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# zbMATH — the first resource for mathematics

Differential geometry and its applications. 2nd ed. (English) Zbl 1153.53001
Washington, DC: Mathematical Association of America (MAA) (ISBN 978-0-88385-748-9/hbk). xxi, 467 p. (2007).
This textbook covers standard topics of differential geometry when presented for undergraduates interested in mathematics or physics, i.e., it deals with curves and surfaces in Euclidean spaces, as well as with initial facts on geodesics and curvature, manifolds and Riemannian geometry. Among less common topics we should mention constant mean curvature surfaces and calculus of variations, incuding Pontryagin maximum principle.
The theoretical treatment is endowed with numerous exercises and examples. The concepts and constructions presented are systematically visualized through the use of computer algebra system Maple. Related procedures are included in the text with necessary explanations. The book is carefully written, and may be recommended to students approaching differential geometry for the first time.
Contents:
1. The geometry of curves (Pages 1–66).
2. Surfaces (67–106).
3. Curvatures (107–159).
4. Constant mean curvature surfaces (161–207).
5. Ggeodesics, metrics and isometries (208–273).
6. Holonomy and the Gauss–Bonnet theorem (275–310).
7. The calculus of variations and geometry (311–395).
8. A glimps to higher dimensions (397–435).

##### MSC:
 53-01 Introductory exposition (textbooks, tutorial papers, etc.) pertaining to differential geometry 53A04 Curves in Euclidean and related spaces 53A05 Surfaces in Euclidean and related spaces 53A10 Minimal surfaces in differential geometry, surfaces with prescribed mean curvature 53A07 Higher-dimensional and -codimensional surfaces in Euclidean and related $$n$$-spaces 49-01 Introductory exposition (textbooks, tutorial papers, etc.) pertaining to calculus of variations and optimal control 49Q05 Minimal surfaces and optimization 53-04 Software, source code, etc. for problems pertaining to differential geometry
Maple


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# Having trouble w/ 2 problems on my probability problem set

For (1), start by making the substitution $y = ux$.

format: plain


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# nLab Gabriel filter

## Definition

A Gabriel filter $G$ is a uniform filter of left ideals in a ring $R$ which is idempotent under Gabriel composition of filters.

## Terminology

In our definition this notion is equivalent to topologizing filter; though for some authors the latter notion slightly differs. Stenstroem says Gabriel topology instead of Gabriel filter, because all Gabriel filters form a basis of nieghborhoods of $0$ for a topology on $R$.

## Properties

If $L$ and $L'$ are left ideals in a Gabriel filter $F$, then the set $L L'$ (of all products $l l'$ where $l\in L, l'\in L'$) is an element on $F$. Any uniform filter $F$ is contained in a minimal Gabriel filter $G$ (said to be generated by $F$), namely the intersection of all Gabriel filters containing $F$. Given a Gabriel filter $G$, the class of all $G$-torsion modules (see uniform filter) is a hereditary torsion class.

## Literature

Revised on December 14, 2010 19:59:50 by Zoran Škoda (161.53.130.104)


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### Gol_D's blog

By Gol_D, history, 5 months ago,

1619A - Square String?

Idea: MikeMirzayanov

Tutorial
Solution

1619B - Squares and Cubes

Idea: MikeMirzayanov

Tutorial
Solution

Idea: MikeMirzayanov

Tutorial
Solution

1619D - New Year's Problem

Tutorial
Solution

1619E - MEX and Increments

Idea: SixtyWithoutExam

Tutorial
Solution

1619F - Let's Play the Hat?

