very hard calculus word problem, lets see what you got!?

Question

this is most likely solved by a series, possibly a power series.
We are given unlimited number of identical, very thin dominoes. The length of a domino is 2 cm and it's mass is neglegable (10 ^ -4314 gram per domino).
Suppose we stack them(assume we are infinitely powerful), one on top of the other, so that the stack advances horizontally in one direction because each is offset a little. the top domino is offset 1cm, and as we move down the stack they are offset progressively less from each other so as to keep the center of gravity. the stack is kept stable only by gravity.
1) show that the stack can advance infinite distance. HINT: numerate the dominoes in a stack from the top. Note that a stack of two dominoes can advance 1cm since the center of the first domino should be above the edge of the second domino. Place the third domino underneath the first two dominoes so that the edge of the third domino is below the center of the stack of the two dominoes that has been built already.

Answer 1

Consider the top domino to be centered at zero. Let d(n) be the maximum x-coordinate of the nth domino from the top. Clearly, the left edge of the domino (which is 1 cm left of the center) must be directly below the center of gravity of the other n dominoes, so d(n) is the average of the previous n-1 values of d, plus 1. Thus we get:

d(1)=0
d(2)=1
d(3)=3/2
d(4)=11/6
d(5)=25/12
d(6)=137/60
d(n)= [k=1, n-1]∑(d(k))/(n-1) + 1

Now, this is not by itself very illuminating. However, consider the differences between these terms:

d(2)-d(1)=1
d(3)-d(2)=1/2
d(4)-d(3)=1/3
d(5)-d(4)=1/4
d(6)-d(5)=1/5

Thus, we begin to suspect that the difference between d(n+1) and d(n) is 1/n, and consequently that d(n) = [k=1, n-1]∑(1/k). We shall prove this rigorously by induction:

First, we note that this is trivially true for d(2). Suppose it is true for all d(x) where 2 ≤ x ≤ n. Then:

d(n+1) = [k=1, n]∑(d(k))/n + 1
= 0+ [k=2, n]∑(d(k))/n + 1
= [k=2, n]∑( [j=1, k-1]∑(1/j))/n + 1
= [k=1, n-1]∑( [j=1, k]∑(1/j))/n + 1
= [k=1, n-1]∑( [j=1, k]∑(1/j))/n + 1
We note that the term 1/1 appears n-1 times in this series, 1/2 appears n-2 times, and so forth, thus:
= [k=1, n-1]∑( (n-k)(1/k))/n + 1
= [k=1, n-1]∑(n/k)/n - [k=1, n-1]∑(k/k)/n + 1
= [k=1, n-1]∑(1/k) - (n-1)/n + 1
= [k=1, n-1]∑(1/k) + 1/n
= [k=1, n]∑(1/k)

Thus d(n+1) = [k=1, n]∑(1/k), and by induction, d(n) = [k=1, n-1]∑(1/k) for all n≥2. Thus, the maximum distance which can be spanned by n dominoes is the sum of the first n-1 terms of the harmonic series. However, since the sum of the harmonic series diverges to infinity, so too does the distance spanned by infinitely many dominoes. Q.E.D.

(to see that the harmonic series diverges, consider that 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8... > 1 + 1/2 + (1/4 + 1/4) + (1/8 + 1/8 + 1/8 + 1/8)... = 1 + 1/2 + 1/2 + 1/2 + 1/2..., which clearly diverges to infinity.

Answer 2

Let me try.

Start with 1 domino. Its center of gravity is in the middle, and it weighs 1.

Now put it on a second one so it just balances. The center of gravity
of those 2 dominoes is 3/4 of the way down the second domino, and it
weighs 2.

Put the 2-domino stack on top of a 3rd domino with the center of
gravity right over the end. This will balance. Where is the center of
gravity of this new stack? Imagine a line coming up from the end of
the bottom domino. Clearly it is to the right of the COG, since the
COG of dominos 1 and 2 is right above it, and all of domino 3 is to
the left, so there is an extra domino's weight on the left.

