Real Analysis – Is a function continuous over a closed and positive interval with every rational positive everywhere?

"If f is continuous on (0,1) and if f (x) is positive for each rational x, then does it follow that f is positive at all x?"

I do not know if I'm missing something major here. My textbook tells me that if f can be zero at some points, it must be nonnegative everywhere. It does not make sense to me.

Can we not define a piecewise function like f (x) = x + 1 if $ x in (0,1), f (x) = | x | if x is a rational outside the interval (0,1), f (x) = – | x | if x is an irrational out of range (0,1)?

Such a function would indeed be continuous over the interval (0,1), would be positive for any rational, and would be negative in an infinite number of points.

Mathematical Optimization – Prove the Positive Semi-finitude of a 6X6 Symbolic Matrix

Specifically, I want to check the positive semi-finiteness of the following 6X6 symbolic matrix

{{-2 (-((11 m1^2 - 24 m1 m3 + 72 m3^2)/(
      144 k^2 m1^2 m3^2 (Sigma)^2)) - 2 (Sigma)^2), 
  I - (m1 - 6 m3)/(6 k m1 m3 (Sigma)^2), (
  3/m1 + 1/m3 + (12 (Sigma)^4)/m2)/(6 k (Sigma)^2), 2 (Sigma)^2, 
  1/(8 k^2 m3^2 (Sigma)^2) + 4 (Sigma)^2, (m1 + 12 m3)/(
  12 k m1 m3 (Sigma)^2)}, {-I - (m1 - 6 m3)/(6 k m1 m3 (Sigma)^2), 
  1/(Sigma)^2, 1/(2 (Sigma)^2), 0, 0, 1/(Sigma)^2}, {(
  3/m1 + 1/m3 + (12 (Sigma)^4)/m2)/(6 k (Sigma)^2), 1/(
  2 (Sigma)^2), 1/(Sigma)^2 + (4 (Sigma)^2)/(k^2 m2^2), 
  I + (4 (Sigma)^2)/(k m2), (1/m3 + (16 (Sigma)^4)/m2)/(
  4 k (Sigma)^2), 1/(Sigma)^2}, {2 (Sigma)^2, 
  0, -I + (4 (Sigma)^2)/(k m2), 4 (Sigma)^2, 4 (Sigma)^2, 
  0}, {1/(8 k^2 m3^2 (Sigma)^2) + 4 (Sigma)^2, 0, (
  1/m3 + (16 (Sigma)^4)/m2)/(4 k (Sigma)^2), 4 (Sigma)^2, 
  1/(8 k^2 m3^2 (Sigma)^2) + 6 (Sigma)^2, 
  I + 1/(4 k m3 (Sigma)^2)}, {(m1 + 12 m3)/(12 k m1 m3 (Sigma)^2), 
  1/(Sigma)^2, 1/(Sigma)^2, 0, -I + 1/(4 k m3 (Sigma)^2), 3/(
  2 (Sigma)^2)}}

What depends on the 5 parameters: $ m_ {1} $, $ m_ {2} $, $ m_ {3} $, k $, $ sigma $, which are all positive, that is to say

$ m_ {1}> 0, m_ {2}> 0, m_ {3}> 0, k> 0, sigma> 0 tag {1} $ .

I try to apply the methodology proposed in (Check if a symbolic matrix is ​​semi-definite positive) but Mathematica continues to perform the calculation and gives no results.

Then, as another way of solving the problem, I tried to numerically inspect the minimum and maximum eigenvalue values ​​of the matrix above, using NMinimize() and NMaximize() subject to the constraints indicated in (1). My code is

s=Simplify(Eigenvalues({{-2 (-((11 m1^2-24 m1 m3+72 m3^2)/(144 k^2 m1^2 m3^2 (Sigma)^2))-2 (Sigma)^2),I-(m1-6 m3)/(6 k m1 m3 (Sigma)^2),(3/m1+1/m3+(12 (Sigma)^4)/m2)/(6 k (Sigma)^2),2 (Sigma)^2,1/(8 k^2 m3^2 (Sigma)^2)+4 (Sigma)^2,(m1+12 m3)/(12 k m1 m3 (Sigma)^2)},{-I-(m1-6 m3)/(6 k m1 m3 (Sigma)^2),1/(Sigma)^2,1/(2 (Sigma)^2),0,0,1/(Sigma)^2},{(3/m1+1/m3+(12 (Sigma)^4)/m2)/(6 k (Sigma)^2),1/(2 (Sigma)^2),1/(Sigma)^2+(4 (Sigma)^2)/(k^2 m2^2),I+(4 (Sigma)^2)/(k m2),(1/m3+(16 (Sigma)^4)/m2)/(4 k (Sigma)^2),1/(Sigma)^2},{2 (Sigma)^2,0,-I+(4 (Sigma)^2)/(k m2),4 (Sigma)^2,4 (Sigma)^2,0},{1/(8 k^2 m3^2 (Sigma)^2)+4 (Sigma)^2,0,(1/m3+(16 (Sigma)^4)/m2)/(4 k (Sigma)^2),4 (Sigma)^2,1/(8 k^2 m3^2 (Sigma)^2)+6 (Sigma)^2,I+1/(4 k m3 (Sigma)^2)},{(m1+12 m3)/(12 k m1 m3 (Sigma)^2),1/(Sigma)^2,1/(Sigma)^2,0,-I+1/(4 k m3 (Sigma)^2),3/(2 (Sigma)^2)}}));
(*The first eigenvalue is 0*)
s((1))

