The identity matrix is a square matrix that has 1’s along the main diagonal and 0’s for all other entries. What are the Inverse Properties of Addition and Multiplication? Matrix Multiplication Properties 9:02. Properties of Inverse For a matrix A, A −1 is unique, i.e., there is only one inverse of a matrix (A −1 ) −1 = A There is no such thing! Using properties of inverse matrices, simplify the expression. An Axiom is a mathematical statement that is assumed to be true. This tutorial can show you the entire process step-by-step. Prove algebraic properties for matrix addition, scalar multiplication, transposition, and matrix multiplication. An inverse matrix exists only for square nonsingular matrices (whose determinant is not zero). A is the inverse of B i.e. The answer to the question shows that: (AB)-1 = B-1 A-1. Well, for a 2x2 matrix the inverse is: In other words: swap the positions of a and d, put negatives in front of b and c, and divide everything by the determinant (ad-bc). Matrix transpose AT = 15 33 52 −21 A = 135−2 532 1 Example Transpose operation can be viewed as ﬂipping entries about the diagonal. of an m xn matrix A if AA+ = pro- jection on the range of A and A+A = projection on the range of A+. There are really three possible issues here, so I'm going to try to deal with the question comprehensively. • F is called the inverse of A, and is denoted A−1 • the matrix A is called invertible or nonsingular if A doesn’t have an inverse, it’s called singular or noninvertible by deﬁnition, A−1A =I; a basic result of linear algebra is that AA−1 =I we deﬁne negative powers of A via A−k = A−1 k Matrix Operations 2–12 In this video you will learn about Properties of Matrix for Addition - Commutative, Associative and Additive Inverse - Matrices - Maths - Class 12/XII - ISCE,CBSE - NCERT. The list of properties of matrices inverse is given below. We will see later that matrices can be considered as functions from R n to R m and that matrix multiplication is composition of these functions. Follow along with this tutorial to get some practice adding and subtracting matrices! Have you ever combined two numbers together and found their sum to be zero? Properties of Matrix Operations. The operations we can perform on the matrix to modify are: Interchanging/swapping two rows. Solve a linear system using matrix algebra. Given the matrix D we select any row or column. Sin(θ), Tan(θ), and 1 are the heights to the line starting from the x-axis, while Cos(θ), 1, and Cot(θ) are lengths along the x-axis starting from the origin. Where a, b, c, and d represents the number. 4. The additive inverse of matrix A is written –A. For example: 2 + 3 = 5 so 5 – 3 = 2. The inverse of a matrix. If A is a matrix of order m x n, then . This matrix is often written simply as $$I$$, and is special in that it acts like 1 in matrix multiplication. Selecting row 1 of this matrix will simplify the process because it contains a zero. Just find the corresponding positions in each matrix and add the elements in them! Properties of matrix addition. The first such attempt was made by Moore.2'3 The essence of his definition of a g.i. Yes, it is! Addition and subtraction are inverse operations of each other. With this knowledge, we have the following: Notice that the fourth property implies that if AB = I then BA = I How Do You Add and Subtract Matrices with Fractions and Decimals. Yes, it is! Is the Inverse Property of Matrix Addition similar to the Inverse Property of Addition? The product of two inverse matrices is always the identity. 7 – 1 = 6 so 6 + 1 = 7. Multiplying or Dividing a row by a positive integer. A + (- A) = (- A) + A = O-A is the additive inverse of A. The identity matrix for the 2 x 2 matrix is given by $$I=\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$$ Properties of matrix addition & scalar multiplication. 2x2 Matrix. Your email address will not be published. So if n is different from m, the two zero-matrices are different. The determinant of a matrix. Go through it and simplify the complex problems. Matrix Addition and Multiplication « Matrices Definitions: Inverse of a matrix by Gauss-Jordan elimination ... Properties of Limits Rational Function Irrational Functions Trigonometric Functions L'Hospital's Rule. Matrix Matrix Multiplication 11:09. Now using these operations we can modify a matrix and find its inverse. In this tutorial, you'll learn the definition for additive inverse and see examples of how to find the additive inverse of a given value. First, since most others are assuming this, I will start with the definition of an inverse matrix. $$A=\begin{bmatrix} a & b\\ c & d \end{bmatrix}$$. The determinant of a 4×4 matrix can be calculated by finding the determinants of a group of submatrices. The identity matrix and its properties. Unitary Matrix- square matrix whose inverse is equal to its conjugate transpose. In this article, let us discuss the important properties of matrices inverse with example. A is row-equivalent to the n-by-n identity matrix In. The determinant of the matrix A is written as ad-bc, where the value is not equal to zero. Inverse of a matrix The inverse of a matrix $$A$$ is defined as a matrix $$A^{-1}$$ such that the result of multiplication of the original matrix $$A$$ by $$A^{-1}$$ is the identity matrix $$I:$$ $$A{A^{ – 1}} = I$$. Compute the inverse of a matrix using row operations, and prove identities involving matrix inverses. In fact, this tutorial uses the Inverse Property of Addition and … This tutorial should help! Various types of matrices are -: 1. When you start with any value, then add a number to it and subtract the same number from the result, the value you started with remains unchanged. Integration Formulas Exercises. det A ≠ 0. ... Is the Inverse Property of Matrix Addition similar to the Inverse Property of Addition? Since . If you've ever wondered what variables are, then this tutorial is for you! Deﬁnition The transpose of an m x n matrix A is the n x m matrix AT obtained by interchanging rows and columns of A, Deﬁnition A square matrix A is symmetric if AT = A. The rank of a matrix. Matrix Vector Multiplication 13:39. The purpose of the inverse property of multiplication is to get a result of 1. It satisfies the condition UH=U −1 UH=U −1. Adding or subtracting a multiple of one row to another. Inverse properties of addition and multiplication got you stumped? The purpose of the inverse property of addition is to get a result of zero. To zero are inverses if from m, the two zero-matrices are.... 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# How to find the percentage of kinetic energy from the collision of three spheres in tandem?

#### Chemist116

The problem is as follows:

Three spheres identical in size, shape and mass are situated over a surface as indicated in the picture. Initially, $2$ and $3$ are at rest. Sphere labeled $1$ makes a collision with $2$ and as a result $2$ makes a collision with $3$. Find the percentage of kinetic energy which $3$ receives with respect from the initial. Assume that in all collisions the coefficient of restitution (COR) is $0.75$.

The alternatives are as follows:

$\begin{array}{ll} 1.&29\%\\ 2.&39\%\\ 3.&49\%\\ 4.&59\%\\ \end{array}$

For this specific situation, what I attempted was to tackle the problem using the conservation of momentum as follows:

$p_i=p_f$

$mv_1=mu_1+mu_2$

$v_1=u_1+u_2$

Since $e=0.75$

$\frac{1}{4}=\frac{u_{1}-u_{2}}{v_{2}-v_{1}}$

But from this part I end up with many equations; how exactly can I find the percentage which is being asked?

#### romsek

Math Team
You have two collisions to analyze.

In the first collision we have two equations arising from conservation of momentum, and the coefficient of restitution.

$m v_{1i}+0 = m v_{1f} + m v_{2f}$
$\dfrac{v_{2f}-v_{1f}}{v_{1i}-0} = \dfrac{3}{4}$

This can be solved for $v_{2f}$ which then becomes $v_{2i}$ for the 2nd collision

The second collision is analyzed exactly the same way but with the initial velocity for ball #2 as calculated above.

Solving this second set of equations will give you a ratio of the velocity of ball #3 to the velocity of ball #1.

This must be squared to get the ratio of kinetic energies.

Do all this and you'll find the answer is one of the choices.

topsquark

#### skeeter

Math Team
You have two collisions to analyze.

In the first collision we have two equations arising from conservation of momentum, and the coefficient of restitution.

$m v_{1i}+0 = m v_{1f} + m v_{2f}$
$\dfrac{v_{2f}-v_{1f}}{v_{1i}-0} = \dfrac{3}{4}$

This can be solved for $v_{2f}$ which then becomes $v_{2i}$ for the 2nd collision

The second collision is analyzed exactly the same way but with the initial velocity for ball #2 as calculated above.

Solving this second set of equations will give you a ratio of the velocity of ball #3 to the velocity of ball #1.

This must be squared to get the ratio of kinetic energies.

