# Double branch $\sqrt x$ or square function turned 90°?

I have this idea for a graph but don't know what function could describe it better.

The idea is something like the "squared" function turned $90$ degrees to the right, so that possible values for $x$ are always positive and $y$ may be both positive and negative.

The graphs of $\sqrt x$ and $-\sqrt x$ combined look good too, but I don't know how to write that as a single function (eh, I'm so bad with these things).

Basically, anything that may look like this will do.

Looking forward to some solution.

What you're looking for can be described as a parabola opening towards the positive $x$ axis. I'm going to refer to it as a "sideways parabola," since the "standard" parabola that people learn opens towards the positive $y$ axis.

The bad news: You cannot express a sideways parabola as a function of $x$. Why? Let's go back and look at a restriction on functions:

Vertical Line Test: For a relation to be a function, it must (colloquially) pass the vertical line test. That is, you must be able to draw a vertical line anywhere on the relation's graph and the line must intersect the relation in at most one point.

In the picture below, I've marked the two intersections that a vertical line makes on a sideways parabola. This shows that the sideways parabola is not a function.

However, the good news is that one may still describe such a graph with mathematical notation. Two such ways are below, but keep in mind that they are not functions of $x$.

$$x=y^2$$ $$y=\pm\sqrt{x}$$

• okay, somehow i didn't know about the vertical line. this is a really good explanation, thank you. – Don Jan 14 '15 at 4:22

The graph you describe will have equation $x=y^2$.

You cannot write it in the form "$y=\langle\hbox{function of$x$}\rangle$" because, just as you pointed out, every $x>0$ will correspond to two $y$ values, not just one.

• Concise. Correct. +1. – MPW Jan 14 '15 at 4:55

If you want the square function $$y = x^2$$ turned 90 degrees to the right, you get the inverse $$x = y^2$$ or written in another way (like you already mentioned) $$y = \pm \sqrt{x}.$$

I'm not actually sure what you mean by writing it as a single function as it has two branches (the positive and the negative one).

• what i meant comes from the fact that i'm stuck with this program which will only accept input like y = sqrt(x), so there's no +/-, and giving it sqrt(x)+(-sqrt(x)) makes no sense of course. – Don Jan 14 '15 at 4:20
• Well, in that case graphing a function with two branches like that seems to be impossible. – 655321 Jan 14 '15 at 4:30
• More accurately, $y = \sqrt{x}$ is the inverse of $y = x^2, x \geq 0$, while $y = -\sqrt{x}$ is the inverse of $y = x^2, x \leq 0$. The equation $x = y^2$ is not a function of $x$ since there are two values of $y$ for each value of $x > 0$. – N. F. Taussig Jan 14 '15 at 12:54

The curve obtained by combining the graphs of $y = \sqrt{x}$ and $y = -\sqrt{x}$ is $x = y^2$. It is not a function of $x$ since there are two values of $y$ for each $x > 0$. However, you can use the parametric equations \begin{align*} x(t) & = t^2\\ y(t) & = t \end{align*} to write both $x$ and $y$ as functions of a third variable $t$. Observe that $x(t) = [y(t)]^2$.

By using parametric equations to express both $x$ and $y$ as functions of $t$, you can describe curves that are not necessarily functions of $x$.


# Prove $A \implies C$

Given, $$(A \lor B) \implies C$$, prove $$A \implies C$$

My Proof:

1 By Conditional Exchange,

$$\neg(A \lor B) \lor C$$

2 By DeMorgan's Law,

$$(\neg A \land \neg B) \lor C$$

3 By Simplification,

$$\neg A \lor C$$

4 By Conditional Exchange,

$$A \implies C$$

My question pertains to steps 2 and 3. I used Demorgan's and Simplification on a subformula of a premise -- can I do that? Usually, I would separate the subformulas, but I don't think I could do so in this case.

Thanks.

• Your steps look valid – J. W. Tanner Jul 26 '19 at 4:13
• Alternative Proof: By disjunction, $A \implies A \vee B$. Since $A \vee B \implies C$, and since $A \implies A \vee B$, then $A \implies C$ by hypothetical syllogism. – JavaMan Jul 26 '19 at 4:24
• My logic is rusty, so take this with a grain of salt: but I do not see the problem. You are asked to prove "If A, then C." so I do not see why you can't start by assuming $A$ is true. In math, to prove a statement of the form $P \implies Q$, it is perfectly valid to assume $P$ is true, since if $P$ is false, then $P \implies Q$ vacuously. If this is a rigorous (philosophical) logic course, then maybe I'm missing some details in logical formalism, so you could/should ask your teacher (or others here). – JavaMan Jul 26 '19 at 4:31
• This isn't circular reasoning. Circular reasoning would be assuming that "$A \implies C$ is true to prove that $A \implies C$ is true. Here, you are really proving that $A \implies C$ using cases: whether $A$ is true or not. – JavaMan Jul 26 '19 at 4:35

\begin{align} &(A\lor B)\to C &&\text{Premise} \\ \iff & \lnot(A \lor B)\lor C&& \text{Conditional Exchange}\\\iff & (\lnot A\land\lnot B)\lor C&&\text{de Morgan's}\\\iff &(\lnot A\lor C)\land(\lnot B\lor C)&&\text{Distribution}\\\implies &\lnot A\lor C&&\text{Simplification}\\ \iff & A\to C&&\text{Conditional Exchange} \end{align}
• The obvious case where substituting a weakening does not work is when the phrase is in an antecedent. $~~(A\land B)\to B$ does not imply $A\to B$. – Graham Kemp Jul 26 '19 at 4:34
Instead, use distribution to get $$(\lnot A\lor C)\land (\lnot B\lor C),$$ and then use simplification to get $$\lnot A\lor C.$$


No. 116

#### Summing up integers to a given limit

 Q: Suppose an arbitrary number n is given e.g n=5. We want to compute the sum of all integers ranging from 1 to 5: 1 + 2 + 3 + 4 + 5 = 15 Implement the following method by using a loop:/** * Summing up all integers starting from 0 up to and including a given limit * Example: Let the limit be 5, then the result is 1 + 2 + 3 + 4 + 5 * * @param limit The last number to include into the computed sum * @return The sum of 1 + 2 + ... + limit */ public static long getSum (int limit) { ... }For the sake of getting used to it write some unit tests beforehand. BTW: Is it possible to avoid the loop completely achieving the same result? A: The solution's class de.hdm_stuttgart.de.sd1.sum.Summing is being contained within: Maven module source code available at sub directory P/Sd1/summing/V1 below lecture notes' source code root, see hints regarding import. Online browsing of API and implementation. We start by defining unit tests beforehand:/** * Testing negative values and zero. */ @Test public void testNonPositiveLimits() { assertEquals(0, Summing.getSum(-5)); assertEquals(0, Summing.getSum(0)); } /** * Testing positive values. */ @Test public void testPositiveLimits() { assertEquals(1, Summing.getSum(1)); assertEquals(3, Summing.getSum(2)); assertEquals(15, Summing.getSum(5)); assertEquals(5050, Summing.getSum(100)); // Carl Friedrich Gauss at school assertEquals(2147450880, Summing.getSum(65535));// Highest Integer.MAX_VALUE compatible value }The crucial part here is determining 1 + 2 + ... + 65535 being equal to 2147450880. Legend has it the young Carl Friedrich Gauss at school was asked summing up 1 + 2 + ... + 99 + 100. His teacher was keen for some peaceful time but the young pupil decided otherwise: $\underbrace{1 + \underbrace{2 + \underbrace{3 + \dots + 98}_{101} + 99}_{101} + 100}_{101}$ Since there are 50 such terms the result is $50 × 101 = 5050$. Generalizing this example we have: $∑ i = 1 n i = n  ( n + 1 ) 2$ For the current exercise we do not make use of this. Instead we implement Summing.getSum(...) by coding a loop: /** * Summing up all integers starting from 0 up to and including a given limit * Example: Let the limit be 5, then the result is 1 + 2 + 3 + 4 + 5 * * @param limit The last number to include into the computed sum * @return The sum of 1 + 2 + ... + limit */ public static long getSum (int limit) { int sum = 0; for (int i = 1; i <= limit; i++) { sum += i; } return sum; } }This passes all tests.

No. 117

#### Summing up, the better way

Q:

The previous solution of Summing up integers to a given limit suffers from a performance issue. When dealing with large values like 65535 looping will take some time:

long start = System.nanoTime();
System.out.println("1 + 2 + ... + 65535" + "=" + getSum(65535));
long end = System.nanoTime();
System.out.println("Elapsed time: " + (end - start) + " nanoseconds");
1 + 2 + ... + 65535=2147450880
Elapsed time: 1169805 nanoseconds

Barely more than one millisecond seems to be acceptable. But using the method for calculations inside some tight loop this might have a serious negative performance impact.

Thus implement a better (quicker) solution avoiding the loop by using the explicit form. When you are finished re-estimate execution time and compare the result to the previous solution.

Provide unit tests and take care of larger values. What is the largest possible value? Test it as well!

