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

$f(x) >0$ whenever $x \neq 0$
$f(0)=0$ 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

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:

October 20, 2014 at 5:52 pm | Reply

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.

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 $p$ .

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 $f(x) = 1.2172336282112216574082793255625$ .

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

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)

prahlad says:

June 22, 2015 at 12:35 pm | Reply

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
hm says:

October 20, 2014 at 5:50 pm | Reply

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:
  
  October 20, 2014 at 5:52 pm | Reply
  
  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.
hossain says:

October 20, 2014 at 5:49 pm | Reply

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.
Fan says:

August 30, 2013 at 5:18 am | Reply

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 $p$ .
Fan says:

August 30, 2013 at 5:13 am | Reply

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 $f(x) = 1.2172336282112216574082793255625$ .
Kavi Rama Murthy says:

July 4, 2013 at 11:52 am | Reply

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

July 2, 2013 at 12:50 pm | Reply

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)

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Problem Set 2

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