Idea: MikeMirzayanov

Tutorial
Solution

1619G - Unusual Minesweeper

Idea: Gol_D

Tutorial
Solution

1619H - Permutation and Queries

Idea: Brovko

Tutorial
Solution

• +73

 » 5 months ago, # |   +13 Really liked Problem D. The observation about how pigeonhole theorem can be used in the check function of binary search was very good.
•  » » 5 months ago, # ^ |   +2 You guys used binary search ? Wow
•  » » » 5 months ago, # ^ |   0 How have you solved it? Would you like to share your approach?
•  » » » » 5 months ago, # ^ |   0 A = maximum of Top-2 by shops B = minimum of Top-1 by friends Result is min(A, B) https://codeforces.com/contest/1619/submission/140109705
•  » » » » » 5 months ago, # ^ | ← Rev. 3 →   0 podtelkin could you please explain why this idea worked? UPD: Got it. Nice idea
•  » » » » » » 5 months ago, # ^ |   +7 你可以在这里看到中文版本(Link)I think I can explain it.(Please forgive my poor English.)my code is hereAccording to the description of the problem, you can easily get the follow inequality: $ans \leq \text{minimum of Top-1 by friends}$I use a[i][j] to describe the j's contribution to friends numbered i.Then let's suppose $n-1 \leq m$.Now I give a strategy for choosing gifts: Choose a friend numbered I, and find his favorite gift J. Let's suppose a[I][J] is also the maximum among a[1][J],a[2][J],...,a[n][J].(That means J's contribution is also the biggest if it's for I.)Let me call this Hypothesis A. Then find the Top-2 among a[1][J],a[2][J],...,a[n][J], and pick the corresponding I'. Now both I and I' will be given the gift J. Joy Point of them is top1ByFriends[I] and top2ByShops[J]. For the rest n-2 friends, we can always pick their favorite gifts. For those friends, their Joy Point is top1ByFriends[i]. So if we try to choose everyone as the lucky I, we will have several $\alpha$ that satisfy the following condition: $max\{\alpha\}\leq \text{maximum of Top-2 by shops}$ After having this consequence, let's think the situation that Hypothesis A doesn't hold. It's quite easy to come to the conclusion that if top1ByFriends[I] < top2ByShops[J], the $\alpha$ will be determined by top1ByFriends[I] so any I' who has a[I'][J] bigger than a[I][J] is OK, because he won't influence the result in this situation. So now we can say: $ans \leq \text{min{minTop1ByFriends[i],maxTop2By Shops[j]}}$You can prove the result is exactly $\text{min{minTop1ByFriends[i],maxTop2By Shops[j]}}$ by category discussion.
•  » » » » » » » 5 months ago, # ^ |   +1 Supplement:If $n-1$ > $m$, everyone can get their favorite gifts.So ans is minTop1ByFriends[i].And you can find that at least two friends will have their favorite gifts in the same shop, so the maxTop2ByShops[j] won't less than minTop1ByFriends[i].
•  » » » » » » » 5 months ago, # ^ |   0 Thank you very much man
•  » » » » » » » 4 months ago, # ^ |   0 Thank you very much :)
•  » » » » » 5 months ago, # ^ |   0 I think one simple way of explaining this approach is:1) we have to pick at least two gifts from the same store (by the pidgeonhole principle).2) we can always choose the store that has the highest second max, and pick the max and second max gifts from this store and give them for the respective friends (lets call this decision opt*), and I argue this choice can always be made and still achieve an optimal solution: if our solution has 2 other items from any same store, then the second highest of these two will be smaller than the second highest of the opt* choice (by definition of highest second max), making it a worse choice. thus, we can always choose opt* and still get a optimal solution.3) now pick the best n-2 for the remaining n-2 friends.
•  » » » » 5 months ago, # ^ | ← Rev. 2 →   +4 What i did in the contest was we know that we can visit n-1 shops ,so the idea is that for every shop we can see the maximum and the 2nd maximum joy values which are attained for any two friends .let say its i and j ,so leaving these two friends apart we can easily take the maximum of joys from any shops for the rest of the friends .i won't suggest you to see my implementation it's really messy and the binary search logic is way more easy to implement and think .my implementation involves map of multiset of pair of int which was the best i could think in the contest
•  » » » » 5 months ago, # ^ | ← Rev. 3 →   +3 well, i have a solution using sort and some magicthis is my solution. I scared to fail the system test, but some how it passed.140057739here are my thoughts while doing this problem btw: if i sort the items by price and when I reach the $i$_th item, if i can delete all the item before $i$ and still albe to take some of the items after, i can just consider $i$ as an answer after that is finding how to check if the remaining item can sastisfy the condition at most n — 1 shop: i'm not so sure about this, but if one item was deleted from all shop you should break the loop, and another case is when the remaining items are the same as the number of remaining shop -> this means you can take two item from any shop so you also need to break loop
•  » » 5 months ago, # ^ |   +2 I also used the pigeonhole principle to solve it but the binary search part is still unclear to me... Like how I can prove that if I can get x units of joy I would be able to get x-1 units of joy? Please care to explain...Thankyou
•  » » » 5 months ago, # ^ |   +3 The binary search function $f(x)$ means I can get $x$ or more units of joy (not necessarily exactly $x$ units)If $f(x)$ is true, then $f(i)$ for $i < x$ is definitely trueIf $f(x)$ is false, then $f(i)$ for $i > x$ is definitely false.So you find the maximum $x$ for which $f(x)$ is true by binary search.
•  » » » » 5 months ago, # ^ |   0 Ohh, thanks a lot... Misunderstood binary search function.
•  » » » 5 months ago, # ^ |   0 Whenever i see maximize the minimum of something my brain just automatically think of binary search. In this problem i struggled with verifying if a particular mid can be achieved.
•  » » 5 months ago, # ^ | ← Rev. 4 →   -7 Solution to problem D without binary search :- Suppose we can visit at most n shops instead of n — 1, since we have n friends, we can select the best shop for each friend and doing so we will be selecting at most n shops. Now, for n — 1 shops, since we have n friends, there must be a shop from which we buy gifts for at least 2 friends. Hence we select the shop with the highest 2nd max. Now we need to select at most n — 2 shops for n — 2 friends. We can simply take the best shop for each remaining friend. Edit: I don't have any proof, but it works
 » 5 months ago, # | ← Rev. 2 →   0 When upsolving G with an unordered_map I got TLE for problem G. When using a map it passes. Took me a long time to realize this was the problem.Gol_D Did you on purpose make a test case that blows up an unordered_map? I doubt this could've happened accidentally. Especially since my code runs quickly on other max cases, it's only Case 14 that is slow.
•  » » 5 months ago, # ^ |   +17 Yes, they do keep such a test case on purpose.
•  » » » 5 months ago, # ^ |   0 I believe it is because someone got hacked using unordered_map during the open hacking phase, then the test case used was included in the main tests
 » 5 months ago, # | ← Rev. 3 →   0 Implementation of C really messed up my contest it was failing on test case 3. Did anyone else face this issue.My Solution Can someone please have a look and point out what I missed.
•  » » 5 months ago, # ^ |   0 It is not an implementation issue, there is a logical error in this part: if (b_last_digit >= 10 or b_last_digit < 0) { b = -1; break; } Please check 140048744 for what should have been in place of the above piece of code.
 » 5 months ago, # |   0 can anyone please explain G in bit more detail ?
•  » » 5 months ago, # ^ |   +17 This is actually a graph problem, let's see it in graph terms.Let each bomb be a node in our graph, there is an edge between two nodes if they share a coordinate and have distance $\leq K$. Now look at its connected components. Clearly if we trigger a bomb, then the entire connected component containing that bomb will explode as well. So a connected component can be triggered by either: we trigger any bomb by ourselves, or we let the earliest bomb explode naturally. Now we have two tasks to solve: Find all the connected components (and the minimum bomb timer in each component), and choose which components to trigger manually (and which will explode naturally).To find connected components: Suppose all bombs are on a straight line, then it's easy. Sort the bombs, add an edge between all pairs of consecutive bombs with distance $\leq K$. For the general problem, use a map> mapx, mapy; or something like that to store bombs on the same x-straight line or y-straight line and solve it as if all bombs are on the straight line. You now have the graph.To choose which components to trigger: Try solving the problem for $K = 0$ (i.e. no bombs trigger one another). Obviously we want to sort the bomb timers, and trigger the highest timers first, until the rest explodes naturally.
•  » » » 5 months ago, # ^ | ← Rev. 2 →   +5 Ahh, sorry I meant H, there are so many problems in contest nowadays, nice explanation, thanks :) I understand sqrt decomp. is applied. I got the idea of query of type 2, but not of type1, why we need inverse indices and that part.
•  » » » » 5 months ago, # ^ |   +3 Think about the permutation as a functional graph where each vertex points to another vertex. In order to answer type 2 queries efficiently, we need a "jump" array which gives the index we will be at if we step forward $Q$ times. When we perform a type 1 query on indices $x$ and $y$, most values do not need to be recomputed--the only indices are $x$ and the $Q-1$ ones that jumped over $x$, so $r_x$, $r_{r_x}$, etc., and similarly for $y$.To recalculate these, start with $x$ and step forward $Q$ times to find $jump_x$. Then step backwards using the reverse permutation to calculate $jump_{r_x}$, $jump_{r_{r_x}}$, etc.
•  » » » » » 5 months ago, # ^ |   0 Thanks !!
•  » » » 5 months ago, # ^ | ← Rev. 2 →   0 140270271 Can you take a look at this submission of G, please? I would appreciate a lot.
 » 5 months ago, # |   0 In problem C's code it should be 18 instead of 19 if(x < y && y >= 10 && y <= 19) b.emplace_back(y — x);
•  » » 5 months ago, # ^ |   0 2 one digit numbers can never sum upto 19. It can be at worse 9+9.
•  » » 5 months ago, # ^ |   0 No in the worst case both a and b contain 9,9. so 9 + 9 = 18
 » 5 months ago, # | ← Rev. 3 →   0 I saw some solutions for problem H that use link cut tree, could anyone help me to understand the idea behind these solutions?
•  » » 4 months ago, # ^ |   0 My understanding is: A permutation can be decomposed into several cycles. If you view each element (1,2,...,n) as a vertex and each (i -> p[i]) as a directed edge, you end up with a directed graph of some simple cycles. A type 1 query "1 x y" is equivalent to cut cycle(s) at the edge (x -> p[x]) and the edge (y -> p[y]), and join the broken cycle(s) with new edges (x -> p[y]) and (y -> p[x]). A type 2 query "2 i k" is equivalent to, (say vertex i is on cycle[i]) find the (k mod length(cycle[i]))-st neighbor of i on cycle[i]. To simulate the above process, all you need is a data structure that represents cycles and supports quick split/merge (for type 1 queries) and find k-th element (for type 2 queries). One obvious choice is the splay tree. Given link-cut tree is mostly based on splay tree, I guess that would work as well.
 » 5 months ago, # |   -7 couldn't get AC on E because of integer overflow. makes me wonder if there are any downsides to always using long long instead of int, I couldn't see any time difference with $2.10^9$ operations
•  » » 5 months ago, # ^ |   0 Maybe it will need more memory
•  » » » 5 months ago, # ^ |   0 yes, it will require double the memory. however, most questions have 256 MB of memory and I don't think I've ever used half of it.Here are the times for $2.10^9$ operationsint : 1107 ms, 1154 mslong long : 1154 ms, 1123 ms
 » 5 months ago, # |   0 In problem E editorial code why sort(a.begin(), a.end()); ? is it a mistake , or something i'm missing ?
•  » » 5 months ago, # ^ |   0 the sorting is not required. you don't even need to store all elements in a[], all you need is the frequency array (cnt[]).
 » 5 months ago, # |   0 Can someone explain this part from Problem G ? Calculating Min From Componentsint ans = res.size() - 1; for (int i = 0; i < res.size(); ++i) { ans = min(ans, max((int)res.size() - i - 2, res[i])); } cout << ans << '\n';
•  » » 5 months ago, # ^ |   0 I guess you forgot to sort res vector, I also did the same mistake
•  » » 6 weeks ago, # ^ |   0 If you are considering bombs till component i to be exploded by timeout, then remaining bombs need to be detonated manually. The count of remaining bombs is (res.size()-i-1). Since you can start detonating from time 0, total time taken to manually detonate remaining bombs is ((res.size()-i-1)-1) i.e. (res.size()-i-2). For a particular i, you would be taking max of time taken to manually detonate and time taken to explode by timeout. Because we can be sure that in that time, all bombs are gone. The minimum such time where all bombs are gone would be the answer.
 » 5 months ago, # | ← Rev. 7 →   +26 Solution for problem D without the log factor:For every person, find $B_i$. $B_i$ means the best shop to choose while buying a gift for person $i$. From now on, I'll rephrase "buying a gift for a person $i$ from shop $s$" as "assigning person $i$ to shop $s$.Case 1: Number of different "optimal" shops (i.e. number of distinct integers $B_i$ for $1 \leq i \leq n$) does not exceed $n - 1$. We can assign person $i$ to $B_i$ and calculate the answer directly. Case 2: Otherwise. Then there has to be at least one person $i$ such that it's not assigned to $B_i$. Let's call such a person "unhappy". Then I claim that there is no more than $2$ unhappy people in an optimal assignment. Proof is left as an exercise for the reader. What we can conclude from here is that in an optimal assignment, we'll choose exactly $2$ people and assign both of them to the same shop. Note that none or one of them can be assigned to their best shop. And every person $i$ in remaining $n - 2$ people are assigned to $B_i$. Call the shop with $2$ people assigned as $s$, and friends to be assigned to $s$ as $x$ and $y$. Without loss of generality, assume that $p_{sx} \leq p_{sy}$. What is the answer for a fixed $x$, $y$, $s$? Turns out it's the minimum of $min_{1 \leq i \leq n, i \neq x, i \neq y}(p_{B_ii}), p_{sx}$ and $p_{sy}$.Note that we can assume as if $y$ is also assigned to $B_y$ because we know that $p_{sy}$ can't minimize the answer while $p_{sx}$ is there. Therefore we don't have to iterate over $y$'s and only take the minimum of $min_{1 \leq i \leq n, i \neq x}(p_{B_ii})$ and $p_{sx}$. So let's iterate over different $x$ and $s$. But we have to find a way to ensure that there is at least one $y$ such that $p_{sx} \leq p_{sy}$ holds for a fixed $x$, $s$. Here is a way to check it: We can store for every shop the value of its most valuable gift and its count. Call it $V_s$ and $C_s$ for shop $s$ respectively. If $p_{sx} < V_s$ or $p_{sx} = V_s$ and $C_s \geq 2$ then it is a valid $(x, s)$ pair.Complexity is $\mathcal{O}(nm)$:In case 1, checking is done in $\mathcal{O}(nm)$. In case 2, at every iteration of $x$, you first find $min_{1 \leq i \leq n, i \neq x}(p_{B_ii})$ in $\mathcal{O}(n)$ and iterate shop $s$ in $\mathcal{O}(m)$. Total runtime is $\mathcal{O}(n(n + m))$, which is bounded by $\mathcal{O}(nm)$ because $n \leq m$ must hold at case 2.See example submission: 140114559. Note that I have swapped the dimensions of $p$ in the problem for implementation simplicity. I also have a vector $d$ to count the number of distinct shops for case 1.
•  » » 5 months ago, # ^ |   +4 Brilliant!
•  » » 5 months ago, # ^ |   0 Can you help me rectify error in this comment, i have described a similar idea but i can't make sense of downvotes
 » 5 months ago, # |   0 I can't figure out what's wrong in my solution? It's failing on test 2. Can anyone please help? my code.
•  » » 5 months ago, # ^ |   0 You should consider the situation that (num2
•  » » » 5 months ago, # ^ |   0 Thank you very much dude, I got my mistake.
 » 5 months ago, # |   0 At first, I think the permutation in problem H can be considered with a graph with n nodes and n edges (A Base loop tree)Finally, I realize that it is just a graph with serval loops.What a pity!
 » 5 months ago, # | ← Rev. 2 →   0 In problem B, replacing sqrt(n) with sqrt(n+ 0.5L) give correct answer. Why is this happening? Thanks in advance.Hacked solution : https://codeforces.com/contest/1619/submission/140041728Accepted solution : https://codeforces.com/contest/1619/submission/140180623
•  » » 5 months ago, # ^ | ← Rev. 2 →   0 Because of representation of real numbers in binary. For example, pow(10^9, 1/3) = 999.9999... and not 1000 as it should be. By adding 0.5 you're avoiding this problem.
 » 5 months ago, # |   0 problem F is actually a joke
•  » » 5 months ago, # ^ |   0 Can you please explain the joke part in it?
•  » » » 5 months ago, # ^ |   0 it's easier than D and E. look at this submission
•  » » » » 5 months ago, # ^ |   0 Only after you realize it/read the editorial :)D/E are like just do whatever said, F needs a lil bit of thinking
•  » » » » » 5 months ago, # ^ |   0 No buddy, he is right, problem F is comparable to C, just a little bit of implementation and knowledge of deque(I used it for putting last to front), D involved more than 5x thinking than F... I am more convinced on this fact also due to the line mentioned in the announcement also that problems may not be in their proper order... According to me, the order could be A B C F E D.
•  » » » » » » 5 months ago, # ^ |   0 Whenever you see a problem like "maximize the minimum" most of the time it turns out to be a binary search problem. That's the reason I said D didn't require much thinking for me.
 » 5 months ago, # |   0 Why need to sort the input array of problem E? As we use the frequency array later, any need to sort the input array?
•  » » 5 months ago, # ^ |   0 you are completely right, filling and sorting of vector a are not required
 » 5 months ago, # |   0 Please help me debugging my code for problem G. It is getting WA on test case 2 line 6. 140296294I think I am committing some mistake in calculating ans at the very end of the code.
 » 5 months ago, # |   +2 I solve B using inclusion and exclusion principle. here's my submission:140032787I was worried about the case where I use both sqrt and cbrt but it passed all testcases!After the contest I found that most of the contestants solved this problem using set or any other data structure. However, I find this solution much easier.
•  » » 5 months ago, # ^ |   0 Brilliant idea!
•  » » 5 months ago, # ^ | ← Rev. 3 →   -18 for someone who can't understand this idea... count numbers of the form x^2 such that x^2 <= n (say A2) count numbers of the form x^3 such that x^3 <= n (say A3) numbers of the form x^6 are counted twice so we subtract them once.ans = A2 + A3 — A6
 » 5 months ago, # |   0 Problem C : I'm getting Memory limit exceeded on test 2. I can't rectify my code. Plz help. My Submission
 » 5 months ago, # |   0 No need to sort the array in problem E.
 » 5 months ago, # |   0 Problem E: If anybody is getting TLE then try multiset.upper_bound(num) rather than upper_bound(multiset.begin(), multiset.end(), num). Reason
 » 5 months ago, # |   0 What is the meaning of res.size()-i-2 in the G tutorial?
• »
»
5 months ago, # ^ |
0

