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https://quant.stackexchange.com/questions/68635/characteristics-of-factor-portfolios/68646
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# characteristics of factor portfolios
In the paper Characteristics of Factor Portfolios (https://papers.ssrn.com/sol3/papers.cfm?abstract_id=1601414), when it discusses pure factor portfolios, it says that simple style factor portfolios have zero exposure to all other style, country, and industry factors. Could someone help me understand the math for why the style factor portfolios have zero exposure to all other style, country, and industry factors?
So, for example, if we are interested in the return of a P/E factor and a P/B factor, we would gather the P/E and P/B for all of our stocks into a matrix of loadings $$B$$. $$B$$ would have two columns – one containing P/E and one containing P/B for all assets. We then regress $$R$$ (a vector containing the returns of all assets) on $$B$$. OLS regression gives us $$f= (B’B)^{-1} B’R$$ = the returns of the style factors for this particular period. The rows of $$(B’B)^{-1} B’$$ are considered to be the factor portfolios.
So, let’s go one step further and look at the loadings of the portfolio on the individual styles by multiplying the factor portfolios with the matrix of loadings. This gives $$(B’B)^{-1} B’B = I$$ - an identity matrix. Hence, the loadings of each factor portfolio are 1 against the particular style and 0 against any other style.
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# characteristics of factor portfolios
In the paper Characteristics of Factor Portfolios (https://papers.ssrn.com/sol3/papers.cfm?abstract_id=1601414), when it discusses pure factor portfolios, it says that simple style factor portfolios have zero exposure to all other style, country, and industry factors. Could someone help me understand the math for why the style factor portfolios have zero exposure to all other style, country, and industry factors? So, for example, if we are interested in the return of a P/E factor and a P/B factor, we would gather the P/E and P/B for all of our stocks into a matrix of loadings $$B$$. $$B$$ would have two columns – one containing P/E and one containing P/B for all assets. We then regress $$R$$ (a vector containing the returns of all assets) on $$B$$. OLS regression gives us $$f= (B’B)^{-1} B’R$$ = the returns of the style factors for this particular period. The rows of $$(B’B)^{-1} B’$$ are considered to be the factor portfolios. So, let’s go one step further and look at the loadings of the portfolio on the individual styles by multiplying the factor portfolios with the matrix of loadings.
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This gives $$(B’B)^{-1} B’B = I$$ - an identity matrix.
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https://math.stackexchange.com/questions/851072/theorem-on-giuga-number/851114
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# Theorem on Giuga number
Giuga number : $n$ is a Giuga number $\iff$ For every prime factor $p$ of $n$ , $p | (\frac{n}{p}-1)$
How to prove the following theorem on Giuga numbers
$n$ is a giuga number $\iff$ $\sum_{i=1}^{n-1} i^{\phi(n)} \equiv -1 \mod {n}$
## 1 Answer
The $\Rightarrow$ part. For first, a giuga number must be squarefree, since, by assuming $p^2\mid n$, we have that $p$ divides two consecutive numbers, $\frac{n}{p}$ and $\frac{n}{p}-1$, that is clearly impossible. So we have: $$n = \prod_{i=1}^{k} p_i$$ that implies: $$\phi(n) = \prod_{i=1}^{k} (p_i-1).$$ By considering the sum $$\sum_{i=0}^{n-1}i^{\phi(n)}$$ $\pmod{p_i}$ we have that all the terms contribute with a $1$, except the multiples of $p_i$ that contribute with a zero. This gives: $$\sum_{i=0}^{n-1}i^{\phi(n)}\equiv n-\frac{n}{p_i}\equiv (n-1)\pmod{p_i}\tag{1}$$ that holds for any $i\in[1,k]$. The chinese theorem now give: $$\sum_{i=0}^{n-1}i^{\phi(n)}\equiv n-1\pmod{\prod_{i=1}^{k}p_i}$$ that is just: $$\sum_{i=0}^{n-1}i^{\phi(n)}\equiv -1\pmod{n}$$ as claimed. For the $\Leftarrow$ part, we have that the congruence $\!\!\!\pmod{n}$ implies the congruence $\!\!\!\pmod{p_i}$, hence $(1)$ must hold, so we must have: $$\frac{n}{p_i}\equiv 1\pmod{p_i}$$ that is equivalent to $p_i\mid\left(\frac{n}{p_i}-1\right).$
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# Theorem on Giuga number
Giuga number : $n$ is a Giuga number $\iff$ For every prime factor $p$ of $n$ , $p | (\frac{n}{p}-1)$
How to prove the following theorem on Giuga numbers
$n$ is a giuga number $\iff$ $\sum_{i=1}^{n-1} i^{\phi(n)} \equiv -1 \mod {n}$
## 1 Answer
The $\Rightarrow$ part. For first, a giuga number must be squarefree, since, by assuming $p^2\mid n$, we have that $p$ divides two consecutive numbers, $\frac{n}{p}$ and $\frac{n}{p}-1$, that is clearly impossible. So we have: $$n = \prod_{i=1}^{k} p_i$$ that implies: $$\phi(n) = \prod_{i=1}^{k} (p_i-1).$$ By considering the sum $$\sum_{i=0}^{n-1}i^{\phi(n)}$$ $\pmod{p_i}$ we have that all the terms contribute with a $1$, except the multiples of $p_i$ that contribute with a zero. This gives: $$\sum_{i=0}^{n-1}i^{\phi(n)}\equiv n-\frac{n}{p_i}\equiv (n-1)\pmod{p_i}\tag{1}$$ that holds for any $i\in[1,k]$. The chinese theorem now give: $$\sum_{i=0}^{n-1}i^{\phi(n)}\equiv n-1\pmod{\prod_{i=1}^{k}p_i}$$ that is just: $$\sum_{i=0}^{n-1}i^{\phi(n)}\equiv -1\pmod{n}$$ as claimed.
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For the $\Leftarrow$ part, we have that the congruence $\!\!\!\pmod{n}$ implies the congruence $\!\!\!\pmod{p_i}$, hence $(1)$ must hold, so we must have: $$\frac{n}{p_i}\equiv 1\pmod{p_i}$$ that is equivalent to $p_i\mid\left(\frac{n}{p_i}-1\right).$
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https://gamedev.stackexchange.com/questions/138165/how-can-i-move-and-rotate-an-object-in-an-infinity-or-figure-8-trajectory/138167
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# How can I move and rotate an object in an “infinity” or “figure 8” trajectory?
I know that the easiest way to move an object with the figure 8 trajectory is:
x = cos(t);
y = sin(2*t) / 2;
but I just don't know how to rotate it, lets says with a new variable r as rotation, how do I merge it into the above formula ? can anyone please advise me on what is the simplest and cheapest way/formula to move and rotate the figure 8 trajectory ?
## 1 Answer
The object should point in the direction of the derivative, which is [-sin(t), cos(2t)]. Its angle is atan2(cos(2t), -sin(t)).
Edit: OP is apparently asking how to rotate the "trajectory," not the object itself.
To rotate the figure, choose an angle, θ, in radians, that you'd like the trajectory to be rotated. The position along this rotated figure is:
x = cos(θ) * cos(t) - sin(θ) * sin(2t)/2
y = sin(θ) * cos(t) + cos(θ) * sin(2t)/2
• so how would I modify the formula to get a rotated figure of 8 ? – user1998844 Mar 3 '17 at 18:24
• That is a completely different question than the one I answered. I'll edit my answer with a solution to this question. – Drew Cummins Mar 3 '17 at 18:31
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# How can I move and rotate an object in an “infinity” or “figure 8” trajectory? I know that the easiest way to move an object with the figure 8 trajectory is:
x = cos(t);
y = sin(2*t) / 2;
but I just don't know how to rotate it, lets says with a new variable r as rotation, how do I merge it into the above formula ? can anyone please advise me on what is the simplest and cheapest way/formula to move and rotate the figure 8 trajectory ? ## 1 Answer
The object should point in the direction of the derivative, which is [-sin(t), cos(2t)]. Its angle is atan2(cos(2t), -sin(t)). Edit: OP is apparently asking how to rotate the "trajectory," not the object itself. To rotate the figure, choose an angle, θ, in radians, that you'd like the trajectory to be rotated. The position along this rotated figure is:
x = cos(θ) * cos(t) - sin(θ) * sin(2t)/2
y = sin(θ) * cos(t) + cos(θ) * sin(2t)/2
• so how would I modify the formula to get a rotated figure of 8 ? – user1998844 Mar 3 '17 at 18:24
• That is a completely different question than the one I answered. I'll edit my answer with a solution to this question.
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– Drew Cummins Mar 3 '17 at 18:31
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https://stats.stackexchange.com/questions/592820/how-can-i-find-the-expectation-value-to-this-problem
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# How can I find the expectation value to this problem?
At a wedding reception on an evening the representative of the host is taking it as an occasion to exercise and explain a classical analytic problem. specifically, he insists that he would start serving the food only when the first table, which is arranged for 12 guests to dine together, has guests born in every twelve months of the year. assume that any given guest is equally likely to be born in any of the twelve months of the year, and that new guests were arriving at every two minutes then. what is the expected waiting time of the first arriving guest before the food gets served eventually?
Since this looked like a Coupon Collector's problem variation, my initial approach was to determine the sum of the expected value of each guests of unique birth months.
X ~ FS(p) [First Success Distribution]
X = time needed until food gets served
$$E[X] = E[X1] + E[X2] + ... + E[X12]$$
$$=> E[X] = 12/12 + 12/11 + ... + 12/1$$
However, this is where i ran into problem, since I don't know how to handle the arrival at every two minutes in my equation. Should I just multiply by 2? Or am i missing something very obvious or basic trivia? Help will be appreciated.
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# How can I find the expectation value to this problem? At a wedding reception on an evening the representative of the host is taking it as an occasion to exercise and explain a classical analytic problem. specifically, he insists that he would start serving the food only when the first table, which is arranged for 12 guests to dine together, has guests born in every twelve months of the year. assume that any given guest is equally likely to be born in any of the twelve months of the year, and that new guests were arriving at every two minutes then. what is the expected waiting time of the first arriving guest before the food gets served eventually? Since this looked like a Coupon Collector's problem variation, my initial approach was to determine the sum of the expected value of each guests of unique birth months.
|
X ~ FS(p) [First Success Distribution]
X = time needed until food gets served
$$E[X] = E[X1] + E[X2] + ... + E[X12]$$
$$=> E[X] = 12/12 + 12/11 + ... + 12/1$$
However, this is where i ran into problem, since I don't know how to handle the arrival at every two minutes in my equation.
|
https://math.stackexchange.com/questions/3003033/show-lim-x-to-x-0-fxx-x-0-0-when-f-mathbbr-subset-mathbbr
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# Show $\lim_{x \to x_0^+} f(x)(x-x_0) =0$ when $f(\mathbb{R}) \subset \mathbb{R}^+$ & monotone increasing.
Show $$\lim_{x \to x_0^+} f(x)(x-x_0) =0$$ when $$f(\mathbb{R}) \subset \mathbb{R}^+$$ & monotone increasing.
Try
I need to show,
$$\forall \epsilon >0, \exists \delta >0 : x \in (x_0, x_0 + \delta) \Rightarrow |f(x) (x-x_0)| < \epsilon$$
I think I could find some upper bound $$M >0$$ such that $$|f(x) (x-x_0)| \le M |x - x_0|$$.
Let $$M = f(x_0 + \epsilon)$$, and let $$\delta = \frac{\epsilon}{\max \{2M, 2 \}}$$, then clearly $$f(x) \le f(x_0 + \epsilon) = M$$
But I'm not sure $$|f(x) (x-x_0)| \le M |x - x_0|$$.
Any hint about how I should proceed?
Hint: Observe \begin{align} |f(x)(x-x_0)|\leq |f(x_0)||x-x_0| \end{align} for all $$x\leq x_0$$.
Use $$M=f(x_0+1)$$ and cosider $$\delta=\min\{\frac{1}{2},\frac{\epsilon}{2M}\}$$.
Fix $$\varepsilon>0$$. Let $$M=f(x_0+1)$$ and choose $$\delta=\mathrm{min}\{1,\frac{\varepsilon}{M}\}$$. For each $$x\in(x_0,x_0+\delta)$$, $$|f(x)|\leq M$$ since $$f$$ is strictly increasing. Thus, $$|f(x)(x-x_0)|\leq M|x-x_0|.
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# Show $\lim_{x \to x_0^+} f(x)(x-x_0) =0$ when $f(\mathbb{R}) \subset \mathbb{R}^+$ & monotone increasing. Show $$\lim_{x \to x_0^+} f(x)(x-x_0) =0$$ when $$f(\mathbb{R}) \subset \mathbb{R}^+$$ & monotone increasing. Try
I need to show,
$$\forall \epsilon >0, \exists \delta >0 : x \in (x_0, x_0 + \delta) \Rightarrow |f(x) (x-x_0)| < \epsilon$$
I think I could find some upper bound $$M >0$$ such that $$|f(x) (x-x_0)| \le M |x - x_0|$$. Let $$M = f(x_0 + \epsilon)$$, and let $$\delta = \frac{\epsilon}{\max \{2M, 2 \}}$$, then clearly $$f(x) \le f(x_0 + \epsilon) = M$$
But I'm not sure $$|f(x) (x-x_0)| \le M |x - x_0|$$. Any hint about how I should proceed? Hint: Observe \begin{align} |f(x)(x-x_0)|\leq |f(x_0)||x-x_0| \end{align} for all $$x\leq x_0$$. Use $$M=f(x_0+1)$$ and cosider $$\delta=\min\{\frac{1}{2},\frac{\epsilon}{2M}\}$$. Fix $$\varepsilon>0$$. Let $$M=f(x_0+1)$$ and choose $$\delta=\mathrm{min}\{1,\frac{\varepsilon}{M}\}$$. For each $$x\in(x_0,x_0+\delta)$$, $$|f(x)|\leq M$$ since $$f$$ is strictly increasing.
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Thus, $$|f(x)(x-x_0)|\leq M|x-x_0|.
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https://matheducators.stackexchange.com/questions/18576/ramanujan-results-for-middle-school/18599
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# Ramanujan results for middle school?
Pls I wonder what Ramanujan's results could be explained to middle school level audience, ie without using integral etc that is up to university curriculum?
For example Ramanujan's infinite radicals could be explained easily
$$3=\sqrt {1+2{\sqrt {1+3{\sqrt {1+\cdots }}}}}$$
• Why Ramanujan specifically? Jul 17, 2020 at 22:52
• just his results have many infinite forms, which sound fun! @ChrisCunningham Jul 18, 2020 at 12:54
## 1 Answer
You can try the Rogers–Ramanujan identities:
• The number of partitions of $$n$$ in which adjacent parts are at least 2 apart is the same as the number of partitions of $$n$$ in which each part ends with 1,4,6,9.
• The number of partitions of $$n$$ without 1 in which adjacent parts are at least 2 apart is the same as the number of partitions of $$n$$ in which each part ends with 2,3,7,8.
For example, taking $$n=10$$:
• Partitions in which adjacent parts are at least 2 apart: $$10 = 10 = 9 + 1 = 8 + 2 = 7 + 3 = 6 + 4 = 6 + 3 + 1$$
• Partitions in which each part ends with 1,4,6,9: $$10 = 9 + 1 = 6 + 4 = 6 + 1 + 1 + 1 + 1 = 4 + 4 + 1 + 1 = 4 + 6\times 1 = 10 \times 1$$
• Partitions without 1 in which adjacent parts are at least 2 apart: $$10 = 10 = 8 + 2 = 7 + 3 = 6 + 4$$
• Partitions in which each part ends with 2,3,5,8: $$10 = 8 + 2 = 7 + 3 = 3 + 3 + 2 + 2 = 2 + 2 + 2 + 2 + 2$$
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# Ramanujan results for middle school? Pls I wonder what Ramanujan's results could be explained to middle school level audience, ie without using integral etc that is up to university curriculum? For example Ramanujan's infinite radicals could be explained easily
$$3=\sqrt {1+2{\sqrt {1+3{\sqrt {1+\cdots }}}}}$$
• Why Ramanujan specifically? Jul 17, 2020 at 22:52
• just his results have many infinite forms, which sound fun! @ChrisCunningham Jul 18, 2020 at 12:54
## 1 Answer
You can try the Rogers–Ramanujan identities:
• The number of partitions of $$n$$ in which adjacent parts are at least 2 apart is the same as the number of partitions of $$n$$ in which each part ends with 1,4,6,9. • The number of partitions of $$n$$ without 1 in which adjacent parts are at least 2 apart is the same as the number of partitions of $$n$$ in which each part ends with 2,3,7,8.
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For example, taking $$n=10$$:
• Partitions in which adjacent parts are at least 2 apart: $$10 = 10 = 9 + 1 = 8 + 2 = 7 + 3 = 6 + 4 = 6 + 3 + 1$$
• Partitions in which each part ends with 1,4,6,9: $$10 = 9 + 1 = 6 + 4 = 6 + 1 + 1 + 1 + 1 = 4 + 4 + 1 + 1 = 4 + 6\times 1 = 10 \times 1$$
• Partitions without 1 in which adjacent parts are at least 2 apart: $$10 = 10 = 8 + 2 = 7 + 3 = 6 + 4$$
• Partitions in which each part ends with 2,3,5,8: $$10 = 8 + 2 = 7 + 3 = 3 + 3 + 2 + 2 = 2 + 2 + 2 + 2 + 2$$
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https://math.stackexchange.com/questions/1575088/find-the-value-of-the-series-sum-limits-n-1-infty-fracn2n
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# Find the value of the series $\sum\limits_{n=1}^ \infty \frac{n}{2^n}$ [duplicate]
Find the value of the series $\sum\limits_{n=1}^ \infty \dfrac{n}{2^n}$
The series on expanding is coming as $\dfrac{1}{2}+\dfrac{2}{2^2}+..$
I tried using the form of $(1+x)^n=1+nx+\dfrac{n(n-1)}{2}x^2+..$ and then differentiating it but still it is not coming .What shall I do with this?
• This might help
– user297008
Dec 14, 2015 at 12:36
• Looks like the derivative of a geometric series to me Dec 14, 2015 at 12:37
• See this for other ideas. Dec 14, 2015 at 12:38
• Just differentiate $\frac{1}{2(1-x)}=\frac12\sum x^n$ and set $x=\frac12$. Dec 14, 2015 at 12:39
$$\sum_{n=1}^{\infty}\frac{n}{2^n}=\lim_{m\to\infty}\sum_{n=1}^{m}\frac{n}{2^n}=\lim_{m\to\infty}\frac{-m+2^{m+1}-2}{2^m}=$$ $$\lim_{m\to\infty}\frac{-2^{1-m}+2-2^{-m}m}{1}=\frac{0+2-0}{1}=\frac{2}{1}=2$$
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# Find the value of the series $\sum\limits_{n=1}^ \infty \frac{n}{2^n}$ [duplicate]
Find the value of the series $\sum\limits_{n=1}^ \infty \dfrac{n}{2^n}$
The series on expanding is coming as $\dfrac{1}{2}+\dfrac{2}{2^2}+..$
I tried using the form of $(1+x)^n=1+nx+\dfrac{n(n-1)}{2}x^2+..$ and then differentiating it but still it is not coming .What shall I do with this? • This might help
– user297008
Dec 14, 2015 at 12:36
• Looks like the derivative of a geometric series to me Dec 14, 2015 at 12:37
• See this for other ideas. Dec 14, 2015 at 12:38
• Just differentiate $\frac{1}{2(1-x)}=\frac12\sum x^n$ and set $x=\frac12$.
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Dec 14, 2015 at 12:39
$$\sum_{n=1}^{\infty}\frac{n}{2^n}=\lim_{m\to\infty}\sum_{n=1}^{m}\frac{n}{2^n}=\lim_{m\to\infty}\frac{-m+2^{m+1}-2}{2^m}=$$ $$\lim_{m\to\infty}\frac{-2^{1-m}+2-2^{-m}m}{1}=\frac{0+2-0}{1}=\frac{2}{1}=2$$
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https://cstheory.stackexchange.com/questions/46185/computing-3d-viewpoint-of-a-set-of-non-intersecting-segments
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# Computing 3D viewpoint of a set of non-intersecting segments
Consider the following problem: we are given a finite set of bounded line-segments in $${\mathbb R}^3$$, and we want to decide whether there exists a point $$p\in {\mathbb R}^3$$ from which no two segments obscure one another.
Can this be done efficiently?
### Problem statement:
More precisely and formally: we are given $$n$$ line segments $$\ell_1,\ldots,\ell_n$$, where each segment is defined as $$\ell_i=\{tu_i+(1-t) v_i: t\in [0,1]\}$$ with $$u_i,v_i\in {\mathbb Q}^3$$ (we assume rational coordinates).
We wish to decide whether there exists a point $$p\in {\mathbb R}^3$$ such the lines connecting $$p$$ with each point on the lines are distinct, and if so, compute it.
Is there an efficient solution? Is there a hardness lower bound?
UPDATE: Given the lack of answers so far, what about the case where the line segments connect two adjacent point in a 3D $$k\times k\times k$$ grid ? Then, they are all parallel to some axis, they are all of length 1, etc. Does this make it significantly easier?
### Inefficient solution:
Observe that for each pair of lines $$\ell_1,\ell_2$$, the points from which the lines do obscure each other can be described as a polyhedron defined as the intersection of 4 half-spaces: for every 3 points in {u_1,v_1,u_2,v_2}, the hyperplane defined by them is a boundary such that on one side of it, the lines do not obscure each other.
Thus, we can represent the set of "bad points" (those from which at least one pair obscure each other) as a union of $$n^2$$ polyhedra (not necessarily disjoint). Then, all we need is to test it's complement for emptiness. This can be done e.g. using Fourier-Motzkin quantifier elimination, whose complexity is quite bad. On top of this, we first need to convert a CNF representation to DNF, which may involve an exponential blowup.
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# Computing 3D viewpoint of a set of non-intersecting segments
Consider the following problem: we are given a finite set of bounded line-segments in $${\mathbb R}^3$$, and we want to decide whether there exists a point $$p\in {\mathbb R}^3$$ from which no two segments obscure one another. Can this be done efficiently? ### Problem statement:
More precisely and formally: we are given $$n$$ line segments $$\ell_1,\ldots,\ell_n$$, where each segment is defined as $$\ell_i=\{tu_i+(1-t) v_i: t\in [0,1]\}$$ with $$u_i,v_i\in {\mathbb Q}^3$$ (we assume rational coordinates). We wish to decide whether there exists a point $$p\in {\mathbb R}^3$$ such the lines connecting $$p$$ with each point on the lines are distinct, and if so, compute it. Is there an efficient solution? Is there a hardness lower bound? UPDATE: Given the lack of answers so far, what about the case where the line segments connect two adjacent point in a 3D $$k\times k\times k$$ grid ? Then, they are all parallel to some axis, they are all of length 1, etc. Does this make it significantly easier? ### Inefficient solution:
Observe that for each pair of lines $$\ell_1,\ell_2$$, the points from which the lines do obscure each other can be described as a polyhedron defined as the intersection of 4 half-spaces: for every 3 points in {u_1,v_1,u_2,v_2}, the hyperplane defined by them is a boundary such that on one side of it, the lines do not obscure each other.
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Thus, we can represent the set of "bad points" (those from which at least one pair obscure each other) as a union of $$n^2$$ polyhedra (not necessarily disjoint).
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https://math.stackexchange.com/questions/3186954/to-find-an-orthonormal-basis-for-the-row-space-of-a
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# To find an orthonormal basis for the row space of $A$.
To find an orthonormal basis for the row space of $$A = \begin{bmatrix} 2 & -1 & -3 \\ -5 & 5 & 3 \\ \end{bmatrix}$$.
Let $$v_1 = (2\ -1 \ -3)$$ and $$v_2 = (-5 \ \ \ 5 \ \ \ 3)$$.
Using the Gram-Schmidt Process, I found an orthonormal basis $$e_1 = \frac{1}{\sqrt{14}} (2\ -1 \ -3)$$ and $$e_2 = \frac{1}{\sqrt{5}} (-1 \ \ \ 2 \ \ \ 0)$$.
So an orthonormal basis for the row space of $$A =\{ e_1,e_2\}$$ .
IS the solution correct?
• Did you try checking if the two vectors you obtained are orthogonal (i.e. their dot product is $0$)? You should also probably show us the steps in your working, so we can see where you went wrong. – Minus One-Twelfth Apr 14 at 2:45
• Even more importantly, have you checked that $v_1$ and $v_2$ are actually elements of the row space? – amd Apr 14 at 3:31
## 1 Answer
Verify your Gram-Schmidt process again.
Note that we have $$V_1=X_1$$ and $$V_2 = X_2-\frac {X_2.V_1}{V_1.V_1}V_1$$
My calculations did not match with yours.
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# To find an orthonormal basis for the row space of $A$. To find an orthonormal basis for the row space of $$A = \begin{bmatrix} 2 & -1 & -3 \\ -5 & 5 & 3 \\ \end{bmatrix}$$. Let $$v_1 = (2\ -1 \ -3)$$ and $$v_2 = (-5 \ \ \ 5 \ \ \ 3)$$. Using the Gram-Schmidt Process, I found an orthonormal basis $$e_1 = \frac{1}{\sqrt{14}} (2\ -1 \ -3)$$ and $$e_2 = \frac{1}{\sqrt{5}} (-1 \ \ \ 2 \ \ \ 0)$$. So an orthonormal basis for the row space of $$A =\{ e_1,e_2\}$$ . IS the solution correct? • Did you try checking if the two vectors you obtained are orthogonal (i.e. their dot product is $0$)? You should also probably show us the steps in your working, so we can see where you went wrong. – Minus One-Twelfth Apr 14 at 2:45
• Even more importantly, have you checked that $v_1$ and $v_2$ are actually elements of the row space? – amd Apr 14 at 3:31
## 1 Answer
Verify your Gram-Schmidt process again.
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Note that we have $$V_1=X_1$$ and $$V_2 = X_2-\frac {X_2.V_1}{V_1.V_1}V_1$$
My calculations did not match with yours.
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https://dsp.stackexchange.com/questions/54772/system-function-h-omega-relationship-to-odd-and-even-components-of-hn
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# system function $H(\omega)$ relationship to odd and even components of h[n]
What qualities of $$h[n]$$ are necessary for:
$$H(e^{j\omega}) = DTFT\{h_{even}[n]\} + j\ DTFT\{h_{odd}[n]\}$$
Do all real / causal h[n] have the property that:
$$H(e^{j\omega}) = DTFT\{h_{even}[n]\} + j\ DTFT\{h_{odd}[n]\}$$
where:
$$h_{even}[n] = \frac{1}{2}(h[n] + h[-n])$$
$$h_{odd}[n] = \frac{1}{2}(h[n] - h[-n])$$
The DTFT relationships
$$x_{even}[n]=\frac12\left(x[n]+x^*[-n]\right)\Longleftrightarrow\textrm{Re}\left\{X(e^{j\omega})\right\}$$
and
$$x_{odd}[n]=\frac12\left(x[n]-x^*[-n]\right)\Longleftrightarrow j\,\textrm{Im}\left\{X(e^{j\omega})\right\}$$
hold for any sequence $$x[n]$$ for which the DTFT exists. There is no assumption about $$x[n]$$ being real-valued or causal (note the complex conjugation $$^*$$ in the definition of even and odd signals). If $$x[n]$$ is real-valued you can leave out the conjugation.
Note that the DTFT of the odd part $$x_{odd}[n]$$ equals $$j$$ times the imaginary part of the DTFT $$X(e^{j\omega})$$, so you have
$$X(e^{j\omega})=\textrm{DTFT}\{x_{even}[n]\}+\textrm{DTFT}\{x_{odd}[n]\}$$
(without a $$j$$ on the right-hand side).
• thanks, makes sense now. any suggestion for title? Jan 12 '19 at 15:55
• @MrCasuality: If your question has been answered you can accept this answer by clicking on the green check mark to its left, thanks. Jan 12 '19 at 17:07
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# system function $H(\omega)$ relationship to odd and even components of h[n]
What qualities of $$h[n]$$ are necessary for:
$$H(e^{j\omega}) = DTFT\{h_{even}[n]\} + j\ DTFT\{h_{odd}[n]\}$$
Do all real / causal h[n] have the property that:
$$H(e^{j\omega}) = DTFT\{h_{even}[n]\} + j\ DTFT\{h_{odd}[n]\}$$
where:
$$h_{even}[n] = \frac{1}{2}(h[n] + h[-n])$$
$$h_{odd}[n] = \frac{1}{2}(h[n] - h[-n])$$
The DTFT relationships
$$x_{even}[n]=\frac12\left(x[n]+x^*[-n]\right)\Longleftrightarrow\textrm{Re}\left\{X(e^{j\omega})\right\}$$
and
$$x_{odd}[n]=\frac12\left(x[n]-x^*[-n]\right)\Longleftrightarrow j\,\textrm{Im}\left\{X(e^{j\omega})\right\}$$
hold for any sequence $$x[n]$$ for which the DTFT exists. There is no assumption about $$x[n]$$ being real-valued or causal (note the complex conjugation $$^*$$ in the definition of even and odd signals). If $$x[n]$$ is real-valued you can leave out the conjugation.
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Note that the DTFT of the odd part $$x_{odd}[n]$$ equals $$j$$ times the imaginary part of the DTFT $$X(e^{j\omega})$$, so you have
$$X(e^{j\omega})=\textrm{DTFT}\{x_{even}[n]\}+\textrm{DTFT}\{x_{odd}[n]\}$$
(without a $$j$$ on the right-hand side).
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http://math.stackexchange.com/questions/99199/solution-of-fredholm-integral-equation-of-the-first-kind-with-symmetric-rational
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solution of Fredholm integral equation of the first kind with symmetric rational kernel
How can be solved this Fredholm first kind integral equation: $$f(x)=\frac{1}{\pi}\int_{0}^{\infty}\frac{g(y)}{x+y}dy$$
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The equation
$$f(x)=\frac{1}{\pi}\int_{0}^{\infty}\frac{g(y)}{x+y}\mathrm{d}y$$
has solution
\begin{align} y(x) &= \frac{1}{2 i} \lim_{\epsilon \to 0^+} \left\{f(-x-i\epsilon)-f(-x+i\epsilon)\right\} \\ &= \frac{1}{\sqrt{x}} \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!} \left(\frac{\pi}{x} \frac{\mathrm{d}}{\mathrm{d}x}\right)^{2k} \left\{\sqrt{x}f(x)\right\}. \end{align}
Source: Polyanin and Manzhirov, Handbook of Integral Equations, section 3.1-3, #17.
Numerous other sources are cited below the entry there.
-
You could try a Mellin transform. Since $\int _{0}^{\infty }\!{\frac {{x}^{s-1}}{x+y}}{dx}={y}^{s-1}\pi \,\csc \left( \pi \,s \right)$ for $y > 0$ and $0 < \Re s < 1$, the Mellin transforms of $f$ and $g$ satisfy $Mf(s) = \csc(\pi s) Mg(s)$ for $0 < \Re s < 1$. You might then try inverting $Mg(s)$ using the inversion formula
$$g(y) = \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} Mf(s) \sin(\pi s)\ ds$$
where $0 < c < 1$, under appropriate convergence assumptions.
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solution of Fredholm integral equation of the first kind with symmetric rational kernel
How can be solved this Fredholm first kind integral equation: $$f(x)=\frac{1}{\pi}\int_{0}^{\infty}\frac{g(y)}{x+y}dy$$
-
The equation
$$f(x)=\frac{1}{\pi}\int_{0}^{\infty}\frac{g(y)}{x+y}\mathrm{d}y$$
has solution
\begin{align} y(x) &= \frac{1}{2 i} \lim_{\epsilon \to 0^+} \left\{f(-x-i\epsilon)-f(-x+i\epsilon)\right\} \\ &= \frac{1}{\sqrt{x}} \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!} \left(\frac{\pi}{x} \frac{\mathrm{d}}{\mathrm{d}x}\right)^{2k} \left\{\sqrt{x}f(x)\right\}. \end{align}
Source: Polyanin and Manzhirov, Handbook of Integral Equations, section 3.1-3, #17. Numerous other sources are cited below the entry there. -
You could try a Mellin transform. Since $\int _{0}^{\infty }\! {\frac {{x}^{s-1}}{x+y}}{dx}={y}^{s-1}\pi \,\csc \left( \pi \,s \right)$ for $y > 0$ and $0 < \Re s < 1$, the Mellin transforms of $f$ and $g$ satisfy $Mf(s) = \csc(\pi s) Mg(s)$ for $0 < \Re s < 1$.
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You might then try inverting $Mg(s)$ using the inversion formula
$$g(y) = \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} Mf(s) \sin(\pi s)\ ds$$
where $0 < c < 1$, under appropriate convergence assumptions.
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http://math.stackexchange.com/questions/435026/algebraic-divison
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# Algebraic Divison
Is there a way to break the left hand side expression such that it takes the the right hand side form?
$(a+b)/(c+d)=a/c+b/d+k$
Where $k$ is some expression.
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Yes, and that expression would be $(a+b)/(c+d) - a/c - b/d$. Are you looking for something less stupid or more specific? – Patrick Da Silva Jul 3 '13 at 3:17
Solve for $k$, as Patrick indicated: \begin{align} k&=\frac{a+b}{c+d}-\frac{a}{c}-\frac{b}{d}\\ &=\frac{cd(a+b)-ad(c+d)-bc(c+d)}{cd(c+d)}\\ &=\frac{acd+bcd-acd-ad^2-bc^2-bcd}{cd(c+d)}\\ &=\frac{-ad^2-bc^2}{cd(c+d)} \end{align} In the words of lots of movie cops over the years, "Move along, folks, there's nothing to see here."
