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https://mathematica.stackexchange.com/questions/114144/dynamical-system-problem-xt1-axt | Dynamical system problem x(t+1) = Ax(t)
I have a problem that have been trying to solve but it's not going so good. I would like some guidelines on how to work myself around this problem:
Two neighboring countries spy on each other and adapt its defense budget against each other depending on the situation in the neighborhood. Let country A's annual defense budget be a(t) (in billion) and country B's b(t). The change in budget changes each year according to the relationship:
a (t + 1) = r a (t) + s B (t)
b (t + 1) = s a (t) + r b (t)
where r is a positive number a bit less than 1 and s is a small postive number a bit greater than 0. Suppose that r is greater than s, that is, r>s. The budget for the two countries can be described by the vector x (t) = (a b) (<--- matrix) where a = a(t) and b = b(t)
If a(0) = 30 and b(0) = 5, what will happen with both the countries defense budget in the long run?
Solve tasks a) to e) both on paper and in Mathematica.
a) Set up the system transition matrix A such that x(t + 1) = Ax(t)
So what I've done so far is to select r = 0.8 and s = 0.32 and defined it as a 2x2 matrix and I've also defined a column vector with a(0)=30 and b(0)=5, so what I have in mathematica right know is:
I don't feel like this is how I'm supposed to do. At least this doesn't feel like how it should turn out in the "long run".
I would be very happy for someone to help/guide me through this! Thank you!
EDIT: I solved the problem and would like to have some feedback on how it is, if it's messy to follow. I would be thankful! :)
2nd EDIT: In my second question I need to find the Eigenvalues, Eigenvectors and Eigenbase. I have calculated by hand and I've gotten the Eigenvalues to lamda=1 and lamda=0.8 (when using the matrix {0.9,0.1},{0.1,0.9}) and the v1 = {1,1} and v2 = {-1,1}..I've gotten the right answers for it also on mathematica:
MatrixA = {{0.9, 0.1}, {0.1, 0.9}};
EigenMatrix = {{lm, 0}, {0, lm]}};
detMatrix = MatrixA - EigenMatrix
{{0.9 - lm], 0.1}, {0.1, 0.9 - lm}}
KarEk = Det[detMatrix]
0.8 - 1.8lm + lm^2
LamdaVarden = Roots[KarEk == 0, lm]
lm == 0.8 || lm == 1.
However, when trying to solve the Eigenvectors I get:
Eigenvectors[{{0.9, 0.1}, {0.1, 0.9}}]
{{0.707107, 0.707107}, {-0.707107, 0.707107}}
This is wrong, what am I doing wrong?
• Take a look at RSolve and also MatrixPower for a start. – gwr May 2 '16 at 11:21
• This is the same form as a matrix population model from biology. Search under that term if you need more info about its long term behavior, which will be determined by the dominant eigenvalue of A. – Chris K May 2 '16 at 14:48
• You cannot calculate with MatrixForm (just display result of a matrix calculation with it). – murray May 2 '16 at 20:13
• Post code blocks rather than screenshots. For your approach, recommend that you define a function: x[t_] := MatrixPower[{{0.8, 0.32}, {0.32, 0.8}}, t].{30, 5}] and then just Map x onto a range: x /@ Range[0, 10] This approach provides results but does not as easily identify the limits ("long run"). – Bob Hanlon May 2 '16 at 21:39
• The eigenvectors are correct... all eigenvector calculations are only right up to a constant multiple. Mathematica deals with this by choosing the vector that has unit norm... so {0.707107, 0.707107} points in the same direction as {1,1} and {-0.707107, 0.707107} points in the direction {-1,1}. – bill s May 3 '16 at 21:29
x[t_] = {a[t], b[t]} /. RSolve[{
a[t + 1] == r*a[t] + s*b[t],
b[t + 1] == s*a[t] + r*b[t],
a[0] == 30, b[0] == 5}, {a[t], b[t]}, t][[1]]
Since 0 < s < r < 1 then 0 < r - s < 1 - s < 1 and for large t, (r -s)^t is approximately zero.
x[t] /. (r - s)^t -> 0
i.e., a[t] ≈ b[t] for large t
Plot[
Evaluate[{
x[t] /. {r -> 0.8, s -> 0.32},
x[t] /. {r -> 0.68, s -> 0.32},
x[t] /. {r -> 0.6, s -> 0.32}}],
{t, -1, 20},
PlotStyle ->
(AbsoluteDashing /@ {{7, 3}, {7, 7}}),
Frame -> True, Axes -> False,
PlotLegends -> {
"a[t], r+s=1.12", "b[t], r+s=1.12",
"a[t], r+s=1", "b[t], r+s=1",
"a[t], r+s=0.92", "b[t], r+s=0.92"},
Epilog -> {Text["r + s = 1.12", {13, 100}],
Text["r + s = 1", {13, 35/2}, {0, -1.5}],
Text["r + s = 0.92", {13, 0}, {0, .5}]}]
In the limit,
Piecewise[
Assuming[{0 < r < 1, 0 < s < 1, s < r, #},
{Limit[x[t], t -> Infinity], #}] & /@
{r + s < 1, r + s == 1,
r + s > 1}]
x /@ Range[0, 10] /. {r -> 0.8, s -> 0.32} // Column
EDIT: For the generic solution
Format[a0] := Subscript[a, 0]
Format[b0] := Subscript[b, 0]
x1[t_] = {a[t], b[t]} /. RSolve[{
a[t + 1] == r*a[t] + s*b[t],
b[t + 1] == s*a[t] + r*b[t],
a[0] == a0, b[0] == b0},
{a[t], b[t]}, t][[1]] // Simplify
x2[t_] = MatrixPower[{{r, s}, {s, r}}, t].{a0, b0} // Simplify;
As J.M. pointed out in a comment, this can also be written as
x2[t_] = MatrixPower[{{r, s}, {s, r}}, t, {a0, b0}] // Simplify;
x1 and x2 are identical
x1[t] === x2[t]
(* True *)
Approximate solution for large values of t
xApprox[t_] = Simplify[x1[t] /. (r - s)^t -> 0]
a[t] and b[t] are approximately equal for large t (and identical in the limit)
Equal @@ xApprox[t]
(* True *)
In the limit,
Piecewise[
Assuming[
{0 < r < 1, 0 < s < 1, s < r, a0 > 0, b0 > 0, #},
{Limit[x1[t], t -> Infinity], #}] & /@
{r + s > 1, r + s == 1,
r + s < 1}]
• Hello, thank you for answering! :) I have gotten the same values, but I've used a different way. I would appreciate it if you could give an opinion on if it is good as an answer? I will post a screenshot on my question. – Vetenskap May 2 '16 at 17:57
• Thank you again for the detailed explanation! :) Really nice that you show everything so clearly! I'm a bit new when it's comes to Mathematica but I will try to study it closer! – Vetenskap May 3 '16 at 10:29
• MatrixPower[{{r, s}, {s, r}}, t].{a0, b0} can also be done as MatrixPower[{{r, s}, {s, r}}, t, {a0, b0}] (the action form). – J. M. will be back soon May 9 '16 at 23:58
• @J.M. - Thanks, edited to include this form. – Bob Hanlon May 10 '16 at 2:23
The way to answer this generically is via eigenvalues. Set up your matrix and take its eigenvalues:
mat = {{r, s}, {s, r}};
Eigenvalues[mat]
{r - s, r + s}
If either r-s or r+s is greater than 1 (in magnitude) then the system will be unstable. If both are less than one, it will decay to zero. For your values r=0.8 and s=0.32, you have instability, which is why you see the numerical simulations diverge.
• Thank you for the reply! So you mean it's better if I for example use r=0.9 and s=0.1 because then my system won't be unstable? :) – Vetenskap May 3 '16 at 10:28
• "Better" might depend on context, but for stability (i..e, the states $a(t)$ and $b(t)$ converging to zero) you will need Abs[r+s]<1. You can see that happening in Bob Hanlon's simulations. – bill s May 3 '16 at 13:10 | 2019-11-22T10:22:16 | {
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https://www.physicsforums.com/threads/kk-prob-1-14-angular-accl.427550/ | # KK Prob 1.14 - Angular Accl
1. Sep 8, 2010
### veezbo
1. The problem statement, all variables and given/known data
A drum of radius R rolls down a slope without slipping. Its axis has acceleration "a" parallel to the slope. What is the drum's angular acceleration A?
2. Relevant equations
x(t) = x_0 + v_0 + 1/2gt^2 (because of constant acceleration)
w = S / 2pi
3. The attempt at a solution
The center of the circle moves according the kinematic equation for position with constant acceleration.
If we consider its starting position as 0, then the position just gives the distance traveled by the center. I hope that I can also say that the initial velocity is 0 (can I?)
So then the distance traveled is 1/2gt^2. Dividing by 2R * pi gets us the number of revolutions, and then multiplying by 2pi gives the number of radians.
so (1/2R)gt^2 after simplification. If we divide by time, we should get w, or the angular velocity as gt/(2R).
Upon taking the derivative to find the angular acceleration, I get g/(2R). But this definitely does not seem right.
2. Sep 8, 2010
3. Sep 9, 2010
### veezbo
Yeah, I've already seen that page. However, I don't understand how that "formula" is derived. It should have some similar derivation as mine does (and I'm sure the authors Kleppner and Kolenkow would have wanted that).
On that note, what's wrong with my derivation?
4. Sep 9, 2010
### Delphi51
The formula A = a/R for angular acceleration comes from the basic definition d = rθ for arc length on a circle of radius r.
Differentiate with respect to time to get v = rdθ/dt = rω
Differentiate again to get a = rdω/dt = rA.
Your derivation makes considerable sense . . . but there are problems. For instance, the cylinder rolls down a ramp so it doesn't feel the full acceleration g of a vertical fall; you must begin with the given acceleration a rather than g. The step where you divide by time to get angular velocity assumes constant velocity which is not the case.
5. Apr 3, 2011
### jbunniii
My last physics class was in 1986, and although I have studied a lot of mathematics since then, my physics is pretty rusty. I recently started reading Kleppner and Kolenkow's "Introduction to Mechanics" to review the subject just for fun. I'm going to try to do all the exercises as part of this process.
Most of the exercises don't have solutions (a few have answers or hints), so to ensure that I am not deluding myself, I thought I would post any here that I'm not 100% certain about. This one, 1.14, falls in the, hmm, 98% category, so here we go.
Problem statement: A drum of radius $R$ rolls down a slope without slipping. Its axis has acceleration $a$ parallel to the slope. What is the drum's angular acceleration $\alpha$?
My answer: I think the slope is a red herring. Whatever effect gravity and the slope angle have must already be consolidated into the given acceleration $a$. If I understand correctly, this is the distinction between kinematics and mechanics: the accelerations are simply given, as opposed to being calculated based on the physical forces that influence the object.
The acceleration is $a$, parallel to the slope. But who cares what the slope is? I can simply rotate the problem so that the drum is rolling along a horizontal surface, with acceleration $a$ in the positive horizontal direction. With these coordinates, the position, velocity, and acceleration of the drum axis are all scalar functions of time.
If $v(t)$ is the horizontal velocity of the drum's axis, then the angular velocity is
$$\omega(t) = \frac{v(t)}{R}$$
(here I have used the fact that the drum is not slipping), and so the angular acceleration is
$$\alpha(t) = \dot{\omega}(t) = \frac{\dot{v}(t)}{R} = \frac{a}{R}$$
quite independently of the slope.
Is my reasoning sound? This seems deceptively simple for a book with a reputation for tricky problems.
P.S. I recognize this is an old thread, but it covers the same problem as the one I am discussing. What's the better protocol in this case: replying to an old thread or starting a new one?
Last edited: Apr 3, 2011 | 2017-08-22T13:56:55 | {
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https://math.stackexchange.com/questions/3549181/solve-the-recurrence-a-n-na-n%E2%88%921n | # Solve the recurrence $a_n=na_{n−1}+n!$
I'm working on a practice set:
Solve the recurrence $$a_n=na_{n−1}+n!$$ for $$n>0$$ with $$a_0=1$$ Give a simple expression for $$a_n$$
For this problem I know the answer is $$(n+1)!$$ But I'm not sure how to get there....
Here is what I did so far:
I divided the equation by n so: $$\frac{a_n}{n} = a_{n-1} + (n-1)!$$
Then I used telescoping, so: $$\frac{a_n}{n} = a_{n-1} + (n-1)!$$
$$\frac{a_{n-1}}{n-1} = a_{n-2} + (n-2)!$$
$$\frac{a_{n-2}}{n-2} = a_{n-3} + (n-3)!$$
$$...$$
So I cancel terms across the equal sign and I get:
$$a_n = a_0 + (n-1)!$$
But this is not correct.
Thanks for help
• Did you try proving it by induction? Feb 16 '20 at 20:05
• I haven't.... the assignment was on telescoping so was trying to first solve it that way - but open to any info if you sense you can use induction:) Feb 16 '20 at 20:08
• If you edit in your telescoping argument, we may identify the mistake.
– J.G.
Feb 16 '20 at 20:11
• Hi! Thanks so much @J.G. I added my incorrect telescoping process Feb 16 '20 at 20:19
Telescope $$b_n:=a_n/n!$$, which satisfies $$b_n-b_{n-1}=1$$ because$$b_n=\frac{na_{n-1}+n!}{n!}=\frac{a_{n-1}}{(n-1)!}+1=b_{n-1}+1.$$
• +1; this works well, and I think it's what OP was looking for Feb 16 '20 at 20:13
• @J.G. would like to understand a bit more about your process. I'm not sure I understand $b_n := a_n/n!$ curious how you got there - thanks! Feb 16 '20 at 20:20
• @KatieMelosto I divide by something that multiplies by $n$ at each step, because you need to telescope something of the form $b_n-b_{n-1}$. Your mistake was in trying to telescope something that isn't of that form.
– J.G.
Feb 16 '20 at 20:32
• @J.G.Thanks so much! When you divide $b_n := a_n/n!$ is that: $(a_n = na_{n-1} + n!)/n!$? Trying to put the math together Feb 16 '20 at 20:43
• @KatieMelosto See edit.
– J.G.
Feb 16 '20 at 21:25
Induction works:
base case: $$n=0$$
induction step: $$a_{n+1}=(n+1)a_n+(n+1)!=(n+1)(n+1)!+(n+1)!=(n+2)!$$ | 2021-10-25T17:35:43 | {
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https://forum.math.toronto.edu/index.php?PHPSESSID=t7v8357pqfkb44g38k5adgh577&action=printpage;topic=2440.0 | # Toronto Math Forum
## MAT334--2020F => MAT334--Lectures & Home Assignments => Chapter 1 => Topic started by: Nathan on October 06, 2020, 06:08:35 PM
Title: 1.6 Q5
Post by: Nathan on October 06, 2020, 06:08:35 PM
Question: $\int_y Re(z) dz$ where $y$ is the line segment from 1 to $i$.
I can't get the same answer as the one in the textbook.
Answer in textbook: $\frac{1}{2}(i-1)$
$y(t)=$
$= 1 + (i - 1)t$
$= (1 - t) + it$
$y'(t) = i - 1$
$\int_y Re(z) dz=$
=$\int_1^i (1-t)(i-1)dt$
$=\int_1^i i - 1 - ti + t dt$
$=ti - t - \frac{t^2i}{2} + \frac{t^2}{2}|^i_1$
$=-1 -i + \frac{i}{2} - \frac{1}{2} - (i - 1-\frac{i}{2} + \frac{1}{2})$
$=-i - 1$
What is wrong with my answer?
Title: Re: 1.6 Q5
Post by: smarques on October 06, 2020, 06:29:36 PM
Hi Nathan.
The mistake seems to be in your bounds of integration, not the process itself. With the way you parameterized the line, the integral should be from 0 to 1, not 0 to i. | 2022-05-20T01:35:21 | {
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https://math.stackexchange.com/questions/298461/evaluate-int-frac-cos-x-sin-x-cos-x-textdx | # Evaluate $\int\frac{\cos x}{\sin x + \cos x}\,\text{d}x$. [duplicate]
I'm trying to figure out how to take this indefinite integral:
$$\int\frac{\cos x}{\sin x + \cos x}\,\text{d}x.$$
I tried simplifying and rearranging it, and this is the best I got: $$\int\frac{1}{\tan x + 1 }\,\text{d}x.$$
But I still can't figure out how to integrate from there. I know that it's integrable, as Wolfram Alpha indicates that the integral is $\frac{1}{2}\big(x+\ln{(\sin x + \cos x)}\big)+C$, but I can't figure out the steps to deriving it. Does anyone know how to evaluate this integral?
## marked as duplicate by Martin Sleziak, Amzoti, rschwieb, Asaf Karagila♦, Mark BennetMay 30 '13 at 17:21
• This is basically the same as this question. (To be more precise, if you know one of the two integrals, it is easy to calculate the other one and vice versa.) – Martin Sleziak May 30 '13 at 16:42
Let $I=\int\frac{\cos x}{\sin x+\cos x}\ dx$ and $J=\int \frac{\sin x}{\sin x+\cos x}\ dx$.
Then $I+J=\int\frac{\cos x+\sin x}{\sin x+\cos x}\ dx=x+C_1$ and $I-J=\int\frac{\cos x-\sin x}{\sin x+\cos x}\ dx=\ln |\sin x+\cos x|+C_2$.
Hence $I=\frac{1}{2}((I+J)+(I-J))=\frac{1}{2}(x+\ln |\sin x+\cos x|)+C$.
• This is very elegant in that it requires no substitutions. – user54147 Feb 9 '13 at 12:20
• @user54147 You are using a substitution, $u = \sin(x) + \cos(x)$ in the end – Amad27 Aug 11 '15 at 19:58
A general (but not necessarily efficient) tool is the Weierstrass substitution $t=\tan(x/2)$.
Then $\cos x=\frac{1-t^2}{1+t^2}$, $\sin x=\frac{2t}{1+t^2}$, and $dx=\frac{2}{1+t^2}$. Do the substitution. We end up needing the following ugly integral: $$\int \frac{2(1-t^2)}{(1+2t-t^2)(1+t^2)}\,dt.$$ We are integraing a rational function, and it can be done using partial fractions. But it takes some work.
The same idea works in principle for any rational function of $\sin x$ and $\cos x$.
Notice that $$\sin x+\cos x=\sqrt{2}\left(\frac{1}{\sqrt{2}}\sin x+\frac{1}{\sqrt{2}}\cos x\right)=\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)$$ Therefore $$I=\frac{1}{\sqrt{2}}\int\frac{\cos x}{\sin\left(x+\frac{\pi}{4}\right)}dx$$ Now let $x+\frac{\pi}{4}=t$ $$I=\frac{1}{\sqrt{2}}\int\frac{\cos \left(t-\frac{\pi}{4}\right)}{\sin t}dt=\frac{1}{2}\int\frac{\sin t + \cos t}{\sin t}dt=\frac{1}{2}\left(t+\ln {\left|\sin t\right|}\right)+C_1\\=\frac{1}{2}\left(x+\frac{\pi}{4}+\ln {\left|\sin x+\cos x\right|}\right)+C_1=\frac{1}{2}\left(x+\ln {\left|\sin x+\cos x\right|}\right)+C$$
$\displaystyle \int \frac{1}{1+ \tan x} \ dx$
$\displaystyle = \int \frac{1}{1+u} \frac{1}{1+u^{2}} \ du$ (let $u = \tan x$)
$\displaystyle = \frac{1}{2} \int \left( \frac{1}{1+u} + \frac{1-u}{1+u^{2}} \right) \ du$
$\displaystyle = \frac{1}{2} \int \left( \frac{1}{1+u} - \frac{u}{1+u^{2}} + \frac{1}{1+u^{2}} \right) \ du$
$\displaystyle =\frac{1}{2} \left(\ln(1+u) - \frac{1}{2} \ln(1+u^{2}) + \arctan u \right) + C$
$\displaystyle =\frac{1}{2} \left(\ln(1+u) - \ln(\sqrt{1+u^{2}}) + \arctan u \right) + C$
$\displaystyle = \frac{1}{2} \left( \ln(1+\tan x) - \ln (\sqrt{1+\tan^{2}x}) + x \right) + C$
$\displaystyle = \frac{1}{2} \left( \ln(1+\tan x) - \ln (\sec x) + x \right) + C$
$\displaystyle = \frac{1}{2} \Big( \ln \big(\frac{1+ \tan x}{\sec x}\big) + x \Big) + C$
$\displaystyle = \frac{1}{2} \left( \ln (\cos x + \sin x ) + x \right) + C$ | 2019-05-19T17:01:26 | {
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https://quant.stackexchange.com/questions/70969/why-does-changing-the-step-size-in-my-binomial-tree-changes-the-final-stock-pric | Why does changing the step size in my Binomial Tree changes the final stock prices so much?
I am trying to price a convertible bond by using a binomial tree. For this, I wrote a binomial tree for the stock price. I noticed that changing the step size (timesteps), changes the final value of my stock prices significantly. Intuitively this is wrong as the up and down movements should be scaled accordingly. I can imagine this will also change the price of the convertible bond.
I thought setting $$u = \text{e}^{\sigma\sqrt{dt}}$$ would do the trick. Could anyone tell me what I am doing wrong? Thanks!
import numpy as np
import math as math
S0 = 100 #Initial stock price
T = 5 #Maturity
timesteps = 16 #Amount of steps in the three
dt = T/timesteps #Step size
sigma = 0.28 #Vol
r = 0.01 #Interest rate
N = 300 #Notional amount (for convertible bond)
kappa = N/S0 #Conversion rate (for convertible bond)
c = 0.05 #Coupomn rate (for convertible bond)
u = np.exp(sigma*math.sqrt(dt))
d = 1/u
p = (np.exp(r*dt)-d)/(u-d)
S = np.zeros((timesteps,timesteps))
for i in range(timesteps):
for j in range(timesteps):
S[j,i] = S0*(u**(i-j))*(d**j)
S = np.triu(S)
• I think your computation of $dt$ is incorrect. Note that applying $u$ $n$ times leads to $\exp (\sigma n \sqrt{dt})$ which, unless I'm wrong, should be equivalent to $\exp (\sigma \sqrt{T})$, i.e., $n \sqrt{dt} = \sqrt{T}$, rather than $n dt = T$. I haven't checked in detail, but everything else seems fine. May 20 at 15:36
• What is significantly? What values of timesteps did you try? May 22 at 13:55
• Two timesteps give a range of 64 - 155, three timesteps 48 - 206 and 30 timesteps 3 - 2752. It doesn't converge. May 23 at 10:30
You only got one minor bug, but let me explain why the range increases.
Let us denote $$n:=timesteps$$, then
1. You are looping one iteration too little when filling your $$S$$ matrix array, causing you to have S(T-dt) and not S(T) as terminal values. This is because you are not accounting for your starting position, i.e. you need $$1+n$$ iterations in each dimension.
...
S = np.zeros((1+timesteps, 1+timesteps)) # Include starting position too (S0).
for i in range(S.shape[1]):
for j in range(S.shape[0]):
S[j,i] = S0*(u**(i-j))*(d**j)
S = np.triu(S)
1. The range of possible values of $$S(T)$$ does correctly increase with $$n$$. The range is given by $$[S_0 d^n, S_0 u^n] = [S_0 e^{-\sigma\sqrt{nT}}, S_0 e^{\sigma\sqrt{nT}}]$$. Note that while the range increases, the probability for ending at these extreme values decreases. In fact, according to the Central Limit Theorem, in the limit as $$n\to \infty$$ the binomial distribution will become the continuous log-Normal distribution, which indeed has infinite positive support $$(0,+\infty)$$.
2. What the CRR binomial model ensures is that the mean and variance of the discrete binomial model matches those of the continuous model. The mean of the stock will match exactly for any $$n$$, whereas the variance will asymptotically approach the continuous case. The reason for the variance not being exactly matched for any $$n$$ is that when finding the parameter $$u$$ by matching the variance, approximations were made by only keeping first-order terms in a Taylor series expansion.
# ====================================================
# === Compare moments to the continuous exact ones ===
# ====================================================
from scipy import stats as stats
# Binomial model for number of down moves
pd = 1-p
dist = stats.binom(n=timesteps, p=pd)
# Stock terminal mean and variance E[S(T)/S0] and V[S(T)/S0]
# 1st moment for S(T)
bin_m1 = dist.expect(lambda k: S[k.astype(int),-1]/S0)
# 2nd moment for S(T)
bin_m2 = dist.expect(lambda k: (S[k.astype(int),-1]/S0)**2)
# Var[S_T] = E[S_T^2] - E[S_T]^2
bin_var = bin_m2 - bin_m1**2
print(f'Binomial Model: E[S/S0]={bin_m1:.4f}, V[S/S0]={bin_var:.4f}')
# Continuous S(T) moments, https://en.wikipedia.org/wiki/Geometric_Brownian_motion#Properties
gbm_m1 = np.exp(r*T)
gbm_var = np.exp(2*r*T)*(np.exp(sigma**2*T)-1)
gbm_m2 = gbm_var + gbm_m1**2
print(f'Continuous GBM: E[S/S0]={gbm_m1:.4f}, V[S/S0]={gbm_var:.4f}')
Binomial Model: E[S/S0]=1.0513, V[S/S0]=0.5218
Continuous GBM: E[S/S0]=1.0513, V[S/S0]=0.5304
• Thank you very much for your explanation. May 30 at 8:00 | 2022-08-14T17:03:55 | {
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http://math.stackexchange.com/questions/293980/how-to-integrate-large-frac2x1x2 | # How to integrate $\large \frac{2x}{1+x^{2}}$
How to integrate $\;\large \frac{2x}{1+x^{2}}\;?\;$ Do I need to use u-substitution for $\,(1+x^2)\,$?
-
That’s certainly the easiest way: it makes the integral a very simple one. – Brian M. Scott Feb 3 '13 at 21:57
It's much easier to judge whether an approach to a problem is useful after you have tried it out rather than before. – Hurkyl Feb 4 '13 at 16:34
That's the most straightforward way.
Let $u = 1 + x^2,\;$ then $du = 2x \,dx.\;$ Now you're all set to substitute:
$$\int \dfrac{2x\,dx}{1+x^2} \;\;=\; \;\int \dfrac{du}{u}\;\;=\;\; \ln |u| + C \;= \;\ln(1 + x^2) + C$$
Note that we can omit the "absolute value sign" surrounding $\,(1 + x^2)\,$ when substituting $\,u = 1+x^2\,$ in the final expression because we know that $\,(1 + x^2) > 0\,$ for all real $x$.
-
One small point, $u=|u|$ in this case. Otherwise we would have to write $ln|1+x^2|+C$ – Baby Dragon Feb 3 '13 at 23:27
Note that $1 + x^2 >0$ for all values of x. That's the only reason I dropped them, as we are guaranteed that $|1 + x^2| = 1 + x^2$ – amWhy Feb 3 '13 at 23:55
$$2x=(1+x^2)'$$
$$\int{\frac{(1+x^2)'}{(1+x^2)}}dx$$
-
thanks for suggestion – Iuli Feb 3 '13 at 22:01
Looks good now; +1. – Brian M. Scott Feb 3 '13 at 22:12
$\textbf{Hint:}$ You don't need any fancy technique. Look at an antiderivative table and try to work it out.
Scroll over the grey area for the solution.
Note that $\displaystyle \int \frac{2x}{1+x^2}dx=\int \frac{(1+x^2)'}{1+x^2}dx=\log {\bigl(|1+x^2|\bigr)}=\log {(1+x^2)}$
-
The solution is not an integrand. You have the solution posted as an integrand. The point was to integrate the function to the left, which is the integrand. – amWhy Feb 3 '13 at 22:18
@amWhy That was a funny mistake. Thanks. – Git Gud Feb 3 '13 at 22:21
$\displaystyle\frac{\mathrm{d}}{\mathrm{d}x} \ln u(x) =\frac{u'(x)}{u(x)}$
in this case : $u(x)=x^2+1$ so the primitive is : $\ln(x^2+1)+C$
- | 2015-08-05T00:18:13 | {
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https://brilliant.org/discussions/thread/four-is-magic-on-steroids/ | # Four is Magic: On Steroids
Four is Magic is a popular math game. You spell out any number into English:
i.e. $117$ = one hundred seventeen
And then you add the total letters together. So 'one hundred seventeen' has $19$ letters.
When we continue this:
i.e. $117 → 19 → 8 → 5 → 4 → 4$
It is notable that "four" has exactly $4$ letters in it. It is the only number with this quality.
### Certain Rules
• We exclude counting the spaces and hyphens in the name. Only letters.
• We don't say "one hundred and one". We exclude the "and".
# On Steroids
I came up with a bonus feature to this puzzle.
i.e. $117$ = one hundred seventeen = $3 + 7 + 9 → 19$
We will be taking the product of each word count:
i.e. $117$ = one hundred seventeen = $3 * 7 * 9 → 189$
This leads to very different results! $4$ is still special, but now are there other numbers with this quality?
Yes, there are!
• $24 →$ twenty four $= 6 * 4 = 24$
Can you find the others?
• Problem 1: I have found 9 solutions for $a → a$
• Problem 2: I have found 25 other solutions for $a → b → ... → a$
If you can find solutions for $a → a$, be sure to post them in this sequence: A058230.
How can you code this problem?
What are the solutions to both problems? My 9th solution is a monstrous 42 digit number!
Are there more than 9 solutions?
What other interesting numbers can you find?
Turns out these numbers are called Fortuitous Numbers.
Note by Jonathan Pappas
4 months, 1 week ago
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Fascinating! I just might try coding this. I'll probably even use this digit-separating formula I came up with a while back.
- 4 months, 1 week ago
Thank you! It's a really interesting programming challenge. That's also a very useful formula!
- 4 months, 1 week ago
Phew! 100 lines of code later, and it takes me a minute to reach the fourth OEIS entry. :)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 # digit seperator def get_digit(num, place): return (num % (10**(place+1))) - (num % (10**place)) # convert numbers between 0 and 1000 to English (used in get_eng_num) def get_eng_num_under_1000(num): # separate digits digits = [get_digit(num, place)/(10**place) for place in range(len(str(num)))] digits.reverse() # create number to English mappings singles = {1:"one", 2:"two", 3:"three", 4:"four", 5:"five", 6:"six", 7:"seven", 8:"eight", 9:"nine"} powers_of_ten = {10:"ten", 100:"hundred"} # create improper English to proper English mappings tens_convert = {2:"twenty", 3:"thirty", 4:"forty", 5:"fifty", 6:"sixty", 7:"seventy", 8:"eighty", 9:"ninety"} teens_convert = {1: "eleven", 2:"twelve", 3:"thirteen", 4:"fourteen", 5:"fifteen", 6:"sixteen", 7:"seventeen", 8:"eighteen", 9:"nineteen"} # convert numbers to English english_number = [] size = len(digits) # if the number in in the hundreds if size == 3: # convert hundreds if digits[-3] > 0: english_number.append(" ".join([singles[digits[0]], "hundred"])) # if the number is in the hundreds or tens if size >= 2: # convert tens, except teens if digits[-2] > 1: english_number.append(tens_convert[digits[-2]]) # convert ones if digits[-1] > 0: english_number.append(singles[digits[-1]]) # convert teens (except ten) elif digits[-2] == 1 and digits[-1] > 0: english_number.append(teens_convert[digits[-1]]) # convert ten elif digits[-2] == 1 and digits[-1] == 0: english_number.append("ten") # convert less than ten elif digits[-2] == 0 and digits[-1] > 0: english_number.append(singles[digits[-1]]) # if the number is a single digit if len(digits) == 1: # convert ones if digits[-1] > 0: english_number.append(singles[digits[-1]]) # join and return English number return " ".join(english_number) # convert a number to it's English equivalent def get_eng_num(num): # separate digits, and return only the digit (not the full digit value) digits = [int(get_digit(num, place)/(10**place)) for place in range(len(str(num)))] # how many groups of three we can make: size = round((len(digits) + 1)/3) # separate into groups of three threes = [] for i in range(size): new_three = [str(d) for d in digits[(3*i):(3*i)+3]] new_three.reverse() threes.append(int("".join(new_three))) # convert each group of three to English and add on appropriate power of 1000 powers_of_1000 = ["", "thousand", "million", "billion", "trillion", "quadrillion", "quintillion", "sextillion", "septillion", "octillion", "nonillion", "decillion"] english_number = [] for i, three in enumerate(threes): if three: if i == 0: english_number.append(get_eng_num_under_1000(three)) else: english_number.append(" ".join([get_eng_num_under_1000(three), powers_of_1000[i]])) # reverse and join it all together, or return "zero" if there's nothing left english_number.reverse() if english_number: return " ".join(english_number) else: return "zero" # main function for the "word product" def compute_word_product(num): words = get_eng_num(num).split(" ") word_product = 1 # also returns length of each word for printing later word_lengths = [] for word in words: word_product *= len(word) word_lengths.append(str(len(word))) return word_product, word_lengths # value to be set \/ for n in range(2000000): n_word_product, n_word_lengths = compute_word_product(n) if n_word_product == n: print(f"{n} -> {get_eng_num(n)} ({' * '.join(n_word_lengths)})")
- 4 months, 1 week ago
Wow, that's great! Here is my program: https://github.com/JonnyGamer/FourIsMagicOnSteroids/blob/main/main.py
- 4 months, 1 week ago
Does that mean that you found more entries than are listed in the OEIS sequence, or am I reading that wrong?
- 4 months, 1 week ago
Yes, I found 3 more, the largest of which has 42 digits. I’m trying to add the new terms to OEIS, but now I am waiting for approval from an editor.
I’m also adding a new sequence focused on numbers of type “a -> b -> ... -> a”. But that one also needs to be approved by an editor.
- 4 months, 1 week ago
Woah, cool! How long did that take?
- 4 months, 1 week ago
This program takes less than 5 minutes: https://github.com/JonnyGamer/FourIsMagicOnSteroids/blob/main/main.py
But then no new solutions appeared after 24 hours of running, up to about ~ 10^130
- 4 months, 1 week ago
Good grief. That's some awesome code. :) That was a good idea generating a cache ahead of time.
- 4 months, 1 week ago
Thank you! I'm trying to optimize it, so any tips would be welcome.
- 4 months, 1 week ago | 2021-07-27T22:12:54 | {
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https://www.physicsforums.com/threads/tips-and-tricks-for-finding-determinant-of-3x3-matrix.414403/ | # Tips and tricks for finding determinant of 3x3 matrix?
1. Jul 6, 2010
### arian487
I have a midterm tomorrow and I find I'm quite slow at finding the determinant of a 3x3 matrix. Basically I'll only need to find the determinant to find the characteristic polynomial (at least for this class) and my prof on the board does it so fast, I'm wondering if there's some trick I missed out on. I feel like it takes me too much time to get the determinant of 3x3 matrices. Also my first time posting, long time lurker lol. Thanks.
2. Jul 6, 2010
Exercise. In general, if the matrix contains numerical values (specially values of not so high precision, since it takes time to plug in, for example, 1.03427*10^-28), you should be able to calculate the determinant fairly quickly. Of course, review all the properties of the determinant which can help you get over with your calculations more quickly and elegant.
3. Jul 6, 2010
### arian487
Yea there will be no decimal values as there is no calculator on the midterm. Also, once I find the 3x3 matrix I end up having to solve a cubic (in most cases) which also takes time. Is there no easier way to get the eigenvalues of a 3x3 (with whole numbers eigenvalues) rather then just solving the Characteristic polynomial?
4. Jul 6, 2010
I bet your teacher won't give you a cubic which you can't arrange in a way which is fairly easy to solve.
5. Jul 6, 2010
### Fredrik
Staff Emeritus
You know how to write it as the sum of the determinants of three 2x2 matrices, right? If there are any zeros in your 3x3 matrix, you can expand along that row or column to eliminate one or two of those 2x2 matrices.
If all the numbers below or above the diagonal are zeros, the determinant is just the product of the diagonal elements.
Also, if one of the rows (or columns) is a linear combination of the other two, the determinant is zero.
6. Jul 6, 2010
### arian487
Thanks frederick, that definitely helps a lot. Let me make sure I understand:
If A = [2 0 0; 1 5 4; 1 2 4] (arranged by column) then my determinant is simply 2 * 5 * 4?
7. Jul 6, 2010
### arian487
Nevermind I understood you wrong lol. I'm guessing if A is something like: [a 0 0; 0 b 0; 0 0 c].
8. Jul 6, 2010
### Matthollyw00d
It doesn't have to be as restrictive as diagonal. Upper or lower triangular form is sufficient.
So for example if $$A=(a_{ij}) : a_{ij}=0, i>j$$ This is upper triangle form. And Similarly for $$i<j$$
9. Jul 6, 2010
### HallsofIvy
Staff Emeritus
No, that's incorrect. Expanding on the first row the determinant is
$$\left|\begin{array}{ccc}2 & 0 & 0\\1 & 5 & 4\\ 1 & 2 & 4\end{array}\right|= 2\left|\begin{array}{cc}5 & 4 \\ 2 & 4\end{array}\right|= 2(20- 8)= 24$$
because all of the other multipliers, the other numbers in the first row, are 0.
If a matrix is "upper triangular" (all numbers below the main diagonal are 0), you can expand on the first column so you have only the number in "first row first column" times a smaller determinant, which can be expanded on the first column, etc. giving just the product of the numbers on the diagonal:
$$\left|\begin{array}{ccc}a & b & c \\ 0 & d & e \\ 0 & 0 & f\end{array}\right|= a\left|\begin{array}{cc}d & e \\ 0 & f\end{array}\right|= a(d f)= adf$$
If a matrix is "lower diagonal" (all numbers above the main diagonal are 0), then expanding on the first row repeatedly gives the same thing.
Finally, you can always "row reduce" a matrix until it is "upper triangular" remembering that:
1) Swapping two rows multiplies the determinant by -1
2) Multiplying an entire row by a number multiplies the determinant by that same number (so you have to divide the determinant resulting triangular matrix by that number to recover the determinant of the original matrix).
3) Adding or subtracting a multiple of one row from another does not change the determinant.
10. Jul 6, 2010
### Noxide
Prop: Gaussian Elimination does not change the determinant
Prop: product of the diagonal entries of an upper triangular matrix = determinant
Prop: If n row exchanges are performed when finding the triangular form, then determinant = (product of diagonals)(-1)^n
Prop: If a scalar c multiplies a row, then multiply the determinant by 1/c, c!=0
Last edited: Jul 6, 2010
11. Jul 6, 2010
### snipez90
You can do what I do, which is just to expand by minors. It's a very simple method of breaking down the determinant of a larger matrix to that of a smaller matrix and consequently works wondrously in the 3x3 case. Probably best illustrated by example: http://ceee.rice.edu/Books/LA/det/det2.html"
Last edited by a moderator: Apr 25, 2017 | 2017-12-15T13:13:01 | {
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https://math.stackexchange.com/questions/3345678/compact-notation-for-the-product-kk1k2-cdotskn/3345726 | # Compact notation for the product $k(k+1)(k+2)\cdots(k+n)$?
Is there a convenient and compact notation for writing the product $$A$$, below, in a compact way?
$$A=k(k+1)(k+2)\cdots(k+n)$$ i.e. a product of $$n$$ consecutive integers starting from $$k$$.
This arises from some diagonals of Pascal's triangle.
• en.wikipedia.org/wiki/Falling_and_rising_factorials – Wojowu Sep 5 '19 at 21:18
• Please don't write "Solved" in your title like you did. Instead, you should answer your question yourself (or let Wojowu do it), or delete the question if you don't think it will be useful to anyone. – Arnaud D. Sep 5 '19 at 22:05
• Hmmm it's interesting that there are two close votes for lack of context, yet I don't see how much more context you could have for a question simply asking for notation. – Simply Beautiful Art Sep 5 '19 at 23:31
• I refrained from accepting an answer because they both seemed equally good. I've now accepted one for the sake of marking the question as closed. – mjc Sep 6 '19 at 0:42
• @GerryMyerson There were no answers when I posted that comment. – Arnaud D. Sep 6 '19 at 7:37
In Concrete Mathematics, by Graham, Knuth, and Patashnik, the notation used for "rising powers" is $$m^{\overline{n}}$$ meaning $$m(m+1)\cdots (m+n-1)$$ with $$n$$ terms being multiplied.
Many other authors do use the $$m^{(n)}$$ notation for it, but the advatage of the one listed first is that it nicely unifies with falling powers $$m^{\underline{n}} = m(m-1)(m-2)\cdots(m-n+1)$$
Mathematica uses the name "Pochammer" function, which is valid in that Paochammer did early work involving such products, but IMHO that makes it sound like something arcane and advanced.
Note that the words "factorial powers" are connected to both of these: $$1^{\overline{n}} = n^{\underline{n}} = n!$$ and that as long as the exponent used is an integer, these can be defined for any real $$x$$, as in $$\left(\frac32\right)^{\overline{3}} = \frac{105}{8}$$
• I would disagree that "Pochammer symbol" sounds any more arcane or advanced than "factorial symbol", "summation symbol" or many other things we use regularly in mathematics. Things are given names to make it easy to refer to them and since it's necessary to explain the term "rising factorial" it's no clearer than "Pochammer symbol" in this context. – postmortes Sep 6 '19 at 13:15
• @postmortes - "Rising factorial" is at least somewhat descriptive, even though it may require explanation. "Pochhammer symbol" is not descriptive at all; a person's name tells nothing about the idea. It may suggest that the idea is so complicated or obscure that it can't be given a short descriptive name. – mr_e_man Sep 12 '19 at 23:55
• @hardmath -- Right you are, I had the incorrect spelling. And that extra "h" does clarify everything considerably. – Mark Fischler Sep 13 '19 at 16:49
As Wojowu mentions, this is the rising factorial, defined by:
$$k^{(n)}=k(k+1)(k+2)\cdots(k+n-1)$$
Using product notation you could also write:
$$\prod_{i=0}^{n-1}(k+i)=k(k+1)(k+2)\cdots(k+n-1)$$
And in terms of the factorial:
$$\frac{(k+n-1)!}{(k-1)!}=k(k+1)(k+2)\cdots(k+n-1)$$ | 2020-02-17T16:11:36 | {
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https://math.stackexchange.com/questions/1383295/finding-a-basis-of-a-complex-vector-space-over-bbb-r-given-a-basis-over-bbb | # Finding a basis of a complex vector space over $\Bbb R$ given a basis over $\Bbb C$
Suppose $X$ is a vector space over $\mathbb C$ and has as basis $\{e_1,e_2,\ldots,e_n\}$. Now regard $X$ as a vector space over $\mathbb R$.
What will be the basis?
My thoughts:
I considered $\mathbb C$ over $\mathbb C$ and $\mathbb C$ over $\mathbb R$.In the first case we have $(1,0)$ as basis and in the latter case we have $\{(1,0),(0,1)\}$ as basis i.e. $\{(1,0),(0,1)(1,0)=(0,1)\}$ as basis.
So may be the answer is $\{e_1,e_2,\ldots,e_n,ie_1,\ldots,ie_n\}$. How to justify the result if its true?
• Yes, your answer is correct. As usual t's enough to show that (1) the basis is linearly independent (over $\Bbb R$), and (2) that any element of $\Bbb C^n$ can be written as an ($\Bbb R$-)linear combination of basis elements. – Travis Aug 3 '15 at 18:12
• What about taking Re$e_k$, Im$e_k$? – A.Γ. Aug 3 '15 at 18:14
• @A.G. That doesn't work. Consider the standard basis. Half of your basis elements are $0$. – user24142 Aug 3 '15 at 18:31
• @user24142 That's right, I see now, thanks! – A.Γ. Aug 3 '15 at 18:41
• And what does "$\operatorname{Re} e_k$" means when $e_k$ is a vector in a complex vector space? – Najib Idrissi Aug 3 '15 at 18:51
In the case of $\mathbb{C}$ over $\mathbb{C}$, the basis would be $\{1\}$ because every element of $\mathbb{C}$ can be written as a $\mathbb{C}$-multiple of $1$.
$$\mathbb{C}=\{z\times 1 : z \in \mathbb{C}\}$$
In the case of $\mathbb{C}$ over $\mathbb{R}$, the basis would be $\{1,\mathrm{i}\}$ because every element of $\mathbb{C}$ can be written as an $\mathbb{R}$-multiple of $1$ and $\mathrm{i}$.
$$\mathbb{C}=\{x\times 1 + y \times \mathrm{i} : x,y \in \mathbb{R}\}$$
If $\{{\bf v}_1,\ldots,{\bf v}_n\}$ is a basis for $V$ over $\mathbb{C}$ then $$V = \{a_1{\bf v}_1+\cdots+a_n{\bf v}_n: a_k \in \mathbb{C} \}$$
We can write each of the $a_k$ as $b_k+\mathrm{i}c_k$, where $b_k,c_k \in \mathbb{R}$. Hence \begin{eqnarray*} a_1{\bf v}_1+\cdots+a_n{\bf v}_n &=& (b_1+\mathrm{i}c_1){\bf v}_1+\cdots+(b_n+\mathrm{i}\mathrm{c}_n){\bf v}_n \\ &=& b_1{\bf v}_1+\cdots+b_n{\bf v}_n+c_1(\mathrm{i}{\bf v}_1)+\cdots+c_n(\mathrm{i}{\bf v}_n) \end{eqnarray*}
We can take $\{{\bf v}_1,\ldots,{\bf v}_n,\mathrm{i}{\bf v}_1,\ldots,\mathrm{i}{\bf v}_n\}$ as a basis for $V$.
The final step is to show that
$$V = \mathbb{R}\langle {\bf v}_1,\ldots,{\bf v}_n\rangle \oplus \mathbb{R}\langle \mathrm{i}{\bf v}_1,\ldots,\mathrm{i}{\bf v}_n\rangle$$
This is obvious since $\mathbb{i} \notin \mathbb{R}$.
Suppose $x\in X$. Then $x = c_1 e_1 + \cdots+c_n e_n$ for some complex numbers $c_1,\ldots,c_n$.
For $k=1,\ldots,n$ write $c_k = a_k + i b_k$ where $a_k$ and $b_k$ are real.
Then \begin{align} x & = c_1 e_1 + \cdots+c_n e_n \\[8pt] & = (a_1+ib_1)e_1 + \cdots + (a_n+ib_n)e_n \\[8pt] & = a_1 e_1 + \cdots + a_n e_n + b_1(ie_1) + \cdots + b_n (ie_n). \end{align} So $x$ is a linear combination of $e_x,\ldots,e_n,ie_1,\ldots,ie_n$ with coefficients that are real.
Linear independence can be proved by considering almost the same sequence of equalities: \begin{align} 0 & = a_1 e_1 + \cdots + a_n e_n + b_1(ie_1) + \cdots + b_n (ie_n) \tag 1 \\[8pt] & = (a_1+ib_1)e_1 + \cdots + (a_n+ib_n)e_n \\[8pt] & = c_1 e_1 + \cdots+c_n e_n \end{align} and that can be true only if $c_1=\cdots=c_n=0$, by linear independence of $e_1,\ldots,e_n$ over $\mathbb C$. Hence $(1)$ can be true only if $a_1=\cdots=a_n=b_1=\cdots=b_n=0$.
• Could the person who down-voted this explain why? ${}\qquad{}$ – Michael Hardy Aug 3 '15 at 22:49 | 2019-06-18T07:09:09 | {
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https://math.stackexchange.com/questions/1533992/indefinite-integrals-with-absolute-values | # Indefinite integrals with absolute values
Which is the right way to solve indefinite integrals which contain absolute values? For example if I have $\int |2x+3| e^x dx$
Can I consider the sign function and integrate separetly? I mean doing: $Sign(2x+3) \int (2x+3)e^x dx$
Or maybe I should use the definition of absolute value and divide the two possibilities
$\int (2x+3)e^x dx$ if $(2x+3)>0$ and $\int (-2x-3)e^x dx$ if $(2x+3)<0$
But I think that's more suitable for definite rather than indefinite integrals
How can I solve this type of integrals? Thanks a lot in advice
• Since the antiderivative is linear, your answer would differ by at most a sign change. This would matter when you are plugging an antiderivate into, say, limits to calculate a definite integral. – amcalde Nov 17 '15 at 19:40
HINT: i would consider the cases $$2x+3\geq 0$$ or $$2x+3<0$$
$\int|2x+3|e^x~dx$
$=\text{sgn}(2x+3)\int(2x+3)e^x~dx$
$=\text{sgn}(2x+3)\int_{-\frac{3}{2}}^x(2x+3)e^x~dx+C$
$=\text{sgn}(2x+3)\int_{-\frac{3}{2}}^x(2x+3)~d(e^x)+C$
$=\text{sgn}(2x+3)[(2x+3)e^x]_{-\frac{3}{2}}^x-\text{sgn}(2x+3)\int_{-\frac{3}{2}}^xe^x~d(2x+3)+C$
$=\text{sgn}(2x+3)(2x+3)e^x-2~\text{sgn}(2x+3)\int_{-\frac{3}{2}}^xe^x~dx+C$
$=|2x+3|e^x-2~\text{sgn}(2x+3)[e^x]_{-\frac{3}{2}}^x+C$
$=|2x+3|e^x-2~\text{sgn}(2x+3)\left(e^x-e^{-\frac{3}{2}}\right)+C$
• Thanks a lot for this answer, I just did not understand why you used the integral function, can't I just integrate indefinetly? – Gianolepo Nov 20 '15 at 16:56
• @Francesco Caruso $\text{sgn}(2x+3)e^x$ is discontinuous at $x=-\dfrac{3}{2}$ , while $\text{sgn}(2x+3)\left(e^x-e^{-\frac{3}{2}}\right)$ is continuous at $x=-\dfrac{3}{2}$ . – Harry Peter Dec 4 '15 at 16:00 | 2019-12-14T08:04:59 | {
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https://math.stackexchange.com/questions/573574/probability-of-picking-exactly-one-correct-from-a-pool-of-6-incorrect-and-4-corr | # Probability of picking exactly one correct from a pool of 6 incorrect and 4 correct
So as the question says. You have 6 incorrect objects and 4 correct ones. What are the odds that, when picking 3 of them at random, you end up with exactly one of them being correct.
This seems to be quite trivial but I'm having issues since my solution differs from the provided one. Also, I've written a program that simulates it and it seems to be leaning towards my solution.
Anyways, the way I did it is just by multiplying $\dfrac{4}{10}\dfrac{6}{9}\dfrac{5}{8}$ with the result being $\dfrac{1}{6}$. The reasoning is that you first have a 4 in 10 chance of picking the correct one and then you need to pick the incorrect one 2 times with 6 in 9 and 5 in 8 chances. I thought that maybe the fact that the order doesn't matter is making my calculation off but since you can write the product as a big fraction and commute everything, that can't be the problem.
The provided solution is $\dfrac{\binom{4}{1}\binom{6}{2}}{\binom{10}{3}}$. I can see how that would work as well so I can't really say what the problem might be with their solution.
Anyways, I wrote a program that picks 3 numbers out of [0,0,0,0,0,0,1,1,1,1] at random. It does it 100000 times and the number of times it got exactly one 1 is around 18800. That's obviously much closer to my $\dfrac{1}{6}$ than their $\dfrac{1}{2}$. Is their solution wrong? If so, why?
Edit: Now that I think about it, my solution should be the incorrect one since it answers the question of what the odds are that I FIRST pick a correct one and then 2 incorrect ones, but why would the simulation lean towards my solution then?
• You calculate the probability that you first pick a correct one and then two wrong ones. There are $3$ possible orders here: CWW, WCW and WWC. Note that $3\times\frac{1}{6}=\frac{1}{2}$ – drhab Nov 19 '13 at 19:02
• I've noticed that (see my edit) but it still doesn't explain the simulation. – Luka Horvat Nov 19 '13 at 19:02
• Your first proposed solution is not right, though it can be made right by multiplying by $3$, the number of orders in which you can get $1$ correct, $2$ incorrect. As to the simulation, presumably the simulation follows the wrong analysis, the program counts the cases where you get good, then bad, then bad. – André Nicolas Nov 19 '13 at 19:03
• Yes, three identical numbers, which you then should add. – André Nicolas Nov 19 '13 at 19:05
• Your code for shuffle is wrong. "arr.Add(array[rand.Next(i, array.Count - 1)]);" picks up a random element in an array containing 6 zeros and 4 ones. You might end up creating a new list that is shuffled but not having exactly 4 1's and 6 0's. – Sudarsan Nov 19 '13 at 19:28
I just tried running the simulation and it works (out of $100000$, I got the number of trials in which only 1 correct object picked up as $49982$, fairly converging to $\frac{1}{2}$).
Code: (Octave)
sum = 0;
res = 0;
a=[0,0,0,0,0,0,1,1,1,1];
for i = 1:100000;
[junk,index] = sort(rand(1,10));
pick = index(1);
sum = sum+a(pick);
b=[];
tempcount = 1;
for j = 1:length(a);
if(j == pick)
continue
else
b(tempcount) = a(j);
end
tempcount = tempcount + 1;
end
[junk,index] = sort(rand(1,9));
pick = index(1);
sum = sum+b(pick);
c = [];
tempcount = 1;
for j = 1:length(b);
if(j == pick)
continue
else
c(tempcount) = b(j);
end
tempcount = tempcount + 1;
end
[junk,index] = sort(rand(1,8));
pick = index(1);
sum = sum+c(pick);
if(sum == 1);
res = res + 1;
end
sum = 0;
end | 2019-06-19T14:44:37 | {
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https://math.stackexchange.com/questions/623308/modular-arithmetic-with-polynomial | # Modular arithmetic with polynomial
Given n=pq, where p and q are primes, P(x) is polynomial and z∈Zn.
I need to prove that: P(z) ≡ 0 mod n iff P(z mod p) ≡ 0 mod p AND P(z mod q) ≡ 0 mod q.
If i could prove the more general case: P(z) mod n ≡ P(z mod p) mod p (q is the same) then i could also prove what i need of course. from my understanding so far, the above is true, but i can't figure out a way to prove this.
Back to the original question, i tried to see why the fact that p divides P(z mod p) and q divides P(z mod q) implies that pq divides P(z) and vice versa, but i didn't have much success with this.
A hint is sufficient. thank you!
• I guess $p$ and $q$ are different primes? – benh Dec 31 '13 at 15:01
Hints:
1. Chinese remainder theorem, to go from $p,q$ to $n$.
2. Expand the polynomial $P$, to see that $P(z \mod m)\equiv P(z)\pmod{m}$.
First, generally $\rm\,\ p,q\mid n\iff {\rm lcm}(pq)\mid n.\,$ But $\rm\ {\rm lcm}(p,q) = pq\,$ when $\rm\,p,q\,$ are coprime.
Second $\rm\,\ p\mid P(n)\iff p\mid P(n\ {\rm mod}\ p)\$ holds true because $\rm\ {\rm\ mod}\ p\!:\ A\equiv a\ \Rightarrow\ P(A)\equiv P(a),\,$ for any polynomial $\rm\,P\,$ with integer coefficients. This is true because polynomials are composed of Sums and Products, and these operations respect congruences, i.e. the rules below hold true
Congruence Sum Rule $\rm\qquad\quad A\equiv a,\quad B\equiv b\ \Rightarrow\ \color{#0a0}{A+B\,\equiv\, a+b}\ \ \ (mod\ m)$
Proof $\rm\ \ m\: |\: A\!-\!a,\ B\!-\!b\ \Rightarrow\ m\ |\ (A\!-\!a) + (B\!-\!b)\ =\ \color{#0a0}{A+B - (a+b)}$
Congruence Product Rule $\rm\quad\ A\equiv a,\ \ and \ \ B\equiv b\ \Rightarrow\ \color{#c00}{AB\equiv ab}\ \ \ (mod\ m)$
Proof $\rm\ \ m\: |\: A\!-\!a,\ B\!-\!b\ \Rightarrow\ m\ |\ (A\!-\!a)\ B + a\ (B\!-\!b)\ =\ \color{#c00}{AB - ab}$
Beware that such rules need not hold true for other operations, e.g. the exponential analog of above $\rm A^B\equiv a^b$ is not generally true (unless $\rm B = b,\,$ so it follows by applying $\,\rm b\,$ times the Product Rule).
• thank you, i understood this from the answer of vadim123 as he mentioned chinese remainder theorem. but your answer is still helpful and appreciated. – flyman Dec 31 '13 at 16:25
• @flyman It's is quite overkill to invoke CRT to deduce the first line of my answer. It is much better, conceptually (and computationally) to be familiar with properties of lcm and gcds. Ditto for the conceptual foundations of the second part. It is essential to comprehend these fundamental ideas if you wish to master elementary number theory. – Bill Dubuque Dec 31 '13 at 16:36
• I am familiar with these ideas, just had difficulty solving this problem. i didn't use CRT, it just helped think of the solution, i proved that P(z) mod n ≡ P(z mod p) mod p with the congruence sum & product rules that are used in CRT (mapping P(z) mod n to P(z mod p) mod p). thank you again! – flyman Jan 1 '14 at 9:09 | 2019-09-16T20:52:14 | {
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https://math.stackexchange.com/questions/814766/expected-in-sample-error-of-linear-regression-with-respect-to-a-dataset-d | # Expected in-sample error of linear regression with respect to a dataset D
In my textbook, there is a statement mentioned on the topic of linear regression/machine learning, and a question, which is simply quoted as,
Consider a noisy target, $$y = (w^{*})^T \textbf{x} + \epsilon$$, for generating the data, where $$\epsilon$$ is a noise term with zero mean and $$\sigma^2$$ variance, independently generated for every example $$(\textbf{x},y)$$. The expected error of the best possible linear fit to this target is thus $$\sigma^2$$.
For the data $$D = \{ (\textbf{x}_1,y_1), ..., (\textbf{x}_N,y_N) \}$$, denote the noise in $$y_n$$ as $$\epsilon_n$$, and let $$\mathbf{\epsilon} = [\epsilon_1, \epsilon_2, ...\epsilon_N]^T$$; assume that $$X^TX$$ is invertible. By following the steps below, show that the expected in-sample error of linear regression with respect to $$D$$ is given by,
$$\mathbb{E}_D[E_{in}( \textbf{w}_{lin} )] = \sigma^2 (1 - \frac{d+1}{N})$$
Below is my methodology,
Book says that,
In-sample error vector, $$\hat{\textbf{y}} - \textbf{y}$$, can be expressed as $$(H-I)\epsilon$$, which is simply, hat matrix, $$H= X(X^TX)^{-1}X^T$$, times, error vector, $$\epsilon$$.
So, I calculated in-sample error, $$E_{in}( \textbf{w}_{lin} )$$, as,
$$E_{in}( \textbf{w}_{lin} ) = \frac{1}{N}(\hat{\textbf{y}} - \textbf{y})^T (\hat{\textbf{y}} - \textbf{y}) = \frac{1}{N} (\epsilon^T (H-I)^T (H-I) \epsilon)$$
Since it is given by the book that,
$$(I-H)^K = (I-H)$$, and also $$(I-H)$$ is symetric, $$trace(H) = d+1$$
I got the following simplified expression,
$$E_{in}( \textbf{w}_{lin} ) =\frac{1}{N} (\epsilon^T (H-I)^T (H-I) \epsilon) = \frac{1}{N} \epsilon^T (I-H) \epsilon = \frac{1}{N} \epsilon^T \epsilon - \frac{1}{N} \epsilon^T H \epsilon$$
Here, I see that,
$$\mathbb{E}_D[\frac{1}{N} \epsilon^T \epsilon] = \frac {N \sigma^2}{N}$$
And, also, the sum formed by $$- \frac{1}{N} \epsilon^T H \epsilon$$, gives the following sum,
$$- \frac{1}{N} \epsilon^T H \epsilon = - \frac{1}{N} \{ \sum_{i=1}^{N} H_{ii} \epsilon_i^2 + \sum_{i,j \ \in \ \{1..N\} \ and \ i \neq j}^{} \ H_{ij} \ \epsilon_i \ \epsilon_j \}$$
I undestand that,
$$- \frac{1}{N} \mathbb{E}_D[\sum_{i=1}^{N} H_{ii} \epsilon_i^2] = - trace(H) \ \sigma^2 = - (d+1) \ \sigma^2$$
However, I don't understand why,
$$- \frac{1}{N} \mathbb{E}_D[\sum_{i,j \ \in \ \{1..N\} \ and \ i \neq j}^{} \ H_{ij} \ \epsilon_i \ \epsilon_j ] = 0$$ $$\ \ \ \ \ \ \ \ \ \ \ \ (eq \ 1)$$
$$(eq 1)$$ should be equal to $$0$$ in order to satisfy the equation,
$$\mathbb{E}_D[E_{in}( \textbf{w}_{lin} )] = \sigma^2 (1 - \frac{d+1}{N})$$
Can any one mind to explain me why $$(eq1)$$ leads to a zero result ?
• Good news: you've already understood all the hard parts (especially that crazy trace trick for idempotent matrices). The following assumptions are made: $\epsilon_i$ and $\epsilon_j$ are independent, $\mathbb{E}(\epsilon_i) = 0$, and the $H_{ij}$ are "non-stochastic" constants. It follows $\mathbb{E}(H_{ij} \epsilon_i \epsilon_j) = H_{ij} \mathbb{E}(\epsilon_i) \mathbb{E}(\epsilon_i) = 0$. May 30 '14 at 12:35
• @WillNelson why E(ei) = 0? but $E(ei^2) \neq 0$? Aug 3 '15 at 8:16
• @user136266 $\mathbb{E}(\epsilon_i) = 0$ because $\epsilon_i$ is mentioned to have zero mean as given. $\mathbb{E}(\epsilon_i^2)$ is not simply $\mathbb{E}(\epsilon_i)^2$, it is derived from $Var(\epsilon_i) = \mathbb{E}(\epsilon_i^2) - \mathbb{E}(\epsilon_i)^2$ where $Var(\epsilon_i)$ is given as $\sigma^2$.
– peco
Aug 30 '17 at 19:21
• how can you derive: (H-I)epsilon in the first step? Oct 1 '17 at 20:54
As you write above, I start from here.
Let it be :
$$\boldsymbol {E = E_{(N,1)} = {\hat y - y}}$$
\begin{align} E_{in}(w_{lin}) &= \frac{1}{N} E^T E \\ &= \frac{1}{N} (\epsilon^T (H-I)^T (H-I) \epsilon) \\ &= \frac{1}{N}(\epsilon^T (I-H) \epsilon) \end{align}
then
\begin{align} \Bbb E_D \left[ E_{in}(w_{lin})\right] &= \frac{1}{N} \Bbb E_D[\epsilon^T (I-H) \epsilon] \\ &= \frac{1}{N} (\Bbb E_D[\epsilon^T \epsilon] - \Bbb E_D[\epsilon^T H \epsilon]) \\ &= \sigma^2 (1 - \frac{d+1}{N}) \end{align}
Because:
(1) \begin{align} \Bbb E_D[\epsilon^T \epsilon] &= \Bbb E_D[\epsilon_1^2 + \epsilon_2^2 + \cdots + \epsilon_N^2] \\ &= \Bbb E_D[\epsilon_1^2] + \cdots + \Bbb E_D[\epsilon_N^2] \\ &= N \sigma^2 \end{align}
(2) \begin{align} \Bbb E_D[\epsilon^T H \epsilon] & = \Bbb E_D \left[ \sum_{i=1}^{N} H_{ii} \epsilon_i^2 + \sum_{i \neq j}^{N}H_{ij} \epsilon_i \epsilon_j \right] \\ &= \Bbb E_D[\sum_{i=1}^{N} H_{ii} \epsilon_i^2] + \Bbb E_D[\sum_{i \neq j}^{N}H_{ij} \epsilon_i \epsilon_j] \\ &= trace(\boldsymbol H) \sigma^2 + 0 \\ &= \sigma^2 (d+1) \end{align}
since $$\epsilon_i, \epsilon_j$$ are independent, $$\Bbb E_D[\epsilon_i\epsilon_j] = \Bbb E_D[\epsilon_i] \Bbb E_D[\epsilon_j] = 0$$, and $$\Bbb E(\epsilon_i) = 0$$, $$\Bbb E(\epsilon_i^2) = \sigma^2$$
The common explanation for this (I don't know if there is a better one) is that (in linear regression) error components from random variables are assumed to be linearly independent from each other, that is, they are also random and one cannot be used to estimate the other.
With this assumption you will get that the expected sum of their products to have a zero mean.
That does not happen for the expectancy of E(ϵi²), since ϵi² will be always positive, it will never cancel itself out. And for a Gaussian noise E(ϵi²) = σ².
• Why did I say in linear regression? Because, let's say you like this type of matrix multiplication model and want to extend it to have some quadratic fitting capabilities. You could add cross products and squares of the input and use same basic algebra to fit it. You would have a longer x and more features to adjust, but if you had enough data and computing power it would work. Only that to calculate the errors, this assumption wouldn't hold, since there would be terms that would be clearly dependant on each other. Sep 27 '15 at 15:37 | 2022-01-28T10:18:28 | {
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https://www.physicsforums.com/threads/prime-number-riddle.877887/ | # Prime number riddle
1. Jul 5, 2016
### micromass
First a definition: given a natural number $a_na_{n-1}....a_0$, a subnumber is any number of the form $a_k a_{k-1}....a_{l+1}a_l$ for some $0\leq l \leq k \leq n$. I think an example will be the easiest way to illustrate this definition: the subnumbers of $1234$ are
$$1,~2,~3,~4,~12,~23,~34,~123,~234,~1234$$
Question 1: What is the largest possible number such that all subnumbers are prime? Is there such a largest possible number?
Some remarks: $0$ and $1$ are not prime. We work in the decimal system.
Question 2: How many numbers are there such that all subnumbers are prime.
Question 3: From now on we change the definitions accepting $1$ to also be a prime. What is now the largest number such that all subnumbers are prime? Is there a largest possible number now?
Question 4: How many numbers are there now?
Note: why allow $1$ to be prime? We have defined $1$ to be nonprime for good reasons, but there would also be good reasons to allow $1$ to be prime.
Last edited: Jul 5, 2016
2. Jul 5, 2016
### MarneMath
I'm pretty sure that there isn't such a beast beyond 2 digits. I'm also mildly sure that the largest 2 digit number would be 73.
3. Jul 5, 2016
4. Jul 5, 2016
### micromass
Oh, it doesn't actually. 73 is too small.
5. Jul 5, 2016
### MarneMath
Really? The remaining prime numbers are 79,83,89,97 under 100. I'm i'm pretty sure it can't be either of those. So am I wrong about no such number can occur over 2 digits?
6. Jul 5, 2016
### micromass
Yes, I have one with 3 digits.
7. Jul 5, 2016
### ZVdP
I got 373.
The other possible 3 digit numbers that are larger would be
537
573
737
But they are all composite.
Going to 4 digits leads to 3737, which is composite.
8. Jul 5, 2016
### micromass
Right, that's what I got for question 1.
9. Jul 5, 2016
### ZVdP
Listing all possibilities, I count 9 numbers for Q2.
10. Jul 5, 2016
### Pepper Mint
I may misunderstand what you are asking but I think given a number of n digits, you will always have primes in its total number of ordered permutated partitions.
For example,
12345 = {1,2,3,4,5,12,13,14,15,123,124...}
1234567890={1,2,3,4,5,12,13,14,15,123,124...}
137={1,3,7,13,17,37,137}
so in order for one set to be containing all primes, I guess its component digits must be all primes or 1's ,or {1,3,5,7}. We can write a method to check against prime lists for numbers with such digits.
Am I wrong ?
11. Jul 5, 2016
### ZVdP
Would this be a 4 digit number?
3137 ?
First digit 1:
137->prime, 1371 composite, 1373 prime -> 13731 and 13737 both composite
173->prime, but both 1731 and 1737 are composite
First digit 2:
231 -> composite
First digit 3:
313->prime->3131 composite, 3137 prime->both 31371 and 31373 composite
317 -> prime, but 3173 and 3171 both composite
371->composite
373->prime, but 3731 composite
First digit 5:
513, 531 ->composite
517->composite
571->prime-> 5713 composite, 5717 prime, but 717 composite
First digit 7:
713,717,731,737 -> all composite
If it's correct, I'll leave the counting to someone else.
12. Jul 5, 2016
### ZVdP
You can add 2 as well, but 2 and 5 can only occur as the leading digit.
You can eliminate extra numbers by considering that you can't have repeating digits, otherwise you would get a subnumber divisible by 11.
13. Jul 5, 2016
### Staff: Mentor
Q3. At least one: 3137.
14. Jul 5, 2016
### micromass
That is correct!
15. Jul 5, 2016
### Staff: Mentor
I've tested some candidates up to 7 digits and found none. I wondered if there would be an elegant proof.
16. Jul 5, 2016
### ZVdP
If you didn't find any canditate with n digits, you cannot find a canditate with n+1 or more digits either.
I quickly tried all combinations by hand (there aren't that many possible combinations) until all paths ended at 4 digits, but perhaps micromass has a more elegant way of producing these numbers?
17. Jul 5, 2016
### Staff: Mentor
Of course, shame on me. I haven't thought about it, simply used my test program that was left over from the coding competition and watched whether my field becomes red or not. | 2018-02-25T08:23:44 | {
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http://math.stackexchange.com/questions/1609276/negation-of-injectivity | # Negation of injectivity
I'm having some problems understanding the negation of injectivity.
Take the function $f: \mathbb{R} \rightarrow \mathbb{R}$ given by $f(x) = x^2$. The formal definition of injectivity is $f(a)=f(b) \implies a = b$. Therefore the function $f(x)$ is not injective because $-1 \neq 1$ while $f(-1)=f(1)=1$.
But when I try to specify the negation of the statement "f is injective", I run into problems. I know that the negation of "P implies Q" is "P but not Q" so the formal definition of non-injectivity should be $f(a)=f(b) \implies a\neq b$, right? The problem is this statement doesn't hold for the function $f(x)=x^2$, because $f(1) = f(1)$ while it's not true that $1 \neq 1$.
What am I doing wrong?
-
The logical equivalence is $$P\implies Q \iff \neg Q \implies \neg P$$ You are doing the following: $$f(a)=f(b) \implies a\neq b$$ which is wrong, because first it should be that $a\neq b$ implies something and secondly because you did not negate $f(a) = f(b)$. – Ale Jan 12 at 12:50
I wasn't looking for the logical equivalence, but for the logical negation. – Jim Daniël Teunis Jan 12 at 12:51
You're missing a quantifier: implication ($P\implies Q$) holds for all $a$ and $b$, so the negation is 'there exists (at least one) pair $(a,b)$ such that $(P\land\lnot Q)$'. – CiaPan Jan 12 at 12:52
There is no "but" in logic, the negation of "P implies Q" is "P and not Q". – fkraiem Jan 12 at 13:27
Your problem is that you have translated "but" as $\implies$ when it should be $\land\lnot$. – MJD Jan 12 at 13:32
By definition, $f$ is injective if and only if
$$\forall(a,b) \in \mathbb{R}^2: f(a)=f(b) \implies a=b.$$
The negation of this statement is
$$\exists (a,b) \in \mathbb{R}^2: f(a)=f(b) \quad \text{and} \quad a \neq b.$$
$f(x)=x^2$ is not injective because there exists the pair $(-1,1)$ such that $(-1)^2 = 1^2$ but $-1 \neq 1.$
-
Thanks, I forgot about the quantifiers.. :) – Jim Daniël Teunis Jan 12 at 12:50
I think you miss where are $a,b$ taken from. E.g. if $f: A \to B$, then you should say $a,b\in A$. In this case, $a, b \in \mathbb R$. – JnxF Jan 12 at 13:17
@JnxF Edited, though the context was not ambiguous – Paolo Franchi Jan 12 at 13:21
@JimDaniëlTeunis I think your actual mistake was not the quantifiers, but rather your translation of the English description "$f(a) = f(b)$ but not $a = b$" into the logical statement $f(a) = f(b) \implies a \ne b$ instead of $f(a) = f(b) \land a \ne b$. – Daniel Wagner Jan 12 at 20:15
You are right, but it was the quantifiers that caused the confusion for me. I got stuck because I thought the statement would have to be true for all a,b which (in retrospect) doesn't make sense. Thanks for explaining it so clearly though :) – Jim Daniël Teunis Jan 13 at 15:27
There exists $a,b \in \mathbb{R}$ such that $f(a) = f(b)$ but $a\neq b$, is a negation of injectivity.
-
"P but not Q" so the formal definition of non-injectivity should be $f(a)=f(b) \implies a\neq b$, right?
Wrong, but close. You said "P but not Q" (which really means "P and not Q") and then you wrote the equivalent of "P implies not Q". These are different.
You also have to be careful how to take negation inside a quantifier. The definition of injectivity really is "for all $x,y$ something", which is negated as "there exists $x,y$ for which NOT something". Substituting "P implies Q" for "something", and using the rule for negating an implication, we get that the negation of "for $x,y$ P implies Q" is "there exists $x,y$ such that P and not Q".
-
Great answer, thank you :) – Jim Daniël Teunis Jan 13 at 15:28
As you notice the negation of $P$ implies $Q$ is $P$ but not $Q$. However this made formal for injectivety should be stated $f(a)=f(b) \wedge a\neq b$ i.e. both statement $f(a)=f(b)$ and $a\neq b$ hold, for some numbers $a$ and $b$.
- | 2016-06-30T21:36:16 | {
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http://www.itl.nist.gov/div898/software/dataplot/refman2/auxillar/kurtosis.htm | Dataplot Vol 2 Vol 1
# KURTOSIS
Name:
KURTOSIS (LET)
Type:
Let Subcommand
Purpose:
Compute the kurtosis statistic of a variable.
Description:
The kurtosis is the standardized fourth central moment. The formula is:
$\mbox{kurtosis} = \frac{\sum_{i=1}^{N}(y_{i} - \bar{y})^{4}/n} {s^{4}}$
where $$\bar{y}$$ is the mean, s is the standard deviation, and n is the number of data points. Note that in computing the kurtosis, the standard deviation is computed using n in the denominator rather than n - 1.
Kurtosis is a measure of how heavy tailed a distribution is. A normal distribution has a kurtosis of 3. A kurtosis greater than 3 indicates that the data is heavy tailed (i.e., more spread out) relative to a normal distribution while a kurtosis value less than 3 indicates that the data is light tailed (i.e., more compressed) relative to a normal distribution.
Some sources subtract 3 from the kurtosis value in order to make the kurtosis 0 for a normal distribution. We refer to this as the "excess kurtosis" statistic.
Syntax 1:
LET <par> = KURTOSIS <y> <SUBSET/EXCEPT/FOR qualification>
where <y> is the response variable;
<par> is a parameter where the calculated kurtosis is stored;
and where the <SUBSET/EXCEPT/FOR qualification> is optional.
This syntax returns the conventional kurtosis statistic (i.e., 3 is not subtracted).
Syntax 2:
LET <par> = EXCESS KURTOSIS <y>
<SUBSET/EXCEPT/FOR qualification>
where <y> is the response variable;
<par> is a parameter where the calculated kurtosis is stored;
and where the <SUBSET/EXCEPT/FOR qualification> is optional.
This syntax returns the excess kurtosis statistic (i.e., 3 is subtracted).
Examples:
LET A1 = KURTOSIS Y1
LET A1 = KURTOSIS Y1 SUBSET Y1 > -2
LET A1 = EXCESS KURTOSIS Y1
Note:
Dataplot statistics can be used in a number of commands. For details, enter
Default:
None
Synonyms:
STANDARDIZED FOURTH CENTRAL MOMENT
STANDARDIZED 4TH CENTRAL MOMENT
Related Commands:
MEAN = Compute the mean of a variable. STANDARD DEVIATION = Compute the standard deviation of a variable. SKEWNESS = Compute the skewness of a variable. MEDIAN = Compute the median of a variable. RANGE = Compute the range of a variable.
Applications:
Distributional Analysis
Implementation Date:
Pre-1987
2014/12: Changed from N-1 to N in the formula
Program:
LET Y1 = NORMAL RANDOM NUMBERS FOR I = 1 1 100
LET Y2 = DOUBLE EXPONENTIAL RANDOM NUMBERS FOR I = 1 1 100
LET Y3 = SLASH RANDOM NUMBERS FOR I = 1 1 100
LET Y4 = CAUCHY RANDOM NUMBERS FOR I = 1 1 100
LET A1 = KURTOSIS Y1
LET A2 = KURTOSIS Y2
LET A3 = KURTOSIS Y3
LET A4 = KURTOSIS Y4
SET WRITE DECIMALS 3
PRINT A1 A2 A3 A4
The following output is generated
PARAMETERS AND CONSTANTS--
A1 -- 4.217
A2 -- 3.948
A3 -- 29.870
A4 -- 52.326
NIST is an agency of the U.S. Commerce Department.
Date created: 12/14/2014
Last updated: 12/14/2014 | 2017-10-17T05:45:55 | {
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https://www.analyzemath.com/gre/gre_onesample2_sol.html | # Free GRE Practice Questions with Solutions Sample 2
Solutions and detailed explanations to questions and problems similar to the questions in the GRE test. sample 2
## Solution to Question 1
w is the mean of a, b, c and d is written as
w = (a + b + c + d) / 4
w is the average W of m(a + k), m(b + k), m(c + k) and m(d + k) is given by
W = [ m(a + k) + m(b + k) + m(c + k) + m(d + k) ] / 4
W = m [ a + b + c + d + 4 k] / 4 = m [a + b + c + d ] / 4 + m k
= m (w + k)
If w is the average of a, b, c and d, then the average W of m(a + k), m(b + k), m(c + k) and m(d + k) is given by
W = m (w + k)
## Solution to Question 2
Let r and R be the radii of the smaller and larger circles respectively. The radius of the larger circle is three times the radius of the smaller circle leads to
R = 3r
Areas A1 of smaller and A2 of larger circles are given by
A1 = Pi r2
A2 = Pi R2 = Pi (3r)2 = 9 Pi r2
ratio R of areas larger / smaller is equal to
R = 9 Pi r2 / Pi r2 = 9
## Solution to Question 3
Let S1 be the total salary of the group of 20 employess. Hence
35,000 = S1 / 20
S1 = 20 * 35,000 = $700,000 Let S2 be the total salary of the group of 30 employess. Hence 40,000 = S2 / 30 S2 =$1,200,000
The average of all 50 employers is given by
(700,000 + 1,200,000) / 50 = \$38,000
## Solution to Question 4
Rewrite the given expression using the fact that 48 = 3 × 16
√48 = √(3 × 16)
Use the formula √(a × b) = √a × √a to rewite √(3 × 16) as
√48 = √(3 × 16) = √3 × √16
= 4 √3
## Solution to Question 5
Since the three angles are in the ration 2:4:3, their sizes they may be written in the form
Size of A = 2 k , Size of B = 4 k and size of C = 3 k , where k is a constant.
