# Evaluating $\lim_{k\to\infty}\sum_{n=1}^{\infty} \frac{\sin\left(\pi n/k\right)}{n}$

Recently, I was asked by a friend to compute the limit of the following series

$$\displaystyle{\lim_{k\to\infty}}\sum_{n=1}^{\infty} \frac{\sin\left(\frac{\pi n}{k}\right)}{n}$$

Having seen a similar problem to this before, Difficult infinite trigonometric series, I used the same complex argument approach as seen in that problem.

Ultimately, for this problem, I obtained $$\frac{\pi}{2}$$ as my answer. However, this limit can also be interpreted as a Riemann Sum, except the answer to the Riemann Sum differs from what I obtained as my answer, and according to Wolfram Alpha, the answer is expressed in terms of $$Si$$, where $$Si$$ is the sine integral.

I'm wondering, does the limit invalidate the argument approach, or is there something else I'm missing, because this limit if I'm not mistaken is a Riemann Sum after all?

• In a Riemann sum we would have both 'variables' tending to infinity simultaneously. In this case we have both $n$ and $k$ tending to infinity seperately and hence the Riemann sum approach should not be possible. This approach seems valid. – Peter Foreman Aug 12 at 21:48
• Nor can you (directly) apply dominated cvg theorem ... – Olivier Aug 12 at 21:51
• You effectively have $\lim_{k\to\infty}\lim_{N\to\infty}f(N,k)$ (where $N$ denotes the maximum summation bound). With a Riemann sum we would have $\lim_{N\to\infty}f(N)$ where $k$ may be used as an index such that the function $f(N)$ is of the form $\frac1N\sum_{k=1}^N g(k/N)\to\int_0^1g(x)\mathrm{d}x$. – Peter Foreman Aug 12 at 21:54
• (Rephrasing of what Peter says) You have $\lim_{k \to \infty} \frac{1}{k} \sum_{n=1}^{k} \frac{\sin(\pi n/k)}{(n/k)} \to \int_{0}^{1} \frac{sin(\pi x)}{x} dx$ by Riemann sum argument; but you have to manage a larger interval here... ($n\ge 1$, not just $n=1...k$) – Olivier Aug 12 at 22:03
• You can do a collection of Riemann sums to deal with n=1...k, then n=k+1...2k, n=2k+1...3k, but you would then need a further argument to put these sums altogether. – Olivier Aug 12 at 22:16

The Riemann Sum would be \begin{align} \lim_{k\to\infty}\sum_{n=1}^\infty\frac{\sin\left(\frac{\pi n}k\right)}{n} &=\lim_{k\to\infty}\sum_{n=1}^\infty\frac{\sin\left(\frac{\pi n}k\right)}{n/k}\frac1k\\ &=\int_0^\infty\frac{\sin(\pi x)}x\,\mathrm{d}x\\ &=\int_0^\infty\frac{\sin(x)}x\,\mathrm{d}x\\[3pt] &=\frac\pi2\tag1 \end{align} However, a cleaner way is to note that \begin{align} \sum_{n=1}^\infty\frac{\sin\left(\frac{\pi n}k\right)}{n} &=-\mathrm{Im}\!\left(\log\left(1-e^{i\pi/k}\right)\right)\\ &=\frac\pi2-\frac\pi{2k}\tag2 \end{align} and the limit is easy.

Take Care

One must be careful with the convergence of the Riemann Sum. Here is one method to control the remainders.

Because $$|\sin(\pi x)|\le1$$, we have $$\int_m^{m+1}\left|\frac{\sin(\pi x)}x\right|\,\mathrm{d}x \le\frac1m\tag3$$ Furthermore, $$\int_m^{m+2}\sin(\pi x)\,\mathrm{d}x=0$$, thus, \begin{align} \left|\int_m^{m+2}\frac{\sin(\pi x)}x\,\mathrm{d}x\right| &=\left|\int_m^{m+2}\sin(\pi x)\left(\frac1x-\frac1{m+1}\right)\mathrm{d}x\right|\\ &\le\frac1{m(m+1)}+\frac1{(m+1)(m+2)}\\[6pt] &=\frac1m-\frac1{m+2}\tag4 \end{align} Therefore, for any $$N\ge m$$, $$\left|\int_m^N\frac{\sin(\pi x)}x\,\mathrm{d}x\right| \le\frac1m\tag5$$ Because $$|\sin(\pi x)|\le1$$, we have $$\sum_{n=mk}^{(m+1)k}\left|\frac{\sin\left(\frac{\pi n}k\right)}{n}\right| \le\frac1m\tag6$$ Furthermore, $$\sum\limits_{n=mk}^{(m+2)k}\sin\left(\frac{\pi n}k\right)=0$$, thus, \begin{align} \left|\sum_{n=mk}^{(m+2)k}\frac{\sin\left(\frac{\pi n}k\right)}{n}\right| &=\left|\sum_{n=mk}^{(m+2)k}\sin\left(\frac{\pi n}k\right)\left(\frac1n-\frac1{(m+1)k}\right)\right|\\ &\le\frac1{m(m+1)}+\frac1{(m+1)(m+2)}\\[6pt] &=\frac1m-\frac1{m+2}\tag7 \end{align} Therefore, for any $$M\ge mk$$, $$\left|\sum_{n=mk}^M\frac{\sin\left(\frac{\pi n}k\right)}{n}\right|\le\frac1m\tag8$$ For any $$\epsilon\gt0$$, let $$m\ge\frac4\epsilon$$. Then Riemann Sums allow us to choose a $$k$$ large enough so that $$\left|\int_0^m\frac{\sin(\pi x)}x\,\mathrm{d}x-\sum_{n=1}^{mk}\frac{\sin\left(\frac{\pi n}k\right)}{n/k}\frac1k\right|\le\frac\epsilon2\tag9$$ Inequalities $$(5)$$ and $$(8)$$ show that for any $$N\ge m$$ and $$M\ge mk$$, $$\left|\int_m^N\frac{\sin(\pi x)}x\,\mathrm{d}x\right|\le\frac\epsilon4 \quad\text{and}\quad \left|\sum_{n=mk}^M\frac{\sin\left(\frac{\pi n}k\right)}{n/k}\frac1k\right|\le\frac\epsilon4\tag{10}$$ Inequalities $$(9)$$ and $$(10)$$ show that, for the $$k$$ chosen for $$(9)$$, $$\left|\sum_{n=1}^\infty\frac{\sin\left(\frac{\pi n}k\right)}{n}-\frac\pi2\right|\le\epsilon\tag{11}$$ Since $$\epsilon\gt0$$ was arbitrary, $$(11)$$ says that $$\lim_{k\to\infty}\sum_{n=1}^\infty\frac{\sin\left(\frac{\pi n}k\right)}{n}=\frac\pi2\tag{12}$$

• Riemann sums works on a finite intervals; what's the additional argument used here to extend this to (0,\infty) ? – Olivier Aug 12 at 22:11
• I've gotta say, that is one sexy solution and elegant approach! +1 :) – Sanjoy Kundu Aug 12 at 22:16
• @Olivier: In this case, we can take the sum from $1$ to $mk$ to get the integral from $0$ to $m$. Then carefully let $m\to\infty$. – robjohn Aug 12 at 22:27
• @Olivier: I believe one can use Abel's Test to estimate the remainder. – robjohn Aug 12 at 22:42
• @Olivier: I have added a section showing one method to control the remainders. It is a bit long, to include a lot of detail, but quite typical for this kind of Riemann Sum. – robjohn Aug 13 at 8:55


Main Content

# Solve System of ODEs with Multiple Initial Conditions

This example compares two techniques to solve a system of ordinary differential equations with multiple sets of initial conditions. The techniques are:

• Use a `for`-loop to perform several simulations, one for each set of initial conditions. This technique is simple to use but does not offer the best performance for large systems.

• Vectorize the ODE function to solve the system of equations for all sets of initial conditions simultaneously. This technique is the faster method for large systems but requires rewriting the ODE function so that it reshapes the inputs properly.

The equations used to demonstrate these techniques are the well-known Lotka-Volterra equations, which are first-order nonlinear differential equations that describe the populations of predators and prey.

### Problem Description

The Lotka-Volterra equations are a system of two first-order, nonlinear ODEs that describe the populations of predators and prey in a biological system. Over time, the populations of the predators and prey change according to the equations

`$\begin{array}{l}\frac{\mathrm{dx}}{\mathrm{dt}}=\alpha \mathit{x}-\beta \mathrm{xy},\\ \frac{\mathrm{dy}}{\mathrm{dt}}=\delta \mathrm{xy}-\gamma \mathit{y}.\end{array}$`

The variables in these equations are

• $\mathit{x}$ is the population size of the prey

• $\mathit{y}$ is the population size of the predators

• $\mathit{t}$ is time

• $\alpha$, $\beta$, $\delta$, and $\gamma$ are constant parameters that describe the interactions between the two species. This example uses the parameter values $\alpha =\gamma =1$, $\beta =0.01$, and $\delta =0.02$.

For this problem, the initial values for $\mathit{x}$ and $\mathit{y}$ are the initial population sizes. Solving the equations then provides information about how the populations change over time as the species interact.

### Solve Equations with One Initial Condition

To solve the Lotka-Volterra equations in MATLAB, write a function that encodes the equations, specify a time interval for the integration, and specify the initial conditions. Then you can use one of the ODE solvers, such as `ode45`, to simulate the system over time.

A function that encodes the equations is

```function dpdt = lotkaODE(t,p) % LOTKA Lotka-Volterra predator-prey model delta = 0.02; beta = 0.01; dpdt = [p(1) .* (1 - beta*p(2)); p(2) .* (-1 + delta*p(1))]; end ```

(This function is included as a local function at the end of the example.)

Since there are two equations in the system, `dpdt` is a vector with one element for each equation. Also, the solution vector `p` has one element for each solution component: `p(1)` represents $\mathit{x}$ in the original equations, and `p(2)` represents $\mathit{y}$ in the original equations.

Next, specify the time interval for integration as $\left[0,15\right]$ and set the initial population sizes for $\mathit{x}$ and $\mathit{y}$ to 50.

```t0 = 0; tfinal = 15; p0 = [50; 50];```

Solve the system with `ode45` by specifying the ODE function, the time span, and the initial conditions. Plot the resulting populations versus time.

```[t,p] = ode45(@lotkaODE,[t0 tfinal],p0); plot(t,p) title('Predator/Prey Populations Over Time') xlabel('t') ylabel('Population') legend('Prey','Predators')```

Since the solutions exhibit periodicity, plot the solutions against each other in a phase plot.

```plot(p(:,1),p(:,2)) title('Phase Plot of Predator/Prey Populations') xlabel('Prey') ylabel('Predators')```

The resulting plots show the solution for the given initial population sizes. To solve the equations for different initial population sizes, change the values in `p0` and rerun the simulation. However, this method only solves the equations for one initial condition at a time. The next two sections describe techniques to solve for many different initial conditions.

### Method 1: Compute Multiple Initial Conditions with `for-`loop

The simplest way to solve a system of ODEs for multiple initial conditions is with a `for`-loop. This technique uses the same ODE function as the single initial condition technique, but the `for`-loop automates the solution process.

For example, you can hold the initial population size for $\mathit{x}$ constant at 50, and use the `for`-loop to vary the initial population size for $\mathit{y}$ between 10 and 400. Create a vector of population sizes for `y0`, and then loop over the values to solve the equations for each set of initial conditions. Plot a phase plot with the results from all iterations.

```y0 = 10:10:400; for k = 1:length(y0) [t,p] = ode45(@lotkaODE,[t0 tfinal],[50 y0(k)]); plot(p(:,1),p(:,2)) hold on end title('Phase Plot of Predator/Prey Populations') xlabel('Prey') ylabel('Predators') hold off```

The phase plot shows all of the computed solutions for the different sets of initial conditions.

### Method 2: Compute Multiple Initial Conditions with Vectorized ODE Function

Another method to solve a system of ODEs for multiple initial conditions is to rewrite the ODE function so that all of the equations are solved simultaneously. The steps to do this are:

• Provide all of the initial conditions to `ode45` as a matrix. The size of the matrix is `s`-by-`n`, where `s` is the number of solution components and `n` is the number of initial conditions being solved for. Each column in the matrix then represents one complete set of initial conditions for the system.

• The ODE function must accept an extra input parameter for `n`, the number of initial conditions.

• Inside the ODE function, the solver passes the solution components p as a column vector. The ODE function must reshape the vector into a matrix with size `s`-by-n. Each row of the matrix then contains all of the initial conditions for each variable.

• The ODE function must solve the equations in a vectorized format, so that the expression accepts vectors for the solution components. In other words, `f(t,[y1 y2 y3 ...])` must return `[f(t,y1) f(t,y2) f(t,y3) ...]`.

• Finally, the ODE function must reshape its output back into a vector so that the ODE solver receives a vector back from each function call.

If you follow these steps, then the ODE solver can solve the system of equations using a vector for the solution components, while the ODE function reshapes the vector into a matrix and solves each solution component for all of the initial conditions. The result is that you can solve the system for all of the initial conditions in one simulation.

To implement this method for the Lotka-Volterra system, start by finding the number of initial conditions `n`, and then form a matrix of initial conditions.

