text:
## Results (1-50 of at least 1000)

Next
Label $\alpha$ $A$ $d$ $N$ $\chi$ $\mu$ $\nu$ $w$ prim arith $\mathbb{Q}$ self-dual $\operatorname{Arg}(\epsilon)$ $r$ First zero Origin
2-483-161.20-c1-0-29 $1.96$ $3.85$ $2$ $3 \cdot 7 \cdot 23$ 161.20 $$1.0 1 -0.486 0 2.29622 Modular form 483.2.r.a.181.4 2-483-161.32-c1-0-25 1.96 3.85 2 3 \cdot 7 \cdot 23 161.32$$ $1.0$ $1$ $0.270$ $0$ $2.04202$ Modular form 483.2.y.a.193.7
2-483-161.2-c1-0-17 $1.96$ $3.85$ $2$ $3 \cdot 7 \cdot 23$ 161.2 $$1.0 1 0.0114 0 1.36183 Modular form 483.2.y.a.163.10 2-483-161.20-c1-0-28 1.96 3.85 2 3 \cdot 7 \cdot 23 161.20$$ $1.0$ $1$ $0.172$ $0$ $2.21032$ Modular form 483.2.r.a.181.14
2-483-161.25-c1-0-13 $1.96$ $3.85$ $2$ $3 \cdot 7 \cdot 23$ 161.25 $$1.0 1 -0.215 0 1.37338 Modular form 483.2.y.b.25.6 2-483-161.32-c1-0-24 1.96 3.85 2 3 \cdot 7 \cdot 23 161.32$$ $1.0$ $1$ $0.0231$ $0$ $1.92195$ Modular form 483.2.y.b.193.11
2-483-161.19-c1-0-2 $1.96$ $3.85$ $2$ $3 \cdot 7 \cdot 23$ 161.19 $$1.0 1 -0.348 0 0.341506 Modular form 483.2.bf.a.19.20 2-483-161.2-c1-0-16 1.96 3.85 2 3 \cdot 7 \cdot 23 161.2$$ $1.0$ $1$ $0.0808$ $0$ $1.14979$ Modular form 483.2.y.b.163.3
2-483-161.20-c1-0-12 $1.96$ $3.85$ $2$ $3 \cdot 7 \cdot 23$ 161.20 $$1.0 1 0.102 0 1.15787 Modular form 483.2.r.a.181.13 2-483-161.20-c1-0-27 1.96 3.85 2 3 \cdot 7 \cdot 23 161.20$$ $1.0$ $1$ $0.343$ $0$ $2.09632$ Modular form 483.2.r.a.181.10
2-483-161.20-c1-0-6 $1.96$ $3.85$ $2$ $3 \cdot 7 \cdot 23$ 161.20 $$1.0 1 0.406 0 0.619048 Modular form 483.2.r.a.181.28 2-483-161.25-c1-0-12 1.96 3.85 2 3 \cdot 7 \cdot 23 161.25$$ $1.0$ $1$ $-0.0599$ $0$ $1.29673$ Modular form 483.2.y.b.25.10
2-483-161.25-c1-0-4 $1.96$ $3.85$ $2$ $3 \cdot 7 \cdot 23$ 161.25 $$1.0 1 0.405 0 0.765021 Modular form 483.2.y.b.25.2 2-483-161.32-c1-0-23 1.96 3.85 2 3 \cdot 7 \cdot 23 161.32$$ $1.0$ $1$ $0.292$ $0$ $1.86365$ Modular form 483.2.y.b.193.5
2-483-161.18-c1-0-7 $1.96$ $3.85$ $2$ $3 \cdot 7 \cdot 23$ 161.18 $$1.0 1 -0.499 0 0.653610 Modular form 483.2.y.b.340.14 2-483-161.19-c1-0-19 1.96 3.85 2 3 \cdot 7 \cdot 23 161.19$$ $1.0$ $1$ $0.298$ $0$ $1.67866$ Modular form 483.2.bf.a.19.1
2-483-161.19-c1-0-31 $1.96$ $3.85$ $2$ $3 \cdot 7 \cdot 23$ 161.19 $$1.0 1 0.448 0 2.59448 Modular form 483.2.bf.a.19.21 2-483-161.2-c1-0-15 1.96 3.85 2 3 \cdot 7 \cdot 23 161.2$$ $1.0$ $1$ $-0.0827$ $0$ $1.12427$ Modular form 483.2.y.b.163.1
2-483-161.2-c1-0-28 $1.96$ $3.85$ $2$ $3 \cdot 7 \cdot 23$ 161.2 $$1.0 1 0.206 0 2.32620 Modular form 483.2.y.a.163.13 2-483-161.20-c1-0-11 1.96 3.85 2 3 \cdot 7 \cdot 23 161.20$$ $1.0$ $1$ $-0.323$ $0$ $1.02197$ Modular form 483.2.r.a.181.25
2-483-161.20-c1-0-23 $1.96$ $3.85$ $2$ $3 \cdot 7 \cdot 23$ 161.20 $$1.0 1 0.0817 0 1.78212 Modular form 483.2.r.a.181.16 2-483-161.20-c1-0-26 1.96 3.85 2 3 \cdot 7 \cdot 23 161.20$$ $1.0$ $1$ $-0.0317$ $0$ $2.09145$ Modular form 483.2.r.a.181.22
2-483-161.20-c1-0-31 $1.96$ $3.85$ $2$ $3 \cdot 7 \cdot 23$ 161.20 $$1.0 1 -0.341 0 2.51316 Modular form 483.2.r.a.181.9 2-483-161.20-c1-0-5 1.96 3.85 2 3 \cdot 7 \cdot 23 161.20$$ $1.0$ $1$ $0.149$ $0$ $0.594556$ Modular form 483.2.r.a.181.3
2-483-161.20-c1-0-9 $1.96$ $3.85$ $2$ $3 \cdot 7 \cdot 23$ 161.20 $$1.0 1 0.0491 0 1.00323 Modular form 483.2.r.a.181.8 2-483-161.25-c1-0-11 1.96 3.85 2 3 \cdot 7 \cdot 23 161.25$$ $1.0$ $1$ $-0.0238$ $0$ $1.28232$ Modular form 483.2.y.a.25.8
2-483-161.25-c1-0-22 $1.96$ $3.85$ $2$ $3 \cdot 7 \cdot 23$ 161.25 $$1.0 1 0.470 0 1.86766 Modular form 483.2.y.a.25.14 2-483-161.25-c1-0-31 1.96 3.85 2 3 \cdot 7 \cdot 23 161.25$$ $1.0$ $1$ $-0.0951$ $0$ $3.38889$ Modular form 483.2.y.a.25.16
2-483-161.32-c1-0-12 $1.96$ $3.85$ $2$ $3 \cdot 7 \cdot 23$ 161.32 $$1.0 1 -0.306 0 1.28001 Modular form 483.2.y.a.193.15 2-483-161.32-c1-0-22 1.96 3.85 2 3 \cdot 7 \cdot 23 161.32$$ $1.0$ $1$ $0.329$ $0$ $1.80580$ Modular form 483.2.y.a.193.10
2-483-161.18-c1-0-29 $1.96$ $3.85$ $2$ $3 \cdot 7 \cdot 23$ 161.18 $$1.0 1 0.369 0 2.07484 Modular form 483.2.y.a.340.6 2-483-161.18-c1-0-6 1.96 3.85 2 3 \cdot 7 \cdot 23 161.18$$ $1.0$ $1$ $0.338$ $0$ $0.587528$ Modular form 483.2.y.a.340.14
2-483-161.19-c1-0-11 $1.96$ $3.85$ $2$ $3 \cdot 7 \cdot 23$ 161.19 $$1.0 1 0.100 0 1.12006 Modular form 483.2.bf.a.19.19 2-483-161.19-c1-0-18 1.96 3.85 2 3 \cdot 7 \cdot 23 161.19$$ $1.0$ $1$ $0.105$ $0$ $1.54812$ Modular form 483.2.bf.a.19.16
2-483-161.19-c1-0-25 $1.96$ $3.85$ $2$ $3 \cdot 7 \cdot 23$ 161.19 $$1.0 1 0.315 0 2.07573 Modular form 483.2.bf.a.19.14 2-483-161.19-c1-0-30 1.96 3.85 2 3 \cdot 7 \cdot 23 161.19$$ $1.0$ $1$ $0.0521$ $0$ $2.51249$ Modular form 483.2.bf.a.19.29
2-483-161.19-c1-0-9 $1.96$ $3.85$ $2$ $3 \cdot 7 \cdot 23$ 161.19 $$1.0 1 -0.152 0 0.858827 Modular form 483.2.bf.a.19.17 2-483-161.2-c1-0-14 1.96 3.85 2 3 \cdot 7 \cdot 23 161.2$$ $1.0$ $1$ $-0.0777$ $0$ $1.09951$ Modular form 483.2.y.b.163.13
2-483-161.2-c1-0-21 $1.96$ $3.85$ $2$ $3 \cdot 7 \cdot 23$ 161.2 $$1.0 1 0.307 0 1.84370 Modular form 483.2.y.b.163.2 2-483-161.2-c1-0-27 1.96 3.85 2 3 \cdot 7 \cdot 23 161.2$$ $1.0$ $1$ $0.425$ $0$ $2.24400$ Modular form 483.2.y.a.163.2
2-483-161.2-c1-0-5 $1.96$ $3.85$ $2$ $3 \cdot 7 \cdot 23$ 161.2 $$1.0 1 -0.368 0 0.518562 Modular form 483.2.y.a.163.4 2-483-161.20-c1-0-10 1.96 3.85 2 3 \cdot 7 \cdot 23 161.20$$ $1.0$ $1$ $-0.283$ $0$ $1.00978$ Modular form 483.2.r.a.181.20
2-483-161.20-c1-0-17 $1.96$ $3.85$ $2$ $3 \cdot 7 \cdot 23$ 161.20 $$1.0 1 -0.106 0 1.34314 Modular form 483.2.r.a.181.18 2-483-161.20-c1-0-22 1.96 3.85 2 3 \cdot 7 \cdot 23 161.20$$ $1.0$ $1$ $0.390$ $0$ $1.77443$ Modular form 483.2.r.a.181.7
2-483-161.20-c1-0-24 $1.96$ $3.85$ $2$ $3 \cdot 7 \cdot 23$ 161.20 $$1.0 1 0.250 0 1.95720 Modular form 483.2.r.a.181.12 2-483-161.20-c1-0-25 1.96 3.85 2 3 \cdot 7 \cdot 23 161.20$$ $1.0$ $1$ $0.190$ $0$ $1.99879$ Modular form 483.2.r.a.181.17
2-483-161.18-c1-0-25 $1.96$ $3.85$ $2$ $3 \cdot 7 \cdot 23$ 161.18 $$1.0 1 0.0409 0 1.94996 Modular form 483.2.y.a.340.8 2-483-161.20-c1-0-30 1.96 3.85 2 3 \cdot 7 \cdot 23 161.20$$ $1.0$ $1$ $-0.318$ $0$ $2.41259$ Modular form 483.2.r.a.181.5
2-483-161.17-c1-0-3 $1.96$ $3.85$ $2$ $3 \cdot 7 \cdot 23$ 161.17 $$1.0 1 -0.172 0 0.322773 Modular form 483.2.bf.a.178.7 2-483-161.20-c1-0-4 1.96 3.85 2 3 \cdot 7 \cdot 23 161.20$$ $1.0$ $1$ $0.104$ $0$ $0.506253$ Modular form 483.2.r.a.181.1


text:
How do you evaluate lne^(5x)?