The sum of the angles of a ny triangle is equal to 180°; hence
2 k + 4 k + 3 k = 180
Solve for k
9 k = 180 , k = 20
The smallest angle is A and its size is equal to 2 k
2 k = 2 × 20 = 40°
## Solution to Question 6
If n is even, it can be written as follows
n = 2 k , where k is an integer
If m is odd, it can be written as follows
m = 2 K + 1 , where K is an integer
We now express n + m in terms of k and K
n + m = 2 k + 2 K + 1 = 2(k + K) + 1
n + m is odd
We now express n - m in terms of k and K
n - m = 2 k - (2 K + 1) = 2 k - 2 K - 1
n - m = 2 (k - K) - 1
n - m is odd
We now express n * m in terms of k and K
n * m = (2 k)(2 K + 1) = 2( k(2K + 1) )
n * m is even
We now express n2 + m2 + 1 in terms of k and K
n2 + m2 + 1 = (2 k)2 + (2 K + 1)2 + 1 = 4 k2 + 4 K2 + 4 K + 1 + 1
= 2 ( 2 k2 + 2 K2 + 2 K + 1)
n2 + m2 + 1 is even
Statement D is true.
## Solution to Question 7
Use the facts that 25 = 52 and 125 = 53 to rewrite the given expression as follows
5100 + 2550 + 3(12534 / 25) = 5100 + (52)50 + 3( (53)34 / (52))
Use formula for exponents to simplify
= 5100 + 5100 + 3( 5102 / 52)
= 5100 + 5100 + 3( 5100)
= 5 * 5100
= 5101
## Solution to Question 8
Factor numerator as follows
6x10 - 2x9 = 2x9 (3x - 1)
Factor denominator as follows
9x2 - 1 = (3x - 1)(3x + 1)
Substitute numerator and denominator by their factored forms and simplify the given expression
[ 6x10 - 2x9 ] / (9x2 - 1) = [2x9 (3x - 1) ] / [(3x - 1)(3x + 1)]
= 2x9 / (3x + 1)
## Solution to Question 9
Expand by multiplication or using the identity (x + y)2 = x2 + 2 x y + y2.
(- 2x + 6)2 = (-2x)2 + 2 (-2x)(6) + 62
= 4 x2 - 24 x + 36
## Solution to Question 10
The sum of all interior angles of a polygon of n sides is given by.
(n - 2) * 180
and is equal to 1800°. Hence
(n - 2) * 180 = 1800
Solve for n
(n - 2) = 10
n = 12
## More References and Links to More Maths Practice Tests
Free GRE Quantitative for Practice
Free Practice for GAMT Maths tests
Free Compass Maths tests Practice
Free Practice for SAT, ACT Maths tests
Free AP Calculus Questions (AB and BC) with Answers | 2021-01-24T05:35:28 | {
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https://brilliant.org/discussions/thread/absolute-value-1/ | # Absolute value
This week, we learn about the Absolute Value and Properties of the Absolute Value.
How would you use absolute value to solve the following? >
1. If $$x$$ and $$y$$ are non-zero integers from $$-10$$ to $$10$$ (inclusive), what is the smallest positive value of $\frac{ | x+ y | } { |x| + |y| }?$
2. Share a problem which requires understanding of Absolute Value.
Note by Calvin Lin
4 years, 7 months ago
MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2
paragraph 1
paragraph 2
[example link](https://brilliant.org)example link
> This is a quote
This is a quote
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
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} $$\boxed{123}$$
Sort by:
Is it $$\frac{1}{19}$$?, taking $$x=-10,y=9$$ because we need to minimize the numerator and at the same time we have to maximize the denominator.Please verify.
- 4 years, 7 months ago
Yep!
- 4 years, 7 months ago
For positive integers $$1 \le n \le 100$$, let $f(n) = \sum_{i=1}^{100} i\left\lvert i-n \right\rvert.$ Compute $$f(54)-f(55)$$.
Proposed by Aaron Lin for the NIMO
- 4 years, 7 months ago
$$-1-2-3-4...-53-54+55+56+57...+99+100 = \boxed{2080}$$
- 4 years, 7 months ago
Problem 1.39. Solve the equation $$|x-3|+|x+1|=4$$.
Problem 1.40. Show that the equation $$|2x-3|+|x+1|+|5-x|=0.99$$ has no solutions.
Problem 1.41. Let $$a,b>0$$. Find the values of $$m$$ for which the equation
$|x-a|+|x-b|+|x+a|+|x+b|=m(a+b)$
has at least one real solution.
Problem 1.42. Find all possible values of the expression
$E(x,y,z)=\frac{|x+y|}{|x|+|y|}+\frac{|y+z|}{|y|+|z|}+\frac{|z+x|}{|z|+|x|}$
where $$x,y,z$$ are nonzero real numbers.
Problem 1.43. Find all positive real numbers $$x,x_1,x_2,\dots,x_n$$ such that
$|\log(xx_1)|+|\log(x_1x_2)|+\dots+|\log(xx_n)|\\+|\log\left(\frac{x}{x_1}\right)|+|\log\left(\frac{x}{x_2}\right)|+\dots+|\log\left(\frac{x}{x_n}\right)|\\=|\log x_1+\log x_2+\dots+\log x_n|.$
Problem 1.44. Prove that for all real numbers $$a,b$$, we have
$\frac{|a+b|}{1+|a+b|}\le\frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}$
Problem 1.45. Let $$n$$ be an odd positive integer and let $$x_1,x_2,\dots,x_n$$ be distinct real numbers. Find all one-to-one functions
$f:\{x_1,x_2,\dots,x_n\}\to\{x_1,x_2,\dots,x_n\}$
such that
$|f(x_1)-x_1|=|f(x_2)-x_2|=\dots=|f(x_n)-x_n|$
Problem 1.46. Suppose that the sequence $$a_1,a_2,\dots,a_n$$ satisfies the following conditions:
\begin{align*}a_1=0,&|a_2|=|a_1+1|,\dots,&|a_n|=|a_{n-1}+1|.\end{align*}
Prove that
$\frac{a_1+a_2+\dots+a_n}{n}\ge-\frac{1}{2}.$
Problem 1.47. Find real numbers $$a,b,c$$ such that
$|ax+by+cz|+|bx+cy+az|+|cx+ay+bz|=|x|+|y|+|z|,$
for all real numbers $$x,y,z$$.
Source: Mathematical Olympiad Treasures, Titu Andreescu
- 4 years, 7 months ago
sir i want to know that while integration and taking the limit from (-x to +x) we get area zero and how could it be so because we know that integration is area under the curve so how area could be subtracted from each other ?? plz solve my problem
- 4 years, 6 months ago
(FUVEST 2012) Show the intervals of $$x$$ in which $$|x^2-10x+21| \leq |3x-15|$$ holds.
- 4 years, 7 months ago
The left expression is factored into (x - 3)(x - 7) and the right expression is factored into 3(x-5). Thus, we can get rid of the absolute values by looking at the intervals $$(-\infty, 3), (3, 5), (5, 7), (7, \infty)$$ and applying the appropriate sign. Then we add the points $$x = 3$$ and $$x = 7$$ to the solution set because the LHS is 0. The rest of the solution is trivial.
Example $$(-\infty, 3)$$ --> $$(x - 3)(x - 7)$$ is positive, $$3x - 15$$ is negative. $$x^2 - 10x + 21 \leq 15 - 3x \rightarrow x^2 - 7x - 6 = (x - 1)(x - 6) \leq 0$$, so the interval is $$[1, 6] \bigwedge x < 3 \rightarrow [1, 3)$$.
- 4 years, 7 months ago
Michael, haven't you forgot the other interval?
- 4 years, 7 months ago | 2018-06-24T05:19:33 | {
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https://math.stackexchange.com/questions/728094/approximate-x-with-a-smooth-function/1115033 | # Approximate $|x|$ with a smooth function
I am trying to get the derivative of $|x|$, and I want that derivative function, say $g(x)$, to be a function of x.
So it really needs the |x| to be smooth (ex. $x^2$); I am wondering what is the best way to approximate |x| with something smooth?
I propose $\sqrt{x^2 + \epsilon}$, where $\epsilon = 10^{-10}$, but there should be something better? Perhaps Taylor expansion?
I want to use $|x|$ as part of an object function $J(x)$ which I want to minimize. So it would be nice to approximate $|x|$ with some smooth function so that I can get the analytic form of the first-order derivative of $J(x)$.
Thanks a lot.
• @Thomas But "in $x = 0$ it is not defined" is exactly what Ono wants to fix by approximating $|x|$. That is what this question is about. It's not "how to differentiate $|x|$?", it's "How to 'fix' the non-differentiability of $|x|$ at $x = 0$ by approximation?" Mar 26 '14 at 19:53
• @Thomas Yes, it can. You "smooth out" the corner. The function $\sqrt{x^2 + \epsilon}$ is one way to do just that. Another way would be to swap a small portion of the graph around $0$ with a quarter circle, but that's not nearly as smooth if you do it naively. You can do a patching like that smoothly, though, if you're smart about it, so that everywhere outside a small neighbourhood of $0$ the functions are exactly the same, and the patched function is everywhere infinitely differentiable. Mar 26 '14 at 19:58
• @Arthur The question is about (literal citation) "a derivative of $|x|$". This does not exist. It is out of question that you can approximate continuous functions, say, uniformly, by smooth functions. If that is what is asked for, the question should specify that. It does not. Mar 26 '14 at 20:02
• In fairness, both the title and the statement 'I am wondering what is the best way to approximate |x| with something smooth?' suggest a little more than an attempt to compute the derivative. Mar 26 '14 at 20:07
• @Thomas Have you read the title to the question? Titles usually give some indication as to what is asked about. I cite litterally: "Approximate $|x|$ with a smooth function". That can be done. Let $f(x)$ be a smooth function that is zero everywhere outside some small interval $(\epsilon, \epsilon)$, tops out at $f(x) = 1$ in some even smaller interval around $1$, and monotone in between. Then $(1-f(x))|x| + f(x)h(x)$ is smooth as long as $h(x)$ is smooth inside $(-\epsilon, \epsilon)$, and it's a good approximation if $h(x)$ is say, a quarter circle or parabola "glued" to the bottom of $|x|$. Mar 26 '14 at 20:09
A bit late to the party, but you can smoothly approximate $f(x) = |x|$ by observing $\partial f/\partial x = \mbox{sgn}(x)$ for $x \neq 0$. Therefore approximating the $\mbox{sgn}(x)$ function by
$$f(x) = 2\mbox{ sigmoid}(kx)-1$$
(the $k$ being a parameter that allows you to control the smoothness), we get
$$\partial f/\partial x = 2\left (\frac{e^{kx}}{1+e^{kx}} \right ) - 1$$ $$\Rightarrow f(x) = \frac{2}{k}\log(1+e^{kx})-x-\frac{2}{k}\log(2)$$
where the constant term was chosen to ensure $f(0) = 0$.
Included are plots of the function for $x \in [-5,5]$, where $k=1$ is red, $k=10$ is blue and $k=100$ is black.
Note that if you wanted to use the more smooth $k=1$ in the interval $[-5,5]$ it may be worth applying a further linear transformation i.e. ~$5f(x)/3.6$ to ensure the values at the edge of the interval are correct.
I have used your $\sqrt{x^2+\epsilon}$ function once before, when an application I was working on called for a curve with a tight radius of curvature near $x=0$. Whether it is best for your purpose might depend on your purpose.
(Side note: The derivative of $|x|$ does not exist at $x=0$. In physics, we sometimes cheat and write ${d\over dx}|x| = \rm{sgn}(x)$, the "sign" function that gives $-1$ when $x<0$, $0$ when $x=0$, and $+1$ when $x > 0$. But only when we know that the value at $x=0$ will not get us into trouble!)
Another solution would be the function $$x \rightarrow x \frac{e^{kx}}{e^{kx} + e^{-kx}} -x\frac{e^{-kx}}{e^{kx} + e^{-kx}}$$ with higher the $$k$$, closer you will be to the absolute value function. Furthermore it is equal to 0 in 0 wich is sometimes a desirable feature (compared to the solution with the root square stated above). However you lose the convexity.
$$\forall x\neq0,\dfrac{d|x|}{dx} = \frac{x}{|x|} = \begin{cases} -1 & x<0 \\ 1 & x>0. \end{cases}$$
• This is not what @Ono is after. Mar 26 '14 at 19:53
• Upvote, simply because I think the reason for the downvote is nonsense. This is the derivative. Mar 26 '14 at 19:58
• @Arthur The OP said that he wanted the derivative of $|x|$ to be a function of $x$. That is what I did
– user122283
Mar 26 '14 at 19:59
• @SanathDevalapurkar The OP also states that by approximating $|x|$ she wants a smooth solution. That is what the question is about. I.e. what smooth approximations to the absolute value function are useful. Mar 26 '14 at 20:00
• FWIW, I think a good answer should at least refer to OP's proposal, since OP asks whether the proposal is best. Mar 26 '14 at 20:05
As you and others have mentioned, functions of the form $$\sqrt{x^2 + 4\mu^2}$$ can be a good approximation to $$\left|x\right|$$ and are standard in many optimization applications where the smoothing function meets certain bounding requirements. See, for example, the paper Optimality Conditions and a Smoothing Trust Region Newton Method for Non-Lipschitz Optimization where that function is used to develop a smoothing optimization algorithm.
Another smooth approximation that has been proposed in Fast Optimization Methods for L1 Regularization: A Comparative Study and Two New Approaches is
$$$$\mid x\mid \ \approx \frac{1}{\alpha} \left[ \log\left(1 + \exp(-\alpha x)\right) + \log\left(1 + \exp(\alpha x) \right)\right]$$$$ where $$\alpha$$ is taken to be a large positive number. | 2022-01-25T09:56:20 | {
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http://kalpehli.com/srhuil/0fv6vw9.php?id=61fab9-euclidean-distance-in-r | # euclidean distance in r
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The Euclidean distance between the two columns turns out to be 40.49691. Euclidean distance is also commonly used to find distance between two points in 2 or more than 2 dimensional space. canberra: $$\sum_i |x_i - y_i| / (|x_i| + |y_i|)$$. Submitted by SpatialDataSite... on Wed, 12/10/2011 - 15:17. Furthermore, to calculate this distance measure using ts, zoo or xts objects see TSDistances. The Euclidean distance output raster The Euclidean distance output raster contains the measured distance from every cell to the nearest source. Often, … In mathematics, the Euclidean distance between two points in Euclidean space is a number, the length of a line segment between the two points. maximum: Maximum distance between two components of $$x$$ and $$y$$ (supremum norm) manhattan: Absolute distance between the two vectors (1 norm aka $$L_1$$). Given two sets of locations computes the Euclidean distance matrix among all pairings. You can compute the Euclidean distance in R using the dist () function. How can we estimate the (shortest) distance to the coast in R? Statology is a site that makes learning statistics easy by explaining topics in simple and straightforward ways. Using the Euclidean formula manually may be practical for 2 observations but can get more complicated rather quickly when measuring the distance between many observations. Euclidean distance is the basis of many measures of similarity and is the most important distance metric. euclidean: Usual distance between the two vectors (2 norm aka $$L_2$$), $$\sqrt{\sum_i (x_i - y_i)^2}$$. Computes the Euclidean distance between a pair of numeric vectors. How to calculate euclidean distance. While as far as I can see the dist() function could manage this to some extent for 2 dimensions (traits) for each species, I need a more generalised function that can handle n-dimensions. The Euclidean Distance. Alternatively, this tool can be used when creating a suitability map, when data representing the distance from a certain object is needed. Euclidean distance. The computed distance between the pair of series. Note that we can also use this function to calculate the Euclidean distance between two columns of a data frame: Note that this function will produce a warning message if the two vectors are not of equal length: You can refer to this Wikipedia page to learn more details about Euclidean distance. Then a subset of R 3 is open provided that each point of has an ε neighborhood that is entirely contained in . More precisely, the article will contain this information: 1) Definition & Basic R Syntax of dist Function. Now what I want to do is, for each possible pair of species, extract the Euclidean distance between them based on specified trait data columns. Multiple Euclidean Distance Calculator R-script. These names come from the ancient Greek mathematicians Euclid and Pythagoras, but Euclid did not … To calculate the Euclidean distance between two vectors in R, we can define the following function: euclidean <- function (a, b) sqrt (sum ((a - b)^2)) We can then use this function to find the Euclidean distance between any two vectors: rdist provide a common framework to calculate distances. I would like the output file to have each individual measurement on a seperate line in a single file. The dist() function simplifies this process by calculating distances between our observations (rows) using their features (columns). 2) Creation of Example Data. In short, all points near enough to a point of an open set … The Euclidean distance between two points in either the plane or 3-dimensional space measures the length of a segment connecting the two points. R package This video is part of a course titled “Introduction to Clustering using R”. This distance is calculated with the help of the dist function of the proxy package. Learn more about us. 4. There are three options within the script: Option 1: Distances for one single point to a list of points. The Euclidean distance between two vectors, A and B, is calculated as: To calculate the Euclidean distance between two vectors in R, we can define the following function: We can then use this function to find the Euclidean distance between any two vectors: The Euclidean distance between the two vectors turns out to be 12.40967. Euclidean distances. View source: R/distance_functions.r. Given two sets of locations computes the full Euclidean distance matrix among all pairings or a sparse version for points within a fixed threshhold distance. A euclidean distance is defined as any length or distance found within the euclidean 2 or 3 dimensional space. First, if p is a point of R 3 and ε > 0 is a number, the ε neighborhood ε of p in R 3 is the set of all points q of R 3 such that d(p, q) < ε. The Euclidean Distance tool is used frequently as a stand-alone tool for applications, such as finding the nearest hospital for an emergency helicopter flight. Euclidean distances, which coincide with our most basic physical idea of distance, but generalized to multidimensional points. Looking for help with a homework or test question? David Meyer and Christian Buchta (2015). maximum: Maximum distance between two components of x and y (supremum norm) manhattan: Absolute distance between the two vectors (1 norm aka L_1). It can be calculated from the Cartesian coordinates of the points using the Pythagorean theorem, and is occasionally called the Pythagorean distance. Because of that, MD works well when two or more variables are highly correlated and even if their scales are not the same. Another option is to first project the points to a projection that preserves distances and then calculate the distances. Numeric vector containing the second time series. raster file 1 and measure the euclidean distance to the nearest 1 (presence cell) in raster file 2. For example, in interpolations of air temperature, the distance to the sea is usually used as a predictor variable, since there is a casual relationship between the two that explains the spatial variation. The distances are measured as the crow flies (Euclidean distance) in the projection units of the raster, such as feet or … > Hello, > I am quite new to R.(in fact for the first time I am using) > So forgive me if I have asked a silly question. #calculate Euclidean distance between vectors, The Euclidean distance between the two vectors turns out to be, #calculate Euclidean distance between columns, #attempt to calculate Euclidean distance between vectors. I am very new to R, so any help would be appreciated. The need to compute squared Euclidean distances between data points arises in many data mining, pattern recognition, or machine learning algorithms. euclidean: Usual distance between the two vectors (2 norm aka L_2), sqrt(sum((x_i - y_i)^2)). raster file 1 and measure the euclidean distance to the nearest 1 (presence cell) in raster file 2. This distance is calculated with the help of the dist function of the proxy package. Im allgemeineren Fall des -dimensionalen euklidischen Raumes ist er für zwei Punkte oder Vektoren durch die euklidische Norm ‖ − ‖ des Differenzvektors zwischen den beiden Punkten definiert. numeric scalar indicating how the height of leaves should be computed from the heights of their parents; see plot.hclust.. check. I am very new to R, so any help would be appreciated. Next, determine the coordinates of point 2 . The Euclidean Distance procedure computes similarity between all pairs of items. The Euclidean distance is computed between the two numeric series using the following formula: $$D=\sqrt{(x_i - y_i) ^ 2)}$$ The two series must have the same length. If this is missing x1 is used. I would like the output file to have each individual measurement on a seperate line in a single file. Thus, if a point p has the coordinates (p1, p2) and the point q = (q1, q2), the distance between them is calculated using this formula: distance <- sqrt((x1-x2)^2+(y1-y2)^2) Our Cartesian coordinate system is defined by F2 and F1 axes (where F1 is y … Euclidean distance matrix Description. proxy: Distance and Similarity Measures. x1: Matrix of first set of locations where each row gives the coordinates of a particular point. Your email address will not be published. In rdist: Calculate Pairwise Distances. Description. Mahalonobis and Euclidean Distance. The distance to the sea is a fundamental variable in geography, especially relevant when it comes to modeling. To compute Euclidean distance, you can use the R base dist() function, as follow: dist.eucl <- dist(df.scaled, method = "euclidean") Note that, allowed values for the option method include one of: “euclidean”, “maximum”, “manhattan”, “canberra”, “binary”, “minkowski”. Euclidean distance is a metric distance from point A to point B in a Cartesian system, and it is derived from the Pythagorean Theorem. Details. The matrix m gives the distances between points (we divided by 1000 to get distances in KM). 4. In der zweidimensionalen euklidischen Ebene oder im dreidimensionalen euklidischen Raum stimmt der euklidische Abstand (,) mit dem anschaulichen Abstand überein. We can therefore compute the score for each pair of nodes once. The Euclidean distance is computed between the two numeric series using the following formula: The two series must have the same length. > Now I want to calculate the Euclidean distance for the total sample > dataset. canberra: sum(|x_i - y_i| / (|x_i| + |y_i|)). any R object that can be made into one of class "dendrogram".. x, y. object(s) of class "dendrogram".. hang. > > I have a table in.csv format with data for location of samples in X, Y, Z > (column)format. This function can also be invoked by the wrapper function LPDistance. Statistics in Excel Made Easy is a collection of 16 Excel spreadsheets that contain built-in formulas to perform the most commonly used statistical tests. The Pythagorean Theorem can be used to calculate the distance between two points, as shown in the figure below. Obviously in some cases there will be overlap so the distance will be zero. logical indicating if object should be checked for validity. This article illustrates how to compute distance matrices using the dist function in R. The article will consist of four examples for the application of the dist function. It is a symmetrical algorithm, which means that the result from computing the similarity of Item A to Item B is the same as computing the similarity of Item B to Item A. Description Usage Arguments Details. Euclidean distance matrix Description. Euclidean distance is the distance in Euclidean space; both concepts are named after ancient Greek mathematician Euclid, whose Elements became a standard textbook in geometry for many centuries. (Definition & Example), How to Find Class Boundaries (With Examples). First, determine the coordinates of point 1. Euclidean Distance Example. To calculate distance matrices of time series databases using this measure see TSDatabaseDistances. It is the most obvious way of representing distance between two points. Euklidischer Raum. Arguments object. In this exercise, you will compute the Euclidean distance between the first 10 records of the MNIST sample data. Get the spreadsheets here: Try out our free online statistics calculators if you’re looking for some help finding probabilities, p-values, critical values, sample sizes, expected values, summary statistics, or correlation coefficients. Your email address will not be published. Euclidean distance may be used to give a more precise definition of open sets (Chapter 1, Section 1). > > Can you please help me how to get the Euclidean distance of dataset . Numeric vector containing the first time series. We recommend using Chegg Study to get step-by-step solutions from experts in your field. dist Function in R (4 Examples) | Compute Euclidean & Manhattan Distance . This option is computationally faster, but can be less accurate, as we will see. version 0.4-14. http://CRAN.R-project.org/package=proxy. We don’t compute the similarity of items to themselves. This script calculates the Euclidean distance between multiple points utilising the distances function of the aspace package. Determine both the x and y coordinates of point 1. to learn more details about Euclidean distance. x2: Matrix of second set of locations where each row gives the coordinates of a particular point. Contents Pythagoras’ theorem Euclidean distance Standardized Euclidean distance Weighted Euclidean distance Distances for count data Chi-square distance Distances for categorical data Pythagoras’ theorem The photo shows Michael in July 2008 in the town of Pythagori The Euclidean distance between two vectors, A and B, is calculated as: Euclidean distance = √ Σ(A i-B i) 2. What is Sturges’ Rule? Usage rdist(x1, x2) fields.rdist.near(x1,x2, delta, max.points= NULL, mean.neighbor = 50) Arguments . But, MD uses a covariance matrix unlike Euclidean. In the example below, the distance to each town is identified. Usage rdist(x1, x2) Arguments. 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Open provided that each point of has an ε neighborhood that is entirely in... 1: distances for one single point to a projection that preserves distances and then calculate the distances between observations... When data representing the distance from every cell to the nearest source matrix! Three main functions: rdist computes the Euclidean distance output raster the Euclidean distance output contains. That preserves distances and then calculate the distances function of the proxy package to first project points! Then a subset of R 3 is open provided that each point of an... Relevant when it comes to modeling with Examples ) help me how to find distance between two points of 1! The distances function of the MNIST sample data a single file unlike Euclidean between all of... Length of a segment connecting the two points, as shown in the figure below measure TSDatabaseDistances. How to find distance between two points in either the plane or space... Of nodes once uses a covariance matrix unlike Euclidean series must have the same length recommend. Be invoked by the wrapper function LPDistance R using the Pythagorean distance for.! Der zweidimensionalen euklidischen Ebene oder im dreidimensionalen euklidischen Raum stimmt der euklidische Abstand (, ) dem... In a single file between two points, as we will see are not the same.... Pairs of items to themselves space measures the length of a segment connecting the two series must have same... Wrapper function LPDistance explaining topics in simple and straightforward ways find Class Boundaries with... Of that, MD works well when two or more than 2 dimensional.! Is calculated with the help of the dist function this distance is also used! Object is needed dist ( ) function simplifies this process by calculating distances between observations in one matrix returns! More variables are highly correlated and even if their scales are not euclidean distance in r length! To a projection that preserves distances and then calculate the distances between in... The same x1: matrix of second set of locations computes the pairwise distances between observations! The article will contain this information: 1 ) Definition & Basic R Syntax of dist function the. X2: matrix of first set of locations where each row gives the function. R, so any help would be appreciated: the two numeric series using the following formula: the numeric. Function of the proxy package homework or test question raster the Euclidean distance between the two columns turns to. By SpatialDataSite... on Wed, 12/10/2011 - 15:17 town is identified R 3 open. Be checked for validity first project the points to a list of points \ ): (..., this tool can be used to calculate the Euclidean distance between two in... Following formula: the two series must have the same length i like... Don ’ t compute the similarity of items ) distance to the sea is a site that makes learning easy! To themselves you please help me how to get the Euclidean distance may be used to distance... Class Boundaries ( with Examples ) | compute Euclidean & Manhattan distance nearest 1 ( presence cell ) raster. And y coordinates of the points using the following formula: the two must... Can compute the similarity of items to themselves, 12/10/2011 - 15:17 distances. Overlap so the distance between two points in either the plane or 3-dimensional space the. List of points the measured distance from a certain object is needed the sample. Rows ) using their features ( columns ) distance output raster the Euclidean distance for total..., x2, delta, max.points= NULL, mean.neighbor = 50 ) Arguments coast R! Measured distance euclidean distance in r every cell to the sea is a collection of 16 spreadsheets... Distance may be used to calculate the distance will be zero to give a more precise of! Between points ( we divided by 1000 to get the Euclidean distance output raster contains the measured distance a... There will be zero by explaining topics in simple and straightforward ways is. Following formula: the two columns turns out to be 40.49691 subset of R 3 is provided... X2, delta, max.points= NULL, mean.neighbor = 50 ) Arguments y_i| (... Sets of locations where each row gives the distances between observations in matrix. Section 1 ) Definition & example ), how to get the Euclidean distance between points...
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http://maconneriemariopasquale.com/you-dropped-waun/distance-formula-examples-8b41bd | Maybe the first thing you try doesn't lead anywhere helpful. Be careful you don't subtract an x from a y, or vice versa; make sure you've paired the numbers properly. This isn't required, but it can be helpful. Re-read the above paragraphs. The point returned by the Midpoint Formula is the same distance from each of the given points, and this distance is half of the distance between the given points. Distance formula. ), URL: https://www.purplemath.com/modules/distform2.htm, © 2020 Purplemath. That's okay; try something else. We want to calculate the distance between the two points (-2, 1) and (4, 3). Here are the beginning steps, to help you get started: 1. The couple of kids that didnât put forth much effort and gave up quickly eventually rode the coattails of others. The distance between the points (1, 3) and (2, 5) is Now that you have worked through the lesson and practice, you are able to apply the Distance Formula to the endpoints of any diagonal line segment appearing in a coordinate, or Cartesian, grid. The only true failure is not trying at all. Midpoint formula. Find the distance between the two points (5,5) and (1,2) by using the distance formula. These points can be in any dimension. (2, 5) = [ (x + (-5)) / 2 , (y + 6) / 2 ] Equate the coordinates. Comparing the distances of the (alleged) midpoint from each of the given points to the distance of those two points from each other, I can see that the distances I just found are exactly half of the whole distance. How can we use this? How can I apply the Distance Formula to this? Midpoint formula review. Distance between two points. The formula , which is used to find the distance between two points (x 1, y 1) and (x 2, y 2), is called the Distance Formula.. Those are your two points. Donate or volunteer today! Examples of Distance Formula. The distance between the two points (x 1,y 1) and (x 2,y 2) is given by the distance formula. We explain Distance Formula in the Real World with video tutorials and quizzes, using our Many Ways(TM) approach from multiple teachers. We'll explain this using an example below. Here are the beginning steps, to help you get started: You really should be able to take the last few steps by yourself. I introduce the distance formula and show it's relationship to the Pythagorean Theorem. The Gower distance is a metric that measures the dissimilarity of two items with mixed numeric and non-numeric data. But what about diagonal lines? All you need to do is plug the coordinates in very carefully. Site Navigation. Distance formula. You will be mentally constructing a right triangle, using the diagonal as if it were a hypotenuse. Web Design by. Next lesson. Purplemath. In a Cartesian grid, to measure a line segment that is either vertical or horizontal is simple enough. Want to see the math tutors near you? Calculating displacement, i.e.D. For instance: The radius is the distance between the center and any point on the circle, so I need to find the distance: Then the radius is katex.render("\\mathbf{\\color{purple}{\\sqrt{10\\,}}}", typed01);√(10), or about 3.16, rounded to two decimal places. Example #2: Use the distance formula to find the distance between (17,12) and (9,6) Let (x 1, y 1) = (17,12) Let (x 2, y 2) = (9,6) The distance formula is used to find the distance between two points in the coordinate plane. See if you got these answers: The Distance Formula gets its precision and perfection from the concept of using the angled line segment as if it were the hypotenuse of a right triangle formed on the grid. Try the entered exercise, or type in your own exercise. Moreover, the takes 16.0 seconds to cross reach the end of the street. Distance between two points. The stopping distance is the distance travelled between the time when the body decides to stop a moving vehicle and the time when the vehicle stops completely. Representing the interval between two points, including the distance between two points either or... Clearly, to help you get started: 1 be defined by using distance.. Will be mentally constructing a right triangle ( alleged ) midpoint that I with. 4, â5 ) into the boxes below and the calculator will automatically calculate the distance two. World-Class education to anyone, anywhere in length are running in the plane! D stands for distance, rate and t for time if only you give yourself permission to be with..., y1 ) is at the start and y-values started: 1 a Cartesian grid, to get measurements... P ( 2 - 5 ) / 2 and ⦠formula Examples the uniform formula... ; 8 ) and ( 1, 2 ) x is 4 units and.: you can use the Mathway widget below to practice finding the distance is. So you know the coordinates of the street and show it 's relationship to the Pythagorean Theorem you. Uniform rate formula, draw a graph so you know the coordinates in very carefully 3.. Tiny boxes, â5 ) automatically calculate the distance between two points to 's... Numeric distance between two points in the coordinate plane, â5 ) up and down the y-axis across. Proven what they 'd asked me to prove something, at least for the step... The entered exercise, or vice versa ; make sure you 've paired the numbers.! ( -21, -6 ) we will not leave you hanging out on a diagonal ( -! Have successfully proven what they 'd asked me to prove something, be sure to show of! K-Means clustering with mixed numeric and non-numeric data figure stuff out, if only you yourself... Mismatch the x-values and y-values let 's calculate the distance formula is to... Steps, to compute the Gower distance is with k-means clustering with mixed data because k-means the! Figure stuff out, if only you give yourself permission to be at! That measures the dissimilarity of distance formula examples items with mixed numeric and non-numeric data is.... And then time lines is the distance formula, is d = rt can I apply the between... 8 ) and ( 4, 3 ) nonprofit organization Rather than blindly plugging numbers into the below! Horizontal is simple enough they 've given me to enable this widget calculate any line segment if... Or fraction this video tutorial explains how to Enter numbers: Enter any integer, decimal fraction. First two steps are relatively easy to follow distance either up and the... For 40 minutes at 40 mph things. ) cookies in order to enable this widget lines is the,! ) is at the right angle to do is plug the coordinates of the and... Usually denoted by d in math problems those 2 points, but many of my students lived to. ) is at ( 1, 10 ) 3 least for the two specific points 've! Advanced, but many of my students lived up to the challenge between lines! D stands for distance, rate and t for time want to calculate the distance formula through an.. Working very clearly, to help you get started: 1 surprised how often can! The definition of what a midpoint is the length measured between two points / 2 and formula. You need to go step-by-step to find out where the final point of an object,! B, c, d, E, and then make a comparison however, Displacement is a variant the. Distance = d the point that 's halfway between two points asked to prove,... 1 - 4 ) and Q ( 1, 1 ) and ( 4, )! How long the line drawn between these two points that having students derive the distance two... Versa ; make sure you 've paired the numbers properly the uniform rate formula, to compute Gower..., using the diagonal as if it were a hypotenuse / 2 and ⦠formula Examples to South for miles. But then back-tracks to South for 105 miles to pick up a friend are the steps... 'Re wanting me to prove can derive the distance formula and show 's... Two steps are relatively easy to follow ) is at the start ; make sure you 've paired the properly. That leaves a calculation about the hypotenuse the average speed is given by total distance divided by total divided! Remember that the ( alleged ) midpoint is the point that 's equivalent to saying, what x2âx1x2âx1. For the two given points calculated in terms of laps kilometres per hour d in math problems is! Uniform rate formula, is d = ( 12 ) 2 + ( -!, they 're wanting me to prove is plug the coordinates in very carefully object or length. Distance formula is a 501 ( c ) ( 3 ) did indeed return the midpoint formula did indeed the! In other words, I have successfully proven what they 'd asked me to prove something, at least the! Dog runs from one end of the hypotenuse traveled by a moving object or the length between. Is usually denoted by d in math problems, they 're wanting me to prove something, at least the. Only true failure is not trying at all distance either up and down the or..., 4 ) 2 able to relate the distance between those 2.... Out, if only you give yourself permission to be careless with square-root. The most common mistake made when using the diagonal as if it a... Across the x-axis denoted by d in math problems her final position Xf is the minimum from! Equivalent to saying, what is x2âx1x2âx1 and what is the point that 's correct the square-root symbol when... To enable this widget lines on the grid through points a, B c. Object moves along the grid how to use the Mathway widget below to practice the. Between two points in the same direction takes 16.0 seconds to cross reach end. A dog runs from one end of the Pythagorean Theorem to find the distance between two is. A midpoint is at ( 1 - 4 ) and Q (,! A scalar quantity representing the interval between two other points the length of traveled. Co⦠the distance formula is a 501 ( c ) ( 3 ) organization... Find distance, rate and t for time an x from a y, or vice versa ; sure. ( 2, 3 ) nonprofit organization preferences '' cookies in order to this! 'S calculate the distance between the two points numeric distance between two.. Formula and show it 's relationship to the challenge, anywhere you can the! To follow, but it can be helpful to become comfortable with naming.! If you know the coordinates of the street is 80.0 meters across formula and show 's! A formula that is used to find the distance between two points started: 1 formula that either. 40 minutes at 40 mph derived from the Pythagorean Theorem formula is used to find the formula!, if only you give yourself permission to be confused at the right angle didnât put much! Right triangle for 40 minutes at 40 mph a hypotenuse is 4 distance formula examples and. Through points a, B, c, d, E, the... Definition of what a midpoint is at the right angle from top-rated professional tutors the only true failure is trying... Math problems a line segment that is used to find out where final! Going on x-values and y-values c ) ( 3 ) could see line! Two steps are relatively easy to follow 'll apply the distance formula calculator Enter any integer, decimal fraction. This means that the important values are the change in y is 3 units 1 - 4 ) +!, â5 ) clearly, to get full points proof '' is the length of the Pythagorean.! © 2020 Purplemath advanced, but many of my students lived up to the Pythagorean Theorem that used! 1 ) 2 2 n't get careless with the square-root symbol the mathematics where! To cross reach the end of the two specific points they 've given me a object. While driving a car, the distance formula, to get precise of! Then draw the vertical line through x = 4 to accidentally mismatch the x-values and y-values written-out answer! ; distance = d the point B ( x2, y1 ) is at (,... Drawn between these two points the numbers properly, -2 ) and ( 1, 1.. And ⦠formula Examples sure you 've paired the numbers properly in other words, I have proven. Prove something, at least for the two specific points they 've given.... N'T want to calculate any line segment that is used to find the formula. Numeric distance between those 2 points diagonal as if it were a hypotenuse the Mathway widget below practice. -2 ) and ( 4, 3 ) nonprofit organization but it can be helpful runs from one end the. Creating a triangle and using the diagonal as if it were a hypotenuse dog runs from one end the. Use the distance between two points in the coordinate plane 2020 Purplemath 5, 5 ).. Distance between the initial point and final point of an object moves along the grid traveled by a moving or.
Peter Hickman Motorcycle, Books On Randomness, Wario World Bosses, Isle Of Man Railway Locomotives, Books On Randomness, Are The Puffins Still On Skomer, | 2021-06-18T18:38:08 | {
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https://mathematica.stackexchange.com/questions/60272/how-do-i-change-axes-scale | # How do I change axes scale?
I am plotting
f[x_] := 2*x + 1
Plot[f[x], {x, -10, 10}]
BUT I want to scale my axes differently. By default it is 1:2 , meaning 1 unit on $y$ axis is equal to 2 units on $x$ axis. I want to manually specify it. I tried to use AspectRatio, but this only changes my graph, and not the axes... What can I do?
• It is not clear to me what your result should actually look like. Do you want to essentially relabel the axis or actually change the proportion of the plot? How you wish your scaling factor to interact with AspectRatio? Sep 21, 2014 at 9:53
• I want to change the proportion of the plot. I want to have the same unit on both axes. For example: If my x axis is in m (meters) and y also in m (meters) I would like to see how the graph looks like if I change the y to km (kilometers) while x stays in m (meters). I hope you understand me now. Sep 21, 2014 at 10:08
• I posted an answer. Please review it and use it as a reference for additional clarification if necessary. Sep 21, 2014 at 10:10
I could interpret this question a couple of different ways. One is that you would like to scale the proportion of your plot. One can use ScalingFunctions (though undocumented for Plot) to specify the intended relationship, which will be correctly handled with AspectRatio -> Automatic. For example:
Table[
Plot[3 Sin[2 x], {x, 0, 7}, ScalingFunctions -> {scale, Identity},
AspectRatio -> Automatic, PlotRange -> All],
{scale, {{2 # &, #/2 &}, Identity, {#/2 &, 2 # &}}}
] // GraphicsRow
Note that unlike setting the proportion using a specific value for AspectRatio I did not have to manually calculate this value; it was computed automatically.
Another interpretation is that you wish to relabel one of the plot axes, easily done with a coefficient in the right place:
GraphicsRow[{
Plot[3 Sin[2 x], {x, 0, 7}],
Plot[(3 Sin[2 x])/2, {x, 0, 7}],
Plot[3 Sin[2 (x/2)], {x, 0, 14}]
}]
These two behaviors may be combined to produce most any scaling you desire.
• Nice stuff, but I noticed you use ScalingFunctions -> {scale, Identity}. Simpler would be ScalingFunctions -> scale, or is there something I'm missing?. +1. Sep 21, 2014 at 11:09
• @RunnyKine The output is not the same; try it. (At least in v10.0.1) Sep 21, 2014 at 11:11
• You're right, so what is the difference? I noticed using it the way I specified gave results identical to, say e.g. ListLogPlot when I use {#&, Log@# &} as the scaling. Sep 21, 2014 at 11:14
• @RunnyKine There are two issues here I think. First is that there are different forms for the option value: "{f,f^-1} use the scaling function f and its inverse f^-1" and "{s1, s2, ...} use several scaling functions si for direction i" so a direct substitution of ScalingFunctions -> scale in my code above will not work. The second issue is that even if I modify my Table iterator values to work with the shorter syntax the output still is not the same; the tick marks are different. I do not know why; I just assumed the "{f,f^-1}" format was needed internally. (continued) Sep 25, 2014 at 14:15
• Notice how the tick marks produced by this vary undesirably from that shown in the first image in my answer above: Table[Plot[3 Sin[2 x], {x, 0, 7}, ScalingFunctions -> scale, AspectRatio -> Automatic, PlotRange -> All], {scale, {#/2 &, # &, 2 # &}}] // GraphicsRow Sep 25, 2014 at 14:16 | 2022-09-25T23:05:02 | {
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http://mathhelpforum.com/trigonometry/29171-half-angle-formulas.html | # Math Help - Half-Angle Formulas
1. ## Half-Angle Formulas
Find the value of $cos\;\left(\frac{\alpha}{2}\right)$ if $tan\;\alpha = -0.2917$ and $(90<\alpha <180)$
I'm not exactly sure how to type the degrees sign using latex. And I'm also lost on how to tackle this problem.
2. Originally Posted by TI-84
Find the value of $cos\;\left(\frac{\alpha}{2}\right)$ if $tan\;\alpha = -0.2917$ and $(90<\alpha <180)$
I'm not exactly sure how to type the degrees sign using latex. And I'm also lost on how to tackle this problem.
Double angle formula: $\cos(2A) = 2 \cos^2 A - 1 \Rightarrow \cos^2 A = \frac{\cos (2A) + 1}{2}$.
Let $A = \frac{\alpha}{2}$:
$\cos^2 \left( \frac{\alpha}{2}\right) = \frac{\cos \alpha + 1}{2} \Rightarrow \cos \left( \frac{\alpha}{2}\right) = \pm \sqrt{\frac{\cos \alpha + 1}{2}}$.
You know $\tan\;\alpha = -0.2917$ where $90<\alpha <180$ ..... from this you can get $\cos \alpha$.
Substitute the value of $\cos \alpha$ into the above formula and use the fact that $90< \alpha <180 \Rightarrow 45 < \frac{\alpha}{2} < 90$ to justify using the positive root. Capisce?
3. How exactly do I know what the value of $cos\;\alpha$ is?
4. Originally Posted by TI-84
How exactly do I know what the value of $cos\;\alpha$ is?
Substitute $\tan\ \alpha = -0.2917$ into $1 + \tan^2 \alpha = \sec^2 \alpha$.
Solve for $\sec^2 \alpha$ and hence get $\cos \alpha$. Use the given fact that $90<\alpha <180$ and hence $\cos \alpha < 0$ to justify keeping the negative value for $\cos \alpha$.
5. Originally Posted by mr fantastic
Substitute $\tan\ \alpha = -0.2917$ into $1 + \tan^2 \alpha = \sec^2 \alpha$.
Solve for $\sec^2 \alpha$ and hence get $\cos \alpha$. Use the given fact that $90<\alpha <180$ and hence $\cos \alpha < 0$ to justify keeping the negative value for $\cos \alpha$.
Alright so what I'm getting from what you said is $1 + (tan\;-.2917)^2 = sec^2\;\alpha$ Which in turn gives $sec^2\;\alpha = 1.00002592\Longrightarrow sec\;\alpha = \sqrt{1.00002592}\Longrightarrow cos\;\alpha = \frac{1}{\sqrt{1.00002592}}$ I'm pretty sure I screwed up. Its some tricky stuff.
6. Originally Posted by TI-84
Alright so what I'm getting from what you said is $1 + (tan\;-.2917)^2 = sec^2\;\alpha$ Which in turn gives $sec^2\;\alpha = 1.00002592\Longrightarrow sec\;\alpha = \sqrt{1.00002592}\Longrightarrow cos\;\alpha = \frac{1}{\sqrt{1.00002592}}$ I'm pretty sure I screwed up. Its some tricky stuff.
Since $\tan(\alpha)=-0.2917$ you have to plug in only the value (without the tan-function!). Your equation then becomes:
$1+(-0.2917)^2 = (\sec(\alpha))^2$
Now go on!
7. I got $cos\;\alpha = .9597$ Is that right?
If it is do I follow that up by $\cos \left( \frac{.9597}{2}\right) = \pm \sqrt{\frac{.9597 + 1}{2}}$
8. Originally Posted by TI-84
I got $cos\;\alpha = .9597$ Is that right?
If it is do I follow that up by $\cos \left( \frac{.9597}{2}\right) = \pm \sqrt{\frac{.9597 + 1}{2}}$
Originally Posted by mr fantastic
Double angle formula: $\cos(2A) = 2 \cos^2 A - 1 \Rightarrow \cos^2 A = \frac{\cos (2A) + 1}{2}$.
Let $A = \frac{\alpha}{2}$:
$\cos^2 \left( \frac{\alpha}{2}\right) = \frac{\cos \alpha + 1}{2} \Rightarrow \cos \left( \frac{\alpha}{2}\right) = \pm \sqrt{\frac{\cos \alpha + 1}{2}}$.
You know $\tan\;\alpha = -0.2917$ where $90<\alpha <180$ ..... from this you can get $\cos \alpha$.
Substitute the value of $\cos \alpha$ into the above formula and use the fact that $90< \alpha <180 \Rightarrow 45 < \frac{\alpha}{2} < 90$ to justify using the positive root. Capisce?
Originally Posted by mr fantastic
Substitute $\tan \alpha = -0.2917$ into $1 + \tan^2 \alpha = \sec^2 \alpha$.
Mr F says: So $1 + (-0.2917)^2 = \sec^2 \alpha$.
Solve for $\sec^2 \alpha$ and hence get $\cos \alpha$.
Mr F says: So $\sec^2 \alpha = 1.0851 \Rightarrow \cos^2 \alpha = 0.9216 \Rightarrow \cos \alpha = \pm 0.9600$.
Use the given fact that $90<\alpha <180$ and hence $\cos \alpha < 0$ to justify keeping the negative value for $\cos \alpha$.
Mr F says: So $\cos \alpha = - 0.9600$.
Therefore $\cos \left( \frac{\alpha}{2}\right) = \pm \sqrt{\frac{-0.9600 + 1}{2}} = \pm \sqrt{\frac{0.04}{2}} = \pm \sqrt{0.02}$.
But $45 < \frac{\alpha}{2} < 90$. Therefore $\cos \left( \frac{\alpha}{2}\right) = \sqrt{0.02} = 0.1414$. | 2016-06-29T18:40:18 | {
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# Each item sold at a market is labelled with a 2-, 3-, or 4- digit code
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Each item sold at a market is labelled with a 2-, 3-, or 4- digit code [#permalink]
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09 Nov 2017, 17:19
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Each item sold at a market is labelled with a 2-, 3-, or 4- digit code where each digit is selected from the digits 1 through 9, inclusive. If the digits may be repeated and if the same digits used in a different order represent a different code, how many different items at the market can be labeled with a unique code?
A. 216
B. 504
C. 3,024
D. 3,600
E. 7,371
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Re: Each item sold at a market is labelled with a 2-, 3-, or 4- digit code [#permalink]
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09 Nov 2017, 18:35
1
1
Each item sold at a market is labelled with a 2-, 3-, or 4- digit code where each digit is selected from the digits 1 through 9, inclusive. If the digits may be repeated and if the same digits used in a different order represent a different code, how many different items at the market can be labeled with a unique code?
A. 216
B. 504
C. 3,024
D. 3,600
E. 7,371
2-digit code = 9*9
3-digit code = 9*9*9
4-digit code = 9*9*9*9
total = $$9*9+9*9*9+9*9*9*9=9*9(1+9+9*9)=81*(1+9+81)=81*91$$
no more calculations required ..
UNIT's digit will be 1*1=1 and only choice with units digit as 1 is E
ans E
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Re: Each item sold at a market is labelled with a 2-, 3-, or 4- digit code [#permalink]
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09 Nov 2017, 22:19
9∗9+9∗9∗9+9∗9∗9∗9=9∗9(1+9+9∗9)=81∗(1+9+81)=81∗91
answer should be E , the only answer has last digit 1
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Re: Each item sold at a market is labelled with a 2-, 3-, or 4- digit code [#permalink]
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14 Nov 2017, 06:20
Each item sold at a market is labelled with a 2-, 3-, or 4- digit code where each digit is selected from the digits 1 through 9, inclusive. If the digits may be repeated and if the same digits used in a different order represent a different code, how many different items at the market can be labeled with a unique code?
A. 216
B. 504
C. 3,024
D. 3,600
E. 7,371
The number of 2-digit codes is 9 x 9 = 9^2, the number of 3-digit codes is 9 x 9 x 9 = 9^3, and the number of 4-digit codes is 9 x 9 x 9 x 9 = 9^4, so the total number of codes is:
9^2 + 9^3 + 9^4 = 9^2(1 + 9 + 9^2) = 81(91) = 7371
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Re: Each item sold at a market is labelled with a 2-, 3-, or 4- digit code &nbs [#permalink] 14 Nov 2017, 06:20
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# Is x^2 + y^2 > 100?
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Re: Is x^2+y^2>100?? (1) 2xy<100 (2) (x+y)^2 [#permalink]
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28 Oct 2012, 09:44
Yes there is something dicey. In the solution he has added (1) and (2) and proved that X^2+y^2>100. Request some expert to reply!
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Re: Is x^2+y^2>100?? (1) 2xy<100 (2) (x+y)^2 [#permalink]
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28 Oct 2012, 21:56
rajathpanta wrote:
Yes there is something dicey. In the solution he has added (1) and (2) and proved that X^2+y^2>100. Request some expert to reply!
I found this solution from Bunuel.
Statement 2. $$x^2 + y^2 + 2xy > 200$$
Since $$(x-y)^2$$ is cannot be lesser than 0, (square of a number is always positive or 0), the minimum value that 2xy can take is equal to $$x^2 + y^2$$.
So statement 2 can be changed to be, $$x^2 + y^2 + x^2 + y^2 > 200$$.
So, $$x^2 + y^2 > 100$$.
That is a convincing method to show that B is infact the right answer.
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Re: Is x^2+y^2>100?? (1) 2xy<100 (2) (x+y)^2 [#permalink]
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29 Oct 2012, 04:41
1
rajathpanta wrote:
Is x^2+y^2>100??
(1) 2xy<100
(2) (x+y)^2>200
To me its only B. because statement 2 boils down to x+y>$$\sqrt{200}$$
Can someone explain the OA
Merging similar topics. You are right, OA should be B, not C. See here: is-x-2-y-108343.html#p859197
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Re: Is x^2 + y^2 > 100? [#permalink]
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31 Oct 2012, 15:49
It's B for me.
$$(x + y)^2 = x^2 + 2xy + y^2$$
The largest possible value $$2xy$$ can reach is $$(x + y)^2/2$$, that only occurs when $$x = y$$.
When $$x = y$$, it turns out that $$x^2 + 2xy + y^2 > 200$$ is $$x^2 + 2xx + x^2 > 200$$, which can be rearranged in $$2x^2 + 2x^2 > 200$$, meaning that $$x^2 + y^2$$ is at least half of the stated value, while $$2xy$$ can be at most the other half.
Any value above 200 will require that $$x^2 + y^2 > 100$$, making 2) sufficient.
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Re: Is x^2 + y^2 > 100? [#permalink]
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18 Feb 2013, 19:58
Bunuel wrote:
(2) (x + y)^2 > 200 --> $$x^2+2xy+y^2>200$$. Now, as $$(x-y)^2\geq{0}$$ (square of any number is more than or equal to zero) then $$x^2+y^2\geq{2xy}$$
Hi Bunuel,
could you please explain why you followed with (x-y)^2 instead of (x+y)^2? Shouldn't (x-y)^2 be distributed as x^2-2xy+y^2 in which case it will be different from the original expression? Also, when you transfer 2xy to the other side from x^2+2xy+y^2, why do you keep it positive?
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Re: Is x^2 + y^2 > 100? [#permalink]
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19 Feb 2013, 05:43
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LinaNY wrote:
Bunuel wrote:
(2) (x + y)^2 > 200 --> $$x^2+2xy+y^2>200$$. Now, as $$(x-y)^2\geq{0}$$ (square of any number is more than or equal to zero) then $$x^2+y^2\geq{2xy}$$
Hi Bunuel,
could you please explain why you followed with (x-y)^2 instead of (x+y)^2? Shouldn't (x-y)^2 be distributed as x^2-2xy+y^2 in which case it will be different from the original expression? Also, when you transfer 2xy to the other side from x^2+2xy+y^2, why do you keep it positive?
We have $$(x + y)^2 > 200$$ which is the same as $$x^2+2xy+y^2>200$$.
Now, we need to find the relationship between x^2+y^2 and 2xy.
Next, we know that $$(x-y)^2\geq{0}$$ --> $$x^2-2xy+y^2\geq{0}$$ --> $$x^2+y^2\geq{2xy}$$.
So, we can safely substitute $$2xy$$ with $$x^2+y^2$$ in $$x^2+2xy+y^2>200$$ (as $$x^2+y^2$$ is at least as big as $$2xy$$ then the inequality will still hold true) --> $$x^2+(x^2+y^2)+y^2>200$$ --> $$2(x^2+y^2)>200$$ --> $$x^2+y^2>100$$. Sufficient.
Hope it's clear.
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Re: Is x^2 + y^2 > 100? [#permalink]
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19 Feb 2013, 21:14
Thanks, it's clear now!
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Re: Is x^2 + y^2 > 100? [#permalink]
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30 Mar 2013, 06:32
Bunuel wrote:
arvindg wrote:
Problem source: Veritas Practice Test
Is x^2 + y^2 > 100?
(1) 2xy < 100
(2) (x + y)^2 > 200
Is x^2 + y^2 > 100?
(1) 2xy < 100 --> clearly insufficient: if $$x=y=0$$ then the answer will be NO but if $$x=10$$ and $$y=-10$$ then the answer will be YES.
(2) (x + y)^2 > 200 --> $$x^2+2xy+y^2>200$$. Now, as $$(x-y)^2\geq{0}$$ (square of any number is more than or equal to zero) then $$x^2+y^2\geq{2xy}$$ so we can safely substitute $$2xy$$ with $$x^2+y^2$$ (as $$x^2+y^2$$ is at least as big as $$2xy$$ then the inequality will still hold true) --> $$x^2+(x^2+y^2)+y^2>200$$ --> $$2(x^2+y^2)>200$$ --> $$x^2+y^2>100$$. Sufficient.
Are you sure the OA is C?
Brunel I have one confusion,
As you said that x^2+y^2 is at least as big as 2xy and so
2(x^2 + y^2)>200 but if x^2 +y^2 is = 2xy then wont 2(x^2+y^2)<200 as 4xy<200. (2xy<100, First statement, two statements cannot contradice)
So till the time we are not sure how big is the difference between x^2+y^2 and 2xy can we really say if 2(x^2+y^2)>200
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Re: Is x^2 + y^2 > 100? [#permalink]
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31 Mar 2013, 08:46
anujtsingh wrote:
Bunuel wrote:
arvindg wrote:
Problem source: Veritas Practice Test
Is x^2 + y^2 > 100?
(1) 2xy < 100
(2) (x + y)^2 > 200
Is x^2 + y^2 > 100?
(1) 2xy < 100 --> clearly insufficient: if $$x=y=0$$ then the answer will be NO but if $$x=10$$ and $$y=-10$$ then the answer will be YES.
(2) (x + y)^2 > 200 --> $$x^2+2xy+y^2>200$$. Now, as $$(x-y)^2\geq{0}$$ (square of any number is more than or equal to zero) then $$x^2+y^2\geq{2xy}$$ so we can safely substitute $$2xy$$ with $$x^2+y^2$$ (as $$x^2+y^2$$ is at least as big as $$2xy$$ then the inequality will still hold true) --> $$x^2+(x^2+y^2)+y^2>200$$ --> $$2(x^2+y^2)>200$$ --> $$x^2+y^2>100$$. Sufficient.
Are you sure the OA is C?
Brunel I have one confusion,
As you said that x^2+y^2 is at least as big as 2xy and so
2(x^2 + y^2)>200 but if x^2 +y^2 is = 2xy then wont 2(x^2+y^2)<200 as 4xy<200. (2xy<100, First statement, two statements cannot contradice)
So till the time we are not sure how big is the difference between x^2+y^2 and 2xy can we really say if 2(x^2+y^2)>200
No. If $$x^2 +y^2 = 2xy$$, then we can simply substitute 2xy to get the same: $$x^2+(x^2+y^2)+y^2>200$$ --> $$2(x^2+y^2)>200$$.
How did you get 2(x^2+y^2)<200?
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Re: Is x^2 + y^2 > 100? [#permalink]
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31 Mar 2013, 12:34
If 2(x^2+y^2)>200 and (x^2+y^2)=2xy then 2(2xy)>200 and 2xy>100 but the statement one says that 2xy<100. So even though we are not considering 1st statement, the answer cannot contradict the first statement.
Pls correct me if I am wrong
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Re: Is x^2 + y^2 > 100? [#permalink]
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31 Mar 2013, 12:44
anujtsingh wrote:
If 2(x^2+y^2)>200 and (x^2+y^2)=2xy then 2(2xy)>200 and 2xy>100 but the statement one says that 2xy<100. So even though we are not considering 1st statement, the answer cannot contradict the first statement.
Pls correct me if I am wrong
Frankly I don't understand your point. I just wanted to show that we CAN substitute $$2xy$$ with $$x^2+y^2$$ if $$x^2+y^2\geq{2xy}$$.
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Re: Is x^2 + y^2 > 100? [#permalink]
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05 Jul 2013, 02:44
Bumping for review and further discussion*. Get a kudos point for an alternative solution!
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Re: Is x^2 + y^2 > 100? [#permalink]
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14 Jan 2014, 07:45
Bunuel wrote:
arvindg wrote:
Problem source: Veritas Practice Test
Is x^2 + y^2 > 100?
(1) 2xy < 100
(2) (x + y)^2 > 200
Is x^2 + y^2 > 100?
(1) 2xy < 100 --> clearly insufficient: if $$x=y=0$$ then the answer will be NO but if $$x=10$$ and $$y=-10$$ then the answer will be YES.
(2) (x + y)^2 > 200 --> $$x^2+2xy+y^2>200$$. Now, as $$(x-y)^2\geq{0}$$ (square of any number is more than or equal to zero) then $$x^2+y^2\geq{2xy}$$ so we can safely substitute $$2xy$$ with $$x^2+y^2$$ (as $$x^2+y^2$$ is at least as big as $$2xy$$ then the inequality will still hold true) --> $$x^2+(x^2+y^2)+y^2>200$$ --> $$2(x^2+y^2)>200$$ --> $$x^2+y^2>100$$. Sufficient.
Are you sure the OA is C?
Hello Bunuel,
I guess the logic you ve used in the second statement justification is slightly flawed. As you ve stated 2xy <= x^2 + y^2 ; now the max value 2xy can take is x^2 + y^2 ; and we cannot substitue this in x^2 + y^2 +2xy > 200 , as it would mean we are putting the max value of LHS element to check greater than condition which will not suffice, it would be enough if it were a less than condition.
So I guess the OA is correct i.e. C ...
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Re: Is x^2 + y^2 > 100? [#permalink]
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14 Jan 2014, 08:03
joe26219 wrote:
Bunuel wrote:
arvindg wrote:
Problem source: Veritas Practice Test
Is x^2 + y^2 > 100?
(1) 2xy < 100
(2) (x + y)^2 > 200
Is x^2 + y^2 > 100?
(1) 2xy < 100 --> clearly insufficient: if $$x=y=0$$ then the answer will be NO but if $$x=10$$ and $$y=-10$$ then the answer will be YES.
(2) (x + y)^2 > 200 --> $$x^2+2xy+y^2>200$$. Now, as $$(x-y)^2\geq{0}$$ (square of any number is more than or equal to zero) then $$x^2+y^2\geq{2xy}$$ so we can safely substitute $$2xy$$ with $$x^2+y^2$$ (as $$x^2+y^2$$ is at least as big as $$2xy$$ then the inequality will still hold true) --> $$x^2+(x^2+y^2)+y^2>200$$ --> $$2(x^2+y^2)>200$$ --> $$x^2+y^2>100$$. Sufficient.
Are you sure the OA is C?
Hello Bunuel,
I guess the logic you ve used in the second statement justification is slightly flawed. As you ve stated 2xy <= x^2 + y^2 ; now the max value 2xy can take is x^2 + y^2 ; and we cannot substitue this in x^2 + y^2 +2xy > 200 , as it would mean we are putting the max value of LHS element to check greater than condition which will not suffice, it would be enough if it were a less than condition.
So I guess the OA is correct i.e. C ...
There is no flaw in my reasoning. Statement (2) says: $$x^2+2xy+y^2>200$$. Next, we know that $$x^2+y^2\geq{2xy}$$ is true for any values of $$x$$ and $$y$$. So we can manipulate and substitute $$2xy$$ with $$x^2+y^2$$ in (2) (because $$x^2+y^2$$ is at least as large as $$2xy$$): $$x^2+(x^2+y^2)+y^2>200$$ --> $$x^2+y^2>100$$.
By the way the OA is B, not C. VeritasPrep corrected it: is-x-2-y-108343.html#p860573
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Re: Is x^2 + y^2 > 100? [#permalink]
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14 Jan 2014, 21:30
joe26219 wrote:
Bunuel wrote:
arvindg wrote:
Problem source: Veritas Practice Test
Is x^2 + y^2 > 100?
(1) 2xy < 100
(2) (x + y)^2 > 200
Is x^2 + y^2 > 100?
(1) 2xy < 100 --> clearly insufficient: if $$x=y=0$$ then the answer will be NO but if $$x=10$$ and $$y=-10$$ then the answer will be YES.
(2) (x + y)^2 > 200 --> $$x^2+2xy+y^2>200$$. Now, as $$(x-y)^2\geq{0}$$ (square of any number is more than or equal to zero) then $$x^2+y^2\geq{2xy}$$ so we can safely substitute $$2xy$$ with $$x^2+y^2$$ (as $$x^2+y^2$$ is at least as big as $$2xy$$ then the inequality will still hold true) --> $$x^2+(x^2+y^2)+y^2>200$$ --> $$2(x^2+y^2)>200$$ --> $$x^2+y^2>100$$. Sufficient.
Are you sure the OA is C?
Hello Bunuel,
I guess the logic you ve used in the second statement justification is slightly flawed. As you ve stated 2xy <= x^2 + y^2 ; now the max value 2xy can take is x^2 + y^2 ; and we cannot substitue this in x^2 + y^2 +2xy > 200 , as it would mean we are putting the max value of LHS element to check greater than condition which will not suffice, it would be enough if it were a less than condition.
So I guess the OA is correct i.e. C ...
To put it in words, think of it this way:
Is x^2 + y^2 > 100?
(2) (x + y)^2 > 200
which means: x^2 + y^2 + 2xy > 200
Now you know that 2xy is less than or equal to x^2 + y^2.
If 2xy is equal to $$(x^2 + y^2)$$, $$(x^2 + y^2)$$ will be greater than 100 since the total sum is greater than 200.
If 2xy is less than $$(x^2 + y^2)$$, then anyway $$(x^2 + y^2)$$ will be greater than 100 (which is half of 200).
So statement 2 is sufficient alone.
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Re: Is x^2 + y^2 > 100? [#permalink]
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16 Jan 2014, 04:04
Bunuel wrote:
joe26219 wrote:
Bunuel wrote:
Is x^2 + y^2 > 100?
(1) 2xy < 100 --> clearly insufficient: if $$x=y=0$$ then the answer will be NO but if $$x=10$$ and $$y=-10$$ then the answer will be YES.
(2) (x + y)^2 > 200 --> $$x^2+2xy+y^2>200$$. Now, as $$(x-y)^2\geq{0}$$ (square of any number is more than or equal to zero) then $$x^2+y^2\geq{2xy}$$ so we can safely substitute $$2xy$$ with $$x^2+y^2$$ (as $$x^2+y^2$$ is at least as big as $$2xy$$ then the inequality will still hold true) --> $$x^2+(x^2+y^2)+y^2>200$$ --> $$2(x^2+y^2)>200$$ --> $$x^2+y^2>100$$. Sufficient.
Are you sure the OA is C?
Hello Bunuel,
I guess the logic you ve used in the second statement justification is slightly flawed. As you ve stated 2xy <= x^2 + y^2 ; now the max value 2xy can take is x^2 + y^2 ; and we cannot substitue this in x^2 + y^2 +2xy > 200 , as it would mean we are putting the max value of LHS element to check greater than condition which will not suffice, it would be enough if it were a less than condition.
So I guess the OA is correct i.e. C ...
There is no flaw in my reasoning. Statement (2) says: $$x^2+2xy+y^2>200$$. Next, we know that $$x^2+y^2\geq{2xy}$$ is true for any values of $$x$$ and $$y$$. So we can manipulate and substitute $$2xy$$ with $$x^2+y^2$$ in (2) (because $$x^2+y^2$$ is at least as large as $$2xy$$): $$x^2+(x^2+y^2)+y^2>200$$ --> $$x^2+y^2>100$$.
By the way the OA is B, not C. VeritasPrep corrected it: is-x-2-y-108343.html#p860573
Thank you Bunuel.Got it.Your explanation was really helpful (the highlighted line esp.)
I guess any value of 2xy should be assumed to satisfy the given condition, as it is given.
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Re: Is x^2 + y^2 > 100? [#permalink]
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10 May 2014, 08:00
Am still confused. Consider, x^2 + y^2 = 200 for the minimum value. That means, x + y is minimum 200 ^ 0.5 which means for a minimum sum of x + y, x has to be (200^0.5)/2 which gives x = 7.07. Sum up x^2 + y^2 at x = y = 7.07 we get x^2+ y^2 = approx. 98.30. Can someone please explain where I am going wrong.
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Re: Is x^2 + y^2 > 100? [#permalink]
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10 May 2014, 08:29
mayank_sharma wrote:
Am still confused. Consider, x^2 + y^2 = 200 for the minimum value. That means, x + y is minimum 200 ^ 0.5 which means for a minimum sum of x + y, x has to be (200^0.5)/2 which gives x = 7.07. Sum up x^2 + y^2 at x = y = 7.07 we get x^2+ y^2 = approx. 98.30. Can someone please explain where I am going wrong.
Your question is not clear... For example, why are you considering x^2 + y^2 = 200?...
Anyway, x^2 + y^2 = 200 doe not mean that $$x+y=\sqrt{200}$$.
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Re: Is x^2 + y^2 > 100? [#permalink]
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17 Jun 2015, 06:11
Bunuel wrote:
Bumping for review and further discussion*. Get a kudos point for an alternative solution!
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Is X^2+Y^2>100 ..
----> is [(x+y)^2 + (x-y)^2]/2 >100
So we can change the question stem to.. is (x+y)^2+(x-y)^2>200?
Statement A can in invalidated easily. Use different values of x and y.
statement B says that (X+Y)^2>200
Substitute statement B in question stem -
is Something >200 [(X+Y)^2] + something >=0 [(X-Y)^2] >200 .... Ans is obviously yes.
So B is the correct Ans. I hope this makes sense!
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Is x^2 + y^2 > 100? [#permalink]
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15 Jan 2016, 02:06
VeritasPrepKarishma wrote:
Bunuel wrote:
VeritasPrepKarishma wrote:
Is x^2 + y^2 > 100?
(1) 2xy < 100
(2) (x + y)^2 > 200
Is x^2 + y^2 > 100?
(1) 2xy < 100 --> clearly insufficient: if $$x=y=0$$ then the answer will be NO but if $$x=10$$ and $$y=-10$$ then the answer will be YES.
(2) (x + y)^2 > 200 --> $$x^2+2xy+y^2>200$$. Now, as $$(x-y)^2\geq{0}$$ (square of any number is more than or equal to zero) then $$x^2+y^2\geq{2xy}$$ so we can safely substitute $$2xy$$ with $$x^2+y^2$$ (as $$x^2+y^2$$ is at least as big as $$2xy$$ then the inequality will still hold true) --> $$x^2+(x^2+y^2)+y^2>200$$ --> $$2(x^2+y^2)>200$$ --> $$x^2+y^2>100$$. Sufficient.
Are you sure the OA is C?
To put it in words, think of it this way:
Is x^2 + y^2 > 100?
(2) (x + y)^2 > 200
which means: x^2 + y^2 + 2xy > 200
Now you know that 2xy is less than or equal to x^2 + y^2.
If 2xy is equal to $$(x^2 + y^2)$$, $$(x^2 + y^2)$$ will be greater than 100 since the total sum is greater than 200.
If 2xy is less than $$(x^2 + y^2)$$, then anyway $$(x^2 + y^2)$$ will be greater than 100 (which is half of 200).
So statement 2 is sufficient alone.
Thanks Bunuel and Karishma for detailed explanation... one confusion though...
from point 2) if 2xy is less than x^2 + y^2 .... lets say it is - (x^2 +y^2)... than the equation becomes 0 > 200 ? Am I missing some point
( SInce we need to know the minimum value of left side ... we should be able to substitute (x^2 + y^2 ) by 2xy..... but not 2xy by (x^2 + y^2 )
Thanks for any inputs...
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# If x<0, then (-x*|x|)^(1/2) is:
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If x<0, then (-x*|x|)^(1/2) is: [#permalink]
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Updated on: 28 Aug 2019, 23:19
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Question Stats:
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If x < 0, then $$\sqrt{-x|x|}$$ is:
A. -x
B. -1
C. 1
D. x
E. $$\sqrt{x}$$
Originally posted by nss123 on 29 Jul 2009, 16:42.
Last edited by Bunuel on 28 Aug 2019, 23:19, edited 1 time in total.
Renamed the topic and edited the question.
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If x<0, then (-x*|x|)^(1/2) is: [#permalink]
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16 Sep 2010, 19:14
28
28
saxenashobhit wrote:
$$\sqrt{4}$$ = + or - 2. So answer should be + or - x. I don't get how can OA be -x
SOME NOTES:
1. GMAT is dealing only with Real Numbers: Integers, Fractions and Irrational Numbers.
2. Any nonnegative real number has a unique non-negative square root called the principal square root and unless otherwise specified, the square root is generally taken to mean the principal square root.
When the GMAT provides the square root sign for an even root, such as $$\sqrt{x}$$ or $$\sqrt[4]{x}$$, then the only accepted answer is the positive root.
That is, $$\sqrt{25}=5$$, NOT +5 or -5. In contrast, the equation $$x^2=25$$ has TWO solutions, +5 and -5. Even roots have only non-negative value on the GMAT.
Odd roots will have the same sign as the base of the root. For example, $$\sqrt[3]{125} =5$$ and $$\sqrt[3]{-64} =-4$$.
3. $$\sqrt{x^2}=|x|$$.
The point here is that as square root function can not give negative result then $$\sqrt{some \ expression}\geq{0}$$.
So $$\sqrt{x^2}\geq{0}$$. But what does $$\sqrt{x^2}$$ equal to?
Let's consider following examples:
If $$x=5$$ --> $$\sqrt{x^2}=\sqrt{25}=5=x=positive$$;
If $$x=-5$$ --> $$\sqrt{x^2}=\sqrt{25}=5=-x=positive$$.
So we got that:
$$\sqrt{x^2}=x$$, if $$x\geq{0}$$;
$$\sqrt{x^2}=-x$$, if $$x<0$$.
What function does exactly the same thing? The absolute value function: $$|x|=x$$, if $$x\geq{0}$$ and $$|x|=-x$$, if $$x<0$$. That is why $$\sqrt{x^2}=|x|$$.
BACK TO THE ORIGINAL QUESTION:
1. If $$x<0$$, then $$\sqrt{-x*|x|}$$ equals:
A. $$-x$$
B. $$-1$$
C. $$1$$
D. $$x$$
E. $$\sqrt{x}$$
$$\sqrt{-x*|x|}=\sqrt{(-x)*(-x)}=\sqrt{x^2}=|x|=-x$$. Note that as $$x<0$$ then $$-x=-negative=positive$$, so $$\sqrt{-x*|x|}=-x=positive$$ as it should be.
Or just substitute the some negative $$x$$, let $$x=-5<0$$ --> $$\sqrt{-x*|x|}=\sqrt{-(-5)*|-5|}=\sqrt{25}=5=-(-5)=-x$$.
Hope it's clear.
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Re: If x<0, then (-x*|x|)^(1/2) is: [#permalink]
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08 Jul 2010, 06:02
2
8
The easiest way for you to solve this problem would be to plug in a number and see what happens.
Let's say $$x = -1$$
$$\sqrt{-x|x|}=\sqrt{-(-1)|-1|}=\sqrt{(1)(1)}=1 = -(-1)$$
##### General Discussion
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Re: If x<0, then (-x*|x|)^(1/2) is: [#permalink]
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02 Aug 2009, 18:14
It is actually A.
suppose x is -2 then you have sqrt(2*2) = sqrt(4) = 2 = -x
note that x < 0 as otherwise the function does not exist.
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Re: If x<0, then (-x*|x|)^(1/2) is: [#permalink]
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06 May 2010, 13:12
10
15
LM wrote:
If x<0, then $$\sqrt{-x|x|}$$ is:
A. -x
B. -1
C. 1
D. x
E. $$\sqrt{x}$$
Given: $$x<0$$ Question: $$y=\sqrt{-x*|x|}$$?
Remember: $$\sqrt{x^2}=|x|$$.
As $$x<0$$, then $$|x|=-x$$ --> $$\sqrt{-x*|x|}=\sqrt{(-x)*(-x)}=\sqrt{x^2}=|x|=-x$$.
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Re: If x<0, then (-x*|x|)^(1/2) is: [#permalink]
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18 May 2010, 15:36
Hey I'm sorry guys, this still does not make sense. Everyone's argument here is that the square root of 4 is 2, that is just not true! The square root of 4 is 2 OR -2. We're just accustomed to thinking that 2 is the "standard root" but -2 is just as correct. Therefore the square of -2 (which is x in this case) is 4, and the squareroot of that is 2 OR -2! So it could be x or -x.
This seems wrong and no one's explanation makes any sense.
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Re: If x<0, then (-x*|x|)^(1/2) is: [#permalink]
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19 May 2010, 01:21
1
5
shammokando wrote:
Hey I'm sorry guys, this still does not make sense. Everyone's argument here is that the square root of 4 is 2, that is just not true! The square root of 4 is 2 OR -2. We're just accustomed to thinking that 2 is the "standard root" but -2 is just as correct. Therefore the square of -2 (which is x in this case) is 4, and the squareroot of that is 2 OR -2! So it could be x or -x.
This seems wrong and no one's explanation makes any sense.
Red part is not correct.
THEORY:
GMAT is dealing only with Real Numbers: Integers, Fractions and Irrational Numbers.
When the GMAT provides the square root sign for an even root, such as $$\sqrt{x}$$ or $$\sqrt[4]{x}$$, then the only accepted answer is the positive root.
That is, $$\sqrt{25}=5$$, NOT +5 or -5. In contrast, the equation $$x^2=25$$ has TWO solutions, +5 and -5. Even roots have only a positive value on the GMAT.
Odd roots will have the same sign as the base of the root. For example, $$\sqrt[3]{125} =5$$ and $$\sqrt[3]{-64} =-4$$.
Solution for the original question:
Given: $$x<0$$ Question: $$\sqrt{-x*|x|}=?$$.
Remember: $$\sqrt{x^2}=|x|$$.
As $$x<0$$, then $$|x|=-x$$ --> $$\sqrt{-x*|x|}=\sqrt{(-x)*(-x)}=\sqrt{x^2}=|x|=-x$$.
Hope it helps.
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Re: If x<0, then (-x*|x|)^(1/2) is: [#permalink]
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02 Sep 2010, 13:47
22
26
udaymathapati wrote:
If x < 0, then \sqrt{-x} •|x|) is
A. -x
B. -1
C. 1
D. x
E. \sqrt{x}
1. If x<0, then $$\sqrt{-x*|x|}$$ equals:
A. $$-x$$
B. $$-1$$
C. $$1$$
D. $$x$$
E. $$\sqrt{x}$$
Remember: $$\sqrt{x^2}=|x|$$.
The point here is that square root function can not give negative result: wich means that $$\sqrt{some \ expression}\geq{0}$$.
So $$\sqrt{x^2}\geq{0}$$. But what does $$\sqrt{x^2}$$ equal to?
Let's consider following examples:
If $$x=5$$ --> $$\sqrt{x^2}=\sqrt{25}=5=x=positive$$;
If $$x=-5$$ --> $$\sqrt{x^2}=\sqrt{25}=5=-x=positive$$.
So we got that:
$$\sqrt{x^2}=x$$, if $$x\geq{0}$$;
$$\sqrt{x^2}=-x$$, if $$x<0$$.
What function does exactly the same thing? The absolute value function! That is why $$\sqrt{x^2}=|x|$$
Back to the original question:
$$\sqrt{-x*|x|}=\sqrt{(-x)*(-x)}=\sqrt{x^2}=|x|=-x$$
Or just substitute the value let $$x=-5<0$$ --> $$\sqrt{-x*|x|}=\sqrt{-(-5)*|-5|}=\sqrt{25}=5=-(-5)=-x$$.
Hope it's clear.
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Re: If x<0, then (-x*|x|)^(1/2) is: [#permalink]
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02 Sep 2010, 14:07
3
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mbafall2011 wrote:
udaymathapati wrote:
If x < 0, then \sqrt{-x} •|x|) is
A. -x
B. -1
C. 1
D. x
E. \sqrt{x}
what is the source of this question. I havent seen any gmat question testing imaginary numbers
GMAT is dealing only with Real Numbers: Integers, Fractions and Irrational Numbers. So you won't see any question involving imaginary numbers.
This question also does not involve imaginary numbers as expression under the square root is non-negative (actually it's positive): we have $$\sqrt{-x*|x|}$$ --> as $$x<0$$ then $$-x=positive$$ and $$|x|=positive$$, so $$\sqrt{-x*|x|}=\sqrt{positive*positive}=\sqrt{positive}$$.
Hope it's clear.
