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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 nth 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.

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8 thoughts on “Problem Set 2

  1. 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. 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.

    • 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. 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. 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. 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. 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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