Oct 30, 2016

$5 x$

Explanation:

Evaluate $\ln {e}^{5 x}$

Recall the log formula $\log {x}^{a} = a \log x$

$\ln {e}^{5 x} = 5 x \ln e$

And because $\ln e = 1$...

$5 x \ln e = 5 x$

format: plain


text:
# So Alice and Bob want to flip a coin… – STOC 2014 Recaps (Part 10)

A major use case for coin flipping (with actual coins) is when you’re with friends, and you have to decide where to eat. This agonizing decision process can be elegantly avoided when randomness is used. But who’s doing the coin flipping? How do you know your friend isn’t secretly choosing which restaurant to go to? Cryptography offers a solution to this, and Sunoo will tell us about how this solution is actually equivalent to one way functions!

In this post, we look at a fundamental problem that has plagued humankind since long before theoretical computer science: if I don’t trust you and you don’t trust me, how can we make a fair random choice? This problem was once solved in the old-fashioned way of flipping a coin, but theoretical computer science has made quite some advances since.

What are the implications of this? In their recent STOC paper, Berman, Haitner, and Tentes show that the ability for two parties to flip a (reasonably) fair coin means that one-way functions exist. This, in turn has far-reaching cryptographic implications.

A function $f$ is one-way if it is “easy” to compute $f(x)$ given any input $x$, and it is “hard”, given the image $y$ of a random input, to find a preimage $x'$ such that $f(x')=y$. The existence of one-way functions imply a wide range of fundamental cryptographic primitives, including pseudorandom generation, pseudorandom functions, symmetric-key encryption, bit commitments, and digital signatures – and vice versa: the seminal work of Impagliazzo and Luby [1] showed that the existence of cryptography based on complexity-theoretic hardness assumptions – encompassing the all of the aforementioned primitives – implies that one-way functions exist.

About coin-flipping protocols, however, only somewhat more restricted results were known. Coin-flipping has long been a subject of interest in cryptography, since the early work of Blum [2] which described the following problem:

“Alice and Bob want to flip a coin by telephone. (They have just divorced, live in different cities, want to decide who gets the car.)”

More formally, coin-flipping protocol is a protocol in which two players interact by exchanging messages, which upon completion outputs a single bit interpreted as the outcome of a coin flip. The aim is that the coin should be (close to) unbiased, even if one of the players “cheats” and tries to bias the outcome towards a certain value. We say that a protocol has constant bias if the probability that the outcome is equal to 0 is constant (in a security parameter).

Impagliazzo and Luby’s original paper showed that coin-flipping protocols achieving negligible bias (that is, they are very close to perfect!) imply one-way functions. Subsequently, Maji, Prabhakaran and Sahai [3] proved that coin-flipping protocols with a constant number of rounds (and any non-trivial bias, i.e. $1/2-o(1)$) also imply one-way functions. Yet more recently, Haitner and Omri [4] showed that the same holds for any coin-flipping protocol with a constant bias (namely, a bias of $\frac{\sqrt{2}-1}{2}-o(1)$). Finally, Berman, Haitner and Tentes proved that coin-flipping of any constant bias implies one-way functions. The remainder of this post will give a flavor of the main ideas behind their proof.

The high-level structure of the argument is as follows: given any coin-flipping protocol $\Pi=(\mathsf{A},\mathsf{B})$ between players $\mathsf{A}$ and $\mathsf{B}$, we first define a (sort of) one-way function, then show that an adversary capable of efficiently inverting that function must be able to achieve a significant bias in $\Pi$. The one-way function used is the transcript function which maps the players’ random coinflips to a protocol transcript. The two main neat insights are these:

• Consider the properties of a coin-flipping protocol when interpreted as a zero-sum game between two players: Alice wins if the outcome is 1, and Bob wins otherwise. If the players play optimally, who wins? It turns out that from the winner, we can deduce that there is a set of protocol transcripts where the outcome is bound to be the winning outcome, no matter what the losing player does: that is, transcripts that are “immune” to the losing player.
• A recursive variant of the biasing attack proposed by Haitner and Omri in [4] is proposed. The new attack can be employed by the adversary in order to generate a transcript that lies in the “immune” set with probability close to 1 – so, this adversary (who has access to an inverter for the transcript function) can bias the protocol outcome with high probability.

The analysis is rather involved; and there are some fiddly details to resolve, such as how a bounded adversary can simulate the recursive attack efficiently, and ensuring that the inverter will work for the particular query distribution of the adversary. Without going into all those details, the last part of this post describes the optimal recursive attack.

Let $\mathsf{Inv}$ be the function that takes as input a pair $(\mathbf{t},b)$, where $\mathbf{t}$ is a random transcript of a partial (incomplete) honest execution of $\Pi$ and $b\in\{0,1\}$, and outputs a random pair of random coinflips $(r_1,r_2)$ for the players, satisfying the following two conditions: (1) they are consistent with $\mathbf{t}$, that is, the coinflips could be plausible next coinflips given the partial transcript $\mathbf{t}$; and (2) there exists a continuation of the protocol after $\mathbf{t}$ and the next-coinflips $(r_1,r_2)$ that leads to the protocol outcome $b$.

It seems that an effective way for the adversary to use $\mathsf{Inv}$ might be to call $\mathsf{Inv}(\mathbf{t},1)$ for each partial transcript $\mathbf{t}$ at which the relevant player has to make a move, and then to behave according to the outputted coins. We call this strategy the biased-continuation attack, which is the crucial attack underlying the result of [4].

The new paper proposes a recursive biased-continuation attack that adds an additional layer of sophistication. Let $\mathsf{A}^{(0)}=\mathsf{A}$ be the honest first player’s strategy. Now, define $\mathsf{A}^{(i)}$ to be attacker which, rather than sampling a random 1-continuation among all the possible honest continuations of the protocol $(\mathsf{A},\mathsf{B})$, instead samples a random 1-continuation among all continuations of $(\mathsf{A}^{(i-1)},\mathsf{B})$. Note that $\mathsf{A}^{(1)}$ is the biased-continuation attacker described above! It turns out that as the number of recursions grows, the probability that the resulting transcript will land in the “immune” set approaches 1 – meaning a successful attack! Naïvely, this attack may require exponential time due to the many recursions required; however, the paper circumvents this by analyzing the probability that the transcript will land in a set which is “almost immune”, and finding that this probability approaches 1 significantly faster.

[1] One-Way Functions Are Essential for Complexity Based Cryptography. Impagliazzo, Luby (FOCS 1989).
[2] Coin Flipping by Telephone: A Protocol for Solving Impossible Problems. Blum (CRYPTO 1981).
[3] On the Computational Complexity of Coin Flipping. Maji, Prabhakaran, Sahai (FOCS 2010).
[4] Coin Flipping with Constant Bias Implies One-Way Functions. Haitner, Omri (FOCS 2011).

# Spaghetti code crypto – STOC 2014 Recaps (Part 6)

We return to our STOC recap programming with Justin Holmgren, who will tell us about an exciting new result in program obfuscation. (However, he doesn’t tell us how to say “indistinguishability obfuscation” 10 times fast).

Justin Holmgren on How to Use Indistinguishability Obfuscation: Deniable Encryption, and More, by Amit Sahai and Brent Waters

One very recent topic of study in cryptography is program obfuscation – that is, how to make a program maximally unintelligible, while still preserving the functionality. We will represent programs as circuits, which for most purposes is equivalent to any other representation. At some level, obfuscation of a program seems like it should be an easy task – any programmer knows that it’s easy to end up with unreadable spaghetti code, but theoretical treatment of the issue was lacking, and finding a construction with provable security properties has proven quite difficult.

In 2001, Barak et al. [1] precisely formulated black-box obfuscation, which says that if you can compute something given the obfuscated program, you could also compute it given black-box access to the program. This is a natural and very strong notion of obfuscation. If we could obfuscate programs according to this definition, we could easily solve many difficult problems in cryptography. For example, a public-key fully homomorphic encryption scheme could be constructed from an ordinary symmetric-key encryption scheme by obfuscating a program which decrypts two input ciphertexts, and outputs the encryption of their NAND. Unfortunately, [1] also showed that black-box obfuscation is impossible to achieve.

One weaker definition of obfuscation that [1] did not rule out is called indistinguishability obfuscation (iO). It requires that if two circuits $C_1$ and $C_2$ are functionally equivalent (for all $x$, $C_1(x) = C_2(x)$), then their obfuscations are indistinguishable. In 2013, Garg et al. [2] gave the first candidate construction for iO. This construction is not proven secure under standard assumptions (e.g. the hardness of factoring), but no attacks on the construction are currently known, and it has been proven that certain types of generic attacks cannot possibly work.

It isn’t immediately clear whether iO is useful. After all, what information is the obfuscation really hiding if it only guarantees indistinguishability from functionally equivalent programs? “How to use Indistinguishability Obfuscation: Deniable Encryption and More”, by Amit Sahai and Brent Waters [3], shows several applications of iO to building cryptographic primitives, and proposes iO as a “hub of cryptography”, meaning that constructions which use iO as a building block would be able to benefit from improvements in the efficiency or underlying computational assumptions of iO constructions.

As a first example of how to use iO, they show how to construct an IND-CPA public key cryptosystem based on one-way functions and iO. This uses a few other basic and well-studied cryptographic primitives (it is known that all of them exist if one-way functions exist). First, a pseudorandom generator is a deterministic function which takes as input a random string $s$, and computes a longer string $r$ which is indistinguishable from truly random. The length of $r$ can be polynomially larger than the length of $s$, but we will let $r$ be twice the length of $s$. A pseudorandom function family is parameterized by a secret key $k$ and an input $x$, and produces outputs of some fixed length. Oracle access to $f_k(\cdot)$ for some random unknown $k$ is computationally indistinguishable from oracle access to a truly random function. The main special property that [2] require is a puncturable pseudorandom function family. That is, for a secret key $k$ and an input $x$, a “punctured” key $k\{x\}$ can be computed. $k\{x\}$ allows one to compute $f_k(y)$ for all $y \neq x$, but essentially reveals no information about $f_k(x)$.

The power of iO comes from combining its indistinguishability guarantees with those of other cryptographic primitives, in what is known as the hybrid method. We will start with the security game played with a real scheme, and give a sequence of other games, each indistinguishable from the previous one to the adversary, such that the adversary obviously has no advantage in the last game. This will prove that the scheme is secure.

For example, a simple public key encryption scheme for one-bit messages (let $G$ be a PRG, and $F$ be a puncturable pseudorandom function family) would be the following.

• The secret key $SK$ is just a puncturable PRF key $k$.
• The public key $PK$ is the obfuscation of an Encrypt program. Encrypt takes a message $m$ and randomness $r$, and outputs $(G(r), F_k(G(r)) \oplus m)$.
• To decrypt a ciphertext $(c_1, c_2)$ given $SK$, one just computes $F_k(c_1) \oplus c_2$.

