matrices - The relationship between matrix rank and its characteristic polynomial coefficients

Given the matrix characteristic polynomial coefficients. Is there a quick way to determine the rank of the matrix?

Answer

In general nothing can be said about rank of the matrix by merely looking at char polynomial. Take $$A=\left[\begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array}\right]$$ Rank$A=1$, though char polynomial is $x^2=0$. But if your matrix is diagonalizable, ''effective degree'' of the characteristic polynomial is equal to the matrix rank, since for a diagonalizable matrix, rank turns out to be number of non zero eigenvalues.

algebra precalculus - Which form is more simplified? $frac{x-2sqrt{xy}-y}{x-y}$ or $1-frac{2sqrt{xy}}{x-y}$

Which of the following is more simplified, for two rational numbers $x$ and $y$?

$$\frac{x-2\sqrt{xy}-y}{x-y} \qquad \text{or} \qquad 1-\frac{2\sqrt{xy}}{x-y}$$

Sorry for the lack of context; not really much context to give...

summation - Limit of triple sum

Suppose one has the following triple sum:

$$S_n=\sum_{s=0}^n\sum_{t=0}^s\sum_{u=0}^sf(n,t)g(n,u)$$

where for all $n$, $-\alpha< S_n <\alpha$ for some real constant $\alpha<\infty$. Since $S_n$ is bounded above and below by a constant may one interchange the limit with the first summand, obtaining
$$\lim_{n\to\infty}S_n=\sum_{s=0}^{\infty}\lim_{n\to\infty}\left(\sum_{t=0}^s\sum_{u=0}^sf(n,t)g(n,u)\right)?$$

Since the limit is now inside the first summand, may one now consider $s$ as a constant and thus bring the limit inside the two other summands to the right of it, yielding

$$\lim_{n\to\infty}S_n=\sum_{s=0}^{\infty}\sum_{t=0}^s\sum_{u=0}^s\left(\lim_{n\to\infty}f(n,t)g(n,u)\right)?$$

If not, why not?

calculus - Is this notation good for the chain rule derivative?

When we take this derivative, for example:

$$y = \log(\sin x)$$
We call $u = \sin x$, so we have:

$$\frac{dy}{dx} = \frac{d y}{du}\frac{du}{dx} = \frac{1}{u}\cos x = \frac{\cos x}{\sin x}$$

But for me, it's better to do:

$$\frac{d\log\color{Blue}{\sin x}}{d\color{Blue}{\sin x}}\frac{d\sin \color{Red}{x}}{d\color{Red}{x}} = \frac{1}{\color{Blue}{\sin x}}\cos \color{Red}{x}$$
It makes easy to do the 'pattern-matching' just by looking at the differentials. No substitution. I know that $\frac{d \log[\mbox{something}]}{d[\mbox{something}]} = \frac{1}{\mbox{something}}$ for example.

However, it looks 'hairy' when I try with larger derivatives, like, for the function:

$$(13x^2-5x+8)^{\frac{1}{2}}$$

we do:
$$\frac{d(13x^2-5x+8)^{\frac{1}{2}}}{dx} = \frac{d\color{Green}{(13x^2-5x+8)}^{\frac{1}{2}}}{d\color{Green}{(13x^2-5x+8)}}\frac{d(13x^2-5x+8)}{dx} = \frac{1}{2\sqrt{\color{Green}{13x^2-5x+8}}}(26x -5)$$
but it's really better for me to do like this, instead of doing the bla bla bla of changing variables and stuff. But I'm afraid my teacher does not accept this. Is this notation/way of doing good for you guys?

One more example: $$\frac{d}{dx}\sqrt{(\sin(7x+\ln(5x)))} = $$

$$\frac{d[\color{Blue}{\sin(7x+\ln(5x))}]^{1/2}}{d[\color{Blue}{\sin(7x+\ln(5x))}]}\frac{d[\sin\color{Red}{(7x+\ln(5x))}]}{d[\color{Red}{7x+\ln(5x)}]}\left[\frac{d[7\color{Purple}{x}]}{d[\color{Purple}{x}]} + \frac{d[\ln(\color{Purple}{5x})]}{d[\color{Purple}{5x}]}\frac{d[5x]}{d[x]}\right] = $$
$$\frac{1}{2}\left[\color{Blue}{\sin(7x+\ln(5x))}\right]^{-1/2}\cdot\cos(\color{Red}{7x+\ln(5x)})\left[7 + \frac{1}{\color{Purple}{5x}}\cdot 5\right]$$
So we get rid of the substitution!

