I think this is arguable since you can reason inductively over partial as well as linear orders.

http://en.wikipedia.org/wiki/Structural_induction

Define the symm. derivative of f at x as lim h->0 [(f(x+h) + [f(x-h) - 2f(x)) / (h)], provided the lim exists.

Prove there is a point x in (0,1) where the ordinary derivative exist.

Note: It's unknown if f is differentiable but f is cont. on [0,1] and the symm. derivative exists everywhere on (0,1). Also, f(0)=f(1)=0.

This is a little beyond my skill level, but here is my attempt.

Take the lim h->0 [(f(x+h) + [f(x-h) - 2f(x)) /(h)].

Rearrange it so that,

lim h->0 [f(x+h) - f(x) / h] + [f(x-h) - f(x) / h]

My guess is that the limits on both side of the equation must exist. For example, if f(x)= 1- x, then f(x+h)= 1- x- h so (f(x+h)-f(x))/h= -h/h which has limit -1. So the left hand side exists, but I am not sure about the right hand side.

Also, I considered a contradiction, letting there not be a point in (0,1) where the ordinary derivative exists and using the Mean Value Theorem. But I am not sure if I can get away with that.

Thank you for any help.

I am doing astronomy homework and have no idea how to solve this problem.

Pretend that galxies are spaced evenly, 2 MPC apart and the average mass of a galaxy is 10 to the 11th solar masses. What is the average density of matter in the universe. (Hints: The volume of a sphere is 4/3 pie r cubed and the mass of the sun is 2x 10 to the 30 kg.)

I know it has something to do with d=m/v but thats where I become lost. I do not know how to find the radius unless its the 2mpc but I am not sure.

You have to consider the sphere packing problem in 3D.

http://en.wikipedia.org/wiki/Sphere_packing

http://en.wikipedia.org/wiki/Close-packing

I think that is going to far in depth this is an introduction to astronomy class and she had worked it out on the board fairly easy but I just do not remember how it was done nor can I find my notes.

You're confusing manifolds with vector spaces. A *vector subspace* certainly has to have this property. A topological subspace, however, doesn't have to be closed under addition or scalar multiplication since it cares only about the topological properties of ℝⁿ (if you look up the definition of manifold you won't see scalar multiplication anywhere, because it has no meaning in this context). Does this answer your question?

Yup.

slipstream -

Ty sir! Very helpful.

Let q = quantity

FC = fixed costs

VC = variable costs

TC = total costs

ATC = average total cost

we have that,

TC = FC + VC

ATC = TC/q

MC = marginal cost

A firm's average cost function is equal to a firm's average total cost function since it includes both variable costs and fixed costs.

If ATC = f(q) = 200 +48q^2

Then ATC = TC/q = 200 + 4q^2

Solving for TC, we get TC = 200q + 4q^3

The marginal cost function is just the derivative of the total cost function.

MC = TC' = 200 + 12*q^2

Originally Posted by PairTheBoard
"Induction is the defining property of the Natural Numbers."



I'm not an expert on foundations, but I think the reason you are allowed to "reason inductively" in these other scenarios is because you have adopted the defining property of the Natural Numbers, namely the legitimacy of inductive reasoning. You are working within an axiomatic system that includes the Natural Numbers and thus includes inductive reasoning which you can apply in various scenarios.

PairTheBoard

Okay, but saying that induction is the defining property of the naturals and saying that the defining property of the naturals (e.g. a certain type of linear order) is that which legitimizes induction are different imo.

I mean, induction is just a bunch of modus ponens in a row. It's the in-a-row-ness that defines the naturals imo.

Edit: To take the domino analogy: you can define a line of dominos without any reference to logic; all induction tells you is that if the first falls then they will all fall.

But if you are building your logical system from scratch you have to decide if you want an axiom that allows you to make that conclusion. It seems like something you should allow. So you adopt it as an axiom. When you adopt that axiom, you have adopted the natural numbers. And if you want to build a system of axioms that includes the natural numbers you adopt your "domino" axiom. That's how you do it.

Your "domino" axiom seems logical. But it doesn't follow from more primitive logical axioms. It is a logical axiom you assume and assuming it is equivalent to assuming the existence of and defining the natural numbers.

PairTheBoard

Domino Axiom: Any amount of dominos arranged into a line will all fall if the first falls.

