Archive for the ‘Infinite Sums’ Category

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!}


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


\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.


On Infinite-Term Functions with Converging Integrals

August 9th, 2013 No comments

Claim.  Take the function  f \colon [0,1] \to \mathbb{R} with rule

 f(x) =\sum_{i=0}^\infty \frac{x^i}{i+1} = 1 + \frac{x}{2} + \frac{x^2}{3} + \ldots


\int_0^1 f(x) \, dx = \frac{\pi^2}{6}

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

\int_0^1 \sum_{i=0}^\infty \frac{x^i}{i+1}\, dx = \sum_{i=0}^\infty\left. \left( \frac{x^{i+1}}{(i+1)^2} \right\vert_0^1 \right) = \sum_{i=0}^\infty \frac{1}{(i+1)^2} = \sum_{i=1}^\infty \frac{1}{i^2}

which is a converging series (to \frac{\pi^2}{6} ). 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  \frac{1}{n^2} 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.  For example,

Claim. The function  g \colon [0,1] \to \mathbb{R} with rule g(x) = 1 + x + x^2 + \ldots = \sum_{i=0}^\infty x^i does not have a converging integral in the interval 0 to 1.

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

 \begin{aligned} 1 & \\ 1 & + 1/2 \\ 1 & + 1/2 + 1/3 \\ etc. & \end{aligned}

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.  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:

1 - x + x^2 - x^3 + \ldots

with alternating coefficients gives the convergent alternating series 1 - 1/2 + 1/3 - \ldots = ln(2)

1 - \frac{2 x}{3} + \frac{3 x^2}{5} - \frac{4 x^3}{7} + \ldots

also with alternating coefficients and odd denominators gives the convergent Leibniz alternating series 1 - 1/3 + 1/5 - 1/7 + \ldots = \frac{\pi}{4}

1 +\frac{2 x}{1} + \frac{3 x^2}{2} + \frac{4 x^3}{3} + \frac{5 x^4}{5} +\ldots

with denominators are the Fibonacci numbers yield the convergent Fibonacci series at the integral 1 + 1 +1/2 + 1/3 + 1/5 + \ldots = \psi

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.  Thus, recall from my previous post that the following infinite sum converges

 \sum_{i=1}^\infty \frac{1}{(2 i)!} = \frac{(e - 1)^2}{2 e}

which implies we can create the infinite term function

 h(x) = \sum_{i=0}^\infty \frac{(i + 1) x^i}{(2 (i+1))!} = \frac{1}{2} + \frac{2 x}{4!} + \frac{3 x^2}{6!} + \ldots

which of course converges to  \frac{(e - 1)^2}{2 e} in area in the interval  [0, 1] .

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.  The manner by which we approach convergently that number describes the shape of the curve of the infinite term function in that interval.  I shall put pretty pictures forthwith to illustrate the concept.

On a Rather Surprising Result

April 14th, 2013 No comments

Hmm... I have figured out something rather surprising.  It is this:

Claim. The infinite sum

\lim_{n \to \infty} s^o(n) = \sum_{i=1}^\infty \frac{1}{(2i)!} = \frac{(e-1)^2}{2e}

 On the other hand the infinite sum

\lim_{n \to \infty} s^e(n) = \sum_{i=0}^\infty \frac{1}{(2i + 1)!} = \frac{e^2 - 1}{2e}

 Thus, both infinite sums are convergent.

One way to prove this is by using hyperbolic sine and hyperbolic cosine Maclaurin expansions.  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.