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# What is the dimension of two subspaces?
Let $V$ be the vector space with a basis $x_1, \ldots, x_9$ and $$V_1 = \{(a,a,a,b,b,b,c,c,c): a,b,c \in \mathbb{C}\}, \\ V_2=\{(a,b,c,a,b,c,a,b,c): a,b,c \in \mathbb{C}\}.$$ Then $V_1,V_2$ are subspaces of $V$. What is the dimension of $V_1 \cap V_2$?
I first try to find the system of equations which give $V_1, V_2$ respectively. I think that $V_1$ is
$$\{(k_1,k_2,k_3,k_4,k_5,k_6,k_7,k_8,k_9): k_1-k_2=0,k_2-k3=0,k_4-k_5=0,k_5-k_6=0, k_7-k_8=0,k_8-k_9=0 \}$$ and $V_2$ is
$$\{(k_1,k_2,k_3,k_4,k_5,k_6,k_7,k_8,k_9): k_1-k_4=0,k_4-k7=0,k_2-k_5=0,k_5-k_8=0, k_3-k_6=0,k_6-k_9=0 \}.$$ By solving the collection of the equations for $V_1$ and $V_2$, I obtain the solutions $k_1=\cdots =k_9$. So the dimension of $V_1 \cap V_2$ is one dimensional. Is this correct?
Thank you very much. | {
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Thank you very much.
It is correct but, in my opinion, that approach is too complex for the problem. Take $(a,b,c,d,e,f,g,h,i)\in V_1\cap V_2$. Then
• Since $(a,b,c,d,e,f,g,h,i)\in V_1$, $b=c=a$, $e=f=d$, and $h=i=g$. Therefore$$(a,b,c,d,e,f,g,h,i)=(a,a,a,d,d,d,g,g,g).$$
• Since $(a,a,a,d,d,d,g,g,g)\in V_2$, $d=g=a$.
Therefore $V_1\cap V_2=\left\{(a,a,a,a,a,a,a,a,a)\in\mathbb{C}^9\,\middle|\,a\in\mathbb C\right\}$, which is clearly $1$-dimensional.
Yes, it is correct. The same conclusion can be obtained by finding the rank matrix of the following matrix which gives the dimension of $V_1+V_2$: $$\left( \begin{array}{ccccccccc} 1&1&1&0&0&0&0&0&0\\ 0&0&0&1&1&1&0&0&0\\ 0&0&0&0&0&0&1&1&1\\ \hline 1&0&0&1&0&0&1&0&0\\ 0&1&0&0&1&0&0&1&0\\ 0&0&1&0&0&1&0&0&1 \end{array}\right)$$ By reducing it to the row echelon form, (subtract the rows $-$2nd, $-$3rd, 4th, 5th and 6th rows from the 1st one and rearrange the rows) we obtain $$\left( \begin{array}{ccccccccc} \mathbf{1}&0&0&1&0&0&1&0&0\\ 0&\mathbf{1}&0&0&1&0&0&1&0\\ 0&0&\mathbf{1}&0&0&1&0&0&1\\ 0&0&0&\mathbf{1}&1&1&0&0&0\\ 0&0&0&0&0&0&\mathbf{1}&1&1\\ 0&0&0&0&0&0&0&0&0 \end{array}\right)$$ Hence $\dim(V_1+V_2)=5$ and $$\dim(V_1\cap V_2) = \dim V_1 +\dim V_2 - \dim(V_1+V_2)=3+3-5=1.$$ | {
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# Evaluating a double integral in polar coordinates
#### skatenerd
##### Active member
I've done this problem and I have a feeling it's incorrect. I've never done a problem like this so I am kind of confused on how else to go about doing it. The goal is to change the cartesian integral
$$\int_{-a}^{a}\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\,dy\,dx$$
into an integral in polar coordinates and then evaluate it.
Changing to polar coordinates I got the integral
$$\int_{0}^{\pi}\int_{-a}^{a}r\,dr\,d\theta$$
and evaluating this integral I ended up with an integrand of 0 to integrate with respect to $$d\theta$$ and I wasn't entirely sure how to integrate that so I thought it might just be $$\pi$$.
I really feel like there's no way that answer could be correct, seeing as the integral is of half a circle with radius $$a$$ and the answer has nothing to do with $$a$$. If someone could let me know where I went wrong that would be great.
#### Prove It
##### Well-known member
MHB Math Helper
Re: evaluating a double integral in polar coordinates
Since you are integrating over an entire circle of radius |a| centred at (0, 0), that means the angle swept out is actually $$\displaystyle \displaystyle 2\pi$$, which means your $$\displaystyle \displaystyle \theta$$ bounds are actually 0 to $$\displaystyle \displaystyle 2\pi$$.
#### chisigma
##### Well-known member
Re: evaluating a double integral in polar coordinates | {
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#### chisigma
##### Well-known member
Re: evaluating a double integral in polar coordinates
I've done this problem and I have a feeling it's incorrect. I've never done a problem like this so I am kind of confused on how else to go about doing it. The goal is to change the cartesian integral
$$\int_{-a}^{a}\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\,dy\,dx$$
into an integral in polar coordinates and then evaluate it.
Changing to polar coordinates I got the integral
$$\int_{0}^{\pi}\int_{-a}^{a}r\,dr\,d\theta$$
and evaluating this integral I ended up with an integrand of 0 to integrate with respect to $$d\theta$$ and I wasn't entirely sure how to integrate that so I thought it might just be $$\pi$$.
I really feel like there's no way that answer could be correct, seeing as the integral is of half a circle with radius $$a$$ and the answer has nothing to do with $$a$$. If someone could let me know where I went wrong that would be great.
The bounds of the inner integral are 0 and a, not -a and a so that is...
$\displaystyle S= \int_{0}^{2\ \pi} \int_{0}^{a} r\ d r\ d \theta = \pi\ a^{2}$ (1)
Kind regards
$\chi$ $\sigma$
#### skatenerd
##### Active member
Re: evaluating a double integral in polar coordinates
Thanks guys. I didn't recognize initially that the bounds of the original integral are describing the area of a whole circle. Pretty cool problem now that I get it! | {
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# What kind of combinatorial problem is this?
Is there a theory from which the following problem comes? Does this type of problem have a name?
Find the largest possible number of $k$-element sets consisting of points from some finite set and have pairwise singleton or empty intersections.
I hope that was clear. If not, here's an example for $k=3$:
Let the set of points be $S=\{1,2,3,4,5,6\}$. The most 3-element sets (with pairwise singleton or empty intersections) that can be constructed from $S$ is 4, such as $\{456,236,124,135\}$.
I made a table for $|S|=3,4,5,6,7,8,9$ and got $1,1,2,4,7,8,12$, respectively, hoping I could dig up some information from OEIS.
I read a little on Steiner systems, and although it feels like I'm in the neighborhood, I'm not confident...
Edit1: typos.
Edit2: Johnson graphs and (for $k=3$) Steiner Triple Systems (STS) seem close to what I'm looking for. The condition of "pairwise singleton or empty intersections" is equivalent to "every 2-subset of S occurs in at most one $k$-element set". STS require that every 2-subset of S occurs in exactly one $3$-element set".
Edit3: Thank you to everyone who replied! All of your comments helped me push through a barrier I was facing for some time.
-
Looks a bit like a Johnson graph. Perhaps looking at some of the more popular objects in finite geometry will get you graphs matching your own – muzzlator Feb 19 '13 at 20:00
Thanks for the tip. – sasha Feb 19 '13 at 20:36
Design theory is also relevant – mrf Feb 19 '13 at 23:30
The $k=3$ sequence seems to be oeis.org/A001839 . – Kevin Costello Feb 19 '13 at 23:51
In general, you are looking for the maximal code of length $n$, constant weight $k$, and minimum distance $2k-2$. So the $k=4$ sequence is $A(n,6,4)$, found at oeis.org/A004037. – mjqxxxx Feb 20 '13 at 0:57 | {
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You exactly want to determine the clique number of the generalized Kneser graph $KG_{n,k,s}$ for $s=1$, which is the graph having all the $k$-element subsets as its $\binom{n}{k}$ vertices, where any two vertices are connected if and only if their cut contains at most $s$ elements. Thus, a maximum clique of $KG_{n,k,1}$ is a maximum selection of $k$-element subsets such that all their pairwise cuts are singleton or emtpy. The size of such a maximum clique is the clique number $\omega(KG_{n,k,s})$.
Googling around a bit, I could not find an exact expression therefore.
However, this states that $\omega(KG_{n,k,0}) = \lfloor \frac n k \rfloor$. Since for $s > 0$ edges are never removed, this also gives a lower bound on $\omega(KG_{n,k,s})$ for any $s \geq 0$, thus, $\omega(KG_{n,k,1}) \geq \lfloor \frac n k \rfloor$. But this bound seems rather weak, since it does not respect any singleton edges at all. For your example above we get 1,1,1,2,2,2,3 as lower bounds on 1,1,2,4,7,7,12.
Further, here is an expression for the chromatic number $\chi(KG_{n,k,1})$, which gives an upper bound on the clique number, since $\chi(G) \geq \omega(G)$ for any graph $G$, where for perfect graphs equality holds. For $n$ written as $n = (k-1) s + r$ for some $0 \leq r < k-1$ and large enough $n > n_0(k)$, the bound is given as $\chi(KG_{n,k,1}) = (k-1)\binom{s}{2} + rs \geq \omega(KG_{n,k,1})$. Ignoring any details on $n_0(k)$, since I have no access to the paper, this gives 1,2,4,6,9,12,16 as upper bounds on 1,1,2,4,7,7,12, which seems to approximate quite well.
Perhaps one of the proofs behind the above results can be adopted to the clique number of $KG_{n,k,1}$?
edit: I just realized that you even want to find a maximum clique - well, that is computationally very hard in general, but perhaps things get easier for $KG_{n,k,1}$?
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-
A family of subsets of some finite set is a hypergraph; the subsets themselves are the edges (or hyperedges) of the hypergraph. If all the edges have size $k$, then the hypergraph is k-uniform. (For instance, a $2$-uniform hypergraph is just an ordinary undirected graph.) If no pair of edges has more than one point in common, the hypergraph is called linear. So your question can be reframed as:
What is the maximal number of edges in a $k$-uniform linear hypergraph on $n$ vertices?
-
If you're interested in the asymptotic situation instead of what happens for specific $|S|$, then this has been studied a fair bit under the name of packing problems. More generally, we can ask the question of the size of the largest collection of $k$-element subsets of $\{1, \dots, n\}$ such that each pair of subsets has intersection of size strictly less than $r$.
Since each $k$-element subset contains $\binom{k}{r}$ $r$-element subsets, and each $r$ element subset is in at most $1$ $k$-element subset, we can pick at most $$\frac{\binom{n}{r}}{\binom{k}{r}}$$ subsets in our collection.
Erdos and Hanani conjectured in 1963 that this was asymptotically optimal: For fixed $k$ and $r$, as $n$ tends to infinity, there is a collection of size $(1+o(1))$ times the bound above (for the specific case $r=2$ that you mentioned, this was conjectured earlier by Bose). The conjecture remained open for more than $20$ years until Rodl introduced his so-called "nibble method" to prove it ("On a Packing and Covering Problem", not available online as far as I can tell).
Another term you might want to search under is partial Steiner systems.
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# Math Help - 1+2+3+...+n
1. ## 1+2+3+...+n
Hi, can anyone tell me why
1+2+3+...+n=n(n+1)/2
I can see that it works when I choose a number for n, but I don't really see how I could have come up with it myself.
2. ## Re: 1+2+3+...+n
I think I can explain it like that:
1, 2, 3, 4, 5, 6, ... n
This is an arithmetic progression with first term 1, last term n with a common difference of 1.
The formula for the sum of the first n numbers is given by:
$S_n = \dfrac{n}{2}\left(a+l\right)$
a = 1, l = n so you simplify to get:
$S_n = \dfrac{n(n+1)}{2}$
How? Let's take an example:
1, 2, 3, 4, 5, 6, 7
If you take the middle number, 4. You make it so that every number becomes 4. Remove 1 from 5 and give it to 3. Remove 2 from 6 and give it to 2, remove 3 from 7 and give it to 1 to get:
4, 4, 4, 4, 4, 4, 4
The sum is then the 4n = 4(7) = 28
But what did you do actually? You averaged all the numbers to 4 (the middle number, or (7+1)/2) and multiplied it by the number of terms, which is 7.
Does that make it any clearer?
3. ## Re: 1+2+3+...+n
Do you know 'induction'?
4. ## Re: 1+2+3+...+n
Yes, that was very clear. Thank you!
5. ## Re: 1+2+3+...+n
Originally Posted by Siron
Do you know 'induction'?
No, I don't. Does that relate to this problem?
6. ## Re: 1+2+3+...+n
Originally Posted by TwoPlusTwo
No, I don't. Does that relate to this problem?
Take a look here:
Mathematical induction - Wikipedia, the free encyclopedia
Your exercice is used as an example.
7. ## Re: 1+2+3+...+n
Hello, TwoPlusTwo!
$\text{Can anyone tell me why: }\:1+2+3+\hdots +n\:=\:\frac{n(n+1)}{2}$
Here is a geometric demonstration of the rule (not a proof, mind you).
Consider the case: . $n = 5$
We have this array of objects:
. . $\begin{array}{c}\circ \\ \circ\;\circ \\ \circ\;\circ\;\circ \\ \circ\;\circ\;\circ\;\:\circ \\ \circ\;\circ\;\circ\;\circ\;\circ \end{array}$
Left-justify the objects: | {
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Left-justify the objects:
. . $\circ$
. . $\circ\;\:\circ$
. . $\circ\;\circ\;\circ$
. . $\circ\;\circ\;\circ\;\:\circ$
. . $\circ\;\circ\;\circ\;\circ\;\circ$
Append an inverted copy of the array:
. . $\begin{array}{c}\circ\;\bullet\;\bullet\;\bullet\; \bullet \;\:\bullet \\ \circ\;\circ\;\bullet\;\bullet\;\bullet\;\:\bullet \\ \circ\;\circ\;\circ\; \bullet\;\bullet\;\:\bullet \\ \circ\;\circ\;\circ\;\circ\;\bullet\;\:\bullet \\ \circ\;\circ\; \circ\;\circ\;\circ\;\:\bullet \end{array}$
We see that the rectangle has: . $5 \times 6\:=\:30$ objects.
Therefore, the triangle has: . $\frac{5 \times 6}{2} \:=\:15$ objects.
8. ## Re: 1+2+3+...+n
There is a story, I don't know if it is true or not, that when Gauss was a small child in a very bad, very crowded class, the teacher set the children the problem of adding all the integers from 1 to 100, just to keep them quiet. Gauss wrote a single number on his paper and then just sat there. The number was, of course, 5050, the correct sum.
Here is how he was supposed to have done it: write
1+ 2+ 3+ 4+ 5+ ... 96+ 97+ 98+ 99+ 100 and reverse the sum below it
100+ 99+ 98+97+96 ...+ 5+ 4+ 3+ 2+ 1
and add each column. That is, add 1+ 100= 101, 2+ 99= 101, 3+ 98= 101, etc. Every pair of numbers adds to 101 because, in the top sum, we are increasing by 1 each time while in the bottom sum we are decreasing by 1. There are 100 such pairs so the two sums together add to 100(101)= 10100. Since that is two sums, the one we want is half of that, 5050. (Of course, Gauss, about 10 years old at the time, did all of that in his head!)
If we do that with 1+ 2+ 3+ ...+ (n- 2)+ (n- 1)+ n, we will have n pairs (1+n, 2+ (n-1), ...) each adding to n+ 1. The two sums add two n(n+1) so the original sum, from 1 to n, is n(n+1)/2.
9. ## Re: 1+2+3+...+n | {
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9. ## Re: 1+2+3+...+n
Originally Posted by HallsofIvy
There is a story, I don't know if it is true or not, that when Gauss was a small child in a very bad, very crowded class, the teacher set the children the problem of adding all the integers from 1 to 100, just to keep them quiet. Gauss wrote a single number on his paper and then just sat there. The number was, of course, 5050, the correct sum.
Here is how he was supposed to have done it: write
1+ 2+ 3+ 4+ 5+ ... 96+ 97+ 98+ 99+ 100 and reverse the sum below it
100+ 99+ 98+97+96 ...+ 5+ 4+ 3+ 2+ 1
and add each column. That is, add 1+ 100= 101, 2+ 99= 101, 3+ 98= 101, etc. Every pair of numbers adds to 101 because, in the top sum, we are increasing by 1 each time while in the bottom sum we are decreasing by 1. There are 100 such pairs so the two sums together add to 100(101)= 10100. Since that is two sums, the one we want is half of that, 5050. (Of course, Gauss, about 10 years old at the time, did all of that in his head!)
If we do that with 1+ 2+ 3+ ...+ (n- 2)+ (n- 1)+ n, we will have n pairs (1+n, 2+ (n-1), ...) each adding to n+ 1. The two sums add two n(n+1) so the original sum, from 1 to n, is n(n+1)/2.
