Sunday, May 26, 2013

Tackling Andrica's conjecture. Part 3

Here is an interesting, but slightly detached from the previous two articles, result ... Let's look at the following two sequences: $$a_{n}=\sqrt{p_{n+1}} - \sqrt{p_{n}}$$ $$b_{n}=ln\left ( \frac{1+\sqrt{p_{n+1}}}{1+\sqrt{p_{n}}} \right )^{1+\sqrt{p_{n+1}}}$$

Here is a short Python code to visualise the sequences:

import math
import matplotlib.pyplot as plt

primes = []

def isPrime(n):
    l = int(math.sqrt(n)) + 1
    for i in xrange(2,l):
        if (n % i) == 0:
            return False
    return True

def calculateLog(sqrt_p1, sqrt_p2):
    sqrt_1_p2 = 1.0 + sqrt_p2
    r = math.log(sqrt_1_p2/(1.0 + sqrt_p1))
    return r * sqrt_1_p2

N = 1500000

print "populate primes ..."
for i in xrange(2, N):
    if isPrime(i):

sqrt_diff = [] # sqrt diffs
diff = []      # simple diffs
log_calcs = [] # log calcs
x = []
for i in xrange(1, len(primes)):
    sqrt_p2 = math.sqrt(primes[i])
    sqrt_p1 = math.sqrt(primes[i-1])
    sqrt_diff.append(sqrt_p2 - sqrt_p1)
    diff.append(primes[i] - primes[i-1])
    log_calcs.append(calculateLog(sqrt_p1, sqrt_p2))

for i in xrange(len(sqrt_diff)):
    print sqrt_diff[i]," = s(",primes[i+1],") - s(",primes[i],")"

plt.plot(x, sqrt_diff)
plt.plot(x, log_calcs)
plt.hist(diff, 1000)

And here is how both sequences look like ($a_{n}$ the first and $b_{n}$ the second): Quite asymptotic, aren't they? Indeed they are ...

Lemma 3. $$\sqrt{p_{n+1}} - \sqrt{p_{n}} \leq ln\left ( \frac{1+\sqrt{p_{n+1}}}{1+\sqrt{p_{n}}} \right )^{1+\sqrt{p_{n+1}}} \leq \left ( \frac{1+\sqrt{p_{n+1}}}{1+\sqrt{p_{n}}} \right )\cdot \left ( \sqrt{p_{n+1}} - \sqrt{p_{n}} \right )$$

Let's look at this function $f_{6}(x)=\frac{\sqrt{p_{n+1}}}{1+x\cdot \sqrt{p_{n+1}}}$. Obviously, ${ln\left ( 1+x\cdot \sqrt{p_{n+1}} \right )}'=f_{6}(x)$. As a result $$\int_{\sqrt{\frac{p_{n}}{p_{n+1}}}}^{1} f_{6}\left ( x \right )dx = ln\left ( 1+x\cdot \sqrt{p_{n+1}} \right )|_{\sqrt{\frac{p_{n}}{p_{n+1}}}}^{1}=ln\left ( \frac{1+\sqrt{p_{n+1}}}{1+\sqrt{p_{n}}} \right )$$

According to Mean Value Theorem, $\exists \mu \in \left (\sqrt{\frac{p_{n}}{p_{n+1}}} ,1 \right )$ such that: $$\int_{\sqrt{\frac{p_{n}}{p_{n+1}}}}^{1} f_{6}\left ( x \right )dx = f_{6}\left ( \mu \right )\cdot \left ( 1- \sqrt{\frac{p_{n}}{p_{n+1}}} \right )$$

Putting all together: $$ln\left ( \frac{1+\sqrt{p_{n+1}}}{1+\sqrt{p_{n}}} \right ) = \frac{\sqrt{p_{n+1}}-\sqrt{p_{n}}}{1+\mu \cdot \sqrt{p_{n+1}}}$$

Because $$\sqrt{\frac{p_{n}}{p_{n+1}}}< \mu < 1 \Rightarrow 1+\sqrt{p_{n}}< 1+\mu \cdot \sqrt{p_{n+1}} < 1+\sqrt{p_{n+1}} $$

And we get $$\frac{\sqrt{p_{n+1}}-\sqrt{p_{n}}}{1+\sqrt{p_{n+1}}}\leq ln\left ( \frac{1+\sqrt{p_{n+1}}}{1+\sqrt{p_{n}}} \right )\leq \frac{\sqrt{p_{n+1}}-\sqrt{p_{n}}}{1+\sqrt{p_{n}}}$$ which proves this lemma.

Noting $\Delta_{n}=\sqrt{p_{n+1}}-\sqrt{p_{n}}$, this becomes: $$\frac{\Delta_{n}}{1+\sqrt{p_{n+1}}}\leq ln\left ( 1 + \frac{\Delta_{n}}{1+\sqrt{p_{n}}} \right )\leq \frac{\Delta_{n}}{1+\sqrt{p_{n}}}$$ or $$\frac{1+\sqrt{p_{n}}}{1+\sqrt{p_{n+1}}}\leq ln\left ( 1 + \frac{\Delta_{n}}{1+\sqrt{p_{n}}} \right )^{\frac{1+\sqrt{p_{n}}}{\Delta_{n}}}\leq 1$$

Is this result of any use? I don't know yet, but it looks like: $$\left ( 1 + \frac{\Delta_{n}}{1+\sqrt{p_{n}}} \right )^{\frac{1+\sqrt{p_{n}}}{\Delta_{n}}} \rightarrow e, n \to \infty$$ using this result.

Tuesday, May 7, 2013

Few cloud based APIs

If you came to the point of being ready to start a new project, here are few links to some cloud based APIs to consider:

1. StormPath - quite good a user management and you get free perks as well.

2. Mandrill - if you need a good notification (email based) system. Again, quite a decent free of charge limit.

3. MongoLab - well, that's MongoDB. And if you don't know what it is by now, then you shouldn't worry about it at all (caught your attention? Google for it then). Free 0.5Gb instance is a good starting point.

4. Google Chart API - Not the best one, but it's free and quite ok(ish) for building dashboards.

5. GeckoBoard - a better dashboard API, with HTML5 support (WebSockets and all that stuff), not for free though ...

And of course, you can test all that with Amazon for free for one year.

This is it for now, I will do my best to keep this list updated.