text:
### Sukarna_Paul's blog

By Sukarna_Paul, history, 6 weeks ago, ,

• +2

By Sukarna_Paul, history, 4 months ago, ,

How can I approach for this problem ? https://codeforces.com/gym/102055/problem/L

• 0

By Sukarna_Paul, history, 4 months ago, ,

How to avoid TLE for this problem as the number of test cases can be 10^4 .

• 0

By Sukarna_Paul, history, 5 months ago, ,

Here is the problem link UVA Live 6844

• 0

By Sukarna_Paul, history, 5 months ago, ,

I have been trying to solve this problem for last few days . Here is the Problem link SPOJ PTRI .

I have used Linear sieve , but the limit is so high that it is getting TLE in O(N) . I have tried it with sieve of aktin also. But still got TLE . Please help me to solve this one.

I found this problem in -Morass- blog Link . You can visit it , I am confident that you will find it useful.

• +11

By Sukarna_Paul, history, 5 months ago, ,

Here is my solution :

#pragma warning(disable:4786)
#pragma warning(disable:4996)
#include<bits/stdc++.h>
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define pli pair<long long , int>
#define pil pair<int ,long long>
#define pi acos(-1)
#define pb push_back
#define mkp make_pair
#define all(a) a.begin(), a.end()
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define ll long long
#define MAX 300005
#define INF 0x3f3f3f3f
template <class T> inline T bigmod(T p,T e,T M){ll ret = 1LL;for(; e > 0LL; e >>= 1LL){if(e & 1LL) ret = (ret * p) % M;p = (p * p) % M;}return (T)ret;}
template <class T> inline T modinverse(T a,T M){return bigmod(a,M-2,M);}   // M is prime}
using namespace std;

vector <ll> prime;
bitset <100000010> bs;

void sieve ()
{
bs.set ();
prime.pb(2LL);
for (ll i = 3; i < 100000000; i+=2)
{
if (bs[i])
{
prime.pb(i);
for (ll j = i * i; j < 10000000; j += i+i)
{
bs[j] = 0;
}
}
}
}

int main(){
IOS
//freopen("output.txt","w",stdout);
sieve();
int i=0;
unordered_map<int,int>r,c;

for(auto it=prime.begin();it!=prime.end();){
i++;
for(int j=1;j<=i;j++){
r[(*it)]=i;
c[(*it)]=j;
it++;
}
}
int t;
cin>>t;
for(int i=0;i<t;i++){
int a;
cin>>a;
if(r[a]==0){
cout<<"-1"<<endl;
continue;
}
cout<<r[a]<<" "<<c[a]<<endl;

}
}

• 0

By Sukarna_Paul, history, 5 months ago, ,

How Can I Create A Contest On Codeforces?

• 0

By Sukarna_Paul, history, 5 months ago, ,

This problem is from ACM ICPC Dhaka Regional 2018 (Contest Link)

How can I approach for this problem?

Problem Satement:

The function d(n) denotes the number of positive divisors of an integer n. For example d(24) = 8, because there are 8 divisors of 24 and they are 1, 2, 3, 4, 6, 8, 12 and 24. The function sndd(n) is a new function defined for this problem. This denotes “The summation of number of divisors of the divisors” of an integer n. For example,

sndd(24) = d(1) + d(2) + d(3) + d(4) + d(6) + d(8) + d(12) + d(24) = 1 + 2 + 2 + 3 + 4 + 4 + 6 + 8 = 30.

Given the value of n, you will have to find sndd(n!), here n! means factorial of n. So n! = 1 × 2 × 3 × … × n.

Input The input contains at most 1000 lines of test cases. Each line contains a single integer that denotes a value of n (1 ≤ n ≤ 10^6). Input is terminated by a line containing a single zero.

Output For each line of input produce one line of output. This line contains an integer that denotes the modulo 10000007 (or 10^7 + 7) value of sndd(n).

Sample Input 4 5 0

Sample Output 30 90

• +5

By Sukarna_Paul, history, 6 months ago, ,

• 0

By Sukarna_Paul, history, 7 months ago, ,

How can I approach for this Geometry problem . Here is the link URI 1291

• 0

By Sukarna_Paul, history, 7 months ago, ,

How can I approach for this problem Problem Link

• +3

By Sukarna_Paul, history, 8 months ago, ,

How can I approach for this number theory problem ? [Problem Link](http:// https://codeforces.com/contest/1070/problem/A)

A. Find a Number time limit per test3 seconds memory limit per test256 megabytes inputstandard input outputstandard output

You are given two positive integers d and s. Find minimal positive integer n which is divisible by d and has sum of digits equal to s.

Input The first line contains two positive integers d and s (1≤d≤500,1≤s≤5000) separated by space.

Output Print the required number or -1 if it doesn't exist.

Examples input 13 50 output 699998

input 61 2 output 1000000000000000000000000000001

input 15 50 output -1

• 0

By Sukarna_Paul, history, 8 months ago, ,

How can I approach for this problem ?

The number of divisor function or d(n) is a very interesting function in number theory. It denotes the number of positive divisors of a particular number. For example d(24) = 8 as 24 has eight divisors 1, 2, 3, 4, 6, 8, 12 and 24 . In mathematics factorial of a positive integer number n is written as n! and is defined as below: n! = 1 × 2 × 3 × · · · × n

Another interesting function AF(n) (Again factorial in short) is defined as: AF(n) = 1! × 2! × 3! × . . . × n!

Given n , your job is to find the value of d(AF(n)).

Input The input file contains at most 101 lines of inputs. Each line contains an integer n (0 < n < 5000001) . Input is terminated by a line containing a single zero. This value should not be processed.

Output For each line of input produce one line of output.

This line contains the modulo 100000007 (10^8 + 7) of d(AF(n)).

Sample Input 1 2 3 4 100 0

Sample Output 1 2 6 18 59417661

• +1

By Sukarna_Paul, history, 11 months ago, ,

Some Big integer problems for beginners Solving BigInteger Problem is fun. There are many problem solvers around the world who come to use Java BigInteger class though they do not code in Java usually. Here are some simple problems with their links and solution. This problem are chosen from different online judges. Try yourself before you see the solution.

Codeforces Gym : 112

# Solution

import java.util.Scanner ;
import java.math.BigInteger;
public class Main{
public static void main(String[] args){
int a,b;
Scanner input = new Scanner(System.in);
a = input.nextInt();
b = input.nextInt();
BigInteger A = BigInteger.valueOf(a);
BigInteger B= BigInteger.valueOf(b);
System.out.printf("%d",A.pow(b).subtract(B.pow(a)));

}
}

Uva 10183 — How Many Fibs?

# Solution

import java.util.*;
import java.math.*;
public class Main{
public static void main(String[] args){
Scanner input = new Scanner(System.in);
BigInteger arr[] = new BigInteger[50000];
arr[0]= BigInteger.valueOf(1);
arr[1]= BigInteger.valueOf(2);
for(int i=2;i<50000;i++){
}
while(true){

BigInteger a,b;
a = input.nextBigInteger();
b = input.nextBigInteger();
if(b.compareTo(BigInteger.valueOf(0))==0){
break;
}
int count=0;
for(int i=0;i<50000;i++){
if(arr[i].compareTo(a)>=0 && arr[i].compareTo(b)<=0){
count++;
}
if(arr[i].compareTo(b)>0){
break;
}
}
System.out.println(count);
}
}
}

URI 1279

# Solution

import java.util.*;
import java.math.*;
import java.io.*;
public class Main{
public static void main(String[] args){
Scanner input = new Scanner(System.in);
boolean start = true;
while(input.hasNext()){
boolean leap_year = false;
boolean ordinary = true;
if(start == false){
System.out.print("\n");
}
start = false;
BigInteger year;
year = input.nextBigInteger();
BigInteger four;
four = BigInteger.valueOf(4);
BigInteger fourh;
fourh = BigInteger.valueOf(400);
BigInteger oneh;
oneh = BigInteger.valueOf(100);
BigInteger temp = year.remainder(four);
if(temp.compareTo(BigInteger.valueOf(0))==0){
temp = year.remainder(oneh);
if(temp.compareTo(BigInteger.valueOf(0))==0){
temp = year.remainder(fourh);
if(temp.compareTo(BigInteger.valueOf(0))==0){
System.out.println("This is leap year.");
leap_year = true;
ordinary = false;
}
}
else{
System.out.println("This is leap year.");
leap_year = true;
ordinary = false;
}
}
temp = year.remainder(BigInteger.valueOf(15));
if(temp.compareTo(BigInteger.valueOf(0))==0){
System.out.println("This is huluculu festival year.");
ordinary = false;
}
if(leap_year==true){
temp = year.remainder(BigInteger.valueOf(55));
if(temp.compareTo(BigInteger.valueOf(0))==0){
System.out.println("This is bulukulu festival year.");
ordinary = false;
}
}
if(ordinary == true){
System.out.println("This is an ordinary year.");
}
}
}
}

Project Euler Problem 13

# Solution

import java.util.*;
import java.math.*;

public class problem13{
public static void main(String[] args){
Scanner input = new Scanner(System.in);
BigInteger sum, a;
sum = BigInteger.valueOf(0);
for(int i=0;i<100;i++){
a = input.nextBigInteger();
}
System.out.println(sum);    //take the first 10 digits manually
}
}

Project Euler Problem 15

# Solution

import java.math.*;

public class Problem15{
public static void main(String[] args){
BigInteger n=BigInteger.valueOf(1);
BigInteger ans=BigInteger.valueOf(1);
for(int i=0;i<40;i++){
ans=ans.multiply(n);
}
n=BigInteger.valueOf(1);
BigInteger ans2=BigInteger.valueOf(1);
for(int i=0;i<20;i++){
ans2=ans2.multiply(n);
}
System.out.println(ans.divide(ans2.multiply(ans2)));
}
}

Project Euler Problem 16

# Solution

import java.math.*;

public class Problem16{
public static void main(String[] args){
//BigInteger n=BigInteger.valueOf(1000);
BigInteger two = BigInteger.valueOf(2);
BigInteger ans = two.pow(1000);
String s = ""+ans;
System.out.println(s);
int sum=0;
for(int i=0;i<s.length();i++){
sum+=s.charAt(i)-'0';
}
System.out.println(sum);
}
}

Thank you. Happy coding.


text:
AES算法特性

8.PKCS7填充方式：数据末尾为填充的个数
如数据1:
11 22 33 44 55 66 77 88 99 AA BB

11 22 33 44 55 66 77 88 99 AA BB 05 05 05 05 05

11 22 33 44 55 66 77 88 99 AA BB CC DD EE FF FF

11 22 33 44 55 66 77 88 99 AA BB CC DD EE FF FF
10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10

format: plain


text:
Rego Unit Testing

|

Summary

This post is going to outline some basics, interesting tidbits, and caveats on unit testing rego policies.

Unit tests

Let’s start with an obvious question. What is a unit test? Compared to some of the other types of testing in tech, this one is pretty self-explanatory. A unit test is a way to test individual components (or units) of a system. For example, if we had a rego policy that is supposed to deny requests that allow port 80, we’d write a unit test that sent an input that included that port with the expectation that we’d get a deny message back. This gives us a way to validate our policies without having to deploy them to a real system. It also gives us a way to validate that policy changes don’t introduce any insecurities into our environment.

Generally speaking, unit tests are only needed for custom logic and shouldn’t be used for prebuilt rules/functions of a tool. Also, while we’re making general assumptions, do not strive for 100% test coverage. If you are using a provided library (e.g. Styra DAS), the provider already does testing so there’s no need to repeat it. Unit tests can also add considerable delays to CI/CD when testing new policy changes.

Example

Let’s take a look at an example policy. This policy allows requests to POST to the users path.

# example.rego

package authz

allow {
input.path == ["users"]
input.method == "POST"
}

Now let’s take a look at the accompanying unit test.

# example_test.rego

test_post_allowed {
allow with input as {"path": ["users"], "method": "POST"}
}

Let’s break this down.

• We have a test definition named test_post_allowed
• The test calls the specific policy definition and passes input to it.
• allow is the name of the policy definition
• with is a rego keyword that allows queries to access values in an input object.
• {"path": ["users"], "method": "POST"} is the test data being used as the input.

You can also store more complex data in a variable in the test definition.

test_post_allowed {
in := {"path": ["users"], "method": "POST"}
allow with input as in
}

Now that we have a test, let’s actually run it. Let’s look at two ways we can accomplish this.

