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Ray Dalio's All Weather Portfolio (Backtest, What Is It, Returns, Performance) [Trading & Investment]
Ray Dalio's All Weather Portfolio is a pretty well-known portfolio, and for good reasons, in our opinion. Ray Dalio and his team at Bridgewater Associates constructed the All Weather Portfolio to withstand every economic environment (and shocks) possible. What exactly is the All-Weather Portfolio? Which assets are included? How has it performed? What was the performance during the financial crisis in 2008/09, Covid-19, and the bear market of 2022? These are some of the questions we answer in this article. We backtest the All Weather Portfolio.
Ray Dalio's All Weather Portfolio was constructed to stand the test of time, no matter the investing climate – be it inflation, deflation, or stagflation – and the portfolio consists of both stocks, bonds, and commodities. We backtest its performance.
The annual returns have been lower than for stocks, but the max drawdowns are substantially smaller. It's a defensive portfolio. Both during the financial crisis in 2008/09, Covid-19 in 2020, and the bear market of 2022 it performed better than stocks.
We show you how you can construct an All-Weather Portfolio and how you can backtest it. Let's get started:
Who is Ray Dalio?
What is the All Weather Portfolio?
The All Weather principles and the risk parity approach
Strategy Shop
What are the assets included in the All Weather Portfolio?
How do you make an All Weather Portfolio yourself?
Why is the All Weather portfolio good?
Is the All Weather portfolio good for retirees or FIREs?
When should you rebalance the All Weather portfolio?
The Pros and cons of the All Weather portfolio
How has the All Weather Portfolio performed? Let's backtest
Backtest of All Weather Portfolio
All Weather Portfolio and drawdowns
How did the All-Weather portfolio perform during the financial crisis in 2008/09?
How did the All-Weather portfolio perform in 2020 during Covid-19?
How did the All-Weather portfolio perform in 2022?
What happens if you leverage the All Weather Portfolio?
Changing the allocations in the All Weather Portfolio
How to code the All Weather portfolio in Amibroker
The All Weather Portfolio investment strategy – conclusions:
Let's start with a very brief introduction about Ray Dalio:
Ray Dalio is one of the most famous money managers on this planet. He founded the hedge fund Bridgewater Associates in 1975, and after a setback in the 1980s, he has managed to create an outstanding track record for his fund: 11.5 %. Considering the huge asset base, this result is really impressive.
As a result, Bridgewater Associates has attracted a lot of money from both private enterprises and public sovereign entities. Bridgewater Associates is currently the world's biggest hedge fund, to our knowledge, with over 155 billion dollars under management.
Ray Dalio's investment approach is mostly top-down and based on macro fundamentals, and the All-Weather Portfolio is a result of this: It's all about asset classes – no stock pickings.
We don't go into details about Ray Dalio in this article, but we recommend our separate article about Ray Dalio.
Now that you briefly know who Ray Dalio is, we go on to explain the All Weather Portfolio, the main purpose of this article:
What I'm trying to say is that for the average investor, what I would encourage them to do is to understand that there's inflation and growth. It can go higher and lower and to have four different portfolios essentially that make up your entire portfolio that gets you balanced…..I think that the first thing is you should have a strategic asset allocation mix that assumes that you don't know what the future is going to hold.
As we have already mentioned in the introduction, Ray Dalio's All Weather Portfolio was constructed to stand the test of time, no matter the investing climate – be it inflation, deflation, or stagflation. Because the portfolio is put together to perform well under ANY economic environment, it consists of many assets.
Why does the All Weather Portfolio consist of many asset classes?
It's all about diversification and correlation. We have diversification to avoid being invested in just one asset class, not to put all our eggs in one basket, and to potentially have a low or no correlation between the asset classes. We have in a previous article explained the benefits of lowly correlated assets or strategies in a portfolio:
Does your trading strategy complement your portfolio of strategies?
Uncorrelated or non-correlated assets and strategies – benefits and advantages
What does correlation mean in trading?
Mark Spitznagel – Safe Haven Investing (explains why you want to mitigate risk)
The aim of Bridgewater's All Weather Portfolio is to use diversification as a tool to smooth returns and lower drawdowns. If the asset classes don't move in tandem, then diversification can help you lower drawdowns and sleep well at night.
Based on backtests and simulations, Ray Dalio's Bridgewater Associates composed the portfolio like this:
55% bonds
30% US stocks (US stocks are about 50% of world market capitalization)
15% hard assets and commodities
That is a lot of exposure to bonds! It's a surprisingly small allocation to stocks, the most successful asset class over the last century (measured in annual returns). But because the portfolio is balanced and constructed to survive and "prosper" in any economic scenario, it consists of several asset classes and not just stocks.
It is about balancing a portfolio's risk exposures to attain a greater chance of investment success than what is offered by traditional, equity-centric approaches to asset allocation. We are very pleased to see investors moving in this direction, as we have been communicating and debating our "All Weather" principles of balanced beta with our clients and with the investment community for the past sixteen years. From these core principles the Risk Parity space was born. We want to take this opportunity to revisit those original principles, and to explain how we apply them to achieve reliable balance, which has proven its value in our real-time management of All Weather since 1996 and in 85 years of back-testing. The best way to achieve reliable balance is to design a portfolio based on a fundamental understanding of the environmental sensitivities inherent in the pricing structure of asset classes. This is the foundation of the All Weather approach.
– Bob Prince
The groundwork for the All Weather investment principles was done in the 1980s and the early 1990s, and in 1996 Dalio decided to use the approach for his own family trust.
Bob Prince, the Co-Chief Investment Officer for Bridgewater Associates, wrote a paper in August 2011 called Risk Parity Is About Balance, where he laid out the foundation of the All Weather investment principles. The main takeaway was that a balanced portfolio achieves the same returns as equities with 1/3 of the risk. Bob Prince presented this backtest in the paper:
All Weather Portfolio historical performance
One takeaway from their findings is their All Weather risk parity approach. We will not get into details about the risk parity approach but in general terms, it's about keeping the risk for each asset class constant based on volatility. Because we don't know future volatility, trailing volatility was used as an estimate. Rising volatility means more risk and thus lower allocation in the portfolio.
In our backtests, we don't take into consideration the volatility, and thus we are simplifying.
The best strategies can be found in our….
Backtested trading strategies
Yes, I'd like to make money!
Let's look at the different asset classes more deeply.
The chart below is taken from the book Triumph of the Optimists and shows the US' returns for stocks and bonds:
Historical performance stocks and bonds
As you can see, stocks have been, by far, a much better asset class than bonds.
Why then allocate so much to bonds and less to stocks?
It's because Ray Dalio and his team do not necessarily look to optimize the total returns, but rather to create a "sleep well at night" portfolio with smaller drawdowns and good risk-adjusted returns. The latter is a very important strategy metric. The portfolio is backtested from the period after WW2, and there is a reason why they implemented a lot of bonds at the expense of less to stocks.
Again, we emphasize that the All Weather Portfolio is an investment portfolio whose purpose is to perform well under different economic environments – preferably ANY environment. Most investors have long forgotten the stagflation period of the 1970s where stocks produced negative real returns, while real assets soared. Real assets had mediocre returns from 2000 until 2021 when they suddenly started rising after being "dormant" for two decades. We all fall prone to the recency bias and quickly forget the past.
One other important issue is that the time horizon for the portfolio is decades. Even though stocks have been the best for over a century, they don't necessarily perform well during shorter time periods of say 1-5 years. We suspect bonds are added to serve as a balancing anchor to the portfolio.
Stocks, bonds, and commodities are included because they have their own separate contributions to the returns of a portfolio:
Stocks are included because they have historically produced good returns. If you own a broad portfolio of stocks, you participate in public companies' wealth creation. Over the last 100 years, stocks have risen about 10% annually with dividends reinvested. This is before inflation, but most research we have seen shows a real return of about 6%. Thus, over the long term, you should have a decent hedge against inflation. Stocks go the same way as the GDP – up.
The drawback with stocks is drawdowns. The market dropped almost 90% from the top in 1929, and in 2008/09 the S&P 500 dropped 55%. These drops are gut-wrenching for most investors!
Can you handle these drawdowns? History shows that retail investors have poor returns, and one of the reasons is they buy market tops and sell market bottoms. They can't stomach losses. They fear a loss will become bigger and sell. When the market recovers they reenter at much higher prices. It's a vicious cycle.
A bond is a loan to a company or sovereign government. As compensation, the lender (the owner of the bond) is paid a coupon at certain intervals. Because the interest rates fluctuate over the life of the bond, the price of the bond goes up and down. We have explained the relationship in our article named what happens to stocks when bonds go up.
Thus, just like stocks, bonds also have drawdowns. However, they are normally of a smaller magnitude than stocks.
Commodities are hard assets like gold, metals, grains, oil, etc. They are real assets, tangible assets that we use in everyday life. Without commodities, the world faces supply shocks, as evidenced after Covid-19 and during the Ukraine-Russian war. It's the grease that keeps the wheel turning and is essential for increasing living standards.
Why are commodities included in the All Weather Portfolio?
We assume it's because they tend to rise when stocks and bonds go down. Rising commodity prices are inflationary, resulting in lower prices for stocks (and bonds). To offset this risk, commodities are included in the portfolio. It's to balance the portfolio – to mitigate risk and neutralize falling stock prices. 2022 is a perfect example of this: stocks and bonds are both down, while commodities are up (a lot). Because of this, trend-following funds are having a fantastic year in 2022.
Now that you know the asset allocations, you might wonder how you can go about and create such a portfolio yourself:
How can you diversify like Ray Dalio and Bridgewater Associates?
We have never seen or read Ray Dalio explain explicitly how "normal" retail investors can create the All Weather Portfolio. However, we see several options on the internet. Due to a wide range of mutual funds and ETFs, it's a lot easier now than before to put together a portfolio.
We suggest the same as many others do and pick these five different ETFs:
40% TLT (long-term Treasuries)
30% SPY (US stocks, S&P 500)
15% IEI (intermediate-term U.S. Bonds)
7.5% GLD (gold)
7.5% DBC (commodities, commodity index tracking fund)
I guess every broker offers these ETF's and thus you can easily compose this portfolio and rebalance whenever you want. You can of course change some of these components yourself but be sure you know what you are doing and that you understand the holdings of the ETF or the fund.
Because of rebalancing, which should be done at least once a year, it's preferable to use a tax-deferred account.
First and foremost the All-Weather Portfolio is a defensive portfolio. It's not made to make you rich quickly (nothing is), and you are almost guaranteed to underperform in a bull market in stocks. So then, why invest in such a portfolio?
Here are some arguments for having an All Weather Portfolio:
It's a defensive portfolio. You'll most likely underperform stocks in a rising market, but outperform in a falling market.
Because of number 1, you'll have smaller drawdowns than in stocks, and thus be less likely to commit behavioral mistakes.
It's about preserving what you have, not going for the jugular.
It's potentially a smart portfolio for retirees or those about to retire in a few years (because of estimated smaller drawdowns).
Because the All-Weather Portfolio is constructed to survive any investment climate, it's potentially a good portfolio for retirees and FIREs. It's a hands-off portfolio that you only need to rebalance 1-2 times per year.
We made a separate article a few months back where we made some simulations of why you need to diversify to different asset classes if you plan to withdraw capital for consumption (or whatever reason):
How likely are you to go broke as retired or FIRE? (Sequence risk, diversification, and withdrawal rate)
There are no definite answers to this. However, we believe at the least you should rebalance once a year, perhaps semi-annually, but not more than quarterly.
But it also depends on your tax situation. If you have a tax-deferred account, something that is strongly recommended, even paramount, you have fewer things to worry about and consider, which means you can balance as often as you want. That said, it most likely makes no sense to rebalance more often than quarterly.
There is no free lunch in the stock market, and lower variability comes at a cost: over the long run, the All Weather Portfolio will most likely underperform both stocks and 60/40 (stock/bonds) as can be seen from the graphs further below in the article.
This means that the All Weather Portfolio is not for everyone. Especially if you are young, for example in your 20s, it would make more sense to be more aggressive (invest more in stocks) as you have time on your side to let your capital compound (and the stomach to handle drawdowns).
This is probably what most readers care about: how has the All Weather Portfolio performed?
We only have data going back to 2007, but Nick Maggiulli has made a backtest going back to 1970 in a blogpost called The Definitive Guide to the All Weather Portfolio on his website ofdollarsanddata.com.
All Weather Portfolio real annualized returns by decade (historical performance)
During the 1970s and 2000s when stocks performed poorly, the All Weather Portfolio outperformed. During bull markets in the 80s, 90s, and the 2010s, it underperformed. This is exactly what you can expect because of the high allocation to bonds.
Below is a backtest we did from July 2007 until today with daily rebalancing. We used daily rebalancing to better capture the ups and downs of the portfolio. The difference in returns compared to monthly and even annual returns are small (about 0.35% lower for annual rebalancing). We used the following ETFs and weightings:
This is what the equity curve looks like:
Backtest of the All Weather Portfolio
The CAGR is a modest 4.5% (4.15% with annual rebalancing) but keep in mind that our results don't include reinvested dividends, perhaps underestimating the results by 1-1.5% annually. For comparison, S&P 500 returned 6.7% (not including dividends).
If we look at each year and month separately we get the following table:
Annual returns of the All Weather Portfolio
Let's compare the equity curves of both the All Weather Portfolio and S&P 500:
All Weather Portfolio vs stocks (S&;P 500)
Clearly, the red line, which is the All Weather Portfolio, has a substantially smoother path, although you end up with less capital. It's a trade-off.
Most backtests scan a test in less than a second and voila! – you have an equity curve. We tend to focus on the end result, and not the journey of the equity curve. That is a shame because most traders and investors give up in the midst of a panic or sell-off in the markets. Most retail investors end up with mediocre results exactly because of this reason.
You don't know how much pain you can suffer until you are in the midst of it. Thus, drawdowns are very important to understand. Because of this, we backtest All Weater Portfolio during the three most dramatic selloffs since 2007:
The red line is the All Weather portfolio and the blue line is S&P 500. Clearly, over many years, the All Weather Portfolio performed much better than stocks with a max drawdown of 17%. It was not until late 2016 that stocks managed to recover the "losses" relative to the All Weather Portfolio.
The All Weather Portfolio during 2008/09
Imagine being a retiree in January 2008 and only being invested in stocks! The sell-off would be heart-ripping.
One of the reasons it took so long for stocks to recover relative to the All Weather Portfolio is the compounding effect. When you suffer a huge drawdown of 50% you need 100% returns to recover. But when drawdowns are smaller you can start compounding at higher levels after the sell-off.
Let's go to the Covid-19 debacle in March 2020.
Stocks sold off 33% while the All Weather Portfolio had a much more modest drawdown of about 6%:
All Weather Portfolio during Covid
Unlike the financial crisis in 2008/09, stocks recovered quickly due to a swift response from central banks with plenty of money printing. By the end of the year, stocks had managed to recover the losses relative to the All Weather Portfolio.
The "crash" of 2022 involves both stocks and bonds. Bonds have often been a safe haven, but not this time. Due to the selloff in bonds, the only place to hide has been in commodities and this has softened the decline somewhat:
The All Weather Portfolio in 2022
The red line is All Weather Portfolio and the blue line is S&P 500. The drawdown for the All Weather Portfolio is about 12%.
You might argue you can leverage the All Weather Portfolio because of its low drawdowns.
Let's run a backtest where you leverage 1.5 times equity. The equity curve looks like this:
The All Weather Portfolio with leverage (backtest)
The CAGR goes up to 6.5% from 4.5% (still not including dividends) and the max drawdown increases to 25% during the financial crisis (but still a pretty modest drawdown).
How does it compare to holding S&P 500?
The All Weather Portfolio with leverage vs stocks (S&;P 500)
Over the 15-year backtest the All Weather Portfolio returns the same as S&P 500 but with significantly lower drawdowns. We have not deducted interest payments for the leverage.
It's a lot of bonds in the portfolio. What happens if use the same ETFs and assets but change the weightings?
Let's change the weightings to this:
7.5% DBC (commodities, commodity index tracking fund, please also read our take on commodity trading strategy)
Without leverage, the portfolio would have fared like this:
The All Weather Portfolio with different weightings (backtest)
The CAGR goes up to 5.2%, but the max drawdown goes up to 29%.
Let's do another last backtest of the All Weather Portfolio with the following weightings:
15% GLD (gold)
10% DBC (commodities, commodity index tracking fund)
The CAGR goes up to 5.3% and we experience slightly reduced drawdowns:
All Weather Portfolio with different weightings (backtest)
If you want to backtest Ray Dalio's All Weather Portfolio yourself or you are struggling to backtest similar strategies, you can buy the Amibroker code for the All Weather Portfolio (plus the code for all the other free trading strategies we have provided). Please click on the green banner below. We have also provided Tradestation/Easy Language code for about 50% of the strategies.
Ray Dalio's All Weather Portfolio offers lower drawdowns but it comes at the expense of lower returns than stocks (in the long term). Thus, the portfolio is not for those who are going for the jugular but is more directed to risk-averse and conservative investors. It could potentially suit those who are retirees, FIREs, or soon to become retired.
However, as with every investment, there is no guarantee that the All Weather Portfolio will perform well in the future, as witnessed in 2022 when bonds sold off heavily.
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Build and Flash with Eclipse IDE
********************************

