number theory - Meaning of equality in zeta regularization

It is known that
$$\sum\limits_{n = 1}^\infty{n = 1 + 2 + 3 + \cdots} = \infty$$
but it is also known that
$$\sum\limits_{n = 1}^\infty{n = 1 + 2 + 3 + \cdots} = -\frac{1}{{12}}$$
which can obtained using the analytic continuation of $\zeta(s)$. My question is: What is the true meaning of equality here? It is a common practice to write $f = {\mathcal O}(g)$ to mean $f \in {\mathcal O}(g)$. Is this something similar?

Answer

One thing that is sometimes lost when talking about series is what the symbol $$\tag{1}\sum_{n=1}^\infty a_n$$ means.

The point is that it is defined as a limit, i.e. the usual definition of the symbol $(1)$ is that it is the limit (if it exists) of the sequence of partial sums. This is already subtle, because the notion of limit depends on the topology. The topology is canonical on $\mathbb R$ or $\mathbb C$, but not in normed vector spaces where one can still consider series. This definition also stumbles with the case of conditionally convergent series, where any other limit can be achieved by reordering.

Anyway, the definition of $(1)$ as the limit of the sequence of partial sums (which is $\infty$ in your example) is not the only possible definition. There are other notions, such as Cesàro-sum and Abel-sum (these two the most noteworthy among others).

Yet another meaning one can given to the $\sum\limits_{n=1}^\infty n$ is the analytic continuation you mention. This is an example of the so called Ramanujan* summation, where the number $-1/12$ is obtained for the series in the question.

In Cesàro, you define
$$
\sum_{n=1}^\infty a_n=\lim_{N\to\infty}\frac{s_1+\cdots+s_N}N

$$
(if it exists), where $s_k=\sum\limits_{n=1}^k a_n$. Summable in the canonical sense implies Cesàro summable, with the same limit, but not vice versa.

The other canonical summability is the Abel one: you define
$$
\sum_{n=1}^\infty a_n=\lim_{x\to1^-}\sum_{n=1}^\infty a_nx^n.
$$
Again, Cesàro summability implies Abel summability to the same limit, but not vice versa.

To see an example, consider $a_n=(-1)^n$. The usual series doesn't exist. But, using Cesàro or Abel, one can say that $\sum\limits_{n=1}^\infty(-1)^n=\frac12$.

elementary number theory - Fermat's little theorem

This is a very interesting word problem that I came across in an old textbook of mine. So I mused over this problem for a while and tried to look at the different ways to approach it but unfortunately I was confused by the problem and I don't really understand how to do it either, hence I am unable to show my own working or opinion. The textbook gave a hint about using Fermat's little theorem but I don't really understand it and I'm really not sure about how to approach it. Any guidance hints or help would be truly greatly appreciated. Thanks in advance :) So anyway, here the problem goes: (It is composed of three parts)

$a)$ Determine the remainder when $2^{2017}+1 $ is divided by $17$.

$b)$ Prove that $30^{99} + 61^{100}$ is divisible by $31$.

$c)$ It is known that numbers $p$ and $8p^2 + 1 $ are primes. Find $p$.

Answer

In problem (a), use Fermat's little theorem, which says (or a rather, a very slightly different version says) that for any prime number $p$, and any integer $n$ that's not divisible by $p$, we have
$$n^{p-1}\equiv 1\bmod p$$
In particular, use $n=2$ and $p=17$. Keep in mind that $2017=(126\times 16)+1$.

In problem (b), note that $30\equiv 61\equiv -1\bmod 31$ (you don't even have to use Fermat's little theorem here).

In problem (c), use Andre's hint above: if $p$ is any prime number other than $3$, then $p^2\equiv 1\bmod 3$ (which you can see is an application of Fermat's little theorem). What does that mean $8p^2$ is congruent to modulo $3$? What does that mean $8p^2+1$ is congruent to modulo $3$? Can a prime number be congruent to that modulo $3$?

calculus - if $int_1^{infty}f(x) mathrm dx$ converges, must $int_1^{infty}f(x)sin x mathrm dx$ converge?

I can't use any of the convergence tests I learned because I have no information on $f(x)$, in particular I don't know if it's continuous or positive.

The only thing I could think of was that if $\displaystyle \int_{1}^{\infty}f(x)\ \mathrm dx$ was absolutely convergent, then $|f(x)\sin x| \leq |f(x)|$ would imply by the comparison test that $\displaystyle \int_{1}^{\infty}f(x)\sin x\ \mathrm dx$ converges.

So if I want to find a counter-example I have to pick $f(x)$ so that $\displaystyle \int_{1}^{\infty}f(x)\ \mathrm dx$ conditionally converges, but I can't think of one.

Answer

Consider $f(x)=\sin(x) / x$.

combinatorics - Proving that $sumlimits_{k=0}^{n} {{m+k} choose{m}} = { m+n+1 choose m+1 }$

I have to prove that:

$$\sum_{k=0}^{n} {{m+k} \choose{m}} = { m+n+1 \choose m+1 }$$

I tried to open up the right side with Pascal's definition that:
$$ { n \choose k} = {n-1 \choose {k}} + {n-1 \choose {k-1}}$$

Here is what I came up with, and I am sure it is wrong because it does not equal the left side:

$$ {m+n+1 \choose m+1} = {m+n \choose m+1} + {m+n \choose m} = ... ={m+n \choose m+1} + {m+n-1 \choose m} + {m+n-2 \choose m-1} + ... + {m \choose m+1} = \sum_{k=0}^{n} {m+k \choose m+k-n+1 } $$

Which, again, probably is wrong because it is not equal $\sum_{k=0}^{n} { m+k \choose m}$.
Any help is appreciated

Answer

Here is a purely combinatorial proof:

Consider picking $m+1$ numbers out of $\{1,2,...,m, \color{ #009900}{m + 1}, \color{ #009900}{m + 1} + 1,...,\color{ #009900}{m + 1} + (n - 1),\color{ #009900}{m + 1}+n\}$.