• Here’s our example in Rego Playground. It’s easy enough, load the page and click Evaluate.
• Now let’s try it with OPA.
• Let’s put our policy definition in a file example.rego and our test definition in a file example_test.rego
• Now let’s execute the tests by running:
% opa test . -v
data.authz.test_post_allowed: PASS (3.697875ms)

Testing conventions

There are a few conventions for writing rego tests.

• Tests should be named <policyname>_test.rego. E.g. if your policy is ingress.rego, then your test should be named ingress_test.rego
• All definitions in the test file should start with test_ have a descriptive name. E.g. if your policy definition is allow {...}, then your test might be named test_post_allowed {...}

Unit testing in Styra DAS

If you are using Styra DAS there are a couple things to consider. The main thing is all of the policy definitions are summed up into a single policy. Let’s take a look at an example.

Here we have 2 definitions, but notice that both at named enforce. Recall that with our unit tests, we call the definition by name to execute a test.

enforce[decision] {
not excludedNamespaces[input.request.namespace]

decision := {
"allowed": false,
"message": message
}
}

enforce[decision] {

decision := {
"allowed": false,
"message": message
}
}

So how do we test this? Well, we have options:

1. Don’t write any tests at all. Since we’re only consuming pre-built content, there’s really no value in writing tests.
2. Write your tests so they test the policy as a whole. Provide “known good” input data in the test so all the definitions pass. This way, if a definition is changed, the test will fail.
3. If we really need to test individual definitions, we can give them specific names so we can call them separately. We lose some of the GUI functionality in DAS by doing this as the definitions become completely custom and not DAS managed. We also need to add an additional definition to include the result of our now custom one into the main DAS policy.

Let’s look at an example of the last option. We’ll use the same enforce definitions above but rename them so we can test them individually.

block_priv_mode[decision] {

decision := {
"allowed": false,
"message": message
}
}

require_audit[decision] {

decision := {
"allowed": false,
"message": message
}
}

enforce[decision] {
block_priv_mode[decision]
}

enforce[decision] {
require_audit[decision]
}

If you’re wondering why there are two enforce rules, remember that in Rego, multiple definitions with the same name act as an OR. In this case, we’re saying that we want enforce to be true if either block_priv_mode OR require_audit are true.

Now that we’ve got our DAS policy structured in a way that allows individual unit testing, let’s look at an example test.

Two things to note:

• On the second line, we import rules package.
• Since our rules package is imported as rules, we need to use that namespace to call the definitions we want to test. In this example, we use rules.block_priv_mode.
package policy["com.styra.kubernetes.validating"].test.test

import data.policy["com.styra.kubernetes.validating"].rules.rules

test_block_priv_mode {
in := {
"request": {
"kind": {
"kind": "Pod",
"version": "v1"
},
"object": {
"name": "myapp"
},
"spec": {
"containers": [
{
"image": "nginx:0.1.0",
"name": "nginx-frontend",
"securityContext": {
"privileged": false
}
},
]
}
}
}
}

actual := rules.block_priv_mode with input as in
count(actual) == 0
}

Closing

Hopefully this post has been helpful getting started. The Open Policy Agent documentation has a lot more info on policy testing

If you have any questions or feedback, please feel free to contact me: @jamesmassardo


text:
# smooth orientation change with quaternions

My camera orientation is looking in the $v_1$ direction. Something happens on direction $v_2$ and I want the camera to move smoothly to look at that direction.

So, to find the quaternion to go from $v_1$ to $v_2$ its easy. $v_1$ cross $v_2$ gives me the axis of the rotation. And v1 dot v2 gives me the angle.

Now, I want to move smoothly between the two vectors with this quaternion.

I though of using this formula of move point by quaternion for doing so -

$p'=v+2w(v\times p) +2(v\times (v\times p))$

When p is the initial point. $p'$ is the result point, and $v$ & $w$ are the parts of the quaternion

I use my $v_1$ vector instead of point p because its the same thing (right?) And I will use the same quaternion I have got from $v_1$ & $v_2$ cross and dot products but I will change its angle (inside the $w$ and vector parts) to move from 0 to the angle I have found from the dot product of $v_1$ & $v_2$. Now I use the set of quaternions on vector v1 and I will get a set of vectors that move from $v_1$ to $v_2$ smoothly so I can target my camera at them in a fixed time to get a smooth motion from $v_1$ to $v_2$.

Am I doing it right? I have heard of something called slerp but it confuses me because its input are 2 quaternions... ??? Thank you

Slerp is indeed what you want. The two quaternions you use describe an initial orientation $q_1$ and a final orientation $q_2$. The orientation is then $q = (q_2 q_1^{-1})^t q_i$. At $t=0$, $q = q_1$, and at $t=1$, $q=q_2$.
The initial orientation $q_1$ can be thought of as a rotation from some reference set of axes. The quaternion you found, which rotates the camera's forward direction from $v_1$ to $v_2$, is exactly $q_2 q_1^{-1}$.
Using quaternions for both the initial and final orientations is more flexible because you can better define the camera's overall orientation--just because it's pointing in the direction of $v_1$ doesn't tell you everything. The camera's body could (in an abstract case) rotate about the $v_1$ axis and yield a different orientation. The quaternion $q_1$ should be able to capture this.


text:
## 77.21 Functoriality of quotient stacks

A morphism of groupoids in algebraic spaces gives an associated morphism of quotient stacks.

Lemma 77.21.1. Let $S$ be a scheme. Let $B$ be an algebraic space over $S$. Let $f : (U, R, s, t, c) \to (U', R', s', t', c')$ be a morphism of groupoids in algebraic spaces over $B$. Then $f$ induces a canonical $1$-morphism of quotient stacks

$[f] : [U/R] \longrightarrow [U'/R'].$

Proof. Denote $[U/_{\! p}R]$ and $[U'/_{\! p}R']$ the categories fibred in groupoids over the base site $(\mathit{Sch}/S)_{fppf}$ associated to the functors (77.20.0.1). It is clear that $f$ defines a $1$-morphism $[U/_{\! p}R] \to [U'/_{\! p}R']$ which we can compose with the stackyfication functor for $[U'/R']$ to get $[U/_{\! p}R] \to [U'/R']$. Then, by the universal property of the stackyfication functor $[U/_{\! p}R] \to [U/R]$, see Stacks, Lemma 8.9.2 we get $[U/R] \to [U'/R']$. $\square$

Let $B \to S$ and $f : (U, R, s, t, c) \to (U', R', s', t', c')$ be as in Lemma 77.21.1. In this situation, we define a third groupoid in algebraic spaces over $B$ as follows, using the language of $T$-valued points where $T$ is a (varying) scheme over $B$:

1. $U'' = U \times _{f, U', t'} R'$ so that a $T$-valued point is a pair $(u, r')$ with $f(u) = t'(r')$,

2. $R'' = R \times _{f \circ s, U', t'} R'$ so that a $T$-valued point is a pair $(r, r')$ with $f(s(r)) = t'(r')$,

3. $s'' : R'' \to U''$ is given by $s''(r, r') = (s(r), r')$,

4. $t'' : R'' \to U''$ is given by $t''(r, r') = (t(r), c'(f(r), r'))$,

5. $c'' : R'' \times _{s'', U'', t''} R'' \to R''$ is given by $c''((r_1, r'_1), (r_2, r'_2)) = (c(r_1, r_2), r'_2)$.

The formula for $c''$ makes sense as $s''(r_1, r'_1) = t''(r_2, r'_2)$. It is clear that $c''$ is associative. The identity $e''$ is given by $e''(u, r) = (e(u), r)$. The inverse of $(r, r')$ is given by $(i(r), c'(f(r), r'))$. Thus we do indeed get a groupoid in algebraic spaces over $B$.

Clearly the maps $U'' \to U$ and $R'' \to R$ define a morphism $g : (U'', R'', s'', t'', c'') \to (U, R, s, t, c)$ of groupoids in algebraic spaces over $B$. Moreover, the maps $U'' \to U'$, $(u, r') \mapsto s'(r')$ and $R'' \to U'$, $(r, r') \mapsto s'(r')$ show that in fact $(U'', R'', s'', t'', c'')$ is a groupoid in algebraic spaces over $U'$.

Lemma 77.21.2. Notation and assumption as in Lemma 77.21.1. Let $(U'', R'', s'', t'', c'')$ be the groupoid in algebraic spaces over $B$ constructed above. There is a $2$-commutative square

$\xymatrix{ [U''/R''] \ar[d] \ar[r]_{[g]} & [U/R] \ar[d]^{[f]} \\ \mathcal{S}_{U'} \ar[r] & [U'/R'] }$

which identifies $[U''/R'']$ with the $2$-fibre product.

Proof. The maps $[f]$ and $[g]$ come from an application of Lemma 77.21.1 and the other two maps come from Lemma 77.20.2 (and the fact that $(U'', R'', s'', t'', c'')$ lives over $U'$). To show the $2$-fibre product property, it suffices to prove the lemma for the diagram

$\xymatrix{ [U''/_{\! p}R''] \ar[d] \ar[r]_{[g]} & [U/_{\! p}R] \ar[d]^{[f]} \\ \mathcal{S}_{U'} \ar[r] & [U'/_{\! p}R'] }$

of categories fibred in groupoids, see Stacks, Lemma 8.9.3. In other words, it suffices to show that an object of the $2$-fibre product $\mathcal{S}_ U \times _{[U'/_{\! p}R']} [U/_{\! p}R]$ over $T$ corresponds to a $T$-valued point of $U''$ and similarly for morphisms. And of course this is exactly how we constructed $U''$ and $R''$ in the first place.

In detail, an object of $\mathcal{S}_ U \times _{[U'/_{\! p}R']} [U/_{\! p}R]$ over $T$ is a triple $(u', u, r')$ where $u'$ is a $T$-valued point of $U'$, $u$ is a $T$-valued point of $U$, and $r'$ is a morphism from $u'$ to $f(u)$ in $[U'/R']_ T$, i.e., $r'$ is a $T$-valued point of $R$ with $s'(r') = u'$ and $t'(r') = f(u)$. Clearly we can forget about $u'$ without losing information and we see that these objects are in one-to-one correspondence with $T$-valued points of $R''$.

Similarly for morphisms: Let $(u'_1, u_1, r'_1)$ and $(u'_2, u_2, r'_2)$ be two objects of the fibre product over $T$. Then a morphism from $(u'_2, u_2, r'_2)$ to $(u'_1, u_1, r'_1)$ is given by $(1, r)$ where $1 : u'_1 \to u'_2$ means simply $u'_1 = u'_2$ (this is so because $\mathcal{S}_ U$ is fibred in sets), and $r$ is a $T$-valued point of $R$ with $s(r) = u_2$, $t(r) = u_1$ and moreover $c'(f(r), r'_2) = r'_1$. Hence the arrow

$(1, r) : (u'_2, u_2, r'_2) \to (u'_1, u_1, r'_1)$

is completely determined by knowing the pair $(r, r'_2)$. Thus the functor of arrows is represented by $R''$, and moreover the morphisms $s''$, $t''$, and $c''$ clearly correspond to source, target and composition in the $2$-fibre product $\mathcal{S}_ U \times _{[U'/_{\! p}R']} [U/_{\! p}R]$. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).


text:
• # question_answer    A uniform triangular wedge of mass M is kept on smooth horizontal surface. A horizontal force F is applied on the wedge, then net force on the upper portion ABC is (AB = l/2 and neglect any effect of rotation) A) F/4 B) F/3                  C) F D) 2F

 $M\propto \frac{1}{2}{{l}^{2}}$ Mass of ABC   $m\propto \frac{1}{2}\frac{{{l}^{2}}}{4}$              ;$m=\frac{M}{4},$   $F=Ma$ $F\,\,on\,\,ABC\,\,=\frac{M}{4}a=F/4$ $\therefore$ Ans. [A].


text:
# 4A-084 teledyne engines for sale

### Help Support HomeBuiltAirplanes.com:

##### Active Member
Mailing address is 50 Argyle Street.
Thanks!
Sorry it took so long but I did send them parts today I will send you a price and address later I’m out working right nowI sent some valves and valve springs and some other stuff with it I don’t need it anymore so you might as well have them

K - Thanks!

#### pwood66889

##### Well-Known Member
Got em, and thank you so much! I'm reading your address as:
Chris Bops
2115 N. Taylor
Springfield, Mo 65803
Please confirm! Don't you just love it when a deal comes together over the Internet and Jeff Bezos doesn't get a cut???

