A Brief History of Infinity – A Paradox
A Brief History of Infinity - by Brian Clegg
A good little read. This book taught me a lot of things about set theory and Georg Cantor. It also taught me about a very interesting paradox within set theory, that was identified by Bertrand Russell.
It goes something like this.
Set theory is something so simple that we take it for granted. It merely points out that certain things, although not identical to one another, are similar enough to be grouped into a “set.”
In fact, set theory is where the working concept of numbers in general finds canon. For instance, the number one looks like this in set theory:
{{Ø}1}
The symbol we as a people have for the concept of the number one (1) is a “set,” which is contained by brackets.
The “Ø” is called the empty set.
Brackets contain the emptiness of the empty set, and it is denoted by {}. The empty set being “Ø,” and Ø being equal to zero for most intents and purposes.
{} = Ø
The number two in set theory could be written like this:
{{{Ø}1}2}
The number two contains the number one, which contains the empty set. The pattern here is that the number three would contain the number two, which contains the number two, which contains the number one, and so on. This can be repeated to infinity, in nested brackets form. Cantor considered numbers this way. And came up with the whole idea of transfinite numbers, which are numbers beyond infinity. The book above goes over this concept quite nicely.
Now that we have some contextual knowledge of set theory, the foundation of all mathematics. Here’s what’s wrong with it!
Consider the set of all <somethings>. <somethings> could be anything you want.
Let’s say “llamas,” the set of all llamas.
I denote this set, the set of all llamas, in red for easier reference. Trust me, this gets kinda twisted.
So please consider the set of all llamas. Now the fact that there is a set of all llamas implies that there is an inverse set, a set of all NOT llamas which shall henceforth be denoted in green. Now lots of things fall into the set of all NOT llamas category: you, me, a planet, electrons, horses, etc. Anything that isn’t a llama falls into this category. You’re not a llama. I’m not a llama. Planets aren’t llamas, and vice versa, even though a llama might exist on a planet. And the same goes for electrons. There are obviously a lot of electrons that might be considered “in” or “part of” a llama, but a llama is definitely NOT an electron and vice versa. The same goes for the set itself.
The what?
The set itself. The container. The thing that we named “set of all llamas.”
A set is a concept, it’s a thought. Isn’t that a thing?
But wait, thoughts and feelings are ephemeral. They’re not real things.
Are then none of your thoughts real things that can be considered part of a set? And how do these non-things keep making and changing real things? I mean, if you want to get technical about it, thoughts are a bio-chemical reaction and that’s definitely about as physical and real as anything.
So anyway,
Please consider the fact that the set of all NOT llamas is a member of itself?
A member?
Yes, an element, a constituent, a part of the whole. The set itself is not a llama, and so that set is a member of itself.
So let’s generalize, and consolidate, what knowledge we’ve gleaned.
There are at least two types of sets. There are sets that ARE members of themselves, and sets that are NOT members of themselves. For instance, the set of all llamas is not a member of itself, but the set of all NOT llamas is a member of itself. And this general rule applies to all sets. That’s why we chose “llamas,” arbitrarily.
So let’s just consider the sets that are NOT members of themselves, for a second. We’ll call this super set the set of all sets that are NOT members of themselves.
Now let’s ask ourselves the same question we asked about the other sets to categorize them. Is this set a member of itself, or not?
Let’s explore both possibilities.
The set of all sets that are NOT members of themselves IS a member of itself.
– Well then it’s actually not a member of itself by definition. If it’s a member of the set of all sets that are NOT members of themselves, then it can’t be a member of itself…if it is a member of itself.
The set of all sets that are NOT members of themselves IS NOT a member of itself.
– Paradoxically, it IS a member of itself since it’s NOT a member of itself.
Bertrand Russell pointed out this paradox within set theory, and proved it mathematically.
He doesn’t propose a new theory of “the way things work,” but merely points out that set theory albeit useful might not be the way things actually work.
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SINGULAR POINTS AND EXCEPTIONAL SINGULARITIES
FOR DIFFERENTIAL COMPLEX EQUATION
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LAGRANGIAN AND ELLIPTIC DIFFERENTIAL EQUATION
ABSTRACT Laplace’s equation describes the equilibrium distribution of energy, species and electromagnetic fields under certain assumptions. It is a second-order, partial-differential equation with closed boundaries. The simplest case to consider is a rectangular region with x and y coordinates such that 0 £ x £ L and 0 £ y £ H. The dependent variable is a potential, u(x,y), which may be temperature, electromagnetic potential, velocity potential or a range of other physical variables. We assume that the potential is known over the complete boundary. In subsequent problems we will consider the case where the potential gradient, rather than the potential, is known over a portion of the boundary. In two dimensions, Laplace’s equation with a simple set of boundary conditions may be written as follows
[1]
Here, the boundary conditions give u = 0 at x = 0, x = L, and y = 0. At y = H, the boundary condition is a known function of x. Note that there may be a discontinuity at edges of the upper boundary (x = 0, y = H) and (x = L, y = H).
