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7.2

#### Lecture 32: Minimum Spanning Trees

Minimum spanning trees: Borůvka’s, Prim’s and Kruskal’s algorithms

Early in the 20th century, the challenge of electrifying towns and cities was one of the pressing issues facing civil engineers. In 1926, a Moravian academic named <NAME> considered the problem and came up with a solution. A workable solution needed to:

• Connect all the towns to the power stations, either directly or indirectly, so that everyone who was supposed to receive electricity actually did.

• At least for an initial deployment, simply ensuring that everyone gets connected is more important than making sure that there are backup connections for each place.

• Building an initial deployment is expensive, so minimizing costs is important.

More impressively, Borůvka expressed the solution not solely in terms of wires and electricity, but in terms of an abstract weighted graph. The lack of backup connections implies that a solution must be a tree (as opposed to a graph that might have multiple paths or cycles); the connectedness requirement means it should be a spanning tree (as opposed to a disconnected forest of multiple trees); and the cost focus means it should be a minimum spanning tree (as opposed to a more expensive tree).

(Unfortunately, Borůvka wrote up his solution in Czech, in a relatively obscure Czech journal, and few European or American computer scientists heard about the result. It was rediscovered several times, and finally became better known when Sollin rediscovered it in 1965 and published it...in French. Coming up with a useful result only matters when people find out about it!)

A fellow Czech researcher, <NAME>, built upon Borůvka’s result and simplified it, and published a related algorithm in 1930. That algorithm, too, was not widely known until 1957, when Robert Prim rediscovered it. Today, the fastest known algorithms for solving the minimum spanning tree problem are a hybrid of Borůvka’s and Prim’s results.

Prim’s algorithm, as we’ll see below, is closely related to a third minimum-spanning-tree algorithm, discovered by <NAME> in 1956: they differ only in the ordering that they choose edges to add to the tree. However, the detailed bookkeeping of the two algorithms is very different, and it’s worth studying both.

##### 32.1Running example

We’ll use the following classes in our examples below:
 class Vertex { ArrayList outEdges; } class Edge { Vertex from, to; int weight; } class Graph { ArrayList vertices; }

Consider the following graph:

The edges are drawn without directional arrows, and the numbers on each edge represents their weight.

Do Now!

List the edges (e.g., AB,BF,EC etc) of three different spanning trees for this graph. Find a minimum spanning tree for this graph. How can you be confident that it’s actually minimal? What strategy did you use to pick the edges?

One possible minimum spanning tree for this graph are the edges AB,BE,BF,EC,CD. Let’s see how to compute that.

Do Now!

Suppose we want to construct a spanning tree; what edge should we start with? Why?

It seems intuitively obvious that we should start with the edge EC, since it’s the cheapest edge in the entire graph. How can we be certain that that edge must be in the final minimum spanning tree? Let’s think about it by contradiction: suppose it wasn’t part of the minimum spanning tree. Then we must have some other spanning tree, that is supposedly minimal, that managed to find some other path connecting E and C. We’re going to improve this spanning tree, and lower its overall weight. That other path connecting E to C has some weight W. Let’s deliberately introduce a cycle, by adding the edge EC to this path. This cycle now has weight W+15. We can break the cycle by removing any of its edges, and all the nodes will remain connected. So we are free to choose whichever edge we like, and any of the other edges will be more expensive than edge EC (why?). Which in turn means that deleting some other edge will result in a spanning tree that has weight W + 15 - (something greater than 15), which is a lower total weight than the tree that we started with, contradicting our assumption that we had a minimum spanning tree.

Let’s make this argument concrete: suppose that path were EA,AB,BC. Then the path weight is 50+30+40 = 120. We add the edge CE to make a cycle of weight 120+15 = 135. Now we can choose to remove any of the four edges in that cycle, and the overall tree is still connected. If we delete AE, the total weight drops to 135 - 50 = 85, which is lower than the 120 that we started with. If we drop AB, it drops to 105, and if we drop BC, it drops to 95. All of those are better than the 120 that we started with, so evidently we should have chosen edge EC to begin with.

Do Now!

Which edge should we select next?

Surely, we should select the next cheapest edge, CD, since any other path that connects C to D will cost more than 25.

Do Now!

Which edge comes next? Why?

There are actually two plausible answers here. We could choose edge BE, because it’s the cheapest edge connected to our current tree that doesn’t create a cycle. Or we could choose edge AB, because it’s the cheapest edge anywhere that doesn’t create a cycle. These two choices lead, respectively, to Prim’s and Kruskal’s algorithms.

Do Now!

How many edges must there be in a spanning tree, in terms of the total number V of vertices and E of edges in the overall graph? Why? Prove it.

There is an interesting duality between the two algorithms. To produce a minimum spanning tree, there are three criteria we need to establish: minimality, spanning-ness and tree-ness. Both Prim’s algorithm and Kruskal’s algorithm below run until their result is spanning. But they choose different importances for the other two considerations:
• Prim’s algorithm essentially decides to build a single tree at all times, and then works to be sure that it is minimal. Every step makes some obvious progress to the end goal, but the proof of correctness is a bit subtle.

• Kruskal’s algorithm guarantees minimality at all times, but builds many trees that eventually link up into a single final answer. This is somewhat reminiscent of how heapsort had two invariants to preserve, and chose to preserve one at all times and then fix up the other one. Here we break the "one tree" invariant, and restore it over time.

##### 32.2Prim’s algorithm

Prim’s approach gows a single tree, starting from the globally cheapest edge, until it produces a spanning tree. At each stage, it chooses the cheapest edge that makes the tree bigger without creating a cycle. We obviously need to keep track of the edges chosen for our tree. But beyond that, Prim’s approach implies that we need to keep track of

• The set of vertices we haven’t connected yet: since all the connected vertices are part of a single tree, then we should only take edges that connect to a new, not-yet-connected vertex.

• The set of candidate edges available, sometimes called the frontier (because it is the boundary between the known part of the graph and the unknown)

• An efficient way to choose the cheapest candidate edge

Do Now!

What data structure do we have that’s really efficient at adding items and keeping track of the minimum (or maximum) item among them?

Let’s keep track of the set of unconnected vertices by a HashMap<Vertex, Boolean>, that initially maps each vertex to false, and as we connect the vertex, we update the map to true to indicate it’s now connected.

To keep track of our candidate edges, we’ll use a priority queue, this time using a min-heap (as we used in the previous lecture). As we connect each vertex, we add its out edges to this priority queue, and then we’ll process edges in order of removal from that min-heap, ensuring that we always choose the cheapest edge available to us at any given point.

Exercise

From the description above, try implementing Prim’s algorithm yourself.

As it turns out, we can generalize the argument made above: we don’t need to start with the globally cheapest edge. We can start from any vertex in the graph, and follow the algorithm sketch above.

The following pseudocode is one possible implementation of Prim’s algorithm:

 // The final resulting tree ArrayList tree = new ArrayList(); // The set of connected vertices HashMap connected = new HashMap(); // The priority queue of candidate edges PriorityQueue frontier = new PriorityQueue(); If the graph's vertices are empty, just return the empty tree Initialize the connected map to map each vertex to false Pick some initial vertex, v. Set its connectedness to true, and add all its edges to the frontier. While(the frontier is not empty) Pick the cheapest edge from the frontier, suppose it connects X to Y. If Y is already connected to the tree: Discard this edge // it would create a cycle Else: Add the edge XY to the tree Mark Y as connected Add all the out-edges of Y to the frontier Return the tree

Exercise

Work through the code above, for the running example graph, and determine what the resulting tree would be.

1. Let’s say we start from vertex A. After the initialization step, we have
Tree:      []
Connected: [A]
Frontier:  [AB(30), AE(50)]

2. We select the cheapest edge from the frontier, and since B is not yet connected, we use this edge and add the out-edges of B to our frontier:
Tree:      [AB(30)]
Connected: [A, B]
Frontier:  [BA(30), BE(35), BC(40), AE(50), BF(50)]

Do Now!

Why do we add BA to the frontier?

3. We select the cheapest edge from the frontier. Since A is already connected, we discard this edge.
Tree:      [AB(30)]
Connected: [A, B]
Frontier:  [BE(35), BC(40), AE(50), BF(50)]

4. We select the cheapest edge from the frontier, and since E is not yet connected, we use this edge and add the out-edges of E to our frontier:
Tree:      [AB(30), BE(35)]
Connected: [A, B, E]
Frontier:  [EC(15), EB(35), BC(40), AE(50), BF(50), EA(50)]

Do Now!

