why isn't it as simple as saying expanding (n+1)^2 to get n^2+2n+1 so that the sum of (n+1) consecutive odd terms is whatever the sum n^2 is plus the term 2n+1 added ? My brains turning to mush. Bed time.

Yes, exactly. That's the proof. You showed that if the sum of the first n odds is n^2 for some n, then for that n the sum of the first n+1 must be n^2 + 2n + 1 = (n+1)^2. So you proved by induction that the sum of the first n odds is n^2 for all n.

Next show inductively that the sum of the first n integers is n(n+1)/2.

And if you're confident, show that the sum of the first n cubes is a square

The numbers that get squared form the 3rd diagonal of Pascal's triangle. 1,3,6,10,15,21,....

The formula for the sum of the first n cubes is 1/4 * (n^2+n)^2. You can prove that inductively.

The 2nd diagonal is 1,2,3,4,... who's squares gave the sum of the first n odds as we saw..

The 4th diagonal is 1,4,10,20,35,... who's squares are 1,16,100,400,1225,.... Those are the sums of the first n values of 1,15,84,300,825,..... Any significance to that?

Also, the nth cube is the sum of n odd integers:

1^3 = 1

2^3 = 3+5

3^3 = 7+9+11

4^3 = 13+15+17+19

etc.

notsureifserious.

I don't understand how that involved induction. Hopefully the following demonstrates where I'm confused:

(1) n^2=1+3+5+...+(2n-1)

Then I think that if that is true the following must be true:
(2) (n+1)^2=1+3+5+...+(2(n+1)-1)

(3) Then, let's say, we "hypothesize" that the sum of the first n+1 odd numbers should be n^2+(2n+1)

In my mind then, that means we need to show (3) is true by showing that (2) can be simplified to look exactly like (1) with an extra (2n+1) added on the end.

But (2) simplifies to 1+3+5...+(2n+1) which is the same as (1) with an extra 2 added on at the end. Not what we were hoping to show.

Induction involves 2 steps. First, show it's true for some n. We did that. Then, show that IF it's true for some n, then it's true for n+1. We did that by showing that if the first n odds sum to n^2, then the first n+1 odds sum to (n+1)^2. That completes the proof.


You don't have to hypothesize (3) or prove (3), you know it's true. We said n^2 is the sum of the first n odds for some n, and the next odd is 2n+1, so obviously the sum of the first n+1 odds is n^2 + (2n+1). But that's just (n+1)^2, so we proved that being true for n makes it true for n+1.


No, (2) is the same as (1) with an extra 2n+1 added to the end. The second to the last term of (2) is 2n-1. The last term if (2) is 2(n+1)-1 = 2n+1.

You don't need (2) though. We showed (1) for some n, and we know (3). That's all you need.

Johann Faulhaber and sums of powers

http://arxiv.org/pdf/math/9207222v1.pdf

was wondering if anyone had a chance to look at this one, it's a page back. Is there going to be some sort of weird piecewise CDF? it looks like when r is between 12 and 32 there is like a semi-oval inside of the rectangle of possible locations... could we really have a problem that difficult to model? it seems pretty tough to me

This is not my area of expertise, but I'll give it a shot.

Introduce a coordinate system which has the lower left corner of the area as its origin, its y-axis going through A, and the x-axis parallel to AB (so it goes through the southern border). Units are kilometers. Our plant is at (50, -12).

Each possible quake is a line segment parallel to the x-axis of length 20. We identify each such line segment with its eastern end; this is a point where 0<=x<=80, and 0<=y<=20. Altogether the possible eastern ends cover an area of 1600. We need to figure out which part of that area is bad for our plant.

Let's consider which point of the quake line is closest to us.

• The eastern end of the quake is closest to us. This happens if x>=50, and (x-50)^2 + (y+12)^2 <= 900.
• The western end is closest to us. This happens if x<=30, and (x-30)^2+(y+12)^2 <= 900
• The closest point lies in the interior of the quake. This happens if 30<x<50, and y <= 18.

Solving for y should give us three integrals. The middle one is simply a rectangle; the other ones are parts of circles which can be solved by trigonometric substitution.

