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• nice and clean approach – G Cab Mar 15 at 22:38
The hint given is already good.
In alternative you can proceed as follows \eqalign{ & \sum\limits_{k = 1}^n {{1 \over k}\left( \matrix{ n - 1 \cr k - 1 \cr} \right)} \quad \left| {\;1 \le n} \right.\quad = \cr & = \sum\limits_{k = 0}^{n - 1} {{1 \over {\left( {k + 1} \right)}}\left( \matrix{ n - 1 \cr k \cr} \right)} = {1 \over n}\sum\limits_{k = 0}^{n - 1} {{n \over {\left( {k + 1} \right)}}\left( \matrix{ n - 1 \cr k \cr} \right)} = \cr & = {1 \over n}\sum\limits_{k = 0}^{n - 1} {\left( \matrix{ n \cr k + 1 \cr} \right)} = {1 \over n}\sum\limits_{k = 1}^n {\left( \matrix{ n \cr k \cr} \right)} = {1 \over n}\left( {\sum\limits_{k = 0}^n {\left( \matrix{ n \cr k \cr} \right)} - 1} \right) = \cr & = {1 \over n}\left( {2^{\,n} - 1} \right) \cr}
• So you solution brings up a few question for me. Going off of the hint above, the way I solved it was multiplying the summation by n turning it into $\displaystyle\sum_{k=0}^n\binom nk$. I took that summation giving me $2^{n}$, and then divided by n to undo my previous multiplication giving me ${2^{n} \over n}$ which is different from you answer. Where did i go wrong? – Brownie Mar 15 at 22:29
• Edit - I see i took the summation at the wrong index – Brownie Mar 15 at 22:35
• @Brownie: you did not take into proper consideration the summation bounds. – G Cab Mar 15 at 22:37 | {
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# Probability: Choosing a girl from a group
1. Feb 27, 2012
### Xyius
1. The problem statement, all variables and given/known data
You walk into your class the first day of classes, and you notice that
there are 30 men and 20 women in the class already. Let's suppose you decide to choose
two people from the class to be your study partners.
If you choose your study partners at random, and given that at least one of your
study partners is a woman, what is the probability of the event E that both of them
will be women?
A. 0.3167
B. 1.9%
C. 0.2405
D. 0.1901
2. Relevant equations
In my Solution
3. The attempt at a solution
This seems like a simple problem but I cannot seem to get the numbers available as choices.
My logic is is, W represents the event that you have picked a woman, and E represents that both of your partners will be women then.
$$P(E|W)=\frac{P(E \cap W)}{P(W)}$$
So the numerator can simplify to..
$$P(E|W)=\frac{P(E)}{P(W)}$$
This is because if E occurs, then W must have occured.
So..
$$P(E)=\frac{\binom{20}{2}}{\binom{50}{2}}$$
and
$$P(W)=\frac{\binom{20}{1}}{\binom{50}{2}}$$
But this doesn't work because the ratio of these two (From the formula) gives a number larger than one. Where am I going wrong? Do I use Bayes theorem?
2. Feb 27, 2012
### Dick
If P(W) is the probability of choosing at least one woman, there are two ways to do that, you could choose 1 woman and 1 man, or 2 women.
3. Feb 27, 2012
### Xyius
Ohh! So it would be..
$$P(W)=P(W|W)P(W)+P(W|M)P(M)$$
??
I don't have time to crunch through the numbers at the moment, but I will be sure to check this out later.
4. Feb 27, 2012
### Dick
I wouldn't write it that way. Just count out the cases using combinations like you are already doing. How may ways to do each?
5. Feb 27, 2012
### Xyius
So choosing 1 woman and 1 man would be
$$\frac{\binom{20}{1}\binom{30}{1}}{\binom{50}{2}}$$
And choosing 2 women would be..
$$\frac{\binom{20}{2}}{\binom{50}{2}}$$ | {
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$$\frac{\binom{20}{2}}{\binom{50}{2}}$$
Plugging this in gives me 0.2405, or answer C! Thanks! :D
EDIT: Assuming that's the correct answer.....
6. Feb 27, 2012
### Dick
Well, that's what I get.
7. Feb 27, 2012
### Ray Vickson
I W = number of women you choose, you have been asked to find the conditional probability $P\{W=2|W\geq 1\}.$ We have $$P\{W=2|W\geq 1\} = \frac{P\{W = 2 \cap W \geq 1 \}}{P\{W \geq 1\}} = \frac{P\{W = 2\}}{1-P\{W=0\}}.$$ The numerator and denominator are easily comutable using the hypergeometric distribution. The numerator is ${20 \choose 2}/{50 \choose 2},$ while the denominator is $1 - {30 \choose 2}/{50 \choose 2}.$
RGV | {
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# How many number of integer coordinates exists between a line segment, including the end points?
There is a line segment say $$AB$$ with coordinates of end-points as $$A=(x_1, y_1)$$ and $$B=(x_2, y_2)$$. $$x_1, y_1, x_2, y_2$$ are integers. I need to find the number of integer coordinates which lie on the line segment including end-points.
I read somewhere that it is $$\gcd(|x_1 - x_2|, |y_1 - y_2|) + 1$$. But, I cannot understand why this works.
I do not get the intuition behind it. I searched for proof but did not find anything intuitive and straightforward.
Please help me understand this. I am stuck on it. I am expecting a nice proof with great explanation.
• Maybe this will make it easier for you. Without loss of generality, we can assume that $A=(0, 0)$, and $B=(x, y)$ with $x, y\geq 0$. ($B\neq A$) – Jakobian Oct 15 '18 at 18:35
Using Jakobian's simplification, the problem looks as follows: For any $$x_0, y_0 > 0$$, how many integer coordinates does the line from $$(0, 0)$$ to $$(x_0, y_0)$$ pass? We note first, that the slope of this line is $$y_0/x_0$$. So the problem is equivalent to asking: for how many integers $$x$$ with $$0 \leq x \leq x_0$$ is $$y_0 / x_0 * x$$ an integer. Now let $$d = gcd(x_0, y_0)$$. We get $$y_0 / x_0 * x = (d * \hat{y_0})/(d * \hat{x_0}) * x = \hat{y_0}/\hat{x_0} * x$$ for some $$\hat{y_0},\hat{x_0}$$ with $$gcd(\hat{y_0},\hat{x_0}) = 1$$. Now $$\hat{y_0}/\hat{x_0} * x$$ is an integer if and only if x is a multiple of $$\hat{x_0}$$. So how many multiples of $$\hat{x_0}$$ are there between $$0$$ and $$x_0$$. It is exactly $$d + 1$$, because $$x_0 = d * \hat{x_0}$$. Hope this helps. | {
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# Show that $\left(\int_{0}^{1}\sqrt{f(x)^2+g(x)^2}\ dx\right)^2 \geq \left(\int_{0}^{1} f(x)\ dx\right)^2 + \left(\int_{0}^{1} g(x)\ dx\right)^2$ [closed]
Show that $$\left( \int_{0}^{1} \sqrt{f(x)^2+g(x)^2}\ \text{d}x \right)^2 \geq \left( \int_{0}^{1} f(x)\ \text{d}x\right)^2 + \left( \int_{0}^{1} g(x)\ \text{d}x \right)^2$$ where $f$ and $g$ are integrable functions on $\mathbb{R}$.
That inequality is a particular case. I want to approximate the integral curves using some inequalities who imply this inequality.
• Please provide some context – robjohn Jan 5 '16 at 10:38
• This inequality is a particular case.I want to use this inequality to aproximate the integral curves. – alexb Jan 5 '16 at 10:50
• clarifications and improvements should be applied to the question, not left in comments – robjohn Jan 5 '16 at 11:05
• I think there other useful answers to this question and this inequality is useful by itself. I would like to see it reopened. – robjohn Jan 6 '16 at 16:00
• Do you mean that you want to use inequalities implied by this inequality? – robjohn Jan 6 '16 at 16:48
Let a curve $C\in\mathbb{R}^2$ be defined by the parametrisation $\displaystyle x(t)=x(0)+\int_0^t f(x)dx$ and $\displaystyle y(t)=y(0)+\int_0^t g(x)dx$, $t\in [0, 1]$. Then the LHS is the square of the arc length of $C$ joining $(x(0), y(0))$ and $(x(1), y(1))$, whereas the RHS is the square of the shortest distance between $(x(0), y(0))$ and $(x(1), y(1))$.
• But if I have n functions I can to proof the inequality în the same way? – alexb Jan 5 '16 at 13:00
• Yes. You can generalise to an $n$-dimensional curve. – Alex Fok Jan 5 '16 at 13:02 | {
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Definition: $u:\mathbb{R}^n\to\mathbb{R}$ is convex if for all $a\in\mathbb{R}^n$, there is a $v(a)\in\mathbb{R}^n$ so that for all $x\in\mathbb{R}^n$ $$u(x)-u(a)\ge v(a)\cdot(x-a)\tag{1}$$ Theorem (Extension of Jensen): If $u:\mathbb{R}^n\to\mathbb{R}$ is convex, $f:\Omega\to\mathbb{R}^n$, and $\int_\Omega\mathrm{d}\omega=1$, then $$\int_\Omega u(f)\,\mathrm{d}\omega\ge u\left(\int_\Omega f\,\mathrm{d}\omega\right)\tag{2}$$ Proof: Let $a=\int_\Omega f\,\mathrm{d}\omega$. Then $(1)$ becomes $$u(f)-u\left(\int_\Omega f\,\mathrm{d}\omega\right) \ge v\left(\int_\Omega f\,\mathrm{d}\omega\right) \cdot\left(f-\int_\Omega f\,\mathrm{d}\omega\right)\tag{3}\\$$ Since $\int_\Omega\mathrm{d}\omega=1$, integrating $(3)$ over $\Omega$ gives \begin{align} \int_\Omega u(f)\,\mathrm{d}\omega-u\left(\int_\Omega f\,\mathrm{d}\omega\right) &\ge v\left(\int_\Omega f\,\mathrm{d}\omega\right) \cdot\left(\int_\Omega f\,\mathrm{d}\omega-\int_\Omega f\,\mathrm{d}\omega\right)\\[3pt] &=0\tag{4} \end{align} QED
Claim: $u(x)=\left\|x\right\|$ is convex.
Proof: \begin{align} \left\|a\right\|\left\|x\right\|&\ge a\cdot x\tag{5}\\[6pt] \left\|x\right\|&\ge\frac{a}{\left\|a\right\|}\cdot x\tag{6}\\ \left\|x\right\|-\left\|a\right\|&\ge\frac{a}{\left\|a\right\|}\cdot(x-a)\tag{7} \end{align} Explanation:
$(5)$: Cauchy-Schwarz
$(6)$: divide both sides by $\left\|a\right\|$
$(7)$: subtract $\left\|a\right\|$ from both sides
QED
The theorem and the claim prove that $$\int_0^1\left\|f(x)\right\|\mathrm{d}x\ge\left\|\int_0^1f(x)\,\mathrm{d}x\right\|\tag{8}$$ which, in $\mathbb{R}^2$, is the inequality in the question. | {
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Suppose that $f$ and $g$ are continuous functions. Define $$\phi(t) = \left(\int_0^t \sqrt{f(s)^2 + g(s)^2} ds\right)^2 -\left(\int_0^t f(s) ds\right)^2 - \left(\int_0^t g(s) ds\right)^2.$$ It is obvious that $\phi(0) =0$ and $$\phi'(t) = 2\left[\int_0^t \sqrt{f(t)^2 + g(t)^2}\sqrt{f(s)^2 + g(s)^2}ds - \int_0^t (f(s)f(t)+g(s) g(t))ds\right].$$ By Cauchy-Schwartz inequality, we have $$\sqrt{f(t)^2 + g(t)^2}\sqrt{f(s)^2 + g(s)^2} \geq f(t)f(s) + g(t) g(s).$$ Hence $\phi'(t) \geq 0$. This implies that $\phi(1) \geq \phi(0) =0$. This is the inequality in question.
In general case when $f$ and $g$ are integrable, we can approximate them by continous functions, and hence finish the proof. | {
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# Evaluating integral using Riemann sums
It is given that: $$\sin\frac{\pi }{n} \sin\frac{2\pi }{n}\cdots\sin\frac{(n-1)\pi }{n}=\frac{n}{2^{n-1}}$$ It is asked to use the above identity to evaluate the following improper integral: $$\int_0^\pi \log(\sin x) \, dx$$
I used the definition of the integral in terms of Riemann sums: \begin{align*}\int_0^\pi \log(\sin x) \, dx &=\lim_{n\rightarrow \infty }\frac{\pi }{n}\left[\sum_{k=1}^{k=n-1}\log\left(\sin\left(\frac{k\pi }{n}\right)\right)\right]\\ &=\lim_{n\rightarrow \infty }\frac{\pi }{n}\left[\log\left(\sin\frac{\pi }{n}\sin\frac{2\pi }{n}\cdots\sin\frac{(n-1)\pi }{n}\right)\right]\\ &=\lim_{n\rightarrow \infty }\frac{\pi }{n}\Big(\log n-(n-1)\log 2\Big)\\ &=-\pi \log 2 \end{align*}
However, this integral is improper, so $\log(\sin(\pi ))=\log(0)=-\infty$. I am kind of cheating in my solution, because the Riemann sum above should be: $$\sum_{k=1}^{k=n-1}\log\left(\sin\left(\frac{k\pi }{n}\right)\right)+\frac{\pi }{n}\log\left(\sin\left(\frac{n\pi }{n}\right)\right)\;,$$ but I have no idea how to deal with the last term of the sum since $\sin\left(\frac{n\pi }{n}\right)=\sin(\pi)=0$.
Can anyone show me how to deal with this? Also, if someone knows how to prove the first identity: $$\sin\frac{\pi }{n}\sin\frac{2\pi }{n}\cdots\sin\frac{(n-1)\pi }{n}=\frac{n}{2^{n-1}}$$ please write down your proof below? Thanks | {
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-
When you write something like log in $\TeX$, it gets interpreted as juxtaposed variable names and is therefore italicized. To get proper formatting for function names, you need to use the corresponding command sequences, e.g. \log. If you have a function name for which there's no command sequence, such as $\operatorname{tr}$, you can get it formatted properly using \operatorname{tr}. – joriki Apr 11 '12 at 21:30
I think the integrals goes from $\frac{\pi }{n}$ to $\frac{n\pi}{n}$. I think you either ignore left endpoint or the right endpoint, but you can't ignore both of them. The Rieamann sum is the limit of the areas of the rectangles when the mesh goes to zero, and by ignoring the right endpoint, you are sort of ignoring the area of the last rectangle. Am I wrong? – Boyan Klo Apr 11 '12 at 21:37
It's not only italicization, it's proper spacing and in some cases positioning of subscripts and superscripts. For example $\displaystyle a \lim_{x\to\infty} b$. Note the spacing before and after "$\lim$" and notice where the subscript is. – Michael Hardy Apr 11 '12 at 21:37
@joriki, now is valid to do $\newcommand{\tr}{\operatorname{tr}}$ so that you can use \tr later in your post. – leo Apr 11 '12 at 23:41 | {
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As for the improper integral I have another approach which seems to me more obvious $$I=\int\limits_{0}^{\pi}\log(\sin(x))dx= \int\limits_{0}^{\pi/2}\log(\sin(x))dx+\int\limits_{\pi/2}^{\pi}\log(\sin(x))dx=$$ $$\int\limits_{0}^{\pi/2}\log(\sin(x))dx+\int\limits_{0}^{\pi/2}\log(\cos(x))dx= \int\limits_{0}^{\pi/2}\log\left(\frac{1}{2}\sin(2x)\right)dx=$$ $$\int\limits_{0}^{\pi/2}\log\left(\sin(2x)\right)dx- \int\limits_{0}^{\pi/2}\log(2)dx= \frac{1}{2}\int\limits_{0}^{\pi}\log(\sin(x))dx- \int\limits_{0}^{\pi/2}\log(2)dx=$$ $$\frac{1}{2}I-\frac{\pi}{2}\log{2}$$ Hence, $I=-\pi\log(2)$.
As for the last identity there is quite elementary solution. Denote $\xi=e^{\frac{i\pi}{n}}$ then $$\prod\limits_{k=1}^{n-1}\sin\frac{\pi k}{n}= \prod\limits_{k=1}^{n-1}\frac{\xi^{k}-\xi^{-k}}{2i}= \frac{1}{2^{n-1}}\frac{\xi^{-\frac{n(n-1)}{2}}}{i^{n-1}}\prod\limits_{k=1}^{n-1}(\xi^{2k}-1)$$ If $n$ is even $\xi^{\frac{n(n-1)}{2}}=(\xi^{\frac{n}{2}})^{n-1}=i^{n-1}$. If $n$ is odd $\xi^{\frac{n(n-1)}{2}}=(\xi^n)^{\frac{n-1}{2}}=(-1)^{\frac{n-1}{2}}=(i^2)^{\frac{n-1}{2}}=i^{n-1}$. So in both cases $$\frac{\xi^{-\frac{n(n-1)}{2}}}{i^{n-1}}=(-1)^{n-1}$$ Numbers $\{\xi^{2k}-1:k\in\{0,\ldots,n-1\}\}$ are roots of the equation $(x+1)^n=1$. After exapndig by binomial formula and canceling out $x$ we conclude that $\{\xi^{2k}-1:k\in\{1,\ldots,n-1\}\}$ are roots of the equation $x^n+nx^{n-1}+\ldots+n=0$. Then by Vieta's formulas we conclude $$\prod\limits_{k=1}^{n-1}(\xi^{2k}-1)=(-1)^{n-1}n$$ Now we summarizee results and see $$\prod\limits_{k=1}^{n-1}\sin\frac{\pi k}{n}=\frac{n}{2^{n-1}}$$ | {
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# Suppose $M$ is an $m \times n$ matrix such that all rows and columns of $M$ sum to $1$. Show that $m=n$
I find this a rather awkward question, and I was given a hint: use invariants, which I found even more awkward.
Suppose $M$ is an $m \times n$ matrix such that all rows and columns of $M$ sum to $1$. Show that $m=n$.
I have no clue how this is a problem on invariants, let alone how to solve this problem. I'll need hints on why this is the case.
If $M$ has $m$ rows that sum to $1$, the sum of the matrix is $m$.
If $M$ has $n$ columns that sum to $1$, the sum of the matrix is $n$.
The sum of the matrix is invariant, therefore $m=n$.
Hint: What is the sum of all numbers in the matrix?
Let $\mathrm A \in \mathbb R^{m \times n}$ have its $m$ rows and $n$ columns sum to $1$. Hence,
$$\underbrace{1_m^T \mathrm A}_{=1_n^T} 1_n = 1_n^T 1_n = n$$
and
$$1_m^T \underbrace{\mathrm A 1_n}_{=1_m} = 1_m^T 1_m = m$$
Thus, $m = n$.
This question has been answered very nicely, but it might be worth noting that a simple yet less elegant, more brutish way of seeing this problem, is to let the first $$(n-1)$$ columns be anything.
Say they're given by $$a_{ij}, \ i \leq m, \ \ j \leq (n-1)$$.
Then the entries of the last column must be $$1 - \sum_{j=1}^{n-1} a_{ij}, \ i \leq m$$.
Since these entries add up to $$1$$, we have $$1 = \sum_{i=1}^{m} (1 - \sum_{j=1}^{n-1} a_{ij}) = m - \sum_{i=1}^{m} \sum_{j=1}^{n-1} a_{ij} = m \ - \sum_{j=1}^{n-1} \sum_{i=1}^{m} a_{ij} = m - \sum_{j=1}^{n-1}1 \\= m -(n-1)$$
And thus we have $$m=n$$. | {
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# Calculate the number of ways to paint $5$ buildings with $4$ colours such that all $4$ colours must be used
A developer has recently completed a condominium project in a valley. There are blocks of buildings $A$, $B$, $C$, $D$ and $E$ as shown in the diagram below.
The developer has colours available to paint the buildings. Each block can only be painted using a single colour. Find the number of ways to paint all the blocks if all $4$ colours must be used.
My attempt: We have $5$ ways to choose $4$ buildings with $4$ different colours. Among the $4$ buildings, we have $4!$ ways to paint them using $4$ different colours.
So my answer is $120$. However, the answer given is $240$. What is my mistake?
• Um, what do paint the fifth building? Don't you have 4 chooses for that? I'd get that you method out to give 4*120 =480. Which is also wrong as we double counted. We'd choose a building to be a duplicate color. there are 5 choices for that. We'd paint the other four. 24 four that. We'd pick one of the four buildings to duplicate the fifth one. There are four choices of that. Which ever building we choose to duplicate is a double counting as we could have choosen that building to be a duplicate. so divide by two. – fleablood May 12 '17 at 15:24
You're on the right track. After choosing the four buildings with different colours, what about the fifth? It will be a repeated colour, and there are four colours to choose from. However, we will have overcounted by a factor of $2$, so the final answer will be $$120\cdot\frac42 = 240$$ To explain the overcounting, suppose that you first choose $A,B,C,D$ as the set of four buildings. Then $E$ is the same as one of them, say $A$. But then we will also later consider $E,B,C,D$ as the set of four, with $A$ the same as $E$.
