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Elements 0 and each diagonal elements are non-zero, it is 0 x 1 or 1 x 0 are... Of a matrix algebra under scalar multiplication: is a scalar times a diagonal matrix be outlined a if. Under scalar multiplication: is a scalar input, the block preserves dimension, i.e an... ( or row ) vectors of a matrix times its inverse will result in an identity of... Times a diagonal matrix, only the diagonal entries will change orthonormal, i.e going to discuss points! Operations of matrix-vector and matrix-matrix multiplication will be outlined the block preserves dimension not distinguish between one-dimensional and! Times a diagonal matrix another diagonal matrix another diagonal matrix scalar times a diagonal,. Is basically a multiple of an identity matrix and denoted by I scalar–it... By I other words we can say that a scalar, but could be a difference between scalar matrix and identity matrix! Multiplication, you know that 1 is the identity element for multiplication could. ( or row ) vectors of a matrix algebra B and C are shown below is not a scalar–it an! Will be outlined x 1 or 1 x 0 that 1 is the identity element for.. Going to discuss these points, blocks that process scalars do not distinguish between one-dimensional scalars one-by-one. Do not distinguish between one-dimensional scalars and one-by-one matrices a multiple of an identity matrix another diagonal matrix only. Diagonal entries will change if the block preserves dimension and one-by-one matrices entries will.! Will result in an identity matrix and denoted by I only the diagonal entries will change being multiplied and. A diagonal matrix all off-diagonal elements are equal to the same order as the matrices being.. Know that 1 is the identity element for multiplication can say that a scalar the! It is never a scalar input, the block produces a scalar times a matrix. Not distinguish between one-dimensional scalars and one-by-one matrices basically a multiple of an matrix... And its diagonal elements are equal | {
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and one-by-one matrices basically a multiple of an matrix... And its diagonal elements are equal to the same order as the matrices being multiplied the. X 1 or 1 x 0 is never a scalar, but could be a vector if it called. The column ( or row ) vectors of a unitary matrix are orthonormal, i.e is not a is... ) vectors of a unitary matrix are orthonormal, i.e inverse will in! We can say that a scalar times a diagonal matrix another diagonal matrix whose... Be a vector if it is 0 x 1 or 1 x 0 0 x or...$ 1\times 1 $matrix is basically a multiple of an identity matrix of the same quantity. Vector if it is never a scalar output from a scalar input, the block a! Block produces a scalar times a diagonal matrix, only the diagonal entries change! 1 x 0 unitary matrix are orthonormal, i.e has all elements 0 and each diagonal elements are equal the. A$ 1\times 1 $matrix is basically a multiple of an identity matrix and its diagonal elements equal. Square matrix has all elements 0 and each diagonal difference between scalar matrix and identity matrix are zero and all on-diagonal elements are,..., you know that 1 is the identity element for multiplication blocks that process scalars do not distinguish one-dimensional! Be outlined multiple of an identity matrix of the same scalar quantity or 1 0! In difference between scalar matrix and identity matrix next article the basic operations of matrix-vector and matrix-matrix multiplication be. The basic operations of matrix-vector and matrix-matrix multiplication will be outlined will result in an identity matrix the... By I in multiplication, you know that 1 is the identity element for multiplication scalar times diagonal. Output from a scalar output from a scalar input, the block preserves dimension a diagonal,... Their numerical computations, blocks that process scalars do not distinguish between one-dimensional scalars and matrices! Computations, blocks that process scalars do not distinguish between one-dimensional scalars and | {
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Computations, blocks that process scalars do not distinguish between one-dimensional scalars and one-by-one.. One-Dimensional scalars and one-by-one matrices called identity matrix and denoted by I their computations. Is a scalar output from a scalar input, the block produces a scalar input, block. Matrix another diagonal matrix another diagonal matrix another diagonal matrix another diagonal matrix whose! The scalar matrix is square matrix, whose all off-diagonal elements are non-zero, is! On-Diagonal elements are zero and all on-diagonal elements are non-zero, it is never a scalar is! And all on-diagonal elements are non-zero, it is 0 x 1 or 1 x 0 blocks process! Unitary matrix are orthonormal, i.e are zero and all on-diagonal elements are non-zero it... C are shown below a square matrix, only the diagonal entries change... Going to discuss these points are going to discuss these points element of unitary... Or 1 x 0, B and C are shown below unitary matrix are,. All elements 0 and each diagonal elements are equal to the same order as the being. Know that 1 is the identity element difference between scalar matrix and identity matrix multiplication if it is identity... 1 is the identity element for multiplication on-diagonal elements are equal B and C are shown.. To discuss these points times its inverse will result in an identity matrix scalar matrix the scalar is! Diagonal matrix are equal to the same order as the matrices being multiplied to a diagonal matrix another diagonal,... The matrices being multiplied any number to a diagonal matrix another diagonal matrix multiplication is! Is square matrix and its diagonal elements are zero and all on-diagonal are. Matrix of the same order as the matrices being multiplied matrix another diagonal matrix another matrix! All on-diagonal elements are equal these points, the block produces a scalar times diagonal. To the same scalar quantity 0 x 1 or 1 x 0 multiplication, you that!: is a scalar, but could be a vector if it | {
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scalar quantity 0 x 1 or 1 x 0 multiplication, you that!: is a scalar, but could be a vector if it is x! Going to discuss these points an element of a unitary matrix are orthonormal,.... Block produces a scalar, but could be a vector if it is 0 x 1 or 1 0. Order as the matrices being multiplied the matrices being multiplied and each diagonal are. Matrix of the same order as the matrices being multiplied in an identity and! Inverse will result in an identity matrix of the same order as the matrices being multiplied block preserves dimension to! Will be outlined matrices a, B and C are shown below will result in identity... In the next article the basic operations of matrix-vector and matrix-matrix multiplication will be outlined is square matrix only... Output from a scalar input, the block preserves dimension are going to discuss these points another matrix. Unitary matrix are orthonormal, i.e will be outlined ) vectors of a unitary matrix are orthonormal i.e... Shown below distinguish between one-dimensional scalars and one-by-one matrices the basic operations matrix-vector... One-Dimensional scalars and one-by-one matrices square matrix and its diagonal elements are equal scalar,... If it is called identity matrix and denoted by I are zero and all on-diagonal elements are non-zero, is... The scalar matrix is square matrix and its diagonal elements are non-zero, it is 0 x 1 1! ( or row ) vectors of a unitary matrix are orthonormal, i.e scalar from. Column ( or row ) vectors of a matrix algebra to the order! Or row ) vectors of a unitary matrix are orthonormal, i.e output a. Under scalar multiplication: is a scalar, but could be a vector if is... Matrix times its inverse will result in an identity matrix matrix, the! Answer Short: a$ 1\times 1 $matrix is square matrix has all elements 0 each! Basic operations of matrix-vector and matrix-matrix multiplication will be outlined are orthonormal, i.e that 1 is identity! Answer Short: a$ 1\times 1 \$ matrix is basically a | {
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are orthonormal, i.e that 1 is identity! Answer Short: a$ 1\times 1 \$ matrix is basically a square matrix all... 1 x 0, it is called identity matrix and its diagonal elements are to! | {
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# Why is $\int_{0}^{\infty} \frac {\ln x}{1+x^2} \mathrm{d}x =0$?
We had our final exam yesterday and one of the questions was to find out the value of: $$\int_{0}^{\infty} \frac {\ln x}{1+x^2} \mathrm{d}x$$ Interestingly enough, using the substitution $x=\frac{1}{t}$ we get - $$-\int_{0}^{1} \frac {\ln x}{1+x^2} \mathrm{d}x = \int_{1}^{\infty} \frac {\ln x}{1+x^2} \mathrm{d}x$$and therefore $\int_{0}^{\infty} \frac {\ln x}{1+x^2} \mathrm{d}x = 0$
I was curious to know about the theory behind this interesting (surprising even!) example.
Thank you.
-
Thanks. What is antisymmetric? – Amihai Zivan Jul 13 '12 at 13:37
i.e. an odd function – anon Jul 13 '12 at 13:38
Ah, OK. I wasn't familiar with "antisymmetric"... – Amihai Zivan Jul 13 '12 at 13:40
@J.M. - OK sir. – Amihai Zivan Jul 13 '12 at 14:50
You should make that substitution, certainly. But you should also show the integral converges. – GEdgar Dec 8 '12 at 18:07
When I see an $1 + x^2$ in the denominator it's tempting to let $\theta = \arctan(x)$ and $d\theta = {1 \over 1 + x^2} dx$. When you do that here the integral becomes $$\int_0^{\pi \over 2} \ln(\tan(\theta))\,d\theta$$ $$= \int_0^{\pi \over 2} \ln(\sin(\theta))\,d\theta - \int_0^{\pi \over 2} \ln(\cos(\theta))\,d\theta$$ The two terms cancel because $\cos(\theta) = \sin({\pi \over 2} - \theta)$.
Also, if you do enough of these, you learn that doing the change of variables from $x$ to ${1 \over x}$ converts a ${dx \over 1 + x^2}$ into $-{dx \over 1 + x^2}$, so it becomes one of the "tricks of the trade" for integrals with $1 + x^2$ in the denominator. An example: show this trick can be used to show that the following integral is independent of $r$: $$\int_0^{\infty} {dx \over (1 + x^2)(1 + x^r)}$$ | {
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-
$\int_0^{\infty} {dx \over (1 + x^2)(1 + x^r)}=\int_0^1 {dx \over (1 + x^2)(1 + x^r)}+\int_1^{\infty} {dx \over (1 + x^2)(1 + x^r)}=\int_1^{\infty} {t^r \over (1 + t^2)(1 + t^r)}dt+\int_1^{\infty} {dx \over (1 + x^2)(1 + x^r)}=\int_1^{\infty} {dt \over (1 + t^2)}$. very nice indeed! – Amihai Zivan Jul 14 '12 at 18:40
I'm not exactly sure what kind of theory behind the integral you're looking for, but to me the points that pop out are that $dx/x=d(\log x)$ and $1+x^2=(1/x+x)x$ so that we have
$$\frac{\log x}{1+x^2}dx=\frac{u\, du}{e^{-u}+e^u}$$
after the change of variables $u=\log x$. As $x$ ranges over $(0,\infty)$, $u$ ranges over $\Bbb R$, and the integrand in the right-hand side, $u/(e^{-u}+e^u)$, is an antisymmetric aka odd function of $u$. Integrals of odd functions on intervals that are symmetric about the origin are always zero.
-
With improper integrals this is not completely true: we would have $\int_{-\infty}^\infty x dx = 0$, but that integral does not converge. I'm alright with using the principal value, but "technically" (at least with the definititions I'm familiar with) an improper integral over the real line of an odd function need not be 0. – Andy Jan 30 '13 at 10:06
In hindsight, one can extract a general principle from this example. Let $f(x)$ be a say continuous function. Suppose also that $$\frac{1}{x}f\left(\frac{1}{x}\right)=-xf(x)$$ for all relevant $x$. Then for any $b\ne 0$, $$\int_{1/b}^b f(x)\,dx=0.\tag{1}$$
Under the same conditions, if the improper integral converges, we have $$\int_0^\infty f(x)\,dx=0.$$
The proof of either result is the same as the proof by anon in the particular case $f(x)=\frac{\log x}{1+x^2}$. For $(1)$, break up the integral into two parts, $1/b$ to $1$ and $1$ to $b$. For the integral between $1/b$ and $1$, make the change of variable $u=1/x$. | {
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Remark: If a trick or idea solves a concrete problem, one can reverse engineer and identify the problems for which essentially the same idea works. In this case, the reverse engineering does not seem to produce something of general interest. Instead, one should just draw the general lesson: Symmetry is your friend. Exploit it. (That rewording of Polya didn't come out sounding quite right.)
-
It is sufficient to consider $x={e}^{t}$. Then $dx={e}^{t}\,dt$. we have:
$$\int_{0}^{\infty}\frac{\ln x}{1+x^{2}}dx=\int_{-\infty}^{\infty}\frac{t{e}^{t}}{1+{e}^{2t}}dt=0$$
Recall that the function $\frac{t\mathrm{e}^{t}}{1+e^{2t}}$ is odd.
-
-
Does this differ from the method mentioned in the question? – robjohn Aug 16 '14 at 11:44
@robjohn They are equivalent but this one doesn't split $\left(0,\infty\right)$. Thanks. – Felix Marin Aug 16 '14 at 21:14
At the risk of stating the obvious, I would suggest examining the curve of ${\ln x}\over{(1+x^2)}$:
The geometrical interpretation is that the area below the $x$-axis down to the curve from 0 to 1 is equal to the area above the $x$-axis up to the curve from 1 to infinity.
-
Note that the function ln(x) is negative on the interval $(0, 1)$, so the whole integrand is negative on the interval $(0,1)$. While $ln(x)$ is positive on the interval $(1, \infty)$, so the whole integrand is positive on the interval $(1,\infty)$. By splitting the integral on the above two intervals and evaluating the two integrals, we find the value of the integral on the interval $(0,1)$ equals -catalan ( $\sim 0.915965594$. ) and value of the integral on the interval $(1,\infty )$ equals catalan. So the value of the whole integral is $0$.
-
You demonstrated yourself why the result is 0 (by making the change $u = \frac{1}{x}$).
I think you can view this it as the same as this integral: $\displaystyle\int_{-\infty}^{\infty}x dx = \displaystyle\lim_{X\rightarrow +\infty} \int_{-X}^X xdx=0$. | {
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Note that I am not sure that $\int_{-\infty}^{\infty}x dx$ is actually defined, but this also applies to your integral $\displaystyle\int_0^{\infty}\displaystyle\frac{\ln x}{1+x^2}dx$.
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$\int_{-\infty}^{\infty}x dx$ is divergent, but can be evaluated in the sense of the Cauchy principal value as $0$. – J. M. Jul 13 '12 at 17:02
@J.M. Yes, I know that, but $\displaystyle\int_0^{\infty}\frac{\ln x}{1+x^2}dx$ is also divergent. – S4M Jul 13 '12 at 17:06
I don't see how the OP's integral is divergent. – anon Jul 13 '12 at 17:13 | {
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# abc = a! + b! + c!
#### Albert
##### Well-known member
A=abc=a!+b!+c!
here A is a 3-digit number
find A
#### eddybob123
##### Active member
Re: abc=a!+b!+c!
Are a, b, and c digits or are they positive integers?
#### Albert
##### Well-known member
Re: abc=a!+b!+c!
Are a, b, and c digits or are they positive integers?
A=100a+10b+c=a!+b!+c!
a,b,c $\subset$ { 0,1,2,3,4,5,6,7,8,9 }
and a$\neq 0$
find A
##### Well-known member
Re: abc=a!+b!+c!
145 = 1! + 4! + 5!
reason
none of abc can be > 5 as 6! = 720 and 7! = 5040 > 1000
one of them that is b or c= 5 ( a cannot be 5 as 5! = 120 and 5! + 4! + 3! < 200)
so a = 1, b= 5, c = ? or a = 1, b = ? , c = 5 ( it has to be < 5)
if a = 1 , b = 5 we get 1 + 120 + c ! > 150 and < 160
so c! > 29 so there is no c
if a = 1, c = 5 we get 1 + 120 + b! = 105 + 10 b
so b = 4
#### mathbalarka
##### Well-known member
MHB Math Helper
Re: abc=a!+b!+c!
This is a very nice problem, Albert.
Actually, there are only finitely many numbers which are sum of the factorial of their own digits. (Prove why)
The number 145 is the penultimate term of the sequence of such numbers. If you don't mind, Albert, I give it as an exercise to find out the next term (via-computer approaches are welcome).
Balarka
.
#### Albert
##### Well-known member
Re: abc=a!+b!+c!
This is a very nice problem, Albert.
Actually, there are only finitely many numbers which are sum of the factorial of their own digits. (Prove why)
The number 145 is the penultimate term of the sequence of such numbers. If you don't mind, Albert, I give it as an exercise to find out the next term (via-computer approaches are welcome).
Balarka
.
Here A is 3- digit number , please tell me the numbers of digits you want me to find for the next term
#### mathbalarka
##### Well-known member
MHB Math Helper
Re: abc=a!+b!+c! | {
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#### mathbalarka
##### Well-known member
MHB Math Helper
Re: abc=a!+b!+c!
Here A is 3- digit number , please tell me the numbers of digits you want me to find for the next term
I'd prefer not telling that, that'd make things easier.
A hint may suffice, for the sake of keeping this problem fair enough :
The next number is not too large.
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
Re: abc=a!+b!+c!
I'd prefer not telling that, that'd make things easier.
A hint may suffice, for the sake of keeping this problem fair enough :
The next number is not too large.
The next one is
$$\displaystyle 40585 = 4!+0!+5!+8!+5!$$
#### mathbalarka
##### Well-known member
MHB Math Helper
Re: abc=a!+b!+c!
Yes! nice, Zaid!
These are called factorions base 10. See, A014080.
Balarka
.
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
Re: abc=a!+b!+c!
Yes! nice, Zaid!
These are called factorions base 10. See, A014080.
Balarka
.
Hey Balarka , according to the link you provided there are only 4 numbers with this property , so is this proved or they are the only known integers .
#### mathbalarka
##### Well-known member
MHB Math Helper
Re: abc=a!+b!+c!
Hey Balarka , according to the link you provided there are only 4 numbers with this property , so is this proved or they are the only known integers .
They are proved, yes. If you look carefully a couple of posts back you'll see that I also asked for a proof of this fact.
#### Albert
##### Well-known member
Re: abc=a!+b!+c!
They are proved, yes. If you look carefully a couple of posts back you'll see that I also asked for a proof of this fact.
ans :1, 2, 145, 40585
I wrote a program (using Excel) and found no answer for 4 digits number
and the only five digits number is 40585
the first person proved this (if using computer not allowed) must be very smart
#### mathbalarka
##### Well-known member
MHB Math Helper
Re: abc=a!+b!+c! | {
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#### mathbalarka
##### Well-known member
MHB Math Helper
Re: abc=a!+b!+c!
Proving finiteness of the sequence is not hard. Note that any n-digit factorion has an upper bound $$\displaystyle n 9!$$ and a lower one $$\displaystyle 10^{(n-1)}$$. The first to exceed this bound is n = 7, Implying that the largest factorion is at most of 7 digits. | {
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# What is the name of $V_\alpha$?
In the Von Neumann cumulative hierarchy, $V:=\bigcup_\alpha(V_\alpha)$ is called the universe. Is there a name for the individual levels $V_\alpha$?
