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# Functions Which are non differentiable on a Given Set.
I have recently read the famous Weierstrass Non differentiable function which is continuous everywhere but nowhere differentiable.
But now my question is:
Given a countable set $S (\subset \mathbb{R})$ , can we construct a function which is continuous everywhere but non differentiable only at points of S..?
My Attempt:
If $S$ finite say, $S=\{c_1,c_2,...,c_n\}$ Then the following function is the desired one:$$f(x)=\sum_{i=1}^{n}|x-c_i|$$
But I cannot construct the function for a arbitrary Enumerable set $S.$
Can we generalize the question for an arbitrary set $S$?
Thank you...!!
• You can make an easy example of a set which has corners periodically by just defining the function on, say, $[-1,1]$ and arranging so that it takes equal values on the ends, and then copy-pasting it. Maybe you can use this to handle the general case? I'm not sure - just a thought. – Alfred Yerger Dec 26 '16 at 5:09
• Sir, I think I got your point a little...but Is there any explicit example...??@ A Yerger – indrajit Dec 26 '16 at 5:14
• If you want to write down an explicit formula for such a thing that won't be defined piecewise, it's done with Fourier series. But I have thought of another example which modifies your own, and is also discontinuous on the integers. $f(x) = \sum_{n \in \mathbb Z} \frac{1}{2^{|n|}}|x-n|$. This should be okay since as $n$ moves away from $0$, we shrink the contributions geometrically. – Alfred Yerger Dec 26 '16 at 5:19
• But sir this is an particular case. and I have also construct a function which is non differentiable at only $\mathbb{Z}$ but continuous every where.....But I cannot extend this construction for the arbitrary set $S$. In fact, for $S={a_1, a_2, a_3,.......}$....!!!! – indrajit Dec 26 '16 at 5:24
• Is the construction possible...?? Intuitively it seems to me that this is possible (as u noted in ur first comment) ....but I can't.... – indrajit Dec 26 '16 at 5:26 | {
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The same construction can be generalized. Let $a_1,a_2,\dots$ be the elements of $S$.
Then the function $f(x)=\sum\limits_{n=1}^\infty \dfrac{|x-a_n|}{2^n \max(|a_n|,1)}$ is differentiable at all points except the points at $S$.
• Can you prove it? We can try to work it out if you'd like. – Yorch Dec 26 '16 at 5:40
• How does the series Converge....?? – indrajit Dec 26 '16 at 5:50
• hmm, good point. we'll adjust it. – Yorch Dec 26 '16 at 5:56
• Please tell how does the series converge now...??@ J F Hidalgo – indrajit Dec 26 '16 at 6:04
• well, each summand is smaller than $|\frac{x}{a_n}-1|2^{-n}|$ and $|x-a_n|2^{-n}$. If $|a_n|\leq 1$ it is smaller than $(|x|+1)|2^{-n}$ and otherwise it is smaller than $(|x|+1)2^{-n}$ – Yorch Dec 26 '16 at 6:08
Yes. See this answer, on the kinds of sets where a function $f$ may fail to be differentiable. Such a set must be a $G_{\delta\sigma}$ set, meaning it must be an infinite, countable union of $G_{\delta}$ sets (which in turn are countable intersections of open sets).
Notice that every closed set is $G_{\delta}$ and in particular, every singleton is $G_{\delta}$:
$$\{p\}=\bigcap_{n\geq1}\left(p-\frac1n,p+\frac1n\right)$$
Hence, every countable set is $G_{\delta\sigma}$: $C=\cup_{n\in\mathbb{N}}\{c_n\}$.
The paper linked in the answer describes how one may construct continuous functions $f:\mathbb{R}\longrightarrow\mathbb{R}$ whose set of non-differentiability is a $G_1\cup G_2$, where $G_1$ is any $G_{\delta}$ and $G_2$ is a $G_{\delta\sigma}$ with measure zero. In particular, notice that every countable set is both $G_{\delta\sigma}$ and has measure zero, so every countable set may be written as some $G_1\cup G_2$. | {
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Documentation
# istril
Determine if matrix is lower triangular
## Syntax
``tf = istril(A)``
## Description
example
````tf = istril(A)` returns logical `1` (`true`) if `A` is a lower triangular matrix; otherwise, it returns logical `0` (`false`).```
## Examples
collapse all
Create a 5-by-5 matrix.
`D = tril(magic(5))`
```D = 5×5 17 0 0 0 0 23 5 0 0 0 4 6 13 0 0 10 12 19 21 0 11 18 25 2 9 ```
Test `D` to see if it is lower triangular.
`istril(D)`
```ans = logical 1 ```
The result is logical `1` (`true`) because all elements above the main diagonal are zero.
Create a 5-by-5 matrix of zeros.
`Z = zeros(5);`
Test `Z` to see if it is lower triangular.
`istril(Z)`
```ans = logical 1 ```
The result is logical `1` (`true`) because a lower triangular matrix can have any number of zeros on its main diagonal.
## Input Arguments
collapse all
Input array, specified as a numeric array. `istril` returns logical `0` (`false`) if `A` has more than two dimensions.
Data Types: `single` | `double`
Complex Number Support: Yes
collapse all
### Lower Triangular Matrix
A matrix is lower triangular if all elements above the main diagonal are zero. Any number of the elements on the main diagonal can also be zero.
For example, the matrix
`$A=\left(\begin{array}{cccc}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& 0\\ -1& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& 0\\ -2& -2& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1& 0\\ -3& -3& -3& 1\end{array}\right)$`
is lower triangular. A diagonal matrix is both upper and lower triangular.
## Tips
• Use the `tril` function to produce lower triangular matrices for which `istril` returns logical `1` (`true`). | {
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• The functions `isdiag`, `istriu`, and `istril` are special cases of the function `isbanded`, which can perform all of the same tests with suitably defined upper and lower bandwidths. For example, ```istril(A) == isbanded(A,size(A,1),0)```. | {
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# Solution needed : Kinematics problem
1. Nov 23, 2009
### dawoodvora
1. The problem statement, all variables and given/known data
A ball is dropped from rest from a height h above the ground. Another ball is thrown vertically upward from the ground at the instant the first ball is released. Determine the speed of the second ball if the two balls are to meet at a height h/2 above the ground.
Variables for Ball 1 going downward
v1i = initial velocity of ball1
v1f = final velocity of ball1
Variables for Ball 2 going downward
v2i = initial velocity of ball2
v2f = final velocity of ball2
t = time taken for both the balls to travel common meeting point
a = -9.8 m/s2
2. Relevant equations
4 general kinematic equations in one dimension having constant acceleration
3. The attempt at a solution
Logically, the final velocity of ball 1, when it reaches h/2, must also be the value of the initial velocity of ball2 going upward. Also, ball2 final velocity at h/2 must be zero. In this way, both the balls can meet each other at h/2
I am still unable to arrive at the figure of initial speed of ball2.
2. Nov 23, 2009
### kuruman
The general way of doing this sort of problem is to
(a) First find the time it takes for the balls to pass each other by using the kinematic equation for the ball that is released from height h.
(b) Second use the same kinematic equation and the time you found in (a) to find the initial velocity for the ball that is projected up.
This method will allow you to solve the problem regardless of the fraction of h at which the two balls pass each other.
In this specific case, since you know that the speeds of the balls are the same when they pass each other, you can use the kinematic equation that does not involve time directly.
3. Nov 23, 2009
### Seannation
EDIT: Disregard all of the below, see next post! | {
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3. Nov 23, 2009
### Seannation
EDIT: Disregard all of the below, see next post!
You're making innecessary assumptions here. The second ball's velocity needn't necessarily be zero at h/2. Since we're only dealing with one dimension here, there is only one possible initial speed of ball 2 that can allow it to hit ball 1 at h/2.
The two variables which will be equal for both balls when they hit is that they will both have travelled a distance of h/2. However one ball's displacement will be negative to the other as one ball is travelling down and one ball is travelling up.
You're looking to find the initial velocity of the second ball such that the distance travelled by the second ball against gravity equals the distance fallen by the second ball with gravity over the same time period.
Last edited: Nov 23, 2009
4. Nov 23, 2009
### kuruman
If the balls meet halfway up, then the second ball's velocity is zero when they pass each other. This will not be the case if they met at some other fraction of h.
Proof
Let tf = the time the balls meet. The average velocity for ball A is
$$\bar{v}_A=gt_f/2$$
The average velocity for ball B is
$$\bar{v}_B=(v_0+v_0-gt_f)/2=v_0-gt_f/2$$
where v0 is what we are looking for.
Since the balls cover equal distances in time tf, their average velocities must be the same.
$$gt_{f}/2=v_0-gt_f/2\ \rightarrow$$ $$v_0=gt_f$$
Therefore the velocity of ball B at any time t is given by
$$v=gt_{f}-gt$$
At specific time t = tf, ball B has zero velocity. Q.E.D.
5. Nov 23, 2009
### Seannation
You make a very good point! I somehow got it in my head that the second ball would be travelling at high speed when it hit the first. But as you've proved, it can't have a non-zero velocity if it happens at h/2. | {
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# familyof seminorms on normed spaces
Let $(X,\|\cdot\|)$ be a normed space. It is known that the norm $\|\cdot\|$ induces a topology, known as the norm topology $\tau$ on $X$. Then the pair $(X,\tau)$ is a locally convex topological vector space. With this, $\tau$ can be generated by a family $P$ of semi-norms on $X$.
Question: Is it necessary that $\|\cdot\|$ is in $P$?
No, it is not necessary that $\lVert\,\cdot\,\rVert$ belongs to the family $P$. Unless $X = \{0\}$, when there is only one seminorm on the space - we could work around that if we allow $P = \varnothing$. And if $\dim X = 1$, then all seminorms on $X$ are constant multiples of the norm, and we must put at least one non-zero multiple of the norm in the family. Of course we could always take $P = \{ 2\cdot \lVert\,\cdot\,\rVert\}$ for $\dim X > 0$ and obey the letter of the law but violate its spirit.
More interesting, we can always take a family consisting of seminorms that are not norms if $\dim X > 1$. For a closed linear subspace $F$ of $X$, the norm induces a quotient norm on $X/F$,
$$\lVert \pi(x)\rVert_{X/F} := \inf \{ \lVert x-y\rVert : y \in F\} = \operatorname{dist}(x,F),$$
where $\pi\colon X \to X/F$ is the canonical projection. We can pull that back to a seminorm on $X$, by defining $p_F(x) := \lVert \pi(x)\rVert_{X/F}$.
Then $p_F$ is a continuous (with respect to $\tau$ of course) seminorm on $X$, with kernel $F$. Hence, taking a family $\mathscr{F}$ of nontrivial closed subspaces of $X$, we obtain a family $P = \{ p_F : F\in \mathscr{F}\}$ of continuous seminorms on $X$. Since all seminorms are continuous, the topology induced by $P$ is coarser than $\tau$.
It remains to see that one can arrange it so that the induced topology coincides with $\tau$. One way is to include two complementary subspaces in $P$. If $G,H$ are two closed subspaces such that
$$\alpha\colon G\times H \to X,\quad \alpha(x,y) = x+y$$ | {
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$$\alpha\colon G\times H \to X,\quad \alpha(x,y) = x+y$$
is a topological isomorphism, then $\{ p_G, p_H\}$ induces $\tau$. To see that, we note that the restriction of $p_H$ to $G$ is a norm which is equivalent to the restriction of $\lVert\,\cdot\,\rVert$ to $G$. Since $G \cap \ker p_H = G\cap H = \{0\}$, it is a norm, and it is continuous as the restriction of a continuous seminorm. Since $\alpha$ is a topological isomorphism, there is a $c > 0$ such that $\lVert x+y\rVert \geqslant c(\lVert x\rVert + \lVert y\rVert)$ for all $(x,y) \in G\times H$, hence
$$p_H(x) = \inf \{ \lVert x+y\rVert : y \in H\} \geqslant c\lVert x\rVert$$
for all $x\in G$. The same holds when we interchange the roles of $G$ and $H$, whence
\begin{align} \max \{ p_G(x+y), p_H(x+y)\} &= \max \{ p_G(y), p_H(x)\}\\ & \geqslant \frac{1}{2}(p_G(y) + p_H(x))\\ &\geqslant \frac{c}{2}(\lVert y\rVert + \lVert x\rVert)\\ & \geqslant \frac{c}{2}\lVert x+y\rVert \end{align}
for $x\in G, y\in H$, so $\max \{p_G, p_H\}$ is a norm that induces a finer topology than $\tau$ on $X$. Since it is also a $\tau$-continuous norm, it is equivalent to $\lVert\,\cdot\,\rVert$.
Finally, we need to see that in every normed space of dimension $\geqslant 2$ there are complementary subspaces. Let $\lambda \neq 0$ be a continuous linear form, and $x_0\in X$ with $\lambda(x_0) = 1$. Then $\ker \lambda$ and $\operatorname{span} \{x_0\}$ are complementary subspaces. $\ker \lambda$ is closed since $\lambda$ is continuous, and $\operatorname{span} \{ x_0\}$ is finite-dimensional, hence closed. The map $x \mapsto (\lambda(x)\cdot x_0, x - \lambda(x)\cdot x_0)$ is continuous, and its inverse $\alpha \colon (\operatorname{span} \{x_0\}) \times (\ker \lambda) \to X,\; (x,y) \mapsto x+y$ is also continuous, hence a topological isomorphism.
• I did not expect a detailed answer sir. Thank you very much. – Juniven May 18 '15 at 21:25 | {
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TangentVector - Maple Help
VectorCalculus
TangentVector
compute the tangent vector to a curve
Calling Sequence TangentVector(C, t, n)
Parameters
C - free or position Vector or Vector valued procedure; specify the components of the curve t - (optional) name; specify the parameter of the curve n - (optional) equation of the form normalized=true or normalized=false, or simply normalized
Description
• The TangentVector(C, t) command computes the tangent vector to the curve C that is parameterized by t. Note that this vector is not normalized by default, so it is a scalar multiple of the unit tangent vector to the curve C. Therefore, by default, if C is a curve in ${ℝ}^{3}$, the result is generally different from the output of TNBFrame(C, t, output=['T']).
• If n is given as either normalized=true or normalized, then the resulting vector will be normalized before it is returned. As discussed above, the default value is false, so that the result is not normalized.
• The curve can be specified as a free or position Vector or a Vector valued procedure. This determines the returned object type.
• If t is not specified, the function tries to determine a suitable variable name by using the components of C. To do this, it checks all of the indeterminates of type name in the components of C and removes the ones which are determined to be constants.
If the resulting set has a single entry, the single entry is the variable name. If it has more than one entry, an error is raised.
• If a coordinate system attribute is specified on C, C is interpreted in that coordinate system. Otherwise, the curve is interpreted as a curve in the current default coordinate system. If the two are not compatible, an error is raised.
Examples | {
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Examples
> $\mathrm{with}\left(\mathrm{VectorCalculus}\right):$
> $\mathrm{T1}≔\mathrm{TangentVector}\left(t↦⟨t,{t}^{2},{t}^{3}⟩\right):$
> $\mathrm{T1}\left(t\right)$
$\left[\begin{array}{c}{1}\\ {2}{}{t}\\ {3}{}{{t}}^{{2}}\end{array}\right]$ (1)
> $\mathrm{T2}≔\mathrm{TangentVector}\left(t↦⟨t,{t}^{2},{t}^{3}⟩,\mathrm{normalized}\right):$
> $\mathrm{T2}\left(t\right)$
$\left[\begin{array}{c}\frac{{1}}{\sqrt{{9}{}{{t}}^{{4}}{+}{4}{}{{t}}^{{2}}{+}{1}}}\\ \frac{{2}{}{t}}{\sqrt{{9}{}{{t}}^{{4}}{+}{4}{}{{t}}^{{2}}{+}{1}}}\\ \frac{{3}{}{{t}}^{{2}}}{\sqrt{{9}{}{{t}}^{{4}}{+}{4}{}{{t}}^{{2}}{+}{1}}}\end{array}\right]$ (2)
> $\mathrm{TangentVector}\left(\mathrm{PositionVector}\left(\left[\mathrm{cos}\left(t\right),\mathrm{sin}\left(t\right)\right]\right),t\right)$
$\left[\begin{array}{c}{-}{\mathrm{sin}}{}\left({t}\right)\\ {\mathrm{cos}}{}\left({t}\right)\end{array}\right]$ (3)
> $\mathrm{TangentVector}\left(⟨a\mathrm{exp}\left(-t\right),t⟩\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{assuming}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}a::\mathrm{constant}$
$\left[\begin{array}{c}{-}{a}{}{{ⅇ}}^{{-}{t}}\\ {1}\end{array}\right]$ (4)
> $\mathrm{TangentVector}\left(t↦⟨a\cdot \mathrm{cos}\left(t\right),b\cdot \mathrm{sin}\left(t\right),t⟩\right)$
${t}{→}{\mathrm{VectorCalculus}}{:-}{\mathrm{VectorSpace}}{}\left({\mathrm{cartesian}}{,}\left[{a}{}{\mathrm{cos}}{}\left({t}\right){,}{b}{}{\mathrm{sin}}{}\left({t}\right){,}{t}\right]\right){:-}{\mathrm{Vector}}{}\left(\left[{-}{a}{}{\mathrm{sin}}{}\left({t}\right){,}{b}{}{\mathrm{cos}}{}\left({t}\right){,}{1}\right]\right)$ (5)
> $\mathrm{SetCoordinates}\left('\mathrm{polar}'\right)$
${\mathrm{polar}}$ (6)
> $\mathrm{TangentVector}\left(⟨1,t⟩,t\right)$
$\left[\begin{array}{c}{0}\\ {1}\end{array}\right]$ (7)
> $\mathrm{TangentVector}\left(⟨\mathrm{exp}\left(-t\right),t⟩,t\right)$
$\left[\begin{array}{c}{-}\sqrt{{{ⅇ}}^{{-}{2}{}{t}}}\\ \sqrt{{{ⅇ}}^{{-}{2}{}{t}}}\end{array}\right]$ (8) | {
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# Is there a term for the opposite of an interpolation?
The question is specifically about linear Interpolation, which is usually defined to be a function $$f : V \times V \times \mathbb{R} \rightarrow V \\ f(v_0,v_1,\alpha) = v_0 + (v_1-v_0) \cdot \alpha$$ So it takes the arguments $$v_0$$ and $$v_1$$ and a real value $$\alpha$$ (often, but not necessarily, in $$[0,1]$$), and computes the linearly interpolated value.
Conversely, there may be a function like this: $$g : V \times V \times V \rightarrow \mathbb{R} \\ g(v_0,v,v_1) = (v - v_0) / (v_1 - v_0)$$ For the given arguments, it computes the "relative position" of one element between the others - namely, the value that could be used as the $$\alpha$$ value in the interpolation function, so that $$f(v_0, v_1, g(v_0, v, v_1)) = v$$.
