Orbits!

OK, this is a really interesting post for me. Just stick with it, OK?

In math, there are two types of orbits:
1) An elliptical path traced by an object due to gravitational pull towards another object. AKA the orbit that you know.
2) A repetition of a function infinitely upon itself. (this term might be outdated- I found the lesson in a precalculus book at my college-aged cousins' house)

For orbit #1, an orbit is elliptical due to two factors:
1) Gravity is imperfect.
2) the way that an object enters (or forms in) the gravitational area.
Anyway, the point is that the object that it orbits around is a focus of that ellipse. The foci are the two points that determine the ellipse-- the sum of the distances from any point on the ellipse to each of the foci is constant.

If you think about it, the Sun isn't at the center of the solar system–– it's at one of the foci.

Here's a good problem about it.

Denying the laws of probability and physics itself, let's say the planets in our solar system are perfectly lined up to one side of the sun. The center of the ellipse is approximately 1.5 million km from the center of the sun. If Earth is at its closest point to the sun (148 million km), and the center of gravity of the solar system remains where it is no matter where the planets travel (which is a lie), what is the furthest point of Earth's orbit from the sun?

In the ellipse above, A and B are the foci, D is the center, and C and E are points on the major and minor axes respectively. 

Anyhow, in this case, AD is equal to 1.5 million kilometers and EA is equal to 148 million kilometers. The sun is at A and Earth is at E. 

Since the furthest point is basically the point on the other side of point B, the formula for the farthest point would be:

( (EA) + (AD) ) * 2 - (EA)

Which is about 151 million km, or an accurate-ish 93.8 million miles.

Another thing you might want to know about ellipses is that in the above example (pretend the scale is off), ED=AC=(AD)^2 + (CD)^2.

~~~~

Soooo, now to the other orbit.

Function orbits are similar to orbits: they go around and around a particular value. Now, to explain them.

Imagine a function f(x) for any value x1. Now imagine f(f(x1)). Then f(f(f(x1))). Continue this pattern until you are 106 years old. The orbit of f(x) for that value of x1 would be the infinite sequence 

x, f(x), f(f(x)).... on until you are 106. Or really, infinity years old. The last (or technically last) term of that sequence would be called the limit of the orbit, because the sequence would get infinitely close to the limit but not to the limit.

The limits are also called 'fixed points', because if x1 is equal to its limit, it goes automatically to the limit. Therefore, the limit becomes 'fixed', as its position never changes after the first term.

If you think about it, the fixed points are all on the line y=x, as shown for f(x)= sqrt(x):

In fact, this is a good function to show you something about.

For f(x)=sqrt(x):

For the x1 values of 0<x1<1, what is the limit?

For the x1 values of x>1, what is the limit?

As you can see, both limits to the domain containing non-fixed points are 1. So, 1 is an attracting limit while 0 is repelling.

There are a couple important uses of orbits.

What is the value of 1+(2/(1+(2/(1+(2/...

This is the limit of the orbit 1+(2/x). Calculating the value of the limit would be:

1+(2/x) = x

x^2 - x - 2 = 0

x= 2, -1

Since all of the signs are positive, the root must be positive 2.

*Note*- The above problem could be thought of as the limit of (2/(1+x)) plus one.

The easier way to explain population is through orbits. Population prediction is really important for people in government, and they use dynamic systems to answer their questions. Here's an example of a model:

f(x+1)= 4 * f(x) * (1 - f(x))

where f(x) is a decimal less than one representing the current population. This function is true because of four things:

1)–– (1-f(x)) is a representation of death, especially by starvation or sickness. This would be true because as population goes up, resources get used quicker and lead to a severe lack of supplies.

2)–– f(x) is a representation of birth, which also increases as the population increases. 

3)–– 4 is needed to multiply because 0.5 is lost twice when the other terms are multiplied.

4)–– The function is recursive because we're trying to find the population next year, not any general year.

~~~~~

Have fun pondering over what I'm trying to explain!

Stay coolio,

John

Your Least Favorite Post

That's right... we're doing

So here are some tips for using the ID's--

1) Memorize those formulae- esp. sum and difference identities.

