number theory - In abstract algebra, what is an intuitive explanation for a field?

Wikipedia has the following to say about fields.

In mathematics, a field is one of the fundamental algebraic structures used in abstract algebra. It is a nonzero commutative division ring, or equivalently a ring whose nonzero elements form an abelian group under multiplication. As such it is an algebraic structure with notions of addition, subtraction, multiplication, and division satisfying the appropriate abelian group equations and distributive law.

What is an intuitive way about thinking about a field?

functions - What's the value of $frac{(x-1)x^2}{x-1}$ when $x=1$?

Consider the function
$$f(x)= \frac{(x-1)x^2}{x-1},\quad x \in \Bbb R\ .$$
This function gives simply gives $f(x)=x^2$ by cancelling the term $x-1$, if I am not wrong. The variable here is $x$ which can take arbitrary value, we just cancel that term as for any $x$, suppose $x=0,2,3,4, \ldots,-1,-2,-3,\ldots$ then the term is going to get cancelled, but what about the value $1$? If we have $x=1$ there then the term would not be defined, so why do we always cancel the similar terms without thinking what values variable takes?
I know this question is really very strange but I want to be clear about my approach.
This question is somewhat which I never asked before.

Answer

The function is undefined at $x = 1$ because $\frac00$ is undefined. If you graphed this the graph would look like the graph of $g(x) = x^2$ with a point simply missing from the graph. We say that the limit of the function as $x$ tends to $x^2$ (=1) from either $x \gt 1$ or $x \lt 1$ tends to 1. ($\lim_{x \rightarrow 1} f(x) = x^2$)

We also refer to $x=1$ as a "removable singularity". If ever you factor and divide by $(x - 1)$ you must specify that you are presuming $x$ does not equal $1$ for the rest of your conclusion.

This is a very common occurrence and plays very heavily in calculus.

It's good that you caught it.

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complex analysis - Can we use analytic continuation to obtain $sum_{n=1}^infty n = b, bneq -frac{1}{12}$

Intuitive question

It is a popular math fact that the sum definition of the Riemann zeta function:
$$\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} $$
can be extended to the whole complex plane (except one) to obtain $\zeta(-1)=-\frac{1}{12}$. The right hand side for the above equation in $-1$ becomes the sum of the natural numbers so in some sense we have obtained a value for it. My question is: is this value depending on the choice of the Riemann zeta function as the function to be analytically continued, or do we always get $-\frac{1}{12}$?

Proper formulation

let $(f_n:D\subset \mathbb{C}\rightarrow \mathbb{C})_{n\in \mathbb{N}}$ be a sequence of functions and $a \in \mathbb{C}$ with $\forall n\in \mathbb{N}: f_n(a) = n$ and
$$f(z):=\sum_{n=0}^\infty f_n(z)$$
convergent on a part of the complex plane, such that it can be analytically continued to a part of the plane that contains $a$. Does it then follow that, under this continuation, $f(a)=-\frac{1}{12}$ and why (or can you give a counterexample)?

Examples

• The case of the Riemann zeta function is the case where $\forall n
  \in \mathbb{N}: f_n(s) = \frac{1}{n^s}$ and $a=-1$


• The case where $\forall n \in \mathbb{N}: f_n(z) = \frac{n}{z^n}$ and $a=1$ does yield the sum of all natural numbers but it's continuation $\frac{z}{(z-1)^2}$ has a pole at $a$.

How to convert from complex number to polar coordinates (solve for r and theta).

How can I convert $ze^z$ to polar coordinates and what are the answers for ($r$ and $\theta$)?
I've been trying to solve the question for so long, but I unfortunately didn't succeed.

sequences and series - How do I evaluate this sum: :$sum_{n=3}^{infty} frac{1}{1-n+n^2-cdots+(-n)^k}$ and $k$ is positive integer?

Wolfram alpha showed after some calculations for evaluation of this series :

$\sum_{n=3}^{\infty} \frac{1}{1-n+n^2-\cdots+(-n)^k}$ for example for $k=10$ ,

I have got this result which it close to $0$.

My question here is : : How do I evaluate this sum: :$$\sum_{n=3}^{\infty} \frac{1}{1-n+n^2-\cdots+(-n)^k}$$ ?

Note: I exclude the singularity points just i would like to Know how do i evaluate it

Thank you for any help

calculus - Find formula of sum $sin (nx)$

I wonder if there is a way to calculate the

$$S_n=\sin x + \sin 2x + … + \sin nx$$

but using only derivatives ?

Answer

Using telescopic sums:

$$ \sin(mx)\sin(x/2) = \frac{1}{2}\left(\cos\left((m-1/2)x\right)-\cos\left((m+1/2)x\right)\right)$$
Hence:
$$ S_n \sin\frac{x}{2} = \frac{1}{2}\left(\cos\frac{x}{2}-\cos\left(\left(n+\frac{1}{2}\right)x\right)\right)=\sin\frac{nx}{2}\cdot\sin\frac{(n+1)x}{2}.$$