5. We select the cheapest edge from the frontier, and since C is not yet connected, we use this edge and add the out-edges of B to our frontier:
Tree:      [AB(30), BE(35), EC(15)]
Connected: [A, B, E, C]
Frontier:  [CE(15), CD(25), BC(40), CB(40), AE(50), BF(50), EA(50)]

6. We select the cheapest edge from the frontier. Since E is already connected, we discard this edge.
Tree:      [AB(30), BE(35), EC(15)]
Connected: [A, B, E, C]
Frontier:  [CD(25), BC(40), CB(40), AE(50), BF(50), EA(50)]

7. We select the cheapest edge from the frontier, and since D is not yet connected, we use this edge and add the out-edges of D to our frontier:
Tree:      [AB(30), BE(35), EC(15), CD(25)]
Connected: [A, B, E, C, D]
Frontier:  [DC(25), BC(40), CB(40), AE(50), BF(50), EA(50), DF(50)]

8. We select the cheapest edge from the frontier. Since C is already connected, we discard this edge.
Tree:      [AB(30), BE(35), EC(15), CD(25)]
Connected: [A, B, E, C, D]
Frontier:  [BC(40), CB(40), AE(50), BF(50), EA(50), DF(50)]

9. We select the cheapest edge from the frontier. Since C is already connected, we discard this edge.
Tree:      [AB(30), BE(35), EC(15), CD(25)]
Connected: [A, B, E, C, D]
Frontier:  [CB(40), AE(50), BF(50), EA(50), DF(50)]

10. We select the cheapest edge from the frontier. Since B is already connected, we discard this edge.
Tree:      [AB(30), BE(35), EC(15), CD(25)]
Connected: [A, B, E, C, D]
Frontier:  [AE(50), BF(50), EA(50), DF(50)]

11. We select the cheapest edge from the frontier. Since E is already connected, we discard this edge.
Tree:      [AB(30), BE(35), EC(15), CD(25)]
Connected: [A, B, E, C, D]
Frontier:  [BF(50), EA(50), DF(50)]

12. We select the cheapest edge from the frontier, and since F is not yet connected, we use this edge and add the out-edges of F to our frontier:
Tree:      [AB(30), BE(35), EC(15), CD(25), DF(50)]
Connected: [A, B, E, C, D, F]
Frontier:  [BF(50), EA(50), DF(50), FB(50), FD(50)]

13. At this point we could stop, since we have connected all the vertices. Or, mechanically, we could work through the remaining edges of the frontier until it empties out, at which point our algorithm terminates with the tree above.

Do Now!

At several points, we added edges to the frontier that had duplicate edge weights. Does it matter which of those edges we take to use in our spanning tree (assuming they don’t create any cycles)?

(Obviously the algorithm sketch above is pointlessly inefficient, since it processes each edge twice. At the cost of some slightly more tedious bookkeeping, we could prevent that: we could choose to add an edge only if its endpoint was not already connected, which would automatically avoid doubling back on any edge. But this isn’t strictly necessary for the overall correctness of the algorithm.)

Exercise

Prove that this algorithm actually works. Use a variant of the reasoning above (about why the globally minimum edge should be used) to argue that whenever we connect a new vertex to the growing tree, we could not possibly have produced a better path to get there, and therefore the tree is indeed minimal so far.

Exercise

Produce a big-O worst-case runtime analysis for Prim’s algorithm.

##### 32.3Kruskal’s algorithm

Prim’s algorithm above relied on a min-heap priority queue to constantly shuffle new edges into the frontier and choose the best-available option at each stage. We might reasonably consider edges in another order: simply sort them from cheapest to most expensive, and use them in increasing order if at all possible. But this loses the guarantee that we only have one tree growing at any given time; we have a forest instead. This implies that we need to keep track of
• Each of the trees being grown so far

• An efficient way to tell if both endpoints of an edge belong to the same tree

That second point is critical: we didn’t have to worry about that with Prim’s algorithm because there was only ever one tree, so asking if both endpoints were in the "same" tree amounted to asking if they were both in the only tree, which could be handled by the connected map. Here, we need to keep track of which tree, which means we’re going to need more than a boolean.

Do Now!

Brainstorm some ideas for how to keep track of this information. How can we tell apart two distinct trees?

We could try to generalize the map a bit, mapping vertices to other vertices somehow. But to which ones? And do we need to update many vertices in the map, each time we join two trees via an edge? This sounds potentially horribly inefficient, but there is a very clever data structure that we can use to make this work to our advantage.

##### 32.3.1The Union/Find data structure

The goal of the union/find data structure is to allow us to take a set of items (such as nodes in a graph) and partition them into groups (such as nodes connected by spanning trees) in such a way that we can easily find whether two nodes are in the same group, and union two disjoint groups together. Intuitively, we accomplish this by naming each group by some representative element, and then two items can be checked for whether they are in the same group by checking if they have the same representative element.

##### 32.3.1.1Example

Suppose each of our vertices was given a string name (as we’ve been labelling the vertices so far). Then the union-find data structure is a HashMap<String, String> that maps (the name of) each node to (the name of) a node that it is connected to. Initially, every node name is mapped to itself, signifying that every node is its own representative element, or equivalently, that it is not connected to anything.

Recall the example from above:

Our HashMap will map every node name to itself:

                                     Representatives, visually:
+---+---+---+---+---+---+    A     B     C     D     E     F
Node:   | A | B | C | D | E | F |
+---+---+---+---+---+---+
Link:   | A | B | C | D | E | F |
+---+---+---+---+---+---+

Spanning tree so far:
Kruskal’s algorithm begins by sorting the list of edges in the graph by edge weight, from shortest to longest:

EC(15)
CD(25)
AB(30)
BE(35)
BC(40)
FD(50)
AE(50)
BF(50)
When we add edge EC(15), nodes E and C are now connected:

                                      Representatives, visually:
+---+---+---+---+---+---+     A     B     D     E     F
Node:   | A | B | C | D | E | F |                       ^
+---+---+---+---+---+---+                       |
Link:   | A | B | E | D | E | F |                       C
+---+---+---+---+---+---+

Spanning tree so far:          (C E)
We next add edge CD(25). Since C’s representative is E, and D’s representative is D, they are currently separate, so adding this edge would not create a cycle. We can therefore union them and set D’s representative’s representative to C’s representative:

                                      Representatives, visually:
+---+---+---+---+---+---+     A     B     E     F
Node:   | A | B | C | D | E | F |                 ^
+---+---+---+---+---+---+                / \
Link:   | A | B | E | E | E | F |               C   D
+---+---+---+---+---+---+

Spanning tree so far:          (C D) (C E)

Do Now!

Careful! Why must we union the representatives of two nodes, and not the nodes themselves?

                                      Representatives, visually:
+---+---+---+---+---+---+     A     E     F
Node:   | A | B | C | D | E | F |     ^     ^
+---+---+---+---+---+---+     |    / \
Link:   | A | A | E | E | E | F |     B   C   D
+---+---+---+---+---+---+

Spanning tree so far:          (A B) (C D) (C E)
We now have three connected components: Nodes B and A form one of them, node F is a singleton, and nodes C, D, and E are in the third component.

We add edge BE(35). That means we add a link from the representative for B (which is A) to the representative for node E (which is E):

                                      Representatives, visually:
+---+---+---+---+---+---+       E     F
Node:   | A | B | C | D | E | F |       ^
+---+---+---+---+---+---+      /|\
Link:   | E | A | E | E | E | F |     A C D
+---+---+---+---+---+---+     ^
|
B

Spanning tree so far:          (A B) (B E) (C D) (C E)
We still have two components. When we try to add the edge BC(40) to the graph, we notice that the representative for node C is the same as the representative for the node B. Therefore adding this edge would create a cycle, so we discard it.