```n = length(y0); p0_all = [50*ones(n,1) y0(:)]';```

Next, rewrite the ODE function so that it accepts `n` as an input. Use `n` to reshape the solution vector into a matrix, then solve the vectorized system and reshape the output back into a vector. A modified ODE function that performs these tasks is

```function dpdt = lotkasystem(t,p,n) %LOTKA Lotka-Volterra predator-prey model for system of inputs p. delta = 0.02; beta = 0.01; % Change the size of p to be: Number of equations-by-number of initial % conditions. p = reshape(p,[],n); % Write equations in vectorized form. dpdt = [p(1,:) .* (1 - beta*p(2,:)); p(2,:) .* (-1 + delta*p(1,:))]; % Linearize output. dpdt = dpdt(:); end ```

Solve the system of equations for all of the initial conditions using `ode45`. Since `ode45` requires the ODE function to accept two inputs, use an anonymous function to pass in the value of `n` from the workspace to `lotkasystem`.

`[t,p] = ode45(@(t,p) lotkasystem(t,p,n),[t0 tfinal],p0_all);`

Reshape the output vector into a matrix with size `(numTimeSteps*s)`-by-`n`. Each column of the output `p(:,k)` contains the solutions for one set of initial conditions. Plot a phase plot of the solution components.

```p = reshape(p,[],n); nt = length(t); for k = 1:n plot(p(1:nt,k),p(nt+1:end,k)) hold on end title('Predator/Prey Populations Over Time') xlabel('t') ylabel('Population') hold off```

The results are comparable to those obtained by the `for`-loop technique. However, there are some properties of the vectorized solution technique that you should keep in mind:

• The calculated solutions can be slightly different than those computed from a single initial input. The difference arises because the ODE solver applies norm checks to the entire system to calculate the size of the time steps, so the time-stepping behavior of the solution is slightly different. The change in time steps generally does not affect the accuracy of the solution, but rather which times the solution is evaluated at.

• For stiff ODE solvers (`ode15s`, `ode23s`, `ode23t`, `ode23tb`) that automatically evaluate the numerical Jacobian of the system, specifying the block diagonal sparsity pattern of the Jacobian using the `JPattern` option of `odeset` can improve the efficiency of the calculation. The block diagonal form of the Jacobian arises from the input reshaping performed in the rewritten ODE function.

### Compare Timing Results

Time each of the previous methods using `timeit`. The timing for solving the equations with one set of initial conditions is included as a baseline number to see how the methods scale.

```% Time one IC baseline = timeit(@() ode45(@lotkaODE,[t0 tfinal],p0),2); % Time for-loop for k = 1:length(y0) loop_timing(k) = timeit(@() ode45(@lotkaODE,[t0 tfinal],[50 y0(k)]),2); end loop_timing = sum(loop_timing); % Time vectorized fcn vectorized_timing = timeit(@() ode45(@(t,p) lotkasystem(t,p,n),[t0 tfinal],p0_all),2);```

Create a table with the timing results. Multiply all of the results by 1e3 to express the times in milliseconds. Include a column with the time per solution, which divides each time by the number of initial conditions being solved for.

```TimingTable = table(1e3.*[baseline; loop_timing; vectorized_timing], 1e3.*[baseline; loop_timing/n; vectorized_timing/n],... 'VariableNames',{'TotalTime (ms)','TimePerSolution (ms)'},'RowNames',{'One IC','Multi ICs: For-loop', 'Mult ICs: Vectorized'})```
```TimingTable=3×2 table TotalTime (ms) TimePerSolution (ms) ______________ ____________________ One IC 1.6558 1.6558 Multi ICs: For-loop 34.333 0.85832 Mult ICs: Vectorized 5.1357 0.12839 ```

The `TimePerSolution` column shows that the vectorized technique is the fastest of the three methods.

### Local Functions

Listed here are the local functions that `ode45` calls to calculate the solutions.

```function dpdt = lotkaODE(t,p) % LOTKA Lotka-Volterra predator-prey model delta = 0.02; beta = 0.01; dpdt = [p(1) .* (1 - beta*p(2)); p(2) .* (-1 + delta*p(1))]; end %------------------------------------------------------------------ function dpdt = lotkasystem(t,p,n) %LOTKA Lotka-Volterra predator-prey model for system of inputs p. delta = 0.02; beta = 0.01; % Change the size of p to be: Number of equations-by-number of initial % conditions. p = reshape(p,[],n); % Write equations in vectorized form. dpdt = [p(1,:) .* (1 - beta*p(2,:)); p(2,:) .* (-1 + delta*p(1,:))]; % Linearize output. dpdt = dpdt(:); end```

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# Integral #4

#### Random Variable

##### Well-known member
MHB Math Helper
Show that for positive parameters $a$, $b$, and $c$,

$$\int_{0}^{\infty} \left( \frac{e^{-ax}-e^{-bx}}{x^{2}} + (a-b) \frac{e^{-cx}}{x} \ \right) \ dx = b-a + a \ln \left(\frac{a}{c} \right) - b \ln \left(\frac{b}{c} \right)$$

#### ZaidAlyafey

##### Well-known member
MHB Math Helper

$$\displaystyle F = \int_{0}^{\infty} \left( \frac{e^{-ax}-e^{-bx}}{x^{2}} + (a-b) \frac{e^{-cx}}{x} \ \right) \ dx$$

$$\displaystyle F_a =\int_{0}^{\infty} \frac{e^{-cx}-e^{-ax}}{x}\ dx = \ln \left( \frac{a}{c}\right)$$

$$\displaystyle F= a \ln \left( \frac{a}{c}\right) - a + C$$

$$\displaystyle F_b = C_b$$

$$\displaystyle F_b = \int_{0}^{\infty} \frac{e^{-bx}-e^{-cx}}{x} dx = -\ln \left( \frac{b}{c}\right)$$

We have $$\displaystyle C_b = -\ln \left( \frac{b}{c}\right)$$ so

$$\displaystyle C = -b\ln \left( \frac{b}{c}\right)+b$$

$$\displaystyle F = b-a +a \ln \left( \frac{a}{c}\right)-b\ln \left( \frac{b}{c}\right)$$

#### Random Variable

##### Well-known member
MHB Math Helper
@ Zaid

I actually didn't even consider differentiating inside of the integral.

I took the boring approach of finding an antiderivative in terms of the exponential integral and then using the expansion of the exponential integral at $x=0$ to evaluate the antiderivative at the lower limit.

When you integrated to find the constant $C$, how did you know that he constant of integration was zero?

Alternatively what you could do to find $C$ is to let $a=b$ in the original integral.

#### ZaidAlyafey

##### Well-known member
MHB Math Helper

When you integrated to find the constant $C$, how did you know that he constant of integration was zero?

Alternatively what you could do to find $C$ is to let $a=b$ in the original integral.
Yeah, I made the evaluations not clear but I actually considered that $C$ as a function of $b$ so I differentiated with respect to $b$. Again putting $a=b$ is by far much clearer.

Can you post your approach ?

#### Random Variable

##### Well-known member
MHB Math Helper
It's easier to work the upper incomplete gamma function.

$$\Gamma(0,a x) = -\text{Ei}(- ax) = \int_{ax}^{\infty} \frac{e^{-t}}{t} \ dt = \int_{x}^{\infty} \frac{e^{-au}}{u} \ du$$

Let's first derive the expansion at $x=0$.

$$\Gamma(0,x) = \int_{x}^{\infty} \frac{e^{-t}}{t} \ dt = \int_{1}^{\infty} \frac{e^{-t}}{t} \ dt - \int_{1}^{x} \frac{e^{-t}}{t} \ dt$$

$$= \int_{1}^{x} \frac{1-e^{-t}}{t} \ dt - \ln x + \int_{1}^{\infty} \frac{e^{-t}}{t} \ dt$$

$$= -\ln x + \Big( -\int_{0}^{1} \frac{1-e^{-t}}{t} \ dt + \int_{1}^{\infty} \frac{e^{-t}}{t} \ dt \Big) + \int_{0}^{x} \frac{1-e^{-t}}{t} \ dt$$

$$= - \ln x - \gamma - \int_{0}^{x} \frac{e^{-t}-1}{t} \ dt$$

$$= - \ln x - \gamma \int_{0}^{x} \Big(-1 + \frac{t}{2} - \frac{t^{2}}{6} + \ldots \Big) \ dt$$

$$= - \ln x - \gamma + x + \mathcal{O}(x^{2})$$

And it's not really needed here, but one can find an asymptotic expansion at $x= \infty$ by integrating by parts.

$$\Gamma(0,x) \sim \frac{e^{-x}}{x} + \mathcal{O}\left( \frac{e^{-x}}{x^{2}} \right) \ \ x \to \infty$$

Then

$$\int \left( \frac{e^{-ax}-e^{-bx}}{x^{2}} + (a-b) \frac{e^{-cx}}{x} \ \right) \ dx = \int \frac{e^{-ax}-e^{-bx}}{x^{2}} \ dx + (a-b) \int \frac{e^{-cx}}{x} \ dx$$

$$= -\frac{e^{-ax} -e^{-bx}}{x} + \int \frac{-ae^{-ax}+be^{-bx}}{x} \ dx -(a-b) \Gamma(0,cx)$$

$$= \frac{e^{-bx} -e^{-ax}}{x} + a \Gamma(0,ax) - b \Gamma(0,bx) - (a-b) \Gamma(0,cx)+C$$

$$\int^{\infty}_{0} \left( \frac{e^{-ax}-e^{-bx}}{x^{2}} + (a-b) \frac{e^{-cx}}{x} \ \right) \ dx$$

$$= 0 - \lim_{x \to 0} \left[ \frac{e^{-bx} -e^{-ax}}{x} + a \left(-\ln ax - \gamma \right) - b \left(-\ln bx - \gamma \right) -(a-b) \left(-\ln cx - \gamma \right) \right]$$

$$= - \lim_{x \to 0} \left[ \frac{e^{-bx} -e^{-ax}}{x} - a \ln \left(\frac{a}{c} \right) + b \left(\frac{b}{c} \right) \right]$$

$$= b-a + a \ln \left(\frac{a}{c} \right) - b \ln \left(\frac{b}{c} \right)$$

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# Why is a Linear Equation Called Linear?

#### (New Question of the Week)

From time to time we get a question that is more about words than about math; usually these are about the meaning or origin of mathematical terms. Fortunately, some of us love words as much as we love math. But the question I want to look at here, which came in last month, is about both the word and the math; in explaining why the word is appropriate, we are learning some things about math itself.

## But it’s not a line …

Here is Christine’s question:

Why is an equation like 2x + 4 = 10 called a linear equation in one variable? Clearly, the solution is a point on one axis, the x-axis, not a line on the two-axis Cartesian coordinate plane? Or are all linear equations in one variable viewed as vertical lines?

Clearly, she is saying, the phrase “linear equation” means “the equation of a line”. And there is a similarity between the linear equation in one variable above, and a linear equation in two variables, such as $$y = 2x + 4$$, which definitely is the equation of a line. But with only one variable, the only way to say that $$2x + 4 = 10$$ is the equation of a line is to plot it on a plane as the vertical line $$x = 3$$. Is that the intent of the term?

No, it goes further than that, because you also have to consider three or more variables. I initially gave just a short answer, to see what response it would trigger before digging in deeper:

The term “linear”, though derived from the idea that a linear equation in two variables represents a line, has been generalized from that to mean that the equation involves a polynomial with degree 1. That is, the variable(s) are only multiplied by constants and added to other terms, with nothing more (squaring, etc.).

So you could say that the term has been taken from one situation that gave it its name, and applied to more general cases with different numbers of variables. A linear equation in three variables, for example, represents a plane.

From my perspective, “linear” means far more than “having a graph that is a line”. My first thought when I see the word (outside of an elementary algebra class) is “first-degree polynomial”. Although one initially connects it to straight lines, when we extend its use (and almost every idea in math is an extension of something simpler), the idea we carry forward is the degree, not the number of dimensions. For example, here is the beginning of the Wikipedia article about linear equations:

A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and (the first power of) a single variable (however, different variables may occur in different terms). A simple example of a linear equation with only one variable, x, may be written in the form: ax + b = 0, where a and b are constants and a ≠ 0.

Linear equations can have one or more variables. An example of a linear equation with three variables, x, y, and z, is given by: ax + by + cz + d = 0, where a, b, c, and d are constants and a, b, and c are non-zero.

But why would we call any polynomial equation with degree 1 “linear”, when it is only in two dimensions that it is related to a line? In the one-variable case, as Christine said, the graph is really a point; and in this three-variable case it is a plane:

## But, still – it’s not a line!

Christine wasn’t quite convinced:

Thank you so much. I am very particular with terminology when I teach mathematics. I have to say, I do not like this generalization of the term. I think it is misleading.

Hmmm … is the term “linear” misleading? Not to mathematicians, and I would hope not to students once they get used to it. Yet it’s true that it doesn’t quite mean what it seems to say. And is generalization bad? I think it’s the essence of what math is – and also an integral part of languages,which are always extending the meaning of words to cover new needs.

Here is my response:

Hi, Christine.

Thanks for writing back with additional thoughts. Let’s think a little more deeply about it.

First, this is standard terminology that has been in use for 200 years by many great mathematicians, so we should be very careful about considering it a bad idea. I don’t think you’re likely to convince anyone to change; the word is used not only of a linear equation in itself, but of “systems of linear equations” (in contrast to “non-linear equations”); of the whole major field of “linear algebra”; and for related concepts like “linear combination”, “linear transformation”, and “linear independence”, all of which apply to any number of variables or dimensions. So the term is very well established with a particular but broad meaning, and (at least after the first year) no one is misled by it. We know what it means, and what it means is more than just “line”.

