Is f necessarily constant?

Let f:\mathbb{R} \rightarrow \mathbb{R} be a differentiable function such that f'(x)=0 forall  x \in \mathbb{Q}. Does it follow that f is necessarily constant?

Solution

Since f'(x)=0 forall x \in \mathbb{Q} it follows that f(x)=c forall x \in \mathbb{Q}. Let x_0 \in \mathbb{R} \setminus \mathbb{Q} and suppose that f(x_0) \neq c. Since \mathbb{Q} is dense there will exist a sequence \{q_n\}_{n \in \mathbb{N}} of rational numbers that it converges to x_0. Thus:

\displaystyle c \neq f\left ( x_0 \right ) = \lim_{x\rightarrow x_0} f(x)= \lim_{n \rightarrow +\infty} f \left ( q_n \right ) = c

contradicting what we had assumed in the first place. Hence f is constant.

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A contour integral

Let a, b \in \mathbb{C} such that |b|<1. Prove that:

\displaystyle \frac{1}{2\pi} \oint \limits_{\left | z \right |=1} \left | \frac{z-a}{z-b} \right |^2 \left | {\rm d} z\right | = \frac{\left | a-b \right |^2}{1- \left | b \right |^2} + 1

Solution

We have successively:

\begin{aligned} \frac{1}{2\pi}\oint \limits_{|z|=1}\left | \frac{z-a}{z-b} \right |^2 \left | {\rm d} z\right | &=\frac{1}{2\pi} \oint \limits_{|z|=1} \frac{\left ( z-a \right )\left ( \bar{z}-\bar{a} \right )}{\left ( z-b \right )\left ( \bar{z}-\bar{b} \right )}\frac{{\rm d}z}{iz}\\ &= \frac{1}{2\pi i }\oint \limits_{|z|=1} \frac{\left ( z-a \right )\left ( \frac{1}{z}-\bar{a} \right )}{\left ( z-b \right )\left ( \frac{1}{z}-\bar{b} \right )}\frac{{\rm d}z}{z}\\ &= \frac{1}{2\pi i}\oint \limits_{|z|=1}\frac{\left ( z-a \right )\left ( 1-\bar{a}z \right )}{b\left ( 1-\bar{b}z \right )}\left ( \frac{1}{z-b}- \frac{1}{z} \right )\, {\rm d}z\\ &=\frac{\left ( z-a \right )\left ( 1-\bar{a}z \right )}{b\left ( 1-\bar{b}z \right )}\bigg|_{z=b}- \frac{\left ( z-a \right )\left ( 1-\bar{a}z \right )}{b\left ( 1-\bar{b}z \right )} \bigg|_{z=0} \\ &= \frac{\left ( a-\bar{b} \right )\left ( \bar{a}-b \right )}{1-b\bar{b}}+1 \\&= \frac{\left | a-b \right |^2}{1-|b|^2}+1 \end{aligned}

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Convergence of series

Examine if the series:

\displaystyle \sum_{n=1}^{\infty}\bigg ( e - \left ( 1+\tfrac{1}{n} \right )^n \bigg )

converges.

Solution

We are using the basic inequality

e^x \geq x +1 \quad \text{forall} \; x \in \mathbb{R}

as well as the Hermite Hadamard inequality. Applying the Hermite – Hadamard inequality we have that

    \begin{align*} n \ln \left ( 1 + \frac{1}{n} \right ) &= n \int_{n}^{n+1} \frac{{\rm d}x}{x} \\ &\leq \frac{n}{2} \left ( \frac{1}{n} + \frac{1}{n+1} \right ) \\ &= 1 - \frac{1}{2n+2} \end{align*}

Exponentiating we get that

    \begin{align*} \left ( 1 + \frac{1}{n} \right )^n & \leq e e^{-1/(2n+2)} \\ &\leq \frac{e}{1+ \frac{1}{2n+2}} \quad \quad (e^x \geq x +1)\\ &= e \left ( 1 -\frac{1}{2n+3} \right ) \end{align*}

and hence

\displaystyle e - \left ( 1 + \frac{1}{n} \right )^n \geq \frac{e}{2n+3}

This leads us to the conclusion that the series diverges.

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Constant sign

Let f:\mathbb{S} \subset \mathbb{R}^3 \rightarrow  \mathbb{R} be a non zero continuous function on a connected surface \mathbb{S}. Prove that f does not change sign on \mathbb{S}.

Solution

Suppose that f(p)<0 and f(q)>0 for some p,q \in \mathbb{S}. Obviously p \neq q because if p=q then

0<f(q)=f(p)<0

and this is an obscurity. Since \mathbb{S} is connected there exists a continuous curve \gamma:[a, b] \rightarrow \mathbb{S} such that

\gamma (a) = p \quad , \quad \gamma(b)=q

Define g(t) = f \left ( \gamma(t) \right ) , \; t \in [a,b]. This function is obviously continuous. Note that g(a)g(b)<0. So by Bolzano’s theorem, there exists a t_0 \in [a, b] such that g(t_0)=0. This, in return means that f\left ( \gamma(t_0) \right ) =0 but this is impossible since f has a constant sign.

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Double series

Evaluate the series:

\displaystyle \sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}\sum_{n=0}^{\infty}\frac{1}{k2^n+1}

(Putnam Competition , 2016)

Solution

Since the double series the series converges absolutely we can interchange the summation. Thus:

\begin{aligned} \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k} \sum_{n=0}^{\infty} \frac{1}{k2^n+1} &= \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k} \sum_{n=0}^{\infty} \int_{0}^{1} x^{k 2^n } \, {\rm d}x\\ &= \sum_{n=0}^{\infty} \int_{0}^{1}\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k} x^{2^n k} \, {\rm d}x\\ &= \sum_{n=0}^{\infty} \int_{0}^{1} \log \left ( 1+x^{2^n} \right ) \, {\rm d}x\\ &= \int_{0}^{1}\sum_{n=0}^{\infty} \log \left ( 1+ x^{2^n} \right ) \, {\rm d}x \\ &= \int_{0}^{1} \log \prod_{n=0}^{\infty} \left ( 1+x^{2^n} \right ) \, {\rm d}x \\ &= - \int_{0}^{1} \log \left ( 1-x \right ) \, {\rm d}x \\ &=1 \end{aligned}

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