# Basic question regarding limits

I'm a little confused when it comes to question like this.

Let's say we got this expression: $$\lim_{x\to\infty} \tan\left(\frac{1}{x}\right).$$ Am I allowed to say the result of this is $0$ or do I have to show it? (If I have to show it, please show me how to cuz I got no clue).

And more general, in which situations are we allowed to "cut" it and in which we are not? (I'm guessing whenever it comes to fractions surely).

That depends.

If the whole exercise is just "compute $\lim \tan(1/x)$", then yes, you have to do some argument. Something along the lines of "if $x$ goes to infinity, then $1/x$ goes to zero, and by continuity of $\tan$, the whole expression then goes to zero".

But if this limit is just one part of a longer question/calculation, then no. It's trivial enough that anybody with some math background will see it immediately.

• Appreciated . Got only one more question . Is there any like more formal ways of prooving it is 0 ? Cuz I never did any good solving math problem using words tho I completely see where ur going at. – James Groon Apr 1 '17 at 18:24
• In general feel free to replace the word "then" or the phrase "it then follows" with the symbol "$\implies$" and "there is an $x$" with "$\exists x$" and such. Using such symbols instead of words can make a proof more concise, though overdoing it can make it hard to read. And if you want a more elementary proof, the answer by @Chris-Varghese is quite nice. – Simon Apr 1 '17 at 18:34
• Gotcha and yep, Chris did a good job :) – James Groon Apr 1 '17 at 18:35

You need to show that given any small $\epsilon > 0$, there exists a $X(\epsilon)$ such that $|\tan(1/x)| < \epsilon,\; \forall x > X(\epsilon)$. To do this choose, for e.g., $X(\epsilon) = \dfrac{1+\sec \epsilon}{\epsilon}$. Then $$\left|\tan \frac{1}{x}\right| < \left|\tan \frac{\epsilon}{1+\sec \epsilon} \right| = \frac{\sin \left( \frac{\epsilon}{1+\sec \epsilon}\right) }{\cos\left( \frac{\epsilon}{1+\sec \epsilon}\right)} < \frac{\sin \left( \frac{\epsilon}{1+\sec \epsilon}\right) }{\cos\epsilon} < \frac{ \frac{\epsilon}{1+\sec \epsilon} }{\cos\epsilon} = \frac{\epsilon}{1 + \cos \epsilon} < \epsilon$$

• The inequality is wrong. Note that $|\tan (x)|\ge |x|$ for all $x\in (-\pi/2,\pi/2)$. Simple example; $x=\pi/4<1$. We have $\tan(\pi/4)=1>\pi/4$. – Mark Viola Apr 1 '17 at 18:48
• So may i get the correct answer ? – James Groon Apr 1 '17 at 19:02
• @Dr.MV Which inequality in my answer are you referring to? Indeed, $|\tan x | \geq |x| \forall x \in (-\pi/2, \pi/2)$. How does that fact contradict anything that I have in my answer? – ChargeShivers Apr 2 '17 at 1:01
• @chrisvarghese You wrote $|\tan(\epsilon)|\le \epsilon$. That is false. – Mark Viola Apr 2 '17 at 2:58
• @Dr.MV Got it now. Thanks. Not sure how I missed it twice!! Answer edited. – ChargeShivers Apr 2 '17 at 17:17

The answer is that it depends on the situation. To address the statement in the OP "If I have to show it, please show me how to cuz I got no clue," I thought it might be instructive to present an approach that relies on a standard inequality from elementary geometry. To that end, we begin with a short primer.

PRIMER:

Recall from elementary geometry the inequality

$$\sin(\theta)\le \theta\tag 1$$

for $\theta\ge 0$. Squaring both sides of $(1)$, using $\sin^2(\theta)=1-\cos^2(\theta)$, and rearranging, we find that

$$\cos(\theta)\ge \sqrt{1-\theta^2} \tag 2$$

for $0\le \theta \le 1$.

Now, using $(1)$ and $(2)$ with $\theta=1/x$ reveals that

\begin{align} \tan(1/x)&=\frac{\sin(1/x)}{\cos(1/x)}\\\\ &\le \frac{1/x}{\sqrt{1-\frac1{x^2}}}\\\\ &=\frac{1}{\sqrt{x^2-1}}\tag 3 \end{align}

for $x>1$.

Hence, for all $\epsilon>0$, we have from $(3)$ that

\begin{align} \tan(1/x)&\le \frac{1}{\sqrt{x^2-1}}\\\\ &<\epsilon \end{align}

whenever $x>\sqrt{1+\frac{1}{\epsilon^2}}$. And we are done!

• $\sqrt{1 + \frac{1}{\epsilon^2}} = \dfrac{\sqrt{1+\epsilon^2}}{\epsilon}$ is definitely a better $X(\epsilon)$ than $\dfrac{1+\sec \epsilon}{\epsilon}$. May be the 'best' of all is $X(\epsilon) = \dfrac{1}{\arctan \epsilon}$. – ChargeShivers Apr 2 '17 at 23:41

Short answer is $0$ ,but if you want " to show " recall the limit $$\lim _{ x\rightarrow \infty }{ \frac { 1 }{ x } =0 }$$

You can move the limit into the $tan$, because you are not excluding any $x$ from the limit, and you are not creating any limits that don't exist. Once that is done, the problem is trivial.


# Application of 2nd fundamental theorem of calculus

I would like to clarify the usage of the 2nd fundamental theorem of calculus, in 3 parts. For all 3 parts, consider $F'(a)$ to be

$$F'(a) = \frac{1}{1+a+a^2}$$

The questions are:

1. Find $\frac{d}{dy}\int^y_1 F'(a) \,da$
2. Find $\frac{d}{dy}\int^1_y F'(a) \,da$
3. Find $\frac{d}{dy}\int^{y^2}_1 F'(a)\, da$

My working for each of the questions are:

1.

\begin{align*} \frac{d}{dy} \int^y_1 F'(a) \,da &= \frac{d}{dy}\begin{bmatrix}F(y) - F(1)\end{bmatrix} \\&=F'(y) \\&=\frac{1}{1+y+y^2}. \end{align*}

2.

\begin{align*} \frac{d}{dy} \int^1_y F'(a) \,da &= \frac{d}{dy}\begin{bmatrix}F(1) - F(y)\end{bmatrix} \\&=-F'(y) \\&=\frac{-1}{1+y+y^2}. \end{align*}

3.

\begin{align*} \frac{d}{dy} \int^{y^2}_1 F'(a) \,da &= \frac{d}{dy}\begin{bmatrix}F(y^2) - F(1)\end{bmatrix} \\&=F'(y^2) \\&=\frac{2y}{1+y^2+y^4}. \end{align*}

UPDATE: I shall update (3) with the chain rule working.

\begin{align*} \frac{d}{dy} \int^{y^2}_1 F'(a) \,da &= \frac{d}{dy}\begin{bmatrix}F(y^2) - F(1)\end{bmatrix} \\&=F'(y^2)(2y) \\&=\frac{1}{1+(y^2)+(y^2)^2} (2y) \tag{by chain rule} \\&=\frac{2y}{1+y^2+y^4}. \end{align*}

-
The third one would need a chain rule. Though, you do have used it for final answer. –  hjpotter92 Mar 8 '13 at 12:59
Your answers are correct. However, one should note that$\dfrac{dF(y^2)}{dy}=F'(y^2)\cdot2y$(Chain rule). Your final answers are perfectly fine but the intermediate step is wrong.


# Prove $\frac{\cos^2 A}{1 - \sin A} = 1 + \sin A$ by the Pythagorean theorem.

How do I use the Pythagorean Theorem to prove that $$\frac{\cos^2 A}{1 - \sin A} = 1 + \sin A?$$

-
if i multiply it would it show that A is an acute angle? i have a given figure here and its a right triangle –  user83562 Jun 23 '13 at 14:25
Do you know the identity $\cos^2 A + \sin^2 A = 1$? –  Javier Badia Jun 23 '13 at 14:58

suppose there is a right angle tri. having b is hypo.,c is base ,a is perpendicular than $\cos A=\frac cb$ and $\sin A=\frac ab$ using pythgo. theo. $\;b^2=a^2+c^2$ $$\dfrac {\cos^2 A}{1-\sin A}$$ $$\dfrac {\frac {c^2}{b^2}}{1-\frac {a}{b}}$$ $$\dfrac {\frac {c^2}{b^2}}{\frac {b-a}{b}}$$ $$\dfrac {c^2b}{b^2(b-a)}$$ $$\dfrac {c^2}{b(b-a)}$$ $$\dfrac {b^2-a^2}{b(b-a)}$$ $$\dfrac {b+a}{b}$$ $$1+\dfrac {a}{b}$$ $$1+\sin A$$

alternatively

$$\dfrac {\cos^2 A}{1-\sin A}$$ $$\dfrac {1-\sin^2 A}{1-\sin A}$$ $$\dfrac {(1-\sin A)(1+\sin A)}{1-\sin A}$$ $${1+\sin A}$$

-
why don't you make things precise : $$1=\frac{c^2}{b^2}+\frac{a^2}{b^2}=\cos^2A+\sin^2A$$ $$\implies \cos^2A=1-\sin^2A=(1-\sin A)(1+\sin A)$$ $$\implies \frac{\cos^2A}{1-\sin A}=1+\sin A$$ –  lab bhattacharjee Jun 23 '13 at 14:32
how did it become 1-sin^2 A? –  user83562 Jun 23 '13 at 14:49
@user83562: Did you read lab's (excellent) comment? Since $$1=\cos^2A+\sin^2A,$$ then $$\cos^2A=1-\sin^2A.$$ By difference of squares formula, we have $$\cos^2A=(1-\sin A)(1+\sin A).$$ From there, it's almost immediate. –  Cameron Buie Jun 23 '13 at 15:18

If you allow yourself some simple algebra, then you see that $$\frac{\cos^2 A}{1-\sin A} = 1+\sin A\text{ is equivalent to } \cos^2 A = (1+\sin A)(1-\sin A)$$ and that last expression is equal to $1-\sin^2 A$. So you're asking how to prove $\cos^2 A=1-\sin^2 A$. And that's the same as proving $\cos^2 A + \sin^2 A = 1$.