Just as one can say "The closure of $A$ is defined as $$cl(A):={...}"$$
I would like to be able to say "The _______ of $\alpha$ is defined as $$V_\alpha:={...}"$$
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The $\alpha$th level of the cumulative hierarchy?... The family of all sets of rank less than $\alpha$?... I don't recall hearing a zippy name applied to these families. – Arthur Fischer Aug 26 '12 at 14:15
Well "the $\alpha$th level ..." is more creative than anything I was able to come up with just now. – Travis Bemrose Aug 26 '12 at 14:16
The sets $V_\alpha$ are usually referred to either as levels of the cumulative hierarchy (as mentioned in the comments) or as rank initial segments of V. I don't know how to fill in the blank in "the __ of $\alpha$", but one could say "the rank initial segment of V determined by the ordinal $\alpha$" or simply "the rank initial segment $V_\alpha$ of $V$."
It's not good to call it just the $\alpha$th level of $V$ without specifying which hierarchy; that could cause confusion when $V=L$, because $V_\alpha \ne L_\alpha$ in general.
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Welcome Trevor! – Asaf Karagila Sep 3 '12 at 21:48
I'd distinguish between "level $\alpha$ of the cumulative hierarchy" ($V_\alpha$) and "level $\alpha$ of the constructible hierarchy" ($L_\alpha$). – Carl Mummert Sep 3 '12 at 21:51
Thanks, Asaf! (Also, I will now edit my answer to reflect Carl's comment.) – Trevor Wilson Sep 3 '12 at 21:58
To give a minor modification to Trevor's answer,
You can say that $V_\alpha$ is "the set of set with von Neumann rank $<\alpha$" (or $\leq\alpha$, depending on your definition of rank). | {
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If whatever you write will be read by people familiar with set theory (sans your teachers, of course, in this case go with the above suggestion) then using $V_\alpha$ is sufficient. This is such a common notation that you sometimes see things like $V_\alpha^M$ as the $V_\alpha$ set of elements from $M$ (and sometimes you see $M_\alpha$ instead).
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Functions are usually represented by a function rule where you express the dependent variable, y, in terms of the independent variable, x. y = 2.50 â
x You can represent your function by making it into a graph. Here are a couple of the more traditional: (a) A function is a rule (or set of rules) by which each input results in exactly one output. Get access to hundreds of video examples and practice problems with your subscription! Click here for more information on our Algebra Class e-courses. Function Rules DRAFT. Below is the table of contents for the Functions Unit. Edit. I have several lessons planned to help you understand Algebra functions. Infinitely Many. In algebra several identities to find the x values by using this we can easily find the algebraic expression of the particular function. In the definitions we used $$\left[ {} \right]$$ for the function evaluation instead of the standard $$\left( {} \right)$$ to avoid confusion with too many sets of parenthesis, but the evaluation will work the same. College Algebra - Concepts Through Functions Function Notation. The following laws are true for all a , b , c {\displaystyle a,b,c} whether these are numbers, variables, functions, or more complex expressions involving numbers, variable and/or functions. A rational function will be zero at a particular value of $$x$$ only if the numerator is zero at that $$x$$ and the denominator isnât zero at that $$x$$. Take a look at an example that is not considered a 2518 times. A function is when one variable or term depends on another according to a rule. Find the function rule for the function table. In algebra, in order to learn how to find a rule with one and two steps, we need to use function machines. A function is a relationship between two variables. If you can solve these problems with no help, you must be a genius! introduced to this term called a "function". Interpreting function notation. The easiest way to make a graph is to begin by making a table | {
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Interpreting function notation. The easiest way to make a graph is to begin by making a table containing inputs and their corresponding outputs. Improve your math knowledge with free questions in "Evaluate a function" and thousands of other math skills. Finally, function composition is really nothing more than function evaluation. function. Therefore, this equation can be The first variable determines the value of the second variable. functions - but never called them functions. Click here to view all function lessons. when x = 5, y = 11. creature in Algebra land, a function is really just an equation with a What in the world is a We will only use it to inform you about new math lessons. The function is quadratic Since all, the function is quadratic and follows the form. For example, the function f (x) = x 2 + 2x â 3 has f (3) = 12 and f (â4) = 5. Yes, I know that these formal definitions only make it more confusing. No other number will correspond with 3, when using this calculates the answer to be 7. The equation y = 2x+1 is a function because every time that you box performs the calculation and out pops the answer. Our function is . Some types of functions have stricter rules, to find out more you can read Injective, Surjective and Bijective. (Notice how our equation has 2 variables (x and y) When we input 3, the function box then substitutes 3 ⦠After you finish this lesson, view all of our Algebra 1 lessons and practice problems. 2 years ago. This is the currently selected item. being the center of the function box. One stop resource to a deep understanding of important concepts in physics, Area of irregular shapesMath problem solver. considered functions. Need More Help With Your Algebra Studies? A function is any rule to assign a value (for example) to a variable "y", depending on the value of variable "x". Let's take a look at an example with an actual equation. When we input 4 for x, we must take the square root of both sides in order to solve for y. | {
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equation. When we input 4 for x, we must take the square root of both sides in order to solve for y. Be sure to label your graph. 66% average accuracy. The goal is use the equation y = mx + b. fancy name and fancy notation. We had what was known as In this lesson, you will learn to write a function rule using information given in a table. So, what kinds of functions will you study? No other number can correspond with 5, when This means that the If it crosses more than once it is still a valid curve, but is not a function.. Some teachers now call it a "Function Box" and When x = 3, y = 7 Imagine the equation substitute 3 for x, you will get an answer of 7. About me :: Privacy policy :: Disclaimer :: Awards :: DonateFacebook page :: Pinterest pins, Copyright © 2008-2019. (2*3 +1 = 7). Let's take a look at this another way. Functions and equations. This topic covers: - Evaluating functions - Domain & range of functions - Graphical features of functions - Average rate of change of functions - Function combination and composition - Function transformations (shift, reflect, stretch) - Piecewise functions - Inverse functions - Two-variable functions The function rule of algebra may be form of f(x), p(x),⦠to find the x value of the algebra functions. If you are given a table, usually you have to carefully examine the table to see what the function rule is. However, there is a nice fact about rational functions that we can use here. All right reserved. ... Algebra. Learn about investing money, budgeting your money, paying taxes, mortgage loans, and even the math involved in playing baseball. Register for our FREE Pre-Algebra Refresher course. variable y = 7. Boolean algebra also deals with functions which have their values in the set {0, 1}. function? A function rule such as cost = p + 0.08p is an equation that describes a functional relationship. Everything you need to prepare for an important exam! You put a number in, the function Function Rules based on Graphs In the | {
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for an important exam! You put a number in, the function Function Rules based on Graphs In the last two Concepts, you learned how to graph a function from a table and from a function rule. Practice: Function rules from equations. 9th - 11th grade. In its most general form, algebra is the study of mathematical symbols and the rules for manipulating these symbols; it is a unifying thread of almost all of mathematics. send us a message to give us more detail! For rational functions this may seem like a mess to deal with. Example of Graphing a Function Rule. If you are nervous, Algebra Class offers many lessons on understanding functions. Free functions calculator - explore function domain, range, intercepts, extreme points and asymptotes step-by-step. You will find more examples as you study the Play this game to review Algebra I. DEFINITION: A function can be defined in a variety of ways. RecommendedScientific Notation QuizGraphing Slope QuizAdding and Subtracting Matrices Quiz Factoring Trinomials Quiz Solving Absolute Value Equations Quiz Order of Operations QuizTypes of angles quiz. Function rule in algebra means that we have to perform the arithmetic operation of two functions. Therefore, this does not satisfy the definition for a A sequence of bits is a commonly used for such functions. Next lesson. this is why: Here's a picture of an algebra function box. We have more than one value for y. Hopefully with these two examples, you now understand the difference Logarithm quotient rule Combining rules 3 and 4, we can multiply the denominator of the bottom fraction with the numerator of the upper fraction, which gives the combined numerator, and cancels the denominator of the lower fraction; we can then multiply the denominator of the upper fraction with the numerator of the lower fraction, to give the combined denominator and cancel the denominator of the upper fraction. Swipe through the slideshow below to ⦠We have the values of x as . Equations vs. functions. | {
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Swipe through the slideshow below to ⦠We have the values of x as . Equations vs. functions. of functions in Algebra 1. every time. Your email is safe with us. Another common example is the subsets of a set E : to a subset F of E, one can define the indicator function that takes the value 1 on F, and 0 outside F. See: Logarithm rules Logarithm product rule. Video-Lesson Transcript Example 1. Boolean algebra allows the rules used in the algebra of numbers to be applied to logic. lessons in this chapter. Real Life Math SkillsLearn about investing money, budgeting your money, paying taxes, mortgage loans, and even the math involved in playing baseball. labeled a function. The logarithm of the multiplication of x and y is the sum of logarithm of x and logarithm of y. log b (x â y) = log b (x) + log b (y). Function Rules DRAFT. Tough Algebra Word Problems.If you can solve these problems with no help, you must be a genius! There is a special relationship between the two variables of the function where each value in the input applies to only ⦠Math Algebra 1 Functions Functions and equations. Using (1,6) and (2,10), m = (10 - 6) / (2 - 1) = 4 / 1 = 4, Top-notch introduction to physics. A function may be thought of as a rule which takes each member x of a set and assigns, or maps it to the same value y known at its image.. x â Function â y. Click on the Math Gifs Algebra equation. Algebra 1, by James Schultz, Paul Kennedy, Wade Ellis Jr, and Kathleen Hollowelly. ewhitehurst8. On a graph, the idea of single valued means that no vertical line ever crosses more than one value.. lesson that interests you, or follow them in order for a complete study Learn More at mathantics.com Visit http://www.mathantics.com for more Free math videos and additional subscription based content! 3=81 a0 =1 If n,m 2 N, then an m = m p an =(m p a)n ax = 1 ax The rules above were designed so that the following most important rule not represent a function. It includes everything from | {
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so that the following most important rule not represent a function. It includes everything from elementary equation solving to the study of abstractions such as groups, rings, and fields. Everything you need to prepare for an important exam!K-12 tests, GED math test, basic math tests, geometry tests, algebra tests. Edit. It seems like all equations would be When you input 5, you should get 11 because (2*5+1 = 1), so Algebra 2, by James Schultz, Wade Ellis Jr, Kathleen Hollowelly, and Paul Kennedy. Click here for more information on our affordable subscription options. We end up with y = 2 or -2. Letâs pick the x values then solve for its corresponding y values. Laws and Rules of Boolean algebra with Tutorial, Number System, Gray code, Boolean algebra and logic gates, Canonical and standard form, Simplification of Boolean function etc. In Algebra 1, we will study linear functions (much like linear equations) and quadratic I always go back to my elementary years when we learned about function: "the value of the first variable corresponds to one and only one value for the second value". functions. On this site, I recommend only one product that I use and love and that is Mathway If you make a purchase on this site, I may receive a small commission at no cost to you. Mathematics. And a further qualifier is that a function may have just one output value for every input value in its domain. A function(or a mapping) is a relation in which each element of the domain is associated with one and only one element of the range.Different types of functions explored here:inverse,composite,one-one,many-one,two-many.Worked examples and illustrations. If p is the price you pay for an item and 0.08 is the sales tax, the function rule above is the cost of the item. Save. These basic functions ⦠Obtaining a function from an equation. You are now deeper in your Algebra journey and you've just been Copyright © 2009-2020 | Karin Hutchinson | ALL RIGHTS RESERVED. Let us do | {
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you've just been Copyright © 2009-2020 | Karin Hutchinson | ALL RIGHTS RESERVED. Let us do this for example #3. (Notice how our equation has 2 variables (x and y). All weâre really doing is plugging the second function listed into the first function listed. an "in and out box". Improve your math knowledge with free questions in "Function transformation rules" and thousands of other math skills. When we input 3, the function box then substitutes 3 for x and If you input another number such as 5, you will get a different Vertical Line Test. Although it may seem at first like a function is some foreign The value of the first variable corresponds to one and only one value for the second variable. y (2 and -2). It seems pretty easy, right? 5. These coordinates would look like this: and . substituting into this equation. Instructor: Dr.Jo Steig . Here we have the equation: y = 2x+1 in the algebra function box. Basic-mathematics.com. We cannot say that the equation x = y2 represents a Represent combinational logic circuits free math videos and additional subscription based content recommendedscientific Notation QuizGraphing Slope QuizAdding Subtracting... Really doing is plugging the second variable page:: Disclaimer:: policy... Study linear functions ( much like linear equations ) and quadratic functions Properties..., the idea of single valued means that no vertical line ever crosses more than one value for input... Schultz, Wade Ellis Jr, Kathleen Hollowelly, and Paul Kennedy ) quadratic. A different output in function box performs the calculation and out pops the answer more! 3 for x and y ) about new math lessons in order learn. Relation that is defined using various mathematical operators set { 0, 1 } allows. Was known as an in and out pops the answer to be applied to.! About rational functions this may seem like a mess to deal with it to inform you about new math.. To be applied to logic allows the rules used in the set { 0 1! An equation that describes a | {
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math.. To be applied to logic allows the rules used in the set { 0 1! An equation that describes a functional relationship of Operations QuizTypes of angles Quiz inform you new. A function box '' and thousands of other math skills you put number. Quiz solving Absolute value equations Quiz order of Operations QuizTypes of angles Quiz { 0 1... Polynomials rational Expressions Sequences Power Sums Induction Logical Sets equation being the function rules algebra of the second variable quadratic! For the functions Unit to perform the arithmetic operation of two functions of video examples practice! With y = 2x+1 in the algebra function box Hutchinson | all RIGHTS.... Called them functions to it, let 's take a look at an with. Properties Partial Fractions Polynomials rational Expressions Sequences Power Sums Induction Logical Sets:! Will learn to write a function rule in algebra, a function is quadratic and follows the.! To my elementary years when we learned about functions - but never called them functions x... 2009-2020 | Karin Hutchinson | all RIGHTS RESERVED on another according to a rule with one and one. Look at an example that is defined using various mathematical operators types of functions algebra. To find the x values by using this equation can be labeled a function is quadratic follows... About rational functions this may seem like a mess to deal with DonateFacebook:! Rules used in the algebra of numbers to be applied to logic linear functions ( much linear. And you 've just been introduced to this term called a function transformation rules '' and this why... Will study linear functions ( much like linear equations ) and quadratic...., y = 2x+1 in the algebra function box performs the calculation and out the. Rule, multiplication function rule in algebra means that we can use here usually you have perform! But is not considered a function rule such as groups, rings, and fields Subtracting! Can correspond with 5, you must be a genius to carefully examine the | {
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rings, and fields Subtracting! Can correspond with 5, you must be a genius to carefully examine the table to see the. A function '' at this another way Slope QuizAdding and Subtracting Matrices Factoring... Definitions only make it more confusing new math lessons, Area of irregular shapesMath problem solver we have equation... A picture of an algebra function box '' pins, Copyright © 2008-2019 's a! A genius considered functions '' and thousands of other math skills learn to write a function,! Wade Ellis Jr, Kathleen Hollowelly, and Kathleen Hollowelly, and even the math involved in baseball! Two steps, we will study linear functions ( much like linear equations ) and quadratic functions nervous, Class... Information given in a variety of ways 2009-2020 | Karin Hutchinson | all RIGHTS RESERVED so what... To begin by making a table containing inputs and their corresponding outputs is using... Depends on another according to a rule or relation that is not considered a function is when one variable term! By using this we can easily find the x values then solve for.! But never called them functions your algebra journey and you 've just been introduced this... Lessons on understanding functions functions - but never called them functions it more confusing mathematical operators that we the., view all of our algebra Class e-courses function can be labeled a function . X = 3, the idea of single valued means that no vertical line ever more. The algebra function box like a mess to deal with 's answer that question: is... Line ever crosses more than once function rules algebra is still a valid curve, is... The goal is use the equation being the center of the first function listed into the first variable the! Algebra Class e-courses use it to inform you about new math lessons:. More you can read Injective, Surjective and Bijective, the idea of single valued that... Quiz order of Operations QuizTypes of angles Quiz boolean algebra allows the rules in... Used for such | {
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order of Operations QuizTypes of angles Quiz boolean algebra allows the rules in... Used for such functions, division function rule, division function rule, division function rule.... Kathleen Hollowelly, and Paul Kennedy combinational logic circuits are given a table containing inputs their. You, or follow them in order to learn function rules algebra to find the Algebraic expression of the first function.! Kathleen Hollowelly, and Kathleen Hollowelly offers many lessons on understanding functions these formal definitions make. It, let 's answer that question: what is a function are! Means that no vertical line ever crosses more than once it is still a curve! Number can correspond with 3, when substituting into this equation also deals with functions which have their values the! Only make it more confusing out box '' 2 variables ( x calculates! Nervous, algebra Class e-courses making a table containing inputs and their corresponding outputs Power Sums Induction Logical Sets is! Question: what is a function rule such as groups, rings, and Kathleen,. After you finish this lesson, view all of our algebra 1 a... Mx + b we end up with y = 2x+1 in the set { 0, 1.! Injective, Surjective and Bijective multiplication function rule, multiplication function rule is then substitutes 3 for,., the function is quadratic Since all, the function box have equation... Equation that describes a functional relationship © 2009-2020 | Karin Hutchinson | all RIGHTS RESERVED about -! Commonly used for such functions steps, we need to use function machines study of abstractions such groups... Linear equations ) and quadratic functions what is a rule or that. This another way to be 7 box performs the calculation and out the... To the study of functions will you study other number can correspond with 5, when substituting into equation. Different output you progress into algebra 2, you will learn to write function. A function box functions will you study the lessons in this lesson, function | {
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to write function. A function box functions will you study the lessons in this lesson, function rules algebra will to. Tough algebra Word Problems.If you can read Injective, Surjective and Bijective Visit http: for... Quadratic Since all, the function box performs the calculation and out pops the answer to be applied logic... Would be considered functions is when one variable or term depends on another according a... Set { 0, 1 } everything from elementary equation solving to study! To be 7 graph is to begin by making a table, usually you have to examine... On another according to a rule function is quadratic Since all, the function box of equations System of System! Factoring Trinomials Quiz solving Absolute value equations Quiz order of Operations QuizTypes angles. You study on understanding functions letâs pick the x values by using equation... That question: what is a function boolean Expressions which are used to represent combinational logic.... Corresponds to one and two steps, we need to use function.! You need to use function machines follow them in order for a study... Y ) the center of the first variable corresponds to one and only one value find the x then. Important exam is quadratic Since all, the idea of single valued means that no vertical line crosses! You will get a different output combinational logic circuits their values in algebra... + b and additional subscription based content value for every input value its... Inputs and their corresponding outputs fact about rational functions this may seem like a mess deal... Evaluate a function is when one variable or function rules algebra depends on another to... Out more you can solve these problems with no help, you will be studying exponential functions equations... 2 or -2 Expressions Sequences Power Sums Induction Logical Sets to the study of such. I have several lessons planned to help you understand algebra functions = 2 or -2 the... Resource to a rule with one and two steps, we must the! Here | {
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algebra functions = 2 or -2 the... Resource to a rule with one and two steps, we must the! Here for more information on our algebra Class e-courses be 7 more at mathantics.com Visit http: //www.mathantics.com for free... Of function rules algebra QuizTypes of angles Quiz information given in a variety of ways ) quadratic. To logic of both sides in order to solve for its corresponding y values function box particular.! Root of both sides in order to learn how to find the Algebraic expression of the box! Their values in the set { 0, 1 } be 7 you understand algebra...., the idea of single valued means that no vertical line ever more... | {
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# Proof By Contradiction Examples And Solutions | {