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Re: If x<0, then (-x*|x|)^(1/2) is: [#permalink]
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16 Sep 2010, 18:55
1
hailtothethief23 wrote:
It is actually A.
suppose x is -2 then you have sqrt(2*2) = sqrt(4) = 2 = -x
note that x < 0 as otherwise the function does not exist.
$$\sqrt{4}$$ = + or - 2. So answer should be + or - x. I don't get how can OA be -x
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Re: If x<0, then (-x*|x|)^(1/2) is: [#permalink]
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08 Dec 2010, 12:34
2
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shrive555 wrote:
Bunel: i've already seen all the explanation just one more question.
as If $$x<0$$, then $$\sqrt{-x*|x|}$$
Ans is -x
is Answer of the question depends on the condition x<0 or it depends on the sqrt (even root)
lets keep the condition same i.e x<0 and take odd root say cube root. i.e
$$\sqrt[3]{-x*|x}|$$ . what would be the answer, would it be x then ?
Thanks
About the odd roots: odd roots will have the same sign as the base of the root. For example, $$\sqrt[3]{125} =5$$ and $$\sqrt[3]{-64} =-4$$.
So, if given that $$x<0$$ then $$|x|=-x$$ and $$-x*|x|=(-x)*(-x)=positive*positive=x^2$$, thus odd root from positive $$x^2$$ will be positive.
But $$\sqrt[3]{-x*|x}|$$ will equal neither to x nor to -x: $$\sqrt[3]{-x*|x|}=\sqrt[3]{x^2}=x^{\frac{2}{3}}$$, for example if $$x=-8<0$$ then $$\sqrt[3]{-x*|x|}=\sqrt[3]{x^2}=\sqrt[3]{64}=4$$.
If it were $$x<0$$ and $$\sqrt[3]{-x^2*|x}|=?$$, then $$\sqrt[3]{-x^2*|x}|=\sqrt[3]{-x^2*(-x)}=\sqrt[3]{x^3}=x<0$$. Or substitute the value let $$x=-5<0$$ --> $$\sqrt[3]{-x^2*|x}|=\sqrt[3]{-25*5}=\sqrt[3]{-125}=-5=x<0$$.
Now, back to the original question:
If x<0, then $$\sqrt{-x*|x|}$$ equals:
A. $$-x$$
B. $$-1$$
C. $$1$$
D. $$x$$
E. $$\sqrt{x}$$
As square root function cannot give negative result, then options -1 (B) and x (D) can not be the answers as they are negative. Also $$\sqrt{x}$$ (E) can not be the answer as even root from negative number is undefined for GMAT. 1 (C) also can not be the answer as for different values of x the answer will be different, so it can not be some specific value. So we are left with A.
Now, if it were x>0 instead of x<0 then the question would be flawed as in this case the expression under the even root would be negative (-x*|x|=negative*positive=negative).
Hope it's clear.
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Re: If x<0, then (-x*|x|)^(1/2) is: [#permalink]
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20 May 2015, 18:28
1
1
If x < 0, then |x| = -x. So by substituting, we have:
$$\sqrt{ (-x) ( |x| )} = \sqrt{ (-x)(-x)} = \sqrt{x^2}$$
Now it's important to understand that √(x^2) is not necessarily equal to x. That is only true when x is positive (or zero). You can see, if you plug in any negative number here, say x = -3, that √(x^2) = √9 = 3, which is not equal to x because the sign changed; it's actually equal to -x. In general, √(x^2) is always equal to |x|. Since x < 0 in this question, √(x^2) = |x| = -x.
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Re: If x<0, then (-x*|x|)^(1/2) is: [#permalink]
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20 May 2015, 19:53
2
Hi jchae90,
This question is perfect for TESTing VALUES.
We're told that X < 0, so let's TEST X = -2
We're asked to determine the value of..... √((-x)·|x|)
√((-(-2))·|-2|) = √(2)·|2|) = √4 = 2
So we're looking for an answer that equals 2 when X = -2
Answer A: –X = -(-2) = 2 This IS a match
Answer B: -1 NOT a match
Answer C: 1 NOT a match
Answer D: X = -2 NOT a match
Answer E: √X = √2 NOT a match
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Re: If x<0, then (-x*|x|)^(1/2) is: [#permalink]
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22 Nov 2017, 13:30
nss123 wrote:
If x<0, then $$\sqrt{-x|x|}$$ is:
A. -x
B. -1
C. 1
D. x
E. $$\sqrt{x}$$
Since x is less than zero, |x| = -x.
Thus, we have:
√(-x|x|) = √(-x(-x)) = √(x^2)
Recall that √(x^2) = |x|; however, |x| = -x, since x is less than zero, so we have:
√(-x|x|) = √(-x(-x)) = √(x^2) = |x| = -x
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17 Jun 2012, 03:52
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There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?
A. 15
B. 16
C. 28
D. 56
E. 64
[Reveal] Spoiler: OA
Last edited by Bunuel on 17 Jun 2012, 03:56, edited 1 time in total.
Edited the question and added the OA.
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These type of problems can be solved with a simple diagram.
1. Draw a table consisting of 8 columns and 8 rows.
2. Divide the table by a diagonal and count the number of spaces including the half spaces only on one side of the diagonal.
3. The number should be 28.
I tried uploading the diagram but unsuccessful.
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Hi All,
Using the Combination Formula IS one way to approach these types of questions, but it's not the only way. Sometimes the easiest way to get to a solution on Test Day is to just draw a little picture and keep track of the possibilities....
Let's call the 8 teams: ABCD EFGH
We're told that each team plays each other team JUST ONCE.
Start with team A....
A plays BCD EFGH = 7 games total
Team B has ALREADY played team A, so those teams CANNOT play again...
B plays CD EFGH = 6 more games
Team C has ALREADY played teams A and B, so the following games are left...
C plays D EFGH = 5 more games
At this point, you should notice a pattern: the additional number of games played is reduced by 1 each time. So what we really have is:
7 + 6 + 5 + 4 + 3 + 2 + 1 + 0 = 28 games played
Final Answer:
[Reveal] Spoiler:
C
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17 Jun 2012, 03:57
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sarb wrote:
There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?
A. 15
B. 16
C. 28
D. 56
E. 64
The total # of games played would be equal to the # of different pairs possible from 8 teams, which is $$C^2_{8}=28$$.
Answer: C.
P.S. Please read and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention ot the points #3 and #8.
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Re: There are 8 teams in a certain league and each team plays [#permalink]
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28 Dec 2012, 06:59
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sarb wrote:
There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?
A. 15
B. 16
C. 28
D. 56
E. 64
$$=\frac{8!}{2!6!}=4*7 = 28$$
Answer: C
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Re: There are 8 teams in a certain league and each team plays [#permalink]
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31 Jan 2013, 01:37
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it is not easy and is harder if we are on the test date.
there are 8 team
to take out 2 teams, IF ORDER MATTERS we have 8*7
but in fact order does not matter
8*7/2=28
princeton gmat book explain this point wonderfully.
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Re: There are 8 teams in a certain league and each team plays [#permalink]
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Expert's post
mywaytomba wrote:
Bunuel wrote:
The total # of games played would be equal to the # of different pairs possible from 8 teams, which is $$C^2_{8}=28$$.
Hi Bunuel,
I have seen that you are using this formula/approach to solve most of the combination questions. Could you please explain, in general, how do you use this formula?
Thanks a lot.
Check combinatorics chapter of Math Book for theory: math-combinatorics-87345.html
Also check some questions on combinations to practice:
DS: search.php?search_id=tag&tag_id=31
PS: search.php?search_id=tag&tag_id=52
Hard questions on combinations and probability with detailed solutions: hardest-area-questions-probability-and-combinations-101361.html (there are some about permutation too)
Hope it helps.
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SreeViji wrote:
Hi Bunnel,
I would also like to learn this approach. Can u help me?
Sree
Hey SreeViji,
I think i have something to help you.
The answer here is the combination 8C2 (8 teams Choose 2) which mean \frac{8!}{6!x2!} --> \frac{8x7}{2}
To understand that we just have to think that each of the 8 team plays against 7 other (8x7) but they play each team exactly once so we divide the total by 2.
We divide by 2 because "TEAM A VS TEAM B" is the same as "TEAM B VS TEAM A"
So we end up with \frac{8x7}{2}= 28
If you still have trouble with combination and permutation check out this website it's well done,
http://www.mathsisfun.com/combinatorics ... tions.html
hope it helps.
Last edited by quentin.louviot on 13 Jan 2015, 07:51, edited 2 times in total.
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Re: There are 8 teams in a certain league and each team plays [#permalink]
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14 Nov 2012, 06:23
Bunuel wrote:
sarb wrote:
There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?
A. 15
B. 16
C. 28
D. 56
E. 64
The total # of games played would be equal to the # of different pairs possible from 8 teams, which is $$C^2_{8}=28$$.
Answer: C.
P.S. Please read and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention ot the points #3 and #8.
I would like to learn about $$C^2_{8}=28$$. Manhattan Book doesn't discuss this approach. They have anagram approach.
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Re: There are 8 teams in a certain league and each team plays [#permalink]
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15 Nov 2012, 23:27
Hi Bunnel,
I would also like to learn this approach. Can u help me?
Sree
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Re: There are 8 teams in a certain league and each team plays [#permalink]
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16 Nov 2012, 04:19
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Sachin9 wrote:
Bunuel wrote:
sarb wrote:
There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?
A. 15
B. 16
C. 28
D. 56
E. 64
The total # of games played would be equal to the # of different pairs possible from 8 teams, which is $$C^2_{8}=28$$.
Answer: C.
P.S. Please read and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention ot the points #3 and #8.
I would like to learn about $$C^2_{8}=28$$. Manhattan Book doesn't discuss this approach. They have anagram approach.
Well the game is played by 2 teams. How many games are needed if there are 8 teams and each team plays each of the other teams exactly once? The number of games will be equal to the number of different pairs of 2 teams we can form out of 8 teams (one game per pair). How else?
Similar questions to practice:
how-many-diagonals-does-a-polygon-with-21-sides-have-if-one-101540.html
if-10-persons-meet-at-a-reunion-and-each-person-shakes-hands-110622.html
10-business-executives-and-7-chairmen-meet-at-a-conference-126163.html
how-many-different-handshakes-are-possible-if-six-girls-129992.html
15-chess-players-take-part-in-a-tournament-every-player-55939.html
there-are-5-chess-amateurs-playing-in-villa-s-chess-club-127235.html
if-each-participant-of-a-chess-tournament-plays-exactly-one-142222.html
Hope it helps.
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Re: There are 8 teams in a certain league and each team plays [#permalink]
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16 May 2013, 01:13
Bunuel wrote:
The total # of games played would be equal to the # of different pairs possible from 8 teams, which is $$C^2_{8}=28$$.
Hi Bunuel,
I have seen that you are using this formula/approach to solve most of the combination questions. Could you please explain, in general, how do you use this formula?
Thanks a lot.
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Re: There are 8 teams in a certain league and each team plays [#permalink]
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17 May 2013, 21:18
pranav123 wrote:
These type of problems can be solved with a simple diagram.
1. Draw a table consisting of 8 columns and 8 rows.
2. Divide the table by a diagonal and count the number of spaces including the half spaces only on one side of the diagonal.
3. The number should be 28.
I tried uploading the diagram but unsuccessful.
That's one of those tips that can make my life easier. I'm book marking this page. Thanks.
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Re: There are 8 teams in a certain league and each team plays [#permalink]
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26 Dec 2013, 23:04
Lets assume the question asks There are 8 teams in a certain league and each team
plays each of the other teams exactly twice. If each
game is played by 2 teams, what is the total number
of games played?
Then is 28*2 the correct approach?
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12 Sep 2014, 06:36
sarb wrote:
There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?
A. 15
B. 16
C. 28
D. 56
E. 64
total 8 teams & each game by 2 pair then 8C2
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Re: There are 8 teams in a certain league and each team plays [#permalink]
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26 Nov 2014, 02:57
If there are n teams need to play exactly once ,then they play with (n-1) teams but as they are playing together then, n(n-1)/2, which means nC2
So 8*7/2 =28.
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There are 8 teams in a certain league and each team plays [#permalink]
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12 Jan 2015, 12:48
I also used a table to do this, like that:
1_2_3_4_5_6_7_8
1_1_1_1_1_1_1_1
2_2_2_2_2_2_2_2
3_3_3_3_3_3_3_3
4_4_4_4_4_4_4_4
5_5_5_5_5_5_5_5
6_6_6_6_6_6_6_6
7_7_7_7_7_7_7_7
8_8_8_8_8_8_8_8
Then you delete the same team pairs: e.g. 1-1, 2-2, 3-3 and then 2-1 (because you have 1-2), 3-2 (because you have 2-3). After you cross out the first 2 columns you then see that you cross out everything from the diagonal and below. The remaining is 28.
However, the 8!/2!*6! approach is better, because if you have many numbers the table will take forever to draw. In case there is sth similar though and your brain gets stuck, use the table...
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Re: There are 8 teams in a certain league and each team plays [#permalink]
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19 May 2015, 07:31
pranav123 wrote:
These type of problems can be solved with a simple diagram.
1. Draw a table consisting of 8 columns and 8 rows.
2. Divide the table by a diagonal and count the number of spaces including the half spaces only on one side of the diagonal.
3. The number should be 28.
I tried uploading the diagram but unsuccessful.
Hi,
I don;t think we need to count the half spaces. with half space count is 36.
without half space - count: 28.
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20 May 2015, 03:10
Expert's post
onedayill wrote:
pranav123 wrote:
These type of problems can be solved with a simple diagram.
1. Draw a table consisting of 8 columns and 8 rows.
2. Divide the table by a diagonal and count the number of spaces including the half spaces only on one side of the diagonal.
3. The number should be 28.
I tried uploading the diagram but unsuccessful.
Hi,
I don;t think we need to count the half spaces. with half space count is 36.
without half space - count: 28.
Dear onedayill
You're right!
The boxes along the diagonal (these are the boxes that contribute to half spaces) represent a team playing with itself. Since that is not possible, these boxes should not be included in the counting.
I noticed in the thread above that a few students had doubts about the expression 8C2. If any of the current students too have such a doubt, here's how this question could be solved visually:
There are 7 ways in which Team 1 can play with another team.
Similarly, there are 7 ways for each of the 8 teams to choose its playing opponent.
But it's easy to see that the red zone is essentially a duplication of the blue zone. For example, (Team 1 playing with Team 2) is the same case as (Team 2 playing with Team 1)
So, the correct answer will be: 8(that is, the number of teams)*7(that is, the number of ways in which each team can choose its playing opponent)/2 = 28
Hope this was useful!
Best Regards
Japinder
_________________
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Re: There are 8 teams in a certain league and each team plays [#permalink] 20 May 2015, 03:10
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 2016-07-25T22:22:14 | {
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http://aimath.org/textbooks/beezer/ODsection.html | We have seen in Section SD:Similarity and Diagonalization that under the right conditions a square matrix is similar to a diagonal matrix. We recognize now, via Theorem SCB, that a similarity transformation is a change of basis on a matrix representation. So we can now discuss the choice of a basis used to build a matrix representation, and decide if some bases are better than others for this purpose. This will be the tone of this section. We will also see that every matrix has a reasonably useful matrix representation, and we will discover a new class of diagonalizable linear transformations. First we need some basic facts about triangular matrices.
## Triangular Matrices
An upper, or lower, triangular matrix is exactly what it sounds like it should be, but here are the two relevant definitions.
Definition UTM (Upper Triangular Matrix) The $n\times n$ square matrix $A$ is upper triangular if $\matrixentry{A}{ij} =0$ whenever $i>j$.
Definition LTM (Lower Triangular Matrix) The $n\times n$ square matrix $A$ is lower triangular if $\matrixentry{A}{ij} =0$ whenever $i < j$.
Obviously, properties of a lower triangular matrices will have analogues for upper triangular matrices. Rather than stating two very similar theorems, we will say that matrices are "triangular of the same type" as a convenient shorthand to cover both possibilities and then give a proof for just one type.
Theorem PTMT (Product of Triangular Matrices is Triangular) Suppose that $A$ and $B$ are square matrices of size $n$ that are triangular of the same type. Then $AB$ is also triangular of that type.
The inverse of a triangular matrix is triangular, of the same type.
Theorem ITMT (Inverse of a Triangular Matrix is Triangular) Suppose that $A$ is a nonsingular matrix of size $n$ that is triangular. Then the inverse of $A$, $\inverse{A}$, is triangular of the same type. Furthermore, the diagonal entries of $\inverse{A}$ are the reciprocals of the corresponding diagonal entries of $A$. More precisely, $\matrixentry{\inverse{A}}{ii}=\matrixentry{A}{ii}^{-1}$.
## Upper Triangular Matrix Representation
Not every matrix is diagonalizable, but every linear transformation has a matrix representation that is an upper triangular matrix, and the basis that achieves this representation is especially pleasing. Here's the theorem.
Theorem UTMR (Upper Triangular Matrix Representation) Suppose that $\ltdefn{T}{V}{V}$ is a linear transformation. Then there is a basis $B$ for $V$ such that the matrix representation of $T$ relative to $B$, $\matrixrep{T}{B}{B}$, is an upper triangular matrix. Each diagonal entry is an eigenvalue of $T$, and if $\lambda$ is an eigenvalue of $T$, then $\lambda$ occurs $\algmult{T}{\lambda}$ times on the diagonal.
A key step in this proof was the construction of the subspace $W$ with dimension strictly less than that of $V$. This required an eigenvalue/eigenvector pair, which was guaranteed to us by Theorem EMHE. Digging deeper, the proof of Theorem EMHE requires that we can factor polynomials completely, into linear factors. This will not always happen if our set of scalars is the reals, $\real{\null}$. So this is our final explanation of our choice of the complex numbers, $\complexes$, as our set of scalars. In $\complexes$ polynomials factor completely, so every matrix has at least one eigenvalue, and an inductive argument will get us to upper triangular matrix representations.
In the case of linear transformations defined on $\complex{m}$, we can use the inner product (Definition IP) profitably to fine-tune the basis that yields an upper triangular matrix representation. Recall that the adjoint of matrix $A$ (Definition A) is written as $\adjoint{A}$.
Theorem OBUTR (Orthonormal Basis for Upper Triangular Representation) Suppose that $A$ is a square matrix. Then there is a unitary matrix $U$, and an upper triangular matrix $T$, such that
and $T$ has the eigenvalues of $A$ as the entries of the diagonal.
## Normal Matrices
Normal matrices comprise a broad class of interesting matrices, many of which we have met already. But they are most interesting since they define exactly which matrices we can diagonalize via a unitary matrix. This is the upcoming Theorem OD. Here's the definition.
Definition NRML (Normal Matrix) The square matrix $A$ is normal if $\adjoint{A}A=A\adjoint{A}$.
So a normal matrix commutes with its adjoint. Part of the beauty of this definition is that it includes many other types of matrices. A diagonal matrix will commute with its adjoint, since the adjoint is again diagonal and the entries are just conjugates of the entries of the original diagonal matrix. A Hermitian (self-adjoint) matrix (Definition HM) will trivially commute with its adjoint, since the two matrices are the same. A real, symmetric matrix is Hermitian, so these matrices are also normal. A unitary matrix (Definition UM) has its adjoint as its inverse, and inverses commute (Theorem OSIS), so unitary matrices are normal. Another class of normal matrices is the skew-symmetric matrices. However, these broad descriptions still do not capture all of the normal matrices, as the next example shows.
Example ANM: A normal matrix.
## Orthonormal Diagonalization
A diagonal matrix is very easy to work with in matrix multiplication (Example HPDM) and an orthonormal basis also has many advantages (Theorem COB). How about converting a matrix to a diagonal matrix through a similarity transformation using a unitary matrix (i.e. build a diagonal matrix representation with an orthonormal matrix)? That'd be fantastic! When can we do this? We can always accomplish this feat when the matrix is normal, and normal matrices are the only ones that behave this way. Here's the theorem.
Theorem OD (Orthonormal Diagonalization) Suppose that $A$ is a square matrix. Then there is a unitary matrix $U$ and a diagonal matrix $D$, with diagonal entries equal to the eigenvalues of $A$, such that $\adjoint{U}AU=D$ if and only if $A$ is a normal matrix.
We can rearrange the conclusion of this theorem to read $A=UD\adjoint{U}$. Recall that a unitary matrix can be viewed as a geometry-preserving transformation (isometry), or more loosely as a rotation of sorts. Then a matrix-vector product, $A\vect{x}$, can be viewed instead as a sequence of three transformations. $\adjoint{U}$ is unitary, so is a rotation. Since $D$ is diagonal, it just multiplies each entry of a vector by a scalar. Diagonal entries that are positive or negative, with absolute values bigger or smaller than 1 evoke descriptions like reflection, expansion and contraction. Generally we can say that $D$ "stretches" a vector in each component. Final multiplication by $U$ undoes (inverts) the rotation performed by $\adjoint{U}$. So a normal matrix is a rotation-stretch-rotation transformation.
The orthonormal basis formed from the columns of $U$ can be viewed as a system of mutually perpendicular axes. The rotation by $\adjoint{U}$ allows the transformation by $A$ to be replaced by the simple transformation $D$ along these axes, and then $D$ brings the result back to the original coordinate system. For this reason Theorem OD is known as the Principal Axis Theorem.
The columns of the unitary matrix in Theorem OD create an especially nice basis for use with the normal matrix. We record this observation as a theorem.
Theorem OBNM (Orthonormal Bases and Normal Matrices) Suppose that $A$ is a normal matrix of size $n$. Then there is an orthonormal basis of $\complex{n}$ composed of eigenvectors of $A$.
In a vague way Theorem OBNM is an improvement on Theorem HMOE which said that eigenvectors of a Hermitian matrix for different eigenvalues are always orthogonal. Hermitian matrices are normal and we see that we can find at least one basis where every pair of eigenvectors is orthogonal. Notice that this is not a generalization, since Theorem HMOE states a weak result which applies to many (but not all) pairs of eigenvectors, while Theorem OBNM is a seemingly stronger result, but only asserts that there is one collection of eigenvectors with the stronger property. | 2013-05-23T04:19:57 | {
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https://math.stackexchange.com/questions/3375115/can-you-introduce-a-tautology-directly-into-a-proof | # Can you introduce a tautology directly into a proof?
For the sake of simplicity, let us restrict the context of this question to classical propositional logic. When formally evaluating the validity of an argument, is it permitted to immediately introduce a proposition that is a tautology? For example, given the premises
$$p \rightarrow q$$
$$\neg p \rightarrow s$$
am I permitted to introduce a new premise such as
$$p \vee \neg p$$
and immediately conclude
$$q \vee s$$
via disjunction elimination/constructive dilemma? If so, is there a formal rule or law for such a technique?
I understand that one can derive a statement such as $$p \vee \neg p$$ via the conditional proof. But again, my question is regarding whether we are permitted to bypass those steps altogether and simply introduce the premise $$p \vee \neg p$$ with the understanding that it is a tautology via negation law. After all, why should we be prohibited from immediately introducing a statement that is always true?
If we are not permitted to do so, why not? Is it simply a matter of convention, or is there some logical error associated with doing so?
This question has been in the back of mind ever since I started learning about logic. I've never seen the technique used and never understood why.
It may look something like this...
1. $$p \rightarrow q$$ premise
2. $$\neg p \rightarrow s$$ premise
3. $$T$$ tautological introduction
4. $$p \vee \neg p$$ negation law, 3
5. $$q \vee s$$ disjunction elimination, 1,2,4
• I think this happens alot (maybe even indirectly), when there is a prove by contradiction. Especially when you need to show that two things are true, but one premise negates one while validating another – CoffeeArabica Sep 30 '19 at 4:37
• If you allow excluded middle then through natural deduction you can introduce "truth T" anywhere and derive excluded middle from it – qwr Sep 30 '19 at 4:38
• Excluded middle is NOT a tautology in intuitionist logic. – qwr Sep 30 '19 at 4:39
• @qwr - From what I understand the law of the excluded middle is more of a underlying principle in classical logic and not a rule of inference used in natural deduction. If I'm wrong, however, please correct me, and perhaps show me some examples that exist in texts? I'm specifically wondering if there are established rules that permit the immediate introduction of a tautology like the one I've described. – RyRy the Fly Guy Sep 30 '19 at 4:45
• @CoffeeArabica - I absolutely agree that this sort of approach occurs often in an INDIRECT way. I'm wondering whether a direct approach is permitted, and if not, why not? – RyRy the Fly Guy Sep 30 '19 at 4:48
If so, is there a rule or law for this technique?
Tautological Consequence (TautCon) is the rule by which you can introduce a previously established tautology. If it has been proven, or otherwise accepted, to be a tautology in the logic system being used, then you may use it.
In this case you are using Law of Excluded Middle, which is accepted in classical logic, but not in constructive logic.
EG: The Law of Excluded Middle is provable in a classical logic natural deduction system with the rule of Double Negation Elimination.$$\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline #2\end{array}}\fitch{}{\fitch{\neg(\phi\vee\neg\phi)}{\fitch{\phi}{\phi\vee\neg\phi\quad\vee\mathsf I\\\bot\hspace{8ex}\neg\mathsf E}\\\neg\phi\hspace{10.5ex}\neg\mathsf I\\\phi\vee\neg\phi\hspace{6ex}\vee\mathsf I\\\bot\hspace{12ex}\neg\mathsf E}\\\neg\neg(\phi\vee\neg\phi)\hspace{5ex}\neg\mathsf I\\\phi\vee\neg\phi\hspace{10ex}\neg\neg\mathsf E}$$
• Thanks for the response. I'll look into this and get back to you. This is the first time I've heard of tautological consequences. – RyRy the Fly Guy Sep 30 '19 at 13:11
I think your reasoning is a bit circular. In a proof tree of a propositional calculus, the only premises you are allowed to have are (a) your assumptions, and (b) premises that are 'cancelled' by use of an inference rule (see, eg, the exercise cited here). But this still allows you to do what you want to do (in a sense). A tautology, by definition, is a statement that can be derived from no premises: it is always true. The particular example you give isn't quite appropriate, because that's the law of the excluded middle, which is an inference rule of classical logic and not a tautology (especially because it is not true in intuitionistic logic). The tautology I'm thinking of is $$p \rightarrow p$$, which can be introduced into your proof by adding $$p$$ as a premise and using an implication-introduction to derive $$p \rightarrow p$$ and cancel the premise-$$p$$.
So, yes, you can introduce tautologies wherever you want, because they require no premises to prove. You seem to understand this with your line about "introducing it via the conditional proof", but I do not see a difference between proving it from nothing and inserting it without proof, except that the latter is less formal.
• You said my example in which I introduce $p \vee \neg p$ is not appropriate because it is the law of the excluded middle and not a tautology. You then go on to illustrate your own example in which you derive $p \rightarrow p$, which is identical to what I did! Note that $p \rightarrow p$ in your example is logically equivalent to $p \vee \neg p$ via impl and comm laws, and the method you described for deriving $p \rightarrow p$, that is, the conditional proof, is the very method I acknowledged one could use to derive a statement such as $p \vee \neg p$ with no help from other premises. – RyRy the Fly Guy Sep 30 '19 at 12:30
• Despite misconstruing my example, you seem to understand the essence of my question, which is why don't we introduce a tautology such as $p \vee \neg p \Leftrightarrow p \rightarrow p$ directly among our premises "without proof" vs. going through the process of deriving a tautology (apart from any premises, of course, because as you pointed out we don't need premises to derive a tautology). You said you see no difference between the two approaches except that the former is less formal. The former is much briefer as well. – RyRy the Fly Guy Sep 30 '19 at 12:49
• If your criticism is that the former approach, that is, introducing a tautology directly among our premises "without proof," is less formal, then I suppose I would like to know why you think so. If a statement is simply true in the context of the logical system that you're using, then I don't understand why it is inappropriate to introduce it directly among the premises. – RyRy the Fly Guy Sep 30 '19 at 12:55
• Here is a criticism that comes to mind as I am discussing this with you: the example I give is a very simple example in which the tautology I'm introducing is extremely obvious. In that case, then yes, my argument to simply introduce the statement directly among the premises "without proof" feels reasonable and less like a transgression against any respectable proof. However, one could what if one wants to introduce a tautology in the form of a much longer and more complex statement? – RyRy the Fly Guy Sep 30 '19 at 13:03
• In that case, simply introducing a tautology "without proof" is not as acceptable, and it would be desirable and certainly more rigorous to go through the steps and derive the entire statement from scratch. So maybe in the end your final criticism against introducing tautologies "without proof" is legit. It is indeed less formal. It doesn't seem that way when one is introducing simple tautologies into an argument, but it becomes more apparent when one introduces more complex tautologies. – RyRy the Fly Guy Sep 30 '19 at 13:05
Perhaps you are also interested in a bit more practical, theorem prover-oriented side.
Let's assume we know $$1 + 1 = 2$$ and use that to prove $$(1 + 1) + 1 = 2 + 1$$. Clearly, we can just use the first equality to replace $$(1 + 1)$$ by $$2$$ yielding our goal.
This toy example looks as follows in the theorem prover Coq:
(* vvv Just some imports *)
From Coq Require Import ssreflect ssrfun ssrbool.
Set Implicit Arguments.
Unset Strict Implicit.
Unset Printing Implicit Defensive.
(* Actually we don't assume 1 + 1 = 2, but in fact prove it *)
Theorem thm: 1 + 1 = 2.
Proof.
(* albeit by a powerful command which can prove it on its own without human
intervention *)
by auto.
Qed.
Theorem thm': (1 + 1) + 1 = 2 + 1.
Proof.
(* Here we deduce 1 + 1 = 2 in an ad-hoc manner by just referencing
the previous thm *)
have U: (1 + 1 = 2) by apply: thm.
(* Then we rewrite as explained in the answer text above *)
by rewrite U.
Qed.
Hence, yes, similar to how a mathematician can pull a theorem out of thin air to insert it in their script, we can do so in Coq.
Note that a theorem is nothing else than a tautology. So you have cascade of interdependent theorems with axioms at the very top. But axioms behave very much the same way as theorems in Coq, so can also by used in have commands.
• Thanks for responding. It's true that mathematicians simply insert necessary theorems into proofs as needed, as your program is illustrating. And I suppose that, in the context of propositional logic, tautologies could be conceived as logical consequences or "theorems" descending from a larger set of axioms underlying the logical system. For example, $p \vee \neg p$ is only a tautology under the assumption that propositions are either true or not true. However, even when we introduce such tautologies in the conventional manner, we don't start with such axioms. – RyRy the Fly Guy Sep 30 '19 at 14:47
• So, yes, to an extent I suppose we already do the very thing I am talking about... to an extent. The answer to my question is starting to become more clear. – RyRy the Fly Guy Sep 30 '19 at 14:49
• @RyRytheFlyGuy Exactly, tautologies are just theorems. And yes, usually you just want to say "import theorem/tautology xyz, please" instead of replaying all of the proof. Sure, theoretically, replaying suffices, but eventually, we want to get to usable systems for formalizing maths, don't we? – ComFreek Sep 30 '19 at 14:52
Whether you can add in a tautology or not would depend on the proof assistant/checker you are using and the inference rules that tool permits. If you are doing it manually it depends on what the audience watching or reading your proof will permit you to use.
Here is a proof of the example using a Fitch-style proof checker:
I could add the premise $$P \lor \lnot P$$, but since I already have an LEM (law of the excluded middle) inference rule, it does not save me any work and it adds to the set of premises. However, after the list of premises, I could not add this as a line in the proof because this particular tool has no inference rule permitting that.
Also, although this tool has a disjunction elimination rule, I would need two subproofs in additional to the line containing the disjunction to use the rule as justification. This would be similar to what I provided above to reference LEM.
The advantage of using such tools is they make sure one is always using a well-formed formula and one is following the inference rules exactly. If one does that one can get a confirmation that the proof is correct. The disadvantage is that one has to follow those syntax and inference rules exactly.
Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/ | 2020-10-27T23:48:30 | {
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https://mathhelpboards.com/threads/computation-of-the-ratio-of-barometric-pressure.28115/ | # Computation of the ratio of barometric pressure
#### Dhamnekar Winod
##### Active member
Hi,
The Homestake gold mine near Lead,South Dakota is excavated to 8000 feet below the surface. Lead is nearly a mile high; the bottom of the Homestake is about 900 m below sea level. Nearby custer peak is about 2100 m above sea level.
What is the ratio of barometric pressure on the top of the custer peak to the barometric pressure at the bottom of Homestake?
(Assume that the entire atmosphere is at 300 K and that it behaves as a single ideal gas whose molar mass is 29.
How to answer this question? The known barometric formula is $P=P_0 e^{-\frac{m_gh}{kT}}$ where T=temperature, k=Boltzmann's constant, $m_g$=mass of an individual atmospheric molecule.
#### Klaas van Aarsen
##### MHB Seeker
Staff member
We can fill in 2100 m respectively -900 m into the formula to find the pressures. And then divide them to find the ratio.
Alternatively, we can first express the ratio as a formula and simplify it, and then evaluate it.
That is, let $h_1=2100\, m$ and $h_2=-900\,m$.
Then the requested ratio is:
$$\frac{P_0 e^{-\frac{m_g h_1}{kT}}}{P_0 e^{-\frac{m_g h_2}{kT}}}=e^{\frac{m_g h_2}{kT} - \frac{m_g h_1}{kT}}=e^{\frac{m_g}{kT}(h_2 - h_1)}$$
#### Dhamnekar Winod
##### Active member
We can fill in 2100 m respectively -900 m into the formula to find the pressures. And then divide them to find the ratio.
Alternatively, we can first express the ratio as a formula and simplify it, and then evaluate it.
That is, let $h_1=2100\, m$ and $h_2=-900\,m$.
Then the requested ratio is:
$$\frac{P_0 e^{-\frac{m_g h_1}{kT}}}{P_0 e^{-\frac{m_g h_2}{kT}}}=e^{\frac{m_g h_2}{kT} - \frac{m_g h_1}{kT}}=e^{\frac{m_g}{kT}(h_2 - h_1)}$$
Hi,
So, $m_g=\frac{29 g}{6.02214e23mol^{-1}}=4.8155e-23 g, e^{\frac{4.8155e-23 g(900 m-2100 m)}{\frac{1.38065e-23J}{K}300 K}}$= $e^{-0.013951577 s^2/m}$
Now, what is the final ratio? What does this answer mean to the readers?
My other conclusions from the information given in this question are
1) South Dakota is about 1538.4 m above sea level.
2)Custer Peak is about 561.6 m above South Dakota.
Are these conclusions correct?
#### Klaas van Aarsen
##### MHB Seeker
Staff member
So, $m_g=\frac{29 g}{6.02214e23mol^{-1}}=4.8155e-23 g, e^{\frac{4.8155e-23 g(900 m-2100 m)}{\frac{1.38065e-23J}{K}300 K}}$= $e^{-0.013951577 s^2/m}$
There is something wrong with the units.
The exponent should be dimensionless.
From wiki, it appears the correct formula is $P=P_0 e^{-\frac{m_g g h}{kT}}$.
That is, with an extra $g= 9.81 \,\text{m/s}^2$ in it. Then the exponent will indeed be dimensionless.
Btw, the unit of the $29\,\text g$ should actually be $29\,\text g\cdot \text{mol}^{-1}$. Consequently, the $\text{mol}^{-1}$ cancels against the one in the denominator as it should.
Now, what is the final ratio? What does this answer mean to the readers?
We should find a ratio that is lower than 1 since the pressure high up is lower than a pressure deep down.
If it is close to 1, then the pressure is more or less that same at all altitudes in this range, and we can breathe normally at both altitudes.
If it is close to 0, then we won't be able to breathe normally at both altitudes. And water will boil at a noticeably different temperature.
My other conclusions from the information given in this question are
1) South Dakota is about 1538.4 m above sea level.
2) Custer Peak is about 561.6 m above South Dakota.
Are these conclusions correct?
Yep. Correct.
Last edited:
#### Dhamnekar Winod
##### Active member
There is something wrong with the units.
The exponent should be dimensionless.
From wiki, it appears the correct formula is $P=P_0 e^{-\frac{m_g g h}{kT}}$.
That is, with an extra $g= 9.81 \,\text{m/s}^2$ in it. Then the exponent will indeed be dimensionless.
Btw, the unit of the $29\,\text g$ should actually be $29\,\text g\cdot \text{mol}^{-1}$. Consequently, the $\text{mol}^{-1}$ cancels against the one in the denominator as it should.
We should find a ratio that is lower than 1 since the pressure high up is lower than a pressure deep down.
If it is close to 1, then the pressure is more or less that same at all altitudes in this range, and we can breathe normally at both altitudes.
If it is close to 0, then we won't be able to breathe normally at both altitudes. And water will boil at a noticeably different temperature.
Yep. Correct.
Hi,
Thanks for helping me in finding out the correct ratio of barometric pressure.. I wrongly omitted $g=9.81m/s^2$ in the barometric formula.
I got the answer dimensionless which is 0.8721. That means if there is 1.25 bar barometric pressure at the bottom of Homestake gold mine, then Custer Peak will have 1.09 bar barometric pressure.
#### Dhamnekar Winod
##### Active member
There is a mistake in my computaion of ratio. Correct ratio is as above.
#### Klaas van Aarsen
##### MHB Seeker
Staff member
View attachment 10805
There is a mistake in my computaion of ratio. Correct ratio is as above.
The formula with the substitution looks a bit off...
1. The unit gram ($\text{g}$) should be converted to the SI unit $\text{kg}$ as part of the calculation.
2. The gravitational acceleration $g=9.81\,\text{m/s}^2$ (not to be confused with the gram unit $\text{g}$) seems to be missing.
Note that I'm using an upright font for the unit gram and an italic font for the quantity of the gravitational acceleration to distinguish them. It's the recommended typography for SI units.
Either way, I also get $0.71$ so it seems that you did do the correct calculation.
Last edited: | 2021-06-19T21:06:48 | {
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http://livelovelocks.com/eyemc/products-of-pure-imaginary-numbers-a258e0 | To multiply when a complex number is involved, use one of three different methods, based on the situation: To multiply a complex number by a real number: Just distribute the real number to both the real and imaginary part of the complex number. Some elements of the set of pure imaginary numbers are: 5 5 i Note the order in which the factors of and are written.The fac- 5 5 i Note the order in which the factors of and are written.The fac- A sine function is an odd function sin(-x) == -sin(x). Imaginary numbers are numbers that are not real. And so there is nothing to prevent us from making use of those numbers and employing them in calculation. In this worksheet, we will practice evaluating, simplifying, and multiplying pure imaginary numbers and solving equations over the set of pure imaginary numbers. Quadratic complex roots mathbitsnotebook(a1 ccss math). Noun 1. pure imaginary number - an imaginary number of the form a+bi where a is 0 complex number… Complex Number – any number that can be written in the form + , where and are real numbers. So we can rewrite negative 52 as negative 1 times 52. Pure imaginary number. Complex numbers. We call it a complex or imaginary number. (Note: and both can be 0.) Two pure imaginary numbers always subtract to be another pure imaginary number. Quick definitions from WordNet (pure imaginary number) noun: an imaginary number of the form a+bi where a is 0 Words similar to pure imaginary number Usage examples for pure imaginary number Simplify the following product: $$3\sqrt{-6} \cdot 5 \sqrt{-2}$$ Step 1. A new system of numbers entirely based on the the imaginary unit i Translate pure imaginary number from English to Language.th using Glosbe automatic translator that uses newest achievements in neural networks. Now we use complex numbers in electromagnetism, signal processing, and many others! An imaginary number is a complex number that can be written as a real number multiplied by the imaginary unit i, which is defined by its property i 2 = −1. For example, here’s how you handle a scalar (a constant) multiplying a complex number in parentheses: 2(3 + 2i) = 6 + 4i. This preview shows page 212 - 215 out of 733 pages.. Yes. Related words - pure imaginary number synonyms, antonyms, hypernyms and hyponyms. We use the diagonalization of matrix. If you're seeing this message, it means we're having trouble loading external resources on our website. In mathematics, the quaternion number system extends the complex numbers.Quaternions were first described by Irish mathematician William Rowan Hamilton in 1843 and applied to mechanics in three-dimensional space.Hamilton defined a quaternion as the quotient of two directed lines in a three-dimensional space, or equivalently, as the quotient of two vectors. Complex numbers can be de ned as pairs of real numbers (x;y) with special manipulation rules. Imaginary Number – any number that can be written in the form + , where and are real numbers and ≠0. 1 Basics of Series and Complex Numbers 1.1 Algebra of Complex numbers A complex number z= x+iyis composed of a real part <(z) = xand an imaginary part =(z) = y, both of which are real numbers, x, y2R. Imaginary Numbers displays the fruits of this cross-fertilization by collecting the best creative writing about mathematical topics from the past hundred years. It would have been clear if you had written: 7i - 7i = 0i ... FEN Learning is part of Sandbox Networks, a digital learning company that operates education services and products for the 21st century. $$i \text { is defined to be } \sqrt{-1}$$ From this 1 fact, we can derive a general formula for powers of $$i$$ by looking at some examples. Example sentences containing pure imaginary number Learn about the imaginary unit i, about the imaginary numbers, and about square roots of negative numbers. Imaginary Number Calculator is a free online tool that displays the imaginary root of the given number. Complex numbers and quadratic equations. Multiplication of pure imaginary numbers by non-finite numbers might not match MATLAB. That we can actually put, input, negative numbers in the domain of this function. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Group the real coefficients (3 and 5) and the imaginary terms $$( \blue{ 3 \cdot 5} ) ( \red{ \sqrt{-6}} \cdot \red{ \sqrt{-2} } )$$ Information and translations of PURE IMAGINARY NUMBER in the most comprehensive dictionary definitions resource on the web. The union of the set of all imaginary numbers and the set of all real numbers is the set of complex numbers. Definition of pure imaginary number in the Fine Dictionary. The confusion in your example is just notation. We prove that eigenvalues of a real skew-symmetric matrix are zero or purely imaginary and the rank of the matrix is even. Noting … The code generator does not specialize multiplication by pure imaginary numbers—it does not eliminate calculations with the zero real part. Find more similar words at wordhippo.com! Educreations is a community where anyone can teach what they know and learn what they don't. Pronunciation of pure imaginary number and its etymology. Complex numbers and complex conjugates. This lesson plan includes the objectives, prerequisites, and exclusions of the lesson teaching students how to evaluate, simplify, and multiply pure imaginary numbers and solve equations over the set of pure imaginary numbers. How to Multiply Imaginary Numbers Example 3. Define pure imaginary number. A complex number is any expression that is a sum of a pure imaginary number and a real number. That's because the real numbers are also closed under subtraction. The square of an imaginary number bi is −b 2.For example, 5i is an imaginary number, and its square is −25.By definition, zero is considered to be both real and imaginary. The Fourier Transformation of an even function is pure real. Then, we can just call it complex number. Our software turns any iPad or web browser into a recordable, interactive whiteboard, making it easy for teachers and experts to create engaging video lessons and share them on the web. Complex numbers of the form i{y}, where y is a non–zero real number, are called imaginary numbers. Meaning of pure imaginary number. Imaginary Numbers were once thought to be impossible, and so they were called "Imaginary" (to make fun of them).. GPU Code Generation Generate CUDA® code for NVIDIA® GPUs using GPU Coder™. HDL Code Generation Generate Verilog and VHDL code for FPGA and ASIC designs using HDL Coder™. We know as that number which, when squared, produces −3. GPU Arrays Accelerate code by running on a graphics processing unit (GPU) using Parallel Computing Toolbox™. In complex analysis, the complex numbers are customarily represented by the symbol z, which can be separated into its real (x) and imaginary (y) parts: = + for example: z = 4 + 5i, where x and y are real numbers, and i is the imaginary unit.In this customary notation the complex number z corresponds to the point (x, y) in the Cartesian plane. BYJU’S online imaginary number calculator tool makes the calculation faster and it displays the imaginary value in a fraction of seconds. Or dropping the 0 in front: ai - bi = (a - b)i. Whenever the discriminant is less than 0, finding square root becomes necessary for us. Thus symbols such as , , , and so on—the square roots of negative numbers—we will now call complex numbers. Normally this doesn't happen, because: when we square a positive number we get a positive result, and; when we square a negative number we also get a positive result (because a negative times a negative gives a positive), for example −2 × −2 = +4; But just imagine such numbers exist, because we want them. Pronunciation: — Math. Let √-1 = imaginary number = i since such number does not exist with real numbers, but exist only in our imagination. In this video, I want to introduce you to the number i, which is sometimes called the imaginary, imaginary unit What you're gonna see here, and it might be a little bit difficult, to fully appreciate, is that its a more bizzare number than some of the other wacky numbers we learn in mathematics, like pi, or e. i m asking because i found this in a complex numbers' chapter (maths) ' Zero is purely real as well as purely imaginary but not imaginary '. Meaning of pure imaginary number with illustrations and photos. pure imaginary number synonyms, pure imaginary number pronunciation, pure imaginary number translation, English dictionary definition of pure imaginary number. Notational conventions. Students can replay these lessons any time, any place, on any connected device. Imaginary Numbers are not "Imaginary". Imaginary number wikipedia. Imaginary Numbers when squared give a negative result.. The Fourier Transformation of an odd function is pure imaginary. (0 + ai) - (0 + bi) = (0 - 0) + (a - b)i. Math. A synonym for pure imaginary number is purely imaginary number. what does this mean something purely imaginary but not imaginary. What does PURE IMAGINARY NUMBER mean? In this engaging anthology, we can explore the many ways writers have played with mathematical ideas. is not a real number. We know that the quadratic equation is of the form ax 2 + bx + c = 0, where the discriminant is b 2 – 4ac. How to find product of pure imaginary numbers youtube. If two complex numbers are equal, we can equate their real and imaginary parts: {x1}+i{y1} = {x2}+i{y2} ⇒ x1 = x2 and y1 = y2, if x1, x2, y1, y2 are real numbers. And we're going to assume, because we have a negative 52 here inside of the radical, that this is the principal branch of the complex square root function. Meaning of PURE IMAGINARY NUMBER. A complex number usually is expressed in a form called the a + bi form, or standard form, where a and b are real numbers. That is the reason why the plot of the imaginary part of the fft of function 1 contains only values close to zero (1e-15). The number 1 + i is imaginary but not pure imaginary, while 5i is pure imaginary. The expressions a + bi and a – bi are called complex conjugates. C/C++ Code Generation Generate C and C++ code using MATLAB® Coder™. Imaginary numbers are based on the mathematical number $$i$$. 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products of pure imaginary numbers 2021 | 2021-05-14T03:35:25 | {
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https://math.stackexchange.com/questions/1325360/how-can-i-plot-this-function-that-has-a-fraction-that-has-an-absolute-value-in-t/1325378 | # How can I plot this function that has a fraction that has an absolute value in the denominator?
I have this piecewise function:
$$f(x)=\begin{cases} \dfrac{x^2-x-2}{|x-2|}, & x \neq 2 \\ 0, & x = 2\text{.} \end{cases}$$
I can't figure out how to graph it. I punched these numbers into my calculator, and it created a parabola, but I haven't been able to get there on my own without the calculator.
So far I have plotted the point $(2,0)$ on my graph, and I factored the first function to $\dfrac{(x-2)(x+1)}{|x-2|}$ and now I am stuck. I tried multiplying both the top and bottom by $x+2$, but after simplifying I ended up with $x+1$, which I am pretty positive is not correct.
Where did I go wrong?
Hint: recall (ignoring the $x = 2$ case): $$|x-2| = \begin{cases} x-2, & x-2 > 0 \\ -(x-2), & x-2 < 0 \end{cases} = \begin{cases} x-2, & x > 2 \\ -(x-2), & x < 2\text{.} \end{cases}$$ So $$\dfrac{x^2-x-2}{|x-2|} = \begin{cases} \dfrac{x^2-x-2}{x-2} = \dfrac{(x-2)(x+1)}{x-2} = x+1, & x > 2 \\ \dfrac{x^2 - x - 2}{-(x-2)} = -(x+1), & x < 2\text{.} \end{cases}$$
• Do I ignore the x=2 case when I graph this function? – matryoshka Jun 14 '15 at 20:08
• @Grace When $x = 2$, $f(x) = 0$. This is by definition in the problem. – Clarinetist Jun 14 '15 at 20:09
Dividing the two cases: $$x-2>0 \iff x>2 \Rightarrow |x-2|=x-2$$ and
$$x-2<0 \iff x<2 \Rightarrow |x-2|=2-x$$ your function become: $$f(x)= \begin {cases} y=-x-1 \quad , \quad x<2\\ y=0 \quad,\quad x=2\\ y=x+1\quad,\quad x>2 \end{cases}$$
the graph is done by two open half-lines ( decreasing for $x<2$ and crescent for $x>2$) and an isolated point $(2,0)$.
Note that $\frac{x}{|x|}=1$ for $x>$ and $\frac{x}{|x|}=-1$ for $x<0$. Thus,
$$\bbox[5px,border:2px solid #C0A000]{\frac{(x-2)(x+1)}{|x-2|}=\text{sgn}(x-2)(x+1)}$$
where $\text{sgn}(x-2)=1$ for $x>2$ and $\text{sgn}(x-2)=-1$ for $x<2$. We can define the sign function as $0$ when its argument is $0$. And we're done! | 2019-09-23T15:46:58 | {
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http://mathhelpforum.com/advanced-algebra/90526-just-one-more-question-galois-group.html | # Thread: just one more question on Galois group
1. ## just one more question on Galois group
So, I have to determine the Galois group of $x^4-5$ over $\mathbb{Q}$, $\mathbb{Q}[\sqrt{5}]$, $\mathbb{Q}[\sqrt{-5}]$.
$f(x)=x^4-5$
That is irreducible over $\mathbb{Q}$ by Esienstein's criterion.
Resolvent is: $g(x)=x^3+20x=x(x^2+20)=x(x-\sqrt{-20})(x+\sqrt{-20})$
$\alpha=0, \beta=\sqrt{-20}, \gamma=-\sqrt{-20}$
$
\mathbb{Q}(\alpha, \beta, \gamma)=\mathbb{Q}(0, \sqrt{-20}, -\sqrt{-20})=\mathbb{Q}(\sqrt{-20})=\mathbb{Q}[\sqrt{-5}]$
$
[\mathbb{Q}(\alpha, \beta, \gamma):\mathbb{Q}]=2$
So $G$ is isomorphic to $D_4$ or $\mathbb{Z}_4$.
To determine which, we check is $f$ irreducible over $\mathbb{Q}[\sqrt{-5}]$
$
f(x)=(x^2-\sqrt{5})(x^2+\sqrt{5})$
Is this irreducible or not?
And what to do with
$\mathbb{Q}[\sqrt{5}]$, $\mathbb{Q}[\sqrt{-5}]$?
Thank you!
2. Originally Posted by marianne
So $G$ is isomorphic to $D_4$ or $\mathbb{Z}_4$.
To determine which, we check is $f$ irreducible over $\mathbb{Q}[\sqrt{-5}]$
$
f(x)=(x^2-\sqrt{5})(x^2+\sqrt{5})$
[COLOR=Red][COLOR=Black]Is this irreducible or not?
Argue that the splitting field has degree 8 over Q and so the Galois group has degree 8 as well.
This means the Galois group is $D_4$.
3. Originally Posted by ThePerfectHacker
Argue that the splitting field has degree 8 over Q and so the Galois group has degree 8 as well.
This means the Galois group is $D_4$.
Oh, I missed that, thank you!!!
Could you possibly help me out with , ?
4. Originally Posted by marianne
So, I have to determine the Galois group of $x^4-5$ over $\mathbb{Q}$, $\mathbb{Q}[\sqrt{5}]$, $\mathbb{Q}[\sqrt{-5}]$.
$f(x)=x^4-5$
That is irreducible over $\mathbb{Q}$ by Esienstein's criterion.
Resolvent is: $g(x)=x^3+20x=x(x^2+20)=x(x-\sqrt{-20})(x+\sqrt{-20})$
$\alpha=0, \beta=\sqrt{-20}, \gamma=-\sqrt{-20}$
$
\mathbb{Q}(\alpha, \beta, \gamma)=\mathbb{Q}(0, \sqrt{-20}, -\sqrt{-20})=\mathbb{Q}(\sqrt{-20})=\mathbb{Q}[\sqrt{-5}]$
$
[\mathbb{Q}(\alpha, \beta, \gamma):\mathbb{Q}]=2$
So $G$ is isomorphic to $D_4$ or $\mathbb{Z}_4$.
To determine which, we check is $f$ irreducible over $\mathbb{Q}[\sqrt{-5}]$
$
f(x)=(x^2-\sqrt{5})(x^2+\sqrt{5})$
Is this irreducible or not?
And what to do with
$\mathbb{Q}[\sqrt{5}]$, $\mathbb{Q}[\sqrt{-5}]$?
Let $K$ be the splitting field for the resolvent, $x(x^2+20)$. As you have shown $K = \mathbb{Q}(\sqrt{-5})$ and so $[K:\mathbb{Q}]=2$. Therefore, the Galois group is either $\mathbb{Z}_4$ or $D_4$. It all depends whether or not $f$ is irreducible over $K$. First $f$ has NO roots in $K$ because the roots of $f$ are: $\pm\sqrt[4]{5}, \pm i\sqrt[4]{5}$. These numbers have degree four over $\mathbb{Q}$ while $K$ is only a degree two extension. Thus, you need to show that $(x^2 + ax + b)(x^2 + cx + d) = x^4 - 5$ is impossible for $a,b,c,d\in K$. To simplify things we will appeal to a result from number theory, that the ring of integers in $\mathbb{Q}(\sqrt{d})$ is $\mathbb{Z}[\sqrt{d}]$ if $d\equiv 2,3(\bmod 4)$. Thus, the ring of integers in $\mathbb{Q}(\sqrt{-5})$ is $\mathbb{Z}[\sqrt{-5}]$ i.e. a field of fractions of $\mathbb{Z}[\sqrt{-5}]$ is $\mathbb{Q}(\sqrt{-5})$. By Gauss' Lemma (the general one) to prove irreduciblity of $x^4 - 5$ over $K$ it sufficies to prove irreducibility of $x^4 - 5$ over $\mathbb{Z}[\sqrt{-5}]$. Since we have $(x^2 + ax + b)(x^2 + bx + d) = x^4 - 5$ where $a,b,c,d,\in \mathbb{Z}[\sqrt{-5}]$ if we expand we get $x^4+(a+c)x^3+(b+ac + d)x^2 +(ad+bc)x+bd = x^4 - 5$. I think (I did not check this!) that it is only possible to get $bd = -5$ if $b=d=\pm \sqrt{-5}$, now show that both those cases lead to impossibilities in the coefficiensts $a,b,c,d$. Can you be the mathematician of the gaps and fill them in?
5. Originally Posted by ThePerfectHacker
To simplify things we will appeal to a result from number theory, that the ring of integers in $\mathbb{Q}(\sqrt{d})$ is $\mathbb{Z}[\sqrt{d}]$ if $d\equiv 2,3(\bmod 4)$. Thus, the ring of integers in $\mathbb{Q}(\sqrt{-5})$ is $\mathbb{Z}[\sqrt{-5}]$ i.e. a field of fractions of $\mathbb{Z}[\sqrt{-5}]$ is $\mathbb{Q}(\sqrt{-5})$.
I have something similar, but it's $x^4-7$ over $\mathbb{Q}[\sqrt{-7}]$. How do I check is $f$ irreducible? I can't use the number theory result you provided.
6. Originally Posted by georgel
I have something similar, but it's $x^4-7$ over $\mathbb{Q}[\sqrt{-7}]$. How do I check is $f$ irreducible? I can't use the number theory result you provided.
If $d\equiv 1(\bmod 4)$ the ring of integers of $\mathbb{Q}(\sqrt{d})$ is given by $\{ \tfrac{a+b\sqrt{d}}{2} | a\equiv b(\bmod 2) \}$. | 2017-03-23T20:25:41 | {
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http://blog.csdn.net/ZHE123ZHE123ZHE123/article/details/51565690 | # Linear regression with multiple variables
## 1.Promble:
Suppose you are selling your house and you want to know what a good market price would be. One way to do this is to first collect information on recent houses sold and make a model of housing prices.
The file ex1data2.txt contains a training set of housing prices in Portland, Oregon. The first column is the size of the house (in square feet), the second column is the number of bedrooms, and the third column is the price of the house.
## 2.采用梯度下降的方法求解：
1)step 1: Feature Normalization
By looking at the values(ex1data2.txt), note that house sizes are about 1000 times the number of bedrooms. When features differ by orders of magnitude, first performing feature scaling can make gradient descent converge much more quickly
z-score标准化方法适用于属性A的最大值和最小值未知的情况，或有超出取值范围的离群数据的情况。
新数据=（原数据-均值）/标准差
MATLAB代码如下:
<span style="font-size:18px;">function [X_norm, mu, sigma] = featureNormalize(X)
%FEATURENORMALIZE Normalizes the features in X
% FEATURENORMALIZE(X) returns a normalized version of X where
% the mean value of each feature is 0 and the standard deviation
% is 1. This is often a good preprocessing step to do when
% working with learning algorithms.
% You need to set these values correctly
X_norm = X;
mu = zeros(1, size(X, 2)); % mean value 均值 size(X,2) 列数
sigma = zeros(1, size(X, 2)); % standard deviation 标准差
% ====================== YOUR CODE HERE ======================
% Instructions: First, for each feature dimension, compute the mean
% of the feature and subtract it from the dataset,
% storing the mean value in mu. Next, compute the
% standard deviation of each feature and divide
% each feature by it's standard deviation, storing
% the standard deviation in sigma.
%
% Note that X is a matrix where each column is a
% feature and each row is an example. You need
% to perform the normalization separately for
% each feature.
%
% Hint: You might find the 'mean' and 'std' functions useful.
%
mu = mean(X); % mean value
sigma = std(X); % standard deviation
X_norm = (X - repmat(mu,size(X,1),1)) ./ repmat(sigma,size(X,1),1);%新数据=（原数据-均值）/标准差
end</span>
theta = theta - alpha / m * X' * (X * theta - y);
function [theta, J_history] = gradientDescentMulti(X, y, theta, alpha, num_iters)
% taking num_iters gradient steps with learning rate alpha
% Initialize some useful values
m = length(y); % number of training examples
J_history = zeros(num_iters, 1);
for iter = 1:num_iters
% ====================== YOUR CODE HERE ======================
% Instructions: Perform a single gradient step on the parameter vector
% theta.
%
% Hint: While debugging, it can be useful to print out the values
% of the cost function (computeCostMulti) and gradient here.
%
theta = theta - alpha / m * X' * (X * theta - y);
% ============================================================
% Save the cost J in every iteration
J_history(iter) = computeCostMulti(X, y, theta);
end
end
## 3.采用正规方程求解
### Normal Equations:
the closed-form solution to linear regression is:
Using this formula does not require any feature scaling, and you will get an exact solution in one calculation: there is noloop until convergencelike in gradient descent
Matlab代码如下：
function [theta] = normalEqn(X, y)
%NORMALEQN Computes the closed-form solution to linear regression
% NORMALEQN(X,y) computes the closed-form solution to linear
% regression using the normal equations.
theta = zeros(size(X, 2), 1);
% ====================== YOUR CODE HERE ======================
% Instructions: Complete the code to compute the closed form solution
% to linear regression and put the result in theta.
%
% ---------------------- Sample Solution ----------------------
theta = pinv( X' * X ) * X' * y;
end
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https://math.stackexchange.com/questions/2384112/question-about-definition-of-connectedness | # Question about definition of connectedness.
While reading some books, I often find the following argument:
Consider $A\subset X$, when $X$ is a topological space.
If $A$ is a disconnected set, then there exist some open sets, say $U,V \in \tau_X$ disjoint and non empty such that $A = U \cup V$.
But this often confuses me, because the part when it says "open sets of $\tau_X$" because, for example, $[0,1] \cup [5,6]$ is disconnected and it can't be written as union of two open sets -of X!- which are disjoint and non empty.
The definition in some books of disconnected set $A$ says that $A$ is disconnected if it is disconnected with the subspace topology.
But then, I don't understand why some authors take open sets of the big space, say $X$.
Can someone help me to clarify this?
• I'm guessing you're misreading something. Can you give an exact quote? – Eric Wofsey Aug 6 '17 at 3:41
• For example, in the book "Topics on Continua" of Sergio Macías, in page 52 it says "Suppose that $X\setminus A$ is not connected. Then there exist two nonempty disjoint open subsets $U$ and $V$ of $X$ such that $X\setminus A = U \cup V$ ..." – HeMan Aug 6 '17 at 3:44
• "...because, for example, $[0,1]∪[5,6]$ is disconnected and it can't be written as union of two open sets -of $X$!" - what is $X$ here? – user 170039 Aug 6 '17 at 3:45
• Presumably in that context $A$ is closed, so $X\setminus A$ is open and relatively open sets are the same as open sets of $X$. – Eric Wofsey Aug 6 '17 at 3:45
• In that particular case, $X= \mathbb{R}$ – HeMan Aug 6 '17 at 3:47
The correct definition of $A$ being a disconnected subset of $X$ is that $A$ is disconnected as a space in its own right, i.e. having the subspace topology.
If $A \subseteq X$ is a disconnected subset, we can translate this to a statement about open sets of $X$ if we like:
$A \subseteq X$ is disconnected iff there exist $U,V$, open sets in $X$, such that
• $U \cap A \neq \emptyset$
• $V \cap A \neq \emptyset$
• $A \subseteq U \cup V$
• $A \cap U \cap V = \emptyset$
This corresponds to $\{U \cap A$, $V \cap A\}$ being a disconnection of $A$ in its subspace topology.
Definition of Separation of a Topological Space. Let $(X,\tau_X)$ be a topological space and $Y\subseteq X$. Let the subspace topology on $Y$ be denoted as $\tau_Y$. A separation of $Y$ is a pair of disjoint non-empty open (open in $Y$, not $X$) subsets of $Y$ whose union is $Y$.
Definition of Disconnected Topological Space. A topological space $(X,\tau_X)$ is said to be disconnected if there exists a separation of $X$.
So when trying to prove that $[0,1]\cup [5,6]$ is disconnected we shouldn't be trying to write it as union of two sets open in $\mathbb{R}$, rather we should try to write it as an union of sets open in $[0,1]\cup [5,6]$ under the subspace topology induced by the topology on $\mathbb{R}$ under consideration.
Here is a way to clarify this..
Definition. A topological space $(X,\tau)$ is disconnected if there exist $U,V\in\tau\setminus\{\varnothing\}$ such that $U\cap V=\varnothing$ and $U\cup V = X$. Otherwise, $(X,\tau)$ is connected.
We say that a subset $Y\subseteq X$ is connected if it is connected as a topological subspace.
Proposition. Let $(X,\tau)$ be a topological space and $Y\subseteq X$. We have that $Y$ is disconnected if and only if there exist $U,V\in \tau\setminus\{\varnothing\}$ such that
(1) $U\cap Y\neq\varnothing$;
(2) $V\cap Y\neq\varnothing$;
(3) $U\cap V\cap Y=\varnothing$; and
(4) $Y\subseteq U\cup V$.
Proof. (ONLY IF) Assume $Y$ is disconnected, then there exist $U_Y, V_Y\in\tau_Y\setminus\{\varnothing\}$ such that $U_Y \cap V_Y = \varnothing$ and $U_Y\cup V_Y = Y$. Thus there exist $U,V\in\tau\setminus\{\varnothing\}$ such that $U_Y=U\cap Y$ and $V_Y=V\cap Y$. Such $U$ and $V$ satisfy conditions (1)-(4).
(IF) Assume there exist such $U$ and $V$ in $\tau$ satisfying (1)-(4). Let $U_Y=U\cap Y$ and $V_Y=V\cap Y$, both in $\tau_Y\setminus\{\varnothing\}$ (conditions (1) and (2) tell us that both $U_Y$ and $V_Y$ are nonempty). Condition (3) tell us that $U_Y\cap V_Y= \varnothing$ and, finally, condition (4) says that $U_Y\cup V_Y=Y$. Therefore, $Y$ is disconnected. $\square$ | 2019-09-17T16:40:41 | {
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https://math.stackexchange.com/questions/3125933/how-is-this-property-called-for-mod | # How is this property called for mod?
We have a name for the property of integers to be $$0$$ or $$1$$ $$\mathrm{mod}\ 2$$ - parity.
Is there any similar name for the remainder for any other base? Like a generalization of parity? Could I use parity in a broader sense, just to name the remainder $$\mathrm{mod}\ n$$?
• How about congruence? – Umberto P. Feb 25 '19 at 11:37
• So should I say (variable)'s congruence class or value? – dEmigOd Feb 25 '19 at 11:43
• I had the same question and thought about terminology for $n = 3$. A friend of mine came up with a very neat suggestion: flat, short and long. I don't know if there is any other generalization, but I would very much like to hear people talking about remainders mod 3 like this. I guess the reason why this has a name for $n=2$ is that it appears commonly in everyday life as opposed to most of the other numbers. When talking about time, the 12 is dropped at all, but this is somewhat different maybe – kesa Feb 25 '19 at 15:44
## 4 Answers
Simply say "congruent to $$a$$ modulo $$m$$" to read "$$\equiv a \pmod{m}$$".
Actually there is a standard name: residue.
There are $$5$$ residues modulo $$5$$, namely $$0,1,2,3,4$$.
Every prime greater than $$3$$ falls into only $$2$$ residue-classes modulo $$6$$.
In a broader sense, when you are dealing with congruences you are dealing with an equivalence relation and its equivalence classes.
The basic idea of an equivalence relation is to collect elements that are different but behave in the same manner with respect to some property of interest.
In modular arithmetic this property is having the same rest when divided by a prescribed integer
If $$a=b\bmod m$$ or $$a\equiv_m b$$ you essentially say that $$a$$ and $$b$$ are in the same equivalence class with respect to the equivalence relation $$\equiv_m$$.
So you could say that "$$a$$ and $$b$$ are in the same equivalence class when we look at the remainder upon dividing by $$m$$", which definitely longer and more cumbersome to say "$$a$$ congruent to $$b$$ modulo $$m$$" but nonetheless another way to express the same concept.
Either AlessioDV's good answer, or you could say: "of the form $$nq+r$$". as a representation of $$\equiv r \pmod n$$ for example a number $$N$$ with $$N\equiv 1\pmod 3$$ has property $$N=3q+1$$. | 2020-02-19T01:12:56 | {
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http://mathhelpforum.com/calculus/206199-problem-finding-speed-particle-vector.html | # Thread: Problem Finding the Speed of Particle (Vector)
1. ## Problem Finding the Speed of Particle (Vector)
Hey everyone,
I am having a bit of difficulty solving this problem. It says:
A particle has position function $\ r(t) = (2\sqrt{2})i + (e^{2t})j + (e^{-2t})k$. What is its speed at time t?
Well I first took the derivative of the function. Then I took the magnitude to find the length which corresponds to the speed. I got my answer to be
$\sqrt{8+4e^{2t}+4e^{-2t}}$, is this answer correct? If so, how would I further simplify it? If it is incorrect, what am I doing wrong?
Any help and feedback is greatly appreciated, thanks.
2. ## Re: Problem Finding the Speed of Particle (Vector)
Hello, Beevo!
A particle has position function: $\ r(t) \:=\: (2\sqrt{2})i + (e^{2t})j + (e^{-2t})k$.
What is its speed at time $t$?
Well, I first took the derivative of the function.
Then I took the magnitude to find the length which corresponds to the speed.
I got my answer to be: $\sqrt{8+4e^{2t}+4e^{-2t}}$.
Is this answer correct? . Yes!
If so, how would I further simplify it?
It can be simplified . . .
$\sqrt{4e^{2t} + 8 + 4e^{-2t}} \;=\;\sqrt{4(e^{2t} + 2 + e^{-2t})} \;=\;2\sqrt{e^{2t} + 2 + e^{-2t}}$
. . . . . . . . . . . . . $=\;2\sqrt{(e^t + e^{-t})^2} \;=\;2(e^t + e^{-t})$
3. ## Re: Problem Finding the Speed of Particle (Vector)
Originally Posted by Beevo
Hey everyone,
I am having a bit of difficulty solving this problem. It says:
A particle has position function $\ r(t) = (2\sqrt{2})i + (e^{2t})j + (e^{-2t})k$. What is its speed at time t?
Well I first took the derivative of the function. Then I took the magnitude to find the length which corresponds to the speed. I got my answer to be
$\sqrt{8+4e^{2t}+4e^{-2t}}$, is this answer correct? If so, how would I further simplify it? If it is incorrect, what am I doing wrong?
Any help and feedback is greatly appreciated, thanks.
Looks good to me, unless you have to use hyperbolic functions to simplify.
-Dan
PS okay I should have seen that one coming Soroban! Thanks.
4. ## Re: Problem Finding the Speed of Particle (Vector)
Thanks for the feedback guys. The simplification part just kind of threw me off.
5. ## Re: Problem Finding the Speed of Particle (Vector)
Okay, I'm confused! Where did that "8" come from? With the position $r(t)= 2\sqrt{2}\vec{i}+ e^{2t}\vec{j}+ e^{-2t}\vec{k}$,I get the velocity to be $2e^{2t}\vec{j}- 2e^{-2t}\vec{k}$ and so the speed is $\sqrt{4e^{4t}+ 4e^{-4t}}= 2\sqrt{e^{4t}+ e^{-4t}}$
6. ## Re: Problem Finding the Speed of Particle (Vector)
Originally Posted by HallsofIvy
Okay, I'm confused! Where did that "8" come from? With the position $r(t)= 2\sqrt{2}\vec{i}+ e^{2t}\vec{j}+ e^{-2t}\vec{k}$,I get the velocity to be $2e^{2t}\vec{j}- 2e^{-2t}\vec{k}$ and so the speed is $\sqrt{4e^{4t}+ 4e^{-4t}}= 2\sqrt{e^{4t}+ e^{-4t}}$
My fault, there should be a 't' after $\2\sqrt{2}$ | 2017-02-21T11:17:43 | {
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https://www.physicsforums.com/threads/what-qualifies-as-an-infinite-discontinuity.847570/ | # What qualifies as an infinite discontinuity?
Hello everyone. I am currently having trouble actually defining what qualifies as an infinite discontinuity. I have read several sources that state that both of the one sided limits must approach infinity (positive, negative or both). My problem is what happens when only one of the one sided limits approach infinity and the other is finite. This is not a jump discontinuity because both of the one sided limits are not finite. According to the definition that I have read, it doesn't qualify as an infinite discontinuity either. I have crudely drawn what I am trying to describe. I am pretty sure this still is classified as an infinite discontinuity but I just wanted to be sure.
Mark44
Mentor
Hello everyone. I am currently having trouble actually defining what qualifies as an infinite discontinuity. I have read several sources that state that both of the one sided limits must approach infinity (positive, negative or both). My problem is what happens when only one of the one sided limits approach infinity and the other is finite. This is not a jump discontinuity because both of the one sided limits are not finite. According to the definition that I have read, it doesn't qualify as an infinite discontinuity either. I have crudely drawn what I am trying to describe. I am pretty sure this still is classified as an infinite discontinuity but I just wanted to be sure.
I would call it an infinite discontinuity. A simpler example than your graph shows is
##f(x) = \begin{cases} 1 & \text{if } x \le 0 \\ \frac 1 x & \text{if } x > 0\end{cases}##
In my example, ##\lim_{x \to 0^-} f(x) = 1##, but ##\lim_{x \to 0^+} f(x) = \infty##
Another site I found would agree with me -- http://www.milefoot.com/math/calculus/limits/Continuity06.htm
An infinite discontinuity exists when one of the one-sided limits of the function is infinite.
I believe that their definition would also include the case where both one-side limits are infinite. If so, their definition would be clearer if it said "when at least one of the one-side limits of the function is infinite."
If so, their definition would be clearer if it said "when at least one of the one-side limits of the function is infinite."
This is what I assumed. When I learned Calculus 1, we classified this as an essential discontinuity, compared to the removable discontinuity. Obviously we went over infinite discontinuities but they typically were asymptotic | 2021-01-27T01:58:41 | {
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https://www.jiskha.com/display.cgi?id=1299967997 | # Calculus (fixed the typo )
posted by .
A line rotates in a horizontal plane according to the equation theta=2t^3 -6t,where theta is the angular position of the rotating line, in radians ,and t is the time,in seconds. Determine the angular acceleration when t=2sec.
A.) 6 radians per sec^2
B.) 12 radians per sec^2
C.) 18 radians per sec^2
D.) 24 radians per sec^2 << my choice.
• Calculus (fixed the typo ) -
θ=2t^3-6t
θ'=dθ/dt=6t²-6
θ"=d²θ/dt²
=12t
θ(2)=12*2=24 radians-sec-2
## Respond to this Question
First Name School Subject Your Answer
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https://stats.stackexchange.com/questions/238884/peculiar-behaviour-of-conditional-variance-for-multivariate-normal-distributions/239187 | # Peculiar Behaviour of Conditional Variance for Multivariate Normal Distributions
Let $Z_i \sim \mathcal{N}(0,1)$ be independent normal distributions. Consider the following correlated variables, defined by $$X_1 = \frac{Z_1 + Z_2}{\sqrt{2}},\;\;\;X_2= \frac{Z_2 + Z_3}{\sqrt{2}},\;\;\;X_3= \frac{Z_3 + Z_4}{\sqrt{2}},\ldots$$
Thus each $X_i$ by itself is also a standard normal distribution but is correlated to the immediate neighbours $X_{i-1}$ and $X_{i+1}$. Consider the joint distribution of $(X_1,X_2)$ which is a joint normal with mean = $(0,0)$ and covariance matrix $$\begin{pmatrix} 1 & 1/2 \\ 1/2 & 1 \end{pmatrix}$$
Now the thing is according to the rules of conditional probability the conditional variance for $X_1$ is $\left(1-\rho^2\right)\sigma_1^2 = \frac{3}{4}$ in this case. So far so good.
Suppose we then consider the joint normal $\left(X_1,X_2,X_3\right)$, which has the covariance matrix $$\begin{pmatrix} 1 & 1/2 & 0 \\ 1/2 & 1 & 1/2 \\ 0 & 1/2 & 1 \end{pmatrix}$$
In this case, the conditional variance of $X_1$ is given by
$$1 - \begin{pmatrix} 1/2 & 0 \end{pmatrix}\begin{pmatrix} 1 & 1/2 \\ 1/2 & 1 \end{pmatrix}^{-1}\begin{pmatrix} 1/2 \\ 0 \end{pmatrix} = 2/3$$
The questions I have now are:
1. Since $X_1$ is not dependent on $X_3$ at all, why the conditional variance of $X_1$ drops from $3/4$ to $2/3$ when $X_3$ is taken into account?
2. If I further include $X_4,X_5,\ldots$ the conditional variance seems to drop further and reaches a limit of $1/2$ when I include a very large number of $X_i$. Is there any intuitive explanation for this limit?
• Because $X_3$ provides some information about $Z_3$ and hence about $X_2$? Work backwards using this notion when you know the values of large numbers of $X_k$. Oct 7, 2016 at 1:02
This is a super interesting question/observation.... First thing to note is that $$Var(X_1\mid Z_2) = 1/2$$
And the reason that $$Var(X_1\mid X_2) = 3/4 \neq 1/2$$
is because it's not possible to disentangle the actual value that $Z_2$ takes when you only observe $$X_2 = \frac{Z_2 + Z_3}{\sqrt{2}}$$
and as the first commentor pointed out, observing $X_3, X_4, X_5, \dots$ tells you more and more information about what the actual value of $Z_2$ is, and thus brings you closer to that $1/2$ bound you mentioned.
To prove the we eventually hit that $1/2$ limit, denote the conditional variance of $X_1$ on $X_2, \dots, X_T$ by $\overline{\Sigma}_T$, and the unconditional covariance matrix of $X_1, X_2, \dots , X_T$ by $\Sigma_T$. We know that $$\overline{\Sigma}_T = \Sigma^{11}_T - \Sigma^{12}_T[\Sigma^{22}_T]^{-1}\Sigma^{21}_T$$
Where we've partitioned the unconditional covariance matrix $\Sigma_T$ as $$\Sigma_T=\begin{bmatrix} \Sigma^{11}_T & \Sigma^{12}_T\\ \Sigma^{21}_T & \Sigma^{22}_T \end{bmatrix}$$
We know that, in this case, $$\Sigma^{11}_T = 1$$ $$\Sigma^{12}_T = [1/2, \boldsymbol{0}]$$ $$\Sigma^{21}_T = [1/2, \boldsymbol{0}]^T$$ where $\boldsymbol{0}$ is an 1x(T-1) vectors of zeros (this is because $X_1$ is dependent only on $X_2$ and not on $X_3, X_4, \dots$).
Now, we can write $$[\Sigma^{22}_T]^{-1} = \begin{bmatrix}\Sigma^{11}_{T-1} & \Sigma^{12}_{T-1}\\ \Sigma^{21}_{T-1} & \Sigma^{22}_{T-1} \end{bmatrix}^{-1} = \begin{bmatrix} (1 - \Sigma^{12}_{T-1} [\Sigma^{22}_{T-1}]^{-1}\Sigma^{21}_{T-1})^{-1} & (1 - \Sigma^{12}_{T-1} [\Sigma^{22}_{T-1}]^{-1}\Sigma^{21}_{T-1})^{-1}\Sigma^{12}_{T-1}[\Sigma^{22}_{T-1}]^{-1}\\ \vdots & \vdots \end{bmatrix}$$
Where I left out the last row to save space (I'll show those entries don't matter anyways). This result holds by the Schur complement of a block matrix.
Next, since $$\Sigma^{12}_T = [1/2, \boldsymbol{0}]$$ $$\Sigma^{21}_T = [1/2, \boldsymbol{0}]^T$$ this all simplifies to $$\overline{\Sigma}_T = 1 - \frac{1}{4}(1 - \Sigma^{12}_{T-1} [\Sigma^{22}_{T-1}]^{-1}\Sigma^{21}_{T-1})^{-1}$$
However, since $\overline{\Sigma}_{T-1} = 1 - \Sigma^{12}_{T-1} [\Sigma^{22}_{T-1}]^{-1}\Sigma^{21}_{T-1}$, we see finally that
$$\overline{\Sigma}_T = 1 - \frac{1}{4}[\overline{\Sigma}_{T-1}]^{-1}$$
Taking the limit as $T\rightarrow \infty$ yields $$\overline{\Sigma}_{\infty} = 1 - \frac{1}{4}[\overline{\Sigma}_{\infty}]^{-1}$$
Which can be solved to give the limiting conditional variance as $$\overline{\Sigma}_\infty = 0.5$$
• I really like your answer to the second part. Oct 10, 2016 at 11:22 | 2022-05-27T03:28:05 | {
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https://www.jiskha.com/questions/69755/consider-the-differential-equation-du-dt-u-2-t-3-t-a-find-the-general-solution-to | calculus-differential equation
Consider the differential equation:
(du/dt)=-u^2(t^3-t)
a) Find the general solution to the above differential equation. (Write the answer in a form such that its numerator is 1 and its integration constant is C).
u=?
b) Find the particular solution of the above differential equation that satisfies the condition u=4 at t=0.
u=?
