text:
# Obtaining the adjacency matrix of Cayley graphs

Is it possible to obtain the adjacency matrix of a Cayley graph of $$Z_3 \times Z_5$$? (Manually or by using a software like GAP).

Will there be a pattern for adjacency matrices of Cayley graphs for a particular type of groups considered (i.e. if we consider Cayley graphs of the groups $$Z_p \times Z_q$$, where p,q are distinct primes, will the adjacency matrices obtained for various choices of p and q be related to each other by some pattern)?

I know we obtain different Cayley graphs for different generating sets chosen to construct the Cayley graph. So if the adjacency matrix is difficult to be taken due to this reason please mention at least for one generating set chosen.

• I am very fond of doing the question by GAP. Maybe we have to arrange sort of certain codes for it. But, by calling simple codes in Maple, you can have the matrix very nice. Are you interested in doing it by Maple which is not a professional abstarct algebra tool? – Mikasa Sep 22 '18 at 20:09
• Yes I am. Please tell me how to. Thanks. – Buddhini Angelika Sep 23 '18 at 4:45
• I'm glad if I can identify a pattern as well (if there is) which will be good to express when you have any prime numbers as p and q. – Buddhini Angelika Sep 23 '18 at 4:46
• Typing ??cayley into GAP shows that the package GRAPE has a method called CayleyGraph which gives you the Cayley graph in some format (specified in the help file). Going from that to an adjacency matrix should be fairly easy. – Tobias Kildetoft Sep 23 '18 at 17:27

In GAP: With the function

AdjacencyMatrixCayleyGraph:=function(elms,gens)
local g,A,i,l;
l:=Length(elms);
A:=NullMat(l,l);
for i in [1..Length(elms)] do
for g in gens do
A[i][Position(elms,elms[i]*g)]:=1;
A[i][Position(elms,elms[i]/g)]:=1; # or -1 if digraph is wanted
od;
od;
return A;
end;

you can call AdjacencyMatrixCayleyGraph with a list of elements of the group and a list of generator.

For example, in the case of a cyclic generator of $$Z_3\times Z_5$$:

gap> g:=AbelianGroup([3,5]);
<pc group of size 15 with 2 generators>
[ [ 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1 ],
[ 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0 ],
[ 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0 ],
[ 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0 ],
[ 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0 ],
[ 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0 ],
[ 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0 ],
[ 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0 ],
[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0 ],
[ 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0 ],
[ 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1 ],
[ 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0 ],
[ 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0 ],
[ 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0 ],
[ 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0 ] ]

• This deserves much + votes. You won't believe how much that old windows friendly version of GAP has helped me to do research. I see what I wanted in screen, delete them and run the codes easily. Thanks for that. Also, as you told me long ago somewhere on this site, the new versions has not constructed as your version unfortunately. – Mikasa Sep 24 '18 at 8:02
• What is that Windows friendly version of GAP? Is it much older version than GAP 4.9.2? – Buddhini Angelika Sep 26 '18 at 11:14
• I also would like to try that since I'm a Windows user. – Buddhini Angelika Sep 26 '18 at 11:23
with(GroupTheory):
with(GraphTheory):
G := DirectProduct(CyclicGroup(3), CyclicGroup(5));
H := CayleyGraph(G):

So the result will be as follows:

Indeed, the output is great but I did like to know the way we could do it by GAP.

• Thanks a lot. Another small question, when giving the command CayleyGraph don't we have to think of the generating sets? Does maple choose a random generating set? – Buddhini Angelika Sep 23 '18 at 7:08
• Thanks for your good question! Honestly, I don't know how Maple does that. I tried to introduce cyclic groups as gens and relations but it didn't get me an exciting result. Maybe I sould be waiting for a programmer which know GAP professionally. – Mikasa Sep 23 '18 at 12:30
• @mrs You can add an option of the form “generators = […]” to the CayleyGraph command to specify a particular sequence of generators for the group. Otherwise, Maple uses the generators returned by “Generators(…)” applied to the group. – James Sep 24 '18 at 1:04
• @James :Thanks so much for notifying that. From years ago, I have thought good points from your side. – Mikasa Sep 24 '18 at 7:40
• Congratulations on you shiny new gold star in group theory!! – amWhy Sep 24 '18 at 22:02

format: plain


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# Why can't I solve for this second derivative?

Here's the equation I have to find the second derivative point for.

$$f(x)=\frac{x+2}{x^{\frac{1}{2}}}$$ $$f'(x) = \frac{x-2}{2x^{\frac{3}{2}}}$$

From here I then calculate the second derivative and set it equal to 0. But it doesn't work.. Take a look:

$$f''(x)=\frac{-x^\frac{3}{2} + 6x^{\frac{1}{2}}}{4x^{3}} = 0$$ FROM first DERIVATIVE TO second:

$$\frac{2x^{\frac{3}{2}}[x-2]'-((x-2)[2x^{\frac{3}{2}}]'}{4x^{3}}$$

$$\frac{2x^{\frac{3}{2}} -3x^{}\frac{3}{2}+6x^{\frac{1}{2}}}{4x^{3}}=0$$ $$-x^{\frac{3}{2}} + 6x^{\frac{1}{2}} = 0$$ $$(-x^{\frac{3}{2}})^{2} + (6x^{\frac{1}{2}})^{2}$$ $$x^{3} + 36x = 0$$ $$x (x^{2} + 36) = 0$$ $$x = 0 \text{ or } x^{2} = - 36 \text{ no solution..}$$

This doesn't seem right.. Yet I have no idea why. It's easier to write the function as a product but I want to solve it using the quotient rule.. What's going on?

• Never mind that. Apr 10 '16 at 21:00
• Why do you say that this "doesn't seem right"? Apr 10 '16 at 21:01
• Why should you expect to be able to solve it? Not every function has an inflection point. Take for example $\frac{1}{2}x^2$ whose second derivative is identically equal to $1$ and is never zero. Apr 10 '16 at 21:01
• Perhaps you could show your work in the quotient rule, your numerator looks suspicious to me. It looks almost like you forgot to multiply by the $-2$. Apr 10 '16 at 21:02
• Please include your work so that we can find where the error is. If you don't include your work, it's hard for us to help you without doing the problem. Apr 10 '16 at 21:06

We proceed via the quotient rule:

$$f''(x)=\frac{2x^{3/2}-(x-2)3x^{1/2}}{4x^3}=\frac{2x^{3/2}-3x^{3/2}+6x^{1/2}}{4x^3}=x^{1/2}\left(\frac{6-x}{4x^3}\right)=\frac{6-x}{4x^{5/2}}$$

We can set $f''(x)=0$ and find that $f''(6)=0$. To find the points of inflection, we simply need to test a point less than and greater than $x=6$.

• Hey Thanks! I got the exact same thing as you did. But what If I do not factor out the $x^{1/2}$ and instead cancel the numerator (the lower part of the fraction) and then do the following: $(-x^{\frac{3}{2}})^{2} + (6x^{\frac{1}{2}})^{2}$ Is this valid? Apr 10 '16 at 21:16
• In general, $a^2+b^2 \neq (a+b)^2$ Apr 11 '16 at 1:32

Not the answer that you expect, but a simple way to get the correct answer.

$$f(x)=x^{1/2}+2x^{-1/2},$$

$$f'(x)=\frac12x^{-1/2}-x^{-3/2},$$

$$f''(x)=-\frac14x^{-3/2}+\frac32x^{-5/2}.$$

Multiplying by $-4x^{-5/2}$, the inflection point is at $x-6=0$.


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# Generalized eigenvector

Generalized eigenvector

In linear algebra, a generalized eigenvector of a matrix "A" is a nonzero vector v, which has associated with it an eigenvalue λ having algebraic multiplicity "k" ≥1, satisfying

: $\left(A-lambda I\right)^kmathbf\left\{v\right\} = mathbf\left\{0\right\}.$

Ordinary eigenvectors are obtained for "k"=1.

For defective matrices

Generalized eigenvectors are needed to form a complete basis of a defective matrix, which is a matrix in which there are fewer linearly independent eigenvectors than eigenvalues. The generalized eigenvectors "do" form a complete basis, as follows from the Jordan form of a matrix.

In particular, suppose that an eigenvalue λ of a matrix "A" has a multiplicity "m" but only a single corresponding eigenvector $x_1$. We form a sequence of "m" generalized eigenvectors $x_1, x_2, ldots, x_m$ that satisfy:

:$\left(A - lambda I\right) x_k = x_\left\{k-1\right\} !$

for $k=1,ldots,m$, where we define $x_0 = 0$. It follows that:

:$\left(A - lambda I\right)^k x_k = 0. !$

The generalized eigenvectors are linearly independent, but are not determined uniquely by the above relations.

Other meanings of the term

* The usage of generalized eigenfunction differs from this; it is part of the theory of rigged Hilbert spaces, so that for a linear operator on a function space this may be something different.

* One can also use the term "generalized eigenvector" for an eigenvector of the "generalized eigenvalue problem"

: $Av = lambda B v.$

* defective matrix
* eigenvector
* Jordan form

Wikimedia Foundation. 2010.

### Look at other dictionaries:

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text:
# How to make mx+n=0 more complicated?

We all know how to solve the equation $mx+n=0$ ($m,n$ are constants).The answer is of course $-\dfrac nm$.

Then what if one day we forget about how to solve $ax+b=0$,but remember the quadratic formula for $ax^2+bx+c=0$?$x=\frac {-b \pm \sqrt {b^2-4ac}}{2a}$

I have 2 ways:

1.$mx+n=0,x(mx+n)=0$.Solve for $mx^2+nx+0=0$,place $a\rightarrow m,b\rightarrow n,c\rightarrow0$,we get $x=\frac{-n\pm\sqrt{n^2-4m\times0}}{2m}=\frac{-n\pm|n|}{2m}=\frac{-n\pm n}{2m}=0\text{ or }-\frac nm$ That $0$ is of course not a root.So the root is $-\dfrac nm$

2.Solve for $0x^2+mx+n=0$,place $a\rightarrow0,b\rightarrow m,c\rightarrow n$.

Wait,$a$ can't be $0$ ! Try using L'Hopital.

