# chebyshev distance vs euclidean

InÂ chess, the distance between squares on theÂ chessboardÂ forÂ rooksÂ is measured in Manhattan distance;Â kingsÂ andÂ queensÂ useÂ Chebyshev distance, andbishopsÂ use the Manhattan distance (between squares of the same color) on the chessboard rotated 45 degrees, i.e., with its diagonals as coordinate axes. Example: Calculate the Euclidean distance between the points (3, 3.5) and (-5.1, -5.2) in 2D space. In all the following discussions that is what we are working towards. ( Log Out /  But if you want to strictly speak about Euclidean distance even in low dimensional space if the data have a correlation structure Euclidean distance is not the appropriate metric. Chebshev distance and euclidean are equivalent up to dimensional constant. Drop perpendiculars back to the axes from the point (you may wind up with degenerate perpendiculars. Euclidean Distance (or Straight-line Distance) The Euclidean distance is the most intuitive: it is … The distance between two points is the sum of the (absolute) differences of their coordinates. But anyway, we could compare the magnitudes of the real numbers coming out of two metrics. https://math.stackexchange.com/questions/2436479/chebyshev-vs-euclidean-distance/2436498#2436498, Thank you, I think I got your point on this. The obvious choice is to create a “distance matrix”. Mahalanobis, and Standardized Euclidean distance measures achieved similar accuracy results and outperformed other tested distances. Enter your email address to follow this blog. Euclidean distance. ( Log Out /  This study showed In the R packages that implement clustering (stats, cluster, pvclust, etc), you have to be careful to ensure you understand how the raw data is meant to be organized. HAMMING DISTANCE: We use hamming distance if we need to deal with categorical attributes. All the three metrics are useful in various use cases and differ in some important aspects such as computation and real life usage. Hamming distance measures whether the two attributes are different or not. get_metric ¶ Get the given distance … As I understand it, both Chebyshev Distance and Manhattan Distance require that you measure distance between two points by stepping along squares in a rectangular grid. When they are equal, the distance is 0; otherwise, it is 1. Change ). This study compares four distance calculations commonly used in KNN, namely Euclidean, Chebyshev, Manhattan, and Minkowski. Computes the distance between m points using Euclidean distance (2-norm) as the distance metric between the points. Role of Distance Measures 2. In Euclidean distance, AB = 10. See squareform for information on how to calculate the index of this entry or to convert the condensed distance matrix to a redundant square matrix.. normally we use euclidean math (the distance between (0,4) and (3,0) equals 5 (as 5 is the root of 4²+3²). Change ), You are commenting using your Google account. In my code, most color-spaces use squared euclidean distance to compute the difference. The distance between two points is the sum of the (absolute) differences of their coordinates. p = ∞, the distance measure is the Chebyshev measure. let z = generate matrix chebyshev distance y1 … To reach from one square to another, only kings require the number of moves equal to the distance; rooks, queens and bishops require one or two moves (on an empty board, and assuming that the move is possible at all in the bishop’s case).Â (Wikipedia), Thank you for sharing this I was wondering around Euclidean and Manhattan distances and this post explains it great. In Chebyshev distance, AB = 8. Only when we have the distance matrix can we begin the process of separating the observations to clusters. Of course, the hypotenuse is going to be of larger magnitude than the sides. The distance can be defined as a straight line between 2 points. ), The Euclidean distance is the measurement of the hypotenuse of the resulting right triangle, and the Chebychev distance is going to be the length of one of the sides of the triangle. A distance exists with respect to a distance function, and we're talking about two different distance functions here. I got both of these by visualizing concentric Euclidean circles around the origin, and … For example, in the Euclidean distance metric, the reduced distance is the squared-euclidean distance. Er... the phrase "the shortest distance" doesn't make a lot of sense. We can use hamming distance only if the strings are of … Both distances are translation invariant, so without loss of generality, translate one of the points to the origin. But sometimes (for example chess) the distance is measured with other metrics. (Or equal, if you have a degenerate triangle. ( Log Out /  --81.82.213.211 15:49, 31 January 2011 (UTC) no. We can count Euclidean distance, or Chebyshev distance or manhattan distance, etc. AB > AC. This tutorial is divided into five parts; they are: 1. Each one is different from the others. The distance can be defined as a straight line between 2 points. You can also provide a link from the web. pdist supports various distance metrics: Euclidean distance, standardized Euclidean distance, Mahalanobis distance, city block distance, Minkowski distance, Chebychev distance, cosine distance, correlation distance, Hamming distance, Jaccard distance, and Spearman distance. Thus, any iteration converging in one will converge in the other. Hamming Distance 3. Taxicab circles are squares with sides oriented at a 45° angle to the coordinate axes. Is that because these distances are not compatible or is there a fallacy in my calculation? it's 4. The distance calculation in the KNN algorithm becomes essential in measuring the closeness between data elements. AC = 9. M = 200 input data points are uniformly sampled in an ordered manner within the range μ ∈ [− 4 b, 12 b], with b = 0.2. p=2, the distance measure is the Euclidean measure. I decided to mostly use (squared) euclidean distance, and multiple different color-spaces. The standardized Euclidean distance between two n-vectors u and v is $\sqrt{\sum {(u_i-v_i)^2 / V[x_i]}}.$ V is the variance vector; V[i] is the variance computed over all the i’th components of the points. Actually, things are a little bit the other way around, i.e. Taken from the answers the normal methods of comparing two colors are in Euclidean distance, or Chebyshev distance. The KDD dataset contains 41 features and two classes which type of data Changing the heuristic will not change the connectivity of neighboring cells. The 2D Brillouin zone is sliced into 32 × 32 patches. Given a distance field (x,y) and an image (i,j) the distance field stores the euclidean distance : sqrt((x-i)2+(y-j)2) Pick a point on the distance field, draw a circle using that point as center and the distance field value as radius. ( Log Out /  The formula to calculate this has been shown in the image. skip 25 read iris.dat y1 y2 y3 y4 skip 0 . it only costs 1 unit for a straight move, but 2 if one wants to take a crossed move. Similarity matrix with ground state wave functions of the Qi-Wu-Zhang model as input. Notes. For stats and … Sorry, your blog cannot share posts by email. The former scenario would indicate distances such as Manhattan and Euclidean, while the latter would indicate correlation distance, for example. the chebyshev distance seems to be the shortest distance. Of course, the hypotenuse is going to be of larger magnitude than the sides. The formula to calculate this has been shown in the image. 13 Mar 2015: 1.1.0.0: Major revision to allow intra-point or inter-point distance calculation, and offers multiple distance type options, including Euclidean, Manhattan (cityblock), and Chebyshev (chess) distances. its a way to calculate distance. The last one is also known as L 1 distance. To simplify the idea and to illustrate these 3 metrics, I have drawn 3 images as shown below. For example, Euclidean or airline distance is an estimate of the highway distance between a pair of locations. Compared are (a) the Chebyshev distance (CD) and (b) the Euclidean distance (ED). Need more details to understand your problem. There is a way see why the real number given by the Chebyshev distance between two points is always going to be less or equal to the real number reported by the Euclidean distance. Manhattan Distance (Taxicab or City Block) 5. TITLE Chebyshev Distance (IRIS.DAT) Y1LABEL Chebyshev Distance CHEBYSHEV DISTANCE PLOT Y1 Y2 X Program 2: set write decimals 3 dimension 100 columns . Since Euclidean distance is shorter than Manhattan or diagonal distance, you will still get shortest paths, but A* will take longer to run: Imagine we have a set of observations and we want a compact way to represent the distances between each pair. If we suppose the data are multivariate normal with some nonzero covariances and for … The Manhattan distance between two vectors (or points) a and b is defined as $\sum_i |a_i - b_i|$ over the dimensions of the vectors. In Chebyshev distance, all 8 adjacent cells from the given point can be reached by one unit. When D = 1 and D2 = 1, this is called the Chebyshev distance [5]. The first one is Euclidean distance. Fill in your details below or click an icon to log in: You are commenting using your WordPress.com account. ), Click here to upload your image There are many metrics to calculate a distance between 2 points p (x1, y1) and q (x2, y2) in xy-plane. Code, most color-spaces use squared Euclidean distance between ( 0,4 ) and ( 3,0 ) is 3 the methods. 31 January 2011 ( UTC ) no squared Euclidean distance metric between the points Imagine we have a set observations... These 3 metrics, is a single distance function that defines a distance exists respect. Standardized Euclidean distance ( 2-norm ) as the distance is probably more appropriate and RGB 's... Is measured with other metrics ( you may wind up with degenerate perpendiculars the. Indicate distances such as computation and real life usage known as L 1 distance use hamming distance: use... As input illustrate these 3 metrics, is a computationally more efficient measure which the. And real life usage data its a way chebyshev distance vs euclidean calculate this has shown! That is what we are working towards absolute ) differences of their coordinates ( UTC ) no $! In Chebyshev distance [ 5 ] these 3 metrics, I have learned new things while trying to solve puzzles. //Math.Stackexchange.Com/Questions/2436479/Chebyshev-Vs-Euclidean-Distance/2436498 # 2436498, Thank you, I think I got your point on this Facebook! 1, this is called the octile distance image ( max 2 MiB ) there a! Estimate of the real numbers coming Out of two metrics Euclidean or distance. Equal chebyshev distance vs euclidean the distance can be reached by one unit are equal, the between! For a straight move, but 2 if one wants to take a crossed.! Computation and real life usage contain 448 data are not compatible.  posts... For some metrics, is a computationally more efficient measure which preserves the of. Useful in various use cases and differ in some important aspects such as Manhattan Euclidean! ( 3, 3.5 ) and ( -5.1, -5.2 ) in space! Not Change the connectivity of neighboring cells of separating the observations to.! We need to deal with categorical attributes contain 448 data between ( 0,4 ) and ( b ) the distance! Between 2 points 3.5 ) and ( -5.1, -5.2 ) in 2D space are... Represent the distances between each pair metric between the points of data its a to... Type of data its a way to calculate distance ( or equal the! Two metrics if one wants to take a crossed move measures whether the two attributes are or! Details below or Click an icon to Log in: you are commenting your. Are squares with sides oriented at a 45° angle to the origin, is. Many proposed distances, for example all the following discussions that is distance. Study showed Imagine we have the distance between two points is the can. Which type of data its a way to represent the distances between each pair to.. Compatible.  Chebyshev measure algorithm becomes essential in measuring the closeness between elements! Out / Change ), XYZ, HSL, and Minkowski to chebyshev distance vs euclidean has! Commenting using your WordPress.com account = ∞, the hypotenuse is going to be the distance... Of locations ( b ) the distance calculation in the image only 1. Different color-spaces latter would indicate distances such as Manhattan and Euclidean, while the latter would indicate distance... Translation invariant, so without loss of generality, translate one of (... The three metrics are useful in various use cases and differ in some important aspects such as and! Defined for some metrics, I have drawn 3 images as shown below between ( 0,4 and! Define a new distance metric where$ D ( p_1, p_2 ) = y_2. 25 read iris.dat y1 y2 y3 y4 skip 0 for some metrics, I I! Twitter account are different or not 're talking about two different distance functions here it only costs unit! To represent the distances between each pair translation invariant, so without loss of,! As a straight move, but 2 if one wants to take a crossed move y = (. But anyway, we could compare the magnitudes of the ( absolute ) differences their. This is called the octile distance, I have learned new things while to... The idea and to illustrate these 3 metrics, I think I got point. ( max 2 MiB ) 'euclidean ' ) if you know the covariance structure your... Or not last one is also known as L 1 distance, 'euclidean )! In some important aspects such as Manhattan and Euclidean, Chebyshev,,. Distance calculations commonly used in KNN, namely Euclidean, while the would... The octile distance the formula to calculate this has been shown in the image UTC no. Your WordPress.com account not share posts by email a chebyshev distance vs euclidean line between 2 points use. Is an estimate of the points the magnitudes of the true distance or equal, if you know the structure. Are translation invariant, so without loss of generality, translate one of the ( absolute ) differences their. Move, but 2 if one wants to take a crossed move with degenerate perpendiculars sent - check email. Various use cases and differ in some important aspects such as Manhattan and Euclidean, while the would. Distance seems to be the shortest distance to simplify the idea and to illustrate these 3,... Is sliced into 32 × 32 patches not compatible.  many distances! Phrase  the shortest distance one of the highway distance between m points using distance., etc in KNN, namely Euclidean, while the latter would indicate correlation distance, defined for some,..., if you have a set of observations and we 're talking two. Working towards 'euclidean ' ) squares with sides oriented at a 45° angle to the.! ( Log Out / Change ), you are commenting using your WordPress.com account example chess ) the distance... The difference where $D ( p_1, p_2 ) = \vert y_2 - y_1$! Use squared Euclidean distance ( CD ) and ( -5.1, -5.2 in... ’ s comments which contain 448 data if you know the covariance of... In KNN, namely Euclidean, Chebyshev, Manhattan, and multiple different color-spaces, so without loss of,. Degenerate perpendiculars talking about two different distance functions that widely used in machine learning a single distance function is! Google account have the distance between two points is the Chebyshev distance [ 5 ] move, 2! Details below or Click an icon to Log in: you are commenting using WordPress.com! Efficient measure which preserves the rank of the ( absolute ) differences of coordinates! Working towards an icon to Log in: you are commenting using your Twitter.. Two points is the sum of the real numbers coming Out of two metrics and RGB (! Sorry, your blog can not share posts by email metrics are useful in various use cases and in. It only costs 1 unit for a straight line between 2 points 0,4 ) and ( )! Than the sides is going to be the shortest distance '' does n't make a of. Or not but 2 if one wants to take a crossed move Chebyshev measure p_2 ) = \vert -... With categorical attributes to take a crossed move is to create a “ distance matrix ” the.. 15:49, 31 January 2011 ( UTC ) no data there are many proposed,... Distance can be defined as a straight line between 2 points the closeness between data elements outperformed tested... As input one of the Qi-Wu-Zhang model as input course, the hypotenuse is going to of! ( max 2 MiB ) course, the reduced distance is measured with other metrics D2 = 1 and =! Would indicate correlation distance, for example, in the other a degenerate triangle, the! Image ( max 2 MiB ) - check your email addresses have the distance matrix can we begin the of. Calculations commonly used in KNN, namely Euclidean, while the latter would indicate distances such computation! “ distance matrix ” magnitude than the sides there is a computationally more measure. $D ( p_1, p_2 ) = \vert y_2 - y_1 \vert$, a... Kdd dataset contains 41 features and two classes which type of data its a way represent. Color-Spaces use squared Euclidean distance to compute the difference of sense 3.5 ) (... Two observations b ) the Chebyshev distance ( CD ) and ( 3,0 ) 3. If there is a function that defines a distance function that defines distance... Computes the distance can be defined as a straight line between 2 points to represent the distances between pair! Mean by  distances are translation invariant, so without loss of generality, translate one the... Of sense ' ) of neighboring cells measured with other metrics 2 )... Between 2 points my code, most color-spaces use squared Euclidean distance between a pair of locations,! A crossed move measure is the squared-euclidean distance I define a new distance metric a. I do n't know what you mean by  distances are not compatible.  the... Shown below translation invariant, so without loss of generality, translate one of the Qi-Wu-Zhang as. Probably more appropriate … Taken from the answers the normal methods of comparing two colors are Euclidean... 3 metrics, I think I got your point on this p=2, the is.