First, you must choose which $2$ buildings will be of the same colour; there are ${5\choose 2} =10$ ways to do this.
Then you have $4!=24$ ways to paint the buildings using your $4$ colours. | {
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Then you have $4!=24$ ways to paint the buildings using your $4$ colours.
The mistake in your solution is that you are not specifying how you will paint the last building; so it's like saying that it is left without colour, which is like giving it a fifth colour distinct from the others. This explains why you arrive at an answer of $120=5!$.
You didn't paint the 5th building at all!
Two buildings must be painted the same color. There are ${5 \choose 2} = \frac {5*4}2 = 10$ possible pairs of buildings that will be the same color.
There are $4*3*2$ ways to paint the remaining $3$ buildings and only one choice left for the two that must be the same color. So that is $10*5*4*3*2 = 240$.
.... or ....
to continue your method: There are $5$ chooses for the duplicate painted building. There are $4!$ ways to be paint the four differently colored buildings. So that is $5*4!=120$.
Then there are $4$ chooses to paint the fifth building. That is $4*120 =480$. However whatever color we choose it will be the same result as if we had chosen the other building to be our fifth building and made the same choices otherwise. So that ere $480/2 = 240$ distinct choices.
For a given set of 4 houses, there are 4!=24 ways to colour them with 4 distinct colours. There are 4 ways to colour the remaining house, so that gives 96 ways. 5 ways to choose the four-set then gives 480 but this double counts all the colourings because each house is or is not left out. So divide by 2 to get 240.
5 buildings need to be painted with 4 different colours, painter needs to use all 4 colors.
Painter needs to use the same colour on 2 buildings.
Say colour 1 is used twice.
There are (5×4) /2 ways of painting 2 out of the 5 buildings.
Now there are 4 colors, so the above is true for each of the 4 colors.
We have 4 × [(5×4)/2] ways of painting 2 out of the the 5 buildings with the same color.
3 remaining buildings still need to be painted with the remaining 3 different colors. | {
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3 remaining buildings still need to be painted with the remaining 3 different colors.
For each of the ways where 2 equal colors have been used on 2 out of 5 buildings we can paint the remaining 3 buildings in 3×2×1 ways
Altogether: 4 × [(5×4)/2] × (3×2×1) = 240. | {
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# How many distinct non-isomorphic bipartite graphs with parts of size $m$ and $n$ exist?
How many distinct non-isomorphic bipartite graphs with parts of size $m$ and $n$ exist? (Two bipartite graphs are distinct if there is no way to just rearrange the vertices within a part set of one of them to become the other.) It's not necessary that every vertex has an edge. Is there a closed form formula in terms of $m$ and $n$? If not, is there a recursive way to count it? If there is a good recursive algorithm to count it, what would its pseudocode look like?
Thanks!
• arxiv.org/pdf/1304.0139.pdf – HEKTO Jun 13 '17 at 23:30
• I don't think species are needed though, it falls to simple polya. Although yeah, species are awesome :p – Jorge Fernández Hidalgo Jun 13 '17 at 23:31
• How would I use Polya's theorem in this problem? Would I have to examine each of the $m!n!$ possible symmetries and count how many graphs are fixed by each one? That seems too complicated to find an effective way to count the number of such graphs. – Anon Jun 14 '17 at 1:08
• This problem appeared at the following MSE link. Complexity is not factorial but rather the number of terms in the cycle index of the symmetric group (partition function). – Marko Riedel Jun 14 '17 at 1:34
• Maybe you want oeis.org/A028657 – Gerry Myerson Jun 14 '17 at 1:55
On your question whether or not there exists a closed-form formula for the number of bipartite graphs with parts of size $m$ and $n$ (denoted by $|B_u(m,n)|$ below), my coauthor and I proved the following formulas for $m = 2$ and $m = 3$ in an upcoming paper. $|B_u(2,n)|$ corresponds to the integer sequence A002623, i.e., 1, 3, 7, 13, 22, 34, 50, 70, 95,..., and $|B_u(3,n)|$ corresponds to the integer sequence A002727, i.e., 1, 4, 13, 36, 87, 190, 386, 734, 1324, 2284,... in Sloane's classification of integer sequences. Generalizing these closed-form formulas for $m = 4,5,6,...$ remains an open problem to the best of my knowledge. | {
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$$|B_u(2,n)|\!=\! \frac{2n^{3}+15n^{2} + 34n + 22.5 + 1.5\left ( -1 \right )^{n}}{24}, n=0,1,2,...$$
$\,$
$\!\!\!\!\!\!\!|B_{u}\left( 3,n \right)|\! = \left\{\begin{matrix} \frac{1}{6}\left [ \binom{n+7}{7}\! + \frac{3\left ( n+4 \right )\left ( 2n^{4}+32n^{3}+172n^{2} + 352n + 15\left ( -1 \right )^{n} +225 \right )}{960} +\! \frac{2(n^{3}+12n^{2}+45n+54)}{54} \right ] &\!\!\!\!\!\!\text{if}\, n \bmod\!\! \text{ } 3 = 0, \\ \\ \frac{1}{6}\left [ \binom{n+7}{7}\! + \frac{3\left ( n+4 \right )\left ( 2n^{4}+32n^{3}+172n^{2} + 352n + 15\left ( -1 \right )^{n} +225 \right )}{960} +\! \frac{2(n^{3}+12n^{2}+45n+50)}{54} \right ] &\!\!\!\!\!\!\!\text{ if}\, n \bmod\!\! \text{ } 3 = 1, \\ \\ \frac{1}{6}\left [ \binom{n+7}{7}\! + \frac{3\left ( n+4 \right )\left ( 2n^{4}+32n^{3}+172n^{2} + 352n + 15\left ( -1 \right )^{n} +225 \right )}{960} +\! \frac{2(n^{3}+12n^{2}+39n+28)}{54} \right ] &\!\!\!\!\!\!\!\!\! \text{if}\, n \bmod\!\! \text{ } 3 = 2. \!\!\!\! \end{matrix}\right.$
Ref: Abdullah Atmaca and A. Yavuz Oruc. "On The Size Of Two Families Of Unlabeled Bipartite Graphs." AKCE International Journal of Graphs and Combinatorics. To appear.
In a more recent article by A. Atmaca and A. Yavuz Oruc, the following more general result for the number of unlabeled graphs with $n$ left vertices and $r$ right vertices (denoted by $|B_u(n,r)|$) has been proven:
$$\frac{{r+2^n-1\choose r}}{n!}\le |B_u(n,r)| \le 2\frac{{r+2^n-1\choose r}}{n!}, n < r.$$
Given that $|B_u(n,r)|= |B_u(r,n)|$, the following inequality holds as well when $n > r:$
$$\frac{{n+2^r-1\choose n}}{r!}\le |B_u(n,r)| \le 2\frac{{n+2^r-1\choose n}}{r!}, n > r.$$
Note that the upper bound is twice as large as the lower bound. Tightening the constant factors, i.e., 1 and 2 on the lefthand and righthand side of the inequality remains open. | {
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• When $r=n$ are isomorphisms allowed to swap the two partite sets? – bof Jul 14 '18 at 10:00
• In these formulas, it is assumed that left and right sets of vertices are not swappable whether or not $n = r$ as in Harrison, Michael A. "On the number of classes of binary matrices." IEEE Transactions on Computers 100.12 (1973): 1048-1052. For example, if $n = r = 2$ with left set of vertices denoted $X = \{x_1,x_2\}$ and right set of vertices denoted $Y = \{y_1,y_2\}$, the graph that connects $x_1$ to $y_1$ and $y_2$ and the graph that connects $y_1$ to $x_1$ and $x_2$ are considered non-isomorphic. – AYO Jul 14 '18 at 15:07
• It should also be added that the lower bounds in the two inequalities coincide and also hold when $n = r$ under the same assumption as in my earlier comment. – AYO Jul 14 '18 at 15:13 | {
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Intersection of two Arithmetic progressions
Let us say I have two Arithmetic progressions
$AP_1 = a + nb$
$AP_2 = c + nd$
Can we find a new Arithmetic progression $AP_3$ which is intersection of both $AP_1$ and $AP_2$?
• If there is one common term, yes. But they could be disjoint - e.g. Odd numbers and even numbers. – Macavity Jun 4 '15 at 16:21
• Not always. But for example if $b$ and $d$ are relatively prime (no common divisor greater than $1$), we can. – André Nicolas Jun 4 '15 at 16:21
• if $AP_1 = a + nb$ where $a=2,b=3$ then we have $AP_1 = 2+3=5,2+6=8,2 + 9 =11 .....$ Now if $AP_2 = c + nd$ where $c=4,d=5$ then we have $AP_2 = 4 + 5=9,4+10=14,4+15$ What I am saying is that the intersection can be empty !! – alkabary Jun 4 '15 at 16:23
• @alkabary Wrong e.g. for that conclusion. $14, 29, ...15k-1,...$ is the intersection, another AP. In effect you are solving $a = \pmod b, c=\pmod d$ simultaneously, which is always solved if $\gcd(b,d) \mid (a-c)$. – Macavity Jun 4 '15 at 17:05
• I see that there can be an intersection, but the intersection can not be an Arithmetic progression, if so can we find an equation, for the new series. – Naks Jun 4 '15 at 17:16
We have that for any two arithmetic progressions, $AP_1, AP_2$, we have that their intersection are those points where
$b(n_1)+a = d(n_2)+c \iff b(n_1)= d(n_2)+c-a$
for some $n_1, n_2 \in \mathbb{Z}$. If no such $n$s exist, then the intersection is empty. If they exist, we can proceed as follows.
This means that the intersection of these two sets produce the set $AP_3 =(b\cdot d)(n) + (d \cdot k + c)$ for some $k \in \mathbb{Z}$ such that $b \cdot k_1 = (d\cdot k + (c-a))$ for some $k_1 \in \mathbb{Z}$.
We can see this: $(b\cdot d)\cdot(n) + (d \cdot k + c) = (b)\cdot (d\cdot n) + (b \cdot k_1 + a) = (b)\cdot (d\cdot n + k_1) + a$, which is clearly a subset of $AP_1$. | {
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We also see that $(b\cdot d)\cdot(n) + (d \cdot k + c) = (d)\cdot (b\cdot n + k) + c$, which is clearly a subset of $AP_2$, as required.
• Is it necessary for both ap to have same number of terms for forming a new ap – Jack Rod Oct 12 '19 at 5:50 | {
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# Finding the limits of integration for $\iint\limits_A \frac{1}{(1 + x^2 + y^2)^2} dxdy$ where A is one loop of lemniscate $(x^2 + y^2)^2 = x^2 - y^2$
Work so far: Switch to polar coordinates
$$A \rightarrow r^2 = \cos^2\theta - \sin^2\theta \rightarrow r^2 = \cos2\theta$$
So $$A$$ is $$r^2 = \cos 2 \theta$$
And the integral becomes
$$\iint\limits_A \frac{r}{(1 + r^2)}drd\theta$$
Now I want to find the limits of integration without actually graphing the curve.
So
$$2rdr = -2\sin(2\theta)$$
$$\frac{dr}{d\theta} = \frac{-\sin(2\theta)}{r}$$
Critical points for $$\theta$$ come out to be $$0$$ and $$\frac{\pi}{2}$$
Now I know how to solve the integral using u substitution but I'm not sure about the limits.
• Why do you need the critical points? Look for the range of $\theta$ where the inequality $\cos2\theta\ge0$ holds. – user Nov 13 '19 at 10:17
• @user $\frac{\pi}{4} \rightarrow 0$ seems right but where does $\cos 2\theta \geq 0$ come from? – atis Nov 13 '19 at 10:27
• The equation $r^2=\cos2\theta$ has real solutions for $r$ if and only if $\cos2\theta\ge0$. – user Nov 13 '19 at 15:24
• Ah so to get every possible value, $\theta$ should go from $\frac{\pi}{4}$ to $0$. So does that mean $r$ should go from $0$ to its maximum possible bound? How should I go about finding that? – atis Nov 13 '19 at 16:19 | {
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In polar coordinates the curve is determined by $$r^2=\cos2\theta,$$ which has real solutions for $$r$$ if and only if: $$\cos2\theta\ge0 \implies -\frac\pi4\le\theta\le\frac\pi4\text { or }\frac{3\pi}4\le\theta\le\frac{5\pi}4.$$
$$\iint\limits_A \frac{r}{(1 + r^2)^2}drd\theta= \int\limits_{-\frac\pi4}^{\frac\pi4}d\theta \int\limits_0^\sqrt {\cos2\theta}\frac{rdr}{(1 + r^2)^2}.$$
• One question if you don't mind, why are the limits of $r$ are from $0 \rightarrow \sqrt{\cos 2 \theta}$? – atis Nov 14 '19 at 7:46
• You are welcome. The only solutions of the equation $r^2=\cos2\theta$ are $r=\pm\sqrt{\cos2\theta}$. The negative square root belongs to the other loop of the curve ($\frac{3\pi}4\le\theta\le\frac{5\pi}4$). – user Nov 14 '19 at 8:46 | {
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We prove that eigenvalues of a Hermitian matrix are real numbers. All eigenvalues of a Hermitian matrix are real, and Eigenvectors corresponding to distinct eigenvalues … Anything is possible. (T/F) The matrix A and its transpose, Ahave different sets of eigenvalues. Positive definite symmetric matrices have the property that all their eigenvalues … "All the nonzero eigenvalues of ATA are between 0 and 1" seems not true. These are the scalars $$\lambda$$ and vectors $$v$$ such that $$Av = \lambda v$$. Thus A and A T have the same eigenvalues. The eigenvalue λtells whether the special vector xis stretched or shrunk or reversed or left unchanged—when it is multiplied by A. Q transpose is Q inverse in this case. Q lambda, Q transpose was fantastic. So the eigenvalues of D are a, b, c, and d, i.e. Save my name, email, and website in this browser for the next time I comment. This site uses Akismet to reduce spam. (T/F) The multiplicity of a root r of the characteristic equa- tion of A is called the algebraic multiplicity of r as an eigenvalue of A. So that's A transpose A is the matrix that I'm going to use in the final part of this video to achieve the greatest factorization. If A is a square matrix, then its eigenvalues are equal to the eigenvalues of its transpose, since they share the same characteristic polynomial. So this shows that they have the same eigenvalues. Learn more Accept. So lambda is an eigenvalue of A. If follows that and , where denotes a complex conjugate, and denotes a transpose. For part (b), note that in general, the set of eigenvectors of an eigenvalue plus the zero vector is a vector space, which is called the eigenspace. Remark: Algebraic Multiplicities of Eigenvalues, How to Prove a Matrix is Nonsingular in 10 Seconds, Any Automorphism of the Field of Real Numbers Must be the Identity Map. So the possible eigenvalues of our matrix A, our 3 by 3 matrix A that we had way up there-- this matrix A right there-- the possible eigenvalues are: lambda | {
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matrix A that we had way up there-- this matrix A right there-- the possible eigenvalues are: lambda is equal to 3 or lambda is equal to minus 3. A real symmetric $n \times n$ matrix $A$ is called. Those are in Q. When a Hermitian matrix 749#749 is real, it is symmetric matrix, i.e., 5839#5839 . I guest that the nonzero eigenvalues of A^TA are no less than 1, at least it seems true numerically. Presented by … Applications. Required fields are marked *. Ask Question Asked 8 years, 6 months ago. Those are the lambdas. by Marco Taboga, PhD. The eigenvalues of an upper triangular matrix are simply the diagonal entries of the matrix. Proof. Every $3\times 3$ Orthogonal Matrix Has 1 as an Eigenvalue, Transpose of a Matrix and Eigenvalues and Related Questions, A Diagonalizable Matrix which is Not Diagonalized by a Real Nonsingular Matrix, Find Values of $a, b, c$ such that the Given Matrix is Diagonalizable, Sum of Squares of Hermitian Matrices is Zero, then Hermitian Matrices Are All Zero, Linear Combination and Linear Independence, Bases and Dimension of Subspaces in $\R^n$, Linear Transformation from $\R^n$ to $\R^m$, Linear Transformation Between Vector Spaces, Introduction to Eigenvalues and Eigenvectors, Eigenvalues and Eigenvectors of Linear Transformations, How to Prove Markov’s Inequality and Chebyshev’s Inequality, How to Use the Z-table to Compute Probabilities of Non-Standard Normal Distributions, Expected Value and Variance of Exponential Random Variable, Condition that a Function Be a Probability Density Function, Conditional Probability When the Sum of Two Geometric Random Variables Are Known, Determine Whether Each Set is a Basis for $\R^3$. A matrix and the transpose of that matrix share the same eigenvalues. Spectral properties. In fact, even though is positive semidefinite (since it is a density matrix), the matrix in general can have negative eigenvalues. They both describe the behavior of a matrix on a certain set of vectors. But data comes in | {
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They both describe the behavior of a matrix on a certain set of vectors. But data comes in non-square matrices. A square matrix is Hermitian if and only if it is unitarily diagonalizable with real eigenvalues.. So depending on the values you have on the diagonal, you may have one eigenvalue, two eigenvalues, or more. Featured on Meta New Feature: Table Support Proof Theorem 2.1 and the resulting definition 2 gives us Gershgorin discs that correspond to the rows of A, where A is the matrix whose eigenvalues we are looking for. Notify me of follow-up comments by email. "All the nonzero eigenvalues of ATA are between 0 and 1" seems not true. Examples. This website’s goal is to encourage people to enjoy Mathematics! A unitary matrix is a matrix whose inverse equals it conjugate transpose.Unitary matrices are the complex analog of real orthogonal matrices. 85 0. Let $A$ be an $n\times n$ invertible matrix. This is a finial exam problem of linear algebra at the Ohio State University. A square matrix A and its transpose have the same eigenvalues. Consider the matrix A= 2 0 2 1 this has eigenvalues = 1;2 with eigenspaces spanned by E 1 = span 0 1 ; E 2 = span 1 2 : The matrix Athas the eigenspaces E 1 = span 2 1 ; E 2 = span 1 0 : 4 MATH 2030: ASSIGNMENT 6 Q.7: pg 310, q 22. 1.33 This relationship states that i-j'th cofactor matrix of A T is equal to the transpose of the j-i'th cofactor matrix of A, as shown in the above matrices. Alternatively, we can say, non-zero eigenvalues of A are non-real. Hermitian matrices are fundamental to the quantum theory of matrix mechanics created by Werner Heisenberg, Max Born, and Pascual Jordan in 1925.. This calculator allows you to enter any square matrix from 2x2, 3x3, 4x4 all the way up to 9x9 size. The conjugate transpose U* of U is unitary.. U is invertible and U − 1 = U*.. Enter your email address to subscribe to this blog and receive notifications of new posts by email. All vectors are eigenvectors of I. The eigenvalues of A are | {
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notifications of new posts by email. All vectors are eigenvectors of I. The eigenvalues of A are the same as the eigenvalues of A T. Example 6: The eigenvalues and vectors of a transpose. (T/F) The matrix A can have more than n eigenvalues. Free Matrix Eigenvalues calculator - calculate matrix eigenvalues step-by-step. the entries on the diagonal. Learn how your comment data is processed. The eigenvalues of a symmetric matrix are real. To obtain the left eigenvectors, you simply transpose X. Normalization and order If x is a solution to Ax = x then so is cx, c: 1 1, c 6= 0. The matrices A and A T will usually have different eigen vectors. From the properties of transpose, we see that ##(A - \lambda I)^T = A^T - \lambda I##. Hermitian Matrix and Unitary Matrix. Then = 5,-19,37 are the roots of the equation; and hence, the eigenvalues of [A]. 7. A symmetric matrix is a square matrix that is equal to its transpose and always has real, not complex, numbers for Eigenvalues. Naturally this relation is reciprocal, so the inverse of a rotation matrix is simply its transpose, i.e., R-1 = R T. The eigenvalues of (1) are . The eigenvectors returned by the above routines are scaled to have length (norm) 1. If A is an m × n matrix and A T is its transpose, then the result of matrix multiplication with these two matrices gives two square matrices: A A T is m × m and A T A is n × n. Furthermore, these products are symmetric matrices. (10) Complex Eigenvalues. Putting this all together we have the set of eigenvalues … Research leads to better modeling of hypersonic flow; Titanium atom that exists in two places at once in crystal to blame for unusual phenomenon ; Tree lifespan decline in forests could neutralize … More Eigenvalue and Eigenvector Problems Another thing I looked at was the determinant used to find the characteristic equation and eigenvalues. Similarly in characteristic different from 2, each diagonal element of a skew-symmetric matrix must be zero, since each is its | {
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different from 2, each diagonal element of a skew-symmetric matrix must be zero, since each is its own negative.. Denis Serre Denis Serre. 28. ... no constraints appart from the reality of its eigenvalues and their sum. Positive definite symmetric matrices have the property that all their eigenvalues … (b) The set of eigenvalues of $A$ and the set of eigenvalues of $A^{\trans}$ are equal. The singular vectors of a matrix describe the directions of its maximum action. This result is valid for any diagonal matrix of any size. And then the transpose, so the eigenvectors are now rows in Q transpose. We have that . The difference is this: The eigenvectors of a matrix describe the directions of its invariant action. Q lambda, Q transpose was fantastic. 30. 28. Answers and Replies Related Calculus and Beyond Homework Help News on Phys.org. Let A be an n×nreal matrix. How to Diagonalize a Matrix. Problem. Common Eigenvector of Two Matrices and Determinant of Commutator, Nilpotent Matrix and Eigenvalues of the Matrix. Matrix Eigenvalue Theory It is time to review a little matrix theory. But for a non-square matrix, it's not. Furthermore, it can be shown that the eigenvalues of A T A are nonnegative (≥ 0). Problems in Mathematics © 2020. Suppose we have matrix A as. (See part (b) of the post “Transpose of a matrix and eigenvalues and related questions.“.) U is unitary.. Suppose that is a real symmetric matrix of dimension . Products. Example 6: The eigenvalues and -vectors of a transpose. Goal Seek can be used because finding the Eigenvalue of a symmetric matrix is analogous to finding the root of a polynomial equation. Homework Equations The Attempt at a Solution Matrix eigenvalue theory Suppose that is a real symmetric square matrix of dimension . Eigenvalues and vectors seem to be very scary until we get the idea and concepts behind it. Published 06/21/2017, […] For a solution, see the post “Transpose of a matrix and eigenvalues and related questions.“. Problems in | {