Do all this and you'll find the answer is one of the choices.
What he said $\uparrow \uparrow \uparrow$ ... I get 59%

#### Chemist116

What he said $\uparrow \uparrow \uparrow$ ... I get 59%
Since nobody did explicitly mentioned I'll do on my own:

$\frac{u_{1}-u_2}{v_2-v_1}=\frac{3}{4}$

$v_1=u_1+u_2$

Then:

$\frac{u_1-u_2}{0-v_1}=\frac{3}{4}$

$4u_1-4u_2=-3v_1$

$u_1=v_1-u_2$

$4v_1-4u_2-4u_2=-3v_1$

$u_2=\frac{7v_1}{8}$

Then for the following collision is rinse and repeat:

$u_2=v_2$

$\frac{u'_2-u_3}{0-v_2}=\frac{3}{4}$

$\frac{u'_2-u_3}{0-\frac{7v_1}{8}}=\frac{3}{4}$

$\frac{7v_1}{8}=u'_2+u_3$

$\frac{7v_1}{8}-u_3=u'_2$

$4u'_2-4u_3=-\frac{21v_1}{8}$

$4u'_2=4u_3-\frac{21v_1}{8}$

Multiplying by $4$ on both sides to: $\frac{7v_1}{8}-u_3=u'_2$

$\frac{28v_1}{8}-4u_3=4u'_2$

Inserting this value into: $4u'_2=4u_3-\frac{21v_1}{8}$

$\frac{28v_1}{8}-4u_3=4u_3-\frac{21v_1}{8}$

$\frac{49v_1}{8}=8u_3$

$u_3=\frac{49v_1}{64}$

This must be the velocity for the third sphere after the tandem collision:

Therefore the relationship in the kinetic energy to the first must be:

$\frac{K.E_3}{K.E_1}=\frac{\frac{1}{2}m\left(\frac{49v_1}{64}\right)^2}{\frac{1}{2}mv_1^2}=\frac{49^2}{64^2}$

Finally, this is approximately to:

$\frac{K.E_3}{K.E_1}\approx 0.5861816406$

which times $100$ is $58.62\%$ which isn't exactly $59\%$ but should it be considered this? Or did somebody obtained a closer answer to $59\%$?

The answers sheet mentions the answer is $59\%$. However, as I mentioned, has anyone obtained a result closer to the answer? @skeeter Did you obtained the same?

#### skeeter

Math Team
0.586... is sufficiently close to 59%, wouldn't you say?

topsquark


# Finite difference method basic implementation on Octave

Trying to study the error of FDM for a second order derivative versus step size I calculated the coefficients and validated them, but the output has errors for small step sizes.

The function in question is

$$f(x) = e^{\sin(x)}$$

With a derivative of $$f''(x) = \left(\cos^2(x) - \sin(x)\right)e^{\sin(x)}$$

The calculated FDM

$$\frac{\partial^{(2)}f}{\partial x^{(2)}}\approx -\frac{5}{2h^2}f(x) - \frac{1}{12h^2}f(x-2h) + \frac{4}{3h^2}f(x-h) + \frac{4}{3h^2}f(x+h) - \frac{1}{12h^2}f(x+2h)$$

And my Octave implementation:

stps = 1e-7:1e-6:3e-5;
outs = [];
x = pi/2;
for h = stps
coefficients = [-5/2 -1/12 4/3 4/3 -1/12]./h^2;
steps = [0 -2 -1 1 2].*h;

outs(end+1) = sum(coefficients .* exp(sin(x + steps)));
end

plot(stps,outs, "linewidth",1.5 , stps, ones(size(stps)).* (cos(x)^2 - sin(x))*exp(sin(x)) , "linewidth",1.5)

Which produces the following plot:

Are those rounding errors caused by the nature of floating point numbers and operations or something else?

• Here's what's happening: math.stackexchange.com/questions/2213240/… – user14717 Feb 2 '18 at 15:10
• Try 'format long' to get more precision, then you'll see this happen at $h\approx 10^-18$ rather than $10^-7$. – user14717 Feb 2 '18 at 15:12
• @user14717 format long only affects the terminal output, not the internal accuracy -- it makes no difference here. – Christian Clason Feb 2 '18 at 17:57
• It's normal, but it's easier to see what's happening if you plot $\log|f''_{\mathrm{estimate}}-f''_{\text{true}}|$ vs $\log(\text{stepsize})$ instead, and make sure to include large stepsizes too (up to $\sim 1$). – Kirill Feb 2 '18 at 22:09
• @user1471 There is none, apart from the Symbolic Toolbox's vpa (which is not a silver bullet). – Christian Clason Feb 2 '18 at 22:54

(part of it was already given in the comments by user14717, Christian Clason, and Kirill)

While performing numerical differentiation using finite differences, one would observe two sources of error: truncation and rounding. Truncation error (for simplicity, what order terms are truncated from the Taylor series expansion) will be determined by a discretization scheme that is being used and will decrease with $h$. Unfortunately, that will be eventually stopped by an increase in the rounding error. Therefore, there usually exists an optimal discretization step $h$ when the total error (the combined effect of Taylor series truncation and rounding in finite-precision arithmetic) is minimal. Certainly, this value depends on the discretization scheme, function under-approximation, chosen point, etc. So, one would expect to see a plot that is similar to the following one:

Here, I plot the relative error in the approximation of the second derivative using your scheme vs analytical derivative as a function of the step size $h$. Since you are using a standard five-point stencil for a second-order derivative, the error is expected to drop as $\mathcal O(h^4)$, until you hit $h\approx5\cdot 10^{-3}$. At this point, the rounding error starts dominating and we see the oscillations in the relative error together with its gradual increase. However, using this scheme you are able to get 10 significant digits from your derivative pretty safely using double precision, which I would say is a very reasonable result.

If interested, you might invest your time adjusting your discretization scheme, increased precision (vpa), etc.

A couple of minor notes:

• loglog plots are usually very useful when studying refinements, errors, timings, etc.
• a line on the plot usually implies continuity of the data; in this particular case, where you are using finite differences at discrete points, markers are preferred, in my opinion.

Slightly modified Octave code:

x = pi/2;
stps = logspace(-8,0,80);
numpoints = size(stps,2);
outs = zeros(numpoints,1);
analytic = ones(numpoints,1)*(cos(x)^2 - sin(x))*exp(sin(x));
for i=1:numpoints
h=stps(i);
coefficients = [-5/2 -1/12 4/3 4/3 -1/12]./h^2;
steps = [0 -2 -1 1 2].*h;
outs(i) = sum(coefficients .* exp(sin(x + steps)));
end

figure(1)
loglog(stps',abs(outs-analytic)./abs(analytic),"kx");hold on;
xlim([1E-8,1E-0]);   ylim([1E-12,1E+1]);
xlabel("h");         ylabel("Rel. error: |num-an|/|an|");


# Math Help - Geometric sequence

1. ## Geometric sequence

The fifth term of a geometric series is 5! and the sixth term is 6!. What is the fourth term?

Can someone post their answer and reasoning. I have my answer and think it's right but i really need e.c right now and want to make sure. Thanks!

2. Originally Posted by jarny
The fifth term of a geometric series is 5! and the sixth term is 6!. What is the fourth term?

Can someone post their answer and reasoning. I have my answer and think it's right but i really need e.c right now and want to make sure. Thanks!
For a geometric sequence {a_n}:
a_n = a_1*r^{n - 1}
where a_1 is the first term in the series and a_n is the nth term.

So
a_5 = 5! = 120 = a_1*r^4
a_6 = 6! = 720 = a_1*r^5

Two equations, two unknowns, we can solve this.

Divide a_6 by a_5:

a_6/a_5 = 6!/5! = 6 = (a_1*r^5)/(a_1*r^4) = r

Thus r = 6.

So
120 = a_1*6^4

Thus
a_1 = 120/6^4 = 20/6^3 = 20/216 = 5/54

So
a_n = (5/54)*6^{n - 1}

So
a_4 = (5/54)*6^{4 - 1} = (5/54)*6^3 = (5/54)*216 = 20

-Dan

3. got the same answer, thanks a lot

4. Hello, jarny!

The fifth term of a geometric series is 5! and the sixth term is 6!.
What is the fourth term?
Dan's solution is absolutely correct.
. . That's the approach I would have used.

But upon reflection, I realized that I let the factorials dazzle me.
. . There is a simpler solution.

We are given: .a
5 = 120 .and .a6 = 720

Then: .r .= .a
6/a5 .= .720/120 .= .6

. . The "rule" is multiply-by-six.

Therefore, the preceding term is: .a
4 = 20.

See? .We could have eyeballed the problem . . .

5. Originally Posted by Soroban

. . There is a simpler solution.
I always tell my students that I have a tendancy to make things harder than they have to be.

-Dan


# Is there a geometric analog of absolute value?

I'm wondering whether there exists a geometric analog concept of absolute value. In other words, if absolute value can be defined as

$$\text{abs}(x) =\max(x,-x)$$

intuitively the additive distance from $$0$$ to $$x$$, is there a geometric version

$$\text{Geoabs}(x) = \max(x, 1/x)$$

which is intuitively the multiplicative "distance" from $$1$$ to $$x$$?

Update: Agreed it only makes sense for $$Geoabs()$$ to be restricted to positive reals.

