Differential inequalities

The other thing I was thinking about recently was the idea of differential inequalities I mentioned before.

I think in the case of: x’ > x if x > 0 we can say something about what x(t) must be, as long as x'(t) is continuous.

I think with careful thinking about this there could be used for a proof that there is no infinitely differentiable f(x) such that its integral and derivative are both less than it for all x > 0

If x'(t) is continuous and x'(t) > x(t) for all t > 0, and both x and x’ are always greater than 0 on that range, then set a = x(1), whatever that positive value is.

You can find an exponential function such that it also is a at t = 1. The function would be:

$g(t) = \frac{a}{e} e^t$

Now notice:
x'(1) > a, because x'(t) > x(t) and x(1) = a

And
$g'(x) = D(\frac{a}{e} e^x) = \frac{a}{e} e^x$
$g'(1) = \frac{a}{e} * e = a$

Call x'(1) = b > a. Even if x'(t) from t = 1 onward is dropping back towards g'(t), there still must be some time h before x'(t) = g'(t), because a continuous function can not have a jump discontinuity, it must smoothly merge back into the curve of g'(t).

That is, for some 1 < t < 1 + h, x'(t) > g'(t).

$\int_{1}^{1+h} x'(t) > \int_{1}^{1+h} g'(t)$
$\int_{1}^{1+h} x'(t) = x(1+h) - x(1) > g(1+h) - g(1)$
$x(1+h) - x(1) > g(1+h) - g(1) \rightarrow x(1+h) > g(1+h) - g(1) + x(1)$

But since g(1) and x(1) are both a:

$x(1+h) > g(1+h) - g(1) + x(1) \rightarrow x(1+h) > g(1+h) - a + a$
$x(1+h) > g(1+h) -a + a \rightarrow x(1+h) > g(1+h)$

This relationship is true for anywhere between $1 < t \leq t + h$ because we can use with some other j < h. So throughout this range, x(t) > g(t), or $x(t) > \frac{a}{e} e^x$

However there is no actual h where g'(t) = x'(t) again, because g”(t) = g'(t), x”(t) > x'(t), x'(t) > g'(t), and so x”(t) > g”(t). x is accelerating away from g, so it cannot rejoin the slope of g(t). Similarly we can show that for all t > 1, \latex x^{(n)} > g^{(n)}.

Since x(t) > 0 for all t > 0, we can pick any arbitrary t > 0 and use this same argument that x(t) > c exp(t) for all t > p and some multiple c. Since we can do so for any p > 0, we can show that there is some c such that x(t) is greater than c exp(t) for all t > 0.

For the other direction, x’ < x, we can use the same argument that from some point p, x(t) could not grow as fast or faster than an exponential that has the same value at p.

If x(2) = 1, then $\frac{1}{e^2} e^x$ has the same value at t=2. x’ is less than 1 there, and up to some distance h from there, so it must fall behind in the same way.

Now my question is how far can you take this line of argument, and what are the pitfalls that look like true statements you can make about inequalities, but you can’t.

It seems like the inequalities largely tell us about the growth rates of a function and nothing about its current value or velocity. We could have a function with x(1) = 1000000, but by the argument it still could not grow exponentially, because the derivative must be less than 1000000 there. It might have a connection with O notation.

Space Beat, The Blackboard and Nanowrimo

I picked the name for this blog based on the idea that I’d be writing the story about the blackboard with the power the change the rules of the universe. I think I’m going to wait on that one and let some more ideas come to me, I don’t seem to have a full novel’s worth yet.

I have now brainstormed three different ideas for my Nanowrimo novel, but I think I’m going to go back to the Hard SF idea I originally had. I backed away from it, because it felt like it may be too confining, but I think I will give it a go, and just skip the boring early parts to get right into the action.

I call it Space Beat, but that’s only a early title. Russia, America and Canada are connected by a massive Siberia-Alaska bridge complex, as well as the ease of trans-Arctic trade in the aftermath of global warming. They have formed the Northern Economic Union more than 150 years before the setting start, and formed the NEU bureau of Law Enforcement to coordinate investigation within the three countries, as well as their off world colonies.

NEU is only one of three major space-faring. The EU and Greater China are the others. All three have colonies and space stations scattered out as far as Jupiter, with two new stations just built on Saturn’s moons a few years before the story’s beginning.

The NEU has set up a Space department charged with tracking suspects who have fled off planet. The NEU’s space department is small, and the cost to outfit just one team of officers in the latest cruiser ship is expensive, especially when fuel costs are taken into account.

Unlike the vast majority of space themed stories, I am taking into account gravity and orbital mechanics, as well as the fact that you can’t really accelerate at more than a few g’s without crushing your officers.