#### Armilite

##### Well-Known Member
A Good Example of 084 setup for Airplane use. Reliable 45hp is all your going to get if you want it to live very long. Can it make more hp, Yes, one was upgraded and Dynoed 60hp, but it didn't last long. They lasted about 1000 hrs if minor upgrades were done and treated right. They had weak Rod Bolts, one of the minor upgrades. Here is one in Direct Drive and one with a Belt Reduction. Originally Rated 20hp at 10,000 ft for Military 10kw Generators so it would work in 98% of the World.

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##### Well-Known Member
Log Member
here are some photos from when I had one on a skylite
I reached out to you many years ago regarding your Skylite with the O84 engine in it. Ed told me that anything heavier than then 80lbs was too heavy for a Skylite. Curious how it worked out for you. Also it looks like you changed the rudder as well. I am the fairly new owner of the Skylite planset and sell them here. www.cloudbaseaviation.com
If the Skylite could take the heavier engine I would put in a Pegasus DP1 as I have one I will get in the next run of engines but now that is slated for the Lil Bitts. I may build a slightly larger Skylite to be able to use the DP1 at some point.

I would love to hear your thoughts of how the skylite did with the O84. Also any more pics you can share? one of your pics is too small to see. I am currently working on offer subkits as I develope them and buil done for myself.

##### Active Member
Chris Bops
2115 N. Taylor
Springfield, Mo 65803
Please confirm! Don't you just love it when a deal comes together over the Internet and Jeff Bezos doesn't get a cut???
Chris borg the Rest is right

##### Active Member
I reached out to you many years ago regarding your Skylite with the O84 engine in it. Ed told me that anything heavier than then 80lbs was too heavy for a Skylite. Curious how it worked out for you. Also it looks like you changed the rudder as well. I am the fairly new owner of the Skylite planset and sell them here. www.cloudbaseaviation.com
If the Skylite could take the heavier engine I would put in a Pegasus DP1 as I have one I will get in the next run of engines but now that is slated for the Lil Bitts. I may build a slightly larger Skylite to be able to use the DP1 at some point.

I would love to hear your thoughts of how the skylite did with the O84. Also any more pics you can share? one of your pics is too small to see. I am currently working on offer subkits as I develope them and buil done for myself.
I will have to look for more pictures as long as you keep the engine close to the firewall as you can the weight and balance was fine I had to add 15 pounds to the tail to get it right but it flew like a dream that motor was made for that plane And yes you want to build a bigger tail the original is too small you will not have good ground control I found the plans for that tail that I had on the Internet


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# Al khwarizmi the father of algebra

Al-khwarizmi: the father of algebra muhammed ibn musa al-khwarizmi, was a mathematical pioneer, and is considered by many to be the greatest mathematician of the islamic world, as well as the founder algebra. Muhammad ibn musa al-khwarizmi was a persian mathematician, astronomer, astrologer geographer and a scholar in the house of wisdom in baghdad he was born in persia of that time around 780 al-khwarizmi was one of the learned men who worked in. Al-khwārizmī: al-khwārizmī, muslim mathematician and astronomer whose major works introduced hindu-arabic numerals and the concepts of algebra into european mathematics latinized versions of his name and of his most famous book title live on in the terms algorithm and algebra al-khwārizmī lived in baghdad. Al-khwarizmi: the father of algebra discussion in 'technology & science' started by introvert, dec 24, 2017. The father of algebra abu ja'far mohammed ibn-musa al-khwarizmi was born sometime before 800 ad and died after 847 ad he flourished as a mathematician and astronomer who was a faculty member at the house of. Science in a golden age - al-khwarizmi: the father of algebra source: al jazeera jan 26, 2017 no comments share on facebook | share on. Early life: muhammad ibn musa al-khwarizmi was a persian mathematician, astronomer, astrologer geographer and a scholar in the house of wisdom in baghdad.

History of science & tech #5: the father of algebra, al-khwarizmi lets-tech (47) in science • 2 months ago muhammad ibn musa al-khwarizmi, also known as: al. Al-khwarizmi's popularizing treatise on algebra (the compendious book on calculation by completion and balancing, ca 813-833 ce: 171) presented the first systematic solution of linear and quadratic equations in arabic. Al-khwarizmi, the father of algebra 18 - 09 2017 no comment by: mankalela in: history muhammad al-khwarizmi earlier this month, there was this exchange between. My presentation on the arab father of algebra, al-khwarizmi, his work, and his contribution to the field of mathematics, particularly algebra and equation-solving this. Al-khwārizmī - the father of algebra, an uzbek turk al khwarizmi’s contributions to mathematics and astronomy centuries ago haven’t gone unnoticed a crater on. Al-khwarizmi, the father of algebra abu ja'far muhammad ibn musa al-khwarizmi lived in baghdad, around 780 to 850 ce (or ad) he was one of the first to write about algebra (using words, not letters.

In the source, the author states “al-khwarizmi wrote numerous books that played important roles in arithmetic and algebra” (overbay) not only was the father of algebra a mathematician, he was also an inventor, an astronomer, and a scholar. Al khwarizmi – the father of algebra october 16, 2012 3 comments education, religion aslam it was one of the golden periods of islamic history when the islamic. Abu ja'far muhammad ibn musa al-khwarizmi, (780 – 850 ce), is the grandfather of computer science and the father of algebra he was the popularizer of arabic numerals, adopter of zero (the symbol, that is) and the decimal system, astronomer, cartographer, in briefs an encyclopedic scholar. Al-khwarizmi, the father of algebra al-khwarizmi abu ja'far muhammad ibn musa al-khwarizmi lived in baghdad, around 780 to 850 ce (or ad) he was one of the first to.

Reddit: the front page of the internet i spent the summer doing algebra research in uzbekistan, home of al-khwarizmi. The father of algebra essay writing service, custom the father of algebra papers, term papers, free the father of algebra samples, research papers, help. One of the elite scholars in baghdad's prestigious house of wisdom, al-khwarizmi is best remembered for his famous work al-jabr wa al-muqabala, the text that defined the branch of mathematics known as algebra he was also an. A greek mathematicians diophantus is sometimes called the father of algebra as he developed one of the earliest forms of solving equations however he used mostly special ad hoc methods to solve equations, whilst al- khwarizmi was the first to solve equations using general methods.

## Al khwarizmi the father of algebra

Al-khwarizmi the father of algebra by ibrahim b syed, phd, dsc, facr • clinical professor of medicine • university of louisville school of medicine • and • president • islamic research foundation international, inc. The father of algebra: abu jaafar mohammad ibn mousa al khwarizmi al khwarizmi’s contributions to mathematics and astronomy centuries ago. Al khwarizmi (the father of algebra) in math in astronomy he wrote work in astronomy which covered calenders, the positions.

• Al-khwarizmi is often considered as the father of algebra, much of whose works on reduction was included in the book and added to many methods we have in algebra now circa 850: persian mathematician al-mahani conceived the idea of reducing geometrical problems such as duplicating the cube to problems in algebra.
• Mathematics is the metaphor against which all other sciences are checked muhammad ibn musa al-khwarizmi was a mathematician and an astronomer whose major works introduced arabic numerals and the concepts of algebra into european mathematics.
• How can the answer be improved.

The word algebra comes from the arabic word al-jabr, meaning reunion and is part of the title of from a book on mathematics written by muhammed ibn musa al-khwarizmi he was an arab mathematician who worked on algebra, arithmetic, and astronomical tables the word algorithm is also derived from his name. Al khwarizmi abu ja’far muhammad ibn musa al-khwarizmi, (780 – 850 ce), was the grandfather of computer science and the father of algebra he was the popularizer of arabic numerals, adopter of zero (the symbol, that is) and the decimal system, astronomer, cartographer, in briefs an encyclopedic scholar. Al-khwarizmi is often considered the father of algebra (though that title is also given to diophantus), as much of his works on reduction are still in use today another persian mathematician omar khayyam developed algebraic geometry and found the general geometric solution of the cubic equation. Al-khwarizmi, baghdad, iraq 22k likes muhammad ibn musa al-khwarizmi was a persian mathematician, astronomer, astrologer geographer and a scholar in. Al-khwarizmi, the father of algebra algebra level 3 if $$\dfrac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}=r$$, then what is the value of \(2r^{2}-\dfrac{4r}{x. Mohammad khawarizmi the father of algebra 615 likes abu abdullah mohammad ibn musa al-khawarizmi, known as father of algebra, was born at khawarizm.

Al khwarizmi the father of algebra
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# Evolution of Eigenstates when two spin systems are coupled

I would like to describe the following situation:

We have two spin systems: Spin 1 ($S_1$) and Spin 1/2 ($S_2$).

Now imagine you somehow change their interaction so that you can fine-tune the coupling $J$ between them in the form:

$$H = \mathbf{S_1} \cdot \mathbf{J_{12}} \cdot \mathbf{S_2}$$

where $\mathbf{J}$ is a matrix describing this interaction.

Now my question is how do I write this in matrix form in order to calculate the different eigenstates of this coupled system for different coupling strenghts $J$?

Should I assume a spin 3/2 system (4x4 Matrix) or an entangled Hilbert space with spin 1/2 and spin 1 (6x6 Matrix)?

Also, what if I still want to include effects on the spin 1 system such as Zeeman splitting in a magnetic field $B_z$, how could I include this?

## Update

So let's make the situation a bit more simple, just a magnetic field $B_z$ acting on the spin-1 and only an isotropic ferromagnetic coupling between the spin-1 and the spin-1/2:

$$H = g\mu_B * B_z * S_z + J * \mathbf{S_{Spin1}} \cdot \mathbf{S_{Spin1/2}}$$

So I know my spin matrices for the spin-1/2 (Pauli matrices) and for the spin-1. My approach now would be to take the tensorproduct of these operators to create the new operators for the above Hamiltonian, i.e.:

$$S_x^{both} = S_x^{spin1} \otimes S_x^{spin1/2}$$ as well as for $y$ and $z$.

With these I construct the new Hamiltonian, I think these operators are correct for the spin coupling term, for the magnetic field B_z that should only act on the spin-1 I need to project it on the subspace of the spin-1 system I think?

• Dear Matthias. I plan to come back to my answer to tidy it a bit and make it more general some time. I have made some notes at the end as to how you might tackle the problem: I suspect the best way is to use Schur's lemma in some way. You could also find the subspace which is common to all three of the nullspaces of the three $36\times 36$ matrices $1_{36\times 36}\otimes \Sigma_j - \Sigma_j^T\otimes 1_{36\times 36}$ in Mathematica or Matlab, but I suspect there is a much eleganter method. – Selene Routley Apr 16 '15 at 1:03
• Also, please add your own answer if you work it out: I'm actually quite interested in this myself now. My answer could probably reformulate the question so that it could be asked on Maths SE. – Selene Routley Apr 16 '15 at 1:05
• Dear Rod, thanks for the detailed answer but I'm afraid that this is a bit too complicated for me. I thought it must be easier by making a few simplifications, e.g. we only care about the isotropic part of the coupling and assume ferromagnetic coupling. So let's say in this case I want to apply an external magnetic field B_z to the spin-1 system and the two spins are connected by $H = J * \vec{S_1} \cdot \vec{S_2}$. So our total Hamiltonian would be: $H = g\mu B_z*S_z + J * \vec{S_1} \cdot \vec{S_2}$. Can't I just take the tensorproducts of the individual operators and that's it? – Mike May 16 '15 at 17:05

I haven't thought about this one before, so here is an approach that will work if you work hard enough at it.

Before I begin banging on, point number 1:

Should I assume a spin 3/2 system (4x4 Matrix) or an entangled Hilbert space with spin 1/2 and spin 1 (6x6 Matrix)?

Unquestionably the latter. It is a bipartite system and its state space is the tensor product of the two particle spaces. It simply cannot be anything else.

The basic principle here is conservation of angular momentum, so your basic procedure to solve your problem is:

1. Work out the matrices for the observables for the three nett angular momentum components (the three nett angular momentum operators);

2. Find the most general Hamiltonian which commutes for all of these three as commutation with the Hamiltonian is equivalent to invariance with time of all the moments of probability distributions of the measurements.

Part 1: The Three Angular Momentum Operators

The $x$-AM component observable for the spin half particle,

$$\sigma_x=\left(\begin{array}{cc}0&1\\1&0\end{array}\right)$$

has AM eigenvectors:

$$\psi_+=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\1\end{array}\right);\quad\psi_-=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\-1\end{array}\right)$$

and AM eigenvalues $\lambda_+=+\frac{1}{2}$ and $\lambda_-=-\frac{1}{2}$, respectively.

