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Mathematical Physics Major at the University of Waterloo.


2d
comment a proof question regarding to eigenvalues and diagonalization
You've shown no attempts to solve the question, so I have no idea what you've tried. In any case, if you've multiplied out the characteristic polynomial you should be able to see why $\lambda_1 \cdots \lambda_n = c_0$ and $\lambda_1 + \cdots + \lambda_n = (-1)^{n-1}c_{n-1}$ at least.
2d
comment a proof question regarding to eigenvalues and diagonalization
Have you tried actually multiplying out the factors of the characteristic polynomial and seeing what you get?
2d
revised a proof question regarding to eigenvalues and diagonalization
added 181 characters in body
Jul
27
comment The sum (or difference) of two irrational numbers
@just1question This is a pretty well known fact. For example, this paper has several proofs. The paper proves the general result, but it wouldn't be too difficult to prove this in the case of a small number of radicals (since it seems you are primarily interested in the case of $a+\sqrt{b}$.
Jul
27
comment The sum (or difference) of two irrational numbers
With your new restriction, your intuition is more or less correct. A more concrete way to say this is the observation that the set of square-free radicals $\{1,\ \sqrt{2},\ \sqrt{3},\ \cdots \}$ is linearly independent over $\mathbb{Q}$.
Jul
25
comment Project sin(x) onto orthonormal basis that span ${(1, x, x^2, x^3, x^4, x^5)}$ on domain $[-\pi, \pi]$
Right now you are calculating the projection onto the subspace spanned by $\{x, x^2\}$. You will continue to contribute to the $x$ term as you project onto larger and larger subspaces. Simply put, you're not done calculating the coefficient of the $x$ term yet. It's the equivalent of going halfway through the Gram-Schmidt procedure in $\mathbb{R}^n$ and wondering why the first entry of your vector doesn't match the answer's. Finish the entire calculation and it should be more clear.
Jul
24
comment Linear Transformation from V to W (bijective) Show that T(v) is a basis of W if B is a basis of V.
I've cleaned up your notation a bit. One point I wish to confirm: In the question, you ask things like "but I don't get why the dimension of w goes up to s as opposed to t." You used the lowercase "w" which I assume meant the vector $\mathbf{w}$ instead of the vector space $W$. Of course, this makes no sense as individual vectors have no dimension. I just want to ask whether this is a typo or a misunderstanding of concepts which an answer might (and should) address?
Jul
24
revised Linear Transformation from V to W (bijective) Show that T(v) is a basis of W if B is a basis of V.
added 533 characters in body
Jul
24
comment Terminology: Alternatives for zero crossing
Zero crossing seems like a perfectly adequate term (even preferred I would say). Is there any reason you don't want to use it?
Jul
23
awarded  Enlightened
Jul
23
awarded  Nice Answer
Jul
23
answered Show that the image of a linear transformation is equal to the kernel
Jul
23
reviewed Approve suggested edit on Show that the image of a linear transformation is equal to the kernel
Jul
14
answered Algebraic Multiplicity of Eigenvalues for a Linear Mapping
Jul
9
awarded  Good Answer
Jul
4
revised Problem on bipartite graphs.
elaboration of notation
Jul
4
comment A better approximation of $({n \over e})^n$ than by $ \Gamma(1+n-1/2)$ ? (Focus is “reverse” to the Stirling approximation)
@GottfriedHelms I've posted my previous comment as an answer. There was actually an error in the comment, which luckily didn't affect the final result. Don't type complicated expressions in comments!
Jul
4
answered A better approximation of $({n \over e})^n$ than by $ \Gamma(1+n-1/2)$ ? (Focus is “reverse” to the Stirling approximation)
Jul
4
comment A better approximation of $({n \over e})^n$ than by $ \Gamma(1+n-1/2)$ ? (Focus is “reverse” to the Stirling approximation)
Well, $(4)$ can be improved in that it's a simplification of Stirling's. Consider the following: $$\Gamma(n+\frac{1}{2}) \sim \sqrt{2\pi \left(n+\frac{1}{2}\right)}\left(\frac{n+\frac{1}{2}}{e}\right)^{n+\frac{1}{2}} = \sqrt{2\pi}\left(\frac{n}{e}\right)^n\frac{\left(1+\frac{1}{2n}\right)^n}{\sqrt{‌​e}}.$$ Now, we recover your $(4)$ by taking $$\left(1+\frac{1}{2n}\right)^n \sim \sqrt{e},$$ but we have a better (albeit more complicated) approximation without making the approximation. In fact, you can improve your asymptotics arbitrarily by taking more and more terms in the asymptotic series.
Jul
4
comment A better approximation of $({n \over e})^n$ than by $ \Gamma(1+n-1/2)$ ? (Focus is “reverse” to the Stirling approximation)
@GottfriedHelms I'm not sure if this helps, but Stirling's approximation doesn't just hold for integer $n$. It's in fact the first term of the asymptotic approximation for $\Gamma(n)$. If you substitute $n+\frac{1}{2}$ into Stirling's formula, you should be able to reduce it (asymptotically) to $(4)$. The fact that the $n$ has moved out of the square root is effectively because $\Gamma(n+\frac{1}{2})$ has introduced another $\sqrt{n}$ to the approximation.