# Modelling precedence relations

I have two tasks $$i$$ and $$k$$ with durations $$d_i$$ and $$d_k$$, where $$d_i$$ and $$d_k$$ are nonnegative variables.

I would like to model that $$i$$ may precede $$k$$ or $$k$$ may precede $$i$$ and that they may not overlap.

So, with $$t_i$$ and $$t_k$$ denoting the start times of $$i$$ and $$k$$, I have to model:

either $$t_i + d_i \le t_k$$ OR $$t_k + d_k \le t_i$$

Introducing a binary variable $$y$$, I can achieve the result with the following two big M constraints:

$$t_i + d_i - t_k \le M y$$

$$t_k + d_k - t_i \le M (1-y)$$

If it is required that $$t_i + d_i \le H$$ and $$t_k + d_k \le H$$ then I can set $$M$$ to be $$M=H$$.

My question is, is what I have done so far correct (what worries me is the variable duration) and can anyone think about a better formulation?

Yes, this is correct and is the classical approach from Manne, On the Job-Shop Scheduling Problem (1960).

In some modeling languages, you can also enforce these implications by using indicator constraints: \begin{align} y = 0 &\implies t_i + d_i \le t_k \\ y = 1 &\implies t_k + d_k \le t_i \\ \end{align}

• Rob, thank you very much for your reply and the suggestion. I will try both the Big-M and the Indicator Constraint approach. Apr 12 at 8:43

Can anyone think about a better formulation?

Another option is to use binary variables $$x_{it}$$ that take value $$1$$ if task $$i$$ starts at time $$t$$. You then need two sets of constraints:

• one start time per task: $$\sum_{t}x_{it} = 1 \quad \forall i$$
• don't overlap tasks: $$\sum_{i}\sum_{k, t+1 - d_i \le k \le t}x_{ik} \le 1 \quad \forall t$$

This formulation is more tight and should solve faster. And it does not require big-Ms.

• This however, will function only in the case of a discrete time formulation? Why is this formulation tighter? Jul 7 at 8:31
• Yes indeed this assumes a discrete time span. That seems reasonable to me. I don't think it is easy to prove that the formulation is tighter - this could be a great separate question. I only have empirical evidence to back this statement. Jul 7 at 9:54

You may also use CPOptimizer within CPLEX that contains scheduling high level concepts. And then you can directly use noOverlap constraints.

In

using  CP;

dvar interval i size 5;
dvar interval k size 4;

dvar sequence seq in append(i,k);

minimize maxl(endOf(i),endOf(k));
subject to
{
noOverlap(seq);
}

the constraint

noOverlap(seq);

makes sure that i and k do not overlap

and in the CPLEX IDE you will see


# Given constraints, in how many ways can actors be chosen for roles?

Given $$13$$ actors and $$6$$ unique roles, in how many ways can the actors be assigned a role if a certain actor (Alan) will not join if another actor (Betty) joins?

My method was to compute total unrestricted number of ways less the number of ways when they're both in it. The first term is $$\binom{13}{6} 6!$$ and the second term is $$\binom{11}{4} 6!$$. The numerical answer I get is $$997920$$ but the correct numerical answer given is $$1116720$$.

• 997920 looks right to me – Hagen von Eitzen Apr 19 at 8:16
• The result 1116720 would fit the task that we only prohibit Betty in a more prominent role than Alan – Hagen von Eitzen Apr 19 at 8:20
• What does being in a more prominent role mean? Also I realise that I get their answer only if I divide my second term by 2 (or 2! ?). – OneGapLater Apr 19 at 8:22
• The given answer will be true if a Alan has problem with Betty but not the other way around, assuming that they join one by one. – SinTan1729 Apr 19 at 9:12
• I think that's what they imply. However, I don't see why my answer doesn't accommodate for that. – OneGapLater Apr 19 at 9:27

Interpretation: Alan and Betty cannot both be cast.

You are correct under this interpretation. We can confirm your answer using a different approach. Observe that either exactly one of Alan or Betty is in the cast or neither Alan nor Betty is in the cast.

Exactly one of Alan or Betty is in the cast: Select whether Alan or Betty is in the cast. Select five of the other eleven actors to be in the cast. Assign roles to the six selected actors. $$\binom{2}{1}\binom{11}{5}6!$$

Neither Alan nor Betty is in the cast: Select six of the other eleven actors to be in the cast. Assign roles to the six selected actors. $$\binom{11}{6}6!$$

Total: Since the two cases above are mutually exclusive and exhaustive, the number of ways of assigning roles to the actors is $$\binom{2}{1}\binom{11}{5}6! + \binom{11}{6}6! = \binom{11}{5}6!\left[\binom{2}{1} + 1\right] = 3\binom{11}{5}6! = 997920$$ as you found.

Interpretation: Alan will not join if Betty has been cast first.

Under this interpretation, Alan and Betty can both be in the cast if Alan is cast first.

Then there are three possibilities:

1. Exactly one of Alan or Betty is in the cast.
2. Neither Alan nor Betty is in the cast.
3. Alan and Betty are both in the cast, because Alan is cast first.

We have covered the first two cases above.

Alan and Betty are both in the cast, because Alan is cast first: Ignoring the order of casting for the moment, if Alan and Betty are both selected, we must select four of the remaining eleven actors. We then assign roles to the six actors. This can be done in $$\binom{11}{4}6!$$ ways. However, by symmetry, in half of these assignments, Betty is cast before Alan. Thus, Alan will only join the cast in half of these assignments. Thus, there are $$\frac{1}{2}\binom{11}{4}6! = 118800$$ ways to assign the roles so that both Alan and Betty are cast.

Total: Since these three cases are mutually exclusive and exhaustive, the roles may be cast if Alan will not join if Betty has already been cast is $$\binom{2}{1}\binom{11}{5}6! + \binom{11}{6}6! + \frac{1}{2}\binom{11}{4}6! = 1116720$$ If this is the intended interpretation, the wording of the question could have been clearer.


# Factorising complex numbers

4 years ago, when I learned about factorising and complex numbers, me and my friend worked on factorising complex numbers.

For example, $4+2i= 3-(-1)+2i = 3-i^2+2i = -(i^2-2i-3) = -(i-3)(i+1)$

The goal was to represent $a+bi$ with product of same form. where $a$ and $b$ are integer.

Another example is, $8+i= 8i^4+i=i(8i^3+1)=i(2i+1)(4i^2-2i+1)=i(2i+1)(-2i-3)=-i(2i+1)(2i+3)$

I showed my teacher, and she said it's useless.

Now I think of it, I don't know why I did this and it looks like same thing just in different form.

Is there any research already done on this or can there be any use of it?

Apparently,

$(n+2)+ni=-(i-(n+1))(i+1)$

$m^3+n^3i=-i(mi+n)(mni+(m^2-n^2))$

• Yes, there's a lot of work done on this. Look up Gaussian Integers. – Angina Seng Jul 26 '18 at 14:07
• And by no means is it useless! – Lubin Jul 26 '18 at 14:11
• There's research already done, but that does not mean there can't be any use of the work you do on this, when you've read up on the existing work. – Henrik supports the community Jul 26 '18 at 14:13
• I was thinking @Henrik, that knowing how to factor Gaussian integers is useful in what might look like other parts of mathematics. – Lubin Jul 26 '18 at 14:15


# How is sample mean divided by sample STD distributed for normal distributions?

Let's assume that we sample N times from a normal distribution with known mean and variance. With a generated sample we calculate sample mean and sample standard deviation. Then we divide sample mean by sample standard deviation (STD) to get the measure of my interest.

The question that I have: Is there an analytical expression for the distribution of the above described measure?

I know that distribution for the sample mean is well known. I have also found an expression for the distribution of sample STD. However, I cannot find the distribution that I need.

• If "STD" is intended to be an abbreviation for standard deviation then the distribution will be a scaled version of a noncentral t distribution. A number of posts on site also discuss this distribution. If I don't locate a sufficiently close duplicate (I didn't with a quick look), and nobody posts a full answer in the meantime, I'll come back and make this more detailed. Nov 15 '18 at 11:33
• @Glen_b Relevant posts are stats.stackexchange.com/a/133274/919 (yours), stats.stackexchange.com/a/17288/919, and stats.stackexchange.com/a/160523/919. The last is very nearly an answer, but it does not explicitly give any expression for the distribution.
– whuber
Nov 15 '18 at 13:22
• @whuber I found a few others a bit like the first two but they didn't seem quite close enough to count as answers for this one (though all were relevant). Nov 15 '18 at 14:54
• This is just to point out that this is the reciprocal of the coefficient of variation. It's also been called signal to noise ratio, although that term doesn't appear to be uniquely defined. Nov 22 '18 at 0:50

Let $$X_1,...,X_n \sim \text{IID N}(\mu, \sigma)$$ be your data points. It is well known from Cochran's theorem that the sample mean and sample variance are independent with distributions:

$$\bar{X}_n \sim \text{N} \Big( \mu, \frac{\sigma^2}{n} \Big) \quad \quad \quad S_n^2 \sim \sigma^2 \cdot \frac{\text{Chi-Sq}(n-1)}{n-1}.$$

Hence, we can form the independent statistics:

$$Z_n \equiv \frac{\bar{X}_n - \mu}{\sigma / \sqrt{n}} \sim \text{N}(0,1) \quad \quad \quad \chi_n \equiv \frac{S_n}{\sigma} \sim \frac{\text{Chi}(n-1)}{\sqrt{n-1}}.$$

With a bit of algebra we then have:

\begin{aligned} \frac{\bar{X}_n}{S_n} &= \frac{\bar{X}_n / \sigma}{S_n / \sigma} \\[6pt] &= \frac{\bar{X}_n / \sigma}{\chi_n} \\[6pt] &= \frac{1}{\sqrt{n}} \cdot \frac{\sqrt{n} \bar{X}_n / \sigma}{\chi_n} \\[6pt] &= \frac{1}{\sqrt{n}} \cdot \frac{\sqrt{n} (\bar{X}_n - \mu)/\sigma + \sqrt{n} \mu/\sigma}{\chi_n} \\[6pt] &= \frac{1}{\sqrt{n}} \cdot \frac{Z_n + \sqrt{n} \mu/\sigma}{\chi_n} \\[6pt] &\sim \frac{1}{\sqrt{n}} \cdot \text{Noncentral T} \big(\sqrt{n} \mu/\sigma, n-1 \big). \\[6pt] \end{aligned}

So you can see that the ratio of the sample mean on the sample standard deviation has a scaled non-central T distribution with non-centrality parameter $$\sqrt{n} \mu/\sigma$$ and degrees-of-freedom $$n-1$$. We can double-check this theoretical result empirically via simulation.