@jessica: one additional thing to note is that you need $c$ and $d$ non-zero. – James Jul 3 '13 at 13:21
@James: and also $c\ne-d$, else the original expression is undefined. – Rick Decker Jul 3 '13 at 13:42
Here's an old chestnut related to your problem. Take $64/16$, cancel the 6s, and you get $4/1$ which happens to be the right answer. Too bad it doesn't work in general. – Rick Decker Jul 4 '13 at 14:05
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# Algebraic Divison
Is there a way to break the left hand side expression such that it takes the the right hand side form? $(a+b)/(c+d)=a/c+b/d+k$
Where $k$ is some expression. -
Yes, and that expression would be $(a+b)/(c+d) - a/c - b/d$. Are you looking for something less stupid or more specific?
|
– Patrick Da Silva Jul 3 '13 at 3:17
Solve for $k$, as Patrick indicated: \begin{align} k&=\frac{a+b}{c+d}-\frac{a}{c}-\frac{b}{d}\\ &=\frac{cd(a+b)-ad(c+d)-bc(c+d)}{cd(c+d)}\\ &=\frac{acd+bcd-acd-ad^2-bc^2-bcd}{cd(c+d)}\\ &=\frac{-ad^2-bc^2}{cd(c+d)} \end{align} In the words of lots of movie cops over the years, "Move along, folks, there's nothing to see here."
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https://math.stackexchange.com/questions/2025090/finding-the-area-bounded-by-two-curves
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# Finding the area bounded by two curves
Find the area of the region bounded by the parabola $$y = 4x^2$$, the tangent line to this parabola at $$(2, 16)$$, and the $$x$$-axis.
I found the tangent line to be $$y=16x-16$$ and set up the integral from $$0$$ to $$2$$ of $$4x^2-16x+16$$ with respect to $$x$$, which is the top function when looking at the graph minus the bottom function. I took the integral and came up with $$\frac{4}{3}x^3-8x^2+16x$$ evaluated between $$0$$ and $$2$$. This came out to be $$\frac{32}{3}$$ but this was the incorrect answer. Can anyone tell me where I went wrong?
Hint: After drawing it, note that you have to calculate $\int_0^1 4x^2\;dx + \int_1^2 4x^2-16x+16\;dx$.
• I got $\frac{8}{3}$. I'm sorry but did you do it right? – Rodrigo Dias Nov 22 '16 at 0:22
• Any time! ${}{}$ – Rodrigo Dias Nov 22 '16 at 0:28
The tangent crosses the $x$ axis at $x=1$, so your integral is including (with the plus sign) also the triangle made by the tangent below the $x$ axis.
The correct way is to integrate only the parabola for $x=0 \cdots 2$ (which is $32/3$ and then subtract the area of the triangle$(1,0),(2,16),(2,0)$, which is $8$, so the net area is $8/3$ .
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# Finding the area bounded by two curves
Find the area of the region bounded by the parabola $$y = 4x^2$$, the tangent line to this parabola at $$(2, 16)$$, and the $$x$$-axis. I found the tangent line to be $$y=16x-16$$ and set up the integral from $$0$$ to $$2$$ of $$4x^2-16x+16$$ with respect to $$x$$, which is the top function when looking at the graph minus the bottom function. I took the integral and came up with $$\frac{4}{3}x^3-8x^2+16x$$ evaluated between $$0$$ and $$2$$. This came out to be $$\frac{32}{3}$$ but this was the incorrect answer. Can anyone tell me where I went wrong? Hint: After drawing it, note that you have to calculate $\int_0^1 4x^2\;dx + \int_1^2 4x^2-16x+16\;dx$. • I got $\frac{8}{3}$. I'm sorry but did you do it right? – Rodrigo Dias Nov 22 '16 at 0:22
• Any time! ${}{}$ – Rodrigo Dias Nov 22 '16 at 0:28
The tangent crosses the $x$ axis at $x=1$, so your integral is including (with the plus sign) also the triangle made by the tangent below the $x$ axis.
|
The correct way is to integrate only the parabola for $x=0 \cdots 2$ (which is $32/3$ and then subtract the area of the triangle$(1,0),(2,16),(2,0)$, which is $8$, so the net area is $8/3$ .
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https://math.stackexchange.com/questions/1735139/how-to-reduce-into-canonical-form/1735960
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# How to reduce into canonical form
Determine the type of the following equation and reduce the PDE to its canonical form $u_{xx} + 4u_{xy} + 4u_{yy} + u = 0$.
We consider pdes in the form $$a_{11}(x,y)u_{xx}+2a_{12}(x,y)u_{xy}+a_{22}(x,y)u_{yy} +F(x,y,u,u_x,u_y)=0$$
Since in our case $a_{12}^2-a_{11}a_{22}=0$, we have that it is parabolic.
Then I think we find $$\frac{dy}{dx}=\frac{a_{12} \pm \sqrt{a_{12}^2-a_{11}a_{22}}}{a_{11}}=2$$ But then what?
Define $\eta \left( x,y \right)=y-2x,\text{ }$ and choose $\xi \left( x,\text{ }y \right)=x$ such that the Jacobian $J:={{\xi }_{x}}{{\eta }_{y}}-{{\xi }_{y}}{{\eta }_{x}}$ does not vanish. Let $v\left( \xi ,\eta \right)=u\left( x,y \right).$ Substituting the new coordinates $\xi$ and $\eta$ into the given equation, we obtain
$$\left( {{v}_{\xi \xi }}-4{{v}_{\xi \eta }}+4{{v}_{\eta \eta }} \right)+4\left( {{v}_{\xi \eta }}-2{{v}_{\eta \eta }} \right)+4{{v}_{\eta \eta }}+v=0\text{ }.$$ Thus,
$${{v}_{\xi \xi }}+v=0,$$ and this is the desired canonical form.
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# How to reduce into canonical form
Determine the type of the following equation and reduce the PDE to its canonical form $u_{xx} + 4u_{xy} + 4u_{yy} + u = 0$. We consider pdes in the form $$a_{11}(x,y)u_{xx}+2a_{12}(x,y)u_{xy}+a_{22}(x,y)u_{yy} +F(x,y,u,u_x,u_y)=0$$
Since in our case $a_{12}^2-a_{11}a_{22}=0$, we have that it is parabolic. Then I think we find $$\frac{dy}{dx}=\frac{a_{12} \pm \sqrt{a_{12}^2-a_{11}a_{22}}}{a_{11}}=2$$ But then what? Define $\eta \left( x,y \right)=y-2x,\text{ }$ and choose $\xi \left( x,\text{ }y \right)=x$ such that the Jacobian $J:={{\xi }_{x}}{{\eta }_{y}}-{{\xi }_{y}}{{\eta }_{x}}$ does not vanish.
|
Let $v\left( \xi ,\eta \right)=u\left( x,y \right).$ Substituting the new coordinates $\xi$ and $\eta$ into the given equation, we obtain
$$\left( {{v}_{\xi \xi }}-4{{v}_{\xi \eta }}+4{{v}_{\eta \eta }} \right)+4\left( {{v}_{\xi \eta }}-2{{v}_{\eta \eta }} \right)+4{{v}_{\eta \eta }}+v=0\text{ }.$$ Thus,
$${{v}_{\xi \xi }}+v=0,$$ and this is the desired canonical form.
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https://math.stackexchange.com/questions/1653694/proving-that-the-sequence-converges
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# Proving that the sequence converges
I would like some help with the following problem. Thanks for any help in advance.
Let $(x_n)$ and $(y_n)$ be convergent sequences of positive real numbers. Let $x_n \xrightarrow[n \to \infty]{} x$ and $y_n \xrightarrow[n \to \infty]{} y$ and suppose that $x > 0$. Prove that the sequence $(x_nn^4 + y_nn^2)^{1/2} - x_n^{1/2}n^2$, $n \geq 1$ converges.
• The universal procedure, multiply the top and missing bottom by $(x_n n^4+y_n^2)^{1/2} +x_n^{1/2}n^2$. – André Nicolas Feb 13 '16 at 21:23
$\begin{array}\\ (x_nn^4 + y_nn^2)^{1/2} - x_n^{1/2}n^2 &=((x_nn^4 + y_nn^2)^{1/2} - x_n^{1/2}n^2) \dfrac{(x_nn^4 + y_nn^2)^{1/2} + x_n^{1/2}n^2}{(x_nn^4 + y_nn^2)^{1/2} + x_n^{1/2}n^2}\\ &=\dfrac{(x_nn^4 + y_nn^2)- x_nn^4}{(x_nn^4 + y_nn^2)^{1/2} + x_n^{1/2}n^2}\\ &=\dfrac{ y_nn^2}{(x_nn^4 + y_nn^2)^{1/2} + x_n^{1/2}n^2}\\ &=\dfrac{ y_n}{(x_n + y_n/n^2)^{1/2} + x_n^{1/2}}\\ \end{array}$
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# Proving that the sequence converges
I would like some help with the following problem. Thanks for any help in advance. Let $(x_n)$ and $(y_n)$ be convergent sequences of positive real numbers. Let $x_n \xrightarrow[n \to \infty]{} x$ and $y_n \xrightarrow[n \to \infty]{} y$ and suppose that $x > 0$. Prove that the sequence $(x_nn^4 + y_nn^2)^{1/2} - x_n^{1/2}n^2$, $n \geq 1$ converges. • The universal procedure, multiply the top and missing bottom by $(x_n n^4+y_n^2)^{1/2} +x_n^{1/2}n^2$.
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– André Nicolas Feb 13 '16 at 21:23
$\begin{array}\\ (x_nn^4 + y_nn^2)^{1/2} - x_n^{1/2}n^2 &=((x_nn^4 + y_nn^2)^{1/2} - x_n^{1/2}n^2) \dfrac{(x_nn^4 + y_nn^2)^{1/2} + x_n^{1/2}n^2}{(x_nn^4 + y_nn^2)^{1/2} + x_n^{1/2}n^2}\\ &=\dfrac{(x_nn^4 + y_nn^2)- x_nn^4}{(x_nn^4 + y_nn^2)^{1/2} + x_n^{1/2}n^2}\\ &=\dfrac{ y_nn^2}{(x_nn^4 + y_nn^2)^{1/2} + x_n^{1/2}n^2}\\ &=\dfrac{ y_n}{(x_n + y_n/n^2)^{1/2} + x_n^{1/2}}\\ \end{array}$
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https://math.stackexchange.com/questions/2022498/area-of-a-circle-from-equation-for-circle
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# Area of a circle from equation for circle
Q. Equation of circle- $2x^2+ \lambda xy+2y^2+( \lambda -4)x+6y-5=0$ find area of the circle.
Attempt- For converting the equation from second degree to first degree $\lambda xy=0$.
Thus, $\lambda =0$ and-
$$(\lambda -4)x = 2gx$$ $$6y=2fy$$ $$c=-5$$ $$g=-2, f=3, c=-5$$
Radius of circle = $\sqrt{4+9+5}=\sqrt{18}$
Area of circle= $\pi *18$
But the answer is $\frac {23}{4} * \pi$
• Your argument would be correct if the coefficients of $x^2$ and $y^2$ were $1$. But they are not. – Leo163 Nov 20 '16 at 11:56
• Do they need to be equal? Correct me but shouldn't just their coefficient be equal? To satisfy $a=b$ where a and b are coefficients of x and y respectively – Akshat Batra Nov 20 '16 at 11:58
• @AkshatBatra Yes, the coefficients of the quadratic expressions for $\;x\,,\,\,y\;$ must be equal if we have a circle (otherwise it is an ellipse), but then you must divide through the whole equation by that common coefficient, and that affects the radius...! – DonAntonio Nov 20 '16 at 11:59
Complete squares after putting $\;\lambda xy=0\implies \lambda =0\;$:
$$0=2x^2+2y^2-4x+6y-5=2(x-1)^2-2+2\left(y-\frac32\right)^2-\frac92-5\implies$$
$$\implies2(x-1)^2+2\left(y-\frac32\right)^2=\frac{23}2\implies(x-1)^2+\left(y-\frac32\right)^2=\frac{23}4$$
and we have a circle of radius $\;\sqrt{\frac{23}4}\;$ , so its area is
$$\pi\sqrt{\frac{23}4}^2=\frac{23\pi}4$$
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# Area of a circle from equation for circle
Q. Equation of circle- $2x^2+ \lambda xy+2y^2+( \lambda -4)x+6y-5=0$ find area of the circle. Attempt- For converting the equation from second degree to first degree $\lambda xy=0$. Thus, $\lambda =0$ and-
$$(\lambda -4)x = 2gx$$ $$6y=2fy$$ $$c=-5$$ $$g=-2, f=3, c=-5$$
Radius of circle = $\sqrt{4+9+5}=\sqrt{18}$
Area of circle= $\pi *18$
But the answer is $\frac {23}{4} * \pi$
• Your argument would be correct if the coefficients of $x^2$ and $y^2$ were $1$. But they are not. – Leo163 Nov 20 '16 at 11:56
• Do they need to be equal? Correct me but shouldn't just their coefficient be equal? To satisfy $a=b$ where a and b are coefficients of x and y respectively – Akshat Batra Nov 20 '16 at 11:58
• @AkshatBatra Yes, the coefficients of the quadratic expressions for $\;x\,,\,\,y\;$ must be equal if we have a circle (otherwise it is an ellipse), but then you must divide through the whole equation by that common coefficient, and that affects the radius...!
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– DonAntonio Nov 20 '16 at 11:59
Complete squares after putting $\;\lambda xy=0\implies \lambda =0\;$:
$$0=2x^2+2y^2-4x+6y-5=2(x-1)^2-2+2\left(y-\frac32\right)^2-\frac92-5\implies$$
$$\implies2(x-1)^2+2\left(y-\frac32\right)^2=\frac{23}2\implies(x-1)^2+\left(y-\frac32\right)^2=\frac{23}4$$
and we have a circle of radius $\;\sqrt{\frac{23}4}\;$ , so its area is
$$\pi\sqrt{\frac{23}4}^2=\frac{23\pi}4$$
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http://math.stackexchange.com/questions/217621/big-omega-proof/217623
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Big Omega Proof
If $f_1(x)$ and $f_2(x)$ are functions from the set of positive integers to the set of positive real numbers and $f_1(x)$ and $f_2(x)$ are both $\Omega(g(x))$, is $(f_1 − f_2)(x)$ also $\Omega(g(x))$?
How do I prove/disprove this?
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No. It is not. For instance, consider $f_1(x) = f_2(x) = x^2$ and $g(x) = x$.
For a slightly non-trivial example, consider $$f_1(x) = x^3 + x^2 + x + 1, f_2(x) = x^3 + x^2 + 1 \text{ and }g(x) = x^2$$ We have $f_1(x),f_2(x) \in \Omega(g(x))$ but $f_1(x) -f_2(x) = x \notin \Omega(g(x))$.
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Big Omega Proof
If $f_1(x)$ and $f_2(x)$ are functions from the set of positive integers to the set of positive real numbers and $f_1(x)$ and $f_2(x)$ are both $\Omega(g(x))$, is $(f_1 − f_2)(x)$ also $\Omega(g(x))$? How do I prove/disprove this? -
No. It is not. For instance, consider $f_1(x) = f_2(x) = x^2$ and $g(x) = x$.
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For a slightly non-trivial example, consider $$f_1(x) = x^3 + x^2 + x + 1, f_2(x) = x^3 + x^2 + 1 \text{ and }g(x) = x^2$$ We have $f_1(x),f_2(x) \in \Omega(g(x))$ but $f_1(x) -f_2(x) = x \notin \Omega(g(x))$.
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http://math.stackexchange.com/questions/180895/subgroup-criterion/180897
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# Subgroup criterion.
I've been reading some stuff about algebra in my free time, and I think I understand most of the stuff but I'm having trouble with the exercises. Specifically, the following:
Prove that a nonempty subset $H$ of a group $G$ is a subgroup if for all $x, y \in H$, the element $xy^{-1}$ is also in H.
Proving that the identity is in $H$ is easy: just take $x=y$, so $x x^{-1} = 1 \in H$. However, I'm having trouble showing that multiplication is closed and that each element in $H$ has an inverse. Can anyone give some hints?
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Hint: For inverses use the fact that the identity is in $H$. Once you have inverses, you can get products using that fact. – Matt Aug 10 '12 at 1:57
1. by given condition for any $x\in H$ we have $xx^{-1}=e$ is in $H$, denote identity element by $e$
2. take any $x\in H$ and $e\in H$ so by the given condition $ex^{-1}=x^{-1}\in H$ so every element of $H$ has inverse in $H$.
3. take any $x,y\in H$ as $y^{-1}\in H$ so by given condition $x(y^{-1})^{-1}=xy\in H$, which proves the closure property.
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So much for hints... – Matt Aug 10 '12 at 2:02
For any $b \in H$, $eb^{-1} = b^{-1} \in H$, so every element has an inverse in $H$. To show closure, note that if $a, b \in H$, then $b^{-1} \in H$ as we have just shown. So $a(b^{-1})^{-1} = ab \in H$. Hence, $H$ is a subgroup.
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# Subgroup criterion. I've been reading some stuff about algebra in my free time, and I think I understand most of the stuff but I'm having trouble with the exercises. Specifically, the following:
Prove that a nonempty subset $H$ of a group $G$ is a subgroup if for all $x, y \in H$, the element $xy^{-1}$ is also in H.
Proving that the identity is in $H$ is easy: just take $x=y$, so $x x^{-1} = 1 \in H$. However, I'm having trouble showing that multiplication is closed and that each element in $H$ has an inverse. Can anyone give some hints? -
Hint: For inverses use the fact that the identity is in $H$. Once you have inverses, you can get products using that fact. – Matt Aug 10 '12 at 1:57
1. by given condition for any $x\in H$ we have $xx^{-1}=e$ is in $H$, denote identity element by $e$
2. take any $x\in H$ and $e\in H$ so by the given condition $ex^{-1}=x^{-1}\in H$ so every element of $H$ has inverse in $H$. 3. take any $x,y\in H$ as $y^{-1}\in H$ so by given condition $x(y^{-1})^{-1}=xy\in H$, which proves the closure property. -
So much for hints... – Matt Aug 10 '12 at 2:02
For any $b \in H$, $eb^{-1} = b^{-1} \in H$, so every element has an inverse in $H$. To show closure, note that if $a, b \in H$, then $b^{-1} \in H$ as we have just shown. So $a(b^{-1})^{-1} = ab \in H$.
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Hence, $H$ is a subgroup.
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https://quantumcomputing.stackexchange.com/questions/5240/whats-my-computational-basis-if-i-want-to-define-a-unitary-operator-that-implem
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What's my computational basis if I want to define a unitary operator that implements a function such as $f(i) = 2^{i+1} \text{mod 21}$?
I know I must define $$U_f$$, the unitary operator, on the computational basis. But what's my computational basis here?
Presumably you want to work with qubits? So the usual computation basis applies: $$|0\rangle$$ and $$|1\rangle$$ for a single qubit, and a composite basis of $$|x\rangle$$ for $$x\in\{0,1\}^n$$ when composing $$n$$ qubits.
What I guess you're asking is how you translate your problem onto qubits. For that, you need to make the decimal values correspond to particular bit values. The conventional way of doing this is using the binary string $$x$$ like a binary number, which corresponds to a decimal value. There are a number of different numbering conventions you can pick from but, for example, you might have $$x=x_0x_1x_2\ldots x_{n-1},$$ meaning that the corresponding decimal value is $$x_{n-1}+2x_{n-2}+4x_{n-3}+\ldots+2^{n-2}x_1+2^{n-1}x_0.$$ $$n$$ bits lets you represent any decimal number 0 to $$2^n-1$$.
It depends on the $$i$$ you want to apply this function, which is to be represented as a bitstring (in this case, I guess an unsigned integer representation) that will be one of the computational basis. It can be a particular one or just a superposition. You need to define the registers (the one containing $$i$$, garbage register containing intermediary results and the one containing the result). Mathematically, your operator will have the following effect if you uncompute intermediary results : $$U_f | i \rangle | 0 \rangle_g | 0 \rangle_f = | i \rangle | 0 \rangle_g | f(i) \rangle_f$$
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What's my computational basis if I want to define a unitary operator that implements a function such as $f(i) = 2^{i+1} \text{mod 21}$? I know I must define $$U_f$$, the unitary operator, on the computational basis. But what's my computational basis here? Presumably you want to work with qubits? So the usual computation basis applies: $$|0\rangle$$ and $$|1\rangle$$ for a single qubit, and a composite basis of $$|x\rangle$$ for $$x\in\{0,1\}^n$$ when composing $$n$$ qubits. What I guess you're asking is how you translate your problem onto qubits. For that, you need to make the decimal values correspond to particular bit values. The conventional way of doing this is using the binary string $$x$$ like a binary number, which corresponds to a decimal value. There are a number of different numbering conventions you can pick from but, for example, you might have $$x=x_0x_1x_2\ldots x_{n-1},$$ meaning that the corresponding decimal value is $$x_{n-1}+2x_{n-2}+4x_{n-3}+\ldots+2^{n-2}x_1+2^{n-1}x_0.$$ $$n$$ bits lets you represent any decimal number 0 to $$2^n-1$$. It depends on the $$i$$ you want to apply this function, which is to be represented as a bitstring (in this case, I guess an unsigned integer representation) that will be one of the computational basis. It can be a particular one or just a superposition. You need to define the registers (the one containing $$i$$, garbage register containing intermediary results and the one containing the result).
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Mathematically, your operator will have the following effect if you uncompute intermediary results : $$U_f | i \rangle | 0 \rangle_g | 0 \rangle_f = | i \rangle | 0 \rangle_g | f(i) \rangle_f$$
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https://math.stackexchange.com/questions/3233705/random-variable-y-following-uniform-distribution-with-parameter-random-x-that-fo
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# Random Variable Y following uniform distribution with parameter Random X that follows geometric.
Random variable X follows geometrical distribution with p=1/4. Random variable Y follows uniform distribution in [-X,X]. I'm looking for P(Y>3/2) and also P(X=2|Y>3/2).I know for a fact that Σ(from k=1 to infinity)zk/k=-log(1-z) for |z|<1.
The probability that $$Y \gt 3/2$$ is actually the probability that $$3/2 \lt Y \lt X$$.
If $$X \lt 1$$ then it is impossible for $$Y$$ to be greater than $$3/2$$.
Therefore $$X$$ must be at least $$2$$ for the probability to even make sense. So we have $$\mathbb{P} \left(Y \gt \frac32 \right) = \sum_{x=2}^\infty \left( \frac14 \right)\left( \frac34 \right) ^{x-1} \left( \frac{x - 3/2}{2x}\right)$$
• No, it is possible for $Y>1.5$ when $X=2$ – Graham Kemp May 20 at 23:19
• @Graham Kemp True, I simply misread it as $Y > 3$ when I solved the problem. Will fix – WaveX May 20 at 23:20
• This sum converges to about $.2159$ – WaveX May 22 at 17:39
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# Random Variable Y following uniform distribution with parameter Random X that follows geometric. Random variable X follows geometrical distribution with p=1/4. Random variable Y follows uniform distribution in [-X,X]. I'm looking for P(Y>3/2) and also P(X=2|Y>3/2).I know for a fact that Σ(from k=1 to infinity)zk/k=-log(1-z) for |z|<1. The probability that $$Y \gt 3/2$$ is actually the probability that $$3/2 \lt Y \lt X$$. If $$X \lt 1$$ then it is impossible for $$Y$$ to be greater than $$3/2$$. Therefore $$X$$ must be at least $$2$$ for the probability to even make sense.
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So we have $$\mathbb{P} \left(Y \gt \frac32 \right) = \sum_{x=2}^\infty \left( \frac14 \right)\left( \frac34 \right) ^{x-1} \left( \frac{x - 3/2}{2x}\right)$$
• No, it is possible for $Y>1.5$ when $X=2$ – Graham Kemp May 20 at 23:19
• @Graham Kemp True, I simply misread it as $Y > 3$ when I solved the problem.
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https://math.stackexchange.com/questions/1732130/discriminant-of-the-cube-quartic
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# Discriminant of the cube, quartic...
I was told the discriminant of the cubic is
$$\Delta=-27q^2-4p^3$$
and that $\Delta>0$ means that there are three real roots. Simply put, why is this the discriminant? I ask this because, looking at Cardano's formula, I thought that we want everything inside the square root to be positive to get real roots(just as in quadratic cases).
Namely, $\frac{q^2}{4}+\frac{p^3}{27}>0$. Which is essentially, $27q^2+4p^3>0$. But the discriminant has a minus on it, and I don't see why. Does the derivation involve resultants and whatnot? Will it be rather complex?
I am wondering if there is a simple explanation as to why this is the case. Similarly, for the quartic, quintic discriminants...will I need to go through resultants for them? Or is there a simpler faster way to determine them?
Your discriminant refers to a cubic equation reduced to the form $$y^3+py+q=0$$ and the Cardano formula says that the solutions are: $$y=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}+\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$$
Note that if $$\Delta=\frac{q^2}{4}+\frac{p^3}{27}>0$$ the arguments of the cubic roots are real numbers, so the roots give only one real value ( and two complex values), so the other solutions of the equation must be complex numbers and we cannot have three real roots.
The proof that for $\Delta \le 0$ we find three real roots is not so simple, you can see here.
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# Discriminant of the cube, quartic... I was told the discriminant of the cubic is
$$\Delta=-27q^2-4p^3$$
and that $\Delta>0$ means that there are three real roots. Simply put, why is this the discriminant? I ask this because, looking at Cardano's formula, I thought that we want everything inside the square root to be positive to get real roots(just as in quadratic cases). Namely, $\frac{q^2}{4}+\frac{p^3}{27}>0$. Which is essentially, $27q^2+4p^3>0$. But the discriminant has a minus on it, and I don't see why. Does the derivation involve resultants and whatnot? Will it be rather complex? I am wondering if there is a simple explanation as to why this is the case. Similarly, for the quartic, quintic discriminants...will I need to go through resultants for them? Or is there a simpler faster way to determine them?
|
Your discriminant refers to a cubic equation reduced to the form $$y^3+py+q=0$$ and the Cardano formula says that the solutions are: $$y=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}+\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$$
Note that if $$\Delta=\frac{q^2}{4}+\frac{p^3}{27}>0$$ the arguments of the cubic roots are real numbers, so the roots give only one real value ( and two complex values), so the other solutions of the equation must be complex numbers and we cannot have three real roots.
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https://math.stackexchange.com/questions/1463140/proof-for-why-a-matrix-multiplied-by-its-transpose-is-positive-semidefinite
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# Proof for why a matrix multiplied by its transpose is positive semidefinite
The top answer to this question says
Moreover if $A$ is regular, then $AA^T$ is also positive definite, since $$x^TAA^Tx=(A^Tx)^T(A^Tx)> 0$$
Suppose $A$ is not regular. It holds that $$x^TAA^Tx=(A^Tx)^T(A^Tx)= \|A^Tx\|^2_2 \ge 0$$ Therefore $AA^T$ is positive semidefinite. Is this argument enough, or am I missing something?
• Yes, that's enough. Oct 3, 2015 at 23:35
• Two comments: 1) Usually, the definition of a positive semidefinite matrix includes the requirement that $A$ is symmetric (or hermitian for complex matrices). You did not check that. 2) Your argument shows that $A^T A$ is positive semidefinite. It does not show that $A^T A$ is not positive definite. Oct 4, 2015 at 10:07
• What is does it mean that "A is regular" in this context?
– Itay
Sep 17, 2016 at 9:32
• It means the same as invertible. So if $A$ is not invertible, then there are $x$ other than $0$ for which $Ax=0$ and thus strict inequality doesn't hold. On the other hand, if $A$ is invertible (thus regular), then $Ax=0$ only holds for $x=0$ and thus strict inequality (definiteness) holds for all $x \ne 0$. I think that more generally in this case regular means that the columns of $A$ are independent. So $A$ doesn't have to be square. Feb 5, 2017 at 20:50
• I think this piece of answer should be added to the top answer of the linked question! Jan 7, 2022 at 15:44
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# Proof for why a matrix multiplied by its transpose is positive semidefinite
The top answer to this question says
Moreover if $A$ is regular, then $AA^T$ is also positive definite, since $$x^TAA^Tx=(A^Tx)^T(A^Tx)> 0$$
Suppose $A$ is not regular. It holds that $$x^TAA^Tx=(A^Tx)^T(A^Tx)= \|A^Tx\|^2_2 \ge 0$$ Therefore $AA^T$ is positive semidefinite. Is this argument enough, or am I missing something? • Yes, that's enough. Oct 3, 2015 at 23:35
• Two comments: 1) Usually, the definition of a positive semidefinite matrix includes the requirement that $A$ is symmetric (or hermitian for complex matrices). You did not check that. 2) Your argument shows that $A^T A$ is positive semidefinite. It does not show that $A^T A$ is not positive definite. Oct 4, 2015 at 10:07
• What is does it mean that "A is regular" in this context? – Itay
Sep 17, 2016 at 9:32
• It means the same as invertible. So if $A$ is not invertible, then there are $x$ other than $0$ for which $Ax=0$ and thus strict inequality doesn't hold.
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On the other hand, if $A$ is invertible (thus regular), then $Ax=0$ only holds for $x=0$ and thus strict inequality (definiteness) holds for all $x \ne 0$.
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https://math.stackexchange.com/questions/1555679/unexpectedly-uniformly-continuous-functions
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# Unexpectedly uniformly continuous functions
The other day in a exam, I was given the following exercise:
Given $f : [0,1] \to \mathbb{R}$ continuous and such that $f(0) = 0, f(1) = 1$, let $g : \mathbb{R} \to \mathbb{R}$ be $g(x) = [x] + f(x - [x])$. Prove that $g$ is uniformly continuous.
I'm looking for more examples of this kind of exercise to practice with (i.e. functions with uniform continuity that are not as straightforward to prove that they are).
Any continuous function $f$ from a closed interval $[a, b]$ to $[a, b]$ is uniformly continuous. See https://en.wikipedia.org/wiki/Heine–Cantor_theorem.
If $f(a) = a$ and $f(b) = b$ then extending $f$ to a function $\Bbb{R} \to \Bbb{R}$ by taking $f(x) = x$ for $x \not\in [a, b]$ still gives a uniformly continuous function.
• The theorem can be applied: you prove $g$ is continuous on $[0, 1]$ and conclude that it is uniformly continuous on $[0, 1]$. Dec 2, 2015 at 1:06
• You are supposed to prove that it is uniformly continuous on $\mathbb{R}$, which is not compact. Dec 2, 2015 at 1:10
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# Unexpectedly uniformly continuous functions
The other day in a exam, I was given the following exercise:
Given $f : [0,1] \to \mathbb{R}$ continuous and such that $f(0) = 0, f(1) = 1$, let $g : \mathbb{R} \to \mathbb{R}$ be $g(x) = [x] + f(x - [x])$. Prove that $g$ is uniformly continuous. I'm looking for more examples of this kind of exercise to practice with (i.e. functions with uniform continuity that are not as straightforward to prove that they are). Any continuous function $f$ from a closed interval $[a, b]$ to $[a, b]$ is uniformly continuous. See https://en.wikipedia.org/wiki/Heine–Cantor_theorem. If $f(a) = a$ and $f(b) = b$ then extending $f$ to a function $\Bbb{R} \to \Bbb{R}$ by taking $f(x) = x$ for $x \not\in [a, b]$ still gives a uniformly continuous function.
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• The theorem can be applied: you prove $g$ is continuous on $[0, 1]$ and conclude that it is uniformly continuous on $[0, 1]$.
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http://math.stackexchange.com/questions/81694/additive-inverse-of-a-nilpotent-element-is-nilpotent
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# Additive inverse of a nilpotent element is nilpotent
An element of a ring $R$ is nilpotent if $a^n=0$ for some $n \ge 1$.
How do I show that additive inverse of $a$ , $-a$ is also nilpotent?
The ring is commutative but may not have a unit element.
-
Using the distributive property, $ab+(-a)b=(a+(-a))b=0\cdot b=0$. Therefore, $$(-a)b=-(ab)\tag{1a}$$ Also, $ab+a(-b)=a(b+(-b))=a\cdot0=0$. Therefore, $$a(-b)=-(ab)\tag{1b}$$ Furthermore, since $a+(-a)=0$, we have $$-(-a)=a\tag{2}$$ Using $(1)$ and $(2)$, it is easy to show by induction that $$(-a)^k=\left\{\begin{array}{}a^k&\text{if }k\text{ is even}\\-(a^k)&\text{if }k\text{ is odd}\end{array}\right.\tag{3}$$ The fact that $a$ is nilpotent and $(3)$ shows that $-a$ is nilpotent.
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I am simply filling in the details of Kb100's suggestion and including $-(-a)=a$, which is needed to show $(3)$. – robjohn Nov 13 '11 at 18:01
Try first proving that $a(-b)=(-a)b=-(ab)$ and then (by induction) that $(-a)^n$ is $a^n$ if $n$ is even and $-(a^n)$ if $n$ is odd.
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# Additive inverse of a nilpotent element is nilpotent
An element of a ring $R$ is nilpotent if $a^n=0$ for some $n \ge 1$. How do I show that additive inverse of $a$ , $-a$ is also nilpotent? The ring is commutative but may not have a unit element. -
Using the distributive property, $ab+(-a)b=(a+(-a))b=0\cdot b=0$. Therefore, $$(-a)b=-(ab)\tag{1a}$$ Also, $ab+a(-b)=a(b+(-b))=a\cdot0=0$. Therefore, $$a(-b)=-(ab)\tag{1b}$$ Furthermore, since $a+(-a)=0$, we have $$-(-a)=a\tag{2}$$ Using $(1)$ and $(2)$, it is easy to show by induction that $$(-a)^k=\left\{\begin{array}{}a^k&\text{if }k\text{ is even}\\-(a^k)&\text{if }k\text{ is odd}\end{array}\right.\tag{3}$$ The fact that $a$ is nilpotent and $(3)$ shows that $-a$ is nilpotent. -
I am simply filling in the details of Kb100's suggestion and including $-(-a)=a$, which is needed to show $(3)$.
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– robjohn Nov 13 '11 at 18:01
Try first proving that $a(-b)=(-a)b=-(ab)$ and then (by induction) that $(-a)^n$ is $a^n$ if $n$ is even and $-(a^n)$ if $n$ is odd.
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https://engineering.stackexchange.com/questions/40847/to-stop-or-to-go-around
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# To stop or to go around?
Let us we are moving in a car. There is a wall in front of us, we need to decide can we go around it or not. It is known that width of the wall is $$w$$ and our speed is $$v = const$$.
What is a simple approach to set formally conditions upon which it is safe to go around?