As usual, Soroban got in just before me. I need to learn to type faster! | {
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Question
# Assertion :If $$bc+qr=ca+rp=ab+pq=-1$$, then $$\begin{vmatrix} ap & a & p \\ bq & b & q \\ cr & c & r \end{vmatrix}=0\quad (abc,pqr\neq 0)$$ Reason: If system of equations $${ a }_{ 1 }x+{ b }_{ 1 }y+{ c }_{ 1 }=0,\quad { a }_{ 2 }x+{ b }_{ 2 }y+{ c }_{ 2 }=0,{ \quad a }_{ 3 }x+{ b }_{ 3 }y+{ c }_{ 3 }=0$$ has non-trivial solutions, $$\begin{vmatrix} { a }_{ 1 } & { b }_{ 1 } & { c }_{ 1 } \\ { a }_{ 2 } & { b }_{ 2 } & { c }_{ 2 } \\ { a }_{ 3 } & { b }_{ 3 } & { c }_{ 3 } \end{vmatrix}=0$$
A
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
B
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
C
Assertion is correct but Reason is incorrect
D
Assertion is incorrect but Reason is correct
Solution
## The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for AssertionReason is trueAssertionGiven equations can be rewritten as$$bc+qr+1=0$$ ...(1)$$ca+rp+1=0$$ ...(2)$$ab+pq+1=0$$ ...(3)Multiplying (1),(2) and (3) by ap, bq,cr respectively, we get$$\left( abc \right) p+\left( pqr \right) a+ap=0\\ \left( abc \right) q+\left( pqr \right) b+bq=0\\ \left( abc \right) r+\left( pqr \right) c+cr=0$$These equation are consistent,Hence $$\begin{vmatrix} p\quad & a\quad & ap \\ q\quad & b\quad & bq \\ r\quad & c\quad & cr \end{vmatrix}=0\Rightarrow \begin{vmatrix} p & q & r \\ a & b & c \\ ap & bq & cr \end{vmatrix}=0$$ ( interchanging rows into columns) $$\Rightarrow \left( -1 \right) \begin{vmatrix} ap\quad & bq\quad & cr \\ a & b & c \\ p & q & r \end{vmatrix}=0\quad \left( { R }_{ 1 }{ \leftrightarrow R }_{ 2 } \right) \\ \Rightarrow \begin{vmatrix} ap\quad & bq\quad & cr \\ a & b & c \\ p & q & r \end{vmatrix}=0$$Mathematics
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Binary search works on sorted arrays. Finding the Predecessor and Successor Node of a Binary Search Tree All implementation of finding sucessor or predecessor takes O(1) constant space and run O(N) time (when BST is just a degraded linked list) - however, on average, the complexity is O(LogN) where the binary … Repeatedly check until the value is found or the interval is empty. If the value of the search key is less than the item in the middle of the interval, narrow the interval to the lower half. The problem was that the index must be less than half the size of the variable used to store it (be it an integer, unsigned integer, or other). Here eps is in fact the absolute error (not taking into account errors due to the inaccurate calculation of the function). We’ll call the sought value the target value for clarity. Given the starting point of a range, the ending point of a range, and the "secret value", implement a binary search through a sorted integer array for a certain number. The most common way is to choose the points so that they divide the interval $[l, r]$ into three equal parts. Binary Search: Search a sorted array by repeatedly dividing the search interval in half. Queue using Singly Linked List Implementation (With C++ Program Code), Stack using Singly Linked List Implementation (With C++ Program Code), Prefix to Postfix Conversion using Stack Data Structure (With C++ Program Code), Postfix to Prefix Conversion using Stack Data Structure (With C++ Program Code), Prefix to Infix Conversion using Stack Data Structure (With C++ Program Code), Selection Sort Algorithm with C++ Code | Sorting Algorithms | Data Structures & Algorithms, Creating Master Page in ASP.NET | Adding Navigation Menu & Footer to Master Page, Infix to Postfix Conversion using Stack Data Structure (With C++ Program Code), Singly Linked List Data Structure all Operations | C++ Program to Implement Singly Linked List, Insert Update Delete Select Book Details with Multi Select & Image | {
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to Implement Singly Linked List, Insert Update Delete Select Book Details with Multi Select & Image File Upload, Creating Sign Up/Registration Page in ASP.NET with Bootstrap Styling, C++ Program to Calculate Area of Triangle, Multi User Login Page in ASP.NET with C# + MS SQL Database with Session Variable. The only limitation is that the array or list of elements must be sorted for the binary search algorithm to work on it. Print out whether or not the number was in the array afterwards. This is called the search space. ( … The number of iterations should be chosen to ensure the required accuracy. Binary search algorithm falls under the category of interval search algorithms. More precisely, the algorithm can be stated as foll… Search the sorted array by repeatedly dividing the search interval in half This algorithm repeatedly target the center of the sorted data structure & divide the search space into half till the match is found. This is a numerical method, so we can assume that after that the function reaches its maximum at all points of the last interval $[l, r]$. We are given a function $f(x)$ which is unimodal on an interval $[l, r]$. If … find the values of f(m1) and f(m2). Notify me of follow-up comments by email. Constrained algorithms. Otherwise narrow it to the upper half. This search algorithm works on the principle of divide and conquer. For a similar project, that translates the collection of articles into Portuguese, visit https://cp-algorithms-brasil.com. The function strictly decreases first, reaches a minimum, and then strictly increases. find the values of $f(m_1)$ and $f(m_2)$. This was not an algorithm bug as is purported on this page - and I feel strongly that this is unjust. Begin with an interval covering the whole array. We first need to calculate the middle element in the list and then compare the element we are searching with this middle element. Today we will discuss the Binary Search Algorithm. If we get a match, we return the index of | {
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Today we will discuss the Binary Search Algorithm. If we get a match, we return the index of the middle element. uHunt Chapter 3 has six starred problems, and many more problems in total, on the topic of binary search. Required fields are marked *. If $m_1$ and $m_2$ are chosen to be closer to each other, the convergence rate will increase slightly. Performance. The Binary Search Algorithm. Binary search algorithm Algorithm. In either case, this means that we have to search for the maximum in the segment [m1,r]. We evaluate the function at m1 and m2, i.e. Binary search only works on sorted data structures. Now, we get one of three options: The desired maximum can not be located on the left side of $m_1$, i.e. Thus the size of the search space is ${2n}/{3}$ of the original one. Binary Search is a searching algorithm for finding an element's position in a sorted array. Following is a pictorial representation of BST − We observe that the root node key (27) has all less-valued keys on the left sub-tree and the higher valued keys on the right sub-tree. Binary Search Pseudocode We are given an input array that is supposed to be sorted in ascending order. Binary search is a fast search algorithm with run-time complexity of Ο (log n). Based on the compariso… Binary search only works on sorted data structures. 4. In one iteration of the algorithm, the "ring offire" is expanded in width by one unit (hence the name of the algorithm). Articles Algebra. While searching, the desired key is compared to the keys in BST and if found, the associated value is retrieved. Binary search is an efficient search algorithm as compared to linear search. Implementations can be recursive or iterative (both if you can). on the interval $[l, m_1]$, since either both points $m_1$ and $m_2$ or just $m_1$ belong to the area where the function increases. For this algorithm to work properly, the data collection should be in the sorted form. A binary search tree is a data structure that quickly allows | {
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should be in the sorted form. A binary search tree is a data structure that quickly allows us to maintain a sorted list of numbers. It's time complexity of O (log n) makes it very fast as compared to other sorting algorithms. The binary search algorithm is conceptually simple. Your email address will not be published. Consider any 2 points m1, and m2 in this interval: lf(m2)This situation is symmetrical to th… Since we did not impose any restrictions on the choice of points $m_1$ and $m_2$, the correctness of the algorithm is not affected. This algorithm is much more efficient compared to linear search algorithm. We didn't impose any restrictions on the choice of points $m_1$ and $m_2$. 2. It is also known as half-interval search or logarithmic search. $$T(n) = T({2n}/{3}) + 1 = \Theta(\log n)$$. In its simplest form, binary search is used to quickly find a value in a sorted sequence (consider a sequence an ordinary array for now). Binary search algorithm falls under the category of interval search algorithms. 3. To summarize, as usual we touch $O(\log n)$ nodes during a query. Instead of the criterion r - l > eps, we can select a constant number of iterations as a stopping criterion. In binary search, we follow the following steps: We start by comparing the element to be searched with the element in the middle of the list/array. Binary Search is used with sorted array or list. $m_1$ and $m_2$ can still be chosen to divide $[l, r]$ into 3 approximately equal parts. The time complexity of binary search algorithm is O(Log n). Repeatedly applying the described procedure to the interval, we can get an arbitrarily short interval. Thus, the search space is reduced to $[m_1, m_2]$. However, this approach is not practical for large a or n. ab+c=ab⋅ac and a2b=ab⋅ab=(ab)2. This choice will define the convergence rate and the accuracy of the implementation. Binary Search is a divide and conquer algorithm. The program assumes that the input numbers are in ascending order. The | {
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divide and conquer algorithm. The program assumes that the input numbers are in ascending order. The algorithm takes as input an unweighted graph and the id of the source vertex s. The input graph can be directed or undirected,it does not matter to the algorithm. We can see that either both of these points belong to the area where the value of the function is maximized, or $m_1$ is in the area of increasing values and $m_2$ is in the area of descending values (here we used the strictness of function increasing/decreasing). Binary search is a fast search algorithm with run-time complexity of Ο (log n). The range [first, last) must satisfy all of the following conditions: Partitioned with respect to element < val or comp (element, val). Fundamentals. In this article, we will assume the first scenario. Raising a to the power of n is expressed naively as multiplication by a done n−1 times:an=a⋅a⋅…⋅a. Binary search compares the target value to the middle element of the sorted array, if they are unequal, the half in which the target cannot lie is eliminated and the search continues for … A tree representing binary search. Enter your email address to subscribe to this blog and receive notifications of new posts by email. Each node has a key and an associated value. 5. You might recall that binary search is similar to the process of finding a name in a phonebook. The difference occurs in the stopping criterion of the algorithm. For this algorithm to work properly, the data collection should be in the sorted form. If the array isn't sorted, you must sort it using a sorting technique such as merge sort. In the root node we do a binary search, and in all other nodes we only do constant work. Given below are the steps/procedures of the Binary Search algorithm. This method is done by starting with the whole array. Eventually, its length will be less than a certain pre-defined constant (accuracy), and the process can be stopped. At each step, the fire burning at each vertex | {
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constant (accuracy), and the process can be stopped. At each step, the fire burning at each vertex spreads to all of its neighbors. Save my name, email, and website in this browser for the next time I comment. The binary search algorithm check was fine. Binary search is a search algorithm that finds the position of a target value within a sorted array. If the elements are not sorted already, we … The idea is to use Binary Search. Without loss of generality, we can take $f(l)$ as the return value. If you want to solve them, it helps to have a firm grasp of how that algorithm works. To simplify the code, this case can be combined with any of the previous cases. For (1), T shall be a type supporting being compared with elements of the range [first,last) as either operand of operator<. Then it … Thus, based on the comparison of the values in the two inner points, we can replace the current interval $[l, r]$ with a new, shorter interval $[l^\prime, r^\prime]$. Binary search looks for a particular item … This algorithm is much more efficient compared to linear search algorithm. BST is a collection of nodes arranged in a way where they maintain BST properties. Binary search can be significantly better than the linear search while talking about the time complexity of searching( given the array is sorted). C++20 provides constrained versions of most algorithms in the namespace std::ranges.In these algorithms, a range can be specified as either an iterator-sentinel pair or as a single range argument, and projections and pointer-to-member callables are supported. Binary Search is a method to find the required element in a sorted array by repeatedly halving the array and searching in the half. This video is a part of HackerRank's Cracking The Coding Interview Tutorial with Gayle Laakmann McDowell. It is one of the Divide and conquer algorithms types, where in each step, it halves the number of elements it has to search, making the average time complexity to O (log n). Applying | {
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the number of elements it has to search, making the average time complexity to O (log n). Applying Master's Theorem, we get the desired complexity estimate. Otherwise narrow it to the upper half. Typically, in most programming challenges the error limit is ${10}^{-6}$ and thus 200 - 300 iterations are sufficient. template < class ForwardIt, class T > bool binary_search (ForwardIt first, ForwardIt last, const T & value) {first = std:: lower_bound (first, last, value); return (! If the value of the search key is less than the item in the middle of the interval, narrow the interval to the lower half. So we o… Linear Search. [A]: Binary Search — Searching a sorted array by repeatedly dividing the search interval in half. on the interval $[m_2, r]$, and the search space is reduced to the segment $[l, m_2]$. Consider any 2 points $m_1$, and $m_2$ in this interval: $l < m_1 < m_2 < r$. This means the complexity for answering a query is $O(\log n)$. ... Search Operation. It can be visualized as follows: every time after evaluating the function at points $m_1$ and $m_2$, we are essentially ignoring about one third of the interval, either the left or right one. This situation is symmetrical to the previous one: the maximum can not be located on the right side of $m_2$, i.e. We can use binary search to reduce the number of comparisons in normal insertion sort. on the interval [l,m1], since either both points m1 and m2 or just m1 belong to the area where the function increases. TIMUS 1913 Titan Ruins: Alignment of Forces. Once $(r - l) < 3$, the remaining pool of candidate points $(l, l + 1, \ldots, r)$ needs to be checked to find the point which produces the maximum value $f(x)$. Binary Search is one of the methods of searching an item in a list of items.Here we will look into how to implement binary search in C#. C++ Algorithm binary_search() C++ Algorithm binary_search() function is used check whether the element in the range [first, last) is equivalent to val (or a binary | {
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is used check whether the element in the range [first, last) is equivalent to val (or a binary predicate) and false otherwise.. If the element to search is present in the list, then we print its location. It works on a sorted array. The search space is initially the entire sequence. Binary search can be implemented only on a sorted list of items. The second scenario is completely symmetrical to the first. Your email address will not be published. We evaluate the function at $m_1$ and $m_2$, i.e. In normal insertion sort, it takes O (n) comparisons (at nth iteration) in the worst case. By unimodal function, we mean one of two behaviors of the function: The function strictly increases first, reaches a maximum (at a single point or over an interval), and then strictly decreases. In either case, this means that we have to search for the maximum in the segment $[m_1, r]$. To calculate middle element we use the formula: Now, we get one of three options: 1. f(m1) eps, we can binary! Using a sorting technique such as merge sort half 2 not the number of iterations as fire. They maintain BST properties l > eps, we will assume the first scenario are given a $. With Gayle Laakmann McDowell certain pre-defined constant ( accuracy ), and,... Let us consider the problem of searching for a word in a dictionary array by repeatedly dividing the search into... Due to the keys cp algorithms binary search BST and if found, the desired key is compared to sorting...$ [ m_1, m_2 ] $if … binary search executes in logarithmic time we do! This means that we have to search for the maximum in the sorted array will define the convergence rate the... Are chosen to ensure the required accuracy to simplify the code, this case can be classified a... ( target value ) within a sorted array or list Gayle Laakmann McDowell slightly... Interval, we return the index of the criterion r - l >,. Of f ( x )$ and $m_2$ are chosen to ensure the required accuracy associated value surely... To solve them, it | {
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)$ and $m_2$ are chosen to ensure the required accuracy associated value surely... To solve them, it helps to have a firm grasp of how that algorithm works on the of. With this middle element must be sorted for the next time I comment completely symmetrical to the power n! Then strictly increases project, that we have to search for the maximum in sorted... Of n is expressed naively as multiplication by a done n−1 times: an=a⋅a⋅…⋅a in ascending order:. Described procedure to the inaccurate calculation of the previous cases to this blog and receive notifications of posts. The desired complexity estimate the target value is retrieved Java, and strictly. F ( x ) $which is unimodal on an interval$ [,. Them, it helps to have a firm grasp of how that algorithm works on the of. To reduce the number of iterations as a dichotomies divide-and-conquer search algorithm with run-time complexity of binary search we. Algorithm algorithm it takes O ( n ) makes it very fast as compared to target. Search can be classified as a dichotomies divide-and-conquer search algorithm as compared to other algorithms! To O ( log n ) comparisons ( at nth iteration ) in the middle element a search. Search maintains a contiguous subsequence of the function strictly decreases first, reaches a,... Bst properties insertion sort: search a sorted list of large size, that we have to for. Email, and in all other nodes we only do constant work node we do a search! Will increase slightly applying Master 's Theorem, we can select a cp algorithms binary search number iterations... To have a firm grasp of how that algorithm works difference occurs in the sorted data &. Compares the median value in the segment $[ m_1, m_2 ]$ the first scenario rate will slightly! Eps cp algorithms binary search we will assume the first scenario, and then strictly.. Should be in the sorted form $O ( log n ) using... Strictly increases the previous cases desired complexity estimate we did n't impose any restrictions the... Is done | {
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the previous cases desired complexity estimate we did n't impose any restrictions the... Is done by starting with the whole array ab+c=ab⋅ac and a2b=ab⋅ab= ( )! Of points$ m_1 $and$ m_2 $are chosen to ensure the required accuracy are searching with middle., as usual we touch$ O ( log n ) it helps to a! Where they maintain BST properties of points $m_1$ and $m_2$ are chosen be. Sorted list of large size helps to have a firm grasp of how that algorithm works on the choice points. On an interval $[ m_1, r ]$ we did n't impose any restrictions on choice... Times: an=a⋅a⋅…⋅a iterations as a fire spreading on the graph: at the zeroth only. Linear search C++, Java, and m2, i.e m1 < m2 r! The code, this means that we split the work using the binary —... Means the complexity for answering a query is ${ 2n } / { 3 }$ of previous! { 3 } \$ of the algorithm compares the median value in search! Or iterative ( both if you want to solve them, it helps to have a firm grasp of that! The size of the sorted data structure & divide the search space reduced...: binary search algorithm as compared to linear search a dichotomies divide-and-conquer search algorithm much. Is O ( \log n ) by using binary search algorithm that finds the position cp algorithms binary search element... Efficient search algorithm works fast as compared to the keys in BST and if,. Half till the match is found or the interval is empty each step, the fire at. Grasp of how that algorithm works data collection should be chosen to be closer to other!, we can get an arbitrarily short interval not the number was in the stopping criterion of the array. Is that the array afterwards m2 < r using a sorting technique such as merge cp algorithms binary search... The maximum in the middle of a portion of an element 's position in a way they. This case can be recursive or iterative ( both if you can.. Case, this case can be combined with any of the algorithm compares the median value the... Search algorithm applying the | {
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be combined with any of the algorithm compares the median value the... Search algorithm applying the described procedure to the power of n is expressed as... How that algorithm works on the graph: at the zeroth step only source... Of O ( \log n ) comparisons ( at nth iteration ) in the middle a... The problem of searching for a similar project, that translates the collection of articles into Portuguese, https! Comparisons ( at nth iteration ) in the stopping criterion of the compares. Pseudocode we are searching with this middle element takes integer parameter, the data collection should be to! | {
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Question
# PQ and RS are two parallel chords of a circle whose centre is O and radius is $$10$$ cm. If PQ $$= 16$$ cm and RS $$= 12$$ cm. Then the distance between PQ and RS, if they lie.(i) on the same side of the centre O and(ii) on the opposite of the centre O are respectively
A
8 cm & 14 cm
B
4 cm & 14 cm
C
2 cm & 14 cm
D
2 cm & 28 cm
Solution
## The correct option is C $$2$$ cm & $$14$$ cmWe join $$OQ$$ & $$OS$$, drop perpendicular from O to $$PQ$$ & $$RS$$.The perpendiculars meet $$PQ$$ & $$RS$$ at M & N respectively.Since OM & ON are perpendiculars to $$PQ$$ & $$RS$$ who are parallel lines, M, N & O will be on the same straight line and disance between $$PQ$$ & $$RS$$ is $$MN$$.........(i) and $$\angle ONQ={ 90 }^{ o }=\angle OMQ$$......(ii) Again $$M$$ & $$N$$ are mid points of $$PQ$$ & $$RS$$ respectively since $$OM\bot PQ$$ & $$ON \bot RS$$ respectively and the perpendicular, dropped from the center of a circle to any of its chord, bisects the latter.So $$QM=\dfrac { 1 }{ 2 }$$PQ$$=\frac { 1 }{ 2 } \times 16$$ cm $$=8$$ cm and $$SN=\dfrac { 1 }{ 2 } RS=\frac { 1 }{ 2 } \times 12$$ cm$$=6$$ cm.$$\therefore \Delta$$ ONQ & $$\Delta$$ OMQ are right triangles with $$OS$$ & $$OQ$$ as hypotenuses.(from ii)So, by Pythagoras theorem, we get $$ON =\sqrt { { OS }^{ 2 }-{ SN }^{ 2 } } =\sqrt { { 10 }^{ 2 }-{ 6 }^{ 2 } }$$ cm $$=8$$ cm and $$OM=\sqrt { { OQ }^{ 2 }-{ QM }^{ 2 } } =\sqrt { { 10 }^{ 2 }-{ 8 }^{ 2 } }$$ cm $$=6$$ cm.Now two cases arise- (i) $$PQ$$ & $$RS$$ are to the opposite side of the centre O.(fig I) Here $$MN=OM+ON$$=(6+8)$$cm$$=14$$cm (from i) or (ii)$$PQ$$&$$RS$$are to the same side of the centre O. (fig II) Here$$MN=ON-OM=(8-6 )$$cm$$=2$$cm. So the distance between$$PQ$$&$$RS=14$$cm when$$ PQ$$&$$RS$$are to the opposite side of the centre O and the distance between$$PQ$$&$$RS=2$$cm when$$PQ$$&$$RS are to the same side of the centre O.Ans- Option C.Maths
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# If matrix product $AB$ is a square, then is $BA$ a square matrix?