If $m>0$,then $\displaystyle\lim_{a\to 0}\frac{-m-\sqrt{m^2-4na}}{2a}$ doesn't exist. $\lim_{a\to 0}\frac{-m+\sqrt{m^2-4na}}{2a}=\lim_{a\to0}\frac{\dfrac{\partial}{\partial a}(-m+\sqrt{m^2-4na}) }{\dfrac{\partial}{\partial a}( 2a)}=\lim_{a\to0}\dfrac{\dfrac{-4n}{2\sqrt{m^2-4na}}}{2}=\frac{-n}{|m|}=-\frac nm$

If $m<0$,then $\displaystyle\lim_{a\to 0}\frac{-m+\sqrt{m^2-4na}}{2a}$ doesn't exist. $\lim_{a\to 0}\frac{-m-\sqrt{m^2-4na}}{2a}=\lim_{a\to0}\frac{\dfrac{\partial}{\partial a}(-m-\sqrt{m^2-4na}) }{\dfrac{\partial}{\partial a}( 2a)}=\lim_{a\to0}\dfrac{\dfrac{4n}{2\sqrt{m^2-4na}}}{2}=\frac{n}{|m|}=-\frac nm$

Can anyone think of other ways of solving $mx+n=0$?

Note by X X
2 years ago

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An alternate spin on your approach would be to start with $mx + n = 0$ and square both sides.

$m^2 x^2 + 2 m n x + n^2 = 0 \\ x = \frac{-2 m n \pm \sqrt{4 m^2 n^2 - 4 m^2 n^2}}{2 m^2} = -\frac{n}{m}$

The advantage of this is that we don't have to artificially neglect any roots

- 2 years ago

Good one!I haven't think of this

- 2 years ago

We could multiply both sides by $mx - n$. Then $(mx + n)(mx - n) = 0(mx - n)$ becomes $m^2x^2 - n^2 = 0$, and using the quadratic formula this solves to $x = \frac{-0 \pm \sqrt{0^2 - 4m^2(-n^2)}}{2m^2}$ or $x = \pm \frac{n}{m}$. Then checking for extraneous solutions, we would eliminate $x = \frac{n}{m}$ and keep $x = -\frac{n}{m}$.

- 2 years ago

Thank you.I haven't think of this one.

- 2 years ago

If we can remember how to graph, we could also just graph y = mx + n and find the x-intercept.

- 2 years ago

Ha Ha,that's true!

- 2 years ago


text:
# Product Profitability Analysis Outdoor Motor Sports Inc. manufactures and sells two styles of All...

Product Profitability Analysis

Outdoor Motor Sports Inc. manufactures and sells two styles of All Terrain Vehicles (ATVs), the 4-wheel drive (4WD), and 2-wheel drive (2WD) from a single manufacturing facility. The manufacturing facility operates at 100% of capacity. The following per unit information is available for the two products:

 4WD 2WD Sales price $6,800$4,400 Variable cost of goods sold 4,280 2,950 Manufacturing margin $2,520$1,450 Variable selling expenses 956 702 Contribution margin $1,564$748 Fixed expenses 740 300 Income from operations $824$448

In addition, the following sales unit volume information for the period is as follows:

 4WD 2WD Sales unit volume 2,900 2,100

Prepare a contribution margin by product report. Calculate the contribution margin ratio for each product as a whole percent.

 Outdoor Motor Sports Inc. Contribution Margin by Product 4WD 2WD    % %


text:
EXODIA
Search…
Codex
Using the website
Protocol Monolith
FAQ

# Why do we need EXODIA in the first place?

Dollar-pegged stablecoins have become an essential part of crypto due to their lack of volatility as compared to tokens such as Bitcoin and Ether. Users are comfortable with transacting using stablecoins knowing that they hold the same amount of purchasing power today vs. tomorrow. But this is a fallacy. The dollar is controlled by the US government and the Federal Reserve. This means a depreciation of dollar also means a depreciation of these stablecoins.
EXODIA aims to solve this by creating a free-floating reserve currency, EXOD, that is backed by a basket of assets. By focusing on supply growth rather than price appreciation, EXODIA hopes that EXOD can function as a currency that is able to hold its purchasing power regardless of market volatility.

# Is EXOD a stablecoin?

No, EXOD is not a stablecoin. Rather, EXOD aspires to become an algorithmic reserve currency backed by other decentralized assets. Similar to the idea of the gold standard, EXOD provides free floating value that its users can always fall back on, simply because of the fractional treasury reserves EXOD draws its intrinsic value from.

# EXOD is backed, not pegged.

Each EXOD is backed by 1 DAI, not pegged to it. Because the treasury backs every EXOD with at least 1 DAI, the DAO would step in and buy back and burn EXOD when it trades below 1 DAI. This has the effect of pushing EXOD price back up to 1 DAI. EXOD could always trade above 1 DAI because there is no upper limit imposed by the protocol. Think pegged == 1, while backed >= 1.
You might say that the EXOD floor price or intrinsic value is 1 DAI. We believe that the actual price will always be 1 DAI + premium, but in the end that is up to the market to decide.

# How does it work?

At a high level, EXODIA consists of its protocol managed treasury, protocol owned liquidity (POL), bond mechanism, and staking rewards that are designed to control supply expansion.
Bond sales generate profit for the protocol, and the treasury uses the profit to mint EXOD and distribute them to stakers. With liquidity bonds, the protocol is able to accumulate its own liquidity.

# What is the deal with (3,3) and (1,1)?

(3,3) is the idea that, if everyone cooperated in EXODIA, it would generate the greatest gain for everyone (from a game theory standpoint). Currently, there are three actions a user can take:
• Staking (+2)
• Bonding (+1)
• Selling (-2)
Staking and bonding are considered beneficial to the protocol, while selling is considered detrimental. Staking and selling will also cause a price move, while bonding does not (we consider buying EXOD from the market as a prerequisite of staking, thus causing a price move). If both actions are beneficial, the actor who moves price also gets half of the benefit (+1). If both actions are contradictory, the bad actor who moves price gets half of the benefit (+1), while the good actor who moves price gets half of the downside (-1). If both actions are detrimental, which implies both actors are selling, they both get half of the downside (-1).
Thus, given two actors, all scenarios of what they could do and the effect on the protocol are shown here:
• If we both stake (3, 3), it is the best thing for both of us and the protocol (3 + 3 = 6).
• If one of us stakes and the other one bonds, it is also great because staking takes EXOD off the market and put it into the protocol, while bonding provides liquidity and DAI for the treasury (3 + 1 = 4).
• When one of us sells, it diminishes effort of the other one who stakes or bonds (1 - 1 = 0).
• When we both sell, it creates the worst outcome for both of us and the protocol (-3 - 3 = -6).

# Why is PCV important?

As the protocol controls the funds in its treasury, EXOD can only be minted or burned by the DAO. This also guarantees that the DAO can always back 1 EXOD with 1 DAI. You can easily define the risk of your investment because you can be confident that the DAO will indefinitely buy EXOD below 1 DAI with the treasury assets until no one is left to sell. You can't trust the FED but you can trust the code.
As the protocol accumulates more PCV, more runway is guaranteed for the stakers. This means the stakers can be confident that the current staking APY can be sustained for a longer term because more funds are available in the treasury.

# Why is POL important?

EXODIA owns most of its liquidity thanks to its bond mechanism. This has several benefits:
• EXODIA does not have to pay out high farming rewards to incentivize liquidity providers a.k.a renting liquidity.
• EXODIA guarantees the market that the liquidity is always there to facilitate sell or buy transaction.
• By being the largest LP (liquidity provider), it earns most of the LP fees which represents another source of income to the treasury.
• All POL can be used to back EXOD. The LP tokens are marked down to their risk-free value for this purpose.

# Why is the market price of EXOD so volatile?

It is extremely important to understand how early in development the EXODIA protocol is. A large amount of discussion has centered around the current price and an expected stable value moving forward. The reality is that these characteristics are not yet determined. The network is currently tuned for expansion of EXOD supply, when paired with the staking, bonding, and yield mechanics of EXODIA, this results in a fair amount of volatility.
EXOD could trade at a very high price because the market is ready to pay a hefty premium to capture a percentage of the current market capitalization. However, the price of EXOD could also drop to a large degree if the market sentiment turns bearish. We would expect significant price volatility during our growth phase so please do your own research as to whether this project suits your own personal goals and investment timelines.

# What is the point of buying it now when EXOD trades at a very high premium?

When you buy and stake EXOD, you capture a percentage of the supply (market cap) which will remain close to a constant. This is because your staked EXOD balance also increases along with the circulating supply. The implication is that if you buy EXOD when the market cap is low, you would be capturing a larger percentage of the market cap.

# What is a rebase?

Rebase is a mechanism by which your staked EXOD balance increases automatically. When new EXOD are minted by the protocol, a large portion of it goes to the stakers. Because stakers only see staked EXOD balance instead of EXOD, the protocol utilizes the rebase mechanism to increase the staked EXOD balance so that 1 staked EXOD is always redeemable for 1 EXOD.

# What is reward yield?

Reward yield is the percentage by which your staked EXOD balance increases on the next epoch. It is also known as rebase rate. You can find this number on the EXODIA staking page.

# What is APY?

APY stands for annual percentage yield. It measures the real rate of return on your principal by taking into account the effect of compounding interest. In the case of EXODIA, your staked EXOD represents your principal, and the compound interest is added periodically on every epoch (28800 Fantom blocks, or around 8 hours) thanks to the rebase mechanism.
One interesting fact about APY is that your balance will grow not linearly but exponentially over time! Assuming a daily compound interest of 2%, if you start with a balance of 1 EXOD on day 1, after a year, your balance will grow to about 1377. That is huge growth!