In the IND-CPA security game, there is a challenger and an adversary. The challenger generates the public and secret key, and gives the public key to the adversary. The challenger also generates an encryption of a random bit $b$, and gives this to the adversary. The adversary wins if he can guess the bit that the challenger chose.

In the first hybrid, the challenger encrypts by choosing a truly random $\tilde{r}$, and generating the ciphertext $(\tilde{r}, F_k(\tilde{r}) \oplus b)$. This is indistinguishable by the security property of a pseudorandom generator.

In the next hybrid, the program Encrypt which is obfuscated and given to the adversary uses a punctured key $k\{\tilde{r}\}$ instead of the full PRF key $k$. This is with high probability functionally equivalent, because $\tilde{r}$ is most likely not in the range of $G$. So by the security of iO, this hybrid is indistinguishable from the previous one.

Next, the ciphertext is generated by choosing a truly random bit $p$, and generating the ciphertext $(\tilde{r}, p \oplus b)$. This is also indistinguishable to the adversary, because the punctured key reveals nothing about $F_k(\tilde{r})$. That is, if these two hybrids were distinguishable, then given only $k\{\tilde{r}\}$, one could distinguish between a truly random bit $p$ and the true value $F_k(\tilde{r})$.

In this hybrid, it is clear (by analogy to a one-time pad) that no adversary can distinguish a ciphertext encrypting $0$ from a ciphertext encrypting $1$. So the original scheme is also secure.

Sahai and Waters achieve more than just public-key encryption – they also achieved the long-standing open problem of publicly deniable encryption. This is an encryption system for which if $Encrypt(m ; r) = c$, one can also compute a “fake” randomness $r'$ for any other message $m'$ such that $Encrypt(m'; r')$ is also equal to c, and $r'$ is indistinguishable from $r$. The motivation is that after a message is encrypted, coercing the encrypter into revealing his message is futile – there is no way of verifying whether what he says is the truth. Note that of course the decrypter could be coerced to reveal his secret key, but there’s nothing that can be done about that.

The basic idea of the deniable encryption scheme is that the encryption public key is again an obfuscated program – this program takes a message and randomness string. If the randomness string has some rare special structure (it is a hidden sparse trigger), it is interpreted as encoding a particular ciphertext, and the program will output that ciphertext. Otherwise, the program encrypts normally. There is also a public obfuscated Explain program, which allows anybody to sample a random hidden sparse trigger, but not test whether or not a string is a hidden sparse trigger.

At present, iO obfuscation is known to imply some very powerful primitives, including for example multi-input functional encryption. It is still open to construct fully homomorphic encryption for iO, or even collision resistant hash functions from iO, or to show that no such reduction is possible.

[1] On the (Im)possibility of Obfuscating Programs, Barak, Goldreich, Impagliazzo, Rudich, Sahai, Vadhan, Yang, 2001.
[2] Candidate Indistinguishability Obfuscation, Garg, Gentry, Halevi, Raykova, Sahai, Waters, 2013.
[3] How to Use Indistinguishability Obfuscation: Deniable Encryption, and More, Sahai, Waters, 2014.

# Can you tell if a bit is random?

Avi Wigderson has this wonderful talk about the power and limitations of randomness in computation, surveying the beautiful area of randomized algorithms, and the surprising connection between pseudorandomness and computational intractability. If you haven’t seen it already, I highly recommend it!

There is a part of the talk when Avi fearlessly engages the age-old question of the meaning of randomness. What does it mean for something to be random? We theoretical computer scientists often speak of “coin tosses”: many of our algorithms toss coins to decide what to do, and Alice and Bob toss coins to foil their mortal enemy Eve. Avi asks, “Is a coin toss random?”

Imagine that I hold in my hand a (fair) coin, and a second after I toss it high in the air, you, as you are watching me, are supposed to guess the outcome when it lands on the floor. What is the probability that you will guess correctly? 50-50 you say? I agree!

(Reproduced from the essay version of his talk). But now, Avi says, imagine that the coin toss was repeated, but this time you had some help: you were equipped with a powerful computing device, and accurate sensors, like an array of high speed video recorders. The coin suddenly starts to look a lot less random now! If your computer and sensors were good enough, you’d likely be able to predict the outcome of the coin toss nearly perfectly. In both cases the coin toss was the same, but the observer changed, and so did the amount of randomness in the situation. Thus, he contends, randomness is in the eye of the beholder.

From here, Avi launches into an excellent discussion of pseudorandomness and other topics. However, I’m going to forge onwards — foolishly, perhaps — with the original theme: what is randomness? As he pointed out, the randomness in the coin toss situation depended on the capabilities of the observer. Thus, one could argue that a coin toss is not truly random, because in principle you could predict the outcome. Is there a situation that is intrinsically random, no matter how powerful or omniscient the observer was? Put another way: is there a process by which I can generate a bit that  no external super-being could possibly predict? Can you tell if a bit is random?

Enter the physics

Do you believe that the universe is governed by quantum mechanics? If so, then it seems that the matter is settled: quantum mechanics guarantees the existence of intrinsically random events. For those who know basic quantum, generating a truly random bit is easy: simply prepare a qubit state $\frac{1}{\sqrt{2}}(|0\rangle + |1 \rangle)$ and measure in the computational basis — done! In fact, there are hardware random number generators that operate on this principle

But how do you actually prepare such a qubit, and perform the measurement? If you were a skilled experimentalist with access to perfectly manufactured mirrors and laser sources and detectors, then maybe. But then, how would you know these mirrors and lasers and detectors were built correctly?

For the rest of us down on Earth who don’t have time to learn experimental quantum physics and personally guarantee that all our laboratory equipment was built to specification, we’d like to be able to take some apparatus like a hardware random number generator and test whether its outputs were genuinely random — without having to open up the device and inspect its insides. Can we test randomness in a black box fashion?

It is easy to see that, if you only had input/output access to a black box device, there’s no test you can perform to reliably tell whether the device’s outputs are random. Intuitively, this is because for every test, you can construct a black box device that deterministically produces outputs that pass the test! Here’s a proof-by-Dilbert-cartoon:

What about statistical tests for randomness? Pseudorandom number generators are usually subject to an extensive battery of tests that check for suspicious patterns or regularities. Again, these are not fool-proof: one can design a fixed string that “diagonalizes” against all these heuristics and passes them all!

The power of non-locality

Black-box randomness testing seems like a hopeless task. But it turns out that, with a tiny extra assumption, it becomes possible! The added ingredient is non-locality.

Imagine that we had a hardware random number generator (HRNG) that we wanted to test, and suppose that the HRNG consisted of two separate pieces — call them Piece A and Piece B. Prevent A and B from communicating, either by surrounding them with Faraday cages or putting them on opposite sides of the galaxy. Now play the following game with them: generate a random bits xy, and give x to A and y to B. Collect output bits a from A and b from B, and check if = Λ y (where b indicates the parity of the two bits). If so, the devices win this curious little game.

Suppose A and B were completely deterministic devices: their outputs a and b were fixed deterministic functions of their inputs x and y, respectively. Then it is not hard to see that, over the random choices of x and y, they can only win this game with probability 3/4.

What if you somehow knew that the devices were winning with probability greater than 3/4? What must you conclude? These devices — and hence their output bits — must be acting randomly. The killer fact is, it’s actually possible to win this game with probability strictly greater than 3/4! This can happen if part A and part B, though isolated, utilize something called quantum entanglement — a phenomenon that Einstein derisively termed “spooky action at a distance”.

We’ve made some headway on randomness testing, it seems. First, this game I’ve described does not require you to peer inside the HRNG. Second, if the HRNG wins the game with high enough probability, you’ve certified that its outputs must contain some randomness. Third, quantum mechanics says it is possible to build a HRNG to win the game. Finally, as a bonus: you don’t need to believe in quantum mechanics in order to trust the conclusion that “If the devices win with probability greater than 3/4, they must be producing randomness”!

What I’ve described to you is a distillation of the famous Bell’s Theorem from physics, which says that quantum mechanics is inconsistent with any so-called “hidden variable” theory of nature that respects locality (i.e. you can separate systems by distances so they cannot communicate or influence each other). This is why people say that quantum mechanics exhibits “non-local” behavior, because games like the one above can be won with probability greater than 3/4 (which has been demonstrated experimentally). Bell’s Theorem has been historically important in establishing a dividing line between classical and quantum theories of nature; but today we have a new interpretation: one can view Bell’s Theorem — published half a century ago — as saying that we can meaningfully perform randomness testing.

Randomness expansion — to infinity and beyond!

“Wait a minute,” you say. “While you’ve presented a test for randomness, this isn’t so useful — to even run the test you need 2 bits of randomness to generate x and y! Why test a HRNG if you already have randomness to begin with?”

True. A HRNG test isn’t very useful when it requires more randomness than is produced. It’s like the current state of fusion reactors: they produce less energy than is required to run them. But researchers in quantum cryptography realized that you could get a randomness efficient test, using less randomness than the amount that is produced by the HRNG. For example, one can start with 100 random bits, run one of these randomness efficient tests, and end up with, say, 200 random bits, a net gain of 100! In a sense, the amount of randomness you now possess has been expanded — hence these tests are called randomness expansion protocols.

Roger Colbeck, in his Ph.D. thesis from 2006, came up with a protocol that uses m bits of initial seed randomness, and certifies cm bits of output randomness using a HRNG that consists of 3 isolated parts, where c is some constant (approximately 3/2). Later in 2010, Pironio et al. gave a protocol that expands m bits of seed randomness to m2 certified bits. In 2011, two simultaneous works — one by Vazirani and Vidick, and one by Fehr, et al. — demonstrated a protocol that attained exponential expansion: starting with m bits, one can certify that a 2-part HRNG produces a whopping 2m certified bits of randomness!

While perhaps 2100 random bits starting from a mere 100 is probably more than you’d ever need, the tantalizing question still stands: is there a limit to how much randomness you can certify, starting with a finite amount of seed randomness? In the upcoming Quantum Information Processing conference, Matt Coudron and I will show that there is no upper limit, and that infinite randomness expansion is possible. We give a protocol where, starting with say 1000 random bits, you can command a finite number of devices (which is eight in our protocol) to produce an unbounded amount of randomness. Furthermore, this protocol is immune to any secret backdoors installed in your HRNG (e.g. by a certain security agency) — the output of your HRNG is guaranteed to be secure against any external eavesdropper.

Classical control of quantum systems

Let’s take a step back from this frenetic expansion of randomness, and appreciate the remarkable nature of these protocols. Note that, this entire time, we know almost nothing about the HRNG being tested. It could have been specially designed by some malicious adversary to try to fool your test as much as possible. It could have at its disposal enormous computational power, like access to the Halting Problem oracle. It could even use quantum entanglement in some yet-unimagined way to try to defeat your test. Yet, by isolating parts of the HRNG from each other, and employing a simple check on their input and outputs, a limited classical being can force the HRNG to produce a long stream of random bits.

In a time when random number generators can be stealthily compromised by altering the dopants in a microchip, or by the intentional weakening of cryptographic standards by government agencies, the ability to rigorously and soundly test randomness has outgrown its philosophical origins in the nature of physical law, and become something of operational significance.

