today I wanted to try and code the Game of Life. So here it is:
Janne
Sunday, October 21, 2018
Friday, October 5, 2018
ILMApilot: Increasing the societal impact of satellite-based atmospheric observations for air quality monitoring
Hi guys,
I just wanted to share Iolanda's pitch from the Finlandia Hall:
See also her new paper on supporting Finnish cleantech sector with satellite data:
Ialongo, I., Fioletov, V., McLinden, C., Jåfs, M., Krotkov, N., Li, C., and Tamminen, J.:
Application of satellite-based sulfur dioxide observations to support the cleantech sector: Detecting emission reduction from copper smelters, Environmental Technology & Innovation, 12, 172-179,
ISSN 2352-1864, https://doi.org/10.1016/j.eti.2018.08.006, 2018.
I just wanted to share Iolanda's pitch from the Finlandia Hall:
See also her new paper on supporting Finnish cleantech sector with satellite data:
Ialongo, I., Fioletov, V., McLinden, C., Jåfs, M., Krotkov, N., Li, C., and Tamminen, J.:
Application of satellite-based sulfur dioxide observations to support the cleantech sector: Detecting emission reduction from copper smelters, Environmental Technology & Innovation, 12, 172-179,
ISSN 2352-1864, https://doi.org/10.1016/j.eti.2018.08.006, 2018.
ILMApilot blog: http://blog.fmi.fi/ILMApilot/
Cheers,
Janne
Wednesday, August 1, 2018
Infinite Magic Squares
Magic squares have interested (recreational) mathematicians for hundreds, if not thousands, of years. A magic square is a $n\times n$ grid filled with integers such that the sum of integers in each row, column and diagonal is equal to a magic constant $M$.
There are various ways to construct magic squares. For odd integers, probably the most famous one is the Siamese method where one also requires that the grid is filled with distinctive positive integers in the range $1, \ldots, n^2$. Below is an example when $n=5$ (Du Royaume de Siam, 1693): \begin{equation*} \begin{array}{|c|c|c|c|c|} \hline 17 & 24 & 1 & 8 & 15 \\ \hline 23 & 5 & 7 & 14 & 16 \\ \hline 4 & 6 & 13 & 20 & 22 \\ \hline 10 & 12 & 19 & 21 & 3 \\ \hline 11 & 18 & 25 & 2 & 9 \\ \hline \end{array} \end{equation*} But what would happen if the grid would be infinite? The simplest “solution” to this problem would be setting all cells to zero \begin{equation*} \begin{array}{c|c|c|c|c} \ddots & \vdots & \vdots & \vdots & \kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.}\\ \hline \cdots & 0 & 0 & 0 & \cdots \\ \hline \cdots & 0 & 0 & 0 & \cdots \\ \hline \cdots & 0 & 0 & 0 & \cdots \\ \hline \kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.} & \vdots & \vdots & \vdots & \ddots \\ \end{array} \end{equation*} but this is not what we are really after here. We can obtain a slightly more interesting solution by subtracting the middle value from a Siamese magic square and adding zeros elsewhere: \begin{equation*} \begin{array}{c|c|c|c|c|c|c|c|c} \ddots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.} \\ \hline \cdots & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\ \hline \cdots & 0 & 4 & 11 & -12 & -5 & 2 & 0 & \cdots \\ \hline \cdots & 0 & 10 & -8 & -6 & 1 & 3 & 0 & \cdots\\ \hline \cdots & 0 & -9 & -7 & \mathbf{0} & 7 & 9 & 0 & \cdots\\ \hline \cdots & 0 & -3 & -1 & 6 & 8 & -10 & 0 & \cdots \\ \hline \cdots & 0 & -2 & 5 & 12 & -11 & -4 & 0 & \cdots\\ \hline \cdots & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\ \hline \kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.} & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{array} \end{equation*} This procedure gives us an infinite magic square where the sum in each row, column and diagonal is equal to zero. This still does not feel quite right as the infinite square has nonzero elements only in the middle. But what about the infinite square below? \begin{equation}\tag{$\infty$}\label{good} \begin{array}{c|c|c|c|c|c|c|c|c} \ddots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.