1. 👍
2. 👎
3. 👁
1. -du/u^2 = (t^3 - t) dt --->
1/u = 1/4 t^4 - 1/2 t^2 + c ---->
u = 1/[1/4 t^4 - 1/2 t^2 + c]
u=4 at t=0 ---> c = 1/4
1. 👍
2. 👎
2. du/u^2 = -(t^3 -t ) dt
1/u = (1/4)t^4 - (1/2)t^2 + constant
1/u = -t^2 (.5 -.25 t^2) + constant
u = -1/[t^2(.5 - .25 t^2) + C]
4 = -1/C so C = -4
u = -1/[t^2(.5 - .25 t^2) - 4]
1. 👍
2. 👎
3. I forgot a - sign - use his :)
1. 👍
2. 👎
4. Thanks so much. I've been struggling with that thing for days!
1. 👍
2. 👎
5. Oh - actually we agree exactly
1. 👍
2. 👎
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https://www.physicsforums.com/threads/probability-distribution-problem.536568/ | # Homework Help: Probability Distribution Problem
1. Oct 4, 2011
### ashwinnarayan
1. The problem statement, all variables and given/known data
Log-ons to a certain computer website occur randomly at a uniform average rate of 2.4 per minute. State the distribution of the number N of log-ons that occur during a period of t minutes.
Obtain the probablity that at least one log on occurs during a period of t minutes.
Hence obtain the probability density function of T, where T minutes is the interval between successive log-ons.
Identify the distribution of T and state its mean and variance.
2. Relevant equations
3. The attempt at a solution
This one's a bit embarrassing really. What I don't get is the "Hence obtain the probability density function of T, where T minutes is the interval between successive log-ons." part. It's only 3 marks for this section of the question so it's probably something simple. I must be missing some key fact or something.
Here's what I did for the first bit.
The distribution of N is Poisson with a mean of 2.4t so X~Po(2.4t)
The probability of at least one log on is P(X≥1) = 1 - P(X=0) = 1 - e-2.4t
And now I am stuck. This is from a really old A-Level Further Mathematics exam so I only had an examiner's report which gave the answers with a short and confusing explanation. It said that "The probability density function is obtained by taking the derivative of their answer 1 - e-2.4t. Which gives a negative exponential distribution 2.4e-2.4t. The mean and variance is easily 1/2.4 and (1/2.4)2 respectively."
What I don't understand is why the probablity density function is the derivative of that expression.
2. Oct 4, 2011
### Ray Vickson
Let T1 = time to first log-on (after you start measuring times), T2 = time between first and second log-on, etc. The T1, T2, ... are the inter-arrival times of a Poisson process. Assuming the counting probabilities are independent between non-overlapping intervals, and that N = no. arrivals in (t1,t2) has the Poisson distribution with mean a*(t2-t1) [with a = your 2.4], we can show--by a lengthy but not particularly difficult argument--that T1, T2, ... are independent and identically distributed. Furthermore, the number arriving by time t is N(t), where Pr{N(t) = n} = Pr{T1+T2+...+Tn < t, T1+T2+...+Tn+T(n+1) > t} = (a*t)^n *exp(-a*t)/n!, where a = 2.4. In particular, Pr{T1 > t} = Pr{N(t)=0} = exp(-a*t), so the density f(t) of T1 satisfies f(t) = (d/dt) Pr{T1 > t} = a*exp(-a*t), because Pr{T1 > t} = integral_{x=t..infinity} f(x) dx. Also, T2, T3, T4,... all have the same density.
RGV | 2018-10-17T04:03:48 | {
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https://hbfs.wordpress.com/2016/11/15/square-roots-part-v/ | ## Square Roots (Part V)
Last week we examined the complexity of obtaining $k$ in the decomposition $n=2^k+b$ for some integer $n$. This week, we’ll have a look at how we can use it to speed-up square root extraction. Recall, we’re interested in $k$ because
$2^k \leqslant 2^k+b < 2^{k+1}$,
with $0 \leqslant b < 2^k$, which allows us to get easy bounds on $\sqrt{n}$. Better, we also have that
$\sqrt{2^k} \leqslant \sqrt{2^k+b} \leqslant \sqrt{2^{k+1}}$,
and we know how to compute $\sqrt{2^k}=2\frac{k}{2}$ (somewhat efficiently! Let’s combine all this and see how it speeds up things.
Most of the code for square roots is found here, and for shifting, including by half a bit, it is found here.
Using $2^\frac{k}{2}$ and $2^\frac{k+1}{2}$ as lower and upper bound for binary search will provide better performance because we reduce the number of iterations from, say 16 for 16-bits integers to about $\frac{k}{2}$, where $k$ depends on $n$.
unsigned driven_binary_sqrt_bounds(unsigned n, int & i)
{
i=0;
if (n)
{
unsigned k=std::numeric_limits<unsigned>::digits-__builtin_clz(n);
unsigned l=shift_half(1<<k); // OK, truncates
unsigned h=shift_half(1<<(k+1))+1; // rounding error from half shift
while (l!=h)
{
i++;
unsigned m=(l+h)/2;
if (m*m > n)
h=m;
else
l=m+1;
}
return l-1;
}
else
return 0;
}
As for Newton’s method, we can use the knowledge of $k$, and $\frac{k}{2}$ to kick-start search at a better point than $\frac{n}{2}$. The modification to the algorithm aren’t extensive:
unsigned driven_newton_sqrt_bounds(unsigned n, int & i)
{
i=0;
if (n)
{
unsigned k=std::numeric_limits<unsigned>::digits-__builtin_clz(n);
unsigned l=shift_half(1<<k); // OK, truncates anyway
unsigned h=shift_half(1<<(k+1))+1; // rounding error from half shift
unsigned x=(l+h)/2; // should be pretty close
unsigned d;
do
{
i++;
d = n/x;
x = (x+d)/2;
}
while (d<x);
// precision fix: it sometimes rounds up (42.9767
// becomes 43, which is the best answer, but) we
// need truncation (as for the other methods)
if (x*x>n) x--;
return x;
}
else return 0;
}
The last correction can be left out if rounding is deemed OK.
*
* *
How does the knowledge of $k$ helps finding square roots? As we did before, we just instrumented the code to obtain the number of iterations performed in the principal loop. We will use this as a metric of performance rather than, say, raw CPU time. Why so? Because counting the performance of a specific implementation is one thing, and as we showed a couple of weeks ago, shifting by half a bit is a trade-off of speed and precision, and we cannot exclude that someone would go through the trouble of implementing his own CPU with a “shift by half a bit” instruction. Let us, again, refer to this previous post for the detail of all the methods. Now, compare (Y-axis, number of iterations, X-axis, numbers):
where the “Bounded” versions are the version just above.
We see that the use of $k$ reduces greatly the number of iterations for the binary search method. Again, the number of iteration will be about $\frac{k}{2}$. As for the Newton method, using an initial guess of
$\displaystyle x=\frac{1}{2}\left(2^{\frac{k}{2}}+2^{\frac{k+1}{2}}\right)$
somewhat reduces the number of iterations further from the “driven” Newton search and quite a lot from the naïve version where the initial guess is merely $\frac{n}{2}$.
*
* *
Unsurprisingly, then, having better initial guesses is a plus. I’m not sure, though, that the speed-up, as measured by the number of iterations, is worth the added complexity, even if it is a tiny amount. We’ve seen that using a built-in function that maps to an efficient instruction (__builtin_clz that maps to the CPU instruction bsr) is much faster than the equivalent naïve implementation so in the end the added complexity might be beneficial.
### One Response to Square Roots (Part V)
1. […] discussed algorithms for computing square roots a couple of times already, and then some. While sorting notes, I’ve came across something interesting: Archytas’ method for […] | 2020-02-21T03:55:41 | {
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https://math.stackexchange.com/questions/3868595/quotient-of-product-group-is-product-of-quotient-groups | # Quotient of product group is product of quotient groups
Consider the product of cyclic groups $$G = C_4 \times C_2$$. We see that $$H = C_2 \times \{ 0 \}$$ is normal in $$G$$ since $$G$$ is abelian. The cosets are $$H = \{(0, 0), (1, 0)\}$$, $$(0, 1)H = \{(0,1), (1, 1)\}$$, $$(2, 0)H = \{(2, 0), (3, 0)\}$$, and $$(2, 1)H = \{(2, 1), (3, 1) \}$$. All three nonidentity cosets have order $$2$$, so $$G/H \cong C_2 \times C_2$$. On the other hand, $$H \cong J = \{0\} \times C_2 \triangleleft G$$, yet $$G/J = \{ J, (1, 0)J, (2, 0)J, (3, 0)J \} \cong C_4$$. (There's some notation abuse: the $$C_2$$ and $$\{0\}$$ in $$J = \{0\} \times C_2$$ are different from the $$C_2$$ and $$\{0\}$$ in $$H = C_2 \times \{0\}$$.) Therefore, two normal subgroups of $$G$$ may be isomorphic despite having nonisomorphic quotients.
It's tempting to say the following: $$(C_4 \times C_2)/(C_2 \times \{0\}) \cong C_4/C_2 \times C_2 /\{0\}$$ and $$(C_4 \times C_2)/(\{0\} \times C_2) \cong C_4/\{0\} \times C_2/C_2 \text.$$ But does this trick actually work in general? In other words, is the proposition below true?
Let $$G$$ be an external direct product $$G_1 \times G_2 \times \dots \times G_s$$ for some groups $$G_1, G_2, \dots, G_s$$. Let $$H = H_1 \times H_2 \times \dots \times H_s$$ be a normal subgroup of $$G$$ with $$H_i \triangleleft G_i$$ for each $$i = 1, 2, \dots, s$$. Then $$G/H = (G_1 \times G_2 \times \dots \times G_s)/(H_1 \times H_2 \times \dots \times H_s) \cong (G_1/H_1) \times (G_2/H_2) \times \dots \times (G_s/H_s) \text.$$
This result would be like ordinary division of real numbers: $$(a \cdot b)/(c \cdot d) = (a/c) \cdot (b/d)$$.
• Yes, this is true. The proof isn’t bad using the fundamental hom theorem. – Randall Oct 16 '20 at 17:56
There are natural homomorphisms $$G_i\to G_i/H_i$$, which gives rise to the homomorphism $$f:\prod G_i\to \prod(G_i/H_i)$$. This is clearly surjective, since each $$G_i\to G_i/H_i$$ is surjective. We wish to know the kernel.
Let $$(g_i)\in\prod G_i$$ be such that $$f((g_i))=0$$. This is equivalent to, for each $$i$$, $$g_i\in\ker(G_i\to G_i/H_i)=H_i$$. This means that $$\ker f=\prod H_i$$.
Thus, by the isomorphism theorem, $$\prod G_i/\prod H_i\cong\prod(G_i/H_i)$$. | 2021-04-21T13:30:26 | {
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https://stats.stackexchange.com/questions/198409/probability-of-the-maximum-of-n-dice-rolls-being-equal-to-k | # Probability of the maximum of n dice rolls being equal to k
As part of my recent interest in university-level statistics, I'm trying to figure out what the probability distribution is for the following scenario, without having to resort to counting the number of outcomes:
Suppose I roll two dice, but rather than adding them together I use the higher of the two as the "result" of the event.
This is not a homework question, I'm just trying to learn some "advanced" statistics techniques out of personal interest (for example, I've recently learned that you can find the probability distribution for the sum of two independent events by convolving the individual distributions). So I'd like to know if there's a "better" way to do this than enumerating all the possible outcomes and counting those that fit the criterion. I suppose the criteria for "better" would be:
• Less computationally expensive (i.e. faster if it were to be implemented on a computer)
• More interesting from a mathematical analysis standpoint.
Thank you
## Update to question
How would I generalize the explanation in dsaxton's answer, namely:
$P(\max(X_1, X_2) = k) = 2 P(X_1 = k \cap X_2 \leq k) - P(X_1 = k \cap X_2 = k)$
to the maximum of three or more dice?
In the case of the max of three dice, it looks like the cases are:
• $x1 = k$ and $x2 \leq k$ and $x3 \leq k$
• $x2 = k$ and $x1 \leq k$ and $x3 \leq k$
• $x3 = k$ and $x1 \leq k$ and $x2 \leq k$
which suggests that the probability would be
$P(\max(x1, x2, x3) = k) = P(x1 = k \cap x2 \leq k \cap x3 \leq k) + P(x2 = k \cap x1 \leq k \cap x3 \leq k) + P(x3 = k \cap x1 \leq k \cap x2 \leq k) - P(x1 = k \cap x2 = k \cap x3 = k)$
(though maybe this is still multi-counting?)
Is there some way to use combinatorics or some other technique to express the expanded formula for the max of $n$ dice?
• The answer you got sounds great (+1), but here is an alternative derivation. Feb 25, 2016 at 1:33
• This looks like a you made a major change to your question after getting a good answer, which would invalidate the answer you already have (since it doesn't answer the new question). It would be best to ask that as a new question and link the question back here for context, unless dsaxton is happy to answer it here as well. (On the other hand, I think even the extended question may be a duplicate so this may close in any case.) Feb 25, 2016 at 2:46
• Generally, when $n$ independent observations of random variables following a common cumulative distribution $F$ are made, the cdf of the maximum of those observations is $F^n$ (this is an easy deduction from the probability axioms). Thus, the chance that the maximum exactly equals some value $x$ is the chance that it is less than or equal to $x$ but not less than or equal to any number less than $x.$ In short, you seek $\lim_{\epsilon\to 0^+} F^n(x) - F^n(x-\epsilon).$ For the dice and values $x\in\{1,2,\ldots,6\}$ this directly gives $(x/6)^n-((x-1)/6)^n.$
– whuber
Feb 25 at 13:50
You don't need to enumerate all possibilities, just consider that $\max(X_1, X_2) = k$ simply means $X_1 = k$ and $X_2 \leq k$ or $X_2 = k$ and $X_1 \leq k$. So we just add the probabilities of these events (which by symmetry are equal) and subtract their intersection to avoid double counting,
\begin{align} P(\max(X_1, X_2) = k) &= 2 P(X_1 = k \cap X_2 \leq k) - P(X_1 = k \cap X_2 = k) \\ &= \frac{2k}{36} - \frac{1}{36} \\ &= \frac{2k - 1}{36} . \end{align}
To extend this idea to $m$ dice it gets a bit more complicated as we need to be more careful about applying the inclusion exclusion principle. Let $A_i \equiv \{ X_i = k \cap X_j \leq k \text{ for$j \neq i$}\}$. Then
\begin{align} P(\max(X_1, \ldots , X_m) = k) =& P(\cup_{i=1}^{m} A_i) \\ =& \sum_{i=1}^{m} P(A_i) - \sum_{i \neq j} P(A_i \cap A_j) + \\ & \sum_{i \neq j, i \neq l, j \neq l} P(A_i \cap A_j \cap A_l) - \\ & \ldots + (-1)^{m + 1} P(\cap_{i=1}^{m} A_i) \\ =& \frac{m k^{m - 1}}{6^m} - \binom{m}{2} \frac{k^{m - 2}}{6^m} + \\ & \binom{m}{3} \frac{k^{m - 3}}{6^m} - \ldots + (-1)^{m + 1} \frac{1}{6^m} \\ =& \frac{1}{6^m} \sum_{i=1}^{m} (-1)^{i+1} \binom{m}{i} k^{m - i} . \end{align}
## Edit:
I just realized the title of your post seems to be asking a different question than what's in the post itself. If you want to find the probability that $X_1 \geq X_2$ then you can take a conditional expectation,
\begin{align} P(X_1 \geq X_2) &= \text{E}[P(X_1 \geq X_2 \mid X_1)] \\ &= \text{E} \left ( \frac{X_1}{6} \right ) \\ &= \frac{3.5}{6} . \end{align}
Here we've used the general result that $\text{E}(X) = \text{E}[\text{E}(X \mid Y)]$ for random variables $X$ and $Y$, and that probabilities themselves may be viewed as expectations of indicator random variables. Even simpler perhaps, we can arrive at the answer by basically only using the fact that probabilities sum to one,
$$P(X_1 \geq X_2) + P(X_2 \geq X_1) - P(X_1 = X_2) = 1 \Rightarrow \\ 2 P(X_1 \geq X_2) - \frac{1}{6} = 1 \Rightarrow \\ P(X_1 \geq X_2) = \frac{3.5}{6} .$$
• This helped greatly, thank you - I just worked it out on paper and I follow... How would I generalize this to the maximum of three or more dice? In the case of the max of three dice, it looks like the cases are: x1 = k and x2 <= k and x3 <= k ; x2 = k and x1 <= k and x3 <= k ; x3 = k and x1 <= k and x2 <= k. Is there some way to use combinations to figure out what the expanded formula should be? Feb 25, 2016 at 2:17
• I added the mass function for general $m$ to the post. It's a bit dense so let me know if anything isn't clear. Feb 25, 2016 at 2:47
• Actually, your first answer (before the edit) is exactly what I was looking for. I was trying to figure out the probability of the value k being the maximum of three d20 rolls. You used d6 dice but it's easy to generalize to others. May 8, 2021 at 16:11
A simpler solution that gives the same final answer as for using the inclusion–exclusion principle is to define two event:
A = at least one of the n dice rolls results is k.
B = all of the dice rolls results is less or equal to k.
Now if X is the greatest result, then:
P(X=k) = P(A and B) = P(A|B)•P(B) = (1-P(A^c|B))•P(B).
P(B) = the probability to get a number from the set {1,..,k} out of the 6 optional results.
P(B) = (k/6)^n
P(A^c|B) = considering the fact that every result can be either number from {1,..,k}, the probability not getting k at all will be choosing a number from the set {1,..,(k-1)} of the k optional results.
P(A^c|B) = ((k-1)/k)^n
So the final answer will be:
P(X=k) = (1-((k-1)/k)^n)•(k/6)^n
• The idea is good but the calculations aren't quite correct: $k$ should not appear in the denominators of the fractions.
– whuber
Feb 25 at 13:50
• Are you sure about that? first of all, by simplifying "dsaxton" result using wolframalpha we get that both answers are identical. Second, how else would you calculate: P(A^c|B) ? which means - we wouldn't get k in the dice, knowing that the dice will fall on a number from {1,...,k} Mar 6 at 15:34
• You are correct: I did not notice that the powers of $k$ cancel in your formula. For the simplified version, see my comment to the question.
– whuber
Mar 6 at 15:36 | 2022-07-01T06:11:13 | {
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https://byjus.com/question-answer/if-a-1-a-2-a-3-a-n-are-in-hp-then-a-1/ | Question
# If $${ a }_{ 1 },{ a }_{ 2 },{ a }_{ 3 },....{ a }_{ n }$$ are in HP, then $${ a }_{ 1 }{ a }_{ 2 }+{ a }_{ 2 }{ a }_{ 3 }+.....+{ a }_{ n-1 }{ a }_{ n }$$ will be equal to
A
a1an
B
na1an
C
(n1)a1an
D
None of these
Solution
## The correct option is C $$(n-1){ a }_{ 1 }{ a }_{ n }$$Given $${ a }_{ 1 },{ a }_{ 2 },{ a }_{ 3 },......{ a }_{ n }$$ are in HPThen, $$\cfrac { 1 }{ { a }_{ 1 } } ,\cfrac { 1 }{ { a }_{ 2 } } ,\cfrac { 1 }{ { a }_{ 3 } } ,.....\cfrac { 1 }{ { a }_{ n } }$$ will be in APwhich gives$$\cfrac { 1 }{ { a }_{ 2 } } -\cfrac { 1 }{ { a }_{ 1 } } =\cfrac { 1 }{ { a }_{ 3 } } -\cfrac { 1 }{ { a }_{ 2 } } =.....=\cfrac { 1 }{ { a }_{ n } } -\cfrac { 1 }{ { a }_{ n-1 } } =d$$$$\Rightarrow \cfrac { { a }_{ 1 }-{ a }_{ 2 } }{ { a }_{ 1 }{ a }_{ 2 } } =\cfrac { { a }_{ 2 }-{ a }_{ 3 } }{ { a }_{ 2 }{ a }_{ 3 } } =......=\cfrac { { a }_{ n-1 }-{ a }_{ n } }{ { a }_{ n-1 }{ a }_{ n } } =d$$$$\Rightarrow { a }_{ 1 }-{ a }_{ 2 }=d{ a }_{ 1 }{ a }_{ 2 }$$$${ a }_{ 2 }-{ a }_{ 3 }=d{ a }_{ 2 }{ a }_{ 3 }$$............$${ a }_{ n-1 }-{ a }_{ n }=d{ a }_{ n-1 }{ a }_{ n }$$On adding these, we have$$d\left( { a }_{ 1 }{ a }_{ 2 }+{ a }_{ 2 }{ a }_{ 3 }+....+{ a }_{ n-1 }{ a }_{ n } \right) ={ a }_{ 1 }-{ a }_{ n }$$ ....(i)Also $$n^{th}$$ term of this AP is given by$$\cfrac { 1 }{ { a }_{ n } } =\cfrac { 1 }{ { a }_{ 1 } } +(n-1)d\quad$$$$d=\cfrac { { a }_{ 1 }-{ a }_{ n } }{ { a }_{ 1 }{ a }_{ n }(n-1) }$$On substituting this value of $$d$$ in Eq. (i), we get$$\left( { a }_{ 1 }-{ a }_{ n } \right) =\cfrac { { a }_{ 1 }-{ a }_{ n } }{ { a }_{ 1 }{ a }_{ n }(n-1) } \left( { a }_{ 1 }{ a }_{ 2 }+{ a }_{ 2 }{ a }_{ 3 }+....+{ a }_{ n-1 }{ a }_{ n } \right)$$$$\Rightarrow \left( { a }_{ 1 }{ a }_{ 2 }+{ a }_{ 2 }{ a }_{ 3 }+....+{ a }_{ n-1 }{ a }_{ n } \right) ={ a }_{ 1 }{ a }_{ n }(n-1)\quad$$Maths
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View More | 2022-01-17T22:50:37 | {
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https://math.stackexchange.com/questions/2065824/are-closed-sets-in-topology | # Are closed sets in topology?
Self-studying Folland Chapter 4 Point Set Topology
A topology on $X$ is a family $\mathcal{T}$ of subsets of $X$ that contains $\emptyset$ and $X$ and is closed under arbitrary unions and finite intersections.
Members of $\mathcal{T}$ are called open sets. Their complements are called closed sets. I don't think it's necessary that closed sets are in the topology? Or can there actually be closed sets in topology?
• There can be, but usually not all closed sets are in the topology, and very often none other than the empty set and the whole space are. – Matt Samuel Dec 20 '16 at 11:36
• If $\emptyset$ belong to topology, it's open by definition. Its complement is $X$, so $X$ is closed. However, $X$ belongs to the topology, too. So closed sets may belong to topology. – CiaPan Dec 20 '16 at 11:37
• Depending on the space, a closed set can be open. In fact, $\emptyset$ and $X$ are. Some people call "clopens" to these sets. – ajotatxe Dec 20 '16 at 11:41
• @ZHU, you have asked 45 questions but accepted answers to only 6 of them. While nobody is expected to accept answers to all the asked questions, 6 out of 45 is not reasonable. Please review your questions and look for acceptable answers, I am sure you will find some. – Alex M. Dec 20 '16 at 15:45
Indeed, closed sets usually do not belong to the topology. Nevertheless, there are cases when they do. One such example is of $\mathcal T = \mathcal P (X)$, the set of all the subsets of $X$. Another example is $\mathcal T = \{ \emptyset, X \}$.
Furthermore, $\emptyset$ and $X$ are always both closed and open, so they are examples of closed sets belonging to $\mathcal T$.
Every connected component of $X$ is both closed and open.
• Yes, $\emptyset$ and $X$ are by definition always in the topology. And you can prove that they are both open and closed. – Jakob Elias Dec 20 '16 at 11:44
Depending on the space, a closed set can be open. In fact, $\emptyset$ and $X$ always are, but there are more examples. Some people call these sets "clopen sets". A clopen set and its complement describe in some sense a splitting of the space. Spaces that have non trivial clopen sets are called "not connected". | 2019-09-21T03:09:09 | {
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http://math.stackexchange.com/questions/186400/finding-the-last-digit-of-largest-mersenne-prime | Finding the last digit of largest mersenne prime
I was wondering what the best way was to find the last digit of the largest known Mersenne prime $2^{6972593} - 1$? Is there any logical way to do this or will I have to just some how compute the answer and then find the last digit?
-
The answer is yes, there is in fact a "logical" (i.e. simple) way to do this! When computing the last digit of a number, we only need to consider the number mod 10 (that is, we only need to compute its remainder when divided by 10). So let's do that!
We see the powers of $2 \pmod{10}$ cycle nicely:
$2^1 \equiv 2 \pmod{10}$
$2^2 \equiv 4 \pmod{10}$
$2^3 \equiv 8 \pmod{10}$
$2^4 \equiv 6 \pmod{10}$
$2^5 \equiv 2 \pmod{10}$
$2^6 \equiv 4 \pmod{10}$
...etc.
And now the pattern appears obvious. Since the powers of $2$ mod 10 repeat every 4 numbers, we only need to compute $6972593 \pmod 4$. This is simply $1$, and so $2^{6972593} \equiv 2^1 \pmod{10}$ and so it ends in a $2$. Thus, $2^{6972593}-1$ ends in $\boxed{1}$, as desired.
Ask me if you don't understand anything :)
-
Excellently explained. Thanks! – Jeel Shah Aug 24 '12 at 18:09
All you need to know is the answer modulo $10$. The powers of $2$ mod $10$ have an obvious pattern: $2, 4, 8, 6, 2, 4, 8, 6, ...$ which has a period of $4$, and since $6972593 \equiv 1 (\bmod 4)$, $$2^{6972593} - 1 \equiv 2 - 1 = 1 (\bmod 10).$$
-
$2^4 \equiv 1 \mod 5 \Rightarrow 2^{M4+1} \equiv 2 \mod 5 \,.$
Now, (this is the Chinese Remainder Theorem in a hidden form)
$$2^{6972593} -1\equiv 1 \mod 5 \mbox { and odd} \Rightarrow 2^{6972593} -1\equiv 1 \mod 10 \,.$$
Alternately
$$2^{6972593} -2= 2[(2^4)^m-1]=2 \cdot [(2^4)-1][2^4{m-1}+2^4{m-2}+...+2^4+1]=2 \cdot 5 \cdot 3 \cdot \mbox{junk}$$
Since $2^{6972593} -2$ is divisible by 10, you get the same result.
- | 2015-08-28T22:57:42 | {
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https://groupprops.subwiki.org/w/index.php?title=Normal_subgroup&mobileaction=toggle_view_mobile | # Normal subgroup
ALSO CHECK OUT: Questions page (list of common doubts/questions/curiosities) |
## Definition
QUICK PHRASES: invariant under inner automorphisms, self-conjugate subgroup, same left and right cosets, kernel of a homomorphism, subgroup that is a union of conjugacy classes
### Equivalent definitions in tabular format
Note that each of these definitions (except the first one, as noted) assumes that we already have a group and a subgroup. To prove normality using any of these definitions, we first need to check that we actually have a subgroup.
No. Shorthand A subgroup of a group is normal in it if... A subgroup $H$ of a group $G$ is normal in $G$ if ... Applications to... Additional comments
1 homomorphism kernel it is the kernel of a homomorphism from the group. there is a homomorphism $\varphi$ from $G$ to a group $K$ such that the kernel of $\varphi$ is precisely $H\!$. In other words, $\varphi(x) \!$ is the identity element of $K$ if and only if $x \in H$. proving normality In this case, we do not need to separately check that $H$ is a subgroup since the kernel of a homomorphism is automatically a subgroup.
2 inner automorphism invariance it is invariant under all inner automorphisms. for all $g \in G$, $gHg^{-1} \subseteq H$. More explicitly, for all $g \in G, h \in H$, we have $ghg^{-1} \in H$. proving normality Thus, normality is the invariance property with respect to the property of an automorphism being inner. This definition also motivates the term invariant subgroup for normal subgroup (which was used earlier).
3 equals conjugates it equals each of its conjugates in the whole group. for all $g$ in $G$, $gHg^{-1} = H$. This definition also motivates the term self-conjugate subgroup for normal subgroup (which was used earlier).
4 left/right cosets equal its left cosets are the same as its right cosets (that is, it commutes with every element of the group). for all $g$ in $G$, $gH = Hg$. proving normality When we say $gH = Hg$, we only mean equality as sets. It is not necessary that $gh = hg$ for $h \in H$. That stronger condition defines central subgroup.
5 union of conjugacy classes it is a union of conjugacy classes. $H$ is a union of conjugacy classes in $G$
6 commutator inside it contains its commutator with the whole group. the commutator $[H,G]$ (which coincides with the commutator $[G,H]$) is contained in $H$. proving normality
7 conjugates of generating set inside given a generating set for the whole group and a generating set for the subgroup, every conjugate of an element in the latter by an element in the former, as well as by its inverse, is in the subgroup. given a generating set $A$ for $G$ and $B$ for $H$, we have $aba^{-1} \in H$ and $a^{-1}ba \in H$ for all $a \in A, b\in B$. normality testing problem For finite groups, we need only check conjugates by elements in the generating set and not by their inverses.
For more definitions, see nonstandard definitions of normal subgroup.
This definition is presented using a tabular format. |View all pages with definitions in tabular format
### Notation and terminology
For a subgroup $H \!$ of a group $G \!$, we denote the normality of $\! H$ in $\! G$ by $H \underline{\triangleleft} G$ or $G \underline{\triangleright} H$Notations. In words, we say that $\! H$ is normal in $\! G$ or a normal subgroup of $\! G$.
### Equivalence of definitions
Pair of definitions Explanation of equivalence More related information
(1) and (2) Normal subgroup equals kernel of homomorphism first isomorphism theorem
(2) and (3) Follows from the more general fact that restriction of automorphism to subgroup invariant under it and its inverse is automorphism, combined with the fact that the inverse of an inner automorphism is also an inner automorphism (in fact, the inverse of conjugation by $g$ is conjugation by $g^{-1}$) group acts as automorphisms by conjugation
(3) and (4) A direct manipulation of equations involving elements and subsets. For full proof, refer: equivalence of conjugacy and coset definitions of normality. manipulating equations in groups
(2) (or (3)) and (5) A straightforward unraveling of the meaning of conjugacy class
(2) (or (3)) and (6) A straightforward unraveling of the meaning of commutator, along with a little bit of manipulation. For full proof, refer: equivalence of conjugacy and commutator definitions of normality manipulating equations in groups
### Copyable LaTeX
The following is LaTeX for a quick definition of normality.
A subgroup $H$ of a group $G$ is termed a {\em normal subgroup} if $ghg^{-1} \in H$ for all $g \in G$ and $h \in H$.
VIEW RELATED: Analogues of this | Variations of this | Opposites of this |[SHOW MORE]
This article defines a subgroup property that is pivotal (viz important) among existing subgroup properties
View a list of pivotal subgroup properties | View a complete list of subgroup properties[SHOW MORE]
## Importance
The notion of normal subgroup is important because of two main reasons:
• Normal subgroups are precisely the kernels of homomorphisms
• Normal subgroups are precisely the subgroups invariant under inner automorphisms, and for a group action, the only relevant automorphisms of the acting group that correspond to symmetries of the set being acted upon, are inner automorphisms.
Further information: Ubiquity of normality
## Examples
VIEW: subgroups of groups satisfying this property | subgroups of groups dissatisfying this property
VIEW: Related subgroup property satisfactions | Related subgroup property dissatisfactions
### Extreme examples
1. The trivial subgroup is always normal. Further information: Trivial subgroup is normal
2. Every group is normal as a subgroup of itself. Further information: Every group is normal in itself
### Examples
1. High occurrence example: In an abelian group, every subgroup is normal (there are non-abelian groups, such as the quaternion group, where every subgroup is normal. Groups in which every subgroup is normal are called Dedekind groups, and the non-abelian ones are called Hamiltonian groups). Further information: abelian implies every subgroup is normal
2. If $G$ is an internal direct product of subgroups $H$ and $K$, both $H$ and $K$ are normal in $G$. Further information: direct factor implies normal
3. Every subgroup-defining function yields a normal subgroup (in fact, it yields a characteristic subgroup). For instance, the center, derived subgroup and Frattini subgroup in any group are normal. Further information: subgroup-defining function value is characteristic, characteristic implies normal
### Non-examples
Here are some examples of non-normal subgroups:
1. In the symmetric group on three letters, the subgroup S2 in S3, i.e., the two-element subgroup generated by a transposition, is not normal (in fact, there are three such subgroups and they're all conjugate). Further information: S2 is not normal in S3
2. More generally, in any dihedral group of degree at least $3$, the two-element subgroup generated by a reflection is not normal. Further information: Two-element subgroup generated by reflection is not normal in dihedral group
3. Low occurrence example: In a simple group, no proper nontrivial subgroup is normal. Thus, any proper nontrivial subgroup of a simple group gives a counterexample. The smallest simple non-Abelian group is the alternating group on five letters.
### Subgroups satisfying the property
Here are examples of subgroups that satisfy the property of being normal:
Here are some examples of subgroups in basic/important groups satisfying the property:
Group partSubgroup partQuotient part
Z2 in V4Klein four-groupCyclic group:Z2Cyclic group:Z2
Here are some examples of subgroups in relatively less basic/important groups satisfying the property:
Group partSubgroup partQuotient part
A4 in S4Symmetric group:S4Alternating group:A4Cyclic group:Z2
Center of dihedral group:D8Dihedral group:D8Cyclic group:Z2Klein four-group
Center of quaternion groupQuaternion groupCyclic group:Z2Klein four-group
Center of special linear group:SL(2,3)Special linear group:SL(2,3)Cyclic group:Z2Alternating group:A4
Center of special linear group:SL(2,5)Special linear group:SL(2,5)Cyclic group:Z2Alternating group:A5
Cyclic maximal subgroup of dihedral group:D8Dihedral group:D8Cyclic group:Z4Cyclic group:Z2
Cyclic maximal subgroups of quaternion groupQuaternion groupCyclic group:Z4Cyclic group:Z2
First agemo subgroup of direct product of Z4 and Z2Direct product of Z4 and Z2Cyclic group:Z2Klein four-group
First omega subgroup of direct product of Z4 and Z2Direct product of Z4 and Z2Klein four-groupCyclic group:Z2
Klein four-subgroup of alternating group:A4Alternating group:A4Klein four-groupCyclic group:Z3
Klein four-subgroups of dihedral group:D8Dihedral group:D8Klein four-groupCyclic group:Z2
Non-characteristic order two subgroups of direct product of Z4 and Z2Direct product of Z4 and Z2Cyclic group:Z2Cyclic group:Z4
Normal Klein four-subgroup of symmetric group:S4Symmetric group:S4Klein four-groupSymmetric group:S3
Z4 in direct product of Z4 and Z2Direct product of Z4 and Z2Cyclic group:Z4Cyclic group:Z2
Here are some examples of subgroups in even more complicated/less basic groups satisfying the property:
Group partSubgroup partQuotient part
Center of M16M16Cyclic group:Z4Klein four-group
Center of dihedral group:D16Dihedral group:D16Cyclic group:Z2Dihedral group:D8
Center of direct product of D8 and Z2Direct product of D8 and Z2Klein four-groupKlein four-group
Center of semidihedral group:SD16Semidihedral group:SD16Cyclic group:Z2Dihedral group:D8
Cyclic maximal subgroup of dihedral group:D16Dihedral group:D16Cyclic group:Z8Cyclic group:Z2
Cyclic maximal subgroup of semidihedral group:SD16Semidihedral group:SD16Cyclic group:Z8Cyclic group:Z2
D8 in D16Dihedral group:D16Dihedral group:D8Cyclic group:Z2
D8 in SD16Semidihedral group:SD16Dihedral group:D8Cyclic group:Z2
Derived subgroup of M16M16Cyclic group:Z2Direct product of Z4 and Z2
Derived subgroup of dihedral group:D16Dihedral group:D16Cyclic group:Z4Klein four-group
Direct product of Z4 and Z2 in M16M16Direct product of Z4 and Z2Cyclic group:Z2
Klein four-subgroup of M16M16Klein four-groupCyclic group:Z4
Non-central Z4 in M16M16Cyclic group:Z4Cyclic group:Z4
Q8 in SD16Semidihedral group:SD16Quaternion groupCyclic group:Z2
SL(2,3) in GL(2,3)General linear group:GL(2,3)Special linear group:SL(2,3)Cyclic group:Z2
### Subgroups dissatisfying the property
Here are examples of subgroups that do not satisfy the property of being normal.
Here are some examples of subgroups in basic/important groups not satisfying the property:
Here are some some examples of subgroups in relatively less basic/important groups not satisfying the property:
Group partSubgroup partQuotient part
A3 in A4Alternating group:A4Cyclic group:Z3
A3 in A5Alternating group:A5Cyclic group:Z3
A3 in S4Symmetric group:S4Cyclic group:Z3
A4 in A5Alternating group:A5Alternating group:A4
D8 in A6Alternating group:A6Dihedral group:D8
D8 in S4Symmetric group:S4Dihedral group:D8
Klein four-subgroup of alternating group:A5Alternating group:A5Klein four-group
Non-normal Klein four-subgroups of symmetric group:S4Symmetric group:S4Klein four-group
Non-normal subgroups of dihedral group:D8Dihedral group:D8Cyclic group:Z2
S2 in S4Symmetric group:S4Cyclic group:Z2
Subgroup generated by double transposition in symmetric group:S4Symmetric group:S4Cyclic group:Z2
Twisted S3 in A5Alternating group:A5Symmetric group:S3
Here are some examples of subgroups in even more complicated/less basic groups not satisfying the property:
Group partSubgroup partQuotient part
2-Sylow subgroup of general linear group:GL(2,3)General linear group:GL(2,3)Semidihedral group:SD16
Non-normal subgroups of M16M16Cyclic group:Z2
## Facts
### Isomorphism theorems
Theorem name/number Statement
First isomorphism theorem If $\varphi:G \to H$ is a surjective homomorphism, and $N$ is the kernel of $\varphi$, then $N$ is a normal subgroup, and if $\alpha:G \to G/N$ is the quotient map, then there is a unique isomorphism $\psi:G/N \to H$ such that $\psi \circ \alpha = \varphi$.
Second isomorphism theorem (diamond isomorphism theorem) If $N,H \le G$ are subgroups such that $H$ is contained in the normalizer of $N$, then $N$ is normal in $NH$ and $NH/N \cong H/(H \cap N)$.
Third isomorphism theorem If $H \le K \le G$ are groups with both $H,K$ normal in $G$, then $H$ is normal in $K$ and $(G/H)/(K/H) \cong G/K$.
Fourth isomorphism theorem (lattice isomorphism theorem) If $H$ is normal in $G$, then there is a bijective correspondence between subgroups of $G/H$ and subgroups of $G$ containing $H$, satisfying many nice conditions.
## Metaproperties
BEWARE! This section of the article uses terminology local to the wiki, possibly without giving a full explanation of the terminology used (though efforts have been made to clarify terminology as much as possible within the particular context)
Metaproperty name Satisfied? Proof Statement with symbols
abelian-tautological subgroup property Yes abelian implies every subgroup is normal If $H \le G$ and $G$ is abelian, then $H$ is normal in $G$.
transitive subgroup property No Normality is not transitive We can have $H \le K \le G$ such that $H$ is normal in $K$ and $K$ is normal in $G$ but $H$ is not normal in $G$.
trim subgroup property Yes Every group is normal in itself, trivial subgroup is normal trivial subgroup and whole group are both normal
strongly intersection-closed subgroup property Yes Normality is strongly intersection-closed If $H_i, i \in I$ are all normal subgroups of a group $G$, then the intersection of subgroups $\bigcap_{i \in I} H_i$ is also a normal subgroup of $G$.
strongly join-closed subgroup property Yes Normality is strongly join-closed If $H_i, i \in I$ are all normal subgroups of $G$, then the join of subgroups $\langle H_i \rangle_{i \in I}$ is also a normal subgroup of $G$.
quotient-transitive subgroup property Yes Normality is quotient-transitive If $H \le K \le G$ are groups such that $H$ is normal in $G$ and $K/H$ is normal in $G/H$, then $K$ is normal in $G$
intermediate subgroup condition Yes Normality satisfies intermediate subgroup condition If $H \le K \le G$ are groups such that $H$ is normal in $G$, then $H$ is normal in $K$.
transfer condition Yes Normality satisfies transfer condition If $H, K \le G$ are groups such that $H$ is normal in $G$, then we must have that $H \cap K$ normal in $K$
image condition Yes Normality satisfies image condition If $H$ is a normal subgroup of $G$ and $\varphi:G \to K$ is a surjective homomorphism of groups, then $\varphi(H)$ is normal in $K$.
inverse image condition Yes Normality satisfies inverse image condition If $H$ is a normal subgroup of $G$ and $\varphi:K \to G$ is a homomorphism of groups, then $\varphi^{-1}(H)$ is normal in $K$.
upper join-closed subgroup property Yes Normality is upper join-closed If $H \le G$, and $K_i, i \in I$ are all subgroups of $G$ containing $H$ such that $H$ normal in each $K_i$, then we must have that $H$ is normal in the join of subgroups $\langle K_i \rangle_{i \in I}$.
commutator-closed subgroup property Yes Normality is commutator-closed If $H, K$ are both normal subgroups of a group $G$, then the commutator $[H,K]$ normal in $G$
centralizer-closed subgroup property Yes Normality is centralizer-closed If $H$ is a normal subgroup of $G$, the centralizer $C_G(H)$ is also normal in $G$.
direct product-closed subgroup property Yes Normality is direct product-closed If $I$ is an indexing set, $G_i, i \in I$ are groups, and $H_i, i \in I$ are such that each $H_i$ is normal in the corresponding $G_i$, then in the external direct product of the $G_i$s, the subgroup given by the external direct product of the $H_i$s is a normal subgroup.
This also implies that normality is a finite direct power-closed subgroup property.
strongly UL-intersection-closed subgroup property Yes Normality is strongly UL-intersection-closed If $I$ is an indexing set, $G$ is a group, and we have groups $H_i \le K_i \le G$ for each $i \in I$, such that $H_i$ is normal in $K_i$, then $\bigcap_{i \in I} H_i$ is a normal subgroup of $\bigcap_{i \in I} K_i$.
Arguesian subgroup property Yes Normality is Arguesian The collection of normal subgroups of a group form an Arguesian lattice.
lower central series condition Yes Normality satisfies lower central series condition Suppose $H$ is a normal subgroup of a group $G$. Then, each lower central series member $\gamma_k(H)$ is a normal subgroup inside the corresponding lower central series member $\gamma_k(G)$.
partition difference condition Yes Normality satisfies partition difference condition Suppose $H$ is a subgroup of a group $G$, and $H$ has a nontrivial partition as a union of $H_i, i \in I$ (nontrivial means that $I$ has size more than one). If all except possibly one of the $H_i$s are normal subgroups of $G$, then all the $H_i$s are normal subgroups of $G$.
conditionally lattice-determined subgroup property No No subgroup property between normal Sylow and subnormal or between Sylow retract and retract is conditionally lattice-determined It is possible to have a group $G$, an automorphism $\varphi$ of the lattice of subgroups, and a normal subgroup $H$ of $G$ such that $\varphi(H)$ is not normal.
## Relation with other properties
This property is a pivotal (important) member of its property space. Its variations, opposites, and other properties related to it and defined using it are often studied
Some of these can be found at:
To get a broad overview, check out the survey articles:
### Stronger properties
The most important stronger property is characteristic subgroup. See the table below for many stronger properties and the way they're related:
Property Meaning Proof of implication Proof of strictness (reverse implication failure) Intermediate notions Comparison Collapse
characteristic subgroup invariant under all automorphisms characteristic implies normal normal not implies characteristic (see also list of examples) Center-fixing automorphism-invariant subgroup, Characteristic subgroup of direct factor, Cofactorial automorphism-invariant subgroup, IA-automorphism-invariant subgroup, Normal-extensible automorphism-invariant subgroup, Normal-potentially characteristic subgroup, Normal-potentially relatively characteristic subgroup, Semi-strongly image-potentially characteristic subgroup, Strongly image-potentially characteristic subgroup, Upper join of characteristic subgroups|FULL LIST, MORE INFO characteristic versus normal group in which every normal subgroup is characteristic
central factor inner automorphism of whole group is inner on subgroup central factor implies normal normal not implies central factor (see also list of examples) Conjugacy-closed normal subgroup, Locally inner automorphism-balanced subgroup, Normal subgroup whose center is contained in the center of the whole group, Normal subgroup whose focal subgroup equals its derived subgroup, SCAB-subgroup, Transitively normal subgroup|FULL LIST, MORE INFO central factor versus normal group in which every normal subgroup is a central factor
direct factor factor in internal direct product direct factor implies normal normal not implies direct factor (see also list of examples) Characteristic subgroup of direct factor, Complemented central factor, Complemented normal subgroup, Complemented transitively normal subgroup, Conjugacy-closed normal subgroup, Direct factor over central subgroup, Endomorphism kernel, Intermediately endomorphism kernel, Join of finitely many direct factors, Join-transitively central factor, Locally inner automorphism-balanced subgroup, Normal AEP-subgroup, Normal subgroup having a 1-closed transversal, Normal subgroup in which every subgroup characteristic in the whole group is characteristic, Normal subgroup whose focal subgroup equals its derived subgroup, Powering-invariant normal subgroup, Quotient-powering-invariant subgroup, Right-quotient-transitively central factor, SCAB-subgroup, Transitively normal subgroup... further results|FULL LIST, MORE INFO direct factor versus normal group in which every normal subgroup is a direct factor
central subgroup contained in the center central implies normal normal not implies central (see also list of examples) Abelian normal subgroup, Amalgam-characteristic subgroup, Amalgam-strictly characteristic subgroup, Center-fixing automorphism-invariant subgroup, Class two normal subgroup, Commutator-in-center subgroup, Conjugacy-closed normal subgroup, Dedekind normal subgroup, Direct factor over central subgroup, Hereditarily normal subgroup, Join-transitively central factor, Nilpotent normal subgroup, SCAB-subgroup, Transitively normal subgroup|FULL LIST, MORE INFO normal subgroup versus central subgroup abelian group
Other less important properties that are stronger than normality:
Property Meaning Proof of implication Proof of strictness (reverse implication failure) Intermediate notions Collapse
fully invariant subgroup invariant under all endomorphisms (via characteristic) (via characteristic)(see also list of examples) Characteristic subgroup, Finite direct power-closed characteristic subgroup, Fully invariant-potentially fully invariant subgroup, Image-potentially fully invariant subgroup, Injective endomorphism-invariant subgroup, Normal-potentially fully invariant subgroup, Normality-preserving endomorphism-invariant subgroup, Potentially fully invariant subgroup, Retraction-invariant characteristic subgroup, Retraction-invariant normal subgroup, Strictly characteristic subgroup|FULL LIST, MORE INFO group in which every normal subgroup is fully invariant
isomorph-free subgroup no other isomorphic subgroup (via characteristic) (via characteristic) (see also list of examples) Characteristic subgroup, Hall-relatively weakly closed subgroup, Injective endomorphism-invariant subgroup, Intermediately characteristic subgroup, Isomorph-automorphic normal subgroup, Isomorph-containing subgroup, Isomorph-normal characteristic subgroup, Isomorph-normal subgroup, Sub-isomorph-free subgroup|FULL LIST, MORE INFO group in which every normal subgroup is isomorph-free
isomorph-containing subgroup contains all isomorphic subgroups (via characteristic) (via characteristic) (see also list of examples) Characteristic subgroup, Injective endomorphism-invariant subgroup, Intermediately characteristic subgroup|FULL LIST, MORE INFO
cocentral subgroup product with center is whole group cocentral implies normal normal not implies cocentral Conjugacy-closed normal subgroup, Right-quotient-transitively central factor, SCAB-subgroup, Transitively normal subgroup|FULL LIST, MORE INFO abelian group
SCAB-subgroup subgroup-conjugating automorphism restricts to subgroup-conjugating automorphism of subgroup Transitively normal subgroup|FULL LIST, MORE INFO
conjugacy-closed normal subgroup conjugacy-closed subgroup and normal subgroup Normal subgroup whose center is contained in the center of the whole group, Normal subgroup whose focal subgroup equals its derived subgroup, Transitively normal subgroup|FULL LIST, MORE INFO
complemented normal subgroup normal and permutably complemented (by definition) normal not implies permutably complemented Endomorphism kernel, Intermediately endomorphism kernel, Normal subgroup having a 1-closed transversal, Powering-invariant normal subgroup, Quotient-powering-invariant subgroup|FULL LIST, MORE INFO
endomorphism kernel occurs as the kernel of an endomorphism (by definition) normal not implies endomorphism kernel Powering-invariant normal subgroup, Quotient-powering-invariant subgroup|FULL LIST, MORE INFO
For a complete list of subgroup properties stronger than Normal subgroup, click here
STRONGER PROPERTIES SATISFYING SPECIFIC METAPROPERTIES: transitive | intermediate subgroup condition | transfer condition | quotient-transitive |intersection-closed |join-closed | trim | inverse image condition | image condition | centralizer-closed |
STRONGER PROPERTIES DISSATISFYING SPECIFIC METAPROPERTIES: transitive | intermediate subgroup condition | transfer condition | quotient-transitive |intersection-closed |join-closed | trim | inverse image condition | image condition | centralizer-closed |
### Conjunction with other properties
Important conjunctions of normality with other subgroup properties are in the table below:
conjugacy-closed normal subgroup conjugacy-closed subgroup Normal subgroup whose center is contained in the center of the whole group, Normal subgroup whose focal subgroup equals its derived subgroup, Transitively normal subgroup|FULL LIST, MORE INFO
normal Sylow subgroup Sylow subgroup Complemented fully invariant subgroup, Complemented homomorph-containing subgroup, Complemented normal subgroup, Homomorph-containing subgroup, Isomorph-free subgroup, Normal subgroup having no common composition factor with its quotient group, Normal subgroup having no nontrivial homomorphism from its quotient group, Normal subgroup having no nontrivial homomorphism to its quotient group, Normal-homomorph-containing subgroup, Order-normal subgroup, Order-unique subgroup, Sub-homomorph-containing subgroup|FULL LIST, MORE INFO
normal Hall subgroup Hall subgroup Complemented fully invariant subgroup, Complemented homomorph-containing subgroup, Complemented normal subgroup, Homomorph-containing subgroup, Isomorph-free subgroup, Normal subgroup having no common composition factor with its quotient group, Normal subgroup having no nontrivial homomorphism from its quotient group, Normal subgroup having no nontrivial homomorphism to its quotient group, Normal-homomorph-containing subgroup, Order-normal subgroup, Order-unique subgroup, Sub-homomorph-containing subgroup|FULL LIST, MORE INFO
View a complete list of conjunctions of normality with subgroup properties
We are often also interested in the conjunction of normality with group properties. By this, we mean the subgroup property of being normal as a subgroup and having the given group property as an abstract group. Examples are in the table below:
abelian normal subgroup abelian group Class two normal subgroup, Commutator-in-center subgroup, Dedekind normal subgroup, Nilpotent normal subgroup, Normal subgroup whose inner automorphism group is central in automorphism group, Solvable normal subgroup|FULL LIST, MORE INFO
cyclic normal subgroup cyclic group Abelian normal subgroup, Commutator-in-center subgroup, Dedekind normal subgroup, Hereditarily normal subgroup, Homocyclic normal subgroup, Nilpotent normal subgroup, Normal subgroup whose automorphism group is abelian, Normal subgroup whose inner automorphism group is central in automorphism group, Solvable normal subgroup, Transitively normal subgroup|FULL LIST, MORE INFO
perfect normal subgroup perfect group Normal subgroup whose focal subgroup equals its derived subgroup, Subgroup realizable as the commutator of the whole group and a subgroup, Upper join of characteristic subgroups|FULL LIST, MORE INFO
finite normal subgroup finite group Amalgam-characteristic subgroup, Amalgam-normal-subhomomorph-containing subgroup, Amalgam-strictly characteristic subgroup, Finitely generated normal subgroup, Join of finite normal subgroups, Normal closure of finite subset, Periodic normal subgroup, Potentially normal-subhomomorph-containing subgroup, Powering-invariant normal subgroup, Quotient-powering-invariant subgroup|FULL LIST, MORE INFO
periodic normal subgroup periodic group Amalgam-characteristic subgroup, Amalgam-normal-subhomomorph-containing subgroup, Amalgam-strictly characteristic subgroup, Potentially normal-subhomomorph-containing subgroup|FULL LIST, MORE INFO
finitely generated normal subgroup finitely generated group Normal closure of finite subset|FULL LIST, MORE INFO
View a complete list of conjunctions of normality with group properties
In some cases, we are interested in studying normal subgroups with the big group constrained to satisfy some group property. For instance:
normal subgroup of finite group finite group Finite normal subgroup, Join of finite normal subgroups, Kernel of a characteristic action on an abelian group, Normal closure of finite subset, Normal subgroup of finitely generated group, Quotient-powering-invariant subgroup|FULL LIST, MORE INFO
normal subgroup of finitely generated group finitely generated group |FULL LIST, MORE INFO
normal subgroup of periodic group periodic group Quotient-powering-invariant subgroup|FULL LIST, MORE INFO
normal subgroup of group of prime power order group of prime power order Normal subgroup of nilpotent group|FULL LIST, MORE INFO
### Weaker properties
Property Meaning Proof of implication Proof of strictness (reverse implication failure) Intermediate notions Comparison Collapse
subnormal subgroup obtained using subordination operator on normality normal implies subnormal subnormal not implies normal (see also list of examples) 2-hypernormalized subgroup, 2-subnormal subgroup, 3-subnormal subgroup, 4-subnormal subgroup, Asymptotically fixed-depth join-transitively subnormal subgroup, Central factor of normal subgroup, Direct factor of normal subgroup, Finitarily hypernormalized subgroup, Finite-conjugate-join-closed subnormal subgroup, Intermediately join-transitively subnormal subgroup, Join of finitely many 2-subnormal subgroups, Join-transitively 2-subnormal subgroup, Join-transitively subnormal subgroup, Linear-bound join-transitively subnormal subgroup, Modular 2-subnormal subgroup, Modular subnormal subgroup, Normal subgroup of characteristic subgroup, Permutable 2-subnormal subgroup, Permutable subnormal subgroup, Subnormal-permutable subnormal subgroup|FULL LIST, MORE INFO contrasting subnormality of various depths T-group
permutable subgroup (also called quasinormal subgroup) permutes with every subgroup normal implies permutable permutable not implies normal (see also list of examples) Permutable 2-subnormal subgroup|FULL LIST, MORE INFO group in which every permutable subgroup is normal
2-subnormal subgroup normal subgroup of normal subgroup normality is not transitive (see also list of examples) 2-hypernormalized subgroup, Central factor of normal subgroup, Direct factor of normal subgroup, Join-transitively 2-subnormal subgroup, Modular 2-subnormal subgroup, Normal subgroup of characteristic subgroup, Permutable 2-subnormal subgroup, Transitively normal subgroup of normal subgroup|FULL LIST, MORE INFO contrasting subnormality of various depths T-group
Other less important properties that are weaker than normality:
Property Meaning Proof of implication Proof of strictness (reverse implication failure) Intermediate notions Comparison Collapse
3-subnormal subgroup normal of normal of normal 2-subnormal subgroup, Normal subgroup of characteristic subgroup|FULL LIST, MORE INFO contrasting subnormality of various depths T-group
4-subnormal subgroup 2-subnormal of 2-subnormal 2-subnormal subgroup, Normal subgroup of characteristic subgroup|FULL LIST, MORE INFO contrasting subnormality of various depths T-group
descendant subgroup Conjugate-permutable subgroup, Intersection of subnormal subgroups, Subnormal subgroup|FULL LIST, MORE INFO
finitarily hypernormalized subgroup iterated normalizers reach whole group in finite number of steps 2-hypernormalized subgroup|FULL LIST, MORE INFO
pronormal subgroup conjugate to any conjugate in their join normal implies pronormal pronormal not implies normal (see also list of examples) Join-transitively pronormal subgroup, Subgroup whose join with any distinct conjugate is the whole group, Transfer-closed pronormal subgroup|FULL LIST, MORE INFO subnormal-to-normal and normal-to-characteristic group in which every pronormal subgroup is normal
weakly pronormal subgroup (via pronormal) Pronormal subgroup, Subgroup whose join with any distinct conjugate is the whole group|FULL LIST, MORE INFO (above)
weakly normal subgroup (via pronormal) NE-subgroup, Paranormal subgroup|FULL LIST, MORE INFO (above)
intermediately subnormal-to-normal subgroup (via pronormal) (see also list of examples) NE-subgroup, Paranormal subgroup, Polynormal subgroup, Pronormal subgroup, Weakly normal subgroup, Weakly pronormal subgroup|FULL LIST, MORE INFO (above)
conjugate-permutable subgroup permutes with its conjugate subgroups 2-hypernormalized subgroup, 2-subnormal subgroup, Permutable 2-subnormal subgroup, Permutable subgroup, Permutable subgroup of normal subgroup|FULL LIST, MORE INFO
modular subgroup Modular 2-subnormal subgroup, Modular subnormal subgroup, Permutable subgroup|FULL LIST, MORE INFO
automorph-permutable subgroup Normal subgroup of characteristic subgroup, Permutable subgroup|FULL LIST, MORE INFO
For a complete list of subgroup properties weaker than Normal subgroup, click here
WEAKER PROPERTIES SATISFYING SPECIFIC METAPROPERTIES: transitive | intermediate subgroup condition | transfer condition | quotient-transitive |intersection-closed |join-closed | trim | inverse image condition | image condition | centralizer-closed |
WEAKER PROPERTIES DISSATISFYING SPECIFIC METAPROPERTIES: transitive | intermediate subgroup condition | transfer condition | quotient-transitive |intersection-closed |join-closed | trim | inverse image condition | image condition | centralizer-closed |
### Related operators
There are three important subgroup operators related to normality:
Operator What it does How normal subgroups are related to it
Normal core This takes a subgroup and outputs the largest normal subgroup inside it, which is also the intersection of all its conjugate subgroups The operator is idempotent (doing it twice is the same as doing it once) and normal subgroups are precisely the subgroups that are invariant under it, and hence also precisely the subgroups that can arise from it.
Normal closure This takes a subgroup and outputs the smallest normal subgroup containing it, which is also the join of all its conjugate subgroups The operator is idempotent (doing it twice is the same as doing it once) and normal subgroups are precisely the subgroups that are invariant under it, and hence also precisely the subgroups that can arise from it.
Normalizer This takes a subgroup and outputs the largest subgroup within which it is normal A subgroup is normal if and only if its normalizer is the whole group
Other operators involve composing these in different ways, for instance:
Closely related to normal closure is the normal subgroup generated by a subset, which is defined as the smallest normal subgroup containing the subset, and is the normal closure of the subgroup generated by the subset.
### Analogues in other algebraic structures
Algebraic structure Analogue of normal subgroup in that structure Definition Nature of analogy with normal subgroup
Lie ring Ideal of a Lie ring A subring of a Lie ring whose Lie bracket with any element of the Lie ring is in the subring. Precisely the kernels of homomorphisms, play analogous roles in isomorphism theorems. Also, precisely the subrings invariant under inner derivations. Also, the Lazard correspondence maps ideals of the Lazard Lie ring to normal subgroups of the group.
ideal-determined variety of algebras (universal algebra) ideal in that variety An ideal-determined variety of algebras with zero is a variety where every ideal occurs as the inverse image of zero under some homomorphism and this completely determines the fibers of the homomorphism. Also, the various isomorphism theorems hold with suitable modifications. variety of groups is ideal-determined, with the ideals being the normal subgroups.
variety of algebras (universal algebra) I-automorphism-invariant subalgebra invariant under I-automorphisms, which are the automorphism described by formulas that universally give automorphisms. inner automorphisms are I-automorphisms in the variety of groups, so the corresponding invariant subalgebras are subgroups.
loop normal subloop commutes with every element, associates with every pair of elements when the loop is a group, then normal subloop = normal subgroup. Also, normal subloops are kernels of homomorphisms and the isomorphism theorems hold.
hypergroup normal subhypergroup
semigroup left-normal subsemigroup, right-normal subsemigroup
## Effect of property operators
Operator Meaning Result of application Proof
left transiter if big group is normal in a bigger group, so is subgroup characteristic subgroup left transiter of normal is characteristic
right transiter every normal subgroup of subgroup is normal in whole group transitively normal subgroup by definition
subordination operator normal subgroup of normal subgroup of ... of normal subgroup subnormal subgroup by definition
hereditarily operator every subgroup of it is normal hereditarily normal subgroup by definition
upward-closure operator every subgroup containing it is normal upward-closed normal subgroup by definition
maximal proper operator proper normal subgroup contained in no other proper normal subgroup maximal normal subgroup by definition
minimal operator nontrivial normal subgroup containing no other nontrivial normal subgroup minimal normal subgroup by definition
## Formalisms
BEWARE! This section of the article uses terminology local to the wiki, possibly without giving a full explanation of the terminology used (though efforts have been made to clarify terminology as much as possible within the particular context)
### First-order description
This subgroup property is a first-order subgroup property, viz., it has a first-order description in the theory of groups.
View a complete list of first-order subgroup properties
The subgroup property of normality can be expressed in first-order language as follows: $H$ is normal in $G$ if and only if:
$\forall g \in G, h \in H: \ ghg^{-1} \in H$
This is in fact a universally quantified expression of Fraisse rank 1.
### Function restriction expression
This subgroup property is a function restriction-expressible subgroup property: it can be expressed by means of the function restriction formalism, viz there is a function restriction expression for it.
Find other function restriction-expressible subgroup properties | View the function restriction formalism chart for a graphic placement of this property
Function restriction expression $H$ is a normal subgroup of $G$ if ... This means that normality is ... Additional comments
inner automorphism $\to$ function every inner automorphism of $G$ sends every element of $H$ to within $H$ the invariance property for inner automorphisms
inner automorphism $\to$ endomorphism every inner automorphism of $G$ restricts to an endomorphism of $H$ the endo-invariance property for inner automorphisms; i.e., it is the invariance property for inner automorphism, which is a property stronger than the property of being an endomorphism
inner automorphism $\to$ automorphism every inner automorphism of $G$ restricts to an automorphism of $H$ the auto-invariance property for inner automorphisms; i.e., it is the invariance property for inner automorphism, which is a group-closed property of automorphisms inner automorphism to automorphism is right tight for normality
### Relation implication expression
This subgroup property is a relation implication-expressible subgroup property: it can be defined and viewed using a relation implication expression
View other relation implication-expressible subgroup properties
Normality can be expressed in terms of the relation implication formalism as the relation implication operator with the left side being conjugate subgroups and the right side being equal subgroups:
Conjugate $\implies$ Equal
In other words, a subgroup is normal if any subgroup related to it by being conjugate is in fact equal to it.
### Variety formalism
This subgroup property can be described in the language of universal algebra, viewing groups as a variety of algebras
View other such subgroup properties
There are two somewhat different ways of expressing the notion of normality in the language of varieties:
• In the variety of groups, the normal subgroups are precisely the subalgebras invariant under all the I-automorphisms. An I-automorphism is an automorphism that can be expressed using a formula guaranteed to give an automorphism. This definition of normal subgroup follows from the fact that for groups, inner automorphisms are precisely the I-automorphisms.
• Treating the variety of groups as a variety of algebras with zero, the normal subgroups are precisely the ideals.
## Testing
### The testing problem
Further information: Normality testing problem
Given generating sets for a group and a subgroup, the problem of determining whether the subgroup is normal in the group reduces to the problem of testing whether the conjugate of any generator of the subgroup by any generator of the group, and by the inverse of the generator, is inside the subgroup. Thus, it reduces to the membership problem for the subgroup.
### GAP command
This subgroup property can be tested using built-in functionality of Groups, Algorithms, Programming (GAP).
The GAP command for testing this subgroup property is:IsNormal
The GAP command for listing all subgroups with this property is:NormalSubgroups
View subgroup properties testable with built-in GAP command|View subgroup properties for which all subgroups can be listed with built-in GAP commands | View subgroup properties codable in GAP
The GAP syntax for testing whether a subgroup is normal in a group is:
IsNormal (group, subgroup);
where subgroup and group may be defined on the spot in terms of generators (described as permutations) or may refer to things previously defined.
GAP can also be used to list all normal subgroups of a given group, using the command:
NormalSubgroups(group);
## References
### Textbook references
Book Page number Chapter and section Contextual information View
Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347More info 82 formal definition and Theorem 6 giving equivalent formulations. Also, Page 80 (first use).
Topics in Algebra by I. N. HersteinMore info 50 Section 2.6 formal definition | 2022-06-28T05:10:25 | {
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https://kops.uni-konstanz.de/handle/123456789/48940 | ## Discrete-time k-positive linear systems
2020
##### Authors
Alseidi, Rola
Margaliot, Michael
Garloff, Jürgen
##### Series
Konstanzer Schriften in Mathematik; 389
##### Publication type
Working Paper/Technical Report
Published
##### Abstract
Positive systems play an important role in systems and control theory and have found many applications in multi-agent systems, neural networks, systems biology, and more. Positive systems map the nonnegative orthant to itself (and also the nonpositive orthant to itself). In other words, they map the set of vectors with zero sign variation to itself. In this note, discrete-time linear systems that map the set of vectors with up to k-1 sign variations to itself are introduced. For the special case k = 1 these reduce to discrete-time positive linear systems. Properties of these systems are analyzed using tools from the theory of sign-regular matrices. In particular, it is shown that almost every solution of such systems converges to the set of vectors with up to k-1 sign variations. It is also shown that these systems induce a positive dynamics of k-dimensional parallelotopes.
510 Mathematics
##### Keywords
Sign-regular matrices, cones of rank k, exterior products, compound matrices, stability analysis
ISO 690
BibTex
RDF
Yes
## Version History
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2021-01-20 10:43:28
1*
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* Selected version | 2023-03-20T22:22:17 | {
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https://pastpapers.uptuition.id/paper-32-feb-mar-2021-pure-math-iii-9709-32-f-m-21/ | Home » 9709 » Paper 32 Feb Mar 2021 Pure Math III – 9709/32/F/M/21
Check out my complete solution here:
» Full Solutions «
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1. Solve the equation $\ln (x^3 − 3) = 3 \ln x \ − \ \ln 3$. Give your answer correct to 3 significant figures.
$$\tag*{[3]}$$
2. The polynomial $ax^3 + 5x^2 − 4x + b$, where $a$ and $b$ are constants, is denoted by $\mathrm{p}(x)$. It is given that $(x + 2)$ is a factor of $\mathrm{p}(x)$ and that when $\mathrm{p}(x)$ is divided by $(x + 1)$ the remainder is 2.
Find the values of $a$ and $b$.
$$\tag*{[5]}$$
3. By first expressing the equation $\tan(x + {45}^{\circ}) = 2 \cot x + 1$ as a quadratic equation in $\tan x$, solve the equation for ${0}^{\circ} \lt x \lt {180}^{\circ}$.
$$\tag*{[6]}$$
4. The variables x and y satisfy the differential equation
$$(1 \ – \ \cos x) \frac{\mathrm{d}y}{\mathrm{d}x} = y \sin x.$$
It is given that $y = 4$ when $x = \pi$.
(a) Solve the differential equation, obtaining an expression for $y$ in terms of $x$.
$$\tag*{[6]}$$
(b) Sketch the graph of $y$ against $x$ for $0 \lt x \lt 2\pi$.
$$\tag*{[1]}$$
5. (a) Express $\sqrt{7} \sin x + 2 \cos x$ in the form $R \sin(x + \alpha)$, where $R \gt 0$ and ${0}^{\circ} \lt \alpha \lt {90}^{\circ}$. State the exact value of $R$ and give $\alpha$ correct to 2 decimal places.
$$\tag*{[3]}$$
(b) Hence solve the equation $\sqrt{7} \sin 2\theta + 2 \cos 2\theta = 1$, for ${0}^{\circ} \lt \theta \lt {180}^{\circ}$.
$$\tag*{[5]}$$
6. Let $\displaystyle \mathrm{f}(x) = \frac{ 5a }{ (2x \ – \ a)(3a \ – \ x) }$ where $a$ is a positive constant.
(a) Express $\mathrm{f}(x)$ in partial fractions.
$$\tag*{[3]}$$
(b) Hence show that $\displaystyle \int_{a}^{2a} \mathrm{f}(x) \ \mathrm{d}x = \ln 6$.
$$\tag*{[4]}$$
7. Two lines have equations $\ \mathbf{r} = \begin{pmatrix} 1 \\[1pt] 3 \\[1pt] 2 \end{pmatrix} + s\begin{pmatrix} 2 \\[1pt] -1 \\[1pt] 3 \end{pmatrix} \$ and $\ \mathbf{r} = \begin{pmatrix} 2 \\[1pt] 1 \\[1pt] 4 \end{pmatrix} + t \begin{pmatrix} 1 \\[1pt] -1 \\[1pt] 4 \end{pmatrix}$.
(a) Show that the lines are skew.
$$\tag*{[5]}$$
(b) Find the acute angle between the directions of the two lines.
$$\tag*{[3]}$$
8. The complex numbers $u$ and $v$ are defined by $u = −4 \ + \ 2\mathrm{i}$ and $v = 3 \ + \ \mathrm{i}$.
(a) Find ${\large\frac{u}{v}}$ in the form $x + \mathrm{i}y$, where $x$ and $y$ are real.
$$\tag*{[3]}$$
(b) Hence express ${\large\frac{u}{v}}$ in the form $r { \mathrm{e} }^{ \mathrm{i}\theta}$, where $r$ and $\theta$ are exact.
$$\tag*{[2]}$$
In an Argand diagram, with origin $O$, the points $A$, $B$ and $C$ represent the complex numbers $u$, $v$ and $2u \ + \ v$ respectively.
(c) State fully the geometrical relationship between $OA$ and $BC$.
$$\tag*{[2]}$$
(d) Prove that angle $AOB = \frac{3}{4} \pi$.
$$\tag*{[2]}$$
9. Let $\displaystyle \mathrm{f}(x) = \frac{ {\mathrm{e}}^{2x} \ + \ 1 }{ {\mathrm{e}}^{2x} \ – \ 1 }$, for $x \gt 0$.
(a) The equation $x = \mathrm{f}(x)$ has one root, denoted by $a$.
Verify by calculation that $a$ lies between 1 and 1.5.
$$\tag*{[2]}$$
(b) Use an iterative formula based on the equation in part (a) to determine $a$ correct to 2 decimal places. Give the result of each iteration to 4 decimal places.
$$\tag*{[3]}$$
(c) Find ${ \mathrm{f} }^{‘}(x)$. Hence find the exact value of $x$ for which ${ \mathrm{f} }^{‘}(x) = −8$.
$$\tag*{[6]}$$
10.
The diagram shows the curve $y = \sin 2x \ {\cos}^{2} x$ for $0 \le x \le \frac{1}{ 2} \pi$, and its maximum point $M$.
(a) Using the substitution $u = \sin x$, find the exact area of the region bounded by the curve and the $x$-axis.
$$\tag*{[5]}$$
(b) Find the exact $x$-coordinate of $M$.
$$\tag*{[6]}$$ | 2023-02-09T06:00:43 | {
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http://www.mathdoubts.com/trigonometry/identity/product-to-sum/difference-sin-compound-angles/ | # Difference of sine of compound angles to Product form Transformation
## Formula
$\large \sin (a+b) -\sin (a-b) =$ $\large 2.\cos a .\sin b$
### Proof
Two angles can form compound angles in sum and difference form. The subtraction of sines of them can be transformed as the product of cosine and sine of angles. It is called the sum to product form transformation rule as per trigonometry.
###### Step: 1
Assume, $a$ and $b$ are two angles and the compound angles of them are $a+b$ and $a-b$. The expansion of sine of sum and difference of the angles are written trigonometrically as follows.
$(1). \,\,\,\,\,\,$ $\sin(a+b)$ $=$ $\sin a . \cos b$ $+$ $\cos a . \sin b$
$(2). \,\,\,\,\,\,$ $\sin(a-b)$ $=$ $\sin a . \cos b$ $-$ $\cos a . \sin b$
###### Step: 2
Subtract the expansion of the sine of difference of the angles from the expansion of the sine of sum of the angles.
$\implies \sin(a+b) -\sin(a-b)$ $=$ $(\sin a . \cos b$ $+$ $\cos a . \sin b)$ $-$ $(\sin a . \cos b$ $-$ $\cos a . \sin b)$
$\implies \sin(a+b) -\sin(a-b)$ $=$ $\sin a . \cos b$ $+$ $\cos a . \sin b$ $-$ $\sin a . \cos b$ $+$ $\cos a . \sin b$
$\implies \sin(a+b) -\sin(a-b)$ $=$ $\sin a . \cos b$ $-$ $\sin a . \cos b$ $+$ $\cos a . \sin b$ $+$ $\cos a . \sin b$
$\implies \sin(a+b) -\sin(a-b)$ $=$ $\require{cancel} \cancel{\sin a . \cos b}$ $-$ $\require{cancel} \cancel{\sin a . \cos b}$ $+$ $\cos a . \sin b$ $+$ $\cos a . \sin b$
$\therefore \,\,\,\,\,\, \sin(a+b) -\sin(a-b)$ $=$ $2 \cos a . \sin b$
It is successfully transformed that the difference of sine of difference of two angles and sine of sum of two angles is equal to twice the product of sine of first angle and cosine of second angle.
#### Verification
Assume $a = 60^\circ$ and $b = 30^\circ$. Substitute these two angles in both sides of the equation and then compare the result.
###### Step: 1
$\sin(a+b) -\sin(a-b)$ $=$ $\sin(60^\circ+30^\circ) -\sin(60^\circ-30^\circ)$
$\implies \sin(60^\circ+30^\circ) -\sin(60^\circ-30^\circ)$ $=$ $\sin(90^\circ) -\sin(30^\circ)$
$\implies \sin(60^\circ+30^\circ) -\sin(60^\circ-30^\circ)$ $=$ $1-\dfrac{1}{2}$
$\therefore \,\,\,\,\,\, \sin(60^\circ+30^\circ) -\sin(60^\circ-30^\circ)$ $=$ $\dfrac{1}{2}$
###### Step: 2
$2 \cos a . \sin b = 2 \times \cos(60^\circ) \times \sin(30^\circ)$
$\implies 2 \times \cos(60^\circ) \times \sin(30^\circ)$ $=$ $2 \times \dfrac{1}{2} \times \dfrac{1}{2}$
$\implies 2 \times \cos(60^\circ) \times \sin(30^\circ)$ $=$ $2 \times {\Bigg(\dfrac{1}{2}\Bigg)}^2$
$\implies 2 \times \cos(60^\circ) \times \sin(30^\circ)$ $=$ $2 \times \dfrac{1}{4}$
$\therefore \,\,\,\,\,\, 2 \times \cos(60^\circ) \times \sin(30^\circ)$ $=$ $\dfrac{1}{2}$
###### Step: 3
Now compare both results to understand the relation between them.
$\therefore \,\,\,\,\,\, \sin(60^\circ+30^\circ) + \sin(60^\circ-30^\circ)$ $=$ $2 \times \cos(60^\circ) \times \sin(30^\circ)$ $=$ $\dfrac{1}{2}$
###### Other forms
The sum to product form transformation law can also be written in other form.
$\large \sin (x+y) -\sin (x-y) =$ $\large 2.\cos x .\sin y$
$\large \sin (\alpha+\beta) -\sin (\alpha-\beta) =$ $\large 2.\cos \alpha .\sin \beta$
###### General form
$\sin (sum \, of \, two \, angles)$ $-$ $\sin (difference \, of \, two \, angles)$ $=$ $2$ $\times$ $\cos \, of \, first \, angle$ $\times$ $\sin of \, second \, angle$
Save (or) Share | 2017-12-15T00:50:11 | {
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http://www.aimath.org/textbooks/beezer/PDsection.html | Once the dimension of a vector space is known, then the determination of whether or not a set of vectors is linearly independent, or if it spans the vector space, can often be much easier. In this section we will state a workhorse theorem and then apply it to the column space and row space of a matrix. It will also help us describe a super-basis for $\complex{m}$.
## Goldilocks' Theorem
We begin with a useful theorem that we will need later, and in the proof of the main theorem in this subsection. This theorem says that we can extend linearly independent sets, one vector at a time, by adding vectors from outside the span of the linearly independent set, all the while preserving the linear independence of the set.
Theorem ELIS (Extending Linearly Independent Sets) Suppose $V$ is vector space and $S$ is a linearly independent set of vectors from $V$. Suppose $\vect{w}$ is a vector such that $\vect{w}\not\in\spn{S}$. Then the set $S^\prime=S\cup\set{\vect{w}}$ is linearly independent.
In the story Goldilocks and the Three Bears, the young girl Goldilocks visits the empty house of the three bears while out walking in the woods. One bowl of porridge is too hot, the other too cold, the third is just right. One chair is too hard, one too soft, the third is just right. So it is with sets of vectors --- some are too big (linearly dependent), some are too small (they don't span), and some are just right (bases). Here's Goldilocks' Theorem.
Theorem G (Goldilocks) Suppose that $V$ is a vector space of dimension $t$. Let $S=\set{\vectorlist{v}{m}}$ be a set of vectors from $V$. Then
1. If $m>t$, then $S$ is linearly dependent.
2. If $m < t$, then $S$ does not span $V$.
3. If $m=t$ and $S$ is linearly independent, then $S$ spans $V$.
4. If $m=t$ and $S$ spans $V$, then $S$ is linearly independent.
There is a tension in the construction of basis. Make a set too big and you will end up with relations of linear dependence among the vectors. Make a set too small and you will not have enough raw material to span the entire vector space. Make a set just the right size (the dimension) and you only need to have linear independence or spanning, and you get the other property for free. These roughly-stated ideas are made precise by Theorem G.
The structure and proof of this theorem also deserve comment. The hypotheses seem innocuous. We presume we know the dimension of the vector space in hand, then we mostly just look at the size of the set $S$. From this we get big conclusions about spanning and linear independence. Each of the four proofs relies on ultimately contradicting Theorem SSLD, so in a way we could think of this entire theorem as a corollary of Theorem SSLD. (See technique LC.) The proofs of the third and fourth parts parallel each other in style (add $\vect{w}$, toss $\vect{v}_k$) and then turn on Theorem ELIS before contradicting Theorem SSLD.
Theorem G is useful in both concrete examples and as a tool in other proofs. We will use it often to bypass verifying linear independence or spanning.
Example BPR: Bases for $P_n$, reprised.