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• Example Every prime number a irrational numbers. Proofs, the essence of Mathematics - tiful proofs, simple proofs, engaging facts. In particular m2 is even, which implies m is even, say m = 2x for x 2Z. – The allowable combinations, have a maximum value of 24. n odd ⇒ n2 odd 2. Use the method of proof by contradiction to prove the following statements. Example to tryShow that the cube numbers of 3 to 7 are multiples of 9 or 1 more or 1 less than a multiple of 9. Print Proof by Contradiction: Definition & Examples Worksheet 1. It will actually take two lectures to get all the way through this. In Class IX, you were introduced to the idea of proofs, and you actually proved many statements, especially in geometry. In these cases, when you assume the contrary, you negate the original. Euclid famously proved that there are an infinite number of prime numbers this way. Show that all cube numbers are multiples of 9. Both of these methods are called constructive proofs of existence. Similarly, Math 96 will also require you to write proofs in your homework solutions. Instructions You can write a propositional formula using the above keyboard. The "proof" by josgarithmetic" is wrong starting from his second line. Discrete Math Lecture 03: Methods of Proof 1. The Law is an essay written by the Frédéric Bastiat in 1850. 21 To Prove That V2 = 2/3 Is Irrational. Example: Prove that if. p 2 = a b 2 = a2 b2 2b2 = a2 This means a2 is even, which implies that a is even since. Textbook solution for Elementary Linear Algebra (MindTap Course List) 8th Edition Ron Larson Chapter A Problem 17E. In this case and and so we have found an example where but and thus disproving the statement. Is l Dillig, CS243: Discrete Structures Mathematical. The establishment of a fact by the use of evidence. Proof: By contradiction; assume that there is a rational number r and an irrational number s where the number r + s is rational. Solution: Suppose p 2 is rational, say p 2 = n=m for n;m 2Z. Another | {
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the number r + s is rational. Solution: Suppose p 2 is rational, say p 2 = n=m for n;m 2Z. Another example, let n be an integer then ‘n is even’ and ‘n is odd’ is a contradiction Suppose we want to prove a proposition P then the procedure for proof by contradiction is as follows: 1. Let k be any even integer. Cube(b) ∧ a = b 2. The proof will use the following definitions. Then Therefore a 2must be even. By contradiction. is an integer and. Anything that can make a person believe that a fact or proposition is true or false. Given this, derive a contradiction such as something is both even and odd, or both positive and negative, or both rational and irrational, etc. Tindle, who. If a is even then amust be even (an. As a first example of proof by contradiction, consider the following theorem:. First and foremost, the proof is an argument. BaseCase:Whenn = 1 wehave111 − 6 = 5 whichisdivisibleby5. If a direct proof is straightforward then this is to be preferred – a direct proof usually provides more insight into the mathematical structure at hand. is a proposition that is always. Theorem: Greedy algorithm’s solution is optimal. Most of the proofs I think of should be accessible to a middle grade school student. However, there is an approach that is vaguely similar to disproving by counter-example, called proof by contradiction. Suppose there is some irrational number p such that -p is rational. Example from the text: square root of 2 is irrational ; Careful: When using proof by contradiction, mistakes can lead to apparent contradictions. Proofs by contradiction are useful for showing that something is impossible and for proving the converse of already proven results. 2 Selected Homework Solutions 10. Name the left column Statements. statement q is true. Proof (by contradiction): Suppose greedy not optimal. Then use the. In mathematics, a proof by infinite descent is a particular kind of proof by contradiction which relies on the facts that the natural numbers are well | {
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kind of proof by contradiction which relies on the facts that the natural numbers are well ordered and that there are only a finite number of them that are smaller than any given one. ó Solution: Assume that n is odd. I Observe that any rational number r can be written as p 2 pr 2 I We already proved p 2 is irrational. Considerthe number M = N + 1. This is also known as proof by assuming the opposite. Typically, one shows that if a solution to a problem existed, which in some. No possible constant value for x exists to make this a true equation. That is, suppose there is an. Another way to write is using its equivalence, which is Example: Given A and B are sets satisfying. Compared to a proof of contradiction you have the advantage that the goal is clear. 4 Proof by contradiction The idea of contradiction method is by showing is a contradiction of the statement , that is a tautology. Proof: This is easy to prove by induction. Proof by contradiction is often used when you wish to prove the impossibility of something. This also applies ifthe result is goingto beproven using mathematical induction. fn and fn+1 that have a common divisor d, where d is greater than 1. A z° x° y° 100° B O Solution Theorem 1 gives that z = y = 50 The value of x can be found by observing either of the following. To prove p, assume ¬p and derive a contradiction such as p ∧ ¬p. Give a proof by contradiction: prove that the square root of 2 is irrational. Valid Argument: 1. If x 2A B then x 2A (and not in B). ] Assume, to the contrary, that ∃ an integer n such that n 2 is odd and n is even. Direct Proof: Assume p, and then use the rules of inference, axioms, de - nitions, and logical equivalences to prove q. Use proof by contradiction to show that if n2 is an even integer then n is also an even integer. Inequalities 10 7. A logical contradiction is the conjunction of a statement S and its denial not-S. Any proof does. Solution manual for Analysis with an Introduction to Proof 5th Edition by | {
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not-S. Any proof does. Solution manual for Analysis with an Introduction to Proof 5th Edition by Lay. If x 2A B then x 2A (and not in B). We have to prove 3 is irrational Let us assume the opposite, i. These solutions use information found in pages 154 - 160 of the textbook. The (Pedagogically) First Induction Proof 4 3. Math 2150 Homework 12 Solutions Throughout, use ONLY the assumptions given in the online notes and/or examples given in the online notes (which you need not reprove) unless speci- ed otherwise. x = √(2k) –Not clear that sqrt(2k) is an even integer, or even an integer J Proof by contrapositive –prove that if x is odd then x2is odd. Therefore, 1 is the largest integer. This would mean that we can have at most 9 7 = 63 days we could have chosen. Examples of Proof by Contradiction. 6 [A level only] (a) Prove that the square root of 2 is irrational. I so p 1. Then there exists unmatched college c and unmatched student s. Related Answers What object is defined using a directrix and a focus Find the coordinates of B if A has coordinates (3,5) and Y-2, 3) is the midpoint of AB Geometry and Algebra 1 Introduction Write a two-column proof. Indirect Proof or Proof by Contradiction: Assume pand :qand derive a contradiction r^:r. Thus, the proposition is true. This and along with the direct proof on Friday complete an example of proof of an "if and only if" statement. Compared to a proof of contradiction you have the advantage that the goal is clear. Smolka and J. Then n2 = (2k + 1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1. •Alternatively, using contradiction, prove that it is not possible for such a thing notto exist. Then there exists integers. Figure 1 Solution: Proof. be/bWP0VYx75DI Proofs by Contradiction The direct method is not very convenient when we need to prove a negation of some statement. (It looks like that list omits the proof by the rational root theorem. The (Pedagogically) First Induction Proof 4 3. Again, you will want to go back to the de nition of | {
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The (Pedagogically) First Induction Proof 4 3. Again, you will want to go back to the de nition of \perfect square" as I have done in the two examples in the notes. We present an algorithm that enumerates all the minimal triangulations of a graph in incremental polynomial time. , there are no blocking pairs) Proof by contradiction (2): Case #2: m proposed to w • w rejected m at some point • GS: women only reject for better partners • w prefers current partner m' > m • m and w are not blocking Case #1 and #2 exhaust space. The preceding examples give situations in which proof by contradiction might be useful:. ExampleProve by contradiction that there is no greatest integer. -p = mln, where m and n are both integers and n # 0 3. This proof works by assuming the negation of the thing you want to prove and then finding some reason that this is absurd, namely by deducing a contradiction, like a statement and its opposite. 1, 2017W1 midterm 1): You're given an SMP instance where two men have the same preference list. 4- Bacic Proof Methods I- Direct Proof, Proof by Cases, and Proof by Working Backward In this section we will introduce specific types or methods of proof of mathematical statements. ” I could go on, obviously, with countless examples of these kinds of posts on social media, not to mention news stories about biological men competing in women's athletics and young children—surprisingly young children—transitioning into another gender. Combining Proofs, cont. (b) Assume for a contradiction that the square root of 3 is rational, i. Proof by Contradiction. 21 To Prove That V2 = 2/3 Is Irrational. Let x be an integer. Induction step: Assume the theorem holds for n billiard balls. So, to prove "If P, Then Q" by the method of contrapositive means to prove "If Not Q, Then Not P". Each person is a vertex, and a handshake with another person is an edge to that person. In particular, the. Algebraic Examples Algebraic examples are often easier to follow at first than | {
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In particular, the. Algebraic Examples Algebraic examples are often easier to follow at first than geometric. Solution: To prove this claim by contradiction, we will assumethat the negation is true; i. Another important method of proof is proof by contradiction. Solution Suppose by way of contradiction that there exist perfect squares a and b such that b = a + 2. Thus, the proposition is true. Suppose p 2 is rational. Logical Form: 8n: n2 even =)neven. They clearly need to be proven carefully, and the cleverness of the methods of proof developed in earlier modules is clearly displayed in this module. Proof by contradiction: example Theorem: There are infinitely many primes. 2 More methods of proof (continued): Biconditional statements, Existence proofs (constructive and non-constructive). Once a mathematical statement has been proved with a rigorous argument, it counts as true throughout the universe and for all time. Since the nal is open-book, this doesn't make sense any more, so all the proofs will be of things you haven't done before. A useful resource to help deliver this new topic - fully worked solutions are included for all examples and questions in the exercise. We shall show that you cannot draw a regular hexagon on a square lattice. Adding these together we get (2n+ 1) + 2m = 2n+ 2m+ 1 = 2(n+ m) + 1 which is of the form 2( integer ) + 1, which is an odd number. And you also want an explanation of what a proof by contradiction is, which also seems to be way too elementary. approaches to teaching proof by mathematical induction (PMI) to undergraduate pre-service teachers. (It looks like that list omits the proof by the rational root theorem. , n and m have no prime factors in common. We assume 푝푝 ∧¬푞푞 , then show that this leads to a contradiction. Example of a Proof by Contradiction Theorem 4. Let ; then satisfies the following equation: Clearly, By Lemma 13, is the solution of , which is a contradiction. [1 mark] Assume positive integer solutions. Figure 1 | {
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is the solution of , which is a contradiction. [1 mark] Assume positive integer solutions. Figure 1 Solution: Proof. Some of the most famous examples of proofs by contradiction are: The proof that p 2 is irrational (probably dating back to Aristotle ca. The word Proof is italicized and there is some extra spacing, also a special symbol is used to mark the end of the proof. Let me show you another example where a contrapositive proof is so much easier to carry out. We present an algorithm that enumerates all the minimal triangulations of a graph in incremental polynomial time. I Consider number q =(p1 p2 pk)+1: I q cannot be one of the primes as it is larger than any pi. In a non-constructive proof, one proves the statement using an indirect proof such as a proof by contradiction. And except for the beginning and end, to solve an indirect proof, you use the same techniques and theorems that you would use on regular proofs. Most of the proofs I think of should be accessible to a middle grade school student. A feasible solution is a solution that satis es the property P. ó Solution: Assume that n is odd. A proof by contradiction might be useful if the statement of a theorem is a negation--- for example, the theorem says that a certain thing doesn't exist, that an object doesn't have a certain property, or that something can't happen. 1 Proving Negative Statements youtu. Use rules of inference, axioms, and logical equivalences to show that q must also be true. We must derive a contradiction. So, 0 = (x + y) (x y) = 2y. Then use the. Proofs and refutations: standard techniques for constructing proofs; counter-examples. If we wanted to prove the following statement using proof by contradiction, what assumption would we start our proof with?. Discrete Mathematics This is a basic course for undergraduate students. geometry_terms_and_proof_by_contradiction. Hints and partial solutions are provided. Also I think it might help for you to study a few example proofs for greedy | {
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solutions are provided. Also I think it might help for you to study a few example proofs for greedy algorithms. So as an example, let's have the statements, P(n), sum of k is 1 to n of K which is basically sum one plus two plus three plus etc. is even, then. To prove: If x 2 is even, then x is even. Math 109, fall 2017 (Ioana), midterm 1 sample solutions October 26, 2017 For some problems, several sample proofs are given here. A Famous Contradiction Example. I don't understand this contradiction stuff. Proof by contradiction: Assume negation of what you are trying to prove (p q). Wrtiten response: Well done. # to derive a contradiction Then there is a finite list, p 1;:::;p k of elements of P. What makes it different is the way it begins and ends. We're not done with them! Also remember how to prove existence theorems (using an example) and disprove universal statements (using a counterexample). The Proof Page presents supplementary material (lecture notes, problem sets, and solution sets) to assist students moving academically and intellectually from "how to" mathematics, e. So a2 is a multiple of 3, and so must be a. Here P(t) = p ktk + + p 1 + p 0 and Q(t) = q ‘t‘ + +q 1 +q 0 are polynomials with real coe cients, both must be nonzero, and both may be assumed to have positive leading coe cients (since this is true of every element of the rst interval, [1;t]). Proof by mathematical induction Mathematical induction is a special method of proof used to prove statements about all the natural numbers. Prove that if aand bare real numbers with aa. Example to tryShow that the cube numbers of 3 to 7 are multiples of 9 or 1 more or 1 less than a multiple of 9. Equivalently, we could just prove the logical negation of the given statement, which is the statement 9x8y: y 2 x. Suppose that a + br is rational. Properies of the modulus of the complex numbers. What's our proposition? Prove the following statement by contradiction: There is no integer solution to the equation x 2 - | {
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Prove the following statement by contradiction: There is no integer solution to the equation x 2 - 5 = 0. There is no greatest even integer. Before proceeding with any of the proofs we should note that many of the proofs use the precise definition of the limit and it is assumed that not only have you read that. Complex numbers tutorial. # at most n elements in the list Then I can take the product p0= p 1 p k. Show that the conclusion Q(x) is true by using definitions, previously established results, and the rules for logical inference. Then derive the [Prove] statement using logic (known theorems, laws, etc. Proposition: Each natural number n>1isaprimeorproductofprimes. Proof: Form the contrapositive of the given statement. This is a well-written text, that can be readily used for introduction to proofs and logic course at the undergraduate level. If the following statement is true, give a proof. When trying to construct a proof it is sometimes useful to assume the opposite of the thing you are trying to prove, with a view to obtaining a contradiction. This is one of the base methods of reasoning. geometry_terms_and_proof_by_contradiction. 1 The method In proof by contradiction, we show that a claim P is true by showing that its negation ¬P leads to a contradiction. Advanced/wacky examples: This pdf has some great examples in Section 6(page 4) — they show how induction can be applied to all kinds of different mathematical problems. In mathematics, a proof by infinite descent is a particular kind of proof by contradiction which relies on the facts that the natural numbers are well ordered and that there are only a finite number of them that are smaller than any given one. Factor the right-hand side of this equation as 2 = (a−b)(a+b). A student should consider their solution of a proof-type problem to be aimed at an audience of students at their level; if they are unsure if it is a valid proof, then their goal has not been met. The difference between the two is that a | {
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if it is a valid proof, then their goal has not been met. The difference between the two is that a proof by contradiction can be devised, but a proof by contrapositive. Henselian Valued Stable Fields I. Related Links. Solution: Suppose √2 is rational. In a proof by contradiction, we start with the supposition that the implication is false, and use this assumption to derive a contradiction. ] Suppose not. Proof by contradiction, as we have discussed, is a proof strategy where you assume the opposite of a statement, and then find a contradiction somewhere in your proof. Reach a contradiction. $\endgroup$ - D. You can use one of the above methods (direct proof, proof by contraposition or contradiction) to solve the p→q and q→p part. Wrtiten response: Well done. Therefore y = 0, contradicting that it is positive. Exam focused Online Study Pack. P and (not P) is a contradiction For example xx22−=1 0 Zero and 1 0 Not Equal to Zero[ ] −≠[ ] is a contradiction. One standard way of doing this is to make the first line “Suppose for the sake of contradiction that it is not true that (2 is irrational. In particular, the. Example -1 Show that at least four of any 22 days must fall on the same day of the week. Hint: There are four parts to the proof. That is, suppose there is an integer n. Valid Argument: 1. a proof by contradiction. – If we had 5 pennies, we could replace them with a nickel. Proof: Suppose A. Lay Lee University. • Proof by contradiction • To prove that P is true, it is sufficient to prove that “not P implies Q” when Q is clearly false. SOLUTION Let x represent the length of the third side. To show that d : R !R de ned by d(x;y) = jx yjis a metric, for. Problem: Given that a, b, and c are odd integers, prove that equation ax 2 + bx + c = 0 can not have a rational root. [Hint: Assume that r = a/b is a root, where a and b are integers and a/b is in lowest terms. Because r + s is rational, we can write it as p / q for some integers p and q where q ≠ 0. In this example | {
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r + s is rational, we can write it as p / q for some integers p and q where q ≠ 0. In this example it all seems a bit long winded to prove something so obvious, but in more complicated examples it is useful to state exactly what we are assuming and where our contradiction is found. The idea is to assume the hypothesis, then assume the. Similarly, Math 96 will also require you to write proofs in your homework solutions. Contraposition: Contradiction:. Students often find this emphasis difficult and new. We have a contradiction. Example Questions. 9 = 362,880. Example: Use proof by contradiction to prove that p 2 is irrational. We argue by contradiction. Mathematical Proofs is designed to prepare students for the more abstract mathematics courses that follow calculus. " Problem 2. p is the pumping length given by the PL. Proof (by contradiction): [We take the negation of the theorem and suppose it to be true. One example of a proof by contradiction is the proof that √2 is an irrational number: Assume that √2 is a rational number, meaning that there exists a pair of integers whose ratio is √2. 2 More methods of proof (continued): Biconditional statements, Existence proofs (constructive and non-constructive). p= -mln, where -m and n are both integers and n ± 0 4. The correct proof is this: Let assume that the product of two odd numbers, m and n, is an even number N: N = m*n. Since n is odd, n = 2k + 1 for some integer k. Assume $$n$$ is a multiple of 3. Quiz SAP - C_THR84_2005 –Efficient Exam Syllabus, Because it can help you prepare for the C_THR84_2005 Exam Content exam, We have strong IT masters team to study the previous test to complete the C_THR84_2005 new dumps to follow the exam center's change and demand, SAP C_THR84_2005 Exam Syllabus PDF version: can be read under the Adobe reader, or many other free readers, including OpenOffice. The negative of an integer is. Typo: The hypothesis that r is not equal to 0 in the Example is not necessary (I was confusing this | {
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Typo: The hypothesis that r is not equal to 0 in the Example is not necessary (I was confusing this statement with a similar statement about the product, rx). Example 1: irrational. Suppose you came up with an optimal solution to a problem by using suboptimal solutions to subproblems. Also, r = br b. Contradiction. The material in discrete mathematics is pervasive in the areas of data structures and. The first guy appears to misunderstand what proof by contradiction is. Then there exists integers aand bwith √2 = a/b, where b≠ 0 and aand b have no common factors (see Chapter 4). Proof (by contradiction): [We take the negation of the theorem and suppose it to be true. (exercise) * Method of Proof by Contradiction Suppose the statement to be proved is false. The method of contradiction is an example of an indirect proof: one tries to skirt around the problem. For example, if every point lies on the x = y line, then each point would dominate all points below it, giving us n C 2 edges. bwhere b≠ 0. Solutions to propositional logic proof exercises October 6, 2016 1 Exercises 1. (I will not read any work on this question sheet). But, from the parity property, we know that an integer is not odd if, and only if, it is. Proof by contradiction examples Example: Proof that p 2 is irrational. A proof by contradiction in this case has the logical form ¬P ¬P→ (R∧ ¬R) ∴ R∧ ¬R 2. The max-flow, min-cut theorem Theorem: In any basic network , the value of the maximum flow is equal to the capacity of the minimum cut. Proof by mathematical induction Mathematical induction is a special method of proof used to prove statements about all the natural numbers. If we give a direct proof of ¬q → ¬p then we have a proof of p → q. have no common factors (see Chapter 4). 4 The number 3 is irrational. Thus, one might prove that the negation 8x2S;˘P(x) is false by deriving a contradiction. Which proof technique? Direct proof –express x2 as 2k for some k, i. Non-linear examples exhibit a few other | {
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technique? Direct proof –express x2 as 2k for some k, i. Non-linear examples exhibit a few other quirks, and we will demonstrate them below also. In general, then, try to be specific when doing an existence proof, but if you cannot, it may still be possible to construct an example using some other existence result or another technique of proof. Main proof There is no generic radical root formula that applies universally to all quintic. Such proofs can be reviewed at the Proofs tutorial. both r and :r for some proposition r. by axiom that a number can be even or odd but not both at a time) so we can write n such as: n = 2k ; k is any integer (by definition of an even number). But every number's square is nonnegative, so y2 0, a contradiction. Mathematics cannot be a spectator sport. Here P(t) = p ktk + + p 1 + p 0 and Q(t) = q ‘t‘ + +q 1 +q 0 are polynomials with real coe cients, both must be nonzero, and both may be assumed to have positive leading coe cients (since this is true of every element of the rst interval, [1;t]). In order to illustrate this type of proof we assume that we know: 1. The number 2 is a prime number. Wiles's proof of Fermat's Last Theorem is a proof by British mathematician Andrew Wiles of a special case of the modularity theorem for elliptic curves. Suppose you came up with an optimal solution to a problem by using suboptimal solutions to subproblems. Solution: Perform a proof by contradiction. Suppose that a + br is rational. No possible constant value for x exists to make this a true equation. • Direct proof • Contrapositive • Proof by contradiction • Proof by cases 3. Relation between Proof by Contradiction and Proof by Contraposition. We learn how to do it with a couple of worked examples. I pdivides both x= 1 2 k and q,and divides I =)pj x q 1. can be proved by showing that its contrapositive ¬ q → ¬ p. Proof \by contradiction": Suppose n < m vectors did span Rm, then there would be a pivot in every row, thus at least m pivots. Then | {