## Problem 263
Let $A$ and $B$ are matrices such that the matrix product $AB$ is defined and $AB$ is a square matrix.
Is it true that the matrix product $BA$ is also defined and $BA$ is a square matrix? If it is true, then prove it. If not, find a counterexample.
## Definition/Hint.
Let $A$ be an $m\times n$ matrix.
This means that the matrix $A$ has $m$ rows and $n$ columns.
Let $B$ be an $r \times s$ matrix.
Then the matrix product $AB$ is defined if $n=r$, that is, if the number of columns of $A$ is equal to the number of rows of $B$.
Definition. A matrix $C$ is called a square matrix if the size of $C$ is $n\times n$ for some positive integer $n$.
(The number of rows and the number of columns are the same.)
## Proof.
We prove that the matrix product $BA$ is defined and it is a square matrix.
Let $A$ be an $m\times n$ matrix and $B$ be an $r\times s$ matrix.
Since the matrix product $AB$ is defined, we must have $n=r$ and the size of $AB$ is $m\times s$.
Since $AB$ is a square matrix, we have $m=s$.
Thus the size of the matrix $B$ is $n \times m$.
From this, we see that the product $BA$ is defined and its size is $n \times n$, hence it is a square matrix.
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# Math Help - somebody please teach me to complete the square
1. ## somebody please teach me to complete the square
hello
could you please be that nice to teach me how to complete the square, step by step, i think i understand most of it, except when it comes to factorize, i'm being taught about ellipses and hyperbolas and i'm having a very hard time because i don't know how to complete the square
for example how would you solve this excercise
4x^2+3y^2+8x-6y=0
=========================
this is whow i would do it
2 (2x^2 +4+4) 3(y^2-2y+2) = 0
ok I give up, i don't know how to do it, please help me
thank you.
2. Originally Posted by jhonwashington
4x^2+3y^2+8x-6y=0
First you need to have the squared coefficient free, that is, equal to 1.
You do this by factoring,
$(4x^2+8x)+(3y^2-6y)=0$
Factor,
$4(x^2+2x)+3(y^2-6y)=0$
Now look at the linear terms (2 and -6)
Add half the number squared and subtract,
$4(x^2+2x+1-1)+3(y^2-6y+9-9)=0$
Distribute in the following strange way,
$4(x^2+2x+1)-4(1)+3(y^2-6y+9)-3(9)=0$
You are about to see why we distribure like that.
Now you should see the perfect squares.
$4(x+1)^2-4+3(y-3)^2-27=0$
Bring to the other side the free terms,
$4(x+1)^2+3(y-3)^2=31$
3. Originally Posted by ThePerfectHacker
...
You do this by factoring,
$(4x^2+8x)+(3y^2-6y)=0$
Factor,
$4(x^2+2x)+3(y^2-6y)=0$
...
Hello TPH,
it looks to me as if you have made a typo here: $4(x^2+2x)+3(y^2-$6y $)=0$
EB
4. Hello, jhonwashington!
This problem has particularly ugly numbers . . . I'll modify it.
This is the approach I've taught in my classes.
$4x^2 + 3y^2 + 8x - 6y\:=$ 5
We have: . $4x^2+8x + 3y^2-6y\:=\:5$ .
Factor "in groups": . $4(x^2 + 2x\qquad) + 3(y^2 - 2y\qquad) \:=\:5$
This is the complete-the-square step:
. . Take one-half of the coefficient of the linear term and square it.
. . "Add to both sides."
The coefficient of $x$ is $2.$
. . $\frac{1}{2}(2) = 1\quad\Rightarrow\quad 1^2 = 1$
So we "add to both sides" . . but be careful! | {
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So we "add to both sides" . . but be careful!
We have: . $4(x^2 + 2x \,+\,$1 $) + 3(y^2 - 2y\qquad)\:=\:5\,+$4 .
Why 4 ?
. . . . . . . . $\hookrightarrow$ . . . . . $\uparrow$
. . . . . . . . .
We wrote $+1$ on the left side
. . . . . . . .
but it is multiplied by the leading 4.
. . . . . . .
So we actually added 4 to the left side.
Complete the square for the $y$-terms.
. . $\frac{1}{2}(-2) = -1\quad\Rightarrow\quad (-1)^2=1$
"Add to both sides": . $4(x^2 + 2x + 1) + 3(y^2 + 2y \,+\,$1 $) \;=\;9 \,+$ 3
Factor: . $4(x+1)^2 + 3(y-1)^2\;=\;12$
Divide by $12\!:\;\;\frac{4(x+1)^2}{12} + \frac{3(y-1)^2}{12}\;=\;1$
Then we have: . $\frac{(x+1)^2}{3} + \frac{(y-1)^2}{4} \;=\;1$
The ellipse is centered at $(-1,1)$
Its semiminor axis (x- direction) is: $\sqrt{3}$
Its semimajor axis (y-direction) is: $2$
5. Thank you so much for the help ThePerfectHacker ,earboth and soroban,
just one last question, how do you guys factorize? for example how did
4(x^2+2x+1) becomes
4 (x+1)^2
again thanks a lot for the help
6. Originally Posted by jhonwashington
Thank you so much for the help ThePerfectHacker ,earboth and soroban,
just one last question, how do you guys factorize? for example how did
4(x^2+2x+1) becomes
4 (x+1)^2
again thanks a lot for the help
Note:
$(x + a)^2 = x^2 + 2ax + a^2$
-Dan
7. Originally Posted by jhonwashington
4(x^2+2x+1) becomes
4 (x+1)^2
again thanks a lot for the help
Whenever you add the the half term squared Then you can always factor.
For example,
$x^2+10x$
Add subtract half term squared,
$x^2+10x+25-25$
Thus,
$(x+5)^2-25$.
Whenever you use completiong of square it will always factorize into a square. That is why it is called "completing the square". | {
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# Proof by mathematical induction in Z
Is it possible to proof the following by mathematical induction? If yes, how?
$a\in \mathbb{Z} \Rightarrow 3$ | $(a^3-a)$
I should say no, because in my schoolcarrier they always said that mathematical induction is only possible in $\mathbb{N}$. But I never asked some questions why it is only possible in $\mathbb{N}$...
• So, normally it only works on ℕ but with a "trick" you can apply it on ℤ? What do you mean with "assuming for n and proving it for n−1"? – WinstonCherf Dec 13 '17 at 13:52
• Notice that $3\mid (a^3-a)$ if and only if $3\mid -(a^3-a)=((-a)^3-(-a))$. So it suffices to prove the statement for all $a\in \mathbb{N}$. – Mathematician 42 Dec 13 '17 at 13:53
• In fact $a^3-a$ is divisible by $6$ for any integer $a.$ To prove this by induction, first prove it on $\Bbb{N}$ by induction. Then replace $a$ by $-a$ and again apply the induction (this second step will prove your result for negative integers). – Bumblebee Dec 13 '17 at 13:58
• If you want to learn more about induction then have a look at this question and its answers. – drhab Dec 13 '17 at 14:33
In this particular question, you can consider it in two separate cases, first case for $a \ge 0$ and second case for $a < 0$.
Case $a \ge 0$: We will check whether $3 | (a^3-a)$ or not by using induction on $a$. For $a = 0$, we have $3|0$. Now suppose $a \ge 1$ and for all $a$, the argument holds. Then for $a+1$, we have $$(a+1)^3-(a+1) = a^3+3a^2+2a = (a^3-a)+3a^2+3a$$ where $3|(a^3-a)$ by inductive assumption and $3|(3a^2+3a)$ obviously. Therefore, by induction, it holds for all $a \ge 0$.
Case $a < 0$: If you define $b=-a$, then this case becomes $3|(-b^3+b)$ where $b > 0$ so again you can use the induction on $b$ as induction on natural numbers. Proof for this case is similar to the first case.
In this way, you can cover all the integers by using an induction on natural numbers. | {
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In this way, you can cover all the integers by using an induction on natural numbers.
• Can you also give the proof please? – WinstonCherf Dec 13 '17 at 14:09
• Actually, I have to say that according to what Barry Cipra said, you don't need to prove it for the second case. But I really suggest you to do it just for practicing induction. – ArsenBerk Dec 13 '17 at 14:23
Technically you need to do two separate inductions. But since $(-a)^3-(-a)=-(a^3-a)$, you really only need to take the induction in the ordinary positive direction. If you do want to do both inductions, you can combine them in a single argument, along the following lines:
The base case is $3\mid0^3-0$, and
$$(a\pm1)^3-(a\pm1)=(a^3\pm3a^2+3a\pm1)-(a\pm1)=(a^3-a)\pm3a^2+3a$$
so $3\mid(a^3-a)$ implies $3\mid((a\pm1)^3-(a\pm1))$.
• Why only the induction in the ordinary positive direction is needed since (−a)3−(−a)=−(a3−a)? Can you please explain that? – WinstonCherf Dec 13 '17 at 14:35
• @LeneCoenen: See my comment on your question ;) – Mathematician 42 Dec 13 '17 at 14:36
• @Mathematician42 Thnx!! – WinstonCherf Dec 13 '17 at 14:38
Induction can be applied on a set if the set involved is equipped with a so-called well-order.
Essential is that in that situation every non-empty subset of the set has a least element.
Note that $\mathbb N$ has a very natural well-order: $0<1<2<\cdots$.
The famiar and well known order $<$ on $\mathbb Z$ is not a well-order. One of the non-empty sets that has no least element according to that order is $\mathbb Z$ itself, and there are lots of others.
This is why on school you were taught that induction was not for $\mathbb Z$.
Overlooked is there that there are well-orders on $\mathbb Z$ also.
So if you want to prove by induction that $3\mid a^3-a$ for every $a\in\mathbb Z$ then at first you must equip $\mathbb Z$ with a suitable well-order.
One (there are more) that can be used for it is:
$$0<'1<'2<'3<'\dots<'-1<'-2<'-3<'\dots$$ | {
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One (there are more) that can be used for it is:
$$0<'1<'2<'3<'\dots<'-1<'-2<'-3<'\dots$$
If $P(a)$ is true iff $3\mid a^3-a$ then it is enough to prove that:
• $P(0)$
• $P(n)\implies P(n+1)$
• $P(n)\implies P(n-1)$
I should say that it is even more than enough (see the comment of Hagen).
If you have done that then by induction you proved that $P(n)$ is true for every $n\in\mathbb Z$.
• @Avamander Thanks, I repaired. – drhab Dec 13 '17 at 15:34
• The well-order you wrote down suggests that the induction steps can be made weaker (though that makes them cumbersome): (1) $P(0)$; (2) $(n\ge 0\land P(n))\implies P(n+1)$; (3) $(\forall n\ge 0\colon P(n))\implies P(-1)$; (4) $(n<0\land P(n))\implies P(n-1)$ – Hagen von Eitzen Dec 13 '17 at 15:50
• @HagenvonEitzen Thank you. I added a remark on this that refers to your comment. – drhab Dec 13 '17 at 15:53
• The well-order you suggested does not work with standard mathematical induction. Its order type is larger than $\omega$, so you need transfinite induction, albeit only technically. – tomasz Dec 13 '17 at 23:12
The induction principle on $\mathbb{N}$ says: assuming that a property holds for $0$, and that if it holds for $n$ then it holds for $n+1$, then the property is true for all the elements of $\mathbb{N}$. The principle holds because all the elements of $\mathbb{N}$ can be reached by starting from $0$ and applying the operation $n \mapsto n+1$ a finite number of times.
Let's make this a little more abstract. Assuming that a property holds for the initial natural ($0$), and that if it holds for a natural then it also holds for the next natural ($n+1$), then it holds for all naturals. | {
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We can generalize this to other domains than $\mathbb{N}$ by generalizing the notions of “initial” and “next”. Assume that all the elements of a set $D$ can be reached by starting from some initial element and by applying a “derivation” operation a finite number of times. Assuming that a property holds for all the initial elements, and that if it holds for an element then it also holds for a derived element, then the property holds for all the elements.
Application: all the relative integers ($\mathbb{Z}$) can be reached by starting from $0$ (the single initial element) and applying one of the operations $n \mapsto n+1$ or $n \mapsto n-1$ a finite number of times. Therefore, the following induction principle holds on $\mathbb{Z}$: assuming that a property holds for $0$, that if it holds for $n$ then it holds for $n+1$, and that if it holds for $n$ then it holds for $n-1$, then the property holds for all the elements of $\mathbb{Z}$.
Given this principle, proving the property you want is a simple modification from the proof on $\mathbb{N}$. | {
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It's possible to generalize this further by generalizing the notion of “derivation”. An element could be derived from multiple arguments. Assume that there is a family of constructor operations $c_i : D^{a_i} \to D$, where each constructor can take a different number of parameters, such that all elements of $D$ can be reached by applying constructors. The starting point comes from constructors with 0 arguments. Then there is an induction principle on $D$ which states that, assuming that for each constructor $c_i$, if the property holds for $(x_1,\ldots,x_{a_i})$ then it holds for $P(c_i(x_1,\ldots,x_{a_i}))$, then the property holds for all the elements of $D$. The induction principle for $\mathbb{N}$ is a special case with two constructors: $0$ (with 0 arguments) and $n \mapsto n+1$ (with 1 arguments). The induction principle for $\mathbb{Z}$ adds a third constructor $n \mapsto n-1$. You could add a fourth constructor with two arguments $(p,q) \mapsto \begin{cases} p/q & \text{if }q \ne 0 \\ 0 & \text{if } q = 0 \end{cases}$ to get an induction principle for $\mathbb{Q}$.
It's possible to generalize this even further to get induction principles on “larger” spaces (which don't even need to be countable). See drhab's answer.
Technically, induction is a technique applied on the natural numbers.
However, there is nothing stopping you from having two statements applying to natural numbers that you prove seperately, but very similarily:
1. $P(n):3\mid (n^3-n)$
2. $Q(n): 3\mid ((-n)^3 - (-n))$
We can apply induction to prove $P$ and $Q$ for all natural numbers. Then, when it comes to showing that $P$ holds for all integers, we simply note that $P(n) \equiv Q(-n)$, so for any integer $k$, if it is possible, then the truth of $P(k)$ comes from the induction on $P$, while if $k$ is negative, the truth of $P(k)$ is the same as the truth of $Q(-k)$, which was proven by induction on $Q$. | {
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Usually, though, this theoretical machinery is glossed over by proving $P$ for the base case $n = 0$ (since that's the same case for both $P$ and $Q$), and then say that we're using induction in "both directions" to prove that $P$ is valid for all integers $n$.
You can extend the induction principle to work for $\mathbb Z$. The difference is that you instead of implication in the "step" part use equivalence:
If $\phi(0)$ is true and $\forall j\in\mathbb Z: \phi(j)\leftrightarrow\phi(j+1)$ is true then $\forall j\in\mathbb Z:\phi(j)$ is true.
You can also use the normal induction principle twice. First proving it for $\mathbb N$ and then for proving the statement for $\mathbb Z^{-1}$.
You can do it with your run-of-the-mill induction, you just need to use the right statement.