### Tip

A:

The solution's class de.hdm_stuttgart.de.sd1.sum.Summing is being contained within:

Since only our implementation changes we reuse our existing unit tests. Our first straightforward implementation attempt reads:

public static long getSum (int limit) {
return limit * (limit + 1) / 2;
}

This fails both unit tests. The first error happens at:

assertEquals(0, Summing.getSum(-5));
java.lang.AssertionError:
Expected :0
Actual   :10

We forgot to deal with negative limit values. Our sum is supposed to start with 0 so negative limit values should yield 0 like in our loop based solution:

public static long getSum(int limit) {
if (limit < 0) {
return 0;
} else {
return limit * (limit + 1) / 2; // Order of operation matters
}
}

This helps but one test still fails:

assertEquals(2147450880, Summing.getSumUsingGauss(65535));
java.lang.AssertionError:
Expected :2147450880
Actual   :-32768

This actually is a showstopper for large limit values: The algebraic value of limit * (limit + 1) / 2 might still fit into an int. But limit * (limit + 1) itself not yet divided by 2 may exceed Integer.MAX_VALUE. Since the multiplication happens prior to dividing by 2 we see this overflow error happen.

Solving this issue requires changing the order of operations avoiding arithmetic overflow. Unfortunately the following simple solution does not work either:

public static long getSum(int limit) {
if (limit < 0) {
return 0;
} else {
return limit / 2 * (limit + 1);
}
}

This is only correct if limit is even. Otherwise division by 2 leaves us with a remainder of 1.

However if limit is uneven the second factor limit + 1 will be even. This observation leads us to the final solution:

public static long getSum(int limit) {
if (limit < 0) {
return 0;
} else if (0 == limit % 2){       // even limit, divide by 2
return limit / 2 * (limit + 1); // Avoiding arithmetic overflow
} else {                          // uneven limit, divide (limit + 1 ) by 2
return (limit + 1) / 2 * limit; // Avoiding arithmetic overflow
}
}

This passes all tests. We finally reconsider execution time:

1 + 2 + ... + 65535=2147450880
Elapsed time: 25422 nanoseconds

Thus execution is roughly 46 times faster compared to the loop based approach. This is not surprising since loop execution is expensive in terms of performance.


# How do I find a flaw in this false proof that $7n = 0$ for all natural numbers?

This is my last homework problem and I've been looking at it for a while. I cannot nail down what is wrong with this proof even though its obvious it is wrong based on its conclusion. Here it is:

Find the flaw in the following bogus proof by strong induction that for all $n \in \Bbb N$, $7n = 0$.

Let $P(n)$ denote the statement that $7n = 0$.

Base case: Show $P(0)$ holds.

Since $7 \cdot 0 = 0$, $P(0)$ holds.

Inductive step: Assume $7·j = 0$ for all natural numbers $j$ where $0 \le j \le k$ (induction hypothesis). Show $P(k + 1)$: $7(k + 1) = 0$.

Write $k + 1 = i + j$, where $i$ and $j$ are natural numbers less than $k + 1$. Then, using the induction hypothesis, we get $7(k + 1) = 7(i + j) = 7i + 7j = 0 + 0 = 0$. So $P(k + 1)$ holds.

Therefore by strong induction, $P(n)$ holds for all $n \in \Bbb N$.

So the base case is true and I would be surprised if that's where the issue is.

The inductive step is likely where the flaw is. I don't see anything wrong with the strong induction declaration and hypothesis though and the math adds up! I feel like its so obvious that I'm just jumping over it in my head.

• Can you really always find natural numbers $i, j$ such that $0 \leq i, j \leq k$ and $i+j = k+1$? – Alex G. Oct 13 '16 at 4:01
• No. The "proof" assumes that for any $k\geq 0$, there exist natural numbers $i, j$ such that $0\leq i, j\leq k$ and $i+j = k+1$ – Alex G. Oct 13 '16 at 4:05
• As a general rule: For fake induction proofs, find the smallest case where the conclusion does not hold, and then do each step in detail with the corresponding numbers inserted, so that it should proof that exact case. That way you will almost always quickly find the problem. In this case, the smallest failing case is $P(1)$, so the number to look at is $k+1=1$, that is, $k=0$. – celtschk Oct 13 '16 at 8:26
• As a matter of English, you can disprove a false theorem (i.e. an untrue statement that is claimed to be a theorem) but not a false proof. The word you need is refute: a false proof (of a real or false theorem) may be refuted. A false theorem may also be refuted, by disproving it (though not by refuting a false proof of it). – John Bentin Oct 13 '16 at 12:48
• en.wikipedia.org/wiki/All_horses_are_the_same_color – v7d8dpo4 Oct 13 '16 at 16:12

The problem is here:

Write $k + 1 = i + j$, where $i$ and $j$ are natural numbers less than $k + 1$.

If $k = 0$, then you are trying to write $1 = i+j$ where $i$ and $j$ are natural numbers less than $1$. The only option for $i$ and $j$ is $0$, but $0+0 \ne 1$.

As a general rule: For fake induction proofs, find the smallest case where the conclusion does not hold, and then do each step in detail with the corresponding numbers inserted, so that it should proof that exact case. That way you will almost always quickly find the problem.

In this case, the smallest failing case is $P(1)$, as that claims $7\cdot 1=0$ which is clearly wrong. Therefore the number to look at is $k+1=1$, that is, $k=0$.

So let's look at the inductive step, and insert $k=0$:

Inductive step: Assume $7\cdot j=0$ for all natural numbers $j$ where $0\le j\le 0$ (induction hypothesis). Show $P(k+1): 7(k+1)=0$.

The only number with $0\le j\le 0$ is $j=0$, so the induction hypothesis is that $7\cdot 0=0$, which clearly is true.

Write $0+1=i+j$, where $i$ and $j$ are natural numbers less than $k+1$.

The only natural number less than $1$ is $0$. Therefore we have to write $0+1 = 0+0$ … oops, that's not right! Error found!

• This is the best answer to the actually posed question, how to find the error in an induction proof of an obviously false statement. The only thing to add is that this method will find at least one problem with the inductive argument, but there could also be other problems that appear at larger values of n. – zyx Oct 13 '16 at 20:46
• @zyx by the well-ordering principle there is a smallest such problematic number... – djechlin Oct 13 '16 at 20:50
• It's an interesting question whether all errors in the induction step can be found by examining a finite set of "problematic numbers" and whether this can be done effectively (given that the conclusion is false for all $n \geq N$ for a known $N$). I am not sure if just considering larger and larger problematic numbers will catch everything. @djechlin – zyx Oct 13 '16 at 20:59
• @zyx you're asking whether a proof can contain an infinite number of errors? I mean each step could also use the fact that $n^2 > n$ for all $n \geq 0$, Then each step would have an error in it, and there would be an infinite number of errors. One could fix the error that $n^2 > n$ is false for $n > 0$ by replacing to the less obviously fallacious claim that $n^2 > n$ only when $n \geq 1$, so that each error would also have a fix that would leave an infinite number of errors. – djechlin Oct 13 '16 at 21:05
• @djechlin, the proof is finite, so has only a finite number of erroneous statements. I'm asking whether there is a form of the "problematic numbers" strategy that is guaranteed to uncover them all. I suspect there is not unless you have some way of computing, for every step of the proof, infinitely many $n$ where the inductive argument requires that step to be true in order for the implication $n \to n+1$ to hold. – zyx Oct 13 '16 at 21:33

Actually, the problem is in the base case — in particular, $P(0)$ isn't enough of a base case.

The inductive step for proving $P(n)$ depends on writing $n$ as a sum of two smaller natural numbers; you can do this when $n \geq 2$, but you can't do this when $n=1$.

If you have both $P(0)$ and $P(1)$ in the base case, that's enough to make the inductive step work.

(of course, you can't prove $P(1)$, so you can't prove the base case)

• This is true but problematic for people learning induction for the first time. You're right that fixing the proof (if it were possible) would require a bigger base case. But the actual error (as in, problematic line in the proof) is in the step case (which has been pointed out several times). – Richard Rast Oct 13 '16 at 15:57

Hint: $1=1+0\neq 0+0$. Study $P(1)$.

Whenever you have to check induction proofs, you should apply the general case in order to prove the first step of the induction. In this particular situation you want to prove P(1): $7*1 = 0$.

Write $k+1=i+j$, where $i$ and $j$ are natural numbers less than $k+1$.

In the first step, this means:

Write $1 = i+j$ where $i,j$ are natural numbers less than $1$.

This statement already shows where the problem is in the induction proof, because the only natural number less than 1 is 0, and 1 cannot be expressed as $0 + 0$.

• This is good advice, but not always sufficient; it's possible to write a bogus inductive "proof" where the general case only fails after two or more steps. For a simple example, consider an attempted proof that all odd numbers greater than 1 are primes, with the base case "3 is prime" and the (obviously false) induction step "$n$ is prime $\implies$ $n+2$ is prime". – Ilmari Karonen Oct 13 '16 at 19:46


# Solving a question by using special products (Students debate to Teacher)

So today,we got back our exam papers,and we found a question marked wrongly and teacher said that it is wrong.We all students do NOT believe this.So here is what happened.

Before reading the next part,this is what we ONLY know (learnt) about rules of special products.

(Under school secondary 2 learning in Singapore)

• Rule $1$: $a^2+2ab+b^2=(a+b)^2$

• Rule $2$ :$a^2-2ab+b^2=(a-b)^2$

• Rule $3$: $a^2-b^2=(a+b)(a-b)$

From the exam paper:

Evaluate $10.2^2-9.8^2$ by using ONLY rules of special products.