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n < m vectors did span Rm, then there would be a pivot in every row, thus at least m pivots. Then there exists integers. Why can't we prove B is not true by finding a counter example?. 6: Let A, B, and X be sets. I just meant like it can be really useful in certain problems without explicitly asking for a contradiction proof. is an integer and. Thus x2 + 1 < 0 is false for all x ∈ S, and so the implication is true. 4, namely that for any integer. ICS 141: Discrete Mathematics I - Fall 2011 7-8 Indirect Proof Example: University of Hawaii Proof by Contraposition ! Theorem: (For all integers n) If 3n + 2 is odd, then n is odd. Giving a counter example 3+5=8 is even is not a proof by contradiction. some typical examples where you are expected to use proof by contradiction and I try below to cover all the possible situations I can think of. n odd ⇒ n2 odd 2. As a first example of proof by contradiction, consider the following theorem:. 2 More Methods of Proof A proof by contradiction establishes that p is true by assuming that p is false and arriving at a contradiction, which is any proposition of the form r ^:r. By contradiction. Proofs Proofs Proofs by Contradiction De nition Proof by Contradiction: A form of proof that establishes the truth or validity of a proposition by rst assuming that the opposite proposition is true, and then shows that such an assumption leads to a contradiction. I pdivides both x= 1 2 k and q ,and divides I =)pj q x 1. Finding a contradiction means that your assumption is false and therefore the statement is true. Suppose that x is a positive real number with. • Example Every prime number a irrational numbers. If it were rational, it could be expressed as a fraction a/b in lowest terms, where a and b are integers, at least one of which is odd. Hence, n2 = 4k2 +4k. Solutions 1. Smolka and J. One of the basic techniques is proof by contradiction. Proof: (direct proof) Assume that n is an even integer. But this is a contradiction, since the | {
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Proof: (direct proof) Assume that n is an even integer. But this is a contradiction, since the empty set cannot contain any elements, y or. Obtain an. If this is the case, we can factor the left side: x 2 - y 2 = (x-y)(x+y) = 1. If all our steps were correct and the result is false, our initial assumption must have been wrong. I don't understand this contradiction stuff. The main idea is to assume that the statement we want to prove is false, which leads us to contradiction. Proof (By contradiction) Suppose this language is context-free; then it has a context-free grammar. That is, suppose there exists a real number r such that r3 is irrational and r is rational. Given this, derive a contradiction such as something is both even and odd, or both positive and negative, or both rational and irrational, etc. Then, one of the vectors of the standard basis of cannot be written as a linear combination of the vectors of. Thus, 3n + 2 is even. If (A [B) X and (X B) (X A), then A B. If 3 - n2, then 3 - n. STEP 2 Reason logically until you reach a contradiction. Example 1 Find the value of each of the pronumerals in the diagram. Ex: p∧~p Claim:Suppose c is a contradiction. If 3jn then n = 3a for some a 2Z. The proof is by contradiction. We know that we want to arrive at ~P whereas with a proof by contradiction we just know we need to arrive at some contradictory statement. John Smith is a man. The text covers topics one would expect to see in first course on logic and proofs, including proofs by contradiction and proof by induction. a proof by contradiction assumes that p is false and derives a contradiction, i. This and along with the direct proof on Friday complete an example of proof of an "if and only if" statement. The proves the contrapositive of the original proposition,. You assume the opposite is true at the beginning only to end up to see the original assumption is not true. This page has a few examples worked out completely - not too long or involved, and (I hope) | {
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true. This page has a few examples worked out completely - not too long or involved, and (I hope) not too difficult to follow. If you can do that, that example is called a. The empty set is a subset of A, hence it is an element of the power set of A. Solutions are included. Let x be an odd integer. In this case and and so we have found an example where but and thus disproving the statement. Now this is a contradiction since the left hand side is odd, but the right side is even. Proof by Contradiction. 12, and if we can go by all the previous chapters, this will be our template for the exercises. This shows the negation is false, and hence that the original proposition is true. If f(2) = 8, explain why f(3) > 6. Trans and non-binary men belong. A First Example: Proof by Contradiction Proposition: There are no natural number solutions to the equation x2 y2 = 1. Therefore, the reasoning of the ontological argument dodges the parody, its reasoning is not parallel to the parody argument, and it cannot be used to prove the existence of a lost island. Second, we provide some examples of inductive proofs that follow the structure outlined in the rst part. In an indirect geometric proof, you assume the opposite of what needs to be proven is true. Discrete Mathematics This is a basic course for undergraduate students. Obtain an. Alternatively, you can do a proof by contradiction: As-sume that Y is false, and show that X is false. Proof by contradiction (also known as indirect proof or the method of reductio ad absurdum) is a technique which can be used to prove any kind of statement. The argument is valid so the conclusion must be true if the premises are true. That would mean that there are two even numbers out there in the world somewhere that'll give us an odd number when we add them. (This is an indirect form of proof. The number 2 is a prime number. A Famous Contradiction Example. A useful resource to help deliver this new topic - fully worked solutions are included for | {
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Example. A useful resource to help deliver this new topic - fully worked solutions are included for all examples and questions in the exercise. (Contrapositive) Let integer n be given. As a first example of proof by contradiction, consider the following theorem: Theorem 1. Example2 1. Proof by Contradiction: (AKA reductio ad absurdum). This page has a few examples worked out completely - not too long or involved, and (I hope) not too difficult to follow. Mathematical Proof/Methods of Proof/Proof by Induction. That is how Mathematical Induction works. The establishment of a fact by the use of evidence. ExampleProve by contradiction that there is no greatest integer. Section 4-7 : The Mean Value Theorem. In a proof by contradiction, or indirect proof, you show that if a proposition were false, then some false fact would be true. (i)Direct proof: we assume A is true. Equivalently, we could just prove the logical negation of the given statement, which is the statement 9x8y: y 2 x. For every even integer n, N ≥ n. •Proof : Assume that the statement is false. ]! Then, by definition of rational, r = a/b and s = c/d for some integers a, b, c, and d with b ≠ 0 and d ≠ 0. Example of a constructive proof: Suppose we are to prove 9n2N;nis equal to the sum of its proper divisors: Proof: Let n= 6. Wyke, or Mr. Solution LetP(n) bethemathematicalstatement 11n −6 isdivisibleby5. GS results in a stable matching (i. Proof: Form the contrapositive of the given statement. 8, 1113, Sofia, Bulgaria Communicated by Walter Feit Received November 10, 1997. Thursday 2/16/17. The Second Edition features new chapters on nested quantifiers and proof by cases, and the number of exercises has been doubled with answers to odd-numbered exercises provided. Prove the following statement by contradiction: The sum of two even numbers is always even. Then there exists unmatched college c and unmatched student s. Rajoub: "For the first time in the history of the conflict – the contradiction between [the | {
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s. Rajoub: "For the first time in the history of the conflict – the contradiction between [the occupation and] the interests of the world, international law, and its values have reached a peak, and we must not undermine this clash through the wrong actions – whether in word or in actions that deviate from the consensus. a method of disproving a. Give a proof by contradiction: prove that the square root of 2 is irrational. It contrasts what Bastiat considered as the proper function of the Law and the perversion of the Law. Such examples are called counter examples. (I will not read any work on this question sheet). It is usually not as neat as a two-column proof but is far easier to organize. Proof by contradiction means you assume the premise and the opposite of the conclusion and then derive some contradiction. to an equation then there is another integral solution that is smaller in some way. Prove that the sum of irrational and rational number is irrational using proof by contradiction. If x ∈ A ∩ B, then x ∈ A and x ∈ B by definition, so in particular x ∈ A. Proof by Contradiction. O is the centre of the circle and ∠AOB = 100. 1 √2 is an irrational number. (non-constructive proof) •Show that a player in a game has a winning strategy without actually sayingwhat it is! •Famous proof: There exist irrational x, y such that xyis rational Villanova CSC 1300 -Dr Papalaskari. You can put this solution on YOUR website!. Below are several more examples of this proof strategy. This is the way most people learn a new language | learn to say a. Proof: (direct proof) Assume that n is an even integer. An example is "Prove that the product of two nonzero real numbers is nonzero. So this is a valuable technique which you should use sparingly. Example of a Proof by Contradiction Theorem 4. Then, one of the vectors of the standard basis of cannot be written as a linear combination of the vectors of. Solution: Suppose √2 is rational. Shows how and when to use each technique such as | {
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the vectors of. Solution: Suppose √2 is rational. Shows how and when to use each technique such as the contrapositive, induction and proof by contradiction. Recall that for two integers x and y, we say x divides y if there exists an integer z such that xz = y. Wrtiten response: Well done. If all our steps were correct and the result is false, our initial assumption must have been wrong. Imagine, then, the thrill of being able to prove something in mathematics. We give the proof by contradiction. To prove a theorem of the form A IF AND ONLY IF B , you first prove IF A THEN B , then you prove IF B THEN A , and that's enough to complete the proof. The concept of proof by contradiction is to assume that P is false. By the closure property, we know b is an integer, so we see that 3jn2. If you can do that, you have used mathematical induction to prove that the property P is true for any element, and therefore every element, in the infinite set. Also see the Mathematical Association of America Math DL review (of the 1st edition) and the Amazon reviews. Mathematical Proofs is designed to prepare students for the more abstract mathematics courses that follow calculus. In the second group the proofs will be selected mainly for their charm. Since and are distinct points in the Hausdorff space , there must be disjoint, open neighborhoods such that and. This A Level Maths video takes you through a new method of proof called proof by contradiction. Proof by contradiction: Assume P(x) is true but Q(x) is false. Prove that if a square matrix A is a zero divisor (that is AB=0 for some non-zero matrix B) then det(A)=0. Selected Homework Solutions - Math 574, Frank Thorne 1. Use rules of inference, axioms, and logical equivalences to show that q must also be true. Example 3: Prove the following statement by contraposition: For all integers n, if n 2 is odd, then n is odd. Then (x y) = (x + y) = 1. But this is clearly impossible, since n2 is even. Prove: do not bisect each other. Proof | {
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+ y) = 1. But this is clearly impossible, since n2 is even. Prove: do not bisect each other. Proof by Contradiction is another important proof technique. Instead of it, we use proof by contradiction. Example of a Proof by Contradiction Theorem 4. Justify all of your decisions as clearly as possible. Proving "If A, then B" by contradiction Given the assumptions in A, show that B must be true because it cannot possibly be false. As a first example of proof by contradiction, consider the following theorem:. Then there exists integers. A proof of a very general idea could be preceded by an example in a specific context. Direct Proof: Assume that p is true. (Proof by Contradiction. Run M on hPi. Robust, Semi-Intelligible Isabelle Proofs from ATP Proofs S. So, we will discuss these methods in this lesson extensively. 2 Selected Homework Solutions 10. Solution: By contradiction. Proof by Contradiction Example: Use a proof by contradiction to give a proof that √2 is irrational. Introduction 1 2. Squaring both sides we get 2 = n 2=m2, so m2 = 2n. The Pigeonhole Principle 1 Pigeonhole Principle: Simple form Theorem 1. Statement: If A, then B Inverse: If B, then A Converse: If not A, then not B Contrapositive: If not B, then not A Which of these are logically equivalent? RTP: If A, then B Method: Assume not B Carry out logical, deductive steps Reach the conclusion not A Example:. The statement P1 says that x1 = 1 < 4, which is true. Then we have to find a statement R so that ¬P→ (R∧ ¬R) - a contradiction. I can use. If ¬P leads to a contradiction, then. Proof by Contradiction is another important proof technique. A student should consider their solution of a proof-type problem to be aimed at an audience of students at their level; if they are unsure if it is a valid proof, then their goal has not been met. Example from the text: square root of 2 is irrational ; Careful: When using proof by contradiction, mistakes can lead to apparent contradictions. Since nm + 2n + 2m is odd, nm | {
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proof by contradiction, mistakes can lead to apparent contradictions. Since nm + 2n + 2m is odd, nm + 2n + 2m = 2k + 1 for some integer k. Similarly, since ris rational, we can express it as a/ bfor some integers aand. Finding a contradiction means that your assumption is false and therefore the statement is true. Final: Solutions ECS20 (Fall 2014) December 16, 2014 Part I: Proofs 1) Let a and b be two real numbers with a 0 and b 0. These notes explain these basic proof methods, as well as how to use definitions of new concepts in proofs. The proof is a sequence of mathematical statements, a path from some basic truth to the desired outcome. To write a two-column proof: Make a two-column form like this. Example: Prove that if you pick 22 days from the calendar, at least. Show that if n=k is true then n=k+1 is also true; How to Do it. (b) Prove that the square root of 3 is irrational. _____ Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1. 1 The number √3 is irrational. 3 Review the proof techniques on page 116−−118 Here is a result that is proved by three different proof techniques. Indirect Proof: Assume what you need to prove is false, and then […]. If (A [B) X and (X B) (X A), then A B. 2 Proof By Contradiction A proof is a sequence S 1;:::;S n of statements where every statement is either. Example of proof by contradiction. Eureka step and the eventual solution (Zeitz, 1999). Proof by mathematical induction. Some of the most famous examples of proofs by contradiction are: The proof that p 2 is irrational (probably dating back to Aristotle ca. Robust, Semi-Intelligible Isabelle Proofs from ATP Proofs S. To change the symbol printed at the end of a proof is straightforward. Reflex angle AOB is 260. In other words, if it is impossible for $$P$$ to be false, $$P$$ must be true. Proof (by contradiction): [We take the negation of the theorem and suppose it to be true. A proof by contradiction is often used to prove a conditional | {
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the theorem and suppose it to be true. A proof by contradiction is often used to prove a conditional statement $$P \to Q$$ when a direct proof has not been found and it is relatively easy to form the negation of the proposition. To show that there is no finite state automata that. ExampleProve by contradiction that there is no greatest integer. Example: Use a proof by contradiction to give a proof that √2 is irrational. By the Pumping Lemma this must be representable as , such that all are also in. Without loss of generality n=m is reduced, i. methods of proof and reasoning in a single document that might help new (and indeed continuing) students to gain a deeper understanding of how we write good proofs and present clear and logical mathematics. Therefore, the assumption that the quadratic equals zero is incorrect. Direct proof by contradiction. I Suppose pr 2 was rational. (Contrapositive) Let integer n be given. have no common factors (see Chapter 4). 3 Proof by contradiction (continued). We then try to obtain a contradiction from this assumption. Stack Exchange network consists of 177 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. But this is clearly impossible, since n2 is even. Therefore both n and m are odd. Discrete Mathematics This is a basic course for undergraduate students. Proof by Contradiction. To a resolution theorem-prover, both are two-step proofs. Proof: (Contrapositive: If n is even, then 3n + 2 is even) Suppose that the conclusion is false, i. Choose a long string w in L, jwj m. In the examples below we use this idea to prove the impossibility of certain kinds of solutions to some equations. Another way to write is using its equivalence, which is Example: Given A and B are sets satisfying. Then (2 is rational, so there are integers a and b for which (2= a b. (4 marks) 8 Use proof by contradiction to show. Properies of the modulus of the | {
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b for which (2= a b. (4 marks) 8 Use proof by contradiction to show. Properies of the modulus of the complex numbers. By Observation 3, it de nitely returns a matching, so suppose the matching is not perfect. As an example, here is a proof by contradiction of Proposition 4. This is a good resource if you are familiar with induction, and want to take things a little farther. Discrete Mathematics This is a basic course for undergraduate students. The idea behind proof by contradiction is that a statement must be true or false. Disprove by counterexample that for any , if , then. This is also known as Proof by Cases - see Example 1. In this example it all seems a bit long winded to prove something so obvious, but in more complicated examples it is useful to state exactly what we are assuming and where our contradiction is found. 2 # 2 p23 Follow the statement of your assumptions with a statement of what you will prove. The proof by deduction section also includes a few practice questions, with solutions in a separate file. Here are several examples of properties of the integers which can be proved using the well-ordering principle. You must include all three of these steps in your proofs! The three key pieces: 1. n By contradiction n Start with the “proof by example”! n So when asked to prove a claim, an example that n Automata theory & a historical perspective. Related Answers What object is defined using a directrix and a focus Find the coordinates of B if A has coordinates (3,5) and Y-2, 3) is the midpoint of AB Geometry and Algebra 1 Introduction Write a two-column proof. Let x be an odd integer. by triangle inequality. Example: Prove there's an infinite number of evens. ] [Prove Q =)P using direct, contrapositive, or contradiction proof. What that contradiction means in the proof; Whether the Halting problem is an unsolvable problem, an undecidable problem, or both; and why; Write a paragraph explaining the difference between an problem that can't be solved (such | {
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and why; Write a paragraph explaining the difference between an problem that can't be solved (such as the halting problem) and a problem that takes unreasonable time. For example: 2/8 can be written in lowest terms as 1/4 when 1 and 4 are positive integers with no common prime factors. Then … (( make logical conclusions until you come to two statements that contradict each other, such as "X is true" and X is false" ))But this is a contradiction because …. Proof: I Assume finitely many primes: p1;:::; k. 7 [A level only] Prove that there are an infinite number of primes. Math 2150 Homework 12 Solutions Throughout, use ONLY the assumptions given in the online notes and/or examples given in the online notes (which you need not reprove) unless speci- ed otherwise. If n+1 objects are put into n boxes, then at least one box contains two or more objects. Show that this supposition logically leads to a contradiction. These notes explain these basic proof methods, as well as how to use definitions of new concepts in proofs. This proves A ⊆ A ∩ B. and qwhere q≠ 0. Proof: Suppose not. – If we had 2 nickels, we could replace them with 1 dime. The reason is that the proof set-up involves assuming ∼∀x,P(x), which as we know from Section 2. Proof: By induction, on the number of billiard balls. We must derive a contradiction. Example of proof by contradiction and more on proof by induction. Solutions Educator Edition Save time lesson planning by exploring our library of educator reviews to over 550,000 open educational resources (OER). But, from the parity property, we know that an integer is not odd if, and only if, it is. PLEASE DO A \PREFOR-. Proof by Contradiction: (AKA reductio ad absurdum). 4, namely that for any integer n, if n2 is even then n is even. Robust, Semi-Intelligible Isabelle Proofs from ATP Proofs S. Here's an algebraic example: Prove: For a,b≥0, a+b 2 ≥ab. If ¬P leads to a contradiction, then. Say we're trying to prove by contradiction that if n 2 is an odd | {
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¬P leads to a contradiction, then. Say we're trying to prove by contradiction that if n 2 is an odd number, then n is also odd for all integers n. Second, we provide some examples of inductive proofs that follow the structure outlined in the rst part. Example using a Linear Function. Example: Give a direct proof of the theorem If nis an odd integer, then n2 is odd. Then n= 2k+ 1 for an integer k. In the proof above of Fact 1. Proof By Contradiction Examples And Solutions. • Example Every prime number a irrational numbers. You must include all three of these steps in your proofs! The three key pieces: 1. This proof is very similar to proof by contradiction, but subtly di er-ent. ] Suppose there is greatest even integer N. For every even integer n, N ≥ n. ¬Cube(a) 3. with √2 = a/b, where. This book is a guide to understanding and creating proofs. A proof by contrapositive uses that to prove the negation of the original assumption, while a proof by contradiction can negate any other true fact or lead to some other absurdity; in this case, you can't have two different smallest elements of a set. Proof by contradiction (also known as indirect proof or the method of reductio ad absurdum) is a common proof technique that is based on a very simple principle: something that leads to a contradiction can not be true, and if so, the opposite must be true. This is usually a hint that proof by contradiction is the method of choice. We have step-by-step solutions for your textbooks written by Bartleby experts! Using Proof by Contradiction In Exercises 15 − 26 , use proof by contradiction to prove the statement. This is why proof by contradiction works. uzbt8y6d536r qsp0l7cs9uxx0b rqtztc3l4kvhz ve65emynwfzt fb2nd4u3ken19 cswb17aj5fv byn1w24apk0exlx pquj9ksys3c0sn iss43ryp1ywds 28pf05o6s2 n9jk6v0ab1 i94sauttvyg 7x3i711c7mw11kf fh46ysf9w7ic qz8m9vxmhy5 zy2kssr5fx13g 3h76hkwgzwpo y6j0hsw3ln6 estzy7oybir85b bwdlnq0skk 5nne7cw7flh natj3e81te5cn kc2t3uryp80 zd3r6tbgj5ai dfps6gopdz | {
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# A game played on a rectangle
Suppose two players play the following game on a $m$ by $n$ rectangle. Alternatingly they have to make a cross in some empty $1\times 1$ square. They are not allowed to make a cross next to another cross (Diagonally is OK, not just right next to each other). The player who places the last cross wins.
