text:
# Whitman has a direct labor standard of 2 hours per unit of output. Each employee has a standard w...

Whitman has a direct labor standard of 2 hours per unit of output. Each employee has a standard wage rate of $23.50 per hour. During July, Whitman paid$105,328 to employees for 4640 hours worked. 2570 units were produced during July. What is the flexible budget amount for direct labor?

a. 120790 b. 202618 c. 218080 d. 226118


text:
×

# $$\sinh$$ and $$\cosh$$

So I recently found out that

$\sin {(a + bi)} = i\sinh {(b - ai)}$

$\cos {(a + bi)} = \cosh {(b - ai)}$

and that

$\sinh {(x)} = \frac {e^x - e^{-x}}{2}$

$\cosh {(x)} = \frac {e^x + e^{-x}}{2}$

Since

$\frac {a}{b} + \frac {b}{a} = \frac {a^2 + b^2}{ab}$

$\frac {a}{b} - \frac {b}{a} = \frac {a^2 - b^2}{ab}$

and

$(\frac {a}{b})^{-x} = (\frac {b}{a})^x$

Re-arranging the equations for $$\sinh{(x)}$$ and $$\cosh {(x)}$$ gives us

$\sinh {(x)} = \frac {e^{2x} - 1}{2e^x}$

$\cosh {(x)} = \frac {e^{2x} + 1}{2e^x}$

Substituting these into $$\sin {(a + bi)}$$ and $$\cos {(a + bi)}$$ when $$x = b - ai$$ gives us

$\sin {(a + bi)} = i\frac {e^{2(b - ai)} - 1}{2e^{b - ai}}$

$\cos {(a + bi)} = \frac {e^{2(b - ai)} + 1}{2e^{b - ai}}$

I'm going to try and expand and substitute by using

$e^{a + bi} = e^a (\cos {(b)} + i\sin {(b)})$

Let $$\cos {\theta} + i\sin {\theta} =$$ $$cis$$ $$\theta$$. So

$e^{a + bi} = e^a \cdot cis (b)$

Next we substitute this new equation in to get

$\sin {(a + bi)} = i\frac {e^{2b} \cdot cis (-2a) - 1}{2e^{b} \cdot cis (-a)}$

$\cos {(a + bi)} = \frac {e^{2b} \cdot cis (-2a) + 1}{2e^{b} \cdot cis (-a)}$

So what does $$cis(a + bi)$$ equal

Substituting in the two equations gives us

$cis(a + bi) = \frac {e^{2b} \cdot cis (-2a) + 1}{2e^{b} \cdot cis (-a)} + i(i\frac {e^{2b} \cdot cis (-2a) - 1}{2e^{b} \cdot cis (-a)})$

That looks unfriendly, let's simplify it

$cis(a + bi) = \frac {1}{e^{b} \cdot cis(-a)}$

(Go through the steps, it works)

This can be simplified even further

$cis(a + bi) = \frac {cis(a)}{e^b}$

P.s. $(\cos {\theta} + i\sin {\theta})^n = \cos {\theta n} + i\sin{\theta n}$

Note by Jack Rawlin
2 years, 11 months ago

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\begin{align*} \text{cis}(a+bi) &= e^{i(a+bi)}\\ &= e^{ia-b}\\ &= \dfrac{e^{ia}}{e^b}\\ &= \dfrac{\text{cis}(a)}{e^b} \end{align*}

- 2 years, 11 months ago

I did consider that but I wanted to make it as complex as possible so it would be fun to work out.

- 2 years, 11 months ago

Lol..Nice note!!

- 2 years, 11 months ago

Very nice note. Learned a lot!

- 2 years, 11 months ago

Although trigonometric functions can operate complex numbers, I think there is no complex angle but only real angles. Just like a doctor who can save lives, not all lives being saved are human being. Do you agree with me?

- 2 years, 10 months ago

format: plain


text:
# If the domain of a coordinate transformation that is reflected over the y-axis is (-3, 2), (4, 5), (2, -3), what is the range? (NEED DONE QUICKLY)

###### Question:

If the domain of a coordinate transformation that is reflected over the y-axis is (-3, 2), (4, 5), (2, -3), what is the range?
(NEED DONE QUICKLY)

### A cubic piece of metal measures 7.50cm on each edge. If the metal is nickel, whose density is 8.90 g/cm3, what is the mass of the cube?

A cubic piece of metal measures 7.50cm on each edge. If the metal is nickel, whose density is 8.90 g/cm3, what is the mass of the cube?...

### A pet fish guidebook says many fish need 1 cubic foot of water in their aquarium for every four fish. Determine whether the aquariums shown below meet that guideline? Aquarium A contains 60 fish. Aquarium B contains 200 fish. Aquarium C contains 50 fish.

A pet fish guidebook says many fish need 1 cubic foot of water in their aquarium for every four fish. Determine whether the aquariums shown below meet that guideline? Aquarium A contains 60 fish. Aquarium B contains 200 fish. Aquarium C contains 50 fish....

### What is the solution to the system of linear equations? These two are not connected in that special brace or bracket you see, these are spaced out, but I think they're connected. 3.5x - 5y = 20 3x + 4y = 13 Please help, I will appreciate it a lot :) Oh, and please show steps. Thank you!

What is the solution to the system of linear equations? These two are not connected in that special brace or bracket you see, these are spaced out, but I think they're connected. 3.5x - 5y = 20 3x + 4y = 13 Please help, I will appreciate it a lot :) Oh, and please show steps. Thank you!...

### If aqueous solutions of pb(no3)2 and nacl are mixed, which ions, if any, are spectator ions?

If aqueous solutions of pb(no3)2 and nacl are mixed, which ions, if any, are spectator ions?...

### In a store the number of dogs is 12 times greater than 3 times of cats. If there are 21 dogs in the store, how many cats are in the store?

In a store the number of dogs is 12 times greater than 3 times of cats. If there are 21 dogs in the store, how many cats are in the store?...

### 4. What is hemolysis, or a hemolytic reaction?

4. What is hemolysis, or a hemolytic reaction?...

### The two small Hebrew states were bordered by the kingdom of Egypt and the kingdoms of __________. A. Persia B. Assyria C. Babylon D. Mesopotamia Please hurry I give 100 points

The two small Hebrew states were bordered by the kingdom of Egypt and the kingdoms of __________. A. Persia B. Assyria C. Babylon D. Mesopotamia Please hurry I give 100 points...

### Which is the meaning of the root -vad- in words such as invade and evasive? put go make speak

Which is the meaning of the root -vad- in words such as invade and evasive? put go make speak...

### Type N over each noun, V over each verb or verb part, and P over each pronoun. ___ ____ __ ___ ___ _ ___ ____ __ ____ ____ The house at the end of the street is being sold.

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### -4 + 4x + 4 = -24 - 8x *

-4 + 4x + 4 = -24 - 8x *...

### . Frosty Snacks sells ice cream in a cylindrical cup and in cones. The small cup of ice cream has a height of 4 inches and a radius of 2 inches. The small cone has a height of 6 inches and a diameter of 5 inches. How much more ice cream will fit into a small cup than a small cone? Use 3.14 for pi.

. Frosty Snacks sells ice cream in a cylindrical cup and in cones. The small cup of ice cream has a height of 4 inches and a radius of 2 inches. The small cone has a height of 6 inches and a diameter of 5 inches. How much more ice cream will fit into a small cup than a small cone? Use 3.14 for pi....

### What does the 6 stand for in 8,375,693.1? 0.6 6 600 6000

What does the 6 stand for in 8,375,693.1? 0.6 6 600 6000...

### Is this true or false?

is this true or false?...

### Help me please! Thank you so much!

Help me please! Thank you so much!...

### Did i draw my triangle right for question 8

did i draw my triangle right for question 8...

### Why does the Moon have a greater effect on the tides than the Sun? A. The Moon is larger than the Sun. B. The Moon is closer than the Sun to Earth. C. The Moon is solid while the Sun is gaseous. D. The Moon rotates more quickly than the Sun. PLEASE HELP ITS DUE TODAY :/

Why does the Moon have a greater effect on the tides than the Sun? A. The Moon is larger than the Sun. B. The Moon is closer than the Sun to Earth. C. The Moon is solid while the Sun is gaseous. D. The Moon rotates more quickly than the Sun. PLEASE HELP ITS DUE TODAY :/...

### HELP ME PLEASE SMART KIDS ONLY

HELP ME PLEASE SMART KIDS ONLY...


text:
# An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass M.

Question:

An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass M. The piston and the cylinder have equal cross sectional area $A$. When the piston is in equilibrium, the volume of the gas is $\mathrm{V}_{0}$ and its pressure is $\mathrm{P}_{0}$. The piston is slightly displaced from the equilibrium position and released. Assuming that the system is completely isolated from its surrounding, the piston executes a simple harmonic motion with frequency.

1. $\frac{1}{2 \pi} \frac{A \gamma P_{0}}{V_{0} M}$

2. $\frac{1}{2 \pi} \frac{V_{0} M_{0}}{A^{2} \gamma}$

3. $\frac{1}{2 \pi} \sqrt{\frac{\mathrm{A}^{2} \gamma \mathrm{P}_{0}}{\mathrm{MV}_{0}}}$

4. $\frac{1}{2 \pi} \sqrt{\frac{\mathrm{MV}_{0}}{\mathrm{A \gamma \textrm {P } _ { 0 }}}}$

Correct Option: , 2

Solution:


text:
Correction to: Scientific Reports https://doi.org/10.1038/s41598-019-45576-3, published online 25 June 2019

The original version of this Article contained an error in the legend of Figure 1.

“Schematic representation of the alternative small-world model as introduced in19 and discussed in this paper. Much like in the original model, we start with $$N$$ nodes placed equidistantly on a ring. However, instead of rewiring, each pair of nodes is connected with distance-based probability $${p}_{d}$$ where $$d$$ is their minimal distance on the ring. Within distance $$d\le k/2$$, nodes are connected with short-range probability $${p}_{S}$$. For larger distances, nodes are connected with long-range probability $${p}_{L}={\beta p}_{S}$$. With increasing redistribution parameter $$0\le \beta \le 1$$ connection probability is redistributed from the short-range regime to the long-range regime while the mean degree $$k$$ i“Acknowledgements” on page 9 s kept constant. Hence at $$\beta =0$$ the short-range probability is unity while the long-range probability is zero which produces a $$k$$-nearest neighbor lattice. With increasing $$\beta$$, long-range “short-cuts” become more probable until at $$\beta =1$$ both connection probabilities are equal and thus the model becomes equal to the Erdős–Rényi model.”