Correct solution: $(10.2+9.8)(10.2-9.8)=(20)(0.4)=8$ (Rule $3$)

Student wrong (Marked as wrong) alternate solution: \begin{align*} (10+0.2)^2-(10-0.2)^2 &=[10^2+2(10)(0.2)+0.2^2]-[10^2-2(10)(0.2)+0.2^2]\\&=100+4+0.04-(100-4+0.04)\\&=104.04-96.04\\&=8\end{align*}(Rules $1$ and $2$)

The question did NOT ask for the easiest and fastest way (and both solution uses ONLY rules of special products) to solve but yet why is student solution wrong? Teacher told us,"Aiya, why need to do so complicated one?" yet she did not answer why is the answer wrong. I debated to her so long but to no avail.

Can anybody think of why the student solution is wrong?

• Your solution is correct. She's butt-hurt because she failed to account for other possible correct solutions. – Git Gud Mar 4 '15 at 16:29
• The teacher is silly ! Both are equivalent things done in two ways! – Shakul Pathak Mar 4 '15 at 16:33
• The unexpected answer is well within the realm of correct. This teacher is out-of-bounds. I teach; I also like unexpected answers that work and strongly encourage this kind of thinking. – ncmathsadist Mar 4 '15 at 19:59
• I think that the second approach is arguably better as it demonstrates knowledge of two of the three rules rather than just one. If you really want to split hairs you could argue that since the question refers to "rules", i.e. plural, the first answer is not even valid as it only uses a single rule! The question should have been disambiguated, e.g. "using exactly one rule", or "using one or more rules". The fact that this was not done indicates a lack of rigour. – Marconius Jul 9 '15 at 12:07
• Whoops,forgot to look back at this post.This is solved.We got our exam marks :D – ministic2001 Oct 18 '15 at 12:45


# Euler function and sums

I've come across a nice little combinatorial problem. Firstly we have vector $(1, 1)$. On each iteration for each two consecutive numbers in the vector we add their sum between them: $$(1, 2, 1)$$ $$(1, 3, 2, 3, 1)$$ $$(1, 4, 3, 5, 2, 5, 3, 4, 1)$$ $$(1, 5, 4, 7, 3, 8, 5, 7, 2, 7, 5, 8, 3, 7, 4, 5, 1)$$ ad infinitum.

Here's the question. How many times number $n$ will be written in the final vector? I'v computed the answers for $1, \dots, 15$: $$2, 1, 2, 2, 4, 2, 6, 4, 6, 4, 10, 4, 12, 6, 8$$

And this values coincide with the values of Euler's totient function. But how to prove that the answer is always $\phi(n)$? And can one simply derive this idea without explicitly writing first values?

• For anyone who wants to explore this with Mathematica, v={1,2,1}; S[v_]:=Table[v[[k]]+v[[k+1]],{k,1,Length[v]-1}]; F[v_]:=Riffle[v,S[v]]; Nest[F,v,n] will give the nth vector – Benedict W. J. Irwin Feb 22 '18 at 14:35
• If we start with the vector $(1,2)$, we seem to end up with $\lfloor n/\tau(n)\rfloor$, i.e. OEIS sequence A078709, where $\tau(n)$ is the number of divisors function. – Benedict W. J. Irwin Feb 22 '18 at 14:58
• Ok, the iteration from $(1,2)$ I just suggested is not true like 1,1,1,1,2,1,3,2,3,2,5,2,6,3,4,4,8,3,9,4, whereas the real sequence A078709 goes like 1,1,1,1,2,1,3,2,3,2,5,2,6,3,3,3,8,3,9,3, so they are not the same, but very close. Your pattern for $\phi(n)$ seems to hold at least to $n=26$ – Benedict W. J. Irwin Feb 22 '18 at 15:11
• These are the denominators in Farey Series. They generate all fractions between $0/1$ and $1/1$. You get the numerators by starting with $0,1$ instead of $1,1$. All these fractions turn out to be in lowest terms, so $\tau(n)$ of them will have $n$ in the denominator. – Michael Feb 22 '18 at 15:52

Let's first name the vectors $v_1, v_2$ and so on.

First of all we can note that consecutive numbers in the vector are coprime by induction. The base case is trivial. Now assume it holds for $v_{n-1}$ and consider $v_n$. Let $a,b$ are consecutive number in $v_n$. Then because of the way numbers are added to the vectors we have $a,|b-a|$ or $b, |a-b|$ appearing as consecutive numbers in $v_{n-1}$. WLOG let it be the first option. But then $\gcd(a,b) = \gcd(a,|b-a|) =1$. Hence the proof.

Now let's notice that a number $n$ can appear in the vector if $a,b$ appear as consecutive integer in a vector and $a+b = n$. But from above we have that neither of $a,b$ can share a common factor of $n$. Now using induction on $n$ we'll prove that such a combination of consecutive integers $(a,b)$ can appear only in a unique vector. The base case for $n=1,2,3$ is true. Now assume that we have consecutive integers $a,b$ appearing in different vectors, namely $v_i$ and $v_j$ such that both sum to $n$. WLOG let $a<b$. Then going back in $v_{i-1}$ and $v_{j-1}$ we have the consecutive integers $a, b-a$ in both vectors. But this contradicts the inductive hypothesis as both pairs sum to $b<n$.

From here we can conclude that an integer $n$ will be at most $\phi(n)$ in the final vector. Now it remains to show that each combination of $a,b$ s.t. $\gcd(a,b) = 1$ appears as consecutive integers at some point.

[UPDATE]

Similar as above we'll prove the last claim by induction on $n$. The base cases $n=1,2,3$ are trivially true. Now assume the claim holds for all numbers less than $n$. Now let $a,b$ be any integers such that $a+b=n$ and $\gcd(a,b)=1$ and WLOG $a<b$. We know that they will appear in $v_i$ if and only if $a,b-a$ appear as consecutive integers in $v_i$. But by the inductive hypothesis we have that $a,b-a$ does appear as consecutive integers in a unique vector. Therfore $a,b$ appear too and in a unique vector. Hence the proof.

SUMMARY: We can count the pairs of the consecutive integers by their left member and summarizing from above we have that if $a<n$ and $\gcd(a,n) = 1$, then eventually the integers $a,n-a$ will appear only once in the vectors. Note that the case $n-a,a$ is counted by using $n-a$. Therefore each integers appears $\phi(n)$ times in the final vector.

• @Igor I just completed my answer, so I would be glad if you could check and possibly point out some mistakes, if any. – Stefan4024 Feb 22 '18 at 15:22
• Thank you for such a detailed answer. – Igor Feb 22 '18 at 23:02


# Expected Value of rolling mean of AR(1) process

Consider a random variable following an AR(1) model:

$$x_t = \mu+\rho x_{t-1} + \epsilon_t$$

The unconditional expected value of this proces is:

$$E(x)=\mu/(1-\rho)$$

Now take a rolling mean of $$x_t$$ of n periods including the current: $$\bar{x} = \frac{1}{n}\sum_{i=0}^{n-1} x_{t-i}$$

Based on Wiki: Sum of normally distributed random variables the expected value is still just the sum of the expected value for each $$x_t$$ which is given above, hence the expected value of $$\bar{x}_t$$ is:

$$E(\bar{x}_t)=1/n \sum_{i=0}^{n-1} E(x_{t-i})=n/n E(x_t)=E(x_t)$$

However, another approach is to roll back each observation in the summand to a common starting point using (this is with thanks to Aleksej at this post here): $$x_t = \mu \sum_{i=0}^{k-1} \rho^i + \sum_{i=0}^{k-1} \rho^i \epsilon_{t-i} + \rho^k x_{t-k}$$

Inserting this into the expression for $$\bar{x}_t$$ yields: $$\bar{x}_t=1/n\sum_{i=0}^{n-1}(\rho^{n-i}x_{t-n}+\mu\sum_{j=0}^{k-i-1}\rho^j+\sum_{j=0}^{k-i-1}\rho^{j}\epsilon_{t-i-j})$$

$$=1/n(\sum_{i=0}^{n-1}x_{t_n}\rho^{n-i}+ \mu\sum_{i=0}^{n-1}\sum_{j=0}^{n-i-1}\rho^{j}+\sum_{i=0}^{n-1}\sum_{j=0}^{n-i-1}\rho^j\epsilon_{t-i-j})$$

Taking the expection to this yields: $$E(\bar{x})=1/n(E[x_{x-t}]\sum_{i=0}^{n-1}\rho^{n-i}+\mu\sum_{i=0}^{n-1}\sum_{j=0}^{n-i-1}\rho^{j})$$

$$=\frac{1}{n}(\frac{\mu}{1-\rho}\sum_{i=0}^{n-1}\rho^{n-i}+\mu\sum_{i=0}^{n-1}\sum_{j=0}^{n-i-1}\rho^{j})$$ But this does not coincide with the previous results using the sum of normal distributed variables.

Can someone point out, if I have made a mistake and which method is the correct one?