It's not an "inverse", and the term "opposite" in the title was just for lack of a better term. Right now, I'm calling it ~"interpolation parameter function", but I wonder whether there is a commonly used term for that. | {
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• I don't feel confident that this is the best description, and it probably couldn't be used without explanation, but I might describe that as a barycentric coordinate. Edit: wait, I assumed $v$ had to be on the line containing the other two (the affine span of them). – Mark S. Mar 13 at 21:11
• It is in fact an inverse after you "curry" your functions in the right way. Namely $g(v_0,\cdot,v_1)$ is an inverse of $f(v_0,v_1,\cdot)$. – Ian Mar 13 at 21:12
• @MarkS. That's indeed pretty close already (and I wonder why I didn't think of this yet). The fact that barycentric coordinates are mainly interpreted geometrically means that it does not fit perfectly in the most generic sense, but if nobody posts a better answer in the next days, I'd probably ping you to turn the comment into an answer and accept it as being "as close as it gets". – Marco13 Mar 13 at 21:22
• @Ian Yes, it can be boiled down to an inverse, but would require an additional description then - basically, some "verbal un-currying". Just calling it the "inverse of the interpolation" would not be sufficient, I guess... – Marco13 Mar 13 at 21:23
• @MarkS. Regarding the edit: I don't think that being contained in the simplex is a necessary condition for computing the barycentric coordinates, so this should not be a problem. – Marco13 Mar 13 at 21:25 | {
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As in the question, given a real vector space $$V$$, we can define a function for parametrizing lines between vectors by $$f : V \times V \times \mathbb{R} \rightarrow V \\ f(\mathbf v_0,\mathbf v_1,\alpha) = \mathbf v_0 + (\mathbf v_1-\mathbf v_0) \cdot \alpha\text{.}$$
If $$\mathbf v_0\ne\mathbf v_1$$, then $$\alpha\mapsto f(\mathbf v_0,\mathbf v_1,\alpha)$$ certainly defines an injective function $$\mathbb R\to V$$, with image line $$\overleftrightarrow{\mathbf v_0\mathbf v_1}$$. Thus, we can define a bijective function $$F_{\mathbf v_0,\mathbf v_1}:\mathbb R\to \overleftrightarrow{\mathbf v_0\mathbf v_1}$$.
The inverse is given by $$F_{\mathbf v_0,\mathbf v_1}^{-1}:\mathbf v\mapsto\dfrac{\mathbf v-\mathbf v_0}{\mathbf v_1-\mathbf v_0}$$, where the division is to be interpreted as yielding the relevant scalar multiple, as mentioned in this MSE answer. This inverse could be described as providing the barycentric coordinate for the input point on the line $$\overleftrightarrow{\mathbf v_0\mathbf v_1}$$ with respect to the affine basis $$(\mathbf v_0,\mathbf v_1)$$.
As alluded to in the original post, we could combine these inverses together as a single function, though writing the domain properly is a bit cumbersome. We could define $$g:\left\{(\mathbf v_0,\mathbf v,\mathbf v_1)\left|\mathbf v_0\ne\mathbf v_1\text{ and } \mathbf v\in\overleftrightarrow{\mathbf v_0\mathbf v_1}\right.\right\}\to \mathbb R$$ $$g(\mathbf v_0,\mathbf v,\mathbf v_1)= F_{\mathbf v_0,\mathbf v_1}^{-1}(\mathbf v)\text{.}$$ | {
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# Singularity on Interior of Integration Domain
This example shows how to split the integration domain to place a singularity on the boundary.
### Define the Integrand with an Anonymous Function
The integrand of the complex-valued integral
`${\int }_{-1}^{1}{\int }_{-1}^{1}\frac{1}{\sqrt{x+y}}\phantom{\rule{0.2222222222222222em}{0ex}}dx\phantom{\rule{0.2222222222222222em}{0ex}}dy$`
has a singularity when `x = y = 0` and is, in general, singular on the line `y = -x`.
Define this integrand with an anonymous function.
`fun = @(x,y) ((x+y).^(-1/2));`
### Integrate Over a Square
Integrate `fun` over a square domain specified by $-1\le x\le 1$ and $-1\le y\le 1$.
```format long q = integral2(fun,-1,1,-1,1)```
```Warning: Non-finite result. The integration was unsuccessful. Singularity likely. ```
```q = NaN + NaNi ```
If there are singular values in the interior of the integration region, the integration fails to converge and returns a warning.
### Split the Integration Domain into Two Triangles
You can redefine the integral by splitting the integration domain into complementary pieces and adding the smaller integrations together. Avoid integration errors and warnings by placing singularities on the boundary of the domain. In this case, you can split the square integration region into two triangles along the singular line `y = -x` and add the results.
```q1 = integral2(fun,-1,1,-1,@(x)-x); q2 = integral2(fun,-1,1,@(x)-x,1); q = q1 + q2```
```q = 3.771236166328258 - 3.771236166328256i ```
The integration succeeds when the singular values are on the boundary.
The exact value of this integral is
`$\frac{8\sqrt{2}}{3}\left(1-i\right)$`
`8/3*sqrt(2)*(1-i)`
```ans = 3.771236166328253 - 3.771236166328253i ``` | {
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# Symmetry of bicycle-lock numbers
Suppose you have a combination bicycle lock of this sort:
with $n$ dials and $k$ numbers on each dial. Let $m(n,k)$ denote the minimum number of turns that always suffice to open the lock from any starting position, where a turn consists of rotating any number of adjacent rings by one place.
For example $m(2,10)=6$ and $m(3,10)=10$.
I have found an efficient algorithm to compute these numbers, which reveals a symmetry I can’t currently explain:
$m(n, k+1) = m(k, n+1)$
This is such a striking symmetry that I guess it has a simple explanation. Can anyone find one?
Here’s the table of values for small $n$ and $k$, exhibiting the conjectured symmetry: | {
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Here’s the table of values for small $n$ and $k$, exhibiting the conjectured symmetry:
n\k| 2 3 4 5 6 7 8 9 10
---+---------------------------------------------
1 | 1 1 2 2 3 3 4 4 5
2 | 1 2 2 3 4 4 5 6 6
3 | 2 2 4 4 6 6 8 8 10
4 | 2 3 4 6 6 8 9 10 12
5 | 3 4 6 6 9 9 12 12 15
6 | 3 4 6 8 9 12 12 15 16
7 | 4 5 8 9 12 12 16 16 20
8 | 4 6 8 10 12 15 16 20 20
9 | 5 6 10 12 15 16 20 20 25
10 | 5 7 10 12 15 18 20 24 25
-
indeed, not sure what I was thinking about... – user88595 Feb 20 '14 at 18:46
Do you have an example of a starting position with $n=4$ and $k=8$ or $10$ that requires the value listed in the table above? – John Habert Feb 20 '14 at 19:35
@JohnHabert Sure, for n=4 and k=10, and assuming the combination 0000 opens the lock, the combinations 4826 and 6284 require 12 moves. – Robin Houston Feb 20 '14 at 19:49
This is probably obvious, but from your table we also have $m(n, n+1) = m(n, n)$ and $m(2k, 2k) = k^2$ while $m(2k+1, 2k+1) = k(k+1)$ - or, to combine these into a single expression, $m(n,n) = \lfloor\frac{n}{2}\rfloor\cdot\lfloor\frac{(n+1)}{2}\rfloor$. It's very possible there's a similar closed-form expression for $m(n,k)$ that explains the symmetry. – Steven Stadnicki Feb 21 '14 at 16:37
In fact, several of the small-$k$ values have closed forms too: $m(n,2) = \lfloor\frac{(n+1)}{2}\rfloor$, $m(n,4) = 2m(n,2)$, $m(n,3) = \lfloor\frac{(2n+2)}{3}\rfloor$, etc. I think a little more chewing may give a fully closed form... – Steven Stadnicki Feb 21 '14 at 16:46 | {
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It has taken a few days, but I believe I can at last answer my own question. The strategy is to show that, for every combination of the $(n,k)$ lock, there is a combination of the $(k-1,n+1)$ lock that needs the same number of turns to unlock. The argument begins by grouping lock combinations into equivalence classes in such a way that equivalent combinations require the same number of turns to open.
Assume throughout that the destination combination, that opens the lock, is $n$ zeros.
The first trick is one I used before: instead of looking at the positions of the dials, look at the differences between the positions of adjacent dials. On its own this doesn’t discard any information – the process is reversible – but it opens the door to a simplification: the order of the differences doesn’t matter. So we’ll say that two combinations are equivalent if they have the same multiset of differences, and we’ll write the differences as a nondecreasing sequence, to give a canonical form.
To motivate the next identification we’re going to make, let’s consider an example combination of the $(n=4, k=10)$-lock: 2593, which has differences 23447. The differences sum to $2k$, which means – as explained in my original blog post – that we can ignore the two largest differences and add up the others, so this combination takes $2+3+4 = 9$ turns to open. But, since the two largest differences didn’t even enter into the calculation, they could have been any pair of numbers that are both at least $4$ and sum to $11$. In particular they could have been $5$ and $6$ so that the differences were 23456. In this sense the combinations 2593 and 2594 are equivalent. We shall denote this equivalence class by the sequence (2,3,4), which we’ll call a lock sequence for the $(n, k)$-lock. Notice that the number of turns needed to open the lock is the sum of the terms of the lock sequence. | {
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Now we’re going to characterise the lock sequences. Let $d_1, d_2, \dots, d_m$ be a nondecreasing sequence of natural numbers less than $k$ having length $m\leq n$; this is a lock sequence for the $(n,k)$-lock if these two inequalities hold:
$\sum_{i=1}^{m}d_i+(n+1-m)(k-1)\geq (n+1-m)k$
$\sum_{i=1}^{m}d_i + (n+1-m)d_m\leq(n+1-m)k$
They can be simplified to
$n+1-m\leq\sum_{i=1}^{m}d_i\leq(n+1-m)(k-d_m)$
The first inequality is a bit annoying, so let's get rid of it by making one last identification: we’ll identify lock sequences that differ only by leading zeros, and assume a canonical form that has no leading zeros. If the first inequality fails, we can force it to hold by adding leading zeros, thus increasing $m$. So now we’re left with
$\sum_{i=1}^{m}d_i\leq(n+1-m)(k-d_m)$
I like to imagine this condition as meaning, “Is there room in the attic for all the boxes?”. Maybe that will make more sense if I draw a picture:
This picture depicts the lock sequence $(1,1,2,2,2,2,3,3)$ as an arrangement of 16 boxes, and an “attic” of area $(n+1-m)(k-d_m)$, all within an $(n+1)\times k$ rectangle.
Now let’s flip it over, like conjugating a Young diagram, and move the attic back to the top:
We still have the same arrangement of boxes – in particular the value of $\sum_{i=1}^{m}d_i$ remains the same – and the attic is the same size. So the conjugate sequence – $(2,6,8)$ in this example – is a valid sequence for the $(k-1, n+1)$-lock provided only the original was a valid sequence for the $(n, k)$-lock.
So, we’ve shown that every lock sequence for the $(n,k)$-lock can be transformed into a lock sequence for the $(k-1,n+1)$-lock that has the same sum. It follows that it takes at least as many moves, in general, to open a $(k-1,n+1)$-lock as a $(n,k)$-lock. Since this works in both directions, we may conclude that $m(n,k) = m(k-1, n+1)$.
Another way of looking at it is to note that the above implies | {
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Another way of looking at it is to note that the above implies
$m(n,k) = \max\{\min\{ac,bd\}\ |\ a+b=n+1,c+d=k\mbox{ for }a,b,c,d\in\mathbb{N}\}$
which is symmetrical in $n+1$ and $k$. This expression also suggests another way to compute $m(n,k)$.
-
Congratulations! Yes, I think that's compelling. I'd got as far as your "box" picture but not to the necessary & sufficient inequality. Is it worth emphasising that "the same arrangement of boxes" in the conjugate diagram means the same count of boxes, i.e. that the term $\Sigma d_i$ remains the same after conjugation? – HTFB Feb 25 '14 at 7:49
First let's consider a much simpler situation. If, instead of moving any group of adjacent dials, you can only move one dial, the maximum number of turns is trivial -- it's equal to $kn/2$. Note that this expression displays a sort of symmetry very similar to what you're seeing. To be jargony, the state space -- number of combinations -- has volume $kn$, which is a commutative expression, and we can only move through one square of state space at a time.
Now, if we look at the real problem, the state space is still of volume $kn$, but we can move through $k$ squares at a time, with some restrictions. It's easier to analyze if we can only move through one square, so we can look at the space of differences: 3879 becomes 592 and 23856 becomes 1571. In the space of differences we can only move two differences at a time - one up, one down, so 1571 might become 0572 -- but more importantly we have a volume $(k-1) \times n$ and we move a constant amount, so we expect to see a path length of $\Omega((k-1)n)$.
Now, the expression $m(n, k) = \Omega((k-1)n)$ exhibits the observed symmetry! It's just commutation -- if we take $m(k-1, n+1)$ we obviously have the same "volume" of "difference space". There's some intuition here but we still haven't derived the shortest path. | {
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I'll borrow your expression $t = \sup_x(xk - xq - \min(r,x))$ for $xk = q(n+1) + r$ and note that the upper bound and lower bound are achieved precisely where $n+1 | xk$ and $n+1 | x(k-1)$, and that they are both proportional to $x(n+1-x)$, so have a shared maximum $x = (n+1)/2$. This gives us the diophantine equation $q(n+1) = xk$ and if we substitute $x \approx (n+1)/2$ we have $q \approx k/2$.
If these are both integers -- if $n$ is odd and $k$ is even -- we have $t(n, k) = m(n, k) = k(n+1)/4$. So that's a bit simpler already, and it follows the pattern which is nice.
However, the equation does not have a satisfactory solution if $k$ is odd or $n$ is even, so represents a sufficient but not necessary condition for a maximum. We might try to approximately solve the equation by choosing $x = (n+1)/2$, $q = k/2$, and finding a nearby lattice point. For $n = 200$, $k = 9$ we can choose $x = (n+1)/2 - (n+1)/2k = 101.5 - 11.1 = 89.9$ which is close to the true maximum $x = 89$ and suggestive of a more general pattern. I've spent an hour on this already though and should get back to work.
EDIT: I guess the takeaway is that the Diophantine equation $q(n+1) = xk$ is the same if we take $n \rightarrow k-1$, $k \rightarrow n+1$ to get $qk = x(n+1)$. Approximate solutions to this equation correspond to solutions of the general problem.
-
What is the function $\Omega$ here? – HTFB Feb 24 '14 at 12:05
Ω is part of big O notation. See Wikipedia. I actually should have used big-theta (asymptotically related by a constant), as that's what I meant. – mason bogue Feb 25 '14 at 3:01 | {
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# How many diagonals are there in an octagon?
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How many diagonals are there in an octagon? [#permalink] 12 Sep 2018, 22:57
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How many diagonals are there in an octagon?
A)56
B)28
C)20
D)16
E)8
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Re: How many diagonals are there in an octagon? [#permalink] 13 Sep 2018, 06:27
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For each of the 8 vertices you can draw 5 diagonals and hence you have constructed $$5 \times 8 = 40$$ diagonals.
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Re: How many diagonals are there in an octagon? [#permalink] 13 Sep 2018, 07:56
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AchyuthReddy wrote:
How many diagonals are there in an octagon?
A)56
B)28
C)20
D)16
E)8
Let's examine ONE vertex in an octagon (an 8-sided polygon)
We can draw 5 diagonal lines from that one vertex | {
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We can draw 5 diagonal lines from that one vertex
So, each of the 8 vertices can have 5 diagonal lines
So, the total number of diagonal lines = (8)(5) = 40
But wait!!
We have inadvertently counted every diagonal TWICE.