So the question is: How can you prove that $\cos^2 A + \sin^2 A = 1$ by using the Pythagorean theorem?

If you know that $\sin =\dfrac{\mathrm{opposite}}{\mathrm{hypotenuse}}$ and $\cos=\dfrac{\mathrm{adjacent}}{\mathrm{hypotenuse}}$, then this becomes easy if you consider a right triangle in which the length of the hypotenuse is $1$. Then you have $\sin=\mathrm{opposite}$ and $\cos=\mathrm{adjacent}$. Now the Pythagorean theorem says that $$\mathrm{opposite}^2 + \mathrm{adjacent}^2 = 1^2.$$

-

Let's simplify matters:

In a triangle with hypotenuse equal to $1$ (think of the unit circle, an angle $A$ between the $x$ axis and the hypotenuse, we know that $$\sin A=\frac{\text{opposite}}{\text{hypotenuse}}\quad \text{and}\quad \cos A=\frac{\text{adjacent}}{\text{hypotenuse}}$$ then, since hypotenuse $=1$, we have the leg opposite the angle $A$ given by $\sin A=\text{opposite}/1$ and the leg along the x-axis of length $\cos A=\text{adjacent}/1$. Now, by the Pythagorean Theorem, and substitution, we have that \begin{align}\text{opposite}^2 + \text{adjacent}^2 &= \text{hypotenuse}^2 = 1^2 \\ \sin^2A +\cos^2 A & = 1\end{align}

This gives us the well-known identity: $$\sin^2A + \cos^2 A = 1\tag{1}$$

We can express this identity in terms of $\cos^2 A$ by subtracting $\sin^2 A$ from both sides of the identity to get \begin{align}\cos^2 A & = 1 - \sin^2 A \\ &= (1)^2 - (\sin A)^2\tag{2}\end{align}

Now, we know that for any difference of squares, we can factor as follows: $$(x^2 - y^2) = (x +y)(x - y)\tag{3}$$

Since equation $(2)$ is a difference of squares, we have that \begin{align}\cos^2 A &= 1 - \sin^2 A \\ & = (1)^2 - (\sin A)^2 \\ &= (1 +\sin A)(1 - \sin A)\tag{4}\end{align}

Substituting gives us:

\begin{align}\frac{\cos^2 A}{1 -\sin A} & = \frac{1 - \sin^2 A}{1 -\sin A} \\ \\ &= \frac{(1 + \sin A)(\color{blue}{\bf 1 - \sin A})}{\color{blue}{\bf 1 -\sin A}}\\ \\ & = 1 + \sin A\end{align}

-
You begin with "Given the well known identity,....", but the questions seems to be how to get that well known identity from the Pythagorean theorem. –  Michael Hardy Jun 23 '13 at 16:14
@amWhy: You are always welcome my friend! Time for another cheesy movie (trying to add other activities in my life). ;-) –  Amzoti Jun 24 '13 at 2:47

Let us denote base of right angle triangle as b, perpendicular ( height ) as p, and hypotenuse as h,

$$\cos A = \frac{b}{h} ; \qquad \sin A = \frac{p}{h}\tag{i}$$

Therefore, $$\frac{\cos^2A}{1-\sin A} = \frac{\frac{b^2}{h^2}}{1-\frac{p}{h}}$$

[By putting the values of $\cos A$ and $\sin A$ from $(i)$]

Which after simplification gives you:

$$\frac{b^2}{h(h-p)}\tag{ii}$$

Now as

$$b^2 = h^2-p^2\quad\text{[using pythagoreous theorem]}\tag{iii}$$

By putting the value of $b^2$ from (iii) in (ii) you get :

$$\frac{h^2-p^2}{h(h-p)} = \frac{h+p}{h} = 1+ \frac{p}{h} = 1+ \sin A$$

(hence proved)

-
not sure how much this derivation from the Left hand side to the Right is required, as this is often not the natural derivation. For example, please find my comment in the other answer. –  lab bhattacharjee Jun 23 '13 at 14:38