2) Simplify to sines and cosines-- using algebra and reciprocal ID's. This is probably the most important step.

3) Use algebra to simplify to the least possible amount of terms.

4) Use a calculator or something to do the rest. (if solving for variable)

Usually, the point of the identities is to

a) prove an identity/theorem or

b) solve for a variable.

Like above, you have to use all of the identities, mostly reciprocals. 

Although it's one of the most annoying parts of math, it's definitely worth knowing.

Another way to solve it is graphing, but that takes away all the coolness of it. Basically, you can find if the graphs of (cot(x)+tan(x)) and (sec(x)csc(x)) are the same.

I might do something with functions and orbits later. Sorry for not posting in a while.

Stay coolio, John

Late Pi Day

Pie? Like the food?

OK, you've probably heard that before, as Pi Day is now more conspicuous in terms of holidays. Happy 6-days-too-late-for-Pi Day!

The approximate ratios for pi are:
3 digits: 22/7
4 digits: 245/78
5 digits: 333/106
6 and 7 digits: 355/113
continued...
38 digits: 2,646,693,125,139,304,345 / 842,468,587,426,513,207
Well, quintillions are pretty accurate, right?

At digit 763, there are 6 nines in a row. This is the Feynman point.

To find the circumference of a circle the size of the universe to within the radius of a single proton, you only need 39 digits. Although that seems like a lot, it really isn't.

Chao Lu from China memorized 67900 digits of pi. Nice use of time, Chao! Especially considering the last fact!
A ten year old actually memorized 410 digits of pi. Either he felt the urge to learn all of them or his parents are gigantic perfectionists.

Pi is equal to...
6 * sqrt(sum (1-inf) (1/n^2) )
4 * (sum (1-inf) ((-1)^n /(2n+1)))

Although pi is determined by a ratio, it cannot be expressed using a ratio. Therefore, the circumference and/or the diameter must be irrational in a circle.

In the first six million digits of pi, five is the most common and zero is the least common. In the first million, five is the most common and seven is the least common.

Pi is technically mentioned in the bible- see 1 Kings 7:23.

Pi has been referenced in a huge amount of books, TV shows, and movies.

Extremely Cool Sites:
The Pi-search page- This finds certain strings of digits from the first 200 million digits of pi. Extremely cool website.
More Links

Keep calm and guess again!

Problems of the Week 2 Answers

1. The answer to this is really obvious. The sum of n (1/n)'s is 1, so you need k=1000 ones to get 1000.

2. By definition, all multiples of x can be represented as Nx mod x, where N is an integer. Working backward, we find out that the code must be a multiple of 7, 8, and 9. The only three digit number with this property is 7*8*9= 504.

3. Let's replace the cost of a cone with x. Al must have (x-24) cents and Bob must have (x-2) cents. If Bob is only two cents short and the sum of their money is less than x, then Al must have 0 cents or 1 cent. Because Al took at least some money from his bank, he took out one cent; solving for x, we find that the cost is 25 cents.

4. This problem is simple replacement. We can take out the one to get 22 as a solution; and, if you take out 2, 3, and 4 and replace them with 24, you get 20. You can't add 25 because only one five follows, so those are your only solutions.

5. Two digit numbers with digits (xy) are written as 10x+y. Therefore the equation for their ages are:
10x+y+(41-14)=10y+x. One of the solutions is already (1, 4), and when we simplify we get
y=x+3. The next positive integer solution would be (2, 5),and therefore the answer is 11 years later.

6. When you graph 0<y<2, 0<x<2, and abs(y-x)<1, you get:

The green area takes up 3/4 of the 2x2 square, so the probability is 3 in 4.

7. After the first repetition, there are three digits. So after two repetitions, there are exactly four possibilities for the number: 033, 123, 213, and 303. However, 033 and 213 are null because no number simplifies to either of the two. 123 and 303 both go to 123 again.

8. For the sake of understanding what I'm saying, let's use the @ symbol in place of 'plus or minus'. abs(x) in this case simplifies to @x, abs(1@x) simplifies to @1@x, and abs(2@1@x) simplifies to @2@1@x=1. The solutions to this are @4, @2, and @0.