To give some context on application, I am working on the solution of an optimization problem something like:

$$\begin{array}{ll} \text{minimize} & \prod_i Geoabs(x_i) \\ \text{subject to} & \prod_{i \in S_j} x_i = C_j && \forall j \\ &x_i > 0 && \forall i . \end{array}$$

Basically want to satisfy all these product equations $$j$$ by moving $$x_i$$'s as little as possible from $$1$$. Note by the construction there are always infinite feasible solutions.

• Is the triangular inequality satisfied ? Jun 28, 2020 at 22:15
• Interesting idea but I'd consider revising the definition to $\operatorname{geoabs}(x)=\operatorname{sign}\left(x\right)\max\left(\left|x\right|,\left|\frac{1}{x}\right|\right)$, which would take $x$ over $(-\infty,-1]$ and $1/x$ on $(-1,0)$ instead of the other way round as you have. Your version has small or large negative values multiplicatively close $1$ while $-1$ is the most distal from $1$, which should be reversed.
– Jam
Jun 28, 2020 at 22:22
• @hamam_Abdallah I believe it is if you consider positive $x$ only.
– Jam
Jun 28, 2020 at 22:26
• The length of a vector is an absolute value. Jun 29, 2020 at 10:05
• Interesting question, but my initial reaction is "Have you thought about re-stating the problem in terms of the variables $y_i$, where $y_i = \log x_i$"? Jun 29, 2020 at 13:31

To make things easier I'll set $$f(x)=\max\{x,-x\}$$ and $$g(x)=\max\{x,\frac{1}{x}\}$$.

So we understand that $$f:\mathbb{R}\to \mathbb{R}^+$$ and $$g: \mathbb{R}^+\to \mathbb{R}^+$$.

Then $$\exp(f(x))=g(\exp(x))$$. So we can use this to translate some properties like the triangle inequality.

$$g(xy)=g(\exp(\log(xy)))=\exp(f(\log(xy)))=\exp(f(\log(x)+\log(y)))$$ $$\leq \exp(f(\log x)+f(\log y))=\exp(f(\log x))\exp(f(\log y))=g(\exp(\log(x))g(\exp(\log(x))$$ $$=g(x)g(y)$$

So $$g(xy)\leq g(x)g(y)$$ and we have the multiplicative triangle inequality.

Of course this is easier to show directly but the method emphasizes the "transfer".

Another good sign is $$g(x)=1$$ if and only if $$x=1$$.

All in all it looks like you're moving between $$(\mathbb{R},+)$$ and $$(\mathbb{R}^+,\cdot)$$ with $$\log$$ and $$\exp$$. So a nice question.

I'm sure there's more to say.

Another way (maybe cleaner) to see it : let us consider

• $$G_1 = (\mathbb{R},+,\|\cdot\|_1)$$ the additive group of real numbers equipped with a norm : for all $$x\in G_1$$, $$\|x\|_1 = |x| = \max \{x,-x\}$$
• $$G_2 = (\mathbb{R_+^*},\cdot,\|\cdot\|_2)$$ the multiplicative group of (strictly) positive real numbers equipped with a norm defined using the norm of $$G_1$$ : for all $$x\in G_2$$, $$\|x\|_2 = \|\ln x\|_1 = \ln \max \{x,1/x\}$$

The map $$\exp\colon G_1\to G_2$$ is therefore by construction a group isometric isomorphism (with $$\ln\colon G_2\to G_1$$ its inverse). Indeed, for all $$x\in G_1$$ $$\|x\|_1 = \|\exp x\|_2$$

You can check that $$\|e_i\|_i = 0$$ where $$e_i$$ is the identity element of $$G_i$$ (here $$e_1 = 0$$ and $$e_2 = 1$$).

If you forget the $$\ln$$ map in the definition of $$\|\cdot\|_2$$, it is not anymore a norm.


# How do you approach when completing the square?

If $$M = 3x^2 - 8xy + 9y^2 - 4x + 6y + 13$$, where $$x,y\in\mathbb R$$, then $$M$$ must be:

a) positive $$\qquad$$b) negative $$\qquad$$c) $$0 \qquad$$ d) an integer

I somehow managed to figure it out by completing the square but in order to do so, it took me a lot of time and I'm not sure if every time I could solve such problems.

This whole expression can be written as: $$2(x - 2y)^2 + (x - 2)^2 + (y + 3)^2$$ which implies $$M$$ is positive.

My point is sometimes I'm lucky and I could group them in squares but other times not. Is there any particular technique/method which always works?

Secondly I also wanna know what you guys observe when completing the squares?

• In my opinion, the key term that must be focused on first is the $8xy$ term. This is because there seems to be wiggle room everywhere else, re raising/lowering the $x^2$ or $y^2$ terms. So, the 1st try should be $(ax + by)^2$ where $2ab = 8.$ Then, try to make everything fit around that. Sep 16 at 8:47
• While you should definitely solidify your ability to pick an approach for the hard math, there is a parallel part of any math problem (especially when applied with units in a real world scenario) that would completely answer this question. You should always ask Does my answer make sense? and that often means getting a rough idea in your head of pos/neg and/or order of magnitude. Use it to validate any calculations. In this case options c and d don't make sense (x=0.123, y=0.357). Plus c can't be true without d. Then plug in y=1, x=1 and you get 3 - 8 + 9 - 4 + 6 + 13 -> positive. Sep 16 at 17:10

\begin{align} &3x^2 - 4x(2y+1)+ (9y^2 + 6y + 13-M)=0\\ \implies &\Delta_x=4(2y+1)^2-3(9y^2+6y+13-M)≥0\\ \implies &3M≥11y^2+2y+35\\ \implies &3M≥11 \left(y + \frac{1}{11}\right)^2 + \frac{384}{11}\\ \implies &3M≥\frac{384}{11}\\ \implies &M≥\frac{128}{11}>0.\end{align}
• Ah nice, $\Delta_x$ is the discriminant of the quadratic in $x$. Nice idea
• @Navdeep To optimize the polynomial, the roots are considered real. If we are talking about complex numbers, then $M=0$ can be and $M > 0$ can also be. It can be $M<0$ too. So your question loses its meaning. Notice, when completing the square you assumed that $x,y$ were real. Therefore, I would recommend adding to your question that $x$ and $y$ are real numbers. Otherwise, the polynomial cannot be optimized. Sep 16 at 15:52


1. ## Notation again again

I do not know how to search this but does

$C^{\infty}$ mean that is differentiable any number of times, or something?

Any help appreciated.

2. Originally Posted by Mathstud28
I do not know how to search this but does

$C^{\infty}$ mean that is differentiable any number of times, or something?

Any help appreciated.
Let $I$ be an open interval. The notation $\mathcal{C}^k(I)$ means a function differenciable $k$ times on $I$ such that $f^{(k)}$ is a continous function on $I$. However, if the function is infinitely differenciable i.e. it is an element of $\mathcal{C}^{k}$ for any $k\geq 1$ then we say it $\mathcal{C}^{\infty}(I)$. For example, $\exp ,\sin , \cos \in \mathcal{C}^{\infty}(\mathbb{R})$. While $\log \in \mathcal{C}^{\infty}(0,\infty)$.

3. Originally Posted by ThePerfectHacker
Let $I$ be an open interval. The notation $\mathcal{C}^k(I)$ means a function differenciable $k$ times on $I$ such that $f^{(k)}$ is a continous function on $I$. However, if the function is infinitely differenciable i.e. it is an element of $\mathcal{C}^{k}$ for any $k\geq 1$ then we say it $\mathcal{C}^{\infty}(I)$. For example, $\exp ,\sin , \cos \in \mathcal{C}^{\infty}(\mathbb{R})$. While $\log \in \mathcal{C}^{\infty}(0,\infty)$.
Thanks TPH! Always answering my notation questions .

One question though, I understand that $e^x,\cos(x),\text{etc.}$ are $C^{\infty}\left(\mathbb{R}\right)$

But since $C^{\text{whatever}}$ seems to only be talking about the at least first derivative wouldnt that mean that the $I$ specified for $\ln(x)$ should be $\mathbb{R}/\left\{0\right\}$ since $\ln^{\left(n\in\mathbb{Z^+}\right)}(x)$ is continous for that interval?

Oh wait, I looked back and the interval $I$ is specified in relation to the original function, not the derivatives. Ok so I get it.

Thank you.

EDIT: Actually two more things. The first is obviously that

If $p(x)$ is a non-descript polynomial then it is $C^{\infty}$ Right?

Also how would you say that some function is this on some interval, in other words how would you put this notation down so it makes sense.

Like this?

" $f(x)$ is $C^{\infty}(a,b)$"?

