284 reputation
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location University of Alabama, Huntsville
age 26
visits member for 2 years, 4 months
seen Dec 11 '12 at 2:08

Sep
9
awarded  Yearling
Dec
11
comment What is the necessary condition for a matrix to have eigenvalue 1?
@berci, that's not technically true, consider I, it has eigenvalue 1, and is most definitely non-singular. Ah missed what you were saying, never mind.
Dec
11
comment What is the necessary condition for a matrix to have eigenvalue 1?
In relation to what Jaime said, think about how matrix multiplication works with a column vector X, that is, what has to be true about the system of equations A represents? This only gets you so far though, really there's only one thing this gives you from the linear algebra properties (the things you already mentioned).
Dec
11
comment A limit to find without using l'Hôpital's rule.
Additionally do you have any information about the growth of functions, and can you find bounds (upper and lower) on sin(7x) that might be helpful with squeeze theorem or a substitution (think about the idea of $a^{\sin(x)}$ when x = $\frac{\pi}{2}$, $0$, and $\frac{3\pi}{2}$)?
Dec
10
comment Differentiability of a Special Function
Amending the above after further consideration: upon looking more closely there is an issue with the derivative, but the sequence of partial sums could still converge even if some of the tail-end points in the sequence don't alone.
Dec
10
comment Differentiability of a Special Function
The issue is that you can pull the differentiation into the series and it becomes a special case of $\frac{d}{dx}a\cos(bx) = -ba\sin(bx)$ for every term in the sequence, and so if that series converges for any x the function is differentiated at some point, and that derivative exists, the function is differentiable. As a result you either have to prove that that series converges nowhere, or that there is some other property that creates an issue with linearity of derivatives in this case. If what you suggested in the edit worked, the initial series would be indeterminate as well.
Dec
10
comment Differentiability of a Special Function
No, as given any n in the natural numbers the derivative exists (provable easily by induction), if it were as set like $\mathbb{N}\cup\infty$ then there would be an issue.
Dec
10
comment Differentiability of a Special Function
His method won't work directly, but the concept behind it will still apply, mainly due to the self-similarity issue this curve will have concerning differentiation. More specifically the curve can at most be differentiated at a countable number of points (and in this case that will not happen due to the lack of smoothness). For an exploration into fractal differentiation see: arxiv.org/pdf/1010.0881.pdf
Dec
10
comment Differentiability of a Special Function
As you've pointed out this isn't mathematically rigorous, and in fact each term in the series does have a proper derivative with respect to x $\frac{-(3^n)\sin(3^nx)}{2^n}$ and the series will converge for some values of x.
Dec
10
awarded  Commentator
Dec
10
comment Need help with simple proof by mathematical induction
I would say that the generalization that "every induction proof is identical" is a very poor generalization, and still doesn't help much with learning how to think about these kinds of problems. I'm guessing that what you've done already and the answers provided you have a template for the proof, and now mainly aren't sure of the inner mechanics of this type of problem. In the real world these mechanics are the primary area variation of inductive proofs themselves. In your class (it would seem) practicing polynomial arithmetic is likely going to be a big push to understanding these mechanics.
Dec
10
answered Differentiability of a Special Function
Dec
10
suggested rejected edit on Differentiability of a Special Function
Nov
8
revised Optimizing response times of an ambulance corp: short-term versus average
deleted 108 characters in body
Nov
8
answered Optimizing response times of an ambulance corp: short-term versus average
Sep
17
comment Prove the Well Ordered Theorem with mathematical induction.
I may be incorrect but I thought you needed the existence well ordering as a property in your set to perform induction over that set first, since you need to define a "first, second third... nth" element before you could act. That said I think you can get the well ordering without induction in this case by simply defining a function from A to N.
Sep
13
comment Showing a sequence is convergent using the $\epsilon$-$N$ definition
It depends on the class and what's expected of you (even down to your professor and their style really). I've had one real analysis class where that wouldn't be a problem, and another with a professor that required us to prove "every detail" of what we were doing (so making a statement f(n) < g(n) required a side proof using the axioms from the start of the course). In most cases though I'd say it should be fine (with the symbolic adjustment as noted made).
Sep
11
comment Is there a “homomorphism/isomorphism” (conceptually) between $\mathbb{C - R}$ and $\mathbb{R - Q}$
@Billy your'e right about that in a lot of ways, but any representation of an irrational number as a split rational/irrational part will have the same algebraic properties, in the same way as you can represent a rational number as a ratio of a (somewhat) arbitrary integer with a natural number. I'm thinking more about the inherent properties that these numbers have regardless of which representation you pick in the context, and not the particulars (I could also represent a complex number as a matrix in a few different ways with similar issues, or with an (infinite) number of polar forms).
Sep
11
awarded  Scholar
Sep
11
accepted Is there a “homomorphism/isomorphism” (conceptually) between $\mathbb{C - R}$ and $\mathbb{R - Q}$