Installing Eclipse IDE
======================

The Eclipse IDE gives you a graphical integrated development environment for writing, compiling and debugging ESP-IDF projects.

* Start by installing the esp-idf for your platform (see files in this directory with steps for Windows, OS X, Linux).

* Download the Eclipse Installer for your platform from eclipse.org_.

* When running the Eclipse Installer, choose "Eclipse for C/C++ Development" (in other places you'll see this referred to as CDT.)

Setting up Eclipse
==================

Once your new Eclipse installation launches, follow these steps:

Import New Project
------------------

* Eclipse makes use of the Makefile support in ESP-IDF. This means you need to start by creating an ESP-IDF project. You can use the skeleton project from github.

* Once Eclipse is running, choose File -> Import...

* In the dialog that pops up, choose "C/C++" -> "Existing Code as Makefile Project" and click Next.

* On the next page, enter "Existing Code Location" to be the directory of your IDF project. Don't specify the path to the ESP-IDF directory itself.

* On the same page, under "Toolchain for Indexer Settings" choose "Cross GCC". Then click Finish.

Project Properties
------------------

* The new project will appear under Project Explorer. Right-click the project and choose Properties from the context menu.

* Click on the "Environment" properties page under "C/C++ Build". Click "Add..." and enter name ``V`` and value ``1``.

* Click "Add..." again, and enter name ``IDF_PATH``. The value should be the full path where ESP-IDF is installed. *Windows users: Use forward-slashes not backslashes for this path, ie C:/Users/MyUser/Development/esp-idf*.

*Windows users only, follow these two additional steps:*

* On the same Environment property page, edit the PATH environment variable. Delete the existing value and replace it with ``C:\msys32\usr\bin;C:\msys32\mingw32\bin;C:\msys32\opt\xtensa-esp32-elf\bin`` (If you installed msys32 to a different directory then you'll need to change these paths to match).

* Click on the "C/C++ Build" top-level properties page then uncheck "Use default build command" and enter this for the custom build command: ``bash ${IDF_PATH}/tools/windows/eclipse_make.sh``.

*All users, continue with these steps:*

Navigate to "C/C++ General" -> "Preprocessor Include Paths" property page:

* Click the "Providers" tab

* In the list of providers, click "CDT Cross GCC Built-in Compiler Settings". Under "Command to get compiler specs", replace the text ``${COMMAND}`` at the beginning of the line with ``xtensa-esp32-elf-gcc``. This means the full "Command to get compiler specs" should be ``xtensa-esp32-elf-gcc ${FLAGS} -E -P -v -dD "${INPUTS}"``.

* In the list of providers, click "CDT GCC Build Output Parser" and type ``xtensa-esp32-elf-`` at the beginning of the Compiler command pattern. This means the full Compiler command pattern should be ``xtensa-esp32-elf-(g?cc)|([gc]\+\+)|(clang)``

* Click OK to close the Properties dialog, and choose Project -> Build to build your project.

Flash from Eclipse
------------------

You can integrate the "make flash" target into your Eclipse project to flash using esptool.py from the Eclipse UI:

* Right-click your project in Project Explorer (important to make sure you select the project, not a directory in the project, or Eclipse may find the wrong Makefile.)

* Select Make Targets -> Create from the context menu.

* Type "flash" as the target name. Leave the other options as their defaults.

* Now you can use Project -> Make Target -> Build (Shift+F9) to build the custom flash target, which will compile and flash the project.