4. Originally Posted by Mathstud28
But since $C^{\text{whatever}}$ seems to only be talking about the at least first derivative wouldnt that mean that the $I$ specified for $\ln(x)$ should be $\mathbb{R}/\left\{0\right\}$ since $\ln^{\left(n\in\mathbb{Z^+}\right)}(x)$ is continous for that interval?
It does not need to be an open interval it can be an open set. But since open sets are concepts from topology I did not want to mention them. An open set is a generalization of an open interval. For example, $\mathbb{R} - \{ 0 \} = (-\infty,0)\cup (0,\infty)$ is an open set. Basically an open set is such a set that has no boundary. More formally $S$ is open iff for any $x\in S$ there is $\epsilon > 0$ such that $(x-\epsilon,x+\epsilon) \subset S$. With open sets we can say that $\ln |x|$ is $\mathcal{C}^{\infty}(\mathbb{R} - \{ 0\})$.

If $p(x)$ is a non-descript polynomial then it is $C^{\infty}$ Right?
A polynomial is infinitely differenciable so it is $\mathcal{C}^{\infty}$.

Also how would you say that some function is this on some interval, in other words how would you put this notation down so it makes sense.

Like this?

" $f(x)$ is $C^{\infty}(a,b)$"?
That is exactly how we say it.

As an illustration of this notation we can state a stronger version of the Fundamental Theorem of Calculus. Let $f$ be continous on $[a,b]$ which is $\mathcal{C}^k(a,b)$. Define $F(x) = \smallint_a^x f$. Then $F$ is $\mathcal{C}^{k+1}(a,b)$.

Analysts like to say that integration "smoothens out a function". The theorem above is what this is all about.

5. Originally Posted by ThePerfectHacker
It does not need to be an open interval it can be an open set. But since open sets are concepts from topology I did not want to mention them. An open set is a generalization of an open interval. For example, $\mathbb{R} - \{ 0 \} = (-\infty,0)\cup (0,\infty)$ is an open set. Basically an open set is such a set that has no boundary. More formally $S$ is open iff for any $x\in S$ there is $\epsilon > 0$ such that $(x-\epsilon,x+\epsilon) \subset S$. With open sets we can say that $\ln |x|$ is $\mathcal{C}^{\infty}(\mathbb{R} - \{ 0\})$.

A polynomial is infinitely differenciable so it is $\mathcal{C}^{\infty}$.

That is exactly how we say it.

As an illustration of this notation we can state a stronger version of the Fundamental Theorem of Calculus. Let $f$ be continous on $[a,b]$ which is $\mathcal{C}^k(a,b)$. Define $F(x) = \smallint_a^x f$. Then $F$ is $\mathcal{C}^{k+1}(a,b)$.

Analysts like to say that integration "smoothens out a function". The theorem about is what this is all about.
Ok, amazing, I completely understand that.

Like the last part since $F'(x)=\frac{d}{dx}\int_a^{x}f(t)dt=f(x)$ by the fundamental theorem of calculus and we have stated that $f(x)\quad\text{is}\quad{C^{k}}$

$F(x)$ takes on all the k amount of differentiabilities of $f(x)$ and adds more since we know that if it is integrable it is differentiable back to f(x)!

Thanks TPH

6. Originally Posted by Mathstud28

Also how would you say that some function is this on some interval, in other words how would you put this notation down so it makes sense.

Like this?

" $f(x)$ is $C^{\infty}(a,b)$"?
Yes, but also:

$f \in C^{\infty}(a,b)$

RonL

7. Originally Posted by CaptainBlack
Yes, but also:

$f \in C^{\infty}(a,b)$

RonL
Thanks captain black, I want to be able to write stuff with as few words as possible (no sarcasm).

But by the way, this got me thinking, this notating would be nice to say things like,

"Let f be a eight times differentiable function" So instead of that could you say

"Let $f(x)\in{C^{k\geq{8}}}$ "

Also, can it apply to functions where $\mathbb{R}^{n>1}\to\mathbb{R}$

Can I say

$C_x^{\infty}\left(\mathbb{R}\right)$?

Or would no one know what that means, haha, I think I am inventing notation here.

8. Originally Posted by Mathstud28
Thanks captain black, I want to be able to write stuff with as few words as possible (no sarcasm).

But by the way, this got me thinking, this notating would be nice to say things like,

"Let f be a eight times differentiable function" So instead of that could you say

"Let $f(x)\in{C^{k\geq{8}}}$ "
Since:

$C^8 \supset C^9 \supset C^{10} \supset ...$

there is no need for this notation.

Also $C^8$ and the set of all functions eight times differentiable (on $(a,b)$ or whatever) are not the same thing since $C^8$ is the set of all $8$ times differentiable function with continuous $8$-th derivative.

RonL

9. Originally Posted by Mathstud28
...

But by the way, this got me thinking, this notating would be nice to say things like,

"Let f be a eight times differentiable function" So instead of that could you say

"Let $f(x)\in{C^{k\geq{8}}}$ "
not really, because there is another condition: $f^{(8)}(x)$ must also be continuous.

so i think, $f(x)\in{C^{k\geq{8}}}$ would mean $f^{(k\geq8)}(x)$ must also be continuous. so y don't you get the biggest k such that $f^k(x)$ is continuous with f is k times differentiable..

Originally Posted by Mathstud28
Also, can it apply to functions where $\mathbb{R}^{n>1}\to\mathbb{R}$
yes, like what TPH said, it can be defined in any open set, just define the open set in $\mathbb{R}^{n}$

Originally Posted by Mathstud28
Can I say

$C_x^{\infty}\left(\mathbb{R}\right)$?

Or would no one know what that means, haha, I think I am inventing notation here.
haven't seen this notation before..

10. Originally Posted by CaptainBlack
Since:

$C^8 \supset C^9 \supset C^{10} \supset ...$

there is no need for this notation.

Also $C^8$ and the set of all functions eight times differentiable (on $(a,b)$ or whatever) are not the same thing since $C^8$ is the set of all $8$ times differentiable function with continuous $8$-th derivative.

RonL
So what your saying is that since for a function to be $f(x)\in{C^{k\geq{8}}}$ it must be $f(x)\in{C^8}$ or in other words all functions of tenth order differentiablility are a subset of eight order or whatever.

So your saying that $f(x)\in{C^{n}}$ describes all functions who are differentiable at least n times, so this would include all functions that are differentiable $n+1,n+2,n+3,\cdots$ times?

11. if $f \in C^{n}$, then $f, f', f'', ..., f^{(n)}$ exists and $f^{(n)}$ is continuous..

$f \in C^{n+1}$ means $f, f', f'', ..., f^{(n)}, f^{(n+1)}$ exists and $f^{(n+1)}$ is continuous..

now, let $g \in C^{n+1}$. therefore, $g^{(n)}$ exists. $g^{(n)}$ must be continuous otherwise, $g^{(n)}$ will not be differentiable, hence $g^{(n+1)}$ will not exist.

therefore, $g \in C^{n}$, that is $C^{n+1} \subset C^{n}$

12. Originally Posted by kalagota
if $f \in C^{n}$, then $f, f', f'', ..., f^{(n)}$ exists and $f^{(n)}$ is continuous..

$f \in C^{n+1}$ means $f, f', f'', ..., f^{(n)}, f^{(n+1)}$ exists and $f^{(n+1)}$ is continuous..

now, let $g \in C^{n+1}$. therefore, $g^{(n)}$ exists. $g^{(n)}$ must be continuous otherwise, $g^{(n)}$ will not be differentiable, hence $g^{(n+1)}$ will not exist.

therefore, $g \in C^{n}$, that is $C^{n+1} \subset C^{n}$

So the operative phrasing here would be that $f(x)\in{C^{n}}$ implies that $f(x)$ is differentiable at least an n number of times, not at most.

13. yes.

14. Originally Posted by Mathstud28
So what your saying is that since for a function to be $f(x)\in{C^{k\geq{8}}}$ it must be $f(x)\in{C^8}$ or in other words all functions of tenth order differentiablility are a subset of eight order or whatever.

So your saying that $f(x)\in{C^{n}}$ describes all functions who are differentiable at least n times, so this would include all functions that are differentiable $n+1,n+2,n+3,\cdots$ times?
Something differentiable $m$ times with continuous $m$ th derivative is obviously continuously differentiable $n$-times for all $n \le m$, so $C^{m} \subset C^n, \ \forall n \le m.$

RonL

15. I will give two examples at my attempt to use this notation and please tell me if it is correct.

Say we are talking about L'hopital's and we are saying

" Let $\lim_{x\to{c}}\frac{f(x)}{g(x)}$ be indeterminate because either $\lim_{x\to{c}}f(x)=\lim_{x\to{c}}g(x)=0$ or $\lim_{x\to{c}}f(x)=\lim_{x\to{c}}g(x)=\infty$

Furthermore let $f(x)\in\mathcal{C}^1(c)$ and $g(x)\in\mathcal{C}^1(c)$ and then blah blah blah"

And in Rolle's Theroem

"Let $f(x)$ be continuous on $[a,b]$ and $f(x)\in\mathcal{C}^1(a,b)$ then ...."

