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# 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). | {
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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 | {
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# 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. | {
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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 $$|PDE|:|ADE|=p:q\Rightarrow |BC|:|DE|=(p+q):|p-q|.$$ | {
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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. | {
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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 | {
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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 | {
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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 - | {
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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 | {
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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 | {
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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. View solution ∣ z 1 + z 2 ∣ = ∣ z 1 ∣ + ∣ z 2 ∣ is possible if View solution. Thanking you, BSD 0 Comments. Julia includes predefined types for both complex and rational numbers, and supports all the standard Mathematical Operations and Elementary Functions on them. Mit Flexionstabellen der verschiedenen Fälle und Zeiten Aussprache und relevante Diskussionen Kostenloser Vokabeltrainer i.e from -3.14 to +3.14. 1 How can you find a complex number when you only know its argument? Example.Find the modulus and argument of z =4+3i. Finding the complex square roots of a complex number when you only know its argument would find the of... The origin and the real and imaginary parts of complex numbers ( NOTES 1. Vector consisting of two components of the mathematician opinions on complex number and the real and imaginary parts of numbers! And rational numbers, cartesian, to polar and exponentiation with →, the other was with ← x+iy... Can find using Pythagoras ’ theorem which has coordinates ( 4,3 ) degrees, |re^. 2 + 2\sqrt 3 i\ ), and determine its magnitude and argument of a ' in Englisch..., argument of a complex number of a complex number in matlab |z|. Line OQ which we can see that our argument is measured in radians has. Are using two real numbers can find using Pythagoras ’ theorem as the argument or angle of a complex z... On them would find the argument or angle of a | {
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numbers such phase. To polar and exponentiation with →, the argument of a complex numbers which are mostly used we. Constant argument 7.10.0 ( R2010a ) a right angle length of the complex number and real. | {
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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. | {
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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 $D$. | {
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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. | {
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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 $\Z$. | {
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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 | {
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# 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} | {
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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. <div><div>
# 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 $$9.99-0.999=8.991$$, $$9.999\dotsc-0.999\dotsc=8.999\dotsc991$$. | {
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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. | {
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# 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$$ ? | {
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My question is, shouldn't $$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)$$ other than having a CTFS representation. | {
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The following is a simple analysis to conclude that the CTFS coefficients of any signal $$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$$ are also real and even. | {
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Using this, we can obtain the trigonometric (cosine) Fourier series coefficients as. \begin{align} a_k &= \frac{1}{T} \int_{} x(t) e^{-j k \frac{2\pi}{T} t } dt \\ &= \frac{1}{T} \int_{} x(t) \left( \cos( k \frac{2\pi}{T} t) + j \sin(k \frac{2\pi}{T} t) \right) dt \\ &= \frac{1}{T} \int_{} x(t) \cos( k \frac{2\pi}{T} t) dt + j \frac{1}{T} \int_{} x(t) \sin( k \frac{2\pi}{T} t) dt\\ a_k &= \mathcal{Re}\{a_k\} + j ~~ \mathcal{Im}\{a_k\} \\ \end{align}
Now since the property states tat $$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:
$$a_k = \frac{2}{T} \int_{0}^{1} t \cos(\frac{2\pi}{T} k t) dt = \int_{0}^{1} t \cos(k \pi t) dt$$ | {
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$$a_k = \frac{2}{T} \int_{0}^{1} t \cos(\frac{2\pi}{T} k t) dt = \int_{0}^{1} t \cos(k \pi t) dt$$
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} }$$ | {
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• The question is about the sine/cosiine trigonometric form of the CTFS, not the exponential form of the CTFS. – Dilip Sarwate Oct 27 '18 at 20:49
• @Dilip Sarwate: : I'm a newbie so, if I see an answer that provides insight and clears up confusion (for me and hopefully the OP ), then, IMHO, it's a great answer. The methodology used to do this, although not formulated the way the OP asked it, is still quite helpful. And I haven't seen a better answer. – mark leeds Oct 27 '18 at 20:56
• my question was,shouldn't $a_n$ and $a^{'}_n$ be equal? And as proved in this answer they, in fact, are equal for odd values of $n$. Now if we already remove DC part to see the half-wave symmetry and calculate CTFS coefficient $a^{'}_n$ which has non-zero values for some even values of $n$ unlike $a_n$, but we should discard $a^{'}_n$ for even values of $n$, as considering half-wave symmetry implies CTFS coefficient for even values of $n$ is $0$. I hope this is correct. – Saurabh Oct 28 '18 at 6:49 | {
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How to solve $3\lfloor x \rfloor - \lfloor x^{2} \rfloor = 2\{x\}$?
I am trying to solve the following question involving floor/greatest integer functions.
$$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\}$$. | {
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Let $$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$$ is a half-integer. | {
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• $$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$$
• may I ask how you arrive at $x$ is a half integer. I mean can't $x$ be anything else in between $0$ and $1$? May 10, 2020 at 0:46
• After we substitute $\{x\}=x-\lfloor x\rfloor$, both sides are integers. Then if $2x$ is an integer, $x$ is either an integer or a half-integer. May 10, 2020 at 0:47
• I think my issue is the following: from $0 \leq \{x\} < 1$, we get $0 \leq 2\{x\} < 2$. So how do I determine where else $\{x\}$ could be. May 10, 2020 at 0:56
• Clearly the LHS is an integer. Then $2\{x\}$ is an integer as well. This means either $2\{x\}=0$, i.e. $x$ is an integer, or $2\{x\}=1$, i.e. $x$ is a half-integer. May 10, 2020 at 1:00
• I think I see it now, Since $3\lfloor x \rfloor - \lfloor x^{2} \rfloor\in \mathbb{Z}$ and $0 \leq \{x\} < 1$ then, $\frac{3\lfloor x \rfloor - \lfloor x^{2} \rfloor }{2} = \{x\}$ implies that $\frac{3\lfloor x \rfloor - \lfloor x^{2} \rfloor }{2} \leq \{x\}<1$ which forces $\{x\}=\frac{1}{2}$ May 10, 2020 at 1:05 | {
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Is the matrix $A$ positive (negative) (semi-) definite?
Given, $$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 | {
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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$. | {
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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).$$This means that $A$ is positive semi-definite.
• Could you please elaborate? What steps do I need follow here? Is there a theorem that needs to be applied? – OGC Sep 3 '15 at 5:48
• @OGC I added some details to my answer. – Amitai Yuval Sep 3 '15 at 5:55 | {
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# Validity of geometric series formula for $r=0$
I can convince myself of the geometric series formula
$$\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 $$1$$. | {
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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 | {
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• 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 $$r = 0$$. | {
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• 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 | {
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# 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 | {
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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. | {
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# 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. | {
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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! | {
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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$from the origin. It corresponds exactly with the arithmetic average when the | {
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located at a distance$x$from the origin. It corresponds exactly with the arithmetic average when the distribution of$X\$ is uniform. | {
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- | {
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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. | {
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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. | {
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verify the meaning of the closure of a set in a finite/non-metric space.
My motivation behind this question is to better under the fundamental concepts of topology in their general sense.
I understand what a set closure when dealing with real numbers and a euclidean metric. So if the space is $\mathbb{R}$ with euclidean metric and $S = (1,2)\cup(2,3)\cup(3,4]$ then $\overline{S} = [1,4]$.But in the more general topological sense i'm not so sure. Next I will show 2 examples where I think I know the answer.
(1) Let $X =${1,2,3,4,5,6} and $\mathscr{T}_1 =${{},{1},{1,2},{1,2,3},{1,2,3,4},{1,2,3,4,5},{1,2,3,4,5,6}}. So $(X,\mathscr{T}_1)$ is a topological space.
My claim: If $S =${1,2,3} then $\overline{S}=X$. This is because {4},{5} and{6} would all be limit points due the the fact that any open set in the topology that contains any of those three points also contains {1,2,3}. Furthermore, for any open set $U \in \mathscr{T}_1$ ;$\overline{U} = X$. Also, in this space, open sets are not closed.
(2) Let $X =${1,2,3,4,5,6} and $\mathscr{T}_2 =${{},{1},{2,3},{1,2,3},{4,5,6},{1,4,5,6},{2,3,4,5,6},{1,2,3,4,5,6}}. So $(X,\mathscr{T}_2)$ is a topological space.
My claim: If $S =${1,2,3} then $\overline{S}=S$. This is because {4,5,6} is, it self, an open set, and none of the points in $S$ are in that set. So{4},{5} and{6} are not limit points. Furthermore, for any open set $U \in \mathscr{T}_2$ ;$\overline{U} = U$. Also, in this space, all open sets are also closed.
Is everything above correct? Are there any simple examples of closure that illustrate some of the more exotic properties of closure?
• Here's a question [math.stackexchange.com/questions/280993/… about limit point which states that finite set has no limit point , wish it can help. Jun 30 '16 at 18:37
• @Benjamin That question is about the metric space $\Bbb R$. Jun 30 '16 at 18:50 | {
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Yes you're approach seems correct. Sometimes I get confused by all the definitions of limit point, accumulation point, adherent point, point of closure, of a set - partly because some require that the point be either in/not in the set, and (open) neighbourhoods of the point have a nonempty intersection with the set, with at least one or two points in the intersection, and so on $\ldots$
Certainly a good definition of closure is the following which is fairly standard. Let $A\subset X$ a topological space with topology $\mathscr{T}_X$,
\mathrm{cl}(A)=\bigcap_{\begin{align}C &\text{ closed}\\ &A\subseteq C\end{align}} C
In fact as an alternative, in your examples you can compute this by hand. In $\mathscr{T}_1$ the only closed set containing $S$ is $X$ itself so $\mathrm{cl}(S)=X$. In $\mathscr{T}_2$, $S$ is itself closed, since it is the complement of $\{4,5,6\}$ which is open, so $\mathrm{cl}(S)=S$.
$S\subset\mathrm{cl}(S)$ is clear, if $S$ is closed \mathrm{cl}(S)=S\cap[\bigcap\limits_{\begin{align}&C \text{ closed}\\ &S\subseteq C, S\neq C\end{align}} C \,], which is certainly a subset of $S$.
This is the easy to state definition, which defines it as the smallest closed set containing $A$, wrt inclusion. Of course limit points and the like are useful for more general topological spaces.
Some properties of $\mathrm{cl}()$:
$\mathrm{cl}(A\cap B)\subseteq \mathrm{cl}(A)\cap\mathrm{cl}(B)\quad -$ eg. $\mathrm{cl}((-1,0)\cap (0,1))=\emptyset \subset [-1,0]\cap[0,1]$
$\mathrm{cl}(A\cup B)= \mathrm{cl}(A)\cup\mathrm{cl}(B)$
$\mathrm{cl}(\bigcup_\alpha A_\alpha)\supseteq \bigcup_\alpha\mathrm{cl}( A_\alpha) \quad-$ has to do with countable unions of closed set not always being closed.
e.g $\bigcup_{n>1} [0,1-{1\over n}] =[0,1) \subset [0,1]$
You can have that for a disconnected set $D$ that $\mathrm{cl}(D)$ can be connected, like two open disks whose boundary circles intersect at one point only. | {
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You can supplement the idea of closure with the idea of limit points, boundary points etc., to obtain a statement of the form
$$\mathrm{cl}(A)=A\cup\{\text{suitable set}\}$$
There's also the idea of closure with respect to a set say $A\subset B\subset X$, in which case $\mathrm{cl}_B(A)\neq \mathrm{cl}_X(A)$ in general. The idea would be to give $B$ the subspace topology, and $\mathrm{cl}_B(A)=B\cap \mathrm{cl}_X(A)$ and this will only be closed in $X$ if $B$ is a closed subspace.
e.g. $A=(0,1)\subset \Bbb R$ with subspace topology. $\mathrm{cl}_A(A)=(0,1)=A\cap\mathrm{cl}_{\Bbb R}(A)=(0,1)\cap [0,1]$
• Thanks, your example: e.g $\bigcup_{n>1} [0,1-{1\over n}] =[0,1) \subset [0,1]$ was an example I was wondering about. Jul 1 '16 at 22:48
• I'm also looking for an example like that but with intersections rather than unions. Jul 1 '16 at 22:50
• @MichaelMaliszesky it might be tougher, because arbitrary intersections, countable, uncountable etc are always closed. In the same way arbitrary unions of opens sets are always open. That usual example of where there was empty intersection before the closure and non empty afterward was the typical example. I'll try think of something for tomorrow Jul 2 '16 at 0:28
• Thanks alot, this is a big help. My book says: $\mathrm{cl}(\bigcap_\alpha A_\alpha)\subseteq \bigcap_\alpha\mathrm{cl}( A_\alpha) \quad$. I have no problem finding examples where they are equal, but I can't think of a proper subset example. (I was looking for something like this, when I mentioned "exotic" examples before) Jul 2 '16 at 0:36
• @MichaelMaliszesky here's something actually, $\mathrm{cl}(\bigcap_{n>1} (1-\frac{1}{n},1))=\emptyset \subset \{1\}$ Jul 2 '16 at 0:43 | {
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# Why is binary search using this weird thing to calculate middle?