“Schematic representation of the alternative small-world model as introduced in19 and discussed in this paper. Much like in the original model, we start with $$N$$ nodes placed equidistantly on a ring. However, instead of rewiring, each pair of nodes is connected with distance-based probability $${p}_{d}$$ where $$d$$ is their minimal distance on the ring. Within distance $$d\le k/2$$, nodes are connected with short-range probability $${p}_{S}$$. For larger distances, nodes are connected with long-range probability $${p}_{L}={\beta p}_{S}$$. With increasing redistribution parameter $$0\le \beta \le 1$$ connection probability is redistributed from the short-range regime to the long-range regime while the mean degree $$k$$ is kept constant. Hence at $$\beta =0$$ the short-range probability is unity while the long-range probability is zero which produces a $$k$$-nearest neighbor lattice. With increasing $$\beta$$, long-range “short-cuts” become more probable until at $$\beta =1$$ both connection probabilities are equal and thus the model becomes equal to the Erdős–Rényi model.”


text:
## The spectrum of the damped wave operator for a bounded domain in $$\mathbb R^2$$.(English)Zbl 1061.35064

The authors consider the damped wave equation with a nonnegative potential on a Riemannian manifold and prove a necessary condition for the real parts of the eigenvalues to accumulate at a point. Then they consider the special case of a sphere and determine an interval which contains the supports of the weak limits of the sequences of probability measures which count the number of eigenvalues in a horizontal strip whose axis is the real axis, as the thickness of the strip tends to infinity. A previous result in this direction is due to J. Sjöstrand. Then some numerical simulations follow.

### MSC:

 35P20 Asymptotic distributions of eigenvalues in context of PDEs 58J45 Hyperbolic equations on manifolds 35B37 PDE in connection with control problems (MSC2000) 93C20 Control/observation systems governed by partial differential equations 34L25 Scattering theory, inverse scattering involving ordinary differential operators 49J20 Existence theories for optimal control problems involving partial differential equations
Full Text:

### References:

 [1] Cornfeld I., Ergodic Theory. (1982) [2] DOI: 10.1137/0151015 · Zbl 0734.35009 [3] DOI: 10.1080/03605309408821015 · Zbl 0818.35072 [4] DOI: 10.1137/S0363012997329445 · Zbl 0999.35010 [5] Freitas, P. 1999. [Freitas 99], Private communication [6] DOI: 10.1215/S0012-7094-93-07122-0 · Zbl 0788.35103 [7] DOI: 10.1137/1026033 · Zbl 0574.65116 [8] DOI: 10.1137/S0363012996291616 · Zbl 0889.35057 [9] DOI: 10.2977/prims/1195142811 · Zbl 0984.35121 [10] Sorenson D. C., Implicitly Restarted Arnoldi/Lanczos Methods for Large Scale Eigenvalue Problems. (1995)
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


text:
# PRIMEGRAPH-Editorial

Author: Mamedov
Tester : Takuki Kurokawa
Editorialist: Kingvarun

Easy Medium

Graph

# PROBLEM:

You need to find maximum number of edges that a graph can have such that a graph such that every node of that graph has prime degree.

# EXPLANATION:

Degree of node is the number of edges connected to that node, and we have to make that prime for the all the nodes.

Maximum number of edges that can be drawn from a single node for a graph containing N nodes is N-1, means each node is connected to all the other remaining nodes.

So, we have to simply find a greatest prime number which is less than equal to N-1 and that will be the number of the edges containing by one single node.

But, if the value of N id odd than it is not possible to create a graph such that every node have same prime degree less than equal to &N-1\$.

Why Not ??

Suppose we have value of N equal to 5 and prime number which is less than or equal to 4 is 3, Can we able to create a graph such that every node has degree 3.

No, because the sum of all the degrees is equal to twice the number of edges. Since the sum of the degrees is even and the sum of the degrees of vertices with even degree is even,

the sum of the degrees of vertices with odd degree must be even. If the sum of the degrees of vertices with odd degree is even, there must be an even number of those vertices.

Here in this case sum of degree of vertices is 5*3 which is odd, and sum of degree of vertices should always be even.

Therefore, what we have to do is simply reduce the degree of one node to 2 and for others keep it as same.

For even no.of number nodes

Maximum edges is (N * P) / 2

For odd no.of number nodes

Maximum edges is (((N-1) * P) / 2) + 1

where P is the maximum prime number less than equal to N-1.

# TIME COMPLEXITY:

O(N * Sqrt(N)) per test case

# SOLUTIONS:

Author's Solution
#pragma GCC optimize ("O3")
#include <bits/stdc++.h>
#define pii pair<int, int>
#define piii pair<pii, int>
#define vi vector<int>
#define vvi vector<vi>
#define vpii vector<pii>
#define vvpii vector<vpii>
#define f first
#define s second
#define oo 1000000001
#define eb emplace_back
#define pb push_back
#define mpr make_pair
#define size(v) (int)v.size()
#define ln '\n'
#define ull unsigned long long
#define ll long long
#define all(v) v.begin(), v.end()

using namespace std;

bool isPrime(int n) {
if (n <= 1) return false;
for (int i = 2; i * i <= n; ++i) {
if (n % i == 0) {
return false;
}
}
return true;
}
void solve() {
int N;
cin >> N;
int P = N - 1;
while (!isPrime(P)) --P;

if (N % 2 == 0) {
cout << (1ll * N * P / 2ll) << ln;
} else {
cout << (1ll * (N - 1) * P / 2ll + 1) << ln;
}
}
int main() {
ios_base::sync_with_stdio(false);

int t = 1;
cin >> t;
while (t--) {
solve();
}
return 0;
}

Tester's Solution
// input validation, not for editorial.
#include <bits/stdc++.h>
using namespace std;

// -------------------- Input Checker Start --------------------

long long readInt(long long l, long long r, char endd) {
long long x = 0;
int cnt = 0, fi = -1;
bool is_neg = false;
while (true) {
char g = getchar();
if (g == '-') {
assert(fi == -1);
is_neg = true;
continue;
}
if ('0' <= g && g <= '9') {
x *= 10;
x += g - '0';
if (cnt == 0)
fi = g - '0';
cnt++;
assert(fi != 0 || cnt == 1);
assert(fi != 0 || is_neg == false);
assert(!(cnt > 19 || (cnt == 19 && fi > 1)));
} else if (g == endd) {
if (is_neg)
x = -x;
if (!(l <= x && x <= r)) {
cerr << l << ' ' << r << ' ' << x << '\n';
assert(false);
}
return x;
} else {
assert(false);
}
}
}

string readString(int l, int r, char endd) {
string ret = "";
int cnt = 0;
while (true) {
char g = getchar();
assert(g != -1);
if (g == endd)
break;
cnt++;
ret += g;
}
assert(l <= cnt && cnt <= r);
return ret;
}

long long readIntSp(long long l, long long r) { return readInt(l, r, ' '); }
long long readIntLn(long long l, long long r) { return readInt(l, r, '\n'); }
string readStringSp(int l, int r) { return readString(l, r, ' '); }
void readEOF() { assert(getchar() == EOF); }

vector<int> readVectorInt(int n, long long l, long long r) {
vector<int> a(n);
for (int i = 0; i < n - 1; i++)
a[n - 1] = readIntLn(l, r);
return a;
}

// -------------------- Input Checker End --------------------

bool is_prime(int n) {
for (int i = 2; i * i <= n; i++) {
if (n % i == 0) {
return false;
}
}
return true;
}

int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int tt = (int) readInt(1, 100, '\n');
while (tt--) {
int n = (int) readInt(3, (int) 1e5, '\n');
int p = n - 1;
while (!is_prime(p)) {
p--;
}
if (n % 2 == 0) {
cout << 1LL * n * p / 2 << '\n';
} else {
cout << 1LL * (n - 1) * p / 2 + 1 << '\n';
}
}
return 0;
}

Editorialist Solution
#include<bits/stdc++.h>
using namespace std;
#define ll long long

bool isprime( ll p)
{
ll count=0;

for(ll i=1; i<=sqrt(p); i++)
{
if(p % i == 0 )
count++;

}
if(count==1)
return true;
else
return false;

}

int main() {

ios_base::sync_with_stdio(false);
cin.tie(NULL);

int t;
cin>>t;
while(t--)
{
ll n , i;
cin>>n;
for(i=n-1; i>=2; i--)
{
if(isprime(i))
break;
}
if(n%2==0)
{
cout << (n * i)/2 << "\n";
}
else
{
cout << ((n-1) * i)/2 +1 << "\n";
}
}
return  0;
}

10 Likes

This problem can also be solved in O(N\cdot \log{N}) time using Sieve of Eratosthenes.

Here is my solution: Solution: 56927008 | CodeChef

Why there is exist graph on N vertexes with prime degree p<=N-1 for each vertex? That is not obvious.
An even sum of vertex degrees is only a necessary condition, but not a sufficient one.

2 Likes

According to Havel Hakimi theorem we simply find that if there is any simple graph exist or not ,
by only knowing the degree of each node ?

From this we can simply proved that there exist a graph with N nodes having Prime degree <= N-1

Solved using Seive Of Eratosthenes in O(N logN).
My solution: Solution: 57011829 | CodeChef

1 Like

Can someone please explain how can we reduce the degree of a single node while keeping the others same, I can only think of reducing degree of two nodes simultaneously. Keeping this in mind the solution I got arived at was (P*(n-2) + 2 + 3)/2 for odd number of nodes. Decreasing the degree of two nodes simultaneously.

We are not reducing the degree of single node by keeping all the same
If the value of N is odd
We are making a graph for N-1 nodes and keeping degree of all N-1 nodes prime and for the last Nth node we connect it between any edges so that degree of that becomes 2 that is also prime

1 Like

Can someone please explain why in for the odd case we are adding 1 in this case as there is one element left in odd to not have its degree prime ((n-1)*prime)/2+1 ??? but why adding 1 only ??

oh I got it, thank you so much bro

Can I know why my Code is getting TLE ??