Second, what would you replace it with? If you don’t want to use the word “linear” except in situations that involve actual lines, what word would you use instead to describe the general class of equations involving variables multiplied only by constants, regardless of the number of variables? We need a word for this bigger idea; that word will either be a familiar word whose meaning is stretched to cover a bigger concept, or some made-up word. The tradition in math has always been to take familiar words and give them new meanings (either more specific, like “group” or “combination” or “function”, or more general, like “number” or “multiplication” or “space”). So what we observe here is found throughout mathematics: a word that has grown beyond its humble beginnings. (This is also true of the entire English language! Most words would be “misleading” if you thought too deeply about their origins.)

I could say that, in a sense, all of mathematics is about generalization (or abstraction). I just mentioned “number”; some people do complain about calling anything other than a natural number a “number”, but the logical development from natural numbers, to integers, to rational numbers, to real and complex numbers, involves repeated broadening of the term, which has been extremely useful. We invent new concepts, and give them old names because they are a larger, more powerful version of the old concept.

I mentioned how common it is in English for a word to grow beyond its original meaning. Sitting here at my computer, I look at the mouse – is it misleading to call it that when it doesn’t have legs, and may not even have a tail? And the computer has a screen; at one time, a screen was a flat surface that hid something that wasn’t to be seen (or that kept out bugs); then it was applied to flat surfaces on which pictures were projected; and then to a surface that shows a picture itself. Is that misleading? It would be if you went back a hundred years …

And thinking again about math terms, I’m reminded of this discussion where I pondered what all the different operations that are called “multiplication” have in common.

## So how do we explain it?

But as I wrote this, I realized that I had gone off in a different direction than Christine, and I wanted to relate my answer to her specific context, linear equations in one variable. The real question was pedagogical: How could she explain this to her students, so that (eventually) “linear” would mean to them what they will need it to mean? I continued:

Now, I’ve been mostly thinking of the “enlarging” development of the word “linear”, taking it to more than two dimensions; you’re thinking specifically of the term used with only one variable. So it may help if we focus on that, to fit your particular context. I’d like to explain linear equations in one variable in a way that should make it clear why we use the word, and that it is not a misnomer.

Consider the equation you asked about, 2x+4=10. The left-hand side is an expression; we call it a linear expression, because if you used it in an equation with two variables, y = 2x+4, its graph would be a straight line. So we call 2x+4 a linear expression (or, later, a linear function). A linear equation in one variable is one that says that two linear expressions are equal (or one is equal to a constant).

Or, to look at it another way, one way to solve this equation would be to graph the related equation y = 2x+4, and find where that line intersects the line y = 10. So the linear equation can be thought of very much in terms of a line.

(If this were a student’s first exposure to linear equations in one variable, she likely wouldn’t have seen graphs of lines yet, and wouldn’t be ready for this discussion; either the word linear would be used without explanation yet, or we would hold off on the word until later.)

Does this help?

It’s often hard to be sure what kind of answer will help; but Christine gave the answer I hoped for:

Yes, your response does help very much. In addition, it gives me a lot to think about. It is true, the grade level I am currently teaching does learn to solve linear equations in one variable before they learn about linear equations and functions. Perhaps this is why I question the use of the term. I must remember that I have had experience in mathematics beyond their years, and therefore, am more thoughtful about what terms they are exposed to and more importantly in what sequence the mathematics is presented to them. I thank you again for such an in depth response and will certainly give this discussion much more thought. It is a pleasure speaking with you.

I imagine if I were introducing these equations to students with no exposure to graphs of lines, I might just mention in passing that these are called “linear equations in one variable”, and that we would soon be seeing why the word “linear” is appropriate. For now, what it means is that all we do with the variable is to multiply it by a constant, and to add things. Kids are used to hearing things they’ll understand later …

### 3 thoughts on “Why is a Linear Equation Called Linear?”

1. Very interesting article, Dr.Peterson!

Just wanted to ask, so then why are polynomials of the second degree quadratics? ‘Quad’ reminds me of 4 (quadrilateral is a shape with 4 sides, a quadrant is 1/4 of a circle)

And what’s the difference between an identity and an expression?

1. Dave Peterson

Hi, Sarah.
First, the “quad” issue is another common question; for an answer, start here: Names of Polynomials
Second, an identity is not an expression, but an equation (that is, a statement that two expressions are equal). In particular, it is an equation that is true for any value of the variable(s) it contains. For instance, a+b = b+a is an identity.


# How can I prove by induction that $9^k - 5^k$ is divisible by 4?

Recently had this on a discrete math test, which sadly I think I failed. But the question asked:

Prove that $9^k - 5^k$ is divisible by $4$.

Using the only approach I learned in the class, I substituted $n = k$, and tried to prove for $k+1$ like this:

$$9^{k+1} - 5^{k+1},$$

which just factors to $9 \cdot 9^k - 5 \cdot 5^k$.

But I cannot factor out $9^k - 5^k$, so I'm totally stuck.

• hint: 9 = (5+4) and distribute. – imranfat Oct 16 '13 at 16:56
• Alternatively you can note that $a^n-b^n=(a-b)\sum _{i=0} ^{n-1} a^i b^{n-1-i}$. – pppqqq Oct 16 '13 at 16:58

\begin{align} 9\cdot 9^k - 5\cdot 5^k & = (4 + 5)\cdot 9^k - 5\cdot 5^k \\ \\ & = 4\cdot 9^k + 5 \cdot 9^k - 5\cdot 5^k \\ \\ & = 4\cdot 9^k + 5(9^k - 5^k)\\ \\ & \quad \text{ use inductive hypothesis}\quad\cdots\end{align}

• And don't forget to include verification of the base case in your proof: $n = 0$ or $n = 1$. – Namaste Oct 16 '13 at 17:10
• Okay this seems to make the most sense and is closest to how my prof. teaches it. Dam I really hate theoretical math and proofs! – krb686 Oct 16 '13 at 17:11
• @amWhy: Nice to get a nice answer +1 – Amzoti Oct 18 '13 at 0:57

${\color{White}{\text{Proof without words.}}}$

• Very nice indeed! – Brian M. Scott Oct 17 '13 at 0:02

$9^{k+1}-5^{k+1}=(8+1)9^k-(4+1)5^k=9^k-5^k+4(2\cdot 9^k-5^k)$ The secret to induction proofs is usually to find a way to relate the $k+1$ case to the $k$ case.

Alternately, just note $9\equiv 1 \pmod 4, 5 \equiv 1 \pmod 4$, so $9^k-5^k \equiv 1^k-1^k \pmod 4$

Note that induction is not actually needed: you can instead use the identity

$$x^n-y^n=(x-y)\sum_{k=0}^{n-1}x^ky^{n-1-k}\;.$$

In this case it becomes

$$9^n-5^n=4\sum_{k=0}^{n-1}x^ky^{n-1-k}\;,$$

and since the summation clearly yields an integer, we have $4\mid 9^n-5^n$. The identity is a standard one, but it’s also not hard to prove:

\begin{align*} (x-y)\sum_{k=0}^{n-1}x^ky^{n-1-k}&=x\sum_{k=0}^{n-1}x^ky^{n-1-k}-y\sum_{k=0}^{n-1}x^ky^{n-1-k}\\\\ &=\sum_{k=0}^{n-1}x^{k+1}y^{n-1-k}-\sum_{k=0}^{n-1}x^ky^{n-k}\\\\ &=\sum_{k=1}^nx^ky^{n-k}-\sum_{k=0}^{n-1}x^ky^{n-k}\\\\ &=x^ny^0+\sum_{k=1}^{n-1}x^ky^{n-k}-\sum_{k=1}^{n-1}x^ky^{n-k}-x^0y^n\\\\ &=x^n-y^n\;. \end{align*}

• My test asked to use induction so that's why I narrowed that down, but thank you for all the in depth info! – krb686 Oct 17 '13 at 3:49
• @krb686: You’re welcome! – Brian M. Scott Oct 17 '13 at 3:50

It sounds like you understand the basic idea, but you're having trouble with the "trick" in the inductive step. This is common for proofs by mathematical induction, and you'll get better at it with more experience.

As you said, you're assuming that $9^k-5^k$ is divisible by four, and you want to prove that $9^{k+1}-5^{k+1}$ is divisible by four. You noticed that you can write the last expression as $9\cdot 9^k-5\cdot 5^k$, and you said you're having trouble applying your assumption to it. What happens if you add $9\cdot 5^k-9\cdot 5^k$ to your expression? This is just zero, so you're not changing anything, but you can write $$9^{k+1}-5^{k+1} =9\cdot 9^k-5\cdot 5^k+9\cdot 5^k-9\cdot 5^k =9\cdot (9^k-5^k)+(9-5)\cdot 5^k.$$ By the assumption, the first term, $9\cdot (9^k-5^k)$, is divisible by four, and by observation, the second term, $(9-5)\cdot 5^k$, is divisible by four, so the sum of the two terms and therefore $9^{k+1}-5^{k+1}$ are divisible by four. This is what you wanted to show.

You can always add zero (or multiply by one), and by writing zero (or one) in a particular way, you can sometimes make an argument easier to prove. Also, since your proof is by mathematical induction, you need to show that $9^n-5^n$ is divisible by four for, say, $n=0$ or $n=1$. You may have already done so and not asked about it in your question, but it's worth mentioning.


# §1.9 Calculus of a Complex Variable

## §1.9(i) Complex Numbers

 1.9.1 $z=x+iy,$ $x,y\in\mathbb{R}$. ⓘ Symbols: $\in$: element of, $\mathbb{R}$: real line and $z$: variable A&S Ref: 3.7.1 Permalink: http://dlmf.nist.gov/1.9.E1 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9 and 1

### Real and Imaginary Parts

 1.9.2 $\displaystyle\Re z$ $\displaystyle=x,$ $\displaystyle\Im z$ $\displaystyle=y.$ ⓘ Defines: $\Im$: imaginary part and $\Re$: real part Symbols: $z$: variable A&S Ref: 3.7.5 3.7.6 Permalink: http://dlmf.nist.gov/1.9.E2 Encodings: TeX, TeX, pMML, pMML, png, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1

### Polar Representation

 1.9.3 $\displaystyle x$ $\displaystyle=r\cos\theta,$ $\displaystyle y$ $\displaystyle=r\sin\theta,$ ⓘ Symbols: $\cos\NVar{z}$: cosine function, $\sin\NVar{z}$: sine function, $r$: radius and $\theta$: angle A&S Ref: 3.7.2 Referenced by: §1.9(ii) Permalink: http://dlmf.nist.gov/1.9.E3 Encodings: TeX, TeX, pMML, pMML, png, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1

where

 1.9.4 $r=(x^{2}+y^{2})^{1/2},$ ⓘ Symbols: $r$: radius A&S Ref: 3.7.3 Permalink: http://dlmf.nist.gov/1.9.E4 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1

and when $z\neq 0$,

 1.9.5 $\theta=\omega,\;\;\pi-\omega,\;\;-\pi+\omega,\mbox{ or }-\omega,$ ⓘ Symbols: $\pi$: the ratio of the circumference of a circle to its diameter and $\theta$: angle Permalink: http://dlmf.nist.gov/1.9.E5 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1

according as $z$ lies in the 1st, 2nd, 3rd, or 4th quadrants. Here

 1.9.6 $\omega=\operatorname{arctan}\left(|y/x|\right)\in\left[0,\tfrac{1}{2}\pi\right].$ ⓘ Symbols: $\pi$: the ratio of the circumference of a circle to its diameter, $[\NVar{a},\NVar{b}]$: closed interval, $\in$: element of and $\operatorname{arctan}\NVar{z}$: arctangent function A&S Ref: 3.7.4 Permalink: http://dlmf.nist.gov/1.9.E6 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1

### Modulus and Phase

 1.9.7 $\displaystyle|z|$ $\displaystyle=r,$ $\displaystyle\operatorname{ph}z$ $\displaystyle=\theta+2n\pi,$ $n\in\mathbb{Z}$. ⓘ Defines: $\operatorname{ph}$: phase Symbols: $\pi$: the ratio of the circumference of a circle to its diameter, $\in$: element of, $\mathbb{Z}$: set of all integers, $z$: variable, $n$: nonnegative integer, $r$: radius and $\theta$: angle Permalink: http://dlmf.nist.gov/1.9.E7 Encodings: TeX, TeX, pMML, pMML, png, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1

The principal value of $\operatorname{ph}z$ corresponds to $n=0$, that is, $-\pi\leq\operatorname{ph}z\leq\pi$. It is single-valued on $\mathbb{C}\setminus\{0\}$, except on the interval $(-\infty,0)$ where it is discontinuous and two-valued. Unless indicated otherwise, these principal values are assumed throughout the DLMF. (However, if we require a principal value to be single-valued, then we can restrict $-\pi<\operatorname{ph}z\leq\pi$.)

 1.9.8 $\displaystyle|\Re z|$ $\displaystyle\leq|z|,$ $\displaystyle|\Im z|$ $\displaystyle\leq|z|,$ ⓘ Symbols: $\Im$: imaginary part, $\Re$: real part and $z$: variable Permalink: http://dlmf.nist.gov/1.9.E8 Encodings: TeX, TeX, pMML, pMML, png, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1
 1.9.9 $z=re^{i\theta},$ ⓘ Symbols: $\mathrm{e}$: base of exponential function, $z$: variable, $r$: radius and $\theta$: angle Permalink: http://dlmf.nist.gov/1.9.E9 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1

where

 1.9.10 $e^{i\theta}=\cos\theta+i\sin\theta;$ ⓘ Symbols: $\cos\NVar{z}$: cosine function, $\mathrm{e}$: base of exponential function, $\sin\NVar{z}$: sine function and $\theta$: angle Permalink: http://dlmf.nist.gov/1.9.E10 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1

see §4.14.