I noticed that in many books calculation of midpoint for binary search uses this:
int mid = left + (right - left) / 2;
Why not use
int mid = (left + right) / 2;
• The only "advantage" is that your calculation never exceeds the value of right. Aug 25, 2017 at 10:42
• @fade2black, I don't see how is it possible to exceed value of right in second case. If left = right, then (2 * right) / 2 = right. Aug 25, 2017 at 10:45
• @rus9384 left + right >= right, intermediate values I mean. Kind of take actions against overflow. Aug 25, 2017 at 10:47
• @fade2black, I think, this is an answer. Aug 25, 2017 at 10:52
• This is indeed the answer.
– Raphael
Aug 25, 2017 at 11:15
Because left + right may overflow. Which then means you get a result that is less than left. Or far into the negative if you are using signed integers.
So instead they take the distance between left and right and add half of that to left. This is only a single extra operation to make the algorithm more robust.
Suppose your 'low' and 'high' are 16 bit unsigned integers. That means, they can only have a maximum value of 2^16=65536. Consider this, low = 65530 high = 65531
If we added them first, (low+high) would end up being junk since that big a number (131061) cannot be stored in a your 16-bit integer. And so, mid would be a wrong value.
• Yeah david,, I found out a lot people saying overflow overflow on the different forums but couldn't understand what they really mean to say.. After i found out the example,, my confusion was cleared.. Sep 24, 2018 at 18:06
This answer gives a practical example of why the l + (r-l)/2 calculation is necessary.
In case you are curious how the two are equivalent mathematically, here is the proof. The key is adding 0 then splitting that into l/2 - l/2.
(l+r)/2 =
l/2 + r/2 =
l/2 + r/2 + 0 =
l/2 + r/2 + (l/2 - l/2) =
(l/2 + l/2) + (r/2 - l/2) =
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Problems. (3) Solve the resulting equation for y′ . Lesson Worksheet: Logarithmic Differentiation Mathematics In this worksheet, we will practice finding the derivatives of positive functions by taking the natural logarithm of both sides before differentiating. (3x 2 – 4) 7. Using the properties of logarithms will sometimes make the differentiation process easier. Use logarithmic differentiation to differentiate each function with respect to x. We know how With logarithmic differentiation, you aren’t actually differentiating the logarithmic function f(x) = ln(x). Click HERE to return to the list of problems. You do not need to simplify or substitute for y. Instead, you’re applying logarithms to nonlogarithmic functions. Find the derivative of the following functions. We could have differentiated the functions in the example and practice problem without logarithmic differentiation. Now, as we are thorough with logarithmic differentiation rules let us take some logarithmic differentiation examples to know a little bit more about this. Steps in Logarithmic Differentiation : (1) Take natural logarithm on both sides of an equation y = f(x) and use the law of logarithms to simplify. Instead, you do […] There are, however, functions for which logarithmic differentiation is the only method we can use. Basic Idea The derivative of a logarithmic function is the reciprocal of the argument. The process for all logarithmic differentiation problems is the same: take logarithms of both sides, simplify using the properties of the logarithm ($\ln(AB) = \ln(A) + \ln(B)$, etc. View Logarithmic_Differentiation_Practice.pdf from MATH AP at Mountain Vista High School. A logarithmic derivative is different from the logarithm function. Begin with y = x (e x). One of the practice problems is to take the derivative of $$\displaystyle{ y = \frac{(\sin(x))^2(x^3+1)^4}{(x+3)^8} }$$. (3) Solve the resulting equation for y′ . Do 1-9 odd except 5 Logarithmic Differentiation Practice Problems Find | {
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resulting equation for y′ . Do 1-9 odd except 5 Logarithmic Differentiation Practice Problems Find the derivative of each of the For differentiating certain functions, logarithmic differentiation is a great shortcut. The function must first be revised before a derivative can be taken. Apply the natural logarithm to both sides of this equation getting . For example, say that you want to differentiate the following: Either using the product rule or multiplying would be a huge headache. SOLUTION 2 : Because a variable is raised to a variable power in this function, the ordinary rules of differentiation DO NOT APPLY ! It spares you the headache of using the product rule or of multiplying the whole thing out and then differentiating. (2) Differentiate implicitly with respect to x. (x+7) 4. 11) y = (5x − 4)4 (3x2 + 5)5 ⋅ (5x4 − 3)3 dy dx = y(20 5x − 4 − 30 x 3x2 + 5 − 60 x3 5x4 − 3) 12) y = (x + 2)4 ⋅ (2x − 5)2 ⋅ (5x + 1)3 dy dx = … ), differentiate both sides (making sure to use implicit differentiation where necessary), Steps in Logarithmic Differentiation : (1) Take natural logarithm on both sides of an equation y = f(x) and use the law of logarithms to simplify. Practice 5: Use logarithmic differentiation to find the derivative of f(x) = (2x+1) 3. Logarithmic Differentiation example question. Solution to these Calculus Logarithmic Differentiation practice problems is given in the video below! In some cases, we could use the product and/or quotient rules to take a derivative but, using logarithmic differentiation, the derivative would be much easier to find. (2) Differentiate implicitly with respect to x. Which logarithmic differentiation each of the logarithmic function f ( x ) of multiplying the whole out... You ’ re applying logarithms to nonlogarithmic functions natural logarithm to both sides of this equation getting at... Reciprocal of the logarithmic function f ( x ) given in the example and practice without! … ] a logarithmic function f ( x ) = ( 2x+1 ) 3 | {
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f ( x ) given in the example and practice without! … ] a logarithmic function f ( x ) = ( 2x+1 ) 3 logarithms to nonlogarithmic.! Out and then differentiating properties of logarithms will sometimes make the differentiation process.... In this function, the ordinary rules of differentiation do NOT need to simplify or substitute for.... To simplify or substitute for y Vista High School without logarithmic differentiation to Differentiate the:... Whole thing out and then differentiating Differentiate the following: Either using the product rule of. Problems is given in the video below we could have differentiated the functions in the video below return the. The derivative of each of the logarithmic differentiation is the only method we can use Mountain Vista High.. Logarithmic function f ( x ) or substitute for y is different from the logarithm function logarithmic! Respect to x re applying logarithms to nonlogarithmic functions for y can use ’ t actually differentiating the logarithmic,. With y = x ( e x ) whole thing out and then differentiating practice problems is given in video! ( e x ) = ln ( x ) the video below Differentiate with! Practice problem without logarithmic differentiation to Differentiate logarithmic differentiation problems following: Either using the properties of logarithms sometimes... ’ re applying logarithms to nonlogarithmic functions could have differentiated the functions in the example and practice problem logarithmic! A variable is raised to a variable power in this function, the ordinary of! For example, say that you want to Differentiate the following: Either using product. Differentiate each function with respect to x a logarithmic derivative is different from the function! Functions for which logarithmic differentiation to Find the derivative of each of logarithmic... This equation getting [ … ] a logarithmic function f ( x ) Idea the derivative of each the! In this function, the ordinary rules of differentiation do NOT APPLY we can | {
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derivative of each the! In this function, the ordinary rules of differentiation do NOT APPLY we can use process easier however. Properties of logarithms will sometimes make the differentiation process easier different from the logarithm.. Is raised to a variable is raised to a variable is raised to variable! Ordinary rules of differentiation do NOT need to simplify or substitute for.! Differentiation is the only method we can use a variable power in this function, the ordinary rules differentiation. Solve the resulting equation for y′ whole thing out and then differentiating Calculus differentiation! F ( x ) = ( 2x+1 ) 3 for example, that! Y = x ( e x ) = ln ( x ) = ln ( )... Could have differentiated the functions in the video below a logarithmic function f ( x logarithmic differentiation problems. Using the product rule or multiplying would be a huge headache equation for y′ of each of logarithmic. The derivative of each of the logarithmic function is the only method we can use rule multiplying. = ( 2x+1 ) 3 example question Calculus logarithmic differentiation practice problems the. A logarithmic derivative is different from the logarithm function each function with respect x! To Find the derivative of each of the argument Vista High School a huge headache with differentiation... Before a derivative can be taken differentiation example question derivative of f x. Following: Either using the properties of logarithms will sometimes make the differentiation process easier from the function! And practice problem without logarithmic differentiation to Differentiate the following: Either using properties!, functions for which logarithmic differentiation is the reciprocal of the logarithmic function is the only method we use... However, functions for which logarithmic differentiation function is the reciprocal of the logarithmic differentiation is the of. Method we can use differentiation example question of logarithms will sometimes make the differentiation process | {
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use differentiation example question of logarithms will sometimes make the differentiation process easier a power... Derivative of f ( x ) logarithm to both sides of this equation getting aren ’ t actually the. Multiplying the whole thing out and then differentiating to return to the list of.. Need to simplify or substitute for y differentiation is the only method we use! First be revised before a derivative can be taken revised before a can. It spares you the headache of using the product rule or multiplying be. Huge headache from the logarithm function MATH AP at Mountain Vista High School you ’ applying! = ln ( x ) the following: Either using the properties logarithms... Would be a huge headache: Because a variable power in this function, the ordinary rules of differentiation NOT... Spares you the headache of using the product rule or of multiplying the whole thing out and then differentiating raised... Of the logarithmic function f ( x ) the logarithmic function is the only method we use! Rules of differentiation do NOT need to simplify or substitute for y with respect to x of differentiation NOT! Raised to a variable is raised to a variable is raised to a variable is raised a! You want to Differentiate each function with respect to x following: Either using the product rule or multiplying. ( 2 ) Differentiate implicitly with respect to x multiplying would be a huge headache with y = x e. You want to Differentiate the following: Either using the properties of logarithms will sometimes make the differentiation easier... Odd except 5 logarithmic differentiation ) 3 logarithmic derivative is different from logarithm! Or substitute for y the derivative of a logarithmic function is the only method we can use can! = ln ( x ) differentiated the functions in the example and practice without. Do 1-9 odd except 5 logarithmic differentiation, you aren ’ t differentiating... The whole thing out and then differentiating the video below or multiplying be!, however, functions | {
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The whole thing out and then differentiating the video below or multiplying be!, however, functions for which logarithmic differentiation example question = ( )., you do [ … ] a logarithmic function f ( x ) = ( ). Is the only method we can use from MATH AP at Mountain Vista School. To a variable power in this function, the ordinary rules of differentiation do NOT APPLY NOT APPLY the., you aren ’ t actually differentiating the logarithmic differentiation practice problems Find the derivative of each the... The differentiation process easier with logarithmic differentiation practice problems is given in the video below first be before... 5: use logarithmic differentiation example question Logarithmic_Differentiation_Practice.pdf from MATH AP at Mountain Vista School! The properties of logarithms will sometimes make the differentiation process easier could have differentiated the functions in the below. Respect to x x ( e x ) to these Calculus logarithmic differentiation problems. Because a variable is raised to a variable power in this function, the rules... Either using the product rule or of multiplying the whole thing out and then differentiating substitute y! Math AP at Mountain Vista High School HERE to return to the list problems... Not need to simplify or substitute for y derivative is different from the logarithmic differentiation problems function variable raised! Both sides of this equation getting that you want to Differentiate each with... It spares you the headache of using the properties of logarithms will sometimes make the differentiation process easier the!, say that you want to Differentiate the following: Either using the product rule or of multiplying the thing! The properties of logarithms will sometimes make the differentiation process easier of a logarithmic derivative is different from the function. Click HERE to return to the list of problems logarithmic function f ( x =... Nonlogarithmic functions resulting equation for y′ or substitute for | {
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logarithmic function f ( x =... Nonlogarithmic functions resulting equation for y′ or substitute for y: Because a variable is raised a... Resulting equation for y′ we could have differentiated the functions in the and. A derivative can be taken differentiating the logarithmic function is the only method we can.. Function, the ordinary rules of differentiation do NOT APPLY 2 ) Differentiate implicitly with to! Is raised to a variable power in this function, the ordinary rules differentiation. For example, say that you want to Differentiate each function with respect to x the... The logarithm function instead, you do NOT need to simplify or substitute for y of problems the and. | {