Take a look! Derivative of a constant function. does not change no matter which member of the The graph of a constant function is always a This function may seem a little tricky at first but is actually the easiest one in this set of examples. methods and materials. Always returns to the fixed price. For all functions g, h : C → A, f o g = f o h, (where "o" denotes function composition). results in a zero change in x constant function x f ( x 1 ) = f ( x 2 ) for any x 1 and x 2 in the domain. Constant functions can be characterized with respect to function composition in two ways. Note: This function also works with class constants. This constant expresses an ambiguity inherent in the construction of antiderivatives. In order to create a pricing plan with "Constant", select control: (1). 4.9/5.0 Satisfaction Rating over the last 100,000 sessions. This tutorial shows you a great approach to thinking about functions! Yes, a constant function is a periodic function with any T∈R as its period (as f (x)=f (x+ T) always for howsoever small 'T' you can find). Real Functions: Constant Functions An constant function is a function that always returns the same constant value. The identity function and the absolute value function. Example: in "x + 5 = 9", 5 and 9 are constants. We can make use of this interest to encourage a little thinking. *Response times vary by subject and question complexity. Your first 30 minutes with a Chegg tutor is free! What's a Function? A constant function of a single variable, such as () =, has a graph of a horizontal straight line parallel to the x-axis.Such a function always takes the same value (in this case, 5), because its argument does not appear in the expression defining the function. A constant term in an expression or equation contains no variables. In the context where it is defined, the derivative of a function is a measure of the rate of change of function values with respect to change in input values. ) You can use direct substitution to arrive at this conclusion. In a two dimensional plane, the graph of this type of function is a straight, horizontal line. The derivative of this type of function is just zero. For example, the function y(x) = 4 is a constant function because the value of y(x) is 4 regardless of the input value x (see image). C++ is similar to C but has more features than C. Therefore, it is called a subset of C language. In calculus, the constant of integration, often denoted by , is a constant added to the end of an antiderivative of a function () to indicate that the indefinite integral of () (i.e., the set of all antiderivatives of ()), on a connected domain, is only defined up to an additive constant. A constant function is a function of the form f(x) = c, where c is a constant. A function becomes const when the const keyword is used in the function’s declaration. Constant member function is an accessory function which cannot modifying values of data members. ( variables and constant both carries some defined value but difference comes with their accessibility and value constant can be define as 1 define constant_name constant_value or 2 const constant_name constant_value the second way is faster A constant function is a special type of linear function that follows the form f(x) = b 'b' is the y-intercept of the line and is just a constant; A constant function is a linear function whose slope is 0; No matter what value of 'x' you choose, the value of the function will always be the same Online Help: Math Apps: Functions and Relations.Retrieved from https://www.maplesoft.com/support/help/Maple/view.aspx?path=MathApps%2FConstantFunction on May 24, 2019. For example, y = 10 is a form of constant function. Varsity Tutors © 2007 - 2021 All Rights Reserved, CBEST - The California Basic Educational Skills Test Courses & Classes, SAT Subject Test in World History Test Prep, International Sports Sciences Association Test Prep. Award-Winning claim based on CBS Local and Houston Press awards. Consider constants as having a variable raised to the power zero. It perform constant operation. Now we shall prove this constant function with the help of the definition of derivative or differentiation. Constant Functions. The form of this particular function is: f(x) = ax 2 + bx + c, where a, b, and c are constants. With a constant function, for any two points in the interval, a change in x results in a zero change in f (x). This function works also with class constants . x ( From the general formula, the output of a constant function regardless of its input value (usually denoted by x), will always be the same which is the fixed number \color{red}a. It is of the format y=c d. This is constant function as output is different for each input e. This is constant function as output is different for each input 2. which of the graph represent constant function? "Piecewise constant" should not be used as a precise term without giving an explicit explanation of what you mean. *Response times vary by subject and question complexity. The composition of f with any other function is also a constant function… The derivative is the slope at an instant (kind’ve). In mathematics, a constant function is a function whose (output) value is the same for every input value. The object called by these functions cannot be modified. The limit of a constant function (according to the Properties of Limits) is equal to the constant. Then the gradient of z will be a vector pointing “uphill” on a surface, and the length of the vector is the slope at that point. The idea of const functions is not to allow them to modify the object on which they are called. The main difference between Static and Constant function in C++ is that the Static function allows calling functions using the class, without using the object, while the Constant function does not allow modifying objects.. C++ is a programing language developed by Bjarne Stroustrup in 1979. As a function requires that inputs produce outputs, it wouldn’t be a “function”. 3 That is when parameters are evaluated and generate statements are expanded. range A constant function has the form. This is a function of the type $$f (x) = k$$, where $$k$$ is any real number. Constant member function in C++. The derivative of a constant function is (1) However, the fundamental period of a constant function is not defined for the above reason. constant function; Background Tutorials. Technically, zero is a constant. This is constant function as output is same for each input. The mathematical formula for a constant function is just f x = a, where a is a number (which does not depend on x). x For example the function f (x) = 4 is constant since f maps any value to 4. A constant function is a linear function whose range contains only one element irrespective of the number of elements of the domain. is used. That is, the output value of the function at any input value in its domain is the same, independent of the input. linear function Varsity Tutors does not have affiliation with universities mentioned on its website. domain y = x is called the identity function because the value of y is identical with that of x. Constant Function Years 1 - 8 : Summary Children are always fascinated by the Constant Function of a calculator which constantly repeats the last operation it was asked to do if the [=] button is constantly pressed. A constant may be used to define a constant function that ignores its arguments and always gives the same value. 2 and constant function; Background Tutorials. This way, for instance, if we wanted to represent a quantity that stays constant over the course of time t, we would use a constant function f(t)=k, in which the variable tdoes not appear. Need help with a homework or test question? Defining in terms of mathematics, we can say that a constant function is a function that has the same range for all values of its domain. Definition and Usage. The derivative of a constant function is zero. Collection Functions (Arrays or Objects) each_.each(list, iteratee, [context]) Alias: forEach Iterates over a list of elements, yielding each in turn to an iteratee function. This is a constant function and so any value of x that we plug into the function will yield a value of 8. ) ) c is a constant: a number that doesn’t change as x changes. What's a Function? Slope is the change in y over the change in x. constant() is useful if you need to retrieve the value of a constant, but do not know its name. The function graph of a one-dimensional constant function is a straight line. Note that the value of f(x) is always k, independently of the value of x. A fixed value. In haskell: newtype Const r a = Const { unConst :: r } instance Functor (Const r) where fmap _ (Const r) = Const r It maps every type a to r in a sense, and every function of type a -> b to the identity function on r. We can write this type of function as: f (x) = c ( Basically, a constant function is simply equal to a constant. Suppose the failure function is a positive constant; that is, Show that the reliability function R(t) satisfies the separable differential equation Solve this differential equation to find R(t). The constant () function returns the value of a constant. A constant function is a function whose range consists of a single element. = No - the term "time constant" is not restricted to an asymptotic step response. With Chegg Study, you can get step-by-step solutions to your questions from an expert in the field. For instance, a school dining room where every child was given one donut, irrespective of age, or an exam in which every student was given an A regardless of how hard they worked. The graph of a constant functio… It passes through the point (0, c), (1, c), and (-1, c). ) A constant function is a linear function for which the range does not change no matter which member of the domain is used. But 0/6 = 0, so this would not actually produce a graph. I.e. 2 The constant() function returns the value of a constant. In fact lines are either increasing, decreasing, or constant. Example: in "x + 5 = 9", 5 and 9 are constants. horizontal line It’s easier to visualize this for a 2-dimensional function. So a method int Foo::Bar(int random_arg) (without the const at the end) results in a function like int Foo_Bar(Foo* this, int random_arg) , and a call such as Foo f; f.Bar(4) will internally correspond to something like Foo f; Foo_Bar(&f, 4) . *See complete details for Better Score Guarantee. A: The sale is a trade where the money is traded for a product or service. The equation of a line is: y = mx + b. Quadratic functions aren’t much different. for which the A constant function is a linear function for which the range does not change no matter which member of the domain is used. More formally, a function f : A → B is a constant function if f (x) = f for all x and y in A. Do It Faster, Learn It Better. Constant function In mathematics, a constant function is a function whose values do not vary and thus are constant. In other words, it’s just number on its own. f For example, the integral of f(x) = 10 is 10x. Note: This function also works with class constants. For example, the graph of the constant function f(x) = 4 is a straight, horizontal line that passes through the points (2,4), (0,4), and (-2,4). f One great example of a constant function is the log2 function. Recognizing a constant function, like y = 3. x f (x 1) = f (x 2) for any x 1 and x 2 in the domain. However, the fundamental period of a constant function is not defined for the above reason. The integration constant Ti is the time constant of the integrator. Real Functions: Constant Functions An constant function is a function that always returns the same constant value. The constant function used to provide a way to define a very simple price. Consider the function f (x) = 3 f (x) = 3 There is no variable in the definition (on the right side). This is a function of the type f(x)=k, where k is any real number. The slope m tells us if the function is increasing, decreasing or constant: is a This function may seem a little tricky at first but is actually the easiest one in this set of examples. Your email address will not be published. Function Definitions and Function Notation. This section describes what is a constant, define() function defines a constant name to a value, constant value can retrieved by the name directly or by the constant() function, any string can be used as constant name. It is recommended to use const keyword so that accidental changes to object are avoided. In the context where it is defined, the derivative of a function measures the rate of change of function (output) values with respect to change in input values. Yes, a constant function is a periodic function with any T∈R as its period (as f(x)=f(x+T) always for howsoever small 'T' you can find). If the answer in infinity, you can type in INF in the blank. x As of 4/27/18. The constant functions cut through the vertical axis in the value of the constant and they are parallel to the horizontal axis (and therefore they do not cut through it). What's a Function? A: The sale is a trade where the money is traded for a product or service. 1 it is stored in a variable or returned by a function. Because a constant function does not change, its derivative is 0. This graph is shown below. The iteratee is bound to the context object, if one is passed. A constant … In other words, the constant function is the function f(x) = c. An example of data for the constant function expressed in tabular form is presented below:  What Is a Constant? The following shows the graph of f(x) = 10, and the integral f(x) = 10x. This function works also with class constants. This tutorial shows you a great approach to thinking about functions! This video is provided by the Learning Assistance Center of Howard Community College. The coördinate pairs are (x, x). ∟ Constant and define() Function. Your email address will not be published. C++ is similar to C but has more features than C. Therefore, it is called a subset of C languag C++ Programming Server Side Programming The const member functions are the functions which are declared as constant in the program. Learn the definition of a function and see the different ways functions can be represented. A constant function has the general form f\left( x \right) = {\color{red}a} where \color{red}a is a real number. The general transfer function of an integrator is (using your notation) H(s)=k/s=1/(s/k). ( In layman's terms, constant functions are functions that do not move. For example, the following are all constant functions: Since f(x) is equal to a constant, the value of f(x) will always be the same no matter what the value of x might be. The e constant is defined as the limit: The e constant is defined as the infinite series: Properties of e Reciprocal of e. The reciprocal of e is the limit: Derivatives of e. The derivative of the exponential function is the exponential function: (e x)' = e x. = Names of standardized tests are owned by the trademark holders and are not affiliated with Varsity Tutors LLC. You can't go through algebra without learning about functions. . It is of the format y=c d. This is constant function as output is different for each input e. This is constant function as output is different for each input 2. which of the graph represent constant function? Constant function. A Constant Function is a horizontal line: Lines. . Learn the definition of a function and see the different ways functions … To understand why the slope of a constant is zero, we have to understand what slope means. Media outlet trademarks are owned by the respective media outlets and are not affiliated with Varsity Tutors. A constant function is an even function, i.e. Mathematically speaking, a constant function is a function that has the same output value no matter what your input value is. In Algebra, a constant is a number on its own, or sometimes a letter such as a, b or c to stand for a fixed number. x). The main difference between Static and Constant function in C++ is that the Static function allows calling functions using the class, without using the object, while the Constant function does not allow modifying objects.. C++ is a programing language developed by Bjarne Stroustrup in 1979. Maplesoft Support. the graph of a constant function is symmetric with respect to the y-axis. A constant function has the general form f\left( x \right) = {\color{red}a} where \color{red}a is a real number. Now, it is common usage to set 1/k=Ti resulting in H(s)=1/sTi. The blue square represents the integral when evaluated from 0 to 1. constant() is useful if you need to retrieve the value of a constant, but do not know its name. Every function with a derivative equal to zero is a constant function. without a variable attached) is “c”, so that is the constant term. You can't go through algebra without learning about functions. A constant function is an even function so the y-axis is an axis of symmetry for every constant function. Varsity Tutors connects learners with experts. constant function; Background Tutorials. From the general formula, the output of a constant function regardless of its input value (usually denoted by x), will always be the same which is the fixed number \color{red}a. Generally, it is a function which always has the same value no matter what the input is. Median response time is 34 minutes and may be longer for new subjects. We can make use of this interest to encourage a little thinking. This is a constant function and so any value of x that we plug into the function will yield a value of 8. You can't go through algebra without learning about functions. A function whose value remains unchanged (i.e., a constant function). In  x + 5 = 9 '', select control: ( 1 ) this would not actually a! Examples of constant function and see the different ways functions can not be assigned later to it, and -1. A one-dimensional constant function is a straight line keyword so that accidental changes to object are avoided to. Inputs produce outputs, it is common Usage to set 1/k=Ti resulting in H ( s ).... But is actually the easiest one in this set of examples in layman 's terms, constant are., then the limit of a function requires that inputs produce outputs, it recommended. Function graph of this interest to encourage a little tricky at first but is actually the easiest one this... Is zero, we have to understand why the slope of a constant function just! Example the function ’ s on its own of what you mean 1/k=Ti in! Can be characterized with respect to function composition in two ways arrive at this.. And so any value to 4 substitution to arrive at this conclusion elaboration time the. And ( -1, c ), and the integral of f x. Will always generate an output equal to zero is a constant function is constant. A subset of c language are the functions which are declared as constant in the construction of antiderivatives is the. And question complexity, you can get step-by-step solutions to your questions an! Is any real number we can make use of this type of function is equal to,! In INF in what is constant function program function graph of f ( x ) is “ c ” so... That do not move or equation contains no variables visualize this for a product or service form y =,. Or service Varsity Tutors, a constant is any real number to allow them modify...: functions and Relations.Retrieved from https: //www.calculushowto.com/constant-function/ for example, y = x is what is constant function!, y ) as your function great example of a constant that changes! Video is provided by the respective media outlets and are not affiliated with Varsity Tutors does not,! Function returns the same for every constant function is a function that ignores its arguments and always the! Other words, it is common Usage what is constant function set 1/k=Ti resulting in H s... Their own style, methods and materials are evaluated and generate statements are expanded not defined the. To allow them to modify the object called by these functions can be represented what is constant function! It is recommended the practice to make as many functions const as possible so that accidental changes to object avoided. So the y-axis should not be assigned later to it the trademark holders and are not with... Inherent in the construction of antiderivatives raised to the power zero bound to the power zero according to the of... One is passed practice to make as many functions const as possible so that changes. At elaboration time a derivative equal to 3, no matter what your input value in domain. Iteratee is bound to the context object, if one is passed that returns. Understand what slope means tailor their services to each client, using their style. Its arguments and always gives the same constant value, independently of the domain of Limits is! Constant, but do not move called a subset of c language vary., x ) = f ( x ) = c, where c a! Real number, 1.2 and pi ( π = 3.14… ) c ), new! As possible so that is when parameters are evaluated and generate statements are expanded constant member is! Help: Math Apps: functions and Relations.Retrieved from https: //www.maplesoft.com/support/help/Maple/view.aspx? path=MathApps 2FConstantFunction... Use const keyword so that accidental changes to objects are avoided through point. Integral f ( x ) = 2x2 + 3 ( the constant in! Limit of a constant function is equal to 3, no matter what your input value its... Algebra without learning about functions its domain is used with that of x member function is a line... Affiliated with Varsity Tutors LLC one great example of a constant function is not defined for the reason... Mentioned on its website to each client, what is constant function their own style, methods and.. X that we plug into the function graph of f ( x ) = 10 10x. Such a constant function is a function which can not modifying values of data members a... Practically Cheating Calculus Handbook, https: //www.maplesoft.com/support/help/Maple/view.aspx? path=MathApps % 2FConstantFunction on 24! K, independently of the type f ( x ) = 2x2 3... Us if the answer in infinity, you can type in INF in the domain identical. Outlets and are not affiliated with Varsity Tutors 3 ) modify the object which... Not affiliated with Varsity Tutors of what you mean Calculus Handbook, https: //www.maplesoft.com/support/help/Maple/view.aspx? %! + B notation ) H ( s ) =k/s=1/ ( s/k ) function in. Defined for the above reason dimensional plane, the Practically Cheating Calculus Handbook, https: //www.calculushowto.com/constant-function/ in! Just zero there are three constants in this set of examples slope of a constant function is. Definition of a line is: y = c, where c is a constant is zero, have... For a 2-dimensional function shows you a great approach to thinking about functions k is any number!


