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Tuesday, June 14, 2005

The order of addition matters!

Now we are ready for our demonstration that the infinite sum:

1 - 1/2 + 1/3 - 1/4 + 1/5....... - 1/(2n) + 1/(2n +1)......
can be rearranged to equal any number that we chose. In other words, the order of addition matters!

The idea is this: choose any number in the real number system that you please. For example, choose, say "2".

Now since we picked a positive number, start with the positive terms.

Start with 1 + 1/3 + 1/5 + 1/7 and keep adding until you first get a number greater than 2.
How do you know that you can do this? Because 1 + 1/3 + ....+ 1/(2k+1) + 1/(2k+3) +....
is an infinite sum by the integral test. Exercise: find how many terms it takes to get bigger than 2.

Next start with the negative terms -1/2 - 1/4 - 1/6....and keep going until you first get a number less than 2. Why can this always be done?

Next start with the first positive term you didn't use in your first step and work your way past 2 again, and then use the negative terms to get less than 2 again, and keep repeating this process.

To see why this process yields a sum of 2, let s1, s2, s3,...denote the following partial sums:

s1 = sum of the first set of positives, s2 = sum of the first set of negative terms, s3 = sum of the second set of positive terms, and so on.

Notice that s1 can exceed 2 by at most some 1/n1. Then when we use s2, note that we only go at most 1/m1 past 2 in the negative direction. Then using s3, we exceed 2 by at most some 1/n2, where 1/n2 < 1/n1 (why does this inequality hold?) and then using s4, we go at most 1/m2 below 2 where 1/m2 < 1/m1.

So, given any interval about our chosen number 2, the partial sums must eventually lie within this interval (this is an "epsilon-delta" argument).

In other words, we have shown that our series has been rearragned to have partial sums which go "back and forth" about 2 and get arbitrarily close to 2.

Infinite Series: integral test

Ok, let's develop an important tool: this is called the integral test.

First of all, remember what a Riemann Sum is; in particular recall the left and right endpoint sums. (Recall all of those little rectangles that you drew when you were first learning about definite integrals).

Next, recall what an improper integral is. This is where you take a limit as one of the limits of the definite integral goes to infinity.

Now suppose f is a function which, on the interval [1, infinity) is decreasing and positive (examples: f = 1/x, f = exp(-x) ). We then have the following:

the sum f(1)+ f(2) +.....+ f(k) + f(k+1) +..... converges if and only if the improper integral
integral(f(x), 1, infinity) converges.

Why? Note that f(1)+ f(2) + .....+f(k) + f(k+1)+...... is a left hand endpoint sum for the integral and therefore is larger than the integral. (why? draw a picture and draw the rectangles for the left hand sum). Note that f(2)+...+f(k) + f(k+1)+.... is a right hand endpoint sum for the improper integral and is smaller than the integral.

Also remember that both sums have a sequence of partial sums which are increasing.

So, if the sum is finite, so is the integral. If the sum is infinite, then so is the sum starting with
f(2) and so is the integral.

That is the integral test, in a nutshell.

As a consequence: sums whose terms look like 1/(ak+b) where a and b are constants are infinite and sums whose terms look like 1/p(k) where p(k) is a polynomial of degree 2 or larger converge.

Monday, June 13, 2005

We have our math-studs too!

We are used to reading about those who excel in many areas, including those in sports, business, acting, politics, as well as the various popular art areas.

But we in mathematics have our stars too; here is one who has achieved the ultimate in excellence in my research area (topology): Michael Freedman.

He was one of our Field's Medal winners and is currently empolyed by Microsoft.

Sunday, June 12, 2005

Infinite Series Part III

So, we see that in order for an infinite sum to make sense, we must use limits.

So, that is how an infinite sum is defined: if Sn is the "n'th partial sum" (what you get when you sum the first "n" terms with ordinary addition, then the infinite sum is the limit of Sn as n goes to infinity, provided the limit exists.

So that is why we need limits and sequences to talk about infinite sums.

Now what about that strange infinite sum 1 - 1/2 + 1/3 - 1/4 + 1/5......

This is an example of an alternating sum because the signs alternate. To find what this sums to, one must study infinite series. But to see that this sum exists:

Let S2n+1 be the odd partial sums. Since each S2n+1 is a finite sum, we can group and sum the terms in any order; we get (1-1/2) + (1/3-1/4).....+ 1/(2n+1) =

1/2 + 1/12 + .....(1/(2n-1) - 1/2n) + 1/(2n+1) =

1/2 + 1/12 + ....((2n - (2n-1))/(2n(2n-1)) + 1/(2n+1) =

1/2 + 1/12 + 1/(4n^2 -2n) + 1/(2n+1)

We need some results about infinite series here, but as n goes to infinity in we get a convergent

series (p test and comparison test). That is why we have all of those tests.

Now here is what is bizarre: if we change the order of the original series, we change the result of the sum! This is not obvious, and to see this we need some infinite series results.

