MISA6zh

2022-11-17

Compute summation of modules expression?

In particular, what I want to look at is the sum

$\sum _{k=1}^{n}(pk\phantom{\rule{1em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}q))$

where $p,q\in {\mathbb{Z}}_{\ge 1}$ can be assumed to be coprime but it would be best if solved in the fullest generality. In the above expression, n is a variable, p,q are fixed, and a(modb) means taking the representative set $\{0,1,2,...,b-1\}$. For example, $7\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}3)=1$ is the only value we agree upon and $7\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}3)\ne -2$.

The problem with this is that the list of representatives are permuted by p and hence the methods presented in the initial link are no longer valid.

It would be nice if we can come up with a closed form, but a really tight upper bound of the expression also works.

In particular, what I want to look at is the sum

$\sum _{k=1}^{n}(pk\phantom{\rule{1em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}q))$

where $p,q\in {\mathbb{Z}}_{\ge 1}$ can be assumed to be coprime but it would be best if solved in the fullest generality. In the above expression, n is a variable, p,q are fixed, and a(modb) means taking the representative set $\{0,1,2,...,b-1\}$. For example, $7\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}3)=1$ is the only value we agree upon and $7\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}3)\ne -2$.

The problem with this is that the list of representatives are permuted by p and hence the methods presented in the initial link are no longer valid.

It would be nice if we can come up with a closed form, but a really tight upper bound of the expression also works.

Patrick Arnold

Beginner2022-11-18Added 21 answers

Step 1

Let the sequence 1p,2p,3p,...qp,...,np. This shows that there are some multiples of qp, or 1qp,2qp,...,rqp. Thus we can assume r be the largest number such that $rq<n$. We can calculate that the number of cycles are $\lfloor \frac{n}{q}\rfloor $. The remaining numbers are rqmodn.

We assume that for some integer ${m}_{1},{m}_{2}$, such that ${m}_{1}p\equiv k\phantom{\rule{thickmathspace}{0ex}}(\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}q)$ and ${m}_{2}p\equiv k\phantom{\rule{thickmathspace}{0ex}}(\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}q)$. By adding up the two equations we have $({m}_{1}-{m}_{2})p\equiv 0\phantom{\rule{thickmathspace}{0ex}}(\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}q)$. We know, for every prime p,q, they are coprime. This means that only for $|{m}_{1}-{m}_{2}|\phantom{\rule{thickmathspace}{0ex}}{\textstyle |}q$. This tells us that in the sets of number 0p,1p,2p,3p,...qp, the modulo of the numbers are always different.

As we take the modulo when $0\le r\le q$, thus for every set of numbers 1p,2p,...,qp, their modulo is $0,1,2,...,q-1$, the total sum of them is

$\lfloor \frac{n}{q}\rfloor \sum _{i=1}^{q-1}i=\frac{1}{2}\lfloor \frac{n}{q}\rfloor q(q-1)$

Step 2

Case 1: $p<q$, thus we can roughly calculate the modulo by hand.

Case 2: $p>q$, we can assume $p=q+a$ for some integer a, we can also know that most of the value of a is even as mostly of the prime numbers are odd. Then, for some integer $\alpha $, $\alpha p=\alpha (q+a)=\alpha q+\alpha a\equiv \alpha a\phantom{\rule{thickmathspace}{0ex}}(\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}q)$. Thus the entire summation can be evaluated as follow:

$\sum _{k=1}^{n}pk\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}q\equiv \frac{1}{2}{\lfloor \frac{n}{q}\rfloor}{q(q-1)}+\sum _{i=1}^{n-rq}ia\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}q=\frac{1}{2}{r}{({q}^{2}-q)}+\sum _{i=1}^{n-rq}ia\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}q$

Feel free to leave a comment if there're mistakes.

Let the sequence 1p,2p,3p,...qp,...,np. This shows that there are some multiples of qp, or 1qp,2qp,...,rqp. Thus we can assume r be the largest number such that $rq<n$. We can calculate that the number of cycles are $\lfloor \frac{n}{q}\rfloor $. The remaining numbers are rqmodn.

We assume that for some integer ${m}_{1},{m}_{2}$, such that ${m}_{1}p\equiv k\phantom{\rule{thickmathspace}{0ex}}(\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}q)$ and ${m}_{2}p\equiv k\phantom{\rule{thickmathspace}{0ex}}(\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}q)$. By adding up the two equations we have $({m}_{1}-{m}_{2})p\equiv 0\phantom{\rule{thickmathspace}{0ex}}(\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}q)$. We know, for every prime p,q, they are coprime. This means that only for $|{m}_{1}-{m}_{2}|\phantom{\rule{thickmathspace}{0ex}}{\textstyle |}q$. This tells us that in the sets of number 0p,1p,2p,3p,...qp, the modulo of the numbers are always different.