Note that you will need to use "make menuconfig" to set the serial port and other config options for flashing. "make menuconfig" still requires a command line terminal (see the instructions for your platform.)

Follow the same steps to add ``bootloader`` and ``partition_table`` targets, if necessary.

.. _eclipse.org: http://www.eclipse.org/

Eclipse Troubleshooting
-----------------------

* ``*** Make was invoked from ... However please do not run make from the sdk or a component directory; ...`` - Eclipse will detect any directory with a Makefile in it as being a possible directory to run "make" in. All component directories also contain a Makefile (the wrong one), so it is important when using Project -> Make Target to always select the top-level project directory in Project Explorer.

redpajama_set_name: RedPajamaGithub


A new postage stamp depicting the rainbow flag to be issued in Sweden on the 4th of May to celebrate Pride, diversity and equal love.
The Swedish postal service has now made sure that you can send proud rainbow greetings all across the world by issuing a new stamp featuring the rainbow flag.
"By issuing a stamp with the Pride Flag, we want to emphasise the equal value of all people and the strength represented by people's unique qualities and differences. To the best of our knowledge, this is the first time the Pride Flag has been reproduced on a stamp – which makes it particularly gratifying that people will now be able to send affectionate greetings with the Pride Flag stamp to all corners of the world," says Per Ljungberg, Head of Communications at PostNord Sweden.
The original idea for the stamp came from Swedish artist Lisa Rydell back in 2014 and was quickly approved by PostNord's Stamp Advisory Committee who then commissioned Rydell to do the design as well.
"For many people, the choice of stamp is an extension of their written message – which will be made especially profound with the new Pride stamp. This motif is also well in line with PostNord's internal diversity initiatives," says Mr Ljungberg from PostNord.


The Crab is a small utility that can help you organize your music albums using various music data sources.
Edit and organize your music files and directories!
TagTuner takes you beyond repetitive typing and speeds up the creation of perfectly organized music library.
The smart KaraFun Player application will help you to transform your computer into a real Karaoke machine.
AllMusicConverter - easily convert any music and movie files at high speed and CD quality.
Determine the chance and enjoy your music.
Clean and simple music player that is beautiful, simple and fun to use.


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Home MMA Bellator Chuck Liddell Was Getting Paid "1 Million a Year"
Photo via UFC.com / Getty Images
Chuck Liddell helped build the UFC but when he lost three fights in a row, all by early KO, his friend league president Dana White offered him a no-show position with the company in return for retiring. That was in 2010.
Last year the UFC sold to WME-IMG, and the new owners cut a number of positions, including Liddell's. Now there are multiple indications that Liddell wants to fight again, and is considering signing with Bellator MMA.
On his Believe You Me podcast, Michael Bisping offered some further detail, noting it was simply something he heard, third-hand information that he does not know for a fact.
"I'm in the circle, where people tell me things," said Bisping, as transcribed by Anton Tabuena for BE. "This is me putting out a disclaimer, if they hear me saying this, I can say 'I said it was third-hand information', when really it might not be third-hand information.
"Somebody at the weekend told me that Chuck Liddell is really pissed off, because he had a pretty sweet deal with the UFC when he left he had this job. He was an ambassador for the UFC. I believe he was on $1 million a year. That's the number I heard. F**in' hell that's a great job.
"It was great that the UFC was doing that, but now of course, new UFC owners came in. I heard that Chuck was really pissed off, so that's why he's possibly going over to Bellator."
"Can he compete? Should he comeback? The answer is no. Can he compete, and should he fight against Tito Ortiz? The answer is probably yes."
Previous articleBREAKING UFC Legend Matt Hughes Hit By Train
Next articleWoodley Opens Up About Relationship With Dana White

redpajama_set_name: RedPajamaCommonCrawl


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Consumption of individual saturated fatty acids and the risk of myocardial infarction in a UK and a Danish cohort
Jaike Praagman, Linda E. T. Vissers, Angela A. Mulligan, Anne Sofie Dam Laursen, Joline W. J. Beulens, Yvonne T. van der Schouw, Nicholas J. Wareham, Camilla Plambeck Hansen, Kay-Tee Khaw, Marianne Uhre Jakobsen, Ivonne Sluijs
ACS - Diabetes & metabolism (VUmc)
APH - Health Behaviors & Chronic Diseases (VUmc)
Clinical Epidemiology, Biostatistics and Bioinformatics (VUmc)
Background: The effect of individual saturated fatty acids (SFAs) on serum cholesterol levels depends on their carbon-chain length. Whether the association with myocardial infarction (MI) also differs across individual SFAs is unclear. We examined the association between consumption of individual SFAs, differing in chain lengths ranging from 4 through 18 carbons, and risk of MI. Methods: We used data from 22,050 and 53,375 participants from EPIC-Norfolk (UK) and EPIC-Denmark, respectively. Baseline SFA intakes were assessed through validated, country-specific food frequency questionnaires. Cox regression analysis was used to estimate associations between intakes of individual SFAs and MI risk, for each cohort separately. Results: During median follow-up times of 18.8 years in EPIC-Norfolk and 13.6 years in Denmark, respectively, 1204 and 2260 MI events occurred. Mean (±SD) total SFA intake was 13.3 (±3.5) en% in EPIC-Norfolk, and 12.5 (±2.6) en% in EPIC-Denmark. After multivariable adjustment, intakes of C12:0 (lauric acid) and C14:0 (myristic acid) inversely associated with MI risk in EPIC-Denmark (HR upper versus lowest quintile: 0.80 (95%CI: 0.66, 0.96) for both SFAs). Intakes in the third and fourth quintiles of C4:0–C10:0 also associated with lower MI risk in EPIC-Denmark. Moreover, substitution of C16:0 (palmitic acid) and C18:0 (stearic acid) with plant proteins resulted in a reduction of MI risk in EPIC-Denmark (HR per 1 energy%: 0.86 (95%CI: 0.78, 0.95) and 0.87 (95%CI: 0.79, 0.96) respectively). No such associations were found in EPIC-Norfolk. Conclusion: The results from the present study suggest that the association between SFA and MI risk depends on the carbon chain-length of the SFA.
International Journal of Cardiology
https://doi.org/10.1016/j.ijcard.2018.10.064
Published - 15 Mar 2019
https://www.scopus.com/inward/record.uri?partnerID=HzOxMe3b&scp=85055468887&origin=inward
Praagman, J., Vissers, L. E. T., Mulligan, A. A., Laursen, A. S. D., Beulens, J. W. J., van der Schouw, Y. T., Wareham, N. J., Hansen, C. P., Khaw, K-T., Jakobsen, M. U., & Sluijs, I. (2019). Consumption of individual saturated fatty acids and the risk of myocardial infarction in a UK and a Danish cohort. International Journal of Cardiology, 279, 18-26. https://doi.org/10.1016/j.ijcard.2018.10.064
Praagman, Jaike ; Vissers, Linda E. T. ; Mulligan, Angela A. et al. / Consumption of individual saturated fatty acids and the risk of myocardial infarction in a UK and a Danish cohort. In: International Journal of Cardiology. 2019 ; Vol. 279. pp. 18-26.
@article{f757b394b8364486a5762dbc54d78299,
title = "Consumption of individual saturated fatty acids and the risk of myocardial infarction in a UK and a Danish cohort",
abstract = "Background: The effect of individual saturated fatty acids (SFAs) on serum cholesterol levels depends on their carbon-chain length. Whether the association with myocardial infarction (MI) also differs across individual SFAs is unclear. We examined the association between consumption of individual SFAs, differing in chain lengths ranging from 4 through 18 carbons, and risk of MI. Methods: We used data from 22,050 and 53,375 participants from EPIC-Norfolk (UK) and EPIC-Denmark, respectively. Baseline SFA intakes were assessed through validated, country-specific food frequency questionnaires. Cox regression analysis was used to estimate associations between intakes of individual SFAs and MI risk, for each cohort separately. Results: During median follow-up times of 18.8 years in EPIC-Norfolk and 13.6 years in Denmark, respectively, 1204 and 2260 MI events occurred. Mean (±SD) total SFA intake was 13.3 (±3.5) en% in EPIC-Norfolk, and 12.5 (±2.6) en% in EPIC-Denmark. After multivariable adjustment, intakes of C12:0 (lauric acid) and C14:0 (myristic acid) inversely associated with MI risk in EPIC-Denmark (HR upper versus lowest quintile: 0.80 (95%CI: 0.66, 0.96) for both SFAs). Intakes in the third and fourth quintiles of C4:0–C10:0 also associated with lower MI risk in EPIC-Denmark. Moreover, substitution of C16:0 (palmitic acid) and C18:0 (stearic acid) with plant proteins resulted in a reduction of MI risk in EPIC-Denmark (HR per 1 energy%: 0.86 (95%CI: 0.78, 0.95) and 0.87 (95%CI: 0.79, 0.96) respectively). No such associations were found in EPIC-Norfolk. Conclusion: The results from the present study suggest that the association between SFA and MI risk depends on the carbon chain-length of the SFA.",
keywords = "Cohort study, Diet, Myocardial infarction, Saturated fatty acids",
author = "Jaike Praagman and Vissers, {Linda E. T.} and Mulligan, {Angela A.} and Laursen, {Anne Sofie Dam} and Beulens, {Joline W. J.} and {van der Schouw}, {Yvonne T.} and Wareham, {Nicholas J.} and Hansen, {Camilla Plambeck} and Kay-Tee Khaw and Jakobsen, {Marianne Uhre} and Ivonne Sluijs",
doi = "https://doi.org/10.1016/j.ijcard.2018.10.064",
journal = "International Journal of Cardiology",
publisher = "Elsevier Ireland Ltd",
Praagman, J, Vissers, LET, Mulligan, AA, Laursen, ASD, Beulens, JWJ, van der Schouw, YT, Wareham, NJ, Hansen, CP, Khaw, K-T, Jakobsen, MU & Sluijs, I 2019, 'Consumption of individual saturated fatty acids and the risk of myocardial infarction in a UK and a Danish cohort', International Journal of Cardiology, vol. 279, pp. 18-26. https://doi.org/10.1016/j.ijcard.2018.10.064
Consumption of individual saturated fatty acids and the risk of myocardial infarction in a UK and a Danish cohort. / Praagman, Jaike; Vissers, Linda E. T.; Mulligan, Angela A. et al.
In: International Journal of Cardiology, Vol. 279, 15.03.2019, p. 18-26.
T1 - Consumption of individual saturated fatty acids and the risk of myocardial infarction in a UK and a Danish cohort
AU - Praagman, Jaike
AU - Vissers, Linda E. T.
AU - Mulligan, Angela A.
AU - Laursen, Anne Sofie Dam
AU - Beulens, Joline W. J.
AU - van der Schouw, Yvonne T.
AU - Wareham, Nicholas J.
AU - Hansen, Camilla Plambeck
AU - Khaw, Kay-Tee
AU - Jakobsen, Marianne Uhre
AU - Sluijs, Ivonne
N2 - Background: The effect of individual saturated fatty acids (SFAs) on serum cholesterol levels depends on their carbon-chain length. Whether the association with myocardial infarction (MI) also differs across individual SFAs is unclear. We examined the association between consumption of individual SFAs, differing in chain lengths ranging from 4 through 18 carbons, and risk of MI. Methods: We used data from 22,050 and 53,375 participants from EPIC-Norfolk (UK) and EPIC-Denmark, respectively. Baseline SFA intakes were assessed through validated, country-specific food frequency questionnaires. Cox regression analysis was used to estimate associations between intakes of individual SFAs and MI risk, for each cohort separately. Results: During median follow-up times of 18.8 years in EPIC-Norfolk and 13.6 years in Denmark, respectively, 1204 and 2260 MI events occurred. Mean (±SD) total SFA intake was 13.3 (±3.5) en% in EPIC-Norfolk, and 12.5 (±2.6) en% in EPIC-Denmark. After multivariable adjustment, intakes of C12:0 (lauric acid) and C14:0 (myristic acid) inversely associated with MI risk in EPIC-Denmark (HR upper versus lowest quintile: 0.80 (95%CI: 0.66, 0.96) for both SFAs). Intakes in the third and fourth quintiles of C4:0–C10:0 also associated with lower MI risk in EPIC-Denmark. Moreover, substitution of C16:0 (palmitic acid) and C18:0 (stearic acid) with plant proteins resulted in a reduction of MI risk in EPIC-Denmark (HR per 1 energy%: 0.86 (95%CI: 0.78, 0.95) and 0.87 (95%CI: 0.79, 0.96) respectively). No such associations were found in EPIC-Norfolk. Conclusion: The results from the present study suggest that the association between SFA and MI risk depends on the carbon chain-length of the SFA.
AB - Background: The effect of individual saturated fatty acids (SFAs) on serum cholesterol levels depends on their carbon-chain length. Whether the association with myocardial infarction (MI) also differs across individual SFAs is unclear. We examined the association between consumption of individual SFAs, differing in chain lengths ranging from 4 through 18 carbons, and risk of MI. Methods: We used data from 22,050 and 53,375 participants from EPIC-Norfolk (UK) and EPIC-Denmark, respectively. Baseline SFA intakes were assessed through validated, country-specific food frequency questionnaires. Cox regression analysis was used to estimate associations between intakes of individual SFAs and MI risk, for each cohort separately. Results: During median follow-up times of 18.8 years in EPIC-Norfolk and 13.6 years in Denmark, respectively, 1204 and 2260 MI events occurred. Mean (±SD) total SFA intake was 13.3 (±3.5) en% in EPIC-Norfolk, and 12.5 (±2.6) en% in EPIC-Denmark. After multivariable adjustment, intakes of C12:0 (lauric acid) and C14:0 (myristic acid) inversely associated with MI risk in EPIC-Denmark (HR upper versus lowest quintile: 0.80 (95%CI: 0.66, 0.96) for both SFAs). Intakes in the third and fourth quintiles of C4:0–C10:0 also associated with lower MI risk in EPIC-Denmark. Moreover, substitution of C16:0 (palmitic acid) and C18:0 (stearic acid) with plant proteins resulted in a reduction of MI risk in EPIC-Denmark (HR per 1 energy%: 0.86 (95%CI: 0.78, 0.95) and 0.87 (95%CI: 0.79, 0.96) respectively). No such associations were found in EPIC-Norfolk. Conclusion: The results from the present study suggest that the association between SFA and MI risk depends on the carbon chain-length of the SFA.
KW - Cohort study
KW - Diet
KW - Myocardial infarction
KW - Saturated fatty acids
UR - https://www.ncbi.nlm.nih.gov/pubmed/30482628
U2 - https://doi.org/10.1016/j.ijcard.2018.10.064
DO - https://doi.org/10.1016/j.ijcard.2018.10.064
JO - International Journal of Cardiology
JF - International Journal of Cardiology
Praagman J, Vissers LET, Mulligan AA, Laursen ASD, Beulens JWJ, van der Schouw YT et al. Consumption of individual saturated fatty acids and the risk of myocardial infarction in a UK and a Danish cohort. International Journal of Cardiology. 2019 Mar 15;279:18-26. Epub 2018. doi: https://doi.org/10.1016/j.ijcard.2018.10.064