### Complex Conjugate

 1.9.11 $\displaystyle\overline{z}$ $\displaystyle=x-iy,$ ⓘ Defines: $\overline{\NVar{z}}$: complex conjugate Symbols: $z$: variable A&S Ref: 3.7.7 Permalink: http://dlmf.nist.gov/1.9.E11 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1 1.9.12 $\displaystyle|\overline{z}|$ $\displaystyle=|z|,$ ⓘ Symbols: $\overline{\NVar{z}}$: complex conjugate and $z$: variable A&S Ref: 3.7.8 Permalink: http://dlmf.nist.gov/1.9.E12 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1 1.9.13 $\displaystyle\operatorname{ph}\overline{z}$ $\displaystyle=-\operatorname{ph}z.$ ⓘ Symbols: $\overline{\NVar{z}}$: complex conjugate, $\operatorname{ph}$: phase and $z$: variable A&S Ref: 3.7.9 Permalink: http://dlmf.nist.gov/1.9.E13 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1

### Arithmetic Operations

If $z_{1}=x_{1}+iy_{1}$, $z_{2}=x_{2}+iy_{2}$, then

 1.9.14 $z_{1}\pm z_{2}=x_{1}\pm x_{2}+\mathrm{i}(y_{1}\pm y_{2}),$ ⓘ Symbols: $z$: variable Permalink: http://dlmf.nist.gov/1.9.E14 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1
 1.9.15 $z_{1}z_{2}=x_{1}x_{2}-y_{1}y_{2}+i(x_{1}y_{2}+x_{2}y_{1}),$ ⓘ Symbols: $z$: variable A&S Ref: 3.7.10 Permalink: http://dlmf.nist.gov/1.9.E15 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1
 1.9.16 $\frac{z_{1}}{z_{2}}=\frac{z_{1}\overline{z}_{2}}{|z_{2}|^{2}}=\frac{x_{1}x_{2}% +y_{1}y_{2}+i(x_{2}y_{1}-x_{1}y_{2})}{x_{2}^{2}+y_{2}^{2}},$ ⓘ Symbols: $\overline{\NVar{z}}$: complex conjugate and $z$: variable A&S Ref: 3.7.13 Permalink: http://dlmf.nist.gov/1.9.E16 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1

provided that $z_{2}\neq 0$. Also,

 1.9.17 $|z_{1}z_{2}|=|z_{1}|\;|z_{2}|,$ ⓘ Symbols: $z$: variable A&S Ref: 3.7.11 Permalink: http://dlmf.nist.gov/1.9.E17 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1
 1.9.18 $\operatorname{ph}\left(z_{1}z_{2}\right)=\operatorname{ph}z_{1}+\operatorname{% ph}z_{2},$ ⓘ Symbols: $\operatorname{ph}$: phase and $z$: variable A&S Ref: 3.7.12 Referenced by: §1.9(i) Permalink: http://dlmf.nist.gov/1.9.E18 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1
 1.9.19 $\left|\frac{z_{1}}{z_{2}}\right|=\frac{|z_{1}|}{|z_{2}|},$ ⓘ Symbols: $z$: variable A&S Ref: 3.7.14 Permalink: http://dlmf.nist.gov/1.9.E19 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1
 1.9.20 $\operatorname{ph}\frac{z_{1}}{z_{2}}=\operatorname{ph}z_{1}-\operatorname{ph}z% _{2}.$ ⓘ Symbols: $\operatorname{ph}$: phase and $z$: variable A&S Ref: 3.7.15 Referenced by: §1.9(i) Permalink: http://dlmf.nist.gov/1.9.E20 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1

Equations (1.9.18) and (1.9.20) hold for general values of the phases, but not necessarily for the principal values.

### Powers

 1.9.21 $z^{n}=\left(x^{n}-\genfrac{(}{)}{0.0pt}{}{n}{2}x^{n-2}y^{2}+\genfrac{(}{)}{0.0% pt}{}{n}{4}x^{n-4}y^{4}-\cdots\right)+i\left(\genfrac{(}{)}{0.0pt}{}{n}{1}x^{n% -1}y-\genfrac{(}{)}{0.0pt}{}{n}{3}x^{n-3}y^{3}+\cdots\right),$ $n=1,2,\dots$. ⓘ Symbols: $\genfrac{(}{)}{0.0pt}{}{\NVar{m}}{\NVar{n}}$: binomial coefficient, $z$: variable and $n$: nonnegative integer A&S Ref: 3.7.22 Permalink: http://dlmf.nist.gov/1.9.E21 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1

### DeMoivre’s Theorem

 1.9.22 $\cos n\theta+i\sin n\theta=(\cos\theta+i\sin\theta)^{n},$ $n\in\mathbb{Z}$.

### Triangle Inequality

 1.9.23 $\left|\left|z_{1}\right|-\left|z_{2}\right|\right|\leq\left|z_{1}+z_{2}\right|% \leq\left|z_{1}\right|+\left|z_{2}\right|.$ ⓘ Symbols: $z$: variable A&S Ref: 3.7.29 Permalink: http://dlmf.nist.gov/1.9.E23 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1

## §1.9(ii) Continuity, Point Sets, and Differentiation

### Continuity

A function $f(z)$ is continuous at a point $z_{0}$ if $\lim\limits_{z\to z_{0}}f(z)=f(z_{0})$. That is, given any positive number $\epsilon$, however small, we can find a positive number $\delta$ such that $|f(z)-f(z_{0})|<\epsilon$ for all $z$ in the open disk $|z-z_{0}|<\delta$.

A function of two complex variables $f(z,w)$ is continuous at $(z_{0},w_{0})$ if $\lim\limits_{(z,w)\to(z_{0},w_{0})}f(z,w)=f(z_{0},w_{0})$; compare (1.5.1) and (1.5.2).

### Point Sets in $\mathbb{C}$

A neighborhood of a point $z_{0}$ is a disk $\left|z-z_{0}\right|<\delta$. An open set in $\mathbb{C}$ is one in which each point has a neighborhood that is contained in the set.

A point $z_{0}$ is a limit point (limiting point or accumulation point) of a set of points $S$ in $\mathbb{C}$ (or $\mathbb{C}\cup\infty$) if every neighborhood of $z_{0}$ contains a point of $S$ distinct from $z_{0}$. ($z_{0}$ may or may not belong to $S$.) As a consequence, every neighborhood of a limit point of $S$ contains an infinite number of points of $S$. Also, the union of $S$ and its limit points is the closure of $S$.

A domain $D$, say, is an open set in $\mathbb{C}$ that is connected, that is, any two points can be joined by a polygonal arc (a finite chain of straight-line segments) lying in the set. Any point whose neighborhoods always contain members and nonmembers of $D$ is a boundary point of $D$. When its boundary points are added the domain is said to be closed, but unless specified otherwise a domain is assumed to be open.

A region is an open domain together with none, some, or all of its boundary points. Points of a region that are not boundary points are called interior points.

A function $f(z)$ is continuous on a region $R$ if for each point $z_{0}$ in $R$ and any given number $\epsilon$ ($>0$) we can find a neighborhood of $z_{0}$ such that $\left|f(z)-f(z_{0})\right|<\epsilon$ for all points $z$ in the intersection of the neighborhood with $R$.

### Differentiation

A function $f(z)$ is differentiable at a point $z$ if the following limit exists:

 1.9.24 $f^{\prime}(z)=\frac{\mathrm{d}f}{\mathrm{d}z}=\lim_{h\to 0}\frac{f(z+h)-f(z)}{% h}.$ ⓘ Symbols: $\frac{\mathrm{d}\NVar{f}}{\mathrm{d}\NVar{x}}$: derivative of $f$ with respect to $x$ and $z$: variable Permalink: http://dlmf.nist.gov/1.9.E24 Encodings: TeX, pMML, png See also: Annotations for 1.9(ii), 1.9(ii), 1.9 and 1

Differentiability automatically implies continuity.

### Cauchy–Riemann Equations

If $f^{\prime}(z)$ exists at $z=x+iy$ and $f(z)=u(x,y)+iv(x,y)$, then

 1.9.25 $\displaystyle\frac{\partial u}{\partial x}$ $\displaystyle=\frac{\partial v}{\partial y},$ $\displaystyle\frac{\partial u}{\partial y}$ $\displaystyle=-\frac{\partial v}{\partial x}$ ⓘ Symbols: $\frac{\partial\NVar{f}}{\partial\NVar{x}}$: partial derivative of $f$ with respect to $x$, $\partial\NVar{x}$: partial differential of $x$, $u(x,y)$: function and $v(x,y)$: function A&S Ref: 3.7.30 Referenced by: §1.9(ii) Permalink: http://dlmf.nist.gov/1.9.E25 Encodings: TeX, TeX, pMML, pMML, png, png See also: Annotations for 1.9(ii), 1.9(ii), 1.9 and 1

at $(x,y)$.

Conversely, if at a given point $(x,y)$ the partial derivatives $\ifrac{\partial u}{\partial x}$, $\ifrac{\partial u}{\partial y}$, $\ifrac{\partial v}{\partial x}$, and $\ifrac{\partial v}{\partial y}$ exist, are continuous, and satisfy (1.9.25), then $f(z)$ is differentiable at $z=x+iy$.

### Analyticity

A function $f(z)$ is said to be analytic (holomorphic) at $z=z_{0}$ if it is differentiable in a neighborhood of $z_{0}$.

A function $f(z)$ is analytic in a domain $D$ if it is analytic at each point of $D$. A function analytic at every point of $\mathbb{C}$ is said to be entire.

If $f(z)$ is analytic in an open domain $D$, then each of its derivatives $f^{\prime}(z)$, $f^{\prime\prime}(z)$, $\dots$ exists and is analytic in $D$.

### Harmonic Functions

If $f(z)=u(x,y)+iv(x,y)$ is analytic in an open domain $D$, then $u$ and $v$ are harmonic in $D$, that is,

 1.9.26 $\frac{{\partial}^{2}u}{{\partial x}^{2}}+\frac{{\partial}^{2}u}{{\partial y}^{% 2}}=\frac{{\partial}^{2}v}{{\partial x}^{2}}+\frac{{\partial}^{2}v}{{\partial y% }^{2}}=0,$

or in polar form ((1.9.3)) $u$ and $v$ satisfy

 1.9.27 $\frac{{\partial}^{2}u}{{\partial r}^{2}}+\frac{1}{r}\frac{\partial u}{\partial r% }+\frac{1}{r^{2}}\frac{{\partial}^{2}u}{{\partial\theta}^{2}}=0$

at all points of $D$.

## §1.9(iii) Integration

An arc $C$ is given by $z(t)=x(t)+iy(t)$, $a\leq t\leq b$, where $x$ and $y$ are continuously differentiable. If $x(t)$ and $y(t)$ are continuous and $x^{\prime}(t)$ and $y^{\prime}(t)$ are piecewise continuous, then $z(t)$ defines a contour.

A contour is simple if it contains no multiple points, that is, for every pair of distinct values $t_{1},t_{2}$ of $t$, $z(t_{1})\neq z(t_{2})$. A simple closed contour is a simple contour, except that $z(a)=z(b)$.

Next,

 1.9.28 $\int_{C}f(z)\mathrm{d}z=\int_{a}^{b}f(z(t))(x^{\prime}(t)+iy^{\prime}(t))% \mathrm{d}t,$ ⓘ Symbols: $\mathrm{d}\NVar{x}$: differential of $x$, $\int$: integral and $C$: closed contour Permalink: http://dlmf.nist.gov/1.9.E28 Encodings: TeX, pMML, png See also: Annotations for 1.9(iii), 1.9 and 1

for a contour $C$ and $f(z(t))$ continuous, $a\leq t\leq b$. If $f(z(t_{0}))=\infty$, $a\leq t_{0}\leq b$, then the integral is defined analogously to the infinite integrals in §1.4(v). Similarly when $a=-\infty$ or $b=+\infty$.

### Jordan Curve Theorem

Any simple closed contour $C$ divides $\mathbb{C}$ into two open domains that have $C$ as common boundary. One of these domains is bounded and is called the interior domain of $C$; the other is unbounded and is called the exterior domain of $C$.

### Cauchy’s Theorem

If $f(z)$ is continuous within and on a simple closed contour $C$ and analytic within $C$, then

 1.9.29 $\int_{C}f(z)\mathrm{d}z=0.$ ⓘ Symbols: $\mathrm{d}\NVar{x}$: differential of $x$, $\int$: integral and $C$: closed contour Permalink: http://dlmf.nist.gov/1.9.E29 Encodings: TeX, pMML, png See also: Annotations for 1.9(iii), 1.9(iii), 1.9 and 1

### Cauchy’s Integral Formula

If $f(z)$ is continuous within and on a simple closed contour $C$ and analytic within $C$, and if $z_{0}$ is a point within $C$, then

 1.9.30 $f(z_{0})=\frac{1}{2\pi i}\int_{C}\frac{f(z)}{z-z_{0}}\mathrm{d}z,$ ⓘ Symbols: $\pi$: the ratio of the circumference of a circle to its diameter, $\mathrm{d}\NVar{x}$: differential of $x$, $\int$: integral and $C$: closed contour Referenced by: §2.3(iii) Permalink: http://dlmf.nist.gov/1.9.E30 Encodings: TeX, pMML, png See also: Annotations for 1.9(iii), 1.9(iii), 1.9 and 1

and

 1.9.31 $f^{(n)}(z_{0})=\frac{n!}{2\pi i}\int_{C}\frac{f(z)}{(z-z_{0})^{n+1}}\mathrm{d}z,$ $n=1,2,3,\dots$,

provided that in both cases $C$ is described in the positive rotational (anticlockwise) sense.