#include <bits/stdc++.h>
using namespace std;

#define fast                    ios_base::sync_with_stdio(false);  cin.tie(NULL);
#define time                    cerr<<"Time taken : " <<(float)clock()/CLOCKS_PER_SEC <<" secs"<<"\n"  ;
#define F                       first
#define S                       second
#define pb                      push_back
typedef long long int           ll  ;

void solve() {

ll n ;
cin >> n;

if (n == 3) {
cout << 3 << "\n"  ;
return ;
}
unordered_map<ll, ll>deg ;
for (ll i = 1; i <= n - 1; i++) {
deg[i]++ ;
deg[i + 1]++  ;
}

deg[n]++ ;
deg[1]++ ;

// for (auto l : deg) {
//     cout << l.F << " " << l.S << "\n"  ;
// }
// cout << "\n" ;

ll edges = 0 ;
if (n % 2 == 0 ) {
ll i = n ;
for (auto k : deg) {
if (i == 0) {
break ;
}
k.S++ ;
edges += k.S;
i-- ;
}

cout << edges / 2 << "\n" ;
}
else {

ll i = n - 1;
for (auto k : deg) {
if (i == 0) {
break ;
}
k.S++ ;
edges += k.S;
i-- ;
}

cout << (edges / 2) + 1 << "\n" ;
// ll p1 = edges ;
// p1 /= 3 ;
// p1++ ;
// p1 += 2;
// cout << p1 << "\n"  ;
}
}

int32_t main() {

fast ; time;

int t = 1;
cin >> t;

while (t--) {
solve()  ;
}

return 0  ;
}


text:
## Log-Sobolev inequalities for subelliptic operators satisfying a generalized curvature dimension inequality

Series:
Stochastics Seminar
Thursday, February 9, 2012 - 15:05
1 hour (actually 50 minutes)
Location:
Skyles 006
,
University of Purdue
Organizer:
Let $\M$ be a smooth connected manifold endowed with a smooth measure $\mu$ and a smooth locally subelliptic diffusion operator $L$ which is symmetric with respect to $\mu$. We assume that $L$ satisfies a generalized curvature dimension inequality as introduced by Baudoin-Garofalo \cite{BG1}. Our goal is to discuss functional inequalities for $\mu$ like the Poincar\'e inequality, the log-Sobolev inequality or the Gaussian logarithmic isoperimetric inequality.


text:
The shuffle algebra and quantum group ${U}_{q}{𝔫}^{-}$

Let $ℱ$ be the free algebra generated by ${f}_{1},\dots ,{f}_{n}$. Define a new product on $ℱ$ by $u∘v = ∑σ∈ Sk+l/ ( Sk×Sl ) qwt(σ,u,v) σ(uv) where k=l(u), ℓ=l(v),$ the sum is over all minimal length coset representatives of the cosets in ${S}_{k+l}/\left({S}_{k}×{S}_{l}\right)$ and $wt(σ,u,v) = ∑ 1≤i σ(j) -⟨ui, vj-k⟩,$ where ${u}_{i}$ is the ${i}^{\mathrm{th}}$letter in the word $u$ and ${v}_{j-k}$is the ${\left(j-k\right)}^{\mathrm{th}}$ letter in $v$. By [Le, prop. 1], $\circ$ is associative. Note that the powers of $q$ in $u\circ v$ depend on the choice of the matrix of the form.

Remark. Leclerc [Le, §2.5 (8)] recursively defines a product by $wa*xb= (w*xb) a+q- ⟨deg(wa), deg(b)⟩ (wa*x)b,$ for words $w$ and $x$ and letters $a$ and $b$. By a straightforward check $wa*xb=xb\circ wa$ so that the product $\circ$ is the opposite of $*$.

The quantum group $Uq𝔫- is the subalgebra of (ℱ,∘) generated by f1,…, fr.$

Let $a\in ℱ$ let ${a|}_{u}$ denote the coefficient of the word $u$ in the expansion of $a$. Then $a\in {U}_{q}{𝔫}^{-}$ if and only if $∑k=0 -⟨ αi, αj∨⟩ +1 (-1)k [ 1-⟨ αi, αj∨⟩ k ] q2/⟨ αi, αi⟩ aTT | z fikfj fi1- ⟨αi, αj∨⟩ -k t=0,$ for all $i\ne j$ with $⟨{\alpha }_{i},{\alpha }_{j}^{\vee }⟩\ne 0$ and all words $z,t$.

Example. (Type ${A}_{2}$) If $C= ( 2-1 -12 ), with α1∨ =α1 and α2∨ =α2,$ then ${U}_{q}{𝔫}^{-}$ contains $1,{f}_{1},{f}_{2}$, $f1∘f2 = f1f2+qf2f1 , f2∘f1 = f2f1+qf1f2$ and $f1∘(f1∘f2) = ( f1f1f2 + q-2 f1f1 f2 + q-1 f1f2 f1) + q( f1f2f1 + q f2f1f1 + q-1 f2f1f1 ) = (1+ q-2) f1f1f2 + (q+q-1) f1f2f1 +(1+q2) f2f1f1 ,$ and the Serre relation for this element (Proposition ??? above) is $[ 1+1 0 ] (1+q-2) - [ 1+1 1 ] (q+q-1) + [ 1+1 2 ] (1+q2) = (1+q-2) - (q+q-1) (q+q-1) + (1+q2) =0.$

Notes and References

These notes are a presentation of the point of view of J.A. Green [Gr], which views the quantum group as a subalgebra of the free associative algebra instead of as a quotient. This version of the quantum group ${U}_{q}{𝔫}^{-}$ is the one which arises naturally as the character ring of the category of finite dimensional graded quiver Hecke algebra modules.

References

[Le] B. Leclerc, Dual canonical bases, quantum shuffles and $q$-characters, Math. Zeitschrift 246 (2004) 691-732 MR2045836 arXiv:math/0209133v3


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# Analytic Geometry | Two Planes and a Angle | Two Solutions

this is me again, I have another problem which I haven't been able to solve, the legend goes like this:
Find the equation of the plane that contains the points $P_1(1,0,-1)$, $P_2(2,0,2)$ and forms a 600 ($\theta$) angle with the other plane $2x-2y+z+6=0$ (Two Solutions).

Here's what I've done:

Since we already know two points, P1(1,0,-1) and P2(2,0,2) they both satisfy a planes equation:
$$Ax+By+Cz+D=0$$ So, we substitute $x,y,z$ respectively for each point, giving use these two equations

1. $A(1)+B(0)+C(-1)+D=0$
2. $A(2)+B(0)+C(2)+D=0$

Simplifying we get:

1. $A-C+D=0$
2. $2A+2C+D=0$

Now, according to the following theorem the angle that is formed by the normals of to planes is the following:

$$\cos{\theta} = \pm \frac{AA'+BB'+CC'}{\sqrt{A^2+B^2+C^2} \sqrt{(A')^2+(B')^2+(C')^2}}$$ Where $[A,B,C]$ Are the director numbers of the normal in the first plane and $[A',B',C']$ Are the director numbers of the normal in the second plane.

Now, given those and taking the known angle of 60o, we substitute the director numbers of $2x-2y+z+6-0:[A'=2,B'=-2,C'=1]$ and the angle in the last formula, giving us: $$\cos{60^\circ} = \pm \frac{A(2)-B(-2)+C(1)}{\sqrt{A^2+B^2+C^2} \sqrt{(2)^2+(-2)^2+(1)^2}}$$

Simplifying, we get(positive because $\theta$ is acute): $$\frac{1}{2}=\frac{2A-2B+C}{3 \sqrt{A^2+B^2+C^2}}$$

Another simplification finally for: $$3(A^2+B^2+C^2)=(4A-4B+2C)^2$$ Now, we have 3 equations:

1. $A-C+D=0$
2. $2A+2C+D=0$
3. $3(A^2+B^2+C^2)=(4A-4B+2C)^2$

If we take 1) and 2) and write them as $D(A,C)$ we get:

1. $D=C-A$
2. $D=-2A-2C$

From which we can obtain A in terms of C $$-2A-2C=C-A$$ $$A=-3C$$ Now we substitute back in 1): $$(-3C)-C+D=0$$ $$C=\frac{1}{4}D$$ Since I want every director number in terms of D we substitute again but in A since A was in function of C. $$A=-3C$$ $$A=-3(\frac{1}{4}D)$$ $$A=\frac{-3}{4}D$$

Now, we have found $A(D)$, $C(D)$ we have yet to find $B(D)$ I'll explain why I think and want everything in terms of D, if we take the general form of a planes equation $Ax+By+Cz+D=0$ and substitute $A$ and $C$ with our known values we get: $$(\frac{-3}{4}D)x+By+(\frac{1}{4}D)z+D=0$$ if we divide by $D \neq 0$ and then multiply by 4 to eliminate fractions: $$\frac{-3}{4}x+\frac{B}{D}y+\frac{1}{4}z+1=0$$ $$-3x+\frac{4B}{D}y+z+4=0$$ From here it's clear that if we find B in terms of D and B is linear we have the planes equation, so, in 3) $3(A^2+B^2+C^2)=(4A-4B+2C)^2$ we substitute A and C: $$3((\frac{-3}{4}D)^2+B^2+(\frac{1}{4}D)^2)=(4(\frac{-3}{4})-4B+2(\frac{1}{4}D))^2$$ $$\vdots$$ $$104B^2+80DB+35D^2=0$$

Here is were I get stuck, I get imaginary parts, since I have to use the quadratic formula, so any help is really appreciated, the problem states that there are two solutions, but this is what I have thinked of if anyone can supply any solution I'll be very gratefull.

Regards.. Tristian

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+1 for showing what you've done. – J. M. Dec 5 '10 at 9:50
@J.M. Hehe thanks I find it fun to use Tex (or LaTeX) syntax, love to see it render.. – Triztian Dec 5 '10 at 9:57

## 2 Answers

The normal vector to the sought plane is orthogonal to the line $P_1 P_2$ and at angle $\pi/3$ to the normal to your given plane.

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 Well, your statement helped me find $A(C)$ using the perpendicularity theorem, but it just gives me a known fact: Director numbers of $l:[1,0,3]$ $$A(1)+B(0)+C(3)=0$$ we get: $$A=-3C$$ – Triztian Dec 5 '10 at 10:17 That follows from orthogonality; now use $60^\circ$. – Robin Chapman Dec 5 '10 at 10:22 That's where I lost you, how can it be orthogonal at $60^\circ$, if in order to be orthogonal it must be a $90^\circ$, where should I use the $60^\circ$?, thank you for your reply and patience by the way. – Triztian Dec 5 '10 at 10:39 @Triztian: It's orthogonal to one vector and at 60 degrees to another vector. There's no contradiction in that. Just to exemplify: the vector (cos 60, sin 60, 0) is at 60 degrees to (1,0,0), and at the same time orthogonal to (0,0,1). – Hans Lundmark Dec 5 '10 at 18:48 @Hans Lundmark, thanks for the explanation,now I get it, but after reading and reading Robin Chapman's statement, it seems that he is saying something that the problem says in it's legend, that is: "That the director numbers of the plane that I'm looking for are orthogonal to the line segment $P_1 P_2$ and that it forms a $60^\circ$ (\frac{\pi}{3}) angle with the known plane's normal... or Maybe I'm missing the point, did anyone look at what I already did? – Triztian Dec 7 '10 at 6:56
show 1 more comment

To flesh out Robin's answer a little: The thing that you really want to find is the unit normal vector $n=(A,B,C)^t$ to the sought plane. Once you have that, it's easy to find the equation of the plane. This vector has to satisfy three conditions:

1. Length one: $A^2+B^2+C^2=1$.

2. Orthogonal to the vector $\overrightarrow{P_1 P_2} = (1,0,3)^t$.

3. 60 degrees to the normal vector of the other plane. If we normalize that vector too, this condition becomes $\frac13 (2,-2,1)^t \cdot (A,B,C) = \cos 60^{\circ}$.

