## V19 of Compendium

I have added version 19 of Compendium! Seems like we've come a long way! Â I feel like up to around page 35ish it is pretty solid. Â Will still work on tying it to QM more directly.

Archive for the ‘Infinite Sums’ Category

August 12th, 2014
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I have added version 19 of Compendium! Seems like we've come a long way! Â I feel like up to around page 35ish it is pretty solid. Â Will still work on tying it to QM more directly.

May 5th, 2014
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This post is a continuation of a previous one.Â I have developed several preparatory (in preparation of other) claims or theorems: Â The first and second claim show that a particular collection of finite polynomial functions have area of 1 in the interval , and are hence also Pasquali patches. Â The third and fourth shows that the finite sum of any such functions actually have converging integrals in the same interval. Â The corollariesÂ show that this is not the case if the sum is infinite. Â 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 alone, and therefore all stationary/limiting/stable surfaces, eigenfunctions or wavevectors (from the quantum mechanics point-of-view).

**Claim 1.** Take with . Â Then .

**Proof by Definition of Integration (Inducing).** Â We show thatÂ . The expression equalsÂ

Â We assume that the th elementÂ although we readily know by the definition of integration that such is true, sinceÂ

The exact same definition argument applies to theÂ th element andÂ

**Claim 2.** The functionsÂ with are Pasquali patches.

**Proof.** Â A Pasquali patch is a function so that . Â Let . Â Since by Claim 1Â Â , then applying the definition meansÂ Â are Pasquali patches .

**Claim 3.** Â The finite polynomial converges in area fromÂ to .

**Proof.** Â We are looking for

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

**Claim 4.** Pick functions from the pool of . Â For example, pick , and . Â Create the function . Â Then .

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

**Corollary 1.** The infinite polynomial diverges in area in the interval fromÂ .

**Proof.** Â Take

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

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

**Corollary 2.** Â The infinite polynomial diverges in area in the interval fromÂ .

**Proof.** Â Take the limit

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

And now sushi.

**Corollary 3.** Â The infinite polynomial Â Â diverges in area in the interval fromÂ with are infinite polynomials constructed by sums of functions picked from the pool and with no repetitions. (Note that the difference of these two infinite polynomials must also be infinite).

**Proof.** Since theÂ is an infinite polynomial, the integral of such will be an infinite string of ones since the functions it containsÂ are and these are Pasquali patches (Claim 2) and there are no repetitions. Â Such infinite sum of ones clearly diverges.

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

Next create the following polynomial differential vector

It is clear that

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

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

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

Â Finally, the corollaries can be summarized in the statement in which we create a matrix 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. Â It is clear then that

**Remark 2.** The cool thing about this notation is that it gives us power to conclude several interesting things. Â For example, scaling of matrices andÂ as by a constant shows convergence at the integral in the interval of every one of the scaled sumsÂ represented by the rows of such matrices. Â Thus:

**Corollary 4.** Let and Â with is a scaling factor. Â Then the area of each of the infinitely many polynomials represented by the matrices dot in the interval from 0 to 1 converge.

**Proof.** Â On the one hand,Â we haveÂ

Â On the other hand,

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

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 thÂ row of . Â We haveÂ

is convergent by definition. Â 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 have converging area in the interval from 0 to 1).

**Corollary 5.** LetÂ Â with is a scaling factor. Â Then the area of each of the infinitely many polynomials represented by the matrix entries of in the interval from 0 to 1 converge.

**Proof.**Â We haveÂ

for all , so this equalsÂ

for all .

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 alone).** Â All Pasquali patches which are functions of alone (and therefore possible limiting surfaces) take the form

**Proof.** We have that, since such is a Pasquali patch, it must conform to the definition. Â ThusÂ

shows this is indeed the case. Â To show that "all" Pasquali patches that are functions of alone are of the form of , we argue by contradiction. Â Suppose that there is a Pasquali patch that is a function of alone which does not take the form of . Â It couldn't possibly be one such that is a finite polynomial, since Â 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. Â But now it couldn't be any infinite polynomial either, by the same definition ofÂ Â which includes infinite sequences so thatÂ is convergent. Â Thus it must be a polynomial formed by dotting divergent sequences (at the sum), but all suchÂ have been happily excluded from the definition ofÂ .

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

**Remark 5.** Â 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 is Maclaurin-expandable so that

Then

**Proof.** Â

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

Â Then the integral

**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.**Â Â Take

Â By applying

**Remark 7.** Now we have a happy way to construct (any and all)Â Pasquali patches which are functions of 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. Â It also gives us extraordinary power to express any number as an infinite sum, for example.

August 9th, 2013
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**Claim.** Â Take the function with rule

Â Then

**Proof.Â **The direct way to see this is by plugging in:Â

which is a converging series (to ). We justify taking the integral inside the sum precisely under the understanding of the convergence of the series (we can prove by induction that the integral of the partial sums of the function are equivalent to the partial sums of the series which is known to converge).

The surprising fact is not the simplicity of the proof, but that an *infinite term function* can have a stable area over a definite interval. If you think about it this may not be so novel, in that Taylor representations converge to a specific (usually elementary) function which may have calculable area (in the interval 0 to 1)... but there are two things to keep in mind: first, some infinite term functions with definite area in 0 to 1 may not be Taylor representations of elementary functions, and, second, a stable area is definitely not a general observation for infinite term functions. Â For example,

**Claim.** The function with ruleÂ does not have a converging integral in the interval 0 to 1.

**Quick Proof.** Â By taking the integral of the function partial sums, we get the sequence

which we can show (through induction) equivalent to the divergent harmonic series.

The point that I'm trying to make here is that we can define infinite-term functions which converge in area over an interval and may or may not be Taylor representations of other elementary functions. Â This observation comes from considerations of function eigenvalues as I've defined them in Compendium.

Here are other convergent-in-area in the interval 0 to 1 infinite-term functions:

with alternating coefficients gives the convergent alternating series

also with alternating coefficients and odd denominators gives the convergent Leibniz alternating series

with denominators are the Fibonacci numbers yield the convergent Fibonacci series at the integral

And in fact, we can construct converging at the integral in the interval 0 to 1 infinite-term functions simply by letting the coefficients of each term be a general index for the term (counting number) times the convergent sequence term. Â Thus, recall from my previous postÂ that the following infinite sum converges

which implies we can create the infinite term function

which of course converges to in area in the interval .

The function eigenvalue idea can be extended to any interval of interest (even infinite ones), but this is a subject of further investigation.

An interesting notion arises when we think of a number as the area under the curve of an infinite term function. Â The manner by which we approach convergently that number describes the shape of the curve of the infinite term function in that interval. Â I shall put pretty pictures forthwith to illustrate the concept.

Categories: Functional Analysis, Infinite Sums, Mathematics

April 14th, 2013
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Hmm... I have figured out something rather surprising. Â It is this:

**Claim.**Â The infinite sum

Â On the other hand the infinite sum

Â Thus, both infinite sums are convergent.

One way to prove this is by using hyperbolic sine and hyperbolic cosine Maclaurin expansions. Â But I argued it differently using (function) eigenvalues.

The proof is detailed in version 11 of "Compendium...", but since there are some ideas that are grossly incomplete (not this proof, I feel it's pretty solid) I haven't gotten around to posting it.

I'm not sure how it fits into the rather big scheme of things yet... but I'm getting there.

Categories: Arithmetic, Functional Analysis, Infinite Sums, Mathematics