### Liouville’s Theorem

Any bounded entire function is a constant.

### Winding Number

If $C$ is a closed contour, and $z_{0}\not\in C$, then

 1.9.32 $\frac{1}{2\pi i}\int_{C}\frac{1}{z-z_{0}}\mathrm{d}z=\mathcal{N}(C,z_{0}),$ ⓘ Defines: $\mathcal{N}(C,z_{0})$: winding number of $C$ (locally) Symbols: $\pi$: the ratio of the circumference of a circle to its diameter, $\mathrm{d}\NVar{x}$: differential of $x$, $\int$: integral and $C$: closed contour Permalink: http://dlmf.nist.gov/1.9.E32 Encodings: TeX, pMML, png See also: Annotations for 1.9(iii), 1.9(iii), 1.9 and 1

where $\mathcal{N}(C,z_{0})$ is an integer called the winding number of $C$ with respect to $z_{0}$. If $C$ is simple and oriented in the positive rotational sense, then $\mathcal{N}(C,z_{0})$ is $1$ or $0$ depending whether $z_{0}$ is inside or outside $C$.

### Mean Value Property

For $u(z)$ harmonic,

 1.9.33 $u(z)=\frac{1}{2\pi}\int^{2\pi}_{0}u(z+re^{i\phi})\mathrm{d}\phi.$

### Poisson Integral

If $h(w)$ is continuous on $|w|=R$, then with $z=re^{i\theta}$

 1.9.34 $u(re^{i\theta})=\frac{1}{2\pi}\int^{2\pi}_{0}\frac{(R^{2}-r^{2})h(Re^{i\phi})% \mathrm{d}\phi}{R^{2}-2Rr\cos\left(\phi-\theta\right)+r^{2}}$

is harmonic in $|z|. Also with $\left|w\right|=R$, $\lim\limits_{z\to w}u(z)=h(w)$ as $z\to w$ within $|z|.

## §1.9(iv) Conformal Mapping

The extended complex plane, $\mathbb{C}\,\cup\,\{\infty\}$, consists of the points of the complex plane $\mathbb{C}$ together with an ideal point $\infty$ called the point at infinity. A system of open disks around infinity is given by

 1.9.35 $S_{r}=\{z\mid|z|>1/r\}\cup\{\infty\},$ $0. ⓘ Symbols: $\cup$: union, $z$: variable, $r$: radius and $S_{r}$: neighborhood Permalink: http://dlmf.nist.gov/1.9.E35 Encodings: TeX, pMML, png See also: Annotations for 1.9(iv), 1.9 and 1

Each $S_{r}$ is a neighborhood of $\infty$. Also,

 1.9.36 $\infty\pm z=z\pm\infty=\infty,$ ⓘ Symbols: $z$: variable Permalink: http://dlmf.nist.gov/1.9.E36 Encodings: TeX, pMML, png See also: Annotations for 1.9(iv), 1.9 and 1
 1.9.37 $\infty\cdot z=z\cdot\infty=\infty,$ $z\not=0$, ⓘ Symbols: $z$: variable Permalink: http://dlmf.nist.gov/1.9.E37 Encodings: TeX, pMML, png See also: Annotations for 1.9(iv), 1.9 and 1
 1.9.38 $z/\infty=0,$ ⓘ Symbols: $z$: variable Permalink: http://dlmf.nist.gov/1.9.E38 Encodings: TeX, pMML, png See also: Annotations for 1.9(iv), 1.9 and 1
 1.9.39 $z/0=\infty,$ $z\neq 0$. ⓘ Symbols: $z$: variable Permalink: http://dlmf.nist.gov/1.9.E39 Encodings: TeX, pMML, png See also: Annotations for 1.9(iv), 1.9 and 1

A function $f(z)$ is analytic at $\infty$ if $g(z)=f(1/z)$ is analytic at $z=0$, and we set $f^{\prime}(\infty)=g^{\prime}(0)$.

### Conformal Transformation

Suppose $f(z)$ is analytic in a domain $D$ and $C_{1},C_{2}$ are two arcs in $D$ passing through $z_{0}$. Let $C^{\prime}_{1},C^{\prime}_{2}$ be the images of $C_{1}$ and $C_{2}$ under the mapping $w=f(z)$. The angle between $C_{1}$ and $C_{2}$ at $z_{0}$ is the angle between the tangents to the two arcs at $z_{0}$, that is, the difference of the signed angles that the tangents make with the positive direction of the real axis. If $f^{\prime}(z_{0})\not=0$, then the angle between $C_{1}$ and $C_{2}$ equals the angle between $C^{\prime}_{1}$ and $C^{\prime}_{2}$ both in magnitude and sense. We then say that the mapping $w=f(z)$ is conformal (angle-preserving) at $z_{0}$.

The linear transformation $f(z)=az+b$, $a\not=0$, has $f^{\prime}(z)=a$ and $w=f(z)$ maps $\mathbb{C}$ conformally onto $\mathbb{C}$.

### Bilinear Transformation

 1.9.40 $w=f(z)=\frac{az+b}{cz+d},$ $ad-bc\not=0$, $c\not=0$. ⓘ Symbols: $z$: variable and $w$: variable Referenced by: §1.9(iv) Permalink: http://dlmf.nist.gov/1.9.E40 Encodings: TeX, pMML, png See also: Annotations for 1.9(iv), 1.9(iv), 1.9 and 1
 1.9.41 $\displaystyle f(-d/c)$ $\displaystyle=\infty,$ $\displaystyle f(\infty)$ $\displaystyle=a/c.$ ⓘ Permalink: http://dlmf.nist.gov/1.9.E41 Encodings: TeX, TeX, pMML, pMML, png, png See also: Annotations for 1.9(iv), 1.9(iv), 1.9 and 1
 1.9.42 $f^{\prime}(z)=\frac{ad-bc}{(cz+d)^{2}},$ $z\not=-d/c$. ⓘ Symbols: $z$: variable Permalink: http://dlmf.nist.gov/1.9.E42 Encodings: TeX, pMML, png See also: Annotations for 1.9(iv), 1.9(iv), 1.9 and 1
 1.9.43 $f^{\prime}(\infty)=\frac{bc-ad}{c^{2}}.$ ⓘ Permalink: http://dlmf.nist.gov/1.9.E43 Encodings: TeX, pMML, png See also: Annotations for 1.9(iv), 1.9(iv), 1.9 and 1
 1.9.44 $z=\frac{dw-b}{-cw+a}.$ ⓘ Symbols: $z$: variable and $w$: variable Permalink: http://dlmf.nist.gov/1.9.E44 Encodings: TeX, pMML, png See also: Annotations for 1.9(iv), 1.9(iv), 1.9 and 1

The transformation (1.9.40) is a one-to-one conformal mapping of $\mathbb{C}\,\cup\,\{\infty\}$ onto itself.

The cross ratio of $z_{1},z_{2},z_{3},z_{4}\in\mathbb{C}\cup\{\infty\}$ is defined by

 1.9.45 $\frac{(z_{1}-z_{2})(z_{3}-z_{4})}{(z_{1}-z_{4})(z_{3}-z_{2})},$ ⓘ Symbols: $z$: variable Permalink: http://dlmf.nist.gov/1.9.E45 Encodings: TeX, pMML, png See also: Annotations for 1.9(iv), 1.9(iv), 1.9 and 1

or its limiting form, and is invariant under bilinear transformations.

Other names for the bilinear transformation are fractional linear transformation, homographic transformation, and Möbius transformation.

## §1.9(v) Infinite Sequences and Series

A sequence $\{z_{n}\}$ converges to $z$ if $\lim\limits_{n\to\infty}z_{n}=z$. For $z_{n}=x_{n}+iy_{n}$, the sequence $\{z_{n}\}$ converges iff the sequences $\{x_{n}\}$ and $\{y_{n}\}$ separately converge. A series $\sum^{\infty}_{n=0}z_{n}$ converges if the sequence $s_{n}=\sum^{n}_{k=0}z_{k}$ converges. The series is divergent if $s_{n}$ does not converge. The series converges absolutely if $\sum^{\infty}_{n=0}|z_{n}|$ converges. A series $\sum^{\infty}_{n=0}z_{n}$ converges (diverges) absolutely when $\lim\limits_{n\to\infty}|z_{n}|^{1/n}<1$ ($>1$), or when $\lim\limits_{n\to\infty}\left|\ifrac{z_{n+1}}{z_{n}}\right|<1$ ($>1$). Absolutely convergent series are also convergent.

Let $\{f_{n}(z)\}$ be a sequence of functions defined on a set $S$. This sequence converges pointwise to a function $f(z)$ if

 1.9.46 $f(z)=\lim_{n\to\infty}f_{n}(z)$ ⓘ Symbols: $z$: variable and $n$: nonnegative integer Permalink: http://dlmf.nist.gov/1.9.E46 Encodings: TeX, pMML, png See also: Annotations for 1.9(v), 1.9 and 1

for each $z\in S$. The sequence converges uniformly on $S$, if for every $\epsilon>0$ there exists an integer $N$, independent of $z$, such that

 1.9.47 $|f_{n}(z)-f(z)|<\epsilon$ ⓘ Symbols: $z$: variable, $n$: nonnegative integer and $\epsilon$: positive number Permalink: http://dlmf.nist.gov/1.9.E47 Encodings: TeX, pMML, png See also: Annotations for 1.9(v), 1.9 and 1

for all $z\in S$ and $n\geq N$.

A series $\sum^{\infty}_{n=0}f_{n}(z)$ converges uniformly on $S$, if the sequence $s_{n}(z)=\sum^{n}_{k=0}f_{k}(z)$ converges uniformly on $S$.

### Weierstrass $M$-test

Suppose $\{M_{n}\}$ is a sequence of real numbers such that $\sum^{\infty}_{n=0}M_{n}$ converges and $|f_{n}(z)|\leq M_{n}$ for all $z\in S$ and all $n\geq 0$. Then the series $\sum^{\infty}_{n=0}f_{n}(z)$ converges uniformly on $S$.

A doubly-infinite series $\sum^{\infty}_{n=-\infty}f_{n}(z)$ converges (uniformly) on $S$ iff each of the series $\sum^{\infty}_{n=0}f_{n}(z)$ and $\sum^{\infty}_{n=1}f_{-n}(z)$ converges (uniformly) on $S$.

## §1.9(vi) Power Series

For a series $\sum^{\infty}_{n=0}a_{n}(z-z_{0})^{n}$ there is a number $R$, $0\leq R\leq\infty$, such that the series converges for all $z$ in $|z-z_{0}| and diverges for $z$ in $|z-z_{0}|>R$. The circle $|z-z_{0}|=R$ is called the circle of convergence of the series, and $R$ is the radius of convergence. Inside the circle the sum of the series is an analytic function $f(z)$. For $z$ in $|z-z_{0}|\leq\rho$ ($), the convergence is absolute and uniform. Moreover,

 1.9.48 $a_{n}=\frac{f^{(n)}(z_{0})}{n!},$ ⓘ Symbols: $!$: factorial (as in $n!$), $z$: variable and $n$: nonnegative integer Permalink: http://dlmf.nist.gov/1.9.E48 Encodings: TeX, pMML, png See also: Annotations for 1.9(vi), 1.9 and 1

and

 1.9.49 $R=\liminf_{n\to\infty}|a_{n}|^{-1/n}.$ ⓘ Defines: $R$: radius of convergence (locally) Symbols: $\liminf$: least limit point and $n$: nonnegative integer Permalink: http://dlmf.nist.gov/1.9.E49 Encodings: TeX, pMML, png See also: Annotations for 1.9(vi), 1.9 and 1

For the converse of this result see §1.10(i).

### Operations

When $\sum a_{n}z^{n}$ and $\sum b_{n}z^{n}$ both converge

 1.9.50 $\sum^{\infty}_{n=0}(a_{n}\pm b_{n})z^{n}=\sum^{\infty}_{n=0}a_{n}z^{n}\pm\sum^% {\infty}_{n=0}b_{n}z^{n},$ ⓘ Symbols: $z$: variable and $n$: nonnegative integer Permalink: http://dlmf.nist.gov/1.9.E50 Encodings: TeX, pMML, png See also: Annotations for 1.9(vi), 1.9(vi), 1.9 and 1

and

 1.9.51 $\left(\sum^{\infty}_{n=0}a_{n}z^{n}\right)\left(\sum^{\infty}_{n=0}b_{n}z^{n}% \right)=\sum^{\infty}_{n=0}c_{n}z^{n},$ ⓘ Symbols: $z$: variable and $n$: nonnegative integer Permalink: http://dlmf.nist.gov/1.9.E51 Encodings: TeX, pMML, png See also: Annotations for 1.9(vi), 1.9(vi), 1.9 and 1

where

 1.9.52 $c_{n}=\sum^{n}_{k=0}a_{k}b_{n-k}.$ ⓘ Symbols: $k$: integer and $n$: nonnegative integer Permalink: http://dlmf.nist.gov/1.9.E52 Encodings: TeX, pMML, png See also: Annotations for 1.9(vi), 1.9(vi), 1.9 and 1

Next, let

 1.9.53 $f(z)=a_{0}+a_{1}z+a_{2}z^{2}+\cdots,$ $a_{0}\not=0$. ⓘ Symbols: $z$: variable Permalink: http://dlmf.nist.gov/1.9.E53 Encodings: TeX, pMML, png See also: Annotations for 1.9(vi), 1.9(vi), 1.9 and 1

Then the expansions (1.9.54), (1.9.57), and (1.9.60) hold for all sufficiently small $|z|$.