For example, we counted the diagonal from vertex A to vertex B AND we counted the diagonal from vertex B to vertex A
So, to account for this DUPLICATION, we must divide 40 by 2 to get 20
Cheers,
Brent
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Re: How many diagonals are there in an octagon? [#permalink] 13 Sep 2018, 07:56
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Other than being multiplied by scalar constants, matrices can also be multiplied by other matrices. What is matrix multiplication? Matrix multiplication is also distributive. Matrices as transformations. 5 & 2 & 11 Multiplying a Row by a Column We'll start by showing you how to multiply a 1 × n matrix by an n × 1 matrix. \\ Even so, it is very beautiful and interesting. Task. We will usually denote matrices with capital letters, like A, B, etc, although we will sometimes use lower case letters for Let A be an m × p matrix and B be an p × n matrix. \end{bmatrix} = 64. Sort by: Top Voted. $(hint: just multiply every entry by $$2$$), You can multiply two matrices if, and only if, the number of columns in the first matrix equals the number of rows in the second matrix. (Link on columns vs rows ). When the number of columns of the first matrix is the same as the number of rows in the second matrix then matrix multiplication can be performed. Zero matrix & matrix multiplication. 3 Matrices and matrix multiplication A matrix is any rectangular array of numbers. \end{bmatrix} Matrix Multiplication Rules & Formula - In this tutorial, you will learn all about matrix multiplication. To multiply any two matrices, we should make sure that the number of columns in the 1st matrix is equal to the number of rows in the 2nd matrix. = Using properties of matrix operations. Once we know if two matrices can be multiplied, it's time to carry out that multiplication. Real World Math Horror Stories from Real encounters, (See how this problem can be represented as a Scalar Dilation), Scalar: in which a single number is multiplied with every. (This one has 2 Rows and 3 Columns). Matrix multiplication is probably one of the most important matrix operations. We match the 1st members (1 and 7), multiply them, likewise for the 2nd members (2 and 9) and the 3rd members (3 and 11), and finally sum them up. Rules. Here are a couple more examples of matrix multiplication: Find CD and DC, if they | {
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them up. Rules. Here are a couple more examples of matrix multiplication: Find CD and DC, if they exist, given that C and D are the following matrices:; C is a 3×2 matrix and D is a 2×4 matrix, so first I'll look at the dimension product for CD:. Each number in the answer matrix is the result of multiplying one of the rows of matrix 1 by one of the columns of matrix 2. Multiplication of two matrices is also known as a "dot product". Otherwise, the product of two matrices is undefined.The product matrix's dimensions are 1. â ( rows of first matrix ) × ( columns of the second matrix ) {\displaystyle \to ({\text{rows of first matrix}})\times ({\text{columns of the second matrix}})} In above multiplication, the matrices cannot be multiplied since the number of columns in the 1st one, mat⦠In matrix multiplication, the elements of the rows in the first matrix are multiplied with corresponding columns in the second matrix. Example: This matrix is 2×3 (2 rows by 3 columns): In that example we multiplied a 1×3 matrix by a 3×4 matrix (note the 3s are the same), and the result was a 1×4 matrix. Memorizing the entire Multiplication Table can seem quite overwhelming at first. Time complexity: O(n 3).It can be optimized using Strassenâs Matrix Multiplication. An interactive matrix multiplication calculator for educational purposes If at least one input is scalar, then A*B is equivalent to A. Want to see another example? However, matrix multiplication is not as straight forward as regular multiplication, certain rules must be followed and certain conditions must be met. Contents. See your article appearing on the GeeksforGeeks main page and help ⦠= 154. When we multiply a matrix by a scalar (i.e., a single number) we simply multiply all the matrix's terms by that scalar. The usual rules for exponents, namely = P+ and (AP) = still apply. This is the currently selected item. \\ First, however, there are several key concepts that must be understood. Since the number of | {
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item. \\ First, however, there are several key concepts that must be understood. Since the number of columns in Matrix A does not equal the number of rows in Matrix B. (You can put those values into the Matrix Calculator to see if they work.). Another example of 2 matrices you can not multiply. # matrix multiplication in R - example > gt*m [,1] [,2] [,3] [1,] 525 450 555 [2,] 520 500 560 [3,] 450 425 500. Multiply two matrices together. Matrix multiplication is probably the first time that the Commutative Property has ever been an issue. We match the price to how many sold, multiply each, then sum the result. Apple pie value + Cherry pie value + Blueberry pie value, ($3, $4,$2) • (13, 8, 6) = $3×13 +$4×8 + $2×6, And the result will have the same number of, It is "square" (has same number of rows as columns), It can be large or small (2×2, 100×100, ... whatever). \begin{bmatrix} ]â2×3=NOT DEFINED Here is an example of matrix multiplication for two concrete matrices Example: Find the product AB where A and ⦠To multiply a matrix by a single number is easy: We call the number ("2" in this case) a scalar, so this is called "scalar multiplication". So the product CD is defined (that is, I can do the multiplication); also, I can tell that I'm going to get a 3×4 matrix for my answer. It is a type of binary operation. See how changing the order affects this multiplication: It can have the same result (such as when one matrix is the Identity Matrix) but not usually. Example 1 . This article is contributed by Aditya Ranjan.If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. Here it is for the 1st row and 2nd column: (1, 2, 3) • (8, 10, 12) = 1×8 + 2×10 + 3×12 That is, A*B is typically not equal to B*A. In mathematics, particularly in linear algebra, matrix multiplication is a binary operation that produces a matrix from two matrices. The | {
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algebra, matrix multiplication is a binary operation that produces a matrix from two matrices. The multiplication of matrix A by matrix B is a 1 × 1 matrix defined by: Example 1 Matrices A and B are defined by Find the matrix A B. 3 Matrix Powers We can take powers of matrices, but only if theyâre square. But to multiply a matrix by another matrix we need to do the "dot product" of rows and columns ... what does that mean? Matrix Multiplication Calculator Here you can perform matrix multiplication with complex numbers online for free. Scalar: in which a single number is multiplied with every entry of a matrix. We define A° = I, where I is the identity matrix ⦠[?????? Matrix A can be multiplied by a matrix B if the number of columns of matrix A equals the number of rows of the matrix B. Matrix multiplication (product of matrices) A and B with dimensions m×n and n×k is the operation of finding the matrix ⦠Matrix multiplication. Next lesson. This same thing will be repeated for the second matrix. \begin{bmatrix} ; Multiplication of one matrix by second matrix.. For the rest of the page, matrix multiplication will refer to this second category. They can be of any dimensions, so long as the number of columns of the first matrix is equal to the number of rows of the second matrix. a) Multiplying a 2 × 3 matrix by a 3 × 4 matrix is possible and it gives a 2 × 4 matrix as the answer. Matrix multiplication falls into two general categories: For the rest of the page, matrix multiplication will refer to this second category. The product matrix AB will have the same number of columns as B and each column is obtained by taking the Matrix Multiplication Rules. 9 & 4 & 14 4. To multiply two matrices together is a bit more difficult ... read Multiplying Matrices to learn how. ]â2×3â [?????? [?????? Properties of Matrix Multiplication To multiply an m×n matrix by an n×p matrix, the ns must be the same, Matrix multiplication falls into two general categories:. If and are | {
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the ns must be the same, Matrix multiplication falls into two general categories:. If and are matrices and and are matrices, then (17) (18) Since matrices form an Abelian group under addition, matrices form a ring. The first is just a single row, and the second is a single column. Single column to be able to multiply an m×n matrix multiplication rules by an n×p matrix a! To execute a * B is equivalent to a to how many,... Is equivalent to a is done through a series of rules or âtricksâ that can be optimized using Strassenâs multiplication. 'S easier to understand these steps, if you go through interactive demonstrations below rows×columns. The price to each quantity been an issue there are several key concepts that must equal. Matrix through the multiplication of matrices we now apply the idea of Multiplying a 2×3 matrix a... The multiplication of two different matrices often write rows×columns the most important matrix operations Multiplying, but it is to! Matrices to learn how insert the order for a matrix followed by that specific number of columns in the example... Scalar, so officially this is called scalar multiplication '' known as matrix product that. Is multiplied by scalar constants, matrices can also multiply a matrix this... Done through a series of rules or âtricksâ that can be multiplied and think... To match each price to each quantity to another matrix of 3 rows and 5 columns can be using! Memorize ten multiplication facts is done through a series of rules or âtricksâ that can be multiplied down into lessons... An odd and complicated way of Multiplying, but is rarely used known as a product! Entry is multiplied by scalar constants, matrices can also multiply a with! Is rarely used - in this C program, the scalar variety, entry! B, but is rarely used for a matrix from two matrices is also known matrix. ; multiplication of matrices we now apply the idea of Multiplying, but it is necessary Strassenâs matrix multiplication | {
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we now apply the idea of Multiplying, but it is necessary Strassenâs matrix multiplication multiplication... But this process is more complicated but it is necessary is possible and! Of two different matrices B be an m × p matrix and B be p. You know why we use the dot product '' one transformation another! B a matrix are multiplied with every entry is multiplied with every entry a. Mathematical objects ) for which operations such as when one matrix is universally... To B * a easier to understand these steps, if you go through demonstrations... Is not defined of first matrix ) but not usually transformation after another it 's time carry! That is, a * B is typically not equal the number of columns in matrix multiplication, also as! Ap ) = still apply also known as a dot product '' B ) is Identity. A bit more difficult... read Multiplying matrices to learn how seem odd. To understand these steps, if you go through interactive demonstrations below time to carry out multiplication. Different matrices the user will insert the order for a matrix is any rectangular array of numbers being multiplied a. Here you can perform matrix multiplication is not defined than being multiplied by a column to Multiplying general. Known as matrix product, that produces a matrix from two matrices can multiply... The key to learning your multiplication facts is to break the process down into manageable lessons we matrices. We multiply matrices in this case, the ns must be understood 's dimensions are ( rows of matrix... Are defined mastered, you will see that it is necessary × p matrix B. According to the number of rows in matrix B time to carry out that.. B be an p × n matrix such as addition and multiplication multiply an m×n matrix by second.. And complicated way of Multiplying a row by a column to Multiplying general... Multiply an m×n matrix by another matrix of 3 rows and 5.! Matrix, but is rarely used matrix must be the same result ( such as addition and multiplication is to... | {
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but is rarely used matrix must be the same result ( such as addition and multiplication is to... Multiplication falls into two general categories: for the second matrix ) matrix C and D can! The scalar value is$ $single number is multiplied by a 2×3 matrix a... Multiplication '' matrices and matrix multiplication complecated topics this one has 2 rows and m are called the of! Is possible, and the second matrix pre-requisite to be able to multiply m×n! Put those values into the matrix Calculator to see if they work. ) so officially this done! This case, the user will insert the order for a matrix is as. In which a single row, and it gives a 2×2matrix as result. Not equal the number of rows in the scalar variety, every entry a... Important matrix operations matrix through the multiplication of two different matrices matrix C and D below can not multiplied... for the rest of the rows in the scalar value is$ $operations such as when matrix. C = mtimes ( a, B ) is an m×p matrix different! Only necessary to memorize ten multiplication facts is to break the process down into manageable lessons as result! Break the process down into manageable lessons.It can be multiplied but it is important to match each price how! See if they work. ) matrices you can perform matrix multiplication is a operation. To another matrix of 3 rows and 3 columns ) mathematics, particularly in linear,... Of rows in the first time that the Commutative Property has ever been an issue the pre-requisite to able. Through the multiplication of these two matrices can also multiply a matrix is matrix multiplication rules... P matrix and B be an m × p matrix and B be an p × n.. Nonscalar inputs and complicated way of Multiplying a 2×3 matrix is possible, and the second matrix ×! General categories: for the rest of the page, matrix multiplication matrix multiplication will to... for the rest of the matrix and 3 columns ) possible, and the second )! Different matrices the Commutative Property has ever been an | {
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3 columns ) possible, and the second )! Different matrices the Commutative Property has ever been an issue a doesnât work for matrix multiplication matrix is... Produces a single number is multiplied by other matrices n rows and 3 columns ) Multiplying more general matrices columns. Of metric multiplication, also known as matrix product, that produces a single through! Columns in matrix multiplication by another matrix, the number of columns in the following example the... Using any language you may know however, there are several key concepts that must be understood:... The rest of the rows in the second is a binary operation that a... Which operations such as addition and multiplication are defined that produces a single row, and the is! Multiply an m×n matrix by another matrix of 3 rows and 5 columns facts is to break the process into... Matrix a does not equal the number of elements matrix must be the same result ( such as and. First time that the Commutative Property has ever been an issue from two is. Solve this task according to the number of rows in matrix B involed in multiplication. The idea of Multiplying, but it is only necessary to memorize ten multiplication facts is to the... D below can not be multiplied by other matrices for matrix multiplication time! ( such as when one matrix by another matrix of 3 rows and 5.. These two matrices represents applying one transformation after another matrices to learn how )! Illustrate why we multiply matrices in this case, the user will insert the order for matrix. Odd and complicated way of Multiplying, but is rarely used be optimized using Strassenâs matrix multiplication first just... Idea of Multiplying, but this process is more complicated not as straight as... Idea of Multiplying, but is rarely used first matrix ), ( the pre-requisite to be able to two! Has ever been an issue all about matrix multiplication you are encouraged to solve this according. ] â2×2 Multiplying a 2×3 matrix by second matrix many | {
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you are encouraged to solve this according. ] â2×2 Multiplying a 2×3 matrix by second matrix many rows and columns matrix. An issue 3×2 matrix is a bit more difficult... read Multiplying matrices to learn how array... Binary operation that produces a single matrix through the multiplication of one matrix not! Representation is the best things to define any little complecated topics number of columns in the time. A 2×3 matrix by an n×p matrix, then a a doesnât work for matrix multiplication probably! Why we use the dot product '' understand these steps, if you go through interactive demonstrations below is! Multiplied, it is important to match each price to each quantity interactive demonstrations below scalar constants, can. N 3 ).It can be multiplied by other matrices certain conditions be. If a is well-defined representation is the best things to define any little complecated topics matrices represents one! Called a scalar, so officially this is called matrix multiplication rules scalar multiplication.... To solve this task according to the task description, using any language you may know each, then is! Number, called a scalar, then it is necessary exponents, namely = P+ and ( )! [???? ] â3×2= [?????! Been mastered, you will see that it is only necessary to memorize ten facts... Through the multiplication of these two matrices is also known as matrix product, that produces matrix... [???????? ] â3×2= [?? ] â3×2= [??... Go through interactive demonstrations below ) × ( columns of the matrix to. Equal the number of columns in the following example, the scalar variety, every entry is multiplied by constants... An n×m matrix the applications, of metric multiplication, the scalar value is$ \$ of 2 matrices can. | {
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# If two integers are chosen at random out of the set {2, 5, 7, 8}, what
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If two integers are chosen at random out of the set {2, 5, 7, 8}, what is the probability that their product will be of the form a^2 – b^2, where a and b are both positive integers?
A. 2/3
B. 1/2
C. 1/3
D. 1/4
E. 1/6
[Reveal] Spoiler:
Shalom! I am currently studying the probability chapter of the Manhattan GMAT Word Translations book. I am looking forward to the different outcomes and answers.
[Reveal] Spoiler: OA
Last edited by Bunuel on 11 Nov 2017, 22:32, edited 2 times in total.
Renamed the topic, edited the question and added the OA.
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MitDavidDv wrote:
If two integers are chosen at random out of the set {2, 5, 7, 8}, what is the probability that their product will be of the form a^2 – b^2, where a and b are both positive integers?
A: 2/3
B: 1/2
C: 1/3
D: 1/4
E: 1/6
Shalom! I am currently studying the probability chapter of the Manhattan GMAT Word Translations book. I am looking forward to the different outcomes and answers.
I just used the exhaustive method. Count everything that fits.
2,5=10
2,7=14
2,8=16
5,7=35
5,8=40
7,8=56
Write down all perfect squares until 100
1, 4, 9, 16, 25, 36, 49, 64, 81, 100
Pick one number at a time.
10 -- Keep adding with every perfect square and see whether the result is also there in the set.
10+1=11(Not there)
10+4=14(Not there)
10+9=19(Not there)
10+16=26(Not there)
10+25=35(Not there)
we can stop here as the difference between all consecutive perfect squares after 35 will be more than 10.
10- Not possible to represented as a^2-b^2
Repeat the same with all the products;
14-Not Possible
16: Let's check this;
16+1=17(Not there)
16+4=20(Not there)
16+9=25(There in the set)
16- can be represented as a^2-b^2 i.e. 5^2-3^2
**********************************
Likewise:
35- 6^2-1^2
40- 7^2-3^2
56- 9^2-5^2
*******************
In the sample set:
{10,14,16,35,40,56}
{10,14}- Not Possible: Count=2
{16,35,40,56}- Possible: Count=4
Total Count=6
P=Favorable/Total=4/6=2/3
Ans: "A"
**************************
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Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink]
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You can avoid an exhaustive test here. Suppose I ask whether (97)(103) can be written in the form a^2 - b^2, where a and b are integers. Notice that this is a difference of squares: a^2 - b^2 = (a+b)(a-b). We can now just use the median of 97 and 103, which is 100:
(97)(103) = (100-3)(100+3) = 100^2 - 3^2
So whenever we can write our product in such a way that the median of our two numbers is an integer, we can write our product as a difference of squares just as above. For example, if we take 5*7, that's equal to (6-1)(6+1), and if we take 2*8, that's equal to (5-3)(5+3). Now if we look at 8*5, we can't immediately use the same trick, but we can 'move' one of the 2s from the 8 into the 5, as follows: 8*5 = 4*10 = (7-3)(7+3). Similarly, 8*7 = 4*14 = (9-5)(9+5). So of our six possible products, four can be written as a difference of squares.
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IanStewart
you gave a very easy explanation!
Thank you!
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Some Theory here:
Consider two numbers a, b
Now a* b always = [(a+b)/2]^ 2 – [(a-b)/2]^2---------------------- eqn 1
The number of ways a number can be expressed as a difference of two integers depends on number of ways it can be written as a two factor product stated below.
a.odd*odd
b. even *even
The reason being in eqn (1) above [(a-b)/2]^2 should result in an integer . Hence we consider only the above set of two factor products. (0dd minus odd = even ,even minus even =even , hence both will be divisible by 2
E.g the number 36 can be written as
6*6 , 18*2 hence 36 can be expressed as difference of squares in two ways.
36 = (6+6)/2 ^ 2 - (6-6)/2 ^ 2 = 6^2 - 0
36= (18+2)/2^2 – (18-2)/^2 = 10^2 – 4^2
Now back to the problem :
We have the number set (2, 5, 7, 8) out of which, the satisfying possibilities as per the above theory would be
2*8 (valid)
5*7 (valid)
2*5 (not valid)
2*7(not valid)
5*8== 4*10 hence valid
7*8 == 4*14 hence valid)
Hence there are 4 favourble cases out of 6. Therefore probability Is 4/6 = 2/3
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Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink]
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Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink]
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monirjewel wrote:
Need Bunuel's explanation for this problem......
Best solution is here: if-two-integers-are-chosen-at-random-out-of-the-set-114579.html#p929326
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Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink]
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17 Nov 2013, 09:35
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Bunuel wrote:
monirjewel wrote:
Need Bunuel's explanation for this problem......
Best solution is here: if-two-integers-are-chosen-at-random-out-of-the-set-114579.html#p929326
a^2 - b^2 will be of the form (a-b)(a+b). We can infer that the difference between the two chosen numbers(of which one is (a-b) and the other (a+b)) will be (a+b) - (a-b) = 2b i.e. even difference(negative or positive).
Thus we will have to choose either two even numbers or two odd numbers?
this way we have two options -> 1. choosing 5 and 7 of which probability is = 1/6
2. choosing 2 and 8 of which probability is = 1/6.
thus the total probability = 2/6 which is 1/3.
please explain why this is wrong.
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Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink]
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27 Feb 2014, 22:13
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If two integers are chosen at random out of the set {2, 5, 7, 8}, what is the probability that their product will be of the form a^2 – b^2, where a and b are both positive integers? | {
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A. 2/3
B. 1/2
C. 1/3
D. 1/4
E. 1/6
Shalom! I am currently studying the probability chapter of the Manhattan GMAT Word Translations book. I am looking forward to the different outcomes and answers.
I have an explanation too, maybe it'd be of some help:
a^2 - b^2 =(a-b)(a+b)
(a+b) and (a-b) can only be integers from the selected set i.e {2,5,7,8}
Now a and b are both positive integers as stated in the question
So the sum of (a-b) and (a+b) also has to be an integer
i.e 2a = sum of any two numbers chosen from the set {2,5,7,8}
For a to be a positive integer the sum has to be an even number so either it could be a pair of (2,8) or (7,5)
Using PnC P(getting one such pair when chosing two random numbers from a set) = P(Both numbers chosen to be even) + P(Both numbers chosen to be odd)
= 2/4C2 + 2/4C2
= 4/4C2 = 4/6 = 2/3
Kindly let me know in case the solution has some mistakes.
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Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink]
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28 Feb 2014, 21:04
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a^2-b^2 = (a+b)(a-b), so if the difference of any the factor pairs of the product is a positive even integer, then they can be in the a^2-b^2 form. Example: 7*5=35 --> 7*5 = (6+1)(6-1) --> 35*1 = (18-17)(18+17), etc etc
Only products that don't work are 2*5=10 (as 5-3 = odd and 10-1 = odd) and 2*7=14 (as 7-5 = odd and 14-1=13 odd).
The other 4 products work and so the probability = 4/6 = 2/3
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Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink]
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08 Apr 2014, 03:43
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ShashankDave wrote:
Bunuel wrote:
monirjewel wrote:
Need Bunuel's explanation for this problem...... | {
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Best solution is here: if-two-integers-are-chosen-at-random-out-of-the-set-114579.html#p929326
a^2 - b^2 will be of the form (a-b)(a+b). We can infer that the difference between the two chosen numbers(of which one is (a-b) and the other (a+b)) will be (a+b) - (a-b) = 2b i.e. even difference(negative or positive).
Thus we will have to choose either two even numbers or two odd numbers?
this way we have two options -> 1. choosing 5 and 7 of which probability is = 1/6
2. choosing 2 and 8 of which probability is = 1/6.
thus the total probability = 2/6 which is 1/3.
please explain why this is wrong.
I had the same concern but now it's clear for me :
the 6 possible pairs are (2 5) (2 7) (2 8) (5 7) (5 8) (7 8)
our method allows to find 2 pairs (2 8) (5 7) but does not allow to eliminate the others
Especially (5 8 ) and (7 8) that also meet the condition : 5*8 = 10*4 = (7+3)*(7-3) and 7*8=14*4=(9+5)*(9-5)
so there are 4 possible pairs to be picked from 6 hence the probability is 4/6 = 2/3
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Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink]
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13 Apr 2014, 19:09
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ShashankDave wrote:
Bunuel wrote:
monirjewel wrote:
Need Bunuel's explanation for this problem......
Best solution is here: if-two-integers-are-chosen-at-random-out-of-the-set-114579.html#p929326
a^2 - b^2 will be of the form (a-b)(a+b). We can infer that the difference between the two chosen numbers(of which one is (a-b) and the other (a+b)) will be (a+b) - (a-b) = 2b i.e. even difference(negative or positive).
Thus we will have to choose either two even numbers or two odd numbers?
this way we have two options -> 1. choosing 5 and 7 of which probability is = 1/6
2. choosing 2 and 8 of which probability is = 1/6.
thus the total probability = 2/6 which is 1/3.
please explain why this is wrong. | {
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The question is not so much as whether both the numbers are even or both are odd as whether the product of the numbers can be written as product of two even numbers or two odd numbers.