} \\ \hline \cdots & \mathbf{-1} & +1 & -1 & +1 & -1 & +1 & \mathbf{-1} & \cdots \\ \hline \cdots & +1 & \mathbf{-1} & +1 & -1 & +1 & \mathbf{-1} & +1 & \cdots \\ \hline \cdots & -1 & +1 & \mathbf{-1} & +1 & \mathbf{-1} & +1 & -1 & \cdots \\ \hline \cdots & +1 & -1 & +1 & \mathbf{-1} & +1 & -1 & +1 & \cdots \\ \hline \cdots & -1 & +1 & \mathbf{-1} & +1 & \mathbf{-1} & +1 & -1 & \cdots \\ \hline \cdots & +1 & \mathbf{-1} & +1 & -1 & +1 & \mathbf{-1} & +1 & \cdots \\ \hline \cdots & \mathbf{-1} & +1 & -1 & +1 & -1 & +1 & \mathbf{-1} & \cdots \\ \hline \kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.} & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{array} \end{equation} It already looks quite magical with only +1 and -1 entries. But where would the series in each row, column and diagonal sum to? One can note that up, down, left and right from each diagonal cell we have Grandi's series $\sum_{n=1}^{\infty} (-1)^{n-1} = 1-1+1-1+1-1+\ldots$ Grandi's series is Cesàro summable, with Cesàro sum $\frac{1}{2}$. One way to justify this value is to set \begin{equation*} S = 1-1+1-1+1-1+\ldots \end{equation*} and then note that $S = 1-S$, and hence $S= \frac{1}{2}$. Now one may calculate \begin{align*} \ldots+1-1+1-1+\ldots &= -1 + \sum_{n=0}^{\infty} (-1)^n + \sum_{n=0}^{\infty} (-1)^n \\ & = -1 + \frac{1}{2}+\frac{1}{2}=0. \end{align*} Thus, the series in every row and column are Cesàro summable, with Cesàro sum $0$. But what about the diagonals? In both diagonals, after the center cell we have $-1-1-1-1-\ldots$ One can recognize this series as a specific case of the Riemann zeta function \begin{equation*} \zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s} \end{equation*} when $s=0$. We have that $\zeta(0)=-\frac{1}{2}$, thus one may write $-1-1-1-1-\ldots = -\zeta(0)=\frac{1}{2}$. In fact, this series is related to Grandi's series via the Dirichlet eta function \begin{equation*} \eta(s) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^s} = (1-2^{1-s})\zeta(s). \end{equation*} Now when $s=0$, we have that \begin{equation*} 1-1+1-1+\ldots = \eta(0)=-\zeta(0) = -1-1-1-1-\ldots \end{equation*} The diagonals are \begin{align*} \ldots-1-1-1-1-\ldots &= -1- \sum_{n=1}^{\infty}\frac{1}{n^0}-\sum_{n=1}^{\infty}\frac{1}{n^0} \\ &=-1-\zeta(0)-\zeta(0)=0. \end{align*} Now the infinite square (\ref{good}) is indeed an infinite magic square as the series in every row, column and diagonal are equal (in above sense) to the magic constant $M=0$. We note that by multiplying the infinite magic square (\ref{good}) with an integer $a$, we obtain another infinite magic square with $M=0$. If we set $a=-1$, we obtain the “evil twin:” \begin{equation*} \begin{array}{c|c|c|c|c|c|c|c|c} \ddots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.} \\ \hline \cdots & \mathbf{+1} & -1 & +1 & -1 & +1 & -1 & \mathbf{+1} & \cdots \\ \hline \cdots & -1 & \mathbf{+1} & -1 & +1 & -1 & \mathbf{+1} & -1 & \cdots \\ \hline \cdots & +1 & -1 & \mathbf{+1} & -1 & \mathbf{+1} & -1 & +1 & \cdots \\ \hline \cdots & -1 & +1 & -1 & \mathbf{+1} & -1 & +1 & -1 & \cdots \\ \hline \cdots & +1 & -1 & \mathbf{+1} & -1 & \mathbf{+1} & -1 & +1 & \cdots \\ \hline \cdots & -1 & \mathbf{+1} & -1 & +1 & -1 & \mathbf{+1} & -1 & \cdots \\ \hline \cdots & \mathbf{+1} & -1 & +1 & -1 & +1 & -1 & \mathbf{+1} & \cdots \\ \hline \kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.} & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{array} \end{equation*} If we square the values of the magic square (\ref{good}), we obtain an infinite square full of ones: \begin{equation*} \begin{array}{c|c|c|c|c} \ddots & \vdots & \vdots & \vdots & \kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.} \\ \hline \cdots & 1 & 1 & 1 & \cdots \\ \hline \cdots & 1 & 1 & 1 & \cdots \\ \hline \cdots & 1 & 1 & 1 & \cdots \\ \hline \kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.} & \vdots & \vdots & \vdots & \ddots \\ \end{array} \end{equation*} In every direction we have \begin{align*} \ldots+1+1+1+1+\ldots &= 1+ \sum_{n=1}^{\infty}\frac{1}{n^0}+\sum_{n=1}^{\infty}\frac{1}{n^0} \\ &=1+\zeta(0)+\zeta(0)=0. \end{align*} Thus, it is also an infinite magic square. This new square, however, seems little bit less magical than the original one (\ref{good}).
