Recent content by karseme

I'm a little stuck here. I should determine the following determinant. I first tried to simplify it a little by using elemntary transformations. And then I did Laplace expansion on the last row. $\begin{vmatrix}2 & 2 & \cdots & 2 & 2 & 1 \\ 2 & 2 & \cdots & 2 & 2 & 2 \\ 2 & 2 & \cdots & 3 & 2 &...

I would say that as x gets large the value of polynomial gets large and as it converges to the infinity the value converges to the infinity...

It is required to determine if there is a non-constant polynomial p with positive coefficients such that function $x \mapsto p(x^2)-p(x)$ is decreasing on $[1,+\infty \rangle$. What should I do here? How should I exactly determine that? What is the right method? My idea was to use somehow the...

Thank you on your effort, but I don't know what L'Hôpital's Rule is. I heard about it, but never used it. And we still haven't done it on my lectures. I will look it up.

How could I calculate: $\displaystyle \lim_{n \rightarrow + \infty}{(\sqrt(n^2+n) - \sqrt[3](n^3+n^2))}$ Everything that I tried I always got infinite forms or 0 in denominator. I don't have any idea what else should I try here.

I need to prove that: $ \arctan{\dfrac{1}{x}}=\dfrac{\pi}{2}- \arctan{x}, \forall x>0$. Now, I assumed $\arctan{\dfrac{1}{x}}=\arccot{x}$. So, I've tried to do this: $\cot{y}=x \implies y=arccot{x} \\ \tan{y}=\dfrac{1}{\cot{y}}=\dfrac{1}{x} \implies y=\arctan{\dfrac{1}{x}} \\ \implies...

But, how can we assume that? It's like let's assume that every number is divisible by 3 for the sake of the convenicence. I don't see how can we assume that here. Maybe it is polynomial, I don't know. Anyway what can we achieve by assuming that, you're still left with $ x \sin{(n...

So, I've got an assignment to prove that f(x)=\cos{(n \cdot \arccos{x})} is a polynomial for \forall n \in \mathbb{N} . Also, we were suggested to use mathematical induction. So, I've tried: Base step: n=1 \implies f(x)=\cos{(\arccos{x})}=x Assumption step: f(x)=\cos{(n \cdot \arccos{x})}...

So, if we assume that both angles are in the second quadrant, then the following must be true: $$ \dfrac{\pi}{2} \leq \pi x \leq \pi \qquad \land \qquad \dfrac{\pi}{2} \leq \pi \sqrt{x} \leq \pi \, \implies \dfrac{1}{2} \leq x \leq 1 $$ Since $$ \pi \sqrt{x} $$ is in the second quadrant we...

Thank you very much. :) It was very helpful.

$$ \sin{(\pi x)}>\cos{(\pi \sqrt{x})} $$ I don't know how to solve this. I would really appreciate some help. I tried to do something, but didn't get anything. If I move cos to the left side, I can't apply formulas for sum. Since arguments of sin and cos have $$ \pi $$, I think there is no way...

Let's say I kind of understand what you have explained. The only definition that I knew for vectors until now was actually that they are line segments that have starting point and ending point, and the properties which characterise them are their modulus(length), direction and orientation. So...

I need some help understanding one task. I know that for some structure to be a vector space all axioms should apply. So if any of those axioms fails then the given structure is not a vector space. Anyway, I have a task where I need to check if $$ \mathbb{C}^n_\mathbb{R} $$ is a vector space...