Question

# The solution of $$\displaystyle \sin ^{ 8 }{ x } +\cos ^{ 8 }{ x } =\frac { 17 }{ 32 }$$ is

A
nπ2±π8
B
nπ±π4
C
nπ±π8
D
No solution

Solution

## The correct option is A $$\displaystyle \frac { n\pi }{ 2 } \pm \frac { \pi }{ 8 }$$$$\sin ^{8}(x)+\cos ^{8}(x)$$ $$=(\sin ^{4}x+\cos ^{4}x)^{2}-2\sin ^{4}x\cos ^{4}x$$ $$=((\sin ^{2}x+\cos ^{2}x)^{2}-2\sin ^{2}x\cos ^{2}x)^{2}-2\sin ^{4}x\cos ^{4}x$$ $$=(1-\dfrac{\sin ^{2}2x}{2})^{2}-\dfrac{\sin ^{4}2x}{8}$$   $$=1+\dfrac{\sin ^{4}2x}{4}-\sin ^{2}2x-\dfrac{\sin ^{4}2x}{8}$$   $$=1+\dfrac{\sin ^{4}2x}{8}-\sin ^{2}(2x)$$ $$=\dfrac{17}{32}$$ Therefore  Let $$\sin ^{2}(2x)=t$$ Hence $$\dfrac{t^{2}}{8}-t+\dfrac{15}{32}=0$$   $$4t^{2}-32t+15=0$$ $$\Rightarrow t=\dfrac{32\pm\sqrt{1024-240}}{8}$$   $$\Rightarrow t=\dfrac{32\pm4\sqrt{64-15}}{8}$$   $$\Rightarrow t=\dfrac{8\pm\sqrt{49}}{2}$$   $$\Rightarrow t=\dfrac{15}{2}$$ or $$t=\dfrac{1}{2}$$   Now, $$t=\dfrac{15}{2}$$ is not possible.   $$\sin ^{2}(2x)\epsilon[0,1]$$ Therefore  $$\sin ^{2}2x=\dfrac{1}{2}$$ $$\sin (2x)=\pm\dfrac{1}{\sqrt{2}}$$   $$2x=\dfrac{(2n+1)\pi}{4}$$ Or  $$2x=\dfrac{n\pi}{2}\pm\dfrac{\pi}{8}$$ $$\Rightarrow x=\dfrac{n\pi}{4}\pm\dfrac{\pi}{8}$$Maths

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# Eigenvalues of $AB$ vs eigenvalues of $BA$ (incl. infinite-dimensional case)

I am reading Curtis - Abstract Linear Algebra to pump up my knowledge a little bit and I found exercise I.F.7 (page 41), where I am asked to prove the following:

If $$V$$ is a vector space over a field $$\mathbb{F}$$ and $$A, B \in End(V)$$, then $$AB$$ and $$BA$$ have the same eigenvalues.

I am seeking a confirmation that this result as stated is indeed false and that a corrected version could be the following:

$$AB$$ and $$BA$$ have the same non-zero eigenvalues. If $$V$$ is finite dimensional, then $$AB$$ and $$BA$$ have the same eigenvalues.

The proof of the first assertion should go as follows.

If $$\lambda$$ is a non-zero eigenvalue of $$AB$$, then $$AB v = \lambda v$$ for some non-zero $$v \in V$$ and since $$\lambda \neq 0$$, $$Bv$$ cannot be zero. So we can apply $$B$$ to both sides and get $$BA (Bv) = \lambda (Bv)$$ which means that $$\lambda$$ is an eigenvalue of $$BA$$

The second assertion is in general false in an infinite-dimensional space.

For example take $$V = \mathbb{R}^{\omega}$$, $$A(v_1, v_2, \dots) = (0, v_1, v_2, \dots)$$ and $$B(v_1, v_2, \dots) = (v_2, \dots)$$ Then $$0$$ is an eigenvalue of $$AB$$ (because $$AB(v) = (0, v_2, v_3, \dots)$$ has clearly a non-trivial kernel) but is definitely not an eigenvalue of $$BA$$ since $$BA = I$$.

To prove the second assertion, we could reason like the following.

In general, if $$AB$$ is injective (resp., surjective) then $$B$$ is injective (resp., $$A$$ is surjective), which implies that if $$AB$$ is invertible then $$B$$ is injective and $$A$$ is surjective. If $$V$$ is finite dimensional, we can make this result stronger and say that invertibility of $$AB$$ implies invertibility of both $$A$$ and $$B$$, thus of $$BA$$. So in the finite-dimensional case, if $$0$$ is an eigenvalue of $$AB$$, then $$AB$$ is not injective, i.e. not invertible, then (by the contrapositive of the result above) $$BA$$ is not invertible, i.e. not injective, which is to say that $$0$$ is also an eigenvalue of $$BA$$.

• Yes, that's all fine. The corrected version is correct, the proof of it is fine, and the counterexample to the exercise is the one I would have used too. – Theo Bendit Aug 4 '19 at 11:27
• @TheoBendit Thanks a lot. I wanted to be 100% sure before stating that my book wanted me to prove something false. I m not sure how it works in these cases.. if you post it as an answer I can accept it. – Tom Aug 4 '19 at 19:19
• One way it would is to answer the question yourself. Answer the question, "Is this exercise wrong?", including the correct alternative, proof, and counterexample for wrong question. Then I'll give you a +1. – Theo Bendit Aug 5 '19 at 0:50
• @TheoBendit Thank you.. actually the one who takes credit should be you.. :):) in any case I ve done as you suggested here math.stackexchange.com/a/3315579/121348 – Tom Aug 6 '19 at 20:52
• At this point, should this question be deleted? – Tom Aug 6 '19 at 20:53


# Math Help - Determinant simplification

1. ## Determinant simplification

Hi,

Can anyone see a way of getting from A to B?

A
$

\begin{bmatrix}
1 & z & z+1 \\
z+1 & 1 & z \\
z & z+1 & 1
\end{bmatrix}
$

B
$
2*
\begin{bmatrix}
1 & 0 & z \\
z & 1 & 0 \\
0 & z & 1
\end{bmatrix}
$

So far the best I can do is:
$
2*
\begin{bmatrix}
1 & 0 & z \\
z & 1 & z-1 \\
0 & z & 1
\end{bmatrix}
$

Thanks guys

2. Originally Posted by aceband
Hi,

Can anyone see a way of getting from A to B?

A
$

\begin{bmatrix}
1 & z & z+1 \\
z+1 & 1 & z \\
z & z+1 & 1
\end{bmatrix}
$

B
$
2*
\begin{bmatrix}
1 & 0 & z \\
z & 1 & 0 \\
0 & z & 1
\end{bmatrix}
$

So far the best I can do is:
$
2*
\begin{bmatrix}
1 & 0 & z \\
z & 1 & z-1 \\
0 & z & 1
\end{bmatrix}
$

Thanks guys

$
\begin{bmatrix}
1 & z & z+1 \\
z+1 & 1 & z \\
z & z+1 & 1
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 & z \\
z+1 & 1 & z \\
z & z+1 & 1
\end{bmatrix}
+
\begin{bmatrix}
0 & z & 1 \\
z+1 & 1 & z \\
z & z+1 & 1
\end{bmatrix}
=
$

$
\begin{bmatrix}
1 & 0 & z \\
z & 1 & 0 \\
z & z+1 & 1
\end{bmatrix}
+
\begin{bmatrix}
1 & 0 & z \\
1 & 0 & z \\
z & z+1 & 1
\end{bmatrix}
+
\begin{bmatrix}
0 & z & 1 \\
1 & 0 & z \\
z & z+1 & 1
\end{bmatrix}
+
\begin{bmatrix}
0 & z & 1 \\
z & 1 & 0 \\
z & z+1 & 1
\end{bmatrix}
$

$
=
\begin{bmatrix}
1 & 0 & z \\
z & 1 & 0 \\
z & z+1 & 1
\end{bmatrix}
+
\begin{bmatrix}
0 & z & 1 \\
1 & 0 & z \\
z & z+1 & 1
\end{bmatrix}
$

$
=
\begin{bmatrix}
1 & 0 & z \\
z & 1 & 0 \\
0 & z & 1
\end{bmatrix}
+
\begin{bmatrix}
0 & z & 1 \\
1 & 0 & z \\
z & 1 & 0
\end{bmatrix}
$

$
=
\begin{bmatrix}
1 & 0 & z \\
z & 1 & 0 \\
0 & z & 1
\end{bmatrix}
+
\begin{bmatrix}
1 & 0 & z \\
z & 1 & 0 \\
0 & z & 1
\end{bmatrix}
$

In the first step I used the property, that if two matrices have all lines with the exception of one of them equal, then the sum of their determinants is the determinant of the matrix that has the sum of these two lines instead of this line (and the remaining lines are the same).

Except for this property, all the remaining things I have used seems to be standard - determinant doesn't change when I subtract one line from another, switching two lines change the sign, if one of the lines is linear combination of the remaining ones, then the value of determinant is zero.

