &, data, {2}], 1]
and then count it like:
data2 = Flatten[Table[{#[[1]], #[[2]]}, {100 #[[3]]}] & /@ data3D, 1];
and then use this function:
FindDistributionParameters[data2,
MultinormalDistribution[{a, b}, {{c, d}, {e, f}}]]
to find the means and coveriance of the data. And then I might do other stuff with the distribution, such as create a distribution object using:
edist = EstimatedDistribution[data2,
MultinormalDistribution[{a, b}, {{c, d}, {e, f}}]]
and then check the distribution against the data, using:
dtest = DistributionFitTest[data2, edist, "HypothesisTestData"]
and create a table of the test results, using:
N[dtest["TestDataTable", All]]
or individually
AndersonDarlingTest[data2, edist, "TestConclusion"]
CramerVonMisesTest[data2, edist, "TestConclusion"]
JarqueBeraALMTest[data2, "TestConclusion"]
MardiaKurtosisTest[data2, "TestConclusion"]
PearsonChiSquareTest[data2, edist, "TestConclusion"]
ShapiroWilkTest[data2, "TestConclusion"]
Now, I'm not a statistician, and I probably only know enough statistics to be dangerous, and I'm not familar with all the tests listed in the table, but I suspect that one of those small numbers is significant in measuring how close the actual data fits the distribution, i.e. how "Gaussian" the set of points is. | {
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"tags": null,
"url": "http://mathematica.stackexchange.com/questions/27642/fitting-a-two-dimensional-gaussian-to-a-set-of-2d-pixels/27754"
} |
Also I might just plot the data as follows and visually judge the gaussiness of the data using these functions (as per VLC's answer to question 2984)
Show[DensityHistogram[d, {.2}, ColorFunction -> (Opacity[0] &),
Method -> {"DistributionAxes" -> "Histogram"}], ListPlot[d]]
or these:
GraphicsColumn[
Table[SmoothDensityHistogram[d, ColorFunction -> "DarkRainbow",
Method -> {"DistributionAxes" -> p}, ImageSize -> 500,
BaseStyle -> {FontFamily -> "Helvetica"},
LabelStyle -> Bold], {p, {True, "SmoothHistogram", "BoxWhisker"}}]]
-
You mentioned in a related question that you would like to do large numbers of these fits quickly.
If each data set has the same dimensions, you can write a fairly fast implementation of Rahul Narain's method by precomputing arrays of x and y coordinates for the data grid, and flattening the data and using Dot to calculate the mean and the elements of the covariance matrix:
x = Table[i, {i, 9}, {j, 9}]//N;
y = Transpose[x];
x = Flatten[x]; y = Flatten[y];
semiaxes[data_] := Module[{min, p, mx, my},
min = Min[data];
p = Flatten[data] - min;
p /= Total[p];
mx = x.p;
my = y.p;
With[{a = (x - mx)^2.p, b = ((x - mx) (y - my)).p, c = (y - my)^2.p},
Sqrt @ Eigenvalues[{{a, b}, {b, c}}]]]
semiaxes[data]
(* {1.86325, 1.50567} *)
This runs in about 340 µs on my PC
Compiling can give you even more speed, but you need to replace Eigenvalues with the explicit symbolic expressions:
semiaxesc =
With[{x = x, y = y},
Compile[{{data, _Real, 2}}, Block[{min, p, mx, my, a, b, c},
min = Min[data];
p = Flatten[data] - min;
p /= Total[p];
mx = x.p;
my = y.p;
a = (x - mx)^2.p;
b = ((x - mx) (y - my)).p;
c = (y - my)^2.p;
{Sqrt[1/2 (a + c - Sqrt[a^2 + 4 b^2 - 2 a c + c^2])],
Sqrt[1/2 (a + c + Sqrt[a^2 + 4 b^2 - 2 a c + c^2])]}],
CompilationTarget -> "C", RuntimeOptions -> "Speed"]];
semiaxesc[data]
(* {1.50567, 1.86325} *)
This runs in about 5.7 µs on my PC. | {
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"lm_q1_score": 0.9643214491222695,
"lm_q1q2_score": 0.868400184465489,
"lm_q2_score": 0.9005297821135384,
"openwebmath_perplexity": 2655.350377263357,
"openwebmath_score": 0.37032657861709595,
"tags": null,
"url": "http://mathematica.stackexchange.com/questions/27642/fitting-a-two-dimensional-gaussian-to-a-set-of-2d-pixels/27754"
} |
semiaxesc[data]
(* {1.50567, 1.86325} *)
This runs in about 5.7 µs on my PC.