Conditions 2 and 3 give a linear system with two equations and three unknowns $A$, $B$, $C$; solve it and you will get a family of solutions depending on one parameter $t$. Condition 1 then gives a quadratic eqation for determining the two possible values of $t$.

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Note

This documents the development version of NetworkX. Documentation for the current release can be found here.

# networkx.algorithms.approximation.clique.large_clique_size¶

large_clique_size(G)[source]

Find the size of a large clique in a graph.

A clique is a subset of nodes in which each pair of nodes is adjacent. This function is a heuristic for finding the size of a large clique in the graph.

Parameters
GNetworkX graph
Returns
k: integer

The size of a large clique in the graph.

Raises
NetworkXNotImplemented

If the graph is directed or is a multigraph.

networkx.algorithms.approximation.clique.max_clique()

A function that returns an approximate maximum clique with a guarantee on the approximation ratio.

networkx.algorithms.clique

Functions for finding the exact maximum clique in a graph.

Notes

This implementation is from [1]. Its worst case time complexity is $$O(n d^2)$$, where n is the number of nodes in the graph and d is the maximum degree.

This function is a heuristic, which means it may work well in practice, but there is no rigorous mathematical guarantee on the ratio between the returned number and the actual largest clique size in the graph.

References

1

Pattabiraman, Bharath, et al. “Fast Algorithms for the Maximum Clique Problem on Massive Graphs with Applications to Overlapping Community Detection.” Internet Mathematics 11.4-5 (2015): 421–448. <https://doi.org/10.1080/15427951.2014.986778>


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# Integrate over $t$ if $dr/dt = \sqrt{1 - r^2}$?

I'm stuck on this step trying to solve a math puzzle. How do I integrate this?

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you cannot integrate this directly but need to first 'separate the variables' in order to solve the equation. Once you've done that, the integral will be a standard result. – Ronald Apr 29 '12 at 23:07
Stuck on what step? Please make the body of your post self-contained. You don't ask readers of a book to start at the spine; you shouldn't ask readers of your post to start at the subject, which is just an indexing item. – Arturo Magidin Apr 29 '12 at 23:07
Please include the question in the question, not just in the title. In any event, it is not clear exactly what integral you are trying to evaluate. – Gerry Myerson Apr 29 '12 at 23:07

You have that

$$\frac{{dr}}{{dt}} = \sqrt {1 - {r^2}}$$

from where

$$\frac{{dr}}{{\sqrt {1 - {r^2}} }} = dt$$

then integrating

$$\arcsin r + C = t$$

You need a initital value to determine $C$.

-

I think you are asking to solve a differential equation. This one is separable, so one has $$\arcsin(r) = t + C$$ or $$r = \sin(t + C).$$

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×

# How do you simplify abs(9)?

Mar 5, 2018

The absolute value function takes any term and transforms it to its non-negative form.

$\left\mid 9 \right\mid = 9$

Mar 7, 2018

Or another way that you could think about the absolute value of a number is how many units away it is from 0. If you think about it that way, you can always remember that the number of units it is from 0 will always have to be positive. Therefore, let's say if you took the absolute value of -9 (which can be written as $| - 9 |$) then it would be 9 also because it, too, is 9 units away from 0.


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My Math Forum Hard logarithm question

 Algebra Pre-Algebra and Basic Algebra Math Forum

January 28th, 2012, 12:58 AM   #1
Newbie

Joined: Jan 2012

Posts: 3
Thanks: 0

Hard logarithm question

Hello!

Attached is a very hard logarithm question that I couldn't answer in my math test. If someone could help me answer it, I would very much appreciate it

Attached Images
 Scan_Pic0002.jpg (166.8 KB, 994 views)

 January 28th, 2012, 01:13 AM #2 Senior Member   Joined: Jul 2011 Posts: 227 Thanks: 0 Re: Hard logarithm question You are given: $2(5^{x+1})=1+\frac{3}{5^x}$ Your first step to make a common denominator for the RHS is a good start: $2(5^{x+1})=\frac{5^x+3}{5^x}$ Now use some logarithm rules: $2(5^{x+1})5^x=5^x+3$ $\Leftrightarrow 2(5^{2x+1})=5^x+3$ $\Leftrightarrow 2(5^{2x}\cdot 5)=5^x+3$ $\Leftrightarrow 10\cdot 5^{2x}-5^x-3=0$ To continue make the substitution $5^x=t$ which will give you a quadratic equation in $t$
 January 28th, 2012, 01:20 AM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Hard logarithm question We are given to solve: $2$$5^{x+1}$$=1+\frac{3}{5^x}$ $10\cdot5^x=1+\frac{3}{5^x}$ Multiply through by $5^x$ as we know $5^x\ne0$ for all real x: $10\cdot5^{2x}=5^x+3$ Arrange as quadratic on $5^x$: $10\cdot$$5^x$$^2-5^x-3=0$ Factor: $$$5\cdot5^x-3$$$$2\cdot5^x+1$$=0$ Assuming x is real, we must discard the negative root, and are left with: $5^x=\frac{3}{5}$ Convert from exponential to logarithmic form: $x=\log_5$$\frac{3}{5}$$=\log_5(3)-\log_5(5)=-1+\log_5(3)$
 January 28th, 2012, 02:23 AM #4 Newbie   Joined: Jan 2012 Posts: 3 Thanks: 0 Re: Hard logarithm question Thank you! I'm really impressed with this forum, the answers are clear and succinct
 January 22nd, 2017, 01:45 AM #5 Newbie   Joined: Jan 2017 From: Somewhere Posts: 1 Thanks: 0 Answer x = log[base 5](3) - 1 Last edited by Wizzy0630; January 22nd, 2017 at 01:47 AM.
 May 31st, 2017, 02:37 AM #6 Newbie   Joined: May 2017 From: China Posts: 1 Thanks: 0 Answer Hey, I know I may be too late to answer you now. You do not need to use logarithm for this question. Set 5^x as y, and rewrite the equation as 2*5*y = 1+3/y. Then continue the equation and factorize the quadratic you get. Keep going until you get your answer. Thanks

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Free Version
Moderate

# Trapezoidal Approximation

APCALC-1BGAVW

Use the Trapezoidal Rule with $4$ trapezoids of equal subintervals to approximate:

$$\int _{ 0 }^{ 4 }{ ( { x }^{ 2 }+2 ) dx }$$

A

$\cfrac{176}{3}$

B

$60$

C

$40$

D

$30$


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### big_aL's blog

By big_aL, 2 months ago,

I don't know if this is a classic problem or not but I can't seem to find a solution for it.

Consider two arrays a = [1, 2, 3, 4, 5] and b = [3, 6, 6]. You are also given two indexes a_ind = 0 and b_ind = 0. In one move you can add 1 to both a_ind and b_ind, if the index for any array goes out of bound then reset that index to 0. The task is to calculate the number of moves it would take such that the value a[a_ind] == b[b_ind] or state that it is impossible.

The brute force method would be to append both arrays to themselves x times. Where x = lcm(lenght(a), lenght(b)) / length(a) for the array a and x = lcm(lenght(a), lenght(b)) / length(b) for the array b, the above arrays would become:

1 2 3 4 5 1 2 3 4 5 1 2 3 4 5

3 6 6 3 6 6 3 6 6 3 6 6 3 6 6

The condition is satisfied at the 12th move. However this method would give TLE on any platform.

What would be an optimal approach for solving it?

• +32

 » 2 months ago, # |   0 Auto comment: topic has been updated by big_aL (previous revision, new revision, compare).
 » 2 months ago, # |   +3 This is from the top of my head but it can be solved in O(NM). In above example assuming answer is at xth position, if its due to first 3 in array A and B coinciding: x=5i+3, x=3j+1. 5i+3=3j+1, 5i-3j=-2, solve with linear diophantine equations. Find minimum x for all equal pairs = O(NM).
•  » » 2 months ago, # ^ |   0 What if the length of a = 10^5 and b = 10^5-1. In such case a linear or logarithmic solution is required.
•  » » » 2 months ago, # ^ |   +3 That's why I said it's from the top of my head.
 » 2 months ago, # | ← Rev. 2 →   +6 This is easy. You'll visit all positions $\mod gcd(len(a),len(b))$. Now, if theres duplicate in those modulo sets, you have the answer YES else NO. The second part is what I am thinking how to do fast.
•  » » 2 months ago, # ^ | ← Rev. 2 →   0 After that, if you find the visited states starting from $i = 0 \mod{gcd(n,m)}$, then, you can use the same visited array to calculate for array b. Since for all the same modulos, they'll visit the same states, except the starting position / number of moves will be different. Simple prefix array /suffix array to find the nearest will work.I'd love to chat for more clarifications.
 » 2 months ago, # |   0 You can use Bézout's identity + (hash)map to get a $O((N + M) \log (N + M))$ :$a[i]$ and $b[j]$ will coincide iff there are $x, y$ such that $i + nx = j + my \iff nx - my = j - i \iff j - i$ is a multiple of $\gcd(n, m)$ so you just need to keep track of values at the indexes mod $\gcd(n, m)$ with a map/hashmap.


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# Self occluding object and alpha blending

Look at the object I've rendered with my app:

It's the same screen twice, above the original and below I've drawn (by hand :P) the shape of the mesh of one of plant's leaves. You can clearly see where the problem is.

From what I understand, this leave is drawn before the other leaves, writing a higher value to the depth buffer but not changing the pixel color (as it's transparent in that particular place). When the other leaves are drawn then their pixels in that place in the buffer are discarded since they're failing the depth test (they're farther away from the camera).

Now while I understand what the problem is I don't know how to solve it. This whole plant is one object so I can't sort by depth. What should I do?

• Possible duplicate of Alpha blending not rendering properly XNA 4.0. See also How to properly implement alpha blending in a complex 3D scene. The short answer is that you'll need to either switch to alpha test (cutout), sort individual leaves, or live with some type of artifacts. – Nathan Reed Mar 29 '14 at 1:14
• Does sorting individual leaves require dividing one object (plant) into multiple ones (leaves)? – NPS Mar 29 '14 at 2:51
• Also, does alpha test/cutout mean discarding pixels (like with HLSL command discard)? – NPS Mar 29 '14 at 3:21
• Yes, it would mean treating each leaf as a separate object for the purposes of sorting. And yes, alpha test means discarding pixels that have too low of an alpha, and rendering the remaining pixels opaque, with no blending. – Nathan Reed Mar 29 '14 at 4:22

While there is some interesting research into order-independent transparency rendering, it's extremely complex to implement. And even sorting individual leaves can still cause artifacts where one leaf overlaps itself. So your safest bet is probably Alpha Testing.

