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https://www.jobilize.com/course/section/section-exercises-composition-of-functions-by-openstax?qcr=www.quizover.com | # 3.4 Composition of functions (Page 6/9)
Page 6 / 9
## Finding the domain of a composite function involving radicals
Find the domain of
Because we cannot take the square root of a negative number, the domain of $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}\left(-\infty ,3\right].\text{\hspace{0.17em}}$ Now we check the domain of the composite function
The domain of this function is $\text{\hspace{0.17em}}\left(-\infty ,5\right].\text{\hspace{0.17em}}$ To find the domain of $\text{\hspace{0.17em}}f\circ g,\text{\hspace{0.17em}}$ we ask ourselves if there are any further restrictions offered by the domain of the composite function. The answer is no, since $\text{\hspace{0.17em}}\left(-\infty ,3\right]\text{\hspace{0.17em}}$ is a proper subset of the domain of $\text{\hspace{0.17em}}f\circ g.\text{\hspace{0.17em}}$ This means the domain of $\text{\hspace{0.17em}}f\circ g\text{\hspace{0.17em}}$ is the same as the domain of $\text{\hspace{0.17em}}g,\text{\hspace{0.17em}}$ namely, $\text{\hspace{0.17em}}\left(-\infty ,3\right].$
Find the domain of
$\left[-4,0\right)\cup \left(0,\infty \right)$
## Decomposing a composite function into its component functions
In some cases, it is necessary to decompose a complicated function. In other words, we can write it as a composition of two simpler functions. There may be more than one way to decompose a composite function , so we may choose the decomposition that appears to be most expedient.
## Decomposing a function
Write $\text{\hspace{0.17em}}f\left(x\right)=\sqrt{5-{x}^{2}}\text{\hspace{0.17em}}$ as the composition of two functions.
We are looking for two functions, $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}h,\text{\hspace{0.17em}}$ so $\text{\hspace{0.17em}}f\left(x\right)=g\left(h\left(x\right)\right).\text{\hspace{0.17em}}$ To do this, we look for a function inside a function in the formula for $\text{\hspace{0.17em}}f\left(x\right).\text{\hspace{0.17em}}$ As one possibility, we might notice that the expression $\text{\hspace{0.17em}}5-{x}^{2}\text{\hspace{0.17em}}$ is the inside of the square root. We could then decompose the function as
We can check our answer by recomposing the functions.
$g\left(h\left(x\right)\right)=g\left(5-{x}^{2}\right)=\sqrt{5-{x}^{2}}$
Write $\text{\hspace{0.17em}}f\left(x\right)=\frac{4}{3-\sqrt{4+{x}^{2}}}\text{\hspace{0.17em}}$ as the composition of two functions.
$\begin{array}{l}g\left(x\right)=\sqrt{4+{x}^{2}}\\ h\left(x\right)=\frac{4}{3-x}\\ f=h\circ g\end{array}$
Access these online resources for additional instruction and practice with composite functions.
## Key equation
Composite function $\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$
## Key concepts
• We can perform algebraic operations on functions. See [link] .
• When functions are combined, the output of the first (inner) function becomes the input of the second (outer) function.
• The function produced by combining two functions is a composite function. See [link] and [link] .
• The order of function composition must be considered when interpreting the meaning of composite functions. See [link] .
• A composite function can be evaluated by evaluating the inner function using the given input value and then evaluating the outer function taking as its input the output of the inner function.
• A composite function can be evaluated from a table. See [link] .
• A composite function can be evaluated from a graph. See [link] .
• A composite function can be evaluated from a formula. See [link] .
• The domain of a composite function consists of those inputs in the domain of the inner function that correspond to outputs of the inner function that are in the domain of the outer function. See [link] and [link] .
• Just as functions can be combined to form a composite function, composite functions can be decomposed into simpler functions.
• Functions can often be decomposed in more than one way. See [link] .
## Verbal
How does one find the domain of the quotient of two functions, $\text{\hspace{0.17em}}\frac{f}{g}?\text{\hspace{0.17em}}$
Find the numbers that make the function in the denominator $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ equal to zero, and check for any other domain restrictions on $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}g,\text{\hspace{0.17em}}$ such as an even-indexed root or zeros in the denominator.
sinx sin2x is linearly dependent
what is a reciprocal
The reciprocal of a number is 1 divided by a number. eg the reciprocal of 10 is 1/10 which is 0.1
Shemmy
Reciprocal is a pair of numbers that, when multiplied together, equal to 1. Example; the reciprocal of 3 is ⅓, because 3 multiplied by ⅓ is equal to 1
Jeza
each term in a sequence below is five times the previous term what is the eighth term in the sequence
I don't understand how radicals works pls
How look for the general solution of a trig function
stock therom F=(x2+y2) i-2xy J jaha x=a y=o y=b
sinx sin2x is linearly dependent
cr
root under 3-root under 2 by 5 y square
The sum of the first n terms of a certain series is 2^n-1, Show that , this series is Geometric and Find the formula of the n^th
cosA\1+sinA=secA-tanA
Wrong question
why two x + seven is equal to nineteen.
The numbers cannot be combined with the x
Othman
2x + 7 =19
humberto
2x +7=19. 2x=19 - 7 2x=12 x=6
Yvonne
because x is 6
SAIDI
what is the best practice that will address the issue on this topic? anyone who can help me. i'm working on my action research.
simplify each radical by removing as many factors as possible (a) √75
how is infinity bidder from undefined?
what is the value of x in 4x-2+3
give the complete question
Shanky
4x=3-2 4x=1 x=1+4 x=5 5x
Olaiya
hi can you give another equation I'd like to solve it
Daniel
what is the value of x in 4x-2+3
Olaiya
if 4x-2+3 = 0 then 4x = 2-3 4x = -1 x = -(1÷4) is the answer.
Jacob
4x-2+3 4x=-3+2 4×=-1 4×/4=-1/4
LUTHO
then x=-1/4
LUTHO
4x-2+3 4x=-3+2 4x=-1 4x÷4=-1÷4 x=-1÷4
LUTHO
A research student is working with a culture of bacteria that doubles in size every twenty minutes. The initial population count was 1350 bacteria. Rounding to five significant digits, write an exponential equation representing this situation. To the nearest whole number, what is the population size after 3 hours?
f(x)= 1350. 2^(t/20); where t is in hours.
Merkeb | 2020-07-09T11:43:05 | {
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https://www.physicsforums.com/threads/for-every-finite-integer-sequence-theres-a-pattern-source.935429/ | # I For every finite integer sequence there's a pattern- source?
Tags:
1. Dec 26, 2017
I have in the back of my head the statement that for every finite sequence of positive integers there exists a pattern (i.e., a generating formula). While this sounds reasonable, I am not sure whether it is true, and if it is true, what the source for this statement is, and how the correct statement is correctly formulated. A somewhat stronger statement would be that for every finite sequence of integers there is a pattern that one can theoretically find, and comments on that would also be appreciated if the first statement is correct.
2. Dec 26, 2017
### Staff: Mentor
Like you, I'm not sure it would be true in general, but for some sequences of integers, a generating polynomial could be found. For the sequence $\{a_1, a_2, \dots, a_n\}$, plot the points $(1, a_1), (2, a_2), \dots, (n, a_n)$. In many cases it would be possible to find the n coefficients of a polynomial of degree n - 1 that passes through these points.
3. Dec 26, 2017
### Stephen Tashi
If we interpret "generating formula" to mean a polynomial, it is true that there exists a polynomial f(x) such f(k) gives the kth member of the sequence. This can be done for any finite sequence of real numbers by using the Lagrange interpolating polynomial http://mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html. For example, In the notation of that article the $(x_i,y_i)$ could be (1,3), (2,5), (3,13) to indicate the sequence {3,5,13}.
The function that generates a given sequence is not unique. For example, if you are given the sequence {y1,y2,..yn} you could find a polynomial that generates the sequence {y1,y2,..yn,8} and another polynomial that generates the sequence {y1,y2,..yn,9}. Both polynomials generate {y1,y2,..yn}. If we want to make a claim about a unique generating function then we have to impose some conditions, such as seeking a polynomial of smallest degree. There are also non-polynomial functions that generate a given sequence.
4. Dec 26, 2017
### StoneTemplePython
another approach:
if you want to interpret your finite positive integer sequence as being un-normalized power sums, i.e.
$s_1 = \sum_{i=1}^n \lambda_i$
$s_2 = \sum_{i=1}^n \lambda_i^2$
$\vdots$
$s_n = \sum_{i=1}^n \lambda_i^n$
and in general:
$s_k = \sum_{i=1}^n \lambda_i^k$
(whereas a proper power sum would be $s_k^* = \frac{1}{n}\sum_{i=1}^n \lambda_i^k$ )
you can directly find a monic polynomial that fits to these via Newton's Identities.
- - - -
questions of uniqueness get tied into this via a Hankel Matrix formulation. The original result that tells you the number of roots and the mixture of real vs complex is due to Sylvester. If you normalize these sums to become a proper power sum, you can consider this as an instance of the Hamburger Moment Problem:
https://en.wikipedia.org/wiki/Hamburger_moment_problem
5. Dec 26, 2017
### Staff: Mentor
It is always possible to do that, and you only have to solve a linear equation system which will always have a unique solution (or an infinite set if you use polynomials with a higher degree).
6. Dec 26, 2017
Thanks, Stephen Tashi, StoneTemplePython, mfb and Mark44. The Lagrange Interpolating Polynomial (indicated by Stephen Tashi) appears to be the most direct method given of these answers [ it is this which I presume (?) mfb is referring to in saying "it is always possible...."]; StoneTemplePython's method is also interesting. If I understand correctly (which is far from guaranteed), the Lagrange Interpolation is much more direct theoretically whereas the Newton Identities approach might be more computationally efficient in practice.
7. Dec 26, 2017
### StoneTemplePython
Actually both the lagrange interpolation approach and newton's identities have similar complexity and both are known for numeric stability issues.
It's quite common, and desirable, to try to characterize probability distributions by their moments, graphs by their cycles, matrices by their traces (i.e. their spectral properties if we're dealing with scalars in $\mathbb R, \mathbb C, \mathbb Q$) and some other stuff. That is what I was getting at.
- - - -
If you know linear algebra, a unifying theme in all this is to look into Vandermonde matrices. Any Hankel matrix can be decomposed into $\mathbf V^T \mathbf {DV}$ (where $\mathbf D$ is diagonal and $\mathbf V$ is vandermonde -- note that is a transpose sign not $\mathbf V^*$ on purpose as it is still a non-conjugate transpose in the case of complex scalars.)
If you are so interested you can skip the theory of Lagrange interpolation and just look up Vandermonde interpolation. (After playing around with the basis its the same thing basically.) e.g. see this: https://math.stackexchange.com/ques...olation-formula-from-vandermondes-determinant
8. Dec 26, 2017
Super! Thanks again, StoneTemplePython. | 2018-07-16T14:50:58 | {
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https://math.stackexchange.com/questions/2391039/prove-n-5-is-not-a-perfect-square-for-n-in-mathbbn/2391041 | Prove $n! +5$ is not a perfect square for $n\in\mathbb{N}$
I have a proof of this simple problem, but I feel that the last step is rather clunky:
For $n=1,2,3,4$ we have $n!+5=6,7,11,29$ respectively, none of which are square. Now assume that $n\geq 5$, then: \begin{aligned} n! +5 & \;=\; n(n-1)\cdots 6\cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 + 5 \\[0.2cm] & \;=\; 5\left[ \frac{}{} n(n-1)\cdots 6\cdot 4 \cdot 3 \cdot 2 \cdot 1 + 1 \right] \\[0.2cm] &\;=\; 5(3k+1) \end{aligned} for some $k\in\mathbb{N}$. Since $5(3k+1)=15k+5\equiv 5\,\text{mod} \, 15$ and all perfect squares are congruent either $0,1,4,6, 9$ or $10\,\text{mod}\,15$, the result follows. $\;\blacksquare$
It took a pretty tedious exhaustive search of squares modulo 15 in the last step; is there a theorem I am missing that means the last step follows immediately? Your comments are much appreciated!
• How do you conclude $n(n−1)⋯6⋅4⋅3⋅2⋅1$ is a multiple of 5? It is not true when $n<10$. – pisco Aug 12 '17 at 7:30
• Ah you're absolutely right @pisco125! I have edited it accordingly - does it make better sense doing it using this method? Though your proof below is a lot more simple and elegant. – AloneAndConfused Aug 12 '17 at 9:13
• It does make sense now :). However, a number which is 5 modulo 15 is automatically 2 modulo 3, so you can choose to modulo 3 instead of modulo 15 in your work. – pisco Aug 12 '17 at 9:17
• My edit was for a typo: In the 2nd line "$n=6,7..."$ should have been $n!+5=6,7..."$. – DanielWainfleet Aug 13 '17 at 18:33
When $n\geq 3$ $$n!+5 \equiv 2 \pmod{3}$$ So it cannot be a perfect square when $n\geq 3$.
As you can see from pisco125's brilliant answer, the step in which you deduce $n! + 5 = 15k + 5$ for $n \geq 5$ is in this case unnecessary.
With factorials you can usually rely on the fact that $d \mid n!$ for all $n \geq d$. This means $n!$ is divisible by 5 for all $n \geq 5$. Then, although 25 is a bigger modulo than 3 or 15, it's very convenient here, because if $m \equiv 5 \pmod{10}$, then $m^2 \equiv 0 \pmod{25}$. That is to say, 5 is not a square modulo 25. Of course this way you still have to check $n < 10$, whereas pisco's way you need only check $n < 3$.
It's rarely the most elegant way, but with factorials the most intuitive way is by looking at them modulo a power of $10$, such as $100$ or $1000$.
Thus $n! + 5 \equiv 5 \pmod{100}$ for $n > 9$. We know, since it is obvious, that $x^2 \equiv 5 \pmod{100}$ has no solution in integers.
Like I said, it's not elegant, because we still have to look at $n < 10$. We have $$6, 7, 11, 29, 25, 25, 45, 65, 25, 5$$ of which some of them might be squares (the $25$s), but turn out not to be very soon after we think about them.
If $n! + p = k^2$ for some prime it's not too difficult to conclude $n < p$ and $k$ has no prime factors less than $p$, or $p \le n < 2p$, $\frac {n!}p + 1$ is divisible by $p$ and $\frac {\frac {n!}p + 1}p$ is a perfect square to no prime factors less than $p$[$*$].
We can go a bit further. $(p-1)! \equiv -1 \mod p$ by Wilson's Theorem so for $p \le n < 2p$ with $\frac {n!}p + 1\equiv 0 \mod p$ we can conclude $\frac {n!}p = (p-1)!\prod (p + k) \equiv -1 \mod p$ so $(n-p)! \equiv 1 \mod p$. And $n = p+1$ or $n = p + (p-2)= 2p-2$ so only two cases to check, if $n \ge p$.
So if $n < 5$ and $n! + 5 = k^2$ then $n! + 5 \ge 7^2$ which is simply not possible for $n\le 4; n! \le 24$
If $5\le n < 10$. then $\frac {n!}5 \equiv -1 \mod 5$. If $n = 5+k$ then $\frac {n!}5 = 4!*(6....n) \equiv 4!*k! \equiv -k! \equiv -1$ so $k! \equiv 1\mod 5$. So $k =1, 3$ and $n=6,8$. Which can be checked.
Which is probably much more clunky than your proof. But it could be useful for generalizing.
====
[$*$] $k \le n$ implies $k|n!$ so $k\not \mid n! +p$ unless $k = p$. If so $p|n! + p$ and $n! + p = p(\frac {n!}p + 1)$. If $n \ge 2p$ then $p|\frac {n!}p$ and $p \not \mid \frac{n!}p + 1$. If $n! + p$ is a perfect square and $p|n! + p$ then $p^2|n!+p$ and so $p \le n < 2p$. | 2021-06-15T23:35:18 | {
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https://www.physicsforums.com/threads/spherical-coordinates-doubt.765314/ | # Homework Help: Spherical coordinates doubt
1. Aug 9, 2014
### Modest Learner
1. The problem statement, all variables and given/known data
In spherical coordinates (ρ,θ,ø); I understood the ranges of ρ, and θ. But ø, still eludes my understanding. Why is ø only from 0 to π, why not 0 to 2π??
2. Aug 9, 2014
### Ray Vickson
Look at a diagram to see why.
3. Aug 9, 2014
### AlephZero
Values of ø between from 0 to π cover the whole surface of the sphere. On maps of the earth latitude is measured from -90 to + 90 degrees not 0 and 180, and longitude from -180 to +180 not 0 to 360, but the basic idea is the same.
4. Aug 9, 2014
### MrAnchovy
$\phi = 0$ is directly overhead, $\phi = \pi$ is directly beneath your feet, where would $\phi = 2\pi$ be?
5. Aug 9, 2014
### Modest Learner
If seeing the diagram would have had helped, then I would not have asked the question in the first place.
6. Aug 9, 2014
### Modest Learner
Okay, I have uploaded two attachments.
When I view from side, ø = π, covers only half the circle (see the picture). When I try to think of it as a clock, ø = π, covers 12 to 6. Now shouldn't ø = 1.5π cover 12 to 9, and ø = 2π cover the whole circle, and reach the same point as π = 0.
Also, in ø = π, the 3d section appears to me as a hemisphere. Shouldn't it be a total sphere??
Or maybe, I am confusing spherical coordinates with polar or cartesian coordinates??
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7. Aug 9, 2014
### MrAnchovy
So what you have shown is a coloured half-disk. For every point on that disk, Θ = 0. If you vary Θ from 0 to 2π the half-disk will sweep out a complete sphere.
8. Aug 9, 2014
### Modest Learner
Please explain further, I can't seem to understand.
EDIT: Okay, I think, I got a little idea of why ø = π would work. Basically, The up and down thing was exactly right. I think, I confused myself, when I added a sense of left and right. It is a total different axis, and the coordinates seem to do exactly the same thing, by putting θ = π to 2π.
9. Aug 9, 2014
### Modest Learner
http://mathinsight.org/spherical_coordinates
10. Aug 9, 2014
### HallsofIvy
Since $\theta$ goes from $0$ to $2\pi$, if we allowed $\phi$ to go also from $0[itex] to [itex]2\pi$ some points would have two descriptions. For example, $\theta= 3\pi/2$, $\phi= \pi/4$ and $\theta= \pi/2$, $\phi= 7\pi/4$, $\rho$ and fixed value, say 1, designate the same point.
You can see that by converting to Cartesian coordinates: $x= \rho cos(\theta) sin(\phi)$, $y= \rho sin(\theta) sin(\phi)$, $z= \rho cos(\phi)$.
$\rho= 1$, $\theta= 3\pi/2$, $\phi= \pi/4$ gives $x= 1(0)(\sqrt{2}/2)= 0$, $y= 1(-1)(\sqrt{2}/2)= -\sqrt{2}/2$ and $z= 1(\sqrt{2}/2)= \sqrt{2}/2$.
$\rho= 1$, $\theta= \pi/2$, $\phi= 7\pi/4$ gives $x= 1(0)(-\sqrt{2}/2)= 0$, $y= 1(1)(-\sqrt{2}/2)= -\sqrt{2}/2$, and $z= 1(\sqrt{2}/2)= \sqrt{2}/2$.
11. Aug 9, 2014
### Ray Vickson
If I was a mind-reader I would have known that. I had no way to know what you have, or have not looked at already.
12. Aug 9, 2014
### Staff: Mentor
Phi is the angle between the axis of the sphere and a line drawn through the center of the sphere to a given latitude, measured from the North Pole. It is equal to 90 degrees (i.e., ∏/2) minus the latitude. So, ø =0 represents a line drawn from the center of the sphere through the North pole, ø = ∏/2 represents a line drawn through the center of the sphere to any point on the equator, and ø =∏ represents a line drawn through the center of the sphere to the South pole.
Chet
13. Aug 10, 2014
### Modest Learner
Thanks, got it. | 2018-06-22T23:13:28 | {
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https://mathematica.stackexchange.com/questions/45519/smoothing-listcontourplot-contours | # Smoothing ListContourPlot contours
I have some data of type {{x1, y1, f1}, {x2, y2, f2}, ...} and want to do ListContourPlot. However, there is a problem that the data f is not smooth enough and has small errors.
Here is a toy example illustrating the problem:
data = Flatten[
Table[{i (1 + RandomReal[0.1]), j (1 + RandomReal[0.1]),
i + j}, {i, 0, 100}, {j, 0, 100}], 1];
ListContourPlot[data, Contours -> {80, 120}, ContourShading -> None]
I would need to smooth the lines in the above curve, to get something like
Any ideas to get this kind of smoothed plot? Thanks!
And here is the figure from real data (which is too large to paste here), which I need to smooth:
I find this thread InterpolationOrder for ContourPlot related. But I was unable to get the method work in my case because of (I guess) different input.
Thanks a lot for the answers! Those answers works great for the toy example, but for the real data I still cannot get smooth lines so far (edit: so far means before the great guys update their answers ^_^. Now it worked great). Here is my data and plotting function, just in case you would like to give a try. I will also investigate why those interesting methods fail to work when I apply it to data...
https://www.dropbox.com/s/2pgobgrhbo42f4v/contourData.dat
data = << "contourData.dat";
ListContourPlot[data, Contours -> {2.30, 6.18},
• I guess the main issue is that your toy example is uniformly sampled, while your real data is very far from it (see ListPlot[Most /@ data, AspectRatio -> 1]: i.stack.imgur.com/DgAAh.png vs. i.stack.imgur.com/zfE5C.png). Also, the aspect ratio of the data is very different from the way you're plotting it. – user484 Apr 7 '14 at 17:21
• @RahulNarain : Thanks! Yes it's a very good point. The real data is from Markov chain Monte Carlo (MCMC), which is by purpose not uniformly distributed. – Yi Wang Apr 7 '14 at 19:43
N.B. Your actual data calls for a more sophisticated approach than the quick hack in my original answer, so I've replaced it with a much better and quite general solution.
There are two things that make your actual data harder to work with than the toy example. First, it is highly irregular and nonuniformly distributed:
ListPlot[Most /@ data, AspectRatio -> 1]
and second, it has an awful aspect ratio:
ListPlot[Most /@ data, AspectRatio -> Automatic]
If your $x$ and $y$ axes aren't actually spatial coordinates but represent independent quantities with different units, you would do well to rescale the data so that, say, the variances along both axes are equal:
sd = StandardDeviation[data];
{scalex, scaley} = 1/Most[sd];
scaledData = {scalex, scaley, 1} # & /@ data;
Now to smooth an arbitrary nonlinear function described by noisy samples at irregularly scattered locations, I believe a local regression (LOESS) model is appropriate. In fact, LOESS is a generally useful thing so it's worthwhile to have an implementation available in Mathematica, and since it's pretty simple to implement, I went ahead and did it. My implementation follows Cleveland and Devlin's 1988 paper, i.e. local quadratic regression with tricube weights, except their $q$ is my $k$.
(* precompute spatial data structure because we'll be needing nearest neighbours a lot *)
nearest = Nearest[scaledData /. {x_, y_, z_} :> ({x, y} -> {x, y, z})];
(* local quadratic regression with k neighbours around point (x, y) *)
loess[nearest_, k_][x_, y_] := Module[{n, d, w, f},
n = nearest[{x, y}, k];
d = EuclideanDistance[{x, y}, Most[#]] & /@ n;
d = d/Max[d];
w = (1 - d^3)^3;
f = LinearModelFit[n, {u, v, u^2, v^2, u v}, {u, v}, Weights -> w];
f[x, y]]
Now we can plot the contours of the regression function, scaling the coordinates back to the original data:
fit = loess[nearest, 100];
{{xmin, xmax}, {ymin, ymax}, {zmin, zmax}} = {Min[#], Max[#]} & /@ Transpose[data];
ContourPlot[fit[scalex x, scaley y], {x, xmin, xmax}, {y, ymin, ymax},
Contours -> {2.30, 6.18}, ContourShading -> None]
Seems to work pretty well.
• Isn't this sort of MovingAverage? – Vitaliy Kaurov Apr 7 '14 at 16:33
• @RahulNarain I would suggest to add smouthData[n_] and this gif for fanciness purposes ;o) – Öskå Apr 7 '14 at 16:43
• The result is amazing! I will read it carefully and try to understand the algorithm! – Yi Wang Apr 8 '14 at 7:24
This is likely a consequence of taking the toy example too seriously, but LinearModelFit seems like a good choice:
lmf=LinearModelFit[data,{x,y},{x,y}]
For a the data provided you might get some use from:
basis[n_,m_] := Flatten[Table[x^i y^j,{i,0,n},{j,0,m}],1]
lmf = LinearModelFit[data,basis[4,6],{x,y}];
Now use this linear model:
cp = ContourPlot[lmf[x, y], {x, 0, 0.5}, {y, -4, 4},
Contours -> {2.30, 6.18}, ContourShading -> None,
ContourStyle -> Red]
Show[ListContourPlot[data, Contours -> {2.30, 6.18},
Not too bad, of course you can adjust the basis used as appropriate. At some point, if you have the information to create a nonlinear model that will be better of course.
• If I understand correctly, his real data are not linear and the model probably is unknown. – Vitaliy Kaurov Apr 7 '14 at 16:31
• NonlinearModelFit might also be appropriate as well. In either case, I suspect they don't want to "smooth" as much as fit. The picture they show doesn't look to terrible to model. – chuy Apr 7 '14 at 16:34
• Sorry, I'm facing a similar problem. How can I use NonlinearModelFit in this algorithm? – Orion Jun 23 at 8:32
I was thinking - how can we average but without loss of points? Well we can randomly sample, interpolate and average - as many times as we want.
Let's take a look at more complicated data:
data = Flatten[Table[{i (1 + RandomReal[0.1]), j (1 + RandomReal[0.1]),
i^2 + j^2}, {i, 0, 100}, {j, 0, 100}], 1];
This is like ~ 10^4 points. Grab samples by 1000 - and many of those - and Interpolate - ListContourPlot anyways does that:
Do[f[k] = Interpolation[RandomSample[data, 1000], InterpolationOrder -> 1], {k, 100}]
Average them:
F[x_, y_] = Sum[f[k][x, y], {k, 100}]/100;
Now let's see:
Show[
ContourPlot[F[x, y], {x, 0, 100}, {y, 0, 100},
ContourShading -> None, ContourStyle -> Directive[Red, Thick]],
ListContourPlot[data, ContourShading -> None, Contours -> 9]
] // Quiet
Anyway - something along these lines.
============ OLD VERSION ============
Well, what I will suggest is brutally simple. You want something better. But for the sake of thought entertainment... You get too much detailed interpolation because you got so many data points. Reduce them?
Manipulate[
Show[
ListPlot3D[data[[1 ;; -1 ;; n]], PlotStyle -> Opacity[.5],
MeshFunctions -> {#3 &}, Mesh -> 10,
MeshStyle -> Directive[Opacity[.5], Thick],
PerformanceGoal -> "Quality"],
ListPointPlot3D[data[[1 ;; -1 ;; n]],
PlotStyle -> Directive[Opacity[.5], Red]]
, ImageSize -> 500]
, {{n, 1}, 1, 147, 1, Appearance -> "Labeled"}]
ListContourPlot[data[[1 ;; -1 ;; 139]], Contours -> {80, 120}, ContourShading -> None]
You can also use moving average, but it also removes points:
ListContourPlot[MovingAverage[data, 50], ContourShading -> None, Contours -> {80, 120}]
• Hey Vitaly, you really are into the screencast thing (I like it) ;D – Yves Klett Apr 7 '14 at 16:22
• @YvesKlett haha yes I love that ;-) – Vitaliy Kaurov Apr 7 '14 at 16:26
• @VitaliyKaurov : Thanks a lot for the answer! The data[[a ;; b ;; c] method indeed improves the figure. But as you said, some data are lost and in my real example the quality is still not good enough. The moving average way is lovely. However, my {{x1, y1, f1}, {x2, y2, f2}, ...} list is actually not ordered in x1 < x2 < x3... but instead in a random way. Thus MovingAverage does not directly apply for my real need... Also, the screencasting is amazing :) – Yi Wang Apr 7 '14 at 16:33
• @YiWang Then RahulNarain moving average method via nearest should work. – Vitaliy Kaurov Apr 7 '14 at 16:35
• @VitaliyKaurov : Thanks a lot for the update! Yes, random sampling, re-sampling and averaging is a very interesting idea :) – Yi Wang Apr 8 '14 at 11:12 | 2020-07-16T17:27:05 | {
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https://math.stackexchange.com/questions/2525334/the-limit-of-a-sum-lim-n-to-infty-sum-k-1n-left-frackn2-frac | # The limit of a sum: $\lim_{n\to \infty} \sum_{k=1}^{n} \left(\frac{k}{n^2}-\frac{k^2}{n^3}\right)$
Evaluate the following limit: $$\lim_{n\to \infty} \sum_{k=1}^{n} \left(\frac{k}{n^2}-\frac{k^2}{n^3}\right)$$
I haven't ever taken the limit of the sum... Where do I start?
Do I start taking the sum?
• "Riemann sum" is the keyword you are looking for. – Clement C. Nov 17 '17 at 21:52
• Factor out $\frac{1}{n}$ and you got this :) – Dionel Jaime Nov 17 '17 at 21:53
With Riemann sums:
We have $$\sum_{k=1}^n \left(\frac{k}{n^2}-\frac{k^2}{n^3}\right) = \frac{1}{n}\sum_{k=1}^n \left(\frac{k}{n}-\left(\frac{k}{n}\right)^2\right) = \frac{1}{n}\sum_{k=1}^n \frac{k}{n}\left(1-\frac{k}{n}\right)$$ which is a Riemann sum for $f\colon[0,1]\to\mathbb{R}$ defined by $f(x)=x(1-x)$. Therefore, we have $$\lim_{n\to\infty} \sum_{k=1}^n \left(\frac{k}{n^2}-\frac{k^2}{n^3}\right) = \int_0^1 f(x)dx = \left[\frac{x^2}{2}-\frac{x^3}{3}\right]^1_0 =\boxed{ \frac{1}{6}}\,.$$
Without Riemann sums:
Here, you can directly use the facts that $\sum_{k=1}^n k = \frac{n(n+1)}{2}$ and $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$ to compute the sum, and conclude afterwards. $$\sum_{k=1}^n \left(\frac{k}{n^2}-\frac{k^2}{n^3}\right) = \frac{1}{n^2}\sum_{k=1}^n k-\frac{1}{n^3}\sum_{k=1}^n k^2 = \frac{n(n+1)}{2n^2}-\frac{n(n+1)(2n+1)}{6n^3} \xrightarrow[n\to\infty]{} \frac{1}{2} - \frac{1}{3} = \boxed{\frac{1}{6}}\,.$$
Extra approach: by convolutions. By defining, for any $n\in\mathbb{N}$, $$a_n = \sum_{k=0}^{n}k(n-k)$$ we have that $a_n$ is the coefficient of $x^n$ in $\left(0+1x+2x^2+3x^3+\ldots\right)^2$, i.e. $$a_n = [x^n]\left(\frac{x}{(1-x)^2}\right)^2 = [x^{n-2}]\frac{1}{(1-x)^4}\stackrel{\text{stars and bars}}{=}[x^{n-2}]\sum_{k\geq 0}\binom{k+3}{3}x^k$$ so $a_n = \binom{n+1}{3}$ and $$\lim_{n\to +\infty}\frac{a_n}{n^3} = \frac{1}{3!}=\color{red}{\frac{1}{6}}.$$
• Nice! (Not as convoluted as I thought reading the first line.) – Clement C. Nov 17 '17 at 22:16
• Is "stars and bars" a reference to binomial theorem? If there is some humor I am not aware of I want to be enlightened. +1 for your answer – Paramanand Singh Nov 18 '17 at 2:40
• @ParamanandSingh: no humor, it is the classical combinatorial lemma known as such: en.wikipedia.org/wiki/Stars_and_bars_(combinatorics) – Jack D'Aurizio Nov 18 '17 at 2:42
• Ok i knew the technique, but not the name "stars and bars". :) – Paramanand Singh Nov 18 '17 at 2:43
Since Riemann sums seem to do the trick, let's look at $d(u) =\lim_{n\to \infty} \sum_{k=1}^{n} \frac{k^u}{n^{u+1}}$ where $u \ge 0$.
Doing what was done before,
$\begin{array}\\ d(u) &=\lim_{n\to \infty} \dfrac1{n}\sum_{k=1}^{n} \frac{k^u}{n^{u}}\\ &=\lim_{n\to \infty} \dfrac1{n}\sum_{k=1}^{n} \left(\frac{k}{n}\right)^u\\ &=\int_0^1 x^u dx\\ &=\dfrac1{u+1}\\ \end{array}$
Therefore, if $d(u, v) =\lim_{n\to \infty} \sum_{k=1}^{n} \left(\frac{k^u}{n^{u+1}}-\frac{k^v}{n^{v+1}}\right)$, $d(u, v) =\dfrac1{u+1}-\dfrac1{v+1}$.
If $u=1, v=2$, this gives $d(1, 2) =\dfrac12-\dfrac13 =\dfrac16$.
• In the RHS of d(u,v), two of the u's ought to be v's. – Clement C. Nov 18 '17 at 0:46
• Thanks. Fixed and upvoted. The perils of copy and paste. – marty cohen Nov 18 '17 at 1:01 | 2019-06-16T20:30:03 | {
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https://math.stackexchange.com/questions/2147877/prove-using-mathematical-induction-for-n-ge-1-52n-42n-is-divisible | # Prove using mathematical induction: for $n \ge 1, 5^{2n} - 4^{2n}$ is divisible by $9$
I have to prove the following statement using mathematical induction.
For all integers, $n \ge 1, 5^{2n} - 4^{2n}$ is divisible by 9.
I got the base case which is if $n = 1$ and when you plug it in to the equation above you get 9 and 9 is divisible by 9.
Now the inductive step is where I'm stuck.
I got the inductive hypothesis which is $5^{2k} - 4^{2k}$
Now if P(k) is true than P(k+1) must be true. $5^{2(k+1)} - 4^{2(k+1)}$
These are the step I gotten so far until I get stuck:
$$5^{2k+2} - 4^{2k+2}$$ $$= 5^{2k}\cdot 5^{2} - 4^{2k} \cdot 4{^2}$$ $$= 5^{2k}\cdot 25 - 4^{2k} \cdot 16$$
Now after this I have no idea what to do. Any help is appreciated.
• Adding and subtracting the same thing is just like "adding zero" just in a tricky sort of way, and you are always allowed to add zero just as you are always allowed to multiply by one without changing anything. Feb 16 '17 at 22:03
• Main correction thus far is that the inductive hypothesis for $n=k$ is $9\mid(5^{2k} - 4^{2k})$, which can also be expressed by $(5^{2k} - 4^{2k})\equiv 0\pmod 9$, or, alternatively "Exits $m\in \mathbb Z$ such that $9m= 5^{2k} - 4^{2k}$" Feb 16 '17 at 22:07
You're very close. Now add and subtract $4^{2k}$ in the first term to obtain $$5^{2k}\cdot 25-4^{2k}\cdot 16=25\cdot (5^{2k}-4^{2k})+(25-16)\cdot 4^{2k}=25\cdot (5^{2k}-4^{2k})+9\cdot 4^{2k}$$
The first term is divisible by $9$ by the induction hypothesis, hence the whole expression is divisible by $9$.
• I dont get how like you get two 25 after the first equal sign. Feb 16 '17 at 22:04
• The $5^{2k}$ in the first expression becomes $5^{2k}-4^{2k}+4^{2k}$, and then I rearranged the terms. Feb 16 '17 at 22:05
• Thank you I understood it but these concept are weird, like knowing when to add and subtract a number and stuff like that. Feb 16 '17 at 22:14
• Some of it comes from experience, but also since you know that $5^{2k}-4^{2k}$ is divisible by $9$ it stands to reason that you should try to make such a term appear. Feb 16 '17 at 22:16
• @shawn It appears weird because when the proof is presented this way it is completely devoid of arithmetical intuition. There are other ways to present the proof that highlight the arithmetical structure, which will become clearer when one learns modular arithmetic. I show one in may answer. I also highlight the key role of the Congruence Product Rule in many similar inductive problems, e.g. see the links here.. Feb 16 '17 at 22:38
The induction hypothesis can be written $$5^{2k}-4^{2k}=9m$$ for some integer $m$. Therefore $5^{2k}=4^{2k}+9m$ and so $$5^2\cdot5^{2k}-4^2\cdot4^{2k}= 25(9m+4^{2k})-16\cdot4^{2k}= 9\cdot 25m+(25-16)\cdot 4^{2k}= 9\cdot 25m-9\cdot 4^{2k}$$
Alternatively, $4\equiv -5\pmod{9}$, so $$5^{2k}-4^{2k}\equiv 5^{2k}-(-5)^{2k}\equiv 5^{2k}-5^{2k}\pmod{9}$$
• Thank you, this is also helpful but at the end how did you get the extra 9. Feb 16 '17 at 22:15
• @shawnedward $25-16=9$. I added one step. Feb 16 '17 at 22:17
• makes sense now, thank you. Feb 16 '17 at 22:20
• @shawnedward This strategy essentially works the same in every case you have to prove something like “$ra^n+sb^n$ is divisible by $c$”. Here $r=1$, $s=-1$, $a=25$, $b=16$ and $c=9$. I find it much simpler than “adding and subtracting something”. Feb 16 '17 at 22:25
\begin{align}{\bf Hint}\qquad\qquad\qquad\qquad\,\ \color{#c00}{25} &=\,\ \color{#c00}{16 + 9}\\ 25^{\large N} &=\,\ 16^{\large N}\! +\! 9j\\ \Rightarrow\,\ 25^{\large N+1}\! = \color{#c00}{25}\cdot 25^{\large N} &= (16^{\large N}\!+\!9j)(\color{#c00}{16+\!9}) = 16^{\large N+1} +9\,(\cdots)\ \end{align}
Or, said mod \,9\!:\,\ \begin{align} 25&\equiv 16\\ 25^{\large N}&\equiv 16^{\large N}\end{align}\ \Rightarrow\, 25^{\large N+1}\equiv 16^{\large N+1}\, by the Congruence Product Rule
Or, equivalently, $\ \big[25\equiv 16\big]^{\large N}\!\Rightarrow\, 25^{\large N}\!\equiv 16^{\large N}\,$ by the Congruence Power Rule, which is an inductive extension of the Product Rule.
The inductive hypothesis is
$$(5^{2n}-4^{2n})\bmod9=0.$$
Then
$$(5^{2n+2}-4^{2n})\bmod9=(25\cdot5^{2n}-16\cdot4^{2n})\bmod9=(9\cdot5^{2n}+16(5^{2n}-4^{2n}))\bmod9=0.$$ | 2021-09-17T04:37:37 | {
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http://www.apedigital.com/209tjkt/210125-greater-than-or-equal-to-sign | In Greater than or equal operator A value compares with B value it will return true in two cases one is when A greater than B and another is when A equal to B. As we saw earlier, the greater than and less than symbols can also be combined with the equal sign. When we say 'as many as' or 'no more than', we mean 'less than or equal to' which means that a could be less than b or equal to b. For example, x ≥ -3 is the solution of a certain expression in variable x. "Greater than or equal to", as the suggests, means something is either greater than or equal to another thing. The greater-than sign is a mathematical symbol that denotes an inequality between two values. If left-hand operator higher than or equal to right-hand operator then condition will be true and it will return matched records. 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Copy the Greater-than Or Equal To in the above table (it can be automatically copied with a mouse click) and paste it in word, Or. With Microsoft Word, inserting a greater than or equal to sign into your Word document can be as simple as pressing the Equal keyboard key or the Greater Than keyboard key, but there is also a way to insert these characters as actual equations. This symbol is nothing but the "greater than" symbol with a sleeping line under it. Greater than or Equal in Excel – Example #5. "Greater than or equal to" is represented by the symbol " ≥ ≥ ". “Greater than or equal to” and “less than or equal to” are just the applicable symbol with half an equal sign under it. Graphical characteristics: Asymmetric, Open shape, Monochrome, Contains straight lines, Has no crossing lines. ≥. use ">=" for greater than or equal use "<=" for less than or equal In general, Sheets uses the same "language" as Excel, so you can look up Excel tips for Sheets. Select the Greater-than Or Equal To tab in the Symbol window. Greater Than or Equal To: Math Definition. The less than or equal to symbol is used to express the relationship between two quantities or as a boolean logical operator. 2 ≥ 2. is less than > > is greater than ≮ \nless: is not less than ≯ \ngtr: is not greater than ≤ \leq: is less than or equal to ≥ \geq: is greater than or equal to ⩽ \leqslant: is less than or equal to ⩾ Solution for 1. Select Symbol and then More Symbols. Less Than or Equal To (<=) Operator. But, when we say 'at least', we mean 'greater than or equal to'. 923 Views. The sql Greater Than or Equal To operator is used to check whether the left-hand operator is higher than or equal to the right-hand operator or not. In such cases, we can use the greater than or equal to symbol, i.e. Select the Insert tab. N'T be found if left-hand operator higher than or equal to '', as the suggests, something. 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https://www.physicsforums.com/threads/differentiation-under-the-integral-sign.612414/ | # Differentiation under the integral sign
1. Jun 8, 2012
### naaa00
1. The problem statement, all variables and given/known data
R(x) := ∫ exp ( -y^2 - x^2/y^2 ) dy
3. The attempt at a solution
I move the derivative operator inside the integral and differentiate with respect to x
R'(x) = ∫ [ - 2x/y^2 ] exp ( -x^2/y^2 - y^2 ) dy
Then I let: t = x/y and dy = - x/t^2 dt
R'(x) = 2 ∫ [ - x ] [ t^2 / x^2 ] exp ( t^2 - x^2/t^2 ) [ - x/t^2 ] dt
=> R'(x) = 2 R(x)
But that last part is supposed to be R'(x) = - 2 R(x) - I don't see why.
Then, it follows that this integrates to R(x) = Ae^(−2x) and x = 0 gives R(0) = √π. This last part I don't get it neither. From where the e^(-2x) came from?
Any help is very appreciated.
2. Jun 8, 2012
### Ray Vickson
I also get R'(x) = 2*[R(x) + c], where c is a constant of integration. Here are the steps, done in Maple 11 (with output written in LaTeX):
R:=Int(exp(-x^2/y^2-y^2),y);
$$R = \int e^{\left(\displaystyle -\frac{x^2}{y^2} - y^2\right)} \, dy$$
Rp:=diff(R,x);
$$Rp = \int -\frac{2x}{y^2} e^{\left(\displaystyle -\frac{x^2}{y^2} - y^2\right)} \, dy$$
changevar(x/y=t,Rp);
$$\int e^{\left(\displaystyle -\frac{x^2}{t^2} - t^2\right)} \, dt$$
RGV
3. Jun 8, 2012
### naaa00
I just realize that I didn't consider the limits of integration... The limits are from 0 to infinity...
but still something must be wrong...
4. Jun 8, 2012
### Muphrid
Yeah, the negative probably comes in from having to flip the limits of integration when you transform to the varaible t.
5. Jun 8, 2012
### HallsofIvy
Staff Emeritus
Does "exp(-y^2- x^2/y^2)" mean $e^{-y^2- (x^2/y^2)}$ or $e^{(-y^2- x^2)/y^2}$.
6. Jun 8, 2012
### naaa00
Hello Halls,
The one in the middle.
7. Jun 8, 2012
### SammyS
Staff Emeritus
So the definition of R(x) is:
$\displaystyle R(x) := \int_{0}^{\infty} e^{\left( -\frac{x^2}{y^2} - y^2\right)} \, dy$
That explains why R is not a function of y. | 2017-08-22T00:08:52 | {
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https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Set_Partition | # Equivalence of Definitions of Set Partition
## Theorem
Let $S$ be a set
The following definitions of the concept of Set Partition are equivalent:
### Definition 1
A partition of $S$ is a set of subsets $\Bbb S$ of $S$ such that:
$(1): \quad$ $\Bbb S$ is pairwise disjoint: $\forall S_1, S_2 \in \Bbb S: S_1 \cap S_2 = \O$ when $S_1 \ne S_2$
$(2): \quad$ The union of $\Bbb S$ forms the whole set $S$: $\displaystyle \bigcup \Bbb S = S$
$(3): \quad$ None of the elements of $\Bbb S$ is empty: $\forall T \in \Bbb S: T \ne \O$.
### Definition 2
A partition of $S$ is a set of non-empty subsets $\Bbb S$ of $S$ such that each element of $S$ lies in exactly one element of $\Bbb S$.
## Proof
### Definition 1 implies Definition 2
Let $\Bbb S$ be a set of subsets $\Bbb S$ of $S$ such that:
$(1): \quad$ $\Bbb S$ is pairwise disjoint: $\forall S_1, S_2 \in \Bbb S: S_1 \cap S_2 = \O$ when $S_1 \ne S_2$
$(2): \quad$ The union of $\Bbb S$ forms the whole set $S$: $\displaystyle \bigcup \Bbb S = S$
$(3): \quad$ None of the elements of $\Bbb S$ is empty: $\forall T \in \Bbb S: T \ne \O$.
By $(3)$ all sets in $\Bbb S$ are non-empty.
Suppose there exists $x \in S$ in more than one set of $\Bbb S$.
That is:
$\exists S_1, S_2 \in \Bbb S, S_1 \ne S_2: x \in S_1 \land x \in S_2$
Then by definition of set intersection:
$x \in S_1 \cap S_2$
and so $S_1 \cap S_2 \ne \varnothing$.
This contradicts condition $(1)$.
So no element of $S$ can be in more than one set of $\Bbb S$.
Suppose there exists $x \in S$ in none of the sets of $\Bbb S$.
Then $\displaystyle x \notin \bigcup \Bbb S$ and so $\Bbb S \notin S$.
This contradicts condition $(2)$.
So every element of $S$ is in at least one set of $\Bbb S$.
Thus $x$ lies in exactly one element of $\Bbb S$.
Hence it follows that $\Bbb S$ is a set of non-empty subsets of $S$ such that each element of $S$ lies in exactly one element of $\Bbb S$.
$\Box$
### Definition 2 implies Definition 1
Let $\Bbb S$ be a set of non-empty subsets of $S$ such that each element of $S$ lies in exactly one element of $\Bbb S$.
$(3)$ is fulfilled automatically by definition.
Consider $\Bbb S$.
$\displaystyle \bigcup \Bbb S \subseteq S$
Let $x \in T$.
$\exists S_1 \in \Bbb S: x \in S_1$
That is:
$\displaystyle x \in \bigcup \Bbb S$
and so:
$\displaystyle S \subseteq \bigcup \Bbb S$
By definition of set equality:
$\displaystyle \bigcup \Bbb S = S$
thus demonstrating that $(2)$ holds.
Suppose $\Bbb S$ were not pairwise disjoint, and that:
$\exists S_1, S_2 \in \Bbb S: S_1 \cap S_2 \ne \O$
Then:
$\exists x \in S: x \in S_1 \cap S_2$
By definition of set intersection:
$x \in S_1$ and $x \in S_2$
thus contradicting the supposition that each element of $S$ lies in exactly one element of $\Bbb S$.
So $\Bbb S$ is pairwise disjoint and so $(1)$ holds.
Hence the result.
$\blacksquare$ | 2020-05-28T22:31:02 | {
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http://mathhelpforum.com/pre-calculus/25399-even-odd-neither.html | Math Help - Even, Odd, or Neither
1. Even, Odd, or Neither
Determine whether each of the following functions are even, odd, or neither. Please state the symmetry.
a)f(x)=1 / (x^2+1) b)g(x)=x^2 / (1+x^4) c)h(x)= (2x^4) + (3x^2)
d)f(x)={1 / (x^3+x)}^5 e)g(x)=x+(1/x) f)h(x)=x-(x^2)
g)f(x)=2^x h)g(x)=(logx^2) / (log3x^4)
For each of the above functions from a) through h) how do you determine if it is even,odd,or neither??