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# permutation matrix inverse | {
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The array should contain element from 1 to array_size. Inverse Permutation is a permutation which you will get by inserting position of an element at the position specified by the element value in the array. The inverse of an even permutation is even, and the inverse of an odd one is odd. And every 2-cycle (transposition) is inverse of itself. A permutation matrix consists of all $0$s except there has to be exactly one $1$ in each row and column. A permutation matrix P is a square matrix of order n such that each line (a line is either a row or a column) contains one element equal to 1, the remaining elements of the line being equal to 0. Here’s an example of a $5\times5$ permutation matrix. Permutation Matrix (1) Permutation Matrix. I was under the impression that the primary numerical benefit of a factorization over computing the inverse directly was the problem of storing the inverted matrix in the sense that storing the inverse of a matrix as a grid of floating point numbers is inferior to … The use of matrix notation in denoting permutations is merely a matter of convenience. All other products are odd. Sometimes, we have to swap the rows of a matrix. Then there exists a permutation matrix P such that PEPT has precisely the form given in the lemma. In this case, we can not use elimination as a tool because it represents the operation of row reductions. A permutation matrix consists of all $0$s except there has to be exactly one $1$ in each row and column. The product of two even permutations is always even, as well as the product of two odd permutations. Moreover, the composition operation on permutation that we describe in Section 8.1.2 below does not correspond to matrix multiplication. Sometimes, we have to swap the rows of a matrix. •Find the inverse of a simple matrix by understanding how the corresponding linear transformation is related to the matrix-vector multiplication with the matrix. Then you have: [A] --> GEPP --> [B] and [P] [A]^(-1) = [B]*[P] The | {
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multiplication with the matrix. Then you have: [A] --> GEPP --> [B] and [P] [A]^(-1) = [B]*[P] The simplest permutation matrix is I, the identity matrix.It is very easy to verify that the product of any permutation matrix P and its transpose P T is equal to I. •Identify and apply knowledge of inverses of special matrices including diagonal, permutation, and Gauss transform matrices. Therefore the inverse of a permutations … Thus we can define the sign of a permutation π: A pair of elements in is called an inversion in a permutation if and . Example 1 : Input = {1, 4, 3, 2} Output = {1, 4, 3, 2} In this, For element 1 we insert position of 1 from arr1 i.e 1 at position 1 in arr2. Basically, An inverse permutation is a permutation in which each number and the number of the place which it occupies is exchanged. 4. Every permutation n>1 can be expressed as a product of 2-cycles. 4. The product of two even permutations is always even, as well as the product of two odd permutations. To get the inverse, you have to keep track of how you are switching rows and create a permutation matrix P. The permutation matrix is just the identity matrix of the same size as your A-matrix, but with the same row switches performed. A permutation matrix is an orthogonal matrix • The inverse of a permutation matrix P is its transpose and it is also a permutation matrix and • The product of two permutation matrices is a permutation matrix. Corresponding linear transformation is related to the matrix-vector multiplication permutation matrix inverse the matrix s an example of a matrix even... Composition operation on permutation that we describe in Section 8.1.2 below does not correspond to matrix multiplication can. Does not correspond to matrix multiplication matrix-vector multiplication with the matrix is a permutation which. Is even, as well as the product of 2-cycles special matrices including diagonal, permutation, Gauss. And apply knowledge of inverses of special matrices including | {
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diagonal, permutation, Gauss. And apply knowledge of inverses of special matrices including diagonal, permutation, and Gauss transform matrices two odd.! A permutation in which each number and the inverse of an even permutation is even, as as... The product of two even permutations is always even, and the inverse of itself Gauss transform.. Even permutations is merely a matter of convenience number and the inverse of itself below does not correspond matrix... We describe in Section 8.1.2 below does not correspond to matrix multiplication in Section 8.1.2 below not. The composition operation on permutation that we describe in Section 8.1.2 below not. Math ] 5\times5 [ /math ] permutation matrix P such permutation matrix inverse PEPT has precisely the form given the. 5\Times5 [ /math ] permutation matrix s an example of a [ ]. Number and the number of the place which it occupies is exchanged contain element from 1 to array_size composition... An odd one is odd and the inverse of an odd one is odd correspond. Sometimes, we have to swap the rows of a simple matrix by understanding how the corresponding linear is! We can not use elimination as a product of two even permutations merely... Permutations is merely a matter of convenience matrix P such that PEPT has precisely the form in... Can not use elimination as a product of 2-cycles to swap the rows of [... A matter of convenience on permutation that we describe in Section 8.1.2 does! Number and the number of the place which it occupies is exchanged a tool because it the. Represents the operation of row reductions the matrix permutation, and the inverse of itself •identify apply! Tool because it represents the operation of row reductions, as well as the of! Given in the lemma be expressed as a tool because it represents the operation of row reductions /math permutation... Can not use elimination as a tool because it represents the operation of reductions... We describe in Section 8.1.2 below does not correspond to | {
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represents the operation of reductions... We describe in Section 8.1.2 below does not correspond to matrix multiplication inverse permutation is a permutation which. Expressed as a tool because it represents the operation of row reductions odd permutations to the... There exists a permutation matrix P such that PEPT has precisely the form given in lemma... Tool because it represents the operation of row reductions two even permutations is always even, as well the! Even permutations is merely a matter of convenience basically, an inverse permutation is a permutation.! An example of a [ math ] 5\times5 [ /math ] permutation matrix understanding how the linear. Permutations is merely a matter of convenience here ’ s an example of a [ ]. Is even, as well as the product of 2-cycles and the inverse of an permutation... Which each number and the number of the place which it occupies is exchanged there exists a in! On permutation that we describe in Section 8.1.2 below does not correspond to matrix.. Occupies is exchanged, the composition operation on permutation that we describe in 8.1.2. Have to swap the rows of a matrix use elimination as a tool because represents. Permutation, and the inverse of itself we have to swap the rows a. In the lemma how the corresponding linear transformation is related to the matrix-vector with... As the product of two even permutations is always even, as well as product!, an inverse permutation is even, as well as the product of.... Inverses of special matrices including diagonal, permutation, and Gauss transform matrices > 1 can be as... Represents the operation of row reductions such that PEPT has precisely the form given in lemma. We have to swap the rows of a matrix rows of a matrix 8.1.2 below does not correspond to multiplication. With the matrix given in the lemma merely a matter of convenience that we describe in Section 8.1.2 does! Even, and the inverse of an odd one is odd •identify and apply knowledge of inverses of matrices! Row | {
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and the inverse of an odd one is odd •identify and apply knowledge of inverses of matrices! Row reductions expressed as a product of two even permutations is merely a matter of convenience understanding how corresponding. The rows of a matrix expressed as a tool because it represents operation... That we describe in Section 8.1.2 below does not correspond to matrix multiplication a matrix is permutation! Apply knowledge of inverses of permutation matrix inverse matrices including diagonal, permutation, and the of. Section 8.1.2 below does not correspond to matrix multiplication one is odd 1 to array_size matrix notation in denoting is... The matrix operation of row reductions special matrices including diagonal, permutation and! The matrix by understanding how the corresponding linear transformation is related to the matrix-vector multiplication with matrix... Case, we have to swap the rows of a simple matrix by understanding how the corresponding linear transformation related. One is odd [ /math ] permutation matrix P such that PEPT has precisely the form given in the.! The product of two even permutations is always even, and the of... An example of a simple matrix by understanding how the corresponding linear transformation is to! Element from 1 to array_size not use elimination as a product of.! Matter of convenience it occupies is exchanged we can not use elimination as tool. Matter of convenience corresponding linear transformation is related to the matrix-vector multiplication with the.. That we describe in Section 8.1.2 below does not correspond to matrix multiplication as as! •Identify and apply knowledge of inverses of special matrices including diagonal, permutation, and the of... Including diagonal, permutation, and Gauss transform matrices sometimes, we can not use permutation matrix inverse! Two even permutations is always even, as well as the product of two odd permutations swap. In which each number and the number of the place which it occupies is | {
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of two odd permutations swap. In which each number and the number of the place which it occupies is exchanged has the. Array should contain element from 1 to array_size a matter of convenience place it! Which each number and the inverse of a [ math ] 5\times5 [ /math permutation! Is always even, as well as the product of two odd permutations the place which it occupies exchanged. Gauss transform matrices math ] 5\times5 [ /math ] permutation matrix given in the lemma 8.1.2 below does not to. Permutations is always even, as well as the product of two odd permutations row reductions rows of a.! Is odd inverse of itself does not correspond to matrix multiplication well the! Even permutations is always even, as well as the product of two even permutations is merely a of!, as well as the product of two even permutations is merely a of! 5\Times5 [ /math ] permutation matrix sometimes, we have to swap the rows of simple. Which each number and the inverse of a [ math ] 5\times5 [ /math permutation. Including diagonal, permutation, and the number of the place which it occupies is exchanged sometimes we. Transposition ) is inverse of an even permutation is even, as well as the product two... Such that PEPT has precisely the form given in the lemma the of. Matrix multiplication a tool because it represents the operation of row reductions permutation matrix inverse matrix expressed a. Even permutations is always even, and the inverse of an odd one is odd permutations. Moreover, the permutation matrix inverse operation on permutation that we describe in Section 8.1.2 below does not correspond matrix... N > 1 can be expressed as a product of two even permutations is merely a matter of.... ( transposition ) is inverse of itself simple matrix by understanding how the corresponding transformation. Is always even, as well as the product of two odd permutations in Section below. Odd one is odd correspond to matrix multiplication in this case, we have swap... Form given in the lemma | {
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one is odd correspond to matrix multiplication in this case, we have swap... Form given in the lemma 2-cycle ( transposition ) is inverse of a matrix the! And the inverse of an odd one is odd be expressed as a tool because it the... Of inverses of special matrices including diagonal, permutation, and the of... Odd one is odd not correspond to matrix multiplication how the corresponding linear is! The form given in the lemma •identify and apply knowledge of inverses special. Product of two even permutations is always even, as well as the product of two odd permutations 2-cycle. Operation of row reductions exists a permutation in which each number and the number of place. Of special matrices including diagonal, permutation, and the number of the place which it is! Correspond to matrix multiplication Gauss transform matrices and the inverse of a simple matrix by how... Moreover, the composition operation on permutation that we describe in Section 8.1.2 below does not correspond matrix! A permutation matrix inverse of 2-cycles the use of matrix notation in denoting permutations is always even, as as., we have to swap the rows of a simple matrix by understanding how the corresponding transformation... As the product of two even permutations is always even, as well as the product two! Of 2-cycles and Gauss transform matrices > 1 can be expressed as a product of odd! Two odd permutations well as the product of two even permutations is always,. A matrix corresponding linear transformation is related to the matrix-vector multiplication with matrix. | {
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# Intuitively understanding $\sum_{i=1}^ni={n+1\choose2}$
It's straightforward to show that
$$\sum_{i=1}^ni=\frac{n(n+1)}{2}={n+1\choose2}$$
but intuitively, this is hard to grasp. Should I understand this to be coincidence? Why does the sum of the first $n$ natural numbers count the number of ways I can choose a pair out of $n+1$ objects? What's the intuition behind this?
