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The hackers within Microsoft must know in their hearts that if the company really cared about users they’d just advise them to switch to OSX.

-Paul Graham

I never thought that a dog could be summarized in such a simple program.

while(true)
{
    dog.DropBall();
 
    Sleep(5);
    dog.GoGetBall();
}

I plan on extending my “holidays” another month. I still have software projects to finish up (which includes rewriting Flog).

If the rod has insulated ends () and obeys the heat equation

,

What is the heat distribution of the rod as a function of time?

Answer:
First assume that the solution to the PDE

,

has the form

Therefore, we can reduce the equation to the following:

Since this equation is true for all x and t, therefore both sides of the equation must equal a constant.

which can be written as two ODEs

Solving the first-order differential is trivial and can easily be shown to have the solution:

Next, we need to solve

The second-order differential has 3 different cases (, , ).

Case
The ODE for this case would be

which has the solution

If we apply the BC , we get

which implies

Case
The ODE for this case would be

which has the solution

Next, we apply the BC .

Since , this implies . Therefore,

If we apply the 2nd BC , we find the following

Therefore, for non-trivial solutions where .

This means,

is a solution.

Case
The ODE for this case would be

where and

which has the solution

If we apply the BC ,we find

Since , we know that which means

is a solution.

If we apply the 2nd BC , we find

Since and , we know that .

This means that there is no solution where .

Since the PDE will satify any linear combination of the above solutions, we find that

If we apply the IC to the solution, we find that and every other is zero.

Therefore, the solution would be

If the rod has insulated ends () and obeys the heat equation

,

What is the heat distribution of the rod as a function of time?

Answer:
First assume that the solution to the PDE

has the form

Therefore, we can reduce the equation to the following:

Since this equation is true for all x and y, therefore both sides of the equation must equal a constant.

This implies that we need to solve two ODEs.

and apply the results to the two ODEs.

There are 3 cases which we need to solve (, , and ).

Case
The ODE for this case would be

which has the solution

Case and
The ODE

which will have the solution

Please note that will be an imaginary number when .

Therefore, if we apply the above solution, we can find the functions that solve the ODEs

which would be

when

when

when

Therefore, the solution would have the form

Any superposition of the above equation will satisfy Laplace’s equation. In order to reduce this solution more, we would need to be given Boundary and Initial Conditions.

Besides solving Differential Equations analytically, I have been looking at numerical approaches to solving these problems. If you use Maple or Matlab, it is fairly straightforward to solve Differential Equations numerically. However, these approaches are not really optimal for larger problems (especially problems that require parallelization). In order to solve these larger problems, it usually requires using a “real” language.

Initially, I started looking for numerical libraries for Perl. The only library that looked decent was Cephes which seems a bit incomplete. The Python library called NumPy seems a lot better, but this would require me to learn Python.

Anyways, while searching the web, I came across a cool little site called “My Physics Lab”. It contains a bunch applets that have physics simulations. The coolest one, I think, is the double pendulum. It is a simple example of chaotic motion.

Also, if you know of any good numerical libraries, please leave a comment (unless it is in C/C++ or Fortran).

Note: Since there are no boundary conditions, there won’t be an exact solution.

Answer: According to Newton’s 2nd Law of Motion, we know that

If we create a force diagram, we can easily see that .

where .

Therefore, we can create the equation of motion as follows:

In order to solve this differential equation, let .

This means:

Therefore, the differential equation becomes

We can solve this using the integrating factor method. According to our differential equation, our integrating factor will be . If we multiple our integrating factor to our ODE, we get

However, since , we know that

If we add the boundary condition, , the equation reduces to

Find the general form for this sequence.

Answer: When analyzing this sequence, the first thing I tried was comparing the difference between each number in the sequence.

Lets investigate this new sequence.

It appears the odd terms are related to each other and so are the even terms. Lets break the odd terms and even terms into two different sequences.

As you can see, the odd terms has the form:

The even terms, likewise, can be seen to have the form:

or

If we merge these results, we can get pattern for the following sequence

which is

We can then use this result to find the solution to

which would be

where .

Find the general form for this sequence.

Find the wavefunction that satisfies Schrödinger’s equation.

Answer:

First assume that can be represented as . This means that Schrödinger’s equation can be manipulated as followed:

Since the this equation holds for all and , both sides of the equation must equal a constant.

Time Component
Next, we will solve for the time component of the above equation.

Time-Independent Component
The time-independent component of the Schrödinger equation would be:

The solution to this problem can be easily solved for the case and . Since , the only way to solve

is to have .