Finally, we add the edge FD(50): after this, every node has the same representative, and therefore all nodes are connected:

                                      Representatives, visually:
+---+---+---+---+---+---+       E
Node:   | A | B | C | D | E | F |       ^
+---+---+---+---+---+---+      /|\
Link:   | E | A | E | E | E | D |     A C D
+---+---+---+---+---+---+     ^   ^
|   |
B   F

Spanning tree so far:          (A B) (B E) (C D) (C E) (D F)
##### 32.3.2Putting the union/find data structure to work

The full Kruskal’s algorithm needs a union/find data structure to handle efficiently connecting components, and also needs a list of the edges used by the algorithm:

 HashMap representatives; List edgesInTree; List worklist = all edges in graph, sorted by edge weights; initialize every node's representative to itself While(there's more than one tree) Pick the next cheapest edge of the graph: suppose it connects X and Y. If find(representatives, X) equals find(representatives, Y): discard this edge // they're already connected Else: Record this edge in edgesInTree union(representatives, find(representatives, X), find(representatives, Y)) Return the edgesInTree

To find a representative: if a node name maps to itself, then it is the representative; otherwise, “follow the links” in the representatives map, and recursively look up the representative for the current node’s parent.

There are additional heuristics for speeding this algorithm up in practice, and they make for a very efficient algorithm. Unfortunately, analyzing these heuristics is beyond the scope of this course, but you can look up the “path-compression” heuristic if you are curious.

To union two representatives, simply set the value of one representative’s representative to the other.

Do Now!

Again, why must we only ever union two representatives, and not two arbitrary nodes?


text: dc.creator WILKINSON, <NAME>AN 2007-05-09T19:37:12Z 2007-05-09T19:37:12Z 1984 https://hdl.handle.net/1911/15874 The definition of the generalized Gauss map is as follows: given an immersed manifold f : M('n) (--->) (//R)('n+k), its Gauss map is the map (gamma)(,f) : M (--->) G(,n,n+k) defined by (gamma)(p) = f(,*)(T(,p)M). As defined the Gauss map is not only invariant under all Euclidean motions but under all affine motions as well. In this thesis we study the problem of determining whether a smooth map (gamma) : M('n) (--->) G(,n,n+k) is the Gauss map for some immersed manifold in (//R)('n+k). Using the method of moving frames, certain necessary and sufficient conditions are found on (gamma) for it to locally be a Gauss map. These conditions specifically involve a tensor induced from the differential of (gamma) called the pseudo affine first fundamental form of (gamma), which is analagous to the affine first fundamental form of an immersed manifold. These conditions are then applied in the study of particular cases. In the case of curves and hypersurfaces in (//R)('N), and surfaces in (//R)('4), the generic Gauss-like map is shown to locally be a Gauss map. For the case of surfaces in (//R)('5), the generic Gauss-like maps are found to satisfy integrability conditions if and only if they are locally Gauss maps. These conditions depend on first and second order information of (gamma). Finally, a certain class of generic Gauss-like maps of codimension two manifolds are studied. For this class of maps, sufficient conditions are found for such analytic maps to be local Gauss maps. These conditions depend on first order information of (gamma). application/pdf eng Mathematics CHARACTERIZING GAUSS MAPS Thesis Text Mathematics Natural Sciences Rice University Doctoral Doctor of Philosophy <NAME>. "CHARACTERIZING GAUSS MAPS." (1984) Diss., Rice University. https://hdl.handle.net/1911/15874.

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## cupcake_queen p+9/5=q, for p one year ago one year ago

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1. jim_thompson5910

Multipliy both sides by 5 Then subtract 9 from both sides $\Large \frac{p+9}{5} = q$ $\Large p+9 = 5q$ $\Large p = 5q - 9$


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1

The source of the magnetic field is the current density $\vec J$. Randomly moving charges have $\vec J =0$ and therefore produce no magnetic field.

0

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0

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2

Yes, a changing magnetic flux will induce a current in a complete conducting loop. If there is a gap in the loop, then current cannot flow and the emf will appear as a voltage difference across the gap. Without a gap, the emf is dissipated by the current in the resistance of the loop and there will be no voltage difference between any two points in a ...

2

Of course yes. Since you know about the Faraday's law, then you may know about the Lenz's law. It says that the polarity of the induced emf is such that the current produced by it, gives rise to magnetic field which opposes the external time varying magnetic field. The emf is the one that is "induced" and it drives the current and not the other way ...

0

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1

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The field (and flux) produced by a bar magnet is always in the same direction along the axis of the bar magnet.

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The 2nd term above, $v\times B\cdot dl$, will be zero in a current-carrying wire since the drift velocity is parallel to the wire segment $dl$. The line integral being non-zero around a loop must be the non-conservative induced electric field, the electrostatic field due to any charges, will have a line integral equal to zero.

0

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They do not care whether your circuit is simple series, parallel, or some complicated circuit. You can use them wherever you want to, but of course, it's useless to use them in a place where it's simple to ... 0 The mistake you are making is similar to if someone concluded that mass (or inertia) is required for something to experience acceleration and motion. It isn't. Similarly, inductance is not required for the production of a magnetic field. It might just seem that way if you are reading about magnetic machines at a high level that does not address the lower ... 1 I would personally define a “steady current” as one that obeys \frac\partial{\partial t} \mathbf J = 0 everywhere. The requirement that the charge density in “magnetostatics” not change with time, \frac\partial{\partial t} \rho = 0, allows the student to use the tools developed during electrostatics to figure out what the electric fields do. 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In the battery there are other forces than macroscopic EM forces (microscopic forces due to chemical reactions) that push the mobile charges against the electrostatic field inside the ... 0 Electrostatic field means that the fields are static. But that is compatible with a static flow of current. According to the Ampére-Maxwell law, that implies a static magnetic field. And a static magnetic field, according to the Maxwell-Faraday law implies that \nabla \times \mathbf E = 0 1 If ρ is linear and zero at the surface, then, ρ =  ρ_o(R – r), where, R, is the radius at the outer surface and,  ρ_o, is your unknown. 0 Search for the patent for an electromagnetic weapon in a 500lb dumb bomb form factor. It will give some insight but not performance parameters. The effectiveness of the weapon is not intrinsic to the weapon. It also depends on the vulnerability of the target. EU technical commissions have been working on IEC performance requirements and measurement methods ... 0 If the magnetic force is not cancelled by the electric force there is no equilibrium, so the charges will move until there is. 3 Your calculation of the gauge covariant derivative is incorrect. The \nabla operator changes depending on what it acts upon, just like the covariant derivative in GR. In general a gauge covariant derivative changes based on the representation under which the object it acts upon transforms. This is actually the same as for the covariant derivative in GR as ... 3 Physics does not have an explanation why electromagnetism exists, but it does describe how it works. One could argue that parallel currents attract because of the Lorentz force and Ampère's law. Conversely, these laws are what they are because parallel currents attract. In the end the laws are as they are to explain, among others, this observation. 1 Faraday might have used an argument like this one... The magnetic fields due to two wires carrying currents in the same direction tend to cancel between the two wires, but re-inforce beyond the two wires (applying right hand grip rule). Magnetic field lines behave as if under tension but exerting lateral pressure, so if you draw the pattern you will see that ... 1 It's certainly possible for there to be both a surface charge present, and a current flowing through a conductor. The surface charge might or might not be the charge that is moving to form the current, though. For example, any time current flows in a wire that also has a DC voltage relative to another nearby conductor, this will be the case. 7 I suspect the reason why you're getting seemingly contradictory answers is that you're asking an "is it A or B?" question where the real answer is "no, it's neither of those things." On one hand, no, an electromagnetic wave is not a "pure mathematical line" — it's a fluctuation in the electromagnetic field, which permeates all ... 4 If I interpret your question correctly, you're asking if the E and B fields constitute some sort of spatial displacement (of a medium in which the wave travels, as with water waves or elastic waves in a solid), on the grounds that they are considered to point in a definite spatial direction. The answer is 'no' - they are better considered to be gradients in ... 2 Update I rephrased a few things and added some additional details after chatting with the OP. Question 1: How can I formally prove this conservative electric field to be \mathbf{E} = -\mathbf{v} \times \mathbf{B} other than with dynamics arguments? The rest of this question isn't really necessary, so I only put in the first part. The electric field ... 0 While this statement is true, it should be stressed that magnetic force does not do work IN REFERENCE OF THE MAGNETIC FIELD. If viewed in other frame of reference( often when the field is moving relative to you), it may do work. 4 A conductor must consist of a union of surfaces and bulk. Therefore, a conducting sheet must have two sides separated by a neutral bulk(However thin that might be). Therefore, within a given volume of an insulating sheet and a conducting sheet, the latter has twice the amount of charge and produces twice the electric field twice. 1 What's particularly bothering me is the equation \nabla \times \mathbf{E}=-\dfrac{\partial \mathbf{B}}{\partial t}. If we were to put it in the integral form, (using Stokes' theorem), there seems to be no reference to V or B, which seems to be incorrect as per definition (1.2).$$ \oint \vec{E} \cdot \vec{dl} = \int_{S} -(\frac{\partial }{\partial t}...