-if we want to find the remaining number of elements after index 'i' , to do so we need to make the difference between the size and the index we are at.

-Indexes are from 0 to res.size()-1 , so the actual difference would be res.size() — (i+1) = res.size()-i-1. (Or you can think it as res.size()- 1 — i too)

-So we found out that the numbers of remaining elements after index 'i' is res.size()-i-1

-Assume we have 'x' remaining elements . Since the seconds begin at 0 , the minimum number of seconds we need to erase the remaining 'x' elements is 'x-1'.

-So we found out that the minimum number of seconds we need to erase the remaining elements after current index 'i' is "the number of remaining elements"-1 = (res.size()-i-1)-1 = res.size()-i-2.

-------------------------------------------------------------------------------------------------

## Below is explanation of the approach if needed

Assume you are at index 'i' in the sorted array 'a' in which you stored the Minimum time u have to wait until 1 bomb from the 'i'th component will explode , erasing the whole component with it.

Since the array is sorted , you wonder : What is the minimum time for all the bombs to explode , if you wait until components 1->i explode (without exploding anything by myself) , and you erase all the remaining components in front of the one from index 'i' by yourself.

Assume it takes 30 seconds for the 'i'th component to explode (all the i-1 components will explode too since the array is sorted) , and u have 40 more components after that . U need to wait 30 seconds , but u needed at least 40 seconds to erase all the remaining ones , so u take the maximum from these 2 values. (If we had 20 more components after this , we needed 20 seconds to erase those by ourselves , so waiting 30 seconds for 'i'th one to explode will be enough , so again we had the maximum from those 2).

•  » » » 5 months ago, # ^ |   0 thank you, I understand.
 » 5 months ago, # | ← Rev. 2 →   0 Did anyone solve prolem H with treap?
•  » » 5 months ago, # ^ |   0 Indeed, it's quite reminiscent of this problem from SecondThread's treap educational contest. Just need to figure out the right treap operations to do, and implement a few new ones.
•  » » » 5 months ago, # ^ |   0 Came up with treap solution as my first approach and became unable to think of the $O((N+Q)\sqrt{N})$ one :(
•  » » » 5 months ago, # ^ |   0 Can you guys leave submission link / code?
•  » » » » 5 months ago, # ^ | ← Rev. 3 →   0 140540828(apologies in advance for the rather rambly Rust code lol... Rust doesn't come with a RNG in its standard library, so there is also code for it)Edit: I also recommend this tutorial video by SecondThread to learn about implementing treaps
 » 5 months ago, # | ← Rev. 2 →   0 Can anyone prove or help in proving the greedy approach for problem D? Thank you.UPD: got it.
 » 5 months ago, # |   0 In the question named "Squares and Cubes".....just use unordered_set rather than ordered_set, it will reduce time complexity. :D
 » 5 months ago, # | ← Rev. 2 →   0 Could someone tell me what is wrong with my code? C Problem/*package whatever //do not write package name here */import java.io.*; import java.util.*;public class GFG { public static Scanner scn=new Scanner(System.in); public static void main (String[] args) { int t=scn.nextInt(); while(t-->0) { long a=scn.nextLong(); long s=scn.nextLong();  ArrayList num1=new ArrayList(); ArrayList sum=new ArrayList(); ArrayList ans=new ArrayList(); while(a>0) { num1.add(a%10); a/=10; } while(s>0) { sum.add(s%10); s/=10; } int i=0; int j=0; boolean flag=false; while(i=two) { ans.add(one-two); i++;j++; } else if(i+119) { flag=true; System.out.println(-1); break; } ans.add(one-two); i+=2; j++; } else{ flag=true; System.out.println(-1); break; } } if(!flag) { if(j=0;k--) { res=res*10+ans.get(k); } System.out.println(res); } } } } }
 » 5 months ago, # |   0 I think I have an O(n + qlogn) solution for problem H: A permutation is a composition of circles (i -> p[i] -> ... -> i). We can represent those circles with treaps, one for each circle, when the nodes are i, p[i], p[p[i]], ... from left to right. Note that we can use split and join to make any element the leftmost. first-type queries can be answered — When we swap p[i], p[j] and i,j are in different circles, we merge two circles with O(1) split & join operations to receive: (i) -> (p[j] -> ... -> j) -> (p[i] -> ... i). When we swap P[i], p[j] and i,j are in the same circle, we split two circles with O(1) split & join operations to receive: (p[j] -> ... i), (p[i] -> ... -> p[j]). Second-type queries can be answered easily — we will keep for each subtree it's size.*We will keep an arrays of pointers, Arr[i] is a pointer for the node that represents i.
 » 4 months ago, # |   0 last problem is crazy!!!
 » 4 months ago, # |   0 I implemented E with a std map instead145083862
 » 2 months ago, # |   0 Could anyone help me Problem E? Why my submission is Memory limit exceeded? submission


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# 90m/s enough for Eve encounter in KSP? (reading delta-v maps)

• I'm trying to understand https://i.imgur.com/yO0bQax.png , in particular I don't get how 90m/s delta-v should be enough to go from a transfer orbit to an Eve encounter:

In my attempt, I go to a transfer orbit by burning straight up against Kerbin's orbit until I get an escape:

This escape burn already gives us about half the required orbital transfer to Eve:

But it's not even close to the 90m/s that is allegedly sufficient for an encounter. For that, I need about 400m/s:

So what's going on?

• The starting point is "elliptical orbit around Kerbin with Pe of 10 km above atmosphere (80km ASL) and Ap at the edge of SOI", not solar orbit. If you try to do the burn while in solar orbit, it's much less efficient (because of Oberth effect).

In theory this means you should be in a (correctly aligned) elliptical orbit around Kerbin, and do the burn of 90m/s at the periapsis.

In practice such an orbit is a pain to align, and it's easier to read as 930+90=1020m/s burn from LKO to Eve intercept.

Experimentally, I see a burn of ~1030m/s from LKO to Eve intercept, which seems in line with the map. (Mechjeb was used to plan this one specifically)

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# How do I divide Laurent polynomials?

I have an example from a paper (listed below) that I cannot figure out. I can divide normal polynomials, but the alternative ways to divide Laurent polynomials is beyond me at the moment. The paper lists two polynomials, and three possible quotient/remainder pairs. The polynomials are,

$$a(z) = z^{-1} + 6 + z, \quad b(z) = 4 + 4z,$$

and the three sets of quotients and remainders are,

$$q(z) = \frac{1}{4}(z^{-1} + 5) , \quad r(z) = -4z,$$

$$q(z) = \frac{1}{4}(z^{-1} + 1), \quad r(z) = 4,$$

$$q(z) = \frac{1}{4}(5z^{-1} + 1), \quad r(z) = -4z^{-1}.$$

I understand how to obtain the last quotient and remainder, but I do not understand how the first two were obtained. A worked example of one or the other would be fantastic, thanks. This is from the Daubechies and Sweldens paper on factoring wavelet transforms.

