# "The world's slowest runner"

• Oct 10th 2007, 10:52 AM
λιεҗąиđ€ŗ
"The world's slowest runner"
Hello everyone, welcome to some reading.

Today our math teacher presented an problem when we were studying Geometric progression. The problem is the following:

A runner covers one meter in his first step, the next step will be 90% of the step before. After how many steps will he have "ran" 100 meters?

...
The problem can be formulated as this: $\frac { 1(0.9^x - 1) } { 0.9 -1 } = 100$

Working some with it we will end up with no good answer, we'll get the $x = \frac { 10^{(\log - 9)} } { 0.9 }$

Negative logarithms? We don't go there, yet.

As putting it as an equation don't work (for us newbies) we did some test-and-see. The calculators loose precision after 10 meters have been covered.

So, my teacher reasoned, the runner will never run 100 meters.

I disputed his answer and compared this problem with one of Zeno's paradoxes about a runner who's step first 1, then a half, then a quarter, and so on.

Will the runner ever cover 100 meters? My answer is yes, he will. But noone would agree with me.

What do you think of this problem?

Zeno's paradoxes - Wikipedia, the free encyclopedia

P.S. I got a russian to agree with me, it's always good to have a russian backing you up.:):cool:
• Oct 10th 2007, 12:02 PM
CaptainBlack
Quote:

Originally Posted by λιεҗąиđ€ŗ
Hello everyone, welcome to some reading.

Today our math teacher presented an problem when we were studying Geometric progression. The problem is the following:

A runner covers one meter in his first step, the next step will be 90% of the step before. After how many steps will he have "ran" 100 meters?

...
The problem can be formulated as this: $\frac { 1(0.9^x - 1) } { 0.9 -1 } = 100$

Working some with it we will end up with no good answer, we'll get the $x = \frac { 10^{(\log - 9)} } { 0.9 }$

Negative logarithms? We don't go there, yet.

As putting it as an equation don't work (for us newbies) we did some test-and-see. The calculators loose precision after 10 meters have been covered.

So, my teacher reasoned, the runner will never run 100 meters.

I disputed his answer and compared this problem with one of Zeno's paradoxes about a runner who's step first 1, then a half, then a quarter, and so on.

Will the runner ever cover 100 meters? My answer is yes, he will. But noone would agree with me.

What do you think of this problem?

Zeno's paradoxes - Wikipedia, the free encyclopedia

P.S. I got a russian to agree with me, it's always good to have a russian backing you up.:):cool:

The runner cannot travel further than:

$\lim_{n \to \infty} \frac { (0.9^n - 1) } { 0.9 -1 }=10$ m

RonL
• Oct 10th 2007, 12:14 PM
λιεҗąиđ€ŗ
Quote:

The runner cannot travel further than: 10 m
How could that be true since he always travels forward?
• Oct 10th 2007, 01:37 PM
CaptainBlack
Quote:

Originally Posted by λιεҗąиđ€ŗ
How could that be true since he always travels forward?

Because he is travelling (on avaerage) more and more slowly, so that in
infinite time he covers but 10m.

RonL
• Oct 14th 2007, 02:13 AM
λιεҗąиđ€ŗ
Quote:

Originally Posted by CaptainBlack
Because he is travelling (on avaerage) more and more slowly, so that in
infinite time he covers but 10m.

RonL

:)

Can you explain how you reason concerning the distance covered affecting the speed?

I.e. He takes smaller steps -> he's slowing down.

Common logic would say the opposite, wouldn't it? That he's actually going faster, since the distance is shrinking?

This is fun :)
• Oct 14th 2007, 09:45 AM
CaptainBlack
Quote:

Originally Posted by λιεҗąиđ€ŗ
:)

Can you explain how you reason concerning the distance covered affecting the speed?

I.e. He takes smaller steps -> he's slowing down.

Common logic would say the opposite, wouldn't it? That he's actually going faster, since the distance is shrinking?

This is fun :)

No its not, you are making a mountain out of a mole hill.

The steps are getting smaller, there is no mention of time in the question
so we can not say anything about speed.

The steps are getting smaller at such a rate that the runner will never
cover more than 10m, however fast he takes them.