This is where you specify a threshold opacity value; anything above that value is rendered 100% opaque (and writes to the depth buffer), and anything below is not rendered at all (and does not write to the depth buffer - ensuring that other, occluded geometry can still be rendered there later).

Older graphics pipelines had a dedicated alpha testing phase for this. Newer ones often just use the discard command under the hood.

The trouble with this is that instead of soft translucent falloff, Alpha Testing gives you sharp aliased edges.

If your texture is relatively crisp-edged and you're usually a long distance from these surfaces, or you're using a post-process antialiasing filter anyway, then these jaggies should not look any worse than typical geometry silhouettes. You can probably stop here.

But if you can get close to the texture or your alpha falloff is very feathery, you'll often see jagged step artifacts.

(Images excerpted from the Wolfire Games Blog. These show the artifacts better than the Unity docs image I was using earlier. Note the fringes in the top-left, incorrect sorting top-right, and aliased edges bottom-left. In the bottom-right, the foreground leaves still show aliasing where they overlap the trunk, but the texture detail makes this less noticeable than aliasing against the sky.)

One way to address these is to use a hybrid 2-pass shader. The first pass renders with alpha testing, to ensure proper z-sorting of the opaque portions. The second pass renders with alpha blending enabled and depth writes disabled, to fill-in the translucent fringes without creating the cutouts you see in the image in the original question, where depth writes from one leaf prevent the others from rendering fully.

The fringes can still visually intersect each other in incorrect ways due to the lack of sorting, but in practice this is often not very noticeable, especially in areas of high texture detail or similar colours overlapping.

A hat tip is due here to the awesome devs at Asteroid Base who introduced me to this hybrid strategy, used to great effect in rendering the enemy characters in Lovers in a Dangerous Spacetime.

• I realize there is no one right apporach and the choice depends on the concrete project. In my case, though, this combo does the trick: alpha blending on, alpha testing on with discard threshold at 0.5f (everything in range 0.5f - 1.0f is alpha blended), both depth testing and writing enabled. It still has some artifact but looks good enough. – NPS Mar 29 '14 at 13:32
• I'm glad you found an approach that meets your needs. One note of caution with using depth writing and alpha blending simultaneously: you can sometimes get translucent fringes where these objects overlap. Eg. unitywars.com/foliage/shot12.jpg – DMGregory Mar 29 '14 at 14:02
• I do but in my case this still looks 10 times better than aliased sharp pixels from traditional cutout. – NPS Mar 29 '14 at 19:42


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location:  Publications → journals → CJM
Abstract view

# Cycle-Level Intersection Theory for Toric Varieties

Published:2004-10-01
Printed: Oct 2004
• Hugh Thomas
 Format: HTML LaTeX MathJax PDF PostScript

## Abstract

This paper addresses the problem of constructing a cycle-level intersection theory for toric varieties. We show that by making one global choice, we can determine a cycle representative for the intersection of an equivariant Cartier divisor with an invariant cycle on a toric variety. For a toric variety defined by a fan in $N$, the choice consists of giving an inner product or a complete flag for $M_\Q= \Qt \Hom(N,\mathbb{Z})$, or more generally giving for each cone $\s$ in the fan a linear subspace of $M_\Q$ complementary to $\s^\perp$, satisfying certain compatibility conditions. We show that these intersection cycles have properties analogous to the usual intersections modulo rational equivalence. If $X$ is simplicial (for instance, if $X$ is non-singular), we obtain a commutative ring structure to the invariant cycles of $X$ with rational coefficients. This ring structure determines cycles representing certain characteristic classes of the toric variety. We also discuss how to define intersection cycles that require no choices, at the expense of increasing the size of the coefficient field.
 Keywords: toric varieties, intersection theory


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# Greatest common divisor

The greatest common divisor (GCD) of two or more integers is the largest integer that is a divisor of all the given numbers.

The GCD is sometimes called the greatest common factor (GCF).

A very useful property of the GCD is that it can be represented as a sum of the given numbers with integer coefficients. From here it immediately follows that the greatest common divisor of several numbers is divisible by any other common divisor of these numbers.

The GCD can be found in two ways. The first way involves factoring the numbers, and the second way uses Euclidean algorithm.

Once the prime factorizations of the given numbers has been found, the greatest common divisor is the product of all common factors of the numbers.

Example:
$270=2\times3^3\times5$
$144=2^4\times3^2$

The common factors are 2 and $3^2$, so the GCD is $2\times3^2=18$

Another Example:
$1200=2^4\times3\times5$
$720=2^4\times3^2\times5$
$288=2^5\times3^2$

The common factors are $2^4$ and 3, making the GCD ${2^4\times3=48}$.

The Euclidean Algorithm is much faster and can be used to give the GCD of any two numbers without knowing their prime factorizations. To find the greatest common divisor of more than two numbers, one can use the recursive formula $\displaystyle GCD(a_1,\dots,a_n)=GCD(GCD(a_1,\dots,a_{n-1}),a_n)$.

Another useful trick is the equation

GCD(x, y) * LCM(x, y) = x * y

Using this, you can find the GCD rather quickly if you know the LCM and the two numbers. This equation is also useful for finding one of the numbers if you know the other number, the GCD and the LCM.


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0

< = less then because the symbol looks like and L

> = greater than because it is not an L :)

Mahesh Godavarti
0

I remember it as the side that is shorter points to the smaller quantity and the side that is longer points to the bigger quantity.

For example, a \lt b , the shorter side of \lt is closer to a and the longer side is closer to b . That means a is less than b .

For example, a \gt b , the shorter side of \gt is closer to b and the longer side is closer to a . That means a is greater than b .


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## Reading the Comics, July 15, 2017: Dawn Of Mathematics Jokes

So I try to keep up with nearly all the comic strips run on Comics Kingdom and on GoComics. This includes some vintage strips: take some ancient comic like Peanuts or Luann and rerun it, day at a time, from the beginning. This is always enlightening. It’s always interesting to see a comic in that first flush of creative energy, before the characters have quite settled in and before the cartoonist has found stock jokes that work so well they don’t even have to be jokes anymore. One of the most startling cases for me has been Johnny Hart’s B.C. which, in its Back To B.C. incarnation, has been pretty well knocking it out of the park.

Not this week, I’m sad to admit. This week it’s been doing a bunch of mathematics jokes, which is what gives me my permission to talk about it here. The jokes have been, eh, the usual, given the setup. A bit fresher, I suppose, for the characters in the strip having had fewer of their edges worn down by time. Probably there’ll be at least one that gets a bit of a grin.

Back To B.C. for the 11th sets the theme going. On the 12th it gets into word problems. And then for the 13th of July it turns violent and for my money funny.

Mark Tatulli’s Heart of the City has a number appear on the 12th. That’s been about as much mathematical content as Heart’s experience at Math Camp has taken. The story’s been more about Dana, her camp friend, who’s presented as good enough at mathematics to be bored with it, and the attempt to sneak out to the nearby amusement park. What has me distracted is wondering what amusement park this could be, given that Heart’s from Philadelphia and the camp’s within bus-trip range and in the forest. I can’t rule out that it might be Knoebels Amusement Park, in Elysburg, Pennsylvania, in which case Heart and Dana are absolutely right to sneak out of camp because it is this amazing place.

Mort Walker’s Beetle Bailey Vintage for the 21st of December, 1960 and rerun the 14th of July, 2017. Wow, I remember when they’d put recipes like this on the not-actual-news segment of the 5:00 news or so, and how much it irritated me that there wasn’t any practical way to write down the whole thing and even writing down the address to mail in for the recipe seemed like too much, what with how long it took on average to find a sheet of paper and a writing tool. In hindsight, I don’t know why this was so hard for me.

Mort Walker’s Beetle Bailey Vintage for the 21st of December, 1960 was rerun the 14th. I can rope this into mathematics. It’s about Cookie trying to scale up a recipe to fit Camp Swampy’s needs. Increasing the ingredient count is easy, or at least it is if your units scale nicely. I wouldn’t want to multiple a third of a teaspoon by 200 without a good stretching beforehand and maybe a rubdown afterwards. But the time needed to cook a multiplied recipe, that gets mysterious. As I understand it — the chemistry of cooking is largely a mystery to me — the center of the trouble is that to cook a thing, heat has to reach throughout the interior. But heat can only really be applied from the surfaces of the cooked thing. (Yes, theoretically, a microwave oven could bake through the entire volume of something. But this would require someone inventing a way to bake using a microwave.) So we must balance the heat that can be applied over what surface to the interior volume and any reasonable time to cook the thing. Won’t deny that at some point it seems easier to just make a smaller meal.

Zach Weinersmith’s Saturday Morning Breakfast Cereal for the 14th goes to the old “inference testing” well again. This comes up from testing whether something strange is going on. Measure something in a sample. Is the result appreciably different from what would be a plausible result if nothing interesting is going on? The null hypothesis is the supposition that there isn’t anything interesting going on: the measurement’s in the range of what you’d expect given that the world is big and complicated. I’m not sure what the physicist’s exact experiment would have been. I suppose it would be something like “you lose about as much heat through your head as you do any region of skin of about the same surface area”. So, yeah, freezing would be expected, considering.

Percy Crosby’s Skippy for the 17th of May, 1930, and rerun the 15th, maybe doesn’t belong here. It’s just about counting. Never mind. I smiled at it, and I’m a fan of the strip. Give it a try; it’s that rare pre-Peanuts comic that still feels modern.

And, before I forget: Have any mathematics words or terms you’d like to have explained? I’m doing a Summer 2017 A To Z and taking requests! Please offer them over there, for convenience. I mean mine.

## Reading the Comics, June 10, 2017: Some Vintage Comics Edition

It’s too many comics to call this a famine edition, after last week’s feast. But there’s not a lot of theme to last week’s mathematically-themed comic strips. There’s a couple that include vintage comic strips from before 1940, though, so let’s run with that as a title.

Glenn McCoy and Gary McCoy’s The Flying McCoys for the 4th of June is your traditional blackboard full of symbols to indicate serious and deep thought on a subject. It’s a silly subject, but that’s fine. The symbols look to me gibberish, but clown research will go along non-traditional paths, I suppose.

Bill Hinds’s Tank McNamara for the 4th is built on mathematics’ successful invasion and colonization of sports management. Analytics, sabermetrics, Moneyball, whatever you want to call it, is built on ideas not far removed from the quality control techniques that changed corporate management so. Look for patterns; look for correlations; look for the things that seem to predict other things. It seems bizarre, almost inhuman, that we might be able to think of football players as being all of a kind, that what we know about (say) one running back will tell us something about another. But if we put roughly similarly capable people through roughly similar training and set them to work in roughly similar conditions, then we start to see why they might perform similarly. Models can help us make better, more rational, choices.

Morrie Turner’s Wee Pals rerun for the 4th is another word-problem resistance joke. I suppose it’s also a reminder about the unspoken assumptions in a problem. It also points out why mathematicians end up speaking in an annoyingly precise manner. It’s an attempt to avoid being shown up like Oliver is.