-

The answer depends on the order in which you try to eliminate the terms. If you start with the $z^{-1}$ term, you can get rid of it by $\frac{1}{4}z^{-1}b(z)$ and you are left with
$$a(z)-\frac{1}{4}z^{-1}b(z) = 5+z$$
For this polynomial you can again chose which one you want to eliminate with a fitting multiple of $b(z)$.
Eliminating the $5$ $$a(z)-\frac{1}{4}z^{-1}b(z)-\frac{5}{4}b(z) = -4z = r(z)$$
eliminating the $z$ $$a(z)-\frac{1}{4}z^{-1}b(z)-\frac{1}{4}b(z) = 4 = r(z)$$


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# Mathematician Solves 30-Year Old Comp Sci Conjecture in Two Pages

Page 1 of 1

## 2 Replies - 1164 Views - Last Post: 26 July 2019 - 12:42 PM

### #1 Skydiver

• Code herder

Reputation: 7132
• Posts: 24,225
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# Mathematician Solves 30-Year Old Comp Sci Conjecture in Two Pages

Posted 25 July 2019 - 09:30 PM

Mathematician Solves Computer Science Conjecture in Two Pages

Quote

The “sensitivity” conjecture stumped many top computer scientists, yet the new proof is so simple that one researcher summed it up in a single tweet.

Quote

I expect that this fall it will be taught — in a single lecture — in every master’s-level combinatorics course.

Quote

“This conjecture has stood as one of the most frustrating and embarrassing open problems in all of combinatorics and theoretical computer science,” wrote Scott Aaronson of the University of Texas, Austin, in a blog post. “The list of people who tried to solve it and failed is like a who’s who of discrete math and theoretical computer science,” he added in an email.

Is This A Good Question/Topic? 2

## Replies To: Mathematician Solves 30-Year Old Comp Sci Conjecture in Two Pages

### #2 macosxnerd101

• Games, Graphs, and Auctions

Reputation: 12657
• Posts: 45,831
• Joined: 27-December 08

## Re: Mathematician Solves 30-Year Old Comp Sci Conjecture in Two Pages

Posted 25 July 2019 - 10:00 PM

The proof of the sensitivity conjecture comes down to the Eigenvalue Interlace Theorem from linear algebra. The application is to the adjacency matrix of the hypercube. I wouldn't be surprised to see Huang's proof as an exercise in Spectral Graph Theory courses, moving forward. The proof is actually that simple.

https://arxiv.org/pdf/1907.00847.pdf

### #3 DarenR

• D.I.C Lover

Reputation: 634
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## Re: Mathematician Solves 30-Year Old Comp Sci Conjecture in Two Pages

Posted 26 July 2019 - 12:42 PM

i stopped at Qn -- to many letters not enough names for me -- ha


text:
## Going back to Germany!

A couple of blog posts ago, I wrote that the decision to go to ESC Boston ended up being a great one for many different reasons.  I came back from the conference energized and really happy that I went.

These feelings were amplified a few days after my return when I received an email from Rolf Segger, the founder of SEGGER Microcontroller (check out their very new website), asking if I would be interested in visiting their headquarters...

## Exact Near Instantaneous Frequency Formulas Best at Peaks (Part 2)

Introduction

This is an article that is a continuation of a digression from trying to give a better understanding of the Discrete Fourier Transform (DFT). It is recommended that my previous article "Exact Near Instantaneous Frequency Formulas Best at Peaks (Part 1)"[1] be read first as many sections of this article are directly dependent upon it.

A second family of formulas for calculating the frequency of a single pure tone in a short interval in the time domain is presented. It...

## Modeling a Continuous-Time System with Matlab

June 6, 2017

Many of us are familiar with modeling a continuous-time system in the frequency domain using its transfer function H(s) or H(jω).  However, finding the time response can be challenging, and traditionally involves finding the inverse Laplace transform of H(s).  An alternative way to get both time and frequency responses is to transform H(s) to a discrete-time system H(z) using the impulse-invariant transform [1,2].  This method provides an exact match to the continuous-time...

## ESC Boston's Videos are Now Up

June 5, 2017

In my last blog, I told you about my experience at ESC Boston and the few videos that I was planning to produce and publish.  Here they are, please have a look and any feedback (positive or negative) is appreciated.

Short Highlight

This is a very short (one minute) montage of some of the footage that I shot at the show & conference.  In future shows, I absolutely need to insert clips here and there of engineers saying a few words about the conference (why they...

## How to Find a Fast Floating-Point atan2 Approximation

Context Over a short period of time, I came across nearly identical approximations of the two parameter arctangent function, atan2, developed by different companies, in different countries, and even in different decades. Fascinated with how the coefficients used in these approximations were derived, I set out to find them. This atan2 implementation is based around a rational approximation of arctangent on the domain -1 to 1:

## Understanding and Relating Eb/No, SNR, and other Power Efficiency Metrics

Introduction

Evaluating the performance of communication systems, and wireless systems in particular, usually involves quantifying some performance metric as a function of Signal-to-Noise-Ratio (SNR) or some similar measurement. Many systems require performance evaluation in multipath channels, some in Doppler conditions and other impairments related to mobility. Some have interference metrics to measure against, but nearly all include noise power as an impairment. Not all systems are...

## Time Machine, Anyone?

Abstract: Dispersive linear systems with negative group delay have caused much confusion in the past. Some claim that they violate causality, others that they are the cause of superluminal tunneling. Can we really receive messages before they are sent? This article aims at pouring oil in the fire and causing yet more confusion :-).

Introduction

## Frequency Dependence in Free Space Propagation

Introduction

It seems to be fairly common knowledge, even among practicing professionals, that the efficiency of propagation of wireless signals is frequency dependent. Generally it is believed that lower frequencies are desirable since pathloss effects will be less than they would be at higher frequencies. As evidence of this, the Friis Transmission Equation[i] is often cited, the general form of which is usually written as:

Pr = Pt Gt Gr ( λ / 4πd )2 (1)

where the...

## Understanding and Preventing Overflow (I Had Too Much to Add Last Night)

December 4, 2013

Happy Thanksgiving! Maybe the memory of eating too much turkey is fresh in your mind. If so, this would be a good time to talk about overflow.

In the world of floating-point arithmetic, overflow is possible but not particularly common. You can get it when numbers become too large; IEEE double-precision floating-point numbers support a range of just under 21024, and if you go beyond that you have problems:

for k in [10, 100, 1000, 1020, 1023, 1023.9, 1023.9999, 1024]: try: ...

## Accurate Measurement of a Sinusoid's Peak Amplitude Based on FFT Data

There are two code snippets associated with this blog post:

Flat-Top Windowing Function for the Accurate Measurement of a Sinusoid's Peak Amplitude Based on FFT Data

and

Testing the Flat-Top Windowing Function

This blog discusses an accurate method of estimating time-domain sinewave peak amplitudes based on fast Fourier transform (FFT) data. Such an operation sounds simple, but the scalloping loss characteristic of FFTs complicates the process. We eliminate that complication by...

## A Differentiator With a Difference

Some time ago I was studying various digital differentiating networks, i.e., networks that approximate the process of taking the derivative of a discrete time-domain sequence. By "studying" I mean that I was experimenting with various differentiating filter coefficients, and I discovered a computationally-efficient digital differentiator. A differentiator that, for low fequency signals, has the power of George Foreman's right hand! Before I describe this differentiator, let's review a few...

## Oscilloscope Dreams

My coworkers and I recently needed a new oscilloscope. I thought I would share some of the features I look for when purchasing one.

When I was in college in the early 1990's, our oscilloscopes looked like this:

Now the cathode ray tubes have almost all been replaced by digital storage scopes with color LCD screens, and they look like these:

Oscilloscopes are basically just fancy expensive boxes for graphing voltage vs. time. They span a wide range of features and prices:...

## Understanding and Implementing the Sliding DFT

Introduction

In many applications the detection or processing of signals in the frequency domain offers an advantage over performing the same task in the time-domain.   Sometimes the advantage is just a simpler or more conceptually straightforward algorithm, and often the largest barrier to working in the frequency domain is the complexity or latency involved in the Fast Fourier Transform computation.   If the frequency-domain data must be updated frequently in a...

## Two jobs

For those of you following closely embeddedrelated and the other related sites, you might have noticed that I have been less active for the last couple of months, and I will use this blog post to explain why. The main reason is that I got myself involved into a project that ended up using a better part of my cpu than I originally thought it would.

I currently have two jobs: one as an electrical/dsp engineer recycled as a web publisher and the other as a parent of three kids. My job...

## Do you like the new Comments System?

I have just finished implementing a new comments system for the blogs.  Do you like it?

I'll wait a few days and make sure it works properly and then I'll port it to the code snippets and papers section.

Thanks!

## DSP Papers, Articles, Theses, etc

March 17, 20111 comment

As you may already know, there is a 'Papers and Theses' section on DSPRelated:http://www.dsprelated.com/documents.phpThere are hundreds of DSP Related documents (articles, papers, theses, dissertations, etc) scattered all around the web, and the goal with this section is to find and list as many of those documents as possible in one place. There are, at the moment, a little over 100 documents listed, which I believe is only a small subset of what is available out there, and I need your help...

## Code Snippets Suggestions

Despite being only a couple of months old, the Code Snippet section ( DSPRelated.com/code.php ) already contains tens of snippets, thanks to the contributors who have taken the time to share their code.

But let's not stop here - there is room for several hundreds more snippets before the database can be said to cover a decent portion of the DSP field.

To keep the momentum going, I will do two things:

First, I am modifying the rewards program.  Instead of...

## Latest DSP Books

December 1, 2010

As you may already know, Rick Lyons has just published a new edition of his highly acclaimed book: "Understanding Digital Signal Processing".   This book has been getting very high ratings and positive reviews from the DSP community since the publication of the first edition.  The 3rd edition seems to contain more than enough new material to justify replacing your old copy.