The story is primarily about the Rookie Jordan Galloway and the Veteran Quintin “Q” Bass, and their pursuit of suspects, especially a group who are connected to a new criminal syndicate.

The “villain”, Ethan “Thunder Man” has adopted the moon Ganymede in orbit around Jupiter as his home, and he has concocted a scheme to divert attention and police resources to issues elsewhere in the system, so he can build up Ganymede’s defenses and declare independence.

While Jordan and Quintin are the protaganists, I intend to make Ethan’s goals at least reasonable, if not his methods.

Pondering higher order functions and derivatives

So I started thinking about other possible alien math systems the past few days and that got me on the topic of higher order functions. These have many names, and a lot of the complexities of things like functional analysis are beyond me, but I do understand a few of the basics, and I understand higher order functions from the perspective of programming.

A higher order function does the same thing with functions that functions do with numbers. It is a function that takes functions as its parameters instead of numbers. The derivative is a classic example:

$Limit_h \rightarrow 0 \frac{f(x+h) - f(x)}{h}$

If we consider x to be a general variable rather than a specific value to be calculated on, than this definition is a higher order function. We provide a function and get another function back. Just like with sin(x) we provide a number and get a number back.

There are tons of higher order functions, but I’m mostly interested in the ones related to the derivative and integral. Of course the integral doesn’t give us back a single value, so I turn it into a higher order function by taking the integral from 0 to x.

You could also have functions that received higher order functions themselves as parameters, such as the following:

$N(O(f(x)), r) = O(O(O(....f(x))$  –r iterations–
$N(D(x^3), 3) = 6$

That’s just a function that iterates a higher order function a given number of times.

The higher order functions I was thinking of both are similar to infinite series in a way. They bring up the same questions of convergence, but its harder for me to say in what sense a function can converge?

$G(f(x), n) = N(D(f(x)), n), n \geq 0$
$G(f(x), n) = N(I(f(x)), -n), n < 0$

So G is a generalized derivative or integrator that gives the nth derivative for positive n, or the -nth  integral for negative n

$P(f(x)) = \sum_{i=0} G(f(X), i)$
$M(f(x)) = \sum_{i=0} G(f(x), -i)$

P represents the sum of all derivatives of a function, and M is the sum of all integrals of a function. P only exists where f(x) is infinitely differentiable, and M exists wherever f(x) is continuous. P seems to “converge” for a lot more functions than M converges for.

Examples of P converging:

1. Any polynomial converges because the derivative will eventually be 0
ex. $x^3 \rightarrow x^3 + 3 x^2 + 6 x + 6$

2. $a^x$ for a < e converges because the derivative decreases like a geometric series

Examples of M converging:

1. $a^x$ for a > e converges in the same kind of geometric series except the multiplier is 1 / log(a) instead of log(a).

The question that I’m thinking about now is whether any function has the property that both P and M converge for it.

Sin(x) might seem to converge for both P and M, since its derivatives cancel each other out. Unfortunately, Sin(x) does not converge to any one value at infinity.

If P and M both converge for a function I think it would be a rather special function. The thing is that, in order for P to converge, the derivatives have to be getting smaller in magnitude. So Abs(G(f(x), n)) has to be decreasing, but that seems to imply that the Integrals are generally increasing in magnitude.

It seems to me that In order for both P and M to converge, there has to be a function that’s the “max” in some sense, so that the derivative and integral of that function are both smaller everywhere than the function itself. That function if found would define a whole range of functions which have converging P and M values.

Does there exist a function f(x) such that for all x > 0 $0 < \int_{0} f(x) < f(x) \land 0 < D(f(x)) < f(x)$ ?

We can think of this, in a way, as a differential inequality. I don’t know if much is known about this sort of thing, but I certainly don’t know how to proceed with it that far. If we consider the integral to be the function g(x) and focus on that we get:

g’ = h
$x>0 \rightarrow g' > g > 0$
$x>0 \rightarrow 0 < h' < h$

Functions that grow very rapidly have larger derivatives and even larger 2nd derivatives and so on. Whereas other functions such as polynomials have their integrals grow such that at some positive a, the higher order terms will dominate and make them larger.

So our f(x) would have to be some strange function, small enough that its derivatives are not any larger, but also strange enough for its integrals to always be smaller. I can imagine this being true on a small interval but not the entire positive real line.

Getting a better idea of what functions have smaller derivatives for all positive x, and which functions have smaller integrals would go a long way to figuring it out, so I will think on that some more.

The Reveal, Part 2

Here are the functions uses in this Alien Math scheme:

D(x) = tanh(x) or the hyberbolic tangent.

