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https://blog.listcomp.com/math/2018/04/10/convex-functions-jensens-inequality-ji-on-expectations-gibbs-inequality | # Convex Functions / Jensen's Inequality / Jensen's Inequality on Expectations / Gibbs' Inequality / Entropy
Yao Yao on April 10, 2018
1. Jensen’s Inequality is the property of convex functions. Convex function 本身就是通过 Jensen’s Inequality 定义的，它们基本就是一回事
2. 在不同的领域，对 Jensen’s Inequality 做不同的展开，可以得到该特定领域的新的 inequality；所以说 Jensen’s Inequality 可以看做是一个总纲
• 基本套路是：该领域有一个 convex function，按 Jensen’s Inequality 展开，得到领域内概念 A 小于概念 B
## 1. Convex Function / Jensen’s Inequality
Let $X$ be a convex set in a real vector space and let $f: X \rightarrow {\mathbb{R}}$ be a function.
• Definition of convex functions
• $f$ is called convex if $f$ statisfies Jensen’s Inequality.
• Jensen’s Inequality
• $\forall x_{1}, x_{2} \in X, \forall t \in [0,1] :\qquad f(t x_{1} + (1-t) x_{2}) \leq t f(x_{1}) + (1-t) f(x_{2})$
• Definition of concave functions
• $f$ is said to be concave if $−f$ is convex.
## 2. Jensen’s Inequality on Expectations
If $X$ is a random variable and $f$ is a convex function:
$f(p_1 x_1 + p_2 x_2 + \dots + p_n x_n) \leq p_1 f(x_1) + p_2 f(x_2) + \dots + p_n f(x_n)$
LHS is essentially $f(\mathrm{E}(X))$ and RHS $\mathrm{E}(f(X))$, which together give
$f(\mathrm{E}(X)) \leq \mathrm{E}(f(X))$
## 3. Gibbs’ Inequality
Let $p = \lbrace p_1, p_2, \dots, p_n \rbrace$ be the true probability distribution for $X$ and $q = \lbrace q_1, q_2, \dots, q_n \rbrace$ be another probability distribution (你可以认为一个假设的 $X$ distrbution). Construct a random variable $Y$ who follows $Y(x) = \frac{q(x)}{p(x)}$. Given $f(y) = -\log(y)$ is a convex function, we have:
$f(\mathrm{E}(Y)) \leq \mathrm{E}(f(Y))$
Therefore:
\begin{aligned} -\log \sum_{i} \big ( p_i \frac{q_i}{p_i} \big ) & \leq \sum_{i} p_i \big (-\log \frac{q_i}{p_i} \big ) \newline -\log 1 & \leq \sum_{i} p_i \log \frac{p_i}{q_i} \newline 0 & \leq \sum_{i} p_i \log \frac{p_i}{q_i} \end{aligned}
$D_{KL}(p \Vert q) \equiv \sum_{i} p_i \log_2 \frac{p_i}{q_i} \geq 0$
\begin{aligned} D_{KL}(p \Vert q) \equiv \sum_{i} p_i \log_2 \frac{p_i}{q_i} &= \sum_{i} p_i \log_2 p_i - \sum_{i} p_i \log_2 q_i \newline &= -H(p) + H(p, q) \geq 0 \end{aligned}
• $H(p)$ is the entropy of distribution $p$
• $H(p, q)$ is the cross entropy of distributions $p$ and $q$
• $H(p) \leq H(p, q) \Rightarrow$ the information entropy of a distribution $p$ is less than or equal to its cross entropy with any other distribution $q$
### Interpretations of $D_{KL}(p \Vert q)$
• In the context of machine learning, $D_{KL}(p \Vert q)$ is often called the information gain achieved if $q$ is used instead of $p$ (This is why it’s also called the relative entropy of $p$ with respect to $q$).
• In the context of coding theory, $D_{KL}(p \Vert q)$ can be construed as measuring the expected number of extra bits required to code samples from $p$ using a code optimized for $q$ rather than the code optimized for $p$.
• In the context of Bayesian inference, $D_{KL}(p \Vert q)$ is amount of information lost when $q$ is used to approximate $p$.
• 简单说，$D_{KL}(p \Vert q)$ 可以衡量两个 distribution $p$ 和 $q$ 的 “接近程度”
• 如果 $p=q$，那么 $D_{KL}(p \Vert q) = 0$
• $p$ 和 $q$ 差异越大，$D_{KL}(p \Vert q)$ 越大
blog comments powered by Disqus | 2021-04-18T11:42:56 | {
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https://std.iec.ch/iev/iev.nsf/17127c61f2426ed8c1257cb5003c9bec/236031e63f7d267cc1257eec00498997?OpenDocument | IEVref:351-50-02ID:
Language:enStatus: Standard
Term: first-order lag element
Synonym1:
Synonym2:
Synonym3:
Symbol:
Definition: linear time-invariant transfer element the transfer function of which has exactly one real-value negative pole and no zero
Note 1 to entry: The transfer function of a first-order lag element is given as
$\frac{V\left(s\right)}{U\left(s\right)}=\frac{{K}_{\text{Ρ}}}{1+{T}_{1}\cdot s}$
where
KP is the proportional action coefficient; T1 is the time constant; s is the complex variable of the Laplace transform; U(s) is the input transform; V(s) is the output transform.
Note 1 to entry: This entry was numbered 351-28-13 in IEC 60050-351:2006.
Publication date:2013-11
Source:
Replaces:
Internal notes:
CO remarks:
TC/SC remarks:
VT remarks:
Domain1:
Domain2:
Domain3:
Domain4:
Domain5: | 2021-05-12T09:25:38 | {
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http://mathhelpforum.com/algebra/221022-functions.html | 1. ## Functions
Functions f and g are defined for x E R by
f:x --->e^x
g:x--->2x-3
(i) Solve the equation fg(x)=7
Function h is defined as gf
(ii)Express h in terms of x and state its range
(iii)Express h inverse in terms of x.
2. ## Re: Functions
(i)x=(ln7+3)/2...its easy. take ln both sides.
(ii)h(x)=2e^x-3
its minimum value will be when the term 2e^x is minimum since 3 is constant ..2e^x can attain a minimum zero(just attain) at x=-infinity.And its maximum value will be infinity.therefore range is (-3,infinity)
(iii)for finding h inverse take log again..
Hope it helps....
3. ## Re: Functions
Note that:
$\displaystyle f \circ g(x) = f(g(x))$
So, in this case, that means:
$\displaystyle f \circ g(x) = e^{g(x)} = e^{2x-3} = (e^{2x})(e^{-3}) = \frac{(e^x)^2}{e^3} = 7$
$\displaystyle e^x = \sqrt{7e^3}$, and taking logs, we get:
$\displaystyle x = \log(\sqrt{7e^3}) = \log((\sqrt{7})(e^{\frac{3}{2}}))$
$\displaystyle = \log(\sqrt{7}) + \log(e^{\frac{3}{2}}) = \log(\sqrt{7}) + \frac{3}{2} = \frac{\log(7) + 3}{2}$
Similarly,
$\displaystyle g \circ f(x) = 2(e^x) - 3$.
Since $\displaystyle e^x$ ranges from (0,∞), it's clear that $\displaystyle 2(e^x)$ also ranges from (0,∞) and thus the range of $\displaystyle g \circ f$ is (-3,∞).
Consequently, $\displaystyle h^{-1} = f^{-1} \circ g^{-1}$ is only defined on (-3,∞). Can you figure out what it is, in terms of x?
4. ## Re: Functions
i solved it maybe it is : (ln(x+3)/2)=h-1(x)
here domain of h-1(x) is (-3,infinity) same as the range .. | 2018-05-22T22:08:45 | {
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https://study.com/academy/answer/find-the-absolute-maximum-and-minimum-values-of-f-on-the-set-d-f-x-y-x-4-plus-y-4-4xy-plus-6-d-x-y-0-less-than-or-equal-to-x-less-than-or-equal-to-3-0-less-than-or-equal-to-y-less-than-or-equal-to-2.html | # Find the absolute maximum and minimum values of f on the set D. f(x,y) = x^{4} + y^{4} - 4xy + 6,...
## Question:
Find the absolute maximum and minimum values of {eq}f {/eq} on the set {eq}D {/eq}.
{eq}f(x,y) = x^{4} + y^{4} - 4xy + 6, \quad\quad D = \{(x,y) \left| \right. 0 \leq x \leq 3,\;\; 0 \leq y \leq 2\} {/eq}
## Comparison of function using Calculus:
If we want to compare {eq}f(x) \ \ \text{and} \ \ g(x) {/eq} consider a function {eq}\phi(x) = f(x) - g(x) {/eq} or {eq}\phi(x) = g(x) - f(x) {/eq} and check whether {eq}\phi(x) {/eq} is increasing or decreasing in given domain of {eq}f(x) \ \ \text{and} \ \ g(x) . {/eq}
{eq}\hspace{30mm} \displaystyle{ f(x,y) = x^4 + y^4 - 4xy + 6 \\ f_{x} = 4x^3 - 4y \\ f_{y} = 4y^3 - 4x } {/eq}
Now to find the critical points solve the system:
{eq}\hspace{30mm} \displaystyle{ 4x^3 - 4y = 0 \\ 4y^3 - 4x = 0 } {/eq}
The real solution of this system is:
{eq}\hspace{30mm} \displaystyle{ (0,0) , \ \ (1,1) \ \ \text{and} \ \ (-1,-1) } {/eq}
Hence the critical points are: {eq}(0,0) , \ \ (1,1) \ \ \text{and} \ \ (-1,-1). {/eq}
To find the absolute maxima we also need to check the function at the boundary points of the given domain. Therefore the region of the given domain {eq}D {/eq} is given below in figure:
Therefore the boundary points are {eq}A \ (0,0) , \ \ B \ (3,0) , \ \ C \ (3,2) \ \ \text{and} \ \ D \ (0,2) {/eq}
Therefore the value of function at these critical and boundary points is given below:
{eq}\text{At} \ \ (0,0) \ \ f(x,y) = 6 \\ \text{At} \ \ (1,1) \ \ f(x,y) = 4 \\ \text{At} \ \ (-1,-1) \ \ f(x,y) = 4 \\ \text{At} \ \ (3,0) \ \ f(x,y) = 87 \\ \text{At} \ \ (3,2) \ \ f(x,y) = 79 \\ \text{At} \ \ (0,2) \ \ f(x,y) = 22 \\ {/eq}
Therefore the given function has absolute maxima at {eq}(3,0) {/eq} in the given region. And has absolute minima at {eq}(1,1) \ \ \text{and} \ \ (-1,-1). {/eq} | 2020-07-05T04:47:47 | {
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http://georgehernandez.com/h/xComputers/JavaScript/Objects/Math.asp | # Math
The Math object in JavaScript
TAGS: JavaScript, Math, Programming, TECH
The Math object in JavaScript. All Math properties and methods are static that are accessed directly from the prototype as opposed to an instance.
## Properties
All the Math properties are in ALL CAPs because they are all constants.
.E. Euler's constant and the base of natural logarithms, approximately 2.718.
.LN10. Natural logarithm of 10, approximately 2.302.
.LN2. Natural logarithm of 2, approximately 0.693.
.LOG10E. Base 10 logarithm of E (approximately 0.434).
.LOG2E. Base 2 logarithm of E (approximately 1.442).
.PI. Ratio of the circumference of a circle to its diameter, approximately 3.14159.
.SQRT1_2. Square root of 1/2; equivalently, 1 over the square root of 2, approximately 0.707.
.SQRT2. Square root of 2, approximately 1.414.
## Methods
.abs(n). Returns the absolute value of a number.
.acos(n). Returns the arccosine (in radians) of a number. The acos method returns a numeric value between 0 and pi radians. If the value of number is outside this range, it returns NaN.
.asin(n). Returns the arcsine (in radians) of a number. The asin method returns a numeric value between -pi/2 and pi/2 radians. If the value of number is outside this range, it returns NaN.
.atan(n). Returns the arctangent (in radians) of a number. The atan method returns a numeric value between -pi/2 and pi/2 radians.
.atan2(y,x). Returns the arctangent of the quotient of its arguments. The atan2 method returns a numeric value between -pi and pi representing the angle theta of an (x,y) point.
.ceil(n). Returns the smallest integer greater than or equal to a number.
.cos(n). Returns the cosine of a number. The cos method returns a numeric value between -1 and 1.
.exp(n). Returns En, where n is the argument, and E is Euler's constant, the base of the natural logarithms.
.floor(n). Returns the largest integer less than or equal to a number.
.log(n). Returns the natural logarithm (base E) of a number.
.max(m,n). Returns the greater of two numbers.
.min(m,n). Returns the lesser of two numbers.
.pow(base,exponent). Returns base to the exponent power, that is, baseexponent.
.random(). Returns a pseudo-random number between 0 and 1. (Seeded from the current time)
.round(n). Returns the value of a number rounded to the nearest integer.
.sin(n). Returns the sine of a number. 1, which represents the sine of the argument.
.sqrt(n). Returns the square root of a number.
.tan(n). Returns the tangent of a number.
GeorgeHernandez.comSome rights reserved | 2017-08-19T10:58:33 | {
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https://math.stackexchange.com/questions/845034/system-of-two-quadratic-equations-with-two-variables | # system of two quadratic equations with two variables
Is there a general way to solve exactly a system of this shape (the $a_i$ are constants):
$$\begin{array}{cc}a_1x^2+a_2x+a_3y^2+a_4y+a_5=0\\ a_6xy+a_7x+a_8y+a_9=0 \end{array}$$
It comes from a geometrical problem: the first equation states two vectors have same length, and the second ones state that they are perpendicular.
It can be reduced to one equation of degree four with one variable, but is it the best we can do?
• In principle, the second equation can be used to solve for $x$ as a rational linear function in $y$. Substituting into the first equation and clearing denominators gives a quartic in $y$, and all quartics can be solved exactly. – davidlowryduda Jun 23 '14 at 19:49
• You ask whether reduction to a quartic is "the best we can do." We know exactly how to solve every quartic exactly, so I think that's pretty good. – davidlowryduda Jun 23 '14 at 20:06
• yes but I was looking for a simpler one, the general solution of quartic equations seems painful to program (I'm looking to implement this). – Denis Jun 23 '14 at 20:41
Assume your $a_1\neq0,\,a_6\neq0$. Dividing by these, you can have the slightly simpler system with two fewer constants,
$$x^2 + b_1 x + b_2 y^2 + b_3 y + b_4 = 0$$
$$x y + b_5 x + b_6 y + b_7 = 0$$
Eliminating $y$ results in a monic quartic of form,
$$x^4+ax^3+bx^2+cx+d=0\tag1$$
The solution of the general quartic is not that bad. It is just,
$$x_{1,2} = -\tfrac{1}{4}a+\tfrac{1}{2}\sqrt{u}\pm\tfrac{1}{4}\sqrt{3a^2-8b-4u+\frac{-a^3+4ab-8c}{\sqrt{u}}}\tag2$$
$$x_{3,4} = -\tfrac{1}{4}a-\tfrac{1}{2}\sqrt{u}\pm\tfrac{1}{4}\sqrt{3a^2-8b-4u-\frac{-a^3+4ab-8c}{\sqrt{u}}}\tag3$$
where,
$$u = \frac{a^2}{4} +\frac{-2b+v_1^{1/3}+v_2^{1/3}}{3}$$
and the $v_i$ are the two roots of the quadratic,
$$v^2 + (-2 b^3 + 9 a b c - 27 c^2 - 27 a^2 d + 72 b d)v + (b^2 - 3 a c + 12 d)^3 = 0$$
P.S. However, one has to be take note that some older CAS have $(-n)^{1/3} = \text{complex}$, even for real $n$. | 2019-11-14T04:12:13 | {
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http://mfleck.cs.illinois.edu/study-problems/basics/logic-3a-sol.html | # Logic: Problem 3a
### Straightforward solution
Here's one solution to the problem, which conveys the two conditions in a straightforward way:
$$(\exists x \in A,\ P(x)) \ \text{ and } \ (\neg \exists y,z \in A, \ P(y)\ \wedge P(z)\ \wedge\ y \not = z )$$
The parentheses aren't strictly necessary. But they are good to include because they make it easier for a reader to understand the expression.
### Minor variation
Notice that the variable from the first half isn't used in the second half. So it's also correct (though perhaps a bit less clear) to re-use the variable name x, as follows:
$$(\exists x \in A,\ P(x)) \ \text{ and } \ (\neg \exists x,y \in A, \ P(x)\ \wedge P(y)\ \wedge\ x \not = y )$$
Notice that the variable x in the first half and the variable x in the second half are two distinct variables.
Here's a rather different sort of answer, equivalent to the above. It's a more hard-core mathematician style of answer: concise but harder to understand. $$\exists x \in A, \ P(x)\ \text{ and } \ ( \forall y \in A,\ P(y) \rightarrow x = y )$$
$$\exists x \in A,\ P(x)\ \text{ and } \ (\neg \exists y \in A, \ P(y)\ \wedge\ x \not = y )$$ | 2018-01-22T00:19:16 | {
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https://stacks.math.columbia.edu/tag/0GEN | Proposition 36.40.5. Let $X$ be a quasi-compact and quasi-separated scheme. Let $G \in D_{perf}(\mathcal{O}_ X)$ be a perfect complex which generates $D_\mathit{QCoh}(\mathcal{O}_ X)$. Let $E \in D_\mathit{QCoh}(\mathcal{O}_ X)$. The following are equivalent
1. $E \in D^-_\mathit{QCoh}(\mathcal{O}_ X)$,
2. $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(G[-i], E) = 0$ for $i \gg 0$,
3. $\mathop{\mathrm{Ext}}\nolimits ^ i_ X(G, E) = 0$ for $i \gg 0$,
4. $R\mathop{\mathrm{Hom}}\nolimits _ X(G, E)$ is in $D^-(\mathbf{Z})$,
5. $H^ i(X, G^\vee \otimes _{\mathcal{O}_ X}^\mathbf {L} E) = 0$ for $i \gg 0$,
6. $R\Gamma (X, G^\vee \otimes _{\mathcal{O}_ X}^\mathbf {L} E)$ is in $D^-(\mathbf{Z})$,
7. for every perfect object $P$ of $D(\mathcal{O}_ X)$
1. the assertions (2), (3), (4) hold with $G$ replaced by $P$, and
2. $H^ i(X, P \otimes _{\mathcal{O}_ X}^\mathbf {L} E) = 0$ for $i \gg 0$,
3. $R\Gamma (X, P \otimes _{\mathcal{O}_ X}^\mathbf {L} E)$ is in $D^-(\mathbf{Z})$.
Proof. Assume (1). Since $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(G[-i], E) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(G, E[i])$ we see that this is zero for $i \gg 0$ by Lemma 36.18.2. This proves that (1) implies (2).
Parts (2), (3), (4) are equivalent by the discussion in Cohomology, Section 20.41. Part (5) and (6) are equivalent as $H^ i(X, -) = H^ i(R\Gamma (X, -))$ by definition. The equivalent conditions (2), (3), (4) are equivalent to the equivalent conditions (5), (6) by Cohomology, Lemma 20.47.5 and the fact that $(G[-i])^\vee = G^\vee [i]$.
It is clear that (7) implies (2). Conversely, let us prove that the equivalent conditions (2) – (6) imply (7). Recall that $G$ is a classical generator for $D_{perf}(\mathcal{O}_ X)$ by Remark 36.17.2. For $P \in D_{perf}(\mathcal{O}_ X)$ let $T(P)$ be the assertion that $R\mathop{\mathrm{Hom}}\nolimits _ X(P, E)$ is in $D^-(\mathbf{Z})$. Clearly, $T$ is inherited by direct sums, satisfies the 2-out-of-three property for distinguished triangles, is inherited by direct summands, and is perserved by shifts. Hence by Derived Categories, Remark 13.36.7 we see that (4) implies $T$ holds on all of $D_{perf}(\mathcal{O}_ X)$. The same argument works for all other properties, except that for property (7)(b) and (7)(c) we also use that $P \mapsto P^\vee$ is a self equivalence of $D_{perf}(\mathcal{O}_ X)$. Small detail omitted.
We will prove the equivalent conditions (2) – (7) imply (1) using the induction principle of Cohomology of Schemes, Lemma 30.4.1.
First, we prove (2) – (7) $\Rightarrow$ (1) if $X$ is affine. Set $P = \mathcal{O}_ X[0]$. From (7) we obtain $H^ i (X, E) = 0$ for $i \gg 0$. Hence (1) follows since $E$ is determined by $R\Gamma (X, E)$, see Lemma 36.3.5.
Now assume $X = U \cup V$ with $U$ a quasi-compact open of $X$ and $V$ an affine open, and assume the implication (2) – (7) $\Rightarrow$ (1) is known for the schemes $U$, $V$, and $U \cap V$. Suppose $E \in D_\mathit{QCoh}(\mathcal{O}_ X)$ satisfies (2) – (7). By Lemma 36.15.1 and Theorem 36.15.3 there exists a perfect complex $Q$ on $X$ such that $Q|_ U$ generates $D_\mathit{QCoh}(\mathcal{O}_ U)$. Let $f_1, \dots , f_ r \in \Gamma (V, \mathcal{O}_ V)$ be such that $V \setminus U = V(f_1, \dots , f_ r)$ as subsets of $V$. Let $K \in D_{perf}(\mathcal{O}_ V)$ be the object corresponding to the Koszul complex on $f_1, \dots , f_ r$. Let $K' \in D_{perf}(\mathcal{O}_ X)$ be
36.40.5.1
$$\label{perfect-equation-detecting-bounded-above} K' = R (V \to X)_* K = R (V \to X)_! K,$$
see Cohomology, Lemmas 20.33.6 and 20.46.10. This is a perfect complex on $X$ supported on the closed set $X \setminus U \subset V$ and isomorphic to $K$ on $V$. By assumption, we know $R\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(Q, E)$ and $R\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(K', E)$ are bounded above.
By the second description of $K'$ in (36.40.5.1) we have
$\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ V)}(K[-i], E|_ V) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(K'[-i], E) = 0$
for $i \gg 0$. Therefore, we may apply Lemma 36.40.1 to $E|_ V$ to obtain an integer $a$ such that $\tau _{\geq a}(E|_ V) = \tau _{\geq a} R (U \cap V \to V)_* (E|_{U \cap V})$. Then $\tau _{\geq a} E = \tau _{\geq a} R (U \to X)_* (E |_ U)$ (check that the canonical map is an isomorphism after restricting to $U$ and to $V$). Hence using Lemma 36.40.3 twice we see that
$\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ U)}(Q|_ U [-i], E|_ U) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(Q[-i], R (U \to X)_* (E|_ U)) = 0$
for $i \gg 0$. Since the Proposition holds for $U$ and the generator $Q|_ U$, we have $E|_ U \in D^-_\mathit{QCoh}(\mathcal{O}_ U)$. But then since the functor $R (U \to X)_*$ preserves $D^-_\mathit{QCoh}$ (by Lemma 36.4.1), we get $\tau _{\geq a}E \in D^-_\mathit{QCoh}(\mathcal{O}_ X)$. Thus $E \in D^-_\mathit{QCoh}(\mathcal{O}_ X)$. $\square$
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). | 2021-04-18T06:00:26 | {
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https://documen.tv/question/when-operated-on-a-household-110-0-v-line-typical-hair-dryers-draw-about-1450-w-of-power-the-cur-15072270-89/ | # When operated on a household 110.0 V line, typical hair dryers draw about 1450 W of power. The current can be modeled as a long, straight wi
Question
When operated on a household 110.0 V line, typical hair dryers draw about 1450 W of power. The current can be modeled as a long, straight wire in the handle. During use, the current is about 2.05 cm from the user’s hand. What is the current in the dryer? What is the resitance of the dryer? What magnetic field does the dryer produce at the users hand?
in progress 0
1 year 2021-08-29T20:11:45+00:00 1 Answers 38 views 0
1. To solve this problem we will start by calculating the resistance through the relationship between it, the voltage and the power. From there we will apply Ohm’s law and later we will find the current. With these values we will have all the variables for the calculation of the magnetic field. So the resistance would be given by
$$P = \frac{V^2}{R}$$
$$R = \frac{V^2}{P}$$
Replacing with our values
$$R = \frac{(110.0V)^2}{1450W}$$
$$R = 8.345\Omega$$
The expression for Ohm’s law is given below,
$$I = \frac{V}{R}$$
$$I = \frac{110.0V}{8.345\Omega}$$
$$I = 13.18A$$
The magnetic field at any point from the current carrying straight wire depends on the current and the distance between the point and the wire, then
$$B = \frac{\mu_0 I}{2\pi r}$$
Here,
$$\mu$$ = Permeability constant
I = Current
r = Distance between the points
Replacing
$$B = \frac{(4\pi*10^{-7} H\cdot m^{-1})(13.18A)}{2\pi (2.05*10^{-2}m)}$$
$$B = 1226.0*10^{-7}T$$
Therefore the magnetic field is given as $$1226.0*10^{-7}T$$ | 2023-02-07T12:34:55 | {
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https://math.stackexchange.com/questions/2225455/rsa-proof-wikipedia-clarification | # RSA proof Wikipedia Clarification?
I have a question involving how this line of the RSA proof from Wikipedia is simplified. Proof
I know that $m^{p-1} \equiv 1(mod p)$ but how does $m^{p-1}$ simplify to 1 from the image?
Modular arithmetic is useful in large part because it respects arithmetic operations like multiplication. As implied here and here on the Wikipedia page for modular arithmetic, if $a\equiv b\pmod p$ then $ac\equiv bc\pmod p$.
In this case, we have $m^{p-1}\equiv1\pmod p$. Therefore, $\left(m^{p-1}\right)^2\equiv 1*m^{p-1}\equiv 1^2\pmod p$. Similarly, $\left(m^{p-1}\right)^3\equiv 1*\left(m^{p-1}\right)^2\equiv 1^3\pmod p$. Assuming $h$ is a nonnegative integer, we can repeat this sort of step until we arrive at $\left(m^{p-1}\right)^h\equiv 1^h\pmod p$. Then we can use the multiplicative property of modular arithmetic one more time to learn that $\left(m^{p-1}\right)^hm\equiv 1^hm\pmod p$. | 2022-08-16T06:14:29 | {
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https://expskill.com/question/if-two-vectors-vecr-vecn_1d_1-and-vecr-vecn_2d_2-are-such-that-vecn_1-vecn_20-then-which-of-the-following-is-true/ | # If two vectors $$\vec{r}.\vec{n_1}=d_1$$ and $$\vec{r}.\vec{n_2}=d_2$$ are such that $$\vec{n_1}.\vec{n_2}$$=0, then which of the following is true?
Category: QuestionsIf two vectors $$\vec{r}.\vec{n_1}=d_1$$ and $$\vec{r}.\vec{n_2}=d_2$$ are such that $$\vec{n_1}.\vec{n_2}$$=0, then which of the following is true?
Editor">Editor Staff asked 11 months ago
If two vectors \vec{r}.\vec{n_1}=d_1 and \vec{r}.\vec{n_2}=d_2 are such that \vec{n_1}.\vec{n_2}=0, then which of the following is true?
(a) The planes are perpendicular to each other
(b) The planes are parallel to each other
(c) Depends on the value of the vector
(d) The planes are at an angle greater than 90°
This question was posed to me in semester exam.
I want to ask this question from Three Dimensional Geometry topic in section Three Dimensional Geometry of Mathematics – Class 12
NCERT Solutions for Subject Clas 12 Math Select the correct answer from above options
Interview Questions and Answers, Database Interview Questions and Answers for Freshers and Experience | 2022-12-05T14:00:28 | {
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https://www.quantumstudy.com/on-a-frictionless-horizontal-surface-assumed-to-be-the-xy-plane-a-small-trolley-a-is-moving-along-a-straight-line-parallel-to-the-y-axis-as-shown-in-figure-with-a-constant-velocity-of/ | # On a frictionless horizontal surface, assumed to be the XY-plane, a small trolley A is moving along a straight line parallel to the Y-axis with a constant velocity of ( √3-1) m/s …
Q: On a frictionless horizontal surface, assumed to be the XY-plane, a small trolley A is moving along a straight line parallel to the Y-axis (as shown in figure) with a constant velocity of ( √3-1) m/s . At a particular instant, when the line OA makes an angle of 45° with the X-axis, ball is thrown along the surface from the origin O. Its velocity makes an angle φ with the X-axis and it hits the trolley.
(a) The motion of the ball is observed from the frame of the trolley. Calculate the angle q made by the velocity vector of the ball with the X0axis in this frame.
(b) Find the speed of the ball with respect of the surface, it φ = 4θ/3. | 2023-03-28T02:34:24 | {
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https://www.shaalaa.com/question-bank-solutions/the-hcf-two-numbers-145-their-lcm-2175-if-one-numbers-725-find-other-euclid-s-division-lemma_40011 | # The Hcf of Two Numbers is 145 and Their Lcm is 2175. If One of the Numbers is 725, Find the Other. - Mathematics
The HCF of two numbers is 145 and their LCM is 2175. If one of the numbers is 725, find
the other.
#### Solution
HCF of two numbers = 145
LCM of two numbers = 2175
Let one of the two numbers be 725 and other be x.
Using the formula, product of two numbers = HCF × LCM
we conclude that
725 × x = 145 × 2175
x = (145 ×2175)/ 725
= 435
Hence, the other number is 435.
Concept: Euclid’s Division Lemma
Is there an error in this question or solution? | 2022-05-18T19:13:13 | {
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https://socratic.org/questions/if-a-sample-of-gas-occupies-6-80-l-at-325-c-what-will-be-its-volume-at-25-c-if-t | If a sample of gas occupies 6.80 L at 325°C, what will be its volume at 25°C if the pressure does not change?
Mar 7, 2016
$\text{3.4 L}$
Explanation:
Even without doing any calculations, you should be able to look at the values given to you and predict that the volume of the gas will decrease as temperature decreases.
When pressure and number of moles of gas are held constant, the volume of a gas and its temperature have a direct relationship - this is known as Charles' Law.
As you know, gas pressure is caused by the collisions that take place between the molecules of gas and the walls of the container.
The more powerful and frequent these collisions are, the higher the pressure of the gas.
Now, temperature is a measure of the average kinetic energy of the gas molecules. When you decrease temperature, you're essentially decreasing the average speed with which these molecules hit the walls of the container.
Because molecules are hitting the walls of the container with less force, you need these collisions to be more frequent in order for pressure to be constant.
This means that the volume of the gas must decrease as well, since the same number of molecules in a smaller volume will result in more frequent collisions with the walls of the container.
So, when temperature decreases, volume decreases as well.
Mathematically, this is written as
$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {V}_{1} / {T}_{1} = {V}_{2} / {T}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, where
${V}_{1}$, ${T}_{1}$ - the volume and temperature of the gas at an initial state
${V}_{2}$, ${T}_{2}$ - the volume and temperature of the gas at a final state
Now, it's very important to remember that you must use absolute temperature, i.e. the temperature expressed in Kelvin.
To go from degrees Celsius to Kelvin, use the conversion factor
$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} T \left[\text{K"] = t[""^@"C}\right] + 273.15 \textcolor{w h i t e}{\frac{a}{a}} |}}}$
So, rearrange the equation for Charles' Law and solve for ${V}_{2}$
${V}_{1} / {T}_{1} = {V}_{2} / {T}_{2} \implies {V}_{2} = {T}_{2} / {T}_{1} \cdot {V}_{1}$
Plug in your values to get
V_2 = ((273.15 + 25)color(red)(cancel(color(black)("K"))))/((273.15 + 325)color(red)(cancel(color(black)("K")))) * "6.80 L" = "3.3895 L"
You need to round this off to two sig figs, the number of sig figs you have for the final temperature of the gas
${V}_{2} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{3.4 L} \textcolor{w h i t e}{\frac{a}{a}} |}}}$ | 2021-07-27T21:43:25 | {
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https://stacks.math.columbia.edu/tag/0H0V | Remark 84.8.4. Assumptions and notation as in Lemma 84.8.3 except we do not require $K$ in $D(\mathcal{C}_{total})$ to be bounded below. We claim there is a natural spectral sequence in this case also. Namely, suppose that $\mathcal{I}^\bullet$ is a K-injective complex of sheaves on $\mathcal{C}_{total}$ with injective terms representing $K$. We have
\begin{align*} R\Gamma (\mathcal{C}_{total}, K) & = R\mathop{\mathrm{Hom}}\nolimits (\mathbf{Z}, K) \\ & = R\mathop{\mathrm{Hom}}\nolimits ( \ldots \to g_{2!}\mathbf{Z} \to g_{1!}\mathbf{Z} \to g_{0!}\mathbf{Z}, K) \\ & = \Gamma (\mathcal{C}_{total}, \mathop{\mathcal{H}\! \mathit{om}}\nolimits ^\bullet ( \ldots \to g_{2!}\mathbf{Z} \to g_{1!}\mathbf{Z} \to g_{0!}\mathbf{Z}, \mathcal{I}^\bullet )) \\ & = \text{Tot}_\pi (A^{\bullet , \bullet }) \end{align*}
where $A^{\bullet , \bullet }$ is the double complex with terms $A^{p, q} = \Gamma (\mathcal{C}_ p, \mathcal{I}^ q_ p)$ and $\text{Tot}_\pi$ denotes the product totalization of this double complex. Namely, the first equality holds in any site. The second equality holds by Lemma 84.8.1. The third equality holds because $\mathcal{I}^\bullet$ is K-injective, see Cohomology on Sites, Sections 21.34 and 21.35. The final equality holds by the construction of $\mathop{\mathcal{H}\! \mathit{om}}\nolimits ^\bullet$ and the fact that $\mathop{\mathrm{Hom}}\nolimits (g_{p!}\mathbf{Z}, \mathcal{I}^ q) = \Gamma (\mathcal{C}_ p, \mathcal{I}^ q_ p)$. Then we get our spectral sequence by viewing $\text{Tot}_\pi (A^{\bullet , \bullet })$ as a filtered complex with $F^ i\text{Tot}^ n_\pi (A^{\bullet , \bullet }) = \prod _{p + q = n,\ p \geq i} A^{p, q}$. The spectral sequence we obtain behaves like the spectral sequence $({}'E_ r, {}'d_ r)_{r \geq 0}$ in Homology, Section 12.25 (where the case of the direct sum totalization is discussed) except for regularity, boundedness, convergence, and abutment issues. In particular we obtain $E_1^{p, q} = H^ q(\mathcal{C}_ p, K_ p)$ as in Lemma 84.8.3.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). | 2022-08-19T05:26:03 | {
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https://math.stackexchange.com/questions/2050281/generating-a-random-variable-from-a-biased-coin/2050290 | Generating a random variable from a Biased Coin
Assume I want to generate a realization of a random variable $X$ with discrete distribution $F_X$. So I know the support of that random variable, but I don't know the distribution. However, if I call a value from the support, someone will tell me probability of that value according to $F_X$.
So in this case, you could call all the values from the support sequentially, then you could know the full distribution $F_X$, then you can use many methods to generate the realization of this random variable. However, I am considering the following procedure, I am not sure whether this procedure will let me generate the realization of $X$:
I first uniformly make a draw ($x$) from the support, then I will be told the probability of that specific value I draw from the support. Let's call this probability $p_x$. Then I flip a biased coin with Head probability $p_x$. If it really turns out to be head, then this value $x$ is generated from distribution $F_X$.
I am not sure whether the above claim is correct or not. Can some one prove it or disprove it mathematically?
Also, what if the biased coin does not turn up to be head? then what should I do? Should I redraw uniformly from the support, then repeat this procedure?
1 Answer
Your process will return $x$ with the proper probability. Now you need to generate all the other possible values with their proper probability. A way in the spirit of your question is to draw a different possible value $y$ and the a priori probability of $y$, $p_y$, so you exclude $x$ from the support for this draw. What you want is the probability of $y$ given that the variable is not $x$, so you want to accept $y$ with probability $\frac {p_y}{1-p_x}$, and you can flip a coin with this probability of heads. If you get heads, return $y$ and you are done. If not, you are still looking. Draw another possible value $z$ from the support except $x,y$ and you want to accept it with probability $\frac {p_z}{1-p_x-p_y}$. As your distribution is discrete, this will terminate within the number of steps equal to the number of possible values. You can even stop one before that, because if you have ruled out all the choices but one, the last choice is it.
• If I want to just generate a series of realizations of $X$ using this method, could I just do the following: uniformly draw and $x$, then I got $p_x$, then I flip this biased coin with H probability being $p_x$. Then if it turns out to be H, I record what I have drawn, which is $x$ and this is one realization of $X$. What if in this flip it turns out to be T, then should keep on flip this coin, until I get an H, or should I redraw a uniform sample $y$ from the support and get $p_y$, then flip a biased coin with H probability being $p_y$? – KevinKim Dec 8 '16 at 22:27
• No, as I said you should draw a new sample from the support less $x$ to get $y$ and $p_y$, then flip a new coin, but the probability of heads is not $p_y$ but $\frac {p_y}{1-p_x}$ The whole procedure above just gets you one realization of $X$. Repeat it as many times as you want realizations. You keep flipping until you get heads, which gives you the realization. When you get tails you don't get one, you keep going. I have edited a bit to make this clearer – Ross Millikan Dec 8 '16 at 22:31 | 2019-06-24T17:20:10 | {
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https://www.math.temple.edu/~hijab/sage/critical_points.html | # Critical Point Finder
Here you choose the function $y= f(x)$ and the interval $[a,b]$, and this page finds and graphs the critical points $y'=0$ and inflection points $y''=0$ in this interval.
Advanced Usage: If the graph extends too far vertically, enter a cutoff ($\infty$ is entered as oo). If the interval is too big, some critical points may be missed. | 2023-02-07T12:37:28 | {
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https://math.stackexchange.com/questions/1028088/calculating-integral-submanifolds | # Calculating Integral Submanifolds
I have the vector fields $v_{1} = x \partial_y - y \partial_x + z \partial_w - w \partial_z$ and $v_{2} = z \partial_x - x \partial_z + w \partial_y - y \partial_w$ on $S^{3} \subset \mathbb{R}^4$.
I have shown that the Lie bracket is zero: $[v_1, v_2] = 0$ and by Frobenius' {$v_1$, $v_2$} form an integral system.
Now I am asked to find the integral submanifolds in $S^3$, but am unsure about how to do this.
I am currently working on the same problem. I believe the idea is to use something akin to the method of characteristics for partial differential equations.
Suppose that $$\gamma(t) = \begin{pmatrix}x(t)\\y(t)\\z(t)\\w(t)\end{pmatrix}$$ is a curve on $S^3$ whose tangent vectors correspond to the vector field $v_1$. Also suppose that $\gamma$ is parameterized so that $x(0) = x_0$, $y(0) = y_0$, $z(0) = z_0$, and $w(0) = w_0$. If this is the case, then the coordinate functions satisfy the following system of ordinary differential equations:
\begin{align} &\dfrac{dx}{dt} = -y\\[0.3cm] &\dfrac{dy}{dt} = x\\[0.3cm] &\dfrac{dz}{dt} = -w\\[0.3cm] &\dfrac{dw}{dt} = z\\ \\ & \text{with $x(0)=x_0$, $y(0) = y_0$, $z(0) = z_0$, and $w(0) = w_0$} \end{align}
Next notice the following:
\begin{align} & \dfrac{\dfrac{dx}{dt}}{\dfrac{dy}{dt}} = \dfrac{dx}{dy} = -\dfrac{y}{x }\\ \\ & \dfrac{\dfrac{dz}{dt}}{\dfrac{dw}{dt}} = \dfrac{dz}{dw} = -\dfrac{w}{z } \end{align}
Solving these ODE's, we arrive at the result: $$\gamma(t) = \begin{pmatrix}\sqrt{A} \cos t\\ \sqrt{A}\sin t\\ \sqrt{B}\cos t\\ \sqrt{B}\sin t\end{pmatrix}$$ Where $A$ and $B$ are determined from the initial conditions to be $x_0^2+y_0^2$ and $z_0^2+y_0^2$ respectively.
Next consider use the same method to flow out from the point $p_t=(x_t, y_t, z_t,w_t)$ by a different curve $\eta(s)$ that satisfies the initial condition $\eta(0) = p_t$ and whose tangent vectors correspond to $v_2$. Following this method should lead you to the complete solution. | 2019-12-12T01:35:44 | {
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https://converter.ninja/volume/imperial-pints-to-dry-quarts/200-imperialpint-to-dryquart/ | 200 imperial pints in dry quarts
Conversion
200 imperial pints is equivalent to 103.20567434887 dry quarts.[1]
Conversion formula How to convert 200 imperial pints to dry quarts?
We know (by definition) that: $1\mathrm{imperialpint}\approx 0.51602837174435\mathrm{dryquart}$
We can set up a proportion to solve for the number of dry quarts.
$1 imperialpint 200 imperialpint ≈ 0.51602837174435 dryquart x dryquart$
Now, we cross multiply to solve for our unknown $x$:
$x\mathrm{dryquart}\approx \frac{200\mathrm{imperialpint}}{1\mathrm{imperialpint}}*0.51602837174435\mathrm{dryquart}\to x\mathrm{dryquart}\approx 103.20567434887\mathrm{dryquart}$
Conclusion: $200 imperialpint ≈ 103.20567434887 dryquart$
Conversion in the opposite direction
The inverse of the conversion factor is that 1 dry quart is equal to 0.00968938971920925 times 200 imperial pints.
It can also be expressed as: 200 imperial pints is equal to $\frac{1}{\mathrm{0.00968938971920925}}$ dry quarts.
Approximation
An approximate numerical result would be: two hundred imperial pints is about one hundred and three point two zero dry quarts, or alternatively, a dry quart is about zero point zero one times two hundred imperial pints.
Footnotes
[1] The precision is 15 significant digits (fourteen digits to the right of the decimal point).
Results may contain small errors due to the use of floating point arithmetic. | 2020-10-29T01:22:49 | {
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http://math.stackexchange.com/questions/534518/whats-the-derivative-fracdefdf-i-of-this-function-ef-fracdfd/534617 | # What's the derivative $\frac{dE[F]}{dF_i}$ of this function $E[F] = |\frac{dF}{dx}|^2$?
The function is
$E[F] = |\frac{dF}{dx}|^2$
where we take $\frac{dF}{dx}$ to be the discrete derivative defined by $F_{i+1} - F_i$.
Could someone walk through why $\frac{dE[F]}{dF_i} = -2F_{i+1}+4F_i-2F_{i-1}$
-
The notation leaves something to be desired. $\frac{dF}{dx}:=F_{i+1}-F_i$.
$$\frac{dE[F]}{dF_i}=\frac{d}{dF_i}|F_{i+1}-F_i|^2=\frac{d}{dF_i}\left(F_{i+1}^2-2F_{i+1}F_i+F_i^2\right)=2F_{i+1}+2F_i^2$$
This doesn't seem right, so perhaps the absolute value really means: take the norm of the vector $\frac{dF}{dx}$ whose $i$'th component is $F_{i+1}-F_i$. Then
$$\left|\frac{dF}{dx}\right|^2=\sum_{i=1}^{n-1}(F_{i+1}-F_i)^2$$
and the set of terms involving $F_i$ are $(F_{i}-F_{i-1})^2+(F_{i+1}-F_i)^2$. Now take derivatives in $F_i$, to get $2F_i-2F_{i-1}-(2F_{i+1}+2F_i)=4F_{i}-2F_{i-1}-2F_{i-1}$. | 2015-08-03T22:16:12 | {
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http://mathhelpforum.com/calculus/87269-solved-find-function-form.html | # Math Help - [SOLVED] Find a function of the form...?
1. ## [SOLVED] Find a function of the form...?
Find a function of the $f(x) = \lambda e^{2x} + \mu e^{x} - \nu x$ where $\lambda, \mu, \nu$ are independent of $x$ and $f(0) = -1$, $f'(\log_{e} 2) = 30$ and $\int_{0}^{\log_{e} 4} \{f(x) + \nu x\} dx = 28.5$
2. Just need to make a system of equations and substitute.
lamda=L mew=M v=V
f(0)=Le^(2*0)+M*e^(0)-v*0=-1
L+M=-1
f'(x)= 2*L*e^(2x)+M*e^(x) -V
f'(ln(2))=8*L+2*M-V=30
integral(L*e^(2*x)+M*e^(x),x,0,ln(4))
solve(L/2*e^(2*X)+M*e^x),x,0,ln(4))
8L+4M-(L/2+M)=28.5
System of equations:
1*L+1*M+0*V=-1
8*L+2*M-1*V=30
15/2*L+3*M=28.5
$7e^{2x} - 8e^{x} - 10x$ | 2014-04-18T22:40:32 | {
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http://www.mathynomial.com/problem/2281 | Problem #2281
2281 Three circles with radius 2 are mutually tangent. What is the total area of the circles and the region bounded by them, as shown in the figure? $[asy] filldraw((0,0)--(2,0)--(1,sqrt(3))--cycle,gray,gray); filldraw(circle((1,sqrt(3)),1),gray); filldraw(circle((0,0),1),gray); filldraw(circle((2,0),1),grey);[/asy]$ $\textbf{(A)}\ 10\pi+4\sqrt{3}\qquad\textbf{(B)}\ 13\pi-\sqrt{3}\qquad\textbf{(C)}\ 12\pi+\sqrt{3}\qquad\textbf{(D)}\ 10\pi+9\qquad\textbf{(E)}\ 13\pi$ This problem is copyrighted by the American Mathematics Competitions.
Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved. | 2018-02-26T03:50:15 | {
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https://www.quantumstudy.com/if-two-charged-particle-each-of-charge-q-mass-m-are-connected-to-the-ends-of-a-rigid-massless-rod-and-is-rotated-about-an-axis/ | # If two charged particle each of charge q mass m are connected to the ends of a rigid massless rod and is rotated about an axis…
Q: If two charged particle each of charge q mass m are connected to the ends of a rigid massless rod and is rotated about an axis passing through the centre and perpendicular to length. Then find the ratio of magnetic moment to angular momentum.
Sol: $\large M = n i A$
$\large M = 2 \times \frac{q \omega}{2 \pi} (\pi (\frac{l}{2})^2)$
$\large M = \frac{q \omega l^2}{4}$
$\large L = 2 (m r^2 \omega) = 2 (m \frac{l^2}{4} \omega)$
$\large L = \frac{m l^2 \omega}{2}$
Hence , $\large \frac{M}{L} = \frac{q}{2 m}$ | 2021-09-22T14:15:53 | {
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http://math.stackexchange.com/questions/147272/hatm-p-cong-prod-i-hatm-q-i?answertab=oldest | # $\hat{M_P}$ $\cong$ $\prod_{i}\hat{M_{Q_i}}$
I think I came up with the following result. But I'm not 100% sure. Is this correct? If yes, how does one prove this?
Theorem? Let $A$ be a discrete valuation ring, $K$ its field of fractions. Let $L$ be a finite separable extension of $K$. Let $B$ be the integral closure of $A$ in $L$. Let P be the maximal ideal of A. Let $Q_i$, $i$ = $1, ..., r$ be the maximal ideals of $B$ lying over $P$. Let $M$ be a finitely generated torsion-free module over $B$. Let $\hat{M_P}$ be the completion of $M$ with respect to $P$-adic topology. Let $\hat{M_{Q_i}}$ be the completion of $M$ with respect to $(Q_i)$-adic topology. Then $\hat{M_P}$ $\cong$ $\prod_{i}\hat{M_{Q_i}}$
EDIT I need this to prove this theorem.
-
As $M$ is free and finite over $B$, the proof is same as in math.stackexchange.com/questions/137888. – user18119 May 20 '12 at 6:51
@Qil Right! I forgot B is a principal ideal domain. :-) Thanks. – Makoto Kato May 20 '12 at 7:22
Your "theorem" holds for any noetherian local ring $A$ and any finitely generated $B$-module $M$. The proof is contained in your answer in the above question. – user18119 May 20 '12 at 21:28
@Qil That's interesting. I guess I need some effort to prove it, though. – Makoto Kato May 22 '12 at 21:37
For any $n\ge 1$, $B/P^nB$ is Artinian with maximal ideals $Q_i/P^nB$, so the canonical map $$B/P^nB \to \prod_{1\le i\le r} B_{Q_i}/P^nB_{Q_i}$$ is an isomorphism. Tensoring by $M$ over $B$ we get a canonical isomorphism $$M/P^n M\simeq \prod_{1\le i\le r} M_{Q_i}/P^nM_{Q_i}.$$ As $Q_i^NB_{Q_i}\subseteq PB_{Q_i}\subseteq Q_iB_{Q_i}$ for some $N\ge 1$, we get $$\widehat{M}=\varprojlim_n M/P^nM\simeq \prod_{1\le i\le r} \widehat{M_{Q_i}}.$$ In fact the theorem you cited is not used here (it is useful if we take inverse limit of the first isomorphism before tensoring by $M$). | 2015-05-24T14:02:11 | {
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http://math.stackexchange.com/questions/732517/one-binomial-equation-sum-i-0z-n-1-choose-in-2-choose-z-i-n-1n | # One Binomial Equation $\sum_{i=0}^{z} {n_1 \choose i}{n_2 \choose z-i} = {n_1+n_2 \choose z}$ [duplicate]
I saw one proof using this formula: $$\sum_{i=0}^{z} {n_1 \choose i}{n_2 \choose z-i} = {n_1+n_2 \choose z}$$ Can anyone help explain it, thank you!
-
## marked as duplicate by Martin Sleziak, Sami Ben Romdhane, mookid, user127.0.0.1, user122283Mar 30 '14 at 17:33
See math.stackexchange.com/questions/337923/… and other posts, which are linked there. – Martin Sleziak Mar 30 '14 at 16:43
A subcommittee of $z$ members will be chosen from a committee consisting of $n_1$ Democrats and $n_2$ Republicans. The number of ways to choose $i$ Democrats and $z-i$ Republicans is $\dbinom{n_1}{i}\dbinom{n_2}{z-i}$. So the total number of ways to choose members of the subcommittee is the expression on the left. But it's also the expression on the right.
This is "Vandermonde's identity" (Google it).
-
$$\left(1+\binom{n_1}{1}x+\binom{n_1}{2}x^2+\cdots\right)=(1+x)^{n_1}$$
Obviously. And same for $n_2$.