EDIT: Simulation study

Based on the comment from mlofton, I have conducted a simulation study, and I find that the expected difference and the simulated difference are indeed very close. Using the formula from the model yields the same as average of the sum of the unconditional means for $$x_t$$

import matplotlib.pyplot as plt
import numpy as np

def simulate_ar_process(initial_value=None, intercept=-0.01, slope=0.5, noise_vol=0.005, obs=100):
# If initial value is None use the unconditional mean
if initial_value is None:
initial_value = intercept / (1-slope)

errors = np.random.normal(0, noise_vol, obs)
pd_levels = []

# Estimate PD-levels
for err in errors:
if len(pd_levels) == 0:
pd_levels.append(initial_value + err)
else:
new_obs = intercept + pd_levels[-1] * slope + err
pd_levels.append(new_obs)

return pd_levels

def calculate_rolling_mean(pd_levels, look_back=20, burn_in=0):
mas = []
for i in range(look_back+burn_in, len(pd_levels)):
current_range = pd_levels[i-look_back:i]
mas.append(np.mean(current_range))

return np.mean(mas)

intercept = -0.01
slope = 0.9
noise_vol = 0.005
obs = 1000
look_back = 20

average_of_rolling_means = []
average_of_x = []
for i in range(100):
x = simulate_ar_process(intercept=intercept, slope=slope, noise_vol=noise_vol, obs=obs)
average_of_rolling_means.append(calculate_rolling_mean(x, look_back=look_back))
average_of_x.append(np.mean(x[20:]))

plt.hist(average_of_x, label="Average of x", alpha=0.3)
plt.hist(average_of_rolling_means, label="Average of rolling mean", alpha=0.3)
plt.legend()

# Calcualte difference
diffs = [x-y for x, y in zip(average_of_x, average_of_rolling_means)]
plt.hist(diffs)
np.average(diffs)

# Theoretical diff
unconditional_mean = intercept / (1-slope)
unconditional_mean_mva = unconditional_mean * np.sum([slope**(look_back-i) for i in range(0, look_back)])
unconditional_mean_mva += intercept * np.sum([slope ** j for i in range(0, look_back) for j in range(0, look_back-i)])
unconditional_mean_mva /= look_back

print("Expected difference: {}".format(unconditional_mean - unconditional_mean_mva))
print("Found difference: {}".format(np.average(diffs)))
#Expected difference: 0.0
#Found difference: 9.324487397407016e-06

However, I cannot see from the formula above, that $$E(\bar{x})=E(x)$$ Can someone point me in the right direction in the deviation of this equality?

• Hi: I wouldn't expect them to be the same because, for the one where you use the result from the Wiki, you're not taking the model into account. The Wiki result would be true for any iid random variable which is normally distributed. The second result-derivation applies due to the model specifics for $x_t$ and $x_t$ is not iid given the model. – mlofton Feb 24 at 18:00
• Hi: Thank you for your comment. I have updated my answer with a simulation study. If you post your original comment and perhaps can give a pointer or two with the remaining part, I'll mark it as the answer. – RVA92 Feb 25 at 7:47
• @mlofton: the linearity of expectation always hold, regardless of the dependence structure between the random variables. – Dayne Feb 26 at 5:50
• Nice derivation and you are absolutely correlation about the correlation structure. My mistake. But when the OP wrote $E(\bar x) = E(x_{i})$, the expectation of the $x_t$ in an AR(1) are not the same unless one refers to the long term unconditional mean which is exactly what you used in your nice derivation. @RVA92: When you wrote the expression for $\bar{x}$, I didn't know that you were referring to the unconditional mean of $x$. The expression does hold in that case. After I send this, I delete my other comments and answer since I was totally confused by the equality. – mlofton Feb 27 at 18:18
• In case, above confuses things even more, I was referring to E(x_t) given the previous $x_t$. That expectation is not constant. because of the dependence structure in the AR(1). For the long term mean, you're absolutely correct. – mlofton Feb 27 at 18:21

There seems to be an error in your calculations. Using a slightly different notation for clarity.

Let the AR(1) process be: $$X_t = \mu + \phi X_{t-1} + e_t$$

Define mean of $$n$$ periods be:

$$\mu_n \equiv \frac{1}{n}\sum\limits_{t=1}^n X_t$$

The expectation operator is linear regardless of dependence structure between the random variables. So, your first expression is correct:

$$\mathbb E(\mu_n) = \mathbb E(X_t) = \frac{\mu}{1-\phi}$$

Second approach, for $$t>1$$:

\begin{align} X_t &= \mu + \phi X_{t-1} + e_t \\ &= \phi^{t-1}X_1+ \sum\limits_{i=0}^{t-2} \Big(\phi^{i}\mu +\phi^{i}e_{t-i}\Big) \tag{1} \end{align}

Using $$(1)$$: \begin{align} n\mu_n &= \sum\limits_{t=1}^n\phi^{t-1}X_1 +\sum\limits_{t=2}^n \sum\limits_{i=0}^{t-2} \Big(\phi^{i}\mu +\phi^{i}e_{t-i}\Big) \\ &= \sum\limits_{t=1}^n\phi^{t-1}X_1 +\sum\limits_{t=2}^n \sum\limits_{i=0}^{t-2} \phi^{i}\mu +\sum\limits_{t=2}^n \sum\limits_{i=0}^{t-2}\phi^{i}e_{t-i} \tag{2} \end{align}

In equation (2) the middle term is non-random and the last term has expected value of $$0$$. So, \begin{align} n\mathbb E(\mu_n) &= \sum\limits_{t=1}^n\phi^{t-1}\mathbb E(X_1) + \sum\limits_{t=2}^n \sum\limits_{i=0}^{t-2} \phi^{i}\mu \\ &= \frac{\mu}{1-\phi}\sum\limits_{t=1}^n\phi^{t-1} + \sum\limits_{t=2}^n \sum\limits_{i=0}^{t-2} \phi^{i}\mu \\ &= \frac{\mu}{1-\phi}\sum\limits_{t=1}^n\phi^{t-1} + \sum\limits_{t=2}^n \bigg( \frac{\mu(1-\phi^{t-1})}{1-\phi}\bigg) \\ &= \frac{\mu}{1-\phi} \bigg( \sum\limits_{t=1}^n\phi^{t-1} + \sum\limits_{t=2}^n (1-\phi^{t-1})\bigg) \\ &=n \frac{\mu}{1-\phi} \end{align}

• Hi Dayne: I got it to match using your approach, so thank you. – RVA92 Feb 26 at 7:46


# Curves in the Complex Plane

Suppose the continuous real-valued functions $x = x(t),$ $y = y(t),$ $a \leq t \leq b,$ are parametric equations of a curve $C$ in the complex plane. If we use these equations as the real and imaginary parts in $z = x+iy$, we can describe the points $z$ on $C$ by means of a complex-valued function of a real variable $t$ called a parametrization of $C$: $$\label{parcurve} z(t) = x(t) + i y(t), \quad a\leq t\leq b.$$

The point $z(a) = x(a) + i y(a)$ or $z_0 = (x(a), y(a))$ is called the initial point of $C$ and $z(b) = x(b) + iy(b)$ or $z_1 = (x(b), y(b))$ is its terminal point. The expression $z(t) = x(t) + iy(t)$ could also be interpreted as a two-dimensional vector function. Consequently, $z(a)$ and $z(b)$ can be interpreted as position vectors. As $t$ varies from $t = a$ to $t = b$ we can envision the curve $C$ being traced out by the moving arrowhead of $z(t)$. This can be appreciated in the following applet with $0 \leq t\leq 1$.

Press Start to animate. You can move the points to change the curve.

For example, the parametric equations $x = \cos t,$ $y = \sin t,$ $0 \leq t \leq 2\pi,$ describe a unit circle centered at the origin. A parametrization of this circle is $z(t) = \cos t + i \sin t,$ or $z(t) = e^{it},$ $0 \leq t \leq 2\pi$.

Press Start to animate.

## Contours

The notions of curves in the complex plane that are smooth, piecewise smooth, simple, closed, and simple closed are easily formulated in terms of the vector function (\ref{parcurve}). Suppose the derivative of (\ref{parcurve}) is $z'(t) = x'(t) + iy'(t).$ We say a curve $C$ in the complex plane is smooth if $z'(t)$ is continuous and never zero in the interval $a \leq t \leq b.$ As shown in Figure 2, since the vector $z'(t)$ is not zero at any point $P$ on $C$, the vector $z'(t)$ is tangent to $C$ at $P$. Thus, a smooth curve has a continuously turning tangent; or in other words, a smooth curve can have no sharp corners or cusps. See Figure 2.

A piecewise smooth curve $C$ has a continuously turning tangent, except possibly at the points where the component smooth curves $C_1, C_2, \ldots, C_n$ are joined together.

A curve $C$ in the complex plane is said to be a simple if $z(t_1) \neq z(t_2)$ for $t_1 \neq t_2,$ except possibly for $t = a$ and $t = b.$ $C$ is a closed curve if $z(a) = z(b).$

$C$ is a simple closed curve if $z(t_1)\neq z(t_2)$ for $t_1\neq t_2$ and $z(a) = z(b).$

In complex analysis, a piecewise smooth curve $C$ is called a contour or path. We define the positive direction on a contour $C$ to be the direction on the curve corresponding to increasing values of the parameter $t$. It is also said that the curve $C$ has positive orientation. In the case of a simple closed countour $C$, the positive direction corresponds to the counter-clockwise direction. For example, the circle $z(t) = e^{it}$, $0 \leq t \leq 2\pi$, has positive orientation. The negative direction on a contour $C$ is the direction opposite the positive direction. If $C$ has an orientation, the opposite curve, that is, a curve with opposite orientation, is denoted by $−C$. On a simple closed curve, the negative direction corresponds to the clockwise direction. For instance, the circle $z(t) = e^{-it}$, $0 \leq t \leq 2\pi$, has negative orientation.

