In this post I want to talk about a thing from the Kryptos universe that are not directly related to the statue. But i think it may be an indirect hint to some Kryptos related methods. The Mayan Symbols in Ed Scheidts driveway I think everyone who knows Kryptos knows Ed Scheidt. The former Chairman of the Cryptographic Center at the CIA and founder of the cryptosystems used around the Kryptos statue. As already shown in Part 4 of my Kryptos series, in the driveway of Ed Scheidts house, there are two symbols: Figure 1 - Garage driveway of Ed Scheidt We denote the left symbol set with $S_1$ and the right one with $S_2$. It took me a while to find his house on Google Maps - Street View. To save you some time, here is the link with a view on the driveway. I you go back in time in Streetview, you can see that the symbols were already there in 2012. But it is impossible to say when they were built. $S_1$ is clearly visible from the street, $S_2$ is hidden in the view. But you can u...
In the current days it seems that people picked up an old discussion and brought it to new attention in their forums. In heated debates they discuss and exchange their point of views, sometimes rather impolite than objective.
They talk about infinite sums of integers, in particular divergent sums, and show how these sums can be manipulated to be equal to a finite value. This causes trouble, since it seems on the first sight rather unintuitive and wrong.
The most used example for their offending object is the simple sum $$S := 1 + 2 + 3 + 4 + ... = \sum^\infty_{i=1} i$$, which can be manipulated such that it is finally equal to $-1/12$. The first one who showed this curiosity was the famous mathematician Srinivasa Ramanujan. And from the fact that a credible mathematician wrote down those things, there should arise some doubt that maybe perhaps there is some truth in these formulas and that the misunderstandings come from the fact, that the formulas are probably not correctly cited and used.
\end{align*} If we double $X$ (writing it ontop of each other for better clarity), we get
\begin{align*} & X + X \\ &= 1 - 2 + 3 - 4 + 5 - 6 \ldots\\ &\;\;\;\;\;\;+ 1 - 2 + 3 - 4 +5 \ldots \;\;\;\;\;\;\;\;\textbf{(*)} \\ & = 1 - 1 + 1 - 1 + 1 - 1 \ldots = Y
\end{align*} Next we subtract $X$ from our target sum $S = 1+2+3+4+5+\ldots$ and we get
\begin{align*} & S - X \\ &1 + 2 + 3 + 4 + \ldots - X \\ = &1 + 2 + 3 + 4 + \ldots - (1 - 2 + 3 - 4 + \ldots )\\ = &4 + 8 + 12 + 16 \ldots \\ = &4(1 + 2 + 3 + 4 + ...) \\ = & 4S \end{align*} hence we have $S-X = 4S \Leftrightarrow = -3S = X$. \begin{equation} -3(1+2+3+4+\ldots) = X = \frac{Y}{2} \end{equation}
so
\begin{equation} 1+2+3+4+\ldots = -\frac{X}{3} = -\frac{Y}{6} \end{equation}
$Y$ is equal to the series $1 - 1 + 1 - 1 \ldots = Y$, so it swaps between $1$ and $0$ and is clearly not infinite large. Some mathematician defines such a series to be equal to the average of the possible outcomes, i.e., $\frac{0+1}{2} = 1/2 = Y$. In that case, we get
\begin{equation} S = 1+2+3+4+\ldots = -\frac{1}{12} \end{equation} which is the results that you will find in most discussion. But if you don't like the idea of the using the average $1/2$, will you agree that the series somehow has to adopt a value between $0$ and $1$? If so, then you get $$-\frac{1}{6} \leq 1+ 2 + 3 + 4 + 5 + \ldots \leq 0$$, which is equally unintuitive as $-1/12$. In the notes of Ramanujan you can find such equation, but only for his type of summation method, i.e., ramanujan summation, which is defined differently from the usual summation method.
The main problem why the usual method does not make sense here, is that the re-alignment at (*) causes problems for divergent series.
This is also crucial for convergent series, if they are only conditional convergent. In that case, swapping the order of the terms can be used to proof other weired things. For example:
&= 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6}...