I understand that there are many possible details such as traction, weight of car etc. and will be glad even for the most simplistic analysis. If you can provide a source, that too is great. Thanks.
• compute max turn radius, as function of w and distance. compute centripetal force as function of that radius and v. compare that force to max lateral force on tires before it slips – Pete W Mar 8 at 17:29
• If you go around it, you will still have a car. – StainlessSteelRat Mar 8 at 18:30
• the calculations for how quickly a car can turn are really too complex unless all you want is a general idea. This would be best accomplished by measuring the car's turning performance empirically. – Tiger Guy Mar 8 at 19:19
Each car depending on its handling has a maximum safe turning speed $$v_{max}$$, and radius $$r$$. Let us say the current speed $$v, then your turning angular velocity is $$\omega= v/r.$$ We need to go an arc of $$\pi/2$$ so the time it takes to turn is $$t=\pi/2\omega= \pi r/2v$$ which will give the decision distance $$x$$ as $$x=t\cdot v$$.
Above was for a wall wider than the cars cornering turn, if it is less, then the arc is smaller and we have $$\theta= arccos(r-\text{car-width})$$.
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# To stop or to go around? Let us we are moving in a car. There is a wall in front of us, we need to decide can we go around it or not. It is known that width of the wall is $$w$$ and our speed is $$v = const$$. What is a simple approach to set formally conditions upon which it is safe to go around? I understand that there are many possible details such as traction, weight of car etc. and will be glad even for the most simplistic analysis. If you can provide a source, that too is great. Thanks. • compute max turn radius, as function of w and distance. compute centripetal force as function of that radius and v. compare that force to max lateral force on tires before it slips – Pete W Mar 8 at 17:29
• If you go around it, you will still have a car. – StainlessSteelRat Mar 8 at 18:30
• the calculations for how quickly a car can turn are really too complex unless all you want is a general idea. This would be best accomplished by measuring the car's turning performance empirically. – Tiger Guy Mar 8 at 19:19
Each car depending on its handling has a maximum safe turning speed $$v_{max}$$, and radius $$r$$. Let us say the current speed $$v, then your turning angular velocity is $$\omega= v/r.$$ We need to go an arc of $$\pi/2$$ so the time it takes to turn is $$t=\pi/2\omega= \pi r/2v$$ which will give the decision distance $$x$$ as $$x=t\cdot v$$.
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Above was for a wall wider than the cars cornering turn, if it is less, then the arc is smaller and we have $$\theta= arccos(r-\text{car-width})$$.
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https://math.stackexchange.com/questions/4525144/when-does-a-limit-function-not-exist
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# When does a limit function not exist??
$$f(x) \begin{cases} \frac{3x-6}{x^4-4}, 0 < x < 2 \\ 0, x=2 \\ \frac{x-2}{\sqrt{3-x} -1} 2 < x < 3 \end{cases}$$
For this expression, can I say that these 3 cases, the limit does not exist because when $$x=2$$ or as $$x$$ approaches $$2$$ from left and right , all of the 3 functions have different values (when I substitute $$x=2$$) thus the limit does not have a finite and unique value. Is that right to say that? If not, when does a limit not exist?
1. $$\lim_{x \to 2^+}$$
2. $$\lim_{x \to 2^-}$$
3. $$\lim_{x \to 2}$$
The value of $$f(x)$$ when $$x=2$$ has nothing to do with the limit. The left hand limit ($$\frac 3{32})$$ and the right hand limit ($$-2$$) are different and that is enough to say that the limit does not exist.
1.$$\lim_{x \to 2^+} f(x) = -2$$ Use LHopital
2.$$\lim_{x \to 2^-} f(x) = \frac{3}{32}$$ Use LHopital
Since $$-2 \not = \frac{3}{32}$$
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# When does a limit function not exist?? $$f(x) \begin{cases} \frac{3x-6}{x^4-4}, 0 < x < 2 \\ 0, x=2 \\ \frac{x-2}{\sqrt{3-x} -1} 2 < x < 3 \end{cases}$$
For this expression, can I say that these 3 cases, the limit does not exist because when $$x=2$$ or as $$x$$ approaches $$2$$ from left and right , all of the 3 functions have different values (when I substitute $$x=2$$) thus the limit does not have a finite and unique value. Is that right to say that? If not, when does a limit not exist? 1. $$\lim_{x \to 2^+}$$
2. $$\lim_{x \to 2^-}$$
3. $$\lim_{x \to 2}$$
The value of $$f(x)$$ when $$x=2$$ has nothing to do with the limit. The left hand limit ($$\frac 3{32})$$ and the right hand limit ($$-2$$) are different and that is enough to say that the limit does not exist.
|
1.$$\lim_{x \to 2^+} f(x) = -2$$ Use LHopital
2.$$\lim_{x \to 2^-} f(x) = \frac{3}{32}$$ Use LHopital
Since $$-2 \not = \frac{3}{32}$$
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https://math.stackexchange.com/questions/1732225/evaluate-the-definite-integral-int-0-infty-fracx-sin-mxx2a2dx/1732246
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Evaluate the definite integral $\int_0^\infty \frac{x\sin mx}{x^2+a^2}dx$
Evaluate the definite integral $$\int_0^\infty \frac{x\sin mx}{x^2+a^2}dx \quad (m,a>0)$$
I tried a trigonometric substitution but did not get anywhere with that I think the multiple variable are throwing me off.
For first, we get rid of one parameter by substituting $x=az$. Then we have to compute: $$I(k) = \int_{0}^{+\infty}\frac{x}{x^2+1}\cdot\sin(kx)\,dx=\frac{1}{2}\int_{\mathbb{R}}\frac{x}{x^2+1}\cdot\sin(kx)\,dx$$ that is half the imaginary part of $\int_{\mathbb{R}}\frac{x e^{ikx}}{x^2+1}\,dx$. By computing the residue of the integrand function at $x=i$ it follows that: $$I(k) = \frac{\pi}{2}\cdot e^{-k},$$ hence:
$$\int_{0}^{+\infty}\frac{x\sin(mx)}{x^2+a^2}\,dx = \color{red}{\frac{\pi}{2}\cdot e^{-am}}.$$
• @Gaffney: the integration path should be a rectangle enclosing $x=i$, indeed. – Jack D'Aurizio Apr 7 '16 at 17:11
• @Gaffney: the situation is almost the same as the usual proof of $\int_{0}^{+\infty}\frac{\sin x}{x}\,dx=\frac{\pi}{2}$. Have a look at robjohn's proof here: math.stackexchange.com/questions/594641/… – Jack D'Aurizio Apr 7 '16 at 17:39
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Evaluate the definite integral $\int_0^\infty \frac{x\sin mx}{x^2+a^2}dx$
Evaluate the definite integral $$\int_0^\infty \frac{x\sin mx}{x^2+a^2}dx \quad (m,a>0)$$
I tried a trigonometric substitution but did not get anywhere with that I think the multiple variable are throwing me off. For first, we get rid of one parameter by substituting $x=az$. Then we have to compute: $$I(k) = \int_{0}^{+\infty}\frac{x}{x^2+1}\cdot\sin(kx)\,dx=\frac{1}{2}\int_{\mathbb{R}}\frac{x}{x^2+1}\cdot\sin(kx)\,dx$$ that is half the imaginary part of $\int_{\mathbb{R}}\frac{x e^{ikx}}{x^2+1}\,dx$. By computing the residue of the integrand function at $x=i$ it follows that: $$I(k) = \frac{\pi}{2}\cdot e^{-k},$$ hence:
$$\int_{0}^{+\infty}\frac{x\sin(mx)}{x^2+a^2}\,dx = \color{red}{\frac{\pi}{2}\cdot e^{-am}}.$$
• @Gaffney: the integration path should be a rectangle enclosing $x=i$, indeed.
|
– Jack D'Aurizio Apr 7 '16 at 17:11
• @Gaffney: the situation is almost the same as the usual proof of $\int_{0}^{+\infty}\frac{\sin x}{x}\,dx=\frac{\pi}{2}$.
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https://matheducators.stackexchange.com/questions/10933/how-to-explain-fractional-terms
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# How to explain "fractional terms"?
as I can see there are mainly two ways to introduce fractional terms. Two examples to demonstrate the two variants:
1. $\frac{a^2+3}{a}; \frac{3}{2c}$
2. $T(a) = \frac{a^2+3}{a}; T(c) = \frac{3}{2c}$.
In fact, the word "function" or "functional term" or "equation with variables" has not been learned at this point, but I think that variant two is able to demonstrate better that any term can be seen as a "number machine", in which you "throw" a number and get a certain result. I could imagine that the first variant leads to young learners being unsure what exactly to do with this experssion.
I made a quick research in some school books here (in Germany) and found those two variants. So even though this is really just a small difference, do you have any opinion or experience?
• What age group do you teach? I use fraction for just numerical and rational function for the functions of polynomials divided by polynomials. Apr 28, 2016 at 15:38
• It is (sometimes) helpful to distinguish "rational expressions" from "rational functions" for precisely the same reason that it is (sometimes) helpful to distinguish between "polynomials" and "polynomial functions". Apr 28, 2016 at 16:04
• Personally, I would not write $T(a) = \frac{a^2+3}{a}$ without first having the notion of "function". Apr 28, 2016 at 16:20
• "I could imagine that the first variant leads to young learners being unsure what exactly to do with this experssion." -- Well they shouldn't think that there's automatically something "to do" on any piece of math without a natural-language direction or question. Nov 2, 2016 at 3:04
• I don't see how fractions are relevant to the question; the same question could be asked about $a+1$.
– user797
Nov 2, 2016 at 12:20
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# How to explain "fractional terms"? as I can see there are mainly two ways to introduce fractional terms. Two examples to demonstrate the two variants:
1. $\frac{a^2+3}{a}; \frac{3}{2c}$
2. $T(a) = \frac{a^2+3}{a}; T(c) = \frac{3}{2c}$. In fact, the word "function" or "functional term" or "equation with variables" has not been learned at this point, but I think that variant two is able to demonstrate better that any term can be seen as a "number machine", in which you "throw" a number and get a certain result. I could imagine that the first variant leads to young learners being unsure what exactly to do with this experssion. I made a quick research in some school books here (in Germany) and found those two variants. So even though this is really just a small difference, do you have any opinion or experience? • What age group do you teach? I use fraction for just numerical and rational function for the functions of polynomials divided by polynomials. Apr 28, 2016 at 15:38
• It is (sometimes) helpful to distinguish "rational expressions" from "rational functions" for precisely the same reason that it is (sometimes) helpful to distinguish between "polynomials" and "polynomial functions".
|
Apr 28, 2016 at 16:04
• Personally, I would not write $T(a) = \frac{a^2+3}{a}$ without first having the notion of "function".
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https://math.stackexchange.com/questions/1518062/let-sn-k-denote-the-signless-stirling-numbers-of-the-first-kind-prove-that
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# Let $s(n,k)$ denote the signless Stirling numbers of the first kind. Prove that…
Let $s(n,k)$ denote the signless Stirling numbers of the first kind. Prove that:
$$s(n,2) = (n-1)!(1 + \frac{1}{2} + \frac{1}{3} +...+ \frac{1}{n-1})$$
-I haven't dealt with Taylor series expansion in a long time and not quite sure how (brand new to)Stirling numbers would play into this proof.Any help is appreciated.
HINT: If you split $[n]$ into two cycles, one of length $k$ and the other of length $n-k$, there are $\binom{n}k$ ways to choose the elements of the $k$-cycle, $(k-1)!$ ways to arrange them in a cycle, and $(n-k-1)!$ ways to arrange the remaining elements in a cycle. That gives you a total of
$$\binom{n}k(k-1)!(n-k-1)!$$
permutations. Now sum over the possible values of $k$. Be careful, though: you’ll be counting every permutation twice.
$$\frac{n}{k(n-k)}=\frac1k+\frac1{n-k}\;.$$
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# Let $s(n,k)$ denote the signless Stirling numbers of the first kind. Prove that…
Let $s(n,k)$ denote the signless Stirling numbers of the first kind. Prove that:
$$s(n,2) = (n-1)! (1 + \frac{1}{2} + \frac{1}{3} +...+ \frac{1}{n-1})$$
-I haven't dealt with Taylor series expansion in a long time and not quite sure how (brand new to)Stirling numbers would play into this proof.Any help is appreciated. HINT: If you split $[n]$ into two cycles, one of length $k$ and the other of length $n-k$, there are $\binom{n}k$ ways to choose the elements of the $k$-cycle, $(k-1)!$ ways to arrange them in a cycle, and $(n-k-1)!$ ways to arrange the remaining elements in a cycle. That gives you a total of
$$\binom{n}k(k-1)! (n-k-1)!$$
permutations. Now sum over the possible values of $k$. Be careful, though: you’ll be counting every permutation twice.
|
$$\frac{n}{k(n-k)}=\frac1k+\frac1{n-k}\;.$$
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https://math.stackexchange.com/questions/724904/a-different-way-to-prove-lim-limits-x-to-x-0gx-with-epsilon-delta
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# A different way to prove $\lim\limits_{x\to x_0}g(x)$ with $\epsilon,\delta$
for $f,g: \mathbb R\to \mathbb R\;\;and\;\; x_0\in \mathbb R$
Assume that $$\lim\limits_{x\to x_0}f(x)=L$$
prove: for every $\epsilon>0$ there is $\delta>0$ so that for every x that sustains $\delta>|x-x_0|>0$ and $\epsilon> |f(x)-g(x)|$ so there is$\lim\limits_{x\to x_0}g(x)=L$
i dont know even how to approach it, thanks in advance for the help
Let $\epsilon > 0$ be arbitrary. Then $\exists \delta_1 > 0$ such that
$$0<|x-x_0|<\delta_1 \Rightarrow |f(x_0) - L| < \epsilon/2$$
and $\exists \delta_2 > 0$ such that
$$0<|x-x_0|<\delta_2 \Rightarrow |f(x) - g(x)| < \epsilon/2.$$
Put $\delta := \min\{\delta_1,\delta_2\}$. Then by the triangle inequality,
$$0<|x-x_0|<\delta \Rightarrow |g(x) - L| \leq |f(x) - L| + |f(x) - g(x)| < \epsilon/2 + \epsilon/2 = \epsilon.$$
NB I hope I have interpreted your question correctly.
• exactly! thanks. i got to $2\epsilon>|f(x)-g(x)|$ but now i understand everything – user137645 Mar 24 '14 at 16:58
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# A different way to prove $\lim\limits_{x\to x_0}g(x)$ with $\epsilon,\delta$
for $f,g: \mathbb R\to \mathbb R\;\;and\;\; x_0\in \mathbb R$
Assume that $$\lim\limits_{x\to x_0}f(x)=L$$
prove: for every $\epsilon>0$ there is $\delta>0$ so that for every x that sustains $\delta>|x-x_0|>0$ and $\epsilon> |f(x)-g(x)|$ so there is$\lim\limits_{x\to x_0}g(x)=L$
i dont know even how to approach it, thanks in advance for the help
Let $\epsilon > 0$ be arbitrary. Then $\exists \delta_1 > 0$ such that
$$0<|x-x_0|<\delta_1 \Rightarrow |f(x_0) - L| < \epsilon/2$$
and $\exists \delta_2 > 0$ such that
$$0<|x-x_0|<\delta_2 \Rightarrow |f(x) - g(x)| < \epsilon/2.$$
Put $\delta := \min\{\delta_1,\delta_2\}$.
|
Then by the triangle inequality,
$$0<|x-x_0|<\delta \Rightarrow |g(x) - L| \leq |f(x) - L| + |f(x) - g(x)| < \epsilon/2 + \epsilon/2 = \epsilon.$$
NB I hope I have interpreted your question correctly.
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https://math.stackexchange.com/questions/1807605/p-1q-1q-2-and-p-1p-2q-1-do-not-have-a-greatest-common-divisor
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# $p_1q_1q_2$ and $p_1p_2q_1$ do not have a greatest common divisor
Question
Let $$p_1,p_2,q_1,q_2$$ be irreducible elements in integral domain $$R$$ such that none are associates to any of the others and $$p_1p_2=q_1q_2$$.
Prove that $$p_1q_1q_2$$ and $$p_1p_2q_1$$ do not have a greatest common divisor.
I have no idea to approach this problem.
It will be started with having gcd and show contradiction. But there are not much theorem to use show the question with integral domain. Help me!!
• Well, $p_1q_1$ is obviously a common divisor. Using $p_1p_2 = q_1q_2$, you get that $p_1q_1q_2 = p_1^2p_2$, so $p_1p_2$ is a common divisor. You could try showing that neither of those are a divisor of the others. May 31, 2016 at 22:57
We have that $p_1q_1$ and $p_1p_2$ are both common divisors, so a greatest common divisor would be divisible by both of these. Suppose there are $a,b,c$ such that $$p_1q_1ac=p_1p_2bc=p_1q_1q_2=p_1^2p_2$$ with $p_1q_1a=p_1p_2b$ being the greatest common divisor. Then $ac=q_2$ and $bc=p_1$. Since these are irreducible, $c$ must be a unit, and it follows that the greatest common divisor is an associate of $p_1q_1q_2$. This implies that $p_1q_1q_2=p_1^2p_2$ divides $p_1p_2q_1$ so there is a $d$ with $p_1p_2q_1=p_1^2p_2d$. It follows that $q_1=p_1d$, which contradicts the fact that $q_1$ is irreducible.
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# $p_1q_1q_2$ and $p_1p_2q_1$ do not have a greatest common divisor
Question
Let $$p_1,p_2,q_1,q_2$$ be irreducible elements in integral domain $$R$$ such that none are associates to any of the others and $$p_1p_2=q_1q_2$$. Prove that $$p_1q_1q_2$$ and $$p_1p_2q_1$$ do not have a greatest common divisor. I have no idea to approach this problem. It will be started with having gcd and show contradiction. But there are not much theorem to use show the question with integral domain. Help me!! • Well, $p_1q_1$ is obviously a common divisor. Using $p_1p_2 = q_1q_2$, you get that $p_1q_1q_2 = p_1^2p_2$, so $p_1p_2$ is a common divisor. You could try showing that neither of those are a divisor of the others. May 31, 2016 at 22:57
We have that $p_1q_1$ and $p_1p_2$ are both common divisors, so a greatest common divisor would be divisible by both of these. Suppose there are $a,b,c$ such that $$p_1q_1ac=p_1p_2bc=p_1q_1q_2=p_1^2p_2$$ with $p_1q_1a=p_1p_2b$ being the greatest common divisor. Then $ac=q_2$ and $bc=p_1$. Since these are irreducible, $c$ must be a unit, and it follows that the greatest common divisor is an associate of $p_1q_1q_2$. This implies that $p_1q_1q_2=p_1^2p_2$ divides $p_1p_2q_1$ so there is a $d$ with $p_1p_2q_1=p_1^2p_2d$.
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It follows that $q_1=p_1d$, which contradicts the fact that $q_1$ is irreducible.
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https://math.stackexchange.com/questions/2897441/orthogonal-family-of-curves-to-the-level-lines-of-fx-y-xy-1
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# Orthogonal family of curves to the level lines of $f(x,y)=xy-1$
Find the orthogonal family of curves to the level lines of $f(x,y)=xy-1$.
My attemp:
The level lines are $$\left\lbrace(x,y)\in\mathbb R^2:xy-1=K\quad\text{for some constant }K\right\rbrace\text.$$
Differentiating both sides: $$\mathfrak F:y+xy'=0\Rightarrow\mathfrak F^\perp:y-\frac x{y'}=0\Rightarrow\mathfrak F^\perp:\frac{y^2}2-\frac{x^2}2=c\Rightarrow\boxed{\mathfrak F^\perp:y^2-x^2=C,\;C\in\mathbb R}\text.$$
Is that correct?
Thanks!
• Thanks! I have plotted some functions and they seem to be orthogonal in the points I chose. Now I think, there is any problem if $K=C=-1$? Should we distinguish cases? Aug 28, 2018 at 16:26
• $K=C=-1$ will give you the coordinate axes and the bisectors and they are also acceptable. They intersect at the origin and are perpendicular. Aug 28, 2018 at 16:29
• $xy-1=K \implies ydx+xdy=0 \quad m_1=\frac{dy}{dx}=-\frac{y}{x}$
• $\frac{y^2}2-\frac{x^2}2=c \implies ydy-xdx=0 \quad m_2=\frac{dy}{dx}=\frac{x}{y}$
and $$m_1\cdot m_2=-\frac{y}{x}\cdot \frac{x}{y}=-1$$
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# Orthogonal family of curves to the level lines of $f(x,y)=xy-1$
Find the orthogonal family of curves to the level lines of $f(x,y)=xy-1$. My attemp:
The level lines are $$\left\lbrace(x,y)\in\mathbb R^2:xy-1=K\quad\text{for some constant }K\right\rbrace\text.$$
Differentiating both sides: $$\mathfrak F:y+xy'=0\Rightarrow\mathfrak F^\perp:y-\frac x{y'}=0\Rightarrow\mathfrak F^\perp:\frac{y^2}2-\frac{x^2}2=c\Rightarrow\boxed{\mathfrak F^\perp:y^2-x^2=C,\;C\in\mathbb R}\text.$$
Is that correct? Thanks! • Thanks! I have plotted some functions and they seem to be orthogonal in the points I chose. Now I think, there is any problem if $K=C=-1$? Should we distinguish cases? Aug 28, 2018 at 16:26
• $K=C=-1$ will give you the coordinate axes and the bisectors and they are also acceptable. They intersect at the origin and are perpendicular.
|
Aug 28, 2018 at 16:29
• $xy-1=K \implies ydx+xdy=0 \quad m_1=\frac{dy}{dx}=-\frac{y}{x}$
• $\frac{y^2}2-\frac{x^2}2=c \implies ydy-xdx=0 \quad m_2=\frac{dy}{dx}=\frac{x}{y}$
and $$m_1\cdot m_2=-\frac{y}{x}\cdot \frac{x}{y}=-1$$
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http://math.stackexchange.com/questions/75803/matrix-norm-identity-derivation
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# Matrix Norm Identity Derivation
I am having trouble figuring out where something in a book I am reading is coming from. (The book is Matrix Computations by Golub and Van Loan, 3rd edition, p.58.) It will probably be obvious once someone points it out to me, but until then I am stuck. It stems from the following lemma (which I understand and can follow):
If $F \in \mathbb{R}^{n \times n}$ and $\|F\|_p < 1$, then $I-F\,\,$ is nonsingular and
$$\left( I-F \right )^{-1} = \sum_{k=0}^{\infty}{F^k}$$
with
$$\|\left( I-F \right )^{-1} \|_p \leq { {1} \over {1-\|F\|_p} }.$$
As a consequence of the above,
$$\| \left( I - F \right )^{-1} - I \|_p \leq { {\|F\|_p} \over {1-\|F\|_p}}.$$
It's the last inequality that I cannot figure out how to derive.
-
Which book? ${}$ – J. M. Oct 25 '11 at 17:40
Golub and Van Loan, Matrix Computations, 3rd edition, p. 58. – anonstudent45678 Oct 25 '11 at 17:46
It's a matrix identity trick + triangle ineq. + submultiplicative norm: $(1-F)^{-1}=I+(1-F)^{-1}F$ – user13838 Oct 25 '11 at 18:26
Thanks to percusse in the comments above. The identity provided makes it rather simple:
$$\left( I - F \right)^{-1} - I = \left( I - F \right)^{-1} F$$
so
$$\|(I-F)^{-1} - I\| \leq \|F\| \cdot \|(I-f)^{-1}\| \leq { {\|F\|} \over {1-\|F\|} }$$
-
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# Matrix Norm Identity Derivation
I am having trouble figuring out where something in a book I am reading is coming from. (The book is Matrix Computations by Golub and Van Loan, 3rd edition, p.58.) It will probably be obvious once someone points it out to me, but until then I am stuck. It stems from the following lemma (which I understand and can follow):
If $F \in \mathbb{R}^{n \times n}$ and $\|F\|_p < 1$, then $I-F\,\,$ is nonsingular and
$$\left( I-F \right )^{-1} = \sum_{k=0}^{\infty}{F^k}$$
with
$$\|\left( I-F \right )^{-1} \|_p \leq { {1} \over {1-\|F\|_p} }.$$
As a consequence of the above,
$$\| \left( I - F \right )^{-1} - I \|_p \leq { {\|F\|_p} \over {1-\|F\|_p}}.$$
It's the last inequality that I cannot figure out how to derive. -
Which book? ${}$ – J. M. Oct 25 '11 at 17:40
Golub and Van Loan, Matrix Computations, 3rd edition, p. 58. – anonstudent45678 Oct 25 '11 at 17:46
It's a matrix identity trick + triangle ineq. + submultiplicative norm: $(1-F)^{-1}=I+(1-F)^{-1}F$ – user13838 Oct 25 '11 at 18:26
Thanks to percusse in the comments above.
|
The identity provided makes it rather simple:
$$\left( I - F \right)^{-1} - I = \left( I - F \right)^{-1} F$$
so
$$\|(I-F)^{-1} - I\| \leq \|F\| \cdot \|(I-f)^{-1}\| \leq { {\|F\|} \over {1-\|F\|} }$$
-
|
https://math.stackexchange.com/questions/4593870/how-to-generate-square-matrices-based-on-constraints-on-their-spectral-radii
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# How to generate square matrices based on constraints on their spectral radii?
After reading the Wikipedia, I am wondering whether there is an algorithm that, given a size $$N$$, can generate an $$N\!\times\! N$$ matrix $$\boldsymbol{W}$$ whose spectral radius satisfies $$\rho(\boldsymbol{W})<1$$, where
$$\rho(\boldsymbol{W}) = \max\limits_{\boldsymbol{W}} \left\{ |\lambda_{1}|, \dots, |\lambda_{n}| \right\}$$
for some natural $$n$$.
• Generate a random matrix $W$ first. Then calculate $\|W\|$ for any submultiplicative norm, such as $\|W\|_\infty=\max_i\sum_j|w_{ij}|$. Now multiply $W$ by $c/\|W\|$, where $c$ is a random number that lies inside $[0,1)$. For this new $W$ we have $\|W\|=c<1$. Hence $\rho(W)<1$. Commented Dec 7, 2022 at 21:23
1. Randomly generate a matrix $$W_0$$ (without a constraint on its spectral radius)
2. Compute $$r_0 = \rho(W_0)$$
3. Generate a random value $$r$$ with $$0 \leq r < 1$$
4. Obtain the desired matrix with $$W = \frac{r}{r_0}W_0$$. We have $$\rho(W) = r < 1$$.
We can make this process faster if we merely require that $$r_0 \geq W_0$$, which means that that the resulting $$W$$ has spectral radius $$\rho(W) \leq r$$. Towards that end, we can simply take $$r_0 = \|W\|$$ for any submultiplicative norm $$\|\cdot\|$$. Note, however, that this makes it relatively unlikely that $$\rho(W)$$ will be close to $$1$$.
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# How to generate square matrices based on constraints on their spectral radii? After reading the Wikipedia, I am wondering whether there is an algorithm that, given a size $$N$$, can generate an $$N\!\times\! N$$ matrix $$\boldsymbol{W}$$ whose spectral radius satisfies $$\rho(\boldsymbol{W})<1$$, where
$$\rho(\boldsymbol{W}) = \max\limits_{\boldsymbol{W}} \left\{ |\lambda_{1}|, \dots, |\lambda_{n}| \right\}$$
for some natural $$n$$. • Generate a random matrix $W$ first. Then calculate $\|W\|$ for any submultiplicative norm, such as $\|W\|_\infty=\max_i\sum_j|w_{ij}|$. Now multiply $W$ by $c/\|W\|$, where $c$ is a random number that lies inside $[0,1)$. For this new $W$ we have $\|W\|=c<1$. Hence $\rho(W)<1$. Commented Dec 7, 2022 at 21:23
1. Randomly generate a matrix $$W_0$$ (without a constraint on its spectral radius)
2. Compute $$r_0 = \rho(W_0)$$
3. Generate a random value $$r$$ with $$0 \leq r < 1$$
4. Obtain the desired matrix with $$W = \frac{r}{r_0}W_0$$. We have $$\rho(W) = r < 1$$. We can make this process faster if we merely require that $$r_0 \geq W_0$$, which means that that the resulting $$W$$ has spectral radius $$\rho(W) \leq r$$.
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Towards that end, we can simply take $$r_0 = \|W\|$$ for any submultiplicative norm $$\|\cdot\|$$.
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https://stats.stackexchange.com/questions/267293/mutual-information-of-two-identical-signals?noredirect=1
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# mutual information of two identical signals [duplicate]
library(infotheo)
> mutinformation(c(1,2,3),c(1,2,3))
[1] 1.098612
> mutinformation(c(1,1,1),c(1,1,1))
[1] 0
Why is there a difference?
I keep reading that this should either be 0 or infinity, yet I get neither
Is this because I have not defined the joint probability?
Why would perfectly similar data have 0 mutual information?
## marked as duplicate by Tim♦, Michael Chernick, kjetil b halvorsen, mdewey, mpiktasJun 23 '17 at 8:06
• The mutual information coefficient measures how related two variables are, but it has a lot to do with their joint and separate entropy. Two variables that are constant have zero entropy, hence zero mutual information. en.wikipedia.org/wiki/Mutual_information – Anna SdTC Mar 14 '17 at 1:49
Entropy and therefore mutual information depend on the probability distribution of your data, not on the values your data takes.
In your case the you have the vectors:
V1 = {1,2,3} and V2 = {1,1,1}
In the first case, the probability of each element is 1/3, in the second case, the probability of each element is 1:
p(V1) = {1/3,1/3,1/3} and p(V2) = {1}
This means that your first vector (V1) has an entropy(total information) of 1.584 bits. You need on average 1.584 bits to describe your vector.
In the case of your second vector, there is no information. You know it always contains a 1. Hence it's entropy equals 0.
When you compute the mutual information between a random variable and itself, you always end up with the total information of your random variable:
I(X,X) = H(X)
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# mutual information of two identical signals [duplicate]
library(infotheo)
> mutinformation(c(1,2,3),c(1,2,3))
[1] 1.098612
> mutinformation(c(1,1,1),c(1,1,1))
[1] 0
Why is there a difference? I keep reading that this should either be 0 or infinity, yet I get neither
Is this because I have not defined the joint probability? Why would perfectly similar data have 0 mutual information? ## marked as duplicate by Tim♦, Michael Chernick, kjetil b halvorsen, mdewey, mpiktasJun 23 '17 at 8:06
• The mutual information coefficient measures how related two variables are, but it has a lot to do with their joint and separate entropy. Two variables that are constant have zero entropy, hence zero mutual information. en.wikipedia.org/wiki/Mutual_information – Anna SdTC Mar 14 '17 at 1:49
Entropy and therefore mutual information depend on the probability distribution of your data, not on the values your data takes. In your case the you have the vectors:
V1 = {1,2,3} and V2 = {1,1,1}
In the first case, the probability of each element is 1/3, in the second case, the probability of each element is 1:
p(V1) = {1/3,1/3,1/3} and p(V2) = {1}
This means that your first vector (V1) has an entropy(total information) of 1.584 bits. You need on average 1.584 bits to describe your vector. In the case of your second vector, there is no information. You know it always contains a 1. Hence it's entropy equals 0.
|
When you compute the mutual information between a random variable and itself, you always end up with the total information of your random variable:
I(X,X) = H(X)
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https://math.stackexchange.com/questions/3201282/finding-a-basis-for-vector-space-of-matrices
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# Finding a basis for vector space of matrices
Find a basis of space of 2×2 matrices $$\{A_1,A_2,A_3,A_4\}$$such that $$A_i^2=A_i$$ for all i. I know how to find basis but I am unable to find such a property for this. Can you help me? Two matrices can be :
[1,0, and [0,0
0,0]. 0,1]
`
• Have you tried brute force? Take an arbitrary matrix $A$ and square it element by element. – amd Apr 25 at 1:43
• Sorry I didn't understand – Tojrah Apr 25 at 1:46
• You seem to be assuming that you need $4$ elements for this basis. Why do you assume this? – The Count Apr 25 at 1:49
• Also, the question phrasing is strange. Do you mean "find a basis for the subspace of $2\times 2$ matrices made up of all matrices $A$ such that $A^2=A$? Or even better, "Find a basis for the subspace $\{A|A^2=A\}$ of $2\times 2$ matrices". Because I doubt that is even a subspace. – The Count Apr 25 at 1:52
• OH OH OH I get it. You need just a basis for the space, and each member of the basis needs to have the property that it is its own square. I see. Well, what is a basis for $V$ that you know? – The Count Apr 25 at 1:54
## 1 Answer
Try adding $$[1,1,0,0]$$ and $$[0,0,1,1]$$.
• Yeah, got it... – Tojrah Apr 25 at 2:02
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# Finding a basis for vector space of matrices
Find a basis of space of 2×2 matrices $$\{A_1,A_2,A_3,A_4\}$$such that $$A_i^2=A_i$$ for all i. I know how to find basis but I am unable to find such a property for this. Can you help me? Two matrices can be :
[1,0, and [0,0
0,0]. 0,1]
`
• Have you tried brute force? Take an arbitrary matrix $A$ and square it element by element. – amd Apr 25 at 1:43
• Sorry I didn't understand – Tojrah Apr 25 at 1:46
• You seem to be assuming that you need $4$ elements for this basis. Why do you assume this? – The Count Apr 25 at 1:49
• Also, the question phrasing is strange. Do you mean "find a basis for the subspace of $2\times 2$ matrices made up of all matrices $A$ such that $A^2=A$?
|
Or even better, "Find a basis for the subspace $\{A|A^2=A\}$ of $2\times 2$ matrices".
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# evaluate the line integral, trouble with setting up and parameterizing and integrating
I tried following the example in my book, and I got stuck when trying to take the integral for the second part with $y = x^2$. I'm not completely confident on what to parameterize with here. I tried replacing $x^2$ with $y$, then in the radical I had $4 + y^3$, but when trying to integrate that with u substitution I couldn't finish it. I'm not sure I'm setting everything up right.