Press Start to animate. You can change the direction of $C$.

Exercise: There is no unique parametrization for a contour $C$. You should verify that \begin{eqnarray*} z(t)&=&e^{it} =\cos t+i\sin t,\; 0\leq t\leq 2\pi \\ z(t)&=& e^{2\pi it} =\cos \left(2 \pi t\right) +i \sin \left(2 \pi t\right),\; 0\leq t\leq 1\\ z(t)&=&e^{\pi/2 it} =\cos \left( \frac{\pi}{2} t\right)+i \sin \left( \frac{\pi}{2} t\right),\; 0\leq t \leq 4 \end{eqnarray*} are all parametrizations, oriented in the positive direction, for the unit circle $|z| = 1$.

NEXT: Complex Integration


# How To Evaluate Nth Roots

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Finding Real nth Roots of a (n > 1 and n is an even integer) Finding Real nth Roots of a (n > 1 and n is an odd integer) Evaluating Expressions with Rational Exponents. Please try again later. 86 21 Take fourth roots of each side. 1 nth Roots and Rational Exponents 401 nth Roots and Rational Exponents EVALUATING NTH ROOTS You can extend the concept of a square root to other types of roots. Section 1-5 : Integrals Involving Roots. I CANNOT use a calculator. A c PA IlqlC Brzi Bg whxtSs K mrSeLs OeJruv Ne7dz. If some numbers in x are negative, n must be odd. Using this formula, we will prove that for all nonzero complex numbers $z \in \mathbb{C}$ there exists \$n. 2 in the text. They can be used instead of the roots such as the square root. notebook March 24, 2015 a. Having an nth root of some number is equivalent to taking that number to the 1/n power. guarantees that a zero or root of f(x) lies in (a,b). Press [2nd][ x 2 ] to select a square root and type the expression you would like to evaluate. Write an explicit or recursive formula for the nth term of an arithmetic sequence, given the value of several of its terms. the group we get is 10 and 648 find the no. We find nth term of the sequence in term of n. It looks quite tedious to do by hand, but the algorithm exists for any root and is similar to the square root one. MATLAB represents polynomials as row vectors containing coefficients ordered by descending powers. Rationalizing the Denominator. 23 = 8 53 = 125 1713 = 5000211 4. "Novel Algorithm for 'Nth Root of Number' using Multinomial Expansion" by Vitthal Jadhav, March 2013 This gives a general algorithm to extract the nth root of any number. Is it possible to calculate n complex roots of a given number using Python? I've shortly checked it, and it looks like Python gives me wrong/incomplete answers: (-27. We’ll start with integer powers of $$z = r{{\bf{e}}^{i\theta }}$$ since they are easy enough. The most common root is the square root. The function thus has a branch cut along the negative half real axis. 12,921 views subscribe 5. Notice that the cursor will stay under the radical sign until you press the right-arrow key (see the last line of the third screen). First we develop the square root method in table form and the see that how it is work after that We have developed division method of square root in formula form. whose cube root is less than or equal to 10. Identify and evaluate square and cube roots. Evaluate all powers from left to right. If the length of p is n+1 then the polynomial is described by:. Algebrator free, 'add subtract multiply divide' worksheets, properties of exponents lesson plan. chebyshev)¶ New in version 1. For example, 4 is a square root of 16 because 4 • 4 or 42 = 16. com - id: 77bdd1-YThlY. Thus, we can think of taking the square root of any positive number, x. Fast Fourier Transformation for poynomial multiplication. Later in this section we will see that using exponent 1 n for nth root is compat- ible with the rules for integral exponents that we already know. Introduction to nth Roots; Radical Form vs. Step 1: Set the expression inside the square root greater than or equal to zero. yaymath 118,922 views. Algebra 2 Standard 12. If n = 2, the root is called square root. In this case, the power 'n' is a half because of the square root and the terms inside the square root can be simplified to a complex number in polar form. This is the video about how to evaluate square roots. 2 Simplify expressions in exponential form. We’ve already seen some integrals with roots in them. homework help on finding nth roots and rational exponents Before you read/watch/listen to “If You Can Read This I Can. Let's look at some: 1. Topical and themed;. This tool is used to calculate the output of almost all the mathematical expressions. And the square root of 25 times 3 is equal to the square root of 25 times the square root of 3. Unfortunately some improper integrals fails to fall under the scope of these tests but we will not deal with them here. How to Find Roots of Unity. Students will simplify nth roots completely, then they will color or draw on the penguin as indicated by their answer. Evaluate the line integral C of F dr, where C is given by the vector function r (t). A square root is defined as a number which when multiplied by itself gives a real non-negative number called a square. Rewrite the power of 16 as products of power of 8 as much as possible. The advantage of using exponents to express roots is that the rules of exponents can be applied to the expressions. Each root gives a particular exponential solution of Each root gives a particular exponential solution of the diﬀerential equation. 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# Metric Spaces: Why $L_\infty$ selects the maximum value

I have a basic question about the metric spaces. There are several metric spaces like $L_1$, $L_2$ to $L_\infty$. The $L_p$ metric is defined by the following equation:

$$d_p(x,y)=\left(\sum_{i=1}^{n}|x_i-y_i|^p \right)^{1/p}$$

Now, when $p = \infty$, then the equation of the matrix is following:

$$d_{\infty}(x,y)=\max_{i=1,2,\ldots, n}|x_i-y_i|$$

Now, I am taking the value of $x$ as $x = [1,2,3]^T$ (Transpose) and compute $L_p$ metric for, $p = 1,2$ and $\infty$.

My question is, why $L_\infty$ metric choose the maximum value.

• i wish you would use $\LaTeX$ so i could copy your equation to save time in answering. please learn the tools. – robert bristow-johnson Mar 3 '16 at 2:53
• I assume you have to go through a mathematical proof of this to fully understand it. Please do not make the mistake and try to explain this result by setting p to higher and higher values - this is an intuitive and often very useful thing to do, but it needs not give you the correct result. – M529 Mar 3 '16 at 9:18
• Shouldn't this migrate to math.SE? – Carl Witthoft Mar 3 '16 at 13:07
• No need for migration, the answer is already on math.SE: math.stackexchange.com/questions/109615/… – Jazzmaniac Mar 3 '16 at 14:19

In the general case, let $$x = (x_1,\dots,x_n)$$ be a finite-length vector (in a finite dimensional space). The finite sequence of absolute values $$|x_i|$$ does attain its maximum (because the sequence is finite), denoted $$M = \max_i |x_i|$$.

Let $$m$$ be the (exact) number of coordinates in $$x = (x_1,\dots,x_n)$$ whose absolute value is equal to $$M$$. Thus, $$1\le m\le n$$. Then, we can lower and upper bound the $$\ell_p$$ norm of $$x$$ as follows: $$(mM^p)^{\frac{1}{p}}\le\ell_p(x)\le (nM^p)^{\frac{1}{p}}\,.$$

Both $$m^\frac{1}{p}$$ and $$n^\frac{1}{p}$$ tend to $$1$$ as $$p\to\infty$$, thus $$\ell_p(x) \to M$$, by the squeeze theorem.

[EDIT] To provide more concrete substance, let us see what happen with your example: $$x=[1,2,3]^T$$ (the transposition does not change the result): $$\|x\|_p = d_p(x,0) = (1^p+2^p+3^p)^{1/p} = 3\times\left(\left(\frac{1}{3}\right)^p+\left(\frac{2}{3}\right)^p+1\right)^{1/p}\,.$$

Both $$\left(\frac{1}{3}\right)^p$$ and $$\left(\frac{2}{3}\right)^p$$ tend to $$0$$ as $$p \to \infty$$. Thus, $$\|x\|_p \to 3$$.

• That's the stuff! :) – Jazzmaniac Mar 3 '16 at 16:53

The $L_p$ norm is

$$d_p(\mathbf{x}, \mathbf{y}) \triangleq \left( \sum\limits_{i=1}^{n} |x_i - y_i|^p \right)^{\frac{1}{p}}$$

there exists a positive value that is the maximum value:

$$M \triangleq \max_{1 \le i \le n} |x_i - y_i|$$

now, suppose you divide both sides of the $L_p$ norm definition by that positive value,

$$\frac{d_p(\mathbf{x}, \mathbf{y})}{M} = \frac{1}{M} \left( \sum\limits_{i=1}^{n} |x_i - y_i|^p \right)^{\frac{1}{p}}$$

\begin{align} \frac{d_p(\mathbf{x}, \mathbf{y})}{M} & = \frac{1}{M} \left( \sum\limits_{i=1}^{n} |x_i - y_i|^p \right)^{\frac{1}{p}} \\ & = \left( \frac{1}{M^p} \right)^{\frac{1}{p}} \left( \sum\limits_{i=1}^{n} |x_i - y_i|^p \right)^{\frac{1}{p}} \\ & = \left( \frac{1}{M^p} \sum\limits_{i=1}^{n} |x_i - y_i|^p \right)^{\frac{1}{p}} \\ & = \left( \sum\limits_{i=1}^{n} \frac{1}{M^p} |x_i - y_i|^p \right)^{\frac{1}{p}} \\ & = \left( \sum\limits_{i=1}^{n} \left( \frac{|x_i - y_i|}{M} \right)^p \right)^{\frac{1}{p}} \\ \end{align}

now ask yourself, what will happen to the right-hand side of this equation as $p \to \infty$? all terms, except for the term that is equal to the maximum, will have value of less than 1. they will go to 0 as $p \to \infty$, but the term that has $|x_i - y_i| = M$, that term is equal to 1 and will remain 1 even as $p \to \infty$.