26

In the image included in your question, one should not view the arrows as literally extending through space or having a spatial length associated to them. The length of these arrows is the magnitude of the field, and they should each be thought of as living inside of their own vector space which is "internal" or "attached" to a given ...

17

While electromagnetic waves do have lateral extension, since the electric and magnetic fields are defined on all of $\mathbb{R}^{3+1}$, most graphical representations, including the one you have included, do not show how an electromagnetic wave extends in all directions. The reason for this is quite simple: representing the entirety of an electromagnetic ...

1

Suppose our object is described by the function $g(x,y,z=0)$. We can write this function as a superposition of plane waves (Fourier decomposition), $$g_{object}(x,y,z=0) = \iint df_x df_y \;G(f_x, f_y,z=0)e^{-2\pi i (f_x x + f_y y)}$$ where $f_x$ and $f_y$ are the spatial frequencies. Hence, the object consists of many different spatial frequencies. Now, ...

6

The plane wave depicted in the image is invariant under translations in $x$ and $y$. It fills all of space.

12

Yes. Often these can be considered plane waves: The vectors represent the electric field of the EM wave. See the description on Commons by Dave3457. Wavefronts can also be spherical, etc.

1

I think you are seeing the problem wrong, the curl of E its not zero, the text, just say that the $z$ component of $\nabla\times \vec{E}$ is zero that means that there is not magnetic field on the direction of the conductor but there exist magnetif field in other dimesions, for example the $y$ component of $\nabla\times \vec{E}$ is $\frac{\partial E_x}{\... 1 This really just comes down to the definition of A. We define A through$\boldsymbol \nabla \times \textbf{A} = \textbf{B}$. Then, using the Maxwell equation$\boldsymbol \nabla \times \textbf{E} = - \frac{\partial\textbf{B}}{\partial t}$we obtain $$\boldsymbol \nabla \times \textbf{E} = - \frac{\partial}{\partial t} \boldsymbol \nabla \times \textbf{A} = -\... 0 The first obvious answer is yes, of course. How could it be otherwise? Momentum is conserved. But it depends on how you divide up the world. Charged particle A is getting accelerated by something. That doesn't just happen for no reason. Charged particle B interacts both with charged particle A and also with whatever is accelerating A. It's the whole system ... 0 Using the integral form of the Ampére-Maxwell law without the current, that is zero across the plates:$$\oint_{\partial \Sigma} \mathbf{B} \cdot \mathrm{d}\boldsymbol{\ell} = \mu_0 \varepsilon_0 \frac{\mathrm{d}}{\mathrm{d}t} \iint_{\Sigma} \mathbf{E} \cdot \mathrm{d}\mathbf{S} $$we can see that around the borders of the plates, the LHS is maximum, ... 0 Assuming that the approximation that the electric field is constant in space but changes with time is reasonable, then since \vec j = 0 between the plates, a possible solution for the \vec \nabla \times \vec B equation which also satisfies \vec \nabla \cdot \vec B = 0 is \vec B = \frac{1}{2c^2} \vec r \times \frac{\partial \vec E}{\partial t}. To see ... 2 In short: There is a lower limit on the spectral width \Delta \omega of unpolarized light. Therefore, strictly monochromatic unpolarized light can not exist, not even mathematically, while spectrally broad unpolarized and polarized light exists. The following holds for classical light which has a well-defined electric field at each point in spacetime. An ... 0 If we start from the charge-current continuity equation for any arbitrary charge distribution, we have:$$ \partial_{t} \rho + \nabla \cdot \mathbf{j} = 0 \tag{0}$$where$\rho$is the charge density,$\mathbf{j}$is the current density (specifically the macroscopic average current density, see pages 248--258 in Jackson [1999] for definition and derivation),... 1 First of all, let me just point out that the radius of the$U(1)$is related to the gauge coupling, so rescaling the radius is essentially the same as rescaling the gauge coupling of the theory. The only place I've seen this pointed out is in references dealing with$S$-duality in electrodynamics since part of the transformation is on the gauge coupling, and ... -1 The answer by Wolpertinger is wrong. Reasoning does not help, so I will use the argument by authority and quote from <NAME>, Waves (Berkeley Physics Course volume 3), 1968. Unpolarized radiation defined. We are now prepared to say what is meant by "unpolarized" light. Unpolarized light is light whose two polarization components ($x\$ and ...

Top 50 recent answers are included


text:
This page captures plans and ideas for improving Metagraph. These items are not in any particular order, and some items on this list may not happen. Feel free to open issues in the issue tracker if you are interested in discussing the details of one of these ideas.

## Project Structure¶

• Split packaging of metagraph into a minimal core package with abstract types and algorithms, and put the basic plugins into separate packages. The metagraph package then becomes a metapackage that installs core and a generally useful set of default plugins.

• The multi-dispatch and automatic translation system in Metagraph is not graph-specific and could be split out into a separate package.

## General Features¶

• Lazy execution of algorithms by returning a Dask task graph from algorithm calls.

• Global preference setting plugin priority. This would be used to more easily switch a workflow from one backend to another without having to modify the calculation itself.

• Allow an abstract algorithm to have a default implementation that only calls other abstract algorithms via the current Resolver.

## Type System and Dispatcher¶

• Nested namespace of properties and validators, that allow common subgroups to be used in multiple types.

• Change unambigous_subcomponents to allowed_translations
• Include which property values can translate to another values

• Clarify what self-translators need to do

• Allow concrete algorithms to specialize on more than the abstract type signature, such as literal values of non-translatable (i.e. “Python”) types.


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# Bounds on trig integrals converted to the complex plane

I am having trouble changing variables and converting real trig integrals into the complex plane. I know these integrals need to be converted to contour integrals along the circle $$|z|=1$$ but I am having trouble changing the bound if the integrals are not already over the interval $$[0,2 \pi]$$.

For instance the book I am using, <NAME> says if the given integral is: $$\int_{0}^{\pi}\frac{8}{5+2 \cos \theta} d \theta$$

The book states since $$\cos(\theta)=\cos (\theta -2 \pi)$$ This implies $$\int_{0}^{\pi}\frac{8}{5+2 \cos \theta} d \theta=\frac{1}{2}\int_{0}^{2\pi}\frac{8}{5+2 \cos \theta} d \theta$$

I know this has something to do with the evenness and periodicity of the cosine function. Can someone explain this to me?

Also another problem in the book is:

$$\int_{- \pi}^{\pi}\frac{1}{1+\sin^{2} \theta} d \theta$$ and this integral is equal to $$\int_{0}^{2\pi}\frac{1}{1+\sin^{2} \theta} d \theta$$

I have also seen the integral: $$\int_{0}^{\frac{\pi}{2}}\frac{\cos 2 \theta}{1+2\cos^{2} \theta} d \theta$$

which equals $$\frac{1}{4} \int_{0}^{2 \pi}\frac{\cos 2 \theta}{1+2\cos^{2} \theta} d \theta$$

I cannot figure out how to easily switch the bounds of the given integral to the interval $$[0,2 \pi]$$ is there any method to easily do this?Can help explain how to do this easily and for any given interval?

For $$\int_0^{\pi} \frac{8}{5+2\cos \theta} d\theta$$ note that that the only variable function is $$\cos \theta$$. Draw the line $$x=\pi$$ on the graph of $$\cos x$$, to notice that it has reflectional symmetry about it. In other words, $$\cos(\pi + x) = \cos(\pi-x)$$. This means the area under the graph in $$[0,\pi]$$ equals that in $$[\pi, 2\pi]$$ and the result follows.
Notice that $$\frac{1}{1+ \sin^2x}$$ is even and so $$\int_{-\pi}^{\pi} \frac{1}{1+\sin^2x}=2 \int_0^{\pi} \frac{1}{1+\sin^2x}$$ Again, by the same reflectional symmetry at $$x=\pi$$ if we expand the interval to $$[0,2\pi]$$, we would have to divide by $$2$$ to preserve the area.
For the last one, you can use a trig identity and the integral becomes$$\int_0^{\frac{\pi}{2}} \frac{\cos 2\theta}{2+\cos 2\theta} d\theta$$
This time, $$\cos 2\theta$$ has reflectional symmetry along $$x=\frac{\pi}{2}$$ and $$x=\frac{3\pi}{2}$$, leading to the areas under the graph in $$[0,\frac{\pi}{2}],\, [\frac{\pi}{2}, \pi],\, [\pi, \frac{3\pi}{2}],\, [\frac{3\pi}{2}, 2\pi]$$ all being equal.