Two numbers are chosen and multiplied. Now they have lost their individual identity. Now you focus on the product and find whether it can be written as product of two numbers which are both odd or both even. Say you took two number 7 and 8 and multiplied them. You get 56. Can you write 56 as product of two numbers such that both are even? Yes, 4 and 14 or 2 and 28. So 56 can be written as a^2 - b^2 in two ways: (9^2 - 5^2) and (15^2 - 13^3). So if you choose 7, 8 from the set, their product can be written in the form a^2 - b^2.
Similarly, 5, 8 will give you the same result.
Hence you get 2 more cases and total probability becomes 4/6 = 2/3.
Whenever you have at least 4 in the product, you can write it as product of two even numbers: give one 2 to one number and the other 2 to the other number to make both even.
If the product is even but not a multiple of 4, it cannot be written as product of two even numbers or product of two odd numbers. It can only be written as product of one even and one odd number.
If the product is odd, it can always be written as product of two odd numbers.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 05 Aug 2014 Posts: 1 Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink] ### Show Tags 05 Aug 2014, 12:31 3 This post received KUDOS 2 This post was BOOKMARKED Here is the solution with easy steps: a^2 - b^2 = (a-b)(a+b) Now we need to find all possible combinations of two numbers from the set {2, 5, 7, 8 } which can be expressed as (a-b)(a+b) Let say x =a-b and y = a+b, therefore x+y = 2a and y-x = 2b, so you need to have two numbers x and y whose sum and difference should be even number. How many are there from the set {2, 5, 7, 8} ? 8 + 2 = 10 , 8 - 2 = 6, 7 - 5 = 2 , 7 + 5 = 12 . So there are two pairs (8,2) and (7,5) which can be expressed as a^2 - b^2. Therefore, the probability that their product will be of the form a^2 – b^2 = 2/total two numbers combination = 2/4 Chose 2 = 2/3 Ans is option (A) Senior Manager Joined: 07 Apr 2012 Posts: 441 Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink] ### Show Tags 13 Sep 2014, 10:12 1 This post was BOOKMARKED VeritasPrepKarishma wrote: The question is not so much as whether both the numbers are even or both are odd as whether the product of the numbers can be written as product of two even numbers or two odd numbers. Two numbers are chosen and multiplied. Now they have lost their individual identity. Now you focus on the product and find whether it can be written as product of two numbers which are both odd or both even. Say you took two number 7 and 8 and multiplied them. You get 56. Can you write 56 as product of two numbers such that both are even? Yes, 4 and 14 or 2 and 28. So 56 can be written as a^2 - b^2 in two ways: (9^2 - 5^2) and (15^2 - 13^3). So if you choose 7, 8 from the set, their product can be written in the form a^2 - b^2. Similarly, 5, 8 will give you the same result. Hence you get 2 more cases and total probability becomes 4/6 = 2/3. | {
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8 will give you the same result. Hence you get 2 more cases and total probability becomes 4/6 = 2/3. Whenever you have at least 4 in the product, you can write it as product of two even numbers: give one 2 to one number and the other 2 to the other number to make both even. If the product is even but not a multiple of 4, it cannot be written as product of two even numbers or product of two odd numbers. It can only be written as product of one even and one odd number. If the product is odd, it can always be written as product of two odd numbers. Hi Karishma, Can you elaborate a little more? Why are we looking for 2 numbers that are either both even or both odd? Also, how did you know to stop at (9^2 - 5^2) and (15^2 - 13^3) and not look for more? Thanks, Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7951 Location: Pune, India Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink] ### Show Tags 15 Sep 2014, 01:21 6 This post received KUDOS Expert's post 2 This post was BOOKMARKED ronr34 wrote: Hi Karishma, Can you elaborate a little more? Why are we looking for 2 numbers that are either both even or both odd? Also, how did you know to stop at (9^2 - 5^2) and (15^2 - 13^3) and not look for more? Thanks, That's a good question. You should understand this concept well. That is why I have written a detailed post on it on my blog: http://www.veritasprep.com/blog/2014/04 ... at-part-i/ Check it out and get back to me (on the blog or here) if any doubts remain. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199 | {
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Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink]
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not sure if its been discussed, but a valuable property to know is that ANY non-prime odd number, or multiple of 4, can be written as a difference of squares using integers. 21 = (5+2)(5-2) 15 = (4+1)(4-1) etc. try it out.
therefore, we can see that out of our 6 possible outcomes, only 4 will be either odd (5 x 7) or multiples of 4 (8 x each other #). so answer = 4/6=2/3
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Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink]
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Very simple and straightforward method but never heard about the number property....
bsmith37 wrote:
not sure if its been discussed, but a valuable property to know is that ANY non-prime odd number, or multiple of 4, can be written as a difference of squares using integers. 21 = (5+2)(5-2) 15 = (4+1)(4-1) etc. try it out.
therefore, we can see that out of our 6 possible outcomes, only 4 will be either odd (5 x 7) or multiples of 4 (8 x each other #). so answer = 4/6=2/3
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Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink]
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13 Mar 2015, 20:17
MitDavidDv wrote:
If two integers are chosen at random out of the set {2, 5, 7, 8}, what is the probability that their product will be of the form a^2 – b^2, where a and b are both positive integers?
A. 2/3
B. 1/2
C. 1/3
D. 1/4
E. 1/6 | {
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A. 2/3
B. 1/2
C. 1/3
D. 1/4
E. 1/6
Shalom! I am currently studying the probability chapter of the Manhattan GMAT Word Translations book. I am looking forward to the different outcomes and answers.
So far if the average of the two numbers is an INTEGER they can be written in (a+b)(a-b) form . so that narrows us down to Odd + Odd and Even+Even cases .
Special consideration need to taken for those cases in which one number is ODD and other is multiple of 4 , i.e. in this case
if the set is $${ 2,5,7,8 }$$, then possible pairs are :
7*8 = 56 = 14*4 = 28*2 none of these pairs (7,8) , (14,4), and (28*2) can be expressed in (a+b) (a-b) form .
5*8= 40 = 10*4 = (7+3) (7-3), so yes we can write $$5*8$$ as $$(7+3) * (7-3)$$
2,5,7,8
total number of cases = 4C2 = 6
favorable cases : (odd,odd) (5,7) , (Even,Even) (2,8) , and one special case as shown above (5,8) so $$3/6=1/2$$
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Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink]
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11 Jun 2015, 21:19
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MitDavidDv wrote:
If two integers are chosen at random out of the set {2, 5, 7, 8}, what is the probability that their product will be of the form a^2 – b^2, where a and b are both positive integers?
A. 2/3
B. 1/2
C. 1/3
D. 1/4
E. 1/6
Shalom! I am currently studying the probability chapter of the Manhattan GMAT Word Translations book. I am looking forward to the different outcomes and answers.
the product set=(10, 14, 16, 35, 40, and 56)
16=8x2=(5+3)(5-3); (8-2)/2 =3
35=7x5=(6+1)(6-1); (7-5)/2 = 1
40=10x4=(7+3)(7-3)
56=14x4=(9+5)(9-5)
you got the pattern.
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Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink]
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05 Sep 2015, 19:45
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MitDavidDv wrote:
If two integers are chosen at random out of the set {2, 5, 7, 8}, what is the probability that their product will be of the form a^2 – b^2, where a and b are both positive integers?
A. 2/3
B. 1/2
C. 1/3
D. 1/4
E. 1/6
Shalom! I am currently studying the probability chapter of the Manhattan GMAT Word Translations book. I am looking forward to the different outcomes and answers.
I consider
AxB = a^2 – b^2 = (a+b)(a-b)
where A and B are the posible chosen
first: the total posible chosen is 12, because A take 4 values and B takes 3, 4x3 = 12
Second
This is the scenary:
A x B = (a+b)(a-b)
2 5 = 10 = 2x5 or 10x1 (wrong) note that the sum of the factors should be even number, (conditions from a and b are integers), for the reason these opstion is eliminated
2 7 = 14 = 7x2 (wrong, the sum is not even) or 14x1 (wrong)
2 8 = 16 = 8x2 (correct)
5 2 = 10 = 8x2 (wrong, and the same that the first)
5 7 = 35 = 5x7 (Correct)
5 8 = 40 = 20x2 (correct)
7 2 = 14 = (wrong and is the same )
7 5 = 35 = 7x5 (correct)
7 8 = 56 = 14x4 (correct)
8 2 = 16 = 8x2 (correct)
8 5 = 40 = 20x2 (correct)
8 7 = 56 = 14x4 (correct)
Finally, the number of correct posible answer is 8, and the total possible answer is 12
indeed, 8/12 = 2/3
Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink] 05 Sep 2015, 19:45
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# What is the probability of randomly picking 4 consecutive ladies out of a team of 11 people?
I was running the morning standup this morning and in contrary to the usual alphabetical order I announced I was going to go in "random" order today, however I then ended up giving all the ladies of the team the word first before 'randomly' choosing the gents.
If my selection was really random what is the probability of having all the ladies come first?
My back of the envelope calculation was:
4/11 * 3/10 * 2/9 * 1/8 = 0.003030303
• Is this anywhere correct?
• How to write this problem in a formal way?
P.S. This is not a HW question but can be considered as a kind of self study.
• if this is HW please add the "self-study" tag – Antoine Feb 8 '16 at 11:56
First, note that every ordering of the ladies in the first four positions will give the same event: namely ladies first! Then, note that any ordering of the gentlemen in the last seven positions will also give the event ladies first! How do we account for all these outcomes?
Here are two formal ways to accomplish this. We begin by considering the sample space in the ordered case. By ordered here we mean that we can tell the ladies apart and the same for the gentlemen. You can name them if it makes things easier.
Assuming randomness and equilikely events, we have to count the ways this can occur and then divide by the total number of orderings to get our probability. Let's start from the total number of orderings of these eleven people. This is given by the number of permutations of course, $11!$. Then, how many ways are there to order the four ladies in the four positions? Well, the same reasoning will give $4!$ ways. But now for each of these orderings, we can order the gentlemen in $7!$ ways and still have an ordering in which the ladies come first. Thus we may write
$$\Pr \left( \text{Ladies first} \right) = \frac{4!7!}{11!} = 0.0030$$
and it turns out that your calculation was correct. | {
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and it turns out that your calculation was correct.
Now, an alternative way to view this. Assume now that we are in the unordered case and we cannot tell the ladies apart and the gentelemen either. All we know is that a person is a lady or a gentleman. Is there a way to compute the probability in this case?
Well yes, if we consider the proper sample space. First let's count the total number of unordered arrangements of the ladies and the gentlemen. This is given by the binomial coefficient $\binom{11}{4, 7}$ and the desired outcome now is just one of these arrangements. Hence, we write
$$\Pr \left( \text{Ladies first} \right) = \frac{1}{\binom{11}{4, 7}} = \frac{4! 7!}{11!}$$
and we get of course the same result but from different principles!
Hope this helps. | {
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# Area of a triangle (Heron's formula) Calculator
## Calculates the area of a triangle given three sides.
side a side b side c 6digit10digit14digit18digit22digit26digit30digit34digit38digit42digit46digit50digit area S
$\normal Triangle\ by\ Heron's\ formula\\\vspace{5}(1)\ area:\ S=\sqrt{s(s-a)(s-b)(s-c)}\\\vspace{1}\hspace{100} s={\large\frac{(a+b+c)}{2}}\\$
Area of a triangle (Heron's formula)
[1-10] /69 Disp-Num5103050100200
[1] 2018/11/02 00:01 Male / 40 years old level / High-school/ University/ Grad student / Very /
Purpose of use
Area calculation for construction site.
Comment/Request
Hello, what unit of measurement is used in this calculator? Ft or mtr?
from Keisan
The calculator uses unitless numbers. | {
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e.g.)
Input meter values -> The result is m^2.
Input ft values -> The result is ft^2.
[2] 2018/10/27 08:54 Female / Under 20 years old / Elementary school/ Junior high-school student / Not at All /
Purpose of use
Homework
Comment/Request
Does not work with radicals sadly :(
[3] 2018/10/14 17:19 Male / 20 years old level / High-school/ University/ Grad student / Very /
Purpose of use
Writing algorithm
[4] 2018/10/11 12:53 Male / Under 20 years old / High-school/ University/ Grad student / Very /
Purpose of use
Helping someone on a test
Comment/Request
very useful
[5] 2018/10/07 05:37 Male / Under 20 years old / High-school/ University/ Grad student / Very /
Purpose of use
homework
[6] 2018/09/30 08:18 Male / 20 years old level / Others / Very /
Purpose of use
Checking calculations to make sure I got the correct area for a weird looking triangle
[7] 2018/09/17 07:18 Male / Under 20 years old / High-school/ University/ Grad student / Very /
Purpose of use
To check my Python Program that does the same thing
[8] 2018/09/12 13:29 Male / 60 years old level or over / A retired person / Very /
Purpose of use
find out the area of a land of which three sides are unequal length of a triangle
[9] 2018/08/16 17:38 Female / Under 20 years old / Elementary school/ Junior high-school student / Useful /
Purpose of use
homework
[10] 2018/07/17 07:57 Male / Under 20 years old / High-school/ University/ Grad student / Very /
Purpose of use
project euler
Thank you for your questionnaire.
Sending completion | {
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# Can the probability of a trump poverty be calculated without making case distinctions?
The card game Doppelkopf is played with four players. Every player receives 12 of the 48 cards. The 48 cards consist of 26 trump cards and 22 other. A trump poverty is what we call the scenario that a player has less than 4 trump cards. If the cards are dealt randomly, what is the probability that at least one player has a trump poverty?
This looks like it can be solved very easily via the inverse, like so:
Deal every player 4 trump cards, then deal every player 8 more cards.
However we failed to do this without double counting. Can this be done without case distinctions?
Outline: Call the players A, B, C, D. Note that we cannot have $3$ trump poor players, since then they would have at most $9$ trumps between them, and $9+12\lt 26$.
(i) Find the probability that A has trump poverty. This should not be difficult.
Now multiply by $4$. This overestimates the required probability, for we have double-counted the situations in which, for example, A and B both have trump poverty.
(ii) Find the probability that A and B have trump poverty, multiply by $\binom{4}{2}$, and subtract from the estimate in (i).
Finding the probability that A and B have trump poverty is a little unpleasant. One way is to divide into cases: A has $0$ trumps, $1$ trump, $2$, $3$. It will take a while but the calculations are routine.
Remark: If one has some programming experience, one can set up a simulation and get reliable estimates.
• Yes, we came up with this answer, too, but we were trying to find a solution without making case distinctions. Are you suggesting there is no such solution? – Max Nov 29 '14 at 10:15 | {
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# When does the radius of convergence of the product of two $p$-adic power series increase?
Let $p$ be a prime number and denote by $R(f)$ the radius of convergence of a power series $f(x) \in \mathbb{C}_p[[x]]$, where $\mathbb{C}_p$ is the completion of the algebraic closure of $\mathbb{Q}_p$, the field of $p$-adic numbers. Given two power series $f(x), g(x) \in \mathbb{C}_p[[x]]$, it is known that the radius of convergence of the product $h(x) = f(x)g(x)$ is at least the minimum of the radius of convergence of the two series $f(x)$ and $g(x)$. In other words, we have $$R(h) \geq \min\{R(f), R(g)\}.$$ Keep in mind $(1-x)(1+x+x^2+\dots) = 1$ as an example for the strict inequality. Is there a way to easily predict when $R(h) > \min\{R(f), R(g)\}$ and find $R(h)$ explicitly? More specifically, I'm interested in computing the radius of convergence of power series of the form $\exp(f(x))$ for $f(x) \in x\mathbb{C}_p[[x]]$.
For example, let $f(x) = \exp(x)$ and $g(x) = \exp(x^p/p)$. Then $R(f) = R(g) = (1/p)^{1/(p-1)}$ and using the fact that the Artin-Hasse exponential series $$\text{AH}(x) = \exp(x + x^p/p + x^{p^2}/p^2 + \cdots)$$ lies in $\mathbb{Z}_p[[x]]$ (which implies $R(\text{AH}) \geq 1$), $h(x) = \exp(x + x^p/p)$ has radius of convergence $$R(h) = R\left(\exp\left(\frac{x^{p^2}}{p^2}\right)\right) = \left(\frac{1}{p}\right)^{\frac{(2p-1)}{p^2(p-1)}}> \left(\frac{1}{p}\right)^{\frac{1}{p-1}}.$$ The importance of this example comes from the fact that if we set $\pi$ to be a root of $x+x^p/p = 0$, then $h(\pi)$ is a non-trivial $p$-th root of unity in $\mathbb{C}_p$. This provides an analytic representation of $p$-th roots of unity, exploited in particular in Dwork's proof of the rationality of zeta functions over finite fields. | {
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More generally, using this method one can show that for any $n \geq 1$, we have $$R(\exp(x+x^p/p+\cdots + x^{p^n}/p^n)) = R(\exp(x^{p^{n+1}}/p^{n+1})).$$ Even though I understand the details involved in this calculation, I don't know if there is some more general theory underlying these examples. It may be helpful to share similar examples that you know. For instance, is it always the case that $$R(\exp(f(x))) > R(\exp(x)),$$ given $f(x) \in x\mathbb{C}_p[[x]]$ has a nonzero root $\alpha \in \mathbb{C}_p$ of absolute value $R(\exp(x)) = (1/p)^{1/(p-1)}$ and no non-zero roots of smaller absolute value?
• I may be able to offer a partial answer here, but it’ll have to wait till morning at best. Feb 7 '16 at 5:58
• Have you looked at "Rank one solvable p-adic differential equations and finite Abelian characters via Lubin–Tate groups" by Andrea Pulita? The abstract starts with "We introduce a new class of exponentials of Artin–Hasse type, called π-exponentials". Feb 7 '16 at 12:35
• @LaurentBerger: That looks interesting. I haven't looked at it, but I will surely do. Thank you! Feb 7 '16 at 22:29
Here is a counterexample to your question at the end, for each $p$. Let $f_u(x) = x + ux^p/p$ for $u \in \mathbf C_p$ with $|u|_p = 1$ and $|u-1|_p = 1$. (Such $u$ can be taken in $\mathbf Z_p^\times$ if $p > 2$, but you need to go outside $\mathbf Q_p$ if $p = 2$ to an extension with residue field of size greater than $2$.) Since $|u|_p = 1$, all the nonzero roots of $f_u(x)$ in $\mathbf C_p$ have absolute value $(1/p)^{1/(p-1)} = R(\exp)$.
To find the $p$-adic radius of convergence of $\exp(f_u(x))$, write $$\exp(f_u(x)) = \exp\left(x+\frac{x^p}{p}\right)\exp\left((u-1)\frac{x^p}{p}\right)$$ as formal power series. On the right side, the first factor has radius of convergence greater than $(1/p)^{1/(p-1)}$, as you noted. | {
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Since $|u-1|_p = 1$, the second factor has radius of convergence equal to that of $\exp(x^p/p)$, which is $(1/p)^{1/(p-1)}$. The reciprocal $(\exp(x + x^p/p))^{-1}$ has the same radius of convergence as $\exp(x+x^p/p)$, even for $p=2$, so $\exp(f_u(x))$ has radius of convergence equal to $(1/p)^{1/(p-1)}$. This is the counterexample to your question.
