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On Infinite Term Functions with Converging Integrals, Part II

May 5th, 2014 No comments

This post is a continuation of a previous one. I have developed several preparatory (in preparation of other) claims or theorems:  The first and second claim show that a particular collection of finite polynomial functions have area of 1 in the interval [0,1], and are hence also Pasquali patches.  The third and fourth shows that the finite sum of any such functions actually have converging integrals in the same interval.  The corollaries show that this is not the case if the sum is infinite.  A fun way to summarize this information is soon after developed, and these observations, though simple, lead us to classify all Pasquali patches which are functions of x alone, and therefore all stationary/limiting/stable surfaces, eigenfunctions or wavevectors (from the quantum mechanics point-of-view).

Claim 1. Take f_i(x) = (i+1) x^i with i = 0 \ldots n.  Then \int_0^1 f_i(x) \, dx = 1, \forall i.

Proof by Definition of Integration (Inducing).  We show that \int_0^1 f_0(x) \, dx = 1. The expression equals 

\int_0^1 x^0 \, dx = \int_0^1 1 \, dx = x \left. \right\vert_0^1 = 1

 We assume that the kth element \int_0^1 f_k(x) \, dx = 1 although we readily know by the definition of integration that such is true, since 

\int_0^1 (k+1) x^k \, dx = x^{k+1} \left. \right\vert_0^1 = 1^{k+1} = 1

The exact same definition argument applies to the k+1th element and 

\int_0^1 (k+2) x^{k+1} \, dx = x^{k+2} \left. \right\vert_0^1 = 1^{k+2} = 1

Claim 2. The functions f_i(x) = (i+1) x^i with i = 0 \ldots n are Pasquali patches.

Proof.  A Pasquali patch is a function p(x,y) so that \int_0^1 p(x,y) dx = 1.  Let p(x,y) = f_i(x).  Since by Claim 1  \int_0^1 f_i(x) \, dx = 1, \forall i = 1 \ldots n, then applying the definition means  f_i(x) = (i+1) x^i are Pasquali patches \forall i= 0 \ldots n .

Claim 3.  The finite polynomial g(x) = \sum_{i=0}^n (i+1) x^{i} converges in area from [0,1] to n+1.

Proof.  We are looking for

\int_0^1 \sum_{i=0}^n (i+1) x^i \, dx

.  The sum is finite so it converges, and there is no issue exchanging the order of the sum and integral. Thus:

\sum_{i=0}^n \int_0^1 (1+i) x^i \, dx =\sum_{i=0}^n \left( x^{i+1} \left. \right\vert_0^1 \right) = \sum_{i=0}^n 1^{i+1} =\sum_{i=0}^n 1 = n+1

Claim 4. Pick n functions from the pool of f_i(x) = (i+1) x^i.  For example, pick f_3(x), f_5(x), and f_7(x).  Create the function h(x) = \sum_i f_i(x).  Then \int_0^1 h(x) \, dx = n.

Proof by induction.  Since by Claim 2 all f_i(x) are Pasquali patches, it follows their integral is 1 in the interval (Claim 1).  Picking 1 function from the pool thus gives an integral of 1 in the interval.  Suppose that picking k functions gives k units at the integral in the interval. Now pick k+1 functions.  The first k functions give k units at the integral in the interval, and the 1 additional function contributes 1 unit at the integral in the interval.  Thus k+1 functions contribute k+1 units at the integral in the interval.

Corollary 1. The infinite polynomial a(x) = \sum_{i=0}^\infty (i+1) x^i diverges in area in the interval from [0,1].

Proof.  Take

\int_0^1 \left( \lim_{n \to \infty} \sum_{i=0}^n (1+i) x^i \right) \, dx =\lim_{n \to \infty} \int_0^1\sum_{i=0}^n (1+i) x^i \, dx

Here exchanging the order of limit and integral is justified by the fact that, term-wise, the integral converges. Next  

\lim_{n \to \infty} n+1 = \infty

Here the second to last step is justified by Claim 3.

Corollary 2.  The infinite polynomial a(x) - h(x) diverges in area in the interval from [0,1].