 1.9.54 $\frac{1}{f(z)}=b_{0}+b_{1}z+b_{2}z^{2}+\cdots,$ ⓘ Symbols: $z$: variable Referenced by: §1.9(vi) Permalink: http://dlmf.nist.gov/1.9.E54 Encodings: TeX, pMML, png See also: Annotations for 1.9(vi), 1.9(vi), 1.9 and 1

where

 1.9.55 $\displaystyle b_{0}$ $\displaystyle=1/a_{0},$ $\displaystyle b_{1}$ $\displaystyle=-a_{1}/a_{0}^{2},$ $\displaystyle b_{2}$ $\displaystyle=(a_{1}^{2}-a_{0}a_{2})/a_{0}^{3},$ ⓘ Permalink: http://dlmf.nist.gov/1.9.E55 Encodings: TeX, TeX, TeX, pMML, pMML, pMML, png, png, png See also: Annotations for 1.9(vi), 1.9(vi), 1.9 and 1
 1.9.56 $b_{n}=-(a_{1}b_{n-1}+a_{2}b_{n-2}+\dots+a_{n}b_{0})/a_{0},$ $n\geq 1$. ⓘ Symbols: $n$: nonnegative integer Permalink: http://dlmf.nist.gov/1.9.E56 Encodings: TeX, pMML, png See also: Annotations for 1.9(vi), 1.9(vi), 1.9 and 1

With $a_{0}=1$,

 1.9.57 $\ln f(z)=q_{1}z+q_{2}z^{2}+q_{3}z^{3}+\cdots,$ ⓘ Symbols: $\ln\NVar{z}$: principal branch of logarithm function, $z$: variable and $q_{j}$: coefficients Referenced by: §1.9(vi) Permalink: http://dlmf.nist.gov/1.9.E57 Encodings: TeX, pMML, png See also: Annotations for 1.9(vi), 1.9(vi), 1.9 and 1

(principal value), where

 1.9.58 $\displaystyle q_{1}$ $\displaystyle=a_{1},$ $\displaystyle q_{2}$ $\displaystyle=(2a_{2}-a_{1}^{2})/2,$ $\displaystyle q_{3}$ $\displaystyle=(3a_{3}-3a_{1}a_{2}+a_{1}^{3})/3,$ ⓘ Symbols: $q_{j}$: coefficients Permalink: http://dlmf.nist.gov/1.9.E58 Encodings: TeX, TeX, TeX, pMML, pMML, pMML, png, png, png See also: Annotations for 1.9(vi), 1.9(vi), 1.9 and 1

and

 1.9.59 $q_{n}=(na_{n}-(n-1)a_{1}q_{n-1}-(n-2)a_{2}q_{n-2}-\cdots-a_{n-1}q_{1})/n,$ $n\geq 2$. ⓘ Symbols: $n$: nonnegative integer and $q_{j}$: coefficients Permalink: http://dlmf.nist.gov/1.9.E59 Encodings: TeX, pMML, png See also: Annotations for 1.9(vi), 1.9(vi), 1.9 and 1

Also,

 1.9.60 $(f(z))^{\nu}=p_{0}+p_{1}z+p_{2}z^{2}+\cdots,$ ⓘ Symbols: $z$: variable, $\nu$: complex and $p_{j}$: coefficients Referenced by: §1.9(vi) Permalink: http://dlmf.nist.gov/1.9.E60 Encodings: TeX, pMML, png See also: Annotations for 1.9(vi), 1.9(vi), 1.9 and 1

(principal value), where $\nu\in\mathbb{C}$,

 1.9.61 $\displaystyle p_{0}$ $\displaystyle=1,$ $\displaystyle p_{1}$ $\displaystyle=\nu a_{1},$ $\displaystyle p_{2}$ $\displaystyle=\nu((\nu-1)a_{1}^{2}+2a_{2})/2,$ ⓘ Symbols: $\nu$: complex and $p_{j}$: coefficients Permalink: http://dlmf.nist.gov/1.9.E61 Encodings: TeX, TeX, TeX, pMML, pMML, pMML, png, png, png See also: Annotations for 1.9(vi), 1.9(vi), 1.9 and 1

and

 1.9.62 $p_{n}=((\nu-n+1)a_{1}p_{n-1}+(2\nu-n+2)a_{2}p_{n-2}+\dots+((n-1)\nu-1)a_{n-1}p% _{1}+n\nu a_{n})/n,$ $n\geq 1$. ⓘ Symbols: $n$: nonnegative integer, $\nu$: complex and $p_{j}$: coefficients Permalink: http://dlmf.nist.gov/1.9.E62 Encodings: TeX, pMML, png See also: Annotations for 1.9(vi), 1.9(vi), 1.9 and 1

For the definitions of the principal values of $\ln f(z)$ and $(f(z))^{\nu}$ see §§4.2(i) and 4.2(iv).

Lastly, a power series can be differentiated any number of times within its circle of convergence:

 1.9.63 $f^{(m)}(z)=\sum_{n=0}^{\infty}{\left(n+1\right)_{m}}a_{n+m}(z-z_{0})^{n},$ $\left|z-z_{0}\right|, $m=0,1,2,\dots$.

## §1.9(vii) Inversion of Limits

### Double Sequences and Series

A set of complex numbers $\{z_{m,n}\}$ where $m$ and $n$ take all positive integer values is called a double sequence. It converges to $z$ if for every $\epsilon>0$, there is an integer $N$ such that

 1.9.64 $|z_{m,n}-z|<\epsilon$ ⓘ Symbols: $z$: variable, $m$: nonnegative integer, $n$: nonnegative integer and $\epsilon$: positive number Permalink: http://dlmf.nist.gov/1.9.E64 Encodings: TeX, pMML, png See also: Annotations for 1.9(vii), 1.9(vii), 1.9 and 1

for all $m,n\geq N$. Suppose $\{z_{m,n}\}$ converges to $z$ and the repeated limits

 1.9.65 $\lim_{m\to\infty}\left(\lim_{n\to\infty}z_{m,n}\right),$ $\lim_{n\to\infty}\left(\lim_{m\to\infty}z_{m,n}\right)$ ⓘ Symbols: $z$: variable, $m$: nonnegative integer and $n$: nonnegative integer Permalink: http://dlmf.nist.gov/1.9.E65 Encodings: TeX, TeX, pMML, pMML, png, png See also: Annotations for 1.9(vii), 1.9(vii), 1.9 and 1

exist. Then both repeated limits equal $z$.

A double series is the limit of the double sequence

 1.9.66 $z_{p,q}=\sum^{p}_{m=0}\sum^{q}_{n=0}\zeta_{m,n}.$ ⓘ Symbols: $z$: variable, $m$: nonnegative integer, $n$: nonnegative integer and $\zeta_{p,q}$: sum Permalink: http://dlmf.nist.gov/1.9.E66 Encodings: TeX, pMML, png See also: Annotations for 1.9(vii), 1.9(vii), 1.9 and 1

If the limit exists, then the double series is convergent; otherwise it is divergent. The double series is absolutely convergent if it is convergent when $\zeta_{m,n}$ is replaced by $|\zeta_{m,n}|$.

If a double series is absolutely convergent, then it is also convergent and its sum is given by either of the repeated sums

 1.9.67 $\sum^{\infty}_{m=0}\left(\sum^{\infty}_{n=0}\zeta_{m,n}\right),$ $\sum^{\infty}_{n=0}\left(\sum^{\infty}_{m=0}\zeta_{m,n}\right).$ ⓘ Symbols: $m$: nonnegative integer, $n$: nonnegative integer and $\zeta_{p,q}$: sum Permalink: http://dlmf.nist.gov/1.9.E67 Encodings: TeX, TeX, pMML, pMML, png, png See also: Annotations for 1.9(vii), 1.9(vii), 1.9 and 1

### Term-by-Term Integration

Suppose the series $\sum^{\infty}_{n=0}f_{n}(z)$, where $f_{n}(z)$ is continuous, converges uniformly on every compact set of a domain $D$, that is, every closed and bounded set in $D$. Then

 1.9.68 $\int_{C}\sum^{\infty}_{n=0}f_{n}(z)\mathrm{d}z=\sum^{\infty}_{n=0}\int_{C}f_{n% }(z)\mathrm{d}z$

for any finite contour $C$ in $D$.

### Dominated Convergence Theorem

Let $(a,b)$ be a finite or infinite interval, and $f_{0}(t),f_{1}(t),\dots$ be real or complex continuous functions, $t\in(a,b)$. Suppose $\sum^{\infty}_{n=0}f_{n}(t)$ converges uniformly in any compact interval in $(a,b)$, and at least one of the following two conditions is satisfied:

 1.9.69 $\int^{b}_{a}\sum^{\infty}_{n=0}|f_{n}(t)|\mathrm{d}t<\infty,$ ⓘ Symbols: $\mathrm{d}\NVar{x}$: differential of $x$, $\int$: integral and $n$: nonnegative integer Referenced by: §1.9(vii) Permalink: http://dlmf.nist.gov/1.9.E69 Encodings: TeX, pMML, png See also: Annotations for 1.9(vii), 1.9(vii), 1.9 and 1
 1.9.70 $\sum^{\infty}_{n=0}\int^{b}_{a}|f_{n}(t)|\mathrm{d}t<\infty.$ ⓘ Symbols: $\mathrm{d}\NVar{x}$: differential of $x$, $\int$: integral and $n$: nonnegative integer Permalink: http://dlmf.nist.gov/1.9.E70 Encodings: TeX, pMML, png See also: Annotations for 1.9(vii), 1.9(vii), 1.9 and 1

Then

 1.9.71 $\int^{b}_{a}\sum^{\infty}_{n=0}f_{n}(t)\mathrm{d}t=\sum^{\infty}_{n=0}\int^{b}% _{a}f_{n}(t)\mathrm{d}t.$ ⓘ Symbols: $\mathrm{d}\NVar{x}$: differential of $x$, $\int$: integral and $n$: nonnegative integer Referenced by: §1.9(vii) Permalink: http://dlmf.nist.gov/1.9.E71 Encodings: TeX, pMML, png See also: Annotations for 1.9(vii), 1.9(vii), 1.9 and 1


# Mod of numbers with large exponents

I've read about Fermat's little theorem and generally how congruence works. But I can't figure out how to work out these two:

• $13^{100} \bmod 7$
• $7^{100} \bmod 13$

I've also heard of this formula:

$$a \equiv b\pmod n \Rightarrow a^k \equiv b^k \pmod n$$

But I don't see how exactly to use that here, because from $13^1 \bmod 7$ I get 6, and $13^2 \bmod 7$ is 1. I'm unclear as to which one to raise to the kth power here (I'm assuming k = 100?)

Any hints or pointers in the right direction would be great.

• Might be useful to recognize that $6=-1 \mod 7$ – Kitter Catter Nov 28 '16 at 0:26
• @KitterCatter Can you explain it a bit more? (possibly as an answer). Is it derived from $n | (a-b)$? – Roshnal Nov 28 '16 at 0:29
• I don't know about using fermat (not a mathematician by trade) but it might be helpful to know that the Euler totient function of a prime,$p$ is $p-1$ and that if $a^{\phi(p)} \equiv 1 \mod p$ – Kitter Catter Nov 28 '16 at 0:31
• Be sure to understand the relation between "mod" as a binary operator vs. ternary relation. See this answer and this one for more on this. If you only know the operator form you will be severely encumbered. – Bill Dubuque Nov 28 '16 at 0:32

The formula you've heard of results from the fact that congruences are compatible with addition and multiplication.

The first power $$13^{100}$$ is easy: $$13\equiv -1\mod 7$$, so $$13^{100}\equiv (-1)^{100}=1\pmod 7.$$

The second power uses Lil' Fermat: for any number $$a\not\equiv 0\mod 7$$, we have $$a^{12}\equiv 1\pmod{13}$$, hence $$7^{100}\equiv 7^{100\bmod12}\equiv 7^4\equiv 10^2\equiv 9\pmod{13}$$

• Very nice answer! When you reference Lil' Fermat did you mean $a\not\equiv0\mod13$? also it looks like you missed a bracket for the exponent in $a^{12} \equiv 1 \mod 13$ – Kitter Catter Nov 28 '16 at 0:47
• Thanks for the answer! I'm still a little unclear on how you got to $10^2$ from $7^4$? – Roshnal Nov 28 '16 at 1:03
• $7^4=(7^2)^2=49^2\equiv 10^2$, that's all. – Bernard Nov 28 '16 at 1:12
• Ah okay, thanks! It's clear now :) – Roshnal Nov 28 '16 at 1:15
• @Kitter Catter: Oh! Yes. I should have re-read my answer before posting. I even missed a pair of brackets. It's fixed now. Thanks for pointing the typos! – Bernard Nov 28 '16 at 1:15

Hint $$\,$$ The key idea is to use modular order reduction on exponents as in the Lemma below. We can find small exponents $$\,e\,$$ such that $$\,a^{\large e}\equiv 1\,$$ either by Euler's totient or Fermat's little theorem (or by Carmichael's lambda generalization), along with obvious roots of $$\,1\,$$ such as $$\,(-1)^2\equiv 1.$$

Theorem $$\ \$$ Suppose that: $$\,\ \color{#c00}{a^{\large e}\equiv\, 1}\,\pmod{\! m}\$$ and $$\, e>0,\ n,k\ge 0\,$$ are integers. Then

$$\qquad n\equiv k\pmod{\! \color{#c00}e}\,\Longrightarrow\,a^{\large n}\equiv a^{\large k}\pmod{\!m}.\$$ Further,  conversely

$$\qquad n\equiv k\pmod{\! \color{#c00}e}\,\Longleftarrow\,a^{\large n}\equiv a^{\large k}\pmod{\!m}\ \,$$ if $$\,a\,$$ has order $$\,\color{#c00}e\,$$ mod $$\,m$$

Proof $$\$$ Wlog $$\,n\ge k\,$$ so $$\,a^{\large n-k} a^{\large k}\equiv a^{\large k}\!\iff a^{\large n-k}\equiv 1\iff n\equiv k\pmod{\!e}\,$$ by here, where we cancelled $$\,a^{\large k}\,$$ using $$\,a^{\large e}\equiv 1\,\Rightarrow\, a\,$$ is invertible so cancellable (cf. below Remark).