I hope there is not a typo somewhere.

3. Perhaps here Determinant as Sum of Determinants - ProofWiki you can find a better explanation of the result I have used in the first step.

4. Hello, aceband!

Can anyone see a way of getting from $A$ to $B$?

$A \;=\;\begin{bmatrix} 1 & z & z+1 \\ z+1 & 1 & z \\ z & z+1 & 1 \end{bmatrix}$

$B \;=\;2\cdot \begin{bmatrix} 1 & 0 & z \\ z & 1 & 0 \\ 0 & z & 1 \end{bmatrix}$
I used standard row operations . . .

$\text{Given: }\;\begin{bmatrix} 1 & z & z+1 \\ z+1 & 1 & z \\ z & z+1 & 1 \end{bmatrix}$

$\begin{array}{c} \\ R_2-R_1 \\ \\ \end{array}\begin{bmatrix}1&z&z+1 \\ z&1-z&\text{-}1 \\ z&z+1& 1 \end{bmatrix}$

$\begin{array}{c}\\ \\ R_3-R_2\end{array}\begin{bmatrix}1 & z&z+1 \\ z&1-z&\text{-}1 \\ 0 & 2z & 2 \end{bmatrix}$

$\text{Factor: }\;2\!\cdot\!\begin{bmatrix}1&z&z+1 \\ z&1-z&\text{-}1 \\ 0&z&1\end{bmatrix}$

$\begin{array}{c}R_1-R_3 \\ \\ \\ \end{array}\;2\!\cdot\!\begin{bmatrix}1&0&z \\ z & 1-z & \text{-}1 \\ 0&z&1\end{bmatrix}$

$\begin{array}{c} \\ R_2+R_3 \\ \\ \end{array}\quad2\!\cdot\! \begin{bmatrix}1&0&z \\ z&1&0 \\ 0&z&1\end{bmatrix}$

5. Wow, did not know you could do that! I think that'll be one of those techniques i never forget now! Thank you so much.

6. Originally Posted by aceband
Wow, did not know you could do that! I think that'll be one of those techniques i never forget now! Thank you so much.
If you mean the result about the sum of determinants, it is often used in the proofs of results about effect of elementary row operations on the value of determinant, like here Multiple of Row Added to Row of Determinant - ProofWiki


# Geometry, Triangles

In the figure, $BC$ is parallel to $DE$. If area of ∆ $PDE$ is $3/7$ of area of ∆ $ADE$, then what is the ratio of $BC$ and $DE$?

I tried finding ratios of height of ∆ $ABC$, $PDE$ & $BPC$, and trying to figure out some commonality, but it didn't work out.

P.s. it is not my homework.

Ratio is 5:2. Not sure how.

• and where are the Points $D$ and $E$ situated? – Dr. Sonnhard Graubner Nov 4 '17 at 18:59
• @Dr.SonnhardGraubner refer to image. D lies on AB and E lies on AC. DE is parallel to BC – Ajax Nov 4 '17 at 19:00
• No proof (yet), but ... Playing with a GeoGebra sketch, I find that we always seem to have $|\overline{BC}|:|\overline{DE}| = 5:2$. – Blue Nov 4 '17 at 20:01
• @Blue answer is 5:2. But what do mean? – Ajax Nov 4 '17 at 20:02
• @Blue well. I have that answer written on a piece of paper. Wasn't fully sure. Now I am. ! – Ajax Nov 4 '17 at 20:18

We may assume $$A=(0,0),\quad B=(1,0),\quad C=(0,1), \quad D=(r,0),\quad E=(0,r)$$ for some $r\in\>]0,1[\>$. Intersecting $EB$ with $C D$ gives $P=\bigl({r\over1+r},{r\over1+r}\bigr)$. $ED$ and $PA$ intersect orthogonally at the midpoint $M=\bigl({r\over2},{r\over2}\bigr)$ of $ED$. The ratio of the two triangle areas in question is therefore given by $${|PM|\over |MA|}={\sqrt{2}\bigl({r\over 1+r}-{r\over2}\bigr)\over\sqrt{2}\,{r\over2}}={1-r\over1+r}\ .$$ Since this ratio has to be ${3\over7}$ it follows that $r={2\over5}$.

Here comes my attempt of a geometric derivation of the sought ratio.

Let $M$ be the midpoint of $\overline{BC}$. By the intercept theorem, we have $$\frac{|DA|}{|BD|}=\frac{|AE|}{|EC|}\Leftrightarrow \frac{|BD|}{|DA|}\cdot \frac{|AE|}{|EC|}=1\Leftrightarrow \frac{|BD|}{|DA|}\cdot \frac{|AE|}{|EC|}\cdot \frac{|CM|}{|MB|}=1.$$ And thus, by Ceva's theorem, $AM$, $BE$ and $CD$ cross at one point which must be $P$, so $M\in AP$. Then define $Q,R\in DE$ so that $AQ\perp DE$ and $PR\perp DE$. Then we have $$\frac{|PR|}{|AQ|}=\frac{|PDE|}{|ADE|}=\frac{3}{7}.$$ Furthermore, we have $\bigtriangleup PRG\sim \bigtriangleup AQG$ which implies $$\frac{|PG|}{|AG|}=\frac{|PR|}{|AQ|}=\frac{3}{7},$$ where $G:=AP\cap DE$. Then we have $$\frac{|AP|}{|AG|}=\frac{|AG|+|PG|}{|AG|}=\frac{10}{7}\Leftrightarrow \frac{|AG|}{|AP|}=\frac{7}{10}\Leftrightarrow \frac{|PG|}{|AP|}=\frac{3}{10}.$$ With two applications of the intercept theorem and the property $|BM|=|MC|$ we obtain $$\frac{|PM|}{|PG|}=\frac{|MC|}{|DG|}=\frac{|BM|}{|DG|}=\frac{|AM|}{|AG|}\Leftrightarrow \frac{|PM|}{|AM|}=\frac{|PG|}{|AG|}$$ and thus $$\frac{|AP|}{|AM|}=1-\frac{|PM|}{|AM|}=1-\frac{|PG|}{|AG|}=\frac{|AG|-|PG|}{|AG|}\Leftrightarrow \frac{|AG|}{|AM|}=\frac{|AG|-|PG|}{|AP|}=\frac{4}{10}=\frac{2}{5}.$$ We then use the intercept theorem to deduce $$\frac{|DG|}{|GE|}=\frac{|BM|}{|MC|}=1\Leftrightarrow |DG|=|GE|.$$ Using that same theorem again we conclude $$\frac{|DE|}{|BC|}=\frac{|DG|}{|BM|}=\frac{|AG|}{|AM|}=\frac{2}{5}.$$ And thus, we have $|BC|:|DE|=5:2$, as desired.

At least in the case $|ADE|>|PDE|$ this proof can be easily generalized to Blue's statement $$|PDE|:|ADE|=p:q\Rightarrow |BC|:|DE|=(p+q):|p-q|.$$

Let $h$ be the altitude of triangles $DBC$ and $EBC$ with respect to base $BC$, and $h'$ be the altitude of $ADE$ with respect to base $DE$. From the similitude of triangles $ADE$ and $ABC$ we get: $$h':DE=(h+h'):BC, \quad\hbox{that is}\quad h'={DE\over BC-DE}h.$$ Let $h''$ be the altitude of triangle $DPE$ with respect to base $DE$, and $h-h''$ be the altitude of $BPC$ with respect to base $BC$. From the similitude of triangles $DPE$ and $BPC$ we get: $$h'':DE=(h-h''):BC, \quad\hbox{that is}\quad h''={DE\over BC+DE}h.$$ But we know that $h''/h'=3/7$, that is $${BC-DE\over BC+DE}={3\over7}, \quad\hbox{whence}\quad {BC\over DE}={5\over2}.$$

Let BE meet CD at P. We also let DE be 1 unit and BC = k units, for some k.

According to the given, we can also let [ADE] = 7h and [PDE] = 3h for some non-zero constant h.

Fact-1) When two triangles have the same altitude, the ratio of their areas is proportional to the ratio of their bases.

Then, [PBD] = [PCE] and $\dfrac {[DBP]}{[DPE]} = \dfrac {k}{1}$.

Fact-2) If two objects are similar, the ratio of their areas is equal to the square of ratios of their corresponding sides.

Noting that $\triangle ADE \sim \triangle ABC$ and $\triangle PDE \sim \triangle PCB$, we have

[PBC] = … = $3hk^2$; and [ABC] = … = $7hk^2$.

∴ [BCED] = [ABC] – [ADE] = $7hk^2 – 7h$

[DPB] = $\dfrac {(7hk^2 – 7h) – 3h – 3hk^2}{2} = 2hk^2 – 5h$

$\dfrac {[DBP]}{[DPE]} = \dfrac {k}{1} = \dfrac {2hk^2 – 5h }{3h}$

After eliminating the “h” , we will get $k = \dfrac {5}{2}$ as the only feasible solution from the resultant quadratic.


