October 25, 2026
If for all $n \in \mathbb{Z}^{+},$ $x_{n} \leq y_{n} \leq z_{n}$ and $\lim_{n \to \infty}x_{n}=u=\lim_{n \to \infty} z_{n},$ then $\lim_{n \to \infty} y_{n}=u.$
Let $\varepsilon_{0} \in \mathbb{R}^{+}$ be arbitrarily chosen. Then we may choose $N_{1}, N_{2} \in \mathbb{Z}^{+}$ such that
$$N_{1} < n \implies |z_{n}-u| < \varepsilon_{0} \text{ and } N_{2} < n \implies |x_{n}-u| < \varepsilon_{0}.$$
Let $N_{0}=\max\{N_{1},N_{2}\}.$ Assume $N_{0} < n.$ By modus ponens,
$$|z_{n}-u| < \varepsilon_{0} \text{ and } |x_{n}-u| < \varepsilon_{0}.$$
By hypothesis, $x_{n}-u \leq y_{n}-u \leq z_{n}-u.$ Thus, since $z_{n}-u \leq |z_{n}-u|,$
$$y_{n}-u \leq z_{n}-u < \varepsilon_{0}.$$
Since $|x_{n}-u| < \varepsilon_{0},$ we have that $-\varepsilon_{0} < x_{n}-u < \varepsilon_{0}.$ Thus,
$$-\varepsilon_{0} < y_{n}-u.$$
Thus,
$$|y_{n}-u| < \varepsilon_{0}.$$
October 3, 2026
Fourier’s Solution to Laplace’s Equation
Currently exploring the history of
analysis
and the “crisis” that preceded its inception. It was apparently
Fourier’s
study of heat flow in 1807. Fourier considered a long and thin plate with
constant temperature at the long sides and heat applied to one of the short
sides. Observe the picture below, taken from Bressoud’s
A Radical Approach to Real Analysis .
Fourier’s scenario can be represented on the $xw$-plane. We can let $z$ represent the temperature at the point $(x,w),$ where $x$ represents the position at the edge of the short side and $w$ represents the distance opposing this short side. We center the short side of the thin plate at the origin so that $x \in [-1,1]$ and $z(-1,0)=z(1,0)=0,$ which would represent the constant temperature at both of the long sides of the plate. The temperature along the bottom edge is described as a function of $x:$
$$z(x,0)=f(x),$$
and Fourier considered the case when $f$ is even, i.e., when $f(-x)=f(x).$ We consider when the temperature along the edge is constant and represent this by
$$z(x,0)=f(x)=1.$$
The problem with this scenario however is that this would then imply that $z(1,0)=0=1.$ This would make the function discontinuous. Fourier observed that the following function satisfies $z \to 0$ as $x \to 1.$
$$f(x)=a_{1}\cos\left(\dfrac{\pi x}{2} \right)+a_{2}\cos\left(\dfrac{3\pi x}{2} \right)+\cdots+a_{k}\cos\left(\dfrac{(2k-1)\pi x}{2}\right),$$
where $k \in \mathbb{Z}^{+}$ and $a_{1},\dots,a_{k}$ are arbitrary constants. He proposed that if this function provides the temperature at the bottom edge of the plate, then at the point $(x,w)$ we have
$$z(x,w)=\sum_{k=1}^{n}a_{k}e^{-(2k-1)\pi w /2}\cos\left(\dfrac{(2k-1)\pi x}{2} \right).$$
In other words, the temperature drops exponentially the further away we are from the edge of the plate. What interests me is how exactly he got this solution! Let’s investigate.Laplace’s equation gives us
$$\dfrac{\partial^{2}z}{\partial x^{2}}+\dfrac{\partial^{2}z}{\partial w^{2}}=0.$$