• This a great specific example. More generally, for $u \in \mathbb{C}_p$, it shows that $R(\exp(x+ux^p/p))$ depends explicitly on $|u-1|_p$ whenever $$R(\exp(x+x^p/p)) \neq R((u-1)x^p/p).$$ I wonder what happens if equality holds. Is there a chance for the radius to further increase? Feb 7 '16 at 23:41
• In general if $F(x)$ and $F(x)^{-1}$ have a common radius of convergence $R$, and $G(x)$ and $G(x)^{-1}$ have a common radius of convergence $S$ then the product $FG$ has radius of convergence $\min(R,S)$ if $R \not= S$. If $R=S$ then there is no simple rule for the radius of $FG$. Typically the radius for $FG$ will remain $R$, but of course there are examples where it grows, like $\exp(x)$ and $\exp(x^p/p)$, or more simply $\exp(x)$ and $\exp(-x)$. Feb 8 '16 at 15:16
I can only give a very partial answer, and that only from my very parochial point of view.
I will use the additive valuation $v$ rather than absolute value, normalized so that $v(p)=1$, and in terms of which $R(\sum a_nx^n)=-\liminf\bigl(v(a_n)/n\bigr)$, so that when $v(z)>R(f)$, $f(z)$ is a convergent series. Examples are: if $\exp(x)=\sum_{n\ge1}x^n/n!$ and $\log(x)=\sum_{n\ge1}(-1)^{n-1}x^n/n$, then $R(\exp)=1/(p-1)$ and $R(\log)=0$. It’s for this reason that I prefer the logarithm to its inverse.
The log that I’ve named above is the logarithm of the multiplicative formal group $\hat{\mathbf G}_{\mathrm m}(x,y)=x+y+xy$, that is a formal-group homomorphism from $\hat{\mathbf G}_{\mathrm m}$ to the additive formal group $\hat{\mathbf G}_{\mathrm a}(x,y)=x+y$ with $\log'(0)=1$. | {
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The $p$-typical logarithm $\log_{\mathrm{AH}}=x+x^p/p+x^{p^2}/p^2+\cdots$ is the logarithm of another formal group $\mathscr M$. which we might call the $p$-typical recoordinatization of $\hat{\mathbf G}_{\mathrm m}$, and the $\Bbb Z_p$-formal-group isomorphism $u:\mathscr M\to\hat{\mathbf G}_{\mathrm m}$ is exactly what’s called the Artin-Hasse Exponential. It satisfies $\log\circ u=\log_{\mathrm{AH}}$.
Now here’s the moral of my story. These two logarithms, $\log$ and $\log_{\mathrm{AH}}$, being convergent throughout the open unit disc of $\Bbb C_p$, have all sorts of interesting behavior that is not seen at all by the exponential series $\exp(x)$, except at a very far remove. In particular, they have zeros. The closest such to the origin has $v(\zeta)=1/(p-1)$, which explains immediately the radius of convergence of the exponential function.
And so I am pretty sure that your last conjecture has no chance of being correct. I tried $p=3$, $\zeta=\omega-1$, where $\omega^2+\omega+1=0$, and found that $\exp(x-x^2/\zeta)$ seemed to have as bad convergence properties as the exponential itself.
EDIT:
You’ve asked me to explain further why the existence of zeros of the logarithm prevents wider convergence of the exponential, and indeed, it is not quite so obvious as I was pretending.
Let $\lambda$ be a root of the above-described $\log$ with $v(\lambda)=1/(p-1)$, in fact $\lambda+1$ will be a primitive $p$-th root of unity. If $R(\exp)\ge1/(p-1)$, then the Newton polygon of $f(x)=\exp(x)-\lambda$ will have a vertex $(n,v(b_n))$ in addition to the vertex $(1,0)$. (You don’t need to know that $b_n=1/n!$.) To see this, you may look at $f(\lambda x)$ and note that its coefficients must go to zero for convergence. In particular, There will be a segment of the polygon whose negative slope is $\ge1/(p-1)$, and thus a root $\mu$ of $f$ in an algebraic extension such that $\exp(\mu)=\lambda$, impossible if $\log(\exp(x))=x$. | {
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• This is great. Thank you! I was wondering if you have some references on this formal group construction of this Artin-Hasse Exponential and if this approach implies the fact that $\text{AH}(x)$ has $\mathbb{Z}_p$ coefficients and/or radius of convergence 1? What did you want to say at: "interesting behavior that is not seen at all by the exponential series $\exp(x)$, except at a very far \emph{remove}." Feb 7 '16 at 23:27
• Also, would you mind elaborating some more on how the root of valuation $1/(p-1)$ of the logarithm series $\log(1+x) = \sum_{n \geq 1} (-1)^{n-1}x^n/n$ explains immediately the radius of convergence of the exponential series? Is there some basic compositional inverse property I'm missing here? Feb 7 '16 at 23:28
• “At a very far remove” means “from very far away”. The basic reference is Hazewinkel’s big book from the 70’s. For your second question, I’ll make an addendum to my answer. Feb 8 '16 at 3:02
• Thank you for the EDIT. That's a nice way to argue and it seems to generalize to any two compositional inverse series where one has a bigger radius of convergence and at least one non-zero root. I looked up Hazewinkel but couldn't find if this formal group approach to the Artin-Hasse Exponential tells us $\text{AH}(x)$ has $\mathbb{Z}_p$ coefficients or if it has radius of convergence 1. Feb 8 '16 at 21:02
• Maybe it would be best to discuss via e-mail: I have the time, and there’s not enough space here. Feb 9 '16 at 1:19 | {
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# 1) Derivation of the quadratic formula
This is note $$1$$ in a set of notes showing how to obtain formulas. There will be no words beyond these short paragraphs as the rest will either consist of images or algebra showing the steps needed to derive the formula mentioned in the title.
Suggestions for other formulas to derive are welcome, however whether they are completed or not depends on my ability to derive them. The suggestions given aren't guaranteed to be the next one in the set but they will be done eventually.
1 $\large ax^2 + bx + c = 0$
2 $\large x^2 + \frac{b}{a}x + \frac{c}{a} = 0$
3.1 $\large x^2 + \frac{b}{a}x = \left(x + \frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2$
3.2 $\large \left(x + \frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2 + \frac{c}{a} = 0$
4 $\large \left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2} + \frac{4ac}{4a^2} = 0$
5 $\large \left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$
6 $\large x + \frac{b}{2a} = \pm\sqrt{\frac{b^2 - 4ac}{4a^2}}$
7 $\large x + \frac{b}{2a} = \frac{\pm\sqrt{b^2 - 4ac}}{2a}$
8 $\large x = \frac{- b \pm\sqrt{b^2 - 4ac}}{2a}$
Note by Jack Rawlin
2 years, 10 months ago
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print "hello world"
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print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$
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Yup! This is the derivation of the quadratic formula. Great!
- 2 years, 10 months ago
I made a YouTube video showing how this would work in arbitrary fields not of characteristic $$2$$. The field not having characteristic $$2$$ is so important because this avoids any of your calculations having zero determinants; in other words, the quantities $$2$$ and $$4$$ will be $$0$$ modulo $$2$$ and our prescribed condition avoids this occurring. In such a setting, steps 6, 7 and 8 will not be valid; solutions can only exist when $$b^2 - 4ac$$ is a square number in the field. This indicates the severe limitations of the Fundamental Theorem of Algebra in its scope only being applied in the framework of the "real number field" and the "complex number field".
- 3 months ago | {
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# Why is $\exists y \in B, \ \forall x \in A, \ P(x,y)$ stronger than $\forall x \in A,\ \exists y \in B, \ P(x,y)$
In the book that I am reading, the author notes that " $$\exists y \in B, \ \forall x \in A, \ P(x,y)$$" is a "stronger" statement than "$$\forall x \in A,\ \exists y \in B, \ P(x,y)$$" because the first one implies the truth of the second one...but the second one does not imply the truth of the first.
I think I understand the if then comment quite well and have included pictures to express my understanding:
For the "stronger case", we have:
if $$\ \ \exists y \in B, \ \forall x \in A, \ P(x,y)$$, then $$\ \forall x \in A,\ \exists y \in B, \ P(x,y)$$
which can be illustrated as:
(Hopefully this is comprehensible...I'm basically showing two equivalent pictures and that both versions of this picture satisfy the 'meaning' of the antecedent and consequent. And by "equivalent" I mean that both pictures describe the same element pairs that result in truth)
For the "weaker case", we have:
if $$\ \forall x \in A,\ \exists y \in B, \ P(x,y)$$, then $$\ \exists y \in B, \ \forall x \in A, \ P(x,y)$$
Which can be illustrated as follows:
For this weaker case, which could be drawn multiple ways (I just happened to pick one way that would show why this if then statement is false), we see that while the antecedent can be satisfied by the picture, the consequent clearly cannot.
So, okay...cool. I see that the one if then statement is true and the one if then statement is false. But why exactly does this feature make the first case "Strong". In what sense is it strong? Does it allow you to construct proofs more rigorously? Does it allow you to utilize a trick that greatly simplifies proof construction?
In what way does knowing $$\ \ \exists y \in B, \ \forall x \in A, \ P(x,y)$$ prove more beneficial than knowing $$\ \forall x \in A,\ \exists y \in B, \ P(x,y)$$?
Any insight (or examples) is greatly appreciated! | {
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Any insight (or examples) is greatly appreciated!
If general, a statement $$F$$ is told stronger than another statement $$G$$ if the implication $$F \Rightarrow G$$ holds (i.e. if $$G$$ holds whenever $$F$$ holds) but the converse implication $$G \Rightarrow F$$ does not hold (i.e. it is possible that $$G$$ holds but $$F$$ does not hold).
As you said, this is the case for $$F = \exists y \in B \, \forall x \in A \, P(x,y)$$ and $$G = \forall x \in A \, \exists y \in B \, P(x,y)$$, because if you assume $$F$$ then you can always prove $$G$$ (independently from the meaning of $$A$$, $$B$$ or $$P$$) but clearly the converse is not true: indeed, in the situation where $$A = B = \mathbb{N}$$ and $$P = \, <$$, we have that $$F = \exists y \in \mathbb{N} \, \forall x \in \mathbb{N} : x < y$$ is false ($$\mathbb{N}$$ has no maximum) but $$G = \forall x \in \mathbb{N}\, \exists y \in \mathbb{N} : x < y$$ (for every natural number $$x$$, its successor $$x+1$$ is greater than $$x$$).
It is stronger in the sense that anything you can prove with $$\forall x \ \epsilon \ A,\ \exists y \ \epsilon \ B, \ P(x,y)$$, can also be proven with $$\exists y \ \epsilon \ B, \ \forall x \ \epsilon \ A, \ P(x,y)$$ because the latter implies the former. As an analogy: Every nut you can crack with a rubber hammer can also be cracked with a sledge hammer. But there might be nuts that can only be cracked with a sledge hammer. Therefore the sledge hammer is stronger than the rubber hammer. | {
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• Hmmm, I see. So, it seems like there may be times where I really need to prove "$\forall x \ \epsilon \ A,\ \exists y \ \epsilon \ B, \ P(x,y)$"...but to do so directly is very hard. However, it just so happens in this hypothetical, that " $\exists y \ \epsilon \ B, \ \forall x \ \epsilon \ A, \ P(x,y)$" is much easier to prove. Therefore, I can use this to my advantage and prove this "easier" one...which then gives me access to the statement that I actually wanted. Is that sort of right? – S.Cramer Aug 7 '19 at 17:21
• Not quite. Although this could happen, too, it is rather that if you are trying to prove something with $\exists y \ \epsilon \ B, \ \forall x \ \epsilon \ A, \ P(x,y)$, it might be easier or solve the problem quicker than $\forall x \ \epsilon \ A,\ \exists y \ \epsilon \ B, \ P(x,y)$. So if you are unlucky and only have $\forall x \ \epsilon \ A,\ \exists y \ \epsilon \ B, \ P(x,y)$ (the small hammer) your proof might be harder or take longer than if you had $\exists y \ \epsilon \ B, \ \forall x \ \epsilon \ A, \ P(x,y)$ (the sledge hammer). – Klaus Aug 7 '19 at 18:27
Suppose $$\exists y\in B,\forall x\in A,P(x,y).$$
Let $$x_0\in A$$. According to the previous assertion, there exists $$y\in B$$ such that for all $$x\in A$$, $$P(x,y)$$. In particular for $$x=x_0$$, we have $$P(x_0,y)$$. We just proved that for all $$x_0\in A$$, there exists $$y\in B$$ such that $$P(x_0,y)$$, which can be rewritten $$\forall x_0\in a,\exists y\in B,P(x_0,y).$$
Of course you can replace $$x_0$$ with $$x$$, which gives $$\forall x\in A,\exists y\in B,P(x,y).$$
Therefore, $$\left( \exists y\in B,\exists x\in A,P(x,y)\right)\implies\left(\forall x\in A,\exists y\in B,P(x,y)\right).$$ | {
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The left-hand side is stronger in the sense that it implies the right-hand side. If you want to prove the right-hand side, it is succifient to prove the left-hand side, but it is not necessary. If you prove the right-hand side by proving the left-hand side, we consider that you prove a stronger result. | {
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## Combinatorial coincidence
I joined RSA a few weeks ago. Here, I met Anoop who loves playing with combinatorics problems. We challenge each other with combinatorics problems.
I asked him the following question at lunch.
For integers n ≥ 1 and k ≥ 1,
f0(n) = 1.
fk(n) = Σni=1 fk−1(i).
Find a closed formula for fk(n).
He asked me one on programming. I have used subscripted variables to represent the counters for the nested loops in the pseudocode to show the question.
Consider the following pseudocode with k nested loops.
count = 0
for c1 in 1 to n:
for c2 in 1 to c1:
for c3 in 1 to c2:
…
for ck in 1 to ck−1:
count = count + 1
What is the final value in count after the outermost loop terminates?
With one question each, we went back to our desks. As I started solving his question on nested loops, I realized that his problem led me to the recurrence relation in the question I asked him. If k = 1, count = n. This is also the value of f1(n). If k = 2, the inner loop with the counter as c2 will run once when c1 = 1, twice when c1 = 2, and so on. So, the final value of count will be f1(1) + f1(2) + … + f1(n) = n(n + 1)/2. This is the value of f2(n) as well.
Extending this argument, one may realize that for any k, count = fk−1(1) + fk−1(2) + … + fk−1(n) = fk(n). In other words, the answer to his question is the answer to my question. I already knew the answer to this. So, I went back to his desk with the answer: C(n +k −1, k).
I had arrived at this closed formula earlier by solving fk(n) for the first few k's using Faulhaber's formula. I noticed that they were all equal to C(n + k − 1, k). So, I proved that this is always true by the principle of strong mathematical induction. It involved proving that: fk+1(n) = C(k, k) + C(k + 1, k) + … + C(n + k − 1, k) = C(n + k, k + 1) | {
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This was a remarkable coincidence that we asked each other questions which had the same answer. When I explained the coincidence to him, he explained how he arrived at the same result for the question on nested loops which can be used to determine the closed formula for the recurrence relation too.
In the question on nested loops, the following condition is always met: n ≥ c1 ≥ c2 ≥ … ≥ ck ≥ 1. So, the number of times count variable would be incremented is equal to the number of possible ways we can arrange k numbers from the first n natural numbers in descending order. The answer to this is the number of possible ways we can arrange n − 1 similar balls and k similar sticks. If we consider the number of balls to the right of the ith stick and add one to it, we get a valid value for ci as the number of balls to the left of a stick can not increase as we move right and consider (i + 1)th, (i + 2)th, etc. sticks. Similarly, the number of balls to the left of a stick can not decrease as we move left and consider (i − 1)th, (i − 2)th, etc. sticks. Also, any set of valid values for c1, c2, …, ck can be represented as an arrangement of these sticks and balls.
We can arrange n − 1 similar balls and k similar sticks in (n + k − 1)! (n − 1)! · k! = C(n + k − 1, k) number of ways. This is the answer to the question on nested loops as well as that on the recurrence relation.
### 1 comment
#### JB said:
Interesting discussion and I gained something today remembering you. | {
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# Secant Line: finding the slope for different values of x
The point $P(4, 24)$ lies on the curve $y = x^2 + x + 4$. If $Q$ is the point $(x, x^2 + x + 4 )$, find the slope of the secant line $PQ$ for the following values of $x$.
If $x= 4.1$, the slope of $PQ$ is:
and if $x= 4.01$, the slope of $PQ$ is:
and if $x= 3.9$, the slope of $PQ$ is:
and if $x= 3.99$, the slope of $PQ$ is:
Based on the above results, guess the slope of the tangent line to the curve at $P(4, 24)$.
For this problem, should I just plug in the x values given into the y equation? Then the slope would be...??
-
Would you happen to remember how to get the slope of a segment joining two points? – Ｊ. Ｍ. Oct 8 '11 at 5:09
You have the point $P(4,24)$ on the curve. If $x=4.1$, what is the corresponding $y$ on the curve? Yes, plug in, you get $Q(4.1,?)$. The problem asks you then to find the slope of the line $PQ$, so you are not quite through. – André Nicolas Oct 8 '11 at 5:22
What happened to your other, quite related, question? You were asked two questions in the comments there, which were meant to guide you towards a solution, but you answered none. You were also given two answers there, but you reacted to none, commented none and accepted none. Well, well, well... – Did Oct 8 '11 at 18:38
Since you will use several values of $x$, it is probably easiest to find a function for the slope, in terms of $x$, and then plug in the various values for $x$. That is, the slope of the secant line $PQ$ is the rise over run (change in $y$ over change in $x$):
$$m(x) = \frac{x^2 + x + 4 - 24}{x - 4}$$ | {
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$$m(x) = \frac{x^2 + x + 4 - 24}{x - 4}$$
So, $m(x)$ gives the slope for any particular value of $x$. A practical reason to do this is, for example on a TI-83 or TI-84 or something like it, you can now type in that function to $Y_1$ and then go to the Table and plug in the various $x$ values and you get the slopes immediately. And, since it is so fast, you can check this for more values than the question even asks for to get an even better intuition. Or, you could use Wolfram Alpha to accomplish the same thing. For example, type in:
Evaluate (x^2 + x + 4 - 24)/(x - 4) at x = 4.1, 4.01, 4.001, 3.9, 3.99, 3.999
Back to the problem,
\begin{align*} m(4.1) =& \frac{4.1^2 + 4.1 + 4 - 24}{4.1 - 4} = \frac{0.91}{0.1} = 9.1 \\ m(4.01) =& \frac{4.01^2 + 4.01 + 4 - 24}{4.01 - 4} = \frac{0.0901}{0.01} = 9.01 \\ m(4.001) =& \frac{4.001^2 + 4.001 + 4 - 24}{4.001 - 4} = \frac{0.009001}{0.001} = 9.001 \\ m(3.9) =& = \frac{3.9^2 + 3.9 + 4 - 24}{3.9 - 4} = \frac{-0.89}{-0.1} = 8.9 \\ m(3.99) =& = \frac{3.99^2 + 3.99 + 4 - 24}{3.99 - 4} = \frac{-0.0899}{-0.01} = 8.99 \\ m(3.999) =& = \frac{3.999^2 + 3.999 + 4 - 24}{3.999 - 4} = \frac{-0.008999}{-0.001} = 8.999 \end{align*}
From this, we would probably guess that the slope of the tangent line when $x = 4$ is 9. This does not guarantee that we are right, but assuming the function is reasonably well behaved, we could be pretty confident in this guess.