$tanh(a+b) = \frac{tanh(a)+tanh(b)}{1+tanh(a) tanh(b)}$

$Z(a,b) = \frac{a+b}{1+ab}$

$C(x) = atanh(x) = \frac{1}{2} log(\frac{1+a}{1-a})$

Z(D(a), D(b)) then is tanh(a + b). So that is why C(Z(D(a), D(b))) is a + b

This is all pretty familiar stuff. You can the expression in Z if you take Einstein’s formula for relativistic velocity addition, and set the units so that c = 1.

In the wikipedia articles on Special Relativity, they mention something called rapidity which is the arctanh of the velocity. All of the relativistic velocity stuff can be turned into normal addition by using this arctanh.

W(a, b) was then the expression for getting the original velocity given the combined velocity and one of the originals.

$W(a,b) = \frac{b^2 - a*b}{a b^2 - b}$

If a = 0, then W(a,b) is -b which is a handy way to get negatives.

The next step was to see if something akin to relativistic multiplication could be done. If you calculate Z(a,a), Z(a,Z(a,a)) and so on, eventually you discover a pattern of odd polynomial terms in the numerator and even terms in the denominator. These are the terms of the binomial expansion of $(a+1)^b$

An example of what this looks like would be:

$Z(a, Z(a,a)) = \frac{a^3+3 a}{3 a^2 + 1}$
$Z(a, Z(a, Z(a,a))) = \frac{4 a^3+4 a}{a^4 + 6 a^2+1}$

I studied these even and odd sequences and learned that you can represent them using two binomials

Even terms: $(a+1)^b + (1-a)^b$
Odd terms: $(a+1)^b - (1-a)^b$

Extending the exponents to all values, then, we get the analogue of multiplication in this system:

$K(a,b) = \frac{(a+1)^b - (1-a)^b}{(a+1)^b + (1-a)^b}$

So D and C are the hyberbolic tangent and its inverse, which can be used to exploit the tanh’s angle addition identity. The same addition identify can be extended to K to prove that C(K(D(a), b)) = a b.

And you need a separate value for 1 for two reasons. If aliens scaled their numbers so that 1 was C, it would make sense for the numeral system to rely on smaller numbers. Also 1 can’t be produced in the system, and the only reciprocal of 1 is 1 itself, so you need a special number E for 1.

Since K is already non-associative, I don’t think any extension to an analogue of exponentiation can be made. K is already an exponential function.

Alien Math: The Reveal, Part 1

In part one, I will give strong hints as to what the Alien math system is, in part 2 I will reveal it all and why I think its a particularly coherent system. While working on the derivative from my first post, I started to get a sense of what thinking in this system is like.

Its all about function D, and its inverse, function C. D is the wrapper, and C is the unwrapper, and certain functions work best if their operands are wrapped, and some work best if they are unwrapped.

The expression for $\frac{dZ}{da}$ is huge so I will build it up, bit by bit.

First, we have K(D(b), b). It seems that K wants the first operand wrapped and the second not. We will name this sub expressions e1, e2, e3 and so on.

e1 = C(K(D(b),b))

So we wrap it up for K to work on, and the unwrap the result.

Reveal #1: C(K(D(a), b) = a * b

So e1 = C(K(D(b), b) = $b^2$

You can chain these values by stacking the repetitions on the left side like in:

e2 = C(K(K(K(D(b),b), a), a)) = $a^2 * b^2$
e3 =  C(K(K(QT, a), b) = 2 a*b

Its not as elegant because K is not associative or commutative, unlike ordinary multiplication.

Reveal #2: The numeral system uses RSTU as 0,1,2,3 in base 4. However, numbers start after the decimal point, so you have to use Q, which is the reciprocal function, to get numbers larger than 1.

For numbers like three that are difficult to get in base 4 using this method, you can use the alternate system. It uses E’s to count up small numbers. E = 1, EE = 2, EEE = 3 and so on.

QT is $\frac{1}{0.2 base 4}$ which is 2.