# How is the APY calculated?

The APY is calculated from the reward yield (a.k.a rebase rate) using the following equation:
$APY = ( 1 + rewardYield )^{1095}$
It raises to the power of 1095 because a rebase happens 3 times daily. Consider there are 365 days in a year, this would give a rebase frequency of 365 * 3 = 1095.
Reward yield is determined by the following equation:
$rewardYield = EXOD_{distributed} / EXOD_{totalStaked}$
The number of EXOD distributed to the staking contract is calculated from EXOD total supply using the following equation:
$EXOD_{distributed} = EXOD_{totalSupply} \times rewardRate$

# Why does the price of EXOD become irrelevant in the long term?

As illustrated above, your EXOD balance will grow exponentially over time thanks to the power of compounding. Let's say you buy an EXOD for $400 now and the market decides that in 1 year time, the intrinsic value of EXOD will be$2. Assuming a daily compound interest rate of 2%, your balance would grow to about 1377 EXODs by the end of the year, which is worth around $2754. That is a cool$2354 profit! By now, you should understand that you are paying a premium for EXOD now in exchange for a long-term benefit. Thus, you should have a long time horizon to allow your EXOD balance to grow exponentially and make this a worthwhile investment.

# What will EXOD's intrinsic value be in the future?

There is no clear answer for this, but the intrinsic value can be determined by the treasury performance. For example, if the treasury could guarantee to back every EXOD with 100 DAI, the intrinsic value will be 100 DAI.

# How does the protocol manage to maintain the high staking APY?

Let’s say the protocol targets an APY of 100,000%. This would translate to a rebase rate of about 0.6328%, or a daily growth of about 2%. Please refer to the equation above to learn how APY is calculated from the rebase rate.
If there are 100,000 of EXOD staked right now, the protocol would need to mint an additional 2000 EXOD to achieve this daily growth. This is achievable if the protocol can bring in at least 2000 DAI daily from bond sales. If the protocol fails to achieve this, the APY of 100,000% cannot be guaranteed.

# Do I have to unstake and stake EXOD on every epoch to get my rebase rewards?

No. Once you have staked EXOD with EXODIA, your staked EXOD balance will auto-compound on every epoch. That increase in balance represents your rebase rewards.

# What will happen if there is a bank run on Exodia?

We're using OHM as an example, as Dune analytics is not available for Fantom.
Fractional reserve banking works because depositors don’t withdraw their funds all at once. A depositor’s faith in the banking system rests on regulations and agencies like Federal Deposit Insurance Corporation (FDIC).
OHM does not have FDIC insurance but it has an incentive structure that protects stakers. Let’s take a look at how it performs during a hypothetical bank run. In this scenario, we assume the majority of stakers would panic and unstake their tokens from Olympus - the staking percentage which stands at 92% now quickly collapses to 3.3%, leaving only 55,000 OHM staked.
Next, we assume the Risk-Free Value (RFV) inflows to the treasury completely dry up. For context, RFV is currently growing at about 1$million every 2 days (source from Olympus queries). However, during a bank run this growth will likely stop. Finally, we assume that those last standing stakers bought in at a price of$500 per OHM. The initial investment of these stakers would be:
As of September 15 2021, the total OHM supply is 2,082,553 and the RFV is $47,041,833. Remember that 1 OHM is backed by 1 USD (DAI or FRAX). By subtracting these two numbers, we know 44,959,280 OHM will eventually get issued to the remaining stakers. In roughly a year, these stakers who are holding 55,000 OHM will have: $55000+44959280 = 45014280$$27.5 million investment made by these stakers will turn into about \$45 million based on cash flow alone if they stay staked (recall that 1 OHM is backed by 1 USD). In this bank run scenario, the stakers who stay staked not only get their money back, but also make some profit. Therefore, (3,3) isn’t just a popular meme, it is actually a dominant strategy.
The above scenario is unlikely to play out because when other people find out that extremely high rewards are being paid to the stakers, they will copy the strategy by buying and staking OHM. This is also why the percentage of OHM staked in Olympus has consistently remained over 90% since launch.
Edit on GitHub
Why do we need EXODIA in the first place?
Is EXOD a stablecoin?
EXOD is backed, not pegged.
How does it work?
What is the deal with (3,3) and (1,1)?
Why is PCV important?
Why is POL important?
Why is the market price of EXOD so volatile?
What is a rebase?
What is reward yield?
What is APY?
How is the APY calculated?
Why does the price of EXOD become irrelevant in the long term?
What will EXOD's intrinsic value be in the future?
How does the protocol manage to maintain the high staking APY?
Do I have to unstake and stake EXOD on every epoch to get my rebase rewards?
What will happen if there is a bank run on Exodia?


text:
# Internal Group Direct Product Isomorphism

## Theorem

Let $G$ be a group.

Let $H_1, H_2$ be subgroups of $G$.

Let $\phi: H_1 \times H_2 \to G$ be the mapping defined by $\map \phi {h_1, h_2} := h_1 h_2$.

If $\phi$ is a (group) isomorphism, then both $H_1$ and $H_2$ are normal subgroups of $G$.

## Proof

$\phi$ is an isomorphism, so in particular a (group) homomorphism.

Thus by Induced Group Product is Homomorphism iff Commutative, every element of $H_1$ commutes with every element of $H_2$.

Now suppose $a \in G$.

As $\phi$ is an isomorphism, it follows that $\phi$ is surjective.

$\exists h_1 \in H_1, h_2 \in H_2: a = h_1 h_2$

Now any element of $a H_1 a^{-1}$ is in the form $a h a^{-1}$ for some $h \in H_1$.

Thus:

 $\displaystyle a h a^{-1}$ $=$ $\displaystyle \paren {h_1 h_2} h \paren {h_1 h_2}^{-1}$ $\displaystyle$ $=$ $\displaystyle h_1 h_2 h h_2^{-1} h_1^{-1}$ $\displaystyle$ $=$ $\displaystyle h_1 h h_2 h_2^{-1} h_1^{-1}$ $h \in H_1$ and $h_2 \in H_2$ commute $\displaystyle$ $=$ $\displaystyle h_1 h h_1^{-1} \in H_1$

Thus $a H_1 a^{-1} \subseteq H_1$, and $H_1$ is therefore a normal subgroup of $G$.

Similarly, any element of $a H_2 a^{-1}$ is in the form $a h a^{-1}$ for some $h \in H_2$.

Thus:

 $\displaystyle a h a^{-1}$ $=$ $\displaystyle \paren {h_1 h_2} h \paren {h_1 h_2}^{-1}$ $\displaystyle$ $=$ $\displaystyle h_1 h_2 h h_2^{-1} h_1^{-1}$ $\displaystyle$ $=$ $\displaystyle h_1 h' h_1^{-1}$ as $h' = h_2 h h_2^{-1} \in H_2$ $\displaystyle$ $=$ $\displaystyle h' h_1 h_1^{-1}$ $h_1 \in H_1$ and $h' \in H_2$ commute $\displaystyle$ $=$ $\displaystyle h' \in H_2$

Thus $a H_2 a^{-1} \subseteq H_2$ and $H_2$ is normal as well.

$\blacksquare$


text:
# Article

Short Communications
2005. V. 43. № 5. P. 796–799
Sagintaeva Zh.I., Kasenova Sh.B., Edil'baeva S.T., Davrenbekov S.Zh., Kasenov B.K., Mustafin E.S., Zhumadilov E.K.
The Heat Capacity and Electrophysical Properties of Neodymium and Lithium Chromite $\text{NdLiCr}_2\text{O}_5$
Annotation
The method of dynamic calorimetry is used to investigate the heat capacity of chromite $\text{NdLiCr}_2\text{O}_5$ in the temperature range from $298$ to $673$ K. A $\lambda$-like peak is observed at $523$ K.The equations of temperature dependence of the heat capacity of this chromite are derived.The temperature dependences of the thermodynamic functions are calculated.The results of electrophysical investigations indicate that this compound may be classed with semiconductors.
Article reference:
Sagintaeva Zh.I., Kasenova Sh.B., Edil'baeva S.T., Davrenbekov S.Zh., Kasenov B.K., Mustafin E.S., Zhumadilov E.K. The Heat Capacity and Electrophysical Properties of Neodymium and Lithium Chromite $\text{NdLiCr}_2\text{O}_5$, High Temp., 2005. V. 43. № 5. P. 796


text:
Why doesn't octahedral d8 system exhibit Jahn-Teller effect?

1. Chemistry LibreTexts stated clearly that (under the section 'Large $$\Delta$$' and 'Small $$\Delta$$') octahedral d$$^8$$ system does not exhibit Jahn-Teller distortions. But then it stated also:

...Some common examples (that exhibit distortions) include Cr$$^{3+}$$, Co$$^{3+}$$, and Ni$$^{2+}$$

Isn't Ni$$^{2+}$$ d$$^8$$(Group 10, +2, so 10 - 2 = 8) and should not exhibit distortions?

1. I believe this question is asking the similar thing. But then when the energy level diagrams drawn out as shown below, I saw that even for d$$^8$$, if singlet favoured over triplet, the distortion does lower the energy (and so should happen).

So what is wrong with the two arguments above?

• Your distortion doesn’t preserve spin. As drawn, you should have one electron in z2 and one in x2-y2. Something is wrong with the sentence that you quoted. Perhaps they meant Ni(III), which does exhibit JT distortions. I also don’t know where they got Co(III) from. Co(III) is nearly always low spin and hence no JT distortion. Likewise, Cr(III) is d3, which should not have JT distortions. – orthocresol Dec 10 '20 at 11:44
• Having looked at the paragraph in more detail, I think it is just poorly written and these are supposed to be examples of ions which don’t exhibit JT distortions, rather than ions which do. They literally picked out d3, d6, and d8 examples. – orthocresol Dec 10 '20 at 11:48
• @orthocresol I just remember the spin must be preserved! But based on that, does that mean that, all d8 axial elongated tetragonal complexes will be paramagnetic (as total spin cannot be 0)? – TheLearner Dec 10 '20 at 12:03
• You can put the electron in the lower orbital, thus getting a singlet state, basically what you've drawn. The only question is whether it's more stable than the alternative, which is to have a triplet state with one electron in $z^2$ and $x^2-y^2$. That would depend on the classical considerations of pairing energy and orbital energy gap: for example, in square planar complexes, the $z^2$ and $x^2-y^2$ are so far apart that both electrons sit inside the $z^2$. Thus, for example, $\ce{[Ni(CN)4]^2-}$ is diamagnetic. But that is no longer the original topic of Jahn–Teller distortions. :-) – orthocresol Dec 10 '20 at 12:13
• I know I shouldn't say thank you here, but I still want to thank you to @orthocresol for often answering my question so pertinently. – TheLearner Dec 10 '20 at 23:33


text:
# Pre Calculus

Let $\bold{w} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}$. There exists a $2 \times 2$ matrix $\bold{P}$ such that
$\text{proj}_{\bold{w}} \bold{v} = \bold{P} \bold{v}$
for all 2-dimensional vectors $\bold{v}$. Find $\bold{P}$.