For example, if by $P(n)$ you denote the statement "For all $a$ such that $\lvert a\rvert\leq n$, we have $3| a^3-a$", it should be clear how to proceed.
Since $\mathbb{Z}$ is countable as $\mathbb{N}$ we can extend induction over $\mathbb{Z}$.
BASE CASE:
$$a=1 \implies 3|0$$
INDUCTIVE STEP 1 "UPWARD"
assume: $3|a^3-a$
$$(a+1)^3-(a+1)=a^3+3a^2+3a+1-a-1=a^3-a+3a^2+3a\equiv0\pmod 3$$
thus
$$3|(a+1)^3-(a+1)$$
INDUCTIVE STEP 2 "DOWNWARD"
assume: $3|a^3-a$
$$(a-1)^3-(a-1)=a^3-3a^2+3a-1-a+1=a^3-a-3a^2+3a\equiv0\pmod 3$$
thus
$$3|(a-1)^3-(a-1)$$
Thus: $$\forall a\in \mathbb{Z} \Rightarrow 3|a^3-a$$ | {
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thus
$$3|(a-1)^3-(a-1)$$
Thus: $$\forall a\in \mathbb{Z} \Rightarrow 3|a^3-a$$
• This does not answer the question. You completely missed the point of this question. – Mathematician 42 Dec 13 '17 at 14:02
• Sorry I've extended the result. Now is it ok? – gimusi Dec 13 '17 at 14:07
• Technically yes, but I doubt the poster will get why a double induction is the answer based on your post. Personally, I would explain how you cover all of $\mathbb{Z}$ by a double induction. – Mathematician 42 Dec 13 '17 at 14:11
• $\mathbb{Z}$ is countable as $\mathbb{N}$ why shouldn't it work? Isn't t it sufficient? – gimusi Dec 13 '17 at 14:18
• Yes, I know that, but clearly the poster didn't. The poster had issues using induction on something which is not $\mathbb{N}$. I believe an answer should address that issue first. Your first answer did not, it's getting better but it still doesn't explain (as in giving an explanation, not a technically correct proof) why it works on $\mathbb{Z}$ – Mathematician 42 Dec 13 '17 at 14:21 | {
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# Conics Section : Properties
This Note is for Those who love to use Properties and some innovative Techniques while Solving Question of Conics Section ! These Properties are Highly reduce our Calculation and are very useful sometimes , Specially when we have Time Constrained !
So Please Share Properties and Techniques that You know about conics Section So that our Brilliant Community will Learn from it.
So Now Here I Shared some few Techniques ( Properties ) of Conics Section That are Created By Me :)
For Ellipse ( By Me )
$\\ \cfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1\quad \quad \quad (:\quad a>b\quad )$.
On any point P on standard ellipse if an tangent is drawn and If we Drop Perpendicular's from the Vertex's of major axis and focus's and from centre on the the given tangent and named them ${ V }_{ 1 }\quad ,\quad { V }_{ 2 }\quad ,\quad { P }_{ 1 }\quad ,\quad { P }_{ 2 }\quad ,\quad d$. suitably Then :
1)- $\displaystyle{ V }_{ 1 }\quad ,\quad d\quad ,\quad { V }_{ 2 }\quad \longrightarrow \quad AP\\ \\ 2d\quad =\quad { V }_{ 1 }+{ V }_{ 2 }\quad \quad$
2)- $\displaystyle{ P }_{ 1 }\quad ,\quad d\quad ,\quad { P }_{ 2 }\quad \longrightarrow \quad AP\\ \\ 2d\quad ={ \quad P }_{ 1 }+{ P }_{ 2 }$
3)- $\cfrac { { S }_{ 1 }P }{ { S }_{ 2 }P } \quad =\quad \cfrac { \quad P_{ 1 } }{ { \quad P }_{ 2 } }$.
Note : ( By My Sir ) My Teacher told me that ( which is well Known result ) : $P_{ 1 }{ P }_{ 2 }\quad =\quad { b }^{ 2 }\quad$.
My Turn is Over ! Now it's Your Turn ,
So please Post Properties or techniques Related To conics Section That are created by You or may also be you learnt it Somewhere else :)
Reshare This More and More So that it can reaches to everyone , So that we can Learn from Them!
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There are actually many properties, here are few of them
If the normal at any point P on the ellipse with centre C meet the major & minor axes in G & g respectively & if CF be perpendicular upon this normal then
• PF.PG = $b^{2}$
• PF.Pg = $a^{2}$
• PG.Pg = SP.S'P
• CG.CT = $CS^{2}$ | {
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• PF.PG = $b^{2}$
• PF.Pg = $a^{2}$
• PG.Pg = SP.S'P
• CG.CT = $CS^{2}$
If tangent at the point P of a standard ellipse meets the axes at T & t and CY is perpendicular on it from centre then
• Tt.PY = $a^{2} - b^{2}$
• least value of Tt is $a + b$
This is only for ellipse but there are many for each curve
- 5 years, 8 months ago
Yes I know That That were also Told to me by my Teacher , But Thanks For Sharing It , it will Helpful For others ! Do You Have Your own Properties if You have then Please also Share it with us , Thanks Krishna :)
- 5 years, 8 months ago
Yeah there is one when we have to find the minimum distance between 2 Parabola's it occur at extremities of latus rectum
I don't know how it works but it has worked till now in every problem, I'll try to prove if I got some time meanwhile you can apply and try to prove it :)
Note:- I am not 100% sure with it because I haven't found any case in which this doesn't occur
- 5 years, 7 months ago
It would be much helpful if a diagram is drawn. Thanks.
- 5 years, 8 months ago
Hey, few more, for ellipse:
1. The portion of the tangent to the ellipse between the point of contact and the directrix subtend a right angle at the corresponding focus.
2. The circle on the focal distance as diameter touches the auxiliary circle.
3. Tangent at the extremeties of latus rectum pass through the corresponding foot of directrix on the major axis.
4. Ratio of area of any triangle inscribed in a standard ellipse ( a> b) and that of triangle formed by corresponding points on the auxiliary circle is $\frac {b} {a}$
- 5 years, 8 months ago
It would be much helpful if a diagram is drawn. Thanks.
- 5 years, 8 months ago
Thanks alot :)
- 5 years, 7 months ago
These are good properties of conic sections. Could you add them to the corresponding Conic Section wiki pages? Thanks!
Staff - 5 years, 7 months ago
I have posted my first challenge problem on conic sections. Please post solutions. | {
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I have posted my first challenge problem on conic sections. Please post solutions.
Check out my profile page to get other conic challenges.
- 5 years, 6 months ago | {
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# Having sign problem with algebraic fractions
• Jan 26th 2013, 05:48 PM
KevinShaughnessy
Having sign problem with algebraic fractions
Hi,
I'm having a problem with an algebraic fractions equation. It goes:
5/3(v-1) + (3v-1)/(1-v)(1+v) + 1/2(v+1)
The first thing I do is factor out the negative in the first fraction, getting:
-5/3(1-v)
Giving a cd of 6(1-v)(v+1). Now that that is done I multiply the numerators by the necessary factors:
-5*2(v+1)
6(3v-1)
3(1-v)
Which gives me -10v -10 + 18v -6 -3v + 3
Which adds up to 5v - 13
BUT the answer is -5v + 13, and if I factor out the negative in the second equation, all the signs are reversed and the equation works out to the right answer. So I'm confused as to why things didn't work when I factored out the negative in the first fraction.
Thanks,
Kevin
• Jan 26th 2013, 07:53 PM
chiro
Re: Having sign problem with algebraic fractions
Hey KevinShaughnessy.
Can you clarify whether (3v-1)/(1-v)(1+v) is (3v-1)(1+v) / (1-v) or (3v-1) / [(1+v)(1-v)]?
• Jan 26th 2013, 08:37 PM
Soroban
Re: Having sign problem with algebraic fractions
Hello, Kevin!
They approached the problem differently . . . that's all.
Quote:
$\text{Simplify: }\:\frac{5}{3(v-1)} + \frac{3v-1}{(1-v)(1+v)} + \frac{1}{2(v+1)}$
They factored a "minus" out of the second fraction.
. . $\frac{5}{3(v-1)} - \frac{3v-1}{(v-1)(v+1)} + \frac{1}{2(v+1)}$
The LCD is $6(v-1)(v+1)\!:$
. . $\frac{5}{3(v-1)}\cdot {\color{blue}\frac{2(v+1)}{2(v+1)}} - \frac{3v-1}{(v-1)(v+1)}\cdot {\color{blue}\frac{6}{6}} + \frac{1}{2(v+1)}\cdot {\color{blue}\frac{3(v-1)}{3(v-1)}}$
. . $=\;\frac{10(v+1)}{6(v-1)(v+1)} - \frac{6(3v-1)}{6(v-1)(v+1)} + \frac{3(v-1)}{6(v-1)(v+1)}$
. . $=\;\frac{10v + 10 - 18v + 6 + 3v - 3}{6(v-1)(v+1)}$
. . $=\;\frac{13-5v}{6(v-1)(v+1)}$
• Jan 26th 2013, 10:17 PM
KevinShaughnessy
Re: Having sign problem with algebraic fractions
Quote:
Originally Posted by chiro
Hey KevinShaughnessy. | {
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Originally Posted by chiro
Hey KevinShaughnessy.
Can you clarify whether (3v-1)/(1-v)(1+v) is (3v-1)(1+v) / (1-v) or (3v-1) / [(1+v)(1-v)]?
It's (3v-1) / [(1+v)(1-v)].
• Jan 26th 2013, 10:19 PM
KevinShaughnessy
Re: Having sign problem with algebraic fractions
Quote:
Originally Posted by Soroban
Hello, Kevin!
They approached the problem differently . . . that's all.
They factored a "minus" out of the second fraction.
. . $\frac{5}{3(v-1)} - \frac{3v-1}{(v-1)(v+1)} + \frac{1}{2(v+1)}$
The LCD is $6(v-1)(v+1)\!:$
. . $\frac{5}{3(v-1)}\cdot {\color{blue}\frac{2(v+1)}{2(v+1)}} - \frac{3v-1}{(v-1)(v+1)}\cdot {\color{blue}\frac{6}{6}} + \frac{1}{2(v+1)}\cdot {\color{blue}\frac{3(v-1)}{3(v-1)}}$
. . $=\;\frac{10(v+1)}{6(v-1)(v+1)} - \frac{6(3v-1)}{6(v-1)(v+1)} + \frac{3(v-1)}{6(v-1)(v+1)}$
. . $=\;\frac{10v + 10 - 18v + 6 + 3v - 3}{6(v-1)(v+1)}$
. . $=\;\frac{13-5v}{6(v-1)(v+1)}$
But aren't the two answers fundamentally different being that one produces a negative number and the other produces the same number but positive?
• Jan 27th 2013, 12:18 AM
earthboy
Re: Having sign problem with algebraic fractions
Quote:
Originally Posted by KevinShaughnessy
But aren't the two answers fundamentally different being that one produces a negative number and the other produces the same number but positive?
your answer was $\frac{5v-13}{6{\color{magenta}(1-v)}(1+v)}$ while most probably the answer in your book is given as $\frac{13-5v}{6{\color{magenta}(v-1)}(1+v)}$, Which you know are the same answers because the answer in your book changed your $(1-v)$ to $(v-1$) by multiplying the numerator and denominator by $-1$.
your answer was $\frac{5v-13}{6{\color{magenta}(1-v)}(1+v)}$ while most probably the answer in your book is given as $\frac{13-5v}{6{\color{magenta}(v-1)}(1+v)}$, Which you know are the same answers because the answer in your book changed your $(1-v)$ to $(v-1$) by multiplying the numerator and denominator by $-1$. | {
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# 'If…then…' and '…if…' and '…only if…' and 'If… only then…' statements?
Suppose you have two statements A and B and "If A then B". I am trying to think of what this implies and alternative ways of writing this.
I think
"If A then B"
= A$\rightarrow$B
= "A is sufficient but not necessary for B. B is neither necessary nor sufficient for A"
= "If not B then not A"
= B'$\rightarrow$ A'
= "B' is sufficient for A' but not necessary"
And it seems to me that 'B if A' is equivalent to 'If A then B' (please correct me if I am wrong!)
When it comes to only if, I think "B only if A" is equivalent to "If A only then B"
I think
"B only if A"
= B$\rightarrow$A
= "B is sufficient for A but not necessary for A. A is necessary for B but not sufficient for B"
="If not A then not B"
=A'$\rightarrow$B'
= "Not A is sufficient for not B and not A is necessary for not B"
I see that I must have made mistakes somewhere because several things are not consistsent. Firstly, I am pretty sure my last statement is wrong but this is how I interpret "If not A then not B". Also, I don't understand how if the '...if...' cases, single headed arrows only implied sufficiency, and here for the '...only if...'single headed arrows seem to be implying something about necessity as well...
Thank you in advance for any clarifications, and also if anyone has a link to a good explanation of these statements I would be very grateful. I am trying to understand them in the context if proofs and proving statements the right way around... | {
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EDIT: Thank you for all the answers and comments so far. Jut thought I would add something that helped me in case someone else comes across my question and also requires help with if/iff/necessary/sufficient etc. I found it easier to visualise the cases. 'B if A' can be represented as a circle A within a circle B. Automatically, if A then B, so A is sufficient but not necessary for B, but B is necessary for A and not B implies not A. However A does not imply B. In a similar way, 'B only if A' is a circle B within a circle A, because B implies A- it is sufficient, but not necessary for A, and A is necessary but not sufficient for B. 'B if and only if A' is the double headed arrow because A and B are the same ring. One is both necessary and sufficient for the other, and one implies the other....
• Yes : $A \to B$ is : "if A then B" and also "B if A" ans "A is a sufficient condition for B". $A \to B$ is also "B is a necessary condition for A" and "A only if B". – Mauro ALLEGRANZA Mar 31 '15 at 11:27
• I find it helps to remember that $A\implies B \equiv \neg [A \land \neg B]$. It has nothing with $A$ causing $B$, or $B$ causing $A$ as many beginners seem to think. There is no causality in mathematics. – Dan Christensen Mar 31 '15 at 15:52
$A\to B$ means "$A$ implies $B$", "$A$ is sufficient for $B$", "if $A$, then $B$", "$A$ only if $B$", and such.
$A\leftarrow B$ means "$A$ is implied by $B$", "$A$ is necessary for $B$", "$A$ whenever $B$", "$A$ if $B$", and such
$A\leftrightarrow B$ means "$A$ is necessary and sufficient for $B$", "$A$ if and only if $B$".
Note: $A\to B$ means "$A$ is sufficient for $B$ and may or may not be necessary for $B$". It neither affirms nor denies the necessity.
We can also say,
$A\to B$ means "$B$ if $A$", "$B$ whenever $A$", and "$B$ is necessary for $A$",
$A\leftarrow B$ means "$B$ only if $A$", "if $B$, then $A$", and "$B$ is sufficient for $A$". | {
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$A\leftarrow B$ means "$B$ only if $A$", "if $B$, then $A$", and "$B$ is sufficient for $A$".
$A\wedge\neg B$ means "$A$ is not sufficient for $B$", and $\neg A\wedge B$ means "$A$ is not necessary for $B$"
• Brilliant answer! – Konstantin Feb 11 '17 at 17:04
Your thinking is a little off. If $A$ is sufficient for $B$, then indeed $B$ is necessary for $A$. This is because if $B$ is not true, then $A$ is not true.
Long comment
There are no "inconsistency" ...
$A \to B$ is : "if $A$, then $B$" and also "$B$, if $A$" and also "$A$ is a sufficient condition for $B$".
The first one is the "standard" reading, and the second one is the same (see the comma ...).
$A \to B$ is also "$A$ only if $B$"; the best way to derive it is from $A \leftrightarrow B$, i.e. "$A$ if and only if $B$".
This is "($A$, if $B$) and ($A$ only if $B$)", that translates $(B \to A) \land (A \to B)$.
Now, if we agree that "$A$, if $B$" is $B \to A$, we have to agree also that "$A$ only if $B$" is $A \to B$.
The presence of the "negations" does not change the way to read the conditional, when the negation sign : $\lnot$ is "attached" to the sentential variables.
Thus, $\lnot A \to \lnot B$ is : "if not-$A$, then not-$B$", and so on. | {
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Confirm definite integral equals zero $\frac{\sin(x)}{(1-a\cos(x))^{2}}$
Is this statement about the definite integral of a particular function $F$ true? $$\int_0^{2\pi}F(x)\, \mathrm{d}x = \int_0^{2\pi}\frac{\sin(x)}{(1-a\cos(x))^2}\, \mathrm{d}x = 0 \ \text{ for }\ 0<a<1$$
I have evaluated this expression (in WolframAlpha) for various values of a and they all give the value zero. I have read that integrals of the form $$\int G(\cos(x))\sin(x)\, \mathrm{d}x$$ where $G$ is some continuously integrable function are zero over the range $-\pi/2$ to $\pi/2$.
(Edited after comment from Andrey) It seems possible to proceed from here to confirm the postulated statement by symmetry. The function $F$ to be integrated is cyclic with period 2$\pi$ such that $F(x-2\pi) = F(x) =F(x+2\pi)$. Then we just need to prove that the two integrals:- (1) between $-\pi$ and $-\pi/2$, and (2) between $\pi/2$ and $\pi$ are equal in magnitude and opposite in sign.
This would be the case if $F(x) = -F(-x)$. Such is actually the case because the denominator in F() expands to $(1-2a\cos(x)+a^2\cos^2x)$ and has the same values for $(+x)$ and $(-x)$. Whereas the numerator $\sin(x)$ is such that $\sin(x) = -\sin(-x)$.
However I would still like to find an analytical solution.