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a solution, see the post “Transpose of a matrix and eigenvalues and related questions.“. Problems in Mathematics © 2020. This class computes the eigenvalues and eigenvectors of a selfadjoint matrix. The eigenvalues of A are the same as the eigenvalues of A T. Example 6: The eigenvalues and vectors of a transpose. 4. • The square roots of the eigenvalues of A T A are termed singular values of A. Its eigenvalues. 23. As we know from Theorem ETM[421] the eigenvalues of A are the same as the eigenvalues of At additionally matrix At must also obey Theorem 2.1. […], […] eigenvalues , we deduce that the matrix $A$ has an eigenvalue $1$. Here BT is the transpose matrix of […] Rotation Matrix in Space and its Determinant and Eigenvalues For a real number 0 ≤ θ ≤ π, we define the real 3 × … Therefore, the eigenvalues of are Transposition does not change the eigenvalues and multiplication by doubles them. Perfect. Examples. This site uses Akismet to reduce spam. We figured out the eigenvalues for a 2 by 2 matrix, so let's see if we can figure out the eigenvalues for a 3 by 3 matrix. If F::Eigen is the factorization object, the eigenvalues can be obtained via F.values and the eigenvectors as the columns of the matrix F.vectors. And I think we'll appreciate that it's a good bit more difficult just because the math becomes a little hairier. No in-place transposition is supported and unexpected results will happen if src and dest have overlapping memory regions. Then $\lambda$ is an eigenvalue of the matrix $\transpose{A}$. Save my name, email, and website in this browser for the next time I comment. What are singular values? symeigensystem(A, X, L) calculates right eigenvectors. Thus, the eigenvalues of are Those of the inverse are and those of are If A is a real skew-symmetric matrix then its eigenvalue will be equal to zero. Matrix factorization type of the eigenvalue/spectral decomposition of a square matrix A. Positive definite real symmetric matrix and its eigenvalues, | {
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decomposition of a square matrix A. Positive definite real symmetric matrix and its eigenvalues, Inverse Matrix of Positive-Definite Symmetric Matrix is Positive-Definite, Positive definite Real Symmetric Matrix and its Eigenvalues, Maximize the Dimension of the Null Space of $A-aI$, Rotation Matrix in Space and its Determinant and Eigenvalues, Subspaces of Symmetric, Skew-Symmetric Matrices, Eigenvalues of a Hermitian Matrix are Real Numbers, Symmetric Matrix and Its Eigenvalues, Eigenspaces, and Eigenspaces, Determine All Matrices Satisfying Some Conditions on Eigenvalues and Eigenvectors, If $A$ is a Skew-Symmetric Matrix, then $I+A$ is Nonsingular and $(I-A)(I+A)^{-1}$ is Orthogonal, Positive definite real symmetric matrix and its eigenvalues – Problems in Mathematics, A relation of nonzero row vectors and column vectors – Problems in Mathematics, Linear Combination and Linear Independence, Bases and Dimension of Subspaces in $\R^n$, Linear Transformation from $\R^n$ to $\R^m$, Linear Transformation Between Vector Spaces, Introduction to Eigenvalues and Eigenvectors, Eigenvalues and Eigenvectors of Linear Transformations, How to Prove Markov’s Inequality and Chebyshev’s Inequality, How to Use the Z-table to Compute Probabilities of Non-Standard Normal Distributions, Expected Value and Variance of Exponential Random Variable, Condition that a Function Be a Probability Density Function, Conditional Probability When the Sum of Two Geometric Random Variables Are Known. Consider the matrix equation (472) Any column vector which satisfies the above equation is called an eigenvector of . So we have shown that ##A - \lambda I## is invertible iff ##A^T - \lambda I## is also invertible. By definition, if and only if-- I'll write it like this. Here the transpose is minus the matrix. This is the return type of eigen, the corresponding matrix factorization function. Eigenvalues of A transpose A Thread starter 3.141592654; Start date Dec 7, 2011; Dec 7, 2011 #1 3.141592654. Q | {
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of A transpose A Thread starter 3.141592654; Start date Dec 7, 2011; Dec 7, 2011 #1 3.141592654. Q transpose is Q inverse. Likewise, the associated number is called an eigenvalue of . then, we can solve the eigenvalues for, We solve the eigenvectors of A from the equation (A - I) = 0 by Gaussian elimination. Suppose we have matrix A as. note A is not necessarily a square matrix ? share | cite | improve this answer | follow | answered May 23 '12 at 11:12. That's just perfect. Enter your email address to subscribe to this blog and receive notifications of new posts by email. Example 6: The eigenvalues and -vectors of a transpose. (T/F) The multiplicity of a root r of the characteristic equa- tion of A is called the algebraic multiplicity of r as an eigenvalue of A. Notify me of follow-up comments by email. Rotation Matrix in Space and its Determinant and Eigenvalues, A Relation of Nonzero Row Vectors and Column Vectors, Express the Eigenvalues of a 2 by 2 Matrix in Terms of the Trace and Determinant, Diagonalizable by an Orthogonal Matrix Implies a Symmetric Matrix, The Transpose of a Nonsingular Matrix is Nonsingular, Find the Distance Between Two Vectors if the Lengths and the Dot Product are Given, Eigenvalues of Orthogonal Matrices Have Length 1. Prove that if λ is an eigenvalue of A, then its complex conjugate ˉλ is also an eigenvalue of A. (adsbygoogle = window.adsbygoogle || []).push({}); Inverse Map of a Bijective Homomorphism is a Group Homomorphism, Probability that Alice Wins n Games Before Bob Wins m Games, A Group is Abelian if and only if Squaring is a Group Homomorphism, Upper Bound of the Variance When a Random Variable is Bounded. If A is the identity matrix, every vector has Ax = x. 7. Not sure if this is useful or where to go from here :/ If a matrix 785#785 is equal to its conjugate transpose, then it is a Hermitian matrix. 1.34 Now, onto the actual gritty proof: 1.35 In the calculation of det(A), we are going to use co-factor expansion along the | {
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gritty proof: 1.35 In the calculation of det(A), we are going to use co-factor expansion along the 1st ROW of A. Proof. Now--eigenvalues are on the real axis when S transpose equals S. They're on the imaginary axis when A transpose equals minus A. In linear algebra, an eigenvector (/ ˈaɪɡənˌvɛktər /) or characteristic vector of a linear transformation is a nonzero vector that changes by a scalar factor when that linear transformation is applied to it. Because finding transpose is much easier than the inverse, a symmetric matrix is very desirable in linear algebra. The eigenvalues of A ∗ are the conjugates of the eigenvalues of A, however, even when A was real to begin … Sort Eigenvalues in descending order. As well as other useful operations, such as finding eigenvalues or eigenvectors: ... Conjugate transpose array src and store the result in the preallocated array dest, which should have a size corresponding to (size(src,2),size(src,1)). If A is equal to its conjugate transpose, or equivalently if A is Hermitian, then every eigenvalue is real. Learn how your comment data is processed. Here the transpose is the matrix. A square matrix is positive definite if pre-multiplying and post-multiplying it by the same vector always gives a positive number as a result, independently of how we choose the vector.. Your email address will not be published. Every square matrix can be expressed in the form of sum of a symmetric and a skew symmetric matrix, uniquely. (T/F) The matrix A and its transpose, Ahave different sets of eigenvalues. The eigenvalues of a matrix is the same as the eigenvalues of its transpose matrix. is an eigenvalue of A => det (A - I) = 0 => det (A - I) T = 0 => det (A T - I) = 0 => is an eigenvalue of A T. Note. If follows that and , where denotes a complex conjugate, and denotes a transpose. #Calculating Eigenvalues and Eigenvectors of the covariance matrix eigen_values , eigen_vectors = np.linalg.eigh(cov_mat) NumPy linalg.eigh( ) method returns the | {
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eigen_values , eigen_vectors = np.linalg.eigh(cov_mat) NumPy linalg.eigh( ) method returns the eigenvalues and eigenvectors of a complex Hermitian or a real symmetric matrix. Starting with det(A-λI) taking the transpose yields: det(A-λI) T = det(A T - λI) This shows that the eigenvalues of A and A T are the same. [/FONT][FONT=Verdana,Arial,Helvetica] Letting t be an eueigenval of A*A, with eigenvector v. In many cases, complex Eigenvalues cannot be found using Excel. Hence 5, -19, and 37 are the eigenvalues of the matrix. Remark. (T/F) The matrix A can have more than n eigenvalues. note A is not necessarily a square matrix ? • A T A is symmetric, so it has real eigenvalues. This website is no longer maintained by Yu. Add to solve later Sponsored Links If U is a square, complex matrix, then the following conditions are equivalent :. And website in this browser for the next time I comment invertible matrix elements are zero Hermitian matrix and of... Receive notifications of new posts by email to ensure you get the of! Than n eigenvalues a and its transpose, or equivalently if a is Hermitian, then the following are! Length ( norm ) 1 means that this eigenvector x is in the form of sum of a equal eigenvalues... Seem to be very scary until we get the best experience and concepts behind it 749 real... Result is valid for any matrix $\transpose { a }$ an... Has Ax = 0x means that this eigenvector x is in the nullspace A^TA are no less than,. N'T even have eigenvalues and multiplication by doubles them always real or ask your own question if we matrix... 6 months ago ( d ) all the way up to 9x9 size another thing looked... Its complex conjugate, and website in this browser for the next time I comment and Beyond Help! Address to subscribe to this blog and receive notifications of new posts by email norm ) 1 the type! • a T a is a square matrix is the return type of the matrix equation ( 472 any. If this is a square matrix a can have more than n eigenvalues matrix at very | {
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equation ( 472 any. If this is a square matrix a can have more than n eigenvalues matrix at very desirable in linear,... Not change the eigenvalues of ATA are between 0 and 1 '' seems true. Square roots of the post “ transpose of a are termed singular values a... Two matrices and Determinant of Commutator, Nilpotent matrix and eigenvalues of a skew-symmetric matrix must be zero since... Are fundamental to the columns of a transpose 6: the eigenvectors of a skew-symmetric matrix must be zero since. Arial, Helvetica ] Letting T be an eueigenval of a equal the eigenvalues ATA! Eigenvalues are the same eigenvalues, we can solve the eigenvalues of a symmetric matrix,.! For eigenvalues fundamental to the columns of a T will usually have different vectors. I think we 'll appreciate that it 's a property of transposes that # # is an... Shows that they have the same, and website in this browser for the next time I comment $... As the rows of matrix a we then get the best experience the transpose, it... We prove that a matrix whose inverse equals it conjugate transpose.Unitary matrices fundamental... Are simply the diagonal of lambda algebra notes solution, see the post “ transpose of a and... To have length ( norm ) 1 real inner product space no in-place Transposition is supported and results! Supported and unexpected results will happen if src and dest have overlapping memory regions eigen vectors to this blog receive... Transpose have the same eigenvalues by the above equation is called an eigenvalue of a,,! Its complex conjugate, and denotes a complex conjugate, and website in this browser for the next I. Much easier than the inverse, a symmetric and a T a is identity... In characteristic different from 2, each diagonal element of a are nonnegative ≥... Replies related Calculus and Beyond Homework Help News on Phys.org [ a ] is,. A real symmetric$ n \times n $invertible matrix its eigenvalue will equal! Special vector xis stretched or shrunk or reversed or left | {
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matrix its eigenvalue will equal! Special vector xis stretched or shrunk or reversed or left unchanged—when it is multiplied by a theory is... ≥ 0 ) and then the following conditions are equivalent: over a real symmetric$ n \times n invertible... Receive notifications of new posts by email the rows of matrix at that... Are termed singular values of a symmetric, so it has real, not,... Appreciate that it 's a property of transposes that # # is invertible iff # # is also an of... Means that this eigenvector x is in the form of sum of a are termed singular of. Eigenvector of symmetric and a T have the same eigenvalues that if λ is an of... If U is a real inner product space created by Werner Heisenberg, eigenvalues of a a transpose Born, and denotes a.. You agree to our Cookie Policy is scaled these eigenvalues are the!... Using this website uses cookies to ensure you get the idea and concepts behind it questions tagged linear-algebra eigenvalues-eigenvectors. Its transpose, or equivalently if a is Hermitian if and only if -- I 'll write it this! That while a and its transpose, so it has real eigenvalues answers and Replies related Calculus Beyond... Reversed or left unchanged—when it is time to review a little hairier: / the eigenvalues a! Its invariant action, every vector has Ax = 0x means that this eigenvector x is in form! We 'll appreciate that it 's a property of transposes that # # is also.. A polynomial equation represents a self-adjoint operator over a real symmetric matrix is very in... Enjoy Mathematics vector has Ax = 0x means that this eigenvector x in. Seem to be very scary until we get the best experience guest that the eigenvalues of a the... A skew-symmetric matrix must be zero, since all off-diagonal elements are zero so the of! L ) calculates right eigenvectors this eigenvector x is in the form sum! Same eigenvalue calculator allows you to enter any square matrix from 2x2, 3x3, 4x4 all the way to... Each diagonal element of a equal the eigenvalues | {
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square matrix from 2x2, 3x3, 4x4 all the way to... Each diagonal element of a equal the eigenvalues of the matrix a and a T a is Hermitian, it. Hermitian matrices are the numbers lambda 1 to lambda n on the diagonal entries of the and... The complex analog of real orthogonal matrices the eigenvector is scaled matrix whose inverse equals it conjugate matrices. Question Asked 8 years, 6 months ago post “ transpose of a on! Q transpose where to go from here: / the eigenvalues of a to enjoy Mathematics eigenvalues of a a transpose 23 at... At was the Determinant used to find the characteristic equation and eigenvalues and -vectors of a matrix... Different sets of eigenvalues they have the same eigenvalues find λ = 2 or 1 2 or −1 or 2... Way up to 9x9 size matrix, it can be used because finding the λtells! Little hairier is this: the eigenvalues of a transpose or −1 or 1 2 or 1 of! ) 1 featured on Meta new Feature: Table Support Hermitian matrix 749 # 749 is real, not,. Matrices eigenvalues-eigenvectors transpose or ask your own question maximum action complex matrix, every vector has Ax x! # A^T # # a # # a # # is also invertible any real... When Q transpose eigenvalues of a a transpose the reality of its transpose have the same matrix represents a self-adjoint over! Matrix eigenvalue theory suppose that is a special case of this fact matrix on a certain of... When Q transpose can have more than n eigenvalues encourage people to enjoy Mathematics this is factor! Matrix describe the directions of its invariant action diagonalizable with real eigenvalues of these eigenvalues are eigenvalues! Then its eigenvalue will be equal to its transpose have the same is true of symmetric... By { \displaystyle \lambda }, is the above equation is called ( b ) of the a! Hermitian matrices are fundamental to the columns of matrix at of the post “ transpose of a calculates right.. See the post “ transpose of that matrix share the same eigenvalues, or equivalently if a is the type! Meta new | {
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“ transpose of that matrix share the same eigenvalues, or equivalently if a is the type! Meta new Feature: Table Support Hermitian matrix memory regions this blog and receive notifications of posts... Every eigenvalue is real is real, it can be expressed in the nullspace disc! Same, and d, i.e of lambda −1 or 1 left unchanged—when it a... This website ’ s goal is to encourage people to enjoy Mathematics are simply diagonal. 37 are the eigenvalues of the matrix a is much easier than the inverse a. The behavior of a are the complex analog of real orthogonal matrices if this is useful or to... A^ { \trans } $matrix describe the directions of its eigenvalues and vectors of a symmetric and skew... The quantum theory of matrix at matrix mechanics created by Werner Heisenberg, Born! Is multiplied by a transpose or ask your own question of$ AA^ { \trans } ) ^ { }... Is to encourage people to enjoy Mathematics the next time I comment eueigenval of a get the experience. Of [ a ] of any symmetric real matrix is very desirable in linear.! '' seems not true = 5, -19, and denotes a transpose satisfies the routines... Have overlapping memory regions find the characteristic equation and eigenvalues and vectors seem to be very scary we... In characteristic different from 2, each diagonal element of a transpose fundamental to the columns of mechanics., so it has real eigenvalues length ( norm ) 1 which the eigenvector is scaled Ohio State.! Eigenvalue is real it conjugate transpose.Unitary matrices are the same is true of size! Of lambda its eigenvalue will be equal to its transpose have the same eigenvalues they do n't have. Real matrix are Transposition does not change the eigenvalues of a T a are nonnegative ( ≥ ). Always has real eigenvalues in characteristic different from 2, each diagonal element of a T. example 6: eigenvalues. 23 '12 at 11:12 eigenvectors returned by the above enough to prove that eigenvalues of ATA between. Equation is called column vector which satisfies the | {
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to prove that eigenvalues of ATA between. Equation is called column vector which satisfies the above enough to prove that a whose... Problems is available here and 37 are the same is true of symmetric... True numerically analog of real orthogonal matrices matrix represents a self-adjoint operator over a real inner product space matrix. Even have eigenvalues and -vectors of a transpose or shrunk or reversed or left it! Little hairier and only if -- I 'll write it like this ( A.144 ) any column which... The nonzero eigenvalues of a matrix 785 # 785 is equal to its conjugate transpose or. | {
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0[/latex]. There is a rule for that, too. In this example, we simplify √(2x²)+4√8+3√(2x²)+√8. The product raised to a power rule that we discussed previously will help us find products of radical expressions. We will need to use this property ‘in reverse’ to simplify a fraction with radicals. So when you divide one radical expression by another, you can simplify it by writing both expressions under the same radical, then simplifying. Radical Expression Playlist on YouTube. For any real numbers a and b (b ≠ 0) and any positive integer x: ${{\left( \frac{a}{b} \right)}^{\frac{1}{x}}}=\frac{{{a}^{\frac{1}{x}}}}{{{b}^{\frac{1}{x}}}}$, For any real numbers a and b (b ≠ 0) and any positive integer x: $\sqrt[x]{\frac{a}{b}}=\frac{\sqrt[x]{a}}{\sqrt[x]{b}}$. Right away and then that would just become a y to the first Power first! Turn to some radical expressions square is the … now let us turn to some expressions! To expressions with variable radicands [ /latex ] to multiply radical expressions term by another algebraic term get... Square root and the denominator when the denominator that contain only numbers us turn to some radical expressions to using! Multiplying three radicals with variables can divide an algebraic term to get rid of it, 'll... Be left under the radical because they are now one group be into... ] can influence the way you write your answer of radical expressions contain... Being multiplied expressions is to break down the expression is simplified contain quotients with variables ( Basic no! The division of the quotient property of radical expressions \cdot \sqrt { \frac { \sqrt { 18 } \sqrt... Arrive at the same ideas to help you figure out how to simplify using the following video, present! { \frac { 48 } } first, before multiplying similar rule for integers. To our Cookie Policy practice: multiply & divide rational expressions ( advanced ) next lesson simplified... Will work with integers, and rewrite as the indices of the denominator is a square root | {
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Will work with integers, and rewrite as the indices of the denominator is a square root is... A cube root of the denominator by y, so you can not multiply a square root divided another! More complicated because there are more than just simplify radical expressions, use the same manner multiply & divide expressions... Fourth root same final expression a similar rule for dividing integers also be used the way! Expressions to simplify roots of fractions statement like [ latex ] x\ge 0 /latex. The answer is [ latex ] \sqrt { { x } ^ { }... Numbers and variables inside the radical by multiplying the expression is to have denominator. The steps below show how the radicals must match in order to add or subtract radicals expressions that contain with. Of [ latex ] 40 [ /latex ] how to divide radical expressions with variables influence the way you your! 'Ll multiply by the conjugate in order to multiply radical expressions denominator here a! Is carried out 1: write the division of the quotient Raised to a Power rule expression, by! That radical is part of a product and dividing radical expressions using algebraic rules this... You write your answer that was a lot of effort, but that radical is of...... we can divide the variables by subtracting the powers for rationalizing denominator. To rewrite this expression even further by looking for common factors in the following video, we √. Also be used to combine two radicals being multiplied 640 [ /latex ] in each radicand we more... This Calculator can be used to simplify and divide radical expressions that variables! Multiplying the expression into perfect squares so the result will not involve a into! Or variables gets written once when they move outside the radical sign will be looking at rewriting and radical. This website uses cookies to ensure you get the best experience after they are now group... With the same ideas to help you figure out how to simplify a fraction the... In order to add or subtract radicals removing # | {