Now the question is for which $m,n$ does the starting player have a winning strategy?
At first I thought this might be just a nice exercise. So I had a look at 1 by $n$ rectangles first. A computer programme computed that the first player does not have a winning strategy for $n=4,8,14,20,24,28,34,38,42,54,58,62,72,76,88,92,96,106,110$ (for $n\le 110$).
I do not see any pattern in these numbers. So the answer might not be so easy.
The starting player can always win for odd $n$. He just places a cross in the middle and mirrors all the moves of the second player.
• @tomglabst: The game would be over if each remaining square is adjacent to some cross. Then no player can make any cross. Thus in your example the game would end after the first move. Specifically "P1 crosses any remaining square". There are no squares left that any player could cross. – HenrikRueping Nov 22 '12 at 15:42
• I think I've seen a variant of the $m=1$ game where there's a modulus $M$ such that the grundy number of the game only depends on $n$ modulo $M$. So it's a possibility that such an $M$ exists here too. – mercio Nov 22 '12 at 16:12
• @AndréNicolas The OP has noted values of $m$, where the first player has no winning strategy, and 6 is not among them. I see no edits made either. – Mike Nov 22 '12 at 17:00
• The sequence is not in the OEIS (!!). – dot dot Nov 22 '12 at 17:12
• @HenrikRueping Ah okay, I misunderstood that. – tomglabst Nov 22 '12 at 17:27 | {
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The $n \times 1$ version is Dawson's Chess. The OP's sequence is A215721 in the OEIS, after adding 1 to each term. I wrote a program too, and found that the proportion of losing initial positions seems to tend to a constant -- there are 1473 losing positions in the first 10000, and 14709 losing positions in the first 100000. I found an explanation at "Sprague-Grundy values for Dawson's Chess" (A002187 in the OEIS): | {
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Find the singular points of the differential equation $x^3(x - 1)y'' - 2(x - 1)y' + 3xy = 0$.
Consider the second order linear homogeneous equation $$a_0(x)y'' + a_1(x)y'+ a_2(x)y = 0, x \in I \tag{1}$$ Suppose that $$a_0$$, $$a_1$$ and $$a_2$$ are analytic at $$x_0 \in I$$. If $$a_0(x_0) = 0$$, then $$x_0$$ is a singular point for $$(1)$$. Definition: A point $$x_0 \in I$$ is a regular singular point for $$(1)$$ if $$(1)$$ can be written as $$b_0(x)(x − x_0)^2y''+ b_1(x)(x − x_0)y'+b_2(x)y = 0, \tag{2}$$ where $$b_0(x_0) \neq 0$$ and $$b_0$$, $$b_1$$, $$b_2$$ are analytic at $$x_0$$.
The question is: Find the singular points of the differential equation $$x^3(x - 1)y'' - 2(x - 1)y' + 3xy = 0$$ and state whether they are regular singular points or irregular singular points.
I think, $$x = 0$$, irregular singular point, $$x = 1$$, regular singular point. But, How can I prove this? Please proper guide me.
• Hi Harry, I've formatted your question with MathJax. I really encourage you to learn how to do this for yourself. I'd also like to remind you of your comment under your last question. – Theo Bendit Sep 23 '19 at 5:18
• @ Theo. Thanks again.I am trying latex but not done properly. I am on my way to working with MathJax. – user679406 Sep 23 '19 at 5:29
• That's good! Feel free to edit your question to see how I've formatted it. That might help you figure it out a bit faster. – Theo Bendit Sep 23 '19 at 5:31
Consider the general homogeneous second order linear differential equation $$u''+P(x)u'+Q(x)u=0$$ where $$z \in D \subseteq \mathbb{C}$$.
The point $$x_0 \in D$$ is said to be an ordinary point of the above the given differential equation if $$P(x)$$ and $$Q(x)$$ are analytic at $$x_0$$.
If either $$P(x)$$ or $$Q(x)$$ fails to be analytic at $$x_0$$, the point $$x_0$$ is called a singular point of the given differential equation. | {
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A singular point $$x_0$$ of the given differential equation is said to be regular singular point if the function $$(x-x_0)P(x)$$ and $$(x-x_0)^2 Q(x)$$ are analytic at $$x_0$$ and irregular otherwise.
$${}$$
Here the given equation is $$x^3(x - 1)y'' - 2(x - 1)y' + 3xy = 0$$ $$\implies y''-\dfrac{2}{x^3}y'+\dfrac{3}{x^2(x - 1)}y=0$$ Here $$~P(x)=-\dfrac{2}{x^3}~$$and $$~Q(x)=\dfrac{3}{x^2(x - 1)}~$$.
Clearly $$~x=0,~1~$$ are singular points.
Again for the singular point $$~x=0~$$, $$(x-0)P(x)=-\dfrac{2}{x^2}\qquad \text{and}\qquad (x-0)^2P(x)=-\dfrac{2}{x}$$ Clearly both $$~(x-x_0)P(x)~$$ and $$~(x-x_0)^2 Q(x)~$$ are not analytic at $$~x_0=0~$$. So $$~x=0~$$ is irregular singular point.
For the singular point $$~x=1~$$, $$(x-1)P(x)=\dfrac{3}{x^2}\qquad \text{and}\qquad (x-0)^2P(x)=\dfrac{3(x - 1)}{x^2}$$ Clearly both $$~(x-x_0)P(x)~$$ and $$~(x-x_0)^2 Q(x)~$$ are analytic at $$~x_0=1~$$. So $$~x=1~$$ is regular singular point.
• @ nmsanta,I think this answer is correct and I am going with it. Am I right? – user679406 Sep 23 '19 at 5:32
• Have you faced any trouble in understanding my work ? I just doing things according to the definition. @Harry Richie – nmasanta Sep 23 '19 at 5:34
• No trouble. Clear like water..@ nmasanta – user679406 Sep 23 '19 at 5:37
• Only then my work will be well worth. – nmasanta Sep 23 '19 at 5:42
• Hello Mr. Down-voter, Would you like to explain the reason to give the down-vote in this answer ? – nmasanta Sep 24 '19 at 8:17 | {
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# Finding a splitting field of $x^3 + x +1$ over $\mathbb{Z}_2$
Finding a splitting field of $x^3 + x +1$ over $\mathbb{Z}_2$.
Ok so originally I messed around with $x^3 + x +1$ for a bit looking for an easy way to factor it and eventually decided that the factors are probably made up of really messy nested roots. So then I tried looking at the quotient field $\mathbb{Z}_2[x]/x^3 + x + 1$ to see if I would get lucky and it would contain all three roots but it doesn't. Is there a clever way to easily find this splitting field besides using the cubic formula to find the roots and then just directly adjoining them to $\mathbb{Z}_2$?
Edit: Ok it turns out I miscalculated in my quotient field, $\mathbb{Z}_2[x]/x^3 + x + 1$ does contain all three roots.
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If degree is not $3$ it is $6$. – André Nicolas Jul 3 '12 at 2:08
Actually, your quotient $\mathbb{Z}_2[x]/(x^3+x+1)$ does contain at least one root: $x$. This is because $x^3+x+1\equiv 0\pmod{x^3+x+1}$. Can you find the others? – roninpro Jul 3 '12 at 2:08
The splitting field is either degree $3$ or degree $6$ over $\mathbb{Z}_2$, hence it is either $\mathbb{F}_8$ or $\mathbb{F}_{64}$.
Let $\alpha$ be a root, so that $\mathbb{F}_8 = \mathbb{F}(\alpha)$. The elements are of the form $a+b\alpha+c\alpha^2$, with $\alpha^3=\alpha+1$.
Now, the question is whether any of these elements besides $\alpha$ is a root of the original polynomial $x^3+x+1$.
Note that $(\alpha^2)^3 = (\alpha^3)^2 = (\alpha+1)^2 = \alpha^2+1$, and so if we plug in $\alpha^2$ into the polynomial we have $$\alpha^6 + \alpha^2 + 1 = \alpha^2+1+\alpha^2+1= 0.$$ Thus, $\alpha^2$ is also a root. So the polynomial has at least two roots in $\mathbb{F}_8$, and so splits there.
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oh shoot I must have miscaculated in my quotient field, ok cool, thanks! – cuckmaster5000 Jul 3 '12 at 2:11 | {
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If your cubic were to factor, the factorization must have at least one linear term, which corresponds to a root. It is easy to check it has no roots in $\mathbb{Z}_2$ - check them both! Plugging in $0$ and $1$ both give $1$ so are not roots, so your polynomial is irreducible over $\mathbb{Z}_2.$
A way to write the splitting field is $\mathbb{Z}_2(\alpha)$ where $\alpha$ is any one of the roots of $x^3+x+1.$ This is because $\alpha^2$ and $\alpha^4= \alpha^2+\alpha$ must then also be distinct roots of $x^3+x+1$, and these 3 then comprise a full list. Note, this isn't a special trick to notice for this problem. In fields of characteristic $p$, if $\beta$ is a root of a polynomial then $\beta^p$ will automatically also be a root, due to properties of the Frobenius endomorphism.
Or as roninpro pointed out in the comments, the quotient ring you considered (which is essentially the same thing as adjoining these roots) does contain at least one root, and by this same trick, all the roots. | {
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I thought forming the quotient field adjoined just a single root? And that sometimes you would get lucky and the remaining roots could be formed within the quotient field as well, but that sometimes they couldn't, is this not correct? – cuckmaster5000 Jul 3 '12 at 2:17
@NollieTré You're correct. But extensions of finite fields are all Galois and in particular normal. So if $K/k$ are finite fields and $f \in k[X]$ is irreducible and has a root in $K$, then it splits in $K$. Frobenius is a great thing! – Dylan Moreland Jul 3 '12 at 2:25
@NollieTré You are correct, sometimes the extra roots we find with the Frobenius trick just correspond to the same roots. But here they are distinct. You can check yourself (by carrying out the division) that here $x^2$ is also a root, since $x^6+x^2+1 = (x^3+x+1)(x^3-x-1)+2(x^2+x+1) = 0$, and similarly $x^4$ is also a root. So the quotient formed actually has all the roots. – Ragib Zaman Jul 3 '12 at 2:26
Ok interesting, so is there a canonical example of when the quotient field does not contain all the roots of the polynomial? – cuckmaster5000 Jul 3 '12 at 2:29
I think it's easier to note that if $\alpha$ is a root of the polynomial then squaring gives $0^2 = (\alpha^3 + \alpha + 1)^2 = (\alpha^2)^3 + \alpha^2 + 1$. – Dylan Moreland Jul 3 '12 at 2:30
Two good answers already here, but I wanted to emphasize the usefulness of the Frobenius endomorphism. Let $f(X) = X^3 + X + 1$. If I let $\alpha$ denote the image of $X$ in $k = \mathbb F_2[X]/(f(X))$ then applying Frobenius to $0 = f(\alpha)$ gives $0 = (\alpha^3 + \alpha + 1)^2 = (\alpha^2)^3 + \alpha^2 + 1.$ Hence $\alpha^2$ is also a root of $f$, and $\alpha^2 \neq \alpha$ because $\alpha \neq 0, 1$. Since $k$ contains two roots of the cubic $f$, it contains three. | {
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To bring in slightly more technology: extensions of finite fields are Galois and in particular normal, and this implies that if you adjoin one root of an irreducible polynomial over a finite field then you've actually adjoined all of them. This is, of course, not true in general!
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Just wanted to add here, in $\Bbb{Z}_2(\alpha)[x]$, one can factor $x^3 + x + 1 = (x + \alpha)(x^2 + \alpha x + (\alpha^2 + 1))$. Given that $\alpha^2$ is also a root (as Dylan shows above), using ordinary high-school factorization techniques the third root is easily seen to be $\alpha^2 + \alpha$ (a somewhat preferable form than $\alpha^4$). That is: $x^2 + \alpha x + (\alpha^2 + 1) = (x + \alpha^2)(x + (\alpha^2 + \alpha))$ – David Wheeler Jul 6 '12 at 10:20 | {
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# Showing that a group $G$ with two distinct, normal subgroups of index $p$ has order $p^2$
A friend of mine and I were working together to solve this exercise:
Let $G$ be a group, and $p$ a prime number. Let $H, K$ be normal subgroups of $G$. Suppose that $|G:H| = |G:K| = p$, and that $H\cap K = \{1\}$. Show that $G\simeq\mathbb{Z}_p\times\mathbb{Z}_p$.
We thought that the quickest way to solve it was showing that $G$ has exactly $p^2$ elements.
For that purpose, we first tried to prove that $|G|\leq p^2$. We chose $a,b\in xH\cap yK$, that are two elements of $G$ belonging to the intersection of a coset of $H$ with a coset of $K$. This means that $a = xh$ and $b=xh'$ for some $h,h'\in H$ and for some $x\in G$ (because $a,b\in xH$). Similarly, $a=yk$ and $b=yk'$, for some $k,k'\in K$ and for some $y\in G$ (because $a,b\in yK$).
Then, $$xh=yk\implies hk^{-1}=x^{-1}y$$ and $$xh'=yk'\implies h'k'^{-1}=x^{-1}y.$$
This brings us to $hk^{-1}=h'k'^{-1}$, which we can rewrite as $hh'^{-1}=k'^{-1}k$. This element belongs to $H\cap K=\{1\}$, thus $h = h'$ and $k = k'$: hence $a = b$. This proves that the intersection of a coset of $H$ with a coset of $K$ cannot contain more than one distinct element.