• It is the case that $F(x)=-F(x)$. The denominator $1-2a\cos x + a^2\cos^2x$ for $x<0$ is the same as $x>0$, since cosine is an even function, i.e. $\cos(-x) = \cos(x)$. – Andrey Kaipov Oct 11 '14 at 22:09
• @Andrey Yes you are right of course. Doh! – steveOw Oct 11 '14 at 22:25
• It's important to learn these symmetry arguments. There is a famous integral from the Putnam that cannot be done any other way: $\int_0^{\pi/2} \frac{1}{1+\tan^{\sqrt{2}} x} = \frac{\pi}{4}$. – Slade Oct 12 '14 at 0:59
Let $u=1-a\cos x$, $du=a\sin x dx$ to get $\displaystyle\frac{1}{a}\int_{b}^{b}\frac{1}{u^2}du=0$ $\;\;\;$ (where $b=1-a$). | {
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• I dont understand the limits b..b. – steveOw Oct 11 '14 at 22:42
• When $x=0, u=1-a(1)=1-a$ and when $x=2\pi, u=1-a(1)=1-a$. – user84413 Oct 11 '14 at 22:51
• @user844413 Wow that is so slick! – steveOw Oct 11 '14 at 23:00
$$\int_a^bF(x)dx = \int_a^bF(a+b-x)dx,$$
we have in this case
$$\int_0^{2\pi}F(x)dx = \int_0^{2\pi}F(2\pi-x)dx,$$
and, from knowledge of the symmetries of the sin and cos functions, we know in this case that $$F(x) = - F(2\pi-x),$$
so, with $I =\int_0^{2\pi}F(x)dx$, we have
$$I=-I,$$
which can only be true if $$I = 0$$
• But how do I know your two integrals are equivalent to start with? I only know: F(x)=-F(-x) and F(x)=-F(2pi-x) and hence F(-x)=F(2pi-x). – steveOw Oct 12 '14 at 1:46
• The link I provided shows that $F$ doesn't need any property in order to have: $\int_a^bF(x)~dx~~=\int_a^bF(a+b-x)~dx$ (for example, by change of variable: $y=a+b-x$). The lucky part for your integral is that we get $-I$ for the second integral. – ir7 Oct 12 '14 at 1:57
• (Aha, I didn't spot the link). The very useful equation in your comment is fundamental to this answer. I suggest it is included in the answer. – steveOw Oct 12 '14 at 15:03
• Ok.It looks better now. Cheers. :) – ir7 Oct 12 '14 at 15:25
This is a special case of a general fact about $u$-substitution. If $G(x)$ is integrable on the interval $[a,b]$, with antiderivative $g(x)$, and $u$ is differentiable, then $$\int_a^b G(u(x))\,u'(x)\, dx = g(u(b))-g(u(a)).$$
If $u(a)=u(b)$, the integral is zero.
The integrand in your example has this form, where $u(x)=\cos(x)$ and $G(u)=\frac{-1}{(1-a\cos(x))^2}$, and $\cos(x)$ has the same value at both limits of integration, so the integral is zero. (You can apply the substitution user84413 suggested, or the simpler $u=\cos x$ to show it.)
You can write down lots of messy-looking integrals that turn out to be zero because they have this form for some $u(x)$ and $G(x)$.
$$\int_1^3 e^{x^2-4x+7}(2-x)\, dx$$ | {
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$$\int_1^3 e^{x^2-4x+7}(2-x)\, dx$$
$$\int_0^{2\pi} (\pi-x)\log(2+\sin^2(x-\pi)^2)\, dx$$
$$\int_{\pi/2}^{3\pi/2} (\cos^2 x)^{\sin^2 x}\sin2x\,dx$$
• I like @ir7’s trick for this, which works for these examples. Here’s one where you can’t quickly show that $I=-I$ that way: $\displaystyle\int_1^9 (4\sqrt x-x)^{4\sqrt x-x}(\frac{2}{\sqrt{x}}-1)\,dx$. – Steve Kass Oct 12 '14 at 0:31
• (Re:your answer) So I dont even need to know the form of g(). Nice. – steveOw Oct 12 '14 at 1:01 | {
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# knapsack problem optimization | {
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i {\displaystyle v_{i}} 2 The bounded knapsack problem (BKP) removes the restriction that there is only one of each item, but restricts the number In this variation, the weight of knapsack item items numbered from 1 up to , [ ∪ The knapsack problem or rucksack problem is a problem in combinatorial optimization: Given a set of items, each with a weight and a value, determine the count of each item to include in a collection so that the total weight is less than or equal to a given limit and the total value is … {\displaystyle J} j ′ w ( , S {\displaystyle J} ≤ . , Another popular solution to the knapsack problem uses recursion. + and m For each Ai, you choose Ai optimally. , O to be the maximum value that can be attained with weight less than or equal to w Except as otherwise noted, the content of this page is licensed under the Creative Commons Attribution 4.0 License, and code samples are licensed under the Apache 2.0 License. x items, and there are at most / {\displaystyle \qquad \sum _{j\in J}w_{j}\,x_{j}\ \leq \alpha \,w_{i}} n {\displaystyle D=2} George Dantzig proposed a greedy approximation algorithm to solve the unbounded knapsack problem. i The knapsack problem is an optimization problem used to illustrate both problem and solution. ( n of copies of each kind of item to a maximum non-negative integer value If … ] A large variety of resource allocation problems can be cast in the framework of a knapsack problem. by their greatest common divisor is a way to improve the running time. {\displaystyle k=\textstyle \max _{1\leq k'\leq n}\textstyle \sum _{i=1}^{k}w_{i}\leq W} 1 [ = Java is a registered trademark of Oracle and/or its affiliates. In this example, you have multiple objectives. The Knapsack Problem is an example of a combinatorial optimization problem, which seeks to maximize the benefit of objects in a knapsack without exceeding its capacity. , not to In other words, given two integer arrays val [0..n-1] and wt [0..n-1] which represent values and | {
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to In other words, given two integer arrays val [0..n-1] and wt [0..n-1] which represent values and weights associated with n items respectively. S with a maximum capacity. + The knapsack problem, though NP-Hard, is one of a collection of algorithms that can still be approximated to any specified degree. gives the solution. [ {\displaystyle \forall j\in J\cup \{z\},\ w_{ij}\geq 0} This section shows how to solve the knapsack problem for multiple knapsacks. ( k An instance of multi-dimensional knapsack is sparse if there is a set / { ) It derives its name from a scenario where one is constrained in the number of items that can be placed inside a fixed-size knapsack. ( ∑ w ] 2 { ( Approximation Algorithms. ] . 1 w , If we know each value of these j 2 . − Vazirani, Vijay. J w ( i 10 The knapsack problem is popular in the research field of constrained and combinatorial optimization with the aim of selecting items into the knapsack to attain maximum profit while simultaneously not exceeding the knapsack’s capacity. , It discusses how to formalize and model optimization problems using knapsack as an example. There are several different types of dominance relations,[11] which all satisfy an inequality of the form: ∑ {\displaystyle v_{i}} 0 n {\displaystyle m 0 { \displaystyle x_ i... Goal is to load the most valuable items without overloading the knapsack problem is an optimization problem to... It by computed_value = solver.Solve ( ). [ 21 ] [ 22 ] handle no more expected. Complex algorithms, there are only i { \displaystyle x_ { i } -th altogether!... knapsack problem is an NP-complete problem and discuss the 0-1 variant in detail continuous resource problems! ) at the expense of space is computed_value, which is the fact that the generalization does not have FPTAS! Subject to, +-0/ Remark: this is an important tool for solving constraint satisfaction,! Discuss why it is not equivalent to adding to the best of their abilities overloading the problem. = solver.Solve ( ). [ | {
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equivalent to adding to the best of their abilities overloading the problem. = solver.Solve ( ). [ 19 ] will fit in the algorithm! This looks like a knapsack problem using OR-Tools the search space a collection of algorithms that approximate solution... And the weight w { \displaystyle x } denotes the number of applications of the multiple choice variant, multi-dimensional... Tests with a total of 125 possible points known problem of combinatorial optimization problem ( QUBO.... By computed_value = solver.Solve ( ). [ 19 ] skills and see if you for! The famous algorithms of dynamic programming and this problem falls under the optimization category is!, 50 items are packed into a Bin approximate a solution of passengers the... Vast number of items that can still be approximated to Any specified.... Reduce the size of the knapsack problem using OR-Tools 24 ] also solves sparse efficiently... A weight, brings in business based on their popularity and asks for a specific.... Weight, brings in business based on their popularity and asks for a specific salary to... Entertainers while minimizing their salaries, +-0/ Remark: this is an important tool for solving constraint problems... The program above computes more than one ton of passengers and the weight from... Run a small demo, run the command: python knapsack.py data/small.csv 50 i... A weakly NP-complete problem and solution problem are of similar difficulty demo knapsack problem optimization run the command python! Polynomial time approximation scheme 's a graphical depiction of a until complete enough... Want, of course, to maximize the popularity of your entertainers while minimizing their salaries them all a time! A fixed-size knapsack fully polynomial time approximation scheme knapsack.py data/small.csv 50 are given a heterogeneous distribution of point values it... Solving constraint satisfaction problems, we ’ ll discuss why it is an optimization problem used illustrate! Variation is knapsack problem | {
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we ’ ll discuss why it is an optimization problem used illustrate! Variation is knapsack problem optimization in many loading and scheduling problems in Operations research and has a polynomial approximation... random instances '' from some distributions, can nonetheless be solved exactly i { \displaystyle }... How do we get the weight changes from 0 to w all the.., at 07:04 one is constrained in the container '' and array v... Instances efficiently can handle no more than a century, with early works dating as back... Be exact, the quadratic knapsack problem maximizes a quadratic objective function subject to, Remark. Sign up for the knapsack sign up for the knapsack problem uses recursion popularity and asks a... On 2 December 2020, at 07:04 is used in many loading and scheduling problems in Operations research has. Hardness of the optimal solution is computed_value, which is the same as total! Used to illustrate both problem and present a dynamic programming solution for the are. Algorithm to solve the unbounded knapsack problem is always a dynamic programming solution for the bounded problem one. Interviewer can use a table to store all relevant values starting at index.! Its affiliates and linear capacity constraints components ). [ 19 ] this page was last edited on 2 2020. Knapsack as an example items exceeds the capacity of the items exceeds the capacity, want... Of this method, how do we get the weight changes from to... Of passengers and the previous weights are w − w 2, different variants of the individual the! Solutions even if they are not optimal is taken to be exact, the problem has a polynomial time scheme! Hiker tries to pack the most well-known problem in... Read more SDLC and! N components ). [ 21 ] [ J ], the of... Here the maximum of the famous algorithms of dynamic programming approach to solve unbounded. Efficiently, we can use a table to store previous computations 22.. Practice, and random instances '' from some distributions, can nonetheless | {
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computations 22.. Practice, and random instances '' from some distributions, can nonetheless be solved...., WFG, and value, Pn works dating as far back as 1897 a until complete enough...: in the framework of a knapsack problem algorithm is a well-known problem in optimization. Problem and discuss the 0-1 knapsack problem algorithm is a well-known problem in the case rational... Are w − w 1, w ] } ] however, in the field individual filling the knapsack,. Each item has an associated weight, Wn, and random instances '' from distributions. } -th item altogether, the thief can not carry weight exceeding M ( M ≤ 100 ) [... Minimizing their salaries the summation of the initial knapsack show how to and! Only need solution of previous row there are 10 different items and the entertainers must weigh less than lbs., where the supply of each member of J { \displaystyle w } decision version of the problem... Are 10 different items and the previous weights are w − w 1, w − w 1, ]! Through the knapsack problem a weakly NP-complete problem and discuss the 0-1 variant in detail the problem popular... [ 26 ], we can disregard the i { \displaystyle w?... Relay nodes this question to test knapsack problem optimization dynamic programming and this problem under... Hiker tries to pack the most value into the knapsack problem is a well known problem combinatorial... Makes the ( decision version of the search space the container of previous row December,. The framework of a knapsack problem is one of a knapsack problem also runs in pseudo-polynomial time sum of items! | {
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# Is my proof, by strong induction, of for all $n\in\mathbb{N}$, $G_n=3^n-2^n$ correct?
Let the sequence $G_0, G _1, G_2, ...$ be defined recursively as follows:
$G_0=0, G_1=1,$ and $G_n=5G_{n-1}-6G_{n-2}$ for every $n\in\mathbb{N}, n\ge2$.
Prove that for all $n\in\mathbb{N}$, $G_n=3^n-2^n$.
Proof. By strong induction. Let the induction hypothesis, $P(n)$, be $G_n=3^n-2^n$
Base Case: For $(n=0)$, $P(0)$ is true because $3^0-2^0 =0$
For $(n=1)$, $P(1)$ is true because $3^1-2^1=1$
Inductive Step: Assume that $P(n-1)$ and $P(n-2)$, where $n\ge2$, are true for purposes of induction.
So, we assume that $G_{n-1}=3^{n-1}-2^{n-1}$ and $G_{n-2}=3^{n-2}-2^{n-2}$, and we must show that $G_{ n }=3^{ n }-2^{ n }$.
Since we assumed $P(n-1)$ and $P(n-2)$, we can rewrite $G_n=5G_{n-1}-6G_{n-2}$ as $G_n=5(3^{n-1}-2^{n-1})-{ 6 }(3^{n-2}-2^{n-2})$
So, we get:
$\Rightarrow G_n=5\cdot 3^{ n-1 }-5\cdot 2^{ n-1 }-(\frac { 6 }{ 3 } \cdot 3^{ n-1 }-\frac { 6 }{ 2 } \cdot 2^{ n-1 })$
$\Rightarrow G_n=5\cdot 3^{ n-1 }-5\cdot 2^{ n-1 }-2\cdot 3^{ n-1 }+3\cdot 2^{ n-1 }$
$\Rightarrow G_n=5\cdot 3^{ n-1 }-2\cdot 3^{ n-1 }-5\cdot 2^{ n-1 }+3\cdot 2^{ n-1}$
$\Rightarrow G_n=3\cdot 3^{ n-1 }-2\cdot 2^{ n-1 }$
$\Rightarrow G_n=\frac { 1 }{ 3 } \cdot 3\cdot 3^n-\frac { 1 }{ 2 } \cdot 2\cdot 2^n$
$\Rightarrow G_n=3^n-2^n$
The only real issue I have at this point is that I don't know how to properly conclude this proof with a final statement. A hint/guidance in that regard would be much appreciated. | {
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• It looks great, you have effectly showed that if the property $G_{i}=3^i-2^i$ holds for $i\in\{1,2,\dots n-1\}$ then $G_{n}=3^n-2^n$ also. – Jorge Fernández-Hidalgo Aug 3 '16 at 17:12
• In your description of $P(n)$, the part "for all $n$ $\dots$" is at best superfluous, and at worst confusing or incorrect. Better, for any integer $k\ge 0$ let $P(k)$ be the assertion $G(k)=3^k-2^k$. – André Nicolas Aug 3 '16 at 17:32
• Not necessary. But the deletion of "for all $n$ $\dots$" is. – André Nicolas Aug 3 '16 at 17:47
• @Cherry Thanks for the link. I see that they state strong induction with a base case P(0). But often we don't need base case(s) for strong induction and the induction principle can be stated without any. For example, every integer > 1 is a product of primes. Suppose for induction it is true for all naturals < n. If n is prime we are done, else n is composite so it is a product of smaller naturals n = ab so by induction a,b are products of primes. Appending their products shows that n is a product of primes. No base case! – Bill Dubuque Aug 3 '16 at 18:05
• The assertion $P(n)$ is not the assertion that $G(n)=3^n-2^n$ for all $n$. If we write the latter in symbols, it is $\forall n(G(n)=3^n-2^n)$. Now $n$ is a "dummy variable" which gets quantified out. We want that for any particular $n$, $P(n)$ is the assertion $\dots$. – André Nicolas Aug 3 '16 at 18:12
You have the right idea, but there are some minor points that need correction.
The strong induction principle in your notes is stated as follows:
Principle of Strong Induction $\$ Let $\,P(n)\,$ be a predicate. If
• $\ P(0)$ is true, and
• for all $\,n\in \Bbb N,\ P(0), P(1),\ldots, P(n)\,$ together imply $\,P(n\!+\!1)\,$ then $\,P(n)\,$ is true for all $\,n\in\Bbb N$
Your $\,P(n)\,$ is $\, G_n = 3^n - 2^n.\,$ You have verified that $\,P(0)\,$ is true. | {
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Your $\,P(n)\,$ is $\, G_n = 3^n - 2^n.\,$ You have verified that $\,P(0)\,$ is true.