Also of possible interest to you, a new DSP book by C. Britton Rorabaugh titled "

## Code Snippets Section Now LIVE

November 2, 20101 comment

The new code sharing section is now live and can be accessed HERE.

Please take a few minutes to rate and/or comment the snippets that you have the expertise to judge.

If you think of some code snippets that you would like to share with the DSP community, please apply to become a contributor HERE.

If you are not aware of the reward program for contributors, your can learn about it HERE.

## New Code Sharing Section & Reward Program for Contributors!

UPDATE (11/02/2010): The code section is now live.

UPDATE 2 (01/31/2011): The reward program has changed.  A flat fee of \$20 per code snippet submitted will now be paid.

_______________

I am very happy to finally announce the imminent launch of the new code sharing section.  My vision for this new section is a rich library of high quality code snippets for the DSP community, from processor specific functions to Matlab or Scilab routines, from the simplest filter...

## 50,000th Member Announced!

January 11, 2010

In my last post, I wrote that DSPRelated.com was about to reach the 50,000 members mark.  Well, I am very happy to announce that it happened during the holidays, and the lucky person is Charlie Tsai from Taiwan.  Charlie is an assistant professor in the Department of Electrical Engineering at the National Central University in Taiwan where he teaches the "Biomedical Signal Processing" class.  He is also the advisor of the

## Almost 50,000 Members!

November 26, 20091 comment

I am very happy to announce that DSPRelated.com will reach the 50,000 registered members mark before the end of 2009. To celebrate this milestone, I will buy a BMW 5 to the 50,000th person to register (please make sure to confirm you email address to activate your registration).  Please read the fine prints after the picture.

I am just having fun here and it's not even April's fool day.  The 50,000th member won't get a BMW (I wish I could offer it!),...


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### Mathematics Class XI

Unit-I: Sets and Functions
Chapter 1: Sets
Unit-II: Algebra
Chapter 5: Binomial Theorem
Chapter 6: Sequence and Series
Unit-III: Coordinate Geometry
Chapter 1: Straight Lines
Chapter 2: Conic Sections
Unit-IV: Calculus
Unit-V: Mathematical Reasoning
Unit-VI: Statistics and Probability
Chapter 1: Statistics
Chapter 2: Probability

# Graphs of Trigonometric Functions

(i)  Graph of

We take the interval , where length of the interval is .

(i) For , decreases from .

(ii) For , increases from .

(iii) For , increases from .

(iv) For , decrease from .

i.e. , its graph lies between the lines   and .

(ii) Graph of

In interval

(i) For , increases from

(ii) For , decreases from

at and

, .

This graph lies between   and .

(iii) Graph of

is not defined at and . It increases from to in and then from to in .

(iv) Graph of

is not defined at and . So, the graph of is  a decreasing graph.

(v) Graph of

(i) decreases from in and is undefined at .

(ii) It again decreases from in and increases from in and is undefined at

(iii) It increases from in .

(vi) Graph of

(i) is undefined at . It increases from in the interval .

(ii) It decreases from in and is undefined at .

(iii) It decreases from in and again increases from   in and is undefined at .

Scroll to Top


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1. ## Induction

Hi there!
This is my problem
Calculate $\sum_{k=1}^n \frac{1}{(2k-1)(2k+1)}$ And then prove it, with help of induction?

Can any one of you help me with this problem?

2. Originally Posted by Kluringen
Hi there!
This is my problem
Calculate $\sum_{k=1}^n \frac{1}{(2k-1)(2k+1)}$ And then prove it, with help of induction?

Can any one of you help me with this problem?
if you get to the expression it should be equal to, can you do the induction?

this is a telescoping sum. meaning, you can write it in a different form that when you expand it, most of the terms will cancel out, leaving you with what the sum actually is.

note that $\frac 1{(2k - 1)(2k + 1)} = \frac 12 \bigg( \frac 1{2k - 1} - \frac 1{2k + 1} \bigg)$ (if you can't expand like that using algebra, just use partial fractions)

so that $\sum_{k = 1}^n \frac 1{(2k - 1)(2k + 1)} = \frac 12 \sum_{k = 1}^n \bigg( \frac 1{2k - 1} - \frac 1{2k + 1} \bigg)$

now expand the sum on the right to see what cancels out. that will get you your formula to do the induction with


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## What is a determinant?

A simple introduction to a determinant is that it’s the area of a box.

Working in two dimensions, I’ll outline

• the geometric picture of a linear transformation
• the geometric picture of a determinant as an area
• how the geometric picture leads to a few important properties of determinants
• how the geometric picture of linear transformations can be expressed with matrices
• how the geometric picture leads us to a formula for the determinant of a matrix

This post is long already. To keep it from becoming even longer, in some places I have had to leave out certain steps in the logic.

We’ll start with the coordinate plane. It’s a grid of points.

Then we scissor it or blow it up or shrink it down. Here are some examples:

To make them, I took the original image and applied the “shear”, “rotate”, and “scale” tools in GIMP (an open-source PhotoShop equivalent). You can try it yourself on any image just by using those tools.

These are called “linear transformations”. To simplify the way we picture them, we can just look at what they do to a box at the origin. I’ll make the box 3×3 so it’s visible, but imagine that each line represents a distance 1/3, so the sides of the box are length 1.

If we wanted, we could use something more complicated:

But since the apple is made from little boxes and all little boxes get treated the same way, we might as well focus on what happens to just one box.

Under any linear transformation, the box turns into a parallelogram.

The area of that parallelogram is called the determinant of the linear transformation.

There’s one extra rule. If the red and blue sides switch (as they would if I used the “flip” tool in GIMP), the determinant is negative. Here’s an example:

Since any area is made from little boxes and each little box’s area gets multiplied by the determinant, the area of any shape at all gets multiplied by the determinant. So for the apple, the determinant is the area of the apple on the right divided by the area of the apple on the left.

So that’s what a determinant is. What remains is to show what it’s about and what it has to do with the matrices you were wondering about.

Let’s look at some properties first. Imagine doing two transformations in a row. We’ll call this “multiplying the transformations”. The result is just another linear transformation.

When we do these sequential transformations, the area of our box gets multiplied by the determinant each time. If the determinant of the first transformation is 3 and the determinant of the second transformation is 5, the area gets multiplied by 15 overall, so the determinant of the combined transformation is 15. Multiplying transformations means multiplying determinants.

Next we’ll think about inverses. An inverse is a transformation that takes you back to where you started. The inverse of a transformation that rotates 45 degrees clockwise and multiplies everything by 2 is a transformation that rotates 45 degrees counterclockwise and cuts everything in half.

The determinant of the first transformation is 4 because each side of the box is doubled. The determinant of the second transformation is 1/4.

This is a general rule. Suppose two transformations are inverses. Then their determinants must multiply to 1, because the area of the box doesn’t change overall.

Next suppose a transformation’s determinant is zero. Then it doesn’t have an inverse because any number times zero is still zero, so there’s no transformation that takes the determinant back to one.

Geometrically, a transformation with zero determinant collapses everything to a line.

The line doesn’t have to be flat like this. It could be at any angle. Also, I didn’t collapse this completely to a line, since then you couldn’t see it. Transformations with zero determinant are bad news.

To review

• Linear transformations are some combination of the “scale”, “rotate”, “shear”, and “flip” tools in Photoshop.
• The determinant of a linear transformation is the factor by which the transformation changes the area.
• The determinants of inverse transformations multiply to 1.
• If the determinant is zero, the matrix doesn’t have an inverse. (The converse of this also holds, although we didn’t discuss it.)

Let’s move on to matrices. Take a linear transformation like this:

If we superimpose the original onto the final, we can see the coordinates of the new parallelogram in terms of the original grid.

We can describe the transformation completely using four numbers, two for the coordinates of the blue side and two for the coordinates of the red side. We’ll call those numbers $a, b, c, d$.

We’ll represent points with column matrices. So the point $(a,b)$ will be represented by the matrix $\left[ \begin{array}{c} a \\ b \end{array} \right]$. (A matrix doesn’t have to be square. This is a 2×1 matrix.)

With this notation, we can represent our linear transformation by

$\left[ \begin{array}{c} 1 \\ 0 \end{array} \right] \to \left[ \begin{array}{c} a \\ b \end{array} \right]$

$\left[ \begin{array}{c} 0 \\ 1 \end{array} \right] \to \left[ \begin{array}{c} c \\ d \end{array} \right]$

This actually represents the entire transformation, even though it looks like we’ve only looked at two points. The reason is that any other point is made up out of the two we’ve already examined. For example

$\left[ \begin{array}{c} 4 \\ 7 \end{array} \right] = \left[ \begin{array}{c} 4 \\ 0 \end{array} \right] + \left[ \begin{array}{c} 0 \\ 7 \end{array} \right] \to \left[ \begin{array}{c} 4a \\ 4b \end{array} \right] + \left[ \begin{array}{c} 7c \\ 7d \end{array} \right] = \left[ \begin{array}{c} 4a + 7c \\ 4b + 7d \end{array} \right]$

There’s a much more convenient way to write all this, which is in the form of a 2×2 matrix. $\left[ \begin{array}{c} a \\ b \end{array} \right]$, which is the blue part of our parallelogram, becomes the first column of the matrix. $\left[ \begin{array}{c} c \\ d \end{array} \right]$ is the second column.

We can view matrix multiplication as

$\left[ \begin{array}{cc} a & c \\ b & d \end{array} \right] \left[ \begin{array}{c} e \\ f \end{array} \right] = e \left[ \begin{array}{c} a \\ b \end{array} \right] + f \left[ \begin{array}{c} c \\ d \end{array} \right] = \left[ \begin{array}{c} ea + fc \\ eb + fd \end{array} \right]$

Check that this works for the example of $\left[ \begin{array}{c} 4 \\ 7 \end{array} \right]$.

You may have learned to do this multiplication one row at a time rather than one column at a time. The result is the same.

This shows how a matrices describe linear transformations. All that remains is to tie in the concept of a determinant.