Fourier decided to show that our function $z$ satisfies this partial differential equation. Since the plate is in thermal equilibrium (neither heating up nor cooling down) the temperature must satisfy Laplace’s equation. To solve this, Fourier sought a solution of the form
$$z(x,w)=g(w)h(x).$$
By Laplace’s equation,
$$g”(w)h(x)+h”(x)g(w)=0.$$
Assuming $g”(w)\neq 0$ and $h”(x)\neq0,$
$$\dfrac{1}{g”(w)h”(x)}\left(g”(w)h(x)+h”(x)g(w) \right)=\dfrac{1}{g”(w)h”(x)} \cdot 0.$$
Thus,
$$\dfrac{h(x)}{h”(x)} + \dfrac{g(w)}{g”(w)}=0.$$
Thus,
$$-\dfrac{h(x)}{h”(x)}=\dfrac{g(w)}{g”(w)}.$$
Notice that the left side is independent of $w$ and the right side is independent of $x.$ Thus, there exists a constant $C$ such that
$$-\dfrac{h(x)}{h”(x)}=C=\dfrac{g(w)}{g”(w)}.$$
Thus, $g(w)=Cg”(w).$ Therefore, $g$ is always the same sign as $g”$ or always the opposite sign. Recall that we want $z$ to be continuous, so we need $g(-1)=g(1)=0.$ Then, $g$ and $g”$ (whose sign describes the concavity) can never be the same, and so $g(w)/g”(w).$ Then we have that
$$\dfrac{g(w)}{g”(w)}=-A$$
and
$$\dfrac{h(x)}{h”(x)}=A,$$
where $A >0.$ Fourier decided to set $A=\dfrac{1}{t^{2}}$ (??) and solved for $g(w)$ and $h(x):$
$$\dfrac{g(w)}{g”(w)}=\dfrac{1}{t^{2}} \implies g(w)=\dfrac{1}{t^{2}}g”(w) \implies \dfrac{1}{t^{2}}u^{2}-1=0,$$
by our study of ordinary differential equations. Then $u=t, -t$. Thus, $g(w)=c_{1}e^{-tw}+c_{2}e^{tw}.$ And,
$$\dfrac{h(x)}{h”(x)}=-\dfrac{1}{t^{2}} \implies \dfrac{1}{t^{2}}h”+h=0 \implies u^{2}=-t^{2} \implies u=\pm it.$$
Thus, $h(x)=c_{3}\cos tx+ c_{4} \sin tx.$
Note that $c_{2}$ must be zero, since the temperature of $z$ must go down as we move away from the edge, i.e., as $w$ increases. And $c_{4}$ must equal 0, since $z$ must be even. Thus, we have the desired solution
$$\alpha e^{-tw}\cos tx,$$
where $t$ is an odd multiple of $\pi/2,$ and $\alpha, t$ are constants. Thus, the general solution is given by
$$z(x,w)=a_{1}e^{-\pi w/2}\cos\left(\dfrac{\pi x}{2} \right) + \cdots + a_{n}e^{-(2n-1)\pi w/2}\cos\left(\dfrac{(2n-1)\pi x}{2} \right)=\sum_{k=1}^{n}a_{k}e^{-(2k-1)\pi w /2}\cos\left(\dfrac{(2k-1)\pi x}{2} \right)$$
The problem here is getting this to satisfy $f(x)=1.$ Fourier decided to let $n \to \infty$, and that’s when the so-called crisis really began. To be continued..
October 2, 2026
Proposition: Suppose the function $\alpha$ increases on the interval $[a,b]$, $a \leq c \leq b$, $\alpha$ is continuous at $c,$ $f(c)=1$, and $f(x)=0$ if $x \neq c.$ Then $f \in \mathscr{R}$.
We wish to prove that for any $\varepsilon > 0,$ there exists a partition $P \in \mathscr{P}(\left [a,b] \right)$ such that $U(P, f, \alpha)-L(P,f,\alpha) < \varepsilon.$ Let
\begin{align*}
M_{k}&=\sup {f(x) : x \in [x_{k-1}, x_{k}] } \\
m_{k}&=\inf {f(x) : x \in [x_{k-1}, x_{k}] }
\end{align*} Let $\varepsilon_{0} > 0$ be given. Note that, by the continuity of $\alpha$ at $c,$ we can choose a $\delta_{0} > 0$ such that, for all $x \in [a,b],$
\[
|x-c|<\delta_{0} \implies
|\alpha(x)-\alpha(c)|<\dfrac{\varepsilon_{0}}{2}.