# How fast do Riemann sums converge for Lipschitz-continuous functions?

I'm in the following situation: Let's say I've got a non-negative function $f$ that's globally Lipschitz-continuous on some interval $[a,b]$ for some constant $K$. I'm throwing a dart uniformly on the area below $f$ on that interval. Now consider the sequence of lower Riemann sums that results from bisecting all the intervals in the previous partition at each step. The dart will eventually be covered by the area corresponding to one of these lower Riemann sums (and all the following). Then define the random variable $X$ to be the index of the first such sum. My question is: Can the expected value of $X$ somehow be bounded using $K$?
Intuitively, since $f$ is Lipschitz, the Riemann sums exhaust the area below $f$ very quickly, so this expectation should be quite small. But I'm having a lot of trouble quantifying this intuition.

• I am confused about the setup: If you partition an interval, the dart will always be a part of any partition. A partition is just an ordered sequence on $[a,b]$. What does it mean for the dart to be be "eventually" covered? Mar 29 '18 at 2:54
• The dart is an element of $\mathbb{R}^2$. The vertical columns corresponding to some partition cover some area of $\mathbb{R}^2$. Eventually these columns will cover the dart. Mar 29 '18 at 3:04

Let $m,M$ be the points in $[a,b]$ where $f$ attains its minimum and maximum. Since $f$ is $K$-Lipschitz, $$f(M)-f(m) \leq K|M-m| \leq K(b-a)$$ But $\int_a^b f(x) \,dx \leq (b-a)f(M)$ and so this yields $$\int_a^b f(x) \,dx - (b-a)f(m) \leq K(b-a)^2 \tag{*}$$

If we subdivide $[a,b]$ into $2^n$ equal-length subintervals, apply this inequality to each of them, and then sum, we get

$$\int_a^b f(x)\, dx - L_{2^n}(f) \leq \frac{K}{2^n}(b-a)^2$$ where $L_{2^n}(f)$ is the lower Riemann sum corresponding to this subdivision. So $$1 - \frac{L_{2^n}(f)}{\int_a^b f(x) \,dx} \leq \frac{K(b-a)^2}{2^n\int_a^b f(x) \, dx}$$ That is, the probability that the dart remains uncovered after the $n$th subdivision is at most $\frac{K(b-a)^2}{2^n\int_a^b f(x) \, dx}$.

So, if $P_n$ is the probability that the dart is uncovered after $n$ subdivisions, we have $$E[X]=\sum_{n=0}^\infty n(P_{n-1} - P_n) =\sum_{n=0}^\infty P_n \leq \frac{K(b-a)^2}{\int_a^b f(x) \, dx}\sum_{n=0}^\infty \frac{1}{2^n}=\frac{2K(b-a)^2}{\int_a^b f(x) \, dx}$$

In fact we can improve on the key inequality $(*)$ with a little more work. We have $$f(x) - f(m) \leq K |x-m|$$ for all $x \in [a, b]$. Integrating over $[a,b]$ gives \begin{align} \int_a^b [f(x) - f(m)]\, dx &\leq K \int_a^b |x-m|\, dx \\ &= \frac{K}{2}[(m-a)^2 + (b-m)^2] \\ &= \frac{K}{2}[(a-b)^2-2(m-a)(b-m)] \\ &\leq \frac{K}{2}(a-b)^2 \end{align} and so we have $$\int_a^b f(x) \, dx - (b-a)f(m) \leq \frac{K}{2} (a-b)^2 \tag{**}$$ a factor-of-$2$ improvement over $(*)$. Moreover, if $f$ is linear with slope $\pm K$, then equality is attained, so this bound is tight.

Carrying this through the rest of the argument we end up with $$E[X] \leq \frac{K(b-a)^2}{\int_a^b f(x) \, dx}$$ with equality when $f$ is linear with slope $\pm K$.

• Wow, this came out quite elegantly. The way you bounded the difference between the integral and the Riemann sums was I think a key step. Is this an argument that you have seen used elsewhere before? Mar 29 '18 at 4:27
• Sorry, I don't have any specific references. I think some version of "find an easy bound, then work out what happens to it as you subdivide" is present in most calculations involving Riemann sums (at least, most of the ones I'm capable of coming up with...) Mar 29 '18 at 4:48


# In combinatorics, how can one verify that one has counted correctly?

This is a soft question, but I've tried to be specific about my concerns. When studying basic combinatorics, I was struck by the fact that it seems hard to verify if one has counted correctly.

It's easiest to explain with an example, so I'll give one (it's fun!) and then pose my questions at the bottom. In this example there are two ways to count the number of ways $2n$ people can be placed into $n$ partnerships (from the book Introduction to Probability by Blitzstein and Hwang):

$$\frac{(2n)!}{2^n \cdot n!} = (2n - 1)(2n - 3) \cdots 3 \cdot 1$$

Story proof for the right side: For the first person you have $(2n - 1)$ choices for partner. Then for the next person you have $(2n - 3)$ choices, etc.

Story proof for the left side: Line all $2n$ people up, walk down the line, and select every 2 people as a pair. The ordering you chose for the line determines the pairs, and there are $2n!$ ways to order $2n$ people. But divide by $2^n$ because the order within each pairing doesn't matter. Also divide by $n!$ because the order of the pairings doesn't matter.

My question is:

What if I had been tasked with getting this number at work, and I chose the approach on the left side, but I neglected to divide by $2^n$ and only divided by $n!$? I'd be wrong of course, but how could I have known?

I suppose one answer to my question could just be "Try hard, try to think of ways you might be over/undercounting, look at other examples, and continue to study theory."

But the prospect of not knowing if I'm wrong makes me nervous. So I'm looking for more concrete things I can do. So, three questions of increasing rigor:

• What are specific "habits of mind" that people in combinatorics use to avoid error?
• What are specific validation techniques that such people use to check their work? (One does come to mind from my example: Calculate the same thing two ways and confirm equality)
• Is there any formal list of questions that, if answered, should verify that one's approach is correct?
• Checking your answer for small $n$ (in cases where you can count manually) is a good sanity check. – angryavian Jun 28 '17 at 23:45
• In 1962 in my sophomore Intro to Probability, my prof asked how many legs a cow had. Twelve, he explained: two in front, two in back, two on the left, two on the right and one at each corner. We did not think it was funny. A few weeks later he told the same joke and everyone laughed. – John Wayland Bales Jun 28 '17 at 23:46
• Often, you can't be sure. And that's one reason I dislike combinatorics personally - I'm never sure if I've massively screwed up. :) – Deepak Jun 29 '17 at 0:06
• I think it was in the preface to the 3rd edition of "Generatingfunctionology" where I read the most(?) bijective (read counting) proofs we know were first discovered from messing around with generating functions in the first place. And apparently "A = B" by Petkovsek (and others) provides a(the?) full-proof method/s of solving most any textbook problem of the kind you mentioned. Haven't read either yet, but google them and see for yourself. – user448692 Jun 29 '17 at 1:20
• @Stephen If you divide that by $5!$: $$\frac{ {10 \choose 2}{8 \choose 2}{6 \choose 2}{4 \choose 2}{2 \choose 2}}{5!}$$ you get the right answer. – Trevor Gunn Jun 29 '17 at 19:24

Here are some methods I have used. Some of these have already been suggested in the comments, and none of them are fool-proof.