I'm also not sure how to tell which integral goes with which segment. Do you just assume the first goes with the first and second with the second?
hint
the first segment can be parametrised as
$$x=0+2t \;\;\;,\;\; y=0+t$$
the integral along this segment is
$$\int_0^1 \Bigl ((2t+2t)(2dt)+4t^2 (dt)\Bigr)=$$
$$4\int_0^1t (2+t)dt=\frac {16}{3}$$
You can do it for the second segment defined by
$$x=2+t \;\;\;, \;\; y=1-t$$
• also can you explain how you parametrized those equations? – 2316354654 Nov 26 '17 at 20:00
• @2316354654 the line AB is parametrized by $x=x_A+t (x_B-x_A)$ and $y=...$ – hamam_Abdallah Nov 26 '17 at 20:14
• why is your integral from 1 to 0? I thought it would be 2 to 0. – 2316354654 Nov 26 '17 at 21:02
• is the integral always from 1 to 0when you parameterize in terms of t? – 2316354654 Nov 26 '17 at 21:15
• @2316354654 Yes. t=0 gives the first point of the segment and t=1 gives the second. – hamam_Abdallah Nov 27 '17 at 11:36
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# evaluate the line integral, trouble with setting up and parameterizing and integrating
I tried following the example in my book, and I got stuck when trying to take the integral for the second part with $y = x^2$. I'm not completely confident on what to parameterize with here. I tried replacing $x^2$ with $y$, then in the radical I had $4 + y^3$, but when trying to integrate that with u substitution I couldn't finish it. I'm not sure I'm setting everything up right. I'm also not sure how to tell which integral goes with which segment. Do you just assume the first goes with the first and second with the second? hint
the first segment can be parametrised as
$$x=0+2t \;\;\;,\;\; y=0+t$$
the integral along this segment is
$$\int_0^1 \Bigl ((2t+2t)(2dt)+4t^2 (dt)\Bigr)=$$
$$4\int_0^1t (2+t)dt=\frac {16}{3}$$
You can do it for the second segment defined by
$$x=2+t \;\;\;, \;\; y=1-t$$
• also can you explain how you parametrized those equations?
|
– 2316354654 Nov 26 '17 at 20:00
• @2316354654 the line AB is parametrized by $x=x_A+t (x_B-x_A)$ and $y=...$ – hamam_Abdallah Nov 26 '17 at 20:14
• why is your integral from 1 to 0?
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https://math.stackexchange.com/questions/485648/transvection-matrices-generate-sl-n-mathbbr
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# Transvection matrices generate $SL_n(\mathbb{R})$
I need to prove that the transvection matrices generate the special linear group $\operatorname{SL}_n \left(\mathbb{R}\right)$.
I want to proceed using induction on n.
I was able to prove the 2x2 case, but I am having difficulty with the n+1 case.
I supposed that the elementary matrices of the first type generate $SL_n(\mathbb{R})$. And I want to show that an elementary matrix of the first type of order n+1 cas generate $SL_{n+1}(\mathbb{R})$
Multiplying a matrix on the left or the right by a transvection matrix corresponds to adding a multiple of one row/column to another row/column. So we just need to prove that we can reduce any matrix in ${\rm SL}_n(K)$ to the identity matrix using row and column operations of this type. This works for any field $K$.
The proof is by induction on $n$ and there is nothing to prove for $n=1$, so assume that $n>1$. Let $A = (a_{ij}) \in {\rm SL}_n(K)$.
1. If $a_{12} = 0$ then, choose some $j$ with $a_{1j} \ne 0$ and add column $j$ to column $2$ to get $a_{12} \ne 0$.
2. Subtract $(a_{11}-1)/a_{12}$ times column $2$ from column $1$ to get $a_{11}=1$.
3. For all $j>1$ subtract $a_{1j}$ times column $1$ from column $j$ to get $a_{1j}=0$. Similarly use row operations to get $a_{j1}=0$ for all $j>1$.
You have now reduced to the case $n-1$.
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# Transvection matrices generate $SL_n(\mathbb{R})$
I need to prove that the transvection matrices generate the special linear group $\operatorname{SL}_n \left(\mathbb{R}\right)$. I want to proceed using induction on n.
I was able to prove the 2x2 case, but I am having difficulty with the n+1 case. I supposed that the elementary matrices of the first type generate $SL_n(\mathbb{R})$. And I want to show that an elementary matrix of the first type of order n+1 cas generate $SL_{n+1}(\mathbb{R})$
Multiplying a matrix on the left or the right by a transvection matrix corresponds to adding a multiple of one row/column to another row/column. So we just need to prove that we can reduce any matrix in ${\rm SL}_n(K)$ to the identity matrix using row and column operations of this type. This works for any field $K$. The proof is by induction on $n$ and there is nothing to prove for $n=1$, so assume that $n>1$. Let $A = (a_{ij}) \in {\rm SL}_n(K)$. 1. If $a_{12} = 0$ then, choose some $j$ with $a_{1j} \ne 0$ and add column $j$ to column $2$ to get $a_{12} \ne 0$. 2. Subtract $(a_{11}-1)/a_{12}$ times column $2$ from column $1$ to get $a_{11}=1$. 3. For all $j>1$ subtract $a_{1j}$ times column $1$ from column $j$ to get $a_{1j}=0$.
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Similarly use row operations to get $a_{j1}=0$ for all $j>1$.
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http://math.stackexchange.com/questions/119357/definition-of-an-extreme-set
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# Definition of an extreme set?
I have an issue with a definition in Rudin's Functional Analysis in the paragraph regarding the Krein-Milman Theorem.
"Let $K$ be a subset of a vector space $X$. A nonempty set $S$ in $K$ is called an extreme set if no point of $S$ is an internal point of a line interval whose end points are in $K$ but not in $S$. Analytically, the condition can be expressed as follows: if $x$ and $y$ are in $K$, if $t$ is in $(0, 1)$, and if $tx + (1 - t)y$ is in $S$, then $x$ and $y$ are in $S$. The extreme points of $K$ are the extreme sets that consist of just one point."
For this condition to be equivalent to the definition, one should replace the conclusion by: "$\dots$ then $x$ is in $S$ or $y$ is in $S$." It turns out that this is indeed equivalent when $S$ consists of a single point, but not in general.
So my question is: what is the good definition for an extreme set?
-
@t.b.: thank you. Rudin's proof does not assume its extreme sets to be convex, though. Anyway, it looks like the condition should be taken to be the definition here. – user26770 Mar 12 '12 at 19:43
The second sentence is convoluted and, to me, hard to understand. The third sentence conforms to several other texts. If you want a fancier definition you can use the following:
Let $\mathsf{con}$ denote the convex hull operator. For $S \subseteq K$, the set $S$ is an extreme subset of $K$ if and only if for all $D \subseteq K$ we have $S \cap \mathsf{con}(D) = S \cap \mathsf{con}(S \cap D)$. If you want to, you only have to consider finite subsets $D$.
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# Definition of an extreme set? I have an issue with a definition in Rudin's Functional Analysis in the paragraph regarding the Krein-Milman Theorem. "Let $K$ be a subset of a vector space $X$. A nonempty set $S$ in $K$ is called an extreme set if no point of $S$ is an internal point of a line interval whose end points are in $K$ but not in $S$. Analytically, the condition can be expressed as follows: if $x$ and $y$ are in $K$, if $t$ is in $(0, 1)$, and if $tx + (1 - t)y$ is in $S$, then $x$ and $y$ are in $S$. The extreme points of $K$ are the extreme sets that consist of just one point." For this condition to be equivalent to the definition, one should replace the conclusion by: "$\dots$ then $x$ is in $S$ or $y$ is in $S$." It turns out that this is indeed equivalent when $S$ consists of a single point, but not in general. So my question is: what is the good definition for an extreme set? -
@t.b. : thank you. Rudin's proof does not assume its extreme sets to be convex, though. Anyway, it looks like the condition should be taken to be the definition here. – user26770 Mar 12 '12 at 19:43
The second sentence is convoluted and, to me, hard to understand. The third sentence conforms to several other texts. If you want a fancier definition you can use the following:
Let $\mathsf{con}$ denote the convex hull operator.
|
For $S \subseteq K$, the set $S$ is an extreme subset of $K$ if and only if for all $D \subseteq K$ we have $S \cap \mathsf{con}(D) = S \cap \mathsf{con}(S \cap D)$.
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Let $X,Y$ be independent exponential RV's with respective pdf's $f(x) = \lambda e^{-\lambda x}$ and $f(y) = \mu e^{-\mu y}$. We want to find the pdf of $Z=X-Y$. I originally tried the convolution ...
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### Difference of two exponential RVs [duplicate]
Let $X,Y$ be independent exponential RV's with respective pdf's $f(x) = \lambda e^{-\lambda x}$ and $f(y) = \mu e^{-\mu y}$. We want to find the pdf of $Z=X-Y$. I originally tried the convolution ...
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### Linear Combination of Exponential Random Variables [duplicate]
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### How to Find CDF of $T_1-T_2$? [duplicate]
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### Find PDF of $Y = X_1-X_2$ [duplicate]
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$Y = X_1-X_2$ and the question asks to find the pdf for $Y$?
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# Characteristic Portfolio for an Attribute [closed]
Given a vector of attributes(eg.E/P ratios, betas) for N assets
$a^T = {a_1,a_2,...,a_N}$ The exposure of portfolio $h_P$ to attribute a is
$a = \sum_{n}a_n h_{P,n}$
Proposition: There is a unique portfolio $h_a$ that has minimum risk and unit exposure to a. The holdings(weights) of the characteristic portfolio $h_a$ are given by
$h_a = \frac{V^{-1}a}{a^TV^{-1}a}$
For the prrof we write:
Minimise $h^TVh$ subject to constriant : $h^Ta=1$
Using Langrange multiplier we get the equations:
a. $h^Ta = 1$
b. $Vh - \lambda a = 0$
Question: How does substituting a in b yields the result of the proposition ?
## closed as off-topic by Alex C, Helin, LocalVolatility, amdopt, Attack68Aug 13 '18 at 17:52
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "Basic financial questions are off-topic as they are assumed to be common knowledge for those studying or working in the field of quantitative finance." – Alex C, Helin, LocalVolatility, amdopt, Attack68
If this question can be reworded to fit the rules in the help center, please edit the question.
From b. we get $Vh = \lambda a$, so $h=\lambda V^{-1}a$ (assuming V is invertible).
Using this to evaluate a. we get $h^Ta = \lambda a^T V^{-1}a=1$ (assuming $V^{-1}$ is symmetric). We can solve this for lambda: $\lambda=\frac{1}{a^T V^{-1}a}$
$$h=\frac{V^{-1}a}{a^T V^{-1}a}$$
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# Characteristic Portfolio for an Attribute [closed]
Given a vector of attributes(eg.E/P ratios, betas) for N assets
$a^T = {a_1,a_2,...,a_N}$ The exposure of portfolio $h_P$ to attribute a is
$a = \sum_{n}a_n h_{P,n}$
Proposition: There is a unique portfolio $h_a$ that has minimum risk and unit exposure to a. The holdings(weights) of the characteristic portfolio $h_a$ are given by
$h_a = \frac{V^{-1}a}{a^TV^{-1}a}$
For the prrof we write:
Minimise $h^TVh$ subject to constriant : $h^Ta=1$
Using Langrange multiplier we get the equations:
a. $h^Ta = 1$
b. $Vh - \lambda a = 0$
Question: How does substituting a in b yields the result of the proposition ? ## closed as off-topic by Alex C, Helin, LocalVolatility, amdopt, Attack68Aug 13 '18 at 17:52
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "Basic financial questions are off-topic as they are assumed to be common knowledge for those studying or working in the field of quantitative finance." – Alex C, Helin, LocalVolatility, amdopt, Attack68
If this question can be reworded to fit the rules in the help center, please edit the question. From b. we get $Vh = \lambda a$, so $h=\lambda V^{-1}a$ (assuming V is invertible). Using this to evaluate a. we get $h^Ta = \lambda a^T V^{-1}a=1$ (assuming $V^{-1}$ is symmetric).
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We can solve this for lambda: $\lambda=\frac{1}{a^T V^{-1}a}$
$$h=\frac{V^{-1}a}{a^T V^{-1}a}$$
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https://math.stackexchange.com/questions/2718899/line-through-a-trapezoids-diagonals
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# Line through a trapezoids diagonals
Given a trapezoid $ABDC$, and line segment $PQ$ where $P$ and $Q$ are points on $AC$ and $BD$, respectively, s.t. $AB||PQ||CD$. Suppose $PQ$ intersects $BC$ at $K$ and $AD$ at $J$, prove that $PK$ and $JQ$ are equal.
Below is an extreme case where $K$ and $J$ are at point $O$. Part of the reason must be because the lines are parallel. Also, I checked on GeoGebra and seen that they are numerically equal,i.e. $PK=JQ$.
• $P$, $K$ and $J$ are collinear. How can $PK=JP$ if $J\ne K$? – CY Aries Apr 2 '18 at 15:35
• Sorry for the confusion, I edited the question. – John Glenn Apr 2 '18 at 15:36
As $\triangle ABC\sim\triangle PKC$, $AB:PK=BC:KC$.
As $\triangle ABD\sim\triangle JQD$, $AB:JQ=BD:QD$.
As $\triangle BCD\sim\triangle BKQ$, $BC:BK=BD:BQ$.
So, $BC:KC=BC:(BC-BK)=BD:(BD-BQ)=BD:QD$
Therefore, $AB:PK=AB:JQ$ and hence $PK=JQ$
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# Line through a trapezoids diagonals
Given a trapezoid $ABDC$, and line segment $PQ$ where $P$ and $Q$ are points on $AC$ and $BD$, respectively, s.t. $AB||PQ||CD$. Suppose $PQ$ intersects $BC$ at $K$ and $AD$ at $J$, prove that $PK$ and $JQ$ are equal. Below is an extreme case where $K$ and $J$ are at point $O$. Part of the reason must be because the lines are parallel. Also, I checked on GeoGebra and seen that they are numerically equal,i.e. $PK=JQ$. • $P$, $K$ and $J$ are collinear. How can $PK=JP$ if $J\ne K$? – CY Aries Apr 2 '18 at 15:35
• Sorry for the confusion, I edited the question. – John Glenn Apr 2 '18 at 15:36
As $\triangle ABC\sim\triangle PKC$, $AB:PK=BC:KC$. As $\triangle ABD\sim\triangle JQD$, $AB:JQ=BD:QD$. As $\triangle BCD\sim\triangle BKQ$, $BC:BK=BD:BQ$.
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So, $BC:KC=BC:(BC-BK)=BD:(BD-BQ)=BD:QD$
Therefore, $AB:PK=AB:JQ$ and hence $PK=JQ$
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https://physics.stackexchange.com/questions/756305/force-problem-with-punching-glove
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# Force Problem with Punching Glove?
Consider a man with a thick padded punching glove who strikes a bag. He then strikes that same bag without a punching glove. Both punches travel at the same velocity. Which one exerts more force? Which one takes longer to stop?
Here's my understanding:
We know by newton's 2nd law that $$F=ma$$. With the glove there is more mass and therefore more inertia. So the glove should take longer to stop.
For force, with the glove there is more mass and once again there should be more force.
The answers state that the force with the glove is smaller and the glove takes longer to stop. I seem to have gotten the force one wrong so what did I do incorrectly?
Edit: I understand the correct solution. However I struggle to understand what's wrong with my solution. Why must I neglect the mass?
• Neglect the mass of the glove - this problem is mainly about impulse. Mar 22 at 4:40
• Given your assumption, you were on the right track. This problem doesn't spell everything out very well. I suspect you are supposed to think of the glove as padding more than extra mass. Mar 22 at 4:59
• @dandan0101 Thanks I understand the solution. However I struggle to understand what's wrong with my solution. Why must I neglect the mass? Mar 22 at 5:05
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# Force Problem with Punching Glove? Consider a man with a thick padded punching glove who strikes a bag. He then strikes that same bag without a punching glove. Both punches travel at the same velocity. Which one exerts more force? Which one takes longer to stop?
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Here's my understanding:
We know by newton's 2nd law that $$F=ma$$.
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https://math.stackexchange.com/questions/3190768/comparing-two-normal-distributions
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# Comparing two normal distributions
Given a normal distribution $$X$$~$$N(60,9^2)$$ with a random variable $$A$$ and a normal distribution $$Y$$~$$N(50,7^2)$$ with a random variable $$B$$, how do I go about finding the probability $$P(B>A)$$?
(Given that A and B are independent events).
If ther are dependent you cannot do this. If they are independent then $$C=B-A$$ has normal distribution with mean $$50-60$$ and variance $$9^{2}+7^{2}$$. You can compute $$P(C>0)$$ by integrating the density function from $$0$$ to $$\infty$$.
• So we get to something like this? $f(x) = \begin{cases} 0, & \text{if$A≥x≥B$} \\ c, & \text{if$A≤x≤B$} \end{cases}$ – sdds Apr 17 at 8:01
• The answer is $\int_0^{\infty} \frac 1 {\sqrt {260 \pi}} e^{-(x+10)^{2}/{260}}dx$. – Kavi Rama Murthy Apr 17 at 8:04
Hint: If $$A$$ and $$B$$ are independent, you have that $$A-B \sim \mathcal{N}(10,7^2+9^2)$$ and $$P(B>A) = P(A-B<0)$$.
If they are dependent and are jointly normally distributed, you need to use the joint distribution.
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# Comparing two normal distributions
Given a normal distribution $$X$$~$$N(60,9^2)$$ with a random variable $$A$$ and a normal distribution $$Y$$~$$N(50,7^2)$$ with a random variable $$B$$, how do I go about finding the probability $$P(B>A)$$? (Given that A and B are independent events). If ther are dependent you cannot do this. If they are independent then $$C=B-A$$ has normal distribution with mean $$50-60$$ and variance $$9^{2}+7^{2}$$. You can compute $$P(C>0)$$ by integrating the density function from $$0$$ to $$\infty$$. • So we get to something like this? $f(x) = \begin{cases} 0, & \text{if$A≥x≥B$} \\ c, & \text{if$A≤x≤B$} \end{cases}$ – sdds Apr 17 at 8:01
• The answer is $\int_0^{\infty} \frac 1 {\sqrt {260 \pi}} e^{-(x+10)^{2}/{260}}dx$.
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– Kavi Rama Murthy Apr 17 at 8:04
Hint: If $$A$$ and $$B$$ are independent, you have that $$A-B \sim \mathcal{N}(10,7^2+9^2)$$ and $$P(B>A) = P(A-B<0)$$.
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https://math.stackexchange.com/questions/3176314/expected-value-of-the-sum
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# Expected value of the sum
Let $$S = \sum_{i=1}^{N} X_i.$$
$$\mathbb EX = 1, \mathbb DX = 2.$$
$$N$$ has a negative binomial distribution with parameters $$k=80$$ and $$p=0.4.$$
$$\mathbb P(N=l)=\begin{pmatrix} k+l-1 \\ l \\ \end{pmatrix} p^kq^l, l=0,1,2,.. .$$ Find $$\mathbb ES = (\mathbb EN)( \mathbb EX).$$
How to find $$(\mathbb EN)$$?
• You could simply recall the formula for the expected value of a Negative Binomial random variable, which is $\color{blue}{kq/p}$ here. This gives $80\times 0.6/0.4 = 120$. Or are you asking for a proof of the formula? – Minus One-Twelfth Apr 5 at 19:34
• What does $\Bbb D$ denote? – J.G. Apr 5 at 19:34
• $\newcommand{\E}{\mathbb{E}}\newcommand{\P}{\mathbb{P}}$If you're asking for a proof of the formula for the expected value of a Negative Binomial random variable, perhaps the easiest way is to recall that such a Negative Binomial random variable $X$ is equal in distribution to a sum of $k$ Geometric$(p)$ random variables, so the expected value is $\E[ X] = k\times \E[Z]$ where $Z\sim\mathsf{Geom}(p)$ (i.e. $\P(Z=l) = q^l p$ for $l=0,1,2,\ldots$). Then recall or prove that $\E[Z] = q/p$. – Minus One-Twelfth Apr 5 at 19:41
• Oh, I forgot you used $N$ for the Negative Binomial random variable. You can replace the $X$ above with $N$ if you want. – Minus One-Twelfth Apr 5 at 19:47
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# Expected value of the sum
Let $$S = \sum_{i=1}^{N} X_i.$$
$$\mathbb EX = 1, \mathbb DX = 2.$$
$$N$$ has a negative binomial distribution with parameters $$k=80$$ and $$p=0.4.$$
$$\mathbb P(N=l)=\begin{pmatrix} k+l-1 \\ l \\ \end{pmatrix} p^kq^l, l=0,1,2,.. .$$ Find $$\mathbb ES = (\mathbb EN)( \mathbb EX).$$
How to find $$(\mathbb EN)$$? • You could simply recall the formula for the expected value of a Negative Binomial random variable, which is $\color{blue}{kq/p}$ here. This gives $80\times 0.6/0.4 = 120$. Or are you asking for a proof of the formula? – Minus One-Twelfth Apr 5 at 19:34
• What does $\Bbb D$ denote? – J.G. Apr 5 at 19:34
• $\newcommand{\E}{\mathbb{E}}\newcommand{\P}{\mathbb{P}}$If you're asking for a proof of the formula for the expected value of a Negative Binomial random variable, perhaps the easiest way is to recall that such a Negative Binomial random variable $X$ is equal in distribution to a sum of $k$ Geometric$(p)$ random variables, so the expected value is $\E[ X] = k\times \E[Z]$ where $Z\sim\mathsf{Geom}(p)$ (i.e. $\P(Z=l) = q^l p$ for $l=0,1,2,\ldots$).
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Then recall or prove that $\E[Z] = q/p$.
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http://math.stackexchange.com/questions/315569/logarithm-its-algebraic-properties-and-algorithmic-complexity-mit-opencoursewa?answertab=active
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# Logarithm, its algebraic properties and algorithmic complexity (MIT OpenCourseware 6.042)
Does $$2^{\log{n}} T(n / 2^{\log n}) = 2^{\log{n}} T(1)?$$ If so, how?
Also, I don't understand how the following equality works:
$$\sum_{i=0}^{\log(n) -1}{2^i} = 2^{\log n} - 1$$
I'm afraid I'm either missing some context or just algebraic properties. It seems like the latter equality is a property that I'm supposed to know or be given? Prof. Leighton performs this calculation in the first lecture (approx. 32:17) of the Mathematics for Computer Science course for OCW, when analyzing the Merge-Sort algorithm.
-
The second equality is just a special case of: $$\sum_{i=0}^n 2^{i} = 2^{n+1} - 1$$ which can be proved using induction. It can also be viewed as the first $n+1$ terms of a geometric series: $$\sum_{i=0}^n x^i = \frac{x^{n+1} - 1}{x -1} \qquad (\text{ for } x \neq 1)$$ – JavaMan Feb 27 '13 at 4:08
Keep in mind that log in CS almost always means log base 2. So in particular, $2^{\log n}=n$ (from the definition of log, obviously), which is how $n/{2^{\log n}}$ becomes 1.
There's all sorts of ways to interpret the second formula. One way is to think of it in terms of binary representation. Note that $\sum_{i=0}^N{2^i}$ is the same as the N-digit binary number that consists of all 1s. E.g. for $N=4$: $1111_2$. Add one to this number and you get $10000_2$, which represents $2^5=2^{N+1}$. Subtract the one to get $2^{N+1}-1$.
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Ah, it was the base 2 that I forgot. Thank you! – user52207 Feb 27 '13 at 13:20
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# Logarithm, its algebraic properties and algorithmic complexity (MIT OpenCourseware 6.042)
Does $$2^{\log{n}} T(n / 2^{\log n}) = 2^{\log{n}} T(1)?$$ If so, how? Also, I don't understand how the following equality works:
$$\sum_{i=0}^{\log(n) -1}{2^i} = 2^{\log n} - 1$$
I'm afraid I'm either missing some context or just algebraic properties. It seems like the latter equality is a property that I'm supposed to know or be given? Prof. Leighton performs this calculation in the first lecture (approx. 32:17) of the Mathematics for Computer Science course for OCW, when analyzing the Merge-Sort algorithm. -
The second equality is just a special case of: $$\sum_{i=0}^n 2^{i} = 2^{n+1} - 1$$ which can be proved using induction. It can also be viewed as the first $n+1$ terms of a geometric series: $$\sum_{i=0}^n x^i = \frac{x^{n+1} - 1}{x -1} \qquad (\text{ for } x \neq 1)$$ – JavaMan Feb 27 '13 at 4:08
Keep in mind that log in CS almost always means log base 2. So in particular, $2^{\log n}=n$ (from the definition of log, obviously), which is how $n/{2^{\log n}}$ becomes 1. There's all sorts of ways to interpret the second formula. One way is to think of it in terms of binary representation. Note that $\sum_{i=0}^N{2^i}$ is the same as the N-digit binary number that consists of all 1s. E.g. for $N=4$: $1111_2$.
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Add one to this number and you get $10000_2$, which represents $2^5=2^{N+1}$.
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http://math.stackexchange.com/questions/89474/a-rough-estimation-for-the-number-of-square-free-integers/89480
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# A Rough Estimation for the number of square free integers
Show by a sieve argument that the number of square free integers not exceeding $x$ is less than $$x\prod_p\left(1-\frac{1}{p^2}\right)+o(x),$$where the product extends over all primes.
I happened to see this exercise this morning, and still fail to prove it. Could you give me a proof?
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Do you mean "sum" in "where the product extends"? – Dimitrije Kostic Dec 8 '11 at 2:42
@DimitrijeKostic Sorry, it should be product. – Kou Dec 8 '11 at 2:51
@AndréNicolas Sorry, you are right. – Kou Dec 8 '11 at 2:52
Rough sketch: Have you seen the Wikipedia section on the distribution of squarefree integers? It essentially gives your solution. All you need to notice is that for large $n$, the "probability" that an integer has $p^2$ as a factor is $\frac{p^2-1}{p^2} = \left( 1 - \frac{1}{p^2}\right)$. Since the probability that $n$ has $p^2$ as a factor is roughly independent of the fact it has $q^2$ where $p$ and $q$ are distinct primes, this implies that
$$\frac{Q(x)}{x} \approx \prod_p \left( 1 - \frac{1}{p^2} \right).$$
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# A Rough Estimation for the number of square free integers
Show by a sieve argument that the number of square free integers not exceeding $x$ is less than $$x\prod_p\left(1-\frac{1}{p^2}\right)+o(x),$$where the product extends over all primes. I happened to see this exercise this morning, and still fail to prove it. Could you give me a proof? -
Do you mean "sum" in "where the product extends"? – Dimitrije Kostic Dec 8 '11 at 2:42
@DimitrijeKostic Sorry, it should be product. – Kou Dec 8 '11 at 2:51
@AndréNicolas Sorry, you are right. – Kou Dec 8 '11 at 2:52
Rough sketch: Have you seen the Wikipedia section on the distribution of squarefree integers? It essentially gives your solution.
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All you need to notice is that for large $n$, the "probability" that an integer has $p^2$ as a factor is $\frac{p^2-1}{p^2} = \left( 1 - \frac{1}{p^2}\right)$.
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https://puzzling.stackexchange.com/questions/24277/find-the-next-term-in-the-number-series
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Find the next term in the number series?
$9, 27, 31, 155, 161, 1127, ?$
I observed that the first term and second term respectively are $3^2$ and $3^3$. But I have no idea about the subsequent terms.
One more observation I did was, the difference in all of these: $$9, +18= 27,+4=31,+124=155,+6=161,+966=1127.$$ We see the after each alternate number, the difference is increasing: $4,6,...$ so next should be $8$, which gives the answer as $1135$. Which is surprisingly correct. But what is the pattern in the middle numbers?
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Find the next term in the number series? $9, 27, 31, 155, 161, 1127, ?$
I observed that the first term and second term respectively are $3^2$ and $3^3$. But I have no idea about the subsequent terms.
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One more observation I did was, the difference in all of these: $$9, +18= 27,+4=31,+124=155,+6=161,+966=1127.$$ We see the after each alternate number, the difference is increasing: $4,6,...$ so next should be $8$, which gives the answer as $1135$.
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https://physics.stackexchange.com/questions/378472/meaning-of-extraction-ratio-e-fracc-ai-c-aoc-ai-c-bi
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# Meaning of extraction ratio $E=\frac{C_{Ai}-C_{Ao}}{C_{Ai}-C_{Bi}}$
My goal is to understand the meaning of the extraction ratio (artificial kidney, not other systems). I have the system in figure ($Q$ volumetric flow, $C$ concentration, the red line is a permeable membrane):
On my book (page 320 Cooney - Biomedical engineering principles) there is the definition of a parameter, the extraction ratio (note that all concentrations are evaluated at the same time):
$$E=\frac{C_{Ai}(t)-C_{Ao}(t)}{C_{Ai}(t)-C_{Bi}(t)}$$ where E is constant ($Q_A$ and $Q_B$ are constants).
In words, it say the extraction ratio "represents the amount of solute concentration change achieved relative to what would result from complete equilibrium with a very large supply" of liquid B "having a concentration $C_{Bi}$.
At page 549 of Saltzman - Biomedical Engineering, it say "The extraction ratio, E, is the solute concentration change" in the liquid A "compared to the theoretical solute concentration change that would occur if the" liquid A and B "came to equilibrium".
Mathematically what does mean equilibrium? Maybe $C_{Ao}=C_{Bo}$ or $C_{Ai}=C_{Bi}$ or $C_{A}=C_{B}$ for all value of $x$ or...?
Morover, how can I see mathematically the two terms of the comparison?
This ratio seem to express the actual concentration gained by A : $\Delta C_A = C_{A}(i) - C_{A}(0)$ divided by the maximum gain this system can have, which is given by : $\Delta^{MAX} C_A = C_{A}(i) - C_{B}(i)$.
To understand why the second is the maximum gain, consider the static case where you let the two fluid for an infinite time from $C_A(i)$ on the A side and $C_B(i)$ and the B side and assume that B is infinitely large, ie its concentration will not change as it is giving particule to the A side. In the end, the A side will be at equilibrium with the B side with a constration of $C_{A}(t=\infty) = C_B(i)$
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# Meaning of extraction ratio $E=\frac{C_{Ai}-C_{Ao}}{C_{Ai}-C_{Bi}}$
My goal is to understand the meaning of the extraction ratio (artificial kidney, not other systems). I have the system in figure ($Q$ volumetric flow, $C$ concentration, the red line is a permeable membrane):
On my book (page 320 Cooney - Biomedical engineering principles) there is the definition of a parameter, the extraction ratio (note that all concentrations are evaluated at the same time):
$$E=\frac{C_{Ai}(t)-C_{Ao}(t)}{C_{Ai}(t)-C_{Bi}(t)}$$ where E is constant ($Q_A$ and $Q_B$ are constants). In words, it say the extraction ratio "represents the amount of solute concentration change achieved relative to what would result from complete equilibrium with a very large supply" of liquid B "having a concentration $C_{Bi}$. At page 549 of Saltzman - Biomedical Engineering, it say "The extraction ratio, E, is the solute concentration change" in the liquid A "compared to the theoretical solute concentration change that would occur if the" liquid A and B "came to equilibrium". Mathematically what does mean equilibrium? Maybe $C_{Ao}=C_{Bo}$ or $C_{Ai}=C_{Bi}$ or $C_{A}=C_{B}$ for all value of $x$ or...? Morover, how can I see mathematically the two terms of the comparison? This ratio seem to express the actual concentration gained by A : $\Delta C_A = C_{A}(i) - C_{A}(0)$ divided by the maximum gain this system can have, which is given by : $\Delta^{MAX} C_A = C_{A}(i) - C_{B}(i)$. To understand why the second is the maximum gain, consider the static case where you let the two fluid for an infinite time from $C_A(i)$ on the A side and $C_B(i)$ and the B side and assume that B is infinitely large, ie its concentration will not change as it is giving particule to the A side.
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In the end, the A side will be at equilibrium with the B side with a constration of $C_{A}(t=\infty) = C_B(i)$
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https://math.stackexchange.com/questions/2142900/describing-linear-transformation
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# Describing Linear transformation
For each of the following linear transformations, write down its matrix and describe the transformation
a) $g(x,y)=(4x,6y)$
b) $h(x,y)=(x+2y,y)$
c) $k(x,y)=(y,x)$
So I have worked out the matrices:
$\begin{bmatrix} 4 & 0 \\ 0 & 6 \end{bmatrix}$
$\begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$
$\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$
Not sure what the transformations would be?
• you are correct! – Arnaldo Feb 13 '17 at 19:16
• For example, for the first case the transformation is $$T\binom xy=\begin{pmatrix}4&0\\0&6\end{pmatrix}\binom xy$$ – DonAntonio Feb 13 '17 at 19:17
• Next time use:meta.math.stackexchange.com/questions/5020/… – Arnaldo Feb 13 '17 at 19:17
The "describe" part is asking you what each transformation does to the input $(x,y)$; think of this as a vector in $\mathbb{R}^2$. For example, a transformation that sends $(x,y)$ to $(-x,y)$ is a reflection over the $y$-axis.
Start with the transformation $k$; that has a nice "symmetry".
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# Describing Linear transformation
For each of the following linear transformations, write down its matrix and describe the transformation
a) $g(x,y)=(4x,6y)$
b) $h(x,y)=(x+2y,y)$
c) $k(x,y)=(y,x)$
So I have worked out the matrices:
$\begin{bmatrix} 4 & 0 \\ 0 & 6 \end{bmatrix}$
$\begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$
$\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$
Not sure what the transformations would be? • you are correct!
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– Arnaldo Feb 13 '17 at 19:16
• For example, for the first case the transformation is $$T\binom xy=\begin{pmatrix}4&0\\0&6\end{pmatrix}\binom xy$$ – DonAntonio Feb 13 '17 at 19:17
• Next time use:meta.math.stackexchange.com/questions/5020/… – Arnaldo Feb 13 '17 at 19:17
The "describe" part is asking you what each transformation does to the input $(x,y)$; think of this as a vector in $\mathbb{R}^2$.
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https://math.stackexchange.com/questions/1151581/prove-that-inta-a-setminus-bda
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# Prove that $int(A)=A\setminus bd(A)$
$A$ is a subset of a metric space $M$.
