continuity - Can unbounded discontinuous functions be locally bounded?

Consider the function $$f(x) = \frac{x^3}{1+x^3}$$

Obviously this function is discontinuous at $x = -1$ therefore discontinuous on $\mathbb{R}$. Moreover, it is unbounded at the same point. Now, I would not say that this function is locally bounded either, as not all sets $f(A)$ are bounded for a neighbourhood $A$ about any $x_0 \in \mathbb{R}$. Is this reasoning correct? Can an unbounded discontinuous function be locally bounded?

Answer

If I understand your definition correctly, the function $$f(x) = \begin{cases}x - 1 & x < 0\\
0 & x = 0\\
x+1 & x>0\end{cases}$$

fits the bill.

If you want a rational function, then no can do, because if $p, q$ are polynomials with no common factors, then $f(x)=\frac{p(x)}{q(x)}$ is continuous if $q(x)\neq 0$ and is unbounded around $x$ if $q(x)=0$.

trigonometry - Limit of $lim limits_{x to frac{5π}{2}^+} frac{5x - tan x}{cos x}$

So I have the following problem:

$$\lim \limits_{x \to \frac{5π}{2}^+} \frac{5x - \tan x}{\cos x}$$

I can't figure out how to get the limit. I tried splitting it up to:

$$\lim \limits_{x \to \frac{5π}{2}^+} \Big(\frac{5x}{\cos x} - \frac{\tan x}{\cos x}\Big)$$

I'm lost and unsure of what to do next. I'm taking a Calc 1 class and we have not yet gotten to L'hopitals and other methods yet (and also I am not sure how I could incorporate those ideas either).

linear algebra - Find a basis of the subspaces of $mathbb{R}^4$ generated by the vectors

Find a basis of the subspaces of $\mathbb{R}^4$ generated by the vectors $$v_1=(1,1,2,0),v_2=(-1,0,1,0),v_3=(2,-2,0,0),v_4=(0,0,-1,2)$$

First of all I wrote these vectors as rows of a matrix then applied the following transformations to reduce the matrix in row echelon form:-
$R_2+R_1$ & $R_3-2R_1$

Then $R_3+4R_2$ and finally $\frac{1}{10}R_3$
Then I wrote the non zero rows in row echelon form as $B={(1,1,2,0),(0,1,3,0),(0,0,1,0)}$ which forms basis .
Am I right here?

Answer

Using the vectors given in the problem, we can define a set: $$A = \{v_1, v_2, v_3, v_4\}$$

Remember the definition of a Basis set:

Given: V is a vector space and B is a subset of V, we say that B is a basis of V IFF $V = span(B)$ and B is linearly independent.

Row-Reducing the matrix form of the set A yields:
$$\begin{bmatrix}1&-1&2&0\\1&0&-2&0\\2&1&0&-1\\0&0&0&2\end{bmatrix}$$
$$ <=>$$
$$\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix}$$
Since each column contains a leading "1", we can say that the set A is linearly independent.

This result implies that the set A itself satisfies the definition of a basis.

$B = \{v_1,v_2,v_3,v_4\}$

Functional equation $f(x+y)-f(x)-f(y)=alpha(f(xy)-f(x)f(y))$ is solvable without regularity conditions

I was reviewing this question and got motivated to solve this general problem:

Find all functions $f:\mathbb R\to\mathbb R$ such that for all real numbers $x$ and $y$,
$$f(x+y)-f(x)-f(y)=\alpha(f(xy)-f(x)f(y))\tag0$$
where $\alpha$ is a nonzero real constant.

I found out that it can be solved in a similar way to what I did in my own answer to that question. It was interesting for me that we don't need any regularity conditions like continuity. As I didn't find any questions about the same functional equation on the site, I thought it might be useful to post my own answer to it.

I appreciate any other ways to solve the problem or any insights helping me understand why there's no need of regularities while some functional equations like Cauchy's need such additional assumptions to have regular solutions.