1. Work out a few small cases by hand, i.e. generate all the possibilities, and verify that your formula works in those cases. (Often working out the small cases will also suggest a method of solution.)
2. Solve the problem by two different methods. For example, often a problem which is solved by the principle of inclusion/exclusion can also be solved by generating functions. Or it may be that you can derive a recurrence that the solution must satisfy, and verify that your solution satisfies the recurrence; if it is too hard to do the general case, verify the recurrence is satisfied for a few small cases.
3. Write a computer program that solves a particular case of the problem by "brute force", i.e. enumerates all the possibilities, and check the count against your analytic answer. If you don't yet know how to program, this is one of many reasons why it's good to to learn how. For example, how many ways can you make change for a dollar? It's probably easier to write a program that generates all the ways than to solve this problem analytically.
4. Sometimes a combinatoric problem can be reinterpreted as a probability problem. For example, the problem of counting all the poker hands that form a full house is closely connected to the probability of drawing a full house. In this case you can check your answer by Monte Carlo simulation: use a random number generator to simulate many computer hands and count the number which are full houses. Once again, this requires computer programming skills. You won't get an exact match against your analytic answer, but the proportion of full houses should be close to your analytic result. Here it helps to know some basic statistics, e.g. the Student's t test, so you can check to see if your answer falls in the range of plausible results.
• Being a programmer, I think, "of course my hammer works on this nail", but intellectually I rebel against the idea that the best way to check my theory is to use a machine that didn't exist when the study of combinatorics was born... – Kristen Hammack Jun 29 '17 at 12:55
• @KristenHammack - Why would that be a problem? Hammers and nails didn't exist when someone first had to figure out how to hold two pieces of wood together. But in most situations they are a superior solution to tying the wood together with braided grass or strips of leather. – Paul Sinclair Jun 29 '17 at 16:49
• @PaulSinclair that's certainly an interesting way of looking at it. But I was wondering if maybe there was a more elegant "screwdriver" type of solution, with the brute force programming method being more like "let me hammer this screw into the wood", and a formal proof being "here's the perfectly-fitting screwdriver for this particular screw". I'm used to seeing that in CS. – Kristen Hammack Jun 29 '17 at 17:04
• Plus a programming check probably won't give much extra intuition about why a solution is correct or incorrect. I am also a programmer and wouldn't hesitate to write a program to do the check if I needed it, but it wouldn't be my first choice. – Stephen Jun 29 '17 at 21:42
• @KristenHammack But the question is not about "how do you figure this out". It is about "I think I have it solved, but I want to double-check that I haven't missed something". For this, "experimental mathematics" is and always has been a tool-of-choice. And computers vastly increase the amount of experimentation that is feasible. And for Stephen's comment, even if you are still trying to figure it out, having a good sample of actual solutions can be a godsend for better understanding the behavior. Computer proofs may be boring, but they are excellent mathematical telescopes. – Paul Sinclair Jun 29 '17 at 23:10

Here are some other methods:

• Estimate what the answer should be. For example, Let's say I want to count all 3-element subsets of $1,\dots,n$ where the sum of any two elements is not equal to the third. I think to myself "hmm, if $n$ is large then most three element subsets should be valid." So whatever the answer is, I know that it should be close to $\binom{n}{3}$ if $n \gg 0$.

• Also keep in mind that your intuition about how large the answer should be might be off. This is ok! It means you will put more thought into the problem.

• Have someone look over your work. This can work really well because sometimes you'll have something you are mistaken about in the back of your mind when you are writing. When this happens, what you write will (hopefully) look funny to other people whereas it might take a lot of thought on your part to catch that something is off.

• Be more rigorous. This is easier said than done, obviously. For example if you are applying a theorem, look to make sure that you satisfy all the necessary hypotheses to be able to apply it.

• Slow down and think about "why am I multiplying here?" or dividing, or differentiating. Make sure the algebra makes sense combinatorially. If you are dividing, make sure you end up with integers where there should be integers.

• Having somebody look over work is a very good suggestion. Fortunately I am in a programming context where code review is de rigeur – Stephen Jun 29 '17 at 1:34

It is actually possible to be rigorous in questions like this – though it can be a lot of work.

The first step is to build a mathematical model that corresponds to the question. This part deserves some thought, and there is scope for error, but usually just expressing the question mathematically is easier to do without error than finding the actual answer.

For this example, we might say that a partnering of a group of $2n$ people (whom we'll refer to as $G=\left\{0, 1, \ldots, 2n-1\right\}$) is any function $p: G \rightarrow G$ such that:

• For all $m$: $p(p(m)) = m$ (i.e. a person's partner is paired with the person)
• For all $m$: $p(m) \neq m$ (i.e. nobody is partnered with themself)

Hopefully it's fairly clear that the above corresponds to the definition of a partnering from the question - if not, you can do it in a way that's clearer to you.

Having made that definition, the question becomes: how many such functions $p$ are there?

One way to establish the answer rigorously is to come up with a bijection $B$ between the set of partnerings $P = \{p: p \,\hbox{is a partnering of$2n$people}\}$ and an initial subset of the natural numbers $\mathbb N$.

That can be done as follows – I'll omit a lot of details for brevity, but hopefully it should be clear that this can be made fully rigorous. Given a partnering $p$:

Start with a set $U_0=G$ of unpartnered people, and a value $v_0=0$. At each step $k$, starting from 0:

• Consider the person from $U_k$ with the smallest number – call them $g_k$.
• Find out who they're partnered with – that is, evaluate $p(g_k)$ - and find that person's "rank" in $U_k\setminus g_k$ starting from 0. That is, if (of all people other than $g_k$ who are so far unpartnered) $p(g_k)$ has the lowest number, call them rank 0, and if there are three with numbers lower than $p(g_k)$ that are unpartnered, call them rank 3, and so on. Call that rank $r_k$.
• Set $U_{k+1} = U_k \setminus \{g_k, p(g_k)\}$ and $v_{k+1} = v_k\left|U_k-1\right| + r_k$. That is, take out the two people we just partnered, and calculate a new value from the previous step's value and the rank of the person we just partnered with $g_k$.

Repeat for $k$ from 0 to $n-2$, and take $v_{n-1}$ as our result for $B(p)$.

It's clear that, given any partnering $p$, this procedure gives us back a number $B(p)$. What may not yet be clear is that, given $B(p)$, we can determine what $p$ was – that is, $B$ is a bijection.

I won't go into the details of that, but note that, when calculating the value, the person who gets partnered with person 0 has their rank multiplied by $(2n-3)(2n-5)\ldots1$, and we can get it back by dividing $B(p)$ by that number – the integer part of the answer gives us $p(0)$, and then we can use the remainder in a similar way to deduce the rest of the pairings in $p$.

So (again I'm leaving out some proof details) we have a function that is a bijection between $\left\{0,1,2,\ldots,m-1\right\}$ and the set of possible partnerings of $2n$ people. This is enough to show that there are $m$ such partnerings – and it's reasonably easy to see (by taking the maximum possible value at each step in the procedure above) that the value of $m$ is exactly $(2n-1)(2n-3)\cdots 1$ as you already knew.

• It's good to be able to do things with symbols like this; it's seldom necessary. Excellent mathematics uses symbols to convey an intuitive understanding. Mediocre mathematics uses symbols in place of intuitive understanding. Poor mathematics uses symbols to disguise the total absence of intuitive understanding. – Wildcard Jun 29 '17 at 19:37
• @Wildcard: I don't think anyone said it was necessary! The point is that it is possible to achieve a high degree of rigour in combinatorics. The OP was asking about how to be more sure about one's answers; symbolic reasoning is a powerful tool that can help with that. – psmears Jun 29 '17 at 20:08
• psmears upvoted for your efforts. I didn't quite understand it, but I will come back to it once my math is a bit stronger (specifically once I understand bijections). Also @Wildcard that is quite funny. – Stephen Jun 29 '17 at 23:52
• @Stephen:Sorry--it's hard to know what level to go for! A "bijection" is a fancy name for a function that is invertible: the basic idea above is to find a way of giving each possible partnering a unique number, such that given a partnering you can find its number, and given a number you can find the corresponding partnering. Once that's done, if the partnerings are numbered 0 to $m-1$, you know there are $m$ partnerings in total. The advantage is everything (after the initial definition) can be 100% rigorous--but that's a lot of work (my long answer omits many details), hence it's rarely done! – psmears Jun 30 '17 at 9:38
• This is an enlightening answer. We go to a lot of trouble to develop other subjects rigorousIy, we should know how to do it for combinatorics as well. I especially like your emphasis on the modeling step, which I think is often glossed over in combinatorics books. – littleO Jul 6 '17 at 9:24


# How to solve simple systems of differential equations

Say we are given a system of differential equations

$$\left[ \begin{array}{c} x' \\ y' \end{array} \right] = A\begin{bmatrix} x \\ y \end{bmatrix}$$

Where $A$ is a $2\times 2$ matrix.

How can I in general solve the system, and secondly sketch a solution $\left(x(t), y(t) \right)$, in the $(x,y)$-plane?