I know I will need to prove $A$ is a subset of $M$. As well as $M$ is a subset of $A$. So for, $A$ is a subset of $A$: $int(A)$ implies that it is a subset of $A$ itself. Thus if $x$ is in $int(A)$ it must also be in $A$. Not sure if I'm off to right track any help is great
Suppose $x\in int(A)$. Then there exists an open neighbourhood $U_{x}$ of $x$ such that $x\in U_{x}\subseteq A$.
So we have $U_{x}\cap A^{c}=\varnothing$.
Now assume that $x\in bd(A)$.
Then for all neighbourhood $U$ of $x$, $U\cap A\neq \varnothing$ and $U\cap A^{c}\neq \varnothing$.
But this is a contradiction, since $U_{x}\cap A^{c}=\varnothing$.
So $x\notin bd(A)$ and hence $x\in A\setminus bd(A)$.
That is $int(A)\subseteq A\setminus bd(A)$.
Now conversely suppose $y\in A\setminus bd(A)$.
Then $y\in A$ and $y\notin bd(A)$.
Then there exist a neighbourhood $U_{y}$ of $y$ such that $U_{y}\cap A^{c}= \varnothing$.
So $y\in U_{y}\subseteq A$ and hence $y\in int(A)$.
Therefore $A\setminus bd(A)\subseteq int(A)$.
Thus $int(A)= A\setminus bd(A)$. $\square$
Hint: $x\in bd (A)$ if for every open neighborhood $V$ of $x$, we have that
$$A\cap V\neq \emptyset \text{ and } (M-A)\cap V\neq\emptyset.$$
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# Prove that $int(A)=A\setminus bd(A)$
$A$ is a subset of a metric space $M$. I know I will need to prove $A$ is a subset of $M$. As well as $M$ is a subset of $A$. So for, $A$ is a subset of $A$: $int(A)$ implies that it is a subset of $A$ itself. Thus if $x$ is in $int(A)$ it must also be in $A$. Not sure if I'm off to right track any help is great
Suppose $x\in int(A)$. Then there exists an open neighbourhood $U_{x}$ of $x$ such that $x\in U_{x}\subseteq A$. So we have $U_{x}\cap A^{c}=\varnothing$. Now assume that $x\in bd(A)$. Then for all neighbourhood $U$ of $x$, $U\cap A\neq \varnothing$ and $U\cap A^{c}\neq \varnothing$. But this is a contradiction, since $U_{x}\cap A^{c}=\varnothing$. So $x\notin bd(A)$ and hence $x\in A\setminus bd(A)$. That is $int(A)\subseteq A\setminus bd(A)$. Now conversely suppose $y\in A\setminus bd(A)$. Then $y\in A$ and $y\notin bd(A)$. Then there exist a neighbourhood $U_{y}$ of $y$ such that $U_{y}\cap A^{c}= \varnothing$. So $y\in U_{y}\subseteq A$ and hence $y\in int(A)$. Therefore $A\setminus bd(A)\subseteq int(A)$.
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Thus $int(A)= A\setminus bd(A)$.
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http://math.stackexchange.com/questions/143265/volume-integral
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# Volume integral
I want to compute with a triple integral the volume of the object that is calculated with $z = \sqrt{y}, z = 0 , x=0$ and $x+y=1$. I managed to draw the object. Then I tried to find the limits of the integrals.
I think $0<z<\sqrt{y}$ , $x^2<y<1-x$ and $0<x<\frac{\sqrt{5}}{2}-\frac{1}{2}$.
But with these limits I end up with a negative result which is not very good when trying to find volume.
Any ideas cause I am really confused right now.
-
First $z$ goes $0$ to $\sqrt{y}$, then $y$ goes $0$ to $1-x$, then $x$ goes $0$ to $1$. – André Nicolas May 9 '12 at 23:12
A couple of things: (1) Do you also have a bound of $y = 0$? (2) I think your first limits $0<z<\sqrt{y}$ were ok, but the next two are funny. Why do you have $x^2<y$? – froggie May 9 '12 at 23:12
Nothing about $y=0$. I did $x^2<y$ cause I read in the book that you must take the projection to $z=0$ and find the limits as you do for a 2-integral. – srimas May 9 '12 at 23:17
I guess I was just worried about the $\sqrt{y}$ when $y$ is negative. I think it must be implicit to take $y>0$. In which case I agree with Andre Nicolas's bounds. – froggie May 9 '12 at 23:19
There is an unstated but implicit assumption that $y \ge 0$, because of the bound on $z$. This forces $x \le 1$. Thus our triple integral can be expressed as the iterated integral $$\int_{x=0}^1\left(\int_{y=0}^{1-x}\left(\int_{z=0}^{\sqrt{y}} dz \right)dy\right)dx.$$ The ensuing computation poses no special difficulties. The result is $\dfrac{4}{15}$.
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# Volume integral
I want to compute with a triple integral the volume of the object that is calculated with $z = \sqrt{y}, z = 0 , x=0$ and $x+y=1$. I managed to draw the object. Then I tried to find the limits of the integrals. I think $0<z<\sqrt{y}$ , $x^2<y<1-x$ and $0<x<\frac{\sqrt{5}}{2}-\frac{1}{2}$. But with these limits I end up with a negative result which is not very good when trying to find volume. Any ideas cause I am really confused right now. -
First $z$ goes $0$ to $\sqrt{y}$, then $y$ goes $0$ to $1-x$, then $x$ goes $0$ to $1$. – André Nicolas May 9 '12 at 23:12
A couple of things: (1) Do you also have a bound of $y = 0$? (2) I think your first limits $0<z<\sqrt{y}$ were ok, but the next two are funny. Why do you have $x^2<y$? – froggie May 9 '12 at 23:12
Nothing about $y=0$. I did $x^2<y$ cause I read in the book that you must take the projection to $z=0$ and find the limits as you do for a 2-integral. – srimas May 9 '12 at 23:17
I guess I was just worried about the $\sqrt{y}$ when $y$ is negative. I think it must be implicit to take $y>0$. In which case I agree with Andre Nicolas's bounds. – froggie May 9 '12 at 23:19
There is an unstated but implicit assumption that $y \ge 0$, because of the bound on $z$. This forces $x \le 1$.
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Thus our triple integral can be expressed as the iterated integral $$\int_{x=0}^1\left(\int_{y=0}^{1-x}\left(\int_{z=0}^{\sqrt{y}} dz \right)dy\right)dx.$$ The ensuing computation poses no special difficulties.
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https://math.stackexchange.com/questions/1061375/equivalence-relation-properties
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# Equivalence Relation Properties
I have to define whether the following relation is symmetric, reflexive, and transitive.
Define a relation R on Z as follows: (x, y) ∈ R if and only if x = |y|:
This is my answer so far:
Is R symmetric? No
Is R reflexive? No
Is R transitive? No
I am not exactly confident with this answer because I am not sure if we should check these properties based on x, y for all x, yR or based on x, y for all x, yZ
Note that $R$ is defined on $\mathbb Z$, so we assume that $x,y \in \mathbb Z$.
You're correct that $R$ is not symmetric:
• Take $x = 3 \in \mathbb Z$ and $y = -3 \in \mathbb Z$. Notice that $3 = |-3|$ but $-3 \neq 3 = |3|$. So $(x,y) \in R$ but $(y,x) \notin R$.
You're correct that $R$ is not reflexive:
• Take $x = -3 \in \mathbb Z$. Notice that $-3 \neq 3 = |-3|$. So $(x,x) \notin R$.
However, $R$ is in fact transitive:
• Choose any $x,y,z \in \mathbb Z$ such that $(x,y),(y,z) \in R$ so that $x = |y|$ and $y = |z|$. Then by substituting, notice that: $$x = |y| = ||z|| = |z|$$ So $(x,z) \in R$, as desired.
• As far as I know, picking values for (x,y),(y,z)∈R requires us to pick 3 distinct values for x, y, and z. However, if we want to satisfy the equation x=|y| and y=|z|, we can only choose at the most 2 distinct values. Or it does not matter? – amundi12 Dec 10 '14 at 21:08
• The three values need not be distinct. Indeed, the identity relation (which only contains ordered pairs of the form $(a,a)$) is transitive. – Adriano Dec 10 '14 at 21:11
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# Equivalence Relation Properties
I have to define whether the following relation is symmetric, reflexive, and transitive. Define a relation R on Z as follows: (x, y) ∈ R if and only if x = |y|:
This is my answer so far:
Is R symmetric? No
Is R reflexive? No
Is R transitive? No
I am not exactly confident with this answer because I am not sure if we should check these properties based on x, y for all x, yR or based on x, y for all x, yZ
Note that $R$ is defined on $\mathbb Z$, so we assume that $x,y \in \mathbb Z$. You're correct that $R$ is not symmetric:
• Take $x = 3 \in \mathbb Z$ and $y = -3 \in \mathbb Z$. Notice that $3 = |-3|$ but $-3 \neq 3 = |3|$. So $(x,y) \in R$ but $(y,x) \notin R$. You're correct that $R$ is not reflexive:
• Take $x = -3 \in \mathbb Z$. Notice that $-3 \neq 3 = |-3|$. So $(x,x) \notin R$. However, $R$ is in fact transitive:
• Choose any $x,y,z \in \mathbb Z$ such that $(x,y),(y,z) \in R$ so that $x = |y|$ and $y = |z|$.
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Then by substituting, notice that: $$x = |y| = ||z|| = |z|$$ So $(x,z) \in R$, as desired.
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https://puzzling.stackexchange.com/questions/85151/fibonacci-cycles-back-figure-out-the-fibonacci-number-fbn
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# Fibonacci Cycles Back...Figure out the Fibonacci Number FBN
$$Given$$:
$$F$$, $$B$$, $$N$$, $$U$$, $$V$$ are all digits that can vary from 0 to 9..but not necessarily distinct.
$$FBN$$, $$NBF$$, $$UV$$, $$VU$$ are all concatenated numbers.
From information given below, what is $$FBN$$?
One possible solution
$$F=U=1$$, $$V=2$$, $$N=B=4$$
Reasoning
$$144 = 12^2$$ and $$441=21^2$$ and $$FBN=144$$ is a Fibonacci number.
Alternative,
It's easy to show that the above is the only solution with $$F>0$$ as there are only five Fibonacci numbers between $$100$$ and $$1000$$ and only one is a square. If we also allow $$F=0$$ then there are two other possible solutions.
$$F=B=U=0, V=N=1$$ $$F=B=U=V=N=0$$ These would satisfy the constraints given that the reversal strictly implies a $$3$$-digit reversal at the top and a $$2$$-digit reversal at the bottom.
• 121 is not a fibonacci
– Uvc
Commented Jun 16, 2019 at 15:02
• @Uvc sorry I missed that requirement, updated now. Commented Jun 16, 2019 at 15:06
• That’s much better...
– Uvc
Commented Jun 16, 2019 at 15:07
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# Fibonacci Cycles Back...Figure out the Fibonacci Number FBN
$$Given$$:
$$F$$, $$B$$, $$N$$, $$U$$, $$V$$ are all digits that can vary from 0 to 9..but not necessarily distinct. $$FBN$$, $$NBF$$, $$UV$$, $$VU$$ are all concatenated numbers. From information given below, what is $$FBN$$? One possible solution
$$F=U=1$$, $$V=2$$, $$N=B=4$$
Reasoning
$$144 = 12^2$$ and $$441=21^2$$ and $$FBN=144$$ is a Fibonacci number. Alternative,
It's easy to show that the above is the only solution with $$F>0$$ as there are only five Fibonacci numbers between $$100$$ and $$1000$$ and only one is a square. If we also allow $$F=0$$ then there are two other possible solutions.
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$$F=B=U=0, V=N=1$$ $$F=B=U=V=N=0$$ These would satisfy the constraints given that the reversal strictly implies a $$3$$-digit reversal at the top and a $$2$$-digit reversal at the bottom.
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https://math.stackexchange.com/questions/3042419/differential-equation-beginner-question
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# differential equation - beginner question [closed]
If I have a differential equation on the form
$$y = y' \cdot c_1$$
can I freely solve for $$y'$$ and use the solution for
$$y' = y \cdot c_2$$
where $$c_2 = \frac{1}{c_1}$$?
## closed as off-topic by RRL, Saad, Cesareo, metamorphy, José Carlos SantosDec 21 '18 at 12:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Saad, Cesareo, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
Yes, of course. Assuming that $$c \in \mathbb R$$ is a constant, then if $$c \neq 0$$ :
$$y = y' \cdot c \Leftrightarrow y' = y \cdot \frac{1}{c} \equiv y \cdot c$$
Since $$c$$ is an arbitrary constant, any expression of it will also be a constant, so you can always "manipulate" it to be just $$c$$. Note that only if you have some certain restrictions for $$c$$, then you will need to take these in mind on how they affect the expression $$1/c$$.
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# differential equation - beginner question [closed]
If I have a differential equation on the form
$$y = y' \cdot c_1$$
can I freely solve for $$y'$$ and use the solution for
$$y' = y \cdot c_2$$
where $$c_2 = \frac{1}{c_1}$$? ## closed as off-topic by RRL, Saad, Cesareo, metamorphy, José Carlos SantosDec 21 '18 at 12:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Saad, Cesareo, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question. Yes, of course.
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Assuming that $$c \in \mathbb R$$ is a constant, then if $$c \neq 0$$ :
$$y = y' \cdot c \Leftrightarrow y' = y \cdot \frac{1}{c} \equiv y \cdot c$$
Since $$c$$ is an arbitrary constant, any expression of it will also be a constant, so you can always "manipulate" it to be just $$c$$.
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https://math.stackexchange.com/questions/858477/graph-with-edge-disjoint-cycles/858510
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# Graph with edge disjoint cycles
If the vertices of graph have a degree of at least $n\geq2$, show that the graph has at least $\frac{n}{2}$ edge disjoint cycles.
Unsure how to approach this, but I understand that edge disjoint cycles are cycles within a graph that don't have the same edge.
• I think you mean it has at least $\lfloor \frac{n}{2} \rfloor$ disjoint cycles since for example: a $K_4$ doesn't have at least 1.5 disjoint cycles. – Jorge Fernández Hidalgo Jul 7 '14 at 2:19
## 1 Answer
Before you do anything, note that the graph must be finite. If the graph weren't finite, a trivial example of a graph with every vertex of degree $2$ without any cycle would be the integer line, where every point is labelled with an integer and each point $n$ would be connected to points $n+1$ and $n-1$.
With that out of the way, it's mostly a matter of actually drawing out these edge-disjoint cycles.
We can use induction. Let $G$ be a finite graph where every vertex is of degree at least $2$. Then, pick any vertex $v_0$, and starting from that vertex, trace a path through the vertices $v_1, v_2, v_3, \ldots$. If at any point along this path we reach a vertex $v_n$ that we already passed earlier in the path, we have a cycle and we're done. But supposing we'd never visited $v_n$ before, then only one edge on $v_n$ is part of the path, and therefore we can continue drawing the path by the another edge on $v_n$ since the degree of $v_n$ is at least 2.
Continue until you do get a vertex that's already in the path – because the graph is finite, this will happen eventually by the pigeonhole principle.
Now that we've proven that, do you see how to approach the problem when each vertex is of degree at least $2n$?
• I understand that this isn't a complete answer, but I'm not sure if you're doing this as a homework question or not. I will fill in the rest eventually. – Joe Z. Jul 7 '14 at 1:49
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# Graph with edge disjoint cycles
If the vertices of graph have a degree of at least $n\geq2$, show that the graph has at least $\frac{n}{2}$ edge disjoint cycles. Unsure how to approach this, but I understand that edge disjoint cycles are cycles within a graph that don't have the same edge. • I think you mean it has at least $\lfloor \frac{n}{2} \rfloor$ disjoint cycles since for example: a $K_4$ doesn't have at least 1.5 disjoint cycles. – Jorge Fernández Hidalgo Jul 7 '14 at 2:19
## 1 Answer
Before you do anything, note that the graph must be finite. If the graph weren't finite, a trivial example of a graph with every vertex of degree $2$ without any cycle would be the integer line, where every point is labelled with an integer and each point $n$ would be connected to points $n+1$ and $n-1$. With that out of the way, it's mostly a matter of actually drawing out these edge-disjoint cycles. We can use induction. Let $G$ be a finite graph where every vertex is of degree at least $2$. Then, pick any vertex $v_0$, and starting from that vertex, trace a path through the vertices $v_1, v_2, v_3, \ldots$. If at any point along this path we reach a vertex $v_n$ that we already passed earlier in the path, we have a cycle and we're done.
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But supposing we'd never visited $v_n$ before, then only one edge on $v_n$ is part of the path, and therefore we can continue drawing the path by the another edge on $v_n$ since the degree of $v_n$ is at least 2.
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https://cs.stackexchange.com/questions/113485/recurrence-relation-not-solvable-by-the-master-theorem?noredirect=1
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# Recurrence relation (not solvable by the master theorem) [duplicate]
Consider the following recursion: $$\begin{cases} T(n) = 2T(\frac{n}{2}) + \frac{n}{\log n} &n > 1 \\ O(1) &n = 1 \end{cases}$$.
The master theorem doesn't work, as the exponent of $$\log n$$ is negative. So I tried unfolding the relation and finally got the equation: $$T(n) = n[1 + \frac{1}{\log(\frac{n}{2})} + \frac{1}{\log(\frac{n}{4})} + ... + \frac{1}{\log(2)}]$$.
I do not know how to simplify (inequalities to use???) from here. A trivial method would be to assume that all reciprocal of the log terms are $$< \frac{1}{\log(2)}$$, and since there are $$\log n$$ terms, the summation of all the reciprocal-log terms is $$< \frac{\log n }{\log(2)} = \log_2 n$$, which gives $$T(n) = O(n \log n)$$. However this is a very poor approximation, as by the master theorem we can check that the time complexity for the recursive relation $$T(n) = 2T(\frac{n}{2}) + n$$ is $$O(n \log n)$$. Can someone find a tighter correct upper bound?
• Check the Wikipedia page. – Yuval Filmus Sep 6 '19 at 21:52
• Note that it's the master theorem (like a master key), not Master's theorem (named after some Professor Master). – David Richerby Sep 7 '19 at 9:11
Wikipedia has a slight extension of the master theorem which covers your case: case 2b here. For the recurrence $$T(n)=aT(n/b)+f(n)$$ where $$f(n)=\Theta(n^{\log_b a}/\log n)$$, it gives $$T(n)=n^{\log_b a}\log\log n$$.
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# Recurrence relation (not solvable by the master theorem) [duplicate]
Consider the following recursion: $$\begin{cases} T(n) = 2T(\frac{n}{2}) + \frac{n}{\log n} &n > 1 \\ O(1) &n = 1 \end{cases}$$. The master theorem doesn't work, as the exponent of $$\log n$$ is negative. So I tried unfolding the relation and finally got the equation: $$T(n) = n[1 + \frac{1}{\log(\frac{n}{2})} + \frac{1}{\log(\frac{n}{4})} + ... + \frac{1}{\log(2)}]$$. I do not know how to simplify (inequalities to use???) from here. A trivial method would be to assume that all reciprocal of the log terms are $$< \frac{1}{\log(2)}$$, and since there are $$\log n$$ terms, the summation of all the reciprocal-log terms is $$< \frac{\log n }{\log(2)} = \log_2 n$$, which gives $$T(n) = O(n \log n)$$. However this is a very poor approximation, as by the master theorem we can check that the time complexity for the recursive relation $$T(n) = 2T(\frac{n}{2}) + n$$ is $$O(n \log n)$$. Can someone find a tighter correct upper bound? • Check the Wikipedia page. – Yuval Filmus Sep 6 '19 at 21:52
• Note that it's the master theorem (like a master key), not Master's theorem (named after some Professor Master). – David Richerby Sep 7 '19 at 9:11
Wikipedia has a slight extension of the master theorem which covers your case: case 2b here.
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For the recurrence $$T(n)=aT(n/b)+f(n)$$ where $$f(n)=\Theta(n^{\log_b a}/\log n)$$, it gives $$T(n)=n^{\log_b a}\log\log n$$.
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https://math.stackexchange.com/questions/228950/find-a-complex-number-geometrically
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# Find a complex number geometrically
Consider the triangle $\Delta ABC$, which $D$ is the midpoint of segment $BC$, and let the point G be defined such that $(GD)= \frac{1}{3}(AD)$. Assuming that $z_A, z_B, z_C$ are the complex numbers representing the points $(A, B, C)$:
a. Find the complex number $z_G$ that represents the point $G$
b. Show that $(CG)= \frac{2}{3}(CF)$ and that $F$ is the midpoint of the segment $(AB)$
How would you go about solving this? I would apply the distance formula but I am not given any actual complex number. I know that a complex number can be represented as a vector connected to origin.
Let $z_1, z_2, z_3$ be the points A,B,C. Then it is clear that $$D=\frac{z_2+z_3}{2}$$
The parametric equation of the line from $D$ to $A$ is $$\gamma(t)=\frac{z_2+z_3}{2}+t (z_1 -\frac{z_2+z_3}{2})$$
Therefore $G=\gamma( \frac{1}{3} )$ which you can simplify.
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# Find a complex number geometrically
Consider the triangle $\Delta ABC$, which $D$ is the midpoint of segment $BC$, and let the point G be defined such that $(GD)= \frac{1}{3}(AD)$. Assuming that $z_A, z_B, z_C$ are the complex numbers representing the points $(A, B, C)$:
a. Find the complex number $z_G$ that represents the point $G$
b. Show that $(CG)= \frac{2}{3}(CF)$ and that $F$ is the midpoint of the segment $(AB)$
How would you go about solving this? I would apply the distance formula but I am not given any actual complex number. I know that a complex number can be represented as a vector connected to origin. Let $z_1, z_2, z_3$ be the points A,B,C.
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Then it is clear that $$D=\frac{z_2+z_3}{2}$$
The parametric equation of the line from $D$ to $A$ is $$\gamma(t)=\frac{z_2+z_3}{2}+t (z_1 -\frac{z_2+z_3}{2})$$
Therefore $G=\gamma( \frac{1}{3} )$ which you can simplify.
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https://math.stackexchange.com/questions/3178301/summation-inequality-limit-for-decreasing-sequences
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# summation inequality/limit for decreasing sequences
Let $$(w_n)_{n=1}^\infty$$ be a decreasing sequence of real numbers satisfying $$\lim_{n\to\infty}w_n=0\;\text{ and }\;\sum_{n=1}^\infty w_n=\infty.$$ Conjecture. For each $$M,N\in\mathbb{N}$$ we have $$\lim_{j\to\infty}\frac{\sum_{n=(M-1)j+1}^{(M-1)j+N}w_n}{\sum_{n=(M-1)j+1}^{Mj}w_n}=0.$$ Obviously this is true if $$M=1$$ and I suspect it is true for general $$M$$, but I'm stuck on proving it. The problem is that $$w_n/w_{n+1}$$ can be arbitrarily large. On the other hand, such large jumps must be spaced out to ensure that $$\sum w_n=\infty$$, and it seems like that might be enough to get the limit expression small. But a proof is eluding me, and maybe the conjecture isn't even true, my intuition notwithstanding.
Any help would be much appreciated. Thanks!
## 1 Answer
Let us take $$M = 2$$, $$N = 1$$.
Let us define $$w_n$$ by steps. Say we have already defined it up to $$m$$. Then let $$w_{m + 1} = w_m$$, $$k > \max(\frac{m}{w_m}, m)$$ and $$w_n = \frac{w_m}{m - 1}$$ for $$n = m + 2, \ldots, m + m$$. Then for $$j = m$$ we have $$\frac{\sum_{n=m+1}^{m+1} w_n}{\sum_{n=m+1}^{2m} w_n} = \frac{w_m}{w_m + \frac{w_m}{m - 1} \cdot (m - 1)} = \frac{1}{2}$$.
Also we have $$\sum_{n=m+1}^{m+m} w_m > 1$$, so our series diverge.
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# summation inequality/limit for decreasing sequences
Let $$(w_n)_{n=1}^\infty$$ be a decreasing sequence of real numbers satisfying $$\lim_{n\to\infty}w_n=0\;\text{ and }\;\sum_{n=1}^\infty w_n=\infty.$$ Conjecture. For each $$M,N\in\mathbb{N}$$ we have $$\lim_{j\to\infty}\frac{\sum_{n=(M-1)j+1}^{(M-1)j+N}w_n}{\sum_{n=(M-1)j+1}^{Mj}w_n}=0.$$ Obviously this is true if $$M=1$$ and I suspect it is true for general $$M$$, but I'm stuck on proving it. The problem is that $$w_n/w_{n+1}$$ can be arbitrarily large. On the other hand, such large jumps must be spaced out to ensure that $$\sum w_n=\infty$$, and it seems like that might be enough to get the limit expression small. But a proof is eluding me, and maybe the conjecture isn't even true, my intuition notwithstanding. Any help would be much appreciated. Thanks! ## 1 Answer
Let us take $$M = 2$$, $$N = 1$$. Let us define $$w_n$$ by steps. Say we have already defined it up to $$m$$. Then let $$w_{m + 1} = w_m$$, $$k > \max(\frac{m}{w_m}, m)$$ and $$w_n = \frac{w_m}{m - 1}$$ for $$n = m + 2, \ldots, m + m$$. Then for $$j = m$$ we have $$\frac{\sum_{n=m+1}^{m+1} w_n}{\sum_{n=m+1}^{2m} w_n} = \frac{w_m}{w_m + \frac{w_m}{m - 1} \cdot (m - 1)} = \frac{1}{2}$$.
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Also we have $$\sum_{n=m+1}^{m+m} w_m > 1$$, so our series diverge.
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http://math.stackexchange.com/questions/70951/1st-order-ode-problem
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# 1st order ODE problem
http://img440.imageshack.us/img440/7352/11unled.jpg
Having some trouble. Im not sure what d_t(u) and :u(with the . above it) means. Would appreciate some help. Thanks
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The "overdot" notation for the derivative with respect to time is due to Newton. The notation now has limited popularity. – André Nicolas Oct 8 '11 at 22:39
The first several things listed are just different notations for the derivative of u with respect to t: $du/dt$, $d_t(u)$, and $\dot{u}$ are all the same thing.
You're asked to solve $u'-g(t)u=h(t)$ with initial condition $u(0)=u_0$.
The trick to solving such an equation is to multiply by some function so that the left hand side is the result of differentiating using the product rule: Say multiply by $X$ and get $Xu'-Xg(t)u = Xh(t)$. We want...
$$Xu'+X'u=\frac{d}{dt}\left[Xu\right] = Xu'-Xg(t)u$$
Thus we need $X'u=-Xg(t)u$ that is $X'=-Xg(t)$ so that $X=exp(-\int_0^t g(x)\,dx)$
Hopefully this will get you started. By the way this kind of equation is a "first order linear ODE". http://en.wikipedia.org/wiki/Linear_differential_equation (see first order)
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$\mathrm d_t u$ and $\dot u$ are both just alternative notations for $\frac{\mathrm du}{\mathrm dt}$.
In fact, that is what the $=:$ symbol means: "The symbol to the right of $=:$ is is hereby defined to be a name for the thing to the left".
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# 1st order ODE problem
http://img440.imageshack.us/img440/7352/11unled.jpg
Having some trouble. Im not sure what d_t(u) and :u(with the . above it) means. Would appreciate some help. Thanks
-
The "overdot" notation for the derivative with respect to time is due to Newton. The notation now has limited popularity. – André Nicolas Oct 8 '11 at 22:39
The first several things listed are just different notations for the derivative of u with respect to t: $du/dt$, $d_t(u)$, and $\dot{u}$ are all the same thing. You're asked to solve $u'-g(t)u=h(t)$ with initial condition $u(0)=u_0$. The trick to solving such an equation is to multiply by some function so that the left hand side is the result of differentiating using the product rule: Say multiply by $X$ and get $Xu'-Xg(t)u = Xh(t)$. We want...
$$Xu'+X'u=\frac{d}{dt}\left[Xu\right] = Xu'-Xg(t)u$$
Thus we need $X'u=-Xg(t)u$ that is $X'=-Xg(t)$ so that $X=exp(-\int_0^t g(x)\,dx)$
Hopefully this will get you started. By the way this kind of equation is a "first order linear ODE". http://en.wikipedia.org/wiki/Linear_differential_equation (see first order)
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$\mathrm d_t u$ and $\dot u$ are both just alternative notations for $\frac{\mathrm du}{\mathrm dt}$.
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In fact, that is what the $=:$ symbol means: "The symbol to the right of $=:$ is is hereby defined to be a name for the thing to the left".
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https://cstheory.stackexchange.com/questions/52238/differing-definitions-of-a-weak-learner
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# Differing definitions of a weak learner
I've been reading about boosting and have come across basically two definitions of a weak learner. Basically for hypothesis $$h$$ and target $$c$$, some definitions says that $$h$$ is a weak learner if $$E[h(x)c(x)] \geq \gamma$$ while others says that it is a weak learner if $$Pr[h(x) \neq c(x)] \leq \frac{1}{2} - \gamma$$. I don't really understand how these two definitions are related, are we made to make assumptions about the weights that the hypothesis function puts on the points in order to calculate expectation from probability? Thanks!
Suppose that the output of $$h,c$$ is either $$+1$$ or $$-1$$. Then $$h(x)c(x)=1$$ iff $$h(x)=c(x)$$. Moreover, if we let $$p=\Pr[h(x)=c(x)]$$, then
\begin{align*} \mathbb{E}[h(x)c(x)] &= 1 \cdot \Pr[h(x)=c(x)] + (-1) \cdot \Pr[h(x)\ne c(x)]\\ &= p - (1-p) = 2p-1. \end{align*}
Now, $$\Pr[h(x)\ne c(x)] \le \frac12 - \gamma$$ is equivalent to $$p \ge \frac12 + \gamma$$, which in turn is equivalent to $$\mathbb{E}[h(x)c(x)] \ge 2 \cdot (\frac12 + \gamma) -1 = 2\gamma$$. So, the two conditions are equivalent, up to a factor of two in $$\gamma$$.
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# Differing definitions of a weak learner
I've been reading about boosting and have come across basically two definitions of a weak learner. Basically for hypothesis $$h$$ and target $$c$$, some definitions says that $$h$$ is a weak learner if $$E[h(x)c(x)] \geq \gamma$$ while others says that it is a weak learner if $$Pr[h(x) \neq c(x)] \leq \frac{1}{2} - \gamma$$. I don't really understand how these two definitions are related, are we made to make assumptions about the weights that the hypothesis function puts on the points in order to calculate expectation from probability? Thanks! Suppose that the output of $$h,c$$ is either $$+1$$ or $$-1$$. Then $$h(x)c(x)=1$$ iff $$h(x)=c(x)$$. Moreover, if we let $$p=\Pr[h(x)=c(x)]$$, then
\begin{align*} \mathbb{E}[h(x)c(x)] &= 1 \cdot \Pr[h(x)=c(x)] + (-1) \cdot \Pr[h(x)\ne c(x)]\\ &= p - (1-p) = 2p-1.
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\end{align*}
Now, $$\Pr[h(x)\ne c(x)] \le \frac12 - \gamma$$ is equivalent to $$p \ge \frac12 + \gamma$$, which in turn is equivalent to $$\mathbb{E}[h(x)c(x)] \ge 2 \cdot (\frac12 + \gamma) -1 = 2\gamma$$.
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https://math.stackexchange.com/questions/3300552/problem-about-an-ellipse-distance-between-two-distance-in-another-view
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# Problem about an ellipse distance between two distance in another view
I have just solved the problem the distance between two points on the circumference of an ellipse following the outer curve after I read the following article.
Is it possible to find the distance between two points on the circumference of an ellipse following the outer curve?
Then I want to solve this problem in another view.
Please watch the example below which link of a question of the owner I have pasted above.
example ←here
This ellipse's equation is $$\frac{x^2}{2^2} + \frac{y^2}{1^2} = 1$$ (Just what original question has mentioned).
So my question is, suppose we know arc-length between two points $$x_A$$ and $$x_B$$ and we know Coordinate of $$x_A$$, how to calculate coordinate of $$x_B$$？If this solution is impossible, we can assume that $$x_A$$ = (0, 1).In this case, the equation has very strong Symmetry, so I think we can solve the problem.
• Both $x_A$ and $x_B$ are located at first quadrant or $x_A = (1, 0)$ and $x_B$ is located at first quadrant. – 石原秀一 Jul 23 '19 at 3:18
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# Problem about an ellipse distance between two distance in another view
I have just solved the problem the distance between two points on the circumference of an ellipse following the outer curve after I read the following article. Is it possible to find the distance between two points on the circumference of an ellipse following the outer curve? Then I want to solve this problem in another view. Please watch the example below which link of a question of the owner I have pasted above. example ←here
This ellipse's equation is $$\frac{x^2}{2^2} + \frac{y^2}{1^2} = 1$$ (Just what original question has mentioned). So my question is, suppose we know arc-length between two points $$x_A$$ and $$x_B$$ and we know Coordinate of $$x_A$$, how to calculate coordinate of $$x_B$$？If this solution is impossible, we can assume that $$x_A$$ = (0, 1).In this case, the equation has very strong Symmetry, so I think we can solve the problem.
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• Both $x_A$ and $x_B$ are located at first quadrant or $x_A = (1, 0)$ and $x_B$ is located at first quadrant.
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https://math.stackexchange.com/questions/4340372/how-to-solve-the-congruence-x30-%E2%89%A1-81x6-pmod269-using-primitive-rootswi
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# How to solve the congruence $x^{30} ≡ 81x^6 \pmod{269}$ using primitive roots(without indices)?
So I know that 3 is a primitive root of 269.
How can I solve $$x^{30} ≡ 81x^6 \pmod{269}$$
Even if I substitute $$x$$ with $$3^y$$, where $$y$$ lies between 0 and 267, I can’t get any solutions.
We can easily see that $$x=0$$ is a solution.
Since $$3$$ is a primite root modulo $$269$$, let $$x \equiv 3^t \pmod{269}$$. hence, we get the congruence:
$$3^{30t} \equiv 3^4 \cdot 3^{6t} \pmod{269}$$ We know that if $$g$$ is a primitive root modulo $$n$$, then $$g^r \equiv g^s \pmod{n} \iff r \equiv s \pmod{\phi(n)}$$ Since $$269$$ is a prime we get that $$\phi(269) = 268$$, and then we get the congruence: $$30t \equiv 4 \cdot 6t \pmod{268}$$ which gives $$2$$ solutions for $$t$$, $$\space$$ $$t\equiv 0 \pmod{268}$$ or $$t\equiv 134 \pmod{268}$$.