There are various ways to construct magic squares. For odd integers, probably the most famous one is the Siamese method where one also requires that the grid is filled with distinctive positive integers in the range $1, \ldots, n^2$. Below is an example when $n=5$ (Du Royaume de Siam, 1693): \begin{equation*} \begin{array}{|c|c|c|c|c|} \hline 17 & 24 & 1 & 8 & 15 \\ \hline 23 & 5 & 7 & 14 & 16 \\ \hline 4 & 6 & 13 & 20 & 22 \\ \hline 10 & 12 & 19 & 21 & 3 \\ \hline 11 & 18 & 25 & 2 & 9 \\ \hline \end{array} \end{equation*} But what would happen if the grid would be infinite? The simplest “solution” to this problem would be setting all cells to zero \begin{equation*} \begin{array}{c|c|c|c|c} \ddots & \vdots & \vdots & \vdots & \kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.}\\ \hline \cdots & 0 & 0 & 0 & \cdots \\ \hline \cdots & 0 & 0 & 0 & \cdots \\ \hline \cdots & 0 & 0 & 0 & \cdots \\ \hline \kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.} & \vdots & \vdots & \vdots & \ddots \\ \end{array} \end{equation*} but this is not what we are really after here. We can obtain a slightly more interesting solution by subtracting the middle value from a Siamese magic square and adding zeros elsewhere: \begin{equation*} \begin{array}{c|c|c|c|c|c|c|c|c} \ddots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.} \\ \hline \cdots & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\ \hline \cdots & 0 & 4 & 11 & -12 & -5 & 2 & 0 & \cdots \\ \hline \cdots & 0 & 10 & -8 & -6 & 1 & 3 & 0 & \cdots\\ \hline \cdots & 0 & -9 & -7 & \mathbf{0} & 7 & 9 & 0 & \cdots\\ \hline \cdots & 0 & -3 & -1 & 6 & 8 & -10 & 0 & \cdots \\ \hline \cdots & 0 & -2 & 5 & 12 & -11 & -4 & 0 & \cdots\\ \hline \cdots & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\ \hline \kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.} & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{array} \end{equation*} This procedure gives us an infinite magic square where the sum in each row, column and diagonal is equal to zero. This still does not feel quite right as the infinite square has nonzero elements only in the middle. But what about the infinite square below? \begin{equation}\tag{$\infty$}\label{good} \begin{array}{c|c|c|c|c|c|c|c|c} \ddots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.} \\ \hline \cdots & \mathbf{-1} & +1 & -1 & +1 & -1 & +1 & \mathbf{-1} & \cdots \\ \hline \cdots & +1 & \mathbf{-1} & +1 & -1 & +1 & \mathbf{-1} & +1 & \cdots \\ \hline \cdots & -1 & +1 & \mathbf{-1} & +1 & \mathbf{-1} & +1 & -1 & \cdots \\ \hline \cdots & +1 & -1 & +1 & \mathbf{-1} & +1 & -1 & +1 & \cdots \\ \hline \cdots & -1 & +1 & \mathbf{-1} & +1 & \mathbf{-1} & +1 & -1 & \cdots \\ \hline \cdots & +1 & \mathbf{-1} & +1 & -1 & +1 & \mathbf{-1} & +1 & \cdots \\ \hline \cdots & \mathbf{-1} & +1 & -1 & +1 & -1 & +1 & \mathbf{-1} & \cdots \\ \hline \kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.} & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{array} \end{equation} It already looks quite magical with only +1 and -1 entries. But where would the series in each row, column and diagonal sum to? One can note that up, down, left and right from each diagonal cell we have Grandi's series $\sum_{n=1}^{\infty} (-1)^{n-1} = 1-1+1-1+1-1+\ldots$ Grandi's series is Cesàro summable, with Cesàro sum $\frac{1}{2}$. One way to justify this value is to set \begin{equation*} S = 1-1+1-1+1-1+\ldots \end{equation*} and then note that $S = 1-S$, and hence $S= \frac{1}{2}$. Now one may calculate \begin{align*} \ldots+1-1+1-1+\ldots &= -1 + \sum_{n=0}^{\infty} (-1)^n + \sum_{n=0}^{\infty} (-1)^n \\ & = -1 + \frac{1}{2}+\frac{1}{2}=0. \end{align*} Thus, the series in every row and column are Cesàro summable, with Cesàro sum $0$. But what about the diagonals? In both diagonals, after the center cell we have $-1-1-1-1-\ldots$ One can recognize this series as a specific case of the Riemann zeta function \begin{equation*} \zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s} \end{equation*} when $s=0$. We have that $\zeta(0)=-\frac{1}{2}$, thus one may write $-1-1-1-1-\ldots = -\zeta(0)=\frac{1}{2}$. In fact, this series is related to Grandi's series via the Dirichlet eta function \begin{equation*} \eta(s) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^s} = (1-2^{1-s})\zeta(s). \end{equation*} Now when $s=0$, we have that \begin{equation*} 1-1+1-1+\ldots = \eta(0)=-\zeta(0) = -1-1-1-1-\ldots \end{equation*} The diagonals are \begin{align*} \ldots-1-1-1-1-\ldots &= -1- \sum_{n=1}^{\infty}\frac{1}{n^0}-\sum_{n=1}^{\infty}\frac{1}{n^0} \\ &=-1-\zeta(0)-\zeta(0)=0. \end{align*} Now the infinite square (\ref{good}) is indeed an infinite magic square as the series in every row, column and diagonal are equal (in above sense) to the magic constant $M=0$. We note that by multiplying the infinite magic square (\ref{good}) with an integer $a$, we obtain another infinite magic square with $M=0$. If we set $a=-1$, we obtain the “evil twin:” \begin{equation*} \begin{array}{c|c|c|c|c|c|c|c|c} \ddots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.} \\ \hline \cdots & \mathbf{+1} & -1 & +1 & -1 & +1 & -1 & \mathbf{+1} & \cdots \\ \hline \cdots & -1 & \mathbf{+1} & -1 & +1 & -1 & \mathbf{+1} & -1 & \cdots \\ \hline \cdots & +1 & -1 & \mathbf{+1} & -1 & \mathbf{+1} & -1 & +1 & \cdots \\ \hline \cdots & -1 & +1 & -1 & \mathbf{+1} & -1 & +1 & -1 & \cdots \\ \hline \cdots & +1 & -1 & \mathbf{+1} & -1 & \mathbf{+1} & -1 & +1 & \cdots \\ \hline \cdots & -1 & \mathbf{+1} & -1 & +1 & -1 & \mathbf{+1} & -1 & \cdots \\ \hline \cdots & \mathbf{+1} & -1 & +1 & -1 & +1 & -1 & \mathbf{+1} & \cdots \\ \hline \kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.} & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{array} \end{equation*} If we square the values of the magic square (\ref{good}), we obtain an infinite square full of ones: \begin{equation*} \begin{array}{c|c|c|c|c} \ddots & \vdots & \vdots & \vdots & \kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.} \\ \hline \cdots & 1 & 1 & 1 & \cdots \\ \hline \cdots & 1 & 1 & 1 & \cdots \\ \hline \cdots & 1 & 1 & 1 & \cdots \\ \hline \kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.} & \vdots & \vdots & \vdots & \ddots \\ \end{array} \end{equation*} In every direction we have \begin{align*} \ldots+1+1+1+1+\ldots &= 1+ \sum_{n=1}^{\infty}\frac{1}{n^0}+\sum_{n=1}^{\infty}\frac{1}{n^0} \\ &=1+\zeta(0)+\zeta(0)=0. \end{align*} Thus, it is also an infinite magic square. This new square, however, seems little bit less magical than the original one (\ref{good}).