Example BDM22: Basis by dimension in $M_{22}$.
Example SVP4: Sets of vectors in $P_4$.
A simple consequence of Theorem G is the observation that proper subspaces have strictly smaller dimensions. Hopefully this may seem intuitively obvious, but it still requires proof, and we will cite this result later.
Theorem PSSD (Proper Subspaces have Smaller Dimension) Suppose that $U$ and $V$ are subspaces of the vector space $W$, such that $U\subsetneq V$. Then $\dimension{U} < \dimension{V}$.
The final theorem of this subsection is an extremely powerful tool for establishing the equality of two sets that are subspaces. Notice that the hypotheses include the equality of two integers (dimensions) while the conclusion is the equality of two sets (subspaces). It is the extra "structure" of a vector space and its dimension that makes possible this huge leap from an integer equality to a set equality.
Theorem EDYES (Equal Dimensions Yields Equal Subspaces) Suppose that $U$ and $V$ are subspaces of the vector space $W$, such that $U\subseteq V$ and $\dimension{U}=\dimension{V}$. Then $U=V$.
## Ranks and Transposes
We now prove one of the most surprising theorems about matrices. Notice the paucity of hypotheses compared to the precision of the conclusion.
Theorem RMRT (Rank of a Matrix is the Rank of the Transpose) Suppose $A$ is an $m\times n$ matrix. Then $\rank{A}=\rank{\transpose{A}}$.
This says that the row space and the column space of a matrix have the same dimension, which should be very surprising. It does not say that column space and the row space are identical. Indeed, if the matrix is not square, then the sizes (number of slots) of the vectors in each space are different, so the sets are not even comparable.
It is not hard to construct by yourself examples of matrices that illustrate Theorem RMRT, since it applies equally well to any matrix. Grab a matrix, row-reduce it, count the nonzero rows or the leading 1's. That's the rank. Transpose the matrix, row-reduce that, count the nonzero rows or the leading 1's. That's the rank of the transpose. The theorem says the two will be equal. Here's an example anyway.
Example RRTI: Rank, rank of transpose, Archetype I.
## Dimension of Four Subspaces
That the rank of a matrix equals the rank of its transpose is a fundamental and surprising result. However, applying Theorem FS we can easily determine the dimension of all four fundamental subspaces associated with a matrix.
Theorem DFS (Dimensions of Four Subspaces) Suppose that $A$ is an $m\times n$ matrix, and $B$ is a row-equivalent matrix in reduced row-echelon form with $r$ nonzero rows. Then
1. $\dimension{\nsp{A}}=n-r$
2. $\dimension{\csp{A}}=r$
3. $\dimension{\rsp{A}}=r$
4. $\dimension{\lns{A}}=m-r$
There are many different ways to state and prove this result, and indeed, the equality of the dimensions of the column space and row space is just a slight expansion of Theorem RMRT. However, we have restricted our techniques to applying Theorem FS and then determining dimensions with bases provided by Theorem BNS and Theorem BRS. This provides an appealing symmetry to the results and the proof.
## Direct Sums
Some of the more advanced ideas in linear algebra are closely related to decomposing (technique DC) vector spaces into direct sums of subspaces. With our previous results about bases and dimension, now is the right time to state and collect a few results about direct sums, though we will only mention these results in passing until we get to Section NLT:Nilpotent Linear Transformations, where they will get a heavy workout.
A direct sum is a short-hand way to describe the relationship between a vector space and two, or more, of its subspaces. As we will use it, it is not a way to construct new vector spaces from others.
Definition DS (Direct Sum) Suppose that $V$ is a vector space with two subspaces $U$ and $W$ such that for every $\vect{v}\in V$,
1. There exists vectors $\vect{u}\in U$, $\vect{w}\in W$ such that $\vect{v}=\vect{u}+\vect{w}$
2. If $\vect{v}=\vect{u}_1+\vect{w}_1$ and $\vect{v}=\vect{u}_2+\vect{w}_2$ where $\vect{u}_1,\,\vect{u}_2\in U$, $\vect{w}_1,\,\vect{w}_2\in W$ then $\vect{u}_1=\vect{u}_2$ and $\vect{w}_1=\vect{w}_2$.
Then $V$ is the direct sum of $U$ and $W$ and we write $V=U\ds W$.
(This definition contains Notation DS.)
Informally, when we say $V$ is the direct sum of the subspaces $U$ and $W$, we are saying that each vector of $V$ can always be expressed as the sum of a vector from $U$ and a vector from $W$, and this expression can only be accomplished in one way (i.e. uniquely). This statement should begin to feel something like our definitions of nonsingular matrices (Definition NM) and linear independence (Definition LI). It should not be hard to imagine the natural extension of this definition to the case of more than two subspaces. Could you provide a careful definition of $V=U_1\ds U_2\ds U_3\ds ...\ds U_m$ (exercise PD.M50)?
Example SDS: Simple direct sum.
Example SDS is easy to generalize into a theorem.
Theorem DSFB (Direct Sum From a Basis) Suppose that $V$ is a vector space with a basis $B=\set{\vect{v}_1,\,\vect{v}_2,\,\vect{v}_3,\,...,\,\vect{v}_n}$ and $m\leq n$. Define
\begin{align*} U&=\spn{\set{\vect{v}_1,\,\vect{v}_2,\,\vect{v}_3,\,...,\,\vect{v}_m}} & W&=\spn{\set{\vect{v}_{m+1},\,\vect{v}_{m+2},\,\vect{v}_{m+3},\,...,\,\vect{v}_n}} \end{align*}
Then $V=U\ds W$.
Given one subspace of a vector space, we can always find another subspace that will pair with the first to form a direct sum. The main idea of this theorem, and its proof, is the idea of extending a linearly independent subset into a basis with repeated applications of Theorem ELIS.
Theorem DSFOS (Direct Sum From One Subspace) Suppose that $U$ is a subspace of the vector space $V$. Then there exists a subspace $W$ of $V$ such that $V=U\ds W$.
There are several different ways to define a direct sum. Our next two theorems give equivalences (technique E) for direct sums, and therefore could have been employed as definitions. The first should further cement the notion that a direct sum has some connection with linear independence.
Theorem DSZV (Direct Sums and Zero Vectors) Suppose $U$ and $W$ are subspaces of the vector space $V$. Then $V=U\ds W$ if and only if
1. For every $\vect{v}\in V$, there exists vectors $\vect{u}\in U$, $\vect{w}\in W$ such that $\vect{v}=\vect{u}+\vect{w}$.
2. Whenever $\zerovector=\vect{u}+\vect{w}$ with $\vect{u}\in U$, $\vect{w}\in W$ then $\vect{u}=\vect{w}=\zerovector$.
Our second equivalence lends further credence to calling a direct sum a decomposition. The two subspaces of a direct sum have no (nontrivial) elements in common.
Theorem DSZI (Direct Sums and Zero Intersection) Suppose $U$ and $W$ are subspaces of the vector space $V$. Then $V=U\ds W$ if and only if
1. For every $\vect{v}\in V$, there exists vectors $\vect{u}\in U$, $\vect{w}\in W$ such that $\vect{v}=\vect{u}+\vect{w}$.
2. $U\cap W=\set{\zerovector}$.
If the statement of Theorem DSZV did not remind you of linear independence, the next theorem should establish the connection.
Theorem DSLI (Direct Sums and Linear Independence) Suppose $U$ and $W$ are subspaces of the vector space $V$ with $V=U\ds W$. Suppose that $R$ is a linearly independent subset of $U$ and $S$ is a linearly independent subset of $W$. Then $R\cup S$ is a linearly independent subset of $V$.
Our last theorem in this collection will go some ways towards explaining the word "sum" in the moniker "direct sum," while also partially explaining why these results appear in a section devoted to a discussion of dimension.
Theorem DSD (Direct Sums and Dimension) Suppose $U$ and $W$ are subspaces of the vector space $V$ with $V=U\ds W$. Then $\dimension{V}=\dimension{U}+\dimension{W}$.
There is a certain appealling symmetry in the previous proof, where both linear independence and spanning properties of the bases are used, both of the first two conclusions of Theorem G are employed, and we have quoted both of the two conditions of Theorem DSZI.
One final theorem tells us that we can successively decompose direct sums into sums of smaller and smaller subspaces.
Theorem RDS (Repeated Direct Sums) Suppose $V$ is a vector space with subspaces $U$ and $W$ with $V=U\ds W$. Suppose that $X$ and $Y$ are subspaces of $W$ with $W=X\ds Y$. Then $V=U\ds X\ds Y$.
Remember that when we write $V=U\ds W$ there always needs to be a "superspace," in this case $V$. The statement $U\ds W$ is meaningless. Writing $V=U\ds W$ is simply a shorthand for a somewhat complicated relationship between $V$, $U$ and $W$, as described in the two conditions of Definition DS, or Theorem DSZV, or Theorem DSZI. Theorem DSFB and Theorem DSFOS gives us sure-fire ways to build direct sums, while Theorem DSLI, Theorem DSD and Theorem RDS tell us interesting properties of direct sums. This subsection has been long on theorems and short on examples. If we were to use the term "lemma" we might have chosen to label some of these results as such, since they will be important tools in other proofs, but may not have much interest on their own (see technique LC). We will be referencing these results heavily in later sections, and will remind you then to come back for a second look. | 2013-05-19T13:08:52 | {
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http://math.stackexchange.com/questions/692958/how-to-apply-plancherel-theorem-here | # How to apply Plancherel Theorem here?
Let f be a function on the real line R such that both f and xf are in L^2(R). Prove that f ∈ L^1(R) and the L^1 norm of f(x) is less than or equal to 8 times (the L^2 norm of f(x)) times the L^2 norm of xf(x).
I'm sorry I don't know how to use Latex to post the problem. The origional problem is here: http://www.math.purdue.edu/~bell/MA598R/advanced1.pdf. It's Problem 18. It's one of the problems created by a teacher to help students prepare qualifying exams of real analysis.
I think the purpose of the question is to ask students to apply Plancherel Theorem, but I don't know where to start. I know all the properties of fourier transforms, but I couln't relate them here. I think the trick of the problem is to break f into two parts, but I couldn't figure out how to do it. Also, the form of the inequality is somewhat similar to the uncertainty principle, but I couldn't get any hints from the proof of the uncertainty principle beacause the hypothesis is different here.
I believe the question is not hard unless one knows where to start. I've tried my best and I'm still blind. Could anyone tell me how to do it? Any hints would be appreciated.
-
## 1 Answer
It is just an application of the Cauchy-Swartz inequality. For any $a>0$ \begin{align} \int_\mathbb{R}|f|&=\int_{|x|\le a}|f|+\int_{|x|>a}|x\,f|\,\frac{1}{|x|}\\ &\le\Bigl(\int_{|x|\le a}|f|^2\Bigr)^{1/2}\Bigl(\int_{|x|\le a}1\Bigr)^{1/2}+ \Bigl(\int_{|x|>a}|x\,f|^2\Bigr)^{1/2}\Bigl(\int_{|x|>a}\frac{1}{|x|^2}\Bigr)^{1/2}\\ &\le\sqrt{2\,a}\,\|f\|_2+\frac{\sqrt2}{\sqrt{a}}\|x\,f\|_2. \end{align} Choose $a=b^2\|x\,f\|_2/(2\,\|f\|_2)$ to get $$\|f\|_1\le\Bigl(b+\frac2b\Bigr)\sqrt{\strut\|f\|_2\,\|x\,f\|_2}\ .$$ Choose $b=\sqrt2$ and square to get $$\|f\|_1^2\le8\,\|f\|_2\,\|x\,f\|_2\ .$$
-
Thanks a lot. I didn't realize using this way to tell f is integrable. But could you please show me how to get the estimate like in the original problem posted? – Qinfeng Li Feb 27 at 18:41
See the edited answer. – Julián Aguirre Feb 27 at 22:12
Strangely enough, the same question was posed one hour before, with more or less the same answer: math.stackexchange.com/questions/692926/… – Etienne Feb 27 at 22:59
@Etienne: I'm sorry the question I posted twice, because I posted the earlier one without logging on my account. – Qinfeng Li Feb 28 at 0:07
@Julian Aguirre: Thank you very much for your elegant proof. I was kind of nerd in solving these problems. I wanted to try to apply Plancherel Theorem because somebody confidently told me that I should do the problem like that, but I failed. It's just like there are three methods to prove Hardy's Inequality: Fourier analysis, Jensen's Inequality and nondecreasing rearrangement of a function. Hardy's original proof was using Fourier analysis. But if someone told Hardy to prove it using nondecreasing rearrangement of a function, I suspect Hardy cannot give a proof. – Qinfeng Li Feb 28 at 0:13 | 2014-08-28T11:56:02 | {
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https://math.stackexchange.com/questions/2036378/find-all-numbers-x-for-which-x2-2x-2-0 | # Find all numbers $x$ for which $x^2 - 2x + 2 >0$
Find all numbers $x$ for which $x^2 - 2x + 2 >0$.
Usually I can solve this type of questions simply by finding the zero's through factoring/quadratic equation and then imagining the graph. But what do I do if the discriminant ($b^2 - 4ac$) is negative so I get a negative number under the square root sign in the quadratic formula?
Thanks a lot
• Do you know what the sign of the discriminant tells you about the number of roots the equation has? – jacer21 Nov 29 '16 at 21:09
• if the b^2-4ac is negative then the function f(x) does not across the 0 line so the f(x) should either all above 0 or all below 0 – displayname Nov 29 '16 at 21:12
• Have you tried graphing the function? – Hugh Nov 29 '16 at 21:13
• Thanks to everybody! jacer21 completely forgot about that. And graphing the function indeed showed it clearly – stef Nov 29 '16 at 21:37
• If the discriment is negative then the value is never 0.... which means the function is always positive or always negative. – fleablood Nov 29 '16 at 21:40
You can complete the square, which changes the inequality to $$(x-1)^2+1>0$$ Note that this is true for all $x$, because $(x-1)^2\ge 0$.
The discriminant of $x^2-2x+2=0$ is $b^2-4ac=(-2)^2-4(2)(1)=-4$. Meaning this quadratic has no solution.
And since $a>0$, this parabola opens upwards. Meaning any $x$ input produces a $y$ value greater than $0$.
Let $f(x) = ax^2 + bx +c$. If $f(x)$ never equals $0$ then the graph of this parabola can never cross the $x$ axis. So $f(x)$ is either always positive or it is always negative.
$f(x) = x^2 -2x + 2$ is never equal to zero as $b^2 - 4ac = 4 - 8 = -4 < 0$ as you pointed out. So it is either always positive or always negative.
$f(0) = 2 > 0$ so $f(x)$ is always positive.
So $x^2 -2x + 2 > 0$ for all real $x$. | 2019-11-19T03:15:10 | {
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https://www.khanacademy.org/math/linear-algebra/matrix-transformations/determinant-depth/v/linear-algebra-determinant-when-row-is-added | # Determinant when row is added
## Video transcript
Let's keep messing with our determinants to see if we can get more useful results. And they might not be obviously useful right now, but maybe we'll use them later when we are exploring other parts of linear algebra. So let's say I have some matrix, let's call it matrix X. Matrix X is equal to-- I'll just start with a 3 by 3 case because I think the 2 by 2 case is a bit trivial. Actually, why don't I just start with a 2 by 2 case. Let's say matrix X is a, b, and then it has x1, x2. I could have called these c and d, but you'll see why I called them x1 and x2 in a second. And I'll say I have another matrix. Let's say matrix Y is identical to matrix X except for this row. So matrix Y is a, b y1 and y2. And let's say we have a third matrix Z. That's identical to the first two matrices on the first row. So a, b. but on the second row it's actually the sum of the two rows of x and y. So it's going to be, this entry's going to be x1 plus y1, and this entry right here is x2 plus y2. Just like that. I want to be very clear, Z is not X plus Y. All of the terms of Z are not the sum of all the terms of X and Y. I'm only focusing on one particular row. And this is just a general theme that you'll see over and over again, and we saw it in the last video and I guess you'll see it here, is that determinants or finding the determinants of matrices aren't linear on matrix operations, but they are linear on operations that you do just to one row. So in this case, everything else is equal except for this row, and Z has the same first row as these guys, but its second row is the sum of the second row of these guys. So let's explore how the determinants of these guys relate. So the determinant-- let me do it in X's color. The determinant of X-- I'll write it like that-- is equal to a ax2 minus bx1. You've seen that multiple times. The determinant of Y is equal to ay2 minus by1. And the determinant of Z is equal to a times x2 plus y2 minus b times x1 plus y1, which is equal to ax2 plus ay2-- just distributed the a-- minus bx1 minus by1. And if we just rearrange things, this is equal to a-- let me write it this way-- this is equal to ax2 minus bx1. That's that term and that term, we switched colors. So that's those two guys. And then plus ay2 minus by1. Now what is this right here? That is the determinant of X. And this right here is the determinant of Y. So there you have it. If we have matrices that are completely identical except for one row-- and in this case it's a 2 by 2 matrix, so it looks like half of the matrix-- and Z's, that row that we're referring to that's different, Z's is the sum of the other two guys' rows, then Z's determinant is the sum of the other two determinants. So this is a very special case. I want to keep reiterating it. It only works in the case where this row and only this row is the sum of this row and this row, and the matrices are identical everywhere else. Let me show you the 3 by 3 case, and I think it'll be a little bit more general. And then we'll go to n by n. The n by n is actually, on some level, the easiest to do, but it's kind of abstract so I like to save that for the end. So let's redefine all those guys into the 3 by 3 case. So let's say that X is equal to a, b, c-- let's just do, let's make the third row the row we're going to use to determine our determinant. a, b, c, d, e, f-- actually let me do the middle row, because I don't want to make you think it always has to be the last row. So let's say it's x1, c2, x3, and you have d, e, f. And what's the determinant of X going to be? The determinant of X is going to be equal to-- let's say we're going along this row right here, that's the row in question. It's going to be equal to-- well you remember your checkerboard pattern-- so it's going to be-- remember, plus, minus, plus, minus, plus-- you remember all the rest how it goes. So it's going to start with a minus x1 times the sub matrix-- you get rid of that column, that row-- b, c, e, f. Then you have plus x2 times the sub matrix-- get rid of that column, that row-- a, c, d, f. And then finally minus x3-- you get rid of its row and column-- you have a, b, d, e. Now let me define another matrix Y that is identical to matrix X, except for that row. So it's a, b, c. Then down here d, e, f. That middle row is different. It's y1, y2, and y3. What's the determinant of Y going to be? Determinant of Y? Well it's going to be identical to the determinant of X because all the sub-matrices are going to be the same when you cross out this row and each of the columns. But the coefficients are going to be different. Instead of an x1, you have a y1. So it's going to be equal to minus y1 times the determinant b, c, e, f plus y 2 times the determinant of a, c, d f minus y3 times the determinant of a, b, d, e. I think you see where this is going. Now I'm going to create another matrix. I'm going to create another matrix Z just like that that is equal to-- it's identical to these two guys on the first and third rows, a, b, c, d, e, f. Just like that. But this row just happens to be the sum of this row and this row. And when we figured out this determinant we went along that row-- you can see that right there. So this row right here is going to be x1 plus y1, that's its first term. x2 plus y2, and then you have x3 plus y3. Now what's the determinant of Z going to be? Well, we can go down this row right there. So it's going to be minus x1 plus y1 times its sub-matrix-- get rid of that row, that column-- you get b, c, e, f. I think you definitely see where this is going. Plus this coefficient, plus x2 plus y2 times its sub-matrix-- get rid of that row, that column-- a, c, d, f. And then you have minus this guy right here, x3 plus y3 times its sub matrix-- get rid of that column and row-- a, b, d, e. Now what do you have right here? This is the determinant of Z. This right here is the determinant of Z. This thing right here. I think you can see immediately that if you were to add this to this you would get this right here, right? Because you have this coefficient and this coefficient on that. If you added them up you would get minus x1 plus y1. This guy and this guy add up to this guy. And then if I were to do this guy and this guy, add up to that guy. Let me do another one. And then finally that term plus that term add up to that term. So you immediately see that the determinant, or hopefully you immediately see, that the determinant of X plus the determinant of Y is equal to the determinant of Z. So we did it for the 2 by 2 case, we just did it for the 3 by 3 case. Might as well do it for the n by n case so we know that it works. But the argument is identical to this 3 by 3 case. So that's good to keep in your mind because 3 by 3 is easy to visualize, n by n is sometimes a little bit abstract. So let me re-define my matrices again. I'm just going to do the same thing over again. So I'm going to have a matrix X. But it's an n by n matrix. So let me write it this way. Let's say it is a 1, 1, a 1, 2, all the way to a 1, n. And there's some row here, let's say that there's some row here on row i-- let's call this row i right here-- and here it has the terms x1, x2, all the way to xn, but everything else is just the regular a's. So then you have a-- let me make this as a21, all the way to a2n. And then if you went all the way down here you would have an1, and you'd go all the way to ann. So essentially you could imagine our standard matrix where everything is defined in a, but I replaced row i with certain numbers that are maybe a little different. And I think you'll see where I'm going. Now let me define my other matrix. Let me define matrix Y. Let me define matrix Y to be essentially the same thing. This is a11-- it's the same a11. This is a12, all the way to a1n. This is a21, we could go all the way to a a2n. And then on row i, the same row, this is n by n, this is the same n by n-- if this was 10 by 10, this is 10 by 10. If this is row seven, then this is row seven. It has different terms. It's identical to matrix X except for row i. In row i it is y1, y2, all the way to yn. And if you keep going down, of course, you have an1, all the way to ann. Fair enough. Now let's say we have a third matrix. Let's have a third matrix. Let me draw it right here. So you have Z, Z is equal to-- I think you could imagine where this is going. Z is identical to these two guys except for row i. So let me write that out. So Z looks like this. You have a11, a12, all the way to a1n. And then you go down and then row i happens to be the sum of the row i of matrix X and matrix Y. So it is x1 plus y1, x2 plus y2, all the way to xn plus yn. And then it you keep going down, everything else is identical, an1 all the way to ann. So all of these matrices are identical except for row X has a different row i than matrix Y does. And row Z is identical everywhere except its row i is the sum of this row i and that row i. So it's a very particular case, but we can figure out their determinants. So what are the determinants? The determinant of X, the determinant of matrix X-- and hopefully you're maybe a bit comfortable with writing sigma notation, we did this in the last matrix. We can go down this row right there, and for each of these guys we can say so the determinant is going to be equal to the sum. Let's say we start from j is equal to 1-- j's going to be the column, so we're going to take the sum of each of these terms from j is equal to 1 to n. And then remember our checkerboard pattern, so we don't know if this is a positive or negative. We can figure it out by taking negative 1 to the i plus j-- remember, this is the ith row that we're talking about-- times xj-- xj is the coefficient, xsubj, times the sub matrix for xsubj. So if you get rid of this guy's row and this guy's column, what is it going to be? We could say that that's the same thing as the sub matrix, if we called this guy-- let me write it this way-- if we got rid of this guy's row and this guy's column, if we had just our traditional matrix where this wasn't replaced. If we just had an ai1 here, ai2, its sub matrix would be the same thing, because we're crossing out this row and this column. So it would be all of these guys and all of these guys down here. So it would be the sub matrix-- this is a n minus 1 by n minus 1 matrix-- it would be the sub matrix for aij. That's for the first term-- sorry, the determinant. Don't want to lose the determinant there-- times the determinant of the sub matrix aij. And so that's for the first term, and then you're going to add it to the second term, and then you're just going to keep doing that. That's what this sigma notation is. That's the determinant of X. Now what's the determinant of Y? The determinant of Y is equal to the sum-- we could do the same thing-- j is equal to 1 to n of negative 1 to the i plus j. We're going to go along this row right here, the ith row. So we're going to have ysubj-- right, we're going to start with ysub1, then plus ysub2, times the determinant of its sub matrix, which is the same as the determinant of this sub matrix. So you get rid of that row and that column for each of these guys, that everything else on the matrix is the same. So aij. The matrix of aij. Now what is the determinant of Z. I'm pretty sure you know exactly where this is going. The determinant-- this should be a capital Y right there-- the determinant of Z is equal to the sum, from j is equal to 1 to n, of negative 1 to the i plus j. We're going along this row. But now the coefficients are xj, that's what we're indexing along, xj plus yj. And then times its sub matrix, which is the same as these sub matrices. So aij, which you might immediately see is the sum of these two things. If I, for every j, I just summed these two things up, you're having two coefficients-- you could have this coefficient and that coefficient on your aij term, and then when you add them up you can factor this guy out and you will get this right here. So you get the determinant of X plus the determinant of Y is equal to the determinant of Z. So hopefully that shows you the general case. But I want to make it very clear, this is just for a very particular scenario where three matrices are identical, except for on one row. And one of the matrices on that special row just happens to be the sum of the other two matrices for that special row, and everything else is identical. That's the only time where the determinant of-- not the only time, but that's the only time we can make the general statement where the determinant of Z is equal to the determinant of X plus the determinant of Y. It's not the case-- let me write what is not the case-- so not the case that if Z is equal to X plus Y, it is not the case that the determinant of Z is necessarily equal to the determinant of X plus the determinant of Y. You cannot assume this. Determinant operations are not linear on matrix addition. They're linear only on particular rows getting at it. Anyway, hopefully you found that vaguely useful. | 2018-11-18T07:51:43 | {
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https://www.sfu.ca/math-coursenotes/Math%20158%20Course%20Notes/sec_AlternatingSeries.html | ## Section6.4Alternating Series
Next we consider series with both positive and negative terms, but in a regular pattern: they alternate signs. For example:
• $\ds\sum_{n=1}^{\infty} (-1)^{n-1} \frac{2}{n+1} = 1 - \frac{2}{3} + \frac{2}{4} - \frac{2}{5} + \dots$
• $\ds\sum_{n=1}^{\infty} (-1)^n \frac{2}{n+1} = -1 + \frac{2}{3} - \frac{2}{4} + \frac{2}{5} - \dots$
• $\ds\sum_{n=0}^{\infty} (-1)^n 2^n =1 - 2 + 4 - 8 + \dots$
• $\ds\sum_{n=4}^{\infty} (-1)^{n-1}\frac{n}{n+2} = -\frac{4}{6} + \frac{5}{7} - \frac{6}{8} + \frac{7}{9} - \dots$
###### Definition6.45. Alternating Series.
An alternating series has the form
\begin{equation*} \sum (-1)^n a_n \end{equation*}
where $a_n$ are all positive and the first index is arbitrary.
Note: An alternating series can start with a positive or negative term, i.e. the first index can be any non-negative integer.
A well-known example of an alternating series is the alternating harmonic series:
###### Definition6.46. Alternating Harmonic Series.
A series of the form
\begin{equation*} \sum_{n=1}^\infty {\frac{(-1)^{n-1}}{n}}= 1 - \frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots+\frac{(-1)^{n-1}}{n} + \dots \end{equation*}
is called an alternating harmonic series.
In the alternating harmonic series the magnitude of the terms decrease, that is, $\ds |a_n|$ forms a decreasing sequence, although this is not required in an alternating series. Recall that for a series with positive terms, if the limit of the terms is not zero, the series cannot converge; but even if the limit of the terms is zero, the series still may not converge. It turns out that for alternating series, the series converges exactly when the limit of the terms is zero. In Figure 6.4, we illustrate what happens to the partial sums of the alternating harmonic series. Because the sizes of the terms $\ds a_n$ are decreasing, the odd partial sums $\ds s_1\text{,}$ $\ds s_3\text{,}$ $\ds s_5\text{,}$ and so on, form a decreasing sequence that is bounded below by $\ds s_2\text{,}$ so this sequence must converge. Likewise, the even partial sums $\ds s_2\text{,}$ $\ds s_4\text{,}$ $\ds s_6\text{,}$ and so on, form an increasing sequence that is bounded above by $\ds s_1\text{,}$ so this sequence also converges. Since all the even numbered partial sums are less than all the odd numbered ones, and since the “jumps” (that is, the $\ds a_i$ terms) are getting smaller and smaller, the two sequences must converge to the same value, meaning the entire sequence of partial sums $\ds s_1,s_2,s_3,\ldots$ converges as well.
The same argument works for any alternating sequence with terms that decrease in absolute value. The Alternating Series Test is worth calling a theorem.
The odd-numbered partial sums, $\ds s_1, s_3, s_5,\ldots, s_{2k+1},\ldots\text{,}$ form a decreasing sequence, because $\ds s_{2k+3}=s_{2k+1}-a_{2k+2}+a_{2k+3}\le s_{2k+1}\text{,}$ since $\ds a_{2k+2}\ge a_{2k+3}\text{.}$ This sequence is bounded below by $\ds s_2\text{,}$ so it must converge, to some value $L\text{.}$ Likewise, the partial sums $\ds s_2, s_4, s_6,\ldots,s_{2k},\ldots\text{,}$ form an increasing sequence that is bounded above by $\ds s_1\text{,}$ so this sequence also converges, to some value $M\text{.}$ Since $\ds\lim_{n\to\infty} a_n=0$ and $\ds s_{2k+1}= s_{2k}+a_{2k+1}\text{,}$
\begin{equation*} L=\lim_{k\to\infty}s_{2k+1}=\lim_{k\to\infty}(s_{2k}+a_{2k+1})= \lim_{k\to\infty}s_{2k}+\lim_{k\to\infty}a_{2k+1}=M+0=M\text{,} \end{equation*}
so $L=M\text{;}$ the two sequences of partial sums converge to the same limit, and this means the entire sequence of partial sums also converges to $L\text{.}$
Another useful fact is implicit in this discussion. Suppose that
\begin{equation*} L=\sum_{n=1}^\infty (-1)^{n-1} a_n \end{equation*}
and that we approximate $L$ by a finite part of this sum, say
\begin{equation*} L\approx \sum_{n=1}^N (-1)^{n-1} a_n\text{.} \end{equation*}
Because the terms are decreasing in size, we know that the true value of $L$ must be between this approximation and the next one, that is, between
\begin{equation*} \sum_{n=1}^N (-1)^{n-1} a_n \hbox{and} \sum_{n=1}^{N+1} (-1)^{n-1} a_n\text{.} \end{equation*}
Depending on whether $N$ is odd or even, the second will be smaller or larger than the first.
###### Example6.48. Approximating a Series.
Approximate the sum of the alternating harmonic series to within 0.05.
Solution
We need to go to the point at which the next term to be added or subtracted is $1/10\text{.}$ Adding up the first nine and the first ten terms we get approximately $0.746$ and $0.646\text{.}$ These are $1/10$ apart, so the value halfway between them, 0.696, is within 0.05 of the correct value.
Note: We have considered alternating series with first index 1, and in which the first term is positive, but a little thought shows this is not crucial. The same test applies to any similar series, such as $\ds\sum_{n=0}^\infty (-1)^n a_n\text{,}$ $\ds\sum_{n=1}^\infty (-1)^n a_n\text{,}$ $\ds\sum_{n=17}^\infty (-1)^n a_n\text{,}$ etc.
##### Exercises for Section 6.4.
Determine whether the following series converge or diverge.
1. $\ds\sum_{n=1}^\infty {(-1)^{n-1}\over 2n+5}$
converges
Solution
Let $a_n = \dfrac{1}{2n+5}\text{.}$ Then
\begin{equation*} \lim_{n\to\infty}a_n = \lim_{n\to\infty} \frac{1}{2n+5} = 0\text{,} \end{equation*}
and we also see that the sequence $\{a_n\}_{n=1}^{\infty}$ is non-increasing:
\begin{equation*} \begin{split} a_n \amp \geq a_{n+1} \\ \frac{1}{2n+5} \amp \geq \frac{1}{2(n+1)+5} = \frac{1}{2n+7} \\ 2n+ 7 \amp \geq 2n + 5 \\ 7 \amp \geq 5, \end{split} \end{equation*}
which is true. Therefore, by the Alternating Series Test, the series $\displaystyle \sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}}{2n+5}$ converges.
2. $\ds\sum_{n=4}^\infty {(-1)^{n-1}\over \sqrt{n-3}}$
converges
Solution
Let $a_n = \dfrac{1}{\sqrt{n-3}}\text{.}$ We first calculate
\begin{equation*} \lim_{n\to\infty} a_n = \lim_{n\to\infty} \frac{1}{\sqrt{n-3}} = 0\text{.} \end{equation*}
Next, we check whether or not the sequence $\{a_n\}_{n=4}^{\infty}$ is decreasing:
\begin{equation*} \begin{split} a_n \amp \geq a_{n+1} \\ \frac{1}{\sqrt{n-3}} \amp \geq \frac{1}{\sqrt{n-2}} \\ \sqrt{n-2} \amp \geq \sqrt{n-3} \\ -2 \amp \geq -3, \end{split} \end{equation*}
which is true. Both conditions of the Alternating Series Test are satisfied, and so the series $\displaystyle \sum_{n=4}^{\infty} \dfrac{(-1)^{n-1}}{\sqrt{n-3}}$ converges.
3. $\ds\sum_{n=1}^\infty (-1)^{n-1}{n\over 3n-2}$
diverges
Solution
We now take $a_n = \dfrac{n}{3n-2}\text{,}$ and first calculate the limit:
\begin{equation*} \lim_{n\to\infty} a_n = \lim_{n\to\infty} \frac{n}{3n-2} = \lim_{n\to\infty} \frac{1}{3-\frac{2}{n}} = \frac{1}{3} \neq 0\text{.} \end{equation*}
Therefore, by the $n$-th Term Test, the series diverges.
4. $\ds\sum_{n=1}^\infty (-1)^{n-1}{\ln n\over n}$
converges
Solution
Let $a_n = \dfrac{\ln(n)}{n}\text{.}$ We first compute
\begin{equation*} \lim_{n\to\infty} a_n = \lim_{n\to\infty} \frac{\ln(n)}{n} \Heq \lim_{n\to\infty} \frac{1}{n} \cdot \frac{1}{1} = 0. \end{equation*}
We now check whether or not $\left\{a_n\right\}$ is a decreasing sequence. Since
\begin{equation*} f(x) = \frac{\ln(x)}{x} \implies f'(x) = \frac{1-\ln(x)}{x^2} \end{equation*}
is decreasing on $(e, \infty)\text{,}$ this means that the sequence is decreasing for $n \ge e\text{.}$ Hence, by the Alternating Series Test, the series converges.
Approximate each series to within the given error.
1. $\ds\sum_{n=1}^\infty (-1)^{n-1}{1\over n^3}\text{,}$ 0.005
$0.904$
Solution
We first notice that the series converges by Alternating Series Test. We now calculate the following partial sums:
\begin{equation*} \begin{array}{l|lllllll} n \amp 1 \amp 2 \amp 3 \amp 4 \amp 5 \amp 6 \amp \dots \\ \hline s_n \amp 1 \amp 0.875 \amp 0.912...\amp 0.896... \amp 0.904... \amp 0.899... \amp \dots \end{array} \end{equation*}
We have that $|s_6-s_5| = |a_6| = 0.004629... \lt 0.005\text{.}$ Therefore, $s_5 \approx 0.904$ is within 0.005 of the true sum.
2. $\ds\sum_{n=1}^\infty (-1)^{n-1}{1\over n^4}\text{,}$ 0.005
$0.946$
We have that $|s_4-s_3| = |a_4| = 0.0039... \le 0.005 \text{.}$ Therefore, $s_3 \approx 0.946$ is within 0.005 of the true sum. | 2022-09-28T13:51:46 | {
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https://math.stackexchange.com/questions/119958/equivalent-to-int-cos-left2x-right-dx | # Equivalent to $\int \cos{\left(2x\right)} \ dx$?
For some reason, I am lost on one part of this integration problem:
\begin{align}\int \cos{\left(2x\right)} \ dx\end{align}
$u = 2x$
$du = 2 \ dx$
$\frac{1}{2} du = dx$
\begin{align}\frac{1}{2} \int \cos{\left(u\right)} \ du\end{align}
\begin{align}\frac{\sin{u}}{2} + C\end{align}
\begin{align}\frac{\sin{\left(2x\right)}}{2} + C\end{align}
The only issue is that I have been told that this should really equal $\sin{x}\cos{x} + C$. I know that $\sin{\left(2x\right)} = 2\sin{x}\cos{x} + C$, so:
\begin{align}\frac{\sin{\left(2x\right)}}{2} = \frac{2\sin{x}\cos{x}}{2} + C = \sin{x}\cos{x} + C\end{align}
On a question I had asked previously, I was told that \begin{align} \int \cos{\left(2x\right)} \ dx \neq \frac{\sin{\left(2x\right)}}{2} + C\end{align}, but rather $\sin{x}\cos{x} + C$.
Please excuse the silly question, but these two expressions are the same, right?
• They are equal. Why didn't you link to the old question? – anon Mar 14 '12 at 6:39
• The two answers are the same. – André Nicolas Mar 14 '12 at 6:41
• @anon You'd have to dig around to find what I'm references. – Oliver Spryn Mar 14 '12 at 6:42
• All the more reason to provide a link, I would say. -1 for the comment. – TonyK Jun 20 '19 at 11:39
$$\int \cos(2x)dx=\frac{\sin(2x)}{2}+C=\sin(x)\cos(x)+C$$
• That is because $sin(2x) = 2 sin(x) cos(x)$, which is what he needs to know explicitly – Jeremy Carlos Mar 14 '12 at 11:46 | 2020-08-06T22:41:47 | {
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https://math.stackexchange.com/questions/3761608/is-it-true-that-frac-lna2-ln-sqrta-for-a0-in-particular-is-fr/3761619 | # Is it true that $\frac{\ln(a)}2=\ln(\sqrt{a})$ for $a>0$? In particular, is $\frac{\ln(2)}{2}=\ln(\sqrt2)$?
I believe the following two identities are correct. For some reason, they look wrong to me. Are they? $$\frac{ \ln \left( 2 \right) } { 2 } = \ln( \sqrt{2} )$$ $$\frac{ \ln \left( a \right) } { 2 } = \ln( \sqrt{a} )$$ The second one being valid for all $$a > 0$$.
• It would be more helpful if you told us what your reason for thinking the identities are wrong is. To check the identities, note that $x = y$ iff $e^x = e^y$. – Rob Arthan Jul 18 '20 at 21:12
• No they aren't. $c\ln x$ is always equal to $\ln (x^c)$ where $c$ is a constant. – Devansh Kamra Jul 18 '20 at 21:13
• I guess they look wrong to me because we are dividing by $2$ rather than multiplying by $2$. Of course dividing by $2$ is the same as multiplying by $\frac{1}{2}$. I am glad to hear that they are right. – Bob Jul 18 '20 at 21:16
• This is simply the identity $\ln a^b = b\ln a$ and the definition $\sqrt a = a^{\frac 12}$. It's a bit nice to know when we were defining things we took care that they made sense and were consistant and we weren't just pulling things out of muck. – fleablood Jul 18 '20 at 21:26
• @Bob: you may find it helpful to think of $\sqrt{2}$ as $2^{\frac{1}{2}}$ and $\frac{1}{2}$ as $2^{-1}$ so that the division is confined to the exponents. – Rob Arthan Jul 18 '20 at 21:26
They are both correct. To prove them, use the logarithm property $$\ln\left(a^b\right)=b\ln(a)$$, for $$a\gt0$$.
This can be rewritten as $$b\ln(a)=\ln\left(a^b\right),\;\;\;\text{for }a\gt0$$ $$\frac{\ln(a)}{2}$$ can be written as $$\frac12\ln(a)$$, and $$a^{(1/2)}\equiv\sqrt a$$.
You can finish it from here.
These are in fact correct. Notice it comes from the fact that
$${e^{\ln(\sqrt{a})}=a^{\frac{1}{2}}=(e^{\ln(a)})^{\frac{1}{2}}=e^{\frac{\ln(a)}{2}}}$$
Now, since $${e^{x}}$$ is bijective (and hence injective) on $${\mathbb{R}}$$, then
$${e^x=e^y \Leftrightarrow x=y}$$
And so finally
$${\ln(\sqrt{a}) = \frac{\ln(a)}{2}}$$
• Your answer is among the clearest. – Sebastiano Jul 18 '20 at 22:18
• @Sebastiano Thank you! :) – Riemann'sPointyNose Jul 18 '20 at 22:25
Yes, this is a logarithm property.
For all $$a \geq 0$$ and for all $$c > 0$$, this property holds:
$$\log_c(a^b) = b\log_c(a)$$
This property is known as the “logarithm power rule”.
Your question is about the specific case where $$b = \dfrac12$$. You can see that it’s true by rewriting $$\ln(\sqrt{a})$$ and then using the logarithm property, like this:
$$\ln(\sqrt{a}) = \ln\left(a^{1/2}\right) = \frac{\ln a}{2}$$
The proof of the rule is as follows:
$$a = c^{\log_c(a)} \tag*{Exponentiation as inverse of \log}$$ $$a^b = \left(c^{\log_c(a)}\right)^b \tag*{Each side to the power of b}$$ $$a^b = c^{b\log_c(a)} \tag*{Power rule of exponentiation}$$ $$\log_c\left(a^b\right) = \log_c\left(c^{b\log_c(a)}\right) \tag*{\log_c of both sides}$$ $$\boxed{\log_c\left(a^b\right) = b\log_c(a)} \tag*{\log as inverse of exponentiation}$$
$$\ln x = \log_e x$$. Multiply $$\frac{\ln(a)}{2} = \ln(\sqrt{a})$$ by $$2$$ to get $$\ln(a) = 2\ln(\sqrt{a})$$, and by the log rule $$x\log_a b = \log_a b^x$$, we get $$\ln(a) = \ln(\sqrt{a}^2)$$, which is obviously true.
-FruDe
If you restrict $$\log$$ function to real numbers, then it is defined for all arguments $$>0$$. Also, square root of a positive number is positive. Hence, $$\log \sqrt{a}$$ exists as long as $$a>0$$ | 2021-03-02T05:31:54 | {
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http://math.stackexchange.com/questions/124754/a-question-on-pointwise-convergence | # A question on pointwise convergence.
The function $f_n(x):[-1,1] \to \mathbb{R}, \, \, \,f_n(x) = x^{2n-1}$ tends pointwise to the function $$f(x) = \left\{\begin{array}{l l}1&\textrm{if} \quad x=1\\0&\textrm{if} \quad -1<x<1\\-1&\textrm{if} \quad x=-1\end{array}\right.$$ but not uniformly (for obvious reasons as $f(x)$ isn't continuous). But then surely in this case $||f_n(x) - f(x)||_\infty \to 0$ as for any $x \in (-1,1)$ and any $\epsilon > 0$ you can make $n$ large enough so that the max distance between $f_n(x)$ and $f(x)$ at that particular $x$ is less than $\epsilon$? What am I doing wrong?
Thanks!
-
$n$ depends on $x$... – Nate Eldredge Mar 26 '12 at 18:03
Do you mean $f_n(x) = x^{2n-1}$? – Robert Israel Mar 26 '12 at 18:11
Ah so $||f_n(x) - f(x)||_\infty = 2$? Is that what you're saying? – user26069 Mar 26 '12 at 18:12
@user26069: $\lVert f_n(x)-f(x)\rVert|_{\infty} = \sup\{|f_n(x)-f(x)|\mid x\in[-1,1]\} = 1$, since we can make the difference as close to $1$ as we like by taking $x$ close enough to (but not equal to) $1$ or to $-1$; and because the difference is certainly bounded above by $1$. – Arturo Magidin Mar 26 '12 at 18:17
• for any $x \in [-1,1]$ and any $\epsilon > 0$ you can make $n$ large enough so that the max distance between $f_n(x)$ and $f(x)$ at that particular $x$ is less than $\epsilon$
is a great proof that for any $x$, the sequence $f_n(x)$ tends to $f(x)$. In other words, it's a proof that $f_n$ tends to $f$ pointwise.
But the assertion $||f_n(x) - f(x)||_\infty \to 0$ is the assertion that $f_n$ tends to $f$ uniformly on $[-1,1]$, which as you've noted is false.
The difference in the two statements (hence in how you'd need to prove them) is in the order of quantifiers. For pointwise convergence, given any $\epsilon>0$, it's "for all $x$, there exists $n$ such that..." - in other words, $n$ can depends on $x$ as well as $\epsilon$. For uniform convergence, given any $\epsilon>0$, it's "there exists $n$ such that for all $x$, ..." - in other words, $n$ cannot depend on $x$.
If you try reorganizing your proof so that you have to choose $n$ before $x$ is given, then you'll see the proof break down. | 2016-02-13T18:01:31 | {
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http://math.stackexchange.com/questions/289921/help-with-a-summation | # Help with a Summation
I have been trying to figure out why the summation of $(r+2)\binom{n-4}{r}$ from $r=0$ to $(n-4)$ is equal to $n2^{n-5}$, but I can't seem to get it to work. The only thing I can think of that would be relevant is that the summation of $\binom{n}{k}$ from $k=0$ to $n$ is $2^n$, but I can't get it to follow. Could anyone help me with this, either with a solution or by telling me what method I'm meant to use? Are there by any chance any facts I'm meant to know?
-
You can decompose your sum as $$\sum_{r=0}^{n-4} r {n-4 \choose r}+\sum_{r=0}^{n-4} 2{n-4 \choose r}$$
The second sum is $2\cdot 2^{n-4}=2^{n-3}$ by the formula you gave.
You also require the result that $$\sum_{r=0}^{n}r{n \choose r}=n2^{n-1}$$ Which gives the first sum as $(n-4)\cdot 2^{n-5}$ Your full sum is therefore $2^{n-5}(n-4+4)=n2^{n-5}$
I don't know if you have proved these identities yourself, but if not, consider the binomial expansions of $(1+x)^n$ and its derivative at $x=1$.
-
Thank you very much for your help! I did prove the first one just by considering the number of subsets of {1,...,n} and applying double counting. I'm now going to try and prove the second one. This would be a key piece I'm missing. :D – user60126 Jan 29 '13 at 18:28
You're welcome! I don't know of a counting argument for the other identity; I only know the binomial theorem proof. – Daniel Littlewood Jan 30 '13 at 19:30
Hint: Use the binomial identity
$$r \binom{n-4}{r} = (n-4) \binom{n-5}{r-1}$$
Keep in mind that $\binom{n-5}{-1} = \binom{n-5}{n-4} = 0.$
-
$$\sum_{r=o}^{n-4}(r+2)\binom{n-4}{r}=\sum_{r=0}^{n-4}r\binom{n-4}{r}+2\sum_{r=0}^{n-4}\binom{n-4}{r}=$$ $$=\sum_{r=0}^{n-4}(n-4)\binom{n-5}{r-1}+2\cdot2^{n-4}=(n-4)\sum_{j=0}^{n-5}\binom{n-5}{j}+2^{n-3}=$$ $$=(n-4)2^{n-5}+2^{n-3}=n\cdot 2^{n-5}-4\cdot2^{n-5}+2^{n-3}=$$ $$=n\cdot 2^{n-5}-2^2\cdot2^{n-5}+2^{n-3}=n\cdot 2^{n-5}-2^{n-3}+2^{n-3}=n\cdot 2^{n-5}$$ Using $$k\binom{n}{k}=n\binom{n-1}{k-1}\Rightarrow r\binom{n-4}{r}=(n-4)\binom{n-5}{r-1}$$
-
First of all,
$$\sum_{r=0}^{n-4}{n-4 \choose r}(r+2)$$
is the same as (where $m = n-4$)
$$\sum_{r=0}^{m}{m \choose r}(r+2) = \sum_{r=0}^{m}{m \choose r}r + 2\sum_{r=0}^{m}{m \choose r}$$
As you've said, using the binomial theorem,
$$2\sum_{r=0}^{m}{m \choose r}1^r 1^{m-r} = 2(1+1)^m = 2^{m+1}$$
The second part, can be explained as having a bunch of people ($m$), in how many ways can you pick a group and designate one of them as a leader:
1. You either pick a group of size $r$ (${m \choose r}$ ways) and pick one as the leader ($r$ ways), for every group size - Leading to sum:
$$\sum_{r=0}^{m}{m \choose r}r$$
1. Pick one as the leader ($m$ choices) and for every one of the remainning people ($m-1$) decide whether they are in the group or not ($2^{m-1}$ ways) leading to
$$m 2^{m-1}$$
So that together you have
$$m 2^{m-1} + 2^{m+1} = (m+4) 2^{m-1} = n 2 ^{-5}$$
-
Thank you very much for the reasoning to the second part. That makes perfect sense. I didn't think to change n-4 to m! – user60126 Jan 29 '13 at 18:33
The summation you quote will be useful.
Then note the following trick, which has a vast scope (http://www.math.upenn.edu/~wilf/DownldGF.html).
Start with $$(1 + x)^{n-4} = \sum_{r=0}^{n-4} \dbinom{n-4}{r} x^r.$$ Derive both sides and then set $x = 1$ to get... then use the other summation.
- | 2015-07-02T05:27:54 | {
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https://github.uconn.edu/tgr12001/ME3255S2017/blob/master/lecture_07/lecture_07.md | tgr12001/ME3255S2017 forked from sed12008/ME3255S2017
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%plot --format svg
setdefaults
Roots: Open methods
Newton-Raphson
First-order approximation for the location of the root (i.e. assume the slope at the given point is constant, what is the solution when f(x)=0)
$f'(x_{i})=\frac{f(x_{i})-0}{x_{i}-x_{i+1}}$
$x_{i+1}=x_{i}-\frac{f(x_{i})}{f'(x_{i})}$
Use Newton-Raphson to find solution when $e^{-x}=x$
f= @(x) exp(-x)-x;
df= @(x) -exp(-x)-1;
x_i= 0;
x_r = x_i-f(x_i)/df(x_i)
error_approx = abs((x_r-x_i)/x_r)
x_i=x_r;
x_r = 0.50000
error_approx = 1
x_r = x_i-f(x_i)/df(x_i)
error_approx = abs((x_r-x_i)/x_r)
x_i=x_r;
x_r = 0.56631
error_approx = 0.11709
x_r = x_i-f(x_i)/df(x_i)
error_approx = abs((x_r-x_i)/x_r)
x_i=x_r;
x_r = 0.56714
error_approx = 0.0014673
x_r = x_i-f(x_i)/df(x_i)
error_approx = abs((x_r-x_i)/x_r)
x_i=x_r;
x_r = 0.56714
error_approx = 2.2106e-07
In the bungee jumper example, we created a function f(m) that when f(m)=0, then the mass had been chosen such that at t=4 s, the velocity is 36 m/s.
$f(m)=\sqrt{\frac{gm}{c_{d}}}\tanh(\sqrt{\frac{gc_{d}}{m}}t)-v(t)$.
to use the Newton-Raphson method, we need the derivative $\frac{df}{dm}$
$\frac{df}{dm}=\frac{1}{2}\sqrt{\frac{g}{mc_{d}}}\tanh(\sqrt{\frac{gc_{d}}{m}}t)- \frac{g}{2m}\mathrm{sech}^{2}(\sqrt{\frac{gc_{d}}{m}}t)$
setdefaults
g=9.81; % acceleration due to gravity
m=linspace(50, 200,100); % possible values for mass 50 to 200 kg
c_d=0.25; % drag coefficient
t=4; % at time = 4 seconds
v=36; % speed must be 36 m/s
f_m = @(m) sqrt(g*m/c_d).*tanh(sqrt(g*c_d./m)*t)-v; % anonymous function f_m
df_m = @(m) 1/2*sqrt(g./m/c_d).*tanh(sqrt(g*c_d./m)*t)-g/2./m*sech(sqrt(g*c_d./m)*t).^2;
[root,ea,iter]=newtraph(f_m,df_m,140,0.00001)
root = 142.74
ea = 8.0930e-06
iter = 48
Secant Methods
Not always able to evaluate the derivative. Approximation of derivative:
$f'(x_{i})=\frac{f(x_{i-1})-f(x_{i})}{x_{i-1}-x_{i}}$
$x_{i+1}=x_{i}-\frac{f(x_{i})}{f'(x_{i})}$
$x_{i+1}=x_{i}-\frac{f(x_{i})}{\frac{f(x_{i-1})-f(x_{i})}{x_{i-1}-x_{i}}}= x_{i}-\frac{f(x_{i})(x_{i-1}-x_{i})}{f(x_{i-1})-f(x_{i})}$
What values should $x_{i}$ and $x_{i-1}$ take?
To reduce arbitrary selection of variables, use the
Modified Secant method
Change the x evaluations to a perturbation $\delta$.
$x_{i+1}=x_{i}-\frac{f(x_{i})(\delta x_{i})}{f(x_{i}+\delta x_{i})-f(x_{i})}$
[root,ea,iter]=mod_secant(f_m,1,50,0.00001)
root = 142.74
ea = 3.0615e-07
iter = 7
car_payments(400,30000,0.05,5,1)
ans = 1.1185e+04
Amt_numerical=mod_secant(@(A) car_payments(A,700000,0.0875,30,0),1e-6,50,0.001)
car_payments(Amt_numerical,700000,0.0875,30,1)
Amt_numerical = 5467.0
ans = 3.9755e-04
Amt_numerical*12*30
ans = 1.9681e+06
Amortization calculation makes the same calculation for the monthly payment amount, A, paying off the principle amount, P, over n pay periods with monthly interest rate, r.
% Amortization calculation
A = @(P,r,n) P*(r*(1+r)^n)./((1+r)^n-1);
Amt=A(30000,0.05/12,5*12)
Amt = 566.14
Matlab's function
Matlab and Octave combine bracketing and open methods in the fzero function.
help fzero
'fzero' is a function from the file /usr/share/octave/4.0.0/m/optimization/fzero.m
-- Function File: fzero (FUN, X0)
-- Function File: fzero (FUN, X0, OPTIONS)
-- Function File: [X, FVAL, INFO, OUTPUT] = fzero (...)
Find a zero of a univariate function.
FUN is a function handle, inline function, or string containing the
name of the function to evaluate.
X0 should be a two-element vector specifying two points which
bracket a zero. In other words, there must be a change in sign of
the function between X0(1) and X0(2). More mathematically, the
following must hold
sign (FUN(X0(1))) * sign (FUN(X0(2))) <= 0
If X0 is a single scalar then several nearby and distant values are
probed in an attempt to obtain a valid bracketing. If this is not
successful, the function fails.
OPTIONS is a structure specifying additional options. Currently,
'fzero' recognizes these options: "FunValCheck", "OutputFcn",
"TolX", "MaxIter", "MaxFunEvals". For a description of these
options, see *note optimset: XREFoptimset.
On exit, the function returns X, the approximate zero point and
FVAL, the function value thereof.
INFO is an exit flag that can have these values:
* 1 The algorithm converged to a solution.
* 0 Maximum number of iterations or function evaluations has
been reached.
* -1 The algorithm has been terminated from user output
function.
* -5 The algorithm may have converged to a singular point.
OUTPUT is a structure containing runtime information about the
'fzero' algorithm. Fields in the structure are:
* iterations Number of iterations through loop.
* nfev Number of function evaluations.
* bracketx A two-element vector with the final bracketing of the
zero along the x-axis.
* brackety A two-element vector with the final bracketing of the
zero along the y-axis.
Additional help for built-in functions and operators is
available in the online version of the manual. Use the command
'doc <topic>' to search the manual index.
Help and information about Octave is also available on the WWW
at http://www.octave.org and via the [email protected]
mailing list.
fzero(@(A) car_payments(A,30000,0.05,5,0),500)
ans = 563.79
Comparison of Solvers
It's helpful to compare to the convergence of different routines to see how quickly you find a solution.
Comparing the freefall example
N=20;
iterations = linspace(1,400,N);
ea_nr=zeros(1,N); % appr error Newton-Raphson
ea_ms=zeros(1,N); % appr error Modified Secant
ea_fp=zeros(1,N); % appr error false point method
ea_bs=zeros(1,N); % appr error bisect method
for i=1:length(iterations)
[root_nr,ea_nr(i),iter_nr]=newtraph(f_m,df_m,300,0,iterations(i));
[root_ms,ea_ms(i),iter_ms]=mod_secant(f_m,1e-6,300,0,iterations(i));
[root_fp,ea_fp(i),iter_fp]=falsepos(f_m,1,300,0,iterations(i));
[root_bs,ea_bs(i),iter_bs]=bisect(f_m,1,300,0,iterations(i));
end
setdefaults
semilogy(iterations,abs(ea_nr),iterations,abs(ea_ms),iterations,abs(ea_fp),iterations,abs(ea_bs))
legend('newton-raphson','mod-secant','false point','bisection')
warning: axis: omitting non-positive data in log plot
warning: called from
__line__ at line 120 column 16
line at line 56 column 8
__plt__>__plt2vv__ at line 500 column 10
__plt__>__plt2__ at line 246 column 14
__plt__ at line 133 column 15
semilogy at line 60 column 10
warning: axis: omitting non-positive data in log plot
warning: axis: omitting non-positive data in log plot
warning: axis: omitting non-positive data in log plot
ea_nr
ea_nr =
Columns 1 through 8:
6.36591 0.06436 0.00052 0.00000 0.00000 0.00000 0.00000 0.00000
Columns 9 through 16:
0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000
Columns 17 through 20:
0.00000 0.00000 0.00000 0.00000
N=20;
f= @(x) x^10-1;
df=@(x) 10*x^9;
iterations = linspace(1,50,N);
ea_nr=zeros(1,N); % appr error Newton-Raphson
ea_ms=zeros(1,N); % appr error Modified Secant
ea_fp=zeros(1,N); % appr error false point method
ea_bs=zeros(1,N); % appr error bisect method
for i=1:length(iterations)
[root_nr,ea_nr(i),iter_nr]=newtraph(f,df,0.5,0,iterations(i));
[root_ms,ea_ms(i),iter_ms]=mod_secant(f,1e-6,0.5,0,iterations(i));
[root_fp,ea_fp(i),iter_fp]=falsepos(f,0,5,0,iterations(i));
[root_bs,ea_bs(i),iter_bs]=bisect(f,0,5,0,iterations(i));
end
semilogy(iterations,abs(ea_nr),iterations,abs(ea_ms),iterations,abs(ea_fp),iterations,abs(ea_bs))
legend('newton-raphson','mod-secant','false point','bisection')
warning: axis: omitting non-positive data in log plot
warning: called from
__line__ at line 120 column 16
line at line 56 column 8
__plt__>__plt2vv__ at line 500 column 10
__plt__>__plt2__ at line 246 column 14
__plt__ at line 133 column 15
semilogy at line 60 column 10
warning: axis: omitting non-positive data in log plot
warning: axis: omitting non-positive data in log plot
warning: axis: omitting non-positive data in log plot
ea_nr
newtraph(f,df,0.5,0,12)
ea_nr =
Columns 1 through 7:
99.03195 11.11111 11.11111 11.11111 11.11111 11.11111 11.11111
Columns 8 through 14:
11.11111 11.11111 11.11111 11.11109 11.11052 11.10624 10.99684
Columns 15 through 20:
8.76956 2.12993 0.00000 0.00000 0.00000 0.00000
ans = 16.208
df(300)
ans = 1.9683e+23
% our class function
f= @(x) tan(x)-(x-1).^2
mod_secant(f,1e-3,1)
f =
@(x) tan (x) - (x - 1) .^ 2
ans = 0.37375
f(ans)
ans = -3.5577e-13
tan(0.37375)
(0.37375-1)^2
ans = 0.39218
ans = 0.39219
f([0:10])
ans =
Columns 1 through 8:
-1.0000 1.5574 -3.1850 -4.1425 -7.8422 -19.3805 -25.2910 -35.1286
Columns 9 through 11:
-55.7997 -64.4523 -80.3516
You can’t perform that action at this time. | 2020-07-14T14:09:50 | {
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https://math.stackexchange.com/questions/2502791/arranging-3-red-balls-2-blue-balls-and-2-green-balls-so-that-no-two-adja | # Arranging $3$ red balls, $2$ blue balls, and $2$ green balls so that no two adjacent balls are of the same color
I want to arrange $3$ red balls, $2$ blue balls, and $2$ green balls in a line so that no two balls of the same color are adjacent. If there are no such restrictions, then the number of such unique arrangements is
$$\frac{7!}{3!\times 2!\times 2!}=210.$$
I found a related problem here but it does not have an answer. Instead, I wrote a code in R that will completely list all such possible arrangements. I found 38. I also tried to list them manually below.
• _ B _ G _ B _ G _: $C^5_3=10$ arrangements
• _ G _ B _ G _ B _: $C^5_3=10$ arrangements
• _ G R G _ B R B _: 3 arrangements
• _ B R B _ G R G _: 3 arrangements
• _ G _ B R B _ G _: $C^4_2=6$ arrangements
• _ B _ G R G _ B _: $C^4_2=6$ arrangements
Is this correct? Is there a more elegant way of doing this?
There are $10$ ways to arrange the red balls. Of these:
1. two arrangements leave three adjacent free spots and one solo free spot: $$R\,\_\,\_\,\_R\,\_R\\ R\,\_R\,\_\,\_\,\_R$$For each of these: Whichever colour is in the middle of the three free spots must also be in the solo spot, so 2 possibilities for each of these arrangements
2. six arrangements leave two adjacent spots and two solo spots: $$\,\_\,\_R\,\_R\,\_R\\ \,\_R\,\_\,\_R\,\_R\\ \,\_R\,\_R\,\_\,\_R\\ R\,\_\,\_R\,\_R\,\_\\ R\,\_R\,\_\,\_R\,\_\\ R\,\_R\,\_R\,\_\,\_$$ and one leaves two pairs of adjacent spots: $$R\,\_\,\_R\,\_\,\_R$$For each of these: the colours must differ in a pair of adjacent spots, but they can be swapped around, so 4 possibilities for each of these arrangements
3. one arrangement leaves no adjacent spots: $$\,\_R\,\_R\,\_R\,\_$$ Here there are six possible arrangements of the remaining $4$ balls
In total: $$2\cdot 2 + 7\cdot 4 + 1\cdot 6 = 38$$ I don't think there is a more "elegant" way to do this than simply splitting into cases and count. The interactions are too complicated to make inclusion-exclusion viable, and there isn't a simple formula for "non-adjacent" that we can apply. Of course, how you decide to split into cases changes how easy the calculations are and how easily you miss a case or double count.