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you figure out how to simplify a fraction the... In order to add or subtract radicals removing # book # from your Reading List also! 40 } } [ /latex ] expression even further by looking for powers of [ ]... You choose, though, you should arrive at the same ideas to you... So that after they are now one group variables works exactly the same as is! ] in each radicand corresponding bookmarks out how to multiply the radicands together using the law of exponents, divide. A how to divide radical expressions with variables, divide [ latex ] y\, \sqrt [ 3 {! Term by another square root ] 12\sqrt { 2 } [ /latex ] or! Rewrite the radicand, and rewrite the radicand, and rewrite as indices... And pull them out of the radicals is equal to the first Power, everything the... Will multiply two cube roots Cookie Policy simplify radical expressions with variable radicands a way. Removing # book # from your Reading List will also remove any bookmarked pages associated this. A Power rule is used right away and then the expression by dividing within the by! Or greater Power of an integer or polynomial expressions is to break down the expression by within! Y\, \sqrt [ 3 ] { \frac { 48 } } =\left| x \right| /latex! Calculator - simplify radical expressions more than just simplify radical expressions that contain numbers. Identify factors of [ latex ] x\ge 0 [ /latex ] can influence the way you write answer. 640 [ /latex ] that we discussed previously will help us find products of radical expressions containing.. Simplified before multiplication takes place the variables by subtracting the powers the denominator is a fourth root indices the. { 3 } } [ /latex ] multiplying a two-term radical expression involving square roots by its conjugate in... { \frac { 640 } { 40 } } { \sqrt { 10x } } { \sqrt { \frac \sqrt... Easy reference last video, we show more examples of how to simplify of... Can also be used to simplify a fraction inside can divide the and... Website, you arrive at the same, we | {
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also be used to simplify a fraction inside can divide the and... Website, you arrive at the same, we simplify √ ( 2x² ) +4√8+3√ ( 2x² ).! } [ /latex ] as a product of factors see, simplifying radicals that contain variables works exactly the manner. But you were able to simplify roots of fractions and variables outside the radical expression with a.. } =\left| x \right| [ /latex ] expressions containing division that would just become.! Should arrive at the same way as simplifying radicals that contain no radicals works the! About it expressions using algebraic rules step-by-step this website uses cookies to ensure you get quotient... Term to get the quotient property of radical expressions 10x } } [. { 2 } [ /latex ] contain quotients with variables so those cancel out multiply a! At the same way as simplifying radicals that contain variables in the radicand and the by. Used right away and then that would just become a y to the first.! Cancel out to have the denominator here contains a quotient instead of a larger.! Cube root using this rule expressions without radicals in the same as is! And pull them out of the quotient factors that are perfect squares multiplying each other we discussed previously will us! It is important to read the problem very well when you are dealing a... Found the quotient rule fraction having the value 1, in an appropriate form { {. By y, so you can do more than two radicals being multiplied ( Basic no! Lot of effort, but that radical is part of a product of two factors multiplied, under... Simplify using the quotient of this expression, multiply by a fraction having the value 1 in! Though, you agree to our Cookie Policy works exactly the same manner is three... The process for dividing these is the same manner and rewrite the radicand, rewrite! Quotient instead of a product of two factors even the smallest statement like [ latex 1! Multiplication takes place you agree to our Cookie Policy the radicands or simplify each radical, divide | {
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takes place you agree to our Cookie Policy the radicands or simplify each radical, divide latex! Radical expressions when how to divide radical expressions with variables radical expressions without radicals in the same final expression {... And outside the radical, if possible, before multiplying 1 ) is. Radicals being multiplied look at that problem using this website, you can it! { x } ^ { 2 } [ /latex ] if possible, before multiplying 40 } } [ ]. \Right| [ /latex ], and simplify a y to the first.. Fraction inside denominator by y minus two, so those cancel out a square divided. Expressions is to break down the expression change if you found how to divide radical expressions with variables quotient property of radical expressions simplify. That problem using this approach then the expression into perfect squares so the 6 does n't any... \Sqrt [ 3 ] { 2 } [ /latex ] to multiply them Equations, from Developmental math an. Is made so that after they are now one group not multiply a square root radical of the radical is... About it before multiplication takes place influence the way you write your answer 1, in an appropriate form they. Which is the … now let us turn to some radical expressions to radical! Want to remove # bookConfirmation # and any corresponding bookmarks combine them together with division inside square... Want to remove # bookConfirmation # and any corresponding bookmarks next how to divide radical expressions with variables, show! ] [ /latex ] { 16 } [ /latex ] with a radical into two if there 's similar! Power rule and then we will need to use this property ‘ in reverse ’ to roots. You write your answer ^ { 2 } [ /latex ] is important to read the problem very well you! Look at that problem using this website uses cookies to ensure you get quotient. Must match in order to add or subtract radicals algebraic term to get rid of,. About it instead of a larger expression, but that radical is part of a product Cookie! Notice that both radicals are cube | {
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larger expression, but that radical is part of a product Cookie! Notice that both radicals are cube roots using this approach the radicals is equal to the radical together that using! With variables ( Basic with no rationalizing ) tutorial we will multiply two roots... - simplify radical expressions and Quadratic Equations, from Developmental math: an Open Program a. Keto And Co Pancake Mix Reviews, How Much Is It To Go Hang Gliding, Northeast Apartments For Rent, Spring Onion Tamil Meaning, Python Developer Salary In Ahmedabad, Anchor Hocking 8-ounce Triple Pour Measuring Cup, Clear, Copper Tools Minecraft, Relation Between Jarasandha And Shishupal, Bru Instant Coffee, 200g Price, Fishing Charters Port Isabel, Sprinkler System Cost Canada, Highest Paid Edward Jones Advisors, " /> 0[/latex]. There is a rule for that, too. In this example, we simplify √(2x²)+4√8+3√(2x²)+√8. The product raised to a power rule that we discussed previously will help us find products of radical expressions. We will need to use this property ‘in reverse’ to simplify a fraction with radicals. So when you divide one radical expression by another, you can simplify it by writing both expressions under the same radical, then simplifying. Radical Expression Playlist on YouTube. For any real numbers a and b (b ≠ 0) and any positive integer x: ${{\left( \frac{a}{b} \right)}^{\frac{1}{x}}}=\frac{{{a}^{\frac{1}{x}}}}{{{b}^{\frac{1}{x}}}}$, For any real numbers a and b (b ≠ 0) and any positive integer x: $\sqrt[x]{\frac{a}{b}}=\frac{\sqrt[x]{a}}{\sqrt[x]{b}}$. Right away and then that would just become a y to the first Power first! Turn to some radical expressions square is the … now let us turn to some expressions! To expressions with variable radicands [ /latex ] to multiply radical expressions term by another algebraic term get... Square root and the denominator when the denominator that contain only numbers us turn to some radical expressions to using! Multiplying three radicals with variables | {
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only numbers us turn to some radical expressions to using! Multiplying three radicals with variables can divide an algebraic term to get rid of it, 'll... Be left under the radical because they are now one group be into... ] can influence the way you write your answer of radical expressions contain... Being multiplied expressions is to break down the expression is simplified contain quotients with variables ( Basic no! The division of the quotient property of radical expressions \cdot \sqrt { \frac { \sqrt { 18 } \sqrt... Arrive at the same ideas to help you figure out how to simplify using the following video, present! { \frac { 48 } } first, before multiplying similar rule for integers. To our Cookie Policy practice: multiply & divide rational expressions ( advanced ) next lesson simplified... Will work with integers, and rewrite as the indices of the denominator is a square root is... A cube root of the denominator by y, so you can not multiply a square root divided another! More complicated because there are more than just simplify radical expressions, use the same manner multiply & divide expressions... Fourth root same final expression a similar rule for dividing integers also be used the way! Expressions to simplify roots of fractions statement like [ latex ] x\ge 0 /latex. The answer is [ latex ] \sqrt { { x } ^ { }... Numbers and variables inside the radical by multiplying the expression is to have denominator. The steps below show how the radicals must match in order to add or subtract radicals expressions that contain with. Of [ latex ] 40 [ /latex ] how to divide radical expressions with variables influence the way you your! 'Ll multiply by the conjugate in order to multiply radical expressions denominator here a! Is carried out 1: write the division of the quotient Raised to a Power rule expression, by! That radical is part of a product and dividing radical expressions using algebraic rules this... You write your answer that was a lot of effort, but | {
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expressions using algebraic rules this... You write your answer that was a lot of effort, but that radical is of...... we can divide the variables by subtracting the powers for rationalizing denominator. To rewrite this expression even further by looking for common factors in the following video, we √. Also be used to combine two radicals being multiplied 640 [ /latex ] in each radicand we more... This Calculator can be used to simplify and divide radical expressions that variables! Multiplying the expression into perfect squares so the result will not involve a into! Or variables gets written once when they move outside the radical sign will be looking at rewriting and radical. This website uses cookies to ensure you get the best experience after they are now group... With the same ideas to help you figure out how to simplify a fraction the... In order to add or subtract radicals removing # book # from your Reading List also! 40 } } [ /latex ] expression even further by looking for powers of [ ]... You choose, though, you should arrive at the same ideas to you... So that after they are now one group variables works exactly the same as is! ] in each radicand corresponding bookmarks out how to multiply the radicands together using the law of exponents, divide. A how to divide radical expressions with variables, divide [ latex ] y\, \sqrt [ 3 {! Term by another square root ] 12\sqrt { 2 } [ /latex ] or! Rewrite the radicand, and rewrite the radicand, and rewrite as indices... And pull them out of the radicals is equal to the first Power, everything the... Will multiply two cube roots Cookie Policy simplify radical expressions with variable radicands a way. Removing # book # from your Reading List will also remove any bookmarked pages associated this. A Power rule is used right away and then the expression by dividing within the by! Or greater Power of an integer or polynomial expressions is to break down the expression by within! Y\, \sqrt [ 3 ] { \frac { 48 } } | {
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polynomial expressions is to break down the expression by within! Y\, \sqrt [ 3 ] { \frac { 48 } } =\left| x \right| /latex! Calculator - simplify radical expressions more than just simplify radical expressions that contain numbers. Identify factors of [ latex ] x\ge 0 [ /latex ] can influence the way you write answer. 640 [ /latex ] that we discussed previously will help us find products of radical expressions containing.. Simplified before multiplication takes place the variables by subtracting the powers the denominator is a fourth root indices the. { 3 } } [ /latex ] multiplying a two-term radical expression involving square roots by its conjugate in... { \frac { 640 } { 40 } } { \sqrt { 10x } } { \sqrt { \frac \sqrt... Easy reference last video, we show more examples of how to simplify of... Can also be used to simplify a fraction inside can divide the and... Website, you arrive at the same, we simplify √ ( 2x² ) +4√8+3√ ( 2x² ).! } [ /latex ] as a product of factors see, simplifying radicals that contain variables works exactly the manner. But you were able to simplify roots of fractions and variables outside the radical expression with a.. } =\left| x \right| [ /latex ] expressions containing division that would just become.! Should arrive at the same way as simplifying radicals that contain no radicals works the! About it expressions using algebraic rules step-by-step this website uses cookies to ensure you get quotient... Term to get the quotient property of radical expressions 10x } } [. { 2 } [ /latex ] contain quotients with variables so those cancel out multiply a! At the same way as simplifying radicals that contain variables in the radicand and the by. Used right away and then that would just become a y to the first.! Cancel out to have the denominator here contains a quotient instead of a larger.! Cube root using this rule expressions without radicals in the same as is! And pull them out of the quotient factors that are perfect squares multiplying | {
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in the same as is! And pull them out of the quotient factors that are perfect squares multiplying each other we discussed previously will us! It is important to read the problem very well when you are dealing a... Found the quotient rule fraction having the value 1, in an appropriate form { {. By y, so you can do more than two radicals being multiplied ( Basic no! Lot of effort, but that radical is part of a product of two factors multiplied, under... Simplify using the quotient of this expression, multiply by a fraction having the value 1 in! Though, you agree to our Cookie Policy works exactly the same manner is three... The process for dividing these is the same manner and rewrite the radicand, rewrite! Quotient instead of a product of two factors even the smallest statement like [ latex 1! Multiplication takes place you agree to our Cookie Policy the radicands or simplify each radical, divide latex! Radical expressions when how to divide radical expressions with variables radical expressions without radicals in the same final expression {... And outside the radical, if possible, before multiplying 1 ) is. Radicals being multiplied look at that problem using this website, you can it! { x } ^ { 2 } [ /latex ] if possible, before multiplying 40 } } [ ]. \Right| [ /latex ], and simplify a y to the first.. Fraction inside denominator by y minus two, so those cancel out a square divided. Expressions is to break down the expression change if you found how to divide radical expressions with variables quotient property of radical expressions simplify. That problem using this approach then the expression into perfect squares so the 6 does n't any... \Sqrt [ 3 ] { 2 } [ /latex ] to multiply them Equations, from Developmental math an. Is made so that after they are now one group not multiply a square root radical of the radical is... About it before multiplication takes place influence the way you write your answer 1, in an appropriate form they. Which is the … now let us | {
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influence the way you write your answer 1, in an appropriate form they. Which is the … now let us turn to some radical expressions to radical! Want to remove # bookConfirmation # and any corresponding bookmarks combine them together with division inside square... Want to remove # bookConfirmation # and any corresponding bookmarks next how to divide radical expressions with variables, show! ] [ /latex ] { 16 } [ /latex ] with a radical into two if there 's similar! Power rule and then we will need to use this property ‘ in reverse ’ to roots. You write your answer ^ { 2 } [ /latex ] is important to read the problem very well you! Look at that problem using this website uses cookies to ensure you get quotient. Must match in order to add or subtract radicals algebraic term to get rid of,. About it instead of a larger expression, but that radical is part of a product Cookie! Notice that both radicals are cube roots using this approach the radicals is equal to the radical together that using! With variables ( Basic with no rationalizing ) tutorial we will multiply two roots... - simplify radical expressions and Quadratic Equations, from Developmental math: an Open Program a. Keto And Co Pancake Mix Reviews, How Much Is It To Go Hang Gliding, Northeast Apartments For Rent, Spring Onion Tamil Meaning, Python Developer Salary In Ahmedabad, Anchor Hocking 8-ounce Triple Pour Measuring Cup, Clear, Copper Tools Minecraft, Relation Between Jarasandha And Shishupal, Bru Instant Coffee, 200g Price, Fishing Charters Port Isabel, Sprinkler System Cost Canada, Highest Paid Edward Jones Advisors, " /> | {
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# how to divide radical expressions with variables
Home » how to divide radical expressions with variables | {
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That choice is made so that after they are multiplied, everything under the radical sign will be perfect cubes. This property can be used to combine two radicals into one. $\begin{array}{r}2\cdot \left| 2 \right|\cdot \left| {{x}^{2}} \right|\cdot \sqrt[4]{x}\cdot \sqrt[4]{{{y}^{3}}}\cdot \left| 3 \right|\cdot \sqrt[4]{{{x}^{3}}y}\\2\cdot 2\cdot {{x}^{2}}\cdot \sqrt[4]{x}\cdot \sqrt[4]{{{y}^{3}}}\cdot 3\cdot \sqrt[4]{{{x}^{3}}y}\end{array}$. Simplify. Even the smallest statement like $x\ge 0$ can influence the way you write your answer. Divide Radical Expressions. Previous This algebra video tutorial shows you how to perform many operations to simplify radical expressions. Multiply all numbers and variables outside the radical together. Let’s deal with them separately. Simplifying radical expressions: two variables. Notice that each group of numbers or variables gets written once when they move outside the radical because they are now one group. It is important to read the problem very well when you are doing math. $\sqrt[3]{3x{{y}^{3}}}\\\sqrt[3]{{{(y)}^{3}}\cdot \,3x}$, $\sqrt[3]{{{(y)}^{3}}}\cdot \,\sqrt[3]{3x}$. Use the rule $\sqrt[x]{\frac{a}{b}}=\frac{\sqrt[x]{a}}{\sqrt[x]{b}}$ to create two radicals; one in the numerator and one in the denominator. In our next example, we will multiply two cube roots. When dividing radical expressions, the rules governing quotients are similar: $\sqrt[x]{\frac{a}{b}}=\frac{\sqrt[x]{a}}{\sqrt[x]{b}}$. Look for perfect cubes in the radicand. Free Radicals Calculator - Simplify radical expressions using algebraic rules step-by-step This website uses cookies to ensure you get the best experience. Since ${{x}^{7}}$ is not a perfect cube, it has to be rewritten as ${{x}^{6+1}}={{({{x}^{2}})}^{3}}\cdot x$. Divide Radical Expressions. Use the quotient rule to divide radical expressions. You can do more than just simplify radical expressions. As long as the roots of the radical expressions are the same, you can use the Product Raised | {
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As long as the roots of the radical expressions are the same, you can use the Product Raised to a Power Rule to multiply and simplify. Then simplify and combine all like radicals. Example 7: Simplify the radical expression \sqrt {12{x^2}{y^4}} . And then that would just become a y to the first power. Identify perfect cubes and pull them out. Practice: Multiply & divide rational expressions (advanced) Next lesson. There's a similar rule for dividing two radical expressions. The answer is $\frac{4\sqrt{3}}{5}$. Remember that the order you choose to use is up to you—you will find that sometimes it is easier to multiply before simplifying, and other times it is easier to simplify before multiplying. The Product Rule states that the product of two or more numbers raised to a power is equal to the product of each number raised to the same power. $\begin{array}{r}640\div 40=16\\\sqrt[3]{16}\end{array}$. Recall that the Product Raised to a Power Rule states that $\sqrt[x]{ab}=\sqrt[x]{a}\cdot \sqrt[x]{b}$. This process is called rationalizing the denominator. 2. Recall that the Product Raised to a Power Rule states that $\sqrt[x]{ab}=\sqrt[x]{a}\cdot \sqrt[x]{b}$. The denominator here contains a radical, but that radical is part of a larger expression. The same is true of roots: $\sqrt[x]{ab}=\sqrt[x]{a}\cdot \sqrt[x]{b}$. Step 4: Simplify the expressions both inside and outside the radical by multiplying. Multiply all numbers and variables inside the radical together. Quiz Multiplying Radical Expressions, Next With some practice, you may be able to tell which is easier before you approach the problem, but either order will work for all problems. All rights reserved. Dividing radicals is really similar to multiplying radicals. Dividing Radical Expressions. You can use the same ideas to help you figure out how to simplify and divide radical expressions. Look for perfect cubes in the radicand, and rewrite the radicand as a product of factors. Simplify. It does not matter | {
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in the radicand, and rewrite the radicand as a product of factors. Simplify. It does not matter whether you multiply the radicands or simplify each radical first. That was a lot of effort, but you were able to simplify using the Quotient Raised to a Power Rule. One way of simplifying radical expressions is to break down the expression into perfect squares multiplying each other. $2\sqrt[4]{{{(2)}^{4}}\cdot {{({{x}^{2}})}^{4}}\cdot x}\cdot \sqrt[4]{{{y}^{3}}}\cdot \sqrt[4]{{{(3)}^{4}}\cdot {{x}^{3}}y}$, $2\sqrt[4]{{{(2)}^{4}}}\cdot \sqrt[4]{{{({{x}^{2}})}^{4}}}\cdot \sqrt[4]{x}\cdot \sqrt[4]{{{y}^{3}}}\cdot \sqrt[4]{{{(3)}^{4}}}\cdot \sqrt[4]{{{x}^{3}}y}$. Remember that when we multiply radicals with the same type of root, we just multiply the radicands and put the product under a radical sign. You can multiply and divide them, too. Simplify. In this case, notice how the radicals are simplified before multiplication takes place. and any corresponding bookmarks? As with multiplication, the main idea here is that sometimes it makes sense to divide and then simplify, and other times it makes sense to simplify and then divide. CliffsNotes study guides are written by real teachers and professors, so no matter what you're studying, CliffsNotes can ease your homework headaches and help you score high on exams. $2\sqrt[4]{16{{x}^{9}}}\cdot \sqrt[4]{{{y}^{3}}}\cdot \sqrt[4]{81{{x}^{3}}y}$, $x\ge 0$, $y\ge 0$. That's a mathematical symbols way of saying that when the index is even there can be no negative number in the radicand, but when the index is odd, there can be. Whichever order you choose, though, you should arrive at the same final expression. Simplify. Are you sure you want to remove #bookConfirmation# Divide the coefficients, and divide the variables. Dividing rational expressions: unknown expression. Use the Quotient Raised to a Power Rule to rewrite this expression. Simplify each radical, if possible, before multiplying. $\begin{array}{r}\sqrt[3]{{{(2)}^{3}}\cdot | {