From the first solution we get that $$x\equiv 1 \pmod{269}$$ which means that $$\big\{ x = 1 +269k \mid k\in \mathbb{Z}\big\}$$ is one set of solutions.
From the second solution we get that $$x\equiv 268 \pmod{269}$$ which means that $$\big\{ x = 268 +269k \mid k\in \mathbb{Z}\big\}$$ is another set of solutions.
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# How to solve the congruence $x^{30} ≡ 81x^6 \pmod{269}$ using primitive roots(without indices)? So I know that 3 is a primitive root of 269. How can I solve $$x^{30} ≡ 81x^6 \pmod{269}$$
Even if I substitute $$x$$ with $$3^y$$, where $$y$$ lies between 0 and 267, I can’t get any solutions. We can easily see that $$x=0$$ is a solution. Since $$3$$ is a primite root modulo $$269$$, let $$x \equiv 3^t \pmod{269}$$. hence, we get the congruence:
$$3^{30t} \equiv 3^4 \cdot 3^{6t} \pmod{269}$$ We know that if $$g$$ is a primitive root modulo $$n$$, then $$g^r \equiv g^s \pmod{n} \iff r \equiv s \pmod{\phi(n)}$$ Since $$269$$ is a prime we get that $$\phi(269) = 268$$, and then we get the congruence: $$30t \equiv 4 \cdot 6t \pmod{268}$$ which gives $$2$$ solutions for $$t$$, $$\space$$ $$t\equiv 0 \pmod{268}$$ or $$t\equiv 134 \pmod{268}$$. From the first solution we get that $$x\equiv 1 \pmod{269}$$ which means that $$\big\{ x = 1 +269k \mid k\in \mathbb{Z}\big\}$$ is one set of solutions.
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From the second solution we get that $$x\equiv 268 \pmod{269}$$ which means that $$\big\{ x = 268 +269k \mid k\in \mathbb{Z}\big\}$$ is another set of solutions.
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http://math.stackexchange.com/questions/130049/error-propagation-and-averages-a-practical-question
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Error propagation and averages, a practical question
Ok, I'm sure this is simple, yet I'm confused.
Lets say my goal is to obtain an average concentration C=N/V so I take a number of N and V readings with errors:
$N_1=50\pm2, V_1=100\pm4$
$N_2=102\pm2, V_2=205\pm4$
$N_3=52\pm2, V_3=99\pm4$
Now if I remember correctly the standard deviation s propagates to C with
$s_{C} = \sqrt{(\frac{\partial{C}}{\partial{N}}s_N)^2+(\frac{\partial{C}}{\partial{V}}s_V)^2}$
But now how do I get the average with a meaningful total standard deviation? Or should I approach this in a different way altogether?
-
Compute $C_1$, $C_2$, $C_3$ as well as $s_{C_1}$, $s_{C_2}$ and $s_{C_3}$ separately, using the formula you have given. Next, compute the mean $\overline{C} = \dfrac{C_1+C_2+C_3}{3}$, and use the formula you have given to determine the standard deviation $s_{\overline{C}}$.
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Error propagation and averages, a practical question
Ok, I'm sure this is simple, yet I'm confused. Lets say my goal is to obtain an average concentration C=N/V so I take a number of N and V readings with errors:
$N_1=50\pm2, V_1=100\pm4$
$N_2=102\pm2, V_2=205\pm4$
$N_3=52\pm2, V_3=99\pm4$
Now if I remember correctly the standard deviation s propagates to C with
$s_{C} = \sqrt{(\frac{\partial{C}}{\partial{N}}s_N)^2+(\frac{\partial{C}}{\partial{V}}s_V)^2}$
But now how do I get the average with a meaningful total standard deviation? Or should I approach this in a different way altogether? -
Compute $C_1$, $C_2$, $C_3$ as well as $s_{C_1}$, $s_{C_2}$ and $s_{C_3}$ separately, using the formula you have given.
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Next, compute the mean $\overline{C} = \dfrac{C_1+C_2+C_3}{3}$, and use the formula you have given to determine the standard deviation $s_{\overline{C}}$.
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http://physics.stackexchange.com/questions/17840/figuring-out-the-force-using-newtons-2nd-law/17842
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# Figuring out the force using Newton's 2nd Law
Here's a problem on my test review:
A 12 kg crate resta on a horizontal surface and a boy pulls on it with a force that is 30° below the horizontal. If the coefficient of static friction is 0.40, the minimum magnitude force he needs to start the crate moving is:
Okay, so I found the equation for the forces in the Y direction: $$\sum F_{Y} = F_{N} - mg - F\sin\theta = 0$$ $$F_{N} = mg + F\sin\theta$$
And the X direction: $$\sum F_{X} = F\cos(\theta) - F_{f} = 0$$ $$F\cos\theta = F_{f}$$
Solving for the force: $$F\cos\theta = u_{s}F_N$$ $$F\cos\theta = u_s (mg + F\sin\theta)$$ $$F\cos{-30°} = 0.40 \left[(12 kg)(9.8 \frac{m}{s^2}) + F\sin{-30°} \right]$$ $$F \left[\cos{-30°} - (.4)(\sin{-30°} \right] = 47.04 N$$ $$F=44N$$
However, it looks like the correct answer was 71. Any ideas where I went wrong here?
-
The expression for $F_Y$ should be $$\displaystyle\sum F_Y=F_N-mg+F \sin\theta\;.$$ All terms in such expressions should start out positive. They can turn negative later depending on the angle at which they're applied. In fact, if you wanted to be super-explicit you could write $$\displaystyle\sum F_Y=F_N\sin(90°)+mg\sin(-90°)+F \sin\theta\;.$$ It's hardly necessary in this case, though.
Working from $F\cos(-30°)=\mu_sF_N=0.4\left[9.807\times 12-F\sin(-30°)\right]$ gives $F=70.7$. Check your math. – rdhs Dec 5 '11 at 5:40
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# Figuring out the force using Newton's 2nd Law
Here's a problem on my test review:
A 12 kg crate resta on a horizontal surface and a boy pulls on it with a force that is 30° below the horizontal. If the coefficient of static friction is 0.40, the minimum magnitude force he needs to start the crate moving is:
Okay, so I found the equation for the forces in the Y direction: $$\sum F_{Y} = F_{N} - mg - F\sin\theta = 0$$ $$F_{N} = mg + F\sin\theta$$
And the X direction: $$\sum F_{X} = F\cos(\theta) - F_{f} = 0$$ $$F\cos\theta = F_{f}$$
Solving for the force: $$F\cos\theta = u_{s}F_N$$ $$F\cos\theta = u_s (mg + F\sin\theta)$$ $$F\cos{-30°} = 0.40 \left[(12 kg)(9.8 \frac{m}{s^2}) + F\sin{-30°} \right]$$ $$F \left[\cos{-30°} - (.4)(\sin{-30°} \right] = 47.04 N$$ $$F=44N$$
However, it looks like the correct answer was 71. Any ideas where I went wrong here? -
The expression for $F_Y$ should be $$\displaystyle\sum F_Y=F_N-mg+F \sin\theta\;.$$ All terms in such expressions should start out positive. They can turn negative later depending on the angle at which they're applied. In fact, if you wanted to be super-explicit you could write $$\displaystyle\sum F_Y=F_N\sin(90°)+mg\sin(-90°)+F \sin\theta\;.$$ It's hardly necessary in this case, though.
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Working from $F\cos(-30°)=\mu_sF_N=0.4\left[9.807\times 12-F\sin(-30°)\right]$ gives $F=70.7$.
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https://physics.stackexchange.com/questions/636671/internal-force-and-centre-of-mass
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Internal force and centre of mass
I have still not yet learn about the centre of mass of a system but I really do need to solve this question that keeps bugging me. I know that internal forces cannot accelerate the centre of mass of a system, but assuming that I have a block-earth system and the block is dropping from a point and considering that gravitational force is an internal force for this system, won't the centre of mass accelerate since the block is accelerating while the earth is not?
Btw, it would be great if someone can explain the concept of centre of mass of a system to me in a much simple way.
Thanks
Actually, from Newton's Third Law of motion, the Earth is attracted towards the block with exactly the same magnitude of force, with which it attracts the block. But we don't see it accelerating simply because its mass is (numerically) much, much greater than the gravitational force. You can also calculate this acceleration, from Newton's law of universal gravitation. We have: $$F=\frac {GMm}{R^2}$$ Hence, $$a_{Earth}=\frac {F}{M}$$ which comes out to be negligibly small in most cases, hence unnoticeable.
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Internal force and centre of mass
I have still not yet learn about the centre of mass of a system but I really do need to solve this question that keeps bugging me. I know that internal forces cannot accelerate the centre of mass of a system, but assuming that I have a block-earth system and the block is dropping from a point and considering that gravitational force is an internal force for this system, won't the centre of mass accelerate since the block is accelerating while the earth is not? Btw, it would be great if someone can explain the concept of centre of mass of a system to me in a much simple way. Thanks
Actually, from Newton's Third Law of motion, the Earth is attracted towards the block with exactly the same magnitude of force, with which it attracts the block. But we don't see it accelerating simply because its mass is (numerically) much, much greater than the gravitational force. You can also calculate this acceleration, from Newton's law of universal gravitation.
|
We have: $$F=\frac {GMm}{R^2}$$ Hence, $$a_{Earth}=\frac {F}{M}$$ which comes out to be negligibly small in most cases, hence unnoticeable.
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https://math.stackexchange.com/questions/3476886/solve-by-direct-integration
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# Solve by direct integration
$$\frac{\partial^2 z}{\partial x^2}+z=0$$, given that at $$x=0, z=e^y$$ and $$\frac{\partial z}{\partial x}=1$$
I am doing it in following way:
Integrating w.r.t x twice
$$\implies \frac{\partial z}{\partial x} +zx=f(y)$$, where f(y) is an arbitrary function ...(1)
$$\implies z +\frac {zx^2}{2}=xf(y)+g(y)$$, where g(y) is an arbitrary function ...(2)
Putting $$x=0, z=e^y$$ and $$\frac{\partial z}{\partial x}=1$$ in (1) and (2)
$$\implies e^y=g(y)$$ and $$f(y)=1$$
$$\implies z(1+\frac {x^2}{2})=x+e^y$$
But the answer to this question is given as $$z=sinx+e^ycosx$$.
Can someone tell me how to solve this PDE by direct integration only?
• $z$ is supposed to be a function of $x$. The integral of $z$ w.r.t. $x$ is not $zx$. Commented Dec 15, 2019 at 5:15
Just pretend that $$y$$ is a constant and solve the equation as if $$z$$ is a function of $$x$$ alone. Then the solution is $$A \cos x+B\sin x$$ where $$A$$ and $$B$$ are constants. Now if you consider $$y$$ as a variable then you have to replace $$A$$ and $$B$$ by functions of $$y$$. The initial conditions easily give you $$A=e^{y}$$ and $$B=1$$ so the solution is $$e^{y} \cos x+\sin x$$.
• The general solution of the ODE $y''+y=0$ is $y=A\cos x+B\sin x$. I am assuming that you know this already since this is only an ODE and only one variable calculus is involved. @ShivaneeGupta Commented Dec 15, 2019 at 5:31
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# Solve by direct integration
$$\frac{\partial^2 z}{\partial x^2}+z=0$$, given that at $$x=0, z=e^y$$ and $$\frac{\partial z}{\partial x}=1$$
I am doing it in following way:
Integrating w.r.t x twice
$$\implies \frac{\partial z}{\partial x} +zx=f(y)$$, where f(y) is an arbitrary function ...(1)
$$\implies z +\frac {zx^2}{2}=xf(y)+g(y)$$, where g(y) is an arbitrary function ...(2)
Putting $$x=0, z=e^y$$ and $$\frac{\partial z}{\partial x}=1$$ in (1) and (2)
$$\implies e^y=g(y)$$ and $$f(y)=1$$
$$\implies z(1+\frac {x^2}{2})=x+e^y$$
But the answer to this question is given as $$z=sinx+e^ycosx$$. Can someone tell me how to solve this PDE by direct integration only? • $z$ is supposed to be a function of $x$. The integral of $z$ w.r.t. $x$ is not $zx$. Commented Dec 15, 2019 at 5:15
Just pretend that $$y$$ is a constant and solve the equation as if $$z$$ is a function of $$x$$ alone. Then the solution is $$A \cos x+B\sin x$$ where $$A$$ and $$B$$ are constants. Now if you consider $$y$$ as a variable then you have to replace $$A$$ and $$B$$ by functions of $$y$$. The initial conditions easily give you $$A=e^{y}$$ and $$B=1$$ so the solution is $$e^{y} \cos x+\sin x$$.
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• The general solution of the ODE $y''+y=0$ is $y=A\cos x+B\sin x$.
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http://math.stackexchange.com/questions/209163/proving-r-ge-2-r-using-synthetic-geometry
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# Proving $R\ge 2 r$ using synthetic geometry
If $R$ and $r$ be the radii of the circumcircle and incircle of a triangle, then how do I prove by synthetic geometry(i.e. without trigonometry) that $R\ge 2r$?
I am aware of a trigonometric proof but I am not quite sure if I can come up with a synthetic one. In case anyone is interested in the trigonometric proof, I can add it if you ask me to.
-
You mean the incircle? – joriki Oct 8 '12 at 11:58
@joriki,Oh yes, I do mean incircle. – Richard Nash Oct 8 '12 at 14:17
Several of the arguments given in these two questions do not involve trigonometry: math.stackexchange.com/questions/170853/… math.stackexchange.com/questions/170813/… – Micah Oct 8 '12 at 14:26
You can prove something stronger:
$$OI^2=R^2-2Rr$$
Let $ABC$ be your triangle, extend $AI$ until it meets the Circle at $A'$.
Extend $OI$ until it meets the circle at $D$ and $E$. By the power of point to the circle,
$$AI \cdot A'I= ID \cdot IE = (R+OI)(R-OI)= R^2-OI^2 \,.$$
Now, oberve that $ICA'$ is isosceles, by proving that the angles are equal, and obtain $A'I=A'C$.
Last but not least, drop the perpendicular $IC'$ from $I$ onto $AB$, and the diametre $A'A''$ from $A'$. By similar triangle $AIC' \sim A'A''C$, you have
$$AI \cdot A'C =2Rr$$
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That is exactly what I was looking for.Thanks . – Richard Nash Oct 8 '12 at 14:38
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# Proving $R\ge 2 r$ using synthetic geometry
If $R$ and $r$ be the radii of the circumcircle and incircle of a triangle, then how do I prove by synthetic geometry(i.e. without trigonometry) that $R\ge 2r$? I am aware of a trigonometric proof but I am not quite sure if I can come up with a synthetic one. In case anyone is interested in the trigonometric proof, I can add it if you ask me to. -
You mean the incircle? – joriki Oct 8 '12 at 11:58
@joriki,Oh yes, I do mean incircle. – Richard Nash Oct 8 '12 at 14:17
Several of the arguments given in these two questions do not involve trigonometry: math.stackexchange.com/questions/170853/… math.stackexchange.com/questions/170813/… – Micah Oct 8 '12 at 14:26
You can prove something stronger:
$$OI^2=R^2-2Rr$$
Let $ABC$ be your triangle, extend $AI$ until it meets the Circle at $A'$. Extend $OI$ until it meets the circle at $D$ and $E$. By the power of point to the circle,
$$AI \cdot A'I= ID \cdot IE = (R+OI)(R-OI)= R^2-OI^2 \,.$$
Now, oberve that $ICA'$ is isosceles, by proving that the angles are equal, and obtain $A'I=A'C$. Last but not least, drop the perpendicular $IC'$ from $I$ onto $AB$, and the diametre $A'A''$ from $A'$.
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By similar triangle $AIC' \sim A'A''C$, you have
$$AI \cdot A'C =2Rr$$
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That is exactly what I was looking for.Thanks .
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https://math.stackexchange.com/questions/708661/russian-roulette-how-many-people-left
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# Russian roulette, how many people left
I have a questian about a game similar to russian roulette. Suppose that we have n people in a room. Every round, everyone shoots a random person. So it can happen that everbody dies, or if everyone shoots the same person only two people die(the unlucky person and the person that hè shot). I want to know what the expected number of surviving persons is after one round. I have no clue how to approach this problem.
• "Russian roulette" normally involves shooting oneself, not shooting other people. – vadim123 Mar 11 '14 at 23:10
Let $X_i=1$ if $i$ survives, and $X_i=0$ if she does not.
Then the number of survivors is $X_1+\cdots+X_n$. By the linearity of expectation the expected number of survivors is $E(X_1)+\cdots +E(X_n)$.
The probability that $i$ survives is the probability nobody shoots at her. This is $\left(\frac{n-2}{n-1}\right)^{n-1}$, since every person other than herself must shoot at someone other than herself.
The expectation is therefore $n\left(\frac{n-2}{n-1}\right)^{n-1}$, approximately $\frac{n}{e}$ unless $n$ is small.
Hint: pick one person and figure the chance he survives. There are $n-1$ people who might shoot him. Then use the linearity of expectation to get the expected number surviving.
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# Russian roulette, how many people left
I have a questian about a game similar to russian roulette. Suppose that we have n people in a room. Every round, everyone shoots a random person. So it can happen that everbody dies, or if everyone shoots the same person only two people die(the unlucky person and the person that hè shot). I want to know what the expected number of surviving persons is after one round. I have no clue how to approach this problem. • "Russian roulette" normally involves shooting oneself, not shooting other people. – vadim123 Mar 11 '14 at 23:10
Let $X_i=1$ if $i$ survives, and $X_i=0$ if she does not. Then the number of survivors is $X_1+\cdots+X_n$. By the linearity of expectation the expected number of survivors is $E(X_1)+\cdots +E(X_n)$. The probability that $i$ survives is the probability nobody shoots at her. This is $\left(\frac{n-2}{n-1}\right)^{n-1}$, since every person other than herself must shoot at someone other than herself.
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The expectation is therefore $n\left(\frac{n-2}{n-1}\right)^{n-1}$, approximately $\frac{n}{e}$ unless $n$ is small.
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https://math.stackexchange.com/questions/1113489/finding-the-nth-term-in-a-recursive-coupled-equation
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# Finding the nth term in a recursive coupled equation.
I'm probably missing something simple, but if I have the recursive sequence: $$a_{i+1} = \delta a_i+\lambda_1 b_i$$ $$b_{i+1} = \lambda_2 a_i + \delta b_i$$
how would I find a formula for $a_n$, $b_n$, or even $\frac{a_n}{b_n}$, given, for example, $a_0 = 1$, $b_0=0$?
I've tried expanding it out and looking for patterns but to no avail - I'm sure there must be an analytic solution to this, I really would rather not do it computationally!
Let $X_i:=[a_i,b_i]^T$ then $$X_{i+1}=AX_i$$ where $$A:=\begin{bmatrix}\delta&\lambda_1\\\lambda_2&\delta\end{bmatrix}$$
Then it is clear that $$X_i=A^{i}X_0$$
To get a nice formula for $A^i$ first write it as $P^{-1}JP$, where $J$ is its Jordan form. Notice that if $\lambda_1\lambda_2\neq0$ $J$ is going to be diagonal (even better).
In any case, the worst that can happen is that $J=D+N$ where $D$ is diagonal and $N^2=0$. The, $$A^i=P^{-1}J^iP=P^{-1}(D^i+iD^{i-1}N)P$$ where the powers $D^i$ and $D^{i-1}$ are easy to write because it is just rising the diagonal elements to $i$.
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# Finding the nth term in a recursive coupled equation. I'm probably missing something simple, but if I have the recursive sequence: $$a_{i+1} = \delta a_i+\lambda_1 b_i$$ $$b_{i+1} = \lambda_2 a_i + \delta b_i$$
how would I find a formula for $a_n$, $b_n$, or even $\frac{a_n}{b_n}$, given, for example, $a_0 = 1$, $b_0=0$? I've tried expanding it out and looking for patterns but to no avail - I'm sure there must be an analytic solution to this, I really would rather not do it computationally! Let $X_i:=[a_i,b_i]^T$ then $$X_{i+1}=AX_i$$ where $$A:=\begin{bmatrix}\delta&\lambda_1\\\lambda_2&\delta\end{bmatrix}$$
Then it is clear that $$X_i=A^{i}X_0$$
To get a nice formula for $A^i$ first write it as $P^{-1}JP$, where $J$ is its Jordan form. Notice that if $\lambda_1\lambda_2\neq0$ $J$ is going to be diagonal (even better). In any case, the worst that can happen is that $J=D+N$ where $D$ is diagonal and $N^2=0$.
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The, $$A^i=P^{-1}J^iP=P^{-1}(D^i+iD^{i-1}N)P$$ where the powers $D^i$ and $D^{i-1}$ are easy to write because it is just rising the diagonal elements to $i$.
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https://math.stackexchange.com/questions/2762320/is-it-possible-to-build-a-unitary-matrix-from-a-diagonal-one
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# Is it possible to build a unitary matrix from a diagonal one?
Let $\Delta$ be a diagonal, non-invertible matrix with complex entries. Is it possible to come up with a matrix $M$ such that
$U\equiv M.\Delta$
is unitary?
Though I don't know about the proof, I heard that there is a theorem stating that any invertible matrix $A$ with complex entries can be written as
$A=U.T$
where $T$ is upper triangular. Since diagonal matrices are a subset of upper triangular, I basically want to know if the converse of this theorem above is true, and if there's an algorithmic way to find $A$ such that $A.T^{-1}$ defines a unitary.
PS.: $M$ has the same dimension as $\Delta$
• Is $M$ any matrix? Then set $M=U\Delta^{-1}$. I guess you need some properties of $M$ that are not mentioned above.
– A.Γ.
May 1, 2018 at 19:20
• You're right, let's assume $\Delta$ does not have inverse.
– dwfa
May 1, 2018 at 20:29
• If $\Delta$ is square then it is not possible ($|\det U|=1$, but $\det M\Delta=0$). In general, $\Delta$ must have full column rank necessarily. But for diagonal matrices it means that neglecting zeros below we get an invertible diagonal submatrix.
– A.Γ.
May 1, 2018 at 20:35
Consider a general case of $n\times m$ matrix $M$ and $m\times n$ matrix $\Delta$ (maybe even non-diagonal). Then we have two cases:
Case 1: $\text{rank}\,\Delta<n$.
Then $\text{rank}\,U=\text{rank}\,M\Delta\le\text{rank}\,\Delta<n$ $\Rightarrow$ impossible for a unitary $U$.
Case 2: $\text{rank}\,\Delta=n$.
Then $\Delta$ has a left inverse, and choosing $M$ to be any left inverse (for example, $M=(\Delta^*\Delta)^{-1}\Delta^*$) gives $U=I$ (unitary).
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# Is it possible to build a unitary matrix from a diagonal one? Let $\Delta$ be a diagonal, non-invertible matrix with complex entries. Is it possible to come up with a matrix $M$ such that
$U\equiv M.\Delta$
is unitary? Though I don't know about the proof, I heard that there is a theorem stating that any invertible matrix $A$ with complex entries can be written as
$A=U.T$
where $T$ is upper triangular. Since diagonal matrices are a subset of upper triangular, I basically want to know if the converse of this theorem above is true, and if there's an algorithmic way to find $A$ such that $A.T^{-1}$ defines a unitary. PS. : $M$ has the same dimension as $\Delta$
• Is $M$ any matrix? Then set $M=U\Delta^{-1}$. I guess you need some properties of $M$ that are not mentioned above. – A.Γ. May 1, 2018 at 19:20
• You're right, let's assume $\Delta$ does not have inverse. – dwfa
May 1, 2018 at 20:29
• If $\Delta$ is square then it is not possible ($|\det U|=1$, but $\det M\Delta=0$). In general, $\Delta$ must have full column rank necessarily. But for diagonal matrices it means that neglecting zeros below we get an invertible diagonal submatrix. – A.Γ. May 1, 2018 at 20:35
Consider a general case of $n\times m$ matrix $M$ and $m\times n$ matrix $\Delta$ (maybe even non-diagonal). Then we have two cases:
Case 1: $\text{rank}\,\Delta<n$. Then $\text{rank}\,U=\text{rank}\,M\Delta\le\text{rank}\,\Delta<n$ $\Rightarrow$ impossible for a unitary $U$. Case 2: $\text{rank}\,\Delta=n$.
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Then $\Delta$ has a left inverse, and choosing $M$ to be any left inverse (for example, $M=(\Delta^*\Delta)^{-1}\Delta^*$) gives $U=I$ (unitary).
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https://math.stackexchange.com/questions/2642498/understanding-the-cartesian-product-pa-times-a
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# Understanding the Cartesian product: $P(A) \times A$
Let $A = \{x,y,z\}$ and $B = \{1,2,3\}$.
The power set of $A$ is $$P(A) = \{∅, \{x\}, \{y\}, \{z\}, \{x, y\}, \{x, z\}, \{y, z\}, \{x, y, z\}\}.$$
When working out the Cartesian product for example $A \times B$ you simply order the pairs in which the first element comes from A and the second elements come from B.
A x B = {$(x,1),(x,2),(x,3),(y,1),(y,2),(y,3),(z,1),(z,2),(z,3)$}
B x A = {$(1,x),(1,y)...(3,z)$}
For the Cartesian product $P(A)\times A$ you can not create a list of ordered pairs as $P(A)$ has a list of lists and $A$ has a list of elements. Is there a way to find the Cartesian product of $P(A)\times A$?
• It seems to me that the set would consist of $$\{(x, y)\mid x\in P(A) \land y \in A\}$$ That is, the ordered pair consists of (set in P(A), element in A). – Namaste Feb 8 '18 at 22:37
• So, for example (∅,x), ({x},x), {{y},x), ({z},x), ({x,y},x). This makes sense. – Danny Sanderson Feb 8 '18 at 22:42
The set would consist of $$P(A)\times A = \{(a, b)\mid a\in P(A) \land b \in A\}$$ That is, the ordered pair consists of (set (element) in $P(A)$, element in $A$).
$$P(A)\times A = \{(\varnothing, x), (\varnothing, y), (\varnothing, z), (\{x\}, x), \ldots (\{y\}, x), \ldots (\{x, y, z\}, z)\}$$
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# Understanding the Cartesian product: $P(A) \times A$
Let $A = \{x,y,z\}$ and $B = \{1,2,3\}$. The power set of $A$ is $$P(A) = \{∅, \{x\}, \{y\}, \{z\}, \{x, y\}, \{x, z\}, \{y, z\}, \{x, y, z\}\}.$$
When working out the Cartesian product for example $A \times B$ you simply order the pairs in which the first element comes from A and the second elements come from B. A x B = {$(x,1),(x,2),(x,3),(y,1),(y,2),(y,3),(z,1),(z,2),(z,3)$}
B x A = {$(1,x),(1,y)...(3,z)$}
For the Cartesian product $P(A)\times A$ you can not create a list of ordered pairs as $P(A)$ has a list of lists and $A$ has a list of elements. Is there a way to find the Cartesian product of $P(A)\times A$? • It seems to me that the set would consist of $$\{(x, y)\mid x\in P(A) \land y \in A\}$$ That is, the ordered pair consists of (set in P(A), element in A). – Namaste Feb 8 '18 at 22:37
• So, for example (∅,x), ({x},x), {{y},x), ({z},x), ({x,y},x). This makes sense. – Danny Sanderson Feb 8 '18 at 22:42
The set would consist of $$P(A)\times A = \{(a, b)\mid a\in P(A) \land b \in A\}$$ That is, the ordered pair consists of (set (element) in $P(A)$, element in $A$).
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$$P(A)\times A = \{(\varnothing, x), (\varnothing, y), (\varnothing, z), (\{x\}, x), \ldots (\{y\}, x), \ldots (\{x, y, z\}, z)\}$$
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https://math.stackexchange.com/questions/3218130/minimizing-univariate-quadratic-via-gradient-descent-choosing-the-step-size/3218189
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# Minimizing univariate quadratic via gradient descent — choosing the step size
I'm learning gradient descent method and I saw different (and opposite) things on my referrals.
I have the following function
$$f(x) = 2x^2 - 5x$$
and I have to calculate some iterations of gradient descent from $$x_0 = 1$$. So, I calculate the function at $$x_0$$, the derivative of the function at $$x_0$$ and now I have to apply the formula
$$x_1 = x_0 - \alpha \cdot f'(x_0)$$
Is $$\alpha$$ randomly chosen or do I have to force the formula to $$0$$ value? I'm quite confused.
The way you choose $$\alpha$$ depends, in general, on the information you have about your function. For example, for the function in your example, it is
$$f'(x) = 4x - 5$$
and $$f''(x) = 4$$, so $$f'$$ is Lipschitz continuous with Lipschitz constant $$L=4$$. You should then choose $$a$$ to be smaller than $$1/L$$, so, in this case, $$a<0.25$$.
In general, you might not know $$L$$. Then you have to resort to a linesearch method (e.g., exact linesearch or Armijo's rule).
You can read Chapter 3 in the book of Nocedal and Wright.
According to the Wikipedia article on gradient descent, $$\alpha$$ is a positive real number. You should choose a small $$\alpha$$, such as $$\alpha = 0.1$$ in your case to avoid going past the minimum value.
• $0.1$ can be quite a large number. – Pantelis Sopasakis May 8 at 9:28
• It still works since the minimum point is at $x = -\frac{-5}{2 \cdot 2} = 1.25$. It really depends on the given function. – Toby Mak May 8 at 9:30
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# Minimizing univariate quadratic via gradient descent — choosing the step size
I'm learning gradient descent method and I saw different (and opposite) things on my referrals. I have the following function
$$f(x) = 2x^2 - 5x$$
and I have to calculate some iterations of gradient descent from $$x_0 = 1$$. So, I calculate the function at $$x_0$$, the derivative of the function at $$x_0$$ and now I have to apply the formula
$$x_1 = x_0 - \alpha \cdot f'(x_0)$$
Is $$\alpha$$ randomly chosen or do I have to force the formula to $$0$$ value? I'm quite confused. The way you choose $$\alpha$$ depends, in general, on the information you have about your function. For example, for the function in your example, it is
$$f'(x) = 4x - 5$$
and $$f''(x) = 4$$, so $$f'$$ is Lipschitz continuous with Lipschitz constant $$L=4$$. You should then choose $$a$$ to be smaller than $$1/L$$, so, in this case, $$a<0.25$$. In general, you might not know $$L$$. Then you have to resort to a linesearch method (e.g., exact linesearch or Armijo's rule). You can read Chapter 3 in the book of Nocedal and Wright. According to the Wikipedia article on gradient descent, $$\alpha$$ is a positive real number. You should choose a small $$\alpha$$, such as $$\alpha = 0.1$$ in your case to avoid going past the minimum value. • $0.1$ can be quite a large number.
|
– Pantelis Sopasakis May 8 at 9:28
• It still works since the minimum point is at $x = -\frac{-5}{2 \cdot 2} = 1.25$.
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http://math.stackexchange.com/questions/186553/test-the-convergence-of-sum-n-0-infty-fracnk1nk-k?answertab=active
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# Test the convergence of $\sum_{n=0}^{\infty} \frac{n^{k+1}}{n^k + k}$
Problem: Test the convergence of $\sum_{n=0}^{\infty} \frac{n^{k+1}}{n^k + k}$, where $k$ is a positive constant.
I'm stumped. I've tried to apply several different convergence tests, but still can't figure this one out.
-
Ugh, I mistyped the series. Should I ask a new question? It should've been $\sum_{n=0}^\infty \frac{n^{k-1}}{n^k+k}$. – Damir Aug 25 '12 at 8:55
Use the theorem "If a series $\sum_{n=0}^{\infty} a_n$ converges, then $\lim_{n\rightarrow \infty} a_n =0$". – Mhenni Benghorbal Aug 25 '12 at 11:34
$$\frac{n^{k+1}}{n^k +k} =n \frac{1}{1+\frac{k}{n^k}}$$
What happens when $n \to \infty$?
$$\frac{n^{k+1}}{n^k+k}\geq\frac{n^{k+1}}{2n^k}=\frac{1}{2}n\xrightarrow [n\to\infty]{}\infty\neq 0$$
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# Test the convergence of $\sum_{n=0}^{\infty} \frac{n^{k+1}}{n^k + k}$
Problem: Test the convergence of $\sum_{n=0}^{\infty} \frac{n^{k+1}}{n^k + k}$, where $k$ is a positive constant. I'm stumped. I've tried to apply several different convergence tests, but still can't figure this one out. -
Ugh, I mistyped the series. Should I ask a new question? It should've been $\sum_{n=0}^\infty \frac{n^{k-1}}{n^k+k}$. – Damir Aug 25 '12 at 8:55
Use the theorem "If a series $\sum_{n=0}^{\infty} a_n$ converges, then $\lim_{n\rightarrow \infty} a_n =0$". – Mhenni Benghorbal Aug 25 '12 at 11:34
$$\frac{n^{k+1}}{n^k +k} =n \frac{1}{1+\frac{k}{n^k}}$$
What happens when $n \to \infty$?
|
$$\frac{n^{k+1}}{n^k+k}\geq\frac{n^{k+1}}{2n^k}=\frac{1}{2}n\xrightarrow [n\to\infty]{}\infty\neq 0$$
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https://math.stackexchange.com/questions/2692806/is-the-number-of-boundary-conditions-sufficient-for-a-unique-solution
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# Is the number of boundary conditions sufficient for a unique solution?