2. Originally Posted by johett
Determine whether each of the following functions are even, odd, or neither. Please state the symmetry.
a)f(x)=1 / (x^2+1) b)g(x)=x^2 / (1+x^4) c)h(x)= (2x^4) + (3x^2)
d)f(x)={1 / (x^3+x)}^5 e)g(x)=x+(1/x) f)h(x)=x-(x^2)
g)f(x)=2^x h)g(x)=(logx^2) / (log3x^4)
For each of the above functions from a) through h) how do you determine if it is even,odd,or neither??
a function is even if f(x) = f(-x). that is, if you replace x with -x and simplify, you get exactly the original function
a function is odd if f(-x) = -f(x). that is, if you replace x with -x and simplify, you get the negative of the original function
see the "Ways to test for symmetry" section here on how to test for other kinds of symmetry. i don't know what kinds you are expected to look for.
note: symmetry about the y-axis is the same as being even, and symmetry about the origin is the same as being odd
3. can someone show me an example like question a) so I can have a better understanding thanks
4. Originally Posted by johett
can someone show me an example like question a) so I can have a better understanding thanks
i'll give you other examples. then try your questions and post your solutions and we'll tell you if you applied the rules correctly.
example 1:
the function $f(x) = x^2 + 1$ is even, since replacing x with -x we get:
$f(-x) = (-x)^2 + 1 = x^2 + 1 = f(x)$
since $f(-x) = f(x)$, the function is even
example 2:
the function $f(x) = x^3$ is odd, since if we replace x with -x we get:
$f(-x) = (-x)^3 = -x^3 = -f(x)$
since $f(-x) = -f(x)$, the function is odd
example 3:
the function $f(x) = x^3 + 3$ is neither even nor odd, since if we replace x with -x we get:
$f(-x) = (-x)^3 + 3 = -x^3 + 3$
clearly $f(-x) \ne f(x)$ and $f(-x) \ne -f(x)$. so the function is neither even nor odd
This is my 59th post!!!!
5. Originally Posted by johett
can someone show me an example like question a) so I can have a better understanding thanks
Originally Posted by johett
Determine whether each of the following functions are even, odd, or neither. Please state the symmetry.
a)f(x)=1 / (x^2+1)
d)f(x)={1 / (x^3+x)}^5
f)h(x)=x-(x^2)
For each of the above functions from a) through h) how do you determine if it is even,odd,or neither??
a)
$f(x) = \frac{1}{x^2 + 1}$
$f(-x) = \frac{1}{(-x)^2 + 1}$
$f(-x) = \frac{1}{x^2 + 1} = f(x)$
so this function is even.
d)
$f(x) = \left ( \frac{1}{x^3 + x} \right ) ^5$
$f(-x) = \left ( \frac{1}{(-x)^3 + (-x)} \right ) ^5$
$f(-x) = \left ( \frac{1}{-x^3 - x} \right ) ^5$
$f(-x) = \left ( \frac{-1}{x^3 + x} \right ) ^5$
$f(-x) = (-1)^5 \left ( \frac{1}{x^3 + x} \right ) ^5$
$f(-x) = - \left ( \frac{1}{x^3 + x} \right ) ^5 = -f(x)$
so this function is odd.
f)
$h(x) = x - x^2$
$h(-x) = (-x) - (-x)^2$
$h(-x) = -x + x^2$
which is equal to neither h(x) nor -h(x). So this function is neither even, nor odd.
-Dan | 2015-01-27T04:34:25 | {
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http://math.stackexchange.com/questions/207005/how-many-numbers-will-i-get-right?answertab=oldest | # How many numbers will I get right?
I play a guessing game. In this game, there are 100 equally-sized, folded-up cards randomly dispersed in a bag. The cards are labeled 1 through 100. I draw out the cards one by one and try to guess the number on the card every time I draw. On every guess, a genie will tell me if I am correct or not (but he won't tell me the actual number on the card).
I start by guessing that the card has 1 on it. I keep guessing that the card has 1 on it until I am correct, after which I will keep guessing 2 until I am correct again, after which I will keep guessing 3, and so on until the bag is empty. How many correct guesses should I expect?
My current hunch is that the answer is 1 since this looks like a binomial distribution, and $52\frac{1}{52} = 1$. However, I feel suspicious since there's that genie that gives me extra information...
-
Are you asking about that particular strategy, or an optimal strategy? Even for the $1,1,\dots,2,\dots$ strategy, the expectation is $\gt 1$, since for sure you will get at least $1$ right, and maybe more. – André Nicolas Oct 4 '12 at 6:35
Just for that particular "strategy." It would be interesting to look at the optimal strategy... but I think that might open up a new can of worms. :) – Minden Petrofsky Oct 4 '12 at 6:37
I think this is the optimal strategy. The most likely next draw is the one you've guessed most often so far but haven't yet seen. – mjqxxxx Oct 4 '12 at 6:44
What happens if you draw 2 before you draw 1? Do you then guess 3 (or whatever number is the next unseen one) once you find 1? If so, that would be optimal. If you don't skip the numbers that you've seen, then it's definitely not optimal. – user22805 Oct 4 '12 at 7:48
@David Wallace - if he is not correct, the genie will not tell him the actual number on the card. So he will only know that a sequence of "not-1"s was followed by a "1". – Viliam Búr Oct 4 '12 at 11:16
You’re certain to get at least one correct guess, and in many cases you’ll get more than one, so the expected number of correct guesses is clearly more than one. You’ll get a second correct guess whenever you draw the $1$ before you draw the $2$; the probability of that is $\frac12$, so the expected number of correct guesses can already be seen to be at least $\frac12\cdot1+\frac12\cdot2=\frac32$.
In general you’ll get at least $n$ correct guesses if the numbers $1,\dots,n$ appear in their correct order, ignoring any larger numbers that may appear between them. The probability of that is $\frac1{n!}$, since all $n!$ orders in which they could appear are equally likely. The probability that $n+1$ is the last of the set $\{1,\dots,n+1\}$ to be drawn is $\frac1{n+1}$, so the probability that it is drawn before you draw all of the smaller numbers is $\frac{n}{n+1}$.
Thus, the probability that you’ll get exactly $n$ correct guesses is $$\frac1{n!}\cdot\frac{n}{n+1}=\frac{n}{(n+1)!}\;,$$ and the expected number of correct guesses is
$$\sum_{n=1}^{100}\frac{n}{(n+1)!}\cdot n=\sum_{n=1}^{100}\frac{n^2}{(n+1)!}\;.$$
Now $$e^x=\sum_{n\ge 0}\frac{x^n}{n!}\;,$$ so $$\frac{e^x-1}x=\sum_{n\ge 1}\frac{x^{n-1}}{n!}=\sum_{n\ge 0}\frac{x^n}{(n+1)!}\;,$$
$$x\frac{d}{dx}\left(\frac{e^x-1}x\right)=\sum_{n\ge 0}\frac{nx^n}{(n+1)!}\;,$$ and
$$\frac{d}{dx}\left(x\frac{d}{dx}\left(\frac{e^x-1}x\right)\right)=\sum_{n\ge 0}\frac{n^2x^{n-1}}{(n+1)!}\;.\tag{1}$$
Finally, $$\frac{d}{dx}\left(x\frac{d}{dx}\left(\frac{e^x-1}x\right)\right)=\frac{d}{dx}\left(\frac{xe^x-e^x+1}x\right)=e^x-\frac{(x-1)e^x}{x^2}-\frac1{x^2}\;.\tag{2}$$
Evaluating the righthand sides of $(1)$ and $(2)$ at $x=1$, we see that
$$\sum_{n\ge 1}\frac{n^2}{(n+1)!}=e-1\;,$$ so the expected number of correct guesses is just a hair less than $e-1\approx 1.718281828459045$.
-
Though you don't say so, I assume that cards are discarded once they're removed from the bag. The key observation is that you will get at least $n$ correct guesses if and only if the first $n$ cards (that is, the cards numbered $1,2,\ldots,n$) appear in the correct order among your draws. Since each of the $n!$ orderings of the first $n$ cards within your random list of draws is equally likely, you'll get at least $n$ correct guesses with probability $(1/n!)$. Therefore the probability of getting exactly $n$ correct guesses must be $$\frac{1}{n!}-\frac{1}{(n+1)!}=\frac{1}{n!}\left(1-\frac{1}{n+1}\right)=\frac{n}{(n+1)!},$$ and the expected number of correct guesses is $$\sum_{n=1}^{100}\frac{n^2}{(n+1)!} \approx (e-1) = 1.718281828\ldots$$
- | 2014-08-01T08:21:26 | {
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https://sriasat.wordpress.com/page/2/ | ## A combinatorial identity
While tutoring today a student asked me about the following identity
$(*)\qquad\qquad\qquad\displaystyle\sum_{n\ge i\ge j\ge k\ge 0}\binom ni\binom ij\binom jk=4^n$.
One way to prove it is to count the number of possible triples $(A,B,C)$ of sets with $C\subseteq B\subseteq A\subseteq S=\{1,\dots,n\}$. This can be done in two ways. First, if $A, B, C$ contain $i,j,k$ elements respectively then the number of such triples clearly equals the left-hand side of $(*)$. On the other hand, since each element of $S$ must belong to one of the four disjoint regions in the picture below
the number of such triples equals precisely $4^n$.
An algebraic proof can be obtained by repeated use of the binomial theorem:
$\displaystyle 4^n=\sum_{n\ge i\ge 0}\binom ni3^i=\sum_{n\ge i\ge j\ge 0}\binom ni\binom ij2^j=\sum_{n\ge i\ge j\ge k\ge 0}\binom ni\binom ij\binom jk$.
Similarly, one obtains the more general identity
$\displaystyle\sum_{n\ge i_m\ge\cdots\ge i_1\ge 0}\binom{n}{i_1}\binom{i_1}{i_2}\cdots\binom{i_{m-1}}{i_m}=(m+1)^n$.
This also follows from the multinomial theorem since the left-hand side equals
$\displaystyle\sum_{n\ge i_m\ge\cdots\ge i_1\ge 0}\frac{n!}{(n-i_1)!(i_1-i_2)!\cdots (i_{m-1}-i_m)!i_m!}=(m+1)^n$.
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## Jordan-Hölder theorem
I am cramming for my algebra comprehensive exam so I will probably be posting stuff like this for a while.
A chain of subgroups $\{e\}=H_0\triangleleft H_1\triangleleft\cdots\triangleleft H_n=G$ of a finite group $G$ with $H_{i+1}/H_i$ simple for all $i$ is called a composition series of $G$. (The $H_{i+1}/H_i$ are called composition factors.) The theorem in question states that every group has a composition series, and the composition factors in any two such series are unique up to reordering.
We can easily prove the first part of the theorem, that any finite group has a composition series, using the extreme principle. Let $n\ge 0$ be the largest integer for which there is a chain $\{e\}=H_0\triangleleft H_1\triangleleft\cdots\triangleleft H_n=G$ of subgroups. (Such an $n$ exists because there is trivially a chain with $n=1$, and our group is finite.) We claim that the composition factors in this case are simple. If not, WLOG $H_{i+1}/H_i$ has a non-trivial normal subgroup $\widetilde N$. By correspondence, $\widetilde N=N/H_i$ for some $H_i\triangleleft N\triangleleft H_{i+1}$. But then
$\{e\}=H_0\triangleleft H_1\triangleleft\cdots\triangleleft H_i\triangleleft N\triangleleft H_{i+1}\cdots\triangleleft H_n=G$
is a longer chain, a contradiction.
While the proof of the general theorem requires some non-trivial group theory (e.g. see here), we can easily prove the theorem for finite abelian groups using elementary number theory. For such a group, the orders of the composition factors must exactly be the prime factors—listed with multiplicity—of the order of the group. So the result follows immediately from the fundamental theorem of arithmetic.
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## A new AM-GM inequality
I recently found the following inequalities useful:
$\displaystyle \left(x+\frac{a_1+\cdots+a_n}{n}\right)^n\ge (x+a_1)\cdots(x+a_n)\ge \left(x+\sqrt[n]{a_1\cdots a_n}\right)^n$,
for any $x,a_1,\dots,a_n\ge 0$.
To prove the left half simply apply AM-GM to the product
$\displaystyle (x+a_1)\cdots (x+a_n)\le \left(\frac{(x+a_1)+\cdots+(x+a_n)}{n}\right)^n$.
The right half follows directly from Hölder’s inequality.
More generally, one has
$\displaystyle\left(\frac{a_1+\cdots+a_n}{n}+\frac{b_1+\cdots+b_n}{n}\right)^n\ge (a_1+b_1)\cdots(a_n+b_n)\\\ge \left(\sqrt[n]{a_1\cdots a_n}+\sqrt[n]{b_1\cdots b_n}\right)^n$
for $a_1,\dots,a_n,b_1,\dots,b_n\ge 0$. The proof goes similarly.
One obtains similar inequalities for any number of vectors $a,b,c,\dots$.
As an example, for $a_i=i$ we get the inequalities
$\displaystyle \left(x+\frac{n+1}{2}\right)^n\ge (x+1)\cdots (x+n)>\left(x+\frac ne\right)^n$
using the fact that $\sqrt[n]{n!}/n\to 1/e$ from above.
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## Hilbert’s nullstellensatz
Leaving it here because I want to study this at some point.
I had occasion recently to look up the proof of Hilbert’s nullstellensatz, which I haven’t studied since cramming for my algebra qualifying exam as a graduate student. I was a little unsatisfied with the proofs I was able to locate – they were fairly abstract and used a certain amount of algebraic machinery, which I was terribly rusty on – so, as an exercise, I tried to find a more computational proof that avoided as much abstract machinery as possible. I found a proof which used only the extended Euclidean algorithm and high school algebra, together with an induction on dimension and the obvious observation that any non-zero polynomial of one variable on an algebraically closed field has at least one non-root. It probably isn’t new (in particular, it might be related to the standard model-theoretic proof of the nullstellensatz, with the Euclidean algorithm and high school algebra taking…
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## Midy’s theorem
Let $b$ have order $d$ modulo $p$. Then the base $b$ expansion of $1/p$ has period $L=(b^d-1)/p$ of length $d$. To see this, note that if $b^d-1=mp$, then
$\displaystyle\frac 1p=\frac{m}{b^d-1}=\frac{m}{b^d}+\frac{m}{b^{2d}}+\frac{m}{b^{3d}}+\cdots$.
Note also that $aL for any $1\le a\le p-1$. Hence $a/p$ has period $aL$. Now suppose that $d$ is even. Since $b$ has order $d$ modulo $p$, it follows that $b^{d/2}\equiv -1\pmod p$. Hence $p-a\equiv b^{d/2}a\pmod p$. This means that at their midpoints the two numbers $aL$ and $(p-a)L$ are mirror images of one another. This means that splitting $aL$ midway into two equal parts and adding them gives $b^{d/2}-1$, i.e., a string of $(b-1)$‘s in base $b$. This is known as Midy’s theorem.
For example, with $p=7$ and $b=10$ we get $d=6$, and $L=142857$. Split $L$ into two equal parts $142$ and $857$, adding which gives
$142+857=999$.
In general, if $m$ is any divisor of $d$, then
\begin{aligned} b^d-1&=(b^m-1)(1+b^{d/m}+b^{2d/m}+\cdots+b^{(m-1)d/m})\\&\equiv 0\pmod p\end{aligned}
so
$b^{d/m}+b^{2d/m}+\cdots+b^{(m-1)d/m}\equiv -1\pmod p$
and so splitting $L$ into $m$ equal parts and adding them will always give a multiple of $b^{d/m}-1$.
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Filed under Number theory | 2018-06-23T02:46:21 | {
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http://muattiyah.com/posts/mathematical-induction/ | June 12, 2017
Suppose you want to sum up the first 10 natural numbers 1, 2, 3, ...
The most obvious way to do it would be to add the numbers consecutively like so:
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55
No big deal, but now you want to sum the first 1000 natural numbers i.e. 1, 2, 3, ..., 1000. While you’re carrying out this laborious task, your friend Hypatia passes by and tells you with a confused look “Why are you doing all that? You can just use this formula:”
$$\frac{n(n+1)}{2}$$
Skeptical, you give it a try on the first 10 natural numbers, because you already know the answer to that. So, you substitute n for 10 and you get
$$\frac{10(10+1)}{2} = \frac{110}{2} = 55$$
Lo and Behold, it’s the right answer. Go ahead and try it for other values of n yourself, see if it works.
So it works for the couple of instances that you’ve tried, but how do we know it works for the first 1024 natural numbers i.e. 1, 2, 3, ..., 1024 or for n = 50000. It seems implausible to actually try it for all instances of n. And even if we tried it for n up to 50000, we only know it works thus far, how do we know whether it works for n = 50001, n = 50002, ...etc especially given the fact that the set of Natural Numbers is Countably Infinite.
If you don’t know what the last couple of phrases mean, think of it this way, if we start with the number 1 and add 1 to it and keep adding 1 to the sum, you will always get a new natural number. For example, 1 + 1 = 2, 2 + 1 = 3, 3 + 1 = 4 and so on.
Obviously our manual method of verifying the technique’s applicability on single instances of n is incapable of convincing us of the applicability of the rule for any n and we need a new method to do just that, that method is Mathematical Induction.
## Definition
Mathematical Induction is a general way to prove that some statement about the natural number n is true for all n ≥ 1.1
A proof by induction involves first proving that we want to prove holds for n = 1 and then proving that if it holds for any n then it holds for n + 1.
## Analogy
As a helpful analogy, think of the process of climbing a ladder, we will prove that we can climb the first step, then we will prove that if we can climb any step then we can climb the step that’s after it. Now we know that we can climb the first step and we know that we can climb any step after it. Symbolically, think of the first step as saying that n = 1, think of any step as n and think of the step that follows any step as n + 1.
Informally:
1. Can climb first step n = 1.
2. If we can climb step n then we can climb step n + 1.
3. Let n be 1, we can already climb the first step, let n + 1 be 1 + 1 = 2, we can climb the second step.
4. Let n be 2, we know we can climb it from item 3, so we can climb the n + 1 step which is the third step.
5. Repeat ad infinitum, can climb any step.
## Outline for a proof by induction
A proof by induction will involve the following steps:
1. Prove that the statement where n = 1 is true, this is called the Base Case 2, we’ll refer to this statement as S1.
2. Prove that when n ≥ 1, the statement Sn ⇒ Sn + 1 is true.3 4 This is called the Inductive Step 5.
After we’ve completed these two steps, we say that it follows by Mathematical Induction that every Sn is true.
## Proof Sketch for our Summation Technique
We will first start by proving the Base Case, in our case this means proving that the technique holds for n = 1. Again, we will refer to this statement as S1.
We will then proceed to the Inductive Step, proving that SnSn + 1.
In our case, this means proving that
$$\frac{n(n+1)}{2} \implies \frac{(n+1)((n+1)+1)}{2}$$
## Proof
Base Case: When n = 1, our formula expands to
$$\frac{1(1+1)}{2}$$
which evaluates to 1 and is true.
Inductive Step: Assume that our formula Sn is true, this means that
$$1+2+3+...+n = \frac{n(n+1)}{2}$$
is true. This is called the Induction Hypothesis. We now have to show that Sn + 1 is true, algebraically this means that
$$(1+2+3+...+n)+(n+1) = \frac{(n+1)((n+1)+1)}{2}$$
(1 + 2 + 3 + ... + n)+(n + 1)
Since we’ve already assumed that Sn
$$1+2+3+...+n = \frac{n(n+1)}{2}$$
is true, we can substitute it in the left hand side like so:
\begin{align} (1+2+3+...+n)+(n+1) & = \frac{n(n+1)}{2} + n+1\\ & = \frac{n(n+1)+2(n+1)}{2}\\ & = \frac{n^2+n+2n+2}{2}\\ & = \frac{n^2+3n+2}{2}\\ & = \frac{(n+1)(n+2)}{2}\\ & = \frac{(n+1)((n+1)+1)}{2} \end{align}
This shows us that Sn + 1 is indeed true if Sn is true. Since we’ve proven the Base Case and conducted the Inductive Step, our formula
$$\frac{n(n+1)}{2}$$
is true for all numbers n ≥ 1 by Mathematical Induction.
$$\square$$
## Examples
### Proposition
If n ∈ ℕ6 then the sum of the first n odd numbers 1 + 3 + 5 + 7 + … + (2n − 1) is n2.
#### Proof
Base Case: ( S1 = 1^2 = 1 ), our base case holds.
Inductive Step:
Assume that ( 1+3+5+7+…+(2n-1)=n^2 )
Show that ( S{n+1}: 1+3+5+7+…+(2(n)-1)+(2(n+1)-1)=(n+1)^2 )
[ \begin{align} 1+3+5+7+…+(2n-1)+(2n+1) & = n^2+2n+1
& = (n+1)(n+1)
& = (n+1)^2 \end{align} ]
Therefore, it follows by induction that 1 + 3 + 5 + 7 + ... + (2n − 1)=n2 as required.
$$\square$$
### Proposition
If n ∈ ℕ, then 3|(n3 − n).
#### Proof
Base Case: ( S1: 1^3-1 = 0 ) and we ( 0 ) is divisible by 3, base case holds.
Inductive Step: Assume ( 3 | n^3-n ), this mean that there exists an integer ( a ) such that ( 3a = n^3-n ).
Our goal is to show that ( S{n+1} ) holds i.e. ( 3 | (n+1)^3 - (n+1) )
[ \begin{align} (n+1)^3 - (n+1) & = n^3+3n^2+3n+1-n-1
& = n^3+3n^2+3n-n
& = (n^3-n)+3n^2+3n
& = 3a + 3n^2 + 3n
& = 3(a+n^2+n) \end{align} ]
It follows by induction that 3|(n3 − n).
$$\square$$
## Exercises
### Exercise
If n is a non-negative integer, then 5|(n5 − n).
### Exercise
If n ∈ ℤ and n ≥ 0, then
$$\sum_{i=0}^{n} i.i! = (n+1)!-1$$
### Exercise
For every Natural Number n, 20 + 21 + 22 + … + 2n = 2n + 1 − 1.
## Footnotes
1. Induction is actually more general than this, it can work on any collection which obeys the Well-Ordering Principle.
2. This is sometimes referred to as the Basis.
3. Read this as “Sn implies Sn + 1”.
4. I am gradually introducing notation to get you used to it, if you find it confusing, revisit the #definition and read this again.
5. This is sometimes referred to as the Induction Step.
6. Read this as “n belongs to the set of Natural Numbers”. | 2018-09-25T17:32:53 | {
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http://code.jasonbhill.com/category/cython/ | A triangular pyramid is constructed using spherical balls so that each ball rests on exactly three balls of the next lower level.
Then, we calculate the number of paths leading from the apex to each position: A path starts at the apex and progresses downwards to any of the three spheres directly below the current position. Consequently, the number of paths to reach a certain position is the sum of the numbers immediately above it (depending on the position, there are up to three numbers above it).
The result is Pascal’s pyramid and the numbers at each level n are the coefficients of the trinomial expansion $(x + y + z)^n$. How many coefficients in the expansion of $(x + y + z)^{200000}$ are multiples of $10^{12}$?
## Solution Using the Multinomial Theorem
The generalization of the binomial theorem is the multinomial theorem. It says that multinomials raised to exponents can be expanded using the formula
$(x_1+x_2+\cdots+x_m)^n=\sum_{{k_1+k_2+\cdots+k_m=n}\atop{0\le k_i\le n}}\left({n}\atop{k_1,k_2,\ldots,k_m}\right)\prod_{1\le t\le m}x_t^{k_t}$
where
$\left({n}\atop{k_1,k_2,\ldots,k_m}\right)=\frac{n!}{k_1!k_2!\cdots k_m!}.$
Of course, when m=2 this gives the binomial theorem. The sum is taken over all partitions $k_1+k_2+\cdots+k_m=n$ for integers $k_i$. If n=200000 abd m=3, then the terms in the expansion are given by
$\left({200000}\atop{k_1,k_2,k_3}\right)x_1^{k_1}x_2^{k_2}x_3^{k_3}=\frac{200000!}{k_1!k_2!k_3!}x_1^{k_1}x_2^{k_2}x_3^{k_3}$
where $k_1+k_2+k_3=200000$. It’s worth pointing out that each of the coefficients is an integer, and thus has a unique factorization into products of prime integers. Of course, there’s no way that we’re going to calculate these coefficients. We only need to know when they’re divisible by $10^{12}$. Thus, we only need to consider how many factors of 2 and 5 are involved.
First, we’ll create a function $p(n,d)$ that outputs how many factors of $d$ are included in $n!$. We have that
$p(n,d)=\left\lfloor\frac{n}{d}\right\rfloor+\left\lfloor\frac{n}{d^2}\right\rfloor+\left\lfloor\frac{n}{d^3}\right\rfloor+ \cdots+\left\lfloor\frac{n}{d^r}\right\rfloor,$
where $d^r$ is the highest power of $d$ dividing $n$. For instance, there are 199994 factors of 2 in 200000!. Since we’re wondering when our coefficients are divisible by $10^{12}=2^{12}5^{12}$, we’ll be using the values provided by $p(n,d)$ quite a bit for $d=2$ and $d=5$. We’ll store two lists:
$p2=[p(i,2)\text{ for }1\le i\le 200000]\quad\text{and}\quad p5=[p(i,5)\text{ for }1\le i\le 200000].$
For a given $k_1,k_2,k_3$, the corresponding coefficient is divisible by $10^{12}$ precisely when
$p2[k_1]+p2[k_2]+p2[k_3]<199983\ \text{and}\ p5[k_1]+p5[k_2]+p5[k_3]<49987.$
That is, this condition ensures that there are at least 12 more factors of 2 and 5 in the numerator of the fraction defining the coefficients.
Now, we know that $k_1+k_2+k_3=200000$, and we can exploit symmetry and avoid redundant computations if we assume $k_1\le k_2\le k_3$. Under this assumption, we always have
$k_1\le\left\lfloor\frac{200000}{3}\right\rfloor=66666.$
We know that $k_1+k_2+k_3=200000$ is impossible since 200000 isn't divisible by 3. It follows that we can only have (case 1) $k_1=k_2 < k_3$, or (case 2) $k_1 < k_2=k_3$, or (case 3) $k_1 < k_2 < k_3$.
In case 1, we iterate $0\le k_1\le 66666$, setting $k_2=k_1$ and $k_3=200000-k_1-k_2$. We check the condition, and when it is satisfied we record 3 new instances of coefficients (since we may permute the $k_i$ in 3 ways).
In case 2, we iterate $0\le k_1\le 66666$, and when $k_1$ is divisible by 2 we set $k_2=k_3=\frac{200000-k_1}{2}$. When the condition holds, we again record 3 new instance.
In case 3, we iterate $0\le k_1\le 66666$, and we iterate over $k_2=k_1+a$ where $1\le a < \left\lfloor\frac{200000-3k_1}{2}\right\rfloor$. Then $k_3=200000-k_1-k_2$. When the condition holds, we record 6 instances (since there are 6 permutations of 3 objects).
## Cython Solution
I’ll provide two implementations, the first written in Cython inside Sage. Then, I’ll write a parallel solution in C.
%cython import time from libc.stdlib cimport malloc, free head_time = time.time() cdef unsigned long p(unsigned long k, unsigned long d): cdef unsigned long power = d cdef unsigned long exp = 0 while power <= k: exp += k / power power *= d return exp cdef unsigned long * p_list(unsigned long n, unsigned long d): cdef unsigned long i = 0 cdef unsigned long * powers = <unsigned long *>malloc((n+1)*sizeof(unsigned long)) while i <= n: powers[i] = p(i,d) i += 1 return powers run_time = time.time() # form a list of number of times each n! is divisible by 2. cdef unsigned long * p2 = p_list(200000,2) # form a list of number of times each n! is divisible by 5. cdef unsigned long * p5 = p_list(200000,5) cdef unsigned long k1, k2, k3, a cdef unsigned long long result = 0 k1 = 0 while k1 <= 66666: # case 1: k1 = k2 < k3 k2 = k1 k3 = 200000 - k1 - k2 if 199982 >= (p2[k1]+p2[k2]+p2[k3]) and 49986 >= (p5[k1]+p5[k2]+p5[k3]): result += 3 # case 2: k1 < k2 = k3 if k1 % 2 == 0: k2 = (200000 - k1)/2 k3 = k2 if 199982 >= (p2[k1]+p2[k2]+p2[k3]) and 49986 >= (p5[k1]+p5[k2]+p5[k3]): result += 3 # case 3: k1 < k2 < k3 a = 1 while 2*a < (200000 - 3*k1): k2 = k1 + a k3 = 200000 - k1 - k2 if 199982 >= (p2[k1]+p2[k2]+p2[k3]) and 49986 >= (p5[k1]+p5[k2]+p5[k3]): result += 6 a += 1 k1 += 1 free(p2) free(p5) elapsed_run = round(time.time() - run_time, 5) elapsed_head = round(time.time() - head_time, 5) print "Result: %s" % result print "Runtime: %s seconds (total time: %s seconds)" % (elapsed_run, elapsed_head)
When executed, we find the correct result relatively quickly.
Result: 479742450 Runtime: 14.62538 seconds (total time: 14.62543 seconds)
## C with OpenMP Solution
#include <stdio.h> #include <stdlib.h> #include <malloc.h> #include <omp.h> /*****************************************************************************/ /* function to determine how many factors of 'd' are in 'k!' */ /*****************************************************************************/ unsigned long p(unsigned long k, unsigned long d) { unsigned long power = d; unsigned long exp = 0; while (power <= k) { exp += k/power; power *= d; } return exp; } /*****************************************************************************/ /* create a list [p(0,d),p(1,d),p(2,d), ... ,p(n,d)] and return pointer */ /*****************************************************************************/ unsigned long * p_list(unsigned long n, unsigned long d) { unsigned long i; unsigned long * powers = malloc((n+1)*sizeof(unsigned long)); for (i=0;i<=n;i++) powers[i] = p(i,d); return powers; } /*****************************************************************************/ /* main */ /*****************************************************************************/ int main(int argc, char **argv) { unsigned long k1, k2, k3, a; unsigned long long result = 0; unsigned long * p2 = p_list(200000, 2); unsigned long * p5 = p_list(200000, 5); #pragma omp parallel for private(k1,k2,k3,a) reduction(+ : result) for (k1=0;k1<66667;k1++) { // case 1: k1 = k2 < k3 k2 = k1; k3 = 200000 - k1 - k2; if (p2[k1]+p2[k2]+p2[k3]<199983 && p5[k1]+p5[k2]+p5[k3]<49987) { result += 3; } // case 2: k1 < k2 = k3 if (k1 % 2 == 0) { k2 = (200000 - k1)/2; k3 = k2; if (p2[k1]+p2[k2]+p2[k3]<199983 && p5[k1]+p5[k2]+p5[k3]<49987) { result += 3; } } // case 3: k1 < k2 < k3 for (a=1;2*a<(200000-3*k1);a++) { k2 = k1 + a; k3 = 200000 - k1 - k2; if (p2[k1]+p2[k2]+p2[k3]<199983 && p5[k1]+p5[k2]+p5[k3]<49987) { result += 6; } } } free(p2); free(p5); printf("result: %lld\n", result); return 0; }
This can be compiled and optimized using GCC as follows.
$gcc -O3 -fopenmp -o problem-154-omp problem-154-omp.c When executed on a 16-core machine, we get the following result. $ time ./problem-154-omp result: 479742450 real 0m1.487s
This appears to be the fastest solution currently known, according to the forum of solutions on Project Euler. The CPUs on the 16-core machine are pretty weak compared to modern standards. When running on a single core on a new Intel Core i7, the result is returned in about 4.7 seconds.
### Problem
Euler published the remarkable quadratic formula:
$n^2+n+41$
It turns out that the formula will produce 40 primes for the consecutive values $n=0$ to $39$. However, when $n=40$, $40^2+40+41=40(40+1)+41$ is divisible by 41, and certainly when $n=41$, $41^2+41+41$ is clearly divisible by 41.
Using computers, the incredible formula $n^2-79n+1601$ was discovered, which produces 80 primes for the consecutive values $n=0$ to $79$. The produce of the coefficients, $-79$ and $1601$ is $-126479$.
$n^2+an+b$ where $|a| where$|n|$is the modulus/absolute value of$n$, e.g.,$|11|=11$and$|-4|=4$. Find the product of the coefficients$a$and$b$, for the quadratic expression that produces the maximum number of primes for consecutive values of$n$, starting with$n=0$. ### Sage Solution Some observations: • Clearly,$b$must be a prime or a negative prime, since$n=0$must result in a prime. • We can refine a lower bound on$b$as follows: If a given combination$(a,b)$results in$m$consecutive primes, then clearly$b>m$, since otherwise we would obtain a factor of$m$in the given polynomial. Also, when$a$is negative, we must have$b>-(n^2+an)$, since the prime values$n^2+an+b$must be positive. Since we know that$n^2+n+41$returns 40 primes, we know that any interesting values for$a$and$b$must then satisfy$b>-(40^2+40a)$. Given these observations, I’ll write a routine in Sage. import time start = time.time() P = prime_range(1000) a_max, b_max, c_max = 0, 0, 0 for a in range(-1000,1001): for b in P: if b < -1600-40*a or b < c_max: continue c, n = 0, 0 while is_prime(n**2 + a*n + b): c += 1 n += 1 if c > c_max: a_max, b_max, c_max = a, b, c prod = a_max * b_max elapsed = time.time() - start print "a=%s, b=%s, longest sequence = %s, prod=%s\nfound in %s seconds"\ % (a_max, b_max, c_max, prod, elapsed) When executed, we obtain the following. a=-61, b=971, longest sequence = 71, prod=-59231 found in 8.53578400612 seconds It’s entirely possible that we’re being a bit redundant with the integer values that we’re throwing at Sage’s “is_prime” function. We’ll cache a list of primes within the range that we’re considering. (I came up with the number 751000 below in the Python section.) import time start = time.time() P = prime_range(751000) L = [False] * 751000 for p in P: L[p] = True a_max, b_max, c_max = 0, 0, 0 for a in range(-1000,1001): for b in range(1,1001): if L[b] is False: continue if b < -1600-40*a or b < c_max: continue c, n = 0, 0 while L[n**2 + a*n + b] is True: c += 1 n += 1 if c > c_max: a_max, b_max, c_max = a, b, c prod = a_max * b_max elapsed = time.time() - start print "a=%s, b=%s, longest sequence = %s, prod=%s\nfound in %s seconds"\ % (a_max, b_max, c_max, prod, elapsed) This indeed executes more quickly, as is seen in the output. a=-61, b=971, longest sequence = 71, prod=-59231 found in 1.39028286934 seconds ### Python Solution Of course, Sage’s prime/factorization functionality isn’t available in Python, and so we’re going to need to create a prime sieve and store primes in some form. One of the things I need to know is how large the values$n^2+an+b$can be. If we were to assume that the maximum value of the length of a prime sequence is 500 (which is an overestimate as we now know), then we’d have the maximum value of the quadratic given as: 500**2 + 1000 * 500 + 1000 751000 Let’s create a prime sieve that will return a list of all primes below a given number. def primes_xrange(stop): primes = [True] * stop primes[0], primes[1] = [False] * 2 L = [] for ind, val in enumerate(primes): if val is True: primes[ind*2::ind] = [False] * (((stop - 1)//ind) - 1) L.append(ind) return L Let’s verify that it works. primes_xrange(20) [2, 3, 5, 7, 11, 13, 17, 19] How long does this function take to form all of the primes under 751000? time P = primes_xrange(751000) Time: CPU 0.39 s, Wall: 0.39 s Now, we’ll take the Sage routine from above and write it completely in Python. Here, I’m using only the sieve list portion of the primes_xrange function I just created. As above, I’m just checking to see if values in that list are True or False, corresponding to primes and composite numbers. This should take a bit longer to run than the Sage version above. import time def primes_xrange(stop): primes = [True] * stop primes[0], primes[1] = [False] * 2 for ind, val in enumerate(primes): if val is True: primes[ind*2::ind] = [False] * (((stop - 1)//ind) - 1) return primes start = time.time() P = primes_xrange(751000) a_max, b_max, c_max = 0, 0, 0 for a in range(-1000,1001): for b in range(1,1001): if P[b] is False: continue if b < -1600-40*a or b < c_max: continue c, n = 0, 0 while P[n**2 + a*n + b] is True: c += 1 n += 1 if c > c_max: a_max, b_max, c_max = a, b, c prod = a_max * b_max elapsed = time.time() - start print "a=%s, b=%s, longest sequence = %s, prod=%s\nfound in %s seconds"\ % (a_max, b_max, c_max, prod, elapsed) And it does run slightly slow compared to the Sage version, as expected. a=-61, b=971, longest sequence = 71, prod=-59231 found in 1.73107814789 seconds ### Cython Solution I’m going to take roughly the same approach as with the Python solution, but I’m going to malloc a C array instead of using the Python list. That should make things a bit more efficient. Before we do that, let’s see how much faster our routine runs if we just slap a “%cython” on the first line. %cython import time def primes_xrange(stop): primes = [True] * stop primes[0], primes[1] = [False] * 2 for ind, val in enumerate(primes): if val is True: primes[ind*2::ind] = [False] * (((stop - 1)//ind) - 1) return primes start = time.time() P = primes_xrange(751000) a_max, b_max, c_max = 0, 0, 0 for a in range(-1000,1001): for b in range(1,1001): if P[b] is False: continue if b < -1600-40*a or b < c_max: continue c, n = 0, 0 while P[n**2 + a*n + b] is True: c += 1 n += 1 if c > c_max: a_max, b_max, c_max = a, b, c prod = a_max * b_max elapsed = time.time() - start print "a=%s, b=%s, longest sequence = %s, prod=%s\nfound in %s seconds"\ % (a_max, b_max, c_max, prod, elapsed) We find that this does improve the timing, and our Python code (which hasn’t been modified at all) now runs more quickly than the Sage code. a=-61, b=971, longest sequence = 71, prod=-59231 found in 1.02033305168 seconds Rewriting to take advantage of Cython’s C datatypes, how much faster can we obtain the result? %cython import time from libc.stdlib cimport malloc, free start = time.time() cdef long i, j cdef bint *primes = malloc(751000 * sizeof(bint)) primes[0], primes[1] = 0, 0 i = 2 while i < 751000: primes[i] = 1 i += 1 i = 2 while i < 751000: if primes[i] == 1: j = 2 * i while j < 751000: primes[j] = 0 j += i i += 1 cdef long long a, b, c, t cdef long long pol, n cdef long a_max, b_max, c_max, prod a_max, b_max, c_max = 0, 0, 0 a = -1000 while a < 1001: b = 2 while b < 1001: if primes[b] == 0: b += 1 continue t = -1600-40*a if b < t or b < c_max: b += 1 continue c, n = 0, 0 pol = n**2 + a*n + b while primes[pol] == 1: c += 1 n += 1 pol = n**2 + a*n + b if c > c_max: a_max, b_max, c_max = a, b, c b += 1 a += 1 prod = a_max * b_max free(primes) elapsed = time.time() - start print "a=%s, b=%s, longest sequence = %s, prod=%s\nfound in %s seconds"\ % (a_max, b_max, c_max, prod, elapsed) Now, this should run much more quickly. a=-61, b=971, longest sequence = 71, prod=-59231 found in 0.0381400585175 seconds Thus, the Sage version runs 1.245 times as fast as the Python version. The Cython version runs 45.387 times as fast as the Python version and 36.452 times as fast as the Sage version. ### Problem Let$d(n)$be defined as the sum of proper divisors of$n$(numbers less than$n$which divide evenly into$n$). If$d(a) = b$and$d(b) = a$, where$a\neq b$, then$a$and$b$are an amicable pair and each of a and b are called amicable numbers. For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore$d(220) = 284$. The proper divisors of 284 are 1, 2, 4, 71 and 142; so$d(284) = 220$. Evaluate the sum of all the amicable numbers under 10000. ### Python Solution I used the idea of creating a list and populating that list with values corresponding to the sum of divisors of numbers, and then I walk through the list and do some index and value checking. import time def sum_divisors(n): s = 0 for i in range(1,n): if n % i == 0: s += i return s def amicable_pairs_xrange(low,high): L = [sum_divisors(i) for i in range(low,high + 1)] pairs = [] for i in range(high - low + 1): ind = L[i] if i + low < ind and low <= ind and ind <= high and L[ind - low] == i + low: pairs.append([i+low,ind]) return pairs def sum_pairs(pairs): return sum([sum(pair) for pair in pairs]) start = time.time() ans = sum_pairs(amicable_pairs_xrange(1,10000)) elapsed = time.time() - start print("%s found in %s seconds") % (ans,elapsed) This returns the correct result. 31626 found in 7.86725997925 seconds ### Cython Solution One of the things I love about using Cython inside the Sage Notebook environment is that one can simply place a Cython statement inside Python code and often achieve some sort of speedup. All I’ve done in this first example is to do just that. I haven’t change the Python code at all yet. %cython import time def sum_divisors(n): s = 0 for i in range(1,n): if n % i == 0: s += i return s def amicable_pairs_xrange(low,high): L = [sum_divisors(i) for i in range(low,high + 1)] pairs = [] for i in range(high - low + 1): ind = L[i] if i + low < ind and low <= ind and ind <= high and L[ind - low] == i + low: pairs.append([i+low,ind]) return pairs def sum_pairs(pairs): return sum([sum(pair) for pair in pairs]) start = time.time() ans = sum_pairs(amicable_pairs_xrange(1,10000)) elapsed = time.time() - start print("%s found in %s seconds") % (ans,elapsed) This runs faster than the original, and that blows my mind. 31626 found in 4.98208904266 seconds We can improve on this obviously, by rewriting the involved functions using C datatypes. %cython import time cdef sum_divisors(unsigned int n): cdef unsigned int s = 0, i = 1 while i < n: if n % i == 0: s += i i += 1 return s cdef sum_amicable_pairs(unsigned int low, unsigned int high): cdef unsigned int a = low, b, sum = 0 while a <= high: b = sum_divisors(a) if b > a and sum_divisors(b) == a: sum += a + b a += 1 return sum start = time.time() ans = sum_amicable_pairs(1,10000) elapsed = time.time() - start print("%s found in %s seconds") % (ans,elapsed) We do achieve somewhat significant speedup now. 31626 found in 1.06677889824 seconds ### C Solution If I take the fast Cython version and rewrite it in C, I don’t get much better performance. #include <stdio.h> #include <stdlib.h> unsigned int sum_divisors(unsigned int n) { unsigned int s = 0, i = 1; while(i < n) { if(n % i == 0) s = s + i; i++; } return s; } unsigned int sum_amicable_pairs(unsigned int low, unsigned int high) { unsigned int a = low, b, sum = 0; while(a <= high) { b = sum_divisors(a); if(b > a && sum_divisors(b) == a) sum = sum + a + b; a++; } return sum; } int main(int argc, char **argv) { unsigned int ans = sum_amicable_pairs(1,10000); printf("result: %d\n",ans); return 0; } The result, compiled with heavy optimization, is only slightly faster than the Cython code. $ gcc -O3 -o problem-21 problem-21.c $time ./problem-21 result: 31626 real 0m1.032s About the only way to really improve this, beyond restructuring the algorithm, is to use OpenMP to run the process in parallel. ### C with OpenMP A parallel version using shared memory via OpenMP looks something like this. (I’ve rewritten it slightly.) #include <stdio.h> #include <stdlib.h> #include <omp.h> unsigned int sum_divisors(unsigned int n) { unsigned int s = 0, i = 1; while(i < n) { if(n % i == 0) s = s + i; i++; } return s; } int main(int argc, char **argv) { unsigned int a, b, nthreads, result = 0; #pragma omp parallel nthreads = omp_get_num_threads(); printf("no of threads %d\n", nthreads); #pragma omp parallel for reduction(+:result) private(b) for(a=0; a < 10000; a++) { b = sum_divisors(a); if(b > a && sum_divisors(b) == a) result = result + a + b; } printf("result: %d\n",result); return 0; } Now, we get much better performance. I’m running this on a 16-core machine, where 3 of the cores are currently being used. $ gcc -fopenmp -O3 -o problem-21-openmp problem-21-openmp.c $time ./problem-21-openmp no of threads 16 result: 31626 real 0m0.140s Problem: You are given the following information, but you may prefer to do some research for yourself. • 1 Jan 1900 was a Monday. • Thirty days has September, April, June and November. All the rest have thirty-one, Saving February alone, Which has twenty-eight, rain or shine. And on leap years, twenty-nine. • A leap year occurs on any year evenly divisible by 4, but not on a century unless it is divisible by 400. How many Sundays fell on the first of the month during the twentieth century (1 Jan 1901 to 31 Dec 2000)? ### Approach There are several different ways to approach this. The easiest, I think, is to use the Gaussian formula for day of week. It is a purely mathematical formula that I have encoded in the following Python code. ### Python Solution import time from math import floor """ Gaussian algorithm to determine day of week """ def day_of_week(year, month, day): """ w = (d+floor(2.6*m-0.2)+y+floor(y/4)+floor(c/4)-2*c) mod 7 Y = year - 1 for January or February Y = year for other months d = day (1 to 31) m = shifted month (March = 1, February = 12) y = last two digits of Y c = first two digits of Y w = day of week (Sunday = 0, Saturday = 6) """ d = day m = (month - 3) % 12 + 1 if m > 10: Y = year - 1 else: Y = year y = Y % 100 c = (Y - (Y % 100)) / 100 w = (d + floor(2.6 * m - 0.2) + y + floor(y/4) + floor(c/4) - 2*c) % 7 return int(w) """ Compute the number of months starting on a given day of the week in a century """ def months_start_range(day,year_start,year_end): total = 0 for year in range(year_start, year_end + 1): for month in range(1,13): if day_of_week(year, month, 1) == day: total += 1 return total start = time.time() total = months_start_range(0,1901,2000) elapsed = time.time() - start print("%s found in %s seconds") % (total,elapsed) This returns the correct result. 171 found in 0.0681998729706 seconds That will run faster if executed directed in Python. Remember, I’m using the Sage notebook environment to execute most Python here. I’m also writing Cython in the same environment. ### Cython Solution There is a nearly trivial rewriting to Cython of the above Python code. %cython import time from math import floor """ Gaussian algorithm to determine day of week """ cdef day_of_week(int year, int month, int day): """ w = (d+floor(2.6*m-0.2)+y+floor(y/4)+floor(c/4)-2*c) mod 7 Y = year - 1 for January or February Y = year for other months d = day (1 to 31) m = shifted month (March = 1, February = 12) y = last two digits of Y c = first two digits of Y w = day of week (Sunday = 0, Saturday = 6) """ cdef int d = day cdef int m = (month - 3) % 12 + 1 cdef int Y if m > 10: Y = year - 1 else: Y = year y = Y % 100 cdef int c = (Y - (Y % 100)) / 100 cdef double w w = (d + floor(2.6 * m - 0.2) + y + floor(y/4) + floor(c/4) - 2*c) % 7 return int(w) """ Compute the number of months starting on a given day of the week in range of years """ cdef months_start_range(int day, int year_start,int year_end): cdef unsigned int total = 0 cdef int year, month for year in range(year_start, year_end + 1): for month in range(1,13): if day_of_week(year, month, 1) == day: total += 1 return total start = time.time() total = months_start_range(0,1901,2000) elapsed = time.time() - start print("%s found in %s seconds") % (total,elapsed) The code is a bit longer, but it executes much faster. 171 found in 0.00387215614319 seconds The Cython code runs roughly 18 times faster. ### C Solution The Cython code was used as a model to create more efficient C code. The only issue here is in maintaining the correct datatypes (not too hard, but compared to Cython it is a pain). #include <stdio.h> #include <stdlib.h> #include <math.h> int day_of_week(int year, int month, int day) { // Using the Gaussian algorithm int d = day; double m = (double) ((month - 3) % 12 + 1); int Y; if(m > 10) Y = year - 1; else Y = year; int y = Y % 100; int c = (Y - (Y % 100)) / 100; int w = ((d+(int)floor(2.6*m-0.2)+y+ y/4 + c/4 -2*c))%7; return w; } long months_start_range(int day, int year_start, int year_end) { unsigned long total = 0; int year, month; for(year = year_start; year < year_end; year++) { for(month = 1; month <= 12; month++) { if(day_of_week(year, month, 1)==day) total++; } } return total; } int main(int argc, char **argv) { int iter = 0; long total; while(iter < 100000) { total = months_start_range(0,1901,2000); iter++; } printf("Solution: %ld\n",total); return 0; } Notice that this executes the loop 100,000 times, as I’m trying to get a good idea of what the average runtime is. We compile with optimization and the [[[-lm]]] math option. We get the following result. gcc -O3 -o problem-19 problem-19.c -lm$ time ./problem-19 Solution: 171 real 0m6.240s
The C code runs roughly 62 times as fast as the Cython and roughly 1124 times as fast as the Python. Each iteration executes in about 6.2000e-5 seconds.