Proof.  Take the limit

 \lim_{n \to \infty} \left[ a(x) - h(x) \right]

Taking n to infinity applies to a(x) only which we know diverges by Corollary 1.  The same limit  has no effect on h(x) as the sum it is composed of is finite and adds up to an integer constant, say m.  We conclude that any infinite collection of terms of f_i(x) diverges, even when a finite number of them may be absent from the sum.

And now sushi.

Corollary 3.  The infinite polynomial  a(x) - b(x) diverges in area in the interval from [0,1] with a(x), b(x) are infinite polynomials constructed by sums of functions picked from the pool f_i(x) = (i+1) x^i and with no repetitions. (Note that the difference of these two infinite polynomials must also be infinite).

Proof. Since the a(x) - b(x) is an infinite polynomial, the integral of such will be an infinite string of ones since the functions it contains are f_i(x) and these are Pasquali patches (Claim 2) and there are no repetitions.  Such infinite sum of ones clearly diverges.

Remark 1.  We can view what we have learned in the claims from a slightly different vantage point.  Create the infinite identity matrix

 I = \left[ \begin{array}{cccc} 1 & 0 & 0 & \ldots \\ 0 & 1 & 0 & \ldots \\ \vdots & \vdots & \vdots & \ddots \end{array} \right]

Next create the following polynomial differential vector

 D =\left[ \begin{array}{c} 1 \\ 2x \\ 3x^2 \\ \vdots \end{array} \right]

It is clear that

 \int_0^1 I_i \cdot D \, dx =1

for all rows  i of  I .  We can omit the little  i because this definition applies to all rows and:

 \int_0^1 I \cdot D \, dx = \int_0^1 D \, dx= \left[ \begin{array}{c} 1 \\ 1 \\ \vdots \end{array} \right] = \bf{1}

This of course summarizes Claims 1 and 2.  Next, define the matrix J consisting of rows which are finite sums of rows of I (so that each row of J consists of a finite number of ones at any position, namely n such coming from n picked rows of I).  Claims 3 and 4 are summarized in the statement  

 \int_0^1 J\cdot D \, dx = S

where S is the vector consisting of the sum of the rows of J, which, since it is made up of a finite number of ones at each row, adds up to a constant integer at each row:

S = \left[ \begin{array}{c} n_1 \\ n_2 \\ \vdots \end{array} \right]

 Finally, the corollaries can be summarized in the statement in which we create a matrix  K consisting of rows with a finite number of zeroes (and an infinite number of ones) or an infinite number of zeroes but an infinite number of ones as well.  It is clear then that

 \int_0^1 K\cdot D \, dx = \infty

Remark 2. The cool thing about this notation is that it gives us power to conclude several interesting things.  For example, scaling of matrices  I and  J as by a constant  t shows convergence at the integral in the interval  \left[ 0,1 \right] of every one of the scaled sums  represented by the rows of such matrices.  Thus:

Corollary 4. Let  I^* = t \cdot I and J^* = t \cdot J with  t is a scaling factor.  Then the area of each of the infinitely many polynomials represented by the matrices I^*, J^* dot D in the interval from 0 to 1 converge.

Proof.  On the one hand, we have 

 \int_0^1 I^* \cdot D \, dx = \int_0^1 t \cdot I \cdot D \, dx =t \left( \int_0^1 I \cdot D \, dx \right) = \bf{t}

 On the other hand,

 \int_0^1 J^* \cdot D \, dx =\int_0^1 t \cdot J\cdot D \, dx = t \left( \int_0^1 J\cdot D \, dx \right) = t \cdot S =\left[ \begin{array}{c} t \cdot n_1 \\ t \cdot n_2 \\ \vdots \end{array} \right]

Remark 3. Next consider the infinite-matrix formed by convergent sequences (at the sum) at each row,

A = \left[ \begin{array}{cccc} \vdots & \vdots & \vdots & \vdots \\ 1 & \frac{1}{2^2} & \frac{1}{3^2} & \ldots \\ \vdots & \vdots & \vdots & \vdots \end{array} \right]

Depicted is the reciprocals of squares which we know converges at the sum (Basel problem), simply for illustration, but all convergent sequences would be in the ith row of A.  We have 

 \int_0^1 A_i\cdot D \, dx = \sum_j a_{i,j}

is convergent by definition.  The cool thing is we can easily prove in one swoop that all sequences that are scaled will also converge at the sum (and the infinite polynomials with coefficients A \cdot D have converging area in the interval from 0 to 1).