Checking the distribution by simulation: In the R code below we create a function to simulate $$m$$ samples of size $$n$$ from the IID normal model and generate the $$m$$ ratio statistics from these samples. We plot the kernel density of these simulated statistics against the theoretical distribution above in order to confirm that the theoretical result is correct.

#Simulate m values of the ratio statistic for samples of size n
SIMULATE <- function(m, n, mu, sigma) {
X <- array(rnorm(n*m, mean = mu, sd = sigma), dim = c(m,n));
R <- rep(0, m);
for (i in 1:m) { R[i] <- mean(X[i,])/sd(X[i,]); }
R; }

#Plot the density of the simulated values against theoretical
PLOTSIM  <- function(m, n, mu, sigma) {
require(stats); require(ggplot2);
RR    <- SIMULATE(m, n, mu, sigma);
DENS  <- density(RR);
DENS$$yy <- dt(DENS$$x*sqrt(n), df = n-1, ncp = sqrt(n)*mu/sigma)*sqrt(n);
DATA <- data.frame(x = DENS$$x, y = DENS$$y, yy = DENS$yy); ggplot(data = DATA, aes(x = x)) + geom_line(aes(y = y), size = 1.2, colour = 'black') + geom_line(aes(y = yy), size = 1.2, colour = 'red', linetype = 'dotted') + theme(plot.title = element_text(hjust = 0.5, size = 14, face = 'bold'), plot.subtitle = element_text(hjust = 0.5)) + ggtitle('Density plot - Simulated Data') + labs(subtitle = paste0('(Sample size = ', n, ', Simulation size = ', m, ')')) + xlab('Sample Mean / Sample Standard Deviation') + ylab('Density'); } #Generate example plot set.seed(1); m <- 10^4; n <- 100; mu <- 12; sigma <- 6; PLOTSIM(m, n, mu, sigma);  • thanks as usual @Ben! I ran the same simulation in Python and got identical results. Could you suggest the analytical formula for$mean(\frac{\mu}{\sigma})$and$stdev(\frac{\mu}{\sigma})$as a function of$\mu$and$\sigma$? I am not familiar with the Noncentral T-distribution, I couldn't figure it out myself. Nov 25 '18 at 20:43 • Just have a look at the moments of the noncentral T distribution. To get the moments for the above ratio, all you need to do is substitute the degrees-of-freedom$n-1$and the noncentrality parameter$\sqrt{n} \mu / \sigma\$. Good luck!
– Ben
Nov 25 '18 at 21:42
• @elemolotiv, would it be possible for you to share your Python code? I am asking since I am also working in Python. Nov 26 '18 at 8:53

@Roman, here is my Python code, of @Ben's example above

• it produces a file "walks.tsv" with N realisations of a random walk with mean=μ and stdev=σ.

• for each realisation there is a row, with the sample mean, sample stdev, and sample mean/stdev.

• I loaded "walk.tsv" in Excel and used Excel to aggregate and plot the data.

Hope it helps 🙂

import math, numpy.random

class RandomWalk:

def __init__(self, step_mean, step_stdev, steps_per_walk):
self.step_mean = step_mean
self.step_stdev = step_stdev
self.steps = steps_per_walk
self.log = open("walk.tsv","w+")
self.log_names()

def realize(self):
total = 0
total_sqr = 0
for i in range(self.steps):
step = float(numpy.random.normal(self.step_mean, self.step_stdev, 1))
total += step
total_sqr += step * step
self.sample_mean = total / self.steps
self.sample_stdev = math.sqrt(total_sqr / self.steps - self.sample_mean * self.sample_mean)
self.log_values()

def log_names(self):
self.log.write("dist_mean\t")
self.log.write("dist_stdev\t")
self.log.write("dist_mean/stdev\t")
self.log.write("steps\t")
self.log.write("sample_mean\t")
self.log.write("sample_stdev\t")
self.log.write("sample_mean/stdev\n")
self.log.flush()

def log_values(self):
self.log.write("{:0.1f}\t".format(self.step_mean))
self.log.write("{:0.1f}\t".format(self.step_stdev))
self.log.write("{:0.2f}\t".format(self.step_mean / self.step_stdev))
self.log.write("{}\t".format(self.steps))
self.log.write("{:0.1f}\t".format(self.sample_mean))
self.log.write("{:0.1f}\t".format(self.sample_stdev))
self.log.write("{:0.2f}\n".format(self.sample_mean / self.sample_stdev))
self.log.flush()

def simulate(step_mean, step_stdev, steps_per_walk, walks):
walk = RandomWalk(step_mean, step_stdev, steps_per_walk)
for i in range(walks):
walk.realize()

simulate(step_mean = 12, step_stdev = 6, steps_per_walk = 100, walks = 10000)


# I Electric field created by point charges and conducting plane

Tags:
1. Apr 2, 2017

### KV71

I came upon this:
http://physics.stackexchange.com/qu...change-if-we-place-a-metal-plat/323006#323006

question on Physics Stackexchange which I found very interesting.

The configuration is basically two positive point charges q and a conducting plane equidistant from both charges. What I found most fascinating in particular, is one answer that claims

"if the plate is a plane that extends to infinity, there will be two image charges -q at the positions of each of the original positive charges so that the electric field everywhere is zero"

My question is, is this true? If so, could someone explain why or perhaps tell me more about this?

2. Apr 2, 2017

Staff Emeritus
If the plate extends to infinity, it can get as many cancelling charges as it needs, "from infinity".

3. Apr 2, 2017

### KV71

So the field everywhere is zero?

4. Apr 2, 2017

Staff Emeritus
I don't think so, because the charge and its image are not in the same place.

5. Apr 2, 2017

### KV71

I think there are two images and the image for the charge below the conductor is sitting at the position of the charge above the conductor and vice versa.

6. Apr 2, 2017

Staff Emeritus
Fields from the other side of the conductor don't penetrate the conductor.

7. Apr 3, 2017

### vanhees71

It's very easy to see, why the field cannot be 0. Just draw a sphere $V$ around one of the postive charges, and use Gauss's Law,
$$\int_{\partial V} \mathrm{d}^2 \vec{F} \cdot \vec{E}=Q.$$
The key to understand this is that the image charges are not really there, but they are used as a mathematical trick to get the total field consisting of the field of the real charge and the influence charges in the plate, i.e., to fulfill the boundary conditions.

For your example you can easily solve the problem indeed by using image charges. You need to treat only one charge first. Say the infinite plane defines the $xy$ plane of a cartesian coordinate system, and let the charge $Q$ sit on the $z$ axis at $(0,0,a)$ with $a>0$.

You have to solve the Laplace equation for the potential
$$\Delta \phi=-Q \delta(x) \delta(y) \delta(z-a)$$
with the boundary condition $\phi=0$ for $z=0$.

Obviously the solution for $z<0$ is $\phi=0$ and for $z>0$ you write down the solution for the field of the true charge $Q$ at $(0,0,a)$ and account for the boundary conditions by substituting the plate by the image charge $-Q$ at $(0,0,-a)$, because then you solve for sure the Laplace equation and the boundary conditions, i.e., you have
$$\phi(\vec{x})=\begin{cases} 0 & \text{for} \quad z<0, \\ \frac{Q}{4 \pi} \left [\frac{1}{\sqrt{x^2+y^2+(z-a)^2}}-\frac{1}{\sqrt{x^2+y^2+(z+a)^2}} \right] &\text{for} \quad z>0. \end{cases}$$
For the other charge, $Q'$ sitting at $(0,0,-b)$ ($b>0$) you have in an analogous way
$$\tilde{\phi}(\vec{x})=\begin{cases} \frac{Q'}{4 \pi} \left [\frac{1}{\sqrt{x^2+y^2+(z+b)^2}}-\frac{1}{\sqrt{x^2+y^2+(z-b)^2}} \right] &\text{for} \quad z<0,\\ 0 & \text{for} \quad z>0. \end{cases}$$
The total field thus is
$$\phi_{\text{tot}}(\vec{x})= \begin{cases} \frac{Q'}{4 \pi} \left [\frac{1}{\sqrt{x^2+y^2+(z+b)^2}}-\frac{1}{\sqrt{x^2+y^2+(z-b)^2}} \right ] &\text{for} \quad z<0,\\ \frac{Q}{4 \pi} \left [\frac{1}{\sqrt{x^2+y^2+(z-a)^2}}-\frac{1}{\sqrt{x^2+y^2+(z+a)^2}} \right] & \text{for} \quad z>0. \end{cases}$$
Now set $Q'=Q$ and $b=a$, and you see that the field is not 0 also in this symmetric case.

That's a great exercise to understand the role of the "mirror charges"!