(づ｡◕‿‿◕｡)づ $\ \ u, v, y$ go away!

Answer

This notation is absolutely acceptable, more than fine, and often used extensively in integral calculus, especially by professional mathematicians.

In this context, it is very convenient because the integration can be seen in all aspects. In other words, you are integrating with respect to a complicated expression and the form of the integrand makes sense with this expression and you want to show that clearly. In essence, it is sometimes used to be more explicit. It expresses steps more clearly at times and has the benefit of being able to visualize equalities more easily.

Be warned that this notation does get overcomplicated at times and you will want to add some substitutions for sanity, but for simpler calculations, it can be useful and looked upon favorably. You should look to study differential calculus as a topic in its own right, where this notation is used as well (in summary, the "denominator" of the derivative $\text{d}[\text{something such as} \,x]$ is eliminated and we just deal with what are called differentials). Your instructor is very intolerant, indeed, if he doesn't allow this.

As a matter of fact, there was a point towards the end of my high school career (at this point, I was studying beyond what I was enrolled in) I used this notation in a "Calculus II" type class (class focusing on expressing calculus with infinite series and basically a structured introduction to approximating evaluations of functions using different types of series), and my teacher (actually a PhD, but taught calculus at my high school in his spare time - shoutout to Dr. Brandell), and my mentor in my early mathematics self-study, commended me and demonstrated this point of view to the rest of my peers. This is very innovative actually for a young mathematician and shows promise. I'd even be willing to talk to to your instructor personally and make an argument in your defense, backed by some of my former and current professors.

Good luck in your pursuit of knowledge in mathematics and keep discovering ways to improve your notation (and, most importantly, keep asking questions)!

calculus - Notation of derivatives...

I asked my teacher the difference between this notations.

(1) $$\frac{dy}{dx}$$

(2) $$\frac{\delta y}{\delta x}$$
(3) $$\frac{\Delta y}{\Delta x}$$

He told me that there is no difference.

I really don't think he's right...

Question:

I think that (1) and (2) is more like the convention expressing the limit of a fraction. (3) instead really represent de ratio of the increments of y and x

Am I right?

Answer

Typically,

1. $\displaystyle\frac{dy}{dx}$ is the derivative (the slope of the tangent line);




2. $\displaystyle\frac{\delta y}{\delta x(t)}$ is a functional derivative where $y=y[x]$ is a functional of $x(t)$;




3. $\displaystyle\frac{\Delta y}{\Delta x}$ is the difference quotient (the slope of the secant line).

elementary set theory - Inverse of bijection proving it is surjective.

I understand that the inverse of a bijection is a bijection. To proof this you need to formally proof it is injective and surjective. I can prove it is injective and i understand what it is in the following way:

Let $f:A\to B$ be a bijection and let $f^{-1}: B\to A$ be its inverse. To show $f^{-1}$ is a bijection, you must show it is an injection: Let $x_1,x_2 \in B$ such that $f^{-1}(x_1)=f^{-1}(x_2)$. The the inverse we have $x_1= f(f^{-1}(x_2))=x_2$. This shows $f^{-1}$ is injective.

I am having a hard time proving this is surjective formally. I understand this is the definition of a surjection but I dont understand how this applies. Basically I dont understand it.

$$\forall b_i \in B~~ \exists a_j \in A \text{ such that } b_i = f(a_j)$$

Could someone explain the inverse of a bijection, to prove it is a surjection please?

Answer

Being a surjection just means you reach all the elements of your target set, here $A$.
It is quite easy to show it here: take $x\in A$, then $x=f^{-1}(f(x))$, so $x$ is reached by $f^{-1}$. Therefore $f^{-1}$ is surjective.

calculus - Limit of $frac{tan(x)-x}{x^3}$ as $x$ approaches $0$ without L'Hospital's Rule

I am trying to find the limit of $\frac{\tan(x)-x}{x^3}$ as $x$ approaches $0$. I know that this can be found by using L'Hospital's Rule 3 times. Is there a way to solve this problem without using L'Hospital's Rule?

Please do not use Taylor series; I consider this to be an equivalent method. I have noticed that the required number of applications of L'Hospital's Rule is precisely the order of the first non-zero derivative, which I think is essentially because a product is $0$ if and only if at least one factor is $0$.

Answer

you can simplify $$\frac{\tan x - x}{x^3}= \frac{\sin x - x \cos x}{x^3\cos x } = \frac{x - \frac{x^3}{6}+\cdots - x\left(1 - \frac{x^2}2+\cdots\right)}{x^3} = \frac 13 \text{ as } x \to 0. $$