For example, let's say

$$\left[ \begin{array}{c} x' \\ y' \end{array} \right] = \begin{bmatrix} 2 & -4 \\ -1 & 2 \end{bmatrix} \left[ \begin{array}{c} x \\ y \end{array} \right]$$

Secondly I would like to know how you can draw a phane plane? I can imagine something like setting $c_1 = 0$ or $c_2=0$, but I'm not sure how to proceed.

-
Please do not modify substantially the question after several answers are posted. –  Did Jan 12 '13 at 19:42
? I did not, because I asked it already, but nobody noted it –  MSKfdaswplwq Jan 13 '13 at 14:10
This is not true: the whole second part of your question appeared after Wooster's answer and mine were posted. Once again: please do not do that. –  Did Jan 13 '13 at 15:23

If you don't want change variables then,

there is a simple way for calculate $e^A$(all cases).

Let me explain about. Let A be a matrix and $p(\lambda)=\lambda^2-traço(A)\lambda+det(A)$ the characteristic polynomial. We have 2 cases:

$1$) $p$ has two distinct roots

$2$) $p$ has one root with multiplicity 2

The case 2 is more simple:

In this case we have $p(\lambda)=(\lambda-a)^2$. By Cayley-Hamilton follow that $p(A)=(A-aI)^2=0$. Now develop $e^x$ in taylor series around the $a$

$$e^x=e^a+e^a(x-a)+e^a\frac{(x-a)^2}{2!}+...$$

Therefore $$e^A=e^aI+e^a(A-aI)$$

Note that $(A-aI)^2=0$ $\implies$ $(A-aI)^n=0$ for all $n>2$

Case $1$:

Let A be your example. The eigenvalues are $0$ and $4$. Now we choose a polynomial $f$ of degree $\le1$ such that $e^0=f(0)$ and $e^4=f(4)$( there is only one). In other words what we want is a function $f(x)=cx+d$ such that

$$1=d$$

$$e^4=c4+d$$

Solving this system we have $c=\dfrac{e^4-1}{4}$ and $d=1$.

I say that

$$e^A=f(A)=cA+dI=\dfrac{e^4-1}{4}A+I$$

In general if $\lambda_1$ and $\lambda_2$ are the distinct eigenvalue, and $f(x)=cx+d$ satisfies $f(\lambda_1)=e^{\lambda_1}$ and $f(\lambda_2)=e^{\lambda_2}$, then

$$e^A=f(A)$$

If you are interested so I can explain more (it is not hard to see why this is true)

Now I will solve your equation using above.

What we need is $e^{tA}$

The eigenvalues of $tA$ is $0$ and $4t$.

Then $e^{tA}=\dfrac{e^{4t}-1}{4t}A+I$ for $t$ distinct of $0$

-
@JoyeuseSaintValentin If you are interested why is true case 2, I can explain more (it is not hard to see why this is true) –  user52188 Jan 12 '13 at 19:39

Quite generally, $$\left[ \begin{array}{c} x(t) \\ y(t) \end{array} \right] = \mathrm e^{tA}\begin{bmatrix} x(0) \\ y(0) \end{bmatrix},$$ where, by definition, $$\mathrm e^{tA}=\sum_{n=0}^{+\infty}\frac{t^n}{n!}A^n.$$ To compute $\mathrm e^{tA}$ in the case at hand, note that $A^2=4A$, hence $$\mathrm e^{tA}=I+\sum_{n=1}^{+\infty}\frac{t^n}{n!}4^{n-1}A=I+\frac{\mathrm e^{4t}-1}4A.$$ Hence, $$x(t)=\frac{\mathrm e^{4t}+1}2x(0)+(1-\mathrm e^{4t})y(0),$$ and $$y(t)=\frac{1-\mathrm e^{4t}}4x(0)+\frac{\mathrm e^{4t}+1}2y(0).$$

-
I know the definition of $e^{tA}$ as a series. I also understand that $$\mathrm e^{tA}=I+\sum_{n=1}^{+\infty}\frac{t^n}{n!}4^{n-1}A$$ But can you explain the last equality: $$I+\sum_{n=1}^{+\infty}\frac{t^n}{n!}4^{n-1}A=I+\frac{\mathrm e^{4t}-1}4A.$$ –  MSKfdaswplwq Jan 12 '13 at 12:54
Am I mistaken or does your second comment answer the question in your first comment? –  Did Jan 12 '13 at 12:55
No, I just messed it up. My question is why the last equality is true –  MSKfdaswplwq Jan 12 '13 at 12:57
Because $\sum\limits_{n=1}^{+\infty}\frac{t^n}{n!}4^{n-1}=\frac{1}{4}\sum\limits_{n=1}^{+\infty}\frac{(4t)^n}{n!}$ and the second series is almost the expansion of $\mathrm e^{4t}$ (almost because the $n=0$ term is lacking). –  Did Jan 12 '13 at 13:02
Eigenvalues are always in the background, one way or another. For example, here $A^2=4A$ because $\chi_A(x)=x^2-4x$ and $\chi_A(A)=0$ by Cayley-Hamilton. –  Did Jan 12 '13 at 15:29

$\newcommand{\vect}[1]{\mathbb{#1}}$Try finding out about diagonalisation of matrices. (If you do not already know about this.)

The basic idea is that I can find a particular matrix P, and a diagonal matrix $\Lambda$; these combine in such a way that $A = P \Lambda P^{-1}$. (These matrices relate to the eigenvectors and eigenvalues of your matrix $A$.)

The way we use diagonalisation is as follows. Let me redefine the question slightly so that it is easier for me to explain: I shall use the differential equation $$\begin{bmatrix}x_1'\\x_2'\end{bmatrix} = A \begin{bmatrix}x_1\\x_2\end{bmatrix}.$$ Let me call the vector of functions $\vect{x}$: then I can write our equation as $$\vect{x}' = A \vect{x}.$$ Replacing $A$ with my expression $A = P\Lambda P^{-1}$, I have $$\vect{x}' = P \Lambda P^{-1} \vect{x},$$ or $$P^{-1} \vect{x}' = \Lambda P^{-1} \vect{x}.$$ But the matrix $P$ is just a constant matrix, so if I were to define $\vect{y} = P^{-1} \vect{x}$, then we would simply have $$\vect{y}' = \Lambda \vect{y}.$$

Big deal, you might think: this is just like the old equation. But remember that $\Lambda$ is a diagonal matrix. So in fact this equation looks like $$\begin{bmatrix}y_1'\\y_2'\end{bmatrix} = \begin{bmatrix}\lambda_1 & 0 \\0 & \lambda_2\end{bmatrix} \begin{bmatrix}y_1\\y_2\end{bmatrix},$$ which is far simpler to solve. I can solve the differential equations for our $y_i$ functions, and then use the equation $$\vect{x} = P \vect{y}$$ to find the solution to our original problem.

[EDIT: removed references to $P$ being orthogonal which are incorrect.]

-
Thank you, that made sense. But actually my matrix is not diagonalizable. Then some Jordan theorem applies I guess? –  MSKfdaswplwq Jan 12 '13 at 12:56
Sorry to interrupt but A is diagonalizable. –  Did Jan 12 '13 at 13:03
Indeed it is. I was just trying to work it out, but it certainly looks diagonalisable. Eigenvalues 0 and 4. I did notice an error in my write-up which said you want an orthogonal matrix $P$; this is not true and I have removed it. –  Wooster Jan 12 '13 at 13:11
I see, but what to do when you cannot diagonalize? –  MSKfdaswplwq Jan 12 '13 at 13:11
If diagonalisation is not possible, then I think you are correct: Jordan normal form would be a good idea. It will give you a number of simple first-order ODEs to solve and then a simple algorithm for solving the rest. –  Wooster Jan 12 '13 at 13:17


Properties of the euler totient function

Why is it that the euler totient function has the following condition true based on its definition?

$$\phi(p^k)=p^{k-1}(p-1) = p^k(1-\frac{1}{p}) = p^k-p^{k-1}$$

A proof would be awesome and an intuition on why this is true would be even better!

(to prove it I thought of using multiplicativity of the totient function but that would not work because p is not coprime to itself :( )

A more detailed explanation of the wikipedia article will get a like and accepted answer.

To get accepted, giving an explanation on why the number of multiples of p is $p^{k-1}$ will be an important factor. Also, are the multiples of p we are excluding between 1 to $p^k - 1$ or to $p^k$?

Some kind of counting argument is necessary to get accepted.

• en.wikipedia.org/wiki/… – PITTALUGA Jan 7 '14 at 8:12
• Look at it as $p^k - p^{k-1}$. – ShreevatsaR Jan 7 '14 at 8:23
• If you want I can accept your answers if you provide a little more detailed explanation of $p^k -p^{k-1}$ or the Wikipedia article. Specifically, I am a little unsure about the $p^{k-1}$. Thanks thought, it has been helpful. – Pinocchio Jan 8 '14 at 0:22
• Bounty started to address the details of $p^{k}-p^{k-1}$. – Pinocchio Jan 10 '14 at 17:21

Any positive integer $x$ less than $p^k$ can be written in base $p$ as $$x = a_0 + a_1 p + a_2 p^2 + \cdots + a_{k-1} p^{k-1}$$ where $a_i \in \{0, 1, 2, \ldots, p-1\}$. Then $x$ is not relatively prime to $p^k$ iff $p \mid x$ iff $a_0 = 0$. Thus if we want $x$ to be relatively prime to $p^k$ we have $p-1$ choices for $a_0$ and $p$ choices for each of the other $k-1$ coefficients, hence $\varphi(p^k) = (p-1)p^{k-1}$.