Which wouldn’t help with Percy Crosby’s Skippy for the 7th of April, 1930, and rerun the 5th. Skippy’s got a smooth line of patter to get out of his mother’s tutoring. You can see where Percy Crosby has the weird trait of drawing comics in 1930 that would make sense today still; few pre-World-War-II comics do.

Niklas Eriksson’s Carpe Diem for the 7th of June, 2017. If I may intrude in someone else’s work, it seems to me that the problem-solver might find a hint to what ‘x’ is by looking to the upper right corner of the page and the $x = \sqrt{13}$ already there.

Niklas Eriksson’s Carpe Diem for the 7th is a joke about mathematics anxiety. I don’t know that it actually explains anything, but, eh. I’m not sure there is a rational explanation for mathematics anxiety; if there were, I suppose it wouldn’t be anxiety.

George Herriman’s Krazy Kat for the 15th of July, 1939, and rerun the 8th, extends that odd little faintly word-problem-setup of the strips I mentioned the other day. I suppose identifying when two things moving at different speeds will intersect will always sound vaguely like a story problem.

George Herriman’s Krazy Kat for the 15th of July, 1939, as rerun the 8th of June, 2017. I know the comic isn’t to everyone’s taste, but I like it. I’m also surprised to see something as directly cartoonish as the brick stopping in midair like that in the third panel. The comic is usually surreal, yes, but not that way.

Tom Toles’s Randolph Itch, 2 am rerun for the 9th is about the sometimes-considered third possibility from a fair coin toss, and how to rig the results of that.

• #### goldenoj 9:01 pm on Sunday, 11 June, 2017 Permalink | Reply

Skippy is fascinating. Had to check if it was really from the 30s http://www.gocomics.com/skippy/2017/06/06 might also be a math comic.

You might want to put your Twitter handle in the sidebar – didn’t realize I had already seen you there via the blog-bot.

Like

• #### Joseph Nebus 11:57 pm on Monday, 12 June, 2017 Permalink | Reply

I didn’t realize I didn’t have my Twitter handle in the sidebar. Thanks, though, I’m glad to do stuff that makes me easier to find or understand, especially if it doesn’t require ongoing work.

Skippy, now, that’s not just a 1930s comic but one of the defining (American) comic strips. Basically every comic strip that stars kids who think is imitating it, either directly or through its influences, particularly Charles Schulz and Peanuts. It’s uncanny, especially when you compare it to its contemporaries, how nearly seamlessly it would fit into a modern comics page. It’s rather like Robert Benchley or Frank Fay in that way; now-obscure humorists or performers whose work is so modern and so influential that a wide swath of the modern genre is quietly imitating them.

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• #### Joseph Nebus 11:58 pm on Monday, 12 June, 2017 Permalink | Reply

Oh yes, and you’re right; I could’ve fit the comic from the 6th of June into a Reading the Comics post if I’d thought a bit more about it. Good eye!

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# FT232R for RS-485 (DMX512) with Isolation - TX Optocoupler

I am trying to make a DMX controller board using the FTDI FT232R (FT232RL) chip for USB to UART and a MAX485 to output data over RS-485. I have a working prototype done, but for the finished product I am designing a PCB and want to add in isolation between the UART side and the RS-485 side.

The design I've come up with for the isolation is to use a DC to DC converter to isolate the VCCs from each other and an optocoupler (6N137) to isolate the FT232R TX pin from the MAX485's driver input.

So now for the question: For driving the optocoupler, can I use the TX pin with a $$\470\Omega\$$ resistor to sink the optocoupler's input cathode? What I calculated is I need a minimum of 5 mA to make sure I exceed the LED threshhold. In the case of the lowest voltage difference of $$\2.6V\$$ ($$\(VCC\rightarrow5V) - (V_{F,max}\rightarrow1.8V) - (V_{ol}=0.6V)\$$), $$\520\Omega\$$ ($$\2.6V/5mA\$$) is the maximum resistance. (Specs below)

I can provide a schematic if that would help makes sense of all that I've said

(Note: The RX pin is not connected at all as I am doing DMX only, not DMX+RDM, so I have the MAX485 constantly driving)

6N137 Optocoupler

FT232RL FTDI USB to Serial

• Have you already seen the FTDI application notes for opto-isolated interfaces? ftdichip.com/Support/Documents/AppNotes/… – nvuono Dec 19 '19 at 18:10
• From where would the power come from to the isolated side? DMX512 transmitter is usually ground referenced so what would the isolation gain? Also are you absolutely sure that the PC software and USB drivers are able to perform their timing within DMX512 specs? Usually the PC would feed data to a microcontroller with non-realtime timing and the microcontroller would send DMX512 frames with realtime timing. Also, FTDI sells RS485 cables, so it would just need the XLR connector really.. – Justme Dec 19 '19 at 19:01
• @Justme The power for the isolated side would be the output of a 5V to 5V Isolated DC/DC Converter (example) Isolation for DMX is recommended since RJ45 connectors are allowed connectors for DMX512-A and some applications using RJ-45 connectors (namely phone lines) have up to 48VDC on the line. Isolation is to limit what gets fried if one were to accidentaly connect a phone line to a RJ-45 DMX port – Matthew Dec 19 '19 at 21:57
• @Justme I am basically imitating the Entec OpenDMX Controller which is based on the FT232R chip. DMX512 only operates at 250kbaud which is well within the capabilities of of the FT232R (3Mbaud) and MAX485 (2.5Mbaud) You are correct that I could just use an FTDI RS485 cable, but I am going to use the a similar isolation design for some DMX lights I have designed (MAX485 and Arduino Nano/ATMega328P) – Matthew Dec 19 '19 at 22:03
• @Matthew Ok, isolated converter is good, but preferred transmitter is still ground referenced while isolated transmitter is only allowed. It is the receivers that have to be isolated. Also the RJ45 connectors are only allowed in fixed installations or locked patch bay areas with limited entry, so that nobody connects phone lines to them accidentally, as that could indeed fry the RS485 transmitter chip and everything else on the bus. But it could also be protected with suitable components. There are also all-in-one isolating tranceivers that don't need external isolated supply. – Justme Dec 19 '19 at 22:14


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A 5000 cruzado banknote featuring Candido Portinari, overstamped as a 5 cruzados novos banknote
ISO 4217
Code BRC
Denominations
Subunit
1/100 centavo
Symbol Cz\$
Banknotes
Freq. used 10, 50, 100, 500, 1000, 5000, 10,000 cruzados
Coins
Freq. used 1, 5, 10 cruzados
Rarely used 1, 5, 10, 20, 50 centavos
Demographics
User(s)  Brazil
Issuance
Central bank Central Bank of Brazil
Website www.bc.gov.br
Printer Casa da Moeda do Brasil
Mint Casa da Moeda do Brasil

The cruzado was the currency of Brazil from 1986 to 1989. It replaced the second cruzeiro (at first called the "cruzeiro novo") in 1986, at a rate of 1 cruzado = 1000 cruzeiros (novos) and was replaced in 1989 by the cruzado novo at a rate of 1000 cruzados = 1 cruzado novo.

This currency was subdivided in 100 centavos and it had the symbol ${\displaystyle \mathrm {CzS} \!\!\!\Vert }$ and the ISO 4217 code BRC.

## Coins

Stainless-steel coins were introduced in 1986 in denominations of 1, 5, 10, 20 and 50 centavos, and 1 and 5 cruzados, with 10 cruzados following in 1987. Coin production ceased in 1988.

100 cruzado coins commemorating the centenary of the abolition of slavery (the Lei Áurea) were produced in 1988. Although rare in circulation, the designs were carried over into the series issued between 1989 and 1991.

## Banknotes

The first banknotes were overprints on cruzeiro notes, in denominations of 10, 50 and 100 cruzados. Regular notes followed in denominations of 10, 50, 100 and 500 cruzados, followed by 1000 cruzados in 1987, 5000 and 10,000 cruzados in 1988.


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# Companion Computer Peripherals

This section contains information about peripherals for connecting Pixhawk (PX4) to a companion computer, and peripherals that might be connected to such a companion computer (and potentially triggered/accessed by PX4).

## Companion/Pixhawk Communication

Typical companion computer work with Pixhawk requires a companion link to transmit/receive the data between the companion computer and Pixhawk hardware (e.g. Intel NUC and Pixhawk 4).

There are a few devices that allow this communication bridge such as FTDI USB breakouts and level shifters (see below).

PX4 configuration for communicating with a companion computer over MAVLink configuration is covered in MAVLink (OSD / Telemetry). Relevant topics/sections in the developer guide include: Companion Computer for Pixhawk Series, Robotics and RTPS/ROS2 Interface: PX4-FastRTPS Bridge.

### FTDI Devices

The FTDI USB adapters are the most common way of communicating between companion computer and Pixhawk. They are usually plug and play as long as the IO of the adapter is set to 3.3v. In order to utilize the full capability/reliability of the serial link offered on the Pixhawk hardware, flow control is recommended.

#### Where To Buy and Features

Device 3.3v IO (Default) Flow Control Tx/Rx LEDs JST-GH
PixDev FTDI JST-GH Breakout Yes Yes Yes Yes
mRo USB FTDI Serial to JST-GH (Basic) Capable Capable No Yes
SparkFun FTDI Basic Breakout Yes No Yes No

### Logic Level Shifters

On occasion a companion computer may expose hardware level IO that is often run at 1.8v or 5v, while the Pixhawk hardware operates at 3.3v IO. In order to resolve this, a level shifter can be implemented to safely convert the transmitting/receiving signal voltage.

## Cameras

Cameras are among the most common form of data collection on a drone whether it be hobbyist, educational, or industrial use.

### Stereo Cameras

Stereo cameras are typically used for depth perception, path planning and SLAM. They are in no way guaranteed to be plug and play with your companion computer.

## Data Telephony (LTE)

An LTE USB module can be attached to a companion computer and used to route MAVLink traffic between the flight controller and the Internet.

There is no "standard method" for a ground station and companion to connect over the Internet. Generally you can't connect them directly because neither of them will have a public/static IP on the Internet.

Typically your router (or the mobile network) has a public IP address, and your GCS computer/vehicle are on a local network. The router uses network address translation (NAT) to map outgoing requests from your local network to the Internet, and can use the map to route the responses back to requesting system. However NAT has no way to know where to direct the traffic from an arbitrary external system, so there is no way to initiate a connection to a GCS or vehicle running in the local network.

A common approach is to set up a virtual private network between the companion and GCS computer (i.e. install a VPN system like zerotier on both computers). The companion then uses mavlink-router to route traffic between the serial interface (flight controller) and GCS computer on the VPN network.