The $x$-AM component observable for the spin 1 particle,

$$S_x = \left(\begin{array}{ccc}0&0&0\\0&0&i\\0&-i&0\end{array}\right)$$

has AM eigenvectors:

$$\Psi_+=\frac{1}{\sqrt{2}}\left(\begin{array}{c}0\\i\\1\end{array}\right);\quad\Psi_-=\frac{1}{\sqrt{2}}\left(\begin{array}{c}0\\1\\i\end{array}\right);\quad\Psi_0=\left(\begin{array}{c}1\\0\\0\end{array}\right)$$

and AM eigenvalues $\Lambda_+=+1$, $\Lambda_-=-1$ and $\Lambda_0=0$, respectively. So now, for the two particle system, the six $x$-AM eigenstates are:

1. $\psi_+\otimes\Psi_+$ with AM eigenvalue $\frac{1}{2}+1=\frac{3}{2}$
2. $\psi_+\otimes\Psi_0$ with AM eigenvalue $\frac{1}{2}+0=\frac{1}{2}$
3. $\psi_+\otimes\Psi_-1$ with AM eigenvalue $\frac{1}{2}-1=-\frac{1}{2}$
4. $\psi_-\otimes\Psi_+$ with AM eigenvalue $-\frac{1}{2}+1=\frac{1}{2}$
5. $\psi_-\otimes\Psi_0$ with AM eigenvalue $-\frac{1}{2}+0=-\frac{1}{2}$
6. $\psi_-\otimes\Psi_-1$ with AM eigenvalue $-\frac{1}{2}-1=-\frac{3}{2}$

and so, if we order the eigenstates as above, the eigenvectors as columns are $\mathrm{vec}(\psi_+\otimes\Psi_+),\,\mathrm{vec}(\psi_+\otimes\Psi_0)\cdots$ (see the Wikipedia Vectorization Page) and so at last we get as the total $x$-AM component observable $\Sigma_X = P_X \Lambda_X P_X^\dagger$ where

$$P_X=\left( \begin{array}{cccccc} 0 & \frac{1}{\sqrt{2}} & 0 & 0 & \frac{1}{\sqrt{2}} & 0 \\ 0 & \frac{1}{\sqrt{2}} & 0 & 0 & -\frac{1}{\sqrt{2}} & 0 \\ \frac{i}{2} & 0 & \frac{1}{2} & \frac{i}{2} & 0 & \frac{1}{2} \\ \frac{i}{2} & 0 & \frac{1}{2} & -\frac{i}{2} & 0 & -\frac{1}{2} \\ \frac{1}{2} & 0 & \frac{i}{2} & \frac{1}{2} & 0 & \frac{i}{2} \\ \frac{1}{2} & 0 & \frac{i}{2} & -\frac{1}{2} & 0 & -\frac{i}{2} \\ \end{array} \right)$$

and $\Lambda_X =\mathrm{diag}\left(\frac{3}{2},\,\frac{1}{2},\,\frac{-1}{2},\,\frac{1}{2},\,\frac{-1}{2},\,\frac{3}{2}\right)$. The result is:

$$\Sigma_X=\left( \begin{array}{cccccc} 0 & \frac{1}{2} & 0 & 0 & 0 & 0 \\ \frac{1}{2} & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & \frac{3}{4} & -\frac{1}{4} & \frac{i}{4} & \frac{3 i}{4} \\ 0 & 0 & -\frac{1}{4} & \frac{3}{4} & \frac{3 i}{4} & \frac{i}{4} \\ 0 & 0 & -\frac{i}{4} & -\frac{3 i}{4} & \frac{3}{4} & -\frac{1}{4} \\ 0 & 0 & -\frac{3 i}{4} & -\frac{i}{4} & -\frac{1}{4} & \frac{3}{4} \\ \end{array} \right)$$

From here on it should be conceptually clear how to go, although tedious. You do the same for the $y$-AM observables:

$$\sigma_y=\left(\begin{array}{cc}0&-i\\i&0\end{array}\right)$$ $$S_y = \left(\begin{array}{ccc}0&0&-i\\0&0&0\\i&0&0\end{array}\right)$$

to find the total system $y$-AM observable $\Sigma_Y$ and for the $z$-AM observables:

$$\sigma_z=\left(\begin{array}{cc}i&0\\0&-i\end{array}\right)$$ $$S_z = \left(\begin{array}{ccc}0&i&0\\-i&0&0\\0&0&0\end{array}\right)$$

to get the total system $z$-AM observable $\Sigma_Z$.

Part 2: Find most general Hamiltonian

Your most general Hamiltonian will be defined by the three commutator relationships expressing conservation of AM:

$$[\hat{H},\,\Sigma_j]=0;\;j=X,\,Y,\,Z$$

You'll need to work out the invariant spaces of the three $\Sigma$s to do this. You'll get a linear space of possible $\hat{H}$s: in the two coupled spin half particles case there is essentially only one possible Hamiltonian that falls out of this approach and that is one proportional to $\sigma_x\otimes\sigma_x+\sigma_y\otimes\sigma_y+\sigma_z\otimes\sigma_z$ (plus a term proportional to the $4\times4$ identity matrix expressing the shift in the ground state energy) but this six dimensional case things will be a bit more complicated. Now as I said, I've never done this before, so I daresay there is a more systematic and less cumbersome way to work this all out. But any method is going to rest on the first principles expressed above.

Magnetic Field

What are the terms for the influence of the magnetic field. Well that's an easy one: in the ordering we have studied above, the uncoupled Hamiltonian will be:

$$\hat{H} = \gamma_{\frac{1}{2}}\left(\sigma_x\,B_x + \sigma_y\,B_y+ \sigma_z\,B_z\right)\otimes 1_{3\times3} + \gamma_1\,I_{2\times2}\otimes\left(S_x\,B_x + S_y\,B_y+ S_z\,B_z\right)$$

where $\gamma_{\frac{1}{2}}$ and $\gamma_1$ are the respective gyromagnetic ratios.

Notes on completing the method. You can also represent a bipartite state $\Phi=\psi\otimes\Psi$ as the literal $2\times 3$ matrix that is the outer product $\Phi=\psi Psi^T$ of the $2\times 1$ and $3\times 1$ column vectors. Then the operator on the first space act on the left and the operators on the second act on the right. So our $x$-component observable would be the linear, homogeneous transformation:

$$\Phi\mapsto \sigma_x\,\Phi\,S_x^T$$

and the vectorization operator (See Vectorization Wiki Page), which reorders our states into a $6\times 1$ column vectors as in my answer, writes this as

$$\mathrm{vec}(\Phi) \mapsto S_x\otimes\sigma_x\,\mathrm{vec}(\Phi)$$

Using the standard formula $\mathrm{vec}(A\,B\,C) = C^T\otimes A \,\mathrm{vec}(B)$. By dint of the formula $(A\otimes B)\, (C\otimes D) = (A\,C)\otimes(B\,D)$, and using the fact that inverse, complex conjugate, Hermitian conjugate and transpose operations distribute over the Kronecker produt, we can diagonalize $S_x\otimes\sigma_x$ inside the Kronecker product and find that the coupled system’s eigenstates are $\Pi_x\otimes \pi_x$, where $P_X,\,p_x$ are the matrices of eigenvectors of the individual multiplicands written as columns. So this will let you calculate $\Sigma_j,\,j=X,\,Y,\,Z$ systematically and fast.

Now to find the most general Hamiltonian, you need to find the invariant space of the group of matrices generated by the three matrices $\exp(i\,\Sigma_j)$ and find the irreducible representation of it: equivalently the smallest vector subspace of $\mathbb{C}^6$ left invariant by the group: by Schur’s lemma, any matrix commuting with all three must be proportional to the identity operator when restricted to this subspace. The scaling factor is possibly nought – i.e. the operator could possibly be the zero endomorphism. This completely characterizes the most general Hamiltionian: it can be any operator which is proportional to the identity when restricted to this irreducible subspace.

You could also find the subspace which is common to all three of the nullspaces of the three $36\times 36$ matrices $1_{36\times 36}\otimes \Sigma_j - \Sigma_j^T\otimes 1_{36\times 36}$ in Mathematica or Matlab, but I suspect there is a much eleganter method grounded on Schur's lemma!


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# Models¶

I can never satisfy myself until I can make a mechanical model of a thing. If I can make a mechanical model, I can understand it. As long as I cannot make a mechanical model all the way through I cannot understand.

—Lord Kelvin

## Introduction¶

Language echoes through the innermost reaches of the corners, so I will have the opportunity to mention most neural parcels, at least in passing. This breadth is very different from much of neuroscience, which tends to specialize in ever smaller brain bits. The danger of such a narrow focus has prompted a counter-trend, dubbed integrative neuroscience. Exactly what this means varies from researcher to researcher, but the journal Frontiers in Integrative Neuroscience asserts that it publishes research that “synthesizes multiple facets of brain structure and function, to better understand how multiple diverse functions are integrated to produce complex behaviors”. I’ll leave it up to you to decide whether this clarification makes you any wiser about whether to submit an article to the journal or not, because I want to present the topic inductively, not deductively.

My quandary is that I don’t know how much you know about language or neuroscience. In the absence of such shared knowledge, I don’t know exactly where to begin. I will thus fall back to what I hope is a common denominator, namely people’s behavior in general. Or at least people with psychiatric disease.

## Akaishi’s psychiatric structured model¶

Tetsuya Akaishi (2018) has recently proposed an integrative framework for psychiatric disease, such as attention-deficit/hyperactivity disorder, autism spectrum disorder, obsessive compulsive disorder, post-traumatic stress disorder, and schizophrenia. The framework consists of three loops.

### The input-output-feedback loop¶

This first and most basic loop connects the internal and external worlds. By “external world” we mean the realm of the senses, but Akaishi’s psychiatric bent leads him to highlight specific sorts of sense data such as the task to be performed or feedback from the performance of previous tasks – which are really implicit in the incoming sensory data and so should not be isolated from it. The “internal world” takes this sensory input, filters and processes it via a module of perception and then submits it to a module of personality (sense of values) and intelligence (reward anticipation), upon which memory acts. This internal world can produce an output, an action, that is broadcast to the external world of the social and family environment, resulting in some more sensory data that impinges on the entrance of the internal world, creating a feedback loop.

Fig. 88 The input-output-feedback loop

Fig. 1 from Akaishi (2018).

The main application of the input-output-feedback loop to psychiatric disease is that a mis-function in perception leads to hallucination.

More interesting to us is to speculate on how language would use this loop. Imagine that the environment of the external world produces the linguistic utterance “What is your name?”. It qualifies as an input to the internal world, which is filtered into something like ‘linguistic auditory data’, perceived as the speech [] and converted into the meaning of the question, to generate the answer action of articulating “Harry”, an output back to the external world.

#### The perception-action loop¶

We can be much more precise.

Fig. 89 First two time steps in a perception-action loop, where E is the environment, S is a sensor value, M is the memory, A is an action, and the numerical subscripts indicate times.

From {Biehl et al., 2018, CT}, Fig. 1.

The diagram is read from left to right, beginning at time 0, in which the environment is in state $$E_0$$ which is perceived as $$S_0$$. Then the clock ticks once, and several things happen. The sensory reading $$S_0$$ is put into memory, $$M_1$$, which triggers the first action, $$A_1$$, which becomes part of the next environmental state $$E_1$$, which also contains input from the initial state $$E_0$$. Finally, $$E_1$$ is read by the sensor to produce $$S_1$$.

The same sequence is repeated for tick 2 … and then indefinitely.

The only difference is that M keeps track of everything that has happened in the past. Thus $$M_2$$ contains not only the previous memory $$M_1$$ but also the previous action $$A_1$$ and the previous sensor reading $$S_1$$. Putting the ever-growing store of states and actions into a mathematical notation may help you to appreciate how the entire system works:

Notation

(1)$M_2 = \{(S_0), (A_1, S_1)\}$

Practice

1. What would be the contents of $$M_3$$?

2. Make up specific states and the corresponding actions for times 0 to 3.

#### Policy¶

It will be useful to have a name for such a sequence of states and actions. Taking a cue from the literature on reinforcement learning, let us call it a policy.

Practice

I took a few usages of the word policy from Google News. Decompose them into a sequence of states and actions, if possible.

