summation - Steps to find the general formula for the series

Look at this series: 1, 2, 4, 6, 9, 12, 16, 20, 25, 30, 36, 42, ...
I've spent two days trying to find the formula for the nth term, but it is too difficult to find a way. Can you help me with the steps to get the general formula for the nth term?. If you can express it as a summation, it would be great.
Thanks in advance, God bless you!

functions - Cardinality of power set and binary sequence

Let $A$ be a set and $P(A)$ be the power set of $A$. Define $B(A)$ as
the set of all functions $F:A\rightarrow\{0,1\}$. For example,
$B(\mathbb{N})$ is the set of all binary sequences. Prove that $P(A)$
has the same cardinality as $B(A)$.

When $A$ is finite, this is easy to prove. I am interested in other cases; for instance when $A$ is countably infinite or uncountable. I am also a bit confused with the definition of $B(A)$. Could anyone help me with this one please?

Answer

The bijection is given by defining the function $F_X:A\to\{0,1\}$ with $X\subseteq A$ as:
\begin{align}
F_X(a)=\begin{cases}
1&\text{if $a\in X$}\\
0&\text{if $a\notin X$}
\end{cases}
\end{align}

The things you have to show is that $G:\mathcal P(A)\to B(A)$ with $G(X)=F_X$ is a bijection.

matrices - Frobenius Norm, Triangle inequality, and complex conjugates

I found a thread that solved the problem I need to turn in (or confirmed that I had done it correctly) but doesn't really resolve some confusion I have regarding norms and inner products.

I need to show that the Frobenius norm obeys the general definition of a matrix norm, and only the triangle inequality is giving me any trouble, but that's been worked to death : Frobenius Norm Triangle Inequality

But I went about it somewhat differently and it's highlighted a few concepts I'm shaky on. Here is my approach:

Starting from the defintion
$$
||A||_F = \left( \sum^m_{i=1} \sum^n_{j=1} |A_{ij}|^2 \right)^{1/2}.
$$

Now consider

$$
||A+B||_F = \left( \sum^m_{i=1} \sum^n_{j=1} |A_{ij}+B_{ij}|^2 \right)^{1/2}.
$$

Noting that each element $A_{ij}, B_{ij}$ can be thought of as vectors in $\Re^2$, I can apply the good old fashioned triangle inequality to the square root of each summand

such that $|A_{ij}+B_{ij}|\leq |A_{ij}|+|B_{ij}|$

Squaring both sides gives me

$|A_{ij}+B_{ij}|^2\leq |A_{ij}|^2+|B_{ij}|^2 +2|A_{ij}||B_{ij}|$

I see that I'm on the right track, but I'm afraid I'm a bit stuck here.

If I sum over all elements, I get back

$||A+B||_F^2 \leq ||A||_F^2 + ||B||_F^2 + 2\sum^m_{i=1} \sum^n_{j=1}|A_{ij}||B_{ij}|$

Is my approach hopelessly flawed, or is there some way I can salvage this?

elementary set theory - Question about cardinality of some set of functions

The question in its original form deals with the problem of deciding whether the set $T$ of all irrational numbers in the set $[0,1]$ such that they have only digits $0$ and $1$ in their decimal expansion is countable or uncountable?

I have been thinking along this lines and this form of the problem deals with the question in the title, if we look at the set $S$ of all functions $f: \mathbb N \to \{ 0,1 \}$ then this set of functions describe all numbers in $[0,1]$ which have $0$ and $1$ in their decimal expansion, and if $S$ is countable then obviously $T$ is also countable as a subset of $S$ but if $S$ is uncountable then $T$ is uncountable because $R=S\setminus T$ is the set of all rational numbers in the $[0,1]$ that have $0$ and $1$ in their decimal expansion so obviously $R$ is countable as a subset of $\mathbb Q$, which is countable.

So how to decide whether $S$ is countable or uncountable?