• Robert, what if there are two terms in the sum that are equal to $M$? Shouldn't the sum be equal to 2 in that case? – MBaz Mar 3 '16 at 3:48
• hi Robert, thanks. It's a very good answer. So, Suppose, I have two matrix: x = [1,2,3]T and Y = 0. And, I want to compute, L1, L2 and L-infinity. So, here, Max value (M according to your explanation) would be 3? Right? Just to make sure. Thanks. – Odrisso Mar 3 '16 at 4:11
• This is a very nice and intuitive explaination. However, I suppose it is not 100% mathematically sound, which is why @MBaz asked the question with the factor of 2, if your vector had two entries corresponding to M. I think the problem stems from not obeying the limit from the start on. Getting the factor 1/M into the brackets of the sum is critical and probably the point where it fails when p goes to infinity. – M529 Mar 3 '16 at 9:14
• You could maybe argue that the sum s is bounded between 1 and n. Hence you have lim p->inf s^(1/p) which is 1. – M529 Mar 3 '16 at 9:25
• @David, it doesn't matter if it's trivial. If you make false assumptions about the nature of something and don't explicitly consider the case that these assumptions fail, your argument is wrong. Yes, mathematics is pedantic. That's why it's so useful. – Jazzmaniac Mar 4 '16 at 19:05


# Number Theory6 Successive numbers no one is prime

#### Amer

##### Active member
is it possible to find a 6 Successive numbers like
x , x+1 , x+2 , x+3 ,x+4 ,x+5 such that one one is prime ?

Thanks

#### chisigma

##### Well-known member
is it possible to find a 6 Successive numbers like
x , x+1 , x+2 , x+3 ,x+4 ,x+5 such that one one is prime ?

Thanks
It is comfortable to verify that $x=5!=120$ satisfies the request... and the reason of that is easy to see...

Kind regards

$\chi$ $\sigma$

#### chisigma

##### Well-known member
Of course $x=5!$ is not the only and neither the 'smallest' solution. Setting $x=90$ You have 7 consecutive non prime numbers...

Kind regards

$\chi$ $\sigma$

#### chisigma

##### Well-known member
Why don't try to generalize the problem: given k, how to compute an n such that n, n+1, n+2,...,n+k are all non prime numbers?...

Kind regards

$\chi$ $\sigma$

#### chisigma

##### Well-known member
Why don't try to generalize the problem: given k, how to compute an n such that n, n+1, n+2,...,n+k are all non prime numbers?...
An easy way to get the result in the particular case where k is prime is based on the consideration that $k|n \implies k|(n+m k)$. Setting $n = 2 \cdot 3 \cdot 5 \cdot ... \cdot k$ we are sure that $n+2,n+3,...,n+k+1$ are all non prime numbers. For example...

$\displaystyle k=11 \implies n=2310 \implies 2312,2313,2314,2315,2316,2317,2318,2319,2320,2321,2322\ \text{are all non prime }$

Although 'easy' this method is often 'excessive' because the effective quantity consecutive non prime numbers can be greater. In the given example 2311 is prime so that the sequence starts at 2312 but 2323,2324,2325,2326,2327,2328,2329,2330,2331 and 2332 are non prime numbers [2333 is prime...] and the effective sequence's length is 20 [not 11]...

Kind regards

$\chi$ $\sigma$

#### Amer

##### Active member
Thanks very much, you are amazing. (f)

#### Amer

##### Active member
An easy way to get the result in the particular case where k is prime is based on the consideration that $k|n \implies k|(n+m k)$. Setting $n = 2 \cdot 3 \cdot 5 \cdot ... \cdot k$ we are sure that $n+2,n+3,...,n+k+1$ are all non prime numbers. For example...

$\displaystyle k=11 \implies n=2310 \implies 2312,2313,2314,2315,2316,2317,2318,2319,2320,2321,2322\ \text{are all non prime }$

Although 'easy' this method is often 'excessive' because the effective quantity consecutive non prime numbers can be greater. In the given example 2311 is prime so that the sequence starts at 2312 but 2323,2324,2325,2326,2327,2328,2329,2330,2331 and 2332 are non prime numbers [2333 is prime...] and the effective sequence's length is 20 [not 11]...

Kind regards

$\chi$ $\sigma$
nice one I get it
$$n +2 = 2.3.4...k +2 = 2(3.4...k +1 )$$ not prime
$$n+3 = 2.3.4...k + 3 = 3(2.4...k+1)$$ not prime
Thanks

#### soroban

##### Well-known member
Hello, Amer!

Is it possible to find a 6 consecutive integers numbers like
x , x+1 , x+2 , x+3 ,x+4 ,x+5 such that not one is prime? . Yes!

One solution is: .$$x \:=\:7!+2$$

. . $$\begin{array}{c}7!+2\text{ is divisible by 2} \\ 7!+3\text{ is divisible by 3} \\ 7!+4\text{ is divisible by 4} \\ 7!+5\text{ is divisible by 5} \\ 7!+6 \text{ is divisible by 6} \\ 7!+7\text{ is divisible by 7} \\ \end{array}$$

There is a simpler (and much longer) list:

. . $$\begin{array}{ccc} 114 &=& 2\cdot57 \\ 115 &=& 5\cdot 23 \\ 116 &=& 2\cdot 58 \\ 117 &=& 3\cdot39 \\ 118 &=& 2\cdot59 \\ 119 &=& 7\cdot17 \\ 120 &=& 2\cdot60 \\ 121 &=& 11\cdot11 \\ 122 &=& 2\cdot61 \\ 123 &=& 3\cdot41 \\ 124 &=& 2\cdot62 \\ 125 &=& 5\cdot25 \\ 126 &=& 2\cdot63 \end{array}$$

#### Amer

##### Active member
Hello, Amer!

One solution is: .$$x \:=\:7!+2$$

. . $$\begin{array}{c}7!+2\text{ is divisible by 2} \\ 7!+3\text{ is divisible by 3} \\ 7!+4\text{ is divisible by 4} \\ 7!+5\text{ is divisible by 5} \\ 7!+6 \text{ is divisible by 6} \\ 7!+7\text{ is divisible by 7} \\ \end{array}$$

There is a simpler (and much longer) list:

. . $$\begin{array}{ccc} 114 &=& 2\cdot57 \\ 115 &=& 5\cdot 23 \\ 116 &=& 2\cdot 58 \\ 117 &=& 3\cdot39 \\ 118 &=& 2\cdot59 \\ 119 &=& 7\cdot17 \\ 120 &=& 2\cdot60 \\ 121 &=& 11\cdot11 \\ 122 &=& 2\cdot61 \\ 123 &=& 3\cdot41 \\ 124 &=& 2\cdot62 \\ 125 &=& 5\cdot25 \\ 126 &=& 2\cdot63 \end{array}$$
Thanks, and Hello
we can make a list with k numbers which are not prime like that
$$k! + i$$ i=1,...,k


# Discrepancy when calculating LTI system output using inverse z-Transform

I'm given a difference equation, $y[n]-0.4y[n-1]=x[n]$, and asked to find the natural response $y_n[n]$, forced response $y_f[n]$ and complete response $y[n]$ if $x[n]=4 (0.25)^nu[n]$ and $y[0]=0$.

By one approach I've read, $$\text{characteristic equation:}\quad z-0.4=0 \quad\therefore y_n[n]=A(0.4)^n \\ \\ y_f[n]=B(0.25)^n \quad\text{substitute back into diffeq:}\\ B(0.25)^n -0.4B(0.25)^{n-1}=B(0.25)^n - \frac{0.4}{0.25}B(0.25)^n= 4(0.25)^nu[n]\\ \therefore B=-\frac{20}{3}, \quad y_f[n]=-\frac{20}{3}(0.25)^nu[n]\\ \therefore y[n] = A(0.4)^n-\frac{20}{3}(0.25)^nu[n]\\ y[0]=0 \rightarrow A=\frac{20}{3} \\ y[n]= \frac{20}{3}(0.4)^n-\frac{20}{3}(0.25)^nu[n]\\$$ Great! Alternatively, I should be able to determine $y[n]$ by taking the inverse z Transform of $Y(z)=H(z)X(z)$, which in this case is (I'm pretty sure) Y(z)= \frac{z}{z-0.4}\cdot 4 \frac{z}{z-0.25}\\ \begin{align} \frac{Y(z)}{z} &= \frac{4z}{(z-0.4)(z-0.25)}=\frac{C}{z-0.4}+\frac{D}{z-0.25}\\ &=\frac{32}{3} \frac{1}{z-0.4} - \frac{20}{3}\frac{1}{z-0.25}\\ \end{align}\\ \therefore Y(z)=\frac{1}{3} \left[\frac{32z}{z-0.4} - \frac{20z}{z-0.25}\right]\\ \therefore y[n]= \left[ \frac{32}{3}(0.4)^n - \frac{20}{3}(0.25)^n \right] u[n] which is almost the same, but not quite.