Answer

First, letting $y=1$ in (0) and rearranging the terms, we get:
$$f(x+1)=f(1)+(1+\alpha-\alpha f(1))f(x)\tag1$$
$$\therefore\:f(2)=(2+\alpha-\alpha f(1))f(1)\tag2$$
Next, substituting $x+1$ for $x$ in (1) and using (1) and (2) we have:
$$f(x+2)=f(2)+(1+\alpha-\alpha f(1))^2f(x)\tag3$$
Next, letting $y=2$ in (0) and using (3) we get:
$$\left((1+\alpha-\alpha f(1))^2-1\right)f(x)=\alpha(f(2x)-f(2)f(x))$$

$$\therefore\:f(2x)=\left(2(1-f(1))+\alpha(1-f(1))^2+f(2)\right)f(x)$$
Thus using (2) we have:
$$f(2x)=(2+\alpha-\alpha f(1))f(x)\tag4$$
Now, substituting $2x$ for $x$ and $2y$ for $y$ in (0) and using (4), we'll get:
$$\beta(f(x+y)-f(x)-f(y))=\alpha\beta^2(f(xy)-f(x)f(y))$$
where $\beta=2+\alpha-\alpha f(1)$. Mutiplying (0) by $\beta$ and subtracting the last equation, we'll have:
$$\beta(\beta-1)(f(xy)-f(x)f(y))=0\tag5$$
If $\beta=0$ then by (4) we conclude that $f$ is the constant zero function. So this case can only happen when $\alpha=-2$.

If $\beta=1$ then by (1) we conclude that $f$ is the constant $1+\frac1\alpha$ function.

If $\beta\neq0$ and $\beta\neq1$ then by (0) and (5) we have:
$$f(xy)=f(x)f(y)\tag6$$
$$f(x+y)=f(x)+f(y)\tag7$$
By letting $y=x$ in (6) we conclude that $f(x)$ is nonnegative for nonnegative $x$. Combining with (7) we find out that $f$ is increasing. An increasing additive function is of the form $f(x)=kx$. So by (6) we conclude that $f$ is the constant zero function or the identity function.

real analysis - If $fcolonmathbb Rtomathbb R$ is continuous and satisfies $f(x+y) = f(x) + f(y)$ and $f(1)=1$, then $f(x)=x$ for all $x$.

Let
$f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function such that:
$$f(x+y) = f(x) + f(y)$$

$$f(1) = 1$$ Show that $$f(x) = x$$ I have been having trouble approaching this problem. I have shown, through a system of equations, that $f(x+y) = x + y$, but that's about as far as I can get. Appreciate any help anyone has to offer!

Answer

See that

$$f(0)=f(0)+f(0)=2f(0)\implies f(0)=0$$

$$\begin{align}f(a+b+c+\dots+z) & =f(a)+f(b+c+\dots+z) \\&=f(a)+f(b)+
f(c+\dots+z)\\&=\dots\\&=f(a)+f(b)+f(c)+\dots+f(z)\end{align}$$

---

For natural numbers $x$, we have

$$f(x)=f(\underbrace{1+1+1+\dots+1}_x)=\underbrace{f(1)+f(1)+\dots+f(1)}_x=1+1+1+\dots+1=x$$

$$f(x)=x\ \forall\ x\in\mathbb N$$

---

For positive rational numbers,

$$1=f(1)=f(\underbrace{\frac1x+\frac1x+\frac1x+\dots+\frac1x}_x)=f(\frac1x)+f(\frac1x)+\dots+f(\frac1x)=xf(\frac1x)$$

$$1=xf(\frac1x)\implies f(\frac1x)=\frac1x\ \forall\ x\in\mathbb N$$

$$f(\frac yx)=f(\frac1x+\frac1x+\dots+\frac1x)=f(\frac1x)+f(\frac1x)+\dots+f(\frac1x)=yf(\frac1x)=\frac yx\\f(\frac yx)=\frac yx\ \forall\ (x,y)\in\mathbb N\times\mathbb N$$

