text:
Please use this identifier to cite or link to this item: http://arks.princeton.edu/ark:/88435/dsp010z708z67d
DC FieldValueLanguage
dc.contributor.authorHall, Jonathan V.-
dc.contributor.authorKrueger, Alan B.-
dc.date.accessioned2015-01-22T15:32:18Z-
dc.date.available2015-01-22T15:32:18Z-
dc.date.issued2015-01-
dc.identifier.urihttp://arks.princeton.edu/ark:/88435/dsp010z708z67d-
dc.description.abstractThis paper provides the first comprehensive analysis of Uber’s driver-partners, based on both survey data and anonymized, aggregated administrative data. Uber has grown at an exponential rate over the last few years, and drivers who partner with Uber appear to be attracted to the platform in large part because of the flexibility it offers, the level of compensation, and the fact that earnings per hour do not vary much with hours worked, which facilitates part-time and variable hours. Uber’s driver-partners are more similar in terms of their age and education to the general workforce than to taxi drivers and chauffeurs. Uber may serve as a bridge for many seeking other employment opportunities, and it may attract well-qualified individuals because, with Uber’s star rating system, driver-partners’ reputations are explicitly shared with potential customers. Most of Uber’s driver-partners had full- or part-time employment prior to joining Uber, and many continued in those positions after starting to drive with the Uber platform, which makes the flexibility to set their own hours all the more valuable. Uber’s driver-partners also often cited the desire to smooth fluctuations in their income as a reason for partnering with Uber.en_US
dc.language.isoen_USen_US
dc.relation.ispartofseriesWorking Papers (Princeton University. Industrial Relations Section) ; 587-
dc.titleAn Analysis of the Labor Market for Uber’s Driver-Partners in the United Statesen_US
dc.typeWorking Paperen_US
pu.projectgrantnumber27400 E0292en_US
Appears in Collections:IRS Working Papers

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# Christoffel symbol and einstein summation convention

## Homework Statement

I know that by definition Γijkei=∂ej/∂xk implies that Γmjk=em ⋅ ∂ej/∂xk (e are basis vectors, xk is component of basis vector). Can I write it in the following form? Γjjk=ej ⋅ ∂ej/∂xk Why or why not?

Orodruin
Staff Emeritus

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# How can I make sure message m ∈ Zp

I am implementing ORUTA and using JPBC library for it.
In one of the algorithms, for hashing a message, it is specified that for message m:

$m \in \mathbb Z_p$ , for some large prime $p$.

I have a message text, so how can I make sure that the above equation holds true?
(If possible someone please reply with example code or any predefined funtion able to do this).

• We won't answer you how you can do this with your favorite library, but rather how it is done in theory. If you want code, you have to ask on StackOverflow. – SEJPM Mar 2 '17 at 11:54

This test is really quite simple (this is the theoretical base, not code because this is not StackOverflow):

1. You convert your byte-array m into the integer $m$ according to your applicable standard
2. $$m\in\mathbb Z_p\iff 0\leq m \land m < p$$ So in order for an integer $m$ to be in the additive group $\mathbb Z_p$ it needs to be non-negative ($\geq 0$) and smaller than $p$.
3. Bonus: If you're looking for membership of the multiplicative group instead of the additive group: $$m\in\mathbb Z^*_p\iff m\in\mathbb Z_p\land m\neq 0\iff 0<m\land m<p$$ So in order for an integer to be in the multiplicative group $\mathbb Z^*_p$ it needs to be in the additive group and not be $0$ or (equivalently) it needs to be positive ($>0$) and smaller than $p$.

Note that this all only applies when $p\in\mathbb P$ is a prime.

• This does depend very much on your representation of $\mathbb{Z}_p$, which doesn't have to be the set $\{0,\ldots,p-1\}$. Although in crypto-world this is usually what people use. – CurveEnthusiast Mar 2 '17 at 12:56

It depends on what exactly you want to achieve. I don't know ORUTA.

But in general: You have to convert your message to a numerical representation somehow. Then the message has to be short enough to be in Zp.

e.g. in Java:

    BigInteger p = [your p value]
String message = "test message";
BigInteger m = new BigInteger(1,message.getBytes(StandardCharsets.UTF_8));
if(m.compareTo(p)>0) {
not good, your message is too big!
}

Usually in signature schemes, you use the hash value of the message instead of the message, because a hash function has a fixed output size and for a suitable p you can be certain it is always < p.