text:
LiveKindly raises an additional $135M to expand to the US
Permission granted by LiveKindly Company
Jessi Devenyns
The multi-brand plant-based platform LiveKindly Collective raised $135 million in a round led by Blue Horizon Corporation, with participation from Trustbridge Partners, EQT and Griffith Foods.
Funds from this round will be used to increase capacity and accelerate a 2021 nationwide launch of its brands into the U.S. The collective will also work toward developing new products, with a particular focus on plant-based chicken and eggs.
This latest capital infusion brings the company's total funding to $335 million since it launched in March. The company, which is led by a former Unilever veteran, has since grown its foothold in the plant-based space with five international deals.
LiveKindly wants to take on the meat industry and the company said it is working toward that by first becoming "one of the highest funded plant-based food companies globally."
With $335 million secured, the collective is now working on expanding the brands it has acquired by increasing their distribution and market penetration. For its 2021 launch into the U.S., LiveKindly plans to bring over its recently acquired portfolio brands, including Fry Family Food Co., which is popular in South Africa; Germany-based LikeMeat; and Sweden-based Oumph.
Plant-based foods have surged in popularity during the pandemic, which has helped LiveKindly amid its launch. LiveKindly CEO Kees Kruythoff told The Wall Street Journal that there was increased demand for its Oumph and Fry Family brands since they are in the frozen section. He also told the Journal that the collective plans to continue expanding its portfolio by backing more acquisitions of companies and brands before the end of this year with this new round of capital.
LiveKindly is focusing on plant-based chicken alternatives. The company said earlier this year that because chicken is the world's most-consumed meat, it is a logical protein to offer consumers an alternative. The company also noted plant-based chicken can be made using a few simple ingredients, unlike the long ingredients list and processing needed for plant-based beef.
The simplicity of chicken production coupled with its popularity makes the protein an attractive target for manufacturers that have largely focused on beef until now. LiveKindly is looking to jump in and fill the demand by more widely rolling out Fry Family Food Co. stateside. The South African family-owned firm offers a variety of vegan chicken products.
In addition to chicken, LiveKindly's brands offer a lot of variety. Oumph sells meat analogue strips and "meat" topped pizzas, for example, while LikeMeat offers kebabs, chicken nuggets, chicken curry and schnitzel.
LiveKindly is also working to introduce its own products with a focus on price parity. According to the National Chicken Council, the retail price per pound of chicken was $1.87 per pound in 2019. Products like Beyond Meat chicken strips, which are now discontinued, retailed for $5.99 for a 9-ounce bag.
As it expands its brands and works on developing more products, LiveKindly has built a team of industry veterans from Nestlé, Unilever and Ferrero that will be an advantage for its launch into the U.S. market. These veterans are able to offer their experience and know-how to the smaller companies it acquires to help make the connections necessary to more rapidly expand these brands in a crowded marketplace with major players.
Business Wire LIVEKINDLY Collective Announces an Additional $135 Million Capital Raise Following a Successful Founders Round of $200 Million Earlier this Year
Filed Under: Corporate


text:
\section{Introduction}

Wave propagation through discrete structures remains an active area of research today.  The triangular lattice is one of the five two-dimensional Bravais lattice types and appears naturally in applications. For example, close-packed planes  occur frequently in some kinds of crystals in the form of triangular lattice \cite{BH54, Bu66}. Besides, one can consider two-dimensional passive propagation media, a host microstrip line network periodically loaded with series capacitors and shunt inductors for signal processing and filtering (as shown in Figure \ref{Fig_1}). This type of inductor-capacitor lattice is referred to a negative-refractive-index transmission-line (NRI-TL) metamaterial \cite{CI} or, simply, left-handed 2D metamaterial.

Motivated by applications of recent interest related to analog circuits, crystalline materials, and metamaterials, we investigate thin-slit diffraction problems for triangular lattices. The peculiar structure allows us to consider the problems on the semi-infinite triangular lattice. For this lattice, we study Dirichlet problems for the two-dimensional discrete Helmholtz equation in a half-plane (mathematically formulated in the next section). When one is interested in the analysis of regular processes in which waves corresponding to the microstructural scales can be neglected, then the continuum limit of corresponding equations can be investigated. In this case, we arrive at the famous problem in applied mathematics – the half-plane diffraction problem. It is well known that the classical continuum model of wave diffraction can be considered only as the slowly-varying approximation of a discrete or structured material. Nowadays the understanding of nanostructure and microstructure phenomena in modern materials and composites in critical conditions is of crucial importance. It turns out that a discrete structure provides an effective way to describe microstructural processes and influences within, cf., e.g., \cite{Br, CI, Do, Sl}. Therefore, there is an obvious necessity to avoid continuum limits and instead directly analyze the discrete Helmholtz diffraction problems. Although the similar problem for square lattice has been studied in \cite{BO}, see also \cite{DK3}, its extension to a triangular lattice model is not direct.

Our main interest lies in the investigation of the problems with wave number $k$ within the pass-band which is an arbitrary non-zero complex number in general. In this paper, we use the well approved approach to provide new results for a real wave number $k\in (0,3)\backslash\{2\sqrt{2}\}$ without passing to a complex wave number.
In order to have the unique solvability results and an exact representation formula for the solution, we use the notion of the radiating solution \cite{DK2} and the method of images to construct the Dirichlet Green's function for the half-plane. Notice that the cases with the wave numbers within the stop-band is mathematically more simple, they do not need the radiation conditions and is not considered here. Finally, we propose a method for numerical calculation. The efficiency of our approach is demonstrated in an example related to the propagation of wave fronts in metamaterials through two small openings.