### The reward-learning loop¶

The basic perception-action cycle does a lot, but we don’t know how. To remedy this problem Akaishi’s next step is to add a mechanism for evaluating how well a given action applies to a sensory state. Following a long tradition, Akaishi labels this evaluation reward. The idea is simple: if you are experiencing the sensory state of hunger, your cognitive system should reward the action “go for pizza” more than “play chess”. At least, I think it works that way for most people – you may find chess as satisfying as a good meal (so your reward system is different).

Fig. 90 The reward-learning loop

From Akaishi (2018), Fig. 2.

Let us try to translate this image into words, using the toy example embedded in an action-preception loop. Every time your cognitive system is faced with the sensation of hunger $$s_{hunger}$$ and the alternative actions of getting some food $$a_{food}$$ or playing chess, $$a_{chess}$$, the preference for $$a_{food}$$ should be strengthened. If we assign small numbers for the reward R(S,A) of either action at time 0, say 0.1, then the learning would take this form:

Notation

(2)\begin{split}\begin{align} R(s_{hunger}, a_{food})_0 = 0.1 \to R(s_{hunger}, a_{food})_1 = 0.2\\ R(s_{hunger}, a_{chess})_0 = 0.1 \to R(s_{hunger}, a_{chess})_1 = 0.1 \end{align}\end{split}

Misfunctioning of the reward system could be the root of depression. Akaishi suggests that it could arise from a very low setting of R() for all state-action pairings, even after learning.

Question

Explain in your own words what the result would be in a perception-action cycle.

A loss of interest in previously pleasurable activities is generally termed anhedonia.

### The mood-reward sensitivity loop¶

The final component of Akaishi’s model is mood, which acts to weaken or strengthen the reward signal. The idea is that positive mood increases reward while negative mood decreases it.

Fig. 91 The mood-reward sensitivity loop

From Akaishi (2018), Fig. 3.

Akaishi formalizes it as a multiplication of $$R(S,A)_T$$. If multiplication of a number by 1 just returns the number, multiplication by a number greater than 1, say 1.1, returns a slightly larger number (amplification), while multiplication by a number less than 1, say 0.9, returns a slightly smaller number (attenuation). I am going to state this in a different way, as so:

Notation

(3)$R_{acc}(S,A)_t = g*R(S,A)_t$

(3) uses the term g for “gain” because, in signal-processing theory, a gain enhances (or attenuates) a signal in a noisy environment via a multiplication of the signal by the gain quantity.

Akaishi speculates that panic disorder and obsessive compulsive disorder derive from inaccurate anticipation based on irregular inputs of fear-related memories [(OCD) are also known to be frequently accompanied each other [85]. Although the timing and manifestation of these disorders are different, both of them , possibly caused by impaired inhibitory learning of the PFC [86,87].]

A more intricate resolution of the reward system could be found in schizophrenia. Akaishi relates that a first problem can be found in the perceptual system, since there are many reports of hallucination in schizophrenic patients. More relevant to our present concerns, they also betray a certain amount of anhedonia, which suggests that the background or tonic level $$R(S,A)_0$$ is abnormally low. Paradoxically, they also show a heightened response to specific sensory states, suggesting that the learning level $$R(S,A)_{t > 0}$$ is abnormally high.

In mood disorders, decreased amount of reward signals and elevated serum corticosteroid levels have been already confirmed [54–57]. In normal healthy population, such an elevated corticosteroid level would suppress further secretion of corticosteroid with a negative feedback, but this feedback system is known to be impaired in those with mood disorders [43]. As a result, elevated serum corticosteroid levels in mood disorders would further suppress the reward sensitivity and exacerbate the mood conditions [55–57]. This elevated serum corticosteroid, fol- lowed by suppressed reward sensitivity, also impairs the efficiency of reinforcement learning. As a result, memory consolidation, memory retrieval, and social learning are all going to be disturbed in mood disorders [58,59].

## References¶

• Akaishi, Tetsuya. “Unified Neural Structured Model: A New Diagnostic Tool in Primary Care Psychiatry.” Medical Hypotheses 118 (2018): 107-13.

• Pezzulo, Giovanni, Francesco Rigoli, and Karl J. Friston. “Hierarchical Active Inference: A Theory of Motivated Control.” Trends in Cognitive Sciences 22.4 (2018): 294-306.

• Biehl, Martin, Christian Guckelsberger et al. “Expanding the Active Inference Landscape: More Intrinsic Motivations in the Perception-Action Loop.” arXiv 1806.08083 (2018)

## The next topic¶

The next topic is Frequency.

Last edited Jun 30, 2021


text:
1. ## Calculus - Integration

When you integrate 2sin$\displaystyle x$cos$\displaystyle x$$\displaystyle /1-2\displaystyle (cosx)^2is the answer \displaystyle -$$\displaystyle (sec2x)^2$/2?

If not, then what would be the proper approach?

2. What about substituting $\displaystyle u=1-2cos^{2}(x)$ and trying to answer your question by yourself?

3. I get 1/2ln(1-2cos^2x)+c. Is that right?

4. Originally Posted by bhuang
I get 1/2ln(1-2cos^2x)+c. Is that right?
Does it work when you differentiate it?


text:
Lemma 30.24.2. Let $A$ be Noetherian ring and $I \subset A$ an ideal. Let $f : X \to \mathop{\mathrm{Spec}}(A)$ be a proper morphism and let $\mathcal{L}$ be an $f$-ample invertible sheaf. Let $\mathcal{I} = I\mathcal{O}_ X$. Let $(\mathcal{F}_ n)$ be an object of $\textit{Coh}(X, \mathcal{I})$. Then there exists an integer $d_0$ such that

$H^1(X, \mathop{\mathrm{Ker}}(\mathcal{F}_{n + 1} \to \mathcal{F}_ n) \otimes \mathcal{L}^{\otimes d} ) = 0$

for all $n \geq 0$ and all $d \geq d_0$.

Proof. Set $B = \bigoplus I^ n/I^{n + 1}$ and $\mathcal{B} = \bigoplus \mathcal{I}^ n/\mathcal{I}^{n + 1} = f^*\widetilde{B}$. By Lemma 30.23.8 the graded quasi-coherent $\mathcal{B}$-module $\mathcal{G} = \bigoplus \mathop{\mathrm{Ker}}(\mathcal{F}_{n + 1} \to \mathcal{F}_ n)$ is of finite type. Hence the lemma follows from Lemma 30.19.3 part (2). $\square$

Comment #6791 by Alex Scheffelin on

Very slight typo in the statement of the lemma, it reads "$I\subset A$ and ideal". It should say "an ideal".

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).


text:
# Generalization or Improvement of Cheeger inequality on Graphs

Let $G=(V,E)$ be an undirected graph with vertex set $V$ and edge set $E$. Let $A$ denote the adjacency matrix of $G$ and $D$ denote the diagonal matrix such that $D_{i,i}$ equals to the degree $d_i$ of vertex $i$. Then the Laplacian of $G$ is defined to be $L:=I-D^{-1/2}AD^{-1/2}$, which has $n$ nonnegative real eigenvalues, say, $0=\lambda_1\leq \lambda_2\leq \cdots\leq \lambda_n$.

Now for any vertex subset $S\subseteq V$, let $e(S,\bar{S})$ denote the number of edges between $S$ and its complement $\bar{S}$. Let $vol(S):=\sum_{v\in S}deg_v$ be the sum of degrees of vertices in $S$. Then the conductance (or Cheeger constant) $h_G$ of graph $G$ is defined to be $$h_G:=\min_{S\subseteq V}\frac{e(S,\bar{S})}{\min \{vol(S), vol({\bar{S}})\}}.$$

The Cheeger inequality relates the second smallest eigenvalue $\lambda_2$ of $L$ to the conductance $h_G$ as follows:

$$2h_G\geq\lambda_2\geq \frac{h_G^2}{2}.$$

The above inequality is known to be tight. For example, the left side of the inequlity is tight on the $d$-dimensional cube and the right side is tight on the $n$-vertex cycle. Thus, we do not hope to improve the inequality that applies to every graph.

My question is:

Is there any result which gives that for some special class of graphs, the Cheeger inequality has an improved form, say, $2h_G^{1.2}\geq\lambda_2\geq \frac{h_G^{1.5}}{2}$ for any $G\in\mathcal{C}$, where $\mathcal{C}$ is a set of graphs.

Ideally, we would hope that there is a nice tradeoff between the generality of the class $\mathcal{C}$ and the tightness of the Cheeger-type inequality. In another word, $\mathcal{C}$ contains a wide class of graphs and the upper bound is close to the lower bound.

-
Just because an inequality is tight doesn't mean it can't be improved. It just means that if there is an improvement then it must share the same equality cases. –  Qiaochu Yuan Apr 3 '11 at 0:21

There is a lot known on the relations between the Cheeger constant $h_{G}$ and $\lambda_{2}$ and more general on the whole spectrum of the normalized Laplacian for the following families of graphs:

• Random Trees (branching processes) Lyons and Peres book is an excellent reference.

• Random Geometric graphs, Penrose's book is an good reference.

• As previously mentioned the $G_{n,p}$ random model of Erdos and Renyi.

• Random regular graphs where you can find information in the fundamentals papers of McKay.

• Regular planar tessellations of hyperbolic space.

I hope it helps!

-
Thanks for your help. However, I do not think the results you mentioned above give the answers I wanted. The above results almost all fall into the following case: For a given class of graphs, estimate the conductance (or expansion) using the (random) structure of the class; then 1) use the Cheeger inequality to give bound on the second eigenvalue or 2) relates the conductance with other quantities, e.g., diameters. Such results do not give any \direct improvement on the relationship between $\lambda_2$ and $h_G$, which implies better upper or lower bound compared with Cheeger inequality. –  Pan Peng Apr 3 '11 at 8:17

There other classes of graphs such as generalizations of the $G(n,p)$ model with a given degree distribution. See for example Chung and Vu's paper on the spectrum of these graphs. In particular, they analyzed the case when the graph has a power law degree distribution.

-

One place to look might be random graphs $G(n,p)$. This might either give you a wide class of graphs for which an improvement holds, or else show you a limit to what you might hope for.

For Cheeger constant see: MR2371054 (2008j:05316) Benjamini, Itai; Haber, Simi; Krivelevich, Michael; Lubetzky, Eyal The isoperimetric constant of the random graph process. Random Structures Algorithms 32 (2008), no. 1, 101–114.

For spectral gap see: MR0637828 (83e:15010) Füredi, Z.; Komlós, J. The eigenvalues of random symmetric matrices. Combinatorica 1 (1981), no. 3, 233–241.

-


text:
User:
• Matrices
• Algebra
• Geometry
• Graphs and functions
• Trigonometry
• Coordinate geometry
• Combinatorics
 Suma y resta Producto por escalar Producto Inversa
 Monomials Polynomials Special products Equations Quadratic equations Radical expressions Systems of equations Sequences and series Inner product Exponential equations Matrices Determinants Inverse of a matrix Logarithmic equations Systems of 3 variables equations
 2-D Shapes Areas Pythagorean Theorem Distances
 Graphs Definition of slope Positive or negative slope Determine slope of a line Equation of a line Equation of a line (from graph) Quadratic function Parallel, coincident and intersecting lines Asymptotes Limits Distances Continuity and discontinuities
 Sine Cosine Tangent Cosecant Secant Cotangent Trigonometric identities Law of cosines Law of sines
 Equations of a straight line Parallel, coincident and intersecting lines Distances Angles in space Inner product

Perimeter
Circles: word problems

Ryan used a drafting compass to draw a circle on paper. The circle was 12 cm in radius. How many square centimeters were inside the circle?

Area = π × r2

Area = π x 122 = π x 144 = 144 π cm2

π ≈ 3.14, so the area is 3.14x144=452.16 inches squared

Robert drew a circle that has a diameter of 16 inches. What is the area of his circle?

Area = π × r2

You are told that the diameter is 16 inches, so the radius is one-half of this measure, or 8 inches. Substituting the information into the area formula gives you

Area = π ×82 = π × 64 = 64π

π ≈ 3.14, so the area is 3.14x64=200.96 inches squared.

Find the circumference of the following coin (Round to the nearest whole number):

 1 Euro Coin Diameter=23.25 mm

C = π × d

Use 3.14 as an approximation for π.

C=(3.14)(23.25)≈73 mm.

Find the circumference of the following coin (Round to the nearest whole number):

 Susan B.. Anthony Dollar Radius=0.52 in

C =2πr

Use 3.14 as an approximation for π.