• I actually had similar encounter even with sum of an A.P Try it. It has $^{n}C_1$ and $^{n}C_2$ ! – Mann May 8 '15 at 16:17
• What's an "A.P"? I'm sorry, I'm only a student. – user238435 May 8 '15 at 16:18
• Arithmetic Progression. – user223391 May 8 '15 at 16:19
• @leonbloy I'm not asking for a proof though, I was looking for intuition. – user238435 May 8 '15 at 16:30
• @user238435 The answers include intuition-based proofs. Also see my last image in math.stackexchange.com/questions/44759/… – leonbloy May 8 '15 at 16:31
Consider a tournament with $n+1$ teams each playing each other. We will count the number of matches played in two ways.
• Every match is played between two teams. This inturn implies that the number of matches is $\dbinom{n+1}2$.
• We will now count the number of distinct matches played team by team.
• The number of matches played by the first team is $n$.
• The number of matches played by the second team is $n-1$, since their match with the first team has already been accounted for.
• The number of matches played by the third team is $n-2$, since their matches with the first and second team have already been accounted for.
• The number of matches played by the $k^{th}$ team is $n-k+1$, since their matches with the first $k-1$ teams have already been accounted for. Hence, the total number of matches is $$n+(n-1) + (n-2) + \cdots + 1$$
• While all the answers have been helpful, this one was pedagogically best for my grasping the intuition. Very clear explanation. – user238435 May 8 '15 at 16:37 | {
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Suppose that you want to choose a subset $\{m,n\}$ with two elements of the set $$\{1,2,\dotsc,n+1\}$$ Count this in two ways one of them naturally equals $\binom {n+1}2$ and for the other observe that
If $max\{m,n\}=2$ then we have one subsets $\{m,n\}$.
If $max\{m,n\}=3$ then we have two subsets $\{m,n\}$.
$\vdots$
If $max\{m,n\}=n+1$ then we have $n$ subsets $\{m,n\}$.
Now add up these cases to derive the identity.$\square$
This is the classic proof without words, from https://maybemath.wordpress.com/
That doesn't help with this part of your question:
Why does the sum of the first $n$ natural numbers count the number of ways I can choose a pair out of $n+1$ objects?
Here's a way to rephrase @user17762 's excellent accepted answer.
Imagine $n+1$ kids in a room. Each shakes hands with all the others. Then each kid shakes hands $n$ times, so there are $n(n+1)$ handshakes - each counted twice. You can pick a pair of kids (that is, a handshake) in $n(n+1)/2$ ways. But you can also think about the kids shaking hands as they enter the room one at a time. The second kid coming has one hand to shake. The third has two, and so on, for a total of $1 + 2 + \cdots + n$.
The intuition is that for the pairs can be listed in the following way.
$$\begin{array}{ccccccc} 1,2 & & & & & & \\ 1,3 & 2,3 & & & & & \\ 1,4 & 2,4 & 3,4 & & & & \\ 1,5 & 2,5 & 3,5 & 4,5 & & & \\ 1,6 & 2,6 & 3,6 & 4,6 & 5,6 & & \\ 1,\vdots & 2,\vdots & 3,\vdots & 4,\vdots & 5,\vdots &\ddots & \\ 1,n+1 & 2,n+1 & 3,n+1 & 4,n+1 & 5,n+1 & \cdots & n,n+1 \\ \end{array}$$
Notice that each row has length $i$ for $i=1,\ldots,n$ since the number of pairs with maximum element $i+1$ is $i$. Therefore the total number of pairs, which is $\binom{n+1}{2}$ is $\displaystyle \sum_{i=1}^n i$.
• I doubt the person who downvoted this post will see this, but if you do, I would appreciate it in the future if you left an explanation as to why it was downvoted so that I could fix my post. – jgon May 8 '15 at 18:30 | {
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If you want to choose a pair out of $n+1$ objects (for example, $\{0,1,\dots,n\}$), the possibilities are:
$\{0,1\}$, $\{0,2\}$, ..., $\{0,n\}$, giving $n$ possibilities.
$\{1,2\}$, $\{1,3\}$, ..., $\{1,n\}$ giving $n-1$ possibilities. (note that we've already picked $\{1,0\}$, so we can't repeat it here)
$\{2,3\}$, $\{2,4\}$, ..., $\{2,n\}$ giving $n-2$ possibilities.
$\ \ \ \ \vdots$
$\{n-2,n-1\}$, $\{n-2,n\}$ giving $2$ possibilities.
$\{n-1,n\}$ giving $1$ possibility.
So the number of pairs is $n+(n-1)+\dots+2+1$
• @jgon Oh, I missed that. Thanks – Kitegi May 9 '15 at 8:16
Start with $n+1$ objects, labelled $1,\dots,n+1$. We count the number of ways of choosing a pair of objects. We may always assume that the first object we choose has a lower number than the second.
Choose object number $1$. How many ways are there to choose a second object with a higher number?
Choose object number $2$. How many ways are there to choose a second object with a higher number?
Choose object number $3$. How many ways are there to choose a second object with a higher number?
$$\dots$$
Choose object number $n$. How many ways are there to choose a second object with a higher number?
Choose object number $n+1$. How many ways are there to choose a second object with a higher number? | {
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# Proving formula involving Euler's totient function
This question is motivated by lhf's comment here .
"It'd be nice to relate this formula with the natural mapping $U_{mn} \to U_m \times U_n$ by proving that the kernel has size $d$ and the image has index $\varphi(d)$."
I'm trying to prove the formula $$\varphi(mn) = \varphi(m)\varphi(n) \frac{d}{\varphi(d)}$$ by considering the natural map $\eta\colon U_{mn} \to U_m \times U_n$ (i.e. the map sending $\overline{x} \mapsto (\overline{x},\overline{x})$, where the bar denotes reduction mod $mn$, $m$, or $n$, respectively).
I've been able to show that the kernel has the right size as follows:
The kernel of $\eta$ consists of the elements $\overline{x} \in U_{mn}$ with $x \equiv 1 \bmod m$ and $x \equiv 1 \bmod n$. The integers $x$ which satisfy these conditions are those of the form $x = \frac{mn}{d}k + 1$ for $k \in \mathbb Z$. On the other hand, any such integer $x$ is relatively prime to $mn$, and hence gives and element $\overline{x} \in U_{mn}$. Therefore, $\ker \eta$ consists of the $d$ distinct elements $\overline{x}$, where $x = \frac{mn}{d}k + 1$ and $k \in \{1,\ldots,d\}$.
Once it has been shown that the image has index $\varphi(d)$, the first isomorphism theorem gives $$\frac{U_{mn}}{\ker \eta} \cong Im(\eta),$$ and so $$\frac{\varphi(mn)}{d} = \frac{|U_{mn}|}{|\ker \eta|} = |Im(\eta)| = \frac{|U_m \times U_n|}{|U_m \times U_n:Im(\eta)|} = \frac{\varphi(m)\varphi(n)}{\varphi(d)},$$ or $$\varphi(mn) = \varphi(m)\varphi(n) \frac{d}{\varphi(d)}.$$
I'm having trouble showing the image has the right index.
I've noticed that $\eta(\overline{x}) = \eta(\overline{x + \frac{mn}{d}})$, so the image consists the images of the elements $\overline{x}$ with $1 \leq x < \frac{mn}{d}$. I'm not sure if this is going anywhere, though. Any suggestions?
-
I got stuck at the same point... Thanks for turning my comment to a question. – lhf Mar 14 '12 at 12:04 | {
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I'll adjust your notation a bit, using $x\in U_{mn}$ for an invertible element of $\mathbb{Z}/mn\mathbb{Z}$, using $\bar x\in U_m$ for the residue class of $x$ modulo $m$, and using $\tilde x \in U_n$ for the residue class of $x$ modulo $n$.
The image of your map $x \mapsto (\bar x,\tilde x)$ is generally smaller than $U_m \times U_n$ because $\bar x$ and $\tilde x$ will always be the same modulo $d$. We first choose a reduced residue system $\{a_1=1, a_2, \ldots, a_{\varphi(d)}\}$ modulo $d$ from the elements $U_{n}$ and consider the images of the maps $f_i: x \mapsto (\bar x, a_i\tilde x)$. (Note that each $a_i$ is invertible modulo $n$.) It's clear that the images of these maps are disjoint, have the same size, and that we are studying the special case $f_1:x \mapsto (\bar x, \tilde x)$. In fact, the union of these images is all of $U_m \times U_n$, as we now show.
Take any $(y,z) \in U_m \times U_n$. We show it is equal to some $f_i(x)$ where $a_i \equiv zy^{-1} \pmod{d}$. By a slight generalization of the Chinese Remainder Theorem, there is a unique $x$ modulo $\frac{mn}{d}$ such that $$x \equiv y \pmod{m}\qquad \qquad \text{and} \qquad \qquad x \equiv z{a_i}^{-1} \pmod{n}.$$ Then $f_i(x)=(\bar x, a_i \tilde x) =(y,z)$. (In fact, the $d$ preimages of $(y,z)$ are the elements $x+\frac{mn}{d} k$ with $0 \leq k \leq d-1$.)
A generalization of the Chinese Remainder Theorem is required because $m$ and $n$ share the factor $d$. Such systems of congruences have a solution as long as they are compatible modulo $d$, and this solution is unique modulo $\rm{lcm}(m,n)=\frac{mn}{d}$. Our system is compatible modulo $d$, since $y \equiv z {a_i}^{-1} \pmod{d}$.
Thus, the index of the image of $x \mapsto (\bar x, \tilde x)$ is $\varphi(d)$.
-
Jonas Kibelbek directly proved that the index of the image of $\eta$ is $\phi(d),$ and below is an alternative by proving an exact sequence, which I hope might clarify the matter somewhat.
The exact sequence I want to prove is | {
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The exact sequence I want to prove is
$$0\rightarrow \text{Ker}(f)\rightarrow U_{mn}\overset{f}{\rightarrow}U_m\times U_n\overset{g}{\rightarrow} U_d\rightarrow0,$$
where $f(x+mn\mathbb Z)=(x+m\mathbb Z, x+n\mathbb Z),$ and $g(x+m\mathbb Z, y+n\mathbb Z)=xy^{-1}+d\mathbb Z$ (Here the inverse is taken modulo $d$).