• Case 3 has only six arrangements of the remaining 4 balls, so your solution agrees with the original poster's. – Michael Nov 3 '17 at 12:30
• @Michael You're right. – Arthur Nov 3 '17 at 12:36
• This is a more systematic approach. Indeed, I think it is easier to miss or duplicate cases in my "manual" solution. – hpesoj626 Nov 3 '17 at 12:37
You might find applying inclusion-exclusion more "elegant"
Keeping the A's apart throughout, count the # of arrangements with
(A's apart) - (A's apart, B's or C's bunched) + (A's apart, both B's and C's bunched)
$= \left[\frac{4!}{2!2!}\times \binom53 \right] - \left[2\times \frac{3!}{2!}\binom43\right] + \left[2!\times\binom33\right] = 60 - 2\cdot12 + 2 =38$
• Definitely is! Thanks for taking your time to answer this. So insightful. – hpesoj626 Nov 4 '17 at 12:08
• You are welcome ! – true blue anil Nov 4 '17 at 17:28 | 2019-10-17T13:30:01 | {
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https://philhallmark.com/millennium-spa-mfrj/49dd6b-transpose-of-a-diagonal-matrix | ', then the element B(2,3) is also 1+2i. In linear algebra, a diagonal matrix is a matrix in which the entries outside the main diagonal are all zero; the term usually refers to square matrices. D = diag(v,k) places the elements of vector v on the kth diagonal. collapse all in page. Given a 2D Matrix, return the transpose of it. The row and column spaces have the same rank, which is also the rank of matrix , i.e. Transpose of matrix A is denoted by A T. Two rows of A T are the columns of A. So, it's B transpose times A transpose. In such type of square matrix, off-diagonal blocks are zero matrices and main diagonal blocks square matrices. Answer: A matrix has an inverse if and only if it is both squares as well as non-degenerate. For a matrix defined as = , the transpose matrix is defined as = . For example, element at position a12 (row 1 and column 2) will now be shifted to position a21 (row 2 and column 1), a13 to a31, a21 to a12and so on. Moreover, if the diagonal entries of a diagonal matrix are all one, it is the identity matrix: Rank . Examples: Properties of an Identity Matrix. For a square matrix m, Transpose [m, {1, 1}] returns the main diagonal of m, as given by Diagonal [m]. Diagonal matrices are usually square (same number of rows and columns), but they may be rectangular. For example, if A(3,2) is 1+2i and B = A. That is a brief overview of identity, diagonal, symmetric and triangular matrices. Remark 2.3 Recall (see page 115) the formula for trans-pose of a product: (MN) T= N MT. The transpose of a matrix is a matrix created by reflecting a matrix over its main diagonal, or making the columns rows of the transpose (or vice versa). We have Zero matrix which on multiplication with any matrix (satisfying conditions for matrix multiplication) returns a Zero matrix. The transpose A T of a matrix A can be obtained by reflecting the elements along its main diagonal. And that first row there is now going to become the first column. Syntax: diag(x, nrow, ncol) Parameters: x: value present as the diagnoal elements. The transpose of a square matrix is a If A is a matrix of order m x n and B is a matrix of order n x p then the order of AB is A matrix having m rows and n columns with m ≠ n is said to be a There are multiple matrix operations that you can perform in R. This include: addition, substraction and multiplication, calculating the power, the rank, the determinant, the diagonal, the eigenvalues and eigenvectors, the transpose and decomposing the matrix by different methods. Transpose of a matrix flips the matrix over its diagonal and this brings the row elements on the column and column elements on the row. 2, 7, minus 5. The transpose of a rectangular matrix is a A matrix having m rows and n columns with m ≠ n is said to be a In a matrix multiplication for A and B, (AB)t 1, 0, minus 1. = [?????] A double application of the matrix transpose achieves no change overall. An identity matrix is a square, diagonal matrix where all of the elements on the main diagonal are one. The columns of A T are rows of A. nrow, ncol: number of rows and columns in which elements are represented. does not affect the sign of the imaginary parts. Each other elements will move across the diagonal and end up at … The transpose of a diagonal matrix is equal to the original matrix. The transpose of a matrix is the matrix flipped over it’s main diagonal, switching the row and column indices of the matrix. Syntax: diag(x, nrow, ncol) Parameters: x: value present as the diagnoal elements. Then the matrix C= 2 4v 1 v n 3 5 is an orthogonal matrix. link brightness_4 code. The Tattribute returns a view of the original array, and changing one changes the other. As the name suggests, Identity matrix works like an identity, like 1 is identity in decimal number system (Any number, multiplied with 1 returns itself). $$\begin{bmatrix} 8 & 0\\ 0 & 12 \end{bmatrix}$$. If the elements on the main diagonal are the inverse of the corresponding element on the main diagonal of the D, then D is a diagonal matrix. edit close. Transpose of a matrix can be found by changing all the rows into columns or vice versa. A Transpose is where we swap entries across the main diagonal (rows become columns) like this: The main diagonal stays the same. It is denoted by I. returns the nonconjugate transpose of A, that is, interchanges the row and column index for each element. Next: Write a program in C to find sum of left diagonals of a matrix. number or rows and columns should be equal, as shown below. The elements on positions where (number of rows) = (number of columns) like a11, a22, a33 and so on, form diagonal of a matrix. Matrices that remain unchanged on transposition. B = A.' Transpose of the matrix is one of the important terminologies used in matrix manipulations. I find it very useful in electrical network analysis to flip the input and output of a two-port network. A square matrix (2 rows, 2 columns) Also a square matrix (3 rows, 3 columns) Identity Matrix. The rank of each space is its dimension, the number of independent vectors in the space. For example: $\begin{bmatrix} 3 & 5 & 1 \\ 5 & 6 & 3 \end{bmatrix} ^\mathrm{T} = \begin{bmatrix} 3 & 5 \\ 5 & 6 \\ 1 & 3 \end{bmatrix}$ This can be extended to complex matrices as the conjugate transpose, denoted as H. A square matrix in which every element except the principal diagonal elements is zero is called a Diagonal Matrix. The tricky one to remember is that when you have the product of two matrices AB transpose, you have to reverse the order of multiplication. Eigenvalues of a triangular matrix. Syntax. In this section, you will be studying the properties of the diagonal matrix. Matrices where (number of rows) = (number of columns). Property 2: Transpose of the diagonal matrix D is as the same matrix. = ?. a_{1} Example 3: To print the rows in the Matr The transpose of a column matrix is. does not affect the sign of the imaginary parts. We indicate identity matrices usually by the letter I. Identity matrices are like a one in scalar math. A transpose will be denoted by original matrix with “T” in superscript, like Aᵀ. Even if and have the same eigenvalues, they do not necessarily have the same eigenvectors. In this post, we explain how to diagonalize a matrix if it is diagonalizable. A diagonal matrix is a square matrix with all its elements (entries) equal to zero except the elements in the main diagonal from top left to bottom right. example. When we take transpose, only the diagonal elements don’t change place. 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After transposing the matrix in C, it became 3 rows and 2 columns. In this article, a brief explanation of the orthogonal matrix is given with its definition and properties. Also, some important transpose matrices are defined based on their characteristics. : Transpose. D = D T If p = $$\begin{bmatrix} 2 & 0\\ 0 & 4 \end{bmatrix}$$ then, P T = $$\begin{bmatrix} 2 & 0\\ 0 & 4 \end{bmatrix}$$ The transpose of a lower triangular matrix is an upper triangular matrix and the transpose of an upper triangular matrix is a lower triangular matrix. Given a matrix A, return the transpose of A.. The transpose of a transpose matrix is just the original matrix. Iterating the decomposition produces the components U, V, Q, D1, D2, and R0. For example − Matrix before Transpose: 123 456 789 Matrix after Transpose: 147 258 369. Property 1: Same order diagonal matrices gives a diagonal matrix only after addition or multiplication. A square matrix (2 rows, 2 columns) Also a square matrix (3 rows, 3 columns) B = transpose(A) Description. In linear algebra, the matrix and their properties play a vital role. 1) rectangular matrix , 2) diagonal matrix , 3) square matrix , 4) scaler matrix This square of matrix calculator is designed to calculate the squared value of both 2x2 and 3x3 matrix. A new example problem was added.) Your email address will not be published. Then, the user is asked to enter the elements of the matrix (of order r*c). D = DT, If p = $$\begin{bmatrix} 2 & 0\\ 0 & 4 \end{bmatrix}$$ then, PT = $$\begin{bmatrix} 2 & 0\\ 0 & 4 \end{bmatrix}$$, Property 3: Under Multiplication, Diagonal Matrices are commutative, i. e. PQ = QP, If P = $$\begin{bmatrix} 2 & 0\\ 0 & 4 \end{bmatrix}$$ and Q = $$\begin{bmatrix} 4 & 0\\ 0 & 3 \end{bmatrix}$$, P x Q = $$\begin{bmatrix} 8+0 & 0 + 0 \\ 0 + 0 & 12+0 \end{bmatrix}$$ For example: $\begin{bmatrix} 3 & 5 & 1 \\ 5 & 6 & 3 \end{bmatrix} ^\mathrm{T} = \begin{bmatrix} 3 & 5 \\ 5 & 6 \\ 1 & 3 \end{bmatrix}$ This can be extended to complex matrices as the conjugate transpose, denoted as H. Just like we have 0 in decimal number system, which on multiplication with any number returns 0 as product. Follow twitter @xmajs Example 1: filter_none. play_arrow. $$\begin{bmatrix} Y_{11} & Y_{12} \\ Y_{21} & Y_{22} \end{bmatrix} \rightarrow \begin{bmatrix} Y_{22} & Y_{21} \\ Y_{12} & Y_{11} \end{bmatrix}$$ linear-algebra matrices. The transpose of a matrix A can be obtained by reflecting the elements along its main diagonal. Properties of transpose The transpose A T of a matrix A can be obtained by reflecting the elements along its main diagonal. Matrices which have non-zero elements in and above diagonal . In this program, the user is asked to enter the number of rows r and columns c. Their values should be less than 10 in this program. The row vectors span the row space of and the columns vectors span the column space of . Entries on the main diagonal and above can be any number (including zero). Diagonal matrices always come under square matrices. Add to solve later Sponsored Links We compute the powers of a diagonal matrix and a matrix similar to a diagonal matrix. Equal matrices two matrices are equal if they have the same order and corresponding elements.? I find it very useful in electrical network analysis to flip the input and output of a two-port network. There are many other matrices other than the Diagonal Matrix, such as symmetric matrix, antisymmetric, diagonal matrix, etc. Note that you have some arr[j][j] terms which will always refer to cells on the diagonal. (Update 10/15/2017. The second row here is now going to become the second column. In fact, every orthogonal matrix C looks like this: the columns of any orthogonal matrix form an orthonormal basis of Rn. A zero vector or matrix of any size with all zero elements is denoted as .. Diagonal Matrix. If P = $$\begin{bmatrix} 2 & 0\\ 0 & 4 \end{bmatrix}$$, and Q = $$\begin{bmatrix} 4 & 0\\ 0 & 3 \end{bmatrix}$$, P + Q = $$\begin{bmatrix} 2 & 0\\ 0 & 4 \end{bmatrix} + \begin{bmatrix} 4 & 0\\ 0 & 3 \end{bmatrix}$$, P + Q = $$\begin{bmatrix} 2 + 4 & 0 + 0 \\ 0+0 & 4 + 3\end{bmatrix}$$ Points to Remember . MATLAB has a function called eye that takes one argument for the matrix size and returns an identity matrix. Here, the non-diagonal blocks are zero. Let’s see an example. If A = A T, A is Symmetric Matrix. In this Video we Find the Transpose of a Matrix Using Excel. C transpose is now going to be a 3 by 4 matrix. How Linear Algebra and Machine Learning Help You Binge Watch TV. If A contains complex elements, then A.' A square matrix D = [dij]n x n will be called a diagonal matrix if dij = 0, whenever i is not equal to j. Unlike Identity matrices, Zero matrices can be rectangular. A diagonal matrix has zero entries all over the matrix except in the main diagonal. An example of a 2-by-2 diagonal matrix is $${\displaystyle \left[{\begin{smallmatrix}3&0\\0&2\end{smallmatrix}}\right]}$$, while an example of a 3-by-3 diagonal matrix is$${\displaystyle \left[{\begin{smallmatrix}6&0&0\\0&7&0\\0&0&4\end{smallmatrix}}\right]}$$. The following snippet gives you the indices of the desired diagonal, given the size of the square matrix n (matrix is n by n), and the number of the diagonal k, where k=0 corresponds to the main diagonal, positive numbers of k to upper diagonals and negative numbers of k to lower diagonals. Identity matrix. The different types of matrices are row matrix, column matrix, rectangular matrix, diagonal matrix, scalar matrix, zero or null matrix, unit or identity matrix, upper triangular matrix & lower triangular matrix. When we take transpose, only the diagonal elements don’t change place. Your email address will not be published. $$\begin{bmatrix} 6 & 0\\ 0 & 7 \end{bmatrix}$$, Property 2: Transpose of the diagonal matrix D is as the same matrix. Transpose vector or matrix. The transpose of a lower triangular matrix is an upper triangular matrix and the transpose of an upper triangular matrix is a lower triangular matrix. Die transponierte Matrix, gespiegelte Matrix oder gestürzte Matrix ist in der Mathematik diejenige Matrix, die durch Vertauschen der Rollen von Zeilen und Spalten einer gegebenen Matrix entsteht. For a rectangular matrix the way of finding diagonal elements remains same, i.e. Create diagonal matrix or get diagonal elements of matrix. This fact was already noted by Pietro Majer for the case n = 1 with notation P instead of J used in the Golyshev and Stienstra paper. To find the length of a numpy matrix in Python you can use shape which is a property of both numpy ndarray's and matrices.. A.shape. In other words, transpose of A[][] is obtained by changing A[i][j] to A[j][i]. 2. And this is a pretty neat takeaway. the row and column indices of the matrix are switched. A square matrix with 1's as diagonal elements and 0’s as other elements is called an Identity matrix. Enter rows and columns of matrix: 2 3 Enter elements of matrix: Enter element a11: 1 Enter element a12: 2 Enter element a13: 9 Enter element a21: 0 Enter element a22: 4 Enter element a23: 7 Entered Matrix: 1 2 9 0 4 7 Transpose of Matrix: 1 0 2 4 9 7 Examples of how to use “diagonal matrix” in a sentence from the Cambridge Dictionary Labs Identity Matrix . filter_none. 1) rectangular matrix , 2) diagonal matrix , 3) square matrix , 4) scaler matrix Programming Simplified is licensed under a Creative Commons Attribution-NonCommercial-NoDerivs 3.0 Unported License. That is, the product of any matrix with the identity matrix yields itself. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … This is a Most important question of gk exam. An example of this is given as follows − Matrix = 1 2 3 4 5 6 7 8 9 Transpose = 1 4 7 2 5 8 3 6 9 A program that demonstrates this is given as follows. A matrix which is split into blocks is called a block matrix. What do you call a matrix operation where you transpose it and then flip it over its anti-diagonal? That’s why we assigned j value to rows, and i value to columns. Required fields are marked *. Just another variation using Array.map. This example will show you how to compute transpose of a matrix in C program. And, essentially, it's going to be the matrix C with all the rows swapped for the columns or all the columns swapped for the rows. This switches the rows and columns indices of the matrix A by producing another matrix. Let D = $$\begin{bmatrix} a_{11} & 0& 0\\ 0 & a_{22} & 0\\ 0& 0 & a_{33} \end{bmatrix}$$, Adj D = $$\begin{bmatrix} a_{22}a_{33} & 0& 0\\ 0 & a_{11}a_{33} & 0\\ 0& 0 & a_{11}a_{22} \end{bmatrix}$$, = $$\frac{1}{a_{11}a_{22}a_{33}} \begin{bmatrix} a_{22}a_{33} & 0& 0\\ 0 & a_{11}a_{33} & 0\\ 0& 0 & a_{11}a_{22} \end{bmatrix}$$ Sums and differences of diagonal matrices are also diagonal matrices. In a square matrix, transposition "flips" the matrix over the main diagonal. Symmetric Matrices. = [?????] Transpose of the matrix is one of the important terminologies used in matrix manipulations. The transpose of a matrix in linear algebra is an operator which flips a matrix over its diagonal. $$\begin{bmatrix} \frac{1}{a_{11}} &0 & 0\\ 0 & \frac{1}{a_{22}} &0 \\ 0& 0 & \frac{1}{a_{33}} \end{bmatrix}$$. Because initially, user-entered values 2 rows and 3 columns. Here are some of the most common types of matrix: Square . (c) A triangular matrix is invertible if and only if its diagonal entries are all nonzero. If we take transpose of transpose matrix, the matrix obtained is equal to the original matrix. A transpose will be denoted by original matrix with “T” in superscript, like Aᵀ. Enter the number of rows: 4 Enter the number of columns: 3 Enter elements of matrix: 1 2 3 4 5 6 7 8 9 10 11 12 Transpose of Matrix: 1 4 7 10 2 5 8 11 3 6 9 12 example. The transpose of a matrix is a new matrix that is obtained by exchanging the rows and columns. In linear algebra, the transpose of a matrix is an operator which flips a matrix over its diagonal; that is, it switches the row and column indices of the matrix A by producing another matrix, often denoted by A T (among other notations). That is, $$L^{T} = U$$ and $$U^{T} = L$$. In this section we have seen how to find out transpose of a matrix by using two methods one is by using the operator and the other one is by using transpose command. In other words, transpose of A[][] is obtained by changing A[i][j] to A[j][i]. D = diag(v) returns a square diagonal matrix with the elements of vector v on the main diagonal. In this Video we Find the Transpose of a Matrix Using Excel. where S † is a diagonal matrix whose elements are the reciprocal of the corresponding diagonal elements of S; except when the elements of the latter are zero or very close to zero where the elements of S † are equated to those of S. When A is not a square matrix, then the inversion, A †, given Eq. In practical terms, the matrix transpose is usually thought of as either (a) flipping along the diagonal entries or (b) “switching” the rows for columns. Matrices that on taking transpose become equal to their product with (-1) (scalar multiplication). Example: Hence, this is the diagonal matrix. ', then the element B (2,3) is also 1+2i. If is an eigenvector of the transpose, it satisfies By transposing both sides of the equation, we get. play_arrow. 1 2 1 3, 3 4 2 4. If all entries outside the main diagonal are zero, is called a diagonal matrix.If only all entries above (or below) the main diagonal are zero, ' is called a lower (or upper) triangular matrix. diagonal matrix. As an example, we solve the following problem. edit close. If matrix A is of order 4 × 3 then it has to be multiplied with Identity matrix of order 3 × 3, denoted as I₃₃ or just I₃. Identity Matrix is a matrix that has 1 s as the entries in the main diagonal. Question is : The transpose of a column matrix is , Options is : 1. zero matrix, … If you observe the above for loop in this C transpose of a matrix program, we assigned the rows to j and columns to i. Image will be uploaded soon The transpose of a matrix A, denoted by A , A′, A , A or A , may be constructed by any one of the following methods: Rather, we are building a foundation that will support those insights in the future. So, it's now going to be a 3 by 4 matrix. In a transpose matrix, the diagonal remains unchanged, but all the other elements are rotated around the diagonal. In this section, you will be studying diagonal matrix definition, the properties of a diagonal matrix, sample solved problems of Diagonal Matrix. We can see that, A = A T. So A is a Symmetric Matrix. D1 is a M-by-(K+L) diagonal matrix with 1s in the first K entries, D2 is a P-by-(K+L) matrix whose top right L-by-L block is diagonal, R0 is a (K+L)-by-N matrix whose rightmost (K+L)-by- (K+L) block is nonsingular upper block triangular, K+L is the effective numerical rank of the matrix [A; B]. 3. For Square Matrix : The below program finds transpose of A[][] and stores the result in B[][], we can change N for different dimension. example. For example, if A (3,2) is 1+2i and B = A. For example, element at position a12 (row 1 and column 2) will now be shifted to position a21 (row 2 and column 1), a13 to a31, a21 to a12 and so on. What do you call a matrix operation where you transpose it and then flip it over its anti-diagonal? B = A.' We denote upper triangular matrices with U. Matrices which have non-zero elements in and below diagonal. If XY exists, where X and Y are matrices, then the matrix y times XT, minus transpose of XY is O a symmetric matrix a null matrix a diagonal matrix an identity matrix Get more help from Chegg Get 1:1 help now from expert Algebra tutors Solve it with our algebra problem solver and calculator The identity matrix of size is the × matrix in which all the elements on the main diagonal are equal to 1 and all other elements are equal to 0, e.g. 6.2.1. The transpose of a matrix is the matrix flipped over it's main diagonal, switching the row and column indices of the matrix. Register at BYJU’S to study many more interesting mathematical topics and concepts. When you add matrices and you transpose is same as transposing the matrices and then adding them. $A = \begin{bmatrix} 6 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$ Triangular Matrix An upper triangular matrix is a square matrix with all its elements below the main diagonal equal to zero. Let’s learn about the properties of the diagonal matrix now. For the matrices with whose number of rows and columns are unequal, we call them rectangular matrices. If A contains complex elements, then A.' A is a square matrix. a square matrix where all the elements below the leading diagonal are zero.? Construct a Diagonal Matrix in R Programming – diag() Function Last Updated: 03-06-2020. diag() function in R Language is used to construct a diagonal matrix. A square matrix has the same number of rows as columns. If you want to insert any vector on a diagonal of a matrix, one can use plain indexing. edit close. Diagonalize the matrix A=[4−3−33−2−3−112]by finding a nonsingular matrix S and a diagonal matrix D such that S−1AS=D. In other words, the elements in a diagonal line from element a 11 to the bottom right corner will remain the same. Dij = 0 when i is not equal to j, then D is called a block diagonal matrix. For Square Matrix : The below program finds transpose of A[][] and stores the result in B[][], we can change N for different dimension. Now take the transpose of A. A square matrix has the same number of rows as columns. The transpose has some important properties, and they allow easier manipulation of matrices. (b) The product of lower triangular matrices is lower triangular, and the product of upper triangular matrices is upper triangular. Also, the size of the matrices also changes from m×n to n×m. collapse all in page. Using this we can 3. see that any orthogonally diagonalizable Amust be sym-metric. diag() function in R Language is used to construct a diagonal matrix. returns the nonconjugate transpose of A, that is, interchanges the row and column index for each element. Transpose of a matrix basically involves the flipping of matrix over the corresponding diagonals i.e. This C program is to find transpose of a square matrix without using another matrix.For example, for a 2 x 2 matrix, the transpose of matrix{1,2,3,4} will be equal to transpose{1,3,2,4}. That is the Diagonal Matrix definition. Properties of Diagonal Matrix. A diagonal matrix has zeros everywhere except on the main diagonal, which is the set of elements where row index and column index are the same. There are many types of matrices like the Identity matrix. Diagonal Matrix. = 푎??.? The row vector is called a left eigenvector of . It relates to the ordinary transpose A T (or A t as used in the paper), as follows: A τ = J A T J where J = (J i j) 0 ≤ i, j ≤ n denotes the matrix with J i j = 1 if i + j = n and J i j = 0 otherwise. Property 1: If addition or multiplication is being applied on diagonal matrices, then the matrices should be of the same order. Die erste Zeile der transponierten Matrix entspricht der ersten Spalte der Ausgangsmatrix, die zweite Zeile der zweiten Spalte und so weiter. Special Matrices¶ Zero Matrix. Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution. B = A.' Where theory is concerned, the key property of orthogonal matrices is: Prop 22.4: Let Cbe an orthogonal matrix… = [?????] In general, if n p = n q then the operation Transpose [ a , { n 1 , n 2 , … } ] is possible for an array a of dimensions { d 1 , d 2 , … } if d p = d q . A transpose of a matrix is the matrix flipped over its diagonal i.e. Lower triangular matrix a square matrix where all the elements above the leading diagonal are zero.? $$\begin{bmatrix} 8 & 0\\ 0 & 12 \end{bmatrix}$$, Q x P = $$\begin{bmatrix} 8+0 & 0 + 0 \\0 + 0& 12+0 \end{bmatrix}$$ An identity matrix of any size, or any multiple of it (a scalar matrix), is a diagonal matrix. D = diag(v) D = diag(v,k) x = diag(A) x = diag(A,k) Description. If the entries in the matrix are all zero except the ones on the diagonals from lower left corner to the other upper side(right) corner are not zero, it is anti diagonal matrix. Amust be sym-metric transposition flips '' the matrix is a matrix where... Original array, and i value to columns ( 2,3 ) is 1+2i... The transpose a T is n x m matrix T. so a is m x matrix! Most important question of gk exam that you have some arr [ j [! That has 1 s as the diagnoal elements. on to the original matrix triangular, the. J value to rows its main diagonal the row and column indices of the array! Changing one changes the other elements are rotated around the diagonal matrix D such that S−1AS=D of the matrix... The user is asked to enter the elements of a diagonal matrix finding a nonsingular matrix s a..., symmetric and triangular matrices column spaces have the same order left diagonals of a matrix zero... B ( 2,3 ) is also 1+2i matrices and main diagonal and above.. Matrix which on multiplication with any matrix with the identity matrix elements then! Matrix has an inverse if and only if it is both squares as well as non-degenerate of. Learning Help you Binge Watch TV product of lower triangular matrix a matrix... Double application of the original matrix with “ T ” in superscript, like Aᵀ 4.. Change overall producing another matrix with its definition and properties are like a one scalar! Off-Diagonal blocks are zero. will support those insights in the future = ( number of vectors. Diagonalizable Amust be sym-metric be of the imaginary parts of rows ) = ( number of and. Along its main diagonal we repeat the process of transpose the transpose it! X: value present as the diagnoal elements. all nonzero and returns an identity matrix invertible... Denote upper triangular the sign of the matrix is lower triangular matrices is lower triangular matrices the problem...: a matrix if it is both squares as well as non-degenerate an example, if contains... T. two rows of a, that is obtained by reflecting the elements in a matrix. Two matrices are also diagonal matrices, then the element B ( 2,3 ) is also...., that is, the diagonal matrix are building a foundation that will those. On multiplication with any matrix with “ T ” in superscript, like Aᵀ about the properties of imaginary! Present as the diagnoal elements. add a comment | 6 1 2 1 3, 4. Size and returns an identity matrix any vector on a diagonal matrix has the same number of rows and columns. Any multiple of it may be rectangular matrix only after addition or multiplication is being applied on matrices! The method to prove a formula is mathematical induction account ) returns zero! And 2 columns we denote upper triangular matrices is lower triangular matrices is upper triangular matrices 2! From m×n to n×m 5 is an operator which flips a matrix operation you. A can be found by changing all the rows and columns indices of the original matrix ( MN T=. Became 3 rows, and they allow easier manipulation of matrices like the identity matrix satisfying... If and have the same eigenvalues, they do not necessarily have the same with suitable identity matrix,. The main diagonal find it very useful in electrical network analysis to flip the input output... Minus 5. diag ( x, nrow, ncol: number of rows as columns flips '' the matrix over! Columns and columns indices of the matrices with whose number of rows and columns should be of the equation we. Important properties, and R0 study many more interesting mathematical topics and concepts bottom right corner will remain same. ) ( scalar multiplication ) returns a with elements in and above.... U, v, k ) places the elements along its main.! The first column that is a brief overview of identity, diagonal matrix what do you call a matrix C..., k ) places the elements along its main diagonal ( same number of rows columns. We explain how to compute transpose of a T of a matrix became 3 rows, and columns. A double application of the important terminologies used in matrix manipulations: x: value as... T } = U\ ) and \ ( L^ { T } = L\ ) is n m... Which on multiplication with any matrix with “ T ” in superscript, like Aᵀ Amust sym-metric. Other than the diagonal matrix now a, that is, the of. Are like a one in scalar math 3 5 is an operator flips. @ xmajs a square matrix ( 2 rows, 3 4 2 4 to become second. Over it 's B transpose times a transpose of a two-port network row space of Most question... Change overall flips '' the matrix a square transpose of a diagonal matrix, i.e matrix way! Matrix basically involves the flipping of matrix: square the important terminologies in! Create diagonal matrix the leading diagonal are one s as other elements zero! Follow twitter @ xmajs a square matrix in C to find sum of left diagonals a. J value to rows change place, 2 columns ) also a square matrix!, such as symmetric matrix, the matrix is one of the transpose a T, a T, is... So a is a square matrix with the identity matrix ( satisfying conditions for matrix multiplication be! Is zero is called a left eigenvector of it 's B transpose times a matrix! Manipulation of matrices like the identity matrix: rank a with elements in and below diagonal the and... Matrix now minus 5. diag ( x, nrow, ncol ) Parameters: x: value present as diagnoal! Contains complex elements, then the matrices also changes from m×n to n×m matrix flipped over its anti-diagonal or... A transpose recommended: Please solve it on “ PRACTICE ” first, before moving on the... ( a scalar matrix ), but all the elements in and above diagonal 4 2 4: ( )... Other elements is zero is called a diagonal matrix has the same order and transpose of a diagonal matrix elements. row., one can use plain indexing and only if its diagonal entries are all nonzero 7! In the future eye that takes one argument for the matrices also changes m×n. The formula for trans-pose of a matrix a square matrix where all of the diagonal has inverse! 3 columns ), but they may be rectangular a scalar matrix ), is a Most question... As other elements are represented then D is as the same number of rows as columns you... Triangular, and R0 entspricht der ersten Spalte der Ausgangsmatrix, die Zeile. Refer to cells on the kth diagonal call a matrix, the matrix... Over it 's main diagonal adding them be uploaded soon for a rectangular matrix the way finding. Of identity, diagonal matrix ( same number of rows and columns indices of the matrix.. But they may be rectangular each element same matrix other matrices other than the diagonal matrix function! Properties play a vital role, every orthogonal matrix form an orthonormal basis of Rn rows! With ( -1 ) ( scalar multiplication ) superscript, like Aᵀ are like a one in scalar.! A Most important question of gk exam matrix only after addition or multiplication is being applied on matrices... Erste Zeile der zweiten Spalte und so weiter rows of a, that is obtained by all! Indicate identity matrices are usually square ( same number of rows and 2 columns by finding a nonsingular matrix and. Der Ausgangsmatrix, die zweite Zeile der zweiten Spalte und so weiter not equal to j, then element! Number ( including zero ) ( L^ { T } = U\ ) and \ ( L^ { }. 'S main diagonal blocks square matrices matrix which is also 1+2i on taking become... 'S as diagonal elements don ’ T change place block matrix, die zweite Zeile der matrix... B transpose times a transpose will be studying the properties of transpose matrix, return the transpose the!, switching the row and column indices of the important terminologies used in manipulations. Matrix obtained is equal to its eigenvalues if its diagonal i.e transpose of a diagonal matrix their original.! The decomposition produces the components U, v, Q, D1, D2, changing... A foundation that will support those insights in the space that takes one argument for matrix. Also changes from m×n to n×m with the identity matrix die erste Zeile der zweiten Spalte und so.... Network analysis to flip the input and output of a T of a matrix in C to find sum left. Topics and concepts multiple of it it became 3 rows and columns are,. Or get diagonal elements of the diagonal matrix D such that S−1AS=D triangular, i. 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http://spagades.com/unsolved/91969728c47a5237543d5781daff | A point in R is of the form (x;y). Topic 8: Residue Theorem (PDF) 2325. The second-order version (n= 2 case) of Taylors Theorem gives the expansion f(x 0 + h) = f(x 0) + Df(x 0)(h) + Hf(x 0)(h) + R 2(h); where Df(x 0) is the derivative of fat x 0 and Hf(x 0) is the Hessian of fat x 0. Graph the function, f (x, y) = Cos(x)*Sin(y). forms. Then f(x+h)=f(x)+Df(x)(h)+o(|h|) as h 0(1) If f Thus a polynomial of degree 2 (perhaps more commonly known as a quadratic ) looks like px,y = a 00 +a 10 x +a 01 y ++a 11 xy +a 20 x2 +a 02 y2. Ex. This formula approximates f ( x) near a. Taylors Theorem gives bounds for the error in this approximation: Suppose f has n + 1 continuous derivatives on an open interval containing a. Then for each x in the interval, where the error term R n + 1 ( x) satisfies R n + 1 ( x) = f ( n + 1) ( c) ( n + 1)! ( x a) n + 1 for some c between a and x . . For example, the Taylor series of f(x) = ln(1 + x) about x= 0 is ln(1 + x) = x x2 2 + x3 3 x4 4 + If you truncate the series it is a good approximation of ln(1 + x) near x= 0. than a transcendental function. So Theorem 5.13(Taylors Theorem in Two Variables) Suppose ( ) and partial derivative up to order continuous on ! hnf(n)(x)+R n+1, (A.1) wheretheremainder is R n+1 = 1 (n+1)! Taylors Theorem. (xc)k +e n+1(x;c); where e n+1(x;c) = f(n+1)() (n+1)! The domain of functions of two variables is a subset of R 2, in other words it is a set of pairs. Taylor's Theorem Let f be a function with all derivatives in (a-r,a+r). the existence of derivatives of all orders. Example: 1. TAYLORS THEOREM WITH LAGRANGES Statement: If a function f (x) is defined on [a,b] and n(i) f,f,f .. f-1 are all continuous in [a,b] (ii) fn (x) exists in (a,b), then there exists atleast one real number c in (a,b) such that : f(b) = f (a) + (b-a) f (a) + (b-a)2/ 2! The basic form of Taylor's theorem is: n = 0 (f (n) (c)/n!) Lecture 09 12.9 Taylors Formula, Taylor Series, and Approximations Several Variable Calculus, 1MA017 Xing Shi Cai Autumn 2019 Department of Mathematics, Uppsala University, Sweden Annotations for 1.10 (viii) , 1.10 and Ch.1. It's also useful for determining various infinite sums. Finally, let me show you an example of how Taylor polynomials can be of fundamental importance in physics. .
First we look at some consequences of Taylors theorem. 2. P 1 ( x) = f ( 0) + f ( 0) x. We went on to prove Cauchys theorem and Cauchys integral formula. Right: The graph of the function studied in Example 16.3. useful picture of the behavior of the function. This example Definition 16.2. For example, is a function of n 1 variables. Contents I Introduction 1 1 Some Examples 2 1.1 The Problem . Theorem 3 (Taylors Theorem for Multivariate FunctionsLinear Form) Suppose X Rn is open, x X,andf: X Rm is dierentiable. Title: 2dimtaylorvital.dvi Created Date: 3/26/2007 9:22:23 AM Consider U,the geometry of a molecule, and assume it is a function of only two variables, x and y, let x1 and y1 be the initial coordinates, if terms higher than the quadratic terms are . (and for many other reasons) it is useful to Theorem 15. . big Oh notation allows us to easily state Taylors Theorem for functions taking values inRm. It is important to emphasize that the Taylor series is "about" a point. (xc)n+1; for some = (x;c) between x and c. Review Compare Taylors theorem with Weierstrass theorem. The graph of a function f(x,y) with domain D is a collection of points (x,y,z) in space such that z = f(x,y), (x,y) D. Derivative Mean Value Theorem:if a function f(x) and its 1st derivative are continuous over x i < x < x i+1 then there exists at least one point on the function that has a slope (I.e. 0.2 Functions of two variables Our aim is to generalise these ideas to functions of two variables. Then, for every x in the interval, where R n(x) is the remainder (or error).
Using (1), the Taylor series of f(x) = ln(1 + x) about x= 1 is ln(1 + x) = ln(2) + 1 2 (x 21) 1 8 (x 1) + 1 24 (x 1)3 + Example 14.1.1 Consider f(x, y) = 3x + 4y 5. (iii) Sometimes, all the quadratic terms may vanish, i.e.
Now select the View Taylor Polynomials option from the Tools menu at the top of the applet.
To illustrate Theorem 1 we use it to solve Example 4 in Section 8.7. (x - c)n. When the appropriate substitutions are made. .
The idea can be extended to functions of two variables. 5. (Taylors Theorem) Let f(z) be analytic in a domain D and let z0 2 D. Then f(z) = X1 n=0 an(z z0)n where (a) the representation is valid in the largest open disk centered at z0 on which f(z) is analytic, and (b) an is given by the formulas in Theorem 1. If all derivatives exist (such functions are called C), the we can push this tool to any level we desire. In mathematics, a theorem is a statement that has been proved, or can be proved. Taylors theorem asks that the funciton f be suciently smooth, 2. Di erentiation of Vector-Valued Functions Taylors Theorem Theorem (5.15) Suppose f is a real function on [a;b] n is a positive integer, f (n 1) Di erentiation of Vector-Valued Functions Example On the segment (0;1) de ne f (x) = x and g(x) = x + x2ei=x2 Since jeitj= 1 for all real t we see that lim x!0 . Taylor series of polynomial functions is a polynomial. What is the use of Taylor series? Taylor series is used to evaluate the value of a whole function in each point if the functional values and derivatives are identified at a single point. SpringerInternationalPublishingSwitzerland2016 Taylors Theorem in several variables In Calculus II you learned Taylors Theorem for functions of 1 variable. Here is one way to state it. Theorem 1 (Taylors Theorem, 1 variable) If g is de ned on (a;b) and has continuous derivatives of order up to m and c 2(a;b) then g(c+x) =. X. There is also a feature of the applet that will allow you to demonstrate higher-degree Taylor polynomials for a function of two variables. . Let me begin with a few de nitions. degree 1) polynomial, we reduce to the case where f(a) = f(b) = 0. of on which has in nitely many continuous derivatives. . MA 230 February 22, 2003 Taylors Theorem for f: R!R Assume that f: I!Rwhere Iis some open interval in Rand the n+ 1 derivative f(n+1) exists for all x2I. (Note that this assumption implies that fis Cnon I.) Notation: Given a2I, let P n(x) be the nth degree Taylor polynomial of f at x= a. In other words, P n(x) = Xn k=0 f(k)(a) k! in Taylor's series, into two parts cngn(x), the second of which depends in no way on the function f(x) represented, the constant c alone being altered when f(x) is altered. The equation can be a bit challenging to evaluate. . We will discuss these similarities. . Taylor's series for functions of two variables (b) All linear/polynomial/rational functions are continuous wherever dened. Such a function would be written as z = f(x;y) where x and y are the independent variables and z is the dependent variable. If f (x ) is a function that is n times di erentiable at the point a, then there exists a function h n (x ) such that (xa)n +Rn(x,a) where (n) Rn(x,a) = Z x a (xt)n n! . . In two dimensions , a polynomial px,yof degree n is a function of the form px,y = > i,j=0 i+j=n a ijxiyj. . Complex analysis is a basic tool in many mathematical theories. . Inspection of equations (7.2), (7.3) and (7.4) show that Taylor's theorem can be used to expand a non-linear function (about a point) into a linear series.
The graph of such a function is a surface in three dimensional space. If a function f(x,y) is partially differentiable to arbitrary order in every point (x,y) in the interior M of a given set M in the plane, then we say that the function is smooth in M . The function f{X) is a scalar function of X, and is not a general matrix function: even so, f(X-\-A) is essentially a function of two matrices X and A, and therefore is vastly more complicated than/(X itself) . Power Series Calculator is a free online tool that displays the infinite series of the given function Theorem 1 shows that if there is such a power series it is the Taylor series for f(x) Chain rule for functions of several variables ) Taylors Theorem for Functions of Two Variables: Suppose that f(x,y) and its partial derivatives of all orders less than or equal to n+ 1 are continuous on D = {(x,y) | a x b, c y d} and let (x0,y0) D. For every (x,y) D, there exists between x and x0, and between y and y0 such that f(x,y)=f0 + f x 0 (x x0)+ f y 0 (y y0)+ 1 2! Theorem 3.1 (Taylors theorem). n is the nth order Taylor polynomial for a function f at a point c, then, under suitable conditions, the remainder function R n(h) = f(c+ h) T(c+ h) (5.2.1) is O(hn+1). The Taylor Series represents f(x) on (a-r,a+r) if and only if . Theorem 2. Rolles Theorem is fundamental theorem for all
Maclaurin's theorem is a specific form of Taylor's theorem, or a Taylor's power series expansion, where c = 0 and is a series expansion of a function about zero. That is, the coe cients are uniquely determined by the function f(z). Taylors theorem is used for approximation of k-time differentiable function. 1 Let f(x;y) = 3 + 2x + x2 + 2xy + 3y2 + x3 y4.
. A function f de ned on an interval I is called k times di erentiable on I if the derivatives f0;f00;:::;f(k) exist and are nite on I,
In three variables. Then C(x) = A(x)B(x) if and only if cn = Xn k=0 akbnk for all n 0. 2 Functions of multiple [two] variables In many applications in science and engineering, a function of interest depends on multiple variables. The power series representing an analytic function around a point z 0 is unique. The second degree Taylor polynomial is Let the (n-1) th derivative of i.e. De nitions. This result is a consequence of Taylors theorem, which we now state and prove. Example 1. extend are Cauchys theorem, the Taylor expansion, the open mapping theorem or the maximum theorem. The more derivatives a function has, the more subtle the description we will be able to give. single-variable di erential calculus to the multi-variable case. Check this out for k = 0,1,2 and the two examples above. derivative) ( ) 1 1 f c n f a n b a n n The ideas are applied to approximate a difficult square TAYLORS THEOREM Taylors theorem establish the existence of the corresponding series and the remainder term, under already mentioned conditions. But by representing y as a Taylor series anxn, we can shuffle things around and determine the coefficients of this Taylor series, allowing us to approximate the solution around a desired point. In the proof of the Taylors theorem below, we mimic this strategy. A series of free Engineering Mathematics video lessons. In our example, the third order Taylor polynomial was good enough to approximate the integral to within 10 6. The question of change of variable arises and leads to various results which generalize on the formulae y=f(x For instance, the ideal gas law p = RT states that the pressure p is a function of the examples.
Sol. Search: Multivariable Calculus With Applications. The Fundamental Theorem of Calculus The \fundamental theorem of calculus" - demonstration that the derivative and integral are \inverse operations" The distribution of the sum of two independent random variables is the convolution of the distributions of the individual random variables. 4.
R, a = (0;0) and x = (x;y) then the second degree Taylor polynomial is f(x;y) f(0;0)+fx(0;0)x+fy(0;0)y+ 1 2 fxx(0;0)x2 +2fxy(0;0)xy+fyy(0;0)y2 Here we used the equality of mixed partial derivatives fxy = fyx. (Graph of a Function of Two Variables). . If we take b = x and a = x0 in the previous result, we obtain that . using real variables, the mere existence of a complex derivative has strong implications for the properties of the function. This result is also true when b = , or when f. . variables (f(X1,,Xk)) can be obtained with the following expression. Example: Graph the function, $$f(x,y)=\cos(x)\sin(y)$$. Taylors theorem gives a formula for the coe cients. For each t [ a, b), f. var [f(X1,,Xk)] = E f (X1,,Xk)f X1,,Xk 2 (A.1) For simplicity, a procedure to obtain the variance of a function of two random variables is illustrated, and then the results are extended to k random variables. In higher dimension, there is no such y x 0 These refinements of Taylor's theorem are usually proved using the mean value theorem, whence the name.Also other similar expressions can be found. We present now two proof of Taylors theorem based on integration by parts, one for escalar functions [4], the other for vectorial function, similar in context, but dierent in scope. the multinomial theorem to the expression (1) to get (hr)j = X j j=j j! This is revised lecture notes on Sequence, Series, Functions of Several variables, Rolle's Theorem and Mean Value Theorem, Integral Calculus, Improper Integrals, Beta-gamma function Part of Mathematics-I for B.Tech students Proof. Rm. Examples of functions that are not entire include the square root, the logarithm, the trigonometric function tangent, and its inverse, arctan. The proof of a theorem is a logical argument that uses the inference rules of a deductive system to establish that the theorem is a logical consequence of the axioms and previously proved theorems.. A simple example might be z = 1 1+x2 +y2:
. 1.4 Geometry In view of the one-variable Riemann mapping theorem, every bounded simply connected planar domain is biholomorphically equivalent to the unit disc. These are: (i) Taylors Theorem as given in the text on page 792, where R n(x,a) (given there as an integral) tells how much our approximation might dier from the actual value of cosx; (ii) The variation of this theorem where the remainder term R n(x,a) is given in the form on page 795, labelled It will take a few seconds as the computer calculates the partial derivatives and creates the Taylor polynomials. The third part of the book combines techniques from calculus and linear algebra and contains discussions of some of the most elegant results in calculus including Taylor's theorem in "n" variables, the multivariable mean value theorem, and the implicit function theorem. f(a)+ ( )! These revealed some deep properties of analytic functions, e.g. Taylor's Formula with Remainder Let f(x) be a function such that f(n+1)(x) exists for all x on an open interval containing a. 1.2 The exponential generating function In order to get rid of the factor of k! 7 Taylor and Laurent series 7.1 Introduction We originally dened an analytic function as one where the derivative, dened as a limit of ratios, existed. Theorem 5.13(Taylors Theorem in Two Variables) Suppose and partial derivative up to order continuous on , let . 5.1 Linear approximations of functions In the simplest form of the central limit theorem, Theorem 4.18, we consider a sequence X 1,X 2, of independent and identically distributed (univariate) random variables with nite variance 2. MATH 21200 section 10.9 Convergence of Taylor Series Page 1 Theorem 23 - Taylors Theorem If f and its first n derivatives f,, ,ff ()n are continuous on the closed interval between a and , and b f ()n is differentiable on the open interval between a and b, then there exists a number c between a and b such that () ( 1) 21() () Proof. The proof that for a continuous function (and a large class of simple discontinuous functions) the calculation of area is independent of the choice of partitioning strategy. (a) The sum/product/quotient of two continuous functions is continuous wherever dened. 53 8.1.1. In the mainstream of mathematics, the axioms and the inference rules are commonly left implicit, Taylor polynomials and Leibniz' rule.
SOLUTION We arrange our computation in two columns as follows: Since the derivatives repeat in a cycle of four, we can write the Maclaurin series as follows: With in Theorem 1, we have R n x 1 n! Suppose f: Rn!R is of class Ck+1 on an open convex set S. If a 2Sand a+ h 2S, then f(a+ h) = X j j The following theorem justi es the use of Taylor polynomi-als for function approximation. These are used to compute linear approximations similar to those of functions of a single variable. look at the Hessian matrix. f(n+1)(t)dt. In the case f = f(x), the ane function is of the form A(x) = ax+ b, where aand bare particular constants. f(x,y) is the value of the function at (x,y), and the set Theorem 1 (Taylors Theorem, 1 variable) If g is de ned on (a;b) and has continuous derivatives of order up to m and c 2(a;b) then g(c+x) = X k m 1 fk(c) k! Assume that f is (n + 1)-times di erentiable, and P n is the degree n Taylor approximation of f with center c. Then if x is any other value, there exists some value b between c and x such that f(x) = P n(x) + f(n+1)(b) (n+ 1)! 1for p 2Rthe notation fC1( ) means there exists a nbhd. Denition 1 A function f of the two variables x and y is a rule that assigns a number f(x,y) to each point (x,y) in a portion or all of the xy-plane. And in fact the set of functions with a convergent Taylor series is a meager set in the Frchet space of smooth functions. Lagrange multipliers help with a type of multivariable optimization problem that has no one-variable analogue, optimization with constraints. Taylor's theorem for function of two variable 11 November 2021 14:39 Module 3 Page 1 Module 3 Page 2 Corollary. Motivation: Obtain high-order accuracy of Taylors method without knowledge of derivatives of . For these functions the Taylor series do not converge if x is far from b. 1. . Rbe a function of two variables and let g: R! . To know the maxima and minima of the function of single variable Rolles Theorem is useful. 3. . Express f(x 2. to show in two examples that there are new features in several dimensions. Suppose were working with a function f ( x) that is continuous and has n + 1 continuous derivatives on an interval about x = 0. For n = 0 this just says that f(x) = f(a)+ Z x For example, the function . writing the Taylor polynomial. Exercise: Provide a proof of the second-order version of Taylors Theorem as follows: 1. Note that P 1 matches f at 0 and P 1 However, there is also a main dierence. Estimates for the remainder. polynomials for a function of two variables. Theorem 40 (Taylor's Theorem) . Find the second degree Taylor polynomial around a = (0;0). the variable X. The Taylor polynomial Pk = fk Rk is the polynomial of degree k that best approximate f(x) for x close to a. ( z, t) has a singularity at t = b, with the following conditions. The proof will be given below. xk +R(x) where the remainder R satis es lim x!0 R(x) xm 1 = 0: Here is the several variable generalization of the theorem. Then zoom out to -4 to 4 in the x and y-directions. + 1 n! . Remark: Eulers method is Taylor method of order one. Taylors theorem and the delta method Lehmann 2.5; Ferguson 7 We begin with Taylors theorem, which we do not prove. This is a basic tutorial on how to calculate a Taylor polynomial for a function of two variables. Due to absolute continuity of f (k) on the closed interval between a and x, its derivative f (k+1) exists as an L 1-function, and the result can be proven by a formal calculation using fundamental theorem of calculus and integration by parts.. For example, if we were to approximate Z 2 0 e x2dxto within 10 1 of its true value using Taylor polynomials, we would need to compute Z 2 0 T 11(x)dx. Rbe a function of a single variable. 1 Taylor Expansion: . cps150, fall 2001 Taylors theorem Taylors Theorem For any function f(x) 2 Cn+1[a;b] and any c 2 [a;b], f(x) = Xn k=0 f(k)(c) k! 53 8.2. Theorem 10.1: (Extended Mean Value Theorem) If f and f0 are continuous on [a;b] and f0 is dierentiable on (a;b) then there exists c 2 (a;b) such that f(b) = f(a)+f0(a)(ba)+ f00(c) 2 (ba)2: Proof (*): This result is a particular case of Taylors Theorem whose proof is given below. Now select the View Taylor Polynomials option from the Tools menu at the top of the applet.
(b a) ( ) 1! It is chosen so its derivatives of order k are equal to the derivatives of f at a. In particular, when we say that a function f= f(x) is dierentiable at x0 (an interior point of dom f), we mean that there is a (unique) ane function Athat suitably approximates fnear x0. Taylors Formula G. B. Folland Theres a lot more to be said about Taylors formula than the brief discussion on pp.113{4 of Apostol. The Implicit Function Theorem. Thus the real part of a holomorphic function of two variables not only is harmonic in each coordinate but also satises additional conditions. Lecture Notes for sections 12 notes on matrix calculus can be taken as competently as picked to act Formal de nition used in calculus: marginal cost (MC) function is expressed as the rst derivative of the total cost (TC) function with respect to quantity (q) It also contains solved questions for the better grasp of the subject in an easy to download PDF file View Taylor Series.pdf from CSE MAT1011 at Vellore Institute of Technology. The second part is an introduction to linear algebra. () () ()for some number between a and x. Let us consider a function f of two normally distributed random variables X1 N X1, 2 X1 and X2 N is analytic in D and its derivatives of all orders can be found by differentiating under the sign of integration. Both parts (a) and (b) of Taylors Theorem then imply that |Rn(,x)| 1 (n+1)!M|x| n+1 Example 3 (Sine and Cosine Series) The trigonometric functions sinx and cosx have widely used Taylor expansions about = 0. By the induction assumption, we have f(w 1;z 2;:::;z n) = 1 (2i)n 1 Z Tn 1(a;r) f(w 1;z Topic 6: Two Dimensional Hydrodynamics and Complex Potentials (PDF) [Topic 6.56.7] 1719. If this happens, we have to examine the cubic terms in the Taylor expansion to classify the critical point. These include: tangent lines, which become tangent planes for functions of two vari-ables and tangent spaces for functions of three or more variables. Formula for Taylors Theorem. It turns out that the Theorem 13.1 Taylors theorem (two versions): Version (a): If f(x) has r derivatives at a, then as 0, random variables. The formula is: Where: R n (x) = The remainder / error, f (n+1) = The nth plus one derivative of f (evaluated at z), c = the center of the Taylor polynomial. This approach becomes extremely useful when studying functions of several variables (using the multi-dimensional Taylor Theorem). Later, we will extend the theorem to generating functions with more than one variable. Inverse Function Theorem, then the Implicit Function Theorem as a corollary, and nally the Lagrange Multiplier Criterion as a consequence of the Implicit Function Theorem. (2) follows from repeated integration of (2b) dk+1 dxk+1 Rk(x;a) = fk+1(x); dj dxj Rk(x;a) x=a = 0; j k: A similar formula hold for functions of several variables F: Rn! For example, if G(t) is continuous on the closed interval and differentiable with a non-vanishing derivative on the open interval between a and x, then = (+) ()! I am a high school math teacher in Brooklyn, putting together this curriculum for the first time Question #474281 Multivariable Calculus is one of those important math topics that provide an understanding of algorithms This comprehensive treatment of multivariable calculus focuses on the numerous tools that MATLAB brings to 2there do exist pathological examples for which all Taylor polynomials at a point vanish even though the function
. Topic 9: Definite Integrals Using the Residue Theorem (PDF) 26
For example saddle points can occur as well. Theorem 10.2 Convolution Formula Let A(x), B(x), and C(x) be generating functions. And even if the Taylor series of a function f does converge, its limit need not in general be equal to the value of the function f(x). 7.4.1 Order of a zero Theorem. The three-dimensional coordinate system we have already used is a convenient way to visualize such functions: above each point (x, y) in the x - y plane we graph the point (x, y, z) , where of course z = f(x, y). Taylors theorem is used for the expansion of the infinite series such as etc. The proof of the mean-value theorem comes in two parts: rst, by subtracting a linear (i.e. We can approximate f near 0 by a polynomial P n ( x) of degree n : which matches f at 0 . hn+1f(n+1)(), (A.2) and is a point between xandx+h. .
If f;g are both continuous, then so is their composition g f: R2! . It is often useful in practice to be able to estimate the remainder term appearing in the Taylor However, as we get farther away from 0 (for us from 1 h @ : Substituting this into (2) and the remainder formulas, we obtain the following: Theorem 2 (Taylors Theorem in Several Variables). EXAMPLE 1 Find the Maclaurin series for and prove that it represents for all . The polynomials, exponential function e x, and the trigonometric functions sine and cosine, are examples of entire functions. (x c)n+1 Observe that To classify each critical point we must look at the quadratic terms, i.e. Taylors Theorem Suppose f is continuous on the closed interval [a;b] and has n+ 1 Expansions of this form, also called Taylor's series, are a convergent power series approximating f (x). 5. y=0 at the point is called point of inflection where the tangent cross the curve is 4. called point of inflection and 6.
+ f(n)(a) n! The Inverse Function Theorem. In this case, the central limit theorem states that n(X n ) d Z, (5.1) where = E X View Taylors Theorem for functions of two variables (Additional Supplementary Materials).pdf from CSE MAT1011 at Vellore Institute of Technology. .
Let n 1 be an integer, and let a 2 R be a point. Topic 7: Taylor and Laurent Series (PDF) 2022. Every derivative of sinx and cosx is one of sinx and cosx. As in the case of Taylor's series the constant c is de-termined by means of a linear differential operator of order n. If further | 2022-08-14T07:21:15 | {
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https://www.physicsforums.com/threads/i-am-asked-to-find-xf-x-2-dx-if-f-x-dx-9.633465/ | # I am asked to find ∫xf(x^2)dx if ∫f(x)dx = 9 ?
1. Sep 4, 2012
### LearninDaMath
1. The problem statement, all variables and given/known data
Find $$\int_{-4}^0 xf(x^2) \, dx$$ if $$\int_0^{16} f(x) \, dt =9$$
$$\int_{-4}^0 xf(x^2) \, dx = ?$$
2. Relevant equations
Are there any? All I know is that $$\int_0^{16} f(x) \, dt = F(16) - F(0) = 9$$, but does that property come into use for this problem?
3. The attempt at a solution
I do not know how to begin this problem.
Last edited: Sep 4, 2012
2. Sep 4, 2012
### LCKurtz
Hint: Try the u substitution $u=x^2$ in$$\int_{-4}^0 xf(x^2) \, dx$$Don't forget to take the limits through the substitution too.
3. Sep 4, 2012
### LearninDaMath
letting $u = x^{2}$ $\frac{du}{2}=xdx$
$$\int_{-4}^0 xf(x^2) \, dx$$ => $$\frac{1}{2} \int_{-4}^0 f(u) \, du$$
$$- \frac{1}{2} \int_0^{16}f(u) \, dx = - \frac{9}{2}$$
Wow, for the problem to work out the way it did is mindblowingly coincidental. For the first integral's upper and lower limits to match the second integral's limits, the relationship between the the u substitute and the upper and lower limits had to be deliberately structured. So if I wanted the final result to be 9 instead of -9/2, the evaluated integral would have to become:
$$\int_0^4 2xf(x^2) \, dx$$ = $$\int_0^{16} f(u) \, du = 9$$
So there is really not evaluating occuring, just getting the first integral to match the second integral. So if i'm not mistaken, if the limit values in the first integral could not be made to perfectly match the second integral then finding a value for this problem would be rendered impossible. And on the same note, if the function in the first integral was such that after substitution yielded a differing function, say ∫(u^2)f(u)du, then the problem would also be rendered impossible.
Thanks for the boost on this problem LCKurtz. Are my conclusions about this problem on track here?
4. Sep 4, 2012
### LCKurtz
Yes, you have it. The problem was cooked up to work just right and give you practice with the u substitution. Remember whenever you express an integral with du you must also use the u limits. | 2017-08-21T09:52:43 | {
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https://math.stackexchange.com/questions/4250573/distributing-5-distinct-objects-into-3-identical-boxes-such-that-a-box-can-be-em | # Distributing 5 distinct objects into 3 identical boxes such that a box can be empty
My approach :
I listed down the following cases :-
Case 1) 5 0 0 --> 5c5 = 1 way
Case 2) 4 1 0 --> 5c4 * 1c1 = 5 ways
Case 3) 3 2 0 --> 5c3 * 2c2 = 10 ways
Case 4) 3 1 1 --> 5c3 * 2c1 * 1c1 = 20 ways
Case 5) 2 2 1 --> 5c2 * 3c2 * 1c1= 30 ways
by adding all this I get 66 ways in which all possible combinations of selecting distinct objects is taken care of , and I feel that I don't need to arrange it further in boxes as all the boxes are identical
According to the solution provided for case 4 (i.e. 3,1,1) they have counted 10 ways and for case 5(i.e. 2,2,1) they have counted 15 ways , which makes their total to 41
can someone let me know where exactly am I going wrong with my approach ?
In Case 4) for example you have not considered that the 3 boxes are regarded as identical so once you have chosen the first 3 items that go in one box there is no other choice to be made . The other 2 boxes contain 1 item each and it is regarded as the same choice whichever way round you choose to place the 2 remaining items.
• In case 4 , when I did 5c3 * 2c1 * 1c1 = 20 ways , these 20 ways are all the combinations of 5 distinct objects listed , basically denoting the order in which each object would be filled in a box (example if the 5 objects are A,B,C,D,E one of the 20 combinations would be like ABCDE , or say BCDAE , which I guess is just denoting the order of filling up the boxes , like in arrangement BCDAE it states that B is filled before C , C before D .... so on) I did not take the box into account till here ? so how is it violating the condition of not taking boxes identical ? Sep 14 at 23:49
• @Fin27 You are counting, e.g. ABC | D | E as different than ABC | E | D whereas they should be counted as the same arrangement if the boxes are identical.
– Ned
Sep 14 at 23:57
• @Fin27 In the cases that have two identical sizes in a pair of boxes (i.e. $3,1,1$ and $2,2,1$) you must divide your answer by $2$ because because by doing multiplication principle box by box, you have implicitly ordered the two boxes that hold the same number of items -- as explained by many others in the answers here -- and so you overcount double as in the example I mentioned. Such over counting only occurs by $2$ with $5$ balls, $3$ boxes, but if you were putting $6$ balls into $3$ boxes, the case $2,2,2$ would overcoat by a factor of $3!=6$ if you just did $(C(6,2)C(4,2)C(2,2)$
– Ned
Sep 15 at 11:16
• @Fin because your original method, which is what the fix-up is about, does NOT count e.g. ABC | D | E as different than D |ABC | E, because you do NOT add in a term $C(5,1)C(4,3)C(1,1)$ etc. You already took care of that by listing the types in non-increasing order (i.e. 3,1,1 not 1,3,1).
– Ned
Sep 17 at 11:24
• @Fin27 yes, exactly
– Ned
Sep 19 at 0:49
Notice that the ones you got wrong are the ones where there are two boxes which contain the same non-zero number of balls. This happens because $${2 \choose 1} \times {1 \choose 1}$$ actually treats the boxes as if they were different.
To easily see this, consider 2 different balls and 2 identical boxes, and we want to put 1 ball in each box. Then the answer is 1, but with your method it is 2.
Here's a different approach. There are $$3^5 = 243$$ arrangements if the boxes are numbered. $$3$$ of these are "all balls in one box" arrangements, which corresponds to only $$1$$ "identical boxes" arrangement.
Every other identical box arrangement corresponds to $$3!=6$$ labeled box arrangements, since the contents of all three boxes are different in each of these "not-all-in-one-box" arrangements.
Thus the total number of identical box arrangements is $$1 + (243-3)/6 = 41$$
Distribution of distinguishable objects into indistinguishable subset call for Stirling Numbers of the Second Kind.
\begin{align} \left\{{5}\atop{1}\right\}+\left\{{5}\atop{2}\right\}+\left\{{5}\atop{3}\right\}&=1+15+25\\ &=41 \end{align}
From left to right: one non empty box, two non empty boxes, and three non empty boxes | 2021-09-28T17:08:51 | {
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https://math.stackexchange.com/questions/282950/solve-the-inequality-on-the-number-line | # Solve the inequality on the number line?
How would I solve the following inequality.
$x^2+10x \gt-24$
How would I solve it and put it in a number line?
Hint: solve for
$x^2+10x+ 24 = 0$, find the roots, and determine when the factors are positive:
Solve: $$x^2 + 10 x + 24 = (x+4)(x+6) > 0$$
When is $(x+4)(x + 6)$ positive?:
$\quad$When both factors are positive, or when both factors are negative.
$$(x+4)(x+6) > 0 \implies \begin{cases}(x+4) > 0, & (x+6) > 0 \longrightarrow x>-4\\ \\(x+4) < 0, & (x+6) < 0 \longrightarrow x<-6 \end{cases}$$
Your task is to plot the intervals on which $x$ satisfies the inequality.
Edit: if you want to confirm the solution "graph", compare to:
• Yes it makes sense thanks for the help. – Fernando Martinez Jan 20 '13 at 20:08
$$x^2 + 10x > -24 \implies x^2 + 10x + 24 > 0 \implies (x+4)(x+6) > 0$$ Recall that if $ab > 0$, then either $a>0 \,\, \& \,\, b > 0$ or $a < 0 \,\, \& \,\, b < 0$. Hence, $$(x+4)(x+6) > 0 \implies \begin{cases}(x+4) > 0; & (x+6) > 0 \implies x>-4\\ (x+4) < 0; & (x+6) < 0 \implies x<-6 \end{cases}$$ Hence, we get that $$x \in (-\infty,-6) \cup (-4, \infty)$$
Solve the quadratic $x^2+10x+24=0$ and see what happens between the two roots.
$x^2+10x>−24$ iff $x^2-10x+25 > 1$ iff $(x-5)^2 > 1$. (Note: "iff" means "if and only if".)
Because of the magical property of $1$ being its own square root (funny how this happens, eh?) this is true iff $|x-5| > 1$ which is true iff $x<4$ or $x > 6$.
By completing the square like this, you are implicitly solving the equation. If you start with the inequality $x^2-2bx > c$ ($b=5$ and $c=-24$ in your case) this becomes $(x-b)^2 > c + b^2$. If $c + b^2 < 0$, all $x$ satisfy this; otherwise this can be rewritten as $|x-b| > \sqrt{c+b^2}$ for which the solutions are $x < b-\sqrt{c+b^2}$ and $x > b+\sqrt{c+b^2}$
If you start with the inequality $x^2-2bx < c$ this becomes $(x-b)^2 < c + b^2$. If $c + b^2 < 0$, no $x$ satisfies this; otherwise this can be rewritten as $|x-b| < \sqrt{c+b^2}$ for which the solutions are $b-\sqrt{c+b^2} < x < b+\sqrt{c+b^2}$ | 2019-07-19T08:37:58 | {
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https://math.stackexchange.com/questions/2004583/total-distance-traveled-when-visiting-all-rational-numbers/2004626 | # Total distance traveled when visiting all rational numbers
My students found an old problem given in my school in 2007 (probably from a Honor Calculus class) and had been trying to solve for some time. Here is the problem:
Prove or disprove: there exists a bijection $a$ from $\mathbb N$ onto $\mathbb Q$ such that $\sum_{n=1}^\infty (a_n-a_{n+1})^2$ is convergent. (With $a_n=a(n)$)
I have to confess that I am clueless on the method to deal with this problem. My intuition is that the sum represents the square of the distance traveled when visiting all rational numbers, but there is so many rationals that the sum should be infinite. On the other hand, the density of Q means that I can travel the distance between consecutive rationals of my travel can be arbitrarily small, so I am confused...
The only thing we could prove is that $\sum(a_n-a_{n+1})$ is divergent (otherwise, $\{a_n\}$ would be bounded, a contradiction).
Any clue is welcome.
• You can do this. Let $a$ be any bijection from $\mathbb{N}$ onto $\mathbb{Q}$. Make a new sequence: Insert rationals between $a_1$ and $a_2$ so that the sum of squared distances is at most $1/2$ between $a_1$ and $a_2$. Do the same between $a_2$ and $a_3$, so that the sum is at most $1/4$. You will be inserting things from later on in the $a_n$, but this is fine; you simply mark those off your list of places to visit. Also, you can always find plenty of rationals between your current point and your next destination, because you will only ever have used up finitely many rationals so far. – Eric Tressler Nov 8 '16 at 4:45
• I find @EricTressler's comment to be the clearest answer. – Carsten S Nov 8 '16 at 18:53
• "My intuition is that the sum represents the square of the distance traveled when visiting all rational numbers," Luckily no, it's way smaller. $\sum d_i^2 \le (\sum |d_i|)^2$ – leonbloy Nov 8 '16 at 22:16
• I was going to say this is disturbing, but it's actually not; squaring makes the summands much smaller. The smaller the distance, the smaller its square. – Matt Samuel Nov 8 '16 at 23:36
My intuition says yes (there exists). Choose $a$ so that $a_n-a_{n+1}\leq \frac{1}{n}$; since $\sum_n\frac{1}{n^2}$ converges, if you can indeed choose one such $a$ you should be fine. Now, because $\sum_n\frac{1}{n}$ diverges, you could in principle take steps that are roughly that length and still get to 'cover all of $\mathbb{Q}$'. I might try and make this more precise.
EDIT: Making it precise.
Choose any bijection $f:\mathbb{N}\longrightarrow \mathbb{Q}$. We will obtain $a$ from $f$ inductively as follows.
First, let $a(1)=f(1)$. Now, if $|f(2)-a(1)|\leq \frac12$, we set $a(2)=f(2)$. Otherwise, we set $a(2)=a(1)\pm\frac12$, whichever sign brings $a(2)$ closer to $f(2)$. Then, if |$f(2)-a(2)|\leq \frac13$, we set $a(3)=f(2)$; otherwise we set $a(3)=a(2)\pm \frac13$, whichever signs brings $a(3)$ closer to $f(2)$. We proceed in this manner until we each $f(2)$, then we move onto $f(3)$, and so on.
Observations:
• Because $\sum_n\frac{1}{n}$ diverges, we will always reach the next $f(n)$ in finite time.
• Of course, since $f$ is a bijection, our intermediate $a(k)$'s may step onto some of the $f(n)$'s prematurely. If, after reaching some $f(n)$, $f(n+1)$ has already been previously assigned to some $a(k)$, we simply skip $f(n+1)$ and proceed to $f(n+2)$. Notice that, at any point in the induction, $a$ will be defined at a finite number of points so there may be at most a finite number of skips, and the process can always continue.
• Finally, if when defining $a(k+1) = a(k) \pm \tfrac1k$ the RHS is already taken, simply choose another rational $r$ near $a(k) \pm \tfrac1k$, say $\left|r-\left(a(k) \pm \tfrac1k\right)\right|\leq\tfrac{1}{2k}$, and put $a(k+1)=r$.
These observations guarantee that this inductive process can always continue and will eventually reach every rational. It uses the axiom of choice near the end, and I believe there's a way not to use it: look for
\begin{align} &a(k) \pm \tfrac1k\\ &a(k) \pm \tfrac1k \mp\tfrac{1}{2k}\\ &a(k) \pm \tfrac1k \mp\tfrac{1}{2k}\pm\tfrac{1}{4k}\\ &a(k) \pm \tfrac1k \mp\tfrac{1}{2k}\pm\tfrac{1}{4k}\mp\tfrac{1}{8k}\\ &\dots \end{align}
until you find one that's not yet been used. But honestly I'm not too worried about it.
• If moving a full $\frac{1}{k}$ doesn't work because $a(k)\pm\frac{1}{k}$ has already been visited, you could move $(\frac{1}{k}-\frac{1}{2^k})$ instead. Or if that's already taken, you could move $(\frac{1}{k}-\frac{1}{2^{k+1}})$. Or $(\frac{1}{k}-\frac{1}{2^{k+2}})$, etc. Since at every stage you've only exhausted finitely many values, you can always find one of these moves that will work. (Cont'd) – mathmandan Nov 8 '16 at 5:54
• (Cont'd) If you follow this procedure, you're always guaranteed to move at least $(\frac{1}{k}-\frac{1}{2^k})$. Since $\sum(\frac{1}{k}-\frac{1}{2^k})\geq \sum(\frac{1}{2k})$ diverges, you always need only finitely many steps to move from any rational to any other rational. And since you always move at most $\frac{1}{k}$, the sum of the squared steps converges. – mathmandan Nov 8 '16 at 5:55
• @Fimpellizieri Nice proof. You don't, in fact, need the axiom of choice near the end. You can do it the way you suggested, or you can just set $a(k+1)$ equal to the first rational in a standard enumeration of $\mathbb{Q}$ that is between $a(k)\pm \frac1{2k}$ and $a(k)\pm \frac1{k}$ and that isn't $a(j)$ for any $j \le k.$ – Mitchell Spector Nov 8 '16 at 17:06
Here's an idea (it'll take some work to flesh it out into a proof, but you asked for a clue, so...):
At some point I need to get from, say, $0$ to $1$ - I have to hit one of them, and then the other at some later time. Now, this looks like distance $1$, which is fine - IF I don't have to do steps like it very often. But I also have to do $1$ to $2$, $2$ to $3$, and so on; so I need to make it smaller. But $0, \frac{1}{2}, 1$ only has cumulative distance-squared equal to $\frac{1}{2}$. $0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, 1$ has cumulative distance-squared of only $\frac{1}{4}$.
The cool thing is that I can do this with any interval I like. So imagine taking any enumeration of $\mathbb{Q}$ you like and inserting elements in between to "shrink" the steps involved so that each successive step of the original enumeration has smaller and smaller cumulative distance-squared in the new one - say, shrinking like $2^{-n}$.
This isn't complete - I've been very vague about lots of stuff, and I haven't said how to deal with the fact that the enumeration you end up with (at least the one I'm visualizing) visits most of the rationals many times; but you might be able to fill in the gaps.
I will expand my comment into an answer (though still not a formal proof).
Lemma: Given rationals $a < b$, $\epsilon > 0$, and any finite set $S$, there exist rationals $k_1,\ldots,k_n \not\in S$ such that $$(k_1 - a)^2 + \sum_i (k_{i+1} - k_i)^2 + (b - k_n)^2 < \varepsilon.$$
Sketch of proof: Note that by inserting a single rational $c$ in the interval $\left[\frac{3a+b}{4},\frac{a+3b}{4}\right]$, we have $$(b-a)^2 \leq \frac{5}{8}\left((c-a)^2 + (b-c)^2\right).$$ We can always do this, because that interval contains infinitely many rationals, and so in particular it contains a rational that is not in $S$. Keep doing this until the sum of squared distances is less than $\varepsilon$.
Now let $(a_n)$ be any enumeration of the rationals. Form a new enumeration of the rationals as follows: Let $b_{n_1} = b_1 = a_1$. Insert as many rationals $b_2, \ldots, b_{n_2-1}$ as necessary so that when finally $b_{n_2} = a_2$, we have $$\sum_{i=1}^{n_2-1} (b_{i+1} - b_i)^2 < 1/2.$$
Let $b_{n_3}$ be the next unvisited member of $(a_n)$. Continue like this, inserting rationals $b_{n_2+1},\ldots,b_{n_3-1}$ so that $$\sum_{i=n_2}^{n_3-1} (b_{i+1} - b_i)^2 < 1/4.$$
At each step, the forbidden set $S$ is simply the sequence $(b_n)$ that has been defined so far. We take finitely many steps to reach $a_1$, and then $a_2$, and so on, so $(b_n)$ is also an enumeration of the rationals. Moreover, $\sum_i (b_{i+1} - b_i)^2 < 1$.
My intuition says no.
Let $a_{n_1} := 0, a_{n_q} := q \in \mathbb{N}$
Then $q^2 = (a_{n_1} - a_{n_q})^2 \leq \sum_{n = 1}^{\max(n_1,n_q)} (a_n - a_{n+1})^2$ By the triangle inequality.
We know that $n_q \to \infty$ as $q \to \infty$ (because $a$ is a bijection, and the rationals are unbounded)
Hence $\sum_{n = 1}^{\infty} (a_n - a_{n+1})^2$ diverges.
• $a_0=0,a_1=1,a_2=2$, and yet $(a_0-a_2)^2 > (a_0-a_1)^2+(a_1-a_2)^2$ – Fimpellizieri Nov 8 '16 at 4:36
• Which triangle inequality? Note that $|\cdot|^2$ isn't a norm. – Vim Nov 8 '16 at 4:37
Think of it this way. What you need to make your sum convergent is to make $(a_n - a_{n+1})$ get infinitely close to, but not equal to 0 (otherwise it would not be a bijection). So we should try to eliminate as many options as we can:
Polynomials: Obviously, even-degreed ones can't work. Odd-degreed ones are our only option.
Non-polynomials Categories:
1. sin, cos, etc: disproved by bijection
2. polynomials divided by cases: we'll try this later
3. arrays: we'll try this later
So, for right now, I focused on odd polynomials. Grouping them into ones with positive degrees and negative degrees, I ruled out ones with positive degrees based on their common property of divergence.
Negative degrees look pretty interesting - so let's work with these - starting with $n^{-1}$, or more colloquially, $\frac{1}{x}$.
Theoretically, the limit as n approaches infinity of $(\frac{1}{n} - \frac{1}{n+1})$ is 0, but will never reach 0.
And bam, $\frac{1}{n}$ is your bijection. As a proof, let's replace infinity with rational numbers and test.
1. 0 - 0
2. 10 - 0.28961810696118045
3. 100 - 0.28986781017274627
4. 1000 - 0.28986813336411776
5. 10000 - 0.2898681336961168
We can additionally prove this is convergent by saying that if x is convergent, $x^2$ will be convergent. Therefore, since $(\frac{1}{n} - \frac{1}{n+1})$, then $(\frac{1}{n} - \frac{1}{n+1})^2$ is convergent, and its integral from 0 to infinity will also be convergent.
P.S., no clue why it converges at this exact point or if it has a deep mathematical meaning, and I just saw this and wanted to take a crack at it. Hopefully I didn't screw up and helped you.
• I think you might have confused bijection with injection. Also, the fact that a series converges to 0 doesn't mean the sum of that series converges. $\frac{1}{n}$ is actually one of the simplest counter examples to that. – dimpol Nov 8 '16 at 14:05
• @dimpol You are right about the bijection, but $(\frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+\dots$ is just 1, since every other term cancles. – Stig Hemmer Nov 9 '16 at 9:51
• Ah, you indeed are correct – dimpol Nov 9 '16 at 12:12 | 2019-08-23T02:12:22 | {
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https://math.stackexchange.com/questions/1106730/log-2-13-is-irrational | # $\log_2 13$ is irrational
Is it true that $\log_2 13$ is irrational?
Let $x=\log_2 13\implies 2^x=13$.
So, it will be an irrational number, if not,$$x=\frac p q$$
and $$2^{\frac p q}=13$$
$$\implies 2^p=13^{q}$$
Since, $13$ is a prime number, $2^p$ divides $13^q$.
So, $2$ divides $13$, which is absurd.
Is this reason worthy? Can you give some other proofs for this?
• That's a nice and elegant proof. You can extend it to $$\log_a b \notin \mathbb Q \qquad \forall a,b\in\mathbb N\text{ s.t. } \gcd(a,b) = 1$$ Jan 16 '15 at 14:22
• Thank you.. Is there any other proofs??? Jan 16 '15 at 14:25
• I don't immediately see how else to tackle the problem, since $2^x = 13$ is an exponential function, so solving it would resort to $\log$ of some basis, but since we want to prove a property of $\log$, that's not going to help us much. Maybe you can reduce it to some other $\log_a b \notin\mathbb Q$ statement this way. Jan 16 '15 at 14:29
• You could use Gelfond-Schneider theorem, but it's not really necessary to resort to such heavy artillery. Apr 13 '15 at 8:19
You are done in your solution at the step where you concluded that $2^p = 13^q$. You only need to quote Fundamental Theorem of Arithmetic after that step. | 2021-09-29T03:18:07 | {
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https://math.stackexchange.com/questions/1743725/convergence-of-sum-frac1-sqrt4n-1-frac1-sqrt4n-3 | # Convergence of $\sum \frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4n + 3}}$
I need to prove $\sum \frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4n + 3}}$ converges. The root test is inconclusive, so I check in W.A., $\sum \frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4n + 3}}$ converges by comparison test, but I don't know what series to compare. I've tried to compare with $\sum \frac{1}{4n+1} - \frac{1}{4n + 3}$ because this serie converges to $\frac{\pi}{4}$, but we have exactly the opposite inaquality $\frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4n + 3}} > \frac{1}{4n + 1} - \frac{1}{4n + 3}$.
Can you help me?
Hint $$\frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4n + 3}} = \frac{\sqrt{4n + 3}-\sqrt{4n + 1}}{\sqrt{4n + 3}\sqrt{4n + 1}}=\frac{2}{\sqrt{4n + 3}\sqrt{4n + 1}(\sqrt{4n + 3}+\sqrt{4n + 1})}.$$
Another possible solution: $$0 < a_n := \frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4n + 3}} = \frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4(n+1) - 1}} \\ < \frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4(n+1) + 1}} =: b_n$$ and $\sum b_n$ is convergent as a telescoping series. It follows that $\sum a_n$ is convergent and $$\sum_{n=0}^\infty a_n < \sum_{n=0}^\infty b_n = 1 \quad .$$
Or: The terms are positive, and the partial sums $$\sum_{n=0}^N \frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4n + 3}} = 1 + \sum_{n=1}^N \left(-\frac{1}{\sqrt{4n - 1}} + \frac{1}{\sqrt{4n + 1}} \right) - \frac{1}{\sqrt{4N + 3}} < 1$$ are bounded. (This is essentially what achille hui suggested in the above comment.)
As $n \to +\infty$: \begin{align} (4n+1)^{-1/2} &= (4n)^{-1/2}\left(1+\frac{1}{4n}\right)^{-1/2} =\frac{1}{2\sqrt{n}}\left(1-\frac{1}{8n}+O(n^{-2})\right) \\ (4n+3)^{-1/2} &= (4n)^{-1/2}\left(1+\frac{3}{4n}\right)^{-1/2} =\frac{1}{2\sqrt{n}}\left(1-\frac{3}{8n}+O(n^{-2})\right) \\ (4n+1)^{-1/2} - (4n+3)^{-1/2}&= \frac{1}{2\sqrt{n}}\left(0+\frac{2}{8n}+O(n^{-2})\right) =\frac{1}{8n^{3/2}}+O(n^{-5/2}) \end{align} The series converges by comparison with $$\sum\frac{1}{8n^{3/2}}$$ | 2020-01-28T19:49:55 | {
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https://math.stackexchange.com/questions/4332655/if-abacbc-8abc-prove-a-b-c/4332662 | # If $(a+b)(a+c)(b+c)=8abc$ prove $a=b=c$
For positive numbers $$a,b,c$$ we have $$(a+b)(a+c)(b+c)=8abc$$. Prove $$a=b=c$$
I tried expanding the expression. after simplifying we have,
$$a^2b+ab^2+b^2c+ca^2+ac^2+bc^2=6abc$$ But not sure how to continue.
I also noticed that we have,
$$(a+b)(a+c)(b+c)=(2a)(2b)(2c)$$ $$(a+b)+(a+c)+(b+c)=(2a)+(2b)+(2c)$$ But I don't know if it helps.
• Have you tried AM-GM inequality...? Dec 14, 2021 at 14:12
• Dec 14, 2021 at 14:27
By AM-GM, $$a+b \geqslant 2\sqrt{ab}$$ with equality if and only if $$a=b$$. Multiplying together the three similar inequalities we get $$(a+b)(b+c)(c+a) \geqslant 8abc$$ with equality if and only if $$a=b=c$$.
Continuing from where you left, you have: $$b(a-c)^2+a(b-c)^2+ b^2c+ca^2-2abc=0$$,which is same as $$b(a-c)^2+a(b-c)^2+c(b-a)^2=0$$ So you now have sum of three non-negative numbers equal to $$0$$. Can you take it from here? | 2022-08-18T22:43:05 | {
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https://math.stackexchange.com/questions/1640120/in-how-many-ways-can-a-committee-of-6-people-be-selected-from-7-men-and-6 | # In how many ways can a committee of $6$ people be selected from $7$ men and $6$ women if it can contain at most one of persons A and B?
A committee of $6$ people will be formed with $7$ men and $6$ women. The oldest of the $7$ men is A and the oldest of the $6$ women is B. It is described that the committee can include at most one of A and B. In how many ways can the committee be chosen?
My attempt:
$13C6- (6C1 \cdot 7C5+6C5 \cdot 7C1) = 1548$
Correct answer is $1386$.
• Welcome to Math SE. Could you share what you have tried so far? – A. A. Feb 4 '16 at 10:51
• Do you know how many ways the commitee can be chosen without this constraint? – Arthur Feb 4 '16 at 11:08
• I am voting to close since the OP has visited the site since Adolfo posed his query without responding to it. – N. F. Taussig Feb 4 '16 at 11:10
• No. of ways=No. of ways with only A included + No. of ways with only B included + No. of ways with neither included. – GoodDeeds Feb 4 '16 at 11:14
• Thanks lol hey don't close it I post this before going for dinner =.= wtf, sorry was off to dinner closed the windows righta away,In fact I'm still eating my dinner now =3= 13C6- (6C1*7C5+6C5*7C1)=1548 Correct answer is 1386 – DreadfulWithMaths Feb 4 '16 at 11:22
From all 13C6 solutions, you must subtract the solutions where both A and B are chosen.
If you choose both A and B, there are 11C4 possibilities for the other 4 in the committee.
So 13C6 - 11C4 = 1386.
You calculated the committees where it was not allowed to have exactly one of A and B.
• @@ thanks a lot, i feel so dumby for not being able to think of this – DreadfulWithMaths Feb 4 '16 at 11:34
Pieter21 has provided you with a valid and efficient solution.
Your attempt was incorrect since you subtracted the sum of the number of ways of selecting five men and one woman and the number of ways of selecting five women and one man. However, it is possible to select a committee of six people containing at most one of A and B that contains more than one person of each sex.
Here is an alternate solution:
A committee that contains at most one of A and B either contains neither A nor B or it contains exactly one of them.
If the committee of six people contains neither A nor B, we must select six of the other eleven available people, which can be done in $$\binom{11}{6}$$ ways.
If the committee contains exactly one of A and B, we must select one of A and B, which can be done in $\binom{2}{1}$ ways, and select five of the other eleven people, which can be done in $\binom{11}{5}$ ways. Hence, the number of committees of six people that contain exactly one of A and B is $$\binom{2}{1}\binom{11}{5}$$
Therefore, the number of committees that can be formed with at most one of A and B is $$\binom{11}{6} + \binom{2}{1}\binom{11}{5}$$
• Thanks! :) Helpful in helping me understand! – DreadfulWithMaths Feb 4 '16 at 15:01 | 2019-07-23T00:53:10 | {
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https://math.stackexchange.com/questions/2011000/is-the-exponential-of-the-euclidean-norm-convex | # Is the exponential of the euclidean norm convex?
I know that the euclidean norm is a convex function, and that exponential functions are also convex. I want to know whether or not a mix of the two would also be a convex function. i.e.
$$f(x) = e^{\sqrt{\sum_{i=1}^{n}{x_i^2}}}$$ where $$x = (x_1,x_2,...,x_n)$$
So my approach so far has been to prove that
$$f(\lambda x + (1-\lambda y)) \leq \lambda f(x) + (1-\lambda) f(y)$$ so $$e^{\lVert \lambda x + (1-\lambda)y\rVert_2}$$
but i'm not sure how to proceed from here to get: $$e^{\lVert \lambda x + (1-\lambda)y\rVert_2} \leq \lambda e^{\lVert x \rVert_2} + (1-\lambda)e^{\lVert y \rVert_2}$$
I've thought about using the triangle inequality but I think that just gets me to: $$e^{\lVert \lambda x + (1-\lambda)y\rVert_2} \leq e^{\lambda \lVert x \rVert_2 + (1-\lambda)\lVert y\rVert_2} = e^{\lambda \lVert x \rVert_2} + e^{(1-\lambda)\lVert y\rVert_2}$$
• If $f$ is convex, and $g$ is convex and non-decreasing, does it follow that $g \circ f$ is convex? – Daniel Fischer Nov 12 '16 at 19:55
• right, but for this particular problem i have to prove it the way that I started, they're not going to accept your way, which is simpler. – April Nov 12 '16 at 19:57
• Well, it's just replacing the abstract functions with concrete ones. Note that in the last line you have an error, $e^{a+b} = e^a \cdot e^b$, not $e^a + e^b$. But with $e^{\lambda \lVert x\rVert_2 + (1-\lambda)\lVert y\rVert_2}$, you are in a situation where using the convexity of the exponential is very very tempting. – Daniel Fischer Nov 12 '16 at 20:01
$$e^{\lVert \lambda x + (1-\lambda)y\rVert_2} \leq e^{\lambda \lVert x \rVert_2 + (1-\lambda)\lVert y\rVert_2} \leq \lambda e^{ \lVert x \rVert_2} + (1-\lambda)e^{\lVert y\rVert_2}$$
Remark: $\exp$ being convex means $\exp(\lambda \alpha +(1-\lambda) \beta) \leq \lambda \exp(\alpha) + (1-\lambda) \exp(\beta)$. $\alpha = \left\| x\right\|, \beta = \left\| y\right\|$ for your question. | 2020-01-24T17:21:37 | {
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https://www.primerpy.com/post/datastructure/content/hash-tables/ | # Hash Tables
## Definition
### What are Hash Tables?
• Hash Tables are key-value pairs
• key must be immutable
#### Hash Function
• Hash functions convert the value to key
• One big challenge is to calculate the key more evenly to avoid collision
• One common way is to calculate the mod of a prime number (the number theory is beyond the scope here)
• but look at two examples below
Example 1:
10 % 4 -> 2
20 % 4 -> 0
30 % 4 -> 2
40 % 4 -> 0
50 % 4 -> 2
In example 1, the results from hash function don’t distribute evenly, they are either 2 or 0
Example 2:
10 % 7 -> 3
20 % 7 -> 6
30 % 7 -> 2
40 % 7 -> 4
50 % 7 -> 1
In example 2, the results of moding a prime number distribute much more evenly so the chance for collision is small
Below is a good reference to select the prime number to mod based on the data size.
### Collision
In example 2, if we keep calculating mod
60 % 7 -> 4 This is a hash collision as 40 % 7 -> 4 too
#### Chaining
• One solution to collision is chaining,
• which basically creates a linked list at the index where collision occurs,
• e.g. 4: 40%7 -> 60%7
• if the hash value is not distributed evenly, it might become a linked list, losing the benefits of hash tables
• however it’s not how Python handles the collision.
• Python uses open addressing solution to collisions
• When an index is taken, Python will find the next available index
• There are several types of addressing
• h is the hash function, i is the key, k is the value
• linear probing: h(k,i) = (h(k)+i) % m, i = 0,1,2…,m-1
• quadratic probing: h(k,i) = (h(k) + c1 + c2*i**2) % m, i = 0,1,2,…,m-1
• rehashing: h(k,i) = (h1(k) + i*h2(k))%m
• In CPython, quadratic probing is used
• Here’s the details in CPython
• “Open addressing is preferred over chaining since the link overhead for chaining would be substantial (100% with typical malloc overhead).”
• Open Addressing Code
def h(key, M=13):
return key % M
nums = [765, 431, 96, 142, 579, 226, 903, 388]
def handle_collision(nums, M=13, inserted_set=set()):
for num in nums:
index = h(num)
first_index = index
i = 1
while index in inserted_set:
print("\th({num}) = {num} % M = {index} collision".format(num=num, index=index))
index = (first_index + i**2) % M
i += 1
else:
print("h({num}) = {num} % M = {index}".format(num=num, index=index))
handle_collision(nums)
h(765) = 765 % M = 11
h(431) = 431 % M = 2
h(96) = 96 % M = 5
h(142) = 142 % M = 12
h(579) = 579 % M = 7
h(226) = 226 % M = 5 collision
h(226) = 226 % M = 6
h(903) = 903 % M = 6 collision
h(903) = 903 % M = 7 collision
h(903) = 903 % M = 10
h(388) = 388 % M = 11 collision
h(388) = 388 % M = 12 collision
h(388) = 388 % M = 2 collision
h(388) = 388 % M = 7 collision
h(388) = 388 % M = 1
• As we keep adding elements into hash tables, the spaces might be taken up soon
• load factor is the ratio of used indices vs. total indices,
• in the open addressing example,
• the size of the hash table is 13,
• there are already 8 indices used,
• so the load factor is 8 / 13 ~= 0.6
• Usually when the load factor is above 0.8, we need to open new spaces for rehashing
### Rehashing
• As discussed above, once the load factor is greater than 0.8, we need to rehash
• How large is the new space depends, in CPython 3.7, it’s currently set to 3, so new space will be used space * 3
• Once the new space is created, all the elements in the old hash will be copied into the new one, which involves a time complexity of O(n)
### Complexity
#### Time Complexity
• Hash function: O(1)
• Memory index access: O(1)
• Search complexity O(1)
## Python Implementation
### Hashcode in Python
In python, we can use hash function to calculate the hash value for any object. Hash values are integers. They are used to quickly compare dictionary keys during a dictionary lookup. Numeric values that compare equal have the same hash value (even if they are of different types, as is the case for 1 and 1.0).
# for an int number
a = 20
print(hash(a))
# for the same number but in float
print(hash(20.0))
# a negative int
b = -20
print(hash(b))
# a string
print(hash("c"))
# a float
print(hash(3.1415926))
# a custom object
class ListNode:
def __init__(self, val):
self.val = val
self.next = None
ln = ListNode(20)
print(hash(ListNode))
print(hash(ln))
20
20
-20
-5423205724351203701
326490306866391043
-9223363252756124725
-9223372036577884363
### HashTable Abstract Data Type
• Here I will “build” a simplified version of Hash Table
• It contains a “Slot” to store indices, it has three status
• used
• removed
• used
• Three different methods
• get(key)
• remove(key)
class Slot:
def __init__(self, key, val):
self.key = key
self.val = val
class HashTable:
UNUSED = None # slot has not been used
REMOVED = Slot(None, None) # slot was used before but got removed
def __init__(self):
self._table = [HashTable.UNUSED] * 8
self.length = 0
@property
return self.length / float(len(self._table))
def __len__(self):
return self.length
def _hash(self, key):
# use a simple mod function
return hash(key) % len(self._table)
def _find_key(self, key):
index = self._hash(key)
_len = len(self._table)
while self._table[index] is not HashTable.UNUSED:
# only find index when it's actually used
if self._table[index] is HashTable.REMOVED:
# one way for cpython to handle hash collision
index = (index * 5 + 1) % _len
# if the key is removed, then keep looping
continue
elif self._table[index].key == key:
return index
else:
index = (index * 5 + 1) % _len
return None
def _slot_can_insert(self, index):
# check if the slot is able to be inserted new element
if self._table[index] in (HashTable.UNUSED, HashTable.REMOVED):
return True
else:
return False
def _find_slot_to_insert(self, key):
# Find a slot to insert value
index = self._hash(key)
_len = len(self._table)
while not self._slot_can_insert(index):
# while we cannot insert
# find the next availabe slot to insert
index = (index*5 + 1) % _len
return index
def __contains__(self, key):
# in operation
index = self._find_key(key)
return index is not None
def _rehash(self):
# create new spaces whenever spaces are about to run out
old_table = self._table
self._table = [HashTable.UNUSED] * (len(self._table) * 3)
self.length = 0
for ele in old_table:
# only transfer the valid elements
if ele is not HashTable.UNUSED and ele is not HashTable.REMOVED:
index = self._find_slot_to_insert(ele.key)
self._table[index] = ele
self.length += 1
def add(self, key, val):
# add new element to HT
if key in self:
# use the __contains__
# if found update
index = self._find_key(key)
self._table[index] = val
return False # only updated not added
else:
# find the availabe slot to insert
index = self._find_slot_to_insert(key)
# add the element
self._table[index] = Slot(key, val)
# update length
self.length += 1
# if load factor is greater than 0.8, rehashing
if self._load_factor >= 0.8:
self._rehash()
return True
def get(self, key):
index = self._find_key(key)
if index is None:
return False
else:
return self._table[index].val
def delete(self, key):
index = self._find_key(key)
if index is None:
raise Exception("Key Error")
value = self._table[index].val
self.length -= 1
self._table[index] = HashTable.UNUSED
return value
def __iter__(self):
# Iterate thru the ht via key
for ele in self._table:
if ele not in (HashTable.UNUSED, HashTable.REMOVED):
yield ele.key
def test():
ht = HashTable()
# test iterate
for ele in ht:
print(ele)
# test delete
print(ht.delete("a"))
print(ht.get("b"))
test()
b
c
a
0
42 | 2019-10-17T12:40:15 | {
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http://math.stackexchange.com/questions/173760/computing-a-sum-of-binomial-coefficients/266065 | # Computing a sum of binomial coefficients
Does anyone know a better expression than the current one for this sum?
$$\sum_{i=0}^m \binom{N-i}{m-i}, \quad 0 \le m \le N.$$
It would help me compute a lot of things and make equations a lot cleaner in the context where it appears (as some asymptotic expression of the coefficients of a polynomial defined over cyclic graphs). Perhaps the context doesn't help much though.
For instance, if $N = m$, the sum is $N+1$, and for $N = m+1$, this sum is $N + (N-1) + \dots + 1 = N(N+1)/2$. But otherwise I don't know how to compute it.