To return to our original question, though: can we tell if a bit is random? We can, but only by employing some some of the deepest ideas drawn from theoretical computer science and quantum physics. More broadly, this paradigm of classically controlling untrusted quantum devices has given rise to what’s called device independent quantum information processing. Not only can we classically manipulate quantum devices to produce genuine randomness, we can do a whole lot more: today we know how to securely exchange secret keys and even perform general quantum computations in a device independent manner! The scope and potential of device independent QIP is enormous, and I think we’ve only gotten started.


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# d20pro

As mentioned in my last post, I’ve been using d20pro to run a traditional table-top roleplaying game. d20pro serves as a virtual tabletop with lots of cool features, displaying maps and character markers, keeping track of character sheets, and even rolling the dice for you. All of your NPCs and monsters are built in. You can share arbitrary files with your players. You can draw custom maps, if you don’t have an existing one to load into the software. Maps can have fog of war, so the players only see what the GM wants them to see.

In addition, d20pro takes care of combat for you, in large part. You choose the type of attack, the target, and any modifiers. d20pro rolls the dice, compares it to the target’s defense, and lets the GM decide whether it was a hit or miss and modify the results if necessary. Then d20pro rolls damage, and again the GM gets a chance to modify and approve it. d20pro also keeps track of initiative, which as always the GM can modify at will. It has built-in functions for ready actions, charging, and a few other common occurrences. You can also do skill rolls, saving throws, and cast spells. Check out this promo video.

d20pro does have some limitations and flaws. Naturally, it only works for d20-based games. The spell system works OK, but needs some major improvement before it becomes really useful. Currently you have to create all of the spells and their effects by hand, and it’s a rather cumbersome process. And since players cannot edit their own character sheets, the GM has to manually select which spells the players choose at the beginning of each day (a rather idiotic rule in both Pathfinder and DnD, I must say). Finally, the overall user interface is a bit awkward at times. Many windows are too small or too large, so either the contents flow out of the window, or it takes a lot of scrolling to see the contents, or the window simply goes off the screen. Some buttons are super tiny, while others are oddly large.

But d20pro is written in Java, which means it works on any platform, and that’s a huge bonus for a group with two Mac users. d20pro is also very affordable compared to other virtual table top programs. It’s just $30 for a judge license (if you want to run games) and$10 for a player license (if all you want to do is play). Judge licenses let you have 2 players connect for free. Licenses are permanent, one time purchases, not subscriptions. Thank all of creation the sensible d20pro team didn’t decide to go with the suck-em-dry software-as-a-service model I hate so much.

Overall, d20pro hasn’t necessarily sped up our games at all. Combat has been somewhat simplified, but you occasionally have to spend time tinkering with the software to get it to do what you want (especially with spells), and communicating with 6 guys on a conference call can get chaotic and confusing at times with no visual cues to help things along. Not to mention the occasional technology glitches that are bound to occur, like people’s internet connections going out, Skype crashes or inexplicable disconnections, mic problems, and so forth. (I should mention that you don’t have to use it for long-distance playing; it can also make a great game aid for in-person games, as well.)

On the other hand, the d20pro team is very responsive to the community, often replying personally to posts in their user forums, unlike most other software companies I know. They are also active about developing the software—they are all gamers and use it for their own games. And it has allowed us to play much more often, since it eliminates travel time, babysitting challenges, and the need for a large, quiet, and dedicated space in which to play. Previously we would play every other month or so at most. Now we can play every other week.


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# (x^2)cosx

• Sep 4th 2006, 11:52 AM
TexasGirl
(x^2)cosx
Calculate the integral from 0 to pi of (x^2)cosxdx
• Sep 4th 2006, 12:08 PM
ThePerfectHacker
Quote:

Originally Posted by TexasGirl
Calculate the integral from 0 to pi of (x^2)cosxdx

I will find the anti-derivative then you can complete the problem.
---
$\int x^2\cos x dx$.
Let $u=x^2$ and $v'=\cos x$
Then,
$u'=2x$ and $v=\sin x$
Thus, by parts,
$x^2\sin x-2\int x \sin xdx$
Now, you do it again,
Let $u=x$ and $v'=\sin x$
Then,
$u'=1$ and $v=-\cos x$
Thus, by parts,
$x^2\sin x-2\left( -x\cos x-\int -\cos xdx \right)$
Simplify,
$x^2\sin x+2x\cos x-2\int \cos xdx$
Thus,
$x^2\sin x+2x\cos x-2\sin x+C$
• Sep 4th 2006, 12:11 PM
Soroban
Hello, TexasGirl!

You need to integrate by parts . . . twice!

Quote:

$I \;= \;\int_0^{\pi} x^2\cos x\,dx$

Let: $u = x^2\qquad dv = \cos x\,dx$

Then: $du = 2x\,dx\quad v = \sin x$

And we have: . $I\;=\;x^2\sin x - 2\int x\sin x\,dx$

Let: $u = x\qquad dv = \sin x\,dx$

Then: $du = dx\quad v = -\cos x$

And we have: . $I \;= \;x^2\sin x - 2\left[-x\cos x - \int(-\cos x)\,dx\right]$
. . . . . . . . . . $I \;= \;x^2\sin x + 2x\cos x - 2\int\cos x\,dx$

Therefore:. . . $\boxed{I \;= \;x^2\sin x + 2x\cos x - 2\sin x + C}$

Edit: Too fast for me, TPHacker!


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# Sine and Cosine are Periodic on Reals/Corollary

{{mergeto|Cosine of Angle plus Straight Angle]] and [[Sine of Angle plus Straight Angle}}

## Corollary to Sine and Cosine are Periodic on Reals

$\map \cos {x + \pi} = -\cos x$
$\map \sin {x + \pi} = -\sin x$

$\cos x$ is strictly positive on the interval $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$ and strictly negative on the interval $\openint {\dfrac \pi 2} {\dfrac {3 \pi} 2}$
$\sin x$ is strictly positive on the interval $\openint 0 \pi$ and strictly negative on the interval $\openint \pi {2 \pi}$

## Proof

$\map \sin {x + \dfrac \pi 2} = \cos x$
$\map \cos {x + \dfrac \pi 2} = -\sin x$

Thus:

$\map \sin {x + \pi} = \map \cos {x + \dfrac \pi 2} = -\sin x$
$\map \cos {x + \pi} = -\map \sin {x + \dfrac \pi 2} = -\cos x$

It follows directly that:

$\forall x \in \closedint {-\dfrac \pi 2} {\dfrac \pi 2}: \cos x \ge 0$

Hence:

$\forall x \in \closedint {\dfrac \pi 2} {\dfrac {3 \pi} 2}: \cos x \le 0$

The result for $\sin x$ follows similarly, or we can use:

$\map \sin {x + \dfrac \pi 2} = \cos x$

$\blacksquare$


text:
# Math PI represented musically

I'm creating a program to output the values of pi as sound. I've seen a few videos and some other representations, and want to try to make my own for fun, except i'm unsure how to proceed.

There are the notes A,B,C,D,E,F, and G.

Unfortunately, there are 10 possible values for a digit, 1,2,3,4,5,6,7,8,9, or 0. I read something about a rule of fifths, but I am unsure how I would map the 10 numbers to sound. I'm not sure if this is the right place to ask this question.

How would you go about mapping a sound to numbers 0-9? I'd like to keep the spread as even as possible, so if every possible sound was on a line, 0-9 would be evenly distributed, if that makes sense.

Sorry if this is in the wrong location.

• It's pretty irrelevant, but I just had to post this, as I love it: m.youtube.com/watch?v=BDMBtQjS1bQ – Bob Broadley Aug 28 '14 at 23:00
• I'm rather surprised this got so many upvotes. As both a musician and a mathematician, I'd reject this question for vapidity! If you want to encode $\pi$ into tones, then map each digit to a note. If you want to create a piece which is representative of $\pi$ , then write something evocative of circles, spheres, and radiuses. – Carl Witthoft Aug 29 '14 at 11:58
• I know little of music, but if you have seven notes, just use pi in base 7 mathematically... turner.faculty.swau.edu/mathematics/materialslibrary/pi/… – Mark Aug 29 '14 at 13:12
• @CarlWitthoft: I agree. Alternatively, one might use a π/4 time signature, or use intervals with frequency ratio π:3 as the melodic step size. But using the digits of π is basically just like using a random stream of numbers. – leftaroundabout Aug 29 '14 at 14:18

One option if you're primarily interested in representing the individual digits of pi is to use a representation in a base other than 10. For example pi base 12 would have an individual digital for each chromatic note.

Here's a website that might help get you started: http://www.virtuescience.com/pi-in-other-bases.html

The number 10 doesn't necessarily map well to values in traditional musical theory. (For instance, there are 12 chromatic pitches per octave, using conventional divisions of the octave; diatonic scales have seven pitches; note durations are related as powers or negative powers of 2). So, for this reason, the world is your oyster! I guess you can choose any 10 values to map to the 10 digits.

This does mean that you are unlikely to get a definitive answer (and so some may consider this post to be off-topic), but here are some of my own suggestions for suitable mappings:

1. Divide an octave by ten (i.e. use 10-TET) although this youtube clip suggests somebody may have beaten you to it…!
2. Use 10 different chords.
3. Use 10 different note durations (which could be either related by powers of 2, or as tenths of a particular value).
4. Use a pair of interlocking pentatonic scales (I like this idea, maybe stereo separated, but it's not my project…)
5. Use a combination of the ideas above, eg. different frequencies each also have their own related duration (and stereo positioning…)

In the end, I came up with these ideas very quickly, the possibilities are vast, and experimentation is the key.

ADDITIONAL INFO: If you are trying to find the frequencies in 10-TET tuning you would multiply a starting frequency by 2 to the power of n/10, where n is integer values in the range 0-9.

• You could use pi in base 7 or base 12. – Cole Johnson Aug 29 '14 at 6:17

Pi can also be expressed through various infinite series. I like François Viète series discovered in 1593:

Square root from 2 is half octave distance. Maybe it is possible to represent the series as some sequence of sounds? Or maybe some other series would fit better? This might reproduce the spirit of Pi even better than replaying its decimal representation.

One Idea I haven't seen mentioned is rhythm. Perhaps you can use some of the spare digits as a change in pace (f.e. switch from eights to quavers). Or you could map the spare digits to pre-conceived rhythmic motives.

Another idea would be to use the digits that are not mapped to a note to switch instrument.

HTH.

Why use base 10? You have to make some compromise somewhere, and since π is already transcendental, there is no rational radix that will accurately represent π. If you use heptary, π ≈ 3.0663651432036134110263402244652226643520650240155443215426431025161154565220002622436103301443233631. These digits map perfectly to the seven pitches in an octave. Using octal would add in either a b3, b7, or #4. Quintary would yield a perfect pentatonic scale.

If you want to make a nice piece of music (which I presume you do, simply encoding pi would seem a bit wasteful), I'd avoid trying to generate the music mechanically, and instead use pieces of pi as inspiration.

For instance:

• Writing it in 22/7 (an approximation of pi)
• Using the first 5 or so digits as a motif in some way, and using the others not as much. You could then have another 50 or so digits as a fast cadenza type section.
• Using cyclical chord progressions (pi being about circles)

There's no reason you have to stay within one octave. You can use, for instance, C-D-E-F-G-A-B-c-d-e for your digits. So the first five notes are E-C-F-C-G-d, for example This has the advantage of being extremely intuitive to any musician, since you'd just be referring to scale degrees in C major (with 10 being 0). I can personally just sit there and read the notes and play them without thinking about it.