So we have our products and our literal numbers.  Now:

Reveal #3: C(Z(D(a), D(b))) = a + b

The wrapping and unwrapping is similar to K except both parameters are wrapped. It can be extended similarly to K.

e4 = C(Z(D(e2), Z(D(e3), D(E))) = e2+e3+1 = $a^2 * b^2 + 2 a b + 1$

Reveal #4: W(R, a) = -a

W(a,b) is not a – b, but when the first parameter is R (0), the result is the negative of the second param. This is the only way to introduce negative numbers in this system.

e5 = C(Z(D(W(R, e1)), D(E))) = -e1 + 1 = $-b^2 + 1$

Using Q as the reciprocal, and the construction in Reveal #1, we can construct the final expression, and expand it using the e’s

dz/da = C(K(D(e5), Q(e4)))
dz/da = C(K(D(C(Z(D(W(R, C(K(D(b),b)))), D(E)))), Q(C(Z(D(C(K(K(K(D(b),b), a), a))), Z(D(C(K(K(QT, a), b)), D(E))))))

You can simplify further by cancelling the C(D(?)) and D(C(?)) expressions.

dz/da = C(K(D(C(Z(D(W(R, C(K(D(b),b)))), D(E)))), Q(C(Z(D(C(K(K(K(D(b),b), a), a))), Z(D(C(K(K(QT, a), b)), D(E)))))) = $\frac{-b^2+1}{a^2 b^2 + 2 a b + 1}$

Correction and A few notes

I had the function C wrong in my notes, which affected example #4 in the Alien math post. It should have been:

4. C(UU) = TRURTSUU…

I kept trying to calculate whether the functions in the sytem could represent all polynomials, all rationals, etc, and I kept coming up with results I knew were wrong. Eventually I realized it was function C which I had set up incorrectly.

#6 is still the same, because its the identity I thought the original C held for, but in fact it was wrong for that C and right for the correct C.

I also realized my numeral system was incapable of representing negative numbers directly, but it can do so using one of the functions.

This system was motivated by the same time of extensions that were used with our ordinary arithmatic, except that the start operator is different so the extensions are different, but they have parallels.

One of the functions / operators is a simple named one in our system. It was carried over because it fit naturally into the system. One of the other operators is named in some sources, but it is somewhat obscure. The rest are expressions, not fundamental functions or operators in our system.

This system makes some limited sense, as far as aliens using it. Certain expressions are a lot simpler in it, but derivatives and other calculations produce very messy expressions in the Alien system. Maybe parts of it are actually used with different names and symbols, in the real world.

I’ll reveal the big clue next week and then a couple days after that show the functions and operators.

The Blackboard: Idea

This blog is half about my upcoming novel “The Blackboard” (name may change), which is about a blackboard that has the power to change reality based on the equations written upon it.

This blog is also half about the fusion of creative and mathematical ideas. I like to write, and though I don’t have a degree in mathematics, I can’t help but think about the ideas that I can understand. The appeal of the “The Blackboard” to me, as a writer is that I can fuse these two things into a story.

My plan is to bring creative mathematics as well as updates on my story’s progress, as well as the progress of things to be written in the future.

There will be plenty of equations, but nothing beyond Undergrad college stuff. It won’t always be mathy stuff, so I hope not to scare anyone away too much.

All the obvious names with blackboard in them were taken, so I had to settle for my user handle + blackboard @ wordpress.com. Maybe I will be able to come up with something that is not taken but still interesting later.

The Blackboard Synopsis

The full equations of reality fill only one blackboard, in Stanford, in a locked classroom that hasn’t been used in two semesters. The blackboard appears just a few years from now, when science has made another small yet crucial step forward.

The classroom has been unlocked. The facilities team forgot to lock up after replacing the broken projector. A grad student from Colorado scrambles to find a blackboard to write down his fading insight on a stumper of a problem his professor gave him. He stumbled into the classroom and sees the blackboard full of equations.

The greek characters are recognizable, but the way they are used in the equations stumps him. After looking at the board for a few seconds, he realizes there is no message to not erase the equations. So he erases them, and frantically writes down his solution to his Astrophysics homework. He takes a picture of the solution with his phone’s camera. He didn’t think to take a picture of the weird equations on there before.

He made a slight mistake, one that would only cost him two points from the professor. It is a typo that would have massive effects. From the moment he left the classroom, the Universe was governed by his equations, including the erroneous bit.

The error begins to cause huge problems, but the light caused by the changes will not arrive for millions of years in most cases. However, in particle accelerators, the change in equations starts to be noticed. It affects only the most high energy collisions at CERN.

The blackboard was placed in the classroom in question for the benefit of the professor teaching the next class in the room. He is one of the world renowned physicists. He is leading the search to unlock the greatest mysteries of physics and cosmology.

The student realizes what he has done after hearing about the CERN’s findings, and tries to fix the equations. Each time he tries, he makes things worse. Only the student and professor are unaffected by the botched equations. And eventually the professor discovers the student and blackboard.

Then the professor has to attempt to write the equations that will restore the Universe to what it once was. He hopes that he can draw more and more out of the student’s memory of the original blackboard scene.

Complicating things is third force in this story: the mysterious being who placed the blackboard in that classroom for the professor to change reality. Is the mysterious being good, bad or cruelly indifferent?