Answer

The set $S$ is often denoted as $2^\Bbb N$, and it is not hard to show that $2^\Bbb N$ has the same cardinality as $\mathcal P(\Bbb N)$, by mapping a subset to its indicator function.

Use Cantor's theorem to conclude that $S$ is uncountable.

One can actually show further that $|[0,1]|=|\mathcal P(\Bbb N)|$, and that in fact $T$ has the same cardinality as $[0,1]$ as well.

Easy functional equation

Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ such that:

$$f(2f(x)+f(y))=2x+f(y)\qquad \forall x,y \in \mathbb{R}.$$

If you put $x=y=0$, you get $f(3f(0))=f(0)$. What deductions about $f(0)$ can you then make?

Clearly from above $f(0)=0$ is a solution . . . so,

Putting $x=0$ gives $f(2f(0)+f(y))=f(y)$

$\rightarrow$ $f(f(y))=f(y)$

So $f(x)=x$ is a solution, but is it the only one?

I think it probably is, but how to prove?

Answer

$$f(2f(x)+f(y))=2x+f(y)\qquad \forall x,y \in \mathbb{R}.$$
Interchaning $x$ and $y$ you get

$$f(f(x)+2f(y))=f(x)+2y \,.$$

Claim 1: $f(x)$ is 1 to 1.

Indeed, if $f(x)=f(y)$ then

$$2x+f(x)=2x+f(y)=f(2f(x)+f(y))=f(f(x)+2f(y))=f(x)+2y $$

This implies that $x=y$.

Now, you can do part of what you did:

$$f(2f(0)+f(y))=f(y)\qquad \forall y \in \mathbb{R}.$$

Since $f$ is 1 to 1 you get

$$2f(0)+f(y)=y \,.$$

Thus

$$f(y)=y-2f(0)\,.$$

Setting $y=0$ you get $f(0)=0$ and thus $f(x)=x$ is the only solution.

real analysis - Is the result for $3sumlimits_{n=1}^inftyfrac{H_nH_n^{(2)}}{n^6}+sumlimits_{n=1}^inftyfrac{H_nH_n^{(3)}}{n^5}$ known in the literature?

I was able to get the following result

$$3\sum\limits_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^6}+\sum\limits_{n=1}^\infty\frac{H_nH_n^{(3)}}{n^5}=11\zeta(3)\zeta(6)+\frac52\zeta(4)\zeta(5)-\frac{13}{6}\zeta^3(3)-2\zeta(2)\zeta(7)-5\zeta(9)$$
where $H_n^{(p)}=1+\frac1{2^p}+\cdots+\frac1{n^p}$ is the $n$th generalized harmonic number of order $p$.

based on a nice identity and some manageable Euler sums. Is this result known in the literature? Can we evaluate the terms separately?

Answer

In answer to your question, can the sums be evaluated separately? Yes they can. The results for each of these two Euler sums can be found in the 2016 paper Euler sums and integrals of polylogarithm functions by Ce Xu et al.

The results are:
$$\sum_{n = 1}^\infty \frac{H_n H^{(2)}_n}{n^6} = \frac{17}{6} \zeta (3) \zeta (6) + \frac{173}{72} \zeta (9) + \frac{1}{4} \zeta (4) \zeta (5) - 3 \zeta (2) \zeta (7) - \frac{2}{3} \zeta^3 (3) \quad \text{(See Eq. 3.18)}$$
and
$$\sum_{n = 1}^\infty \frac{H_n H^{(3)}_n}{n^5} = \frac{679}{24} \zeta (9) - 11 \zeta (2) \zeta (7) - \frac{1}{2} \zeta (3) \zeta (6) - \frac{29}{4} \zeta (4) \zeta (5) - \frac{1}{6} \zeta^3 (3).$$

elementary number theory - What is the best algorithm for finding the last digit of an enormous exponent?

I found most answers here not clear enough for my case such as

$$
123155131514315^{4515131323164343214547}
$$

I wrote the $n\bmod10$ in Python and execution time ran out. So I need a faster algorithm or method. Sometimes, the result is incorrect as it failed the test case.