1. 👍 0
2. 👎 0
3. 👁 260
1. can you skip the LaTex? It appears that you are saying that

w = (2,1) and there is a 2x2 P such that
w•v = Pv

or something like that. Correct?

1. 👍 0
2. 👎 0
posted by Steve

## Similar Questions

Find the $2 \times 2$ matrix $\bold{A}$ such that $\bold{A} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ and $\bold{A} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -7 \\ 4 asked by Zheng on February 25, 2017 2. ### Pre Calculus For a vector \bold{v}, let \bold{r} be the reflection of \bold{v} over the line \[\bold{x} = t \begin{pmatrix} 2 \\ -1 \end{pmatrix}.$ [asy] unitsize(1 cm); pair O, V, R; O = (0,0); V = (3,1); R = reflect(O,(2,-1))*(V);

asked by Bob on September 20, 2017

Find the $2 \times 2$ matrix $\bold{A}$ such that $\bold{A} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ and $\bold{A} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -7 \\ 4 asked by Zheng on February 22, 2017 4. ### Math Find the 2 \times 2 matrix \bold{A} such that \[\bold{A} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ and $\bold{A} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -7 \\ 4 asked by Zheng on February 23, 2017 5. ### Precalculus Compute the distance between the parallel lines given by \[\begin{pmatrix} 1 \\ 4 \end{pmatrix} + t \begin{pmatrix} 4 \\ 3 \end{pmatrix}$ and $\begin{pmatrix} -5 \\ 6 \end{pmatrix} + s \begin{pmatrix} 4 \\ 3 \end{pmatrix}.$

asked by Zheng on February 13, 2017
6. ### Precalculus

The vector $\begin{pmatrix} k \\ 2 \end{pmatrix}$ is orthogonal to the vector $\begin{pmatrix} 3 \\ 5 \end{pmatrix}$. Find $k$. I thought the answer should be 10/4/

asked by Zheng on February 8, 2017
7. ### Precalculus

Latex: The vector $\begin{pmatrix} k \\ 2 \end{pmatrix}$ is orthogonal to the vector $\begin{pmatrix} 3 \\ 5 \end{pmatrix}$. Find $k$. Regular: The vector , is orthogonal to the vector . Find k. I can't seem to figure it out, I

asked by Zheng on February 9, 2017
8. ### Pre Calculus

Find all $2 \times 2$ matrices $\bold{A}$ that have the property that for any $2 \times 2$ matrix $\bold{B}$, $\bold{A} \bold{B} = \bold{B} \bold{A}.$

asked by Bob on September 20, 2017
9. ### Algebra

Round each number to the place value indicated by the digit in bold. A: 3587 5 is bold. B: 148213 the first one, one is bold. C: 23785 3 is bold. D: 2357 5 is bold.

asked by Kali G on September 10, 2014
10. ### Language Arts

Directions: replace the bold faced word with a contraction. 1. We will(bold faced) see you t the English class next week! Answer: We'll 2. There is (bold faced) no milk in the refrigerator. Answer: There's 3. There will (bold

asked by Patrick on March 2, 2016

More Similar Questions


text:
First convert them to fractions (the answer choices are fractions): a) 0.333... = 1/3 b) 0.555... = 5/9 Now add a and b: 1/3 (or 3/9 for a common denominator) + 5/9 = 8/9 That fraction is in simplest form, so leave it as it is. Hope this helps!

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text:
e02 Chapter Contents
e02 Chapter Introduction
NAG C Library Manual

# NAG Library Function Documentnag_2d_cheb_eval (e02cbc)

## 1  Purpose

nag_2d_cheb_eval (e02cbc) evaluates a bivariate polynomial from the rectangular array of coefficients in its double Chebyshev series representation.

## 2  Specification

 #include #include
 void nag_2d_cheb_eval (Integer mfirst, Integer mlast, Integer k, Integer l, const double x[], double xmin, double xmax, double y, double ymin, double ymax, double ff[], const double a[], NagError *fail)

## 3  Description

This function evaluates a bivariate polynomial (represented in double Chebyshev form) of degree $k$ in one variable, $\stackrel{-}{x}$, and degree $l$ in the other, $\stackrel{-}{y}$. The range of both variables is $-1$ to $+1$. However, these normalized variables will usually have been derived (as when the polynomial has been computed by nag_2d_cheb_fit_lines (e02cac), for example) from your original variables $x$ and $y$ by the transformations
 $x-=2x-xmax+xmin xmax-xmin and y-=2y-ymax+ymin ymax-ymin .$
(Here ${x}_{\mathrm{min}}$ and ${x}_{\mathrm{max}}$ are the ends of the range of $x$ which has been transformed to the range $-1$ to $+1$ of $\stackrel{-}{x}$. ${y}_{\mathrm{min}}$ and ${y}_{\mathrm{max}}$ are correspondingly for $y$. See Section 8). For this reason, the function has been designed to accept values of $x$ and $y$ rather than $\stackrel{-}{x}$ and $\stackrel{-}{y}$, and so requires values of ${x}_{\mathrm{min}}$, etc. to be supplied by you. In fact, for the sake of efficiency in appropriate cases, the function evaluates the polynomial for a sequence of values of $x$, all associated with the same value of $y$.
The double Chebyshev series can be written as
 $∑i=0k∑j=0laijTix-Tjy-,$
where ${T}_{i}\left(\stackrel{-}{x}\right)$ is the Chebyshev polynomial of the first kind of degree $i$ and argument $\stackrel{-}{x}$, and ${T}_{j}\left(\stackrel{-}{y}\right)$ is similarly defined. However the standard convention, followed in this function, is that coefficients in the above expression which have either $i$ or $j$ zero are written $\frac{1}{2}{a}_{ij}$, instead of simply ${a}_{ij}$, and the coefficient with both $i$ and $j$ zero is written $\frac{1}{4}{a}_{0,0}$.
The function first forms ${c}_{i}=\sum _{j=0}^{l}{a}_{ij}{T}_{j}\left(\stackrel{-}{y}\right)$, with ${a}_{i,0}$ replaced by $\frac{1}{2}{a}_{i,0}$, for each of $i=0,1,\dots ,k$. The value of the double series is then obtained for each value of $x$, by summing ${c}_{i}×{T}_{i}\left(\stackrel{-}{x}\right)$, with ${c}_{0}$ replaced by $\frac{1}{2}{c}_{0}$, over $i=0,1,\dots ,k$. The Clenshaw three term recurrence (see Clenshaw (1955)) with modifications due to Reinsch and Gentleman (1969) is used to form the sums.

## 4  References

Clenshaw C W (1955) A note on the summation of Chebyshev series Math. Tables Aids Comput. 9 118–120
Gentleman W M (1969) An error analysis of Goertzel's (Watt's) method for computing Fourier coefficients Comput. J. 12 160–165

## 5  Arguments

1:     mfirstIntegerInput
2:     mlastIntegerInput
On entry: the index of the first and last $x$ value in the array $x$ at which the evaluation is required respectively (see Section 8).
Constraint: ${\mathbf{mlast}}\ge {\mathbf{mfirst}}$.
3:     kIntegerInput
4:     lIntegerInput
On entry: the degree $k$ of $x$ and $l$ of $y$, respectively, in the polynomial.
Constraint: ${\mathbf{k}}\ge 0$ and ${\mathbf{l}}\ge 0$.
5:     x[mlast]const doubleInput
On entry: ${\mathbf{x}}\left[\mathit{i}-1\right]$, for $\mathit{i}={\mathbf{mfirst}},\dots ,{\mathbf{mlast}}$, must contain the $x$ values at which the evaluation is required.
Constraint: ${\mathbf{xmin}}\le {\mathbf{x}}\left[i-1\right]\le {\mathbf{xmax}}$, for all $i$.
6:     xmindoubleInput
7:     xmaxdoubleInput
On entry: the lower and upper ends, ${x}_{\mathrm{min}}$ and ${x}_{\mathrm{max}}$, of the range of the variable $x$ (see Section 3).
The values of xmin and xmax may depend on the value of $y$ (e.g., when the polynomial has been derived using nag_2d_cheb_fit_lines (e02cac)).
Constraint: ${\mathbf{xmax}}>{\mathbf{xmin}}$.
8:     ydoubleInput
On entry: the value of the $y$ coordinate of all the points at which the evaluation is required.
Constraint: ${\mathbf{ymin}}\le {\mathbf{y}}\le {\mathbf{ymax}}$.
9:     ymindoubleInput
10:   ymaxdoubleInput
On entry: the lower and upper ends, ${y}_{\mathrm{min}}$ and ${y}_{\mathrm{max}}$, of the range of the variable $y$ (see Section 3).
Constraint: ${\mathbf{ymax}}>{\mathbf{ymin}}$.
11:   ff[mlast]doubleOutput
On exit: ${\mathbf{ff}}\left[\mathit{i}-1\right]$ gives the value of the polynomial at the point $\left({x}_{\mathit{i}},y\right)$, for $\mathit{i}={\mathbf{mfirst}},\dots ,{\mathbf{mlast}}$.
12:   a[$\mathit{dim}$]const doubleInput
Note: the dimension, dim, of the array a must be at least $\left(\left({\mathbf{k}}+1\right)×\left({\mathbf{l}}+1\right)\right)$.
On entry: the Chebyshev coefficients of the polynomial. The coefficient ${a}_{ij}$ defined according to the standard convention (see Section 3) must be in ${\mathbf{a}}\left[i×\left(l+1\right)+j\right]$.
13:   failNagError *Input/Output
The NAG error argument (see Section 3.6 in the Essential Introduction).