Corollary 5. Let  A^* = t \cdot A with  t is a scaling factor.  Then the area of each of the infinitely many polynomials represented by the matrix entries of A^* \cdot D in the interval from 0 to 1 converge.

Proof.  We have 

 \int_0^1 A^*_i\cdot D \, dx = \sum_j a_{i,j}

for all i, so this equals 

 \int_0^1 t \cdot A_i\cdot D \, dx = t \left(\int_0^1 A_i\cdot D \, dx \right) = t \cdot\sum_j a_{i,j}

for all i.

All of these small and obvious observations lead to this:

Claim 5. The Grand Classification Theorem of Limiting Surfaces (A General and Absolutely Complete Classification of Pasquali patches which are functions of x alone).  All Pasquali patches which are functions of x alone (and therefore possible limiting surfaces) take the form

 p(x) = \frac{A_i \cdot D}{\sum_j a_{i,j}}

Proof. We have that, since such  p(x) is a Pasquali patch, it must conform to the definition.  Thus 

 \int_0^1 p(x) \, dx = \int_0^1 \frac{A_i \cdot D}{\sum_j a_{i,j}} \, dx = \frac{\int_0^1 A_i \cdot D \, dx}{\sum_j a_{i,j}} = \frac{\sum_j a_{i,j}}{\sum_j a_{i,j}} = 1

shows this is indeed the case.  To show that "all" Pasquali patches that are functions of x alone are of the form of p(x), we argue by contradiction.  Suppose that there is a Pasquali patch that is a function of x alone which does not take the form of p(x).  It couldn't possibly be one such that is a finite polynomial, since  A_i was defined to be that matrix formed by all convergent sequences at the sum at each row and it can be scaled any which way we like, and this includes sequences with a finite number of nonzero coefficients.  But now it couldn't be any infinite polynomial either, by the same definition of  A_i which includes infinite sequences so that \sum_j a_{i,j} is convergent.  Thus it must be a polynomial formed by dotting divergent sequences (at the sum), but all such have been happily excluded from the definition of A.

Remark 4.  Thus, EVERY convergent series has an associated Pasquali patch (which is solely a function of  x), and vice versa, covering the totality of the Pasquali patch functions of x universe and the convergent series universe bijectively.

Remark 5.  Notice how the definition takes into account Taylor polynomial coefficients (thus all analytic functions are included) and those that are not (even those that are as yet unclassified), and all sequences which may be scaled by a factor as well.

Claim 6. Let f(x) is Maclaurin-expandable so that

 f(x) = \sum_{n=0}^\infty \frac{f^n(0) x^n}{n!}

Then

 \sum_{n=0}^\infty\frac{f^n(0)}{(n+1)!} = \int_0^1 f(x) \, dx

Proof.  

\int_0^1 f(x) \, dx = \int_0^1 A_i \cdot D \, dx

for some i row of A.  Such a row would have to be of form

 A_i = \left[ \begin{array}{cccc} f(0) & \ldots & \frac{f^n(0)}{n! (n+1)} & \ldots \end{array} \right]

 Then the integral

\int_0^1 A_i \cdot D \, dx = \sum_j a_{i,j} =\sum_{n=0}^\infty \frac{f^n(0)}{n! (n+1)} = \sum_{n=0}^\infty \frac{f^n(0)}{(n+1)!}

Remark 6. Notice that all Maclaurin-expandable functions converge in area (have stable area) in the interval from 0 to 1, a remarkable fact.

Example 1.  Take

f(x) = e^x = \sum_{n=0}^\infty \frac{x^n}{n!}

 By applying Claim 6, it follows that

 \sum_{n=0}^\infty \frac{1}{(n+1)!} = \int_0^1 e^x \, dx = e - 1

Remark 7. Now we have a happy way to construct (any and all) Pasquali patches which are functions of x alone, merely by taking a sequence which is convergent at the sum.

Remark 8. Quantum mechanically, we now know all possible shapes that a stationary (limiting) eigen wavevector can take.

Remark 9. This gives us extraordinary power to calculate convergent sums via integration, as the next examples show.  It also gives us extraordinary power to express any number as an infinite sum, for example.