Multiply these two equations, we get,
$$(1+x)^{n_1+n_2} = \left(1+\binom{n_1}{1}x+\binom{n_1}{2}x^2+\cdots\right)\left(1+\binom{n_2}{1}x+\binom{n_2}{2}x^2+\cdots\right)$$
Your given summation is the coefficient of $x^z$ in the RHS. That is equal to $\binom{n_1+n_2}{z}$ by the Binomial Theorem.
-
This is the Vandermonde identity. Choosing $z$ elements from a total of $n_1+n_2$, you must choose some number $i$ with $0\leq i\leq z$ from the first $n$, and the remaining $z-i$ from the last $n_2$. Whence the formula.
- | 2015-07-06T16:30:24 | {
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https://mathoverflow.net/questions/370909/an-elementary-inequality | # An “elementary” inequality
The following is left unproven in a monograph. The author refers to it as "elementary exercise" but I am unable to prove it. Any insight is appreciated. $$\int f \log f d\mu \le 2 \left[\int|f-1|^p d\mu\right]^{1/p}+\frac{2}{p-1}\int |f-1|^p d\mu,\quad p>1$$ where $$f$$ is a probability density.
• You really mean $f$ a probablity density? not $\mu$? Also please quote the monograph. – YCor Sep 5 at 7:03
• @YCor I rather expect that $\mu$ is probability measure. Because the inequality is not homogeneous with respect to $\mu$. – Fedor Petrov Sep 5 at 9:04
• This follows from $(1+x) \log (1+x)\le 2x +2/(p-1)x^p$ for $x \ge 0$ writing $0 \le f=1+g$ and integrating only where $g \ge 0$. – Giorgio Metafune Sep 5 at 10:38
• $\mu$ is indeed a probability measure in the context of the monograph. f is probability density with respect to $\mu$. But seems that it is true in general by Metafune's proof? The inequality is from (13.6) in "Dynamic to random matrix" by Erdos and Yau. – Daniel Li Sep 5 at 17:20
• I used that $\mu$ is a probability measure to have $\|g\|_1 \le \|g\|_p$ and $f \ge 0$. – Giorgio Metafune Sep 5 at 17:37 | 2020-10-24T17:28:31 | {
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https://www.jobilize.com/online/course/2-5-geometric-mean-descriptive-statistics-by-openstax?qcr=www.quizover.com | # 2.5 Geometric mean
Page 1 / 1
The mean (Arithmetic), median and mode are all measures of the “center” of the data, the “average”. They are all in their own way trying to measure the “common” point within the data, that which is “normal”. In the case of the arithmetic mean this is solved by finding the value from which all points are equal linear distances. We can imagine that all the data values are combined through addition and then distributed back to each data point in equal amounts. The sum of all the values is what is redistributed in equal amounts such that the total sum remains the same.
The geometric mean redistributes not the sum of the values but the product of multiplying all the individual values and then redistributing them in equal portions such that the total product remains the same. This can be seen from the formula for the geometric mean, $\stackrel{~}{x}$ : (Pronounced x-tilde)
$\stackrel{~}{x}={\left(\prod _{i=1}^{n}{\pi }_{i}\right)}^{\frac{1}{n}}=\sqrt[n]{{x}_{1}*{x}_{2}\bullet \bullet \bullet {x}_{n}}={\left({x}_{1}*{x}_{2}\bullet \bullet \bullet {x}_{n}\right)}^{\frac{1}{n}}$
where $\pi$ is the mathematical symbol that tells us to multiply all the ${x}_{i}$ numbers in the same way capital Greek sigma tells us to add all the ${x}_{i}$ numbers. Remember that a fractional exponent is calling for the nth root of the number thus an exponent of 1/3 is the cube root of the number.
The geometric mean answers the question, "if all the quantities had the same value, what would that value have to be in order to achieve the same product?” The geometric mean gets its name from the fact that when redistributed in this way the sides form a geometric shape for which all sides have the same length. To see this, take the example of the numbers 10, 51.2 and 8. The geometric mean is the product of multiplying these three numbers together (4,096) and taking the cube root because there are three numbers among which this product is to be distributed. Thus the geometric mean of these three numbers is 16. This describes a cube 16x16x16 and has a volume of 4,096 units.
The geometric mean is relevant in Economics and Finance for dealing with growth: growth of markets, in investment, population and other variables the growth of which there is an interest. Imagine that our box of 4,096 units (perhaps dollars) is the value of an investment after three years and that the investment returns in percents were the three numbers in our example. The geometric mean will provide us with the answer to the question, what is the average rate of return: 16 percent. The arithmetic mean of these three numbers is 23.6 percent. The reason for this difference, 16 versus 23.6, is that the arithmetic mean is additive and thus does not account for the interest on the interest, compound interest, embedded in the investment growth process. The same issue arises when asking for the average rate of growth of a population or sales or market penetration, etc., knowing the annual rates of growth. The formula for the geometric mean rate of return, or any other growth rate, is:
${r}_{s}={\left({x}_{1}*{x}_{2}\bullet \bullet \bullet {x}_{n}\right)}^{\frac{1}{n}}-1$
Manipulating the formula for the geometric mean can also provide a calculation of the average rate of growth between two periods knowing only the initial value ${a}_{0}$ and the ending value ${a}_{n}$ and the number of periods, $n$ . The following formula provides this information:
${\left(\frac{{a}_{n}}{{a}_{0}}\right)}^{\frac{1}{n}}=\stackrel{~}{x}$
Finally, we note that the formula for the geometric mean requires that all numbers be positive, greater than zero. The reason of course is that the root of a negative number is undefined for use outside of mathematical theory. There are ways to avoid this problem however. In the case of rates of return and other simple growth problems we can convert the negative values to meaningful positive equivalent values. Imagine that the annual returns for the past three years are +12%, -8%, and +2%. Using the decimal multiplier equivalents of 1.12, 0.92, and 1.02, allows us to compute a geometric mean of 1.0167. Subtracting 1 from this value gives the geometric mean of +1.67% as a net rate of population growth (or financial return). From this example we can see that the geometric mean provides us with this formula for calculating the geometric (mean) rate of return for a series of annual rates of return:
${r}_{s}=\stackrel{~}{x}-1$
where ${r}_{s}$ is average rate of return and $\stackrel{~}{x}$ is the geometric mean of the returns during some number of time periods. Note that the length of each time period must be the same.
As a general rule one should convert the percent values to its decimal equivalent multiplier. It is important to recognize that when dealing with percents, the geometric mean of percent values does not equal the geometric mean of the decimal multiplier equivalents and it is the decimal multiplier equivalent geometric mean that is relevant.
## Formula review
The Geometric Mean: $\stackrel{~}{x}={\left(\prod _{i=1}^{n}{\pi }_{i}\right)}^{\frac{1}{n}}=\sqrt[n]{{x}_{1}*{x}_{2}\bullet \bullet \bullet {x}_{n}}={\left({x}_{1}*{x}_{2}\bullet \bullet \bullet {x}_{n}\right)}^{\frac{1}{n}}$
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
how did you get the value of 2000N.What calculations are needed to arrive at it
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https://www.vedantu.com/question-answer/the-sum-of-n-terms-of-two-arithmetic-class-11-maths-cbse-5f5fa86368d6b37d16353ce1 | Question
# The sum of ‘n’ terms of two arithmetic progressions is in ratio $5n+4:9n+6$. find the ratio of their ${18}^{\text{th}}$ terms:
Hint: In this problem, we will be using the concept of the sum of an arithmetic progression (A.P). In order to solve this question, we divide the sum of n terms of the two arithmetic progression(A.P) with each other and equate it with $5n+4:9n+6$.
Now we will try to get the equation in the form of the ${\text{n}}^{\text{th}}$ term of an A.P and will find the values for ‘n’. Upon putting that value of ‘n’ in $5n+4:9n+6$. We can calculate the ratio of the ${18}^{\text{th}}$ term of two given A.P.
Complete step by step solution: As mentioned in the question, there are two arithmetic progressions with different first terms and different common differences.
For the first A⋅P:-
Let the first term of A⋅P be = a, and the common difference be = d;
So, Sum of ‘n’ terms of an A⋅P is;
${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$
and the nth term of an A⋅P is;
${{a}_{n}}=a+\left( n-1 \right)d$
For the second A⋅P:-
Let the first term of A⋅P. be = A and the common difference be = D;
So, the sum of its ‘n’ terms will be;
${{S}_{n}}=\dfrac{n}{2}\left[ 2A+\left( n-1 \right)D \right]$
And the nth term of an A⋅P is;
${{A}_{n}}=A+\left( n-1 \right)D$
It’s given in the question that the ratio of the sum of ‘n’ terms of the two AP is $5n+4:\ 9n+6;$
$\Rightarrow \dfrac{\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]}{\dfrac{n}{2}\left[ 2\text{A}+\left( n-1 \right)\text{D} \right]}=\dfrac{5n+4}{9n+6}$
$\Rightarrow \dfrac{\left[ 2a+\left( n-1 \right)d \right]}{\left[ 2A+\left( n-1 \right)d \right]}=\dfrac{5n+4}{9n+6}$
Taking L.H.S;
When we take ‘2’ common in numerator and denominator;
$=\dfrac{2\left[ a+\left( \dfrac{n-1}{2} \right)d \right]}{2\left[ A+\left( \dfrac{n-1}{2} \right)D \right]}$
$=\dfrac{\left[ a+\left( \dfrac{n-1}{2} \right)d \right]}{\left[ A+\left( \dfrac{n-1}{2} \right)D \right]}$
So, now;
$\dfrac{\left[ a+\left( \dfrac{n-1}{2} \right)d \right]}{\left[ A+\left( \dfrac{n-1}{2} \right)D \right]}=\dfrac{5n+4}{9n+6}$ (1)
We need to find the ratio of the 18th term of Arithmetic progression:
$=\dfrac{18\text{th}\ \text{term}\ \text{of}\ \text{1st}\ \text{A}\cdot \text{P}}{18\text{th}\ \text{term}\ \text{of}\ \text{2nd}\ \text{A}\cdot \text{P}}$
$\dfrac{{{a}_{18}}\ \text{of}\ 1\text{st}\ \text{A}\cdot \text{P}}{{{A}_{18}}\ \text{of}\ 2\text{nd}\ \text{A}\cdot \text{P}}$
$=\dfrac{a+\left( 18-1 \right)d}{A+\left( 18-1 \right)D}$
$=\dfrac{a+17d}{A+17D}$ (2)
Comparing equation (2) with equation (1); $a+17d=a+\left( \dfrac{n-1}{2} \right)d$
$\Rightarrow 17=\dfrac{n-1}{2}$
$\Rightarrow n-1=17\times 2$
$\Rightarrow n-1=34$
$\Rightarrow n=34+1$
$\Rightarrow n=35$
Now, putting $n=35$ in equation (1);
$\Rightarrow \dfrac{\left[ a+\left( \dfrac{n-1}{2} \right)d \right]}{\left[ A+\left( \dfrac{n-1}{2} \right)D \right]}=\dfrac{5n+4}{9n+6}$
$\Rightarrow \dfrac{a+\left( \dfrac{35-1}{2} \right)d}{A+\left( \dfrac{35-1}{2} \right)D}=\dfrac{5\left( 35 \right)+4}{9\left( 35 \right)+6}$
$\Rightarrow \dfrac{a+\left( \dfrac{34}{2} \right)d}{A+\left( \dfrac{34}{2} \right)D}=\dfrac{175+4}{315+6}$
$\Rightarrow \dfrac{a+17d}{A+17D}=\dfrac{179}{321}$
Therefore, $\dfrac{18\text{th}\ \text{term}\ \text{of}\ 1\text{st}\ \text{A}\cdot \text{P}}{18\text{th}\ \text{term}\ \text{of}\ 2\text{nd}\ \text{A}\cdot \text{P}}=\dfrac{179}{321}$
Hence, the ratio of ${18}^{\text{th}}$ term of ${1}^{\text{st}}$ A⋅P and ${18}^{\text{th}}$ term of ${2}^{\text{nd}}$ A⋅P is 179: 321.
Note: The sum of the ‘n’ terms of any A⋅P is ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ and ${n}^{\text{th}}$ term of an A⋅P is ${{a}_{n}}=a+\left( n-1 \right)d$ where ‘a’ is the first term of an A.P and ‘d’ is common difference of an A.P. | 2020-09-29T04:02:38 | {
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http://samjshah.com/tag/calculus/ | # An expanded understanding of basic derivatives – graphically
The guilt that I feel for not blogging more regularly this year has been considerable, and yet, it has not driven me to post more. I’ve been overwhelmed and busy, and my philosophy about blogging it is: do it when you feel motivated. And so, I haven’t.
Today, I feel a slight glimmer of motivation. And so here I am.
Here’s what I want to talk about.
In calculus, we all have our own ways of introducing the power rule for derivatives. Graphically. Algebraically. Whatever. But then, armed with this knowledge…
that if $f(x)=x^n$, then $f'(x)=nx^{n-1}$
…we tend to drive forward quickly. We immediately jump to problems like:
take the derivative of $g(x)=4x^3-3x^{-5}+2x^7$
and we hurdle on, racing to the product and quotient rules… We get so algebraic, and we go very quickly, that we lose sight of something beautiful and elegant. This year I decided to take an extra few days after the power rule but before problems like the one listed above to illustrate the graphical side of things.
Here’s what I did. We first got to the point where we comfortably proved the power rule for derivatives (for n being a counting number). Actually, before I move on and talk about the crux of this post, I should show you what we did…
Okay. Now I started the next class with kids getting Geogebra out and plotting on two graphics windows the following:
and they saw the following:
At this point, we saw the transformations. On the left hand graph, we saw that the function merely shifted up one unit. On the right hand graphs, we saw a vertical stretch for one function, and a vertical shrink for the other.
Here’s what I’m about to try to illustrate for the kids.
Whatever transformation a function undergoes, the tangent lines to the function also undergoes the exact same transformation.
What this means is that if a function is shifted up one unit, then all tangent lines are shifted up one unit (like in the left hand graph). And if a function undergoes vertical stretching or shrinking, all tangent lines undergo the same vertical stretching or shrinking.
I want them to see this idea come alive both graphically and algebraically.
So I have them plot all the points on the functions where $x=1$. And all the tangent lines.
For the graph with the vertical shift, they see:
The original tangent line (to $f(x)=x^2$) was $y=2x-1$. When the function moved up one unit, we see the tangent line simply moved up one unit too.
Our conclusion?
Yup. The tangent line changed. But the slope did not. (Thus, the derivative is not affected by simply shifting a function up or down. Because even though the tangent lines are different, the slopes are the same.)
Then we went to the second graphics view — the vertical stretching and shrinking. We drew the points at $x=1$ and their tangent lines…
…and we see that the tangent lines are similar, but not the same. How are they similar? Well the original function’s tangent line is the red one, and has the equation $y=2x-1$. Now the green function has undergone a vertical shrink of 1/4. And lo and behold, the tangent line has also!
To show that clearly, we did the following. The original tangent line has equation $y=2x-1$. So to apply a vertical shrink of 1/4 to this, you are going to see $y=\frac{1}{4}(2x-1)$ (because you are multiplying all y-coordinates by 1/4. And that simplifies to $y=0.5x-0.25$. Yup, that’s what Geogebra said the equation of the tangent line was!
Similarly, for the blue function with a vertical stretch of 3, we get $y=3(2x-1)=6x-3$. And yup, that’s what Geogebra said the equation of the tangent line was.
What do we conclude?
And in this case, with the vertical stretching and shrinking of the functions, we get a vertical stretching and shrinking of the tangent lines. And unlike moving the function up or down, this transformation does affect the slope!
I repeat the big conclusion:
Whatever transformation a function undergoes, the tangent lines to the function also undergoes the exact same transformation.
I didn’t actually tell this to my kids. I had them sort of see and articulate this.
Now they see that if a function gets shifted up or down, they can see that the derivative stays the same. And if there is a vertical stretch/shrink, the derivative is also vertically stretched/shrunk.
The next day, I started with the following “do now.” We haven’t learned the derivative of $\sin(x)$, so I show them what Wolfram Alpha gives them.
For (a), I expect them to give the answer $g'(x)=3\cos(x)$ and for (b), $h'(x)=-\cos(x)$.
The good thing here is now I get to go for depth. WHY?
And I hear conversations like: “Well, g(x) is a transformation of the sine function which gives a vertical stretch of 3, and then shifts the function up 4. Well since the function undergoes those transformations, so does the tangent lines. So each tangent line is going to be vertically stretched by 3 and moved up 4 units. Since the derivative is only the slope of the tangent line, we have to see what transformations affect the slope. Only the vertical stretch affects the slope. So if the original slope of the sine function was $\cos(x)$, then we know that the slope of the transformed function is $3\cos(x)$.
That’s beautiful depth. Beautiful.
For (b), I heard talk about how the negative sign is a reflection over the x-axis, so the tangent lines are reflected over the x-axis also. Thus, the slopes are the opposite sign… If the original sine functions slope of the tangent lines was $\cos(x)$, then the new slopes are going to be $-\cos(x)$.
This isn’t easy for my kids, so when I saw them struggling with the conceptual part of things, I whipped up this sheet (.docx).
And here are the solutions
And here is a Geogebra sheet which shows the transformations, and the new tangent line (and equation), for this worksheet.
Now to be fair, I don’t think I did a killer job with this. It was my first time doing it. I think some kids didn’t come out the stronger for this. But I do feel that the kids who do get it have a much more intuitive understanding of what’s going on.
I am much happier to know that if I ask kids what the derivative of $q(x)=6x^9$ is, they immediately think (or at least can understand) that we get $q'(x)=6*9x^8$, because…
our base function is $x^9$ which has derivative (aka slope of the tangent line) $9x^8$… Thus the transformed function $6x^9$ is going to be a vertical stretch, so all the tangent lines are going to be stretched vertically by a factor of 9 too… thus the derivative of this (aka the slope of the tangent line) is $q'(x)=6*9x^8$.
To me, that sort of explanation for something super simple brings so much graphical depth to things. And that makes me feel happy.
# Do They Get It? The Instantaneous Rate of Change Exactly
Today in calculus I wanted to check if students really understood what they were doing when they were finding the instantaneous rate of change. (We haven’t learned the word derivative yet, but this is the formal definition of the derivative.)
So I handed out this worked out problem.
And I had them next to each of the letters write a note answering the following individually (not as a group):
A: write what the expression represents graphically and conceptually
B: write what the notation $\lim_{h\rightarrow0}$ actually means. Why does it need to be there to calculate the instantaneous rate of change. (Be sure to address with h means.)
C: write what mathematical simplification is happening, and why were are allowed to do that
D: write what the reasoning is behind why were are allowed to make this mathematical move
E: explain what this number (-1) means, both conceptually and graphically
It was a great activity. I had them do it individually, but I should have had students (after completing it) discuss in groups before we went to the whole group context. Next time…
Anyway, the answers I was looking for (written more drawn out):
A: the expressions represents the average rate of change between two points, one fixed, and the other one defined in relation to that first point. The average rate of change is the constant rate the function would have to go at to start at one point and end up at the second. Graphically, it is the slope of the secant line going through those two points.
B: the $\lim_{h\rightarrow0}$ is simply a fancy way to say we want to bring h closer and closer and closer to zero (infinitely close) but not equal zero. That’s all. The expression that comes after it is the average rate of change between two points. As h gets closer and closer to 0, the two points get closer and closer to each other. We learned that if we take the average rate of change of two points super close to each other, that will be a good approximation for the instantaneous rate of change. If the two points are infinitely close to each other, then we are going to get an exact instantaneous rate of change!
C: we see that $\frac{h}{h}$ is actually 1. We normally would not be allowed to say that, because there is the possibility that h is 0, and then the expression wouldn’t simplify to 1. However we know from the limit that h is really close to 0, but not equal to 0. Thus we can say with mathematical certainty that $\frac{h}{h}=1$
D: as we bring h closer and closer to 0, we see that $h-1$ gets closer and closer to -1. Thus if we bring h infinitely close to 0, we see that $h-1$ gets infinitely close to -1.
E: the -1 represents the instantaneous rate of change of $x^2-5x+1$ at $x=2$. This is how fast the function is changing at that instant/point. It is graphically understood as the slope of the tangent line drawn at $x=2$.
I loved doing this because if a student were able to properly answer each of the questions, they really truly understand what is going on.
# Starting Calculus with Area Functions
So I decided to try a new beginning to (non-AP) calculus this year. Instead of doing an algebra bootcamp and diving into limits, I decided to teach kids a new kind of function transformation. I’d say this is something that makes my classroom uniquely mine (this is my contribution to Mission 1 of Explore the MTBoS). I don’t think anyone else I know does something like this.
You see, I was talking with a fellow calculus teacher, and we had a big realization. Yes, calculus is hard for kids because of all the algebra. But also, calculus involves something that students have never seen before.
It involves transformations that morph one graph into another graph. And not just standard up, down, left, right, stretch, shrink, reflect transformations. Although they do transform functions, they don’t make them look too different from the original. Given a function and a basic up, left, reflect, shrink transformation of it, you’d be able to pair them up and say they were related… But in calculus, students start grappling with seriously weird and abstract transformations. For example: if you hold an f(x) graph and an f’(x) graph next to each other — they don’t look alike at all. You would never pair them up and say “oh, these are related.”
So I wanted to start out with a unit on abstract and weird function transformations. Turns out, even though the other teacher and I had brainstormed 5 different abstract function transformations, I got so much mileage out of one of them that I didn’t have to do anything else. You see: I introduced my kids to integrals, without ever saying the word integrals. Well, to be fair, I introduced them to something called the area transformation and the only difference between this and integrals is that we can’t have negative area. [1]
You can look at this geogebra page to see what I mean by area functions.
Here’s the packet I created (.docx)
That packer was just the bare backbones of what we did. There was a lot of groupwork in class, a lot of conceptual questions posed to them, and more supplemental documents that were created as I started to realize this was going to morph into a much larger unit because I was getting so much out of it. (I personally was finding so much richness in it! A perfect blend of the concrete and the abstract!)
Here are other supplemental documents:
2013-09-16 Abstract Functions 1.5
2013-09-17 Abstract Functions 1.75
2013-09-20 Area Function Concept Questions
2013-09-23 Abstract Functions 1.9375
• It’s conceptual, so those kids who aren’t strong with the algebraic stuff gain confidence at the start of the year
• Kids start to understand the idea of integration as accumulation (though they don’t know that’s what they are doing!)
• Kids understand that something can be increasing at a decreasing rate, increasing at a constant rate, or increasing at an increasing rate. They discovered those terms, and realized what that looks like graphically.
• Kids already know why the integral of a constant function is a linear function, and why the integral of a linear function is a quadratic function.
• Kids are talking about steepness and flatness of a function, and giving the steepness and flatness meaning… They are making statements like “because the original graph is close to the x-axis near x=2, not much area is being added as we inch forward on the original graph, so the area function will remain pretty flat, slightly increasing… but over near x=4, since the original function is far from the x-axis, a lot of area is being added as we inch forward on the original graph, so the area function shoots up, thus it is pretty steep”
• Once we finish investigating the concept of “instantaneous rate of change” (which is soon), kids will have encountered and explored the conceptual side of both major ideas of calculus: derivatives and integrals. All without me having used the terms. I’m being a sneaky teacher… having kids do secret learning.
I mean… I worked these kids hard. Here is a copy of my assessment so you can see what was expected of them.
I love it.
Love. It.
LOVE.
IT.
I’m going to put a picture gallery below of some things from my smartboards.
This slideshow requires JavaScript.
[1] To be super technical, I am having kids relate $f(x)$ and $\int_{0}^{x} |f(t)|dt$
# Quotes from Calculus
Seniors are done with classes. (The rest of the Upper School is preparing for final exams this week, and finals are administered next week.) Yesterday one of my calculus students gave me this 12-page booklet she prepared. All year, she had been writing down quotations from class — from students and from me. This was her final product.
I don’t think it would be right to include the student quotations, but below here are some that are attributed to me. I remember some of them, and some of them I am clueless! Most of them won’t make any sense to you, gentle reader. Oh well!
“Derivatize!” — Mr. Shah
“Laughing is the only thing we can do, otherwise we would cry” — Mr. Shah
“Does this make the diddy [ditty] make more sense?” “P. Diddy” — Mr. Shah and Stu
“There’s still 2 minutes left, keep working” — Mr. Shah
“Fish, fish, fish, fish, fish, fish. 6 fish!” — Mr. Shah
“You can harangue him” — Mr. Shah
“It’s just depressing as a teacher when students admire clocks” — Mr. Shah
“I have 3 declarations, is that okay?” “No” — Student and Mr. Shah
“Sorry that you’re so sensitive” — Mr. Shah
“Anyone taking Latin here? Too bad. Ha! It’s in Greek.” — Mr. Shah
“Crust” — Mr. Shah
“”You need parentheses or else you’re gonna die” — Mr. Shah
“Oh no he didn’t!” — Mr. Shah
“Are you having special difficulties?” — Mr. Shah to Student
“Uh-uh boo boo” — Mr. Shah
“Jesus!” “Jesus!” “Hey, let’s keep religion out of this” — Student, Student, and Mr. Shah
“You’re a whack sharpener” — Mr. Shah to Student
“What if I put formaldehyde in this? And then spit in it?” — Mr. Shah to Student
“What’s the point in the spit? After the formaldehyde she’d already be dead” — Student
“That was my fault for listening to anyone but my brain” — Mr. Shah
“I have hearing” — Mr. Shah
“How was your weekend Mr. Shah?” *silence* “Oh, okay” — Student
“I got 99 problems and they’re all problematic” — Mr. Shah
“Who wants to volunteer to factor out these 100 terms?” *silence* “No one?” — Mr. Shah
“What… what’s the derivative of tan(x)?” “This isn’t happening” — Student and Mr. Shah
“Make your life easiah!” — Mr. Shah
“Yes sir” “I prefer your majesty” — Student and Mr. Shah
“Hey! Hey! Hey! This doesn’t sound mathy” — Mr. Shah
“Draw the boxes” “Why?” ” Because I order it” — Mr. Shah and Student
“When I see these things, I get like heart palpitations” — Mr. Shah
“Let’s come up with our own definition of genius” — Mr. Shah
“The baby mama rule, ugh! You guys have me calling it this instead of the inception rule” — Mr. Shah
“I pick one kid in every class to blame for everyone getting sick. I blame Student” — Mr. Shah
“Student die!” “Did you just tell Student to die?” “No I said duck!” — Mr. Shah and Student
“A long, long time ago… in a classroom right here” — Mr. Shah
“So what’s the derivative?” “With the letter? I can’t do it with letters” “Yo, pass it over here” — Mr. Shah, Student, and Student
“Doing it all at once is a little cray cray” — Mr. Shah
“We’re so close to being done” “We’re not done yet?” — Mr. Shah and Student
“Where is my pencil honey boo boo child” — Mr. Shah
“That’s bad news bears” — Mr. Shah
“Hush! No questions. We’re imagining” — Mr. Shah
“My favorite flowers are ranunculus” — Mr. Shah
“Do we have this sheet?” “Yes… but I don’t want you to take it out” “So how are we gonna do it?” — Student and Mr. Shah
“Do you have your phone in your hand?” “Never have I ever” — Mr. Shah and Student
“A baby, in a baby, in a momma” — Mr. Shah
“Student, I’m asking you this because you’re snarky” — Mr. Shah
“Derp!” — Mr. Shah
“What if I just say give me the Riemann Sum?” “You won’t” — Mr. Shah and Student
“Did I do well?” “No coach, you didn’t” — Student and Mr. Shah
“I put a little doo-hickey on the right side” — Mr. Shah
“I have a QQ Mr. Shah” — Mr. Shah
“Hush yourself child” — Mr. Shah
“Repetitious and tedious” — Mr. Shah
“Hey, fight me!” “Don’t tempt us” — Student and Mr. Shah
“Can’t you read it? More a exact!” — Mr. Shah
“He’s doing his thing” “What’s his thing?” “He’s running” “Attempting to run” — Student, Mr. Shah, and Student
“We should look at this and say…” “That ain’t right” — Mr. Shah and Student
“Holy Mother… Superior” — Mr. Shah
“They’re full of hogwash” — Mr. Shah
# Some Random Things I Have Liked
## The Concept of Signed Areas
In calculus, after first introducing the concept of signed areas, I came up with the “backwards problem” which really tested what kids understood. (This was before we did any integration using calculus… I always teach integration of definite integrals first with things they draw and calculate using geometry, and then things they do using the antiderivatives.)
I made this last year, so apologies if I posted it last year too.
[.d0cx]
Some nice discussions/ideas came up. Two in particular:
(1) One student said that for the first problem, any line that goes through (-1.5,-1) would have worked. I kicking myself for not following that claim up with a good investigation.
(2) For all problems, only a couple kids did the easy way out… most didn’t even think of it… Take the total signed area and divide it over the region being integrated… That gives you the height of a horizontal line that would work. (For example, for the third problem, the line $y=\frac{2\pi+4}{7}$ would have worked.) If I taught the average value of a function in my class, I wouldn’t need to do much work. Because they would have already discovered how to find the average value of a function. And what’s nice is that it was the “shortcut”/”lazy” way to answer these questions. So being lazy but clever has tons of perks!
## Motivating that an antiderivative actually gives you a signed area
I have shown this to my class for the past couple years. It makes sense to some of them, but I lose some of them along the way. I am thinking if I have them copy the “proof” down, and then explain in their own words (a) what the area function does and (b) what is going on in each step of the “proof,” it might work better. But at least I have an elegant way to explain why the antiderivative has anything to do with the area under a curve.
Note: After showing them the area function, I shade in the region between $x=3$ and $x=4.5$ and ask them what the area of that bit is. If they understand the area function, they answer $F(4.5)-F(3)$. If they don’t, they answer “uhhhhhh (drool).” What’s good about this is that I say, in a handwaving way, that is why when we evaluate a definite integral, we evaluate the antiderivative at the top limit of integration, and then subtract off the antiderivative at the bottom limit of integration. Because you’re taking the bigger piece and subtracting off the smaller piece. It’s handwaving, but good enough.
## Polynomial Functions
In Precalculus, I’m trying to (but being less consistent) have kids investigate key questions on a topic before we formal delve into it. To let them discover some of the basic ideas on their own, being sort of guided there. This is a packet that I used before we started talking formally about polynomials. It, honestly, isn’t amazing. But it does do a few nice things.
[.docx]
Here are the benefits:
• The first question gets kids to remember/discover end behavior changes fundamentally based on even or odd powers. It also shows them that there is a difference between $x^2$ and $x^4$… the higher the degree, the more the polynomial likes to hang around the x-axis…
• The second question just has them list everything, whether it is significant seeming or not. What’s nice is that by the time we’re done with the unit, they will have a really deep understanding of this polynomial. But having them list what they know to start out with is fun, because we can go back and say “aww, shucks, at the beggining you were such neophytes!”
• It teaches kids the idea of a sign analysis without explaining it to them. They sort of figure it out on their own. (Though we do come together as a class to talk through that idea, because that technique is so fundamental to so much.)
• They discover the mean value theorem on their own. (Note: You can’t talk through the mean value theorem problem without talking about continuity and the fact that polynomials are continuous everywhere.)
## The Backwards Polynomial Puzzle
As you probably know, I really like backwards questions. I did this one after we did So I was proud that without too much help, many of my kids were really digging into finding the equations, knowing what they know about polynomials. A few years ago, I would have done this by teaching a procedure, albeit one motivated by kids. Now I’m letting them do all the heavy lifting, and I’m just nudging here and there. I know this is nothing special, but this course is new to me, so I’m just a baby at figuring out how to teach this stuff.
[.docx]
# Related Rates, Yet Another Redux
I posted in 2008 how I didn’t actually find related rates all that interesting/important in calculus. The problems that I could find were contrived, and I didn’t quite get the “bigger picture.” In 2011, I posted again about something I found from a conference that used Logger Pro, was pretty interesting, and helped me get at something less formulaic.
I still don’t know how I feel about related rates. I’m torn. Part of me wants to totally eliminate them from the curriculum (which means I can also possibly eliminate implicit differentiation, because right now I see one of the main purposes of implicit differentiation is to prime students for related rates). Part of me feels there is something conceptually deeper that I can get at with related rates, and I’m missing it.
I still don’t have a good approach, but this year, I am starting with the premise that students need to leave with one essential truth:
Often times, as we change one thing, it affects a number of other things. However, the way that the other things are affected can vary greatly.
Right now, to me, that’s the heart of related rates. (To be honest, it took some conversation with my co-teacher before we were able to stumble upon this essential understanding.)
In order to get at this, we are starting our related rates unit with these two worksheets. A nice bonus is that it gets students to think about the shape of a graph, which is what we’ll be embarking on next.
The TD;DR for the idea behind the worksheets: Students study a circle which has it’s radius increase by 1 cm each second, and see how that changes the area and circumference. Then students study a circle which has it’s area increase by 10 cm^2 each second, and see how that changes the radius and circumference. The big idea is that even though one thing is changing, that one thing affects a number of different things, and it changes them in different ways.
[.docx] [.docx]
(A special thanks to Bowman for making the rocket and camera problem dynamic on Geogebra.)
It’s not like this is a deep investigation or they come out knowing anything super special. But the main takeaway that I want them to get from it becomes pretty apparent. And what’s really powerful (for me, as a teacher trying to illustrate this essential understanding) is seeing the graphs of how the various thing change.
***
I had students finish the first packet one night. Before we started going over it, or talking about it, I started today’s class asking for a volunteer to blow up balloons. (We got a second volunteer to tie the balloons.) While he practiced breathing even breaths, I tied and taped an empty balloon to the whiteboard.
Then I asked our esteemed volunteer to use one breath to blow up the first balloon. Taped it up. Again, for two breaths. Taped. Et cetera until we got a total of six balloons taped.
Then I asked what things are measurable in the balloons.
Bam. List.
(We should have listed more. Color. What it’s made of. Thickness of rubber.]
Then I asked what we did to the balloon.
Added volume. A constant volume (ish) in each balloon.
Which of the other things changed as a result?
How did they change?
This five minute start to class reinforced the main idea (hopefully). We changed one thing. It changed a bunch of other things. But just because one thing changed in one particular way doesn’t mean that everything changed in that same way. For example, just because the volume increased at a constant rate doesn’t mean the radius changed at a constant rate.
***
This is about all I got for now. I’m going to teach the rest of the topic the way I always do. It’s not up to my personal standards, but I still am struggling to get it there. I suppose to do that, I’ll have to see a more nuanced bigger picture with related rates, or find something that approaches what’s happening more visually, dynamically, or conceptually.
PS. The more I mull it over, the more I think that geogebra has to be central to my approach next year… teaching students to make sliders to change one parameter, and having them develop something that dynamically illustrates how a number of other things change. And then analyzing how those things change graphically and algebraically.
(A simple example: Have a rectangle where the diagonal changes length… what gets affected? The sides, the angle between the diagonal and the sides of the rectangle, the area, the perimeter, etc. How do each of these things get affected as the diagonal changes?)
# What does it mean to be going 58 mph at 2:03pm?
That’s the question I asked myself when I was trying to prepare a particular lesson in calculus. What does it mean to be going 58 mph at 2:03pm? More specifically, what does that 58 mean?
You see, here’s the issue I was having… You could talk about saying “well, if you went at that speed for an hour, you’d go 58 miles.” But that’s an if. It answers the question, but it feels like a lame answer, because I only have that information for a moment. That “if” really bothered me. Fundamentally, here’s the question: how can you even talk about a rate of change at a moment, when rate of change implies something is changing. But you have a moment. A snapshot. A photograph. Not enough to talk about rates of change.
And that, I realized, is precisely what I needed to make my lesson about. Because calculus is all about describing a rate of change at a moment. This gets to the heart of calculus.
I realized I needed to problematize something that students find familiar and understandable and obvious. I wanted to problematize that sentence “What does it mean to be going 58 mph at 2:03pm?”
And so that’s what I did. I posed the question in class, and we talked. To be clear, this is before we talked about average or instantaneous rates of change. This turned out to be just the question to prime them into thinking about these concepts.
Then after this discussion, where we didn’t really get a good answer, I gave them this sheet and had them work in their groups on it:
I have to say that this sheet generated some awesome discussions. The first question had some kids calculate the average rate of change for the trip while others were saying “you can’t know how fast the car is moving at noon! you just can’t!” I loved it, because most groups identified their own issue: they were assuming that the car was traveling at a constant speed which was not a given. (They also without much guidance from me discovered the mean value theorem which I threw in randomly for part (b) and (c)… which rocked my socks off!)
As they went along and did the back side of the sheet, they started recognizing that the average rate of change (something that wasn’t named, but that they were calculating) felt like it would be a more accurate prediction of what’s truly going on in the car when you have a shorter time period.
In case this isn’t clear to you because you aren’t working on the sheet: think about if you knew the start time and stop time for a 360 mile trip that started at 2pm and ended a 8pm. Would you have confidence that at 4pm you were traveling around 60 mph? I’d say probably not. You could be stopping for gas or an early dinner, you might not be on a highway, whatever. But you don’t really have a good sense of what’s going on at any given moment between 2pm and 8pm. But if I said that if you had a 1 mile trip that started at 2pm and ended at 2:01pm, you might start to have more confidence that at around 2pm you were going about 60 mph. You wouldn’t be certain, but your gut would tell you that you might feel more confident in that estimate than in the first scenario. And finally if I said that you had a 0.2 mile trip that started at 2pm and ended at 2:01:02pm, you would feel more confident that you were going around 72mph at 2pm.
And here’s the key… Why does your confidence in the prediction you made (using the average rate of change) increase as your time interval decreases? What is the logic behind that intuition?
And almost all groups were hitting on the key point… that as your time interval goes down, the car has less time to fluctuate its speed dramatically. In six hours, a car can change up it’s speed a lot. But in a second, it is less likely to change up it’s speed a lot. Is it certain that it won’t? Absolutely not. You never have total certainty. But you are more confident in your predictions.
Conclusion: You gain more certainty about how fast the car is moving at a particular moment in time as you reduce the time interval you use to estimate it.
The more general mathematical conclusion: If you are estimating a rate of change of a function (for the general nice functions we deal with in calculus), if you decrease a time interval enough, the function will look less like a squiggly mess changing around a lot, and more and more like a line. Or another way to think about it: if you zoom into a function at a particular point enough, it will stop looking like a squiggly mess and more and more like a line. Thus your estimation is more accurate, because you are estimating how fast something is going when it’s graph is almost exactly a line (indicating a constant rate of change) rather than a squiggly mess.
I liked the first day of this. The discussions were great, kids seemed to get into it. After that, I explicitly introduced the idea of average rate of change, and had them do some more formulaic work (this sheet, book problems). And then finally, I tried exploiting the reverse of the initial sheet. I gave students an instantaneous rate of change, and then had them make predictions in the future.
It went well, but you could tell that the kids were tired of thinking about this. The discussions lagged, even though the kids actually did see the relationships I wanted them to see.
My Concluding Thoughts: I came up with this idea of the first sheet the night before I was going to teach it. It wasn’t super well thought out — I was throwing it out there. It was a success. It got kids to think about some major ideas but I didn’t have to teach them these ideas. Heck, it totally reoriented the way I think about average and instantaneous rate of change. I usually have thought of it visually, like
But now I have a way better sense of the conceptual undergirding to this visual, and more depth/nuance. Anyway, my kids were able to start grappling with these big ideas on their own. However, I dragged out things too long. We spent too long talking about why we have to use a lot of average rates of changes of smaller and smaller time intervals to approximate the instantaneous rate of changes, instead of just one average rate of change over a super duper small time interval. The reverse sheet (given the instantaneous rate of change) felt tedious for kids, and the discussion felt very similar. It would have been way better to use it (after some tweaking) to introduce linear approximations a little bit later, after a break. There were too much concept work all at once, for too long a period of time.
The good news is that after some more work, we finally took the time to tie these ideas all together, which kids said they found super helpful. | 2013-12-10T12:23:00 | {
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https://codeforces.com/blog/entry/17126 | Endagorion's blog
By Endagorion, 5 years ago, ,
Somehow I don't feel sleepy at all at 6am, which is kinda weird. I wonder if writing an editorial will help.
SquareScores (250)
The standard trick is to use the linearity of expectation: for every substring count the probability that it consists of the same characters, and sums all these probabilities. There are O(n 2) substrings, which is a reasonable amount in the given limitations. However, iterating over all symbols in the substring yields O(n 3) solution, so we should optimize something.
My way was as follows: fix the beginning of the substring, and process all substrings with this beginning altogether. Count q c — the probability that current substring consists only of characters c. If we append one symbol from the right, q c changes in a simple way: if the symbol is a letter x, then all q c with c ≠ x become zeros, while q x does not change; else, if the symbol is question mark, every q c gets multiplies by p c. So, we process all substrings with fixed beginning in O(mn) (where m is the number of different symbols), for a O(mn 2) solution.
Do you have a more effective/simple solution for this problem?
SuccessiveSubstraction (450)
When we open all the brackets, every number adds up with a plus or a minus sign by it. Which configurations are possible? It is not difficult to prove that all possible configurations are as follows: the first number is always a plus, the second is always a minus, while all other numbers with a plus sign form no more than two consecutive segments. So, we just have to find two subsegments of the array with the greatest total sum. There are lots of ways to do it in O(n), the most straightforward (IMO) being DP (how many elements are processed; the number of consecutive segments already closed; is a segment currently open). Simply recalculate the DP after every change in the array, for a solution of O(n 2) complexity.
TwoEntrances (850)
What if there is only one entrance in the tree? The problem becomes a classic one: count the number of topological orderings of rooted tree with edges directed from the root. There is a simple subtree DP solution; also, a closed formula is available: , where n is the total number of vertices, w(v) is the number of vertices in the subtree of v.
Now, two entrances s 1 and s 2 are present. Consider vertices that lie on the path connecting s 1 and s 2 in a tree; every vertex of the path can be associated with a subtree consisting of vertices not on the path. The last vertex to cover will be either s 1 or s 2 (WLOG, assume it's s 1), the second-to-last will be either s 2 or the neighbour of s 1 in the path, etc. The observation is that at any moment there is a consecutive subsegment of the s 1- s 2 path which is covered.
Count dp l, r — the number of ways to cover the subsegment from l-th to r-th vertex in the path with all the hanging subtrees. The number of ways such that the l-th vertex is the last to cover is , where ord l is the number of topological orderings of the subtree of l-th vertex (which can be calculated as described above), t l is the size of this subtree, and T l, r is the total number of vertices in the subtrees of l-th, ..., r-th vertices. The intuition behind this formula is that we independently deal with the segment [l + 1;r] and the l-th subtree, and the last vertex is always the l-th vertex of the path. The number of ways that cover r-th vertex in the last place is obtained symmetrically; dp l, r is simply the sum of those two quantities. The base of DP is clearly dp l, l = ord l.
Clearly, this subsegment DP is calculated in O(n 2), with all the preliminaries being simple DFS'es and combinatorial primitives which also can be done in O(n 2).
• +74
» 5 years ago, # | +16 OIerRobbin in my room coded a different solution to 450, which I think is quite cool.He finds the maximum consecutive segment in the data, inverts signs on that segment and then simply finds the maximum consecutive segment again. In case he finds two overlapping segments that way, it is still a valid solution, since one could achieve the same result by choosing two non-overlapping parts.
» 5 years ago, # | 0 Thanks for this, I didn't see it before, and I had just made a comment on the topcoder editorial about an O(mn) solution — please excuse the (sortof) duplication. Without wildcards, you can move through the string left to right. If you come across an 'a', that gives you +1 for the single character substring, then another +1 if there was an 'a' before it due to the extra 'aa', then another +1 if there was an 'aa' before it, up to a total of n+1 for n matching characters before it. So you can get O(mn) by storing the previous character and a counter for how many in a row that character has been.Dealing with expectation values, the 'previous character' becomes an array of doubles. If an 'a' comes along, you get a score of 1 + previous['a'], previous['a'] gets incremented, all other previous[] go to zero. If '?' comes along, you get a score of +1, a score of previous[i] for all i, and previous[i] gets +1 at a rate of p[i]. I wonder if I can figure out how to post code... double score = 0; double[] previous = new double[26]; for (int i = 0; i < s.Length; i++) { if (s[i] == '?') { score += 1; for (int j = 0; j < 26; j++) { score += previous[j] * p[j]; previous[j] = (previous[j] + 1) * p[j]; } } else { int index = s[i] - 'a'; score += 1 + previous[index]; for (int j = 0; j < 26; j++) { if (j == index) previous[j] += 1; else previous[j] = 0; } } } return score;
» 5 years ago, # | 0 How do we care about vertices in subtrees rooted not in [l,r]? We can partially fill some of them. | 2020-05-26T00:44:53 | {
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https://www.elitedigitalstudy.com/10617/a-circular-disc-of-mass-10-kg-is-suspended-by-a-wire-attached-to-its-centre | A circular disc of mass 10 kg is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be 1.5 s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant α is defined by the relation J = –α θ, where J is the restoring couple and θ the angle of twist).
Asked by Abhisek | 1 year ago | 155
##### Solution :-
Mass of the circular disc = 10 kg
Period of torsional oscillation = 1.5 s
Radius of the disc = 15 cm = 0.15 m
Restoring couple, J = –α θ
Moment of inertia, I = ($$\dfrac{1}{2}$$) mR2
= ($$\dfrac{1}{2}$$) x 10 x (0.15)2
= 0.1125 kgm2
Time period is given by the relation
T = $$2\pi \sqrt{\dfrac{I}{\alpha }}$$
So,$$\alpha =\dfrac{4\pi^{2}I }{T^{2}}α= \dfrac{4\times (3.14)^{2}\times 0.1125}{(1.5)^{2}}$$
$$\dfrac{4.44}{2.25}$$
Answered by Pragya Singh | 1 year ago
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An air chamber of volume V has a neck area of cross-section into which a ball of mass m just fits and can move up and down without any friction. Show that when the ball is pressed down a little and released, it executes SHM. Obtain an expression for the time period of oscillations assuming pressure-volume variations of air to be isothermal [see Figure] | 2022-11-30T16:21:50 | {
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http://mathramble.wordpress.com/2011/05/ | # The spectrum of a direct sum of rings
The following exercise is taken from Atiyah-Macdonald, chapter 1.
Let $A$ be a commutative ring. Show that the following statements are equivalent:
(i) $X=\mbox{Spec }A$ is disconnected;
(ii) $A \cong A_1 \oplus A_2$, where neither $A_1$ or $A_2$ is the zero ring;
(iii) $A$ contains an idempotent $\neq 0, 1$.
Proof : first we do the easiest parts. It’s clear that $(ii) \Rightarrow (iii)$; for example, the elements $(1, 0)$ and $(0,1)$ are such idempotents. To show that $(iii) \Rightarrow (ii)$, let $r \in A$ be an idempotent which is not $0$ or $1$. Let $r'=1-r$. Then $r'^2 = 1-2r+r^2 = 1-2r+r = 1-r$, so $r'$ is also an idempotent which is not $0$ or $1$. Now since $1=r+r'$, we have $a=ra+r'a$ for every $a \in A$; so $A=(r)+(r')$. Since $rr'=r(1-r)=r-r^2=r-r=0$, the sum is actually direct (as a sum of $A$-modules); indeed, if $ra+r'b=0$, then, multiplying by $r$ we have $r^2a=ra=0$. Similarily $r'b=0$. So the sum is direct. Give $(r)$ an internal ring structure by defining its unit to be $r$; this works, since $r(ra)=ra$ for every $ra \in (r)$. Then $A \cong (r) \oplus (r')$ as a direct sum of rings.
Now clearly $(ii) \Rightarrow (i)$; indeed, every prime ideal of $A$ contains one of $(0,1)$ or $(1,0)$, but none contains both.
Now we show that $(i) \Rightarrow (iii)$. Let $V(R_1)$ and $V(R_2)$ be closed sets in $X$ such that $X = V(R_1) \sqcup V(R_2) = V(R_1 \cap R_2)$. Suppose without loss of generality that $R_1$ and $R_2$ are ideals. Since the union is disjoint, we must have $R_1+R_2=A$, so there exists $r_1 \in R_1$ and $r_2 \in R_2$ with $r_1+r_2=1$. We must show that we can take $r_1$ and $r_2$ so that $r_1r_2=0$; this will cause the sum $R_1+R_2=A$ to be direct. Since $V(R_1 \cap R_2) = X$, and since $r_1r_2 \in R_1\cap R_2$, we have $V(r_1r_2) \supset V(R_1R_2)=X$. Thus every prime ideal in $A$ contains $r_1r_2$. Thus $r_1r_2$ is contained in the nilradical of $A$, so it is nilpotent.
Let $n$ be such that $r_1^nr_2^n=0$. Then, by the binomial theorem, $r_1^n+r_2^n = 1 +r_1r_2s$ for some $s \in A$. Since $r_1r_2$ is nilpotent, so is $r_1r_2s$; hence $r_1^n+r_2^n$ is a unit, say $\alpha(r_1^n+r_2^n)=1$. Let $r_1'=\alpha r_1^n$ and $r_2'=\alpha r_2^n$. Then $r_1'+r_2'=1$ and $r_1'r_2'=0$, as promised. This shows that $A=r_1' A \oplus r_2' A$ as $A$-modules. We can, as before, define a ring structure on $r_i'A$ be letting $r_i'$ be the unit, since we have $r_1'(r_1'+r_2')=r_1'=r_1'^2$ since $r_1r_2=0$. Thus $A \cong r_1' A \oplus r_2' A$ as rings.
# A cute problem
I came up with this little problem last night. It’s not very difficult to prove but still fun (I think). Here it is : let $A$ be a commutative ring, and let $h(u,v) \in A[u,v]$. Suppose that, for any polynomials $f(u), g(u) \in A[u]$, we have $h(f(u), g(u))=h(g(u), f(u))$. Then $h(u,v)=h(v,u)$.
I’ll post my solution in a couple of days to see if anyone can come up with an alternative solution in the meantime. :)
# A nice problem in Galois theory
This problem was given to me by my research supervisor. This is the problem and my solution:
Let $k$ be a field of characteristic zero, and $k(x)$ the rational function field in one variable over $k$. Suppose $F_1$ and $F_2$ are subfields of $k(x)$ such that $[k(x):F_1]$ and $[k(x):F_1]$ are finite. Is it possible for $[k(x):F_1\cap F_2]$ to be infinite? Indeed, it is. Note that $\mbox{Gal }(k(x)/k) = \mbox{PSL}_2(k)$, the projective special general linear group over $k$ (we identify its elements with Möbius transformations). Let $S, T$ denote the generators of the modular group (or rather, their image in $\mbox{PSL}_2(k)$). Note that they are of finite order. Hence, taking $F_1=k(x)^{}$ and $F_2=k(x)^{}$, we have $[k(x):F_i]<\infty$ for $i=1,2$. However, $F_1 \cap F_2$ is fixed by the whole modular group, which is infinite since $k$ has characteristic zero. Hence $k(x)$ cannot be of finite index over $F_1 \cap F_2$.