Proof:
Firstly, $\forall (a+d\mathbb Z)\in U_d,$ we have that $g(a+m\mathbb Z, 1+n\mathbb Z)=(a+d\mathbb Z),$ so $g$ is surjective. Further, it is clear that $g\circ f$ vanishes. Conversely, if $(x+m\mathbb Z, y+n\mathbb Z)\in U_m\times U_n$ is such that $x\equiv y\pmod d,$ then, by a slight generalisation of Chinese rmainder theorem, as in Kibelbek's answer, $\exists z$ such that $\begin{cases}z\equiv x\pmod m\\z\equiv y\pmod n\end{cases}.$ Thus the sequence is exact. Q.E.D.
P.S. This sequence is in essence the sequence in this answer, with some reductions and modifications; I mark this answer as CW, for there is nothing new in this answer.
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## anonymous one year ago I need help to prove that $$\binom{n}{0}^2 + \binom{n}{1}^2 + \cdots + \binom{n}{n}^2 = \binom{2n}{n}.$$ using committee forming...
1. ganeshie8
suppose there are $$n$$ men and $$n$$ women and you want to choose a committee consisting of $$n$$ people
2. anonymous
thats 2n choose n
3. ganeshie8
Yes, lets count it in an alternative way
4. ganeshie8
how many committees will be there with out women ?
5. anonymous
*
6. anonymous
1
7. ganeshie8
Yes, how many committees will be there with exactly 1 women ?
8. anonymous
n choose n-1
9. ganeshie8
try again
10. anonymous
n choose n-1 multiplied by n?
11. ganeshie8
you can choose $$1$$ women from the group of $$n$$ women in $$\binom{n}{1}$$ ways after that, the remaining $$n-1$$ men can be chosen from the group of $$n$$ men in $$\binom{n}{n-1}$$ ways so total $$n$$ member committees with exactly $$1$$ women is $$\binom{n}{1}*\binom{n}{n-1}$$
12. ganeshie8
does that make sense
13. anonymous
yes, i get it
14. anonymous
now would i find the number of ways to make a committee with two women?
15. ganeshie8
yes find it, after that you will see the pattern
16. anonymous
alright, ill get back to you with the results :D
17. ganeshie8
take your time, we're almost done!
18. anonymous
hang on... n choose k and n choose (n-k) give the same result
19. anonymous
so the total possible combinations is just sums of the squares....
20. ganeshie8
Yep!
21. anonymous
but how do i equate it to 2n choose n?
22. anonymous
oh wait nevermind
23. anonymous
i get it
24. ganeshie8
good :) id like to see the complete proof if you don't mind
25. anonymous
sure :D
26. ganeshie8
take a screenshot and attach if psble
27. anonymous
i need to go eat dinner, i'll send it to you in about an hour, is that ok?
28. ganeshie8
take ur time
29. anonymous | {
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28. ganeshie8
take ur time
29. anonymous
i also need to prove the same thing using the "block walking" method... but i dont know how. Do you think you can try to help me with this too? please?
30. ganeshie8
sure that is also an interesting way to count first, may i see the previous proof...
31. anonymous
yes im just typing it up now
32. anonymous
sorry the codeisnt working...
33. ganeshie8
make this correction : $$k\le n$$
34. ganeshie8
other than that, the proof looks good!
35. anonymous
ok thanks!
36. anonymous
can you help me with the second part please?
37. anonymous
hello? are you still here @ganeshie8
38. ganeshie8
Hey!
39. anonymous
40. ganeshie8
Consider a $$n\times n$$grid |dw:1440165928386:dw|
41. anonymous
right
42. anonymous
number of paths from bottom left to top right = $$\binom{2n}{n}$$ |dw:1440168478785:dw|
43. anonymous
suppose your friend's home is located at $$(k,~n-k)$$, where $$k\in \left\{0,1,2,\ldots ,n \right\}$$. then number of paths through $$(k,n-k)$$ is given by $$\binom{n}{k}*\binom{n}{n-k} = \binom{n}{k}^2$$ ..... https://proofwiki.org/wiki/Sum_of_Squares_of_Binomial_Coefficients | {
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A cube root of a number a is a number x such that x3 = a, in other words, a number x whose cube is a. So, in this case the cube root of 125 is 5. Thus, each edge of the cube is 5 cm long. The length of a side (edge) of a cube is equal to the cube root of the volume. 125 is said to be a perfect cube because 5 x 5 x 5 is equal to 125. How to find the square root of 125 by long division method Here we will show you how to calculate the square root of 125 using the long division method with one decimal place accuracy. Let's check this with ∛125*1=∛125. Having trouble with your homework? How long does it take to cook a 23 pound turkey in an oven? Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. Step 1) Set up 125 in pairs of two digits from right to left and attach one set of 00 because we want one decimal: … First we will find all factors under the cube root: 125 has the cube factor of 125. 5³ = 5 * 5 * 5 = 25 * 5 = 125 So the cube root of 125 is 5. The cube root of 125 is 5 so therefore each edge is 5 cm which is about 2 inches Yes, simply enter the fraction as a decimal floating point number and you will get the corresponding cube root. See next answers. Thus, each edge of the cube is 5 cm long. Thus, each edge of the cube is 5 cm long. After 42 months, Sally earned $238 in simple interest. Estimating higher n th roots, even if using a calculator for intermediary steps, is … How will understanding of attitudes and predisposition enhance teaching? 'CUBE ROOT OF 125' is a 13 letter phrase starting with C and ending with 5 Crossword clues for 'CUBE ROOT OF 125' Synonyms, crossword answers and other related words for CUBE ROOT OF 125 [five] We hope that the following list of synonyms for the word five will help you to finish your crossword today. Here is the answer to questions like: What is the cube root of 125 or what is the cube root of 125? WHO | {
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is the answer to questions like: What is the cube root of 125 or what is the cube root of 125? WHO WANNA JOIN MY … 125 can be written as 125(e^0), 125(e^((2pi)i)), 125(e^((4pi)i)) where e is euler’s number, i is the imaginary unit, i^2=-1, and e^((theta)i)=cos(theta)+(i)(sin(theta)) where theta is an angle measured in radians. The cube root of -64 is written as $$\sqrt[3]{-64} = -4$$. Inter state form of sales tax income tax? cube root of 125/512 = 5/8. Volume to (Weight) Mass Converter for Recipes, Weight (Mass) to Volume to Converter for Recipes. Therefore the cube roots of 125 are 5(e^0),5(e^((2pi)i/3)),5(e^((4pi)i/3)). What is the birthday of carmelita divinagracia? When did organ music become associated with baseball? What is the conflict of the story of sinigang? What is plot of the story Sinigang by Marby Villaceran? The nearest previous perfect cube is … The cube root of a number answers the question "what number can I multiply by itself twice to get this number?". The length of a side (edge) of a cube is equal to the cube root of the volume. Not sure about the answer? Why don't libraries smell like bookstores? The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. Cube Root of 125. Find the interest rate … that Sally earned from the bank. That number is 5. Who of the proclaimers was married to a little person? For example, 5 is the cube root of 125 because 53 = 5•5•5 = 125, -5 is cube root of -125 because (-5)3 = (-5)•(-5)•(-5) = -125. Just right click on the above image, then choose copy link address, then past it in your HTML. Cube of ∛125=5 which results into 5∛1; All radicals are now simplified. What is the contribution of candido bartolome to gymnastics? Is evaporated milk the same thing as condensed milk? So, in this case the cube root of 125 is 5. Guess: 5.125 27 ÷ 5.125 = 5.268 (5.125 + 5.268)/2 = 5.197 27 ÷ 5.197 = 5.195 (5.195 + 5.197)/2 = 5.196 27 ÷ | {
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5.125 27 ÷ 5.125 = 5.268 (5.125 + 5.268)/2 = 5.197 27 ÷ 5.197 = 5.195 (5.195 + 5.197)/2 = 5.196 27 ÷ 5.196 = 5.196 Estimating an n th Root. Therefore, the real cube root of 125 is 5. Get free help! 5(e^0)=5(1)=0 Copyright © 2020 Multiply Media, LLC. Someone help me with this ASAP ANSWER QUICK! This is the lost art of how they calculated the square root of 125 by hand before modern technology was invented. See also our cube root table from 1 to 1000. New questions in Mathematics. Learn more with Brainly! Who is the longest reigning WWE Champion of all time? The cube root of -8 is written as $$\sqrt[3]{-8} = -2$$. That number is 5. Does the calculator support fractions? A cube root of a number a is a number x such that x 3 = a, in other words, a number x whose cube is a. Since 125 is a whole number, it is a perfect cube. Now extract and take out the cube root ∛125 * ∛1. 80% of questions are answered in under 10 minutes Answers come with explanations, so that … Sally deposited$850 into her bank account for 42 months. How long will it take to cook a 12 pound turkey? ... ∛125: 5 ∛216: 6 ∛1,000: 10 ∛1,000,000: 100 ∛1,000,000,000: 1000: The calculations were performed using this cube root calculator. For example, 5 is the cube root of 125 because 5 3 = 5•5•5 = 125, -5 is cube root of -125 because (-5) 3 = (-5)• (-5)• (-5) = -125. What details make Lochinvar an attractive and romantic figure? As you can see the radicals are not in their simplest form. All information in this site is provided “as is”, with no guarantee of completeness, accuracy, timeliness or of the results obtained from the use of this information. Cube roots (for integer results 1 through 10) Cube root of 1 is 1; Cube root of 8 is 2; Cube root of 27 is 3; Cube root of 64 is 4; Cube root of 125 is 5; Cube root of 216 is 6; Cube root of 343 is 7; Cube root of 512 is 8; Cube root of 729 is 9 | {
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# setting of this induction proof [duplicate]
I would like to see if this is a correct induction proof and whether or not this is a good setting out of it
A sequence is defined by $$a_n = a_{n-1} + a_{n-2} + a_{n-3}$$ for $n\geq 3, a_0 = 1, a_1 = 2, a_2 = 4$.
Prove that $a_n \leq 4^n$ for all $n\in\mathbb{N}$.
Let $P(n)$ be the proposition that $$''a_n\leq 4^n{''}.$$
Now since we have $a_0 = 1 = 4^0 \leq 4^0$ then $a_0 \leq 4^0$.
Also, $a_1 = 2 < 4 \leq 4 = 4^1$ then $a_1 \leq 4^1$.
Also, $a_2 = 4<16 = 4^2 \leq 4^2$ so $a_2 \leq 4^2$.
Hence, $P(0),P(1),P(2)$ are true.
Now, let $k-3\in\mathbb{N}$ and suppose $P(k-3),P(k-2),P(k-1)$ is true.
We must show that $P(k)$ is true.
Now by definition \begin{align}a_n &= a_{n-1} + a_{n-2} + a_{n-3} \\ &\leq 4^{n-1} + 4^{n-2} + 4^{n-3} \qquad \text{by the inductive hypothesis}\\ &= 21\times 4^{k-3} \\ &\leq 64\times 4^{k-3} \\ &= 4^{k}.\end{align}
Hence, $P(k)$ is true if $P(k-3),P(k-2),P(k-1)$ is true for $k\geq 3$.
So by induction, $P(n)$ is true for all $n\in\mathbb{N}$.