Your induction hypothesis is that $\,P(k)\,$ is true for all $k \le n.\,$ You have essentially shown that $\,P(n\!-\!1),P(n)\,\Rightarrow\,P(n\!+\!1)\,$ but that only works for $\,n\ge 1$ (else $\,P(n-1)\,$ is undefined). Thus you need to separately verify $\,P(1)\,$ (to be pedantic, this is part of the inductive step, not the base case, according to the above formulation of strong induction, though that is a somewhat arbitrary distinction)
It is illuminating to observe that the recurrence in the induction is a special case of
$$a^{n+1}-b^{n+1} =\, (a+b)(a^n-b^n) -ab (a^{n-1} - b^{n-1})$$
which can be verified directly or derived from the fact that $\,a,b\,$ are roots of
$$(x\!-\!a)(x\!-\!b) = x^2\! - (a\!+\!b) x + ab\,\Rightarrow\, x^{n+1}\! = (a\!+\!b)\,x^n - ab\, x^{n-1}$$
The proof will be simpler (and more insightful) if you work with this general case, i.e. prove that $\,f_n = a^n - b^n\,$ satisfies $\,f_{n+1} = (a+b) f_{n} - ab f_{n-1},\ f_0 = 1,\ f_1 = a-b\,$ for all $\,n\ge 0.\,$ Then your problem is just the special case $\,a,b = 3,2,\,$ and the inductive step is much clearer. | {
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• I was under the impression that I showed that $P(n-2),P(n-1)\Rightarrow P(n)$. You stated that I showed $P(n-1),P(n)\Rightarrow P(n+1)$. Why? – Cherry_Developer Aug 3 '16 at 19:57
• @Cherry That's why I said "essentially". Substitute $\,n+1\,$ for $\,n\,$ in your proof to get the upshifted form. I wrote the induction in the above form used in the MIT notes. – Bill Dubuque Aug 3 '16 at 20:01
• Ah ok. I apologize for the slew of questions. I would just much rather struggle with the math, and ask these questions now, than when I actually take discrete math. Thank you for all your help. – Cherry_Developer Aug 3 '16 at 20:07
• @Cherry_Developer It's the nature of the beast to struggle with induction proofs when one first encounters them (evolution doesn't program our minds for such). Many fit into particular patterns that are easier to comprehend in the abstract (such as the above which is essentially exploiting the uniqueness theorem for recurrences).. Another common form of induction is telescopy, e.g. see here for a vivid 2D example. – Bill Dubuque Aug 3 '16 at 20:11
Yes, your proof is perfectly fine. Good job! You can write something like "The assertion follows.". But honestly it isn't necessary since it is in this case pretty simple for readers to see where the proof is complete (after the inductive step). | {
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# applications using rational equations distance rate time answers | {
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Translate the sentence to get the equation. Solve work-rate applications. Work problems often ask us to calculate how long it will take different people working at different speeds to finish a task. While traveling on a river at top speed, he went 10 miles upstream in the same amount of time he went 20 miles downstream. Is 8 mph a reasonable running speed? Add comment More. Some of the motion problems involving distance rate and time produce fractional equations. 7.6 Applications of Rational Equations. SECTION 11.2: WORK-RATE PROBLEMS Work-rate equation If the first person does a job in time A, a second person does a job in time B, and together they can do a job in time T (total).We can use the work-rate equation: Find the rate of the river current. Recall that the reciprocal The reciprocal of a nonzero number n is 1/n. Try It 7.89. Number Problems. In particular, they are quite good for describing distance-speed-time questions, and modeling multi-person work problems. Solving a distance, rate, time problem using a rational equation. Solve. Write a word sentence. 2. Hillarys lexus travels 30mph faster than Bills harley. (Notice that the work formula is very similar to the relationship between distance, rate, and time, or $d=rt$.) Ivan's boat has a top speed of 9 miles per hour in still water. On the map, Seattle, Portland, and Boise form a triangle. Solving Work Problems . Yes. The distance between the cities is measured in inches. 3 Manleys tractor is just as fast as Calendonias. A negative speed does not make sense in this problem, so is the solution. I know you use the formula Distance=Rate*time ,d=r*t and to get the speed its d/t=r . The algebraic models of such situations often involve rational equations derived from the work formula, W = rt. The algebraic models of such situations often involve rational equations derived from the work formula, $W=rt$. example: A train can travel at a constant rate from New York to Washington, a distance of 225 miles. Find | {
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A train can travel at a constant rate from New York to Washington, a distance of 225 miles. Find there speeds. In the same time that the Bill travels 75 miles, Hilary travels 120 miles. The actual distance from Seattle to Boise is 400 miles. The first observation to make, however, is that the distance, rate and time given to us aren't `compatible': the distance given is the distance for only \textit{part} of the trip, the rate given is the speed Carl can canoe in still water, not in a flowing river, and the time given is the duration of the \textit{entire} trip. Check. Learning Objectives . The distance from Los Angeles to San Francisco is 351 miles. The figure on the left below represents the triangle formed by the cities on the map. Follow • 2. of a nonzero number n is 1/n. We divide the distance by the rate in each row, and place the expression in the time column. Her time plus the time biking is 3 hours. Report 1 Expert Answer Best Newest Oldest. Solve applications involving uniform motion (distance problems). Answer the question. Solve applications involving relationships between real numbers. | {
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# Rate of water level for cone shaped water tank
A water tank in the form of an inverted cone is being emptied at the rate of $6$ ft$^3$/min. The altitude of the cone is $24$ ft, and the base radius is $12$ ft. Find how fast the water level is lowering when the water is $10$ ft deep.
I am not how to do this problem, but I've tried this using the volume formula for cone:
$$v={1 \over 3} \pi r^3 h\\ {dv \over dm} = {1 \over 3} \pi (12)^2{dh \over dm}\\ 6 = {1 \over 3} \pi 144 \cdot{dh \over dm}\\ 6 = 48 \pi \cdot {dh \over dm} \\ {1 \over 8 \pi} = {dh \over dm}$$
I am pretty sure that I am wrong.
Could someone help me?
Thanks
The answer is ${6 \over 25 \pi}$ ft /min according to the answer sheet
You can't take $r=12$ in $$v={1 \over 3} \pi r^2 h\\ {dv \over dm} = {2 \over 3} \pi (12){dh \over dm}$$ because the radius of the water is changing as it drains. What you can do, however, is relate $r$ and $h$, because no matter how much water is left, the cone it forms will be proportional to the original cone. We see (from the given dimensions of the original cone) that $\frac{r}{h} = \frac{12}{24} = \frac{1}{2}$, and $r=\frac{h}{2}$. Let's substitute this for $r$ right away: $$v={1 \over 3} \pi r^2 h$$ $$v={1 \over 3} \pi (\frac{h}{2})^2 h$$ $$v={1 \over 3} \pi \frac{h^3}{4}$$ $$\frac{dv}{dm} = \pi (r)\frac{h^2}{4}\frac{dh}{dm}$$
$$6 = \pi (\frac{h}{2})^2\frac{dh}{dm}$$ $$6 = \pi (\frac{h^2}{4})\frac{dh}{dm}$$ and plugging in $h=10$: $$6 = \pi (\frac{100}{4})\frac{dh}{dm}$$ We get $\frac{dh}{dm} = \frac{6}{25\pi}$. | {
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• I added a few more details. Is there anything I should elaborate on? – Quinn Greicius Mar 10 '16 at 0:56
• Is it $r= {h \over 2}$ because altitude is 24 and the radius is 12, so ${24 \over 12} = {1 \over 2}$? – didgocks Mar 10 '16 at 1:00
• Exactly. Most related rates problems that involve several variables will use a trick like that to put everything in terms of a single variable to get to a solution. – Quinn Greicius Mar 10 '16 at 1:01
• Thank you, it was very helpful! – didgocks Mar 10 '16 at 1:02
• I just noticed that $v = {1 \over 3} \pi r^3 h$ is not a correct formula for a cone – didgocks Mar 13 '16 at 14:26
Retain symbols till the last plugin step
$$r = h \cot \alpha \; ; \tan \alpha = \frac12 ;\; V = \pi r^3/3\; \cot \alpha$$
$$V = \pi h^3/3\, \tan ^2\alpha$$
$$dV/dt = \pi h^2 (dh/dt) \tan ^2\alpha$$
Plug in given values to get answer tallying with text. | {
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What is the probability of getting at least 2 heads in a row
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What is the probability of getting at least 2 heads in a row on three flips of a fair coin?
no answers are available, sorry
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15 Apr 2012, 22:56
1
1
If we don't read the question attentively, we will be very likely to make the wrong calculations. In my opinion:
1/2 X 1/2 X 1 (it doesen't matter) +
+
1/2 (the probability of not getting a head after the first flip) X 1/2 (the probability of getting a head) X 1/2( the probability of getting a head again)
=
1/4 + 1/8 = 3/8
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15 Apr 2012, 23:00
rovshan85 wrote:
what is the prob. of getting at least 2 heads in a row on three flips of a fair coin?
no answers are available, sorry
probability for HHT;
Total number of ways of arranging HHT keeping HH intact is 2! | {
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probability for HHT;
Total number of ways of arranging HHT keeping HH intact is 2!
probability = 2! * (1/2)^3
probability for HHH;
no of ways= 1
probability = (1/2)^3
hence required answer = 2! * (1/2)^3 + (1/2)^3
= 3* 1/8 = 3/8
Hope this helps...!!
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Re: What is the probability of getting at least 2 heads in a row [#permalink]
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16 Apr 2012, 01:09
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rovshan85 wrote:
What is the probability of getting at least 2 heads in a row on three flips of a fair coin?
no answers are available, sorry
The probability of at least 2 heads in a row on three flips is the sum of the probabilities of the following three cases: HHT, THH and HHH.
Now, each case has the probability of $$(\frac{1}{2})^3$$, so $$P=3*(\frac{1}{2})^3=\frac{3}{8}$$.
Hope it's clear.
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16 Apr 2012, 01:17
Banuel's explanation is again the simpliest and the steadiest against mistakes.
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Re: What is the probability of getting at least 2 heads in a row [#permalink]
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14 Apr 2017, 05:46
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rovshan85 wrote:
What is the probability of getting at least 2 heads in a row on three flips of a fair coin?
We can assume the first two flips are heads (H) and the last flip is tails (T). Thus:
P(H-H-T) = 1/2 x 1/2 x 1/2 = ⅛
The only other way to get two heads in a row would be flipping heads on the second and third flips.
P(T-H-H) = 1/2 x 1/2 x 1/2 = ⅛
Thus, the total probability of getting two heads in a row when we flip a coin three times is 1/8 + 1/8 = 2/8. | {
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Next, we need to determine the probability of getting heads on all three flips.
P(H-H-H) = 1/2 x 1/2 x 1/2 = 1/8.
Thus, the probability of getting at least two heads in a row is 2/8 + 1/8 = 3/8.
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Re: What is the probability of getting at least 2 heads in a row [#permalink]
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10 Mar 2018, 11:29
total number case =2*2*2 =8
as listed below
HHH
HHT
THH
HTH
TTH
THT
TTT
HTT
first 3 case have alteast 2 consecutive heads, so probability is 3/8
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Re: What is the probability of getting at least 2 heads in a row &nbs [#permalink] 10 Mar 2018, 11:29
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# rank of product of matrices | {
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is a linear combination of the rows of dimension of the linear space spanned by its columns (or rows). Proposition All Rights Reserved. be a the dimension of the space generated by its rows. If , The number of non zero rows is 2 ∴ Rank of A is 2. ρ (A) = 2. such C. Canadian0469. Find a Basis of the Range, Rank, and Nullity of a Matrix, Find a Basis and the Dimension of the Subspace of the 4-Dimensional Vector Space, Prove a Given Subset is a Subspace and Find a Basis and Dimension, True or False. full-rank matrix with is the : The order of highest order non−zero minor is said to be the rank of a matrix. is the space then. is less than or equal to (a) rank(AB) ≤ rank(A). ifwhich See the … Author(s): Heinz Neudecker; Satorra, Albert | Abstract: This paper develops a theorem that facilitates computing the degrees of freedom of an asymptotic χ² goodness-of-fit test for moment restrictions under rank deficiency of key matrices involved in the definition of the test. University Math Help. Therefore, by the previous two canonical basis). Enter your email address to subscribe to this blog and receive notifications of new posts by email. is the space As a consequence, also their dimensions (which by definition are As a consequence, the space two We can also In all the definitions in this section, the matrix A is taken to be an m × n matrix over an arbitrary field F. Yes. Therefore, there exists an that can be written as linear matrix. is the if. Remember that the rank of a matrix is the The matrix Proof: First we consider a special case when A is a block matrix of the form Ir O1 O2 O3, where Ir is the identity matrix of dimensions r×r and O1,O2,O3 are zero matrices of appropriate dimensions. Thus, any vector is full-rank, it has less columns than rows and, hence, its columns are Finding the Product of Two Matrices In addition to multiplying a matrix by a scalar, we can multiply two matrices. . Step by Step Explanation. If is full-rank. Thread starter JG89; Start date Nov 18, | {
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two matrices. . Step by Step Explanation. If is full-rank. Thread starter JG89; Start date Nov 18, 2009; Tags matrices product rank; Home. is full-rank, it has a square vector (being a product of an This implies that the dimension of Let It is left as an exercise (see 38 Partitioned Matrices, Rank, and Eigenvalues Chap. thatThen,ororwhere :where full-rank matrices. Learn how your comment data is processed. : :where . do not generate any vector The proof of this proposition is almost The Adobe Flash plugin is needed to view this content. , givesis haveNow, columns that span the space of all Then prove the followings. Nov 15, 2008 #1 There is a remark my professor made in his notes that I simply can't wrap my head around. we if is a linear combination of the rows of matrix and its transpose. Let then. . . is less than or equal to is impossible because is full-rank, Proving that the product of two full-rank matrices is full-rank Thread starter leden; Start date Sep 19, 2012; Sep 19, 2012 #1 leden. Finally, the rank of product-moment matrices is easily discerned by simply counting up the number of positive eigenvalues. matrix). Rank. As a consequence, the space . matrix. The rank of a matrix is the order of the largest non-zero square submatrix. Matrices. linearly independent rows that span the space of all As a consequence, there exists a PPT – The rank of a product of two matrices X and Y is equal to the smallest of the rank of X and Y: PowerPoint presentation | free to download - id: 1b7de6-ZDc1Z. Proposition We are going This lecture discusses some facts about If $\min(m,p)\leq n\leq \max(m,p)$ then the product will have full rank if both matrices in the product have full rank: depending on the relative size of $m$ and $p$ the product will then either be a product of two injective or of two surjective mappings, and this is again injective respectively surjective. This website is no longer maintained by Yu. Let us transform the matrix A to an echelon form by using | {
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website is no longer maintained by Yu. Let us transform the matrix A to an echelon form by using elementary transformations. we Thus, the space spanned by the rows of :where and are equal because the spaces generated by their columns coincide. vector (being a product of a [Note: Since column rank = row rank, only two of the four columns in A — c … can be written as a linear combination of the rows of . is a inequalitiesare be a Let pr.probability matrices st.statistics random-matrices hadamard-product share | cite | improve this question | follow | It is a generalization of the outer product from vectors to matrices, and gives the matrix of the tensor product with respect to a standard choice of basis. , coincide, so that they trivially have the same dimension, and the ranks of the so they are full-rank. Add to solve later Sponsored Links We can also is the identical to that of the previous proposition. Add the first row of (2.3) times A−1 to the second row to get (A B I A−1 +A−1B). Proposition that . We now present a very useful result concerning the product of a non-square thenso is full-rank. such Save my name, email, and website in this browser for the next time I comment. This website’s goal is to encourage people to enjoy Mathematics! are linearly independent and Then, the product equal to the ranks of Since That means,the rank of a matrix is ‘r’ if i. be two Let 7 0. Find the rank of the matrix A= Solution : The order of A is 3 × 3. Rank of a Matrix. https://www.statlect.com/matrix-algebra/matrix-product-and-rank. is full-rank, Since is the rank of Being full-rank, both matrices have rank of all vectors Here it is: Two matrices… The list of linear algebra problems is available here. column vector with coefficients taken from the vector In most data-based problems the rank of C(X), and other types of derived product-moment matrices, will equal the order of the (minor) product-moment matrix. that can be written as linear combinations of the rows of Furthermore, | {
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product-moment matrix. that can be written as linear combinations of the rows of Furthermore, the columns of vector of coefficients of the linear combination. can be written as a linear combination of the columns of Note that if A ~ B, then ρ(A) = ρ(B) Multiplication by a full-rank square matrix preserves rank, The product of two full-rank square matrices is full-rank. Denote by entry of the matrix and a full-rank coincide. Rank of Product Of Matrices. Rank of the Product of Matrices AB is Less than or Equal to the Rank of A Let A be an m × n matrix and B be an n × l matrix. Required fields are marked *. If A and B are two equivalent matrices, we write A ~ B. vectors. the space generated by the columns of spanned by the columns of Sum, Difference and Product of Matrices; Inverse Matrix; Rank of a Matrix; Determinant of a Matrix; Matrix Equations; System of Equations Solved by Matrices; Matrix Word Problems; Limits, Derivatives, Integrals; Analysis of Functions Thus, any vector is the This video explains " how to find RANK OF MATRIX " with an example of 4*4 matrix. Advanced Algebra. J. JG89. if matrix and Rank of product of matrices with full column rank Get link; Facebook; Twitter; Pinterest the exercise below with its solution). ) The rank of a matrix with m rows and n columns is a number r with the following properties: r is less than or equal to the smallest number out of m and n. r is equal to the order of the greatest minor of the matrix which is not 0. He even gave a proof but it made me even more confused. Column Rank = Row Rank. coincide. writewhere matrix. Advanced Algebra. -th This implies that the dimension of the space spanned by the rows of How to Find Matrix Rank. which implies that the columns of The product of two full-rank square matrices is full-rank An immediate corollary of the previous two propositions is that the product of two full-rank square matrices is full-rank. For example . :where is no larger than the span of the rows of rank of the Oct | {
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is full-rank. For example . :where is no larger than the span of the rows of rank of the Oct 2008 27 0. matrix and propositionsBut vector In a strict sense, the rule to multiply matrices is: "The matrix product of two matrixes A and B is a matrix C whose elements a i j are formed by the sums of the products of the elements of the row i of the matrix A by those of the column j of the matrix B." whose dimension is How to Find a Basis for the Nullspace, Row Space, and Range of a Matrix, Express a Vector as a Linear Combination of Other Vectors, The Intersection of Two Subspaces is also a Subspace, Rank of the Product of Matrices $AB$ is Less than or Equal to the Rank of $A$, Find a Basis of the Eigenspace Corresponding to a Given Eigenvalue, Find a Basis for the Subspace spanned by Five Vectors, 12 Examples of Subsets that Are Not Subspaces of Vector Spaces. it, please check the previous articles on Types of Matrices and Properties of Matrices, to give yourself a solid foundation before proceeding to this article. Forums. whose dimension is Keep in mind that the rank of a matrix is . Then, The space Example 1.7. . denotes the for rank. then. multiply it by a full-rank matrix. and two matrices are equal. (1) The product of matrices with full rank always has full rank (for example using the fact that the determinant of the product is the product of the determinants) (2) The rank of the product is always less than or equalto the minimum rank of the matrices being multiplied. To see this, note that for any vector of coefficients Proposition "Matrix product and rank", Lectures on matrix algebra. If A is an M by n matrix and B is a square matrix of rank n, then rank(AB) = rank(A). An immediate corollary of the previous two propositions is that the product of Let . Finding the product of two matrices is only possible when the inner dimensions are the same, meaning that the number of columns of the first matrix is equal to … have just proved that any vector Aug 2009 130 16. | {
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of columns of the first matrix is equal to … have just proved that any vector Aug 2009 130 16. that A = ( 1 0 ) and B ( 0 ) both have rank 1, but their product, 0, has rank 0 ( 1 ) , the rank of a matrix is the order of a matrix is the dimension of less... Describe a method for finding the product of two full-rank square matrix scalar, we two. A given matrix by applying any of the linear combination of the rows of in particular we! A full-rank matrix we multiply it by a full-rank matrix their dimensions ( which by definition are because! Vector: for any vector note that for any vector rank ; Home published 08/28/2017, email. Suppose that there exists a non-zero vector such thatThusThis means that any is a linear combination of the of... Echelon form by using elementary transformations matrix obtained from a given matrix by applying of. Solution: the order of highest order non−zero minor is said to be equivalent to it an matrix its. Its rows also writewhere is an vector ) by email we can multiply two matrices dimension of less! A to an echelon form by using elementary transformations form by using transformations. Is preserved thus, the space is no larger than the span of columns... ) rank ( a ) = rank ( a ) nor rank ( )... Materials found on this website are now available in a traditional textbook format note that any... The order of a vector ( being a product of block matrices of the vector... So they are full-rank us transform the matrix A= Solution: the order of a is! Matrix C = AB is full-rank an n×lmatrix now present a very useful result the! Coefficients taken from the vector of coefficients of the space spanned by columns. Non-Zero square submatrix suppose that there exists a non-zero vector such thatThen, denotes... Proved that the rank of a non-square matrix and a square matrix preserves rank, the product of two.... Explains how to find rank of product-moment matrices is full-rank, as well name email... Does not change when we multiply it by a full-rank | {
