This post has a pretty weird title, but the problem is easy to state and uses a few interesting mathematical concepts.  It’s worth going through.  Let’s start with the basics.


Problem 1.  Let S_{\alpha}(n) = 1^{\alpha} + 2^{\alpha} + \dots + n^{\alpha}.  Show that S_{\alpha}(n) is a polynomial for each \alpha\in {\mathbb N} and that the degree of the polynomial is \alpha + 1.


Indeed, for example, we have  that 1 + 2 + \dots + n = \frac{n(n+1)}{2}, as we learned in Calculus, and this is a polynomial of degree 2.  Similarly, 1^{2} + 2^{2} + \dots + n^{2} = \frac{n(n+1)(2n+1)}{6}, which is a polynomial of degree 3.  In the same respect, 1^{3} + 2^{3} + \dots + n^{3} = \frac{(n(n+1))^{2}}{4}, which is a polynomial of degree 4.

The associated polynomials in this case are given by Faulhaber’s formula:


Theorem (Faulhaber).  For \alpha\in {\mathbb N} we have \displaystyle \sum_{k=1}^{n}k^{\alpha} = \frac{1}{1 + \alpha}\sum_{j=0}^{\alpha}(-1)^{j}\binom{\alpha + 1}{j}B_{j}n^{\alpha + 1 - j}.


This formula looks terrifying, but it is not hard to apply in practice.  You may be wondering, though, what the B_{j}‘s in this formula stand for.  These are the strange and wonderful Bernoulli numbers, of course!  I always enjoy seeing these creatures, because they unexpectedly pop up in the strangest problems.  There are a number of ways to define these numbers, one of which is to just write them out sequentially, starting with B_{0}:


1, -\frac{1}{2}, \frac{1}{6}, 0, -\frac{1}{30}, 0, \frac{1}{42}, 0, -\frac{1}{30}, 0, \dots


But in this case it is not so easy to guess the next value.  The clever reader will notice that all of the odd numbered Bernoulli numbers (except the first) are zero, but other than that there does not seem to be a clear pattern.  Fortunately, we can construct a function which generates the values as coefficients; we’ll call this function (surprise!) a generating function.


Definition.  We define the sequence \{B_{j}\}_{j=0}^{\infty} by

\displaystyle \frac{t}{e^{t} - 1} = \sum_{j=0}^{\infty}B_{j}\frac{t^j}{j!}.


Notice that this will, in fact, generate the B_{j} as coefficients times \frac{1}{j!}.  Neat.  In practice, you can use a program like Mathematica to compute B_{j} for pretty large values of j; but, of course, there are lists available.  We can now use Faulhaber’s formula above, which gives us (assuming we have proven that the formula holds!) that the sums of powers of natural numbers form polynomials of degree \alpha + 1.


But something else happens that’s pretty interesting.  Let’s look at some of the functions.


\begin{array}{|c|c|}\hline \alpha & \mbox{Polynomial}\\\hline &\\ 1 & \frac{1}{2}n^{2} + \frac{1}{2}n\\ &\\ 2 & \frac{1}{3}n^{3} + \frac{1}{2}n^{2} + \frac{1}{6}n\\ & \\ 3 & \frac{1}{4}n^{4} + \frac{1}{2}n^{3} + \frac{1}{4}n^{2} \\ & \\ 4 & \frac{1}{5}n^{5} + \frac{1}{2}n^{4} + \frac{1}{3}n^{3} - \frac{1}{30}n\\ & \\ \hline \end{array}


Look at the coefficients in each of these polynomials.  Anything strange about them?  Consider them for a bit.


Problem.  Look at the coefficients.  What do you find interesting about them?  Note that, in particular, for a fixed \alpha, the coefficients of the associated polynomial sum to 1.  Convince yourself that this is probably true (do some examples!) and then prove that it is true.  Do this before reading the statements below.


Anecdote.  I spent quite a while trying to write down the "general form" of a polynomial with elementary symmetric polynomials and roots to try to see if I could prove this fact using some complex analysis and a lot of terms canceling out.  This morning, I went into the office of the professor to ask him about what it means that these coefficients sum up to 1.  He then gave me a one-line (maybe a half-line) proof of why this is the case. 