\end{align*}This definition is from the textbook. If we now rearrange the terms, we can build
\begin{align*}
\log(2) &= \sum^\infty_{i=1} \frac{(-1)^{i+1}}{i} \\
&= 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6}...\\
&= \left(1-\frac{1}{2}\right) - \frac{1}{4} + \left(\frac{1}{3} - \frac{1}{6}\right) - \frac{1}{8} + \left(\frac{1}{5}-\frac{1}{10}\right) ...\\
& = \frac{1}{2} - \frac{1}{4} + \frac{1}{6} - \frac{1}{8} + \frac{1}{10} ...\\
& = \frac{1}{2}\left(1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} ... \right) \\
& = \frac{1}{2}\log(2)
\end{align*} which is obviously non-sense. Since from this you can derive $$\log(2) = 1/2\log(2) \Leftrightarrow 1 = 0$$
\begin{align*}
&\left(1 + \frac{1}{2}\right)\log(2)\\
&= \left(1 + \frac{1}{2}\right)\left(1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6}...\right)\\
& = \left(1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} - \frac{1}{8}...\right)\\
&+ \left(\frac{1}{2} - \frac{1}{4} + \frac{1}{6} - \frac{1}{8} + \frac{1}{10} - \frac{1}{12} + \frac{1}{14} - \frac{1}{16}...\right)
\end{align*}
One could rearrange the last two sums again in a similar way as shown in the example above to create the statement $\log(2) = \frac{3}{2}\log(2)$.
\begin{align*}
& = \left(1 + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} + \frac{1}{7} - \frac{1}{8}...\right)\\&+ \left( - \frac{1}{4} - \frac{1}{8} + \frac{1}{10} - \frac{1}{12} + \frac{1}{14} - \frac{1}{16}...\right)\\
& = 1 + \frac{1}{3} - \frac{2}{4} + \frac{1}{5} + \frac{1}{7} - \frac{2}{8} + \frac{1}{10} - \frac{1}{12} + \frac{1}{14} - \frac{1}{16} ... \\
& = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} ...
\end{align*}
What we cut of from the last line are the terms $\frac{1}{7} + \frac{1}{10} - \frac{1}{12} + \frac{1}{14} - \frac{1}{16}$, which does not yet "fit" into the final series, because from the first series more terms are used than from the second series.
If you keep track of this, the supply terms, we can correct this formula. We focus on the first of the two sums, i.e., $1-1/2+1/3-1/4...$ which is known to be $\log(2)$. We pick $m=4m'=8$ terms of this sum and show how many terms of that sum arise again. Finally, we count what remains.
\begin{align*}
&\left(1 + \frac{1}{2}\right)\log(2)\\
& = \left(1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} - \frac{1}{8}...\right)\\
&+ \left(\;\;\;\;\;\;\frac{1}{2} \;\;\;\;\;\;\;-\frac{1}{4} \;\;\;\;\;\;\;+\frac{1}{6} \;\;\;\;\;\;\;- \frac{1}{8} + \frac{1}{10} - \frac{1}{12} + \frac{1}{14} - \frac{1}{16}...\right) \\
&= 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} \\
&+ \left(\frac{1}{5}+\frac{1}{7}\right) //\text{remaining terms from the first sum} \\
&+ \left(\frac{1}{10}-\frac{1}{12}+\frac{1}{14}-\frac{1}{16}\right) //\text{remaining terms from the second sum}
\end{align*}
Every odd term of the second sum cancels with terms on positions $2+4k$ of the first sum and every even term of the second sum yields with every term at $4k$ of the first sum, the missing $2k$-th term of the final sum. Hence if i take $m'=4m$ terms from sum $A$, one needs $2m$ from sum $B$. It remains from $B$
