Blopa

Hello forum, I'm trying to create a program that will securily delete a file, as you know, even if you delete a file (be it manually or through a program with the remove() function) the file always stays there until its deleted by something written over it. My question is, anyone has any idea of how could you securely delete a file? I have thought of for example first using <fstream> to overwrite the file (suppose a .txt file, change all whats written on it converted to 0's) and then finally using remove() Is this method secure enough? What could I use? Thanks for the help

I know how to do that, but, there isn't any code I can put inside the program to make it copy/paste in an easier way? D= Thanks by the way

Exactly, I meant the console, but in windows that one doesnt works, also to copy paste you must right click and make many process to copy and paste, I want to make something like that the user inputs "/c" "/v" and the program automaticly copies or pastes the information

Hello there forum, I have recently been working on encryption software (I'm an amateur in everything, just starting my physics studies, but I'm doing my best). I've been building it on C++ (specially with Dev-C++), I'm using "the black screen" for everything, and I was wondering if any of you knew an useful script that allows the black screen to "paste" a string copied from somewhere else (like when you copy a text from somewhere and then simply 'paste' it on Word) and also another one to "copy" a desired string so it can be easily pasted somewhere else. -Thanks for your time, Blopa

Hello forum, I'm just starting using C++, but I've got a problem, I'm working with Dev-C++ and supposedly a code like this: #include <stdio.h> int main() { printf("whatever"); return 0; } would make an output window appear saying "whatever", I checked in many other sites and it should work, nevertheless, the window opens on my computer and quickly closes up... don't know why is it happening... By the way, I'm working with version 4.9.9.2 Thanks

My question is, if I have the KOH in water, will the second reaction take place? or is the KOH to stable to get into another reaction?

Hey there forum, the other day I was "exploding" some potassium with water, I know becase of what I saw that the reaction is: 2 K + 2 H2O ---> 2 KOH + H2 My question is, considering there's extra water, the reaction would continue as following? KOH + H2O ---> KOOH + H2 Does that happens? And the water is still caustic? Does another reaction take place after that one?

Yeah, I knew the answer but... I was wondering of the formal way to do it, testing the numbers isnt really a good way to solve it, even though, I admire you for testing it with a program (I use Actionscript (kids programming) to program), and thanks for helping ^^

I understand your analysis I did the same, but still couldnt solve it... how do you think you can solve it with that? And yeah the key is in the 200<N<300 but I cant figure out how to use it... Thanks by the way And I really dont remember how he worked the thing about they noticing the missing coins, I just paraphrased the problem, I read the book a year ago and remember the problem today

Hey there forum, I have been reading this awesome book called The Man Who Calculated (really recommended, it's a good recreational book), in this book the main character faces different mathematical problems which he solves easily, in the book it appears one which he only answers but doesnt explains how he solved it, so I in my curiosity tried to solve the problem through math, but it became troublesome, I would really appreaciate if someone here helps me. Here's the story of the problem: 3 sailors got some money for doing a very good work, the amount of golden coins they were going to get was between 200 and 300. They were going to split it in the morning. During the night one of them woke up and decided he was going to take his share before the morning, so he went to where the money was kept and divided the coins in 3, there was one coin left, so he thought "we are going to fight over this coin in the morning", so he decided to throw it to the sea. He took his coins and left the rest in there. Later in the night, another one of the sailors woke up with the same idea, but without knowing the other one had already taken his part, so he went and splited the coins in 3, and there was also one coin left, so after thinking the same as the other sailor, he threw it to the sea, he took his share and leaved the rest. After that the story repeats with the third sailor. And later, another sailor assigned by the captain in the morning went to help them divide it, finding out there was (again) one coin left, which he took for himself for the trouble. That's the story, I tried to solve it, but I can't, I wonder if one of you guys can solve it, I would be thankful for the instruction.

