Recent content by Euge

It was assumed in the proof that ##Df(a)## is the identity, from which the inequality ##2|f(x_1) - f(x_2)| \ge |x_1 - x_2|## was obtained. Your function has derivative ##1/3##, not ##1##.

Observe $$\lim_{x\to 1} \frac{\sqrt[n]{x}-1}{\sqrt[m]{x}-1} = \lim_{x\to 1} \frac{\sqrt[n]{x} -1}{x-1} \cdot \lim_{x\to 1} \frac{x-1}{\sqrt[m]{x}-1}$$ Use the identity $$t^n - 1 = (t-1)(t^{n-1} + t^{n-2} + \cdots + t + 1)$$ with ##t = \sqrt[n]{x}## to simplify ##\lim_{x\to 1}...

Hi @issacnewton, in your inductive step you assumed ##k \ge 3## and the claim holds whenever ##2 \le m + n \le k##. So it is unnecessary to prove ##P(m,n)## is true whenever ##m + n = 3##. Also, in your inductive step, you assume ##k = m + n##, but it should be ##k + 1 = m + n##, since you are...

This was buried in my answer, but for a short, non-inductive proof of the result, observe that ##(p+1)\binom{p}{q} = (q+1)\binom{p+1}{q+1}## is even because ##q + 1## is; since ##p+1## is odd, we deduce ##\binom{p}{q}## is even.

The answer depends on what properties of the 3-dimensional cross product you want preserved in higher dimensions. For definiteness, let's suppose ##n \ge 3## and there is a "cross product" on ##\mathbb{R}^n##, which we assume to be a continuous map ##\times : \mathbb{R}^n \times \mathbb{R}^n \to...

The dual matrix in the first sense is ill-defined. For example, if ##A## is a ##3\times 2## matrix, there does not exist an ##m\times n## matrix ##B## such that ##AB## is a number. In the second sense you mentioned, I suppose the dual of a complex square matrix ##A## is the adjugate of ##A##...

When ##m + n = 2##, ##m =1##, ##n = 1## and ##\binom{2\cdot 1}{2\cdot 1-1} = \binom{2}{1} = 2##, so ##P(m,n)## is true in this case. Now assume ##k > 2## and ##P(m,n)## is true whenever ##m + n < k##. If ##m + n = k##, then by the induction hypothesis ##\binom{2(m-1)}{2(n-1)-1} =...

Let ##(X,d_\infty)## be the space of all bounded functions ##f : E \to \mathbb{C}##, equipped with the sup metric ##d_\infty##. Convergence of a sequence of functions with respect to ##d_\infty## is the same as uniform convergence of the sequence. So the condition that ##\mathcal{B}## is the...

Yes, this is a correct approach. I thought you wanted to prove the original statement with standard two-variable induction, in which you induct along the horizontal and vertical directions. Sorry about that. :smile: You can also induct as @Orodruin suggested or induct on along the lines ##m + n...

You need to show separately that if ##P(m-1, n)## is true, then ##P(m,n)## is true, and if ##P(m,n-1)## is true, then ##P(m,n)## is true. First, suppose ##P(m-1,n)## is true. Then, as you claimed, ##\binom{2m-2}{2n-1}## is even. Continue the recurrence: $$\binom{2m-1}{2n-1} = \binom{2m-2}{2n-1}...

If ##H## is a separable Hilbert space and ##T## is a trace-class operator on ##H## (i.e., a bounded linear operator on ##H## whose sequence of singular values is summable), then one may define the determinant of ##I + T## as the product ##\prod_{j = 1}^\infty (1 + \lambda_j(T))## where the...

Yes, I meant a ring. Fixed.

This does not define a representation. A representation of ##G## (by complex matrices) is a group homomorphism ##\pi : G \to GL(n,\mathbb{C})## for some ##n \ge 1##. So, not only is ##\pi(g)## an invertible ##n\times n## complex matrix for every ##g\in G##, but also ##\pi(g)\pi(h) = \pi(gh)##...

Let ##n \ge 1## and ##1 \le m \le n##. Suppose ##\lambda_1,\ldots, \lambda_m : \mathbb{C}^n \to \mathbb{C}## are ##\mathbb{C}##-linearly independent linear functionals. For each abelian group ##G##, determine the cohomology ring $$H^*(\mathbb{C}^n \setminus \bigcup_{k = 1}^m \ker \lambda_k ; G)$$