I don’t know whether such a construction is possible if $k$ has prime characteristic. I’d be interested to know if you find out! | 2014-04-21T15:02:57 | {
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https://socratic.org/questions/how-to-add-and-subtract-radical-equations-3sqrt2-9-1-2sqrt32-sqrt9-8 | # How to add and subtract radical equations (3sqrt2)/9 +1/(2sqrt32)+sqrt9/8?
Apr 3, 2015
We first simplify the individual terms as much as possible
$\frac{3 \sqrt{2}}{9} = \frac{\sqrt{2}}{3}$ (divide upper and lower by $3$)
$32 = 2 \cdot 16 = 2 \cdot {4}^{2} \to \sqrt{32} = 4 \sqrt{2} \to$
1/(2sqrt32)=1/(8sqrt2
$9 = {3}^{2} \to \frac{\sqrt{9}}{8} = \frac{3}{8}$
Our problem has now evolved into:
$\frac{\sqrt{2}}{3} + \frac{1}{8 \sqrt{2}} + \frac{3}{8}$
We need to take care of the second term by multiplying (above and below) by $\sqrt{2}$
$\frac{1}{8 \sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{8 \cdot 2} = \frac{\sqrt{2}}{16}$
Now we need to get all numerators alike; we'll have to get them to $48$ (= smallest common multiple)
$\frac{16 \cdot \sqrt{2}}{16 \cdot 3} + \frac{3 \cdot \sqrt{2}}{3 \cdot 16} + \frac{6 \cdot 3}{6 \cdot 8} =$
$\frac{16 \sqrt{2} + 3 \sqrt{2} + 18}{48} = \frac{19 \sqrt{2} + 18}{48} = \frac{19}{48} \sqrt{2} + \frac{3}{8}$ | 2021-09-28T12:35:59 | {
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https://stacks.math.columbia.edu/tag/04XV | Lemma 41.7.2. Let $f : X \to S$ be a morphism of schemes. The following are equivalent:
1. $f$ is a closed immersion,
2. $f$ is a proper monomorphism,
3. $f$ is proper, unramified, and universally injective,
4. $f$ is universally closed, unramified, and a monomorphism,
5. $f$ is universally closed, unramified, and universally injective,
6. $f$ is universally closed, locally of finite type, and a monomorphism,
7. $f$ is universally closed, universally injective, locally of finite type, and formally unramified.
Proof. The equivalence of (4) – (7) follows immediately from Lemma 41.7.1.
Let $f : X \to S$ satisfy (6). Then $f$ is separated, see Schemes, Lemma 26.23.3 and has finite fibres. Hence More on Morphisms, Lemma 37.40.1 shows $f$ is finite. Then Morphisms, Lemma 29.44.15 implies $f$ is a closed immersion, i.e., (1) holds.
Note that (1) $\Rightarrow$ (2) because a closed immersion is proper and a monomorphism (Morphisms, Lemma 29.41.6 and Schemes, Lemma 26.23.8). By Lemma 41.7.1 we see that (2) implies (3). It is clear that (3) implies (5). $\square$
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). | 2021-12-01T21:55:07 | {
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http://texgraph.tuxfamily.org/aide/TeXgraphse96.html | ### 5.60 Mtransform
• Mtransform( <list>, <matrix> ).
• Description: that function applies the <matrix> to the <list> and returns the result. If the <list> contain the constant jump, it is returned in the result without been modified. The <matrice> represents the analytic expression of an affine plane transformation, this is a three complexes list $\left[z1,z2,z3\right]$: $z1$ is the translation vector affix, $z2$ is the affix of the first column vector of the matrix of the linear part in the base (1,i), and $z3$ is the affix of the second column vector of the matrix of the linear part. For example, the identity matrix is written like this: [0,1,i] (This is the default matrix). | 2017-07-26T00:40:20 | {
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https://www.projectrhea.org/rhea/index.php/HW5ECE38F15 | # Homework 5, ECE438, Fall 2015, Prof. Boutin
Hard copy due in class, Wednesday September 30, 2015.
The goal of this homework is to get an intuitive understanding on how to DT signals with different sampling frequencies in an equivalent fashion.
## Question 1
Downsampling and upsampling
a) What is the relationship between the DT Fourier transform of x[n] and that of y[n]=x[4n]? (Give the mathematical relation and sketch an example.)
b) What is the relationship between the DT Fourier transform of x[n] and that of
$z[n]=\left\{ \begin{array}{ll} x[n/4],& \text{ if } n \text{ is a multiple of } 4,\\ 0, & \text{ else}. \end{array}\right.$
(Give the mathematical relation and sketch an example.)
## Question 2
Downsampling and upsampling
Let $x_1[n]=x(Tn)$ be a sampling of a CT signal $x(t)$. Let D be a positive integer.
a) Under what circumstances is the downsampling $x_D [n]= x_1 [Dn]$ equivalent to a resampling of the signal with a new period equal to DT (i.e. $x_D [n]= x(DT n)$)?
b) Under what circumstances is it possible to construct the sampling $x_3[n]= x(\frac{T}{D} n)$ directly from $x_1[n]$ (without reconstructing x(t))?
## Question 3
Define System 1 as the following LTI system
$x(t)\rightarrow \left[ \begin{array}{c} \text{ LPF} \\ \text{ no gain} \\ \text{cutoff at 1000Hz} \end{array}\right] \rightarrow \left[ \begin{array}{ccc} & & \\ & H(f) & \\ & & \end{array}\right] \rightarrow y(t)$
where the frequency response H(f) corresponds to a band-pass filter with no gain and cutoff frequencies f1=200Hz and f2=600Hz.
a) Sketch the graph of the frequency response H(f) of System 1.
b) Sketch the graph of the frequency response $H_1(\omega)$ that would make the following system equivalent to System 1.
$x(t) \rightarrow \left[ \begin{array}{c} \text{LPF} \\ \text{ no gain }\\ \text{ cutoff at 1000Hz} \end{array}\right] \rightarrow \left[ \begin{array}{c} \text{C/D Converter} \\ \text{6000 samples per second} \end{array}\right] \rightarrow \left[ \begin{array}{c} H_1(\omega) \end{array}\right] \rightarrow \left[ \begin{array}{c} \text{D/C Converter} \\ \text{6000 samples per second} \end{array}\right] \rightarrow y(t)$
## Question 4
Define System 2 as the following LTI system
$x[n]\rightarrow \left[ \begin{array}{ccc} & & \\ & H_1(\omega) & \\ & & \end{array}\right] \rightarrow y[n]$
where the frequency response $H_1(\omega)$ is the one you obtained in Question 3. Is it possible to implement System 2 as follows? Answer yes/no. If you answered yes, sketch the graph of the required LPF1 and frequency response H2. If you answered no, explain why not. (Hint: the first two parts of the system correspond to an "interpolator".)
$x[n] \rightarrow \left[ \begin{array}{ccc} & & \\ & \text{Upsample by factor 2} & \\ & & \end{array}\right] \rightarrow \left[ \begin{array}{ccc} & & \\ & \text{LPF1 } & \\ & & \end{array}\right] \rightarrow \left[ \begin{array}{ccc} & & \\ & H_2(\omega) & \\ & & \end{array}\right] \rightarrow \left[ \begin{array}{ccc} & & \\ & \text{Downsample by factor 2} & \\ & & \end{array}\right] \rightarrow y([n]$
## Question 5
Define System 3 as the following LTI system
$x[n] \rightarrow \left[ \begin{array}{c} \text{LPF} \\ \text{ no gain }\\ \text{ cutoff at} \frac{\pi}{2} \end{array}\right] \rightarrow \left[ \begin{array}{ccc} & & \\ & H_1(\omega) & \\ & & \end{array}\right] \rightarrow y[n]$
where the frequency response $H_1(\omega)$ is the one you obtained in Question 3.
a) Is it possible to implement System 3 as follows? Answer yes/no. If you answered yes, sketch the graph of the required LPF2 and frequency response H3. If you answered no, explain why not. (Hint: the last two parts of the system correspond to an "interpolator".)
$x[n] \rightarrow \left[ \begin{array}{c} \text{LPF} \\ \text{ no gain }\\ \text{ cutoff at} \frac{\pi}{2} \end{array}\right] \rightarrow \left[ \begin{array}{ccc} & & \\ & \text{Downsample by factor 2} & \\ & & \end{array}\right] \rightarrow \left[ \begin{array}{ccc} & & \\ & H_3(\omega) & \\ & & \end{array}\right] \rightarrow \left[ \begin{array}{ccc} & & \\ & \text{Upsample by factor 2} & \\ & & \end{array}\right] \rightarrow \left[ \begin{array}{c} \text{LPF2} \end{array}\right] \rightarrow y([n]$
Hand in a hard copy of your solutions. Pay attention to rigor!
## Presentation Guidelines
• Write only on one side of the paper.
• Use a "clean" sheet of paper (e.g., not torn out of a spiral book).
• Staple the pages together.
• Include a cover page. | 2020-08-13T01:57:16 | {
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https://blog.plover.com/math/seventh-root.html | # The Universe of Discourse
Sat, 18 Feb 2023
This Math SE question seems destined for deletion, so before that happens I'm repatriating my answer here. The question is simply:
Let !!n!! be an integer, and you are given that !!n^7=19203908986159!!. How would you solve for !!n!! without using a calculator?
If you haven't seen this sort of thing before, you may not realize that it's easier than it looks because !!n!! will have only two digits.
The number !!n^7!! is !!14!! digits long so its seventh root is !!\left\lceil \frac {14}7\right\rceil=2!! digits long. Let's say the digits of !!n!! are !!p!! and !!q!! so that the number we seek is !!n=10p+q!!.
The units digit of !!n^7!! is odd so !!q!! is odd. Clearly !!q\ne 1!! and !!q\ne 5!!. Units digits of numbers ending in !!3,7,9!! repeat in patterns:
$$\begin{array}{rl} 3 & 3, 9, 7, 1, 3, 9, \mathbf 7, \ldots \\ 7 & 7, 9, 3, 1, 7, 9, \mathbf 3, \ldots \\ 9 & 9, 1, 9, 1, 9, 1, \mathbf 9, \ldots \end{array}$$
so that !!(10p+q)^7!! can end in !!9!! only if !!q=9!!. Let's rewrite !!10p+9!! as !!10(p+1)-1!!.
Expanding !!(10(p+1)-1)^7!! with the binomial theorem, we get
$$10^7(p+1)^7 - 7\cdot10^6(p+1)^6 + \binom72 10^5(p+1)^5 - \cdots$$
The first term here is by far the largest; it is more than !!\frac{10}7p!! times as large as the second largest. Ignoring all but the first term, we want $$(p+1)^7 \approx 1920390.$$
!!2^7!! is only !!128!!, far too small. !!5^7!! is only !!78125!!, also too small. But !!8^7 = 2^{21} \approx 2000000!! because !!2^{10}\approx 1000!!. This is just right, so !!p+1=8!! and the final answer is $$79.$$
If the !!n!! were a three digit number, these kinds of tricks wouldn't be sufficient, and I would use more systematic and general methods. But I think it's a nice example of how far you can get with mere tricks and barely any theory.
This post was brought to you by the P.D.Q. Bernoulli Intitute of Lower Mathematics. It is dedicated to Gian Pietro Farina. I told him I planned to blog more, and he said he was especially looking forward to my posts about math. And then I posted like eleven articles in a row about Korean dog breed names and goose snot.
[ Addendum 20230219: Roger Crew pointed out that I forgot the binomial coefficients in the binomial theorem expansion. I have corrected this. ]
[ Addendum 20230228: Roger Crew also pointed out a possibly simpler way to find !!p!!. ] | 2023-03-26T09:41:03 | {
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https://math.stackexchange.com/questions/1238987/how-to-use-cauchys-integral-formula-with-more-than-one-pole | # How to use Cauchy's integral formula with more than one pole?
$$\int_{\gamma} \frac{z^2}{z(z-2)}, \quad \gamma(\theta) = 3e^{i\theta}, 0 \leq \theta \leq 2\pi$$
Cauchy's integral formula is given by:
$$\int\limits_{\gamma} \frac{f(z)}{(z-a)^{n+1}} = \frac{2\pi i}{n!} f^{(n)}(a)$$
And I can choose my holomorphic $$f(z) = z^2$$. But it doesn't seem like I can get my integral into a form like $$(z - a)^n$$ in the denominator. Am I missing some algebraic trick to do this?
Also, if $$\gamma(\theta)$$ was $$e^{i\theta}$$, then I could choose my holomorphic function to be $$\frac{z^2}{z-2}$$?
• en.m.wikipedia.org/wiki/Residue_theorem – Ant Apr 17 '15 at 12:38
• Isn't $\frac{z^2}{z(z-2)}=\frac{z}{z-2}$? (with $z=0$ being a removal singularity) – kennytm Apr 17 '15 at 12:39
• $\frac{z^2}{z\left(z-2\right)}=\frac{z}{z-2}$ so that you can take $f\left(z\right)=z$. – Nicolas Apr 17 '15 at 12:40
• Oh, I didn't know we're allowed to cancel out fractions inside an integral. Thanks – mr eyeglasses Apr 17 '15 at 12:40
We have $$\dfrac{z^2}{z(z-2)} = \dfrac{z}{z-2} = 1 + \dfrac2{z-2}$$ | 2020-04-09T08:48:22 | {
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https://faculty.math.illinois.edu/Macaulay2/doc/Macaulay2-1.19.1/share/doc/Macaulay2/NormalToricVarieties/html/_is__Cartier_lp__Toric__Divisor_rp.html | # isCartier(ToricDivisor) -- whether a torus-invariant Weil divisor is Cartier
## Synopsis
• Function: isCartier
• Usage:
isCartier D
• Inputs:
• D, ,
• Outputs:
• , that is true if the divisor is Cartier
## Description
A torus-invariant Weil divisor $D$ on a normal toric variety $X$ is Cartier if it is locally principal, meaning that $X$ has an open cover $\{U_i\}$ such that $D|_{U_i}$ is principal in $U_i$ for every $i$.
On a smooth variety, every Weil divisor is Cartier.
i1 : PP3 = toricProjectiveSpace 3; i2 : assert all (3, i -> isCartier PP3_i)
On a simplicial toric variety, every torus-invariant Weil divisor is $\QQ$-Cartier, which means that every torus-invariant Weil divisor has a positive integer multiple that is Cartier.
i3 : W = weightedProjectiveSpace {2,5,7}; i4 : assert isSimplicial W i5 : assert not isCartier W_0 i6 : assert isQQCartier W_0 i7 : assert isCartier (35*W_0)
In general, the Cartier divisors are only a subgroup of the Weil divisors.
i8 : X = normalToricVariety (id_(ZZ^3) | -id_(ZZ^3)); i9 : assert not isCartier X_0 i10 : assert not isQQCartier X_0 i11 : K = toricDivisor X; o11 : ToricDivisor on X i12 : assert isCartier K | 2022-07-01T04:31:11 | {
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https://stacks.math.columbia.edu/tag/0CRB | ## 109.11 Ext groups
Ext groups are defined in Algebra, Section 10.70.
Exercise 109.11.1. Compute all the Ext groups $\mathop{\mathrm{Ext}}\nolimits ^ i(M, N)$ of the given modules in the category of $\mathbf{Z}$-modules (also known as the category of abelian groups).
1. $M = \mathbf{Z}$ and $N = \mathbf{Z}$,
2. $M = \mathbf{Z}/4\mathbf{Z}$ and $N = \mathbf{Z}/8\mathbf{Z}$,
3. $M = \mathbf{Q}$ and $N = \mathbf{Z}/2\mathbf{Z}$, and
4. $M = \mathbf{Z}/2\mathbf{Z}$ and $N = \mathbf{Q}/\mathbf{Z}$.
Exercise 109.11.2. Let $R = k[x, y]$ where $k$ is a field.
1. Show by hand that the Koszul complex
$0 \to R \xrightarrow { \left( \begin{matrix} y \\ -x \end{matrix} \right) } R^{\oplus 2} \xrightarrow {(x, y)} R \xrightarrow {f \mapsto f(0, 0)} k \to 0$
is exact.
2. Compute $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(k, k)$ where $k = R/(x, y)$ as an $R$-module.
Exercise 109.11.3. Give an example of a Noetherian ring $R$ and finite modules $M$, $N$ such that $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(M, N)$ is nonzero for all $i \geq 0$.
Exercise 109.11.4. Give an example of a ring $R$ and ideal $I$ such that $\mathop{\mathrm{Ext}}\nolimits ^1_ R(R/I, R/I)$ is not a finite $R$-module. (We know this cannot happen if $R$ is Noetherian by Algebra, Lemma 10.70.9.)
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). | 2021-01-19T08:43:31 | {
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http://mathhelpforum.com/advanced-statistics/133944-gamma-distribution-mgf.html | # Math Help - Gamma distribution mgf
1. ## Gamma distribution mgf
Y has gamma distribution with α=n/2 for some positive integer n and β equal to some specified value. Using method of mgf, show that W=2Y/β has a chi-square dist. with n df.
I know what the mgfs of gamma and chi-square are. The part that is confusing me is the W=2Y/β and how to plug in the mgf's and such.
2. Obtain the MGF of W.
$M_W(t)= E(e^{Wt})=E(e^{2Yt/\beta})=M_Y(2t/\beta)$
So, insert $2t/\beta$ where you have t in the MGF of Y. | 2014-03-15T19:36:06 | {
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http://angrystatistician.blogspot.com/2012/06/infamous-monty-hall.html | ## Wednesday, June 27, 2012
### The Infamous Monty Hall
A (Brief) History:
The Monty Hall problem is arguably the most infamous probability puzzle in recent history. It was originally proposed in its current form in 1975, but only really surged into the public spotlight in 1990 when in appeared in a Parade column written by Marilyn vos Savant. For more details on the history of this problem see Wikipedia: Monty Hall problem.
The Original:
As given in Marilyn's column the problem read:
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1 [but the door is not opened], and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
The phrasing here is ambiguous. Does Monty always show you a goat? Does he only show you a goat when switching would lead to a win? Does he only show you a goat when switching would lead to a loss? In fact, Monty could assign you any probability $$p$$ of winning when switching that he wanted. All Monty needs to do is with probability $$p$$ only show you a goat if switching wins; otherwise, he'd only show you a goat if switching loses. Some of the unstated assumptions undoubtedly led to additional confusion in what's already a conceptually difficult problem for many people.
Unambiguous Phrasing (Krauss and Wang 2003):
Suppose you're on a game show and you're given the choice of three doors [and will win what is behind the chosen door]. Behind one door is a car; behind the others, goats [unwanted booby prizes]. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one [uniformly] at random. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door. Imagine that you chose Door 1 and the host opens Door 3, which has a goat. He then asks you "Do you want to switch to Door Number 2?" Is it to your advantage to change your choice?
Solution:
Standing, of course, gives us a $$1/3$$ probability of winning. But what if we switch? Well, the car is behind one of the doors, and if it's behind Door 1 with probability $$1/3$$ it must be behind one of the remaining two doors with probability $$2/3$$. But it's not behind Door 3 because Monty just showed us a goat. Aha, so the probability that the prize is behind Door 2 must be $$2/3$$ now, so we should switch! Another way of seeing that this is correct is to pretend that you close your eyes and cover your ears right before Monty Hall reveals the goat. Now by switching you'll win the prize if it's behind either Door 2 or Door 3, so your probability of winning increases from $$1/3$$ to $$2/3$$ by switching.
My next entry will be a solution using Bayes' Theorem. | 2017-03-24T12:06:33 | {
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https://encyclopediaofmath.org/index.php?title=Liouville_normal_form&diff=prev&oldid=27240 | # Difference between revisions of "Liouville normal form"
2010 Mathematics Subject Classification: Primary: 34A30 Secondary: 34B24 [MSN][ZBL]
The Liouville normal form is a way of writing a second-order ordinary linear differential equation $$\label{eq1} \derivn{y}{x}{2} + p(x)\deriv{y}{x} + \left( q(x) + \lambda r(x) \right) y = 0,$$ in the form $$\label{eq2} \derivn{\eta}{\xi}{2} + \left( \lambda + \phi(\xi) \right) \eta = 0,$$ where $\lambda$ is parameter. If $p(x) \in C^1$, $r(x) \in C^2$ and $r(x) > 0$, then equation \ref{eq1} reduces to the Liouville normal form \ref{eq2} by means of the substitution $\eta(\xi) = \Phi(x)y(x),\quad \xi = \int_\alpha^x \sqrt{r(t)}\,\mathrm{d}t, \quad \Phi(x) = r(x)^{1/4} \exp\left( \frac{1}{2}\int_\alpha^x p(t)\,\mathrm{d}t \right),$ which is called the Liouville transformation (introduced in [Li]). The Liouville normal form plays an important role in the investigation of the asymptotic behaviour of solutions of \ref{eq1} for large values of the parameter $\lambda$ or the argument, and in the investigation of the asymptotics of eigenfunctions and eigenvalues of the Sturm–Liouville problem (see [Ti]).
#### References
[In] E.L. Ince, "Ordinary differential equations", Dover, reprint (1956) [Ka] E. Kamke, "Differentialgleichungen: Lösungen und Lösungsmethoden", 1. Gewöhnliche Differentialgleichungen, Chelsea, reprint (1947) [Li] J. Liouville, J. Math. Pures Appl., 2 (1837) pp. 16–35 [Ti] E.C. Titchmarsh, "Eigenfunction expansions associated with second-order differential equations", 1–2, Clarendon Press (1946–1948)
How to Cite This Entry:
Liouville normal form. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Liouville_normal_form&oldid=27240
This article was adapted from an original article by M.V. Fedoryuk (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article | 2021-10-15T22:45:03 | {
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https://math.stackexchange.com/questions/3374298/solving-a-functional-equation-using-even-and-odd-properties | # Solving a functional equation using even and odd properties
I need help solving this functional equations problem.
Find all $$\ f : \mathbb{R} \rightarrow \mathbb{R}$$ such that for all $$x,y\ \epsilon \ \mathbb{R}$$, the two following equations hold:
$$f(3x)=f\left(\frac{x+y}{(x+y)^2+1}\right) + f\left(\frac{x-y}{(x-y)^2+1}\right)$$
$$f(x^2-y^2)=(x+y)f(x-y)+(x-y)f(x+y)$$
What I did was using the second equation to get that $$f(0)=0$$ by setting $$x=y=0$$.
then setting $$x=0, \ y$$ arbitrary in the first equation to get:
$$0=f(0)=f\left(\frac{y}{y^2+1}\right) + f\left(\frac{-y}{y^2+1}\right)$$ $$\implies f\left(\frac{y}{y^2+1}\right)=-f\left(\frac{-y}{y^2+1}\right) \implies f$$ is odd.
But then I got stuck here, I tried many ways but always end up with a variation
of $$f(-y^2)=2yf(-y)=-2yf(y)$$
Any suggestions how to go on from here?
I'm going to disregard the first equation altogether, not because it is ugly (though it is), but because one equation is just enough for a problem. Another one does not feel like it belongs here; for all I know, it could have been pasted by mistake.
Now to the point. $$f(x^2-y^2)=(x+y)f(x-y)+(x-y)f(x+y)\tag1$$
Let $$a=x+y,\;b=x-y$$ $$f(ab)=af(b)+bf(a)\tag2$$
Plugging $$b=a$$, we get $$f(a^2)=2af(a)$$. Now plugging $$b=a^2$$, we get $$f(a^3)=3a^2f(a)$$, and continuing in this manner, we end up with $$f(a^n)=na^{n-1}f(a)$$.
Now let's apply that in reverse: $$f(a)=na^{n-1\over n}f(a^{1/n})$$, hence $$f(a^{1/n})={1\over n}f(a)/a^{n-1\over n}$$.
Now let's put the two together: $$f(a^{p\over q}) = {1\over q}f(a^p)/a^{p(q-1)\over q}={p\over q}f(a)\cdot a^{p-1-{p(q-1)\over q}}={p\over q}f(a)\cdot a^{{p\over q}-1}\tag3$$ or, in other words, for any rational $$r$$ we have $$f(a^r)=r\cdot f(a)\cdot a^{r-1}\tag4$$
Well, rationals are dense in $$\mathbb R$$, but what's between them? The discontinuous solutions are plenty. If you know what the Hamel basis is, then you know where to find them, and if you don't know, then trust me, you don't want to know. But if we stick to $$f$$ being continuous everywhere (or anywhere, for that matter), the only way to achieve that is for (4) to hold at any real $$r$$.
Say we fix $$a$$ and looks at some $$x$$ (not the same $$x$$ as in the beginning):
$$x=a^{\log_ax}\implies f(x)=\log_ax\cdot f(a)\cdot a^{\log_ax-1}=C\cdot x\cdot\ln x\tag5$$
So this is the only family of continuous solution to your second equation. If you are still interested, you may go and check how well does this sit with the first equation (hint: not well, except the trivial case of $$C=0$$).
That's it.
• I know this is late but at that time I thought this was above my level. But now I understand everything except the last equation (5). where does x.lnx come from? Mar 1 '20 at 15:59
• Why, it comes from what is right before it. $a^{\log_ax}$ is just $x$, and $\log_ax$ is basically $\ln x$ times stuff. Mar 1 '20 at 22:11 | 2021-11-27T14:07:16 | {
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"openwebmath_perplexity": 146.52780637760148,
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"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9817357253887771,
"lm_q2_score": 0.6654105454764746,
"lm_q1q2_score": 0.6532573045446887
} |
http://mathhelpforum.com/calculus/13321-differentiate-function-problem.html | Math Help - Differentiate and Function problem
1. Differentiate and Function problem
Hello Friends,
I have 2 problems,
Here are they:
1) Differentiate y w.r.t. x where
y=e^(log tan x^4)^2
2) For the following function, find a point of maxima and a point of minima if these exist:
f(x)=17x^5-14x^3+44
Thanks.
harshal.
2. 1) Let v = tan(x^4) and u = log(v), then y = e^u².
Then by the chain rule: dy/dx = dy/du . du/dv . dv/dx
2) Solve: df/dx = 0, check whether its zeroes are min/max. | 2015-03-04T12:00:02 | {
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"url": "http://mathhelpforum.com/calculus/13321-differentiate-function-problem.html",
"openwebmath_score": 0.832008421421051,
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"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.981735725388777,
"lm_q2_score": 0.6654105454764747,
"lm_q1q2_score": 0.6532573045446887
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https://ch.mathworks.com/help/stats/repeatedmeasuresmodel.mauchly.html | # mauchly
Class: RepeatedMeasuresModel
Mauchly’s test for sphericity
## Syntax
``tbl = mauchly(rm)``
``tbl = mauchly(rm,C)``
## Description
example
````tbl = mauchly(rm)` returns the result of the Mauchly’s test for sphericity for the repeated measures model `rm`. It tests the null hypothesis that the sphericity assumption is true for the response variables in `rm`.For more information, see Mauchly’s Test of Sphericity.```
````tbl = mauchly(rm,C)` returns the result of the Mauchly’s test based on the contrast matrix `C`.```
## Input Arguments
expand all
Repeated measures model, returned as a `RepeatedMeasuresModel` object.
For properties and methods of this object, see `RepeatedMeasuresModel`.
Contrasts, specified as a matrix. The default value of `C` is the Q factor in a QR decomposition of the matrix M, where M is defined so that Y*M is the difference between all successive pairs of columns of the repeated measures matrix Y.
Data Types: `single` | `double`
## Output Arguments
expand all
Results of Mauchly’s test for sphericity for the repeated measures model `rm`, returned as a `table`.
`tbl` contains the following columns.
Column NameDefinition
`W`Value of Mauchly’s W statistic
`ChiStat`Chi-square statistic value
`DF`Degrees of freedom of the Chi-square statistic
`pValue`p-value corresponding to the Chi-square statistic
Data Types: `table`
## Examples
expand all
`load fisheriris`
The column vector `species` consists of iris flowers of three different species: setosa, versicolor, and virginica. The double matrix `meas` consists of four types of measurements on the flowers: the length and width of sepals and petals in centimeters, respectively.
Store the data in a table array.
```t = table(species,meas(:,1),meas(:,2),meas(:,3),meas(:,4),... 'VariableNames',{'species','meas1','meas2','meas3','meas4'}); Meas = dataset([1 2 3 4]','VarNames',{'Measurements'});```
Fit a repeated measures model, where the measurements are the responses and the species is the predictor variable.
`rm = fitrm(t,'meas1-meas4~species','WithinDesign',Meas);`
Perform Mauchly’s test to assess the sphericity assumption.
`mauchly(rm)`
```ans=1×4 table W ChiStat DF pValue _______ _______ __ __________ 0.55814 84.976 5 7.6149e-17 ```
The small $p$-value (in the `pValue` field) indicates that the sphericity, hence the compound symmetry assumption, does not hold. You should use epsilon corrections to compute the $p$-values for a repeated measures anova. You can compute the epsilon corrections using the `epsilon` method and perform the repeated measures anova with the corrected $p$-values using the `ranova` method. | 2022-01-22T07:37:01 | {
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"url": "https://ch.mathworks.com/help/stats/repeatedmeasuresmodel.mauchly.html",
"openwebmath_score": 0.8735157251358032,
"openwebmath_perplexity": 3769.407618925084,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9817357243200245,
"lm_q2_score": 0.6654105454764747,
"lm_q1q2_score": 0.6532573038335295
} |
https://www.gradesaver.com/textbooks/math/calculus/calculus-8th-edition/chapter-3-applications-of-differentiation-3-4-limits-at-infinity-horizontal-asymptotes-3-4-exercises-page-241/6 | ## Calculus 8th Edition
$$\lim_{x\to \infty} f(x) \approx 0.135$$
(a) Given $$f(x) =\left(1-\frac{2}{x}\right)^x$$ From the figure , it is clear that as $x\to\infty$ $f(x)\to 0.135$ (b) by using values of $x$ , we get $f(10)=0.107374$, $f(100) =0.135064$, $f(1000) =0.1353327$, and $f(10000) =0.1353352$ , then $$\lim_{x\to \infty} f(x) \approx 0.135$$ | 2019-11-22T11:02:57 | {
"domain": "gradesaver.com",
"url": "https://www.gradesaver.com/textbooks/math/calculus/calculus-8th-edition/chapter-3-applications-of-differentiation-3-4-limits-at-infinity-horizontal-asymptotes-3-4-exercises-page-241/6",
"openwebmath_score": 0.9875689148902893,
"openwebmath_perplexity": 127.71694422614358,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9817357243200245,
"lm_q2_score": 0.6654105454764747,
"lm_q1q2_score": 0.6532573038335295
} |
https://groupprops.subwiki.org/wiki/Endomorphism_ring_of_an_abelian_group | # Endomorphism ring of an abelian group
Suppose $A$ is an abelian group. The endomorphism ring of $A$, denoted $\operatorname{End}(A)$ is defined as follows:
1. As a set, it is the set of all endomorphisms of $A$.
2. The addition is pointwise addition in the target group. In other words, for endomorphisms $f,g$ of $A$, we define $f + g$ as the map $a \mapsto f(a) + g(a)$. Thus, the additive identity is the zero map (the map sending everything to zero) and the negation is the pointwise negation.
3. The multiplication is given by function composition. In other words, $fg$ is the map sending $a$ to $f(g(a))$. The identity for multiplication is the identity map. | 2021-03-08T00:28:52 | {
"domain": "subwiki.org",
"url": "https://groupprops.subwiki.org/wiki/Endomorphism_ring_of_an_abelian_group",
"openwebmath_score": 0.9937881231307983,
"openwebmath_perplexity": 77.4991703874908,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9817357237856482,
"lm_q2_score": 0.6654105454764747,
"lm_q1q2_score": 0.6532573034779499
} |
http://immersivemath.com/ila/ch02_vectors/ch02.html | © 2015-2019 Jacob Ström, Kalle Åström, and Tomas Akenine-Möller Forum
# Chapter 2: Vectors
One of the most important and fundamental concepts in linear algebra is the vector. Luckily, vectors are all around us, but they are, in general, not visible. The common ways to introduce a vector is either to begin with the strict mathematical definition, or to discuss examples of vectors, such as velocities, forces, acceleration, etc. For a more intuitive and hopefully faster understanding of this important concept, this chapter instead begins with an interactive demonstration and a clear visualization of what a vector can be. In this case, a ball's velocity, which consists of a direction (where the ball is going) and a speed (how fast it is going there), is shown in Interactive Illustration 2.1.
Interactive Illustration 2.1: This little breakout game shows the concept of a vector. Play along for an interactive introduction. Control the paddle with left/right keys, or touch/swipe.
Interactive Illustration 2.1: This little breakout game shows the concept of a vector. Play along for an interactive introduction. Control the paddle with left/right keys, or touch/swipe.
In this book, we denote points by capital italic letters, e.g., $A$, $B$, and $Q$. For most of the presentation in the early chapters, we will use two- and three-dimensional points, and some occasional one-dimensional points. We start with a definition of a vector.
Definition 2.1: Vector
Let $A$ and $B$ be two points. A directed line segment from $A$ to $B$ is denoted by:
$$\overrightarrow{AB}.$$ (2.1)
This directed line segment constitutes a vector. If you can move the line segment to another line segment with the same direction and length, they constitute the same vector.
$A$
$B$
$\overrightarrow{AB}$
$C$
$D$
$\overrightarrow{CD} = \vc{v}$
For instance, the two line segments $\overrightarrow{AB}$ and $\overrightarrow{CD}$ in Interactive Illustration 2.2 constitute the same vector as can be seen when pushing the "forward" button.
We say that $\overrightarrow{AB}$ is a vector and that
$$\overrightarrow{AB} = \overrightarrow{CD}.$$ (2.2)
A shorter notation for vectors is to use a single bold face characters, such as $\vc{v}$. As is shown in the illustration, $\vc{v} = \overrightarrow{AB} = \overrightarrow{CD}$. Some books make a difference between directed line segments and vectors, and reserve the short hand variant $\vc{v}$ for true vectors and the longer $\overrightarrow{AB}$ for directed line segments. While this may be mathematically more stringent, this difference is ignored for the purposes of this book, and we use vectors and directed line segments as one and the same thing.
We also use the terms tail point and tip point of a vector when this is convenient, where the tip point is where the arrowhead is, and the tail point is the other end.
A vector is completely defined by its
1. direction, and
2. its length
Note that a starting position of a vector is missing from the list above. As long as the direction and length is not changed, it is possible to move it around and have it start in any location. This is illustrated in Interactive Illustration 2.3.
Interactive Illustration 2.3: A vector does not have a specific starting position. This vector is drawn at a certain position, but even when it is moved to start somewhere else, it is still the same vector. Click/touch Forward to move the vector.
Interactive Illustration 2.3: A vector does not have a specific starting position. This vector is drawn at a certain position, but even when it is moved to start somewhere else, it is still the same vector. Click/touch Forward to move the vector.
$\vc{u}$
$\vc{u}$
$\vc{u}$
$\vc{u}$
$\vc{u}$
$\vc{u}$
$\vc{u}$
$\vc{u}$
$\vc{u}$
$\vc{u}$
$\vc{u}$
$\vc{u}$
$\vc{u}$
$\vc{u}$
$\vc{u}$
The length of a vector is denoted by $\ln{\overrightarrow{AB}}$, or in shorthand by $\ln{\vc{v}}$.
$$\text{length of vector:}\spc\spc \ln{\vc{v}}$$ (2.3)
The length of a vector is a scalar, which just means that it is a regular number, such as $7.5$. The term scalar is used to emphasize that it is just a number and not a vector or a point. Exactly how the length of a vector can be calculated will be deferred to Chapter 3.
Note that the order of the points is important, i.e., if you change the order of $A$ and $B$, another vector, $\overrightarrow{BA}$, is obtained. It has opposite direction, but the same length, i.e., $\ln{\overrightarrow{AB}} = \ln{\overrightarrow{BA}}$. Even $\overrightarrow{AA}$ is a vector, which is called the zero vector, as shown in the definition below.
Definition 2.2: Zero Vector
The zero vector is denoted by $\vc{0}$, and can be created using a directed line segment using the same point twice, i.e., $\vc{0}=\overrightarrow{AA}$. Note that $\ln{\vc{0}}=0$, i.e., the length of the zero vector is zero.
Two vectors, $\vc{u}$ and $\vc{v}$, are parallel if they have the same direction or opposite directions, but not necessarily the same lengths. This is shown to the right in Figure 2.4. Note how you can change the vectors in the figure, some can be changed by grabbing the tip, others by grabbing the tail. The notation
$$\vc{u}\, ||\, \vc{v}$$ (2.4)
means that $\vc{u}$ is parallel to $\vc{v}$. The zero vector $\vc{0}$ is said to be parallel to all other vectors. Next, we will present how two vectors can be added to form a new vector, and then follows scalar vector multiplication in Section 2.3.
There are two fundamental vector operations in linear algebra, namely, vector addition and scalar vector multiplication, where the latter is sometimes called vector scaling. Most of the mathematics in this book build upon these two operations, and even the most complex operations often lead back to addition and scaling. Vector scaling is described in Section 2.3, while vector addition is described here. Luckily, both vector addition and vector scaling behave as we would expect them to.
The sum, $\vc{u}+\vc{v}$, of two vectors, $\vc{u}$ and $\vc{v}$, is constructed by placing $\vc{u}$, at some arbitrary location, and then placing $\vc{v}$ such that $\vc{v}$'s tail point coincides with $\vc{u}$'s tip point, and $\vc{u}+\vc{v}$ is the vector that starts at $\vc{u}$'s tail point, and ends at $\vc{v}$'s tip point.
Exactly how the vector sum is constructed is shown in Interactive Illustration 2.5 below.
Interactive Illustration 2.5: Two vectors, $\vc{u}$ and $\vc{v}$, are shown. These will be added to form the vector sum $\vc{u} + \vc{v}$. Note that the vectors can be changed as usual by dragging their tips. Click/press Forward to continue to the next stage of the illustration.
Interactive Illustration 2.5: Two vectors, $\hid{\vc{u}}$ and $\hid{\vc{v}}$, are shown. These will be added to form the vector sum $\hid{\vc{u} + \vc{v}}$. Note that the vectors can be changed as usual by dragging their tips. Click/press Forward to continue to the next stage of the illustration.
$\vc{u}+\vc{v}$
$\vc{v}$
$\vc{v}$
$\vc{v}$
$\vc{u}$
$\vc{u}$
$\vc{u}$
$\vc{v}$
So far, we have only illustrated the vector addition in the plane, i.e., in two dimensions. However, it can also be illustrated in three dimensions. This is done below in Interactive Illustration 2.6. Remember that you can rotate the figure by moving the mouse while right clicking or by using a two-finger swipe.
Interactive Illustration 2.6: Two vectors, $\vc{u}$ and $\vc{v}$, are shown. These will be added to form the vector sum, $\vc{u} + \vc{v}$. Note that the vectors can be changed as usual by dragging their tip points. If you do so, you will move the points in the plane of the screen. Click/press Forward to continue to the next stage of the illustration.
Interactive Illustration 2.6: In this final stage, we have added some dashed support lines to make it easier to see the spatial relationships. Recall that you can press the right mouse button, keep it pressed, and move the mouse to see the vector addition from another view point. For tablets, the same maneuver is done by swiping with two fingers. Note that by changing the point of view like this, you can verify that $\hid{\vc{u}}$, $\hid{\vc{v}}$, and $\hid{\vc{u}+\vc{v}}$ all lie in the same plane. Try also to move the vectors so that the projected points no longer end up on a straight line.
$\vc{u}$
$\vc{u}+\vc{v}$
$\vc{v}$
$\vc{v}$
$\vc{v}$
As we saw in the Breakout Game 2.1, the speed of the ball was increased by 50% after a while. This is an example of vector scaling, where the velocity vector simply was scaled by a factor of $1.5$. However, a scaling factor can be negative as well, and this is all summarized in the definition below, and instead of the term vector scaling, we also use the term scalar vector multiplication.
Definition 2.4: Scalar Vector Multiplication
When a vector, $\vc{v}$, is multiplied by a scalar, $k$, the vector $k\vc{v}$ is obtained, which is parallel to $\vc{v}$ and its length is $\abs{k}\,\ln{v}$. The direction of $k\vc{v}$ is opposite $\vc{v}$ if $k$ is negative, and otherwise it has the same direction as $\vc{v}$. If $k=0$, then $k\vc{v}=\vc{0}$.
A corollary to this is that if the two vectors $\vc{u}$ and $\vc{v}$ satisfy $\vc{u} = k \vc{v}$ for some scalar $k$, then $\vc{u}$ and $\vc{v}$ are parallel.
Scalar vector multiplication is shown in Interactive Illustration 2.7 below. The reader is encouraged to play around with the illustration.
Interactive Illustration 2.7: Here, we show how a vector, $\vc{v}$, can be multiplied by a scalar, $k$, so that $k\vc{v}$ is generated. The reader can move the vector, $\vc{v}$, and also manipulate the value of $k$ by dragging the slider below the illustration. Note what happens to $k\vc{v}$ when $k$ is negative. As an exercise, try to make the tip point of $\vc{v}$ meet coincide with the tip point of $k\vc{v}$.
Interactive Illustration 2.7: Here, we show how a vector, $\hid{\vc{v}}$, can be multiplied by a scalar, $\hid{k}$, so that $\hid{k\vc{v}}$ is generated. The reader can move the vector, $\hid{\vc{v}}$, and also manipulate the value of $\hid{k}$ by dragging the slider below the illustration. Note what happens to $\hid{k\vc{v}}$ when $\hid{k}$ is negative. As an exercise, try to make the tip point of $\hid{\vc{v}}$ meet coincide with the tip point of $\hid{k\vc{v}}$.
$k=$
$\vc{v}$
$k\vc{v}$
Now that we can both add vectors, and scale vectors by a real number, it is rather straightforward to subtract two vectors as well. This is shown in the following example.
Example 2.1: Vector Subtraction
Note that by using vector addition (Definition 2.3) and scalar vector multiplication (Definition 2.4) by $-1$, we can subtract one vector, $\vc{v}$, from another, $\vc{u}$ according to
$$\underbrace{\vc{u} + (\underbrace{-1\vc{v}}_{\text{scaling}})}_{\text{addition}} = \vc{u}-\vc{v},$$ (2.5)
where we have introduced the shorthand notation, $\vc{u}-\vc{v}$, for the expression to the left of the equal sign. Vector subtraction is illustrated below.
Interactive Illustration 2.8: Vector subtraction, $\vc{u}-\vc{v}$, is illustrated here. First, only the two vectors, $\vc{u}$ and $\vc{v}$, are shown.
Interactive Illustration 2.8: Finally, we see that $\hid{\vc{u}-\vc{v}}$ is the vector from $\hid{\vc{v}}$'s tip point to $\hid{\vc{u}}$'s tip point. The reader can move the red ($\hid{\vc{u}}$) and green ($\hid{\vc{v}}$) vectors, and as an exercise, try out what happens if one of $\hid{\vc{u}}$ or $\hid{\vc{v}}$ is set to the zero vector, and also try setting $\hid{\vc{u}=\vc{v}}$.
$\vc{u}$
$\vc{v}$
$-\vc{v}$
$-\vc{v}$
$-\vc{v}$
$\vc{u}-\vc{v}$
$\vc{u}-\vc{v}$
$\vc{u}-\vc{v}$
Example 2.2: Box
In this example, we will see how a box can be created by using three vectors that all make a right angle with each other.
Interactive Illustration 2.9: In this example, we have one red, one green, and one blue vector. These all make right angles with each other, and they are constrained to be like that. The length of the vectors can be changed interactively, though. In the following, we will show how a box can be built from these vectors.
Interactive Illustration 2.9: In this example, we have one red, one green, and one blue vector. These all make right angles with each other, and they are constrained to be like that. The length of the vectors can be changed interactively, though. In the following, we will show how a box can be built from these vectors.
There are a number of different rules for using both vector addition and scalar vector multiplication. This is the topic of the next section.
Using vectors in calculations with vector addition and scalar vector multiplication is fairly straightforward. They behave as we might expect them to. However, rules such as $\vc{u}+(\vc{v}+\vc{w})=(\vc{u}+\vc{v})+\vc{w}$, must be proved. The rules for vector arithmetic are summarized in Theorem 2.1.
Theorem 2.1: Properties of Vector Arithmetic
Assuming that $\vc{u}$, $\vc{v}$, and $\vc{w}$ are vectors of the same size, and that $k$ and $l$ are scalars, then the following rules hold:
\begin{gather} \begin{array}{llr} (i) & \vc{u}+\vc{v} = \vc{v}+\vc{u} & \spc\text{(commutativity)} \\ (ii) & (\vc{u}+\vc{v})+\vc{w} = \vc{u}+(\vc{v}+\vc{w}) & \spc\text{(associativity)} \\ (iii) & \vc{v}+\vc{0} = \vc{v} & \spc\text{(zero existence)} \\ (iv) & \vc{v}+ (-\vc{v}) = \vc{0} & \spc\text{(negative vector existence)} \\ (v) & k(l\vc{v}) = (kl)\vc{v} & \spc\text{(associativity)}\\ (vi) & 1\vc{v} = \vc{v} & \spc\text{(multiplicative one)} \\ (vii) & 0\vc{v} = \vc{0} & \spc\text{(multiplicative zero)} \\ (viii) & k\vc{0} = \vc{0} & \spc\text{(multiplicative zero vector)} \\ (ix) & k(\vc{u}+\vc{v}) = k\vc{u}+k\vc{v} & \spc\text{(distributivity 1)} \\ (x) & (k+l)\vc{v} = k\vc{v}+l\vc{v} & \spc\text{(distributivity 2)} \\ \end{array} \end{gather} (2.6)
While most (or all) of the rules above feel very natural and intuitive, they must be proved nevertheless. The reader is encouraged to look at the proofs, and especially at the interactive illustrations, which can increase the feeling and intuition for many of the rules.
$(i)$ This rule (commutativity) has already been proved in the figure in Definition 2.3. Another way to prove this rule is shown below in Interactive Illustration 2.10.
Interactive Illustration 2.10: This interactive illustration shows commutativity of vector addition. This means that $\vc{u}+\vc{v}=\vc{v}+\vc{u}$. Click/touch Forward to continue.
Interactive Illustration 2.10: Finally, we also show the other translated vectors both to the left and right. As can be seen, the resulting vector sum is the same, regardless of the order of the operands. Recall that the vectors can be moved around.
$\textcolor{#aa0000}{\vc{u}}$
$\textcolor{#aa0000}{\vc{u}}$
$\textcolor{#aa0000}{\vc{u}}$
$\textcolor{#aa0000}{\vc{u}}$
$\textcolor{#00aa00}{\vc{v}}$
$\textcolor{#00aa00}{\vc{v}}$
$\textcolor{#00aa00}{\vc{v}}$
$\textcolor{#00aa00}{\vc{v}}$
$\textcolor{#0000aa}{\vc{u}+\vc{v}}$
$\textcolor{#0000aa}{\vc{v}+\vc{u}}$
$(ii)$ The proof of this rule (associativity) is shown in Interactive Illustration 2.11.
Interactive Illustration 2.11: Consider the three vectors, $\vc{u}$, $\vc{v}$, and $\vc{w}$. Since vectors do not have a specific starting point, we have arranged them so that $\vc{v}$ starts where $\vc{u}$ ends, and $\vc{w}$ starts where $\vc{v}$ ends.
Interactive Illustration 2.11: Consider the three vectors, $\hid{\vc{u}}$, $\hid{\vc{v}}$, and $\hid{\vc{w}}$. Since vectors do not have a specific starting point, we have arranged them so that $\hid{\vc{v}}$ starts where $\hid{\vc{u}}$ ends, and $\hid{\vc{w}}$ starts where $\hid{\vc{v}}$ ends.
$\textcolor{#aa0000}{\vc{u}}$
$\textcolor{#00aa00}{\vc{v}}$
$\textcolor{#0000aa}{\vc{w}}$
$\textcolor{#00aaaa}{\vc{v}+\vc{w}}$
$\vc{u}+(\vc{v}+\vc{w})$
$\textcolor{#aaaa00}{\vc{u}+\vc{v}}$
$(\vc{u}+\vc{v})+\vc{w}$
$\vc{u}+\vc{v}+\vc{w}$
$(iii)$ Since, there zero vector has length zero, the definition of vector addition gives us that $\vc{v}+\vc{0}$ is the same as $\vc{v}$.
$(iv)$ Since $-\vc{v}$ is exactly $\vc{v}$ with opposite direction, the sum will be zero.
$(v)$ The approach to this proof is to start with the left hand side of the equal sign, and find out what the direction and length is. Then the same is done for the right hand side of the equal sign. The details are left as an exercise for the reader.
$(vi)$ Since $1$ is a positive number, we know that $1\vc{v}$ and $\vc{v}$ have the same direction, so it only remains to control that they have the same length. The length of the left hand side of the equal sign is $\abs{1}\,\ln{\vc{v}}=\vc{v}$, and for the right hand side it is $\ln{\vc{v}}$, i.e., they are the same, which proves the rule.
$(vii)$ and $(viii)$ First, note the difference between these. In $(vii)$, we have a scalar zero times $\vc{v}$ equals a zero vector, and in $(viii)$, we have a scalar, $k$ times a zero vector, which is equals the zero vector. $(vii)$ is actually defined in Definition 2.4, so only $(viii)$ needs to be proved. The length of both $k\vc{0}$ and $\vc{0}$ are zero, which proves the rule.
$(ix)$ First, we refer the reader to Interactive Illustration 2.12. Be sure to press Forward until the last stage of the illustration. The formal proof (of distributivity) follows after the illustration.
Interactive Illustration 2.12: This illustration helps show the rule $k(\vc{u}+\vc{v}) = k\vc{u}+k\vc{v}$. First, we simply show two vectors, $\vc{u}$ and $\vc{v}$, and their sum, $\vc{u}+\vc{v}$. Press Forward to continue.
