Mathematics

next values : I : 2 + 2 ^ 2 II : 2 + 2 ^ 3 III : 2 ^ 1 + 2 ^ 2 + 2 ^ 3

area of square inside the circle is undimensial, because if we get the side of this square according to ray of circle we will get the undimensial area am i correct ?

I noticed a sort of probability problem in which different sequences of events have the same probability despite the events being dependent. I've proven an example below, but I still don't fully grasp why. Here is the problem I use to demonstrate. There is a bag of tubes. Testing for bad tubes, you test the tubes one at a time. Tubes aren't returned to the bag after testing, so there is one less tube each time (event dependence). Despite this, the probability of a good then a bad is equal to the probability of a bad then a good. variables: x = bad tubes / total tubes y = 1 / total tubes. derived: If x is all the bad tubes, 1-x is all the good tubes. y is 1 single tu…

high of triangle same-side around the circle is 3 r, to 2r ( 2 ray of circle ) 2 ray of circle to diagonal of square inside the circle is 2 to 2 / sqrt(2)

( 2 ^ 2 ) ^ 2 is not like 2 ^ 3 in dimenosion for examp

picture : http://s6907.chomikuj.pl/ChomikImage.aspx?e=BNis-wCuwcNDkpjatdNzsgfVQtxAral5imWMh2pcl5i3we1zEJnLAb22thcIwURy0BZgsnZ7wrNEOE4LLgkqslbSC1Gc0yINw0Y-KxmPKgc&pv=2

so i remember, logarith is dependence between a to c in a^b = c so if we have 5 ^ 3 = 125 between 125 and 5 is dependence 125 / 5 = 25, so squart from 2 = 25

suggestion : 2r is diagonal of square inside the oval

is it a dependence between d ( diagonal of square from equal of area of circle ) to r ( ray of this circle ) d / r pi * r ^ 2, pi * sqrt(2) ^ 2, d = sqrt(2) * d ( for third value ) ....

I’m working on a financial problem about budget of households. Households in a state fill a form about their net budget in every year and our insurance company investigate their financial status and find the exact amount of their budget. The net budget can be positive or negative. I’m designing a system with neural network that we can find the households that Falsify their net budget, So my output is a binary [ 0 and 1] which 1 is false net budget and 0 is true net budget. Suppose that households’ net budget is P and our investigated outcome of net budget is R. As you know: Now we have this: R>=P If P=R then : Criterion = 0 (true net budget)If P>0 and R>0 …

if we have function like this y = ( x - 2 ) ^ 2 + 1, what we will get if we do : y = - ( ( x - 2) ^ 2 + 1 ) , and more to find the top of function ( highest value ) doesnt matter the ( x - 2) ^ 2 ..... for x-2 beacause theres allways is 0 to find the top of function

In mathematics, two variables are proportional if a change in one is always accompanied by a change in the other, and if the changes are always related by use of a constant. i.e. y=kx Doesn’t the above definition apply to Y=k+x

Hello I just disovered some new functions: sinIntegral(x) (Si(x)) and cosIntegral(x) (Ci(x)) and just played around with them on GeoGebra. Now, I found something kind of 'beautiful': the average area between those functions, limited by [math]a-1[/math] and [math]a[/math] (so just basically any 'area-block' with width 1), equals [math]\frac{\pi}{2}[/math] when the number of elements of which the average is taken reaches [math]\infty[/math]: [math]\lim_{n\to\infty}{\left[\frac{\int_{1}^{n}{Si(x) dx}-\int_{1}^{n}{Ci(x) dx}}{n-1}\right]}=\frac{\pi}{2}[/math]. In order to prove it, I worked out the left side, but I don't really know how to do it further: [m…

about the highest and lowest of function, if we get the lowest value of function like sinus and cosinus ( exmaple ) , od minus them we get the medium of function, but the doing it on back, we maybe get the lowest and hightest of this function and if we get y = 2x + 1, and change it to 2x - 1, we can do 2x + 1 / 2x - 1, and dont know if it will make something

Hello, I have a question. So we all know that, in the real number system, you can only have positive and negative numbers, right? Because The numbers cycle twice because there it is one dimension. And you can only have right or left. 1*-1=-1, -1*-1=1. Then you have imaginary numbers, right. i*i=-1. -1*i= -i. -i*i = 1 1*i=i They cycle because on the complex plane, there is four quadrants, thus can cycle four times. So, now, lets go to three dimensions shall we. Is there a number that cycles 8 times? Please answer... And if there isn't one, lets try making one up, shall we? Oh, and when we make it up, how would it be used? Josh