\section{Basic notations and formulation of the problem}

Following the customary notation in mathematics, let $\bZ$, $\bZ^+$, $\bZ^-$, $\bN$, $\bR$, and $\bC$ denote the set of integers, positive integers, negative integers, non-negative integers, real numbers and complex numbers, respectively. We denote by $e_1=(1,0)$, $e_2=(0,1)$ the standard base of $\bZ^2$ ($=\bZ\times \bZ$).

Consider a two-dimensional infinite triangular lattice $\mathfrak{T}$ defined as a periodic simple graph $\{\mathcal{V}, \mathcal{E}\}$, where
\[
\mathcal{V}=\{(x_1+x_2/2,\sqrt{3}x_2/2)\subset \mathbb{R}^2 : (x_1,x_2)\in\mathbb{Z}^2\}
\]
is a vertex set, and $\mathcal{E}$ is an edge set, whose endpoints $(a,b)\in \mathcal{V}\times \mathcal{V}$ are adjacent points, i.e., $\mid a-b\mid=1$, cf. Figure \ref{Fig_1}. The time-harmonic discrete waves in $\mathfrak{T}$ can be described by solutions of the discrete Helmholtz equation

\begin{figure}[t!]
\subfigure{\includegraphics[scale=0.42]{Fig_1}}
\subfigure{\raisebox{18mm}{\includegraphics[scale=0.50]{Triangle_LC}}}
\caption{Triangular lattice. Connection between $x=(x_1,x_2)\in \bZ^2$ and the Euclidean coordinates of the vertexes (black dots) is established via  $(x_1,x_2) \to (x_1+x_2/2,\sqrt{3}x_2/2)$. The nearest neighbor interactions based on the triangular lattice are shown with thick blue lines. Triangular lattice can be viewed as a left-handed 2D inductor-capacitor metamaterial. A host transmission-line is loaded periodically with series capacitors and shunt inductors.}
\label{Fig_1}
\end{figure}

\begin{equation}\label{eq:Helmholtz}
(\Delta_d + k^2)u(x)=g(x), \quad x=(x_1,x_2)\in \bZ^2.
\end{equation}
Here, $\Delta_d$ denotes the discrete (a 7-point) Laplacian defined as follows
\begin{equation*}
\begin{aligned}
\Delta_d u(x)=&u(x+e_1)+u(x-e_1)+u(x+e_2)+u(x-e_2)\\
&+u(x+e_1-e_2)+u(x-e_1+e_2)-6u(x),
\end{aligned}
\end{equation*}
where $g$ is a function with bounded support, i.e., $g\in C_0(\bZ^2)$.

 Let $\Omega=\mathring{\Omega}\cup \Gamma$ be a discrete half-plane in $\bZ^2$, where
$\mathring{\Omega}:=\mathbb{Z}\times \mathbb{\bZ^+}$ and $\Gamma := \partial \Omega =\{(x_1,0) : x_1\in \mathbb{Z}\}$.
We consider the Dirichlet poblem for the discrete Helmholtz equation on the semi-infinite lattice:

\begin{subequations}
\begin{align}
(\Delta_d +k^2)u(x) &= 0,\quad\quad\  \ \textup{in}\  \mathring{\Omega}, \label{eq:H1} \\
u(y)&= f(y),\quad \textup{on}\  \Gamma. \label{eq:H2}
\end{align}
\end{subequations}
Here, $f$ is a given function supported on a finite subset of $\Gamma$.

Thus, we are interested in studying the problem of the existence and uniqueness of a function $u:\Omega\to \bC$ such that $u$ satisfies the discrete Helmoltz equation \eqref{eq:H1} with $k \in (0,3)\backslash \{2\sqrt{2}\}$ and the boundary condition \eqref{eq:H2}. From now on we will refer to this problem as Problem $\mathcal{P}_{\mathrm{D}}$.

\section{Lattice Green's function}

Denote by $\cG(x;y)$ the Green's function for the discrete Helmholtz equation \eqref{eq:Helmholtz} centered at $y$ and evaluated at $x$. Then the function $\cG(x;y)$ satisfies
\begin{equation}\label{eq:greens}
(\Delta_d+k^2)\cG(x;y)=\delta_{x,y},
\end{equation}
where $\delta_{x,y}$ is the Kronecker delta. For brevity, we use the notation $\cG(x)$ for $\cG(x;0)$. Notice that
$\cG(x;y)=\cG(x-y)$.

Using the discrete Fourier transform and the inverse Fourier transform
we get
\begin{equation}\label{eq:gmn}
\cG(x)=\frac{1}{4\pi^2}\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}\frac{e^{\iota(x\cdot \xi)}}{\sigma(\xi;k)}d\xi, \quad \xi=(\xi_1,\xi_2),
\end{equation}
where
\begin{equation}
\begin{aligned}
\sigma(\xi;k^2)&=e^{\iota\xi_1}+e^{-\iota\xi_1}+e^{\iota\xi_2}+e^{-\iota\xi_2}+e^{\iota\xi_1}e^{-\iota\xi_2}+e^{-\iota\xi_1}e^{\iota\xi_2}-6+k^2\\
&=k^2-6+2\cos\xi_1+2\cos\xi_2+2\cos(\xi_1-\xi_2).
\end{aligned}
\end{equation}

The lattice Green's function $\mathcal{G}$ is quite well known when $k^2\in \bC\backslash[0, 9]$ (cf., e.g., \cite{Ho}).
Notice that if $k^2\in \bC\backslash[0, 9]$ then $\sigma\neq 0$ and, consequently, $\mathcal{G}$ in \eqref{eq:gmn} is well defined. In this case
$\mathcal{G}(x)$ decays exponentially when $\mid x \mid\to\infty$. Moreover, we have
\begin{equation}\label{eq:lateq}
\mathcal{G}(x_1,x_2)=\mathcal{G}(x_2,x_1)=\mathcal{G}(-x_1,-x_2)=\mathcal{G}(x_1+x_2,-x_2)
\end{equation}
for all $x=(x_1,x_2) \in \bZ^2$.

Let us show $\mathcal{G}(x_1,x_2)=\mathcal{G}(x_1+x_2,-x_2)$ while other identities are trivial.
According to \eqref{eq:gmn}, we have
\begin{eqnarray*}
\mathcal{G}(x_1+x_2,-x_2) & = & \frac{1}{4\pi^2}\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}\frac{e^{\iota x_1\xi_1 +\iota x_2(\xi_1-\xi_2)}}{\sigma(\xi_1,\xi_2;k)}d\xi_1 d\xi_2 \\
& = &  \frac{1}{4\pi^2}\int_{-\pi}^{\pi}\int_{\eta_1-\pi}^{\eta_1+\pi}\frac{e^{\iota x_1\eta_1 +\iota x_2\eta_2}}{\sigma(\eta_1,\eta_2;k)}d\eta_2 d\eta_1
\end{eqnarray*}
with $\eta_1=\xi_1, \, \eta_2=\xi_1-\xi_2$.
The following equality can be easily obtained by changing the variable $\eta_2$ to $\eta_2-2\pi$
\[
\int_{0}^{\pi}\int_{\pi}^{\eta_1+\pi}\frac{e^{\iota x_1\eta_1 +\iota x_2\eta_2}}{\sigma(\eta_1,\eta_2;k)}d\eta_2 d\eta_1=
\int_{0}^{\pi}\int_{-\pi}^{\eta_1-\pi}\frac{e^{\iota x_1\eta_1 +\iota x_2\eta_2}}{\sigma(\eta_1,\eta_2;k)}d\eta_2 d\eta_1.
\]
 The factor $e^{\iota x_2 2\pi}$ is equal to $1$ since $x_2\in \mathbb{Z}$. Similar arguments give us
\[
\int_{-\pi}^{0}\int_{\eta_1-\pi}^{-\pi}\frac{e^{\iota x_1\eta_1 +\iota x_2\eta_2}}{\sigma(\eta_1,\eta_2;k)}d\eta_2 d\eta_1=
\int_{-\pi}^{0}\int_{\eta_1+\pi}^{\pi}\frac{e^{\iota x_1\eta_1 +\iota x_2\eta_2}}{\sigma(\eta_1,\eta_2;k)}d\eta_2 d\eta_1.
\]
Taking into account the last two equalities, we get
\[
\mathcal{G}(x_1+x_2,-x_2) = \frac{1}{4\pi^2}\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}\frac{e^{\iota x_1\eta_1 +\iota x_2\eta_2}}{\sigma(\eta_1,\eta_2;k)}d\eta_2 d\eta_1 = \mathcal{G}(x_1,x_2).
\]

In the case $k^2\in (0, 9)\backslash\{8\}$ the expression \eqref{eq:gmn} is understood as follows: we replace $k^2$ by $k^2+\iota\varepsilon$ with $0<\varepsilon\ll 1$ and let $\varepsilon\to 0$ at the end of the calculation, cf. \cite{DK2}. Thus, we define the lattice Green's function for $k\in (0, 3)\backslash\{2\sqrt{2}\}$ as a pointwise limit of
\[
(R_{\lambda+\iota\varepsilon}\delta_{x,0})(x):=\frac{1}{4\pi^2}\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}\frac{e^{\iota x\cdot\xi }d\xi_1 d\xi_2}{\sigma(\xi;k^2+\iota\varepsilon)}
\]
as $k^2+\iota\varepsilon \to k^2+\iota 0$ and denote it again by $\mathcal{G}(x)$, i.e., $\mathcal{G}(x)=(R_{\lambda+\iota 0}\delta_{x,0})(x) $. Clearly, $\mathcal{G}(x)$ is a solution to equation \eqref{eq:greens} and satisfies equalities  \eqref{eq:lateq}.

\section{Unique solvability result}

In order to simplify further arguments let us introduce following vectors:
\begin{equation*}
\begin{aligned}
e_1=(1,0),\quad e_2=(0,1),\quad e_3=e_1-e_2,\quad e_4=-e_1,\quad e_5=-e_2,\quad e_6=-e_3.
\end{aligned}
\end{equation*}
For any point $x\in\bZ^2$ we define the 6-neighbourhood $F^0_x$ as the set of points $\{x+e_j: j=1,...,6\}$ and the neighbourhood $F_x$ as $F^0_x\bigcup\{x\}$. We say that $R\subset \bZ^2$ is a region if there exist disjoint nonempty subsets $\Rint$ and $\partial R$ of $R$ such that
\begin{itemize}
\item[(a)] $R=\Rint\cup \partial R$,
\item[(b)] if $x\in \mathring{R}$ then $F_x\subset R$,
\item[(c)] if $x\in\partial R$ then there is at least one point $y\in F^0_x$ such that $y\in \Rint$.
\end{itemize}
Clearly, the subsets $\Rint$ and $\partial R$ are not defined uniquely by $R$, but henceforth, it will be always assumed that $\Rint$ and $\partial R$ are also given and fixed for a given region $R$ in $\bZ^2$. We also say that $x$ is an interior (boundary) point of $R$ if $x\in \Rint$ ($x\in \partial R$).