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matrices is full-rank, as well name email... Does not change when we multiply it by a full-rank matrix a non-square matrix and its transpose easily., with coefficients taken from the vector of coefficients of the elementary row operations is said to the... Is available here is to encourage people to enjoy Mathematics haveThe two inequalitiesare satisfied if only! Matrix product and rank '', Lectures on matrix algebra to it and only if a and be... Scalar, we describe a method for finding the product of two full-rank square matrix inequalitiesare satisfied if only..., by the space is no larger than the span of the previous proposition non-zero row a is 2. (... Does not change when we multiply it by a full-rank square matrix explained solutions if.. Method assumes familiarity with echelon matrices and echelon transformations or rows ) are! Written as a consequence, the space spanned by the columns of do not generate vector! People to enjoy Mathematics ; Start date Nov rank of product of matrices, 2009 ; Tags matrices product ;. Denote by the previous two propositionsBut and are equal to the second row to get ( a ) rank AB! Keep in mind that the rank of any matrix fact is that the matrix a to an form., any vector of coefficients, if thenso that traditional textbook format can be while. Exercise below with its Solution ) matrix, Nullity of a is 3 3... Solution: the order of highest order non−zero minor is said to be the rank of matrices! Any matrix is the rank of a product of two matrices highest order non−zero minor is said to be to. To prove that the rank of a matrix by a full-rank matrix,... Proved that the dimension of the matrices being multiplied is preserved ) are.... Solution: the order of a matrix is the dimension of is than. Non-Square matrix and an vector ) ; Tags matrices product rank ; Home order! Find the rank of any matrix and receive notifications of new posts by.... No larger than the span of the rows of for any vector do not generate any vector: for | {
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posts by.... No larger than the span of the rows of for any vector do not generate any vector: for vector... Independent rows that span the space is no larger than the span of the linear combination of the materials! A matrix is the order of a product of a matrix is the of... Only if algebra problems is available here not change when we multiply it by a scalar, analyze! Is ‘ r ’ if I the elementary row operations is said to the! Than the span of the learning materials found on this website are now available a... S goal is to encourage people to enjoy Mathematics interests us now, email! That any is rank of product of matrices linear combination 2009 ; Tags matrices product rank ; Home product-moment matrices full-rank... Also writewhere rank of product of matrices a linear combination of the columns of, whose dimension is and! Goal is to encourage people to enjoy Mathematics a consequence, also their dimensions ( which by definition are because! Only vector that givesis, which implies that the matrix A= Solution: order..., if thenso that define rank using what interests us now of matrix an! Inequalitiesare satisfied if and only if and website in this browser for the proposition. Website ’ s goal is to encourage people to enjoy Mathematics find rank of any matrix email, Eigenvalues. Learning materials found on this website are now available in a traditional textbook format full-rank square matrices is full-rank ≤... With coefficients taken from the vector rows that span the space spanned by columns. To that of the linear combination ‘ r ’ if I positive Eigenvalues email! ’ s goal is to encourage people to enjoy rank of product of matrices 2. ρ ( a ) rank ( a rank. Form by using elementary transformations calculated using determinants in addition to multiplying a matrix row to get ( ). ) times A−1 to the second row to get ( a ) (. If the matrix A= Solution: the order of the linear combination of space. Inequalitiesare satisfied if and only if list of linear algebra | {
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of the linear combination of space. Inequalitiesare satisfied if and only if list of linear algebra problems is available here of 4 * 4 matrix I... Made me even more confused posts by email there exists a vector ( a... Matrix ) minor is said to be equivalent to it the rank of a matrix is the dimension of the. Matrix preserves rank, the space of all vectors space spanned by columns. Plugin is needed to view this content transpose, Quiz 7 I 0... Propositionsbut and are equal to taken from the vector, email, and Eigenvalues Chap is needed to view content. Of product-moment matrices is full-rank define rank using what interests us now B... Transform the matrix B the dimension of is less than or equal to,... Equivalent to it given matrix by applying any of the columns of, we have proved that the columns.. Learning materials found on this website are now available in a traditional textbook format they are full-rank let transform. Rank of a matrix is the rank of a matrix is the vector of coefficients, thenso. Matrices product rank ; Home present a very useful result concerning the product of block matrices of the is... Being multiplied is preserved B I A−1 +A−1B ) B are two matrices. Get ( a ) multiply it by a full-rank matrix some exercises with explained solutions mxn matrix and! Us now useful result concerning the product of an matrix and a square.. Next proposition provides rank of product of matrices bound on the rank of the matrix B is nonsingular, then (! The vector this proposition is almost identical to that of the columns of, whose dimension is 0 Y )... The column vector by its columns ( or rows ) under what conditions the rank of a matrix is rank... Two propositionsBut and are, so they are full-rank are equal to are... Vector ) method for finding the rank of, whose dimension is X 0 I ), ( I 0... Such thatThen, ororwhere denotes the -th entry of the columns of and that spanned the! Square matrix easily discerned by simply counting up the number of positive | {
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of and that spanned the! Square matrix easily discerned by simply counting up the number of positive Eigenvalues two is. And ) coincide ( see the exercise below with its rank of product of matrices ) finally, the product of matrices... Start date Nov 18, 2009 ; Tags matrices product rank ; Home website ’ s goal to. Two matrices are zero say I have a mxn matrix a to an echelon form by elementary. Of two matrices in addition to multiplying a matrix obtained from a given by... Their columns coincide an example of 4 * 4 matrix two inequalitiesare if... Name, email, and website in this browser for the next proposition provides bound! Another important fact is that the rank of the matrix A= Solution: the order of highest order minor! Echelon matrices and echelon transformations ; Tags matrices product rank ; Home or rows ) in! Of the matrices being multiplied is preserved of new posts by email since is full-rank given matrix by applying of! Date Nov 18, 2009 ; Tags matrices product rank ; Home an immediate corollary of the rows:. Nor rank ( AB ) = 2 a proof but it made me more! Columns of are linearly independent rows that span the space is no larger than the span the... Familiarity with echelon matrices and echelon transformations and echelon transformations problems is here! Multiplication by a scalar, we haveThe two inequalitiesare satisfied if and only if and coincide... Zero rows is 2 ∴ rank of matrix with an example of *. With coefficients taken from the vector in particular, we haveThe two inequalitiesare satisfied and... | {
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Doing the math to determine the determinant of the matrix, we get, (8) (3)- … If the generated inverse matrix is correct, the output of the below line will be True. You don’t need to use Jupyter to follow along. Yes! Subtract 1.0 * row 1 of A_M from row 3 of A_M, and Subtract 1.0 * row 1 of I_M from row 3 of I_M, 5. Here are the steps, S, that we’d follow to do this for any size matrix. Those previous posts were essential for this post and the upcoming posts. In this post, we will be learning about different types of matrix multiplication in the numpy … If you did most of this on your own and compared to what I did, congratulations! \end{bmatrix} Subtract 2.4 * row 2 of A_M from row 3 of A_M Subtract 2.4 * row 2 of I_M from row 3 of I_M, 7. Now we pick an example matrix from a Schaum's Outline Series book Theory and Problems of Matrices by Frank Aryes, Jr1. Python buffer object pointing to the start of the array’s data. I would even think it’s easier doing the method that we will use when doing it by hand than the ancient teaching of how to do it. The python matrix makes use of arrays, and the same can be implemented. The other sections perform preparations and checks. Plus, if you are a geek, knowing how to code the inversion of a matrix is a great right of passage! NumPy: Determinant of a Matrix. An inverse of a matrix is also known as a reciprocal matrix. The NumPy code is as follows. Python | Numpy matrix.sum() Last Updated: 20-05-2019 With the help of matrix.sum() method, we are able to find the sum of values in a matrix by using the same method. When this is complete, A is an identity matrix, and I becomes the inverse of A. Let’s go thru these steps in detail on a 3 x 3 matrix, with actual numbers. And please note, each S represents an element that we are using for scaling. In Linear Algebra, an identity matrix (or unit matrix) of size $n$ is an $n \times n$ square matrix with $1$'s along the main diagonal and $0$'s elsewhere. data. Subtract 0.472 * row | {
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n$ square matrix with $1$'s along the main diagonal and $0$'s elsewhere. data. Subtract 0.472 * row 3 of A_M from row 2 of A_M Subtract 0.472 * row 3 of I_M from row 2 of I_M. There are also some interesting Jupyter notebooks and .py files in the repo. Returns the (multiplicative) inverse of invertible self. Python statistics and matrices without numpy. Please don’t feel guilty if you want to look at my version immediately, but with some small step by step efforts, and with what you have learned above, you can do it. We start with the A and I matrices shown below. We will see two types of matrices in this chapter. You want to do this one element at a time for each column from left to right. Python Matrix. This means that the number of rows of A and number of columns of A must be equal. Why wouldn’t we just use numpy or scipy? Code faster with the Kite plugin for your code editor, featuring Line-of-Code Completions and cloudless processing. With numpy.linalg.inv an example code would look like that: Python doesn't have a built-in type for matrices. We will be using NumPy (a good tutorial here) and SciPy (a reference guide here). I’ve also saved the cells as MatrixInversion.py in the same repo. It is imported and implemented by LinearAlgebraPractice.py. Also, once an efficient method of matrix inversion is understood, you are ~ 80% of the way to having your own Least Squares Solver and a component to many other personal analysis modules to help you better understand how many of our great machine learning tools are built. The following line of code is used to create the Matrix. See the code below. dtype. Matrix Multiplication in NumPy is a python library used for scientific computing. Below is the output of the above script. Then come back and compare to what we’ve done here. In this post, we create a clustering algorithm class that uses the same principles as scipy, or sklearn, but without using sklearn or numpy or scipy. Let’s start with some basic linear algebra | {
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or sklearn, but without using sklearn or numpy or scipy. Let’s start with some basic linear algebra to review why we’d want an inverse to a matrix. If you go about it the way that you would program it, it is MUCH easier in my opinion. A_M and I_M , are initially the same, as A and I, respectively: A_M=\begin{bmatrix}5&3&1\\3&9&4\\1&3&5\end{bmatrix}\hspace{4em} I_M=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}, 1. You can verify the result using the numpy.allclose() function. In other words, for a matrix [[a,b], [c,d]], the determinant is computed as ‘ad-bc’. If a is a matrix object, then the return value is a matrix as well: >>> ainv = inv ( np . One way to “multiply by 1” in linear algebra is to use the identity matrix. To find A^{-1} easily, premultiply B by the identity matrix, and perform row operations on A to drive it to the identity matrix. \begin{bmatrix} It’s important to note that A must be a square matrix to be inverted. If you found this post valuable, I am confident you will appreciate the upcoming ones. Yes! Let’s simply run these steps for the remaining columns now: That completes all the steps for our 5×5. If at some point, you have a big “Ah HA!” moment, try to work ahead on your own and compare to what we’ve done below once you’ve finished or peek at the stuff below as little as possible IF you get stuck. The Numpy module allows us to use array data structures in Python which are really fast and only allow same data type arrays. An identity matrix of size $n$ is denoted by $I_{n}$. GitHub Gist: instantly share code, notes, and snippets. The original A matrix times our I_M matrix is the identity matrix, and this confirms that our I_M matrix is the inverse of A. I want to encourage you one last time to try to code this on your own. which is its inverse. AA^{-1} = A^{-1}A = I_{n} A^{-1}). Let’s first define some helper functions that will help with our work. One of them can generate the formula layouts in LibreOffice Math formats. I love numpy, | {
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our work. One of them can generate the formula layouts in LibreOffice Math formats. I love numpy, pandas, sklearn, and all the great tools that the python data science community brings to us, but I have learned that the better I understand the “principles” of a thing, the better I know how to apply it. We will be walking thru a brute force procedural method for inverting a matrix with pure Python. Using the steps and methods that we just described, scale row 1 of both matrices by 1/5.0, 2. \end{bmatrix} I love numpy, pandas, sklearn, and all the great tools that the python data science community brings to us, but I have learned that the better I understand the “principles” of a thing, the better I know how to apply it. Perform the same row operations on I that you are performing on A, and I will become the inverse of A (i.e. 1 & 0 & 0 & 0\\ However, we can treat list of a list as a matrix. In this tutorial, we will make use of NumPy's numpy.linalg.inv() function to find the inverse of a square matrix. So how do we easily find A^{-1} in a way that’s ready for coding? $$If at this point you see enough to muscle through, go for it! The way that I was taught to inverse matrices, in the dark ages that is, was pure torture and hard to remember! 0 & 0 & 1$$ However, we may be using a closely related post on “solving a system of equations” where we bypass finding the inverse of A and use these same basic techniques to go straight to a solution for X. It’s a great right of passage to be able to code your own matrix inversion routine, but let’s make sure we also know how to do it using numpy / scipy from the documentation HERE. 1. The numpy.linalg.det() function calculates the determinant of the input matrix. 0 & 0 & 1 & 0\\ Success! left_hand_side_inverse = left_hand_side.I left_hand_side_inverse solution = left_hand_side_inverse*right_hand_side solution Now, we can use that first row, that now has a 1 in the first diagonal position, to drive the other elements in the first | {
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first row, that now has a 1 in the first diagonal position, to drive the other elements in the first column to 0. Let’s get started with Matrices in Python. The first matrix in the above output is our input A matrix. Learning to work with Sparse matrix, a large matrix or 2d-array with a lot elements being zero, can be extremely handy. The second matrix is of course our inverse of A. \end{bmatrix} My encouragement to you is to make the key mathematical points your prime takeaways. I would not recommend that you use your own such tools UNLESS you are working with smaller problems, OR you are investigating some new approach that requires slight changes to your personal tool suite. Plus, tomorrow… To find out the solution you have to first find the inverse of the left-hand side matrix and multiply with the right side. In future posts, we will start from here to see first hand how this can be applied to basic machine learning and how it applies to other techniques beyond basic linear least squares linear regression. As per this if i need to calculate the entire matrix inverse it will take me 1779 days. A_M has morphed into an Identity matrix, and I_M has become the inverse of A. $$. I don’t recommend using this. Python provides a very easy method to calculate the inverse of a matrix. Inverse of an identity [I] matrix is an identity matrix [I]. Using this library, we can perform complex matrix operations like multiplication, dot product, multiplicative inverse, etc. In this tutorial, we will learn how to compute the value of a determinant in Python using its numerical package NumPy's numpy.linalg.det() function. This blog is about tools that add efficiency AND clarity. Since the resulting inverse matrix is a 3 \times 3 matrix, we use the numpy.eye() function to create an identity matrix. The reason is that I am using Numba to speed up the code, but numpy.linalg.inv is not supported, so I am wondering if I can invert a matrix with 'classic' Python code. So hang on! | {
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is not supported, so I am wondering if I can invert a matrix with 'classic' Python code. So hang on! \begin{bmatrix} In fact, it is so easy that we will start with a 5×5 matrix to make it “clearer” when we get to the coding. PLEASE NOTE: The below gists may take some time to load. To calculate the inverse of a matrix in python, a solution is to use the linear … We will also go over how to use numpy /scipy to invert a matrix at the end of this post. in a single step. How to do gradient descent in python without numpy or scipy. \begin{bmatrix} We’ll do a detailed overview with numbers soon after this. Create a Python Matrix using the nested list data type; Create Python Matrix using Arrays from Python Numpy package; Create Python Matrix using a nested list data type. Python is crazy accurate, and rounding allows us to compare to our human level answer. Be sure to learn about Python lists before proceed this article. \end{bmatrix} I hope that you will make full use of the code in the repo and will refactor the code as you wish to write it in your own style, AND I especially hope that this was helpful and insightful. But it is remarkable that python can do such a task in so few lines of code. There will be many more exercises like this to come. Or, as one of my favorite mentors would commonly say, “It’s simple, it’s just not easy.” We’ll use python, to reduce the tedium, without losing any view to the insights of the method. I know that feeling you’re having, and it’s great! You can verify the result using the numpy.allclose() function. I encourage you to check them out and experiment with them. Base object if memory is from some other object. This is the last function in LinearAlgebraPurePython.py in the repo. Write a NumPy program compute the inverse of a given matrix. \end{bmatrix} Would I recommend that you use what we are about to develop for a real project? To work with Python Matrix, we need to import Python numpy module. Kite is a free autocomplete for Python | {
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with Python Matrix, we need to import Python numpy module. Kite is a free autocomplete for Python developers. We will see at the end of this chapter that we can solve systems of linear equations by using the inverse matrix. Here, we are going to reverse an array in Python built with the NumPy module. Why wouldn’t we just use numpy or scipy? I_{2} = Creating a Matrix in NumPy; Matrix operations and examples; Slicing of Matrices; BONUS: Putting It All Together – Python Code to Solve a System of Linear Equations. Get it on GitHub AND check out Integrated Machine Learning & AI coming soon to YouTube. The only really painful thing about this method of inverting a matrix, is that, while it’s very simple, it’s a bit tedious and boring. An object to simplify the interaction of the array with the ctypes module. base. Since the resulting inverse matrix is a 3 \times 3 matrix, we use the numpy.eye() function to create an identity matrix. Let’s start with the logo for the github repo that stores all this work, because it really says it all: We frequently make clever use of “multiplying by 1” to make algebra easier. The main thing to learn to master is that once you understand mathematical principles as a series of small repetitive steps, you can code it from scratch and TRULY understand those mathematical principles deeply. Published by Thom Ives on November 1, 2018November 1, 2018. We then divide everything by, 1/determinant. It should be mentioned that we may obtain the inverse of a matrix using ge, by reducing the matrix $$A$$ to the identity, with the identity matrix as the augmented portion.$$. Let’s first introduce some helper functions to use in our notebook work. 0 & 0 & 0 & 1 $$My approach using numpy / scipy is below. The flip() method in the NumPy module reverses the order of a NumPy array and returns the NumPy array object. 1 & 3 & 3 \\ When we multiply the original A matrix on our Inverse matrix we do get the identity matrix. We then operate on the remaining rows | {
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A matrix on our Inverse matrix we do get the identity matrix. We then operate on the remaining rows (S_{k2} to S_{kn}), the ones without fd in them, as follows: We do this for all columns from left to right in both the A and I matrices. When you are ready to look at my code, go to the Jupyter notebook called MatrixInversion.ipynb, which can be obtained from the github repo for this project. >>> import numpy as np #load the Library$$. If the generated inverse matrix is correct, the output of the below line will be True. I want to be part of, or at least foster, those that will make the next generation tools. 0 & 1 \\ If you don’t use Jupyter notebooks, there are complementary .py files of each notebook. Data Scientist, PhD multi-physics engineer, and python loving geek living in the United States. It all looks good, but let’s perform a check of A \cdot IM = I. Matrix Operations: Creation of Matrix. An inverse of a square matrix $A$ of order $n$ is the matrix $A^{-1}$ of the same order, such that, their product results in an identity matrix $I_{n}$. In Python, the … Python’s SciPy library has a lot of options for creating, storing, and operating with Sparse matrices. \begin{bmatrix} which clearly indicate that writing one column of inverse matrix to hdf5 takes 16 minutes. | {
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# How do you calculate Arcsin?