This is also quite simple, but if you say you want to compute a hash, it is probably not what you want to do?

• Both answers solved my doubt.. but unfortunately, I can mark only one as answer. Thank you so much. – Kashyap Kotak Mar 3 '17 at 3:42


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×

# Conjecture about Continued Square Roots

Let $$(s_n)_{n = 0}^{\infty}$$ be a sequence of the form $$s_0 = \sqrt{a}, s_1 = \sqrt{a + \sqrt{a}}, s_2 = \sqrt{a + \sqrt{a + \sqrt{a}}}, ...$$

As $$n$$ approaches infinity, the sequence approaches a "continued square root", a number of the following form:

$$x = \sqrt{a + \sqrt{a + \sqrt{a + ...}}}$$

I'll leave it as an exercise to show that if $$a > 0$$, then this continued square root always converges, and the result is the positive root of the equation $$a = x^2 - x$$.

Let's take $$a = 2$$ as an example. By the formula I just mentioned, we have $$\sqrt{2+\sqrt{2+\sqrt{2+...}}} = 2$$.

One question I asked myself is, how fast do these continued square roots converge to $$2$$?

To put it more precisely, let's consider the sequence $$2 - \sqrt{2}, 2 - \sqrt{2 + \sqrt{2}}, 2 - \sqrt{2 + \sqrt{2 + \sqrt{2}}}$$, and so on. The decimal approximations for these numbers are 0.5858, 0.1522, 0.03843, 0.0096305, etc. Do you notice the pattern? The numbers seem to be getting about $$4$$ times smaller each time.

Let's make this more rigorous. In the original sequence, we have $$s_{n+1} = \sqrt{2 + s_n}$$. Therefore, $\lim_{n \to \infty} \dfrac{2 - s_{n+1}}{2 - s_n} =\lim_{n \to \infty} \dfrac{2 - \sqrt{2 + s_n}}{2 - s_n} = \lim_{n \to \infty} \dfrac{(2 - \sqrt{2 + s_n})(2 + \sqrt{2 + s_n})}{(2 - s_n)(2 + \sqrt{2 + s_n})}$$= \lim_{n \to \infty} \dfrac{4 - (2 + s_n)}{(2 - s_n)(2 + \sqrt{2 + s_n})} = \lim_{n \to \infty} \dfrac{1}{2 + \sqrt{2 + s_n}} = \dfrac{1}{4}$

So, the differences between the continued square roots and their limit (2) do, in fact, approach a geometric sequence with $$\dfrac{1}{4}$$ as the common ratio. Hence, $$s_n$$ can be asymptotically approximated by a function $$s_n \approx 2 - \dfrac{b}{4^n}$$.

Now, the million dollar question: What is this constant $$b$$?

In other words, what is $$\displaystyle\lim_{n\to\infty} 4^n(2 - s_n)$$?

I won't tell you the answer in case you want to solve it on your own. If you succeed, I'd love to see how you solved it so feel free to share it in the comments. If you'd like to see my solution, you can find it here.

Now let's generalize the previous result for the real number $$a$$. Let's replace $$2$$ with $$x^2 - x$$, so that our sequence looks like this: $s_n = \left\{\sqrt{x^2 - x}, \sqrt{(x^2 - x) + \sqrt{(x^2 - x)}}, \sqrt{(x^2 - x) + \sqrt{(x^2 - x)+\sqrt{x^2 - x}}} ...\right\}$ You can show (using a similar proof to mine above) that $\lim_{n \to \infty} \dfrac{x - s_{n+1}}{x - s_n} = \dfrac{1}{2x}$

Hence, the generalized sequence can be approximated by the function $$s_n \approx x - \dfrac{L_x}{(2x)^n}$$ where $$L_x$$ is the following limit: $L_x = \lim_{n \to \infty} (2x)^n (x - s_n)$

Now, I have a conjecture for you. I haven't been able to prove/disprove it yet; see if you can:

Let $$s_n$$ be a sequence defined recursively by $$s_0 = \sqrt{x^2 - x}$$ and $$s_{n+1} = \sqrt{(x^2 - x) + s_n}$$ for $$n \in \mathbb{N}$$, for some $$x > 1$$. Let $$L_x = \lim_{n \to \infty} (2x)^n (x - s_n)$$. Then,

$\lim_{x \to \infty} L_x = \dfrac{1}{2}$

Note by Ariel Gershon
2 years, 3 months ago

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Just to confirm, is $$b=\dfrac{\pi^2}{16}$$ ?

