Recent content by nuuskur

Why is it a problem that eigenvalues are not real? A fundamental matrix of the system is a matrix-valued function ##\Phi## such that the columns of ##\Phi(t)## are linearly independent solutions of the given system for all ##t##. Simply write how ##\Phi(t)## is parametrised with respect to ##t##...

Typos in your exponents. Right now you have zero at the very end. Explain the 5th equality from the bottom.

Shooting from hip, I'd say choice is not necessary for this to occur. To prove choice, we would have an arbitrary family of nonempty sets and we must find a choice function. The condition (**) is formulated in only finite terms. I don't see how it provides an angle to tackle with infinite...

Recall that ## 2\sin^2x = 1-\cos 2x ## and consider the behaviour of ##\frac{\sin t}{t}## as ##t\to 0##.

Second derivative positive implies ##f## is convex for ##x<a##. Take anything that is convex and decreasing for (some) ##x<a## as counterexample to your claim of ##f'(x)>0##. As for the initial claim, think again of convexity. The slope at ##x=a## must be as large as it gets as ##x\to a-##...

Existence of Schauder basis implies separable (e.g here)

As you say, an ##m\times n## matrix over ##\mathbb R##, say, represents a linear map ##\mathbb R^n \to\mathbb R^m##. Hence, a row would correspond to a linear map ##\mathbb R^n\to\mathbb R##. A projection, for instance.

Summation is a binary operation, hence you can extend it to only finite sums. This ## \sum _{k=0}^\infty c_k x^k ## is not a polynomial. The sum of two polynomials is a polynomial. That is indeed how vector spaces are defined. There is a generating set (of arbitrary cardinality) and the space...

Fair play on your last point. As for the first one, optimisation can be done later. First and foremost, it is important the argument is correct.

It is not wrong to check the claim up to ##m## manually. Sometimes that's even helpful. The induction hypothesis then is that the claim is true for some ##n\geqslant m## (weak induction); for all ##k\leqslant n##, where ##n\geqslant m## (strong induction). In both cases the task is to prove...

You can formulate more precisely. Instead of saying "consider ##n=m## case", we can say "assume that ## B^nx = \alpha ^nx ## for some ##n\geqslant 2##". Then for the case ##n+1## we have the equalities B^{n+1}x = BB^nx = B\alpha ^nx = \alpha ^nBx = \alpha ^n\alpha x = \alpha ^{n+1}x. Right now...

The determinant of an upper or lower triangular matrix is equal to the product of the elements on the leading diagonal. An upper triangular matrix is a square matrix whose entries below the leading diagonal are zero. The claim follows quickly provided you are familiar with the Laplace...

It's a power series at the point ##a=9## whose radius of convergence ##R## is given by \frac{1}{R} = \limsup _n \frac{1}{\sqrt[n]{n9^n}} = \frac{1}{9}. Hence, interval of convergence contains ##(0,18)##. For ##x=18## we get ## \sum \frac{(-1)^n}{n} ##, which converges. For ##x=0## we get...

Your reasoning is strange. On top of what's already mentioned, you write something like "convergent, therefore absolutely convergent". This is not true. For sufficiently large indices it holds that \left\lvert \frac{\cos f(n)}{n!}\right\rvert \leqslant \frac{1}{n^2}. What do we conclude?

Please avoid using some ambiguous word soup as a title. None of us know what Hard or Medium hard means and which Q7 of which tutorial it is. The only helpful part of the title currently is "MVT".