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# How to dissolve chalk (CaCO3)?

I've read that vinegar can dissolve $$\ce{CaCO3}$$, I know that extreme heat can decompose it, but how to dissolve $$\ce{CaCO3}$$ at home? I need to dissolve chalk in my home because of repairing something. Can someone help me do it without heating $$\ce{CaCO3}$$? Any replacement for that?

• As you noted, how about vinegar? That is a fairly common household item. – <NAME> Mar 14 '17 at 12:52
• as I know the reaction process (CaCO3 + Vinegar) is very slow, do you know another thing which can make the reaction faster? Can heating Vinegar help speed up the reaction? As I know heating can affect reactions, but I'm not sure about this reaction... – <NAME> Mar 14 '17 at 12:57
• If you want to repair something with $\ce{CaCO3}$, than it sounds like using glazing putty; especially near to wood. Vinegar however is an acid that will start to decompose the carbonate in chalk, yielding CO2. Instead I recommend either a mixture of (finely ground $\ce{CaCO3}$ and water) -- like a paste, or lineseed oil. – Buttonwood Mar 14 '17 at 13:23
• There are products on the market for defurring the inside of kettles, you need one of these. They mostly contain sulfamic acid – Waylander Mar 14 '17 at 13:28
• Chalk is not noted for being particularly strong - scrape the bulk of it off and finish with some vinegar on a sponge or cloth. – <NAME> Mar 14 '17 at 13:36

$$\ce{CaCO3}$$ is sedimenting when mixed with water, and therefore is not easy to remove with hot water. Nevertheless several methods exist that actually dissolve it that are better than vinegar:

• You could add hydrochloric acid again to convert it into soluble Calcium chloride, according to equilibrium: $$\ce{CaCO3 (s) + 2 HCl (aq) <=> CaCl2 (aq) + H2CO3(aq)}$$
• You could use carbonic acid instead $$\ce{CaCO3 (s) + H2CO3 <=> Ca(HCO3)2}$$ this is the reason shells are getting dissolved by an increase of the amount of carbon dioxyde in solution: $$\ce{CaCO3 (s) + CO2 (aq) + H2O <=> Ca(HCO3)2}$$
• But I recommend Sulfamic acid ($$\ce{H3NSO3}$$), is frequently used for removing rust and limescale, replacing the more volatile and irritating hydrochloric acid, it may be a little more expensive but is certainly not toxic.
• Sulfamic acid is an eye irritant, however. – <NAME> Mar 29 at 14:07
• Lots of things are eye irritants, including orange oil, so we have to live with that with any chemical method. – <NAME> Apr 26 at 16:40

Calcium carbonate will dissolve quickly in a strong acid, like hydrochloric, even if it is dilute and therefore not so dangerous.

An organic acid has a problem in that it may have a relatively low pH at first, but after it dissolves a bit of the $$\ce{CaCO3}$$, a significant amount of the acid anion is formed in the solution, which stifles the ionization of the acid, so the rate of dissolution slows down. So, vinegar is OK, but you need to use a lot.

Perhaps the problem is with the other materials, which may be binders (as Buttonwood suggested). Perhaps an organic solvent could dissolve the binder and thus remove the chalk.

However, it may be that the "chalk" is not $$\ce{CaCO3}$$, but rather a prepared mixture with clay or gypsum (https://en.wikipedia.org/wiki/Chalk). In that case, trying to dissolve traces remaining from your project will not be easy. Try to dissolve some of your "chalk" starting material - if it can't be dissolved easily, you may be directed toward mechanical methods.

Basically $\ce{CaCO3}$ is not soluble in water but it can disperse in water by using a dispersing agent. The solution looks like milk . The viscosity of solution up to solid content or quantity of $\ce{CaCO3}$ in solution. That is my opinion.


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# Difference between Ljung Box and McLeod Li Test?

I'm a little confused. What is the difference between the Ljung-Box test and the McLeod-Li test?

The Ljung-Box test is a test for linearity in the mean and the McLeod-Li is a test for nonlinearity on the mean and/or in variance?

Ljung-Box test is a test for autocorrelation in either raw data or model residuals. The test is described quite clearly in a Wikipedia article. I will add that alternatives to Ljung-Box test exist and in some situations are preferred; see this thread comparing Ljung-Box test with Breusch-Godfrey test.

McLeod-Li test is a test for autoregressive conditional heteroskedasticity in either raw data or residuals from a conditional mean model (but not for residuals from a GARCH model; there Li-Mak test should be used instead).

The similarity between the two tests is that McLeod-Li test takes the form of a Ljung-Box test applied on squared raw data or squared model residuals. Despite the similarity in their shapes, the purposes of the tests are different, as explained above.

Also, what you state in your second paragraph is incorrect.

"Li-Mak test should be used instead". I'm doing a Time Series Financial Course and we won't see this test. Is [this/there] a better test?

When applied on standardized residuals from a GARCH model, McLeod-Li test does not have the usual null distribution, thus you do not have a benchmark to compare the test statistic to. Consequently, you cannot obtain a p-value or tell whether the test statistic is above or below a 5% (or whatever %) critical value. Meanwhile, Li-Mak test is especially suited for this situation. See <NAME>. and <NAME>. (1994) "On the squared residual autocorrelations in non-linear time series with conditional heteroscedasticity. Journal of Time Series Analysis 15, 627–36.

• Thanks Richards! Interesting your observation: "Li-Mak test should be used instead". Im doing a Time Series Financial Course and we wont see this test. Is a better test? – digsydinner Oct 2 '15 at 19:09
• When applied on standardized residuals from a GARCH model, McLeod-Li test does not have the usual null distribution, thus you do not have a benchmark to compare the test statistic to, and consequently you cannot obtain a $p$-value or tell whether the test statistic is above or below a 5% (or whatever %) critical value. Meanwhile, Li-Mak test is especially suited for this situation. See <NAME>. and <NAME>. (1994) On the squared residual autocorrelations in non-linear time series with conditional heteroscedasticity. Journal of Time Series Analysis 15, 627–36. – <NAME> Oct 2 '15 at 19:23


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# Calculate the limit $\lim \limits_{n \to \infty} |\sin(\pi \sqrt{n^2+n+1})|$

Calculate

$$\lim \limits_{n \to \infty} |\sin(\pi \sqrt{n^2+n+1})|$$

-
Is $n$ here required to be an integer, or a real number? –  MJD Apr 2 '13 at 22:10
As MJD hints, if n is a real number, then your limit doesn't exist, since you're bouncing between 0 and 1. –  <NAME> Apr 2 '13 at 22:14
The answers all take $n$ to be restricted to the integers, their answers would equally apply if $n$ was restricted to the naturals as well, right? –  user66807 Apr 23 '13 at 0:20

Note that $$\sqrt{n^2+n+1}-n=\frac{n+1}{\sqrt{n^2+n+1}+n}\to 1/2$$ as $n\to\infty$.

For even $n$, $\sin(\sqrt{n^2+n+1}\pi)=\sin(\sqrt{n^2+n+1}\pi-n\pi)\to \sin(\pi/2)=1$ as $n\to\infty, n$ even.

For odd $n$, $\sin(\sqrt{n^2+n+1}\pi)=-\sin(\sqrt{n^2+n+1}\pi-n\pi)\to -\sin(\pi/2)=-1$ as $n\to\infty, n$ odd.

Therefore, $$|\sin(\sqrt{n^2+n+1}\pi)|\to 1$$ as $n\to\infty$.

-

Edit: This assumes that $n$ is restricted to the integers.

Firstly, since $\mid \sin \theta \mid$ is periodic with period $\pi$, we want to look at the value of $\pi \sqrt{n^2 + n +1} \pmod{\pi}$.

Secondly, convince yourself that $\lim_{n\rightarrow \infty} \pi \sqrt{n^2+n+1} \pmod{\pi} = \frac {1}{2} \pi$.

