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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

 Then

\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.