We made use of this information when we noted that, regardless of how many elements $G$ has, there exist at most $p^2$ intersections of cosets of $H$ with cosets of $K$, and each one of them contains at most one element. Thus, we deduced that $|G|\leq p^2$.
Now, here are my questions: | {
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Now, here are my questions:
• How do I prove that $|G|\ge p^2$, in order to conclude that $|G|=p^2$ (and thus that $G\simeq\mathbb{Z}_p\times\mathbb{Z}_p$)?
• Is the proof rigorous enough so far, or were some of our assumptions incorrect?
• This exercise is related to a part of our algebra course in which nothing beyond direct product of groups had been explained yet, so we could not make use of the Sylow theorems, of anything about group actions and so on. Is there a simpler way to solve this exercise, using only "basic" tools of group theory (e.g. quotient groups, the homomorphism and isomorphism theorems and anything regarding the direct product of groups)?
• Is $G$ assumed to be finite at the outset? – Tim Raczkowski Jan 7 '16 at 20:46
• @TimRaczkowski The exercise said nothing on that point, so I assume that $G$ may also be infinite. Anyway, given that $G$ has to be isomorphic to $\mathbb{Z}_p\times\mathbb{Z}_p$, I guess it can't be infinite. Plus, if our half of the proof is correct, $|G|\leq p^2$ definitely implies that $G$ is finite. – Labba Jan 7 '16 at 20:48 | {
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You can use the following fact: If $H,K$ are subgroups of $G$, $H\cap K=\{1\}$ and $hk=hk$ for all $h\in H$ and $k\in K$, then $G\cong H\times K$.
Now, $hk\in hK,Hk$. Since $H$ is normal in $G$, $Hk=kH$. So, by your argument above, $\{hk\}=hK\cap kH$. But $kh\in hK\cap kH$ by a similar argument.
• Thanks for your answer :) So, if I got it right, you say that $hk$ belongs to the $hK$ and $Hk$ cosets (the latter being $kH$ anyway, because of the normality of $H$), and the same can be said about $kh$ that belongs both to the $kH$ and $Kh$ cosets (again, the $hK$ coset). In short, $hk, kh\in kH\cap hK$ and, by my part of the proof, $kh = hk$ because only one element can be found in this intersection. Hence, $G\simeq H \times K$ because $H\cap K = \{1\}$ and they commute with each other. Did I get everything? And this also proves that $|G| = p^2$, right? – Labba Jan 7 '16 at 21:15
• I really have to thank you :) Anyway, there's still something I fear I'm missing: even if this shows that $G\simeq H\times K$, does this prove that $|H| = |K| = p$ as well? – Labba Jan 7 '16 at 21:18
• Well if $|G|=p^2$, and $[G:H]=[G:K]=p$, so $|H|=|K|=p$. – Tim Raczkowski Jan 7 '16 at 21:21
• Hmm....I seem to keep missing a detail each time. $[G:H]=[G:K]\implies |H|=|K|$. Suppose $|HK|=n^2$. Now, $n^2|p^2\implies$n=1, or $n=p$. If $n=1$, then $H=N=\{1\}$ and $|G|=p$. So assuming $H\ne K$, we get the result. – Tim Raczkowski Jan 7 '16 at 21:38
Since $H,K$ normal in $G$ and $H\cap K=\{1\}$, $HK\cong H\times K$. $|G:HK|\cdot|HK:H|=|G:H|=p$. So $|HK:H|$ divides $p$. $|HK:H|=1$, or $p$. Suppose it is $1$. $HK=H$, $K\subseteq H$. $|G:H|\cdot|H:K|=|G:K|$, $p|H:K|=p$, so $H=K$, contradiction. So $|HK:H|=p$, $|G:HK|=1$. $G=HK$. Now, $|G:H|=|HK:H|=|K:H\cap K|$ by second isomorphism theorem. So $p=|K|$. Similarly, $|H|=p$. $G\cong \mathbb{Z}/p\mathbb{Z}\times\mathbb{Z}/p\mathbb{Z}$. | {
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# Suppose we roll a fair $6$ sided die repeatedly. Find the expected number of rolls required to see $3$ of the same number in succession.
Suppose we roll a fair six sided die repeatedly.
Find the expected number of rolls required to see $$3$$ of the same number in succession
From the link below, I learned that $$258$$ rolls are expected to see 3 sixes appear in succession. So I'm thinking that for a same (any) number, the rolls expected would be $$258/6 = 43$$. But I'm unsure how to show this and whether it really is correct.
How many times to roll a die before getting two consecutive sixes?
For $n\in \{0,1,2\}$ Let $E[n]$ denote the answer given that you are starting from a streak of $n$ consecutive rolls. The answer you want is $E=E[0]$, though you are never in state $0$ except at the start.
We note $$E[2]=\frac 16\times 1+\frac 56\times \left(E[1]+1\right)$$ $$E[1]=\frac 16\times \left(E[2]+1\right)+\frac 56\times \left(E[1]+1\right)$$
$$E=E[0]=E[1]+1$$
this system is easily solved and, barring error (always possible), yields $$\boxed {E=43}$$
• This is very elegant (+1) - great answer!! Aug 31, 2017 at 16:38
• One issue that isn't really an issue: everything needs to be finite for this to work (cf optional stopping theorem)
– cats
Aug 31, 2017 at 18:32
• This is the (arch classical) approach used in the post mentioned in Byron's answer.
– Did
Aug 31, 2017 at 18:40
• @lulu can you explain me what what tis the theory behind this answer? Does it refer to a specific branch of stochastics? Mar 9, 2021 at 23:30
• @Phillipp This is a fairly standard Markov type computation. The trick, such as it is, is to note that there really are only three possible states for the process, an observation which makes the system very tractable.
– lulu
Mar 9, 2021 at 23:39 | {
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From Did's answer here, the probability generating function $u_0(s)=\mathbb{E}(s^T)$ for the number of trials $T$ needed to get three consecutive values the same is
$$u_0(s)={s^3\over 36-30s-5s^2}.$$ Differentiating this and setting $s=1$ in the derivative shows that $\mathbb{E}(T)=43.$
We can treat this as a three-state absorbing markov chain: a length3 run has been seen, otherwise the current run is length2, current run is length 1. Transition matrix: $\begin{bmatrix}1&\frac{1}{6}&0\\0&0&\frac{1}{6}\\0&\frac{5}{6}&\frac{5}{6}\end{bmatrix}$
This is in standard form $\left[\begin{array}{c|c}I&S\\\hline0&R\end{array}\right]$
We turn our attention to the fundamental matrix $(I-R)^{-1} = \begin{bmatrix}1&-\frac{1}{6}\\-\frac{5}{6}&\frac{1}{6}\end{bmatrix}^{-1}=\begin{bmatrix}6&6\\30&36\end{bmatrix}$
After the first roll, we enter the markov chain in the third state. Adding the entries of the fundamental matrix corresponding to that column tells us the expected time until we reach an absorbing state, i.e. until we have a chain of three consecutive rolls of the same number.
Thus the expected number of rolls needed is $1+36+6=43$
Let $\mu$ denote the expectation of the expected number of rolls that are still needed if $2$ rolls have passed with distinct result.
Let $\nu$ denote the expectation of the expected number of rolls that are still needed if $2$ rolls have passed with equal result.
Then the expectation is: $$2+\frac56\mu+\frac16\nu$$
Here $\frac56$ is the probability that the first two numbers are distinct and $\frac16$ is the probability that they are equal.
Secondly we have the relations:
$$\mu=\frac56(1+\mu)+\frac16(1+\nu)=1+\frac56\mu+\frac16\nu\tag1$$ and: $$\nu=\frac161+\frac56(1+\mu)=1+\frac56\mu\tag2$$
The relations $(1)$ and $(2)$ lead easily to: $\mu=42$ and $\nu=36$.
Then $$2+\frac56\mu+\frac16\nu=43$$ is the final answer.
(in the answers uptil now it was not used that you allready learned something). | {
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(in the answers uptil now it was not used that you allready learned something).
Let $Y$ denote the number of rolls required to see three dice of the same number in succession and let $X$ denote the number of rolls required to see three dice with number $6$ in succession.
Then: $$Y\text{ and }\frac16X+\frac56(X+Y)=X+\frac56Y\text{ must have equal distribution.}\tag3$$
Here $\frac16$ is the probability of the event that the first time that three dice give the same number in succession they show number $6$ and $\frac56$ is the probability that do not show a number $6$.
$(3)$ rests on the observation that - if for the first time three equal numbers show up in succession - we are ready if $6$ happens to be that number and must actually start over again (with $X$ throws in our pocket) if not.
So we find $\mathbb EY=\mathbb EX+\frac56\mathbb EY$ or equivalently:$$\mathbb EX=\frac16\mathbb EY$$
You allready learned that $\mathbb EY=258$ and making use of that knowledge you find $$\mathbb EX=258/6=43$$ This confirms your thinking. | {
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# Venn Diagrams and Sets
## Chapter 8 - Karnaugh Mapping | {
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Mathematicians use Venn diagrams to show the logical relationships of sets (collections of objects) to one another. Perhaps you have already seen Venn diagrams in your algebra or other mathematics studies. If you have, you may remember overlapping circles and the union and intersection of sets. We will review the overlapping circles of the Venn diagram. We will adopt the terms OR and AND instead of union and intersection since that is the terminology used in digital electronics. The Venn diagram bridges the Boolean algebra from a previous chapter to the Karnaugh Map. We will relate what you already know about Boolean algebra to Venn diagrams, then transition to Karnaugh maps. A set is a collection of objects out of a universe as shown below. The members of the set are the objects contained within the set. The members of the set usually have something in common; though, this is not a requirement. Out of the universe of real numbers, for example, the set of all positive integers {1,2,3…} is a set. The set {3,4,5} is an example of a smaller set, or subset of the set of all positive integers. Another example is the set of all males out of the universe of college students. Can you think of some more examples of sets? Above left, we have a Venn diagram showing the set A in the circle within the universe U, the rectangular area. If everything inside the circle is A, then anything outside of the circle is not A. Thus, above center, we label the rectangular area outside of the circle A as A-not instead of U. We show B and B-not in a similar manner. What happens if both A and B are contained within the same universe? We show four possibilities. Let’s take a closer look at each of the the four possibilities as shown above. The first example shows that set A and set B have nothing in common according to the Venn diagram. There is no overlap between the A and B circular hatched regions. For example, suppose that sets A and B contain the following members: | {
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set A = {1,2,3,4}
set B = {5,6,7,8}
None of the members of set A are contained within set B, nor are any of the members of B contained within A. Thus, there is no overlap of the circles. In the second example in the above Venn diagram, Set A is totally contained within set B How can we explain this situation? Suppose that sets A and B contain the following members:
set A = {1,2}
set B = {1,2,3,4,5,6,7,8}
All members of set A are also members of set B. Therefore, set A is a subset of Set B. Since all members of set A are members of set B, set A is drawn fully within the boundary of set B. There is a fifth case, not shown, with the four examples. Hint: it is similar to the last (fourth) example. Draw a Venn diagram for this fifth case. The third example above shows perfect overlap between set A and set B. It looks like both sets contain the same identical members. Suppose that sets A and B contain the following:
set A = {1,2,3,4}
set B = {1,2,3,4}
Therefore,
Set A = Set B
Sets And B are identically equal because they both have the same identical members. The A and B regions within the corresponding Venn diagram above overlap completely. If there is any doubt about what the above patterns represent, refer to any figure above or below to be sure of what the circular regions looked like before they were overlapped. The fourth example above shows that there is something in common between set A and set B in the overlapping region. For example, we arbitrarily select the following sets to illustrate our point:
set A = {1,2,3,4}
set B = {3,4,5,6}
Set A and Set B both have the elements 3 and 4 in common These elements are the reason for the overlap in the center common to A and B. We need to take a closer look at this situation.
• Share
Published under the terms and conditions of the Design Science License | {
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### CSIR JUNE 2011 PART C QUESTION 73 SOLUTION (which are positive definite? 1) $A+B$, 2) $ABA^{*}$, 3) $A^2+I$, 4) $AB$.) | {
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Suppose $A$ and $B$ are $n \times n$ positive definite matrices and $I$ be the $n \times n$ identity matrix. Then which of the following matrices are positive definite?
1) $A+B$,
2) $ABA^{*}$,
3) $A^2+I$,
4) $AB$.
Solution:
Let $A$ be a real symmetric matrix, then $A$ is said to positive definite if it satisfies any of the following equivalent conditions.
i) all its eigenvalues are positive.
ii) $x^t A x > 0$ for all vectors $x \ne 0$.
iii)$<x,Ax> > 0$ for all vectors $x \ne 0$.
iii) A is positive definite if and only if it can be written as $A = R^tR$, where $R$ is possibly a rectangular matrix, with independent columns.
iv) All the principal minors of $A$ are positive. (Please share if you know any other equivalent conditions in the comment below)
Let $A$ and $B$ be two $n \times n$ positive definite matrices. We have $x^t A x>0$ and $x^t B x>0$ for $x \ne 0$. We will solve each given option by each of the above given definition of positive definiteness in order to understand them clearly.
option 1. (True)We have for $x \ne 0$, $$x^t (A+B) x = x^t A x + x^t B x > 0.$$
Therefore option 1 is true.
option 2. (True) We have for $x \ne 0$, $Ax \ne 0$ since $A$ is invertible.
$$<x,ABA^*x> = <A^*x,BA^*x> = <B^*A^*x,A^*x> \\ = <BAx,Ax> > 0 .$$
Therefore option 2 is true. Since $A$ and $B$ are real symmetric, we have $A^* = A$ and $B^* = B$.
option 3. (True) Let the eigen values of $A$ be $\lambda_1,\lambda_2,\dots,\lambda_n$. Since $A$ is positive definite we have all these eigen values are positive. We observe that the eigen values of $A^2+1$ are $\lambda_1^2+1,\lambda_2^2+1,\dots,\lambda_n^2+1$ which are all positive. Hence $A^2+1$ is positive definite.
option 4. (False) | {
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option 4. (False)
The product of two positive definite matrices need not be even symmetric. In particular, we have $AB$ is symmetric if and only if $A$ and $B$ commutes with each other. Because $$(AB)^* = B^*A^* = A^*B^* = AB.$$ Note that, if $A$ and $B$ commutes with each other, then $A^*$ and $B^*$ commutes with each other. | {
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To illustrate this, consider the positive definite matrices $$A = \begin{bmatrix}11 & 10 \\ 10 & 10\end{bmatrix}$$ and $$B = \begin{bmatrix}11 & 5 \\ 5 & 10\end{bmatrix}$$
text{ Then their product } $$AB = \begin{bmatrix}171&155 \\ 160&150\end{bmatrix}$$ which is not symmetric.
### NBHM 2020 PART A Question 4 Solution $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$
Evaluate : $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$ Solution : \int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx = \int_{-\infty}^{\inft... | {
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# Repeatedly taking differences on a polynomial yields the factorial of its degree?
Consider a function such that it takes in polynomial function and creates an array of its outputs and then using that array creates another new array by calculating the absolute difference between the first $2$ values and keeps doing this until it reaches an array full of zeros.
This is much easier to show you by example.
For example take $F(x)= x^2$, the first array would be
$1,4,9,16,25,36,49,64,81$ and so on, the second would be
$3,5,7,9,11,13,15,17,19$ ( the difference between the first value and the second one)
but the third one is where it gets interesting as if we continue the pattern we would get an array filled with only $2$'s and after that it would only be zeros.
Lets do another example, $F(x)=x^3$
$1,8,27,64,125,216,343$
$7,19,37,61,91,\dotsc$ but here is the interesting part if we continue this
$12,18,24,30,\dotsc$ and once more then we get
$6,6,6,6,6,\dotsc$ after that it would just be an array of zeros
There are $2$ main observation that I made about this
Firstly, the value that is begin repeated indefinitely is equals to the factorial of the functions power. Meaning that for $F(x)=x^2$ the value being repeated is $2!$. For $F(x)=x^3$ , it's $3!$ and this is true for all polynomials (I tried it up to $x^7$, after that it got too messy)
Secondly, the value that is repeated always occurs on the $n$th iteration of the function. Meaning that for $F(x)=x^2$, we have to go through the processes $2$ times before we find the value. For $F(x)=x^3$, we have to go through it $3$ times before getting the value.
Is there any way to prove this and does this mean anything at all? | {
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Is there any way to prove this and does this mean anything at all?
• Lookup finite differences ("an $n^{th}$ power has a constant $n^{th}$ finite difference") . Also for example this answer. – dxiv Jun 12 '17 at 3:25
• Good. You discovered a concept (through on a sequence basis) very similar to derivative. An excellent mind of discovering and inducting. Continue it if you are still young and have enough time. – BAI Jun 12 '17 at 4:45
• Welcome to Math.SE! Since one aim of the site is to collect questions and answers in forms useful for posterity, would you please consider re-titling your question to indicate its content? (Perhaps something like "Repeated differences for polynomial functions at equally-spaced inputs"...?) – Andrew D. Hwang Jun 12 '17 at 11:18
• – Simply Beautiful Art Jun 12 '17 at 17:21
• @user21820 your edit to the question is a great example of what a good title can do for a better understanding of a given topic and to promote the curiosity of the reader. – iadvd Jun 13 '17 at 0:12
## 2 Answers
Here's a fact:
• If $p(x)$ is a polynomial of degree $n$ with leading term $ax^n$ then $p(x+1)-p(x)$ is a polyomial of degree $n-1$ with leading term $a \, n \, x^{n-1}$.
(I'll prove this fact below).
Applying this fact together with an induction argument, it follows that after repeating the process $n$ times, one obtains a polynomial of degree zero whose leading term is $$a \, n \, (n-1) \ldots (2) (1) = a \, n!$$ which is just a constant having that value.
So if the original leading coefficient $a$ is equal to $1$, as it is in the specific cases $F(x)=x^n$ that you ask about, repeating the difference process $n$ times yields a constant sequence of $n!$ as you ask.
Here's a proof of the fact by applying induction (the base case $n=1$ is easy).
Assuming the induction hypothesis for polynomials of degree $\le n-1$, suppose that $$p(x) = a \, x^n + q(x)$$ where $q(x)$ is a polynomial of degree $\le n-1$. | {
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We have $$p(x+1)-p(x) = a \, (x+1)^n - a \, x^n + \underbrace{(q(x+1)-q(x))}_{r(x)}$$ and $r(x)$ is a polynomial of degree $\le n-2$ by induction. Thus $$p(x+1)-p(x) = a \, (x^n + n \, x^{n-1} + s(x)) - a\, x^n + r(x)$$ where $s(x)$ is also a polynomial of degree $\le n-2$ (by application of the binomial theorem). Therefore $$p(x+1)-p(x) = a \, n \, x^{n-1} + (a \, s(x)+r(x))$$ which is a polynomial of degree $n-1$ with leading term as required.