Denote by $(\partial R)_j$, $j=1,...,6$, a set of all boundary points $y\in\partial R$ such that $y-e_j\in \Rint$ and call it the sides of the boundary $R$. Note that a boundary point $x$ can simultaneously belong to all six sides of $R$. Thus, these sides may overlap each other. Clearly, $\partial R$ is the union of its six sides, $\partial R=\cup_{j=1}^6 (\partial R)_j$.

As it was already mentioned above, $y\in\partial R$ may be a point of intersection of several sides of $\partial R$. However, in our arguments presented below it will be always clear which side is needed to be considered. Under this condition, we define the discrete derivative in the outward normal direction $e_j$, $j=1,\dots,6$,
\[
Tu(y)=u(y)-u(y-e_j), \quad y\in(\partial R)_j.
\]

Let us introduce the following set $H_0=\{(0,0)\}$ and then define $H_{N}$, $N\in\mathbb{N}$, with the help of recurrence formula
\[
H_{N}:=\bigcup_{x\in H_{N-1}}F_{x}
\]
with $\mathring{H}_N:=H_{N-1}$ and $(\partial H)_{N}:=H_{N}\backslash \mathring{H}_N$.

Let $R$ be a finite region. Representing $R=\bigcup_{x\in \Rint}{F_x}$, we can easily derive a discrete analogue of Green's second identity
\begin{equation}\label{eq:GRsecond}
\sum_{x\in \Rint}(u(x)\Delta_d v(x)-v(x)\Delta_d u(x))=\sum_{y\in \partial R} (u(y)Tv(y)-v(y)Tu(y)).
\end{equation}

Now let us give a definition of a radiating solution on the discrete half-plane. First, we consider the case $k^2\in (0,8)$.
We say that $u:\Omega \to\bC$ satisfies the radiation condition at infinity if
\begin{equation}
\label{eq:radcond}
\left\{\begin{aligned} u(x)&=O(\mid x\mid^{-\frac12}),\\
u(x+e_j)&=e^{\iota \xi^*_j(\alpha,k)} u(x)+O(\mid x\mid^{-\frac32}),\quad j=1,2,
\end{aligned}
\right.
\end{equation}
with the remaining term decaying uniformly in  all directions $x/\mid x\mid$, where $x$ is characterized as $x_1=\mid x\mid\cos\alpha$,  $x_2=\mid x\mid\sin\alpha$, $0\le \alpha \le \pi$. Here, $\xi^*_j(\alpha,k)$ is the $j$th coordinate of the point $\xi^*(\alpha, k)$.

For $k^2\in (8,9)$ we say that $u:\Omega\to\bC$ satisfies the radiation condition at infinity if it can be represented as a sum of functions $u_1$ and $u_2$ such that each of them satisfies
\begin{equation}
\label{eq:radcond2}
\left\{\begin{aligned} u_i(x)&=O(\mid x\mid^{-\frac12}),\\
u_i(x+e_j)&=e^{\iota \xi^{*,i}_j(\alpha,k)} u_i(x)+O(\mid x\mid^{-\frac32}),\quad i,j=1,2,
\end{aligned}
\right.
\end{equation}
with the remaining term decaying uniformly in  all directions $x/\mid x\mid$, where $x$ is characterized as $x_1=\mid x\mid\cos\alpha$,  $x_2=\mid x\mid\sin\alpha$, $0\le \alpha \le \pi$. Here, $\xi^{*,i}_j(\alpha,k)$ is the $j$th coordinate of the point $\xi^{*,i}(\alpha, k)$, $i=1,2$.

It is shown in \cite{DK2} that $\mathcal{G}(x)$ satisfies the radiation conditions introduced above. Now we are ready to introduce a notion of a radiating solution to the discrete Helmholtz equation \eqref{eq:Helmholtz}.

\begin{definition} Let $k^2\in (0,8)$. A solution $u$ to the discrete Helmholtz equation \eqref{eq:H1} is called radiating if it satisfies the radiation condition \eqref{eq:radcond}.

Let $k^2\in (8,9)$. A solution $u$ to the discrete Helmholtz equation \eqref{eq:H1} is called radiating if it satisfies the radiation condition \eqref{eq:radcond2}.
\end{definition}

We represent the Dirichlet Green's function for the half-plane as follows:
\[
\mathcal{G}^+(x;y)=\mathcal{G}(x;y)-\mathcal{G}(\hat{x};y), \quad y\in \bZ^2,
\]
with $x=(x_1,x_2)\in \Omega$ and $\hat{x}=(x_1+x_2,-x_2)$. Notice that  $\mathcal{G}^+(x;y)$ can be represented equivalently as
\[
\mathcal{G}^+(x;y)=\mathcal{G}(x;y)-\mathcal{G}(x;\hat{y}).
\]
Indeed, using the property \eqref{eq:lateq}, we have
\begin{equation}\label{eq:Greenyhat}
\mathcal{G}(\hat{x};y)=\mathcal{G}(x_1+x_2-y_1,-x_2-y_2)=\mathcal{G}(x_1-y_1-y_2, x_2+y_2)=\mathcal{G}(x; \hat{y}).
\end{equation}
Since $\widehat{(y_1,0)}=(y_1,0)$, we immediately get $\mathcal{G}^+(x;y)=0$ for $y \in \Gamma$. Moreover,
\[
(\Delta_d+k^2)\mathcal{G}^+(x;y)=(\Delta_d+k^2)\mathcal{G}(x;y)-(\Delta_d+k^2)\mathcal{G}(\hat{x};y)=\delta_{x,y}-\delta_{\hat{x},y}=\delta_{x,y}
\]
for all $x\in \mathring{\Omega}$ and $y\in \Omega$. From \eqref{eq:Greenyhat} we see that $\mathcal{G}^+(x;y)$ satisfies the radiation condition.  Indeed,
for any fixed $y\in \Gamma$ the angle $\alpha$ from the radiation conditions \eqref{eq:radcond} and \eqref{eq:radcond2} is the same for $\mathcal{G}(x;y)=\mathcal{G}(x-y)$ and $\mathcal{G}(x;\hat{y})=\mathcal{G}(x-\hat{y})$ when $\mid x\mid \to \infty$.

\begin{theorem}\label{th:repr}
Let $k \in (0,3)\backslash \{2\sqrt{2}\}$, and $u$ be a given function $\Omega \to \bC$ that satisfies the radiation condition. The function $u\mid_{\Gamma}$ has a finite support. Then, for any point $x\in \mathring{\Omega}$, we have the following representation formula
\[
u(x)=\sum_{y\in \Gamma}\big(u(y)T\cG^+(x;y)-\cG^+(x;y)Tu(y)\big)+\sum_{y\in \mathring{\Omega}}\cG^+(x;y)(\Delta_d+k^2) u(y).
\]
In particular, if $u$ is a solution to the discrete Helmholtz equation
\[
(\Delta_d+k^2)u(x)=0\quad \textup{in}\  \mathring{\Omega},
\]
then
\begin{equation}\label{eq:repr}
u(x)=\sum_{y\in \Gamma}u(y)T\cG^+(x;y).
\end{equation}
\end{theorem}

\begin{proof}
Denote by $\tilde{u}$ the extension of $u$ to $\bZ^2$ by zeros, i.e,
$\tilde{u}(x)=u(x)$ if $x\in \Omega$ and $\tilde{u}(x)=0$ if $x\in \bZ\times\bZ^-$. Then, for any finite region $H_N$, $N\in\mathbb{N}$, we apply Green's second identity \eqref{eq:GRsecond} where we take $v(y)=\cG^+(x;y)$. Here, note that $\cG^+$ is a fundamental solution for $y\in \mathring{\Omega}$, and
$\tilde{u}(y)=0$ for $y\in \bZ\times\bZ^-$. Thus, $u(y)(\Delta_d+k^2) v(y)$ disappears in the following identity
\[
u(y)\Delta_d v(y)-v(y)\Delta_d u(y)=u(y)(\Delta_d+k^2) v(y)-v(y)(\Delta_d+k^2) u(y)
\]
as far as $y\in \bZ\times\bZ^-$. For $x\in \mathring{H}_N$, we get
\begin{align*}
\tilde{u}(x)&=\sum_{y\in \mathring{H}_N}\tilde{u}(y)\delta_{x,y}\\
       &=\sum_{y\in \partial H_N} \big(\tilde{u}(y)T\cG^+(x;y)-\cG^+(x;y)T\tilde{u}(y)\big)+\sum_{y\in \mathring{H}_N}(\cG^+(x;y)(\Delta_d+k^2) \tilde{u}(y)).
\end{align*}

Passing to the limit $N\to \infty$, we use exactly the same arguments as in the proof of Theorem 5.2 from \cite{DK2} and, consequently, we get
\[
\tilde{u}(x) = \sum_{y\in \bZ^2}\cG^+(x;y)(\Delta_d+k^2) \tilde{u}(y).
\]
For the function $u$, we obtain
\[
u(x)=\sum_{y\in \Gamma}\big(u(y)T\cG^+(x;y)-\cG^+(x;y)Tu(y)\big)+\sum_{y\in \mathring{\Omega}}\cG^+(x;y)(\Delta_d+k^2) u(y).
\]
Taking into the account that $\cG^+(x;y)=0$ when $y\in \Gamma$, for a solution $u$ to the discrete Helmholtz equation we get the following quality
\[
u(x)=\sum_{y\in \Gamma}u(y)T\cG^+(x;y), \quad x\in \mathring{\Omega}.
\]
\end{proof}

Now we are ready to prove the unique solvability result for the discrete Helmholtz equation on the semi-infinite triangular lattice.
\begin{theorem}
Let $k \in (0,3)\backslash \{2\sqrt{2}\} $ then the Problem $\mathcal{P}_{\mathrm{D}}$ has a unique radiating solution $u$ which
can be represented as
\begin{equation}\label{eq:HPsolution}
u(x)=\sum_{y\in \Gamma}(\delta_{x,y}-\mathcal{G}^+(x;y+e_2)-\mathcal{G}^+(x,y+e_2-e_1))f(y).
\end{equation}
\end{theorem}
\begin{proof} To prove the uniqueness result, it is sufficient to show that the corresponding homogeneous problem has only
the trivial solution. Let $u$ be a radiating solution to the homogeneous problem $\mathcal{P}_D$. Then Theorem \ref{th:repr} immediately implies $u(x)=0$ for all $x\in \mathring{\Omega}$.