Consider the following partial differential equation in the bounded domain $\Omega\subset \mathbb{R}^2$ $$\Delta^2 - \nabla\cdot(k(x,y)\nabla u) + \lambda u = f,\text{ in }\Omega$$ where $\lambda >0$ is a constant and $k(x,y) >0$ is a given function of the position. Further, $f(x,y)$ is a given function. The boundary conditions are given by $$u =0,\,\Delta u = 0,\text{ on }\partial\Omega.$$
I need to find the minimisation problem corresponding to this boundary value problem and explain whether the number of boundary conditions is sufficient for a unique solution. I have the following theorem:
Let $L:\sum(\Omega)\to\sum'(\Omega)$, where $\sum(\Omega)$ is a linear space, and suppose that $L$ is linear, self-adjoint, positive and coercive, and let $u_0\in\sum:Lu_0 = f$. Then $u_0\in\sum(\Omega)$ minimises $$F(u) = \int_\Omega\dfrac{1}{2}uLu - ufd\Omega.$$
Hence, after applying integration by parts twice on the first term and once on the second term and using Gauss' theorem, the minimisation problem corresponding to the boundary value problem is $$F(u) = \int_\Omega \dfrac{1}{2}\nabla\cdot\|\nabla u\|^2 + \dfrac{1}{2}k(x,y)\|\nabla u\|^2 + \lambda u^2 - uf\,d\Omega.$$
My Question is: How do I show whether or not the number of boundary conditions is sufficient for a unique solution? I don't have a clue how I should show this.
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# Is the number of boundary conditions sufficient for a unique solution? Consider the following partial differential equation in the bounded domain $\Omega\subset \mathbb{R}^2$ $$\Delta^2 - \nabla\cdot(k(x,y)\nabla u) + \lambda u = f,\text{ in }\Omega$$ where $\lambda >0$ is a constant and $k(x,y) >0$ is a given function of the position. Further, $f(x,y)$ is a given function. The boundary conditions are given by $$u =0,\,\Delta u = 0,\text{ on }\partial\Omega.$$
I need to find the minimisation problem corresponding to this boundary value problem and explain whether the number of boundary conditions is sufficient for a unique solution. I have the following theorem:
Let $L:\sum(\Omega)\to\sum'(\Omega)$, where $\sum(\Omega)$ is a linear space, and suppose that $L$ is linear, self-adjoint, positive and coercive, and let $u_0\in\sum:Lu_0 = f$.
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Then $u_0\in\sum(\Omega)$ minimises $$F(u) = \int_\Omega\dfrac{1}{2}uLu - ufd\Omega.$$
Hence, after applying integration by parts twice on the first term and once on the second term and using Gauss' theorem, the minimisation problem corresponding to the boundary value problem is $$F(u) = \int_\Omega \dfrac{1}{2}\nabla\cdot\|\nabla u\|^2 + \dfrac{1}{2}k(x,y)\|\nabla u\|^2 + \lambda u^2 - uf\,d\Omega.$$
My Question is: How do I show whether or not the number of boundary conditions is sufficient for a unique solution?
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http://math.stackexchange.com/questions/555908/such-of-sum-of-any-two-adjacent-numbers-is-equal-to-the-sum-of-the-opposite-numb?answertab=oldest
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such of sum of any two adjacent numbers is equal to the sum of the opposite numbers
Give the numbers 1 to 10 on the edges of the diametric chords for the image given below such that such of sum of any two adjacent numbers is equal to the sum of the opposite numbers
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Going clockwise
$$10, 1, 4, 5, 8, 9, 2, 3, 6, 7$$
is a solution.
The pairs to compare are:
$$10 + 1 = 9 + 2 \\ 1 + 4 = 2 + 3 \\ 4 + 5 = 3 + 6 \\ 5 + 8 = 6 + 7 \\ 8 + 9 = 7 + 10$$
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For how I actually got that, I started with the guess that (1,10) and (2,9) would be pairs, and worked my way clockwise, going up. – Dennis Meng Nov 7 '13 at 18:43
Is there any logic you have used – sai kiran grandhi Nov 7 '13 at 18:46
Well, as an example, while I was working clockwise, I knew that 1 and 2 would be in another set of pairs, where 1's partner is once again one higher than 2's partner (and similar reasoning while working around the circle). Why I chose 3 and 4 next mostly amounts to intuition and a bit of luck. – Dennis Meng Nov 7 '13 at 18:48
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such of sum of any two adjacent numbers is equal to the sum of the opposite numbers
Give the numbers 1 to 10 on the edges of the diametric chords for the image given below such that such of sum of any two adjacent numbers is equal to the sum of the opposite numbers
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Going clockwise
$$10, 1, 4, 5, 8, 9, 2, 3, 6, 7$$
is a solution.
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The pairs to compare are:
$$10 + 1 = 9 + 2 \\ 1 + 4 = 2 + 3 \\ 4 + 5 = 3 + 6 \\ 5 + 8 = 6 + 7 \\ 8 + 9 = 7 + 10$$
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For how I actually got that, I started with the guess that (1,10) and (2,9) would be pairs, and worked my way clockwise, going up.
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https://economics.stackexchange.com/questions/13566/prove-if-succsim-is-rational-then-if-x-succ-y-succsim-z-then-x-succ-z
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# Prove if $\succsim$ is rational then: if $x \succ y \succsim z$, then $x \succ z$
Prove if $\succsim$ is rational then: if $x \succ y \succsim z$, then $x \succ z$
By definition of $\succ$,
$$\tag{1} x \succ y \iff x \succsim y, \; \neg \; y \succsim x$$
Where $\neg$ is the negation symbol.
So we want to show both
$$x \succsim z \; and \; \neg z \succsim x$$
We are given
$$\tag{2} y \succsim z$$
By $(1)$ and $(2)$ and transitivity of $\succsim$ (since $\succsim$ is rational),
$$\tag{3} x \succsim y \succsim z \Rightarrow x \succsim z$$
We also need $\neg \; z \succsim x$
Suppose the contrary that $z \succsim x$, but then
$$z \succsim x \succ y \succsim z$$
Such that
$$\tag{4} z \succ z$$
is a contradiction (as $\succ$ is irreflexive) so it must be that
$$\tag{5} \neg \; z \succsim x$$
Applying $(3)$ and $(5)$ we have the desired property
$$x \succsim z, \; \neg \; z \succsim x \Rightarrow x \succ z$$
My question is if I am allowed to arrive at the statement $(4)$ from the line above it?
• No. Getting to $(4)$ the way you do requires the sort of "transitivity" on $\succ$ that you are trying to show in the first place. Commented Sep 28, 2016 at 22:12
By contradiction, assume $z \succsim x$. However, we already know that $y \succsim z$, which is a contradiction (after a couple more steps). Can you now see this?
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# Prove if $\succsim$ is rational then: if $x \succ y \succsim z$, then $x \succ z$
Prove if $\succsim$ is rational then: if $x \succ y \succsim z$, then $x \succ z$
By definition of $\succ$,
$$\tag{1} x \succ y \iff x \succsim y, \; \neg \; y \succsim x$$
Where $\neg$ is the negation symbol. So we want to show both
$$x \succsim z \; and \; \neg z \succsim x$$
We are given
$$\tag{2} y \succsim z$$
By $(1)$ and $(2)$ and transitivity of $\succsim$ (since $\succsim$ is rational),
$$\tag{3} x \succsim y \succsim z \Rightarrow x \succsim z$$
We also need $\neg \; z \succsim x$
Suppose the contrary that $z \succsim x$, but then
$$z \succsim x \succ y \succsim z$$
Such that
$$\tag{4} z \succ z$$
is a contradiction (as $\succ$ is irreflexive) so it must be that
$$\tag{5} \neg \; z \succsim x$$
Applying $(3)$ and $(5)$ we have the desired property
$$x \succsim z, \; \neg \; z \succsim x \Rightarrow x \succ z$$
My question is if I am allowed to arrive at the statement $(4)$ from the line above it? • No. Getting to $(4)$ the way you do requires the sort of "transitivity" on $\succ$ that you are trying to show in the first place. Commented Sep 28, 2016 at 22:12
By contradiction, assume $z \succsim x$.
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However, we already know that $y \succsim z$, which is a contradiction (after a couple more steps).
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https://math.stackexchange.com/questions/2099864/misconception-of-the-surface-element-of-an-hyperboloid
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# Misconception of the surface element of an hyperboloid
I am trying the compute the surface of a given hyperboloid: $x^2+y^2-z^2=1$ between the planes $z=-4$ and $z=2$.
1. I have found that the surface element is $$d\vec{S}=(\sqrt{1+z^2}cos\theta,\sqrt{1+z^2}sin\theta,-z),$$ so in order to compute the surface I should use: $$dS=||d\vec{S}||=\sqrt{1+z^2+z^2}=\mathbf{\sqrt{1+2z^2}}d\theta dz.$$ So my doubt it is now:
2. Why if I use the cilindrical surface element $dS=\rho d\theta dz$, where $\rho=\sqrt{1+z^2}$, giving $$dS=\mathbf{\sqrt{1+z^2}}d\theta dz,$$ I don't get the same answer? It is that $2$ in front of the $z$ that bothers me.
wiritnig our surface under the form $f(x,y) = z$ if $z \ge 0$ then $z = \sqrt {{x^2} + {y^2} - 1}$ and if $z \le 0$ then $z = - \sqrt {{x^2} + {y^2} - 1}$. now we pass to polar cordinates to getting V=$2\pi (\int\limits_1^{\sqrt {17} } {\sqrt {{r^2} - 1} rdr + } \int\limits_1^{\sqrt 5 } {\sqrt {{r^2} - 1} rdr)}$
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# Misconception of the surface element of an hyperboloid
I am trying the compute the surface of a given hyperboloid: $x^2+y^2-z^2=1$ between the planes $z=-4$ and $z=2$. 1. I have found that the surface element is $$d\vec{S}=(\sqrt{1+z^2}cos\theta,\sqrt{1+z^2}sin\theta,-z),$$ so in order to compute the surface I should use: $$dS=||d\vec{S}||=\sqrt{1+z^2+z^2}=\mathbf{\sqrt{1+2z^2}}d\theta dz.$$ So my doubt it is now:
2. Why if I use the cilindrical surface element $dS=\rho d\theta dz$, where $\rho=\sqrt{1+z^2}$, giving $$dS=\mathbf{\sqrt{1+z^2}}d\theta dz,$$ I don't get the same answer? It is that $2$ in front of the $z$ that bothers me. wiritnig our surface under the form $f(x,y) = z$ if $z \ge 0$ then $z = \sqrt {{x^2} + {y^2} - 1}$ and if $z \le 0$ then $z = - \sqrt {{x^2} + {y^2} - 1}$.
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now we pass to polar cordinates to getting V=$2\pi (\int\limits_1^{\sqrt {17} } {\sqrt {{r^2} - 1} rdr + } \int\limits_1^{\sqrt 5 } {\sqrt {{r^2} - 1} rdr)}$
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https://math.stackexchange.com/questions/803707/cuboid-for-which-the-volume-the-surface-area-and-the-perimeter-are-numerically
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# Cuboid for which the volume, the surface area and the perimeter are numerically equal
How to show that it is impossible to have a cuboid for which the volume, the surface area and the perimeter are numerically equal ? The perimeter of a cuboid is the sum of the lengths of all its twelve edges.
• Does your definition of "cuboid" require the faces to be rectangular? May 21, 2014 at 5:00
Then you are asking for a solution to the equations $$4(l + w + h) = 2(lw + wh + hl) = lwh.$$ Setting $c = l + w + h \in \mathbb{R}$, we construct the polynomial $$(x - l)(x - w)(x - h) = x^3 - c x^2 + 2c x - 4c,$$ and we wish to know if it has three positive real roots for any real $c$. The discriminant of this polynomial is $$-432 c^2+112 c^3-12 c^4 = -4c^2(3c^2 - 28c + 108),$$ which is strictly negative for all $c$, implying that there are two nonreal roots, and no solution to the equations exists for real numbers $l, w, h$.
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# Cuboid for which the volume, the surface area and the perimeter are numerically equal
How to show that it is impossible to have a cuboid for which the volume, the surface area and the perimeter are numerically equal ? The perimeter of a cuboid is the sum of the lengths of all its twelve edges. • Does your definition of "cuboid" require the faces to be rectangular? May 21, 2014 at 5:00
Then you are asking for a solution to the equations $$4(l + w + h) = 2(lw + wh + hl) = lwh.$$ Setting $c = l + w + h \in \mathbb{R}$, we construct the polynomial $$(x - l)(x - w)(x - h) = x^3 - c x^2 + 2c x - 4c,$$ and we wish to know if it has three positive real roots for any real $c$.
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The discriminant of this polynomial is $$-432 c^2+112 c^3-12 c^4 = -4c^2(3c^2 - 28c + 108),$$ which is strictly negative for all $c$, implying that there are two nonreal roots, and no solution to the equations exists for real numbers $l, w, h$.
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http://math.stackexchange.com/questions/124485/finding-taking-off-speed-help?answertab=votes
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# Finding taking-off Speed? Help
An athlete jumps a horizontal distance of $7.18$m. He was airborne for $2.29$ seconds. The acceleration due to gravity is taken as $9.81$ms$^{-2}$. Assume that air resistance is negligible. Calculate the take-off speed in ms$^{-1}$
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What have you tried? How are you attempting to solve the problem? Where are you getting stuck? – Matthew Conroy Mar 26 '12 at 3:00
What do you know about this kind of problem? What facts and equations have you seen that deal with this kind of situation? – Gerry Myerson Mar 26 '12 at 5:37
The horizontal component $v_1$ of the velocity is an unchanging $\dfrac{7.18}{2.29}$ metres per second.
Maximum height is reached after $\dfrac{2.29}{2}$ seconds. If $v_2$ is the initial vertical component of the velocity, then the vertical component of the velocity, after $t$ seconds, is $v_2-9.81t$ (for $t\le 2.29$). This vertical component of the velocity reaches $0$ at time $\frac{2.29}{2}$. It follows that $$v_2-9.81\frac{2.29}{2}=0.$$ Now we know $v_1$ and $v_2$. The initial speed is $\sqrt{v_1^2+v_2^2}$.
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# Finding taking-off Speed? Help
An athlete jumps a horizontal distance of $7.18$m. He was airborne for $2.29$ seconds. The acceleration due to gravity is taken as $9.81$ms$^{-2}$. Assume that air resistance is negligible. Calculate the take-off speed in ms$^{-1}$
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What have you tried? How are you attempting to solve the problem? Where are you getting stuck? – Matthew Conroy Mar 26 '12 at 3:00
What do you know about this kind of problem? What facts and equations have you seen that deal with this kind of situation? – Gerry Myerson Mar 26 '12 at 5:37
The horizontal component $v_1$ of the velocity is an unchanging $\dfrac{7.18}{2.29}$ metres per second. Maximum height is reached after $\dfrac{2.29}{2}$ seconds. If $v_2$ is the initial vertical component of the velocity, then the vertical component of the velocity, after $t$ seconds, is $v_2-9.81t$ (for $t\le 2.29$). This vertical component of the velocity reaches $0$ at time $\frac{2.29}{2}$.
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It follows that $$v_2-9.81\frac{2.29}{2}=0.$$ Now we know $v_1$ and $v_2$.
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https://math.stackexchange.com/questions/2110936/what-does-it-mean-to-evaluate-the-product-of-elements-of-a-group
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# What does it mean to evaluate the product of elements of a group?
Consider $G$, a finite abelian group where $\forall a\in G, a\neq e$, we have $a^2 \neq e$. Given that the list $a_1,...,a_n$ lists out all elements of $G$ without repetition, evaluate $a_1...a_n$ (their product).
To me this seems like a bit of an oddly phrased question, unless my intuition is correct.
My intuition: because $G$ is abelian, rearrange the elements in the list so that each element is adjacent to its own inverse. Then, it is clear that $a_1...a_n=e$.
Could it really be this simple? Is there a more rigorous way to phrase my argument?
• You're almost there. But why is there no element which is it's own inverse? In that case your reasoning doesn't work. Jan 23, 2017 at 22:15
• math.stackexchange.com/questions/53185/… Jan 23, 2017 at 22:21
This argument is correct. The assumption $a^2\ne e$ for all $a\ne e$ tells us that the inverse of every $a\ne e$ differs from $a$, so the construction of your list is possible.
Obviously this assumption is essential. Indeed, in $\Bbb Z_4=\{0,1,2,3\}$ (with addition) we have $0+1+2+3=2$. In this group we have $2+2=0$.
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# What does it mean to evaluate the product of elements of a group? Consider $G$, a finite abelian group where $\forall a\in G, a\neq e$, we have $a^2 \neq e$. Given that the list $a_1,...,a_n$ lists out all elements of $G$ without repetition, evaluate $a_1...a_n$ (their product). To me this seems like a bit of an oddly phrased question, unless my intuition is correct. My intuition: because $G$ is abelian, rearrange the elements in the list so that each element is adjacent to its own inverse. Then, it is clear that $a_1...a_n=e$. Could it really be this simple? Is there a more rigorous way to phrase my argument? • You're almost there. But why is there no element which is it's own inverse? In that case your reasoning doesn't work. Jan 23, 2017 at 22:15
• math.stackexchange.com/questions/53185/… Jan 23, 2017 at 22:21
This argument is correct. The assumption $a^2\ne e$ for all $a\ne e$ tells us that the inverse of every $a\ne e$ differs from $a$, so the construction of your list is possible. Obviously this assumption is essential. Indeed, in $\Bbb Z_4=\{0,1,2,3\}$ (with addition) we have $0+1+2+3=2$.
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In this group we have $2+2=0$.
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https://physics.stackexchange.com/questions/206146/potential-energy-problem
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# Potential Energy Problem
We know that, the potential energy of a particle is $U= mgh$ where $h$ is the height from the zero potential point.
But what I have found in Resnick's Fundamental of physics book is that they wrote the potential energy for the sphere at height h is equal$= Mgh$ .
Will not it be $= Mg(h+R)$? I have added the $R$ to the height $h$ because, from the ground the height of the center of mass of the sphere is $(h+R)$.
• The diagram is a bit confusing; at the top point, the center point of the ball is not $h+R$ above the top of the ramp, but instead $h+Rcos(\theta)$. That might work if the ramp were floating so the ball could roll to touching the bottom of the ramp, so that the center point ended up at $Rcos(\theta)$... – Daniel Griscom Sep 9 '15 at 22:16
Well, it depends on what you call zero potential energy. If you say "ball sitting on the floor" is zero potential energy, then $Mgh$ is the correct formula for the potential energy of the ball on the ramp. If you say "ball material flattened onto the floor" is zero, then $Mg(h+R)$ is indeed the correct formula. If you say "ball sitting at the bottom of a 30 meter pit" is zero, then the formula would be $Mg(h+30)$.
None of these zero points is "wrong"; potential energy is always determined relative to some arbitrary zero point. As the problem is written, though, it implies that zero is "ball sitting on the floor", hence $Mgh$ is indeed the expected answer.
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# Potential Energy Problem
We know that, the potential energy of a particle is $U= mgh$ where $h$ is the height from the zero potential point. But what I have found in Resnick's Fundamental of physics book is that they wrote the potential energy for the sphere at height h is equal$= Mgh$ . Will not it be $= Mg(h+R)$? I have added the $R$ to the height $h$ because, from the ground the height of the center of mass of the sphere is $(h+R)$. • The diagram is a bit confusing; at the top point, the center point of the ball is not $h+R$ above the top of the ramp, but instead $h+Rcos(\theta)$. That might work if the ramp were floating so the ball could roll to touching the bottom of the ramp, so that the center point ended up at $Rcos(\theta)$... – Daniel Griscom Sep 9 '15 at 22:16
Well, it depends on what you call zero potential energy. If you say "ball sitting on the floor" is zero potential energy, then $Mgh$ is the correct formula for the potential energy of the ball on the ramp. If you say "ball material flattened onto the floor" is zero, then $Mg(h+R)$ is indeed the correct formula. If you say "ball sitting at the bottom of a 30 meter pit" is zero, then the formula would be $Mg(h+30)$. None of these zero points is "wrong"; potential energy is always determined relative to some arbitrary zero point.
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As the problem is written, though, it implies that zero is "ball sitting on the floor", hence $Mgh$ is indeed the expected answer.
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https://stats.stackexchange.com/questions/12094/how-to-add-two-random-variables-pdf
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# How to add two random variable's pdf? [closed]
X,Y are independent random variables. X's pdf = f(x) Y's pdf = g(x) if Z= X+Y what is the Z's pdf? Can it be calculated?
• The first hit upon Googling the very title of this question produces a good answer: please read our faq about the wisdom of doing a little bit of research before asking a question.
– whuber
Jun 19, 2011 at 16:35
• @whuber but now this question is ;) Jun 10, 2013 at 1:59
If $X$ and $Y$ are two independent, continuous random variables, then you can find the distribution of $Z=X+Y$ by taking the convolution of $f(x)$ and $g(y)$:$$h(z)=(f*g)(z)=\int_{-\infty}^{\infty}f(x)g(z-x)dx$$If $X$ and $Y$ are two independent, discrete random variables, then you can find the distribution of $Z=X+Y$ by taking the discrete convolution of $X$ and $Y$:$$\mbox{P}(Z=k)=\sum_{i=-\infty}^{\infty}\mbox{P}(X=i)\cdot\mbox{P}(Y=k-i)$$
• It's pretty much the same as what I've written above. $X-Y$ is the same as $X+(-Y)$. The distribution of $-Y$ is the same as the distribution of $Y$ except that it is mirrored over the vertical line at 0. Jun 19, 2011 at 18:15
If $A$ is the domain for $f(x)$ then $h(z)=\int_{x\in A}f(x)g(z-x)\,dx$.
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# How to add two random variable's pdf? [closed]
X,Y are independent random variables. X's pdf = f(x) Y's pdf = g(x) if Z= X+Y what is the Z's pdf? Can it be calculated? • The first hit upon Googling the very title of this question produces a good answer: please read our faq about the wisdom of doing a little bit of research before asking a question. – whuber
Jun 19, 2011 at 16:35
• @whuber but now this question is ;) Jun 10, 2013 at 1:59
If $X$ and $Y$ are two independent, continuous random variables, then you can find the distribution of $Z=X+Y$ by taking the convolution of $f(x)$ and $g(y)$:$$h(z)=(f*g)(z)=\int_{-\infty}^{\infty}f(x)g(z-x)dx$$If $X$ and $Y$ are two independent, discrete random variables, then you can find the distribution of $Z=X+Y$ by taking the discrete convolution of $X$ and $Y$:$$\mbox{P}(Z=k)=\sum_{i=-\infty}^{\infty}\mbox{P}(X=i)\cdot\mbox{P}(Y=k-i)$$
• It's pretty much the same as what I've written above. $X-Y$ is the same as $X+(-Y)$. The distribution of $-Y$ is the same as the distribution of $Y$ except that it is mirrored over the vertical line at 0.
|
Jun 19, 2011 at 18:15
If $A$ is the domain for $f(x)$ then $h(z)=\int_{x\in A}f(x)g(z-x)\,dx$.
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https://math.stackexchange.com/questions/2853598/restriction-of-interval-while-proving-an-inequality
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# Restriction of interval while proving an inequality
In many books and online , the inequality $\sin x<x$ is defined only over the interval $[0, \frac{\pi}{2}]$ . However it is easy to check that the inequality holds good over the interval $[\frac{\pi}{2} , \pi]$ as well. We can check it graphically . In fact it holds good for all positive numbers. Is there a special reason for proving this inequality over the restricted interval of $[0, \frac{\pi}{2}]$ ?
• You probably mean $\left(0, \frac\pi2\right]$. At $x = 0$ we have equality. – mechanodroid Jul 16 '18 at 17:08
• @mechanodroid right . I didn’t pay attention to that. Thanks for correcting me. – Aditi Jul 16 '18 at 17:15
You are correct. The inequality holds for all positive $x$. Easiest way to see this is probably to compare derivative.
In regards to the restriction of the inequality: It is possible that in your case, you only need the fact that $\sin x<x$ on $[0,\pi/2]$, rather than on the whole positive real line.
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# Restriction of interval while proving an inequality
In many books and online , the inequality $\sin x<x$ is defined only over the interval $[0, \frac{\pi}{2}]$ . However it is easy to check that the inequality holds good over the interval $[\frac{\pi}{2} , \pi]$ as well. We can check it graphically . In fact it holds good for all positive numbers. Is there a special reason for proving this inequality over the restricted interval of $[0, \frac{\pi}{2}]$ ? • You probably mean $\left(0, \frac\pi2\right]$. At $x = 0$ we have equality. – mechanodroid Jul 16 '18 at 17:08
• @mechanodroid right . I didn’t pay attention to that. Thanks for correcting me. – Aditi Jul 16 '18 at 17:15
You are correct. The inequality holds for all positive $x$. Easiest way to see this is probably to compare derivative.
|
In regards to the restriction of the inequality: It is possible that in your case, you only need the fact that $\sin x<x$ on $[0,\pi/2]$, rather than on the whole positive real line.
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https://math.stackexchange.com/questions/3131965/coordinates-of-line-in-sphere-with-x-y-rotation
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# Coordinates of line in sphere with x,y rotation
Lets say that I have a line with one end fixed to the center of a sphere, and the other end can freely rotate. If I were to rotate the line around the x and y axes, what would the coordinates be for the freely-rotating end?
Here's what I need this for:
I have a rectangle, and need to find the coordinates of each vertex when the rectangle is rotated (the rectangle will be rotated from its center at (0,0,0), and can have any width and height). I only need to rotate the rectangle around the x,y (or any pair really, but x,y are just a nice looking pair), because rotating the unincluded axis will be like rolling a pencil back and forth in your thumb.
I have found some helpful resources for questions similar to mine, but they don't solve my problem. For example, this here I've tried to understand, but I have minimal experience with matrices, and this doesn't exactly solve my problem.
2. You say we have a line with one end fixed at the origin; say that the point on this line which is distance 1 from the origin is $$v =(a,b,c).$$ When I rotate $$v$$ around the $$z$$ axis, I won't change it's $$z$$ coordinate--this will still be $$c.$$ If I rotate by an angle of $$\theta,$$ the output vector will be $$(cos(\theta)a - sin(\theta)b, sin(\theta)a + cos(\theta)b, c).$$
• If you would like to first rotate $(a,b,c)$ by an angle of $\theta$ around the $z$ axis and then rotate the result by an angle of $\alpha$ around the $x$ axis, the output would be $(a cos(\theta) - b sin(\theta), a sin(\theta) cos(\alpha) + b cos(\theta) cos(\alpha) - c sin(\alpha), a sin(\theta) sin(\alpha) + b cos(\theta) sin(\alpha) + c cos(\alpha).$ Commented Mar 2, 2019 at 1:20
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# Coordinates of line in sphere with x,y rotation
Lets say that I have a line with one end fixed to the center of a sphere, and the other end can freely rotate. If I were to rotate the line around the x and y axes, what would the coordinates be for the freely-rotating end? Here's what I need this for:
I have a rectangle, and need to find the coordinates of each vertex when the rectangle is rotated (the rectangle will be rotated from its center at (0,0,0), and can have any width and height). I only need to rotate the rectangle around the x,y (or any pair really, but x,y are just a nice looking pair), because rotating the unincluded axis will be like rolling a pencil back and forth in your thumb. I have found some helpful resources for questions similar to mine, but they don't solve my problem. For example, this here I've tried to understand, but I have minimal experience with matrices, and this doesn't exactly solve my problem. 2.
|
You say we have a line with one end fixed at the origin; say that the point on this line which is distance 1 from the origin is $$v =(a,b,c).$$ When I rotate $$v$$ around the $$z$$ axis, I won't change it's $$z$$ coordinate--this will still be $$c.$$ If I rotate by an angle of $$\theta,$$ the output vector will be $$(cos(\theta)a - sin(\theta)b, sin(\theta)a + cos(\theta)b, c).$$
• If you would like to first rotate $(a,b,c)$ by an angle of $\theta$ around the $z$ axis and then rotate the result by an angle of $\alpha$ around the $x$ axis, the output would be $(a cos(\theta) - b sin(\theta), a sin(\theta) cos(\alpha) + b cos(\theta) cos(\alpha) - c sin(\alpha), a sin(\theta) sin(\alpha) + b cos(\theta) sin(\alpha) + c cos(\alpha).$ Commented Mar 2, 2019 at 1:20
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https://math.stackexchange.com/questions/2805398/how-is-equation-of-a-circle-in-argand-plane-given-by-z-bar-z-a-bar-z-bar
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How is equation of a circle in Argand plane given by $z\bar z + a \bar z + \bar a z+ b = 0$?
My book gives the equation $z\bar z + a \bar z + \bar a z+ b = 0$ for general equation of circle with centre -a and radius $\sqrt {a \bar a - b}$. I can understand that general equation of a circle can be $|z-z_0|=r$ or $z \bar z - z_0 \bar z -\bar z_0 z + z_0 \bar z_0 - r^2 = 0$, but how does that give rise to this equation? I am referring to the formula list in the Arihant 39 years JEE solved papers.
• Compare coefficients of $z$ and $\bar z$ Commented Jun 2, 2018 at 12:44
Note that\begin{align}z\overline z+a\overline z+\overline az+b=0&\iff(z+a)\overline{(z+a)}=a\overline a-b\\&\iff|z+a|^2=a\overline a-b.\\&\iff\bigl|z-(-a)\bigr|^2=a\overline a-b.\end{align}
In
$$Z \bar Z = r^2$$
making $Z = z + a$
$$(z+a)(\bar z + \bar a) = r^2$$
or
$$z\bar z+z \bar a+ \bar z a + a\bar a -r^2 = 0$$
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How is equation of a circle in Argand plane given by $z\bar z + a \bar z + \bar a z+ b = 0$? My book gives the equation $z\bar z + a \bar z + \bar a z+ b = 0$ for general equation of circle with centre -a and radius $\sqrt {a \bar a - b}$. I can understand that general equation of a circle can be $|z-z_0|=r$ or $z \bar z - z_0 \bar z -\bar z_0 z + z_0 \bar z_0 - r^2 = 0$, but how does that give rise to this equation? I am referring to the formula list in the Arihant 39 years JEE solved papers.
|
• Compare coefficients of $z$ and $\bar z$ Commented Jun 2, 2018 at 12:44
Note that\begin{align}z\overline z+a\overline z+\overline az+b=0&\iff(z+a)\overline{(z+a)}=a\overline a-b\\&\iff|z+a|^2=a\overline a-b.\\&\iff\bigl|z-(-a)\bigr|^2=a\overline a-b.\end{align}
In
$$Z \bar Z = r^2$$
making $Z = z + a$
$$(z+a)(\bar z + \bar a) = r^2$$
or
$$z\bar z+z \bar a+ \bar z a + a\bar a -r^2 = 0$$
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http://math.stackexchange.com/questions/127529/problem-on-the-mean-value-theorem
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# Problem on the Mean Value Theorem
Given that $\epsilon > 0$ prove that $\ln(x_2) - \ln(x_1) < \epsilon(x_2 - x_1)$ if $1 < x_1 < x_2$ and $x_1$ is sufficiently large. How large must $x_1$ be for the inequality to be guaranteed true?
-
It was a long time ago, but I couldn't restrain myself)
Solution 1 (without MVT):
Denote $f_1 (x) = \log x, \ f_2 (x) = \epsilon x$. At the lower bound $f_1(1)=0 < f_2(1)= \epsilon$. Hence, $f_2(1) >f_1(1)$. Taking the derivatives of both functions we see that the rate of growth of $f_{2}(x) >f_{1}(x)$ for $\delta > \frac{1}{\epsilon}$, which certainly includes $x_2$.
Solution 2 (with MVT):
Consider the function $h(x)= \epsilon x -\log x$. Then
$$\frac{h(x_2)-h(x_1)}{x_2-x_1} = \frac{\epsilon(x_2 - x_1) - \log x_2 + \log x_1}{x_2 - x_1}= \epsilon -\frac{\log x_2 - \log x_1}{x_2 -x_1}=\epsilon - \frac{1}{\xi}$$ with $x_1 \leq \xi \leq x_2$. The last step is due to MVT and the derivative of the $\log$ function. Now we want this to be positive so that $h(x) >0$, so this holds for $\xi >\frac{1}{\epsilon}$, so the smallest possible value for $x_1$ is again $\frac{1}{\epsilon}$.
-
Haha, thanks, Alex!!! – user9292 Feb 18 '13 at 21:56
Hint: What does the mean value theorem tell you? The derivative of $ln(x)$ is what? When is $1/y$ smaller than epsilon? Write everything out explicitly and I am sure you will get it.
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# Problem on the Mean Value Theorem
Given that $\epsilon > 0$ prove that $\ln(x_2) - \ln(x_1) < \epsilon(x_2 - x_1)$ if $1 < x_1 < x_2$ and $x_1$ is sufficiently large. How large must $x_1$ be for the inequality to be guaranteed true? -
It was a long time ago, but I couldn't restrain myself)
Solution 1 (without MVT):
Denote $f_1 (x) = \log x, \ f_2 (x) = \epsilon x$. At the lower bound $f_1(1)=0 < f_2(1)= \epsilon$. Hence, $f_2(1) >f_1(1)$. Taking the derivatives of both functions we see that the rate of growth of $f_{2}(x) >f_{1}(x)$ for $\delta > \frac{1}{\epsilon}$, which certainly includes $x_2$. Solution 2 (with MVT):
Consider the function $h(x)= \epsilon x -\log x$.
|
Then
$$\frac{h(x_2)-h(x_1)}{x_2-x_1} = \frac{\epsilon(x_2 - x_1) - \log x_2 + \log x_1}{x_2 - x_1}= \epsilon -\frac{\log x_2 - \log x_1}{x_2 -x_1}=\epsilon - \frac{1}{\xi}$$ with $x_1 \leq \xi \leq x_2$.
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https://math.stackexchange.com/questions/1265557/triangles-sine-and-cosine-problem/1816710
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# Triangles, sine and cosine problem
Hi everyone I tried solving this countless times but I always get the wrong answer! what I did first is 600/tan(46) - 600/tan(40) and that sounded reasonable to find the answer! but I keep getting it wrong :( there was a similar question but I got it right.
Thank you.