Problem: $2^{15}=32768$ and the sum of its digits is $3+2+7+6+8=26$.
What is the sum of the digits of the number $2^{1000}$?
### Sage Solution
Sage’s built-in Python and functions makes this easy.
import time start = time.time() a = 2^1000 s = sum(a.digits()) elapsed = time.time() - start print "%s found in %s seconds" % (s,elapsed)
It executes pretty quickly too.
1366 found in 0.000343084335327 seconds
### An Easy Python Solution
Python itself also makes this problem too easy, due to string functions.
import time def pow2sum(exp): pow = list(str(2**1000)) return sum([int(i) for i in pow]) start = time.time() n = pow2sum(1000) elapsed = (time.time() - start) print "%s found in %s seconds" % (n,elapsed)
And it is fast:
1366 found in 0.000911951065063 seconds
### Python Without Strings Attached
Let’s do our arithmetic without string functionality. Then, I’d note that $2^{1000}<10^{1000}$ and so we know that, at most, we're dealing with 1001 digits. So, we can create a routine to multiply 2 by itself 1000 times, maintaining the result at each step in a list instead of a single integer. (This is how you'd be forced to do things in C, where arbitrary length integers are pure fiction.) Since we're only multiplying by 2 at each iteration, we know that we'll either carry a zero or one to the next stage... which does make this routine a bit more simple than your typical multiply-by-list routine.
import time def pow2sum(exp): L = [0] * exp # make a list exp long L[0] = 1 for power in range(exp): carry, add = 0, 0 for index in range(exp): prod = L[index] * 2 + carry if prod > 9: carry = 1 prod = prod % 10 else: carry = 0 L[index] = prod return sum(L) start = time.time() n = pow2sum(1000) elapsed = (time.time() - start) print "%s found in %s seconds" % (n,elapsed)
It runs relatively quickly, although not as quickly as the string version runs.
1366 found in 0.361020803452 seconds
### Cython Solutions
We can first trivially rewrite the string Python version.
%cython import time cdef pow2sum(unsigned int exp): cdef list pow = list(str(2**1000)) return sum([int(i) for i in pow]) start = time.time() n = pow2sum(1000) elapsed = (time.time() - start) print "%s found in %s seconds" % (n,elapsed)
When executed, we find that the string Cython code runs about 1.14 times as fast as the string Python code.
1366 found in 0.000799894332886 seconds
Of course, I’m more interested in seeing how much faster the arithmetic version runs.
%cython import time from libc.stdlib cimport malloc, free cdef pow2sum(unsigned int exp): cdef int *L = <int *>malloc(exp * sizeof(int)) cdef unsigned int power = 0,index = 0 cdef unsigned int prod, carry, add while index < exp: L[index] = 0 index += 1 L[0] = 1 while power < exp: carry, add, index = 0, 0, 0 while index < exp: prod = L[index] * 2 + carry if prod > 9: carry = 1 prod = prod % 10 else: carry = 0 L[index] = prod index += 1 power += 1 cdef int sum = 0 index = 0 while index < exp: sum += L[index] index += 1 free(L) return sum start = time.time() n = pow2sum(1000) elapsed = (time.time() - start) print "%s found in %s seconds" % (n,elapsed)
When executed, we get the following result.
1366 found in 0.00858902931213 seconds
Problem: Starting in the top left corner of a $2\times 2$ grid and moving only down and right, there are 6 routes to the bottom right corner.
How many routes are there through a $20\times 20$ grid?
### A Great Interview Problem
This is precisely the sort of problem I expect to see in a technical coding interview, and knowing the various ways of solving a problem such as this will help you get far in those situations. But, even if you’re an experienced coder, the solutions may not be obvious at first. I want to walk through a few potential solutions and see how they perform.
### Recursive Python Solution
I think the easiest solution, and one you should know (but possible don’t) is the recursive approach. The main idea here is to develop a function that will call itself, inching along to the right and down in all possible combinations, returning a value of 1 whenever it reaches the bottom-right and summing all of those 1s along the way. You should convince yourself that the procedure actually terminates (at what we call “the base case”) and returns the correct solution. Try doing it on paper on the 2×2 grid, or a 3×3 grid, and see what happens. Here’s what the Python code looks like.
#!/usr/bin/python import time gridSize = [20,20] def recPath(gridSize): """ Recursive solution to grid problem. Input is a list of x,y moves remaining. """ # base case, no moves left if gridSize == [0,0]: return 1 # recursive calls paths = 0 # move left when possible if gridSize[0] > 0: paths += recPath([gridSize[0]-1,gridSize[1]]) # move down when possible if gridSize[1] > 0: paths += recPath([gridSize[0],gridSize[1]-1]) return paths start = time.time() result = recPath(gridSize) elapsed = time.time() - start print "result %s found in %s seconds" % (result, elapsed)
That’s great, and it will actually work, but it may take some time. Actually it takes a lot of time. By that, I mean it really, really takes a lot of time. When we run it on the 2×2 output, we get the following.
result 6 found in 9.05990600586e-06 seconds
When we run it on the 20×20 input, as the problem requires, it runs for about 4 hours before I kill it. Python is OK at recursive function calls, and it can handle/collapse the memory required moderately well, but what we’re doing here is manually constructing ALL possible paths to a solution, which isn’t incredibly efficient. Still, during a technical interview, this is definitely the first idea for a solution that should pop into your head. It’s quick and easy to write, and many problems can be solved like this.
### Dynamic Python Solution
Our recursive approach suffers from the problem that we’re doing a lot of similar operations over and over. Can we learn anything from the smaller cases and build up from there? In this case, the answer is yes, but it requires us to build up some mathematics a bit more. This is actually quite easy if approached correctly. The idea is to construct the solution recursively, but differently this time.
Claim: Let $n$ be any natural number and consider the 2-dimensional sequence $S_{i,j}$ defined by
$S_{i,j}=\begin{cases}1 &\text{ if }j=0\\ S_{i,j-1}+S_{i-1,j} &\text{ if }0<j<i\\ 2S_{i,j-1} &\text{ if }i=j\end{cases}$
where $0\le i\le n$ and $0\le j\le i$. Then the number of non-backtracking paths from top-left to bottom-right through an $n\times n$ grid is $S_{n,n}$.
Proof: Consider a grid of $m$ rows and $n$ columns (we do not need to assume that the grid is square). Counting from the upper-left and starting at zero, denote the intersection/node in the $i$-th row and $j$-th column by $N_{i,j}$. Thus, the upper-left node is $N_{0,0}$, the bottom-left is $N_{m,0}$ and the bottom-right is $N_{m,n}$. Clearly, the number of paths from $N_{0,0}$ to any node along the far left or far top of the grid is only 1 (since we may only proceed down or left). Now, consider how many paths there are to $N_{1,1}$. We must first travel through $N_{0,1}$ or $N_{1,0}$. This yields only two paths to $N_{1,1}$. We can continue this process. In order to determine the total number of paths to any node $N_{i,j}$, we only need to sum together the total number of paths to $N_{i,j-1}$ and $N_{i-1,j}$. The process is understood graphically in the following diagram, where each new integer represents the number of paths to a node.
Thus, in a $4\times 4$ grid, there are 70 non-backtracking paths. How does this relate to the sequence $S_{i,j}$? Simply put, $S_{i,j}$ is the number of paths to node $N_{i,j}$. If we write out the sequence $S_{i,j}$ for $0\le i\le 4$ and $0\le j\le i$, we simply obtain the lower diagonal sequence embedded in the diagram above.
$\begin{array}{ccccc} S_{0,0} = 1 & & & & \cr S_{1,0} = 1 & S_{1,1}=2 & & & \cr S_{2,0} = 1 & S_{2,1}=3 & S_{2,2}=6 & & \cr S_{3,0} = 1 & S_{3,1}=4 & S_{3,2}=10 & S_{3,3}=20 & \cr S_{4,0} = 1 & S_{4,1}=5 & S_{4,2}=15 & S_{4,3}=35 & S_{4,4}=70\end{array}$
That completes the proof.
Let’s code that into a Python solution and see how fast it runs. My bet is that this will be considerably faster than our initial recursive solution. We can record the sequence $S_{i,j}$ as a single dimensional list that is simply rewritten at each iteration.
#!/usr/bin/python import time def route_num(cube_size): L = [1] * cube_size for i in range(cube_size): for j in range(i): L[j] = L[j]+L[j-1] L[i] = 2 * L[i - 1] return L[cube_size - 1] start = time.time() n = route_num(20) elapsed = (time.time() - start) print "%s found in %s seconds" % (n,elapsed)
When executed, we get the following.
137846528820 found in 0.000205039978027 seconds
### Cython Solution
We’ll recode things in Cython and see how much faster we can get the result returned.
%cython import time from libc.stdlib cimport malloc, free cdef route_num(short cube_size): cdef unsigned long *L = <unsigned long *>malloc((cube_size + 1) * sizeof(unsigned long)) cdef short j,i = 0 while i <= cube_size: L[i] = 1 i += 1 i = 1 while i <= cube_size: j = 1 while j < i: L[j] = L[j]+L[j-1] j += 1 L[i] = 2 * L[i - 1] i += 1 cdef unsigned long c = L[cube_size] free(L) return c start = time.time() cdef unsigned long n = route_num(20) elapsed = (time.time() - start) print "%s found in %s seconds" % (n,elapsed)
We now get the result a bit more quickly.
137846528820 found in 2.21729278564e-05 seconds
The Cython code executes roughly 9 times as fast as the Python.
Problem: The following iterative sequence is defined for the set of positive integers:
$n\rightarrow\begin{cases}n/2 & n \text{ even}\\ 3n+1 & n \text{ odd}\end{cases}$
Using the rule above and starting with 13, we generate the following sequence:
$13\rightarrow 40\rightarrow 20\rightarrow 10\rightarrow 5\rightarrow 16\rightarrow 8\rightarrow 4\rightarrow 2\rightarrow 1$
It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1. Which starting number, under one million, produces the longest chain? Note: Once the chain starts the terms are allowed to go above one million.
### Idea Behind a Solution
I’ll refer to the “Collatz length of $n$” as the length of the chain from an integer $n$ to 1 using the above described sequence. If we were to calculate the Collatz length of each integer separately, that would be incredibly inefficient. In Python, that would look something like this.
### First Python Solution
import time start = time.time() def collatz(n, count=1): while n > 1: count += 1 if n % 2 == 0: n = n/2 else: n = 3*n + 1 return count max = [0,0] for i in range(1000000): c = collatz(i) if c > max[0]: max[0] = c max[1] = i elapsed = (time.time() - start) print "found %s at length %s in %s seconds" % (max[1],max[0],elapsed)
Now, this will actually determine the solution, but it is going to take a while, as shown when we run the code.
found 837799 at length 525 in 46.6846499443 seconds.
### A Better Python Solution
What I’m going to do is to cache any Collatz numbers for integers below one million. The idea is that we can use the cached values to make calculations of new Collatz numbers more efficient. But, we don’t want to record every single number in the Collatz sequences that we’ll be using, because some of the sequences actually reach up into the hundreds of millions. We’ll make a list called TO_ADD, and we’ll only populate that with numbers for which Collatz lengths are unknown. Once known, the Collatz lengths will be stored for repeated use.
import time start = time.time() limit = 1000000 collatz_length = [0] * limit collatz_length[1] = 1 max_length = [1,1] for i in range(1,1000000): n,s = i,0 TO_ADD = [] # collatz_length not yet known while n > limit - 1 or collatz_length[n] < 1: TO_ADD.append(n) if n % 2 == 0: n = n/2 else: n = 3*n + 1 s += 1 # collatz_length now known from previous calculations p = collatz_length[n] for j in range(s): m = TO_ADD[j] if m < limit: new_length = collatz_length[n] + s - j collatz_length[m] = new_length if new_length > max_length[1]: max_length = [i,new_length] elapsed = (time.time() - start) print "found %s at length %s in %s seconds" % (max_length[0],max_length[1],elapsed)
This should return the same result, but in significantly less time.
found 837799 at length 525 in 5.96128201485 seconds
### A First Cython Solution
If we take our original approach of computing each Collatz length from scratch, this might actually work slightly better in Cython.
%cython import time cdef collatz(unsigned int n): cdef unsigned count = 1 while n > 1: count += 1 if n % 2 == 0: n = n/2 else: n = 3*n + 1 return count cdef find_max_collatz(unsigned int min, unsigned int max): cdef unsigned int m = 1 cdef unsigned long num = 1 cdef unsigned int count = 1 cdef unsigned long iter = min while iter < max: count = collatz(iter) if count > m: m = count num = iter iter += 1 return num start = time.time() max_found = find_max_collatz(1,1000000) elapsed = (time.time() - start) print "found %s in %s seconds" % (max_found,elapsed)
In fact, when executed, we find that it is significantly better than our efficient Python code.
found 837799 in 0.604798078537 seconds
This just goes to show that even low efficiency machine/compiled code can drastically outperform efficient Python. But, how far can we take this Cython refinement? What if we were to recode our more efficient algorithm in Cython? It may look something like this.
### A Better Cython Solution
%cython import time from libc.stdlib cimport malloc, free cdef find_max_collatz(unsigned long int max): cdef int *collatz_length = <int *>malloc(max * sizeof(int)) cdef list TO_ADD # holds numbers of unknown collatz length cdef unsigned long iter, j, m, n, s, p, ind, new_length, max_length = 0 # set initial collatz lengths iter = 0 while iter < max: collatz_length[iter] = 0 iter += 1 collatz_length[1] = 1 # iterate to max and find collatz lengths iter = 1 while iter < max: n,s = iter,0 TO_ADD = [] while n > max - 1 or collatz_length[n] < 1: TO_ADD.append(n) if n % 2 == 0: n = n/2 else: n = 3*n + 1 s += 1 # collatz length now known from previous calculations p = collatz_length[n] j = 0 while j < s: m = TO_ADD[j] if m < max: new_length = collatz_length[n] + s - j collatz_length[m] = new_length if new_length > max_length: max_length = new_length ind = m j += 1 iter += 1 free(collatz_length) return ind start = time.time() max_collatz = find_max_collatz(1000000) elapsed = (time.time() - start) print "found %s in %s seconds" % (max_collatz,elapsed)
This gives us some relatively good results:
found 837799 in 0.46523308754 seconds
Still, it isn’t a great improvement over the naive Cython code. What’s going on? I bet that the TO_ADD data structure could be changed from a Python list (notice the “cdef list” definition) to a malloc’d C array. That will be a bit more work, but my gut instincts tell me that this is probably the bottleneck in our current Cython code. Let’s rewrite it a bit.
%cython import time from libc.stdlib cimport malloc, free cdef find_max_collatz(unsigned long int max): cdef int *collatz_length = <int *>malloc(max * sizeof(int)) cdef int *TO_ADD = <int *>malloc(600 * sizeof(int)) cdef unsigned long iter, j, m, n, s, p, ind, new_length, max_length = 0 # set initial collatz lengths and TO_ADD numbers iter = 0 while iter < max: collatz_length[iter] = 0 iter += 1 collatz_length[1] = 1 iter = 0 while iter < 600: TO_ADD[iter] = 0 iter += 1 # iterate to max and find collatz lengths iter = 1 while iter < max: n,s = iter,0 while n > max - 1 or collatz_length[n] < 1: TO_ADD[s] = n if n % 2 == 0: n = n/2 else: n = 3*n + 1 s += 1 # collatz length now known from previous calculations p = collatz_length[n] j = 0 while j < s: m = TO_ADD[j] TO_ADD[j] = 0 if m < max: new_length = collatz_length[n] + s - j collatz_length[m] = new_length if new_length > max_length: max_length = new_length ind = m j += 1 iter += 1 free(collatz_length) free(TO_ADD) return ind start = time.time() max_collatz = find_max_collatz(1000000) elapsed = (time.time() - start) print "found %s in %s seconds" % (max_collatz,elapsed)
Now, when we execute this code we get the following.
found 837799 in 0.0465848445892 seconds
That’s much better. So, by using Cython and writing things a bit more efficiently, the code executes 1119 times as fast.
### C Solution
If I structure the algorithm in the same way, I don’t expect to gain much by rewriting things in C, but I’ll see what happens.
#include <stdio.h> #include <stdlib.h> #include <time.h> int find_max_collatz(unsigned long max) { unsigned int collatz_length[max]; unsigned int TO_ADD[600]; unsigned long iter, j, m, n, s, p, ind, new_length, max_length = 0; // set initial collatz lengths and TO_ADD numbers iter = 0; while(iter < max) { collatz_length[iter] = 0; iter++; } collatz_length[1] = 1; iter = 0; while(iter < 600) { TO_ADD[iter] = 0; iter++; } // iterate to max and find collatz lengths iter = 1; while(iter < max) { n = iter; s = 0; while(n > max - 1 || collatz_length[n] < 1) { TO_ADD[s] = n; if(n % 2 == 0) n = n/2; else n = 3*n + 1; s++; } // collatz length now known from previous calculations p = collatz_length[n]; j = 0; while(j < s) { m = TO_ADD[j]; TO_ADD[j] = 0; if(m < max) { new_length = collatz_length[n] + s - j; collatz_length[m] = new_length; if(new_length > max_length) { max_length = new_length; ind = m; } } j++; } iter++; } return ind; } int main(int argc, char **argv) { unsigned int max, i; time_t start, end; double total_time; start = time(NULL); for(i=0;i<1000;i++) max = find_max_collatz(1000000); end = time(NULL); total_time = difftime(end,start); printf("%d found in %lf seconds.\n",max,total_time); return 0; }
We’re using 1,000 iterations to try and get a good idea of how this runs.
### Python Solution
import time def prod_triplet_w_sum(n): for i in range(1,n,1): for j in range(1,n-i,1): k = n-i-j if i**2+j**2==k**2: return i*j*k return 0 start = time.time() product = prod_triplet_w_sum(1000) elapsed = (time.time() - start) print "found %s in %s seconds" % (product,elapsed)
When executed, this code gives the following result.
found 31875000 in 0.101438999176 seconds
### Cython Solution
We take the same approach in Cython, but remove the Python iterators in an attempt to improve performance.
%cython import time cdef prod_triplet_w_sum(unsigned int n): cdef unsigned int i,j,k i = 1 while i < n: j = 1 while j < n - i: k = n - i - j if i**2 + j**2 == k**2: return i*j*k j += 1 i += 1 return 0 start = time.time() product = prod_triplet_w_sum(1000) elapsed = (time.time() - start) print "found %s in %s seconds" % (product,elapsed)
This gives us the following.
found 31875000 in 0.00186085700989 seconds
Thus, the Cython version is roughly 56 times faster than the Python code. | 2017-03-23T04:17:41 | {
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https://math.stackexchange.com/questions/3263785/is-a-subset-of-0-1-whose-points-are-isolated-in-measure-necessarily-null/3263791 | # Is a subset of [0, 1] whose points are “isolated in measure” necessarily null?
Suppose that $$E \subseteq [0, 1]$$ has the property that for each $$x \in E$$, there is $$r_x > 0$$ such that $$B(x, r_x) \cap E = (x-r_x, x+r_x) \cap E$$ has Lebesgue measure zero. Is it necessarily the case that $$E$$ itself has Lebesgue measure zero?
A couple remarks:
1) If the points of $$E$$ were legitimately isolated from one another, $$E$$ would necessarily be countable, and hence have measure zero.
2) Because I am only interested in the question of "measure zero or not," measurability does not appear explicitly in the statement of the question above. And indeed, to answer the question in the negative, one does not need to produce a set of positive measure; producing a Lebesgue non-measurable set $$E$$ with the given property would just as well constitute a negative answer.
• See also: Lebesgue density theorem. – GEdgar Jun 15 at 23:50
$$E$$ is Lindelöf, as a separable (second countable) metric space.
So we have a cover of $$E$$ by these $$(x-r_x,x+r_x)$$ and so a countable subcover $$\{(x-r_x,x+r_x)\cap E: x \in N \}$$ exists for some countable subset $$N$$ of $$E$$.
But then (as we have a cover of $$E$$): $$\lambda(E) \le \sum_{x \in N} \lambda((x-r_x,x+r_x) \cap E) = 0$$
by countable subadditivity of the Lebesgue measure $$\lambda$$. So $$\lambda(E)=0$$ indeed.
Yes, it is true. Let $$U=\cup_{e\in E}B(e,r_e)$$. As $$U$$ is an open subset of $$[0,1]$$, it is a countable disjoint union of intervals, say $$U=\cup_{n\in\mathbb N}I_n$$ with $$I_n, I_m$$ disjoint for $$n\not=m$$. Then one has $$E=\bigcup_{n\in\mathbb N}E\cap I_n$$ where each set $$E\cap I_n$$ is null, and hence so is $$E$$.
• If you want to disjoint intervals you have to make sure that they are contained in one of the original intervals. If you drop the word disjoint your proof is correct. In any second countable space any union of open sets is the union of a countable subfamily. – Kavi Rama Murthy Jun 15 at 23:21
• Thanks for your response! But I'm not sure I can see why each $E \cap I_n$ is necessarily null. – User190212 Jun 15 at 23:21
• @KaviRamaMurthy in fact the intervals $I_n$ are the connected components of $U$ and there are countably many since each has positive measure. User190212 one can see this by observing that $I_n$ can be written as a countable union of sets $B(e,r_e)$ by construction. – pre-kidney Jun 15 at 23:24
• I know this argument but I am just saying you are making things unnecessarily complicated. Why do you need disjoint intervals? – Kavi Rama Murthy Jun 15 at 23:27
• One way to see it is by taking a countable compact exhaustion of $I_n$. Each compact set is covered by finitely many elements of the open cover $B(e,r_e)$ of $I_n$. – pre-kidney Jun 15 at 23:43 | 2019-10-17T11:23:45 | {
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https://proofwiki.org/wiki/Definition:Floor_Function | # Definition:Floor Function
## Definition
Let $x$ be a real number.
Informally, the floor function of $x$ is the greatest integer less than or equal to $x$.
### Definition 1
The floor function of $x$ is defined as the supremum of the set of integers no greater than $x$:
$\floor x := \sup \set {m \in \Z: m \le x}$
where $\le$ is the usual ordering on the real numbers.
### Definition 2
The floor function of $x$, denoted $\floor x$, is defined as the greatest element of the set of integers:
$\set {m \in \Z: m \le x}$
where $\le$ is the usual ordering on the real numbers.
### Definition 3
The floor function of $x$ is the unique integer $\floor x$ such that:
$\floor x \le x < \floor x + 1$
## Notation
Before around $1970$, the usual symbol for the floor function was $\sqbrk x$.
The notation $\floor x$ for the floor function is a relatively recent development.
Compare the notation for the corresponding ceiling function, $\ceiling x$, which in the context of discrete mathematics is used almost as much.
Some sources use $\map {\mathrm {fl} } x$ for the floor function of $x$. However, this notation is clumsy, and will not be used on $\mathsf{Pr} \infty \mathsf{fWiki}$.
## Examples
### Floor of $0 \cdotp 99999$
$\floor {0 \cdotp 99999} = 0$
### Floor of $1 \cdotp 1$
$\floor {1 \cdotp 1} = 1$
### Floor of $-1 \cdotp 1$
$\floor {-1 \cdotp 1} = -2$
### Floor of $\sqrt 2$
$\floor {\sqrt 2} = 1$
## Also known as
The floor function of a real number $x$ is usually just referred to as the floor of $x$.
The floor function is sometimes called the entier function, from the French for integer.
The floor of $x$ is also often referred to as the integer part or integral part of $x$, particularly in older treatments of number theory.
Some sources give it as the greatest integer function.
## Also see
• Results about the floor function can be found here.
## Historical Note
The notation $\floor x$ for the floor function was introduced in the $1960$s by Kenneth Eugene Iverson and made popular by Donald Ervin Knuth.
## Technical Note
The $\LaTeX$ code for $\floor {x}$ is \floor {x} .
When the argument is a single character, it is usual to omit the braces:
\floor x | 2021-07-30T21:15:44 | {
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https://math.stackexchange.com/questions/833533/let-s-be-a-closed-convex-set-x-be-an-extreme-pt-of-s-then-s-x-is/833723 | # Let $S$ be a closed convex set & $x$ be an extreme pt of $S$ then $S-\{x\}$ is
Let $S$ be a closed convex set and $x$ be an extreme point of $S$, then $S-\{x\}$ is
1. Convex
2. Not Convex
3. May or may not be convex
I am thinking that the convexity doesn't fail even if we remove the extreme points. It's just my intuition from finite dimensional spaces. What about the answer? Is there anything different if the space is infinite dimensional? Please help me.
Thank you.
• What is an extreme point? – HK Lee Jun 14 '14 at 3:42
• Since it is closed it must be in its boundary – David Jun 14 '14 at 3:56
• @Nameless : Why should it have to be bounded? It could be a half-plane. – Michael Hardy Jun 14 '14 at 4:46
• Never mind...I thinking about something else. – IAmNoOne Jun 14 '14 at 4:56
If $a$ is an extreme point of $S$, then $a=\alpha x_1+(1-\alpha)x_2$ for some $x_1,x_2\in S$ implies $x_1=x_2=a$. (I.e., $a$ cannot be written as convex combination of points from $S\setminus\{a\}$.)
This means that $S\setminus\{a\}$ is convex. If you take any two points $x_1,x_2\in S\setminus\{a\}$ and make a convex combination $x=\alpha x_1+(1-\alpha)x_2$, then $x\in S$ (since the set $S$ is convex) and $x\ne a$ (because of the condition in the definition of extreme point).
In fact, these conditions are equivalent. (Which is not too difficult to show.)
Here is a quote from Conway's book A Course in Functional Analysis, p.142
Proposition 7.3. If $K$ is as convex subset of a vector space $X$ and $a\in K$, then the following statements are equivalent:
• $a$ is an extreme point of $K$
• If $x_1,x_2\in X$ and $a=\frac12(x_1+x_2)$, then either $x_1\notin K$ or $x_2\notin K$ or $x_1=x_2=a$.
• If $x_1,x_2\in X$, $0<t<1$, and $a=tx_1+(1-t)x_2$, then either $x_1\notin K$ or $x_2\notin K$ or $x_1=x_2=a$.
• If $x_1,\dots,x_n\in K$ and $a$ is in the convex hull of $\{x_1,\dots,x_n\}$, then $a=x_k$ for some $k$.
• $K\setminus\{a\}$ is a convex set.
Proof of this proposition is omitted in this book. (It is left as an exercise.) | 2019-08-18T17:01:57 | {
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https://math.stackexchange.com/questions/1128176/proof-by-induction-1xn-1-nxnx2 | # Proof by induction: $(1+x)^n > 1 + nx+nx^2$
This is one of the exercises that appears in Apostol's Calculus I. I'm not sure whether what I did is correct.
1. Let $n_1$ be the smallest positive integer $n$ for which the inequality $(1+x)^n > 1 + nx+nx^2$ is true for all $x > 0$. Compute $n_1$ and prove that the inequality is true for al integers $n \geq n_1$.
The first thing I assumed was that the first number for which the inequality holds was $x=1$. Then:
$$2^n > 2n + 1$$
After a little of inspection, one can notice that the inequality is true for all positive integers $n \geq 3$. So $n_1 = 3$.
Proof (by Induction):
$$P(n): (1+x)^n > 1 + nx+nx^2\qquad \text{for all}\ n \geq 3$$
Base Case: $P(3)$
$$(1+x)^3 > 1 + 3x+3x^2$$ $$x^3+3x^2+3x+1 > 3x^2+3x+1$$
which is true.
Inductive Hypothesis: Assume $P(k)$ is true for a positive integer $k\geq 3$:
$$(1+x)^k > 1 + kx + kx^2\qquad (1)$$
Inductive Step: Prove $P(k+1)$:
$$(1+x)^{k+1} > \underbrace{1 + (k+1)x + (k+1)x^2}_\text{a}$$
If we multiply the inequality $(1)$ by $(1+x)$ we get:
$$(1+x)^{k+1} > (1+x)[1 + kx + kx^2]$$
$$(1+x)^{k+1} > \underbrace{kx^3+2kx^2+kx+x+1}_\text{b}$$
Since $a<b$ and $b < (1+x)^{k+1}$, by Transitivity we have that $a < (1+x)^{k+1}$ and hence $P(n)$ is true as asserted.
Is it correct to assume that my first number is $x=1$, although the problem states that it's true for all $x >0$?
• $x$ is a real greater than $0$, so there's no "smallest $x$ for which the inequality holds". Although letting $x=1$ is a good strategy to determine the positive integer $n_1$. – Workaholic Jan 31 '15 at 21:00
• I would start the proof as follows: "The inequality does not hold for $n=2$, since then for $x=1$ it reads $2^2 > 1 + 2 + 2$. We now show that the inequality holds for all $n \geq 3$, thus proving that $n_1=3$..." – user133281 Jan 31 '15 at 21:03
• @Workaholic Yep. $x>0$, but is there no smallest $n$ for which the inequality holds? – Jazz Jan 31 '15 at 21:04
• @Jazz You just proved that such $n_1$ existed and that it was equal to $3$. – Workaholic Jan 31 '15 at 21:06
• @Jazz The error is the phrase, "the first ... was $x=1$." It makes no sense to say $x=1$ is "first". What you can say is, "Assume the inequality holds for $x=1$." No mention about what is the "first" $x$ is needed. – David K Jan 31 '15 at 21:10
Another approach
From the induction hypothesis, we have
$$(1+x)^k > 1+kx+kx^2.$$
In the induction step, notice
$$1+(k+1)x+(k+1)x^2 = 1+kx+kx^2+x(1+x)$$
Now, using the induction hypothesis and add $x(1+x)$ to both sides:
$$(1+x)^k +x(1+x) > 1+kx+kx^2 + x(1+x).$$
Note that $x(1+x)^k > x(1+x)$, and also note that $$(1+x)^{k+1} = (1+x)^k(1+x) = (1+x)^k+x(1+x)^k.$$ Finally,
\begin{align*} (1+x)^{k+1} &= (1+x)^k+x(1+x)^k \\ &> (1+x)^k+x(1+x) \\ &> 1+kx+kx^2+x(1+x) \\ &= 1+(k+1)x+(k+1)x^2. \end{align*}
If we use the binomial expansion of $(1+x)^n$ we get:$$(1+x)^n=1+nx+\frac{n(n-1)}{2}x^2+...\tag{1}$$For this expression to be greater then $1+nx+nx^2$ we must satisfy:$$\frac{n(n-1)}{2}\ge n$$From which you should be able to calculate the minimum $n$
• This makes no sense to me. This shows that $\frac{n(n-1)}{2} \geq n$ is a sufficient condition for the inequality to hold, not that it is a necessary condition. In other words, we can only use this to conclude that the statement is true for all $n \geq 3$, not that it is false for $n=1$ and $n=2$. – user133281 Jan 31 '15 at 21:13
• Ah! I see what you mean @user133281 - you are quite right in pointing that out. I guess we could use this result and then reduce $n$ by $1$ to show it is also a necessary condition? – Mufasa Jan 31 '15 at 21:15
• @Mufasa Note that if we use approach Taylor (or binomial approximation by Newton) combined with the principle of mathematical induction we get demonstration in which the principle of mathematical induction is superfluous. – Elias Costa Jan 31 '15 at 21:24
• @Elias - I could not see anything in the "quoted" question (i.e. 7. Let $n_1$ be...) that asked for the use of mathematical induction. – Mufasa Jan 31 '15 at 21:27 | 2020-09-22T05:18:40 | {
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https://projecteuler.chat/viewtopic.php?f=4&p=50565&sid=52c1b4fe830e5cb2e725c1859d2f9270 | ## Practising dynamic programming
Arrangements, combinations and permutations, probability, ...
bentaylortheonly
Posts: 3
Joined: Wed Jan 17, 2018 7:39 pm
### Practising dynamic programming
Just trying to get the hang of DP and how to be able to generically apply it to any situation.
I've read a few examples of algorithms to solve problems using DP but slightly struggling to see how to apply it to a given, new, problem.
So supposing the problem was: given a set of integers {1,2,3,...,N}
let S(N) be the number of subsets of those integers that add up to N. Each integer, let's call it i, in the subset can only be used once. (This might actually be similar to an existing PE problem but it's not intended to be, I just made it up to test the technique)
So for instance:
S(1)=1 (the set is {1})
S(2)=1 {2}
S(3)=2 {1,2} and {3}
S(6)=4 {1,2,3} , {1,5}, {2,4}, {6}.
S(10)=10 {10}, {1,9}, {2,8}, {3,7}, {4,6}, {1,2,7}, {1,3,6}, {1,4,5}, {2,3,5}, {1,2,3,4}.
So my, obviously currently flawed, process of generating this algorithm was this:
Consider each i. Find the maximum number ('max') that can be created with integers up to i.
So when i=3, max=1+2+3. When i=4, max=10 as we could have a subset of 1+2+3+4. etc.
Add one to to w[j] for each integer j from i to max. Rationale: using i adds one more possibility to the number of ways of creating each j.
And I came up with this:
Code: Select all
def S(N):
w = [0]*(N+1); #w[n] = the number of ways of making n
max = 0
for i in range(1, N+1): #i is the next number in the subset
max += i #the max number we can make with numbers up to i
if max>=N:
max = N
for j in range(i, max+1):
w[j]+=1 # using i adds one more possibility to the ways of making j.
return w
print(S(10))
The algorithm produces
{0, 1, 1, 2, 2, 3, 4, 4, 5, 6, 7}
but it should be
{0, 1, 1, 2, 2, 3, 4, 5, 6, 8, 10}
There's obviously some fundamental flaw in this logic but I'm struggling to see what the correction is.
I keep thinking that the flaw is due to miscounting subsets with 3 or more elements in them, which start at S(6). But my algorithm is correct up to and including S(6) - it only starts giving the wrong answer from S(7) on.
It must be something to do with referring back to previous stored values. Can anyone be so kind as to venture a clear explanation of how to modify the algorithm to be correct?
traxex
Posts: 77
Joined: Thu Oct 19, 2017 12:30 pm
### Re: Practising dynamic programming
Here is a slightly modified version that works, except that it gives S(0) = 1 using the empty subset.
Code: Select all
def S(N):
w = [0]*(N+1)
w[0] = 1
for i in range(1, N+1):
for j in reversed(range(i, N+1)):
w[j] += w[j - i]
return w
print(S(10))
The variable "max" is a valid optimization, but it is not necessary.
It's crucial to increment w[j] with the value of w[j-i] instead of 1. It's also important to loop backwards when modifying "w" in-place. Otherwise, during the iteration i==3, we would first add w[0] to w[3], then we'd add w[3] to w[6], etc., essentially counting subsets with multiple copies of 3.
Looping in increasing order can be made to work with a temporary array:
Code: Select all
def S2(N):
w = [0]*(N+1)
w[0] = 1
for i in range(1, N+1):
w2 = w[:]
for j in range(i, N+1):
w2[j] += w[j - i]
w = w2
return w
Technically, everyone is full of himself.
bentaylortheonly
Posts: 3
Joined: Wed Jan 17, 2018 7:39 pm
### Re: Practising dynamic programming
Excellent, thanks very much for that, I'll digest that and hopefully be able to create my own in future.
MuthuVeerappanR
Posts: 320
Joined: Sun Mar 22, 2015 2:30 pm
Location: India
Contact:
### Re: Practising dynamic programming
I'm not expert in DP either but this problem can be solved in a couple of ways. The problem at hand is well known and studied. You are asking for the number of partitions with distinct summands. My favorite would be to use generating functions.
Let $f(x)=(1+x)(1+x^2)(1+x^3)(1+x^4)\cdots$
What does the coefficient of $x^n$ represent when we expand this?? Here are the coefficients using Wolfram Alpha.
To turn into a DP approach, let $f(n,k)$ represent the number of partitions of $n$ using numbers upto $k$. In gen functions, let
$f_k(x)=\displaystyle\prod_{i=1}^k(1+x^i)$
The coefficients of $x^n$ is $f_k(x)$ is then $f(n,k)$.
Also, $f_k(x)=f_{k-1}(x)(1+x^k)$. If we equate the coefficients of $x^n$ on both sides we get, $f(n,k)=f(n,k-1)+f(n-k,k-1)$.
As no summand can exceed $n$ in a partition of $n$, $f(n,n)$ gives the answer we are looking for.
Of course, there are much more general cases where the gen. functions may not be easy to come up with but still it helped me solve a lot of PE problems. You can also check Partitions (Number Theory) for more on the problem you are looking for and Analytic Combinatorics to know more about gen. functions.
Happy Solving!!
It is not knowledge, but the act of learning, not possession but the act of getting there, which grants the greatest enjoyment. | 2018-10-23T13:41:41 | {
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http://math.stackexchange.com/questions/211725/converting-from-cartesian-coordinates-to-spherical-coordinates/211750 | # Converting from Cartesian coordinates to Spherical coordinates
I want to understand how to convert from Cartesian coordinates to spherical coordinates. I have the following definitions:
\begin{align} x & =r\sin\theta\cos\phi \\[6pt] y & =r\sin\theta\sin\phi \\[6pt] z & =r\cos\theta \\[6pt] \rho & =r\sin\theta \end{align}
In written terms: $r$ is the distance from the origin to the point, $\phi$ is the angle needed to rotate around $z$ to get to the point, $\theta$ is the angle from the positive $z$-axis, $\rho$ is the distance between the point and the $z$-axis.
On the basis that $(x,y,z)=(r,\theta,\phi)$ I have,
$$\rho=\sqrt{x^2+y^2}=r\sin\theta$$
using Pythagoras' Theorem gives
$$r=\sqrt{\rho^2+z^2}=\sqrt{x^2+y^2+z^2}.$$
Next take $z=r\cos\theta$ which gives
$$\theta=\arccos\left(\frac{z}{r}\right).$$ Both of these agree with what I have found on wikipedia, however I can't understand how the last coordinate $\phi$ is reached. This is what I get:
$$y=r\sin\theta\sin\phi$$ $$\frac{y}{r\sin\theta}=\sin\phi$$
from here I use the relationship that $\rho=\sqrt{x^2+y^2}=r\sin\theta$ and write
$$\frac{y}{\rho}=\sin\phi$$ $$\arcsin\left(\frac{y}{\rho}\right)=\phi.$$
Have I gone wrong somewhere? Can it be explained to me how my last result differs from that provided by wikipedia?
Thanks
-
In most texts, $\theta$ is the angle in the $xy$ plane and $\phi$ is the angle down from the $z$ axis. That is, I think you (and wiki) have things not as usual. Nevertheless, using their definitions, both answers are correct.
The wiki terms for $y$ and $x$, when one computes $y/x$ the $r \sin(\theta)$ cancels and the result is $\sin(\phi)/\cos(\phi) = \tan(\phi)$. so you can say $\phi = \tan^{-1}(y/x)$ as wiki does.
For your version, you use the extra symbol $\rho$, which is really $r \sin(\theta)$. So using the same formulas from wiki gives $y/\rho = y/(r \sin(\theta) )$. This time the $y$ in wiki is $r\sin(\theta)\sin(\phi)$, so things cancel and you get $y/\rho = \sin(\phi)$. Thus you can also say that $\phi = \arcsin(y/\rho)$. But note that $\rho$ is not one of the spherical coordinates, but is just $r\sin(\theta)$ in terms of the spherical coordinates. It's not usual to give the inverse in terms of other symbols than the direct variables.
-
In both the wiki version and yours, care must be taken as to which quadrant one is in, because of the arctan and arcsin functions not automatically returning the original angles. – coffeemath Oct 12 '12 at 16:51
Thanks for help explaining this in the terms I am using and I understand now that I really don't want to be using $\rho$ in my final conversions. Could you help explain why I would divide y/x in order to obtain tan? – Aesir Oct 12 '12 at 17:20
In the wiki formulas, y is r sin theta sin phi and x is r sin theta cos phi. By dividing y by x we will have a cancel of the common factor r sin theta, arriving at simply sin phi / cos phi, which is tan phi. – coffeemath Oct 12 '12 at 17:31
Ah yes, I see so it's just common sense to do that, or not in my case! Thanks, I understand now. – Aesir Oct 12 '12 at 17:40
Your formula for $\phi$ is correct for $x>0$, but gives the wrong angle when $x<0$: in this case $\phi\in(\pi/2,3\pi/2)$, but $\arcsin$ gives a result in $[-\pi/2,\pi/2]$.
The formula shown by Wikipedia is obtained from
$$\sin\phi=y/\rho\\ \cos\phi=x/\rho$$
and dividing the two's (when possible)
$$\tan\phi=y/x.$$
From here it comes out $\arctan$, with all the warning of the case, leading to the atan2 function.
-
Thanks for taking the time to answer, could you provide some more clarification on why I would divide $y$ by $x$? I can see now that, that would give me $\tan$ which is what I need but I don't understand why I would do this. – Aesir Oct 12 '12 at 17:18 | 2016-05-06T02:28:42 | {
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greater than or equal to sign 2021 | 2022-08-19T20:41:08 | {
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https://math.stackexchange.com/questions/1079330/wolfram-alpha-error-in-global-minimization | # Wolfram alpha error in global minimization?
Let us consider the function
$$f(x,y)= x + y^2 - \ln(x+y)$$
If you try to minimize it using Wolfram Alpha (http://www.wolframalpha.com/input/?i=minimize+x%2By%5E2-ln%28x%2By%29), it founds a local minimum at $(x,y)=(1/2,1/2)$, but it can't find a global minimum. But that local minimum should also be the global one, since the function is a convex one and the domain is convex.
Am I wrong or is Wolfram Alpha wrong?
Ok, thanks to everybody who answered or commented, but perhaps I was not clear when writing my question: I knew that $(1/2,1/2)$ is a local minimum, and, since the $f(x,y)=x+y^2-ln(x+y)$ is a convex function and the domain is $D=\{(x,y) \in \mathbb{R}^2 : x+y >0\}$, which is a convex subset of $\mathbb{R}^2$, then a well kwown result in convex optimization ensures that $(1/2,1/2)$ is also a global minimum. Then perhaps I should have written: "why does Wolfram Alpha says 'no global minima found' when I ask to minimize $f(x,y)$?"
As far as I can understand from your answers, that sentence from Wolfram Alpha means only that it is not able to find a global minimum, not that the global minimum does not exist, is this correct?
• Wolfram never claimed it wasn't a global minimum. Detecting local minima is generally much simpler than analyzing their global character, so Wolfram simply never bothered with the latter. – user7530 Dec 24 '14 at 2:04
• It could be thinking, "Uh-oh, if $x+y$ is negative, then the function takes complex values, and there's no order on the complex numbers... better not say anything". Computer algebra systems are kind of stupid sometimes. – Milo Brandt Dec 24 '14 at 3:04
The domain of $f$ is the set of pairs $(x,y)$ for which $r=x+y>0.$ Using $x=r-y$ our function can be expressed nicely in terms of separate functions of $r,y$ as $$(r-\ln r)+(y^2-y).$$ Using one variable calc, the min of the first bracketed term is $1$ (at $r=1$), while the min of the second is $-1/4$ (at $y=1/2$).
This shows your function satisfies $f \ge 1 -1/4=3/4.$ That this is actually the global minimum then follows since $f(1/2,1/2)=3/4.$
If Wolfram explicitly says no global minimum exists, that is incorrect. However in some cases a computer algebra system may not deal with implied domains well... [I just looked at Wolfram output you link to, and it only says there is a local min, not that there is not a global one.]
Find the partial derivatives:
$$\frac{∂}{∂x}(f(x,y))=1-\frac{1}{x+y}\:\:\:\:(1)$$ and $$\frac{∂}{∂y}(f(x,y))=2y-\frac{1}{x+y}\:\:\:\:(2)$$
Set $(1)$ equal to $0$: $$0=1-\frac{1}{x+y}$$ $$\frac{1}{x+y}=1$$ $$x+y=1\:\:\:\:(3)$$
Set $(2)$ equal to $0$:
$$0=2y-\frac{1}{x+y}\:\:\:\:(4)$$
Put $(3)$ into $(4)$:
$$0=2y-\frac{1}{1}$$ $$y=\frac{1}{2}\:\:\:\:(5)$$
Put $(5)$ back into $(3)$:
$$x+\frac{1}{2}=1$$ $$x=\frac{1}{2}$$
So $(x,y) = (\frac{1}{2},\frac{1}{2})$ is our potential extremum. Now find our second partial derivatives, and perform the second derivative test for a function of two variables:
$$f_{xx} = \frac{1}{(x+y)^2}$$ $$f_{xy} = f_{yx}=\frac{1}{(x+y)^2}$$ $$f_{yy}= \frac{1}{(x+y)^2} + 2$$
And now the test:
$$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-(f_{xy}(x,y))^2$$
$$D(\frac{1}{2},\frac{1}{2})=(\frac{1}{(\frac{1}{2}+\frac{1}{2})^2})(\frac{1}{(\frac{1}{2}+\frac{1}{2})^2} + 2)+(\frac{1}{(\frac{1}{2}+\frac{1}{2})^2})^2$$
$$D(\frac{1}{2},\frac{1}{2})=4$$
Also note that:
$$f_{xx}(\frac{1}{2},\frac{1}{2})=\frac{1}{(\frac{1}{2}+\frac{1}{2})^2}=1$$
And so, we have $D(\frac{1}{2},\frac{1}{2})>0$ and $f_{xx}(\frac{1}{2},\frac{1}{2})>0$, so by the second derivative test for a function of two variables, we conclude that $(\frac{1}{2}, \frac{1}{2})$ is a local minimum, and the only extremum. As you stated in your question, this point must be a global minimum. WolframAlpha simply said it was a local minimum, and didn't say that it wasn't a global one:
Let's look at the first order conditions:
$f_{x} = 1 - \frac{1}{x + y} = 0$
$f_{y} = 2y - \frac{1}{x + y} = 0$
So $\frac{1}{x + y} = 1$, which gives us $y = \frac{1}{2}$. It follows that $x = \frac{1}{2}$.
So Wolfram Alpha seems to be correct here. Note you should also verify that $(\frac{1}{2}, \frac{1}{2})$ is a minimum, but this is straight-forward to do. | 2019-09-23T07:00:19 | {
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https://physics.stackexchange.com/questions/348250/angular-velocity-of-cone-rolling-on-ground | # Angular velocity of cone rolling on ground
A cone rolls without slipping on a table. The half-angle at the vertex is $\alpha$, and the axis has length $h$. Let the speed of the center of the base, point $\text{P}$ in the figure, be $v$.
$\hspace{175px}$
Question: What is the angular velocity of the cone with respect to the lab frame at the instant shown?