Remembering that a determinant is the area of a box, we can find a formula for the determinant by looking at some properties of area.

The area of the original 1×1 box is 1. That means

$\left| \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right| = 1$

because that’s the identity matrix. It’s the linear transformation that does nothing. (The vertical lines around the matrix indicate that we’re taking a determinant.)

When we switch the blue and red sides of the box, the determinant is -1. The matrix that does this is

$\left| \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right| = -1$

When we multiply the blue side by two, the determinant gets multiplied by that same factor. Since this is represented in the matrix by multiplying the first column by two, we have

$\left| \begin{array}{cc} 2 & 0 \\ 0 & 1 \end{array}\right| = 2$

and similarly

$\left| \begin{array}{cc} 2 & 0 \\ 0 & 3 \end{array}\right| = 6$

$\left| \begin{array}{cc} 1 & 1 \\ 0 & 0 \end{array}\right| = ?$

This matrix is not invertible. It collapse everything onto the x-axis, making a “box” of zero area, so its determinant is zero. Similarly,

$\left| \begin{array}{cc} 0 & 0 \\ 1 & 1 \end{array}\right| = 0$

The final property we need of determinants/areas is linearity. Check out this picture:

It requires a little explanation. There are three linear transformations here, all sharing the same red side. The first two have the blue and purple sides. These are smaller. When we add them up, we get the third one with the gray side, so this picture represents adding linear transformations (which is different than multiplying them.) The green area is the area of the big transformation with the gray side.

The two smaller ones, with the blue and purple sides, have a total area equal to the green area. We can see this because there is a triangle of stuff that’s outside the green area, and therefore not counted. However, there’s also a triangle of extra stuff in the green area that’s not part two smaller parallelograms. These two triangles have the same area and cancel each other out, so that the small parallelograms have the same total area as the single big one.

Translating this into matrices means we can add determinants when one column is shared. This is called linearity in a column. For example

$\left| \begin{array}{cc} a & 0 \\ b & 1 \end{array}\right| + \left| \begin{array}{cc} c & 0 \\ d & 1 \end{array}\right| = \left| \begin{array}{cc} a+c & 0 \\ b + d & 1 \end{array}\right|$

So the properties we found are

• The determinant of the identity is one.
• The determinant of the matrix that switches horizontal and vertical is -1.
• Multiplying a column by a number multiplies the determinant by that number.
• The determinant is linear in a column.

These properties combined let us find the determinant of any matrix. Start with

$\left| \begin{array}{cc} a & c \\ b & d \end{array}\right|$

use linearity in the first column to write this as

$\left| \begin{array}{cc} a & c \\ 0 & d \end{array}\right| + \left| \begin{array}{cc} 0 & c \\ b & d \end{array}\right|$

now use linearity in the second column to make it

$\left| \begin{array}{cc} a & c \\ 0 & 0 \end{array}\right| + \left| \begin{array}{cc} a & 0 \\ 0 & d \end{array}\right| + \left| \begin{array}{cc} 0 & c \\ b & 0 \end{array}\right| + \left| \begin{array}{cc} 0 & 0 \\ b & d \end{array}\right|$

We have already set up the tools to evaluate each of these individually. The determinant is

$\left| \begin{array}{cc} a & c \\ b & d \end{array}\right| = 0 + ad - cb - 0$

That’s the area of the parallelogram. You could find it by other geometrical means, too, but knowing the formula for the determinant makes it easy.

### 8 Responses to “What is a determinant?”

1. Quora Says:

What is a determinant of a matrix?…

The simpler the English, the more of it there will have to be. Some main points are * A matrix is a transformation which can be thought of as any combination of the “scale”, “rotate”, “shear”, and “flip” buttons in Photoshop. * As Vaibhav Mally…

2. Jim Says:

In general a linear transformation in Rn multiplies k-dimensional
volume in any k-flat by a constant factor which depends on the k-flat. To calculate this factor for any k-flat take any basis of the
linear subspace corresponding to the k-flat and arrange them as
column vectors forming an n x k array, Calculate all k x k minors
of this array , square them and add up the squares.
Next apply the matrix of the given linear transformation to the k x k
array and calculate all k x k minors of the resultant n x k array and
again square them and add them up.
Divide the second sum of squares of k x k minors by the first and
then take the square root of the result.
This the factor by which the given linear transformation multiplys
all k-volumes in the given k-flat.

3. Jim Says:

I meant to say in the seventh line “apply the matrix of the given
linear matrix to the n x k array of column vectors”
Sorry.

4. Jim Says:

I meant to say in the above correction “apply the matrix of the given linear transformation to the n x k array of column vectors”
Sorry.

5. Jim Says:

In the second to last line “multipys” should be “multiplies”.
Sorry

6. simplethebest1 Says:

In what tool have you make the designs?., thanks

7. Voislav Sauca Says:

With what tools are making the designs?., thanks

8. Mark Eichenlaub Says:

They were made with GIMP.


text:
# Peeter Joot's (OLD) Blog.

• ## Archives

 Adam C Scott on avoiding gdb signal noise… Ken on Scotiabank iTrade RESP …… Alan Ball on Oops. Fixing a drill hole in P… Peeter Joot's B… on Stokes theorem in Geometric… Exploring Stokes The… on Stokes theorem in Geometric…

• 317,406

## Linear transformations that retain two by two positive definiteness.

Posted by peeterjoot on October 4, 2009

# Motivation

Purely for fun, lets study the classes of linear transformations that retain the positive definiteness of a diagonal two by two quadratic form. Namely, the Hamiltonian

\begin{aligned}H = P^2 + Q^2\end{aligned} \quad\quad\quad(1)

under a change of variables that mixes position and momenta coordinates in phase space

\begin{aligned}\mathbf{z}' =\begin{bmatrix}p \\ q\end{bmatrix}=\begin{bmatrix}\alpha & \beta \\ a & b \\ \end{bmatrix}\begin{bmatrix}P \\ Q\end{bmatrix}= A \mathbf{z}\end{aligned} \quad\quad\quad(2)

We want the conditions on the matrix $A$ such that the quadratic form retains the diagonal nature

\begin{aligned}H = P^2 + Q^2 = p^2 + q^2\end{aligned} \quad\quad\quad(3)

which in matrix form is

\begin{aligned}H = \mathbf{z}^\text{T} \mathbf{z} = {\mathbf{z}'}^\text{T} \mathbf{z}' \end{aligned} \quad\quad\quad(4)

So the task is to solve for the constaints on the matrix elements for

\begin{aligned}I = A^\text{T} A = \begin{bmatrix}\alpha & a \\ \beta & b \\ \end{bmatrix}\begin{bmatrix}\alpha & \beta \\ a & b \\ \end{bmatrix}\end{aligned} \quad\quad\quad(5)

Strictly speaking we can also scale and retain positive definiteness, but that case is not of interest to me right now so I’ll use this term as described above.

# Guts

The expectation is that this will neccessarily include all rotations. Will there be any other allowable linear transformations? Written out in full we want the solutions of

\begin{aligned}\begin{bmatrix}1 & 0 \\ 0 & 1 \\ \end{bmatrix}=\begin{bmatrix}\alpha & a \\ \beta & b \\ \end{bmatrix}\begin{bmatrix}\alpha & \beta \\ a & b \\ \end{bmatrix}= \begin{bmatrix}\alpha^2 + a^2 & \alpha \beta + a b \\ \alpha \beta + a b & \beta^2 + b^2\end{bmatrix}\end{aligned} \quad\quad\quad(6)

Written out explicitly we have three distinct equations to reduce

\begin{aligned}1 = \alpha^2 + a^2 \end{aligned} \quad\quad\quad(7)

\begin{aligned}1 = \beta^2 + b^2 \end{aligned} \quad\quad\quad(8)

\begin{aligned}0 = \alpha \beta + a b \end{aligned} \quad\quad\quad(9)

Solving for $a$ in 9 we have

\begin{aligned}a &= -\frac{\alpha \beta}{b} \\ \implies \\ 1 &= \alpha^2 \left( 1 + \left(-\frac{\beta}{b} \right)^2 \right) \\ &= \frac{\alpha^2}{b^2} \left( b^2 + \beta^2 \right) \\ &= \frac{\alpha^2}{b^2} \end{aligned}

So, provided $b \ne 0)$, we have a first simplifying identity

\begin{aligned}\alpha^2 = b^2\end{aligned} \quad\quad\quad(10)

Written out to check, this reduces our system of equations

\begin{aligned}\begin{bmatrix}1 & 0 \\ 0 & 1 \\ \end{bmatrix}=\begin{bmatrix}\alpha & a \\ \beta & \pm \alpha \\ \end{bmatrix}\begin{bmatrix}\alpha & \beta \\ a & \pm \alpha \\ \end{bmatrix}= \begin{bmatrix}\alpha^2 + a^2 & \alpha \beta \pm a \alpha \\ \alpha \beta \pm a \alpha & \beta^2 + \alpha^2\end{bmatrix}\end{aligned} \quad\quad\quad(11)

so our equations are now

\begin{aligned}1 = \alpha^2 + a^2 \end{aligned} \quad\quad\quad(12)

\begin{aligned}1 = \beta^2 + \alpha^2\end{aligned} \quad\quad\quad(13)

\begin{aligned}0 = \alpha (\beta \pm a )\end{aligned} \quad\quad\quad(14)

There are two cases to distinguish here. The first is the more trivial $\alpha = 0$ case, for which we find $a^2 = \beta^2 = 1$. For the other case we have

\begin{aligned}\beta = \mp a\end{aligned} \quad\quad\quad(15)

Again, writing out in full to check, this reduces our system of equations

\begin{aligned}\begin{bmatrix}1 & 0 \\ 0 & 1 \\ \end{bmatrix}=\begin{bmatrix}\alpha & a \\ \mp a & \pm \alpha \\ \end{bmatrix}\begin{bmatrix}\alpha & \mp a \\ a & \pm \alpha \\ \end{bmatrix}= \begin{bmatrix}\alpha^2 + a^2 & 0 \\ 0 & a^2 + \alpha^2\end{bmatrix}\end{aligned} \quad\quad\quad(16)