• the two distances should be added; not subtracted. – abel May 4 '15 at 0:32
• That actually worked, thanks! – AliAlM May 4 '15 at 0:37
• you are welcome. what do you mean actually? were you expecting not to work? – abel May 4 '15 at 0:39
$tan40^\circ=\frac{600}{x}$
$tan46^\circ=\frac{600}{y}$
So,Distance = $(x+y)=\frac{600}{tan40^\circ}+\frac{600}{tan46^\circ}$
$=6oo(\frac{1}{tan40^\circ}+\frac{1}{tan46^\circ})$
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# Triangles, sine and cosine problem
Hi everyone I tried solving this countless times but I always get the wrong answer! what I did first is 600/tan(46) - 600/tan(40) and that sounded reasonable to find the answer! but I keep getting it wrong :( there was a similar question but I got it right. Thank you. • the two distances should be added; not subtracted. – abel May 4 '15 at 0:32
• That actually worked, thanks! – AliAlM May 4 '15 at 0:37
• you are welcome. what do you mean actually? were you expecting not to work?
|
– abel May 4 '15 at 0:39
$tan40^\circ=\frac{600}{x}$
$tan46^\circ=\frac{600}{y}$
So,Distance = $(x+y)=\frac{600}{tan40^\circ}+\frac{600}{tan46^\circ}$
$=6oo(\frac{1}{tan40^\circ}+\frac{1}{tan46^\circ})$
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http://math.stackexchange.com/questions/305622/tensor-product-proof
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# Tensor Product proof
I have to prove something for my matrix-algebra course. it's the following proof:
I have to prove that $A\otimes B$ is invertible, if and only if $B\otimes B$ is invertible.
Please explain this in simple language, I'm only a first year econometrics student.
-
Are $A$ and $B$ square matrices? – Pete L. Clark Feb 16 '13 at 17:27
Could be, it's not given.. – Sjoerd Smaal Feb 16 '13 at 17:29
This might help give you some intuition: math.mcgill.ca/msnarski/tensors1.PDF – snarski Feb 16 '13 at 18:03
I think $B\otimes B$ should be corrected to $B\otimes A$. See,Invertibility of a Kronecker Product – M.Sina Feb 16 '13 at 18:43
This is wrong. Take the zero matrix as $A$ and for $B$ any invertible matrix. Then the Kronecker-product of $B$ with itself is invertible with inverse $B^{-1} \otimes B^{-1}$ (can be proven by matrix multiplication), but $A \otimes B$ is zero, and therefore not invertible.
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# Tensor Product proof
I have to prove something for my matrix-algebra course. it's the following proof:
I have to prove that $A\otimes B$ is invertible, if and only if $B\otimes B$ is invertible. Please explain this in simple language, I'm only a first year econometrics student. -
Are $A$ and $B$ square matrices? – Pete L. Clark Feb 16 '13 at 17:27
Could be, it's not given.. – Sjoerd Smaal Feb 16 '13 at 17:29
This might help give you some intuition: math.mcgill.ca/msnarski/tensors1.PDF – snarski Feb 16 '13 at 18:03
I think $B\otimes B$ should be corrected to $B\otimes A$. See,Invertibility of a Kronecker Product – M.Sina Feb 16 '13 at 18:43
This is wrong. Take the zero matrix as $A$ and for $B$ any invertible matrix.
|
Then the Kronecker-product of $B$ with itself is invertible with inverse $B^{-1} \otimes B^{-1}$ (can be proven by matrix multiplication), but $A \otimes B$ is zero, and therefore not invertible.
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https://math.stackexchange.com/revisions/048fd15c-723b-443a-aa21-0fe93632f450/view-source
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Here's a very simple yet quite good approximation to $\sum\nolimits_{k = 1}^{n - 1} {\frac{{x^k }}{{n - k}}}$, suitable for $|x|>1$ and $n$ sufficiently large (which is the interesting case, in my opinion).
First write
$$S(n,x) := \sum\limits_{k = 1}^{n - 1} {\frac{{x^k }}{{n - k}}} = x^n \sum\limits_{k = 1}^{n - 1} {\frac{{u^k }}{k}},$$
where $u=1/x$. If $|x|>1$, then $|u|<1$; hence $\sum\nolimits_{k = 1}^\infty {\frac{{u^k }}{k}} = - \log (1 - u)$.
So, noting that $- \log (1 - \frac{1}{x}) = \log (\frac{x}{{x - 1}})$, this gives rise to the approximation
$$S(n,x) \approx x^n \log \bigg(\frac{x}{{x - 1}}\bigg),$$
provided that $n$ is sufficiently large. It turns out that this simple approximation is quite good. (You can check by yourself.) See the table for an example.
----------
Rounded values of $S(n,x)$ and its approximation, for $n=10$ and several values of $x$.
$$\begin{array}{ccc}x&S(n,x)&x^n \log\left(\frac{x}{x-1}\right)\\-5&-1780484.04&-1780483.95\\-3&-16987.42&-16987.34\\2&709.598&709.783\\ 4&301656.39&301656.52\\6&1.102428722 \times 10^7&1.102428734 \times 10^7\end{array}$$
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Here's a very simple yet quite good approximation to $\sum\nolimits_{k = 1}^{n - 1} {\frac{{x^k }}{{n - k}}}$, suitable for $|x|>1$ and $n$ sufficiently large (which is the interesting case, in my opinion). First write
$$S(n,x) := \sum\limits_{k = 1}^{n - 1} {\frac{{x^k }}{{n - k}}} = x^n \sum\limits_{k = 1}^{n - 1} {\frac{{u^k }}{k}},$$
where $u=1/x$. If $|x|>1$, then $|u|<1$; hence $\sum\nolimits_{k = 1}^\infty {\frac{{u^k }}{k}} = - \log (1 - u)$. So, noting that $- \log (1 - \frac{1}{x}) = \log (\frac{x}{{x - 1}})$, this gives rise to the approximation
$$S(n,x) \approx x^n \log \bigg(\frac{x}{{x - 1}}\bigg),$$
provided that $n$ is sufficiently large. It turns out that this simple approximation is quite good. (You can check by yourself.) See the table for an example.
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Rounded values of $S(n,x)$ and its approximation, for $n=10$ and several values of $x$.
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http://math.stackexchange.com/questions/247953/proof-for-fermats-little-theorem
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# Proof for Fermat's Little Theorem
The proof was given as:
1. Let $p\in\mathbb{N}^{+}$
2. Let $a\in\mathbb{N}$
3. Let $p$ be prime
4. We know that $p\,\vert\,a^p-a$ by proposition of Chapter 1
5. Therefore $a^p-a\equiv 0\pmod p$ by Definition 8.4.1
6. Therefore $a^p\equiv a\pmod p$ by Proposition 8.4.7
Q.E.D.
But how do I get from 4 to 5 and 5 to 6?
Is 4 to 5 trying to say, since $p$ divides $(a^p - a)$ then the remainder is always 0? Then 5 - 6, how do I just move the $a$ over?
1 (original question)
-
Step 4. is critical. If you understand that, you know the essence. Step 5. and 6. are trivial by definition of $\equiv$, and goes exactly as you wrote at the end. – Berci Nov 30 '12 at 10:31
The step from 4 to 5 is just the definition of congruence. By definition $a\equiv b\pmod m$ means that $m\mid a-b$, so if you know from step 4 that $p\mid a^p-a$, then you know that $p\mid(a^p-a)-0$ and hence by definition that $a^p-a\equiv 0\pmod p$. In fact step 5 is unnecessary, since you can go directly from 4 to 6: if you know that $p\mid a^p-a$, then you know that $a^p\equiv a\pmod p$ by the definition of congruence.
$$a\equiv b\pmod m\quad\text{iff}\quad a-b\equiv 0\pmod m\;.$$
By definition $a\equiv b\pmod m$ means that $m\mid a-b$, but that’s certainly true if and only if $m\mid(a-b)-0$, which means exactly that $a-b\equiv 0\pmod m$.
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# Proof for Fermat's Little Theorem
The proof was given as:
1. Let $p\in\mathbb{N}^{+}$
2. Let $a\in\mathbb{N}$
3. Let $p$ be prime
4. We know that $p\,\vert\,a^p-a$ by proposition of Chapter 1
5. Therefore $a^p-a\equiv 0\pmod p$ by Definition 8.4.1
6. Therefore $a^p\equiv a\pmod p$ by Proposition 8.4.7
Q.E.D. But how do I get from 4 to 5 and 5 to 6? Is 4 to 5 trying to say, since $p$ divides $(a^p - a)$ then the remainder is always 0? Then 5 - 6, how do I just move the $a$ over? 1 (original question)
-
Step 4. is critical. If you understand that, you know the essence. Step 5. and 6. are trivial by definition of $\equiv$, and goes exactly as you wrote at the end. – Berci Nov 30 '12 at 10:31
The step from 4 to 5 is just the definition of congruence.
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By definition $a\equiv b\pmod m$ means that $m\mid a-b$, so if you know from step 4 that $p\mid a^p-a$, then you know that $p\mid(a^p-a)-0$ and hence by definition that $a^p-a\equiv 0\pmod p$.
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http://math.stackexchange.com/questions/64646/find-angledac
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# find $\angle{DAC}$
In $\triangle{ABC}$, given $\angle{A}=80^\circ$, $\angle{B}=\angle{C}=50^\circ$, D is a point in $\triangle{ABC}$, which $\angle{DBC}=20^\circ,\angle{DCB}=40^\circ$. Then how to find find $\angle{DAC}$?
thanks.
-
I wonder if this isn't the same kind of problem as math.stackexchange.com/questions/63819/… – Gerry Myerson Sep 15 '11 at 1:35
(didn't do a careful check, but...) possible duplicate of Finding an angle within an 80-80-20 isosceles triangle – Aryabhata Sep 15 '11 at 2:15
I tried the geometric method but could not succeed. However the trigonometric route yields the result. Here it is: Extend $CD$ to meet $AB$ at $G$. Angle $B$ is $50$. Angle $BCD$ is $40$. So $CG$ is perpendicular to $AB$. Let $AH$ be the right bisector of angle $A$ resting on $BC$ at $H$. Let the required angle $DAC$ be $k$ degrees. $$GD = AG\tan(80-k) = BG\tan(30),$$ $$\tan(80-k) = \frac{GD}{AG} = {BG\tan(30)}{AG}.$$ $$BG = BC\cos(50)$$ $$BC = 2BH = 2AB\cos(50).$$ So, $$BG = 2AB\cos(50) \cos(50)$$ $$AG = AC \cos(80) = AB \cos(80)$$ So, $$\tan(80-k) = \frac{BG\tan(30)}{AG} = \frac{2\cos )(50) \cos (50) \tan (30)}{\cos(80)}$$ $$80-k = \arctan\left(\frac{2\cos (50) \cos (50) \tan (30)}{\cos (80)}\right) = 70^\circ$$ Thus the required $\angle{DAC} = 10^\circ$
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# find $\angle{DAC}$
In $\triangle{ABC}$, given $\angle{A}=80^\circ$, $\angle{B}=\angle{C}=50^\circ$, D is a point in $\triangle{ABC}$, which $\angle{DBC}=20^\circ,\angle{DCB}=40^\circ$. Then how to find find $\angle{DAC}$? thanks. -
I wonder if this isn't the same kind of problem as math.stackexchange.com/questions/63819/… – Gerry Myerson Sep 15 '11 at 1:35
(didn't do a careful check, but...) possible duplicate of Finding an angle within an 80-80-20 isosceles triangle – Aryabhata Sep 15 '11 at 2:15
I tried the geometric method but could not succeed. However the trigonometric route yields the result. Here it is: Extend $CD$ to meet $AB$ at $G$. Angle $B$ is $50$. Angle $BCD$ is $40$. So $CG$ is perpendicular to $AB$. Let $AH$ be the right bisector of angle $A$ resting on $BC$ at $H$. Let the required angle $DAC$ be $k$ degrees.
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$$GD = AG\tan(80-k) = BG\tan(30),$$ $$\tan(80-k) = \frac{GD}{AG} = {BG\tan(30)}{AG}.$$ $$BG = BC\cos(50)$$ $$BC = 2BH = 2AB\cos(50).$$ So, $$BG = 2AB\cos(50) \cos(50)$$ $$AG = AC \cos(80) = AB \cos(80)$$ So, $$\tan(80-k) = \frac{BG\tan(30)}{AG} = \frac{2\cos )(50) \cos (50) \tan (30)}{\cos(80)}$$ $$80-k = \arctan\left(\frac{2\cos (50) \cos (50) \tan (30)}{\cos (80)}\right) = 70^\circ$$ Thus the required $\angle{DAC} = 10^\circ$
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https://math.stackexchange.com/questions/890062/inequality-involving-traces-and-matrix-inversions
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# Inequality involving traces and matrix inversions
The following question kept me wondering for some weeks:
Given the symmetric matrices $A,B,C\in\mathbb{R}^{n\times n}$ where $A$ and $C$ are positive definite (hence invertible), and $B$ is positive semidefinite (hence not necessarily invertible) with $\operatorname{trace}(B)\neq 0$, prove that
$$\operatorname{trace}\left\{C^{-1/2}BC^{-1/2}(A^{-1}+C^{-1/2}BC^{-1/2})^{-1}(A+\frac{n}{\operatorname{trace}(B)}C)^{-1}\right\}\geq \operatorname{trace}\left\{(A+\frac{n}{\operatorname{trace}(B)}C)^{-2}A \right\}.$$
If it would help, one can also consider the simpler version with $C=I$: prove that
$$\operatorname{trace}\left\{B(A^{-1}+B)^{-1}(A+\frac{n}{\operatorname{trace}(B)}I)^{-1}\right\}\geq \operatorname{trace}\left\{(A+\frac{n}{\operatorname{trace}(B)}I)^{-2}A \right\}.$$
Please note that the matrix inversion lemma is not applicable at first as $B$ is positive semidefinite. Although I'm not sure, it seems like $\operatorname{trace}(B)=\operatorname{trace}\{(\operatorname{trace}(B)/n)I\}$ should be utilized in some way, and it may also help to interpret the trace operator in terms of the Frobenius norm. I would highly appreciate if anyone can provide some help or suggestions on this.
• Crossposting: mathoverflow.net/questions/178089/…. Aug 8, 2014 at 11:55
• This inequality was shown to be incorrect at Mathoverflow. Therefore, people can disregard my question. Aug 16, 2014 at 22:28
M. Lin's answer at Math Overflow show that the inequality doesn't need to hold even in the case $C=I$.
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# Inequality involving traces and matrix inversions
The following question kept me wondering for some weeks:
Given the symmetric matrices $A,B,C\in\mathbb{R}^{n\times n}$ where $A$ and $C$ are positive definite (hence invertible), and $B$ is positive semidefinite (hence not necessarily invertible) with $\operatorname{trace}(B)\neq 0$, prove that
$$\operatorname{trace}\left\{C^{-1/2}BC^{-1/2}(A^{-1}+C^{-1/2}BC^{-1/2})^{-1}(A+\frac{n}{\operatorname{trace}(B)}C)^{-1}\right\}\geq \operatorname{trace}\left\{(A+\frac{n}{\operatorname{trace}(B)}C)^{-2}A \right\}.$$
If it would help, one can also consider the simpler version with $C=I$: prove that
$$\operatorname{trace}\left\{B(A^{-1}+B)^{-1}(A+\frac{n}{\operatorname{trace}(B)}I)^{-1}\right\}\geq \operatorname{trace}\left\{(A+\frac{n}{\operatorname{trace}(B)}I)^{-2}A \right\}.$$
Please note that the matrix inversion lemma is not applicable at first as $B$ is positive semidefinite. Although I'm not sure, it seems like $\operatorname{trace}(B)=\operatorname{trace}\{(\operatorname{trace}(B)/n)I\}$ should be utilized in some way, and it may also help to interpret the trace operator in terms of the Frobenius norm. I would highly appreciate if anyone can provide some help or suggestions on this. • Crossposting: mathoverflow.net/questions/178089/…. Aug 8, 2014 at 11:55
• This inequality was shown to be incorrect at Mathoverflow. Therefore, people can disregard my question.
|
Aug 16, 2014 at 22:28
M. Lin's answer at Math Overflow show that the inequality doesn't need to hold even in the case $C=I$.
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https://math.stackexchange.com/questions/2519203/prove-if-g-fixes-an-element-in-x-then-it-fixes-more-than-one-element
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# Prove if $G$ fixes an element in $X$ then it fixes more than one element
Let $p$ be a prime and let $X$ be a finite set whose cardinality is divisible by $p$. Assume a group $G$ of order $p^n$ for $n>1$ acts on $X$. Show that if $G$ fixes an element in $X$ then it fixes more than one element.
No idea with this problem. Maybe there is some useful key facts about such kind of group. May I please ask to pointing them out or prove it directly? Many thanks!
Use the fact that $X$ can be decomposed into disjoint union of orbits. Assume that there is exactly one element fixed in $X$ by $G$. Now, the other orbits must have length dividing $p^{n}$ and are strictly bigger than $1$, and hence must be divisible by $p$. Hence $X$ is a disjoint union of sets, one of length one, and the others of length divisible by $p$, so $|X|=1+kp$ for some $k >0$. Contradiction. We must have another fixed point.
If $x \in X$, then the orbit stabilizer theorem tells us that $|O(x)| = (G:G_x)$, where $|O(x)|$ is the order of the orbit containing $x$ and $G_x$ is the stabilizer of $x$. If no other element is fixed, then there exists $y \in X$ such $|O(y)| = (G:G_y) \neq 1$ and hence $O(y) = p^m$ but $X$ has only $p$ elements so $|O(y)| = p$. Thus $O(y) = X$ and we no longer have $O(x) \cap O(y) = \emptyset$
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# Prove if $G$ fixes an element in $X$ then it fixes more than one element
Let $p$ be a prime and let $X$ be a finite set whose cardinality is divisible by $p$. Assume a group $G$ of order $p^n$ for $n>1$ acts on $X$. Show that if $G$ fixes an element in $X$ then it fixes more than one element. No idea with this problem. Maybe there is some useful key facts about such kind of group. May I please ask to pointing them out or prove it directly? Many thanks! Use the fact that $X$ can be decomposed into disjoint union of orbits. Assume that there is exactly one element fixed in $X$ by $G$. Now, the other orbits must have length dividing $p^{n}$ and are strictly bigger than $1$, and hence must be divisible by $p$. Hence $X$ is a disjoint union of sets, one of length one, and the others of length divisible by $p$, so $|X|=1+kp$ for some $k >0$. Contradiction. We must have another fixed point. If $x \in X$, then the orbit stabilizer theorem tells us that $|O(x)| = (G:G_x)$, where $|O(x)|$ is the order of the orbit containing $x$ and $G_x$ is the stabilizer of $x$. If no other element is fixed, then there exists $y \in X$ such $|O(y)| = (G:G_y) \neq 1$ and hence $O(y) = p^m$ but $X$ has only $p$ elements so $|O(y)| = p$.
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Thus $O(y) = X$ and we no longer have $O(x) \cap O(y) = \emptyset$
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https://math.stackexchange.com/questions/1897981/circle-centre-point-from-two-angles-and-circle-overal
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# Circle centre point from two angles and circle overal
There are two overlapping circles. Point A is known (it lies on the y axis) and the y value is given. This point makes a tangent to a circle with an unknown centre point and radius (centre C on the diagram)Diagram. A tangent angle at an overlap is known at point B and diameter is specified for this circle.
From this how can you calculate the Centre point and radius of the unknown circle?
The two reference circles are a distraction. You have two points $A$ and $B$ on the unknown circle and the directions (after a little bit of computation) of the tangents to the unknown circle at those two points. Its center $C$ is at the intersection of the perpendiculars to those two tangent lines at the two points of tangency.
The slope of the tangent to either of the reference circles at the point $(x,y)$ is simply $-x/y$, and you can then find the slope of the tangent to the unknown circle via the formula for the tangent of a sum.
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# Circle centre point from two angles and circle overal
There are two overlapping circles. Point A is known (it lies on the y axis) and the y value is given. This point makes a tangent to a circle with an unknown centre point and radius (centre C on the diagram)Diagram. A tangent angle at an overlap is known at point B and diameter is specified for this circle. From this how can you calculate the Centre point and radius of the unknown circle? The two reference circles are a distraction. You have two points $A$ and $B$ on the unknown circle and the directions (after a little bit of computation) of the tangents to the unknown circle at those two points. Its center $C$ is at the intersection of the perpendiculars to those two tangent lines at the two points of tangency.
|
The slope of the tangent to either of the reference circles at the point $(x,y)$ is simply $-x/y$, and you can then find the slope of the tangent to the unknown circle via the formula for the tangent of a sum.
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https://math.stackexchange.com/questions/1908498/can-a-equiv-x-cdot-b-mod-m-be-solved-for-x
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# Can $a \equiv x \cdot b$ (mod $m$) be solved for $x$?
Suppose I have $a \equiv x \cdot b$ (mod $m$) and I know the value of $a, b, m$.
The value of $m$ is of the form $2^n$ (I don't know if that even helps)...
I want to know if there exist a unique value of $x$ in $[0, m -1)$ that satisfies the equation. And if so, then how do I obtain it?
This is what I've done so far:
First I considered $a$ and $b$ to be between $0$ and $m-1$
$a \equiv x \cdot b$ (mod $m$)
$$m | (bx - a)$$ $$bx - a = km$$ $$x = \frac{km + a}{b}$$ Since $x$ ranges from $0$ to $m-1$, $k$ must be from $0$ to $b-1$
I'm confused how to proceed from here...
• Do you know the extended Euclidean algorithm and the Bézout indetity? – quid Aug 30 '16 at 14:18
This equation has a unique solution if and only if $b$ is a unit modulo $m$, which happens if and only if $b$ and $m$ are coprime.
To solve $$a\equiv xb\pmod{m}\tag{1}$$ we can solve $$xb+ym=a\tag{2}$$ which requires $\gcd(b,m)\mid a$. Thus, we would like to solve $$x\frac{b}{\gcd(b,m)}+y\frac{m}{\gcd(b,m)}=\frac{a}{\gcd(b,m)}\tag{3}$$ To solve $(3)$, we can use the Extended Euclidean Algorithm. One implementation is detailed in this answer.
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# Can $a \equiv x \cdot b$ (mod $m$) be solved for $x$? Suppose I have $a \equiv x \cdot b$ (mod $m$) and I know the value of $a, b, m$. The value of $m$ is of the form $2^n$ (I don't know if that even helps)... I want to know if there exist a unique value of $x$ in $[0, m -1)$ that satisfies the equation. And if so, then how do I obtain it? This is what I've done so far:
First I considered $a$ and $b$ to be between $0$ and $m-1$
$a \equiv x \cdot b$ (mod $m$)
$$m | (bx - a)$$ $$bx - a = km$$ $$x = \frac{km + a}{b}$$ Since $x$ ranges from $0$ to $m-1$, $k$ must be from $0$ to $b-1$
I'm confused how to proceed from here...
• Do you know the extended Euclidean algorithm and the Bézout indetity? – quid Aug 30 '16 at 14:18
This equation has a unique solution if and only if $b$ is a unit modulo $m$, which happens if and only if $b$ and $m$ are coprime. To solve $$a\equiv xb\pmod{m}\tag{1}$$ we can solve $$xb+ym=a\tag{2}$$ which requires $\gcd(b,m)\mid a$.
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Thus, we would like to solve $$x\frac{b}{\gcd(b,m)}+y\frac{m}{\gcd(b,m)}=\frac{a}{\gcd(b,m)}\tag{3}$$ To solve $(3)$, we can use the Extended Euclidean Algorithm.
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http://math.stackexchange.com/questions/294813/compute-the-expected-number-of-descents-in-a-random-permutations-in-s-n
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# Compute the expected number of descents in a random permutations $\in$ $S_n$
Let $\sigma$ be a permutation of [n]. There is a descent at i, $1\leq i\leq n$, if $\sigma(i)>\sigma(i+1)$. Compute the expected number of descents in a random permutations $\in$ $S_n$
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What are your thoughts? Can you say something about the relation between the expected number of descents and the expected number of ascents? What is the expected number of places where $\sigma$ changes ins some direction? – Henning Makholm Feb 4 '13 at 20:37
Linearity of expectation. Randomly choose a location, then compute the expectation. Then sum them up. – Kimmi Feb 4 '13 at 20:40
Sounds good. Which answer does that lead to? – Henning Makholm Feb 4 '13 at 20:43
Alternatively, you can pair up a permutation with another permutation so that the total 'descents' is a constant. – Thomas Andrews Feb 4 '13 at 20:43
Linearity of expectation works. Also works to say that there are $n-1$ places where $\sigma(i) \neq \sigma(i+1)$, and note that by symmetry you should expect equal numbers of ascents and descents. – mjqxxxx Feb 4 '13 at 21:08
There are $n-1$ places for comparison, either descents or ascents, and they are symmetric. So the expected number of descents in permutation is $\frac{n-1}{2}$
-
The correct answer is $\frac{n+1}{2}$.
Let $I_i$ be a random variable, indicating the i-th element in a permutation being the beginning of a descent.
$P(I_i)=\frac{1}{2}$, as any possible pair is either ascending or descending (for $i\ne 1$)
$P(I_i)=1$, as the first element is always a beginning.
$A=\sum_{i=1}^nI_i=1+\sum_{i=2}^nI_i$
$E(A)=1+\sum_{i=2}^n P(I_i) = \frac{n+1}{2}$
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# Compute the expected number of descents in a random permutations $\in$ $S_n$
Let $\sigma$ be a permutation of [n]. There is a descent at i, $1\leq i\leq n$, if $\sigma(i)>\sigma(i+1)$. Compute the expected number of descents in a random permutations $\in$ $S_n$
-
What are your thoughts? Can you say something about the relation between the expected number of descents and the expected number of ascents? What is the expected number of places where $\sigma$ changes ins some direction? – Henning Makholm Feb 4 '13 at 20:37
Linearity of expectation. Randomly choose a location, then compute the expectation. Then sum them up. – Kimmi Feb 4 '13 at 20:40
Sounds good. Which answer does that lead to? – Henning Makholm Feb 4 '13 at 20:43
Alternatively, you can pair up a permutation with another permutation so that the total 'descents' is a constant. – Thomas Andrews Feb 4 '13 at 20:43
Linearity of expectation works. Also works to say that there are $n-1$ places where $\sigma(i) \neq \sigma(i+1)$, and note that by symmetry you should expect equal numbers of ascents and descents. – mjqxxxx Feb 4 '13 at 21:08
There are $n-1$ places for comparison, either descents or ascents, and they are symmetric. So the expected number of descents in permutation is $\frac{n-1}{2}$
-
The correct answer is $\frac{n+1}{2}$. Let $I_i$ be a random variable, indicating the i-th element in a permutation being the beginning of a descent. $P(I_i)=\frac{1}{2}$, as any possible pair is either ascending or descending (for $i\ne 1$)
$P(I_i)=1$, as the first element is always a beginning.
|
$A=\sum_{i=1}^nI_i=1+\sum_{i=2}^nI_i$
$E(A)=1+\sum_{i=2}^n P(I_i) = \frac{n+1}{2}$
-
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https://mathematica.stackexchange.com/questions/36836/efficient-matrix-multiplication
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# Efficient matrix multiplication
I have a list of vectors vecs = {vec1, vec2, ..., vecN} where veci is a list with length $N$.
Now I have a matrix $N\times N$ called mat.
I would like to efficiently get all the numbers veci.mat.veci for $i=1$ to $N$ in a list. How do I do it?
• Perhaps, MapThread[Dot,{vecs.mat,vecs}]? (untested). Commented Nov 11, 2013 at 21:26
• @LeonidShifrin Is your matrix multiplication engine broken? Commented Nov 11, 2013 at 21:31
• @belisarius Let's say it is just busy :). Actually, even just a simplest sample input with expected output would make this question way more attractive, and I guess not just for me. Commented Nov 11, 2013 at 21:33
• @LeonidShifrin It works. Thanks! Commented Nov 11, 2013 at 21:50
• All right, I will then post this as an answer, to not keep this among the unanswered questions. Commented Nov 11, 2013 at 22:12
This is one way:
MapThread[Dot,{vecs.mat,vecs}]
Here's another way:
Total[vecs.mat * vecs, {2}]
If the vectors happen to come naturally as the columns of vecs,
rather than its rows, then this will get what you want:
Total[vecs * mat.vecs]
One way to calculate is to leave everything in matrix form
v = RandomReal[{-1, 1}, {m=5, 10}];
mat = RandomReal[{-1, 1}, {10, 10}];
Diagonal[v.mat.Transpose[v]]
Since we're looking for fast ways, this seems faster:
v = RandomReal[{-1, 1}, {m=1000, 10}];
mat = RandomReal[{-1, 1}, {10, 10}];
v[[#]].mat.v[[#]] & /@ Range[m]
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# Efficient matrix multiplication
I have a list of vectors vecs = {vec1, vec2, ..., vecN} where veci is a list with length $N$. Now I have a matrix $N\times N$ called mat.
|
I would like to efficiently get all the numbers veci.mat.veci for $i=1$ to $N$ in a list.
|
https://math.stackexchange.com/questions/4953206/within-a-space-which-is-a-3-sphere-what-the-is-surface-area-of-a-2-sphere
| 1,723,116,616,000,000,000
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# Within a space which is a 3-sphere, what the is surface area of a 2-sphere?
If an ant lives on a ball (2-sphere) with radius $$r$$, draws a circle and measures the radius of the circle as $$R$$ as it walks along the ball's curved area, it will find that the length of the circle is $$U = 2\pi r \sin (R/r)$$.
Question: raising all dimensions by one, what is the area of a 2-sphere if it lives in a 3-sphere? By raising all dimensions by one, I mean
• space to live in: 2-sphere -> 3-sphere
• geometrical object created: 1-sphere (circle) -> 2-sphere (ball)
• length of circle $$U$$ -> area of 2-sphere $$A$$
• measured radius of of the object created: $$R$$ in both cases
• radius of space we live in: $$r$$ in both cases
What is $$f$$ in $$A = f(R, r)$$?
An $$n$$-sphere of radius $$R$$ in an $$(n+1)$$-spherical space of radius $$r$$ (with $$R$$ measured inside the spherical space and $$r$$ measured in $$\Bbb R^{n+2}$$ where the $$(n+1)$$-sphere is defined) is an $$n$$-sphere of radius $$r\sin\frac Rr$$ in a flat (i.e. Euclidean) $$(n+1)$$-dimensional cross section of $$\Bbb R^{n+2}$$. (This is easiest to visualize with $$n=0$$.) The surface measure is the same in both spaces (unlike the interior volume!), so we can plug $$r\sin\frac Rr$$ into the appropriate standard formula to compute it. For $$n=2$$ this gives $$4\pi (r\sin\frac Rr)^2$$.
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# Within a space which is a 3-sphere, what the is surface area of a 2-sphere? If an ant lives on a ball (2-sphere) with radius $$r$$, draws a circle and measures the radius of the circle as $$R$$ as it walks along the ball's curved area, it will find that the length of the circle is $$U = 2\pi r \sin (R/r)$$. Question: raising all dimensions by one, what is the area of a 2-sphere if it lives in a 3-sphere? By raising all dimensions by one, I mean
• space to live in: 2-sphere -> 3-sphere
• geometrical object created: 1-sphere (circle) -> 2-sphere (ball)
• length of circle $$U$$ -> area of 2-sphere $$A$$
• measured radius of of the object created: $$R$$ in both cases
• radius of space we live in: $$r$$ in both cases
What is $$f$$ in $$A = f(R, r)$$? An $$n$$-sphere of radius $$R$$ in an $$(n+1)$$-spherical space of radius $$r$$ (with $$R$$ measured inside the spherical space and $$r$$ measured in $$\Bbb R^{n+2}$$ where the $$(n+1)$$-sphere is defined) is an $$n$$-sphere of radius $$r\sin\frac Rr$$ in a flat (i.e. Euclidean) $$(n+1)$$-dimensional cross section of $$\Bbb R^{n+2}$$. (This is easiest to visualize with $$n=0$$.) The surface measure is the same in both spaces (unlike the interior volume! ), so we can plug $$r\sin\frac Rr$$ into the appropriate standard formula to compute it.
|
For $$n=2$$ this gives $$4\pi (r\sin\frac Rr)^2$$.
|
https://cs.stackexchange.com/questions/41097/how-to-solve-a-simple-linear-equation-using-a-binary-tree-data-structure
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# How to solve a Simple Linear Equation using a binary tree data structure
i am currently working on a school project that takes in a simple linear equation and has to return the value of x, the code i have transforms x + 3 = 3x - 2 into a binary tree format like so:
=
/ \
+ -
/ \ / \
x 3 * 2
/ \
3 x
now that i have the expression in this format could someone please explain how can i obtain the value of x, any help is appreciated and if you have an alternative method that may make it easier i would love to hear it thank you
• Interesting question (+1). I suspect that there might be a way to go from the parse tree to a solution (if one exists), but it's not obvious to me. For example, you need to ensure that both the left and right subtrees produce a linear expression. That would mean that the subtrees of any * node can't both be expressions involving $x$. Given these constraints, as I said, I suspect that you could manipulate the parse tree to get the solution, but I'll have to look at it later. For practical purposes, I'd just use @Yuval's idea. Apr 7, 2015 at 1:02
You don't really want to solve the equation using this representation. You want to convert it into a normal form such as $ax+b = 0$, from which you can read the solution. The first step would be to replace $P = Q$ with $P - Q = 0$. The second step would be to normalize $P - Q$ to some standard form, say a polynomial (or in you case, perhaps $ax+b$). One way to do this is to convert each subexpression into normal form recursively, using rules for handling leaves such as numerical constants and $x$, and mathematical operations such as $+,-,\times$.
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# How to solve a Simple Linear Equation using a binary tree data structure
i am currently working on a school project that takes in a simple linear equation and has to return the value of x, the code i have transforms x + 3 = 3x - 2 into a binary tree format like so:
=
/ \
+ -
/ \ / \
x 3 * 2
/ \
3 x
now that i have the expression in this format could someone please explain how can i obtain the value of x, any help is appreciated and if you have an alternative method that may make it easier i would love to hear it thank you
• Interesting question (+1). I suspect that there might be a way to go from the parse tree to a solution (if one exists), but it's not obvious to me. For example, you need to ensure that both the left and right subtrees produce a linear expression. That would mean that the subtrees of any * node can't both be expressions involving $x$. Given these constraints, as I said, I suspect that you could manipulate the parse tree to get the solution, but I'll have to look at it later. For practical purposes, I'd just use @Yuval's idea. Apr 7, 2015 at 1:02
You don't really want to solve the equation using this representation. You want to convert it into a normal form such as $ax+b = 0$, from which you can read the solution.
|
The first step would be to replace $P = Q$ with $P - Q = 0$.
|
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