With the minor key you are using, you could also borrow from the major, ala the harmonic minor. Then you could use A-B-C-D-E-F-F#-G-G#-a. This isn't as intuitive, but it produces some interesting effects, without seeming completely chromatic. The first few notes become C-A-D-A-E-G#-B-F-E-C-E G-G#-F#-G#-C-B-C-G-D-F B-F-D-C-C-G-C-B-F#-G#-E-a. That last bit sounds like an actual cadence.

As a programmer, I love this idea and of course I thought about this as well already but didn't have any time yet to try this out :).

Basically, I believe your line with notes is incorrect. You should start by choosing a key in which you want to write it. I believe your key would be Am, or is it a coincidence? I would work with something like this:

char notes[] = {'A', 'B', 'C', 'D', 'E', 'F', 'G'};
String pi = String.valueOf(Math.PI);
for(int i = 0; i < pi.length(); i++) {
char currentChar = pi.charAt(i);
if(currentChar != '.') {//negate the decimal didigt
System.out.println(currentChar);
char currentNote = notes[(Integer.parseInt(currentChar + "") - 1) % notes.length];
System.out.println(currentNote);
}
}

Note that this is Java.

EDIT: This code is designed so that if you have a number that is larger than the amount of notes provided, it will start from the first note again. So 8 would be back note 0 (A).

Other answers have suggested using different bases. For an event in the Physics department, I did play pi in quintal, and there is a video. The sheet was generated using a script and Lilypond.

Bonus: also in octal, but this one is not annotated.

An option which no one has really mentioned is to use those extra digits for special purposes (ie Change tempo, another instrument). If the primary instrument is a piano, I'd imagine that simply assigning a digit to the snare, bass and cymbal would add a lot of flair to your final music.

In fact, adding new instruments will open you up to a bunch of new ways to do this. You could let each digit represent an operation. Each operation could represent an instrument, tempo, or effect change, etc... Also, each operation would then read in as many more digits as it needed in order to satisfy it's parameter count. This will allow for your beats to have things like sustain, accent, etc... for your notes.

This will actually make the problem of having more data space than note space even more of a problem except for the fact that it was never really a problem to begin with. Just pad the extra space with the next higher and lower octaves. This will result in a slight imbalance of notes but we don't really want a balance anyway...

Which does lead us to another problem. I think your best bet is to use patterns found in popular music to ensure that your note dispersion is pleasant. I found this site which has the data we'd want for determining that:

Your application will be more likely to sound better when it produces notes with a similar dispersion pattern to this. You may find that some instruments may work vastly differently though so (as with all of this) experimentation will be important.

Once you have this system working, I'd suggest trying to think of another song which it sounds similar to what you have and doing a more exact note dispersion graph which actually matches a song in the same key. Read the page at the link above to see what I mean.

If you wanted to step it up another notch, you'll have the ability to add as many functions as you want. You just need to decide how large an op code is (1,2,3 chars) depending on how many different functions you have and handle them all (even if large blocks of them do the same thing). If you really want this to be a complex symphony, I would suggest that you separate the processing step from the playback step to eliminate any timing issues you'll have due to the variable data rate inherent to such a design. Fortunately, there are standards. One popular standard is called MusicXML. If you make your program to simply generate those files. Then you can later play those files back using a MusicXML player.

Update: If you'd like to see my experiments with this you may get them here.

• hooktheory.com/blog/… It says nothing about "instruments sounding better" in a specific key. It just presents the data as "there is a general trend favoring key signatures with less sharps and flats but this is not universal." – Lyd Aug 29 '14 at 19:57
• That's not how it works, at all. That's now what the article claims, at all. The Garage Band application (the "smart instruments" app) shows and uses the diatonic chords of the harmony you select, you can select any key. It sounds good because of the diatonicism of the possible chord progressions, not because it is in a specific key. Your understanding of the article is completely broken, distorted, misleading, and completely wrong. You and the article are claiming and talking about very different things. It's evident that you don't know about what you are talking about. – Lyd Aug 30 '14 at 0:18
• I've got to back @JCPedroza up on this one. Unless you have perfect pitch you won't know the difference between C major and Db major when you listen to piece. It's completely wrong to think that this half step accounts for a 20% drop in popularity because C major sounds better. C major is popular because it's all natural notes (i.e. white keys on the piano) which are easier to play and write, especially for people without formal training. Your'e 100% wrong, krowe, in the conclusions you've drawn from this article. – MarkM Aug 31 '14 at 3:56
• @krowe, fine, but if you don't care WHY a key is more popular why do you keep asserting things like: "At the end of the day these keys are the most common because they sound better." That is a (unsupported) answer to the question WHY the key is more popular that you say you don't care about. – MarkM Sep 1 '14 at 18:59
• @krowe - I'm definitely on the side of JC and Mark. Those keys are most common because they are easier to read and write. A lot of instruments were designed around these keys, as early music was Modal and the use of #/b were not common or necessary. Other instruments are transposed and have certain tendencies, such as music written for horns having more b's than #'s. So more Jazz stuff is in flat keys and more rock stuff is in # keys (due to guitars). The piece of your argument (specific to the topic of keys) that is agreeable is that certain keys are more common but I'd say that is all. – Basstickler Sep 2 '14 at 12:49

Here's another idea, bizarre even by my standards. For those not familiar w/ the 12-tone composition rule (as originally stated; probably changed many times), it says you cannot repeat any tone until the other 11 have been played. Serially or in chords is allowed.

So here's the "12-tone pi" composition rule: For each digit of pi, you're allowed to skip that many tones in the next 12-cycle. E.g., first cycle only needs 9 (12-3) tones before repeating; the next cycle needs 11 (12-1) tones, and so on.

Bonus points if you can make the piece NOT sound like Webern or Berio :-)

• Dudes: downvoting sans reason is considered poor form. Other than perhaps allowing too much interpretability by the composer, what's wrong with this approach? – Carl Witthoft Sep 2 '14 at 14:47


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## Trigonometry (11th Edition) Clone

(a) $f(2)=5$ (b) $f(-1)=11$
(a) Here we require the y-value of the ordered pair when $x=2$. As indicated by the ordered pair $(2,5)$, when $x=2$, $y=5$. Therefore, $f(2)=5$. (b) Here we require the y-value of the ordered pair when $x=-1$. As indicated by the ordered pair $(-1,11)$, when $x=-1$, $y=11$. Therefore, $f(-1)=11$.


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## “For” in Spanish Depends Upon What It’s For!

It sounds so simple. You once see that para is the equivalent of the English for. So you start using it as an exact translation every time you want to say for. Sorry…it just doesn’t work that way. I’ve many times had to squelch the teacher in me and not correct someone who is saying Gracias para todo (it should be Gracias por todo) or when in a restaurant someone orders and says por mí (it should be para mí).  Para and por cannot simply be thought of as the English “for”, as they are both very rich in meaning. To use them successfully you must think beyond the surface word of for and think of the meaning. Para is the easier, more limited one. Think of it as handing something to someone, as destination of an object. Este regalo es para ti.

Other uses are:

1. For the purpose of or in order to

2. Directed to (here’s the dinner order – las enchiladas para mí, por favor).

3. In the direction of when talking about a place. Voy para México este fin de semana.

4. Meaning by when it’s a time in the future. Necesito las camisas para el viernes.

In a very broad sense you can think of por as being more inward and para with an arrow pointing outward. A really frequent thing to say is Gracias por todo. In this sense, it means in exchange for. Other uses with this exchange idea are in buying things. ¿Cuánto por el libro? Pagué 10 dólares por el regalo.      Think of por as having these extended meanings: Movement along, through, around, by or about: Caminé por toda la ciudad. Háblame por teléfono. Time or duration when something occurs. Estudié por 2 horas. Cause (not the purpose) of an action: La compañía fracasó por la economía. Per: Diez por ciento. 10%  Supporting or in favor of: Voté por él como presidente. For the sake of: Se sacrificó por el honor. Agent of an action meaning by. Fué escrito por Shakespeare.

Means of transportation: Me gusta viajar por tren.

There are many idioms with por: por cierto – certainly, por primera vez – for the first time, por todas partes – everywhere, por fin – at last, por eso – therefore, por lo general – generally, as a rule, por dentro – inside, palabra por palabra – word for word, por supuesto – of course, por casualidad – by chance, por mi parte – as for me, por lo menos – at least, por allí– around there, that way, por adelantado – in advance

And don’t forget to distinguish between Por qué, meaning why and Para qué, meaning for what purpose.

## Use of the word Por

Para and por cannot simply be thought of as the English “for”, as they are both very rich in meaning.  To use them successfully you must think beyond the surface words.  Para is the easier, more limited one.  Think of it as handing something to someone, as destination of an object.  Este regalo es para ti.

Another use is for the purpose of – Este ejercicio es para aumentar (increase) el vocabulario.   In a very broad sense you can think of por as being more inward and para with an arrow pointing outward.  A really frequent thing to say is Gracias por todo.  In this sense, it means in exchange for.  (Using para here is never correct, but it’s a common mistake we see frequently here in the office when people write us.)

Other uses with this exchange idea are in buying things.  ¿Cuánto por el libro?  Pagué 10 dólares por el regalo. Think of por as having these extended meanings:             By (times), per:  Dos por cuatro son ocho.  Voy a la clase dos veces por semana.  Prefiero viajar por tren.  Necesito hablar por teléfono con mi amigo.  El libro fue escrito por Carlos Fuentes.              Through (along), throughout, during   Caminamos por el parque.  Voy a hacerlo por la mañana (an unspecified time).

For the sake of, in favor of, on behalf of     Lo hice por el amor.  Voté por el mejor candidato.

Contrast the following: Compré el regalo para ti.  (It’s going outward to you as a recipient.)  Compré el regalo por ti.  (I bought it on your behalf or for your sake.)

Trabajo para la compania, y trabajo por mi familia.  I work for the company, and I work for the sake of my family.

For a period of time – Leí el libro por dos horas anoche.  But here be careful.  Para is used for a time deadline or specific time in the future.  Necesito el taxi para el lunes.  Te amo para siempre.

There are many idioms with por:  por cierto – certainly, por primera vez  – for the first time, por todas partes – everywhere, por fin – at last, por eso – therefore, por lo general – generally, as a rule, por dentro – inside, palabra por palabra – word for word, por supuesto – of course, por casualidad – by chance, por mi parte – as for me, por lo menos – at least, por allí– around there, that way, por adelantado – in advance

And don’t forget to distinguish between Por qué, meaning why and Para qué, meaning for what purpose.


text:
# Force Diagram

(a) Assume that the Earth is a uniform sphere with a radius r which rotates such that the speed of the ground at the equator is v which is constant.

(i) Draw a diagram showing the forces acting on a person standing on the equator.

There would be the following forces acting on the person:

Gravity (mg)
Normal (-mg)
Friction at the point of contact to the earth
Centripetal force (?)

(ii) Explain why a person in this situation must experience a net force even though their speed is constant.

Does this have to do with the centripetal force? So they need to have a net force towards the center of the earth??

Any help with these would be greatly appreciated.

CompuChip
Homework Helper
Since you are talking about force diagrams, I assume you know about vectors. In (ii) the question says that $|\vec v|$ is constant, but is $\vec v$ constant as well?

so the magnitude of v is constant but the direction is changing hence you are accelerating and if you are accelerating your are experiencing a net force?