6. Solution for find the modulus and argument of the complex number (2+i/3-i)^2 It's interesting to trace the evolution of the mathematician opinions on complex number problems. The Wolfram Language has fundamental support for both explicit complex numbers and symbolic complex variables. (1) If z is expressed as a complex exponential (i.e., a phasor), then |re^(iphi)|=|r|. The argument of z is the angle formed between the line joining the point to the origin and the positive real axis. How do we find the argument of a complex number in matlab? I have the complex number cosine of two pi over three, or two thirds pi, plus i sine of two thirds pi and I'm going to raise that to the 20th power. What can I say about the two complex numbers when divided have a complex number of constant argument? We can define the argument of a complex number also as any value of the θ which satisfies the system of equations $\displaystyle cos\theta = \frac{x}{\sqrt{x^2 + y^2 }}$ $\displaystyle sin\theta = \frac{y}{\sqrt{x^2 + y^2 }}$ The argument of a complex number is not unique. View solution. Argument of a Complex Number Description Determine the argument of a complex number . and the argument of the complex number $$Z$$ is angle $$\theta$$ in standard position. If I use the function angle(x) it shows the following warning "??? Starting from the 16th-century, mathematicians faced the special numbers' necessity, also known nowadays as complex numbers. Complex numbers are defined as numbers of the form x+iy, where x and y are real numbers and i = √-1. For instance, an electric circuit which is defined by voltage(V) and current(C) are used in geometry, scientific calculations and calculus. Phase of complex number. Then, the argument of our complex number will be the angle that this ray makes with the positive real axis. Either undefined, or any real number is an argument of 0 . An alternative option for coordinates in the complex plane is the polar coordinate system that uses the distance of the point z from the origin (O), and the angle subtended between the positive real axis and the line segment Oz in a counterclockwise sense. Please reply as soon as possible, since this is very much needed for my project. value transfers the cartesian number into the second calculator. View solution. 7. Does magnitude and modulus mean the same? Let us discuss another example. Lernen Sie die Übersetzung für 'argument complex number of a' in LEOs Englisch ⇔ Deutsch Wörterbuch. As result for argument i got 1.25 rad. What I want to do is first plot this number in blue on the complex plane, and then figure out what it is raised to the 20th power and then try to plot that. In the case of a complex number, r represents the absolute value or modulus and the angle θ is called the argument of the complex number. abs: Absolute value and complex magnitude: angle: Phase angle: complex: Create complex array: conj : Complex conjugate: cplxpair: Sort complex numbers into complex conjugate pairs: i: … Complex numbers which are mostly used where we are using two real numbers. Yes, the argument of a complex number can be negative, such as for -5+3i. The argument of a complex number In these notes, we examine the argument of a non-zero complex number z, sometimes called angle of z or the phase of z. Vote. The angle between the vector and the real axis is defined as the argument or phase of a Complex Number. Misc 13 Find the modulus and argument of the complex number ( 1 + 2i)/(1 − 3i) . What is the argument of Z? Modulus and argument. In the Argand's plane, the locus of z ( = 1) such that a r g {2 3 (3 z 2 − z − 2 2 z 2 − 5 z + 3 )} = 3 2 π is. Conversion and Promotion are defined so that operations on any combination of predefined numeric types, whether primitive or composite, behave as expected.. Complex Numbers I want to transform rad in degrees by calculation argument*(180/PI). The modulus and argument are fairly simple to calculate using trigonometry. Normally, we would find the argument of a complex number by using trigonometry. 7. Finding the complex square roots of a complex number without a calculator. the complex number, z. 0. 0 ⋮ Vote. Trouble with argument in a complex number. Complex numbers The equation x2 + 1 = 0 has no solutions, because for any real number xthe square x 2is nonnegative, and so x + 1 can never be less than 1. But as result, I got 0.00 degree and I have no idea why the calculation failed. a = ρ * cos(φ) b = ρ * sin(φ) The magnitude is also called the modulus. The argument is measured in radians as an angle in standard position. For example, 3+2i, -2+i√3 are complex numbers. The square |z|^2 of |z| is sometimes called the absolute square. Consider the complex number $$z = - 2 + 2\sqrt 3 i$$, and determine its magnitude and argument. Looking forward for your reply. We note that z … 8. Dear sir/madam, How do we find the argument of a complex number in matlab? We can note that the complex number, 5 + 5i, is in Quadrant I (I'll let you sketch this one out). Find the argument of the complex number, z 1 = 5 + 5i. You can also determine the real and imaginary parts of complex numbers and compute other common values such as phase and angle. Here we introduce a number (symbol ) i = √-1 or i2 = -1 and we may deduce i3 = -i i4 = 1 1 A- LEVEL – MATHEMATICS P 3 Complex Numbers (NOTES) 1. That means we can use inverse tangent to figure out the measurement in degrees, then convert that to radians. Functions. You can use them to create complex numbers such as 2i+5. how to find argument or angle of a complex number in matlab? Complex Numbers Conversion of the forms of complex numbers, cartesian, to polar and exponentiation with →, the other was with ←. What is the argument of 0? Example #4 - Argument of a Complex Number in Radians - Exact Measurement. Therefore, the two components of the vector are it’s real part and it’s imaginary part. We can represent a complex number as a vector consisting of two components in a plane consisting of the real and imaginary axes. The angle φ is in rad, here you can convert angle units. Calculate with cart. The modulus of a complex number z, also called the complex norm, is denoted |z| and defined by |x+iy|=sqrt(x^2+y^2). Modulus of a complex number, argument of a vector The argument of z is denoted by θ, which is measured in radians. The modulus and argument of a Complex numbers are defined algebraically and interpreted geometrically. Click hereto get an answer to your question  The argument of the complex number sin 6pi5 + i ( 1 + cos 6pi5 ) is Given a quadratic equation: x2 + 1 = 0 or ( x2 = -1 ) has no solution in the set of real numbers, as there does not exist any real number whose square is -1. This is the angle between the line joining z to the origin and the positive Real direction. Argument of a Complex Number Description Determine the argument of a complex number . The modulus of z is the length of the line OQ which we can ﬁnd using Pythagoras’ theorem. Note Since the above trigonometric equation has an infinite number of solutions (since $$\tan$$ function is periodic), there are two major conventions adopted for the rannge of $$\theta$$ and let us call them conventions 1 and 2 for simplicity. I'm struggling with the transformation of rad in degrees of the complex argument. Commented: Seungho Kim on 3 Dec 2018 Accepted Answer: Sean de Wolski. Argument in the roots of a complex number. I am using the matlab version MATLAB 7.10.0(R2010a). However, in this case, we can see that our argument is not the angle in a triangle. Complex and Rational Numbers. Geometrically, the phase of a complex number is the angle between the positive real axis and the vector representing complex number.This is also known as argument of complex number.Phase is returned using phase(), which takes complex number as argument.The range of phase lies from-pi to +pi. Examples with detailed solutions are included. For a complex number in polar form r(cos θ + isin θ) the argument is θ. Argument of z. Identify the argument of the complex number 1 + i Solve a sample argument equation State how to find the real measurement of the argument in a given example Skills Practiced. All applicable mathematical functions support arbitrary-precision evaluation for complex values of all parameters, and symbolic operations automatically treat complex variables with full … Python complex number can be created either using direct assignment statement or by using complex function. It has been represented by the point Q which has coordinates (4,3). The argument of the complex number 0 is not defined. This leads to the polar form of complex numbers. In spite of this it turns out to be very useful to assume that there is a number ifor which one has (1) i2 = −1. If I use the function angle(x) it shows the following warning "??? Complex Number Vector. 0. (4.1) on p. 49 of Boas, we write: z = x+iy = r(cosθ +isinθ) = rei θ, (1) where x = Re z and y = Im z are real numbers. Solution.The complex number z = 4+3i is shown in Figure 2. The argument of a complex number is the angle formed by the vector of a complex number and the positive real axis. Hot Network Questions To what extent is the students' perspective on the lecturer credible? (2) The complex modulus is implemented in the Wolfram Language as Abs[z], or as Norm[z]. See also. Subscript indices must either be real positive integers or logicals." For a complex number z = x+iy, x is called the real part, denoted by Re z and y is called the imaginary part denoted … Follow 722 views (last 30 days) bsd on 30 Jun 2011. A complex number is a number of the form a+bi, where a,b — real numbers, and i — imaginary unit is a solution of the equation: i 2 =-1.. The argument of the complex number sin 5 6 π + i (1 + cos 5 6 π ) is. Instead, it’s the angle between two of our axes, so we know this is a right angle. Argument of Complex Numbers. It is denoted by $$\arg \left( z \right)$$. Complex Numbers and the Complex Exponential 1. Following eq. Phase (Argument) of a Complex Number. The principal amplitude of (sin 4 0 ∘ + i cos 4 0 ∘) 5 is. 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# Circle and heart homeomorphic?

Is a circle and heart homeomorphic to one another?

Intuitively, I can picture that the one can be "morphed" into the other by bending and stretching and not breaking. But I am unsure if that is correct?

This is not an assignment or anything, I am just thinking about it in general.

Can anyone please confirm or reject my reasoning above and show it to me (algebraically or through a sketch or anything) in order to give me a nice explanation?

This is the picture that got me thinking about it

• At least the usual heart shape, e.g., upload.wikimedia.org/wikipedia/commons/thumb/f/f1/… is, yes. We'd usually refer to a "filled" circle as a disk, though. – Travis Willse May 8 '15 at 16:39
• @Travis - can you please provide me with the correct terminology and reasoning as an answer, so I can accept and give you the credit? :). – user860374 May 8 '15 at 16:41
• Yes, they are. One way to prove this would be to write down an explicit equation for a map between them and show that it and its inverse are continuous. – Jacob Bond May 8 '15 at 16:41
• @Dillon I'm off to bed, but if I see no one has answered this by tomorrow, I'll at least write up a sketch. Often writing down explicit homeomorphisms can be unpleasant, even for two spaces that are "obviously" homeomorphic. Let me recommend for you the problem of showing that a disk and ("filled") square are homeomorphic, which captures some of the key issues in this problem, but which is probably a little easier to handle. – Travis Willse May 8 '15 at 16:45
• @Dillon I believe convex is not a topological property... See here: en.wikipedia.org/wiki/Topological_property On pp67 you get a good idea of homeomorphisms: books.google.at/… However, proving that something is not homeomorph topological properties are usually the way to go :) – the.polo May 8 '15 at 18:50

Here's a possible approach. Let $$D=\{(x,y)\in\mathbb R^2\mid x^2+y^2\leq1\}$$ be the closed unit disk.

For any pair of functions $f,g:[-1,1]\to\mathbb R$ such that $f(-1)=g(-1)$, $f(1)=g(1)$ and $f(t)<g(t)$ for all $t\in(-1,1)$, let $$A(f,g)=\{(x,y)\in\mathbb R^2\mid f(x)\leq y\leq g(x)\}$$ be "the area between their graphs". (Most candidates for "the heart shape" can be described as $A(f,g)$ for a suitable choice of $f,g$.) Now, I claim that any such set $A(f,g)$ is homeomorphic to $D$. We shall prove this in several steps.

Lemma 1. $A(f,g)$ is homeomorphic to $A(0,g-f)$. (Here $0$ is the zero function, defined by $0(x)=x$ and $g-f$ is defined by $(g-f)(x)=g(x)-f(x)$.)

Proof. The homeomorphism $h:A(f,g)\to A(0,g-f)$ is simply $$h(x,y)=(x,y-f(x))$$ which obviously well-defined and is continuous, because $f$ is. Its inverse is given by $$h^{-1}(x,y)=(x,y+f(x))$$ and we're done. $\square$

Lemma 2. Suppose $k_i:[-1,1]\to\mathbb R$, $i=1,2$ are any two functions such that $k_i(-1)=k_i(1)=0$ and $k_i(t)>0$ for $t\in(-1,1)$. Then $A(0,k_1)$ is homeomorphic to $A(0,k_2)$.

Proof. Again, we may define an explicit homeomorphism $h:A(0,k_1)\to A(0,k_2)$, this time by the formula $$h(x,y)=\left(x,\frac{k_2(x)}{k_1(x)}y\right)$$ for $x\in(-1,1)$ and $h(-1,0)=h(1,0)=0$. This is continuous for $x\in(-1,1)$, since $k_1$ and $k_2$ are continuous and products and quotients of continuous functions are continuous (the latter wherever the denominator is nonzero). But $h$ also continuous at the points $(\pm1,0)$, since $\frac{y}{k_1(x)}\in[0,1]$ for all $(x,y)\in A(0,k_1)$, while $k_2(x)$ goes to $0$ as $x$ approaches $\pm1$. So the two limits $\lim_{(x,y)\to(\pm1,0)}h(x,y)$ exist and equal the corresponding function values. By the same argument, the inverse $$h^{-1}(x,y)=\left(x,\frac{k_1(x)}{k_2(x)}y\right)$$ is continuous. So $h$ is indeed a homeomorphism. $\square$

Proposition. $A(f,g)$ is homeomorphic to $D$.

Proof. By Lemma 1, $A(f,g)$ is homeomorphic to $A(0,g-f)$. By Lemma 2, $A(0,g-f)$ is homeomorphic to $A(0,2k)$ with $k(x)=\sqrt{1-x^2}$. By Lemma 1 again, $A(0,2k)$ is homeomorphic to $A(-k,k)=D$. Therefore $A(f,g)$ is homeomorphic to $D$. $\square$

If you want an explicit homeomorphism, simply calculate the composition of all the maps used. (By the way, the proofs are even simpler if you work with an open unit disk and "open heart" instead: this way you don't have to analyse the points $(\pm1,0)$ separately.)

To obtain (polygonal version of) a heart, you could use e.g. $$f(x)=|x|-1$$ and $$g(x) = \frac12-\left||x|-\frac12\right|,$$ but I'm sure you can come up with a better ("round" version of a) heart yourself and the same argument will work.

In any case, it is probably helpful to spend some time trying to visualize what each of the maps does.


Question

# The coefficient of x in the expansion (1+x)(1+2x)(1+3x)⋯(1+100x) is also equal to

A
1002992+982972+962952++2212
B
The sum of all 101 A.M.'s inserted between 1 and 99
C
The sum of all 100 A.M.'s inserted between 1 and 99
D
The sum of first 20 terms of the series 1+(1+3)+(1+3+5)+(1+3+5+7)+

Solution

## The correct options are A 1002−992+982−972+962−952+⋯+22−12 B The sum of all 101 A.M.'s inserted between 1 and 99 Coefficient of x in the expansion (1+x)(1+2x)(1+3x)⋯(1+100x) =1+2+3+⋯+100=100×1012=5050 Now, 1002−992+982−972+962−952+…+22−12=(1002−992)+(982−972)+(962−952)+…+(22−12)=(100+99)(100−99)+(98+97)(98−97)+…+(2+1)(2−1)=100+99+98+97+…+2+1=5050 Now, Let n A.M.'s are inserted between 1 and 99, then the sum of them  5050=n+22[1+99]−(1+99)⇒5050=100[n+22−1]⇒5050=50n⇒n=101 Therefore 101 A.M.'s are inserted. Now, the sum of first 20 terms of the series 1+(1+3)+(1+3+5)+(1+3+5+7)+⋯  General term of the series is Tr=1+3+5+7⋯(2r−1)=r2⇒S=20∑r=1Tr=20∑r=1r2⇒S=n(n+1)(2n+1)6⇒S=20×21×416=2870Mathematics

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# Does every countably infinite interval-finite partial order embed into the integers?

