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Problem Set 2

In this problem set we present four questions:

(Middle School Students)
The irrational number $0.1234012340012340001234\dots$ is formed
by using alternating blocks of $1234$ and zeros, where the $n$th block
of zeros following the decimal contains $n$ zeros. What is the digit
in the 2550th place following the decimal?

(High School Students) Consider the function $f : \mathbb{R} \rightarrow \mathbb{R}$ given by

$f(x) = \left \{ \begin{array}{ll} 1-\frac{\sin(x)}{x} & x \neq 0 \cr 0 & \mbox{otherwise} \end{array} \right.$

Find the maximum and minimum value of the function.

(Undergraduate Students) Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be an infinitely differentiable function, such that

1. $f(x) >0$ whenever $x \neq 0$
2. $f(0)=0$ and $f^{(n)}(0) = 0$ where $n \geq 1$ and $f^{(n)}$ is the $n$-th derivative of $f.$

Let $g : \mathbb{R} \rightarrow \mathbb{R}$ is given by

$g(x) = \sqrt{f(x)},\,\,\forall x \in {\mathbb R}$

Decide whether $g$ is infinitely differentiable on $\mathbb{R}$.

(Masters Students and above) If $A$ is a measurable subset of $\mathbb{R}$ such that

$a\in A,\,\,\,b\in A,\,\,\, a\neq b\,\,\, \Rightarrow \,\,\, \frac{a+b}{2}\notin A$

then $A$ has measure $0.$

8 thoughts on “Problem Set 2”

Undergraduate : No, it may not be.Consider the function $$f(x) = \left\{ \begin{array}{ll} \sin^2{\left(\frac{1}{x^2}\right)}e^{-\frac{1}{x^2}}+e^{-\frac{2}{x^2}} & \quad x \neq 0 \\ 0 & \quad x = 0 \end{array} \right.$$Since $e^{-\frac{1}{x^2}}=o(x^n)$ as $x\to0$ for each $n$, it can be shown inductively that its k-th derivative $f^{(k)}(x)=o(x^n)$ as $x\to0$ and $f^{(k+1)}(0)=\lim_{x\to0}\frac{f^{(k)}(x)}{x}=\lim_{x\to0}o(x^{n-1})=0$ for each $k\geq 0$.Hence $f$ is a $C^{\infty}$ function and satisfies the hypothesis of the question.Now we show that the second derivative of the function $g(x)=\sqrt{f(x)}$ is discontinuous at $0$.For $x\neq0$, $g''(x)=\frac{f''(x)}{2\sqrt{f(x)}}-\frac{f'(x)^2}{4f(x)^{\frac{3}{2}}}$. At $x_k=\frac{1}{\sqrt{k\pi}},k\in\mathbb{N}$ the second term $\frac{f'(x_k)^2}{4f(x_k)^{\frac{3}{2}}}=\frac{4e^{-\frac{1}{x_k^2}}}{x_k^6}\to0$ as $k\to\infty$. But the first term $\frac{f''(x_k)}{2\sqrt{f(x_k)}}=\frac{4}{x_k^6}+\frac{8}{x_k^6}e^{-\frac{1}{x_k^2}}-\frac{6}{x_k^4}e^{-\frac{1}{x_k^2}}\to\infty$ as $k\to\infty$. Hence $g''(x_k)\to\infty$ as $k\to\infty$ but $x_k\to0.$ So $g(x)$ is not a $C^2$ function.

On the other hand, the $C^\infty$ function $$f(x) = \left\{ \begin{array}{ll} e^{-\frac{1}{x^2}} & \quad x \neq 0 \\ 0 & \quad x = 0 \end{array} \right.$$ is such that $f^{(n)}(0)=0$ for each $n\in\mathbb{N}$ and clearly its square root $g(x)=\sqrt{f(x)}$ is also a $C^\infty$ function

2. hm says:

the digit in the 2550th place is 4. The first 71 blocks consist of 2546 digits. the 4th digit of the 72 th block is 4, which is the 2550 digit. we can guess that 2550 th digit occured in the 72 th block by using 5+6+…….+n=2550, which gives n=71.05.

• hm says:

the digit in the 2550th place is 4. The first 71 blocks consist of 2546 digits. the 4th digit of the 72 th block is 4, which is the 2550 digit. we can guess that 2550 th digit occured in the 72 th block by using 5+6+…….+n=2550, which gives n=71.05.

3. hossain says:

the digit in the 2550th place is 4. The first 71 blocks consist of 2546 digits. the 4th digit of the 72 th block is 4, which is the 2550 digit. we can guess that 2550 th digit occured in the 72 th block by using 5+6+…….+n=2550, which gives n=71.05.

4. Fan says:

For the undergraduate problem, I guess something surprising is going to happen…
For the graduate problem, if $m(A) > 0$, take a rectangle almost occupied by $A$. Choose a point $p \in A$ sufficiently close to its center and consider reflection about $p$.

5. Fan says:

For the high school problem, the maximum of $f$ should be attained at the solutions of $\tan x = x$ in the interval $[\pi/2, 3\pi/2]$. Numerically it is
$x = 4.4934094579090641753078809272803$ with $f(x) = 1.2172336282112216574082793255625$.

6. Kavi Rama Murthy says:

The value at 4.6 (slightly smaller than 3*pi/2 exceeds the value at 3*pi/2.
Kavi Rama Murthy

7. mihir says:

I am trying to give solution to the problem for high school students
f(x)=1-(sinx/x) for x not equal to 0
=0 otherwise
as x-sinx>0, 1>sinx/x
therefore 1-sinx/x will be positive for x belongs to (0,pi/2] & as x takes values greater the numerator of sinx/x will be greater hence the 1-sinx/x will be positive
therefore minimum value of f(x)=0
in [pi,3pi/2]
sinx is negative
hence at 3pi/2 the valiue of sinx/x will be -2/(3pi)
hence value of f(x) becomes 1-(-2/(3pi)) i.e. 1+2/3(pi)
as x takes value greater the value of sinx/x will be very very small
hence at x=3pi/2 f(x) is maximum
hence maximum of f(x)=1+2/3(pi)