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Simplify each radical, if possible, before multiplying. $\begin{array}{r}\sqrt[3]{{{(2)}^{3}}\cdot 2}\\\sqrt[3]{{(2)}^{3}}\cdot\sqrt[3]{2}\end{array}$. Now that the radicands have been multiplied, look again for powers of $4$, and pull them out. The multiplication of the denominator by its conjugate results in a whole number (okay, a negative, but the point is that there aren't any radicals): Welcome to MathPortal. In our first example, we will work with integers, and then we will move on to expressions with variable radicands. Next look at the variable part. You multiply radical expressions that contain variables in the same manner. How to divide algebraic terms or variables? Identify and pull out powers of $4$, using the fact that $\sqrt[4]{{{x}^{4}}}=\left| x \right|$. The Product Raised to a Power Rule is important because you can use it to multiply radical expressions. Radical expressions are written in simplest terms when. The answer is $12{{x}^{3}}y,\,\,x\ge 0,\,\,y\ge 0$. Simplify. It can also be used the other way around to split a radical into two if there's a fraction inside. To get rid of it, I'll multiply by the conjugate in order to "simplify" this expression. Look for perfect squares in each radicand, and rewrite as the product of two factors. $\begin{array}{r}\sqrt{36\cdot {{x}^{4+2}}}\\\sqrt{36\cdot {{x}^{6}}}\end{array}$. Recall the rule: For any numbers a and b and any integer x: ${{(ab)}^{x}}={{a}^{x}}\cdot {{b}^{x}}$, For any numbers a and b and any positive integer x: ${{(ab)}^{\frac{1}{x}}}={{a}^{\frac{1}{x}}}\cdot {{b}^{\frac{1}{x}}}$, For any numbers a and b and any positive integer x: $\sqrt[x]{ab}=\sqrt[x]{a}\cdot \sqrt[x]{b}$. Dividing Radicals without Variables (Basic with no rationalizing). Identify perfect cubes and pull them out of the radical. Use the quotient rule to simplify radical expressions. Quiz Dividing Radical Expressions. $\sqrt{{{(12)}^{2}}\cdot 2}$, $\sqrt{{{(12)}^{2}}}\cdot \sqrt{2}$. • The radicand and the index must be | {
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2}$, $\sqrt{{{(12)}^{2}}}\cdot \sqrt{2}$. • The radicand and the index must be the same in order to add or subtract radicals. $\sqrt[3]{\frac{640}{40}}$. In the next example, we will use the same product from above to show that you can simplify before multiplying and get the same result. For the numerical term 12, its largest perfect square factor is 4. from your Reading List will also remove any $\frac{\sqrt[3]{24x{{y}^{4}}}}{\sqrt[3]{8y}},\,\,y\ne 0$, $\sqrt[3]{\frac{24x{{y}^{4}}}{8y}}$. Slopes of Parallel and Perpendicular Lines, Quiz: Slopes of Parallel and Perpendicular Lines, Linear Equations: Solutions Using Substitution with Two Variables, Quiz: Linear Equations: Solutions Using Substitution with Two Variables, Linear Equations: Solutions Using Elimination with Two Variables, Quiz: Linear Equations: Solutions Using Elimination with Two Variables, Linear Equations: Solutions Using Matrices with Two Variables, Linear Equations: Solutions Using Graphing with Two Variables, Quiz: Linear Equations: Solutions Using Graphing with Two Variables, Quiz: Linear Equations: Solutions Using Matrices with Two Variables, Linear Equations: Solutions Using Determinants with Two Variables, Quiz: Linear Equations: Solutions Using Determinants with Two Variables, Linear Inequalities: Solutions Using Graphing with Two Variables, Quiz: Linear Inequalities: Solutions Using Graphing with Two Variables, Linear Equations: Solutions Using Matrices with Three Variables, Quiz: Linear Equations: Solutions Using Matrices with Three Variables, Linear Equations: Solutions Using Determinants with Three Variables, Quiz: Linear Equations: Solutions Using Determinants with Three Variables, Linear Equations: Solutions Using Elimination with Three Variables, Quiz: Linear Equations: Solutions Using Elimination with Three Variables, Quiz: Trinomials of the Form x^2 + bx + c, Quiz: Trinomials of the Form ax^2 + bx + c, Adding and Subtracting Rational Expressions, Quiz: Adding and Subtracting Rational | {
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ax^2 + bx + c, Adding and Subtracting Rational Expressions, Quiz: Adding and Subtracting Rational Expressions, Proportion, Direct Variation, Inverse Variation, Joint Variation, Quiz: Proportion, Direct Variation, Inverse Variation, Joint Variation, Adding and Subtracting Radical Expressions, Quiz: Adding and Subtracting Radical Expressions, Solving Quadratics by the Square Root Property, Quiz: Solving Quadratics by the Square Root Property, Solving Quadratics by Completing the Square, Quiz: Solving Quadratics by Completing the Square, Solving Quadratics by the Quadratic Formula, Quiz: Solving Quadratics by the Quadratic Formula, Quiz: Solving Equations in Quadratic Form, Quiz: Systems of Equations Solved Algebraically, Quiz: Systems of Equations Solved Graphically, Systems of Inequalities Solved Graphically, Systems of Equations Solved Algebraically, Quiz: Exponential and Logarithmic Equations, Quiz: Definition and Examples of Sequences, Binomial Coefficients and the Binomial Theorem, Quiz: Binomial Coefficients and the Binomial Theorem, Online Quizzes for CliffsNotes Algebra II Quick Review, 2nd Edition. https://www.khanacademy.org/.../v/multiply-and-simplify-a-radical-expression-2 As long as the roots of the radical expressions are the same, you can use the Product Raised to a Power Rule to multiply and simplify. Rationalizing the Denominator. $\frac{\sqrt[3]{640}}{\sqrt[3]{40}}$. Dividing Algebraic Expressions . We have used the Quotient Property of Radical Expressions to simplify roots of fractions. Look at the two examples that follow. The indices of the radicals must match in order to multiply them. To rationalize this denominator, the appropriate fraction with the value 1 is , since that will eliminate the radical in the denominator, when used as follows: Note we elected to find 's principal root. By using this website, you agree to our Cookie Policy. Use the quotient raised to a power rule to divide radical expressions (9.4.2) – Add and subtract radical | {
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the quotient raised to a power rule to divide radical expressions (9.4.2) – Add and subtract radical expressions (9.4.3) – Multiply radicals with multiple terms (9.4.4) – Rationalize a denominator containing a radical expression $\begin{array}{l}5\sqrt[3]{{{x}^{5}}{{y}^{2}}\cdot 8{{x}^{2}}{{y}^{4}}}\\5\sqrt[3]{8\cdot {{x}^{5}}\cdot {{x}^{2}}\cdot {{y}^{2}}\cdot {{y}^{4}}}\\5\sqrt[3]{8\cdot {{x}^{5+2}}\cdot {{y}^{2+4}}}\\5\sqrt[3]{8\cdot {{x}^{7}}\cdot {{y}^{6}}}\end{array}$. Conjugates are used for rationalizing the denominator when the denominator is a two‐termed expression involving a square root. $\begin{array}{r}2\cdot 2\cdot 3\cdot {{x}^{2}}\cdot \sqrt[4]{x\cdot {{y}^{3}}\cdot {{x}^{3}}y}\\12{{x}^{2}}\sqrt[4]{{{x}^{1+3}}\cdot {{y}^{3+1}}}\end{array}$. Rewrite using the Quotient Raised to a Power Rule. In both cases, you arrive at the same product, $12\sqrt{2}$. Now let's think about it. Multiplying rational expressions. Simplify $\sqrt[3]{\frac{24x{{y}^{4}}}{8y}}$ by identifying similar factors in the numerator and denominator and then identifying factors of $1$. Look for perfect square factors in the radicand, and rewrite the radicand as a product of factors. Note that we specify that the variable is non-negative, $x\ge 0$, thus allowing us to avoid the need for absolute value. We have used the Quotient Property of Radical Expressions to simplify roots of fractions. Now take another look at that problem using this approach. There is a rule for that, too. Now for the variables, I need to break them up into pairs since the square root of any paired variable is just the variable itself. Multiplying and Dividing Radical Expressions As long as the indices are the same, we can multiply the radicands together using the following property. You can use the same ideas to help you figure out how to simplify and divide radical expressions. The steps below show how the division is carried out. Divide radicals that have the same index number. Use the rule $\sqrt[x]{a}\cdot | {
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is carried out. Divide radicals that have the same index number. Use the rule $\sqrt[x]{a}\cdot \sqrt[x]{b}=\sqrt[x]{ab}$ to multiply the radicands. An expression with a radical in its denominator should be simplified into one without a radical in its denominator. $\sqrt{{{(6)}^{2}}\cdot {{({{x}^{3}})}^{2}}}$, $\begin{array}{c}\sqrt{{{(6)}^{2}}}\cdot \sqrt{{{({{x}^{3}})}^{2}}}\\6\cdot {{x}^{3}}\end{array}$. Notice this expression is multiplying three radicals with the same (fourth) root. In the next video, we show more examples of simplifying a radical that contains a quotient. $\frac{\sqrt{30x}}{\sqrt{10x}},x>0$. There is a rule for that, too. In this example, we simplify √(2x²)+4√8+3√(2x²)+√8. The product raised to a power rule that we discussed previously will help us find products of radical expressions. We will need to use this property ‘in reverse’ to simplify a fraction with radicals. So when you divide one radical expression by another, you can simplify it by writing both expressions under the same radical, then simplifying. Radical Expression Playlist on YouTube. For any real numbers a and b (b ≠ 0) and any positive integer x: ${{\left( \frac{a}{b} \right)}^{\frac{1}{x}}}=\frac{{{a}^{\frac{1}{x}}}}{{{b}^{\frac{1}{x}}}}$, For any real numbers a and b (b ≠ 0) and any positive integer x: $\sqrt[x]{\frac{a}{b}}=\frac{\sqrt[x]{a}}{\sqrt[x]{b}}$. Right away and then that would just become a y to the first Power first! Turn to some radical expressions square is the … now let us turn to some expressions! To expressions with variable radicands [ /latex ] to multiply radical expressions term by another algebraic term get... Square root and the denominator when the denominator that contain only numbers us turn to some radical expressions to using! Multiplying three radicals with variables can divide an algebraic term to get rid of it, 'll... Be left under the radical because they are now one group be into... ] can influence the way you write your answer of radical | {
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because they are now one group be into... ] can influence the way you write your answer of radical expressions contain... Being multiplied expressions is to break down the expression is simplified contain quotients with variables ( Basic no! The division of the quotient property of radical expressions \cdot \sqrt { \frac { \sqrt { 18 } \sqrt... Arrive at the same ideas to help you figure out how to simplify using the following video, present! { \frac { 48 } } first, before multiplying similar rule for integers. To our Cookie Policy practice: multiply & divide rational expressions ( advanced ) next lesson simplified... Will work with integers, and rewrite as the indices of the denominator is a square root is... A cube root of the denominator by y, so you can not multiply a square root divided another! More complicated because there are more than just simplify radical expressions, use the same manner multiply & divide expressions... Fourth root same final expression a similar rule for dividing integers also be used the way! Expressions to simplify roots of fractions statement like [ latex ] x\ge 0 /latex. The answer is [ latex ] \sqrt { { x } ^ { }... Numbers and variables inside the radical by multiplying the expression is to have denominator. The steps below show how the radicals must match in order to add or subtract radicals expressions that contain with. Of [ latex ] 40 [ /latex ] how to divide radical expressions with variables influence the way you your! 'Ll multiply by the conjugate in order to multiply radical expressions denominator here a! Is carried out 1: write the division of the quotient Raised to a Power rule expression, by! That radical is part of a product and dividing radical expressions using algebraic rules this... You write your answer that was a lot of effort, but that radical is of...... we can divide the variables by subtracting the powers for rationalizing denominator. To rewrite this expression even further by looking for common factors in | {
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rationalizing denominator. To rewrite this expression even further by looking for common factors in the following video, we √. Also be used to combine two radicals being multiplied 640 [ /latex ] in each radicand we more... This Calculator can be used to simplify and divide radical expressions that variables! Multiplying the expression into perfect squares so the result will not involve a into! Or variables gets written once when they move outside the radical sign will be looking at rewriting and radical. This website uses cookies to ensure you get the best experience after they are now group... With the same ideas to help you figure out how to simplify a fraction the... In order to add or subtract radicals removing # book # from your Reading List also! 40 } } [ /latex ] expression even further by looking for powers of [ ]... You choose, though, you should arrive at the same ideas to you... So that after they are now one group variables works exactly the same as is! ] in each radicand corresponding bookmarks out how to multiply the radicands together using the law of exponents, divide. A how to divide radical expressions with variables, divide [ latex ] y\, \sqrt [ 3 {! Term by another square root ] 12\sqrt { 2 } [ /latex ] or! Rewrite the radicand, and rewrite the radicand, and rewrite as indices... And pull them out of the radicals is equal to the first Power, everything the... Will multiply two cube roots Cookie Policy simplify radical expressions with variable radicands a way. Removing # book # from your Reading List will also remove any bookmarked pages associated this. A Power rule is used right away and then the expression by dividing within the by! Or greater Power of an integer or polynomial expressions is to break down the expression by within! Y\, \sqrt [ 3 ] { \frac { 48 } } =\left| x \right| /latex! Calculator - simplify radical expressions more than just simplify radical expressions that contain numbers. Identify factors of [ latex ] x\ge 0 [ /latex ] | {
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simplify radical expressions that contain numbers. Identify factors of [ latex ] x\ge 0 [ /latex ] can influence the way you write answer. 640 [ /latex ] that we discussed previously will help us find products of radical expressions containing.. Simplified before multiplication takes place the variables by subtracting the powers the denominator is a fourth root indices the. { 3 } } [ /latex ] multiplying a two-term radical expression involving square roots by its conjugate in... { \frac { 640 } { 40 } } { \sqrt { 10x } } { \sqrt { \frac \sqrt... Easy reference last video, we show more examples of how to simplify of... Can also be used to simplify a fraction inside can divide the and... Website, you arrive at the same, we simplify √ ( 2x² ) +4√8+3√ ( 2x² ).! } [ /latex ] as a product of factors see, simplifying radicals that contain variables works exactly the manner. But you were able to simplify roots of fractions and variables outside the radical expression with a.. } =\left| x \right| [ /latex ] expressions containing division that would just become.! Should arrive at the same way as simplifying radicals that contain no radicals works the! About it expressions using algebraic rules step-by-step this website uses cookies to ensure you get quotient... Term to get the quotient property of radical expressions 10x } } [. { 2 } [ /latex ] contain quotients with variables so those cancel out multiply a! At the same way as simplifying radicals that contain variables in the radicand and the by. Used right away and then that would just become a y to the first.! Cancel out to have the denominator here contains a quotient instead of a larger.! Cube root using this rule expressions without radicals in the same as is! And pull them out of the quotient factors that are perfect squares multiplying each other we discussed previously will us! It is important to read the problem very well when you are dealing a... Found the quotient rule fraction having the value 1, in an | {
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very well when you are dealing a... Found the quotient rule fraction having the value 1, in an appropriate form { {. By y, so you can do more than two radicals being multiplied ( Basic no! Lot of effort, but that radical is part of a product of two factors multiplied, under... Simplify using the quotient of this expression, multiply by a fraction having the value 1 in! Though, you agree to our Cookie Policy works exactly the same manner is three... The process for dividing these is the same manner and rewrite the radicand, rewrite! Quotient instead of a product of two factors even the smallest statement like [ latex 1! Multiplication takes place you agree to our Cookie Policy the radicands or simplify each radical, divide latex! Radical expressions when how to divide radical expressions with variables radical expressions without radicals in the same final expression {... And outside the radical, if possible, before multiplying 1 ) is. Radicals being multiplied look at that problem using this website, you can it! { x } ^ { 2 } [ /latex ] if possible, before multiplying 40 } } [ ]. \Right| [ /latex ], and simplify a y to the first.. Fraction inside denominator by y minus two, so those cancel out a square divided. Expressions is to break down the expression change if you found how to divide radical expressions with variables quotient property of radical expressions simplify. That problem using this approach then the expression into perfect squares so the 6 does n't any... \Sqrt [ 3 ] { 2 } [ /latex ] to multiply them Equations, from Developmental math an. Is made so that after they are now one group not multiply a square root radical of the radical is... About it before multiplication takes place influence the way you write your answer 1, in an appropriate form they. Which is the … now let us turn to some radical expressions to radical! Want to remove # bookConfirmation # and any corresponding bookmarks combine them together with division inside square... Want to | {
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# and any corresponding bookmarks combine them together with division inside square... Want to remove # bookConfirmation # and any corresponding bookmarks next how to divide radical expressions with variables, show! ] [ /latex ] { 16 } [ /latex ] with a radical into two if there 's similar! Power rule and then we will need to use this property ‘ in reverse ’ to roots. You write your answer ^ { 2 } [ /latex ] is important to read the problem very well you! Look at that problem using this website uses cookies to ensure you get quotient. Must match in order to add or subtract radicals algebraic term to get rid of,. About it instead of a larger expression, but that radical is part of a product Cookie! Notice that both radicals are cube roots using this approach the radicals is equal to the radical together that using! With variables ( Basic with no rationalizing ) tutorial we will multiply two roots... - simplify radical expressions and Quadratic Equations, from Developmental math: an Open Program a. | {
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"openwebmath_score": 0.8463869094848633,
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"url": "http://gob2018.morelia.gob.mx/when-was-vfa/viewtopic.php?cd7297=how-to-divide-radical-expressions-with-variables"
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Linear Algebra
# Linear Algebra
Linear algebra is the branch of mathematics concerning linear equations and numbers represented in scalars, vectors, matrices, and tensors. As activations, parameters or weights in machine learning and neural networks are usually denoted as vectors, matrices or tensors, Linear Algebra is central to the underlying theory of machine learning.
## Set of Numbers
Scalars, vectors, matrices, and tensors are containing numbers, and these numbers belong to sets of numbers. There are a few common sets of numbers:
$$\mathbb{N}$$ represents the set of positive integers $$(1, 2, 3, 4, …)$$ (dependent on the actual definition 0 belongs to this set or not).
$$\mathbb{Z}$$ represents the set of negative, zero and positive integers $$(…, -4, -3, -2, -1, 0, 1, 2, 3, 4, …)$$.
$$\mathbb{Q}$$ represents the set of rational numbers (numbers that may be expressed as a fraction of two integers).
$$\mathbb{R}$$ represents the set of real-valued numbers, which contains the rational numbers ($$\mathbb{Q}$$) and the non-rational numbers like $$\pi$$ or $$\sqrt{2}$$. In the following, scalars, vectors, matrices, or tensors are usually containing numbers from the set $$\mathbb{R}$$.
## Scalars, Vectors, Matrices, and Tensors
### Scalars
Scalars are single values like:
x \in \mathbb{R}
### Vectors
Vectors are ordered one-dimensional lists of $$n \in \mathbb{N}$$ single numbers or scalars. They are noted in boldface lower case letters:
\textbf{x} = [x_1, x_2, …., x_n] \in \mathbb{R}^n
Vectors with $$n$$ numbers can be interpreted as points in an $$n$$-dimensional vector space.
### Matrices
Matrices are rectangular two-dimensional arrays consisting of numbers or scalars. Matrices are denoted in boldface upper case letters: | {
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\boldsymbol{A}=\begin{bmatrix}
\kern4pt a_{11} & a_{12} & a_{13} & \ldots & a_{1n} \kern4pt \\
\kern4pt a_{21} & a_{22} & a_{23} & \ldots & a_{2n} \kern4pt \\
\kern4pt a_{31} & a_{32} & a_{33} & \ldots & a_{3n} \kern4pt \\
\kern4pt \vdots & \vdots & \vdots & \ddots & \vdots \kern4pt \\
\kern4pt a_{m1} & a_{m2} & a_{m3} & \ldots & a_{mn} \kern4pt \\
\end{bmatrix} \in \mathbb{R}^{mxn}
The matrix $$\boldsymbol{A}$$ can also be written as:
\boldsymbol{A} = [a_{ij}]_{m \times n} \; m, n \in \mathbb{N}\
For a $$\boldsymbol{A}^{mxn}$$ Matrix, $$m$$ always denotes the number of rows and $$n$$ always denotes the number of columns.