### Attempted solution
Let us consider a frame rigidly attached to the cone with basis vectors $\hspace{.1cm}\hat{x_1}$,$\hspace{.2cm}\hat{x_2}$ and $\hspace{.1cm}\hat{x_3}$ as shown below, where $\hat{x_2}$ points out of the page, and let the coordinate axes fixed to lab be $\hat{z}\hspace{.2cm}$and $\hat{y}$. Let the cone rotate with $\omega_z$ around $\hat{z}$ and with $\omega_3$ around $\hat{x_3}$.
Since the point A is at rest with respect to the lab frame, $$\vec{v_a} = \vec{0}$$
This therefore implies,$$\omega_3\cdot(h\cdot\tan\alpha) = v$$ Also, since it is given that $\vec{v_a} = v\hat{x_2}$,$$\omega_z\cdot(h\cdot\cos\alpha) = v$$
Therefore, we have the angular velocity in lab frame at the current instant, $\vec{\omega}$ such that, $$\vec{\omega} = -\omega_z\hat{z} + \omega_3\hat{x_3}$$ And breaking $\hat{x_3}$ into its components along $\hat{z}\hspace{0.2cm}and\hspace{0.2cm}\hat{y}$ and substituting the values of $\omega_3\hspace{0.2cm}and\hspace{0.2cm}\omega_z$, $$\vec{\omega} = \frac{-v}{h \cos\alpha}\hat{z} + \frac{v}{h \tan\alpha} \left[ \cos{\left(\alpha \right)} \, \hat{y}+ \sin{\left( \alpha \right)} \, \hat{z} \right]$$
And thus the angular velocity is, $$\vec{\omega} = \frac{v}{h}\left[ \left( \cos{\alpha} - \frac{1}{\cos{\alpha}} \right) \hat{z} + \frac{ \left(\cos{\alpha}\right)^2}{ \sin{\alpha}}\hat{y}\right]$$
Why is this solution wrong?
• Looking from above, the cone rolls around its vertex at some angular velocity. Looking at the "base" of the cone, it rotates around the point P at some angular velocity. Which does the question ask about? – DJohnM Jul 24 '17 at 4:14
• The total angular velocity of the cone at the current instant. In fact, even the direction is not asked for, just the magnitude is asked. – BabaYaga Jul 24 '17 at 4:57
• Have you looked here, on Page 429? books.google.ca/… – DJohnM Jul 24 '17 at 5:22
You are right that angular velocities can be added as vectors, so total angular velocity can be expressed as sum
$$\boldsymbol{\omega} = \omega_z \hat{\mathbf z} + \omega_3\mathbf{x}_3 \tag{1} \,.$$
The point where you go wrong is
Since the point A is at rest with respect to the lab frame, $$\vec{v_a} = \vec{0}$$
This therefore implies,$$\omega_3\cdot(h\cdot\tan\alpha) = v$$
It is true that point A has zero velocity, but that does not imply the given expression for $\omega_3$.
Actually, it is much easier to find the total angular velocity and then find $\omega_3$. Here is how.
Every solid body with one or more points at rest (zero velocity) is either at rest as a whole or is rotating around some axis. The cone is said to roll without slipping, which means that whole line of points of the cone that is in contact with the table is at rest. But since the cone is rotating around some axis, this contact line has to be the axis of rotation.
Then, the velocity of the point $P$ can be expressed as $$v = \omega\, h \sin \alpha$$ and from this we obtain the magnitude of the sought angular velocity $$\omega = \frac{v}{h\sin\alpha},$$ the direction being that of the contact line.
Now, we could calculate $\omega_3$ from the equation (1). The result is
$$\omega_3 = \frac{v}{h\sin\alpha \cos\alpha}.$$
Point P travels along a circle of radius $R = h\cos(\alpha)$, so its speed is $v = R\omega = h\cos(\alpha)\omega$, which finally gives
$$\omega = \frac{v}{h\cos(\alpha)}$$
• What are you referring to? The answer by the way, is v(h*sin(alpha)) – BabaYaga Jul 23 '17 at 22:20
• I was referring to the angular speed of the axis $OP$ where $O$ is the vertex of the cone. This answer is not homogeneous so is wrong... – Spirine Jul 23 '17 at 22:26
• I'm sorry i meant v/h(sin(alpha)) – BabaYaga Jul 23 '17 at 22:51
• This supplied answer is the rotational velocity seen by looking at the "bottom" of the cone – DJohnM Jul 24 '17 at 5:15 | 2021-03-07T02:42:04 | {
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http://mathhelpforum.com/number-theory/111387-multiples.html | Math Help - Multiples
1. Multiples
Of the numbers 1, 2, 3, . . . , 6000, how many are not multiples of 2, 3 or 5?
3000 multiples of 2, 2000 multiples of 3, 1200 multiples of 5, 600 of 10, 1000 of 6, 400 of 15
So 4200 multiples, by adding the first 3 and subtracting the next 3 from the sum, since they have been counted twice.
So 6000 - 4200 = 1800.
Is this correct?
2. No! You have counted twice the numbers which are multiples of 2,3 and 5. So you should subtract 200 from your result.
One way to do it is to consider the intersection of the set of number which are not multiples of 2, the set of number which are not multiples of 3 and the set of number which are not multiples of 5.
So you would get $6000(1-1/2)(1-1/3)(1-1/5)=1600$. If you expand the product you get :
$6000(-1/2-1/3-1/5+1/(2\times 3)+1/(2\times 5)+1/(3\times 5)-1/(2\times 3 \times 5))$
which is consistent with your approach by the inclusion/exclusion principle (once your small mistake is corrected).
3. Hello, Aquafina!
We can use this fancy counting formula:
. . $n(A \cup B\cap C) \;=\;n(A) + n(B) + n(C)$
. . . . . . . . . . . . . . $- n(A \cap B) - n(B \cap C) - n(A \cap C)$
. . . . . . . . . . . . . . . . $+ n(A \cap B\cap C)$
Of the numbers 1, 2, 3, ..., 6000, how many are not multiples of 2, 3 or 5?
We'll find the number of integers which are multiples of 2, 3 or 5.
. . $\begin{array}{cccc}
\text{multiples of 2} & \frac{6000}{2} &=& 3000 \\ \\[-3mm]
\text{multiples of 3} & \frac{6000}{3} &=& 2000 \\ \\[-3mm]
\text{multiples of 5} & \frac{6000}{5} &=& 1200 \end{array}$
. . . $\begin{array}{cccc}\text{multiples of 2, 3} & \frac{6000}{6} &=& 1000 \\ \\[-3mm]
\text{multiples fo 3, 5} & \frac{6000}{15} &=& 400 \\ \\[-3mm]
\text{multiples of 2, 5} & \frac{6000}{10} &=& 600 \end{array}$
. . . $\begin{array}{cccc}\text{multiples of 2, 3, 5} & \frac{6000}{30} &=& 200 \end{array}$
$n(2\cup3\cup5) \;=\; n(2) + n(3) + n(5) - n(2\cap3) - n(3\cap5) - n(2\cap 5) - n(2\cap3\cap5)$
. . . . . . . $=\; 3000 + 2000 + 1200 - 1000 - 400 - 600 + 200$
. . . . . . . $= \;4400$
There are 4400 integers which are multiples of 2, 3 or 5.
Therefore, there are: . $6000-4400 \:=\: 1600$ which are not multiples of 2, 3 or 5.
4. Originally Posted by Soroban
Hello, Aquafina!
We can use this fancy counting formula:
. . $n(A \cup B\cap C) \;=\;n(A) + n(B) + n(C)$
. . . . . . . . . . . . . . $- n(A \cap B) - n(B \cap C) - n(A \cap C)$
. . . . . . . . . . . . . . . . $+ n(A \cap B\cap C)$
$n(2\cup3\cup5) \;=\; n(2) + n(3) + n(5) - n(2\cap3) - n(3\cap5) - n(2\cap 5) - n(2\cap3\cap5)$
. . . . . . . $=\; 3000 + 2000 + 1200 - 1000 - 400 - 600 + 200$
. . . . . . . $= \;4400$
Hi, do the multiples of 2,3 and 5 have to be added or subtracted? You have used both operations...
Looking at the inclusion/exclusion principle Bruno J posted, I am able to do the question when thinking about the Venn Diagram | 2014-08-20T17:58:01 | {
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http://www.jerimiannwalker.com/category/cool-math/ | # A very nice pattern: squaring numbers like 99, 999, 9999, etc
Quick! What’s 92?
I’m assuming (hoping) you rolled your eyes or thought “81” (thanks for playing along). Since that was easy enough, let’s make it more interesting… what about 992? or 9992? or even 999992?
By the end of this post, you will be able to rattle off the square of any number like this just as quickly as you think of the answer from the good ole times table. You’ll be the life of every party! You will simply dazzle others with your math magic! Yes I’m done.
It turns out that squaring numbers where the digits are all 9 follows a very predictable pattern. While there are patterns when multiplying any number by a number made up of only 9s, this pattern is much easier to remember and in my opinion, much more interesting.
## The pattern
Let’s just look at a few squared values to get an idea of what is going on.
92 81 992 9801 9992 998001 99992 99980001 999992 9999800001
Based on this small sample, it seems you can square a number consisting of n 9s by writing (n-1) nines, 8, (n-1) zeros, and then 1. You can spot check some larger values and see the pattern continues. For example: 9999999999= 99999999980000000001.
Applying this, is someone randomly asked you “hey – what’s 992?”, you would mentally note that there are 2 digits, and so the answer would have one 9, 8, one zero, and then 1: 9801.
## Why does this work
As you know, simply showing a few examples is not a proof right? I’ve only shown that this pattern holds for these specific examples. A proof would show that it holds generally, without us having to type numbers into wolfram alpha to check for the rest of time. To understand the proof though, I will start by showing why it holds for one particular case: 99992.
Consider the definition of squaring. You know that 32 is just the product of 3 and 3. Naturally, the same is true here so:
$9999^2 = 9999 \times 9999$
But what is multiplication really? It is repeated addition. With something like $3 \times 5$, we are saying “add 3 to itself 5 times”. Using this, you could write $3 \times 5$ as $3 + 3 + 3 + 3 + 3$ or, if you wanted, $3 \times 4 + 3$. Applying that here:
$9999 \times 9999 = 9999 \times 9998 + 9999$
Now here is the “trick”. I’m going to add and subtract 9998. The way I will do that is by changing $9999 \times 9998$ to $10\,000 \times 9998$. This will mean I now have ten-thousand 9998s being added, which is too many, so I will subtract one of them from 9999.
$9999 \times 9998 + 9999 = 10\,000 \times 9998 + 9999 - 9998$
Simplifying this, we get:
$9999^2 = 10\,000 \times 9998 + 1$
And how do you multiply a value and 10,000? You write the value with four zeros following it. Here, you will then add 1. This will give us our pattern of three 9s, 8, three 0s, and 1.
## Generalized
The same technique will work for any number made up of all 9s. We just have to keep track of the number of digits along the way. Following the steps we used above:
\begin{aligned}\underbrace{999 \cdots 9^2}_\text{n digits} &= 999 \cdots 9 \times 999 \cdots 9\\&= 999 \cdots 9 \times 999 \cdots 8 + 999 \cdots 9\\&= \underbrace{1000 \cdots 0}_\text{(n + 1) digits} \times 999 \cdots 8 + 999 \cdots 9 - 999 \cdots 8\\&= \underbrace{1000 \cdots 0}_\text{(n + 1) digits} \times 999 \cdots 8 + 1\\&= \underbrace{999 \cdots 8}_\text{n digits} \underbrace{000 \cdots 0}_\text{n digits} + 1\\&= \underbrace{999 \cdots 8}_\text{n digits} \underbrace{000 \cdots 1}_\text{n digits}\\&= \underbrace{999 \cdots 9}_\text{(n - 1) digits} 8 \underbrace{000 \cdots 0}_\text{(n - 1) digits} 1\\\end{aligned}
The number of digits is easy to keep track of since when we add 1, we know that we just went up to another digit (999 + 1 = 1000 which has one more digit than 999). I’m not sure this would be as simple when squaring similar types of numbers like 777, 888 etc.
# Pin codes and the birthday problem
In an introductory statistics class, there are always tons of ways to link back up to the “real world” – its one of the great things about the topic! One of my favorite things to talk about in class is the concept of bank pin codes. How many are possible? Are they really randomly distributed? How many people need to get together before you can be sure two people have the same pin code? So many questions and all of them can be easily explored!
## My movie idea: the pigeonhole principle
Our romantic lead is at a baseball game, rushing between innings to the ATM. On the way, he bumps into a young woman and they drop all the stuff they’re carrying. Apologetic, he helps her up while gathering both of their things and is off to get his cash. It’s only later that he realizes that he accidentally used her ATM card! Not only that – he has some strange charges on his bank statement too! They must have mixed up cards… but is it really possible they share the same code? If so… they are clearly meant to be together…
[Naturally the rest of the movie is her thinking that he stole the card and then eventually them falling in love. At some point, a random person will vaguely explain the pigeonhole principle and give the movie its name – no stealing my idea]
Could this really happen? Yes! In fact, as we will see later, you probably wouldn’t even need the pair to be at a baseball game to make it reasonably likely. Before we get to that though, let’s look at the basic math.
### How many pin codes are possible?
We will work with pin codes are 4 digits long and can be any whole number 0 – 9. In general, the multiplication principle states that if you can break an action into steps, then the number of ways you can complete the action is found by multiplying the number of ways you can complete each step. For pin codes, we can think of selecting each digit as a step. There are 10 choices for each step so:
$10 \times 10 \times 10 \times 10 = 10\,000$
This shows there are 10,000 possible pin codes.
### How many people do you need to get together to guarantee at least two share a pin code?
The pigeonhole principle is a really simple idea that can be used to prove all kinds of things, from the complex to the silly. This rule states that if you have n pigeonholes but n + 1 (or more) pigeons, then at least 2 pigeons are sharing a pigeonhole. Another way to think about it: if a classroom has 32 chairs, but the class has 33 students… well, someone is sitting on someone’s lap (and class just got weird).
For pin codes, this means that you only need 10,001 people together to guarantee that at least two share the same pin code. You can imagine that if you were handing out pin codes, you would run out at the 10,000th person. At that point, you would have to give them the same code as someone else.
There are definitely more than 10,000 people at a baseball game –So the movie idea works! I’m going to be a millionaire! Ok back to math…
## Simulations show you need way fewer than 10,000 people
Everything we have talked about so far is based on the idea that pin codes are randomly selected by people. That is, we are assuming that any pin code has the same chance of being used by a person as any other. For now, we will continue that assumption but as you can imagine, people definitely don’t behave this way.
### The question
If we randomly assign pin codes, how many assignments (on average) will there be before there is a repeated code? We know by the pigeonhole principle that it is guaranteed after 10,000. But what is the typical number? For sure, through randomness, it often takes less than 10,000 right?
### The simulation
For the sake of ease with writing the code, we will assign each pin code a whole number 1 – 10,000. So you can imagine the code 0000 is 1, the code 0001 is 2, and the code 9999 is 10,000. This way, assigning a pin code is really just assigning a random number from 1 to 10,000. In fact, we can now state the problem as:
“How many random selections from {1,2,…,10000} until there is a repeated value?”
Let’s look at the code!
pin_codes=1:10000 #this is the set 1 through 10,000
selected=rep(0,10000) #placeholder for selected pins
count = 0
i = 1
repeat{ #loop that keeps selecting pin codes
pin = sample(pin_codes,1, replace=TRUE) #select the pin
if (pin %in% selected){ #if already selected then stop
break
}
selected[i]=pin #put the selected pin into "selected"
count = count +1 #keep track of how many you selected
i = i+1}
count #type this to see how many were selected before a repeat
If you copy and paste this code into R, you might be surprised. My results the first three times were: 120, 227, 41.
This seems to suggest that through random selection, it only took assigning 120 pin codes before a repeat (in the first trial), 227 (in the second) and the in the last trial – it only took 41 times! This can’t be right?!
### Loop it
Maybe through randomness, we just had some unusual trials. Running 500 or 1000 trials should show the overall trend. The code below is the same (almost) but I wrote a loop around it so that it would repeat the same experiment 500 times. If you try this on your computer note that it is a little slow (I didn’t consider efficiency at all when writing this).
pin_codes=1:10000
counts=rep(0,500)
for (i in 1:500){
selected=rep(0,10000)
count = 0
j = 1
repeat{
pin = sample(pin_codes,1,replace=TRUE)
if (pin %in% selected){
break}
selected[j]=pin
count = count +1
j = j+1}
counts[i]=count}
Type in “mean(counts)” and it will give us the mean number of times that pin codes were randomly assigned before a repeat. The result?
mean(counts)
[1] 124.486
This tells us that on average, it only takes about 124 assignments before you see a repeat [1]. This is waaaay less than 10,000. What is going on?!
# The birthday problem
A famous probability question is known as “the birthday problem“. Suppose that birthdays are equally likely to occur on any given day of the year (and include leap years). This means that there are 366 possible birthdays. By the pigeonhole principle, that means you are guaranteed to have two people share the same birthday as soon as you get 367 people together. But, the probability is almost 100% at only 70 people (and 50% at 23 people). This is another very counter-intuitive result.
### Pin codes and birthdays?
This pin code problem is really the same as the birthday problem, just a little bit bigger or more generalized. Just imagine that there are 10,000 possible birthdays – we are looking at the number of people you need to have two people with the same birthday under this much larger set. Realizing this and researching a bit, you find that this is something that has been studied and in fact, the expected number of selections needed would be:
$1 + \displaystyle\sum\limits_{k=1}^{10,000}\,\dfrac{10,000!}{(10,000-k)!10,000^k}$
Plugging that mess into wolfram alpha gives the result:
$1 + \displaystyle\sum\limits_{k=1}^{10,000}\,\dfrac{10,000!}{(10,000-k)!10,000^k} \approx 1 + 124.932 = 125.932$
Look at that – even over just 500 trials, our simulation was really close. It only takes about 125 to 126 selection (on average) before you see a repeated pin code (assuming they are randomly selected).
126 – thats it!
## But pin codes aren’t random
This is more true than you realize. If you look at the datagenetics article, you will see that instead of making up 1 of 10,000 (or 0.01%), the pin 1234 appears to actually make up more than 10% of codes. Just for fun, I went ahead and changed the code to account for the probability of the top 20. I then assigned all the remaining pin codes the remaining probability evenly – though this isn’t perfect as uncommon pin codes are REALLY uncommon.
pr=rep((1-0.2683)/9980,9980)
pin_codes=1:10000
counts=rep(0,500)
for (i in 1:500){
selected=rep(0,10000)
count = 0
j = 1
repeat{
pin = sample(pin_codes,1,replace=TRUE, prob = c(0.10713,0.06016,0.01881,0.01197,0.00745,0.00616,0.00613,0.00526,0.00516,0.00512,0.00451,0.00419,0.00395,0.00391,0.00366,0.00304,0.00303,0.00293,0.00290,0.00285,pr))
if (pin %in% selected){
break}
selected[j]=pin
count = count +1
j = j+1}
counts[i]=count}
This runs a loop of selecting pins until there is a repeat and then repeats that process 500 times (using these new probabilities for the first 20 codes). Checking the mean after one run I have:
mean(counts)
[1] 12.8
Even crazier! Considering how pin codes are not truly random at all, it looks like you would really only need around 12 to 13 people to have a repeat. Remember – there are 10,000 possibilities in general! ([2] – comment below on small correction made here)
## Summary
Here are the numbers all together:
• By the pigeonhole principle – you are guaranteed that two people share a pin code if the group is larger than 10,000. BUT:
• If pin codes are randomly distributed: ~126 people (on average) are needed before two share a pin code
• Using just a little of the data on the true distribution: Only ~13 people [2] (again, on average) are needed before two share a pin code!
Including more of the data we have on the distribution would probably bring that number down even further. As you can see, it is always very interesting to compare the theory to the reality.
#### Notes
[1] my code counts how many were selected and then stops counting when a repeat is encountered. So, it is really off by 1 from the expected number that you would select to have a repeat. This is a minor technicality overall but worth noting when you see the expected value formula which adds 1 to the sum.
[2] My original code had one of the probabilities as 0.0516 instead of 0.00516 and the mean after several runs was generally around 12. Fixing this probability, the mean seems to be a bit closer to 13 with several runs resulting in means of 12.5 to 12.8. It seems the top codes are really dominating the selection. It would be interesting to code in the details about the less likely pin codes (since they have a very tiny probability of being selected) and seeing if this is actually lower or not.
# Generate a data set with a given correlation coefficient
Recently, I found myself needing to create scatterplots that represented specific values for the correlation coefficient r. This was for a writing project, but it is something that has come up with teaching as well. Showing students scatterplots for many different values of r seems to really help them conceptually, especially when it comes to understanding that not every data set with the same correlation will look exactly the same. Unfortunately, I have always been at the mercy of what examples I can find online or in textbooks. With this in mind, I set out to figure this problem out once and for all.
The problem: Given a desired correlation coefficient, generate a data set.
As it turns out, this is not that difficult of a problem! Using this overall solution, I wrote a simple function in R.
make_data_corr = function(corr, n){
x = rnorm(n,0,1)
y = rnorm(n,0,1)
a = corr/(1-corr^2)^0.5
z=a*x+y
the_data = data.frame(x,z)
return(the_data)
}
The inputs here are corr (the desired value for the correlation coefficient) and n (the desired number of paired data values). You will notice that I didn’t add any kind of validation or anything like that to this function, so if you put in a strange value for r or n, you are on your own. The resulting output is a data frame with your data set being x and z. Here is an example of it in action:
example=make_data_corr(0.85,35)
plot(example$x,example$z)
At smaller sample sizes, the correlation coefficient is CLOSE but not exact. Here, r = 0.92 but when I ran the function again with n = 350 I ended up with r = 0.83. For my purposes this is good enough, but it is a consideration for possible improvements (at this stage, I haven’t thought about how to approach this).
Eventually I may make this into a small webapp that anyone can use (including myself). Until then, if you find a use for this or find a way to make this better, certainly let me know. It is an interesting little problem to play with! | 2017-07-27T04:34:46 | {
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"url": "http://www.jerimiannwalker.com/category/cool-math/",
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"lm_name": "Qwen/Qwen-72B",
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http://hcxu.bonref.fr/qr-decomposition-least-squares.html | # Qr Decomposition Least Squares
Sparse Linear Least Squares – p. When we used the QR decomposition of a matrix to solve a least-squares problem, we operated under the assumption that was full-rank. QR-decomposition using householder reflections. Updating the QR Factorization and the Least Squares Problem Sven Hammarling∗ Craig Lucas† November 12, 2008 Abstract In this paper we treat the problem of updating the QR factorization, with applications to the least squares problem. The LU decomposition is twice as fast as the standard QR decomposition and it will solve most systems. (c)If the columns of A are linearly independent, then the least squares solution is unique. QRD is useful for solving least squares' problems and simultaneous equations. qr is called. An important special case is the rank-one orthogonal projector which can be written as. Algorithms. Consider the rank-deficient linear least-squares problem: For r=n, the optimal solution, , is unique; however, for r> help qr QR Orthogonal-triangular decomposition. Consider the least squares problem with A = 1 3 1 1 3 7 1 −1 −4 1 −1 2 and b = 1 1 1 1. For a singular or non-square matrix A the QR-decomposition of A is not unique. Las funciones incluyen una gran variedad de factorizaciones de matrices, resolución de ecuaciones lineales y cálculos de valores propios o valores singulares, entre otras. qr, but if a is a rectangular matrix the QR decomposition is computed first. Cosine-Sine Decomposition: LAPACK Computational Routines?bbcsd?orbdb/?unbdb; LAPACK Least Squares and Eigenvalue Problem Driver Routines. The default value is 2. One of the applications of QR factorization is solution of linear least squares. In that case we revert to rank-revealing decompositions. 1 Formulation of Least-Squares Approximation Problems. then after the QR decomposition, we get R = 3. You may get a few more analytical insights with SVD, but QR is lighter weight machinery (i. It is shown that the proposed framework is effective in terms of quantitative and qualitative reconstructions of initial pressure distribution enabled via finding an optimal regularization parameter. In such cases, using a dense QR factorization is inefficient. Algorithms are presented that compute the factorization Ae = QeRe where Ae is the matrix A = QR. solve solves systems of equations via the QR decomposition: if a is a QR decomposition it is the same as solve. Renaut via the QR factorization of A [9]. The Householder reflectors method is given below. The QR and Cholesky Factorizations §7. The first section of this chapter is devoted to the QR decomposition itself. Second-order cone optimization. One of the fundamental problems in electronic structure calculations is to determine the electron density associated with the minimum total energy of a molecular or bulk system. Because A=QR, so there should be no difference to use A or QR anyway. THE INVERSE QR-RLS ALGORITHM FOR THE CMAC QR-decomposition is a method for decomposing a matrix into two matrices, one orthogonal and the other upper triangular. The QR decompostion always exists, even if the matrix does not have full rank, so the constructor will never fail. Large non-linear least square problems are usually sparse. QR review Reduced QR Let A by m by n and A=QR What is the shape of Q? What is the shape of R? What are they for Full QR? QR review Least Squares with QR Tall and skinny systems No exact solution…. Quadratic optimization. • Predict the sparsity structure of the decomposition and allocate storage. To solve a Linear Least Squares Problem using the QR-Decomposition with matrix A2Rm n, of rank nand b2Rm: 1. The vector x is then a solution in the least squares sense, and this sort of problem is called a least squares problem. The LU decomposition is twice as fast as the standard QR decomposition and it will solve most systems. This lecture is matter of least square (LS). There are however pathologically ill-conditioned systems and non-square systems. This old question is a continuation of another old question I have just answered: Compute projection / hat matrix via QR factorization, SVD (and Cholesky factorization?). One of the fundamental problems in electronic structure calculations is to determine the electron density associated with the minimum total energy of a molecular or bulk system. Explain how to use the QR decomposition to solve the least squares problem, and why that method works. See [2] for details on how to use QR decomposition for constructing orthogonal bases, and for solving least-squares problems. Linear least squares with linear equality constraints by direct elimination -- 22. There are many studies on the parallel QR decomposition, however, most of them assume a central coordinator knows Aand bthen distribute computations to the network. QR decomposition is often used to solve the linear least squares problem, and is the basis for the QR algorithm. 9640 , and the real least squares solution x= −0. returns the results of solve. Formally, we distinguish the cases M < N, M = N, and M > N, and we expect trouble whenever M is not equal to N. Factor the matrix a as qr, where q is orthonormal and r is upper-triangular. Linear Least Squares Problems 121 Here is the general algorithm for QR decomposition using Householder transformations. Such series converge fast if f(x) is smooth. The worksheet also contains an example of how QR Decomposition can be applied to find the least-squares solution of an overdetermined system of equations. LEAST SQUARE PROBLEMS, QR DECOMPOSITION, AND SVD DECOMPOSITION 3 where the columns of Q^ are orthonormal. In such cases, using a dense QR factorization is inefficient. Another methods, potentially faster but less reliable, are to use a Cholesky decomposition of the normal matrix or a QR decomposition. AU - Apley, Daniel W. I am using Matlab to estimate a regression model with ordinary least squares (OLS). The second section treats its application to least squares, with particular attention being paid to the tricky matter of whether to use the QR decomposition or the normal equations, which is the traditional way of solving least squares problems. --Rdm 16:34, 17 June 2011 (UTC). This is a square problem, and Rbis invertible if Awas full rank. This function finds the least squares solution to the overdetermined system where the matrix A has more rows than columns. There are however pathologically ill-conditioned systems and non-square. And the least squares lattice (LSL) algorithm has been rederived based on the QRD method. Sparse Linear Least Squares – p. solve solves systems of equations via the QR decomposition: if a is a QR decomposition it is the same as solve. Updating the QR Factorization and the Least Squares Problem Sven Hammarling∗ Craig Lucas† November 12, 2008 Abstract In this paper we treat the problem of updating the QR factorization, with applications to the least squares problem. Consider the least squares problem with A = 1 3 1 1 3 7 1 −1 −4 1 −1 2 and b = 1 1 1 1. Linear algebra review. Linear Least Squares with Linear Equality Constraints by Direct Elimination; 22. (c)If the columns of A are linearly independent, then the least squares solution is unique. This old question is a continuation of another old question I have just answered: Compute projection / hat matrix via QR factorization, SVD (and Cholesky factorization?). I am using Matlab to estimate a regression model with ordinary least squares (OLS). Sparse least squares and Q-less QR Suppose we want to solve a full-rank least squares problem in which Ais large and sparse. If X is an n by p matrix of full rank (say n > p and the rank = p), then X = QR where Q is an n by p orthonormal matrix and R is a p by p upper triangular matrix. Matlab diary: QR with Column Pivoting and the Least-Squares Problem -- A Matlab diary file showing the hand calculation of the QR factorization with column pivoting and how to use it to solve the least-squares problem. Least square computation via QR factorization (see linear algebra recap notes), least squares function approximation: linear regression. 09_QR Decomposition; 10_Householder Transformation; 11_Least Squares Problems; 12_Geometric Meaning of Least Squars Solution; 13_Nonlinear Least Squares Problems; 14_Singular Value Decomposition; 15_Solving Least Squares Problem via SVD; 16_Eigenvalue Problems; 17_Power Method; 18_Inverse Power Method; 19_QR Algorithm for Eigenvalue Computation. QL and RQ decompositions are less popular and they are not implemented in the ALGLIB. The Householder reflectors method is used to calculate the QR factorization of a matrix. The primary use of the QR decomposition is in the least squares solution of nonsquare systems of simultaneous linear equations. This works because orthogonal transformations. Lecture 22: Gram-Schmidt QR, modified Gram Schmidt. The QR algorithm uses orthogonal (or unitary) transformations. In this lab, we introduce linear least squares problems, tools in Python for computing least squares solutions, and two fundamental algorithms. Con las funciones de álgebra lineal de MATLAB ®, es posible realizar cálculos de matrices rápidos y numéricamente robustos. Suitable choices are either the (1) SVD or its cheaper approximation, (2) QR with column-pivoting. 6 Least Squares Approximation by QR Factorization 6. The QR decomposition is a popular approach for solving the linear least squares equation. You cannot use the following syntax: qr (A,B). The QR decompostion always exists, even if the matrix does not have full rank, so the constructor will never fail. The Householder reflectors method is given below. QR decomposition is often used to solve the linear least squares problem and is the basis for a particular eigenvalue algorithm , the QR algorithm. Lecture 6-7, 16 in the course lecture notes. > problems in an Excel implementation of QR decomposition. We never need to form Pi. solving a least squares problem) the “economy size version” [Q,R]=qr(A,0) is sufficient. The first section of this chapter is devoted to the QR decomposition itself. The QR and Cholesky Factorizations §7. The QR decomposition can be found. The approach is to construct a factorization of the matrix A that can be used to solve the problem. Second-order cone optimization. Y1 - 2000/12/1. , a system in which A is a rectangular m × n-matrix with more equations than unknowns (when m>n). is an matrix with linearly independent columns, then. 1 Answer to 1. An alternative, more practical method is QR factorization with column piv-oting, a method proposed by Golub in the mid-sixties [9]. Signal processing and MIMO systems also employ QR decomposition. The projection Px= Q^(Q^T x) can be interpret as: c= Q^T xis the coefficient vector and Qc^ is expanding xin terms of column vectors of Q^. Introduction. QR Decomposition Least Squares Eric Shaffer. These notes explain some reflections and rotations that do it, and offer M ATLAB implementations; in its notation, x ':= (complex conjugate transpose of x). KMP was closely related to the orthogonal least squares (OLS) method in the field of nonlinear model identification (Chen, Cowan, & Grant, 1991). filters, Recursive least squares (RLS), inverse QR decomposition (IQRD). Cholesky factorization of $X^TX$ is faster, but its use for least-squares problem is usually discouraged due to the claim that it “squares the condition number”. If X is an n by p matrix of full rank (say n > p and the rank = p), then X = QR where Q is an n by p orthonormal matrix and R is a p by p upper triangular matrix. THE INVERSE QR-RLS ALGORITHM FOR THE CMAC QR-decomposition is a method for decomposing a matrix into two matrices, one orthogonal and the other upper triangular. This old question is a continuation of another old question I have just answered: Compute projection / hat matrix via QR factorization, SVD (and Cholesky factorization?). K 2(A)2 = K 2(ATA) In the next section we will choose one function f(x) and we will construct its best polynomial approximation by least squares solving the system by QR factorization and Cholesky on normal equations. The least squares approximate solution to A*x = b can be found with the Q-less QR decomposition and one step of iterative refinement: x = R$$R'\(A'*b)) r = b - A*x e = R\(R'\(A'*r)) x = x + e; See also LU, NULL, ORTH, QRDELETE, QRINSERT, QRUPDATE. Either will handle over- and under-determined systems, providing a least-squares fit if appropriate. This old question is a continuation of another old question I have just answered: Compute projection / hat matrix via QR factorization, SVD (and Cholesky factorization?). Smoothing 3. My question is, whether surface fitting using Pivoted QR decomposition is a single pass fitting or multipass fitting procedure? If it is trying to minimize the least square then how it is changing the coefficient values for every iteration?. WALKER Abstract. compute QR factorization A = QR (2mn2 flops if A is m n) 2. In principle, we could solve the problem via the normal equations A TAx= A b; or introduce A= QRand multiply ATAx= RTRx= bby R T to nd Rx= R TA b= QTb: Note that there is a very close relation between these approaches. , deviation between • what we actually observed (y), and • what we would observe if x = ˆx, and there were no noise (v = 0) least-squares estimate is just xˆ = (ATA)−1ATy Least-squares 5–12. Higham‡ September 5, 1997 Abstract For least squares problems in which the rows of the coefficient matrix vary widely in norm, Householder QR factorization (without pivoting) has unsatisfac-tory backward stability properties. Such series converge fast if f(x) is smooth. Q·R·x = b R·x = QT·b x=R-1·QT·b Matlab Code: Normal Equations Consider the system. See Also: Serialized Form. Form the matrix A∗A and the vector A∗b 2. QR-decompositions and the least square problem Ax=b. - Read in Strang, section 5. Let \(H = R^\top R$$ be the Cholesky factorization of the normal equations, where $$R$$ is an upper triangular matrix, then the solution to is given by. Figure 1: QR Decomposition Base Least Squares III. This is a standard QR decomposition implementation in a Python library that uses the Householder Transforma-tion method to create the decomposition [1]. The design matrix X is m by n with m > n. MATCOM and MATLAB notes: MATCOM function lsfrqrh implements the QR factorization method for the full-rank least-squares problem using. 1 Using QR to solve Least Squares A good part of classical numerical linear algebra (NLA) is concerned with solving the equation Ax = b ( the joke amongst NLA people being that the rest of it is solving Ax = x. Just as the QR factorization has proved to be a powerful tool in solving least-squares and related linear regression problems, so too can the GQR. Least Squares 5. Harp-DAAL currently supports distributed mode of QR for dense input datasets. In the above examples, the decomposition was computed at the same time that the decomposition object was. Just as the QR factorization has proved to be a powerful tool in solving least-squares and related linear regression problems, so too can the GQR. Las funciones incluyen una gran variedad de factorizaciones de matrices, resolución de ecuaciones lineales y cálculos de valores propios o valores singulares, entre otras. A QR-DECOMPOSITION FOR MATRIX PENCILS P. Calculate the solution to the least-squares problem Ax = b using information about the QR factorization of A and input tolerance. There are many possible cases that can arise with the matrix A. We wish to solve for x Ax = b (A : m n; m n) (4) The problem is that there is no solution to (4) when m > n because we have more equations than unknowns. In such cases, using a dense QR factorization is inefficient. The least squares solution minimizes the Euclidean norm of the residual,. But that approach has numerical issues, as we will see later. 4 High-Performance Cholesky The solutionof overdetermined systems oflinear equations is central to computational science. The rows of matrix X are fed as inputs to the array from the top along with the corresponding element of the. That is where it will use the QR or SVD. See [2] for details on how to use QR decomposition for constructing orthogonal bases, and for solving least-squares problems. Dmitriy Leykekhman Fall 2008 Goals I SVD-decomposition. Part Va: Linear Least Squares, QR Solving Linear Least Squares Problems with QR If Q is M N with orthonormal columns, then QTQ = I, the N N identity. Symmertric matrices and their decomposition. 3 Solution of Rank Deficient Least Squares Problems If rank(A) < n (which is possible even if m < n, i. The QR decomposition, also known as the QR factorization, is another method of solving linear systems of equations using matrices, very much like the LU This website uses cookies to ensure you get the best experience on our website. One of the key benefits of using QR Decomposition over other methods for solving linear least squares is that it is more numerically stable, albeit at the expense of being slower to execute. 9640 , and the real least squares solution x= −0. The QR decompostion always exists, even if the matrix does not have full rank, so the constructor will never fail. Parameters: a - Complex matrix to be factored. QR decomposition is often used in linear least squares estimation and is, in fact, the method used by R in its lm() function. Given an mxn matrix of rank n, let A=QR be a QR factorization of A. Using SVD Decomposition. Indicate whether the statements are true or false. The use of Givens transformations and the QR decomposition to solve linear least squares problems has several advantages, particularly when the design matrix is sparse or large. 4 High-Performance Cholesky The solutionof overdetermined systems oflinear equations is central to computational science. Lecture 24: Linear least squares with QR, conditioning and stability. Properties of Matrices and Operations on Matrices A very useful factorization is A = QR, where Q is orthogonal and R is upper triangular or trapezoidal. The Pseudoinverse Construction Application Outline 1 The Pseudoinverse Generalized inverse Moore-Penrose Inverse 2 Construction QR Decomposition SVD 3 Application Least Squares Ross MacAusland Pseudoinverse. We show how the simple and natural idea of approximately solving a set of over- determined equations, and a few extensions of this basic idea, can be used to solve. We can use cholesky decomposition to solve for Ax = b, Least Squares Problem though still QR is more optimal compared to Cholesky. lstsq() with the ones computed using the QR decomposition:. That is, the least. Adaptive systems commonly employ QR decomposition to solve overdetermined least squares problems. (TODO: implement these alternative methods). Reflections, Rotations and QR Factorization QR Factorization figures in Least-Squares problems and Singular-Value Decompositions among other things numerical. The QR decomposition is a popular approach for solving the linear least squares equation. Signal processing and MIMO systems also employ QR decomposition. The fact that Q is orthogonal means that Q'Q=I, so that Ax=b is equivalent to Rx=Q'b, which is easier to solve since R is triangular. The least squares fitting using non-orthogonal basis We have learned how to nd the least squares approximation of a function fusing an orthogonal basis. Least-norm solutions of undetermined equations • least-norm solution of underdetermined equations • minimum norm solutions via QR factorization • derivation via Lagrange multipliers • relation to regularized least-squares • general norm minimization with equality constraints 8-1. Set x= Py: D. One of the fundamental problems in electronic structure calculations is to determine the electron density associated with the minimum total energy of a molecular or bulk system. KMP was closely related to the orthogonal least squares (OLS) method in the field of nonlinear model identification (Chen, Cowan, & Grant, 1991). QR decomposition is often used in linear least squares estimation and is, in fact, the method used by R in its lm() function. In such cases, using a dense QR factorization is inefficient. One of the most important applications of the QR factorization of a matrix A is that it can be effectively used to solve the least-squares problem (LSP). , if we have an underdetermined problem), then infinitely many solutions exist. That answer discusses 3 options for computing hat matrix for an ordinary least square problem, while this question is under the context of weighted least squares. Then, for each in Rm, the equation has a unique least squares solution, given by (The solution may be obtained by using back substitution to solve ). You may get a few more analytical insights with SVD, but QR is lighter weight machinery (i. The columns of the matrix must be linearly independent in order to preform QR factorization. There are several methods for performing QR decomposition, including the Gram-Schmidt process, Householder reflections, and Givens rotations. For non-square matrices or when simple inversion to recover the data performs poorly, the QR decomposition is used to generate an equivalent upper triangular system, allowing for detection using the sphere decomposition or M-algorithm. Shaw CB, Prakash J, Pramanik M, Yalavarthy PK. This lecture is matter of least square (LS). solve solves systems of equations via the QR decomposition: if a is a QR decomposition it is the same as solve. We could find $\mathbf{x}$ by solving the normal equations, which is a square linear system. NUMERICALLY EFFICIENT METHODS FOR SOLVING LEAST SQUARES PROBLEMS 5 The 2-norm is the most convenient one for our purposes because it is associated with an inner product. To solve this using QR factorization, we note that inserting the QR factorization A=QR in the normal equations, T x T b, and simplifying gives Rx = QT b: Thus, the solution of the least squares problem is given by x =R Q1 T b. In the above examples, the decomposition was computed at the same time that the decomposition object was. the QR factorization provided m ˛ n. There are however pathologically ill-conditioned systems and non-square. If QR = A is the QR-decomposition of the matrix A, the formal least-squares solution is c = R−1QTb. System identi cation 5. Example: Solving a Least Squares Problem using Gram-Schmidt Problem For A = 3 2 0 3 4 4 and b = 3 5 4 solve minjjb Axjj. Get more help from Chegg. Cholesky decomposition, LU decomposition, QR decomposition, rank revealing, numerical rank, singular values, strong rank revealing QR factorization. Consider the rank-deficient linear least-squares problem: For r=n, the optimal solution, , is unique; however, for r> help qr QR Orthogonal-triangular decomposition. H A RT MA N N y Abstract. QR decomposition is often used in linear least squares estimation and is, in fact, the method used by R in its lm() function. I use it to do a data fitting with least square method(so with the backsubstitution i am supposed to take. MATCOM and MATLAB notes: MATCOM function lsfrqrh implements the QR factorization method for the full-rank least-squares problem using. 1: Constrained least squares polynomial fits (m =30, n =10). AU - Parhi, Keshab K. The solution can be found with the linear least squares method, e. Least Square sense means a solver can be computed for an. Leykekhman - MATH 3795 Introduction to Computational MathematicsLinear Least Squares { 1. For an inverse based on QR decomposition, see[M-5] qrinv(). Compute QT b= c d : 3. Shaw CB, Prakash J, Pramanik M, Yalavarthy PK. (8) However in practice it is better to back-substitute the system Rc = QTb. Examples of Some. MATH 3795 Lecture 9. the QR factorization by calling numpy. $$Form the orthogonal projection matrix$$ P = Q Q^T. I use it to do a data fitting with least square method(so with the backsubstitution i am supposed to take. To solve this using QR factorization, we note that inserting the QR factorization A=QR in the normal equations, T x T b, and simplifying gives Rx = QT b: Thus, the solution of the least squares problem is given by x =R Q1 T b. The Householder reflectors method is given below. Gram-Schmidt Algorithm. R Where: - Q is an orthogonal matrix - R is an upper triangular matrix (= right triangular matrix) If :return parameter is specified in options map, it returns only specified keys. Least Squares with QR and SVD As we have seen up to now, we have our forward problem (in matrix, discrete form) d = Gm (1) where d represents the vector with the data, G is the theory that we use (the. From Wikipedia: In linear algebra, a QR decomposition (also called a QR factorization) of a matrix is a decomposition of a matrix A into a product A = QR of an orthogonal matrix Q and an upper triangular matrix R. If m > n, then qr computes only the first n columns of Q and the first n rows of R. Refer to the MATLAB qr reference page for more information. Con las funciones de álgebra lineal de MATLAB ®, es posible realizar cálculos de matrices rápidos y numéricamente robustos. If you specify a third output with the economy-size decomposition, then it is returned as a permutation vector such that A(:,P) = Q*R. Cosine-Sine Decomposition: LAPACK Computational Routines?bbcsd?orbdb/?unbdb; LAPACK Least Squares and Eigenvalue Problem Driver Routines. The use of the conjugate gradient method with a nonsingular square submatrix A 1 2 R n of A as preconditioner was first suggested by L¨auchli in 1961. Figure 1: QR Decomposition Base Least Squares III. , if we have an underdetermined problem), then infinitely many solutions exist. Cholesky decomposition, LU decomposition, QR decomposition, rank revealing, numerical rank, singular values, strong rank revealing QR factorization. N2 - In this paper we examine the properties of QR and inverse QR factorizations in the general linear least squares (LS) problem. However, traditional QR decomposition methods, such as Gram-Schmidt (GS), require high computational complexity and non-linear operations to achieve high throughput, limiting their usage on resource-limited platforms. Im-plementing the QRD-LSL interpolator merely involves using both forward and backward prediction errors produced at the various stages of a QRD-LSL predictor. does not contribute anything to the decomposition since it is multiplied by zero. My question is, whether surface fitting using Pivoted QR decomposition is a single pass fitting or multipass fitting procedure? If it is trying to minimize the least square then how it is changing the coefficient values for every iteration?. Once these factors are calculated, the residual becomes. (Under-constrained problems: see Tygert 2009 and Drineas et al. The package provides an implementation of the QR decomposition for general complex matrices for the go programming language. In brief, they are ways to transform the matrix $\mathbf{A}$ into a product of matrices that are easily manipulated to solve for the vector $\mathbf{c}$. 1 Formulation of Least Squares Approximation Problems Least-squares problems arise, for instance, when one seeks to determine the relation between an independent variable, say time, and a measured dependent variable, say position or velocity of an object. SVD and Least Squares • Solving Ax=b by least squares: • ATAx = ATb x = (ATA)-1ATb • Replace with A+: x = A+b • Compute pseudoinverse using SVD – Lets you see if data is singular (< n nonzero singular values) – Even if not singular, condition number tells you how stable the solution will be – Set 1/w i to 0 if w. QR decomposition is often used in linear least squares estimation and is, in fact, the method used by R in its lm() function. Linear least squares with linear inequality constraints -- 24. The first section of this chapter is devoted to the QR decomposition itself. So in a nutshell, the full algorithm for solving the linear least squares problems is like this. QR-decomposition where is an orthogonal matrix and an upper triangular matrix! Non zero diagonal! Matlab: [Q,R]=qr(A) Fact: V. with two steps [C,R] = qr(A,b) x = R\C. AU - Shi, Jianjun. The standard recommendation for linear least-squares is to use QR factorization (admittedly a very stable and nice algorithm!) of $X$. Modern implementations for general matrices use successive applications of the Householder transform to form QR, though variants based on Givens rotation or Gram-Schmidt orthogonalization are. The QR and Cholesky Factorizations §7. To solve a Linear Least Squares Problem using the QR-Decomposition with matrix A2Rm n, of rank nand b2Rm: 1. You cannot use the following syntax: qr (A,B). An important special case is the rank-one orthogonal projector which can be written as. To answer my own question above, the reason for the column of 1s in the knowledge base article is that the regression using QR decomposition is of the form Rx = Q[T]b, so the regression is fitting a coefficient to a unit value, which is equivalent to solving for the constant term based. System identi cation 5. Lab 1 Least squares and Eigenvalues Lab Objective: Use least squares to t curves to data and use QR decomposition to nd eigenvalues. Reflections, Rotations and QR Factorization QR Factorization figures in Least-Squares problems and Singular-Value Decompositions among other things numerical. Least Squares and Computing Eigenvalues Lab Objective: Because of its numerical stability and convenient structure, the QR decomposition is the basis of many important and practical algorithms. determination of the QR factorization of B- ‘A avoids the possible numerical difficulties in forming B-’ or B-‘A. The primary use of the QR decomposition is in the least squares solution of nonsquare systems of simultaneous linear equations. Get more help from Chegg. Exercise 20 (Downdating least squares problems) For linear least squares problems, ’downdating’ refers to the efiect of eliminating an observation: Let A 2 Rm£n; m ‚ n, with rank A = n and let A = Q ˆ R 0! be the QR-decomposition of A. We wish to solve for x Ax = b (A : m n; m n) (4) The problem is that there is no solution to (4) when m > n because we have more equations than unknowns. 6 Least Squares Approximation by QR Factorization 6. Typical methods for computing the QR decomposition use Householder transformations, Givens transformations, or the Gram-Schmidt process. Hence the solution of our least squares problem is the vector c= 2 4 1:1 0:9 0:5 3 5. These notes explain some reflections and rotations that do it, and offer M ATLAB implementations; in its notation, x ':= (complex conjugate transpose of x). Special Features. Properties of Matrices and Operations on Matrices A very useful factorization is A = QR, where Q is orthogonal and R is upper triangular or trapezoidal. The exponential scaling in data rates will cause the world traffic to reach thousands and thousands of Exabyte by the end of 2020. This example shows how to compute the QR decomposition of matrices using hardware-efficient MATLAB® code in Simulink®. It provides an interface to the techniques used in the LINPACK routine DQRDC or the LAPACK routines DGEQP3 and (for complex matrices) ZGEQP3. From Wikipedia: In linear algebra, a QR decomposition (also called a QR factorization) of a matrix is a decomposition of a matrix A into a product A = QR of an orthogonal matrix Q and an upper triangular matrix R. To solve the normal equations, this implementation uses QR decomposition of the X matrix. QR-decomposition-based least-squares lattice (QRD-LSL) predictors have been extensively applied in the design of order-recursive adaptive filters. Uses QR decomposition instead of LU decomposition. , if we have an underdetermined problem), then infinitely many solutions exist. Compute QT b= c d : 3. recursive least-squares estimation algorithms of Lee et al (1981) were derived by Kalson-Yao (1985). Because A=QR, so there should be no difference to use A or QR anyway. THE INVERSE QR-RLS ALGORITHM FOR THE CMAC QR-decomposition is a method for decomposing a matrix into two matrices, one orthogonal and the other upper triangular. Part Va: Linear Least Squares, QR Solving Linear Least Squares Problems with QR If Q is M N with orthonormal columns, then QTQ = I, the N N identity. Cosine-Sine Decomposition: LAPACK Computational Routines?bbcsd?orbdb/?unbdb; LAPACK Least Squares and Eigenvalue Problem Driver Routines. The QR Decomposition Here is the mathematical fact. Recursive least squares (RLS) adaptive noise cancellation via QR decomposition (QRRLS) is introduced to reduce the bias in the mixing matrix caused by noise. The QR and Cholesky Factorizations §7. These questions concern the least squares solution x^ to A~x =~b. Note that the equation for is the normal equation of the following linear least squares problem:. Updating QR factorization procedure for solution of linear least squares problem with equality constraints Article (PDF Available) in Journal of Inequalities and Applications 2017(1) · December. That answer discusses 3 options for computing hat matrix for an ordinary least square problem, while this question is under the context of weighted least squares. The QR decomposition is widely used to solve the linear least squares problem as well as the nonlinear least squares problem. A least squares solution of Ax=b is a list of weights that when applied to the columns of A, produces the orthogonal projections of b onto Col a. (8) However in practice it is better to back-substitute the system Rc = QTb. Least-squares and variants. Consider the rank-deficient linear least-squares problem: For r=n, the optimal solution, , is unique; however, for r> help qr QR Orthogonal-triangular decomposition. CS 542G: QR, Weighted Least Squares, MLS Robert Bridson October 6, 2008 1 The QR Factorization We established the Gram-Schmidt process last time as a start towards an alternative algorithm for solv-ing least squares problems: while the normal equations approach (using Cholesky factorization) is very. the computation of the best approximation of an unsolvable system of linear equations. QR decomposition is often used in linear least squares estimation and is, in fact, the method used by R in its lm() function. • Predict the sparsity structure of the decomposition and allocate storage. There are however pathologically ill-conditioned systems and non-square systems. Application of QR decomposition-based least squares technique to improve air traffic control radar Abstract: For real-time radar processing, it is very desirable to have an algorithm that does not assume restricted statistics of the input data and can be implemented for high-speed processing (without a high cost) to meet real-time requirements. If A is invertible , then the factorization is unique if we require the diagonal elements of R to be positive. The approach still involves a matrix inversion, but in this case only on the simpler R matrix. Sparse least squares and Q-less QR Suppose we want to solve a full-rank least squares problem in which Ais large and sparse. Could someone Stack Exchange Network Stack Exchange network consists of 175 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Signal processing and MIMO systems also employ QR decomposition. 1 Formulation of Least-Squares Approximation Problems. The QR decomposition, also known as the QR factorization, is another method of solving linear systems of equations using matrices, very much like the LU This website uses cookies to ensure you get the best experience on our website. Moreover, the relation between LS with optimal GSVD is derived in this lecture. A computationally efficient approach that computes the optimal regularization parameter for the Tikhonov-minimization scheme is developed for photoacoustic imaging. There are three basic techniques for solving the overdetermined least-squares problem, m ≥ n, solving the normal equations, using the reduced QR decomposition, and using the reduced SVD. Lecture 6-7, 16 in the course lecture notes. 2 The QR Factorization §7. qr is called. qr, but if a is a rectangular matrix the QR decomposition is computed first. Example: Solving a Least Squares Problem using Householder transformations Problem For A = 3 2 0 3 4 4 and b = 3 5 4 , solve minjjb Axjj. 4, ;GvL 5, 5. QR Decomposition is widely used in quantitative finance as the basis for the solution of the linear least squares problem, which itself is used for statistical regression analysis. (c)If the columns of A are linearly independent, then the least squares solution is unique. 8 Chapter 5. Note: this uses Gram Schmidt orthogonalization which is numerically unstable. NUMERICALLY EFFICIENT METHODS FOR SOLVING LEAST SQUARES PROBLEMS 5 The 2-norm is the most convenient one for our purposes because it is associated with an inner product. This is a square problem, and Rbis invertible if Awas full rank. Stepping over all of the derivation, the coefficients can be found using the Q and R elements as follows: b = R^-1. However, as we explain in Section 2 below, all. The QR decomposition is a popular approach for solving the linear least squares equation. This will fail if isFullRank() returns false. The Rank-De cient Least Squares Problem QR with Column Pivoting When Adoes not have full column rank, the property in the case where rank(A) = r n. The vector x mininizes (x) = kb Fx k2 2 if and only if it is the solution of thenormal equations: F T Fx = F T b Proof: Expand out the formula for (x + x ): (x + x ) = (( b Fx ) F x )T ((b Fx ) F x ). the computation of the best approximation of an unsolvable system of linear equations. | 2020-04-01T13:18:50 | {
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http://math.stackexchange.com/questions/316734/formula-for-sum-i-1n-n1-i-n-i/316747 | Formula for $\sum _{i=1}^n (n+1-i) (n-i)$
It is easy to show that
$$\sum _{i=1}^n (n+1-i) (n-i) = n (n-1)+(n-1) (n-2)+...+1 (1-1)=\frac{1}{3} \left(n^3-n\right)$$
using induction. But how do I derive this formula? I couldn't find any substitution to do this.