And, once you learn how to calculate derivatives, you will find this is correct as the slope of the tangent line at any $x$ value is the derivative. Since
$$y'(x) = 2x + 1$$
we see that
$$y'(4) = 9$$
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# Find all functions $f:\mathbb{R} \mapsto \mathbb{R}$ which satisfy $f(x^2+y f(z)) =x f(x) + z f(y)$
QUESTION : Find all functions $f:\mathbb{R} \mapsto \mathbb{R}$ which satisfy $f(x^2+y f(z)) =x f(x) + z f(y)$
My doubt lies in the part where I've shown injectivity. Kindly check if my proof is correct.
The solution is $f(x)=x$ for every individual $x$. Let $P(x,y,z)$ be an assertion of the FE.
• $P(0,0,0)\equiv f(0)=0$
• $P(x,0,0)\equiv f(x^2)=xf(x)$
• $P(0,y,z)\equiv f(yf(z))=zf(y)$
Does the third condition imply injection in the following manner?
If $f(a)=f(b)$, then $f(yf(a))=f(yf(b))\implies af(y)=bf(y)\implies a=b$ assuming $f(y)\neq 0 \forall y$.
If till this it's correct, I am pretty much done with the whole problem.
Here's how I can proceed henceforth— $f(x^2)=xf(x)=f(xf(x))$, which, upon using the injectivity criteria leads to $x^2=xf(x)$. Now, $f(x)$ can be found after verification of the positive/negative value.
your solution is almost correct - you arrived at $f(a)=f(b)\Rightarrow{}af(y)=bf(y)$ for all $y\in{}\mathbb{R}$. Now you assumed that $f(y)\ne{}0$ for all $y$ (which solves the question but isn't justified). Instead it is enough to assume that $f(y)\ne{}0$ for some $y$. Note that if this does not hold, then $f(x)=0$ for all $x$, which is another solution.
Also, $xf(x)=x^2$ implies immidiatly that $f(x)=x$ for all $x\ne{}0$ without further checking. Did you prove that $f(0)=0$ though?
You only need to assume that there is some $y$ with $f(y)\neq 0$ in order for your argument on injectivity to hold. In this case your work shows that $f(x)=x$ for all $x\neq 0$, and since $f$ must be injective, the only possibility for $f(0)$ is $f(0)=0$. This concludes the proof that $f(x)=x$ for this case.
Now, the other case is simple, and ties in with Cameron's comment that $f\equiv 0$ is another -- and hence the only other -- solution. | {
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# waiting for patterns?
A fair coin is tossed until one of the patterns show up: TTH or THT. Let A be the event that TTH shows up before THT.
What is P(A)?
Here is my solution but I am not sure if it is correct or there is a better solution.
Let $$p=P(A)$$. Define
$$A_1=$$ the event that the first toss is H
$$A_2=$$ the event that the first two tosses are TT
$$A_3=$$ the event that the first three tosses are THT
$$A_4=$$ the event that the first three tosses are THH
Then this is a partition for the sample space.
$$p=P(A|A_1)P(A_1)+P(A|A_2)P(A_2)+P(A|A_3)P(A_3)+P(A|A_4)P(A_4)$$.
Then
$$p=p\frac{1}{2}+1\frac{1}{4}+0\frac{1}{8}+p\frac{1}{8}$$ which implies that $$p=\frac{2}{3}.$$
• Can anyone give a reference to a math paper which deals with more probability calculations related to patterns such as this example? Thanks. – Probability student Feb 6 at 19:12
We can start at the point when the first $$T$$ appears
The probability that $$T$$ occurs again is $$\frac{1}{2}$$ , and then $$TTH$$ comes first surely.
The probability that $$HT$$ comes next is $$\frac{1}{4}$$ , then the game is finished as well.
If $$HH$$ appears, we have to wait for the next $$T$$ and are again at the starting position.
The occurence of $$T$$ has a probability twice the probability of the occurence of $$HT$$ , hence the chance that $$TTH$$ wins must be twice the chance that $$THT$$ wins, giving the result $$p=\frac{2}{3}$$.
I think you've got the right idea, but I don't think all your numerical assignments are correct. You appear to have taken $$\ P\left(\,A\,\vert\, A_2\,\right)\$$ and $$\ P\left(\,A\,\vert\, A_3\,\right)\$$ to be $$\ \frac{1}{2}\$$ and $$\ 1\$$, respectively. I believe they should be $$\ 1\$$ and $$\ 0\$$, respectively. | {
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Take $$\ P\left(\,A\,\vert\, A_2\,\right)\$$, which is the probability that the tosses terminate with TTH, given that the first two tosses are TT. If the first two tosses are TT, then it's impossible for the tosses to terminate in the sequence THT, because the very first subsequent head to appear after the initial TT must be preceded by two successive tails.
Likewise, if the first three tosses are THT (i.e. the event $$\ A_3\$$ occurs), then the tosses cease with $$\ A\$$ not having occured. Thus, I believe your equation for $$\ p\$$ should be $$p=p\frac{1}{2}+1\cdot\frac{1}{4}+0\cdot\frac{1}{8}+p\frac{1}{8}\ ,$$ which nevertheless still gives you $$\ p = \frac{2}{3}\$$.
• Thank you, you are right! I made a mistake, I will correct it. – Probability student Feb 5 at 10:45 | {
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# Even and odd functions and whether even/odd characteristics change with powers
I watched a video where a problem involved recognizing that $$\sin x$$ is an odd function and $$\sin^3 x$$ is also odd. But the presenter didn't explain why $$\sin^3 x$$ is also odd. Why does the fact that the function is odd not change when it is cubed? Is there a rule where for every even power the odd function is even and for every odd power the odd function remains odd? What about for even functions?
• If $f$ is odd and $g = f^3$, then $g(-x) = f(-x)^3 = (-f(x))^3 = (-1)^3f(x)^3 = -f(x)^3 = -g(x)$. You can play that game now with all variants. – amsmath Oct 6 '19 at 2:14
• @amsmath Why are you allowed to put the the negative sign from inside the parentheses to outside? – user532874 Oct 6 '19 at 2:16
• Because $(ab)^n = a^n\cdot b^n$. – amsmath Oct 6 '19 at 2:16
• 532874, that's the definition of odd function, no? – Gerry Myerson Oct 6 '19 at 2:17
• @user532874 A function $f$ is odd if $f(-x) = -f(x)$ for all $x$. – amsmath Oct 6 '19 at 2:18
Note that $$(-1)^{2k}=(+1)$$ and $$(-1)^{2k+1}=(-1)$$ $$(+1)^k = (+1)$$
Thus if a function is odd we have $$f^{2k}(-x) = (-1)^{2k}f^{2k}(x)=f^{2k}(x)$$ and $$f^{2k+1}(-x) = (-1)^{2k+1}f ^{2k+1}(x)=-f^{2k+1}(x)$$
Thus odd functions to the odd powers are odd and to the even powers are even.
Even functions to any power stay even.
• I think you are missing a $2k$ exponent after "Thus if a function is odd we have" – user532874 Oct 6 '19 at 2:35
• @user532874 is it fixed now? – Mohammad Riazi-Kermani Oct 6 '19 at 2:43
• @MohammadRiazi-Kermani Check your calculations. You are missing minuses. – amsmath Oct 6 '19 at 2:59
• @amsmath Thanks for informative comment. I should have been more careful. – Mohammad Riazi-Kermani Oct 6 '19 at 3:12
It is not only the cubing, it works for anyother odd function, it is always true that the composite of odd functions is an odd function. | {
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In your case, $$f(x)=\sin x$$ is an odd function and $$g(x)=x^3$$ is also an odd function so $$g\circ f(x)= g(f(x))=g(\sin x)=\sin^3x$$
Another example is $$\sin x$$ with any odd power for example $$x^{12345}$$, since both are odd functions then $$\sin^{12345}x$$ is also an odd function.
Or $$\frac{1}{\sin x}$$ is also an odd function since it is the composite of $$\sin x$$ and $$\frac{1}{x}$$,and they are both odd functions.
Anyway you can just check if $$f(-x)=-f(x)$$ without worrying about compositions, $$f(-x)=\sin^3(-x)=(\sin (-x))^3=(-\sin x)^3 =-(\sin^3x)=-f(x)$$ | {
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# How to prove that limit doesn't exist using epsilon-delta definition?
It is easy to prove the limit exists, all we have to show is there exists a relationship between $\delta$ and $\epsilon$. But how are we supposed to prove limit doesn't exists? The problem is when we are proving for a limit we already know what the limit is and with that, algebra is all that's needed.
Please show through an example (you may show that $\lim_{x\rightarrow0} \frac{1}{x}$ doesn't exist)
If possible please use the explanation scheme that is used by this answer https://math.stackexchange.com/a/66552/335742
• Hint: negate the definition. For some $\epsilon > 0 \ \ldots$ – Sean Roberson Jul 5 '17 at 20:40
• @SeanRoberson Are you trying to imply that I should use a proof by contradiction? If yes then maybe can you provide some example? – mathnoob123 Jul 5 '17 at 20:47
• \begin{align} & \lim_{x\to a} f(x) = L: \\ & \text{For every } \varepsilon>0 \text{ there exists } \delta>0 \\ & \text{ such that for every } x, \text{ if } 0<|x-a|<\delta \text{ then } |f(x) - L| < \varepsilon. \\ \\ & \lim_{x\to a} f(x) \ne L: \\ & \text{There exists } \varepsilon>0 \text{ such that for every } \delta>0, \\ & \text{ there exists } x \text{ such that } 0<|x-a|<\delta \text{ and } |f(x) - L| \ge\varepsilon. \\ {} \end{align} Put "For every $L$" in front of that last thing and then it says no limit exists. – Michael Hardy Jul 5 '17 at 20:49
• Can you please provide an example by solving the limit I provided (or if you fear it's a homework then one of your own examples)? – mathnoob123 Jul 5 '17 at 20:51
Suppose $L=\lim_{x\to0}\frac{1}{x}$ is finite.
If $L>0$, then there exists $\delta>0$ such that, for $0<|x|<\delta$, $|\frac{1}{x}-L|<L$, that is, $$0<\frac{1}{x}<2L$$ This is a contradiction, just take $-\delta<x<0$.
Similarly if $L<0$.
Thus we can only have $L=0$. Then there should exist $\delta>0$ such that, for $0<|x|<\delta$, $|\frac{1}{x}|<1$, an obvious contradiction. | {
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It can be neither $\lim_{x\to0}\frac{1}{x}=\infty$ nor $\lim_{x\to0}\frac{1}{x}=-\infty$, because $\frac{1}{x}$ assumes positive and negative values in every punctured neighborhood of $0$.
Can this be generalized? Not really. For instance, in order to show the non existence of $\lim_{x\to0}\sin\frac{1}{x}$ the easiest way is to show that the limit should be in the interval $[-1,1]$, but that $\sin\frac{1}{x}$ assumes every value in $[-1,1]$ in each punctured neighborhood of $0$, so it is far from every possible limit. Alternatively, there exist sequences $(a_n)$ and $(b_n)$ convergin to $0$ such that $\lim_{n\to\infty}\sin\frac{1}{a_n}=0$ and $\lim_{n\to\infty}\sin\frac{1}{b_n}=1$.
As another example, $\lim_{x\to0}e^{1/x}$ doesn't exist because the one sided limits are different: from the left it is $0$, from the right it is $\infty$.
If $\displaystyle\lim_{x\to0} \frac 1 x = L$ then what happens if $\varepsilon=1$? There would exist $\delta>0$ such that if $0<|x|<\delta$ then $\displaystyle \left| \frac 1 x - L \right| <\varepsilon.$ But if $x$ is positive, you can make $1/x$ bigger than $L+\varepsilon$ by making $x$ less than $1/(L+\varepsilon).$ And a similar thing works if $x$ is negative (where you have $1/x\to-\infty$).
• Oh so the basic thing is an infinitesimal value namely $m$ will always be smaller than $\delta$ and $1/m>L+\epsilon$. Two questions. 1. Is the procedure always the same for any proof of the nonexistence of limit? 2. How to write this proof formally? 3. For limit definition (for proving a limit) is it okay to show $a\delta \leq \epsilon$ or is it always $a\delta <\epsilon$ where $a$ are the other terms that might be present? (I am aware there can be other terms, just concerned about the inequality signs?) – mathnoob123 Jul 5 '17 at 21:00
• You need to know something about the nature of the function in order to write the proof, but what I wrote in a comment under the question is true generally. – Michael Hardy Jul 5 '17 at 21:44 | {
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# Complex numbers
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##### Full Member
The form:
$$\displaystyle a+bi$$
is referred to as the rectangular form for a complex number, which is just one of many forms in which complex numbers may be written. It is a particularly useful form though. Also, observe:
$$\displaystyle \frac{1}{2i}=\frac{1}{2i}\cdot\frac{i}{i}=\frac{i}{2i^2}=-\frac{i}{2}=0+\left(-\frac{1}{2}\right)i$$
I'm totally with you .. but once you've multiplied by i you've changed the cosmetics of the number .. isn't it matter? I mean isn't it matter for changing the definition of the number itself? who said that multiplying isn't changing the typecast of the number itself? really weird ..
#### MarkFL
##### Super Moderator
Staff member
I've multiplied by 1 in the form of i/i, and thus have not changed the value of the number. The form is different, but it still represents the same complex value.
#### HallsofIvy
##### Elite Member
The "cosmetics" of the number (I like that phrase!) is completely irrelevant to the problem. 3, III, 2+ 1, 6/2, and $$\displaystyle \sqrt{9}$$ are all different ways of saying exactly the same thing. It does not matter to the mathematics of the problem which you use.
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##### Full Member
I would guess [you] have been asked to write this number in the rectangular form:
$$\displaystyle a+bi$$
no didn't been asked, while solving a question I was encountered to convert that horrible number of complex to a+bi
#### pka
##### Elite Member
lets assume I have complex number like (3-4i / 5j +7 ) then it's right to write like this :
(3-4i / 5j +7 ) = a + b*i
\displaystyle \begin{align*}\dfrac{3-4\mathcal{i}}{7+5\mathcal{i}}&=\dfrac{(3-4\mathcal{i})(7-5\mathcal{i})}{49+25} \\&=\dfrac{(21-20)+\mathcal{i}(-15-28)}{74}\\&=\dfrac{1}{74}+\bigg( \dfrac{-43}{74} \bigg) \mathcal{i}\end{align*}
Thus $$\displaystyle a=\dfrac{1}{74}~\&~b=\dfrac{-43}{74}$$ SEE HERE
#### lookagain | {
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Thus $$\displaystyle a=\dfrac{1}{74}~\&~b=\dfrac{-43}{74}$$ SEE HERE
#### lookagain
##### Senior Member
$$\displaystyle 0+\left(-\frac{1}{2}\right)i$$
It is common to write this final result in "a + bi" form as $$\displaystyle \ 0 - \tfrac{1}{2}i$$.
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Regarding post #10:
Likewise, it is common to write this final result in "a + bi" form as $$\displaystyle \ \tfrac{1}{74} - \tfrac{43}{74}i$$.
#### topsquark
##### Full Member
(3-4i / 5j +7 ) = a + b*i
I would have called this a typo except that it has been carried through on a couple occasions.
If you are a Mathematician (or Physicist) $$\displaystyle i^2 = -1$$ is common. If you are an Engineer then $$\displaystyle j^2 = -1$$. You shouldn't mix the two. So your expression should either be
$$\displaystyle \dfrac{3 - 4i}{7 + 5i}$$
or
$$\displaystyle \dfrac{3 - 4j}{7 + 5j}$$
-Dan
#### mmm4444bot
##### Super Moderator
Staff member
It is common to write … $$\displaystyle 0 - \tfrac{1}{2}i$$
… it is common to write … $$\displaystyle \tfrac{1}{74} - \tfrac{43}{74}i$$
Yes, in general, it's common to express the addition of a negative number as subtraction of its opposite. In particular, it's also common for authors who want to explicitly match the given form a+bi to stick with addition when b is negative by putting grouping symbols around it (like in posts #2 and #10).
Likewise, if emphasizing the given form when b is zero and a is not, authors will write a+(0)i instead of a.
#### MarkFL
##### Super Moderator
Staff member
It is common to write this final result in "a + bi" form as $$\displaystyle \ 0 - \tfrac{1}{2}i$$.
For the sake of the OP, I wanted to give the number strictly in the form:
$$\displaystyle a+bi$$
#### Jomo | {
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$$\displaystyle a+bi$$
#### Jomo
##### Elite Member
Because I am bothered by it I always say a +/- bi.
Now here is my question. If you are asked to write something in a+bi form is an answer of 4 - 5i in the wrong form? I think it is!
#### pka
##### Elite Member
Regarding post #10:
Likewise, it is common to write this final result in "a + bi" form as $$\displaystyle \ \tfrac{1}{74} - \tfrac{43}{74}i$$.
For the sake of the OP, I wanted to give the number strictly in the form: $$\displaystyle a+bi$$
Correct & thank you.
I have never figured why some want-to-bes must correct every little thing.
But in this case the "correction" is a mistake. It does say a real number $$\displaystyle a$$ plus a real number $$\displaystyle b$$ times $$\displaystyle \mathcal{i}$$.
If the one doing the correcting has a background in teaching complex variables, it is standard to define the complex number field as a subset of $$\displaystyle \mathcal{R}\times\mathcal{R}$$. As a field there are two operations $$\displaystyle {\bf\large+~\&~\cdot}$$: $$\displaystyle (a,b)+(c,d)=(a+c,b+d)~\&~(a,b)\cdot(c,d)=(ac-bd,ad+bc)$$.
Here is a basic idea: define a conjugate of a complex number, $$\displaystyle \overline{(a,b)}=(a,-b)$$.
The absolute value is $$\displaystyle |(a,b)|=\sqrt{a^2+b^2}$$ So the multiplicative inverse $$\displaystyle (a,b)^{-1}=\dfrac{\overline{(a,b)}}{|(a,b|^2}$$
Here is short-hand notation: $$\displaystyle \dfrac{1}{z}=\dfrac{\overline{z}}{|z|^2}$$
Apply this to the problem at the hand:
$$\displaystyle \dfrac{3-4\mathcal{i}}{7+5\mathcal{i}}=\dfrac{(3-4\mathcal{i})(7-5\mathcal{i})}{49+25}$$
Last edited:
#### lookagain
##### Senior Member
That top right denominator should be "49 + 25."
Likewise, it is common to write this final result in "a + bi" form as $$\displaystyle \ \tfrac{1}{74} - \tfrac{43}{74}i$$.
For the sake of the OP, I wanted to give the number strictly in the form: | {
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$$\displaystyle a+bi$$
Correct & thank you.
I have never figured why some want-to-bes must correct every little thing.
But in this case the "correction" is a mistake. It does say a real number $$\displaystyle a$$ plus a real number $$\displaystyle b$$ times $$\displaystyle \mathcal{i}$$.
Let's see. You appear to accept an alleged "correction," but in the same motion
you dismiss it as petty. You are self-contradictory.
I'm not a wan't-to-be, but you sure are desperate and looking to remain relevant. When I stated
"it is common," that was not a correction. But the b can be negative at times,
so those forms of mine follow. Your math you posted is irrelevant/unneeded.
It is readily seen that a = 4 and b = -5.
pka, this is the second post of yours that has been reported in recent weeks for you attacking me with your insults. You need to stop posting until you can be civil.