To show the existence results, let us first check that $u(x)$ satisfies the boundary condition \eqref{eq:H2}. Since $\mathcal{G}^+(x;y)=0$ for all $x\in \Gamma$ and any $y$ we get
\[
u(x_1,0)=\sum_{y_1\in \mathbb{R}}\delta_{x_1,y_1}\delta_{0,0}f(y)=f(x_1).
\]
Now let us check that $u(x)$ satisfies the Helmholtz equation \eqref{eq:H1}. For $x_2>1$ all terms $(\Delta_d+k^2)\delta_{x,y}$,  $(\Delta_d+k^2)\mathcal{G}^+(x;y+e_2)=\delta_{x,y+e_2}$ and $(\Delta_d+k^2)\mathcal{G}^+(x;y+e_2-e_1)=\delta_{x,y+e_2-e_1}$ are equal to zero. Consequently, $(\Delta_d+k^2)u(x)=0$  for points $x=(x_1,x_2)$ with $x_2>1$.
It remains to consider the case $x_2=1$. By the direct calculation we have
\begin{eqnarray*}
  (\Delta_d+k^2)\delta_{x,y}f(y)&=&f(x_1)+f(x_1+1),\\
  (\Delta_d+k^2)\mathcal{G}^+(x;y+e_2)f(y)&=&f(x_1),\\
  (\Delta_d+k^2)\mathcal{G}^+(x;y+e_2-e_1)f(y)&=&f(x_1+1),
\end{eqnarray*}
and as a result $(\Delta_d+k^2)u(x)=0$. Thus, $u(x)$ is a solution to \eqref{eq:H1}. Since $\mathcal{G}^+$ satisfies the radiation condition and $f$ is supported on the finite subset of $\Gamma$ it follows that $u(x)$ is the unique radiating solution to the problem \eqref{eq:H1}, \eqref{eq:H2}.
\end{proof}

\section{Numerical results}

The main issue for numerical evaluation of the solution \eqref{eq:HPsolution} is to compute the lattice Green's function. For this purpose we apply  the method developed in \cite{BC}. Using 8-fold symmetry, we need only to compute the lattice Green's function $\cG(i,j)$ with $i\ge j \ge 0$. Following to \cite{BC}, let us introduce the vectors
$\cV_{2p}=(\cG(2p,0),\cG(2p-1,1),\dots,\cG(p,p))^\top$ and $\cV_{2p+1}=(\cG(2p+1,0),\cG(2p,1),\dots,\cG(p+1,p))^\top$ that collect all distinct Green's functions $\cG(i,j)$ with ``Manhattan distances" $\mid i \mid + \mid j \mid$ of $2p$ and $2p+1$, respectively. For any Manhattan distance larger than 1, equation
\[
(\Delta_d+k^2)\cG(x)=\delta_{x,0}
\]
can be written in the matrix form $\gamma_n(k)\cV_n=\alpha_n(k)\cV_{n-1}+\beta_n(k)\cV_{n+1}$
where $\alpha_n(k)$, $\beta_n(k)$ and $\gamma_n(k)$ are sparse matrices (cf., Appendix A). Notice that only the
dimensions of these matrices depend on $n$. It is shown in \cite{BC} that, for any $n\ge 1$, we have
\[
\cV_n=A_n(k)\cV_{n-1},
\]
where the matrices $A_n(k)$ are defined by the following recurrence formula
\[
A_n(k)=[\gamma_n(k)-\beta_n(k)A_{n+1}]^{-1}\alpha_n(k).
\]
They can be computed starting from a sufficiently large $N$ with $A_{N+1}(k)=0$. Here, it is worth mentioning, that for $k=2$ we need to choose a better ``initial guess" than $A_{N+1}(k)=0$, since in this case $\mathrm{det}\gamma_n(k)=0$ and the matrix $\gamma_n(k)-\beta_n(k)A_{n+1}$ is not invertible.

Once $A_n(k)$ are known, we have $\cV_n=A_n(k)\dots A_1(k)\cV_{0}$, where
$\cV_{0}=\cG(0,0)$. In particular, $\cV_1=\cG(1,0)=A_1(k)\cG(0,0)$ which, together with
$6\cG(1,0)-(6-k^2)\cG(0,0)=1$, gives $\cG(0,0)=1/[6A_1(k)-6+k^2]$. This completes the calculation of the Green's function using elementary operations and no integrals. Notice also one more important advantage of this method. The $A_n(k)$ matrices are calculated coming down from asymptotically large Manhattan distances.  As they are propagated towards smaller Manhattan distances, it definitely gives us the physical solution.

Finally, we demonstrate our theoretical and numerical approaches for the diffraction problem on triangular lattice. For this purpose we take $k=\sqrt{2}$ and consider the wave $U(x_1,x_2,t)=\exp(\iota x_2 \pi/3-\iota\omega t)$ on the semi-infinite lattice $\mathbb{Z}\times \mathbb{Z}^-$ which encounters an obstacle at $x_2=0$ with two small openings formed by four nodes $\{(-11,0),(-10,0), (10,0), (11,0)\}$. It is easy to check that $\exp(\iota x_2 \pi/3)$ satisfies the discrete Helmholtz equation. Thus, our goal is to evaluate numerically a radiating solution of Problem $\mathcal{P}_{\mathrm{D}}$ where, for the given data, we take $f(\pm 10)=f(\pm 11)=1$ and  $f(m)=0$, $m\in\bZ^+\backslash \{-11,-10,10,11\}$.

\begin{figure}[t!]
\subfigure[The density plot of $\mathrm{Re}\,\mathcal{G}$.]{\includegraphics[scale=0.45]{Green_1}}
\subfigure[The density plot of $\mathrm{Re}\,u$.]{\includegraphics[scale=0.45]{10_11_twopoints}}
\begin{center}
\subfigure[The graph of $\mathrm{Re}\,u(\cdot,25\sqrt{3})$.]{\includegraphics[scale=0.6]{2_sqrt2_50}}
\end{center}
\caption{Plots are represented in original coordinates of the triangular lattice $\mathfrak{T}$. $k=\sqrt{2}$.}
\label{Fig_2}
\end{figure}

The results of numerical evaluations are plotted in Figure \ref{Fig_2} in original coordinates of $\mathfrak{T}$, where (a) shows the density plots of the Green's function $\mathrm{\Re e}\,\mathcal{G}$, (b) shows the density plots of $\cos(\iota x_2 \pi/3)$ on the lower half-plane and
$\mathrm{\Re e}\,u$ for the case $k=\sqrt{2}$ on the upper half-plane, and (c) presents the graph of $\mathrm{Re}\,u(\cdot,25\sqrt{3})$.  Some key features of the numerical solution can be immediately observed, namely, as expected, the symmetry of $\mathrm{\Re e}\,u$ and contributions from the wavefront that create a variable intensity.

\section{Discussion}

In this paper, we have constructed the discrete scattering theory for the two-dimensional discrete Helmholtz equation with a real wave number $k\in (0,3)\backslash\{2\sqrt{2}\}$ for the semi-infinite triangular lattices. The main objective was to prove the unique solvability result and derive an exact formula for the solution to a Dirichlet problem. For simplicity we restricted ourselves to compact boundary data. For non-compact boundary data we should require some extra conditions at infinity, cf. \cite{DK3}.

Similarly to the continuum theory, we used the notion of the radiating solution for the continuous Helmholtz equation
\begin{equation}\label{eq:Helmc}
\Delta u(x)+ \kappa^2 u(x)=0,\quad \kappa\in \mathbb{R}\backslash \{0\},\  x\in \mathbb{R}^2.
\end{equation}
Recall that a solution $u$ to the equation \eqref{eq:Helmc} is called radiating if it satisfies the Sommerfeld radiation condition
\begin{equation}\label{eq:zomm}
  \frac{\partial}{\partial \mid x \mid}u(x) - \iota \mid\kappa\mid u(x)=o(\mid x \mid ^{-\frac12})
\end{equation}
for $\mid x \mid\to \infty$ uniformly in all directions $x/\mid x \mid$. There are evident similarities and differences between conditions \eqref{eq:radcond}, \eqref{eq:radcond2} and \eqref{eq:zomm}.
For example, we see the same decay at infinity, but, in contrast to \eqref{eq:zomm}, the second condition of \eqref{eq:radcond} or \eqref{eq:radcond2} is anisotropic: the factor $e^{\iota \xi^*_j(\alpha,k)}$ is not a constant. Moreover, it is well known that the radiating solutions $u$ to the continuous Helmholtz equation automatically satisfy the Sommerfeld's finiteness condition, however this fact is not shown for the discrete case so far. Therefore, the first condition of \eqref{eq:radcond} or \eqref{eq:radcond2} is included in our definition.

In the present paper, the problems under consideration have an infinite boundary. Within the continuum framework it is well-known that, in general, when the surface is unbounded, we cannot neglect the contribution of that surface waves at infinity. In this case, the Sommerfeld radiation condition is no longer appropriate and a proper modification is needed. To the best of the authors' knowledge, different radiation conditions are provided only for the half-plane (and locally perturbed half-plane) problems, cf. \cite{CW, CW1, CWZ, CWZ1, DMN, DMN1}, and finding radiation conditions for arbitrary wedge-shaped regions remains open. In case of the square lattice we have proposed sufficient conditions for the given boundary data at infinity, which ensures to have an unique radiation solution to the corresponding problem, cf. \cite{DK3}. Here, in order to avoid further technical difficulties, we restricted ourselves with compactly supported data on the boundary.

Finally, let us note that comparing the results of problems in the continuous framework and results obtained in the continuum limit of discrete problems deserves the high interest of scientists, however, it is beyond the scope of this paper.