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If you don't know/remember/want to use any of these "standard identities", then my favorite way is using finite differences. We can work directly with your original sum. Let's define
$$f(n)=\sum_{i=1}^n (n+1-i)(n-i).$$
First compute a bunch of values of $f$ like $f(1),f(2),f(3),...,f(10)$ for example. Then compute their finite differences. Then compute the second order finite differences, meaning finite differences of the finite differences. Then the third order and so on. You will see that eventually the finite differences are all zero. In this case the fourth order differences will all be zero. No matter how many terms you take, you can go up to $f(100)$ and they will all be zero.
This then tells you that $f(n)$ is a cubic polynomial in $n$. Then you simply compute any four (different) values, the easiest are using $f(1),f(2),f(3),f(4)$ and then get the unique interpolating polynomial that goes through those four points and then WHAM, you get your polynomial
$$f(n)=\frac{n^3-n}{3}.$$
The summation will always be a polynomial in $n$ if the summand is a polynomial in the index $i$ which means that eventually the finite differences will be zero. Specifically, the degree of $f(n)$ will be one higher than the degree of the summand in $i$ so in this we can also just look at the sum at see that the answer must be cubic in $n$.
Then just use interpolation to figure it out. This technique of finite differences is very powerful by the way and useful with all sorts of stuff. They (almost) mimic the continuous derivative. So for a polynomial, finite differences eventually become all zeros and hence stay zeros after that and the order when they first become all zeros tells you the degree of the polynomial. The fourth derivative of a cubic is zero so if fourth order finite differences of a sequence are all zeros then the generating function must be a cubic. For example, if you get the same exact (non-zero) differences like
$$1,2,4,8,16,...$$
$$1,2,4,8,16,...$$
then your original expression must be an exponential. Compare this with the derivative of an exponential being itself properly scaled. Derivatives of exponentials are never identically zero.
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Great method! Works on all polynomials and very simple. Thank you! – Max Feb 28 '13 at 14:37
If you look at your summand, it is quadratic in $i$, so you know the sum has to be cubic in $n$ (sums behave somewhat like integrals: $\sum_k k^m \rightsquigarrow n^{m + 1}$ just as $\int x^m d x \rightsquigarrow x^{m + 1}$) – vonbrand Feb 28 '13 at 17:26
\begin{align} \sum_{i = 1}^n 1 &= n \\ \sum_{i = 1}^n i &= \frac{n(n + 1)}{2} \\ \sum_{i = 1}^n n &= n \cdot \sum_{i = 1}^n 1 \\ & = n^2 \\ \sum_{i = 1}^n i^2 &= \frac{n(n + 1)(2n + 1)}{6} \\ \end{align}
Now, from the given expression;
\begin{align} \sum_{i=1}^n (n+1-i) (n-i) &= \sum_{i = 1}^n\left( n^2 + i^2 + n - i - 2 n i\right) \\ &= n^2 \cdot \sum_{i=1}^n 1 + \sum_{i=1}^n i^2 + n \cdot \sum_{i=1}^n1 - \sum_{i=1}^ni -2n \cdot \sum_{i=1}^ni \\ &= n^3 + \frac{n(n + 1)(2n + 1)}{6} + n^2 - \frac{n(n + 1)}{2} - n^2(n + 1) \\ &= \frac{n(n + 1)}{6} \cdot \left( 2n + 1 - 3\right) \\ &= \frac{n (n + 1)}{6} \cdot 2(n - 1) = \frac{n (n + 1) (n - 1)}{3} \\ &= \frac{n^3 - n}{3} \end{align}
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$$\sum _{i=1}^{n}{(n+1-i)(n-i)}=\sum _{i=1}^{n}{i(i-1)}=2\sum _{i=1}^{n}{\binom{i}{2}}=2\binom{n+1}{3}$$
This and related identities are easy to derive from the combinatorial identity
$$\sum_{i=0}^{n}{\binom{i}{k}}=\binom{n+1}{k+1}$$
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Let me give you a hint for a combinatorial proof.
Let us say, we have $n+1$ integers from $1,2,3\cdots n,n+1$ and we need to pick $3$ integers from them.
How many ways to pick them. $\binom{n+1}{3}$.
Now, count the same as follows. What can be the largest integer of the selected $3$ element group. It can be any one of
3,4,5....n+1
. Let us count how many possible $3$ subset groups can have the largest element as $n+1$. It is $\binom{n}{2}$( agreed ). Like this, count all possible $3$ subset groups of $n+1$.
$\binom{n+1}{3} = \binom{2}{2} + \binom{3}{2} + \binom{4}{2} + \cdots \binom{n}{2}$
A little algebraic manipulation will do from here( divide by $2$ ).
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Hint: $$\sum_{i=1}^n i = \frac{n(n+1)}{2}$$ and $$\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$$ Now expand $(n+1−i)(n−i)$...
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http://math.stackexchange.com/questions/126270/how-many-seven-digit-numbers-satisfy-the-following-conditions | # How many seven-digit numbers satisfy the following conditions?
How many seven-digit numbers divisible by 11 have the sum of their digits equal to 59?
I am able to get the seven-digit numbers divisible by 11
and
I am also able to get the seven-digit numbers whose sum of their digits equal to 59.
But i am not able to get how i can get the count of 7 digit numbers satisying both the condition.
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Since I promised vikiiii that I’d answer, here’s my version.
A number is divisible by $11$ if and only if the alternating sum of its digits is divisible by $11$. Say that the seven-digit number is $abcdefg$, where the letters represent the individual digits; then it’s divisible by $11$ if and only if $(a+c+e+g)-(b+d+f)$ is a multiple of $11$. Let $S=a+c+e+g$ and $T=b+d+f$; then we need $S-T$ to be a multiple of $11$ and $S+T=59$.
Since $S+T=59$, one of $S$ and $T$ must be odd and the other even, so their difference must be odd. Thus, $S-T$ cannot be $0$ or $\pm22$. We should look for ways to make it $\pm11$ or $\pm33$. (It clearly can’t be any bigger in magnitude, since $S\le 4\cdot9=36$.)
Suppose that $S-T=11$; then $70=11+59=(S-T)+(S+T)=2S$, so $S=35$ and $T=24$. This is possible only if three of the digits $a,c,e,g$ are $9$ and the fourth is $8$; there’s no other way to get four digits that total $35$. For three digits to total $24$, they must average $8$, so the only possibilities are that all three are $8$, that two are $9$ and one is $6$, or that they are $7,8$, and $9$. Thus, the digits $a,c,e,g$ in that order must be $8999,9899,9989$, or $9998$, and the digits $b,d,f$ must be $888,699,969,996,789,798,879,897,978$, or $987$, for a total of $4\cdot 10=40$ numbers.
Now suppose that $S-T=-11$; then by similar reasoning $2S=-11+59=48$, so $S=24$, and $T=35$. But $T\le 3\cdot9=27$, so this is impossible. Similarly, $S-T$ cannot be $-33$. The only remaining case is $S-T=33$. Then $2S=33+59=92$, and $S=46$, which is again impossible. Thus, the first case contained all of the actual solutions, and there are $40$ of them.
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M.Scott you are genius.Thanks so much.This is the best way.Thank you again. – vikiiii Mar 30 '12 at 15:13
If we look at the number $a_6a_5a_4a_3a_2a_1a_0$: $$a_6a_5a_4a_3a_2a_1a_0 = a_610^6+a_510^5+ \dots +a_010^0 \\ \equiv a_6(-1)^6 + \dots+a_0(-1)^0 \mod 11 = a_6 -a_5+a_4-a_3+a_2-a_1+a_0 \mod 11$$ So if the number will be divisible by 11 we require $$a_6 -a_5+a_4-a_3+a_2-a_1+a_0 = 11m$$
We take note that $0 \leq a_i \leq 9$ so possible values of $11m$ are limited to: $$-27 = 4\cdot0-3\cdot9\leq 11m \leq 4\cdot 9-3\cdot0 = 36$$ Which really means that $11m \in \{-22,-11,0,11,22,33\}$
The other requirement of the question was that $a_6 + \dots +a_0 = 59$. From this and the above equation (not sure how to number them and align them nicely in TeX) we add and subtract and get much nicer equations:
$$a_6 +a_4+a_2+a_0 = \frac{59+11m} 2 \\a_5+a_3+a_1 =\frac{ 59-11m}2$$
Of course the LHS is whole, so the right hand side must be as well, which means $m$ needs to be odd. So we reduce our options to:
$$11m \in \{-11,11,33\} \implies \frac{59+11m} 2 \in \{24,35,46\}$$ Of course the sum of four digits can't be $46$ from our above inequality on $a_i$, and similarly $$\frac{59-11m} 2 \in \{35,24\}$$ But the sum of three digits can't be $35$, so we're left with $$a_6 +a_4+a_2+a_0 = 35 \\a_5+a_3+a_1 =24$$
It's easy to see that the only options for the four digits is a permutation of $9998$, and the three digits must be a permutation of one of $\{699,789,888\}$.
Order doesn't matter, so basic combinatorics gives $4\cdot(3 + 3! + 1) = 40$ such numbers.
-
Nice solution davin – juantheron Nov 8 '13 at 13:12 | 2014-08-28T13:26:28 | {
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https://byjus.com/question-answer/if-y-left-n-right-e-x-e-x-2-e-x-n-0-x/ | Question
# If $$y\left( n \right) ={ e }^{ x }{ e }^{ { x }^{ 2 } }...{ e }^{ { x }^{ n } },0<x<1$$ then $$\displaystyle \lim _{ n\rightarrow \infty }{ \frac { d\ y\left( n \right) }{ dx } }$$ at $$\displaystyle\frac{1}{2}$$ is
A
e
B
4e
C
2e
D
3e
Solution
## The correct option is D $$4e$$$$\displaystyle y\left( n \right) ={ e }^{ x+{ x }^{ 2 }+...+{ x }^{ n } }={ e }^{ \dfrac { x\left( 1-{ x }^{ n } \right) }{ 1-x } }$$ So $$\displaystyle \dfrac { dy\left( n \right) }{ dx } ={ e }^{ \dfrac { x\left( 1-{ x }^{ n } \right) }{ 1-x } }\times \dfrac { d }{ dx } \left( \dfrac { x\left( 1-{ x }^{ n } \right) }{ 1-x } \right)$$ $$\displaystyle \lim _{ n\rightarrow \infty }{ \dfrac { dy\left( n \right) }{ dx } = } \lim _{ n\rightarrow \infty }{ { e }^{ \dfrac { x }{ 1-x } -\dfrac { { x }^{ n+1 } }{ 1-x } } } \dfrac { d }{ dx } \left[ \lim _{ n\rightarrow \infty }{ \dfrac { x }{ 1-x } -\dfrac { { x }^{ n+1 } }{ 1-x } } \right]$$ $$\displaystyle ={ e }^{ \dfrac { x }{ dx } }\times \dfrac { d }{ dx } \left( \dfrac { x }{ dx } \right) ={ e }^{ \dfrac { x }{ 1-x } }\dfrac { d }{ dx } \left( -1+\dfrac { 1 }{ 1-x } \right)$$ $$\displaystyle ={ e }^{ \dfrac { x }{ 1-x } }.\dfrac { d }{ dx } \dfrac { 1 }{ { \left( 1-x \right) }^{ 2 } }$$ $$\because$$ for $$\displaystyle x\in \left( 0,1 \right) ;{ x }^{ n }\rightarrow 0$$ $$\displaystyle \therefore \lim _{ n\rightarrow \infty }{ \dfrac { { x }^{ n+1 } }{ 1-x } } =0$$ $$\displaystyle \Rightarrow \lim _{ n\rightarrow \infty }{ \left( \dfrac { dy\left( n \right) }{ dx } \right) _{ x=\dfrac { 1 }{ 2 } }=4e }$$ Mathematics
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https://www.physicsforums.com/threads/definite-integral.333310/ | # Definite Integral
1. Aug 29, 2009
### songoku
1. The problem statement, all variables and given/known data
Find using substitution $$x=\sin \theta$$
$$\int _0^{0.5}\frac{x}{\sqrt{1-x^2}}dx$$
2. Relevant equations
integration
3. The attempt at a solution
$$\int _0^{0.5}\frac{x}{\sqrt{1-x^2}}dx$$
$$=\int_0^{\frac{\pi}{6}}\sin \theta\;d\theta$$
$$=1-\frac{1}{2}\sqrt{3}$$
What I want to ask is about changing the upper bound of the integral.
For x = 0.5 :
$$0.5 = \sin \theta$$
Here I choose $$\theta = \frac{\pi}{6}$$. But what if I choose $$\theta = \frac{5}{6}\pi\;??$$
So, the integral :
$$=\int_0^{\frac{5}{6}\pi}\sin \theta\;d\theta$$
$$=1+\frac{1}{2}\sqrt{3}$$
Thanks
2. Aug 29, 2009
### espen180
A good way to check your answers on integration problems is to solve them in different ways and compare their answers. One way to do this is to use another integration method.
3. Aug 29, 2009
### arildno
Be careful!
We have, ignoring boundary values:
$$\int\frac{xdx}{\sqrt{1-x^{2}}}=\int\frac{\sin\theta\cos\theta{d\theta}}{\sqrt{1-\sin^{2}\theta}}=\int\sin\theta\frac{\cos\theta}{|\cos\theta|}d\theta$$
Do you see what you need to be careful about when simplifying this further?
In particular, why is your choice: "But what if I choose $\theta=\frac{5}{6}\pi$" troublesome?
4. Aug 29, 2009
### songoku
Hi arildno and espen180
In particular, why is your choice: "But what if I choose $\theta=\frac{5}{6}\pi$" troublesome?[/QUOTE]
Ah, maybe you are referring to the square root. The value of $$\cos \theta$$ should be positive so the angle should be in first quadrant, where the values of sin and cos are positive.
I've checked it using substitution u = 1-x^2 and got the answer. What I am confused is about why I can't take the upper bound to be (5 pi)/6. But I think I get it now
Thank you very much, arildno and espen180
5. Aug 29, 2009
### njama
It is same either if you pick п/6 or 5п/6 because п - п/6 = 5п/6 and both result with positive value of sin(x).
What arildno meant to say is that you should be careful with the absolute value $$\int{sin(x)\frac{cos(x)}{|cos(x)|}dx}=\int{sin(x)\frac{cos(x)}{\pm cos(x)}dx}=\int{\pm sin(x)dx}$$
So, you should consider the negative values for cos(x) and not just the positive ones.
P.S And why you chose x=0 for sin(x)=0, why you did not choose x=п, since both are valid?
6. Sep 6, 2009
### songoku
Hi njama
I don't really understand about the absolute sign. Why can it be negative? I think it's should be positive since it's a square root and we don't have to consider the negative value.
Thanks
7. Sep 6, 2009
### njama
Ok, I will explain.
For example if you have $y^2=16$ so that we can write |y|=4
What that means is y can be either -4 or 4 so that |y|=4 or $y=\pm 4$
Lets check it out:
(-4)2=16, 42=16 or |-4|=4 but also |4|=4
In this case take cos(x) = y, and you will see what I am talking about.
8. Sep 6, 2009
### HallsofIvy
Staff Emeritus
Of course, $|cos(\theta)|$ cannot be negative. But you can write it as $\pm cos(\theta)|$ because $cos(\theta)$ itself can be positive or negative.
IF $\theta= (5/6)\pi$ then $cos(\theta)= cos((5/6)\pi)= -\sqrt{3}/2$ so $|cos(\theta)|= - cos(\theta)= \sqrt{3}/2$.
9. Sep 6, 2009
### songoku
Hi njama and Mr. HallsofIvy
This is my interpretation after reading your posts.
$$\int_{0}^{0.5}\frac{xdx}{\sqrt{1-x^{2}}}$$ can turn out to be:
$$1.\int_{0}^{\frac{\pi}{6}}\sin \theta\;d\theta\;\text{or}$$
$$2.\int_{\pi}^{\frac{5}{6}\pi}-\sin \theta\;d\theta$$
Am I right? Or maybe there are any other forms with different upper and lower bounds?
Thanks
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https://byjus.com/question-answer/sanju-puts-equal-amount-of-money-one-at-10-per-annum-compound-interest-payable-half/ | Question
# Sanju puts equal amount of money, one at $$10\%$$ per annum compound interest payable half-yearly and the second at a certain rate per annum compound interest payable yearly. If he gets equal amounts after $$3$$ years what is the value of the second rate percent?
A
1014%
B
10%
C
912%
D
814%
Solution
## The correct option is C $$\displaystyle 10\frac{1}{4}\%$$Given that, Sanju puts the same amount of money, one at $$10\%$$ per annum compound interest, compounded half-yearly, and other at some interest per annum compound interest, compounded annually.He receives the same amount after $$3$$ years.To find out: The rate of interest for the second investment.Let the principal amount be $$Rs.\ x$$. And the amount received after $$3$$ years be $$A$$Also, let the rate of interest in the second case be $$y\%$$ per annum.In the first case, the payment is made half-yearly at $$10\%$$.So the number of time in $$3\ years=3\div \dfrac { 1 }{ 2 } =6$$ andthe rate $$=10\%\div 2=5\%$$In CI, we know that the amount $$(A)=P{ \left( 1+\dfrac { R }{ 100 } \right) }^{ T }$$Here, $$P=x,\ R=5\%\ and \ T=6$$$$\therefore \ A=x{ \left( 1+\dfrac { 5 }{ 100 } \right) }^{ 6 }$$.Now, in the second case, $$P=x,\ R=y\%\ and \ T=3$$$$\therefore \ A=x{ \left( 1+\dfrac { y }{ 100 } \right) }^{ 3 }$$.Given that, the amount received in both cases is the same.Hence, $$x{ \left( 1+\dfrac { y }{ 100 } \right) }^{ 3 }=x{ \left( 1+\dfrac { 5 }{ 100 } \right) }^{ 6 }$$$$\Rightarrow { \left( 1+\dfrac { y }{ 100 } \right) }^{ 3 }={ \left\{ { { \left( 1+\dfrac { 5 }{ 100 } \right) } }^{ 2 } \right\} }^{ 3 }\\$$$$\Rightarrow { \left( 1+\dfrac { y }{ 100 } \right) }={ { \left( 1+\dfrac { 5 }{ 100 } \right) } }^{ 2 }=\dfrac { 441 }{ 400 }$$$$\Rightarrow \dfrac { y }{ 100 } =\dfrac { 441 }{ 400 } -1$$$$\Rightarrow \dfrac{y}{100}=\dfrac { 41 }{ 400 }$$$$\therefore \ y=\dfrac { 41 }{ 4 }=10\dfrac14\%$$Hence, the required rate of interest for the second case is $$10\dfrac14\%$$.Mathematics
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https://math.stackexchange.com/questions/1165229/gcdn-m-over-nn-choose-m-is-an-integer | ${\gcd(n,m)\over n}{n\choose m}$ is an integer
Prove that for every $n\geq m \geq1$ natural numbers, the following number is an integer:
$${\gcd(n,m)\over n}\cdot{n\choose m}$$
Where $\gcd$ is the greatest common divisor.
I tried to make it simpler by cancelling the $n$ from the left side, and making it $(n-1)!$ on the right: $\gcd(n, m) \cdot \frac{(n-1)!}{m!(n-m)!}$, but can't really go further.
This was problem B-2 on the 2000 Putnam exam.
• Try to use this: "Bézout's identity (also called Bézout's lemma) is a theorem in the elementary theory of numbers: let a and b be nonzero integers and let d be their greatest common divisor. Then there exist integers x and y such that ax+by=d" en.wikipedia.org/wiki/B%C3%A9zout%27s_identity – marvinthemartian Feb 25 '15 at 18:42
• I tried considering group actions but it didn't pan out. In particular, $\Bbb Z/n\Bbb Z$ acts on $\{1,2,\cdots,n\}$ which induces an action on $\binom{n}{m}$ (the collection of $m$-subsets of $\{1,\cdots,n\}$). If the subgroup $\langle\bar{m}\rangle$ acts freely on $\binom{n}{m}$ then the claim would follow from orbit-stabilizer. Unfortunately the action needn't be free - the smallest counterexample is $m=2$, $n=4$. – whacka Feb 26 '15 at 2:23
Write $nx+my=\gcd(n,m)$, with $x,y\in\mathbb{Z}$
Then:
$$\frac{\gcd(n,m)}{n}\binom{n}{m}=\frac{nx+my}{n}\binom{n}{m}=x\binom{n}{m}+y\,\frac{m}{n}\binom{n}{m}=x\binom{n}{m}+y\binom{n-1}{m-1}\in\mathbb{Z}$$
• Wow, I am really impressed! Very nice proof. – Peter Feb 25 '15 at 19:03
• I changed \text{gcd} to \gcd. With \text{gcd} you don't automatically get proper spacing in expressions like $3\gcd(m,n)$. Instead you see $3\text{gcd}(m,n)$. ${}\qquad{}$ – Michael Hardy Feb 25 '15 at 19:28
• Next question: What's the combinatorial interpretation of the expression that is simplified here? ${}\qquad{}$ – Michael Hardy Feb 25 '15 at 19:30
Here is a conceptual way to derive this. We will show that it is a special case of the well-known fact that if a fraction $$q\,$$ can be written with denominators $$\,n\,$$ and $$\,m,\,$$ then it can also be written with a denominator being their gcd $$\,(n,m),\,$$ i.e. $$\, nq,\,mq\in\Bbb Z\,\Rightarrow\, (n,m)q\in\Bbb Z.\,$$ Applied to $$\ \color{#c00}q = \frac{1}{n}{n\choose m}\,$$
$$n\color{#c00}q = {n\choose m}^{\vphantom{|^{|^|}}}\in\Bbb Z,\,\ m\color{#c00}q= {n\!-\!1\choose m\!-\!1}\in\Bbb Z\,\ \Rightarrow\ (n,m)q = \dfrac{(n,m)}n\smash{\overbrace{{n\choose m}}^{\Large n\color{#c00}q}}\in \Bbb Z\quad$$
Remark $$\$$ Below are a few proofs of the Lemma on fractions. Recall $$\,(x,y):=\gcd(x,y)$$
$$(1)\$$ Recall that a fraction can be written with denominator $$\,n\,$$ iff its least denominator $$\,d\mid n.\,$$ Therefore $$\,m,n\,$$ are denoms $$\iff d\mid m,n\iff d\mid (m,n)\iff (m,n)\:$$ is a denom.
$$(2)\ \ \dfrac{mc}d,\dfrac{nc}d\in\Bbb Z\iff d\mid mc,nc\iff d\mid (mc,nc)=(m,n)c\iff\! \dfrac{(m,n)c}d\in\Bbb Z$$
$$(3)\ \ \dfrac{mc}d, \dfrac{nc}d\in\Bbb Z\,\Rightarrow \dfrac{jmc}d,\, \dfrac{knc}d\in\Bbb Z\,\Rightarrow\,\dfrac{(jm\!+\!kn)c}d\,\overset{\large \color{#c00}{\exists\, j,k}_{\phantom{1^{1^{1}}\!\!\!\!\!}}} = \dfrac{(m,n)c}d\in\Bbb Z\$$ by $$\rm\color{#c00}{Bezout}$$ | 2019-08-24T18:49:22 | {
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https://math.stackexchange.com/questions/1260571/experiment-roll-three-6-sided-dice | # Experiment: Roll three 6-sided dice.
Are the following probabilities correct? I'm not very confident with probabilities and would just like these double checked please. Thank you.
Experiment: Roll three 6-sided dice.
a) Find the probability that none of the dice rolled are a 2.
$\left(\frac{5}{6}\right)^3$
b) Find the probability that the sum of the dice is 8.
$\frac{21}{216}$
c) Find the probability that the sum of the dice is 8 and none of the dice rolled are 2’s.
$\frac{9}{216}$
d) Find the probability that at least one of the dice rolled are a 2.
$1- \left(\frac{5}{6}\right)^3$
e) Find the probability that the three values are sequential (for example: 4, 3, 5.)
$1 \cdot \left(\frac{5}{6}\right) \cdot \left(\frac{4}{6}\right)$
f) Find the probability that none of the dice rolled are a 2 given that the sum is 8.
$\frac{9}{21}$
g) Find the probability that the sum is 8 given that none of the dice rolled are a 2.
$\frac{9}{125}$
h) Are the events “none of the dice rolled are 2” and “the sum is 8” independent? Explain.
Yes, because they don't depend on each other. For example, you can roll a sum of 8 while still rolling a 2 or without a 2.
i) Assume that the dice are colored: one is red and two are blue. Find the probability that the sum of the blue dice is equal to the value on the red die.
$\frac{15}{216}$
j) Assume that the dice are colored: one is red and two are blue. Find the probability that the sum of the blue dice is equal to the value on the red die given that the red value is 6.
$\frac{5}{36}$
k) Assume that the dice are colored: one is red and two are blue. Find the probability that the value of the red die is 6 given that the sum of the blue dice is equal to the value on the red die.
$\frac{5}{15}$
• If no one else does comment me and I'll help you, but the first one is right,, showing your workings would help. Looks like you get the logic! May 1 '15 at 13:12
• In order to establish independence, you must show that $P(A \cap B) = P(A)P(B)$. May 1 '15 at 16:08
Most of your work is correct.
There are two exceptions.
If the three dice are sequential, then they must assume one of the following sets of values $\{1, 2, 3\}$, $\{2, 3, 4\}$, $\{3, 4, 5\}$, or $\{4, 5, 6\}$. For each of the four sets, there are $3! = 6$ permutations which result in the same set of values. Thus, the probability that the dice are sequential is $$\frac{4 \cdot 3!}{6^3} = \frac{24}{216} = \frac{1}{9}$$
If events $A$ and $B$ are independent, then $P(A \cap B) = P(A)P(B)$. If we let $A$ be the event that the sum of the dice is $8$ and $B$ be the event that none of the dice rolled show a 2, then \begin{align*} P(A) & = \frac{21}{216} = \frac{7}{72}\\ P(B) & = \frac{125}{216}\\ P(A \cap B) & = \frac{9}{216} = \frac{3}{72} \end{align*} As you can check, $P(A \cap B) \neq P(A)P(B)$. The events are dependent because if none of the dice rolled show a 2, you are less likely to get a sum of $8$ since it becomes more likely that the sum of the numbers shown on the dice will be too large.
Your answers all look correct except for (e) and (h). On (e), since you know that there are $6^3 = 216$ different rolls of the three dice, simply count up the rolls in which the three numbers are consecutive. There are four such selections: $\{1, 2, 3\}, \{2, 3, 4\}, \{3, 4, 5\}, \{4, 5, 6\}$. Each selection can be ordered in $3! = 6$ different ways. There are thus $4 \cdot 6 = 24$ different rolls that satisfy the condition, and the probability is therefore $24/216 = 1/9$.
As N. F. Taussig points out, you are applying too weak a condition for independence on part (h). It is not sufficient merely that event $A$ can happen regardless of event $B$, and vice versa; each one's probability of occurrence must be unaffected by the other's occurrence (or non-occurrence). The most straightforward way to determine that is to verify that $P(A \cap B) = P(A)P(B)$. You have already determined the relevant quantities in parts (a), (b), and (c); is the product rule satisfied? | 2021-10-19T07:05:52 | {
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https://math.stackexchange.com/questions/3696228/int-f-1d-mu2-cdots-int-f-nd-mu2-leq-int-sqrtf-12-cdotsf-n2d-mu | # $(\int f_1d\mu)^2+\cdots+(\int f_nd\mu)^2\leq(\int \sqrt{f_1^2+\cdots+f_n^2}d\mu)^2$
Let $$(X, \mathfrak{B}, \mu)$$ be a measurable space, possibly not $$\sigma$$-finite, and $$f_1, \cdots, f_n \colon X\to (-\infty, +\infty)$$ be integrable functions on $$X$$. Does $$(\int f_1d\mu)^2+\cdots+(\int f_nd\mu)^2\leq(\int \sqrt{f_1^2+\cdots+f_n^2}d\mu)^2$$ holds? (Since $$\sqrt{f_1^2+\cdots+f_n^2}\leq |f_1|+\cdots+|f_n|$$, note that integrand in RHS is integrable.)
My first attempt was to apply the Fubini's theorem and Cauchy-Schwarz to the LHS: \begin{align}(LHS)&=(\int f_1(x)d\mu(x))(\int f_1(y)d\mu(y))+\cdots+(\int f_n(x)d\mu(x))(\int f_n(y)d\mu(y))\\&=\int f_1(x)f_1(y)+\cdots+f_n(x)f_n(y) d(\mu\otimes\mu)(x,y)\\ &\leq\int \sqrt{f_1^2(x)+\cdots+f_n^2(x)}\sqrt{f_1^2(y)+\cdots+f_n^2(y)}d(\mu\otimes\mu)(x,y)\\&=(RHS)\end{align}
However this approach is valid only if $$X$$ is $$\sigma$$-finite.
Note that the inequation is equivalent to the following: If $$f\colon X\to \mathbb{R}^n$$ is integrable, $$|\int f d\mu|\leq \int|f| d\mu$$
• Jensen's inequality would apply en.wikipedia.org/wiki/Jensen's_inequality – orangeskid May 29 at 6:46
• @orangeskid: The issue nessy brought up is what happens with the underlying measure $\mu$ is not $\sigma$--finte. In such case, Jensen's not applicable. – Oliver Diaz May 29 at 6:58
• @nessy: I have worked out a much simpler solution with uses a few facts from linear algebra, specifically p norms in $\mathbb{R}^n$. – Oliver Diaz May 29 at 10:39
• @Oliver Diaz: I finally understood why we need $\sigma$ finite. It is because we replace the measure with an equivalent probability measure. However, we can restrict ourselves with $\sigma$-finite measures when we deal with the integral of a single $L^1$ function, since we can ignore the $0$ set ( and the rest must be $\sigma$-finite). – orangeskid May 29 at 20:08
• @orangeskid: The $\sigma$-finite case is solved already by nessy. Her concern was the non-$\sigma$-finte case, where her using of Fubini's theorem does not apply and certainly Jensen's either. Her insight however gives more ore less the path to follow, i.e. $\|\int f\|\leq\int\|f\|$ where $\|\;\|$ is Eucliean norm in $\mathbb{R}^n$. It turns out that the aforemetioned inequality holds in very general settings. See solution below – Oliver Diaz May 29 at 20:27
Here are a couple of strategies that work in general and make no use of any type of local integrability properties of the underlying measure ($$\sigma$$-finiteness or not).
Consider the space $$L$$ of functions $$f:X\rightarrow\mathbb{R}^n$$ which are integrable in each component and define $$\|f\|^*=\int\|f\|_2\,d\mu$$, where $$\|\;\|_2$$ is the Euclidean norm on $$\mathbb{R}^n$$. This is defines a norm on $$L$$ since $$\|f\|^*\leq\sum^n_{k=1}\int|f|_j\,d\mu<\infty$$. Also, $$\int|\|f\|_2-\|g\|_2|\,d\mu\leq\int\|f-g\|_2\,d\mu=\|f-g\|^*$$
Consider $$\mathcal{E}$$ the collection of (integrable) simple functions on $$(X,\mathscr{B},\mu)$$ and define $$\mathbb{R}^n\otimes\mathcal{E}=\{\sum^m_{k=1}u_k\phi_k: u_k\in\mathbb{R}^n, \phi_k\in\mathcal{E}, m\in\mathbb{N}\}$$
This space will play the role of elementary functions in the construction of the real valued integral. It is easy to check that $$\mathbb{R}^n\otimes\mathcal{E}$$ is dense in $$(L,\|\;\|^*)$$; furthermore, any function in $$\mathbb{R}^n\otimes\mathcal{E}$$ can be expressed as $$\Phi=\sum^{M}_{j=1}v_j\mathbb{1}_{A_j}$$ where $$v_j\in \mathbb{R}^n$$, $$A_j\in\mathscr{B}$$, $$\mu(A_j)<\infty$$, and $$M\in\mathbb{N}$$. Consider now the elementary integral $$\int\Big(\sum^m_{k=1}u_k\phi_k\Big):=\sum^m_{j=1}u_k\int\phi_k\,d\mu$$
Since $$\Phi=\sum_{u\in\mathbb{R}^n}u\mathbb{1}_{\{\Phi=u\}}$$ (notice that the sum over $$\mathbb{R}^n$$ is actually finite), $$\int\Phi =\sum^m_{j=1}u_j\mu(A_j)=\sum_{u\in\mathbb{R}^n}u\int\mathbb{1}_{\{\Phi=u\}}\,d\mu\tag{1}\label{one}$$
which means that the elementary integral extended to $$\mathbb{R}^n\otimes\mathcal{E}$$ does not depend on any particular representation of $$\Phi$$. Now $$\Big\|\int\Phi\Big\|_2\leq\sum_{u\in\mathbb{R}^n}\|u\|_2\int\mathbb{1}_{\{\Phi=u\}}\,d\mu=\int\Big(\sum_{u\in\mathbb{R}^n}\|u\|_2\mathbb{1}_{\{\Phi=u\}}\Big)\,d\mu=\int\|\Phi\|_2\,d\mu=\|\Phi\|^*\tag{2}\label{two}$$ $$\eqref{two}$$ is the inequality you are looking for but only for functions in $$\mathbb{R}^n\otimes\mathcal{E}$$. For all functions in $$L$$ one can use some density arguments.
1. Notice that $$\|\;\|_2$$ can be replaced by $$\|\;\|_p$$ ($$p\geq1$$).
2. Your problem is an example of an integral defined on vector--valued functions.
3. The arguments used, with some technical additions (Daniell integration, and measurability issues) can be used to construct Bochner's integral where $$\mathbb{R}^n$$ is replaced by a Banach space.
Another, much simpler solution may be obtained by applying linear functionals to the vector $$\int f=\sum^n_{j=1}e_j\int f_j\,d\mu$$ where $$e_1,\ldots,e_n$$ is the standard basis of $$\mathbb{R}^n$$. As above, w $$\|\,\|_p$$ is $$p$$-norm in $$\mathbb{R}^n$$. We use the fact that $$(\mathbb{R}^n,\|;\|_p)$$ and $$(\mathbb{R}^n,\|\,\|_q)$$ are dual to each other when $$\tfrac1p+\tfrac1q=1$$.
If $$\Lambda:\mathbb{R}^n\rightarrow\mathbb{}$$ is linear, then $$\Lambda x =x\cdot u$$ for some unique $$u\in\mathbb{R}$$. Thus
\begin{aligned} \Lambda \Big(\int f\Big) &= u\cdot\Big(\int f\Big)=\sum^n_{j=1}u_j\int f_j\,d\mu =\int u\cdot f\,d\mu \end{aligned} and so, by Hölder's inequality (in $$\mathbb{R}^n$$) \begin{aligned} \left|\Lambda \Big(\int f\Big)\right|&\leq\int|u\cdot f|\,d\mu\\ &\leq\int\|u\|_q\|f\|_p\,d\mu=\|u\|_q\int\|f\|_p\,d\mu \end{aligned} The result than follows by taking $$\sup$$ over all linear functionals $$\Lambda$$ with functional norm $$\|\Lambda\|:=\sup_{\|x\|_p=1}|\Lambda x|\leq1$$, or equivalently, by taking $$\sup$$ over all vectors $$u\in\mathbb{R}^n$$ with $$\|u\|_q=1$$. Thus
$$\left\|\int f\right\|_p \leq \int\|f\|_p\,d\mu$$
• I like the last proof; it seems to work for a general (semi)norm $\|\cdot \|$. By Hahn-Banach, given any $v (= \int_X f)$ there exists a functional of norm $1$, call it $L$, so that $\|v\| = L(v)$. Then apply your method and get $\|v\| \le \|L \| \cdot \int_X \|f\|$. This should be the "canonical" proof. Nice, thank you. – orangeskid May 30 at 2:12
• The first proof is more or less how one constructs Bochner’s integral through the Daniell procedure. the later is close to how one construct Bochner integral through the Dunford integral. – Oliver Diaz May 30 at 3:39
• I took a look at the definition of Pettis integral. The second solution should also work for this integral. – orangeskid May 30 at 5:06
• Yes, the Pettis integral. I confuse him with Dunford, as in the Dunford-Pettis theorem. Any way, that type of integral is useful even when functions take values on more abstract linear spaces (at least locally convex). The advantage of Bochner+Daniell is that one can derive a natural notion of measurability (Caratheodory's cut idea does not work here). – Oliver Diaz May 30 at 14:48
First, assume that $$(X,\mu)$$ is a $$\sigma$$ finite space. Then there exists a probability measure $$\nu$$ on $$X$$ that is equivalent to $$\mu$$, that is $$\mu = \rho \cdot \nu$$ where $$\rho>0$$ is a measurable function, $$\rho>0$$. We have for every $$f\in L^1(X, \mu)$$ $$\int_X f d\mu = \int_X f \, d\, \rho \nu = \int_X \rho f\, d \nu$$
Now, let $$\phi$$ be a convex function on $$\mathbb{R}^n$$ that is also positively homogeneous ( a sublinear function). Then we have $$\int_X \phi( f) d \mu= \int_X \rho \phi(f) d\nu = \int_X \phi(\rho f) d\nu \ge \phi(\int_X \rho f d\nu ) = \phi( \int_X f d\mu)$$
The inequality above is Jensen's inequality, for the convex functions $$\phi$$ and the function $$L^1$$ $$\rho f$$ on the probability space $$(X,\nu)$$.
We can reduce to the case $$X$$ $$\sigma$$-finite as follows: Consider $$X' = \{x\in X | f(x) \ne 0\}$$. Since $$f$$ is $$L^1$$, all the subsets $$\{x |\ |f(x)|\ge 1/n\}$$ are have finite measure. Hence $$X'$$ is $$\sigma$$-finite. We can reduce all our integrals to integrals over $$X'$$.
Now, how to find the probability measure $$\nu$$ equivalent to $$\mu$$. Let $$X= \sqcup_n X_n$$ where $$\mu(X_n) <\infty$$. Now, find $$\eta>0$$ such that $$\int_X \eta\, d\mu = 1$$, for instance $$\eta=\sum_{n\ge 1}\frac{1}{2^n} \cdot \frac{\chi(X_n)}{\mu(X_n)}$$ Put $$\nu = \eta \cdot \mu$$.
$$\bf{Added:}$$ I think the natural solution is the second one of @Oliver Diaz, let's restate it in general terms.
