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i love mathematical puzzles. i have included a few of my favorites, with answers (spoilers) though i suggest you try them out yourself. sometimes i give hints. i think more people should get interested in math before the school system beats it out of them, though that is a rant for another time

• olympiad proofs and inequalities
  [hint: some of the inequalities from this document may be helpful to you!]
  
  
U697 from Mathematical Reflections 2025 Issue 2
{Problem U697} Let \(f : [a, b] \to \mathbb{R}\) be a continuous function. Prove the inequality: \begin{equation} \left( \int_a^b f(x) \, dx \right)^4 \leq (b-a)^2 \left( \int_a^b e^x f^2(x) \, dx \right) \left( \int_a^b \frac{f^2(x)}{e^x} \, dx \right). \end{equation}
Spoilers
Define the functions: \begin{align*} u(x) &= f(x) \sqrt{e^x}, \\ v(x) &= f(x) \frac{1}{\sqrt{e^x}}. \end{align*} Then: \begin{align*} \int_a^b u(x) v(x) \, dx &= \int_a^b f(x) \sqrt{e^x} \cdot f(x) \frac{1}{\sqrt{e^x}} \, dx = \int_a^b f^2(x) \, dx. \end{align*} Also, \begin{align*} \|u\|^2 &= \int_a^b u^2(x) \, dx = \int_a^b f^2(x) e^x \, dx, \\ \|v\|^2 &= \int_a^b v^2(x) \, dx = \int_a^b \frac{f^2(x)}{e^x} \, dx. \end{align*} By the Cauchy-Schwarz inequality, \begin{equation} \left( \int_a^b f^2(x) \, dx \right)^2 \leq \left( \int_a^b f^2(x) e^x \, dx \right) \left( \int_a^b \frac{f^2(x)}{e^x} \, dx \right). \label{eq:cs1} \end{equation} Applying Cauchy-Schwarz again: \begin{equation} \left( \int_a^b f(x) \cdot 1 \, dx \right)^2 \leq \left( \int_a^b f^2(x) \, dx \right)(b-a). \end{equation} Squaring both sides: \begin{equation} \left( \int_a^b f(x) \, dx \right)^4 \leq (b-a)^2 \left( \int_a^b f^2(x) \, dx \right)^2. \label{eq:cs2} \end{equation} Combining equations 1 and 2 gives: \begin{align*} \left( \int_a^b f(x) \, dx \right)^4 &\leq (b-a)^2 \left( \int_a^b f^2(x) \, dx \right)^2 \\ &\leq (b-a)^2 \left( \int_a^b f^2(x) e^x \, dx \right) \left( \int_a^b \frac{f^2(x)}{e^x} \, dx \right). \end{align*}

U698 from Mathematical Reflections 2025 Issue 2

{Problem U698} Let \( f(z) = u(x, y) + iv(x, y) \) be a holomorphic function on \( \mathbb{C} \), where \( z = x + iy \), and the real part of \( f \) is given by: \( u(x, y) = 2x^2 - 3xy - 2y^2. \)

Spoilers

Since \( f \) is holomorphic, it satisfies the Cauchy-Riemann equations: \[ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}. \] We now have: \[ \frac{\partial v}{\partial y} = 4x - 3y, \quad \frac{\partial v}{\partial x} = 3x + 4y. \] Integrating \( \frac{\partial v}{\partial y} \) with respect to \( y \): \[ v(x, y) = \int (4x - 3y)\,dy = 4xy - \frac{3}{2}y^2 + C(x). \] Differentiate with respect to \( x \) to determine \( C(x) \): \[ \frac{\partial v}{\partial x} = 4y + C'(x). \] Set this equal to \( \frac{\partial v}{\partial x} = 3x + 4y \), yielding: \[ C'(x) = 3x \Rightarrow C(x) = \frac{3}{2}x^2. \] Thus, \[ v(x, y) = 4xy - \frac{3}{2}y^2 + \frac{3}{2}x^2. \] For a holomorphic function, the derivative is: \[ f'(z) = \frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x}. \] From earlier: \[ \frac{\partial u}{\partial x} = 4x - 3y, \quad \frac{\partial v}{\partial x} = 3x + 4y, \] So, \[ f'(z) = (4x - 3y) + i(3x + 4y). \] Computing the modulus: \begin{align*} |f'(z)|^2 &= (4x - 3y)^2 + (3x + 4y)^2 \\ &= 16x^2 - 24xy + 9y^2 + 9x^2 + 24xy + 16y^2 \\ &= 25x^2 + 25y^2 = 25(x^2 + y^2). \end{align*} Therefore: \[ |f'(z)| = 5\sqrt{x^2 + y^2}. \] Since \( |z| = \sqrt{x^2 + y^2} \), we have: \[ \left| \frac{f'(z)}{z} \right| = \frac{|f'(z)|}{|z|} = \frac{5\sqrt{x^2 + y^2}}{\sqrt{x^2 + y^2}} = \boxed{5}. \]

• dandy pdf/epub
  a book about traveling through time (only forward, though) to get over that one really nasty break-up you had
  
  

• non-fiction
• can we please stop acting like it's not porn if it's also oil on canvas?
  an article about lady parts
  
  
• marking molecules
  an article about synapse formation
  
  

• other
• short blog-type posts that don't fit anywhere else
  [this is arguably the part of the site I update the most frequently. I have many interesting opinions, internet!]
  
  
• poetry & spoken word
  don't expect me to meter, please.
  
  
• CV
  yep. doing that.
  
  
• the dietary choices of gwen
  a small bitsy game about your new friend gwen

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