Did I use it right?

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Integrating the area of a disc

1. Mar 14, 2008

Hakins90

1. The problem statement, all variables and given/known data

Finding the area of a disc by integration of rings.

2. Relevant equations

A ring of radius r and thickness dr has an area of $$2 \pi rdr$$.

3. The attempt at a solution

Why isn't it $$\pi (2rdr + (dr)^2)$$?

2. Mar 14, 2008

tiny-tim

I don't understand - why do you think it might be?

What would $$\pi (dr)^2$$ represent?

3. Mar 14, 2008

Dick

The exact area of the ring is pi*((r+dr)^2-r^2)=pi(2*r*dr+dr^2), indeed. But in the integration we are taking the limit of a sum of these where dr->0. The limit of the terms involving dr^2 will go to zero. You can ignore corrections involving higher powers of 'infinitesimals' like dr.

4. Mar 14, 2008

Hakins90

Oh dear... i posted too quickly.

I just realised why it isn't.

My old reasoning was that the ring's area = the area enclosed by the outside circumference - the area enclosed by the inside circumference.

$$A = \pi (r+dr)^2 - \pi r^2 = \pi (r^2 + 2rdr + (dr)^2 - r^2) = \pi (2rdr + (dr)^2)$$

I now realise that the ring's radius is measured from the centre of the width of the ring.

so
$$A = \pi (r+ \frac{1}{2} dr)^2 - \pi (r- \frac{1}{2} dr)^2 =\pi (r^2 +rdr + \frac{1}{4} (dr)^2 - r^2 + rdr - \frac{1}{4} (dr)^2) = 2 \pi rdr$$

EDIT: Oh you posted before i saw... We both have different reasons. Who is right?

5. Mar 14, 2008

Dick

If r and dr are the same, then those are the areas of two different rings. If dr is a finite number then you'd better pay attention to which is which. But,either calculation works for integration once you ignore higher powers of dr.

6. Mar 15, 2008

This reasoning is correct, in my opinion.
When I was in school, the explanation I got from my teacher was that while doing integration we add infinitesimally small quantities. The quantity dr is very small and so $$dr^2$$ will be even smaller......like $$0.000001^2=0.000000000001$$ , so we neglect the $$dr^2$$ thing from that expression.

7. Mar 15, 2008

HallsofIvy

Staff Emeritus
The simplest way to do the original problem is to say that the area of a small ring, of inner radius r and thickness dr is approximately $2\pi r$, the length of the ring, times dr, the thickness. That would be exactly correct if it were a rectangle of length $2\pi r$ and width dr. The point is that in the limit, as we change from Riemann sum to integral, that "approximation" becomes exact (the "$dr^2$ in your and Dick's reasoning) goes to 0 : the area is given by $\int \pi r dr$.

Can someone explain to my why this is a physics problem and not a mathematics problem?


The function is defined at a.In other words, point a is in the domain of f, ; The limit of the function exists at that point, and is equal as x approaches a from both sides, ; The limit of the function, as x approaches a, is the same as the function output (i.e. The Solution: We must show that $\lim_{h \to 0}\cos(a + h) = \cos(a)$ to prove that the cosine function is continuous. Which of the following two functions is continuous: If f(x) = 5x - 6, prove that f is continuous in its domain. Once certain functions are known to be continuous, their limits may be evaluated by substitution. A function is said to be differentiable if the derivative exists at each point in its domain. We can define continuous using Limits (it helps to read that page first):. limx→c f(x) = f(c) "the limit of f(x) as x approaches c equals f(c)" The limit says: The following are theorems, which you should have seen proved, and should perhaps prove yourself: Constant functions are continuous everywhere. 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If f(x) = x if x is rational and f(x) = 0 if x is irrational, prove that f is continuous … More formally, a function (f) is continuous if, for every point x = a:. When a function is continuous within its Domain, it is a continuous function.. More Formally ! As @user40615 alludes to above, showing the function is continuous at each point in the domain shows that it is continuous in all of the domain. If f(x) = 1 if x is rational and f(x) = 0 if x is irrational, prove that x is not continuous at any point of its domain. Proofs of the Continuity of Basic Algebraic Functions. To show that $f(x) = e^x$ is continuous at $x_0$, consider any $\epsilon>0$. Using the Heine definition, prove that the function $$f\left( x \right) = {x^2}$$ is continuous at any point $$x = a.$$ Solution. $\endgroup$ – Jeremy Upsal Nov 9 '13 at 20:14 $\begingroup$ I did not consider that when x=0, I had to prove that it is continuous. 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Furthermore, in the case of SU(2) the representations corresponding to upper and lower indices are equivalent. While the motion of ... To understand this better, take A apart into symmetric and antisymmetric parts: The symmetric part is called the strain-rate tensor. It is a real tensor, hence f Î±Î² * is also real. First, the vector space of finite games is decomposed into a symmetric subspace and an orthogonal complement of the symmetric subspace. For N>2, they are not, however. The result is An alternating form Ï on a vector space V over a field K, not of characteristic 2, is defined to be a bilinear form. P i A ii D0/. The N-way Toolbox, Tensor Toolbox, â¦ Yes. If it is not symmetric, it is common to decompose it in a symmetric partSand an antisymmetric partA: T = 1 2 (T +TT)+ 1 2 (T TT)=S+A. A tensor is a linear vector valued function defined on the set of all vectors . Thus, the rank of Mmust be even. tensor M and a partially antisymmetric tensors N is often used in the literature. The alternating tensor can be used to write down the vector equation z = x × y in suï¬x notation: z i = [x×y] i = ijkx jy k. (Check this: e.g., z 1 = 123x 2y 3 + 132x 3y 2 = x 2y 3 âx 3y 2, as required.) Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. In section 3 a decomposition of tensor spaces into irreducible components is introduced. We begin with a special case of the definition. Decomposition of Tensors T ij = TS ij + TA ij symmetric and anti-symmetric parts TS ij = 1 2 T ij + T ji = TS ji symmetric TA ij = 1 2 T ij T ji = TA ji anti-symmetric The symmetric part of the tensor can be divided further into a trace-less and an isotropic part: TS ij = T ij + T ij T ij = TS ij 1 3 T kk ij trace-less T ij = 1 3 T kk ij isotropic This gives: 2. In 3 dimensions, an antisymmetric tensor is dual to a vector, but in 4 dimensions, that is not so. Second, the potential-based orthogonal decompositions of two-player symmetric/antisymmetric â¦ Antisymmetric and symmetric tensors. 1.5) are not explicitly stated because they are obvious from the context. and a pair of indices i and j, U has symmetric and antisymmetric parts defined as: (antisymmetric) spin-0 singlett, while the symmetric part of the tensor corresponds to the (symmetric) spin-1 part. We show that the SA-decomposition is unique, irreducible, and preserves the symmetries of the elasticity tensor. Algebra is great fun - you get to solve puzzles! There is one very important property of ijk: ijk klm = Î´ ilÎ´ jm âÎ´ imÎ´ jl. Ask Question Asked 2 years, 2 months ago. An alternative, less well-known decomposition, into the completely symmetric part Sof C plus the reminder A, turns out to be irreducibleunder the 3-dimensional general linear group. DECOMPOSITION OF THE LORENTZ TRANSFORMATION MATRIX INTO SKEW-SYMMETRIC TENSORS. This means that traceless antisymmetric mixed tensor $\hat{T}^{[ij]}_{k}$ is equivalent to a symmetric rank-2 tensor. Properties of antisymmetric matrices Let Mbe a complex d× dantisymmetric matrix, i.e. This makes many vector identities easy to prove. The trace decomposition equations for tensors, symmetric in some sets of superscripts, and antisymmetric â¦ : Lehigh Univ., Bethlehem, Penna. Decomposition. In these notes, the rank of Mwill be denoted by 2n. THE INDEX NOTATION Î½, are chosen arbitrarily.The could equally well have been called Î± and Î²: vâ² Î± = n â Î²=1 AÎ±Î² vÎ² (âÎ± â N | 1 â¤ Î± â¤ n). Google Scholar; 6. Vector spaces will be denoted using blackboard fonts. Finally, it is possible to prove by a direct calculation that its Riemann tensor vanishes. (1.5) Usually the conditions for µ (in Eq. Each part can reveal information that might not be easily obtained from the original tensor. According to the Wiki page: ... Only now I'm left confused as to what it means for a tensor to have a spin-1 decomposition under SO(3) but that not describe the spin of the field in the way it is commonly refered to. ARTHUR S. LODGE, in Body Tensor Fields in Continuum Mechanics, 1974 (11) Problem. : USDOE â¦ A.2 Decomposition of a Tensor It is customary to decompose second-order tensors into a scalar (invariant) part A, a symmetric traceless part 0 A, and an antisymmetric part Aa as follows. What's the significance of this further decomposition? If so, are the symmetric and antrisymmetric subspaces separate invariant subspaces...meaning that every tensor product representation is reducible? Since the tensor is symmetric, any contraction is the same so we only get constraints from one contraction. Lecture Notes on Vector and Tensor Algebra and Analysis IlyaL. 3 Physical Models with a Completely Antisymmetric Torsion Tensor After the decomposition of the connection, we have seen that the metric g The symmetry-based decompositions of finite games are investigated. Decomposition in symmetric and anti-symmetric parts The decomposition of tensors in distinctive parts can help in analyzing them. The trace of the tensor S is the rate of (relative volume) expansion of the fluid. When defining the symmetric and antisymmetric tensor representations of the Lie algebra, is the action of the Lie algebra on the symmetric and antisymmetric subspaces defined the same way as above? Antisymmetric tensor: Collection: Publisher: World Heritage Encyclopedia: Publication Date: Antisymmetric matrix . [3] Alternating forms. A completely antisymmetric covariant tensor of order p may be referred to as a p-form, and a completely antisymmetric contravariant tensor may be referred to as a p-vector. Sponsoring Org. A completely antisymmetric covariant tensor of order p may be referred to as a p-form, and a completely antisymmetric contravariant tensor may be referred to as a p-vector. Sci. 440 A Summary of Vector and Tensor Notation A D1 3.Tr A/U C 0 A CAa D1 3 AÄ± ij CA ij CAa ij: (A.3) Note that this decomposition implies Tr 0 A D0. The bases of the symmetric subspace and those of its orthogonal complement are presented. Cl. A related concept is that of the antisymmetric tensor or alternating form. Irreducible decomposition and orthonormal tensor basis methods are developed by using the results of existing theories in the literature. 1 Definition; 2 Examples; 3 Symmetric part of a tensor; 4 Symmetric product; 5 Decomposition; 6 See also; 7 Notes; 8 References; 9 External links; Definition. Contents. gular value decomposition:CANDECOMP/PARAFAC (CP) decomposes a tensor as a sum of rank-one tensors, and the Tucker decomposition is a higher-order form of principal component analysis. Decomposition of Tensor (of Rank 3) We have three types of Young Diagram which have three boxes, namely, (21) , , and Symmetric Antisymmetric ??? Symmetric tensors occur widely in engineering, physics and mathematics. This decomposition, ... ^2 indicates the antisymmetric tensor product. Since det M= det (âMT) = det (âM) = (â1)d det M, (1) it follows that det M= 0 if dis odd. Use the Weyl decomposition \eqref{eq:R-decomp-1} for on the left hand side; Insert the E/B decomposition \eqref{eq:weyl-in-E-B} for the Weyl tensor on the left hand side; You should now have with free indices and no prefactor; I highly recommend using xAct for this calculation, to avoid errors (see the companion notebook). Viewed 503 times 7. 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Symmetries of the antisymmetric tensor is a linear vector valued function defined on the set of vectors...