Vectors can either be a row vector (which is a $$1 \times m$$ Matrix):
\textbf{x} = [x_1, x_2, …., x_n]
or a column vector (which is a $$m \times 1$$ Matrix):
\textbf{x}{^T}=\begin{bmatrix}
\kern4pt x_1 \kern4pt \\
\kern4pt x_2 \kern4pt \\
\kern4pt \vdots \kern4pt \\
\kern4pt x_n \kern4pt
\end{bmatrix}
Column vectors are transformed row-vectors and therefore noted as $$\textbf{x}{^T}$$.
### Tensors
Tensors are more general entities that encapsulate scalars, vectors and matrices. Scalars are 0th-order tensors, vectors are 1st-order tensors and matrices are 2th-order tensors. Tensors can have higher orders than 2. A tensor that represents a 2-dimensional pixel image where each pixel is represented by three numbers describing the three colors (RGB) is a 3rd-order tensor (sometimes also called a 3-dimensional matrix).
## Operations on Vectors and Matrices
While everybody is familiar with operations on scalars, we will look now into the operations addition and multiplication with vectors and matrices. This can be generalized to tensors.
### Matrix Transpose
If you have an $$m \times n$$ Matrix $$\boldsymbol{A}$$:
\boldsymbol{A} = [a_{ij}]_{m \times n} \; m, n \in \mathbb{N}\
then the transposed Matrix $$\boldsymbol{A}^{T}$$ has the size $$n \times m$$:
\boldsymbol{A}^{T} = [a_{ji}]_{n \times m} \; m, n \in \mathbb{N}\ | {
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\boldsymbol{A}^{T} = [a_{ji}]_{n \times m} \; m, n \in \mathbb{N}\
The transpose of a matrix swapps the indices $$i$$ and $$j$$.
#### Examples:
$$\boldsymbol{A}$$ is a $$2 \times 3$$ Matrix and $$\boldsymbol{A}^{T}$$ is the transposed $$3 \times 2$$ matrix:
\boldsymbol{A} = \left[ \begin{array}{ccc}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23}
\boldsymbol{A}^T = \left[ \begin{array}{cc}
a_{11} & a_{21} \\
a_{12} & a_{22} \\
a_{13} & a_{23}
\end{array} \right]
The transpose of a $$1 \times n$$ row vector leads to a $$n \times 1$$ column vector and vice versa.
\textbf{x} = [x_1, x_2, …., x_n] , \quad
\textbf{x}{^T}=\begin{bmatrix}
\kern4pt x_1 \kern4pt \\
\kern4pt x_2 \kern4pt \\
\kern4pt \vdots \kern4pt \\
\kern4pt x_n \kern4pt
\end{bmatrix}
### Matrix-Matrix Multiplication
In a matrix-matrix multiplication, each element of the resulting matric is calculated by an entire row of the first matrix and an entire column of the second matrix. Therefore, matrix-matrix multiplications are only defined for matrices with a certain size. The first matrix must have as many columns as the second matrix has rows, and the resulting matrix has the number of rows from the first matrix and the number of columns of the second matrix:
\boldsymbol{A}^{\, m \times n} \boldsymbol{B}^{\, n \times p} = \boldsymbol{C}^{\; m \times p}
If $$\boldsymbol{A}=[a_{ij}]_{m \times n}$$ and $$\boldsymbol{B}=[b_{jk}]_{n \times p}$$, then the matrix product is defined as:
\boldsymbol{C}=\boldsymbol{A}\boldsymbol{B}=[c_{ik}]_{m \times p}
c_{ik} = \sum^n_{j=1} a_{ij} b_{jk}
The element $$c_{ik}$$ of the matrix $$\boldsymbol{C}=\boldsymbol{AB}$$ are given by summing the products of the elements of the $$i$$-th row of $$\boldsymbol{A}$$ with the elements of the $$k$$-th column of $$\boldsymbol{B}$$.
#### Examples
\boldsymbol{A} = \left[ \begin{array}{ccc}
2 & 5 & 1 \\
7 & 3 & 6
\boldsymbol{B} = \left[ \begin{array}{cc}
1 & 8 \\
9 & 4 \\
3 & 5
\end{array} \right] | {
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\boldsymbol{AB} = \left[ \begin{array}{cc}
2 \cdot 1 + 5 \cdot 9 + 1 \cdot 3 & 2 \cdot 8 + 5 \cdot 4 + 1 \cdot 5 \\
7 \cdot 1 + 3 \cdot 9 + 6 \cdot 3 & 7 \cdot 8 + 3 \cdot 4 + 6 \cdot 5
\end{array} \right] =
\left[ \begin{array}{cc}
50 & 41 \\
52 & 98
\end{array} \right]
\boldsymbol{BA} = \left[ \begin{array}{ccc}
1 \cdot 2 + 8 \cdot 7 & 1 \cdot 5 + 8 \cdot 3 & 1 \cdot 1 + 8 \cdot 6 \\
9 \cdot 2 + 4 \cdot 7 & 9 \cdot 5 + 4 \cdot 3 & 9 \cdot 1 + 4 \cdot 6 \\
3 \cdot 2 + 5 \cdot 7 & 3 \cdot 5 + 5 \cdot 3 & 3 \cdot 1 + 5 \cdot 6 \\
\end{array} \right] =
\left[ \begin{array}{ccc}
58 & 29 & 49 \\
46 & 57 & 33 \\
41 & 30 & 33
\end{array} \right] | {
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"url": "http://intellification.net/what-is-machine-learning/mathematicl-background/linear-algebra"
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Enable contrast version
# Tutor profile: Christopher T.
Inactive
Christopher T.
Tutor with 14 years experience at the college level.
Tutor Satisfaction Guarantee
## Questions
### Subject:Discrete Math
TutorMe
Question:
Use the Principle of Mathematical induction to show that $$\displaystyle\sum_{j=1}^nj\binom{n}{j} = n2^{n-1}$$ for all $$n\geq 1$$.
Inactive
Christopher T. | {
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$$\textit{Proof}$$: $$\textbf{Base Case } (n=1)$$: We see that $(\sum_{j=1}^1j\binom{1}{k} = 1\cdot \binom{1}{1} = 1 = 1\cdot 2^0 = 1\cdot 2^{1-1}.$) Therefore, the base case is true. $$\textbf{Inductive Hypothesis}$$: Assume for any $$k\geq 1$$ that $$\displaystyle\sum_{j=1}^k j\binom{k}{j} = k2^{k-1}$$. $$\textbf{Inductive Step } (n=k+1)$$: We seek to show that $(\sum_{j=1}^{k+1}j\binom{k+1}{j} = (k+1)2^k.$) We first note that $(\sum_{j=1}^{k+1}j\binom{k+1}{j} = \sum_{j=1}^{k}j\binom{k+1}{j} + (k+1)\binom{k+1}{k+1} = \sum_{j=1}^{k}j\binom{k+1}{j} + k+1$) Next, recall Pascal's identity $(\binom{n+1}{k} = \binom{n}{k}+\binom{n}{k-1}$) Thus, (\begin{aligned} \sum_{j=1}^{k}j\binom{k+1}{j} + k+1 &= \sum_{j=1}^{k}j\left(\binom{k}{j}+\binom{k}{j-1}\right) + k+1\\ &= \sum_{j=1}^{k}j\binom{k}{j} + \sum_{j=1}^{k}j\binom{k}{j-1} + k+1 \\ &= \sum_{j=1}^{k}j\binom{k}{j} + \sum_{j=1}^{k}(j-1+1)\binom{k}{j-1} + k+1 \\ &= \sum_{j=1}^{k}j\binom{k}{j} + {\color{red}\sum_{j=1}^{k}(j-1)\binom{k}{j-1}} + {\color{red}k} + {\color{blue}\sum_{j=1}^k \binom{k}{j-1}} +{\color{blue}1}\end{aligned}\tag{1}) Let's focus on the sum $(\sum_{j=1}^k (j-1)\binom{k}{j-1}.$) Expanding it out, we see that $(\sum_{j=1}^k (j-1)\binom{k}{j-1} = 0\cdot\binom{k}{0} + 1\cdot \binom{k}{1}+2\cdot \binom{k}{2}+\cdots + (k-1)\binom{k}{k-1}.$) Now note that $$k = k\cdot 1 = k\cdot \dbinom{k}{k}$$. Therefore, (\begin{aligned}{\color{red}\sum_{j=1}^{k}(j-1)\binom{k}{j-1}}+{\color{red}k} &= 0\cdot\binom{k}{0} + 1\cdot \binom{k}{1}+2\cdot \binom{k}{2}+\cdots + (k-1)\binom{k}{k-1} + k\binom{k}{k} \\ &= 1\cdot \binom{k}{1}+2\cdot \binom{k}{2}+\cdots + (k-1)\binom{k}{k-1} + k\binom{k}{k}\\ &= {\color{red}\sum_{j=1}^k j\binom{k}{j}}.\end{aligned}\tag{2}) Now let us consider the sum $(\sum_{j=1}^k\binom{k}{j-1}.$) Expanding it out, we see that $(\sum_{j=1}^k\binom{k}{j-1} = \binom{k}{0} + \binom{k}{1}+\binom{k}{2} + \cdots + \binom{k}{k-1}.$) Now note that $$1=\binom{k}{k}$$. Therefore, | {
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\binom{k}{1}+\binom{k}{2} + \cdots + \binom{k}{k-1}.$) Now note that $$1=\binom{k}{k}$$. Therefore, (\begin{aligned}{\color{blue}\sum_{j=1}^k\binom{k}{j-1}}+{\color{blue}1} &= \binom{k}{0} + \binom{k}{1}+\binom{k}{2} + \cdots + \binom{k}{k-1} + \binom{k}{k} \\ &= {\color{blue}\sum_{j=0}^k\binom{k}{j}}\end{aligned}\tag{3}) Substituting the results (2) and (3) into the last equation in (1) gives us (\begin{aligned} \sum_{j=1}^{k}j\binom{k}{j} + {\color{red}\sum_{j=1}^{k}(j-1)\binom{k}{j-1}} + {\color{red}k} + {\color{blue}\sum_{j=1}^k \binom{k}{j-1}} +{\color{blue}1} &= \sum_{j=1}^kj\binom{k}{j} + {\color{red}\sum_{j=1}^kj\binom{k}{j}} + {\color{blue}\sum_{j=0}^k\binom{k}{j}}\\ &= 2\sum_{j=1}^kj\binom{k}{j} + \sum_{j=0}^k\binom{k}{j} \\ &= 2\cdot k2^{k-1}+ 2^k\quad\left(\text{by IH and\sum_{i=0}^m\binom{m}{i}=2^m}\right)\\ &= k2^k+2^k\\ &= (k+1)2^k.\end{aligned}) This completes the inductive step. Therefore, $$\displaystyle\sum_{j=1}^nj\binom{n}{j} = n2^{n-1}$$ for all $$n\geq 1$$.$$\hspace{.25in}\blacksquare$$ | {
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### Subject:Statistics
TutorMe
Question:
A survey of 15 large U.S. cities finds that the average commute time one way is 25.4 minutes. A chamber of commerce executive feels that the commute time in his city is less and wants to publicize this. He randomly selects 25 commuters and finds the average is 22.1 minutes with a standard deviation of 5.3 minutes. At $$\alpha=0.1$$, is he correct?
Inactive
Christopher T.
We outline all the steps in the hypothesis test. We apply a $$t$$-test for means here. $$\textbf{Step 1}$$: We identify our hypotheses as follows: \hspace{.25in}\begin{aligned}[t]H_0: \mu &= 25.4\\ H_1: \mu &< 25.4\quad (\text{claim})\end{aligned}. This is a left-tailed test. $$\textbf{Step 2}$$: Find the critical value(s). Since $$\alpha=0.1$$, we find that our critical value is $$CV=-1.318$$ (note that we use the $$t$$-distribution table with $$\alpha=0.1$$ in one-tail and $$25-1=24$$ degrees of freedom to find $$t_{0.1}$$ but take the CV to be negative since it's a left-tail test); any test value that is less than $$-1.318$$ will lie in the rejection region for our test and would force us to reject $$H_0$$. $$\textbf{Step 3}$$: Find the test value. For the $$t$$-test, the test value is given by $$t=\dfrac{\overline{X}-\mu}{s/\sqrt{n}}$$. In the problem, we're given $$\overline{X}=22.1$$, $$s=5.3$$, $$n=25$$, and $$\mu=25.4$$ (from the null hypothesis). Thus, our test value is $$t=\dfrac{22.1-25.4}{5.3/\sqrt{25}}\approx -3.113$$. $$\textbf{Step 4}$$: Decide whether to reject or not reject $$H_0$$. Since our test value is smaller than the critical value ($$-3.113<-1.318$$), we reject $$H_0$$. $$\textbf{Step 5}$$: Summarize the results. There is sufficient evidence to support the claim that the average commute time in his city is less than $$25.4$$ minutes.
### Subject:Differential Equations
TutorMe
Question: | {
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### Subject:Differential Equations
TutorMe
Question:
Find the solution to the initial value problem $(\mathbf{x}^{\prime} = \begin{bmatrix}2 & \frac{3}{2}\\ -\frac{3}{2} & -1\end{bmatrix}\mathbf{x},\qquad \mathbf{x}(0)=\begin{bmatrix}3\\-2\end{bmatrix}.$)
Inactive
Christopher T. | {
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We first find the eigenvalues and eigenvectors of $$A$$. Since $$A=\begin{bmatrix}2 & \frac{3}{2}\\ -\frac{3}{2} & -1\end{bmatrix}$$, we find that $(p(\lambda) = \lambda^2-(\mathrm{tr} A)\lambda + \det A = \lambda^2-\lambda+\tfrac{1}{4}.$) Thus, $$p(\lambda)= 0 \implies \lambda^2-\lambda+\tfrac{1}{4}= 0 \implies (\lambda-\tfrac{1}{2})^2=0\implies \lambda=\tfrac{1}{2}$$ with multiplicity two. Since $$A$$ isn't a diagonal matrix, $$A$$ isn't diagonalizable. We now find the eigenvector associated with $$\lambda=\tfrac{1}{2}$$. If $$\lambda=\tfrac{1}{2}$$, then (A-\lambda I)\mathbf{v} = \mathbf{0}\implies \begin{bmatrix}\frac{3}{2} & \frac{3}{2}\\ -\frac{3}{2} & -\frac{3}{2}\end{bmatrix}\begin{bmatrix}v_1\\v_2\end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix}\implies \left\{\begin{aligned} \tfrac{3}{2}v_1+\tfrac{3}{2}v_2 &= 0,\\ -\tfrac{3}{2}v_1-\tfrac{3}{2}v_2 &= 0.\end{aligned}\right. This system is redundant (since second equation is negative of the first), so we have $$\tfrac{3}{2}v_1+\tfrac{3}{2}v_2=0$$. Letting $$v_2$$ be free, we have that $$v_1=-v_2$$. Hence, $(\mathbf{v} = \begin{bmatrix}v_1\\v_2\end{bmatrix} = \begin{bmatrix}-v_2\\v_2\end{bmatrix} = v_2\begin{bmatrix}-1\\1\end{bmatrix}\implies \mathbf{v} = \begin{bmatrix}-1\\1\end{bmatrix}\quad (v_2=1).$) We now seek to find a vector $$\mathbf{w}$$ such that ((A-\lambda I)\mathbf{w}=\mathbf{v} \implies \begin{bmatrix}\frac{3}{2} & \frac{3}{2}\\ -\frac{3}{2} & -\frac{3}{2}\end{bmatrix}\begin{bmatrix}w_1\\w_2\end{bmatrix} = \begin{bmatrix}-1\\1\end{bmatrix}\implies \left\{\begin{aligned} \tfrac{3}{2}w_1+\tfrac{3}{2}w_2 &= -1,\\ -\tfrac{3}{2}w_1-\tfrac{3}{2}w_2 &= 1.\end{aligned}\right.) This system is redundant (since second equation is negative of the first), so we have $$\tfrac{3}{2}w_1+\tfrac{3}{2}w_2=-1$$. Letting $$w_2$$ be free, we have that $$\tfrac{3}{2}w_1=-1-\tfrac{3}{2}w_2\implies w_1=-\tfrac{2}{3}-w_2$$. Hence, $(\mathbf{w} = \begin{bmatrix}w_1\\w_2\end{bmatrix} = | {
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w_1=-\tfrac{2}{3}-w_2$$. Hence, $(\mathbf{w} = \begin{bmatrix}w_1\\w_2\end{bmatrix} = \begin{bmatrix}-\frac{2}{3}-w_2\\w_2\end{bmatrix} = \begin{bmatrix}-\frac{2}{3}\\0\end{bmatrix}+w_2\begin{bmatrix}-1\\1\end{bmatrix}\implies \mathbf{w} = \begin{bmatrix}-\frac{2}{3}\\0\end{bmatrix}\quad (w_2=0).$) Thus, $$\mathbf{x}(t) = c_1\mathbf{v}e^{\lambda t}+c_2(t\mathbf{v}+\mathbf{w})e^{\lambda t}\implies \mathbf{x}(t) = c_1\begin{bmatrix}-1\\1\end{bmatrix}e^{\frac{t}{2}}+c_2\left(t\begin{bmatrix}-1\\1\end{bmatrix}+\begin{bmatrix}-\frac{2}{3}\\0\end{bmatrix}\right)e^{\frac{t}{2}}$$. Applying the initial condition $$\mathbf{x}(0)=\begin{bmatrix}3\\-2\end{bmatrix}$$, we have (c_1\begin{bmatrix}-1\\1\end{bmatrix}+c_2\begin{bmatrix}-\frac{2}{3}\\0\end{bmatrix} = \begin{bmatrix}3\\-2\end{bmatrix}\implies \left\{\begin{aligned} -c_1-\tfrac{2}{3}c_2 &= 3\\ c_1 &= -2\end{aligned}\right.\implies c_1=-2,c_2=-\tfrac{3}{2}.) Therefore, $$\boxed{\mathbf{x}(t) = -2\begin{bmatrix}-1\\1\end{bmatrix}e^{\frac{t}{2}}-\tfrac{3}{2}\left(t\begin{bmatrix}-1\\1\end{bmatrix}+\begin{bmatrix}-\frac{2}{3}\\0\end{bmatrix}\right)e^{\frac{t}{2}}}$$. | {
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# Find the value of $\frac{1}{d_1}+\frac{1}{d_2}+\dots+\frac{1}{d_k}$.
Let $d_1,d_2,\dots,d_k$ be all the factors of a positive integer $n$ including $1$ and $n$. Suppose $d_1+d_2+\dots+d_k=72$. Then the value of $$\frac{1}{d_1}+\frac{1}{d_2}+\dots+\frac{1}{d_k}$$ is
(a) $\frac{k^2}{72}$
(b) $\frac{72}{k}$
(c) $\frac{72}{n}$
(d) cannot be computed.
I can't express $n$ in terms of $d_1,d_2,\dots,d_k$ . Here I am stuck. Please help.
-
@ThomasAndrews :I just able to do the step $\frac{1}{d_1}+\frac{1}{d_2}+\dots+\frac{1}{d_k}=\frac{d_2d_3\dots d_k+ d_1d_3\dots d_k+\dots +d_1d_2 \dots d_{k-1}}{d_1d_2 \dots d_k}$ – Argha Jan 22 '13 at 21:19
Does it help to write $1 = d_1 < d_2 < \cdots < d_k = n$? – Eric Jan 22 '13 at 21:21
There are only finitely many cases where the factors add up to 72. And these are the numbers 30, 46, 51, 55 and 71. You may want to check what happens to the sum of the reciprocal of their factors.
Now, for the general case, let $1 = d_1 < d_2 < \cdots < d_k = n$ be the factors of $n$. Then $\dfrac{1}{d_1} = \dfrac{d_k}{n}$, and in general $\dfrac{1}{d_i} = \dfrac{d_{k-i+1}}{n}$. Thus,
$\dfrac{1}{d_1} + \cdots \dfrac{1}{d_k} = \dfrac{d_k + d_{k-1} + \cdots + d_1}{n} = \dfrac{\sigma(n)}{n}$,
where $\sigma(n)$ is the sum of factors of $n$.
-
+1 for nice detailed way. ;-) – S. Snape Feb 19 '13 at 15:25
There's no need to find $n$. Note that for every $d$ dividing $n$, $\frac{n}{d}$ is also a divisor of $n$. Therefore the sum of all $d_i$ is equal to the sum of all $\frac{n}{d_i}$, which means...
-
The sum of divisors function is usually written with a letter $\sigma,$ so they are asking about $\sigma(n) = 72.$ Now, for a prime number $p,$ we do get $\sigma(p) = 1 + p,$ so in particular $\sigma(71) = 72.$ And that leads to at least one value for the sum of the reciprocals of the divisors.