Where did I go wrong? Also, in the second approach, how do you take into account the initial condition of $y[0]=0$? And thirdly, in the first approach, is the first term of $y[n]$ multiplied by $u[n]$ or not? If so, how do you show this?

(for anyone wondering, this is not homework for a course, just self study...)

• Don't you mean $y[-1]=0$ instead of $y[0]=0$? – Matt L. Aug 20 '15 at 6:54
• The question says y[0]=0 but I guess its mistaken. But its not clear to me why y[0] can't be zero... – Westerley Aug 20 '15 at 17:02

The way to solve such problems is to use the unilateral $\mathcal{Z}$-transform, which allows you to take initial conditions into account. Note that the unilateral $\mathcal{Z}$-transform of $y[n-1]$ is

$$\mathcal{Z}\{y[n-1]\}(z)=z^{-1}Y(z)+y[-1]\tag{1}$$

where $Y(z)$ is the $\mathcal{Z}$-transform of $y[n]$. Using $(1)$ you can transform the given difference equation $y[n]-\frac25 y[n-1]=x[n]$, resulting in

$$Y(z)=\frac{X(z)}{1-\frac{2}{5}z^{-1}}+\frac{\frac{2}{5}\cdot y[-1]}{1-\frac{2}{5}z^{-1}}\tag{2}$$

With $X(z)$ given by

$$X(z)=\frac{4}{1-\frac14z^{-1}}\tag{3}$$

Eq. $(2)$ becomes (after partial fraction expansion)

$$Y(z)=\left(\frac{32}{3}+\frac{2y[-1]}{5}\right)\frac{1}{1-\frac{2}{5}z^{-1}}-\frac{20}{3}\frac{1}{1-\frac14z^{-1}}\tag{4}$$

From $(4)$, the resulting output sequence is

$$y[n]=\left(\frac{32}{3}+\frac{2y[-1]}{5}\right)\left(\frac25\right)^nu[n]-\frac{20}{3}\left(\frac14\right)^nu[n]\tag{5}$$

Eq. $(5)$ is the general form of the output sequence of the given system with the given input sequence. The value $y[-1]$ is the initial condition of the system, i.e. the state the system is in right before the input signal starts. With $y[-1]=0$ you obtain the same solution as you obtained using the inverse $\mathcal{Z}$-transform (neglecting any initial conditions by using $y[n-1]\Longleftrightarrow z^{-1}Y(z)$).

With the given constraint $y[0]=0$, you can obtain the required value of $y[-1]$ by considering the difference equation at $n=0$:

$$y[0]-\frac25 y[-1]=x[0]=4$$

which gives $y[-1]=-10$. Plugging that value into Eq. $(5)$ finally gives the result

$$y[n]=\frac{20}{3}\left(\frac25\right)^nu[n]-\frac{20}{3}\left(\frac14\right)^nu[n]\tag{6}$$

As Matt L pointed out, your initial condition is wrong. y[0] is NOT 0, it's 4. This is obvious from the difference equation. Once you use y[0] = 4, both results come out to be the same.

• As mentioned in my comment to @matt, the question does say y[0]=0 but it's likely wrong... but its not clear to me why y[0] CAN'T be zero... I see that x[0] is 4, but why can't the natural response at n=0 cancel out that? – Westerley Aug 20 '15 at 17:03
• Note that $y[0]$ can be zero by an appropriate choice of the initial condition $y[-1]$ (see my answer). – Matt L. Aug 21 '15 at 7:48


# Number of ways to divide $10$ children into two teams of $5$?

Consider the following two examples from A First Course in Probability by Sheldon Ross,

EXAMPLE 5b Ten children are to be divided into an A team and a B team of 5 each. The A team will play in one league and the B team in another. How many different divisions are possible?

Solution by the Book itself There are $$\frac{10!}{5!5!} = 252$$ possible divisions.

My Solution and Reasoning We first choose five children which can be done in $$C(10, 5)$$ possible ways. Then we decide which team(either A or B) to assign these children to, which can be done in $$2!$$ ways. Finally we assign the remaining five children to the remaining team, depending on the second level(if the first five children were assigned to A, then the remaining five should be assigned to B, and if the first five were assigned to B, the remaining to A). Therefore this level has only $$1$$ possible way, and hence the final answer is $$C(10, 5) \times 2! \times 1 = 252 \times 2 \times 1 = 504$$.

EXAMPLE 5c In order to play a game of basketball, 10 children at a playground divide themselves into two teams of 5 each. How many different divisions are possible?

Solution by the Book itself Note that this example is different from Example 5b because now the order of the two teams is irrelevant. That is, there is no A and B team, but just a division consisting of 2 groups of 5 each. Hence, the desired answer is

$$\frac{10!/(5!5!)}{2!} = 126$$

My Solution and Reasoning First we choose 5 children for one team which can be done is $$C(10, 5)$$ ways and then we put the remaining children in the other team, which is done in $$1$$ way. Thus the answer is $$C(10, 5) \times 1 = 252$$, which can also be obtained by dividing the answer to Example 5b by 2, or in other words, removing the order of the groups in Example 5b.

There seems to be a conceptual misunderstanding in my way of thinking, but what is it?

PARTICULAR PROBLEM

For your particular problem, for labeled teams

• label the children $$0\; through\;$$9
• choose $$5$$ for team $$A$$, say $$13579$$, team $$B$$ automatically $$02468$$
• This is different from $$02468$$ in Team $$A$$ and $$13579$$ for team $$B$$, so total $$\binom{10}5$$ choices

But if teams are unlabeled $$[13579-02468] \equiv [02468-13579]$$
hence division by $$2$$

GENERAL GUIDANCE

It is important to have a clear concept over combinatorics of teams depending on whether they are labeled or unlabeled

To take a more complex example for greater clarity, suppose you are to choose $$4$$ teams of $$3$$ each from $$12$$ players

• If the teams are labeled, eg Horses, Greyhounds , Panthers, etc
the answer is $$\binom{12}3\binom93\binom63\binom33$$
which can be variously written as $$\binom{12}{3,3,3,3}\; or\; \dfrac{12!}{3!3!3!3!}$$

• If the teams are unlabeled, we shall have to divide by $$4!$$, to remove permutations between identical teams

• An important point to note is that unlabeled teams in effect become labeled if sizes differ, or composition differs (eg boys' team or girls' team)

Note that in both casees your answer is doubled from the book's solution so you should question whether you really ought to multiply by $$2!=2$$.

Let's consider a simpler case of 4 children $$\{1,2,3,4\}$$ and the teams are of size 2. Then all possible options for Team A are $$\{1,2\},\{1,3\},\{1,4\},\{2,3\},\{2,4\},\{3,4\}$$ giving a total of $$\frac{4!}{2!2!}=6$$ options. What you are doing is counting each option twice; Say when you count $$\{1,2\}$$, you count once for this being assigned to $$A$$ and once for $$B$$. However, note that the event of $$\{1,2\}$$ belonging to $$B$$ is exactly the event that $$\{3,4\}$$ belongs to $$A$$ which we already count. Hence we should not multiply by 2.


# How to evaluate the sum $\sum_{n=1}^{\infty}\frac{1}{(2n+1)(2n+2)}\left(1+\frac{1}{2}+…+\frac{1}{n}\right)$

How to evaluate the sum: $$S=\sum_{n=1}^{\infty}\frac{1}{(2n+1)(2n+2)}\left(1+\frac{1}{2}+...+\frac{1}{n}\right)$$ Can anyone help me,I really appreciate it.