 

Latest version of Compendium

February 9th, 2014 No comments

I am including here the latest version of the Compendium (v17).

Part I v17

A Remark based on False Claim 1

January 26th, 2014 No comments

So I was thinking that, the statistical description of a dynamical system (as one described by a generator Pasquali patch) really does give us a lot of power in computing the probable "position" of a particle (photon, electron) moving in space at different (integer) time intervals.  If the quantum mechanical supposition of time having a minimum discreteness (Planck-time) is correct, we can find the "finest" Pasquali patch generator that will give a complete description of the dynamical system.  Any Pasquali patch generator descriptive of the system which is not this "first" will generate an accurate, yet less refined ("coarser") version of the system (this is what we mean by Claim 3 of the previous post, in that such Pasquali patch will be "contained" in the finest description, yet is not the finest), and in fact either system of course converges to the same steady state (this is what is meant by Claim 4).  If we are able to find a continuous description (like the Shrodinger equation, via a "Pasqualian") of such system then we are in luck (this description would be the finest, though non-discrete, description), and I speculate though I cannot be sure yet that either discrete descriptions will be contained in such.

Whatever the description of the dynamical system via a generator Pasquali patch (or a Pasqualian), each Pasquali patch represents the transition (position) probabilities of a particle (photon, electron) moving within that system.  If we suppose that the particle moves with same velocity (take photons in vacuum as an example), then each Pasquali patch power is descriptive of the transition-position probability at equally spaced spacial or distance intervals.  Though I've remarked about this before, if it were the case that, for a particle with a particular (steady) velocity, the Pasquali patch power is not exactly equally spaced in distance intervals, it must mean that the arrow of time is bent (time is moving faster for smaller-spaced intervals, slower for longer-spaced).  We have not yet described accelerating particles but at present that is not of our interest.

However, we can tell if time is passing so long as each Pasquali patch description is different at each (equal or unequal) interval.  If a single Pasquali patch were to describe the system at EACH distance interval, there is no way to know if time is moving at all.  Take for example the False Claim 1 of the previous post.  We had the collection \mathbb{Q} = \{Q^1, Q^2, \ldots, Q^k, \ldots\}_{k \in \mathbb{Z}^+} with Q = q(x,y) being an explicit function of y and converging to Q^\infty = 1.  We could track the time-distance interval via the Pasquali patch power, so that 1 was the first distance interval from start (we take it as given that time is not being bent, so that a fixed distance implies the passage of 1 unit of time), 2 was the second distance interval (2 units of time), and so on, and we could tell if time were bent if each power were descriptive of different distance intervals.  Furthermore, since each Q \in \mathbb{Q} is different, this implies each position transition probability is different and the system is in movement.

This is definitely not the case with the collection \mathbb{P} = \{P^1 = 1, P^2 = 1, \ldots, P^j, \ldots \}_{j \in \mathbb{Z}^+} which also converges to P^\infty = 1.  Since at each distance interval the movement probability is the same (uniform), one cannot be convinced that each power represents a distance interval equal to equally spaced time intervals or different-spaced time intervals.  Where we could with \mathbb{Q} ascertain that time was moving, we cannot with \mathbb{P}.  The statistical description cannot tell if the system is frozen.

When a system has reached the steady state (which, is the highest entropy state!), there is no way to tell if time flows, as the statistical description is and forever will be unchanging.  Recall that the steady state for a Pasquali patch is always a function of x alone, say p(x).  Furthermore recall that any power of p(x) is always p(x) itself (see Compendium).  We reach an impasse: is time flowing normally, faster than what is conventional, slower? At such a point it is impossible to say, at least from the statistical point of view.  We would have to track particles individually in order to ascertain if they have deviated their path at all (it could be the case that they shifted to all positions with equal probability or in the shape of p(x), e.g., but we cannot be sure of either situation).

Compendium of Claims and Proofs, Including New Ones, Part I

December 3rd, 2012 No comments

I've condensed this exceptional mathematical wisdom here, which is still transforming as I organize and jot down ideas.

Part I v16 (latest, but very unorganized after @Dynamics)

Part I v15

Part I v14

Part I v13

Part I v12

Part I v11

Part I v10

Part I v9

Part I v8

Part I v7

Part I v6

Part I v5

Part I v4

Part I v3