This method has the benefit that the GCS computer address can be static within the VPN, so the configuration of the mavlink router does not need to change over time. In addition, the communication link is secure because all VPN traffic is encrypted (MAVLink 2 itself does not support encryption).

You can also choose to route to the VPN broadcast address (i.e. x.x.x.255:14550, where 'x' depends on the VPN system). This approach means that you do not need to know the IP address of the GCS computer, but may result in more traffic than desired (since packets are broadcast to every computer on the VPN network).

Some USB modules that are known to work include:

• Huawei E8372 and Huawei E3372
• The E8372 includes WiFi which you can use to configure the SIM while it is plugged into the companion (making the development workflow a little easier). The E3372 lacks WiFi, so you have to configure it by plugging the stick into a laptop.


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• Home
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• What is an Equilateral Triangle – Definition, Properties & Examples

# What is an Equilateral Triangle – Definition, Properties & Examples

December 26, 2022

This post is also available in: हिन्दी (Hindi)

A triangle is a 2D shape having three sides, three vertices, and three angles. Although the sum of three angles of a triangle is always $180^{\circ}$, there are various types of triangles classified according to their length of sides – equilateral, isosceles, or scalene, or measures of angles – acute, right or obtuse.

Let’s understand what is an equilateral triangle and its properties.

## What is an Equilateral Triangle? – Equilateral Triangle Definition

An equilateral triangle is a triangle in which all three sides are equal and the angles are also equal. The value of each angle of an equilateral triangle is $60^{\circ}$ therefore, it is also known as an equiangular triangle. The equilateral triangle is a regular polygon or a regular triangle as angles are equal and sides are also equal.

In the above figure, $\triangle \text{ABC}$ is an equilateral. The length of all its sides is equal, i.e., $\text{AB} = \text{BC} = \text{CA} = a$. Also, the measure of all three angles is equal, i.e., $\angle \text{A} = \angle \text{B} = \angle \text{C} = 60^{\circ}$.

## Shape of Equilateral Triangle

The shape of an equilateral triangle is regular. The word ‘Equilateral’ is formed by the combination of two words, i.e., ‘Equi’ meaning equal, and ‘Lateral’ meaning sides. An equilateral triangle is also called a regular polygon or regular triangle since all its sides are equal.

## Properties of Equilateral Triangle

The following are the important properties of an equilateral triangle that helps to distinguish it from other types of triangles.

• The sum of all the angles of an equilateral triangle is equal to $180^{\circ}$.
• In an equilateral all three sides are equal.
• All three angles are congruent(or qual) to $60^{\circ}$ in a equilateral triangle.
• An equilateral triangle is a regular polygon with three sides.
• The perpendicular drawn from the vertex of the equilateral triangle to the opposite side bisects it into equal halves.
• The perpendicular drawn from the vertex of the equilateral triangle divides the angle from where the perpendicular is drawn into two equal angles, i.e., $30^{\circ}$ each.
• The orthocentre, incentre, and centroid are at the same point.
• In an equilateral triangle, the median, the angle bisector, and the altitude for all sides are the same.
• The area of an equilateral triangle is $\frac{\sqrt{3}a^2}{4}$
• The perimeter of an equilateral triangle is $3a$.

## Each Angle of an Equilateral Triangle is $60^{\circ}$

Let $\triangle \text{ABC}$ be an equilateral triangle, then $\angle \text{A} = \angle \text{B} = \angle \text{C} = 180^{\circ}$.

Let’s prove the above statement.

In a triangle the sum of three angles is $180^{\circ}$ (Angle Sum Property of Triangle)

Therefore, $\angle \text{A} + \angle \text{B} + \angle \text{C} = 180^{\circ}$ ————– (1)

Also, $\angle \text{A} = \angle \text{B} = \angle \text{C}$ ($\triangle \text{ABC}$ is an equilateral triangle) ———— (2)

From (1) and (2),  $\angle \text{A} + \angle \text{A} + \angle \text{A} = 180^{\circ}$

$=> 3\angle \text{A} = 180^{\circ} => \angle \text{A} = 60^{\circ}$

From (2) $\angle \text{A} = \angle \text{B} = \angle \text{C} = 60^{\circ}$.

## Perpendicular Drawn From the Vertex Divides an Equilateral Triangle into Two Equal Halves

Let $\triangle \text{ABC}$ be an equilateral triangle and a perpendicular $\text{AD}$ is drawn the vertex $\text{A}$ on the opposite side $\text{BC}$, then $\triangle \text{ABD} \cong \triangle \text{ADC}$.

Let’s prove the above statement.

In $\triangle \text{ABD}$ and $\triangle \text{ADC}$

$\text{AB} = \text{AC}$ (Sides of an equilateral triangle)

$\text{AD} = \text{AD}$ (Common Side)

$\angle \text{ADB} = \angle \text{ADC} = 90^{\circ}$ ($\text{AD} \perp \text{BC}$)

Therefore, $\triangle \text{ABD} \cong \triangle \text{ADC}$ (RHS Congruence Criterion)

Thus, $\triangle \text{ABD}$ and $\triangle \text{ADC}$ are equal.

## Perpendicular Drawn From the Vertex Divides The Angle into Two Equal Angles

Let $\triangle \text{ABC}$ be an equilateral triangle and a perpendicular $\text{AD}$ is drawn the vertex $\text{A}$ on the opposite side $\text{BC}$, then $\angle \text{BAD} = \angle \text{DAC}$.

Let’s prove the above statement.

In $\triangle \text{ABD}$ and $\triangle \text{ADC}$

$\text{AB} = \text{AC}$ (Sides of an equilateral triangle)

$\text{AD} = \text{AD}$ (Common Side)

$\angle \text{ADB} = \angle \text{ADC} = 90^{\circ}$ ($\text{AD} \perp \text{BC}$)

Therefore, $\triangle \text{ABD} \cong \triangle \text{ADC}$ (RHS Congruence Criterion)

Thus, $\angle \text{BAD} = \angle \text{DAC}$ (Corresponding Parts of Congruent Triangles)

## Equilateral Triangle Theorem

Let $\triangle \text{ABC}$ be an equilateral triangle and $\text{P}$ be a point on the arc $\widehat{\text{BC}}$ of the circumcircle of the $\triangle \text{ABC}$, then according to equilateral triangle theorem $\text{PA} = \text{PB} + \text{PC}$.

Let’s prove the above statement.

For a cyclic quadrilateral $\text{ABPC}$, we have $\text{PA}⋅\text{BC}=\text{PB}⋅\text{AC}+\text{PC}⋅\text{AB}$

Also, for an equilateral triangle ABC, $\text{AB} = \text{BC} = \text{AC}$

Therefore, $\text{PA}.\text{AB} = \text{PB}.\text{AB}+\text{PC}.\text{AB}$

Taking $\text{AB}$ as a common, we get $\text{PA}.\text{AB}=\text{AB} \left(\text{PB}+\text{PC} \right)$

$=>\text{PA} = \text{PB} + \text{PC}$.

## Difference Between Equilateral, Isosceles, and Scalene Triangle

The following are the main differences between equilateral, isosceles, and scalene Triangle

## Practice Problems

1. What is meant by an equilateral triangle?
2. Prove that the sum of each angle of an equilateral triangle is $60^{\circ}$
3. Prove that a perpendicular drawn from the vertex of the equilateral triangle to the opposite side bisects it into equal halves.
4. Prove that a perpendicular drawn from the vertex of the equilateral triangle divides the angle from where the perpendicular is drawn into two equal angles.

## FAQs

### What is meant by an equilateral triangle?

An equilateral triangle is a triangle in which all sides are equal and angles are also equal. The value of each angle of an equilateral triangle is $60^{\circ}$ therefore, it is also known as an equiangular triangle.

### What is an equilateral triangle and its formula?

An equilateral triangle is a triangle in which all sides are equal and angles are also equal. For an equilateral triangle, $\text{ABC}$ formulas are
a) $\angle \text{A} = \angle \text{B} = \angle \text{C}$
b) $\text{Perimeter} = 3a$ ($a$ is the length of each side)
c) $\text{Area} = \frac{\sqrt{3}}{4}a^2$ ($a$ is the length of each side)

### Why is it called an equilateral?

The word ‘Equilateral’ is formed by the combination of two words, i.e., ‘Equi’ meaning equal, and ‘Lateral’ meaning sides. An equilateral triangle is also called a regular polygon or regular triangle since all its sides are equal.

## Conclusion

An equilateral triangle is a triangle in which all three sides are equal and the angles are also equal. The value of each angle of an equilateral triangle is $60^{\circ}$ therefore, it is also known as an equiangular triangle. In an equilateral triangle, the medians, altitudes, and angle bisectors are the same lines and meet at the same point.


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## What is Bond Yield?

Bond yield is the return an investor realizes on a bond. The bond yield can be defined in different ways. Setting the bond yield equal to its coupon rate is the simplest definition. The current yield is a function of the bond's price and its coupon or interest payment, which will be more accurate than the coupon yield if the price of the bond is different than its face value. More complex calculations of a bond's yield will account for the time value of money and compounding interest payments. These calculations include yield to maturity (YTM), bond equivalent yield (BEY) and effective annual yield (EAY). (Discover the difference between Bond Yield Rate vs. Coupon Rate).

1:56

## Overview of Bond Yield

When investors buy bonds, they essentially lend bond issuers money. In return, bond issuers agree to pay investors interest on bonds through the life of the bond and to repay the face value of bonds upon maturity. The simplest way to calculate a bond yield is to divide its coupon payment by the face value of the bond. This is called the coupon rate.

$\text{Coupon Rate}=\frac{\text{Annual Coupon Payment}}{\text{Bond Face Value}}$

If a bond has a face value of $1,000 and made interest or coupon payments of$100 per year, then its coupon rate is 10% ($100 /$1,000 = 10%). However, sometimes a bond is purchased for more than its face value (premium) or less than its face value (discount), which will change the yield an investor earns on the bond.

## Bond Yield Vs. Price

As bond prices increase, bond yields fall. For example, assume an investor purchases a bond that matures in five years with a 10% annual coupon rate and a face value of $1,000. Each year, the bond pays 10%, or$100, in interest. Its coupon rate is the interest divided by its par value.

If interest rates rise above 10%, the bond's price will fall if the investor decides to sell it. For example, imagine interest rates for similar investments rise to 12.5%. The original bond still only makes a coupon payment of $100, which would be unattractive to investors who can buy bonds that pay$125 now that interest rates are higher.

If the original bond owner wants to sell her bond, the price can be lowered so that the coupon payments and maturity value equal yield of 12%. In this case, that means the investor would drop the price of the bond to $927.90. In order to fully understand why that is the value of the bond, you need to understand a little more about how the time value of money is used in bond pricing, which is discussed later in this article. If interest rates were to fall in value, the bond's price would rise because its coupon payment is more attractive. For example, if interest rates fell to 7.5% for similar investments, the bond seller could sell the bond for$1,101.15. The further rates fall, the higher the bond's price will rise, and the same is true in reverse when interest rates rise.