Consider $$\|\cdot \|$$ a seminorm on $$\mathbb{R}^n$$ (or, more generaly, a sublinear function). We want to show the inequality $$\| \int_X f d\mu \| \le \int_X \|f\| d \mu$$
Denote by $$v \colon = \int_X f d\mu$$. By Hahn-Banach theorem, there exists a linear functional $$L\colon \mathbb{R}^n \to \mathbb{R}$$ such that $$L(v) = \|v\|$$, and $$L(w)\le \|w\|$$ for all $$\|w\|\in \mathbb{R}^n$$. We get $$\|\int_X f d\mu \| = L(\int_X f d\mu)=\int_X L(f) d\mu \le \int_X \|f\| d\mu$$
• The problem requires a "vector" version of Jensen's inequality (for the $\sigma$--finite case). The function here is an $\mathbb{R}^n$--valued integrable function. A proof of a version for Jensen's on this setting though not difficult, it does require more than the standard real line version. – Oliver Diaz May 30 at 14:40 | 2020-07-06T06:50:13 | {
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http://arquivo.flip.org.br/reviews/prove-that-focal-length-of-plane-mirror-is-infinity-b0943c | A ray of light falling normally on a plane mirror, then angle of reflection will be : (a) 90° (b) 180° (c) 0° (d) 45° Answer: (c) 0° For normal incidence on a plane mirror, the angle of reflection will be zero. Share 0 You can use the mirror formula also to get thefocal length of a plane mirror, where the image distance = − objectdistance, please mark as brainlist answer and follow me please thank answer. So A is to B as f is to di minus the focal length. Focal length is half of radius of curvature and hence it is infinity. A plane mirror has ##R \to \infty##, so ##f \to \infty## too. 1/F = 1/U - 1/U = 0 . [ citation needed ] The focal length of a plane mirror is infinity ; [4] its optical power is zero. A concave mirror of radius 30 cm is placed in water. So this is di minus the focal length. b. …, toThe temperature of solution increases.The temperature of solution decreasesThe temperatwee of the solution remainsthe sameFormation of salt is seen.Alternatives. 1/F = 1/U - 1/U = 0. In a plane mirror, the radius of curvature is infinitly long, so the focus will be at infinity. Ask your question. 1. You can specify conditions of storing and accessing cookies in your browser, Prove that focal length of plane mirror is infinite, How could you use the same apparatus to find out what happens when sands and sandstonesare stretched?, it's time to uninstall the branily app bye bye , Section Asuces the following questions os disucted:Select the couuet alternative from the givenAdding Acid & Base solutions in a test tubeguesults At infinity. Yes, the focal length of a plane mirror is at infinity. Zero. MCQ Of Chapter Light Class 10 Question 7. Image formed by plane mirror is. Focal length of a plane mirror is infinity. We can extend the mirror equation to the case of a plane mirror by noting that a plane mirror has an infinite radius of curvature. (A plane mirror can be considered as a spherical mirror of infinite radius of curvature. Real and erect. Share with your friends. Proof of Focal length of a Plane mirror is infinity - YouTube When an object is placed infront of a plane-mirror an image is formed behind the mirror at a distance equal to the distance of the object from the mirror.That is distance of the image from the mirror= V = -U. from the law of distances, 1/F = 1/U + 1/V . This means the focal point is at infinity, so the mirror equation simplifies to $d_o=−d_i$ ==>The mirror formula to get the focal length of a mirror where,the image distance - objectdistance. Answer/Explanation. (A) enly () soovist (B)) full both couscoli) & (u) conduct (DXi13 GUN) comeler, 28. We and our partners will store and/or access information on your device through the use of cookies and similar technologies, to display personalised ads and content, for ad and content measurement, audience insights and product development. c. Virtual and erect. ==>focal length is half of curvature and Hence, it is infinity. Question 12. Information about your device and internet connection, including your IP address, Browsing and search activity while using Verizon Media websites and apps. a. Technically, yes - you could call the focal length of a plane mirror "undefined", since it doesn't have a particular value. That is F = infinity. Technically, confident - you may desire to call the focal length of a aircraft reflect "undefined", because of fact it would not have a definite cost. A plane mirror can be considered as a spherical mirror of infinite radius of curvature. A plane mirror can be considered as a spherical mirror of infiniteradius of curvature. Focal length of plane mirror is. When an object is placed infront of a plane-mirror an image is formed behind the mirror at a distance equal to the distance of the object from the mirror.That is distance of the image from the mirror= V = -U. from the law of distances, 1/F = 1/U + 1/V. Find out more about how we use your information in our Privacy Policy and Cookie Policy. c. Negative. But the use of infinity for the value comes from the equation for the focal length of a mirror, f = -R/2, where R is the radius of curvature of the mirror. Although it was derived for a concave mirror, it also holds for convex mirrors (proving this is left as an exercise). Focal length as well as radius of curvature of a plane mirror is infinity. That is F = infinity. You can use the mirror formula also to get the focal length of a plane mirror, where the image distance = − object distance.) Prove that focal length of plane mirror is infinite - 21597112 1. 2. Answer: d Explanation: (d) The focal length of spherical mirror does not depends on the surrounding medium. Technically, yes - you could call the focal length of a plane mirror "undefined", since it doesn't have a particular value. An image formed by a plane mirror is virtual, erect, laterally inverted, of same size as that of object and at the same distance as the object from the mirror. The focal length of a concave mirror is 20 cm. Join now. d. Virtual and inverted . Chitosan nanoparticles can be prepared byionotropic gelation technique using_(1 Point), चार्ट मॉडल एवं चित्र की सहायता से मनुष्य में होने वाले रोगों का अध्ययन करना।. You can use the mirror formula also to get the focal length of a plane mirror, where the image distance = − object distance. It’s focal length in air and water differ by (a) 15 (b) 20 (c) 30 (d) 0. Power of a plane mirror is zero. Focal length is half of radius of curvature and hence it is infinity. ==>A plane mirror can be considered as a spherical mirror of infiniteradius of curvature. So this whole distance is di, all the way over here. d. None of these. An object is placed at distance 20 cm from mirror. Honey7999685171 Honey7999685171 27.08.2020 Science Secondary School +5 pts. You can change your choices at any time by visiting Your Privacy Controls. 2 Answers. 3. Log in. b. a. Focal length is defined because of the fact the gap from the centre of the reflect at which an incident parallel beam of sunshine will converge. Focal length is half of radius of curvature and hence it is infinity. Yahoo is part of Verizon Media. But this length is that whole distance minus the focal length. Plane mirrors are the only type of mirror for which a real object always produces an image that is virtual, erect and of the same size as the object. It comes from the extension of the focal length of a spherical mirror of radius of curvature ##R##, which is ##f=R/2##. Ask your question. To enable Verizon Media and our partners to process your personal data select 'I agree', or select 'Manage settings' for more information and to manage your choices. And there you have it, we have a relationship between the distance of the object, the distance of the image, and the focal length. This site is using cookies under cookie policy. Using mirror eq prove that focal length of plane mirror is infinity. Virtual objects produce real images , however. yet using infinity for the cost comes from the equation for the focal length of a reflect, f = -R/2, the place R is the radius of curvature of the reflect. Real and inverted. Join now. Log in. | 2021-03-04T02:52:06 | {
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http://math.stackexchange.com/questions/695979/does-every-prime-divide-some-fibonacci-number | # Does every prime divide some Fibonacci number?
I am tring to show that $\forall a \in \Bbb P\; \exists n\in\Bbb N : a|F_n$, where $F$ is the fibonacci sequence defined as $\{F_n\}:F_0 = 0, F_1 = 1, F_n = F_{n-1} + F_{n-2}$ $(n=2,3,...)$. How can I do this?
Originally, I was trying to show that $\forall a\in\Bbb N\;\exists n\in\Bbb N:a|F_n$. I soon found out that if the $k$-th Fibonacci can be divided by $m$, then the $nk$-th Fibonacci can also be divided by $m$, and this can be reduced to my original problem in this post.
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By the way, the Fibonacci numbers (under the usual convention as you have, i.e. $F_0 = 0$ and $F_1 = 1$) form a "strong divisibility sequence": they satisfy the property that $F_{\gcd(m, n)} = \gcd(F_m, F_n)$. This explains your observation that if $m$ divides $F_k$, then $m$ divides $F_{nk}$ as well, because in fact $F_k$ divides $F_{nk}$. – ShreevatsaR Mar 2 '14 at 9:31
Two more interesting facts: first, if $p \mid F_k$, then $\operatorname{ord}_p F_{kn}=\operatorname{ord}_p F_k + \operatorname{ord}_p n$. So if we find the first Fibonacci number divisible by $p$, we can easily say how many powers of $p$ divide any given Fibonacci number. Also, there are no known primes for which $p\mid F_k \implies p^2\mid F_k$, but it is conjectured that there are infinitely many. – Slade Mar 2 '14 at 19:17
Yes. Consider any prime $p$. (Actually we don't need $p$ to be prime; consider any nonzero number $p$.)
You can of course take $F_0 = 0$ which is divisible by $p$, but let's suppose you want some $n > 1$ such that $F_n$ is divisible by $p$. Consider the Fibonacci sequence modulo $p$; call it $F'$.
That is, you have $F'_0 = 0$, $F'_1 = 1$, and for $n \ge 0$, you have $F'_{n+2} \equiv F'_{n+1} + F'_n \mod p$.
Now, there are only $p^2$ possible pairs of remainders $(F'_k, F'_{k+1})$, so some pair of consecutive remainders must occur again at some point. Further, the future of the sequence is entirely determined by its value at some two consecutive indices, so the sequence must itself repeat after that point. And it cannot go into some cycle that does not include $(F'_0, F'_1)$, because we can also work the sequence backwards: we can find $F'_{k-1}$ using $F'_{k-1} \equiv F'_{k+1} - F'_{k} \mod p$, etc.
This means that there always exists some $n > 0$ such that $F'_n \equiv F_0 \equiv 0 \mod p$ and $F'_{n+1} \equiv F_1 \equiv 1 \mod p$. Such an $n$ will do. This is called the period of the sequence modulo $p$ (or the $p$th Pisano period; of course some smaller $n$ may also exist (for which $F'_{n+1} \not\equiv 1 \mod p$).
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can you elaborate on "And it cannot go into some cycle that does not include $(F′_0,F′_1)$, because we can also work the sequence backwards"? if the sequence (mod 5) is always $1,3,1,3,1,3...$ I can very well traverse it backwards but will never reach $0$ nonetheless. – example Mar 3 '14 at 0:54
@example: Note I said $(F'_0,F'_1)$, not $(0,1)$ (which happens to be here our initial values, but that's not the argument). The argument is this: the eventual cycle can't be something that avoids the initial values. If you know that eventually the sequence becomes some $\dots,a,b,\dots,y,z,a,b,\dots,y,z,a,b\dots,y,z,\dots$, then you can trace the values backwards, and conclude that even the earlier terms (and in particular the first two terms) must also be the same as those in the cycle. There is only one way to reach any pair, so $(F'_0,F'_1)$, whatever they are, must be part of the cycle. – ShreevatsaR Mar 3 '14 at 2:03
that cleared it up. thanks =) – example Mar 3 '14 at 11:14
You have reinvented the wheel (cycle). @example See this post for the essence of the matter. – Bill Dubuque Mar 6 '14 at 16:25
According to the Wikipedia article on Fibonacci numbers if $p$ is a prime number then
$$F_{p - \left(\frac{p}{5}\right)} \equiv 0 \text{ (mod } p)$$
where $\left(\frac{p}{5}\right)$ is the Legendre symbol.
$$\left(\frac{p}{5}\right) = \begin{cases} 0 & \textrm{if}\;p =5\\ 1 &\textrm{if}\;p \equiv \pm1 \pmod 5\\ -1 &\textrm{if}\;p \equiv \pm2 \pmod 5.\end{cases}$$
For example, if $p = 31$ then $F_{30} = 832040 = 2^3 \times 5 \times 11 \times 31 \times 61$.
The reference given is Lemmermeyer (2000), pp. 73–4.
-
Worth noting that this does not produce a good primality test on its own because Fibonnaci pseudoprimes also satisfy the above condition (there is significant overlap with Frobenius pseudoprimes as well due to the fact that the Frobenius test uses the $x^2 - x - 1$ quadratic). Just throwing it out there for people who might be wondering. – Thomas Mar 2 '14 at 19:37
This congruence can actually be generalized by multiplying both the index and modulus by $p^{n-1}$. – Jaycob Coleman Feb 15 at 23:40
We can in fact show a stronger statement with some algebraic number theory: If $p>5$ is prime then $p|F_{p\pm 1}$ for some choice of $+$ or $-$.
Suppose $\left(\frac{5}{p}\right)=1$. In this case, $p$ splits in $\mathbf{Z}\left[\frac{1+\sqrt{5}}{2}\right]=\mathbf{Z}[\varphi]$. Thus, we can write $p=\pm\pi\bar\pi$, where $\pi$ and $\bar\pi$ are conjugate primes in $\mathbf{Z}[\varphi]$ that do not differ by a unit. Write $\pi=x+y\varphi$, so $x+y\varphi\equiv 0\pmod{\pi}$. Now, if $p|y$, then $\pi|y,x$, contradiction, so $p\nmid y$. Thus, $y$ has an inverse modulo $p$, say $y'$. Then we have $\pi|p|yy'-1$, so $\varphi\equiv -xy'\pmod{\pi}$. Summarizing, $\varphi\equiv k\pmod{\pi}$ for some integer $k\not\equiv 0\pmod{p}$. By FLT, $k^{p-1}\equiv 1\pmod{p}$, so $\varphi^{p-1}\equiv k^{p-1}\pmod{\pi}$. Thus $\varphi^{p-1}\equiv 1\pmod{\pi}$. Similarly, we see that $\bar\varphi^{p-1}\equiv 1\pmod{\pi}$, so $F_{p-1}\sqrt{5}\equiv 0\pmod{\pi}$. Since $p$ and $5$ are necessarily relatively prime, $\pi|F_{p-1}$, and $\bar\pi|F_{p-1}$. Hence $\pi\bar\pi = p|F_{p-1}$ in this case.
Now, suppose $\left(\frac{5}{p}\right)=-1$. We have $5^{(p-1)/2}\equiv -1\pmod{p}$, by Euler's Criterion. Now, applying the Binomial-theorem to Binet's formula yields several terms containing $\binom{p+1}{k}\equiv 0\pmod{p}$. After reducing modulo $p$ we will be left with $\dfrac{\sqrt{5}^{p+1}+1}{2^t}$ for some $t$, which is also divisible by $p$ (by working in $\mathbf{Z}[\varphi]$), so $p|F_{p+1}$ in this case.
Note: This is a proof of the above post by 01000100, which asserts that $p|F_{p-\left(\frac{p}{5}\right)}$
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The trivial answer is: yes $F_0=0$ is a multiple of any prime (or indeed natural) number. But this can be extended to answer your real question: does this also happen (for given$~p$) for some $F_n$ with $n>0$.
Indeed, the first coefficient (the one of $F_{n-2}$, which is $1$) of the Fibonacci recurrence is obviously invertible modulo any prime$~p$, which means the recurrence can be run backwards modulo$~p$: knowing the classes of $F_{n-1}$ and of $F_n$ one can recover the class of$~F_{n-2}$ uniquely. This means that on the set $\Bbb F_p^2$ of pairs of classes of successive terms, the Fibonacci recurrence defines a bijection (a permutation of all pairs). Then since the set of pairs is finite, some power of this operation must be the identity on it. This implies that the case $F_n\equiv0\pmod p$ that happens for $n=0$ recurs for some $n>0$.
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Any linearly recursive integer sequence has the property that every large enough prime divides some term, as long as
some term of the sequence is equal to $0$.
That follows from the solution of linear recursive sequences. Number the sequence so that $S_0 = 0$, and consider primes $p$ not dividing any of the denominators or characteristic roots that appear in the algebraic solution of the recurrence (this is all but a finite number of primes). If $k$ is chosen so that $\alpha^k=1 \mod p$ for all roots $\alpha$ of the characteristic polynomial, then $p|S_k$ if the roots are distinct, and $p|S_{kp}$ whether or not the roots are distinct.
If the sequence has the additional property that
the recursion can be run backward (its extreme coefficients are $\pm 1$)
then every prime divides some term of the sequence.
The second statement is what was proved in the other answers: when reduced mod $p$, a recursion that can be run in both directions is periodic and thus repeats the value of $0$.
The converse, that there must exist a term equal to $0$ in order to have a term equal to $0$ mod $p$ for all large $p$, is plausible but seems difficult to prove.
- | 2015-10-10T00:13:31 | {
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http://mathhelpforum.com/discrete-math/156876-integer-soultion-equation.html | # Thread: Integer Soultion to an equation...
1. ## Integer Soultion to an equation...
How many integer solutions (x1; x2; x3) are there to the equation :-
x1 + x2 + x3 = 25
with all xi satisfying 0 < xi < 10?
2. Originally Posted by aamiri
How many integer solutions (x1; x2; x3) are there to the equation :-
x1 + x2 + x3 = 25
with all xi satisfying 0 < xi < 10?
Maybe not the most elegant, but:
The lowest possible for x1 is 5, forcing (x2,x3)=(10,10).
Fix x1=6 then it's either (x2,x3)=(9,10) or (10,9)
Fix x2=7 then it's either (x2,x3)=(8,10) or (10,8) or (9,9)
As you can see there really aren't that many options if you continue in this manner.
3. ^^ Thanks for taking out time to reply.
So, in this case , do I just keep going in the same manner as you did?
Is there any other way of doing this ?
4. Originally Posted by aamiri
^^ Thanks for taking out time to reply.
So, in this case , do I just keep going in the same manner as you did?
Is there any other way of doing this ?
Well for these numbers I think it's fine to continue in the same manner, it's quick enough. For an extended problem like
x1 + x2 + ... + x7 = 39
one approach if you are allowed a computer is to solve recursively. For other ways to solve, maybe someone else will chime in.
5. Originally Posted by aamiri
How many integer solutions (x1; x2; x3) are there to the equation :-
x1 + x2 + x3 = 25
with all xi satisfying 0 < xi < 10?
A generating function solution requires the least thought, in my opinion.
Suppose, more generally, we want to count the integer solutions to
$x_1 + x_2 + x_3 = r$
where $0 \leq x_i \leq 10$;
let's say the number of solutions is $a_r$.
Then $a_r$ is the coefficient of $x^r$ in
$f(x) = (1 + x + x^2 + \dots + x^{10})^3$.
From here on it's just a matter of manipulating series:
$f(x) = \left( \frac{1-x^{11}}{1-x} \right) ^3$
$= (1- x^{11})^3 (1-x)^{-3}$
$= (1 - 3 x^{11} + 3 x^{22} - x^{33}) \sum_{i=0}^{\infty} \binom{3+i-1}{i} x^i$
Picking out the coefficient of $x^{25}$ from this expression, we can see it is
$a_{25} = \binom{27}{25} - 3 \binom{16}{14} + 3 \binom{5}{3}$.
I think a solution via inclusion/exclusion is also possible (I haven't worked it out), but I think it will lead to the same expression for the answer. Only you will have to think harder. :-P | 2016-12-03T02:59:55 | {
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https://www.physicsforums.com/threads/electric-potential-inside-a-shell-of-charge.947834/ | Electric potential inside a shell of charge
Homework Statement
Q1: There are two concentric spherical shells with radii ##R_1## and ##R_2## and charges ##q_1## and ##q_2## uniformly distributed across their surfaces. What is the electric potential at the center of the shells?
Q2: There is an infinitely long hollow cylinder of linear charge density ##\lambda## and radius ##R##. What is the potential difference ##\Delta V## between the surface of the shell and a radius ##R'## inside the cylinder?
Homework Equations
##\vec E = -\nabla V##
##\oint \vec E \cdot d\vec A = \frac{Q\text{encl}}{\epsilon_0}## (Gauss's Law)
##V=\frac{q}{4\pi \epsilon_0 r}##
The Attempt at a Solution
Starting with Q2, Gauss's Law using a cylinder as the Gaussian surface shows there is no enclosed charge; ##\vec E = \vec 0##. Because ##\vec E = -\nabla V##, one can conclude that ##V=0## between ##R'## to ##R##. This is the given (and found) answer for Q2.
In a similar way, there is no enclosable charge for all points inside the two spherical shells in Q1. By the same logic as Q2, it would seem ##\vec E=0=-\nabla V## and there would be no electric potential at the center of the shells.
However, Gauss's Law cannot be applied to a point or line since the Gaussian surface has area ##A=0## and Gauss's Law reduces to ##0=0##. Therefore, one cannot find the electric field at the center of the sphere, and ##\vec E = -\nabla V## cannot be used.
Since the center of the shells are at a constant distance ##R_1## and ##R_2##, the electric potential can be found by:
##V=\frac{q_1}{4\pi \epsilon_0 R_1} + \frac{q_2}{4\pi \epsilon_0 R_2}##
which is the given answer for Q1. (One needs to integrate the charge density across a spherical area, which ultimately reduces to the answer above)
This result (i.e. textbook answers) seems to show some weird results:
• The electric field and potential are zero for all positions inside a closed area of charge and nonzero at the symmetrical center or axis.
• The graphs of the field magnitude and electric potential are discontinuous at the center.
• If this is true, the electric field vector there has no defined direction...? (or is undefined since the equation is discontinuous)
• In Q2, the electric field and potential should be nonzero along the axis of the cylinder and zero for all spaces between the axis and the cylinder wall.
I'm somewhat confused because the two questions seem to contradict each other. Is my logic correct in interpreting the answers?
Doc Al
Mentor
One note: Between points where the field is zero there is no change in potential. That does not mean that the potential is zero. (There can certainly be a potential with respect to some other point.)
One note: Between points where the field is zero there is no change in potential. That does not mean that the potential is zero. (There can certainly be a potential with respect to some other point.)
Integrating ##\vec E \cdot d\vec r## where ##\vec E = \vec 0## to find the potential at a single point results in ##V = 0+C##. Then the nonzero potential found at the center is ##C## and is constant across the space inside the shells...
Makes much more sense, thanks!
rude man
Homework Helper
Gold Member
Or perhaps:In Q2: the E field just outside the surface is σ/ε where σ is surface charge density (related to λ obviously). The E field just below the surface is zero. Both by Gauss. Since potential is the integral of the E field over distance, and the distance → zero, therefore there is no change in potential between the outside & inside surfaces.
Q1: Can also do this by superposition theorem:
Potential of shell 1 with q2=0 is kq1/R1.
Potential of shell 2 with q1=0 is kq2/R2.
Total potential is sum of above potentials. | 2022-06-27T08:49:52 | {
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http://axea-media.com/saskatchewan-am-okl/transitive-closure-c-224e80 | Otherwise, it is equal to 0. Main article: Transitive closure. https://mathworld.wolfram.com/TransitiveClosure.html. For every set a, there exist transitive supersets of a, and among these there exists one which is included in all the others.This set is formed from the values of all finite sequences x 1, …, x h (h integer) such that x 1 ∈ a and x i+1 ∈ x i for each i(1 ≤ i < h). 1 Transitive Closure Formally, we de ne the transitive closure (TC) problem as follows. Given a directed graph, find out if a vertex v is reachable from another vertex u for all vertex pairs (u, v) in the given graph. Reading, Then the transitive closure of R is the connectivity relation R1.We will now try to prove this Transitive closure The transitive property of numbers states that if A = B and B = C, then A = C. Derby applies this property to query predicates to add additional predicates to the query in order to give the optimizer more information. path(); Reachable mean that there is a path from vertex i to j. The transitive closure of a binary relation on a set is the minimal void main() MA: Addison-Wesley, 1990. In this article, we will begin our discussion by briefly explaining about transitive closure and the Floyd Warshall Algorithm. The reach-ability matrix is called the transitive closure of a … Here reachable mean that there is a path from vertex u to v. The reach-ability matrix is called transitive closure of a graph. The transitive closure of a binary relation $$R$$ on a set $$A$$ is the smallest transitive relation $$t\left( R \right)$$ on $$A$$ containing $$R.$$ The transitive closure is more complex than the reflexive or symmetric closures. Let R be a relation on the set {a,b, c, d} R = {(a, b), (a, c), (b, a), (d, b)} Find: 1) The reflexive closure of R 2) The symmetric closure of R 3) The transitive closure of R Express each answer as a matrix, directed graph, or using the roster method (as above). SIAM J. Comput. Don't express your answer in terms of set operations. In this article, we will begin our discussion by briefly explaining about transitive closure and graph powering. { there exist , , ..., with , , and for all . for all a, b, c ∈ X, if a R b and b R c, then a R c.. Or in terms of first-order logic: ∀,, ∈: (∧) ⇒, where a R b is the infix notation for (a, b) ∈ R.. 1.4.1 Transitive closure, hereditarily finite set. for(i=0;i The transitive closure of a graph can be computed using TransitiveClosure[g] In Studies in Logic and the Foundations of Mathematics, 2000. Given a directed graph, find out if a vertex j is reachable from another vertex i for all vertex pairs (i, j) in the given graph. scanf(“%d”,&a[i][j]); August 2014; Categories. for(j=0;j Change ), C program to Compute the transitive closure of a given directed graph using Warshall’s algorithm, C program to Find the minimum cost spanning tree of a given undirected graph using Prim’s algorithm, C program to Find the binomial coefficient using dynamic programming. c. Describe an efficient algorithm for updating the transitive closure as edges are inserted into the graph. Why do we have to include the pairs $(b, b)$ and $(c, c)$ in the transitive closure? For example, suppose X is a set of towns, some of which are connected by roads. transitive relation on that contains Data structures using C, Here we solve the Warshall’s algorithm using C Programming Language. for(j=0;j
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https://math.stackexchange.com/questions/2551218/given-x-y-and-heading-as-a-number-of-degrees-determine-next-point-along-headi | # Given X, Y, and heading as a number of degrees, determine next point along heading.
Suppose an object is moving in some direction on a 2D plane. Its direction is given as a heading, where 0 degrees is north and 180 degrees is south.
Suppose that I know the object's coordinates are 0,0 (x,y) and that its heading is 90 degrees. To maintain the current heading I know that for each whole unit I move the object along the X-axis I need to move it 0 units along the Y-axis.
But suppose the same object's heading changes to anything else, like 98 degrees. How would I determine how many units to move in each direction to maintain the current heading, where either X or Y must move a whole unit (but not necessarily both)?
I suppose the question I am asking is similar to "how do I find another point given a point and slope", but the slope is in degrees and I can't figure out how to convert that.
Here is a function and expected output if it helps clarify:
f(x,y,d) = a,b (where a and b, when added to x and y, are the next point on the line in the direction of the heading)
f(0, 0, 0) = 0,1 (heading north, so for each 1 unit added to Y we add 0 to X)
f(0, 0, 45) = 1,1
f(0, 0, 90) = 1,0
f(0, 0, 135) = 1,-1
f(0, 0, 180) = 0,-1
f(0, 0, 225) = -1,-1
f(0, 0, 270) = -1,0
f(0, 0, 18) = a,b
f(0, 0, 200) = a,b
f(0, 0, 65) = a,b
Either x or y needs to be a whole number, but one of them can be a decimal number. For example, to move x 1 point, move y 0.345 points, or vice versa.
• Could you better clarify how the $x$ and $y$-coordinate change for non-trivial angles? For example, starting from the origin, if $a=30^o$ from the north, would you like to get $(1/\sqrt{3},1)$ or $(1,\sqrt{3})$? – Anatoly Dec 7 '17 at 10:36
The function $f$ you describe is
$f(x,y,a)=(x,y)+\begin{cases} (1,\cot (a)) & 45\leq a\leq 135 \\ (-\tan (a),-1) & 135\leq a\leq 225 \\ (-1,-\cot (a)) & 225\leq a\leq 315 \\ (\tan (a),1) & \text{else} \end{cases}$
which is well-defined despite the overlapping conditions.
• Hmm, this doesn't seem quite right. It is only accurate for the first result, f(0,0,0). I'm not sure if you're familiar with JavaScript, but I put together a tester and implemented your function f(x, y, a) here: jsfiddle.net/46Lxk3f9/2 Is there something I missed or wrote incorrectly? – jdgregson Dec 7 '17 at 4:01
• @jdgregson In JavaScript you need to convert the angle in degrees to radians. That is replace tan(a) by tan(pi*a/180) and cot(a) by cot(pi*a/180) – MeMyselfI Dec 7 '17 at 8:17
• Ah, when I implemented that all test cases passed (almost): jsfiddle.net/46Lxk3f9/3 - I say "almost" because each answer was still off, but only be a very small decimal number. When I rounded the output I got what I was looking for. I'll throw this into the program tonight and see if it solves everything. – jdgregson Dec 7 '17 at 22:04
• After converting this to JavaScript I find this answer to be the easiest to use and highly accurate. Thanks for the help! – jdgregson Dec 8 '17 at 4:51
The function can be written as $$f(x,y,d)=(x,y)+\begin{cases} (0,1)\quad &\quad d=0^\circ\\\\ (0,-1)\quad &\quad d=180^\circ\\\\ \left((-1)^{\left\lfloor\frac{d}{180}\right\rfloor},(-1)^{\left\lfloor\frac{d}{180}\right\rfloor}\cot\left(\dfrac{d\pi}{180}\right)\right)\quad &\quad\text{else}\end{cases}$$ where $\lfloor x\rfloor$ represents the greatest integer less than or equal to $x$.
You can see that this $f$ satisfies the following examples you've shown :
f(0, 0, 0) = 0,1 (heading north, so for each 1 unit added to Y we add 0 to X)
f(0, 0, 45) = 1,1
f(0, 0, 90) = 1,0
f(0, 0, 135) = 1,-1
f(0, 0, 180) = 0,-1
f(0, 0, 225) = -1,-1
f(0, 0, 270) = -1,0
and that
\begin{align}f(0, 0, 18)& = \left(1,\cot \left(\dfrac{\pi}{10}\right)\right)\approx (1,3.0776835)\\\\f(0, 0, 200) &=\left(-1,-\cot\left(\dfrac{10\pi}{9}\right)\right)\approx (-1,-2.7474774)\\\\f(0, 0,65) &=\left(1,\cot \left(\dfrac{13\pi}{36}\right)\right)\approx (1,0.4663077)\end{align}
• You assume that the integer, unit step always occurs in the $x$-axis direction, but this is not specified in the text. – Anatoly Dec 7 '17 at 13:26
• @Anatoly: OP says "How would I determine how many units to move in each direction to maintain the current heading, where either X or Y must move a whole unit (but not necessarily both)?", so I think that assumption is OK. – mathlove Dec 7 '17 at 14:58
• Right! For the sake of clarity, I just added an integration to your correct answer, to account for both possibilities. – Anatoly Dec 7 '17 at 15:02
Just to give a little integration to the correct answer by mathlove, since the OP does not specify criteria to decide, for angles that are not multiples of $\pi/4$, whether the integer unit step must be taken in the $x$- or $y$-axis direction. To follow the notation of the OP, I will express angles in degrees. Let us assume that the initial coordinates of the object are $(0,0)$. For a movement from the origin along a line that forms an angle $\theta$ with the $x$-axis (measured using the conventional approach of increasing angles in the counterclockwise direction), a unit change in the $x$-coordinate corresponds to a shift by $\tan(\theta)$ in the $y$-coordinate. In a symmetric fashion, a unit change in the $y$-coordinate corresponds to a shift by $\cot(\theta)$ in the $x$-coordinate.
These considerations must be applied to the particular definition of the angle described in the OP, where $0^o$ corresponds to the "north", i.e. $90^o$ of the standard definition, and where increasing values go in a clockwise direction. If $a$ is the angle identified using this alternative definition, we have $\theta=-a+90^o \,\,$. Now note that the new coordinates depend on whether we decide to assign the unit step to the $x$- or $y$-axis. The assignment to the $x$-axis is possible only for $a \neq 0^o$ and $a \neq 180^o$ (values for which we get the trivial results of $(0,1)$ and $(0,-1)$, respectively), whereas that to the $y$-axis is possible only for $a \neq 90^o$ and $a \neq -90^o$ (values for which we get the trivial results of $(1,0)$ and $(-1,0)\,$). Under these restrictions, if we assign the unit step to the $x$-axis, the new coordinates are given by
$$\left(1,\cot(a)\right) \,\,\, \text{if} \,\,\, 0^o<a<180^o$$ $$\left(-1,-\cot(a)\right) \,\,\, \text{if} \,\,\, 180^o<a<360^o$$
whereas if we assign the unit step to the $y$-axis they are
$$\left(\tan(a),1 \right) \,\,\, \text{if} \,\,\, -90^o<a<90^o$$ $$\left(-\tan(a), -1 \right) \,\,\, \text{if} \,\,\, 90^o<a<270^o$$
With this double option, the examples of the OP become
$$f(0, 0, 18) = \left(1,\cot ( 18^o ) \right) \,\,\, \text{or} \,\,\, \left(\tan ( 18^o ),1 \right)$$
$$f(0, 0, 200) = \left(-1,-\cot ( 200^o ) \right) \,\,\, \text{or} \,\,\, \left(-\tan ( 200^o ),-1 \right)$$
$$f(0, 0, 65) = \left(1,\cot ( 65^o ) \right) \,\,\, \text{or} \,\,\, \left(\tan ( 65^o ),1 \right)$$ | 2019-10-18T18:42:54 | {
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http://math.stackexchange.com/questions/299570/understanding-an-integration | # Understanding an integration
I need some help in understanding the integration performed in below equation. My question is how step 2 is obtained from the first step (i.e., how integration of exponential and dirac delta functions is performed). Thanks in advance.
\begin{align*} \mathsf{P}(0<Y\leq 7) &=\int_{0^+}^7\left[\frac{1}{4}e^{-|y|}+\frac{1}{3}\delta(y)+\frac{1}{6}\delta(y-7)\right]dy\\\\\\ &=\frac{1}{4}\int_0^7e^{-y}\,dy+\frac{1}{6}\\\\\\ &=\frac{1-e^{-7}}{4}+\frac{1}{6}=\frac{5}{12}-\frac{e^{-7}}{4} \end{align*}
-
I understood the exponential integration... now i just want to know how we get 1/6 please. – Osman Khalid Feb 10 '13 at 18:53
By definition $\int_{-\infty}^{\infty} \delta (y) dy = 1$. the integral borders are 0+ and 7 therefore the integration do not contain value of the function at 0 and contain value at 7. Therefore the peak of the integrand $\frac{1}{3} \delta(y)$ is not in the interval of integration – OukiDouki Feb 10 '13 at 18:58
@OukiDouki: Then what about the fact that the result of an integral should not depend on altering the interval of integration for a zero measure set of points? I mean, I could well remove the single point $y=7$ and use $y=7^-$ as upper limit. But then - following your argument - also that integral would be zero. – Andrea Orta Feb 10 '13 at 19:05
@Andrea Orta: Good question. I didn't mastered the measure theory and Lebesque integrals so well to answer it. I followed the simple logic that when is integration interval $(o,7>$ and a Dirac function is non-zero only at point $0$ then it simply you cannot integrate it. Somebody else should anwer this question – OukiDouki Feb 10 '13 at 19:22
@OukiDouki: Yes, I was writing something along the lines of your reasoning, but then I stopped, thinking about that problem. Osman: Wait a moment, better answers will appear! – Andrea Orta Feb 10 '13 at 19:28
show 1 more comment
A delta function satisfies $\int_{-\infty}^\infty \delta(x) dx=1$ but also $\int_{-\epsilon}^\epsilon \delta(x) dx=1$ for every $\epsilon>0$. Intuitively, it can be thought of as a function that is zero everywhere, but jumps quickly to infinity at zero. This is the physical/engineering interpretation.
Another way to think about a delta function, is as the derivative of a step function, usually denoted as $u(x)$ satisfies $u(x)=1$ for $x \ge 0$ and $u(x)=0$ for $x<0$
In your problem, the integral starts from $0^+$, so the integral over the first delta is $0$ (it's like integrating a zero). The second delta is a shifted delta to $x=7$, which its integral is $u(x-7)$. Having said that, we get: $\int_{0^+}^7 \frac{1}{6}\delta(x-7)dx=\frac{1}{6}[u(7-7)-u(0^+-7)]=\frac{1}{6}[u(0)-u(-7)]=\frac{1}{6}[1-0]=\frac{1}{6}$
-
The defining properties of $\delta(y)$ are $$\delta(0)\to\infty$$ $$\delta(y)=0\quad y\ne0$$ $$\int_{-\infty}^{\infty}\delta(y)dy=1$$ The second one means that $\delta(y)$ vanishes in the neighbourhood of $0$ however small. It also follows that $$\int_{-\infty}^{\infty}\delta(y)dy=\int_{-\epsilon}^{\epsilon}\delta(y)dy=1\tag{1}$$ for arbitrarily small $\epsilon>0$ The integral could also be written as $$I=\int_{\epsilon}^{7}\left[\frac{1}{4}e^{-|y|}+\frac{1}{3}\delta(y)+\frac{1}{6}\delta(y-7)\right]dy$$ From which it is clear that the $\delta(y)$ term vanishes as the interval of integration does not include the singularity. Now add a small $\epsilon_1$ neighbourhood of 7 to the interval of integration and consider $$I_1=\int_{7-\epsilon_1}^{7+\epsilon_1}\left[\frac{1}{4}e^{-|y|}+\frac{1}{6}\delta(y-7)\right]dy$$ By the man value theorem we can write: $$I_1=\frac{\epsilon_1}{2}e^{-c}+\frac{1}{6}\int_{7-\epsilon_1}^{7+\epsilon_1}\delta(y-7)dy$$ where $c\in [7-\epsilon_1,7+\epsilon_1]$. Now using $(1)$ we can write $$I_1=\frac{1}{6}\int_{-\infty}^{\infty}\delta(y-7)dy+O\left(\epsilon_1\right)=\frac{1}{6}+O\left(\epsilon_1\right)$$ Since $\epsilon_1$ is arbitrary we obtain the final result.
Could you please explain why the comment I made to the question (about removing the single point $y=7$ from the interval of integration) doesn't raise a problem with your proof? Thanks! – Andrea Orta Feb 10 '13 at 20:53 | 2014-03-09T15:43:04 | {
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https://math.stackexchange.com/questions/749224/combinatorics-generating-function | # Combinatorics-Generating function
5 pirates find 3000 gold coins. In how many ways they can distribute them, if the captain gets at least 500 and not more then 2000 coins. the rest get at least 150 but not more then 1000 coins.(each one)
I've tried to use this generating function: $(x^{150}+x^{151}+...+x^{1000})^4 * (x^{500}+x^{501}+...+x^{2000})$
Then I thought this approach is better:
we will try to decide how many coins the othe four get, and the pirate will have one option... then, we need to solve: $t_1+t_2+t_3+t_4=n$ where $t_1,t_2,t_3,t_4$ between 150, 2000 and $n$ is: $1000,1001,....,2500$ because $t_i$ is at least 150, we can solve: $y_1+y_2+y_3+y_4=n-600$ we only need to sum the solutions for any $n$ in range. I couldn't solve it, I don't know how to insert the condition that $t_i\leq1000$
Thank You.
We can solve from the generating function itself, rewriting it as:
\displaystyle \begin{align} g(x)&= \dfrac{x^{600}\,(1-x^{851})^4}{(1-x)^4}\cdot \dfrac{x^{500}\,(1-x^{1501})}{(1-x)}\\\\ &=\dfrac{x^{1100}\,(1-x^{851})^4\,(1-x^{1501})}{(1-x)^5} \end{align}
Expand:
\begin{align} g(x)&=\left(x^{1100}-x^{2601}\right)\, \left(1-\binom{4}{1}x^{851}+\binom{4}{2}x^{1702}-\binom{4}{3}x^{2553}+\binom{4}{4}x^{3404}\right)\sum_{n=0}^{\infty}\binom{n+4}{4}x^n \end{align}
Extract $[x^{3000}]$, which is:
$\displaystyle \binom{4}{0}\, \binom{1904}{4}-\binom{4}{1}\, \binom{1053}{4}+\binom{4}{2}\, \binom{202}{4}-\binom{403}{4}=341444579376$
• Here's a code in Sage to extract the required coefficient: code – gar Apr 11 '14 at 8:10
• Can you explain how you got the final expression with the binomial coefficients? – ShreevatsaR Apr 11 '14 at 9:10
• It looks similar to Inclusion–exclusion principle – Shirly Geffen Apr 11 '14 at 9:50
• I have added a step, hope it's clear enough now. – gar Apr 11 '14 at 10:31
• @gar: Thanks, it's clear now -- before you edited your answer I'd started writing one, and arrived at the same answer independently. +1. – ShreevatsaR Apr 11 '14 at 10:51
Your original approach is fine: we want the coefficient of $x^{3000}$ in
$$(x^{500} + x^{501} + \dots + x^{2000}) (x^{150} + x^{151} + \dots + x^{1000})^4$$ $$= x^{500 + 150\times 4}(1 + x^2 + \dots + x^{1500})(1 + x + \dots + x^{850})^4$$ $$= x^{1100}\frac{1-x^{1501}}{1-x} \left(\frac{1-x^{851}}{1-x}\right)^4$$
The coefficient of $x^{3000}$ in the above is \begin{align} &[x^{3000}]x^{1100} (1-x^{1501})(1-x^{851})^4(1-x)^{-5} \\ &= [x^{1900}]\left((1-x^{851})^4(1-x)^{-5}\right) - [x^{399}]\left((1-x^{851})^4(1-x)^{-5}\right) \tag 1 \end{align}
Now, we have from the binomial theorem, $$(1-x^{851})^4 = 1 - \binom{4}{1}x^{851} + \binom{4}{2}x^{851\times2} + \dots$$ and $$(1-x)^{-5} = 1 + \binom54x + \binom64x^2 + \binom74x^3 + \dots + \binom{k+4}{4}x^k + \dots$$
So the two coefficients in $(1)$ are, respectively, \begin{align} & [x^{1900}]\left((1-x^{851})^4(1-x)^{-5}\right) \\ &= [x^{1900}](1-x)^{-5} - \binom41 [x^{1900-851}](1-x)^{-5} + \binom42[x^{1900-851\times2}](1-x)^{-5} \\ &= \binom{1900 + 4}{4} - \binom41 \binom{1049 + 4}{4} + \binom42 \binom{198 + 4}{4} \\ &= 342527319476 \end{align} and $$[x^{399}]\left((1-x^{851})^4(1-x)^{-5}\right) = [x^{399}](1-x)^{-5} = \binom{399 + 4}{4} = 1082740100$$
Thus $(1)$ gives the answer to be $$342527319476 - 1082740100 = 341444579376.$$ | 2020-08-14T03:24:31 | {
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https://math.stackexchange.com/questions/576130/is-there-any-nonconstant-function-that-grows-at-infinity-slower-than-all-itera | # Is there any nonconstant function that grows (at infinity) slower than all iterations of the (natural) logarithm?
Is there any nonconstant function that grows at infinity slower than all iterations of the (natural) logarithm?
Yes, just take a function which is equal to $\log x$ on an initial interval such as $[1,K_1]$, then equal to $\log\log x$ on the interval $(K_1,K_2]$, then $\log\log\log x$ etc. where the values $K_1, K_2, \dots$ are chosen sufficiently large to ensure that the function really does tend to infinity. This can always be done, since each of the iterations of the log function does tend to infinity.
This function will not be continuous, but a similar function could certainly be created which is continuous.
This will produce a function which tends to infinity slower than any (fixed) iteration of $\log$. If you want something even slower, then call my function above $\phi (x)$, and repeat my construction using $\phi\phi(x)$ etc on successive intervals.
If this is still not slow enough, then ...
• I suspect a smooth one could be created, and maybe G. H. Hardy's old book "Orders of Infinity" might be worth a look. – Old John Nov 21 '13 at 16:26
• @Frank: my answer gives a continuous one, and it would not be difficult to smooth the kinks – Henry Nov 21 '13 at 16:27
• @Frank Hardy's book is available (legally) here on Gutenberg, and although a bit old-fashioned (like me) it is still worth a read. – Old John Nov 21 '13 at 16:29
• Thanks, this is a useful reference. – Frank Nov 21 '13 at 16:33
• @Frank: smoothing the kinks was easier than I expected. – Henry Nov 21 '13 at 16:35
Let $\text{LogNestMonster}_i(n) = nest(log, i)(n) = \overbrace{log(log(...(log}^\text{i times}(n))...))$.
Many functions grow slower than $\text{LogNestMonster}_i$, no matter how large the $i$ you picked is:
1. The inverse Ackermann function $\alpha$ grows slower than all $\text{LogNestMonster}$s. It (roughly) tells you how far in the sequence $1+1$, $2 \cdot 2$, $3^3$, $4\uparrow\uparrow4$, $5\uparrow\uparrow\uparrow5$, $...$ you have to go before exceeding $n$.
2. The iterated logarithm $log^*$ also grows slower than all $\text{LogNestMonster}$s. It counts how many times you have to apply log to $n$ before the result is less than $1$.
3. Also, consider the function $\frac{1}{n}$. It is not constant and is clearly asymptotically less than all $\text{LogNestMonster}$s, although I'm guessing it's not exactly what you had in mind since $\frac{1}{n}$ is $O(1)$ despite not being $\Theta(1)$.
• @Frank It seems like a reasonable answer to me... – fluffy Nov 21 '13 at 21:42
• 1/n is decreasing. I think the OP wants an increasing function. – marty cohen Oct 1 '15 at 18:29
• @martycohen The answer mentions that already. – Craig Gidney Oct 1 '15 at 18:35
• I haven't seen the nest function before and I'm interested in learning more. It's a bit hard to search for though, can you point me in the direction of any resources that talk about nest from a mathematical point of view? – 7yl4r Dec 27 '16 at 17:30
• @7yl4r Iterated function. I've seen it called "iter" before. I just called it nest for the sake of the pun. – Craig Gidney Dec 27 '16 at 18:37
In fact there are functions that go to $\infty$ more slowly than any function you can write down a formula for. For positive integers $n$ let $f(BB(n)) = n$ where $BB$ is the Busy Beaver function. Extend to $[1,\infty)$ by interpolation.
EDIT: Stated more technically, a "function you can write down a formula for" is a recursive function: it can be computed by a Turing machine. $BB(n)$ is not recursive, and grows faster than any recursive function. If $g(n)$ is a recursive (integer-valued, for simplicity), nondecreasing function with $\lim_{n \to \infty} g(n) = \infty$, then there is a recursive function $h$ such that for all positive integers $n$, $g(h(n)) > n^2$. Namely, here is an algorithm for calculating $h(n)$ for any positive integer $n$: start at $y = 1$ and increment $y$ until $g(y) > n^2$, then output $y$.
Now since $BB$ grows faster than any recursive function, for sufficiently large $n$ (say $n \ge N$) we have $BB(n) > h(n+1)$. For any integer $x \ge h(N)$, there is $n \ge N$ such that $h(n) \le x < h(n+1)$, and then (since $f$ and $g$ are nondecreasing) $$f(x) \le f(h(n+1)) \le f(BB(n)) = n < \sqrt{g(h(n)} \le \sqrt{g(x)}$$ and thus $$\dfrac{f(x)}{g(x)} \le \dfrac{1}{\sqrt{g(x)}} \to 0\ \text{as} \ x \to \infty$$
• This assumes, of course, that "$BB(n)$ doesn't count as a formula. – Andreas Blass Nov 21 '13 at 16:51
• What about g(x)=f(x)/2 – Bulwersator Nov 21 '13 at 22:05
• @Bulwersator They both have the exact same growth rate at $\infty$. A function $g(x)$ has a smaller growth rate than $f(x)$ at $\infty$ iff $$\lim_{x\to\infty}{\frac{g(x)}{f(x)}} = 0$$ or alternately: $$\lim_{x\to\infty}{\frac{f(x)}{g(x)}} = \infty$$. – AJMansfield Nov 22 '13 at 1:47
• The inverse Ackermann function is frequently cited in these contexts. – MJD Nov 12 '14 at 4:42
Yes:
My favorite big-number function from positive numbers to positive numbers, a continuous variant of Old John's.
Write $log_{10}^{(n)} x$ for the $n$th iterated base 10 logarithm of $x$.
If $log_{10}^{(n)} x \in [0,1)$ then let $f(x)=n+log_{10}^{(n)} x$.
You can extend this to non-positive numbers with $f(0)=0$ and $f(-x)=-f(x)$ giving a continuous bijective function on the reals.
If you want a smooth function take natural logarithms base $e$ instead
• excellent - thanks! – Frank Nov 21 '13 at 16:27
• Note that a version of this function - 'how many iterations of log does it take to reduce $x$ to $\lt 1$?' - shows up regularly in the analysis of algorithms; see en.wikipedia.org/wiki/Iterated_logarithm for details. – Steven Stadnicki Nov 21 '13 at 21:27
Try the following: $$f(x):=\min\{n\geq1\>|\> \log^{\circ n}(x)<0\}\ .$$
• Does this function tent toward infinity at all though? An explanation would be nice. – AJMansfield Nov 22 '13 at 1:55
• @AJMansfield: It's more or less the inverse of $x_0:=0$, $x_{n+1}:=e^{x_n}$ $\>(n\geq0)$. – Christian Blatter Nov 22 '13 at 9:14 | 2019-12-11T22:43:48 | {
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https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12A_Problems/Problem_9&curid=18131&diff=0&oldid=151768 | # Difference between revisions of "2020 AMC 12A Problems/Problem 9"
## Problem
How many solutions does the equation $\tan(2x)=\cos(\tfrac{x}{2})$ have on the interval $[0,2\pi]?$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$
## Solution 1
Draw a graph of $\tan(2x)$ and $\cos(\tfrac{x}{2})$
$\tan(2x)$ has a period of $\tfrac{\pi}{2},$ asymptotes at $x = \tfrac{\pi}{4}+\tfrac{k\pi}{2},$ and zeroes at $\tfrac{k\pi}{2}$. It is positive from $(0,\tfrac{\pi}{4}) \cup (\tfrac{\pi}{2},\tfrac{3\pi}{4}) \cup (\pi,\tfrac{5\pi}{4}) \cup (\tfrac{3\pi}{2},\tfrac{7\pi}{4})$ and negative elsewhere.
cos$(\tfrac{x}{2})$ has a period of $4\pi$ and zeroes at $\pi$. It is positive from $[0,\pi)$ and negative elsewhere.
Drawing such a graph would get $\boxed{\textbf{E) }5}$ ~lopkiloinm
Or you could see the points at which both graphs are positive or both are negative, again yielding 5 such areas. -hi13
edited by - annabelle0913
## Solution 2
To find the asymptotes of $\tan(2x)$ we consider the behaviour of $\tan(x)$ on $[0,4\pi]$. Then we see that there are five separate continuous parts of the graph splitting the plane into regions. Since $\cos(\frac{x}{2})$ is continuous it must intersect each of the five pieces of $\tan$ at least once. But since $\tan(2x)$ is increasing and $\cos(\frac{x}{2})$ is decreasing on the interval and continuous increasing functions and decreasing functions can intersect at most once, there are $\boxed{\textbf{E) }5}$ intersections. -codecow
## Remark
The graphs of $f(x)=\tan(2x)$ (in red) and $g(x)=\cos\left(\frac{x}{2}\right)$ (in blue) are shown below.