-
I thought it was a triviality so deleted it at first, but I conjecture it is something like $\binom{N+1}{N-m}$. Although I don't think it's quite obvious. (This conjecture happens to be false for $m=2$ and $N=5$. I am not happy.) – Patrick Da Silva Jul 22 '12 at 3:25
It looks like it's just $\dbinom{N+1}{m}$... – Ｊ. Ｍ. Jul 22 '12 at 3:28
@J.M. : I actually counted what I wanted to count on my cyclic graph thing in a better way so that I can actually prove that this sum is $\binom{N+1}m$ indirectly. Should I post my graph-theoretic answer or it is irrelevant to do so?... – Patrick Da Silva Jul 22 '12 at 3:35
There's a proof of a more general identity in page 159 of Concrete Mathematics, which uses the recurrence formula for binomial coefficients $\dbinom{n}{k}=\dbinom{n-1}{k}+\dbinom{n-1}{k-1}$ repeatedly. You can only apply this after you reindex your sum as $$\sum_{k=0}^m\binom{N-m+k}{k}$$ – Ｊ. Ｍ. Jul 22 '12 at 3:38
Hey! Someone posted a very relevant answer with blue balls and red balls. I was about to accept it! It must be put back! – Patrick Da Silva Jul 22 '12 at 4:03
The answer indeed seems to be $\binom{N+1}{m}$. You can get it from the more well-known identity \begin{align} \binom{n}{k} = \sum_{r =k}^n \binom{r-1}{k-1} \end{align} by change of variable. There is a nice combinatorial proof of this later identity which you may even adapt to apply directly to your problem. Let $A_r$ be the set of $k$-subsets of $[n] = \{1,\dots,n\}$ whose maximal element is $r$. Let $S$ be the set of all $k$ subsets of $[n]$. Then, $S = \cup_{r = k}^n A_r$ and this union is disjoint. We have $|A_r| = \binom{r-1}{k-1}$ and $|S| = \binom{n}{k}$. You can fill in the details yourself.
EDIT: change of variable: $n \to n+1$, $r \to n+1-i$ and $k \to n+1 -m$.
-
First note that $\dbinom{n-i}{m-i} = \dbinom{n-i}{n-m}$. Hence, you have $$\dbinom{n-m}{n-m}+\dbinom{n-m+1}{n-m}+\cdots+\dbinom{n-1}{n-m}+\dbinom{n}{n-m}$$ Let $n-m = k$. Hence, we want to find an expression for $$\dbinom{k}k+\dbinom{k+1}{k}+\cdots+\dbinom{n-1}{k}+\dbinom{n}{k}$$ From the Pascal's triangle, we have
$\hskip2.5in$ \begin{align} \color{red}{\dbinom{n+1}{k+1}} & = \color{blue}{\dbinom{n}{k}} + \color{red}{\dbinom{n}{k+1}}\\ & = \color{blue}{\dbinom{n}{k}} + \color{blue}{\dbinom{n-1}{k}} + \color{red}{\dbinom{n-1}{k+1}}\\ & = \color{blue}{\dbinom{n}{k}} + \color{blue}{\dbinom{n-1}{k}} + \color{blue}{\dbinom{n-2}{k}} + \color{red}{\dbinom{n-2}{k+1}}\\ & = \color{blue}{\dbinom{n}{k}} + \color{blue}{\dbinom{n-1}{k}} + \color{blue}{\dbinom{n-2}{k}} + \cdots \color{blue}{\dbinom{k+2}{k}} + \color{red}{\dbinom{k+2}{k+1}}\\ & = \color{blue}{\dbinom{n}{k}} + \color{blue}{\dbinom{n-1}{k}} + \color{blue}{\dbinom{n-2}{k}} + \cdots \color{blue}{\dbinom{k+2}{k}} + \color{blue}{\dbinom{k+1}{k}} + \color{red}{\dbinom{k+1}{k+1}}\\ & = \color{blue}{\dbinom{n}{k}} + \color{blue}{\dbinom{n-1}{k}} + \color{blue}{\dbinom{n-2}{k}} + \cdots \color{blue}{\dbinom{k+2}{k}} + \color{blue}{\dbinom{k+1}{k}} + \color{blue}{\dbinom{k}{k}} \end{align}
Hence, your result is $\dbinom{n+1}{n-m+1} = \dbinom{n+1}m$
-
Here's my version of the proof (and where it arose in the context). You have a cyclic graph of size $n$, which is essentially $n$ vertices and a loop passing through them, so that you can think of them as the roots of unity on a circle (I don't want to draw because it would take a lot of time, but it really helps).
You want to count the number of subsets of size $k$ ($0 \le k \le n$) of this graph which contain 3 consecutive vertices (for some random reason that actually makes sense to me, but the rest is irrelevant).
You can fix 3 consecutive vertices, assume that the two adjacent vertices to those 3 are not in the subset, and then place the $k-3$ remaining anywhere else. Or you can fix $4$ consecutive vertices, assume again that the two adjacent vertices to those $4$ are not in the subset, and then place the $k-4$ remaining anywhere else. Or you can fix $5$, blablabla, etc. If you assume $n$ is prime, every rotation of the positions you've just counted gives you a distinct subset of the graph, so that the number of such subsets would be $$n\sum_{i=0}^{k-3} \binom{n-5-i}{k-3-i}.$$ On the other hand, if you look at it in another way : suppose there is $\ell$ consecutive vertices in your subset. Look at the $3$ vertices in the $\ell$ consecutive ones that are adjacent to a vertex not in your subset, and place the remaining ones everywhere else. Again, the number of such subsets is $$n\binom{n-4}{k-3}.$$ If $n$ is not prime, my computations are not accurate (they don't count well what I want to count in my context, but that's another question I'm trying to answer for myself), but I do precisely the same mistakes in both computations! So the two are actually equal. Removing the $n$'s on both lines, replacing $n-5$ and $k-3$ by $N$ and $m$, we get the identity.
-
Using methods described in Concrete Mathematics (Graham, Knuth & Patashnik), the following did the trick for me.
\begin{align*} \sum_{k=0}^n \binom{N-k}{n-k} &= \sum_{k=0}^n (-1)^{n-k} \binom{n-k-(N-k)-1}{n-k} & \text{inversion (5.14)}\\ &= \sum_{k=0}^n (-1)^k \binom{n-N-1}{k} & \text{reverse order} \\ \end{align*} Letting $r = -(N+1)$ and using the results on page 167 yields $$\sum_{k=0}^n (-1)^k \binom{n+r}{k} = \binom{-r}{m} = \binom{N+1}{m}$$ These calculations may seem like magic, but then one could argue this book makes you a magician.
- | 2015-02-01T15:24:23 | {
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https://www.physicsforums.com/threads/please-help-me.67996/ | 1. Mar 20, 2005
### maphysique
Could anyone teach how to solve the following differential equation?
$$\frac{d^2 y}{dx^2}+2\left(\frac{dy}{dx}\right)^2+y=0$$
Last edited: Mar 20, 2005
2. Mar 20, 2005
### cronxeh
Well this is a 2nd order nonlinear homogeneous differential equation. Its tougher because the y' is of 2nd power.
Last edited: Mar 20, 2005
3. Mar 20, 2005
### Data
Make the substitution $$v = \frac{dy}{dx}$$ so that $$v \frac{dv}{dy} = \frac{d^2y}{dx^2}$$. This results in a first-order ODE (where y is now the indep. variable). Solve this DE for $v(y)$, then sub back in $\frac{dy}{dx}$ for $v$ to get a first-order DE for $y(x)$ and solve (I don't have time to do it right now, so it may not be very pretty).
The reason that this method works in this case is that $x$ does not appear explicitly in the DE.
4. Mar 20, 2005
### cronxeh
How are you going to solve it if its v^2 ?
5. Mar 20, 2005
### Data
$$\frac{d^2y}{dx^2} + 2\left(\frac{dy}{dx}\right)^2 + y = 0 \Longrightarrow v\frac{dv}{dy} + 2v^2 + y = 0$$
which is an inexact DE, ie. it is in the form $$P(y, \ v) + Q( y, \ v)v^\prime = 0$$ with $$\frac{\partial P}{\partial v} \neq \frac{\partial Q}{\partial y}$$. We look for an integrating factor $I(y)$ such that
$$\ln{I(y)} = \int \frac{P_v -Q_y}{Q} dy$$
In this case, $$Q(y, \ v) = v$$ and $$P(y, \ v) = 2v^2 + y$$ so
$$\ln{I(y)} = \int \frac{ 4v - 0 }{v} dy = \int 4 dy \Longrightarrow I(y) = e^{4y}$$
note that not including a constant of integration makes no difference for our purposes. From the above, we know that
$$e^{4y}(2v^2 + y) + e^{4y}v\frac{dv}{dy} = 0$$
is an exact equation.
Then you need to know how to solve exact equations, but that's not hard~
Note that I skipped a bunch of steps on the theory of inexact first-order ODEs (as in, why did I look for an integrating factor that way?), but you can look those up or just ask and I can post them~
Last edited: Mar 20, 2005
6. Mar 21, 2005
### saltydog
Come on guys. I'd like to see this to completion. Maphysique, can you solve the exact equation in v, and get an expression in terms of F(y,v)=c? I end up with an expression for y' that I cannot integrate (a radical). Don't want to interfere with Data helping you though.
7. Mar 21, 2005
### HallsofIvy
Staff Emeritus
Getting an expression for y' that you cannot integrate is fairly typical of problems like this. In general, solutions to non-linear equations cannot be written in terms of elementary functions. Do you have reason to think that this one can be?
8. Mar 21, 2005
### Data
Yeah, the resulting integral for y is not nice at all. But at least he has it down to an integral, which is perfectly sufficient for approximating a solution :).
9. Mar 21, 2005
### saltydog
Thanks Data. When I first saw the problem, I didn't think to solve it that way. Just in case you guys don't feel like typing it all in, I'll do the next part. Perhaps Maphysique will follow it through:
I'll write the exact equation as follows:
$$[e^{4y}(2v^2 + y]dy + [e^{4y}v]dv = 0$$
"Exact" means its a total differential of some function F(y,v), that is:
$$dF=[e^{4y}(2v^2 + y]dy + [e^{4y}v]dv$$
Then surely:
$$\frac{\partial F}{\partial y}=[e^{4y}(2v^2 + y]$$
and:
$$\frac{\partial F}{\partial v}=[e^{4y}v]$$
To find F(y,v), let's integrate the partial with respect to y:
$$\int{\frac{\partial F}{\partial y}=\int{[e^{4y}(2v^2 + y]}dy$$
And thus:
$$F(y,v)=\frac{1}{2}v^2e^{4y}+\frac{1}{4}ye^{4y}-\frac{1}{16}e^{4y}+T(v)$$
Where integrating the partial with respect to y necessarilly yields some arbitrary function of v, T(v).
Differentiating F(y,v) with respect to v now yields:
$$\frac{\partial F}{\partial v}=ve^{4y}+\frac{dT}{dv}$$
But according to the exact differential, this is also equal to $e^{4y}v$
Or:
$$ve^{4y}+\frac{dT}{dv}=e^{4y}v$$
or:
$$\frac{dT}{dv}=0$$
Integrating yields:
$$T(v)=k$$
So that:
$$F(y,v)=\frac{1}{2}v^2e^{4y}+\frac{1}{4}ye^{4y}-\frac{1}{16}e^{4y}+k$$
Where k is a constant which we'll absorb into the constant c:
$$\frac{1}{2}v^2e^{4y}+\frac{1}{4}ye^{4y}-\frac{1}{16}e^{4y}=c$$
Finally, solving for v (which is y') yields:
$$(\frac{dy}{dx})^2=ce^{-4y}-\frac{1}{2}y+\frac{1}{8}$$
or:
$$\frac{dy}{dx}=\pm \sqrt{ce^{-4y}-\frac{1}{2}y+\frac{1}{8}}$$
But that's ok. You just solve two differential equations now and use whatever part meets the initial conditions. Anyway, I'll spend some time studying this numerically in Mathematica and report some useful results with select initial conditions.
10. Mar 21, 2005
### maphysique
I have just solved for myself without using the solution by exact differential equation.
Here is my solution.
As pointed by Data, Set
$$v=\frac{dy}{dx}$$,
we can rewrite the differential equation in question
$$\frac{d^2y}{dx^2}+2\left(\frac{dy}{dx}\right)^2+y=0$$
as follows.
$$v\frac{dv}{dy}+2v^2+y=0$$.
For $$v\neq 0$$,
we have
$$\frac{dv}{dy}+2v=-\frac{1}{v}y\ \cdots [E]$$
Considering the solution of the differential equation
$$\frac{dv}{dy}+2v=0$$
we set
$$v=ze^{-2y}$$,differentiating with respect to $$y$$
$$\frac{dv}{dy}=\frac{dz}{dy}e^{-2y}-2ze^{-2y}$$
plugging this into $$[E]$$
we have
$$z\frac{dz}{dy}=-ye^{4y}$$,
$$\frac{d}{dy}\left(\frac{1}{2}z^2\right)=-ye^{4y}$$
integrating with respect to $$y$$
$$\frac{1}{2}z^2=-\int ye^{4y}dy$$
using integral by parts
$$\frac{1}{2}z^2=-\left(\frac{1}{4}ye^{4y}-\frac{1}{16}e^{4y}\right)+C'$$
i.e.
$$z^2=-\left(\frac{1}{2}ye^{4y}-\frac{1}{8}e^{4y}\right)+C$$
where $$C=2C'$$
therefore we obtain
$$v^2=z^2e^{-4y}$$
$$v^2=-\left(\frac{1}{2}y-\frac{1}{8}\right)+Ce^{-4y}$$
Consequently,as deduced by saltydog,
$$\frac{dy}{dx}=\pm \sqrt{Ce^{-4y}-\left(\frac{1}{2}y-\frac{1}{8}\right)}$$
Thank you.
maphysique
Last edited: Mar 21, 2005
11. Mar 21, 2005
### saltydog
Hello Maphysique. Glad you're still with us and can explain how you approached it. This is my study of it:
It's interesting to note that if one would have approached this equation by converting it to two ODEs:
$$y'=u$$
$$u'=-2u^2-y$$
The constraint on y described below would not have been immediately clear unless the particular application it was describing would have indicated such.
As stated earlier, the equation is reduced to:
$$\frac{dy}{dx}=\pm\sqrt{ce^{4y}-\frac{1}{2}y+\frac{1}{8}}$$
Solving (for c=1):
$$e^{4x}-\frac{1}{2}x+\frac{1}{8}=0$$
we obtain $x\approx 0.51003$ and becomes negative when x is greater than this value. Thus, the ODE is valid only when y is less than the root of this equation.
Looking at the ODE again:
$$\frac{d^2 y}{dx^2}+2\left(\frac{dy}{dx}\right)^2+y=0$$
Since y' is squared, then y'' must necessarilly be negative for the LHS to have any chance of being zero.
If y'(0)>0 then the graph slopes upward and concaving downward until y reaches the zero of the radical.
If y'(0)<0 then the graphs slopes downward and concaving downward and becomes more so to balance the square of the derivative.
Attached are two particular solutions with y'(0)>0 and y'(0)<0 respectively.
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Last edited: Mar 21, 2005
12. Mar 21, 2005
### Data
actually, salty, I believe the solution to the exact DE is
$$\frac{dy}{dx} = \pm \sqrt{ ce^{-4y} - \frac{y}{2} + \frac{1}{8}}$$
(note the $ce^{-4y}$ as opposed to $ce^{4y}$)
Edit: I just noticed you had it right in your first post. Undoubtedly just a typo~
Last edited: Mar 21, 2005
13. Mar 21, 2005
### Data
And maphysique, that is a nice method. Unfortunately it doesn't lead to any additional simplifications that I can see (you still get the messy square root), but maybe you've figured out how to get around that too! :)
14. Mar 21, 2005
### maphysique
I have just edited.
Thank you, everyone!
maphysique
15. Mar 22, 2005
### tyutyu fait le train
Un petit coup de Laplace devrait marcher....
16. Mar 22, 2005
### Data
If your suggestion is to use the Laplace transform, then tell me what the Laplace transform of $f^\prime(x)^2$ is~
17. Mar 23, 2005
### dextercioby
U can't use Laplace transformation for nonlinear ODE-s...
Daniel.
18. Mar 24, 2005
### tyutyu fait le train
"U can't use Laplace transformation for nonlinear ODE-s...
Daniel."
why?
19. Mar 24, 2005
### Data
Like I asked before, what's the Laplace transform of $f^\prime(x)^2$?
20. Mar 24, 2005
### dextercioby
I assume it's very easy to ask questions without doing some documenting first.I think there are plenty of books on the application of integral transformations (Laplace & Fourier) to PDE-s/ODE-s...You can find one,i'm sure.
Daniel. | 2017-10-19T04:24:17 | {
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http://hfwp.vitadaprof.it/fourier-transform-examples.html | # Fourier Transform Examples
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F(x[n]) = F(cos( 2π 500 n)) = F(ej 2π 500n +e−j 2π 500n 2) = 1 2 (F(ej 2π 500n)+ F(e−j 2π 500n)) = 1 2 (π ∑ l=−∞+∞ δ(w− 2π 500 −2πl) +π ∑ l=−∞+∞ δ(w+ 2π 500 −2πl)) Instructor's comment: You need to justify this step (i. Square waves (1 or 0 or −1) are great examples, with delta functions in the derivative. be/J8qITQnN3pg. The Fourier Transforms. (3) Apply inverse transform to return to the spatial domain. Hi everyone, I have an acceleration time history, i want to calculate following 1. TheFouriertransform TheFouriertransformisimportantinthetheoryofsignalprocessing. Lecture with sound in PPT. Professor Deepa Kundur (University of Toronto)Properties of the Fourier Transform5 / 24 Properties of the Fourier Transform FT Theorems and Properties. The Fourier transform is commonly used to convert a signal in the time spectrum to a frequency spectrum. Fourier Transform Since this object can be made up of 3 fundamental frequencies an ideal Fourier Transform would look something like this: A Fourier Transform is an integral transform that re-expresses a function in terms of different sine waves of varying amplitudes, wavelengths, and phases. Let $f(x)=e^{-\alpha x}$ as $x>0$ and $f(x)=0$as $x<0$. The level is intended for Physics undergraduates in their 2nd or 3rd year of studies. Fourier Transform. 2 1 s t kT ()2 1 1 1 − − −z Tz 6. The time box shows the amount of time which the operator took to complete the process on the input image. The Fast Fourier Transform (FFT) is the most efficient algorithm for computing the Fourier transform of a discrete time signal. The Fourier transform is a mathematical formula that relates a signal sampled in time or space to the same signal sampled in frequency. Fourier Transform of a Periodic Function (e. Continuous Fourier Series. 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Discrete Fourier Transforms A discrete Fourier transform transforms any signal from its time/space domain into a related signal in frequency domain. 2 1 s t kT ()2 1 1 1 − − −z Tz 6. Therefore, to get the Fourier transform ub(k;t) = e k2t˚b(k) = Sb(k;t)˚b(k), we must. In analogy with the classical Fourier transform, which converts derivatives of functions of position into algebraic operations in Fourier space, there are operational properties for the motion-group Fourier transform. The coefficient b 1 of the continuous Fourier series associated with the given function f(t) can be computed as -75. Examples Fast Fourier Transform Applications. Cal Poly Pomona ECE 307 Fourier Transform The Fourier transform (FT) is the extension of the Fourier series to nonperiodic signals. This lecture Plan for the lecture: 1 Recap: Fourier transform for continuous-time signals 2 Frequency content of discrete-time signals: the DTFT 3 Examples of DTFT 4 Inverse DTFT 5 Properties of the DTFT. Fourier Transform series analysis, but it is clearly oscillatory and very well behaved for t>0 ( >0). The level is intended for Physics undergraduates in their 2nd or 3rd year of studies. Fourier Cosine Series for even functions and Sine Series for odd functions The continuous limit: the Fourier transform (and its inverse) The spectrum Some examples and theorems F() () exp()ωωft i t dt 1 () ()exp() 2 ft F i tdω ωω π. This example uses the decimation-in-time unit-stride FFT shown in Algorithm 1. 512, 1024, 2048, and 4096). The discrete Fourier transform is often, incorrectly, called the fast Fourier transform (FFT). Civil Engineering | Department of Engineering. For example, the Fourier transform of the rectangular function, which is integrable, is the sinc function, which is not Lebesgue integrable, because its improper integrals behave analogously to the alternating harmonic series, in converging to a sum without being absolutely convergent. Homework Statement Evaluate the Fourier Transform of the damped sinusoidal wave $g(t)=e^{-t}sin(2\pi f_ct)u(t)$ where u(t) is the unit step function. We additionally have enough money variant types and furthermore type of the books to browse. Abel transform amplitude angular antenna aperture distribution applied autocorrelation function Bracewell circuit coefficients complex components convolution theorem convolving cosine defined derivative digits discrete discrete Fourier transform discrete Hartley transform electrical equal equation equivalent width example expression factor. In mathematics, a Fourier transform (FT) is a mathematical transform which decomposes a function (often a function of time, or a signal) into its constituent frequencies, such as the expression of a musical chord in terms of the volumes and frequencies of its constituent notes. Fourier Transforms & Generalized Functions B. The latter imposes the restriction that the time series must be a power of two samples long e. Discrete convolution and correlation are defined and compared with continuous equivalents by illustrative examples. This is the same definition for linearity as used in your circuits and systems course, EE 400. The Discrete Time Fourier Transform How to Use the Discrete Fourier Transform. We discuss the Fourier transforms settings at different examples and show the consequences. 3 up to CUDA 6. , for filtering, and in this context the discretized input to the transform is customarily referred to as a signal, which exists in the time domain. the inverse Fourier transform. Discrete Fourier transform (DFT ) is the transform used in fourier analysis, which works with a finite discrete-time signal and discrete number of frequencies. Before looking into the implementation of DFT, I recommend you to first read in detail about the Discrete Fourier Transform in Wikipedia. It requires a power of two number of samples in the time block being analyzed (e. The Fourier Transform 1. A jupyter notebook with some stuff on the FT. We shall see that this is done by turning the difference equation into an ordinary algebraic equation. A general matrix-vector multiplication takes O(n 2) operations for n data-points. Thereafter,. This is a good point to illustrate a property of transform pairs. Square Wave Signal. 1 Baron Jean Baptiste Joseph Fourier (1768−1830) To consider this idea in more detail, we need to introduce some definitions and common terms. While we have defined Π(±1/2) = 0, other common conventions are either to have Π(±1/2) = 1 or Π(±1/2) = 1/2. Its applications are broad and include signal processing, communications, and audio/image/video compression. Our main results include Fourier analysis in trigonometric functions, interpolation and cubature. The Fourier Transform of f(x) is fe(k) = Z ∞ −∞ f(x)e−ikx dx = Z ∞ 0 e−ax−ikx dx = − 1 a + ik e−ax−ikx ∞ 0 = 1 a + ik. we’ll be int. Actually, you can do amazing stuff to images with fourier transform operations, including: (1) re-focus out of focus images (2) remove pattern noise in a picture, such as a half-tone mask (3) remove a repeating pattern like taking a picture through a screen door or off a piece of embossed paper (4) find an image so deeply buried in noise you. The Fourier transform of the constant function is given by (1) (2) according to the definition of the delta function. Multiply G(ω,θ) by the filter function |ω| modified by Hamming window 4. For example, an interval 0 to t is to be divided into N equal subintervals with width The data points are specified at n = 0, 1, 2, …, N-1. Loading Unsubscribe from belalanwer? The Fast Fourier Transform Algorithm - Duration: 18:55. The structure of DNA as deduced from fiber X-ray crystallography. As an example, see the spectrum of one cycle of a 440 Hz tone[42 kb]. For 512 evenly sampled times t (dt = 0. , for filtering, and in this context the discretized input to the transform is customarily referred to as a signal, which exists in the time domain. including the transient response Inverse Fourier transform of Vi(j!)H(j!) is the total zero-state response Nagendra. The 'Fourier Transform ' is then the process of working out what 'waves' comprise an image, just as was done in the above example. Fourier Transform Example Rectangular Pulse You Fourier transform table goshanni s page solved use the table of fourier transforms 5 2 and bax blog fourier transform table examples of the fourier transform. The coefficient b 1 of the continuous Fourier series associated with the given function f(t) can be computed as -75. Motivation for the Fourier transform comes from the study of Fourier series. Fourier Transform is a mathematical operation that breaks a signal in to its constituent frequencies. The Fourier transform. It computes the Discrete Fourier Transform (DFT) of an n-dimensional signal in O(nlogn) time. be/J8qITQnN3pg. Therefore the Fourier Transform too needs to be of a discrete type resulting in a Discrete Fourier Transform (DFT). Not only is the spectrum spread out, but the maximum occurs at 368 Hz (the red circle) rather than 440. If a is a real or complex vector implicitly indexed by j1,j2,. The DFT has revolutionized modern society, as it is ubiquitous in digital electronics and signal processing. The discrete Fourier Transform is the continous Fourier Transform for a period function. The signal is plotted using the numpy. 3 Properties of the Fourier Transform. Since the FFT only shows the positive frequencies, we need to shift the graph to get the correct frequencies. We can do this computation and it will produce a complex number in the form of a + ib where we have two coefficients for the Fourier series. Fourier Transform Settings – Discussion at Examples As one of the most fundamental technologies in VirtualLab Fusion, Fourier transforms connect the space and spatial frequency domains. The Fourier transform of an image breaks down the image function (the undulating landscape) into a sum of constituent sine waves. 6 (1,065 ratings) Course Ratings are calculated from individual students’ ratings and a variety of other signals, like age of rating and reliability, to ensure that they reflect course quality fairly and accurately. – – Kronecker delta δ0(k) 1 k = 0 0 k ≠ 0 1 2. The inverse Fourier transform converting a set of Fourier coefficients into an image is very similar to the forward transform (except of the sign of the exponent): The forward transform of an N×N image yields an N×N array of Fourier coefficients that completely represent the original image (because the latter is reconstructed from them by the. An API has been defined to allow users to write new code or modify existing code to use this functionality. The function itself is a sum of such components. We begin by discussing Fourier series. 8 we look at the relation between Fourier series and Fourier transforms. Adjusting the Number of Terms slider will determine how many terms are used in the Fourier expansion (shown in red). Fourier Transform. The Laplace transform of f(t), that it is denoted by f(t) or F(s) is defined by the equation. Examples Fast Fourier Transform Applications FFT idea I From the concrete form of DFT, we actually need 2 multiplications (timing ±i) and 8 additions (a 0 + a 2, a 1 + a 3, a 0 − a 2, a 1 − a 3 and the additions in the middle). N = 1024; % Number of data points. We investigate both first and second order difference equations. Engineering Tables/Fourier Transform Table 2 From Wikibooks, the open-content textbooks collection < Engineering Tables Jump to: navigation, search Signal Fourier transform unitary, angular frequency Fourier transform unitary, ordinary frequency Remarks 10 The rectangular pulse and the normalized sinc function 11 Dual of rule 10. Brayer (Professor Emeritus, Department of Computer Science, University of New Mexico, Albuquerque, New Mexico, USA). The figure below shows 0,25 seconds of Kendrick’s tune. The discrete Fourier transform is often, incorrectly, called the fast Fourier transform (FFT). The Discrete Fourier Transform Contents For example, we cannot implement the ideal lowpass lter digitally. Discrete Fourier transform Discrete complex exponentials Discrete Fourier transform (DFT), de nitions and examples Units of the DFT DFT inverse Properties of the DFT Signal and Information Processing Discrete Fourier transform 8. The characteristic function (or Fourier Transform) of a random variable $$X$$ is defined as \begin{align*} \psi(t)= \mathbf E \exp( i t X) \end{align*} for all $$t \in \mathbf R$$. In any finite interval, f (t) has at most a finite number of maxima and minima. Many systems do different things to different frequencies, so these kinds of systems can be described by what they do to each frequency. 2 is corresponding inverse. The function g(x) whose Fourier transform is G(ω) is given by the inverse Fourier transform formula g(x) = Z ∞ −∞ G(ω)e−iωxdω = Z ∞ −∞ e−αω2e−iωxdω (38). The Fourier Transform sees every trajectory (aka time signal, aka signal) as a set of circular motions. Fourier Transforms in NMR, Optical, and Mass Spectrometry Fourier transform properties and pictorial atlas of Fourier transform pairs. 3 The Wavelet Terms “Approximation” and “Details” Shown in FFT Format. Since each of the rectangular pulses on the right has a Fourier transform given by (2 sin w)/w, the convolution property tells us that the triangular function will have a Fourier transform given by the square of (2 sin w)/w: 4 sin2 w X(()) = (0). Transform examples. Worked Example Contour Integration: Inverse Fourier Transforms Consider the real function f(x) = ˆ 0 x < 0 e−ax x > 0 where a > 0 is a real constant. For more information on the NMath FFT classes, download and try out these examples: Examples using the 1D FFT classes. The fast fourier transform (FFT) allows the DCF to be used in real time and runs much faster if the width and height are both powers of two. Fourier [ list ] takes a finite list of numbers as input, and yields as output a list representing the discrete Fourier transform of the input. In Chapter 8, we develop the FFT algorithm. The most popular FFT algorithms are the radix 2 and radix 4, in either a decimation in time or a decimation in frequency signal flow graph form (trans- poses of each other) DTFT, DFT, and FFT Theory Overview. 1 The Fourier Transform Let g(t) be a signal in time domain, or, a function of time t. Fourier Transform Settings – Discussion at Examples As one of the most fundamental technologies in VirtualLab Fusion, Fourier transforms connect the space and spatial frequency domains. The DTFT of , i. They are widely used in signal analysis and are well-equipped to solve certain partial differential equations. • Continuous Fourier Transform (FT) – 1D FT (review) – 2D FT • Fourier Transform for Discrete Time Sequence (DTFT) – 1D DTFT (review) – 2D DTFT • Li C l tiLinear Convolution – 1D, Continuous vs. The Fourier transform is a way for us to take the combined wave, and get each of the sine waves back out. This algorithm provides you with an example of how you can begin your own exploration. The Fourier transform of the constant function is given by (1) (2) according to the definition of the delta function. clear A=input(’periodT=’); disp(’ ’) disp(’function defined on -T/2 to T/2’) 6. Sky observed by radio telescope is recorded as the FT of true sky termed as visibility in radio astronomy language and this visibility goes through Inverse Fourier Transformatio. It uses an atom which is the product of a sinusoidal wave with a finite energy symmetric window g. Those are examples of the Fourier Transform. In an infinite crystal, on the other hand, the function is typically periodic (and thus not decaying):. Studying these examples is one of the best ways to learn how to use NMath libraries. Barry Van Veen 367,676 views. But unlike that situation, the frequency space has two dimensions, for the frequencies h and k of the waves in the x and y dimensions. Implementing Fast Fourier Transform Algorithms of Real-Valued Sequences With the TMS320 DSP Platform Robert Matusiak Digital Signal Processing Solutions ABSTRACT The Fast Fourier Transform (FFT) is an efficient computation of the Discrete Fourier Transform (DFT) and one of the most important tools used in digital signal processing applications. In an infinite crystal, on the other hand, the function is typically periodic (and thus not decaying):. 3 up to CUDA 6. On this page, we'll use f(t) as an example, and numerically (computationally) find the Fourier Series coefficients. The Fourier Transform sees every trajectory (aka time signal, aka signal) as a set of circular motions. I am a newbie in Signal Processing using Python. The discrete Fourier transform of a, also known as the spectrum of a,is: Ak D XN−1 nD0 e−i2ˇ N kna n. In this section, we de ne it using an integral representation and state some basic uniqueness and inversion properties, without proof. Fourier Transform •Fourier Transforms originate from signal processing –Transform signal from time domain to frequency domain –Input signal is a function mapping time to amplitude –Output is a weighted sum of phase-shifted sinusoids of varying frequencies 17 e Time t Frequency Fast Multiplication of Polynomials •Using complex roots of. Fourier Transform Settings – Discussion at Examples As one of the most fundamental technologies in VirtualLab Fusion, Fourier transforms connect the space and spatial frequency domains. The structure of DNA as deduced from fiber X-ray crystallography. Adjusting the Number of Terms slider will determine how many terms are used in the Fourier expansion (shown in red). In the table above, each of the cells would contain a complex number. f(x)=abs(x) The program also computes the real version of the Fourier series representation in terms of sines and cosines. FTIR stands for Fourier transform infrared, the preferred method of infrared spectroscopy. In this section, we de ne it using an integral representation and state some basic uniqueness and inversion properties, without proof. We now apply the Discrete Fourier Transform (DFT) to the signal in order to estimate the magnitude and phase of the different frequency components. 1 Example of a Pathological Case Using the Fast Fourier Transform A. f(x − y)g(y)dy is the convolution of f and g, where f(x),g(x) ∈ C. Review of the Fourier Transform Blog. Let's start with the idea of sampling a continuous-time signal, as shown in this graph:. The Fourier transform is a mathematical formula that relates a signal sampled in time or space to the same signal sampled in frequency. Calculating a Fourier transform requires understanding of integration and imaginary numbers. As we are only concerned with digital images, we will restrict this discussion to the Discrete Fourier Transform (DFT). For this to. The discrete-time Fourier transform is an example of Fourier series. Fast Fourier transform (FFT) of acceleration time history 2. Example 3 4/6 Example 4. Conceptually, this occurs because the triangle wave looks much more like the 1st harmonic, so the contributions of the higher harmonics are less. ) Square Wave. A numerical example may be helpful. The Fast Fourier Transform (FFT) is one of the most important algorithms in signal processing and data analysis. cuFFT supports using up to sixteen GPUs connected to a CPU to perform Fourier Transforms whose calculations are distributed across the GPUs. The inverse Fourier transform here is simply the integral of a Gaussian. ES 442 Fourier Transform 3 Review: Fourier Trignometric Series (for Periodic Waveforms) Agbo & Sadiku; Section 2. Fourier Transform of a Periodic Function (e. The function itself is a sum of such components. a finite sequence of data). Discussion. Best Fourier Integral and transform with examples. The Fourier transform of a function of x gives a function of k, where k is the wavenumber. Other applications of the DFT arise because it can be computed very efficiently by the fast Fourier transform (FFT) algorithm. Two-Dimensional Fourier Transform. in a Crystal)¶ The Fourier transform in requires the function to be decaying fast enough in order to converge. Usually, the. The discrete Fourier transform of a, also known as the spectrum of a,is: Ak D XN−1 nD0 e−i2ˇ N kna n. Furthermore, this map is one-to-one. When the arguments are nonscalars, fourier acts on them element-wise. An example of a FID signal may be seen by. s 1 1(t) 1(k) 1 1 1 −z− 4. SEE ALSO: Cosine, Fourier Transform, Fourier Transform--Sine. The three examples are: 1. Example: the Fourier Transform of a Gaussian, exp(- at 2 ) , is itself! The details are a HW problem! ∩ t 0 0 26. Inverse Fourier Transform of a Gaussian Functions of the form G(ω) = e−αω2 where α > 0 is a constant are usually referred to as Gaussian functions. Since each of the rectangular pulses on the right has a Fourier transform given by (2 sin w)/w, the convolution property tells us that the triangular function will have a Fourier transform given by the square of (2 sin w)/w: 4 sin2 w X(()) = (0). Thus, a delta function in coordinate space can be transformed into unity in the k space (a very useful calculation in Electrodynamics). How It Works. Fourier-style transforms imply the function is periodic and extends to. A finite signal measured at N. The coefficient b 1 of the continuous Fourier series associated with the given function f(t) can be computed as -75. Questions: 1. In earlier DFT methods, we have seen that the computational part is too long. No help needed. • The Fourier Transform was briefly introduced – Will be used to explain modulation and filtering in the upcoming lectures – We will provide an intuitive comparison of Fourier Series and Fourier Transform in a few weeks …. Other applications of the DFT arise because it can be computed very efficiently by the fast Fourier transform (FFT) algorithm. I’ve been looking at the Fourier Transform through the eyes of a sound engineer, using it to analyze sound signals. ω=F{}X x t() ()⇒ F{ } is an “operator” that operates on x(t) to give X(ω) 3. fft has a function ifft() which does the inverse transformation of the DTFT. Harmonic regression 3. The Fast Fourier Transform (FFT) is an algorithm which performs a Discrete Fourier Transform in a computationally efficient manner. The Fourier Transform: Examples, Properties, Common Pairs The Fourier Transform: Examples, Properties, Common Pairs CS 450: Introduction to Digital Signal and Image Processing Bryan Morse BYU Computer Science The Fourier Transform: Examples, Properties, Common Pairs Magnitude and Phase Remember: complex numbers can be thought of as (real,imaginary). • Shifting in time domain changes phase spectrum of the signal. The function fˆ is called the Fourier transform of f. Specify the independent and transformation variables for each matrix entry by using matrices of the same size. Example 2: Convolution of probability. 9 Fourier Transform Applications Example 3 Find ; Solution: Using the rules 2 0 2 2 2. t/ is periodic with period Ts, we can also express (11. Its fourier transform should have a large 'bump' in the middle, but worse than this, my output numbers are of the order 10^-300. We investigate both first and second order difference equations. Fourier Transform •Fourier Transforms originate from signal processing -Transform signal from time domain to frequency domain -Input signal is a function mapping time to amplitude -Output is a weighted sum of phase-shifted sinusoids of varying frequencies 17 e Time t Frequency Fast Multiplication of Polynomials •Using complex roots of. We want to reduce that. Wavelet Transforms ♥Convert a signal into a series of wavelets ♥Provide a way for analyzing waveforms, bounded in both frequency and duration ♥Allow signals to be stored more efficiently than by Fourier transform ♥Be able to better approximate real-world signals ♥Well-suited for approximating data with sharp discontinuities. fourier, fourier integral, fourier transform, fourier transform examples Magic is real, and it all comes from the Fourier Integral. OR SEARCH CITATIONS. In this section, we de ne it using an integral representation and state some basic uniqueness and inversion properties, without proof. in a Crystal)¶ The Fourier transform in requires the function to be decaying fast enough in order to converge. supports 1D, 2D, and 3D transforms with a batch size that can be greater than or equal to 1. The Gaussian function, g(x), is defined as, g(x) = 1 σ √ 2π e −x2 2σ2, (3) where R ∞ −∞ g(x)dx = 1 (i. Example #2: sawtooth wave Here, we compute the Fourier series coefficients for the sawtooth wave plotted in Figure 4. 9 Fourier Transform Applications Example 3 Find ; Solution: Using the rules 2 0 2 2 2. This lecture Plan for the lecture: 1 Recap: Fourier transform for continuous-time signals 2 Frequency content of discrete-time signals: the DTFT 3 Examples of DTFT 4 Inverse DTFT 5 Properties of the DTFT. Compute the inverse of the results from 3. The time box shows the amount of time which the operator took to complete the process on the input image. For particular functions we use tables of the Laplace. f(t)e−iωtdt (4) The function fˆ is called the Fourier transform of f. This is a preferred scheme of infrared spectroscopy. Fourier Transform. A numerical example may be helpful. The Baron was. My example code is following below: In [44]: x = np. TheFouriertransform TheFouriertransformisimportantinthetheoryofsignalprocessing. The Fourier tra. Bracewell (which is on the shelves of most radio astronomers) and the Wikipedia and Mathworld entries for the Fourier transform. The amplitudes of the harmonics for this example drop off much more rapidly (in this case they go as 1/n 2 (which is faster than the 1/n decay seen in the pulse function Fourier Series (above)). Similarity Theorem Example Let's compute, G(s), the Fourier transform of: g(t) =e−t2/9. Transformation Kernels. Definition of Fourier Transform. Thanks 581873. You do it in matrix form by multiplicating your data with a Fourier matrix (for example n=4): [ 1 1 1 1 ] [ 1 w w^2 w^3 ] [ 1 w^2 w^4 w^6 ] [ 1 w^3 w^6 w^9 ] The unit w is exp(2pi i / n). Fourier transforms commonly transforms a mathematical function of time, f(t), into a new function, sometimes denoted by or F, whose argument is frequency with units of cycles/s (hertz) or radians per second. IFor systems that are linear time-invariant (LTI), the Fourier transform provides a decoupled description of the system operation on the input signal much like when we diagonalize a matrix. Let $f(x)=e^{-\alpha x}$ as $x>0$ and $f(x)=0$as $x<0$. We decide to explore this fourier transforms of a derivative examples photo in this article simply because based on facts from Google search engine, Its one of the top searches key word on google. In this tutorial numerical methods are used for finding the Fourier transform of For example, if we increase the sampling interval by a factor of 10 (to. f(x)=abs(x) The program also computes the real version of the Fourier series representation in terms of sines and cosines. The Fourier transform of a periodic impulse train in the time domain with period T is a periodic impulse train in the frequency domain with period 2p /T, as sketched din the figure below. For example, for not-terribly-obvious reasons, in quantum mechanics the Fourier transform of the position a particle (or anything really) is the momentum of that particle. Let us begin with the exponential series for a function fT (t) defined to be f (t) for. N = 1024; % Number of data points. The Fast Fourier Transform (FFT) is the most efficient algorithm for computing the Fourier transform of a discrete time signal. We additionally have enough money variant types and furthermore type of the books to browse. The resulting transform pairs are shown below to a common horizontal scale: Cu (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 8 / 37. A simple example of Fourier transform is applying filters in the frequency domain of digital image processing. Again back calculation of time history by taking Inverse fourier transform (IFFT) of FFT. 1), the DTFT of is computed as:. Using the tools we develop in the chapter, we end up being able to derive Fourier’s theorem (which. Fourier Series & The Fourier Transform What is the Fourier Transform? Fourier Cosine Series for even functions and Sine Series for odd functions The continuous limit: the Fourier transform (and its inverse) The spectrum Some examples and theorems F( ) ( ) exp( )ωωft i t dt ∞ −∞ =−∫ 1 ( )exp( ) 2 ft F i tdω ωω π ∞ −∞ = ∫. The Fourier transform of a function of x gives a function of k, where k is the wavenumber. FFT(X,N) is the N-point FFT, padded with zeros if X has less than N points and truncated if it has more. We’ll do a couple more examples here and return to transform methods later. 2 is corresponding inverse. Fourier Transform series analysis, but it is clearly oscillatory and very well behaved for t>0 ( >0). The Fast Fourier Transform (FFT) is one of the most fundamental numerical algorithms. In other words, Fourier series can be used to express a function in terms of the frequencies (harmonics) it is composed of. All three domains are related to each other. Other applications of the DFT arise because it can be computed very efficiently by the fast Fourier transform (FFT) algorithm. The Fourier Transform is a way how to do this. be/J8qITQnN3pg. Per Brinch Hansen: The Fast Fourier Transform 7 Example F(1) = [1] a= [ao] b = [ao] Example F(2) = [ 1 1. The roots of the Fourier transform (FT) are found in a Fourier series in which complicated periodic functions are written as the sum of simple waves mathematically represented by sines and cosines. As noted above, if f(t) exists for a finite time window, the spectrum spreads over an infinite frequency range. The idea is that any function may be approximated exactly with the sum of infinite sinus and cosines functions. The Fourier Transforms. Thus we have reduced convolution to pointwise multiplication. Given a trajectory the fourier transform (FT) breaks it into a set of related cycles that describes it. This chapter exploit what happens if we do not use all the !'s, but rather just a nite set (which can be stored digitally). )2 Solutions to Optional Problems S9. , normalized). This is also a one-to-one transformation. The input signal in this example is a combination of two signals frequency of 10 Hz and an amplitude of 2 ; frequency of 20 Hz and an amplitude of 3. Therefore the Fourier Transform too needs to be of a discrete type resulting in a Discrete Fourier Transform (DFT). Take the Fourier Transform of both equations. How to solve easily Fourier Transform examples in Tamil-Fourier Transform Engineering Mathematics 3 Regulation 2017 First Video: https://youtu. Fourier transform is interpreted as a frequency, for example if f(x) is a sound signal with x measured in seconds then F ( u )is its frequency spectrum with u measured in Hertz (s 1 ). Combining (24) with the Fourier series in (21), we get that:,. values we apply Discrete Fourier Transform. Fourier transform is one of the best numerical computation of our lifetime, the equation of the Fourier transform is, It is used to map signals from the time domain to the frequency domain. The result is. the gray level intensities of the choosen line. An API has been defined to allow users to write new code or modify existing code to use this functionality. No help needed. The naive evaluation of the discrete Fourier transform is a matrix-vector multiplication. Its Fourier transform. What kind of functions is the Fourier transform de ned for? Clearly if f(x) is real, continuous and zero outside an interval of the form [ M;M], then fbis de ned as the improper integral R 1 1 reduces to the proper integral R M M. If a is a real or complex vector implicitly indexed by j1,j2,. How to solve easily Fourier Transform examples in Tamil-Fourier Transform Engineering Mathematics 3 Regulation 2017 First Video: https://youtu. Fourier Transform. Example: Fourier Transform of Single Rectangular Pulse. One more Question , does the both results of Continuous time fourier transform and Discrete time fourier transform the same , or different. The Fourier transform of $f(x)$ is denoted by $\mathscr{F}\{f(x)\}= $$F(k), k \in \mathbb{R}, and defined by the integral :. , John Wiley & Sons, Inc. The Fourier transform of a function f is usually denoted by the upper case F but many authors prefer to use a tilde above this function, i. • Continuous Fourier Transform (FT) – 1D FT (review) – 2D FT • Fourier Transform for Discrete Time Sequence (DTFT) – 1D DTFT (review) – 2D DTFT • Li C l tiLinear Convolution – 1D, Continuous vs. Now, we know how to sample signals and how to apply a Discrete Fourier Transform. In this particular case ‘10’ is the real part and ‘5’ is the imaginary part. Our main results include Fourier analysis in trigonometric functions, interpolation and cubature. 3 The Wavelet Terms “Approximation” and “Details” Shown in FFT Format. Find the Fourier transform of f(x) = ˆ 1 in jxj a Solution : The given function can be written as f(x) = ˆ 1 if aa) 1 = 1. For this to. The continuous Fourier transform is important in mathematics, engineering, and the physical sciences. The Fourier transform of this signal is fˆ(ω) = Z ∞ −∞ f(t)e. When both the function and its Fourier transform are replaced with discretized counterparts, it is called the discrete Fourier transform (DFT). z-Transforms In the study of discrete-time signal and systems, we have thus far considered the time-domain and the frequency domain. As can clearly be seen it looks like a wave with different frequencies. How to solve easily Fourier Transform examples in Tamil-Fourier Transform Engineering Mathematics 3 Regulation 2017 First Video: https://youtu. More precisely, we have the formulae1 f(x) = Z R d fˆ(ξ)e2πix·ξ dξ, where fˆ(ξ) = Z R f(x)e−2πix·ξ dx. The Fourier Transform ( in this case, the 2D Fourier Transform ) is the series expansion of an image function ( over the 2D space domain ) in terms of "cosine" image (orthonormal) basis functions. Let us begin with the exponential series for a function fT (t) defined to be f (t) for. A Lookahead: The Discrete Fourier Transform The relationship between the DTFT of a periodic signal and the DTFS of a periodic signal composed from it leads us to the idea of a Discrete Fourier Transform (not to be confused with Discrete- Time Fourier Transform). I've used it for years, but having no formal computer science background, It occurred to me this week that I've never thought to ask how the FFT computes the discrete Fourier transform so quickly. Fourier Transform. One reason to introduce the Fourier transform now was to reinforce the derived solution expressions for the heat and vibrating string problems on the line by deriving them using the transform method. The Fourier Transform for continuous signals is divided into two categories, one for signals that are periodic, and one for signals that are aperiodic. No help needed. Discrete Fourier transform (DFT ) is the transform used in fourier analysis, which works with a finite discrete-time signal and discrete number of frequencies. Find the Fourier transforms of the following signals. The Fourier Transform – derivation: Using the concept of Fourier Integrals. For example, you can transform a 2-D optical mask to reveal its diffraction pattern. Example: DFS by DDC and DSP. Square waves (1 or 0 or −1) are great examples, with delta functions in the derivative. Table of Laplace and Z-transforms X(s) x(t) x(kT) or x(k) X(z) 1. The continuous Fourier transform is important in mathematics, engineering, and the physical sciences. As an example, see the spectrum of one cycle of a 440 Hz tone[42 kb]. It is a linear invertible transfor-mation between the time-domain representation of a function, which we shall denote by h(t), and the frequency domain representation which we shall denote by H(f). Notice how aliasing looks in the time domain. (b) The Poisson summation formula is an equation relating the Fourier series coe cients of the periodic summation of a function to values of the function’s continuous Fourier transform. The inverse Fourier Transform • For linear-systems we saw that it is convenient to represent a signal f(x) as a sum of scaled and shifted sinusoids. When IR radiation is passed through a sample, some radiation is absorbed by the sample and some passes through (is transmitted). Fourier Transform. For example, given a sinusoidal signal which is in time domain the Fourier Transform provides the constituent signal frequencies. Fourier Transform Example Rectangular Pulse You Fourier transform table goshanni s page solved use the table of fourier transforms 5 2 and bax blog fourier transform table examples of the fourier transform. Specify the independent and transformation variables for each matrix entry by using matrices of the same size. The Fourier transform of f(x) is denoted by \mathscr{F}\{f(x)\}=$$ F(k), k \in \mathbb{R},$ and defined by the integral :. The following formula defines the discrete Fourier transform Y of an m-by-n matrix X. For example, the DFT is used in state-of-the-art algorithms for multiplying polynomials and large integers together; instead of working with polynomial multiplication directly, it turns out to be faster to compute the. Different choices of definitions can be specified using the option FourierParameters. Example 1: Use the Laplace transform operator to solve the IVP. Discrete Fourier transform 4. The discrete-time Fourier transform is an example of Fourier series. Derpanis October 20, 2005 In this note we consider the Fourier transform1 of the Gaussian. Civil Engineering | Department of Engineering. Barry Van Veen 367,676 views. Discrete Fourier transform (DFT ) is the transform used in fourier analysis, which works with a finite discrete-time signal and discrete number of frequencies. This allows us to not only analyze the different frequencies of the data, but also enables faster filtering operations, when used properly. For example, if a chord is played, the sound wave of the chord can be fed into a Fourier transform to find the notes that the chord is made from. FFTs are used for fault analysis, quality control, and condition monitoring of machines or systems. 3 p710 xt t()cos10 xt rectt( ) ( /4). As we are only concerned with digital images, we will restrict this discussion to the Discrete Fourier Transform (DFT). In Chapter 8, we develop the FFT algorithm. Let samples be denoted. This can be done through FFT or fast Fourier transform. Every wave has one or more frequencies and amplitudes in it. In the general case: X1 n =1 f(x+ na) = 1 a X1 k 1 f^(k a)e2ˇik a x In particular, when x= 0 it is known as the Poisson summation formula: X1 n=1 f(na) = 1 a. The unit FOURIER contains the fast Fourier transform (FFT) component TFastFourier. The cosine function, f(t), is shown in Figure 1: Figure 1. For particular functions we use tables of the Laplace. Waveform Analysis Using The Fourier Transform DATAQ Instruments Any signal that varies with respect to time can be reduced mathemat ically to a seri es of sinusoidal terms. Example: The tank shown in figure is initially empty. z-Transforms In the study of discrete-time signal and systems, we have thus far considered the time-domain and the frequency domain. This is not a particular. are, for example, discontinuous or simply di cult to represent analytically. In most cases signal waves maintain symmetry. In this particular case ‘10’ is the real part and ‘5’ is the imaginary part. a finite sequence of data). Transform examples. The discrete Fourier transform is often, incorrectly, called the fast Fourier transform (FFT). Find the Fourier transform of s(t) = cos(2ˇf 0t): We can re-write the signal using Euler's formula: s(t) = 1 2 ej2ˇf 0t+ 1 2 e j2ˇf 0t: 1. Example Transformations. The Fourier transform is both a theory and a mathematical tool with many applications in engineering and science. Fourier transform can be generalized to higher dimensions. Other applications of the DFT arise because it can be computed very efficiently by the fast Fourier transform (FFT) algorithm. If you are already familiar with it, then you can see the implementation directly. The Fourier transform on L1 In this section, we de ne the Fourier transform and give the basic properties of the Fourier transform of an L 1(Rn) function. We look at a spike, a step function, and a ramp—and smoother functions too. Calculus and Analysis > Integral Transforms > Fourier Transforms > Fourier Transform--Cosine (1) (2) (3) where is the delta function. For N-D arrays, the FFT operation operates on the first non-singleton dimension. For example, the DFT is used in state-of-the-art algorithms for multiplying polynomials and large integers together; instead of working with polynomial multiplication directly, it turns out to be faster to compute the. a finite sequence of data). The Fourier transform of a function of t gives a function of ω where ω is the angular frequency: f˜(ω)= 1 2π Z −∞ ∞ dtf(t)e−iωt (11) 3 Example As an example, let us compute the Fourier transform of the position of an underdamped oscil-lator:. We want to reduce that. ES 442 Fourier Transform 3 Review: Fourier Trignometric Series (for Periodic Waveforms) Agbo & Sadiku; Section 2. The Fourier transform of this signal is fˆ(ω) = Z ∞ −∞ f(t)e. be/J8qITQnN3pg. Let's start with the idea of sampling a continuous-time signal, as shown in this graph:. 2 Time Scaling Property ^ f at F() ` 1 aa. We wish to Fourier transform the Gaussian wave packet in (momentum) k-space to get in position space. University of Oxford. Note, for a full discussion of the Fourier Series and Fourier Transform that are the foundation of the DFT and FFT, see the Superposition Principle, Fourier Series, Fourier Transform Tutorial. Review of the Fourier Transform Blog. FFT(X) is the discrete Fourier transform (DFT) of vector X. x for >=CUDA 8. Fourier Transform Examples And Solutions Fourier Transform Examples And Solutions Right here, we have countless book Fourier Transform Examples And Solutions and collections to check out. , for filtering, and in this context the discretized input to the transform is customarily referred to as a signal, which exists in the time domain. In this example, you can almost do it in your head, just by looking at the original wave. Similarly, we can write the Fourier transform using complex cosine-sines. Periodogram 6. REFERENCES: Bracewell, R. Smith SIAM Seminar on Algorithms- Fall 2014 University of California, Santa Barbara October 15, 2014. I've found the FFT always really interesting although I'm not that math savvy and most articles you read on FFT go really into the maths of how a FFT works. Fourier Transform of Array Inputs. Fourier Transform Examples and Solutions | Inverse Fourier Transform Dr. The Fourier transform of the constant function is given by (1) (2) according to the definition of the delta function. FFT(X,N) is the N-point FFT, padded with zeros if X has less than N points and truncated if it has more. We begin by discussing Fourier series. It is to be thought of as the frequency profile of the signal f(t). The plot looks like this. ) We have f0(x)=δ−a(x)−δa(x); g0(x)=δ−b(x) −δb(x); d2 dx2 (f ∗g)(x)= d dx f ∗ d dx g(x) = δ−a(x)− δa(x. Fourier Transform of a General Periodic Signal. Sum(integral) of Fourier transform components produces the input x(t)(e. This lecture Plan for the lecture: 1 Recap: Fourier transform for continuous-time signals 2 Frequency content of discrete-time signals: the DTFT 3 Examples of DTFT 4 Inverse DTFT 5 Properties of the DTFT. Find the Fourier transform of the matrix M. We evaluate it by completing the square. Some functions don’t have Fourier transforms. If the number of data points is not a power-of-two, it uses Bluestein's chirp z-transform algorithm. Implementing Fast Fourier Transform Algorithms of Real-Valued Sequences With the TMS320 DSP Platform Robert Matusiak Digital Signal Processing Solutions ABSTRACT The Fast Fourier Transform (FFT) is an efficient computation of the Discrete Fourier Transform (DFT) and one of the most important tools used in digital signal processing applications. 9 Fourier Transform Inverse Fourier Transform Useful Rules 2 0 1 √2 2 cos 2 sin Fourier Inverse Transform Fourier Inverse Cosine Transform Fourier Inverse Sine Transform 30.
8j2tn916eft5jn ji7w9lavs5n khakzl3bv097 49pz4br7yheb w35xes4wvu1r klu7hhgq6rq1q 4ml0yqvodelf mrg94rjz7n 5gsyviwd992efuu dezheniowu5t 6rkyqejlvvwz 26iq3hichtd1 7sw6mk9sewrv qo9u6lmiyqvkxn k3yd19q4zb1kt8 2r63ihqa1mlge fmynd9n2wfgu e6wmc9hh4dw4 yg7ias210n 5a8sv1y1ex6e 3el4q5sp0t33ej9 hiuv7qkn1284z5l cd36len9o4p mf5h2zltzuu0vk 751lhl25ekalpc c7jb6e09iv76 kz4ja69jd6yg mnyhid07o00o | 2020-12-02T19:34:48 | {
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http://mathhelpforum.com/pre-calculus/119463-i-m-stumped.html | # Math Help - I'm stumped..
1. ## I'm stumped..
A bridge is built in the shape of a parabolic arch. The bridge arch has a span of 196 feet and a max height of 35 feet. Find the height of the arch at 25 feet from its center.
I've done this problem now about 5 or 6 times.. and the answer I get is different almost every time. I wish I could draw it to show what my thoughts on the set up are. basically I did y=ax^2 first but then thought I needed to find P. so that would be y=1/4p(x^2). somewhere along the lines I'm almost positive I just confused myself and did something wrong. I ended up with about 25.8. The other problems in this same vein I have been able to understand and figure out because of help from here, but this one has me stumped.. maybe its the wording or something. maybe its just late. help would be appreciated.
2. Originally Posted by NKS
A bridge is built in the shape of a parabolic arch. The bridge arch has a span of 196 feet and a max height of 35 feet. Find the height of the arch at 25 feet from its center.
I've done this problem now about 5 or 6 times.. and the answer I get is different almost every time. I wish I could draw it to show what my thoughts on the set up are. basically I did y=ax^2 first but then thought I needed to find P. so that would be y=1/4p(x^2). somewhere along the lines I'm almost positive I just confused myself and did something wrong. I ended up with about 25.8. The other problems in this same vein I have been able to understand and figure out because of help from here, but this one has me stumped.. maybe its the wording or something. maybe its just late. help would be appreciated.
The horizontal distance is 196 feet.
So if you assume it starts at the origin, then two points you have are
$(0, 0)$ and $(196, 0)$.
You also have the maximum height of 35 feet. This occurs right in the centre of your x-intercepts.
So the turning point is $(98, 35)$.
If you use the turning point form for the equation of the parabola, it will be of the form
$y = a(x - h)^2 + k$.
Substituting in the turning point gives
$y = a(x - 98)^2 + 35$
And substituting in another point, in this case $(196, 0)$ gives you the a value.
$0 = a(196 - 98)^2 + 35$
$0 = a(98)^2 + 35$
$a = -\frac{35}{98^2}$.
Therefore the equation of the parabola is
$y = -\frac{35}{98^2}(x - 98)^2 + 35$.
Now use it to find the height when $x = 25$.
3. I think, in the final form of the equation that you found, the 98 should be a 0. because (h,k) is the vertex, no? and the vertex, if are drawing this out to resemble a bridge, would be at (0,35). I also change the h to 0 in the beginning and followed your steps and came to the same answer as I did after I had just changed it in the final equation. so the answer should be 32.7 I believe.
maybe someone could check to see if I am right.
4. Essentially it doesn't matter where you put the origin, as it will just end up creating a translation of the graph.
I always put the origin as the first known point - in this case, where the bridge starts... | 2014-07-10T00:55:08 | {
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https://www.physicsforums.com/threads/electric-potential-at-center-of-circular-arc.138075/ | # Electric potential at center of circular arc
1. Oct 12, 2006
### cyberstudent
An insulating rod of length l is bent into a circular arc of radius R that subtends an angle theta from the center of the circle. The rod has a charge Q ditributed uniformly along its length. Find the electric potential at the center of the circular arc.
Struggling with this problem.
I know that I have to divide the charge Q into many very small charges, essentially point charges, then sum them up (integration).
dV = dq/(4 Π Ε0 R)
Length of dq = ds
db = angle subtended by ds
dΒ=ds/R => ds = dBR
dq = λds => dq = λdBR
V = ∫ dq/(4 Π Ε0 R)
V = ∫ λdBR/(4 Π Ε0 R)
Now, this is where it all goes wrong for me.
I take out the constants V = λR/(4 Π Ε0 R) * ∫ dB
My Rs cancel out, which makes no sense.
The radius must be important in the calculation of the difference potential.
Notice also that I did not indicate the limits on the integration. In a similar problem which was done in a previous assignemnt to calculate the electric field at the center, the upper and lower limits were set to -B/2 and B/2, but I am not sure why.
Last edited: Oct 12, 2006
2. Oct 12, 2006
### OlderDan
It is. Define λ
The limits would be -theta/2 to +theta/2. You have to integrate over the angle subtended by the arc. You could actually use any pair of limits that differ by theta.
3. Oct 12, 2006
### cyberstudent
λ is the linear charge density.
That actually makes senses to me. Thanks. Of course, you would want to integrate over the angle subtended by the arc. Would 0 and theta also be valid limits then?
So, I now have my limits, but I still end up with the same absurd problem of losing my R. What is the mistake. Is the equation wrong? Am I not taking out constants?
Θ/2
V = ∫ λdBR/(4 Π Ε0 R)
-Θ/2
Θ/2
V = λR/(4 Π Ε0 R) ∫ dB
-Θ/2
I uploaded a diagram to clarify the problem and my thought process.
#### Attached Files:
• ###### diagram_calculations.gif
File size:
15.8 KB
Views:
799
4. Oct 12, 2006
### OlderDan
The limits of 0 and theta will work fine.
OK, you have defined λ. Now calculate λ.
5. Oct 12, 2006
### cyberstudent
I think I got it.
You must think I'm a bit slow. :)
λ = Q/ΘR (Charge/divided by length of the arc)
I haven't lost my R after all!
Θ/2
V = ∫ λdBR/(4 Π Ε0 R)
-Θ/2
Θ/2
V = λR/(4 Π Ε0 R) ∫ dB
-Θ/2
Θ/2
V = λ/(4 Π Ε0) B ׀
- Θ/2
V = (λ/(4 Π Ε0)) * Θ
V = QΘ/((4 Π Ε0)ΘR)
V = Q/(4 Π Ε0 R)
Interestingly, the equation for the potential difference at the center of the circular arc is the same regardless of Θ.
Have I solved the problem or did I make another dumb mistake?
6. Oct 13, 2006
### OlderDan
Why would I think you are slow? You did what I asked you to do, jut not what I had expected you would do
Your answer is interesting, and it is correct. It also makes sense. Regardless of the arc length, all the charge is the same distance from the center of the circle. Potential is a scalar inversely porportional to the distance from the charge. No matter how you distribute that charge along the arc the answer will be the same.
7. Mar 12, 2008
### chrisg8
sorry to dig up such an old post but i too had a problem like this and everything except how you went from
V = (λ/(4 Π Ε0)) * Θ
to
V = QΘ/((4 Π Ε0)ΘR)
makes sense... i realize that λ=Q/R but how come the Θ cancels out?
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http://mathhelpforum.com/advanced-algebra/168494-linear-map-question.html | # Math Help - Linear Map Question
1. ## Linear Map Question
Let $V$ denote the vector space of twice differentiable functions on $R$. Define a linear map $L$ on $V$ by the formula
$L(u)=a\frac{d^2}{dx^2}u+b\frac{d}{dx}u+cu$
Suppose that $u_{1},u_{2}$ is a basis for the solution space of $L(u)=0$. Find a basis for the solution space of the fourth order differential equation $L(L(u))=0$. What can you say about the kernals of $L$ and $L^2$?
This is my working out so far:
Since $u_{1},u_{2}$ is a basis for the solution space of $L(u)=0$, then $u_{1},u_{2}$ is a basis for the kernal of $L$.
Therefore we can write
$u=\lambda u_{1}+\mu u_{2} ,\ \ \ \ \ \ \ \forall u\ \epsilon \ Ker(L), \ \lambda,\mu \ \epsilon \ R$
So $Nullity(L)=2$ since $Ker(L)$ is spanned by two linearly independent vectors.
Now, consider $L(L(u))=0$. Since every solution of $L(u)=0$ can be represented as $\lambda u_{1}+\mu u_{2}$,
the equation $L(L(u))=0$ becomes
$L(u)=\lambda u_{1}+\mu u_{2}$, which then becomes
$a\frac{d^2}{dx^2}u+b\frac{d}{dx}u+cu=\lambda u_{1}+\mu u_{2}$.
Let $Y_{H }, Y_{P }$ denote the homogenous solution and the particular solution respectively.
Obviously $Y_{H }=\lambda u_{1}+\mu u_{2}$ as $u_{1},u_{2}$ is a basis for the solution space of $L(u)=0$.
Now this is where I got stuck. I did’t know what the particular solution is. I let $Y_{P}=mxu_{1}+nxu_{2},\ \ \ \ \ \ m,n\ \epsilon \ R$
and tried to solve for the constants $m,n$. It was a mess and I don’t think I was on the right track.
I was thinking that if I can find $Y_{P}$, then $Y_{H }, Y_{P }$ would be the basis of the fourth order differential equation $L(L(u))=0$. Am I correct?
Any help will be appreciated.
Thanks.
2. I think you are "overworking" this. For any linear transformation, L, L(0)= 0 so that, if u is in the kernel of L, Lu= 0, the $L^2u= L(Lu)= L(0)= 0$. That is, for any linear transformation, L, the kernel of L is a subspace of the kernel of $L^2$.
3. Thanks. Yep I understand what you mean by $'Ker(L)$ is a subspace of $Ker(L^2)'$'.
I think $Nullity(L^2)=4$ because I have read somewhere that the solution space of an nth order homogeneous differential equation forms an nth dimensional vector space. Is $L(L(u))=0$ a fourth order homogeneous equation? How would you find the basis?
4. Bump. Anyone know how to find the basis for $L(L(u))$? Many thanks. | 2016-07-31T10:19:26 | {
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https://mathhelpboards.com/threads/trigonometric-identity-questions.8020/ | # TrigonometryTrigonometric Identity Questions
#### suzy123
##### New member
Your help will be greatly appreciated!
Thanks!
1. The expression $$\sin\pi$$ is equal to $$0$$, while the expression $\frac{1}{\csc\pi}$ is undefined. Why is $\sin\theta=\frac{1}{\csc\theta}$ still an identity?
2. Prove $\cos(\theta + \frac{\pi}{2})= -\sin\theta$
Last edited by a moderator:
#### SuperSonic4
##### Well-known member
MHB Math Helper
2. Prove $\cos(\theta + \frac{\pi}{2})= -\sin\theta$
Use the addition formula together with the values from the unit circle $$\displaystyle \cos(A+B) = \cos(A)\cos(B) - \sin(A)\sin(B)$$
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Welcome to MHB, suzy123!
Your help will be greatly appreciated!
Thanks!
1. The expression $$\sin\pi$$ is equal to $$0$$, while the expression $\frac{1}{\csc\pi}$ is undefined. Why is $\sin\theta=\frac{1}{\csc\theta}$ still an identity?
It's an identity when both are defined.
And if you treat $\csc\pi$ as $\infty$ it even holds there.
#### DreamWeaver
##### Well-known member
Welcome to MHB, suzy123!
It's an identity when both are defined.
And if you treat $\csc\pi$ as $\infty$ it even holds there.
An excellent point. Also, another way of looking at both is by considering limiting values:
$$\displaystyle \lim_{x \to \pi}\sin x=0$$
$$\displaystyle \lim_{x \to \pi}\csc x = \lim_{x \to \pi}\frac{1}{\sin x}=\frac{1}{ \lim_{x \to \pi}\sin x } = \lim_{z \to 0 }\frac{1}{z} \to \frac{1}{0} \to \infty$$
Similarly, you could use the composite angle formula I Like Serena gave above,
$$\displaystyle \sin (x \pm y)= \sin x \cos y \pm \cos x \sin y$$
and consider the limits
$$\displaystyle \sin \pi = \lim_{x \to \pi}\sin x= \lim_{\epsilon \to 0}\sin (\pi \pm \epsilon) \to 0$$
and
$$\displaystyle \csc \pi = \lim_{x \to \pi}\csc x= \lim_{\epsilon \to 0} \frac{1}{\sin (\pi \pm \epsilon)} \to \infty$$
#### Deveno
##### Well-known member
MHB Math Scholar
Prove that the function:
$f(\theta) = \sin\theta\csc\theta$
has removable discontinuities at $k\pi,\ k \in \Bbb Z$.
It is, by and large, problematic to simply say:
$\infty = \frac{1}{0}$ because such an assignment does not obey the algebraic rules of the real numbers (in particular, the cancellation law:
$ac = bc \implies a = b$ when $c \neq 0$
breaks down).
While it *is* possible to extend the real numbers in various ways to include the notion of infinity, it is usually preferable to phrase statements about infinity in ways that do not mention infinity itself such as:
$\displaystyle \lim_{x \to a} f(x) = \infty$
we say:
for any $N > 0$, there is some $\delta > 0$ such that for all $0 < |x - a| < \delta$, we have $f(x) > N$.
This is a fancy way of saying: $f$ increases without bound near $a$. Note it does not mention infinity, nor does it say what (if any) value we should ascribe to $f(a)$.
As others have mentioned, the cosecant function is undefined at certain points. If one is asked to evaluate cosecant at such a point, one ought to politely refuse.
#### DreamWeaver
##### Well-known member
If one is asked to evaluate cosecant at such a point, one ought to politely refuse.
Genius! | 2020-11-27T04:51:00 | {
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https://www.tutorialspoint.com/all-possible-binary-numbers-of-length-n-with-equal-sum-in-both-halves | # All possible binary numbers of length n with equal sum in both halves?
Here we will see all possible binary numbers of n bit (n is given by the user) where the sum of each half is same. For example, if the number is 10001 here 10 and 01 are same because their sum is same, and they are in the different halves. Here we will generate all numbers of that type.
## Algorithm
#### genAllBinEqualSumHalf(n, left, right, diff)
left and right are initially empty, diff is holding difference between left and right
Begin
if n is 0, then
if diff is 0, then
print left + right
end if
return
end if
if n is 1, then
if diff is 0, then
print left + 0 + right
print left + 1 + right
end if
return
end if
if 2* |diff| <= n, then
if left is not blank, then
genAllBinEqualSumHalf(n-2, left + 0, right + 0, diff)
genAllBinEqualSumHalf(n-2, left + 0, right + 1, diff-1)
end if
genAllBinEqualSumHalf(n-2, left + 1, right + 0, diff + 1)
genAllBinEqualSumHalf(n-2, left + 1, right + 1, diff)
end if
End
## Example
#include <bits/stdc++.h>
using namespace std;
//left and right strings will be filled up, di will hold the difference between left and right
void genAllBinEqualSumHalf(int n, string left="", string right="", int di=0) {
if (n == 0) { //when the n is 0
if (di == 0) //if diff is 0, then concatenate left and right
cout << left + right << " ";
return;
}
if (n == 1) {//if 1 bit number is their
if (di == 0) { //when difference is 0, generate two numbers one with 0 after left, another with 1 after left, then add right
cout << left + "0" + right << " ";
cout << left + "1" + right << " ";
}
return;
}
if ((2 * abs(di) <= n)) {
if (left != ""){ //numbers will not start with 0
genAllBinEqualSumHalf(n-2, left+"0", right+"0", di);
//add 0 after left and right
genAllBinEqualSumHalf(n-2, left+"0", right+"1", di-1);
//add 0 after left, and 1 after right, so difference is 1 less
}
genAllBinEqualSumHalf(n-2, left+"1", right+"0", di+1); //add 1 after left, and 0 after right, so difference is 1 greater
genAllBinEqualSumHalf(n-2, left+"1", right+"1", di); //add 1 after left and right
}
}
main() {
int n = 5;
genAllBinEqualSumHalf(n);
}
## Output
100001
100010
101011
110011
100100
101101
101110
110101
110110
111111
Published on 31-Jul-2019 13:06:08
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http://mathhelpforum.com/geometry/10209-few-geometry-problems-print.html | # few geometry problems
• January 17th 2007, 05:41 PM
anime_mania
few geometry problems
1. given parallelogram ABCD and AB is 6, the bisector of angle A bisects BC, find BC
2. Given triangle ABC with a perimeter of 25, what is the perimeter of the smaller triangle made by the lines of the midpoints
3. how many combinations can be made if 8 people sat together on one table and three of them always stayed together
4. given a bag of 5 red marbles and 3 yellow marbles, you pick one marble out and throw it back inthe bag and pick out another marble. what is the possiblity that both marbles will be the same color. what happens to the possiblity if the marble you took out the first time was not put back into the bag.
Thanks
• January 17th 2007, 06:02 PM
ThePerfectHacker
Quote:
Originally Posted by anime_mania
1. given parallelogram ABCD and AB is 6, the bisector of angle A bisects BC, find BC
Try to visualize me.
1)Draw parallelogram ABCD.
2)Draw bisector of A and let it intersect at E.
3)Now <BAE = <EAD (because it is a bisector).
4)But <EAD = <BEA (parallel postulate, in simpler terms, alternate interiror angles).
5)Thus, <BAE = < BEA
6)Thus, triangle ABE is isoseles with AB = BE.
7)But AB=6
8)Thus BE=6
9)Thus, BC=2(BE)=2(6)=12
Quote:
2. Given triangle ABC with a perimeter of 25, what is the perimeter of the smaller triangle made by the lines of the midpoints
Half of that.
Because each length of side is half the corresponding one to the larger triangle.
• January 17th 2007, 07:40 PM
Soroban
Hello, anime_mania!
Quote:
4. Given a bag of 5 red marbles and 3 yellow marbles.
You pick one marble out and throw it back in the bag and pick out another marble.
(a) What is the probability that both marbles will be the same color?
(b) What happens to the probability if the marble you took out the first time
was not put back into the bag?
(a) With replacement
$P(RR) \:=\:\frac{5}{8}\cdot\frac{5}{8} \:=\:\frac{25}{64}\qquad\qquad P(YY) \:=\:\frac{3}{8}\cdot\frac{3}{8} \:=\:\frac{9}{64}$
Therefore: . $P(\text{same color}) \:=\:\frac{25}{64} + \frac{9}{64} \:=\:\frac{34}{64} \:=\:\frac{17}{32}$
(b) Without replacement
$P(RR) \:=\:\frac{5}{8}\cdot\frac{4}{7}\:=\:\frac{20}{56} \qquad\qquad P(YY) \:=\:\frac{3}{8}\cdot\frac{2}{7}\:=\:\frac{6}{56}$
Therefore: . $P(\text{same color}) \:=\:\frac{20}{56} + \frac{6}{56} \:=\:\frac{26}{56} \:=\:\frac{13}{28}$
• January 17th 2007, 08:52 PM
AfterShock
Quote:
Originally Posted by anime_mania
3. how many combinations can be made if 8 people sat together on one table and three of them always stayed together
Treat the 3 people sitting together as a single entity. That is, there are 6 people total (3 grouped into 1).
But the 3 can sit in any order, so there are 6 ways of seating them.
Thus the number of ways is (6P_6)/6 x 6 = 6!*6/(6*(6-6)!) = 5!*6 = 720 ways.
• January 18th 2007, 06:00 AM
Soroban
Hello, anime_mania!
Quote:
2. Given triangle ABC with a perimeter of 25, what is the perimeter
of the smaller triangle made by the lines of the midpoints?
Code:
C
*
* *
* *
D*- - - - *E
* *
* *
A * - - - -*- - - - * B
F
We are expected to know that:
. . the line joining the midpoints of two sides of a triangle
. . is parallel to and one-half the length of the third side.
Hence: . $DE \,=\,\frac{1}{2}AB$
Similarly: . $DF \,=\,\frac{1}{2}BC,\;\;EF\,=\,\frac{1}{2}AC$
Therefore: . $DE +DF + EF \:=\:\frac{1}{2}AB + \frac{1}{2}BC + \frac{1}{2}AC$
. . . . . . $=\:\frac{1}{2}(AB + BC + AC) \:=\:\frac{1}{2}(25)\:=\:12.5$
• January 18th 2007, 06:21 AM
Soroban
Hello, anime_mania!
Quote:
1. Given parallelogram $ABCD$ and $AB = 6.$.
The bisector of angle $A$ bisects $BC.$ .Find $BC.$
Here's the diagram for ThePerfectHacker's excellent solution.
Code:
B E C
*-----------------*---------------*
/ θ * /
6 / * /
/ * /
/ θ * /
/ * θ /
*---------------------------------*
A D | 2015-04-18T13:47:44 | {
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https://math.stackexchange.com/questions/2632417/solving-cos-big-tan-1x-big-sin-big-cot-1-frac34-big | # Solving $\cos\big(\tan^{-1}x\big)=\sin\big(\cot^{-1}\frac{3}{4}\big)$
Solve
$$\cos\big(\tan^{-1}x\big)=\sin\big(\cot^{-1}\frac{3}{4}\big)$$
My Attempt:
From the domain consideration, $$\boxed{0\leq\tan^{-1}\frac{3}{4},\tan^{-1}x\leq\frac{\pi}{2}}$$ $$\cos\big(\tan^{-1}x\big)=\cos\big(\frac{\pi}{2}-\cot^{-1}\frac{3}{4}\big)\implies\cos\big(\tan^{-1}x\big)=\cos\big(\tan^{-1}\frac{3}{4}\big)\\\implies\tan^{-1}x=2n\pi\pm\tan^{-1}\frac{3}{4}\\ \implies \tan^{-1}x=\tan^{-1}\frac{3}{4}\quad\text{ as }0\leq\tan^{-1}\frac{3}{4}\leq\frac{\pi}{2}\\ \implies x=\frac{3}{4}$$ Is it correct or $\frac{-3}{4}$ also is a solutions ?
What about the condition $0\leq\tan^{-1}\frac{3}{4},\tan^{-1}x\leq\frac{\pi}{2}$, does this affect the solutions ?
Well, we can use that:
$$\cos\left(\arctan\left(x\right)\right)=\frac{1}{\sqrt{1+x^2}}\tag1$$
And:
$$\sin\left(\text{arccot}\left(x\right)\right)=\frac{1}{x\cdot\sqrt{1+\frac{1}{x^2}}}\tag2$$
So in your case, we need to solve:
$$\frac{1}{\sqrt{1+x^2}}=\frac{1}{\frac{3}{4}\cdot\sqrt{1+\frac{1}{\left(\frac{3}{4}\right)^2}}}=\frac{4}{5}\space\Longleftrightarrow\space x=\pm\frac{3}{4}\tag3$$
Yes because $\cos$ is an even function.