Hint.  What value would we plug in to a polynomial to find the sum of the coefficients?  What does plugging in this value mean in terms of the sum?


An interesting question came up during my studying today which made me think about some of the different ways we can think about polynomials.  The solution to this problem was not immediately obvious to me, and, in fact, it wasn’t until I looked up a completely unrelated problem (in a numerical methods book!) that some solution became clear.

Question:  Suppose that \{f_{n}\} is a sequence of polynomials with f_{n}:{\mathbb R}\to {\mathbb R} each of degree m > 1 and suppose f_{n}\to f pointwise.  Show that f is also a polynomial of degree no more than m


Some interesting points come up here.  First is that we only have pointwise convergence — it wasn’t even immediately obvious to me how to prove the resulting limit was continuous, let alone a polynomial of some degree.  Second, we know very little about the polynomials except for what degree they are.  This should be an indication that we need to characterize them with respect to something degree-related.

Indeed, polynomials can be represented in a few nice ways.  Among these are:


  • In the form f(x) = a_{0} + a_{1}x + \cdots + a_{n}x^{n} where it is usually stated that a_{n}\neq 0.
  • In terms of their coefficients.  That is, if we have a list of polynomials of degree 3 to store on a computer, we could create an array where the first column is the constant, the second is the linear term, and so forth.  This is sort of like decimal expansion.
  • Where they send each point.  That is, if we know what f(x) is equal to for each x, we could recover it.
  • If a polynomial is of degree m then, somewhat surprisingly, we can improve upon the previous statement: if we know the value of f for m + 1 distinct points, then we can find f, and f is the unique such polynomial of that degree which has those values.  (Note that if we were to have m+1 points and a polynomial of degree k > m, then many polynomials of this degree could fit the points.  Consider, for example, m = 0 and k = 1.  Then we have one points and we want to fit line through it.  Clearly this can be done in infinitely many ways.)


  This last one is going to be useful for us.  So much so that it might be a good idea to prove it.


Lemma.  Let x_{1}, \dots, x_{m+1} be m + 1 distinct points in {\mathbb R}, and let y_{1}, \dots, y_{m+1} be distinct points in {\mathbb R}.  Then there is a unique polynomial of degree at most m such that f(x_{i}) = y_{i} for each i considered here.


Proof.  This is an exercise in linear algebra.  We need to solve the system of linear equations

\displaystyle \sum_{n = 1}^{m} a_{n}x_{i}^{n} = y_{i}

where i spans 1, \dots, m+1, for the constants a_{n}\in {\mathbb R}.  Notice that this is simply plugging x_{i} into a general polynomial of degree m.  Notice that the matrix that this forms will be a Vandermonde matrix.  Since each x_{i} is distinct, the determinant of this matrix is nonzero, which implies that there is a unique solution.  This gives us our coefficients, and note that this is a polynomial not necessarily of degree exactly m, since some coefficients may be 0, but it is at most m\diamond


[Note: For those of you who forgot your linear algebra, the end of this goes like this: if we let our coefficients be denoted by the column matrix b and our Vandermonde matrix is denoted by A, then we want to solve Ab = Y where Y is the column vector with entries y_{i}.  If A has non-zero determinant, then it is invertible, and so we have that b = A^{-1}y gives us our coefficients.]


Neato.  But now we need to specialize this somewhat for our proof. 


Corollary.  Let the notation and assumptions be as in the last lemma.  For I\in \{1, \dots, m+1\}, let g_{i} be the unique polynomial of degree at most m with g_{i}(x_{j}) = \delta_{i,j} (where \delta_{i,j} = 0 if i\neq j and \delta_{i,i} = 1).  Then every polynomial f of degree at most m is of the form \displaystyle f(x) = \sum_{i = 1}^{m+1}f(x_{i})g_{i}(x) for each x\in {\mathbb R}.