9. Drawing out the hexagon, it looks like this:

Notice that the corners are equilateral triangles of side lengths 9, 14, and 10. If you think about it, the other equilateral triangle will have corner triangles of 7, 11, and 6. The side length of the large triangle will consist of the side lengths of two smaller triangles and the side inbetween them. That would make the triangle's side length equal 7+14+6=6+10+11=11+9+7= 27 units.

10. Draw this out too.

Because FA is tangent to circle C at point F, radius FC is perpendicular to FA. This forms right triangle CAF, with hypotenuse CA and legs FA and FC. We know that FC is 3 units, and AC is 5 units, so therefore FA is 4 units long.

Sorry for the kind-of bad explanations I made. I tried, at least!

Stay coolio,
John

PS: Credit to SketchBookExpress and GeoGebra for the illustrations.

Problems of the Week (or General Period of Time) 2

Here're some problems from the Math Contests book I got for winning a contest–– see if you can do them!

1. (Easy) In the ordered sequence of positive integers, {1, 2, 2, 3, 3, 3, 4, 4, 4, 4....}, each positive integer n occurs in a block of n terms. For what value of k is the sum of the reciprocals of the first k terms equal to 1000?

2. (Easy) My area code is a positive three-digit number. Add 7 to it and the result is divisible by y. Add 8 to it and the result is divisible by 8. Add 9 to it and the result is divisible by 9. What is my area code?

3. (Easy-Moderate) Al and Bob each took some money from their piggy bank to get ice cream. Al was 24 cents short of a cone and Bob was 2 cents short. They pooled their money just to find out that they were still short. How much money did Bob take out of his bank? (sorry the names are boring, the original ones were Rufus and Dufus. I thought those would be stupid enough to take away the point of the problem...)

4. (Easy-Moderate) The factorial 23! can be written as the product of n consecutive integers such that 1< n <23. What are the possible values of n?

5. (Easy-Moderate) On his birthday, Brian was fourteen and his father was 41. What's the next time their ages will be the 'reverse' of each other? (Don't trial and error!)

6. (Moderate-Hard) Both x and y are positive numbers less than 2. Every positive number less than 2 is equally likely to be x and ditto y. What is the probability that x and y differ by less than one?

 7.  (Moderate) Form any positive integer n with less than 10 digits. Form the three digit number x where the hundreds digit is equal to the number of even digits in that number, the tens is equal to the number of odd digits, and the ones digit is equal to the sum of the other two. Then replace the original number with the new one. If this process is repeated infinitely, what number remains?

 8. (Moderate-Hard) What are the real values of x that satisfy   abs(2 - abs(1- abs(x))) = 1?

9. (Hard) An equiangular hexagon with side-lengths 9, 11, 10, 6, 14, and 7 (going clockwise) can be inscribed in an equilateral triangle with side-length 30. What is the side length of the other possible triangles?

10. (Hard) Two externally tangent circles have radii 2 and 3 units. If there is a segment whose endpoints are the center of the small circle and the point at which it is tangent to the larger circle, what is the segment's length?

The answers will be posted tomorrow evening!

Stay coolio,
John

Analysis of Solving Mazes (logic edition)

How do you solve a simple maze without knowing what the maze looks like from above?

Well, let's first define the maze. Let's say the maze (I'll call it the Johnesian maze) is a rectangular 'grid' with 2 openings on edges of the grid (for start and exit), and enough walls taken out as to provide at least one path between the openings. Remember that these features define the maze–– the grid is a Johnesian maze only if it has exactly two openings on the edge, and enough grid-walls are removed to provide a path to each.

To show you the first and easiest method, let's think about the lines (note: lines are connected strips walls, not the walls themselves) in the maze. There must be two lines, because the two lines end at the start and finish and cannot be connected in the middle (or else there is no solution path). There can also be island lines, or lines that touch neither the start nor finish.
By that logic, there are two lines that touch both the start and the finish. Therefore, you can just follow one of the walls beside the start until you eventually reach the finish. This is called the left (or right)-hand rule.