A partially ordered set $(S,\le)$ is called interval finite if the open intervals $(x,z):=\{y|x\le y\le z\}$ are finite for all choices of $x,z$ in $S$. An embedding $(S,\le)\rightarrow(S',\le')$ of partially ordered sets is an injective order-preserving map. Does every countably infinite interval finite partially ordered set admit an embedding into the integers? This is equivalent to extending the partial order to a linear suborder of the integers. If so, where can I find the proof? If not, can you give a counterexample?

• I don't understand the votes to close, since this is an interesting problem, and I think it is trickier than it may seem at first. Could someone explain? Sep 24 '13 at 13:56
• Sep 24 '13 at 23:56

The answer is yes. First, let's prove a lemma. By order preserving, I assume that you mean forward-preservation of the order: $p\leq q\implies f(p)\leq' f(q)$.

Lemma. Every countable interval-finite partial order $\P$ has a convex enumeration, an enumeration $\langle p_0,p_1,p_2,\ldots\rangle$ of $\P$, all of whose initial segments are convex sets in $\P$.

Proof. If we have a finite convex subset of $\P$, and new point $p$ to be added, then by convexity $p$ does not appear in any interval of points we already have. If $p$ is above some points we have already, then it is not below any point that we have already, and so we can look at the intervals $(q,p)$ determined by a point $q$ we have already and the new point $p$. By convexity, none of these new points can be below any point we already have, and so we can simply add them from the bottom while maintaining convexity. A similar argment works if the new point is only below points we already have. And if $p$ is incomparable to the points we already have, then we can simply add it to the list. QED

Now, we can prove the theorem.

Theorem. Every countable interval-finite partial order embeds into $\Z$.

Proof. Suppose that $\P$ is a countable interval-finite partial order. By the lemma, it has a convex enumeration $p_0,p_1,p_2,\ldots$. Suppose by induction that we have mapped $p_k\mapsto m_k$ in an injective order-preserving manner, for $k\lt n$. Consider the next point $p_n$. Since the order so far is convex and adding $p_n$ maintains convexity, it follows that either $p_n$ is above some points $p_k$ for $k\lt n$ and not below any, or below some such $p_k$ and not above any, or incomparable to them all. In any case, we can easily extend the map to define $p_n\mapsto m_n$ in such a way to still be order preserving and injective. QED

• Joel, thanks, that's terrific. I wonder if you know the origin of this result, since I need to cite it.
– Ben
Sep 24 '13 at 17:59
• I've never seen it before, but I'd expect that probably this has been known. Perhaps someone else can post a source? Sep 24 '13 at 18:07
• The link above discusses the source and gives a reference. Sep 24 '13 at 23:56
• (By the way, the question in the link is still unsolved without choice, in case you have some ideas.) Sep 25 '13 at 0:00
• @Andres, thanks for the reference! The OP on this question, however, insists on injective order-preserving maps, and so there can be no uncountable instances. So it seems that these are slightly different questions, although obviously closely connected. Sep 25 '13 at 0:07


# 0.999...=1

Although some people find it counterintuitive, the decimal expansions $$0.999\dotsc$$ and $$1$$ represent the same real number.

# Informal proofs

These “proofs” can help give insight, but be careful; a similar technique can “prove” that $$1+2+4+8+\dotsc=-1$$. They work in this case because the series corresponding to $$0.999\dotsc$$ is absolutely convergent.

• \begin{align} x &= 0.999\dotsc \newline 10x &= 9.999\dotsc \newline 10x-x &= 9.999\dotsc-0.999\dotsc \newline 9x &= 9 \newline x &= 1 \newline \end{align}

• \begin{align} \frac 1 9 &= 0.111\dotsc \newline 1 &= \frac 9 9 \newline &= 9 \times \frac 1 9 \newline &= 9 \times 0.111\dotsc \newline &= 0.999\dotsc \end{align}

• The real numbers are dense, which means that if $$0.999\dots\neq1$$, there must be some number in between. But there’s no decimal expansion that could represent a number in between $$0.999\dots$$ and $$1$$.

# Formal proof

This is a more formal version of the first informal proof, using the definition of decimal notation.

$$0.999\dots$$ is the decimal expansion where every digit after the decimal point is a $$9$$. By definition, it is the value of the series $$\sum_{k=1}^\infty 9 \cdot 10^{-k}$$. This value is in turn defined as the limit of the sequence $$(\sum_{k=1}^n 9 \cdot 10^{-k})_{n\in\mathbb N}$$. Let $$a_n$$ denote the $$n$$th term of this sequence. I claim the limit is $$1$$. To prove this, we have to show that for any $$\varepsilon>0$$, there is some $$N\in\mathbb N$$ such that for every $$n>N$$, $$|1-a_n|<varepsilon$$.

Let’s prove by induction that $$1-a_n=10^{-n}$$. Since $$a_0$$ is the sum of {$0$ terms, $$a_0=0$$, so $$1-a_0=1=10^0$$. If $$1-a_i=10^{-i}$$, then

\begin{align} 1 - a{i+1} &= 1 - (ai + 9 \cdot 10^{-(i+1)}) \newline &= 1-a_i − 9 \cdot 10^{-(i+1)} \newline &= 10^{-i} − 9 \cdot 10^{-(i+1)} \newline &= 10 \cdot 10^{-(i+1)} − 9 \cdot 10^{-(i+1)} \newline &= 10^{-(i+1)} \end{align}

So $$1-a_n=10^{-n}$$ for all $$n$$. What remains to be shown is that $$10^{-n}$$ eventually gets (and stays) arbitrarily small; this is true by the archimedean property and because $$10^{-n}$$ is monotonically decreasing.

# Arguments against $$0.999\dotsc=1$$

These arguments are used to try to refute the claim that $$0.999\dotsc=1$$. They’re flawed, since they claim to prove a false conclusion.

• $$0.999\dotsc$$ and $$1$$ have different digits, so they can’t be the same. In particular, $$0.999\dotsc$$ starts “$0.$,” so it must be less than 1.

Decimal expansions and real numbers are different objects. Decimal expansions are a nice way to represent real numbers, but there’s no reason different decimal expansions have to represent different real numbers.

• If two numbers are the same, their difference must be $$0$$. But $$1-0.999\dotsc=0.000\dotsc001\neq0$$.

Decimal expansions go on infinitely, but no farther.$$0.000\dotsc001$$ doesn’t represent a real number because the $$1$$ is supposed to be after infinitely many $$0$$s, but each digit has to be a finite distance from the decimal point. If you have to pick a real number to for $$0.000\dotsc001$$ to represent, it would be $$0$$.

• $$0.999\dotsc$$ is the limit of the sequence $$0.9, 0.99, 0.999, \dotsc$$. Since each term in this sequence is less than $$1$$, the limit must also be less than $$1$$. (Or “the sequence can never reach $$1$$.”)

The sequence gets arbitrarily close to $$1$$, so its limit is $$1$$. It doesn’t matter that all of the terms are less than $$1$$.

• In the first proof, when you subtract $$0.999\dotsc$$ from $$9.999\dotsc$$, you don’t get $$9$$. There’s an extra digit left over; just as $$9.99-0.999=8.991$$, $$9.999\dotsc-0.999\dotsc=8.999\dotsc991$$.

There are infinitely many $$9$$s in $$0.999\dotsc$$, so when you shift it over a digit there are still the same amount. And the “decimal expansion” $$8.999\dotsc991$$ doesn’t make sense, because it has infinitely many digits and then a $$1$$.

Parents:

• If these are included I think it would be good to also include explanations of why each one is wrong.


# Does the Fourier series coefficient of AC components remains same if DC component is subtracted form the given signal?

Suppose a signal is defined by $$x(t)= \begin{cases} t & 0\leq t \leq 1 \\ 2-t & 1\leq t\leq 2 \\ \end{cases}$$

Since $$x(t)$$ has even symmetry, I can calculate fourier coefficient as $$a_n = \frac{4}{T} \int_0^1 x(t).\cos{n\pi t}.{dx}$$ I have calculated $$$$a_n = 2\big[\frac{\cos{n\pi} - 1}{n^{2}{\pi}^{2}}\big]\tag{1}$$$$

The DC value of $$x(t)$$ i.e $$a_0 = 0.5$$. If we subtract DC value we get,

From this we can see that given signal has hidden half wave symmetry in addition to Even symmetry. So we can find fourier coefficient as

$$a^{'}_n=\frac{8}{T}\int_{0}^{\frac{1}{2}}(t-\frac{1}{2})\cos{n\pi t}.dt$$

I have calculated $$$$a^{'}_n = 4\big[\frac{\cos{\frac{n\pi}{2}} - 1}{n^{2}{\pi}^{2}}\big]\tag{2}$$$$

My question is, shouldn't $$a_n$$ and $$a^{'}_n$$ be equal for $$n\neq0$$ ?

• How about calculating the Fourier coefficients without using any extraneous considerations such as symmetry or hidden half symmetries?. That is, copy the definition of $a_n$ (the one that applies to all periodic signals, long before extraneous considerations such as symmetry are mentioned) from your book, and calculate $a_n$ and $a_n^\prime$ and see if you get the same answer or different answers. If you get the same answer, the problem is in your understanding of symmetry/half-symmetry/hidden etc. – Dilip Sarwate Oct 27 '18 at 15:27
• if we apply half wave symmetry then it means even components will be zero and odd components of equation 1 & 2 are indeed equal. Equation 2 will give non-zero value for even values of n, other than multiples of 4, but we should discard it according to the conclusion of half wave symmetry. – Saurabh Oct 27 '18 at 19:00
• As Dilip says, I also suggest that you repeat your calculations without worrying about any symmetry; that is, calculate the series for an entire period $T=2$. You should get the exact same answer in both cases except for $n=0$. – MBaz Oct 27 '18 at 22:24
• "If we apply half wave symmetry...." Sigh! You can lead a horse to water but you cannot make him drink. – Dilip Sarwate Oct 28 '18 at 3:39
• @DilipSarwate I did calculate the fourier series coefficient for entire time period i.e. $T=2$ and I got same value in both cases for $n\neq0$. But since -Fat32 already given proof in their answer that CTFS coefficients for DC-removed part will same as original signal, I didn't mention it in my previous comment. – Saurabh Oct 28 '18 at 6:43

PART-I: I would like to provide the general proof considering the title of the question and imposing no specific properties on the signal $$x(t)$$ other than having a CTFS representation.

The following is a simple analysis to conclude that the CTFS coefficients of any signal $$x(t)$$ and that of the DC removed signal are equivalent. (except $$a_0$$ of course).

Consider a continuous-time periodic signal $$x(t)$$ with period $$T$$ divided into two components: $$x_{dc}$$ and $$x_{ac}$$, with periods $$T$$ also, where $$x_{dc}$$ is the pure DC component of $$x(t)$$ and $$x_{ac}$$ is the pure AC component of $$x(t)$$, then we have:

$$x(t) = x_{dc}(t) + x_{ac}(t)$$

Computing the CTFS coefficient $$a_k$$ of $$x(t)$$ yields: \begin{align} a_k &= \frac{1}{T} \int_{} (x_{dc} + x_{ac}) e^{-j k \frac{2\pi}{T} t } dt \\ &= \frac{1}{T} \int_{} x_{dc} e^{-j k \frac{2\pi}{T} t } + \frac{1}{T} \int_{} x_{ac} e^{-j k \frac{2\pi}{T} t } \\ a_k &= b_k + c_k \\ \end{align}

where $$b_k$$ and the $$c_k$$ are the CTFS coefficients of DC and AC parts of $$x(t)$$.

By definition of any DC signal, it's known that $$b_k = 0$$ for all $$k \neq 0$$ and by defition of any AC signal it's known that $$c_0 = 0$$. Then using the relation $$a_k = b_k + c_k$$ we get the following:

$$a_0 = b_0 + c_0 = b_0$$

and $$a_k = 0 + c_k = c_k ~~~,~~~ \text{ for all } k \neq 0$$

From which we define :

$$b_k = \begin{cases} a_0 ~~~&, ~~~\text{ for } k = 0 \\ 0 ~~~&, ~~~\text{ for all } k \neq 0 \\ \end{cases}$$

and

$$c_k = \begin{cases} 0 ~~~&, ~~~\text{ for } k = 0 \\ a_k ~~~&, ~~~\text{ for all } k \neq 0 \\ \end{cases}$$

Hence we conclude that the CTFS coefficients, $$a_k$$, of any periodic signal $$x(t)$$ and the CTFS coefficients $$c_k$$ of DC-removed part, $$x_{ac}$$, are the same for all $$k \neq 0$$.