8. Apr 4, 2017

### KV71

Thanks for the very informative and well-written post @vanhees71 . Indeed, I liked this problem too because it has a solution that does not strike me immediately but is actually obvious when I gave it some more thought. Also , the owner of the answer that I referenced in my intro post deleted their answer (must have realized it is wrong) and has posted a new answer--you may look at it (it uses essentially the same method as yours but talks in terms of the field and superposition). After reading it, I am amazed that iafter an infinite conducting plate is placed between two charges, "the force on the charges changes sign". Very interesting indeed.

Let me know what you think!

9. Apr 4, 2017

### KV71

@Vanadium 50 I didn't understand what you meant earlier by
but now I understand. The link in my last post #8 helped me. Thanks so much! It turns out you were absolutely right!


# Find all triplets $(a,b,c)$ less than or equal to 50 such that $a + b +c$ be divisible by $a$ and $b$ and $c$.

Find all triplets $$(a,b,c)$$ less than or equal to 50 such that $$a + b +c$$ be divisible by $$a$$ and $$b$$ and $$c$$.(i.e $$a|a+b+c~~,~~b|a+b+c~~,~~c|a+b+c$$) for example $$(10,20,30)$$ is a good triplet. ($$10|60 , 20|60 , 30|60$$).

Note: $$a,b,c\leq 50$$ and $$a,b,c\in N$$.

In other way the question says to find all $$(a,b,c)$$ such that $$lcm(a,b,c) | a+b+c$$

After writing different situations, I found that if $$gcd(a,b,c) = d$$ then all triplets are in form of $$(d,2d,3d)$$ or $$(d,d,d)$$ or $$(d,d,2d)$$ are answers. (of course the permutation of these like $$(2d,3d,d)$$ is also an answer). It gives me $$221$$ different triplets. I checked this with a simple Java program and the answer was correct but I cannot say why other forms are not valid. I can write other forms and check them one by one but I want a more intelligent solution than writing all other forms. Can anyone help?

My java code: (All of the outputs are in form of $$(d,d,d)$$ or $$(d,2d,3d)$$ or $$(d,d,2d)$$ and their permutations.)

import java.util.ArrayList;
import java.util.Collections;

public class Main {
public static void main(String[] args) {
int count = 0;
for (int i = 1; i <= 50; i++) {
for (int j = 1; j <= 50; j++) {
for (int k = 1; k <= 50; k++) {
int s = i + j + k;
if (s % i == 0 && s % j == 0 && s % k == 0 && i != j && j != k && i != k) {
ArrayList<Integer> array = new ArrayList<Integer>();
array.clear();
int g = gcd(gcd(i, j), k);
Collections.sort(array);
int condition = 4; //To find out whether it is (d,d,d) or (d,d,2d) or (d,2d,3d)
if (array.get(0) == 1 && array.get(1) == 1 && array.get(2) == 1) {
condition = 1;
}
if (array.get(0) == 1 && array.get(1) == 1 && array.get(2) == 2) {
condition = 2;
}
if (array.get(0) == 1 && array.get(1) == 2 && array.get(2) == 3) {
condition = 3;
}
System.out.printf("%d %d %d ::: Condition: %d\n", i, j, k, condition);
count++;
}
}
}
}
System.out.println(count);
}

public static int gcd(int a, int b) {
if (b == 0) {
return a;
} else
return gcd(b, a % b);
}
}

• ... I recall seeing this question yesterday... – Servaes Apr 2 at 16:14
• @Servaes I used the search and didn't find this question. But as you said,I checked and found it. I am not the same person. Maybe his source and I was the same because I was investigating homework of a discrete mathematics course of a university and I found this question and he/she stated that it is his homework.I found the question interesting and asked it here. I completely checked the conditions with a Java program and find tested my hypothesis but I don't know how to prove it without checking all different forms. for better clarification, I'll add my java code to the problem. – amir na Apr 2 at 16:47
• What does a triplet being less than $50$ mean? That each term is less than 50? are you assuming each term is non-negative? – fleablood Apr 2 at 16:53
• Also what does "m is divisible to k" mean? Does that mean $\frac km$ is an integer? Or that $\frac mk$ is an integer? Or something else? I usually hear "m is divisible by k" to mean $\frac mk$ is an integer. – fleablood Apr 2 at 16:55
• @Servaes math.stackexchange.com/questions/3170626/… I didn't find this in search because the title of that question is not very good and don't have the actual question and I think stack Exchange only search by title. – amir na Apr 2 at 16:57

If $$a\leq b\leq c$$ then $$c\mid a+b+c$$ implies $$c\mid a+b$$ and so $$a+b=cz$$ for some $$z\in\Bbb{N}$$. Then $$cz=a+b\leq2b\leq2c,$$ and so $$z\leq2$$. If $$z=2$$ then the inequalities are all equalities and so $$a=b=c$$. Then the triplet $$(a,b,c)$$ is of the form $$(d,d,d)$$.

If $$z=1$$ then $$c=a+b$$, and then $$b\mid a+b+c$$ implies that $$b\mid 2a$$. As $$b\geq a$$ it follows that either $$b=a$$ or $$b=2a$$. If $$b=a$$ then $$c=2a$$ and the triplet $$(a,b,c)$$ is of the form $$(d,d,2d)$$. If $$b=2a$$ then $$c=3a$$ and the triplet $$(a,b,c)$$ is of the form $$(d,2d,3d)$$.

This allows us to count the total number of triplets quite easily;

1. The number of triplets of the form $$(d,d,d)$$ is precisely $$50$$; one for each positive integer $$d$$ with $$d\leq50$$.
2. The number of triplets of the form $$(d,d,2d)$$ is precisely $$25$$; one for each positive integer $$d$$ with $$2d\leq50$$. Every such triplets has precisely three distinct permutations of its coordinates, yielding a total of $$3\times25=75$$ triplets.
3. The number of triplets of the form $$(d,2d,3d)$$ is precisely $$16$$; one for each positive integer $$d$$ with $$3d\leq50$$. Every such triplets has precisely six distinct permutations of its coordinates, yielding a total of $$6\times 16=96$$ triplets.

This yields a total of $$50+75+96=221$$ triplets.

• Simple code finds $221$. – David G. Stork Apr 4 at 4:56
• @DavidG.Stork A simple count shows the same ;) – Servaes Apr 4 at 13:29


Puzzle of gold coins in the bag

At the end of Probability class, our professor gave us the following puzzle:

There are 100 bags each with 100 coins, but only one of these bags has gold coins in it. The gold coin has weight of 1.01 grams and the other coins has weight of 1 gram. We are given a digital scale, but we can only use it once. How can we identify the bag of gold coins?

After about 5 minutes waiting, our professor gave us the solution (the class had ended and he didn't want to wait any longer):

Give the bags numbers from 0 through 99, then take 0 coins from the bag number 0, 1 coin from the bag number 1, 2 coins from the bag number 2, and so on until we take 99 coins from the bag number 99. Gather all the coins we have taken together and put them on the scale. Denote the weight of these coins as $W$ and the number of bag with gold coins in it as $N$, then we can identify the bag of gold coins using formula $$N=100(W-4950)$$ For instance, if the weight of all coins gathered is $4950.25$ grams, then using the formula above the bag number 25 has the gold coins in it.

My questions are:

1. How does the formula work? Where does it come from?
2. Do we have other ways to solve this puzzle? If yes, how?
3. If the digital scale is replaced by a traditional scale, the scale like symbol of libra or the scale in Shakespeare's drama: The Merchant of Venice (I don't know what is the name in English), then how do we solve this puzzle?
• If you can pull coins out of the bag, maybe you can see the difference instead of weighing. ;) – jpmc26 Nov 13 '14 at 19:54
• @jpmc26 One of friends also said like that but our prof only smiled. :-D – Venus Nov 14 '14 at 9:54
• This puzzle appeared on one of the episodes of Columbo: imdb.com/title/tt0075864 – Stephen Montgomery-Smith Nov 18 '14 at 23:04
• @StephenMontgomery-Smith That's very old movie, I haven't been born yet when the movie aired so I don't watch it. Anyway, you're the one who also reviews Prof. Otelbaev's work, how is it going? Is it correct so far? – Venus Nov 19 '14 at 6:29
• @Venus You can see that episode on Netflix. Otelbaev said there was a mistake in his work. math.stackexchange.com/questions/634890/… – Stephen Montgomery-Smith Nov 19 '14 at 14:46

To understand the formula, it would be easiest to explain how it works conceptually before we derive it.

Let's simplify the problem and say there are only 3 bags each with 2 coins in them. 2 of those bags have the 1 gram coins and one has the 1.01 gram gold coins. Let's denote the bags arbitrarily as $Bag_0$, $Bag_1$, and $Bag_2$. Similarly to your problem, let's take 0 coins from $Bag_0$, 1 coin from $Bag_1$, and 2 coins from $Bag_2$. We know that the gold coins must be in one of those bags, so there are three possibilities when we weigh the three coins we removed:

Gold Coins in $Bag_0$: So the weight of the 3 coins on the scale are all 1 gram. So the scale will read 3 grams.

Gold Coins in $Bag_1$: So the weight of 1 of the coins is 1.01 grams and 2 of the coins are 2 grams. So the scale will read 3.01 grams.

Gold Coins in $Bag_2$: So the weight of 2 of the coins is 2.02 grams and 1 of the coins is 1 gram. So the scale will read 3.02 grams.

So each possibility has a unique scenario. So if we determine the weight, we can determine from which bag those coins came from based on that weight.

We can generalize our results from this simplified example to your 100 bag example.

Now for deriving the formula. Say hypothetically, of our 100 bags, all 100 coins in each of the 100 bags weigh 1 gram each. In that case, when we remove 0 coins from $Bag_0$, 1 from $Bag_1$, up until 99 coins from $Bag_{99}$, we'll have a total of 4950 coins on the scale, which will equivalently be 4950 grams. Simply put, if $n$ is our Bag number (denoted $Bag_n$), we've placed $n$ coins from each $Bag_n$ onto the scale for $n = 0, 1, 2, ... 99$.