In either scenario, the coupon rate no longer has any meaning for a new investor. However, if the annual coupon payment is divided by the bond's price, the investor can calculate the current yield and get a rough estimate of the bond's true yield.

$\text{Current Yield}=\frac{\text{Annual Coupon Payment}}{\text{Bond Price}}$

The current yield and the coupon rate are incomplete calculations for a bond's yield because they do not account for the time value of money, maturity value or payment frequency. More complex calculations are needed to see the full picture of a bond's yield.

## Yield to Maturity

A bond's yield to maturity (YTM) is equal to the interest rate that makes the present value of all a bond's future cash flows equal to its current price. These cash flows include all the coupon payments and its maturity value. Solving for YTM is a trial and error process that can be done on a financial calculator, but the formula is as follows:

\begin{aligned} &\text{Price}=\sum^T_{t-1}\frac{\text{Cash Flows}_t}{(1+\text{YTM})^t}\\ &\textbf{where:}\\ &\text{YTM}=\text{ Yield to maturity} \end{aligned}

In the previous example, a bond with $1,000 face value, five years to maturity and$100 annual coupon payments was worth $927.90 in order to match a YTM of 12%. In that case, the five coupon payments and the$1,000 maturity value were the bond's cash flows. Finding the present value of each of those six cash flows with a discount or interest rate of 12% will determine what the bond's current price should be.

## Bond Equivalent Yield – BEY

Bond yields are normally quoted as a bond equivalent yield (BEY), which makes an adjustment for the fact that most bonds pay their annual coupon in two semi-annual payments. In the previous examples, the bonds' cash flows were annual, so the YTM is equal to the BEY. However, if the coupon payments were made every six months, the semi-annual YTM would be 5.979%.

The BEY is a simple annualized version of the semi-annual YTM and is calculated by multiplying the YTM by two. In this example, the BEY of a bond that pays semi-annual coupon payments of \$50 would be 11.958% (5.979% X 2 = 11.958%). The BEY does not account for the time value of money for the adjustment from a semi-annual YTM to an annual rate.

## Effective Annual Yield – EAY

Investors can find a more precise annual yield once they know the BEY for a bond if they account for the time value of money in the calculation. In the case of a semi-annual coupon payment, the effective annual yield (EAY) would be calculated as follows:

\begin{aligned} &\text{EAY}=\left(\frac{1+\text{YTM}}{2}\right)^2-1\\ &\textbf{where:}\\ &\text{EAY}=\text{ Effective annual yield}\\ \end{aligned}

If an investor knows that the semi-annual YTM was 5.979%, then he or she could use the previous formula to find the EAY of 12.32%. Because the extra compounding period is included, the EAY will be higher than the BEY.

## Complications Finding a Bond's Yield

There are a few factors that can make finding a bond's yield more complicated. For instance, in the previous examples, it was assumed that the bond had exactly five years left to maturity when it was sold, which would rarely be the case.

When calculating a bond's yield, the fractional periods can be dealt with simply; the accrued interest is more difficult. For example, imagine a bond has four years and eight months left to maturity. The exponent in the yield calculations can be turned into a decimal to adjust for the partial year. However, this means that four months in the current coupon period have elapsed and there are two more to go, which requires an adjustment for accrued interest. A new bond buyer will be paid the full coupon, so the bond's price will be inflated slightly to compensate the seller for the four months in the current coupon period that have elapsed.

Bonds can be quoted with a "clean price" that excludes the accrued interest or the "dirty price" that includes the amount owed to reconcile the accrued interest. When bonds are quoted in a system like a Bloomberg or Reuters terminal, the clean price is used.

## Bond Yield Summary

A bond's yield is the return to an investor from the bond's coupon and maturity cash flows. It can be calculated as a simple coupon yield, which ignores the time value of money and any changes in the bond's price or using a more complex method like yield to maturity. The yield to maturity is usually quoted as a bond equivalent yield (BEY), which makes bonds with coupon payment periods less than a year easy to compare. A classic strategy is to use a bond ladder technique to maximize profits with multiple bonds coming into maturity at different times.

Bonds can be purchased through a variety of different sources. A common way to go about purchasing some bond types is to use an investment account through a broker. (For related reading, see "What Do Constantly Low Bond Yields Mean for the Stock Market?")


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# Is FPV a specific category of drones or just a way to use them?

I'm approaching the drone world and I'm studying the basics before buying the first one or attending a course.

I would use them for personal pleasure, not for work nor for racing or group flying. But I'd love the FPV way to fly them. Most of the drones on the market have a smartphone holder on the remote so you can see the video stream in real-time when flying.

I'm also aware that you can put the smartphone inside a "headset" (like the ones for VR). Hence, if I'm not wrong almost any drone can be used with an "FPV" approach.

What I'm asking here is if my assumption is correct, or perhaps there are drones specifically built to be flown in FPV. Again, I'm not interested in racing or acrobatic features.

I'm thinking for example about the latency of the video streaming: if I fly keeping an eye on the drone I can live with some latency, but if I rely on the camera only I guess the latency should be as low as possible.

Bottom line: can I pick up any drone out there or should I pay attention to some specific features?

## UPDATE

After the very useful information I received now I better understand the scenario, so I have to update my question - hoping to not going off-topic.

My final use of the drone is for surveillance on my farmland and helping with farm works. I thought FPV was the best choice because I would "feel" flying above, but now I'm not sure if I was right.

I'm not sure I entirely understand what you're specifically asking about when you say "is FPV a category of drones", but here's my answer:

Fundamentally, all FPV (First Person View) means is that the pilot maneuvers by using some sort of camera feed from the drone. It is true that this often requires that the drone have special hardware (e.g. camera and VTX) to support this feature, but I wouldn't say that this means FPV represents a special subset of RC aviation.

For example, FPV flight is one of the core tenets of RC multirotors (including quadcopters) and it would be difficult to find a modern commercial or hobbyist multirotor aircraft without the ability to easily add hardware to support FPV. (assuming it isn't already supported "out of the box" and excluding ultra-cheap models for <$50 USD) I don't think you've really described whether you're interested in purchasing a Ready-to-Fly (RTF) model or assembling your own from parts, so it's hard to make more specific recommendations as to what to look for in product listings. Generally, though, FPV is a major selling point for any drone, so it likely isn't supported if not explicitly mentioned in the product listing. • Sorry for the missing information. I was going to find an RTF model but as an electronic engineer I'm very interested in assembling my own as well. My budget is >$500 so I guess there are chances to find a suitable model that supports FPV.
– Mark
Dec 29, 2021 at 8:15

ifconfig's answer is probably what you're looking for, but I thought this might help answer your question: FPV is more a category of drones than a way to fly them.

For example, you can fly any drone (speaking generally) smooth and slow, or perform sharp maneuvers, depending on what you feel like. Regardless of how a drone is built or what type of components you have, you can fly it in a great variety of styles.

The category of drone, on the other hand, is something that is specific to the drone and is not dependent on how you fly. A cinematography platform has a professional camera and is usually large, a helicopter has fewer rotors than a quadcopter, and an FPV drone has the capability to stream video to a first-person view monitor. None of these things necessarily limit how you can fly, they only categorize the drone by what components it has.

The fact that a drone has FPV capability doesn't dictate whether you can fly it around trees or just in a straight line, it only means that the drone has certain components that put it into the category of "FPV". Hence why FPV is more a category of drones than a way to use a drone.

I'd say that yes, it is a category (although you can also FPV a plane or car too)

IMHO, FPV (or race) drones have fixed cameras and are flown entirely by the camera.

The alternative are photography drones, which have a gimbaled camera. The gimbal isolates the camera from the movement and attitude of the drone, which gives steadier video and the choice of camera direction, but it means that half the pilot's effort goes into aiming the camera and the attitude of the drone is not intuitive. As as result these drones need a GPS and barometer, and a flight controller that positions the drone relative to these inputs. That makes them easy enough to fly that you can concentrate on aiming the camera and framing the shot.

This gives a very different flying experience. With an FPV drone you tell the drone "pitch forward at 10 degrees per second" until you judge that it's at the right angle, then you tell it to hold it's attitude while it speeds up, and you adjust the throttle to maintain height and the other controls to steer. With a photography drone you just tell it "fly north at 10mph" and it works out what angle is required and compensates for cross winds. The experience is more like being a movie director than a pilot.

That different flying experience has led to two very different hobbies. FPV for the fun of flying, and aerial photography.

Of course, you can fix a gimbled camera and fly it FPV, but it's an expensive and fragile way to do it. Similarly lots of FPV drones have HD cameras and take nice video, but they are limited in the angles they can take, and typically built for speed and power (and strength) rather than duration.

Your question might not be too much related to the aircraft, but more about the level of the gimbal movement freedom and controllment.

Here is the phenogram diagram for your reference.

UAV -- Hardware -- Quadcopter -- With Gimbal & Camera -- Gimbal mode: Free
\_ Gimbal mode: Follow
\_ Gimbal mode: FPV (Locked mode)

The most freedom of camera operation is the first type. To operate this type of drone, you may need a second operator, whose role is to be responsible for the operation of the camera movement.

The type of drones is also the most expensive one. It is also a standard to define the professional drone or consumer drone.

Almost all of the consumer drones use follow-mode gimbal in the current market.

And, if you lock the gimbal with the aircraft, it will be like what people say the "FPV mode".

-- Update:

Refer to DJI's terminologies:

Free Mode: Pitch, roll and yaw are all controllable, meaning the gimbal can move independently of the product's yaw. In this mode, even if the product yaw changes, the camera will continue pointing in the same world direction.

FPV (First Person View) Mode: Only pitch is controllable. Yaw and roll will be fixed relative to the product while pitch remains controllable.

Illustration:

As you can see in the illustration, the drones can all be exactly the same, but the camera's gimbal makes different.

A drone that can provide Yaw follow mode can also provide FPV mode, but a drone with FPV mode gimbal cannot provide the Yaw follow mode.

This is because the drone with FPV mode gimbal has only one degree of movement freedom in its gimbal.

Just a quick comparison

• Interesting, thanks! May you just describe a little bit more the "follow" mode? I'm not sure to fully understand how it works.
– Mark
Jan 16 at 8:12
• I've updated with DJI's terminologies as above. Jan 16 at 10:05
• The difference between them is the degree of control of the gimbal. Jan 16 at 10:07

FPV is a specific build and is NOT a good choice for your application. The primary use for FVPV drones is chase, which could be for pleasure, could be business, etc.

If you want to chase animals, perfect fit.

A typical race or freestyle drone is ALL POWER and short battery life. Even adjusting pitch and KV you will still have shorter flights than a non-FPV. Your best bet is a sub-250 gram DJI unit.