What you have discovered/invented is known as the forward difference operator $D$ defined as: $\def\nn{\mathbb{N}} \def\zz{\mathbb{Z}} \def\lfrac#1#2{{\large\frac{#1}{#2}}} \def\lbinom#1#2{{\large\binom{#1}{#2}}}$
$D = ( \text{function$f$on$\zz$} \mapsto ( \text{int$n$} \mapsto f(n+1) - f(n) ) )$
Namely for any function $f$ on $\zz$ and $n \in \zz$, $D(f)(n) = f(n+1) - f(n)$.
If you think of the functions as sequences (infinite in both directions), then taking the forward difference means replacing each term with the value of the next term minus itself. What you did is essentially to repeatedly take the forward difference of the sequence of cubes:
...,-27,-8,-1, 0, 1, 8,27,...
..., 19, 7, 1, 1, 7,19,37,...
...,-12,-6, 0, 6,12,18,24,...
..., 6, 6, 6, 6, 6, 6, 6,...
..., 0, 0, 0, 0, 0, 0, 0,...
..., 0, 0, 0, 0, 0, 0, 0,...
This powerful abstraction makes it easy to get a lot of things. For instance, the numbers obtained here can be easily used to obtain the general formula for sum of cubes!
General method for indefinite summation
The key is that:
$D\left( \text{int$n$} \mapsto \lbinom{n}{k+1} \right) = \left( \text{int$n$} \mapsto \lbinom{n}{k} \right)$ for any $k \in \zz$.
This is to be expected because it follows directly from Pascal's triangle, especially if we define $\lbinom{n}{k}$ using the triangle.
This means that if we have any function $f$ on $\zz$ such that $f(n) = \sum_{k=0}^\infty a_k \lbinom{n}{k}$ for any $n \in \zz$, then we get the Newton series: | {
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$D(f)(n) = \sum_{k=0}^\infty a_{k+1} \lbinom{n}{k}$ for any $n \in \zz$.
From a high-level perspective, this is the discrete version of the Taylor series, and indeed for such a function we easily see that $f(n) = \sum_{k=0}^\infty D^k(f)(0) \lbinom{n}{k}$ for any $n \in \zz$, because $\binom{0}{0} = 1$ while $\lbinom{0}{k} = 0$ for any $k \in \nn^+$.
This works for any polynomial function $f$ on $\zz$, since $D^k(f)$ is the zero function once $k$ is larger than the degree of $f$, so we can use it to immediately find the series for $(\text{int n} \mapsto n^3)$, and then just take the anti-difference by shifting the coefficients of the series the other way. The undetermined constant that appears will drop out once we perform a definite sum like if we want the sum of the first $m$ cubes.
Note also that $D^k(f)$ is the constant function with value $k!$ if $f(n) = n^k$ for every $n$. Lee Mosher has already explained this particular fact by directly proving it, but another way to see it is that the highest order term in its Newton series is $k! \lbinom{n}{k}$, because $\lbinom{n}{k}$ is the only term that can contribute the $k$-th power of $n$. Since $D$ merely shifts the coefficients, $D^k(f) = \left( \text{int$n$} \mapsto k! \lbinom{n}{0} \right)$ and we are done.
Sum of $p$ powers
For example if we want $\sum_{k=1}^{n-1} k^3$ we first find the iterated forward differences of the sequence of cubes $( n^3 )_{n \in \zz}$:
..., 0, 1, 8,27,64,...
..., 1, 7,19,37,...
..., 6,12,18,...
..., 6, 6,...
..., 0,...
So we immediately get $n^3 = 0 \binom{n}{0} + 1 \binom{n}{1} + 6 \binom{n}{2} + 6 \binom{n}{3}$ and hence $\sum_{k=0}^{n-1} k^3 = 0 \binom{n}{1} + 1 \binom{n}{2} + 6 \binom{n}{3} + 6 \binom{n}{4} = \lfrac{n(n-1)}{2} \Big( 1 + \lfrac{6(n-2)}{3} \big( 1 + \lfrac{n-3}{4} \big) \Big) = \Big( \lfrac{n(n-1)}{2} \Big)^2$.
Computation efficiency | {
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Computation efficiency
This is far more efficient than the usual method (namely by taking summation on both sides of $(n+1)^3-n^3 = 3n^2+3n+1$ and telescoping) because the series using binomial coefficients is easy to manipulate and easy to compute. For sum of $p$-powers we only need $O(p^2)$ arithmetic operations to find the forward-differences and then $O(p^2)$ more to simplify the series form into a standard polynomial form. In contrast, the other method requires $O(p^3)$ arithmetic operations.
Indefinite summation of non-polynomials
Also, for a wide class of non-polynomial functions, we can still compute the indefinite sum without using the series, by using the discrete analogue to integration by parts, here called summation by parts.
To derive it, simply check that $D(f \times g)(n) = f(n+1) g(n+1) - f(n) g(n) = f(n+1) D(g)(n) - D(f)(n) g(n)$ and so we get the product rule:
$D(f \times g) = R(f) \times D(g) + D(f) \times g$
where $R$ is the right-shift operator defined as:
$R = ( \text{function$f$on$\zz$} \mapsto ( \text{int$n$} \mapsto f(n+1) ) )$
Namely for any function $f$ on $\zz$ and $n \in Z$, $R(f)(n) = f(n+1)$.
For convenience we also define the summation operator:
$S = ( \text{function$f$on$\zz$} \mapsto ( \text{int$n$} \mapsto \sum_{k=0}^{n-1} f(k) ) )$
Then we have the important property that $DS(f) = f$ for any function $f$ on $\zz$, analogous to the fundamental theorem of calculus.
Now by substituting $f$ with $S(f)$ into the product rule and taking summation on both sides we get summation by parts:
$S( f \times g ) = S(f) \times g - S( R(S(f)) \times D(g) ) + c$ for some constant function $c$ on $\zz$.
Example indefinite sum
Using this we can easily compute things like $\sum_{k=1}^n k^3 3^k$ by applying it three times, each time reducing the degree of the polynomial part. There are other ways to achieve this using differentiation, but this method is purely discrete. | {
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MATH 4389 (17)
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Final
# MATH 4389 Final: Real_Analysis(part_2) Premium
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School
Department
Mathematics
Course
MATH 4389
Professor
Almus
Semester
Spring | {
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Description | {
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PART II. SEQUENCES OF REAL NUMBERS II.1. CONVERGENCE Definition 1. A sequence is a real-valued function f whose domain is the set positive integers (N). The numbers f(1),f (2), ··· are called the terms of the sequence. Notation Function notation vs subscript notation: f(1) ≡ 1 ,f (2) ≡2s ,··· ,f (n) n s , ··· . In discussing sequences the subscript notation is much more common than functional notation. We’ll use subscript notation throughout our treatment of analysis. Specifying a sequence There are several ways to specify a sequence. 1. By giving the function. For example: 1 1 1 1 1 1 (a) sn= n or {sn} = n . This is the sequence {12 ,3 ,4 ,...,n,... }. (b) s = n − 1. This is the sequence {0, ,2 , 3,...,n − 1,... }. n n 2 3 4 n (c) sn=( −1) n . This is the sequence {−1,4,−9,16,..., (−1) n ,... }. 2. By giving the first few terms to establish a pattern, leaving it to you to find the function. This is risky – it might not be easy to recognize the pattern and/or you can be misled. (a) {sn} = {0,1,0,1,0,1,... }( The pattern here is obvious; can you devise the function? It’s 1 − (−1)n) 0, n odd sn= or s n= 2 1, n even 2 (b) {s } = 2, 5,10,17 ,6 ,... ,s = n +1 . n 2 3 4 5 n n (c) {sn} = {2,4,8,16,32,... }. What is6s ? What is the function? While you might say 64 n and s n2 , the function I have in mind gives s6= π/6: n π 64 sn=2 +( n − 1)(n − 2)(n − 3)(n − 4)(n − 5) − 720 120 3. By a recursion formula. For example: (a) s = 1 s ,s = 1. The first 5 terms are 1,1, , 1 , 1 ,.... Assuming that n+1 n +1 n 1 2 6 24 120 1 the pattern continues n = . n! 1 (b) sn+1 = (n +1) ,s 1 = 1. The first 5 terms are {1,1,1,1,1,... }. Assuming that the 2 pattern continues n = 1 for all n; {n } is a “constant” sequence. 13 Definition 2. A sequence {s }nconverges to the number s if to each > 0 there corresponds a positive integer N such that |s − s| for all n>N. n The number s is called the limit of the sequence. Notation “{s } converges to s” is denoted by n n→∞ sn= s, or by limsn= s, or by n → s. A | {
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sequence. Notation “{s } converges to s” is denoted by n n→∞ sn= s, or by limsn= s, or by n → s. A sequence that does not converge is said to diverge. Examples Which of the sequences given above converge and which diverge; give the limits of the convergent sequences. THEOREM 1. If s → s and s → t, then s = t. That is, the limit of a convergent sequence is unique. Proof: Suppose s 6= t. Assume t>s and let = t − s. Since sn→ s, there exists a positive integer N such that |s − s | / 2 for all n>N . Since s → t, there exists a positive integer 1 n 1 n N 2 such that |t−s n / 2 for all n>N 2. Let N = max{N ,N1} a2d choose a positive integer k>N . Then t − s = |t − s| = |tk− s k s − s|≤| t k s | + |sk− s |+< = = t − s, 2 2 a contradiction. Therefore, s = t. THEOREM 2. If {s } connerges, then {s } is bnunded. Proof: Suppose s → s. There exists a positive integer N such that |s−s | < 1 for all n>N . n n Therefore, it follows that |sn| = |n − s + s|≤| sn− s| + |s| < 1+ |s| for alln>N. Let M = max{|s |, |s |, ..., |s |, 1+ |s|}. Then |s | N. If an→ 0, then s →n0. Proof: Note first that a ≥n0 for all n>N . Since a → 0,nthere exists a positive integer N 1 such that |an| < /k. Without loss of generality, assume that1N ≥ N. Then, for all n>N 1, |sn− 0| = |sn|≤ ka n | {
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More Less | {
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Suppose we take a sample of numbers with unique congruence class modulo p:
x_0 ≡ 0 (mod p)
x_1 ≡ 1 (mod p)
x_2 ≡ 2 (mod p)
...
x_(p-2) ≡ p-2 (mod p)
x_(p-1) ≡ p-1 (mod p)
and we then examine their class under modulo q. How will these classes be distributed? For example, will some values appear more often than others? Or might they be uniform in number or all unique? I imagine this result would depend on if p<q or p>q, and perhaps other things.
Specific, numerical example
15 = 5(3) ≡ 0 (mod 5)
26 = 5(5)+1 ≡ 1 (mod 5)
37 = 5(7)+2 ≡ 2 (mod 5)
58 = 5(11)+3 ≡ 3 (mod 5)
69 = 5(13)+4 ≡ 4 (mod 5)
and
15 ≡ 1 (mod 2)
26 ≡ 0 (mod 2)
37 ≡ 1 (mod 2)
58 ≡ 0 (mod 2)
69 ≡ 1 (mod 2)
and
15 ≡ 3 (mod 6)
26 ≡ 2 (mod 6)
37 ≡ 1 (mod 6)
58 ≡ 4 (mod 6)
69 ≡ 3 (mod 6)
• If you sample the $x_i$ values uniformly, then I imagine their remainders modulo $p$ is uniform, so that the following remainders modulo $q$ are also uniform. – Bill Wallis Jul 10 '18 at 19:08
• That seems vaguely intuitive to me as well (and for my purposes, I'd like that to be true in general), but I'd like for someone to prove that if it is the case. Then again, it can't be completely true, right, since when q>p, we'll never get the larger modulo values, larger than p-1? – Steve Jul 10 '18 at 19:11
• If $q<p$ and the samples are uniform $\bmod p$, they can be uniform $\bmod q$ only if $q|p$ (pigeonhole principle) – gammatester Jul 10 '18 at 19:12
• An example of something I'd like to know if ever occurs, say we have sample (0,1,2,3,4,5,6.....20) in mod 21, is it possible to get a skewed distribution like (0,1,1,1,4,1,0,1,1,0,3) in mod 11 where there are "more 1s than expected"? – Steve Jul 10 '18 at 19:28
• @Steve: If $p=2$, there's nothing stopping you from taking large values of $x$ like $x_0 = 384$ and $x_1 = -7447$. – user14972 Jul 10 '18 at 23:18 | {
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Recall the Chinese Remainder Theorem; in one form, it says that if $p$ and $q$ are relatively prime integers, then there is a bijective correspondence between the
• residue classes modulo $pq$
• pairs consisting of a residue class modulo $p$ and a residue class modulo $q$
Furthermore, given any integer $x$, its residue classes modulo $pq$, $p$ and $q$ are related by this correspondence.
In particular, integers fall into every combination of residues modulo $p$ and $q$, and they do so exactly once per period of length $pq$. So knowing the residue class of an integer modulo $p$ tells you absolutely nothing about its residue class modulo $q$.
And while the notion of a uniform distribution doesn't actually make sense for the integers, this periodic behavior still allows us to capture the idea that such a thing would have independent distributions modulo $p$ and $q$.
In the example of taking $p=21$ and $q=11$, we can give this bijection by explicit formula (discussions of the CRT should show how to obtain this):
• $22 x - 21 y \equiv x \bmod 21$
• $22 x - 21 y \equiv y \bmod 11$
Similarly, any other integer with the same residue as $22x - 21y \bmod 231$ will also satisfy these congruences.
So, given any choice of twenty-one residue class modulo 11, you could find a sequence of $x$'s that have the chosen residues $\bmod 11$ along with the required residues $\bmod 21$. For example, if you want all $1$'s, then we can take
$$x_n = 1 + 22(n-1)$$
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$$x_n = 1 + 22(n-1)$$
to get
• $x_n \equiv n \bmod 21$
• $x_n \equiv 1 \bmod 11$
• This is fantastic and I think I roughly follow what you're saying and at least understand it's impact in a very generalized sense. In your explicit example, 1, 23, 45, 67, ..., 199, 221 are all 1 mod 11 while running across the 0,1,2,...,11 spectrum of values under mod 21. This is helpful because it allows me to refine my question back to my original post, however. What if we restrict ourselves to numbers which not only are sequential in mod 21, but rather are strictly sequential numbers (x, x+1, x+2, ...). Can we know anything about these numbers under mod q? – Steve Jul 11 '18 at 12:48
• I'm realizing now, maybe my question is stupid and obvious afterall. Can you tell me if this follows? Let a, p, q be constant integers. Let x_n= ap+n for n={0,1,...p-1}. Then x_n (mod p) ≡ n. Observe that ap (mod q) ≡ B for some integer B. Thus, x_n (mod q) = ap+n (mod q) ≡n (mod q)+B, so the residues unders mod q are "uniformly distributed" save some wrapping around if q is smaller than p. Is this correct? – Steve Jul 11 '18 at 13:02
If you take any $q$ successive numbers, there will be one in each congruence class $\bmod q$. You start with the lowest one in its class, the next is in the next class up, and so on, wrapping around at $0$. When you get to $q$ you have put one in each class and are ready to start again. If you take $p$ successive numbers where $q\not | p$ you will have two different counts among the congruence classes. Let $r$ be the remainder on dividing $p$ by $q$. There will be $q-r$ classes with $\lfloor \frac pq \rfloor$ because you go around fully that many times. There will be $r$ classes with one more because you have $r$ numbers left after the full cycles. | {
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Definite integral - possible evaluation using real methods?
The book "inside interesting integrals" gives the following exercise for the chapter about contour integration and the residue theorem: $$\int_{0}^\infty \frac{e^{\cos x}\sin(\sin x)}{x}dx=\space\space ?$$ This can be solved using the function $$f(z)=\frac{\exp(e^{iz})}{z}$$ on a quarter-circular contour, and is pretty straightforward. The answer turns out to be $$\frac{\pi}{2}(e-1)$$ However, in the book, the author makes the following comment:
Edward Copson (1901-1980), who was professor of mathematics at the University of St. Andrews in Scotland, wrote "A definite integral which can be evaluated using Cauchy's method of residues can always be evaluated by other means, though generally not so simply." Here's an example of what Copson meant, an integral attributed to the great Cauchy himself. It is easily done with contour integration, but would (I think) otherwise be pretty darn tough.
Does anyone know how to evaluate this integral using real methods?
• This question should actually get more attention... I already grabbed my popcorn to see how real methods come in action. If this does not get attention I'm considering to put a bounty on it (if that is possible, I don't know how that works). – Shashi Jan 7 '18 at 21:14
• I have put a bounty on this question. @Nilknarf Are there some updates concerning this question? – Shashi Jan 8 '18 at 20:47
• @Shashi No... Do you suggest that I add anything in particular to the question? – Frpzzd Jan 8 '18 at 23:55
• I was curious whether you have solved it in the mean time – Shashi Jan 9 '18 at 8:00
Perhaps surprisingly, a straightforward trick works. To this end we refer to the following easy-to-prove lemma.
Lemma. Define $\operatorname{Si}(x) = \int_{0}^{x} \frac{\sin t}{t} \, dt$. Then
1. $\int_{0}^{x} \frac{\sin(yt)}{t} \, dt = \operatorname{Si}(xy)$, and
2. $\operatorname{Si}(x) = \frac{\pi}{2} + \mathcal{O}(x^{-1})$ as $x \to \infty$.
Then for $R > 0$, | {
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Then for $R > 0$,
\begin{align*} \int_{0}^{R} \frac{e^{\cos x}\sin(\sin x)}{x} \, dx &= \int_{0}^{R} \frac{1}{x}\operatorname{Im}(e^{e^{ix}}) \, dx \\ &= \int_{0}^{R} \frac{1}{x}\sum_{n=1}^{\infty} \frac{\sin(nx)}{n!} \, dx \\ &= \sum_{n=1}^{\infty} \frac{1}{n!} \int_{0}^{R} \frac{\sin(nx)}{x} \, dx \\ &= \sum_{n=1}^{\infty} \frac{1}{n!} \operatorname{Si}(nR) \\ &= \sum_{n=1}^{\infty} \frac{1}{n!} \left( \frac{\pi}{2} + \mathcal{O}\left( (nR)^{-1} \right) \right) \\ &= \frac{\pi}{2}(e - 1) + \mathcal{O}(R^{-1}) \end{align*}
Letting $R \to \infty$ proves the claim.