## 6  Error Indicators and Warnings

NE_ALLOC_FAIL
Dynamic memory allocation failed.
On entry, argument $〈\mathit{\text{value}}〉$ had an illegal value.
NE_INT
On entry, ${\mathbf{k}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{k}}\ge 0$.
On entry, ${\mathbf{l}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{l}}\ge 0$.
NE_INT_2
On entry, ${\mathbf{k}}=〈\mathit{\text{value}}〉$ and ${\mathbf{l}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{k}}\ge 0$ and ${\mathbf{l}}\ge 0$.
On entry, ${\mathbf{mfirst}}=〈\mathit{\text{value}}〉$ and ${\mathbf{mlast}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{mfirst}}\le {\mathbf{mlast}}$.
NE_INTERNAL_ERROR
An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance.
Unexpected failure in internal call to nag_1d_cheb_eval (e02aec).
NE_REAL_2
On entry, ${\mathbf{xmin}}=〈\mathit{\text{value}}〉$ and ${\mathbf{xmax}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{xmin}}<{\mathbf{xmax}}$.
On entry, ${\mathbf{y}}=〈\mathit{\text{value}}〉$ and ${\mathbf{ymax}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{y}}\le {\mathbf{ymax}}$.
On entry, ${\mathbf{y}}=〈\mathit{\text{value}}〉$ and ${\mathbf{ymin}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{y}}\ge {\mathbf{ymin}}$.
On entry, ${\mathbf{ymin}}=〈\mathit{\text{value}}〉$ and ${\mathbf{ymax}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{ymin}}<{\mathbf{ymax}}$.
NE_REAL_ARRAY
On entry, $i=〈\mathit{\text{value}}〉$, ${\mathbf{x}}\left[i-1\right]=〈\mathit{\text{value}}〉$ and ${\mathbf{xmax}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{x}}\left[i-1\right]\le {\mathbf{xmax}}$.
On entry, $i=〈\mathit{\text{value}}〉$, ${\mathbf{x}}\left[i-1\right]=〈\mathit{\text{value}}〉$ and ${\mathbf{xmin}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{x}}\left[i-1\right]\ge {\mathbf{xmin}}$.

## 7  Accuracy

The method is numerically stable in the sense that the computed values of the polynomial are exact for a set of coefficients which differ from those supplied by only a modest multiple of machine precision.

The time taken is approximately proportional to $\left(k+1\right)×\left(m+l+1\right)$, where $m={\mathbf{mlast}}-{\mathbf{mfirst}}+1$, the number of points at which the evaluation is required.
This function is suitable for evaluating the polynomial surface fits produced by the function nag_2d_cheb_fit_lines (e02cac), which provides the array a in the required form. For this use, the values of ${y}_{\mathrm{min}}$ and ${y}_{\mathrm{max}}$ supplied to the present function must be the same as those supplied to nag_2d_cheb_fit_lines (e02cac). The same applies to ${x}_{\mathrm{min}}$ and ${x}_{\mathrm{max}}$ if they are independent of $y$. If they vary with $y$, their values must be consistent with those supplied to nag_2d_cheb_fit_lines (e02cac) (see Section 8 in nag_2d_cheb_fit_lines (e02cac)).
The arguments mfirst and mlast are intended to permit the selection of a segment of the array x which is to be associated with a particular value of $y$, when, for example, other segments of x are associated with other values of $y$. Such a case arises when, after using nag_2d_cheb_fit_lines (e02cac) to fit a set of data, you wish to evaluate the resulting polynomial at all the data values. In this case, if the arguments x, y, mfirst and mlast of the present function are set respectively (in terms of arguments of nag_2d_cheb_fit_lines (e02cac)) to x, ${\mathbf{y}}\left(S\right)$, $1+\sum _{i=1}^{s-1}{\mathbf{m}}\left(i\right)$ and $\sum _{i=1}^{s}{\mathbf{m}}\left(i\right)$, the function will compute values of the polynomial surface at all data points which have ${\mathbf{y}}\left[S-1\right]$ as their $y$ coordinate (from which values the residuals of the fit may be derived).

## 9  Example

This example reads data in the following order, using the notation of the argument list above:
 $N k l a[i-1], for i=1,2,…,k+1×l+1 ymin ymax y[i-1] Mi-1 xmin[i-1] xmax[i-1] X1i XMi, for i=1,2,…,N.$
For each line ${\mathbf{y}}={\mathbf{y}}\left[i-1\right]$ the polynomial is evaluated at $M\left(i\right)$ equispaced points between $\mathrm{X1}\left(i\right)$ and $\mathrm{XM}\left(i\right)$ inclusive.

### 9.1  Program Text

Program Text (e02cbce.c)

### 9.2  Program Data

Program Data (e02cbce.d)

### 9.3  Program Results

Program Results (e02cbce.r)


text:
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# Composite BOC (CBOC)

Fundamentals
Title Composite BOC (CBOC)
Author(s) J.A Ávila Rodríguez, University FAF Munich, Germany.
Year of Publication 2011

## CBOC Implementation

The CBOC implementation is a particular case of the CBCS Modulation. The CBOC modulation can be expressed as follows

where we can recognize that in this particular case the BCS component is BOC(6,1) with a percentage of power of 1/11. Furthermore the high frequency signal is placed on both the data and pilot channels and equal power for both channels is assumed. In addition, the total Galileo Open Service (OS) power is equal to that of the PRS and the modulation parameters are the following:

Table 1: Interplex parameters of the CBOC(6,1,1/11) modulation.

## Optimal Tracking of CBOC

As one can imagine, until the final Galileo implementation of MBOC was selected, different CBOC implementations were under consideration. In the next lines we summarize the main results of the analyses carried out to optimally track CBOC each of the proposed solutions.

Following the results from [O. Julien et al., 2006][1] we will show in this chapter the tracking performance of the CBOC modulation when only one channel of the Galileo E1 OS (data or pilot) is used. As we have seen above, there are many ways of implementing MBOC and even if we would constrain our analysis to CBOC, it can be shown that the variety of solutions is still broad. Indeed, the high frequency BOC(6,1) component could go in principle on both data and pilot channels, only on the pilot, only on the data, the power splits could change, and they all would still fulfil the MBOC definition that we gave above. To simplify thus our analysis, the three main cases are exposed next. These correspond basically to the three CBOC implementations that were considered for the Galileo E1 OS:

• The use of a CBOC(6,1,1/11) where both the data and pilot channels have a BOC(6,1) component in anti-phase. In this case, the power of the BOC(6,1) part is 1/11 of the channel total power. This is the implementation finally selected for Galileo E1 OS.
• The use of a CBOC(6,1,2/11) where only one of the channels, the pilot or the data, has a BOC(6,1) component with alternating sign, while the other channel is a pure BOC(1,1). In this case, the power of the BOC(6,1) part is 2/11 of the pilot channel total power. Moreover, it is important to note that the alternation of BOC(6,1) is necessary to get rid of the cross spectral terms that would result otherwise. This will be shown in the following lines.
• CBOC(6,1,1/11) is used on both the data and pilot channels, being BOC(6,1) in both channels with alternating sign. In this case, the data and pilot channels have a BOC(6,1) power of 1/11 of the channel total power.

Recalling now the general CBCS definition from:

and substituting BCS by BOC(6,1) we can define the general CBOC model as follows:

and looking only at the pilot channel for the three cases described above, we have:

where $\rho$ indicates the amount of power on the high frequency BOC(6,1) component:

The last model to introduce is the alternating model, which is shown to be:

for even chips, and

for odd chips. The autocorrelation function of the three models is shown next:

As we can see, the two first auto-correlations present an additional cross-term due to the existence of cross-correlation between BOC(1,1) and BOC(6,1). As we mentioned above, this cross-term is not desirable and must be eliminated to generate a spectrum according to the MBOC definition. This is possible if data and pilot are in anti-phase (each with a different sign) or if the BOC(6,1) component alternates its sign. Nonetheless, we will keep these signals for reference in the following figures to show the performance. We show next the autocorrelation functions of the different analyzed solutions compared with the TMBOC solution for a receiver bandwidth of 24 MHz:

Figure 1: CBOC and TMBOC Autocorrelation Function.

As we can recognize in the figure above, the shape of the auto-correlation function highly depends on the amount of power on BOC(6,1). Indeed, the higher the value of $\rho$, the more ripples the auto-correlation will present and the better the potential performance will be. Furthermore, it is interesting to see that the sharpest slope around the origin is shown by TMBOC and the version while TMBOC(6,1,2/11) and perform similar. MBOC is analyzed next regarding the following three criteria:

• False tracking points
• Code tracking noise
• Multipath-induced code tracking error

### False Tracking Points

If we take a closer look at the auto-correlation function of Figure 1 above, it is clear to recognize the existence of secondary peaks that could lead to stable false locks. Indeed, the more ripples the function presents and the more accentuated the undulations are, the higher the probability that we lock on a non-desired but stable tracking point. Fortunately, not all the solutions are equally susceptible to suffering from this effect, since this depends actually on how the auto-correlation function looks like. In fact, while the selected CBOC(1/11) for Galileo E1 OS is not likely to lead to stable false lock points as shown in [J.-A. Avila-Rodriguez et al., 2006][2], the existence of false lock points would be nearly unavoidable for the implementation, implying thus a higher complexity to detect the right peak. This is in fact the prize for allocating more BOC(6,1) power on the channel. Finally, it is important to note that no matter which of the implementations we look at, since BOC(1,1) is the dominant signal in all of them, a false lock detector is still necessary in order to make sure that the receiver is tracking the signal based on the correct autocorrelation main peak, and not the secondary of BOC(1,1).

### Thermal Noise-Induced Code Tracking Error

In order to understand how the code tracking noise behaves for the different implementations of MBOC, we present in the next figure the Cramér Rao Lower Bound.

Figure 2: BOC(1,1), CBOC and TMBOC Cramér Rao Lower Bound with 1 Hz Loop Bandwidth, 1/12 Chip E-L Spacing, 4 ms Integration and 12 MHz One-Sided Filter.

