calculus - Calculate $lim_{ntoinfty}(sqrt{n^2+n}-n)$.

Introduction:

An exercise from "Principles of mathematical Analysis, third edition" by Rudin, page 78.

---

Exercise:

Calculate $\lim_{n\to\infty}(\sqrt{n^2+n}-n)$.

---

Explanation:

I have a hard time to grasp how to handle limits like this. I don't know how to start and what to look for. I've checked with mathematica and the answer should be $\frac{1}{2}$, and of course i've got the wrong answer. I find limits unintuitive. In the book, they proved "the limits of some sequences which occur frequently". The limits they proved were:

(a) If $p>0$ then $\lim_{n\to\infty}\frac{1}{n^p}=0$

(b) If $p>0$ then $\lim_{n\to\infty}\sqrt[n]{p}=1$

(c) $\lim_{n\to\infty}\sqrt[n]{n}=1$

(d) If $p>0$ and $\alpha$ is real, then $\lim_{n\to\infty}\frac{n^\alpha}{(1+p)^n}=0$

(e) If $|x|<1$, then $\lim_{n\to\infty}x^n=0$.

When they proved all of the above theorems it felt like they used the fact that they knew the limits. For example:

---

Proof of (b): If $p>1$, put $x_n=\sqrt[n]{p}-1$. Then, $x_n>0$ and by the binomial theorem,

$$1+nx_n\leq(1+x_n)^n=p$$

so that $$0

Hence $x_n\to 0$. And so on...

---

That is, I think they used the fact that the limit were 1 when they did put $x_n=\sqrt[n]{p}-1$. Before I compute a limit do I have to guess one? How can I do that when I don't think this is intuitive? Have you any tips how you do when you shall tackle a problem like this? How do you start when you want to compute a limit?

---

Solution:

This is how I did it:

$\sqrt{n^2+n}-n=\sqrt{n}\sqrt{n-1}-n=\sqrt{n}(\sqrt{n-1}-\sqrt{n})$.

Since $$(\sqrt{n-1}-\sqrt{n})\to 0\text{ when }n\to\infty.$$

The product approaches $0$. Which is obviously not true. I did realise this after a while. Since one of the factor grows really big while the other gets really small and I guess they tend to take each other out, so it's pretty clear it shouldn't approach 0, but I don't think it's clear that it should approach $\frac{1}{2}$ either. Thanks for your help.

Answer

Hint: multiply and divide for $(\sqrt{n^2+n}+n)$.

functions - Why is the domain of x raised to x (0,infinity)?

My math teacher was explaining how to draw graphs of given functions. For $f(x)=x^x$ he put the domain as $(0,\infty)$. Why is this true (if it is)?
For $x= -2$, $(-2)^{-2}$ is $1/4$, and so the function is defined at $x= -2$. Isn't it?

elementary number theory - Divisibility test for $4$

Claim: A number is divisible by $4$ if and only if the number formed by the last two digits is divisible by $4$.

Here's where I've gotten so far.

Let $x$ be an $(n+1)$-digit number. So $x= a_na_{n-1} \dots a_2a_1a_0$. If $a_1 = 0$ and $a_0 =0$, then $x$ is a multiple of $100$ and therefore clearly divisible by $4$. So we must deal with the case when $(a_1 \neq 0 \lor a_0 \neq 0)$.

Then if $10a_1 + a_0 \equiv 0 \mod 4$ is true, then $x$ is divisible by $4$.

Do I need to do anything else or is this done? I feel like it's not quite complete, but I'm not sure how to proceed.

Answer

Pick $h, j \in \{0,1,\dots 9\}$ then $$100k + 10h + j \equiv 10h+j \mod 4$$ because $4 \mid 100$

So we have $$100k + 10h + j \equiv 0 \mod 4 \ \Leftrightarrow \ 10h + j \equiv 0 \mod 4$$

linear algebra - Can a positive definite matrix have complex eigenvalues?

We know that a positive definite matrix has positive eigenvalues. I wanted to know if there is any result that shows whether a positive definite matrix can have complex eigenvalues.
I am currently calculating a covariance matrix which has real entries and is symmetric. In order to find the accuracy of the calculation, I tried to find the eigenvalues of the matrix and then generate c-code using MATLAB coder to use in my Kalman filter equations. However, when the coder gives error that the eigenvalues are complex. It is to be noted that the first two eigenvalues of the matrix are close two zero.

general topology - Order type between two sets and bijection?

I want to show that $$ \{1,2\}\times Z_+\ \text{and} \ Z_+ \times\{1,2\}\ \text{have different order type}$$

If we define $$f(i,j)=(j,i)\ \text{for}\ i\ \text{in }\{1,2\}\ \text{and} \ j\ \text{in}\ Z_+$$

It seems like that this is bijective map between two sets.

However, to show that they are not order isomorphic, how shall I start to show that bijection does not preserve ordering?

It seems like that the way I defined the bijection is not the only way.

I am wondering if there exists any bijection between two sets and that bijection does not preserve order, can I conclude that they have different order type?

combinatorics - Simplify a combinatorial expression

I came a cross a kind of combinatorial expression in my research. I'm wondering if there is a way to simplify or rewrite it. The expression is pretty simple. So I'm posting it here instead of MO. It is the following.

$\displaystyle \sum\limits_{i=0}^n (-1)^i{n \choose i} {x-i \choose l}, $

where in my case $x,l$ are some positive integers. It's not hard to show when $l< n$, the expression is $0$. But I would like to know about any possible formula for $l\geq n$. I tried to search in some combinatorial identity book. There are a lot of similar expressions, but none of the identities seems to apply. Any idea or answers will be greatly appreciated.

Answer

First let $x=m$ and $l=n+j$ & use $\binom{m-i}{n+j} =[y^{n+j}]: (1+y)^{m-i}$
\begin{eqnarray*}
S= \sum_{i=0}^{n} (-1)^i \binom{n}{i} \binom {m-i}{n+j} = [y^{n+j}]: \sum_{i=0}^{n} (-1)^i \binom{n}{i} (1+y)^{m-i} \\
= [y^{n+j}]: (1+y)^m (1- \frac{1}{1+y})^n \\
=[y^{n+j}]: (1+y)^{m-n} y^n =\binom{m-n}{j}.
\end{eqnarray*}

calculus - Dirichlet integral.

I want to prove $\displaystyle\int_0^{\infty} \frac{\sin x}x \,\mathrm{d}x = \frac \pi 2$, and $\displaystyle\int_0^{\infty} \frac{|\sin x|}x \,\mathrm{d}x \to \infty$.

And I found in wikipedia, but I don't know, can't understand. I didn't learn differential equation, laplace transform, and even inverse trigonometric functions.

So tell me easy, please.

Answer

About the second integral: Set $x_n = 2\pi n + \pi / 2$. Since $\sin(x_n) = 1$ and
$\sin$ is continuous in the vicinity of $x_n$, there exists $\epsilon, \delta > 0$ so that $\sin(x) \ge 1 - \epsilon$ for $|x-x_n| \le \delta$. Thus we have:
$$\int_0^{+\infty} \frac{|\sin x|}{x} dx \ge 2\delta\sum_{n = 0}^{+\infty} \frac{1 - \epsilon}{x_n} = \frac{2\delta(1-\epsilon)}{2\pi}\sum_{n=0}^{+\infty} \frac{1}{n + 1/4} \rightarrow \infty $$