So the weight of the coins will be $Weight = 1 + 2 + 3 + ... + 99 = 4950$

But we actually have one bag with gold coins weighing 1.01 grams. And we know that those 1.01 gram coins must be from some $Bag_n$. In our hypothetical example, all of our coins were 1 gram coins, so we must replace the $n$ coins weighed from $Bag_n$ with $n$ gold coins weighing 1.01 grams. Mathematically, we would have: $Weight = 4950 - n + 1.01n = 4950 + .01n = 4950 + n/100$

Rearranging the formula to solve for n, we have: $100(Weight-4950) = n$, where $Weight$ is $W$ and $n$ is $N$ in your example.

I have no knowledge of an alternative answer to this puzzle, but perhaps another member's answer may be enlightening if there is. Technically speaking, you could have denoted the bags from 1 to 100 and gone through a similar process as above, but the method is still the same, so I wouldn't treat it as a new answer.

If our electric scale is replaced by a scale of libra, I don't believe it would be possible to answer this puzzle with only one measurement of weight. But again, perhaps another answer may be enlightening on that.

• To see why it's not possible to find the bag with just one "libra" weighing, consider that one weighing gives as 3 possible states (left, right, balanced), yet we have 100 bags to choose from. In other words we have a trit of information (trit: ternary equivalent of the bit) with each weighing. We will need at least 5 trits to distinguish between 100 possible bags. – Thanassis Oct 2 '18 at 5:29

For #1: imagine for a moment that all the coins are fake. If we took 0 coins from bag 0, 1 coin from bag 1, 2 coins from bag 2... we'd have $99\times100/2=4,950$ coins, and those 4,950 coins would weigh a total of 4,950 grams. But now, say that bag 25 were the one with real coins that are slightly heavier: 0.01 grams heavier, in fact. So the total weight of the coins is $W=4950+0.01N$, where $N$ is the number of the bag with the real coins. But -- we have the weight, not the bag number. So let's invert the equation: we want to find N given W, not the other way around.

\begin{align}W&=4950+0.01N\\W-4950&=0.01N\\100(W-4950)&=N\end{align}

For #2, aside from renumbering the bags, there isn't a different way to do this; no matter what, we have to have a different number coming from the scale for each different possible result.

For #3, you need $\lceil\log_3k\rceil$ weighings to discover the odd coin out; for 100 coins, that's 5 weighings: the first splits the coins into groups of (up to) 34; the second into groups of (up to) 12; the third into groups of (up to) 4; the fourth into groups of (up to) 2; the fifth finds it guaranteed.

Why $\lceil\log_3k\rceil$? Each use of the balance scales actually compares three different groups of coins: the one on the left scale, the one on the right scale, and the one not on the scale at all. If one of the two groups on the scale is heavier, then the gold coin is in that group; if neither, then the gold coin is in the group not on the scale. Thus, each weighing can distinguish between 3 states, and $n$ weighings can distinguish between $3^n$ states. We need an integer solution to $3^n\ge k$, thus $n\ge\log_3k$, thus $n=\lceil\log_3k\rceil$.

• +1 for #3, but how to get this formula: $\lceil\log_3k\rceil$? – Venus Nov 13 '14 at 14:32
• Each use of the balance scales actually compares three different groups of coins: the one on the left scale, the one on the right scale, and the one not on the scale at all. If one of the two groups on the scale is heavier, then the gold coin is in that group; if neither, then the gold coin is in the group not on the scale. Thus, each weighing can distinguish between 3 states, and $n$ weighings can distinguish between $3^n$ states. We need an integer solution to $3^n\ge k$, thus $n\ge\log_3k$, thus $n=\lceil\log_3k\rceil$ – Dan Uznanski Nov 13 '14 at 14:37

According to the given values, the gold coins have essentially the same weight (down to a single percent) as the base ones. Since gold is heavy this means that each gold coin is significantly smaller.

Forget about weighing and just take the bag whose coins are much smaller than the coins in the other bags.

• As much as this isn't the intended strategy, I really can't say this is wrong, because this would totally work. However, in the context of Math.se, it seems that this approach is a bit less mathematical than it was supposed to be. Perhaps it would be better suited for puzzling.se – Asimov Nov 13 '14 at 14:38
• The fake ones could be made of tungsten; tungsten has a density remarkably similar to gold's. – Dan Uznanski Nov 13 '14 at 14:44
• Perhaps the others are all gold-plated platinum or iridium, so they are very nearly the same size as the gold coins. – David K Nov 13 '14 at 14:44
1. If all coins would have equal weight ($1.00$ gram), then the total weight of taking $0$ coins from bag $0$, $1$ coin from bag $1$, etc. would be $4950$ grams. Verify: $1 + 2 + \cdots + 99 = 4950$. To work with the example: if you take $25$ coins from bag $25$, then the total offset in weight is $0.25$ grams.
2. Not that I know of.
3. When the rest of the rules are the same (weighing only once), you cannot solve the puzzle.
• Re 3: With a balance that can only give three different "answers" (left is heavier, right is heavier, or equality), it takes at least five weighings (because $3^4<100$). – Hagen von Eitzen Nov 13 '14 at 14:26

The total weight of coins on the digital scale is $$W=\sum_{m=0}^{99} mw_m$$ where $w_m$ is the weight of each coin in bag number $m.$ But if the gold coins are actually in bag number $N,$ then $$w_m = \begin{cases} 1 + 0.01 && \text{if}\ m = N, \\ 1 && \text{if}\ m \neq N. \end{cases}$$ Therefore the only term of the sum that is not equal to $m$ is the term $Nw_N,$ which is equal to $N + 0.01N.$ So $$W=\sum_{m=0}^{99} mw_m = \left(\sum_{m=0}^{99} m\right) + 0.01N = 4950 + 0.01N.$$ Knowing $W$, we solve for $N.$

Are there other solutions? Certainly! We could choose $m+1$ coins from each bag numbered $m$, or in fact any number of coins as long as we take a different number of coins from each bag and know which bag contributed what number of coins. The same calculation as before tells us how many gold coins are on the scale, and we then deduce which bag they must have come from.

In fact you can find the bag of gold coins among up to $101$ bags this way. That's because there are just $101$ different numbers of coins we can draw from a bag. If any of the bags have more coins, we can solve this for more bags. But we cannot solve the puzzle for so many bags that there would have to be two bags that contribute the same number of coins to the weighing, because if we found that that was the number of gold coins on the scale then we would not know which of the two bags they came from.

Now if you have only a two-pan balance that does not give a reading, it is no longer possible to determine where the gold coins are in one weighing. It will take multiple weighings, as in the solutions to this problem and this problem.


# Math Help - Compounding interest problem help

1. ## Compounding interest problem help

Hello everyone, I am working on a math project relating to logarithms, exponential growth and decay, and I have the following problem to work on:

8. The final amount for $5000 invested for 25 years at 10% annual interest compounded semiannually is$57,337.
a. What is the effect of doubling the amount invested?

b. What is the effect of doubling the annual interest rate?

c. What is the effect of doubling the investment period?

d. Which of the above has the greatest effect on the final amount of the investment?

I can't seem to figure out how to start...

Any suggestions? Thanks!

2. Originally Posted by qcom
Hello everyone, I am working on a math project relating to logarithms, exponential growth and decay, and I have the following problem to work on:

8. The final amount for $5000 invested for 25 years at 10% annual interest compounded semiannually is$57,337.
a. What is the effect of doubling the amount invested?

b. What is the effect of doubling the annual interest rate?

c. What is the effect of doubling the investment period?

d. Which of the above has the greatest effect on the final amount of the investment?

I can't seem to figure out how to start...

Any suggestions? Thanks!
You should know that

$A = P(1 + r)^n$, where P is the principal, r is the percentage interest rate, n is the number of time periods, and A is the final amount.

3. Using Proveit's:
A = 5000(1.05^50) = 57337

25 years, so there are 50 semiannual periods; .10/2 = .05 is rate per period; kapish?

4. Hey thanks a lot 'Prove It' and Wilmer, I think I was mainly confused about the semiannual part, and so you just double the number of years because semiannual means that it is collected twice in one year, right?

Also, when you divide the rate .1 (10% in decimal form) by 2, is that also due to the fact that the interest is collected semiannually?

Otherwise, I know how to proceed and do a-d.

BTW, not much of a concern but I knew the formula as Y = a(1 + r)^t

But it doesn't really make a difference,

5. Correct.
And this is what "happens" during the 25 years:
Code:
         Interest               Balance
0                               5000.00
1          250.00               5250.00
2          262.50               5512.50
3          275.62               5788.12
....
49                             54606.67
50        2730.33              57337.00

6. Alright, now I got it for sure, thanks.

Not sure if you want to help with another problem, but I was also confused with this one, which deals with similar solving techniques, I think...

6. Consider a \$1000 investment that is compounded annually at three different interest rates: 5%, 5.5%, and 6%.

a. Write and graph a function for each interest rate over a time period from 0 to 60 years.

b. Compare the graphs of the three functions.

c. Compare the shapes of the graphs for the first 10 years with the shapes of the graphs between 50 and 60 years.

I think the functions are (for problem 'a'):

y = 1000(1 + .05)^60
y = 1000(1 + .055)^60
y = 1000(1 + .06)^60

Does that look correct?

Now for problem 'b', if I'm not mistaken, just look like horizontal lines waaay up on the y-axis.