Thanks for the help... but I'm not sure how this gives the cumulative density function for the area source as a function of distance "r" from the plant

What do you need that for? The problem only asks for the probability.

the probability of an EQ being less than 30 meters will be

CDF of Fault 1 evaluated at r = 30 * (0.2/0.6) +
CDF of Fault 2 evaluated at r = 30 * (0.3/0.6)
CDF of Area Fault evaluated at r = 30 * (0.1/0.6)

for example, the CDF of Fault 2 is

(sqrt(r^2-8^2) - 16)/(106-20-16), for : sqrt(8^2+16^2) < r < sqrt(8^2+(16+90-20)^2)

Why not compute the probability directly? The fixed value of 30 is given, so you don't need the CDF. I showed how to compute the probability for the area; for Fault 1 it is clearly
(sqrt(30^2-10^2)-5)/(80-5),
and for Fault 2 we can do it similarly. It is just elementary geometry, using only the fact that for a uniform distribution, the probability of hitting a certain subarea (or interval) equals the ratio of subarea/(total area).

As I said, this is not my strongest area and I'm not entirely sure how you define CDFs, but I'm pretty sure that this is not needed here.

Your way probably works but I most likely need to find the CDFs for the class =\

Although for now I will work it out geometrically, as that will still give me the correct answer

edit: for fault 1, wouldnt it be

15/75 + sqrt(30^2-10^2)/75

How does this seem for the area fault? Rectangle plus two quarter ellipses:

(20*18+2*pi*18*sqrt(30^2-12^2))/4/(20*80)

Yes, my mistake. I was inconsistent in using the left and right endpoint.

They're not exactly ellipses, but parts of circles. I think for the area I get something like

could you maybe explain the terms in the integral, i am having a hard time picturing it

The 20*18 is the rectangle in the middle. We have two parts of circles on the left and on the right; I moved them together by removing the rectangle, so we have only one integral. Moreover, I shifted the circle in such a way that the center is at x=0, y=-12. The integral boundaries are where the circle meets the x-axis; the integrand is the equation x^2 + (y+12)^2 = 30^2, solved for y.

If Σ a_n is a convergent series with positive terms, is it true that Σsin(a_n) is also convergent?

My intuition is that it is true. If Σ a_n converges then the limit as n-> infinity of a_n must be zero. This would indicate that as n gets very large the value of sin(a_n) would start to change by a negligible amount so that eventually Σsin(a_n) converges.

The solution manual shows this using the limit comparison test but I think I must be applying the test wrong.
If we let b_n=sin(a_n) then we have

lim_n->infinity sin(a_n)/a_n,

but if the bottom goes to zero as it must, how can we say that this limit goes to a finite value? Couldn't it go to infinity?

You ninja deleted?

For whatever reason I didn't think I could differentiate a_n. I see why l'hospital works.

First note that by symmetry we can assume the earthquake occurred (had midpoint in) the west most half of the area WLOG. Define an earthquakes position (x,y) as its eastmost point and let the point of the area closest to the plant (the south midpoint) be (0,0). Then x in (-10,30) and y in (0,20).

For x < 0, the earthquake is 'within r' if y <= (r-12). These two conditions form a rectangle with area 10(r-12) for 12 <= r <= 32.

For x > 0, we want to find the intersection between a 20x30 rectangle with bottom left corner at (0,0) and the circle of radius r centered at (0,-12). For 12 <= r <= 32, this is the top corner or a quarter circle (it helps to draw it).

This next part is probably possible without a diagram but I'm on phone so here it goes...

Consider the top right quarter of a circle ('quadrant one' in high school trig). Consider a rectangle of height 12 inscribed into this quarter. The area of the quarter circle above this rectangle is the area we wish to find. Now, the diagonal of this rectangle is a radius of our circle, so the rectangle has length sqrt(r^2-144). The diagonal also makes an angle of arccos (12/r) from the vertical axis. So the area above the diagonal is r^2 arccos (12/r) /2. But the area above the diagonal is the area we are after PLUS half the rectangle. So the area we are after is

(1/2) r^2 arccos (12/r) - 6 sqrt (r^2 - 144)

=> the probability that an earthquake is within r of the plant is

(1/800)*(10(r-12) + (1/2) r^2 arccos (12/r) - 6 sqrt (r^2 - 144)

For r = 30:

(1/800) * (450 arccos (2/5) + 36(5-sqrt(21)))

The limit of sin(a_n)/a_n as a_n goes to 0 is 1 by L'Hospital's rule. You can also use the direct comparison test. sin(a_n) < a_n for 0 < a_n < pi. That's because the derivative of sin(x) is cos(x) <= 1, so y = sin(x) never gets higher than y = x for x > 0. We know that the sum of a_n converges, so the sum of sin(a_n) must converge.

It's back now. I had added some stuff about absolute convergence to handle the fact that sin(a_n) can be negative, but you don't need it if you just consider 0 < a_n < pi.


You can't since a_n is discrete, but you can still say that sin(a_n)/a_n goes to 1 as a_n goes to 0 for the same reason that sin(x)/x goes to 1 as x goes to 0.


Did you figure out induction?