-
Fantastic. However, I'm getting a "Eigenvalues::matsq : Argument" which is saying that don't have a square matrix. Is there some preprocessing I need to do for the data array I posted? – Bob Jun 30 '13 at 9:15
Would you mind posting the full script you're running? I'm sure there's some formatting error on my end, but I can't pin it down. Everything works and makes sense, but there seems to be a problem with the Eigenvalue computation on my end. – Bob Jun 30 '13 at 12:33
That is the full script, apart from the definition of data which I copied from the question. I suggest you step through the code one line at a time and make sure that mx, my, a, b, and c are all single numbers. I'm away from the computer this week, so that's about all the help I can offer at the moment. You could try asking in chat if anyone else can get it working. – Simon Woods Jun 30 '13 at 14:58
Very good, as long as it's working on your end, I'll be able to fix it on mine. – Bob Jun 30 '13 at 16:50
The problem was that the dot products were not being recognized as such. Switching a.b for Dot[a,b] did the trick. Very odd. – Bob Jun 30 '13 at 18:19 | {
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"openwebmath_score": 0.37032657861709595,
"tags": null,
"url": "http://mathematica.stackexchange.com/questions/27642/fitting-a-two-dimensional-gaussian-to-a-set-of-2d-pixels/27754"
} |
# Product of two monotonic real functions
I'm confused about the possibility to say that the product of two monotonic real functions $f,g:\mathbb{R} \to \mathbb{R}$ is monotonic.
I found the following proposition:
A. If $g$ and $f$ are two increasing functions, with $f(x)>0$ and $g(x)>0$ $\forall x$, then $f \, g$ is increasing.
B. If $g$ and $f$ are two increasing functions, with $f(x)<0$ and $g(x)<0$ $\forall x$, then $f \, g$ is decreasing.
Does something similar hold also for decreasing functions? That is: is the following sentence correct?
A'. If $g$ and $f$ are two decreasing functions, with $f(x)>0$ and $g(x)>0$ $\forall x$, then $f \, g$ is decreasing.
B'. If $g$ and $f$ are two decreasing functions, with $f(x)<0$ and $g(x)<0$ $\forall x$, then $f \, g$ is increasing.
Moreover, can something be said about the product of a increasing and a decreasing functions?
For example if $f$ is increasing and $g$ is decreasing under what conditions can I say something about the monotony of the product $f g$?
Besides these two practical questions I would like to ask some suggestions on how to prove statement A.
I tried in the following way
$Hp:$
$x_1 >x_2 \implies f(x_1)>f(x_2)>0 \,\,\, \forall x_1,x_2$
$x_1 >x_2 \implies g(x_1)>g(x_2)>0 \,\,\, \forall x_1,x_2$
$Th:$
$x_1 >x_2 \implies f(x_1) g(x_1)>f(x_2) g(x_2)>0 \,\,\, \forall x_1,x_2$
$Proof:$
$f(x_1)>f(x_2)>0 \,\,\, , g(x_1)>g(x_2)>0 \,\,\, \forall x_1,x_2 \implies f(x_1) g(x_1)>f(x_2) g(x_2)>0 \,\,\, \forall x_1,x_2$
Which seems obvious if one thinks about some numbers but I don't really know how I could prove the last implication in rigourous way. So any help in this proof is highly appreciated. | {
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"openwebmath_score": 0.8550617098808289,
"tags": null,
"url": "https://math.stackexchange.com/questions/1996917/product-of-two-monotonic-real-functions"
} |
To prove statement A, let $x > y$. Then, as we know, $f(x) > f(y)>0$ and $g(x) > g(y)>0$. Hence, the following chain of statements proves the claim: \begin{gather} f(x) > f(y) \implies g(x)f(x)> g(x)f(y) \quad(\because g(x)>0)\\ g(x) > g(y) \implies g(x)f(y) > g(y)f(y) \quad(\because f(y)>0)\\ g(x)f(x)> g(x)f(y) > g(y)f(y) \implies fg(x) > fg(y) \end{gather}
An analogous proof would follow for part B if $f$ and $g$ were increasing, with a caveat:let $x > y$. Then, as we know, $f(x) > f(y)$ and $g(x) > g(y)$. Hence, the following chain of statements proves the claim: \begin{gather} f(x) > f(y) \implies g(x)f(x)< g(x)f(y) \quad(\because g(x)<0)\\ g(x) > g(y) \implies g(x)f(y) < g(y)f(y) \quad(\because f(y)<0)\\ g(x)f(x)< g(x)f(y) < g(y)f(y) \implies fg(x) < fg(y) \end{gather}
Now, let us see if the same logic could work with part A':let $x > y$. Then, as we know, $f(x) < f(y)$ and $g(x) < g(y)$. Hence, the following chain of statements proves the claim: \begin{gather} f(x) < f(y) \implies g(x)f(x)< g(x)f(y) \quad(\because g(x)>0)\\ g(x) < g(y) \implies g(x)f(y) < g(y)f(y) \quad(\because f(y)>0)\\ g(x)f(x)< g(x)f(y) < g(y)f(y) \implies fg(x) < fg(y) \end{gather}
That's brilliant, so great intuition for anticipating part A'. Now we will check part B':let $x > y$. Then, as we know, $f(x) < f(y)$ and $g(x) < g(y)$. Hence, the following chain of statements proves the claim: \begin{gather} f(x) < f(y) \implies g(x)f(x)> g(x)f(y) \quad(\because g(x)<0)\\ g(x) < g(y) \implies g(x)f(y) > g(y)f(y) \quad(\because f(y)<0)\\ g(x)f(x)> g(x)f(y) > g(y)f(y) \implies fg(x) > fg(y) \end{gather}
And therefore part B' is also done. Note the above logic carefully, I think all steps are equally important.