Thirdly, by the continuity of $\sin \theta$, $\mid \lim_{n \rightarrow \infty} \sin (\pi \sqrt{n^2+n+1} ) \mid = \mid \sin \lim_{n\rightarrow \infty} \pi \sqrt{n^2+n+1}\mid = \sin \frac {1}{2} \pi$.

-
Can you help us convince ourselves? –  levitopher Apr 2 '13 at 22:16
So, it's oscillating around $\pm1$, no? –  Berci Apr 2 '13 at 22:17
Because of the multiplying by $n$, to the front of the square root. You should be approximating your square root by $1 + \frac {1}{2n}$ instead. Moreover, note that $lim_{n\rightarrow \infty} n f(n) \neq n \lim_{n \rightarrow \infty} f(n)$. –  <NAME> Apr 2 '13 at 22:24

The function $|\sin x|$ is periodic with period $\pi$. Hence

$$\lim \limits_{n \to \infty}|\sin \pi \sqrt{n^2+n+1}|= \lim \limits_{n \to \infty}|\sin \pi (\sqrt{n^2+n+1}-n)|$$

But

$$\lim \limits_{n \to \infty} (\sqrt{n^2+n+1}-n)=\lim \limits_{n \to \infty} \dfrac{n^2+n+1-n^2}{\sqrt{n^2+n+1}+n}=\dfrac{1}{2}$$

It follows that the limit we are computing is equal to $\left |\sin \dfrac{\pi}{2}\right |$, which is 1.

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Yes, that's a neat elementary way. But don't avoid Taylor too long...One day you'll have no choice! –  1015 Apr 2 '13 at 22:35

If you know a little bit about Taylor expansions and big O notation, you can do that easily. We will use $$\sqrt{1+u}=1+\frac{u}{2}+O(u^2)\qquad \mbox{when } u\longrightarrow 0.$$ Now $$\pi\sqrt{n^2+n+1}=\pi n\sqrt{1+\frac{1}{n}+\frac{1}{n^2}}=\pi n\left(1+\frac{1}{2n}+O\left(\frac{1}{n^2} \right)\right)=\pi n+\frac{\pi}{2}+O\left(\frac{1}{n} \right)$$ So $$\sin (\pi\sqrt{n^2+n+1})=\sin\left(\pi n+\frac{\pi}{2}+O\left(\frac{1}{n} \right) \right)=(-1)^n\sin\left(\frac{\pi}{2}+O\left(\frac{1}{n} \right) \right).$$ It follows that $$|\sin (\pi\sqrt{n^2+n+1})|=\left| \sin\left(\frac{\pi}{2}+O\left(\frac{1}{n} \right) \right)\right|\longrightarrow \left| \sin\left(\frac{\pi}{2}\right)\right|=1.$$

-

We have $$\sqrt{n^2+n+1}=n\sqrt{1+\frac{1}{n}+\frac{1}{n^2}}=n\left(1+\frac{1}{2n}+o(\frac{1}{n})\right)=n+\frac{1}{2}+o(1)$$ hence $$\lim \limits_{n \to \infty} |\sin(\pi \sqrt{n^2+n+1})|=\lim \limits_{n \to \infty}|\sin(n\pi+\frac{\pi}{2}+o(1))|=\lim \limits_{n \to \infty}|\cos(o(1))|=1.$$

-


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# Can the RSA accumulator scheme be converted to Elliptic Curve math?

Is it possible to translate the RSA accumulator scheme directly to EC without requiring bilinear pairings?

In RSA we have:

$$A_{n+1} = A_n^c$$ st. $$\{c \: \textrm{prime} \: | \: c \in [\mathcal{A}, \mathcal{B}]\}$$

$$W = A_n$$

$$A_{n+1} \stackrel{?}{=} W^c$$

Would this work in EC like this?

$$A_{n+1} = c A_n$$

$$W = A_n$$

$$A_{n+1} \stackrel{?}{=} c W$$

Where $$A_0$$ is a generator point on the curve and $$c$$ is prime.

If not, then why?


text:
Multiplication and Division
Temperature
Geometry
Area and Perimeter
Random

17 x 9
17 x 9 = 153

### 100

What is the change in temperature from 5°F to 45°F?
40°F

### 100

What is a polygon with 5 sides called?
a pentagon

### 100

A rectangle has a length of 5m and width of 10m. What is the perimeter?
5m + 5m + 10m + 10m = 30m

### 100

n x 9 = 18. Solve for n.
n = 2 because 2 x 9 = 18

112 ÷ 8
112 ÷ 8 = 14

### 200

What is the change in temperature from 8°C to 23°C?
15°C

### 200

What is an angle with a 90° measure called?
a right angle

### 200

What is the perimeter of a square with 6in sides?
6in + 6in + 6in + 6in = 24in

### 200

$61.50 +$ 39.78
$61.50 +$ 39.78 = \$101.28

53 x 6
53 x 6 = 318

### 300

What is boiling point of water in °F?
212°F

### 300

What is an angle with 180° measure called?
a straight angle

### 300

What is the area of a square with 6 in sides?
36 sq. in

### 300

7 decades, 5 years = ____ years
75 years because 7 decades + 5 years is 70 years + 5 years

407 ÷ 2
407 ÷ 2 = 203 r1

### 400

What is the change in temperature from 15°F to -10°F?
25°F (NOT -25°F)

### 400

What is a polygon called if its sides are all the same length and its angles are all the same measure?
a regular polygon

### 400

If the area of a rectangle is 36 sq. mi, and the length is 9 mi, what is the measure of the width?
4mi because 36 ÷ 9 = 4

### 400

Order the numbers from least to greatest: 875, 857, 957, 807
807, 857, 875, 957

47 x 56
47 x 56 = 2,632

### 500

What is the freezing point of water in °C?
0°C

### 500

What are two lines called if intersect and make 4 right angles?
perpendicular lines

### 500

Mr. Phil's closet is 3 feet long and 5 feet wide. What is the area of his closet? If he doubles the length and the width of the closet, what will be the new area?
The area is 15 sq. ft. If the length and width are doubled, the new area will be 60 sq. ft.

### 500

What is one-half plus one-half?
one, or one whole


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<meta http-equiv="refresh" content="1; url=/nojavascript/">
You are reading an older version of this FlexBook® textbook: CK-12 Geometry - Second Edition Go to the latest version.

# 5.4: Medians and Altitudes in Triangles

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Define median and find their point of concurrency in a triangle.
• Apply medians to the coordinate plane.
• Construct the altitude of a triangle and find their point of concurrency in a triangle.

## Review Queue

1. Find the midpoint between (9, -1) and (1, 15).
2. Find the equation of the line between the two points from #1.
3. Find the equation of the line that is perpendicular to the line from #2 through (-6, 2).

Know What? Triangles are frequently used in art. Your art teacher assigns an art project involving triangles. You decide to make a series of hanging triangles of all different sizes from one long piece of wire. Where should you hang the triangles from so that they balance horizontally?

You decide to plot one triangle on the coordinate plane to find the location of this point. The coordinates of the vertices are (0, 0), (6, 12) and (18, 0). What is the coordinate of this point?

## Medians

Median: The line segment that joins a vertex and the midpoint of the opposite side (of a triangle).

Example 1: Draw the median $\overline{LO}$ for $\triangle LMN$ below.

Solution: From the definition, we need to locate the midpoint of $\overline{NM}$. We were told that the median is $\overline{LO}$, which means that it will connect the vertex $L$ and the midpoint of $\overline{NM}$, to be labeled $O$. Measure $NM$ and make a point halfway between $N$ and $M$. Then, connect $O$ to $L$.

Notice that a median is very different from a perpendicular bisector or an angle bisector. A perpendicular bisector also goes through the midpoint, but it does not necessarily go through the vertex of the opposite side. And, unlike an angle bisector, a median does not necessarily bisect the angle.

Example 2: Find the other two medians of $\triangle LMN$.

Solution: Repeat the process from Example 1 for sides $\overline{LN}$ and $\overline{LM}$. Be sure to always include the appropriate tick marks to indicate midpoints.

Example 3: Find the equation of the median from $B$ to the midpoint of $\overline{AC}$ for the triangle in the $x-y$ plane below.

Solution: To find the equation of the median, first we need to find the midpoint of $\overline{AC}$, using the Midpoint Formula.