- 2 years, 3 months ago

Yes, very good!

- 2 years, 3 months ago

If we can solve the functional equation $$g(x)=a g^2\left(\dfrac{kx}{a}\right) - 1$$ where $$a$$ is a fixed constant and $$k$$ is arbitarary, we are almost done solving the conjecture.

- 2 years, 3 months ago

I'm sorry, how would this help?

- 2 years, 3 months ago

Let $$f_{n}(y)=\underbrace{\sqrt{a+\sqrt{a+\ldots+\sqrt{a+y}}}}_{n \ \text{times}}$$, substitute $$y=a\cdot g(x)$$ to simplify $$f_{n}(y)$$. We can then replace $$a$$ by $$t^2-t$$ and calculate $$L_{t}$$ by letting $$n \to \infty$$. Then we can find the conjectured limit by letting $$t \to \infty$$. For example if we let $$a=2$$, $$k=1$$ we will get $$g(x)=2g^2(\frac{x}{2})-1$$. One solution is $$g(x)= \cos (x)$$. From this we can calculate $$b=\dfrac{\pi^2}{16}$$.

- 2 years, 3 months ago

I'm missing quite a few steps in your logic.... What are $$g(x)$$ and $$k$$?

- 2 years, 3 months ago

$$g(x)$$ is an unknown function which satisfies $$g(x)=a g^2\left(\dfrac{kx}{a}\right) - 1$$. If we substitute $$y=a\cdot g(x)$$ in $$f_{n}(y)=\underbrace{\sqrt{a+\sqrt{a+\ldots+\sqrt{a+y}}}}_{n \ \text{times}}$$, we have $$f_{n} (y) = a \cdot g\left(\dfrac{k^nx}{a^n}\right)$$. And if we replace $$a$$ by $$t^2-t$$, we can then calculate $$L_{t}$$ as a function of $$t$$. This will simplify the conjectured limit $$L_{t}$$. If my reasoning is missing any steps, you can ask and I'll try to fix it.

- 2 years, 3 months ago

OK, it makes more sense now as I was able to finish your argument. This is a very interesting approach! If you don't mind though, I'm going to simplify the equation for $$g$$ to make it $$g(x) = a g^2\left(cx\right) - 1$$ (since $$k$$ is an arbitrary constant, let $$c = \dfrac{k}{a}$$).

There are two more stipulations for $$g$$ and $$c$$ which you didn't mention, but they are important. The proof by induction of this approach requires that the following be true:

(1) there must exist a real number $$z$$ such that $$g(z) = \dfrac{\sqrt{a}}{a}$$

(2) $$g(c^n z) \ge 0$$ for all $$n \in \mathbb{N}$$

Also, we can write $$f_n(a) = s_n$$. Then by induction we can show that $$s_n = a * g(c^n z)$$. If we replace $$a$$ with $$t^2 - t$$, then $$L_t$$ will have the following formula: $L_t = \lim_{n\to\infty} (2t)^n \left(t - (t^2 - t) g(c^n z)\right)$

In the case of $$t = 2$$, we can have $$c = \dfrac{1}{2}$$ and $$g(x) = \cos(x)$$, since, as you mentioned, $$\cos(x) = 2\cos^2\left(\dfrac{x}{2}\right) - 1$$. We have $$z = \dfrac{\pi}{4}$$ since $$\cos\left(\dfrac{\pi}{4}\right) = \dfrac{\sqrt{2}}{2}$$. Then we get $L_2 = \lim_{n\to\infty} 4^n \left(2 - 2\cos\left(\dfrac{\pi}{4*2^n}\right)\right)$ This limit does, in fact, evaluate to $$\dfrac{\pi^2}{16}$$.

Interesting as this is, though, I think it won't be as easy to find $$g, c, z$$ for $$t$$ in general.

- 2 years, 3 months ago

I want to point out that it only applies for $$a > 0$$. (For $$a = 0$$, it converges to the lower root of $$a = x^2-x$$, namely $$x = 0$$ instead of $$x = 1$$.)

- 2 years, 3 months ago

Fair point. I'll fix it

- 2 years, 3 months ago