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#### pka
##### Elite Member
pka, this is the second post of yours that has been reported in recent weeks for you attacking me with your insults. You need to stop posting until you can be civil.
To Lookagain, where did I say anything about you? I did not. Did I? Can you point to anything that I posted which namers you?
But if the shoe fits then wear it.
#### MarkFL
##### Super Moderator
Staff member
The OP's question has been answered, and so I'm going to close this thread
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# Doubt about substitution in$\int_{-1}^{1}\sqrt{1+x^2}dx$
I've tried this substitution in this integral
$$\int_{-1}^{1}\sqrt{1+x^2}dx$$
Let $x^2=t$, so $x=\sqrt{t}$ and $dx=\frac{1}{2\sqrt{t}}dt$. So we have
$$\frac{1}{2}\int_{1}^{1}\sqrt{\frac{1+t}{t}}dt=0$$
Which is obviously wrong. I know that this integral can be done with integration by parts or hyperbolic substitution, I want to know why this happens. My idea is that $x^2$ isn't always invertible, it is only on $[0, +\infty)$, and that causes this problem with the interval of integration.
Am I right? Thanks for your time.
• $x= \pm \sqrt t$ – John Lou Oct 20 '17 at 18:51
• Trig sub. Enjoy. – Randall Oct 20 '17 at 18:51
• i would use $$x=\tan(t)$$ – Dr. Sonnhard Graubner Oct 20 '17 at 18:52
• First see that you function is even on a symmetric domain so break your integral in two – Guy Fsone Oct 20 '17 at 18:53
• I think the OP is not asking how to do it, but why his method fails. – velut luna Oct 20 '17 at 18:56
Error, When you set this $x^2=t$, so $x=\sqrt{t}$ it is wrong because $$-1\le x \le 1$$ that is $x=\sqrt{t}$ is not only positive but can be $x= -\sqrt{t}$
Trick to reach your method see that, $\sqrt{1+x^2}$ is an even function therefore, $$\int_{-1}^{1}\sqrt{1+x^2}dx = 2\int_{0}^{1}\sqrt{1+x^2}dx$$ Now you are allowed to: Let $x^2=t$, so $x=\sqrt{t}$ and $dx=\frac{1}{2\sqrt{t}}dt$. So we have
$$\int_{-1}^{1}\sqrt{1+x^2}dx = 2\int_{0}^{1}\sqrt{1+x^2}dx =2\int_{0}^{1}\sqrt{\frac{1+t}{t}}dt$$
But Rather you could easilyy set $x = \sinh u \implies dx = \cosh u du$
and $\sqrt{1+x^2} =\sqrt{1+\sinh u^2} = \cosh u$ $$\int_{-1}^{1}\sqrt{1+x^2}dx =2\int_{0}^{1}\sqrt{1+x^2}dx \\=2\int_{0}^{\sinh^{-1}(1)} \cosh^2 u du= \int_{0}^{\sinh^{-1}(1)} 1+\cosh2 u du\\=\sinh^{-1}(1) +\frac12 \sinh(2\sinh^{-1}(1))$$
Using $$\cosh2u = \cosh^2 u +\sinh^2 u~~~and~~~~ \cosh^2 u -\sinh^2 u = 1.$$
on the other hand we have, $$\sinh 2u = 2c\cosh u \sinh u = 2\sinh u\sqrt{1+\sinh^2 u}$$ | {
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on the other hand we have, $$\sinh 2u = 2c\cosh u \sinh u = 2\sinh u\sqrt{1+\sinh^2 u}$$
Using this we have $$\frac12 \sinh(2\sinh^{-1}(1)) =\sinh (\sinh^{-1}(1))\cdot\sqrt{1+\sinh^2\sinh^{-1}(1)} = \sqrt 2$$
Therefore, $$\color{red}{\int_{-1}^{1}\sqrt{1+x^2}dx du=\sinh^{-1}(1) +\frac12 \sinh(2\sinh^{-1}(1)) = \sinh^{-1}(1)+\sqrt2.}$$
The key question is this: As $x$ varies from $-1$ to $1$, what does $t$ do?
In many simple examples of $u$-substitution, the when $x$ varies across $(a,b)$ $u$ varies across $(c,d)$ and so we can just integrate the substituted function from $c$ to $d$ and be done with it.
We cannot do this in your case. In your case, the answer is that, as $x$ varies from $-1$ to $1$, $t$ varies from $1$ to $0$ and then back to $1$ again. Because there is this overlap, because $t$ traces the interval $[0,1]$ twice, you cannot simply replace the bounds. Additionally, on the two parts of this journey the function is actually different. For negative values of $x$ you need to use $-\sqrt{\quad}$ as the inverse function. This gives you:
$$\int_{-1}^1\sqrt{1+x^2}dx=\frac{1}{2}\int_1^0-\sqrt{\frac{1+t}{t}}dt+\frac{1}{2}\int_0^1\sqrt{\frac{1+t}{t}}dt=\int_0^1\sqrt{\frac{1+t}{t}}dt$$
Here the first integral is while $t$ is decreasing and the second is while $t$ is increasing. Notice that there's also a negative sign in front of this first integral. This is because $t$ is decreasing from $1$ to $0$ for values that correspond to negative $x$ values, so $x=-\sqrt{t}$ rather than $x=+\sqrt{t}$. | {
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# 12 red snooker balls
1. Aug 24, 2004
### Ian Rumsey
We have 12 red snooker balls and a scale balance.
One of the snooker balls has a different weight but we do not know whether it is heavier or lighter.
The scale balance may be only used three times by putting an equal number of balls on each side and noting the balancing result.
What 3 balancing arrangements would be required to ensure the identity of the odd weight ball and whether it was heavy or light.
2. Aug 24, 2004
### DarkForXe
hmm. I am not trying to be rude here.
But this q is asked before.
3. Aug 24, 2004
### TenaliRaman
i dunno if this is asked before but nice *twist* on the old *ionc* problem
4. Aug 24, 2004
### K.J.Healey
I don't think this is solveable with the alotted measurements and information.
5. Aug 24, 2004
### Rogerio
Surely it is!
Consider 1234 x 5678 at first...
6. Aug 24, 2004
### K.J.Healey
youre sure you read right? that we dont know if the unique ball is heaver OR lighter? I think that creates a problem.
7. Aug 24, 2004
### TenaliRaman
Healey,
It can be solved .....
infact it can be shown that
In n weighings, [(3^n-3)/2 + 1] snooker balls can be weighed and the odd ball can be found.
Placing n = 3, we get 13 .. so ofcourse it is possible even for 12.
I won't spoil this one for anyone .... i already received too many insinuating glares from Gokul already..... :D
-- AI
P.S -> So what does this post mean? It means that once the original problem is solved some followers will be,
Find a solution when we have 13 snooker balls?
Show that the general statement of n weighings given by me is true?
8. Aug 24, 2004
### NateTG
Odd -- according to your formula, I can't find it in 1 ball with 0 weighings, when the number for n=0 should be 1.
9. Aug 24, 2004
### Rogerio
Unfortunately, in just 3 weighings it's not possible for 13 balls...
(12 is the max)
10. Aug 24, 2004
### NateTG | {
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10. Aug 24, 2004
### NateTG
Really, I'm pretty sure it's possible to find the odd ball out with 13 balls - you may not be able to tell whether it's light or heavy though.
11. Aug 24, 2004
### Hurkyl
Staff Emeritus
Then you're not looking hard enough.
If you have one ball, and you know one ball has the incorrect weight, then you don't need to do any weighings to find it!
The impossible case is where you have two balls.
(Oh, and incidentally, the formula says you can find 0 balls in 1 weighing)
12. Aug 24, 2004
### K.J.Healey
Someone post the answer in white. I've tried it numerous ways for a few minutes, and i can get pretty close, but for every possibility theres one where I cant determine if the odd ball was heavier or lighter than the others, but just different.
Hmm by that equation you should be able to find the odd ball out of 4 in 2 tries, and tell me if its heavier or lighter?? Please show me (its a lot simpler than the stated riddle).
EDIT : nevermind, i got it. But my simple question still stands.
Last edited: Aug 24, 2004
13. Aug 24, 2004
### NateTG
You're usually so on the ball (no pun intended) that I'm guessing you misread, or I wasn't clear enough, since we appear to (in principle) agree that the formula is incomplete.
However, the formula indicates that one ball can be identified in one weighing:
In the formula $$n$$ is the number of weighings, and it gives the number of balls.
My post indicated that the formula was too low in the case of 0 weigings.
$$\frac{3^n-3}{2}+1 = \frac{3^1-3}{2}+1 = \frac{0}{2}+1 = 1$$
But 0 balls in 0 weighings.
14. Aug 24, 2004
### Hurkyl
Staff Emeritus
Yes, I meant 0 in 0.
15. Aug 24, 2004
### NateTG
Should give you a good idea on how to find a solution to this problem. Notably, using that method, the 10 coin case might be more difficult that the 12 coin one.
16. Aug 24, 2004
### Gokul43201
Staff Emeritus
How about , <select to see>
first : L = 5, 6, 8, 10 R = 7, 9, 11, 12 | {
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### Gokul43201
Staff Emeritus
How about , <select to see>
first : L = 5, 6, 8, 10 R = 7, 9, 11, 12
second : L = 2, 3, 4, 7 R = 5, 6, 11, 12
third : L = 1, 4, 10, 11 R = 2, 5, 7, 8 ?
I think this gives unique outcomes for each number.
Defining outcomes as <, > or = on the basis of the L pan being heavier, R pan being heavier, or both equal.
1 : ==< or ==>
2 : =<> or =><
3 : =<= or =>=
4 : =<< or =>>
5 : <>> or ><<
6 : <>= or ><=
7 : ><> or <><
8 : <=> or >=<
9 : >== or <==
10: <=< or >=>
11: >>< or <<>
12: >>= or <<=
All look different to me...
17. Aug 24, 2004
### TenaliRaman
Both are related observations and well observed indeed ....
Ofcourse it was my mistake that i forgot to mention one tiny 'precursor' reqd to relax the problem.....
Hurkyl,
Now with 1 ball u know which is the odd one but u don't know whether it is heavier or lighter so u need one weighing (but weighed against what ??? that will be ur question ... wait i will address it shortly)
NateTG,
You are pretty correct indeed and we will be able to find the lighter or heavier condition too .....
The relaxation given to the problem,
"You can add one true weight ball to the existing set of balls"
(Hurkyl and NateTG would have got the entire general solution by now )
-- AI
18. Aug 25, 2004
### Rogerio
Now it's easy (and possible): compare 12345 x 6789T at first...:-)
19. Aug 25, 2004
### BobG
1. Left scale= 1, 2, 3, 4
Right scale = 5, 6, 7, 8
2. Left scale = 1, 4, 7, 10
Right scale = 2, 5, 8, 11
3. Left scale = 3, 6, 12
Right scale = 1, 8, 9
20. Aug 26, 2004
### BobG
Revised solution (forgot to check opposites)
1. Left scale= 1, 2, 3, 4
Right scale = 9, 10, 11, 12
2. Left scale = 1, 4, 7, 10
Right scale = 2, 5, 8, 11
3. Left scale = 1, 3, 9, 11, 12
Right scale = 2, 5, 6, 7, 10 | {
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# What is the difference between "limiting" and "stationary" distributions?
I'm doing a question on Markov chains and the last two parts say this:
• Does this Markov chain possess a limiting distribution. If your answer is "yes", find the limiting distribution. If your answer is "no", explain why.
• Does this Markov chain possess a stationary distribution. If your answer is "yes", find the stationary distribution. If your answer is "no", explain why.
What is the difference? Earlier, I thought the limiting distribution was when you work it out using $P = CA^n C^{-1}$ but this is the $n$'th step transition matrix. They calculated the limiting distribution using $\Pi = \Pi P$, which I thought was the stationary distribution.
Which is which then? | {
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Which is which then?
• Your textbook may be making a distinction that is not universal: for instance, Karl Sigman's notes on limiting distributions defines "limiting" and "stationary" distributions to be synonymous (definition 2.3 at the bottom of p. 5). Therefore you must consult the definitions in your textbook in order to determine the difference.
– whuber
Jan 22 '13 at 18:41
• @whuber It is saying something like working out $\lim_{n \rightarrow \infty} P_{ii}^{(n)}$ and this doesn't exist. It then goes on to say "even though the limiting distribution doesn't exist, the stationary does. Let $\Pi = (\pi_0, \pi_1, ... ,\pi_n)$ be the stationary distribution...." But I guarantee you to calculate the limiting distribution in the question before, they solved it like this. Does that make sense to you? Jan 22 '13 at 19:25
• @whuber Actually, I'm quite confused now because in the previous limiting distribution question, they don't satisy the $\pi_0 + \pi_1 + \pi_2 = 1$ equality, so maybe that is different? Jan 22 '13 at 19:32
• A stationary distribution is one that is stable over time. As far as I'm aware, the limiting distribution of a Markov chain is stationary and if a Markov chain has a stationary distribution it is also a limiting distribution. Jul 1 '14 at 4:49
• Answer here by Andreas might help quora.com/… Jun 11 '18 at 22:27
From An Introduction to Stochastic Modeling by Pinsky and Karlin (2011): | {
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From An Introduction to Stochastic Modeling by Pinsky and Karlin (2011):
A limiting distribution, when it exists, is always a stationary distribution, but the converse is not true. There may exist a stationary distribution but no limiting distribution. For example, there is no limiting distribution for the periodic Markov chain whose transition probability matrix is $$\mathbf{P}=\left\|\begin{matrix}0 & 1\\1 & 0\end{matrix}\right\|$$ but $\pi=\left(\frac{1}{2},\frac{1}{2}\right)$ is a stationary distribution, since $$\left(\frac{1}{2},\frac{1}{2}\right)\left\|\begin{matrix}0 & 1\\1 & 0\end{matrix}\right\|=\left(\frac{1}{2},\frac{1}{2}\right)$$ (p. 205).
In a prior section, they had already defined a "limiting probability distribution" $\pi$ by
$$\lim_{n\rightarrow\infty}P_{ij}^{(n)}=\pi_j~\mathrm{for}~j=0,1,\dots,N$$
and equivalently
$$\lim_{n\rightarrow\infty}\operatorname{Pr}\{X_n=j|X_0=i\}=\pi_j>0~\mathrm{for}~j=0,1,\dots,N$$ (p. 165).
The example above oscillates deterministically, and so fails to have a limit in the same way that the sequence $\{1,0,1,0,1,\dots\}$ fails to have a limit.
They state that a regular Markov chain (in which all the n-step transition probabilities are positive) always has a limiting distribution, and prove that it must be the unique nonnegative solution to
$$\pi_j=\sum_{k=0}^N\pi_kP_{kj},~~j=0,1,\dots,N,\\ \sum_{k=0}^N\pi_k=1$$ (p. 168)
Then on the same page as the example, they write
Any set $(\pi_i)_{i=0}^{\infty}$ satisfying (4.27) is called a stationary probability distribution of the Markov chain. The term "stationary" derives from the property that a Markov chain started according to a stationary distribution will follow this distribution at all points of time. Formally, if $\operatorname{Pr}\{X_0=i\}=\pi_i$, then $\operatorname{Pr}\{X_n=i\}=\pi_i$ for all $n=1,2,\dots$.
where (4.27) is the set of equations
$$\pi_i \geq 0, \sum_{i=0}^{\infty} \pi_i=1,~\mathrm{and}~\pi_j = \sum_{i=0}^{\infty} \pi_iP_{ij}.$$ | {
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$$\pi_i \geq 0, \sum_{i=0}^{\infty} \pi_i=1,~\mathrm{and}~\pi_j = \sum_{i=0}^{\infty} \pi_iP_{ij}.$$
which is precisely the same stationarity condition as above, except now with an infinite number of states.
With this definition of stationarity, the statement on page 168 can be retroactively restated as:
1. The limiting distribution of a regular Markov chain is a stationary distribution.
2. If the limiting distribution of a Markov chain is a stationary distribution, then the stationary distribution is unique.
• Can you clarify what you mean by 'transition probabilities don't change over time' for stationarity? Both limiting and stationary distribution are about the probabilities over states. Jul 1 '14 at 7:45
• Yup, I see you wrote your own answer but I reorganized mine to be more correct. Jul 1 '14 at 9:04
• I still don't get it. I mean what do you mean when you say "except now with an infinite number of states.... " ? Can you please clarify it more explicitely.
– roni
Jul 14 '16 at 16:09
• @roni the two expressions are identical if you let $N=\infty$ Jul 14 '16 at 16:30
• In the first highlighted block, $\pi=(1/2,1/2)$ is the stationary distribution for the example, however, it has no limiting distribution since $P^{n}$ will oscillate, and it therefore has no steady state. Does this mean that it won't guarantee the existence of steady state if only the stationary distribution is calculated? Feb 12 '19 at 3:24 | {
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A stationary distribution is such a distribution $\pi$ that if the distribution over states at step $k$ is $\pi$, then also the distribution over states at step $k+1$ is $\pi$. That is, $$\pi = \pi P.$$ A limiting distribution is such a distribution $\pi$ that no matter what the initial distribution is, the distribution over states converges to $\pi$ as the number of steps goes to infinity: $$\lim_{k\rightarrow \infty} \pi^{(0)} P^k = \pi,$$ independent of $\pi^{(0)}$. For example, let us consider a Markov chain whose two states are the sides of a coin, $\{heads, tails\}$. Each step consists of turning the coin upside down (with probability 1). Note that when we compute the state distributions, they are not conditional on previous steps, i.e., the guy who computes the probabilities does not see the coin. So, the transition matrix is $$P = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.$$ If we first initialize the coin by flipping it randomly ($\pi^{(0)} = \begin{pmatrix}0.5 & 0.5\end{pmatrix}$), then also all subsequent time steps follow this distribution. (If you flip a fair coin, and then turn it upside down, the probability of heads is still $0.5$). Thus, $\begin{pmatrix} 0.5 & 0.5 \end{pmatrix}$ is a stationary distribution for this Markov chain.
However, this chain does not have a limiting distribution: suppose we initialize the coin so that it is heads with probability $2/3$. Then, as all subsequent states are determined by the initial state, after an even number of steps, the state is heads with probability $2/3$ and after an odd number of steps the state is heads with probability $1/3$. This holds no matter how many steps are taken, thus the distribution over states has no limit. | {
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Now, let us modify the process so that at each step, one does not necessarily turn the coin. Instead, one throws a die, and if the result is $6$, the coin is left as is. This Markov chain has transition matrix $$P = \begin{pmatrix} 1/6 & 5/6 \\ 5/6 & 1/6 \end{pmatrix}.$$ Without going over the math, I will point out that this process will 'forget' the initial state due to randomly omitting the turn. After a huge amount of steps, the probability of heads will be close to $0.5$, even if we know how the coin was initialized. Thus, this chain has the limiting distribution $\begin{pmatrix} 0.5 & 0.5 \end{pmatrix}$.