Interactive Illustration 2.12: Finally the vector $\hid{k(\vc{u}+\vc{v})}$ is shown as well. Note that the smaller triangle $\hid{\triangle O A_1 B_1}$ is similar to the larger triangle $\hid{\triangle O A_2 B_2}$ since it has the same angle at $\hid{A_1}$ and $\hid{A_2}$ and since the two edges $\hid{\vc{v}}$ and $\hid{\vc{u}}$ is proportional to $\hid{k\vc{v}}$ and $\hid{k\vc{u}}$. Thus it is clear that by adding $\hid{k\vc{u}}$ and $\hid{k\vc{v}}$, we reach $\hid{k\vc{u}+k\vc{v}}$, which is the same as $\hid{k(\vc{u}+\vc{v})}$. Recall that you can press the right mouse button, keep it pressed, and move the mouse to see the vector addition from another perspective. For tablets, the same maneuver is done by swiping with two fingers.
$k=$
$\vc{u}$
$\vc{v}$
$\vc{u} + \vc{v}$
$k\vc{u}$
$k\vc{v}$
$k(\vc{u} + \vc{v})$
$O$
$A_1$
$A_2$
$B_1$
$B_2$
It follows from scalar vector multiplication (Definition 2.4) that
\begin{align} \ln{k\vc{u}} &= \abs{k}\,\ln{\vc{u}}, \\ \ln{k\vc{v}} &= \abs{k}\,\ln{\vc{v}}, \end{align} (2.7)
and if $k>0$ then $\vc{u}$ and $k\vc{u}$ have the same direction, and so do $\vc{v}$ and $k\vc{v}$. On the other hand, if $k<0$ then $\vc{u}$ and $k\vc{u}$ have opposite directions, and so do $\vc{v}$ and $k\vc{v}$. This implies that the triangle, formed by the following set of three points: $\{O$, $O+\vc{u}$, $O+\vc{u}+\vc{v}\}$, is similar to the triangle formed by $\{O$, $O+k\vc{u}$, $O+k\vc{u}+k\vc{v}\}$. That those two triangles are similar also means that $O$, $O+\vc{u}+\vc{v}$ and $O+k\vc{u}+k\vc{v}$ lie on a straight line. Furthermore, since the triangles are similar, and due to (2.7), we know that
$$\ln{k(\vc{u}+\vc{v})} = \abs{k}\,\ln{\vc{u}+\vc{v}}.$$ (2.8)
If $k>0$ then $k(\vc{u}+\vc{v})$ has the same direction as $\vc{u}+\vc{v}$, and if $k<0$ then the have opposite directions. Hence, it follows that $k(\vc{u}+\vc{v}) = k\vc{u}+k\vc{v}$. The rule is trivially true if $k=0$, which concludes the proof of this rule.
$(x)$ This is somewhat similar to $(ix)$, but simpler, and so is left for the reader.
This concludes the proofs for Theorem 2.1.
$\square$
Example 2.3: Vector Addition of Three Vectors
To get an understanding of how vector addition works for more than two vectors, Interactive Illustration 2.13 below shows the addition of three vectors. Recall that vector addition is associative, so we may write $\vc{u}+\vc{v}+\vc{w}$ without any parenthesis.
Interactive Illustration 2.13: This interactive illustration shows the addition of three vectors, shown to the left. The vectors can be moved around as usual, and the interactive illustrated may be advanced by clicking/touching Forward.
Interactive Illustration 2.13: Finally, the black vector is shown, which is the sum of the three vectors. Recall that the vectors to the left can be moved around by clicking close to the tip of the vectors and moving the mouse while pressing. As an exercise, try to make the three vectors sum to zero so a triangle appears to the right.
$\vc{u}$
$\vc{v}$
$\vc{w}$
It is often useful to be able to calculate the middle point of two points. This is described in the following theorem.
Theorem 2.2: The Middle Point Formula
$A$
$B$
$M$
$O$
Assume that $M$ is the middle point of the line segment that goes between $A$ and $B$ as shown in the illustration to the right. Assume $O$ is another point. The vector $\overrightarrow{OM}$, i.e., from $O$ to $M$, can be written as
$$\overrightarrow{OM} = \frac{1}{2}(\overrightarrow{OA} + \overrightarrow{OB}).$$ (2.9)
The vector $\overrightarrow{OM}$ is the sum of $\overrightarrow{OA}$ and $\overrightarrow{AM}$
$$\overrightarrow{OM} = \overrightarrow{OA} + \overrightarrow{AM}.$$ (2.10)
Another way of saying this is that if you start in $O$ and want to end up in $M$, you can either go first from $O$ to $A$ and then from $A$ to $M$ (right hand side of the equation) or go directly from $O$ to $M$ (left hand side of equation).
By going via $B$ instead we get
$$\overrightarrow{OM} = \overrightarrow{OB} + \overrightarrow{BM}.$$ (2.11)
Summing these two equations together gives
$$2\overrightarrow{OM} = \overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{AM} + \overrightarrow{BM}.$$ (2.12)
Since $\overrightarrow{BM}$ is equally long as $\overrightarrow{AM}$ but has opposite direction it must hold that $\overrightarrow{BM} = -\overrightarrow{AM}$. Inserting this in the equation above and dividing by two gives
$$\overrightarrow{OM} = \frac{1}{2}(\overrightarrow{OA} + \overrightarrow{OB}).$$ (2.13)
Sometimes you also see the shorter notation
$$M = \frac{1}{2}(A + B)$$ (2.14)
$\square$
Example 2.4: Sierpinski Triangles using the Middle Point Theorem
We will now show how the middle point formula can be used to generate a geometrical figure called the Sierpinski triangle. Assume we have a triangle consisting of three points, $A$, $B$, and $C$. Using Theorem 2.2, the midpoints of each edge can now be computed. These midpoints can be connected to form four new triangles, where the center triangle is empty. If this process is repeated for each new non-empty triangle, then we arrive at the Sierpinski triangle. This is shown in Interactive Illustration 2.15 below.
Interactive Illustration 2.15: In this illustration, we will show how a geometrical figure, called the Sierpinski triangle, is constructed. We start with three (moveable) points, $A$, $B$, and $C$, connected to form a triangle.
Interactive Illustration 2.15: In this illustration, we will show how a geometrical figure, called the Sierpinski triangle, is constructed. We start with three (moveable) points, $\hid{A}$, $\hid{B}$, and $\hid{C}$, connected to form a triangle.
$A$
$B$
$C$
$M_1 = \frac{1}{2}(A + B)$
$M_2 = \frac{1}{2}(B + C)$
$M_3 = \frac{1}{2}(A + C)$
Example 2.5: Center of Mass Formula
$A$
$B$
$C$
$A'$
$O$
$M$
$2$
$1$
$B'$
$2$
$1$
$M$
In the triangle $ABC$, the point $A'$ is on the mipoint between $B$ and $C$. The line segment from $A$ to $A'$ is called the median of $A$. Let $M$ be the point that divides the median of $A$ in a proportion 2 to 1, shown in the illustration to the right.
The center of mass formula states that
$$\pvec{OM} = \frac{1}{3}(\pvec{OA} + \pvec{OB} + \pvec{OC}).$$ (2.15)
This formula can be proved as follows. We can go from $O$ to $M$ either directly or via $A$, hence
$$\pvec{OM} = \pvec{OA} + \pvec{AM}.$$ (2.16)
It is also possible to arrive at $M$ via $A’$. This gives
$$\pvec{OM} = \pvec{OA'} + \pvec{A'M}.$$ (2.17)
One of the assumptions is that $\pvec{A'M}$ is half the length of $\pvec{AM}$ and of opposite direction, and therefore, it holds that $\pvec{A'M} = -\frac{1}{2}\pvec{AM}$. Inserting this in the equation above gives
$$\pvec{OM} = \pvec{OA'} -\frac{1}{2}\pvec{AM}.$$ (2.18)
We can eliminate $\pvec{AM}$ by adding Equation (2.16) to two times Equation (2.18)
$$\pvec{3OM} = \pvec{OA} + 2\pvec{OA'}.$$ (2.19)
Since $A'$ is the mid point of $B$ and $C$, we known from the middle point formula that $\pvec{OA'} = \frac{1}{2}(\pvec{OB} + \pvec{OC})$. Inserting this in the equation above gives
$$\pvec{3OM} = \pvec{OA} + 2\frac{1}{2}(\pvec{OB} + \pvec{OC}),$$ (2.20)
which simplifies as
$$\pvec{OM} = \frac{1}{3}(\pvec{OA} + \pvec{OB} + \pvec{OC}).$$ (2.21)
This completes the proof.
Note that since the formula is symmetric, it works just as well on the median to $B$. The same point, $M$, will divide the median going from $B$ to $B'$ in proportion $2:1$. This can be seen by pressing forward in the interactive illustration.
This point is also called the center of mass. If the triangle was cut out of cardboard, this is the point where it would balance of the top of a pencil, which exaplains the name "center for mass". Also, if equal point masses were placed in the points $A$, $B$ and $C$, $M$ is the point where they would balance out.
Most of you are probably familiar with the concept of a coordinate system, such as in the map in the first step of Interactive Illustration 2.17 below. In this first step, the axes are perpendicular and of equal length, but this is a special case, as can be seen by pressing Forward. This section will describe the general coordinate systems, and the interaction between vectors, bases, and coordinates.
Interactive Illustration 2.17: A map with an ordinary coordinate system. The center of the map is marked as the origin, and we show the $x$-axis as a horizontal arrow, and the $y$-axis as a vertical arrow. These axes are locally similar to the longitude and latitude, but not on a global scale since the earth is not flat. Press/click Forward to continue to the next stage.
Interactive Illustration 2.17: A map with an ordinary coordinate system. The center of the map is marked as the origin, and we show the $\hid{x}$-axis as a horizontal arrow, and the $\hid{y}$-axis as a vertical arrow. These axes are locally similar to the longitude and latitude, but not on a global scale since the earth is not flat. Press/click Forward to continue to the next stage.
$x$
$y$
$x$
$y$
$x$
$y$
$x$
$y$
$\vc{e}_1$
$\vc{e}_2$
Next, we define how coordinates can be described in one, two, and three dimensions. This is done with the following set of theorems.
Theorem 2.3: Coordinate in One Dimension
$\vc{e}$
$\textcolor{#aa0000}{\vc{v}}$
Let $\vc{e}$ be a non-zero vector on a straight line. For each vector, $\vc{v}$, on the line, there is only one number, $x$, such that
$$\vc{v} = x \vc{e}.$$ (2.22)
(The vector, $\vc{v}$, in the figure to the right can be moved around.)
If $\vc{e}$ and $\vc{v}$ have the same direction, then choose $x=\ln{\vc{v}}/\ln{\vc{e}}$, and if $\vc{e}$ and $\vc{v}$ have opposite directions, then set $x=-\ln{\vc{v}}/\ln{\vc{e}}$. Finally, if $\vc{v}=\vc{0}$, then $x=0$. It follows from the definition of scalar vector multiplication, $x\vc{e}$, that $x$ is the only number that fulfils $\vc{v} = x\vc{e}$.
$\square$
Note that we say that $\vc{e}$ is a basis vector, and that $x$ is the coordinate for $\vc{v}$ in the basis of $\{\vc{e}\}$.
So far, this is not very exciting, but the next step makes this much more useful.
Theorem 2.4: Coordinates in Two Dimensions
$\vc{e}_1$
$\textcolor{#aa0000}{\vc{v}}$
$\vc{e}_2$
$O$
$P_1$
$P_2$
Let $\vc{e}_1$ and $\vc{e}_2$ be two non-parallel vectors (which both lie in a plane). For every vector, $\vc{v}$, in this plane, there is a single coordinate pair, $(x,y)$, such that
$$\vc{v} = x\vc{e}_1 + y\vc{e}_2.$$ (2.23)
(The vectors $\vc{v}$, $\vc{e}_1$, and $\vc{e}_2$ can be moved around in the figure.)
For this proof, we will use Interactive Illustration 2.19. As can be seen, $P_1$ was obtained by drawing a line, parallel to $\vc{e}_2$, from the tip point of $\vc{v}$ until it collides with the line going through $\vc{e}_1$. Similarly, $P_2$ is obtained by drawing a line, parallel to $\vc{e}_1$, from the tip point of $\vc{v}$ until it collides with the line going through $\vc{e}_2$. It is clear that
$$\vc{v} = \overrightarrow{O P_1} + \overrightarrow{O P_2}.$$ (2.24)
Now, let us introduce $\vc{u} = \overrightarrow{O P_1}$ and $\vc{w} = \overrightarrow{O P_2}$. Using Theorem 2.3 on $\vc{u}$ with $\vc{e}_1$ as basis vector, we get $\vc{u} = x \vc{e}_1$. Similarly, for $\vc{w}$ with $\vc{e}_2$ as basis vector, $\vc{w} = y \vc{e}_2$ is obtained. Hence, the vector $\vc{v}$ can be expressed as
$$\vc{v} = \vc{u} + \vc{w} = x \vc{e}_1 + y \vc{e}_2.$$ (2.25)
It remains to prove that $x$ and $y$ are unique in the representation of $\vc{v}$. If the representation would not be unique, then another coordinate pair, $(x',y')$, would exist such that
$$\vc{v} = x' \vc{e}_1 + y' \vc{e}_2.$$ (2.26)
Combining (2.25) and (2.26), we get
\begin{gather} x \vc{e}_1 + y \vc{e}_2= x' \vc{e}_1 + y' \vc{e}_2 \\ \Longleftrightarrow \\ (x-x') \vc{e}_1 = (y'-y) \vc{e}_2. \end{gather} (2.27)
The conclusion from this is that if another representation, $(x',y')$, would exist, then $\vc{e}_1$ and $\vc{e}_2$ would be parallel (bottom row in (2.27)). For instance, if $x'$ is different from $x$, then $(x-x') \neq 0$ and both sides can be divided by $(x-x')$, which gives us
\begin{gather} \vc{e}_1 = \frac{(y'-y)}{(x-x')} \vc{e}_2, \end{gather} (2.28)
which can be expressed as $\vc{e}_1 = k \vc{e}_2$ with $k = \frac{(y'-y)}{(x-x')}$. However, according to the corollary to Definition 2.4 this means that $\vc{e}_1$ and $\vc{e}_2$ are parallel, contradicting the assumption in Theorem 2.4. The same reasoning applies if $y' - y \neq 0$. Hence, we have shown that there is only one unique pair, $(x,y)$, for each vector, $\vc{v}$, by using a proof by contradiction.
$\square$
Note that we say that $\vc{e}_1$ and $\vc{e}_2$ are basis vectors, and that $x$ and $y$ are the coordinates for $\vc{v}$ in the basis of $\{\vc{e}_1,\vc{e}_2\}$.
Next, we will extend this to three dimensions as well.
Theorem 2.5: Coordinates in Three Dimensions
Let $\vc{e}_1$, $\vc{e}_2$, and $\vc{e}_3$ be three non-zero basis vectors, and that there is no plane that is parallel with all three vectors. For every vector, $\vc{v}$, in the three-dimensional space, there is a single coordinate triplet, $(x,y,z)$, such that
$$\vc{v} = x\vc{e}_1 + y\vc{e}_2 + z\vc{e}_3.$$ (2.29)
Start by placing all the vectors $\vc{v}$, $\vc{e}_1$, $\vc{e}_2$ and $\vc{e}_3$ so that they start in the origin according to Interactive Illustration 2.20. Let $\pi_{12}$ be the plane through $O$ that contains $\vc{e}_1$ and $\vc{e}_2$, and let $P$ be the point at the tip of $\vc{v}$, i.e., $\vc{v} = \overrightarrow{OP}$.
$O$
$P_{12}$
$\vc{e}_1$
$\vc{e}_2$
$\vc{e}_3$
$P$
$\pi_{12}$
Interactive Illustration 2.20: Starting with the three vectors $\vc{e}_1$, $\vc{e}_2$ and $\vc{e}_3$, all placed with their tails in a point $O$.
Interactive Illustration 2.20: In summary, going from $\hid{O}$ to $\hid{P}$ can be done by first going to $\hid{P_{12}}$: $\hid{\overrightarrow{OP} = \overrightarrow{OP_{12}} + \overrightarrow{P_{12}P}}$. These two terms can in turn be exchanged using $\hid{\overrightarrow{OP_{12}} = x\vc{e}_1 + y \vc{e}_2}$ and $\hid{\overrightarrow{P_{12}P} = z\vc{e}_3}$. Thus $\hid{\overrightarrow{OP} = \overrightarrow{OP_{12}} + \overrightarrow{P_{12}P} = x\vc{e}_1 + y \vc{e}_2 + z\vc{e}_3}$.
Draw a line from $P$ parallel with $\vc{e}_3$ that intersects the plane $\pi_{12}$ in the point $P_{12}$. It is now clear that we can write $\vc{v}$ as the sum
$$\vc{v} = \overrightarrow{OP} = \overrightarrow{OP_{12}} + \overrightarrow{P_{12}P}.$$ (2.30)
However, according to the Theorem 2.4 (two dimensions), $\overrightarrow{OP_{12}}$ can be written as $\overrightarrow{OP_{12}} = x \vc{e}_1 + y \vc{e}_2$, and according to Theorem 2.3 (on dimension), $\overrightarrow{P_{12}P}$ can be written as $\overrightarrow{P_{12}P} = z \vc{e}_3$. Hence, there exist three numbers $x$, $y$ and $z$, such that
$$\vc{v} = x \vc{e}_1 + y \vc{e}_2 + z \vc{e}_3.$$ (2.31)
We must now prove that $x$, $y$ and $z$ are the only numbers for which this is possible. Assume that there is another set of numbers, $x'$, $y'$ $z'$, that also generates the same vector, $\vc{v}$, that is
$$\vc{v} = x' \vc{e}_1 + y' \vc{e}_2 + z' \vc{e}_3.$$ (2.32)
Combining (2.31) and (2.32) gives
$$x \vc{e}_1 + y \vc{e}_2 + z \vc{e}_3 = x' \vc{e}_1 + y' \vc{e}_2 + z' \vc{e}_3.$$ (2.33)
This can be rearranged to
$$(x-x') \vc{e}_1 + (y-y') \vc{e}_2 + (z-z') \vc{e}_3 = 0.$$ (2.34)
If the new set ($x'$, $y'$, $z'$) is to be different from the other ($x$, $y$, $z$), at least one of the terms must now be different from zero. Assume it is $(x-x')$ (or else, rename the vectors and scalars so that it becomes this term). This means that we can divide by $(x-x')$ to obtain
$$\vc{e}_1 = - \frac{(y-y')}{(x-x')} \vc{e}_2 - \frac{(z-z')}{(x-x')}\vc{e}_3,$$ (2.35)
which also can be expressed as
$$\vc{e}_1 = \alpha \vc{e}_2 + \beta \vc{e}_3,$$ (2.36)
where $\alpha = - \frac{(y-y')}{(x-x')}$ and $\beta = - \frac{(z-z')}{(x-x')}$. However, this means that $\vc{e}_1$ lies in the same plane as $\vc{e}_2$ and $\vc{e}_3$ (see Theorem 2.4), which contradicts the assumption that there is no plane that is parallel to $\vc{e}_1$, $\vc{e}_2$ and $\vc{e}_3$. Thus there cannot exist any other set of values, $x'$, $y'$, $z'$, that satisfies the equation and therefore the proof is complete.
$\square$
Similarly as before, we say that $\vc{e}_1$, $\vc{e}_2$, and $\vc{e}_3$ are basis vectors, and that $x$, $y$, and $z$ are the coordinates for $\vc{v}$ in the basis of $\{\vc{e}_1,\vc{e}_2,\vc{e}_3\}$.
Now, we can finally see where the vector representation using coordinates comes from. If we assume that a certain basis, $\{\vc{e}_1, \vc{e}_2, \vc{e}_3\}$, is used, then we can write a three-dimensional vector, $\vc{v}$, as
$$\vc{v} = v_x \vc{e}_1 + v_y \vc{e}_2 + v_z \vc{e}_3= \begin{pmatrix} v_x \\ v_y \\ v_z \end{pmatrix},$$ (2.37)
where we have used $v_x$ instead of $x$, $v_y$ instead of $y$, and $v_z$ instead of $z$. This is to make it simpler to mix several different vectors, and still be able to access the individual components. Note that the right-hand expression shows the vector as a column of three numbers, the $x$-coordinate on top, the $y$-coordinate in the middle, and the $z$-coordinate at the bottom. This is such an important notation, so we have summarized it into the following definition:
Definition 2.5: Column Vector Notation
Given a basis, a column vector, $\vc{v}$, in $n$ dimensions (we have used $n\in [1,2,3]$) is a column of $n$ scalar values. These scalar components, sometimes called vector elements, of the vector can either be numbered, i.e., $v_1$, $v_2$, and $v_3$, or we can use $x$, $y$, and $z$ as subscripts when that is more convenient. The notation is:
\begin{gather} \underbrace{ \vc{u} = \begin{pmatrix} u_x \end{pmatrix} = \begin{pmatrix} u_1 \end{pmatrix}}_{\text{1D vector}}, \spc\spc \underbrace{ \vc{v} = \begin{pmatrix} v_x \\ v_y \end{pmatrix} = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix}}_{\text{2D vector}}, \spc\spc \\ \underbrace{ \vc{w} = \begin{pmatrix} w_x \\ w_y \\ w_z \end{pmatrix} = \begin{pmatrix} w_1 \\ w_2 \\ w_3 \end{pmatrix}}_{\text{3D vector}}, \end{gather} (2.38)
where $\vc{u} = u_x \vc{e}_1$, $\vc{v} = v_x \vc{e}_1 + v_y \vc{e}_2$, and $\vc{w} = w_x \vc{e}_1 + w_y \vc{e}_2 + w_z \vc{e}_3$.
We also use a more compact way of writing vectors, which is convenient when writing vectors in text, for example: $\vc{w} = \bigl(w_1,w_2,w_3\bigr)$, which means the same as above (notice the commas between the vector elements).
Column vectors, per the definition above, is the type of vectors that we use mostly throughout this book. Hence, when we say "vector", we mean a "column vector". However, there is also another type of vectors, namely, row vectors. As can be deduced from the name, it is simply a row of scalar values, instead of a column of scalar values. An example of a row vector is:
$$\bigl(1\spc 2\spc 5 \bigr).$$ (2.39)
Any vector, be it row or column, can be transposed, which means that a row vector turns into a column vector, and a column vector turns into a row vector. The notation for a transposed vector is: $\vc{v}^T$. An example is shown below:
$$\vc{v} = \begin{pmatrix} 1\\ 2\\ 5 \end{pmatrix}, \spc\spc\spc \vc{v}^\T = \bigl(1\spc 2\spc 5 \bigr).$$ (2.40)
We summarize the transposing of a vector in the following definition:
Definition 2.6: Transpose of a Vector
The transpose of a vector, $\vc{v}$, is denoted by $\vc{v}^\T$, and turns a column vector into a row vector, and a row vector into a column vector. The order of the vector components is preserved.
Note that with this definition, we can transpose a vector twice, and get back the same vector, i.e., $\bigl(\vc{v}^T\bigr)^T = \vc{v}$. Next, we also summarize the row vector definition below:
Definition 2.7: Row Vector Notation
A row vector is expressed as a transposed column vector, as shown below:
$$\underbrace{ \vc{v}^\T = \bigl( v_x \spc v_y \bigr) }_{\text{2D row vector}}, \spc \spc \underbrace{ \vc{w}^\T = \bigl( w_x \spc w_y \spc w_z \bigr) }_{\text{3D row vector}}.$$ (2.41)
Notice that a row vector never has any commas between the vector elements. This is reserved for the compact notation for column vector (see Definition 2.5).
Now, let us assume that we have two vectors, $\vc{u}$ and $\vc{v}$, in the same basis, i.e.,
$$\vc{u} = u_x \vc{e}_1 + u_y \vc{e}_2 + u_z \vc{e}_3= \begin{pmatrix} u_x \\ u_y \\ u_z \end{pmatrix} \spc\spc \text{and} \spc\spc \vc{v} = v_x \vc{e}_1 + v_y \vc{e}_2 + v_z \vc{e}_3= \begin{pmatrix} v_x \\ v_y \\ v_z \end{pmatrix}.$$ (2.42)
The addition, $\vc{u}+\vc{v}$, becomes:
\begin{align} \vc{u}+\vc{v} &= u_x \vc{e}_1 + u_y \vc{e}_2 + u_z \vc{e}_3 + v_x \vc{e}_1 + v_y \vc{e}_2 + v_z \vc{e}_3 \\ &=(u_x+v_x)\vc{e}_1 + (u_y+v_y)\vc{e}_2 + (u_z+v_z)\vc{e}_3 \\ &= \begin{pmatrix} u_x+v_x \\ u_y+v_y \\ u_z+v_z \end{pmatrix}. \end{align} (2.43)
As can be seen, the vector addition boils down to simple component-wise scalar addition. For scalar vector multiplication, $k\vc{v}$, we have:
\begin{align} k\vc{v} &= k (v_x \vc{e}_1 + v_y \vc{e}_2 + v_z \vc{e}_3) \\ &= (k v_x) \vc{e}_1 + (k v_y) \vc{e}_2 + (k v_z) \vc{e}_3\\ &= \begin{pmatrix} k v_x \\ k v_y \\ k v_z \end{pmatrix}, \end{align} (2.44)
and here we see that each component of the vector is multiplied by $k$.
Example 2.6: Vector Addition and Scalar Multiplication using Coordinates
Assume we have the following vectors in the same basis:
$$\vc{u} = \left( \begin{array}{r} 3 \\ -4 \\ 7 \end{array} \right), \spc \spc \vc{v} = \left( \begin{array}{r} 1 \\ 2 \\ 5 \end{array} \right), \spc \spc \text{and} \spc \spc \vc{w} = \left( \begin{array}{r} 2 \\ -1 \\ 6 \end{array} \right),$$ (2.45)
and that we now want to evaluate $\vc{u} + \vc{v} - 2\vc{w}$. As we have seen above, vector addition is simply a matter of adding the vector elements:
$$\vc{u}+\vc{v} = \left( \begin{array}{r} 3 \\ -4 \\ 7 \end{array} \right) + \left( \begin{array}{r} 1 \\ 2 \\ 5 \end{array} \right) = \left( \begin{array}{r} 3+1 \\ -4+2 \\ 7+5 \end{array} \right) = \left( \begin{array}{r} 4 \\ -2 \\ 12 \end{array} \right).$$ (2.46)
We can also scale a vector by a scalar value, e.g., $k=2$:
$$2\vc{w} = 2 \left( \begin{array}{r} 2 \\ -1 \\ 6 \end{array} \right) = \left( \begin{array}{c} 2\cdot 2 \\ 2\cdot (-1) \\ 2\cdot 6 \end{array} \right) = \left( \begin{array}{r} 4 \\ -2 \\ 12 \end{array} \right),$$ (2.47)
which means that $\vc{u} + \vc{v} - 2\vc{w} = \vc{0}$.
In many calculations, one uses a simple and intuitive basis called the standard basis, which is defined as follows.
Definition 2.8: Standard Basis
The standard basis in this book is as follows for two and three dimensions, that is,
\begin{gather} \underbrace{ \vc{e}_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix},\ \vc{e}_2 = \begin{pmatrix} 0 \\ 1 \end{pmatrix} }_{\mathrm{two-dimensional\ standard\ basis}} \ \ \mathrm{and} \\ \ \\ \underbrace{ \vc{e}_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix},\ \vc{e}_2 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix},\ \vc{e}_3 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}. }_{\mathrm{three-dimensional\ standard\ basis}} \end{gather} (2.48)
In general, for an $n$-dimensional standard basis, the basis vectors, $\vc{e}_i$ have vector elements which are all zeroes, except the $i$:th element, which is a one.
In Chapter 3, we will discuss different types of bases, and we will see that the standard basis is, in fact, an orthonormal basis (Section 3.3).
Example 2.7: Addition in the Standard Basis
In this example, we will illustrate how vector addition is done in the standard basis in order to increase the reader's intuition about addition. See Interactive Illustration 2.21. Recall that the standard basis vector in two dimensions are $\vc{e}_1=(1,0)$ and $\vc{e}_2=(0,1)$.
Interactive Illustration 2.21: Here, two vectors are shown in the standard basis. Click/touch Forward to continue.
Interactive Illustration 2.21: The coordinates of the blue vector are simply the sum of the respective coordinates of the red and green vectors. For example, the $\hid{x}$-coordinate of the blue vector is simply the addition of the $\hid{x}$-coordinates of the red and green vectors. Recall that the vector can be moved around by clicking/touching close to the tip of the red or green vector, and the drag.
$\vc{u}$
$\vc{v}$
$\vc{u}+\vc{v}$
Next, two intuitive examples will be given on the topics of coordinate systems, basis vectors, uniqueness, and coordinates.
Example 2.8: Same Point Expressed In Different Bases
Note that the same point will have different coordinates when different basis vectors are used, as shown in Interactive Illustration 2.22. Note in the illustration that when the basis vectors change, the coordinates change too, but the point stays at the same place all the time.
Interactive Illustration 2.22: Here, we show how the same point, $P$, can be expressed in different coordinate systems. In the first step, we have an ordinary coordinate system where the first basis vector, $\vc{e}_1$ (thick red arrow) and the second basis vector, $\vc{e}_2$ (green thick arrow) make a right angle, and they are of equal length. The coordinates $(2,1)$ mean that if we start from the origin, go two steps along $\vc{e}_1$ and one step along $\vc{e}_2$, with the result that we end up in $P$.
Interactive Illustration 2.22: Here is another example, where the coordinates for $\hid{P}$ equals $\hid{(3,1)}$. Note that you can move the two basis vectors and $\hid{P}$, while the coordinates will adjust accordingly. Note also that if you place the two basis vectors so that they become almost parallel, then the coordinates start to rise dramatically and erratically. This makes sense, since if the vectors were indeed completely parallel, you would only be able to represent points on the line going from the origin along the first basis vector (or the second, which would be equivalent). If they are slightly different, this small difference must be enhanced by a large number in order to reach $\hid{P}$.
$O$
$P =$
$\vc{e}_1$
$\vc{e}_2$
$\vc{e}_2$
$\vc{e}_2$
Going back to stage two in Interactive Illustration 2.22, it is obvious that adding the two basis vectors together brings us the vector $\overrightarrow{OP}$ exactly, so $\overrightarrow{OP} = 1.0 \vc{e}_1 + 1.0 \vc{e}_2$ must hold. Thus, $(1, 1)$ is a valid coordinate pair for the point $P$. However, one may ask whether there are any other coordinates that will also describe the point $P$, now that the basis vectors no longer need to make a right angle. The answer to this is no, as we have seen in the proof to Theorem 2.4. A bit more intuition about why this is so, can be obtained from Interactive Illustration 2.23.
Interactive Illustration 2.23: In this interactive figure, the fat arrows represent the basis vectors, $\vc{e}_1$ and $\vc{e}_2$. The point, $P$, has the coordinates $(2.5, 1.0)$ since, if you start in the origin, you need to go $2.5$ steps along $\vc{e}_1$ (thin red arrow) and one step along $\vc{e}_2$ (thin green arrow) to get to $P$. But could other coordinates also work? Press/click Forward to advance the illustration.
Interactive Illustration 2.23: In this interactive figure, the fat arrows represent the basis vectors, $\hid{\vc{e}_1}$ and $\hid{\vc{e}_2}$. The point, $\hid{P}$, has the coordinates $\hid{(2.5, 1.0)}$ since, if you start in the origin, you need to go $\hid{2.5}$ steps along $\hid{\vc{e}_1}$ (thin red arrow) and one step along $\hid{\vc{e}_2}$ (thin green arrow) to get to $\hid{P}$. But could other coordinates also work? Press/click Forward to advance the illustration.
In this chapter, we have introduced the notion of vectors. The definitions of a vector in Section 2.5 and the basic operations, such as vector addition (Definition 2.3) and multiplication with a scalar (Definition 2.4), have been defined geometrically. We then showed that these two operations fulfill a number of properties in Theorem 2.1. This definition works for $\R^1$, $\R^2$, and $\R^3$. For higher dimensions, it is difficult for us to use the geometric definition. The notion of a two-dimensional and three-dimensional vector is in itself very useful, but geometric vectors are also a stepping stone for understanding general linear spaces or vector spaces. This more general theory is extremely useful for modeling and understanding problems when we have more than three unknown parameters. The reader may want to skip the following section, and revisit it later depending on his/her needs.
In this section, we will first give a definition of a $\R^n$.
Definition 2.9: Real Coordinate Space
The vector space $\R^n$ is defined as $n$-tuples $\vc{u} = (u_1, u_2, \ldots, u_n)$, where each $u_i$ is a real number. It is a vector space over the real numbers $\R$, where vector addition $\vc{u}+\vc{v}$ is defined as $\vc{u}+\vc{v} = (u_1+v_1, u_2+v_2, \ldots, u_n+v_n)$ and scalar-vector multiplications is defined as $k\vc{v} = (k v_1, k v_2, \ldots, k v_n)$, where $k\in \R$.
Note that using these definitions for vector addition and scalar-vector multiplication, the properties of Theorem 2.1 all hold.
Example 2.10:
Let $\vc{u}=(1,2,3,4,5)$ and $\vc{v}=(5,4,3,2,1)$ be two vectors in $\R^5$. What is $\vc{u}+\vc{v}$, $3\vc{u}$ and $3\vc{u}+3\vc{v}$?
$$\vc{u}+\vc{v} = (1,2,3,4,5) + (5,4,3,2,1) = (1+5,2+4,3+3,4+2,5+1) = (6,6,6,6,6)$$ (2.49)
$$3\vc{u}= 3(1,2,3,4,5) = (3 \cdot 1,3 \cdot 2,3 \cdot 3,3 \cdot 4,3 \cdot 5) = (3,6,9,12,15)$$ (2.50)
$$3\vc{u}+3\vc{v} = 3 (\vc{u}+\vc{v}) = 3 (6,6,6,6,6) = (18,18,18,18,18)$$ (2.51)
In this last step, the result ($\vc{u}+\vc{v}$) from Equation (2.49) was used.
Definition 2.10: Basis in $\R^n$
A basis in $\R^n$ is a set of vectors $\{\vc{e}_1, \ldots, \vc{e}_m\}$ in so that for every vector $\vc{u}\in\R^n$, there is a unique set of coordinates $(u_1, \ldots, u_m)$ so that
$$\vc{u} = \sum_{i=1}^m u_i \vc{e}_i.$$ (2.52)
Example 2.11: Canonical Basis in $\R^n$
The canonical basis in $\R^n$ is the following set of basis vectors
$$\begin{cases} \begin{array}{ll} \vc{e}_1 &= (1, 0, \ldots, 0), \\ \vc{e}_2 &= (0, 1, \ldots, 0), \\ \vdots & \\ \vc{e}_n &= (0, 0, \ldots, 1). \end{array} \end{cases}$$ (2.53)
#### 2.6.1 The General Definition
We will now present an abstract definition of a vector space. Then we will show that any finite-dimensional vector space over $\R$ is in fact 'the same as' $\R^n$ that we defined earlier in Definition 2.9.
Definition 2.11: Vector space
A vector space consists of a set $V$ of objects (called vectors) and a field $F$, together with a definition of vector addition and multiplication of a scalar with a vector, in such a way that the properties of Theorem 2.1 holds.
A vector space consists of a set $V$ of objects. As we shall see in one example, the vector space is the set of images of size $m \times n$ pixels. In another example, the vector space is a set of polynomials up to degree $5$. The elements of the field $F$ are called scalars. A field is a set of objects where addition, subtraction, multiplication and division is well defined and follows the usual properties. Most often the field used is the set of real numbers $\R$ or the set of complex numbers $\mathbb{C}$, but one could use more exotic fields, such as integers modulu a prime number, e.g., $\mathbb{Z}_3$.
Example 2.12: Polynomials up to degree 2
Polynomials in $x$ up to degree 2 with real coefficients is a vector space over $\R$. Here if $u = u_0 + u_1 x + u_2 x^2$ and $v = v_0 + v_1 x + v_2 x^2$, where each coefficient $u_i$ and $v_i$ is a real number. Here vector addition $u+v$ is defined as $u+v = (u_0+v_0) + (u_1+v_1) x + (u_2+v_2) x^2$ and scalar-vector multiplications is defined as $ku = k u_0 + k u_1 x + k u_2 x^2$.
Example 2.13: Gray-scale images
Gray-scale images, where each pixel intensity is a real number is a vector space over $\R$. Here if the pixel of the image $u$ at position $(i,j)$ has intensity $u_{i,j}$ and similarily if the pixel of the image $v$ at position $(i,j)$ has intensity $v_{i,j}$, then vector addition is defined as an image $u+v$ where the intensity of the pixel at position $(i,j)$ is $u_{i,j}+v_{i,j}$. The scalar-vector multiplications is defined as the image $ku$ where the pixel at position $(i,j)$ has intensity $k u_{i,j}$,
Example 2.14: $\mathbb{Z}_3$ Coordinate Space
The vector space $\mathbb{Z}_3^n$ is defined as n-tuples $\vc{u} = (u_1, u_2, \ldots, u_n)$, where each $u_i$ is one of the integers $0$, $1$, or $2$. It is a vector space over the integers $0$, $1$, and $2$. Vector addition $\vc{u}+\vc{v}$ is defined as $\vc{u}+\vc{v} = (u_1+v_1, u_2+v_2, \ldots, u_n+v_n)$ and scalar-vector multiplications is defined as $k\vc{v} = (k u_1, k u_2, \ldots, k u_n)$. Here the multiplications and additions of two scalars are done moduli 3.
Definition 2.12: Basis in Vector Space
A basis in a finite dimensional vector space $V$ over $F$ is a set of vectors $\{\vc{e}_1, \ldots, \vc{e}_m\}$ in so that for every vector $\vc{u} \in V$, there is a unique set of coordinates $(u_1, \ldots, u_m)$ with $u_i \in F$, so that
$$\vc{u} = \sum_{i=1}^m u_i \vc{e}_i.$$ (2.54)
The number, $m$, of basis vectors is said to be the dimension of the vector space. We will later show that this is a well-defined number for a given vector space.
Theorem 2.6: Vector in Vector Space
Let $V$ be a vector space over $\R^m$ and let $\{\vc{e}_1, \ldots, \vc{e}_m\}$ be a basis. Then each vector $\vc{u}$ can be identified with its coordinates $(u_1, \ldots, u_m)$.
In this way, one can loosely say that each $m$-dimensional vector space over $\R$ is 'the same thing' as $\R^m$.
The vector concept has been treated in this chapter, and the vector addition and scalar vector multiplication operations have been introduced. In addition, we have seen that these operations behave pretty much as expected, i.e., similar to how we calculate with real numbers. To make the vectors a bit more practical, the basis concept was introduced, and we saw how a, e.g., three-dimensional vector can be represented by three scalar numbers with respect to a certain basis. Finally, we also introduced the concept of a higher-dimensional vector space $\R^n$ very briefly. In Chapter 3, the dot product operation will be introduced. It is useful when measuring length and angles.
Chapter 1: Introduction (previous) Chapter 3: The Dot Product (next) | 2019-09-16T00:28:59 | {
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https://byjus.com/question-answer/the-adjacent-sides-of-a-parallelogram-have-length-20-cm-and-12-cm-the-distance/ | Question
# The adjacent sides of a parallelogram have length $$20\ cm$$ and $$12\ cm$$. The distance between the two shorter sides is $$5\ cm$$. Find the area of parallelogram and the distance between the longer sides.
Solution
## Area$$=b\times h$$$$=5\times 12=60㎠$$$$=20\times x$$$$\Rightarrow x=3㎝$$$$\therefore$$Area$$=60 \; {cm}^{2}$$; Distance between longer sides$$=3 \; cm$$Mathematics
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View More | 2022-01-21T14:02:36 | {
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https://benvitalenum3ers.wordpress.com/2015/04/13/triangular-number-system-of-equations/ | ## Triangular number – System of equations
An integer of the form $T_{n} = n(n+1)/2$ is called a Triangular number.
System of equations:
$T_{A} \; + \; T_{B} \; = \; T_{C}$
$T_{A} \; - \; T_{B} \; = \; T_{D}$
one solution is the pair $T_6, \; T_5$
$T_6 \; + \; T_5 \; = \; 21 \; + \; 15 \; = \; 36 \; = \; T_7$
$T_6 \; - \; T_5 \; = \; 21 \; - \; 15 \; = \; 6 \; = \; T_2$
Find other pairs.
Paul found:
$T_{18} \; + \; T_{14} \; = \; 171 \; + \; 105 \; = \; 276 \; = \; T_{23}$
$T_{18} \; - \; T_{14} \; = \; 171 \; - \; 105 \; = \; 66 \; = \; T_{11}$
$T_{37} \; + \; T_{27} \; = \; 703 \; + \; 378 \; = \; 1081 \; = \; T_{46}$
$T_{37} \; - \; T_{27} \; = \; 703 \; - \; 378 \; = \; 325 \; = \; T_{25}$
$T_{44} \; + \; T_{39} \; = \; 990 \; + \; 780 \; = \; 1770 \; = \; T_{59}$
$T_{44} \; - \; T_{39} \; = \; 990 \; - \; 780 \; = \; 210 \; = \; T_{20}$
$T_{86} \; + \; T_{65} \; = \; 3741 \; + \; 2145 \; = \; 5886 \; = \; T_{108}$
$T_{86} \; - \; T_{65} \; = \; 3741 \; - \; 2145 \; = \; 1596 \; = \; T_{56}$
$T_{91} \; + \; T_{54} \; = \; 4186 \; + \; 1485 \; = \; 5671 \; = \; T_{106}$
$T_{91} \; - \; T_{54} \; = \; 4186 \; - \; 1485 \; = \; 2701 \; = \; T_{73}$
$T_{116} \; + \; T_{104} \; = \; 6786 \; + \; 5460 \; = \; 12246 \; = \; T_{156}$
$T_{116} \; - \; T_{104} \; = \; 6786 \; - \; 5460 \; = \; 1326 \; = \; T_{51}$
$T_{132} \; + \; T_{125} \; = \; 8778 \; + \; 7875 \; = \; 16653 \; = \; T_{182}$
$T_{132} \; - \; T_{125} \; = \; 8778 \; - \; 7875 \; = \; 903 \; = \; T_{42}$
$T_{247} \; + \; T_{242} \; = \; 30628 \; + \; 29403 \; = \; 60031 \; = \; T_{346}$
$T_{247} \; - \; T_{242} \; = \; 30628 \; - \; 29403 \; = \; 1225 \; = \; T_{49}$
$T_{278} \; + \; T_{209} \; = \; 38781 \; + \; 21945 \; = \; 60726 \; = \; T_{348}$
$T_{278} \; - \; T_{209} \; = \; 38781 \; - \; 21945 \; = \; 16836 \; = \; T_{183}$
$T_{392} \; + \; T_{374} \; = \; 77028 \; + \; 70125 \; = \; 147153 \; = \; T_{542}$
$T_{392} \; - \; T_{374} \; = \; 77028 \; - \; 70125 \; = \; 6903 \; = \; T_{117}$
math grad - Interest: Number theory
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### 2 Responses to Triangular number – System of equations
1. Paul says:
I think you may have mislabelled the resultant T values, 36 is T8 and 6 is T3.
Here are a few I found, with the first T number <500
T6 + T5 = 21 + 15 = 36 = T8 &
T6 – T5 = 21 – 15 = 6 = T3
T18 + T14 = 171 + 105 = 276 = T23 &
T18 – T14 = 171 – 105 = 66 = T11
T37 + T27 = 703 + 378 = 1081 = T46 &
T37 – T27 = 703 – 378 = 325 = T25
T44 + T39 = 990 + 780 = 1770 = T59 &
T44 – T39 = 990 – 780 = 210 = T20
T86 + T65 = 3741 + 2145 = 5886 = T108 &
T86 – T65 = 3741 – 2145 = 1596 = T56
T91 + T54 = 4186 + 1485 = 5671 = T106 &
T91 – T54 = 4186 – 1485 = 2701 = T73
T116 + T104 = 6786 + 5460 = 12246 = T156 &
T116 – T104 = 6786 – 5460 = 1326 = T51
T132 + T125 = 8778 + 7875 = 16653 = T182 &
T132 – T125 = 8778 – 7875 = 903 = T42
T247 + T242 = 30628 + 29403 = 60031 = T346 &
T247 – T242 = 30628 – 29403 = 1225 = T49
T278 + T209 = 38781 + 21945 = 60726 = T348 &
T278 – T209 = 38781 – 21945 = 16836 = T183
T392 + T374 = 77028 + 70125 = 147153 = T542 &
T392 – T374 = 77028 – 70125 = 6903 = T117
Paul. | 2018-01-20T22:49:51 | {
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https://www.esaral.com/q/find-the-area-bounded-by-the-curve-69140/ | Find the area bounded by the curve
Question:
Find the area bounded by the curve $y=\sin x$ between $x=0$ and $x=2 \pi$
Solution:
The graph of y = sin x can be drawn as
$\therefore$ Required area = Area OABO + Area BCDB
$=\int_{0}^{\pi} \sin x d x+\left|\int_{\pi}^{2 \pi} \sin x d x\right|$
$=[-\cos x]_{0}^{\pi}+\left|[-\cos x]_{\pi}^{2 \pi}\right|$
$=[-\cos \pi+\cos 0]+|-\cos 2 \pi+\cos \pi|$
$=1+1+|(-1-1)|$
$=2+|-2|$
$=2+2=4$ units | 2022-12-06T17:16:19 | {
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https://www.fact-archive.com/encyclopedia/Kepler%27s_laws_of_planetary_motion | Search
# Kepler's laws of planetary motion
Johannes Kepler's primary contributions to astronomy/astrophysics were the three laws of planetary motion. Kepler derived these laws, in part, by studying the observations of Brahe. Isaac Newton would later design his laws of motion and universal gravitation and verify that Kepler's laws could be derived from them. The generic term for an orbiting object is "satellite".
Contents
## Kepler's laws of planetary motion
• Kepler's second law (1609): A line joining a planet and its star sweeps out equal areas during equal intervals of time.
## Kepler's first law
The orbit of a planet about a star is an ellipse with the star at one focus.
There is no object at the other focus of a planet's orbit. The semimajor axis, a, is half the major axis of the ellipse. In some sense it can be regarded as the average distance between the planet and its star, but it is not the time average in a strict sense, as more time is spent near apocentre than near pericentre.
### Connection with Newton's laws
Newton proposed that "every object in the universe attracts every other object along a line of the centers of the objects proportional to each objects mass, and inversely proportional to the square of the distance between the objects."
This section proves that Kepler's first law is consistent with Newton's laws of motion. We begin with Newton's law F=ma:
$m\frac{d^2\mathbf{r}}{dt^2} = f(r)\widehat{\mathbf{r}}$
Here we express F as the product of its magnitude and its direction. Recall that in polar coordinates:
$\frac{d\mathbf{r}}{dt} = \dot r\widehat{\mathbf{r}} + r\dot\theta\widehat{\theta}$
$\frac{d^2\mathbf{r}}{dt^2} = (\ddot r - r\dot\theta^2)\widehat{\mathbf{r}} + (r\ddot\theta + 2\dot r \dot\theta)\widehat{\theta}$
In component form we have:
$m(\ddot r - r\dot\theta^2) = f(r)$
$m(r\ddot\theta + 2\dot r\dot\theta) = 0$
Now consider the angular momentum:
$\mathbf{L} = \left|\mathbf{r} \times m\frac{d\mathbf{r}}{dt}\right| = \left|mr^2\dot\theta\right|$
So:
$r^2\dot\theta = \ell$
where $\ell=L/m$ is the angular momentum per unit mass. Now we substitute. Let:
$r = \frac{1}{u}$
$\dot r = -\frac{1}{u^2}\dot u = -\frac{1}{u^2}\frac{d\theta}{dt}\frac{du}{d\theta}= -\ell\frac{du}{d\theta}$
$\ddot r = -\ell\frac{d}{dt}\frac{du}{d\theta} = -\ell\dot\theta\frac{d^2u}{d\theta^2}= -\ell^2u^2\frac{d^2u}{d\theta^2}$
The equation of motion in the $\hat{\mathbf{r}}$ direction becomes:
$\frac{d^2u}{d\theta^2} + u = - \frac{1}{m\ell^2u^2}f\left(\frac{1}{u}\right)$
Newton's law of gravitation states that the central force is inversely proportional to the square of the distance so we have:
$\frac{d^2u}{d\theta^2} + u = \frac{k}{m\ell^2}$
where k is our proportionality constant.
This differential equation has the general solution:
$u = A\cos(\theta-\theta_0) + \frac{k}{m\ell^2}.$
Replacing u with r and letting θ0=0:
$r = \frac{1}{A\cos\theta + \frac{k}{m\ell^2}}$.
This is indeed the equation of a conic section with the origin at one focus. Q.E.D.
## Kepler's second law
A line joining a planet and its star sweeps out equal areas during equal intervals of time.
This is also known as the law of equal areas. Suppose a planet takes 1 day to travel from points A to B. During this time, an imaginary line, from the Sun to the planet, will sweep out a roughly triangular area. This same amount of area will be swept every day.
As a planet travels in its elliptical orbit, its distance from the Sun will vary. As an equal area is swept during any period of time and since the distance from a planet to its orbiting star varies, one can conclude that in order for the area being swept to remain constant, a planet must vary in velocity. Planets move fastest when at perihelion and slowest when at aphelion.
This law was developed, in part, from the observations of Brahe that indicated that the velocity of planets was not constant.
This law corresponds to the angular momentum conservation law in the given situation.
### Proof of Kepler's second law:
Assuming Newton's laws of motion, we can show that Kepler's second law is consistent. By definition, the angular momentum $\mathbf{L}$ of a point mass with mass m and velocity $\mathbf{v}$ is :
$\mathbf{L} \equiv m \mathbf{r} \times \mathbf{v}$.
where $\mathbf{r}$ is the position vector of the particle.
Since $\mathbf{v} = \frac{d\mathbf{r}}{dt}$, we have:
$\mathbf{L} = \mathbf{r} \times m\frac{d\mathbf{r}}{dt}$
taking the time derivative of both sides:
$\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F} = 0$
since the cross product of parallel vectors is 0. We can now say that $|\mathbf{L}|$ is constant.
The area swept out by the line joining the planet and the sun, is half the area of the parallelogram formed by $\mathbf{r}$ and $d\mathbf{r}$.