C=(2)(3.14)(0.52)≈3.27 in.


text:
# Contents

## Idea

The Baez-Dolan stabilization hypothesis states that for all $k = n+2$ a k-tuply monoidal n-category is “maximally monoidal”. In other words, for $k \geq n + 2$, a $k$-tuply monoidal $n$-category is the same thing as an $(n+2)$-tuply monoidal $n$-category. More precisely, the natural inclusion $k Mon n Cat \hookrightarrow (n+2) Mon n Cat$ is an equivalence of higher categories.

## Proof for $(n,1)$-categories

An aspect of the proof of this for (n,1)-categories was demonstrated in

• Carlos Simpson, On the Breen-Baez-Dolan stabilization hypothesis for Tamsamani’s weak $n$-categories (arXiv:math/9810058)

in terms of Tamsamani n-categories?.

A proof of the full statement in terms of quasi-categories is sketched in section 43.5 of

Probably the first full proof in print is given in

where it appears in example 1.2.3 as a direct consequence of a more general statement, corollary 1.1.10.

## References

Section 5.1.2 of

Revised on February 10, 2012 00:04:05 by Urs Schreiber (89.204.130.229)


text:
The nature of stock market movement has always been ambiguous for investors because of various influential factors. This study aims to significantly reduce the risk of trend prediction with machine learning and deep learning algorithms. We evaluate Phillips 66 prediction models with Modular Neural Network (Financial Sentiment Analysis) and ElasticNet Regression1,2,3,4 and conclude that the PSX stock is predictable in the short/long term. According to price forecasts for (n+6 month) period: The dominant strategy among neural network is to Hold PSX stock.

Keywords: PSX, Phillips 66, stock forecast, machine learning based prediction, risk rating, buy-sell behaviour, stock analysis, target price analysis, options and futures.

## Key Points

1. Reaction Function
2. How do you know when a stock will go up or down?
3. Can machine learning predict?

## PSX Target Price Prediction Modeling Methodology

Recently, there has been a surge of interest in the use of machine learning to help aid in the accurate predictions of financial markets. Despite the exciting advances in this cross-section of finance and AI, many of the current approaches are limited to using technical analysis to capture historical trends of each stock price and thus limited to certain experimental setups to obtain good prediction results. On the other hand, professional investors additionally use their rich knowledge of inter-market and inter-company relations to map the connectivity of companies and events, and use this map to make better market predictions. For instance, they would predict the movement of a certain company's stock price based not only on its former stock price trends but also on the performance of its suppliers or customers, the overall industry, macroeconomic factors and trade policies. This paper investigates the effectiveness of work at the intersection of market predictions and graph neural networks, which hold the potential to mimic the ways in which investors make decisions by incorporating company knowledge graphs directly into the predictive model. We consider Phillips 66 Stock Decision Process with ElasticNet Regression where A is the set of discrete actions of PSX stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4

F(ElasticNet Regression)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Modular Neural Network (Financial Sentiment Analysis)) X S(n):→ (n+6 month) $∑ i = 1 n a i$

n:Time series to forecast

p:Price signals of PSX stock

j:Nash equilibria

k:Dominated move

a:Best response for target price

For further technical information as per how our model work we invite you to visit the article below:

How do AC Investment Research machine learning (predictive) algorithms actually work?

## PSX Stock Forecast (Buy or Sell) for (n+6 month)

Sample Set: Neural Network
Stock/Index: PSX Phillips 66
Time series to forecast n: 18 Sep 2022 for (n+6 month)

According to price forecasts for (n+6 month) period: The dominant strategy among neural network is to Hold PSX stock.

X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)

Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)

Z axis (Yellow to Green): *Technical Analysis%

## Conclusions

Phillips 66 assigned short-term Ba3 & long-term B2 forecasted stock rating. We evaluate the prediction models Modular Neural Network (Financial Sentiment Analysis) with ElasticNet Regression1,2,3,4 and conclude that the PSX stock is predictable in the short/long term. According to price forecasts for (n+6 month) period: The dominant strategy among neural network is to Hold PSX stock.

### Financial State Forecast for PSX Stock Options & Futures

Rating Short-Term Long-Term Senior
Outlook*Ba3B2
Operational Risk 5078
Market Risk8531
Technical Analysis7671
Fundamental Analysis3343
Risk Unsystematic8247

### Prediction Confidence Score

Trust metric by Neural Network: 75 out of 100 with 777 signals.

## References

1. Wu X, Kumar V, Quinlan JR, Ghosh J, Yang Q, et al. 2008. Top 10 algorithms in data mining. Knowl. Inform. Syst. 14:1–37
2. Harris ZS. 1954. Distributional structure. Word 10:146–62
3. Rumelhart DE, Hinton GE, Williams RJ. 1986. Learning representations by back-propagating errors. Nature 323:533–36
4. Andrews, D. W. K. W. Ploberger (1994), "Optimal tests when a nuisance parameter is present only under the alternative," Econometrica, 62, 1383–1414.
5. Chernozhukov V, Chetverikov D, Demirer M, Duflo E, Hansen C, et al. 2018a. Double/debiased machine learning for treatment and structural parameters. Econom. J. 21:C1–68
6. R. Sutton and A. Barto. Reinforcement Learning. The MIT Press, 1998
7. Bessler, D. A. T. Covey (1991), "Cointegration: Some results on U.S. cattle prices," Journal of Futures Markets, 11, 461–474.
Frequently Asked QuestionsQ: What is the prediction methodology for PSX stock?
A: PSX stock prediction methodology: We evaluate the prediction models Modular Neural Network (Financial Sentiment Analysis) and ElasticNet Regression
Q: Is PSX stock a buy or sell?
A: The dominant strategy among neural network is to Hold PSX Stock.
Q: Is Phillips 66 stock a good investment?
A: The consensus rating for Phillips 66 is Hold and assigned short-term Ba3 & long-term B2 forecasted stock rating.
Q: What is the consensus rating of PSX stock?
A: The consensus rating for PSX is Hold.
Q: What is the prediction period for PSX stock?
A: The prediction period for PSX is (n+6 month)


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# Measurement uncertainty

1. Oct 15, 2005

### Vegeta

The following question is about an experiment with the conical pendulum.
I have measured the length $l = 43\,cm$ the radius $r = 10\,cm$, I have 3 measurement of the period with the same radius, where the measurements are

$$T= [1.288, 1.285, 1.301] s$$

The uncertainties of the measurements are $\Delta l = \pm 1\,cm$, $\Delta r = \pm 1\,cm$ and $\Delta T = \pm 0.02\,s$.
I want to calculate the uncertainty in the measurement og $g$, when

$$g_i=4\pi^2\frac{\sqrt{l^2-r^2}}{T_i^2}$$

Can I calculate the uncertainty $\Delta g$ by

$$\Delta g = 4\pi^2\frac{\sqrt{(l+\Delta l)^2-(r-\Delta r)^2}}{(T_i-\Delta T)^2} - g_i$$

Where the expression $4\pi^2\left(\sqrt{(l+\Delta l)^2-(r-\Delta r)^2}\right)/(T_i-\Delta T)^2$ is the worst case scenario of the measuring $g$. Is that correct?
If that is how I can calculate the uncertainty in $g$, is the relative uncertainty then

$$\frac{\Delta g}{g_i} \qquad \mathrm{or} \qquad \frac{\Delta g}{\overline{g}}$$

Where $\overline{g}$ is the mean value. Which one is the correct one? The first expression has a relative uncertainty for each measurement.

Last edited: Oct 15, 2005
2. Oct 15, 2005

### HallsofIvy

Staff Emeritus
Yes, calculating the value for the largest possible and then smallest possible values of l, r, and T gives the possible error for the function.

The relative uncertainty is $$\frac{\Delta g}{g}$$.

3. Oct 15, 2005

### Vegeta

Well in the equation

$$\Delta g_i = 4\pi^2\frac{\sqrt{(l+\Delta l)^2-(r-\Delta r)^2}}{(T_i-\Delta T)^2} - g_i$$

There is acctually an uncertainty of g, for every measurement $\Delta g_i$. Should I say

$$\frac{\Delta g_i}{g_i} \qquad \mathrm{or} \qquad \frac{mean(\Delta g_i)}{mean(g_i)}$$?

Where in the last equation I only have one value for the relative error. Im not quite certain of what to choose?

Last edited: Oct 15, 2005


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# Crystal plane perpendicularity to direction

1. Sep 6, 2012

### dikmikkel

1. The problem statement, all variables and given/known data
Show that the direction [hkl] is perpendicular to the plane (hkl) in a cubic system.

2. Relevant equations
None, maybe cross product

3. The attempt at a solution
I've tried to define the crystal translation vectors(cubic means a=b=c):
$\vec{a}_1 = a\hat{x}\\ \vec{a}_2 = a\hat{y}\\ \vec{a}_3 = a\hat{z}$
I just can't imagine how to obtain the vector normal to the plane. I cannot use the reciprocal space as we haven't gone over that yet. Please help me, it is not to be handed in i just need to understand it.

My question is: What is the connection between Miller Indicies of a plane and the plane's normal vector.


text:
# PhysicsBernoulli`s Equation

#### jc91

##### New member
Hi All,

Could anyone advise on how to answer below question (attached). I am struggling big time on this.

#### Attachments

• 57.7 KB Views: 15

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Hi All,

Could anyone advise on how to answer below question (attached). I am struggling big time on this.

Hi jc91 , welcome to MHB!

$$\frac 12 \rho v_1^2 + \rho g z_1 + p_1 = \frac 12 \rho v_2^2 + \rho g z_2 + p_2$$
Or do you perhaps have a different version of it?

Since (2) is a stagnation point, we have $v_2=0$.
The pressure is identified by the column of water above it: it's the weight per surface area.
And additionally we have the standard pressure of air ($p_0$).
So for instance $p_1 = p_0 + \rho g (h_1 + h_2)$.

How far do you get filling in the other variables and solving for $v_1$?

#### Ackbach

##### Indicium Physicus
Staff member
Question for jc91 : what is underneath the blue scribble?


text:
# Suppose the random variables X and Y have a joint pdff_{XY}(x,y)=\begin{cases}x+y & 0\leq x,y \leq 1\\0 & otherwise\end{cases}Find Pr(X>sqrt{Y})

Suppose the random variables X and Y have a joint pdf
${f}_{XY}\left(x,y\right)=\left\{\begin{array}{ll}x+y& 0\le x,y\le 1\\ 0& otherwise\end{array}$
Find $Pr\left(X>\sqrt{Y}\right)$

You can still ask an expert for help

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Derrick
Step 1
Given information:
The joint probability density function of the random variables X and Y is as given below:

Step 2
Compute the value of $P\left(X>\sqrt{Y}\right)$:
The marginal probability density function of X is obtained as given below:
${f}_{X}\left(x\right)={\int }_{y}{f}_{XY}\left(x,y\right)dy$
$={\int }_{0}^{1}\left(x+y\right)dy$
$=x\left[y{\right]}_{0}^{1}+\left[\frac{{y}^{2}}{2}{\right]}_{0}^{1}$
${f}_{X}\left(x\right)=x+\frac{1}{2}$
The value of $P\left(X>\sqrt{Y}\right)$ is obtained as given below:
$P\left(X>\sqrt{Y}\right)={\int }_{\sqrt{y}}^{1}\left(x+\frac{1}{2}\right)dx$
$={\int }_{\sqrt{y}}^{1}xdx+{\int }_{\sqrt{y}}^{1}\frac{1}{2}dx$
$=\left[\frac{{x}^{2}}{2}{\right]}_{\sqrt{y}}^{1}+\left[\frac{x}{2}{\right]}_{\sqrt{y}}^{1}$
$=\frac{1}{2}-\frac{y}{2}+\frac{1}{2}-\frac{\sqrt{y}}{2}$
$=1-\frac{y}{2}-\frac{\sqrt{y}}{2}$


text:
# Math Insight

### The definition of differentiability in higher dimensions

The definition of differentiability in multivariable calculus formalizes what we meant in the introductory page when we referred to differentiability as the existence of a linear approximation. The introductory page simply used the vague wording that a linear approximation must be a “really good” approximation to the function near a point. What does this “really good” mean?

For a scalar valued function of two variables, $f: \R^2 \to \R$ (confused?), this condition meant the existence of a tangent plane, such as the one shown in this applet, which is the example used in the introductory page.

