Will I ever hit the wall?

[problemchild]

I am standing 10 feet from a wall. I will begin to walk toward the wall. My first step is 4 feet. Each succeeding step is exactly one half the distance of the prior step. So my next step would be 2 feet then one etc. How many steps will it take me to reach the wall? What might the equation that asks this question look like and what would the proof of the answer look like? I don't know.
Thanks- Chris

 

[DrPhil]

I am standing 10 feet from a wall. I will begin to walk toward the wall. My first step is 4 feet. Each succeeding step is exactly one half the distance of the prior step. So my next step would be 2 feet then one etc. How many steps will it take me to reach the wall? What might the equation that asks this question look like and what would the proof of the answer look like? I don't know.
Thanks- Chris

Write the sequence of the steps and sum of steps:
Step size: 4, 2, 1, 1/2, 1/4, . . .
Sum:...... 4, 6, 7, 7.5, 7.75, . . .

Can you see where the Sum is going in the limit? Is it greater than 10?
[I wonder if the problem really said 10', or was it perhaps 8' ?]

Have you learned about geometric series? There is a formula for the sum.

In considering "when" you might hit the wall, you need to say something about how long it takes you to make each step. Usually we say that the time is proportional to size a step .. that is, you are keeping constant speed.

 

[JeffM]

I am standing 10 feet from a wall. I will begin to walk toward the wall. My first step is 4 feet. Each succeeding step is exactly one half the distance of the prior step. So my next step would be 2 feet then one etc. How many steps will it take me to reach the wall? What might the equation that asks this question look like and what would the proof of the answer look like? I don't know.
Thanks- Chris

You will not hit the wall.

\(\displaystyle n\ an\ integer > 1 \implies \left(1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8}\ +\ ...\ \dfrac{1}{2^n}\right) = 1 + \dfrac{2^n - 1}{2^n}.\)

But \(\displaystyle n\ an\ integer > 1 \implies 0 < \dfrac{1}{2^n} < 1 \implies\)

\(\displaystyle - 1 < - \dfrac{1}{2^n} < 0 \implies\)

\(\displaystyle 0 < 1 - \dfrac{1}{2^n} < 1 \implies\)

\(\displaystyle 0 < \dfrac{2^n - 1}{2^n} < 1 \implies\)

\(\displaystyle 1 < 1 + \dfrac{2^n - 1}{2^n} < 2.\)

Now lets say you step forward n + 1 times toward the wall (where n is finite). How far do you get?

\(\displaystyle 4 + \dfrac{4}{2} + \dfrac{4}{4} + \dfrac{4}{8} ...\ \dfrac{1}{2^n} = 4\left(1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8}\ +\ ...\ \dfrac{1}{2^n}\right) = 4\left(1 + \dfrac{2^n - 1}{2^n}\right) < 4 * 2 = 8.\)

Not quite 8 feet.

Edit: If you consider moving forward an infinite number of time, you move exactly 8 feet. I shall not attempt a proof, but you may find that you are very old before you have completed the task.

 

[lookagain]

\(\displaystyle n\ an\ integer > 1 \implies \left(1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8}\ +\ ...\ \dfrac{1}{2^n}\right) = 1 + \dfrac{2^n - 1}{2^n}. \ \ \ \ \ \ \) A plus sign is missing in front of \(\displaystyle \ \dfrac{1}{2^n}.\)



\(\displaystyle 4 + \dfrac{4}{2} + \dfrac{4}{4} + \dfrac{4}{8} ...\ \dfrac{1}{2^n} = 4\left(1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8}\ +\ ...\ \dfrac{1}{2^n}\right) = 4\left(1 + \dfrac{2^n - 1}{2^n}\right) < 4 * 2 = 8. \ \ \ \ \ \ \)Two plus signs are missing in the far left sum, and one plus sign is missing in the next sum.

\(\displaystyle n\ an\ integer > 1 \implies \left(1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8}\ +\ ...\ + \dfrac{1}{2^n}\right) = 1 + \dfrac{2^n - 1}{2^n}.\)

\(\displaystyle 4 + \dfrac{4}{2} + \dfrac{4}{4} + \dfrac{4}{8} \ + \ ...\ + \ \dfrac{1}{2^n} = 4\left(1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8}\ + \ ...\ + \ \dfrac{1}{2^n}\right) = 4\left(1 + \dfrac{2^n - 1}{2^n}\right) < 4 * 2 = 8.\)

.

 

[HallsofIvy]

IF you are not a "point mass"- that is, if your body is real and not a "mathematical fiction", so it extends any distance at all horizontally, then your body will extend past the point directly over your feet and will hit the wall!

 

[JeffM]

IF you are not a "point mass"- that is, if your body is real and not a "mathematical fiction", so it extends any distance at all horizontally, then your body will extend past the point directly over your feet and will hit the wall!

Only if he has a beer belly two feet deep. If he gets to eight feet the wall is still two feet away.

 

[lookagain]

IF you are not a "point mass"- that is, if your body * is real and not a "mathematical fiction", so it extends any distance at all horizontally, then your body will extend past the point directly over your feet and will hit the wall!

But the question didn't mention about a "body" * hitting the wall. The headline for the initial post for this thread is
"Will I ever hit the wall?" So, it's not necessary that someone's body extend past the end of their
feet in order to count for them hitting the wall.

If the leading toe juts out further than the belly, chest, etc. above the feet, then the time to hit the wall would
count when (or if) the leading toe reaches the wall.

That is, a person would be correct in stating that they "hit the wall" at the moment that the frontmost part of a
foot reached/touched the wall first.

 

[lookagain]

Hope you hit it with your nose, lookagain

If I am Pinocchio and I have just told a lie, then I would expect to hit the wall with my nose.