$dA = \begin{matrix}\frac{1}{2}\end{matrix} |\mathbf{r} \times d\mathbf{r}| = \begin{matrix}\frac{1}{2}\end{matrix} \left|\mathbf{r} \times \frac{d\mathbf{r}}{dt}dt\right| = \frac{\mathbf{|L|}}{2m}dt$
Since $|\mathbf{L}|$ is constant, the area swept out by is also constant. Q.E.D.
## Kepler's third law (harmonic law)
The square of the sidereal period of an orbiting planet is directly proportional to the cube of the orbit's semimajor axis.
P2 ~ a3
P = object's sidereal period in years
a = object's semimajor axis, in AU
Thus, not only does the length of the orbit increase with distance, also the orbital speed decreases, so that the increase of the sidereal period is more than proportional.
See the actual figures: attributes of major planets.
Newton would modify this third law, noting that the period is also affected by the orbiting body's mass, however typically the central body is so much more massive that the orbiting body's mass may be ignored. (See below.)
## Applicability
The laws are applicable whenever a comparatively light object revolves around a much heavier one because of gravitational attraction. It is assumed that the gravitational effect of the lighter object on the heavier one is negligible. An example is the case of a satellite revolving around Earth.
## Application
Assume an orbit with semimajor axis a, semiminor axis b, and eccentricity ε. To convert the laws into predictions, Kepler began by adding the orbit's auxiliary circle (that with the major axis as a diameter) and defined these points:
• c center of auxiliary circle and ellipse
• s sun (at one focus of ellipse); $\mbox{length }cs=a\varepsilon$
• p the planet
• z perihelion
• x is the projection of the planet to the auxiliary circle; then $\mbox{area }sxz=\frac ba\mbox{area }spz$
• y is a point on the circle such that $\mbox{area }cyz=\mbox{area }sxz=\frac ba\mbox{area }spz$
and three angles measured from perihelion:
• true anomaly $T=\angle zsp$, the planet as seen from the sun
• eccentric anomaly $E=\angle zcx$, x as seen from the centre
• mean anomaly $M=\angle zcy$, y as seen from the centre
Then
area cxz = area cxs + area sxz = area cxs + area cyz
$\frac{a^2}2E=a\varepsilon\frac a2\sin E+\frac{a^2}2M$
giving Kepler's equation
$M=E-\varepsilon\sin E$.
To connect E and T, assume r = length sp then
$a\varepsilon+r\cos T=a\cos E$ and rsinT = bsinE
$r=\frac{a\cos E-a\varepsilon}{\cos T}=\frac{b\sin E}{\sin T}$
$\tan T=\frac{\sin T}{\cos T}=\frac ba\frac{\sin E}{\cos E-\varepsilon}=\frac{\sqrt{1-\varepsilon^2}\sin E}{\cos E-\varepsilon}$
which is ambiguous but useable. A better form follows by some trickery with trigonometric identities:
$\tan\frac T2=\sqrt\frac{1+\varepsilon}{1-\varepsilon}\tan\frac E2$
(So far only laws of geometry have been used.)
Note that area spz is the area swept since perihelion; by the second law, that is proportional to time since perihelion. But we defined $\mbox{area }spz=\frac ab\mbox{area }cyz=\frac ab\frac{a^2}2M$ and so M is also proportional to time since perihelion—this is why it was introduced.
We now have a connection between time and position in the orbit. The catch is that Kepler's equation cannot be rearranged to isolate E; going in the time-to-position direction requires an iteration (such as Newton's method) or an approximate expression, such as
$E\approx M+\left(\varepsilon-\frac18\varepsilon^3\right)\sin M+\frac12\varepsilon^2\sin 2M+\frac38\varepsilon^3\sin 3M$
via the Lagrange reversion theorem. For the small ε typical of the planets (except Pluto) such series are quite accurate with only a few terms; one could even develop a series computing T directly from M.[1]
## Kepler's understanding of the laws
Kepler did not understand why his laws were correct; it was Isaac Newton who discovered the answer to this more than fifty years later. Newton, understanding that his third law of motion was related to Kepler's third law of planetary motion, devised the following:
$P^2 = \frac{4\pi^2}{G(m_1 + m_2)} \cdot a^3$
where:
Astronomers doing celestial mechanics often use units of years, AU, G=1, and solar masses, and with m2<<m1, this reduces to Kepler's form. SI units may also be used directly in this formula. | 2021-09-28T10:46:15 | {
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http://openstudy.com/updates/508197d8e4b041b051b8dd79 | ## Ilovemeowmeow 3 years ago What is the slope of a line that passes through the point (−2, 3) and is parallel to a line that passes through (3, 7) and (−2, −8)?
1. 3psilon
Use the slope formula with the two last points $\frac{ y_{2 } - y_{1}}{ x_{2}-x_{1} }$
2. 3psilon
That will give you the Slope for the last two points
3. Ilovemeowmeow
Okay let me work it out tell me if its correct..
4. 3psilon
So the slope would be $\frac{ -8-7 }{ -2-3} = -15/-5 ==$
5. 3psilon
Slope = 3
6. 3psilon
And we know that parallel lines have the same slope
7. Ilovemeowmeow
I did it and got 3! thank you!
8. 3psilon
Did you plug in the slope in y = mx + b form? and plug the first coordinates into that and solve for b? | 2016-02-08T21:32:00 | {
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https://groupprops.subwiki.org/wiki/Endomorphism_structure_of_general_linear_group_over_a_finite_field | # Endomorphism structure of general linear group over a finite field
This article gives specific information, namely, endomorphism structure, about a family of groups, namely: general linear group. This article restricts attention to the case where the underlying ring is a finite field.
View endomorphism structure of group families | View other specific information about general linear group | View other specific information about group families for rings of the type finite field
This article describes the endomorphism structure of the general linear group $GL(n,q)$ of degree $n$ over a finite field $\mathbb{F}_q$ of size $q$, where $q$ is a prime power. We will denote by $p$ the characteristic of the field, i.e., the prime of which $q$ is a power, and by $r$ the value $\log_pq$.
## Particular cases
Value of $n$ Comments Endomorphism structure page
1 $GL(1,q) \cong \mathbb{F}_q^*$, which in turn is isomorphic to the cyclic group $\mathbb{Z}/(q - 1)\mathbb{Z}$. Infer from endomorphism structure of cyclic groups
2 The transpose-inverse map is in the subgroup generated by the inner automorphisms and radial automorphisms. Specifically, it is the composite of the map $A \mapsto A/(\det A)$ and conjugation by $\begin{pmatrix} 0 & -1 \\ 1 & 0 \\\end{pmatrix}$. endomorphism structure of general linear group of degree two over a finite field
3 -- endomorphism structure of general linear group of degree three over a finite field
## Endomorphism structure
### Automorphism structure
The automorphism group can be described as a semidirect product in these equivalent ways. This applies only in the case $n > 2$. In the case $n = 2$, the transpose-inverse map is already in the product of the inner automorphism group and the radial automorphism group, and need not be introduced separately. See endomorphism structure of general linear group of degree two over a finite field. The case $n= 1$ is also qualitatively different.
Format Description of base Description of acting group
(Inner automorphism group) $\rtimes$ (Radial automorphism group $\times$ Field automorphism group $\times$ Cyclic group of order two generated by transpose-inverse map) inner automorphism group (the projective general linear group) outer automorphism group
(Inner automorphism $\rtimes$ Cyclic group of order two generated by transpose-inverse map) $\rtimes$ (Radial automorphism group $\times$ Field automorphism group). projective outer linear group no specific name
(Inner automorphism $\rtimes$ Field automorphism group) $\rtimes$ (Radial automorphism group $\times$ Cyclic group of order two generated by transpose-inverse map) general semilinear group
(Inner automorphism group $\times$ Radial automorphism group) $\rtimes$ (Field automorphism group $\times$ Cyclic group of order two generated by transpose-inverse map) no specific name no specific name
The components are described below briefly and then in detail (by default, $n > 2$, see right columns for explanation of how $n = 2$ and $n = 1$ differ).
Construct Value Order Comment Case $n = 2$ Case $n = 1$
automorphism group (Inner automorphism group) $\rtimes$ (Radial automorphism group $\times$ Field automorphism group $\times$ Cyclic group of order two generated by transpose-inverse map) $\frac{2r\varphi(n(q - 1))}{\varphi(n)}q^{\binom{n}{2}} \prod_{i=0}^{n-2} (q^{n-i} - 1)$ The subgroup generated by the transpose-inverse map does not get included as a separate factor because it is a composite of an inner automorphism and a radial automorphism. Thus, the factor of 2 in the numerator must be removed. The factor of $2r$ in the numerator needs to be removed, because both the transpose-inverse and the field automorphisms get included in the radial automorphisms.
inner automorphism group projective general linear group $PGL(n,q)$ $q^{\binom{n}{2}} \prod_{i=0}^{n-2} (q^{n-i} - 1)$ See order formulas for linear groups. Applies as is. Applies as is, though the group becomes a trivial group.
radial automorphism group Kernel of the natural homomorphism $(\mathbb{Z}/n(q - 1)\mathbb{Z})^* \to (\mathbb{Z}/n\mathbb{Z})^*$ $\frac{\varphi(n(q - 1))}{\varphi(n)}$ Here, $\varphi$ denotes the Euler totient function. See description of radial automorphism group below. Applies as in, though the denominator becomes trivial, so it simplifies to $\varphi(2(q - 1))$ Applies as is, though the denominator becomes trivial, so it simplifies to $\varphi(q - 1)$
field automorphism group (Galois group) $\operatorname{Aut}(\mathbb{F}_q) \cong \mathbb{Z}/r\mathbb{Z}$, generated by Frobenius $x \mapsto x^p$ $r$ See description below. Applies as is. Applies as is, but already included in radial automorphism group, so should not count it separately.
cyclic group generated by transpose-inverse map $\mathbb{Z}/2\mathbb{Z}$ 2 Applies as is, but is in direct product of inner automorphism group and radial automorphism group, so should not count it separately. Applies as is, but already included in radial automorphism group, so should not count it separately. Exception: Case that $q = 2$, in which case the group is a trivial group.
semidirect product of inner automorphism group and field automorphism group projective semilinear group $P\Gamma L(n,q)$ $rq^{\binom{n}{2}} \prod_{i=0}^{n-2} (q^{n-i} - 1)$ Applies as is. Applies as is.
semidirect product of inner automorphism group and cyclic group generated by transpose-inverse map projective outer linear group $POL(n,q)$ $2q^{\binom{n}{2}} \prod_{i=0}^{n-2} (q^{n-i} - 1)$ Applies as is. Applies as is, except in the case that $q = 2$, in which case the group is trivial.
outer automorphism group direct product of radial automorphism group, Galois group, and cyclic group of order two generated by transpose-inverse map $\frac{2r\varphi(n(q - 1))}{\varphi(n)}$ The subgroup generated by the transpose-inverse map does not get included as a separate factor because it is a composite of an inner automorphism and a radial automorphism. Thus, the factor of 2 in the numerator must be removed. The factor of $2r$ in the numerator needs to be removed, because both the transpose-inverse and the field automorphisms get included in the radial automorphisms.
#### Inner automorphism group
The inner automorphism group is the quotient of $GL(n,q)$ by its center, which is isomorphic to $\mathbb{F}_q^*$ (explicitly, it is the scalar matrices). This quotient group is the projective general linear group $PGL(n,q)$.
The radial automorphism group is a group of automorphisms of the form:
$A \mapsto A(\det A)^m$
where $m$ is considered modulo $q - 1$, and has the property that $mn + 1$ is invertible modulo $q - 1$. The latter condition is necessary for invertibility. Values of $m$ that fail the condition define endomorphisms that are not automorphisms.
The radial automorphism group is a group of automorphisms of the form:
$A \mapsto A (\det A)^m$
where $m$ is considered modulo $q - 1$, and has the property that $2m + 1$ is relatively prime to $q - 1$. The latter condition is necessary for invertibility. Values of $m$ that fail the condition define endomorphisms that are not automorphisms.
Since $m$ is defined modulo $q - 1$, $mn + 1$ is defined modulo $n(q - 1)$. The set of possible values for $mn + 1$ is the subgroup of the multiplicative group modulo $n(q - 1)$ comprising elements that are 1 modulo $n$. Further, the composition of radial automorphisms corresponds to multiplication of the corresponding $(mn + 1)$ values. Therefore, the radial automorphism group is isomorphic to the subgroup of the multiplicative group modulo $n(q - 1)$ comprising the elements that are 1 modulo $n$. We can see that this is the same as the kernel of the natural surjective homomorphism obtained by going modulo $n$:
$(\mathbb{Z}/n(q - 1)\mathbb{Z})^* \to (\mathbb{Z}/n\mathbb{Z})^*$
By Lagrange's theorem and the fundamental theorem of group homomorphisms, the order of the kernel is the quotient of the order of the group on the left by the group on the right, and hence, is given as follows, where $\varphi$ is the Euler totient function:
$\frac{\varphi(n(q - 1))}{\varphi(n)}$
Note that these calculuations are valid in the cases $n = 1$ and $n = 2$. In both cases, the denominator $\varphi(n)$in the expression for the order becomes 1.
### Field automorphism group (Galois group)
The group of field automorphisms of the field $\mathbb{F}_q$ is the same as its Galois group over its prime subfield $\mathbb{F}_p$, because any automorphism fixes the prime subfield pointwise by definition. This Galois group is a cyclic group of order $r = \log_p q$ (for instance, it is generated by the Frobenius $x \mapsto x^p$, an automorphism of order $r$). It is thus isomorphic to $\mathbb{Z}/r\mathbb{Z}$.
The Galois group acts on the direct product of the inner automorphism group and radial automorphism group by conjugation, but does not preserve inner automorphisms pointwise: a Galois automorphism $\sigma$ acting by conjugation on conjugation by a matrix $B$ gives conjugation by the matrix $\sigma(B)$. The Galois group does commute with the radial automorphism group.
This also applies in the cases $n = 1$ and $n= 2$. The case $n = 1$, however, is anomalous in the following respect: the Galois group is a subgroup of the radial automorphism group, i.e., field automorphisms are radial automorphisms. Note that for $n \ge 2$, the groups intersect trivially (and in fact, form an internal direct product inside the whole automorphism group) so $n = 1$ is very anomalous in this respect.
### Cyclic group of order two generated by the transpose-inverse map
The transpose-inverse map $A \mapsto (A^T)^{-1}$ is an automorphism of order two of the general linear group. This generates a cyclic subgroup of order two inside the automorphism group (the exception is the case $GL(1,2)$, where the transpose-inverse map is trivial).
The transpose-inverse map commutes with the radial automorphism group as well as with the field automorphism group. It acts nontrivially by conjugation on the inner automorphism group. Conjugating the inner automorphism by a matrix $A$ by the transpose-inverse map gives the inner automorphism by the matrix $(A^T)^{-1}$.
The cases $n = 1$ and $n= 2$ are anomalous:
• For $n = 1$, the transpose-inverse map is in the radial automorphism group.
• For $n = 2$, the transpose-inverse map is the composite of the radial automorphism $A \mapsto A/(\det A)$ and conjugation by $\begin{pmatrix} 0 & 1 \\ -1 & 0 \\\end{pmatrix}$. Therefore, it is in the subgroup generated by the inner automorphism group and the radial automorphism group. | 2020-12-05T12:01:13 | {
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https://web2.0calc.com/questions/geometry-help_110 | +0
# Geometry help
0
218
6
Here is the questions that I am stuck on,
1. Point Y lies on line segment $$\overline{XZ}$$ such that XY = 2 and YZ = 4. Semicircles are constructed with diameters $$\overline{XY}, \overline{XZ}$$, and $$\overline{YZ}$$. Find the area of the blue region.
I dont even know how or where to start.
2. A sector of a circle has a central angle of 100$$^\circ$$. If the area of the sector is 250 $$\pi$$, what is the radius of the circle?
I try to put my thoughts and ideas into words because I want to learn, but what I tried has no matamatical words.
If someone could help me with these two questions that would be great,
Thanks in advance
Jun 7, 2020
### 6+0 Answers
#1
0
1. The area of the shaded region is 4*pi.
2. The radius of the circle is 25.
Jun 7, 2020
#2
0
hmmm I am still confused... I dont understand how you got those answers nor they make sence... I am sorry but these answers with no explantions did not help me...
Guest Jun 7, 2020
#3
+30940
+2
1. The distance XZ = XY + YZ = 6, so the radius of the largest semicircle is 3. Hence its area is (1/2)pi32 or A1 = 9pi/2.
The area of the next largest semicircle is A2 = (1/2)pi22 or A2 = 2pi
The area of the smallest semicircle is A3 = (1/2)pi12 or A3 = pi/2
Hence the area of the shaded region = A1 - A2 - A3 = 9pi/2 - 2pi - pi/2 = 2pi
Jun 7, 2020
#4
+980
0
1)
The area for a circle is πr^2.
Thus the large circle is π((2+4)/2)^2 /2 = 9π/2.
The smaller ones are 1π/2 = π/2 and 4π/2 = 2π. Thus we have 9π/2 - π/2 - 2π = 4π-2π = 2π. as the area of blue.
2)
The formula for circle are is πr^2. You have 100/360 = 50/180 = 5/18 part of the area.
5/18 * πr^2 = 250π
5πr^2/ 18 = 250 π
5r^2/18 = 250
5r^2 = 250 * 18
r^2 =250*18/5 = 50*18 = 900
r = 30
If you don't understand anything feel free to ask!
Jun 7, 2020
edited by hugomimihu Jun 7, 2020
#5
+30940
0
2. Area of circle A = pi.r2 Area of circle/Area of segment = A/250pi. But this must equal 360/100.
I'll leave you to take it from here.
I see hugo... has already done it!
Jun 7, 2020
edited by Alan Jun 7, 2020
#6
0
WOW!! thx! I read through your guys explantions and I understand now! I was really stuck on those...
Thank you agian!!
Jun 7, 2020 | 2020-09-29T22:43:25 | {
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http://openstudy.com/updates/560a6dece4b032660b211aa7 | ## mathmath333 one year ago How many different numbers smaller than 2.10^8 can be formed using the digits 1 and 2 only ?
1. mathmath333
\large \color{black}{\begin{align} & \normalsize \text{ How many different numbers smaller than}\ 2\cdot 10^8 \hspace{.33em}\\~\\ & \normalsize \text{ can be formed using the digits 1 and 2 only} \hspace{.33em}\\~\\ & a.)\ 766 \hspace{.33em}\\~\\ & b.)\ 94 \hspace{.33em}\\~\\ & c.)\ 92 \hspace{.33em}\\~\\ & d.)\ 126 \hspace{.33em}\\~\\ \end{align}}
2. ganeshie8
Using two digits, How many 1 digit numbers are possible ? How many 2 digit numbers are possible ? How many 3 digit numbers are possible ? . . . How many 8 digit numbers are possible ?
3. mathmath333
is this correct \large \color{black}{\begin{align} 2^{1}+2^{2}+2^{3}+2^{4}+2^{5}+2^{6}+2^{7}+2^{8}=510\hspace{.33em}\\~\\ \end{align}}
4. ganeshie8
right! Next, look at 9 digit numbers. The left most digit can only be 1. If you fix the left most digit at 1, how many numbers are possible by changing remaining 8 digits ?
5. mathmath333
Why is this site lagging | 2017-01-19T19:54:16 | {
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https://math.stackexchange.com/questions/954049/why-can-strong-law-of-large-numbers-be-applied-in-this-question | # Why can strong law of large numbers be applied in this question?
Let $(X_n)_{n\in\mathbb{N}}$ be i.i.d. random variables taking values in the set of natural number $\mathbb{N}$. Assume that $\mathbb{P}(X_1=i)=p_i>0$ for $i\in\mathbb{N}$. Let $D_n$ denote the carinality of the set $\{X_1,X_2,...,X_n\}$. Prove $(i)$, $D_n\rightarrow\infty$ a.s.; $(ii)$, $D_n/n\rightarrow 0$ in probability. Can one strengthen $(ii)$ to a.s. convergence?
$\left\{ D_{n}\rightarrow\infty\right\} ^{c}=\bigcup_{k=1}^{\infty}\bigcap_{n=1}^{\infty}\left\{ X_{n}\leq k\right\}$ so that $P\left(\left\{ D_{n}\rightarrow\infty\right\} ^{c}\right)\leq\sum_{k=1}^{\infty}P\left(\bigcap_{n=1}^{\infty}\left\{ X_{n}\leq k\right\} \right)$.
For a fixed $k$ we find $P\left(\bigcap_{n=1}^{\infty}\left\{ X_{n}\leq k\right\} \right)\leq P\left(\bigcap_{n=1}^{m}\left\{ X_{n}\leq k\right\} \right)=\left(p_{1}+\cdots+p_{k}\right)^{m}$ for any $m$.
Here $p_{1}+\cdots+p_{k}<1$ so $m\rightarrow\infty$ makes clear that $P\left(\bigcap_{n=1}^{\infty}\left\{ X_{n}\leq k\right\} \right)=0$.
Actually the second convergence can be almost surely
Fix $\epsilon >0$. Let $\tau$ be the first moment when the total probability of elements in $D_n$ is greater than $1 - \frac{\epsilon}{2}$. By the fact that every element will eventually fall into $D_n$, we know $\tau$ is finite almost surely. To strictly prove it, begin with the element with biggest probability, then that with second biggest probability etc.
By the strong law of large number $\limsup\dfrac{D_n - D_\tau}{n - \tau} < \dfrac{\epsilon}{2}$, since after $\tau$, for each step the probability that $D_n$ increase by 1 is less than $\dfrac{\epsilon}{2}$
The above limit is also limsup of $\dfrac{D_n}{n}$.
My question: Why strong law of large number can be applied to $\limsup\dfrac{D_n - D_\tau}{n - \tau} < \dfrac{\epsilon}{2}$?? I don't understand.
For every $n$, consider the random set $Y_n=\mathbb N\setminus\{X_k\mid k\leqslant n\}$, then $Y_{\tau+k-1}\subseteq Y_{\tau}$ for every $k\geqslant1$, hence, for every $n\geqslant\tau$, $$D_n-D_\tau=\sum_{k=1}^{n-\tau}\mathbf 1_{X_{\tau+k}\in Y_{\tau+k-1}}\leqslant\sum_{k=1}^{n-\tau}U_k^\tau,\qquad U_k^\tau=\mathbf 1_{X_{\tau+k}\in Y_\tau}.$$ Conditionally on $Y_\tau$, the sequence $$\left(U_k^\tau\right)_{k\geqslant1}$$ is i.i.d. Bernoulli with probability of success $p(Y_\tau)$ where, for every $B\subseteq\mathbb N$, $p(B)=\sum\limits_{i\in B}p_i$. Thus, by the strong law of large numbers conditionally on $Y_\tau$, one has, almost surely, $$\limsup_{n\to\infty}\frac{D_n-D_\tau}{n-\tau}\leqslant\lim_{n\to\infty}\frac1{n-\tau}\sum_{k=1}^{n-\tau}U^\tau_k=\lim_{n\to\infty}\frac1n\sum_{k=1}^{n-\tau}U^\tau_k=p(Y_\tau).$$ Finally, note that, by definition of $\tau$, $p(Y_\tau)\leqslant\frac12\epsilon$. | 2019-06-17T23:18:08 | {
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https://socratic.org/questions/how-to-apply-the-remainder-theorem | # How to apply the remainder Theorem?
## () Can someone please explain to me how to do 3 c and 3.7? Thank you so much!
Mar 26, 2017
#### Explanation:
According to remainder theorem if $f \left(x\right)$ is divided by $\left(x - a\right)$, then remainder is $f \left(a\right)$. Therefore if $\left(x - a\right)$ is a factor of $f \left(x\right)$, $f \left(a\right) = 0$.
Now coming to questions raised by you solution is given seriatim.
(a) As $f \left(x\right) = {x}^{n} - {a}^{n}$, dividing by $x - a$ gives a remainder $f \left(a\right) = {a}^{n} - {a}^{n} = 0$. Hence ${x}^{n} - {a}^{n}$ is divisible by $x - a$.
(b.I) As $f \left(x\right) = {x}^{n} + {a}^{n}$, dividing by $x + a$ gives a remainder $f \left(a\right) = {\left(- a\right)}^{n} + {a}^{n} = {\left(- 1\right)}^{n} {a}^{n} + {a}^{n}$. This will be $0$ only if $n$ is odd. Hence the condition for $x + a$ to be a factor of ${x}^{n} - {a}^{n}$ is $n$ is odd.
(b.II) As $f \left(x\right) = {x}^{n} - {a}^{n}$, dividing by $x + a$ gives a remainder $f \left(a\right) = {\left(- a\right)}^{n} - {a}^{n} = {\left(- 1\right)}^{n} {a}^{n} - {a}^{n}$. This will be $0$ only if $n$ is even. Hence condition for $x + a$ to be a factor of ${x}^{n} - {a}^{n}$ is $n$ is even.
3 c As $f \left(x\right) = 2 {x}^{3} + {x}^{2} - 5 x + 2$, the factors are of the type $x - a$, where $a$ is a factor of $\frac{2}{2}$ ($\frac{p}{q}$ where $p$ is constant term and $q$ is coefficient of highest power of $x$) i.e. $\pm 1$ or $\pm 2$ or $\pm \frac{1}{2}$. As $f \left(1\right) = 0$, $f \left(2\right) = 0$ and $f \left(\frac{1}{2}\right) = 0$, hence factors are $x - 1$, $x - 2$ and $2 x - 1$ i.e. $2 {x}^{3} + {x}^{2} - 5 x + 2 = \left(x - 1\right) \left(x - 2\right) \left(2 x - 1\right)$.
3.7 When $P \left(x\right)$ is divided by $\left(x - 1\right)$, remainder is $2$, hence $P \left(1\right) = 2$ and as when $P \left(x\right)$ is divided by $\left(x - 2\right)$, remainder is $3$, hence $P \left(2\right) = 3$.
(a) As $P \left(x\right) = \left(x - 1\right) \left(x - 2\right) Q \left(x\right) + a x + b$,
$P \left(1\right) = 2$ gives us $a + b = 2$ and $P \left(2\right) = 3$ gives us $2 a + b = 3$.
Subtracting former from latter, we get $a = 1$ and hence $b = 1$
(b.I) As $P \left(x\right) = \left(x - 1\right) \left(x - 2\right) Q \left(x\right) + a x + b$ and $P \left(x\right)$ is a cubic polynomial, with coefficient of ${x}^{3}$ as $1$, it is of the type
$P \left(x\right) = \left(x - 1\right) \left(x - 2\right) \left(x - k\right) + x + 1$. Now as $- 1$ is a solution to $P \left(x\right) = 0$, we have
$P \left(- 1\right) = \left(- 1 - 1\right) \left(- 1 - 2\right) \left(- 1 - k\right) - 1 + 1 = - 6 - 6 k = 0$
i.e. $k = - 1$ and $P \left(x\right) = \left(x - 1\right) \left(x - 2\right) \left(x + 1\right) + x + 1$ or
$P \left(x\right) = {x}^{3} - 2 {x}^{2} - 4 x - 1$
(b.II) It is apparent that as $P \left(x\right) = \left(x - 1\right) \left(x - 2\right) \left(x + 1\right) + x + 1$, $\left(x + 1\right)$ is a factor of $P \left(x\right)$ and
$P \left(x\right) = \left(x - 1\right) \left(x - 2\right) \left(x + 1\right) + x + 1$
= $\left(x + 1\right) \left({x}^{2} - 3 x + 2 + 1\right) = \left(x + 1\right) \left({x}^{2} - 3 x + 3\right)$
As discriminant in ${x}^{2} - 3 x + 3$ is ${3}^{2} - 4 \times 1 \times 3 = 9 - 12 = - 3$,
there is no other real factor. | 2020-09-20T06:03:12 | {
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https://socratic.org/questions/how-do-you-solve-the-system-4x-y-1-x-2y-7-using-matrix-equation | # How do you solve the system 4x-y=1, x+2y=7 using matrix equation?
Feb 14, 2017
$\left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}1 \\ 3\end{matrix}\right)$
#### Explanation:
In matrix form it's this:
$\left(\begin{matrix}4 & - 1 \\ 1 & 2\end{matrix}\right) \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}1 \\ 7\end{matrix}\right)$
You can the go through the formality of getting the inverse of the matrix (assuming of course it has one). If we start with the determinant:
$\det \left(\begin{matrix}4 & - 1 \\ 1 & 2\end{matrix}\right) = 4 \left(2\right) - 1 \left(- 1\right) = 9$ so the matrix is invertible!
And we know that the inverse is:
${\left(\begin{matrix}4 & - 1 \\ 1 & 2\end{matrix}\right)}^{- 1} = \frac{1}{9} \left(\begin{matrix}2 & 1 \\ - 1 & 4\end{matrix}\right)$
And we can say that:
$\textcolor{red}{\frac{1}{9} \left(\begin{matrix}2 & 1 \\ - 1 & 4\end{matrix}\right)} \textcolor{g r e e n}{\left(\begin{matrix}4 & - 1 \\ 1 & 2\end{matrix}\right)} \left(\begin{matrix}x \\ y\end{matrix}\right) = \textcolor{red}{\frac{1}{9} \left(\begin{matrix}2 & 1 \\ - 1 & 4\end{matrix}\right)} \left(\begin{matrix}1 \\ 7\end{matrix}\right)$
And because the red and the green terms combine as the identity matrix:
$\implies \left(\begin{matrix}x \\ y\end{matrix}\right) = \frac{1}{9} \left(\begin{matrix}2 & 1 \\ - 1 & 4\end{matrix}\right) \left(\begin{matrix}1 \\ 7\end{matrix}\right)$
$\implies \left(\begin{matrix}x \\ y\end{matrix}\right) = \frac{1}{9} \left(\begin{matrix}9 \\ 27\end{matrix}\right) = \left(\begin{matrix}1 \\ 3\end{matrix}\right)$
Generally speaking solving systems is easier using Row Reduction but in the case of a 2x2 it is also pretty easy to work through this method. | 2020-01-21T15:34:24 | {
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https://books.google.gr/books?qtid=837e78b2&lr=&id=EwgUw45FrQkC&hl=el&sa=N&start=110 | ÁíáæÞôçóç Åéêüíåò ×Üñôåò Play YouTube ÅéäÞóåéò Gmail Drive Ðåñéóóüôåñá »
Åßóïäïò
Âéâëßá Âéâëßá 111 - 120 áðü 175 ãéá Reduce compound fractions to simple ones, and mixt numbers to improper fractions....
Reduce compound fractions to simple ones, and mixt numbers to improper fractions ; then multiply the numerators together for a new numerator, and the denominators for. a new denominator.
The New Federal Calculator, Or, Scholar's Assistant: Containing the Most ... - Óåëßäá 134
ôùí Thomas Tucker Smiley - 1825 - 180 óåëßäåò
ÐëÞñçò ðñïâïëÞ - Ó÷åôéêÜ ìå áõôü ôï âéâëßï
## Felter's New Intermediate Arithmetic: Containing Oral and Written Problems ...
Stoddard A. Felter, Samuel Ashbel Farrand - 1875 - 258 óåëßäåò
...each part is -llj. IIIIIIT 1 I. ' ii I 2. Since * of £ is iV- i of f = ,V f of f = ^=i. 3. RULE. — Multiply the numerators together for a new numerator, and the denominators for a new denominator. KOTE. — Mixed numbers may be changed to improper fractions when more convenient, or when the operation...
## Manual of Algebra
William Guy Peck - 1875 - 331 óåëßäåò
...ciple 2°); this gives for the product, ^; that is, ac ac bxd " M' Hence, we have the following RULE. Multiply the numerators together, for a new numerator, and the denominators for a new denominator. EXAMPLES. 1. Multiply |, by g. The product of the numerators is 21aae, and of the denominators ЗOcy...
## Written Arithmetic
George Augustus Walton - 1876
...becomes j;Tgi after cancelling, ~=. ^, Ans. Hence the RULE. To multiply a fraction by a fraction ; — Multiply the numerators together for a new numerator, and the denominators for a new denominator. EXAMPLES. Multiply 4. ft by H. 5- If by |?f. 7. ^ X 2£ = what ? NOTE. — Reduce mixed numbers to...
## New Elementary Arithmetic, Embracing Mental and Written Exercises ...
Henry Bartlett Maglathlin - 1876 - 224 óåëßäåò
...of a fraction called ? What is the multiplying of a fraction by a fraction equivalent to ? Rule. — Multiply the numerators together for a new numerator, and the denominators for a new denominator. Examples, Multiply 2. £by|. Ans. A3. T\byi. Ans. TV. 4. £byf. Ans. §. 6. I- of f by £. Ans. 7....
## Bradbury's Elementary Algebra: Designed for the Use of High Schools and ...
William Frothingham Bradbury - 1877 - 269 óåëßäåò
...and -r—— y = 7— (Art. 93) must be ~h ~^~ y -- i~ o " ' oy v ^ the product sought. Hence, RULE. Multiply the numerators together for a new numerator, and the denominators for a new denominator. .NOTE 1. — Common factors in the numerators and denominators may be cancelled before multiplication....
## Complete Arithmetic: Theoretical and Practical
William Guy Peck - 1877 - 341 óåëßäåò
...similar cases ; hence, the following RULE. Reduce the factors to the form of simple fractions ; then multiply the numerators together for a new numerator, and the denominators for a new denominator. EXAMPLES. 3 1. Multiply 7i by f Ans. ^ x | = | = 4f afa 8. Multiply 2J by | of f Ans. \ x \ XJ = \....
## Complete Arithmetic: Comprising the New Intermediate Arithmetic and Part of ...
Stoddard A. Felter, Samuel Ashbel Farrand - 1877 - 471 óåëßäåò
...ANALYSIS. — By the rule for multiplication of „ .„ fractions, the numerators must be multiplied together for a new numerator, and the denominators ' for a new denominator. 972 As the denominator of a decimal fraction is 729 always a power of ten, the denominator of the result...
## Elementary Arithmetic, Oral and Written
William Guy Peck - 1878 - 232 óåëßäåò
...we deduce tbe following RULE. Reduce the fractions (when necessary) to equivalent simple ones ; then multiply the numerators together for a new numerator, and the denominators for a new denominator. EXAMPLES FOR OBAL WORK. 2. Multiply | by 4. Ans. -fa. 3. Multiply | by -f. 6. Multiply | by \$. 4. Multiply...
## Annual report of the superintendent of public instruction. 31st ..., Ôåý÷ïò 32
...reverse of multiplication. He finds the rule for the multiplication of one fraction by another to be : Multiply the numerators together for a new numerator, and the denominators for a new denominator. When he comes to the division of one fraction by another, he naturally expects, and has good reason...
## New Elementary Algebra: Designed for the Use of High Schools and ..., Âéâëßï 1
Benjamin Greenleaf - 1879 - 336 óåëßäåò
...the operation might have been performed in the same manner as in the second example. Hence the RULE. Multiply the numerators together for a new numerator, and the denominators for a new denominator. NOTE 1. When either of the factors is an entire or mixed quantity, it may be best to reduce it to an... | 2020-02-25T22:12:58 | {
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https://encyclopediaofmath.org/wiki/Branching_process | Branching process
2010 Mathematics Subject Classification: Primary: 60J80 [MSN][ZBL]
A stochastic process describing a wide circle of phenomena connected with the reproduction and transformation of given objects (e.g. of particles in physics, of molecules in chemistry, of some particular population in biology, etc.). The class of branching processes is singled-out by the fundamental assumption that the reproductions of individual particles are mutually independent.
A time-homogeneous branching process $\mu (t)$ with one type of particles is defined as a Markov process with a countable number of states $0, 1 \dots$ the transition probabilities $P _ {ij} (t)$ of which satisfy the additional branching condition:
$$\tag{1 } P _ {ij} (t) = \ \sum _ {j _ {1} + \dots + j _ {i} = j } P _ {1 j _ {1} } (t) \dots P _ {1 j _ {i} } (t).$$
The states $0, 1 \dots$ of a branching process are interpreted as numbers of particles. The probability $P _ {ij} (t)$ is equal to the probability
$${\mathsf P} \{ \mu (t + t _ {0} ) = j \mid \mu (t _ {0} ) = i \}$$
of transition from state $i$ to state $j$ during a time interval of length $t$. The principal analytical tools of branching processes are the generating functions (cf. Generating function)
$$\tag{2 } F (t; s) = \ \sum _ {n = 0 } ^ \infty {\mathsf P} \{ \mu (t) = n \mid \mu (0) = 1 \} s ^ {n} .$$
The equality
$$\tag{3 } F (t + \tau ; s) = F (t; F ( \tau ; s))$$
follows from the branching condition. In branching processes with discrete time, $t$ assumes non-negative integral values, and it follows from (3) that $F(t; s)$ is a $t$- fold iteration of the function $F(s) = F(1; s)$. Such a process is sometimes called a Galton–Watson process. In a continuous-time branching process it is assumed that $t \in [0, \infty )$ and that the right-side derivative
$$\left . \frac{\partial F (t; s) }{\partial t } \right | _ {t=0} = f (s)$$
exists. It follows from (3) that $F(t; s)$ satisfies the differential equation
$$\tag{4 } \frac{\partial F (t; s) }{\partial t } = \ f (F (t; s))$$
and the initial condition $F(0; s) = s$.
If $A = F ^ { \prime } (1)$ and $a = f ^ { \prime } (1)$ are finite, the mathematical expectation ${\mathsf E} \mu (t)$ of the number of particles $\mu (t)$( under the condition $\mu (0) = 1$) is $A ^ {t}$ for a discrete-time branching process and $e ^ {at}$ for a continuous-time branching process. Depending on the values of the parameters $A$ or $a$, branching processes are subdivided into subcritical ( $A < 1$, $a < 0$), critical ( $A=1$, $a=0$) and supercritical ( $A > 1$, $a > 0$). The main characteristic governed by this classification is the behaviour of ${\mathsf E} \mu (t)$ as $t \rightarrow \infty$.
In what follows, the trivial cases $F(s) \equiv s$ and $f(s) \equiv 0$, when
$${\mathsf P} \{ \mu (t) = 1 \mid \mu (0) = 1 \} \equiv 1 ,$$
will not be discussed.
The probability of extinction is one in subcritical and critical branching processes and is less than one in supercritical processes. If ${\mathsf E} \mu (t) \cdot \mathop{\rm ln} \mu (t) < \infty$, then, in the subcritical range, the probability ${\mathsf P} \{ \mu (t) > 0 \}$ of survival of the process behaves asymptotically, as $t \rightarrow \infty$, as $K {\mathsf E} \mu (t)$, where $K$ is a positive constant. In the case of critical branching processes with finite ${\mathsf E} \mu ^ {2} (t)$, the asymptotic behaviour as $t \rightarrow \infty$ is
$${\mathsf P} \{ \mu (t) > 0 \} \approx { \frac{2}{ {\mathsf D} \mu (t) } } ,$$
where ${\mathsf D} \mu (t) = B t$ in discrete-time branching process and ${\mathsf D} \mu (t) = bt$ in continuous-time branching process and where $B = F ^ { \prime\prime } (1)$, $b = f ^ { \prime\prime } (1)$. A more detailed study of the asymptotic behaviour of the distribution of $\mu (t)$ as $t \rightarrow \infty$ shows that the conditional distribution law
$$\tag{5 } S _ {t} (x) = {\mathsf P} \left \{ \frac{\mu (t) }{ {\mathsf E} \{ \mu (t) \mid \mu (t) > 0 \} } \leq x \mid \mu (t) > 0 \right \}$$
converges weakly as $t \rightarrow \infty$ towards a limit distribution $S(x)$ if certain moments of $\mu (t)$ are finite. In subcritical branching processes the limit law $S(x)$ is discrete, while in other cases it is absolutely continuous. Of special interest is the case of critical branching processes for which the limit law $S(x)$ is exponential:
$$\tag{6 } S (x) = 1 - e ^ {-x} ,\ x \geq 0.$$
The distribution (6) is also a limit distribution for near-critical branching processes. More exactly, if one considers the class of generating functions $F(s)$ or $f(s)$ with a bounded third derivative $F ^ { \prime\prime\prime } (1)$, $f ^ { \prime\prime\prime } (1)$ and $F ^ { \prime\prime } (1) \geq B _ {0} > 0$, $f ^ { \prime\prime } (1) \geq b _ {0} > 0$, then
$$\lim\limits _ {\begin{array}{c} t \rightarrow \infty \\ A \rightarrow 1 \end{array} } \ S _ {t} (x) = \ 1 - e ^ {-x} ,\ \ x \geq 0,$$
where $S _ {t} (x)$ is defined by formula (5). The phenomena occurring in near-critical processes at $t \rightarrow \infty$ are said to be transient.
Another model of branching processes is the Bellman–Harris process, in which each particle has a random lifetime with distribution function $G(t)$. At the end of its lifetime, the particle leaves behind $n$ daughter particles with probability $q _ {n}$;
$$\sum _ {n = 0 } ^ \infty q _ {n} = 1.$$
The lifetimes and the number of daughter particles produced by individual particles are independent. Consider a particle of age zero at the initial moment of time $t=0$; then the generating function $F(t; s)$ of the number of particles $\mu (t)$ at the moment of time $t$, given by formula (2), satisfies the non-linear integral equation
$$\tag{7 } F (t; s) = \ \int\limits _ { 0 } ^ { t } h (F (t - u; s)) dG (u) + s (1 - G (t)),$$
where
$$h (s) = \ \sum _ {n = 0 } ^ \infty q _ {n} s ^ {n} .$$
If $G(t)$ is a degenerate distribution function, the Bellman–Harris process is a discrete-time branching process; if $G(t)$ is an exponential distribution function, one obtains a continuous-time branching process. In the general case, the Bellman–Harris process is a non-Markov branching process.
A branching process may also be complicated by the dependence of the particles on their location in space. For instance, let each particle execute an independent Brownian motion in an $r$- dimensional domain $G$ with an absorbing boundary $\partial G$. A particle located inside the domain $G$ will be transformed within time $\Delta t \rightarrow 0$ with probability
$$p _ {n} \Delta t + o ( \Delta t),\ \ n \neq 1,$$
into $n$ particles, and each such particle begins to execute an independent motion along Brownian trajectories starting from its point of genesis. Let $\mu _ {x,t} (A)$ be the number of particles in a set $A$ at the moment of time $t$, that number having been produced by one particle located at the point $x \in G$ at the initial moment of time 0. The generating function
$$H (t, x, s( \cdot )) = \ {\mathsf E} \mathop{\rm exp} \left [ \int\limits _ { G } \mathop{\rm ln} s (y) \mu _ {x,t} (dy) \right ]$$
satisfies the quasi-linear parabolic equation
$$\tag{8 } \Delta H + f (H) = 0$$
with the initial condition $H(0, x, s( \cdot )) = s(x)$ and the boundary condition $H(t, x, s ( \cdot )) \mid _ {x \rightarrow \partial G } = 0$, where $\Delta = {\sum _ {i=1} ^ {r} } \partial ^ {2} / \partial x _ {i} ^ {2}$ is the Laplace operator, and $f(s) = {\sum _ {n} } p _ {n} s ^ {n}$, $p _ {1} = - \sum _ {n \neq 1 } p _ {n}$.
It is assumed in general branching processes that the propagating particles can be characterized by certain parameters, which may be interpreted as age, location of the particle in space, type, size or energy of the particle, etc. Such processes are studied with the aid of generating functions or functionals, for which non-linear differential and integral equations, generalizing equations (4), (7) and (8), are deduced. One may give the following general descriptions of such models of branching processes. Let particles move, independently of each other, in accordance with Markov's law in some phase space $X$. It is assumed that the random lifetime of a particle is a Markov time (cf. Markov moment) which depends on its trajectory. At the end of its lifetime the particle generates new particles, which become distributed over the phase space $X$ in accordance with some probability law. The new particles evolve in a similar manner, independently of each other. In the space of integer-valued measures determined by the number of particles present in subsets of $X$, the branching process is Markovian. However, branching processes are often studied in simpler reduced spaces as well. In such a case many of them become non-Markovian.
In most of the models discussed above the subdivision of processes into subcritical, critical and supercritical processes retains its meaning. Many properties of simple branching processes described by equation (4) are also displayed by more complex systems. In particular, the limiting distribution of (5), for critical processes, usually turns out to be an exponential distribution (6) (in an appropriate norm).
Branching processes are used in calculations of various real processes — biological, genetic, physical, chemical or technological. In real processes the condition of independent motion of each particle is often not met, and the generated daughter particles may interact with each other. This is true of many biological reproduction processes, in processes of propagation of epidemics (cf. Epidemic process), in bimolecular chemical reactions, etc. However, the initial stages of such processes may be calculated with the aid of suitably chosen models of branching processes. This is done in situations in which the number of active particles in the space is relatively small; at such low concentrations their mutual collisions are practically non-existent, and the state of the system evolves by the collisions between active particles and the particles of the medium. In epidemic processes, for example, sick individuals may be considered as such "active particles" . Phenomena related to genetic mutations, for example, may be calculated with the aid of branching processes. The branching process with a finite number of particle types may serve as a model in computing chain reactions; the branching process with diffusion may serve as a model of neutron processes in nuclear reactors. Phenomena occurring in showers of cosmic particles may also be studied with the aid of branching processes. In telephony, the computation of certain expectation systems may be reduced to branching process models.
References
[S] B.A. Sewastjanow, "Verzweigungsprozesse" , Akad. Wissenschaft. DDR (1974) (Translated from Russian) MR0408018 Zbl 0291.60039 [AN] K.B. Athreya, P.E. Ney, "Branching processes" , Springer (1972) MR0373040 Zbl 0259.60002 | 2021-08-01T01:33:00 | {
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http://clay6.com/qa/51446/show-that-the-statement-if-x-is-real-number-such-that-x-3-4x-0-then-x-is-0- | # Show that the statement " If $x$ is real number such that $x^3+4x=0$, then $x$ is 0." is true using contradiction method.
" If $x$ is real number such that $x^3+4x=0$, then $x$ is 0." | 2018-06-23T17:26:47 | {
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http://karagila.org/2017/stationary-preserving-permutations-are-the-identity-on-a-club/ | Asaf Karagila
I don't have much choice...
This is not something particularly interesting, I think. But it's a nice exercise in Fodor's lemma.
Theorem. Suppose that $\kappa$ is regular and uncountable, and $\pi\colon\kappa\to\kappa$ is a bijection mapping stationary sets to stationary sets. Then there is a club $C\subseteq\kappa$ such that $\pi\restriction C=\operatorname{id}$.
Proof. Note that the set $\{\alpha\mid\pi(\alpha)<\alpha\}$ is non-stationary, since otherwise by Fodor's lemma there will be a stationary subset on which $\pi$ is constant and not a bijection. This means that $\{\alpha\mid\alpha\leq\pi(\alpha)\}$ contains a club. The same arguments shows that $\pi^{-1}$ is non-decreasing on a club. But then the intersection of the two clubs is a club on which $\pi$ is the identity. $\square$
This is just something I was thinking about intermittently for the past few years, but now I finally spent enough energy to figure it out. And it's cute. (Soon I will post more substantial posts, on far more exciting topics! Don't worry!)
### There are 5 comments on this post.
By
(Apr 10 2017, 02:13)
One can formulate this theorem in a category theoretic context. The category of filters in the category whose objects are pairs $(X,\mathcal{F})$ where $X$ is a set and $\mathcal{F}$ is a filter on $X$. If $\mathcal{F}$ is a filter on $X$ and $\mathcal{G}$ is a filter on $Y$, then let $\mathcal{A}_{(X,\mathcal{F}),(Y,\mathcal{G})}$ be the set of all functions $f:X\rightarrow Y$ where $f^{-1}[R]\in\mathcal{F}$ whenever $R\in\mathcal{G}$. Let $\simeq$ be the equivalence relation on $\mathcal{A}_{(X,\mathcal{F}),(Y,\mathcal{G})}$ where $f\simeq g$ iff $\{x\in X|f(x)=g(x)\in\mathcal{F}\}$. Then $\mathcal{A}_{(X,\mathcal{F}),(Y,\mathcal{G})}/\simeq$ is the set of all morphisms from $(X,\mathcal{F})$ to $(Y,\mathcal{G})$ in the category of filters. This result that you have proven therefore states that if $\kappa$ is a regular cardinal and $\mathcal{F}$ is the club filter, then the object $(\kappa,\mathcal{F})$ has no non-trivial automorphisms. As all set theorists are aware, the pairs $(X,\mathcal{U})$ where $\mathcal{U}$ is an ultrafilter do not contain any non-trivial endomorphisms. It seems like this property of no non-trivial endomorphisms should be quite prevalent and that there should be many different kinds of normal filters on regular cardinals with no non-trivial automorphisms. Have you found any other examples of filters with no non-trivial automorphisms?
By Asaf Karagila
(Apr 10 2017, 08:35 In reply to Joseph Van Name)
That's interesting. I haven't thought about other examples, though. The motivation for this, actually, comes from symmetric extensions. If one adds a subset, say of $\omega_1$, and you want to use permutations of $\omega_1$ to induce automorphisms of the forcing, what this right here shows, is that if your conditions are stationary sets, then your approach leads to some sort of pseudo-rigidity. (There might be automorphisms, but not ones induced by permutations of $\omega_1$.)
I guess we can formulate this notion as the filter being rigid. Then one of the first obvious question would be, are there rigid filters on $\omega$ which are not ultrafilters?
By Ashutosh
(Apr 16 2017, 00:00)
This has an interesting "application" due to Komjath: Suppose X is a set of reals such that the ideal of meager subsets of X is isomorphic to the non stationary ideal on omega_1 (Komjath showed that such X can consistently exist). Then X^2 is a non meager subset of plane each of whose non meager subsets contains three collinear points. Shelah and I proved that a similar result holds for the null ideal.
By Asaf Karagila
(Apr 16 2017, 00:27 In reply to Ashutosh)
How do you mean that this is an application of this?
By Ashutosh
(Apr 16 2017, 01:23)
Showing that every subset Y of X^2 that does not contain three collinear points is meager boils down to showing that Y can be covered by the diagonal and two other meager sets and this uses the rigidity of the non-stationary ideal.
This is not really an application but just that this fact was exploited here.