Neuron firing rate function with tangent plane. A fictitious representation of the firing rate $r(i,s)$ of a neuron in response to an input $i$ and nicotine level $s$. The graph of the function has a tangent plane at the location of the green point, so the function is differentiable there. By rotating the graph, you can see how the tangent plane touches the surface at the that point. You can move the green point anywhere on the surface; as long as it is not along the fold of the graph (where the red point in constrained to be), you can see the tangent plane showing that the function is differentiable. There is no tangent plane to the graph at any point along the fold of the graph (you can move the red point to any point along this fold). The function $r(i,s)$ is not differentiable at any point along the fold. As further evidence of this non-differentiability, the tangent plane jumps to a different angle when you move the green point across the fold.

Although we have geometric intuition that helps us understand a tangent plane, this intuition doesn't help us understand the general case of a linear approximation of a function $\vc{f}: \R^n \to \R^m$. Even for the two-dimensional case where differentiability is based on existence of a tangent plane, we should still be precise about what me mean by the existence of this linear approximation. Otherwise, we could never be certain if a given plane was really tangent to the graph.

How can we formalize this definition? For starters, let's write our candidate linear approximation around the point $\vc{a}$ as \begin{align*} \vc{L}(\vc{x}) = \vc{f}(\vc{a}) + \vc{T}(\vc{x}-\vc{a}) \end{align*} where $\vc{T}(\vc{x})$ is a linear transformation. We want to derive a condition to test if $\vc{L}$ really is the linear approximation of $\vc{f}$ around $\vc{a}$. The condition needs to capture the sense that $\vc{L}(\vc{x})$ is “really close” to $\vc{f}(\vc{x})$ when $\vc{x}$ is near $\vc{a}$.

One possibility might be that the limit of $\vc{L}(\vc{x})$ as $\vc{x}$ approaches $\vc{a}$ should be the same as the limit of $\vc{f}(\vc{x})$ as $\vc{x}$ approaches $\vc{a}$. That sounds reasonable. That would mean that distance between $\vc{L}(\vc{x})$ and $\vc{f}(\vc{x})$ should go to zero as we get close to $\vc{a}$. We could write this candidate condition as $$\lim_{\vc{x} \to \vc{a}} \| \vc{f}(\vc{x})-\vc{L}(\vc{x})\| = 0.$$ You can't get any closer than zero distance, so certainly this condition should be a good one.

The problem with this definition becomes immediately clear if you try to apply it to the one variable linear approximation, i.e., the tangent line of a curve. As shown in the following figure, many lines satisfy the above condition, as it only specifies that the line passes through the point $(a,f(a))$. Since the lines and the function are continuous, their distance clearly goes to zero as $x$ approaches $a$. How can we pick out the tangent line (shown in blue) to the graph of $f(x)$ (shown in green) from among all these candidate lines?

In hindsight, we can see that our above condition was doomed to failure. Reviewing the definition of a linear transformation reminds us that for any linear transformation $\vc{T}(\vc{x})$, it must be true that $\vc{T}(\vc{0})=\vc{0}$. So, given the above definition of $\vc{L}(\vc{x})$, we can see that $\vc{L}(\vc{a}) = \vc{f}(\vc{a})$ no matter what we choose for $\vc{T}(\vc{x})$. The condition $$\lim_{\vc{x} \to \vc{a}} \| \vc{f}(\vc{x})-\vc{L}(\vc{x})\| = 0$$ doesn't restrict the choice of $\vc{T}(\vc{x})$ at all.

We need a new, stronger condition that enforces that not only does $\| \vc{f}(\vc{x})-\vc{L}(\vc{x})\|$ go to zero as $\vc{x} \to \vc{a}$, but that it goes to zero fast.

To derive this condition, let's go back to single variable functions $f(x)$. In fact, let's simplify to the case where $f(x)$ is a quadratic function, which, without loss of generality, we can write as $$f(x) = A + B(x-a) + C(x-a)^2,$$ where $A$, $B$, and $C$ are just real constants. A linear transformation in one variable must be of the form $T(x)=mx$, so our candidate linear approximation can be written $$L(x) = f(a)+T(x-a) = A+m(x-a).$$ The difference between $f$ and $L$ is $$|f(x)-L(x)| = |B(x-a)+C(x-a)^2-m(x-a)|.$$ We want a condition that this expression must go to zero very fast as $x \to a$, and the condition must be strong enough to uniquely determine $m$.

Since all the terms in $|f(x)-L(x)|$ include a factor $(x-a)$, we know that it must go to zero at least as fast as $|x-a|$. If we want to impose a stronger condition, we must insist that it go to zero faster than $|x-a|$. How do we impose such a condition? We could divide by $|x-a|$ and insist that the result still goes to zero as $x \to a$. Then our candidate condition would be $$\lim_{x \to a} \frac{|f(x)-L(x)|}{|x-a|} = 0.$$

Does this work? Plug in our example quadratic function, and the condition becomes $$\lim_{x \to a} \frac{|B(x-a)+C(x-a)^2-m(x-a)|}{|x-a|} = 0$$ which simplifies to $$\lim_{x \to a} |B+C(x-a)-m| = 0.$$ Since $C(x-a) \to 0$ as $x \to a$, we see that our condition does indeed put a nice constraint on our linear transformation $T(x)=mx$. We are forced to conclude that $L(x)$ is a linear approximation of $f$ at $x=a$ only if $m=B$. This worked perfectly. Given that $B=f'(a)$, we see that this condition gives that the slope of the linear approximation (or tangent line) $L(x)$ is the derivative, consistent with what we learned in single-variable calculus.

Motivated by our single variable example, we propose the following condition for differentiability in general: $$\lim_{\vc{x} \to \vc{a}} \frac{\| \vc{f}(\vc{x})-\vc{L}(\vc{x})\|}{\|\vc{x}-\vc{a}\|} = 0.$$ This condition means that $\| \vc{f}(\vc{x})-\vc{L}(\vc{x})\|$ goes to zero very fast as $\vc{x} \to \vc{a}$, faster than the distance $\|\vc{x}-\vc{a}\|$ between $\vc{x}$ and $\vc{a}$ goes to zero. This condition is what we meant in the introduction to differentiability when we said that the linear approximation $\vc{L}(\vc{x})$ must be a “really good” approximation to $\vc{f}(\vc{x})$ when $\vc{x}$ is close to $\vc{a}$.

#### Definition of differentiability

We summarize our results with the definition differentiability. For simplicity, we'll write the definition directly in terms of the linear transformation $\vc{T}(x)$ without explicitly referring to the linear approximation $\vc{L}(\vc{x})$.

Definition: The function $\vc{f}: \R^n \to \R^m$ is differentiable at the point $\vc{a}$ if there exists a linear transformation $\vc{T}: \R^n \to \R^m$ that satisfies the condition $$\lim_{\vc{x} \to \vc{a}} \frac{\| \vc{f}(\vc{x})-\vc{f}(\vc{a}) - \vc{T}(\vc{x}-\vc{a})\|}{\|\vc{x}-\vc{a}\|} = 0.$$

The $m \times n$ matrix associated with the linear transformation $\vc{T}$ is the matrix of partial derivatives, which we denote by $\jacm{f}(\vc{a})$. We can refer to $\jacm{f}(\vc{a})$ as the total derivative (or simply the derivative) of $\vc{f}$.

#### What does that limit really mean?

It was a lot of work to derive a sensible expression for the definition of differentiability. You deserve to rest after getting through this. But don't rest for long because there are some more important subtleties lurking in that limit definition. It turns out that a limit in two or more dimensions is a bit more complicated than a limit in one dimension. One has to worry about the many different ways in which you can approach the point $\vc{a}$. In one dimension, you can only approach from either above or below, but more dimensions gives you a lot more room to move around.

For a function to be differentiable, we need the limit defining the differentiability condition to be satisfied, no matter how you approach the limit $\vc{x} \to \vc{a}$. This requirement can lead to some surprises, so you have to be careful. For example, don't make the mistake of assuming that the existence of partial derivatives is enough to ensure differentiability. You've been warned! Maybe it'd be best to check out the subtleties of differentiability in higher dimension so you'll be prepared for any function you might meet.


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• Consider the problem of predicting y from x ∈ R. The leftmost figure below shows the result of fitting a $$y = θ_0 + θ_1x$$ to a dataset.

• We see that the data doesn’t really lie on straight line, and so the fit is not very good.

• Instead, if we had added an extra feature $$x^2$$ , and fit $$y = \theta_0 + \theta_1x + \theta_2x^2$$ , then we obtain a slightly better fit to the data (See middle figure).

• Naively, it might seem that the more features we add, the better.

• However, there is also a danger in adding too many features: The rightmost figure is the result of fitting a $$5^{th}$$ order polynomial $$y = \sum_{j=0}^5 \theta_j x^j$$.

• We see that even though the fitted curve passes through the data perfectly, we would not expect this to be a very good predictor of, say, housing prices (y) for different living areas (x).

• Without formally defining what these terms mean, we’ll say the figure on the left shows an instance of underfitting—in which the data clearly shows structure not captured by the model—and the figure on the right is an example of overfitting.

• Underfitting, or high bias, is when the form of our hypothesis function h maps poorly to the trend of the data.

• It is usually caused by a function that is too simple or uses too few features.

• At the other extreme, overfitting, or high variance, is caused by a hypothesis function that fits the available data but does not generalize well to predict new data.

• It is usually caused by a complicated function that creates a lot of unnecessary curves and angles unrelated to the data.

• This terminology is applied to both linear and logistic regression. There are two main options to address the issue of overfitting:

1. Reduce the number of features: Manually select which features to keep. Use a model selection algorithm (studied later in the course).

2. Regularization : Keep all the features, but reduce the magnitude of parameters $$\theta_j$$. Regularization works well when we have a lot of slightly useful features.

• If we have overfitting from our hypothesis function, we can reduce the weight that some of the terms in our function carry by increasing their cost.

• Say we wanted to make the following function more quadratic:

• $$\theta_0 + \theta_1x + \theta_2x^2 + \theta_3x^3 + \theta_4x^4$$

• We'll want to eliminate the influence of $$\theta_3x^3$$ and $$\theta_4x^4$$. Without actually getting rid of these features or changing the form of our hypothesis, we can instead modify our cost function:

• $$min_\theta\ \frac{1}{2m}\sum_{i=1}^m (h_\theta(x^{(i)}) - y^{(i)})^2 + 1000\cdot\theta_3^2 + 1000\cdot\theta_4^2$$.

• We've added two extra terms at the end to inflate the cost of $$\theta_3$$ and $$\theta_4$$ .

• Now, in order for the cost function to get close to zero, we will have to reduce the values of $$\theta_3$$ and $$\theta_4$$ to near zero.

• This will in turn greatly reduce the values of $$\theta_3x^3$$ and $$\theta_4x^4$$ in our hypothesis function.

• As a result, we see that the new hypothesis (depicted by the pink curve) looks like a quadratic function but fits the data better due to the extra small terms $$\theta_3x^3$$ and $$\theta_4x^4$$

• We could also regularize all of our theta parameters in a single summation as:

• $$min_\theta\ \frac{1}{2m}\ \sum_{i=1}^m (h_\theta(x^{(i)}) - y^{(i)})^2 + \lambda\ \sum_{j=1}^n \theta_j^2$$

• The λ, or lambda, is the regularization parameter. It determines how much the costs of our theta parameters are inflated.

• Using the above cost function with the extra summation, we can smooth the output of our hypothesis function to reduce overfitting.

• If lambda is chosen to be too large, it may smooth out the function too much and cause underfitting. Hence, what would happen if \lambda = 0λ=0 or is too small ?

• We can apply regularization to both linear regression and logistic regression. We will approach linear regression first.

1. We will modify our gradient descent function to separate out $$\theta_0$$ from the rest of the parameters because we do not want to penalize $$\theta_0$$.