I agree that this is an axiom in that every single case cannot be checked: it has to be assumed true. Also agree that it doesn't follow from more primitive axioms and also that it can be used to define the naturals. My point is that there are equivalent formulations of what is a "line" of dominos, which are geometrical rather than logical. Maybe what we're disagreeing on is which of these two should be emphasized.

I just got this question for a prop trading interview today:

There are two cities, located 1000 miles apart. There are 3000 apples in City A, and we want to send as many as we can to City B. The only method of transportation is a truck, which can carry a maximum of 1000 apples at a time. For each mile the truck travels, one apple falls out of the back. Assume that at any number of points in the truck's journey, you can stop the truck, unload the apples and leave them on the side of the road (nothing will happen to them). You can pick these apples up later, if desired. What's the maximum number of apples we can transport to city B?

I can do about 833. I have no idea what's best possible, and didn't try to prove it. Are you supposed to answer straight away?

I am pretty sure that the max is 800 this is the way I did it and my reasoning. If its too long sorry:

Basically we are going to be making 3 trips of a thousand apples and dump at the 100 mile mark. So the first trip would be dumping 900, the second 900, and the third 900. We then have 2700 left, we make three trips again 900, 900 and 600 are dumped which leaves 2400. I made this chart.

1. 900 900 900 2700
2. 900 900 600 2400
3. 900 900 300 2100
4. 900 900 x 1800
5. 900 700 1600
6. 900 500 1400
7. 900 300 1200
8. 900 100 1000
9. 900 900
10. 800 800

Now my reason for why this is the max may be flawed. I tried lowering the amount of miles we travel for each trip to ten. So the chart looks like this

990 990 990 2970
990 990 960 2940 etc.

The funny thing is that at the tenth trip doing it the second way, it leaves the exact same number as the first trip doing it the first way: 2700. Therefore I deduce that it will be the same for the rest, but I may be wrong. I hope this helps a little bit.

I don't suppose you could unload and reload every half mile thereby never losing any apples?


PairTheBoard

has anyone here used this book for quantum mechanics before? taking it next semester and wondering if it is any good, so far physics text books have been a bit hit and miss for me.

Quantum Physics of Atoms, Molecules, Solids, Nuclei, and Particles

~ Robert Eisberg (Author), Robert Resnick

I dont think you would get the job with that kind of an answer

My interview was over the phone, but I guess they give you about 2-3 minutes to work out an answer. They're more interested in your thought process than the actual answer I think (or at least, I hope).

I think 833 is optimal.

The key to the puzzle is the observation that it's always best to start a leg of the journey with a full truck. This is because you're paying one apple per mile regardless of how many apples you're transporting.

For the solution, first note that we can get 2000 apples to the 333.3-mile marker by driving 3 groups of 1000 apples to that point. A bit of thought shows we can never get more than 2000 apples past this point, since any such strategy necessitates driving the first 333.3 miles 3 times, which leaves you with at most 2000 apples.

So the first step is to transport the apples 333.3 miles from point A, when we're reduced to 2000 apples. We can then consolodate this into two separate groups of 1000 each. We drive each of those 500 further miles (a total of 333.3+500 = 833.3 miles from the start) unti each group is down to 500 apples. We consolidate those two to one package of 1000 apples, which we then drive the remaining 167.7 miles, leaving us with 833.3 apples. Then we can sell the apples and use the proceeds to fix the back of the truck to prevent produce from rolling off.

I could do 888.88 apples. Beat that.

You can transport 2000 apples at the one-third mark.
1333.33 apples at the two-third mark.
Finally, you can transport 888.88 apples at the final mark.

I'll leave it up to someone else to fill in the details.

i'm with you up until the last step, i don't see how it is possible to get 1333 from the 2/3 mark to the final mark without losing another 666 apples, i'm fairly sure it is impossible to do your last step.

I think you need to at least supply the detail of how you get around the argument that every mile traveled past the 1/3 mark requires double hauls until you're down to 1000 apples. That includes being at the 2/3 mark with 1333.33 apples. Each mile forward from there requires double hauls resulting in a 2 apples/mile cost until down to 1000. That would be an additional 333.33/2 miles which brings you to the same 333.33 + 333.33 + 333.33/2 = 333.33 + 500 = 833.33 mile mark with 1000 apples left as before. So you lose an additional 166.66 apples with a full haul the last 166.66 miles leaving you with 833.33 as before.

PairTheBoard