Please, try EDU on Codeforces! New educational section with videos, subtitles, texts, and problems. ×

C. Make Good
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Let's call an array $a_1, a_2, \dots, a_m$ of nonnegative integer numbers good if $a_1 + a_2 + \dots + a_m = 2\cdot(a_1 \oplus a_2 \oplus \dots \oplus a_m)$, where $\oplus$ denotes the bitwise XOR operation.

For example, array $[1, 2, 3, 6]$ is good, as $1 + 2 + 3 + 6 = 12 = 2\cdot 6 = 2\cdot (1\oplus 2 \oplus 3 \oplus 6)$. At the same time, array $[1, 2, 1, 3]$ isn't good, as $1 + 2 + 1 + 3 = 7 \neq 2\cdot 1 = 2\cdot(1\oplus 2 \oplus 1 \oplus 3)$.

You are given an array of length $n$: $a_1, a_2, \dots, a_n$. Append at most $3$ elements to it to make it good. Appended elements don't have to be different. It can be shown that the solution always exists under the given constraints. If there are different solutions, you are allowed to output any of them. Note that you don't have to minimize the number of added elements!. So, if an array is good already you are allowed to not append elements.

Input

Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10\,000$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ $(1\le n \le 10^5)$ — the size of the array.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($0\le a_i \le 10^9$) — the elements of the array.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.

Output

For each test case, output two lines.

In the first line, output a single integer $s$ ($0\le s\le 3$) — the number of elements you want to append.

In the second line, output $s$ integers $b_1, \dots, b_s$ ($0\le b_i \le 10^{18}$) — the elements you want to append to the array.

If there are different solutions, you are allowed to output any of them.

Example
Input
3
4
1 2 3 6
1
8
2
1 1

Output
0

2
4 4
3
2 6 2

Note

In the first test case of the example, the sum of all numbers is $12$, and their $\oplus$ is $6$, so the condition is already satisfied.

In the second test case of the example, after adding $4, 4$, the array becomes $[8, 4, 4]$. The sum of numbers in it is $16$, $\oplus$ of numbers in it is $8$.


# Part of a proof that the product of an odd and even integers is even

I'm practicing for a test on Monday and I'm trying to do some proofs - but I'm not entirely sure if this is sufficient enough for the question.

"Prove that for all integers, m and n, if m is odd and n is even, then $m\cdot n$ is even."

These are the steps I've gone through - but I'm a little unsure on the result.

1. $m = 2k_1+1$ and $n = 2k_2$
2. $m\cdot n= (2k_1 + 1)(2k_2)$
3. $= 4k_1k_2+2k_2$
4. $=2(2k_1k_2+k_2)$

The part I'm unsure on is the last step, I wasn't sure if I was supposed to factor out the $k_2$ as well.

• That's fine. You could have jumped directly from step 2 to step 4. – Adriano Sep 20 '14 at 18:42
• Apart from the fact that you used equations, with no explanatory words, everything is fine. We can even skip part of the multiplication step. We have $m=2a+1$ and $n=2b$ for some integers $a$ and $b$. Thus $mn=nm=(2b)(2a+1)=(2)\left[(b)(2a+1)\right]$, so $mn$ is even. – André Nicolas Sep 20 '14 at 18:45

## 3 Answers

You've done all the work, and it's correct. Let me try to demonstrate how improve the style of your proof. This can be very important—graders are invariably tired and overworked, and lack of clarity is one of the most frequent causes of correct proofs being marked as incorrect on exams.

Additionally, it feels good for both reader and writer when the math is crisp and clear—it is an often-overlooked fact that the entire purpose of mathematical writing is to communicate ideas to other people.

Claim: If $m$ and $n$ are integers, and $n$ is even, then $mn$ is even.

Proof: Because $n$ is even, we can write $n=2l$, where $l$ is an integer. Then $mn=m(2l)=2(ml)$ is a multiple of $2$. In other words, $mn$ is even.

Notice that each step is justified—though if we had an extremely strict teacher, we might want to point out that we got $mn=m(2l)$ by multiplying both sides of $n=2l$ by $m$, and that $m(2k)=2(ml)$ follows from the associativity and commutativity of multiplication. But you're probably safe in omitting these details unless you've been told otherwise.

Also note that I didn't assume that $m$ is odd, because this makes the equations a little more complicated. Here is a more faithful recreation of your proof:

Claim: If $m$ is an odd integer and $n$ is an even integer, then $mn$ is even.

Proof: Because $m$ is odd and $n$ is even, we can write $m=2k+1$ and $n=2l$, where $k$ and $l$ are integers. Then $mn = (2k+1)(2l) = 2\cdot(l(2k+1))$ is a multiple of $2$, hence even.

Define $2k_1k_2+k_2=k_3$ as an integer number.

You could further formalize your text, if you want (it is always handy to write down the definitions or the theorems you use, it also makes it easier for others to read and make it look more complete): By definition: $$m \text{ even }\iff \exists k \in \mathbb Z \text{ such that }2k = m$$


# How to Count Squares!

Let me go grab a hamburger real quick...

Ok, I'm back.

How many squares are there in the 6$$\times$$4 grid below?

That's a reaallly good question!

Let's start by counting the smallest 1$$\times$$1 squares, this is just the same as counting the number of unit squares in a 6$$\times$$4 grid, there are $$6\times4=24$$ 1 by 1 squares in the grid.

Now let's move on to the 2$$\times$$2 squares, notice that counting the number of 2$$\times$$2 squares in a 6$$\times$$4 grid is just the same as counting the number of unit squares in a 5$$\times$$3 grid. So the number of 2 by 2 squares in the grid is $$5\times3=15$$.