Want to comment? Send me an email! | 2019-01-24T00:05:51 | {
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https://math.stackexchange.com/questions/2773118/general-form-of-picks-theorem | # General Form of Pick's Theorem
I recently derived the most general possible version of Pick's Theorem which works for any shape consisting of the union of any number of polygons (these don't have to be simple polygons either) with any number of holes inside of them (the holes also don't have to be simple polygons). The generalization states that:
$A=I+\frac{B}{2}-\frac{P+P_s}{2}+\frac{H+H_s}{2}$
where $I$ is the total number of interior integer coordinate points of the shape, $B$ is the total number of boundary integer coordinate points of the shape, $P$ is the total number of path-connected domains of finite area only containing boundary and interior points of the shape, $P_s$ is the total number of non-intersecting polygons only containing boundary and interior points of the shape, $H$ is the total number of path-connected domains of finite area only containing boundary and exterior points of the shape, and $H_s$ it the total number of non-intersecting polygons only containing boundary and exterior points of the shape.
If the shape in question is itself a single simple polygon then we have $P=1$, $P_s=1$, $H=0$, $H_s=0$, and the generalization reduces to the standard form of Pick's Theorem:
$A=I+\frac{B}{2}-1$
If the shape in question is the union of $P$ non-intersecting polygons with no holes then the generalization reduces to:
$A=I+\frac{B}{2}-P$
If the shape in question is a single polygon with $H$ holes, all of which are non-intersecting polygons, then the generalization reduces to:
$A=I+\frac{B}{2}-1+H$
My question is this: has this generalization been published anywhere? Is there a simpler version of it somewhere to be found?
• Could you rewrite your formula in terms of Euler characteristic? – Lord Shark the Unknown May 9 '18 at 3:35
• Yes, somebody had already done so back in 1985 here: faculty.math.illinois.edu/~reznick/496-2-6-17.pdf – The Riddler May 9 '18 at 14:20
• The 1985 article ("Pick's Theorem revisited" by Varberg) concludes with the remark: "We have recently learned of an even more general form of Pick's theorem". The referenced article (in German) is "Neuere Studien über Gitterpolygone" by Hadwiger and Wills; an online citation is here. (Sadly, the displayed initial page doesn't include the formula.) – Blue May 12 '18 at 18:09 | 2019-05-19T18:56:53 | {
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http://www.wikidoc.org/index.php/Kumaraswamy_distribution | # Kumaraswamy distribution
For other uses, see Kumaraswamy (disambiguation).
Parameters Probability density functionProbability density function Cumulative distribution functionCumulative distribution function $a>0\,$ (real)
$b>0\,$ (real) $x \in [0,1]\,$ $abx^{a-1}(1-x^a)^{b-1}\,$ $[1-(1-x^a)^b]\,$ $\frac{b\Gamma(1+1/a)\Gamma(b)}{\Gamma(1+1/a+b)}\,$ $\left(1-\left(\frac{1}{2}\right)^{1/b}\right)^{1/a}$ $\left(\frac{a-1}{ab-1}\right)^{1/a}$ (complicated-see text) (complicated-see text) (complicated-see text)
In probability and statistics, the Kumaraswamy's double bounded distribution is a family of continuous probability distributions defined on the interval [0,1] differing in the values of their two non-negative shape parameters, a and b.
It is similar to the Beta distribution, but much simpler to use especially in simulation studies due to the simple closed form of both its probability density function and cumulative distribution function. This distribution was originally proposed by Poondi Kumaraswamy for variables that are lower and upper bounded.
## Characterization
### Probability density function
The probability density function of the Kumaraswamy distribution is
$f(x; a,b) = a b x^{a-1}{ (1-x^a)}^{b-1}.$
### Cumulative distribution function
The cumulative distribution function is therefore
$F(x; a,b)=1-(1-x^a)^b.$
### Generalizing to arbitrary range
In its simplest form, the distribution has a range of [0,1]. In a more general form, we may replace the normalized variable x with the unshifted and unscaled variable z where:
$x = \frac{z-z_{\mathrm{min}}}{z_{\mathrm{max}}-z_{\mathrm{min}}} , \qquad z_{\mathrm{min}} \le z \le z_{\mathrm{max}}. \,\!$
The distribution is sometimes combined with a "pike probability" or a Dirac delta function, e.g.:
$g(x|a,b) = F_0\delta(x)+(1-F_0)a b x^{a-1}{ (1-x^a)}^{b-1}.$
## Properties
The raw moments of the Kumaraswamy distribution are given by:
$m_n = \frac{b\Gamma(1+n/a)\Gamma(b)}{\Gamma(1+b+n/a)} = bB(1+n/a,b)\,$
where B is the Beta function. The variance, skewness, and excess kurtosis can be calculated from these raw moments. For example, the variance is:
$\sigma^2=m_2-m_1^2.$
## Relation to the Beta distribution
The Kuramaswamy distribution is closely related to to Beta distribution. Assume that Xa,b is a Kumaraswamy distributed random variable with parameters a and b. Then Xa,b is the a-th root of a suitably defined Beta distributed random variable. More formally, Let Y1,b denote a Beta distributed random variable with parameters $\alpha=1$ and $\beta=b$. One has the following relation between Xa,b and Y1,b.
$X_{a,b}=Y^{1/a}_{1,b},$
with equality in distribution.
$\operatorname{P}\{X_{a,b}\le x\}=\int_0^x ab t^{a-1}(1-t^a)^{b-1}dt= \int_0^{x^a} b(1-t)^{b-1}dt=\operatorname{P}\{Y_{1,b}\le x^a\} =\operatorname{P}\{Y^{1/a}_{1,b}\le x\} .$
One may introduce generalised Kuramaswamy distributions by considering random variables of the form $Y^{1/\gamma}_{\alpha,\beta}$, with $\gamma>0$ and where $Y_{\alpha,\beta}$ denotes a Beta distributed random variable with parameters $\alpha$ and $\beta$. The raw moments of this generalized Kumaraswamy distribution are given by:
$m_n = \frac{\Gamma(\alpha+\beta)\Gamma(\alpha+n/\gamma)}{\Gamma(\alpha)\Gamma(\alpha+\beta+n/\gamma)}.$
Note that we can reobtain the original moments setting $\alpha=1$, $\beta=b$ and $\gamma=a$. However, in general the cumulative distribution function does not have a closed form solution.
## Example
A good example of the use of the Kumaraswamy distribution is the storage volume of a reservoir of capacity zmax whose upper bound is zmax and lower bound is 0 (Fletcher, 1996).
## References
• Kumaraswamy, P. (1980). "A generalized probability density function for double-bounded random processes". Journal of Hydrology 46: 79–88.
• Fletcher, S.G., and Ponnambalam, K. (1996). "Estimation of reservoir yield and storage distribution using moments analysis". Journal of Hydrology 182: 259–275.
it:variabile casuale di Kumaraswamy | 2013-05-25T16:45:57 | {
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https://planetmath.org/LaplacianMatrixOfAGraph | # Laplacian matrix of a graph
Let $G$ be a finite graph with $n$ vectices and let $D$ be the incidence matrix (http://planetmath.org/ IncidenceMatrixWithRespectToAnOrientation) of $G$ with respect to some orientation. The Laplacian matrix of $G$ is defined to be $DD^{T}$.
If we let $A$ be the adjacency matrix of $G$ then it can be shown that $DD^{T}=\Delta-A$, where $\Delta=\textrm{diag}(\delta_{1},\ldots,\delta_{n})$ and $\delta_{i}$ is the degree of the vertex $v_{i}$. As a result, the Laplacian matrix is independent of what orientation is chosen for $G$.
The Laplacian matrix is usually denoted by $L(G)$. It is a positive semidefinite singular matrix, so that the smallest eigenvalue is 0.
Title Laplacian matrix of a graph LaplacianMatrixOfAGraph 2013-03-22 17:04:40 2013-03-22 17:04:40 Mathprof (13753) Mathprof (13753) 8 Mathprof (13753) Definition msc 05C50 Laplacian matrix | 2019-03-25T23:48:27 | {
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https://socratic.org/questions/58498b8b11ef6b14f5145160 | Question #45160
Feb 13, 2017
See explanation.
Explanation:
To solve this task you need to use the folloing feature of polynomial:
If $a$ is a zero of polynomial with multiplicity $n$, then the polynomial is divisible by ${\left(x - a\right)}^{n}$ and not divisible by ${\left(x - a\right)}^{n + 1}$
As stated above the polynomial would have to be divisible by ${\left(x - 3\right)}^{3}$, and ${\left(x - 0\right)}^{2}$. The only polynomial of degree $5$ fulfilling theses conditions is:
$P \left(x\right) = {\left(x - 3\right)}^{3} \cdot {x}^{2} = \left({x}^{3} + 3 {x}^{2} + 3 x + 1\right) \cdot {x}^{2}$
$= {x}^{5} + 3 {x}^{4} + 3 {x}^{3} + {x}^{2}$ | 2019-11-19T19:59:12 | {
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https://fungrim.org/entry/9fbe4f/ | # Fungrim entry: 9fbe4f
$\left\{{n + 1 \atop k}\right\} = k \left\{{n \atop k}\right\} + \left\{{n \atop k - 1}\right\}$
Assumptions:$n \in \mathbb{Z}_{\ge 0} \;\mathbin{\operatorname{and}}\; k \in \mathbb{Z}_{\ge 1}$
TeX:
\left\{{n + 1 \atop k}\right\} = k \left\{{n \atop k}\right\} + \left\{{n \atop k - 1}\right\}
n \in \mathbb{Z}_{\ge 0} \;\mathbin{\operatorname{and}}\; k \in \mathbb{Z}_{\ge 1}
Definitions:
Fungrim symbol Notation Short description
StirlingS2$\left\{{n \atop k}\right\}$ Stirling number of the second kind
ZZGreaterEqual$\mathbb{Z}_{\ge n}$ Integers greater than or equal to n
Source code for this entry:
Entry(ID("9fbe4f"),
Assumptions(And(Element(n, ZZGreaterEqual(0)), Element(k, ZZGreaterEqual(1))))) | 2021-06-19T07:35:21 | {
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https://www.notion.so/Feynman-s-Trick-Differentiation-under-the-integral-sign-a37c936f79d7442a87314124d71768c0 | C. Hinsley
14 September 2020
Note: In Surely You're Joking, Mr. Feynman!, Feynman mentions the two books from which he learned calculus. The former is "Calculus for the Practical Man" by J.E. Thompson and the latter (the one from which he learned Liebniz' integral rule) is "Advanced Calculus" by F.S. Woods. Both of these texts are intended to prepare the reader for the use of calculus in an applied setting, and I think each of these have a distinct clarity missing from modern texts on the subject. It may be worthwhile to read these texts, even for one who has already studied calculus. The basis for technique in particular can be found on page 141 of the 1934 edition of Woods' text.
These exercises lend themselves to solution by Feynman's technique, which you can see worked out for example problems at various other sources. I found that these sources only had a handful of problems each, so I decided to compile all I could find in one location.
$$(1).\int_0^1 \frac{t^3-1}{\ln t}\ \text{d}t$$
$$(2). \int_0^1 (x\ln x)^{50}\ \text{d}x$$
$$(3). \int_0^\infty e^{-\frac{x^2}{2}}\ \text{d}x$$
$$(4).\int_0^{2\pi} e^{\cos \theta}\cos(\sin\theta)\ \text{d}\theta$$
$$(5). \int_0^\infty \frac{\sin x}{x}\ \text{d}x$$
$$(6). \int_0^{\frac{\pi}{2}} \frac{x}{\tan x}\ \text{d}x$$
$$(7). \int_0^\infty \frac{\ln(1+x^2)}{1 + x^2}\ \text{d}x$$
$$(8). \int_0^1 \frac{x-1}{\ln x}\ \text{d}x$$
$$(9). \int_0^1 \frac{\ln(x+1)}{x^2+1}\ \text{d}x$$ | 2022-05-24T10:09:22 | {
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http://math.stackexchange.com/questions/117926/finding-mode-in-binomial-distribution | # finding mode in Binomial distribution
suppose $X$ has the Binomial distribution with parameters $n,p$ . how can show that if $((n+1)p)$ is integer then $X$ has two mode that is $((n+1)p)$ or $((n+1)p-1)?$
-
Hint: compute the ratio $b(n,p;k+1)/b(n,p;k)$ and check that this ratio is $\gt1$ for every $k\lt k^*$ and $\leqslant1$ for every $k\geqslant k^*$, for some integer $k^*$. – Did Mar 8 '12 at 16:27
Does your acceptance rate mean that you were satisfied by none of the answers you received on the site? – Did Mar 8 '12 at 16:29
## 1 Answer
Let $a_k=P(X=k)$. We have $$a_k=\binom{n}{k}p^kq^{n-k},\qquad\text{and}\qquad \binom{n}{k+1}p^{k+1}q^{n-k- 1},$$ where as usual $q=1-p$. We calculate the ratio $\dfrac{a_{k+1}}{a_k}$. Note that $$\frac{\binom{n}{k+1}}{\binom{n}{k}}$$ simplifies to $$\frac{n-k}{k+1},$$ and therefore $$\frac{a_{k+1}}{a_k}=\frac{n-k}{k+1}\frac{p}{q}=\frac{n-k}{k+1}\frac{p}{1-p}.$$ This is $\ge 1$ if $k \le np+p-1$. Thus if $k< np-p+1$, then $a_{k+1}>a_k$, and if $k>np-p+1$, then $a_{k+1}<a_k$.
The calculation (almost) says that we have equality of two consecutive probabilities precisely if $a_{k+1}=a_k$, that is, if $k=np+p-1$. Note that $k=np+p-1$ implies that $np+p-1$ is an integer.
So unless $k=np+p-1$, there is a single mode, and if $k=np+p-1$ there are two modes, at $np+p-1$ and at $np+p$.
Not quite! We have been a little casual in our algebra. We have not paid attention to whether we might be multiplying or dividing by $0$. We also have casually accepted what the algebra seems to say, without doing a reality check.
Suppose that $p=0$. Then $np+p-1$ is an integer, namely $-1$. But whatever $n$ is, there is a single mode, namely $k=0$. In all other situations where $np+p-1$ is an integer, the $k$ we have identified is non-negative.
However, suppose that $p=1$. Again, $np+p-1$ is an integer, and again there is no double mode. The largest $a_k$ occurs at one place only, namely $k=n$, since $np+p$ is in this case beyond our range.
That completes the analysis when $np+p-1$ is an integer. When it is not, the analysis is simple. There is a single mode, at $\lfloor np+p\rfloor$.
-
I am solving a similar exercise and I have some doubts: why taking the ratio $\frac{p_X(k+1)}{p_X(k)}$ gives you the mode (most probable value), which is defined as $\sup_{x \in R_X} p_X(x)$? – user16924 Sep 7 at 23:26
The sup is in this case a max, since the random variable takes on integer values. Apart from a couple of "degenerate" cases pointed out in the answer, the probabilities rise and then fall. The ratio of consecutive terms is therefore $\gt 1$ for a while, then $\lt 1$, except that in somewhat unusual cases we can have ratio $1$, so two consecutive values each qualify as a mode. Looking at the ratios tells us when the probability has reached a maximum. – André Nicolas Sep 8 at 1:38
Thanks very much for the explanation. – user16924 Sep 8 at 2:47
You are welcome. That was an overview. The detail is in the answer above. – André Nicolas Sep 8 at 2:48 | 2014-11-27T04:29:58 | {
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http://mathoverflow.net/questions/130121/maximum-number-of-vertices-of-hypercube-covered-by-ball-of-radius-r | # Maximum number of Vertices of Hypercube covered by Ball of radius R
Let $R>0$ be given and let $H^n$ be the unit hypercube in $\mathbb{R}^n$. The problem I am facing is to find the maximum number of vertices of $H^n$ which can be covered by a closed $n$-dimensional ball of radius $R$.
I would greatly appreciate any references to literature or ideas on how to approach the problem. I strongly suspect that a solution for any $R$ comes from placing the center of the ball at $(\frac{1}{2},\frac{1}{2},\ldots,\frac{1}{2},0,\ldots,0)$ with the number of nonzero coordinates being $\lfloor 4R^2 \rfloor=\lfloor D^2 \rfloor$, where $D$ is the diameter (for $D \leq n$).
Any references or ideas are welcome.
Thanks!
-
That seems like a good guess to me. You could try using the fact that any set of more than $2^k$ vertices of $H^n$ cannot be contained in a $k$-dimensional affine subspace of $\mathbb R^n$. – Greg Martin May 9 '13 at 0:52
This is not always true. Already if n=2 (we have a square), then if R is a little less than $\sqrt(2)/2$, then the best we can do is to cover 3 vertices. – domotorp Jun 25 '13 at 15:12
@domotorp Indeed, it is not always true. However, if one can cover 3 vertices then one can cover the whole square. I believe the conjecture first breaks down at dimension 16 – user21277 Jun 26 '13 at 19:42
I think you are mistaken, in my example with the given radius we can cover 3 vertices but not the whole square. – domotorp Jun 27 '13 at 15:14
@domotorp Consider 3 vertices of a square. These form an isosceles right triangle. The circumcenter of this triangle is the midpoint of the hypotenuse, which is also the center of the square. – user21277 Jun 28 '13 at 2:10
Let $n=8$ and $R=\sqrt{\frac{7}{8}}$. Then the ball with center $(\frac{1}{8}, \frac{1}{8}, \dots, \frac{1}{8})$ and radius $R$ covers exactly $9$ vertices (those with at most one coordinate equal to $1$), but a ball with radius $R$ cannot cover a $4$-dimensional unit hypercube, which requires radius at least $1$.
There is even no upper bound on the number of points a ball with radius $1$ can cover, provided the dimension is not bounded. Generalizing previous construction, a ball with center $(\frac{1}{n}, \frac{1}{n}, \dots, \frac{1}{n})$ and radius $\sqrt{\frac{n-1}{n}}$ covers $n+1$ vertices of the $n$-dimensional hypercube. As user21277 points out, for $n=16$ this is better than placing the ball in the center of a $4$-dimensional face.
In general, for $n>2i$, a ball with center $(\frac{i}{n}, \frac{i}{n}, \dots, \frac{i}{n})$ and radius $\sqrt{\frac{i(n-i)}{n}}$ covers ${n \choose i} + {n \choose i-1} + \cdots +{n \choose 0}$ vertices: those with at most $i$ coordinates equal to $1$. A ball with the same radius cannot cover all $2^{4i}$ vertices of a $4i$-dimensional unit hypercube. | 2014-10-31T14:56:33 | {
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http://mathhelpforum.com/calculus/63837-help-arc-length-parametric-curves.html | # Math Help - Help with arc length of parametric curves
1. ## Help with arc length of parametric curves
x(t) = 2t - 2sin(t)
y(t) = 2 - 2cos(t)
t is from 0 to 2Pi
Okay, I am able to set up the problem correctly:
x'(t) = 2 - 2cos(t)
y'(t) = 2sin(t)
Length = Integral from 0 to 2Pi of (Sqrt[[2 - 2cos(t)]^2 + [2sin(t)]^2])
= Integral from 0 to 2Pi of (Sqrt[4[1-cos(t)]^2 + [2sin(t)]^2])
I know I can pull the 4 out (making it a 2), but I'm stuck after that. What good does making a substitution of 1 - cos(t) = 2sin(t/2)^2 do?
Help would be greatly appreciated!
2. Dear Drumiester,
next step: bring out 4 from root
next step: break down the bracket under the root and simplify
next step: you can bring out again 2 from root
you get $\sqrt{1-cos(t)}$
Now one trick: $cos(t) = cos(2*t/2) = cos^2(t/2) - sin^2(t/2)$
and
$1 = cos^2(t/2) + sin^2(t/2)$
So
$\sqrt{1-cos(t)} = \sqrt{cos^2(t/2)+sin^2(t/2) - cos^2(t/2) + sin^2(t/2)} = \sqrt{2}*sin(t/2)$
Now it's not so wild, isn't it?
3. Thanks man! I just discovered this site, so I'll be sure to come here for help | 2015-05-29T10:35:03 | {
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https://mathhelpboards.com/threads/angles-in-a-triangle.28085/ | # [SOLVED]Angles in a triangle
#### anemone
##### MHB POTW Director
Staff member
All the angles in triangle $ABC$ are less than $120^{\circ}$. Prove that
$\dfrac{\cos A+\cos B+\cos C}{\sin A+\sin B+\sin C}>-\dfrac{\sqrt{3}}{3}$.
#### anemone
##### MHB POTW Director
Staff member
\begin{tikzpicture}
\coordinate[label=left:A] (A) at (0,0);
\coordinate[label=right:B] (B) at (6, 0);
\coordinate[label=above:C] (C) at (2.4,3);
\coordinate[label=above: $A_1$] ($A_1$) at (7,3);
\coordinate[label=below: $B_1$] ($B_1$) at (7.2,0);
\coordinate[label=below: $C_1$] ($C_1$) at (12,2);
\draw (A) -- (B)-- (C)-- (A);
\draw (7,3) -- (7.2,0)-- (12,2)-- (7,3);
\end{tikzpicture}
Consider the triangle $A_1B_1C_1$ where $\angle A_1=120^{\circ}-\angle A,\,\angle B_1=120^{\circ}-\angle B$ and $\angle C_1=120^{\circ}-\angle C$. The given condition guarantees the existence of such a triangle.
Applying the triangle inequality in triangle $A_1B_1C_1$ gives $B_1C_1+C_1A_1>A_1B_1$, i.e.
$\sin A_1+\sin B_1>\sin C_1$ by applying the law of sines to triangle $A_1B_1C_1$.
It follows that
$\sin (120^{\circ}-A)+\sin (120^{\circ}-B)>\sin (120^{\circ}-C)$ or
$\dfrac{\sqrt{3}}{2}(\cos A+\cos B+\cos C)+\dfrac{1}{2}(\sin A+\sin B+\sin C)>0$.
Taking into account that $a+b>c$ implies $\sin A+\sin B-\sin C>0$, the above inequality can be rewritten as
$\dfrac{\sqrt{3}}{2}\cdot \dfrac{\cos A+\cos B+\cos C}{\sin A+\sin B+\sin C}+\dfrac{1}{2}>0$, from which the conclusion follows. | 2021-01-24T08:52:30 | {
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https://math.stackexchange.com/questions/1720050/solve-the-equation-x3-117y3-5-over-the-integers | # Solve the equation $x^3 + 117y^3 = 5$ over the integers.
Solve the equation $x^3 + 117y^3 = 5$ over the integers.
I have tried solving this. It is clear that one of $x$ or $y$ must be negative. $117$ seemed a strange number. So I found out that $117 = 125 - 8 = 5^3 - 2^3$. I don't know if this is useful but still I'm adding it. So the equation becomes:
$$x^3 + (5y)^3 - (2y)^3 = 5$$
I don't know how to proceed further. I need some hints. Any help would be appreciated.
• Take $\pmod{3^2}$ – lab bhattacharjee Mar 30 '16 at 8:57
• @labbhattacharjee How did you know we had to take a $\pmod 9$? – TheRandomGuy Mar 30 '16 at 9:20
• @Dhrub, As the power of $x,y$ are $3$ – lab bhattacharjee Mar 30 '16 at 9:32
• @Dhruv By Binomial theorem $(3k\pm 1)^3\equiv \pm 1\pmod 9$. Similarly, $(pk+a)^p\equiv a^p\pmod{p^2}$ for any prime $p$, which gives that there are at most $p$ $p$'th powers mod $p^2$ (in this case, there are only $3$ $3$'th powers mod $3^2$, so using mod $3^2$ makes sense). – user236182 Mar 30 '16 at 14:56
• Using mod $7$ would also make sense, because $x^3\equiv \{0,\pm 1\}\pmod{7}$, but unfortunately it doesn't solve it. By Fermat's Little Theorem $x^6\equiv \{0,1\}\pmod{7}$, so $x^3\equiv \{0,\pm 1\}\pmod{7}$. More generally, $x^{\frac{p-1}{2}}\equiv \{0,\pm 1\}\pmod{p}$ for any odd prime $p$ (see Euler's Criterion for a stronger result, namely $x^{\frac{p-1}{2}}\equiv \left(\frac{x}{p}\right)\pmod{p}$ for any odd prime $p$). – user236182 Mar 30 '16 at 15:17
Hint: $$x^3 \equiv 0,1,8 (\bmod 9)$$ $$117y^3 \equiv 0 (\bmod 9)$$ $$5 \equiv 5 (\bmod 9)$$
• Haha, and here I was asking Sage to calculate the class group of $\mathbb Q(\sqrt[3]{-117})$. +1 – RKD Mar 30 '16 at 8:58
• @Dhruv: I saw that $9|117$ and remembered that $x^3 \equiv 0,1,8 (\bmod 9)$ – Roman83 Mar 30 '16 at 10:08
• To prove that $x^3\equiv \{0,\pm 1\}\pmod{9}$, notice that by Binomial Theorem it's easy to see that $(3k\pm 1)^3\equiv \pm 1\pmod{9}$. – user236182 Mar 30 '16 at 14:48 | 2021-04-21T10:09:00 | {
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https://quant.stackexchange.com/tags/implied-volatility/hot | # Tag Info
43
The Black-Scholes option pricing model provides a closed-form pricing formula $BS(\sigma)$ for a European-exercise option with price $P$. There is no closed-form inverse for it, but because it has a closed-form vega (volatility derivative) $\nu(\sigma)$, and the derivative is nonnegative, we can use the Newton-Raphson formula with confidence. Essentially, ...
30
Brenner and Subrahmanyam (1988) provided a closed form estimate of IV, you can use it as the initial estimate: $$\sigma \approx \sqrt{\cfrac{2\pi}{T}} . \cfrac{C}{S}$$
28
Let $t_0, t_1, \ldots, t_n$ be observation dates, where $0=t_0 < \cdots < t_n = T$, and $\{S_t \mid t \geq 0\}$ be the equity price process without dividend payments. Then the realized variance is defined by \begin{align*} \frac{252}{n}\sum_{i=1}^n \ln^2 \frac{S_{t_i}}{S_{t_{i-1}}}. \end{align*} Note that, for sufficiently small $x$, \begin{align*} \...
19
Taking away all frictions and incomplentess of the market, the theory says that European Call and Puts do have the same implied volatility unless there is an arbitrage opportunity by put call parity $$C(t,K) - P(t,K) = DF_t(F_t - K)\ .$$ If you plug the Black-Scholes formula here for the prices of the call and the put, you will see that the equality only ...
19
There is no "plain Black Scholes implied surface" because implied volatilities come from options market prices (calls and put). If you had a whole continuum of call prices $C : \mathbb{R}_+ \times \mathbb{R}_+ \to \mathbb{R}_+$, $(T,K) \mapsto C(T,K)$ you would get a implied volatility function $\sigma_I : \mathbb{R}_+ \times \mathbb{R}_+ \to \mathbb{R}_+$ ...
18
It is a very simple procedure and yes, Newton-Raphson is used because it converges sufficiently quickly: You need to obviously supply an option pricing model such as BS. Plug in an initial guess for implied volatility -> calculate the the option price as a function of your initial iVol guess -> apply NR -> minimize the error term until it is sufficiently ...
17
I generally agree with @dm63's answer: A convex (concave) smile around the forward usually indicates and leptokurtic (platykurtic) implied risk-neutral probability density. Both situations can or cannot admit arbitrage. I provide you with two counterexamples to your statements. A volatility smile that is concave around the forward does not necessarily ...
17
The value of a call option does not go to infinity as the volatility goes to infinity. It tends to the discounted value of the forward $F=S_0 e^{(r-q)T}$, which when the dividend yield is zero, corresponds to the current value of the stock price $S_0$. Let me explain why. The value of a call option increases with volatility as the upside to the option is ...
16
You may want to first broadly categorize volatility models before comparing between them within each class, it does not make sense to compare standard deviation models with an implied vol model. I would broadly classify as follows: Historical realized volatility: Those include standard deviation (sum of squared deviations), realized range volatility ...
16
For an option with price $C$, the P$\&$L, with respect to changes of the underlying asset price $S$ and volatility $\sigma$, is given by \begin{align*} P\&L = \delta \Delta S + \frac{1}{2}\gamma (\Delta S)^2 + \nu \Delta \sigma, \end{align*} where $\delta$, $\gamma$, and $\nu$ are respectively the delta, gamma, and vega hedge ratios. Then it is clear ...
15
Setting aside, that it's not pure riskless arbitrage, but rather statistical arbitrage: You can extract the profit by performing continuous delta hedging. If you constantly adjust your hedge position you gain/lose money by delta hedging. Being long option (gamma long), you sell at higher prices and buy at lower ones. Over the course of time you realize ...
15
Partly because it's hard to get a hold of, the Arslan et. al. paper is starting to assume mythical proportions. As said by Dimitri Vulis, the general idea of the paper is set out in (one or two of) Peter Carr's papers. For the benefit of the OP and others I will try to summarize the most salient points of the paper below and also point out the assumptions ...
14
Along with Gatheral's book, I'd recommend reading Lorenzo Bergomi's "Stochastic Volatility Modelling". The first 2 chapters are available for download on his website. That being said, let me try to give you the basic picture. Below we assume that the equity forward curve $F(0,t)=\Bbb{E}_0^\Bbb{Q}[S_t]$ is given for all $t$ smaller than some relevant ...
13
It seems that you are thinking of the volatility as some sort of standard deviation of your stock price. It is not. In the BS model, $\sigma\sqrt{T}$ is the standard deviation of the log-return $\log(\frac{S_T}{S_0})$. There is no mathematical upper bound to its standard deviation. There is also no mathematical problem with returns being negative either. ...
13
Upon close reading, this appears to be 3 (interesting) questions, not one. I'm not sure if the mods have the tools needed to split it up, so I'm just going to write down the three questions as I see them and then deal with them one by one. Note, it is simpler for me to talk about variance instead of volatility. This has no material impact on the answer. ...
13
Yes it is a better way. Just take a look to figure 3, from Buss and Vilkov (2012, RFS):
13
I think it's interesting to look at this problem graphically also. I get a different answer, depending on whether the option is ITM, ATM, or OTM. In the plot below, all options have 1-year expiry, rates are set to 0.01 and spot is 100. The ITM call has strike 80, the ATM call has strike 100 and the OTM call has strike 150. I added a linear function (y = 40* ...
12
My try to answer this question with some other questions: Is the BS model right? No. Is it useful: yes. Taking a traded price and the BS Model there is only one input factor that is not given by the market: the implied volatility. It is a measure to compare options across time and strike. Are there better models? yes. Those that you mention: The local vol ...
12
Note that total implied variance defined as $$V(T,K) = T\Sigma(T,K)^2$$ should be an increasing function of $T$. Otherwise you have a calendar arbitrage (sell the call with shorter expiry and buy the cheap longer one). If you interpolate linearly your implied volatility is $$\Sigma(T,K) = w\Sigma(T_i,K) + (1-w)\Sigma(T_{i+1},K)$$ with weight $w = \... 12 I'll outline how you can estimate the (implied) real-world density function from (observed) option prices. Having found this real-world density, you can then compute all sorts of probabilities and quantify the market's expectation of future prices. Recall firstly that (European-style) options are priced as risk-neutral expectation of the discounted payoff. ... 10 What they gave you is Newton's formula. If you have a function$f(x)$then you can find the value$x_0$such that$f(x_0) = 0$by this method. It uses the derivative$f'$which in your case is the vega. Your function is: $$f(x) = BS(x) - M$$ where$BS$is the theoretical price with volatility$x$and$M$is the marketprice. Then$f'(x)$is the ... 10 Some Notations It's easy to get lost so let's introduce some notations and let $$\sigma : (t, S, K, \tau) \to \sigma(K,\tau; S, t)$$ denote the implied volatility smile prevailing at time$t$when the spot price is$S_t=S$for an option with strike level$K$and time to expiry$\tau=T-t$. From here onward, we drop the$t$argument to keep notations ... 10 From an equities perspective, there are two concepts that should not be confused in my opinion and context should make the distinction self-explicit: Forward variance swap volatility (A) Forward implied volatility smile (B) I really recommend reading Bergomi's "Stochastic Volatility Modeling" which is an excellent book for equity practitioners. The topics ... 10 Great question. Let me try provide some insights and thoughts regarding your points and questions raised. It may not be a full answer but hopefully it helps to connect the contents in the paper/book with some trading intuition: From a theoretical perspective, I don't see any mistake in your thinking regarding skew decay but two questions arise on my end: ... 9 A very popular choice for mean reversion is the Ornstein–Uhlenbeck process (here in discretized form): $$L_{t+1}-L_t=\alpha(L^*-L_t)+\sigma\epsilon_t$$ Here you see that the level change is governed by some parameter$\alpha$, the mean reversion rate (or speed), and the distance between the long run mean$L^*$and the actual level$L_t\$ plus some noise. A ...
9
The idea of regime switching in volatility is rooted in the observation that volatility is usually fairly consistent and "mild", and occasionally very high, say during a market crash. The concept goes further, though. Not only does the volatility level differ markedly in different regimes, but the behavior of volatility does as well (degree of mean reversion,...
9
Regime switching is another way to describe structural changes in a data series. For example, an inflation timeseries may change states from ARMA to linear as the economy moves from a period of cyclical growth to prolonged recession. A stock price may, say, be determined by and correlated to the main equity index when it has a large market capitalisation ...
9
Consider what happens when IV is lower than realised vol. The person long the IV would make money. So there would ideally be no one selling IV if it's lower than realised vol on an average. Next if IV is equal to RV, then the guy selling the option has no incentive to sell since he won't make money on average. Also he has considerable risk in case RV ...
8
If you look at tick data, you will probably get an even better analysis. However, vix correlation tends to be negative with spx but remember that this is generally more true for when spx tanks. When spx goes up, the correlation isn't as strong. Why? People panic after a drop, therefore leading to people buying options. They don't care about black scholes ...
8
Implied volatility does not have to be equal (so yes, it can be different) for a call and put of same underlying, underlying borrow rates, time to expiration, strike if: If the underlying is a stock and the underlying cannot be easily borrowed for short selling If there are dividends or other costs of carry involved If there is not unlimited liquidity in ...
Only top voted, non community-wiki answers of a minimum length are eligible | 2021-05-08T05:23:41 | {
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https://mathematica.stackexchange.com/posts/103222/revisions | 2 added 403 characters in body edited Jan 2 '16 at 19:12 Stelios 1,33111 gold badge99 silver badges1313 bronze badges One approach to this kind of problems is to discretize the continuous domain to a finite (but sufficiently large) number of (not necessarily equispaced) samples and then approximate the original problem on this grid. The resulting formulation is a linear programming problem for which efficient solvers are available. f[x_] := Exp[x]^(1/2) + 2 - Exp[x] + x^5; l = 5000; (*number of samples of the [0,1] interval*) xRange = Range[0., 1., 1./(l - 1)]; (*grid points*) (*auxiliary matrix; note that {a,b,c}.mA is ax^2+bx+c evaluated on the grid points*) mA = {xRange^2, xRange, Table[1., {l}]}; (*auxiliary matrix; note that {a,b,c}.mB is the derivative of ax^2+bx+c evaluated on the grid points*) mB = {2 xRange, Table[1., {l}]}; sol = (* note introduction of an additional optimization variable delta *) FindMinimum[{delta, And @@ Thread[-delta <= {a, b, c}.mA - f[xRange] <= delta] && And @@ Thread[{a, b}.mB <= 0]}, {delta, a, b, c}] {0.181843, {delta -> 0.181843, a -> 0.0695606, b -> -0.139121, c -> 1.81816}} Plot[{f[x], Evaluate[a x^2 + b x + c /. sol[[2]]]}, {x, 0, 1}] The optimal polynomial approximation in this case is pretty far from close to the actual function due to the non-increasing constraint. Actually, generalizing this approach to larger order polynomials shows that the approximation gets only slightly better. Of course, removing this constraint results in much better approximations that rapidly converge to the actual function with increasing order. One approach to this kind of problems is to discretize the continuous domain to a finite (but sufficiently large) number of (not necessarily equispaced) samples and then approximate the original problem on this grid. The resulting formulation is a linear programming problem for which efficient solvers are available. f[x_] := Exp[x]^(1/2) + 2 - Exp[x] + x^5; l = 5000; (*number of samples of the [0,1] interval*) xRange = Range[0., 1., 1./(l - 1)]; (*grid points*) (*auxiliary matrix; note that {a,b,c}.mA is ax^2+bx+c evaluated on the grid points*) mA = {xRange^2, xRange, Table[1., {l}]}; (*auxiliary matrix; note that {a,b,c}.mB is the derivative of ax^2+bx+c evaluated on the grid points*) mB = {2 xRange, Table[1., {l}]}; sol = (* note introduction of an additional optimization variable delta *) FindMinimum[{delta, And @@ Thread[-delta <= {a, b, c}.mA - f[xRange] <= delta] && And @@ Thread[{a, b}.mB <= 0]}, {delta, a, b, c}] {0.181843, {delta -> 0.181843, a -> 0.0695606, b -> -0.139121, c -> 1.81816}} Plot[{f[x], Evaluate[a x^2 + b x + c /. sol[[2]]]}, {x, 0, 1}] One approach to this kind of problems is to discretize the continuous domain to a finite (but sufficiently large) number of (not necessarily equispaced) samples and then approximate the original problem on this grid. The resulting formulation is a linear programming problem for which efficient solvers are available. f[x_] := Exp[x]^(1/2) + 2 - Exp[x] + x^5; l = 5000; (*number of samples of the [0,1] interval*) xRange = Range[0., 1., 1./(l - 1)]; (*grid points*) (*auxiliary matrix; note that {a,b,c}.mA is ax^2+bx+c evaluated on the grid points*) mA = {xRange^2, xRange, Table[1., {l}]}; (*auxiliary matrix; note that {a,b,c}.mB is the derivative of ax^2+bx+c evaluated on the grid points*) mB = {2 xRange, Table[1., {l}]}; sol = (* note introduction of an additional optimization variable delta *) FindMinimum[{delta, And @@ Thread[-delta <= {a, b, c}.mA - f[xRange] <= delta] && And @@ Thread[{a, b}.mB <= 0]}, {delta, a, b, c}] {0.181843, {delta -> 0.181843, a -> 0.0695606, b -> -0.139121, c -> 1.81816}} Plot[{f[x], Evaluate[a x^2 + b x + c /. sol[[2]]]}, {x, 0, 1}] The optimal polynomial approximation in this case is pretty far from close to the actual function due to the non-increasing constraint. Actually, generalizing this approach to larger order polynomials shows that the approximation gets only slightly better. Of course, removing this constraint results in much better approximations that rapidly converge to the actual function with increasing order. 1 answered Jan 2 '16 at 18:47 Stelios 1,33111 gold badge99 silver badges1313 bronze badges One approach to this kind of problems is to discretize the continuous domain to a finite (but sufficiently large) number of (not necessarily equispaced) samples and then approximate the original problem on this grid. The resulting formulation is a linear programming problem for which efficient solvers are available. f[x_] := Exp[x]^(1/2) + 2 - Exp[x] + x^5; l = 5000; (*number of samples of the [0,1] interval*) xRange = Range[0., 1., 1./(l - 1)]; (*grid points*) (*auxiliary matrix; note that {a,b,c}.mA is ax^2+bx+c evaluated on the grid points*) mA = {xRange^2, xRange, Table[1., {l}]}; (*auxiliary matrix; note that {a,b,c}.mB is the derivative of ax^2+bx+c evaluated on the grid points*) mB = {2 xRange, Table[1., {l}]}; sol = (* note introduction of an additional optimization variable delta *) FindMinimum[{delta, And @@ Thread[-delta <= {a, b, c}.mA - f[xRange] <= delta] && And @@ Thread[{a, b}.mB <= 0]}, {delta, a, b, c}] {0.181843, {delta -> 0.181843, a -> 0.0695606, b -> -0.139121, c -> 1.81816}} Plot[{f[x], Evaluate[a x^2 + b x + c /. sol[[2]]]}, {x, 0, 1}] | 2019-08-21T23:40:28 | {
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https://math.stackexchange.com/questions/1309859/question-about-analytic-function-for-series-solutions | # Question about Analytic function for series solutions
Hello all I am having a bit of confusion in regard to the following;
I understand that when we are working with ODE of the form
$$p_o(x)y''(x)+p_1(x)y'(x)+p_2(x)y(x)=0$$ and we are considering some singular point say $x_o$.
I know then for if all p polynomials, we can use the limit method to determine if it is regular or irregular singular point.
But also I know it must hold in general for all analytic functions and the limit method can only be used with polynomials.
So my question is, how can I know if something is analytic. I know the definition of it, i.e. , if it has a convergent power series at $x_o$, but surely we are not supposed to construct a taylor series for each function we see to do this?
The same methods should be usable here for analytic functions (or more generally, meromorphic functions) as for polynomials. For example, the Wikipedia page "Regular singular point" deals with the coefficients being meromorphic functions. In your case, you want $p_1(z)/p_0(z)$ to have at most a pole of order $1$ and $p_2(z)/p_0(z)$ at most a pole of order $2$ at $x_0$, which will be true if $(z-x_0) p_1(z)/p_0(z)$ and $(z - x_0)^2 p_2(z)/p_0(z)$ are analytic and bounded in a deleted neighbourhood of $x_0$.
Consider $p_{1}(x)/p_{0}(x)$ in the complex $z$ plane. This can then be broken into real and imaginary components $p_{1}(z)/p_{0}(z) = u(x,y)+iv(x,y)$.
Then inside some region where $u(x,y)$ and $v(x,y)$ satisfy the Cauchy Riemann equations:
$p_{1}(z)/p_{0}(z)$ will be analytic. So if your point of interest $z_0=x_0$ lies inside such region then $p_{1}(x)/p_{0}(x)$ will be analytic at the point $x_0$. | 2019-09-16T08:14:44 | {
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https://stacks.math.columbia.edu/tag/0871 | Lemma 69.11.5. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume
1. $f$ is locally of finite type and quasi-affine, and
2. $Y$ is quasi-compact and quasi-separated.
Then there exists a morphism of finite presentation $f' : X' \to Y$ and a closed immersion $X \to X'$ over $Y$.
Proof. By Morphisms of Spaces, Lemma 66.21.6 we can find a factorization $X \to Z \to Y$ where $X \to Z$ is a quasi-compact open immersion and $Z \to Y$ is affine. Write $Z = \mathop{\mathrm{lim}}\nolimits Z_ i$ with $Z_ i$ affine and of finite presentation over $Y$ (Lemma 69.11.1). For some $0 \in I$ we can find a quasi-compact open $U_0 \subset Z_0$ such that $X$ is isomorphic to the inverse image of $U_0$ in $Z$ (Lemma 69.5.7). Let $U_ i$ be the inverse image of $U_0$ in $Z_ i$, so $U = \mathop{\mathrm{lim}}\nolimits U_ i$. By Lemma 69.5.12 we see that $X \to U_ i$ is a closed immersion for some $i$ large enough. Setting $X' = U_ i$ finishes the proof. $\square$
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). | 2023-03-30T01:02:00 | {
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https://physics.stackexchange.com/questions/444817/is-the-maximum-ability-of-a-rope-to-withstand-the-tension-force-constant | # Is the maximum ability of a rope to withstand the tension force constant?
Suppose, I have a rope which has a length of 10 meters. I know that it can handle up to 100 N centrifugal force in circular velocity condition. Then I attached a 10 gram spherical object to the end of the rope and I started rotating the object at 10 revolutions per second without the rope breaking.
If I want to rotate this again with the same angular velocity and object but change the rope length to 100 m. Will the rope be strong enough?
• Have you calculated the centripetal force for the 10 m length of rope? If you then increase the length to 100 m, how does the force change? – Mick Dec 3 '18 at 12:23
• The strength or the rope is related to the total number of twists, and not the rate of twisting. – ja72 Dec 3 '18 at 23:28
Example: Let's assume that the probability that a $$1$$ meter rope is having one or more points that will break at less than $$10$$ N is equal to $$p$$. This would mean that the same probability (of having at least one point which will break at less than $$10$$ N), is going to be $$(1-(1-p)^{10})$$ for a $$10$$ meter rope. | 2020-03-31T22:30:03 | {
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http://clay6.com/qa/512/determine-p-e-f-mother-father-and-son-line-up-at-random-for-a-family-pictur | Browse Questions
# Determine P(E|F) . Mother, father and son line up at random for a family picture E : son on one end, F : father in middle
$\begin{array}{1 1} 0 \\ 1 \\ \frac{1}{2} \\ \frac{1}{3} \end{array}$
Toolbox:
• In any problem that involves determing the outcomes, we can write down the sample space and cout the number of favorable outcomes. Then we find P(E), P(F), P(E$\cap$F) using the set of outcomes.
• Given P(E), P(F), P(E $\cap$ F), P(E/F) $= \large \frac{P(E \;\cap \;F)}{P(F)}$
Given the mother M, father F and son S line up for a family picture, the sample space will be S: (MFS, MSF, FMS, FSM, SMF, SFM) and the total number of outcomes = 6.
For the event E: where the son is on one end, E = (MFS, FMS, SMF, SFM). The total number of outcomes = 4.
$\Rightarrow P(E) = \large \frac{\text{Number of favorable outcomes in E}}{\text{Total number of outcomes in S}} = \frac{4}{6} = \frac{2}{3}$
For the event F: where the father is in the middle, F = (MFS, SFM). The total number of outcomes = 2.
$\Rightarrow P(F) = \large \frac{\text{Number of favorable outcomes in F}}{\text{Total number of outcomes in S}} = \frac{2}{6} = \frac{1}{3}$
For our events, E $\cap$ F = (MFS, SFM). The total number of outcomes = 2.
$\Rightarrow P(E \cap F) = \large \frac{\text{Number of favorable outcomes in E$\cap$F}}{\text{Total number of outcomes in S}} = \frac{2}{6} = \frac{1}{3}$
Given P(E), P(F), P(E $\cap$ F), P(E/F) $= \large \frac{P(E \;\cap \;F)}{P(F)}$
$\Rightarrow P(E/F) = \Large\frac {\Large \frac{1}{3}}{\Large\frac{1}{3}}$ = 1.
edited Jun 18, 2013 | 2016-10-22T08:57:00 | {
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http://openstudy.com/updates/4f1a2038e4b04992dd22891f | ## anonymous 4 years ago the rectangular box has a 2x2 base and height of 1. cosine of angle between AC and AB is...
1. anonymous
2. anonymous
hints please, trying to solve on my own...
3. Mertsj
We need to see the rest of the problem
4. anonymous
i posted attachment
5. Mertsj
I know and it's not on that either
6. anonymous
yes it is
7. anonymous
you sure? i'm pretty sure it's solveable, just can't remember how...i believe the angle between the vector and AB (i'll label x axis) would be roh(?)
8. anonymous
divide the problem into two right triangles
9. anonymous
elica85. First, project the line down onto the base square of the triangle. Find the hypotenuse of the triangle that is formed by doing this. This is our adjacent quantity. We know the opposite quantity. Now we can use the tangent function.
10. anonymous
whoops. project the line down onto the base rectangle of the prism! sorry
11. anonymous
or use dot product of AC with AB
12. Mertsj
The cosine of the angle is 2/3
13. anonymous
ok, the dot product gives me the cosine? anything else i need to do after? and why dot produvct?
14. anonymous
because it gives you the cos of the angle, which is what you are trying to solve for
15. anonymous
|dw:1327112719359:dw|the dot product of A and B is C
16. anonymous
so,$A*B=|AC||AB|\cos \theta$
17. anonymous
let me know if you need more help from here
18. anonymous
Sorry I should have wrote $AC*AB=|AC||AB|\cos \theta$
19. anonymous
thx, still trying to understand...first how do i get AC...it would be sqrt(1^2+?^2)
20. anonymous
no
21. anonymous
o, sqrt(1+8)=3
22. anonymous
so AC=3
23. anonymous
AC=<2,2,1> so |AC|=(2^2+2^2+1^2)^1/2
24. anonymous
ok...i saw that dot product formula quite a few times but i don't get it. <AB>dot<AC>=|AB||AC|...and multiplied by cos...
25. anonymous
yeah, exactly. |AB||AC|cosa, where a i the angle between vectors AB and AC
26. anonymous
do you remember how to calculate the dot product part?
27. anonymous
yes. mult component wise and then add
28. anonymous
yep :)
29. anonymous
ok so have that on the left side...and then solve for cosa
30. anonymous
exactly
31. anonymous
ok ok. so use sina if the question asks to find the sine of the angle?
32. anonymous
no..the dot product only gives you cosines
33. anonymous
so if i'm asked for sine...?
34. anonymous
then you need to use another relation. if you have corresponding angles for example, you can relate the cos of one angle to the sin of another.
35. anonymous
ok, thx you closed the gap a lot for me just from this problem
36. anonymous
or the question might involve the cross-product of two vectors which involve the sin of the angle between the two vectors..depends on the problem
37. anonymous
np
38. anonymous
i get .84 but the closest out of the choices i have is 2/sqrt(5)...or none of the above
39. anonymous
I get cos a = 2/3 | 2016-10-23T08:13:00 | {
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https://math.stackexchange.com/questions/453203/definition-of-sheaf-using-equalizer | # Definition of sheaf using equalizer
Wikipedia give sheaf property using equalizer diagram by saying sheaf property means for any open cover $\{U_i\}$ of $U$
$$F(U) \rightarrow \prod_{i} F(U_i) {{{} \atop \longrightarrow}\atop{\longrightarrow \atop {}}} \prod_{i, j} F(U_i \cap U_j)$$
is an equalizer. What means two arrows? If $i = 1,2,3$ then what means
$$F(U) \rightarrow F(U_1) \times F(U_2) \times F(U_3) {{{} \atop \longrightarrow}\atop{\longrightarrow \atop {}}} F(U_1 \cap U_2) \times F(U_2 \cap U_3) \times F(U_1 \cap U_3)?$$
• There are textual descriptions above and below the indicated diagram at the cited Wikipedia article. Have you read those? – Zhen Lin Jul 27 '13 at 1:44
• @ZhenLin I read it but I still don't understand. I am confused what is meant there. – user23086 Jul 27 '13 at 2:14
• Is there a typo when you say: Similarly , fo each of $i,j \in I$, we have a map $b_{i,j}$: $$res^{U_j}_{U_i \cap U_j} \circ p_j : \prod_{i \in I} F(U_i) \rightarrow F(U_i \cap U_j)$$. Shouldn't it be: $$res^{U_j}_{U_i \cap U_j} \circ p_j : \prod_{i \in I} F(U_j) \rightarrow F(U_i \cap U_j)$$ because you are restricting $U_j$ to $U_i \cap U_j$ unless I am not understanding something. Sorry i couldnt put that in comments because my reputation is not 50 above!! – Avadutta Jun 12 '18 at 6:04
Given maps $f:X\to Y$ and $g:X\to Y$, a natural way of writing them together in the same diagram would be like this: $$X{{f\atop {\Large\longrightarrow}}\atop{{\Large\longrightarrow}\atop g}}Y$$ A map $\mathrm{eq}:E\to X$ is an equalizer (Wikipedia) of the maps $f$ and $g$ if it is final in the category of maps to $X$ that equalize $f$ and $g$. Depicting it together with the maps $f$ and $g$, we say that $$E\xrightarrow{\;\mathrm{eq}\;} X{{f\atop {\Large\longrightarrow}}\atop{{\Large\longrightarrow}\atop g}}Y$$ is an equalizer diagram.