As we can see, the best performance corresponds to the alternating version with 2/11 of power on BOC(6,1) or to its equivalent TMBOC version, since these can take advantage of the high frequency components of the signal the best. Moreover, among the different CBOC solutions with the 1/11 of average power that results from considering data and pilot together, the worst performance is shown to be that of the in-phase solution while the anti-phase is the best. This was to expect since its autocorrelation function’s main peak has the steepest slope. We can also see this if we compare both phase and anti-phase versions in the time domain.

Figure 3: CBOC data chip with BOC(1,1) and BOC(6,1) in-phase.

Equally, the pilot channel would present the following shape for a chip.

Figure 4: CBOC pilot chip with BOC(1,1) and BOC(6,1) in anti-phase.

From the previous figures it is easy to recognize that the pilot channel will have more components at higher frequencies since at 0.5 chips the amplitude variation is higher. Furthermore, the alternating CBOC solution with 1/11 of power on BOC(6,1) performs as the average of the phase and anti-phase solutions shown above. To avoid any confusion, it is important to underline that the solutions studied above correspond to MBOC implementations that were object of study during the design of Galileo E1 OS and GPS L1C. The finally selected CBOC implementation of Galileo has BOC(6,1) on both data and pilot, being these in anti-phase with respect to each other. According to the previous notation, this means that the pilot channel of Galileo E1 OS will be and the data channel .

### Multipath Induced Tracking Error

The multipath performance of a signal highly depends on the shape of the auto-correlation function. Indeed, this was one of the aspects that were more seriously taken into consideration during the optimization of the Galileo Open Service and GPS Civil signals in E1/L1. We show in the next figures the multipath performance of different MBOC implementations by means of the multipath envelopes and the multipath running average error as done in [G.W. Hein et al., 2006a][3] and [G.W. Hein et al., 2006b][4]. Note that TMBOC(6,1,1/11) and perform the same and therefore are overlapped in the next figures.

Figure 5: MBOC Multipath Envelopes with an E-L Spacing of 0.1 chips and a Double-Sided Filter of 24 MHz.
Figure 6: MBOC Multipath Running Average Error with an E-L Spacing of 0.1 chips and a Double-Sided Filter of 24 MHz

As we can see, for an early-late spacing of 0.1 chips and a one-sided front-end filter of 12 MHz, the best performance is exhibited by , followed by , and . Moreover, it is clear to see that all the solutions clearly outperform BOC(1,1). In addition, it is important to note that and TMBOC(6,1,1/11) perform the same in average when both data and pilot are processed in the receiver.

### Conclusions on Optimal CBOC Tracking

As we have seen in previous lines, the best CBOC implementation of MBOC in terms of performance is the anti-phase solution. Nevertheless, as demonstrated in (12), this solution presents the great inconvenient of showing a non desired cross-term that can only be eliminated if the other channel adopts the counterpart . This is indeed the case in the finally selected CBOC implementation of Galileo E1 OS. Then, the total performance of this CBOC implementation, seen as data and pilot together, is equivalent to that of the alternating solution.

Another studied option was to use directly the solution on both the data and pilot channels, in which case both channels would have the same tracking performance. This implementation would perfectly fulfil the MBOC spectrum too but the drawback was be the extra complexity required for the implementation of the alternation. Equally, the option of using a on one channel and a pure BOC(1,1) on the other channel would permit to have one channel with excellent tracking performance, while the other channel would just use a pure BOC(1,1) modulation. This signal would be of interest for some types of receivers with a philosophy behind very much in line with how TMBOC is conceived.

All the figures that we have shown above rely on the assumption that a CBOC replica signal has to be locally generated by the receiver. Indeed, CBOC is a linear combination of two sub-carriers and has thus more than two levels, so that 2 bits are needed to encode the signal. While this should not be a great challenge as some works have already shown [P.G. Mattos, 2007][5], we will study in the next chapter a suboptimal tracking of CBOC using local replicas encoded with only 1 bit, as shown in [J.-A. Avila-Rodriguez et al., 2006] [2] and [O. Julien et al., 2006] [1]. One could of course think of separating the correlation of the incoming CBOC signal with a pure BOC(1,1) replica and, on the other side, with a pure BOC(6,1) replica. Then a simple linear combination of the outputs of both correlators would be sufficient to obtain the same result as if we had correlated directly with the CBOC replica of two bits. However, this processing requires twice as many correlators as the traditional CBOC tracking. In the next lines we present a solution that avoids to have to pay this prize [O. Julien et al., 2006][1] and [J.-A. Avila-Rodriguez et al., 2006] [2].

## Suboptimal Tracking of CBOC: TMBOC like approach

As shown in [J.-A. Avila-Rodriguez et al., 2006] [2], the idea behind the 1-bit receiver technique is to correlate the incoming CBOC signal with a locally generated signal obtained by time-multiplexing a BOC(1,1) sub-carrier and a BOC(6,1) sub-carrier. We can model this replica as follows:

where $S_1$ is the union of the segments of time when a BOC(1,1) sub-carrier is used. Equally, $S_2$, the complement of $S_1$ in the time domain, is the union of the segments of time when a BOC(6,1) sub-carrier is generated. In addition, we write $TM_{61}$ in order to distinguish this technique from the typical CBOC replica approach studied above. The advantage of this approach is that by using such a replica we only need to encode with 1 bit reducing thus the complexity of the receiver. Moreover, time-multiplexing reduces the number of correlators.

Let us define $\alpha$ to designate the percentage of time that the BOC(6,1) sub-carrier is used in one code length, and $\beta = 1 - \alpha$ to represent the amount of time reserved for the BOC(1,1) local sub-carrier part. Furthermore, we will assume that the sign of the BOC(6,1) local sub-carrier in the local replica is taken according to the sign of the BOC(6,1) used in the CBOC signal. This means that for the case the CBOC channel is in-phase, we have a positive sign and thus the percentage of BOC(6,1) time is given by $\alpha^+ = 1 -\beta^+$. Equally, for the anti-phase CBOC component the percentage of BOC(6,1) time is given by $\alpha^- = 1 -\beta^-$. It must be noted that $\alpha^+$ and $\alpha^-$ can be different in general. As a result, the cross-correlation that results from using the $TM_61$ replica with the different CBOC implementations of previous chapter is given by the following expressions:

As we can see in the expressions above, the correlation functions of BOC(1,1) and BOC(6,1) are also weighted as was also the case for the CBOC replica, with the difference that the weights are in this case controlled by the factors $\alpha$ and $\beta$ and the cross-correlation between $TM_{61}$ and CBOC induces additional correlation losses as shown in [O. Julien et al., 2006] [1] and [J.-A. Avila-Rodriguez et al., 2006c].

It is also interesting to see in [O. Julien et al., 2006][1] and [J.-A. Avila-Rodriguez et al., 2006][2] that if the Dot Product (DP) discriminator is employed, different TM61 replicas could be used for the prompt correlator and for the early and late correlators. Indeed, for the particular case of the DP discriminator, the prompt correlator only affects the tracking noise squaring losses and is thus recommendable to choose a local prompt replica that minimizes the correlation loss with the incoming CBOC signal. As shown in [J.-A. Avila-Rodriguez et al., 2006][2], this can be achieved by selecting $\alpha=0$ or what is equivalent, by using a pure BOC(1,1) sub-carrier as proposed in [J.-A. Avila-Rodriguez et al., 2006][2].

On the other hand, the early and late correlator outputs determine the gain of the discriminator as well as the noise correlation between the early and late correlators’ output. Next table shows the tracking noise degradation that results from the use of the TM61 technique for the different studied CBOC cases and we compare it with the optimal TMBOC tracking for the same BOC(6,1) power [O. Julien et al., 2006][1].

Table 2: $TM_{61}\left(\alpha\right)$ Tracking Noise Degradation with respect to TMBOC in Terms of Equivalent $C/N_0$ for Different CBOC Configurations.

As we can see, for CBOC(1/11) the most interesting values are either a high or a low value of , since for those cases the equivalent C/N0 degradation of the tracking noise is the lowest with a value of 1.9 dB for the , of 3 dB for the and of 4 dB for the . It is important to note that $\alpha=0$ implies to use a pure BOC(1,1) replica and $\alpha=1$ a pure BOC(6,1) replica.

In the same manner, the best option for and TMBOC(6,1,2/11) would be to take a value of $\alpha$as high as possible. This means, a pure BOC(6,1) replica for the E-L correlators would be the ideal choice in this case. As a conclusion, in order to have a common architecture for CBOC(6,1,1/11), CBOC(6,1,2/11) and TMBOC(6,1,2/11), the best would be to select a $TM_{61}$ tracking technique that would only use pure local replicas: BOC(1,1) for the prompt correlator and BOC(6,1) for the E-L correlators. Moreover, no time-multiplexing would be needed any more, simplifying thus the complexity of the receiver. As one can imagine, if this approach is valid for the different MBOC implementations discussed in previous pages, it is also valid for the particular CBOC and TMBOC solutions adopted by Galileo and GPS.

Another aspect that we have not touched yet is the multipath rejection of the different discussed TM61 options. As shown in [O. Julien et al., 2006][1]for both the CBOC(1/11) and the CBOC(2/11) cases, a value of $\alpha$ around 0.5 would lead to the best results. Nevertheless, values of $\alpha$ close to 1 would also deliver performances close to the optimum. In addition, using $\alpha=0$ is shown to be suboptimal in terms of multipath rejection since the performance is then that of BOC(1,1) [J.-A. Avila-Rodriguez et al., 2006][2].

As a conclusion, it seems that a sub-optimum implementation of a CBOC-TMBOC receiver based on a 1 bit architecture could use a pure local BOC(1,1) replica for the prompt correlator and a pure local BOC(6,1) replica for the early and late correlators. Moreover, we have seen that this scheme provides with relatively low degradation in terms of code tracking noise compared to the optimal TMBOC tracking.

The preceding derivations pursued to show that processing CBOC with a1 bit receiver is possible already today at the cost of some degradation. However, 2-bit receivers are already reality as shown in [P.G. Mattos, 2007] [5] and an optimal processing with 2 bits would be thus preferred. This would additionally imply some superiority of CBOC with respect to the TMBOC implementation since no blanking would be needed any more. Indeed, blanking of the BOC(6,1) pulses to avoid correlation losses is equivalent to reducing the equivalent code length by a factor 29/33 of the total length (10230). It must be noted though that this supposes a minimum additional complexity.