One of the graphs, for the first equation I gave, looks like this: http://www.wolframalpha.com/input/?i=graph:+y+%3D+1000(1+%2B+.05)^60

Now for 'c', would I just draw the graphs of the equations but replace the 't' value with 10 giving us new equations:

y = 1000(1 + .05)^10
y = 1000(1 + .055)^10
y = 1000(1 + .06)^10

And then the equations for the last ten years, would we need to find the new 'a' value for our equation, instead of just 1000, and then write a new equation for that?

Just please tell me if I'm way off or something!

7. Originally Posted by qcom
> I think the functions are (for problem 'a'):
> y = 1000(1 + .05)^60
> y = 1000(1 + .055)^60
> y = 1000(1 + .06)^60
> Does that look correct?

Not familiar with graphing programs; but above correct as the FINAL
values (60 years later); seems to me this is really what's needed:
y = 1000(1 + .05)^t where t= 0 to 60

> Now for problem 'b', if I'm not mistaken, just look like horizontal lines
> waaay up on the y-axis.

Well, you'd have t (0 to 60) along x-axis, and the y values along y-axis

> Now for 'c', would I just draw the graphs of the equations but replace
> the 't' value with 10 giving us new equations:
> y = 1000(1 + .05)^10
> y = 1000(1 + .055)^10
> y = 1000(1 + .06)^10

Yes, but: y = 1000(1 + .05)^t where t=1 to 10

> And then the equations for the last ten years, would we need to find
> the new 'a' value for our equation, instead of just 1000, and then write > a new equation for that?

Simply this way: y = 1000(1 + .05)^t where t = 51 to 60
.

8. Ok I think I got you, so we would set up equations like this:

y = 1,000(1.05)^10 y = 1,629
y = 1,000(1.05)^20 y = 2,653
y = 1,000(1.05)^30 y = 4,322
y = 1,000(1.05)^40 y = 7,040
y = 1,000(1.05)^50 y = 11,467
y = 1,000(1.05)^60 y = 18,679

And that's just for one of them and you would do that with the other two as well right?

Alright now I get what you meant by 't' goes along the x-axis as it is basically our 'x' value in these functions, right? And then the 'y' would of course go along the 'y' axis.

Now for c, I think I got that as well.

Does this look good? My only concern is that I may need to do an equation like I did for part 'a' for every single number from 1-60 3 times, because there is 3 different interest rates, or is it fine to only do it in intervals of 10?

9. Originally Posted by qcom
....or is it fine to only do it in intervals of 10?
Dunno. I'd assume it is.
Isn't there a way in whatever graphing program you're using
ti give an instruction, like:
Do for t = 1 to 60 : y = (1 + .05)^t ?


# Alternate solution to circular permutation problem with restrictions

Ten chairs are evenly spaced around a round table and numbered clockwise from $1$ through $10$. Five married couples are to sit in the chairs with men and women alternating, and no one is to sit either next to or across from his/her spouse. How many seating arrangements are possible?

$\mathrm{(A)}\ 240\qquad\mathrm{(B)}\ 360\qquad\mathrm{(C)}\ 480\qquad\mathrm{(D)}\ 540\qquad\mathrm{(E)}\ 720$

Solution For the first man, there are $10$ possible seats. For each subsequent man, there are $4$, $3$, $2$, and $1$ possible seats. After the men are seated, there are only two possible arrangements for the five women. The answer is $10\cdot 4\cdot 3\cdot 2\cdot 1\cdot 2 = \boxed{480}$.

I am not satisfied with this solution. Is there another way to do this? (I just don't like this solution's..solution. There are many ways to proof the Pythagorean Theorem, and while you have no logical objections to others, surely there's some you like better or less than others? It's like that. I don't think I "think" like this solution. In any case, I just want to explore some other solutions that make better sense in my mind.)

• Understood. I am asking for another solution to the problem is all though. I will edit it and try to express my intentions better. – mathflair Jan 2 '16 at 1:24
• Even if this were not a request for a more satisfactory solution of some kind, I would hope that you would include more context, such as your own thoughts about solving it. After all, a multiple choice problem will often have some short cuts, and hearing your thoughts of one kind or another may better inform a Reader's responses toward solutions that you find more pleasing. – hardmath Jan 2 '16 at 1:43
• Perhaps you could give an example of how you "think" to a similar problem so we have a better idea of what you are after. Personally that answer is exactly how I think about the problem. – Ian Miller Jan 2 '16 at 2:07
• I just wanted to see if there were other ways to do this problem, so no one had to tailor a solution to my mind! I may have come across that point confusingly. ^^; – mathflair Jan 2 '16 at 5:45

This is not very different from your solution but maybe it is more clear: Let the men be $m_1, m_2, ..., m_5$ and their arrangement be the (ordered) vector $A=(a_1, ..., a_5)$ Since the seats are labelled, the arrangement $(1,3,5,7,9)$ is different than $(3,5,7,9,1)$ etc. There are 10 ways to pick $a_1$ for man $m_1$. For each arrangement vector $A$ there are $4!$ permutations for men $m_2$ to $m_5$ and all the men arrangements are now $10\times 4!$

Without losing generality let's consider one men's arrangement $(1,3,5,7,9)$ For this case let's name the women as $w_i$, where $i$ is the seat number where their spouse is seated. Women can go to the even seats. In seat 2 you cannot have $w_1$ and $w_3$ and not $w_7$ either (across her spouse in seat 7). That leaves two choices for seat 2: $w_5$ or $w_9$. If you pick $w_5$ there is only one possible arrangement for the rest of the even seats to give: $(w_5, w_7,w_9,w_1,w_3)$. Same as you pick $w_9$, there is only one arrangement. Working the same way you can confirm that there are only two possible arrangements for the women as your solution states and get the answer $2\times10\times4!$

Another way

Permissible positions of males in relation to their spouses is either $+3$ or $-3$, e.g. $1-4$ or $1-8$

The chart below with females at odd positions makes it obvious that if the $+3$ option is chosen for one couple, it needs to be so for all couples, (and so, too for $-3$)

$1 - \color{green}4-\color{red}{8}\quad\;\; 1-\color{red}4-\color{green}8$
$3 - \color{green}6-\color{red}{10}\quad 3-\color{red}6-\color{green}{10}$
$5 - \color{green}8-\color{red}{2}\quad\;\; 5-\color{red}8-\color{green}{2}$
$7 - \color{green}{10}-\color{red}4\quad 7-\color{red}{10}-\color{green}4$
$9 - \color{green}2-\color{red}6\quad\;\; 9-\color{red}2-\color{green}6$

Thus $10\cdot4!$ with the alpha female at odd positions, ditto at even positions to finally yield

ans = $2\cdot 10\cdot4!$


# Solving the heat equation on the semi-infinite rod

Cross posted in scicomp.SE.

I want to test the solution which is given below is right by Mathematica.

Please look the post in mathstackexhange

or

Question: Solve the following heat equation on the semi-infinite rod

$\qquad u_t=ku_{xx}$

where $x,t>0$ and

$\qquad u_x(0,t) =0$ and $u(x,0)=\begin{cases} 1, & 0 < x <2 \\ 0, & 2\leq x \end{cases}$

with proper Fourier transform.

$\qquad u(x,t) = \frac{2}{\pi}\int_{0}^{\infty}e^{-s^2 t}\frac{1-\cos(2s)}{s}\cos(sx)ds.$

Code

But I am not sure the solution is right. I am not capable of testing it in Mathematica.

Could you help me?

• I don't we can do much to help without having the code you used to get the answer. If you didn't use Mathematica to solve your problem, then this question is inappropriate -- I would mean you are asking us to both write the code and verify the solution for you. Oct 17 '17 at 17:34
• Well, solving this in Mathematica is quite straightforward, just check the document of DSolve. Anyway, the solution is 1/2 (-Erf[(-2 + x)/(2 Sqrt[k t])] + Erf[(2 + x)/(2 Sqrt[k t])]). A quick test shows the solution in your question seems to be wrong. Oct 18 '17 at 4:44
• Further check shows that, if one wants to express the solution as integration, then it should be $\int_0^{\infty } \frac{\sqrt{\frac{2}{\pi }} \sin (2 w) e^{-k t w^2} \cos (w x)}{w} \, dw$ Oct 18 '17 at 12:08
• Guys, personally I suggest not to close this post, though it's a… "give me the code" question, the problem is interesting, I think. Oct 18 '17 at 12:13
• Please do not cross-post within SE sites. Choose one site and delete the questions from the others. meta.stackexchange.com/q/64068/164803 Oct 19 '17 at 12:17

Personally I think the problem is interesting, so let me extend my comments to an answer. First of all, DSolve can solve OP's problem straightforwardly (in Mathematica 10.3 or higher, if I remember correctly):

With[{u = u[t, x]},
eq = D[u, t] == k D[u, x, x];
ic = u == Piecewise[{{1, 0 < x < 2}}] /. t -> 0;
bc = D[u, x] == 0 /. x -> 0;]

asol = DSolveValue[{eq, ic, bc}, u, {t, x}, Assumptions -> {x > 0, k > 0}];
asol[t, x]
(* 1/2 (-Erf[(-2 + x)/(2 Sqrt[k] Sqrt[t])] + Erf[(2 + x)/(2 Sqrt[k] Sqrt[t])]) *)

Remark

There seems to be a bug in DSolve in v11.2.0.

DSolve[{eq, ic, bc}, u[t, x], {t, x}]

will return unevaluated.

As one can see, DSolve expresses the solution with Erf, so it's not immediately clear whether OP's solution is correct or not, and Mathematica's functions for simplifying also doesn't work well in this case, so let's obtain the analytic solution with another approach, that is, making use of Fourier cosine transform to eliminate the derivative of $x$:

fct = FourierCosTransform[#, x, s] &;

tset = Map[fct, {eq, ic}, {2}] /. Rule @@ bc /.
HoldPattern@FourierCosTransform[a_, __] :> a

tsol = u[t, x] /. DSolve[tset, u[t, x], t][[1]]
(* (E^(-k s^2 t) Sqrt[2/π] Sin[2 s])/s *)

Remark

I've made the transform on the PDE in a quick way, for a more general approach, check this post.