Let $X$ be a topological space, let $F$ be a presheaf on $X$, and let $\{U_i\}_{i\in I}$ be an open cover of $X$. The presheaf $F$ comes with restriction maps $\mathrm{res}_{V}^{U}:F(U)\to F(V)$ for each pair of open sets $U,V$ with $V\subseteq U$. In particular, for each pair of $i,j\in I$, we have restriction maps $$\large\mathrm{res}_{U_i\cap U_j}^{U_i}:F(U_i)\to F(U_i\cap U_j)$$ and $$\large\mathrm{res}_{U_i\cap U_j}^{U_j}:F(U_j)\to F(U_i\cap U_j).$$ By definition, for each $i\in I$, the product (Wikipedia) $\prod_{i\in I}F(U_i)$ comes with a projection map $$p_i:\prod_{i\in I}F(U_i)\longrightarrow F(U_i).$$ Moreover, $F(U_i)$ comes with a restriction map $\mathrm{res}_{U_i\cap U_j}^{U_i}:F(U_i)\to F(U_i\cap U_j)$ for each $j\in I$. Composing them, we have for each pair of $i,j\in I$, a map $a_{i,j}$, as follows: $$\underbrace{(\mathrm{res}_{U_i\cap U_j}^{U_i}\circ p_i)}_{\Large a_{i,j}}:\prod_{i\in I}F(U_i)\longrightarrow F(U_i\cap U_j)$$ By the definition of a product, these $a_{i,j}$'s induce a map (let's call it $\alpha$) from $\prod_{i\in I}F(U_i)$ to the product of all of the $F(U_i\cap U_j)$'s together: $$\alpha:\prod_{i\in I}F(U_i)\longrightarrow \prod_{i,j\in I}F(U_i\cap U_j).$$ Similarly, for each pair of $i,j\in I$, we have a map $b_{i,j}$: $$\underbrace{(\mathrm{res}_{U_i\cap U_j}^{U_j}\circ p_j)}_{\Large b_{i,j}}:\prod_{i\in I}F(U_i)\longrightarrow F(U_i\cap U_j)$$ which come together to form a single map $$\beta:\prod_{i\in I}F(U_i)\longrightarrow \prod_{i,j\in I}F(U_i\cap U_j).$$ These maps $\alpha$ and $\beta$ are the top and bottom arrows in the diagram $$F(U) \longrightarrow \prod_{i\in I} F(U_i) {{{}\atop {\Large\longrightarrow}}\atop{{\Large\longrightarrow}\atop {}}}\prod_{i, j\in I} F(U_i \cap U_j)$$ We say that $F$ is a sheaf (Wikipedia) if, for any open cover $\{U_i\}_{i\in I}$ of our space $X$, this diagram is an equalizer diagram.
Note that $\alpha$ and $\beta$ are not the same map; they take different "paths". Let's trace out what happens: the top-right path is $a_{i,j}$, and the bottom-left path is $b_{i,j}$. $\require{AMScd}$ $$\require{AMScd} \begin{CD} \prod_{i\in I}F(U_i) @>{\Large p_i}>> F(U_i);\\ @V{\Large p_j}VV @VV{\Large\mathrm{res}_{U_i\cap U_j}^{U_i}}V \\ F(U_j) @>>{\Large\mathrm{res}_{U_i\cap U_j}^{U_j}}> F(U_i\cap U_j); \end{CD}$$ The two maps $\alpha$ and $\beta$ to the product $\prod_{i,j\in I}F(U_i\cap U_j)$ agree on a specific element $(s_i)_{i\in I}$ of the product $\prod_{i\in I}F(U_i)$ if and only if $a_{i,j}$ and $b_{i,j}$ agree on it for all pairs of $i,j\in I$. But $a_{i,j}$ and $b_{i,j}$ are clearly going to be different maps in general, so this is not a trivial condition.
• @Zee: The comment below is wrong, the product $$\prod_{i\in I}F(U_j)$$ is just $F(U_j)$ to the power $|I|$ (the cardinality of $I$), a very different object than $$\prod_{i\in I}F(U_i)$$ – Zev Chonoles Apr 5 '19 at 14:22
• It looks to me like one either has to distinguish between $F(U_i \cap U_j)$ and $F(U_j \cap U_i)$, or that one has to specify for each unordered pair $(i,j)$ which of the two maps $\prod_{i} F(U_i) \to F(U_i \cap U_j)$ one is choosing for $\alpha$ and which for $\beta$. – Improve May 13 '19 at 22:45 | 2020-02-19T19:59:24 | {
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https://ijc.fd.uc.pt/h5hmc/b3520f-hypothesis-testing-calculator | Hypothesis Testing Calculator. Binomial Coefficient Calculator and we cannot reject the hypothesis. If the z score is above the critical value, this means that it is is in the nonrejection area, Here are some more working examples of hypothesis testing with detail explanation: accidents a year and the company's claim is inaccurate. Hypothesis. ... Calculator for Statistics. The Test for one proportion in the Tests menu can be used to test the hypothesis that an observed proportion is equal to a pre-specified proportion. The alternative hypothesis is the hypothesis that we believe it actually is. Type in the values from the two data sets separated by commas, for example, 2,4,5,8,11,2. 5%, the 2 ends of the normal This means we want to see if the sample mean is less than the hypothesis mean of $40,000. Chebyshev's Theorem Calculator mean is much lower than what the real mean really is. The right tail method, just like the left tail, has a critical value. This means that the null hypothesis claim is false. Area Under the Curve Calculator This means that if we obtain a z score below the critical value, A hypothesis test for the difference in samples means can help you make inferences about the relationships between two population means. Statistic Math Testing. That is if one is true, the other one must be false and vice versa. An important part of inferential statistics is hypothesis testing. Select the appropriate test for your application, punch in the parameters and receive a comprehensive solution as it would be calculated by hand. mean is much higher than what the real mean really is. Steve Earth. There are 5 main steps in hypothesis testing: State your research hypothesis as a null (H o) and alternate (H a) hypothesis. If you’re talking about conversion-rate AB testing, your hypothesis may involve adding a button, image, or some copy to a page to see if it affects conversion rates. This test can be applied to any univariate dataset. that most likely it receives much more. The mean daily return of the sample is 0.1% and the standard deviation is 0.30%. Left tail hypothesis testing is illustrated below: We use left tail hypothesis testing to see if the z score is above the significance level critical value, in which case we cannot reject the This t-test, unlike the z-test, does not need to know the population standard deviation $$\sigma$$. You can us… Variation B's observed conversion rate was higher than variation A's conversion rate ().You can be 95 % confident that this result is a consequence of the changes you made and not a result of random chance. p value calculator. Step 3 Compute the test value. 4. This online hypothesis testing calculator for population mean helps you to perform the two-tailed and one-tailed statistical hypothesis testing. Step 2 Find the critical value(s) from the appropriate table. HIOX Softwares Pvt Ltd Education. Use this free sample and population statistics calculator to perform a statistical hypothesis test for the given population mean. 3. s = 1. Is there a significant difference in the correlation of both cohorts? alpha beta power hypothesis testing. I greet you this day: First: Read the notes. Notes For Hypothesis Testing In Statistics 1. Testing Mean (known Variance) - Critical Value: computes the critical value for one- and two-sided hypothesis tests about the mean. Hypothesis testing for the mean Calculator: 100 randomly selected items were tested. An important part of inferential statistics is hypothesis testing. This means that the hypothesis is false. We also provide Hypothesis Testing calculator with downloadable excel template. For example, if three outcomes measure the effectiveness of a drug or other intervention, you will have to adjust for these three analyses. Online calculator for computing and testing correlations, calculation of confidence intervals, testing against fixed values, averaging correlations and interpretation of their magnitude ... All hypothesis tests on this page are based on Eid et al. Please select the null and alternative hypotheses, type the hypothesized mean, the significance level, the sample mean, the population standard deviation, and the sample size, and the results of the z-test will be displayed for you Suppose we want to know that the mean return from a portfolio over a 200 day period is greater than zero. ... Null Hypothesis. hypothesis as true. This is because the z score will because the real mean is really greater than the hypothesis mean. An academic study states that the cookbook method of teaching introductory statistics leaves no time for history, philosophy or controversy. Bring it to Science: A hypothesis is an unproven theory about a Scientific problem. Then hit "Calculate" and the test statistic and p-Value will be calculated for you. To use this calculator, a user simply enters in the hypothesis testing method (either one-tail or two-tail), the sample size, and the significance level and clicks the 'Calculate' button. Then hit "Calculate" and the test statistic and p-Value will be calculated for you. This is a classic left tail hypothesis test, where the Significant Figures (Sig Fig) Calculator, Sample Correlation Coefficient Calculator. Everyone. Learn how to perform hypothesis testing with this easy to follow statistics video. This example was to conduct a p-test for a population proportion. H0 is Null Hypothesis. Solve the test statistic, p-value and critical value. This Hypothesis Testing Calculator determines whether an alternative hypothesis is true or not. This means that if we obtain a z score above the critical value, Find P and T score value using the hypothesis testing for population … Hypothesis Testing. hypothesis. With the hypothesis testing method, sample size, and significance level, we can find the t-value for a given data set. of 1%, you are choosing a normal standard distribution that has a rejection area of 1% of the total 100%. Conducting a Hypothesis Test for the Difference in Means. Testing Mean (unknown Variance) - p-value: computes the 2-sided p-value for the statistical hypothesis test about the mean when the population variance is unknown. That is if one is true, the other one must be false and vice versa. In this case, the alternative hypothesis is true. This is the alternative hypothesis. you increase the significance level, the greater area of rejection there is. The significance level represents If the calculated z score is between the 2 ends, we cannot reject the null hypothesis and we reject the alternative hypothesis. Enter your sample means, sample standard deviations, sample sizes, hypothesized difference in means, test type, and significance level to calculate your results. Video transcript. In this case, where you need tofind the critical values for the t-distribution for a given sample size? z score is below the critical value, this means that we cannot reject the null hypothesis and we reject the alternative hypothesis It was found that the average of the sample was 980. Hypothesis Test, also known as Statistical Hypothesis Test is a method of statistical inference. A free online hypothesis testing calculator for population mean to find the Hypothesis for the given population mean. For example, let's say that Hypothesis testing represents a very important part of Statistics, and it is usually misunderstood in terms of the the objectives and methodology. The null hypothesis is the hypothesis that is claimed and that we will test against. rejection area. This really means there are fewer than 400 worker accidents a year and the company's claim is T-value Calculator Bring it to Logic: A hypothesis is defined as the premise of a conditional sattement. Statistical hypothesis testing is considered a mature area within statistics, but a limited amount of development continues. Enter your sample proportions, sample sizes, hypothesized difference in proportions, test type, and significance level to calculate your results. In this case, the null hypothesis which the researcher would like to reject is that the mean daily return for the portfolio is zero. Activity. Large sample proportion hypothesis testing. Parent topic: Statistics. This calculator lets the user input: 'sample size', 'null probability' 'number of successful trials' and 'significance level', and generates a generic test with conclusion, based on these values. because the real mean is actually less than the hypothesis mean. alternative hypothesis is that the mean is greater than 400 accidents a year. So, for example, it could be used to determine whether the mean diastolic blood pressure of a particular group differs from 85, a value determined by a previous study. The following examines an example of a hypothesis test, and calculates the probability of type I and type II errors. How to use this t-test calculator for One Sample. Hypothesis testing for a proportion is used to determine if a sampled proportion is significantly different from a specified population proportion. Third: Solve the questions/solved examples. Hypothesis test need an analyst to state a null hypothesis and an alternative hypothesis. Hypothesis testing helps identify ways to reduce costs and improve quality. A calculator aimed at automating hypothesis tests using the binomial distribution. Variance Calculator If you do a large number of tests to evaluate a hypothesis (called multiple testing), then you need to control for this in your designation of the significance level or calculation of the p-value. Confidence Interval Calculator Expected Value Calculator So the greater the significance level, the smaller or narrower the nonrejection area. z score is above the critical value, this means that we cannot reject the null hypothesis and we reject the alternative hypothesis It is most often used by scientists to test specific predictions, called hypotheses, that arise from theories. than the hypothesis mean of 400. Test for one proportion free online statistical calculator. This calculator calculates the p-value for a given set of data based on the test statistic, sample size, hypothesis testing type (left-tail, right-tail, or two-tail), and the significance level. A significance value (P-value) and 95% Confidence Interval (CI) of the observed mean is reported. Any value Proportion hypothesis testing is applied for making inferences around a proportion, like for election results.The test holds an assumed proportion up against an alternative claim, like a new sample mean.. Therefore, the Containing nine different tests including one and two sample options as well as one-tailed and two-tailed tests for each option. This means we want to see if the sample mean is greater Hypothesis testing is just a method for testing a claim or hypothesis about a parameter in a population, using data measured in a sample. The two tail method has 2 critical values (cutoff points). If you choose a significance level of 5%, you are increasing Enter the size of the sample n, sample mean m, population standard deviation s. 1. n = 1. and we cannot reject the hypothesis. chance you have of accepting the hypothesis, since the nonrejection area decreases. few years. Recall: Ask students to list the steps in the Scientific Method Is Forming a Hypothesis one of the methods? Hypothesis testing is the use of statistics to determine the probability that a given hypothesis is true. The null hypothesis, in this case, is a two-t… There is a wide range of methods for this. the critical value. If you choose a significance level of 20%, you increase the rejection area of the standard normal curve to 20% of the 100%. Hypothesis Testing: Contains Ads. 2. m = 0. Her test statistic, I can never say that right, was t is equal to negative 1.9. For tests involving means or standard deviations, the statistics may either be supplied directly in hypothesis testing software through methods such as setSample1Size or calculated from the data by specifying variable names. We accept true hypotheses and reject false hypotheses. The more whether we accept or reject the hypothesis. In this method, we test some hypothesis by determining the likelihood that a sample statistic could have been selected, if the hypothesis regarding the population parameter were true. When two populations are related, you can compare them by analyzing the difference between their means. On the basis of the data collected and entered into the statistics calculator, DATAtab will then suggest which hypothesis test you need. You will find a description of how to conduct a two sample t-test below the calculator. A neurologist is testing the effect of a drug on response time by injecting 100 rats with a unit dose of the drug, subjecting each to neurological stimulus and recording its response time. If the z score is below the critical value, this means that it is is in the nonrejection area, The degrees of freedom for at-distribution can be derived from the sample size - just subtract one. Sample Size Calculator If the z score is below the critical value, this means that we reject the hypothesis, Hypothesis Testing Calculator is free, comprehensive and intuitive. This means that the null hypothesis is 400. The level of statistical significance is often expressed as the so-called p-value. Hypothesis testing is a statistical method which is used to make decision about entire population, with the help of only sample data. Here we discuss how to calculate Hypothesis Testing along with practical examples. To use the Hypothesis Test Calculator in a meaningful way, you must first formulate your hypothesis and collect your data. Power of a hypothesis test. The results are mutually exclusive. This critical values calculatoris designed to accept your p-value (willingness to accept an incorrect hypothesis) and degrees of freedom. Hypothesis testing has been taught as received unified method. Hypothesis Testing Calculators. In hypothesis testing, the calculated value of Z-statistic (Z 0), Student's t-statistic (t 0), F-statistic (F 0) or χ²-statistic (χ² 0) is compared with the table (critical) values of one or two tailed normal distribution (Z e), t-distribution (t e), F-distribution (F e) or Chi-squared distribution (χ² e) to check if the results of experiments are statistically significant. Step 5 Summarize the results. Two dependent Samples with data Calculator. If the z score is outside of this range, then we reject the null hypothesis and accept the alternative hypothesis rejection area. A population parameter is a numerical constant that represents o characterizes a distribution. getcalc.com's statistic calculator & formulas to estimate Z0 for Z-test, t0 for student's t-test, F0 for F-test & (χ²)0 for χ² test for sample mean, proportion, difference between two means or proportions hypothesis testing in statistics & probability experiments. 1. Fill in the sample size, n, the number of successes, x, the hypothesized population proportion $$p_0$$, and indicate if the test is left tailed, <, right tailed, >, or two tailed, $$\neq$$. P-value > -10 ( Significance level). This test is not performed on data in the data table, but on statistics you enter in a dialog box. This calculator will conduct a complete one-sample t-test, given the sample mean, the sample size, the hypothesized mean, and the sample standard deviation. If the P-value is less than 0.05, the hypothesis that the observed proportion is equal to the pre-specified proportion value is rejected, and the alternative hypothesis that there is a significant difference between the two proportions can be accepted. These functions are very simple to run; beign able to use and interpret them correctly is the hard part. Fourth: Check your solutions with my thoroughly-explained solutions. and whether the differences are statistically significant or not. The alternative hypothesis is that μ > 20, which To make this decision, we come up with a value called as p-value… Hypothesis testing helps the businesses and researchers, to make better data-based decisions. In is common, if not standard, to interpret the results of statistical hypothesis tests using a p-value. Activity. The null hypothesis is that the mean is 400 worker accidents per year. This is a classic right tail hypothesis test, where the Depending on the statistical test you have chosen, you will calculate a probability (i.e., the p-value) of observing your sample results (or more extreme) given that the null hypothesis is true. There are 3 types of hypothesis testing that we can do. There is left tail, right tail, and two tail hypothesis testing. Install. So, our alternative hypothesis is true that more than 52% of parents today believe that electronics and social media are the cause of their teenager’s lack of sleep. Standard Deviation Calculator Hence, H0 is not rejected. If the The usual process of hypothesis testing consists of four steps. Not all implementations of statistical tests return p-values. So if the hypothesis mean is claimed to be 100. One Proportion Z Test is a hypothesis test to make comparison between a group to specified population proportion. Hypothesis testing is a set of formal procedures used by statisticians to either accept or reject statistical hypotheses. Therefore, we want to determine if this number of accidents is greater than what is being claimed. This t-test calculator allows you to use either the p-value approach or the critical regions approach to hypothesis testing! The hypothesis testing for population mean analyses the results of the null hypothesis and the alternative hypothesis of a population. determines One Proportion Z Test Calculator. For example, let's say that Two tail hypothesis testing is illustrated below: We use the two tail method to see if the actual sample mean is not equal to what is claimed in the hypothesis mean. Enter the sample mean, population mean, sample standard deviation, population size and the significance level to know the T score test value, P value and result of hypothesis. Is Testing the Hypothesis also one of the methods? Hypothesis testing helps identify ways to reduce costs and improve quality. The smaller the significance level, the greater the nonrejection area. One-sample t-test free online statistical calculator. The p-value represents the probability of a null hypothesis being true. For example, let's say that a company claims it only receives 20 consumer complaints on average a year. The results are mutually exclusive. As with learning anything related to mathematics, it is helpful to work through several examples. the total rejection area of a normal standard curve. Therefore, it is false and we reject the hypothesis. Add to Wishlist. Planning a statistical experiment and trying to estimate what results you need to accept a hypothesis? If you answered "yes", then you are right! above this critical value in the right tail method represents the rejection area. 2. m = 0. In addition, critical values are used when estimating the expected intervals for observations from a population, such as in tolerance intervals. Enter in the sample sizes and number of successes for each sample, the tail type and the confidence level and hit Calculate and the test statistic, t, the p-value, p, the confidence interval's lower bound, LB, and the upper bound, UB will be shown. because it is outside the range. If the z score calculated is above the critical value, this means There is no point where it says it is an excusive list of hypothesis tests. Z-Hypothesis Testing (stats) Z-Hypothesis Testing (stats) Log InorSign Up. To use this calculator, a user selects the null hypothesis mean (the mean which is claimed), the sample mean, the standard deviation, the sample size, hypothesis test for a population Proportion calculator Fill in the sample size, n, the number of successes, x, the hypothesized population proportion p 0, and indicate if the test is left tailed, <, right tailed, >, or two tailed, ≠. Right tail hypothesis testing is illustrated below: We use right tail hypothesis testing to see if the z score is below the significance level critical value, in which case we cannot reject the null (2011). that we reject the null hypothesis and accept the alternative hypothesis, because the hypothesis Is Testing the Hypothesis also one of the methods? However, we suspect that is has much more accidents than this. When you're working with data, the numbers of the data itself is not very meaningful, because it's not standardized. below this critical value in the left tail method represents the rejection area. If you answered "yes", then you are right! Correlations, which have been retrieved from different samples can be tested against each other. To calculate the p-value, this calculator needs 4 pieces of data: the test statistic, the sample size, the hypothesis testing type (left tail, right tail, or two-tail), and the significance level (α). That flowchart was created to explain what we were teaching. As with learning anything related to mathematics, it is helpful to work through several examples. 4. Sample Correlation Coefficient Calculator H0: u1 - u2 = 0, where u1 is the mean of first population and u2 the mean of the second. Any value Single Sample T-Test Calculator A single sample t-test (or one sample t-test) is used to compare the mean of a single sample of scores to a known or hypothetical population mean. Activity. Therefore, if you choose to calculate with a significance level Therefore, it is false and the alternative hypothesis is true. The Test for one mean can be used to test the hypothesis that a sample mean is equal to a given mean (with unknown standard deviation) or certified value. hypothesis test for a population Proportion calculator. sample mean is actually different from the null hypothesis mean, which is the mean that is claimed. Hypothesis testing can be used for any type of science to show whether we reject or accept a hypothesis based on quantitative computing. An alternative hypothesis typically tries to prove that a relationship exists and is the statement you’re trying to back up. Hypothesis testing is the use of statistics to determine the probability that a given hypothesis is true. Example: Imagine, you want to test, if men increase their income considerably faster than women. To use this calculator, a user simply enters in the hypothesis testing method (either one-tail or two-tail), the sample size, and the significance level and clicks the 'Calculate' button. be in the nonrejection area. Enter the sample mean, population mean, sample standard deviation, population size and the significance level to know the T score test value, P value and result of hypothesis. Step 4 Make the decision to reject or not reject the null hypothesis. However, we believe Fifth: Check your solutions with the calculators as applicable. More about the t-test for one mean so you can better interpret the results obtained by this solver: A t-test for one mean is a hypothesis test that attempts to make a claim about the population mean ($$\sigma$$). Use the calculator below to analyze the results of a difference in two proportions hypothesis test. This Hypothesis Testing Calculator determines whether an alternative hypothesis is true or not. Z-Hypothesis Testing (stats) Log InorSign Up. Paired t-test Calculator Hypothesis. Hypothesis testing calculator with steps. Bring it to Science: A hypothesis is an unproven theory about a Scientific problem. The observed difference in conversion rate isn't big enough to declare a significant winner.There is no real difference in performance between A and B or you need to collect more data. yes '', then you are increasing the rejection area will be in the could... Asks the question: are two or more sets of data the same or different, statistically aimed... Flowchart was created to explain what we were teaching often used by scientists to test specific,... Use this free sample and population statistics calculator, DATAtab will then suggest which hypothesis test procedure ( method. 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Selected items were tested and vice versa within statistics, but a limited amount of development continues containing different. =.38 in the left tail, has a critical value, the z score will be in parameters! Then suggest which hypothesis test to make decision about entire population, with an explanation to. It was found that the mean Scientific method is Forming a hypothesis is the hypothesis mean and two-sided tests... Fewer than 400 worker accidents a year and the company, that the mean is greater than what is claimed... Itself is not performed on data in the correlation of both cohorts to Logic: a hypothesis is or. | 2021-04-15T23:03:24 | {
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http://mathoverflow.net/revisions/55515/list | 4 deleted 1 characters in body
Given a lattice $L$. Can one classify all functions $f:L\rightarrow \mathbb{R}$, that satisfy
$f(a \wedge b)+f(a\vee b) = f(a)+f(b)$.
Some examples are the the set of all finite subsets of a given set $S$. Then every such function is uniquely determined by the element $(f(\{s\}))_{s\in S}\in \prod_S\mathbb{R}$ plus the value on the empty set. Indeed this gives a vector space isomorphism from the set of all such functions to $\mathbb{R}^{|S|+1}$.
For other lattices there are also other additive functions arising naturally. For example if one considers the set of all natural numbers (without $0$) ordered by divisibility and assigns to a natural number (for a fixed prime $p$) the biggest $n$, so that $p^n$ divides this number. The logarithm is another example for such a function.
More general a ultrafilter on the underlysing underlying poset can also be viewed as such a function. So there are a lot of interesting examples. There are even more interesting examples like lattices of measurable sets and their measures, Euler-characteristic of subcomplexes and so on. So my question is: Can one classify the set of all those functions, probably in terms of filters / ultrafilters on the underlying poset?
Given a lattice $L$. Can one classify all functions $f:L\rightarrow \mathbb{R}$, that satisfy
$f(a \wedge b)+f(a\vee b) = f(a)+f(b)$.
Some examples are the the set of all finite subsets of a given set $S$. Then every such function is uniquely determined by the element $(f(\{s\}))_{s\in S}\in \prod_S\mathbb{R}$ plus the value on the empty set. Indeed this gives a vector space isomorphism from the set of all such functions to $\mathbb{R}^{|S|+1}$.
For other lattices there are also other additive functions arising naturally. For example if one considers the set of all natural numbers (without $0$) ordered by divisibility and assigns to a natural number (for a fixed prime $p$) the biggest $n$, so that $p^n$ divides this number. The logarithm is another example for such a function.
More general a ultrafilter on the underlysing poset can also be viewed as such a function. So there are a lot of interesting examples. There are even more interesting examples like lattices of measurable sets and their measures, Euler-characteristic of subcomplexes and so on. So my question is: Can one classify the set of all those functions, probably in terms of filters / ultrafilters on the underlying poset?
2 typo
Given a lattice $L$. Can one classify all functions $f:L\rightarrow \mathbb{R}$, that satisfy
$f(a \wedge b)+f(a\vee b) = f(a)+f(b)$.
Some examples are the the set of all finite subsets of a given set $S$. Then every such function is uniquely determined by the element $(f(\{s\}))_{s\in S}\in \prod_S\mathbb{R}$ plus the value on the empty set. Indeed this gives a vector space isomorphism from the set of all such functions to $\mathbb{R}^{|S|+1}$.
For other lattices there are also other additive functions arising naturally. For example if one considers the set of all natural numbers (without $0$) ordered by divisibility and assigns to a natural number (for a fixed prime $p$) the biggest $n$, so that $p^n$ divides this number. The logarithm is another example for such a function.
More general a ultrafilter on the underlysing poset can also be viewed as such a function. So there are a lot of interesting examples. There are even more interesting examples like lattices of measurable sets and their measures, Euler-characteristic of subcomplexes and so on. So my question is: Can one classify the set of all those functions, probably in terms of filters / ultrafilters on the underlying poset?
1 | 2013-05-22T09:29:47 | {
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https://cn.maplesoft.com/support/help/view.aspx?path=MultiZeta | MultiZeta - Maple Help
MultiZeta
the multiple zeta function
Calling Sequence MultiZeta(${m}_{1},{m}_{2},...,{m}_{n})$
Parameters
${m}_{1},{m}_{2},...,{m}_{n}$ - positive integers
Description
MultiZeta is an implementation of multiple zeta values, also known as the generalized Euler sums over
$\mathrm{MultiZeta}\left({m}_{i}$\mathrm{=}\left(i,1..n\right)\right)=\mathrm{%sum}\left(\mathrm{Multiply}\left(\mathrm{/}\left(1,\mathrm{^}\left({i}_{j},{m}_{j}\right)\right)$\mathrm{=}\left(j,1..n\right)\right),i\right)$
The sum converges for all positive integer arguments, except when the first argument equals one, for instance as in MultiZeta(1,2,3), in which case the function diverges.
With no arguments, MultiZeta() is defined as equal to 1.
Examples
For one argument, MultiZeta reduces to the Riemann Zeta function:
> $\mathrm{%MultiZeta}\left(43\right)=\mathrm{MultiZeta}\left(43\right)$
${\mathrm{%MultiZeta}}{}\left({43}\right){=}{\mathrm{\zeta }}{}\left({43}\right)$ (1)
The more relevant special cases are computed automatically, such as that of two identical arguments, here using a more compact input syntax
> $\left(\mathrm{%MultiZeta}=\mathrm{MultiZeta}\right)\left(27,27\right)$
${\mathrm{%MultiZeta}}{}\left({27}{,}{27}\right){=}\frac{{{\mathrm{\zeta }}{}\left({27}\right)}^{{2}}}{{2}}{-}\frac{{\mathrm{\zeta }}{}\left({54}\right)}{{2}}$ (2)
and of two arguments summing to an odd number
> $\left(\mathrm{%MultiZeta}=\mathrm{MultiZeta}\right)\left(11,8\right);$
${\mathrm{%MultiZeta}}{}\left({11}{,}{8}\right){=}{-}\frac{{75583}{}{\mathrm{\zeta }}{}\left({19}\right)}{{2}}{+}\frac{{9724}{}{{\mathrm{\pi }}}^{{2}}{}{\mathrm{\zeta }}{}\left({17}\right)}{{3}}{+}\frac{{4433}{}{{\mathrm{\pi }}}^{{4}}{}{\mathrm{\zeta }}{}\left({15}\right)}{{90}}{+}\frac{{286}{}{{\mathrm{\pi }}}^{{6}}{}{\mathrm{\zeta }}{}\left({13}\right)}{{315}}{+}\frac{{121}{}{{\mathrm{\pi }}}^{{8}}{}{\mathrm{\zeta }}{}\left({11}\right)}{{9450}}{+}\frac{{8}{}{{\mathrm{\pi }}}^{{10}}{}{\mathrm{\zeta }}{}\left({9}\right)}{{93555}}$ (3)
All Multiple Zeta values of weight less than or equal to seven, can be written solely in terms of classical Zeta values:
> $\left(\mathrm{%MultiZeta}=\mathrm{MultiZeta}\right)\left(2,1,4\right)$
${\mathrm{%MultiZeta}}{}\left({2}{,}{1}{,}{4}\right){=}\frac{{7}{}{{\mathrm{\pi }}}^{{4}}{}{\mathrm{\zeta }}{}\left({3}\right)}{{360}}{-}\frac{{11}{}{{\mathrm{\pi }}}^{{2}}{}{\mathrm{\zeta }}{}\left({5}\right)}{{12}}{+}\frac{{61}{}{\mathrm{\zeta }}{}\left({7}\right)}{{8}}$ (4)
The multiple Zeta values are a special case of the multiple polylogarithm:
> $\left(\mathrm{%MultiPolylog}=\mathrm{MultiPolylog}\right)\left(\left[2,3,4,5\right],\left[1,1,1,1\right]\right);$
${\mathrm{%MultiPolylog}}{}\left(\left[{2}{,}{3}{,}{4}{,}{5}\right]{,}\left[{1}{,}{1}{,}{1}{,}{1}\right]\right){=}{\mathrm{MultiZeta}}{}\left({2}{,}{3}{,}{4}{,}{5}\right)$ (5)
The multiple zeta values obey a large number of identities, primarily the stuffle relation:
> $\mathrm{MultiZeta}\left(7,9\right)\mathrm{MultiZeta}\left(6\right)$
$\frac{{\mathrm{MultiZeta}}{}\left({7}{,}{9}\right){}{{\mathrm{\pi }}}^{{6}}}{{945}}$ (6)
> $\mathrm{MultiZeta}\left(7,9,6\right)+\mathrm{MultiZeta}\left(7,6,9\right)+\mathrm{MultiZeta}\left(6,7,9\right)+\mathrm{MultiZeta}\left(13,9\right)+\mathrm{MultiZeta}\left(7,15\right)$
${\mathrm{MultiZeta}}{}\left({7}{,}{9}{,}{6}\right){+}{\mathrm{MultiZeta}}{}\left({7}{,}{6}{,}{9}\right){+}{\mathrm{MultiZeta}}{}\left({6}{,}{7}{,}{9}\right){+}{\mathrm{MultiZeta}}{}\left({13}{,}{9}\right){+}{\mathrm{MultiZeta}}{}\left({7}{,}{15}\right)$ (7)
Up to 5 digits,
> $\mathrm{evalf}\left[5\right]\left(=\right)$
${0.0084952}{=}{0.0084952}$ (8)
and the duality
> $\mathrm{MultiZeta}\left(2,3,4\right)$
${\mathrm{MultiZeta}}{}\left({2}{,}{3}{,}{4}\right)$ (9)
> $\mathrm{MultiZeta}\left(2,1,1,2,1,2\right)$
${\mathrm{MultiZeta}}{}\left({2}{,}{1}{,}{1}{,}{2}{,}{1}{,}{2}\right)$ (10)
> $\mathrm{evalf}\left(=\right)$
${0.06781184623}{=}{0.06781184623}$ (11)
References
[1] J. Bluemlein, D.J. Broadhurst, J.A.M. Vermaseren. "The Multiple Zeta Value Data Mine", Comput.Phys.Commun. Vol. 181 (2010): p. 582-625.
Compatibility
• The MultiZeta command was introduced in Maple 2018. | 2023-01-30T05:31:42 | {
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https://crypto.stackexchange.com/questions/95789/is-it-possible-to-give-a-definition-for-point-multiplication-on-elliptic-curve | # Is it possible to give a definition for point multiplication on elliptic curve?
As we know that at least in cryptography, the group operation on elliptic curve is just the point addition(https://en.wikipedia.org/wiki/Elliptic_curve_point_multiplication), which is defined on $$E:y^{2}=x^{3}+a x+b$$ as: $$\left(x_{p}, y_{p}\right)+\left(x_{q}, y_{q}\right)=\left(x_{r}, y_{r}\right)$$, $$\lambda=\frac{y_{q}-y_{p}}{x_{q}-x_{p}}$$, $$x_{r}=\lambda^{2}-x_{p}-x_{q}$$, $$y_{r}=\lambda\left(x_{p}-x_{r}\right)-y_{p}$$. My question is: can we give a meaningful definition for point multiplication?
• There is a Math theory behind this: Z-Module; That is, every abelian group is a module over the ring of integers Z in a unique way... Oct 27 '21 at 9:13
• Oct 29 '21 at 8:05
• @kelalaka Thank you so much! Your answer also helps me understand this better. Yeah, I just find it useful to my research if there is one such technique. But now I know it is not the case. Thanks! Oct 31 '21 at 7:17
On an elliptic curve we have
• point addition $$C:=A+B$$ defined for any two points $$A$$ and $$B$$ of the curve (often with special rules for $$A=B$$ or some special points, depending on the coordinate system).
• neutral $$\infty$$ such that $$A+\infty=\infty+A=A$$ for all $$A$$ on the curve (including $$\infty$$)
• opposite $$-A$$ of any $$A$$ on the curve such that $$A+(-A)=(-A)+A=\infty$$ (with $$\infty$$ it's own opposite).
Point addition is associative, and commutative.
From this we can define point multiplication by an integer $$i\in\mathbb Z$$ (also known as scalar multiplication), as $$i\times A\underset{\text{def}}=\begin{cases} \infty&\text{if }i=0\\ ((i-1)\times A)+A&\text{if }i>0\\ (-i)\times (-A)&\text{if }i<0 \end{cases}$$
From this it follows that for all $$A$$ and $$B$$ on the curve (including $$\infty$$) and all integers $$i$$, $$j$$, it holds \begin{align} (i+j)\times A&=(i\times A)+(j\times A)\\ i\times(A+B)&=(i\times A)+(i\times B)\\ (i\times j)\times A&=i\times (j\times A)\\ \end{align} where in the above, the top left addition and bottom left multiplication are in $$\mathbb Z$$, and all the other operations are point addition or point multiplication by an integer.
When we talk about multiplication in Elliptic Curve cryptography, that's most often this multiplication by an integer.
In order to define multiplication of points, we need to designate a particular point $$G$$ and restrict to points $$A$$ that can be obtained as $$A=a\times G$$ for some integer $$a\in\mathbb Z$$. They form a subgroup of the curve. Many groups used in Elliptic Curve Cryptography are cyclic, meaning there exists $$G$$ such that any point of the group can be obtained in this way. For some curves (those with a prime number of point including $$\infty$$, e.g secp256k1 or secp384r1), any point $$G$$ other than $$\infty$$ can be used and all points of the curve are of this form $$A=a\times G$$.
For elliptic curves on a finite field as used in cryptography, there is some minimal strictly positive integer $$n$$ such that $$n\times G=\infty$$ (the order of $$G$$), and that's also the order (the number of elements) of said subgroup. For any $$A$$ in this subgroup, there is a uniquely defined $$a\in[0,n)$$ with $$A=a\times G$$.
We can then define the product of point $$A=a\times G$$ and $$B=b\times G$$ with $$a,b\in[0,n)$$ as the point $$A\times B\underset{\text{def}}=(a\times b\bmod n)\times G$$ That product of elliptic curve points inherits associativity, commutativity, neutral $$G$$, from the corresponding properties of multiplication in $$\mathbb Z_n$$. Distributivity w.r.t. point addition holds. Also, $$(i\times A)\times B=i\times(A\times B)$$ holds for all points $$A$$, $$B$$ which product is defined, and all integers $$i$$.
When $$n$$ is prime (which holds for most curves and generators $$G$$ used in ECC), any point $$A$$ except $$\infty$$ has inverse $$A^{-1}$$ such that $$A\times A^{-1}=A^{-1}\times A=G$$. If $$A=a\times G$$, then $$A^{-1}=(a^{-1}\bmod n)\times G$$.
Notice that this definition of multiplication depends on the choice of $$G$$, and is for the whole curve only when the elliptic curve group is cyclic.
Also, we can compute $$C=A\times B$$ efficiently if we know $$a$$ with $$A=a\times G$$ (as $$C:=a\times B$$) or know $$b$$ with $$B=b\times G$$ (as $$C:=b\times A$$). But otherwise, the best known algorithms have cost $$\Theta(\sqrt n)$$ on standard computers, thus are not polynomial time w.r.t. the bit size of $$n$$.
• that's what I want! Thanks! Oct 28 '21 at 9:55
• Since you went to an educative answer, you should mention scalar multiplication. The theory about this is the Modules, and ECs are Z-Modules since they are forming an abelian group under addition. Modules are relaxation from Vector spaces. The multiplication is not well defined, maybe you should write it as $\times_G$ indicating the action of $G$ on the operation. Oct 28 '21 at 17:44
A group by definition has only one operation. You would need at least a semiring with the new “multiplication” operation also being compatible with the Elliptic Curve addition.
To the best of my knowledge no such meaningful definition exists.
• I see. I just think if it is possible to give one. But anyway, thanks! Oct 27 '21 at 4:32
• @user77340 No, it is not possible to define a group operation as multiplication. EC forms a Z-module, that is. What Fgriue defined is not multiplication as we know from the group theory. What is defined as is the action of the selected base element. If already had we could be talking about EC is a ring! Oct 29 '21 at 8:07 | 2022-01-20T09:13:16 | {
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http://mathhelpforum.com/business-math/153550-selling-price-computer.html | # Thread: selling price of computer
1. ## selling price of computer
The selling price of a computer is $1760 excluding GST. (a) Calculate the amount of GST (12½%) which needs to be added on. (b) Calculate also the final selling price of the computer when GST is added on. (c) What fraction is the GST of the final selling price? I got a) /8 x 1760 =$220
b) 1760 + 220 = \$1980
c) 1/9 = 220
2. I agree. | 2017-07-23T11:07:44 | {
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https://math.stackexchange.com/questions/1047065/if-lim-n-to-infty-f-n-f-almost-everywhere-then-lim-n-to-infty-f-n | # If $\lim_{n \to \infty} f_n=f$ (Almost everywhere) then $\lim_{n \to \infty} f_n=f$ (in measure on $E$)
Suppose $$E$$ is measurable subset of $$\Bbb R$$, $$(f_n)$$ is sequence of measurable functions from $$E$$ to $$[-\infty, \infty]$$ , $$f$$ is function from $$E$$ to $$[-\infty , \infty]$$.
If $$\lim\limits_{n \to \infty} f_n=f$$ (Almost everywhere) then $$\lim\limits_{n \to \infty} f_n=f$$ ( in measure on$$E$$)
Is this Statement True?
$$\lim\limits_{n \to \infty} f_n=f$$ (in measure on $$E$$) i.e. $$\forall \epsilon \gt 0$$ ,$$\exists N \in \Bbb N s.t. \forall n \ge N :$$ $$m^*(\{x \in E |$$ $$|f_n(x)-f(x)|\ge \epsilon\}) \lt \epsilon$$
$$\lim\limits_{n \to \infty} f_n=f$$ (Almost everywhere) i.e. $$m^*(\{ x \in E | \lim\limits_{n \to \infty} f_n\not =f\}=0$$
• It's not necessarily true if $E$ has infinite measure. – David Mitra Dec 1 '14 at 20:20
• @DavidMitra in fact this question is (True\False). Do you have any Counterexample? – agustin Dec 1 '14 at 20:23
• $E=\Bbb R$, $f_n=\chi_{[n,n+1]}$. (You would also have to assume $f_n$ and $f$ to have finite values a.e. for the statement to hold.) – David Mitra Dec 1 '14 at 20:25
• @DavidMitra It's Ok, that's work. – agustin Dec 1 '14 at 20:32
You need to show $$\lim_{n\to\infty}m(x:|f_{n}(x)-f(x)|\geq\epsilon)=0$$ for all $\epsilon>0$. From pointwise a.e. convergence, we have $$m(x:\lim_{n\to\infty}|f_{n}(x)-f(x)|\geq\epsilon)=0.$$
For some reason I thought the OP was working on a probability space (or a space with finite measure). If the $m(E)=+\infty$, the claim is false as exhibited by a moving bump function out to horizontal infinity where $E=\mathbb{R}$ with Lebesgue measure $dx$. | 2020-11-29T20:22:43 | {
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https://metric2011.wordpress.com/2011/02/14/notes-of-urs-langs-lecture-nr-2/ | ## Notes of Urs Lang’s lecture nr 2
Lee: If a space ${X}$ is an absolute ${C}$-Lipschitz retract, with ${C>1}$, can you say anything of its injective hull ? Answer: probably no. For instance, the injective hull of Euclidean ${{\mathbb R}^n}$ is ${\ell_{\infty}^{2^{n-1}}}$ which seems very far from ${{\mathbb R}^n}$ although ${{\mathbb R}^n}$ is an absolute ${\sqrt{n}}$-Lipschitz retract.
1. Injective hulls: basics
1.1. Extremal functions
Let ${X}$ be a metric space. Leopoldo Nachbin, in the 1950’s, already considered the set
$\displaystyle \begin{array}{rcl} \Delta(X)=\{f:X\rightarrow{\mathbb R}\,;\,f(x)+f(y)\geq d(x,y)\}. \end{array}$
Definition 1 The minimal elements of ${\Delta}$ are called extremal functions. The set of extremal functions is denoted by ${E(X)}$.
Note that if ${X}$ is compact, ${f\in\Delta(X)}$ is extremal iff for all ${x\in X}$, there exists ${y\in X}$ such that ${f(x)+f(y)=d(x,y)}$. In general,
${f\in\Delta(X)}$ is extremal iff for all ${x\in X}$, ${f(x)=\sup_{y\in X}d(x,y)-f(y)}$.
Note that all ${d_x :x'\mapsto d(x,x')}$ belong to ${E(X)}$. This yields a map ${e:X\rightarrow E(X)}$.
Lemma 2 Extremal functions are ${1}$-Lipschitz.
Proof: Given ${x\in X}$, define ${g:X\rightarrow{\mathbb R}}$ by ${g(y)=f(x)+d(x,y)}$ and ${g(x')=f(x')}$ for ${x'\not=y}$. Then ${g\in\Delta(X)}$, so
$\displaystyle \begin{array}{rcl} f(y)\leq g(y)=f(x)+d(x,y). \end{array}$
$\Box$
Let
$\displaystyle \begin{array}{rcl} \Delta_{1}(X)=\{f\in\Delta(X)\,;\, f\textrm{ is }1\textrm{-Lipschitz}\}. \end{array}$
Then the ${L_{\infty}}$-norm defines a metric on ${\Delta_1 (X)}$.
Proposition 3 (Dress) There is a map ${p:\Delta(X)\rightarrow E(X)}$ such that
1. ${p(f)\leq f}$.
2. ${\|p(f)-p(g)\|_{\infty}\leq \|f-g\|_{\infty}}$.
Proof: For ${f\in\Delta(X)}$, set ${f^{*}(x)=\sup_{y\in X}d(x,y)-f(y)}$, so ${f}$ is extremal iff ${f=f^*}$. Note that ${f^* \leq f}$. By definition, for all ${x}$, ${y\in X}$,
$\displaystyle \begin{array}{rcl} f^* (x)+f(y)\geq d(x,y), \quad f(x)+f^* (y)\geq d(x,y), \end{array}$
so ${q(f):=\frac{1}{2}(f+f^* \in\Delta(X)}$, and ${q(f)\leq f}$. For every ${g\in \Delta(X)}$,
$\displaystyle \begin{array}{rcl} g^* (x)\leq f^* (x)+\|f-g\|_{\infty}. \end{array}$
Thus
$\displaystyle \begin{array}{rcl} \|q(f)-q(g)\|_{\infty} &\leq& \frac{1}{2}\|f-g\|_{\infty}+\frac{1}{2}\|f^* -g^* \|_{\infty}\\ &\leq& \|f-g\|_{\infty} \end{array}$
Define ${p(f):=\lim_{n\rightarrow\infty}q^{n}(f)}$ (pointwise limit). Then ${p(f)\leq f}$. For all ${n}$, ${p(f)^*\geq q^n (f)^*}$, so
$\displaystyle \begin{array}{rcl} 0\leq p(f)-p(f)^* \leq q^n (f)-q^n (f)^* =2(q^n (f)-q^{n+1} (f)) \end{array}$
which tends to ${0}$, so ${p(f)=p(f)^*}$. $\Box$
Remark 1 If ${|X|\geq 3}$, an infinite iteration is indeed needed to obtain ${p}$.
1.2. Injectivity
Proposition 4 ${\Delta_1 (X)}$ is injective.
Proof: Let ${A\subset B}$ be metric spaces, let ${F:A\rightarrow \Delta_1 (X)}$ be ${1}$-Lipschitz. For ${b\in B}$, ${x\in X}$, put
$\displaystyle \begin{array}{rcl} f_b (x)=\inf_{a\in A}F(a)(x)+d(a,b). \end{array}$
Then ${f_b \geq 0}$. and ${f_b}$ is ${1}$-Lipschitz. For ${a\in A}$, ${f_a =F(a)}$. We have
$\displaystyle \begin{array}{rcl} f_b (x)+f_b (y)&\geq& \inf_{a,\,a'\in A}F(a)(x)+F(a')(y)+d(a,a')\\ &\geq& \inf_{a A}F(a)(x)+F(a)(y)\geq d(x,y) \end{array}$
thus ${f_b \in \Delta_1 (X)}$. Extend ${F}$ to ${B}$ by setting ${F(b)=f_b}$. Then the extended ${F}$ is ${1}$-Lipschitz. $\Box$
Theorem 5 (Isbell)
1. ${E(X)}$ is injective.
2. If ${f:E(X)\rightarrow E(X)}$ is ${1}$-Lipschitz, and fixes ${e(X)}$ pointwise, then ${F}$ is the identity.
3. If ${G:E(X)\rightarrow Y}$ is ${1}$-Lipschitz and ${G\circ e}$ is an isometric embedding, then ${G}$ is an isometric embedding.
4. If ${I:X\rightarrow Y}$ is an isometric embedding and ${Y}$ is injective, then there is an isometric embedding ${G:E(X)\rightarrow Y}$ such that ${G\circ e=I}$.
Proof: 2. ${F(f)(x)=\|F(f)-d_x \|_{\infty}\leq \|f-d_x \|_{\infty}=f(x)}$, so ${F(f)=f}$ by minimality. 3. follows from 2. and 4. follows from 3. $\Box$
2. First examples
2.1. Polyhedral structure on ${E(X)}$, ${X}$ finite
If ${X}$ is bounded, then ${diameter (E(X))\leq diameter(X)}$. If ${X}$ is compact, so is ${E(X)}$ (Arzela-Ascoli).
If ${X}$ is finite, ${E(X)}$ is a polyhedron in ${\ell_{\infty}(X)}$. Note that ${\Delta(X)}$ and ${\Delta_1 (X)}$ are convex, but ${E(X)}$ is not in general. The polyhedral structure can be detected by looking at “equality graphs”. For ${f\in\Delta(X)}$, let ${\mathcal{E}(f)}$ be the set of pairs ${\{x,y\}}$ such that ${f(x)+f(y)=d(x,y)}$. Then ${G(f)=(X,\mathcal{E}(f))}$ is a graph with loops: ${\{x,x\}\in\mathcal{E}(X)}$ iff ${f(x)=0}$. Also
${f\in E(X)}$ iff ${G(f)}$ has no isolated vertices.
Say a graph ${G=(X,\mathcal{E})}$ is admissible if there exists ${f\in E(X)}$ such that ${G=G(f)}$. Every admissible graph corresponds to a polyhedral cell ${P(G)}$ in ${E(X)}$, whose interior points are precisely the functions ${f\in E(X)}$ such that ${G(f)=G}$. ${P(G')}$ is a face of ${P(G)}$ iff ${G'}$ contains ${G}$.
2.2. Dimension of ${E(X)}$
If two functions ${f}$, ${g\in E(X)}$ have the same graph, then for all ${\{v,v'\}\in\mathcal{E}}$, ${f(v)+f(v')=d(v,v')=g(v)+g(v')}$, thus
$\displaystyle \begin{array}{rcl} f(v')-g(v')=-(f(v)-g(v)). \end{array}$
Hence, if there is a path from ${x}$ to ${y}$ in ${G}$ of length ${L}$, then
$\displaystyle \begin{array}{rcl} f(y)-g(y)=(-1)^{L}(f(x)-g(x)). \end{array}$
On connected components of ${G}$ containing an odd cycle, ${f}$ and ${g}$ coincide. On other (even) components, there is exactly one degree of freedom to vary ${f}$ without changing the graph.