We have now only one constraint left, and have reduced things to a single degree of freedom

\begin{aligned}1 = \alpha^2 + a^2 \end{aligned} \quad\quad\quad(17)

Or

\begin{aligned}\alpha = (1 - a^2)^{1/2}\end{aligned} \quad\quad\quad(18)

We’ve already used $\pm$ to distinguish the roots of $\alpha = \pm b$, so here lets imply that this square root can take either positive or negative values, but that we are treating the sign of this the same whereever seen. Our transformation, employing $a$ as the free variable is now known to take any of the following forms

\begin{aligned}A &= \begin{bmatrix}(1 - a^2)^{1/2} & \mp a \\ a & \pm (1 - a^2)^{1/2} \\ \end{bmatrix} \\ &= \begin{bmatrix}0 & \pm 1 \\ (1)^{1/2} & 0 \\ \end{bmatrix} \\ &=\begin{bmatrix}\alpha & \beta \\ a & 0 \\ \end{bmatrix} \end{aligned} \quad\quad\quad(19)

The last of these (the $b=0$ case from earlier) was not considered, but doing so one finds that it produces nothing different from the second form of the transformation above. That leaves us with two possible forms of linear transformations that are allowable for the desired constraints, the first of which screams for a trigonometric parameterization.

For ${\left\lvert{a}\right\rvert} \le 0$ we can parameterize with $a = \sin\theta$. Should we allow complex valued linear transformations? If so $a = \cosh(\theta)$ is a reasonable way to parameterize the matrix for the $a > 0$ case. The complete set of allowable linear transformations in matrix form are now

\begin{aligned}A &= \begin{bmatrix}1^{1/2} \cos\theta & \mp \sin\theta \\ \sin\theta & \pm 1^{1/2} \cos\theta \\ \end{bmatrix} \\ A &= \begin{bmatrix}(-1)^{1/2} \sinh\theta & \mp \cosh\theta \\ \cosh\theta & \pm (-1)^{1/2} \sinh\theta \\ \end{bmatrix} \\ A &= \begin{bmatrix}0 & \pm 1 \\ 1^{1/2} & 0 \\ \end{bmatrix}\end{aligned} \quad\quad\quad(22)

There are really four different matrixes in each of the above. Removing all the shorthand for clarity we have finally

\begin{aligned}A \in \{&\begin{bmatrix}\cos\theta & - \sin\theta \\ \sin\theta & \cos\theta \\ \end{bmatrix}, \begin{bmatrix}-\cos\theta & - \sin\theta \\ \sin\theta & -\cos\theta \\ \end{bmatrix}, \begin{bmatrix}\cos\theta & \sin\theta \\ \sin\theta & - \cos\theta \\ \end{bmatrix}, \begin{bmatrix}-\cos\theta & \sin\theta \\ \sin\theta & \cos\theta \\ \end{bmatrix}, \\ & \begin{bmatrix}i \sinh\theta & -\cosh\theta \\ \cosh\theta & i \sinh\theta \\ \end{bmatrix}, \begin{bmatrix}-i \sinh\theta & -\cosh\theta \\ \cosh\theta & -i \sinh\theta \\ \end{bmatrix}, \begin{bmatrix}i \sinh\theta & \cosh\theta \\ \cosh\theta & - i \sinh\theta \\ \end{bmatrix}, \begin{bmatrix}-i \sinh\theta & \cosh\theta \\ \cosh\theta & i \sinh\theta \\ \end{bmatrix}, \\ &\begin{bmatrix}0 & 1 \\ 1 & 0 \\ \end{bmatrix},\begin{bmatrix}0 & 1 \\ -1 & 0 \\ \end{bmatrix},\begin{bmatrix}0 & -1 \\ 1 & 0 \\ \end{bmatrix},\begin{bmatrix}0 & -1 \\ -1 & 0 \\ \end{bmatrix}\}\end{aligned}

The last four possibilities are now seen to be redundant since they can be incorporated into the $\theta = \pm \pi/2$ cases of the real trig parameterizations where $\sin\theta = \pm 1$, and $\cos\theta = 0$. Employing a $\theta' = -\theta$ change of variables, we find that two of the hyperbolic parameterizations are also redundant and can express the reduced solution set as

\begin{aligned}A \in \{\pm \begin{bmatrix}\cos\theta & - \sin\theta \\ \sin\theta & \cos\theta \\ \end{bmatrix}, \pm \begin{bmatrix}\cos\theta & \sin\theta \\ \sin\theta & - \cos\theta \\ \end{bmatrix}, \pm \begin{bmatrix}i \sinh\theta & -\cosh\theta \\ \cosh\theta & i \sinh\theta \\ \end{bmatrix}, \pm \begin{bmatrix}i \sinh\theta & \cosh\theta \\ \cosh\theta & - i \sinh\theta \\ \end{bmatrix}\}\end{aligned} \quad\quad\quad(25)

I suspect this class of transformations has a name in the grand group classification scheme, but I don’t know what it is.


text:
# Minkowski's theorem

A set in 2 satisfying the hypotheses of Minkowski's theorem.

In mathematics, Minkowski's theorem is the statement that any convex set in ${\displaystyle \mathbb {R} ^{n}}$ which is symmetric with respect to the origin and which has volume greater than ${\displaystyle 2^{n}}$ contains a non-zero integer point. The theorem was proved by Hermann Minkowski in 1889 and became the foundation of the branch of number theory called the geometry of numbers. It can be extended from the integers to any lattice ${\displaystyle L}$ and to any symmetric convex set with volume greater than ${\displaystyle 2^{n}\,d(L)}$, where ${\displaystyle d(L)}$ denotes the covolume of the lattice (the absolute value of the determinant of any of its bases).

## Formulation

Suppose that L is a lattice of determinant d(L) in the n-dimensional real vector space n and S is a convex subset of n that is symmetric with respect to the origin, meaning that if x is in S then x is also in S. Minkowski's theorem states that if the volume of S is strictly greater than 2n d(L), then S must contain at least one lattice point other than the origin. (Since the set S is symmetric, it would then contain at least three lattice points: the origin 0 and a pair of points ±x, where xL \ 0.)

## Example

The simplest example of a lattice is the integer lattice n of all points with integer coefficients; its determinant is 1. For n = 2, the theorem claims that a convex figure in the Euclidean plane symmetric about the origin and with area greater than 4 encloses at least one lattice point in addition to the origin. The area bound is sharp: if S is the interior of the square with vertices (±1, ±1) then S is symmetric and convex, and has area 4, but the only lattice point it contains is the origin. This example, showing that the bound of the theorem is sharp, generalizes to hypercubes in every dimension n.

## Proof

The following argument proves Minkowski's theorem for the specific case of L = ℤ2. It can be generalized to arbitrary lattices in arbitrary dimensions.

Consider the map

${\displaystyle f:S\to \mathbb {R} ^{2}/2L,\qquad (x,y)\mapsto (x{\bmod {2}},y{\bmod {2}})}$

Intuitively, this map cuts the plane into 2 by 2 squares, then stacks the squares on top of each other. Clearly f(S) has area less than or equal to 4, because this set lies within a 2 by 2 square. Assume for a contradiction that f could be injective, which means the pieces of S cut out by the squares stack up in a non-overlapping way. Because f is locally area-preserving, this non-overlapping property would make it area-preserving for all of S, so the area of f(S) would be the same as that of S, which is greater than 4. That is not the case, so the assumption must be false: f is not injective, meaning that there exist at least two distinct points p1, p2 in S that are mapped by f to the same point: f(p1) = f(p2).

Because of the way f was defined, the only way that f(p1) can equal f(p2) is for p2 to equal p1 + (2i, 2j) for some integers i and j, not both zero. That is, the coordinates of the two points differ by two even integers. Since S is symmetric about the origin, p1 is also a point in S. Since S is convex, the line segment between p1 and p2 lies entirely in S, and in particular the midpoint of that segment lies in S. In other words,

${\displaystyle {\tfrac {1}{2}}\left(-p_{1}+p_{2}\right)={\tfrac {1}{2}}\left(-p_{1}+p_{1}+(2i,2j)\right)=(i,j)}$

is a point in S. But this point (i,j) is an integer point, and is not the origin since i and j are not both zero. Therefore, S contains a nonzero integer point.

## Applications

An application of this theorem is the result that every class in the ideal class group of a number field K contains an integral ideal of norm not exceeding a certain bound, depending on K, called Minkowski's bound: the finiteness of the class number of an algebraic number field follows immediately.

Minkowski's theorem is also useful to prove Lagrange's four-square theorem, which states that every natural number can be written as the sum of the squares of four natural numbers.