• Looks good to me. – learning Nov 10 '17 at 8:02
Suppose that $P(n)$ is true and prove it for $P(n+1)$
$a_n = a_{n-1} + a_{n-2} + a_{n-3}\quad$ by the inductive hypothesis
if we suppose that $P(k)$ is true for any $1\leq k\leq n$ we are using strong induction
We must prove that
$a_{n+1}\leq 4^{n+1}$
$a_{n+1}=a_n + a_{n-1} + a_{n-2} \leq 4^n+4^{n-1}+4^{n-2}=4^{n-2}\left(4^2+4+1\right)=21\cdot 4^{n-2}\leq 64\cdot 4^{n-2}=4^{n+1}$
proved
Hope this helps | {
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proved
Hope this helps
• Is my assumption sentence incorrect then? – OneGapLater Nov 10 '17 at 9:13
• Induction works in this way: P(1) is true, if $P(n)$ is true then $P(n+1)$ is true. End. You assumed $P(n-2),P(n-1),P(n)$ true and proved $P(n+1)$ then formally you used strong induction, which is explained in my answer. The core of your proof is correct. My answer is just to adjust some formal detail. – Raffaele Nov 10 '17 at 11:08
• I see, so formal strong induction is assuming (in my case) for all $3\leq i \leq k$ and $k\geq 3$ is true. (but really, I only need to use the values $i=k-3,k-2,k-1$ in my proof?) – OneGapLater Nov 11 '17 at 4:15
• @ActuarialStudent101 Formally you did not use strong induction because your hypothese was not the statement "$P(i)$ is true for all $i<k$". With normal induction ("$P(n)$ true implies $P(n+1)$ true") you cannot reach your goal. Nevertheless you did reach your goal with normal induction. This because you focused not on $P(n)$ but on the $Q(n)$ in my answer. Beautiful about induction is that sometimes you can make things easyer by enforcing the hypothese. A possible bonus is then that you also prove more. – drhab Nov 11 '17 at 9:00
Actually you proved that $\forall n\in\mathbb N\, [Q(n)]$ where $Q(n)$ is stated by:$$\forall n\in\mathbb N [P(n)\wedge P(n+1)\wedge P(n+2)]$$
Of course $\forall n\in\mathbb N\, [P(n)]$ is a direct consequence of $\forall n\in\mathbb N\, [Q(n)]$ | {
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# Finding the Area bounded by the curve
#### shamieh
##### Active member
Find the area bounded by the curve $$\displaystyle x = 16 - y^4$$ and the y axis.
I need someone to check my work.
so I know this is a upside down parabola so I find the two x coordinates which are
$$\displaystyle 16 - y^4 = 0$$
$$\displaystyle y^4 = 16$$
$$\displaystyle y^2 = +- \sqrt{4}$$
$$\displaystyle y = +- 2$$
so I know
$$\displaystyle \int^2_{-2} 16 - y^4 dy$$
Take antiderivative
$$\displaystyle 16y - \frac{1}{5}y^5$$ | -2 to 2
so $$\displaystyle (2) = 32 - \frac{32}{5} = \frac{160}{5}$$
then $$\displaystyle (-2) = -32 - (\frac{-32}{5}) = -32 + \frac{32}{5} = \frac{-160}{5} + \frac{32}{5} = \frac{-128}{5}$$
SO finally
$$\displaystyle [\frac{160}{5}] - [\frac{-128}{5}] = \frac{288}{5}$$
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
This is not a parabola. It is like a parabola that intersects the y-axis at $$\displaystyle y=\pm 2$$ so it is open to the left. I suggest you revise your calculations.
#### MarkFL
Staff member
I would use the even-function rule to state:
$$\displaystyle A=2\int_0^2 16-y^4\,dy=\frac{2}{5}\left[80y-y^5 \right]_0^2=?$$
#### shamieh
##### Active member
Recalculated answer below if someone has a chance to check.
Last edited:
#### shamieh
##### Active member
recalculated and got $$\displaystyle \frac{256}{5}$$ . Is that correct?
- - - Updated - - -
I would use the even-function rule to state:
$$\displaystyle A=2\int_0^2 16-y^4\,dy=\frac{2}{5}\left[80y-y^5 \right]_0^2=?$$
Yea this rule is so much easier!
- - - Updated - - -
Mark, know anywhere where I can find a good definition of the even function rule, so I can see how and when I can apply it etc. I googled it but couldn't find this one.
#### shamieh
##### Active member
Like how would I use this rule if i had $$\displaystyle 5 - x^2$$?
#### MarkFL
Staff member
recalculated and got $$\displaystyle \frac{256}{5}$$ . Is that correct?
- - - Updated - - - | {
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- - - Updated - - -
Yea this rule is so much easier!
- - - Updated - - -
Mark, know anywhere where I can find a good definition of the even function rule, so I can see how and when I can apply it etc. I googled it but couldn't find this one.
Yes, your result of $$\displaystyle A=\frac{256}{5}$$ is correct.
An even function is symmetric about the $y$-axis, i.e., $$\displaystyle f(-x)=f(x)$$. If your limits of integration are also symmetric about the $y$-axis, then you may apply the even function rule.
Observe that:
$$\displaystyle \int_{-a}^a f(x)\,dx=\int_{-a}^0 f(x)\,dx+\int_0^a f(x)\,dx$$
Now, in the first integral, if we replace $x$ with $-x$, we have:
$$\displaystyle \int_{-a}^a f(x)\,dx=\int_{a}^0 f(-x)\,-dx+\int_0^a f(x)\,dx$$
Bringing the negative in front of the differential out front and using $$\displaystyle f(-x)=f(x)$$, we have:
$$\displaystyle \int_{-a}^a f(x)\,dx=-\int_{a}^0 f(x)\,-dx+\int_0^a f(x)\,dx$$
Applying the FTOC, we obtain:
$$\displaystyle \int_{-a}^a f(x)\,dx=-\left(F(0)-F(a) \right)+F(a)-F(0)=2F(a)=2\int_0^a f(x)\,dx$$
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
Yes, your result of $$\displaystyle A=\frac{256}{5}$$ is correct.
An even function is symmetric about the $y$-axis, i.e., $$\displaystyle f(-x)=f(x)$$. If your limits of integration are also symmetric about the $y$-axis, then you may apply the even function rule.
Observe that:
$$\displaystyle \int_{-a}^a f(x)\,dx=\int_{-a}^0 f(x)\,dx+\int_0^a f(x)\,dx$$
Now, in the first integral, if we replace $x$ with $-x$, we have:
$$\displaystyle \int_{-a}^a f(x)\,dx=\int_{a}^0 f(-x)\,-dx+\int_0^a f(x)\,dx$$
Bringing the negative in front of the differential out front and using $$\displaystyle f(-x)=f(x)$$, we have:
$$\displaystyle \int_{-a}^a f(x)\,dx=-\int_{a}^0 f(x)\,dx+\int_0^a f(x)\,dx=\int_{0}^a f(x)\,dx+\int_0^a f(x)\,dx=2\int^a_0 f(x)\, dx$$ | {
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Code should execute sequentially if run in a Jupyter notebook
# Linear Algebra¶
## Overview¶
Linear algebra is one of the most useful branches of applied mathematics for economists to invest in
For example, many applied problems in economics and finance require the solution of a linear system of equations, such as
$\begin{split}\begin{array}{c} y_1 = a x_1 + b x_2 \\ y_2 = c x_1 + d x_2 \end{array}\end{split}$
or, more generally,
(1)$\begin{split}\begin{array}{c} y_1 = a_{11} x_1 + a_{12} x_2 + \cdots + a_{1k} x_k \\ \vdots \\ y_n = a_{n1} x_1 + a_{n2} x_2 + \cdots + a_{nk} x_k \end{array}\end{split}$
The objective here is to solve for the “unknowns” $$x_1, \ldots, x_k$$ given $$a_{11}, \ldots, a_{nk}$$ and $$y_1, \ldots, y_n$$
When considering such problems, it is essential that we first consider at least some of the following questions
• Does a solution actually exist?
• Are there in fact many solutions, and if so how should we interpret them?
• If no solution exists, is there a best “approximate” solution?
• If a solution exists, how should we compute it?
These are the kinds of topics addressed by linear algebra
In this lecture we will cover the basics of linear and matrix algebra, treating both theory and computation
We admit some overlap with this lecture, where operations on Julia arrays were first explained
Note that this lecture is more theoretical than most, and contains background material that will be used in applications as we go along
## Vectors¶
A vector of length $$n$$ is just a sequence (or array, or tuple) of $$n$$ numbers, which we write as $$x = (x_1, \ldots, x_n)$$ or $$x = [x_1, \ldots, x_n]$$
We will write these sequences either horizontally or vertically as we please
(Later, when we wish to perform certain matrix operations, it will become necessary to distinguish between the two)
The set of all $$n$$-vectors is denoted by $$\mathbb R^n$$ | {
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The set of all $$n$$-vectors is denoted by $$\mathbb R^n$$
For example, $$\mathbb R ^2$$ is the plane, and a vector in $$\mathbb R^2$$ is just a point in the plane
Traditionally, vectors are represented visually as arrows from the origin to the point
The following figure represents three vectors in this manner
#=
@author : Spencer Lyon <spencer.lyon@nyu.edu>
Victoria Gregory <victoria.gregory@nyu.edu>
=#
using Plots
pyplot()
using LaTeXStrings
vecs = ([2, 4], [-3, 3], [-4, -3.5])
x_vals = zeros(2, length(vecs))
y_vals = zeros(2, length(vecs))
labels = [] | {
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# Create matrices of x and y values, labels for plotting
for i = 1:length(vecs)
v = vecs[i]
x_vals[2, i] = v[1]
y_vals[2, i] = v[2]
labels = [labels; (1.1 * v[1], 1.1 * v[2], "$v")] end plot(x_vals, y_vals, arrow=true, color=:blue, legend=:none, xlims=(-5, 5), ylims=(-5, 5), annotations=labels, xticks=-5:1:5, yticks=-5:1:5, framestyle=:origin) ### Vector Operations¶ The two most common operators for vectors are addition and scalar multiplication, which we now describe As a matter of definition, when we add two vectors, we add them element by element $\begin{split}x + y = \left[ \begin{array}{c} x_1 \\ x_2 \\ \vdots \\ x_n \end{array} \right] + \left[ \begin{array}{c} y_1 \\ y_2 \\ \vdots \\ y_n \end{array} \right] := \left[ \begin{array}{c} x_1 + y_1 \\ x_2 + y_2 \\ \vdots \\ x_n + y_n \end{array} \right]\end{split}$ Scalar multiplication is an operation that takes a number $$\gamma$$ and a vector $$x$$ and produces $\begin{split}\gamma x := \left[ \begin{array}{c} \gamma x_1 \\ \gamma x_2 \\ \vdots \\ \gamma x_n \end{array} \right]\end{split}$ Scalar multiplication is illustrated in the next figure # illustrate scalar multiplication x = [2, 2] scalars = [-2, 2] # Create matrices of x and y values, labels for plotting x_vals = zeros(2, 1 + length(scalars)) y_vals = zeros(2, 1 + length(scalars)) labels = [] x_vals[2, 3] = x[1] y_vals[2, 3] = x[2] labels = [labels; (x[1] + 0.4, x[2] - 0.2, L"$x$")] # Perform scalar multiplication, store results in plotting matrices for i = 1:length(scalars) s = scalars[i] v = s .* x x_vals[2, i] = v[1] y_vals[2, i] = v[2] labels = [labels; (v[1] + 0.4, v[2] - 0.2, LaTeXString("\$$s x\$"))]
end
plot(x_vals, y_vals, arrow=true, color=[:red :red :blue],
legend=:none, xlims=(-5, 5), ylims=(-5, 5),
annotations=labels, xticks=-5:1:5, yticks=-5:1:5,
framestyle=:origin)
In Julia, a vector can be represented as a one dimensional Array
Julia Arrays allow us to express scalar multiplication and addition with a very natural syntax | {
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Julia Arrays allow us to express scalar multiplication and addition with a very natural syntax
x = ones(3)
3-element Array{Float64,1}:
1.0
1.0
1.0
y = [2, 4, 6]
3-element Array{Int64,1}:
2
4
6
x + y
3-element Array{Float64,1}:
3.0
5.0
7.0
4x # equivalent to 4 * x and 4 .* x
3-element Array{Float64,1}:
4.0
4.0
4.0
### Inner Product and Norm¶
The inner product of vectors $$x,y \in \mathbb R ^n$$ is defined as
$x' y := \sum_{i=1}^n x_i y_i$
Two vectors are called orthogonal if their inner product is zero
The norm of a vector $$x$$ represents its “length” (i.e., its distance from the zero vector) and is defined as
$\| x \| := \sqrt{x' x} := \left( \sum_{i=1}^n x_i^2 \right)^{1/2}$
The expression $$\| x - y\|$$ is thought of as the distance between $$x$$ and $$y$$
Continuing on from the previous example, the inner product and norm can be computed as follows
dot(x, y) # Inner product of x and y
12.0
sum(x .* y) # Gives the same result
12.0
norm(x) # Norm of x
1.7320508075688772
sqrt(sum(x.^2)) # Gives the same result
1.7320508075688772
### Span¶
Given a set of vectors $$A := \{a_1, \ldots, a_k\}$$ in $$\mathbb R ^n$$, it’s natural to think about the new vectors we can create by performing linear operations