Proposition 6 Define the rank ${rk(G)}$ as the number of even components. Then ${\mathrm{dim}(P(G))=rk(G)}$.
Example 1 ${X=K_2}$.
Then the possible graphs are an edge (rank one) and one edge with one loop (rank zero). This produces an interval.
Example 2 ${X=K_3}$.
Then ${E(X)}$ is a tripod. Indeed, admissible graphs of extremal functions can have at most one even component. The complete graph is admissible but odd, producing a vertex. Two-edge admissible graphs contribute three segments. Hinges with a loop contribute three vertices.
The tripod lies in ${\ell_{\infty}^{3}}$ on a triangular face (standard simplex) at the bottom of polyhedron ${\Delta_1}$. It takes infinitely many steps (iterates of ${q}$) to map a constant function ${1/2}$ to ${E(X)}$. It is a bit stupid.
Lee: why don’t you consider the optimal ${t}$ such that ${(1-t)f+tf^*}$ belongs to ${\Delta}$ ?
Example 3 ${X}$ has 4 points with one distance equal to ${2}$ and all other equal to ${1}$ or ${0}$.
The graph with two disjoint edges (one of length ${2}$) is admissible and has rank ${2}$, it contributes a ${2}$-cell, a square sitting diagonally in ${\ell_{\infty}^{2}}$. The resulting polyhedron is the union of the ${2}$-cell with two opposite protruding edges, each of length ${1/2}$. | 2017-04-24T18:58:29 | {
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https://www.studysmarter.us/textbooks/physics/college-physics-urone-1st-edition/uniform-circular-motion-and-gravitation/q63-29-pe-a-large-centrifuge-like-the-one-shown-in-figure-a-/ | Suggested languages for you:
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Q6.3-29 PE
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### College Physics (Urone)
Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000
# A large centrifuge, like the one shown in Figure (a), is used to expose aspiring astronauts to accelerations similar to those experienced in rocket launches and atmospheric re-entries. (a) At what angular velocity is the centripetal acceleration $$10g$$ if the rider is $$15.0{\rm{ m}}$$ from the centre of rotation? (b) The rider’s cage hangs on a pivot at the end of the arm, allowing it to swing outward during rotation as shown in Figure (b). At what angle $$\theta$$ below the horizontal will the cage hang when the centripetal acceleration is $$10g$$? (Hint: The arm supplies centripetal force and supports the weight of the cage. Draw a free body diagram of the forces to see what the angle $$\theta$$ should be.)*Figure (a) NASA centrifuge used to subject trainees to accelerations similar to those experienced in rocket launches and re-entries. (Credit: NASA) (b) Rider in cage showing how the cage pivots outward during rotation. This allows the total force exerted on the rider by the cage to be along its axis at all times.
1. The angular velocity is $$2.55{\rm{ rad}}/{\rm{s}}$$.
2. The angle below the horizontal is $$5.71^\circ$$.
See the step by step solution
## Step 1: Definition of Angular velocity
Angular velocity may be defined as the speed at which an item rotates or revolves around an axis. The SI unit for angular velocity is radians per second because it is expressed as an angle per unit of time.
## Step 2: Calculating centripetal acceleration
The centripetal acceleration is calculated as follows:
$${a_c} = 10g$$ The acceleration owing to gravity is denoted by $$g$$.
Substitute $${\rm{9}}{\rm{.8 m/}}{{\rm{s}}^{\rm{2}}}$$ for $$g$$,
\begin{aligned}{}{a_c} = 10 \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right)\\ = 98{\rm{ m}}/{{\rm{s}}^2}\end{aligned}
The centripetal acceleration is,
$${a_c} = {\omega ^2}r$$
Here, $${a_c}$$ is the centripetal acceleration, $$\omega$$ is the angular velocity, and $$r$$ is the distance of the rider from the center.
To generate an expression for the angular velocity, rearrange the previous equation.$$\omega = \sqrt {\frac{{{a_c}}}{r}}$$
Substitute $$98{\rm{ m}}/{{\rm{s}}^2}$$ for $${a_c}$$, and $$15.0{\rm{ m}}$$ for $$r$$,
\begin{aligned}{}\omega = \sqrt {\frac{{98{\rm{ m}}/{{\rm{s}}^2}}}{{15.0{\rm{ m}}}}} \\ = 2.55{\rm{ rad}}/{{\rm{s}}^2}\end{aligned}
Hence, the angular velocity is $$2.55{\rm{ rad}}/{\rm{s}}$$.
## Step 3: Calculating Angle below the horizontal
The vertical component of the force on the arm will counter the weight i.e.,
\begin{aligned}{}{F_{arm}}\sin \theta = W\\ = mg\end{aligned} $${\rm{(1}}{\rm{.1)}}$$
Here, $${F_{arm}}$$ is the force on the arm, $$\theta$$ is the angle below the horizontal, $$W$$ is the weight, $$m$$ is the mass, and $$g$$ is the acceleration due to gravity.
The horizontal component of the force on the arm will provide necessary centripetal force i.e.,
\begin{aligned}{}{F_{arm}}\cos \theta = {F_c}\\ = m{a_c}\end{aligned} $${\rm{(1}}{\rm{.2)}}$$
Here, $${F_c}$$ is the centripetal force, and $${a_c}$$ is the centripetal acceleration.
Dividing equation $${\rm{(1}}{\rm{.1)}}$$ by equation $${\rm{(1}}{\rm{.2)}}$$,
\begin{aligned}{}\frac{{{F_{arm}}\sin \theta }}{{{F_{arm}}\cos \theta }} = \frac{{mg}}{{m{a_c}}}\\\tan \theta = \frac{g}{{{a_c}}}\end{aligned}
Rearranging the above equation to get an expression angle below the horizontal.
$$\theta = {\tan ^{ - 1}}\left( {\frac{g}{{{a_c}}}} \right)$$
Substitute $$9.8{\rm{ m}}/{{\rm{s}}^2}$$ for $$g$$, and $$98{\rm{ m}}/{{\rm{s}}^2}$$ for $${a_c}$$,
\begin{aligned}{}\theta = {\tan ^{ - 1}}\left( {\frac{{9.8{\rm{ m}}/{{\rm{s}}^2}}}{{98{\rm{ m}}/{{\rm{s}}^2}}}} \right)\\ = 5.71^\circ \end{aligned}
Hence, the angle below the horizontal is $$5.71^\circ$$. | 2023-03-27T23:52:01 | {
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https://math.stackexchange.com/questions/3435119/proof-a-n1-geq-a-n-with-a-n-11-nn-for-n-in-mathbbn?noredirect=1 | # Proof $a_{n+1} \geq a_n$ with $a_n := (1+1/n)^n$ for $n \in \mathbb{N}$ [duplicate]
Let $$a_n := (1+1/n)^n$$ for $$n \in \mathbb{N}$$
How can one prove that $$a_{n+1} \geq a_n$$ for all $$n \in \mathbb{N}$$ with Bernoulli's inequality?
I know that the inequality states that $$(1+x)^r \geq 1+rx$$ for every integer $$r \geq 0$$ and every real number $$x \geq -2$$. And if the exponent $$r$$ is even, then the inequality is valid for all real numbers x.
So I have to use induction and for $$n=1$$ we would get $$(1+1/1)^1 = (1+1/(1+1))^2$$, and that would give $$2 < 2,25$$. But wouldn't that imply that all numbers $$> 1$$ would make $$a_{n+1} > a_n$$? At which case is it equal? Can someone show me where I can find an induction proof for this?
• There is never equality. Note $a_n \to e$ – Henry Nov 14 '19 at 8:42
• @Henry So $a_{n+1} > a_n$ for all $n \in \mathbb{N}$? – Ramanujan Taylor Nov 14 '19 at 8:44
• Yes $\,\,\,\,\,\,\,\,$ – Henry Nov 14 '19 at 8:45
• In particular there is an answer (among many other answers) that user Bernoulli's inequality – Maximilian Janisch Nov 14 '19 at 8:58
## 1 Answer
$$a_{n+1}\ge a_n \quad\Longleftrightarrow\quad \frac{(n+2)^{n+1}}{(n+1)^{n+1}}\ge\frac{(n+1)^n}{n^n} \quad\Longleftrightarrow\quad \frac{n+2}{n+1}\ge\left(\frac{n^2+2n+1}{n^2+2n}\right)^n \\ \quad\Longleftrightarrow\quad \left(1-\frac{1}{n^2+2n+1}\right)^n\ge 1-\frac{1}{n+2}$$ The last inequality holds since $$\left(1-\frac{1}{n^2+2n+1}\right)^n\ge 1-\frac{n}{n^2+2n+1}$$ and $$\frac{1}{n+2}\ge\frac{n}{n^2+2n+1}$$ | 2020-04-03T21:30:10 | {
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http://www.physicsforums.com/showthread.php?t=264405 | ## Numbers in Triangle Rows of 3 are equal rows of 4 are equal
Ok I'm new here and I don't know how to find answers to puzzles.
I've got a few and my brain isn't working.
Picture a triangle with circles in it. Place the numbers 1-10 in the circles on the triangle so that the sum of Rows of 3 are equal and the sum of rows of 4 are equal
O
O O
O O O
O O O O
(If the picture of the triangle doesn't come out it 1 circle on top, then 2, then 3 then 4. The perimeter has 4 circles on all three sides.
PhysOrg.com science news on PhysOrg.com >> 'Whodunnit' of Irish potato famine solved>> The mammoth's lament: Study shows how cosmic impact sparked devastating climate change>> Curiosity Mars rover drills second rock target
Just to help with the picture, try to code tag Code: o o o o o o o o o o
I believe there are 12 solutions to this, which are variants on 2 different patterns. You can rotate by 120 degrees and get a "new" solution effectively, or flip it along one of the 3 axes of symmetry-- giving you 6 variants on a single solution, and 12 patterns all together, since there are 2 "core" solutions. Spoiler Two of the variants on the core solutions are: 2 7 8 3 4 10 9 5 6 1 2 8 6 4 7 5 9 3 1 10 DaveE | 2013-05-21T22:46:54 | {
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https://math.stackexchange.com/questions/3267604/re-writing-div-and-grad-with-differential-operators | # Re-writing Div and Grad with Differential Operators
I have some expressions with grad and div operators and would like to re-write the expressions so that there are no grads or divs. Basically, I have been told that $$\nabla \phi$$, where $$\phi = 1/4 \pi |\textbf{r}|$$, can be re-written as
$$\nabla \phi = \frac{\textbf{r}}{|\textbf{r}|} \frac{\text{d}\phi}{\text{d}|\textbf{r}|}$$.
I might need to refresh my knowledge of vector calculus, but I suppose this just follows from the chain rule?
Edit:
$$\nabla \nabla \phi = \frac{\nabla (\textbf{r})}{|\textbf{r}|}\frac{\partial}{\partial r}\phi + \hat{r} \nabla \bigg( \frac{\partial}{\partial r} \phi \bigg)=\frac{\textbf{r}}{|\textbf{r}|} \frac{\text{d} \phi}{\text{d} |\textbf{r}|} + \frac{\textbf{r}}{|\textbf{r}|}\frac{\textbf{r}}{|\textbf{r}|} \frac{\partial}{\partial r} \bigg( \frac{\partial}{\partial} \phi \bigg) = \frac{\delta_{ij}}{|\textbf{r}|}\frac{\text{d} \phi}{\text{d}|\textbf{r}|} + \frac{\textbf{r} \otimes \textbf{r}}{|\textbf{r}|^2} \frac{\text{d}^2 \phi}{\text{d} |\textbf{r}|^2} ,$$
$$\nabla \nabla \nabla \phi = \nabla \bigg( \frac{\delta_{ij}}{|\textbf{r}|}\frac{\text{d} \phi}{\text{d}|\textbf{r}|} + \frac{\textbf{r} \otimes \textbf{r}}{|\textbf{r}|^2} \frac{\text{d}^2 \phi}{\text{d} |\textbf{r}|^2} \bigg)=\nabla \bigg(\frac{\delta_{ij}}{|\textbf{r}|} \bigg) \frac{\text{d}\phi}{\text{d} |\textbf{r}|} + \frac{\delta_{ij}}{|\textbf{r}|} \nabla \bigg(\frac{\partial \phi}{\partial r} \bigg) + \nabla \bigg(\frac{\textbf{r} \otimes \textbf{r}}{|\textbf{r}|^2} \bigg) \frac{\partial^2 \phi}{\partial r^2} +\frac{\textbf{r} \otimes \textbf{r}}{|\textbf{r}|^2} \nabla \bigg( \frac{\partial^2 \phi}{\partial r^2} \bigg) = \nabla \bigg(\frac{\delta_{ij}}{|\textbf{r}|} \bigg) \frac{\text{d}\phi}{\text{d} |\textbf{r}|} + \frac{\delta_{ij} \otimes \textbf{r}}{|\textbf{r}|^2} \frac{\text{d}^2 \phi}{\text{d} |\textbf{r}|^2} + \frac{\delta_{ij} \otimes \textbf{r}}{|\textbf{r}|^2} \frac{\text{d}^2 \phi}{\text{d} |\textbf{r}|^2} + \frac{\textbf{r} \otimes \delta_{ij} }{|\textbf{r}|^2} \frac{\text{d}^2 \phi}{\text{d} |\textbf{r}|^2} + \frac{\textbf{r} \otimes \textbf{r} \otimes \textbf{r}}{|\textbf{r}|^3} \frac{\text{d}^3 \phi}{\text{d} |\textbf{r}|^3} ,$$
$$\nabla \cdot \nabla \phi = \nabla^2 \phi = \frac{\text{d}^2 \phi}{\text{d} |\textbf{r}|^2}$$.
Your expression for the gradient will work with any function that is a function purely of the radius (i.e., no angular dependencies). In spherical coordinates:
$$\displaystyle \nabla\Phi = \hat{r}\frac{\partial}{\partial r}\Phi + \hat{\theta}\frac{\partial}{\partial \theta}\Phi+\hat{\phi}\frac{1}{\sin{\theta}}\frac{\partial}{\partial \phi}\Phi$$
If $$\Phi$$ is a function of $$r$$ only,
$$\displaystyle \nabla\Phi = \hat{r}\frac{\partial}{\partial r}\Phi$$
But $$\hat{r}=\bf{r}/|\bf{r}|$$ and $$\partial/\partial r = d/d|\bf{r}|$$. Make those substitutions and you get your expression.
• And for example, if I then want to $\nabla \nabla \phi$ it's just repeat the same rule? What would $\nabla \cdot \nabla \phi$ be? – Tom Jun 19 '19 at 17:25
• @Tom Again, if $\Phi$ depends only on $r$, it would just be $\partial^2\Phi/\partial r^2=d^2\Phi/dr^2$ – bob.sacamento Jun 19 '19 at 17:28
• I've tried a few more as practice, I'm not sure if these are correct. One thing that I'm not sure about in the expression is that you end up with $\nabla \delta_{ij}$, what happens when you have the grad operator act on a second rank tensor (obviously I know it will have to be a third rank tensor). – Tom Jun 19 '19 at 19:35 | 2020-01-24T13:59:51 | {
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https://math.stackexchange.com/questions/2289630/can-someone-tell-me-how-this-is-a-quadratic-equation | # Can someone tell me how this is a quadratic equation?
I'm not sure how to complete this- Could someone help describe to me how exactly this is a quadratic equation? Or how can I manipulate the variables to make it one? It has been a while since I've worked with quadratics, also, I've never seen one like this before. Thanks for looking.
$$\frac{1}{200-s} + \frac{1}{s} = \frac{1}{32}$$
• Multiply everything by the common denominator. – Crostul May 20 '17 at 21:18
• @pstumps Get common denominator. $$\frac{1}{200-s}+\frac1s = \frac{1}{200-s}\cdot \frac{s}{s} + \frac{1}{s}\cdot\frac{200-s}{200-s} = \frac{s}{(200-s)s} + \frac{200-s}{(200-s)s} = \frac{s+200-s}{(200-s)s} = \frac{200}{200s-s^2}.$$ – Eff May 20 '17 at 21:32 | 2020-06-05T06:27:16 | {
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http://mathhelpforum.com/algebra/19735-solving-log-equation.html | # Math Help - Solving a Log Equation
1. ## Solving a Log Equation
log6(x+1) + log6(x+2) = 1
The 6 in both equations is the base.
I know that...
log6(x+1) + log6(x+2) = log6(x+1)(x+2)
So I get to here.
log6(x+1)(x+2) = 1
I assume we have to make it equal to 0.
log6(x^2+3x+2) = 1
log6(x^2+3x+1) = 0
Tell me if I'm doing something wrong here.
The answer is supposed to be 1, but I don't understand how to get to the answer.
2. Originally Posted by Jeavus
log6(x+1) + log6(x+2) = 1
The 6 in both equations is the base.
I know that...
log6(x+1) + log6(x+2) = log6(x+1)(x+2)
So I get to here.
log6(x+1)(x+2) = 1
I assume we have to make it equal to 0.
log6(x^2+3x+2) = 1
log6(x^2+3x+1) = 0
that's wrong:
remember the definition of a logarithm... If $\log_a b = c$ then $a^c = b$
so you have: $\log_6 (x^2 + 3x + 2) = 1$
$\Rightarrow 6^1 = x^2 + 3x + 2$
now continue
3. Thanks again.
4. How do I go about solving this equation?
log5(2x+2) - log5(x-1) = log5(x+1)
The 5 in each is the base.
5. Use the formula $\displaystyle\log_ab-\log_ac=\log_a\frac{b}{c}$.
6. Hmmm...
log5(2x+2) - log5(x-1) = log5(x+1)
log5(2x+2/x-1) = log5(x+1)
log5(2x+2/x-1)(x+1) = 0
Where do I go from here?
7. Originally Posted by Jeavus
Hmmm...
log5(2x+2) - log5(x-1) = log5(x+1)
log5(2x+2/x-1) = log5(x+1)
log5(2x+2/x-1)(x+1) = 0
Where do I go from here?
how did you get $\log_5 \left( \frac {2x + 2}{x - 1} \right)(x + 1)$ exactly? you had to subtract $\log_5 (x + 1)$ correct?
when you find what the log should be, follow what i told you in the first post
...or, you could have simply dropped all the logs...
$\log_5 \left( \frac {2x + 2}{x - 1}\right) = \log_5 (x + 1)$
$\Rightarrow \frac {2x + 2}{x - 1} = x + 1$ ............do you see why?
8. $\displaystyle\log_5\frac{2x+2}{x-1}=\log_5(x+1)\Rightarrow\frac{2x+2}{x-1}=x-1$
Now, solve for x.
9. Thanks very much. | 2014-10-20T08:03:24 | {
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http://mathhelpforum.com/algebra/188475-sove-simple-equation-doubt.html | # Math Help - Sove simple equation doubt
1. ## Sove simple equation doubt
I'm solving this:
Code:
3 – 7 * (1 - 2x) = 5 – (x + 9)
And am trying this way:
Code:
3- 7 * (1 - 2x) = 5 - (x + 9)
3 - 7 ..?? ?? = 5 - x - 9)
What should be done in the part I put those question marks?
I'm really lost.
Thanks in advance for any help.
2. ## Re: Sove simple equation doubt
Do you know the distributive law?
In general: $a\cdot (b+c)=a\cdot b+a\cdot c$
Use it to expand $-7\cdot (1-2x)=...$
3. ## Re: Sove simple equation doubt
So, I should do:
$3 - 7 . (1 - 2x) = 5 - (x + 9)$
$3 - 7 . 1 - 7 . 2x = 5 - x - 9)$
$3 - (-7) - 14x = 5 - x - 9$
$-10x = ...$
I'm lost...
4. ## Re: Sove simple equation doubt
You have to pay attention with the signs:
$3-7\cdot (1-2x)=5-(x+9)$
$\Leftrightarrow 3-7\cdot 1 -7\cdot (-2x)=5-x-9$
$\Leftrightarrow 3-7+14x=5-x-9$
$\Leftrightarrow 3-7-5+9=-x-14x$
$\Leftrightarrow 0=-15x$
$\Leftrightarrow x=0$
5. ## Re: Sove simple equation doubt
I was doing:
Code:
3 - 7 * (1 - 2x) = 5 - (x + 9)
3 - 7 * 1 - 7 * (-2x) = ...
3 - (-7) - 14x = ...
When I should be doing:
Code:
3 - 7 * (1 - 2x) = 5 - (x + 9)
3 - 7 * 1 - 7 * (-2x) = ...
3 - 7 - 14x = ...
Why don't I put -7 inside parentheses?
6. ## Re: Sove simple equation doubt
If you expand:
$7(1-2x)=7-14x$
And the LHS is:
$3-[7(1-2x)]=3-(7-14x)=3-7+14x$
If you expand:
$-7(1-2x)=-7+14x$
Now, the LHS is:
$3+[-7(1-2x)]=3-7+14x$ | 2014-09-17T14:03:30 | {
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https://jp.maplesoft.com/support/help/Maple/view.aspx?path=fourier_in_maple | Fourier Transforms in Maple - Maple Help
Home : Support : Online Help : Mathematics : Calculus : Transforms : Fourier Transforms in Maple
Fourier Transforms in Maple
Fourier transforms in Maple can be categorized as either transforms on expressions or transforms on signal data.
To compute the Fourier transform of an expression, use the inttrans[fourier] command. For more details on this command, see the inttrans[fourier] help page.
To compute the Fourier transform of signal data, the following commands are available:
• SignalProcessing[DFT] : This command computes the discrete Fourier transform of an Array of signal data points. The SignalProcessing[DFT] command works for any size Array. For more information, see SignalProcessing[DFT].
• SignalProcessing[FFT] : Similar to the SignalProcessing[DFT] command, SignalProcessing[FFT] computes the discrete Fourier transform of an Array of signal data points. The difference between the two commands is that the SignalProcessing[FFT] command uses the fast Fourier transform algorithm. Note: SignalProcessing[FFT] requires that the size of the Array must be a power of 2, greater than 2. If the Array passed to SignalProcessing[FFT] does not meet this requirement, the SignalProcessing[DFT] command is used instead. Similarly, SignalProcessing[InverseFFT] calls SignalProcessing[InverseDFT] when the passed Array does not meet this requirement. For more information, see SignalProcessing[FFT].
• DiscreteTransforms[FourierTransform] : The DiscreteTransforms[FourierTransform] provides similar functionality to that of SignalProcessing[DFT]. There are some options available in DiscreteTransforms[FourierTransform], such as padding, that are not available in SignalProcessing[DFT]. For more information, see DiscreteTransforms[FourierTransform].
Note: Typically, SignalProcessing[DFT] and SignalProcessing[FFT] are slightly more efficient than DiscreteTransforms[FourierTransform].
The table below provides a summarized comparison of the discrete Fourier transform commands mentioned above.
Feature SignalProcessing[FFT] SignalProcessing[DFT] DiscreteTransforms[Fourier Transform] input single rtable yes yes yes input two rtables (Re/Im) yes yes yes higher-dimensional transforms yes yes yes specify single dim for higher-dimensional transforms no no yes output single rtable yes yes yes output two rtables (Re/Im) yes yes yes padding no no yes apply transform only to initial segment no no yes in place yes yes yes specify output rtable yes yes no specify working storage no no yes size of Array: power of 2 yes yes yes size of Array: other yes (dispatch to DFT) yes yes
Examples
> $\mathrm{signal}≔\mathrm{sin}\left(t\right)\mathrm{exp}\left(-\frac{{t}^{2}}{100}\right)$
${\mathrm{signal}}{≔}{\mathrm{sin}}{}\left({t}\right){}{{ⅇ}}^{{-}\frac{{{t}}^{{2}}}{{100}}}$ (1)
> $\mathrm{plot}\left(\mathrm{signal},t=-30..30\right)$
> $\mathrm{transform}≔\mathrm{inttrans}\left[\mathrm{fourier}\right]\left(\mathrm{signal},t,s\right)$
${\mathrm{transform}}{≔}{-}{10}{}{I}{}\sqrt{{\mathrm{\pi }}}{}{\mathrm{sinh}}{}\left({50}{}{s}\right){}{{ⅇ}}^{{-}{25}{}{{s}}^{{2}}{-}{25}}$ (2)
The transform is purely imaginary:
> $\mathrm{evalc}\left(\mathrm{\Re }\left(\mathrm{transform}\right)\right)$
${0}$ (3)
This is what the imaginary part looks like:
> $\mathrm{plot}\left(\mathrm{\Im }\left(\mathrm{transform}\right),s=-3..3\right)$
> $\mathrm{inttrans}\left[\mathrm{invfourier}\right]\left(\mathrm{transform},s,t\right)$
${\mathrm{sin}}{}\left({t}\right){}{{ⅇ}}^{{-}\frac{{{t}}^{{2}}}{{100}}}$ (4)
Turn the original signal into data by sampling:
> $\mathrm{data}≔\mathrm{Array}\left(1..80,i↦\mathrm{evalf}\left(\mathrm{eval}\left(\mathrm{signal},t=\frac{i}{4}\right)\right)\right)$
$\left[\begin{array}{cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc}0.2472493801& 0.4782284717& 0.6778153055& 0.8330982086& 0.9342719766& 0.9753019572& 0.9543081382& 0.8736433648& 0.7396636850& 0.5622125499& 0.3538622664& 0.1289739763& -0.09734987013& -0.3103399929& -0.4965810793& -0.6449045459& -0.7470908727& -0.7983356335& -0.7974529355& -0.7468109761& -0.6520150601& -0.5213720678& -0.3651855797& -0.1949415782& -0.02245018091& 0.1409909856& 0.2853511479& 0.4024873310& 0.4865936932& 0.5344562763& 0.5455063352& 0.5216811081& 0.4671149321& 0.3876949535& 0.2905236192& 0.1833342324& 0.07390606951& -0.03047787547& -0.1234939721& -0.2001341823& -0.2569380786& -0.2920941320& -0.3054141956& -0.2981943590& -0.2729827805& -0.2332802508& -0.1832019433& -0.1271290188& -0.06937673750& -0.01390182491& 0.03593343608& 0.07752902163& 0.1091418791& 0.1299085975& 0.1398024388& 0.1395353879& 0.1304188346& 0.1141980569& 0.09287587668& 0.06853983412& 0.04320521664& 0.01868354009& -0.003517056003& -0.02225629117& -0.03679098198& -0.04677160933& -0.05221092953& -0.05343083141& -0.05099444295& -0.04563063549& -0.03815763932& -0.02941158576& -0.02018456742& -0.01117540149& -0.002954832805& 0.004054455326& 0.009583845211& 0.01351257059& 0.01585170961& 0.01672117554\end{array}\right]$ (5)
> $\mathrm{SignalProcessing}:-\mathrm{SignalPlot}\left(\mathrm{data}\right)$
> $\mathrm{tdata}≔\mathrm{SignalProcessing}:-\mathrm{DFT}\left(\mathrm{data}\right)$
$\left[\begin{array}{cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc}0.4497659902145549+0.0{}I& 0.5052107976574163+0.03906636874940997{}I& 0.8711587723414446+0.07448652662727094{}I& 1.1182760567766632-1.6032057898841618{}I& -0.8339749012551638-0.6640820060944201{}I& -0.3159645805618081-0.12387884553360848{}I& -0.16663098558056866-0.07884710782613985{}I& -0.10535201213814539-0.059562131612666616{}I& -0.07240929585556202-0.048279430331170076{}I& -0.05222881581212239-0.040706219532542554{}I& -0.03882429156840338-0.035192046627913816{}I& -0.029408211717446228-0.030953615257113244{}I& -0.022514831311322547-0.027566764942940535{}I& -0.01730473009676248-0.024779825343899695{}I& -0.013265506494983152-0.022433066673257066{}I& -0.010068426817571105-0.020419723054675615{}I& -0.007493983240128059-0.018665435725976523{}I& -0.005390693235827055-0.01711665405828176{}I& -0.0036510898568611646-0.015733725210829757{}I& -0.002197117052166677-0.014486588000811401{}I& -0.0009709198499760314-0.013351987698461087{}I& 0.00007114657962727471-0.012311614766041167{}I& 0.0009625191227188982-0.011350826945342068{}I& 0.0017291645449145677-0.010457750572218883{}I& 0.0023914911868130355-0.009622634659130522{}I& 0.0029657088319229798-0.00883737750161765{}I& 0.003464811917516336-0.008095174947573733{}I& 0.0038993009931257852-0.007390253615141137{}I& 0.004277719195538448-0.006717666840270438{}I& 0.004607055731170729-0.0060731353106488785{}I& 0.004893052577030947-0.005452922509981582{}I& 0.005140439896249555-0.004853734699977014{}I& 0.005353117624996999-0.004272640914574546{}I& 0.005534296909190233-0.003707006932207723{}I& 0.005686610182988066-0.0031544417156110626{}I& 0.005812197276695991-0.0026127514510301993{}I& 0.0059127721776296585-0.002079901359869103{}I& 0.0059896751052553494-0.0015539818621112334{}I& 0.006043910803150312-0.0010331800204289058{}I& 0.006076178185685466-0.0005157520706090525{}I& 0.006086888632727769-7.757919228897728{}{10}^{-18}{}I& 0.00607617818568548+0.0005157520706090301{}I& 0.006043910803150335+0.0010331800204288989{}I& 0.0059896751052553685+0.0015539818621112117{}I& 0.005912772177629706+0.0020799013598689917{}I& 0.0058121972766957825+0.002612751451030189{}I& 0.0056866101829880335+0.0031544417156110895{}I& 0.005534296909190161+0.0037070069322077104{}I& 0.005353117624996999+0.004272640914574546{}I& 0.005140439896249608+0.004853734699976993{}I& 0.004893052577031012+0.00545292250998158{}I& 0.004607055731170938+0.006073135310648669{}I& 0.004277719195538379+0.006717666840270417{}I& 0.0038993009931257636+0.007390253615141087{}I& 0.0034648119175163176+0.008095174947573721{}I& 0.0029657088319229863+0.008837377501617644{}I& 0.002391491186813019+0.009622634659130522{}I& 0.0017291645449145827+0.010457750572218876{}I& 0.0009625191227188866+0.01135082694534206{}I& 0.00007114657962730806+0.01231161476604118{}I& -0.0009709198499760225+0.013351987698461087{}I& -0.0021971170521666713+0.014486588000811377{}I& -0.003651089856861188+0.015733725210829764{}I& -0.005390693235826993+0.017116654058281736{}I& -0.007493983240128059+0.018665435725976523{}I& -0.010068426817571098+0.020419723054675668{}I& -0.01326550649498307+0.02243306667325704{}I& -0.017304730096762463+0.024779825343899664{}I& -0.022514831311322574+0.027566764942940525{}I& -0.02940821171744621+0.03095361525711325{}I& -0.03882429156840338+0.03519204662791382{}I& -0.052228815812122374+0.04070621953254256{}I& -0.07240929585556202+0.048279430331170076{}I& -0.10535201213814538+0.05956213161266663{}I& -0.16663098558056866+0.07884710782613985{}I& -0.3159645805618081+0.12387884553360849{}I& -0.8339749012551637+0.6640820060944201{}I& 1.1182760567766632+1.6032057898841616{}I& 0.8711587723414448-0.07448652662727094{}I& 0.5052107976574162-0.039066368749409955{}I\end{array}\right]$ (6)
The following calls the FFT command, which in turn calls the DFT command (since the size of the data is not a power of 2):
> $\mathrm{tdata2}≔\mathrm{SignalProcessing}:-\mathrm{FFT}\left(\mathrm{data}\right)$
$\left[\begin{array}{cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc}0.4497659902145549+0.0{}I& 0.5052107976574163+0.03906636874940997{}I& 0.8711587723414446+0.07448652662727094{}I& 1.1182760567766632-1.6032057898841618{}I& -0.8339749012551638-0.6640820060944201{}I& -0.3159645805618081-0.12387884553360848{}I& -0.16663098558056866-0.07884710782613985{}I& -0.10535201213814539-0.059562131612666616{}I& -0.07240929585556202-0.048279430331170076{}I& -0.05222881581212239-0.040706219532542554{}I& -0.03882429156840338-0.035192046627913816{}I& -0.029408211717446228-0.030953615257113244{}I& -0.022514831311322547-0.027566764942940535{}I& -0.01730473009676248-0.024779825343899695{}I& -0.013265506494983152-0.022433066673257066{}I& -0.010068426817571105-0.020419723054675615{}I& -0.007493983240128059-0.018665435725976523{}I& -0.005390693235827055-0.01711665405828176{}I& -0.0036510898568611646-0.015733725210829757{}I& -0.002197117052166677-0.014486588000811401{}I& -0.0009709198499760314-0.013351987698461087{}I& 0.00007114657962727471-0.012311614766041167{}I& 0.0009625191227188982-0.011350826945342068{}I& 0.0017291645449145677-0.010457750572218883{}I& 0.0023914911868130355-0.009622634659130522{}I& 0.0029657088319229798-0.00883737750161765{}I& 0.003464811917516336-0.008095174947573733{}I& 0.0038993009931257852-0.007390253615141137{}I& 0.004277719195538448-0.006717666840270438{}I& 0.004607055731170729-0.0060731353106488785{}I& 0.004893052577030947-0.005452922509981582{}I& 0.005140439896249555-0.004853734699977014{}I& 0.005353117624996999-0.004272640914574546{}I& 0.005534296909190233-0.003707006932207723{}I& 0.005686610182988066-0.0031544417156110626{}I& 0.005812197276695991-0.0026127514510301993{}I& 0.0059127721776296585-0.002079901359869103{}I& 0.0059896751052553494-0.0015539818621112334{}I& 0.006043910803150312-0.0010331800204289058{}I& 0.006076178185685466-0.0005157520706090525{}I& 0.006086888632727769-7.757919228897728{}{10}^{-18}{}I& 0.00607617818568548+0.0005157520706090301{}I& 0.006043910803150335+0.0010331800204288989{}I& 0.0059896751052553685+0.0015539818621112117{}I& 0.005912772177629706+0.0020799013598689917{}I& 0.0058121972766957825+0.002612751451030189{}I& 0.0056866101829880335+0.0031544417156110895{}I& 0.005534296909190161+0.0037070069322077104{}I& 0.005353117624996999+0.004272640914574546{}I& 0.005140439896249608+0.004853734699976993{}I& 0.004893052577031012+0.00545292250998158{}I& 0.004607055731170938+0.006073135310648669{}I& 0.004277719195538379+0.006717666840270417{}I& 0.0038993009931257636+0.007390253615141087{}I& 0.0034648119175163176+0.008095174947573721{}I& 0.0029657088319229863+0.008837377501617644{}I& 0.002391491186813019+0.009622634659130522{}I& 0.0017291645449145827+0.010457750572218876{}I& 0.0009625191227188866+0.01135082694534206{}I& 0.00007114657962730806+0.01231161476604118{}I& -0.0009709198499760225+0.013351987698461087{}I& -0.0021971170521666713+0.014486588000811377{}I& -0.003651089856861188+0.015733725210829764{}I& -0.005390693235826993+0.017116654058281736{}I& -0.007493983240128059+0.018665435725976523{}I& -0.010068426817571098+0.020419723054675668{}I& -0.01326550649498307+0.02243306667325704{}I& -0.017304730096762463+0.024779825343899664{}I& -0.022514831311322574+0.027566764942940525{}I& -0.02940821171744621+0.03095361525711325{}I& -0.03882429156840338+0.03519204662791382{}I& -0.052228815812122374+0.04070621953254256{}I& -0.07240929585556202+0.048279430331170076{}I& -0.10535201213814538+0.05956213161266663{}I& -0.16663098558056866+0.07884710782613985{}I& -0.3159645805618081+0.12387884553360849{}I& -0.8339749012551637+0.6640820060944201{}I& 1.1182760567766632+1.6032057898841616{}I& 0.8711587723414448-0.07448652662727094{}I& 0.5052107976574162-0.039066368749409955{}I\end{array}\right]$ (7)
tdata and tdata2 are the same, up to tiny float inaccuracies:
> $\mathrm{verify}\left(\mathrm{tdata},\mathrm{tdata2},'\mathrm{Array}'\left('\mathrm{float}'\left(10,\mathrm{test}=2\right)\right)\right)$
${\mathrm{true}}$ (8)
> $\mathrm{tdata3}≔\mathrm{DiscreteTransforms}:-\mathrm{FourierTransform}\left(\mathrm{data}\right)$
$\left[\begin{array}{cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc}0.4497659902145549+0.0{}I& 0.5052107976574163+0.039066368749409955{}I& 0.8711587723414446+0.07448652662727094{}I& 1.1182760567766632-1.603205789884162{}I& -0.8339749012551638-0.6640820060944201{}I& -0.3159645805618081-0.12387884553360849{}I& -0.16663098558056866-0.07884710782613988{}I& -0.1053520121381454-0.05956213161266657{}I& -0.07240929585556202-0.04827943033117009{}I& -0.05222881581212236-0.040706219532542574{}I& -0.038824291568403356-0.03519204662791384{}I& -0.029408211717446116-0.03095361525711323{}I& -0.022514831311322495-0.027566764942940535{}I& -0.01730473009676253-0.024779825343899796{}I& -0.013265506494983152-0.022433066673257066{}I& -0.010068426817571093-0.02041972305467564{}I& -0.007493983240128061-0.018665435725976523{}I& -0.005390693235827055-0.01711665405828174{}I& -0.0036510898568611677-0.015733725210829785{}I& -0.002197117052166725-0.014486588000811432{}I& -0.0009709198499759817-0.013351987698461092{}I& 0.00007114657962733987-0.012311614766041179{}I& 0.0009625191227189075-0.011350826945342072{}I& 0.0017291645449145458-0.010457750572218868{}I& 0.0023914911868130415-0.009622634659130522{}I& 0.0029657088319229924-0.008837377501617671{}I& 0.0034648119175163237-0.008095174947573733{}I& 0.00389930099312586-0.0073902536151410625{}I& 0.004277719195538548-0.0067176668402704875{}I& 0.004607055731170927-0.0060731353106488785{}I& 0.004893052577030996-0.005452922509981607{}I& 0.005140439896249604-0.004853734699977054{}I& 0.005353117624997005-0.004272640914574546{}I& 0.005534296909190279-0.0037070069322076974{}I& 0.005686610182988094-0.0031544417156110804{}I& 0.005812197276695914-0.0026127514510301112{}I& 0.005912772177629639-0.002079901359869095{}I& 0.005989675105255347-0.0015539818621112295{}I& 0.0060439108031502976-0.001033180020428902{}I& 0.006076178185685449-0.0005157520706089898{}I& 0.0060868886327277725-7.757919228897728{}{10}^{-18}{}I& 0.006076178185685475+0.0005157520706090549{}I& 0.00604391080315032+0.0010331800204289082{}I& 0.005989675105255343+0.00155398186211128{}I& 0.005912772177629716+0.002079901359869018{}I& 0.005812197276695808+0.0026127514510301112{}I& 0.005686610182988059+0.00315444171561106{}I& 0.005534296909190154+0.0037070069322076944{}I& 0.005353117624997005+0.004272640914574546{}I& 0.005140439896249597+0.004853734699976988{}I& 0.004893052577031059+0.0054529225099815915{}I& 0.0046070557311709195+0.006073135310648729{}I& 0.004277719195538376+0.006717666840270428{}I& 0.0038993009931257133+0.007390253615141106{}I& 0.0034648119175163003+0.008095174947573737{}I& 0.0029657088319229767+0.00883737750161764{}I& 0.0023914911868130207+0.009622634659130528{}I& 0.0017291645449145692+0.01045775057221889{}I& 0.0009625191227188874+0.011350826945342079{}I& 0.00007114657962720333+0.012311614766041179{}I& -0.0009709198499760143+0.013351987698461097{}I& -0.0021971170521666618+0.01448658800081144{}I& -0.003651089856861185+0.015733725210829736{}I& -0.005390693235826972+0.017116654058281704{}I& -0.007493983240128061+0.018665435725976523{}I& -0.010068426817571101+0.020419723054675647{}I& -0.01326550649498307+0.022433066673257063{}I& -0.01730473009676242+0.02477982534389959{}I& -0.022514831311322602+0.027566764942940518{}I& -0.029408211717446266+0.03095361525711326{}I& -0.03882429156840338+0.03519204662791383{}I& -0.052228815812122444+0.040706219532542554{}I& -0.07240929585556204+0.048279430331170076{}I& -0.1053520121381454+0.05956213161266663{}I& -0.16663098558056866+0.07884710782613985{}I& -0.3159645805618082+0.12387884553360848{}I& -0.8339749012551638+0.6640820060944201{}I& 1.1182760567766632+1.6032057898841618{}I& 0.8711587723414446-0.07448652662727087{}I& 0.5052107976574163-0.03906636874940996{}I\end{array}\right]$ (9)
Moreover, tdata and tdata3 are the same, up to tiny float inaccuracies:
> $\mathrm{verify}\left(\mathrm{tdata},\mathrm{tdata3},'\mathrm{Array}'\left('\mathrm{float}'\left(10,\mathrm{test}=2\right)\right)\right)$
${\mathrm{true}}$ (10)
> $\mathrm{SignalProcessing}:-\mathrm{SignalPlot}\left({⟨\mathrm{~}\left[\mathrm{\Re }\right]\left(\mathrm{tdata}\right),\mathrm{~}\left[\mathrm{\Im }\right]\left(\mathrm{tdata}\right)⟩}^{\mathrm{%T}},'\mathrm{compactplot}'\right)$
> $\mathrm{original}≔\mathrm{SignalProcessing}:-\mathrm{InverseFFT}\left(\mathrm{tdata}\right)$
$\left[\begin{array}{cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc}0.24724938010000005-5.788806807038671{}{10}^{-17}{}I& 0.4782284717-4.6306843959555885{}{10}^{-18}{}I& 0.6778153055-8.813190380294797{}{10}^{-17}{}I& 0.8330982086000001-6.514854332728859{}{10}^{-17}{}I& 0.9342719765999999-3.849326999814238{}{10}^{-17}{}I& 0.9753019572-4.372165814963856{}{10}^{-17}{}I& 0.9543081381999999-1.0713724607539688{}{10}^{-16}{}I& 0.8736433648+1.1462631311827254{}{10}^{-17}{}I& 0.7396636849999999+7.27845345235421{}{10}^{-18}{}I& 0.5622125499-4.186869669466468{}{10}^{-17}{}I& 0.35386226639999996+8.639718032697117{}{10}^{-17}{}I& 0.12897397629999988-3.1013698720138586{}{10}^{-17}{}I& -0.09734987012999996+3.5982754599275804{}{10}^{-17}{}I& -0.3103399928999999+1.0523039104519782{}{10}^{-16}{}I& -0.4965810792999999-3.1136855829179936{}{10}^{-17}{}I& -0.6449045459000001+4.2494308227418165{}{10}^{-17}{}I& -0.7470908727+3.35757149293384{}{10}^{-17}{}I& -0.7983356335+1.3239115740348083{}{10}^{-18}{}I& -0.7974529354999998+5.249906607569585{}{10}^{-17}{}I& -0.7468109761+1.8871850934219758{}{10}^{-17}{}I& -0.6520150600999999-4.1077204126344286{}{10}^{-17}{}I& -0.5213720677999999+5.063547722021758{}{10}^{-17}{}I& -0.36518557969999993-2.6385105329169736{}{10}^{-17}{}I& -0.19494157819999997-5.034994399150584{}{10}^{-17}{}I& -0.02245018091000007+4.520793273779095{}{10}^{-17}{}I& 0.14099098560000015-3.2724974232940717{}{10}^{-17}{}I& 0.2853511479+3.907489070614328{}{10}^{-17}{}I& 0.402487331+8.469533485416189{}{10}^{-18}{}I& 0.48659369319999984+9.698082920104402{}{10}^{-18}{}I& 0.5344562762999999-8.932425093008308{}{10}^{-17}{}I& 0.5455063352-2.1762828718436853{}{10}^{-19}{}I& 0.5216811080999999+2.6482383785783187{}{10}^{-17}{}I& 0.4671149320999999+7.212827541007565{}{10}^{-17}{}I& 0.3876949535+8.14606194735691{}{10}^{-17}{}I& 0.2905236192-3.3022244070640787{}{10}^{-17}{}I& 0.18333423239999988+1.3039786540038024{}{10}^{-16}{}I& 0.07390606950999985-2.5043265028372553{}{10}^{-17}{}I& -0.030477875470000006+1.1462895400597365{}{10}^{-16}{}I& -0.12349397209999992+3.961109875421382{}{10}^{-17}{}I& -0.2001341822999999+5.8889948737155596{}{10}^{-18}{}I& -0.2569380785999999+2.3740175388370566{}{10}^{-17}{}I& -0.29209413199999995+2.457843447149258{}{10}^{-17}{}I& -0.3054141955999998-1.8190162000157572{}{10}^{-17}{}I& -0.2981943589999999-5.7721481393800585{}{10}^{-18}{}I& -0.27298278049999997+2.5840450106396485{}{10}^{-18}{}I& -0.2332802507999998+3.2618811210719874{}{10}^{-18}{}I& -0.1832019433+4.9607848690802386{}{10}^{-17}{}I& -0.12712901879999977-1.2903000941256319{}{10}^{-16}{}I& -0.06937673750000004-1.886823916508005{}{10}^{-17}{}I& -0.013901824909999944-4.5812523542074915{}{10}^{-17}{}I& 0.03593343608000013-4.031822888859644{}{10}^{-17}{}I& 0.0775290216300001-1.0226046617704087{}{10}^{-16}{}I& 0.10914187909999994-9.030053709139358{}{10}^{-17}{}I& 0.12990859750000008-1.0343740793553614{}{10}^{-16}{}I& 0.13980243880000012-5.483011258062978{}{10}^{-17}{}I& 0.13953538790000003+1.7036267749938947{}{10}^{-17}{}I& 0.1304188345999999-8.184830882269641{}{10}^{-17}{}I& 0.11419805689999987-4.6252474029640126{}{10}^{-17}{}I& 0.09287587667999987+5.420140833553328{}{10}^{-18}{}I& 0.06853983411999992-1.1345784577491752{}{10}^{-17}{}I& 0.04320521663999989+2.4471999464506593{}{10}^{-17}{}I& 0.01868354008999981-4.7606716955327633{}{10}^{-17}{}I& -0.003517056003000017-4.908741794575067{}{10}^{-18}{}I& -0.022256291169999994+2.37790804659256{}{10}^{-17}{}I& -0.03679098197999992-8.071254299971714{}{10}^{-18}{}I& -0.04677160933000001+6.568608741830469{}{10}^{-18}{}I& -0.05221092953000003+4.06939146588404{}{10}^{-17}{}I& -0.053430831409999796+6.477669952037744{}{10}^{-17}{}I& -0.05099444294999982+6.45112800847227{}{10}^{-17}{}I& -0.04563063549000001+1.2015062864008614{}{10}^{-16}{}I& -0.038157639320000006+3.1982773214475455{}{10}^{-17}{}I& -0.029411585760000056+7.327520661516035{}{10}^{-17}{}I& -0.02018456741999993-2.004933932523007{}{10}^{-17}{}I& -0.011175401490000025+1.1050934674793172{}{10}^{-17}{}I& -0.0029548328049999457+2.448292107023198{}{10}^{-17}{}I& 0.00405445532599999-5.334304181791691{}{10}^{-17}{}I& 0.009583845211000034+4.511353717067075{}{10}^{-17}{}I& 0.01351257058999993-6.604288736668821{}{10}^{-17}{}I& 0.015851709609999994+1.7161159567782773{}{10}^{-17}{}I& 0.016721175539999883+6.255571657698161{}{10}^{-17}{}I\end{array}\right]$ (11)
> $\mathrm{verify}\left(\mathrm{data},\mathrm{original},'\mathrm{Array}'\left('\mathrm{float}'\left(10,\mathrm{test}=2\right)\right)\right)$
${\mathrm{true}}$ (12)
> | 2021-12-05T22:58:17 | {
"domain": "maplesoft.com",
"url": "https://jp.maplesoft.com/support/help/Maple/view.aspx?path=fourier_in_maple",
"openwebmath_score": 0.8218860626220703,
"openwebmath_perplexity": 1876.8787169509344,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9715639669551474,
"lm_q2_score": 0.6723317123102955,
"lm_q1q2_score": 0.6532132655219376
} |
https://brilliant.org/discussions/thread/proof-please/ | ×
Today my friend gave me a formula which: $$\sqrt{a+2\sqrt{bc}} = \sqrt{b} + \sqrt{c}$$ Where $$b+c=a$$. Can you give me the proof? She got it from her teacher, yet she didn't ask how her teacher got the formula.
Note by Kenny Indrajaya
2 years, 11 months ago
Sort by:
Square both sides, you'll end up with a = b + c · 2 years, 11 months ago
$$\sqrt{ a + 2\sqrt{bc}}$$
$$= \sqrt{b + c + 2\sqrt{bc}}$$
$$= \sqrt{ (\sqrt{b})^2 + (\sqrt{c})^2 + 2\sqrt{bc}}$$
$$= \sqrt{ (\sqrt{b} + \sqrt{c})^2 }$$
$$= \sqrt{b} + \sqrt{c}$$ · 2 years, 11 months ago | 2017-01-22T12:27:51 | {
"domain": "brilliant.org",
"url": "https://brilliant.org/discussions/thread/proof-please/",
"openwebmath_score": 0.9436385035514832,
"openwebmath_perplexity": 1679.559554607689,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.971563966131786,
"lm_q2_score": 0.6723317123102956,
"lm_q1q2_score": 0.6532132649683657
} |
https://math.libretexts.org/Under_Construction/Purgatory/Remixer_University/Username%3A_junalyn2020/Book%3A_Introduction_to_Real_Analysis_(Lebl)/1%3A_Real_Numbers/1.2%3A_The_set_of_real_numbers | # 1.2: The set of real numbers
This page titled 1.2: The set of real numbers is shared under a not declared license and was authored, remixed, and/or curated by Jiří Lebl. | 2022-10-05T19:10:46 | {
"domain": "libretexts.org",
"url": "https://math.libretexts.org/Under_Construction/Purgatory/Remixer_University/Username%3A_junalyn2020/Book%3A_Introduction_to_Real_Analysis_(Lebl)/1%3A_Real_Numbers/1.2%3A_The_set_of_real_numbers",
"openwebmath_score": 0.9999414682388306,
"openwebmath_perplexity": 292.71815743858224,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.971563966131786,
"lm_q2_score": 0.6723317123102956,
"lm_q1q2_score": 0.6532132649683657
} |
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