## References

1. ^ a b c d e f g [O. Julien et al., 2006] O. Julien, C. Macabiau, L. Ries, J.-L. Issler, 1 - Bit processing of Composite BOC (CBOC) Signals, First CNES Workshop on Galileo Signals and Signal Processing, 12-13 October 2006, IAS ( Institut Aero Spatial ) Toulouse, France.
2. ^ a b c d e f g h [J.-A. Avila-Rodriguez et al., 2006] J.-A. Avila-Rodriguez, S. Wallner, G.W. Hein, E. Rebeyrol, O. Julien, C. Macabiau, L. Ries, A. DeLatour, L. Lestarquit, J.-L. Issler, CBOC – An Implementation of MBOC, First CNES Workshop on Galileo Signals and Signal Processing, 12-13 October 2006, IAS ( Institut Aero Spatial ) Toulouse, France.
3. ^ [G.W. Hein et al., 2006a] G. W. Hein, J.-A. Avila-Rodriguez, S. Wallner, A.R. Pratt, J.I.R. Owen, J.-L Issler, J.W. Betz, C.J. Hegarty, S. Lenahan, J.J. Rushanan, A.L. Kraay, T.A. Stansell, MBOC: The New Optimized Spreading Modulation Recommended for Galileo L1 OS and GPS L1C, Proceedings of the International Technical Meeting of the Institute of Navigation, IEEE/ION PLANS 2006, 24-27 April, 2006, Loews Coronado Bay Resort, San Diego, California, USA.
4. ^ [G.W. Hein et al., 2006b] G. W. Hein, J.-A. Avila-Rodriguez, S. Wallner, A.R. Pratt, J.I.R. Owen, J.-L Issler, J.W. Betz, C.J. Hegarty, S. Lenahan, J.J. Rushanan, A.L. Kraay, T.A. Stansell, MBOC: The New Optimized Spreading Modulation Recommended for Galileo L1 OS and GPS L1C, Inside GNSS – Working Papers, Vol.1, No.4, May/June 2006, pp.57-65.
5. ^ a b [P.G. Mattos, 2007] P.G. Mattos, Hardware and Algorithm Implications of Adding Galileo Capability to an SPS GPS Receiver, Proceedings of the National Technical Meeting of the Institute of Navigation, ION-NTM 2007, January 2007, San Diego, California, USA.

## Credits

The information presented in this NAVIPEDIA’s article is an extract of the PhD work performed by Dr. Jose Ángel Ávila Rodríguez in the FAF University of Munich as part of his Doctoral Thesis “On Generalized Signal Waveforms for Satellite Navigation” presented in June 2008, Munich (Germany)


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# defining a proof

## What are proofs? (informal)

### Example 1

The following submission from an anonymous engineer to the January, 1902 edition of Popular Mechanics caught my eye. Seems like something every Boy/Girl Scout and Architect should know.

“HOW TO USE THE WATCH AS A COMPASS: Very few people are aware of the fact that in a watch they are always provided with a compass, with which, when the sun is shining, the cardinal points can be determined. All one has to do is to point the hour hand to the sun and south is exactly half way between the hour and the figure 12 on the watch. This may seem strange to the average reader, but it is easily explained. While the sun is passing over 180 degrees (east to west) the hour hand of the watch passes over 360 degrees (from 6 o'clock to 6 o'clock). Therefore the angular movement of the sun in one hour corresponds to the angular movement of the hour hand in half an hour; hence, if we point the hour hand toward the sun the line from the point midway between the hour hand and 12 o'clock to the pivot of the hands will point to the south. — Engineer.”

They give an argument of correctness; is that really a proof? Well, there are some ambiguities: Do I hold the watch vertically, or, in the plane of the sun's arc? Certainly I can't hold it up-side down, even though this isn't explicitly stated. Furthermore, the correctness of the reasoning relies on some unstated assumptions. E.g., the sun is at its highest (northernmost) point of its transit at noon. Is this actually true? Does it depend on the time of year? I'm not exactly sure (and will have to sit down and scratch my head and draw pictures of orbits, to convince myself). Certainly there are at least a couple of caveats: even beyond account for Daylight Savings Time, the solar-time and clock-time only align at time-zone boundaries, and they drift up to an hour apart, before the next boundary rectifies the difference. Is this presuming I'm in the northern hemisphere? What if I'm on the equator?

To be fair, the intent of this anecdote was to give enough evidence to convince you, not necessarily to be a complete, stand-alone self-contained proof. But in writing out a careful proof, one is forced to consider all the points just made; being forced to understand these can lead you to better understand the procedure yourself. But be careful to distinguish between something which sounds reasonable, and something that you're certain of.

### An argument by form

How can we tell true proofs from false ones? What, exactly, are the rules of a proof? These are the questions which will occupy us.

Proofs are argument by form. We'll illustrate this with three parallel examples of a particular proof form called syllogism.

#### Example 2

 1 All people are mortal. Premise 2 Socrates is a person. Premise 3 Therefore, Socrates is mortal. Syllogism, lines 1,2

#### Example 3

 1 All [substitution ciphers] are [vulnerable to brute-force attacks]. Premise 2 The [Julius Caesar cipher] is a [substitution cipher]. Premise 3 Therefore, the [Julius Caesar cipher] is [vulnerable to brute-force attacks]. Syllogism, lines 1,2

Note that you don't need to know anything about cryptography to know that the conclusion follows from the two premises. (Are the premises indeed true? That's a different question.)

#### Example 4

 1 All griznoxes chorble happily. Premise 2 A floober is a type of griznox. Premise 3 Therefore, floobers chorble happily. Syllogism, lines 1,2

You don't need to be a world-class floober expert to evaluate this argument, either.

#### Aside:

Lewis Carroll, a logician, has developed many whimsical examples of syllogisms and simple reasoning. (Relatedly, note how the social context of Carroll's examples demonstrates some feminist issues in teaching logic.)

As you've noticed, the form of the argument is the same in all these. If you are assured that the first two premises are true, then, without any true understanding, you (or a computer) can automatically come up with the conclusion. A syllogism is one example of a inference rule — that is, a rule form that a computer can use to deduce new facts from known ones.

### Some non-proofs

Of course, not all arguments are valid proofs. Identifying invalid proofs is just as interesting as identifying valid ones.

#### Logic in (?)Culture:

Homer: Ah, not a bear in sight.  The Bear Patrol must be working.
Homer: Thank you, honey.
Lisa:  By your logic, this rock keeps tigers away.
Homer: Oh?  How does it work?
Lisa:  It doesn't work.
Homer: Uh-huh.
Lisa:  It's just a stupid rock.
Homer: Uh-huh.
Lisa:  But I don't see any tigers around here, do you?
[pause]
[A moment's hesitation … and money changes hands.]

(From The Simpsons Much Apu About Nothing.)

If Lisa isn't around, who will identify specious reasoning for us? We can certainly use her approach of finding other particular examples that follow the same argument, yet lead to a clearly erroneous conclusion.

#### Example 5

Suppose that my friend makes the following argument:

 1 Warm cola tastes bad. Premise 2 Warm salt-water tastes bad. Premise 3 Therefore, mixing them together tastes bad. “Common-sense conclusion”, lines 1,2
I'm skeptical, so I have a sip; sure enough, the conclusion is indeed true. But is the proof correct — does the “common-sense conclusion” rule actually hold? In order to refute the form of the argument, we can try similar arguments which have the same form but a false conclusion (as Lisa did).
 1 Ice-cold coke tastes good. Premise 2 Ice coffee tastes good. Premise 3 Therefore, mixing them together tastes good. “Common-sense conclusion”, lines 1 and 2.
After another unfortunate sip, I verify that this conclusion is not true, and therefore my friend's reasoning is at fault.

My friend responds by claiming that the “common-sense conclusion” is too valid; the rule is that bad-taste is preserved upon mixing, not that any taste is preserved. While I'm inclined to believe that, we realize we can still test this more refined rule: can you come up with an instance of mixing together bad-tasting things and ever getting a yummy result? (Say, salt and flour, which can be mixed and baked to get delicious saltines! The argument continues, about whether the form of the argument precludes baking, and so on.)

The end result (after I take some antacid) is that we have a clearer understanding of the initially vague “common-sense conclusion”, and stricter rules about when it applies. Thus, refining the argument has led us to a greater understanding.

The above examples are a bit frivolous, but the procedure of looking for counterexamples applies to many real-world dilemmas. It also highlights the difference between a correct proof, and a faulty proof that might still happen to lead to a true result. (By the way, this is the exact same skill used when trying to come up with an algorithm for a problem: “well, the algorithm works for this input, but can I find a something that makes one of the steps fail?” If so, you then try refining your algorithm “well, I can add a test to take care of that problem; is that enough so that it always works?”)

#### Exercise 1

Solve this statement for [X]: It is wrong to ban [X]. Such a ban would punish those reasonable citizens who would use [X] responsibly, while those who really want to abuse [X] will be able to get it anyway, through a black market which will only subsidize other criminal activities.

##### Solution

This argument is or has been commonly used for varying topics —

• marijuana,
• alcohol,
• all drugs,
• handguns,
• birth control,
• prostitution,
• encryption technology.
The interesting part, is that the traditional Left and Right political positions each use this argument for some of these items, while rejecting the argument when used for other items.

A more rational response is to either accept all the above, or none of the above, or to realize that the stated argument wasn't everything — that there might be implicit assumptions or arguments which actually do distinguish between these cases (the different interpretations of “[X]”). Being able to articulate the differences is essential. The more refined arguments may be more nuanced, and less able to fit into a sound-bite, but lead to a better understanding of one's own values. And sometimes, upon reflection, one may realize that some of the implicit values or premises are things they actually disagree with, once they are precisely spelled out.

In real-world issues, there are often many subtleties, and short arguments that sound airtight might be glossing over factors which are important in practice.