InverseFourierCosTransform has difficulty in transforming tsol, but it doesn't matter because the integral form is just what we want. By checking the formula of inverse Fourier cosine transform, we find the solution should be

$$u(t,x)=\sqrt{\frac{2}{\pi }} \int_0^{\infty } \frac{e^{-k s^2 t} \sqrt{\frac{2}{\pi }} \cos (s x) \sin (2 s)}{s} \, ds$$

It's apparently different from the one in your question, and numeric calculation shows this solution is the same as the one given by DSolve, so the one in your question is wrong.

Finally, a illustration for the solution:

Plot3D[asol[t, x] /. k -> 1 // Evaluate, {x, 0, 4}, {t, 0, 10}]

# Update

Inspired by Ars3nous' comment below, I noticed InverseFourierCosTransform can actually transform tsol. We just need a proper assumption:

InverseFourierCosTransform[tsol, s, x, Assumptions -> k > 0]
(* 1/2 (-Erf[(-2 + x)/(2 Sqrt[k t])] + Erf[(2 + x)/(2 Sqrt[k t])]) *)

Apparently it's the same as asol.

• Following @xzczd amazing answer using cosine transform, with k=1 (you can always scale out k in time) In Mathematica 9.0, I get using InverseFourierCosineTransform, $u(x,t)=\frac{1}{2} \left(-2 x \text{erfc}\left(\frac{x}{2 \sqrt{t}}\right)+\text{erfc}\left(\frac{x+2}{2 \sqrt{t}}\right)+\text{erfc}\left(-\frac{x-2}{2 \sqrt{t}}\right)+\frac{4 \sqrt{t} e^{-\frac{x^2}{4 t}}}{\sqrt{\pi }}-2\right)$ which has two different terms along the mentioned answer. But this solution also does not satisfy boundary conditions. I wonder what is the discrepency for. Nov 5 '17 at 12:18
• @Ars3nous Are you in v9.0.0 or v9.0.1? I just tested in v9.0.1, Win10 64bit, with InverseFourierCosTransform[tsol /. k -> 1, s, x] I got 1/2 (Erf[(2 - x)/(2 Sqrt[t])] + Erf[(2 + x)/(2 Sqrt[t])]), which is consistent with asol. Nov 5 '17 at 12:38


# Find the sum of the series of $\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}+…$

Find the sum of the series $$\frac1{1\cdot3}+\frac1{3\cdot5}+\frac1{5\cdot7}+\frac1{7\cdot9}+\frac1{9\cdot11}+\cdots$$ My attempt solution: $$\frac13\cdot\left(1+\frac15\right)+\frac17\cdot\left(\frac15+\frac19\right)+\frac1{11}\cdot\left(\frac19+\frac1{13}\right)+\cdots$$ $$=\frac13\cdot\left(\frac65\right)+\frac17\cdot \left(\frac{14}{45}\right)+\frac1{11}\cdot\left(\frac{22}{117}\right)+\cdots$$ $$=2\cdot\left(\left(\frac15\right)+\left(\frac1{45}\right)+\left(\frac1{117}\right)+\cdots\right)$$ $$=2\cdot\left(\left(\frac15\right)+\left(\frac1{5\cdot9}\right)+\left(\frac1{9\cdot13}\right)+\cdots\right)$$ It is here that I am stuck. The answer should be $\frac12$ but I don't see how to get it. Any suggestions?

Also, a bit more generally, are there good books (preferably with solutions) to sharpen my series skills?

• @ParclyTaxel Gah! My inability to read signs when strikes again. Retracted. – Xander Henderson Dec 8 '17 at 3:22
• Your question suggests an Infinite sum. Am I right? – Ravi Prakash Dec 8 '17 at 5:26

This is a general approach to evaluate the sum of series, like these.

First find $n^{th}$ term of series.

Let $T_n$ denote the $n^{th}$ term.

We see that,

$T_1 = \frac{1}{\color{green}{1} \cdot \color{teal}{3}}$

$T_2 = \frac{1}{\color{green}{3} \cdot \color{teal}{5}}$

And so on. Let the numbers in $\color{green}{green}$ be $$\color{green}{X_1,X_2,X_3,X_4,..=1,3,5,7...}$$

Clearly they form an A.P. with common difference $=2$

So, $n^{th}$ term of this AP is $1 + (n-1) × 2 = \color{green}{2n-1}$

Similarly, Let the numbers in $\color{teal}{teal}$ be $$\color{teal}{Y_1,Y_2,Y_3,Y_4,..=3,5,7,9...}$$ Clearly they form an A.P. with common difference $=2$

So, $n^{th}$ term of this AP is $3 + (n-1) × 2 =\color{teal}{ 2n+1 }$

So, the $n^{th}$ term of the main question is just

$$T_n = \frac{1}{\color{green}{(2n-1)} \cdot \color{teal}{(2n+1)}}$$

Now, taking summation from $1$ to $n$ , we have,

$$\sum_{n=1}^n \frac{1}{(2n-1) \cdot (2n+1)}$$

$$= \sum_{n=1}^n = \frac{1}{2} \cdot \frac{(2n+1)-(2n-1)}{(2n-1)(2n+1)}$$

$$= \sum_{n=1}^n = \frac{1}{2} \cdot \frac{1}{(2n-1)} - \frac{1}{(2n+1)}$$

$$= \sum_{n=1}^n = \frac{1}{2} \cdot ( 1 - \frac{1}{(2n+1)} )$$

While $n = ∞$,

$$\sum_{n=1}^∞ = \frac{1}{2} \cdot ( 1 - \frac{1}{(2(∞)+1)} )$$

$$= \sum_{n=1}^∞ = \frac{1}{2} \cdot ( 1 - \frac{1}{∞} )$$

Since $\frac{1}{∞} = 0$,

$$\sum_{n=1}^∞ = \frac{1}{2} \cdot ( 1 - 0 )$$

Which is

$$\sum_{n=1}^∞ = \frac{1}{2}$$

This is a classic telescoping series. $$\frac1{n\cdot(n+2)}=\frac12\left(\frac1n-\frac1{n+2}\right)$$ Thus $$\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}+\frac{1}{9\cdot 11}+\cdots$$ $$=\frac12\left(\frac11-\frac13+\frac13-\frac15+\frac15-\frac17+\frac17-\frac19+\frac19-\frac1{11}+\cdots\right)$$ $$=\frac12$$

\begin{align*} \sum_{n=1}\dfrac{1}{(2n-1)(2n+1)}&=\dfrac{1}{2}\sum_{n=1}\left(\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\\ &=\dfrac{1}{2}\sum_{n=1}\left(\dfrac{1}{2n-1}-\dfrac{1}{2(n+1)-1}\right)\\ &=\dfrac{1}{2}\sum_{n=1}\left(\dfrac{1}{f(n)}-\dfrac{1}{f(n+1)}\right)\\ &=\dfrac{1}{2}\dfrac{1}{f(1)}, \end{align*} where $f(n)=2n-1$, and note that $f(n)^{-1}\rightarrow 0$ as $n\rightarrow\infty$.

$$\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}+\frac{1}{9\cdot 11}...=\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}\right)+\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}\right)+\dfrac{1}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}\right)+\cdots=\dfrac12$$

Let me give a general method which is useful for this sum $\frac1{1\cdot3}+\frac1{3\cdot5}+\frac1{5\cdot7}+\frac1{7\cdot9}+\frac1{9\cdot11}+\cdots=\sum_{n=1}^\infty\dfrac{1}{(2n-1)(2n+1)}$ we have

$\sum_{n=1}^\infty\dfrac{1}{(2n-1)(2n+1)}=\sum_{n=0}^\infty\dfrac{1}{(2n+1)(2n+3)}$

we can use this method

$$\sum_{n\geq 0}\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b}\tag{1}$$

where $\psi$ is digamma function.

now we can write

$\sum_{n=1}^\infty\dfrac{1}{(2n-1)(2n+1)}= \frac{1}{4}\sum_{n=1}^\infty\dfrac{1}{(n+\frac{1}{2})(n+\frac{3}{4})}=\frac{1}{2}$

since $\psi(\frac{1}{2})-\psi(\frac{3}{2})=-2$


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The staff here would really like to help, and it would be informative for use if you could get a screen shot of the error and email it to me, and I will forward it to the staff so we can see exactly what the error looks like at your end.

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# Alternating sum of squares of binomial coefficients

I know that the sum of squares of binomial coefficients is just $${2n}\choose{n}$$ but what is the closed expression for the sum $${n\choose 0}^2 - {n\choose 1}^2 + {n\choose 2}^2 + \cdots + (-1)^n {n\choose n}^2$$?

• Can you do it with generating functions? – Nikhil Ghosh Aug 8 '12 at 2:33
• I thought I had something clever. I have deleted my post until I have a chance to think on the case when $n$ is even. $n$ being odd still yields 0, unless I am totally mistaken. – Emily Aug 8 '12 at 2:41
• Wolfram|Alpha gives this closed form. – joriki Aug 8 '12 at 2:45
• I don't really understand why combinatorial proof went more or less unnoticed (while standard application of generating functions is heavily upvoted). – Grigory M Nov 30 '13 at 13:28

$$(1+x)^n(1-x)^n=\left( \sum_{i=0}^n {n \choose i}x^i \right)\left( \sum_{i=0}^n {n \choose i}(-x)^i \right)$$

The coefficient of $$x^n$$ is $$\sum_{k=0}^n {n \choose n-k}(-1)^k {n \choose k}$$ which is exactly your sum.