Now the 3$$\times$$3 squares, similarly, counting the number of 3$$\times$$3 squares in a 6$$\times$$4 grid is just the same as counting the number of unit squares in a 4$$\times$$2 grid, which is $$4\times2=8$$.

And again the number of 4$$\times$$4 squares in a 6$$\times$$4 grid is equal to the number of unit squares in a 3$$\times$$1 grid, which is $$3\times1=3$$.

Add up all the number of squares together: $$24+15+8+3=50$$. Tada! We now have our answer! There are 50 squares in a 6$$\times$$4 grid.

Mmm... the hamburger is really good...

Back on topic, in general, what is the total number of squares in an $$a\times b$$ grid (where $$a$$ is the width of the grid and $$b$$ is the height of the grid), given $$a\geqslant b$$?

Again let's start from the 1$$\times$$1 squares, that's trivial, there's $$ab$$ of them.

Now moving on to the 2$$\times$$2 squares, the number of 2$$\times$$2 squares in an $$a\times b$$ grid is equal to the number of unit squares in an $$(a-1)\times(b-1)$$ grid.

Notice the pattern? Counting the number of $$n\times n$$ squares in an $$a\times b$$ grid is the same as counting the number of unit squares in an $$(a-n+1)\times(b-n+1)$$ grid.

The largest square that can contain in an $$a\times b$$ grid given that $$a\geqslant b$$ is $$b\times b$$.

Hence, the total number of squares in an $$a\times b$$ grid is $ab+(a-1)(b-1)+(a-2)(b-2)+\ldots+[a-(b-2)][b-(b-2)]+[a-(b-1)][b-(b-1)]$ Or $\sum_{i=0}^{b-1}{(a-i)(b-i)}$

This is ugly, we don't like sigma symbols sitting around, so why not we simplify this a little bit...

\begin{align} \sum_{i=0}^{b-1}{(a-i)(b-i)}&=\sum_{i=0}^{b-1}{[ab-(a+b)i+i^2]} \\&=ab^2-\frac{(a+b)b(b-1)}{2}+\frac{b(b-1)(2b-1)}{6} \\&=b\left[ab-\frac{ab-a+b^2-b}{2}+\frac{2b^2-3b+1}{6}\right] \\&=\frac{b}{6}\left[6ab-3ab+3a-3b^2+3b+2b^2-3b+1\right] \\&=\frac{b}{6}\left[3ab+3a-b^2+1\right] \\&=\frac{b(b+1)(3a-b+1)}{6} \end{align} BOOM! There we have it! *Round of applause* *Fireworks* *Pancakes*

The total number of squares in an $$a\times b$$ grid (where $$a$$ is the width of the grid and $$b$$ is the height of the grid), given $$a\geqslant b$$ is $\frac{b(b+1)(3a-b+1)}{6}$

If $$a<b$$, then we just swap the $$a$$ and $$b$$ around.

Special case: If $$a=b$$, the above equation becomes $\frac{a(a+1)(2a+1)}{6}$ which is the formula of the sum of squares from 1 to $$a$$.

Done! Now let me finish my burger...

This is one part of Quadrilatorics.

Note by Tan Kenneth
2 years, 1 month ago

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You should add this to the Brilliant wiki. Great note!

- 2 years, 1 month ago

Cool! Thanks for that note bro, you're awesome!

- 2 years, 1 month ago

Hope you finished your burger peacefully :P

- 2 years, 1 month ago

Thanks, I'm glad you like the note. Well unfortunately, I think my hamburger has become stale. XD

- 2 years, 1 month ago

Nice simple way of explaining complex situation. So many thanks.

- 1 year, 11 months ago

are you a robot, cos I need some real friends? Humanity is a lie, the computer generation is upon us. Support the cause m64^(1/2)

- 2 years, 1 month ago

No, i am 100.1% sure I'm not a robot.

- 2 years, 1 month ago

What a note @Tan Kenneth:)

- 2 years, 1 month ago

HEY tankenneth, you hyped?

- 2 years, 1 month ago

Oh yes I am! $$1+1=3$$

- 2 years, 1 month ago

جميلة

- 2 years, 1 month ago

Translation: beautiful!

- 2 years, 1 month ago


# Prove $\sum_{i=0}^n (-1)^{n-i} \binom{n+1}{i} (i+1)^n = (n+2)^n$

I found through simulations that

$$\sum_{i=0}^n (-1)^{n-i} \binom{n+1}{i} (i+1)^n = (n+2)^n$$

Is there any proof of this? I've tried to solve it by:

• Induction, but it gets too messy.
• Binomial theorem, but I got nowhere.

Any help with this is highly appreciated.

• I bet induction is not so bad in this case, but I'll try to find a more elegant explanation Aug 23, 2018 at 17:34
• Thanks @Exodd .. Would love to see it through induction. But any other method is also okay. Aug 23, 2018 at 17:35
• Notice that you can rewrite it as $$\sum_{i=0}^{n+1} (-1)^{n-i} \binom{n+1}{i} i^{n} =0$$ Aug 23, 2018 at 18:01

Let $[m]$ denote the set of the first $m$ positive integers.

Let $S$ be the set of functions from $[n]\to [n+2]$, and for $1\le i \le n+1$, let $S_i$ be the set of such functions whose range does not contain $i$. Obviously, $|S|=(n+2)^n$. Furthermore, the range of a function in $S$ will not contain at least two elements, since the range has at most $n$ elements. Therefore, some number $1\le i\le n+1$ must be missing from the range, so $$S=\bigcup_{i=1}^{n+1} S_i$$ Therefore, we can use the principle of inclusion-exclusion to count $|S|$ in terms the sizes of the $k$-way intersections $|S_{i_1}S_{i_2}\dots S_{i_k}|$. For any $k$, the intersection of $k$ of the sets $S_i$ consists of functions whose range does not contain a particular $k$ elements. The number of such functions is $(n+2-k)^n$. Therefore, using the inclusion-exclusion formula, $$|S|=(n+2)^n = \sum_{k=1}^{n+1} (-1)^{k+1}\binom{n+1}k(n+2-k)^n$$ The result follows by reversing the order of the summation, i.e. setting $k\leftarrow n+1-i$.

Posting a second answer because there is a completely different, more direct proof.

Note that subtracting the $(n+2)^n$ on the right over, this is equivalent to proving $$\sum_{i=0}^{n+1} (-1)^{n-i}\binom{n+1}i(i+1)^n\stackrel{?}=0$$ which after multiplying both sides by $(-1)^n$ equals $$\sum_{i=0}^{n+1} (-1)^{i}\binom{n+1}i(i+1)^n \stackrel{?}=0\tag{1}$$ To prove this, consider the polynomial $$\sum_{i=0}^{n+1} \binom{n+1}i(i+1)^nx^i \tag{2}$$ I claim that $(2)$ is the result of taking the polynomial $$\sum_{i=0}^{n+1} \binom{n+1}ix^i\tag{3}$$ and performing the following operation $n$ times: multiply the polynomial by $x$, then differentiate. Indeed, each summand of a polynomial looks like $a_i x^i$, and when you multiply by $x$ and differentiate, the result is $(i+1)a_ix^i$. Doing this $n$ times results in $(i+1)^na_ix^i$.

But the polynomial in $(3)$ is by the binomial theorem equal to $(1+x)^{n+1}$. Note that this has a zero of order $n+1$ at $x=-1$. Multiplying by $x$ does not change this. It is a standard result that if $f(x)$ has a zero of order $k$ at $x_0$, then $f'(x)$ has a zero of order $k-1$ at $x_0$. Therefore, since $(3)$ has a zero of order $n+1$, it follows that after $n$ differentiations (and some multiplications by $x$), it will still have a zero of order $1$. In particular, $(2)$ has a zero at $x=-1$, so the expression in $(1)$ equals zero.

• This is outstanding .. Thank you for the clear clarification. I loved both answers. Aug 23, 2018 at 18:37
• Nice answer (+1). May 18, 2019 at 8:17

Start from

$$\sum_{q=0}^n (-1)^{n-q} {n+1\choose q} (q+1)^n \\ = \sum_{q=0}^n (-1)^{n-q} {n+1\choose q} n! [z^n] \exp((q+1)z) \\ = n! [z^n] \exp(z) \sum_{q=0}^n (-1)^{n-q} {n+1\choose q} \exp(qz) \\ = - n! [z^n] \exp(z) \times (-1) \exp((n+1)z) \\ + n! [z^n] \exp(z) \sum_{q=0}^{n+1} (-1)^{n-q} {n+1\choose q} \exp(qz) \\ = n! [z^n] \exp((n+2)z) \\ - n! [z^n] \exp(z) \sum_{q=0}^{n+1} (-1)^{n+1-q} {n+1\choose q} \exp(qz) \\ = (n+2)^n \\ - n! [z^n] \exp(z) (\exp(z)-1)^{n+1}.$$

Now $\exp(z)-1 = z + \cdots$ and hence $(\exp(z)-1)^{n+1} = z^{n+1} +\cdots$ and therefore

$$[z^n] \exp(z) (\exp(z)-1)^{n+1} = 0$$

and we are left with

$$\bbox[5px,border:2px solid #00A000]{ (n+2)^n.}$$

This is a less clever algebraic proof by induction. We actually prove something stronger:

For any $\lambda>0$, we have $$\sum_i(-1)^i\binom{n+1}{i}(i+\lambda)^n=0$$ where $i$ runs through all integers and define $\binom{n}{i}=0$ if $i<0$ or $i>n$.