Use this logic, and see why in most cases, one increasing and one decreasing function doesn't tell you much about the product itself. | {
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"openwebmath_score": 0.8550617098808289,
"tags": null,
"url": "https://math.stackexchange.com/questions/1996917/product-of-two-monotonic-real-functions"
} |
For A' and B', apply A and B. For A': If $f,g$ are decreasing and positive then $-f$ and $-g$ are increasing and negative so by B, the function $(-f)(-g)=fg$ is decreasing. Similarly, apply A to B'.
$e^x$ is increasing and $e^{-x}+1$ is decreasing, and their product $1+e^x$ is increasing.
$e^x+1$ is increasing and $e^{-x}$ is decreasing, and their product $1+e^{-x}$ is decreasing.
The product of a positive increasing and a positive decreasing function can also fail to be monotonic. | {
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"lm_q1q2_score": 0.8683916281290693,
"lm_q2_score": 0.8824278710924296,
"openwebmath_perplexity": 994.7676560267981,
"openwebmath_score": 0.8550617098808289,
"tags": null,
"url": "https://math.stackexchange.com/questions/1996917/product-of-two-monotonic-real-functions"
} |
# Fair and Unfair coin Probability
I am stuck on this question.
A coin with $P(H) = \frac{1}{2}$ is flipped $4$ times and then a coin with $P(H) = \frac{2}{3}$ is tossed twice. What is the probability that a total of $5$ heads occurs?
I keep getting $\frac{1}{6}$ but the answer is $\frac{5}{36}$.
Attempt: $P($all heads on the four coins$)P($either one of the tosses is heads on the two coins$)+P(3$ heads on the four coins$)P($both coins are heads$)$
$P($all heads on the four coins$) = \left(\frac{1}{2}\right)^4 = \frac{1}{16}$.
$P($either one of the tosses is heads on the two coins$) = 1-P($no heads on both tosses$) = 1-\frac{1}{3}\cdot\frac{1}{3} = \frac{8}{9}$.
$P($exactly $3$ heads on the four tosses$) = \frac{1}{4}$.
$P($both coins are heads$) = \frac{2}{3}\cdot\frac{2}{3} = \frac{4}{9}$.
Final Equation: $\frac{1}{16}\cdot\frac{8}{9}+\frac{1}{4}\cdot\frac{4}{9} = \frac{1}{6}$.
Why am I off by $\frac{1}{36}$?
-
Because "either one of..." $\not \equiv$ " exactly one of..." – Ehsan M. Kermani Jun 9 '13 at 4:31
You claim that the problem is with $P($either one of the tosses is heads on the two coins$)$. You need to calculate the probability of there being precisely one occurrence of a head in the tossing of the second coin. In your calculation, you subtracted the probability of no heads from one. The remaining probability covers one head or both heads, but you want to exclude the latter. If you make this adjustment, you will get the correct answer. – Michael Albanese May 30 '15 at 19:38
The required probability will be
$P($exactly $4$ heads from the $4$ flips of $1$st coin$)\cdot P($exactly $1$ head from the $2$ flips of $2$nd coin $)+$
$P($exactly $3$ heads from the $4$ flips of $1$st coin$)\cdot P($exactly $2$ heads from the $2$ flips of $2$nd coin $)$
Using Binomial Distribution, the required probability $$\binom44\left(\frac12\right)^4\left(1-\frac12\right)^{4-4} \cdot\binom21\left(\frac23\right)^1\left(1-\frac23\right)^{2-1}$$ | {
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"lm_q1_score": 0.9877587247081928,
"lm_q1q2_score": 0.8683848945980446,
"lm_q2_score": 0.8791467722591728,
"openwebmath_perplexity": 909.9909338425831,
"openwebmath_score": 0.7746295928955078,
"tags": null,
"url": "http://math.stackexchange.com/questions/415179/fair-and-unfair-coin-probability"
} |
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