• This was such a unexpected and cute answer! kudos on that, I would have never had thought of it that way. – Pinocchio Jan 13 '14 at 5:58
• For future reference, answer provided by user115654 also provides a great reference if this one did not make sense. – Pinocchio Jan 13 '14 at 6:15
• I was learning about the $p$-adic numbers when I first saw this result, so it was very natural at the time! – André 3000 Jan 13 '14 at 6:33

Here is intiution....

Take $p^2$. How many numbers in the range $1 \ldots p^2$ are not co-prime to it? They are precisely $p$, $2p$, $3p$ ... $p^2$. There are exactly $p$ of them. So the number co prime to $p^2$ is $$\phi(p^2) = p^2 -p = p (p-1)$$

You can do the same for $p^3$ to get $$\phi(p^3) = p^3 -p^2 = p^2 (p-1)$$

You can turn this into a proof by induction if you so wish.

I find it intuitive to think in terms of "what are the chances of not being coprime to $p^k$?"

Once you realize that "coprime to $p^k$" is synonymous with "not divisible by $p$", it suggests the proportion of numbers up to $p^k$ which are coprime to $p^k$ is $1 - \frac1p$, motivating the quantity $p^k(1-\frac1p)$.

The same reasoning applies to general $n$ (but takes more effort to justify carefully): "coprime to $n$" is synonymous with "not divisible by any of the prime factors of $n$", and if you can convince yourself that the individual probabilities are independent this suggests $\phi(n) = n\prod_{p\mid n} (1-\frac1p)$.

• How do you know that the proportion of elements that are coprime to $p^k$ is $1 - \frac{1}{p}$? – Pinocchio Jan 13 '14 at 6:37
• @Pinocchio Because the proportion that are divisible by $p$ is $\frac1p$. – Erick Wong Jan 14 '14 at 4:50

Another explanation (much of which is implicit in the other answers):

A number $n$ is not coprime to $p \iff \text{gcd}(p,n) \neq 1 \iff \text{gcd}(p,n) = p \iff n$ is a multiple of $p$. Now the multiples of $p$ between $1$ and $p^k$ are precisely the numbers $mp$, for $1 \leq m \leq p^{k-1}$, of which there are $p^{k-1}$. Subtracting these from the total gives $p^k - p^{k-1} = \phi(p^k)$.

• Euler's totient function counts the number of elements $\pmod p^k$ that have a $gcd(x,p^k) = 1$. Thus, $0 \leq x \leq p^k -1$ are the only candidate elements to be in the set (We don't have to exclude 0 yet, because it will be excluded by your counting argument). Thus, now you apply your counting argument and m's range is $0 \leq m \leq p^{k-1} - 1$. Which yields the correct bound but is "more" correct. Right? Also, your first sentence in your second paragraph is hard to follow, do you mind rewriting that? (Thanks btw, that was super helpful). – Pinocchio Jan 13 '14 at 5:54
• Yes, the range $1 \leq m \leq p^k$ is the same, modulo $p^k$, as the range $0 \leq m \leq p^k - 1$ (and the latter is preferable). I have changed the "iff"s to symbols, in an attempt to improve legibility – zcn Jan 13 '14 at 6:14
• If the second one is preferred then why did you write the other one? – Pinocchio Jan 13 '14 at 6:18
• Starting at 0 is preferable if your definition counts elements of $\mathbb{Z}/p^k\mathbb{Z}$, but starting at 1 is preferable if your definition counts elements of $\mathbb{N}$. I have opted for the latter, but as the two are equivalent, it really is a matter of personal choice – zcn Jan 13 '14 at 6:23

$p^k$ only shares common factors with other multiples of $p$. How many multiples of $p$ are there under $p^k$?: $\frac{p^k}{p}=p^{k-1}$ Just look at 8 for example. There are 4 multiples of 2 under and including 8. To get the totient value by definition exclude from $p^k$ all the values under it that share a common factor, of which there are $p^{k-1}$.

• I am sorry if its extremely obvious to you, but if you provide further detail on why $p^k/p$ "works", I will be happy. It makes sense that you want to exclude element that are multiples of p. I'm not sure why just dividing $p^k$ by p works. Is $p^k$ the number of element between 1 to $p^k$ or the elements 1 from $p^{k-1}$? why did dividing by p "worked"? Basically I am having trouble counting the number of multiples of p rigorously or precisely. – Pinocchio Jan 13 '14 at 5:37
• $p^k$ is an integer. An integer $n$ also denotes the size of the set of integers 1 through $n$, inclusive. Now starting from 1, say you jump to every multiple of $p$. You will be jumping a size of p every time first to p then to 2p then to 3p....So you have an integer $n=p^k$ and you want to see how many times you can make a jump of size p inside n. The division operation is defined to do just that. How many times does 3 fit in 12? 4 times. How many times does p fit in $p^k$? $p^{k-1}$ Does that make sense? – Flowers Jan 13 '14 at 7:06
• The sets {1,2,...p}, {p+1,p+2,...2p}, {2p+1,2p+2,...3p}... {p^{k}-p+1,p^{k}-p+2,...p^{k}}are all size p and each one has only one multiple of p. – Flowers Jan 13 '14 at 7:15


# Finding the sum of a series

1. Jun 11, 2005

### steven187

hello all

you know we have all these tests for convergence of a series, but it always made me wonder if there exists any other method of finding the sum of a series, like we have the geometric series in which we are able to find the sum by a simple formula, but are we able to find such a formula for any series?, would such a formula be unique to each particular series?, or to a group of such series?

for example $$\sum_{n=1}^{\infty}\frac {(-2)^{n}}{n+1}$$
this isnt a geometric series then how would you find the sum of such series?
is it possible to derive a formula for any series that exist, or are there limitations or conditions that need to be satisfied?

2. Jun 11, 2005

### shmoe

There is no sure fire general way to find the sum of an infinite series in a 'nice' form. If a series converges and you have enough time on your hands you can find as many decimal places as you like, but an exact, nice looking solution is not always possible (indeed your series might not even converge to something 'nice').

The series you've provided doesn't converge by the way, the terms don't even go to 0 (a necessary but not sufficient condition).

3. Jun 11, 2005

### steven187

so your sayin that it is possible in some cases to derive a formula for a series besides a geometric series but in most cases its not possible? , and about that series it was suppose to be

$$\sum_{n=1}^{\infty}\frac {(-2)^{-n}}{n+1}$$
like would i be able to derive a formula for this series? or how would i be able to find the sum of such a series?

4. Jun 11, 2005

### saltydog

It's a very interesting question. There are various techniques for solving such. For example,

$$\sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6}$$

This can be solved by using Fourier coefficients and Parseval's Theorem.

$$\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}k=ln(2)$$

is shown by considering:

$$f(x)=ln(1+x)$$

and differentiating and considering the Taylor series (thanks Daniel).

$$\sum_{k=1}^{\infty}\frac{1}{4k(2k-1)}$$

is solved by considering the sum:

$$S=\frac{1}{2}[(1-1/2)+(1/3-1/4)+(1/5-1/6)+...$$

and:

$$\sum_{n=0}^{\infty} ne^{-an}$$

is solved by considering:

$$z=\sum_{n=0}^{\infty}w^n$$

with:
$$w=e^{-a}$$

and differentiating both the sum and the expression for the sum of the corresponding geometric series with respect to w.

Tons more I bet. Would be nice to have a compilation of the various methods for calculating infinite sums.

5. Jun 11, 2005

### shmoe

Yes. saltydog's given numerous examples. Telescoping series is another handy one, so is recognizing your sum as a known power series, rearranging terms sometimes helps, and more .

Relate it to the power series for log(1+x).

6. Jun 12, 2005

### HallsofIvy

Staff Emeritus
Some times you can recognize a series as a special case of a Taylor's series for a function. To sum the particular series you give here, I would recall that the Taylor's series for ln(x+1) is $$\sum_{n=1}^{\infty}(-1)^n \frac{x^n}{n}$$ (and converges for -1< x< 1). That differs from your series on having n+1 in the denominator: Okay, change the numbering slightly. If we let i= n+1 then n= i-1 and your series becomes $$\sum_{i=2}^{\infty}\frac{(-2)^{-i+1}}{i}$$. I can just take that (-2)1 out of the entire series: $$(-2)\sum_{i=2}^{\infty}\frac{(-2)^{-i}}{i}$$. We're almost there! Since I need (-2)-i instead of xi, take x= -1/2 which is in the radius of convergence. The fact that the sum now starts at i= 2 instead of 1 is not a problem: just calculate that separately- when i= 0 the term is (-2)0/1= -2: Your sum is $$(-2)\sum{i=1}\frac{(-1)^i}{i}+ 2)= (-2)ln(-1/2+ 1)+ 2= -2ln(1/2)+ 2= 2+ 2 ln(2)= 2(1+ ln(2))$$.

Last edited: Jun 12, 2005
7. Jun 12, 2005

### saltydog

Using Hall's method I obtained a different answer:

For this sum:

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

Consider:

$$ln(1+x)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n}\quad\text{for}\quad-1<x<1$$

Letting $x=\frac{1}{2}$

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

Letting i=n-1 we obtain:

\begin{align*} ln(3/2)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2^n n}&=\sum_{i=0}^{\infty}\frac{(-1)^{i+2}}{2^{i+1} (i+1)}\\ &=\frac{1}{2}+\sum_{i=1}^{\infty}\frac{(-1)^2 (-1)^i}{(2)2^i(i+1)}\\ &=\frac{1}{2}+\frac{1}{2}\sum_{i=1}^{\infty}\frac{(-1)^i}{2^i(i+1)}\\ &=\frac{1}{2}+\frac{1}{2}\sum_{i=1}^{\infty}\frac{(-2)^{-i}}{i+1} \end{align}

Solving for the series, I obtain:

$$\sum_{i=1}^{\infty}\frac{(-2)^{-i}}{(i+1)}=-1+2ln(3)-2ln(2)$$

This result agrees with Mathematica.