For the case when , we know that . Therefore, the time-independent equation will become:

This equation should look rather familiar because it is the differential equation describing a simple harmonic oscillator. Therefore, the general solution to this differential equation would be:

Therefore, the solution would be:

If we add a requirement that needs to be piecewise-smooth, we can add boundary conditions and to the problem:

Therefore, when , we can see that because

Now that we know , we can apply the second boundary condition to the remaining portion of the equation.

where

This implies

Therefore, we can see that the general solution can be reduced to:

The last step is to normalize this wave-function

Therefore

Therefore, the final solution would be

where

If you want to see the presentation, you can see it on YouTube.

Find the wavefunction that satisfies Schrödinger’s equation.

Answer: In order to balance the chemical equations, we need to find the five unknowns (ab,c,d,e).

Since the number of chemical elements are conserved, we know that:

we need to solve the following linear system.

This system of equations can easily be solved with Maple.

> with(Student[LinearAlgebra]):
> A := Matrix([[1, 1, 0, -1, 0, 0],
		[1, 2, 0, 0, -1, 0],
		[0, 0, 2, -4, -2, 0]]):
> ReducedRowEchelonForm(A)

According to maple, we can reduce the linear system to:

According to the above equation, we don’t have a unique solution. However, we know that the linear system will have a solution where

Since we know that (where ), we can add additional constraints to our linear system. If we look at the above equations, it is easy to see that the following constraint must hold:

Using the linear system and the above constraint, we find the following solutions…

…
…

As you can see, there is still an infinite amount of solutions for this chemical equations. Some can be reduced, however majority of the equations can’t be reduced

Answer: In order to solve this problem, I am going to define the questions a little more formally.

Prove that there a where given

and

I will be solving this problem using induction.

Base Case:
First we need show that the problem is true for the base case ()

Therefore, there a (which would be ) where .

Induction Step (for positive values):
In order to prove that the problem holds for , we need to show:

If where where

In order to prove this statement, we need to reduce .

Since , we have shown if it implies that where .

Induction Step (for negative values):
Since we already proved by induction that the problem holds for all natural numbers, all we need to do to extend this proof to all integers is to show that:

If where where

In order to prove this statement, we need to reduce .

Since , we have shown if it implies that where .

Therefore, we have proven by induction that the sum of the cubes of any three consecutive integers is divisible by 9.

Now, I want you to look at the new Shreddies and then the old one. If for some reason you failed Kindergarten (and didn’t bother to repeat it), I want you to rotate your head 45 degrees. Wow, look at that, the old Shreddies look like the new improved Diamond Shreddies.

I don’t understand how they can justify advertising for this product. Do they think the general public are absolute morons! Don’t you think people will clue in when their bowl of Shreddies Diamonds looks identical to the regular Shreddies? They didn’t even bother changing the grid, so at least something would be different.

I refuse to buy this stuff. If I really want to eat Shreddies Diamonds, I will just buy regular Shreddies and rotate my coordinate system by 45 degrees.

Question: Calculate the current and voltage difference between points D and B in the following circuit.

Answer: The easiest way to solve this problem is to use Kirchhoff’s Laws. We will start off by using Kirchhoff’s first law which is defined as:

The current entering a junction must equal the current out of the junction.

In order to use this law, we first need to define the direction the current is traveling. It really doesn’t matter how you define the current direction. If you picked the wrong direction, the current will just be a negative value which means that it flows the opposite way. This is how I defined the current direction for the circuit.

If we apply Kirchhoff’s first law to junction B, we get the following:

If we this law again on junction D, we get:

Next, we will need to use Kirchhoff’s Second law which is defined as:

The potential differences across all elements around any closed circuit loop must be zero.

If we apply Kirchhoff’s 2nd law and Ohm’s Law to the loop ADB, we get:

because

If we do the same thing to the loop DBC, we get:

because

Now, we will use these laws on the loop ADC, we get:

because

Therefore, we have the following system of linear equations.

which we can put in matrix form.

This system of equations can easily be solved with Maple.

> with(Student[LinearAlgebra]):
> A := Matrix([[0, 1, 1, 0, -1, 0],
		[1, 0, -1, -1, 0, 0],
		[1, -3, 1, 0, 0, 0],
		[0, 0, 1, -2, 1, 0],
		[1, 0, 0, 2, 0, 5]])


> ReducedRowEchelonForm(A)

Therefore, this means that the current going through the point D to B is . If we use Ohm’s law, we can also determine the voltage difference between the points D and B to be .