Undefined Behavior

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Problem. Shanille O'Keal shoots free throws on a basketball court. She hits the first and misses the second, and thereafter the probability that she hits the next shot is equal to the proportion of the shots she has hit so far. What is the probability she hits exactly 50 of her first 100 shots?

Solution. Let $S(N)$ denote the number of successful free throws she has made in her first $N$ attempts. Thus, $S(2) = 1$. The problem is to compute the probability that $S(100) = 50$.

We will consider the more general problem of finding the probability that $S(N) = k$, that is, the problem of finding the probability that she made $k$ shots in her first $N$ attempts. Denote this probability by $P(N,k)$.

Clearly, if she made $k$ shots in $N$ attempts, then she either made $k$ shots in $N - 1$ attempts and missed the $N$th shot or she made $k - 1$ shots in $N - 1$ attempts and made the $N$th shot. By the conditions put forth in the problem, this yields

\[
P(N,k) = P(N - 1, k) \left(1 - \frac{k}{N - 1} \right) + P(N - 1, k - 1) \frac{k - 1}{N - 1}.
\]

Problem. Let $k$ be a fixed positive integer. The $n$-th derivative of $1/(x^k-1)$ has the form $\frac{P_n(x)}{(x^k-1)^{n+1}}$ where $P_n(x)$ is a polynomial. Find $P_n(1)$.

Solution. We take advantage of the information the problem has given us. We simply differentiate the given formula to obtain

\[
\frac{P_{n+1}(x)}{(x^k-1)^{n+2}} = P'_n(x) (x^k - 1)^{-n-1} - (n + 1)P_n(x) (x^k - 1)^{-n-2} kx^{k-1}.
\]

We can solve for $P_{n+1}(x)$; hence

\[
P_{n+1}(x) = P'_n(x) (x^k - 1) - (n + 1) kx^{k-1} P_n(x).
\]

Now setting $x = 1$ yields

\[
P_{n+1}(1) = -(n+1)k P_n(1).
\]

By direct computation, $P_1(x) = -kx^{k-1}$, so $P_1(1) = -k$. Then induction suffices to yield the final result

\[
P_n(1) = (-1)^n(n!)(k^n).
\]

Problem. Given any five points on a sphere, show that some four of them must lie on a closed hemisphere.

Solution. Pick any two distinct points among the five and consider a great circle containing them. This circle divides the sphere's surface into two hemispheres. Since three points remain and there are two hemispheres, the pigeonhole principle shows that at least two of the three remaining points must lie in the same hemisphere. Hence this (closed) hemisphere contains at least four points including the initial two that lie on its boundary.

In a multivariable calculus course, one traditionally studies the major theorems of vector calculus. Vector calculus, of course, is the study of vector fields and their analytic properties. The classic pillars of this theory are the following intimately related results:

• The normal form of Green's theorem. The integral of the divergence of a field over a region of the plane equals the net outward flux of the field over the region's bounding curve:

  \[
  \iint_R \nabla \cdot \mathbf{F} dA = \oint_{\partial R} \mathbf{F} \cdot \mathbf{n} ds\]

• The divergence theorem. The integral of the divergence of a field over a region of Euclidean 3-space equals the net outward flux of the field over the volume's bounding surface:

  \[
  \iiint_V \nabla \cdot \mathbf{F} dV = \iint_{\partial V} \mathbf{F} \cdot \mathbf{n} dS
  \]

• The tangential form of Green's theorem. The integral of the curl of a field over a region of the plane equals the net circulation of the field along the region's bounding curve:

  \[
  \iint_R \nabla \times \mathbf{F} \cdot \mathbf{k} dA = \oint_{\partial R} \mathbf{F} \cdot d\mathbf{r}
  \]

• Stokes's theorem. The net flux of the curl of a field over a surface is equal to the net circulation of the field over the surface's bounding curve:

  \[
  \iint_S \nabla \times \mathbf{F} \cdot \mathbf{n} dS = \oint_{\partial S} \mathbf{F} \cdot d\mathbf{r}
  \]

Observe how these theorems naturally fall into pairs. The normal form of Green's theorem is a two-dimensional version of the divergence theorem, and the tangential form of Green's theorem is Stokes's theorem restricted to the plane.

Here's an interesting problem. Let $F$ represent either $R$ or $C$, the sets of real and complex numbers, respectively. Define $F^{\infty}$ to be the set of all (infinite) sequences of elements of $F$. You should verify that $F^{\infty}$ is a vector space with addition and scalar multiplication defined in the obvious ways. The problem is to show that $F^{\infty}$ is infinite dimensional.

Recall that a vector space is finite dimensional if there is a tuple of vectors that spans the space, and a vector space is infinite dimensional if no such tuple exists. Our task is then naturally to show that in the case of $F^{\infty}$ no spanning tuple exists.

To establish this, we first put forth the following lemma. All vector spaces are over $F$.

Lemma. A vector space $V$ is infinite dimensional if and only if there is a sequence of vectors $v_1,v_2,\dots$ in $V$ such that $(v_1,\dots,v_n)$ is linearly independent for every positive integer $n$.

Proof. A standard result is that the length of any linearly independent tuple is less than or equal to the length of any tuple that spans $V$. If such a sequence existed, then for every $n$ we would have a linearly independent tuple of length $n$. Thus no spanning tuple could exist; otherwise, a linearly independent tuple would be longer than the spanning tuple. Thus $V$ is infinite dimensional (by definition).