• is there a source for this sum? Or, just a practice problem? – karakfa Oct 20 '17 at 15:05
• @MarkViola Mathematica gives the sum of the series. I don't know what is the trick to evaluate it. I think this is a good problem. :) – Ixion Oct 20 '17 at 15:26
• Partial fraction decomposition – JohnColtraneisJC Oct 20 '17 at 15:27
• Since $\displaystyle\frac {\pi^2}{12}=\sum_{r=1}^\infty \frac 1{2r^2}$, and $\displaystyle\ln 2=\sum_{r=1}^\infty \frac {(-1)^{r+1}}{r}$ is it possible to transform the given summation $\displaystyle \sum_{n=1}^\infty \sum_{r=1}^n \frac 1{(2n+1)(2n+2)r}$into $\displaystyle\sum_{r=1}^\infty \frac 1{2r^2}-\left(\sum_{r=1}^\infty \frac {(-1)^{r+1}}{r}\right)^2$? If so, then we are done. – hypergeometric Oct 21 '17 at 15:35
• I am voting for reopening this question since, despite a lack of efforts from the OP, it led to a number of collateral questions, and I think it is correct to leave this original question as a reference. – Jack D'Aurizio Oct 26 '17 at 15:57

A different view on the same problem: $$\sum_{n\geq 1}\frac{x^n}{n}=-\log(1-x),\qquad \sum_{n\geq 1}H_n x^n = \frac{-\log(1-x)}{1-x} \tag{A}$$ $$\sum_{n\geq 1}H_n x^{2n} = \frac{-\log(1-x^2)}{1-x^2}\tag{B}$$
$$\begin{eqnarray*}\sum_{n\geq 1}H_n\left(\frac{1}{2n+1}-\frac{1}{2n+2}\right) &=& \int_{0}^{1}\frac{-\log(1-x^2)}{1+x}\,dx\\&=&-\tfrac{1}{2}\log^22+\int_{0}^{1}\frac{-\log(1-x)}{1+x}\,dx\\&=&-\tfrac{1}{2}\log^22+\int_{0}^{1}\frac{-\log(x)}{2-x}\,dx\tag{C}\end{eqnarray*}$$ and by differentiation under the integral sign, the last integral is related to the series $$\sum_{n\geq 1}\frac{1}{n^2 2^n}=\text{Li}_2\left(\tfrac{1}{2}\right)\stackrel{(*)}{=}\tfrac{\pi^2}{12}-\tfrac{\log^2 2}{2} \tag{D}$$ where $(*)$ follows from the dilogarithm reflection formula, proved here.
It is evident that $$S=\sum_{n=1}^{\infty}\frac{1}{(2n+1)(2n+2)}\left(1+\frac{1}{2}+...+\frac{1}{n}\right) = \frac{\pi^2}{12} - \ln^{2}2$$ and can be evaluated by following the pattern:
Consider the series $$S(x) = \sum_{n=1}^{\infty}\frac{H_{n} \, x^{2n+2}}{(2n+1)(2n+2)}$$ which upon differentiation leads to $S(0) = 0$, $S'(0) = 0$, \begin{align} S(x) &= \sum_{n=1}^{\infty}\frac{H_{n} \, x^{2n+2}}{(2n+1)(2n+2)} \\ S'(x) &= \sum_{n=1}^{\infty}\frac{H_{n} \, x^{2n+1}}{(2n+1)} \\ S''(x) &= \sum_{n=1}^{\infty} H_{n} \, x^{2n} = - \frac{\ln(1- x^2)}{1-x^2}. \end{align} Now, $$- 2 \, S''(x) = \frac{\ln(1-x)}{1-x} + \frac{\ln(1-x)}{1+x} + \frac{\ln(1+x)}{1-x} + \frac{\ln(1+x)}{1+x}$$ which, upon integration, leads to \begin{align} - 4 \, S'(x) &= \ln^{2}(1 + x) - \ln^{2}(1-x) + 2 \, Li_{2}\left(\frac{1-x}{2}\right) - 2 \, Li_{2}\left(\frac{1+x}{2}\right) + \ln4 \, \ln\left(\frac{1+x}{1-x}\right). \end{align} Integrating again leads to $S(x)$. The integrals \begin{align} \int_{0}^{x} \ln^{2}(1-t) \, dt &= (x-1) \, (\ln^{2}(1-x) - 2 \ln(1-x) + 2) + 2 \\ \int_{0}^{x} \ln^{2}(1+t) \, dt &= (x+1) \, (\ln^{2}(1+x) - 2 \ln(1+x) + 2) - 2 \\ \int_{0}^{x} \ln\left(\frac{1+t}{1-t}\right) \, dt &= x \, \ln\left(\frac{1+x}{1-x}\right) + \ln(1-x^2) \\ \int_{0}^{x} Li_{2}\left(\frac{1+t}{2}\right) \, dt &= (1+x) \, Li_{2}\left(\frac{1+x}{2}\right) + x \, \ln\left(\frac{1-x}{2}\right) - \ln(1-x) -x - Li_{2}\left(\frac{1}{2}\right) \\ \int_{0}^{x} Li_{2}\left(\frac{1-t}{2}\right) \, dt &= (x-1) \, Li_{2}\left(\frac{1-x}{2}\right) + (x+1) \, \ln\left(\frac{1+x}{2}\right) -x - Li_{2}\left(\frac{1}{2}\right) + \ln2 \end{align} are needed for the evaluation. Once $S(x)$ is determined set $x=1$ to obtain $$S(1) = \sum_{n=1}^{\infty}\frac{H_{n}}{(2n+1)(2n+2)} = \frac{\pi^2}{12} - \ln^{2}2$$
Following Leucipus' answer, I will use different approach to evaluate $S(1)$. Note \begin{eqnarray} S=S(1)&=&\int_0^1\int_0^xS''(x)dxt\\ &=&-\int_0^1\int_0^x\frac{\ln(1-x^2)}{1-x^2}dxdt\\ &=&-\int_0^1\int_x^1\frac{\ln(1-x^2)}{1-x^2}dtdx\\ &=&-\int_0^1(1-x)\frac{\ln(1-x^2)}{1-x^2}dx\\ &=&-\int_0^1\frac{\ln(1-x^2)}{1+x}dx\\ &=&-\int_0^1\frac{\ln(1-x)}{1+x}dx-\int_0^1\frac{\ln(1+x)}{1+x}dx\\ &=:&-I_1-I_2 \end{eqnarray} Now $$I_2=\int_0^1\ln(1+x)d\ln(1+x)=\frac{1}{2}\ln^2(1+x)\bigg|_0^1=\frac12\ln^22.$$ For $I_1$, under $t=1-x$, one has \begin{eqnarray} \int\frac{\ln t}{2-t}dt&=&-\int_0^1\ln xd\ln(2-t)\\ &=&-\ln x\ln(2-t)+\int\frac{\ln(2-t)}{t}dt\\ &=&-\ln x\ln(2-t)+\int\frac{\ln2+\ln(1-\frac12t)}{t}dt\\ &=&-\ln x\ln(2-t)+\ln2\ln t+Li_2(\frac t2)+C \end{eqnarray} \begin{eqnarray} I_1&=&\int_0^1\frac{\ln t}{2-t}dt=-\int_0^1\ln xd\ln(2-t)\\ &=&-\ln x\ln(2-t)+\ln2\ln t+Li_2(\frac t2)\bigg|_0^1 &=&-\frac{\pi^2}{12}+\frac12\ln^22. \end{eqnarray} Thus $$S=\frac{\pi^2}{12}-\ln^22.$$


# Why is $\{0, 2\}$ a subgroup of $\mathbb Z_4$?

I feel like this should be obvious but why is $$\{0, 2\}$$ a subgroup of $$\mathbb Z_4$$? So, $$\langle 2\rangle=\{0,2\}$$. Shouldn't this set contain the inverse ($$-2$$)? Or does it have to do with the fact that $$(-2)(-2)=4=0$$? Please advise.

• What is the difference between $2$ and "$-2$"? Recall, the minus sign merely means "the additive inverse of" which does not necessarily need to appear in other representations of the element. – JMoravitz Feb 19 '20 at 17:36
• Welcome to math SE. Have a look at mathjax for your mathematical expressions. – Alain Remillard Feb 19 '20 at 17:37
• As for "does it have to do with the fact that $(-2)(-2)=4=0$" No, it doesn't. It has to do with the fact that $2 + 2 = 0$. Addition is what is important here, not multiplication. You will find that the additive inverse of $0$ is $0$, the additive inverse of $1$ (which you might when convenient decide to notate as $-1$) is equal to $3$, the additive inverse of $2$ (which you might when convenient decide to notate as $-2$) is equal to $2$, and so on... – JMoravitz Feb 19 '20 at 17:37
• There's a problem with the title of the question. 2 is not a subgroup of $\mathbb Z_4$, it's an element. I guess you mean the subgroup generated by 2. – Shatabdi Sinha Feb 19 '20 at 17:40
• After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $\checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?. – Shaun Feb 19 '20 at 18:46

$$\langle 2\rangle = \{0, 2\}$$ is a subgroup of the group $$\mathbb Z_4 = \{0, 1, 2, 3\}$$ under modular arithmetic, modulo $$4$$.

The identity of this group is $$0$$, and because $$2+2 \equiv 0 \pmod 4$$, it has order two, and hence $$2$$ generates a group (subgroup) of order 2. In fact, the additive inverse of $$2$$ is $$2$$.

That is, $$\langle 2 \rangle = \{0, 2\} \leq \{0, 1, 2, 3\} = \mathbb Z_4$$.

• Good approach to teach! – Mikasa Mar 20 '20 at 8:36

The elements of $$\Bbb Z_4$$ are not technically $$0$$, $$1$$, $$2$$ and $$3$$; rather, they are equivalence classes of integers with respect to the divisibility of their differences by $$4$$, like so: $$[a]_4:=\{b\in\Bbb Z: 4\mid a-b\}.$$ The operation of the group is defined by $$[x]_4+_4[y]_4=[x+y]_4$$.

Thus, since $$4\mid (-2)+2=0$$, we have $$[-2]_4=[2]_4$$.

• Why the downvote? – Shaun Feb 19 '20 at 19:00
• No, it isn't, @amWhy. – Shaun Mar 11 '20 at 16:58
• But $-2\equiv 2\pmod{4}$. – Shaun Mar 11 '20 at 17:02