Corollary $$\ \ \bbox[7px,border:1px solid #c00]{\!\bmod m\!:\,\ \color{#c00}{a^{\large e}\equiv 1}\,\Rightarrow\, a^{\large n}\equiv a^{\large n\bmod \color{#c00}e}}\,\$$ by $$\ n\equiv n\bmod e\,\pmod{\!e}$$

Remark  If you are familiar with modular inverses then it is not necessary to restrict to nonnegative powers of $$\,a\,$$ above since $$\,a^{\large e}\equiv 1,\ e> 0\,\Rightarrow\,$$ $$a$$ is invertible by $$\,a a^{\large e-1}\equiv 1\,$$ so $$\,a^{\large -1}\equiv a^{\large e-1}$$.

• Often it proves simpler to first reduce $\,e\,$ using Euler's Criterion or quadratic reciprocity, e.g. see here.. – Bill Dubuque Jun 8 at 13:31

Quick answer: $13 = 2\cdot 7-1$ so $13\equiv-1\mod 7$ and therefore $13^{100} \equiv (-1)^{100} \mod 7$

Other one is fairly quick: \begin{eqnarray} \phi(13) = 12\\ \gcd(7,13)=1\\ 7^{100}\equiv7^{4} \mod13\\ 7\rightarrow10\rightarrow5\rightarrow9 \end{eqnarray} Probably a nicer way to do that.

• Thanks for the answer. But can you please explain how you arrived to step 3? (in your four-step part of the answer), and why you have calculated the gcd(7, 13)? – Roshnal Nov 28 '16 at 0:58


Finding prime factors by taking the square root

I'm trying to solve the third Project Euler problem and I'd like a little help understanding a mathematical concept underlying my tentative solution.

The prime factors of 13195 are 5, 7, 13, and 29.

What is the largest prime factor of the number 600851475143 ?

As a caveat, in accordance with the wishes of Project Euler I won't be providing any code, my question is centered on why a mathematical concept works.

Failed Attempt

My first algorithm looked at all the numbers from 1 through 600851475143, but was unable to complete the subsequent computations (concerning primes and factorization) due to memory constraints.

Successful Attempt

My next algorithm looked at all the numbers from 1 through $\sqrt{600851475143}$ and completed the computation successfully.

My Question

Why does evaluating $\sqrt{600851475143}$ work in this instance when I really want to evaluate up to 600851475143 ? How can I be sure this approach won't miss some factor like $2 \cdot n$ or $3 \cdot n$, when $n$ is some number between $\sqrt{600851475143}$ and 600851475143 ?

• If a number $N$ has a prime factor larger than $\sqrt{N}$ , then it surely has a prime factor smaller than $\sqrt{N}$. – barak manos Nov 26 '14 at 11:46
• @barakmanos You may as well write this as the answer. – DanielV Nov 26 '14 at 11:47
• You won't miss $2\cdot n$ or $3\cdot n$ because you have checked for $2$ and $3$. – gammatester Nov 26 '14 at 11:48
• We have to be a little careful in writing the program, since some plausible approaches will miss "large" prime factors when a small prime factor occurs with multiplicity $\gt 1$. – André Nicolas Nov 26 '14 at 11:59
• You don't have to check all the way to $\sqrt{600851475143}$. Once you identify 71 as a factor you know that the largest prime factor of 600851475143, is also the largest prime factor of 600851475143/71=8462696833. So at that point you can limit the search to $\sqrt{8462696833}$ - and so on. Once you hit the square root you know, by the argument given in the answers, that the number you are taking the square root of is prime, and also that it's the largest prime factor of your original number. – Taemyr Nov 26 '14 at 15:22

If a number $N$ has a prime factor larger than $\sqrt{N}$ , then it surely has a prime factor smaller than $\sqrt{N}$.

So it's sufficient to search for prime factors in the range $[1,\sqrt{N}]$, and then use them in order to compute the prime factors in the range $[\sqrt{N},N]$.

If no prime factors exist in the range $[1,\sqrt{N}]$, then $N$ itself is prime and there is no need to continue searching beyond that range.

• As mentioned by André Nicolas, prime factors with >1 multiplicity will result in the larger factor being non-prime. To solve this, keep dividing the larger factor of a pair by the smaller one until it no longer divides. – Mark K Cowan Nov 26 '14 at 14:00
• As an example: For 28, factors are 2,2,7, here 7 is larger than sqroot(28), but there is no single prime number that can combine to form 28 (i.e. x*7=28, where x is prime, this does not exist, since x is 4 which is not prime), so dividing the primes from beginning with 2 multiple time will help, 28/2= 14, 14/2 = 7, then we know the 2 & 2 are prime factors, also take the remaining one 7 is also another prime number completing the prime factor list. best of luck. – Manohar Reddy Poreddy Dec 2 '15 at 0:25

If you do not find a factor less than $\sqrt{x}$, then $x$ is prime for the following reason. Consider the opposite, you find two factors larger than $\sqrt{x}$, say $a$ and $b$. But then $a\cdot b> \sqrt{x}\sqrt{x} = x$. Therefore, if there is a factor larger than $\sqrt{x}$, there must also exist a factor smaller than $\sqrt{x}$, otherwise their product would exceed the value of $x$.

Regarding the last question. You will not miss some factor like $2\cdot n$ because you have already checked if $2$ is a factor.

What you're describing is a prime-testing algorithm known as the sieve of Eratosthenes.

It seems like you already understand the concept:

• Cross out $1$

• Circle $2$ as the first prime, then cross out all of the multiples of $2$ in your list.

• The next not-crossed-out number is the next prime (in this case $3$).

If your list has $N$ numbers, you only need to test until you get to a prime that's bigger than $\sqrt{N}$.

Other users have mentioned the reason for $\sqrt{N}$ being the maximum you need to test until, but I'll add an alternative explanation. If you consider listing out all of the factors of $N$, in order from least to greatest, you will either get an even number or an odd number. $N$ has an odd number of factors if and only if it is a perfect square.

In any case, $N$ always has the same number of factors (strictly) less than its square root as it does (strictly) greater than its square root.

In fact, there's an explicit bijection between the set of factors less than $\sqrt{N}$ and the set of factors greater than $\sqrt{N}$ given by $$f(a) = \frac{N}{a}$$

(where $a$ is a factor of $N$ less than $\sqrt{N}$)


# What is the negation of this sentence

What is the negation of the following case:

Case A: The function $$f$$ has a unique rational fixed point $$a$$.

I would say the function has:

• multiple rational fixed points or
• no fixed point at all or
• at least one irrational fixed point

Not sure if we can make this more "compact".

• @China: I don't agree with this answer. My point of view is explained in my answer. – Taroccoesbrocco Aug 3 at 10:59
• Well, my answer matches better to your 2. than your 1. So in fact we agree on the negation... – zwim Aug 3 at 13:30
• @Taroccoesbrocco: Thank you. "Unique" refers to "rational fixed point" – Germany Aug 3 at 14:46
• @zwim - I'm not sure to understand your comment: If $f$ has an irrational fixed point and a rational fixed point, then your version of the negation is true (because of point 3) but my version of the negation (see point 2 in my answer) is false. So, your negation and my negation are not equivalent, and only one of them is correct. – Taroccoesbrocco Aug 3 at 15:35
• @China - Then the correct negation is in point $(2)$ of my answer, and not in point $(1)$ (or in zwim's answer). – Taroccoesbrocco Aug 3 at 15:45

The sentence can be interpreted in two ways, depending on whether "unique" refers to "fixed point" or "rational fixed point":

1. "Unique" refers to "fixed point": The function $$f$$ has a unique fixed point $$a$$, and moreover $$a$$ is rational. The logical form of this sentence is \begin{align} \exists a : a \in \mathbb{Q} \land a = f(a) \land \forall z (z = f(z) \to z = a) \end{align} In this case, the negation is: the function $$f$$ either has a unique irrational fixed point, or several fixed points, or no fixed points at all (this is equivalent to what @zwim said in his/her answer).

2. "Unique" refers to "rational fixed point": The function $$f$$ has exactly one rational fixed point (and possibly many irrational fixed points). The logical form of this sentence is: \begin{align} \exists a : a \in \mathbb{Q} \land a = f(a) \land \forall z ((z \in \mathbb{Q} \land z = f(z)) \to z = a) \end{align} In this case the negation is: the function $$f$$ has either several rational fixed points or no rational fixed points at all (but it might have irrational fixed points).

Note that the negation of the interpretation $$(2)$$ is more restrictive than the negation of the interpretation $$(1)$$: for instance, if $$f$$ has a irrational fixed point and a rational fixed point, then the negation of $$(1)$$ is true but the negation of $$(2)$$ is false.

Unlike @zwim, in my opinion the intended meaning of the sentence should be $$(2)$$, and not $$(1)$$, unless contextual information (which is missing in the question of the OP) says differently.


# Showing $f(x)=x^4$ is not uniformly continuous

I am looking at uniform continuity (for my exam) at the moment and I'm fine with showing that a function is uniformly continuous but I'm having a bit more trouble showing that it is not uniformly continuous, for example:

show that $x^4$ is not uniformly continuous on $\mathbb{R}$, so my solution would be something like:

Assume that it is uniformly continuous then:

$$\forall\epsilon\geq0\exists\delta>0:\forall{x,y}\in\mathbb{R}\ \mbox{if}\ |x-y|<\delta \mbox{then} |x^4-y^4|<\epsilon$$

Take $x=\frac{\delta}{2}+\frac{1}{\delta}$ and $y=\frac{1}{\delta}$ then we have that $|x-y|=|\frac{\delta}{2}+\frac{1}{\delta}-\frac{1}{\delta}|=|\frac{\delta}{2}|<\delta$ however $$|f(x)-f(y)|=|\frac{\delta^3}{8}+3\frac{\delta}{4}+\frac{3}{2\delta}|$$

Now if $\delta\leq 1$ then $|f(x)-f(y)|>\frac{3}{4}$ and if $\delta\geq 1$ then $|f(x)-f(y)|>\frac{3}{4}$ so there exists not $\delta$ for $\epsilon < \frac{3}{4}$ and we have a contradiction.

So I was wondering if this was ok (I think it's fine) but also if this was the general way to go about showing that some function is not uniformly continuous? Or if there was any other ways of doing this that are not from the definition?

Thanks very much for any help

• So, is this an exam question? – user21436 Apr 22 '12 at 12:17
• No, I'm just practicing for my exam where questions like this (not this one though) will come it. This is from one of the past papers that are for revision – hmmmm Apr 22 '12 at 12:29
• Just trying to ensure we are not taken for a ride. Hope you don't mind. :) – user21436 Apr 22 '12 at 12:33
• @KannappanSampath no its fine- it annoys me when I see people posting assessment questions on forums :) – hmmmm Apr 22 '12 at 12:36

To show that it is not uniformly continuous on the whole line, there are two usual (and similar) ways to do it:

1. Show that for every $\delta > 0$ there exist $x$ and $y$ such that $|x-y|<\delta$ and $|f(x)-f(y)|$ is greater than some positive constant (usually this is even arbitrarily large).
2. Fix the $\varepsilon$ and show that for $|f(x)-f(y)|<\varepsilon$ we need $\delta = 0$.

First way:

Fix $\delta > 0$, set $y = x+\delta$ and check $$\lim_{x\to\infty}|x^4 - (x+\delta)^4| = \lim_{x\to\infty} 4x^3\delta + o(x^3) = +\infty.$$

Second way:

Fix $\epsilon > 0$, thus $$|x^4-y^4| < \epsilon$$ $$|(x-y)(x+y)(x^2+y^2)| < \epsilon$$ $$|x-y|\cdot|x+y|\cdot|x^2+y^2| < \epsilon$$ $$|x-y| < \frac{\epsilon}{|x+y|\cdot|x^2+y^2|}$$

but this describes a necessary condition, so $\delta$ has to be at least as small as the right side, i.e.

$$|x-y| < \delta \leq \frac{\epsilon}{|x+y|\cdot|x^2+y^2|}$$

so if either of $x$ or $y$ tends to infinity then $\delta$ tends to $0$.

Hope that helps ;-)

Edit: after explanation and calculation fixes, I don't disagree with your proof.