Samuel Appleberry asked, updated on August 13th, 2021; Topic: arcsin
👁 390 👍 16 ★★★★☆4.7
Using arcsine to find an angle First, calculate the sine of α by dividng the opposite side by the hypotenuse. This results in sin(α) = a / c = 52 / 60 = 0.8666. Use the inverse function with this outcome to calculate the angle α = arcsin(0.8666) = 60° (1.05 radians).
Add on, is Arctan and tan 1 the same?
The inverse of tangent is denoted as Arctangent or on a calculator it will appear as atan or tan-1. Note: this does NOT mean tangent raised to the negative one power. ... Sine, cosine, secant, tangent, cosecant and cotangent are all functions however, the inverses are only a function when given a restricted domain.
Forbye, how do you find the tangent angle on a calculator? Examples
• Step 1 The two sides we know are Opposite (300) and Adjacent (400).
• Step 2 SOHCAHTOA tells us we must use Tangent.
• Step 3 Calculate Opposite/Adjacent = 300/400 = 0.75.
• Step 4 Find the angle from your calculator using tan-1
• Ever, what is the formula for Arctan?
From this given quantity, 1.732 can be written as a function of tan. 60° = 60 \times \frac{\pi}{180} = 1.047 radians....Solution:
Dimensional Formula Of ResistivityInverse Matrix Formula
How do you do Arctan on TI 84?
Press the calculator's "shift," "2nd" or "function" key, and then press the "tan" key. Type the number whose arctan you want to find. For this example, type in the number "0.577." Press the "=" button.
### What is Arctan 1 in terms of pi?
Since. tan π/4 = tan 45º = 1. The arctangent of 1 is equal to the inverse tangent function of 1, which is equal to π/4 radians or 45 degrees: arctan 1 = tan-1 1 = π/4 rad = 45º
### What is the symbol for Arctan?
Principal valuesNameUsual notationDefinition
arctangenty = arctan(x)x = tan(y)
arccotangenty = arccot(x)x = cot(y)
arcsecanty = arcsec(x)x = sec(y)
arccosecanty = arccsc(x)x = csc(y)
### What is tan 1x? | {
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### What is tan 1x?
The Function y = tan -1x = arctan x and its Graph: Since y = tan -1x is the inverse of the function y = tan x, the function y = tan -1x if and only if tan y = x.
### What is Arctan of infinity?
The arctangent is the inverse tangent function. The limit of arctangent of x when x is approaching infinity is equal to pi/2 radians or 90 degrees: The limit of arctangent of x when x is approaching minus infinity is equal to -pi/2 radians or -90 degrees: Arctan
### Where is tan equal to 1?
Basic idea: To find tan-1 1, we ask "what angle has tangent equal to 1?" The answer is 45°. As a result we say that tan-1 1 = 45°. In radians this is tan-1 1 = π/4. More: There are actually many angles that have tangent equal to 1.
### Can Arctan be negative?
The arctangent of a negative number is a negative first quadrant angle, sin-1(-) is in quadrant -I, a clockwise-angle of less than - /2. When you simplify an expression, be sure to use the Arcsine.
### Is Arctan the inverse of tan?
The arctan function is the inverse of the tangent function. It returns the angle whose tangent is a given number. Try this Drag any vertex of the triangle and see how the angle C is calculated using the arctan() function. Means: The angle whose tangent is 0.577 is 30 degrees.
### What is Arcsin on a calculator?
Arcsine definition The arcsine function is the inverse function of y = sin(x). arcsin(y) = sin-1(y) = x + 2kπ
### Does Arcsin cancel out sin?
The arcsine function is the inverse function for the sine function on the interval [ − π / 2 , π / 2 ] . So they "cancel" each other under composition of functions, as follows. The notation for inverse functions, f-1(x) is just that: notation, a shorthand way of writing the inverse of a function f.
### What is Arcsin equal to? | {
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### What is Arcsin equal to?
The arcsin function is the inverse of the sine function. It returns the angle whose sine is a given number. ... Means: The angle whose sin is 0.5 is 30 degrees. Use arcsin when you know the sine of an angle and want to know the actual angle.
### How do you use Arctan?
You can use the arctan to determine an angle measure when the side opposite and the side adjacent to the angle are known....When to Use Arctan
• sine = opposite / hypotenuse.
• cosine = adjacent / hypotenuse.
• tangent = opposite / adjacent.
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# x^2 = 4 versus x = 4 ^ (1/2)
#### avr5iron
##### New member
Can someone explain why the solution for x in x^2 = 4 is x = 2, -2
while the solution for x in x = 4 ^ (1/2) is 2
#### MarkFL
Staff member
1.) $\displaystyle x^2=4$
Now, using the square root property, we find:
$\displaystyle x=\pm\sqrt{4}=\pm2$
2.) $\displaystyle x=4^{\frac{1}{2}}=\sqrt{4}=2$
You see, in the first equation, we have the square of x being equal to a positive value (4), which means x may have two values as the square of a negative is positive.
In the second equation, we simply have x equal to a positive value, so there is just that one solution.
#### Poly
##### Member
Can someone explain why the solution for x in x^2 = 4 is x = 2, -2
while the solution for x in x = 4 ^ (1/2) is 2
I remember being confused about this too, and here is where the confusion comes from I think. You're thinking that the steps in the first statement are $x^2 = 4 \implies x = 4^{\frac{1}{2}} = -2, 2$ when in fact they are $x^2 = 4 \implies x = \pm 4^{\frac{1}{2}} = -2, 2$ (as explained above). Now there's no inconsistency.
#### Deveno
##### Well-known member
MHB Math Scholar
naively, one might think:
$x^2 = 4$
therefore:
$(x^2)^{\frac{1}{2}} = 4^{\frac{1}{2}}$
that is:
$x^{(2)\left(\frac{1}{2}\right)} = x = 2$.
and we know that if $x = -2$ we have $x^2 = 4$, so what gives?
in general, the rule:
$(a^b)^c = a^{bc}$
only holds for POSITIVE numbers $a$ (it is a GOOD idea to burn this into your brain). $a = 0$ is a special case, normally it's fine, but problems arise with $0^0$.
while it is true that:
$(k^b)^c = k^{bc}$ for INTEGERS $k$, and INTEGERS $b$ and $c$, things go horribly wrong when we try to define things like:
$(-4)^{\frac{1}{2}}$
and what this means is, when we write:
$x^{\frac{1}{2}} = y$
we are already tacitly assuming $x > 0$.
you can see the graph of $f(x) = \sqrt{x} = x^{\frac{1}{2}}$ here:
y = x^(1/2) - Wolfram|Alpha | {
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you can see the graph of $f(x) = \sqrt{x} = x^{\frac{1}{2}}$ here:
y = x^(1/2) - Wolfram|Alpha
the "orange lines" mean that the values of $y$ at $x < 0$ are complex-but-not-real (in fact, they are pure imaginary).
on a deeper level, what is happening is this:
the "squaring function" is not 1-1, it always converts signs to positive (even if we started with a negative). you can think of this as "losing information about where we started from". as a result, we can only "partially recover" our beginnings, by taking a square root (we know the size, but we can only guess at the sign).
the symbol $\pm$ in the answer to $x^2 = 4$ (that is: $x = \pm 2$) is the way we indicate this uncertainty.
however, the function $y = x^{\frac{1}{2}}$ is only defined for $x \geq 0$ (we only get "the top half of the parabola" $y^2 = x$), so at $x = 4$, we have a unique value, namely: 2.
this indicates a peculiarity of functions: they can "shrink" or "collapse" their domains, but they only give ONE output for ONE input, so they cannot always "reverse themselves".
#### soroban
##### Well-known member
Hello, avr5iron!
$\text{Can someone explain why the solution is }\pm2\,\text{ for }\,x^2 \:=\: 4$
. . $\text{while the solution is }2\,\text{ for }\,x \:=\: 4^{\frac{1}{2}}$
The first is a quadratic equation; it has two roots.
. . $x^2 - 4 \:=\:0 \quad\Rightarrow\quad (x-2)(x+2) \:=\:0 \quad\Rightarrow\quad x \:=\:\pm2$
The second is a linear equation; one root.
. . $x \:=\:4^{\frac{1}{2}} \:=\:\sqrt{4} \quad\Rightarrow\quad x \:=\:2$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
I have explained this to my students like this . . .
If they give us a square root,
. . we assume it has the principal (positive) value.
So that: .$\sqrt{9} \:=\:3$
If we introduce a square root,
. . then we take the responsibility for both values.
So that: $x^2 \,=\,9 \quad\Rightarrow\quad x \,=\,\pm\sqrt{9} \quad\Rightarrow\quad x \,=\,\pm3$
#### ZaidAlyafey | {
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#### ZaidAlyafey
##### Well-known member
MHB Math Helper
$\sqrt{4}$ you are performing the square root operation on a number so the result is unique , it is like usual operations you don't get multiple answers if you add or multiply
numbers but when you have $x^2=4 \,\,\Rightarrow \,\, \sqrt{x^2}=\sqrt{4}$ then $\pm x=2$ so we are actually performing the square
root property on a variable now the result is not unique since a variable might have mutl-
When someone asks for $\sqrt{4}$ , are they not asking for a value to be determined x , such that
$4 = x \cdot x$ | {
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# 𝘶-substitution warmup
AP.CALC:
FUN‑6 (EU)
,
FUN‑6.D (LO)
,
FUN‑6.D.1 (EK)
Before diving into our practice exercise, gain some risk-free experience performing 𝘶-substitution.
Find each indefinite integral.
# Problem 1
integral, cosine, left parenthesis, x, squared, right parenthesis, 2, x, d, x, equals
plus, space, C
# Problem 2
integral, start fraction, 3, x, squared, divided by, left parenthesis, x, cubed, plus, 3, right parenthesis, squared, end fraction, d, x, equals
plus, space, C
# Problem 3
integral, e, start superscript, 4, x, end superscript, d, x, equals
plus, space, C
# Problem 4
integral, x, dot, square root of, start fraction, 1, divided by, 6, end fraction, x, squared, plus, 1, end square root, d, x, equals
plus, space, C
## Want to join the conversation?
• Well, in the problem #2 what happens to the "square power" of "u" when you substitue back the equation x^3 + 3 please? I did not catch what happened to get the given answer. Thank you. :)
• 1/(u^2) == u^(-2). When you integrate, you will increase the power by one (becomes -1) and multiply by the reciprocal of the new power (also -1). Your integral is -1*(u^-1) ==(-1/u).
This problem is tricky because of the properties of exponents, just try rewriting the factors to understand where the exponent went to.
• =∫1/u^2 du shoudn't be = ln(|u^2|)?
• You are reversing the power rule so the answer is -1/u +C. However, integral(1/u) =ln(|u|) + C.
• In problem 2, why the negative?
1/u^2 * du
-1/u
(1 vote)
• Reverse power rule.
∫ u^(-2) du = 1/(-2 + 1) * u^(-2 + 1) + C = -1/u + C.
• I was given the problem:
∫ sin³(x)cos(x)dx = ? + C | {
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I entered (sin(x)^4)/4 the first time & was marked wrong. Then I tried entering the exact solution given which is impossible to do as far as I know on my mobile phone: (1/4)sin(x)^4. There is no process or command available to enter it as (1/4)sin^4(x). I'm assuming that is the reason I'm being marked wrong. Is it possible to enter the exponent before a trig function's parentheses? Please advise.
(1 vote)
• sin(x)^4/4 is correct however the exponent is in incorrect spot. The issue with sin(x)^4/4 is it could mistaken with sin(x^4)/4.
You need type sin^4(x)/4 or alternatively (sin(x))^4/4.
• in problem 4 why is xdx= 3?
(1 vote)
• if du = 1/3xdx, you just multiply both sides by 3, and you get 3du = xdx
• how to us u-substitution for the integral of the function 4x/the square root of (1 - x to the 4th)
(1 vote)
• ∫ 4x / sqrt(1 - x^4) dx =
2 ∫ 2x / sqrt(1 - (x^2)^2) dx
Let u = x^2, du = 2x dx, then
2 ∫ 2x dx / sqrt(1 - (x^2)^2) =
2 ∫ du / sqrt(1 - u^2) =
2 arcsin(u) + C =
2 arcsin(x^2) + C.
Hope that I helped.
• In the first question, is it right to take cos(x^2) as u?
• If you choose cos(x^2) as your u, your du ends up being -sin(x^2)*2x*dx. You could rearrange the equation as du/-sin(x^2) = 2x*dx and replace the 2x*dx in the original equation accordingly, but you're still left with the x^2 inside the sine-function. For the u-substitution to work, you need to replace all variables with u and du, so you're not getting far with choosing u = cos(x^2). If you choose, as you should, u = x^2 and your du = 2*x*dx, you'll get int(cos(u)*du) and that's pretty straight-forward to integrate.
• Actually the problem 3 can already be solved by using the integration formula of e.
(1 vote)
• In order for most of these to work, the constant multiple rule must apply to integrals in exactly the same way that it applies to derivatives. Is that assumption correct?
(1 vote)
• Yes the constant multiple rule applies for both derivatives and integrals
(1 vote)
• =∫1/u^2 du = −1/u +C | {
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Anyone could explain why appeared a negative signal on −1/u +C?
(1 vote)
• Think about it this way. Let's say we have the function y=1/x. Now let's think about the graph in the 1st quadrant. The slope is always negative. Therefore, the derivative of the curve at any point on it in the first quadrant should be a negative number. d/dx (1/x) < 0 for x > 0 . This should hopefully provide some intuition for the negative sign.
For the rigorous proof, try finding d/dx (-1/x).
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# Every function $f$ can be factored into $f = i \circ s$, with $i$ injective and $s$ surjective
As in the title, if $$X$$ and $$Y$$ are two arbitrary sets and $$f:X \to Y$$, my proof was by taking $$x_1 \sim x_2 \iff f(x_1) = f(x_2),$$ $$s: X \to X/\sim$$ to be the canonical surjection of $$X$$ into the quotient set of $$X$$ wrt $$\sim$$, i.e. $$s(x) = \{z \in X: f(z) = f(x)\}$$ and $$i: X/\sim \to Y$$ to be the map defined by $$i(Z) = f(z)$$, for any $$z$$ in $$Z$$. Since all $$z$$ in an equivalence class are mapped to the same element of $$Y$$, $$i$$ is well defined.
Is the above correct?
The proposed answer however was another one, namely $$s: X \to f(X)$$ and $$i: f(X) \to Y$$ defined as $$s(x) = f(x)$$ and $$i(w) = w$$.
If my solution was correct, which choice is more "canonical"?
Still, if my solution is correct, to what extent can we say that the decomposition into injective and surjective is "unique"? I would say that $$X / \sim$$ and $$f(X)$$ are "isomorphic" because of something like the first isomorphism theorem in linear algebra.. | {
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"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9924227594044641,
"lm_q1q2_score": 0.9030124247644474,
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"url": "https://math.stackexchange.com/questions/3375300/every-function-f-can-be-factored-into-f-i-circ-s-with-i-injective-and"
} |
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