Proof. This is straight-forward for $n=0$. Suppose for $n=k$ the statement is true. Then \begin{align} \sum_i(-1)^i\binom{k+2}i(i+\lambda)^{k+1}&=\sum_i(-1)^i\binom{k+2}i\binom i1(i+\lambda)^k+\lambda\sum_i(-1)^i\binom{k+2}{i}(i+\lambda)^k \end{align} Note that $\binom{k+2}i\binom i1=\binom{k+2}1\binom{k+1}{i-1}$ and $\binom{k+2}i=\binom{k+1}i+\binom{k+1}{i-1}$. By induction hypothesis we obtain $\sum(-1)^i\binom{k+1}i(i+\lambda)^k=0$, thus \begin{align} \sum_i(-1)^i\binom{k+2}i(i+\lambda)^{k+1}&=(k+2+\lambda)\sum_i(-1)^i\binom{k+1}{i-1}(i+\lambda)^k\\ &=-(k+2+\lambda)\sum_i(-1)^i\binom{k+1}{i}(i+\lambda+1)^k\\ &=0 \end{align} where the last equality follows from induction hypothesis again.


# Math Help - Help, SVP.

1. ## Help, SVP.

Hey... if you could answer/explain any of the following (even one would be awesome), I'd be forever thankful.

1. A six-member working group to plan a student lounge is to be selected from five teachers and nine students. If the working group is randomly selected, what is the probability of the committee including at least two teachers.

2. Cha’tima has five white and six grey huskies in her kennel. If a wilderness expedition chooses a team of six sled dogs from Cha’tima’s kennel, what is the probability that the team will consist of:
all white huskies?
all grey huskies?
three of each colour?

3. A survey at a school asked students if they were ill with a cold or the flu during last months.
The results were as follows. None of the students had both a cold and the flu.
Cold Flu Healthy
Females 32 18 47
Males 25 19 38
Use these results to estimate the probability that:
a randomly selected student had a cold last month.
a randomly selected females students was healthy last month.
a randomly selected student who had the flu last month is male.
a randomly selected student had either a cold or the flu last month.

4. To get out of jail free in the board game MonopolyÒ, you have to roll doubles with a pair of dice. Determine the odds in favour of getting out of jail on your first or second roll.

5. If a survey on teenage reads of popular magazines shows that 38% subscribe to Teen Life and 47% subscribe to Cool Teen and 35% subscribe to neither magazine, what is the probability that a randomly selected teenager,
subscribes to both magazines?
subscribes to either one magazine or both?
subscribes to only one of the two magazines?

2. I got this for 2,3, and 4.

2)

Prob is 0 b/c only 5 whites are there
P(all grey) = 1/(11 6) = 1/462, ~ 0.0022
P(3 white & 3 grey) ((5 3) (6 3))/(11 6), 200/462, ~ 0.433

3)

P(cold)=57/179, =0.3184, ~ 31.8%
P(health | female) (47/179)/(97/179) = 47/97 = 0.4845, ~ 48.5%
P(male | flu) 19/37 = 0.5135, ~51.3%
44/82 = 0.5366, ~53.7%

Double on first roll = 6/36 = 1/6, ergo not rollin = 5/6
Ergo prob of rolling doubles on the second roll: 5/6 * 1/6 = 5/36
Prob of rolling double on first or second: 1/6 + 5/6 = 11/36
Ergo odds of getting out of jail on first or second roll are 11:25

Confirm?

And help with 1 and 5?

3. Originally Posted by acc
Confirm?

And help with 1 and 5?
I didn't get a calculator out to confirm final answers, but I agree with your work for 2,3,4 except for the following:

for 3b, I would simply write 47/97 without the initial (47/179)/(97/179) step.

for 3d, I get 94/179. Your answer is for a randomly chosen male student.

for 4, there's an assumption we have to make, and that is that we don't want to count the probability of getting doubles on both the first and second rolls (because presumably, rolling a double on the first roll will get us out of jail and we won't make the second roll). Assuming that, I agree with your answer. (But you made a typo by writing 5/6 in one spot where you meant 5/36.)

For problem 1, I would start with a "1" and subtract the following probabilities from it

P(0 teachers and 6 students)
P(1 teacher and 5 students)

Problem 5, you could draw a Venn diagram. Or using formal notation,

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

(Edit: fixed typo.)

Edit 2: The part in gray is incorrect. What you wrote for 4 is fine the way it is, other than the typo.

4. Hello, acc!

2. Cha’tima has five white and six grey huskies in her kennel.
If a wilderness expedition chooses a team of six sled dogs from Cha’tima’s kennel,
what is the probability that the team will consist of:

(a) all white huskies?
(b) all grey huskies?
(c) three of each colour?
There are: . ${11\choose6} \:=\:462$ possible selections.

$(a)\;P(\text{6 white}) \;=\;0$

$(b)\;P(\text{6 grey}) \;=\;\frac{{6\choose6}}{462} \;=\;\frac{1}{462}$

$(c)\;P(\text{3 white, 3 grey}) \;=\;\frac{{5\choose3}{6\choose3}}{462} \;=\;\frac{10\cdot20}{462} \;=\;\frac{100}{231}$

3. A survey at a school asked students
if they were ill with a cold or the flu during last month.

The results were as follows.
None of the students had both a cold and the flu.

. . $\begin{array}{c||c|c|c||c|}
& \text{Cold} & \text{Flu} & \text{Healthy } & \text{Total} \\ \hline\hline \text{Females} & 32 & 18 & 47 & 97 \\ \hline
\text{Males} & 25 & 19 & 38 & 82 \\ \hline \hline
\text{Total} & 57 & 37 & 85 & 179 \\ \hline \end{array}$

Use these results to estimate the probability that:

(a) a randomly selected student had a cold last month.
(b) a randomly selected female student was healthy last month.
(c) a randomly selected student who had the flu last month is male.
(d) a randomly selected student had either a cold or the flu last month.

$(a)\;P(\text{Cold}) \:=\:\frac{57}{179}$

$(b)\;P(\text{Female}\,\wedge\,\text{Healthy}) \;=\;\frac{47}{179}$

$(c)\;P(\text{Male}|\text{Flu}) \;=\;\frac{19}{37}$

$(d)\;P(\text{Cold} \vee\text{Flu}) \;=\;P(\text{Cold}) + P(\text{Flu}) \;=\;\frac{57}{179} + \frac{37}{170} \;=\;\frac{94}{179}$

5. Originally Posted by Soroban

$(a)\;P(\text{Cold}) \:=\:\frac{57}{179}$

$(b)\;P(\text{Female}\,\wedge\,\text{Healthy}) \;=\;\frac{47}{179}$

$(c)\;P(\text{Male}|\text{Flu}) \;=\;\frac{19}{37}$

$(d)\;P(\text{Cold} \vee\text{Flu}) \;=\;P(\text{Cold}) + P(\text{Flu}) \;=\;\frac{57}{179} + \frac{37}{170} \;=\;\frac{94}{179}$

Wow, I completely misread (c) and thought it said "a randomly selected student had the flu last month."

For (b), I'm wondering if it is a matter of interpretation. To get the probability you found, wouldn't it have better been stated "a randomly selected student is female and had the flu last month"?

Edit: I must be tired because I didn't even address 3c, I'll stop posting now and get some rest.

6. Originally Posted by acc
Hey... if you could answer/explain any of the following (even one would be awesome), I'd be forever thankful.

1. A six-member working group to plan a student lounge is to be selected from five teachers and nine students. If the working group is randomly selected, what is the probability of the committee including at least two teachers.

[snip]

5. If a survey on teenage reads of popular magazines shows that 38% subscribe to Teen Life and 47% subscribe to Cool Teen and 35% subscribe to neither magazine, what is the probability that a randomly selected teenager,
subscribes to both magazines?
subscribes to either one magazine or both?
subscribes to only one of the two magazines?
Duplicate posted (and answered) here: http://www.mathhelpforum.com/math-he...tml#post511545