$\left(\frac{-6+6}{2}, \frac{-4+(-4)}{2}\right)=\left(\frac{0}{2}, \frac{-8}{2}\right)=(0,-4)$

Now, we have two points that make a line, $B$ and the midpoint. Find the slope and $y-$intercept.

$m &= \frac{-4-4}{0-(-2)}=\frac{-8}{2}=-4\\y &= -4x+b\\-4 &= -4(0)+b\\-4 &= b$

The equation of the median is $y=-4x-4$

## Point of Concurrency for Medians

From Example 2, we saw that the three medians of a triangle intersect at one point, just like the perpendicular bisectors and angle bisectors. This point is called the centroid.

Centroid: The point of concurrency for the medians of a triangle.

Unlike the circumcenter and incenter, the centroid does not have anything to do with circles. It has a different property.

Investigation 5-3: Properties of the Centroid

Tools Needed: pencil, paper, ruler, compass

1. Construct a scalene triangle with sides of length 6 cm, 10 cm, and 12 cm (Investigation 4-2). Use the ruler to measure each side and mark the midpoint.

2. Draw in the medians and mark the centroid.

Measure the length of each median. Then, measure the length from each vertex to the centroid and from the centroid to the midpoint. Do you notice anything?

3. Cut out the triangle. Place the centroid on either the tip of the pencil or the pointer of the compass. What happens?

From this investigation, we have discovered the properties of the centroid. They are summarized below.

Concurrency of Medians Theorem: The medians of a triangle intersect in a point that is two-thirds of the distance from the vertices to the midpoint of the opposite side. The centroid is also the “balancing point” of a triangle.

If $G$ is the centroid, then we can conclude:

$AG &= \frac{2}{3} AD, CG=\frac{2}{3} CF, EG=\frac{2}{3} BE\\DG &= \frac{1}{3} AD, FG=\frac{1}{3} CF, BG=\frac{1}{3} BE$

And, combining these equations, we can also conclude:

$DG=\frac{1}{2} AG, FG=\frac{1}{2} CG, BG=\frac{1}{2} EG$

In addition to these ratios, $G$ is also the balance point of $\triangle ACE$. This means that the triangle will balance when placed on a pencil (#3 in Investigation 5-3) at this point.

Example 4: $I, K$, and $M$ are midpoints of the sides of $\triangle HJL$.

a) If $JM = 18$, find $JN$ and $NM$.

b) If $HN = 14$, find $NK$ and $HK$.

Solution:

a) $JN$ is two-thirds of $JM$. So, $JN =\frac{2}{3} \cdot 18=12$. $NM$ is either half of 12, a third of 18 or $18 - 12$. $NM = 6$.

b) $HN$ is two-thirds of $HK$. So, $14 =\frac{2}{3} \cdot HK$ and $HK = 14 \cdot \frac{3}{2} = 21$. $NK$ is a third of 21, half of 14, or $21-14$. $NK = 7$.

Example 5: Algebra Connection $H$ is the centroid of $\triangle ABC$ and $DC = 5y - 16$. Find $x$ and $y$.

Solution: $HF$ is half of $BH$. Use this information to solve for $x$. For $y$, $HC$ is two-thirds of $DC$. Set up an equation for both.

$\frac{1}{2} BH &= HF \ \text{or} \ BH=2HF && HC=\frac{2}{3} DC \ \text{or} \ \frac{3}{2} HC=DC\\3x+6 &= 2(2x-1) && \frac{3}{2} (2y+8)=5y-16\\ 3x+6 &= 4x-2 && \quad 3y+12=5y-16\\8 &= x && \qquad \quad \ 28=2y$

## Altitudes

The last line segment within a triangle is an altitude. It is also called the height of a triangle.

Altitude: A line segment from a vertex and perpendicular to the opposite side.

Here are a few examples.

As you can see, an altitude can be a side of a triangle or outside of the triangle. When a triangle is a right triangle, the altitude, or height, is the leg. If the triangle is obtuse, then the altitude will be outside of the triangle. To construct an altitude, use Investigation 3-2 (constructing a perpendicular line through a point not on the given line). Think of the vertex as the point and the given line as the opposite side.

Investigation 5-4: Constructing an Altitude for an Obtuse Triangle

Tools Needed: pencil, paper, compass, ruler

1. Draw an obtuse triangle. Label it $\triangle ABC$, like the picture to the right. Extend side $\overline{AC}$, beyond point $A$.
2. Using Investigation 3-2, construct a perpendicular line to $\overline{AC}$, through $B$.

The altitude does not have to extend past side $\overline{AC}$, as it does in the picture. Technically the height is only the vertical distance from the highest vertex to the opposite side.

As was true with perpendicular bisectors, angle bisectors, and medians,the altitudes of a triangle are also concurrent. Unlike the other three, the point does not have any special properties.

Orthocenter: The point of concurrency for the altitudes of triangle.

Here is what the orthocenter looks like for the three triangles. It has three different locations, much like the perpendicular bisectors.

Acute Triangle Right Triangle Obtuse Triangle
The orthocenter is inside the triangle. The legs of the triangle are two of the altitudes. The orthocenter is the vertex of the right angle. The orthocenter is outside the triangle.

Know What? Revisited The point that you should put the wire through is the centroid. That way, each triangle will balance on the wire.

The triangle that we wanted to plot on the $x-y$ plane is to the right. Drawing all the medians, it looks like the centroid is (8, 4). To verify this, you could find the equation of two medians and set them equal to each other and solve for $x$. Two equations are $y=\frac{1}{2} x$ and $y=-4x+36$. Setting them equal to each other, we find that $x = 8$ and then $y = 4$.

## Review Questions

Construction Construct the centroid for the following triangles by tracing each triangle onto a piece of paper and using Investigation 5-3.

1. Is the centroid always going to be inside of the triangle? Why?

Construction Construct the orthocenter for the following triangles by tracing each triangle onto a piece of paper and using Investigations 3-2 and 5-4.

1. What do you think will happen if the triangle is equilateral? What can we say about the incenter, circumcenter, centroid, and orthocenter? Why do you think this is?
2. How many lines do you actually have to “construct” to find any point of concurrency?

For questions 10-13, find the equation of each median, from vertex $A$ to the opposite side, $\overline{BC}$.

1. $A(9, 5), B(2, 5), C(4,1)$
2. $A(-2, 3), B(-3, -7), C(5, -5)$
3. $A(-1, 5), B(0, -1), C(6, 3)$
4. $A(6, -3), B(-5, -4), C(-1, -8)$

For questions 14-18, $B, D$, and $F$ are the midpoints of each side and $G$ is the centroid. Find the following lengths.

1. If $BG = 5$, find $GE$ and $BE$
2. If $CG = 16$, find $GF$ and $CF$
3. If $AD = 30$, find $AG$ and $GD$
4. If $GF = x$, find $GC$ and $CF$
5. If $AG = 9x$ and $GD = 5x - 1$, find $x$ and $AD$.

Write a two-column proof.

1. Given: $\triangle ABC \cong \triangle DEF$ ${\;}$ $\overline{AP}$ and $\overline{DO}$ are altitudes Prove: $\overline{AP} \cong \overline{DO}$
2. Given: Isosceles $\triangle ABC$ with legs $\overline{AB}$ and $\overline{AC}$ $\overline{BD} \bot \overline{DC}$ and $\overline{CE} \bot \overline{BE}$ Prove: $\overline{BD} \cong \overline{CE}$

Use $\triangle ABC$ with $A(-2, 9), B(6, 1)$ and $C(-4, -7)$ for questions 21-26.

1. Find the midpoint of $\overline{AB}$ and label it $M$.
2. Write the equation of $\overleftrightarrow{CM}$.
3. Find the midpoint of $\overline{BC}$ and label it $N$.
4. Write the equation of $\overleftrightarrow{AN}$.
5. Find the intersection of $\overleftrightarrow{CM}$ and $\overleftrightarrow{AN}$.
6. What is this point called?

Another way to find the centroid of a triangle in the coordinate plane is to find the midpoint of one side and then find the point two thirds of the way from the third vertex to this point. To find the point two thirds of the way from point $A(x_1, y_1)$ to $B(x_2, y_2)$ use the formula: $\left(\frac{x_1+2x_2}{3}, \frac{y_1+2y_2}{3}\right)$. Use this method to find the centroid in the following problems.