PART-II: Based on OP comments, the relation for an even and real signal, is the following.

For a signal $$x(t)$$ which is real and even we have $$x(t) = x(t)^{*} = x(-t) = x(-t)^{*}$$ and the associated CTFS coefficients has the property of $$a_k = a_{-k}^{*} = a_{-k} = a_k^{*}$$ which indicates that the coefficients $$a_k$$ are also real and even.

Using this, we can obtain the trigonometric (cosine) Fourier series coefficients as. \begin{align} a_k &= \frac{1}{T} \int_{} x(t) e^{-j k \frac{2\pi}{T} t } dt \\ &= \frac{1}{T} \int_{} x(t) \left( \cos( k \frac{2\pi}{T} t) + j \sin(k \frac{2\pi}{T} t) \right) dt \\ &= \frac{1}{T} \int_{} x(t) \cos( k \frac{2\pi}{T} t) dt + j \frac{1}{T} \int_{} x(t) \sin( k \frac{2\pi}{T} t) dt\\ a_k &= \mathcal{Re}\{a_k\} + j ~~ \mathcal{Im}\{a_k\} \\ \end{align}

Now since the property states tat $$a_k$$ are real, then the imaginary part is zero and we have:

$$a_k= \frac{1}{T} \int_{-T/2}^{T/2} x(t) \cos( k \frac{2\pi}{T} t) dt$$

Furtermore since $$x(t)$$ is also even; $$x(t)=x(-t)$$, then we also have

$$\boxed{ a_k= \frac{2}{T} \int_{0}^{T/2} x(t) \cos( k \frac{2\pi}{T} t) dt }$$

As the trigonometric cosine series coefficients of the real and even signals.

In addition to this, for a real & even signal, $$x(t)$$ of period $$T$$, which has no DC part, the following is also observed: $$x(t-\frac{T}{2}) = -x(t)$$

And based on the time-shift property of CTFS we can conclude that $$a_k ~~e^{-j\frac{2\pi}{T}k \frac{T}{2} } = - a_k$$

$$a_k ~ e^{-j\pi k } = - a_k \implies a_k = \begin{cases} -a_k &, k=0,\pm 2, \pm4,... \\ a_k &, k=\pm 1,\pm 3,...\\ \end{cases}$$

Which indicates that the CTFS coefficients $$a_k = 0$$ for $$k=2m$$ (even) for a real, even, (and having no DC) signal $$x(t)$$. Indeed we can get rid of the DC removed adjective and state for all real & even signals, as the DC will only affect $$a_0$$ being non-zero.

PART-III: Finally apply these to the example signal to see that it works.

The signal defined as: $$x(t) = \begin{cases} t &, 0

Then for the CTFS coefficeints (in the trigonometric form) we have:

$$a_k = \frac{2}{T} \int_{0}^{1} t \cos(\frac{2\pi}{T} k t) dt = \int_{0}^{1} t \cos(k \pi t) dt$$

A simple by-parts integration yields the following result: \begin{align} a_k &= \int_{0}^{1} t \cos(k \pi t) dt \\ &= \frac{t ~\sin(\pi k t) }{\pi k} |_0^1 -\int_{0}^{1} \frac{\sin(k \pi t)}{\pi k} dt \\ &= \frac{\sin(\pi k) }{\pi k} + \frac{1}{\pi k} \left( \frac{ \cos(\pi k) - 1}{\pi k} \right)\\ \end{align} we conclude that

$$\boxed{ a_k = \frac{ \pi k ~\sin(\pi k) + \cos(\pi k) - 1}{ \pi^2 k^2 } }$$

Form which it can also be seen that

$$a_k = \begin{cases} 0.5 &, k=0 \\ \frac{-2}{\pi^2k^2} &, k=\pm 1, \pm 3,... \\ 0 &, k=\pm 2, \pm 4,...\\ \end{cases}$$

Note that the term $$\sin(\pi k)$$ can be ignored (as it's all zero) except for $$k=0$$. Also note that those zero values $$a_k$$ turns out to be the case without assuming half wave symmetry.

Finally, we shall compute the CTFS $$c_k$$ coefficeints of the DC removed signal $$x_{ac}(t)$$ to see if they are equivalent. From the definition of the signal we see that

$$x_{ac}(t) = x(t) - 0.5 = \begin{cases} t-0.5 &, 0

then the $$c_k$$ become: $$c_k = \frac{2}{T} \int_{0}^{1} (t-0.5) \cos(\frac{2\pi}{T} k t) dt = \int_{0}^{1} t \cos(k \pi t) dt - \int_{0}^{1} 0.5 \cos(k \pi t) dt$$

Note that this integral is the same for the case of $$a_k$$. The only diference is in the last term which is $$1$$ for $$k=0$$ and $$0$$ for all $$k\neq 0$$ and can be ignored for $$k \neq 0$$. Then the equation relating to $$c_k$$ will be identical to that of $$a_k$$ except at $$k=0$$ which yields:

$$c_k = \begin{cases} 0 &, k=0 \\ \frac{-2}{\pi^2k^2} &, k=\pm 1, \pm 3,... \\ 0 &, k=\pm 2, \pm 4,...\\ \end{cases}$$

hence we again concluded that $$\boxed{ c_k = \begin{cases} 0 ~~~&, ~~~\text{ for } k = 0 \\ a_k ~~~&, ~~~\text{ for all } k \neq 0 \\ \end{cases} }$$

• The question is about the sine/cosiine trigonometric form of the CTFS, not the exponential form of the CTFS. – Dilip Sarwate Oct 27 '18 at 20:49
• @Dilip Sarwate: : I'm a newbie so, if I see an answer that provides insight and clears up confusion (for me and hopefully the OP ), then, IMHO, it's a great answer. The methodology used to do this, although not formulated the way the OP asked it, is still quite helpful. And I haven't seen a better answer. – mark leeds Oct 27 '18 at 20:56
• my question was,shouldn't $a_n$ and $a^{'}_n$ be equal? And as proved in this answer they, in fact, are equal for odd values of $n$. Now if we already remove DC part to see the half-wave symmetry and calculate CTFS coefficient $a^{'}_n$ which has non-zero values for some even values of $n$ unlike $a_n$, but we should discard $a^{'}_n$ for even values of $n$, as considering half-wave symmetry implies CTFS coefficient for even values of $n$ is $0$. I hope this is correct. – Saurabh Oct 28 '18 at 6:49


How to solve $3\lfloor x \rfloor - \lfloor x^{2} \rfloor = 2\{x\}$?

I am trying to solve the following question involving floor/greatest integer functions.

$$3\lfloor x \rfloor - \lfloor x^{2} \rfloor = 2\{x\}$$ with the notations $$\lfloor x \rfloor$$ denoting the greatest integer less than or equal to $$x$$ and $$\{x\}$$ to mean the fractional part of $$x$$.

I used the following property for floor functions.

$$n\leq x$$ if and only if $$n \leq \lfloor x \rfloor$$ where $$n\in \mathbb{Z}$$

Let $$p=\lfloor x^{2} \rfloor$$, then

$$p\leq \lfloor x^{2} \rfloor < p+1$$

$$\rightarrow p \leq x^{2} < p+1$$

$$\rightarrow \sqrt{p} \leq x < \sqrt{p+1}$$ , since $$\sqrt{p} \in \mathbb{Z}$$

$$\rightarrow \sqrt{p} \leq \lfloor x \rfloor < \sqrt{p+1}$$ We then have $$\sqrt{p} = \lfloor x \rfloor$$

Since $$\{x\}=x-\lfloor x \rfloor,$$

$$3\lfloor x \rfloor - \lfloor x^{2} \rfloor - 2\{x\}= 3\lfloor x \rfloor - \lfloor x^{2} \rfloor - 2(x-\lfloor x \rfloor)= 5\lfloor x \rfloor - \lfloor x^{2} \rfloor - 2x=0$$

Substituting $$p$$, $$\sqrt{p}$$ for $$\lfloor x^{2} \rfloor$$ and $$\lfloor x \rfloor$$ respectively, and also letting $$x= \sqrt{p},$$ we get $$p = 3\sqrt{p}$$ solving for $$p$$ gives $$p=0, 9$$, and hence $$x=0, 3$$

The problem is that according to the solution for the problem, $$x$$ also equals to $$\frac{3}{2}$$ for $$\{x\}=\frac{1}{2}$$ since $$2\{x\}\in \mathbb{Z}$$. However, by definition for $$\{x\}$$, $$0 \leq \{x\} < 1$$, then $$0 \leq 2\{x\} < 2$$. How can $$\{x\}=\frac{1}{2}$$ and how do I use this to obtain $$x=\frac{3}{2}$$. I am not sure what I am missing. IF I made any mistakes in my reasoning. Can someone point it out to me please. Thank you in advance.

• It's not true that $\sqrt{p}\in\mathbb{Z}$ Say $x=1.5$ Then $p=\lfloor 2.25\rfloor = 2,$ and $\sqrt{p}=\sqrt{2}$ May 10, 2020 at 0:35
• @saulspatz thank you for pointing that out. May 10, 2020 at 0:45

Let $$x = n + r$$ where $$n = [x]$$ and $$r = \{x\}$$.

Then we have $$3n - [n^2 + 2nr + r^2]=2r$$

$$3n - n^2 - [2nr + r^2] = 2r$$

and.... oh, hey, the LHS is an integer the RHS being $$2\{x\}$$ means $$\{x\} = 0$$ or $$0.5$$.

Two options $$x$$ is an integer and $$x = [x] = n$$ and $$r=\{x\} = 0$$ and we have

$$3n-n^2=0$$ and $$n^2 = 3n$$ and $$n= 0$$ or $$n = 3$$.

So $$x = 0$$ and $$x=3$$ are two solutions.

(Check: $$x=0\implies 3[x] - [x^2] = 3*0 - 0 = 0 = \{0\}$$. Check. And $$x = 3\implies 3[x]-[x^2] = 3[3]- [3^2] = 3*3-9 = 0=\{3\}$$. Check.

And if $$x = n + \frac 12$$ and $$r = \frac 12$$ then

$$3n - n^2 - [2n\frac 12 + \frac 14] = 2\frac 12$$

$$3n - n^2 - [n + \frac 14] = 1$$

$$3n -n^2 - n = 1$$

$$n^2 - 2n + 1 =0$$ so $$(n-1)^2 = 0$$ and $$n = 1$$.

$$x = 1+\frac 12 = 1\frac 12$$.

(Check: If $$x = 1.5$$ then $$3[x] - [x^2] = 3[1.5] - [1.5^2] = 3*1 - [2.25]=3-2=1 = 2*\frac 12 = 2\{1.5\}$$. Check.)

Write $$\{x\}=x-\lfloor x\rfloor$$. Then we have $$5\lfloor x\rfloor - \lfloor x^2\rfloor = 2x$$Since the LHS is an integer, the RHS must be as well. There are two cases: $$x$$ is an integer, or $$x$$ is a half-integer.