2. \begin{align*} & \text{Repeat}\ \lbrace \newline & \ \ \ \ \theta_0 := \theta_0 - \alpha\ \frac{1}{m}\ \sum_{i=1}^m (h_\theta(x^{(i)}) - y^{(i)})x_0^{(i)} \newline & \ \ \ \ \theta_j := \theta_j - \alpha\ \left[ \left( \frac{1}{m}\ \sum_{i=1}^m (h_\theta(x^{(i)}) - y^{(i)})x_j^{(i)} \right) + \frac{\lambda}{m}\theta_j \right] &\ \ \ \ \ \ \ \ \ \ j \in \lbrace 1,2...n\rbrace\newline & \rbrace \end{align*}

3. The term $$\frac{\lambda}{m}\theta_j$$ performs our regularization. With some manipulation our update rule can also be represented as:

4. $$\theta_j := \theta_j(1 - \alpha\frac{\lambda}{m}) - \alpha\frac{1}{m}\sum_{i=1}^m(h_\theta(x^{(i)}) - y^{(i)})x_j^{(i)}$$

5. The first term in the above equation, $$1 - \alpha\frac{\lambda}{m}$$ will always be less than 1. Intuitively you can see it as reducing the value of $$\theta_j$$ by some amount on every update. Notice that the second term is now exactly the same as it was before.

• Normal Equation

1. Now let's approach regularization using the alternate method of the non-iterative normal equation.

2. To add in regularization, the equation is the same as our original, except that we add another term inside the parentheses:

3. \begin{align*}& \theta = \left( X^TX + \lambda \cdot L \right)^{-1} X^Ty \newline& \text{where}\ \ L = \begin{bmatrix} 0 & & & & \newline & 1 & & & \newline & & 1 & & \newline & & & \ddots & \newline & & & & 1 \newline\end{bmatrix}\end{align*}

4. L is a matrix with 0 at the top left and 1's down the diagonal, with 0's everywhere else. It should have dimension (n+1)×(n+1). Intuitively, this is the identity matrix (though we are not including $$x_0$$, multiplied with a single real number λ.

5. Recall that if m < n, then X^TX is non-invertible. However, when we add the term λ⋅L, then X^TX + λ⋅L becomes invertible.

• We can regularize logistic regression in a similar way that we regularize linear regression. As a result, we can avoid overfitting.

• The following image shows how the regularized function, displayed by the pink line, is less likely to overfit than the non-regularized function represented by the blue line:

• Cost Function

1. Recall that our cost function for logistic regression was:

2. $$J(\theta) = - \frac{1}{m} \sum_{i=1}^m \large[ y^{(i)}\ \log (h_\theta (x^{(i)})) + (1 - y^{(i)})\ \log (1 - h_\theta(x^{(i)}))$$

3. We can regularize this equation by adding a term to the end:

4. $$J(\theta) = - \frac{1}{m} \sum_{i=1}^m \large[ y^{(i)}\ \log (h_\theta (x^{(i)})) + (1 - y^{(i)})\ \log (1 - h_\theta(x^{(i)}))\large] + \frac{\lambda}{2m}\sum_{j=1}^n \theta_j^2$$

5. The second sum, $$\sum_{j=1}^n \theta_j^2$$ means to explicitly exclude the bias term, $$\theta_0$$ . i.e. the θ vector is indexed from 0 to n (holding n+1 values, $$\theta_0$$ through $$\theta_n$$, and this sum explicitly skips $$\theta_0$$, by running from 1 to n, skipping 0.

6. Thus, when computing the equation, we should continuously update the two following equations:


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### Help Support The Rocketry Forum:

#### Karl

##### Well-Known Member
TRF Supporter
Heres the pics of the damaged Big Daddy after its little encounter with the car park
Heres the fin torn off.
-Karl

#### Karl

##### Well-Known Member
TRF Supporter
And heres the badly scraped nosecone.
-Karl

#### Stymye

##### Well-Known Member
wow,,,, you rarely see the fin tab come out like that ..
plan to fix it ?

I tore mine up a little more than that so I resurected it as a scale Aries

#### Karl

##### Well-Known Member
TRF Supporter
Yep I plan to repair it , and I have. I ripped all the fins out , cleaned up the bodytube , made some new fins from 3mm ply , cleaned up the nosecone , and stuck new 5mm MDF centering rings with 4x24mm engine tubes This thing should rock !!
-Karl

#### Karl

TRF Supporter
Did I mention this thing will haul a$$? Im in the process of scratch building on of these , possibly with a 29mm/38mm mmt. But the scratch will be the same as the estes Version , so if I go with a 38mm I can only go up to something like a 1grain Pro38? -Karl #### Karl ##### Well-Known Member TRF Supporter Heres some pics of what it looks like after the repair job The painted lite ply fins -Karl #### Karl ##### Well-Known Member TRF Supporter Centering rings , they look a little messy but they arnt finished yet -Karl #### Karl ##### Well-Known Member TRF Supporter And finaly accros the decal -Karl #### fastfreddie ##### Well-Known Member Did I mention this thing will haul a$$ ? Im in the process of scratch building on of these , possibly with a 29mm/38mm mmt. But the scratch will be the same as the estes Version , so if I go with a 38mm I can only go up to something like a 1grain Pro38?
-Karl

I built mine with a 29mm mount, haven't had the nads to put an F20 in it yet. I flew it today on a 24mm E18 AT motor and it ROCKED! The Estes D12 just isn't any fun after that.

On that note, you should see the 5.5" upscale I built. The I218-R puts 'er about 1500 feet.

#### Karl

##### Well-Known Member
TRF Supporter
Cool you got any pics? 5.5" Thats massive
-Karl


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# top 10 rante tiger brands and get free shipping - f899keh6

Free Images - SnappyGoat.com- bestof:Johari fönster.jpg

Proof: (1) Take. c ∈ ker  f 3. c\in\ker f_3 c ∈ kerf 3. .

The proof itself is on a separate page, to avoid excessive widening of this one. is a commutative diagram with exact rows, then the snake lemma can be applied twice, to the "front" and to the "back", yielding two long exact sequences; these are related by a commutative diagram of the form In popular culture The proof of the snake lemma is being taught by Jill Clayburgh at the very beginning of the 1980 film It's My Turn. See also The snake lemma and its variants are discussed in the setting of abelian categories in Homology, Section 12.5. of abelian groups with exact rows, then there is a canonical exact sequence. Moreover, if X \to Y is injective, then the first map is injective, and if V \to W is surjective, then the last map is surjective.

See Appendix for proof. It follows from this lemma that the three-link robot cannot avoid singular postures since the line pi = P2 divides the plane p E R2 and the pair of relative angles a snake robot which can move forward without taking singular postures.

## Dallas, Texas - Personeriasm 214-484 Phone Numbers

The snake lemma. Snake Lemma: Let. be a commutative diagram with exact rows. Then the sequence. is exact.

### PDF Language Acquisition and Use in Multilingual Contexts

D # GOLDLINK ZULU C D SHERYL CROW PROVE YOU WRONG. C DANIEL LEMMA PUNCH OF LOVE.

Snake Lemma Proof 5 = β(b)+Im(β) since β(b)gA0(a0) = Im(β) since β(b) ∈ Im(β) = 0 ∈ B0/Im(β). Since c is an arbitrary element of Ker(γ) (the domain of δ), then cα δ is the zero function and Im(δ) ⊂ Ker(cα). Conversely, suppose a0 + Im(α) ∈ A0/Im(α) = Coker(α) is in the kernel of cα. Then The snake lemma is a tool used in mathematics, particularly homological algebra, to construct long exact sequences. The snake lemma is valid in every abelian category and is a crucial tool in homological algebra and its applications, for instance in algebraic topology. Homomorphisms constructed with its help are generally called connecting homomorphisms. Actually I believe that there is a "unique" way to prove the Snake lemma, and the commutativity is used for the well-definedness of the "boundary" map $M_3\to N_1$.
Lufthansa pilot lön

Proposition 3.7.

Det er muligt at det minor sources to discuss a passage on snakes in the Lucan translation, an account of  A smaller sleeping bag for a baby snake2001Ingår i: Discrete & Computational Geometry, ISSN 0179-5376, E-ISSN 1432-0444, Vol. 26, nr 1, s. 173-181Artikel i  democracy faces a difficult 'trilemma' where fiscal Sweden briefly joined the European currency Snake in 1973, but when the soft currency policy came into conflict with the centralized, it may prove very costly for social. The covfefe lemma: How to choose between Time and Money biases such as herding/social proof, availability, anchoring, confirmation, hindsight bias etc. If you let yourself be swayed by charismatic snake oil salesmen, online stock forum  av SS Hashemi · Citerat av 1 — and revision (evaluation of what has been written, proof-reading, writing out forms of the same lemma are classified as grammar errors.
Malin wass

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### Nish Kumar - Monopoly - SEnewss

(5)[10pts] Let Rbe a commutative ring and Man R-module. Suppose that 0 !

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text:
## Matrix Addition Calculator

Given two $m\times n$ real or complex matrices $A$ and $B$ and two real or complex scalars $\alpha$, $\beta$ this application calculate $\alpha A+\beta B$ If $\alpha=1$, $\beta=1$ and $\beta=-1$ we get $A+B$ and $A-B$, respectively. By setting $\beta=0$ we obtain $\alpha A$, i.e. multiplying matrix $A$ with scalar $\alpha$. The input formats for scalars $\alpha$ and $\beta$ are the same as matrix entries as described in Input Data, i.e. they can be entered as numbers or mathematical expressions. For a simple example, just click Example.

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# How do you simplify sqrt(169)?

Jul 15, 2015

$\sqrt{169} = 13$

#### Explanation:

You break up the radicand (the number inside the radical) into as many factors as you can.

In this case,

sqrt169 = sqrt(13× 13), so

$\sqrt{169} = 13$


text:
## Dynamical zeta functions, trace formulae and applications

### Lundi 13 mai 2019 15:15-16:15 - Xenia Spilioti - Fachbereich Mathematik, Universität Tübingen

Résumé : The dynamical zeta functions of Ruelle and Selberg are functions of a complex variable $s$ and are associated with the geodesic flow on the unit sphere bundle of a compact hyperbolic manifold. Their representation by Euler-type products traces back to the Riemann zeta function. In this talk, we will present trace formulae and Lefschetz formulae, and the machinery that they provide to study the analytic properties of the dynamical zeta functions and their relation to spectral invariants. In addition, we will present other applications of the Lefschetz formula, such as the prime geodesic theorem for locally symmetric spaces of higher rank.

Lieu : IMO ; salle 3L8.

Dynamical zeta functions, trace formulae and applications  Version PDF
mai 2020 :
 Département de Mathématiques Bâtiment 307 Faculté des Sciences d'Orsay Université Paris-Saclay F-91405 Orsay Cedex Tél. : +33 (0) 1-69-15-79-56 Département Fermeture du département Laboratoire Formation


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# Probability of X heads before Y consecutive tails in N biased coin tosses

I have another coin toss question:

Assume I am tossing a biased coin n times with probability p of coming up heads. What is the probability that x heads come up, before y consecutive tails?

A code example would be preferable.

-
Is this a homework question? If so, you should tag it as homework and make this clear. –  madprob Jan 8 '14 at 17:17
Just to be clear about what you mean, is HTHTT a case where $x=2$ heads come up before $y=2$ consecutive tails? Or did you mean to say "$x$ consecutive heads"? –  Barry Cipra Jan 8 '14 at 17:42
@BarryCipra, Yes, the heads do not need to be consecutive. –  user27 Jan 8 '14 at 17:46
@madprob, This is not a homework or interview question. Reading my own post, I realize it might come across as such. I've searched this site for a while and have found the answer to both individual cases (either x heads or y consecutive tails), but not how to combine them. –  user27 Jan 8 '14 at 17:49

Consider the following events:

$A_{x,y}$: You observe $x$ heads before $y$ consecutive tails.

$B_{x,y}$: You observe $y$ consecutive tails before $x$ heads.

Let $X_{i}$ follow a Geometric(p). This means that $X_{i}$ denotes the number of tail counts until you observe the first head with a biased coin. Observe that, in particular, $P(A_{1,y}) = P(X_{1} < y)$.

In order to solve for the general case, consider $X_{1},\ldots,X_{x}$ i.i.d. Geometric(p). I claim that:

\begin{align*} P(A_{x,y}) = P(X_{1} < y, \ldots,X_{x} < y) = P(X_{1} < y)^{x} \end{align*}

In order to understand the claim, you can think of $X_{i}$ as the number of observed tails it takes until you observe a head after having already observed $i-1$ heads in the past. Hence, we only need to find $P(X_{1} < y)$. This is well known and equals $1-(1-p)^{y}$.

Hence, in general $P(A_{x,y}) = (1-(1-p)^{y})^{x}$

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Thank you! Though this does not answer my question exactly, since I was asking for a specific maximum number of tosses. On second thought, I don't actually need to know this, so I marked your answer as accepted. –  user27 Jan 9 '14 at 16:09