Hello all I'm new to this forum, I don't know if this topic or related ones have been discussed somewhere else, but a quick search did not bring me any relevant result. I'm investigating on geometric progressions with initial value 1 and ratio p/(p-1), where p is a prime number. For two different values of p, it's trivial that the two progressions will have no common term (apart of the initial 1). For example powers of 2, 3/2, 5/4 ... are all distinct. My question is : given two such progressions, can they contain terms arbitrarily close to each other? Put in more formally: Let un(p)=(p/(p-1))n, where p is a prime integer and n a positive integer (n >1) …

Hello everyone I was wondering if there was a general formula (sum, product, ...) to define the n-th derivative of the m-th power of a function f(x) So far, I've found that for [math]n[/math] up to 3, the following should be right (if I made no mistakes): [math]\frac{d^{n}\left[f(x)\right]^m}{dx^n}=\frac{d^{n}f}{dx^n}m\left[f(x)\right]^{m-1}+\left(\frac{df}{dx}\right)^n\left[f(x)\right]^{m-n}\cdot\prod_{i=0}^{m-1}{(m-i)}+n\left[f(x)\right]^{m-n+1}\cdot\prod_{i=1}^{m-1}{\frac{d^{i}f}{dx^i}}\cdot\prod_{i=0}^{m-2}{(m-i)}[/math] Can anyone tell me if it's even possible to make such formula, and if yes, what that formula is exactly? Thanks. Function

Well, with my fascination of the Collatz conjecture, I might as well test other things with it. When I combined some concepts to make one. So, here is what I am presenting: Pretty much what this notation is presenting is that using the Hailstone sequence, sum the reciprocals of all the numbers of the hailstone sequence, besides 1, to a get a solution. My conjecture here is that the summation will always have a high bound 3(unsure about a lower bound). I have tested with many numbers and haven't found a counter example, though larger numbers may present different results. Another part of the conjecture is if the solution can get close to 3, but never re…

You thought it was difficult? They did it... In a set of p*q elements, which is not a field, and where the prime p and q are chosen big (like 500 bits), computing ax is quick, but the reverse operation called discrete logarithm is long - that is, no quick method was known. So much that some methods for computer security rely on that, for instance some passports. http://en.wikipedia.org/wiki/Discrete_logarithm That was before. On May 12 at Eurocrypt 2014, Razvan Barbulescu and his mates have described a method in quasi-polynomial time: http://ec14.compute.dtu.dk/program.html they claim as an example that number sizes that would have needed 2128 operations to crack go in …

I was thinking about the quote "Nothing is impossible" and it brought up the idea of carrying out a specific task the same, but with a different set of restrictions or rules applied to that task needed to be done. Yes, certain things are impossible because of the rules of physics and the rules of mathematics, but some how we find a way to accomplish that task another way. I find it interesting how this can be done even within a different set of rules and restrictions. So, it got me thinking that these tasks being done in a different set of restrictions can be generalized by a function that if you have found the process of completing a task within one set of restrictio…

Hello everyone I'd just like to known if the formulas below are correct. Normally I wouldn't post such a short/maybe dumb question, but since Wolfram|Alpha won't generate these formulas, which are, according to me, somewhat obvious, I'd just like to know if there are correct. [math]a^n-b^n=(a-b)\cdot \sum_{i=1}^n{\left(a^{n-i}\cdot b^{i-1}\right)}[/math] [math]a^n+b^n=(a+b)\cdot \sum_{i=1}^n{\left((-1)^{i-1}\cdot a^{n-i}\cdot b^{i-1}\right)}[/math] Thanks. Function

Given a function f(x) that is not a regular polynomial equation(x^n +/- x^n-1 +/- x-2...), how would one determine if a function is smooth over a curve or not? For example, let us say there is a given function that has a curve involved. Given the conditions above, how would one determine if all the parts of the curve are smooth in the sense that there are no other irregular curves on that curve even at the most minuscule spot of the curve?

This is a problem faced within another topic and I made this topic separate because I wanted to focus on this specific problem, which would deviate from the other topic. However, if moderators feel otherwise then it will be fine if the topic is moved to the other topic. So the problem(which I still haven't solved) is dealing with finding an equation to predict the amount of matrix solutions for a given Collatz-Matrix equation, which is defined by as the following: [math]C(x)_{k\times d}\begin{Bmatrix} a_{f} &b_{f} \\ u_{f}&v_{f} \end{Bmatrix},s(k_{p},d_{p})[/math] Now, how these work is there could be multiple or just 1 matrix solution for a given Coll…