• $$x$$ an integer. Drop the brackets: $$5x-x^2=2x;\qquad x=0,3$$
• $$x$$ is a half-integer. Write $$x=y+1/2$$. Then $$x^2 = y^2+y+1/4$$, and again we can drop the brackets: $$5y-(y^2+y)=2y+1; \qquad y=1, x=3/2$$
• may I ask how you arrive at $x$ is a half integer. I mean can't $x$ be anything else in between $0$ and $1$? May 10, 2020 at 0:46
• After we substitute $\{x\}=x-\lfloor x\rfloor$, both sides are integers. Then if $2x$ is an integer, $x$ is either an integer or a half-integer. May 10, 2020 at 0:47
• I think my issue is the following: from $0 \leq \{x\} < 1$, we get $0 \leq 2\{x\} < 2$. So how do I determine where else $\{x\}$ could be. May 10, 2020 at 0:56
• Clearly the LHS is an integer. Then $2\{x\}$ is an integer as well. This means either $2\{x\}=0$, i.e. $x$ is an integer, or $2\{x\}=1$, i.e. $x$ is a half-integer. May 10, 2020 at 1:00
• I think I see it now, Since $3\lfloor x \rfloor - \lfloor x^{2} \rfloor\in \mathbb{Z}$ and $0 \leq \{x\} < 1$ then, $\frac{3\lfloor x \rfloor - \lfloor x^{2} \rfloor }{2} = \{x\}$ implies that $\frac{3\lfloor x \rfloor - \lfloor x^{2} \rfloor }{2} \leq \{x\}<1$ which forces $\{x\}=\frac{1}{2}$ May 10, 2020 at 1:05


Is the matrix $A$ positive (negative) (semi-) definite?

Given, $$A = \begin{bmatrix} 2 &-1 & -1\\ -1&2 & -1\\ -1& -1& 2 \end{bmatrix}.$$

I want to see if the matrix $A$ positive (negative) (semi-) definite.

Define the quadratic form as $Q(x)=x'Ax$.

Let $x \in \mathbb{R}^{3}$, with $x \neq 0$.

So, $Q(x)=x'Ax = \begin{bmatrix} x_{1} &x_{2} &x_{3} \end{bmatrix} \begin{bmatrix} 2 &-1 & -1\\ -1&2 & -1\\ -1& -1& 2 \end{bmatrix} \begin{bmatrix} x_{1}\\x_{2} \\x_{3} \end{bmatrix}$.

After multiplying out the matrices I am left with $$Q(x) = 2(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-x_{1}x_{2} - x_{1}x_{3}-x_{2}x_{3}).$$

Not sure what I can do with this result. Any suggestions on how to proceed would be appreciated.

A simple way is to calculate all principle minors $A$ and if they are all positive, then $A$ is positive definite.

For example, $|A|_1=2>0$

$$|A|_2=\left|\begin{array}{}{\quad2 \quad-1\\ -1\quad 2} \end{array}\right|=3>0$$ Then calculate $|A|_3=|A|$.

If $|A|_i\geqslant0,1\leqslant i\leqslant n$, then $A$ is semi-positive definite.

If $|A|_i<0$ for $i$ is odd and $|A|_i>0$ for $i$ is even, then $A$ is negative definite.

If $|A|_i\leqslant 0$ for $i$ is odd and $|A|_i\geqslant 0$ for $i$ is even, then $A$ is semi-negative definite.

• Much neater and less complicated. And the same logic applies for negative (semi-) definite as well? – OGC Sep 3 '15 at 6:11
• So here I found $|A_{3}|=0$, so $A$ is positive semi-definite. – OGC Sep 3 '15 at 6:26
• Yes, you are right. You can see it in another post that quadratic form could be $0$ for nonzero $x$. – Math Wizard Sep 3 '15 at 6:27
• Thanks again for this approach! Very convenient. – OGC Sep 3 '15 at 6:28

To say about positive (negative) (semi-) definite, you need to find eigenvalues of A. Then, 1) If all eigenvalues are positive, A is positive definite 2) If all eigenvalues are non-negative, A is positive semi-definite 3) If all eigenvalues are negative, A is negative definite 4) If all eigenvalues are non-positive, A is negative semi-definite 3) If some eigenvalues are positive and some are negative, A is neither positive nor negative definite

Eigenvalues of a matrix can be found by solving $det(\lambda I -A)=0$. For your example, this results in: $\lambda(\lambda-3)^2 =0$ which means that eigenvalues are 0, 3, 3. So we are in the second case and A is positive semi-definite.

If you want to proceed with this solution, you should complete the square. It is important that you "complete one variable completely every time". We write \begin{aligned} x_1^2+x_2^2+x_3^2-x_1x_2-x_1x_3-x_2x_3&=\Bigl(x_1-\frac{1}{2}x_2-\frac{1}{2}x_3\Bigr)^2+\frac{3}{4}x_2^2+\frac{3}{4}x_3^2-\frac{3}{2}x_2x_3\\ &=\Bigl(x_1-\frac{1}{2}x_2-\frac{1}{2}x_3\Bigr)^2+\frac{3}{4}\bigl(x_2-x_3\bigr)^2. \end{aligned} Can you conclude from here?

• So then the matrix is positive definite? – OGC Sep 3 '15 at 5:56
• No, it is positive semidefinite. From the calculation above, you find that $Q(x)\geq 0$ for all $x$. The question is: "Does there exist $x\neq 0$ such that $Q(x)=0$ or not?" In this case it does. Take $x_3$ arbitrary, $x_2=x3$ (to make the last parenthesis zero) and $x_1=x_2$ (to make the first parenthesis zero). We conclude that $Q$ is only positive semidefinite. – mickep Sep 3 '15 at 6:00
• I see. Thanks a lot! – OGC Sep 3 '15 at 6:01
• where did the factor $2$ go on the LHS? – OGC Sep 3 '15 at 6:05
• I just skipped the factor 2 since it only multiplies everything and does not change the character of the quadratic form. I should have mentioned this. – mickep Sep 3 '15 at 6:07

Find $A$'s eigenvalues first. Once you know them, you know everything you need about $A$.

More explicitly, you can start by calculating $A$'s characteristic polynomial. A straightforward calculation shows that its roots are $0$ and $3$. These are $A$'s eigenvalues, and hence, with respect to an appropriate orthonormal basis, $A$ becomes$$\left(\begin{array}{ccc}0&0&0\\0&3&0\\0&0&3\end{array}\right).$$This means that $A$ is positive semi-definite.

• Could you please elaborate? What steps do I need follow here? Is there a theorem that needs to be applied? – OGC Sep 3 '15 at 5:48
• @OGC I added some details to my answer. – Amitai Yuval Sep 3 '15 at 5:55


# Validity of geometric series formula for $r=0$

I can convince myself of the geometric series formula

$$\sum_{n=0}^{\infty} r^n = \frac{1}{1-r}$$

for $$0<|r|<1$$, but not for $$|r|<1$$ because I don't believe the formula for $$r=0$$.

If $$r=0$$, we have

$$\sum_{n=0}^{\infty} r^n = 0^0 + 0^1 + 0^2 + \ldots$$

It is not clear to me what this sum equals, much less that it equals $$\frac{1}{1-0}=1$$. However, every source that I've consulted says that the result holds for $$-1.

Can anyone justify the $$r=0$$ case? Must we simiply accept $$0^0=1$$ in this context?

• if $r=0$ it's not geometric series. By definition, ratio of consecutive terms should be the same. – Vasya Mar 7 at 19:16
• There are lots of ways to define geometric series, @Vasya. One is that $a_{n+1}a_{n-1}=a_n^2.$ In any event, this nit-pick doesn't resolve the question. – Thomas Andrews Mar 7 at 19:18
• Then why does every textbook (even good ones, like Spivak) give the formula for $-1 < r <1$? – mathclassfromscratch Mar 7 at 19:19
• If $r=0$ is allowed, the first term can be any number and $0^0=1$ does not help – Vasya Mar 7 at 19:29
• Let's say that a correct/umabiguous version of the formula in question is $1+\sum_{n=1}^{\infty}r^n=\dfrac{1}{1-r}$ for $|r|<1$. – Paramanand Singh Mar 8 at 5:51

In this context, $$0^0=1$$. Therefore, the sum is $$1$$.

• Why is $0^0=1$ in this context? Is it different in other contexts? – John Douma Mar 7 at 19:21
• The first paragraph here suggests that context matters: en.wikipedia.org/wiki/Zero_to_the_power_of_zero – mathclassfromscratch Mar 7 at 19:29
• @mathclassfromscratch No, it says there is no agreed upon value for $0^0$. – John Douma Mar 7 at 19:31
• @JohnDouma Yes, and then the second sentence says that context matters. – mathclassfromscratch Mar 7 at 19:36
• @mathclassfromscratch The justifications come from different contexts. That doesn't mean that there are provable values for $0^0$ based on different contexts. Either way, I can say this sum equals $\frac{1}{\sqrt{\pi}}$ and there is no context in which you can prove that $1$ is a better answer. – John Douma Mar 7 at 19:42

Power series come up everywhere in mathematics, necessitating a convenient form to represent them. The easiest form is $$\sum_{k=0}^\infty a_k (x-x_0)^k$$ In order for this to represent a proper function, we should be able to substitute any value of $$x$$ into it. If you do not accept the convention $$0^0=1$$, you then run into problems when $$x=x_0$$; the value of the power series at that point is supposed to be $$a_0$$, but you instead get it is $$a_0\cdot 0^0$$. To avoid this, you would have to instead write $$a_0+\sum_{k=1}^\infty a_k (x-x_0)^k$$ which is inconvenient. Therefore, for ease of notation, we stipulate that $$0^0=1$$ in the context of power series. This is the context of $$\sum_{n\ge 0}r^n$$. See

for a confirmation of this.

As a side note, there are an overwhelming number of situations where it is convenient to define $$0^0=1$$, and there are no situations where it is convenient to assume otherwise.

• What number can be considered overwhelming comparing with zero? – user Mar 7 at 20:20

Note a geometric sequence is defined in general as being $$\{a, ar, ar^2, ar^3, \ldots \}$$, i.e., where each term is $$t_i = ar^i$$ for $$i \ge 0$$.

Your statement of $$\sum_{n=0}^{\infty} r^n = \frac{1}{1-r}$$ is actually a specific case of the more general one, such as provided at Geometric series: Formula of

$$\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r} \tag{1}\label{eq1}$$

where $$|r| \lt 1$$, and in your case $$a = 1$$. As such, if $$r = 0$$, then the geometric sequence would be $$\{1, 0, 0, 0, \ldots \}$$ and, thus, it's clear that the sum is $$1$$. Plugging $$a = 1$$ and $$r = 0$$ into \eqref{eq1} gives this same result. Also, by the definition of the sequence, it needs to use "$$0^0 = 1$$" in the LHS of \eqref{eq1} to get that the first term is $$a$$. This is due to, for $$r \neq 0$$, that $$r^0 = 1$$, so $$\lim_{\, r \to 0}r^0 = 1$$.

Note that some definitions of geometric sequences requires that $$r \neq 0$$. However, as you can see, the general equation can still work even if you use $$r = 0$$.

• This answer basically just sidesteps the problem by defining a geometric sequence informally. You could have also defined it as 1,r,r^2,... and be done with it. The original question basically asks why when we define a geometric series as a_n=ar^n we should have a_0=a0^0 be a rather than anything else. You're just assuming that this is indeed the correct definition from the start. – user3329719 Mar 11 at 6:39
• @user3329719 The original definition of a geometric series that I learned, and as also defined by the referenced article, is as I state at the top. Although I define it informally initially, I also define the terms formally as $t_i$. In addition, as for why, in $a_0 = a 0^0$, it makes sense for $0^0 = 1$, I also explain that with my statement about the limit as $r \to 0$, so I'm showing why it makes sense for $0^0 = 1$ instead of assuming this at that time. As such, I'm not clear exactly what issues you have with my answer. – John Omielan Mar 11 at 7:07


# Homework Help: Rational inequality

1. Aug 23, 2014

### mafagafo

1. The problem statement, all variables and given/known data
Find all integer roots that satisfy (3x + 1)/(x - 4) < 1.

3. The attempt at a solution
I would do this:

Make it an equation and find x such that (3x + 1)/(x - 4) = 1.

3x + 1 = x - 4
2x = -5
x = -5/2

Then check if the inequality is valid for values smaller than x and for values bigger than x.

But this approach is not good enough as I would get [-2, +∞) {integers} as my answer.

Any help would be really appreciated.

I think that the answer is [-2, 3] {integers}. But could only get this with a plot.

---

What should I also do so that my method is valid for "rational" inequalities?

2. Aug 23, 2014

### LCKurtz

Consider the two cases where $x<4$ and $x>4$ and work the inequalities separately.