On another hand:

$$(1+x)^n(1-x)^n=(1-x^2)^n=\left( \sum_{i=0}^n {n \choose i}(-1)^ix^{2i} \right)$$

Thus, the coefficient of $$x^n$$ is $$0$$ if $$n$$ is odd or $$(-1)^{\frac{n}2}{n \choose n/2}$$ if $$n$$ is even.

• ty fixed it. I hope that was the only one :) – N. S. Aug 8 '12 at 2:51

Here's a combinatorial proof.

Since $\binom{n}{k} = \binom{n}{n-k}$, we can rewrite the sum as $\sum_{k=0}^n \binom{n}{k} \binom{n}{n-k} (-1)^k$. Then $\binom{n}{k} \binom{n}{n-k}$ can be thought of as counting ordered pairs $(A,B)$, each of which is a subset of $\{1, 2, \ldots, n\}$, such that $|A| = k$ and $|B| = n-k$. The sum, then, is taken over all such pairs such that $|A| + |B| = n$.

Given $(A,B)$, let $x$ denote the largest element in the symmetric difference $A \oplus B = (A - B) \cup (B - A)$ (assuming that such an element exists). In other words, $x$ is the largest element that is in exactly one of the two sets. Then define $\phi$ to be the mapping that moves $x$ to the other set. The pairs $(A,B)$ and $\phi(A,B)$ have different signs, and $\phi(\phi(A,B)) = (A,B)$, so $(A,B)$ and $\phi(A,B)$ cancel each other out in the sum. (The function $\phi$ is what is known as a sign-reversing involution.)

So the value of the sum is determined by the number of pairs $(A,B)$ that do not cancel out. These are precisely those for which $\phi$ is not defined; in other words, those for which there is no largest $x$. But there can be no largest $x$ only in the case $A=B$. If $n$ is odd, then the requirement $\left|A\right| + \left|B\right| = n$ means that we cannot have $A=B$, so in the odd case the sum is $0$. If $n$ is even, then the number of pairs is just the number of subsets of $\{1, 2, \ldots, n\}$ of size $n/2$; i.e., $\binom{n}{n/2}$, and the parity is determined by whether $|A| = n/2$ is odd or even.

Thus we get $$\sum_{k=0}^n \binom{n}{k}^2 (-1)^k = \begin{cases} (-1)^{n/2} \binom{n}{n/2}, & n \text{ is even}; \\ 0, & n \text{ is odd}.\end{cases}$$

• Wonderful proof, Mike! I always prefer combinatorial proofs as they also offer motivation and explanation to the identity and not only a formal proof. What really interests me is whether you can construct a sign-reversing involution to prove the following special case of Dixon's Identity: $\sum_{k=0}^{n} \binom{3n}{k}^{3}(-1)^{k} = (-1)^n\binom{3n}{n,n,n}$. – Ofir Jan 18 '13 at 16:31
• @Ofir: I'm glad you like it! I'm a big fan of combinatorial proofs myself. That's an interesting question about Dixon's identity; you should ask it as a question on the site. – Mike Spivey Jan 18 '13 at 18:56
• I can't edit my original comment, but in the LHS it should be $\binom{2n}{k}^{3}$ instead of $\binom{3n}{k}^{3}$. Reading an article by Zeliberger, I found out there's a combinatorial proof for Dixon's identity by Foata. It should be in the following French book: www-irma.u-strasbg.fr/~foata/paper/ProbComb.pdf . I think pages 37-40 generalize it but I don't know any French, can anyone help out? – Ofir Jan 19 '13 at 15:33
• A belated comment: Your beautiful proof can be easily extended to the more general result that $\sum\limits_{k=0}^n \dbinom{m}{k} \dbinom{m}{n-k} \left(-1\right)^k = \begin{cases} \left(-1\right)^{n/2} \dbinom{m}{n/2} , & \text{ if } n \text{ is even}; \\ 0 , & \text{ if } n \text{ is odd} \end{cases}$ for any nonnegative integers $m$ and $n$. – darij grinberg Dec 20 '17 at 0:45

$$\sum_{m=0}^n (-1)^m{n \choose m}^2= (n!)^2\sum_{m=0}^n \frac{(-1)^m}{(m!)^2((n-m)!)^2}$$

This function feels hypergeometric, so we take the quotient of $c_{m+1}$ and $c_m$ where

$$c_m=\frac{(n!)^2}{(m!)^2((n-m)!)^2}$$

so,

$$\frac{c_{m+1}}{c_{m}}=\frac{((m+1)!)^2((n-m-1)!)^2}{(m!)^2((n-m)!)^2}=\frac{(m-n)^2}{(m+1)^2}$$

after some simplification, confirming that this can be expressed in terms of a hypergeometric function. The previous result gives us the parameters of the function so we find $\sum_{m=0}^\infty c_m x^m = {_2}F_1(-n, -n; 1;-1)$.

$${_2}F_1(-n, -n; 1;-1)= \sum_{m=0}^\infty \frac{((-n)_m)^2}{(1)_k} \frac{(-1)^k}{k!} = \sum_{m=0}^\infty (-1)^m{n \choose m}^2$$

Where $(x)_n=x(x+1)\cdots(x+n-1)=\frac{\Gamma(x+n)}{\Gamma(x)}$ is Pochhammer's symbol.

An identity for ${_2}F_1$ gives an elegant "closed form:"

$${_2}F_1(-n, -n; 1;-1)= \frac{2^{n} \sqrt{\pi} \Gamma(-n+n+1)}{\Gamma\left(\frac{1-n}{2}\right)\Gamma\left(\frac{-n}{2}+n+1\right)}= \frac{2^{n} \sqrt{\pi}}{\Gamma\left(\frac{1-n}{2}\right)\Gamma\left(\frac{n+2}{2}\right)}$$

Now, nothing that if $a$ is a positive integer, ${n \choose n+a}=0$, so

$$\sum_{m=0}^\infty (-1)^m{n \choose m}^2 = \sum_{m=0}^n (-1)^m{n \choose m}^2$$

and finally we get the answer

$$\sum_{m=0}^n (-1)^m{n \choose m}^2=\frac{2^{n} \sqrt{\pi}}{\Gamma\left(\frac{1-n}{2}\right)\Gamma\left(\frac{n+2}{2}\right)}$$

that holds for real numbers as well.


# linear equation graph problems

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# Infinite Ways to an Infinite Geometric Sum

One of my students, K, and I were reviewing Taylor Series last Friday when she asked for a reminder why an infinite geometric series summed to $\displaystyle \frac{g}{1-r}$ for first term g and common ratio r when $\left| r \right| < 1$.  I was glad she was dissatisfied with blind use of a formula and dove into a familiar (to me) derivation.  In the end, she shook me free from my routine just as she made sure she didn’t fall into her own.

STANDARD INFINITE GEOMETRIC SUM DERIVATION

My standard explanation starts with a generic infinite geometric series.

$S = g+g\cdot r+g\cdot r^2+g\cdot r^3+...$  (1)

We can reason this series converges iff $\left| r \right| <1$ (see Footnote 1 for an explanation).  Assume this is true for (1).  Notice the terms on the right keep multiplying by r.

The annoying part of summing any infinite series is the ellipsis (…).  Any finite number of terms always has a finite sum, but that simply written, but vague ellipsis is logically difficult.  In the geometric series case, we might be able to handle the ellipsis by aligning terms in a similar series.  You can accomplish this by continuing the pattern on the right:  multiplying both sides by r

$r\cdot S = r\cdot \left( g+g\cdot r+g\cdot r^2+... \right)$

$r\cdot S = g\cdot r+g\cdot r^2+g\cdot r^3+...$  (2)

This seems to make make the right side of (2) identical to the right side of (1) except for the leading g term of (1), but the ellipsis requires some careful treatment. Footnote 2 explains how the ellipses of (1) and (2) are identical.  After that is established, subtracting (2) from (1), factoring, and rearranging some terms leads to the infinite geometric sum formula.

$(1)-(2) = S-S\cdot r = S\cdot (1-r)=g$

$\displaystyle S=\frac{g}{1-r}$

STUDENT PREFERENCES

I despise giving any formula to any of my classes without at least exploring its genesis.  I also allow my students to use any legitimate mathematics to solve problems so long as reasoning is justified.

In my experiences, about half of my students opt for a formulaic approach to infinite geometric sums while an equal number prefer the quick “multiply-by-r-and-subtract” approach used to derive the summation formula.  For many, apparently, the dynamic manipulation is more meaningful than a static rule.  It’s very cool to watch student preferences at play.

K’s VARIATION

K understood the proof, and then asked a question I hadn’t thought to ask.  Why did we have to multiply by r?  Could multiplication by $r^2$ also determine the summation formula?

I had three nearly simultaneous thoughts followed quickly by a fourth.  First, why hadn’t I ever thought to ask that?  Second, geometric series for $\left| r \right|<1$ are absolutely convergent, so K’s suggestion should work.  Third, while the formula would initially look different, absolute convergence guaranteed that whatever the “$r^2$ formula” looked like, it had to be algebraically equivalent to the standard form.  While I considered those conscious questions, my math subconscious quickly saw the easy resolution to K’s question and the equivalence from Thought #3.

Multiplying (1) by $r^2$ gives

$r^2 \cdot S = g\cdot r^2 + g\cdot r^3 + ...$ (3)

and the ellipses of (1) and (3) partner perfectly (Footnote 2), so K subtracted, factored, and simplified to get the inevitable result.