\]
We prove only the interesting case. Choose a partition $P_{0}=\{x_{0},\dots,x_{N}\}$ with $c \notin P_{0}$ such that $[x_{k-1}, x_{k}] < \delta_{0}$ for all $k \in \{1,\dots,N \}.$ Then, $c \in (x_{i-1}, x_{i})$ for some $i\in \{1,\dots,N \},$ and we have that
\[
U(P_{0}, f, \alpha)-L(P_{0}, f, \alpha)=\alpha(x_i)-\alpha(x_{i-1})=\alpha(x_{i})-\alpha(c)+ \alpha(c)-\alpha(x_{i-1}) < \epsilon_{0}.
\]
September 11, 2026
Darboux sums, the Stieltjes integral, and Chanel
Consider the interval $[a,b] \subseteq \mathbb{R}.$ We define a partition $P=\{x_{0}, x_{1},…,x_{N-1},x_{N} \}$ such that $a=x_{0} < x_{1} < \cdots < x_{N-1} < x_{N}=b.$ Let the function $f:[a,b] \rightarrow \mathbb{R}$ be bounded, and let $\alpha:[a,b] \rightarrow \mathbb{R}$ be non-decreasing. If
\begin{align*}
m_{k} &:= \inf \{f(x) : x \in [x_{k-1}, x_{k}] \}, \\
M_{k} &:= \sup \{f(x) : x \in [x_{k-1}, x_{k}]\},
\end{align*} then we define the upper sum and lower sum of $f$, respectively, by
\begin{align*}
U(P,f,\alpha) &:= \sum_{k=1}^{N} M_{k}(\alpha (x_{k}) - \alpha (x_{k-1})), \\ L(P,f,\alpha) &:= \sum_{k=1}^{N} m_{k}(\alpha (x_{k}) - \alpha (x_{k-1})).
\end{align*} We call these sorts of sums Darboux sums, and we can use them to define the Stieltjes integral.
The lower and upper integrals are defined, respectively, by
\begin{align*}
\int_{\underline{a}}^{b}f(x) d \alpha(x) &:= \sup\{L(P,f,\alpha) : P \in \mathcal{P}([a,b]) \},\\
\int_{a}^{\overline{b}}f(x) d \alpha(x) &:= \inf\{U(P,f,\alpha) : P \in \mathcal{P}([a,b]) \},
\end{align*}
where $\mathcal{P}([a,b])$ denotes the set of all partitions of $[a,b].$ When
$$\int_{\underline{a}}^{b}f(x) d \alpha(x)=I=\int_{a}^{\overline{b}}f(x) d \alpha(x),$$
we just represent this value $I$ by
$$\int_{a}^{b}f(x) d \alpha(x).$$ Nice and all, but there's an interesting property of Darboux sums that merits proof. First, we need a definition: If $P$ and $\tilde{P}$ are partitions of $[a,b],$ we say that $\tilde{P}$ is a refinement of $P$ if $P \subseteq \tilde{P}.$
If $\tilde{P}$ is a refinement of $P,$ then $U(\tilde{P}, f, \alpha) \leq U(P, f, \alpha)$ and $L(\tilde{P}, f, \alpha) \geq L(P, f, \alpha).$
First, we’ll prove that $U(\tilde{P}, f, \alpha) \leq U(P, f, \alpha)$. It suffices to prove that $$U(P, f, \alpha)-U(\tilde{P}, f, \alpha) \geq 0.$$ We consider the case when $\tilde{P}= P \cup \{c\}$. Then, there exists $\ell \in \mathbb{Z}^{+}$ such that $x_{\ell -1}< c < x_{\ell}.$ Observe that $$U(\tilde{P}, f, \alpha) = \sum_{k=1}^{\ell-1}M_{k}(\alpha(x_{k})-\alpha(x_{k-1}))+\sup \{ f(x):x \in [x_{\ell -1}, c]\}(\alpha(c)-\alpha(x_{\ell - 1})) +\sup \{ f(x):x \in [c,x_{\ell}]\}(\alpha(x_{\ell})-\alpha(c)) + \sum_{k=\ell+1}^{N}M_{k}(\alpha(x_{k})-\alpha(x_{k-1})).$$ Thus, it must be the case that $$U(P, f, \alpha)-U(\tilde{P}, f, \alpha) = M_{\ell}(\alpha(x_{\ell})-\alpha(x_{\ell-1}))-\sup \{ f(x):x \in [x_{\ell -1}, c]\}(\alpha(c)-\alpha(x_{\ell - 1})) -\sup \{ f(x):x \in [c,x_{\ell}]\}(\alpha(x_{\ell})-\alpha(c)) =M_{\ell}(\alpha(x_{\ell})-\alpha(c) +\alpha(c)-\alpha(x_{\ell - 1})) -\sup \{ f(x):x \in [x_{\ell -1}, c]\}(\alpha(c)-\alpha(x_{\ell - 1})) -\sup \{ f(x):x \in [c,x_{\ell}]\}(\alpha(x_{\ell})-\alpha(c)) =(M_{\ell}-\sup \{ f(x):x \in [c,x_{\ell}]\})(\alpha(x_{\ell})-\alpha(c)) +(M_{\ell}-\sup \{ f(x):x \in [x_{\ell -1}, c]\})(\alpha(c)-\alpha(x_{\ell-1})) \geq 0,$$ since $\alpha$ is non-decreasing and $M_{\ell}$ is the supremum of $f$ over $[x_{\ell -1}, x_{\ell}].$
Proving $L(\tilde{P},f,\alpha)\geq L(P,f,\alpha)$ is analogous. We need only observe that $$L(\tilde{P},f,\alpha)- L(P,f,\alpha) = \inf\{f(x):x\in[x_{\ell - 1},c] \}(\alpha(c)-\alpha(x_{\ell - 1})) + \inf\{f(x):x\in[c,x_{\ell}] \}(\alpha(x_{\ell})-\alpha(c)) -m_{\ell}(\alpha(x_{\ell})-\alpha(c)+\alpha(c)-\alpha(x_{\ell - 1})).$$ And we are done.
Chanel!!
September 7, 2026
Dropped AIME Problem
Consider $x,y \in \mathbb{R}$ such that $$ |x| \neq |y|, \ x^{3}=15x+4y, \text{ and } y^{3}=4x+15y.$$ Find $x^{2}+y^{2}.$
Solution: Well, observe that $$x^{3} + y^{3} = 19x+19y=19(x+y)=(x+y)(x^{2}-xy+y^{2}).$$ Thus, since $|x| \neq |y|,$ $$x^{2}+y^{2}=19+xy.$$ We also observe that $$x^{3}-y^{3}=11x-11y=11(x-y)=(x-y)(x^{2}+xy+y^{2}).$$ Thus, since $|x|\neq|y|,$ we have that $$x^{2}+y^{2}=11-xy.$$ Thus, $$19+xy=11-xy \implies 2xy=-8 \implies xy = -4.$$ Thus, $x^{2}+y^{2} = 15.$
Demystifying the AC Method
Let’s explore the AC method for factoring quadratic expressions, particularly when the leading coefficient $a$ is unequal to $1$. We consider the anatomy of $$ax^{2}+bx+c,$$ where $a \neq 0$. If this expression is factorable, then it can be written as a product of two linear factors, like so: $$ax^{2}+bx+c = (mx+p)(nx+q).$$ Expanding the right-hand side gives us $$ax^{2}+bx+c = mnx^{2} + (mq+np)x+pq.$$ Then, it must be the case that $a=mn$, $b=mq + np,$ and $c=pq.$ We recall that the AC method states that one would have to multiply $a$ and $c$, then find factors of $ac$ such that their sum is equal to $b.$ From above, we see that $ac=mnpq=mqnp=(mq)(np)$, and so $b$ is indeed the sum of two factors of $ac$.