Now imagine a vertical line that will represent the new center of
gravity. Start this line at the end of domino 3. Clearly the line will
need to be moved to the left, since it passes through the COG of the
stack of dominos 1 and 2, and all of domino 3, a mass of 1 domino, is
to its left. To get the COG we need to move the line past the mass of
half a domino. As lhe line moves parts of all 3 dominos move from one
side to the other, so when it moves 1/6 of the way the two sides are
balanced.

Continue doing the same thing. For the next domino put the center of
mass over the end and compute the new center of mass. Once again, it
starts with the excess of 1 domino on the left, and to move the line
to the COG you need to move left past the mass of 1/2 domino. Moving a unit this time is multiplied by 4 (4 dominos in the stack), so to move
half a domino you need to move 1/4*1/2.

For n, you have a stack of n dominos you put on the n+1st. Start on
the end, you need to move the COG past the mass of 1/2 domino, and
every unit you move it is multiplied by the (n + 1) dominos, so the
new COG moves 1/(2(n + 1))

So the overhang is measured by the series 1/2 + 1/4 + 1/6 + 1/8 +
... = 1/2(1 + 1/2 + 1/4 +...

The series 1/1 + 1/2 + 1/3 + ... is called the harmonic series. It diverges, but is in some ways the slowest series to diverge.

there are a few ways of estimating it. Since ln(x) is the integral of 1/x, you can get a general estimate from looking at ln(x). So if you want

1/2H(x) = 500 (10 meters is 500 domino-lengths), or when the harmonic series exceeds 1000. So e^1000 would be an order-of-magnitude estimate.

Another way of estimating is by comparing to a power series. Split terms up into sets of increasing powers of 2:

H(2^n) = 1 + 1/2 + 1/3 + 1/4 + 1/5 + ... + 1/2^n
H(2^n) = 1 + (1/2 + 1/3) + (1/4 + 1/5 + 1/6 + 1/7) + ... + (1/2^(n - 1) + ... + 1/2^n)
H(2^n) < 1 + (1/2 + 1/2) + (1/4 + 1/4 + 1/4 + 1/4) + ... + (1/2^(n - 1) + ...)
H(2^n) < 1 + 1 + 1 + ... + 1 = n


H(2^n) = 1 + 1/2 + 1/3 + 1/4 + 1/5 + ... + 1/2^n
H(2^n) = 1 + 1/2 + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + ... + (1/2^(n - 1) + ... + 1/2^n)
H(2^n) > 1 + 1/2 + 1/2 + 1/2 + ... + (1/2^(n - 1) + ...)
H(2^n) < 1/2 + 1/2 + 1/2 + ... + 1/2 = n/2

So it is on the order of 2^1000

Answer 3

I solved this problem about a week ago involving book stacking instead of dominoes, but the concept is the same. Just start with the domino farthest out and find it's center of mass. Put that center of mass right above the edge of the domino underneath it and then calculate the center of mass of the two domino system. If I remember correctly, the length it can extend out is equal to 2 times the sum of the harmonic series from 1 to n, and we all know that the sum of the harmonic series diverges as n approaches infinity.

Answer 4

i admire those style of issues. So the backyard is a rectangle. the portion of a rectangle is L*W. the section is to be ninety 8. So L*W=ninety 8. the quantity of fencing is yet in any different case of exclaiming the fringe of the rectangle. through fact one facet is already secure by potential of a barn the fringe (P) or fencing is P=2L+W. to locate the minimum volume of fencing we would desire to cut back the equation P=2L+W. yet first we would desire to re-show that equation with in easy terms one variable, we could randomly decide for L. it rather is the place the 1st equation is supplied in: L*W=ninety 8 W=ninety 8/L Now we could plug that in the time of to the 2d equation: P=2L+(ninety 8/L) to cut back it we would desire to take the spinoff and set it equivalent to 0: dP/dL=2-(ninety 8/L^2)=0 Now we resolve for L: 2-(ninety 8/L^2)=0 2=ninety 8/L^2 a million/2=L^2/ninety 8 ninety 8/2=L^2 40 9=L^2 L=7 Now that we've L, we use the 1st equation to locate W: L*W=ninety 8 7W=ninety 8 W=14 ****** So our answer is L=7 ft and W=14 ft ******* (the quantity of fencing we choose is 2L+W=2*7+14=28 ft while you're curious)