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[Skyrunner]

Here's a non-college, not-calculus question!
 
Say a < b < c.
For
f(x) = | x - a | + | x - b | + | x - c |,
why is f(x)'s minimum always when x = b? :c The graph always seems to look like a V, no matter what numbers I plug in for a, b, and c.

[da_nang]

Easy, first let's partition f(x) into its four cases:

f(x) = | x - a | + | x - b | + | x - c | =

{a-x + b-x + c-x = (a+b+c) - 3x, x < a
{x-a + b-x + c-x = (b+c-a) - x, a <= x < b
{x-a + x-b + c-x = (c-b-a) + x, b <= x < c
{x-a + x-b + x-c = 3x - (a+b+c), x >= c

a < b < c

Trivially noticeable, f(x) is continuous for all x∈ℝ, but not differentiable at x=a, x=b, x=c

Now, what are the minimum points of these cases? The first two are monotonically decreasing, while the latter two are increasing. Thus we get the minimum points f(a), f(b), f(c). Then we determine which one is the smallest.

f(a) = b+c-2a
f(b) = c-a
f(c) = 2c-b-a

Rewrite them:

f(a) = (c-a) + (b-a)
f(b) = (c-a)
f(c) = (c-a) + (c-b)

However, since a < b < c, all these terms are positive, making f(b) the smallest and the global minimum of f(x).

Q.E.D.

[Skyrunner]

Oh, huh. That just required a small bit of thinking. Thanks!

[Loud Whispers]

I have recently been working on a very important physics equation, proposed in an XKCD comic.

Spoiler (click to show/hide)

First I wanted to see how long it took the velociraptor to reach max speed. This stuff is important. Makes later maths easier if the chase drags on.
To find the time I used the equation:
Time = (Final velocity - Initial velocity) / Acceleration
= 6.25 seconds.
With this I wanted to see how much distance the velociraptor could cover in those run up seconds:
Distance = Initial (velocity)(time) + 1/2(acceleration)(time)²
= 78.125 Meters.

78 meters covered in 6.25 seconds. Terrifying speed.


_____________________________________

Next I wanted to check my speed, if I got to 6mph quickly and kept at constant speed, so used:
Velocity x time = Distance traveled
= 37.5 meters.
Then I set about calculating how far I could run against this predator before getting mercilessly butchered, using this formula:
Time = √(distance / acceleration)
= 3.06 seconds, to 3 significant figures.


But this is if you've just popped out of a hiding spot beneath the raptor or jumped down from a tree or something; which would give you 3.06 seconds to form a plan, run or fight.
If you get the 40m head start, it changes things slightly.
Same equation, but with the 40 meters taken into consideration:
4.40 seconds, to 3 significant figures.

This dubious and not the most accurate test will suffice to gather a rough idea of my average reaction time in a calm environment (not filled with velociraptors) : 259 milliseconds.
This means that upon spotting it, I would have 4.141 (+/- 5 milliseconds) to run to safety, plan or fight. That second could mean the difference between life and cruel death.

1. Is that all correct?
2. Cheers to XKCD who made my life 1/3^rd more prepared for a velociraptor attack.

[MonkeyHead]

Seems fine to me, if the assumed values of raptor speed and reactions are reasonable. The equations assume a constant acceleration - a reasonable assumption if air resistance isn't a big deal, which on the scale you are working on shouldn't make too much difference either way.

[Another]

xkcd link is not correct. Should be 135 instead of 137.

Also if your acceleration is capped at higher level than the raptor's than don't run in straight lines and you will be OK assuming the raptor is dumb and has zero reach. For real-world example look at what matadors are doing.

[Loud Whispers]

xkcd link is not correct. Should be 135 instead of 137.

Also if your acceleration is capped at higher level than the raptor's than don't run in straight lines and you will be OK assuming the raptor is dumb and has zero reach. For real-world example look at what matadors are doing.

A fatal mistake Another has made... In the 3rd person.

[Virex]

You are assuming that the raptor needs time to accelerate, while you are at your top speed immediately...

[Zrk2]

You are assuming that the raptor needs time to accelerate, while you are at your top speed immediately...

How long does it take a human to reach max speed? I don't know that it's terribly long, and thus might not be worth accounting for.

[Another]

That assumption was stated in the original problem.

[FearfulJesuit]

Thread necro! I've been bored, recently, and I've been looking at three types of functions, all of which are similar. This class of functions, a↑^bc, where any two of a, b, c are constants and the other is represented by x, classes together both polynomials and exponential functions, and generalizations of them to addition, multiplication, tetration, pentation, etc., and also lumps them in together with functions where both the base and the power are constants, but the form of the function itself varies with respect to x. Two questions:

a. Can the power of the arrow be something other than a whole number- say, could I have the function a↑^3.5b, where the function is sort of halfway between exponentiation and tetration, whatever that would look like?

b. Can one find generalized equations for the derivatives and antiderivatives of
d(a↑^bx)/dx, ∫a↑^bx dx
d(x↑^ab)/dx, ∫x↑^ab dx

and, if the answer to a) is yes,

d(a↑^xb)/dx, ∫a↑^xb dx

where a and b are any given constants and all a, b, x greater than 0? (Just as, for example, ∫xⁿdx=xⁿ⁺¹/n+1 fails when n=-1, I wholly expect there might be some cases where a generalization wouldn't work. But these should be well-defined.)

EDIT: Because a single arrow represents exponentiation, consider also the equations where the arrow is raised to the power of 0 (which I assume would represent multiplication) and -1 (addition). This means that the equations we are really looking at are of the form
a↑^b-2c.

[da_nang]

a. Considering Knuth's up-arrow notation has the power as a positive non-zero integer, you'd need to extend the definition and domain of n.

a↑ⁿb = a^b if n = 1, 1 if b = 0, a↑^n-1(a↑ⁿ(b-1)) otherwise for integers a,b,n where b>=0 and n>=1

b. Per the definition, it's discontinuous everywhere so it isn't differentiable. As for antiderivative, if we assume that the function is zero for any non-integer x then we could consider it to be C, if it exists.

[MagmaMcFry]

Naturally, there are also smooth extensions of a↑^bc into IR³, but there are many of them, and interpolation, differentiation and integration depend on which one you pick. So unless you specify some more constraints that allow the extension to be unique, your question is not well-defined.

[FearfulJesuit]

I think I have stated a question which I understand only superficially...

Dammit, why can't I know all the mathematics?

[da_nang]

Because Gödel said so.