Friday, June 10, 2005

Infinite Series Part II

We've seen that an infinite sum can sometimes fail to exist for "obvious" reasons. Examples 2, 4, and 5 are more subtle.

Let's look at examples 2 and 4: can these sums exist? Note that one does have an infinite number of terms in each sum. But also note that each term is getting progressively smaller. So, there might be a chance that these sums exist!

It turns out that the sum of example 4 indeed exists, but the sum of example 2 does not. Here is how we can see this:

In example 1, recall from calculus the idea of an indefinite integral. Integrate the function "1/x" from x = 1 to x = "3". The value of this integral is "ln(3)". Notice what "1 + 1/2" represents: it represents the left hand Riemann sum with n = 2 and delta x = 1 for this integral and remember that because the function 1/x is decreasing, the left hand sum is greater than the integral.

Now keep doing this; note that 1+ 1/2 + ....+1/n represents the left hand sum with delta x = 1, n = n for the function 1/x. Notice that this partial sum exceeds the value of the integral of 1/x as x goes from 1 to n; that is, 1+1/2 +....1/n exceeds ln(n).

So we see that the n'th partial sum exceeds ln(n) and we know from calculus that ln(n) increases without bound as n increases. Hence these partial sums will eventually exceed and finite number. This sum cannot make sense.

On the other hand, if 1 + 1/2 + ....+(1/2)^n = Sn
Then we can multiply both sides by "1/2" to get
1/2 + 1/4 + ....(1/2)^(n+1) = (1/2)*Sn
Now subtract: Sn - (1/2)Sn = 1 - (1/2)^(n+1) (subtract the two sums and note that the leading 1 doesn't get subtracted off and the last term (1/2)^(n+1) isn't cancelled either.)

Now solve for Sn: Sn = (1 - (1/2)^(n+1))/(1 - (1/2))

Now notice that as n gets bigger (as we take a limit as n goes to infinity) the right hand side becomes 1/(1-(1/2))= 1/(1/2) = 2 because (1/2)^(n+1) goes to zero as n goes to infinity.

So, you see that a sequence of partial sums is determined by the limit as n goes to infinity! Here, that limit is 2 and that is the value of this infinite sum.

Example 5 is trickier. It turns out that this sum exists, but we need some theory of sequences to see this. Also example 5 is just downright weird.

Remember back in algebra where you were told that addition is commutative; that is, you could add things in different order and still get the same answer? This is false with infinite sums! To see why, keep reading.

What is an infinite series?

This entry is designed to help a student understand what an infinite series is and why we need to be fussy.

  • Regular Addition

Not a problem; we "know" what, say, "3 + 4" means, or even "4 + 7 + 5" means, etc. In a similar way, we have an intuitive idea of what "x + 2x^2" means. A sum with any finite number of terms is always well defined.

  • Infinite Addition: where the problem begins.

So, now what does it mean to add an infinite number of terms together? Let's look at the following examples:

  1. 1 + 2 + 3 + ....+ n + (n + 1) +....
  2. 1 + 1/2 + 1/3 +.....+ 1/n + 1/(n+1) +......
  3. 1 - 1 + 1 - 1 + 1 - 1....... + (-1)^(n)+ (-1)^(n+1) +....
  4. 1 + 1/2 + 1/4 + 1/8 + .....+ 1/2^(n) + 1/2^(n+1) + .....
  5. 1 - 1/2 + 1/3 - 1/4 + 1/5 +.....+ (-1)^n)*(1/(n+1))+ (-1)^(n+1)*(1/(n+2))

The first thing to notice is that an infinite sum need not make sense at all! First note that it should be clear that the first sum not make sense. After all, it is clear that as one is adding an infinite number of terms which are getting infinitely large.

To see this we introduce the concept of the partial sum; the first partial sum is 1, the second is 1+2 = 3, the third is 1+2+3 = 6. Exercise: what is the fourth partial sum?

So, we look at the partial sums as a sequence: 1, 3, 6, 10, 15, 21....

and we see that the partial sums are getting bigger without bound. Clearly this infinite sum can't make sense.

Now look at example three: the first partial sum is 1, the second is zero, the third is 1, the fourth is zero, and so on. Exercise: what is the (2k+1)'st partial sum? The 2k'th?

Here the sequence of partial sums (1, 0, 1, 0, 1, 0...) are never going to "steady out" on anything. So this sum can't make sense either.

Ok, what about examples 2, 4, and 5? That is the topic of the next blog entry.

Tuesday, June 07, 2005

Math Journal

Here I am hoping to record my progress in mathmetics research.

My summer projects are to learn about hyperbolic geometry, review some knot theory and to get current, and to

1) work on how classes of curves in the plane detect discontinuities of two variable functions
2) work on polynomial invariants of proper knots (proper embeddings of the real line into 3 space)
3) explore wild knots that have symmetries of all orders (are set wise fixed under peroidic homeomorphisms of S^3 which have S^1 as a fixed point set.