Doc Al
Mentor
so the magnitude of v is constant but the direction is changing hence you are accelerating and if you are accelerating your are experiencing a net force?
Absolutely. So redo your analysis of the forces acting on the person.


text:
## Conjugacy Classes of non-abelian group of order p^3

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Let p be a prime, and let G be a non-abelian group of order $p^3$. We want to find the number of conjugacy classes of G.

First we prove a lemma: Z(G) has order p.

Proof: We know that since G is a non-trivial p-group, then $Z(G)\neq 1$. Since $Z(G)\trianglelefteq G$, by Lagrange’s Theorem, $|Z(G)|=p,p^2,\text{or }p^3$.

Case 1) $|Z(G)|=p$. We are done.

Case 2) $|Z(G)|=p^2$. Then $|G/Z(G)|=p^3/p=p$. Thus $G/Z(G)$ is cyclic which implies that G is abelian. (contradiction).

Case 3) $|Z(G)|=p^3$. This means that the entire group G is abelian. (contradiction).

Next, let $O(x_1),\dots, O(x_n)$ be the distinct conjugacy classes of G.

$O(x_i)=\{gx_i g^{-1}:g\in G\}$, where $C_G(x_i)=\{g\in G:gx_i=x_ig\}$.

Then by the Class Equation, we have $\displaystyle p^3=|G|=\sum_{i=1}^n [G:C_G(x_i)]$.

If $x_i\in Z(G)$, then $C_G(x_i)=G$, which means $[G:C_G(x_i)]=1$.

If $x_i \notin Z(G)$, then $C_G(x_i)\neq G$. Since $x_i\in C_G(x_i)$, thus $Z(G)\subsetneq C_G(x_i)$. Thus we have $p=|Z(G)|<|C_G(x_i)|<|G|=p^3$. Since $C_G(x_i)$ is a subgroup of $G$, Lagrange’s Theorem forces $|C_G(x_i)|=p^2$. Thus $[G:C_G(x_i)]=p^3/p^2=p$.

By the Class Equation, we thus have $p^3=p+(n-p)p$, which leads us to $\boxed{n=p^2+p-1}$.


text:
## Publicacions Matemàtiques

### Stabilization in $H^\infty_{\mathbb{R}}(\mathbb{D})$

Brett D. Wick

#### Abstract

It is shown that for $H^\infty_\mathbb{R}(\mathbb{D})$ functions $f_1$ and $f_2$ with

$\inf_{z\in\mathbb{D}}(\vert f_1(z)\vert+\vert f_2(z)\vert)\geq\delta>0$

and $f_1$ being positive on the real zeros of $f_2$, then there exists $H^\infty_\mathbb{R}(\mathbb{D})$ functions $g_2$ and $g_1$, $g_1^{-1}$ with norm controlled by a constant depending only on $\delta$ and

$g_1f_1+g_2f_2=1\quad\forall\; z\in\mathbb{D}$.

These results are connected to the computation of the stable rank of the algebra $H^\infty_\mathbb{R}(\mathbb{D})$ and to results in Control Theory.

#### Article information

Source
Publ. Mat., Volume 54, Number 1 (2010), 25-52.

Dates
First available in Project Euclid: 8 January 2010

https://projecteuclid.org/euclid.pm/1262962131

Mathematical Reviews number (MathSciNet)
MR2603587

Zentralblatt MATH identifier
1193.46033

#### Citation

Wick, Brett D. Stabilization in $H^\infty_{\mathbb{R}}(\mathbb{D})$. Publ. Mat. 54 (2010), no. 1, 25--52. https://projecteuclid.org/euclid.pm/1262962131

#### References

• H. Bass, $K$-theory and stable algebra, Inst. Hautes Études Sci. Publ. Math. 22 (1964), 5\Ndash60.
• J. Bourgain, On finitely generated closed ideals in $H\sp \infty(D)$, Ann. Inst. Fourier (Grenoble) 35(4) (1985), 163\Ndash174.
• L. Carleson, Interpolations by bounded analytic functions and the corona problem, Ann. of Math. (2) 76 (1962), 547\Ndash559.
• G. Corach and F. D. Suárez, Stable rank in holomorphic function algebras, Illinois J. Math. 29(4) (1985), 627\Ndash639.
• P. L. Duren, “Theory of $H\sp{p}$ spaces”, Pure and Applied Mathematics 38, Academic Press, New York-London, 1970.
• J. B. Garnett, “Bounded analytic functions”, Pure and Applied Mathematics 96, Academic Press, Inc. [Harcourt Brace Jovanovich, Publishers], New York-London, 1981.
• P. W. Jones, D. Marshall, and T. Wolff, Stable rank of the disc algebra, Proc. Amer. Math. Soc. 96(4) (1986), 603\Ndash604.
• R. Mortini, An example of a subalgebra of $H\sp \infty$ on the unit disk whose stable rank is not finite, Studia Math. 103(3) (1992), 275\Ndash281.
• R. Mortini and B. D. Wick, The bass and topological stable ranks of $H^\infty_\mathbb{R}(\mathbb{D})$ and $A_\mathbb{R}(\mathbb{D})$, J. Reine Angew. Math. (to appear).
• R. Rupp, Stable rank of subalgebras of the disc algebra, Proc. Amer. Math. Soc. 108(1) (1990), 137\Ndash142.
• S. Treil, Hankel operators, embedding theorems and bases of co-invariant subspaces of the multiple shift operator, (Russian), Algebra i Analiz 1(6) (1989), 200\Ndash234; translation in: Leningrad Math. J. 1(6) (1990), 1515\Ndash1548.
• S. Treil, The stable rank of the algebra $H\sp \infty$ equals $1$, J. Funct. Anal. 109(1) (1992), 130\Ndash154.
• M. Vidyasagar, “Control system synthesis”, A factorization approach, MIT Press Series in Signal Processing, Optimization, and Control 7, MIT Press, Cambridge, MA, 1985.

• See also: Brett D. Wick. Corrigendum: "Stabilization in \boldmath$H^\infty_{\mathbb{R}}(\mathbb{D})$". Publ. Mat. Volume 55, Number 1 (2011), 251-260.


text:
# American Institute of Mathematical Sciences

2007, 2007(Special): 704-712. doi: 10.3934/proc.2007.2007.704

## Blow up and decay bounds in guasi linear parabolic problems

 1 Dipartimento di Matematica e Informatica, Viale Merello 92, 09123 Cagliari, Italy, Italy

Received  September 2006 Revised  May 2007 Published  September 2007

Aim of this paper is to investigate a class of quasilinear parabolic problems whose solutions may blow up at some finite time. We establish conditions on data sufficient to preclude blow up and to insure that the solution and its spatial gradient decay exponentially for all $t > 0$.
Citation: Monica Marras, Stella Vernier Piro. Blow up and decay bounds in guasi linear parabolic problems. Conference Publications, 2007, 2007 (Special) : 704-712. doi: 10.3934/proc.2007.2007.704
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Impact Factor:


text:
## b.reichard Group Title The male to female ratio in taiwan is 52 to 48 write this ratio as a fraction in reduced form one year ago one year ago

1. cwrw238 Group Title

52 = 2 * 2* 13 48 = 2*2*2*2*3 cancel ut the factors which are common

2. AnimalAin Group Title

$\frac{52}{48}=\frac{13}{12}$

3. rcmagnolia Group Title

thanks


text:
# Can one choose multiple values for a param in YAML for parameterized markdown?

#1

The YAML header allows us to set parameters using the params and choices variables. However, it seems that we can only choose one value from the choices list. Is there a way to select multiple (and hence to have a multi-item vector parameter)? For example, in each analysis report, I would like to compare two fruits (apple vs orange, apple vs pear, etc).

Since I don't know of a way to multi-select from choices, I tried:

params:
fruits:
value: !r c("apple","orange")
choices:
- !r c("apple","orange")
- !r c("apple","pear")
- !r c("pear","orange")

But I got this error " The following parameters cannot be customized: fruits". So I don't even get the choices list to choose from. But it actually works for the default value if I ignore the error and click "Knit".

0 Likes

#2

I haven't used the choices option, but couldn't you pass a vector of two values as a single parameter? For example, to knit an rmarkdown file for every pair of fruits you could do:

library(rmarkdown)
library(tidyverse)

x = c("apple", "pear", "orange", "cherry")

map(combn(x, 2, simplify=FALSE),
~ render(input="myfile.rmd",
output_file = paste("myfile", .x, ".pdf"),
params=list(fruits=.x)))

Then, in myfile.rmd you would presumably have code that would compare each pair of fruits that was passed in as params$fruits. 0 Likes #3 Thanks @joels! This will not output the user interface to choose, but most of the time I run all combinations anyway. So this will be very helpful. I tried it out, but got a "Error: pandoc document conversion failed with error 1". It's a separate question (perhaps some pandoc setting?), but thought you maybe able to point out the issue. Here is the output when running the knit: R-3.5.1> map(combn(x, 2, simplify=FALSE), + ~ render(input="<full_path>/test_markdown3.Rmd", + output_file = paste("test", .x, ".html"), + params=list(fruits=.x))) processing file: test_markdown3.Rmd |............. | 20% inline R code fragments |.......................... | 40% label: setup (with options) List of 5$ include: symbol F
$echo : logi FALSE$ results: chr "hide"
$message: logi FALSE$ warning: logi FALSE

|.......................................                          |  60%
ordinary text without R code

|....................................................             |  80%
label: unnamed-chunk-1
|.................................................................| 100%
ordinary text without R code

output file: test_markdown3.knit.md

/<some_path>/apps/pandoc/1.19.2.1-goolf-1.7.20/bin/pandoc +RTS -K512m -RTS test_markdown3.utf8.md --to html4 --from markdown+autolink_bare_uris+ascii_identifiers+tex_math_single_backslash --output pandoc2f01f5c4dacfa.html pandoc2f01f29532b9d.html --smart --email-obfuscation none --self-contained --standalone --section-divs --template /<some_path>/apps/R/3.5.1-goolf-1.7.20/lib64/R/library/rmarkdown/rmd/h/default.html --no-highlight --variable highlightjs=1 --variable 'theme:bootstrap' --include-in-header /tmp/RtmpCNYbIW/rmarkdown-str2f01f2ce20a4c.html --mathjax --variable 'mathjax-url:https://mathjax.rstudio.com/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML' --variable code_folding=hide --variable code_menu=1
pandoc: pandoc2f01f29532b9d.html: openFile: does not exist (No such file or directory)
Error: pandoc document conversion failed with error 1
In if (nzchar(file_ext)) paste(".", file_ext, sep = "") else "" :
the condition has length > 1 and only the first element will be used

Note that I do need to put in the full path of the rmarkdown file even though they are in the same path. Not sure why. Also, what does the "." in ".x" mean?
Thanks!

0 Likes

#4

I'm not sure what specifically is going wrong. Could you post a self-contained reproducible example (i.e., a minimal rmarkdown document and perhaps a small data sample, if necessary (though you could probably build the data sample into the rmarkdown document), that I can run on my machine and that reproduces the error you're getting?