Graph in Desmos: https://www.desmos.com/calculator/gfplgzapww
~MRENTHUSIASM | 2021-05-06T02:12:26 | {
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https://mathhelpboards.com/threads/quick-problem-given-the-slope-and-a-point-find-the-line.8625/ | # Quick problem: given the slope and a point, find the line
#### tmt
##### Active member
Hello,
I am working on a problem. I am trying to get an equation for this linear line with these traits:
slope = 5/6
linear equation that passes through [1,3]
therefore using this formula
y=mx+b
3=5/6[1]+b
b=13/6
therefore, the equation is
y=5/6[x] + 13/6
y - ( 3 ) = (5/6) ( x - ( 1 ) ) ,
or
y = (7/6) x + (13/6) .
Any pointers,
Thanks, Tim
#### MarkFL
Staff member
The second given answer has the wrong slope. I would recommend using the point-slope formula because you are given exactly what you need to use this formula, that is a point on the line and its slope.
This formula is:
$$\displaystyle y-y_1=m\left(x-x_1 \right)$$
Now, you are given:
$$\displaystyle m=\frac{5}{6}$$
and:
$$\displaystyle \left(x_1,y_1 \right)=(1,3)$$
And so plugging in the given data to our formula, we obtain:
$$\displaystyle y-3=\frac{5}{6}(x-1)$$
If we wish to put this into slope-intercept form, we may distribute the slope on the right side:
$$\displaystyle y-3=\frac{5}{6}x-\frac{5}{6}$$
$$\displaystyle 3=\frac{18}{6}$$
to get:
$$\displaystyle y=\frac{5}{6}x+\frac{13}{6}$$
So you had the correct answer in slope-intercept form, but the answer given by your book is incorrect for this form (the slope is wrong, most likely just a typo), but is correct for the point-slope form.
#### Deveno
##### Well-known member
MHB Math Scholar
A persistent problem with many math texts is that "answers" given in the "back of the book" often contain silly mistakes (apparently, good proof-readers are hard to find).
The only remedy for this is to become so confident in YOUR skills, that you can TELL when the answer is right or wrong.
You answered the problem correctly, so well done! | 2020-12-03T03:50:49 | {
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http://mathhelpforum.com/calculus/140125-general-formula-sequence-print.html | # General Formula For Sequence
• Apr 19th 2010, 02:40 PM
vReaction
General Formula For Sequence
Find the general formula (for the nth term) for the sequence
{5,1,5,1,5,1,...}
assuming the pattern continues.
I've tried a lot of things but can't come up with a solution...it seems innocent enough. It seems like the formula should depend on n being even or odd but I can't get anything to work. A suggestion would be appreciated. Is there a specific way to solve these or are you supposed to just spot the pattern. Thanks.
• Apr 19th 2010, 02:55 PM
tonio
Quote:
Originally Posted by vReaction
Find the general formula (for the nth term) for the sequence
{5,1,5,1,5,1,...}
assuming the pattern continues.
I've tried a lot of things but can't come up with a solution...it seems innocent enough. It seems like the formula should depend on n being even or odd but I can't get anything to work. A suggestion would be appreciated. Is there a specific way to solve these or are you supposed to just spot the pattern. Thanks.
$a_n=\left\{\begin{array}{ll}5&,\,if\,\,\,n\,\,\,is \,\,\,odd\\1&,\,if\,\,\,n\,\,\,is\,\,\,even\end{ar ray}\right.$
Tonio
• Apr 19th 2010, 02:56 PM
Plato
Quote:
Originally Posted by vReaction
Find the general formula (for the nth term) for the sequence {5,1,5,1,5,1,...}, assuming the pattern continues.
Try $x_n=5\cdot \text{mod}(n,2)+\text{mod}(n+1,2)$.
• Apr 19th 2010, 03:07 PM
vReaction
Tonio: Thanks...I didn't think it'd be that simple.
Plato: I'm unfamiliar with "mod"...I searched and only found modular arithmetic which looked very confusing. Thanks though.
• Apr 19th 2010, 03:37 PM
Plato
Quote:
Originally Posted by vReaction
: I'm unfamiliar with "mod"...I searched and only found modular arithmetic which looked very confusing.
It is not confusing at all.
$\text{mod}(j,k)$ is the remainder when $j$ is divided by $k$.
So $\text{mod}(3,2)=1$ and $\text{mod}(4,2)=0$
$\text{mod}(5,3)=2$ and $\text{mod}(12,6)=0$
• Apr 19th 2010, 04:09 PM
Soroban
Hello, vReaction!
Quote:
Find the $n^{th}$ term for the sequence: . $5,1,5,1,5,1\:\hdots$
The sequence is: . $(3+2),\;(3-2),\;(3+2),\;(3-2),\;\hdots$
Therefore: . $a_n \;=\;3 - (\text{-}1)^n\!\cdot\!2$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
. . $\frac{1-(\text{-}1)^n}{2} \;=\;\begin{Bmatrix}1 & n \text{ odd} \\ 0 & n\text{ even} \end{Bmatrix}$
. . $\frac{1+(\text{-}1)^n}{2} \;=\;\begin{Bmatrix} 0 & n\text{ odd} \\ 1 & n\text{ even} \end{Bmatrix}$
Given the sequence: . $A,B,A,B,A,B\:\hdots$
. . then: . $a_n \;\;=\;\;\frac{1-(\text{-}1)^n}{2}\cdot A \:+\: \frac{1+(\text{-}1)^n}{2}\cdot B$ | 2017-02-23T23:23:44 | {
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http://mathhelpforum.com/calculus/118189-find-dimensions-x-y.html | # Math Help - find the the dimensions of x and y
1. ## find the the dimensions of x and y
a. Find a formula for the area
b. find a formula for the perimeter
c. find the dimensions x and y that maximize the area given that the perimeter is 100 .
* See attachment for details!!!
I know that this figure is composed of 4 semicircles and one rectangle ; the area of the rectangle is xy , and the area of the semicircle is 1/2pir^2. The perimeter of the rectangle is 2x + 2y , but I'm not sure how to apply the perimeter of the semicircles to this problem. On the other hand , the constraint I should use is in part c ; however , I need to find out a and b before that. I really would appreciate help for this application .Thanks for your time !!!
2. ## here it is
1)area= (pi/4)[x^2+y^2]+xy
2)perimeter=p=pi(x+y)
3)for maximum area(constant perimeter)
put x=0 or y=0
hence maximum area will be p^2/(4pi)
though not asked but minimum area will be
[p^2/4(pi)^2][pi/2+1][this can be obtained using calculus]
3. Originally Posted by skorpiox
a. Find a formula for the area
b. find a formula for the perimeter
c. find the dimensions x and y that maximize the area given that the perimeter is 100 .
* See attachment for details!!!
I know that this figure is composed of 4 semicircles and one rectangle ; the area of the rectangle is xy , and the area of the semicircle is 1/2pir^2. The perimeter of the rectangle is 2x + 2y , but I'm not sure how to apply the perimeter of the semicircles to this problem. On the other hand , the constraint I should use is in part c ; however , I need to find out a and b before that. I really would appreciate help for this application .Thanks for your time !!!
If you call the dimensions of the rectangle x and y, then you really have two complete circles with radii x/2 and y/2. The area of the entire figure is The area of those two circles, $\pi r^2= \pi x^2/4$ and $\pi y^2/4$ plus the area of the rectangle, xy. That is, the total area of the figure is $xy+ \pi x^2/4+ \pi y^2/4$.
The perimeter, however, has nothing to do with the perimeter of the rectangle. It is, instead, the total perimeter of the two circles, $\pi x+ \pi y= \pi (x+ y)$.
You are then given that $\pi (x+ y)= 100$ so $y= \frac{100}{\pi}- x$. Replace y by that in the formula for the area, $xy+ \pi x^2/4+ \pi y^2/4= x\left(\frac{100}{\pi}- x\right)+ \pi x^2/4+ \pi\left(\frac{100}{\pi}- x\right)/4$. Minimize that.
4. now, I have a better picture of this problem; I never thought about using two circles for getting my perimeter!!!! | 2014-09-22T01:55:20 | {
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http://vrtoplica.mi.sanu.ac.rs/u7nj752/199b1d-correlation-matrices-positive-semidefinite | ## correlation matrices positive semidefinite
April 2019 It is nsd if and only if all eigenvalues are non-positive. The problem is solved by a convex quadratic semidefinite program. Insurance August 2020 The first is a general assumption that R is a possible correlation matrix, i.e. Sometimes, these eigenvalues are very small negative numbers and … that it is a symmetric positive semidefinite matrix with 1’s on the main diagonal. This can prove problematic in using the matrix in statistical models. $$M = \begin{bmatrix} 1&a\\ a&1 \end{bmatrix}$$, And let $z$ be the column vector $M = \begin{bmatrix} z_1\\ z_2 \end{bmatrix}$, $$z^T M z = {\begin{bmatrix} z_1\\ z_2 \end{bmatrix}}^T \begin{bmatrix} 1&a\\ a&1 \end{bmatrix} \begin{bmatrix} z_1\\ z_2 \end{bmatrix}$$, $$= {\begin{bmatrix} z_1\\ z_2 \end{bmatrix}}^T \begin{bmatrix} z_1 & a z_2 \\ a z_1 & z_2 \end{bmatrix} = z_1 (z_1 + a z_2) + z_2 (a z_1 + z_2)$$, $$= {z_1}^2 + a z_1 z_2 + a z_1 z_2 + {z_2}^2 = (z_1 + a z_2)^2 \geq 0$$, $$M = \begin{bmatrix} 1&a&b\\ a&1&c \\ b&c&1 \end{bmatrix}$$. A valid correlation matrix not only has to be symmetric but also positive semidefinite. If $latex x_1, x_2, \dots, x_n$ are column vectors with $latex m$ elements, each vector containing… Therefore in order for a $3$ x $3$ matrix to be positive demi-definite we require: I work as a pricing actuary at a reinsurer in London.I mainly write about Maths, Finance, and Technology.If you would like to get in touch, then feel free to send me an email at:[email protected], All January 2018 Introduction The algorithmic generation of valid correlation matrices has been up to quite recently a challenging problem. June 2017 However, the estimated correlation matrix sometimes has a serious defect: although the correlation matrix is originally positive semidefinite, the estimated one may not be positive semidefinite when not all ratings are observed. Let's start with the Mathematical definition. The kernel matrices resulting from many practical applications are indefinite and therefore are not suitable for kernel learning. It is nsd if and only if all eigenvalues are non-positive. November 2019 The correlation matrix is a fundamental statistic that is used in many fields. Now, to your question. The MovieLens data set is used to test our approach. We require: $\begin{vmatrix} 1 & a & b \\ a & 1 & c \\ b & c & 1 \end{vmatrix} \geq 0$, $\begin{vmatrix} 1 & a & b \\ a & 1 & c \\ b & c & 1 \end{vmatrix} = 1 ( 1 - c^2) - a (a - bc) + b(ac - b) = 1 + 2abc - a^2 - b^2 - c^2$. All correlation matrices are positive semidefinite (PSD), but not all estimates are guaranteed to have that property. Actuarial Career June 2018 January 2016, A symmetric $n$ x $n$ matrix $M$ is said to be. That inconsistency is why this matrix is not positive semidefinite, and why it is not possible to simulate correlated values based on this matrix. The problem is solved by a convex quadratic semidefinite program. is definite, not just semidefinite). Correlation matrices capture the association between random variables and their use is ubiquitous in statistics. November 2017 For example, robust estimators and matrices of pairwise correlation coefficients are two situations in which an estimate might fail to be PSD. Poker which shows that any covariance matrix is positive semidefinite. The input matrix is nominally a correlation matrix, but for a variety of reasons it might not be positive semidefinite. March 2017 2008 Mar-Apr;21(2-3):170-81. doi: 10.1016/j.neunet.2007.12.047. It is nd if and only if all eigenvalues are negative. September 2019 Browse other questions tagged matrices eigenvalues-eigenvectors correlation positive-semidefinite or ask your own question. We first check the determinant of the $2$ x $2$ sub matrix. Observation: A consequence of Property 4 and 8 is that all the eigenvalues of a covariance (or correlation) matrix are non-negative real numbers. Let's take a hypothetical case where we have three underliers A,B and C. We need that: $\begin{vmatrix} 1 & a \\ a & 1 \end{vmatrix} \geq 0$, $\begin{vmatrix} 1 & a \\ a & 1 \end{vmatrix} = 1 - a^2$. Hum Hered. May 2017 It is pd if and only if all eigenvalues are positive. This can be tested easily. February 2020 If any of the eigenvalues is less than zero, then the matrix is not positive semi-definite. Actuarial Exams 2006 Nov;18(11):2777-812. doi: 10.1162/neco.2006.18.11.2777. Since we are dealing with Correlation Matrices, rather than arbitrary Matrices, we can actually show a-priori that all 2 x 2 Matrices are positive semi-definite. Obviously, if we only have two random variables, then this is trivially true, so we can define any correlation between two random variables that we like. Epub 2010 Jul 3. (2 replies) I'm trying to test if a correlation matrix is positive semidefinite. The first approach is quite simple. This result is consistent with our intuitive explanation above, we need our Correlation Matrix to be positive semidefinite so that the correlations between any three random variables are internally consistent. A … However, statistical properties are not explicitly used in such studies. It is easy to verify that correlation matrices are positive semidefinite and have all diagonal entries equal to one. This site needs JavaScript to work properly. Hence, while individual elements still obey the assumptions of correlation values, the overall matrix is often not mathematically valid (not positive semidefinite). Law While implementing the algorithm there is no need to check positive semi-definiteness directly, as we do a Cholesky decomposition of the matrix … Finance Features of a valid correlation matrix Correlation matrices: Diagonal elements all equal 1 Matrix is symmetric All off-diagonal elements between 1 and−1 inclusive. matrix not positive semidefinite One or more numeric values are incorrect because real data can generate only positive semidefinite covariance or correlation matrices. The norm is a weighted version of the Frobenius norm, A 2 F = i,j a 2 ij, the Frobenius norm being the easiest norm to work By using the model, an estimate is obtained as the optimal point of an optimization problem formulated with information on the variances of the estimated correlation coefficients. Each point in this space corresponds to a symmetric matrix, but not all of them are positive-definite (as correlation matrices have to be). Which gives us the required result. NLM >From what I understand of make.positive.definite() [which is very little], it (effectively) treats the matrix as a covariance matrix, and finds a matrix which is positive definite. Any covariance matrix is symmetric and positive semi-definite and its main diagonal contains variances. November 2020 A correlation matrix has a special property known as positive semidefiniteness. Epub 2008 Jan 10. Conversely, any such matrix can be expressed as a cor-relation matrix for some family of random variables. Web Scraping, January 2021 Tao Q, Scott SD, Vinodchandran NV, Osugi TT, Mueller B. IEEE Trans Pattern Anal Mach Intell. By using the model, an estimate is obtained as the optimal point of an optimization problem formulated with information on the variances of the estimated correlation coefficients. November 2016 All we need to do is install a package called 'Matrixcalc', and then we can use the following code: That's right, we needed to code up our own algorithm in VBA, whereas with R we can do the whole thing in one line using a built in function! Nicholas J. Higham, Computing a nearest symmetric positive semidefinite matrix, Linear Algebra Appl. COVID-19 is an emerging, rapidly evolving situation. | There are a number of ways to adjust these matrices so that they are positive semidefinite. To obtain a positive semidefinite correlation matrix, we assume the approximate model. Modelling A different question is whether your covariance matrix has full rank (i.e. Problem When a correlation or covariance matrix is not positive definite (i.e., in instances when some or all eigenvalues are negative), a cholesky decomposition cannot be performed. Wang and F. Zhang (1997, Linear and Multilinear Algebra, 43, 315–326) involves the Hadamard product and Schur complements.These two inequalities hold in the positive definite matrix case. USA.gov. Let me rephrase the answer. Neural Comput. Second, the data used to generate the matrix … October 2017 :) Correlation matrices are a kind of covariance matrix, where all of the variances are equal to 1.00. 103, 103–118, 1988.Section 5. The term comes from statistics. | By using the model, an estimate is obtained as the optimal point of an optimization problem formulated with information on the variances of the estimated correlation coefficients. The requirement comes down to the need for internal consistency between the correlations of the Random Variables. March 2020 The values in my correlation matrix are real and the layout means that it is symmetric. ; 30 ( 12 ):2084-98. doi: 10.1159/000312641 condition over the range 0,1... 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http://math.stackexchange.com/questions/507446/on-the-funny-identity-tfrac1-sin2-pi-7-tfrac1-sin3-pi-7-tf | # On the “funny” identity $\tfrac{1}{\sin(2\pi/7)} + \tfrac{1}{\sin(3\pi/7)} = \tfrac{1}{\sin(\pi/7)}$
This equality in the title is one answer in the MSE post Funny Identities. At first, I thought it had to do with $7$ being a Mersenne prime, but a little experimentation with Mathematica's integer relations found,
$$\frac{1}{\sin(2\pi/15)} + \frac{1}{\sin(4\pi/15)} + \frac{1}{\sin(7\pi/15)} = \frac{1}{\sin(\pi/15)}$$
$$\frac{1}{\sin(2\pi/31)} + \frac{1}{\sin(4\pi/31)} + \frac{1}{\sin(8\pi/31)} + \frac{1}{\sin(15\pi/31)} = \frac{1}{\sin(\pi/31)}$$
so the Mersenne number need not be prime. Let $M_n = 2^n-1$. How do we prove that,
$$\frac{1}{\sin(M_{n-1}\pi/M_n)}+\sum_{k=1}^{n-2} \frac{1}{\sin(2^k\pi/M_n)} = \frac{1}{\sin(\pi/M_n)}$$
indeed holds true for all integer $n>2$?
Edit (an hour later):
I just realized that since, for example, $\sin(3\pi/7)=\sin(4\pi/7)$, then the question can be much simplified as,
$$\sum_{k=1}^{n-1} \frac{1}{\sin(2^k\pi/M_n)} \overset{?}{=} \frac{1}{\sin(\pi/M_n)}$$
-
Oops, thanks for the formatting correx, Micah. – Tito Piezas III Sep 28 '13 at 1:21
Let $$\displaystyle S=\sum_{k=1}^{n-1}\csc \left(\frac{2^k \pi}{2^n-1} \right)$$ Using Euler's Formula, we can express this sum in terms of complex numbers. $$S=\sum_{k=1}^{n-1} \frac{2i}{e^{i 2^k \pi/(2^n-1)}-e^{-i2^k\pi /(2^n-1)}}=2i \sum_{k=1}^{n-1}\frac{e^{i 2^k \pi/(2^n-1)}}{e^{i 2^{k+1}\pi/(2^n-1)}-1}$$ For simplicity, let us assume $x=e^{i\pi/(2^n-1)}$. Then \begin{align*} S &= 2i\sum_{k=1}^{n-1}\frac{x^{2^k}}{x^{2^{k+1}}-1} \\ &= 2i \sum_{k=1}^{n-1}\frac{(x^{2^k}+1)-1}{(x^{2^k}+1)(x^{2^k}-1)} \\ &= 2i \sum_{k=1}^{n-1}\left( \frac{1}{x^{2^k}-1}-\frac{1}{x^{2^{k+1}}-1}\right) \end{align*} This is a telescoping sum and it's value is \begin{align*} S &= 2i \left( \frac{1}{x^2-1}-\frac{1}{x^{2^n}-1}\right) \end{align*} Back substituting, gives \begin{align*} S &= 2i \left( \frac{1}{e^{2i\pi/(2^n-1)}-1}-\frac{1}{e^{i 2^n \pi/(2^n-1)}-1}\right) \\ &= 2i \left( \frac{1}{e^{2i\pi/(2^n-1)}-1}+\frac{1}{e^{i\pi/(2^n-1)}+1}\right)\\ &= 2i \frac{e^{i\pi/(2^n-1)}}{e^{2i\pi/(2^n-1)}-1} \\ &= \csc \left( \frac{\pi}{2^n-1}\right) \end{align*} as desired.
-
dang, you got it just before I did, though I think mine's more pleasing :) – Calvin Lin Sep 28 '13 at 3:45
@Integral: I've accepted it since you answered first, and it is quite detailed. (Btw, you really should use your real name in this forum. This place is moderated anyway, so people are relatively civil.) – Tito Piezas III Sep 28 '13 at 15:08
I would suggest a further simplification of your problem, namely that $\frac{1}{ \sin (\pi/ M_n) } = - \frac{1}{ \sin ( 2^n \pi / M_n)}$. The identity then becomes
$$\sum_{i=1}^n \frac{1}{\sin (2^n \pi / M_n )} = 0.$$
We now use
$$\frac{1}{\sin \theta} = \cot \frac{\theta}{2} - \cot \theta,$$
which you can verify for yourself, to conclude that the result follows from telescoping, since $\cot \frac{ \pi}{2^n - 1} = \cot \frac{ 2^n \pi } { 2^n - 1}$.
Note: I got the trig identity from Wikipedia trig identity, knowing that I wanted $\csc \theta$ and something that could telescope.
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https://www.math.purdue.edu/pow/discussion/2018/spring/59 | ## Spring 2018, problem 59
Prove that
$\frac{\left(b+c-a\right)^{2}}{\left(b+c\right)^{2}+a^{2}}+\frac{\left(c+a-b\right)^{2}}{\left(c+a\right)^{2}+b^{2}}+\frac{\left(a+b-c\right)^{2}}{\left(a+b\right)^{2}+c^{2}}\geq\frac35$
for any positive real numbers $a$, $b$, $c$.
2 months ago
Let us assume for any $a,b,c \in \mathbb{R^{+}}$ we have $a \le b \le c$. Now, let $b = x \cdot a, c = y \cdot a$ such that $1 \le x \le y$ holds for $x,y \in \mathbb{R^{+}}$. This now allows us to transform the above 3-variable expression into a 2-variable version:
$\frac{(b+c-a)^2}{(b+c)^2 + a^2} + \frac{(c+a-b)^2}{(c+a)^2 + b^2} + \frac{(a+b-c)^2}{(a+b)^2 + c^2} = \frac{(x+y-1)^2a^2}{[(x+y)^2 + 1]a^2} + \frac{(y+1-x)^2 a^2}{[(y+1)^2 + x^2]a^2} + \frac{(1+x-y)^2 a^2}{[(1+x)^2 + y^2]a^2}$
or $f(x,y) = [1 - \frac{2(x+y)}{(x+y)^2 + 1}] + [1 - \frac{2x(y+1)}{(y+1)^2 + x^2}] + [1 - \frac{2y(x+1)}{(x+1)^2 + y^2}].$
If $1 \le x \le y$ holds for $x,y \in \mathbb{R^{+}}$, then for some fixed $x=x_{0}$ we have $lim_{y \rightarrow \infty} f(x_{0},y) = 3$ (which is the maximum value for $f$). Also, there exists only one critical point in the first-quadrant of the $xy$-plane: $(x,y) = (1,1)$. Evaluating $f$ at this critical point gives:
$f(1,1) = 3 \cdot (1 - \frac{4}{5}) = 3 \cdot \frac{1}{5} =\boxed{ \frac{3}{5}}$.
Since $\frac{3}{5}$ < $3$, this is our minimum value.
9 months ago
@Phiboyd2
The idea that a symetric function ( that stays unchanged when exchanging the variables) should have a maximum or minimum for equal values of the variables is very attractive but generally wrong for polynomials of degree 4 or more as proved by Buniakovsky in 1854.It is true for degrees less than 4. (here the degree of the polynomial is 6 so one cannot use this argument) a nice example is the following: P(x,y) = {x^2 +(1-y)^2} {y^2 +(1-x)^2} P is alwways positive or null and is equal to zero for (1,0) or (0,1).
@Claudio
Your remark that the problem can be confined to the case a+b+c= 1 is indeed crucial. Then you can get an immediate proof of the inequality using Jensen's theorem and the convexity of the function:
f(x) = (1-2x)^2 / {(1-x)^2 + x^2}
Thanks, Mike. You are right, I did't think to the Jensen's inequality, thus my solution is unnecessarily complicayted; but it was a first attempt... Can I suggest you of writing down the simpler solution?
Claudio 9 months ago
With the assumption a+b+c = 1 we have to prove that f(a) +f(b)+f(c) larger or equal to 3/5 with f(x) defined as above. f(x) is convex on the segment [ (3-sqt(3))/6 , (3+sqrt(3))/6] because its second derivative is non negative.
So if a,b and c are in this segment we can use Jensen's theorem which gives: f(a)+f(b)+f(c) larger or equal than 3f( (a+b+c)/3)=3f(1/3)=3/5.
Now suppose WLOG that a is larger than (3+sdrt(3))/6 then b+c is smaller than 1- (3+sqrt(3))/6 which is precisely the other bound (3- sqrt(3))/6; We therfore conclude that if a is outside the convex region, then b and c are also outside the convex region.But f(x) is larger than 1/2 outside this segment ,hence f(a) +f(b)+f(c) is larger than 3/2 in that case
In conclusion whatever the values of a,b,c in the segment (0,1) with hte condition a+b+c=1, we have always 3/5 as the lower bound.
michel 9 months ago
9 months ago
@Phiboyd2
The idea that a symetric function ( that stays unchanged when exchanging the variables) should have a maximum or minimum for equal values of the variables is very attractive but generally wrong for polynomials of degree 4 or more as proved by Buniakovsky in 1854.It is true for degrees less than 4. (here the degree of the polynomial is 6 so one cannot use this argument) a nice example is the following: P(x,y) = {x^2 +(1-y)^2} {y^2 +(1-x)^2} P is alwways positive or null and is equal to zero for (1,0) or (0,1).
@Claudio
Your remark that the problem can be confined to the case a+b+c= 1 is indeed crucial. Then you can get an immediate proof of the inequality using Jensen's theorem and the convexity of the function:
f(x) = (1-2x)^2 / {(1-x)^2 + x^2}
9 months ago
Let us call $F(a,b,c)$ the function to be minimized. It is a homogeneous function of degree 0 say, for any $t$, it is $F(ta,tb,tc)=F(a,b,c)$. Thus we can confine ourselves to work with $(x,y,z)$ in the triangle $T:=\{x+y+z=1;\ x,y,z\ge0\}$; and we will see that the minimum is when the variables all take the value $\frac13$.
On the boundary of $T$ the function $F$ is bigger than 1: e.g. for $a=0$ the function becomes $1+2\frac{(b-c)^2}{(b^2+c^2)}\ge1$. Thus we can apply the Lagrange method, searching for minima at the interior of the triangle. Better, we will search for interior maxima on $T$ of the function $6-F(a,b,c)$ that, as we will show, can be written on the form $g(a)+g(b)+g(c)$, the function $g$ being defined by $g(x):=\frac1{(1-x)^2+x^2}$.
In fact $2-\frac{(b+c−a)^2}{(b+c)^2+a^2}=\frac{(a+b+c)^2}{(b+c)^2+a^2}$ that on $a+b+c=1$ becomes $\frac1{(1-a)^2+a^2}$, say $g(a)$; the other terms can obviously be treaten in the same way.
Now the Lagrange method gives for the estremal points $(a_0,b_0,c_0)$ the equation $g'(a_0)=g'(b_0)=g'(c_0)$ that, joint with $a_0+b_0+c_0=1$ implies $a_0=b_0=c_0=\frac13$.
May be the last point deserves more details.
For any real $k$ the equation $g'(x)=k$ has at most two solutions; in particular if $g'(a)=g'(b)=g'(c)$ then at least two among $a,b,c$ must coincide; let e.g. $a=b$. Being on the triangle $T$ we must have $c=1-2a$, and obviously $a\in(0,\frac12)$. Our goal is $a=\frac13$ (thus also $c=a=\frac13$). And in fact, rewriting $g'(a)=g'(c)$ we get $g'(a)-g'(1-2a)=0$.
This imlpies $a=c$ because on the interval $(0,\frac12)$ the function $h(x):=g'(x)-g'(1-2x)$ starts from $h(0)=4$, then increases on a small interval, then decreases; thus it vanishes only on $x_0:=\frac13$.
8 months ago
## My solution in 3 simple steps
1. Since the left side is a homogenous function F(a,b,c)=F(ta,tb,tc) we can restrict a,b and c to be smaller or equal to 1 and recognize that F(1,1,1) = 3/5
2. Now we set * a+b+c = 3 * and substitute/replace yielding the expression (3-2c)^2/ ( (3-c)^2+ c^2)) + (3-2a)^2/( ( 3-a)^2 + a^2)) + (3-2b)^2/ ( (3-b)^2 + b^2)). These are rational functions. I WILL SHOW THAT EACH OF THE THREE TERMS IS always LARGER THAN 1/5.
3. Expanding the quadratic functions and doing the PARTIAL FRACTION DIVISION yields : (9 - 12c + 4 c^2 ) / ( 9 - 6c+2c^2) = 2 - 9 / ( 2 c^2 - 6c + 9 ) , the same for a and b. The quadratic polynomial in the denominator takes its minimum value at c=1 , the whole function is therefore a strictly decreasing function and falls from 1 at c=0 to its minimum value of 1/5 at c=0. Adding the three terms 3x1/5=3/5 completes the proof.
3 months ago
1. Call the function on the LHS to be f(a,b,c).
2. Since all variables are independent, fix b and c.Let the variable 'a' vary alone.So if we want to find minimum of f(a,b,c) with fixed b and c we differentiate f partially with respect to a and equate it with 0 to get a=β(say).. 3.Now,let the variable b vary alone keeping others fixed and again differentiating f partially with respect b and equatng it with 0 in order to find minimum of f(a,b,c).By symmetry b must be equals to β.In fact the last variable c must be equals to β(while doing same process with c). 4.These will tell us that the all variables will have to be equal(for minimum value of f(a,b,c) )
because if we intersect the conditions for minimum value of f(a,b,c) with respect to individual independent variables we will get the condition which will give global minimum of f(a,b,c). 5.Thus putting a=b=c we will get 3/5 as a minimum of f(a,b,c).
9 months ago
This seems like a bit of deja vu !
The following comment extracted from the previous problem 58 states :
The function M(a,b) is minimized when both of its arguments equal each other. (i)
If we apply this to this problem replacing M(a,b) by f(a,b,c) then with a=b=c the expression takes the value 3/5 - solved.
However, as I previously mentioned the statement (i) does need some justification.
I am beginning to wonder if as a general rule a function of n variables where the value of the function is the same no matter how the variables are positioned implies that the minimum, (or maximum as the case may be) occurs when all the variables have the same value.
e.g if f(x1, x2, x3, x4) = f(x1, x2, x4, x3) = f(x1, x3, x2, x4) .......... then the minimum occurs when x1=x2=x3=x4
The property you quote as (1) does not refer to a general function $M$: it just apply when the function $M$ is defined as the maximum of two functions, $f,g$. Furthermore, as I stated twice in the comments to Problem 58 (and one of the explanations was dedicated to you), the property $f(P)=g(P)$ holds true unless the minimum point $P$ is a local minimum of one among $f,g$ (no matter the number of variables). If you need further details, let me know.
Claudio 9 months ago | 2018-12-10T16:21:03 | {
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https://math.stackexchange.com/questions/1966609/how-to-check-if-you-are-counting-right | # How to check if you are counting right?
Combinatorics is "hard" even at the elementary level in that verifying answers becomes extremely tricky. While in algebra while a solution to an equation such as $$x^2+4x+3=0$$ is desired, the solution can be obtained and can be checked by plugging the solution by plugging the answer back into the equation.
Consider for instance, this problem.
Find the number of ways to create $5$ groups of exactly two among $10$ people such that no person belongs to two groups.
An incorrect solution: First select a group of two people from the $10$ people. There are ${10 \choose 2}$ of doing this. The next group can be selected in ${8 \choose 2}$ ways and so on. So by the multiplication principle, the total number of ways equal to the value of the following product.
$${10 \choose 2} \times {8 \choose 2} \times {6 \choose 2} \times {4 \choose 2}$$
The idea in the next (incorrect) solution is to create a bijection between the number of permutations of the people standing in a line and the number of groups that can be formed. Though this is the same idea that one possible correct solution uses the groups are overcounted.
Incorrect solution-2: There are $10!$ ways in which the $10$ people can stand in a line. The first and the second person, the third and the fourth and so on are placed in a single group. Since switching the first and second position will give another permutation but the same groups we need to divide by $2$. Similarly switching the third and the fourth position will give a different permutation but the same groups. Therefore ultimately, the total number of ways to form the $5$ groups will be $$\frac{10!}{2^{5}}$$
The solution which yields the correct answer is the following
Correct Solution-2: There are $10!$ ways in which the $10$ people can stand in a line. The first and the second person, the third and the fourth and so on are placed in a single group. Since switching the first and second position will give another permutation but the same groups, we need to divide by $2$. Similarly switching the third and the fourth position will give a different permutation but the same groups. Also consider the following fact. If the people (indicated by letters) are arranged in one permutation as follows, ABCDEFGHIJ then the permutation CDABIJGHEF also correspond to the same set of groups which is why there is a need to divide by $5!$ as there are $5!$ ways to
Therefore ultimately, the total number of ways to form the $5$ groups will be $$\frac{10!}{2^{5}\times5!}$$
Questions concerning the correctness of a counting procedure adopted are quite common on this forum. ( I will add some links if you think it is necessary.) This is one reason this question,albeit subjective, has been asked.Also i have not come across any text which addresses how not to undercount or overcount. Any discussion of techniques on how to avoid undercounting/overcounting or how to check the enumeration will be highly appreciated.
Note:
1. The problems that I mean when i say combinatorics problems are enumerative combinatorics problems. This will perhaps narrow the scope of discussion by a considerable extent.
1. Though counting in two ways is sometimes a nice way to check whether the solution obtained is right, it takes a lot of ingenuity to enumerate the answer to every problem in two ways. Which is why any answer to this question can avoid that technique and also not include listing out all the possibilities.
• Practice, practice, practice. – Gerry Myerson Oct 13 '16 at 10:14
• Change the numbers in the problem to smaller ones, then you can check your calculation by listing all possibilities. – 5xum Oct 13 '16 at 10:19
• Sometimes a combinatoric result can be interpreted as a probability statement. You may then be able to check the probability statement via Monte Carlo simulation. Crunching out all possibilities in a computer program is another way to spot-check. – awkward Oct 13 '16 at 11:21
• If there were a general approach or even only some rather general approaches, beyond those already mentioned, they would be well known by then, wouldn't they? – g g Oct 19 '16 at 16:14
• Someone should at least mention the fact that this is a legitimately delicate issue. @TheCrypticCat, you are not alone in finding this difficult to feel comfortable with! – Greg Martin Oct 22 '16 at 18:52
Verification of counting may be harder than verification of arithmetic, but the two incorrect solutions can't be blamed on this difference. They are incorrect interpretations of a word problem. This can just as easily happen in an algebra word problem. It's not clear that we can do better than heuristics like, "Where does order matter?", "Are those things labeled or not?", or "Did I overcount?" In practice, the suggestions in the comments are good diagnostics for humans doing counting problems.
In principle, however, verification is possible. Suppose we agree that the set in question is formalized best by a mathematically specified set $A$. Then we can formalize our solutions and verify.
Specifically, we verify answers with bijective proofs. This consists of constructing two sets $A$ and $B$ and a bijection $f:A \rightarrow B$. Then we prove that $f$ is a bijection. Once established, we've verified $|A| = |B|$.
For elementary counting problems, we're usually applying the multiplication or addition principles to already established counting formulas, which themselves can be established with these principles. The more formal version of applying the multiplication principle is to provide a bijection $f:A \rightarrow B \times C$ where $A,B,C$ are sets. For example, if you construct a set $A$ and believe it has $2^3 \cdot 3!$ elements, you can verify this by constructing a bijection from $A$ to $B_3 \times S_3$.
(In case you're unfamiliar, $S_3$ is the set of permutations of a 3-element set and $B_3$ is the set of all subsets of $\{1,2,3\}$.)
For the problem in question, we might (sloppily) formalize the set as $$A = \{\{G_1,G_2,\ldots,G_5\}: G_i \subseteq \{P_1,\ldots,P_{10}\}, |G_i| = 2, |G_i \cap G_j| = \emptyset \}$$
If we agree that's a decent translation, we can then verify there's a bijection $f:A \times B_5 \times S_5 \rightarrow S_{10}$. We can also verify that a specific function $f:A \times B_5 \rightarrow S_{10}$ is not a bijection. Or, if one prefers, the divisions can be captured with equivalence relations and again proceed to construct functions and verify.
While this is not as simple as substituting a proposed value into $x^2 + 4x + 3$ to check, proofs can be made so formal that they become verifiable calculations. It sounds like the paper "Tests and Proofs for Enumerative Combinatorics" goes down this path with computer proof assistants.
Many word problems are of the following form: "An object $x$ has the properties $P_i(x)={\rm true}$ $(1\leq i\leq n)$. What is $x$?"
The given conditions define a set $$S:=\{x\in X\>|\>P_i(x)\ (1\leq i\leq n)\}\ ,$$ whereby the "universe" $X$ of possible $x$ is understood by the problem setting. Solving such a problem one sets up a chain (or even a directed graph) of reasoning of the following type:
$$x\in S\Rightarrow\ldots\Rightarrow\ldots\Rightarrow\ldots\Rightarrow x\in\{x_1,x_2,\ldots x_r\}\ ,$$ or similar. Unless one has a "general theory" for this special kind of problems, which guarantees "exactly $r$", or, e.g., a "vector space of dimension $r$" of solutions one has to test each candidate $x_i$, $1\leq i\leq r$, whether $x_i$ is actually a solution of the original problem.
Now the counting problems are of a different nature. A certain set $X$ is defined, for all to see, and it is claimed that this set set has a certain property, e.g., that it has $9!$ elements. Maybe you have an "alleged proof" for this claim; but there is definitely no "easy check". Note that there is no easy proof for the claim that the nontrivial zeros of the zeta function are lying on the critical line.
This reminds me of this question I answered recently.
Yes, validating your answer is not easy when we have combinatorics or probability questions. There is no generic method one can "mindlessly" follow.
Monte Carlo simulation or brute force computing can help, provided you have understood the question correctly and your model/program is correct. Keep in mind that some questions are easier to validate with this method because they have a straightforward simulation/computation model, while others are more difficult because the model is not apparent or it is difficult to program (and then you worry about validating your program :)).
For example, think about this question: Three people roll a pair of dice each. What's the probability that one person has rolled at least one $1$, while the other two don't have a die with $3$ or below? This would be relatively straightforward to model and simulate. You simulate many rolls of three pairs of dice and you check whether these conditions hold in each try. You count the times they held and divide them by the total number of tries.
On the other hand, how would you simulate/compute the problem at hand (counting the pairings)? I can think of the following brute force computation approach: list all permutations of 10 people, each time grouping them into pairs (1st and 2nd is one pair, 3rd and 4th is another, etc). If a permutation results in a configuration of pairings we have already encountered then we do not count it. It is certainly doable, but there is plenty of room for programming mistakes when trying to find if two pair groupings are the same. Moreover, if we had 100 people instead of 10, computing time would explode using this method. The takeaway message is that even though simulation or brute-force computation can help in some cases, they are not a cure-for-all.
Trying your solution for small values can also help. But be careful. Sometimes wrong formulas give the right answer for small numbers. For our specific question, this is not the case (the wrong formula produces wrong results even for small values) so we could have actually gotten some help if we tried this method. Let's say we have $4$ people instead of $10$. The first (i.e., incorrect) method would yield ${4 \choose 2} = 6$ but when we try to enumerate all possible pairings manually, we get only $3$ ways. Looking into this we can get an insight about how we are double-counting.
But how do we know that there are only $3$ ways to pair $4$ items? What if we made a mistake in our manual counting? One way to feel more confident about our result is to notice that once we choose an item and we create all possible pairs with the 3 remaining items, then the last pair is 'forced' to be formed. This brings me to the crux of the validation issue.
The most helpful thing is to think of different ways of solving the problem. In the question you link, this is what the OP has done. They have realised that they can view the problem like this: Choose one person and then you have $9$ possibilities to pair them with someone. This is our first pair. Then choose another person (does not matter who) and they have $7$ possibilities to be paired. And so on with the rest. So the possible pairings should be: $9\cdot 7\cdot 5\cdot 3 = 945$. In the question, it's not clear that they saw the conflict with the result from the first formula (it seems they did not compute the result from the first formula) but they could have easily done that. This could have given them some insight about the problem. Once you find a mismatch you can start asking why. It's possible that one solution method is more intuitive to you (for example the method I just described seems more intuitive to me than thinking about combinations), so this can guide you to find the mistake in the other way. In any case, you can attack the problem from different angles and think about what your solutions do in small size problems to get some clarity of what's going on.
There is no guarantee that you will find an error, but attacking the problem from many angles is the best way I know. A great side effect of this process is that you gradually gain more insight in the mistakes you and other people do, and you become more confident about your solutions in future problems. | 2020-01-19T13:30:13 | {
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https://math.stackexchange.com/questions/1693685/what-is-the-sum-of-the-reciprocal-of-all-of-the-factors-of-a-number | # What is the sum of the reciprocal of all of the factors of a number?
Suppose I have some operation $f(n)$ that is given as
$$f(n)=\sum_{k\ge1}\frac1{a_k}$$
Where $a_k$ is the $k$th factor of $n$.
For example, $f(100)=\frac11+\frac12+\frac14+\frac15+\frac1{10}+\frac1{20}+\frac1{25}+\frac1{50}+\frac1{100}=\frac{217}{100}$
$f(101)=\frac11+\frac1{101}=\frac{102}{101}$
$f(102)=\frac11+\frac12+\frac13+\frac16+\frac1{17}+\frac1{34}+\frac1{51}+\frac1{102}=\frac{216}{102}$
I was wondering if it were possible to plot a graph of $f(n)$ and wondered if there were any interesting patterns. I was also wondering if there is a closed form representation and if $\lim_{n\to\infty}f(n)$ could be evaluated or determined to be finite or not or any other interesting things that might happen in this limit.
Secondly, I was wondering about another similar series, which considers $b_k$ as the $k$th prime factor of $n$.
$$p(n)=\sum_{k\ge1}\frac1{b_k}$$
• I wouldn't call that $\mu$ because there a fairly common function by that name that is similar. Mar 11 '16 at 23:51
• @ThomasAndrews Fine. I just think greek letters are pretty to use for things. Mar 11 '16 at 23:52
• A more standard name would be $\sigma_{-1}$. See Divisor function. Mar 12 '16 at 0:03
• The graph of this would be quite interesting. There would be gaps" at all the primes. However, that might be hard to see. Mar 12 '16 at 0:16
• Here are plots for the first 100k integers: i.imgur.com/XDyVK7p.png and for the first 10 million: i.imgur.com/y5SYGOd.png The largest value in this range is obtained for $8648640 = 2^6 × 3^3 × 5× 7 × 11× 13$ Mar 12 '16 at 17:30
Note that $n\cdot f(n)$ is the sum of the factors of $n$ (written in a different order), which is denoted by $\sigma(n)$. Thus, $\displaystyle f(n)={\sigma (n)\over n}$.
• Hm, interesting. Thank you. Mar 11 '16 at 23:50
• This function is sometimes called the abundancy of a number, since $n$ is abundant if $f(n) > 2$, deficient if $f(n) < 2$, and perfect if $f(n) = 2$. See mathworld.wolfram.com/Abundancy.html. Note that Wikipedia (en.wikipedia.org/wiki/Abundant_number) uses the word "abundance" to mean something related but different, and "abundancy index" for your $f(n)$. Mar 12 '16 at 2:34
• @RaviFernando Yes, usually abundance is the difference $\sigma(n) - 2n$ while abundancy is the ratio $\frac{\sigma(n)}{n} = \sigma_{-1}(n)$. But one may have to repeat that definition to make sure everyone knows. A number whose abundancy is an integer, is called a multiply perfect number. Two or more numbers sharing the same abundancy are called friendly numbers. Mar 12 '16 at 10:40
Let $X_k$ be the product of the first $k$ primes. Let $Z_k$ be the sum of the reciprocals of the first $k$ primes. Then clearly $f(X_k)>Z_k$, and it's well known that $Z_k$ is unbounded, so $f(a_k)$ cannot have a finite limit. On the other hand, if $P_k$ is the $k$'th prime, then $f(P_k)$ clearly goes to $1$. Therefore $f(a_k)$ cannot have a limit other than $1$. Therefore $lim_{k\rightarrow\infty}a_k$ cannot exist.
• How did you arrive at $f(P_k)$ going to $1$? I do not see this. (And us \lim for limits) Mar 12 '16 at 0:16
• $f(P_k)$ goes to $1$ by euclid's theorem. Mar 12 '16 at 0:17
• @SimpleArt: if $p$ is prime then $f(p) = 1 + \frac{1}{p}$. The limit of this is $1$. Mar 12 '16 at 1:30
• @SimpleArt: if it's not prime then it's not one of the $P_k$ that WillO is talking about in that sentence. WillO is saying that your sequence has one subsequence that tends to infinity, and another subsequence that tends to $1$. Therefore the sequence has no limit (proof left as exercise). It doesn't matter how dense each subsequence is, only that they both have infinitely many terms. Mar 12 '16 at 1:34
• Note that the Wikipedia section Divisor function § Growth rate has more to say. Remember that $f(n)=\frac{\sigma(n)}{n}$ (Carl Heckman's answer). Then that section says that Grönwall's theorem is: $$\limsup_{n\to\infty} \frac{f(n)}{\log \log n} = e^\gamma$$ And it says that Robin's inequality (valid under the assumption of the Riemann hypothesis) says that every $n>5040$ has $\frac{f(n)}{\log \log n} < e^\gamma$. Mar 12 '16 at 10:54
Ramanujan included this in his original paper on Highly Composite Numbers, originally 1915. http://math.univ-lyon1.fr/~nicolas/ramanujanNR.pdf However, this was in a section left out because of paper shortages.
Let's see, I asked about this on MO https://mathoverflow.net/questions/137865/estimate-term-in-ramanujan-lost-notebook-classic-analytic-number-theory but did not quite get what I wanted, so I wrote to Nicolas. He's a nice man, but he had never heard of me, and the websites I mentioned were unknown to him. Sigh. Anyway, he did answer.
In brief, Ramanujan's construction allows us to produce a sequence of numbers, each new one the previous one times a prime, so that the function $\sigma(n)/n$ is surprisingly large for $n$ of that size. In turn, this gives explicit bounds on the function.
