Here's an extension of a list posted earlier. If anybody can think of any additions to the list, please post :D!
Perspectives of the world:
-------------------------------
Optimist – The glass is half-full.
Pessimist – The glass is half-empty.
Existentialist – The glass is.
Fatalist – The...
Here's an extension of a list posted earlier. If anybody can think of any additions to the list, please post :D!
Perspectives of the world:
-------------------------------
Optimist – The glass is half-full.
Pessimist – The glass is half-empty.
Existentialist – The glass is.
Fatalist – The...
In the pyramid integration, x varies from 0 to 3. However, each cross section of a rectangular prism has the same base. So you would have instead:
\int_0^3 (3)^2 dx
First of all, there is no such thing as the "volume of a rectangle." If you mean a rectangular prism, then the volume will be different. A rectangular prism with the same square base and height would have a volume of 27 cubic units.
V_{rectangular \ prism} = bh.
V_{pyramid} = \frac{bh}{3}...
I believe we would want (e) and not (f). A thicker string would result in a greater affinity for the string to return back to its equilibrium position. Think of it as: It takes more effort to pull a thicker string, so it'll have a 'stronger desire' to return back to its original position.
Almost...your injection proof is fine. But, when proving a mapping is surjective, we need to show that the domain maps all of the range, i.e., show that for each y, there's an x such that f(x) = y.
You wrote "for y E N," but that's not true. y E B = set of all odd integers greater than 13, not...
phillyolly, you have (k^3 + 5k) + 3(k(k+1) + 2). We want to show this is divisible by 6, right? Well we already know (k^3 + 5k) is divisible by 6 because we assumed so in our induction proof. So now we just have to prove 3(k(k+1) + 2) is divisible by 6=3*2, i.e., prove it's divisible by 3 AND 2...
Stop at = \frac{4k^3 - k + 3(2k + 1)^2}{3} and expand the numerator completely. You know you want 4(k+1)^3. So do as hunt_mat suggested and expand 4(k+1)^3 as an aside (not in the proof) so you know what it is expanded. Subtract this expanded form from your expanded numerator. It should work.
k^3 + 3k^2 + 8k + 6 = (k^3 + 5k) + 3k^2 + 3k + 6.
= (k^3 + 5k) + 3(k^2 + k + 2).
From our induction hypothesis, we know the first term is divisible by 6. So it remains to show that k^2 + k + 2 is divisible by 2 for all k. It's a mini-induction proof within your main induction proof. Can...
You're finding the limit as x approaches a value. For your problem, x --> 3. Hence, you set c=3 in the generalized inequality 0<x-c< \delta. To put in more formal mathematical words, we want to prove that (as jgens said above):
" \forall \epsilon > 0, \ \exists \delta > 0 \ \ni (\forall x)(0 <...
Oh, you just need to know what you want to end up. Well you want to show in the end that y^n > x leads to contradiction. So we would want an h that will give us (y - h)^n > x, thus contradicting the fact that y is the least upper bound. We've already seen that the identity b^n - a^n < (b -...
nope, the unique x part makes it a bijection. It's subtle but it's there! A surjection only guarantees there exists an x in A. A bijection tells us it's a unique x.