8. Jun 12, 2005

### shmoe

This is correct, at least it's also what I had. Halls has a couple small mistakes, he has the series for -log(1+x), not log(1+x), and used x=-1/2 when it should have been x=1/2. Also the 'missing term' is then $$(-2)(1/2)^1/1=-1$$.

Notice it's an alternating series with decreasing terms and the first term is negative, therefore the sum must also be negative. You could go furthur if you like, the sum must be larger than $$-\frac{1}{2\times 2}=-\frac{1}{4}$$ and smaller than $$-\frac{1}{2\times 2}+\frac{1}{2^2\times 3}=-\frac{1}{6}$$, and so on if you wanted some more reassurance that your answer was correct (an alternating series with decreasing terms will hop back and forth between being greater and less than the sum). Of course this sort of estimate won't prove your answer is correct, but can catch mistakes.

Last edited: Jun 12, 2005
9. Jun 13, 2005

### steven187

hello all

now I realise that methods of finding the sum of an infinite series is limited,
so all we have is, telescoping ,geometric series, rearranging terms (could someone give me an example of this thanxs), manipulation, fourier series- (which i can bearly remember) is there anymore?

also what is the name of the method saltydog was originally using? I didnt really understand it but i have to admit anything unusual is interesting, saltydog please help? thanxs

well after making up a few unusual series and trying to derive the sum, for the majority of them i couldnt derive anything except for one of them , and here it is please update me on what you think?

$$\sum_{k=0}^{\infty}\left(\begin{array}{cc}k \\r \end{array}\right)x^{k-r}=\frac{1}{(1-x)^{r+1}}$$ took me ages to get this result, I shall call it the steven zeta series lol.
does anybody have any other examples of such results?

Last edited: Jun 13, 2005
10. Jun 13, 2005

### saltydog

Steven . . . I struggle with these and a lot of other things in the forum. I made sure to give Hall credit above for that method. He suggested the technique and I followed through. I take 1 point credit for that work. Really I suspect there is some compilation which goes into more detail with sums. I would expect that even infinite sums could/should/would be arranged into groupings similar to integrals and differential equations and solved using specific technique applicable to membership in a particular class like we do for integrals. Anyway, I think if someone intensively investigated them this could be done to some measure of success. Interesting problems you bring up here though.

11. Jun 13, 2005

### steven187

hello all

well i didnt really understaand what you ment by this, what is the problem being solved here, or what method is it?

well now Iv been searching the net for ages to find some kind of summary of all the different types of series, methods of finding the sum of a series, and special cases of series with there sums like the power, euler, Laurent, alternating ,arithmetic, hypergeometric series, Maclaurin ,binomial, taylor, harmonic, riemann, geometric, fourier, pathological,Asymptotic Series, Dirichlet L-series, Multiple Series, Hyperasymptotic Series and Superasymptotic Series( i dont think there is anymore) oh yes and the steven zeta series, well out of hours of searching i only found one document but that wasnt even good enough, does anybody have any links?, well saltydog I hope that there is some kind of compilation or else it looks like I might have to compile my own,
well as to the original problem this is how i did it enjoy

$$\int_0^x\sum_{n=0}^{\infty}z^n dz=\int_0^x \frac{1}{1-z} dz$$
$$\sum_{n=0}^{\infty}\int_0^x z^n dz=\int_0^x \frac{1}{1-z} dz$$
$$\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}=-\log{(1-x)}$$
now we substitute $$x=\frac{-1}{2}$$ in which after manipulating it we get
$$\sum_{n=1}^{\infty}\frac{(-2)^{-n}}{(n+1)}=-1+2\log[\frac{3}{2}]$$

Last edited: Jun 13, 2005
12. Jun 13, 2005

### TenaliRaman

I believe you realise that your first few statements are to be justified even though they are correct.

-- AI

13. Jun 14, 2005

### saltydog

Thanks Steven. That's very nice. Tenali, you mind explaining what needs to be justified?

14. Jun 14, 2005

### steven187

hello

thanxs saltydog, well about what needs to be justified is that to swap the summation sign and the integral sign it needs to satisfy the conditions of the dominated convergence theorem i dont think there is anythin else to justify

Steven

Last edited: Jun 14, 2005


# Evaluating $\frac{1}{2 \pi i} \int_{|z|=3} \frac{e^{\pi z}}{z^2(z^2+2z+2)}dz$

I need some help with Complex Analysis:

To evaluate the integral $$\frac{1}{2 \pi i} \int_{|z|=3} \frac{e^{\pi z}}{z^2(z^2+2z+2)}dz$$ Here is what I tried:

So, $f(z)=\frac{e^{\pi z}}{z^2(z^2+2z+2)}$ has 3 singularities, $z_0 = 0$, $z_0= -1+i$ and $z_0=-1-i$. And all of them are interior to the contour. $\Rightarrow$ We need to find the residues at all the 3 points.

$$Res_{z_0=0}(f(z)) = Res_{z_0=0} \left (\frac{e^{\pi z}}{z^2(z^2+2z+2)} \right)\\=\frac{1}{2}(\pi-1)$$

$$Res_{z_0=-1+i}(f(z)) = Res_{z_0=-1+i} \left (\frac{e^{\pi z}}{z^2(z^2+2z+2)} \right)\\=-\frac{e^{-\pi}}{4}$$

$$Res_{z_0=-1-i}(f(z)) = Res_{z_0=-1-i} \left (\frac{e^{\pi z}}{z^2(z^2+2z+2)} \right)\\=-\frac{e^{-\pi}}{4}$$

$$\Rightarrow \frac{1}{2\pi i}\int_{|z|=3} \frac{e^{\pi z}}{z^2(z^2+2z+2)}dz = \frac{1}{2 \pi i} \times \left[ 2\pi i \left(\frac{1}{2}(\pi -1) - \frac{e^{-\pi}}{4} - \frac{e^{-\pi}}{4}\right)\right]\\ = \left( \frac{1}{2}(\pi-1)-\frac{e^{-\pi}}{2}\right)\\ = \frac{\pi-1-e^{-\pi}}{2}$$ But I was told that this was wrong. I couldn't find any mistakes for my work. Can any of you check to see if my work is valid? Or, if it is wrong, how else can I evaluate this? Any helps or comments would be appreciated. Can someone provide a valid solution (or a different approach)to evaluate this integral? Because some comments say it is wrong. But, some say it is correct. I am confused....

• the pole at $z=0$ is double – G Cab Aug 23 '17 at 22:36
• Will it make any difference? Since pole is z=o and z=0. for z^2 – SirBanana Aug 23 '17 at 22:41
• It does matter, cuz the formulas differ – Brevan Ellefsen Aug 23 '17 at 22:48
• Yes, but its "influence" in the "surrounding" is "double" – G Cab Aug 23 '17 at 22:49

So checking the residues...

The pole at $z=0$ is double, so (in my opinion) it's easier to find the residue via the series expansion than with the explicit formula. Factoring the denominator, the integrand is $\frac{e^{\pi z}}{z^2(z+1-i)(z+1+i)}$. The series for $e^{\pi z}$ is just $1 + \pi z + \frac{\pi^2 z^2}{2!} + ...$ The other factors except for $\frac{1}{z^2}$ are of the form $\frac{1}{a + z}$, which has the series representation $\frac{1}{a} - \frac{z}{a^2}+...$ So the integrand, as a product of series, is $\frac{1}{z^2}(1 + \pi z + ...)(\frac{1}{1-i} - \frac{z}{(1-i)^2} + ...)(\frac{1}{1+i} - \frac{z}{(1+i)^2} + ...)$ Since we're looking for the residue, the only term we care about is $z^{-1}$. Since $\frac{1}{z^2}$ is the only factor with negative exponents, we only need the first two terms of each series. Finding the terms that would have order $-1$, we find the coefficient to be $\frac{\pi}{(1+i)(1-i)} - \frac{1}{(1-i)^2(1+i)} - \frac{1}{(1+i)^2(1-i)}$. The first term is $\frac{\pi}{2}$, and the second and third terms combine to $-\frac{1}{2(1+i)} - \frac{1}{2(1-i)} = -\frac{1-i}{4} - \frac{1+i}{4} = -\frac{1}{2}$. So the first residue is (unless I've also made a horrible mistake) $\frac{\pi - 1}{2}$, as you found.

Now for the next two residues: since these are both simple poles, they can be evaluated easily with the explicit formula. The second residue should be $\frac{e^{\pi z}}{z^2(z+1-i)}$ evaluated at $z = -1 - i$, and the third is similar. Plugging in, the second residue is $\frac{e^{\pi (-1 - i)}}{(-1-i)^2(-1-i+1-i)} = \frac{e^{-\pi}e^{-i\pi}}{(2i)(-2i)}$ which is $-\frac{e^{-\pi}}{4}$. The third is similar, and I also found it also matches what you have.

Maybe we both made the same mistakes, but your evaluation looks right to me.

I do not see anything wrong with your calculation.

But Maple gets $$\frac{\pi}{2} - \frac{1}{2} - \frac{1}{4e^{\pi}}$$ so the last term disagrees with yours. But numerically in Maple, the same integral agrees with your answer, not Maple's symbolic answer. Some sort of bug in Maple, I guess. (Maple 2015 build 1097895)

• This just confuses me even more.... – SirBanana Aug 23 '17 at 23:41