The right hand side of your equation is clearly equal to the number of ways of doing this.

Now for any given choice of $m+1$ numbers, the highest number chosen must be some $k$ with $\color{ #009900}{m + 1} \leq k \leq \color{ #009900}{m + 1}+n$. In each of these cases, we must select the remaining $m$ numbers to be chosen from the $k-1$ numbers smaller than $k$.

For $k = m +1$, must pick $m$ numbers to the left of $m + 1$, out of $\{\color{ #0073CF}{1, 2, ..., m}, m+1\}$.
Since there are $ \color{#0073CF}{m}$ such numbers, so $\color{#0073CF}{m}$ possible choices for $m$.
Thus the total number of choices for $m$ numbers $= \binom{\color{#0073CF}{m}}{m}$.

For $k = m +2$, must pick $m$ numbers to the left of $m + 2$, out of $\{\color{ #0073CF}{1, 2, ..., m, m +1}, m+2\}$.
Since there are $ \color{#0073CF}{m + 1}$ such numbers, so $\color{#0073CF}{m + 1}$ possible choices for $m$.
Thus the total number of choices for $m$ numbers $= \binom{\color{#0073CF}{m + 1}}{m}$.

...
For $k = m + 1 + n$, must pick $m$ numbers to the left of $m + 1 + n$, out of $\{\color{ #0073CF}{1, 2, ..., m, m +1, ..., m + n}, m+ n + 1\}$.
Since there are $ \color{#0073CF}{m + n}$ such numbers, so $\color{#0073CF}{m + n}$ possible choices for $m$.
Thus the total number of choices for $m$ numbers $= \binom{\color{#0073CF}{m + n}}{m}$.

Summing up the number of ways of doing this for $k=m+1,...,m+n+1$ yields the LHS of your equation.

calculus - Find the limit of $lim_{xto 0} (frac{1+tan x}{1+sin x})^{csc^3x}$

I failed to find the limit of:lim(x->0) $(\frac{1+tan(x)}{1+sinx})^{\frac{1}{sin^3(x)}}$?
as X approches 0
How do I find the answer for this?

Thanks in advance. the answer supposed to be sqr(e). but my answer was 1.
Can anyone please help me find my mistake?
I DID:

$lim_{x \to 0} (\frac{1+tan(x)}{1+sin(x)})^{\frac{1}{sin^3(x)}}$

$lim_{x \to 0} (\frac{((1+tan(x))^{1/sin(x)}}{((1+ sin(x))^{1/sin(x)}})^{1/sin^2(x)}$

now I look inside:

$lim_{x \to 0} ((1+tan(x))^{1/sin(x)}$ is e
$lim_{x \to 0} ((1+sin(x))^{1/sin(x)}$ is also e

so we get:

$lim_{x \to 0} (\frac{e}{e})^{\frac{1}{sin^2(x)}}$

$lim_{x \to 0} (1)^{\frac{1}{sin^2(x)}}$ = 1

real analysis - Convergence in $L^{infty}$ norm implies convergence in $L^1$ norm

Let $\{f_n\}_{n\in \mathbb{N}}$ be a sequence of measurable functions on a measure space and $f$ measurable. Assume the measure space $X$ has finite measure. If $f_n$ converges to $f$ in $L^{\infty}$-norm , then $f_n$ converges to $f$ in $L^{1}$-norm.

This is my approach:

We know $||f_n-f||_{\infty} \to 0 $ and by definition $||f_n-f||_{\infty} =\inf\{M\geq 0: |f_n-f|\leq M \}.$ Then
\begin{align}
||f_n-f||_1\
&=\int |f_n-f| dm\

&\leq \int|f_n|dm+\int|f|dm\
\end{align}

I don't know how to proceed after that, any help would be appreciated.

Answer

For any function $g$, $||g||_1 = \int_X|g(m)|dm \leq \int_X||g||_\infty dm = \mu(X)*||g||_\infty$ (as $|g(m)| \leq ||g||_\infty$ almost everywhere); $||g||_\infty \geq \frac{||g||_1}{\mu(X)}$, so if $||f_n-f||_\infty$ tends to zero, then $||f_n-f||_1$ tends to zero as well.

matrices - what is the geometry behind the matrix multiplication?

What is the geometry behind the matrix multiplication?

The questions that I am having is the follows.

$\bullet$ I accept that we are viewing $\mathbb R^4$ as either in $\begin{pmatrix} a_{11},a_{12},a_{13},a_{14}\end{pmatrix}$ or $\begin{pmatrix} a_{11}&a_{12}\\a_{13}&a_{14}\end{pmatrix}$.

so, matrix addition makes sense that it gives another vector in that space.

$\bullet$ But, Matrix multiplication does not convince me in this role.

I strucked with,

Like in the matrix addition, (addition of two vectors is nothing but the diagonal of the parallelogram in which the two vectors are adjustcent sides)

is there any vector space diagramatic representation for matrix multiplication??

Note:
I am aiming to teach this factacy to my grade 11 students who are studying their matrices now only.(Means to say, this is the first time they gonna meet matrices)