1. (-1, 3), (5, -2) and (-1, -4)
2. (1, -2), (-5, 4) and (7, 7)
3. Use the coordinates $(x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$ and the method used in the last two problems to find a formula for the centroid of a triangle in the coordinate plane.
4. Use your formula from problem 29 to find the centroid of the triangle with vertices (2, -7), (-5, 1) and (6, -9).

1. $midpoint=\left (\frac{9+1}{2},\frac{-1+15}{2}\right ) = (5,7)$
2. $m=\frac{15+1}{1-9}=\frac{16}{-8}=-2 \qquad \qquad 15=-2(1)+b \qquad \qquad y=-2x+17\!\\{\;}\qquad \qquad \qquad \qquad \qquad \qquad \ \ 17=b$
3. $y=\frac{1}{2} x+b\!\\2= \frac{1}{2} (-6) + b\!\\2=-3+b\!\\5=b\!\\y=\frac{1}{2} x+5$

Feb 23, 2012

Feb 16, 2015


text:
# Reproducing figures4

Posted by <NAME>,

I’m taking a break this week from MATLAB R2010a features to show you a feature that’s been a part of MATLAB for years: generating MATLAB code from a figure. Last week I had the opportunity to visit some of our customers and observe what they do with MATLAB. I was surprised by the amount of effort one person spent on plotting and setting up the figures for series of data. His data sets represented different runs of repeated experiments, and as you can imagine, the figures all looked pretty similar. We suggested that he could save time by writing a function that took in a set of data and created and plot, that way he would only have to write and debug the various commands (such as plot, legend, axis, colormap, etc.) once.

The workflow I suggested to that user is as follows: (A) plot some data, (B) spiffy up the plot from the command line, (C) save those commands to a function file, (D) repeat on additional sets of data. I further suggested that step “C” could be sped up using code generation from the figure. Every figure has a “Generate M-File” command under it’s File menu. This command takes the Handle Graphics objects in the figure and recreates them in a function, where all you have to do is pass in the plot data. What I like about this feature is that the generated function is readable and can teach you about new properties of plot objects.

Here is an example of how it works, using the Scatter Plot from the graf2d demo:

load count.dat
scatter(count(:,1),count(:,2),'r*')
xlabel('Number of Cars on Street A');
ylabel('Number of Cars on Street B');

This code generates the following scatter plot:

From the figure’s “File -> Generate M-File” menu item we get the following function created in the editor:

function createfigure(X1, Y1, S1, C1)
%CREATEFIGURE(X1,Y1,S1,C1)
%  X1:  scatter x
%  Y1:  scatter y
%  S1:  scatter s
%  C1:  scatter c

%  Auto-generated by MATLAB on 16-Apr-2010 11:07:47

% Create figure
figure;

% polar currently does not support code generation, enter 'doc polar' for correct input syntax

% polar(...);

% Create scatter
scatter(X1,Y1,S1,C1,'Marker','*');

% Create xlabel
xlabel('Number of Cars on Street A');

% Create ylabel
ylabel('Number of Cars on Street B');

The quality and re-usefulness of the generated function will depend on the data, as well as the functions used to create the original plot. The generate file command uses low-level objects and basic plot types, so if the original plot was generated by a function that performs a lot of set up or massaging of data, there may be a decent amount of set-up you’ll have to perform in order to use the generated function. For example, in original code from graf2d the plot was performed by the 3-input form of scatter and our generated file uses the 4-input form, whose inputs we have to supply when calling generated createfigure like so:

createfigure(count(:,1),count(:,2),40,'r')

To get back my original plot, I had to intuit what the appropriate inputs need to be. In my simple example, I no longer need to supply the axes labels, but now have to define the marker size (input 3, “40″), and while the marker type, “*,” is taken care of, I still have to tell it “r” (4th input) to make it red. Using a different type of plot could mean that there are fewer things to specify, but they will still likely be different than the original set of inputs.

I hope you can imagine that with a more complicated figure with subplots, multiple titles and legends, etc that this feature could save a bit of work. Sometimes I use this feature in the same way I use GUIDE: I use the interactive mode (Plot Tools) to get the figure how I want it, generate the code, and then copy the appropriate sections to my own hand-written file that is more readable.

## 4 CommentsOldest to Newest

While the generated m-code is readable, I find that saving figures as a *.fig file is more reliable in preserving all aspects of the figure (inc. controls and axes).

As the example that you have provided above shows, polar plots are not currently supported and there are other quirks. Sometimes when I run a generated m-file, I get a run-time error (exception) – I never run into such problems with fig files.

Under the hood, a FIG file is really a simple MAT file that contains all the figure’s properties in a hierarchical structure. It is therefore slightly disappointing that Matlab doesn’t simply recursively loop over the structure’s elements when generating m-files.

I only use the feature “Generate M-file…” after opening plot tools (from an existing figure of course)
It is so easy to change a font , a colour or … whatever from plottols w/o knowing the exact parameter (fontsize, fontcolor…)
But like most of matlab developers , you are quickly limited:
For example , try to generate code from the example in PIE function (already sent last year to French Matlab support):

x = [1 3 0.5 2.5 2];
explode = [0 1 0 0 0];
pie(x,explode)

W/O link, I definetely like the R2010a.
How many customers have reported the error ??? Undefined function or method ‘mikehowareyou’ for input arguments of type ‘double’.
I know that there is a technical note about this common message, but I would prefer a message which would say “Check that ‘mikehowareyou’ is on your path.
With R2010a, a simple right click on any folder allow you to update MATLAB path (no need to know addpath, pathtool , File-> Set Path) …wow great !!

Jason G replied on : 3 of 4

I have to say that “Generate M-file” is one of the most disappointing Matlab commands because it has the opportunity to be so great but falls woefully short. This could be a stellar, useful option to create real m-code that would reproduce the plot exactly when called from the command line and then would allow tinkering with the plot style by editing the m-file directly, or would allow easy regeneration of the plot style with new data. But instead, it lacks potency and is difficult to use by Mike from Mathworks’ own admission. How about starting by saving all of the handlegraphics settings?

The purpose of any command in Matlab should be to save time and/or complexity. Please make “Generate M-file” live up to its potential in your next revision! I am a big Matlab fan (since 1994)

Mike replied on : 4 of 4

@Jason,

Thanks for those comments. We’re working on a whole suite of figure functionality that will replace the current Generate M-code functionality. However although it is seemingly underpowered by its name, there are a few workflows where this is useful, such as the one I highlighted or @Aurélien mentioned.

These postings are the author's and don't necessarily represent the opinions of MathWorks.


text:
# Halfspaces minimise nonlocal perimeter: a proof via calibrations

created by pagliari on 02 May 2019
modified on 26 Nov 2019

[BibTeX]

Accepted Paper

Inserted: 2 may 2019
Last Updated: 26 nov 2019

Journal: Annali di Matematica Pura ed Applicata
Year: 2019

Abstract:

We consider a nonlocal functional $J_K$ that may be regarded as a nonlocal version of the total variation. More precisely, for any measurable function $u\colon \mathbb{R}^d \to \mathbb{R}$, we define $J_K(u)$ as the integral of weighted differences of $u$. The weight is encoded by a positive kernel $K$, possibly singular in the origin. We study the minimisation of this energy under prescribed boundary conditions, and we introduce a notion of calibration suited for this nonlocal problem. Our first result shows that the existence of a calibration is a sufficient condition for a function to be a minimiser. As an application of this criterion, we prove that halfspaces are the unique minimisers of $J_K$ in a ball, provided they are admissible competitors. Finally, we outline how to exploit the optimality of hyperplanes to recover a $\Gamma$-convergence result concerning the scaling limit of $J_K$.


text:
## Thomas' Calculus 13th Edition

$-\dfrac{2}{t^2}-\dfrac{2}{3t^{3/2}}+C$
As we know $\int x^n dx=\dfrac{x^{n+1}}{n+1}+C$ Thus, we have$\int \dfrac{4 +\sqrt t}{t^3} dt=\int \dfrac{4}{t^{3}}+\dfrac{t^{1/2}}{t^3} dt$ or, $=4(\dfrac{1}{-2}t^{-2}+\dfrac{1}{-3/2}t^{-3/2}+C$ or, $=-\dfrac{2}{t^2}-\dfrac{2}{3t^{3/2}}+C$