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Adam is trying to sell Bob a bike for $a$ dollars.

Bob does not agree on the price $b$ ($0 < b < a$).

Adam does not agree on this price but does lower his initial price to $\dfrac{a+b}{2}$ dollars.

Bob responds by offering $\dfrac{b+\frac{a+b}{2}}{2}$ dollars.

They continue haggling this way, each time taking the average of the previous two amounts.

On what price will they converge? Express this price in terms of the two initially proposed amounts $a$ and $b$.

Ѕᴀᴀᴅ
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  • Please edit the question, seems unclear – kayush Mar 15 '18 at 05:33
  • The information underneath the line was an attempt i made how could i make it more clear? –  Mar 15 '18 at 05:37
  • (b+(a+b/2))/2 is this correct? if yes then bob is offering more than Adam. I guess you mean (b+(a+b)/2)/2 – kayush Mar 15 '18 at 05:40
  • The explanations about what you tried are exceedingly difficult to understand. Which geometric series are you considering? And how are the ratios (a/4+3b/4)/(a/2+b/2) and (3a/8+5b/8)/(a/4+3b/4) related to the problem? Are you trying to show that the sequence of prices proposed by Adam and Bob is geometric? Why would it be in the first place? Of course it is not (every geometric sequence is monotone and the sequence of suggested prices is not). – Did Mar 15 '18 at 07:34
  • And you tagged this with the tag (fibonacci-numbers). The tag is accurate but that you use it seems to indicate you have information about, and/or understanding of, the problem that you did not share. – Did Mar 15 '18 at 07:38
  • well they are taking the average of the previous two amounts which is similar like fibonacci numbers except that you are dividing it by 2 –  Mar 15 '18 at 07:58

2 Answers2

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Let $a_n$ be the $n^{\rm th}$ bid of $A$, and let $b_n$ be the $n^{\rm th}$ bid of $B$, with $a_0=a$, $b_0=b<a$. Then $$a_n={1\over2}(a_{n-1}+b_{n-1}),\quad b_n={1\over2}(a_n+b_{n-1})={1\over4}a_{n-1}+{3\over4}b_{n-1}\qquad(n\geq1)\ .\tag{1}$$ It follows that $$[b_n,a_n]\subset[b_{n-1},a_{n-1}]\>,\quad a_n-b_n={1\over4}(a_{n-1}-b_{n-1})\qquad(n\geq1)\ .$$ This shows that $\lim_{n\to\infty} a_n=\lim_{n\to\infty}b_n=\alpha$ for a certain $\alpha\in[b,a]$. Furthermore it follows from $(1)$ that $$a_n+2b_n=a_{n-1}+2b_{n-1}\qquad (n\geq1)\ ,$$ so that $a_n+2b_n=a+2b$ for all $n\geq0$, and this implies $$\alpha={1\over3}\lim_{n\to\infty}(a_n+2b_n)={a+2b\over3}\ .$$

Christian Blatter
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  1. We first write the series of bids: Start with $a$ and then $$ \tfrac{b}{1} \to \tfrac{a+b}{2} \to \tfrac{2b+(a+b)}{4}=\tfrac{a+3b}{4} \to \tfrac{2(a+b)+(a+3b)}{8}=\tfrac{3a+5b}{8} \to \tfrac{2(a+3b)+(3a+5b)}{16}=\tfrac{5a+11b}{16}\to\dotsb \tag{*} $$

  2. The denominators in (*) are the series $\{ 2^0, 2^1, 2^2, 2^3, 2^4,\dotsc\}$.

  3. The nominators in (*) are as follows: $$ 0a+1b \to 1a+1b \to 1a+3b \to 3a+5b \to 5a+11b \to 11a+21b \to \dotsb $$

  4. Let us rewrite the coefficients of $a$ and $b$ as sequences: $$ \begin{align*} \text{coefficients of $a$}: \{a_n\mid n\geq0\}=\{ 0, 1, 1, 3, 5, 11, 21, 43, \dotsc \} \\ \text{coefficients of $b$}: \{b_n\mid n\geq0\}=\{ 1, 1, 3, 5, 11, 21, 43, 85, \dotsc \} \end{align*} $$

  5. Two sequences are the same: $a_n=s_n$ and $b_n=s_{n+1}$ for $n\geq0$, where $$ s_0=0, \quad s_1=1, \quad s_n = 2s_{n-2} + s_{n-1} \quad\text{for $n\geq2$} $$ Note that $a_n+b_n=s_n+s_{n+1}=2^n$ for all $n\geq0$. Moreover, $$ s_n = 2s_{n-2} + s_{n-1} = 2(2^{n-2}-s_{n-1})+s_{n-1} = 2^{n-1}-s_{n-1} \quad\text{for $n\geq2$} $$ so that $$ \frac{1}{2}\left(\frac{s_{n-1}}{2^{n-1}}\right) + \frac{s_n}{2^n}=\frac{s_{n-1} + s_n}{2^n} = \frac{2^{n-1}}{2^n}=\frac{1}{2} \tag{**} $$

  6. If the sequence $\dfrac{s_n}{2^n}$ converges to $p$, then the equation (**) follows that $\frac{1}{2}p+p=\frac{1}{2}$ so that $$ \lim_{n\to\infty} \frac{s_n}{2^n} = p = \frac{1}{3} \tag{***} $$

  7. Let us rewrite the bids in (*) as an equation: $$ \frac{a_na+b_nb}{2^n} = \frac{s_na+s_{n+1}b}{2^n} = \frac{s_na+(2^n-s_n)b}{2^n} = \frac{s_n}{2^n}(a-b) + b \quad\text{for $n\geq0$} $$

On what price will they converge? Express this price in terms of the two initially proposed amounts a and b.

Now from the limit in (***), the final answer is: $$ \lim_{n\to\infty} \left[ \frac{s_n}{2^n}(a-b) + b \right] = \frac{1}{3}(a-b)+b $$

ChoF
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