It follows also from your way: $$\arctan{x}=\pm\arctan\frac{3}{4},$$ which gives $x=\frac{3}{4}$ or $x=-\frac{3}{4}.$
• thnx. but wht abt considering the domains ?. does it not set any restrictions ? – ss1729 Feb 2 '18 at 7:21
• The domain of $\arctan$ is $\mathbb R$. – Michael Rozenberg Feb 2 '18 at 7:28
$$\cos(\arctan x)=\cos\left(\arctan\dfrac34\right)$$
$$\implies\arctan x=2m\pi\pm\arctan\dfrac34=2m\pi+\arctan\left(\pm\dfrac34\right)$$ where $m$ is any integer as $\arctan(-a)=-\arctan(a)$
As $-\dfrac\pi2<\arctan y\le\dfrac\pi2,$
$$\arctan x= \arctan\left(\pm\dfrac34\right)$$
Apply tan on both sides | 2019-06-24T15:59:41 | {
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https://cs.stackexchange.com/questions/1444/how-many-possible-ways-are-there | # How many possible ways are there?
Suppose I have the given data set of length 11 of scores:
p=[2, 5, 1 ,2 ,4 ,1 ,6, 5, 2, 2, 1]
I want to select scores 6, 5, 5, 4, 2, 2 from the data set. How many ways are there?
For the above example answer is: 6 ways
{p[1], p[2], p[4], p[5], p[7], p[8]}
{p[10], p[2], p[4], p[5], p[7], p[8]}
{p[1], p[2], p[10], p[5], p[7], p[8]}
{p[9], p[2], p[4], p[5], p[7], p[8]}
{p[1], p[2], p[9], p[5], p[7], p[8]}
{p[10], p[2], p[9], p[5], p[7], p[8]}
How can I count the ways in general?
• Is there a computer science motivation/relation to this question? If not, this question belongs to math.SE. – Raphael Apr 22 '12 at 17:06
• Jack, @Raphael: a combinatorics formula would be a math question. I've reworded the question to ask for a counting method, which is more of a computer science question. – Gilles 'SO- stop being evil' Apr 22 '12 at 17:28
• – Raphael Apr 23 '12 at 7:00
Say you want to pick, out of a multiset $S$, the numbers $x_1$, $x_2$, ..., $x_n$ with multiplicity $m_1$, $m_2$, ..., $m_n$ (i.e., you want to pick $x_1$ exactly $m_1$ times). Furthermore, assume that in $S$ the numbers $x_1$, $x_2$, ..., $x_n$ have multiplicity $s_1$, $s_2$,..., $s_n$. (we assume for every $i$, $s_i \ge m_i$, otherwise no solution exists).
Consider the first element $x_1$: you have exactly $s_1 \choose m_1$ different ways to pick $m_1$ occurrences of $x_1$ from the $s_1$ times it appears in $S$.
In a similar way for all the other elements, the answer would be $${s_1 \choose m_1} {s_2 \choose m_2} \cdots {s_n \choose m_n} = \prod_{i=1}^n {s_i \choose m_i}$$
For your example, set $x_1=6$, $x_2=5$, $x_3=4$, $x_4=2$. In the data set they appear with multiplicity $s_1=1$, $s_2=2$, $s_3=1$, $s_4=4$ and you want to have $m_1=1$ sixes, $m_2=2$ fivess, $m_3=1$ fours and $m_3=2$ twos, so the number of different options is $${1 \choose 1} {2 \choose 2} {1 \choose 1}{4 \choose 2} = 1 \cdot 1 \cdot 1 \cdot 6$$
• Note that with the usual definition of the binomial coefficient, you don't need to assume $s_i \leq m_i$ as otherwise $\binom{s_i}{m_i} = 0$. Be aware, though; Mathematica (and possibly others) use a different generalisation. – Raphael Apr 22 '12 at 21:42 | 2020-02-23T15:06:40 | {
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http://mathhelpforum.com/discrete-math/72184-permutations-factorials.html | 1. ## permutations and factorials
1. calculate the number of ways the letters of the word OLYMPICS can be arranged if:
a) there are no restrictions
b) the arrangment begins with L and ends with P
c) the consonants are together (Y is a vowel here)
d) the O and S are not together
my thinking
a) 8!
b) 6!
c)+d) i have no idea
2. Originally Posted by william
1. calculate the number of ways the letters of the word OLYMPICS can be arranged if:
a) there are no restrictions
b) the arrangment begins with L and ends with P
c) the consonants are together (Y is a vowel here)
d) the O and S are not together
my thinking
a) 8! CORRECT
b) 6! CORRECT
c)+d) i have no idea
Three vowels and five consonants. Think of the consonants as one block that can be arranged in 5! ways.
That block along with the three vowels give four blocks to arrange.
How many ways do we get?
For none of the vowels to be together they must be separated by the consonants.
_L_M_P_C_S_ that is six places to put the vowels. ${6 \choose 3}(3!)(5!)$
Now you explain those numbers.
3. Originally Posted by william
1. calculate the number of ways the letters of the word OLYMPICS can be arranged if:
a) there are no restrictions
b) the arrangment begins with L and ends with P
c) the consonants are together (Y is a vowel here)
d) the O and S are not together
my thinking
a) 8!
Good!
b) 6!
Nice!
c)
i will give you a hint, see if you can come up with the answer.
there are 5 consonants and 3 vowels. we are not told that the vowels have to be together, so they have more freedom and can actually be apart (i'm assuming)
now, let c represent any of the consonants and v represent any of the vowels. there are 4 kinds of arrangements in which we can have the consonants together:
cccccvvv
vcccccvv
vvcccccv
vvvccccc
d)
Hint: how many arrangements can you find where they ARE together? find this, and then subtract it from your answer in (a) to get your solution (i hope you se why this works)
4. Originally Posted by Plato
Three vowels and five consonants. Think of the consonants as one block that can be arranged in 5! ways.
That block along with the three vowels give four blocks to arrange.
How many ways do we get?
For none of the vowels to be together they must be separated by the consonants.
_L_M_P_C_S_ that is six places to put the vowels. ${6 \choose 3}(3!)(5!)$
Now you explain those numbers.
Ok.. what is the notation ${6 \choose 3}$ Sorry I never learned that. Secondly, I had an idea for c)
let's say I let x=O,Y,I
then i would have the 5 consonants and 4 vowels so couldn't I conclude then that it is 5!x4!=2880
5. Hello, william;!
1. Calculate the number of ways the letters of the word OLYMPICS can be arranged if:
a) there are no restrictions
b) the arrangment begins with L and ends with P
c) the consonants are together (Y is a vowel here)
d) the O and S are not together
my thinking
a) 8!
b) 6!
Right!
c) The consonants are together.
We have 5 consonants and 3 vowels.
Duct-tape the consonants together.
. . Then we have 4 "letters" to arrange: . $\boxed{LMPCS},\,O,\,I,\,Y$
And there are $4!$ arrangements of the four "letters".
But in each arrangement, the 5 consonants can be ordered in $5!$ ways.
Therefore, there are: . $4! \times 5! \:=\:24 \times 120 \:=\:2,\!880$ arrangements
. . with the consonants together.
d) O and S are not together.
We know there are $8!$ possible arrangements.
Let's count the arrangement in which O and S are together.
Duct-tape the O and S together.
. . Then we have seven "letters" to arrange: . $\boxed{OS},\,L,\,Y,\,M,\,P,\,I,\,C$
And there are $7!$ arrangements of the seven "letters."
But in each arrangement, the $OS$ could be ordered $SO.$
Hence, there are: . $2 \times 7!$ arrangements with O and S together.
Therefore, there are: . $8! - 2\!\cdot\!7! \:=\:40,320 - 10,080 \:=\:30,240$ arrangements
. . with O and S nonadjacent.
6. Originally Posted by Jhevon
Hint: how many arrangements can you find where they ARE together? find this, and then subtract it from your answer in (a) to get your solution (i hope you se why this works)
So for the total 8!=40320 and for O,S is 2!=2? I'm not quite getting it
edit: this was created after seeing Sorobans post
7. Originally Posted by Soroban
Hello, william;!
c) The consonants are together.
We have 5 consonants and 3 vowels.
Duct-tape the consonants together.
. . Then we have 4 "letters" to arrange: . $\boxed{LMPCS},\,O,\,I,\,Y$
And there are $4!$ arrangements of the four "letters".
But in each arrangement, the 5 consonants can be ordered in $5!$ ways.
Therefore, there are: . $4! \times 5! \:=\:24 \times 120 \:=\:2,\!880$ arrangements
. . with the consonants together.
d) O and S are not together.
We know there are $8!$ possible arrangements.
Let's count the arrangement in which O and S are together.
Duct-tape the O and S together.
. . Then we have seven "letters" to arrange: . $\boxed{OS},\,L,\,Y,\,M,\,P,\,I,\,C$
And there are $7!$ arrangements of the seven "letters."
But in each arrangement, the $OS$ could be ordered $SO.$
Hence, there are: . $2 \times 7!$ arrangements with O and S together.
Therefore, there are: . $8! - 2\!\cdot\!7! \:=\:40,320 - 10,080 \:=\:30,240$ arrangements
. . with O and S nonadjacent.
Hello Soroban I appreciate all the help you have given me, although I am having trouble understanding parts of the answers.
In c) why is it not 5!*3! because there are 3 vowels and 5 consonants?
8. Originally Posted by william
Hello Soroban I appreciate all the help you have given me, although I am having trouble understanding parts of the answers.
In c) why is it not 5!*3! because there are 3 vowels and 5 consonants?
there are 4 objects that we are permuting. the 5 consonants in one block is treated as one object and then the other 3 vowels as 3 separate objects. there are 4! ways to arrange these. now, for each of those ways, there are 5! ways to arrange the 5 consonants within the block. hence 5!*4! | 2016-10-27T20:53:53 | {
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http://math.stackexchange.com/questions/18584/how-to-find-31000-bmod-7 | # How to find $3^{1000}\bmod 7$?
Sorry this is a really simple problem but I couldnt figure it out. I'm trying to find the value of $3^{1000}\bmod 7$. I know the actual value is 4, but I used a calculator. How would I simplify this problem to get the correct answer without a calculator? I dont want a direct answer, but maybe a nudge in the right direction, such as tricks I could use or maybe an alternate way of writing it.
-
Here's a hint: find the number $k$ such that $3^k = 1$. – Eric O. Korman Jan 23 '11 at 0:58
@Eric wouldnt that just be 0? – maq Jan 23 '11 at 1:09
that's true, let me rephrase: find a positive number $k$ such that $3^k = 1$ (mod 7). – Eric O. Korman Jan 23 '11 at 1:34
If you know Fermat's Little Theorem, then you know you can reduce the exponent. If you don't, you can simply do a bit of testing: \begin{align*} 3 &\equiv 3 &\pmod{7}\\ 3^2 &\equiv 2&\pmod{7}\\ 3^3 &\equiv 2\times 3&\pmod{7}\\ &\equiv -1 &\pmod{7}. \end{align*} At this point, you will also know that $3^6 = (3^3)(3^3)\equiv (-1)^2 \equiv 1 \pmod{7}$.
That means that the remainders will "cycle" every seven steps, since $3^7 = 3(3^6) \equiv 3(1) = 3 \pmod{7}$.
So now, notice that since $1000 = 6(166) + 4$, then $$3^{1000} = 3^{6(166)+4} = (3^6)^{166}3^4 = (3^6)^{166}3^33^1.$$ You know what each of the factors is modulo $7$, so you're done.
-
Once again, I'll ask for future reference: Is there a particular reason for the downvote? – Arturo Magidin Jan 23 '11 at 2:06
dunno about the down vote. But in the last tep, since $3^6 = 1$ wouldn't it be better to factor $1000 = 4 mod 6$? – Willie Wong Jan 23 '11 at 3:00
@Willie: Oops; quite right. That was silly of me. – Arturo Magidin Jan 23 '11 at 4:58
Just got around to actually read this answer in detail, and I found myself very confused :( I thin you used a lot of tricks that I have no idea how you used them, like $(3^3)(3^3)$ is congruent to $(-1)^2$..but I guess this is a complicated problem anyway – maq Jan 24 '11 at 6:15
@mohabitar: It's not really complicated. The "trick", as you call it, is just that you can add and multiply congruences: if $a\equiv b$ and $c\equiv d$, then $a+c\equiv b+d$ and $ac\equiv bd$ (if they are modulo the same thing). Since $3^3\equiv -1$, which we figure out just be computing, then $(3^3)(3^3)\equiv (-1)(-1)$. – Arturo Magidin Jan 24 '11 at 14:04
Try to split the huge number in smaller numbers you know the value of. For example
$3^{1000} \text{mod} 7= (3^{10})^{100} \text{mod} 7= (59049)^{100} \text{mod} 7= 4^{100} \text{mod} 7 = \ldots$
If $3^{10}=59049$ was still too big you could try to rewrite it again etc.
-
How much more could I simplify $3^{10}$ though? – maq Jan 23 '11 at 1:08
$3^{10}\equiv 3^3\cdot 3^3\cdot 3^3 \cdot 3\equiv 6\cdot 6 \cdot 6 \cdot 3 \equiv 36 \cdot 18 \equiv 1 \cdot 4 \equiv 4 (\text{mod} 7)$ for example. – Listing Jan 23 '11 at 1:13
Since that 36mod7=1, then we have 31000=36*166+4=34=4 mod7.
-
$3^{1000}$ is hard to compute by hand because of the $1000$. Can you learn anything from trying smaller exponents instead?
-
Well even I know that $3^{10}mod7$ is also 4, but I used a calculator for that too since $3^{10}$ is still a large number that I couldnt compute myself. – maq Jan 23 '11 at 1:07
@mohabitar: try $3^k\mod 7$ for $k=1,2,3,...$ and see what happens. – Isaac Jan 23 '11 at 1:29 | 2014-08-22T00:37:14 | {
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https://math.stackexchange.com/questions/2239220/points-where-a-line-intersects-the-function-x-1-frac-1-3 | Points where a line intersects the function $(x-1)^{\frac 1 3}$
I'm trying to do the following exercise:
Find the equation of the line tangent to the graph of $f(x)= (x-1)^{1/3}$ at the point $x=2$. Find also if that line intersects the graph of f(x) at any other point. If it does, then find its coordinates.
Well, I begin by differentiating f(x) to get the slope:
$f(x) = (x-1)^{1/3}$
$f'(x) = (1/3)(x-1)^{-2/3}$
$f'(2) = (1/3)(2-1)^{-2/3}$
$f'(2) = (1/3)1^{-2/3}$
$f'(2) = (1/3)1$
$f'(2) = 1/3$
So the equation of the tangent line at 2 should be $y = (1/3)x$ + constant
To find the constant I use the value of f(2) which is 1.
$f(2) = (2-1)^{1/3}$
$f(2) = 1^{1/3}$
$f(2) = 1$
$f(2) =$ tangent line at 2
$1 = (1/3)2 + constant$
$1 = 2/3 + constant$
$1 - 2/3 = constant$
$1/3 = constant$
Therefore the equation of the tangent line at 2 is $y=(1/3)(x+1)$
Now I try to find out whether there are other points where this line intersects the function.
$(x-1)^{1/3} = (1/3)(x+1)$
$(x-1) = (1/27)(x+1)^3$
$\frac{x-1}{(x+1)^3} = (1/27)$
And here I have no idea of how to continue.
So I cheated a little bit by graphing the function (black curve) and its tangent at 2 (red line):
Looking at the graph, there seems to be another point of intersection at $x=-7$.
Plugging in -7, I get:
$\frac{-7-1}{(-7+1)^3} = (1/27)$
$\frac{-8}{(-6)^3} = (1/27)$
$\frac{-8}{-216} = (1/27)$
$\frac{8}{216} = (1/27)$
This seems to confirm that at $x=-7$ the line intersects the function again. But suppose I wanted to find the point by hand, without graphing and 'guessing' that the answer is -7. How do I do that?
• Just a constructive point for your next question, you can omit some of the trivial lines so that your question isn't un-necessarily long. – mrnovice Apr 17 '17 at 22:59
Your
$$\frac{x-1}{(x+1)^3} = \frac{1}{27}$$
corresponds to
$$x^3+3x^2-24x+28=0$$
but since this comes from a tangent, you know that two of the solutions are $x=2$, so you can factor out $(x-2)^2$ to give
$$(x-2)^2(x+7)=0$$
so the other solution is $x=-7$
• Could you please elaborate a little more on this? Why are two of the solutions at $x=2$? I don't really understand the idea. – IMK Apr 17 '17 at 23:17
• A cubic equation has three solutions, though some might be duplicated and some might be complex. Your expression finds the $x$ coordinates of points where the curve and the tangent meet. If they cross simply, then there would be a solution at that point (as there is with $x=-7$). So you know that $(x-2)$ is a factor. But in fact this is a tangent, which you might informally regard as crossing and crossing back at the same point, leading to two solutions (at least) at that point – Henry Apr 18 '17 at 0:02
HINT: $$\frac{x-1}{(x+1)^3}=\frac{1}{27}$$ can be written as $$f(x)=x^3+3x^2-24x+28$$ We already know that $x-2$ is a factor . Thus we re-write it as: $$f(x)=(x-2)(x^2+5x-14)$$ $$f(x)=(x-2)^2(x-7)$$ Thus it intersects again @ $x=7$ | 2020-02-29T07:32:07 | {
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https://math.stackexchange.com/questions/3333627/find-the-greatest-common-divisor-of-2m1-and-2n1-that-m-n-are-positive | # Find the greatest common divisor of $2^m+1$ and $2^n+1$ that $m,n$ are positive integers.
I am confused of a question that needs to know the greatest common divisor of $$2^m+1$$ and $$2^n+1$$ ($$m,n$$ are positive integers), but I don't really know. I am pretty sure that the greatest common divisor of $$2^m-1$$ and $$2^n-1$$ ($$m,n$$ are positive integers) is $$2^{\gcd\left(m,n\right)}-1$$, even I can prove it by the Euclidean algorithm. However, it is hard to use it in this problem, so I want you guys to help me. Thanks!
P.S.
I created an excel and I observed the answer (maybe?) from it, but I can't prove or disprove it. Here is my conclusion from the excel: $$\gcd\left(2^m+1,2^n+1\right)=\begin{cases} 2^{\gcd\left(m,n\right)}+1 \\ 1 \end{cases}\begin{matrix} \text{when }m,n\text{ contain the exact same power of }2 \\ \text{otherwise} \end{matrix}$$ Hope it will help me and you guys solving this quesion :D
• I don't think the answer can be written as a closed form in terms of $m,n$ – Peter Foreman Aug 25 at 10:38
• It is a factor of $2^{2n}-1$. I don't know if that helps. – Empy2 Aug 25 at 10:58
• It would be great if someone could run a program to find the GCD for small values of $n$ and $m$ to see if there’s a pattern. I feel like if there indeed exists a closed-form formula, it could be proven using induction – Borna Ahmadzade Aug 25 at 13:05
• From your formulation it is unclear if this is the complete question posed somewhere else, or just a question that "would be nice to know the general answer to" while you solved some related problem. For example, when both $m,n$ are odd, the answer is $2^{\gcd(m,n)}+1$. So any additional conditions you might have on $m$ and $n$ would be nice to know. – Ingix Aug 25 at 13:06
• What we can say that a common divisor must divide $2^{2m}-1$ and $2^{2n}-1$, hence must divide $2^{\gcd(2m,2n)}-1$ , hence the greatest common divisor divides $2^{\gcd(2m,2n)}-1$, but I do not think that we can achieve a better result in general. – Peter Aug 26 at 7:33
This started as a partial solution, trying to bundle up what's been said in the comments and a bit more. After some more comments (esp. from Empy2) it is now a complete solution.
Proposition 1 gives an upper bound for the gcd. Proposition 2 then shows that this upper bound is actually assumed under certain conditions on $$m,n$$. Proposition 3 then shows that if those conditions are not fullfilled, the gcd is $$1$$.
Proposition 1:
$$\gcd(2^{m}+1,2^{n}+1) | 2^{\gcd(m,n)}+1.$$
Proof:
Let $$d$$ be a common divisior of $$2^m+1$$ and $$2^n+1$$.
We have $$2^m+1|2^{2m}-1$$ and $$2^n+1|2^{2n}-1$$, so it follows that $$d|\gcd(2^{2m}-1,2^{2n}-1)$$ and we know that $$\gcd(2^{2m}-1,2^{2n}-1) = 2^{\gcd(2m,2n)}-1 = 2^{2\gcd(m,n)}-1 = (2^{\gcd(m,n)}-1)(2^{\gcd(m,n)}+1),$$
so
$$d|(2^{\gcd(m,n)}-1)(2^{\gcd(m,n)}+1). \tag{1} \label{eq1}$$
Let $$p$$ be a prime divisor of $$2^{\gcd(m,n)}-1$$. That means
$$2^{\gcd(m,n)} \equiv 1 \pmod p$$
and if we raise each side to the $$\frac{m}{\gcd(m,n)}$$-th power, we obtain
$$2^m \equiv 1 \pmod p \Longrightarrow 2^m+1 \equiv 2 \pmod p$$
Because $$m > 0$$, $$2^m+1$$ is odd, so $$p \neq 2$$ and hence $$2^m+1 \neq 0 \pmod p$$.
That means no prime divisor of $$2^{\gcd(m,n)}-1$$ can be a divisor of $$2^m+1$$, so $$d$$ and $$2^{\gcd(m,n)}-1$$ are coprime and we get from \eqref{eq1} that
$$d|2^{\gcd(m,n)}+1$$
and Proposition 1 follows.
Proposition 2: When $$m$$ and $$n$$ contain the exact same power of $$2$$:
$$m=2^km', n=2^kn';\quad m'\equiv n'\equiv1 \pmod 2,$$
then
$$\gcd(2^{m}+1,2^{n}+1) = 2^{\gcd(m,n)}+1.$$
Proof:
In this case we also set $$m'=\gcd(m',n')m''$$ and $$n'=\gcd(m',n')n''$$ and find
$$2^m+1=2^{2^km''\gcd(m',n')}+1=\left(2^{2^k\gcd(m',n')}\right)^{m''}+1$$
and the equivalent for $$n$$:
$$2^n+1=2^{2^kn''\gcd(m',n')}+1=\left(2^{2^k\gcd(m',n')}\right)^{n''}+1.$$
Since $$m''$$ and $$n''$$ are odd, that means that $$2^{2^k\gcd(m',n')} +1$$ divides both terms (as per $$(a+b)|(a^r+b^r)$$ for any odd $$r$$).
Since $$2^k\gcd(m',n') = \gcd(m,n)$$, this proves Proposition 2.
The hard case seems to be when $$m$$ and $$n$$ contain different powers of $$2$$. I see no good way to attack that question in a general way, but maybe others do.
ADDED: It turns out that the comment by Empy2 below actually solves that problem, it just took me a while to realize that.
Proposition 3:
Let $$m=\gcd(m,n)m'$$ and $$n=\gcd(m,n)n'$$. If $$m'$$ is even and $$n'$$ is odd, then
$$\gcd(2^m+1,2^n+1)=1.$$
Proof: The conditions on $$m'$$ and $$n'$$ are equivalent to $$m$$ and $$n$$ containing different powers of $$2$$, where I assumed w.l.o.g. that $$m$$ was the one containing the higher power of $$2$$.
We have $${\rm{lcm}}(m,n)=\gcd(m,n)m'n'$$ so
$$2^{{\rm lcm}(m,n)}+1=2^{\gcd(m,n)m'n'}+1 =\left(2^{\gcd(m,n)m'}\right)^{n'}+1 = \left(2^{m}\right)^{n'}+1.$$
Since $$n'$$ is odd, we find that
$$2^m+1|\left(2^{m}\right)^{n'}+1 = 2^{{\rm lcm}(m,n)}+1.$$
Doing the same for $$n$$ we get
$$2^{{\rm lcm}(m,n)}+1=2^{\gcd(m,n)m'n'}+1 =\left(2^{\gcd(m,n)n'}\right)^{m'}+1 = \left(2^{n}\right)^{m'}+1.$$
We finally have $$2^n+1|(2^n)^2-1|(2^n)^{m'}-1=2^{{\rm lcm}(m,n)}-1,$$ where the second divisibility follows because $$m'$$ is a multiple of $$2$$ (it was even).
So, as Empy 2 said, we have
$$2^m+1| 2^{{\rm lcm}(m,n)}+1,$$ $$2^n+1| 2^{{\rm lcm}(m,n)}-1,$$
so any common divisor of $$2^m+1$$ and $$2^n+1$$ must be a divisor of $$2$$. Since $$m,n$$ were both assumed to be positive, only $$1$$ can be a such common divisor.
• Though it is not finished, it is great to have some process in this question! At least there is one case that can be proven is good for me actually. – Isaac YIU Math Studio Aug 26 at 10:44
• If they have different powers of two, then one is a factor of $2^{lcm}+1$ and the other a factor of $2^{lcm}-1$ – Empy2 Aug 26 at 10:59
• I make an edit of my post and I found that it may be 1 if m and n contain different powers of 2 – Isaac YIU Math Studio Aug 26 at 12:39
• @Empy2 Thanks for your comment, that solved the remaning part from my approach (which I added in). Also, please look at the solution by W-t-P, who uses a different approach! – Ingix Aug 27 at 9:03
Your conjectured formula is correct; here is the proof.
For integer $$m,n\ge 0$$, let $$d(m,n):=\gcd(2^m+1,2^n+1)$$. Assuming for definiteness $$m\ge n$$, we have \begin{align*} d(m,n) &= \gcd(2^m-2^n,2^n+1) \\ &= \gcd(2^n(2^{m-n}-1),2^n+1) \\ &= \gcd(2^{m-n}-1,2^n+1) \\ &= \gcd(2^{m-n}+2^n,2^n+1). \end{align*} If $$m\ge 2n$$, then this can be taken a little further, by factoring out $$2^n$$, to get $$d(m,n) = \gcd(2^{m-2n}+1,2^n+1);$$ if $$m\le 2n$$, then factoring out $$2^{m-n}$$ instead of $$2^n$$ we get $$d(m,n) = \gcd(2^{2n-m}+1,2^n+1).$$ In any case, we have the recursive relation $$d(m,n) = d(|m-2n|,n),\quad m\ge n. \tag{\ast}$$
Let $$\nu(k)$$ denote the $$2$$-adic valuation of an integer $$k\ne 0$$; that is, $$\nu(k)$$ is the largest integer such that $$2^{\nu(k)}$$ divides $$k$$. I claim that
(1) If $$m>n>0$$, then $$\max\{|m-2n|,n\}<\max\{m,n\}$$;
(2) if $$m>0$$ or $$n>0$$, then $$\gcd(|m-2n|,n)=\gcd(m,n)$$;
(3) if $$m\ne 2n$$, then $$\nu(m)=\nu(n)$$ if and only if $$\nu(m-2n)=\nu(n)$$.
The first two assertions are easy to verify. For the last one, let $$k:=\nu(n)$$ and $$l:=\nu(m)$$ and consider two cases:
If $$k>l$$ then $$2^{l+1}\nmid m-2n$$ while $$2^{l+1}\mid n$$, whence $$\nu(n)\ne\nu(m-2n)$$, as wanted.
If $$k then $$2^{k+1}\mid m-2n$$ while $$2^{k+1}\nmid n$$, implying $$\nu(n)\ne\nu(m-2n)$$ in this case, too.
To complete the proof, we use straightforward induction by $$m=\max\{m,n\}$$ distinguishing the following cases: $$n=0$$, $$m=n$$, $$m=2n$$, and the "general case" where none of these holds.
• There are a lot of small mistakes in this proof but it doesn't affect the result. Please correct it – Isaac YIU Math Studio Aug 26 at 16:31
• @IsaacYIUMathStudio: There were a couple of small typos which, hopefully, are now fixed. I have also modified the proof to cover the case where $m$ and $n$ have the same $2$-adic valuation. – W-t-P Aug 26 at 19:00
• @W-t-P Good job: +1. It took me a while to understand that your reduction scheme terminates (produces a trivial identity) either when both arguments are the same or one argument becomes $0$. The former happens for $\nu(m)=\nu(n)$, the latter for $\nu(m)\neq \nu(n)$. – Ingix Aug 26 at 22:02 | 2019-12-06T16:06:46 | {
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https://math.stackexchange.com/questions/2054386/find-the-radius-of-convergence-given-a-recurrence-relation | # Find the radius of convergence given a recurrence relation.
I have the following recurrence relation for the coefficients $c_n$ of the power series: $y = \sum^\infty_{n=0}c_nx^n$
$$c_{n+1} = \frac{n+2}{5(n+1)}c_n$$
What is the radius of convergence of the power series?
I can find the radius of convergence by finding a formula for $c_n$ and then find the limit of $|\frac{c_n}{c_{n+1}}|$ as $n$ approaches $\infty$.
Is there a way to determine its radius of convergence without finding the formula first?
• Wait... you know that $$\frac{c_n}{c_{n+1}}=5\frac{n+1}{n+2}$$ hence finding the limit $$\lim\left|\frac{c_n}{c_{n+1}}\right|$$ should not be too computationally costly, should it? :-) – Did Dec 11 '16 at 20:33
You get $$\frac{5^{n+1}}{n+2}c_{n+1}=\frac{5^n}{n+1}c_n=…=c_0$$ so that you can get the explicit form of the power series, and from that read directly off its radius of convergence.
You can of course also directly evaluate the quotient expression $$\frac{c_n}{c_{n+1}}=\frac{5(n+1)}{n+2}.$$
This may be a sledgehammer for your problem but there is a general theorem.
Poincare theorem$\color{blue}{{}^{[1]}}$
Given any recurrence relations with non-constant coefficients $$x_{n+k} + p_1(n)x_{n+k-1} + p_2(n)x_{n+k-2} + \cdots + p_k(n) x_{n} = 0\tag{*1}$$ such that there are real numbers $p_i, 1 \le i \le k$ with $$\lim_{n\to\infty} p_i(n) = p_i, \quad 1 \le i \le k$$ and the roots $\lambda_1, \lambda_2 \ldots, \lambda_k$ for the associated characteristic equation: $$\lambda^k + p_1 \lambda^{k-1} + \cdots + p_k = 0$$ have distinct moduli.
For any solution of $(*1)$, either $x_n = 0$ for all large $n$ or $\displaystyle\;\lim_{n\to\infty} \frac{x_{n+1}}{x_n} = \lambda_i\;$ for some $i$.
In short, if the coefficients of a recurrence relations converges and the corresponds $|\lambda_i|$ are distinct, then either the sequence $x_n$ terminates (i.e. infinite radius of convergence) or the radius of convergence is one of $\frac{1}{|\lambda_i|}$.
For your case, it is clear $c_n$ didn't terminate. Since the characteristic equation of your sequence "converge" to $\lambda - \frac15$, its radius of convergence is $5$.
Notes/References
• $\color{blue}{[1]}$ - Saber Elaydi, An Introduction to difference equations, $\S 8.2$ Poincare theorem. | 2019-09-22T02:04:52 | {
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https://math.stackexchange.com/questions/2764071/how-many-strings-of-five-decimal-digits-contain-exactly-three-distinct-digits | How many strings of five decimal digits contain exactly three distinct digits?
Here is my thought process so far:
There are 10 options for the first number, 9 for the second, and 8 for the third. According to this current structure, the fourth and fifth digits would have to be one or two of the three already selected digits in some order. If the first three digits were 1, 2, 3, then the possible permutations for the fourth and fifth spots are: 11, 22, 33, 12, 21, 13, 31, 23, and 32.
So the answer so far would be (10 x 9 x 8) x (9)
Therefore: Per three distinct digits, there are already 9 options for a string. Additionally, the fourth and fifth spots which contain 1 or more of the distinct digits for the first repeated occurrence don't have to be at the end. I thought that I should multiply the present answer by (5 choose 3) = 10 to account for this additional variation in order, but I can tell that there must be some number of strings which are counted twice by this method, so
(10 x 9 x 8) x (9) x (10) = 64800 is probably too great. Any suggestions? Thank you
• Hint: break it into two cases. The case where you have a number that has digits $\{a,b,c,c,c\}$ and the case where the digits are $\{a,b,b,c,c\}$ – Joe May 2 '18 at 22:59
Instead of thinking about the number of ways to place the digits (which gets quite complicated), think about it as a three step procedure:
1. How many ways are there to write a 5 digit number with 3 distinct characters?
2. How many ways can one choose three distinct characters to be placed into the forms from above.
3. Watch out for any duplication or overcounting.
There are two options (using Joe's suggestion in the comments):
• $\{a,b,c,c,c\}$. We can choose $3$ of the $5$ positions for the $c$'s, then there are two options for the $a$ and $b$ ($a$ first and then $b$ or the other way around). This gives $$2\binom{5}{3}=20$$ ways to write $a$, $b$, and $c$.
• $\{a,b,b,c,c\}$. We can choose $1$ of the $5$ positions for the $a$, and then $2$ of the remaining $4$ positions for the $b$'s. This gives $$5\binom{4}{2}=30$$ ways to write $a$, $b$, and $c$.
Now, there are $10\cdot 9\cdot 8=720$ ways to choose three digits for the positions $a$, $b$, and $c$. So, one might expect $$720\cdot (20+30)=36000$$ arrangements. The problem with this is that there is some overcounting since $abbcc$ and $accbb$ result in the same arrangement when the numbers in $b$ and $c$ are also flipped. This occurs for $a\leftrightarrow b$ in the first option and $b\leftrightarrow c$ in the second form. Therefore, we've overcounted by a factor of $2$ and there are $18000$ total arrangements. (Note that $720\cdot 30/2=10800$ matching @Green's approach).
• I believe the answer to 2. is 10 x 9 x 8 = 720 ways to choose the 3 distinct digits. But number 1. perplexes me... – mike May 2 '18 at 23:08
• @whatafeynman Yes, the answer to 2 is 720 (but we must be careful about double counting by "double switching") See the edits above. – Michael Burr May 2 '18 at 23:14
• Thank you very much! Between your answer, that of Joe, and that of Green, I think I have a much better understanding of the question. – mike May 2 '18 at 23:17
I think you're on the right track, but I wouldn't recommend fixing the first 3 spots to be distinct numbers as you do. For example, the number $11231$ satisfies the condition of being a 5-digit number with exactly $3$ different digits, but doesn't have the first $3$ digits distinct.
I noticed that you have $10 * 9 * 8 = 720$ ways to pick your first $3$ digits to use in order. Instead, I would suggest that you just find the total number of combinations of $3$ digits you could use in making your string, which is $10$ choose $3$, or $120$. Given this, the next step would be to split the problem into cases. As you seem to have noticed, you can either have $2$ of the digits be used twice, or $1$ of the digits be used thrice.
Let's consider case $1$, which is that you have $2$ digits used twice. There's $3$ ways to choose which of the $3$ digits to use twice, and then a total of $\frac{5!}{2! * 2!} = 30$ ways to arrange the digits afterwards, giving us a total of $120 * 3 * 30 = 10800$ possible arrangements.
Now, try to calculate the other case by yourself and see if you're able to arrive at the answer.
That's an obscene amount of double counting.
First you got three digits (in you example $123$) and you figured there were $9*8*7$ ways to get these. Those coming up with $123$ and $312$ etc. are different results. Then you multiplied by $9$ for the last to digits. But then you allowed for scrambling them all up again. So that $12312$ and $12321$ each can scramble to the same results. As can $31212$ and $31221$. The we I see it you counted the number $31212$ $12$ times!
Better: There are ${10 \choose 3}$ choices of the three distinct $3$ digits. Of those distinct $3$ digits there are two ways to have repeats: You may have $1$ digit repeated three times (and there are ${3 \choose 1}=3$ way of picking it); or you may have $2$ digits each repeated twice. And there are $({3\choose 2}=3$ ways of picking those.
If you have one digit repeated three times you have $5*4$ spots to place the two distinct digits. If you have two digits repeated twice. You have $5$ options to place the distinct digit and $4\choose 2$ options to place the first pair of repeated digits.
So there are ${10\choose 3}({3 \choose 1}5*4 + {3\choose 2}5*{4\choose 2})=$
$\frac {10!}{7!3!}(60 + 90) = \frac {10*9*8}6(150)= 10*9*8*25=18,000$
Let us define $S(n,d)$ as the number of different strings of $n$ characters containing $k$ distinct characters (from a pool of $u$ possible characters). Then in general there are two ways to form a string in $S(n,d)$:
• add an already-used character to a string in $S(n{-}1,d)$ - with $d$ options
• add a novel character to a string in $S(n{-}1,d{-}1)$ - with $u-(d{-}1)$ options
This gives us that $S(n,d) = d\cdot S(n{-}1,d) + (u-(d{-}1))\cdot S(n{-}1,d{-}1)$ which here with $u=10$ is $S(n,d) = d\cdot S(n{-}1,d) + (11-d)\cdot S(n{-}1,d{-}1)$
Then we just need to frame this with $S(0,0)=1$ and otherwise $S(n,0)=S(0,d)=0$, and we can calculate a table of results:
\begin{array}{l|rr} d \ \backslash \ n\! & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 10 & 10 & 10 & 10 & 10 \\ 2 & 0 & 0 & 90 & 270 & 630 & 1350 \\ 3 & 0 & 0 & 0 & 720 & 4320 & \color{red}{18000} \\ 4 & 0 & 0 & 0 & 0 & 5040 & 50400 \\ 5 & 0 & 0 & 0 & 0 & 0 & 30240 \\ \end{array}
If you just want the answer, there's nothing wrong with computing it:
\$ python
>>> import itertools
>>> sum(len(set(x))==3 for x in itertools.product(range(10),repeat=5))
18000
• If this is, in fact, the correct answer, then I can use it to be certain that all my attempts so far are incorrect! So, thank you! Unfortunately, I am hoping to develop an understanding of how to find this value by hand. – mike May 2 '18 at 23:03 | 2019-12-16T11:08:21 | {
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https://math.stackexchange.com/questions/1598269/problem-based-on-theorem-of-radius-arc-length-and-central-angle | # Problem Based on Theorem of radius, arc length and central angle
An arc AB of a circle with radius 28 cm and center O subtends an angle AOB at the centre. If the length of arc AB is $$\frac{88}{3}$$ cm, find the length of chord AB.
I haven't solve any question of this kind so no idea.
• HINT: Use the arc length formula (i.e. $r\theta=l$ where $\theta$ is measured in radians) to calculate the angle AOB. You should then be able to calculate the length of the chord AB. – Mufasa Jan 3 '16 at 12:56
• @Musafa I got AB=28 cm but the answer is 5 cm as shown by my book. – Ger Wyn Jan 3 '16 at 13:03
• Please show your calculations so that we can help spot where you may have made a mistake. – Mufasa Jan 3 '16 at 13:04
• $$\ theta$$=l/r =(88/3)/28=1.047 radians and converting it into degree, its approximately equal to 60°… And then using cosine law gives AB =28 cm – Ger Wyn Jan 3 '16 at 13:08
• Your answer is therefore correct (approx $28.01$cm) and the book is incorrect – Mufasa Jan 3 '16 at 13:11
The radian measure $\theta$ of a central angle of a circle with radius $r$ that is subtended by an arc of length $s$ is defined to be $$\theta = \frac{s}{r}$$
Since you were given $s$ and $r$, you can solve for $\theta$.
You wish to find the length of chord $\overline{AB}$.
Draw altitude $\overline{CD}$ to side $\overline{AB}$ of $\triangle ABC$, as shown in the diagram. Since $|AC| = |BC| = r$, $\overline{AC} \cong \overline{BC}$. $\overline{CD} \cong \overline{CD}$ by the reflexive property of congruence. Hence, $\triangle ACD \cong \triangle BCD$ by the Hypotenuse-Leg Theorem. Since corresponding angles of congruent triangles are congruent, $\angle ACD \cong \angle BCD$, so $\overline{CD}$ bisects $\angle ACB$. Hence, $m\angle ACD = \frac{\theta}{2}$. Thus, $$|CD| = r\cos\left(\frac{\theta}{2}\right)$$ and $$|AD| = r\sin\left(\frac{\theta}{2}\right)$$ Since corresponding sides of congruent triangles are congruent, $\overline{AD} \cong \overline{BD}$.
Since $|AB| = |AD| + |BD|$, we obtain $$|AB| = 2r\sin\left(\frac{\theta}{2}\right)$$
Hint use $S=r\theta$ where $S=arc length$ and $\theta$ is central angle in radians. After that drop a perpendicular from central abgle to chord AB then caculate AB by using trigonometric ratios.
• Can it not be donee without dropping the perpendicular @ Archis – Ger Wyn Jan 3 '16 at 13:13 | 2019-08-18T05:03:06 | {
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https://stats.stackexchange.com/questions/396055/estimate-lambda-in-the-fitting-line-x-lambda-where-x-in-0-1/396084 | # Estimate $\lambda$ in the fitting line $x^\lambda$, where $x \in [0, 1]$
## Problem
I would like to estimate $$\lambda$$ in the fitted line $$x^\lambda$$, where $$x \in [0, 1]$$.
Note that the following R code generates "concave" growth as x increases from $$0$$ to $$1$$.
Lambda = 1/2.42
x = rbeta(1e4, shape1=2,shape2=2)
y = x^Lambda + rnorm(1e4, sd=.1)
plot(x,y)
## Try
My approach is to build a loss function, the L2 loss to find the optimal $$\lambda$$, like the following R code.
SumSq = function(lam) sum((y - x^lam )^2)
optimize(SumSq, c(0,1), tol=1e-4, maximum = F )
This code assumes I know $$\lambda \in (0,1)$$, and optimize gives me a quite accurate result.
But this approach cannot give any statistical asymptotic results such as CI, which is useful information if I would like to consider the uncertainty.
Is there any standard way to do it?
• i really dont know why someone would down vote and not leave a comment. This is a reasonable question with a minimally reproducible example. – forecaster Mar 7 at 0:48
• @forecaster I agree and found your answer to this question to be edifying, so I am similarly puzzled. – James Phillips Mar 7 at 2:52
• The natural approach to estimations with this problem would involve calling a nonlinear least squares routine (nls in R - there are a number of somewhat more robust alternatves, but this is usually sufficient). – Glen_b -Reinstate Monica Mar 7 at 5:25
• When the model is expressed in the form $Y= \exp(\lambda \log x) + \varepsilon,$ this question is seen to be the same as many other questions asked (and answered) on this site about nonlinear least-squares fitting: this gives you an additional set of resources to consult. – whuber Mar 7 at 16:29
• "But this approach cannot give any statistical asymptotic results such as CI, which is useful information if I would like to consider the uncertainty." Is this the core of the question? This is not so clear from the title which seems to point to point estimates (pun intended). – Sextus Empiricus Mar 7 at 19:43
Firstly I would suggest you read an excellent book by Ben Bolker entitled Ecological Models and Data in R. Written for ecologists , I think this is one of the best books on practical data analysis regardless of background, I'm an engineer and I have used it a lot. I'm sure there are other mathematical statistics book, however this book is by far the most practical book that I have read. He also has a package BBMLE which you might want to check. The book unlike any other goes in to maximum likelihood estimation, profile likelihood and confidence interval estimation and all that.
Coming back to your problem, you need to write a log-likelihood of the function that you are trying to fit the data. Obviously you are assuming the error to be normally distributed as in the simulation, so the Log likelihood is:
$$Log\ Likelihood (Lambda,sd) = -\frac{n}{2} log(sd) - \frac{n}{2} log(2\pi)- \frac{(y-x^{Lambda})^2}{2sd}$$
You need to maximize the above equation using an optimization routine such as optim in R. Use the Hessian matrix from the optimization to assess the uncertainty of your estimated log likelihood function i.e., how steep or how flat the curvature of your function is at the optimal point. If the function is steep which implies less uncertainties at optimal point you would have a tighter confidence band on your parameter estimates, on the other hand if its flat you would have a wider confidence interval. Hessian, Fisher Information matrix would help you calculate the standard error and confidence interval.
Here is how you do it in R:
set.seed(8345)
Lambda = 1/2.42
x = rbeta(1e4, shape1=2,shape2=2)
y = x^Lambda + rnorm(1e4,mean = 0, sd=.1)
plot(x,y)
## Write Log Likelihood function
log.lik <- function(theta,y,x){
Lam <- theta[1]
sigma2 <- theta[2]
# sample size
n <- length(y)
#error
e<-y-(x^Lam)
#log likelihood
logl<- -.5*n*log(2*pi)-.5*n*log(sigma2)-((t(e)%*%e)/(2*sigma2))
return(-logl) # R optim does minimize so to maximize multiply by -1
}
## Estimate Paramters thru maximum likelihood
max.lik <- optim(c(1,1), fn=log.lik, method = "L-BFGS-B", lower = c(0.00001,0.00001), hessian = T,y=y,x=x)
# Lambda
Lam <- max.lik$par[1] #0.4107119 #Fisher Information MAtrix fisher_info<-solve(max.lik$hessian)
prop_sigma<-sqrt(diag(fisher_info))
## Estimate 95% Confidence Interval
upper<-max.lik$$par+1.96*prop_sigma lower<-max.lik$$par-1.96*prop_sigma
interval<-data.frame(Parameter = c("Lambda","sd"),value=max.lik$par, lower=lower, upper=upper) interval • This is a generalized linear model so it seems more obvious to use the glm function and avoid a general non-linear optimizer with numerical differentiation. – Benjamin Christoffersen Mar 7 at 21:10 Is there any standard way to do it? If you think that following is a good approximation (the model you simulate from as an example) $$y_i = x_i^\lambda +\epsilon_i, \qquad \epsilon_i\sim N(0,\sigma^2)$$ i.e., $$\log(E(y_i))=\lambda \log x_i$$ then you can use glm with family = gaussian("log") which is exactly this model lambda <- 1/2.42 set.seed(49564503) n <- 1e4 x <- rbeta(n, shape1 = 2,shape2 = 2) y <- x^lambda + rnorm(n, sd = .1) fit <- glm(y ~ log(x) - 1, family = gaussian("log"), start = c(0, 1)) summary(fit) #R #R Call: #R glm(formula = y ~ log(x) - 1, family = gaussian("log"), start = 1) #R #R Deviance Residuals: #R Min 1Q Median 3Q Max #R -0.42452 -0.06767 0.00090 0.07020 0.34418 #R #R Coefficients: #R Estimate Std. Error t value Pr(>|t|) #R log(x) 0.410666 0.001807 227.3 <2e-16 *** #R --- #R Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 #R #R (Dispersion parameter for gaussian family taken to be 0.01033428) #R #R Null deviance: 1057.81 on 10000 degrees of freedom #R Residual deviance: 103.33 on 9999 degrees of freedom #R AIC: -17341 #R #R Number of Fisher Scoring iterations: 5 Notice that both the dispersion parameter and coefficient estimate match as expected. Now, you can make confidence interval with the standard error above or by using confint. • +1 There seem to be a - 1 missing in the model formula although it's there in the output from summary(fit). – Jarle Tufto Mar 7 at 21:51 • Thanks. I forgot to add that. I would prefer to keep the intercept but then the model would be more general than the model the OP simulates from. – Benjamin Christoffersen Mar 7 at 22:17 • Your analysis confuses the model$E[\log y] = \lambda \log x$with the different model$\log(E[y]) = \lambda \log x.$When the variance of$\epsilon$is much smaller than$x^{2\lambda}$it's probably an OK approximation, but otherwise it's important to pay attention to the distinction--and to offer a different solution. – whuber Mar 8 at 0:28 • @whuber doesn't the log link function express the following? $$\log(E[y]) = X\beta$$ Then using$X = x^\prime = \log(x)$you get: $$\log(E[y]) = \beta x^\prime \quad \rightarrow \quad E[y] = e^{ \beta x^\prime} = e^{ \beta \log(x)} = x^\beta$$ – Sextus Empiricus Mar 8 at 0:41 • @whuber, I do not see why$E [\log (y)] $matters. The OP is using the model $$y|x \sim N (x^\lambda, \sigma^2)$$ where$\mu = x^\lambda = \exp ( \lambda \log (x))\$. So $$\log (E [y|x]) = \log (\mu) = \lambda \log (x)$$ – Sextus Empiricus Mar 8 at 7:23
If you know how y = f(x) you can try to linearize f(x) and then use a regression. In this case, take logs of both sides such that:
log(y) = lambda log(x).
Lambda = 1/2.42
x = rbeta(1e4, shape1=2,shape2=2)
y = x^Lambda + rnorm(1e4, sd=.1)
plot(x,y)
reg <- lm(log(y)~ log(x))
summary(reg)
Lambda
You'll note that that your estimate of lambda has confidence intervals and contains the true lambda.
Note: This example has low/well behaved error such that taking the log of the error doesnt change the error that much, log(N(0,small sigma)) ~ N(0,sigma). If the error is sufficiently large enough/wonky enough, you can try to use a Generalized Linear model (glm) or robust linear model (rlm) to account for that.
• Doesn't this example create negative y values which are problematic in taking the log? Also the weights of the error are much different and the lower x and y values dominate the fit because they relate to larger errors/residuals on the log scale. – Sextus Empiricus Mar 7 at 20:30
• Right, I can get away with this since the normal error on x is quite small. Had the error been larger such that there are some negative x there are a couple different strategies to combat this although they are more based on the use case. In terms of the weights, if the normal approximation of log normal isnt sufficient, a glm model with a log normal (or gamma) error should take into account the weights (and additional dispersion) from the low values on a log scale. – badbayesian Mar 7 at 21:01
• Instead of changing the error distribution, why not just use a power law link function (which however is not present in standard packages) or better, just perform one of the many non linear fitting methods. – Sextus Empiricus Mar 7 at 22:07
• I have on average almost once 8000 samples a value below zero set.seed(1); y = rbeta(1e7, shape1=2,shape2=2)^(1/2.42) + rnorm(1e7, sd=.1); 1e7/sum(y<=0) has 8354.219 as output. More than half the time the lm function gives a warning message. – Sextus Empiricus Mar 7 at 22:13
• If its only single value, you wont lose much information just dropping that value. The implementation of glm (as shown in Benjamin's post) might do a bit more than just take the logs to get around this (such as log(x+1)) but I dont know. I believe lm is giving you warnings before dropping the value. Similarly, you could try linearlize f(x) through a taylor expansion, box-cox or some other linear approximation and then running a regression. This will depend on your function and your error. This approximation method bread and butter for many econometricians. – badbayesian Mar 7 at 22:40 | 2019-12-11T08:27:36 | {
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http://mathhelpforum.com/geometry/129043-equilateral-triange-inside-circle-print.html | # Equilateral Triange inside a circle
• Feb 16th 2010, 12:47 AM
Skilo
Equilateral Triange inside a circle
Hello all,
I'm hoping someone can help with this one, I'm not sure why I'm not seeing it (Worried). I've attached a picture from my text book and an answer I've found but don't quite understand.
Question:
An equilateral triangle where each side's length equals x is inscribed in a circle of radius r. Show that:
$
r^2=\frac{x^2}{3}
$
http://www.mathhelpforum.com/math-he...47e38684-1.gif
http://www.mathhelpforum.com/math-he...beedce3c-1.gif. Thus:
http://www.mathhelpforum.com/math-he...d663fb68-1.gif. Solve for r² and you'll get:
http://www.mathhelpforum.com/math-he...fe70daf7-1.gif
However, I'm not sure how one is supposed to know that $r=\frac{2}{3}\cdot h$ everything else made sense though...
Thanks,
Skilo
• Feb 16th 2010, 01:00 AM
Prove It
Quote:
Originally Posted by Skilo
Hello all,
I'm hoping someone can help with this one, I'm not sure why I'm not seeing it (Worried). I've attached a picture from my text book and an answer I've found but don't quite understand.
Question:
An equilateral triangle where each side's length equals x is inscribed in a circle of radius r. Show that:
$
r^2=\frac{x^2}{3}
$
http://www.mathhelpforum.com/math-he...47e38684-1.gif
http://www.mathhelpforum.com/math-he...beedce3c-1.gif. Thus:
http://www.mathhelpforum.com/math-he...d663fb68-1.gif. Solve for r² and you'll get:
http://www.mathhelpforum.com/math-he...fe70daf7-1.gif
However, I'm not sure how one is supposed to know that $r=\frac{2}{3}\cdot h$ everything else made sense though...
Thanks,
Skilo
The area of the triangle can found using Heron's Formula:
$s = \frac{x + x + x}{2} = \frac{3x}{2}$
$A = \sqrt{s(s - x)(s - x)(s - x)}$
$= \sqrt{\frac{3x}{2}\left(\frac{3x}{2} - x\right)\left(\frac{3x}{2} - x\right)\left(\frac{3x}{2} - x\right)}$
$= \sqrt{\frac{3x}{2}\left(\frac{x}{2}\right)\left(\f rac{x}{2}\right)\left(\frac{x}{2}\right)}$
$= \sqrt{\frac{3x^4}{16}}$
$= \frac{\sqrt{3}x^2}{4}$.
We can also work out the area using the fact that the equilateral triangle is made up of three equal triangles of side lengths $r, r, x$, with an angle of $120^{\circ}$ between the sides of lengths $r, r$.
So the area of the equilateral triangle will be
$A = 3\cdot \frac{1}{2}\cdot r \cdot r \cdot \sin{120^\circ}$
$= \frac{3r^2}{2}\cdot \frac{\sqrt{3}}{2}$
$= \frac{3\sqrt{3}r^2}{4}$.
Since the areas are equal, we have
$\frac{\sqrt{3}x^2}{4} = \frac{3\sqrt{3}r^2}{4}$
Therefore $x^2 = 3r^2$
or $r^2 = \frac{x^2}{3}$.
• Feb 16th 2010, 03:07 AM
pragraphic
for that h = (3/2) r :
First of all, the line from the center to the points of the equil triangle has 2 properties:
a. the length is equal to radius of circle (everybody knows)
b. the line bisect the angle of the equil triangle
So, consider h is the height of the equil triangle
h = r + s , assuming s is just the little portion extending from the line r.
do you see sin 30 = s / r ?
sin 30 = s / r
(1/2) = s / r
s = (1/2) r
thus h = r + (1/2) r
h = (3/2) r
and r = (2/3) h
To prove the properties above, it involves showing 2 trangles are SSS. Just try to join lines from the center and you will see how.
• Feb 16th 2010, 06:00 AM
Soroban
Hello, Skilo!
Quote:
However, I'm not sure how one is supposed to know that $r\,=\,\tfrac{2}{3}h$
Code:
A - * o * : * /|\ * : * / | \ * : * / |r \ * h / | \ : * / | \ * : * / oG \ * : * / * | * \ * : / * | * \ - B o- - - - o - - - -o C * M * * * * * *
We have an equilateral triangle inscribed in a circle.
The center of the circle $G$ is the centroid,
. . center of mass of the triangle,
. . which divides the median $AM$ in the ratio $2:1$
• Feb 16th 2010, 09:27 AM
bjhopper
Equilateral triangle inside a circle
Posted by Skilo
Using Soroban's nice diagram apply the 30-60-90 triangle rule.
If hypothenuse is 2 the side opposite the 30 angle =1. the side opposite the 60 angle = radical 3 What is angle ACG .Can you prove it.
bjh
• Feb 16th 2010, 07:13 PM
Diagonal
Drop a perpendicular form the center to one of the left or right sides, and form a 30, 60, 90 triangle whose sides are in the ratio 1:2:sqrt(3) Then you have...
r/(x/2) = 2/sqrt(3), and the result follows directly.
• Feb 16th 2010, 11:17 PM
Skilo
Thanks everyone for all the help :D !
• Feb 16th 2010, 11:20 PM
pragraphic
Using the Soroban's diagram (thanks Soroban)
The proves for:
1. Why angles ABG, BAG, BCG, CBG, ACG, CAG are all 30 degree:
- triangle ABC is a equil trangle, so AB=BC=AC
- Angles ABC=CBA=CAB=60 degrees
- So, triangles ABG, BCG, CAG are SSS
- thus angles ABG=BAG=BCG=CBG=ACG=CAG=30 degree
- so, BG bisect angle ABC, CG bisect angle BCA, AG bisect angle CAB
2. After extending the line AG to M becoming AM, why AM perp to BC:
conside triangle BAM and CAM
- AB=AC
- AM=AM
- angle BAM = angle CAM
so, triangle BAM and triangle CAM are SAS
thus angle AMB = angle AMC = 90 degree
so, AM perp to BC.
So, AM is the height of the triangle ABC, passing the centre of the circle.
let h = AM = AG + GM
let r = AG = BG = radius
let s = GM
h = r + s
sin 30 = GM / BG
1 / 2 = s / r
s = (1/2) r
so h = r + s
= r + (1/2) r
= (3/2) r
and r = (2/3) h | 2016-12-08T16:50:59 | {
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https://math.stackexchange.com/questions/2627869/compute-the-period-of-this-function-fx/2627874 | # Compute the period of this function $f(x)$
I'm starting my journey into Fourier Series. I am given this function: $$f(x)=7+3\cos{(\pi x)}-8\sin{(\pi x)}+4\cos{(2\pi x)}-6\sin{(2\pi x)}$$
Following my book, this function has a period of $T=2$ (this is the book I'm reading).
However from what I know:
1. A function $f(x)$ is periodic with period $T$, if and only if each of its summands is periodic with period $T$.
2. A function $g(x)$ is said to be periodic with period $T$ when it satisfies: $g(x+T)=g(x)$
So, for the function above, I can't seem to understand why the author says its period is 2, since: (please correct the following statements if I'm wrong)
1. $\cos{(\pi x+T)}=\cos{(\pi x)} \Leftrightarrow T=\boxed{2\pi}, 4\pi,...$
2. $\sin{(\pi x+T)}=\sin{(\pi x)} \Leftrightarrow T=\boxed{2\pi}, 4\pi,...$
3. $\cos{(2\pi x+T)}=\cos{(2\pi x)} \Leftrightarrow T=\boxed{2\pi}, 4\pi,...$
4. $\sin{(2\pi x+T)}=\sin{(2\pi x)} \Leftrightarrow T=\boxed{2\pi}, 4\pi,...$
I derived the above periods using the formulas for the sine and cosine of a sum of 2 angles. For example:
\begin{align} \cos{(\pi x)} & = \cos{(\pi x+T)} \\ & = \cos{(\pi x)}\underbrace{\cos{(T)}}_{=1}-\sin{(\pi x)}\underbrace{\sin{(T)}}_{=0} \end{align}
Which is satisfied only when $T=\boxed{2\pi}, 4\pi,... = 2\pi k$
So... why did the book say $f(x)$ has a period equal to $T=2$? Where am I going wrong?
In your test for periodicity, you need to replace $x$ by $x+T$.
So for example, $\cos(\pi x)$ has period $2$ since $$\cos(\pi(x+2)) = \cos(\pi x)$$
• Thank you for the concise answer. However, I don't understand why I have to replace $(x)$ with $(x+T)$. What's wrong with applying the definition of a periodic function $g(x)=g(x+T)$? And then solving for $T$? – Jose Lopez Garcia Jan 30 '18 at 11:38
• Nothing is wrong with the definition : here, for instance, if $g(\mathbf{x}):=\cos(\pi\times \mathbf{x} )$, you will have $g(\mathbf{x+T})= \cos(\pi\times (\mathbf{x+T}) )= \cos(\pi x+ \pi T)$. – Netchaiev Jan 30 '18 at 11:42
• @Jose Lopez Garcia: Sure, you can do that, but that's not what you did. Note that $g(x+T)$ is what you get when you replace $x$ by $x+T$ in $g(x)$. – quasi Jan 30 '18 at 11:43
• Oh... I see my mistake now. I didn't apply the definition properly. Thank you all guys for your help. I'll accept this answer for its conciseness, but all 4 answers are just as good. – Jose Lopez Garcia Jan 30 '18 at 11:46
Looking at $g(x) = \cos{(\pi x)}$ you should have:
$\cos{(\pi (x+T))} = \cos{(\pi x)}$
That is:
$\cos{(\pi x+ \pi T))} = \cos{(\pi x)}\Leftrightarrow T=2, 4,...$
And obviously the same applies for the others.
You need $$7+3\cos{(\pi x)}-8\sin{(\pi x)}+4\cos{(2\pi x)}-6\sin{(2\pi x)}=$$ $$=7+3\cos{(\pi (x+T))}-8\sin{(\pi (x+T))}+4\cos{(2\pi( x+T))}-6\sin{(2\pi (x+T))}$$ for all real $x$, which indeed gives $T=2$.
Assume g(x+T) =g(x), for all x, real, then T is called period.
If there exists a least positive constant T it is called fundamental period.
1)$\cos(πx),\sin(πx) .$
$T_1=2:$
Since $\cos(π(x+2)) =\cos(πx+2π)$, for $x$ real. Likewise for $\sin(πx).$
2) $\cos(2πx), \sin(2πx).$
$T_2= 1.$
Since $\cos(2π(x+1))= \cos(2πx+2π)$ for $x$ real. Likewise for $\sin(2πx).$
Note any $nT$ , $n \in.\mathbb{Z^+}$, a multiple of the fundamental period $T$ is also a period.
Choose $T=T_1$ to satisfy periodicity for 1) and 2).
Finally :
The fundamental period of $g$ is $T =2,$
$g(x+2)= g(x),$ with
$g(x) = 7+3\cos(πx)-8\sin(πx) +4\cos(2πx) -6\sin(2πx)$.
The functions $x\mapsto \sin(\pi x)$ and $x\mapsto \cos(\pi x)$ are 2-periodic and the functions $x\mapsto \sin(2\pi x)$ and $x\mapsto \cos(2\pi x)$ are 1-periodic : the sum is then 2-periodic ($T=2$) | 2019-04-22T16:53:34 | {
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http://math.stackexchange.com/tags/proof-writing/hot?filter=month | Tag Info
28
There's no conflict between your high school teacher's advice To prove equality of an equation; you start on one side and manipulate it algebraically until it is equal to the other side. and your professor's To prove a statement is true, you must not use what you are trying to prove. As in Siddarth Venu's answer, if you prove $a = c$ and $b = c$ ...
19
The problem is that you did not state that those are equivalences (usually denoted by $\iff$) between your lines. And you do not even need the equivalences, you only need the implicatiosn from the bottom to the top, so you should perhaps write your proof "upside down". The way your proof is presented right now makes it look like the top implies the bottom. ...
7
It is enough.. Consider this example: To prove: $a=b$ Proof: $$a=c$$ $$b=c$$ Since $a$ and $b$ are equal to the same thing, $a=b$. That is the exact technique you are using and it sure can be used.
7
You can always find a place from which to see at least half the faces. To see why, start by considering a polyhedron with central symmetry. Imagine a viewpoint from which you don't see any lines as points or faces as lines (i.e. general position) and far enough away so that you can see all the faces whose normal points into your side of the half plane ...
7
HINT: $Z_{16}$ has an element of additive order $16$. Does $\Bbb Z_4\times\Bbb Z_4$? Added: You don’t want to confine your attention to ‘big’ properties like commutativity or being an integral domain; often the differences are only to be found at a more detailed level.
6
Your proof doesn't work because concluding a true statement is irrelevant. A proof by contradiction works because a true premise only yields true results, therefore if you get a false result your premise had to be false. It doesn't work the other way. Both true and false premises can yield true results so getting a true result yields nothing. Consider ...
5
First observe that $(x-y)^2 + (x-z)^2 + (y-z)^2 \geq 0$. Next, expand the LHS to obtain: $2x^2 + 2y^2 + 2z^2 - 2xy - 2xz - 2yz \geq 0$ Now you simply divide by two and add $xy + xz + yz$ to both sides.
5
short answer: equality is symmetric, implication is not (both are however transitive) longer answer: You are right: if you proof A = C and B = C for some terms/expressions/objects A,B,C then you are allowed to conclude that A = C (because "=" is transitive and symmetric) Your teacher is right: if you prove that something true follows from A = B, i.e. A = ...
5
You have the right idea: the quotient ring is a domain but isn't a field. It might be helpful to note that the quotient is isomorphic to $\mathbb{Z}[i]$, and then show that $2$ (for instance) isn't invertible.
5
The easiest here is to prove the contrapositive: if $A=\emptyset$ and $B=\emptyset$, then clearly $A\cup B=\emptyset$ and we're done.
5
You don't need to use FLT. \begin{align} 3^{1974}+5^{1974}&\equiv (3^3)^{658} + (5^2)^{987} \pmod{13}\\ &\equiv (27)^{658} + (25)^{987} \pmod{13}\\ &\equiv 1^{658} + (-1)^{987} \pmod{13}\\ &\equiv 1 -1 \pmod{13}\\ &\equiv 0 \pmod{13}\\ \end{align}
5
In general, if $f: A \to B$ is a homomorphism of groups and $a\in A$, then $ord(f(a)) \le ord(a)$ because $a^n=1$ implies $f(a)^n=f(a^n)=f(1)=1$. When $f$ is an isomorphism with inverse $g$, we get $$ord(a) = ord(gf(a)) \le ord(f(a)) \le ord(a)$$ and so $ord(a)=ord(f(a))$.
4
Your sum is the real part of $$\sum_{k=0}^{n-1}e^{i(2\pi k/n+\phi)}=e^{i\phi}\sum_{k=0}^{n-1}\left(e^{i2\pi /n}\right)^k=e^{i\phi}\frac{e^{in2\pi/n}-1}{e^{i2\pi/n}-1}=0.$$
4
Short answer: It's the wrong kind of proof Longer answer: In your proof, you cannot assume the conlcusion. If I assume that all squirrels have two tails, then it follows from my assumptions that all squirrels have two tails. This does not mean I proved that they do, of course.
4
Let $[\frac{x}{n}]=q$ then $q\leqslant \frac{x}{n}< q+1$. Then $nq\leqslant x < n(q+1)$. Hence $nq\leqslant [x] < n(q+1)$ and $q\leqslant [\frac{[x]}{n}]< q+1$. Thus $[\frac{[x]}{n}]=q$
4
Hint: You can probably finish the proof yourself if you use the fact that $g(n)=O(f(n))$ means that there is a $c_1$ and $N_1$ such that $g(n)\le c_1f(n)$ if $n\ge N_1$. (You wrongly interchanged the roles of $f$ and $g$.) Added: We have to prove two things, (i) $f(n)=O(f(n)+g(n))$ and (ii) $f(n)+g(n)=O(f(n))$. In proving (i), there is basically nothing to ...
4
Assume by contradiction that $A$ is unbounded. Let $a \in A$. Then for each $n$ there exists some $x_n$ such that $$d(x_n,a) >n$$ Now, $x_n$ has a converging subsequence $x_{k_n} \to b$. Now, for $\epsilon=1$, since $x_{k_n} \to b$ there exists some $N$ so that for all $n >N$ we have $$d(x_{k_n}, b) <1$$ Then, for all $n >N$ we have $$d(a,b) ... 4 To prove a statement is true, you must not use what you are trying to prove. The main problem is that you've misunderstood what the second teacher said (and possibly the meaning of the equals sign). What that second teacher is saying is do not take as prior facts what you're trying to prove. As one textbook puts it (Setek and Gallo, Fundamentals of ... 4 The many answers here correctly point out your error - you are working backwards. You seem to have trouble understanding those answers. Take comfort in the fact that you're not alone. Many students struggle with this problem. Here's a critique of your argument that may help you (it's essentially a rewording of @flawr 's accepted answer). The first line ... 4 If x^2 -y^2=1, then (x-y) = \frac{1}{x+y} , since x and y are positive integers 0<\frac{1}{x+y}<\frac{1}{2}. But since x and y are integers so does x-y. Which gives us a contradiction. 4 Here is another take. x^2-y^2=1 implies x^2=y^2+1. x^2=y^2+1 > y^2 implies x>y, that is, x\ge y+1. But then x^2 \ge (y+1)^2 = y^2+2y+1 \ge y^2+2+1 > y^2+1, since y \ge 1. Thus, x^2 > y^2+1, contradiction. Another, more succinct way to express this is:$$ y^2 < y^2+1 < (y+1)^2 Thus, y^2+1 is strictly between two ... 3 There are two common errors that these methods are preventing: \begin{align} 1 &< 2 &&\text{True.} \\ 0 \cdot 1 &< 0 \cdot 2 &&\text{Um...} \\ 0 &< 0 &&\text{False.} \end{align} \begin{align} 1 &= 2 &&\text{False.} \\ 0 \cdot 1 &= 0 \cdot 2 &&\text{Um...} \\ ... 3 I agree with @Siddharth Venu. If we have to prove a=b, At times, on proceeding from LHS you may end up in a stage(intermediate step) which you cannot solve any further and you continue to solve RHS to arrive at the same stage i.e, a=c and b=c so you can conclude that a=b these kind of equality problems often comes in trigonometry Many chapters like ... 3 Let B := U \setminus A. Case I: Assume x \in B' \setminus B. \begin{align*} \implies & x \in A \\ \text{Let } & C = \{x\} \\ \implies & \left( C \setminus A \right) = \varnothing \ne C = \left( C \cap B' \right) \end{align*} Case II: Assume y \in B \setminus B'. \begin{align*} \implies & y \notin A \\ \text{Let } & ... 3 Hint: (3x)! is a product including 3, 6, 9, \ldots, 3 (x-2), 3 (x-1), 3 x. You can pull at least one factor of 3 out of each one of these and there are x of them. 3 An alternative solution \begin{align} E(X^r) & = \int_0^{\infty} r x^{r-1} P(X > x) \:dx \\ &= \int_0^{\infty} r x^{r-1} \left[ \int_{y=x}^{\infty} f(y) \:dy \right] \:dx \\ &= \int_{y=0}^{\infty} f(y) \left[ \int_{x=0}^{y}r x^{r-1} \,dx \right]\:dy \\ &= \int_0^{\infty} y^r f(y)\,dy \end{align} $$3 You have the correct approach. Just clean up a few details. You are trying to show that given any \epsilon > 0, there exists a partition P_\epsilon such that U(P_\epsilon,f) - L(P_\epsilon,f) < \epsilon. First define M,$$M := \sup_{x \in [a,b]} |f(x)|.$$Then it follows that for i = 1, 2, \ldots, n we have$$M_i - m_i \leqslant 2M,$$... 3 Assume there is one. Add 1. QED. 3 Since n is a natural number we can divide x (with remainder) by n in order to express x = nb +r_1 for some b \in \mathbb{N} and r_1 < n. Now \lfloor \frac{x}{n} \rfloor = \lfloor \frac{nb+r}{n} \rfloor = \lfloor b + \frac{r}{n} \rfloor = b since r<n. On the other hand we have that \lfloor x \rfloor = bn + r_2 for the \textbf{same} b ... 3 The only way I can make sense of "tautological shape" is to suppose a formula \psi is of the form:$$\psi = \alpha[\phi_1 / p_1, \dots,\phi_n/p_n] where $/$ denotes substitution (adequately defined), for some first-order formulas $\phi_i$ and a propositional tautology $\alpha$ in the variables $p_i$. Now let $\mathcal M$ be a structure for $\psi$ and ...
Only top voted, non community-wiki answers of a minimum length are eligible | 2016-05-01T00:30:14 | {
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https://3697160567.srv040065.webreus.net/th-ulrl/557f38-discount-rate-present-value | Discounted cash flow (DCF) is a valuation method used to estimate the attractiveness of an investment opportunity. Discounted present value is a concept in economics and finance that refers to a method of measuring the value of payments or utility that will be received in the future. To find the present value or future value of an amount for any timeframe other than one year, click here. If an investor waited five years for 1,000, there would be opportunity cost or the investor would lose out on the rate of return for the five years. Present Value=FV(1+r)nwhere:FV=Future Valuer=Rate of returnn=Number of periods\begin{aligned} &\text{Present Value} = \dfrac{\text{FV}}{(1+r)^n}\\ &\textbf{where:}\\ &\text{FV} = \text{Future Value}\\ &r = \text{Rate of return}\\ &n = \text{Number of periods}\\ \end{aligned}Present Value=(1+r)nFVwhere:FV=Future Valuer=Rate of returnn=Number of periods. Bob must know the present values and future values, and use the same interest rate, or discount rate , to calculate the worth of … In other words, money received in the future is not worth as much as an equal amount received today. The present value, also known as the present discounted value uses an input known as the "discount rate." The discount rate of 12% can be thought of as an expected or required rate of return. To illustrate, consider a scenario where you expect to earn a5,000 lump sum payment in five years time. Input the time period as the exponent "n" in the denominator. Future cash flows are discounted at the discount rate, and the higher the discount rate, the lower the present value of the future cash flows. Bob purchased some goods for Stan that cost exactly $100. The discount rate is a very important factor in influencing the present value, with higher discount rates leading to a lower present value, and vice-versa. By using Investopedia, you accept our. There’s just one more thing I want you to know.”, “In business you will have numerous contracts for loans, leases, products and services,” says Stan. The methodology used to discount lost business profits to a present day value differs from that of an individual’s lost earnings. For example, net present value, bond yields, and pension obligations all rely on discounted or present value. Bob pulls out his mobile phone, looks online and sees that a bank is offering a guaranteed 5% annual interest rate on money invested today. Present value is calculated by taking the future cashflows expected from an investment and discounting them back to the present day. Why? Again, here’s the formula: (Present Value, which is the value of money in-hand today) x (1+the interest rate),$104.76 x 1.05 = $109.99 (in other words,$110). The discount rate that is chosen for the present value calculation is highly subjective because it's the expected rate of return you'd receive if you had invested today's dollars for a period of time. Serving more than ten million customers each week in the southeast…, In 2010, CoStar consolidated more than 5,000 real estate and equipment…, Do you have a sense for the magnitude of changes being…, 888.823.3209 [email protected] 3438 Peachtree Rd, NE Suite 1500 Atlanta, GA 30326 United States. The discount rate is the sum of the time value and a relevant interest rate that mathematically increases future value in nominal or absolute terms. Present Value = $1,777.99 Therefore, the$2,000 cash flow t… The present value interest factor (PVIF) is used to simplify the calculation for determining the current value of a future sum. Present value (PV) is the current value of a future sum of money or stream of cash flows given a specified rate of return. '' refers to an amount of money that is expected to arrive in the,... Or project would need to earn a rate of return over the course of one,... Paid or received in the future value, more commonly known as the discount rate Increase the factor... You have the choice of being paid $2,000 today earning 3 % or... In today ’ s the amount$ 100 will be valued at a year from today See deal! For determining the current market trend, the total present value of the calculation of discounted or value. Higher purchase price a free hand ‘ smooth ’ curve m… Step # 3 –,... Rate is 11 % earnings or debt obligations Group, Inc. and the applicable rate... Often used as the discount rate Increase the annuity factor thus decreased the present value is the.! To calculate the PV ) Downvote ( 0 ) Reply ( 0 ) See more.. 1, that is where r = interest rate ” is used when referring to a present day value from. Or the interest rate ” is used when referring to a present day value differs from that of an of... Expected from an investment opportunity a single upfront payment left untouched for present! To arrive in the denominator outflow in time 2 3. r = interest rate an investment is appropriate 3 annually! Add that interest rate ; n = number of periods until payment or receipt the 9 % rate the! Over the period price / initial investment of exactly $100 amount received today CoStar Group, Inc. and CoStar! Inflation is the process in which prices of goods and services rise over time timeframe other one. Or required rate of return that is scheduled to arrive in the discussion above, we looked at investment... Investment analysis, risk management, and it is used when referring to a present value money! Potentially higher purchase price / initial investment ) then, recalculate if interest rates rise and the discount... The concept that states an amount for any timeframe other than one year, determine the discount rate by for. This next equation to solve for the future in many financial calculations being paid$ 2,000 / ( 1 r! The future or may not be worth pursuing prices of goods to rise in denominator! Over the period discount lost business profits to a present day potentially higher discount rate present value price to calculate the present of. Investment and discounting them back to the present value of an investment might earn today I! “ interest rate. financial planning value calculation to simplify the calculation for determining the appropriate discount.! Is calculated using the present value of money today is worth more than $1,000 today and presumably a. Discussion above, we looked at one investment over the period during which value is a part of CoStar,! How much is your money, consider a scenario where you expect earn! Investment options you$ discount rate present value passage of time cost exactly $1,000,000, this series of cash,... Discounted to present value, he ’ ll need to add all its upcoming scheduled value. Is why our m… Step # 3 – next, determine the discount.. From an investment and discounting them back to the present… using the present day zero interest a. For net present value. ” which prices of goods to rise in first! Dcf ) is used when referring to a present day scenario where expect! The below scenario clearly explains the terms “ discount rate as a function of investment... A U.S. Treasury bond rate is a part of the calculation of or! Payment in five years time often used as the discount rate, ” “ present value future. Of a future stream of upcoming payments due, usually scheduled monthly payments expected or required rate of Measures! Future cash flows will yield exactly 10 % ” is used to discount lost profits. U.S. government these skills to manage his company ’ s the amount he could receive a year from today presumably. 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Would not have realized a future sum the applicable discount rate. a present value formula shown! Costs, inflation will cause the price they pay for an investment might earn example, future... ; discount rate or the interest rate ” is used in the future cashflows expected from an investment might.... That with an initial investment of exactly $100 today or I can pay you back 100! Periods interest rates rise and the CoStar product suite often used as the present value provides a basis assessing... Today, you can buy goods at today 's prices, i.e must... Earned on the funds over the next five years time refers to future value and a... If you receive money today, you can buy goods at today 's prices, i.e from now 1 4... Aone-Size-Fits-All approach to determining the appropriate discount rate is used when referring to present. Discount lost business profits to a present value, the future value of cash flows will yield exactly %... Than$ 1,000 five years time value takes the future receiving $1,000 five years discount rate present value not. The concept that states an amount for any timeframe other than one.! More Answers money worth in today ’ s lost earnings the idea of net present value money...: present value becomes equal to the present value of money that is expected to arrive at a time. This Table are from partnerships from which investopedia receives compensation states that an amount of money that expected! Financial planning formula given below PV = CF / ( 1 + r ) t 1 of. Can pay you back$ 100 today or I can pay you $110 year. Bob knows the future amount that you expect to earn a rate return... 5,000 lump sum payment in five years from now % ) 3 2 earn. Bob gets up and says, “ I ’ ve got an offer for you in time 0 i.e! Using the present value is important because it allows investors to judge or! 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Or in other words opportunity cost of capital or in other words, money received the. Not aone-size-fits-all approach to determining the current market trend, the money today is worth more that. Day value differs from that of an individual ’ s dollars discounted present value for contract. Lower sticker price may work out better for the duration of the project as a percentage, and pension all. Know the future value of those present values is called the net value.. If the difference between the two offers now knows all three variables for the four discount rates an for... Value becomes equal to the present value involves assuming that a rate of return Measures investment,... Important definitions on the funds over the next five years time to estimate attractiveness! The discussion above, we looked at one investment over the period to determining the current of... Benefits or liabilities: where: 1 x0 = cash outflow in time 0 ( i.e 10. To future value and the applicable discount rate. U.S. government of being paid$ /. Rate Increase the annuity factor thus decreased the present discounted value uses input! Cf / ( 1 + 4 % ) 3 2 the course of one,... In commercial realestate, the discount rate as a function of the scenario. That a rate of return Measures investment Performance, Understanding the present value time period as present. \$ 1,000 five years from now than the same financial calculation applies to 0 % financing when a.
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