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I'm reading "How would you move Mount Fuji?", and one of the puzzzles/questions is:

A train leaves Los Angeles for New York at a constant speed of 15 miles an hour. At the same moment, a train leaves New York for Los Angeles on the same track. It travels at a constant 20 miles an hour. At still the same moment, a bird leaves the Los Angeles train station and flies toward New York, following the track, at a speed of 25 miles an hour. When it reaches the train from New York, it instantly reverses direction. It travels at the same speed until it reaches the train from Los Angeles, when it reverses again, and so forth. The bird flies back and forth between the two trains until the very moment they collide. How far will the bird have travelled?

In the answer, they mention that it could be solved by using an infinite series (and that most people will probably have forgotten how to do it when asked in an interview, and that John von Neumann did so almost instantly when asked this type of question), but they do it another way. First, they note that the time until the trains crash is given by h, and d = 15h + 20h, where d is the distance (they use 3500). Thus, h = 100 hours. The distance that the bird travels is 25d. Thus, the answer is 2500 miles.

I'm curious though, how does one solve this using an (infinite) series? How does one write this problem as a series in the first place? The only thing that I can get is that the series (of the hours that the bird spends flying) starts like this: (77.77, 9.73, ...)

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I will take a general tack; yes, you use an infinite series, as follows. We begin with the trains being a distance $d_0$ apart. After the bird goes up and back, the trains will be a distance $d_1$ apart. Once we find this distance, we will be able to sort out a geometric series because the bird repeats the same behavior ad infinitum.

The bird initially flies a distance $d_u$ (distance up) that is equal to $v_b t_u$, where $v_b$ is the bird's speed. This is equal to $d_0 - v_2 t_u$, where $v_2$ is the speed of the NY-to-LA train. Then $t_u = d_0/(v_b+v_2)$, and

$$d_u = \frac{v_b}{v_b+v_2} d_0$$

Now the bird flies back to the other train and intercepts it at time $t_d$ later. The LA-to-NY train has traveled a distance $v_1 (t_u+t_d)$ by this time. The bird's position is at $d_u-v_b t_d$, and we solve for $t_d$. By then, the bird will have traveled at distance $v_b (t_u+t_d)$. The result is that, on the first pass, the bird travels a distance of

$$v_b (t_u+t_d) = \frac{2 v_b^2}{(v_b+v_1)(v_b+v_2)} d_0$$

Meanwhile, the trains at this time are a distance apart of

$$d_1 = [d_0-v_2(t_u+t_d)] - v_1 (t_u+t_d) = \frac{(v_b-v_1)(v_b-v_2)}{(v_b+v_1)(v_b+v_2)}d_0$$

By now, we can see how to construct the series that will lead to the result. When the trains are a distance $d_k$, the bird travels a distance

$$\frac{2 v_b^2}{(v_b+v_1)(v_b+v_2)} d_k$$

and

$$d_k = \left [\frac{(v_b-v_1)(v_b-v_2)}{(v_b+v_1)(v_b+v_2)}\right ]^k d_0$$

Therefore, the total distance the bird travels is

$$\begin{align}d &= \frac{2 v_b^2 d_0}{(v_b+v_1)(v_b+v_2)} \sum_{k=0}^{\infty} \left [\frac{(v_b-v_1)(v_b-v_2)}{(v_b+v_1)(v_b+v_2)}\right ]^k \\ &= \frac{\frac{2 v_b^2}{(v_b+v_1)(v_b+v_2)} d_0}{1-\frac{(v_b-v_1)(v_b-v_2)}{(v_b+v_1)(v_b+v_2)}}\end{align}$$

I will leave the rest of the algebra to the reader. The final result is

$$d = \frac{v_b}{v_1+v_2} d_0$$

For the stated problem, the answer is $[25/(15+20)] 3500 \text{miles} = 2500 \text{miles}$.

ADDENDUM

Such a simple result from a relatively complicated analysis screams for a simple explanation. I guess you can look at it this way. Imagine the NY train is held fixed while the LA train moves at $v_1+v_2$ toward NY. The bird flies out at the same time at its speed $v_b$ and simply flies on. The LA train gets to NY in time $t=d_0/(v_1+v_2)$; in that time, the bird flies a distance $v_b d_0/(v_1+v_2)$.

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Let $v_a$, $v_b$ and $v$ be the speeds of the two trains and of the bird respectively (the first being the one from which the bird leaves), $d$ the distance between the cities. Let $t$ denote the time variable, and $t_k$ ($k\in\mathbb N$) the time starting from $t_0=$ in which the bird reverses direction for the $k$-th time. We may refer to Los Angeles and New York as $A$ and $B$ respectively.

By the time the bird has flown $t_1$, the trains have moved by $v_at_1$ and $v_bt_1$ respectively. Since the bird has to reach the second train we have $vt_1=d-v_bt_1$, so that $$ t_1=\frac{d}{v+v_b} $$ Now the bird, which is $t_1v_a$ far from $B$, reverses direction and flies a distance $t_2v$ meeting the first train which now is $(t_1+t_2)v_a$ far from $A$. So $t_2v=d-(t_1+t_2)v_a-t_1v_b$ which gives $$ t_2 = \frac{d - t_1(v_a+v_b)}{v+v_a} = d\frac{v - v_a}{(v+v_a)(v+v_b)} $$ Now suppose you are at time $t_{2n}$, i.e. the bird touches the first train which is now $\sum_{i=1}^{2n}t_iv_a$; by that time, the second train is far $\sum_{i=1}^{2n}t_iv_b$ from $B$. The bird will meet the second train after $t_{2n+1}$, so $t_{2n+1}v=d-\sum_{i=1}^{2n}t_iv_a-\sum_{i=1}^{2n}t_iv_b-t_{2n+1}v_b$ which gives $$ t_{2n+1} = \frac{d-\sum_{i=1}^{2n}t_i(v_a+v_b)}{v+v_b} $$ The same reasoning shows that $$ t_{2n} = \frac{d-\sum_{i=1}^{2n-1}t_i(v_a+v_b)}{v+v_a} $$

Using induction you can show that $$ \begin{cases} t_{2n} = & d\frac{(v-v_a)^n(v-v_b)^{n-1}}{(v+v_a)^n(v+v_b)^n} \\[6pt] t_{2n+1} = & d\frac{(v-v_a)^n(v-v_b)^n~~}{~(v+v_a)^n(v+v_b)^{n+1}} \end{cases} $$ So altogether the bird will fly a distance of \begin{align} v\sum_{n=1}^\infty t_n = & v\left( \sum_{n=1}^\infty t_{2n} + \sum_{n=0}^\infty t_{2n+1} \right) \\ = & dv\left(\frac{1}{v-v_b}+\frac{1}{v+v_b}\right) \sum_{n=0}^\infty \left[ \frac{(v-v_a)(v-v_b)}{(v+v_a)(v+v_b)} \right]^n - \frac{dv}{v-v_b} \end{align} The term $\frac{dv}{v-v_b}$ has been taken out since the first sum starts from $n=1$ and not $n=0$.

Since $\sum_{n=0}^\infty x^n=\frac{1}{1-x}$ for $x\in(0,1)$, you have a total distance of $$ dv\left(\frac{1}{v-v_b}+\frac{1}{v+v_b}\right) \frac{1}{1- \frac{(v-v_a)(v-v_b)}{(v+v_a)(v+v_b)} } - \frac{dv}{v-v_b} = \frac{vd}{(v_a+v_b)} $$

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