Okey, so you will see, 1 radian is the amount of degrees the radius "takes" when its put in the circumference, pi radians = 180º therefore: 2pi radians = 360º And then Cap'n Refsmmat is right, the result would be [math]\frac{2 \pi}{5}\, \mbox{rad/sec} [/math]

I see, and sorry, I didnt knew i posted it thrice (I've been having some connection problems =S, seriously I'm really ashamed for the spam and the trouble, sorry forum) and, oh I see, thank you for the help, it seems I'm late for this discovery as well -.- Anyway thanks for your help, and it seems I was correct =D What I did: I was thinking of a kind off "infinite circular ladder" in which each "step" became half of the last one (or also and "empty disc" which I started filling), and thought of the S 1/2^i to solve it, I later thought that it will aproximate infinitly to one, but always would be 1 - (1/2^n) which would become infinitely small, it would always be missing "one of the same size as the last step" so I wrote: S 1/2^i = 1 - 1/2^i the next day I wrote the formula and thought "hey, what if instead of a 2 I use a 3?" I made 1/3 + 1/9 + 1/27 = (9+3+1)/27 = 13/27 and I was like "oh -.-", I tried it again adding one more "step", 1/3 + 1/9 + 1/27 + 1/81 = (27+9+3+1)/81 = 40/81 And then I noticed something, the nominators (upper part of fraction) were equivalent to (3^n - 1)/2; so I wrote: S 1/3^i = (3^n - 1)/(2 · 3^n) = 1/2 - 1/(2 · 3^n) I thought "this looks kinda similar, lets work with the 4" S 1/4^i 1/4 + 1/16 + 1/64 = (16+4+1)/64 = 21/64 this time I was like "what is this?" 1/4 + 1/16 + 1/64 + 1/256 = (64+16+4+1)/256 = 85/256 So I tried to find a relationship between 21/64 and 85/256 I noticed they were similar, and then I divided (actually I dont know why I did it, luck?) 256/3 and I found out it was incredibly close to 85, so I made (256-1)/3 = 85, I was impressed and did the same with the other one (64-1)/3 = 21, I had found the relationship S 1/4^i = (4^n-1)/(3 · 4^n) = 1/3 - 1/(3·4^n) Then I noticed the 3 formulas were similar, and after analyzing how they "changed" I got to the conclusion: S 1/c^i = 1/(c-1) - 1/(c-1)(c^n) I tested it this morning with a program I wrote, it worked on more than 99% of the times with different values (16 decimals, mistakes were due to computer approximation; I put her to calculate it by "adding" and meanwhile by using my formula to see if they were the same) Anyway thanks for your time forum =)

I really dont know how to call those: n E i i=1 in english, so forgive me for the title, anyway, I have come with a big question for you forum, I have lately been working with n E 1/c^i i=1 where: c > 1, c is natural working with those I think I may have found a general formula for any c (that satisfies the past requirements), I was wondering if you have ever heard of it and also if you have it, I would appreaciate. Later I will post my thinking, procedure and conclusion, meanwhile, thank you for your help.

Hey there, mmm, you will see, I'm a calculus student, today as I was coming back home I remembered the logarythm properties, the thing is, I found out I just accepted them as they told me, and never questioned them, so as soon as I got home I tried to analyze the properties from a deeper perspective, I was wondering, there's someone that can explain me and demonstrate me the reason of the properties in logarythms? I would really appreciate it Also, as I was experimenting I came with a log 9, but I (feeling guilty of not having questioned the properties told to me before) decided to solve them using differential calculus instead of a calculator so I made like this: f(x) = log x f'(x) = 1/(x · ln 10) f(9) = f'(10)·dx + f(10) log 9 = -1/(10·ln 10) + 1 So I found another logarythm, ln 10 g(x) = ln x g'(x) = 1/x g(e + (10-e)) = g'(e)·dx + g(e) ln 10 = (10-e)/e + 1 So... I decided to solve them, mmm... the second one is a total fail, real approximation is 2.303, my approx is 3.679 I also tested the first one as its (using the calculator to solve the ln 10) and it gave me 0.9565, while the real approx is 0.9542 (not so much difference 0,002) My question is, what am I doing wrong? The reason of why my second differential failed is because the differential (10-e) is to big? Thanks for your time by the way

Oh, ok =S that sounds scary (everything blowing up) haha... it seems i will need to find another way xD thank you guys ^^

Oh ok, thank you a lot guys, I will try with the hydrogen peroxide soon and tell you how it did -Thanks, Blopa ^^

Hello there community, I'm still a high school student, but I'm much into math, chemistry and physics, so I was wondering if someone could "illuminate" me in a problem. I'm needing H2SO4 (sulfuric acid) but it doesnt seems so easy to get, so I got the wonderful idea of producing it. I know that to make that oxoacid i would need H2O + SO3; my question is, how to make SO3? I suppose that if I burn sulfur (easy to get at the drugstore) it would end being something like these: S + O2 ---> SO2 But that still isn't what I need, someone knows if I can use a catalyst to get SO3? and which one? Any recommendations? Anyway, thanks for your time