\section*{Acknowledgments}

This work was supported by Shota Rustaveli National Science Foundation of Georgia (SRNSFG) [FR-21-301]

\noindent

\section*{Appendix A Sparse matrices}\label{secA1}

The sparse matrices $\alpha_n(k)$, $\beta_n(k)$ and $\gamma_n(k)$ are defined as follows:
if $n=2p$  then  $\alpha_{2p}(k)$ is a $(p+1)\times p$ matrix such that
$\alpha_{2p}(k)\mid_{i,i}=1$, $i=\overline{1,p}$,
$\alpha_{2p}(k)\mid_{i,i-1}=1$, $i=\overline{2,p}$,
while $\alpha_{2p}(k)\mid_{p+1,p}=2$,
and all other matrix elements are zero. The $\beta_{2p}(k)$ is a $(p+1)\times (p+1)$ matrix
such that $\beta_{2p}(k)\mid_{i,i}=1$, $i=\overline{1,p}$,
$\beta_{2p}(k)\mid_{i,i+1}=1$, $i=\overline{2,p}$,
while $\beta_{2p}(k)\mid_{p+1,p+1}=\beta_{2p}(k)\mid_{1,2}=2$, and all other matrix elements are zero.
The $\gamma_{2p}(k)$ is a $(p+1)\times (p+1)$ matrix such that $\gamma_{2p}(k)\mid_{i,i}=6-k^2$, $i=\overline{1,p+1}$,
$\gamma_{2p}(k)\mid_{i,i+1}=\gamma_{2p}(k)\mid_{i,i-1}=-1$, $i=\overline{2,p}$, and $\gamma_{2p}(k)\mid_{1,2}=\gamma_{2p}(k)\mid_{p+1,p}=-2$.

If $n=2p+1$ then $\alpha_{2p+1}(k)$ is a $(p+1)\times (p+1)$ matrix such that
$\alpha_{2p+1}(k)\mid_{i,i}=1$, $i=\overline{1,p+1}$,
$\alpha_{2p+1}(k)\mid_{i,i-1}=1$, $i=\overline{2,p+1}$,
and all other matrix elements are zero. The $\beta_{2p+1}(k)$ is a
$(p+1)\times (p+2)$ matrix such that $\beta_{2p+1}(k)\mid_{i,i}=1$, $i=\overline{1,p+1}$,
$\beta_{2p+1}(k)\mid_{i,i+1}=1$, $i=\overline{2,p+1}$,
while $\beta_{2p+1}(k)\mid_{1,2}=2$, and all other matrix elements are zero.
The $\gamma_{2p}(k)$ is a $(p+1)\times (p+1)$ matrix such that $\gamma_{2p+1}(k)\mid_{i,i}=6-k^2$, $i=\overline{1,p}$,
$\gamma_{2p+1}(k)\mid_{p+1,p+1}=5-k^2$, while $\gamma_{2p+1}(k)\mid_{i,i+1}=-1$, $i=\overline{2,p}$, $\gamma_{2p+1}(k)\mid_{i,i-1}=-1$, $i=\overline{2,p+1}$, and $\gamma_{2p+1}(k)\mid_{1,2}=\gamma_{2p+1}(k)\mid_{p+1,p}=-2$. Finally, $\gamma_1(k)$ is a $1\times 1$ matrixs with an element $4-k^2$.

redpajama_set_name: RedPajamaArXiv


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Two Jenn Air Range Stove Model FCG2042W Oven racks. OEM Jenn Air original parts Jenn Air Stove Model #. Jenn Air part number #. 23 wide x 17 1/4 inches deep. We will also be listing the knobs, and burner grates.
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The seller is "pinkelephantstoo" and is located in Melbourne, Florida. This item can be shipped to United States, to Canada, to United Kingdom, DK, RO, SK, BG, CZ, FI, HU, LV, LT, MT, EE, to Australia, GR, PT, CY, SI, to Japan, to China, SE, KR, ID, to Taiwan, TH, to Belgium, to France, to Hong Kong, to Ireland, to Netherlands, PL, to Spain, to Italy, to Germany, to Austria, RU, IL, to Mexico, to New Zealand, PH, SG, to Switzerland, NO, SA, UA, AE, QA, KW, BH, HR, MY.


Showing 1 - 15 of 22 articles tagged "David Benioff"
Game of Thrones began with so much character and foreshadowing, but it has devolved into a "dragon-delivery service"
The first of six Iron Thrones was spotted in Gloucestershire, England days after the contest was announced.
"We want people to love it," says co-creator D.
King Tommen's beloved cat, who was introduced in Season 4, was never seen again after his owner's death.
On Sunday, HBO released an "official tease" containing new footage, but not from the final season.
Game of Thrones creators explain why the final season is taking so long: "It's the biggest thing we've ever done"
David Benioff, Dan Weiss and Peter Dinklage recorded the Blu-ray commentary for the standout "Pickle Rick" episode.


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Posted byhdeuropixonline February 8, 2020 August 7, 2021 Posted inAction Movies, Crime movies, Drama movies, Horror movies, UncategorizedTags:best James Bond Movies Till 2020, daniel craig james bond movies, james bond latest movie on HDEeuropix, james bond movies list after 2000, new james bond movie 2020
The best James Bond Movies Till 2020
Now Download Full The Best James Bond HDeuropix Movies online in 720p or 1080p HD quality for free. Watch All James Bond movies for free On these website Until now
These are the top best James Bond Movies till 2020:
Casino Royale (2006): Casino royale keeps track of the beginning of James Bond's career. His first mission as "007" takes him to Le Chiffre, the banker of terrorists from around the world. To stop him, and dismantle the terrorist network, Bond must defeat Le Chiffre in a high stake's poker game at Casino Royale. At first, Bond is upset when a beautiful Treasury officer, Vesper Lynd, is assigned to deliver the best for the game and monitor the government's money. But, while Bond and Vesper survive a series of deadly attacks by Le Chiffre and his henchmen, a mutual attraction develops that leads them to greater dangers and events that will mark Bond's life forever. Download Latest Hollywood Movies Online on HDeuropix For Free with no sign up.
Skyfall (2012): The loyalty of James Bond (Daniel Craig), the best agent of the British secret services, for his superior M (Judi Dench) will be put to the test when episodes of her past return to torment her. At the same time, MI6 suffers an attack and 007 will have to locate and destroy the grave danger that villain Silva represents. To achieve this, he will have the help of Agent Eve (Naomie Harris). Now Watch Hollywood Afdah Movies 2020 Online free with no Registration.
The Spy Who Loved Me (1977): Russians and British feel equally threatened after the disappearance of two submarines belonging to their respective fleets, leaving the only microfilm as a trail. While the British assign agent 007 James Bond the mission to find out who is behind the kidnappings, the Russians send spy Anya Amasova, also known as Triple X agent. After a first meeting in which they fight for possession of the microfilm, both agents decide to collaborate and work together against the real enemy: Carl Stromberg. It is a multimillionaire navigation tycoon who plans to use nuclear submarines to destroy Humanity.
Goldeneyes (1995): The seventeenth part of the "Bond" was called the Golden Eye (Goldeneye). This time, the main role was played by Pierce Brosnan. Critics immediately identified him as the best James Bond. Spectacular fights, elegance, and laconicism of the hero do their job.
The action takes place in post-Soviet Russia. James Bond, armed with new tricks of the super-agent, flies to Russia to fulfill a dangerous mission. The criminal group seized control of the Goldeneye secret satellite complex. This poses a threat to the security of highly developed countries. The brilliant Pierce Brosnan with real English charm during the operation manages to seduce several attractive girls and, of course, save the world. Stream 2020 HDeuropix Movies Online without ads or cost
From Russia with love (1963): The theft of a device capable of deciphering complicated communications is jeopardizing important investigations regarding the Russian government. James Bond, Agent 007 of the British Secret Services at the service of His Majesty, will travel to the Soviet Union to find his target.
This is the second film of the well-known spy James Bond. Released in 1963, the tape was directed by writer and director Terence Young. Nominated in the Golden Globes for her soundtrack in 1965.
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Pet Fooled
Steve Jobs the Lost Interview
What Happened Before The Big Bang
The Nazis, A Warning From History. Episode 1
Our Secret Universe: The Hidden Life of the Cell
Fittest on Earth
March of the Penguins
Urban Jungles
Walking with Monsters
"Archaeology" Sort by
4000 Year Old Cold Case: The Body in the Bog
A 4000-year-old body is found preserved in an Irish peat bog, in Cashel, Ireland. To scientists and historians, it could offer brand new clues to solve an ancient mystery: the hundreds of bodies found mummified in the boglands of Northern Europe. Now, will Cashel Man help prove the theory these Irish victims were ancient kings? And what clues can the bog bodies of Europe offer to explain our ancestors' most macabre tradition - ritual murder? Meanwhile, that question could be answered by the bog itself.
Unlocking the Great Pyramid
When ancient architects completed construction on the Great Pyramid at Giza, they left behind the greatest riddle of the engineering world—how did builders lift limestone blocks weighing an average of two and a half tons 480 feet up onto the top of the Pyramid? For centuries, adventurers and Egyptologists have crawled through every passageway and chamber of the Pyramid, measuring and collecting data in an attempt to determine how it was built. For the first time, a revolutionary theory argues that the answer may be inside the Pyramid. Architect Jean-Pierre Houdin and Egyptologist Bob Brier use 3-D software to unlock the secret.
Did God Have a Wife
2011 Culture
Dr Francesca Stavrakopoulou asks whether the ancient Israelites believed in one God as the Bible claims. She puts the Bible text under the microscope, examining what the original Hebrew said, and explores archaeological sites in Syria and the Sinai which are shedding new light on the beliefs of the people of the Bible. Was the God of Abraham unique? Were the ancient Israelites polytheists? And is it all possible that God had another half?
Series: Bible's Buried Secrets
Magical Egypt
In this documentary we moved to Ancient Egypt to learn their vision of the world, life and death, gods and magic. For the ancient Egyptians, the Nile was the main connection between their lives and an infinite supply of provisions. But its waters hid an underworld of darkness and mystery for its first settlers. The Nile was one the spiritual paths that combined life and death, the real world and the underworld.
Ancient Aliens Debunked: Anunnaki
The word Anunnaki means "princely seed" or "princely blood". The idea is that the Anunnaki were direct creations of Anu, who was regarded as the father and king of the gods. Ancient Aliens: "In the ancient texts of Sumeria we have descriptions of these being descending from the sky called the Anunnaki. The term Anunnaki means 'those who from the heavens came."
Series: Ancient Aliens
The Nazis, A Warning From History
The Story of God
How Art Made the World
2006 Art
One Strange Rock
2018 Science
The Universe Season 8
Life In Cold Blood
2008 Nature


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Yalvigi is a city in Karnataka, India.

Cities and towns in Gadag district

redpajama_set_name: RedPajamaWikipedia


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