## References

• Enrico Bombieri and Walter Gubler (2006). Heights in Diophantine Geometry. Cambridge U. P.
• J. W. S. Cassels. An Introduction to the Geometry of Numbers. Springer Classics in Mathematics, Springer-Verlag 1997 (reprint of 1959 and 1971 Springer-Verlag editions).
• John Horton Conway and N. J. A. Sloane, Sphere Packings, Lattices and Groups, Springer-Verlag, NY, 3rd ed., 1998.
• Hancock, Harris (1939). Development of the Minkowski Geometry of Numbers. Macmillan. (Republished in 1964 by Dover.)
• Hazewinkel, Michiel, ed. (2001) [1994], "Geometry of numbers", Encyclopedia of Mathematics, Springer Science+Business Media B.V. / Kluwer Academic Publishers, ISBN 978-1-55608-010-4
• Edmund Hlawka, Johannes Schoißengeier, Rudolf Taschner. Geometric and Analytic Number Theory. Universitext. Springer-Verlag, 1991.
• C. G. Lekkerkerker. Geometry of Numbers. Wolters-Noordhoff, North Holland, Wiley. 1969.
• Wolfgang M. Schmidt. Diophantine approximation. Lecture Notes in Mathematics 785. Springer. (1980 [1996 with minor corrections])
• Wolfgang M. Schmidt.Diophantine approximations and Diophantine equations, Lecture Notes in Mathematics, Springer Verlag 2000.
• Siegel, Carl Ludwig (1989). Lectures on the Geometry of Numbers. Springer-Verlag.
• Rolf Schneider, Convex bodies: the Brunn-Minkowski theory, Cambridge University Press, Cambridge, 1993.
• "Minkowski's theorem". PlanetMath.
• Stevenhagen, Peter. Number Rings.
• Malyshev, A.V. (2001) [1994], "Minkowski theorem", in Hazewinkel, Michiel, Encyclopedia of Mathematics, Springer Science+Business Media B.V. / Kluwer Academic Publishers, ISBN 978-1-55608-010-4


text:
OpenStudy (anonymous):

PLEASE HELLP! Find the perimeter of this trapezoid. E(4, 2), F(2, 5), G(-2, 5), H(-4, 2) P = + 2

6 years ago
OpenStudy (anonymous):

did you draw a picture?

6 years ago
OpenStudy (anonymous):

theres no picture : )

6 years ago
OpenStudy (anonymous):

ok lets try that again now that i have a nice picture and not a wrong one|dw:1314297808715:dw|

6 years ago
OpenStudy (anonymous):

k : ]

6 years ago
OpenStudy (anonymous):

the distance from (-2,5) to (2,5) the top horizontal line, is 4 the distance form (-4,2 ) to (4,2) the bottom horizontal line, is 8

6 years ago
OpenStudy (anonymous):

|dw:1314297956701:dw|

6 years ago


text: Sequence ${a_n}$ has a limit $A$ in $\mathbb{R}$ and a limit $B$ in $\mathbb{Q}_p$. Then $A$ is rational iff $B$ is rational.


text:
Groups St Andrews 2017 in Birmingham

C. M. Campbell, C. W. Parker, M. R. Quick, E. F. Robertson, and C. M. Roney-Dougal, editors
Publisher:
Cambridge University Press
Publication Date:
2019
Number of Pages:
508
Format:
Paperback
Series:
London Mathematical Society Lecture Note Series
Price:
115.00
ISBN:
9781108728744
Category:
Proceedings
[Reviewed by
Charles Traina
, on
06/24/2019
]
Every four years leading researchers in group theory meet to survey recent developments in all aspects of their field.  The present volume is the tenth in this series and consists of twenty-one papers on the following topics: generation of finite simple groups, block theory, fusion systems, algebraic groups, one-relator groups, geometric group theory, and Beauville groups.

We give a brief description of one of the articles, "One-Relator Groups: An Overview" by Gilbert Baumslag, Benjamin Fine, and Gerhard Rosenberger.  The article was selected by the reviewer because it is based on work in one relator group theory of Dr. Wilhelm Magnus, the reviewer's thesis advisor. It also indicates the importance of the work of Max Dehn, Magnus' thesis advisor.

In the 1985 meeting of Groups St. Andrews, Gilbert Baumslag gave a course on one-relator groups which provided a view of the subject up to that point in time. This paper looks at the large amount of work done over the past three decades. Connections with surface groups and elementary theory, and a description of the surface group conjecture and the Gromov conjecture on surface groups are discussed. The authors discuss the solution by Wise of Baumslag's residual finiteness conjecture and also a new Baumslag conjecture on virtually free-by-cyclic groups. Various amalgam decompositions of one-relator groups and the Baumslag-Shalen conjectures are additionally examined. A number of open problems in one-relator group theory and their statuses are discussed. The final discussion is an introduction to a concept called plainarity based on the Magnus breakdown of a one-relator group. This may provide a systematic approach to the problems in the setting of one-relator groups and indicates the importance of the fundamental work done by Wilhelm Magnus in one-relator group theory.

This volume does what the editors intended: it provides a snapshot of the state of the art in group theory." It can serve as a reference work for group theorists, and since many of the articles include open problems, can serve as a source of research topics as well.
Charles Traina is a professor of mathematics at St. John's University, Jamaica N.Y. His research areas are group theory and measure theory.


text:
### Theory:

The measurement is the base of all experiments in a world of science and technology.

The quality of a measurement can be described using terms such as :
• Error.
• Accuracy.
• Precision.
• Approximation.
• Roundoff.
The value of every measurement contains some uncertainty. These uncertainties are called errors.

Accuracy is the closeness of a measured value to the actual value or true value.

Example:
In a shop, you purchased 3.5 kg apple measured using traditional balance, on the way you weight the same pack of apple ina different shop which is digitals balnce and found that it actually weight only 3.3 then, the measurement is considered inaccurate. If the purchased apple measures the same 3.5 kg then the measurement is considered as accurate.

Precision: is the closeness of two or more measurements to each other.
The pack of apple is weighted for five or more time,  and get 3.5 kg for each time, then your measurement is very precise. If the measurement varies then it is considered as imprecise.

Approximation: is the process of finding a number, which is acceptably close to the exact value of the measurement of a physical quantity.
Continuing the same example, if the apple weighs differently in different balances but most of the time (or all the time) the value is above 3.3 kg then we can conclude that the pack of apple is approximately 3.3 kg.

Try an activity:

Calculate the approximate ‘heartbeat’ of a man in a day.
Consider the heart beats per minute as 72 times.
Given:  $$1$$ day ; Heart beats rate $$=$$ $$72$$ per minute approximately.

To find: The approximate  heart beat rate of man in $$1$$ day.

Approximate heart beat rate of man in $$1$$ days:

Approximate heart beat rate in one day $$=$$ $$1 day$$ $$×$$ $$heart\ beat\ rate$$ $$×$$ $$60 seconds$$ $$×$$ $$24 hours$$.

$$=$$ $$1$$ $$×$$ $$72$$ $$×$$ $$60$$ $$×$$ $$24$$

$$=$$ 103680

Thus, the approximation of heart beat rate of man in $$1$$ day is $$103680$$.

Rounding off numbers:  While rounding off a number, if a last digit is greater than 5 then it should be changed or increased by one. The following are a few steps to be followed in rounding off numbers.

Rule for Round off:

Step 1: Identify the last digit to be kept.

Step 2: The following digit is less than $$5$$. So, retain it as the same number. (or) The following digit is equal to or greater than $$5$$. Then, the number is increased by one.

Example 1: Round off the number $$1.344$$.

Given: $$1.344$$.

To find: The round off the number $$1.344$$ to two decimal places.

Round off the number $$1.344$$:

Step: 1 Identify the last digit to be kept.$$4$$ is the last digit to be kept.

Step: 2 The following digit, because $$4$$ is less than $$4$$. So, retain it as $$4$$.

Thus, the round off the number is $$1.34$$.

Example 2: Round off the number $$1.456$$.

Given: $$1.456$$.

To find: The round off the number $$1.456$$ to two decimal places.

Round off the number $$1.456$$:

Step: 1 Identify the last digit to be kept. $$5$$ is the last digit to be kept.

Step: 2. The following digit, because $$6$$ is greater than $$5$$. So, increase $$5$$ by one as $$6$$.

Thus, the round off the number is $$1.46$$.


text:
# Four-photon polarisation tensor in QED

Let's consider the four-photon polarisation tensor $\Pi^{\mu\nu\lambda\rho}$ in QED. It follows from Ward identity that $$k_1^\mu \Pi_{\mu\nu\lambda\rho}(k_1,k_2,k_3,k_4) = 0.$$ After applying the derivative $\frac{\partial}{\partial k_1^\sigma}$ to both sides of the above equality we obtain $$\Pi_{\sigma\nu\lambda\rho}(k_1,k_2,k_3,k_4) = -k_1^\mu \frac{\partial}{\partial k_1^\sigma}\Pi_{\mu\nu\lambda\rho}(k_1,k_2,k_3,k_4).$$ Thus, $\Pi_{\sigma\nu\lambda\rho}$ vanishes for $k_1=k_2=k_3=k_4=0$.

I cannot reproduce the above result by direct calculaition in one-loop approximation. There are six contributing diagrams in this order, all with one fermionic loop. The three depicted below and another three witch differ only in the arrow direction (they give the same integrals as the diagrams above). I'm interested only in a value of the polarisation tensor for vanishing external momenta of photons. We have $$\Pi^{\mu\nu\lambda\rho} = T^{\mu\nu\lambda\rho} + T^{\mu\nu\rho\lambda} + T^{\mu\lambda\nu\rho},$$ where $$T^{\mu\nu\lambda\rho}=2(-ie)^4 i^4 \int\frac{d^4 p}{(2\pi)^4}\frac{\text{tr}\left((m+\gamma \cdot p)\gamma ^{\mu }(m+\gamma \cdot p)\gamma ^{\nu }(m+\gamma \cdot p)\gamma ^{\lambda }(m+\gamma \cdot p)\gamma ^{\rho }\right)}{(p^2-m^2+i0)^4}.$$ The integral in the definition of $T^{\mu\nu\lambda\rho}$ is logarithmically divergent but the divergent contribution to $\Pi^{\mu\nu\lambda\rho}$ vanishes. Finally I get finite answer for $\Pi_{\sigma\nu\lambda\rho}(k_1=k_2=k_3=k_4=0)$ which is unfortunatelly different from $0$?

Does $\Pi_{\sigma\nu\lambda\rho}(k_1=k_2=k_3=k_4=0)$ really vanish?

Yes, at $k_1 = k_2 = k_3 = k_4 = 0$, the amplitude vanishes. And, I think this can be understood in terms of F. Low's soft photon theorem.
Use LoopIntegrate to initiate the computation of the Feynman integrals:
Then after replacing all external momenta to zero, use LoopRefine to obtain an explicit analytic representation of that integral: