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Sameer K.
Math Ph.D Student, Graduate of UChicago
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Linear Algebra
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Question:

Suppose $$V$$ is a finite dimensional vector space over $$\mathbb{R}$$, and $$f: V \to V$$ is a linear map satisfying $$(f\circ f)(v) = -v$$. Prove that the dimension of $$V$$ must be even.

Sameer K.

Since $$f^2 = -I$$, we see that the minimal polynomial of $$f$$ over $$\mathbb{R}$$ has to be $$x^2 + 1 \in \mathbb{R}[x]$$. The minimal polynomial and characteristic polynomial share their root sets, so the characteristic polynomial of $$f$$ must be $$(x^2 + 1)^k$$ for some $$k\ge 1$$. But the degree $$2k$$ of the characteristic polynomial is the dimension of $$V$$, hence $$V$$ is even-dimensional. Key idea: minimal polynomials and characteristic polynomials of linear transformations are intimately related, and information about one tells us information about the other.

Number Theory
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Question:

Find all solutions to the equation $(p^q - q^p = p+q$) where $$p$$ and $$q$$ are both prime numbers.

Sameer K.

Suppose $$p$$ and $$q$$ are primes such that $$p^q - q^p = p+q$$. We can assume $$p\neq q$$, since otherwise we get $$p+q = 0$$. Taking the equation mod $$p$$ gives $(-q^p \equiv q \pmod{p}.$)Since $$p\neq q$$, by Fermat's little theorem we also have $(q^{p} \equiv q \pmod{p} \implies q \equiv -q \pmod{p} \implies 2q \equiv 0 \pmod{p}.$)Hence it follows that $$p = 2$$. The original equation is therefore $$2^q = q^2+ q+2$$. But for any $$x>5$$, we have $$2^x > x^2 + x+ 2$$. It follows that $$q \le 5$$. Checking all the possibilities for $$q$$ now reveals that $$(p,q) = (2,5)$$ is the only possible solution. Key idea: using modular arithmetic and Fermat's little theorem to pare down the possibilities for our primes $$p$$ and $$q$$.

Calculus
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Question:

Evaluate the integral $(\int_{0}^{1} \ln(x) \ln(1-x) \, dx$)

Sameer K.

We can do this in a few steps-- here is a general outline, Step 1: First, prove that for each $$n\in \mathbb{N}$$, $(\int_{0}^{1} x^n \log(x) \, dx = -\frac{1}{(n+1)^2}$)Starting with the identity $(\int_{0}^{1} x^n \, dx = \frac{1}{n+1}$)differentiate both sides with respect to $$n$$ to get the desired result. Key idea: differentiating under the integral sign. Step 2: Expanding $$\log(1-x)$$ as its Taylor series, we have $(\log(x)\log(1-x) = -x\log(x) - \frac{x^2 \log(x)}{2} - \frac{x^3 \log(x)}{3} - \cdots$)Sweeping issues of convergence under the rug, integrating both sides over the interval $$[0,1]$$ gives $(\int_{0}^{1} \log(x) \log(1-x) \, dx = \frac{1}{2^2} + \frac{1}{2\cdot 3^2} + \frac{1}{3\cdot 4^2} + \cdots$)Key idea: turning an integral into a sum using power series expansion. Step 3: We must evaluate the sum $(\sum_{k=1}^{\infty} \frac{1}{k(k+1)^2} = \sum_{k=1}^{\infty} \left(\frac{1}{k} - \frac{1}{k+1} - \frac{1}{(k+1)^2} \right)$) $( = \sum_{k=1}^{\infty} \left(\frac{1}{k} - \frac{1}{k+1}\right) - \sum_{k=1}^{\infty} \frac{1}{(k+1)^2} = 1 - \left( \frac{\pi^2}{6} - 1\right) = 2-\frac{\pi^2}{6}$)Here, we have used Euler's famous Basel identity $$\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$$. Key idea: evaluating a sum using the telescoping technique, recognizing a famous fact we know.

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