For the second, the answer is much less confusing. If one can tell the difference between the start and the finish, and discern therefore that the start is not the finish, you can just pick completely random paths and eventually (after infinity) it is guaranteed that the finish is found. Of course that solution assumes that there is unlimited food and water. This solution is the Random Mouse.

The third is more difficult. As you walk down a passage, draw a line behind you to mark your path. When you hit a dead end turn around and go back the way you came. When you encounter a junction you haven't visited before, pick a new passage at random. If you're walking down a new passage and encounter a junction you have visited before, treat it like a dead end and go back the way you came so that you won't go around in circles or missing passages. If walking down a passage you have visited before (i.e. marked once) and you encounter a junction, take any new passage if one is available, otherwise take an "old" one. Every passage will be either empty (if not visited yet), marked once, or marked twice (if you were forced to backtrack). When you reach the solution, the paths which were marked exactly once will indicate the direct way back to the start. If the maze has no solution, you'll find yourself back at the start with all passages marked twice. It's kind of complicated, but I'm sure it works. BTW, this is called the Tremaux algorithm.

If you can look from above, a good strategy is to fill out all dead ends or nooses (eg, places where you must backtrack to find a solution). Then you fill in the places that lead only to dead ends and nooses. And etc, until the only paths not filled in are solutions.

The last one is the Pledge Algorithm. The idea is to try to always go in one chosen direction, if you do not turn more than 359 degrees in any direction. 
So let's measure our angles- a left turn is worth -1 and a right turn is worth 1, and you add each on to your 'turn count' whenever you turn. Choose a direction that goes towards the finish, and calibrate it to turn count=0, -4, 4, and so on. Just know that no multiple of 4 that is not 0 is equal to 0, just for this algorithm. In the algorithm:
1) If turn count = 0 and path ahead is open, then go forward.
2) If turn count is 0 and there is a wall directly ahead, then follow that wall in a random direction until turn count=0.
This algorithm is confusing and difficult, but it works.

NOTE: Only one of these methods is mine, and a ton of them I don't understand at all. If I have left something out, sorry!

Good sites:
Labyrinth.htm
Wikipedia- Maze solving algorithms

I bet your brain exploded a while ago, so I might as well just shut up now. See ya!

Stay coolio,
John

Squares!

Well, by squares, I obviously mean square numbers, right?

Welcome back to your favorite site! I'm going to do a rather short post today, as I can't really think of much to do.

Here's something I found out a couple of days ago.

For any integer, then x^2 is either a multiple of 4 or one more than a multiple of 4.
Proof: This theory depends on whether that integer is odd or even. Let's represent the set of any integer as N.
odd– (2N+1)^2=4(N^2)+4(N)+1=4(N^2+N)+1=4(N)+1
even– (2N)^2=4(N^2)=4(N)
As one can see, an odd integer squared is one more than four times another integer, and an even integer squared is four times another integer. Isn't that weird?

Another odd thing is that the sums of the first few cubes is always the square of the sum of their base numbers. For example,
1^3 + 2^3 = (1+2)^2
1^3 + 2^3 + 3^3 = (1+2+3)^2
1^3 + 2^3 + 3^3 + 4^3 = (1+2+3+4)^2

Speaking of shape--ish numbers, every square is the sum of two consecutive triangles. I wonder if the sum of two consecutive squares is a pentagonal? Or if you can figure out hyper-cubic numbers? Probably you can-- I'll update that later.

The difference between perfect squares is always odd. (n+1)^2 - n^2 = 2n+1, so if n is an integer, the difference is odd (and the squares are perfect squares).

There's a weird pattern in the squares of all of the integers from 969 to 1000. What is it? (Look closely; no, it's not the thousands digits)

An easy way to approximate a square root without a calculator is by:
1) guessing the root
2) dividing your guess into the original
3) finding the mean of the result and the guess
4) and repeating steps 2 to 4 with what you get.

All square numbers end with the digits 1, 4, 6, 9, 25, or 00.

Stay coolio,
John

PS: If you have a Google account, please follow me through the button on the right! Enter your email into the box to get notifications about new posts! Preferably both please!