This might be a bit more cryptic, so let’s do an example.  Let’s let m = 1 so that we have two points.  Let’s say x_{1} = 0 and x_{2} = 1.  Then we have g_{1} is the unique polynomial of degree at most 1 such that g_{1}(0) = 1 and g_{2}(1) = 0.  Of course, this function will be g_{1}(x) = -x + 1.  Now g_{2}(0) = 0 and g_{2}(1) = 1; this gives us that g_{2}(x) = x.  The theorem now states that any polynomial of degree at most 1 can be written in the form

f(x) = f(0)g_{1}(x) + f(1)g_{2}(x)

= f(0)(-x+1)+ f(1)x = (f(1) - f(0))x + f(0).

For example, let f(x) = 2x + 1.  Then the lemma says f(x) = (3(1) - 1)x + 1 = 2x + 1, as we’d expect.  The power of this lemma will become clear when we use this in the solution.  The proof of this corollary is just a specialization of the previous lemma, so we exclude it.


Solution.  Recall, just for notation, that our sequence \{f_{k}\}\to f pointwise.  Let’s let x_{1}, \dots, x_{m+1} be our distinct points, as usual.  In addition, let’s let g_{1}, \dots, g_{m+1} be defined as in the corollary above. Represent each f_{k} as follows:

\displaystyle f_{k}(x) = \sum_{j = 1}^{m+1}f_{k}(x_{j})g_{j}(x)

for each x\in {\mathbb R}.  Here comes the magic: let k\to\infty and note that f_{n} \to f at every point, so, in particular, on each x_{i} and on x.  We obtain

\displaystyle f(x) = \sum_{j = 1}^{m+1}f(x_{j})g_{j}(x)

for each x\in {\mathbb R}.  But this is the sum of polynomials of degrees at most m, which gives us that f is itself a polynomial of degree at most m\Box


I’ll admit, I did a bit of digging around after finding the lemma above; in particular, this corollary representation of polynomials seems to be a nice way to represent a polynomial if we do not know the coefficients but do know the values at a certain number of points and have that its degree is bounded below that number of points. 


Exercise: Try to write out this representation for m = 2 and m = 3.  If you’re a programmer, why not make a program that allows you to input some points and some values and spits out the polynomial from the corollary above?

Factorials and Exponentials.

November 6, 2011

I’ve been working on a problem (here is a partial paper with some ideas) that’s really easy for any calculus student to understand but quite difficult for even wolfram alpha to work out some cases.  Here’s the idea:

We know that \lim_{n\to\infty}\dfrac{e^n}{n!} = 0.  It’s not hard to reason this out (there are some relatively obvious inequalities, etc.), but I wanted to know what happened if we considered something like:


It turns out, this goes to infinity.  Maybe this is not so surprising.  But, to balance this out, I thought maybe I could add another factorial on the bottom.  What about


where this double factorial is just n! with another factorial at the end.  It turns out, this one goes to 0. 


The problem here is that after ((n!)!)!, Mathematica doesn’t seem to be able to handle the sheer size of these numbers.  Consequently, I only have a few values for this.  I’ve included everything I have in a google-doc PDF (the only way I can think to share this PDF), and I’m looking for suggestions.  Here’s some things I thought of:


  • Stirling’s formula.  Unfortunately, this starts to get very complicated very quickly, and if you consider subbing it in for even ((n!)!)! it can take up a good page of notes.  It also doesn’t reduce as nicely as I’d like.
  • Considering the Gamma function.  It may be easier to work with compositions of the gamma function since it is not discrete and we may be able to use some sort of calculus-type things on it.
  • Number Crunching.  For each of these cases, it seems like there is a point where either the numerator or the denominator "clearly" trumps over the other; this is not the "best" method to use, but it will give me some idea of which values potentially go to infinity and which go to zero.
  • Asymptotics.  I’m not so good at discrete math or asymptotics, so there may be some nice theorems (using convexity, maybe?) in that field that I’ve just never seen before.  Especially things like: if f\sim g then f\circ f \sim g\circ g under such-and-such a condition.


Feel free to comment below if you think of anything.