(*Maximum and Minimum of second eigenvalue*)
NMinimize(s((2)),m1>0&&m2>0&&m3>0&&k>0&&(Sigma)>0,{m1,m2,m3,m3,k,(Sigma)})
NMaximize(s((2)),m1>0&&m2>0&&m3>0&&k>0&&(Sigma)>0,{m1,m2,m3,m3,k,(Sigma)})

(*Maximum and Minimum of third eigenvalue*)
NMinimize(s((3)),m1>0&&m2>0&&m3>0&&k>0&&(Sigma)>0,{m1,m2,m3,m3,k,(Sigma)})
NMaximize(s((3)),m1>0&&m2>0&&m3>0&&k>0&&(Sigma)>0,{m1,m2,m3,m3,k,(Sigma)})

(*Maximum and Minimum of fourth eigenvalue*)
NMinimize(s((4)),m1>0&&m2>0&&m3>0&&k>0&&(Sigma)>0,{m1,m2,m3,m3,k,(Sigma)})
NMaximize(s((4)),m1>0&&m2>0&&m3>0&&k>0&&(Sigma)>0,{m1,m2,m3,m3,k,(Sigma)})

(*Maximum and Minimum of fifth eigenvalue*)
NMinimize(s((5)),m1>0&&m2>0&&m3>0&&k>0&&(Sigma)>0,{m1,m2,m3,m3,k,(Sigma)})
NMaximize(s((5)),m1>0&&m2>0&&m3>0&&k>0&&(Sigma)>0,{m1,m2,m3,m3,k,(Sigma)})

(*Maximum and Minimum of sixth eigenvalue*)
NMinimize(s((6)),m1>0&&m2>0&&m3>0&&k>0&&(Sigma)>0,{m1,m2,m3,m3,k,(Sigma)})
NMaximize(s((6)),m1>0&&m2>0&&m3>0&&k>0&&(Sigma)>0,{m1,m2,m3,m3,k,(Sigma)})

I have especially tried this method because of the following reasoning:

If some eigenvalues ​​become purely negative, their maximum and minimum values ​​will also be negative and, in this situation, the matrix defined above will not be a semi-defined matrix.

Then, in my code, I find that the second eigenvalue (s((2))) therefore has a negative maximum and a minimum, which proves that the matrix has negative eigenvalues ​​and is therefore a non-semi-defined matrix. But the problem is that Mathematica displays the following error at the output of Max and Min calculation.

minimize::cvmit: Failed to converge to the requested accuracy or precision within 100 iterations. >>

So, I have three questions

  1. How to remove the above error?
  2. Is my methodology / reasoning correct for testing the positive semi-finiteness of the matrix?
  3. Do you have any other suggestions to prove the semi-definitive positive nature of the aforementioned matrix?

BigBoxHost Cheapest Offer – Positive Comodo SSL @ $ 4.72 / year

BigBoxHost – Company BigBox LLC

BigBox Host, a leading provider of web hosting, web design and services, is part of the BigBox LLC group, which has been providing quality web hosting solutions to businesses and individuals since 2010. Our goal is to provide exceptional services at the best possible price. To achieve these goals, we innovate and constantly upgrade our services at no additional cost to our customers.

Get the cheapest SSL certificate from Comodo, Geotrust, Symantec, Thawte and RapidSSL at the best price only from BigBox Host.