The solution of Laplace’s equation proceeds by a method known as the separation of variables. In this method we postulate a solution that is the product of two functions, X(x) a function of x only and Y(y) a function of the y only. With this assumption, our solution becomes.
u(x,t) = X(x)Y(y) [2]
We do not know, in advance, if this solution will work. However, we assume that it will and we substitute it for u in equation [1]. Since X(x) is a function of x only and Y(y) is a function of y only, we obtain the following result when we substitute equation [2] into equation [1].
[3]
If we divide the final equation through by the product X(x)Y(y), and move the y derivative to the other side of the equal sign, we obtain the following result.
[4]
The left hand side of equation [6] is a function of x only; the right hand side is a function of y only. The only way that this can be correct is if both sides equal a constant. This also shows that the separation of variables solution works. In order to simply the solution, we choose the constant to be equal to -l2. This gives us two ordinary differential equations to solve.
[5]
Equation [5] shows that we have two separate differential equations, each of which has a known general solution. These equations and their general solutions are shown below.
[6]
[7]
From the solutions in equations [6] and [7], we can write the general solution for u(x,t) = X(x)T(t) as follows.
[8]
We now apply the boundary conditions shown with the original equation [1] to evaluate the constants A, B, C, and D. If we substitute the boundary condition at x = 0 into equation [8], get the following result.
[9]
Because sin(0) = 0 and cos(0) = 1, equation [9] will be satisfied for all y only if B = 0. Thus, we set B = 0. Next we apply the solution in equation [8] (with B = 0) to the boundary condition at y = 0
[10]
Since sinh(0) = 0 and cosh(0) = 1, this boundary condition will be satisfied only if D = 0. The third boundary condition that u = 0 occurs at x = L. At this point we have the following result, using solution in [8] with B = 0 and D = 0 as found previously.
[11]
Equation [11] can only be satisfied if the sine term is zero. This will be true only if lL is an integral times p. If n denotes an integer, we must have
[12]
Since any integral value of n gives a solution to the original differential equations, with the boundary conditions that u = 0 at both boundaries, the most general solution is one that is a sum of all possible solutions, each multiplied by a different constant. In the general solution for one value of n, which we can now write as Asin(lnx)Csinh(lny), with ln = npx/L, we can write the product of two constants, AC, as the single constant, Cn., which may be different for each value of n. The general solution which is a sum of all solutions with different values of n is written as follows
[13]
The final boundary condition, the one at y = H, states that u(x,H) is a given function of x, uN(x). Setting u(x,H) = uN(x) in equation [13] gives the following result.
[14]
In reviewing the solution for u(x,y), we see that the differential equation for X(x) is a Sturm-Liouville problem. This is a result of both the form of the differential equation for X(x) and the boundary conditions for X(x), implied by the boundary conditions that u(0,y) = 0 and u(L,y) = 0. These boundary conditions can be satisfied only if X(0) = 0 and X(L) = 0, giving the homogenous boundary conditions required for a Sturm-Liouville problem.
Because we have a Sturm-Liouville problem for X(x), we know that the solutions sin(npx/L) form a complete orthogonal set over the region 0 ≤ x ≤ L. We can use this fact in equation [14] to get a solution for Cm. If we multiply both sides by sin(mpx/L), where m is another integer, and integrate from a lower limit of zero to an upper limit of L, we get the following result.
[15]
In the second row of equation [15] we can reverse the order of summation and integration because these operations commute. We then recognize that the integrals in the summation all vanish unless m = n, leaving only the sine-squared integral to evaluate. Solving for Cm and evaluating the last integral in equation [15] gives the following result.
[16]
For any initial condition, then, we can perform the integral on the right hand side of equation [16] to compute the values of Cm and substitute the result into equation [13] to compute u(x,y). For example, consider the simplest case where uN(x) = UN, a constant. In this case we find Cm from the usual equation for the integral of sin(ax).
[17]
Since cos(mp) is 1 when m is even and –1 when m is odd, the factor of 1 – cos(mp) is zero when m is even and 2 when m is odd. This gives the final result for Cm shown below.
[18]
Substituting this result into equation [13] gives the following solution to the diffusion equation when u(x,H) = uN(x) = UN, a constant, and all other boundary values of u are zero.
[19]
If we substitute the equation for ln into the summation terms we get the following result.