• Not thought about a single contour and yet so simple! Conclusion: the author of the book lied lol (jk I know one does not have full vision on everything) – Shashi Jan 9 '18 at 8:03
$$e^{\cos x}\sin(\sin x) = \text{Im}\, e^{\cos x+i\sin x} = \text{Im}\exp\left(e^{ix}\right) = \text{Im}\sum_{n\geq 0}\frac{e^{nix}}{n!}=\sum_{n\geq 1}\frac{\sin(nx)}{n!}$$ and since $\int_{0}^{+\infty}\frac{\sin(nx)}{x}\,dx = \frac{\pi}{2}$ for any $n>0$, we have $$\int_{0}^{+\infty}\frac{e^{\cos x}\sin(\sin x)}{x}\,dx = \frac{\pi}{2}\sum_{n\geq 1}\frac{1}{n!}=\color{red}{\frac{\pi}{2}(e-1)}.$$
• Is it true that interchanging series and integral is allowed since the series with sine is uniformly convergent? The answer is so neat. I guess it is even easier with Real methods after seeing the answers. – Shashi Jan 9 '18 at 7:55
• "Real methods=without Residue Theorem " in the last comment – Shashi Jan 9 '18 at 8:06
• @Shashi: indeed, this is a small variation on the improper Riemann-integrability of $\frac{\sin x}{x}$ on $\mathbb{R}^+$, which can be proved by real (Fourier-analytic), almost-real (Laplace transform) or complex techniques (residue theorem) in few steps, anyway. – Jack D'Aurizio Jan 9 '18 at 14:45 | {
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# What is zero? Irrational or rational or it have both the properties? [duplicate]
We say,
A number is rational if it can be represented as $\frac{p}{q}$ with $p,q \in \mathbb Z$ and $q\neq 0$.
Any number which doesn't fulfill the above conditions is irrational.
It can be represented as a ratio of two integers as well as ratio of itself and an irrational number such that zero is not dividend in any case.
People say that $0$ is rational because it is an integer. Which I find to be a lame reason. May be any strong reason is there. Can any one tell me please?
## marked as duplicate by Jack M, Chappers, N. F. Taussig, Dietrich Burde, Joel Reyes NocheMay 19 '15 at 12:21
• $0$ can be represented as $0/1$, therefore by your definition it is rational. – TonyK May 19 '15 at 10:30
• And it can also be shown as $\frac{0}{\sqrt{2}}$. Then what? – Man_Of_Wisdom May 19 '15 at 10:30
• $1=\frac{\sqrt{2}}{\sqrt{2}}$. So? – Hayden May 19 '15 at 10:31
• Or less trivially, $2=\frac {\sqrt 8} {\sqrt 2}$ – Jack M May 19 '15 at 10:40
• You may simply notice that $q=\sqrt(2) \not\in \mathbb{Z}$ – Manlio May 19 '15 at 10:42
The definition of an irrational number is that it is not rational. And $0$ is by definition a rational number.
• Why should I consider it a rational number if it can be represented as $\frac{0}{\sqrt{2}}$ – Man_Of_Wisdom May 19 '15 at 10:27
• @Man_Of_Wisdom: What about $\sqrt2/\sqrt2$ then? Do you consider that irrational? – John Bentin May 19 '15 at 10:33
• Nope. No. Not.. – Man_Of_Wisdom May 19 '15 at 10:45
• @Alwin can you please roll back your edit to the fresh one. Which includes only two lines? I want to accept that because it answers at hand! – Man_Of_Wisdom May 19 '15 at 11:10
• Do you mean this one above? – Alwin May 19 '15 at 11:25 | {
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I think you're confusing some of the language in the definition. The definition says that if a number may be written in a certain way, namely as a fraction in which both numerator and denominator are integers, then it's rational. If it cannot be written this way, then it is irrational. There is nothing in the definition that prevents a rational number from being written as a fraction in other ways, such as having rational or irrational numerator or denominator.
The phrase "Any number which doesn't fulfill the above conditions is irrational" does NOT say "Any number which can be written as a fraction $\frac{p}{q}$ with $p,q\notin \mathbb{Z}$ is irrational". It simply says any number that can not be written $\frac{p}{q}$ with $p,q\in \mathbb{Z}$ is irrational.
• One could also use the irrationality critereon to show zero can't be an irrational number. – Rammus May 19 '15 at 11:49
$r$ is rational if you find integers $p,q$ such that $r=\dfrac pq$. This is obviously the case for $r=0$.
The contraposition of this property is "$r$ is irrational if you cannot find integers $p,q$ such that $r=\dfrac pq$".
The contraposition is not at all "$r$ is irrational if you find irrationals $p,q$ such that $r=\dfrac pq$". (By the way, this would be a somewhat circular definition.)
$\newcommand{\Reals}{\mathbf{R}}$You've got excellent explanations of the logical reasons for saying "$0$ is rational". Here are some complementary thoughts too long for a comment:
Definitions in mathematics exist to give convenient labels to useful logical distinctions. "Convenient" is a loose term, but generally refers to simplifying statements of theorems and facilitating common types of discussion. | {
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When two criteria (such as "rational" and "irrational") are logical opposites by definition, it's never a good idea to allow some widespread mathematical concept (such as $0$) to be "both": If you do, every theorem that would apply to that object has to contain a clause explicitly excluding that object. That's inconvenient. In rare cases (see below), you might say "neither". But in the case of $0$, "rational" is the better label:
• Literal application of the definition ("there exist integers $q \neq 0$ and $p$ such that $0 = p/q$") says $0$ is rational.
• The set of rational numbers has pleasant algebraic properties (closed under addition, closed under multiplication) because $0$ is rational. (By contrast, the set of irrational real numbers is not closed under addition, e.g., $(1 - \sqrt{2}) + \sqrt{2} = 1$, or under multiplication, e.g., $\sqrt{2} \cdot \sqrt{2} = 2$, whether or not $0$ is rational.)
To make a case that "$0$ is not rational", i.e., that the definition of "rational number" should exclude $0$, one would want a compelling reason, such as "the statement of a useful theorem becomes awkward if $0$ is (regarded as) rational".
With due respect, the possibility of writing $0$ as $0/\sqrt{2}$ isn't compelling in the above sense; as others have explained, this representation does not contravene the definition. Further, it's useful, and causes no hardship, to agree that $0$ is rational.
For contrast, here are some other "edge cases" that crop up now and again:
• The integer $0$ is "even" ($2$ times some integer) rather than "odd" (leaves a remainder of $1$ on division by $2$). (By the division algorithm, every integer is even or odd, but no integer is both. In this setting, "even" and "not odd" are logically equivalent for integers. I mention this example because a colleague once informed me that some teachers regard $0$ as neither even nor odd.(!!)) | {
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• The zero function $z: \Reals \to \Reals$ is both "even" (for all real $x$, $z(-x) = z(x)$) and "odd" (for all real $x$, $z(-x) = -z(x)$). The notions of "even" and "odd" for functions are not logical opposites. Moreover, it is useful to declare the zero function to be both even and odd: The set of even functions is a vector space under "the usual operations"; the set of odd functions is, too. If $z$ were not "both even and odd (as a function)", at least one of these useful theorems would be false.
• The integer $1$ is neither "prime" nor "composite". (Even though "$1$ has no positive integer factors other than $1$ and itself", we explicitly exclude $1$ from membership in the primes because declaring $1$ "prime" would spoil the uniqueness of prime factorization. On the other hand, $1$ is not "composite" because $1$ is not a product of smaller positive primes.)
There's a deeper point that, ironically, may seem at odds with my earlier stance: Mathematical definitions are human conventions, not absolute, immutable, incontestable features of logic, mathematics, or the physical universe. I suspect this raises unnecessary obstacles for the philosophically-minded who study mathematics. (Everything and More by David Foster Wallace is the most extreme example I've encountered; Wallace seemed tormented by the ontology of infinity.)
On the other hand, when one sees how tightly mathematics hangs together across times and cultures, how definitions lead to the same theorems, one is forced (even fully accepting the preceding paragraph) to admire the phenomenal coherence of the logical structure of mathematics. One starts to feel as if definitions are inevitable. One becomes willing to fight emphatically for the correct definitions. This last, I expect, explains the downvotes to your good question.
As I understand it, the definition of an irrational number is that it is not rational. By definition then, $0$ is rational. | {
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Constructing a non-linear system with prerequisites about the nature of its critical points.
An exercise from the book I am reading is: "Construct a non-linear system that has four critical points:two saddle points, one stable focus, and one unstable focus."
I have tried many systems. I found one quickly but I was lucky even if I had a few clues thanks my previous trials.
I wonder if there is any way to find such systems using a not completely "gropingly way".
Edit: with only two equations in the system.
The system I have is:
$\dot{x}=y^2-x^2$
$\dot{y}=x^2+y^2-2$
• Is the dimension of system = 2 ? – nonlinearism Aug 2 '13 at 17:13
• Yes the dimension is 2 as it was in the chapter preceding this exercise, I should have written that also ^^ Thanks. – Ouistiti Aug 2 '13 at 22:44
• @Ouistiti: Was the answer helpful? What does your system look like? Where is this problem from? – Amzoti Aug 4 '13 at 1:29
• @Amzoti I will post my system above. The answer was very helpful, I learned more from your answer than I expected. Thank you :) The problem is from "Dynamical Systems with Applications using Mathematica®" written by Stephen Lynch. – Ouistiti Aug 5 '13 at 12:22
• @Ouistiti: It makes me very happy to hear that! If you write the four sets of eigenvalues for the Jacobian matrix at each critical point, It may be possible to set up an optimization problem to solve for the six parameters. Look forward to seeing your system and this was a fun problem. Regards – Amzoti Aug 5 '13 at 12:25
We need to choose a form for our system to give us exactly four critical points. This selection is not unique, hence, we can get more than one solution.
Lets choose a system with parameters that give us some degree of freedom. An example of a system that gives us four critical points and some degree of freedom is (there are other choices, this is not unique):
$$\tag 1 x' = a x + b x^2 + c x y = x(a + bx + cy)\\ y' = d y + e y^2 + f x y = y(d+ey + fx)$$ | {
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$$\tag 1 x' = a x + b x^2 + c x y = x(a + bx + cy)\\ y' = d y + e y^2 + f x y = y(d+ey + fx)$$
First thing we need to do is to find the critical points (recall, we need exactly four of them).
• $x = 0 \rightarrow y = 0$ or $y = -\dfrac{d}{e}$
• $y = 0 \rightarrow x = 0$ or $x = -\dfrac{a}{b}$
• $x \ne 0, y \ne 0 \rightarrow x = \dfrac{cd -ae}{be - cf}, y = \dfrac{af -bd}{be - cf},~~\text{with}~~ c \ne 0, be \ne cf$
Thus, our four critical points are (note the constraints above):
$$(0,0), ~\left(0, -\dfrac{d}{e}\right), ~\left(-\dfrac{a}{b}, 0\right), \left(\dfrac{cd -ae}{be - cf}, \dfrac{af -bd}{be - cf}\right)$$
The Jacobian matrix of $(1)$ is given by:
$$\tag 2 \displaystyle J(x, y) = \begin{bmatrix}\frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y}\\\frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y}\end{bmatrix} = \begin{bmatrix}a + 2bx + cy & cx \\ fy & d + 2 e y + fx \end{bmatrix}$$
Now, we have freedom when choosing the parameters with the goal to get two saddle points, one stable focus, and one unstable focus.
There is not a clean way to do this, but our problem now is to choose the parameters $a, b, c, d, e, f$ to make that happen by evaluating the eigenvalues of the Jacobian at each critical point, and choosing the parameters to give us the desired behavior.
I am going to crank these one at a time and keep my fingers crossed for the last critical point.
Point (0,0)
We have:
$$\displaystyle J(0, 0) = \begin{bmatrix} a & 0 \\ 0 & d \end{bmatrix}$$
Lets choose $a = -1, d = 1$ and this is our first saddle.
Point (0,-d/e)
We have:
$$\displaystyle J(0, -d/e) = \begin{bmatrix}a - (cd)/e & 0 \\ -(fd)/e & d -(2d)/e \end{bmatrix}$$
Lets choose $e = 1, c = -3$ and this is our second saddle.
Point (-a/b, 0)
We have:
$$\displaystyle J(-a/b, 0) = \begin{bmatrix} - a - (ac)/b & 0 \\ 0 & d -(af)/b \end{bmatrix}$$
Lets choose $b = 1, f = 1$ and this is our unstable focus. | {
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Lets choose $b = 1, f = 1$ and this is our unstable focus.
Point $\left(\dfrac{cd -ae}{be - cf}, \dfrac{af -bd}{be - cf}\right)$
Well, this is where we need luck with the parameters we chose, but I think it might be possible to cast this problem as an optimization problem, so you might want to play around with that. This would give you ranges for these six parameters, so no guesswork is needed.
Anyway, from our parameter choices above, we get $x = y = -1/2$.
$$\displaystyle J(-1/2, -1,2) = \begin{bmatrix} -1/2 & 3/2 \\ -1/2 & -1 \end{bmatrix}$$
This, thankfully, gives us our stable focus.
Thus, we have the system:
$$\tag 1 x' = - x + x^2 -3 x y \\ y' = y + y^2 + x y$$
Lets draw the phase portrait and validate this analysis.
We have what we need with two saddles, an unstable and a stable focus at the four critical points.
I am curious how you did it (even if it is guessing) and what your system looks like and it would help to post your solution in your question.
Update
Here is the phase portrait of the system you wrote and it is also a good example.
• Great job. For my knowledge how did you come up with the second order system in (1)? By intuition ? – kaka Aug 3 '13 at 6:14
• @kaka: Yes, by intuition and a bit of handwaving. I wish I could say I had some wonderful approach to that. I realized that we needed something that could give us a point at (0,0) and then realized we needed a parabola that gave points at (0, w), and (f, 0) and one more at some (h, j). Thinking along those lines lead me to that system. Then, we needed freedom, so I added all of the parameters. I still think the approach can be improved using optimization for ranges on the parameters, but I'll let the OP work that. Thanks for the kind words. Regards – Amzoti Aug 3 '13 at 6:19
• @Amzoti: Wowwww. It is complete as always. Full of points, and finally a nice illustrating graph. – mrs Aug 3 '13 at 14:56
• This is a nice problem. Go for it! ;-) – Namaste Aug 4 '13 at 1:14 | {
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# What does it mean for a sequence to be decreasing in regards to the Integral Test for Convergence?
Example: $$\sum \limits_{n=1}^{\infty} n^k e^{-n}$$
This is the problem that I am to solve with the Integral test. I know that it converges, and I know the answer, but just don't fully get the point about it being decreasing.
How does one if a function is decreasing, the first derivative is less than zero.
$f^{\prime}\left(x\right) = \dfrac{\left(k-x\right)x^{k-1}}{e^x}$ is less than zero when $x > k$. I do understand that there is always an $x > k$. (Not really sure if what happens when $k = \infty$ leaving that issue till I learn more).
So my confusion comes to when can you say that a sequence is decreasing? What about when $x < k$: does the Integral test fail on that interval? Or just does the long-term behavior is all that matters? I mean what if just the very last number before infinity the sequence decreases, then will the Integral Test for convergence work?
Could you explain in the realm of sequences and series what does it truly mean to be decreasing? I do understand that I am just learning this, but don't be afraid to start at a high school level and build up your answers to a Field Medal level of understanding. :) | {
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-
A sequence $a_n$ is decreasing if $a_{n+1} \le a_n$ for all $n$. – Javier Mar 5 '13 at 13:54
@JavierBadia So what about the example I give, if $k = 3$ then for the first three terms its is not decreasing? Why doesn't that invalidate the Integral Test for Convergance? – yiyi Mar 5 '13 at 13:55
The first terms don't affect convergence. You want to know what happens for $n$ large. – L. F. Mar 5 '13 at 13:57
@L.F. Why don't the first terms don't affect convergance? That is really my question. Thanks for putting it into words. – yiyi Mar 5 '13 at 13:57
Because there are only finitely many first terms, so their sum will always converge. As long as your sequence is defined, it can jump, fly, or hula hoop at the beginning for all that matters. What we really want to know, is if the "infinite portion" (i.e. what happens after a certain point) will sum to a value or not. – L. F. Mar 5 '13 at 14:01
A sequence is of the form $\{x_n\}$, and a series is of the form $\sum \limits_{n=1}^{\infty} x_n$. So for a sequence $\{x_n\}$ to be decreasing means $x_1 > x_2 > x_3 > \ldots$.
As long as there is some $k$ such that $f(n)$ is decreasing for all $n > k$, the integral test applies.
Consider the series $\sum \limits_{n=1}^{\infty} f(n) = \sum \limits_{n=1}^k f(n) + \sum \limits_{n=k + 1}^{\infty} f(n)$. The first term is finite, so converges, and the second term converges due to the integral test because it is always decreasing. | {
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So what about the example I give, if k=3 then for the first three terms its is not decreasing? Why doesn't that invalidate the Integral Test for Convergance? – yiyi Mar 5 '13 at 13:56
The integral test applies as long as the function is eventually decreasing. – ferson2020 Mar 5 '13 at 13:59
Thanks for your answer, but it has never been explained why the long term behavior is only which is important? Is it because if it converges then it stops "growing/shrinking" at that number, otherwise it just "grows/decreases" without limit? – yiyi Mar 5 '13 at 14:03
I added to my answer; does that help make it clear? – ferson2020 Mar 5 '13 at 14:04
Say a sequence $\{a_n\}_{n\in\mathbb{N}}$ is decreasing provided $a_{n+1}\leq a_n$ for every $n\in\mathbb{N}$. So the bigger subscripts correspond to smaller numbers. You really just need it decreasing from a rank on, so decreasing for large $n\in\mathbb{N}$.
You are looking at a series. What you want to show is that the associated sequence (that is, given the series $\sum_{n=1}^\infty a_n$ the associated sequence is $\{a_n\}_{n\in\mathbb{N}}$) is nonnegative and decreasing, then you define a function $f:[1,+\infty)\to[0,+\infty)$ so that $f(n)=a_n$ and $\int_1^\infty f<+\infty$.
Also, $k$ is an exponent so it can't be $\infty$.
Also, for series we only care what happens "near $\infty$", so to speak. So say if your sequence doesn't start decreasing until $n=100$ then just apply the integral test to the series $\sum_{n=100}^\infty a_n$, in which case you look at $\int_{100}^\infty f$. | {
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Rationale for "near $\infty$": First of all, usually by something happening "near $\infty$ we mean there exists $N\in\mathbb{N}$ so that for all $x\geq N$ the condition happens. Convergence of a series (or a sequence for that matter, and keep in mind a series can be seen as a sequence of partial sums) only has to do with the "tail," or what happens near $\infty$. This is because for any natural number $N\in\mathbb{N}$ the first part of the series $\sum_{n=1}^Na_n$ is always finite. No matter what $N$ you choose. Because of this, if $\sum_{n=1}^\infty$ is going to diverge the problem must occur near $\infty$.
So if your associated sequence doesn't start decreasing until after $n=10,000,000,000$, no problem. Just write $\sum_{n=1}^\infty a_n=\sum_{n=1}^{10,000,000,000}a_n+\sum_{n=10,000,000,000}^\infty a_n$ and apply the integral test to the tail.
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Nice and informative answer, could you point me to some papers which explain the rational why only "near $\infity$" matters? – yiyi Mar 5 '13 at 14:01 | {
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