The .x is a "pronoun" that refers back to the first argument of map (combn(x, 2, simplify=FALSE), which is the list we're iterating over. Each .x is one of the elements of combn(x, 2, simplify=FALSE)). Also, in case it was confusing, the x and the .x have nothing to do with each other. I should have called the x vector fruits or something like that to avoid confusion. See the help for map for more details on the .x argument.

0 Likes

#5

Thanks @joels for the explanation. Here is the minimal Rmd:

---
title: "Compare r params$compareFruits" params: compareFruits: !r c("apple","orange") output: html_document: code_folding: hide --- {r} cat("calling params$compareFruits: \n", params\$compareFruits)

And the R script to knit the Rmd:

library(rmarkdown)
fruits = c("apple", "pear", "orange")

map(combn(fruits, 2, simplify=FALSE),
~ render(input="~/test_markdown3.Rmd",
output_file = paste("test", .x, ".html"),
params=list(compareFruits=.x)))

And some of the sessionInfo (the full sessionInfo maybe too long for here):

rmarkdown_1.10
knitr_1.20
pandoc/1.19.2.1-goolf-1.7.20

0 Likes

#6

It seems the problem is in the file name, since combn returns a list for each combination which causes paste to vectorize over it.

For example, the first element in the list created by combn is [1] "apple" "pear" so paste will create this:

paste("test", combn(fruits, 2, simplify=FALSE)[[1]], ".html")
[1] "test apple .html" "test pear .html"

You can collapse the elements into each other and create the filenames that way, which you can see below:

library(rmarkdown)

fruits = c("apple", "pear", "orange")

purrr::map(combn(fruits, 2, simplify=FALSE),
~ render(input="params.Rmd",
output_file = paste0( paste0(.x, collapse = "_"), ".html"), # File name as fruit combination
params=list(compareFruits=.x)))

1 Like

#7

Ah, yes! Thanks @jdb ! It works now.

0 Likes

#8

I forgot to account for the two-element vector represented by .x, treating it as if it were a single element. Thank you @jdb for the correction!

0 Likes


text:
# Does a target have total cover from a ranged attacker positioned on an opposite corner of a hallway?

Consider the diagram below:

 N    *******   * = Wall
W E   .......   . = Open Space
S    .A***T.   A = Attacker
T = Target

# Does the target have total cover from the attacker?

The following quote has the relevant rules:

To determine whether your target has cover from your ranged attack, choose a corner of your square. If any line from this corner to any corner of the target’s square passes through a square or border that blocks line of effect or provides cover, or through a square occupied by a creature, the target has cover (+4 to AC).

[...]

Total Cover: If you don’t have line of effect to your target (that is, you cannot draw any line from your square to your target’s square without crossing a solid barrier), he is considered to have total cover from you. You can’t make an attack against a target that has total cover.

Source: Core Rulebook pg.195-196

My interpretation:

• Every corner of the attacker's square has at least one line to one of the target's square's corners passing through the wall. The target has cover.
• A line can be traced from either of the northern corners of the attacker to either of the northern corners of the target. This line runs parallel to the wall but does cross it. The target does not have total cover.

# Is this interpretation of the rules correct?

• Welcome to rpg.se! Take the tour if you haven't already and check out this meta for other places you can find help. This is a great first question, well explained with great diagrams. Thanks for participating and happy gaming! – linksassin Jun 25 '19 at 3:33

# The target has total cover from the attack

The Core Rulebook doesn't cover this exact situation very well. As you say, you can draw lines parallel to, but not passing through, the wall. So it is possible this only provides cover, not total cover. My ruling is that the wall provides total cover in this situation and I will justify my reasons for it.

### RAW Support from the Hero's Handbook

There is a diagram on page 57 of the Hero's Handbook (from the Beginner's Box) that covers this exact situation. I will try to translate their pretty diagram here:

      ........   * = Wall
N    2.......   . = Open Space
W E   *TT***3.   1,2,3 = Attacker Number
S    *TT****.   T = Target
*1.*****

Paraphrasing their annotations for each attack.

• Target does not have cover from attacker 1
• Target has cover from attacker 2 because at least one line passes through the wall.
• The target has total cover from attacker 3

Attacker 3 in this scenario is identical to your example. The book goes on to state the following about total cover:

If you are completely hidden behind a wall or other obstacle, enemies can't attack you. If every imaginary line you can draw from your corner to the enemy's corner goes through the cover, you have total cover from that enemy and you can't attack each other.

Now you could argue that the Beginner's Box is an overly simplified version of the rules and it doesn't apply to the full game. It does use very similar wording about the imaginary lines from corners however, and it is intended to serve as an introduction to the core rules. So it is certainly RAI for this situation to provide total cover even if RAW is slightly unclear.

### RAW Support from D&D 3.5

3.5 edition D&D is the immediate predecessor to Pathfinder and its ruleset is mostly compatible. In the Player's Handbook for 3.5e on page 139 we have the following text (emphasis mine):

Two creatures can see each other if they can trace at least one clear straight line from any part of one creature’s space to any part of the other creature’s space. The line is clear if it doesn’t cross or even touch squares that block line of sight.

The qualification that the lines cannot touch squares that block line of sight clears this up. Also note that this does not mean that the creatures have partial cover if they both move 1 square north. The rule is any part to any part not all parts to all parts. The lines can go diagonally to avoid touching the wall.

Thanks to HeyICanChan in the comments for this one.

### My interpretation of Core Rulebook RAW

With the Core Rulebook RAW situation being unclear the best I can do it provide you with my interpretation of the rules. This entire situation hinges on the single line which runs from the north edge of the attack to the north edge of the target. The argument that this does not provide total cover it because the line passes by the cover but does not "cross" it.

I would argue that if a line could be consider to not "cross" cover by only passing along the intersection between the cover square and the next, then the line also does not intersect with the targets square by the same logic. Therefore the wall should provide total cover, is the creatures are considered to be filling their entire square so should the cover.

### Use DM discretion

As there is no clear determination either way you can use some situational discretion. If the target is hiding around the corner, attempting to use the wall as cover they are unlikely to leave an arm or a leg sticking out. In that situation I would rule that the wall provides total cover. Conversely, if the target is ducking in and out of cover to fire back at the attacker then you could easily rule that it only provides cover not total cover.

Use logic and common sense to determine a ruling that feels right based on the situation. You mentioned in the comments that the Hero's Handbook example depicts a jagged wall that passes beyond the grid lines. That is the sort of thing you can use to help determine the correct ruling here. Perhaps jagged cave walls provide total cover where smooth castle bricks would not? However you rule is unlikely to make a major difference as the creatures can easily move slightly further into or out of cover to correct the situation.

• Comments are not for extended discussion; this conversation has been moved to chat. – V2Blast Jun 26 '19 at 2:00
• While I was originally hoping for an answer clearly defined by the core RAW, realistically it seems that there simply isn't one. Therefore I'm accepting this answer for citing as many sources as possible as well as offering a very reasonable interpretation and advice for use in-game. Thanks @linksassin! – David Jun 27 '19 at 3:29

## Yes

Assuming the wall does not touch/cross the line at any point. If the wall does touch the line than the target has total cover. Whether the wall touches the line depends on the convention that you use when drawing your maps - the wall can go up to the line or on to the line - if the former it has cover, if the latter total cover.


text:
Concerning
>**Puzzle 287.** Show that if \$$x\$$ is an object in a compact closed category, \$$(x^\ast)^\ast\$$ is isomorphic to \$$x\$$.

To show isomorphism we need a bijection between these 2 objects.
Since \$$x^\ast\$$ are just objects in the same closed compact category, they also have cap and cup (natural) isomorphisms:
\$$\cap: I \to x^\ast \otimes (x^\ast)^\ast\$$ and \$$\cup: (x^\ast)^\ast \otimes x^\ast \to I\$$. Also we have symmetry \$$\sigma: x \otimes y \to y \otimes x\$$. Let's try to construct a bijection between \$$x\$$ and \$$(x^\ast)^\ast\$$:

\$$x \to x\otimes I \to x \otimes (x^\ast \otimes (x^\ast)^\ast) \to (x \otimes x^\ast) \otimes (x^\ast)^\ast \to (x^\ast \otimes x) \otimes (x^\ast)^\ast \to I \otimes (x^\ast)^\ast \to (x^\ast)^\ast\$$

All arrows here are natural isomorphisms: right unitor, cap, associator, symmetry, cup, left unitor. Bijection between \$$(x^\ast)^\ast\$$ and \$$x\$$ might be constructed in a similar way.

Here comes a question: if a category isn't symmetric, \$$(x^\ast)^\ast\$$ is not isomorphic to \$$x\$$ in general?


text:
# Homework Help: Do i add these in paralell?

1. Oct 27, 2005

### mr_coffee

Hello everyone i have the follwoing circuit:
http://www.webassign.net/hrw/hrw7_27-34.gif I need to find the Potential Difference between a and b. I found the current of R1 and R2, and the questions says:
R1 = 130 , R2 = 75 , and the ideal batteries have emfs E1 = 6.0 V, E2 = 5.5 V, E3 = 4.0 V.

R1 current = .042A
R2 current = .0467

Im confused on what I should add u using kricofs rule? like do i just say
Va = -E2 -E3+ R1*i2 + Vb
Va -Vb = -E2 -E3+ R1*i2
But there is a resistor that looks like paraellel, but is that R1 parellel to R2?

2. Oct 27, 2005

### Tom Mattson

Staff Emeritus
Not to me they don't!

Certainly not. To be in parallel they have to have the same potential difference across them no matter what the values of $\mathcal{E}_1$, $\mathcal{E}_2$, and $\mathcal{E}_1$ which is clearly not the case.

Also, note that $R_1$ is in parallel with $\mathcal{E}_2$, but that $R_2$ is not in parallel with $\mathcal{E}_2$. Since they are not both in parallel with $\mathcal{E}_2$, there is no way that they could possibly be in parallel with each other.

3. Oct 27, 2005

### Tom Mattson

Staff Emeritus
That's not right. If you follow the path from $a$ to $b$ you do not go through $R_1$.


text:
# Difference between revisions of "2015 AMC 10A Problems/Problem 17"

## Problem

A line that passes through the origin intersects both the line $x = 1$ and the line $y=1+ \frac{\sqrt{3}}{3} x$. The three lines create an equilateral triangle. What is the perimeter of the triangle?

$\textbf{(A)}\ 2\sqrt{6} \qquad\textbf{(B)} \ 2+2\sqrt{3} \qquad\textbf{(C)} \ 6 \qquad\textbf{(D)} \ 3 + 2\sqrt{3} \qquad\textbf{(E)} \ 6 + \frac{\sqrt{3}}{3}$

## Solution 1

Since the triangle is equilateral and one of the sides is a vertical line, the triangle must have a horizontal line of symmetry, and therefore the other two sides will have opposite slopes. The slope of the other given line is $\frac{\sqrt{3}}{3}$ so the third must be $-\frac{\sqrt{3}}{3}$. Since this third line passes through the origin, its equation is simply $y = -\frac{\sqrt{3}}{3}x$. To find two vertices of the triangle, plug in $x=1$ to both the other equations.

$y = -\frac{\sqrt{3}}{3}$

$y = 1 + \frac{\sqrt{3}}{3}$

We now have the coordinates of two vertices, $\left(1, -\frac{\sqrt{3}}{3}\right)$ and $\left(1, 1 + \frac{\sqrt{3}}{3}\right)$. The length of one side is the distance between the y-coordinates, or $1 + \frac{2\sqrt{3}}{3}$.