\begin{equation}
\sum^{2m}_{i=1} \frac{(-1)^{i+1}}{2(2m+i)}
\end{equation} and from $A$ the odd terms
\begin{equation}
\sum^{m}_{i=1} \frac{1}{2m+2i-1}
\end{equation} Since the final sum resembles the original sum for $\log(2)$, this remaining terms must be equal to $1/2 \log(2)$ if we let $m$ go to infinity
\begin{equation}
\frac{1}{2}\log(2) = \lim_{m \rightarrow \infty} \left( \sum^{2m}_{i=1} \frac{(-1)^{i+1}}{4m+2i} + \sum^{m}_{i=1} \frac{1}{2m+2i-1} \right)
\end{equation}
If you plot the first values, you see that this indeed converges to $1/2\log(2)$
\begin{array}{| c | c | c | c | c | c | c |} \hline \\ 1/2\log(2) & \textbf{m=1}&\textbf{m=2}&\textbf{m=3}&\textbf{m=4} &\textbf{m=100} &\textbf{m=10000}\\\hline
0.3465736 & 0.3750000 & 0.3684524 & 0.3631494 & 0.3597840 & 0.3471947 & 0.3465798 \\\hline
\end{array}
Proof [That $1+2+3+... = -1/12$]
Take the alternating sum: \begin{align*} X := 1 - 2 + 3 - 4 + 5 - 6 \ldots\end{align*} If we double $X$ (writing it ontop of each other for better clarity), we get
\begin{align*} & X + X \\ &= 1 - 2 + 3 - 4 + 5 - 6 \ldots\\ &\;\;\;\;\;\;+ 1 - 2 + 3 - 4 +5 \ldots \;\;\;\;\;\;\;\;\textbf{(*)} \\ & = 1 - 1 + 1 - 1 + 1 - 1 \ldots = Y
\end{align*} Next we subtract $X$ from our target sum $S = 1+2+3+4+5+\ldots$ and we get
\begin{align*} & S - X \\ &1 + 2 + 3 + 4 + \ldots - X \\ = &1 + 2 + 3 + 4 + \ldots - (1 - 2 + 3 - 4 + \ldots )\\ = &4 + 8 + 12 + 16 \ldots \\ = &4(1 + 2 + 3 + 4 + ...) \\ = & 4S \end{align*} hence we have $S-X = 4S \Leftrightarrow = -3S = X$. \begin{equation} -3(1+2+3+4+\ldots) = X = \frac{Y}{2} \end{equation}
so
\begin{equation} 1+2+3+4+\ldots = -\frac{X}{3} = -\frac{Y}{6} \end{equation}
$Y$ is equal to the series $1 - 1 + 1 - 1 \ldots = Y$, so it swaps between $1$ and $0$ and is clearly not infinite large. Some mathematician defines such a series to be equal to the average of the possible outcomes, i.e., $\frac{0+1}{2} = 1/2 = Y$. In that case, we get
\begin{equation} S = 1+2+3+4+\ldots = -\frac{1}{12} \end{equation} which is the results that you will find in most discussion. But if you don't like the idea of the using the average $1/2$, will you agree that the series somehow has to adopt a value between $0$ and $1$? If so, then you get $$-\frac{1}{6} \leq 1+ 2 + 3 + 4 + 5 + \ldots \leq 0$$, which is equally unintuitive as $-1/12$. In the notes of Ramanujan you can find such equation, but only for his type of summation method, i.e., ramanujan summation, which is defined differently from the usual summation method.
The main problem why the usual method does not make sense here, is that the re-alignment at (*) causes problems for divergent series.
This is also crucial for convergent series, if they are only conditional convergent. In that case, swapping the order of the terms can be used to proof other weired things. For example:
Proof [That $1=0$]
\begin{align*} \log(2) &:= \sum^\infty_{i=1} \frac{(-1)^{i+1}}{i} \\&= 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6}...