Global XCO2 anomalies as seen by Orbiting Carbon Observatory-2
Dear readers,
I would like to advertise our latest work "Global XCO2 anomalies as seen by Orbiting Carbon Observatory-2." It was published yesterday as discussion paper in Atmos. Chem. Phys. Discuss. (ACPD). Here's a little preview:
Hakkarainen, J., Ialongo, I., Maksyutov, S., and Crisp, D.: Global XCO2 anomalies as seen by Orbiting Carbon Observatory-2, Atmos. Chem. Phys. Discuss., https://doi.org/10.5194/acp-2018-649, in review, 2018.
Janne
I would like to advertise our latest work "Global XCO2 anomalies as seen by Orbiting Carbon Observatory-2." It was published yesterday as discussion paper in Atmos. Chem. Phys. Discuss. (ACPD). Here's a little preview:
Hakkarainen, J., Ialongo, I., Maksyutov, S., and Crisp, D.: Global XCO2 anomalies as seen by Orbiting Carbon Observatory-2, Atmos. Chem. Phys. Discuss., https://doi.org/10.5194/acp-2018-649, in review, 2018.
Janne
Monday, April 30, 2018
Pitch like a pro!
As you might remember I was in the Skolar Award finals at Slush 2017. This week we will have a little alumni meeting organized by the Kaskas Media. The 2016 Skolar award winner Virpi Virjamo listed her 5 tips for pitching your research idea like a winner.
So here are my 5+2 tips for pitching like a pro!
1. Don’t forget your science!
I think this was the best advice that I got. Scientists tend to oversimplify when they talk about their work to non-experts. But people are smart and they will understand. If you leave your science out, what is left in your pitch!? You are not a salesman!
2. Keep it simple!
Now this second point seems to contradict what I just said in the first point. It is not. You should keep it simple, but do not oversimplify! Easy as that.
3. Practice, practice, practice!
This may sound self evident, but it is important. You should learn your pitch by heart. Be a pro. I still remember my pitch! Practicing is a good advice also to your scientific presentations. (When I gave my pitch at Slush, I made a little mistake and skipped one slide. Since I knew the pitch, I could do it on autopilot and think how to back it up in the end. It was surreal feeling).
4. Take the advice!
When people try to help you, you should listen. And make your pitch better. They will have good ideas. But remember, it is still your pitch and you have to deliver.
5. The structure! Problem, solution, vision.
This is a technical advice, but still very important. I think the structure should be a) problem, b) solution, and c) vision. So first tell what is the problem that you are solving, not the latest thing in your own research. Then give Your solution to this problem. Finally, tell how your solution will change the world.
6. Be yourself?
This is an advice that you are like to get for someone else but me. I say: Don’t be yourself, be a better version of yourself! In my daily life, I’m quite calm and low-energy guy. In a good way. In my pitch, I wanted to give 120% and be super energetic. It worked out for me.
7. Have fun!
Don’t think about $$ or the victory. It will show and you’ll regret it later. Have fun! This is a unique opportunity. Make the most of it!
Friday, January 5, 2018
2018: Centre of Excellence of Inverse Modelling and Imaging
Happy
New Year 2018!
They are plenty of new things in my academic life. I am
officially back at Finnish Meteorological Institute, now with the title “Senior
Research Scientist.” Because of the large organization change at FMI, the name
of my group is now entitled “Greenhouse Gases and Satellite Methods” in Earth
Observation Research Unit.
 |
| Centre of Excellence of Inverse Modelling and Imaging: applications. |
Probably the biggest news story is that now also our FMI team is part of the new 2018–2025 “Centre of Excellence of Inverse Modelling and Imaging.” The University of Helsinki team, the team where I had the pleasure of visiting for the last seven months, coordinates this Centre! We will have the kick-off meeting next week at Uunisaari. It will be fun!
Regarding
this site, I updated my Biographical Sketch. Enjoy!
In the next
post will recap my experiences from Slush 2017 Skolar Award.
Stay tuned!
Janne
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