The perimeter of the triangle is thus $3\left(1 + \frac{2\sqrt{3}}{3}\right)$, so the answer is $\boxed{\textbf{(D) }3 + 2\sqrt{3}}$

## Solution 2

Draw a line from the y-intercept of the equation $y=1+ \frac{\sqrt{3}}{3} x$ perpendicular to the line $x=1$. There is a square of side length 1 inscribed in the equilateral triangle. The problem becomes reduced to finding the perimeter of an equilateral triangle with a square of side length 1 inscribed in it. The side length is $2\left(\frac{1}{\sqrt{3}}\right) + 1$. After multiplying the side length by 3 and rationalizing, you get $\boxed{\textbf{(D) }3 + 2\sqrt{3}}$.

~savannahsolver

## See Also

Video Solution:

 2015 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 16 Followed byProblem 18 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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# Math Help - applying chain rule to natural log

1. ## applying chain rule to natural log

Hi, I'm very confused about these:
1. y=2^ln x
2. y=ln2^x

for #1, wouldn't it be:
y=2^ln x
y'=ln x (2^ln x) (ln 1/x) ?

but the answer in the book says it's y'=ln x (2^ln x) (1/x).

and I have absolutely no idea how to begin #2.

edit:
sorry, but these too:
3. f(x) = Ae^-Bx
book says f'(x) = Ae^-Bx (-B). I don't understand where would the -Bx go
4. f(x) = L/(1+Ae^-Bx)

thank you so much in advance!

2. Originally Posted by colloquial
Hi, I'm very confused about these:
1. y=2^ln x
2. y=ln2^x

for #1, wouldn't it be:
y=2^ln x
y'=ln x (2^ln x) (ln 1/x) ?

but the answer in the book says it's y'=ln x (2^ln x) (1/x).

and I have absolutely no idea how to begin #2.

thank you so much in advance!
For number 1 use the property that $\frac{d}{du}a^u = a^u\,ln(u)$

a = 2, u = ln(x)

$\frac{d}{dx}(2^{ln(x)}) = 2^{ln(x)}\,ln(ln(x)).$

Perhaps 1/ln(x) is obtained from the chain rule but I don't see it.

For number 2 use the property of logs that says $ln(b^a) = a\,ln(b)$

$y = ln(2^x) = x\,ln(2)$

This should be easy to differentiate. Remember ln(2) is a constant

3. Originally Posted by e^(i*pi)
For number 1 use the property that $\frac{d}{du}a^u = a^u\,ln(u)$
I would suggest you NOT use that property as it isn't true! The correct property is $\frac{a^u}{du}= ln(a) a^u$. Write $a^u= e^{ln(a^u)}= e^{uln(a)}$ and use the chain rule: the derivative is $e^{u ln(a)}$ times the derivative of u ln(a) with respect to u which is ln(a), not ln(u).

a = 2, u = ln(x)

$\frac{d}{dx}(2^{ln(x)}) = 2^{ln(x)}\,ln(ln(x)).$

Perhaps 1/ln(x) is obtained from the chain rule but I don't see it.
The derivative of $2^{ln(x)}$ is, by the chain rule, again, $ln(2) 2^{ln(x)}$ times the derivative of ln(x) which is 1/x. The derivative of $2^{ln(x)}$ is $ln(2)\frac{2^{ln(x)}}{x}$ so your book (believe it or not!) is correct.

For number 2 use the property of logs that says $ln(b^a) = a\,ln(b)$

$y = ln(2^x) = x\,ln(2)$

This should be easy to differentiate. Remember ln(2) is a constant
Exactly right!


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Home OALib Journal OALib PrePrints Submit Ranking News My Lib FAQ About Us Follow Us+
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 Michel Lassalle Mathematics , 1999, Abstract: We present a conjecture about partitions, with a very elementary formulation.
 Narutaka Ozawa Mathematics , 2003, Abstract: This is a detailed survey on the QWEP conjecture and Connes' embedding problem. Most of contents are taken from Kirchberg's paper [Invent. Math. 112 (1993)].
 Carolin Z？belein Mathematics , 2013, Abstract: I want to show one possibility to proof the Collatz conjecture, also called 3n+1 conjecture, for any natural number N. For this, I limit my analysis on the direct odd follower of every natural odd number and show the connections between the already by one reached numbers and their followers, to find an recurrence over all ranges [1,N_{i}], to proof the conjecture.
 Vered Moskowicz Mathematics , 2014, Abstract: The well-known Dixmier conjecture asks if every algebra endomorphism of the first Weyl algebra over a characteristic zero field is an automorphism. We bring a hopefully easier to solve conjecture, called the $\gamma,\delta$ conjecture, and show that it is equivalent to the Dixmier conjecture. Up to checking that in the group generated by automorphisms and anti-automorphisms of $A_1$ all the involutions belong to one conjugacy class, we show that every involutive endomorphism from $(A_1,\gamma)$ to $(A_1,\delta)$ is an automorphism ($\gamma$ and $\delta$ are two involutions on $A_1$), and given an endomorphism $f$ of $A_1$ (not necessarily an involutive endomorphism), if one of $f(X)$,$f(Y)$ is symmetric or skew-symmetric (with respect to any involution on $A_1$), then $f$ is an automorphism.
 BMC Public Health , 2009, DOI: 10.1186/1471-2458-9-431 Abstract: Data on the public's cancer beliefs were collected by means of a postal survey (anonymous questionnaire). Two general practices based in Nottingham and in Mansfield, in east-central England, sent questionnaires to registered patients aged 30 to 70 years. 1,808 completed questionnaires were returned for analysis, 56.5 per cent from women.Women were less likely to underestimate overall cancer incidence, although each sex was more likely to cite a sex-specific cancer as being amongst the most common cancer site. In terms of risk factors, men were most uncertain about the role of stress and sexually-transmitted diseases, whereas women were more likely to rate excessive alcohol and family history as major risk factors. The majority of respondents believed the public health care system should provide cancer screening, but significantly more women than men reported having benefiting from the nationally-provided screening services. Those who were older, in better health or had longer periods of formal education were less worried about cancer than those who had illness experiences, lower incomes, or who were smokers. Actual or potential participation in bowel screening was higher amongst those who believed bowel cancer to be common and amongst men, despite women having more substantial worries about cancer than men.Our results suggest that men's and women's differential knowledge of cancer correlates with women's closer involvement with screening. Even so, men were neither less positive about screening nor less likely to express a willingness to participate in relevant screening in the future. It is important to understand gender-related differences in knowledge and perceptions of cancer, if health promotion resources are to be allocated efficiently.In England, enthusiasm for cancer screening has led to the establishment of organised national programmes, with direct costs being met by the publicly-funded National Health Service (NHS). Since the late-1980s, all women within p
 Physics , 2002, DOI: 10.1016/S0370-2693(03)00127-8 Abstract: We conjecture that in the chiral limit of QCD the spectrum of hadrons is comprised of decoupled, reducible chiral multiplets. A simple rule is developed which identifies the chiral representations filled out by the ground-state hadrons. Our arguments are based on the algebraic structure of superconvergence relations derived by Weinberg from the high-energy behavior of pion-hadron scattering amplitudes.
 Mathematics , 2007, Abstract: In this work we use the number classification in families of the form 6n+1, and 6n+5 with n integer (Such families contain all odd prime numbers greater than 3 and other compound numbers related with primes). We will use this kind of classification in order to attempt an approach to Goldbach strong conjecture. By means of a geometric method of binary bands of numbers we conceive a new form of study of the stated problem.
 P. F. Tupper Mathematics , 2006, Abstract: An open problem in numerical analysis is to explain why molecular dynamics works. The difficulty is that numerical trajectories are only accurate for very short times, whereas the simulations are performed over long time intervals. It is believed that statistical information from these simulations is accurate, but no one has offered a rigourous proof of this. In order to give mathematicians a clear goal in understanding this problem, we state a precise mathematical conjecture about molecular dynamics simulation of a particular system. We believe that if the conjecture is proved, we will then understand why molecular dynamics works.
 Narutaka Ozawa Mathematics , 2012, Abstract: This is an expanded lecture note for "Masterclass on sofic groups and applications to operator algebras" (University of Copenhagen, 5-9 November 2012). It is about algebraic aspects of the Connes Embedding Conjecture. It contains new proofs of equivalence of the Connes Embedding Conjecture, Positivstellensatze for trace positive polynomials, Kirchberg's Conjecture, and Tsirelson's Problem.


text:
# Parameterizing A Circle Projected onto a Plane

## Homework Statement

Find a vector function that parameterizes a curve C which lies in the plane x-y+z=2 and directly above the circle x2 + (y-1)2 = 9

## The Attempt at a Solution

So, in order to parameterize the circle, I simply use x=cos(t), y = sin(t) with some adjustments. Namely, I let x=3cos(t) and y=3sin(t)+1. Then, to determine how the z-coordinates of this curve change according to x,y I use the definition of the plane. So I have z=2-x+y, which I convert using my parametric defintions for x,y.

I have, z=3-3cos(t)+3sin(t). Therefore, my parameterized curve is,

<3cos(t),3sin(t)+1,3-3cos(t)+3sin(t)>

Right?

Related Calculus and Beyond Homework Help News on Phys.org
I like Serena
Homework Helper
Looks good.
But... is it always above the circle?

So I just drew myself a little sketch. Basically I have plane and, beneath it, a cylinder which extends upwards, whose base represents the circle. The intersection of the cylinder and the plane should be the projection of the circle onto the plane, right? In this case, I can't see how the projection could be anywhere else besides over the circle.

As an aside, the vector projection of a onto b is given as follows:

((b [dot] a)/|a|) * a/|a|.

Is it possible to use the idea of projections along with dot products to solve this problem?

I like Serena
Homework Helper
So I just drew myself a little sketch. Basically I have plane and, beneath it, a cylinder which extends upwards, whose base represents the circle. The intersection of the cylinder and the plane should be the projection of the circle onto the plane, right? In this case, I can't see how the projection could be anywhere else besides over the circle.
Consider t=-pi/4.
Is the corresponding point above or below the circle?

As an aside, the vector projection of a onto b is given as follows:

((b [dot] a)/|a|) * a/|a|.

Is it possible to use the idea of projections along with dot products to solve this problem?
This is the formula for projection onto a line.
It doesn't work for projection onto a plane.

Anyway, you still need a parametrization.
The only catch is that the curve is not entirely above the circle (assuming the problem does make this fine distinction).

OH! I see what you mean now. So presumably I need only calculate a z-coordinate for above/below orientation:

z=3-3cos(t)+3sin(t), where t=-pi/4
z=3-3sqrt(2).

So I understand that this point is below the xy-plane, but I don't understand how this impacts the parametrization. Does it change how I am allowed to define t? If the intersection of my cylinder with the plane is still the projection, what difference does "above" or "below" make?

I like Serena
Homework Helper
Well, as I interpret the problem (and I may be wrong), it asks for a parametrization that is "above" the circle.
So I think you're supposed to limit the range of t, such that the curve is indeed "above" the circle.
But perhaps I'm being too nit-picky. You tell me.
Still, it is good practice...