As we take the modulo when $0\le r\le q$, thus for every set of numbers 1p,2p,...,qp, their modulo is $0,1,2,...,q-1$, the total sum of them is

$\lfloor \frac{n}{q}\rfloor \sum _{i=1}^{q-1}i=\frac{1}{2}\lfloor \frac{n}{q}\rfloor q(q-1)$

Step 2

Case 1: $p<q$, thus we can roughly calculate the modulo by hand.

Case 2: $p>q$, we can assume $p=q+a$ for some integer a, we can also know that most of the value of a is even as mostly of the prime numbers are odd. Then, for some integer $\alpha $, $\alpha p=\alpha (q+a)=\alpha q+\alpha a\equiv \alpha a\phantom{\rule{thickmathspace}{0ex}}(\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}q)$. Thus the entire summation can be evaluated as follow:

$\sum _{k=1}^{n}pk\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}q\equiv \frac{1}{2}{\lfloor \frac{n}{q}\rfloor}{q(q-1)}+\sum _{i=1}^{n-rq}ia\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}q=\frac{1}{2}{r}{({q}^{2}-q)}+\sum _{i=1}^{n-rq}ia\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}q$

Feel free to leave a comment if there're mistakes.

Show that the sequence $a}_{n$ is an solution of the recurrence relation [a^n = a^n−1 + 2a^n−2 + 2n − 9 if a] if

a^n = −n + 2

Why does a magnet always attract iron, not wood?

Why is $f(n)={n}^{3}$ not an onto function?

I was doing an example in a book where it asked which of these functions are one to one, the answer in the back said for $f(n)={n}^{3}$ that it is a one to one function. Then it asked which of the functions from the previous example are onto and $f(n)={n}^{3}$ was not included in the list of onto functions.

In a later example, a question asked which of these functions is a bijection, the answer included $f(x)={x}^{3}$

This is confusing because doesn't a function have to be both an onto and one to one to be a bijection? Why would the book say it was not a onto in a previous example yet declare it to be a bijection? Is the book wrong?What is the meaning of l.h.s and r.h.s?

Showing that if n is a natural number larger than 3, then $n!>{2}^{n}$

Showing that if n is a natural number larger than $3$, then $n!>{2}^{n}$

My try:

Base Case:

If $n=4$, then $4!>{2}^{4}$

$24>16$

So, the base case is true.

Assuming $P(k)$ is true.

$k!>{2}^{k}$

Now we need to show that $P(k+1)$ is true.

$(k+1)!={2}^{k+1}$

Proof:

$(k+1)!>(k+1)k!$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}(k+1){2}^{k}$

After this I have no idea how to solve further.

Can anyone explain how to continue.What type of material allows electrons to flow freely?

Conductor

Resistor

Hestitation

RegulatorA set X with cardinality X has how many elements in its power set (the set of all subsets of a set)?

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b. $|X\times X|$

c. ${2}^{|x|}$

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Can a dot product of a permutation of n (1,-1) with a sequence of primes generate unique numbers?

According to Fundamental Theorem of Arithmetic any positive whole number is the product of primes. Therefore, I can create unique numbers by multiplying n primes.

If I have a list of length n generated by a finite sequence in the set 1,-1, e.g., (1,1,-1,-1,1,...), and I do a dot product with a sequence of n primes starting with 3, e.g., (3,5,7,11,13,...), do I have any guarantee that I will generate unique numbers by doing different permutations of 1s and -1s?

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In this example, you could use:

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$\{1,3\},\{2,4\}\to 1\cdot 3+2\cdot 4=3+8=11$

$\{1,4\},\{2,3\}\to 1\cdot 4+2\cdot 3=4+6=10$

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Other than that, no restrictions apply: any amount of distinct letters may be used throughout the entire word, and any letter can be used as many times as we like, as long as every substring inside the length n word complies with the rules above.

There is the obvious case of $k=2$ which results in 2 words, for every n, because you can only start with either letter, and they alternate.

For the larger case, I have come up with a recursive formula:

$C(n,k)=f(0,0,k)$

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At every step, we can either use a letter other than the previous d letters, in which case the length is increased by one, and the chain-length d of distinct consecutive letters is also increased by one. In this case, there are $(k-d)$ such letters we can use at the stage, each subsequently resulting in the same contribution.

Or, we can use a letter already in the last d letters. In this case, the position of the letter matters. If we use the last of the d letters, a whole new chain begins. If instead we use the second last d letter, then a chain of length 2 of consecutive distinct letters begins. The 3rd last would result in a chain of length 3 and so on.

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