$(1)-(3) = S-S\cdot r^2 = g+g\cdot r$

$S\cdot \left( 1-r^2 \right) = g\cdot (1+r)$

$\displaystyle S=\frac{g\cdot (1+r)}{1-r^2} = \frac{g\cdot (1+r)}{(1+r)(1-r)} = \frac{g}{1-r}$

That was cool, but this success meant that there were surely many more options.

EXTENDING

Why stop at multiplying by r or $r^2$?  Why not multiply both sides of (1) by a generic $r^N$ for any natural number N?   That would give

$r^N \cdot S = g\cdot r^N + g\cdot r^{N+1} + ...$ (4)

where the ellipses of (1) and (4) are again identical by the method of Footnote 2.  Subtracting (4) from (1) gives

$(1)-(4) = S-S\cdot r^N = g+g\cdot r + g\cdot r^2+...+ g\cdot r^{N-1}$

$S\cdot \left( 1-r^N \right) = g\cdot \left( 1+r+r^2+...+r^{N-1} \right)$  (5)

There are two ways to proceed from (5).  You could recognize the right side as a finite geometric sum with first term 1 and ratio r.  Substituting that formula and dividing by $\left( 1-r^N \right)$ would give the general result.

Alternatively, I could see students exploring $\left( 1-r^N \right)$, and discovering by hand or by CAS that $(1-r)$ is always a factor.  I got the following TI-Nspire CAS result in about 10-15 seconds, clearly suggesting that

$1-r^N = (1-r)\left( 1+r+r^2+...+r^{N-1} \right)$.  (6)

Math induction or a careful polynomial expansion of (6) would prove the pattern suggested by the CAS.  From there, dividing both sides of (5) by $\left( 1-r^N \right)$ gives the generic result.

$\displaystyle S = \frac{g\cdot \left( 1+r+r^2+...+r^{N-1} \right)}{\left( 1-r^N \right)}$

$\displaystyle S = \frac{g\cdot \left( 1+r+r^2+...+r^{N-1} \right) }{(1-r) \cdot \left( 1+r+r^2+...+r^{N-1} \right)} = \frac{g}{1-r}$

In the end, K helped me see there wasn’t just my stock approach to an infinite geometric sum, but really an infinite number of parallel ways.  Nice.

FOOTNOTES

1) RESTRICTING r:  Obviously an infinite geometric series diverges for $\left| r \right| >1$ because that would make $g\cdot r^n \rightarrow \infty$ as $n\rightarrow \infty$, and adding an infinitely large term (positive or negative) to any sum ruins any chance of finding a sum.

For $r=1$, the sum converges iff $g=0$ (a rather boring series). If $g \ne 0$ , you get a sum of an infinite number of some nonzero quantity, and that is always infinite, no matter how small or large the nonzero quantity.

The last case, $r=-1$, is more subtle.  For $g \ne 0$, this terms of this series alternate between positive and negative g, making the partial sums of the series add to either g or 0, depending on whether you have summed an even or an odd number of terms.  Since the partial sums alternate, the overall sum is divergent.  Remember that series sums and limits are functions; without a single numeric output at a particular point, the function value at that point is considered to be non-existent.

2) NOT ALL INFINITIES ARE THE SAME:  There are two ways to show two groups are the same size.  The obvious way is to count the elements in each group and find out there is the same number of elements in each, but this works only if you have a finite group size.  Alternatively, you could a) match every element in group 1 with a unique element from group 2, and b) match every element in group 2 with a unique element from group 1.  It is important to do both steps here to show that there are no left-over, unpaired elements in either group.

So do the ellipses in (1) and (2) represent the same sets?  As the ellipses represent sets with an infinite number of elements, the first comparison technique is irrelevant.  For the second approach using pairing, we need to compare individual elements.  For every element in the ellipsis of (1), obviously there is an “partner” in (2) as the multiplication of (1) by r visually shifts all of the terms of the series right one position, creating the necessary matches.

Students often are troubled by the second matching as it appears the ellipsis in (2) contains an “extra term” from the right shift.  But, for every specific term you identify in (2), its identical twin exists in (1).  In the weirdness of infinity, that “extra term” appears to have been absorbed without changing the “size” of the infinity.

Since there is a 1:1 mapping of all elements in the ellipses of (1) and (2), you can conclude they are identical, and their difference is zero.

### One response to “Infinite Ways to an Infinite Geometric Sum”

1. Here’s a rather involved comment on your first Footnote about regarding restricting r (as a video link: http://tinyurl.com/riemannzeta)


# If $\{x_n\}$ is a sequence s.t. $x_{2k} \rightarrow x$ and $x_{2k-1} \rightarrow x$, then $x_k \rightarrow x$

If $$\{x_n\}$$ is a sequence s.t. $$x_{2k} \rightarrow x$$ and $$x_{2k-1} \rightarrow x$$, then $$x_k \rightarrow x$$

Just looking for feedback on my proof attemtpt

Proof Attempt

given that the odd and even terms of the sequence both individually converge that means: $$1) \ \exists \ N_1 \ s.t \ \forall \ k \leq N_1 \ |x_{2k} - x| < \frac{\epsilon}{2} \\ 2) \ \exists \ N_2 \ s.t \ \forall \ k \leq N_2 \ |x_{2k-1} - x| < \frac{\epsilon}{2}$$

If we choose $$N = max\{N_1,N_2\}$$

Then: $$|x_{2k} - x_{2k-1}| \leq |x_{2k} - x| + |x_{2k-1} - x| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$

I don't know if it is a big leap, but I'm assuming that $$|x_{2k} - x_{2k-1}|$$ represent consecutive sequential terms so in essence I am bringing together the odd sequential terms and the even sequential terms. Is this the right idea?

• What you've shown is that the sequence is Cauchy. For sequences in $\Bbb R$ this is equivalent to convergence but it is not immediately obvious that that's the case. – Lukas Kofler Oct 21 '18 at 22:40
• I think you meant $k\geq N_1$ and $l\geq N_2$, not the other way around. – Mee Seong Im Oct 21 '18 at 22:40
• @LukasKofler that is true.....and this is supposed to be in $\mathbb{R}$, but since I wasn't thinking about it in those terms, what could I do to prove it? – dc3rd Oct 21 '18 at 22:43
• If you are in $\mathbb R$, you are done, since $\mathbb R$ is complete – Don Thousand Oct 21 '18 at 22:45
• You can just say that for all $n>N$ that $|x_n-x|<\epsilon$ because $n$ is either of the form $2k$ or $2k-1$ for some $k>N$. – kingW3 Oct 21 '18 at 22:47

so in essence I am bringing together the odd sequential terms and the even sequential terms. Is this the right idea?

Ooooh. Ouch. No. That is not the right idea. I'd doesn't matter how close sequential terms get. The classic counter example is the harmonic series in which $$a_n = \sum_{k = 1}^n \frac 1n$$. $$|x_n - x_{n-1}|=\frac 1n \to 0$$ but $$\{a_n\}$$ does not converge.

In this case you are lucky in that you actually have and $$x$$ which $$x_{2k}$$ and $$x_{2k - 1}$$ converge to.

So for any $$\epsilon$$ you have an $$N_1$$ so that $$n > N_1 \implies |x_{2n} - x| < \epsilon$$ and you have an $$N_2$$ so that $$n > N_2 \implies |x_{2n-1} - x| < \epsilon$$.

So if you have $$m > 2n > 2n-1$$ where $$n \ge \max (N_1, N_2)$$ then if $$m$$ is odd then $$m = 2k - 1$$ for $$k > n> N_2$$ so $$|x_m - x|= |x_{2k -1} - x| < \epsilon$$. But if $$m$$ is even then $$m = 2k$$ for $$k > n > N_1$$ so $$|x_m - x| = |x_{2k} - x| < \epsilon$$.

In other words, let $$M = 2\max (N_1, N_2)$$. Then if $$m > M$$ then if $$m = 2k$$we have $$k > N_1$$ and if $$m = 2k -1$$ then we have $$k > N_2$$. ANd either way $$|x_m - x| < \epsilon$$.

.....

Now if you hadn't been given that $$x_{2k}, x_{2k-1}\to x$$ and where given that $$x_{2k}$$ and $$x_{2k-1}$$ were Chauchy and needed to prove $$x_m$$ was cauchy you would have had the right idea only you don't prove it only for the subsequent terms you must prove it for any TWO terms $$m_1, m_2 > N$$.

And we'd do this by taken cases.

Case 1: if $$m_1, m_2$$ are both even then $$m_1, m_2 \ge N_1$$ (um, why did you write $$k \le N_1$$? That was a typo I assume) so $$|x_{m_1} - x_{m_2}| < \frac {\epsilon}2 < \epsilon.$$

Case 2: if $$m_1, m_2$$ are both odd... some thing but with $$N_2$$.

Case 3: If $$m_1$$ is even , $$m_2$$ is odd are opposite parity then $$|x_{m_1} - x_{m_2} \le |x_{m_1} - x_{2k}| + |x_{2k} - x_{2k -1}| + |x_{2k-1} - x_{m_2}| < \frac \epsilon 2 + \frac \epsilon 2 + \frac \epsilon 2 = \frac 32 \epsilon$$.

So you'd have to modify for $$\frac \epsilon 3$$ instead.

• Your solution is very interesting. I haven't been able to remove the notion from my head, but I'm always under the impression that since these are "exercises" or questions from an assignment they will always have an all encompassing solution that captures everything at once. But your solution had to break it down into cases. If we were to expand on this, say that there were 100 or 1000 cases, I guess we would have to just explicitly show them all and possibly over time refine it. Thanks @fleablood for the thorough explanation. – dc3rd Oct 22 '18 at 1:34
• If there were a thousand case we just need to consider the maximum of the resulting $N_i$s If $n > \max N_i$ then $n$ will qualify no matter what. – fleablood Oct 22 '18 at 1:36