#### Example 6

During daylight, there is no need to have headlights (or running lights) on: there's already plenty of light for everybody to see each other by. Even during the day, headlights slightly increase how quickly other drivers see you during (say) a routine, tenth-of-a-second glance in their mirror.

#### Example 7

When in a turn-only lane, there is absolutely no need to signal — since there's only one way to turn, a signal can't communicating any information to other drivers! Glib, but not true: Other defensive drivers presumably know you have only one legal option, but they don't know that you know that, and they are planning reactions in case you surprise them with a sudden illegal maneuver. By signaling, you give them information which helps them better plan for yet other contingencies. Furthermore, it also gives you more confidence that other drivers are expecting your turn, reducing your suspicion that they're about to pull a surprise maneuver on you. (True, these are all low-probability events which almost always turn out to be unnecessary. But avoiding accidents is all about minimizing risks for the one moment events do spiral out of control.)

#### Example 8

“You'll lose weight if and only if you burn more calories than you take in. All those diet-plan books can never get around this, and all their details are pointless.”

True, calorie intake and expenditure solely determine weight loss/gain. But after some thought, we can get examples where the above logic overlooks some relevant differences: If your friend told you they were switching from a diet of 2000 calories of balanced short-term and long-term energy sources (sugars, proteins, and carbs) to a diet of 2000 calories worth of Pixy Stix at breakfast plus a Flintstones multivitamin, would you be optimistic that they would have the willpower to strictly follow the new plan? The two plans are equal when counting calories, but in actuality one really is a better plan. (Even more exaggeratedly, consider a daily plan of 2000 calories of sugar while never drinking any water—since water has no calories, it can't affect your calorie count, according to the above claim.)

These contrived counterexamples help illustrate that it's conceivable that there can be a difference between diet plans, so the initial claim isn't technically true.

The point illustrated is that often real-world arguments incorrectly imply that their result follows from the form of the argument, when in fact the form is not valid in the way a syllogism is. This fallacy can be illuminated by finding a different domain in which the argument fails. The practice of searching for domains which invalidate the argument can help both sides of a debate hone in on bringing the unspoken assumptions to light. The original argument, if its conclusion is indeed true, must be patched either by adding the unspoken assumptions or fixing the invalid form.

#### Exercise 2

Mistakes in syllogisms are hard to make: what are the only two ways to have an error in a syllogism?

##### Solution
1. The argument isn't actually in syllogism form. For example, the following is an incorrect syllogism:
 1 All people don't know my file's password. Premise (Equivalent to “Nobody knows my file's password”, but reworded to be of the required form “All somethings have some property.”.) 2 All hackers are people. Premise 3 Therefore, my file is secure from hackers. Incorrect syllogism, lines 1,2
To be a syllogism, the conclusion would have to be “all hackers don't know my file's password.” The file might or might not be secure, but the above doesn't prove it.
2. One of the two premises is wrong.
 1 All people don't know my file's password. Premise, but possibly false 2 All hackers are people. Premise, but possibly false 3 Therefore, all hackers don't know my file's password. Syllogism, lines 1,2
This proof fails, of course, if some hackers are non-people (e.g., programs), or if some people know the password. (In fact, presumably you know the password!)

Of course, even if a proof fails, the conclusion might be true for other reasons. An incorrect argument doesn't prove the conclusion's opposite!

### Other Inference Rules

Of course, there are more ways to deduce things, beyond a syllogism.

• Who decides what the valid inference rules are?
• Is it always clear when people have used the inference rules correctly?

Consider the following argument about WaterWorld boards:

 1 (A) is next to exactly onepirate. Premise, from either subfigure 2 (A) has only one unexplored neighbor. Premise, from either subfigure 3 If you are an unexpected location next to (A), then you contain apirate. Incorrect conclusion
This conclusion is not valid; while it is correct for the first board shown, it is incorrect for the second. (I make this mistake all the time when playing WaterWorld too quickly, arrggh! — The Author.)

The problem is that the author of the argument presumably meant to conclude “all explored neighbors of (A) contain a pirate”.

Before we can study exact proofs, we need a way of writing exactly what we mean. This will occupy us for the next section.

### The need for a precise language

These previous glitches in the WaterWorld arguments both arise, of course, because we were sloppy about what each sentence meant exactly. We used informal English — a fine language for humans, who can cope with remarkable amounts of ambiguity -- but not a good language for specifying arguments.

#### Aside:

Laws and contracts are really written in a separate language from English — legalese — full of technical terms with specific meanings. This is done because, while some ambiguity is tolerable in 99% of human interaction, the remaining 1% can be very problematic. Even so, legalese still contains intentionally ambiguous terms: When, exactly, is a punishment “cruel and unusual”? What exactly is the “community standard” of indecency? The legal system tries to simultaneously be formal about laws, yet also be flexible to allow for unforeseen situations and situation-specific latitude. (The result of this tension is the position of Judge.)

#### Aside:

Court decisions, while dense reading, are often the model of well-presented arguments.

Consider, from a previous example, the statement “…[this is something] every Boy/Girl Scout and Architect should know”. Does this mean all people who are both a scout and architect, or everybody who is at least one or the other? Genuinely ambiguous, in English! (Often, “and/or” is used to mean “one or the other or possibly both”.)

We'll next look at a way to specify some concepts non-ambiguously, at least for WaterWorld. We need to be more careful about how we state our facts and how we use these known facts to deduce other facts. Remember, faulty reasoning might not just mean losing a silly game. Hardware and software bugs can lead to significant bodily harm (Imagine software bugs in an airplane autopilot or surgical robot system), security loopholes (e.g., in Mozilla or IE), or expensive recalls.

One reaction to the above arguments is “Well, big deal — somebody made a mistake (mis-interpreting or mis-stating a claim); that's their problem. (And sheesh, they sure are dolts!)” But as a programmer, that's not true: Writing large systems, human programmers will err, no matter how smart or careful or skilled they are. Type-checkers catch some errors upon compilation, and test suites catch their share of bugs, but many still remain in real-world software. Thus we are looking for systemic ways to reduce and catch errors, with the ultimate ideal of being able to prove programs correct.

#### Aside:

Other professions have checklists, protocols, and regulations to minimize human error; programming is no different, except that the industry is still working on exactly what the checklists or training should be. Someday, a license will be required for practicing software, at least for software involved with life-safety.

In our study of formal logic, we'll need three things:

• Syntax (language) — a precise syntax and vocabulary for expressing concepts without ambiguity,
• Propositional logic,
• First-order logic (propositional logic, plus relations and quantifiers)
• Semantics (meaning) and modeling — how to connect these formal languages to whatever topic we want to reason about (including our software).
• Reasoning (proofs) — methods of deducing new facts from old. We'll see three types of reasoning, and how to use them for each of our two logics:
• Truth tables
• Boolean Algebra
• Inference Rules
We'll visit these topics in an interleaved manner — first propositional logic (immediately with its semantics) and three methods of reasoning for it; then first-order logic and an in-depth look at its interpretations, and finally the methods of reasoning for first-order logic.

We'll begin with a particular syntax — propositional logic for the game of WaterWorld — before using this syntax to formally deduce safe moves.

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text:
## Return to Question

4 added some background

The question is more serious than it looks!

Recall the list of irreducible simply connected inner symmetric spaces of compact type in dimension $4k+2$:

Hermitian symmetric spaces are of course complex manifolds, and real Grassmannians carrying almost complex structures were classified by P. Sankaran (PAMS, 1991) and Z. Tang (PAMS, 1994). Therefore Thus my question:

anybody know whether $\mathrm{E}_7/(\mathrm{SU}_8/(\mathbb{Z}/2))$ carries carry an almost complex structure?

Note that in dimension $4k$ it is known that an irreducible simply connected inner symmetric space of compact type carries an almost complex structure iff it is Hermitian symmetric.

Edit concerning the dimensions multiple of 4: It is on purpose that I did not mention our paper http://fr.arxiv.org/abs/1003.5172 when I first asked the question. Indeed, the method used in that paper to treat the $4k$-dimensional situation (an index-theoretical obstruction) seems uneffective in the $4k+2$-dimensional cases so I expected other methods to tackle this problem.

Let me nevertheless give a short account on how things work in dimension $4k$. The idea is to use the following criterion:

If a compact Riemannian manifold $(M,g)$ carries a self-dual vector bundle $E$ such that the index of the twisted Dirac operator on $\Sigma M \otimes E\otimes TM^{\mathbb{C}}$ is odd, then $M$ has no almost complex structure (in fact $TM$ is not even stably isomorphic to a complex bundle).

Note that the complex spin bundle $\Sigma M$ and $E$ can be locally defined, only their tensor product has to be globally defined. On then checks that this criterion applies to all irreducible inner symmetric spaces of compact type which are neither Hermitian symmetric nor spheres.

Returning to the dimensions $4k+2$, it is easy to check that there exists a priori no bundle satisfying the above criterion. Nevertheless, one can show real Grassmannians and $\mathrm{E}_7/(\mathrm{SU}_8/(\mathbb{Z}/2))$ do carrysuch a bundle $E$ which satisfies all conditions in the criterion except that it is not self-dual. I don't know if this can lead anywhere...

3 corrected notation

# Does $\mathrm{F}_7/(\mathrm{SU}_8/(\mathbb{Z}/2))$\mathrm{E}_7/(\mathrm{SU}_8/(\mathbb{Z}/2))$carry an almost complex structure? The question is more serious than it looks! Recall the list of irreducible simply connected inner symmetric spaces of compact type in dimension$4k+2$: 1. Hermitian symmetric spaces (one can write them down explicitly); 2. Grassmannians of oriented real$p$-planes in$\mathbb{R}^{p+q}$, with$pq=4k+2$; 3. The 70-dimensional exceptional symmetric space$\mathrm{F}_7/(\mathrm{SU}_8/(\mathbb{Z}/2))$.\mathrm{E}_7/(\mathrm{SU}_8/(\mathbb{Z}/2))$.

Hermitian symmetric spaces are of course complex manifolds, and real Grassmannians carrying almost complex structures were classified by P. Sankaran (PAMS, 1991) and Z. Tang (PAMS, 1994). Therefore my question:

1