Possible Algorithmic Applications

The complex arithmetic used in the demonstration of the theorem can be used, not only to prove the convergence of the series, but also as an algorithm to accomplish an efficient numerical evaluation of the limit of the series. This can be very useful, because series $S(\theta)$ of the type examined here may converge extremely slowly. Let us recall that we have for the partial sums of the series $(1-v)S(\theta)$

$$(1-v)S_{n} = a_{0} + \sum_{k=1}^{n}(a_{k}-a_{k-1})v^{k} - a_{n}v^{n+1}.$$

By our hypotheses, in the $n\to\infty$ limit the last term vanishes, and therefore we may write for the series

$$(1-v)S(\theta) = a_{0} + \sum_{k=1}^{\infty}(a_{k}-a_{k-1})v^{k}.$$

Observe that all the terms in the sum in the right-hand side are negative, due to the fact that the coefficients decrease monotonically. Unlike the original series for $S(\theta)$, the series in the right-hand side in this formula is always absolutely convergent, as we saw during the proof of the theorem. This series converges therefore much faster than the original one, when that original one is not absolutely convergent. We may define new symbols $b_{k}$ for its coefficients,

$$b_{k} = a_{k}-a_{k-1},\mbox{ for }k=1,\ldots,\infty,$$

while for $k=0$ we adopt the definition $b_{0}=a_{0}$, in terms of which we may write that

$$S(\theta) = \frac{1}{1-v} \sum_{k=0}^{\infty}b_{k}v^{k},$$

which holds so long as $v\neq 1$. In order to identify the real and imaginary parts of this expression, which are related respectively to the series of cosines and the series of sines, we write it back in terms of $\theta$, obtaining

$$S(\theta) = \frac{1}{1-e^{\mbox{\boldmath$\imath$}\theta}} \sum_{k=0}^{\infty}b_{k}e^{\mbox{\boldmath$\imath$}k\theta}.$$

The first factor in the right-hand side can be written as

$$\frac{1}{1-e^{\mbox{\boldmath$\imath$}\theta}} = \frac{\m... ... \frac{e^{-\mbox{\boldmath$\imath$}\theta/2}}{\sin(\theta/2)}.$$

If we substitute this in the expression for the series, we get

$$S(\theta) = \frac{\mbox{\boldmath$\imath$}}{2\sin(\theta/... ...}^{\infty}b_{k}e^{\mbox{\boldmath$\imath$}(k\theta-\theta/2)}.$$

From this we may easily identify the series of sines and of cosines, and in this way we obtain for them the relations

$$\begin{eqnarray*} \sum_{k=0}^{\infty}a_{k}\cos(k\theta) & = & \frac{-1}{2\sin... ...\sin(\theta/2)} \sum_{k=0}^{\infty}b_{k}\cos(k\theta-\theta/2). \end{eqnarray*}$$

Separating once again the $k=0$ terms, we get

$$\begin{eqnarray*} \sum_{k=0}^{\infty}a_{k}\cos(k\theta) & = & \frac{a_{0}}{2}... ...}{2\sin(\theta/2)} \sum_{k=1}^{\infty}b_{k}\cos[(k-1/2)\theta]. \end{eqnarray*}$$

The initial terms of these two series can be recognized as the coordinates $a_{0}/2$ and $a_{0}h$ of the initial center of rotation. Note that the other terms in the right-hand sides have half-integers in their arguments, rather than the usual integers of the Fourier series. They may be interpreted as well as series over only odd indices $(2k-1)$, with the half-angle $\theta/2$ as argument.