Comodo certificates:

Comodo SSL Trial – 90 days – FreeOrder now
Comodo PositiveSSL – from $ 4.72 per year
Comodo InstantSSL – From $ 23.09 per year
Comodo InstantSSL Pro – From $ 34.39 per year
Comodo EV SSL – from $ 115.95 per year
Comodo EV SGC SSL – from $ 124.97 per year
Comodo Multi-Domain EV SSL – $ 205.47 per year
Comodo Essential SSL – from $ 8.79 per year
Comodo SGC SSL – from $ 72.96 per year
Comodo SGC Wildcard SSL – From $ 199.67 per year
Comodo Code Signing SSL – From $ 67.09 per year
Comodo PositiveSSL UCC / SAN – From $ 45.65 per year
Comodo PositiveSSL Wildcard – From $ 65.99 per year
Comodo Essential Wildcard SSL – From $ 68.99 per year
Comodo Premium Wildcard SSL – From $ 111.43 per year
Comodo InstantSSL Premium (IP) – From $ 41.75 per year

Buy now

GeoTrust certificates:

GeoTrust QuickSSL Premium – From $ 40.56 per year
GeoTrust TrueBusinessID – From $ 59.98 per year
GeoTrust TrueBusinessID Wildcard – From $ 267.48 per year
GeoTrust TrueBusinessID EV – from $ 132.97 per year
GeoTrust TrueBusinessID SAN – From $ 150.23 per year
GeoTrust TrueBusinessID EV SAN – From $ 284.97 per year

Buy now

Symantec certificates:

Symantec Code Signing SSL – From $ 386.65 per year
Symantec Safe Site – From $ 219.98 per year
Symantec Secure Site – From $ 248.31 per year
Symantec Secure Site PRO – From $ 593.31 per year
Symantec Secure Site PRO EV – From $ 892.47 per year
SAN Symantec Secure Site EV – From $ 614.97 per year
Symantec Secure Site Wildcard – From $ 1457.31 per year

Buy now

Thawte certificates:

Thawte SSL Signing Code for Individuals – From $ 124.97 per year
Thawte Code Signing SSL – From $ 122.47 per year
Thawte SGC SuperCerts SAN – From $ 124.98 per year
Thawte Web Server SAN SSL – From $ 63.98 per year
Thawte SSL 123 – from $ 28.48 per year
Thawte Web Server SAN EV – from $ 149.97 per year
Thawte Wildcard SSL Certificate – From $ 254.97 per year

Buy now

RapidSSL certificates:

RapidSSL Standard – From $ 7.93 per year
RapidSSL WildcardSSL – From $ 81.23 per year

Buy now

Visit us today!
www.BigBoxHost.com

Canadian Writer | 100% positive feedback | Quick return | Bulk order accepted

I offer 100% unique, high quality items on all high quality topics and topics. I offer articles, ebooks, rewrites, press releases, blog posts, product reviews, and more.

As you will see, I have the reputation of offering only the best quality to my clients. If you buy from me, I guarantee that you will receive the best content you will find on this forum, or elsewhere.

Prices start at 1.6 cents per word and I offer discounts on bulk …

Canadian Writer | 100% positive feedback | Quick return | Bulk order accepted

Positive – Black Friday Special Web Hosting Deals | 50% off flat | bodHOST | NewProxyLists

Bodhost is a provider of business solutions. It offers solutions ranging from managed dedicated servers, custom server solutions, data centers, VPS hosting, web hosting, backups and spam experts.

Exclusive Black Friday offers bodHOST on web hosting.

Save big on this black Friday with bodHOST. Redeem your discount code now!

R1 Soft Backup – 50% off

• Hassle-free use
• Multiple replication sites
• High resistance storage
• full encryption

Coupon Code – BFBACK50 | Order now

Spam Experts – 25% discount

• Escape the blacklist
• one-click backup
• No emails sent back
• Easily configure

Coupon Code – BF25SPAM | Order now

Dedicated Servers – 20% off

• Managed servers
• Secure environment
• Optimized application
• Optional firewall

Coupon Code- BFSAVE20DS | Order now

VPS Hosting – 25% off

• 99.95% uptime
• regular monitoring
• 24×7 Support
• Rapid provisioning

Coupon Code – BFGET25VPS | Order now

WordPress Hosting – 30% off

• WordPress pre-installed
• Optimized security
• 24×7 Pro Support
• Performance equipment

Coupon Code – BFBOD30WP | Order now

Hurry up! All offers are valid until the end of December 2, 2019.

If you have any questions, you can contact our sales department by starting a chat or by sending an email to [protected email] Or call us on 8443245054.

group dynamics – Practical use of map O, or how to measure positive consent on the fly

I myself am a strong supporter of the TTRPG Security Toolkit, which has been useful to me in a myriad of campaigns as an average player and player.

You mentioned that you have very little time for any kind of "session 0" or experience with the players. The best thing to do is to give each player a small piece of cardboard or paper and ask him to write any "no-go". "subjects, in a method similar to the" Lines and Sails ", which I will quote here:

Lines are strict limits to content, things that the GM or players do not want to engage. Setting up a line means that the content will not appear at all in the game.