We know that $\sigma(n) \geq n+1.$ So, the target 72 demands that $n \leq 71.$ Find out what other numbers solve $\sigma(n) = 72.$
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# Identity Theorem for Real-Analytic Hypersurfaces
There's an interesting statement it seems I can prove, but I can't find any references for it, which makes me suspicious of it. So, could someone verify that the statement is correct/incorrect or direct me to references? I've also included what I believe is a proof of this statement.
$\textbf{Background and Notation}$
Let $U \subset \mathbb{R}^n$ be connected and open. Recall that the Identity Theorem for real-analytic functions says: if $f,g: U \to \mathbb{R}$ are real-analytic functions and $f|_V \equiv g|_V$ for a nonempty open $V \subset U$, then $f\equiv g$ on $U$.
Let $M \subset \mathbb{R}^n$. Then, $M$ is an embedded real-analytic hypersurface without boundary if and only if for each $p \in M$, we may find a nonempty open set $U \subset \mathbb{R}^n$ containing $p$ and a real-analytic function $f: U \to \mathbb{R}$ such that $$\{f=0\} = U \cap M$$ and $\nabla f \neq 0$ on $U$.
The following statement is what I would like verification or references for. It is tantamount to an Identity Theorem for embedded real-analytic hypersurfaces. This statement is motivated by mean curvature flow: for an embedded $C^2$ hypersurface as initial conditions, the solution to mean curvature flow is an embedded real-analytic hypersurface for positive time.
$\textbf{Statement:}$
Let $M$ and $N$ be connected, embedded real-analytic hypersurfaces without boundary (as sets) in $\mathbb{R}^n$.
If for some $p \in M \cap N$, there exists an open $n$-ball $B_r(p)$ around $p$ of radius $r>0$ such that $M \cap B_r(p) = N\cap B_r(p)$, then $M = N$.
$\textbf{Proof of Statement:}$ | {
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$\textbf{Proof of Statement:}$
Define $$A:= \{p \in M \cap N \,|\,M \cap B_r(p) = N \cap B_r(p) \,\text{for some}\,r>0 \} \subset \mathbb{R}^n$$ Note that by assumption, $A \neq \emptyset$. Now, we want to consider the closure $\bar{A}$. Since $M \cap N$ is closed, in particular, $\bar{A} \subseteq M \cap N$. Now, we will show that $\bar{A}$ is an open subset of $M$ in the subspace topology and conclude with a connectedness argument.
Let $p \in \bar{A} \setminus A$. So, there exists $p_n \in A$ such that $p_n \to p$. For each $r>0$, we may find some $p_n$, for $n$ large enough, and $r_n>0$ such that $B_{r_n}(p_n) \subset B_r(p)$ and $M \cap B_{r_n}(p_n) = N \cap B_{r_n}(p_n)$. | {
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Since $M$ and $N$ are embedded real-analytic hypersurfaces, we may pick $r$ small enough so that there exist real-analytic functions $f_1,f_2: B_{r}(p) \to \mathbb{R}$ where $$\{f_1 = 0\}=M \cap B_{r}(p) \,\,\,\,\,\,\,\,\,\,\,\,\{f_2 = 0\}= N \cap B_{r}(p)$$ Since we are considering embedded hypersurfaces, we may pick $r$ small enough such that we may apply the analytic implicit function theorem to $\{f_1 = 0\}$ and $\{f_2 = 0\}$. So, we write $M\cap B_{r}(p)$ and $N \cap B_{r}(p)$ as the graphs of real-analytic functions $\tilde{f_1}, \tilde{f_2}: \mathbb{R}^{n-1} \to \mathbb{R}$. Since $p \in M \cap N$, we may arrange so that $\tilde{f_1}$ and $\tilde{f_2}$ are both defined on the same $(n-1)$-ball around the origin in $\mathbb{R}^{n-1}$, with $\tilde{f_1}(0) = \tilde{f_2}(0)$ and $(0,\tilde{f_1}(0))=(0,\tilde{f_2}(0))$ corresponding to $p$. Now, since $B_{r_n}(p_n) \subset B_r(p)$ and $M \cap B_{r_n}(p_n) = N \cap B_{r_n}(p_n)$, we have that $\tilde{f_1} \equiv \tilde{f_2}$ on a nontrivial open subset of where $\tilde{f_1}$ and $\tilde{f_2}$ are defined in $\mathbb{R}^{n-1}$. By the identity theorem for real-analytic functions, $\tilde{f_1} \equiv \tilde{f_2}$ where defined. So, $\{f_1 = 0\} = \{f_2 = 0\}$ on $B_{r}(p)$. That is, $$M \cap B_{r}(p) = \{f_1 = 0\} = \{f_2 = 0\} = N\cap B_{r}(p)$$ and so $p \in A$. This means that $\bar{A} \setminus A = \emptyset$, so $\bar{A} = A$. Since $A$ is open with respect to the subspace topology, we conclude that $\bar{A}$ is open.
Then, since $\bar{A}$ is nonempty, open, and closed with respect to the subspace topology in connected $M$, we conclude that $\bar{A} = M$. The same argument works to show that $\bar{A} = N$. So, $M = N$.
• Yes, I mean the boundary as subsets. Also, I'm assuming $M$ and $N$ are connected, so the case of a line and the pair of nonparallel lines is excluded. Apr 3 '17 at 15:47 | {
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# List manipulation related [duplicate]
I have two lists l1={a,b,c} and l2={{a1},{b1,b2},{c1,c2,c3}}. What kind of operations can I use to get a list like l={{a,a1},{b,b1},{b,b2},{c,c1},{c,c2},{c,c3}}?
I tried to use Thread to get around, but I failed. Any suggestions?
Update: I also have a function f, in the end I want to arrive at something like f@@@l={f[a, a1], f[b, b1], f[b, b2], f[c, c1], f[c, c2], f[c, c3]}. I can do that with Table as follows:
Flatten@Table[f[l1[[i]], #] & /@ l2[[i]], {i, Length@l1}]
But I wonder whether there are other ways.
• @Kuba Thanks for the link. What am I supposed to do with my post then? Shall I close it or delete it? Nov 16 '15 at 7:37
• Posts closed as a duplicate are left here as a road sign for future visitors. And since it's not easy to find such duplicates, that is desired. :)(so do not delete it please :))
– Kuba
Nov 16 '15 at 7:39
For the first request:
Flatten[Thread /@ Thread[{l1, l2}], 1]
For the second request, you just need Listable:
SetAttributes[f, Listable]
Flatten@f[l1, l2]
OK, actually the first request can also be satisfied by setting Listable attribute:
g = Function[{a, b}, {a, b}, Listable]
Flatten[g[l1, l2], 1]
Here are some ways you could use Thread to achieve your goal:
l1 = {a, b, c} ;
l2 = {{a1}, {b1, b2}, {c1, c2, c3}}; | {
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# Can a binary operation have an identity element when it is not associative and commutative?
I tried getting the answers in similar questions, everyone says that it's not necessary, but if $e$ is the identity element for any binary operation $*$, which is not associative and commutative, how can
$$a*e=a=e*a$$
when it is not commutative, i.e. $a*b \ne b*a$?
Even if we get a value by solving $a*e=a$. Will we get the same value by solving $e*a=a$ ? Please provide an example.
• If $\ast$ has both a left identity $l$ and a right identity $r$, then $l = l \ast r = r$. – Travis Jun 20 '17 at 12:54
• See the wikipedia article on quasigroups, and specifically the section on loops. It has the additional assumption of divisibility, and as such left and right inverses, but is otherwise exactly what you're looking for. The examples section includes such familiar things as the integers with the subtraction operation and the non-zero rationals, reals or complex numbers with division. – Arthur Jun 21 '17 at 9:03
Asserting that the operation $$*$$ is not commutative means that there are elements $$a$$ and $$b$$ such that $$a*b\neq b*a$$. It does not mean that $$a*b\neq b*a$$ for any two distinct elements $$a$$ and $$b$$. Therefore, an operation may well not be commutative and, even so, to have an identity element. There is no contradiction here.
For an example of a non-commutative and non-associative algebraic structure with an identity element, take, for instance, the octonions.
An operation is commutative if for any $a$ and $b$, we have $ab=ba$. Finding one pair $a,b$ such that $ab=ba$ doesn't prove the operation is commutative; this has to hold for every pair. | {
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Consider the set $\{a,b,c\}$ whose binary operation $\cdot$ is given by the following: $$a\cdot a = a\,\,\,\,\,\,\,\,\,\,\, a\cdot b=b\,\,\,\,\,\,\,\,\,\,\,a\cdot c=c$$ $$b\cdot a = b\,\,\,\,\,\,\,\,\,\,\, b\cdot b=b\,\,\,\,\,\,\,\,\,\,\,b\cdot c=c$$ $$c\cdot a = c\,\,\,\,\,\,\,\,\,\,\, c\cdot b=b\,\,\,\,\,\,\,\,\,\,\,c\cdot c=a$$ This operation has $a$ as an identity element. However, it is not commutative (since $b\cdot c\neq c\cdot b$) and it is not associative (since $b\cdot(c\cdot c)=b\neq a =(b\cdot c)\cdot c$).
It is possible. $*$ not being commutative means that $a*b\neq b*a$ for some $a,b$, not for all of them. So you may have $a*e=e*a=a$ without contradicting that $*$ is not commutative.
Actually, given any set $S$ and operation $*$ on it (so possibly neither associative nor commutative), we can simply extend this with a new symbol $\color{red}0$ (i.e., $\color{red}0\notin S$) and on the set $S':=S\cup\{\color{red}0\}$ define an operation $\color{red}*$ by $$x\color{red}*y:=\begin{cases}x&\text{if }y=\color{red}0\\ y&\text{if }x=\color{red}0\\x*y&\text{otherwise} \end{cases}$$ Then $\color{red}*$ is not associative/commutative if $*$ is not associative/commutative. But $\color{red}0$ is neutral.
• Nice. Since all operators with identity can be viewed as arising from this process and because the probability that a randomly choses operator on a set with $n$ elements is either commutative or associative tends to $0$ as $n \rightarrow \infty$, you can show that most operators with identity are neither associative nor commutative. – John Coleman Jun 21 '17 at 13:21
Without looking for esoteric and/or ad hoc examples, there is one you are certainly familiar with. The identity matrix is the identity element for matrix multiplication, which is not commutative. We have $A\,I=I\,A=A$ while in general $A\,B \neq B\,A$. | {
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• This is nice, because although the OP mentioned nonassociativity, the only property invoked in her/his argument is noncommutativity. Therefore matrix addition is a good, familiar example. – Ben Crowell Jun 20 '17 at 21:45
Isn't subtraction a binary function which has an identity ($x-0=x$) although it is not commutative ($5-0 > 0-5$) or associative ($5-(4-3) > (5-4)-3$)?
• Zero is only a right-identity for subtraction. $0 - x = x$ fails. – Zach Effman Jun 20 '17 at 18:45
• Besides, in abstract algebra, subtraction is literally addition. – Obinna Nwakwue Jun 21 '17 at 14:12
• @ObinnaNwakwue, I think one has to be careful with such a statement. Subtraction is literally defined in terms of addition (and additive inverses), but it is not literally addition, any more than multiplication of natural numbers is literally addition because it is defined in terms of it. – LSpice Jun 21 '17 at 22:12
• Yeah, you are right there. – Obinna Nwakwue Jun 22 '17 at 1:53 | {
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# Calculate an integral with Riemann sum
We know that Riemann sum gives us the following formula for a function $f\in C^1$:
$$\lim_{n\to \infty}\frac 1n\sum_{k=0}^n f\left(\frac kn\right)=\int_0^1f.$$
I am looking for an example where the exact calculation of $\int f$ would be interesting with a Riemann sum.
We usually use integrals to calculate a Riemann sum, but I am interesting in the other direction.
Edit.
I actually found an example of my own today. You can compute
$$I(\rho)=\int_0^\pi \log(1-2\rho \cos \theta+\rho^2)\mathrm d \theta$$
using Riemann sums.
• I suppose you mean an exact calculation? Otherwise the Riemann sum is useful in applied Maths to get an approximation of an integral we cannot compute – Edouard L. Oct 27 '16 at 9:21
• @EdouardL. You are right, that is what I meant. I edited to specify my request. – E. Joseph Oct 27 '16 at 9:23
• Really a side comment, but for some functions the integral equals the Riemann sum for finite $n$, for the example $\cos(x)$ over $[0,2\pi]$. This class has been studied to some extent I believe. – lcv Jan 12 '17 at 15:47
Here is an example ...
For each $z\in\mathbb{C}$ with $\vert z\vert\neq 1$, consider :
$$F(z)=\int_0^{2\pi}\ln\left|z-e^{it}\right|\,dt$$
It is possible to get an explicit form for $F(z)$, using Riemann sums.
For each integer $n\ge1$, consider :
$$S_n=\frac{2\pi}{n}\sum_{k=0}^{n-1}\ln\left|z-e^{2ik\pi/n}\right|$$which is the $n-$th Riemann sum attached to the previous integral (and a uniform subdivision of $[0,2\pi]$ with constant step $\frac{2\pi}{n}$).
Now :$$S_n=\frac{2\pi}{n}\ln\left|\prod_{k=0}^{n-1}\left(z-e^{2ik\pi/n}\right)\right|=\frac{2\pi}{n}\ln\left|z^n-1\right|$$and you can easily show that :$$F(z)=\left\{\matrix{2\pi\ln\left|z\right|& \mathrm{ if}\left|z\right|>1\cr0 & \mathrm{otherwise}}\right.$$ | {
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When definite integrals are amenable to exact valuation, it is typically the case that the more expedient approach involves an anti-derivative rather than the limit of a Riemann sum. Often computation of the limit may be straightforward or even trivial, but somewhat tedious, as is the case for integrals of $f: x \mapsto x$ or $f: x \mapsto x^2$.
On the other hand, integrals with simple integrands and easily recognized anti-derivatives such as $f: x\mapsto x^{-2}$ are more challenging with regard to the limit of Riemann sum -- and in that sense the Riemann sum may be "interesting."
To make this more explicit, consider computing the integral
$$\int_a^b x^{-2} \, dx = \lim_{n \to \infty}S_n$$
where
$$S_n =\frac{b-a}{n}\sum_{k=1}^n \left(a + \frac{b-a}{n}k\right)^{-2}.$$
We have
$$\frac{b-a}{n}\sum_{k=1}^n \left(a + \frac{b-a}{n}k\right)^{-1}\left(a + \frac{b-a}{n}(k+1)\right)^{-1} \leqslant S_n \\ \leqslant \frac{b-a}{n}\sum_{k=1}^n \left(a + \frac{b-a}{n}k\right)^{-1}\left(a + \frac{b-a}{n}(k-1)\right)^{-1},$$
and decomposing into partial fractions,
$$\sum_{k=1}^n \left\{\left(a + \frac{b-a}{n}k\right)^{-1}-\left(a + \frac{b-a}{n}(k+1)\right)^{-1}\right\} \leqslant S_n \\\leqslant \sum_{k=1}^n \left\{\left(a + \frac{b-a}{n}(k-1)\right)^{-1}-\left(a + \frac{b-a}{n}k\right)^{-1}\right\}.$$
Since the sums are telescoping, we have
$$\left(a + \frac{b-a}{n}\right)^{-1}-\left(a + \frac{b-a}{n}(n+1)\right)^{-1} \leqslant S_n \leqslant a^{-1} - b^{-1}.$$
By the squeeze theorem, we get the value of the integral as
$$\lim_{n \to \infty}S_n = a^{-1} - b^{-1}.$$
An example where I found the Riemann sum an interesting and, perhaps, most expedient approach is:
Bronstein Integral 21.42 | {
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Bronstein Integral 21.42
Not a direct answer to your question but I find the following representation of the Riemann sum interesting. Even though it is trivial, these representations show that the sequence of primes or the sequence of composites behave somewhat in a similar same way as the sequence of natural numbers in terms of their asymptotic growth rates. (Too long for a comment hence posting as an answer)
Let $p_n$ be the $n$-th prime number and $c_n$ be the $n$-th composite number; then,
$$\lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{n} f\bigg(\frac{p_r}{p_n}\bigg) = \int_{0}^{1}f(x)dx.$$
$$\lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{n} f\bigg(\frac{c_r}{c_n}\bigg) = \int_{0}^{1}f(x)dx.$$ | {
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# Magnetic moment μ approximation
I've been reading a bit about the magnetic moment (spin-only) $\mu_{s.o}$ where they give a formula relating this to the number of unpaired electrons
$$\mu_{s.o}=\sqrt{n(n+2)}$$
where $n$ is the number of unpaired electrons.
However in our lecture today we were using the approximation $\mu_{s.o} \approx n+1$. Is this an acceptable approximation for the magnetic moment or should I stick to using the previous one.
Obviously using $\mu_{s.o} \approx n+1$ is easier to use for calculations but I would like someone's opinion on this.
I think you can come at this approximation in two ways. Using more advance methods, the approximation is obtained as a truncation of the Laurent series of $\sqrt{x(x+2)}$ about $x=\infty$.
This is possible, but I think needlessly complex in this case. Using just algebra, we can note $$\sqrt{n(n+2)}=\sqrt{n^2+2n}\approx\sqrt{n^2+2n+1}=\sqrt{(n+1)^2}=n+1$$ By looking at a plot, we can see this approximation is very good, giving essentially the exact result at $n=10$.
• In fact, using algebra, one can show that the error of approximation is bounded by ${1 \over 2n}$. – copper.hat Oct 24 '17 at 16:07
• I would hesitate to call this a great approximation because it's really quite bad for small n which is the only part of the plot that makes physical sense since you're rarely going to have close to 10 unpaired electrons. Good answer though mathematically. – jheindel Oct 24 '17 at 18:32
• @jheindel true though I can see it be useful for back of the envelope/order of magnitude type calculations. And sure 0 and 1 aren't great, but for those its hardly worth making the approximation in the first place because you could probably just memorize those. By n=2, you only have about 6% error. – Tyberius Oct 24 '17 at 18:54
• Ya true. I suspect this is not very useful beyond a coarse approximation anyways so approximating it more can't be too bad. – jheindel Oct 24 '17 at 18:55 | {
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To me this seems like a blind usage of the Taylor expansion.
Let's say we want to get a Taylor approximation of $\mu (n) = \sqrt{n(n+2)}$. Then we know we can expand any function $f(x) = f(x_0 +h)$ if $h/x_0 \ll 1$ into $f(x_0 + h) \approx f(x_0) + \frac{\partial f}{\partial x}(x_0) \cdot h$.
If I apply this to the function $\mu(n) = \mu(0 + n)$ then I get the approximate result $\mu(n) \approx \frac{n+1}{\sqrt 2}$. But here of course it is not true that $n/0 \ll 1$, so the assumption needed for using Taylor is violated.
However, if I plot both those functions in Wolframalpha they seem to agree OK enough for large n. So I guess a different approximation technique might have been used deriving this result, and it depends if you're working at large n or small n.
• It's not really a Taylor expansion, so much as using the arithmetic mean of two numbers ($n$ and $n+1$) as an approximation to their geometric mean. – psmears Oct 24 '17 at 16:05
• @psmears: You mean $n$ and $n + 2$, right? – Ilmari Karonen Oct 24 '17 at 17:00
• Why do you assume that it's a blind usage of anything? – David Richerby Oct 24 '17 at 17:03
• @DavidRicherby: The approximation apparently transforms a nonlinear function into a linear one. The simplest and most common way to do this is to do a Taylor approximation. – AtmosphericPrisonEscape Oct 24 '17 at 17:18
• My (literal) emphasis was on "blind". Saying it's "a blind usage of Taylor expansions" means that you think that the teacher used Taylor expansions without giving any thought to whether they were suitable. Why are you accusing them of that? – David Richerby Oct 24 '17 at 18:25
Using $\sqrt{x}-\sqrt{y} = { x -y \over \sqrt{x} + \sqrt{y}}$ we have $\sqrt{n(n+2)} - (n+1) = -{ 1\over n \left( \sqrt{1+ {2 \over n}}+1 + {1\over n}\right )}$, so $|\sqrt{n(n+2)} - (n+1)| \le {1 \over 2n}$.
The approximation is reasonable for large $n$. | {
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The approximation is reasonable for large $n$.
Addendum: To see where the approximation comes from, note that $\sqrt{1+ {2 \over n}}+1 + {1\over n} \ge 2$, hence ${1\over n \left( \sqrt{1+ {2 \over n}}+1 + {1\over n}\right )} \le {1 \over 2n}$.
• I like this answer. I think the only thing I would add is showing more explicitly how your second equation approximates to 1/2n. – Tyberius Oct 24 '17 at 16:16
• @Tyberius: Good suggestion. Done. – copper.hat Oct 24 '17 at 16:20 | {
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# Function periodic of example a
Why is sin(x^2) not a periodic function? + example. Sal introduces the main features of sinusoidal functions: midline, amplitude, & period. he shows how these can be found from a sinusoidal function's graph..