[20]
If we compare this equation to equation [19] in the notes on the solution of the diffusion equation, we see that the sine terms are the same. The exponential time term, , in the diffusion equation, has been replaced by the hyperbolic sine terms in equation [20]. This similarity arises from the fact that we have the same ordinary differential equation, with the same zero boundary conditions, for the variable X(x) in each problem. This is an important feature of separation of variable solutions. The contribution to the solution from a term like
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LAGRANGIAN AND CONSEQUENCES
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INTEGRATION AND MULTIPLICATION
ABSTRACT Average value of a function
To find the average value of a function of two variables, let’s start by looking at the average value of a function of one variable. Note that, over the interval , the integral gives the total area of the region. We could also get the total area of the region by treating the region as a rectangle of length b-a and height equal to the average value of the function.
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FUNDAMENTAL THEOREMS AND LAGRANGIAN
ABSTRACT
Theorem 13 (Rolle’s theorem):
Let be continuous on and differentiable on . If , then there exists at least one number c in at which .
[Intuitions:]
(1)
(2)
(3)
If (3) (figure), in . Thus, . If
(1) or (2), suppose takes on some positive values in . Intuitive, there is a number in , such that , where M is the maximum value of in . Then, .
◆
Theorem 14 (mean-value theorem):
Let be continuous on and differentiable on . If , then there exists at least one number c in at which
.
[justifications of theorem 14:]
.
Then, let . Since
,
by Rolle’s theorem, there is a number c such that
◆
The fundamental theorem of calculus is the core of calculus. The following example provides the intuition of the theorem.
Motivating Example (continue):
The area, A, bounded by over is . Note that the antiderivative of is and . As the interval is divided into subintervals with equal length , the
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The domain for this trigonometric function is all real numbers. There are no discontinuities at all. So the interval of continuity is .
ONE-SIDED LIMITS
We have already suggested that it is necessary to examine a neighborhood both to the left and to the right of “c”, when evaluating a limit. Now we will give it a formal name and formal notation.
“Limit from the right”
Examine the function for values greater than c … or the “positive side of” c.
“Limit from the left”
Examine the function for values less than c … or the “negative side of” c. and these lead to infinity [roblem in case of integration as Riemann integration and Lebesgue integration
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O INFINITY POINT IN CASE OF INTEGRATION
ABSTRACT
To find out the dependence of pressure on equilibrium temperature when two phases coexist.
Along a phase transition line, the pressure and temperature are not independent of each other, since the system is univariant, that is, only one intensive parameter can be varied independently.
When the system is in a state of equilibrium, i.e., thermal, mechanical and chemical equilibrium, the temperature of the two phases has to be identical, the pressure of the two phases has to be equal and the chemical potential also should be the same in both the phases.
Representing in terms of Gibbs free energy, the crite
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ABSTRACT The number of independent variables associated with a multi component, multiphase system is given by the Gibbs Phase Rule, expressed as,
F=C+2-P
Where,
F= The number of independent variables
C= The number of components
P= The number of phases present in the equilibrium
• For a single component (C=1) two phase (P=2) system, one independent intensive property needs to be specified (F=1).
• At the triple point, for C=1, P=3 and thus F=0. None of the properties of a pure substance at the triple point can be varied.
• Two independent intensive properties need to be specified to fix the equilibrium state of a pure substance in a single phase.
Phase diagram for a single component system is given in figure.
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LAGRANGIAN MECHANICS
ABSTRACT
In Lagrangian mechanics, the trajectory of a system of particles is derived by solving the Lagrange equations in one of two forms, either the Lagrange equations of the first kind,[1] which treat constraints explicitly as extra equations, often using Lagrange multipliers;[2][3] or the Lagrange equations of the second kind, which incorporate the constraints directly by judicious choice of generalized coordinates.[1][4] In each case, a mathematical function called the Lagrangian is a function of the generalized coordinates, their time derivatives, and time, and contains the information about the dynamics of the system find as the our person and some prof dr doc stefan i . gheorghita
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MEDICINE AND LAGRANGIAN APPLICATIONS
ABSTRACT
the Korteweg–de Vries equation (KdV equation for short) is a mathematical model of waves on shallow water surfaces. It is particularly notable as the prototypical example of an exactly solvable model, that is, a non-linear partial differential equation whose solutions can be exactly and precisely specified. KdV can be solved by means of the inverse scattering transform. The mathematical theory behind the KdV equation is a topic of active research. The KdV equation was first introduced by Boussinesq (1877, footnote on page 360) and rediscovered by Diederik Korteweg and Gustav de Vries (1895).[2]A linear source driven non-linear dynamical evolution system governed by the d-KdV equation of our present concern arises in the following form with all usual notations in the theoretical description of the non-linear normal mode behaviour of plasma ion acoustic mode through transonic plasma of an assumed finite extension. The ion acoustic wave-induced lowest order perturbed potential in transonic plasma equilibrium configuration can be written in a normalized form as the following
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suppose the sample space S is continuous and uncountable. Such a sample space arises when the outcomes of an experiment are numbers. For example, such sample space occurs when the experiment consists in measuring the voltage, the current or the resistance and observedcved by