RonL
• Oct 14th 2007, 12:25 PM
Soroban
Hello, λιεҗąиđ€ŗ!

Quote:

A runner covers one meter in his first step, the next step will be 90% of the step before.
After how many steps will he have "ran" 100 meters?

The problem can be formulated as: . $\frac{1(0.9^x - 1)}{ 0.9 -1 } = 100$

Working some with it, we end up with: . $x = \frac {10^{(\log(-9)}}{0.9}$ . . . . no
Negative logarithms? We don't go there yet.

So, my teacher reasoned: the runner will never run 100 meters. . . . . Correct!

The total distance is: . $S \:=\:\frac{0.9^x - 1}{0.9 - 1} \:=\:100\quad\Rightarrow\quad \frac{0.9^x - 1}{-0.1} \:=\:100
$

. . $0.9^x - 1\:=\:-10\quad\Rightarrow\quad 0.9^x \:=\:-9$

Take logs: . $\ln(0.9^x) \:=\:\ln(\text{-}9)\quad\Rightarrow\quad x\ln(0.9) \:=\:\ln(\text{-}9)\quad\Rightarrow\quad x \:=\:\frac{\ln(\text{-}9)}{\ln(0.9)}$

This is not a real number . . . There is no solution.
The runner will never run 100 meters.

CaptainBlack is correct.
The total possible distance is: . $d \;=\;\frac{1}{1 - 0.9} \:=\:\frac{1}{0.1} \:=\: 10$ meters.

The runner's speed is assumed to be constant,
. . neither speeding up nor slowing down.

Suppose he runs at 1 meter per second.

He runs 1 meter.
. . This takes 1 second.

Then he runs 0.9 meters.
. . This takes 0.9 seconds.

Then he runs 0.9² = 0.81 meters.
. . This takes 0.81 seconds.

Assuming he could run an infinite number of these intervals:
. . his total distance is: . $d \;=\;1 + 0.9 + 0.9^2 + 0.9^3 + \cdots \;=\;10$ meters.
. . his total time is: . $t \;=\;1 + 0.9 + 0.9^2 + 0.9^3 + \cdots \;=\;10$ seconds.

His average speed is: . $\frac{10\text{ meters}}{10\text{ seconds}} \:=\:1$ meter per second . . . see?

• Oct 14th 2007, 12:41 PM
λιεҗąиđ€ŗ
This problem is really wierd isn't it?

I still have faith in that our runner will run 100 meters...and more if needed!:)
---

There is only one thing I cannot comprehend. It is that the proof of the dichtonomy paradox of zeno can't be applied to this problem. (look at wikipedias article)
• Oct 21st 2007, 08:20 AM
Thomas154321
Think of it this way:

The runner runs 10% of 10m in his first step, then 10% of the remaining 9m in his second step, then 10% of the remaining 8.1m in his next step etc.

Each time he runs 10% of the distance to the 10m line. If he is always running a percentage of the gap, he will never close the gap. So he certainly won't reach over 10m.
• Dec 27th 2007, 01:22 AM
JoFaSs
distance covered: 1m in first step, 1m+0.9(1)m in total of first and second step...Now as the steps become smaller there will be such a point in which a step covers very little distance(almost no distance at all).

Sum of an infinite geometric sequence: $S_{\infty}=\frac{a}{1-r}$
$a=First \ Term\rightarrow1$
$r=Common \ Ratio\rightarrow0.9$

Now we have $S_{\infty}=\frac{1}{1-0.9}$
$S_{\infty}=10m$

:D
• Dec 27th 2007, 01:26 AM
Isomorphism
Quote:

Originally Posted by λιεҗąиđ€ŗ
This problem is really wierd isn't it?
I still have faith in that our runner will run 100 meters...and more if needed!:)

If you still have "faith", then you most be treating maths like a religion. "I have faith in Zeno" - you can say that without any remorse in religion.
But Alas! in mathematics,moreover, after Soroban's detailed proofs, you should have no doubt. In case you have a doubt in his idea, people here can help you. If you just have faith in that fact without reason, then may GOD bless you.

By the way what is the problem with the dichotomy argument??