$$f(x)=x\ \forall\ x\in\mathbb Q^+$$

---

For positive real numbers, every real number is the cauchy sequence of rational numbers, which, since $f(x)$ is continuous, gives us

$$f(x)=x\ \forall\ x\in\mathbb R^+$$

---

Finally, for all negative numbers, we have

$$0=f(0)=f(x-x)=f(x)+f(-x)$$

$$0=f(x)+f(-x)\implies f(-x)=-f(x)=-x$$

real analysis - Does there exist a scalar function $g({bf{x}})$ that satisfies $g({bf{x}} +,{bf{f}}({bf{x}}))= g ({bf{x}})det(I+,{bf{f}}'({bf{x}}))$?

Given a vector valued function $\bf{f}:\mathbb{R}^n\rightarrow\mathbb{R}^n$, what is a scalar function $g:\mathbb{R}^n\rightarrow\mathbb{R}$ that satisfies the following $$g({\bf x} +\mathbf{f}(\mathbf{x}))=g(\mathbf{x})\det(I+\mathbf{f}'(\mathbf{x})),$$ where $\bf{f}'(\bf{x})$ is the Jacobian matrix of $\bf{f}(\bf{x})$ and $I$ is the $n\times n$ identity matrix.

If a solution can't be found for arbitrary $\bf{f}$, what structure can one impose on $\bf{f}$ for there to exist a particular function $g$ that satisfies the above condition.

One particular example of a function $g$ that satisfies the above is all I'm after (i.e., I don't need the most general solution).

Alternatively: Can one prove that there does not exist a function $g$ that satisfies the above?

estimation - a limit about exponential function

$\lim_{n\rightarrow\infty}\frac{1+\frac{n}{1!}+\cdot+\frac{n^n}{n!}}{e^n}=\frac12$

Taking the first $n$ terms of the Taylor series of $e^n$ as the numerator, the limit is true or false? How to prove?

Answer

Assuming that we work

$$a_n=e^{-n}\sum_{k=0}^n \frac{n^k}{k!} $$by the definition of the incomplete gamma function
$$a_n=\frac{\Gamma (n+1,n)}{n \Gamma (n)}$$
We have the relation $$\Gamma (n+1,n)=n \,\Gamma (n,n)+e^{-n}\, n^n$$ which makes
$$a_n=\frac{ n^{n-1}}{e^n\,\Gamma (n)}+\frac{\Gamma (n,n)}{\Gamma (n)}$$ The first term tends to $0$ when $n$ becomes large; to prove it, take its logarithm and use Stirling approximation to get
$$\log\left(\frac{ n^{n-1}}{e^n\,\Gamma (n)} \right)=-\frac{1}{2} \log \left({2 \pi n}\right)-\frac{1}{12
n}+O\left(\frac{1}{n^{5/2}}\right)$$

For the second term, if you look here, you will notice the asymptotics
$$\Gamma(n,n) \sim n^n e^{-n} \sqrt{\frac{\pi}{2 n}}$$ So, neglecting the first term, we have, for large $n$
$$a_n\sim \frac{ n^n e^{-n} }{\Gamma(n)}\sqrt{\frac{\pi}{2 n}}$$ Take logarithms and use Stirling approximation to get

$$\log(a_n)=-\log (2)-\frac{1}{12 n}+O\left(\frac{1}{n^{5/2}}\right)$$ Continue with Taylor
$$a_n=e^{\log(a_n)}=\frac{1}{2}-\frac{1}{24 n}+O\left(\frac{1}{n^{2}}\right)$$
If you use the better asymptotics given in the link $$\Gamma(n,n) = n^n e^{-n} \left [ \sqrt{\frac{\pi}{2 n}} - \frac{1}{3 n} + O\left ( \frac{1}{n^{3/2}} \right ) \right ]$$ doing the same, you should end with
$$a_n=\frac 12-\frac{1}{3 \sqrt{2 \pi n} }-\frac{1}{24 n}+O\left(\frac{1}{n^{3/2}}\right)$$