Putnam 2020 A2

For $k$ a non-negative integer, evaluate
\sum_{j=0}^k 2^{k-j} \binom{k +j}{j}.

Apparently, I only know how to do induction anymore now but this problem is straightforward using it. We suppose the inductive hypothesis, and assume that for a fixed $k$ that
\sum_{j=0}^k 2^{k-j} \binom{k +j}{j} = 2^{2k}.
We will proceed to show
\sum_{j=0}^{k + 1} 2^{k + 1-j} \binom{k + 1 +j}{j} = 2^{2k + 2}.

By a simple binomial identity, we have
\sum_{j=0}^{k + 1} 2^{k + 1-j} \binom{k + 1 +j}{j} &= \sum_{j=0}^{k + 1} 2^{k + 1-j} \binom{k+j}{j} + \sum_{j=0}^{k + 1} 2^{k + 1-j} \binom{k +j}{j – 1}.
Let $A$, $B$ be the first and second term respectively on the right hand side, then
A &= 2\sum_{j=0}^{k} 2^{k-j} \binom{k+j}{j} + \binom{2k + 1}{k+1} \\
&= 2^{2k + 1} + \binom{2k + 1}{k+1}.
For the other term,
B &= \sum_{j=1}^{k + 1} 2^{k + 1-j} \binom{k +j}{j – 1} \\
&= \sum_{j=0}^{k} 2^{k-j} \binom{k +1 +j}{j} \\
&= \frac{1}{2} \sum_{j=0}^{k + 1} 2^{k + 1-j} \binom{k +1 +j}{j} – \frac{1}{2} \binom{2k + 2}{k + 1}.
Thus, we rearranging, we have our result
\frac{1}{2}\sum_{j=0}^{k + 1} 2^{k + 1-j} \binom{k + 1 +j}{j} &= 2^{2k+1} + \binom{2k + 1}{k+1} – \frac{1}{2} \binom{2k + 2}{k + 1} \\
&= 2^{2k+1}.

Scaling Arguments

This is a pretty important concept in PDEs and its numerical approximations. Specifically, tt shows up in Bramble-Hilbert lemma, and domain decomposition analysis.
Most of this post is pretty much written right after reading Toselli and Widlund’s book, so there are a lot of resemblance.

Let $\Omega$ be a bounded domain in $\mathbb{R}^n$ which is ‘nice’ (say Lipschitz boundary) with radius $h$. Now let $u, v \in H^1(\Omega)$ such that
|v|_{H^1(\Omega)} \le C||u||_{H^1(\Omega)}
and we wish to obtain the $h$ dependence from $C$.

What we do is to first consider a scaled domain $\hat \Omega$ which is just $\Omega$ scaled to be of radius 1, with the change of basis $x = h\hat x$.
If we find the corresponding inequality on $\hat \Omega$, then the constant $C$ will not depend on $h$.
Let $\hat v(\hat x) := v(h\hat x)$, then we note that $\hat \nabla \hat v(\hat x) = h\hat \nabla v(h\hat x)$ where $\hat \nabla $ is the gradient with respect to $\hat x$.
|v|^2_{H^1(\Omega)} &= \int_\Omega |\nabla v(x)|^2 \, dx \\
&= \int_{\hat \Omega} |\hat\nabla v(h \hat x)|^2 h^n \, d\hat x \\
&= \int_{\hat \Omega} |\hat\nabla \hat v(\hat x)|^2 h^{-2} h^n \, d\hat x = h^{n-2}|\hat v|_{H^1(\hat \Omega)}^2

But for $L^2$ norm, there is no $h^2$ scaling, hence
||u||_{L^2(\Omega)}^2 &= \int_\Omega |u(x)|^2 \, dx \\
&= \int_{\hat \Omega} |u(h \hat x)|^2 h^n \, d\hat x = h^n ||\hat u||_{L^2(\hat \Omega)}^2.
This is why derivatives mixing causes scaling issues.

Putnam 2003 A2

Let $a_1, \ldots, a_n$ and $b_1, \ldots, b_n$ be non-negative real numbers. Show that $$(a_1\ldots a_n)^{1/n} + (b_1\ldots b_n)^{1/n} \le [(a_1 + b_1) \cdots (a_n + b_n)]^{1/n}.$$

Solution: we will use the generalized Holder’s inequality which states that
$$||f_1\ldots f_n ||_1 \le ||f_1||_{\lambda_1} \cdots ||f_n||_{\lambda_n}$$
for lambda weights $\lambda_1^{-1} + \cdots + \lambda_n^{-1} = 1$ all greater than 1.

Assuming this is true, let $f_i = (a_i^{1/n}, b_i^{1/n})$ and the norms be the discrete $l^p$ norm. This will give us $||f_1 \ldots f_n||_1 = (a_1\ldots a_n)^{1/n} + (b_1\ldots b_n)^{1/n}$ as everything is non-negative. The weight will be uniform $\lambda_i = n$, then the right hand side will be
$$||f_i||_{n} = (a_i + b_i)^{1/n}$$
and we have our inequality.

The sole remaining thing to prove is the generalized Holder’s inequality. We will assume the famous base case of the two element case. In the inductive case, we have
||f_1\cdots f_{n+1}||_1 &\le ||f_1 \cdots f_n||_{\lambda_{n+1}/(\lambda_{n+1} – 1)} ||f_{n+1}||_{\lambda_{n+1}} \\
&= ||(f_1 \cdots f_n)^{\lambda_{n+1}/(\lambda_{n+1} – 1)}||_1^{(\lambda_{n+1} – 1)/\lambda_{n+1}} ||f_{n+1}||_{\lambda_{n+1}}.
From here, just change the weights and use the inductive case and we are done.

Putnam 2003 A1

Let n be a fixed positive integer. How many ways are there to write n as a sum of positive integers, K_n to be the set of tuples of We are not done here, as we would need to show that there exists no other tuples in After spending a good amount of timing trying to prove this, I realized that this is in general not true (in fact, the result I was suppose to be chasing would’ve been disproven if the above statement was true). As a counter example, consider the following counterexample from a Bernstein basis application:

Let The book actually gave quite a lot of hints for this one.

For example, in Python it would be literally one line:

return optimize.fsolve(lambda u: u - self.u0(x - u*t), 1)

Spectacular Books

Mathematicians suck at writing. This is part of the reasons why they went into math in the first place, because they suck at writing. Sucking at writing doesn’t mean that mathematicians don’t write books though; in fact there are tons of math books written by mathematicians.

The problem is most of them suck.

Half of them are for geniuses written by geniuses.

The other half are for geniuses written by borderline geniuses.

By far my two favorite books are William’s Probability with Martingales and Trefethen’s NLA book. Both British authors. Both written in a highly colloquial style.

It’s really too bad academics have this ego-stroking urge to write to the highest denominator rather than to, say grad students or undergrad.

God damn.