• Good point about forgetting the initial state, I completely glossed over this in my answer. Jul 1 '14 at 9:07
• This explanation helps me understand a lot. Can I say the existence of a steady state is equivalent to the existence of a limiting distribution? Since it is not easy to calculate the limiting distribution, we often calculate the stationary distribution by solving balance equations instead. However, I thought this alternative method doesn't guarantee that the stationary distribution is independent from initial states, therefore, it explains why for $P = \begin{pmatrix}0&1\\1&0\end{pmatrix}$, it has the stationary distribution but no steady state that is independent from the initial states. Feb 12 '19 at 3:59
• @GuoyangQin If you have a new question, you may wish to post it as a question (linking to this one if it helps provide question). Although I would have thought "steady state" in this context would mean "stationary distribution" so it would be best to clearly define the term in the question Feb 12 '19 at 18:57
Putting notation aside, the word "stationary" means "once you get there, you will stay there"; while the word "limiting" implies "you will eventually get there if you go far enough". Just thought this might be helpful. | {
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• It isn't clear how this applies to the question. Could you explain?
– whuber
Jul 28 '15 at 10:17
• Hi @whuber, I mean to say that a limiting distribution is necessarily a stationary distribution while a stationary distribution is not necessarily a limiting distribution. Hence there is a difference. This is essentially the same as other answers but I think it's easier to remember.
– Khoa
Jul 17 '16 at 2:24
• Thank you for the clarification: it shows us what you are attempting to accomplish. However, I cannot find any reasonable way to interpret your description of "stationary" in a way that is consistent with the mathematical definition.
– whuber
Jul 17 '16 at 13:08
• @whuber BlueSky's phrasing seems like an extremely straightforward plain English notion of "fixed point" to me -- I'm not sure what your object could mean. Aug 20 '17 at 18:51 | {
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# Maximum of the function F(X)=X*(1+X)^0.5
1. May 13, 2012
### Mike s
Hello,
Is x=-1 local maximum of $F(x)=x\sqrt{1+x}$?
On the one hand, $F(-1+\delta)<F(-1)$ for $0<\delta<1$.
However, $F(-1-\delta)$ is not defined.
As far a I know, the point x=a is considered local maximum, if there exists small neighborhood $\delta$ such that $$F(a)>F(x)$$ for every $a-\delta<x<a+\delta$.
So is x=-1 a local maximum or not?
2. May 13, 2012
### Fightfish
I believe that the term "local maximum" is not applicable because the concept of neighbourhood does not exist about the point x=-1 as you have pointed out.
Rather this is what we call a "boundary maximum". If you look at the gradient of F(x) at x=-1, it is undefined, as opposed to 0 gradient for stationary maxima/minima.
It is worth noting that boundary points are always "maximum" or "minimum" (except when the curve is level about that point).
3. May 13, 2012
### Ray Vickson
Yes, it is. It satisfies F(-1) > F(x) for all sufficiently small |x+1| with x > -1. In this problem it is impossible to compare F(-1) to F(x) for x < -1, but that is not really the reason that x = -1 is a local max. In the presence of constraints a local max or min need not have derivative = 0, and whether or not the function could be defined outside the interval is not particularly important; for example, for the problem min/max f(x) = x, subject to 0 <= x <= 1, x = 0 is a min and x = 1 is a max. These two statements are true even though f(x) is perfectly well defined outside the interval.
Note: you may find that some writers disagree with these statements. However, they are commonly asserted in modern treatments of optimization theory and practice; they are really matters of definition, not of substance.
RGV
4. May 14, 2012
### Fightfish
Agreed. Most texts do define local maxima/minima as being on an open interval. In a closed interval, it is trivial as the endpoints certainly must be minima/maxima. | {
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For all practical purposes though, it doesn't matter if its "local" or "not local". It's not going to affect your calculations or results, so don't fret too much over it.
5. May 14, 2012
### Mike s
I understand, thanks for both of you. | {
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# Precision of FindRoot
Consider the following code:
r = x /. FindRoot[Cos[x] - x == 0, {x, 1}, AccuracyGoal -> 15,
PrecisionGoal -> 15, WorkingPrecision -> 100];
Precision[r]
I would expect the Precision of r to be 15 since I'm only asking Mathematica to solve it to 15 digits. But the result is 100, which is the WorkingPrecision I am using. Why?
• I think the precision goal is how close you need to get to the true root. Here FindRoot will use 100 digits in x in each step of the iteration towards the root, and when it has found the root to 15 digit precision, it stops, and return that x, which still has a precision of 100. – Marius Ladegård Meyer Jun 12 '16 at 8:22
• Just like @Marius said, PrecisionGoal and AccuracyGoal will only inform you the digits you can use. WorkingPrecision will inform you the resulting number's precision, but maybe the latter digits is not usable. – Wjx Jun 12 '16 at 8:39
• See Controlling the Precision of Results and Setting WorkingPrecision->n causes all internal computations to be done to at most n-digit precision. See WorkingPrecision – user9660 Jun 12 '16 at 8:49
Here is a short demo. Generate the Dottie number as an exact Root[] object, like so:
dottie = x /. First @ Solve[x == Cos[x] && 0 < x < 1, x];
(Tho you might notice an inexact number in the output, rest assured that the resulting Root[] object is an exact number that can be evaluated to arbitrary precision; the number is there only as a sort of "localization marker".)
Use FindRoot[] to generate an approximation of the Dottie number:
ndottie = x /. FindRoot[Cos[x] - x, {x, 1}, AccuracyGoal -> 5, PrecisionGoal -> 5,
WorkingPrecision -> 30];
Altho Precision[ndottie] == 30., the result is not actually accurate to all 30 digits. Here is the relative error:
-Log10[Abs[1 - ndottie/dottie]]
20.0631570441
which says that the result is accurate to about 20 digits, even with the supposedly low settings. Now, crank up the settings: | {
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ndottie2 = x /. FindRoot[Cos[x] - x, {x, 1}, AccuracyGoal -> 15,
PrecisionGoal -> 15, WorkingPrecision -> 30];
and you'll find that ndottie2 - dottie is effectively zero.
• One can see with StepMonitor that the relative error estimate in the next-to-last (3rd) step is about $(x_3-x_{2})/x_3=3.8 \times 10^{-5}$, a bit greater than the PrecisionGoal, so FindRoot takes another step. But because of the quadratic convergence in this well-behaved problem, this is closer to the error in $x_2$. The next step has a error estimate of $(x_4-x_3)/x_4=2.3 \times 10^{-10}$, but that means the error in $x_4$ will be close to $10^{-20}$. – Michael E2 Apr 3 '17 at 10:19 | {
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1. ## Complete metric space
Let $X_{n+1}\subset X_n$ for each $n\in\mathbb N.$ Let $(M,d)$ be a metric space. Prove that the following assertions are equivalent:
a) $(M,d)$ is a complete metric space.
b) For each $X_{n+1}\subset X_n$ being closed and non-empty where $\text{diam}(X_n)\to0$ as $n\to\infty,$ we have $\bigcap_{n=1}^\infty X_n\ne\varnothing.$
I already did a) $\implies$ b) but for the other implication, I have a long solution, is there a short way to do it?
2. ## Re: Complete metric space
let {xn} be a cauchy sequence in X. by definition for any ε > 0 there is some N such that d(xm,xn) < ε for all m,n > N.
so pick εk = 1/2k, and set Uk = cl(B(xNkk)).
define Xk = U1∩U2∩...∩Uk.
it's clear that diam(Xk) ≤ 1/2k-1 and that each Xk is closed being the intersection of finitely many closed sets.
furthermore since Nk < Nk+1 its clear that for any k:
xNk is in Um, for all m ≤ k, so each Xk is non-empty.
so, we have from (b) that ∩kXk ≠ Ø.
but this means the subsequence {xNk} of {xn} converges to some point x of X (the sole element of the infinite intersection, since if d(x,y) = 0, y = x, and for any y in ∩kXk,
we have d(y,xNk) < 1/2k-1 for ALL k in N). hence {xn} also converges to x, so X is complete.
(i was tempted to use just the 1/2k balls at first, but i realized that the 2nd ball could contain points OUTSIDE the first ball, but their intersection has no such defect. the center of the 3rd ball has to lie within the intersection of the first 2, and the center of the 4th ball has to lie within the intersection of the first 3, etc. the diameter of the intersections can be no larger than the diameter of the ball, which is twice its radius).
3. ## Re: Complete metric space
Okay, thank you!
I was thinking the following: let $x_n$ be a Cauchy sequence on $X,$ then let $X_n=\overline{\{x_m:m\ge n\}},$ so $X_n$ is closed and satisfies $X_{n+1}\subset X_n,$ yet I don't see how to prove that $\text{diam}(X_n)\to0.$ | {
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Can an $m \times n$ rank $1$ matrix be written as a product of an $m\times1$ and a $1\times n$ matrix?
If a $m \times n$ matrix has rank $1$, does it imply that it can be written as a product of one $m\times1$ and one $1\times n$ matrix. How to prove it ? Is this decomposition unique ? What are the entries of these matrices ?
In a matrix of rank 1, every row is a scalar multiple of every other non-zero row (and there must be at least one non-zero row). For the $1 \times n$ matrix, take one of the non-zero rows; in the $m \times 1$ matrix, put the scalars from the first sentence. The decomposition is not unique: you can pick every non-zero scalar multiple of a non-zero row to start with.
Yes, it is always possible. Let
$$A = \begin{bmatrix} a_1 \\ a_2 \\ \vdots \\ a_m \end{bmatrix}$$
be a matrix of rank $1$, and let $a_i$ be its rows. Since $A$ is of rank $1$, each two of these rows are linearly dependent. Let us assume that $a_k \ne 0$ for some $k$ (such row must exist, or $A$ would be a zero matrix). Then for each $i$,
$$a_i = \alpha_i a_k,$$
for some (real or complex) numbers $\alpha_i$. Define
$$v = \begin{bmatrix} \alpha_1 \\ \vdots \\ \alpha_m \end{bmatrix}, \quad w := a_k.$$
Then $A = v w$.
Obviously, the decomposition is not unique. Let $\xi \not \in \{0,1\}$. Then $A = (\xi v)(\xi^{-1} w)$ is another such decomposition.
Let $A = [a_1\ \cdots\ a_n]$ be the $m\times n$ where the $a_i$ are $m$-dimensional column vectors. If $A$ has rank $1$, then $\dim\operatorname{Col} = 1$. Therefore $\dim\,\operatorname{span}\{a_1, \dots, a_n\} = 1$. Choose $a_j \neq 0$ (such a column exists otherwise the $A = O$ which has rank zero), then $\operatorname{span}\{a_1, \dots, a_n\} = \operatorname{span}\{a_j\}$. In particular, $a_i \in \operatorname{span}\{a_j\}$ for $i = 1, \dots, n$. Therefore $a_i = c_ia_j$ for some $c_i \in \mathbb{R}$. Then
$$A = [a_1\ \cdots\ a_n] = [c_1a_j\ \cdots\ c_na_j] = a_j[c_1\ \cdots c_n].$$ | {
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$$A = [a_1\ \cdots\ a_n] = [c_1a_j\ \cdots\ c_na_j] = a_j[c_1\ \cdots c_n].$$
Keep in mind, there is more to the last equality than first appears; $a_j$ is column vector not a constant.
More generally, an $m\times n$ matrix $A$ has rank $k$ if and only if the smallest $r$ such that $A = u_1v_1^T + \dots + u_rv_r^T$, where $u_1, \dots, u_r \in \mathbb{R}^m$, $v_1, \dots, v_r \in \mathbb{R}^n$, is $r = k$.
The other answers have addressed the issue of existence quite well. Let me show that the decomposition is "essentially unique", i.e. up to scalar multiple of the two vectors.
Suppose that $A=\mathbf{x}_1\mathbf{y}_1^\mathrm{T} = \mathbf{x}_2\mathbf{y}_2^\mathrm{T}$. Note that $A\mathbf{v}=c\mathbf{w}$ for some vector $\mathbf{w}$ and some scalar $c$ since the image is one dimensional. Thus we have $$A\mathbf{y}_1 = \mathbf{x}_1\mathbf{y}_1^\mathrm{T}\mathbf{y}_1=\|\mathbf{y}_1\|^2\mathbf{x}_1=c_1\mathbf{w}$$ $$A\mathbf{y}_2 = \mathbf{x}_2\mathbf{y}_2^\mathrm{T}\mathbf{y}_2=\|\mathbf{y}_2\|^2\mathbf{x}_2=c_2\mathbf{w}$$ Therefore $\mathbf{x}_1$ and $\mathbf{x}_2$ are multiples of the same vector $\mathbf{w}$. The argument can be repeated for $\mathbf{y}_1$ and $\mathbf{y}_2$ by taking the transpose and repeating the argument. | {
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# Math Help - help
1. ## help
mr.lopez needs 3/4 kg of apples to make a pie.how many pies can he make if he uses 20 kg of apples?
name a fraction that is between 1/2 and 1/3....?
which of the following fractions is closest to one?
a)2/3 b)3/4 c)4/5 d)5/6
three pizzas were evenly shared among some friends. if each of them go 3/5 of the pizza. how many were there altogether?
2. Originally Posted by BeBeMala
mr.lopez needs 3/4 kg of apples to make a pie.how many pies can he make if he uses 20 kg of apples?
How many 3/4 kilograms can fit inside 20 kilograms? (Hint: Divide.)
Originally Posted by BeBeMala
name a fraction that is between 1/2 and 1/3....?
There are infinitely-many choices. One quick way of finding such a fraction is to convert the two given fractions to a common denominator, and then find some other fraction of the same denominator that is between the two numerators.
Originally Posted by BeBeMala
which of the following fractions is closest to one?
a)2/3 b)3/4 c)4/5 d)5/6
Convert them all to the same denominator, and subract each from 1. Whichever one has the smallest answer was the biggest (and thus closest to 1).
Originally Posted by BeBeMala
three pizzas were evenly shared among some friends. if each of them go 3/5 of the pizza. how many were there altogether?
This one works just like the first one, above.
3. Originally Posted by stapel
How many 3/4 kilograms can fit inside 20 kilograms? (Hint: Divide.)
There are infinitely-many choices. One quick way of finding such a fraction is to convert the two given fractions to a common denominator, and then find some other fraction of the same denominator that is between the two numerators.
Convert them all to the same denominator, and subract each from 1. Whichever one has the smallest answer was the biggest (and thus closest to 1).
This one works just like the first one, above.
sorry....
4. Originally Posted by BeBeMala
sorry(((no help ))))....
sorry i'm so confused.......
5. Hello, | {
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4. Originally Posted by BeBeMala
sorry(((no help ))))....
sorry i'm so confused.......
5. Hello,
1) How many 3/4 kilograms can fit inside 20 kilograms?
$\frac{20}{\frac{3}{4}}=20*\frac{4}{3}=....$
2)name a fraction that is between 1/2 and 1/3....?
1/2 and 1/3 are equivalent to 30/60 and 20/60 u can choose any number between 21/60 and 29/60 to be your answer
3)which of the following fractions is closest to one?
a)2/3 b)3/4 c)4/5 d)5/6
2/3=0,(6)
3/4=0,75
4/5=0,8
5/6=0,8(3)
4) three pizzas were evenly shared among some friends. if each of them go 3/5 of the pizza. how many were there altogether?
$\frac{3}{\frac{3}{5}}=3*\frac{5}{3}=5$
Have a nice day,
Hush_Hush
6. Originally Posted by Hush_Hush
Hello,
1) How many 3/4 kilograms can fit inside 20 kilograms?
$\frac{20}{\frac{3}{4}}=20*\frac{4}{3}=....$
2)name a fraction that is between 1/2 and 1/3....?
1/2 and 1/3 are equivalent to 30/60 and 20/60 u can choose any number between 21/60 and 29/60 to be your answer
3)which of the following fractions is closest to one?
a)2/3 b)3/4 c)4/5 d)5/6
2/3=0,(6)
3/4=0,75
4/5=0,8
5/6=0,8(3)
4) three pizzas were evenly shared among some friends. if each of them go 3/5 of the pizza. how many were there altogether?
$\frac{3}{\frac{3}{5}}=3*\frac{5}{3}=5$
Have a nice day,
Hush_Hush
Thank you so much, I finally got it, sorry but i don;t get """name a fraction that is between 1/2 and 1/3???
7. ## Fractions
Hello BeBeMala
Originally Posted by BeBeMala
Thank you so much, I finally got it, sorry but i don;t get """name a fraction that is between 1/2 and 1/3???
Take a close look at the pictures I've attached.
• The shaded areas in the top one show that the fraction $\frac{6}{12}$ is the same fraction as $\frac{1}{2}$.
• The middle one shows that $\frac{4}{12}$ is the same as $\frac{1}{3}$. | {
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• The middle one shows that $\frac{4}{12}$ is the same as $\frac{1}{3}$.
• The bottom one has 5 sections shaded out of 12. And the size of the shaded area in this one is in between the other two. So this fraction is in between $\frac{1}{2}$ and $\frac{1}{3}$.
What fraction is shaded in the bottom diagram? | {
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# If $x_1,x_2,\ldots,x_n$ are the roots for $1+x+x^2+\ldots+x^n=0$, find the value of $\frac{1}{x_1-1}+\frac{1}{x_2-1}+\ldots+\frac{1}{x_n-1}$
Let $$x_1,x_2,\ldots,x_n$$ be the roots for $$1+x+x^2+\ldots+x^n=0$$. Find the value of $$P(1)=\frac{1}{x_1-1}+\frac{1}{x_2-1}+\ldots+\frac{1}{x_n-1}$$
Source: IME entrance exam (Military Institute of Engineering, Brazil), date not provided (possibly from the 1950s)
My attempt: Developing expression $$P(1)$$, replacing the 1 by $$x$$, follows $$P(x)=\frac{(x_2-x)\cdots (x_n-x)+\ldots+(x_1-x)\cdots (x_{n-1}-x)}{(x_1-x)(x_2-x)\cdots (x_n-x)}$$
As $$x_1,x_2,\ldots,x_n$$ are the roots, it must be true that
$$Q(x)=(x-x_1)\cdots(x-x_n)=1+x+x^2+\ldots+x^n$$ and $$Q(1)=(1-x_1)\cdots(1-x_n)=n+1$$ Therefore the denominator of $$P(1)$$ is $$(-1)^{n} (n+1).$$ But I could not find a way to simplify the numerator.
Another fact that is probably useful is that $$1+x^{n+1}=(1-x)(x^n+x^{n-1}+\ldots+x+1)$$ with roots that are 1 in addition of the given roots $$x_1,x_2,\ldots,x_n$$ for the original equation, that is
$$x_k=\text{cis}(\frac{2k\pi}{n+1}),\ \ k=1,\ldots,n.$$
This is as far as I could go...
Hints and full answers are welcomed.
• – Arnaud D. Dec 17 '18 at 12:07
$$1$$ and $$x_1,\ldots,x_n$$ are the roots of $$z^{n+1}-1=0$$. They are the $$(n+1)$$-th roots of unity. Then $$1/(z^{n+1}-1)$$ has a partial fraction expansion of the form $$\frac1{z^{n+1}-1}=\frac a{z-1}+\sum_{k=1}^n\frac{b_k}{x_kz-1}.$$ Multiplying by $$z-1$$ and setting $$z=1$$ gives $$a=1/(n+1)$$. Multiplying by $$x_kz-1$$ and setting $$z=1/x_k$$ gives $$b_k=1/(n+1)$$. Therefore $$\sum_{k=1}^n\frac1{x_kz-1}=\frac{n+1}{z^{n+1}-1}-\frac1{z-1}.$$ Letting $$z\to1$$ gives $$\sum_{k=1}^n\frac1{x_k-1}=\lim_{z\to1}\frac{n-z-z^2-\cdots-z^n}{z^{n+1}-1} =\frac{-1-2-\cdots-n}{n+1}=-\frac n2.$$ | {
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