Using lines alone can be tricky, but if you do not have time to discuss party sails, you are right to fall back on the "O / X" continuous consent method. My experience with the X Card has been excellent, both in real life and in live games, it usually goes well with a good party, someone privately reports me an "X" or "Hey, I I am not comfortable with that. "And I'm informing the party that we are taking a step back on this topic." I had to use it myself when a trio of particularly toxic gamers attempted to get away with it. assault an NPC for no other reason than sadistic laughter out of the character.

For the use of the "O" card, I used it only when approaching a veil or a subject that, I know, could be a problem (all that I have encountered before as a veil, violence extreme, claustrophobia, body horror, etc.) to see if the players are doing well. What I will usually do is put an O card on the table or send each player an "O" directly and wait for him to place an X or O card on the table in response or answer to my private message For an X or an O, the decision must be unanimous so that we continue with the thread of the plot. If a person is uncomfortable, I will respect that and we will not continue. This puts a player in a public situation, but I would not say more than the X Card. That's why I usually take 10 minutes to explain the X / O Cards at the beginning of a game and make it clear "If you're pressuring a player to feel uncomfortable with a scene, you risk being excluded from the table for inappropriate behavior. "as simple as that.

To conclude, my interpretation of map O is less a method initiated by the player, but rather a method initiated by the GM. I suggested to my players that they could also trigger a vote if they thought that someone could feel uncomfortable with a subject, but since that time that I used these rules, there was only one vote triggered by a player, which ended a thread of intrigue as it was turned out that this player was friendly enough with another to know a phobia that I had not been aware of, and they were a little shy and did not want to interrupt a story themselves, so the other player was triggered a vote on their behalf, which allowed them to have a safer space to vote with an X, knowing that someone had concerns.

algorithms – Find $ n th $ perfect number, where perfect number is a positive integer whose sum of digits is $ 10 $

I'm going to describe a method being run in $ O (| n |) = O ( log n) $ or $ | n | $ is the length of the number written in decimal representation. We will follow a dynamic programming scheme. For each $ k in {0, dots 10 $ and every positive integer $ j leq n $ we will calculate the number of integers in length k $whose sum of digits equals k $. This can be calculated using dynamic programming where $$ C (k, j) = sum limits_ {i in {0, k points}} C (k-i, j-1), $$
since we can add $ i $ to the left of a number of length $ j-1 $ and sum $ k-i $ build a number of length $ j and sum k $.

Let $ P $ to be a sum prefix on $ C (10, j) $ for $ j = 0, 1 points $clearly $ P (i) $ indicates the whole number of lengths at the most $ i $ summarize up to 10.
Now, the solution to your problem can be searched recursively by finding the biggest $ i $ such as $ P (i) leq n $ and subtract from $ n $.

I suspect that the solution can be found in $ O ( log | n |) = O ( log log n) $ by only building $ C (k, 2 ^ r) $ for integers $ r $ and binary search for the greatest value of $ r $ or $ C (10, 2 ^ r) $ is at most $ n $.

With an array of positive numbers, look for the sum of maximum subsequences <= sum given, so that no element is adjacent.

That's my code

It goes for few cases but not for all,
for example.

number of elements n = 6

elements a (n) = {5,5,10,100,10,5}

and sum = 25

the output is 15, but the output must be 25 (5 + 10 + 10)

click here to see how it works in the IDE

#include
#define ll long long int
using namespace std;
int main(){
    int n;
    ll sum;
    vector v;
    cin>>n;
    ll a(n);
    for(int i=0;i>a(i);
    }
    cin>>sum;
    ll ans = 0;
    ll inc = a(0);
    ll exc = 0;
    ll prev = 0;
    for(int i=1;i

Numerical Analysis – How to solve a semidimensionally positive quadratic matrix equation

I have the following quadratic matrix equation:

$ XAX + X = B $

or $ A $ and $ B $ are all positive defined matrix.

The constraint here is that $ X $ is in fact a covariance matrix and should therefore be positive defined.

All I have to say is that when there is no constraint, the equation can be solved via the Bernoulli iteration in the following form:

$ X_ {k + 1} = -A ^ {- 1} (I-BX_k ^ {- 1}) $

However, this does not seem to be able to preserve the constraint.

Any orientation would be appreciated, thank you.

Why do poor white people complain of positive actions instead of going to university?

Positive action is a myth. Racist whites in power continue to discretely discriminate against us, including refusing to interview someone for an ethnic-sounding name, and so on.