• Thanks for the reply, I think that I use that we are considering all of $\mathbb{R}$ when i choose $x=\delta+\frac{1}{\delta}$ and $y=\frac{1}{\delta}$ as these would not be valid for small $\delta$ in bounded interval? – hmmmm Apr 22 '12 at 13:38
• @hmmmm Ok, I misunderstood what you were saying there. If the calculations are alright, then your proof is fine. – dtldarek Apr 22 '12 at 13:44
• Is it correct to fix $\delta>0$ and choosing $y = x + \delta$ specifically? Since the distance between x and y will be exactly $\delta$, and we want $|x-y|<\delta$. – DrHAL May 2 '16 at 8:08
• @dtldarek Is it possible that one method will fail to show and other will succeed. For example, for $f(x)=1/x$ on $(0,\infty)$, first method leads the limit to 0? – chandresh Feb 3 at 10:18
• @chandresh Function $f(x)=1/x$ is uniformly continuous on $(1, \infty)$, or even on $(a, \infty)$ for any fixed $a > 0$. On the other hand, $f$ is very steep near zero, and so the limit you want to calculate is $$\lim_{x \to 0^+}\left|\frac{1}{x}-\frac{1}{x+\delta}\right|.$$ – dtldarek Feb 5 at 8:04

I will comment on your solution after writing another approach. For any $x,y\in\mathbb{R}$ we have: \begin{align*} |x^{4}-y^{4}|=|(x^{2}-y^{2})(x^{2}+y^{2})|=|(x-y)(x+y)(x^{2}+y^{2})|=|x-y|\cdot |x+y|\cdot |x^{2}+y^{2}| \end{align*}

So what you can see is that even if you take arbitrarily close $x$ and $y$, you can grow the distance of $x^{4}$ and $y^{4}$ as much as you want by taking them far enough away from zero. You can easily conclude from here that the function is not uniformly continuous by a contraposition for example.

Alright, then to your solution. If the calculations would be correct, then it would be fine. You could assume at first that such $\delta>0$ exists for $0<\varepsilon<3$ and conclude with a contradiction. However, I got a bit different calculations than you. Using the above equation we see that: \begin{align*} |f(\frac{\delta}{2}+\frac{1}{\delta})-f(\frac{1}{\delta})|&=|(\frac{\delta}{2}+\frac{1}{\delta})^{4}-\frac{1}{\delta^{4}}|=|\frac{\delta}{2}(\frac{\delta}{2}+\frac{2}{\delta})((\frac{\delta}{2}+\frac{1}{\delta})^{2}+\frac{1}{\delta^{2}})| \\ &= |(\frac{\delta^{2}}{4}+1)(\frac{\delta^{2}}{4}+2\cdot \frac{\delta}{2}\cdot \frac{1}{\delta}+\frac{1}{\delta^{2}}+\frac{1}{\delta^{2}})| \\ &=|(\frac{\delta^{2}}{4}+1)(\frac{\delta^{2}}{4}+1+\frac{2}{\delta^{2}})| \\ &= |\frac{\delta^{4}}{16}+\frac{\delta^{2}}{4}+\frac{1}{2}+\frac{\delta^{2}}{4}+1+\frac{2}{\delta^{2}}| \\ &= |\frac{\delta^{4}}{16}+\frac{\delta^{2}}{2}+\frac{2}{\delta^{2}}+\frac{3}{2}|\\ &= \frac{\delta^{4}}{16}+\frac{\delta^{2}}{2}+\frac{2}{\delta^{2}}+\frac{3}{2} \end{align*} If you're able to find a lower bound for this (which is quite easy) as you did previously, then by choosing an epsilon smaller than that fixed number you may conclude as you did in your original post by contradiction.

• Hey sorry about that, I edited it-hopefully it's right now? – hmmmm Apr 22 '12 at 13:39
• I also edited now my calculation which still differs abit from your new one. Could you show the steps of how you got this answer for $|f(x)-f(y)|$? – T. Eskin Apr 22 '12 at 13:47
• yeah I messed that up quite a bit sorry (I had the wrong power and the wrong delta's) – hmmmm Apr 22 '12 at 13:50
• It should be $|(\frac{\delta}{2}+\frac{1}{\delta})^4-\frac{1}{\delta^4}|$ which would give $|\frac{\delta^4}{16}+\frac{\delta^2}{2}+\frac{2}{\delta}+\frac{3}{2}|$ I think I could conclude a similar thing from here? – hmmmm Apr 22 '12 at 13:52
• Except that is the last $-\frac{1}{\delta^{4}}$ missing from there? Otherwise it looks close to mine. – T. Eskin Apr 22 '12 at 13:57

(1). If $f:(0,\infty)\to \Bbb R$ is differentiable and $\lim_{x\to \infty}f'(x)=\infty$ then $f$ is not uniformly continuous: For any $\delta >0$ and any $x>0$, the MVT implies there exists $y\in (x,x +\delta)$ such that $\frac {f(x+\delta)-f(x)}{\delta}= f'(y).$ Now if $x$ is large enough that $\forall y>x\;(f'(y)>1/\delta)$ then $f(x+\delta)-f(x)=\delta f'(y)>1.$

(2). Given $\delta >0$ take $x>\max (1,1/\delta).$ Then $$(x+\delta)^4-x^4= 4x^3\delta+6x^2\delta^2+4x\delta^3+\delta^4>$$ $$>4x^3\delta=4(x^2)x\delta>4x\delta>4.$$

By (1) or by (2) we have $\sup_{x\in \Bbb R} \{|(x+x')^4-x^4|: |x'-x|<\delta\}> 1$ for every $\delta>0.$ (In fact the $\sup$ is $\infty$.)

Uniform continuity of $f:D\to \Bbb R$ for some (any) domain $D\subset \Bbb R$ means $$0=\lim_{\delta \to 0^+}\sup \{|f(x')-f(x)|:x,x'\in D\land |x'-x|\leq\delta\}.$$

I think you should make this a little simpler (and for uniform continuity in general) All you need to do to show $f:X \to Y$ is not uniformly continuous on $X$ (let's suppose there both subsets of $\Bbb R$), then just give me a SINGLE epsilon such that, NO MATTER HOW SMALL delta is chosen, there will be x and y closer than delta for which the difference in function values exceed epsilon. Thus for instance $|(N+\theta)^4- N^4| \ge 4\theta N^3$ so if you choose $x$ really big (with respect to $\delta, x=N+\theta\,\,\,\ \text{and}\,\,\, y = N,$ then if $0 < \delta/2 < \theta < \delta$ you have $|x-y| < \delta$ yet you still have the variable N to play with to make the difference in function values as large as you like (in particular the difference in function values can always be made bigger than 3 regardless of how small $\delta$ is). Nevertheless, I think your proof is an accurate job!

## ** Simple Trick:**

Given $a>0$ if we take $$x = \frac{1}{a}+2 ~~~and ~~~y = \frac{1}{a}+2+\frac{a}{2} \color{red}{= x +\frac{a}{2}}$$ we have $x,y\in [2,\infty)~~~ and~~~|x-y| =\frac{a}{2}<a$

But since $a\cdot x = 1+ 2a~~~$ and $~~x,y>2\implies x^2+y^2 >8$ we get,

$$|f(x)-f(y)| =(x^2+y^2) |y^2-x^2|\ge 8|y^2-x^2|\\~~~~~~~~~~~~~~~=8| x^2 + 2\frac{a}{2}x +\frac{a^2}{4} - x^2 | \\= 8(ax +\frac{a^2}{4})= 8(1+2a+\frac{a^2}{4} )>8$$

That is $$|f(x)-f(y)| >8$$

Thus $$\color{blue}{\exists \varepsilon_0 =8,\forall~a>0, \exists ~x,y\in \mathbb R: |x-y|< a~~and ~~|f(x)-f(y)|>8}$$

just take $$x = \frac{1}{a}+2 ~~~and ~~~y = \frac{1}{a}+2+\frac{a}{2} \color{red}{= x +\frac{a}{2}}$$


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# The amounts of time that three secretaries worked on a

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07 Feb 2012, 02:12
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The amounts of time that three secretaries worked on a special project are in the ratio of 1 to 2 to 5. If they worked a combined total of 112 hours, how many hours did the secretary who worked the longest spend on the project?

A) 80
B) 70
C) 56
D) 16
E) 14

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Re: The amounts of time that three secretaries worked on a  [#permalink]

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07 Feb 2012, 02:28
8x = 112
=> x = 14

Therefore the secretary who worked the longest spent 14 x 5 = 70 hours on the project

Option (B)
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07 Feb 2012, 05:05
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fiendex wrote:
The amounts of time that three secretaries worked on a special project are in the ratio of 1 to 2 to 5. If they worked a combined total of 112 hours, how many hours did the secretary who worked the longet spend on the project?

A) 80
B) 70
C) 56
D) 16
E) 14

Since, amounts of time that secretaries worked in the ratio of 1 to 2 to 5, then $$\frac{\frac{1x}{2x}}{5x}$$, for some number x --> x+2x+5x=8x=112 --> x=14 --> the secretary who worked the longet spent 5x=70 hours.

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07 Feb 2012, 06:53
1
All three secretaries worked 112h and with a ratio of 1to2to5

8x = 112h
x = 14h

5*14h = 50h --> B
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23 May 2017, 02:48
fiendex wrote:
The amounts of time that three secretaries worked on a special project are in the ratio of 1 to 2 to 5. If they worked a combined total of 112 hours, how many hours did the secretary who worked the longest spend on the project?

A) 80
B) 70
C) 56
D) 16
E) 14

Similar question: https://gmatclub.com/forum/the-amount-o ... 01139.html
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Re: The amounts of time that three secretaries worked on a  [#permalink]

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25 May 2017, 15:48
1
fiendex wrote:
The amounts of time that three secretaries worked on a special project are in the ratio of 1 to 2 to 5. If they worked a combined total of 112 hours, how many hours did the secretary who worked the longest spend on the project?

A) 80
B) 70
C) 56
D) 16
E) 14

The ratio of the amounts of time that the secretaries worked on the project is x : 2x : 5x or a total of 8x.

Since the total number of hours is 112:

8x = 112

x = 14

The secretary who worked the longest worked for 5x = 5(14) = 70 hours.

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08 Dec 2017, 19:36
$$\frac{5}{8ths}$$of the time was spent by the longest working secretary. So the answer is more than 56. An eighth of
112 is 14. 56+14=70
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# Is $X^T X$ invertible if $p > n$?

I am checking other's regression analysis work on a $p > n$ data. I can only see the results but not the process how he did it.

I believe he made mistakes. Since $p > n$, $X^T X$ is not full rank it is not invertible. So we cannot find the OLS coefficient.

However, I am not sure my reasoning is correct. I have not used linear algebra for a long time.

Update:

1. No penalized methods involved.
2. He used stepwise regression for variable selection. A new question: would such algorithm stop if number of variables in the model equal to the number of sample points?
3. His goal is to find out which variables are important. Doesn't care about the prediction power.
• Yes you are correct in that $X^TX$ is non-invertible. However, perhaps your colleague performed some penalized regression like LASSO to get coefficients? Oct 20 '16 at 17:44
• What are $p$ and $n$? Oct 20 '16 at 18:21
• A bit more context may be useful. Is the goal forecasting? Is it consistently estimating some effect $b_i$? Do you know if your colleague performed LASSO or ridge regression? Oct 20 '16 at 18:30
• @RodrigodeAzevedo $p$ and $n$ are classically the number of covariates and of independent observations. IE, $X$ is $n \times p$. Oct 20 '16 at 19:07
• If he's using forward stepwise regression, he'd never have more than $n$ variables in the actual regression matrix used (starting from 1 = the intercept, finding the best variable to add incrementally.) That's not to say forward stepwise regression (or any stepwise regression) is good, it's just to say it would have avoided the identifiability problem. Oct 20 '16 at 19:44

If $\mathrm X$ is $n \times p$ and $p > n$, then it is fat and, thus,
$$\mbox{rank} (\mathrm X) = \mbox{rank} (\mathrm X^T \mathrm X) \leq n < p$$
Hence, $\mathrm X^T \mathrm X$ does not have full rank and, thus, it is not invertible.


## Never Decreasing Digits

#### Problem

A number has strictly increasing digits if each digit exceeds the digit to its left and is said to be a strictly increasing number; for example, 1357.
A number has increasing digits if each digit is not exceeded by the digit to its left and is said to be an increasing number; for example, 45579.

There are exactly 219 increasing numbers below one-thousand.

How many numbers below one million are increasing?

#### Solution

We will begin by deriving, from first principles, the information given in the problem relating to numbers below one-thousand.

It is possible to use binary strings to produce increasing numbers. Consider the following algorithm.

Let S be binary string
Let C = 0
Let K = 1
Label A
Let D be value of Kth digit of S
If D = 1 then output C
If D = 0 then increment c by 1
Increment K by 1
If K has not exceeded the length of S then goto A
Stop

For example, the binary string $001101001$ would produce 2235.

In general it can be seen that an $n$-digit increasing number will be produced by a binary string containing $n$ $1$'s, and the maximum digit will be represented by the number of $0$'s, as each zero causes the "counter" to increase by 1.

Using this idea we can represent all the 3-digit increasing numbers made up of the digits 0, 1, and 2 using 5-digit binary strings consisting of three $1$'s and two $0$'s:

\begin{align}11100 &= 000\\11010 &= 001\\11001 &= 002\\10110 &= 011\\10101 &= 012\\10011 &= 022\\01110 &= 111\\01101 &= 112\\01011 &= 122\\00111 &= 222\end{align}

But we must subtract one to remove $000$, so there are $\dbinom{5}{3} - 1 = 9$ such numbers.

In the same way a 3-digit increasing number using each of the digits 0 to 9 requires a binary string containing three 1's and nine 0's; that is, there are exactly $\dbinom{12}{3} - 1 = 219$ increasing numbers below one-thousand.

Now we can return to the problem: increasing numbers below one million.

As the numbers contain six digits the binary string will contain six 1's and nine 0's. Hence there are $\dbinom{15}{6} - 1 = 5004$ increasing numbers below one million.

Check out the related (strictly) Increasing Digits problem.

Problem ID: 263 (29 Jan 2006)     Difficulty: 4 Star

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