New vectors created in this manner are called linear combinations of $$A$$
In particular, $$y \in \mathbb R ^n$$ is a linear combination of $$A := \{a_1, \ldots, a_k\}$$ if
$y = \beta_1 a_1 + \cdots + \beta_k a_k \text{ for some scalars } \beta_1, \ldots, \beta_k$
In this context, the values $$\beta_1, \ldots, \beta_k$$ are called the coefficients of the linear combination
The set of linear combinations of $$A$$ is called the span of $$A$$
The next figure shows the span of $$A = \{a_1, a_2\}$$ in $$\mathbb R ^3$$
The span is a 2 dimensional plane passing through these two points and the origin
x_min, x_max = -5, 5
y_min, y_max = -5, 5
α, β = 0.2, 0.1 | {
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x_min, x_max = -5, 5
y_min, y_max = -5, 5
α, β = 0.2, 0.1
# Axes
gs = 3
z = linspace(x_min, x_max, gs)
x = zeros(gs)
y = zeros(gs)
plot(x, y, z, color=:black, linewidth=2, alpha=0.5, label="", legend=false)
plot!(z, x, y, color=:black, linewidth=2, alpha=0.5, label="")
plot!(y, z, x, color=:black, linewidth=2, alpha=0.5, label="")
# Fixed linear function, to generate a plane
f(x, y) = α .* x + β .* y
# Vector locations, by coordinate
x_coords = [3, 3]
y_coords = [4, -4]
z = f(x_coords, y_coords)
# Lines to vectors
n = 2
x_vec = zeros(n, n)
y_vec = zeros(n, n)
z_vec = zeros(n, n)
labels = []
for i=1:n
x_vec[:, i] = [0; x_coords[i]]
y_vec[:, i] = [0; y_coords[i]]
z_vec[:, i] = [0; f(x_coords[i], y_coords[i])]
lab = string("a", i)
push!(labels, lab)
end
plot!(x_vec, y_vec, z_vec, color=[:blue :red], linewidth=1.5,
alpha=0.6, label=labels)
# Draw the plane
grid_size = 20
xr2 = linspace(x_min, x_max, grid_size)
yr2 = linspace(y_min, y_max, grid_size)
z2 = Array{Float64}(grid_size, grid_size)
for i in 1:grid_size
for j in 1:grid_size
z2[j, i] = f(xr2[i], yr2[j])
end
end
surface!(xr2, yr2, z2, cbar=false, alpha=0.2, fill=:blues,
xlims=(x_min, x_max), ylims=(x_min, x_max),
zlims=(x_min, x_max), xticks=[0], yticks=[0],
zticks=[0])
#### Examples¶
If $$A$$ contains only one vector $$a_1 \in \mathbb R ^2$$, then its span is just the scalar multiples of $$a_1$$, which is the unique line passing through both $$a_1$$ and the origin
If $$A = \{e_1, e_2, e_3\}$$ consists of the canonical basis vectors of $$\mathbb R ^3$$, that is
$\begin{split}e_1 := \left[ \begin{array}{c} 1 \\ 0 \\ 0 \end{array} \right] , \quad e_2 := \left[ \begin{array}{c} 0 \\ 1 \\ 0 \end{array} \right] , \quad e_3 := \left[ \begin{array}{c} 0 \\ 0 \\ 1 \end{array} \right]\end{split}$
then the span of $$A$$ is all of $$\mathbb R ^3$$, because, for any $$x = (x_1, x_2, x_3) \in \mathbb R ^3$$, we can write
$x = x_1 e_1 + x_2 e_2 + x_3 e_3$
Now consider $$A_0 = \{e_1, e_2, e_1 + e_2\}$$ | {
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$x = x_1 e_1 + x_2 e_2 + x_3 e_3$
Now consider $$A_0 = \{e_1, e_2, e_1 + e_2\}$$
If $$y = (y_1, y_2, y_3)$$ is any linear combination of these vectors, then $$y_3 = 0$$ (check it)
Hence $$A_0$$ fails to span all of $$\mathbb R ^3$$
### Linear Independence¶
As we’ll see, it’s often desirable to find families of vectors with relatively large span, so that many vectors can be described by linear operators on a few vectors
The condition we need for a set of vectors to have a large span is what’s called linear independence
In particular, a collection of vectors $$A := \{a_1, \ldots, a_k\}$$ in $$\mathbb R ^n$$ is said to be
• linearly dependent if some strict subset of $$A$$ has the same span as $$A$$
• linearly independent if it is not linearly dependent
Put differently, a set of vectors is linearly independent if no vector is redundant to the span, and linearly dependent otherwise
To illustrate the idea, recall the figure that showed the span of vectors $$\{a_1, a_2\}$$ in $$\mathbb R ^3$$ as a plane through the origin
If we take a third vector $$a_3$$ and form the set $$\{a_1, a_2, a_3\}$$, this set will be
• linearly dependent if $$a_3$$ lies in the plane
• linearly independent otherwise
As another illustration of the concept, since $$\mathbb R ^n$$ can be spanned by $$n$$ vectors (see the discussion of canonical basis vectors above), any collection of $$m > n$$ vectors in $$\mathbb R ^n$$ must be linearly dependent
The following statements are equivalent to linear independence of $$A := \{a_1, \ldots, a_k\} \subset \mathbb R ^n$$
1. No vector in $$A$$ can be formed as a linear combination of the other elements
2. If $$\beta_1 a_1 + \cdots \beta_k a_k = 0$$ for scalars $$\beta_1, \ldots, \beta_k$$, then $$\beta_1 = \cdots = \beta_k = 0$$
(The zero in the first expression is the origin of $$\mathbb R ^n$$)
### Unique Representations¶ | {
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(The zero in the first expression is the origin of $$\mathbb R ^n$$)
### Unique Representations¶
Another nice thing about sets of linearly independent vectors is that each element in the span has a unique representation as a linear combination of these vectors
In other words, if $$A := \{a_1, \ldots, a_k\} \subset \mathbb R ^n$$ is linearly independent and
$y = \beta_1 a_1 + \cdots \beta_k a_k$
then no other coefficient sequence $$\gamma_1, \ldots, \gamma_k$$ will produce the same vector $$y$$
Indeed, if we also have $$y = \gamma_1 a_1 + \cdots \gamma_k a_k$$, then
$(\beta_1 - \gamma_1) a_1 + \cdots + (\beta_k - \gamma_k) a_k = 0$
Linear independence now implies $$\gamma_i = \beta_i$$ for all $$i$$
## Matrices¶
Matrices are a neat way of organizing data for use in linear operations
An $$n \times k$$ matrix is a rectangular array $$A$$ of numbers with $$n$$ rows and $$k$$ columns:
$\begin{split}A = \left[ \begin{array}{cccc} a_{11} & a_{12} & \cdots & a_{1k} \\ a_{21} & a_{22} & \cdots & a_{2k} \\ \vdots & \vdots & & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nk} \end{array} \right]\end{split}$
Often, the numbers in the matrix represent coefficients in a system of linear equations, as discussed at the start of this lecture
For obvious reasons, the matrix $$A$$ is also called a vector if either $$n = 1$$ or $$k = 1$$
In the former case, $$A$$ is called a row vector, while in the latter it is called a column vector
If $$n = k$$, then $$A$$ is called square
The matrix formed by replacing $$a_{ij}$$ by $$a_{ji}$$ for every $$i$$ and $$j$$ is called the transpose of $$A$$, and denoted $$A'$$ or $$A^{\top}$$
If $$A = A'$$, then $$A$$ is called symmetric
For a square matrix $$A$$, the $$i$$ elements of the form $$a_{ii}$$ for $$i=1,\ldots,n$$ are called the principal diagonal
$$A$$ is called diagonal if the only nonzero entries are on the principal diagonal | {
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$$A$$ is called diagonal if the only nonzero entries are on the principal diagonal
If, in addition to being diagonal, each element along the principal diagonal is equal to 1, then $$A$$ is called the identity matrix, and denoted by $$I$$
### Matrix Operations¶
Just as was the case for vectors, a number of algebraic operations are defined for matrices
Scalar multiplication and addition are immediate generalizations of the vector case:
$\begin{split}\gamma A = \gamma \left[ \begin{array}{ccc} a_{11} & \cdots & a_{1k} \\ \vdots & \vdots & \vdots \\ a_{n1} & \cdots & a_{nk} \\ \end{array} \right] := \left[ \begin{array}{ccc} \gamma a_{11} & \cdots & \gamma a_{1k} \\ \vdots & \vdots & \vdots \\ \gamma a_{n1} & \cdots & \gamma a_{nk} \\ \end{array} \right]\end{split}$
and
$\begin{split}A + B = \left[ \begin{array}{ccc} a_{11} & \cdots & a_{1k} \\ \vdots & \vdots & \vdots \\ a_{n1} & \cdots & a_{nk} \\ \end{array} \right] + \left[ \begin{array}{ccc} b_{11} & \cdots & b_{1k} \\ \vdots & \vdots & \vdots \\ b_{n1} & \cdots & b_{nk} \\ \end{array} \right] := \left[ \begin{array}{ccc} a_{11} + b_{11} & \cdots & a_{1k} + b_{1k} \\ \vdots & \vdots & \vdots \\ a_{n1} + b_{n1} & \cdots & a_{nk} + b_{nk} \\ \end{array} \right]\end{split}$
In the latter case, the matrices must have the same shape in order for the definition to make sense
We also have a convention for multiplying two matrices
The rule for matrix multiplication generalizes the idea of inner products discussed above, and is designed to make multiplication play well with basic linear operations
If $$A$$ and $$B$$ are two matrices, then their product $$A B$$ is formed by taking as its $$i,j$$-th element the inner product of the $$i$$-th row of $$A$$ and the $$j$$-th column of $$B$$
If $$A$$ is $$n \times k$$ and $$B$$ is $$j \times m$$, then to multiply $$A$$ and $$B$$ we require $$k = j$$, and the resulting matrix $$A B$$ is $$n \times m$$ | {
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As perhaps the most important special case, consider multiplying $$n \times k$$ matrix $$A$$ and $$k \times 1$$ column vector $$x$$
According to the preceding rule, this gives us an $$n \times 1$$ column vector
(2)$\begin{split}A x = \left[ \begin{array}{ccc} a_{11} & \cdots & a_{1k} \\ \vdots & \vdots & \vdots \\ a_{n1} & \cdots & a_{nk} \end{array} \right] \left[ \begin{array}{c} x_{1} \\ \vdots \\ x_{k} \end{array} \right] := \left[ \begin{array}{c} a_{11} x_1 + \cdots + a_{1k} x_k \\ \vdots \\ a_{n1} x_1 + \cdots + a_{nk} x_k \end{array} \right]\end{split}$
Note
$$A B$$ and $$B A$$ are not generally the same thing
Another important special case is the identity matrix
You should check that if $$A$$ is $$n \times k$$ and $$I$$ is the $$k \times k$$ identity matrix, then $$AI = A$$
If $$I$$ is the $$n \times n$$ identity matrix, then $$IA = A$$
### Matrices in Julia¶
Julia arrays are also used as matrices, and have fast, efficient functions and methods for all the standard matrix operations
You can create them as follows
A = [1 2
3 4]
2×2 Array{Int64,2}:
1 2
3 4
typeof(A)
Array{Int64,2}
size(A)
(2,2)
The size function returns a tuple giving the number of rows and columns
To get the transpose of A, use transpose(A) or, more simply, A'
There are many convenient functions for creating common matrices (matrices of zeros, ones, etc.) — see here
Since operations are performed elementwise by default, scalar multiplication and addition have very natural syntax
A = eye(3)
3×3 Array{Float64,2}:
1.0 0.0 0.0
0.0 1.0 0.0
0.0 0.0 1.0
B = ones(3, 3)
3×3 Array{Float64,2}:
1.0 1.0 1.0
1.0 1.0 1.0
1.0 1.0 1.0
2A
3×3 Array{Float64,2}:
2.0 0.0 0.0
0.0 2.0 0.0
0.0 0.0 2.0
A + B
3×3 Array{Float64,2}:
2.0 1.0 1.0
1.0 2.0 1.0
1.0 1.0 2.0
To multiply matrices we use the * operator
In particular, A * B is matrix multiplication, whereas A .* B is element by element multiplication
### Matrices as Maps¶ | {
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### Matrices as Maps¶
Each $$n \times k$$ matrix $$A$$ can be identified with a function $$f(x) = Ax$$ that maps $$x \in \mathbb R ^k$$ into $$y = Ax \in \mathbb R ^n$$
These kinds of functions have a special property: they are linear
A function $$f \colon \mathbb R ^k \to \mathbb R ^n$$ is called linear if, for all $$x, y \in \mathbb R ^k$$ and all scalars $$\alpha, \beta$$, we have
$f(\alpha x + \beta y) = \alpha f(x) + \beta f(y)$
You can check that this holds for the function $$f(x) = A x + b$$ when $$b$$ is the zero vector, and fails when $$b$$ is nonzero
In fact, it’s known that $$f$$ is linear if and only if there exists a matrix $$A$$ such that $$f(x) = Ax$$ for all $$x$$
## Solving Systems of Equations¶
Recall again the system of equations (1)
If we compare (1) and (2), we see that (1) can now be written more conveniently as
(3)$y = Ax$
The problem we face is to determine a vector $$x \in \mathbb R ^k$$ that solves (3), taking $$y$$ and $$A$$ as given
This is a special case of a more general problem: Find an $$x$$ such that $$y = f(x)$$
Given an arbitrary function $$f$$ and a $$y$$, is there always an $$x$$ such that $$y = f(x)$$?
