1

Define a sequence where;

$I_s$ are positive integer values,

$I_1$ is a constant value and

$I_2k_1=I_1$

$I_3k_2=I_1+I_2$

$I_4k_3=I_1+I_2+I_3$

this goes on for some $k_i$ value

All variable "$I$" are bounded such that

$I \in\{1,20\}$

and need to minimize $(1/k_1+1/k_2+1/k_3+...+1/k_s)$

With What strategy do i choose $I_2,I_3,I_4 ... I_s,I_{s+1}$ that sum of inverses of $k$ are minimized?

HeroZhang001
  • 2,201
NewBeginner
  • 49

1 Answers1

1

$Z = \frac{1}{k_1} + \frac{1}{k_2} + \frac{1}{k_3}+ ...... + \frac{1}{k_n}$

$\implies Z = \frac{I_2}{I_1} + \frac{I_3}{I_1 + I_2} + \frac{I_4}{I_1 + I_2 + I_3} ... \frac{I_{n+1}}{I_1 + I_2 + I_3 + ....+I_n}$

The following two statements are crucial to the solution -

  • $I_1$ must have the highest possible value in the set {1,2,3,..20} i.e $I_1 = 20$
  • Using Chebyshev's inequality, $n(a_1b_1 + a_2b_2 +... + a_nb_n) \ge (a_1 + a_2 + ... + a_n)(b_1 + b_2 + ... +b_n)$ if either one of the a or b series is increasing and the other is decreasing. Since we know $\frac{1}{I_1} \ge \frac{1}{I_1 + I_2}... \ge ... \frac{1}{I_1+I_2+ ... + I_n} $. Thus for our objective function(Z) to be at its minimum $I_2 \le I_3\le I_4 \le ... \le I_{n+1}$

Further, let us consider this such $I_2$ and $I_3$ that satisfy this, $\frac{1}{I_1 + 1} \ge \frac{I_3}{I_1 + I_2}$

It becomes obvious that I2 = I3 = 1 is the only solution for this, hence by induction we may prove all the integers from $I_2$ to $I_{n+1}$ to be equal to 1

$\therefore I_2 = I_3 = I_4 = .. I_{n+1} = 1$

hence the lowest value of our objective function $(Z)$ is

Z = $\frac{1}{20} +\frac{1}{21} + \frac{1}{22} + ..... + \frac{1}{19 +n}$

Aadi
  • 804