Thanks!

Is there any sense in which (x^0)/0 implies ln|x|?

Why can we integrate 1/x^n for any n except n=1 ? How is that consistent aside from the obvious fact that we would have to divide by zero in order to integrate 1/x the way we'd integrate 1/x^n?

No. That would be a very bad way of thinking about it.

If you work out the details of the power rule for differentiation, you get a certain formula. Start with f(x) = x^n and use f'(x) = lim_{h \to 0} [ f(x + h) - f(x) ] / h. You have to use some binomial theorem stuff to make it work for non-integer values of n, but it can be worked out.

Integration formulas come from looking at these derivative formulas and using the fundamental theorem of calculus to work backwards. Unfortunately, there's no real good way of starting from the definition of an integral and working forward to get a general formula for \int x^n dx for any value of n.

The reason for the special case comes from the fact that the behavior of x^0 is an outlier. It has... issues. One of the issues is that it's not defined at x = 0. Or at least, it's not as simple as most people want to think it is. When thinking about polynomial function of x, we take x^0 = 1 when x = 0 for the sake of continuity. But if we were to think about exponential functions, 0^x should be 0 at 0 for the sake of continuity.

So there's clearly something "different" that happens when n = 0 in that formula. And this quirk is what gives rise to the fact that integrating 1/x gives something that falls outside of the normal pattern. Looking at the derivative rules for power functions, we just don't see anything that we can use to work backwards to get 1/x.

Thanks for the reply, Aaron.

I'm trying to pick up a bit of intuition about induction in my spare time. Is the following inductively sound (if that's the appropriate nomenclature)?

Want to prove that n/(2n^3+1) < n/(2n^3) for all k >=1

Base case
1/(2(1)^3+1) < 1/(2(1)^3)
1/3 < 1/2

Inductive hypothesis:
k/(k^3+1) < k/(2k^3)

(and here is the first spot I wonder if I'm breaking a rule)
if k=1 then
1/3 < 1/2

Prove from inductive hypothesis that it is also true for k+1
To save time: I sub k+1 in for k and then again assume k=1 in this case it simplifies to 2/17 < 2/16, which is obviously true.

The part where I let k = 1 seems like I'm cheating though.

You already did that. You know it's true for some k, namely k=1, so you're done with that. Now you have to show that

k/(2k^3+1) < k/(2k^3)

leads to

(k+1)/[2(k+1)^3+1] < (k+1)/[2(k+1)^3]

For arbitrary k>1, not just k=1.


That one is obvious from the beginning. Try proving that the sum of the first n odd integers is n^2.

1 = 1^2
1 + 3 = 2^2
1+ 3 + 5 = 3^2
1 + 3 + 5 + 7 = 4^2
etc.

You already did that. You know it's true for some k, namely k=1, so you're done with k=1. Now you have to show that

k/(2k^3+1) < k/(2k^3)

leads to

(k+1)/[2(k+1)^3+1] < (k+1)/[2(k+1)^3]

For arbitrary k>1, not just k=1.


That one is obvious from the beginning. Try proving that the sum of the first n odd integers is n^2.

1 = 1^2
1 + 3 = 2^2
1+ 3 + 5 = 3^2
1 + 3 + 5 + 7 = 4^2
etc.

But I feel like its obvious in only a "hand wavy" kind of way. I see that it should be true, but I want to be able to prove its true inductively, and I don't know how to do so using what we've assumed for the inductive hypothesis. Aren't I supposed to be able to use the hypothesis to somehow move to the case where it's true for k+1? Is it simply a case where I just need to figure out the correct algebraic manipulations that lead to what I am trying to prove?

k/(2k^3 + 1) = k/2k^3 * 2k^3 / (2k^3 + 1) < k/2k^3 * 1 = k/2k^3 QED.

Yes, that's the case with any proof by induction.

You want a straightforward case to understand the method, not one with a lot of algebra. My example is very clean. Clearly it works for the first n=1,2,3,4 odd numbers as I already showed. Then you need to show that if the sum of the first n odd numbers is n^2, then the sum of the first n+1 odd numbers is (n+1)^2. That's easy. Hint: the n+1st odd number is 2n+1.

This problem is driving me crazy.



I have the CDF for both of the faults, but I'm stuck on the area one. I'm thinking it's probably going to be piecewise?

I know for example that if it was just another fault, the CDF would be

0, r < 12
(2*sqrt(r^2-12^2)+20)/80. , 12 < r < sqrt(12^2+(50-20)^2)
1, r > sqrt(12^2+(50-20)^2)

I'm going to be pedantic just because I'm trying to make sure I get what's going on:

You just proved it for the Kth case, but aren't you supposed to show it for the K+1 case? (Don't bother doing it because I can see that it's trivially easy to do the same thing for the K+1 case, I just want to make sure that it's the k+1 case we're actually supposed to prove).

Also, haven't you only showed that the above is true in the case that (2k^3/2k^3+1) is less than 1? Wouldn't it also be necessary to prove that that is also the case?

I proved it for all k. I didn't prove it by induction.


2k^3 / (2k^3 + 1) is always less than 1. That's because

2k^3 + 1 > 2k^3.

Now just reciprocate that and multiply by k to get

k/(2k^3 + 1) < k/2k^3.

If you really want to prove it by induction, rewrite the inductive hypothesis as

2k^3 + 1 > 2k^3.

Then show that implies

2(k+1)^3 + 1 > 2(k+1)^3.

That's obvious if you multiply the second one out. You'll get the first one with the same terms added to both sides.

I have a poor intuition for this.

I don't know how to find the formula for the nth partial sum in the n^2 series.

I see that for every n, the last number in the series is 2n-1.

n^2=1+3+5+7+...+(n-3)+(n-1)

The way i see it, the 1 from the front will cancel the one at the end, then the 3 etc etc until on the right we have n numbers of n terms or n^2=n^2.

That works out nicely. Clever too.

Yeah, but proving that 2k^3 + 1 > 2k^3 is idiotic. That's the same as proving 1 > 0. If you know 1 > 0, then add 2k^3 to both sides.

Well sure, in that case it's sort of trivial. It was more an exercise in understanding induction than it was proving anything about that specific series.

You have everything you need. You know the sum of the first n odds is n^2 for some n. You know that the last one added was 2n-1, so the next one that we haven't added yet is 2n+1. So what is the sum of the first n+1 odds?

If this is the right answer I made it way too difficult, but if it is the right answer I'm not sure what I've proved:

n^2+2

Not n^2 + 2. The sum of the first n odd numbers is n^2. The n+1st odd number is 2n+1. What is the sum of the first n+1 odd numbers?

Are yo saying that the first number in the n+1 series is 2n+1?

2n+1 is the n+1st odd number. The 1st odd number is 2*0+1 = 1. The 2nd odd number is 2*1+1 = 3. The 3rd odd number is 2*2+1 = 5. The 4th odd number is 2*3+1 = 7. Etc.

I've been using 1 as an index so that sort of clarifies a bit.

Problems.

Why doesn't this work?
(n+1)^2=1+3+...+(2n+1)
n^2=1+3+...+(2n-1)
(n+1)^2-n^2=2
(n+1)^2=2+n^2

Because writing "..." is dangerous...

You have n terms in row 2, but n+1 terms in row 1.