\end{align*}This definition is from the textbook. If we now rearrange the terms, we can build
\begin{align*}
\log(2) &= \sum^\infty_{i=1} \frac{(-1)^{i+1}}{i} \\
&= 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6}...\\
&= \left(1-\frac{1}{2}\right) - \frac{1}{4} + \left(\frac{1}{3} - \frac{1}{6}\right) - \frac{1}{8} + \left(\frac{1}{5}-\frac{1}{10}\right) ...\\
& = \frac{1}{2} - \frac{1}{4} + \frac{1}{6} - \frac{1}{8} + \frac{1}{10} ...\\
& = \frac{1}{2}\left(1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} ... \right) \\
& = \frac{1}{2}\log(2)
\end{align*} which is obviously non-sense. Since from this you can derive $$\log(2) = 1/2\log(2) \Leftrightarrow 1 = 0$$
Supply-Sum; doing it right
If you keep track of the terms that need to be processed you can rescue the proof. To show this, we build\begin{align*}
&\left(1 + \frac{1}{2}\right)\log(2)\\
&= \left(1 + \frac{1}{2}\right)\left(1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6}...\right)\\
& = \left(1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} - \frac{1}{8}...\right)\\
&+ \left(\frac{1}{2} - \frac{1}{4} + \frac{1}{6} - \frac{1}{8} + \frac{1}{10} - \frac{1}{12} + \frac{1}{14} - \frac{1}{16}...\right)
\end{align*}
One could rearrange the last two sums again in a similar way as shown in the example above to create the statement $\log(2) = \frac{3}{2}\log(2)$.
\begin{align*}
& = \left(1 + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} + \frac{1}{7} - \frac{1}{8}...\right)\\&+ \left( - \frac{1}{4} - \frac{1}{8} + \frac{1}{10} - \frac{1}{12} + \frac{1}{14} - \frac{1}{16}...\right)\\
& = 1 + \frac{1}{3} - \frac{2}{4} + \frac{1}{5} + \frac{1}{7} - \frac{2}{8} + \frac{1}{10} - \frac{1}{12} + \frac{1}{14} - \frac{1}{16} ... \\
& = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} ...
\end{align*}
What we cut of from the last line are the terms $\frac{1}{7} + \frac{1}{10} - \frac{1}{12} + \frac{1}{14} - \frac{1}{16}$, which does not yet "fit" into the final series, because from the first series more terms are used than from the second series.
\begin{align*}
&\left(1 + \frac{1}{2}\right)\log(2)\\
& = \left(1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} - \frac{1}{8}...\right)\\
&+ \left(\;\;\;\;\;\;\frac{1}{2} \;\;\;\;\;\;\;-\frac{1}{4} \;\;\;\;\;\;\;+\frac{1}{6} \;\;\;\;\;\;\;- \frac{1}{8} + \frac{1}{10} - \frac{1}{12} + \frac{1}{14} - \frac{1}{16}...\right) \\
&= 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} \\
&+ \left(\frac{1}{5}+\frac{1}{7}\right) //\text{remaining terms from the first sum} \\
&+ \left(\frac{1}{10}-\frac{1}{12}+\frac{1}{14}-\frac{1}{16}\right) //\text{remaining terms from the second sum}
\end{align*}
Every odd term of the second sum cancels with terms on positions $2+4k$ of the first sum and every even term of the second sum yields with every term at $4k$ of the first sum, the missing $2k$-th term of the final sum. Hence if i take $m'=4m$ terms from sum $A$, one needs $2m$ from sum $B$. It remains from $B$
\begin{equation}
\sum^{2m}_{i=1} \frac{(-1)^{i+1}}{2(2m+i)}
\end{equation} and from $A$ the odd terms
\begin{equation}
\sum^{m}_{i=1} \frac{1}{2m+2i-1}
\end{equation} Since the final sum resembles the original sum for $\log(2)$, this remaining terms must be equal to $1/2 \log(2)$ if we let $m$ go to infinity
\begin{equation}
\frac{1}{2}\log(2) = \lim_{m \rightarrow \infty} \left( \sum^{2m}_{i=1} \frac{(-1)^{i+1}}{4m+2i} + \sum^{m}_{i=1} \frac{1}{2m+2i-1} \right)
\end{equation}
If you plot the first values, you see that this indeed converges to $1/2\log(2)$
\begin{array}{| c | c | c | c | c | c | c |} \hline \\ 1/2\log(2) & \textbf{m=1}&\textbf{m=2}&\textbf{m=3}&\textbf{m=4} &\textbf{m=100} &\textbf{m=10000}\\\hline
0.3465736 & 0.3750000 & 0.3684524 & 0.3631494 & 0.3597840 & 0.3471947 & 0.3465798 \\\hline
\end{array}
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