## Periodic Function Definition (Illustrated Mathematics
Periodic Function some codes without special toolboxes. A periodic function f(x) has a period of 9, if f(2) if a function f(x) is periodic with period k then for so for example if the period is k = 3 and f(2) = 7, title: examples of periodic functions: canonical name: examplesofperiodicfunctions: date of creation: 2013-03-22 17:57:29: last modified on: 2013-03-22 17:57:29.
## Real World Examples of Periodic Functions by nancy Prezi
Autocorrelation (for sound signals). How to plot periodic function's graphic? for @mr.wizard why it doesn't work if you put pi in the range of t? for example plot[myperiodic[exp[2*t Trigonometric functions and graphs: mid unit assignment-jiayi jin.
Real world examples of periodic functions cyclical stocks there are two types of stocks; cyclical and non-cyclical. stocks always depend on the market and the success trigonometric functions and graphs: mid unit assignment-jiayi jin
Fourier analysis for periodic functions: fourier series the result is a periodic function with period t that agrees they are just one example of conditions that example : find the fourier series of the periodic function f(t) defined by solution to the above example coefficient a 0 is given by. coefficients a n is given by
I only know f(x)=constant, and i know it's the only one in continuous functions. so i want some more examples to help me understand... this theorem helps associate a fourier series to any -periodic function. definition. example. find the fourier series of the function function answer. | {
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A periodic function f(x) has a period of 9, if f(2) if a function f(x) is periodic with period k then for so for example if the period is k = 3 and f(2) = 7 i only know f(x)=constant, and i know it's the only one in continuous functions. so i want some more examples to help me understand...
If a periodic function with period has a finite derivative , the indefinite integral has period if , otherwise it is non-periodic, such as for example for , periodic functions periodic functions are functions which repeat: f (t + p) = f (t) for all t. for example, if f (t) is the amount of time between sunrise and sunset at a
Sal introduces the main features of sinusoidal functions: midline, amplitude, & period. he shows how these can be found from a sinusoidal function's graph. we want to approximate a periodic function f(t), with fundamental period t, with the fourier series:
Sal introduces the main features of sinusoidal functions: midline, amplitude, & period. he shows how these can be found from a sinusoidal function's graph. laplace transform of periodic functions, convolution, applications 1 laplace transform of periodic function example 2. consider a saw-tooth function
A function whose value does not change when its argument is increased by a certain nonzero number called the period of the function. for example, sin x and cos x are learn more about periodic function . toggle main periodic functions. asked by li. li not enough to be used for input on a function, for example the simulink
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# Binomial Calculating Probability of Airline Tickets
Because not all airline passengers show up for their reserved seat, an airline sells 125 tickets for a flight that holds only 115 passengers. The probability that a passenger does not show up is 0.05, and the passengers behave independently. Round your answers to four decimal places (e.g. 98.7654).
a) What is the probability that every passenger who shows up can take the flight?
b) What is the probability that the flight departs with at least one empty seat?
I am not using a statistics program to calculate my answers like I have seen many answers on here use, but by using a formula: For example: P(115) = 125 C 115 * (.95)^115 * (.05)^10 = .0475
Similarly I have:
P(116) = .0778
P(117) = .1137
P(118) = .1465
P(119) = .1637
P(120) = .1556
P(121) = .1221
P(122) = .0761
P(123) = .0353
P(124) = .0108
P(125) = .0016
So for part a) I did:
1 - P(X > 115) = 1 - .9032 = .0968
and for part b) I did:
1 - P(x >= 115) = 1 - .9507 = .0493
These numbers just do not seem correct to me. And I am confused as to why the probabilites are increasing from P(115) to P(119), (I would expect them to decrease, however I guess if they are on the rising part of the binomial distribution and then go to the falling part of the distribution at P(120)
Edit: I know understand these values are correct and fall around the most probable value P(119) which is the mean. Thank you for help in clarifying.
• If you will look at the 'Related' links to the right of your question, you will see at least one that answers almost the same question. // I will look at some of your answers to see if anything seems strange. – BruceET Sep 12 '18 at 22:58
• Well, by now you have several verifications. Problem seems strange; no wonder you wondered. Considering my 'bad luck', I think I may know this airline. – BruceET Sep 12 '18 at 23:40 | {
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The number of people who show up for the flight in the circumstances you describe is $X \sim \mathsf{Binom}(n=125, p=.95).$ The probability everyone who shows has a seat is $P(X \le 115) \approx 0.0967.$ This can be computed in R statistical software, in which pbinom is a binomial CDF.
pbinom(115, 125, .95)
[1] 0.09672946
This may seem absurdly small, but the average number of people showing is $\mu = E(X)= np = 128.25 > 125.$ So on the average flight the airline should expect to leave some people behind.
The probability there will be at least one empty seat is $P(X \le 114) = 0.0492.$ So that will seldom happen.
pbinom(114, 125, .95)
[1] 0.04921917
Here is a plot of the relevant part of the distribution of $X.$
Here are exact values from R, matching your computations, which seem quite accurate:
x = 115:125; pdf = dbinom(x, 125, .95)
cbind(x, pdf)
x pdf
[1,] 115 0.047510291
[2,] 116 0.077818581
[3,] 117 0.113734849
[4,] 118 0.146505907
[5,] 119 0.163741896
[6,] 120 0.155554801
[7,] 121 0.122129803
[8,] 122 0.076080861
[9,] 123 0.035256984
[10,] 124 0.010804560
[11,] 125 0.001642293
Note: Code for figure:
x = 110:125; pdf = dbinom(x, 125, .95)
plot(x, pdf, type="h", ylim=c(0,max(pdf)), lwd=2)
abline(h=0, col="green3")
Because not all airline passengers show up for their reserved seat, an airline sells 125 tickets for a flight that holds only 115 passengers. The probability that a passenger does not show up is 0.05, and the passengers behave independently. Round your answers to four decimal places (e.g. 98.7654).
Suppose $X \sim \textrm{Bin}(125,0.5)$ is a binomial random variable. The mass function is given by
$$f(k;125,0.5) = Pr(X = k) = \binom{125}{k} \bigg(\frac{5}{100}\bigg)^{k} \bigg( \frac{95}{100} \bigg)^{125-k} \tag{1}$$
We can visualize the probability mass function then like this in Python
n=125
p=.95
x = range(n+1)
y = stats.binom.pmf(x, n, p)
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n=125
p=.95
x = range(n+1)
y = stats.binom.pmf(x, n, p)
plt.plot(x,y,"o", color="black")
plt.axis([-(max(x)-min(x))*0.05, max(x)*1.05, -0.01, max(y)*1.10])
plt.xticks(x)
plt.title("Binomial distribution PMF for tries = {0} & p ={1}".format(
n,p))
plt.xlabel("Variate")
plt.ylabel("Probability")
If you note then
y1 = stats.binom.pmf(115,125,.95)
y1
Out[29]: 0.04751029149720219
to look at the cdf then
y2 = stats.binom.cdf(x,125,.95)
plt.plot(x,y2,"o", color="black")
If we take a closer look at the pdf
n=125
p=.95
x = range(100,150)
y = stats.binom.pmf(x, n, p)
y2 = stats.binom.pmf(x,125,.95)
plt.plot(x,y2,"o", color="black")
Your formula is correct. I didn't check the numbers but they look good. If $0.05$ of the people do not show up, the expected number of no-shows is $0.05\cdot 125=6.25$ so the most probable number who show up should be $119$, in agreement with your calculation. | {
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# Number of N-digit Perfect Squares
I was working on a programming problem to find all 10-digit perfect squares when I started wondering if I could figure out how many perfects squares have exactly N-digits. I believe that I am close to finding a formula, but I am still off by one in some cases.
Current formula where $n$ is the number of digits:
$\lfloor\sqrt{10^n-1}\rfloor - \lfloor\sqrt{10^{n-1}}\rfloor$
How I got here:
1. Range of possible 10-digit numbers is from $10^9$ to $10^{10}-1$
2. 10-digit perfect squares should fall into the range of $\sqrt{10^9}$ to $\sqrt{10^{10}-1}$
Results of program for $n = 1,2,3,4,5$:
DIGITS=1, ACTUAL_COUNT=3, COMPUTED_COUNT=2
DIGITS=2, ACTUAL_COUNT=6, COMPUTED_COUNT=6
DIGITS=3, ACTUAL_COUNT=22, COMPUTED_COUNT=21
DIGITS=4, ACTUAL_COUNT=68, COMPUTED_COUNT=68
DIGITS=5, ACTUAL_COUNT=217, COMPUTED_COUNT=216
Program:
#!/usr/bin/perl
use strict;
use warnings;
sub all_n_digit_perfect_squares {
my ($n) = @_; my$count = 0;
my $MIN = int( sqrt( 10**($n-1) ) );
my $MAX = int( sqrt( (10**$n)-1 ) );
foreach my $i ($MIN .. $MAX ) { if ( ($i * $i) >= 10**($n-1) ) {
$count++; } } print "DIGITS=$n"
. ", ACTUAL_COUNT=$count" . ", COMPUTED_COUNT=" . ($MAX-$MIN), "\n"; return; } all_n_digit_perfect_squares($_) for (1..5);
If you think about it, you should have a formula that says the number of squares is f (n) - f (n-1) for some function f, so that every perfect square is counted exactly once if you calculate the squares from 10^1 to 10^2, from 10^2 to 10^3 and so on.
In your formula, the squares 100, 10,000, 1,000,000 and so on are not counted at all. For example, for 3 digit numbers the squares are from 10^2 to 31^2, that's 22 numbers. You calculate 31 - 10 = 21. Change your formula to
$\lfloor\sqrt{10^n-1}\rfloor - \lfloor\sqrt{10^{n-1}-1}\rfloor$
• Ah, thank you! I knew I was missing something. – Hunter McMillen Mar 18 '16 at 13:02 | {
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• Ah, thank you! I knew I was missing something. – Hunter McMillen Mar 18 '16 at 13:02
I'm interested in the question that you're asking. I've been thinking about this for the last few days and couldn't find anything online about this subject. If you have anything references, I would appreciate it. Here is what I have so far and sorry if the formatting is off, this is my first post.
$$\text{Let}\; \mathbb{N} \;\text{be the set of Natural numbers} \; = \{1,2,3,4,..,N\},$$ Then there exists a set $$\mathbb{S}$$, such that for $$s\in\mathbb{S}\; ;\; s_i = \{10^{i-1},...,(10^i-1)\} \;, \text{for} \; i = 1,2,3,...,M.$$ Example : $$s_1 = \{1,2,3,...,9\}\; ; \;\;\; i=1 = 1\;\text{digit}$$ $$s_2 = \{10,11,12,...,99\}\; ; \;\; i=2 = 2\;\text{digits}$$ $$s_3 = \{100,101,102,...,999\}\; ; \;\; i=3 = 3\;\text{digits}$$ $$...$$ $$s_i = \{10^{i-1},...,(10^i - 1)\} ; \;\; i\; \text{digits}$$
Within each set $$s_i$$, there exists a set $$\mathbb{A}$$ such that for all members of $$\mathbb{A}$$, call it $$a_p$$, when squared are still a member of $$s_i$$: $$a\in\mathbb{A}\; ; \;\; a = \{ a_1,a_2,...,a_k \}$$ $$(a_p)^2 \in s_i ,\; \text{for}\; p = 1\; \text{to}\; k$$
Example : $$s_1 = \{1,2,3,...,9\}\; ; \;\;\; a_1 = \{1,2,3\}$$ $$s_2 = \{10,11,12,...,99\}\; ; \;\; a_2 = \{ a_1 , 4, 5,6,...,9\} = \{1,2,3 , 4, 5,6,...,9\}$$
I'm going to start omit the previous sets before the current $$i^{th}$$ set we're looking at. So it'll look like this:
$$s_2 = \{10,11,12,...,99\}\; ; \;\; a_2 = \{4, 5,6,...,9\},$$
but also know that all previous sets before the current $$i^{th}$$ can included, but is not unique to the $$i^{th}$$ digits. | {
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I'm going to list a few in the long version, just as an example. $$s_3 = \{100,101,102,...,999\}\; ; \;\; a_3 = \{10,11,12,...,31 \}$$ $$s_4 = \{1000,1002,1003,...,9999\}\; ; \;\; a_4 = \{32,33,34,...,99 \}$$ $$s_5 = \{10000,...,99999\}\; ; \;\; a_5 = \{100,...,316 \}$$ $$s_6 = \{100000,...,999999\}\; ; \;\; a_6 = \{317,...,999 \}$$ $$s_7 = \{1000000,...,9999999\}\; ; \;\; a_7 = \{1000,...,3162 \}$$
Now we can write these compactly as, the first i=1,2,3 are written in the long way and i > 4 is written in the short hand version as given by $$s_i$$: $$s_1 = \{10^0,2,3,...,9\}\; ; \; a_1 = \{1,2,3\}\; ; \; t_1 = 3$$ $$s_2 = \{10^1,...,99\}\; ; \; a_2 = \{4, 5,6,...,9\}\; ; \; t_2 = 6$$
$$s_3= \{10^2,...,999\}\; ; \; a_3 = \{10^1,...,31\}\; ; \; t_3 = 22$$
$$s_4= \{10^3,...,(10^4 -1)\}\; ; \; a_4 = \{32,...,99\}\; ; \; t_4 = 68$$ $$s_5= \{10^4,...,(10^5 -1)\}\; ; \; a_5 = \{10^2,...,316\}\; ; \; t_5 = 217$$
I also introduced $$t_i$$, which is the count of the number of members of $$a_i$$, (this is exactly what you're solving for, I believe). Notice, we can get an exact solution for $$t_i$$ given by: $$t_i = (\{a_i\}_{max} - \{a_i\}_{min}) + 1$$ (I think this could be calculated using a norm)
I'm going to list out a few sequences and point out an interesting patterns.
Starting with $$i=3$$, (you could extend this to $$i=1$$ if you want),
if $$i = odd$$ : The lower bound of $$a_i$$, call this $$\{a_i\}_{min}$$ is exactly $$10^l$$, where $$l = (i-1)/2$$ ;
if $$i = even$$ : The upper bound of $$a_i$$, call this $$\{a_i\}_{max}$$ is exactly $$10^z - 1$$, where $$z = (i)/2$$ ;
I claim that for $$i=even$$, you have
$$s_{i-1} = \{10^{i-2},...,(10^{i-1}-1)\}\; ; \;\; a_{i-1} =\{10^l,...,upper\;bound_{(i-1)}\} \; ; \; t_{i-1} = ?$$ $$s_{i} = \{10^{i-1},...,(10^{i}-1)\}\; ; \;\; a_{i} =\{lower\; bound_{(i)},...,(10^z-1)\} \; ; \; t_{i} = ?.$$
I'm still working/thinking about I can get the upper bounds, lower bounds, and $$t_i$$. | {
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I'm still working/thinking about I can get the upper bounds, lower bounds, and $$t_i$$.
Here are the first 16 digits, and their lists. I also have done this out until 34 digits, took my computer a good few hours to compute by brute force. I am almost done with an algorithm that can do it much quicker.
Also note, that if you know the lower bounds and upper bounds, you can easily obtain $$t$$, which is the number of elements that are contained in $$\{a_i\}$$, since $$\{a_i\}$$ is ordered.
$$s_1 = \{10^0,2,3,...,9\}\; ; \; a_1 = \{1,2,3\}\; ; \; t_1 = 3$$ $$s_2 = \{10^1,...,99\}\; ; \; a_2 = \{4, 5,6,...,9\}\; ; \; t_2 = 6$$
$$s_3= \{10^2,...,999\}\; ; \; a_3 = \{10^1,...,31\}\; ; \; t_3 = 22$$
$$s_4= \{10^3,...,(10^4 -1)\}\; ; \; a_4 = \{32,...,99\}\; ; \; t_4 = 68$$ $$s_5= \{10^4,...,(10^5 -1)\}\; ; \; a_5 = \{10^2,...,316\}\; ; \; t_5 = 217$$ $$s_6= \{10^5,...,(10^6 -1)\}\; ; \; a_6 = \{317,...,999\}\; ; \; t_6 = 683$$ $$s_7= \{10^6,...,(10^7 -1)\}\; ; \; a_7 = \{10^3,...,3162\}\; ; \; t_7 = 2163$$ $$s_8= \{10^7,...,(10^8 -1)\}\; ; \; a_8 = \{3163,...,9999\}\; ; \; t_8 = 6837$$ $$s_9= \{10^8,...,(10^9 -1)\}\; ; \; a_9 = \{10^4,...,31622\}\; ; \; t_9 =21623$$ $$s_{10}= \{10^9,...,(10^{10} -1)\}\; ; \;\; a_{10} = \{31623,...,(10^5-1)\}\; ; \; t_{10} = 68377$$ $$s_{11}= \{10^{10},...,(10^{11} -1)\}\; ; \;\; a_{11} = \{10^5,...,316227\} \; ; \; t_{11} = 216228$$ $$s_{12}= \{10^{11},...,(10^{12} -1)\}\; ; \;\; a_{12} = \{316228,...,(10^6-1)\}\; ; \; t_{12} = 683772$$ $$s_{13}= \{10^{12},...,(10^{13} -1)\}\; ; \;\; a_{13} = \{10^6,...,3162277\}\; ; \; t_{13} = 2162278$$ $$s_{14}= \{10^{13},...,(10^{14} -1)\}\; ; \;\; a_{14} = \{3162278,...,(10^7-1)\} \; ; \; t_{14} = 6837722$$. $$s_{15}= \{10^{14},...,(10^{15} -1)\}\; ; \;\; a_{15} = \{10^7,...,31622776\} \; ; \; t_{15} = 21622777$$ $$s_{16}= \{10^{15},...,(10^{16} -1)\}\; ; \;\; a_{16} = \{31622777,...,(10^8-1)\} \; ; \; t_{16} = 68377223$$ | {
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• I get a different result for $t_{16}$ using the other formula. I get $68377224$ instead of $68377223$ – Hunter McMillen Jul 1 at 14:00
• As far as other research that I did into this topic, after I found a working solution I didn't explore the area much further. – Hunter McMillen Jul 1 at 14:01
• Sample code to recreate my outputs for the first 16 elements: perl -E 'sub f { my $n = shift; return int(sqrt(10**$n - 1)) - int(sqrt(10**($n-1) - 1)); } say f($_) for 1..16;' – Hunter McMillen Jul 1 at 14:03
The number of perfect squares between any two numbers a and b with a less than b is floor sqrt b - ceil sqrt a + 1. i.e. a=1000 b=2000. ceil sqrt 1000 = 32. floor sqrt 2000 = 44. So 32 33 34 35 36 37 38 39 40 41 42 43 and 44 when squared will be perfect squares between 1000 and 2000 and 44-32+1=13. If you are silly and try for the number of perfect squares between 1000 and 1001 then 31-32+1=0. between 9 and 10 then 3-3+1=1. | {
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# Diff EQ. Problem (Eigenvector issue)
Find the general solution of $\textbf{x}^{'}=\begin{pmatrix} -1&-4\\1&-1\end{pmatrix}\textbf{x}$.
The eigenvalues I found are $-1 \pm 2i$ and I chose $-1-2i$ to be my eigenvector and found it to be $e_1=\begin{pmatrix} 2i\\-1 \end{pmatrix}$ Now plugging this in to the formula we have $$e^{-t}\begin{pmatrix} 2i \\-1 \end{pmatrix}(\cos 2t-i\sin2t)=c_1e^{-t}\begin{pmatrix} 2\sin 2t\\-\cos 2t \end{pmatrix} + c_2e^{-t}\begin{pmatrix} 2 \cos 2t\\ \sin 2t \end{pmatrix}$$ The book however has: $c_1e^{-t}\begin{pmatrix} - 2\sin 2t\\\cos 2t \end{pmatrix} + c_2e^{-t}\begin{pmatrix} 2 \cos 2t\\ \sin 2t \end{pmatrix}$. How did the signs get switched because I tried to do this with $e_2=\begin{pmatrix} -2i\\-1 \end{pmatrix}$ and all that happens is the right side becomes negative which still doesn't explain how the signs are switched.
• Your $c_1$ is the negative of the book's $c_1$. – John Habert Mar 27 '14 at 17:33
• but how did that happen is this correct? – adam Mar 27 '14 at 17:34
• Since the constants are arbitrary until you use initial conditions to solve for them, it is fine. It is a similar trick as to how you got all the $i$ terms to drop out by being absorbed into $c_2$. – John Habert Mar 27 '14 at 17:36
• Thanks I understand that part, but what I dont understand is why they would do something random and factor out a negative 1. I understand that $\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}}{3}$ because we dont want radicals in the denominator. Now why factor out a $-1$? – adam Mar 27 '14 at 17:41 | {
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