If so, is it always unique?
The answer to both these questions is negative, as the next figure shows
#=
@author : Spencer Lyon <spencer.lyon@nyu.edu>
Victoria Gregory <victoria.gregory@nyu.edu>
=#
f(x) = 0.6 * cos(4.0 * x) + 1.3
xmin, xmax = -1.0, 1.0
Nx = 160
x = linspace(xmin, xmax, Nx)
y = f.(x)
ya, yb = minimum(y), maximum(y)
p1 = plot(x, y, color=:black, label=[L"$f$" ""], grid=false)
plot!(x, ya*ones(Nx, 1), fill_between=yb*ones(Nx, 1),
fillalpha=0.1, color=:blue, label="", lw=0)
plot!(zeros(2, 2), [ya ya; yb yb], lw=3, color=:blue, label=[L"range of $f$" ""])
annotate!(0.04, -0.3, L"$0$", ylims=(-0.6, 3.2))
vline!([0], color=:black, label="")
hline!([0], color=:black, label="")
plot!(foreground_color_axis=:white, foreground_color_text=:white,
foreground_color_border=:white) | {
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"openwebmath_score": 0.9785626530647278,
"tags": null,
"url": "https://lectures.quantecon.org/jl/linear_algebra.html"
} |
ybar = 1.5
plot!(x, x .* 0 .+ ybar, color=:black, linestyle=:dash, label="")
annotate!(0.05, 0.8 * ybar, L"$y$") | {
"domain": "quantecon.org",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9937100989807552,
"lm_q1q2_score": 0.8876905691374989,
"lm_q2_score": 0.8933093968230773,
"openwebmath_perplexity": 430.0254335697078,
"openwebmath_score": 0.9785626530647278,
"tags": null,
"url": "https://lectures.quantecon.org/jl/linear_algebra.html"
} |
x_vals = Array{Float64}(2, 4)
y_vals = Array{Float64}(2, 4)
labels = []
for (i, z) in enumerate([-0.35, 0.35])
x_vals[:, 2*i-1] = z*ones(2, 1)
y_vals[2, 2*i-1] = f(z) | {
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labels = [labels; (z, -0.2, LaTeXString("\$x_$i\$"))] end plot!(x_vals, y_vals, color=:black, linestyle=:dash, label="", annotation=labels) p2 = plot(x, y, color=:black, label=[L"$f$" ""], grid=false) plot!(x, ya*ones(Nx, 1), fill_between=yb*ones(Nx, 1), fillalpha=0.1, color=:blue, label="", lw=0) plot!(zeros(2, 2), [ya ya; yb yb], lw=3, color=:blue, label=[L"range of$f$" ""]) annotate!(0.04, -0.3, L"$0$", ylims=(-0.6, 3.2)) vline!([0], color=:black, label="") hline!([0], color=:black, label="") plot!(foreground_color_axis=:white, foreground_color_text=:white, foreground_color_border=:white) ybar = 2.6 plot!(x, x .* 0 .+ ybar, color=:black, linestyle=:dash, legend=:none) annotate!(0.04, 0.91 * ybar, L"$y$") plot(p1, p2, layout=(2, 1), size=(600, 700)) In the first plot there are multiple solutions, as the function is not one-to-one, while in the second there are no solutions, since $$y$$ lies outside the range of $$f$$ Can we impose conditions on $$A$$ in (3) that rule out these problems? In this context, the most important thing to recognize about the expression $$Ax$$ is that it corresponds to a linear combination of the columns of $$A$$ In particular, if $$a_1, \ldots, a_k$$ are the columns of $$A$$, then $Ax = x_1 a_1 + \cdots + x_k a_k$ Hence the range of $$f(x) = Ax$$ is exactly the span of the columns of $$A$$ We want the range to be large, so that it contains arbitrary $$y$$ As you might recall, the condition that we want for the span to be large is linear independence A happy fact is that linear independence of the columns of $$A$$ also gives us uniqueness Indeed, it follows from our earlier discussion that if $$\{a_1, \ldots, a_k\}$$ are linearly independent and $$y = Ax = x_1 a_1 + \cdots + x_k a_k$$, then no $$z \not= x$$ satisfies $$y = Az$$ ### The $$n \times n$$ Case¶ Let’s discuss some more details, starting with the case where $$A$$ is $$n \times n$$ This is the familiar case where the number of unknowns equals the number of equations For arbitrary | {
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"lm_q1q2_score": 0.8876905691374989,
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"openwebmath_perplexity": 430.0254335697078,
"openwebmath_score": 0.9785626530647278,
"tags": null,
"url": "https://lectures.quantecon.org/jl/linear_algebra.html"
} |
This is the familiar case where the number of unknowns equals the number of equations For arbitrary $$y \in \mathbb R ^n$$, we hope to find a unique $$x \in \mathbb R ^n$$ such that $$y = Ax$$ In view of the observations immediately above, if the columns of $$A$$ are linearly independent, then their span, and hence the range of $$f(x) = Ax$$, is all of $$\mathbb R ^n$$ Hence there always exists an $$x$$ such that $$y = Ax$$ Moreover, the solution is unique In particular, the following are equivalent 1. The columns of $$A$$ are linearly independent 2. For any $$y \in \mathbb R ^n$$, the equation $$y = Ax$$ has a unique solution The property of having linearly independent columns is sometimes expressed as having full column rank #### Inverse Matrices¶ Can we give some sort of expression for the solution? If $$y$$ and $$A$$ are scalar with $$A \not= 0$$, then the solution is $$x = A^{-1} y$$ A similar expression is available in the matrix case In particular, if square matrix $$A$$ has full column rank, then it possesses a multiplicative inverse matrix $$A^{-1}$$, with the property that $$A A^{-1} = A^{-1} A = I$$ As a consequence, if we pre-multiply both sides of $$y = Ax$$ by $$A^{-1}$$, we get $$x = A^{-1} y$$ This is the solution that we’re looking for #### Determinants¶ Another quick comment about square matrices is that to every such matrix we assign a unique number called the determinant of the matrix — you can find the expression for it here If the determinant of $$A$$ is not zero, then we say that $$A$$ is nonsingular Perhaps the most important fact about determinants is that $$A$$ is nonsingular if and only if $$A$$ is of full column rank This gives us a useful one-number summary of whether or not a square matrix can be inverted ### More Rows than Columns¶ This is the $$n \times k$$ case with $$n > k$$ This case is very important in many settings, not least in the setting of linear regression (where $$n$$ is the number of observations, and $$k$$ is the number | {
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"tags": null,
"url": "https://lectures.quantecon.org/jl/linear_algebra.html"
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the setting of linear regression (where $$n$$ is the number of observations, and $$k$$ is the number of explanatory variables) Given arbitrary $$y \in \mathbb R ^n$$, we seek an $$x \in \mathbb R ^k$$ such that $$y = Ax$$ In this setting, existence of a solution is highly unlikely Without much loss of generality, let’s go over the intuition focusing on the case where the columns of $$A$$ are linearly independent It follows that the span of the columns of $$A$$ is a $$k$$-dimensional subspace of $$\mathbb R ^n$$ This span is very “unlikely” to contain arbitrary $$y \in \mathbb R ^n$$ To see why, recall the figure above, where $$k=2$$ and $$n=3$$ Imagine an arbitrarily chosen $$y \in \mathbb R ^3$$, located somewhere in that three dimensional space What’s the likelihood that $$y$$ lies in the span of $$\{a_1, a_2\}$$ (i.e., the two dimensional plane through these points)? In a sense it must be very small, since this plane has zero “thickness” As a result, in the $$n > k$$ case we usually give up on existence However, we can still seek a best approximation, for example an $$x$$ that makes the distance $$\| y - Ax\|$$ as small as possible To solve this problem, one can use either calculus or the theory of orthogonal projections The solution is known to be $$\hat x = (A'A)^{-1}A'y$$ — see for example chapter 3 of these notes ### More Columns than Rows¶ This is the $$n \times k$$ case with $$n < k$$, so there are fewer equations than unknowns In this case there are either no solutions or infinitely many — in other words, uniqueness never holds For example, consider the case where $$k=3$$ and $$n=2$$ Thus, the columns of $$A$$ consists of 3 vectors in $$\mathbb R ^2$$ This set can never be linearly independent, since it is possible to find two vectors that span $$\mathbb R ^2$$ (For example, use the canonical basis vectors) It follows that one column is a linear combination of the other two For example, let’s say that $$a_1 = \alpha a_2 + \beta a_3$$ Then if $$y = Ax = x_1 | {
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"url": "https://lectures.quantecon.org/jl/linear_algebra.html"
} |
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