For coefficients $a_{k}$ that satisfy the hypotheses of our theorem and that approach zero too slowly for the original series to be absolutely convergent, the series in the right-hand sides converge much faster than the corresponding ones in the left-hand sides, and therefore can be used to calculate approximations to the same limits in a much more efficient way when $v\neq 1$. This constitutes therefore a numerical technique for the calculations of these limits. By using this technique one approaches the limit by following the drift of the center $C(\theta)$, rather than the propagation of the chain of the original series $S(\theta)$. In general, even if the coefficients $a_{k}$ do not satisfy our hypothesis of monotonicity, whenever the series

$$\sum_{k=0}^{\infty}b_{k}v^{k}$$

converges, so does the original series $S(\theta)$, so long as $a_{k}$ tends to zero as $k\to\infty$. Hence, this numerical technique may be useful even in cases when not all our hypotheses are satisfied.

Each one of the extensions of the theorem to other cases, which were discusses previously, also comes with corresponding summation formulas similar to the ones presented above, which in each case may be used for the efficient numerical estimation of the limits. For example, we saw that the result can be extended to series with non-zero coefficients only for even $k$, that is for $k=2j$, and if we consider the angle $\theta'=2\theta$ we may write such a series as

$$S_{n}(\theta') = \sum_{j=0}^{n}a_{j}e^{\mbox{\boldmath$\imath$}j\theta'},$$

which has exactly the same structure as the series just discussed, so that we may write at once that, for $\theta'\neq 2n\pi$ for all integers $n$,

$$\begin{eqnarray*} \sum_{j=0}^{\infty}a_{j}\cos(j\theta') & = & \frac{a_{0}}{2... ...2\sin(\theta'/2)} \sum_{j=1}^{\infty}b_{j}\cos[(j-1/2)\theta'], \end{eqnarray*}$$

where $b_{j}=a_{j}-a_{j-1}$, for $j=1,\ldots,\infty$, and where once more the two initial terms are the coordinates $a_{0}/2$ and $a_{0}h$ of the initial center of rotation. Writing directly in terms of $\theta$ we have

$$\begin{eqnarray*} \sum_{j=0}^{\infty}a_{j}\cos(2j\theta) & = & \frac{a_{0}}{2... ...c{1}{2\sin(\theta)} \sum_{j=1}^{\infty}b_{j}\cos[(2j-1)\theta], \end{eqnarray*}$$

where we have now the condition that $\theta\neq n\pi$ for all integers $n$. Note that in this case, while the series on the left are Fourier series over only the even indices, the ones on the right are over only odd indices. The case in which the series has non-zero coefficients only for odd $k$, that is for $k=2j+1$, can be treated in a similar way. As we saw before, in this case we may write the series as

$$e^{-\mbox{\boldmath$\imath$}\theta'/2} S_{n}(\theta') = \sum_{j=0}^{n}a_{j}e^{\mbox{\boldmath$\imath$}j\theta'},$$

which once more has the same structure as before, except for the additional overall exponential factor. We get therefore in this case, for the series written in terms of $b_{j}$,

$$\begin{eqnarray*} e^{-\mbox{\boldmath$\imath$}\theta'/2} S(\theta') & = & e^... ...)} \sum_{j=0}^{\infty}b_{j}e^{\mbox{\boldmath$\imath$}j\theta}, \end{eqnarray*}$$

which then leads to

$$\begin{eqnarray*} \sum_{j=0}^{\infty}a_{j}\cos[(j+1/2)\theta'] & = & 0 - \f... ...ac{1}{2\sin(\theta'/2)} \sum_{j=1}^{\infty}b_{j}\cos(j\theta'), \end{eqnarray*}$$

where $b_{j}=a_{j}-a_{j-1}$, for $j=1,\ldots,\infty$, and where we see that the two coordinates of the initial center of rotation are now $0$ and $a_{0}R$. Writing directly in terms of $\theta$ we have

$$\begin{eqnarray*} \sum_{j=0}^{\infty}a_{j}\cos[(2j+1)\theta] & = & -\,\frac{1... ...\frac{1}{2\sin(\theta)} \sum_{j=1}^{\infty}b_{j}\cos(2j\theta), \end{eqnarray*}$$

where we have the same condition that $\theta\neq n\pi$ for all integers $n$. Note that in this case, while the series on the left are Fourier series over only the odd indices, the ones on the right are over only even indices.