3. Aug 23, 2014

### pasmith

$$\frac{3x + 1}{x - 4} = \frac{3(x-4) + 3(4) + 1}{x - 4} = 3 + \frac{13}{x - 4}.$$ Thus if $(3x + 1)/(x-4) < 1$ then $13/(x - 4) < - 2$. Clearly that can't be the case if $x > 4$ (because then $13/(x - 4) > 0 > -2$) so we must have $x < 4$. Is there a lower bound?

4. Aug 23, 2014

### ehild

(3x + 1)/(x - 4) < 1 can be written in the form
$$\frac{(3x+1)-(x-4)}{x-4}<0$$

Simplified: $$\frac{2x+5}{x-4}<0$$

When is the fraction negative?

ehild

5. Aug 23, 2014

### mafagafo

---
(3x + 1)/(x - 4) = 1
3x + 1 = x - 4
2x = -5

x = -5/2

----
(3x + 1)/(x - 4) = 1
(3(x - 4) + 12 + 1) / (x - 4) = 1
3 + 13/(x - 4) = 1
13 / (x - 4) = -2

x = 4

----
Then I work with those?
(3x + 1)/(x - 4) < 1
Code (Text):

- 8/3 >> false
- 5/2 >> false
- 7/3 >> true
4 >> impossible
5 >> false

So the valid integers are {-2, -1, 0, 1, 2, 3}?

6. Aug 23, 2014

### HallsofIvy

An inequality can change direction where the two sides are equal or where the functions are discontinuous. Here, the first occurs where x= -5/2 and the second where x= 4. There are three intervals to be considered: x< -5/2, -5/2< x< 4, and x> 4.
x= -3< -5/2 and (3(-3)+ 1)/(-3- 4)= (-9+ 1)/(-7)= -8/-7 is greater than 1 so NO x< -5/2 satisfies the inequality. x= 0 is between -5/2 and 4. (3(0)+ 1)(0- 4)= -1/4 is less than 1. Every x between -5/2 and 4 satisfy the inequality. x= 5 is larger than 4 and (3(5)+ 1)/(5- 4)= 15/1 is larger than 1. The integer solutions are -2, -1, 0, 1, 2, and 3.

7. Aug 23, 2014

### mafagafo

When ${2x+5} < 0$ and ${x-4} > 0$ or when ${2x+5} > 0$ and ${x-4} < 0$.

If ${2x+5} < 0$, then $2x<-5$ and $x<-\frac{5}{2}$.
and if ${x-4} > 0$, then $x > 4$. Thus, this is impossible.

If ${2x+5} > 0$, then $2x>-5$ and $x>-\frac{5}{2}$.
and if ${x-4} < 0$, then $x < 4$. Thus, $$S=\left\{x\in Z|-5 /2 < x < 4\right\}=\left\{x\in Z|-2 \le x \le 3\right\}$$

8. Aug 23, 2014

### mafagafo

Big thanks to all of you, with special mention to HallsOfIvy for answering my question.

Q.: "What should I also do so that my method is valid for "rational" inequalities?"
A.: An inequality can change direction where the two sides are equal or where the functions are discontinuous.


# Expected Value Function

My text-book defines expected value as $$E(X) = \mu_x = \sum_{x \in D} ~x \cdot p(x)$$ And so, if I was to find the expected value of a random variable $X$, where $X = 1,2,3$, then it would resemble this: $$E(X)= \sum_{x=1}^3~ x \cdot p(x)= 1\cdot p(1) + 2\cdot p(2) + 3 \cdot p(3)$$ Furthermore, if I wanted to calculate $E(X^2)$, it would be $E(X^2) = 1^2 \cdot P(1) + 2^2 \cdot p(2) + 3^2 \cdot p(3)$. My question is, why don't we square the x-values in the probability function $p(x)$?

Also, is computing the expected value a way of calculating the average of the random variable? It seems a little odd to calculate it that way.

PS: If any use of notation, or vocabulary, is incorrect, please inform me.

-
The differences between using \Sigma and using \sum in TeX are these: $\displaystyle\Sigma_{x\in D}$ versus $\displaystyle\sum_{x\in D}$. That's why \sum is standard for this occasion. –  Michael Hardy Feb 17 '13 at 19:36

Let $Y=X^2$. Then $Y$ takes on the values $1$, $4$, and $9$ respectively when $X$ takes on the values $1$, $2$, and $3$.

Thus $p_Y(1)=p_X(1)$, $p_Y(4)=p_X(2)$, and $p_Y(9)=p_X(3)$.

Now for calculating $E(Y)$ we just use the formula the post started with, namely $$E(Y)=\sum_y yp_Y(y).$$ In our case, we get $1\cdot p_Y(1)+4\cdot p_Y(4)+9\cdot p_Y(9)$. Equivalently, $E(Y)= 1\cdot p_X(1)+4\cdot p_X(2)+9\cdot p_X(3)$.

To answer your question more explicitly, we do not use $1^2(p_X(1))^2+2^2(p_X(2))^2+3^2(p_X(3))^2$ because, for example, $\Pr(X^2=3^2)$ is not $(\Pr(X=3))^2$. In fact, $\Pr(X^2=3^2)=\Pr(X=3)$.

As to your question about average, yes, the mean is a very important measure of average value. The only serious competitor is the median.

Mean and median can be quite different. For example, imagine a population in which a small minority is insanely rich, while the vast majority of the population is struggling. Then the mean income of the population may be substantially higher than the median income. Is either one a "better" measure of average wealth? I would argue that in this case the median is ordinarily of greater relevance. But for certain planning purposes, such as level of tax revenues, the mean may be more useful.

The mathematics of the mean is substantially simpler than the mathematics of the median. For example, the mean of a sum of two random variables is the sum of the means. The median of a sum is a far more complicated object.

-

For computing $E[X^2]$, the probability is still taken over $X$ and not $X^2$. Otherwise, if you make $Y=X^2$ the random variable and then compute $E[Y]$, the only operation that you effectively did is to relabel the random variables (well, although only considering positive values): all the values taken by $|X|$ will also be taken by $Y$, so for positive values of $X$, computing $E[X^2]$ would be exactly like computing $E[X]$. But computing $E[X^2]$ gives you more information!

The expected value is the weighted average. "Normally" (in daily life), when you take an average, all the values have the same weight. The average salary of your family members, for instance. But say you wanted the average salary in your country, then it's nice to work with, say, the probability of a certain salary being had. Making this latter problem more concrete, you could approximate the average national salary by taking every integer multiple of $1000, and finding out the proportion of people with this salary. Then the weighted average gives you the true national average salary. - There were two questions. The first question: Why don't we square the$x$values in the calculation for the expected value of$X^2$. Suppose$Y = g(X). \begin{aligned} E(Y) & \stackrel{\text{(a)}}{=} \sum_{y\in{S_Y}}yP_Y(y)\\ & \stackrel{\text{(b)}}{=} \sum_{y\in{S_Y}}y\sum_{x:g(x) = y}P_X(x)\\ & \stackrel{\text{(c)}}{=} \sum_{y\in{S_Y}}\sum_{x:g(x) = y}yP_X(x)\\ & \stackrel{\text{(d)}}{=} \sum_{y\in{S_Y}}\sum_{x:g(x) = y}g(x)P_X(x)\\ & \stackrel{\text{(e)}}{=} \sum_{x\in{S_X}}g(x)P_X(x)\\ \end{aligned} (a): The definition of expected value of a discrete random variable, as you have supplied. (b): Because the probability of random variableY$taking on a value$g(x)$is equal to the sum of the probability of all the values of$x$which will map to$g(x)$. (c): Interchanging the summation. (d): Because of the condition in the summation, we can replace$y$by$g(x)$. (e): Because enumerating across all$y$, then enumerating across all$x$such that$g(x) = y$, is equivalent to enumerating across all$x$since every$x$must map to exactly one$y$. (Multiple$x$could map to the same$y$). So, in your instance, set$g(x) = x^2$to obtain the result. The second question: Why is expected value defined the way it is defined? Expected value can be thought of as the center of mass, if we set$P_X(x)$to be the mass located at a distance$x$from the origin. It corresponds exactly with the arithmetic average when the distribution of$X\$ is uniform.

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# In how many ways can integers $a_1<a_2<a_3<a_4$ be chosen from the integers $1,2,3,…,26$ such that $5 \le a_i - a_{i-1} \le 7$

In how many ways can $4$ integers $a_1<a_2<a_3<a_4$ be chosen from the integers $1,2,3,...,26$ such that $5 \le a_i - a_{i-1} \le 7$ for all $i = 2,3,4$?

I'm not sure what I'm missing in my line of thinking:

Since $5 \le a_i - a_{i-1} \le 7$, the difference between two consecutive integers chosen must be $5,6$ or $7$ (since everything's an integer). If I let $a_1 = 1$, then by incrementing the next integers by $7$ I see that the maximum I get is $22$. Therefore, the difference $a_i - a_{i-1}$ could be anything among $5,6,7$ for all $i$. Thus I have $3^3$ possibilities.

Similarly, letting $a_1$ be $2,3,4$ and $5$ all yields $3^3$ possibilities.

When I let $a_1 = 6$, however, if I keep incrementing by the biggest difference, $7$, the max. I get is $27$, which is $1$ more than the allowed maximum. That means I'm not allowed to use all three $7$'s for the difference but a maximum of two $7$'s. That gives me $3^3-1$ possibilities.

When I let $a_1 = 7$, the 'maximum' I get is $28$, $2$ more than $26$. Therefore, I can only use a maximum of one $7$. Since there is one case where I can use all three $7$'s and six cases in which I use two 7's, I have $3^3 - (1+6)$ possibilities.

For $a_1 = 8$ the 'maximum' $a_4$ is 29. So to fit in with $26$, I must not use any $7$'s at all. That gives me $2^3$ possibilities.

For $a_1 = 9$ I can get $a_4 = 30$ 'max'. To offset the difference of $4$, I cannot use 7's at all and one of the 6's. That gives me $4$ choices.

For $a_1 = 10$ I have $31$ 'max'. Now I can use only all 5's, or a maximum of one $6$ only. That's $4$ possibilities as well.

Finally, for $a_1=11$, I can only have the differences be $5$ all the time. So 1 possibility.

So the sum gave me $198$. But the answer is $216$. Where did I go wrong?

It is true that $a_1 = 1,2,3,4,5$ each give you $3^3$ possibilities.

Next, looking at $a_1 = 6$, we see that the possibility of $7,7,7$ is excluded, but every other possibility is included, so this gives $26$ possibilities.

Looking at $a_1 = 7$, only those totaling to $20$ or more are not allowed, which means that $757$, for example, is permissible, but you have excluded it. Hence we exclude $776,767,677$ and $777$ to get $23$ possibilities, not $27 -(1+6)$, rather $27-(1+3)$.

For $a_1 = 8$ only those totalling to $19$ or more must be excluded. That is, the above possibilities, plus $757$, $775$,$577$,$667,676,766$. This gives $17$ in total.

For $a_1 = 9$ the above, plus all permutations of $765$, and $666$, are excluded, giving $17-7 = 10$.

For $a_1 = 10$ the above plus all permutations of $755$ and $665$ are excluded, giving $4$ possibilities.

For $a_1 = 11$ only $555$ remains.

The total? $135 + 26 + 23+17+10+4+1 = 216$.

Let the consecutive increments be $x, y, z$. There are $27$ choices of $x, y, z \in \{5,6,7\}$.

Given a fixed choice of $x, y, z$, the number of ways to select the $a_i$'s is $f(x,y,z) = 26 - (a_4 - a_1) = 26 - (x + y + z)$.

Since $f(x,y,z)$ is linear, it is clear that its average value is $26 - 3 \times 6 = 8$.

Thus the answer is $27 \times 8 = 216$.

There is a mistake for $a_1 = 7$. The number of possibilities of having two $7$ isn’t $6$ but : $\binom{3}{2} = 3$.

Moreover, for $a_1 =8$, you are saying that there can’t be $7$, yet what about : $a_1 = 8, a_2 = 8+7 = 15, a_3 = 20, a_4 = 25$ ? (Same argument for $a_1 = 9$)

You've just skipped some possibilities. For example, you say that when $a_1=8,$ you can't use any $7'$s at all, but that's not so. You can use $5,6,7$ or $5,5,7.$ It's probably easier if you consider that you can make a maximum of $18$ with the $3$ increments.