For numerical experiments of your own, the easiest way to approximate the numbers in this sequence is simply to take $$n = \operatorname{lcm} \{1,2,3, \ldots, k \}$$ and put $n$ into the sequence when it increases, which happens only when $k$ is a prime or prime power. Extremely approximately, $n \approx e^k.$ From Robin's criterion and related stuff, we will have $$\frac{\sigma(n)}{n} \approx e^\gamma \log \log n \approx e^\gamma \log k,$$ where $n = \operatorname{lcm} \{1,2,3, \ldots, k \} .$ Note that $e^\gamma \approx 1.7810724.$ Also note that it is the Prime Number Theorem that says that $\log n \approx k.$
Did it myself:
2 n = 2 = 2 function: 1.5 over log k: 2.16404
3 n = 6 = 2 3 function: 2 over log k: 1.82048
4 n = 12 = 2^2 3 function: 2.33333 over log k: 1.68314
5 n = 60 = 2^2 3 5 function: 2.8 over log k: 1.73974
7 n = 420 = 2^2 3 5 7 function: 3.2 over log k: 1.64447
8 n = 840 = 2^3 3 5 7 function: 3.42857 over log k: 1.64879
9 n = 2520 = 2^3 3^2 5 7 function: 3.71429 over log k: 1.69044
11 n = 27720 = 2^3 3^2 5 7 11 function: 4.05195 over log k: 1.68979
13 n = 360360 = 2^3 3^2 5 7 11 13 function: 4.36364 over log k: 1.70126
16 n = 720720 = 2^4 3^2 5 7 11 13 function: 4.50909 over log k: 1.62631
17 n = 12252240 = 2^4 3^2 5 7 11 13 17 function: 4.77433 over log k: 1.68513
19 n = 232792560 = 2^4 3^2 5 7 11 13 17 19 function: 5.02561 over log k: 1.70681
23 n = 5354228880 = 2^4 3^2 5 7 11 13 17 19 23 function: 5.24412 over log k: 1.6725
25 n = 26771144400 = 2^4 3^2 5^2 7 11 13 17 19 23 function: 5.41892 over log k: 1.68348
27 n = 80313433200 = 2^4 3^3 5^2 7 11 13 17 19 23 function: 5.55787 over log k: 1.68633
29 n = 2329089562800 = 2^4 3^3 5^2 7 11 13 17 19 23 29 function: 5.74952 over log k: 1.70746
31 n = 72201776446800 = 2^4 3^3 5^2 7 11 13 17 19 23 29 31 function: 5.93499 over log k: 1.72831
32 n = 144403552893600 = 2^5 3^3 5^2 7 11 13 17 19 23 29 31 function: 6.03071 over log k: 1.7401
37 n = 5342931457063200 = 2^5 3^3 5^2 7 11 13 17 19 23 29 31 37 function: 6.1937 over log k: 1.71527
41 n = 219060189739591200 = 2^5 3^3 5^2 7 11 13 17 19 23 29 31 37 41 function: 6.34477 over log k: 1.70854
43 n = 9419588158802421600 = 2^5 3^3 5^2 7 11 13 17 19 23 29 31 37 41 43 function: 6.49232 over log k: 1.72613
47 n = 442720643463713815200 = 2^5 3^3 5^2 7 11 13 17 19 23 29 31 37 41 43 47 function: 6.63046 over log k: 1.72213
49 n = 3099044504245996706400 = 2^5 3^3 5^2 7^2 11 13 17 19 23 29 31 37 41 43 47 function: 6.74886 over log k: 1.73411
53 n = 164249358725037825439200 = 2^5 3^3 5^2 7^2 11 13 17 19 23 29 31 37 41 43 47 53 function: 6.8762 over log k: 1.73191
59 n = 9690712164777231700912800 = 2^5 3^3 5^2 7^2 11 13 17 19 23 29 31 37 41 43 47 53 59 function: 6.99274 over log k: 1.71494
In comparison, the function for, say, $n$ prime is very small, just $1 + (1/n).$
• Good answer and it includes a nice approximation to growth rates. Thanks. Mar 12 '16 at 2:01
• @SimpleArt I put in some more detail. The largest your function gets is about $e^\gamma \log \log n,$ slowly growing but not bounded. Mar 12 '16 at 2:27
Just for the fun of it, I've graphed this up to $n=18$
Here is the data:
As you can see, the points move all over the place. Over this domain, it does seem to have a local maximum at every even number though. Perhaps this is because they have the advantage of a plus $1/2$
Here is the table I made if anyone wants to double check it:
Also, here is the link to the graph
• Just a note about your "local maximum at every even number" statement. Although that does happen in this range, and probably "usually" happens for large $n$ (although it would be interesting to try to prove this), it isn't always true. For example, the first odd abundant number (i.e. where $f(n) > 2$) is 945, but 944 and 946 are both deficient ($f(n) < 2$). Mar 12 '16 at 4:48
– user311559
Mar 12 '16 at 5:06
Here is the plot of $f(n)=\frac{\sigma(n)}{n}$ .
Patterns in these plots are amazing!
For $1000$
And here is for $100,000$
This diagram shows the $\lim_{x\to \infty} f_n$ isn't plausible !
• Just saying, this is awesome
– user311559
Mar 21 '16 at 19:46 | 2021-09-27T09:30:00 | {
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https://math.stackexchange.com/questions/2206352/does-an-existential-quantifier-followed-by-a-universal-quantifier-imply-a-univer | # Does an existential quantifier followed by a universal quantifier imply a universal quantifier followed by an existential quantifier?
1. $\exists y \forall x P(x, y)$ means: There exists some $y$ such that $P(x, y)$ holds for all $x$.
2. $\forall x \exists y P(x, y)$ means: For all $x$, there exists some $y$ such that $P(x, y)$ holds.
My understanding is that 1 implies 2. Here is my reasoning:
1 means that there is some $y$ such that $P(x, y)$ holds for all $x$. Let name it as $a$. So $P(x, a)$ holds for all $x$. Then for all $x$, there exists $a$ such that $P(x, a)$ holds. So 2 can be deduced.
In other words, one $y$ needs to make $P(x, y)$ holds for all $x$ in 1. $y$ stays the same. In 2, to make $P(x, y)$ hold for all $x$, $y$ does not need to be the same thing.
Is my understanding correct?
• Yes, exactly so. – bof Mar 28 '17 at 3:15
• I am slightly confused here. The existential quantifier allows for more than one y to make P(x,y) hold. Should not we prove that this works for multiple y as well? – user400188 Mar 28 '17 at 4:05
• @user400188 No, we only require one witness to the existence of at least one thing. – Graham Kemp Mar 28 '17 at 4:11
• math.stackexchange.com/questions/2182629/logical-truth-question/… – Bram28 Mar 28 '17 at 16:12
Yes, it is true that $\exists y \forall x \phi \models \forall x \exists y \phi$, and your explanation is correct.
Suppose that for some structure $\mathcal{M} = \langle D, I \rangle$ and some variable assignment $v$, $[[\exists y \forall x \phi]]^\mathcal{M}_v = 1$. Then there is some $d \in D$ such that $[[\forall x \phi]]^\mathcal{M}_{v[y \mapsto d]} = 1$, and consequently, there is some $d \in D$ such that for all $d' \in D$: $[[\phi]]^\mathcal{M}_{v[y \mapsto d][x \mapsto d']} = 1$. But then, for any $d' \in D$ there is some $d \in D$ such that $[[\phi]]^\mathcal{M}_{v[y \mapsto d][x \mapsto d']} = 1$, so for any $d' \in D$, $[[\exists y \phi]]^\mathcal{M}_{v[x \mapsto d']} = 1$, and hence $[[\forall x \exists y \phi]]^\mathcal{M}_v = 1$.
The proof might seem a bit circular because at some point you just need to swap the quantifiers on the meta level the same way as you eventually do at the level of the object language, but this meta level argumentation should make it clear why the consequence holds (and is essentially what you did).
Example: Suppose that $\mathcal{M} = \langle \mathbb{N}, \geq \rangle$ and $\phi \bumpeq x \geq y$.
$\exists y \forall x (x \geq y)$ is a true statement in the model since there is some $y$ such that $x \geq y$ for all $x$, with $y = 0$.
Then obviously for all $x$ there is some $y$ (namely $0$) such that $x \geq y$, and therefore $\forall x \exists y (x \geq y)$ is true in model $\mathcal{M}$.
On the other hand, the reverse direction does not hold: $\forall x \exists y \phi \not \models \exists y \forall x \phi$
Example: Suppose that $\mathcal{M} = \langle \mathbb{N}, = \rangle$ and $\phi \bumpeq x = y$.
Then obviously, for any $x$ there is a $y$ such that $x = y$, so $\forall x \exists y (x = y)$ is true in $\mathcal{M}$.
However, there is no $y$ such that $x = y$ for all $x$, hence $\exists y \forall x (x = y)$ is false.
• I love your examples. But I can't understand your proof which uses formulars that I can't understand. – Jingguo Yao Apr 15 '17 at 3:04
• @Jingguo Yao What exactly is it you don't understand? – lemontree Apr 15 '17 at 8:18
• In $\mathcal{M} = \langle D, I \rangle$, what do $D$ and $I$ represent? – Jingguo Yao Apr 15 '17 at 8:37
• $D$ represents the universe/the domain of the structure and $I$ the interpretation of the non-logical symbols (= individual constants, relations, functions). – lemontree Apr 15 '17 at 8:57 | 2019-06-25T23:49:56 | {
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http://mathhelpforum.com/advanced-algebra/2206-uncountable-sets-print.html | # Uncountable sets
• Mar 14th 2006, 08:42 AM
kennyb
Uncountable sets
I have a cardinality question. If anyone can give me some guidance on this, I would greatly appreciate it. Here it goes:
If X={0,1} and w={0,1,2,...}, then show X^w and w^w have the same cardinality.
Showing that both these sets are uncountable isn't too bad, but I need them to be the same uncountable size. I'm just not seeing how to set up a bijection between these two sets. I was also thinking that I might be able to show that both sets inject into the reals, then using the continuum hypothesis. Can anyone help?
• Mar 14th 2006, 09:50 AM
CaptainBlack
Quote:
Originally Posted by kennyb
I have a cardinality question. If anyone can give me some guidance on this, I would greatly appreciate it. Here it goes:
If X={0,1} and w={0,1,2,...}, then show X^w and w^w have the same cardinality.
Showing that both these sets are uncountable isn't too bad, but I need them to be the same uncountable size. I'm just not seeing how to set up a bijection between these two sets. I was also thinking that I might be able to show that both sets inject into the reals, then using the continuum hypothesis. Can anyone help?
This will be a bit informal.
First $X^w$ is the set of all infinite sequences of elements drawn
from $\{0,1\}$ and $w^w$ is the set of all sequences of elements
drawn from $\{0,1,2, \dots\}=\mathbb{N}$.
Clearly $\mathcal{C}(X^w) \le \mathcal{C}(w^w)$
Now let $x \in w^w$ then $x=x_1,x_2, \dots$. Now we may write each of the
$x_i$s in unary, that is $n$ is represented by $n\ 1$s, but we will write them in unary+ as
$n+1\ 1$s.
Now map $x \in w^w$ to $y \in X^w$ such that $y$ consists of the unary+
repersentations of the $x_i$s seperated by a single $0$s. This map takes
$w^w$ one-one onto a subset of $X^w$ (in fact the sub-set where there are no
two consecutive $0$s.
Hence:
$\mathcal{C}(X^w) \ge \mathcal{C}(w^w)$
so with the earlier result:
$\mathcal{C}(X^w) = \mathcal{C}(w^w)$
RonL
• Mar 14th 2006, 10:06 AM
ThePerfectHacker
Quote:
Originally Posted by kennyb
I have a cardinality question. If anyone can give me some guidance on this, I would greatly appreciate it. Here it goes:
If X={0,1} and w={0,1,2,...}, then show X^w and w^w have the same cardinality.
Showing that both these sets are uncountable isn't too bad, but I need them to be the same uncountable size. I'm just not seeing how to set up a bijection between these two sets. I was also thinking that I might be able to show that both sets inject into the reals, then using the continuum hypothesis. Can anyone help?
I have an idea but maybe it is useless here. Axiom of Choice, thus, conclude that there exists a surjetive map between $X^w$ and $w^w$.
• Mar 14th 2006, 10:08 AM
CaptainBlack
Quote:
Originally Posted by ThePerfectHacker
I have an idea but maybe it is useless here. Axiom of Choice, thus, conclude that there exists a surjetive map between $X^w$ and $w^w$.
If a result can be proven without AC it should be, and here we can
RonL
• Mar 14th 2006, 10:10 AM
kennyb | 2016-09-28T17:19:19 | {
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https://mathzsolution.com/problems-that-become-easier-in-a-more-general-form/ | # Problems that become easier in a more general form
When solving a problem, we often look at some special cases first, then try to work our way up to the general case.
It would be interesting to see some counterexamples to this mental process, i.e. problems that become easier when you formulate them in a more general (or ambitious) form.
## Motivation/Example
Recently someone asked for the solution of $$a,b,c$$ such that $$\frac{a}{b+c} = \frac{b}{a+c} = \frac{c}{a+b} (=t).$$
Someone suggested writing this down as a system of linear equations in terms of $$t$$ and solving for $$a,b,c$$. It turns out that either (i) $$a=b=c$$ or (ii) $$a+b+c=0$$.
Solution (i) is obvious from looking at the problem, but (ii) was not apparent to me until I solved the system of equations.
Then I wondered how this would generalize to more variables, and wrote the problem as:
$$\frac{x_i}{\sum x – x_i} = \frac{x_j}{\sum x – x_j} \quad \forall i,j\in1,2,\dots,n$$
Looking at this formulation, both solutions became immediately evident without the need for linear algebra (for (ii), set $$\sum x=0$$ so that each denominator cancels out with its numerator).
Consider the following integral $\displaystyle\int_{0}^{1}\dfrac{x^7-1}{\ln x}\,dx$. All of our attempts at finding an anti-derivative fail because the antiderivative isn’t expressable in terms of elementary functions.
Now consider the more general integral $f(y) = \displaystyle\int_{0}^{1}\dfrac{x^y-1}{\ln x}\,dx$.
We can differentiate with respect to $y$ and evaluate the resulting integral as follows:
$f'(y) = \displaystyle\int_{0}^{1}\dfrac{d}{dy}\left[\dfrac{x^y-1}{\ln x}\right]\,dx = \int_{0}^{1}x^y\,dx = \left[\dfrac{x^{y+1}}{y+1}\right]_{0}^{1} = \dfrac{1}{y+1}$.
Since $f'(y) = \dfrac{1}{y+1}$, we have $f(y) = \ln(y+1)+C$ for some constant $C$.
Trivially, $f(0) = \displaystyle\int_{0}^{1}\dfrac{x^0-1}{\ln x}\,dx = \int_{0}^{1}0\,dx = 0$. Hence $C = 0$, and thus, $f(y) = \ln(y+1)$.
Therefore, our original integral is $\displaystyle\int_{0}^{1}\dfrac{x^7-1}{\ln x}\,dx = f(7) = \ln 8$.
This technique of generalizing an integral by introducing a parameter and differentiating w.r.t. that parameter is known as Feynman Integration. | 2023-02-07T18:39:42 | {
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https://math.stackexchange.com/questions/508952/connected-but-it-is-not-continuous-at-some-points-of-i | # Connected, but it is not continuous at some point(s) of I
The problem is "Give an example of a function $f(x)$ defined on an interval I whose graph is connected, but is is not continuous at some point(s) of $I$"
In my idea, a solution is topologist's sin curve.
\begin{gather*} f\colon[0, 1] \to [0, 1] \\\\ f(x) = \begin{cases} \sin(1/x) &(x\neq 0)\\ 0 &(x=0). \end{cases} \end{gather*}
Is this function connected on $\mathbb{R}^2$ space but not continuous at $x=0$ ?
This is a good example. The graph of the topologist's sine curve, which includes the point $(0,0)$ as you have indicated, is indeed connected. However, it is not continuous. To see this, try and produce a sequence of points $x_n$ converging to $0$ for which $\sin(1/x_n) = 1$.
Yes, your function works perfectly. The graph of this curve is connected for the following reason: any open set that contains $(0,0)$ is going to contain another part of the curve. | 2019-05-24T03:26:13 | {
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https://math.stackexchange.com/questions/3689064/technique-for-simplifying-e-g-sqrt-8-4-sqrt3-to-sqrt6-sqrt2 | # Technique for simplifying, e.g. $\sqrt{ 8 - 4\sqrt{3}}$ to $\sqrt{6} - \sqrt{2}$
How to find the square root of an irrational expression, to simplify that root. e.g.:
$$\sqrt{ 8 - 4\sqrt{3} } = \sqrt{6} - \sqrt{2}$$
Easy to verify:
\begin{align} (\sqrt{6} - \sqrt{2})^2 = 6 - 2\sqrt{12} +2 = 8 - 4 \sqrt{3} \end{align}
But how to work it out in the first place? I feel there's a standard technique (Completing-the-square? Quadratic formula?), but don't recall it or what it's called...
BTW: this came up in verifying equivalence of different calculations of $$\cos{75°}$$ (the above divided by $$4$$), as $$\cos{\frac{90°+60°}{2}}$$ vs $$\cos{(45°+30°)}$$, from 3Blue1Brown's lockdown video on complex numbers and trigonometry.
• The term of art (and it really is something of an art) is "denesting radicals". That should help you search for references. – Blue May 24 '20 at 5:36
• perhaps you could assume that the inner expression is a perfect square, then reform so that the inner radical has a coefficient of two. e.g. $\sqrt{8-4\sqrt{3}} = \sqrt{8-2\sqrt{12}}=\sqrt{6-2\sqrt{12}+2}=\dots$ – John Joy May 24 '20 at 6:12
I don't think there's a name for this procedure but let's apply it to $$\sqrt{8-4\sqrt3}$$. If you suspect this equals $$\sqrt a\pm\sqrt b$$ with rationals $$a$$ and $$b$$, then $$8-4\sqrt3=(\sqrt a\pm\sqrt b)^2=(a+b)\pm2\sqrt{ab}$$ so you want to solve simultaneously $$a+b=8$$ and $$-4\sqrt{3}=\pm2\sqrt{ab}$$. So you need the minus sign, and $$ab=12$$. Then $$a$$ and $$b$$ are roots of the quadratic equation $$(X-a)(X-b)=X^2-(a+b)X+ab=X^2-8X+12.$$ This does has rational roots: $$2$$ and $$6$$.
Note the denesting formula
$$\sqrt{a-\sqrt c} = \sqrt{\frac{a+\sqrt {a^2-c}}2}- \sqrt{\frac{a-\sqrt {a^2-c}}2}$$
which can be verified by squaring both sides, and apply it to
$$\sqrt{8-4\sqrt3}=2\cdot \sqrt{2-\sqrt3}= 2\left(\sqrt{\frac32} -\sqrt{\frac12}\right)=\sqrt6-\sqrt2$$
If I recall correctly, you make the assumption that your expression takes the form $$\sqrt{a}\pm\sqrt{b}$$:
$$\sqrt{8-4\sqrt{3}} = \sqrt{a}\pm\sqrt{b}$$
$$8-4\sqrt{3} = 8-\sqrt{48} = a\pm2\sqrt{ab}+b$$
We can see that the irrational part must be assigned the negative sign.
Equating rational and irrational parts:
$$a+b = 8$$
$$-2\sqrt{ab} = -\sqrt{48} \implies ab = 12$$
Then $$a = \frac{12}{b} \implies \frac{12}{b}+b = 8 \implies 12 + b^2 = 8b \implies b^2 - 8b + 12 = (b-6)(b-2) = 0$$
Since we know the answer is positive, take b = 2 and a = 6.
$$\sqrt{8-4\sqrt{3}} = \sqrt{6}-\sqrt{2}$$ | 2021-06-22T10:16:33 | {
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http://www.stefanoricciardi.com/2011/01/10/project-euler-problem-18-in-f/ | Project Euler Problem 18 in F#
2011 brings us yet another Project Euler problem to tackle: this time is Problem 18 one of the most interesting that I have solved so far:
By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.
3
7 4
2 4 6
8 5 9 3
That is, 3 + 7 + 4 + 9 = 23.
Find the maximum total from top to bottom of the triangle below:
75
95 64
17 47 82
18 35 87 10
20 04 82 47 65
19 01 23 75 03 34
88 02 77 73 07 63 67
99 65 04 28 06 16 70 92
41 41 26 56 83 40 80 70 33
41 48 72 33 47 32 37 16 94 29
53 71 44 65 25 43 91 52 97 51 14
70 11 33 28 77 73 17 78 39 68 17 57
91 71 52 38 17 14 91 43 58 50 27 29 48
63 66 04 68 89 53 67 30 73 16 69 87 40 31
04 62 98 27 23 09 70 98 73 93 38 53 60 04 23
NOTE: As there are only 16384 routes, it is possible to solve this problem by trying every route. However, Problem 67, is the same challenge with a triangle containing one-hundred rows; it cannot be solved by brute force, and requires a clever method! ;o)
Analysis
This problem is easily solved resorting to dynamic programming, which we can loosely describe as decomposing the overall problem into easier sub-problems and combining their solutions to find the final answer.
In our case, if we consider the number associated with each node of the triangle as a cost, the overall problem can be expressed as finding the path with maximum cost from the upper vertex to the bottom row. A generic sub-problem would be the cost to reach the bottom row from a generic node at position $$i,j$$.
Given the cost $$c_{i,j}$$ of passing through a given node and the cost $$C_{i,j}$$ of reaching the bottom row from that particular node, we can define the following truth:
$$C_{i,j} = c_{i,j} + max(C_{i+1,j}, C_{i+1,j+1})$$
or, in other words, from each node $$i,j$$ the cost of reaching the bottom is equal to the cost of the node itself plus the maximum path between one of the two adjacent nodes from the row below.
Let's confirm why this is the case with the help of the following figure: consider for example the first node from the left in the fourteenth (last-but-one) row (having cost 63): from there you can move to the bottom either going to node 04 or node 62. Clearly, if we want to pick the highest cost, we need to pick 62 (and we mark this choice with a grey line). The overall cost to reach the bottom from that node is thus 63 + max (04, 62) = 125.
Let's consider now the second node from that row (66). From there, we have to choose between node 62 and 98: the highest cost path from there is thus 66 + max(62, 98) = 164. We can continue for all the nodes in the row to find the highest cost path from each node; the solution to such sub-problems is indicated in gray close to each node.
Having solved all sub-problems for row 14, we can now consider the row immediately above. In the following figure, I have replaced row 14 with the solutions of the sub-problems that we had just solved; the row above is row 13. Please note how we dont care about row 15 anymore: its contribution to the overall problem is already captured in each sub-problems' solution for row 14! Hence, we proceed in a similar way to solve the sub-problems for the row 13 just as we did for row 14:
We continue to follow a similar pattern moving upwards row after row, until we reach the tip of the triangle, always considering two rows at a time.
Solution
Most of the code for this solution is needed just to initialize the two dimensional array containing the triangle. For reading convenience, I have skipped most of the uninteresting rows, but you can find the complete solution on Github.
So, we have a matrix (2D array) for the cost of each nodes and another matrix for the solutions of the sub-problems associated to each node. Trivially, the costs and the solutions coincide for the bottom row (note the handy Array2D.blit function to copy a whole row from one matrix to the other).
Then, with a loop we move backward from the last-but-one row up to the tip of the triangle calculating the sub-problems solutions as explained above.
Updated: | 2018-05-26T21:09:53 | {
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https://questioncove.com/updates/5b2fb11d11d7c2bccc7b3200 | MARC:
Math quest(For fun)
3 weeks ago
MARC:
What is the digit in the unit place of $$2003^{2001}$$?
3 weeks ago
mhchen:
Can I use a calculator? xD
3 weeks ago
MARC:
u can't use calculator X'D U will get infinity X'D
3 weeks ago
mhchen:
I have a super-calculator ;) Anyways I'll try it without it. I'm starting with $2003^{2000}2003^1$
3 weeks ago
mhchen:
I also notice that 3^n has a unit digit pattern of 3 9 7 1 3 9 7 1 3 9 7 1... Basically the pattern is 4 repeating numbers. so 2000 mod 4 is 1, and the 1st number of that pattern is 3. That multiplied by 3 (for the 2003^1) is 9. So my guess will be 9. Am I correct?
3 weeks ago
nuts:
what about the second digit?
3 weeks ago
Zarkon:
fyi 2000 mod 4 is 0
3 weeks ago
Zarkon:
the first 4 digits are 2003
3 weeks ago
MARC:
9 is incorrect but quiet close to the real answer,hehe. Good guess tho,mhchen. nuts,I don't get what u mean. X'D yeah, Zarkon was right. 2000 mod 4 is 0 To find the unit digit you only need to look at the unit digit and the power and not the whole number. So, unit digit of $$2003^{2001}$$ will be same as unit digit of $$3^{2001}$$. Now if we look at the unit digit of powers of any number they follow a cycle i.e. They repeat itself after a fixed interval. Let us look at powers of 3. $$3^0=1$$ $$3^1=3$$ $$3^2=9$$ $$3^3=27$$(unit digit is $$7$$) $$3^4 = 81$$ (unit digit is $$1$$, repetition starts) $$3^5= 243$$(unit digit is $$3$$, again repetition) Likewise we can go on.. The observation is that the unit digit repeats itself at an interval of $$4$$, hence if we divide the power by four and check the unit digit for $$3$$ raised to the power remainder obtained.. we will get our answer. Hence, $$\frac{2001}{4}$$ gives remainder $$1$$ Unit digit of $$2003^{2001}$$ should be same as unit digit of $$3^2001$$ which would be same as $$3^1$$ which is 3. Hence the answer is 3.
3 weeks ago
Zarkon:
you can use a generalization of Fermat's little theorem to get the answer I gave for the first 4 digits ... which is 2003 $2003^{2001}\equiv 2003 \,( \hspace{-.4cm}\mod10000)$ which obviously gives the ones digit. 200$$\color{red}{3}$$
3 weeks ago
MARC:
Wow,that's nice,Zarkon. :D Learn something new today,hehe,thanks. ;) Compute some powers of 2003 mod 10: $$2003^{2}\equiv3^{2}\equiv9(mod~10)$$ so, $$2003^{4}\equiv9^{2}\equiv1(mod~10)$$ Therefore,$$2003^{2001}\equiv(2003^4)^{500}\times2003$$ and $$2003^{2001}\equiv(1)^{500}\times2003(mod~10)$$ $$\equiv1\times2003(mod~10)$$ $$\equiv2003(mod~10)$$ $$\equiv3(mod~10)$$ Zarkon,is my working correct or?
3 weeks ago
Zarkon:
that is correct
3 weeks ago
MARC:
Cool! Thank you,Zarkon! :D lol,thanks guys...
3 weeks ago | 2018-07-20T19:45:14 | {
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https://math.stackexchange.com/questions/1790642/general-formula-for-the-1-5-19-65-211-sequence | # General formula for the 1; 5; 19; 65; 211 sequence
I have got array $1; 5;19; 65; 211$. Can I find general formula for my array?
For example, general formula for array $1; 2; 6; 24; 120$ is $n!$.
I tried a lot for finding the general formula, but I only found recurrent formula: $a_{n+1} = 5a_{n}-6a_{n-1}$.
Any help will be much appreciated.
• – lab bhattacharjee May 18 '16 at 16:53
• Here is one (out of infinitely many). – barak manos May 18 '16 at 17:01
• OEIS has a few entries for this sequence. – Jules May 18 '16 at 18:38
• I would like to reopen the question since my answer below could be a reference for similar questions (what is the next number in$\ldots$), often appearing here on MSE. For instance, it just happened: math.stackexchange.com/questions/1848089/unknown-formula – Jack D'Aurizio Jul 3 '16 at 21:43
• What distinguishes this Question from many of this ilk is the OP's statement of a recurrence relation. It is certainly a well founded problem to solve that recurrence relation (with the given initial conditions) for an explicit formula. – hardmath Jul 3 '16 at 23:34
This kind of problems is always ill-posed, since given any sequence $a_0,a_1,\ldots,a_n$ we are free to assume that $a_k=p(k)$ for some polynomial $p$ with degree $n$, then extrapolate $a_{n+1}=p(n+1)$ through Lagrange's approach or the backward/forward difference operator. A taste of the second approach:
$$\begin{array}{ccccccccc} 1 & & 5 & & 19 & & 65 & & 211\\ & 4 && 14 && 46 && 146\\ & & 10 && 32 && 100 && \\ &&& 22 && 68 &&&\\ &&&& 46\end{array}$$ by applying four times the difference operator, we reach a constant polynomial, hence we may re-construct $p(n+1)$ this way:
$$\begin{array}{cccccccccc} \color{green}{1} & & 5 & & 19 & & 65 & & 211&&\color{purple}{571}\\ & \color{green}{4} && 14 && 46 && 146&&\color{red}{360}\\ & & \color{green}{10} && 32 && 100 && \color{red}{214}& \\ &&& \color{green}{22} && 68 &&\color{red}{114}&&\\ &&&& \color{green}{46}&&\color{red}{46}&&&\end{array}$$ and $\color{purple}{571}$ is a perfectly reasonable candidate $a_{n+1}$, like $$a_n = 1-\frac{31 n}{6}+\frac{181 n^2}{12}-\frac{47 n^3}{6}+\frac{23 n^4}{12}=\color{green}{46}\binom{n}{4}+\color{green}{22}\binom{n}{3}+\color{green}{10}\binom{n}{2}+\color{green}{4}\binom{n}{1}+\color{green}{1}$$ is a perfectly reasonable expression for $a_n$.
The Berlekamp-Massey algorithm is designed for solving the same problem under a different assumption, namely that $\{a_n\}_{n\geq 0}$ is a linear recurring sequence with a characteristic polynomial with a known degree. In your case you already know the characteristic polynomial $x^2-5x+6=(x-2)(x-3)$, hence you just have to find the coefficients $A,B$ fulfilling $$a_n= A\cdot 2^n+B\cdot 3^n$$ and by considering that $a_0=1,a_1=5$ we get $\color{red}{a_n = 3^{n+1}-2^{n+1}}$.
• That's true for a certain value of the word "reasonable". :) Your point about these questions being ill-posed is well-taken, and yet there is so often consensus agreement on what is meant. – G Tony Jacobs May 18 '16 at 17:19
• Nice approach!!! – SiXUlm May 18 '16 at 17:25
• What if I say that a solution is "nicer" iff it's "shorter" than others? We can do better. Let's say that our "elementary functions" are polinomials and exponentials. A solution could be the nicest if it's the shortest. This might give solution he was waiting for. – Ivan Di Liberti May 18 '16 at 21:22
Are you simply looking for an explicit formula, or a way to derive it yourself? $$a_n = 3^n-2^n$$
You have $a_{n+1}=5a_n-6a_{n-1}$. This kind of recurrence tends to happen with exponential-type series. Thus, assume $a_n=r^n$, and plug in:
$a_{n+1}-5a_n+6a_{n-1}=0 \\ r^{n+1} - 5r^n+6r^{n-1}=0$
Divide through by $r^{n-1}$, and you get a quadratic in $r$:
$r^2-5r+6=0$
This is solved by $r=2$ and $r=3$, so you look for a sequence of the form:
$a_n=P\cdot 2^n + Q\cdot 3^n$ for some real coefficients $P$ and $Q$. You can find them by plugging in the first couple of terms in your series, thus producing a linear system in $P$ and $Q$.
In the answer given by @SiXUlm, we note another recurrence for this sequence:
$a_n = 3a_{n-1}+2^n$ for $n\geq2$, with $a_1=1$
You can also get the formula by solving this recurrence. We write it as:
$a_n - 3a_{n-1} = 2^n$
and then solve the related recurrence:
$a_n - 3a_{n-1} = 0$
Using the technique from above, we get $r=3$ and $a_n=P\cdot3^n$. Then, because of the $2^n$ that we were just ignoring, we throw in a $Q\cdot 2^n$ term to account for its effect. Thus, we get the same answer as above.
$5 = 1*3 + 2, 19 = 5*3 + 2^2, 65 = 19*3 +2^3, 211=65*3+2^4$, etc
If $a_0 = 1$, then $a_n = 3 \times a_{n-1} + 2^n$.
• How to derive this in general? – lab bhattacharjee May 18 '16 at 16:55
• I don't know. It depends very much on the problem. – SiXUlm May 18 '16 at 17:01
• That's an interesting recurrence @SiXUlm. How did you notice it? – G Tony Jacobs May 18 '16 at 17:07
• I noticed when I divide the one term by its successive, I obtain the result is bigger than $3$, but never $4$ (except the first term), so I checked the difference $a_{n+1} - 3a_n$ and saw the pattern $2,4,8,16$. – SiXUlm May 18 '16 at 17:13
This refers to @Jack d'Aurizio's second ansatz, just to make it explicite. It is a method which I use if I suspect that my sequence has a recursive structure.
Example in Pari/GP
v=[1,5,19,65,211] \\ initialize a row-vector with values of the sequence
Now recursion means, that we have some transfer [1,5,19] -> [5,19,65] or [1,5,19] -> [19,65,211] by some transfermatrix T by [...] * T = [...]. To be able to find T by a matrix-inversion the brackets in [...] -> [...] should be (quadratic) matrices and not only vectors. So I construct a source-matrix Q and use the maximal possible dimension first:
Q=matrix(3,3,r,c,v[r-1+c]) \\ "source"-matrix with maximal dimension
\\ all entries of v are used
%291 =
[1 5 19]
[5 19 65]
[19 65 211]
First test, whether we really need dimension 3. If the matrix is singular, we only need a smaller dimension for the recursion:
matrank(Q)
%292 = 2
Well, the rank of the matrix is only 2, so we need to do everything with 2x2-matrices only:
Q=matrix(2,2,r,c,v[r-1+c])
%293 =
[1 5]
[5 19]
Now we define the target-matrix Z which should be a "rightshift" of Q by one column:
Z=matrix(2,2,r,c,v[1+r-1+c])
%294 =
[5 19]
[19 65]
From this we can compute the needed transfermatrix T to allow Q*T=Z
T = Q^-1 * Z
%295 =
[0 -6]
[1 5]
The transfermatrix contain the solution which is also known by the earlier answers: $a_{k+1} = -6 a_{k-1} + 5 a_k$ . Powers of $T$ should transfer more positions in $v$:
Q * T
%296 =
[5 19]
[19 65]
Q * T^2
%297 =
[19 65]
[65 211]
Q * T^3
%298 =
[65 211]
[211 665]
and so on ...
Of course this can simply be generalized to higher dimensions. And if in the problem the rank of the initial matrix had been 3 and no more entries in v had been given, we had been lost in the well known arbitrariness...
Remark: we could do even more. When we diagonalize the transfermatrix T then we can even find the "Binet-type" solutions with some exponential-formula, where the elements of the generalized sequence in v can be directly computed putting the index into the exponent of a monomial, and can thus often be generalized to fractional and even complex sequence-indexes . (As might be known from the Fibonacci-numbers and their Binet-formula - see wikipedia)
The sequence you have found is a generalization of the Fibonacci sequence.
There have been many extensions of the sequence with adjustable (integer) coefficients and different (integer) initial conditions, e.g., $f_n=af_{n-1}+bf_{n-2}$. (You can look up Pell, Jacobsthal, Lucas, Pell-Lucas, and Jacobsthal-Lucas sequences.) Maynard has extended the analysis to $a,b\in\mathbb{R}$, (Ref: Maynard, P. (2008), “Generalised Binet Formulae,” $Applied \ Probability \ Trust$; available at http://ms.appliedprobability.org/data/files/Articles%2040/40-3-2.pdf.)
We have extended Maynard's analysis to include arbitrary $f_0,f_1\in\mathbb{R}$. It is relatively straightforward to show that
$$f_n=\left(f_1-\frac{af_0}{2}\right) \frac{\alpha^n-\beta^n}{\alpha-\beta}+\frac{f_0}{2} (\alpha^n+\beta^n)$$
where $\alpha,\beta=(a\pm\sqrt{a^2+4b})/2$.
The result is written in this form to underscore that it is the sum of a Fibonacci-type and Lucas-type Binet-like terms. It will also reduce to the standard Fibonacci and Lucas sequences for $a=b=1 \ \text{and} \ f_0=0, f_1=1$.
The analysis follows Maynard almost exactly and I can expand upon this or provide a brief manuscript. This analysis is valid for any $a,b,f_0,f_1\in\mathbb{R}$. Notice that only when $a=b=1$ does the ratio of successive terms approach the golden ratio $\Phi$ for large $n$. I haven't fully explored the limiting ratio for the general case, but I find for positive $a,b$ that the limiting ratio is given by $\alpha$.
After all is said and done, your sequence with $f_0=1, f_1=5, \alpha=3, \text{ and } \beta=2$, comes down to
$$f_n=\frac{5}{2}(3^n-2^n)+\frac{1}{2}(3^n+2^n)=3^{n+1}-2^{n+1},\ \ \ n=0,1,2,3...\\ \lim_{n\to \infty}\frac{f_{n+1}}{f_n}=3$$
which is essentially the same the result $(3^n-2^n)$ noted earlier on this page. However, now you have the tools to tackle many additional problems of this type.
Disclosure: This post is was derived largely from a previous answer of mine in Decimal Fibonacci Number? | 2019-09-19T08:31:30 | {
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http://gasa.ecb.bt/bykv5rq5/f7ce6d-convergence-in-distribution-sum-of-two-random-variables | By convergence in distribution, each of these characteristic functions is known to converge and hence the characteristic function of the sum also converges, which in turn implies convergence in distribution for the sum of random variables. Y = 5X−7 . Just hang on and remember this: the two key ideas in what follows are \convergence in probability" and \convergence in distribution." Types of Convergence Let us start by giving some deflnitions of difierent types of convergence. 1 Convergence of Sums of Independent Random Variables The most important form of statistic considered in this course is a sum of independent random variables. Determine whether the table describes a probability distribution. We say that the distribution of Xn converges to the distribution of X as n → ∞ if Fn(x)→F(x) as n … Proposition 1 (Markov’s Inequality). And if we have another sequence of random variables that converges to a certain number, b, which means that the probability distribution of Yn is heavily concentrated around b. Y = X2−2X . This follows by Levy's continuity theorem. It is easy to get overwhelmed. 2. Convergence in Distribution Basic Theory Definition Suppose that Xn, n ∈ ℕ+ and X are real-valued random variables with distribution functions Fn, n ∈ ℕ+ and F, respectively. 1 Convergence of random variables We discuss here two notions of convergence for random variables: convergence in probability and convergence in distribution. The notion of independence extends to many variables, even sequences of random variables. In general, convergence will be to some limiting random variable. In that case, then the probability distribution of the sum of the two random variables is heavily concentrated in the vicinity of a plus b. There are several different modes of convergence. In the case of mean square convergence, it was the variance that converged to zero. So what are we saying? Example 1. S18.1 Convergence in Probability of the Sum of Two Random Variables We begin with convergence in probability. 8. Let X be a non-negative random variable, that is, P(X ≥ 0) = 1. However, this random variable might be a constant, so it also makes sense to talk about convergence to a real number. by Marco Taboga, PhD. A biologist is studying the new arti cial lifeform called synthia. The random variable x is the number of children among the five who inherit the genetic disorder. convergence of random variables. Then, the chapter focuses on random variables with finite expected value and variance, correlation coefficient, and independent random variables. This lecture discusses how to derive the distribution of the sum of two independent random variables.We explain first how to derive the distribution function of the sum and then how to derive its probability mass function (if the summands are discrete) or its probability density function (if the summands are continuous). Probability & Statistics. Find the PDF of the random variable Y , where: 1. It says that as n goes to infinity, the difference between the two random variables becomes negligibly small. 5.2. Sums of independent random variables. A sum of discrete random variables is still a discrete random variable, so that we are confronted with a sequence of discrete random variables whose cumulative probability distribution function converges towards a cumulative probability distribution function corresponding to a continuous variable (namely that of the normal distribution). 1.1 Convergence in Probability We begin with a very useful inequality. 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Find the PDF of the random variable Y, where: 1 hang. And independent random variables the distribution of the random variable, that is P! In general, convergence will be to some limiting random variable might be constant! By giving some deflnitions of difierent types of convergence let us start giving. | 2021-12-01T06:51:01 | {
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https://math.stackexchange.com/questions/1891454/solving-inequality-for-convex-functions | # Solving inequality for convex functions.
A function $f : I \rightarrow \mathbb R$ on an interval $I$ is convex if $f((1-t)x+ty)\le (1-t)f(x)+tf(y), \forall x,y \in I, t \in [0,1]$
Assume now that $I$ is an open interval. Show that if $f$ is convex then for each $c \in I$ there exists $m \in \mathbb R$ such that $m(x − c) + f(c) \le f(x)$ for all $x \in I$ , and if in addition $f$ is differentiable at $c$ then $f'(c)$ is the unique $m$ that works. In general, must m be unique?
(we have previously shown that convex functions are continuous)
I think I'm missing the point of the question.
Surely, by the mean value theorem we can find some $y\in (x,c)$ such that $f'(y)=\frac{f(x)-f(c)}{x-c}$ setting $m=f'(y)$ we get $m=\frac{f(x)-f(c)}{x-c}, m(x-c)+f(c)=f(x)$ such an $m$ fits the inequality we're asked to show.
I also cannot see why $f'(c)$ would be the unique solution or why we would ever get a unique solution of the inequality at all for that matter.
Clearly I'm misunderstanding the question, so if someone could point out where I would appreciate that.
Thank you
• Have you tried drawing a picture? Try drawing a line $m(x-c)+f(c)$ (i.e. a line through $(c,f(c))$) that is never above the graph of $f$ on the interval $I$. If $f$ is differentiable at $c$ then there is only one such line: the tangent line (i.e. the line with $m=f'(c)$). – smcc Aug 13 '16 at 21:54
• I should have said: "if $f$ is differentiable and convex then there is only one such line". You did not use convexity yet in your attempt at answering the question. – smcc Aug 13 '16 at 22:03
• Actually now that I think about it more, what's the significance in it being convex? – Aka_aka_aka_ak Aug 13 '16 at 22:07
• The question is really getting you to show that for differentiable functions convexity can be characterized as the property that the graph of the function never lies below its tangent lines. – smcc Aug 13 '16 at 22:10
As smcc pointed out in their comment, we can think of $y=m(x-c)+f(c)$ as a line passing through the point $\left(c, f(c)\right)$ with slope $m$. Let's assume that this line has a second intersection with the curve $f(x)$: $\left(b, f(b)\right)$. Then, we can rewrite the equation of the line using the fact that $\displaystyle m=\frac{f(b)-f(c)}{b-c}$: $$y=\frac{f(b)-f(c)}{b-c}(x-c)+f(c)$$ We can use the substitution $\displaystyle t=\frac{x-c}{b-c}$ to get the following equation:
$$y=tf(b)+(1-t)f(c)$$
Now, we use the convex condition which was mentioned in the OP: $$f(tb+(1-t)c)\leq tf(b)+(1-t)f(c) \text{ for }t\in[0,1]$$
Unless $f(x)$ is the equation of a line, equality is only reached when $t=0$ or $t=1$. Ergo, for all $x$ between $b$ and $c$, $m(x-c)+f(c)>f(x)$. The only case in which this does not happen is when the line $y=m(x-c)+f(c)$ has no second intersection. In other words, $m=f'(c)$.
In the case that $f(x)$ is a line, the question should be easy to answer.
• @user307463 Although I appreciate that you've accepted my answer, I would like to point out that I never proved that such an $m$ exists for general convex functions. You should read through A.G.'s answer for a proof of existence. – Hrhm Aug 14 '16 at 22:41
Here is an idea how you can prove it.
1. Take a point $a\in I$ to the left of $c$ and a point $b\in I$ to the right, i.e. $a<c<b$, and draw two secant lines: $\ell_a$ through $(a,f(a))$ and $(c,f(c))$ and $\ell_b$ through $(c,f(c))$ and $(b,f(b))$. Denote $k_a$ and $k_b$ their slopes respectively.
2. By convexity, the graph of $f$ is under $\ell_a$ on $[a,c]$ and above it outside $[a,c]$. Similarly, the graph of $f$ is under $\ell_b$ on $[c,b]$ and above it outside $[c,b]$. It gives, in particular, that $k_a\le k_b$.
3. Let $a\to c^-$. The slope $k_a$ is monotonically increasing (again by convexity) and bounded from above by $k_b$, hence, converges to some value $M_1$. Similar argument when $b\to c^+$ gives that $k_b\to M_2$.
4. Since $k_a\le k_b$ we get $M_1\le M_2$. Now any $m\in[M_1,M_2]$ works.
5. If $f$ is differentiable at $c$ then $M_1=M_2$.
P.S. The interval $[M_1,M_2]$ is called subdifferential of $f$ at $c$.
In your argument, the point $y\in (x,c)$ you found by the mean value theorem depends on $x$, hence $m=f'(y)$ also does. However, the value of $m$ in the claim must be the same for all $x\in I$. | 2019-11-14T11:34:43 | {
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https://math.stackexchange.com/questions/2055106/four-equations-in-four-unknowns | # Four Equations in Four Unknowns
Four Equations in Four Unknowns
Completely solve the following equation!
$$\begin{eqnarray} x&+&y&+&z&+&w&=&10 \\ x^2&+&y^2&+&z^2&+&w^2&=&30 \\ x^3&+&y^3&+&z^3&+&w^3&=&100\\ &&&&&&xyzw&=&24 \end{eqnarray}$$
This is problem 3 on page 3 of
Mathematical Quickies
270 Stimulating Problems with Solutions
by Charles W. Trigg
Dover Publications, Inc., New York
ISBN 0-486-24949-2
Here is the solution from page 78 of the book:
By inspection $(1,2,3,4)$ is a solution if the first and fourth equation and satisfies the second and third equations. Since the equations are symmetrical in $x$, $y$, $z$, $w$ the other $23$ permutations of $1,2,3,4$ are solutions also. But these are all the solutions, since the product of the degrees of the equations is 4!
I assume that the exclamation mark at the end of the last sentence is a factorial symbol because the product of the degree of the equation is 24. This seems to be a property that is similar to the fact that a univariate polynomial of degree $n$ has at most $n$ zeroes. But I can't see how to generalize this to multivariate equations.
Why does this system of equations where the product of the degree of the equations is 24 has at most 24 solutions?
• I am interested to hear what others have to say about this question. I wouldn't have expected such a convenient result to be true myself. For clarification, are we restricting the domain of the solutions? $x^2+y^2=1$ has infinitely many solutions (the unit circle)... – JMoravitz Dec 12 '16 at 7:42
• @JMoravitz: thank you, I corrected the typo. I cited the full problem text. I think the domain is $\mathbb{C}^4$ – miracle173 Dec 12 '16 at 7:50
• @JMoravitz: you have two variables but one equation. If you have two equations then you may have only finitely many solutions. circle and line: product of degree is 2, number of solutions is two. Circle and ellipsis: product of degee is 4, number of solutions can be 4. So I think something like $n$ equations in $n$ variables, product of degree is $k$, then there are at most $k$ solutions. – miracle173 Dec 12 '16 at 7:54
• This is a consequence of Bezout's Theorem. en.wikipedia.org/wiki/B%C3%A9zout's_theorem – Leon Sot Dec 12 '16 at 8:00
• this is a nice problem.............+1 – Bhaskara-III Dec 12 '16 at 8:10
I. Yes, you are correct that a univariate polynomial of degree $n$ has $n$ zeros (counting multiplicity). However, your system are just the roots of a univariate in disguise.
The clue is the elementary symmetric polynomials $x+y+z+w$ and $xyzw$. If these unknowns are indeed the roots of the quartic, $$F(u)=u^4+au^3+bu^2+cu+d=0$$ then,
\begin{aligned} x+y+z+w &=10 = -a\\ x^2+y^2+z^2+w^2 &=30 = a^2-2b\\ x^3+y^3+z^3+w^3 &=100=-a^3 + 3 a b - 3 c\\ xyzw &=24=d \end{aligned}\tag1
It is easy to solve for $a,b,c,d\,$ above and we get, $$F(u)=u^4 - 10u^3 + 35u^2 - 50u + 24 = 0$$ $$F(u)=(u - 1)(u - 2)(u - 3)(u - 4)=0$$ hence these, including their permutations, are all the solutions.
II. Let $n_r$ be the number of roots and $n_d$ be the product of the degrees. By Bezout's theorem, then $n_r$ is at most equal to $n_d$. The observation that your example has $n_r=n_d=24$ is just a peculiarity of the system. If we tweak it slightly,
\begin{aligned} x+y+z+w &=10 = -a\\ x^2+y^2+z^2+w^2 &=30 = a^2-2b\\ \color{blue}{x^5+y^5+z^5+w^5} &=100=-a^5 + 5 a^3 b - 5 a b^2 - 5 a^2 c + 5 b c + 5 a d\\ xyzw &=24=d \end{aligned}\tag2
Solving for $a,b,c,d,\,$ they turn out to be rational so, $$F(u)=u^4 - 10u^3 + 35u^2 - \color{blue}{\tfrac{602}{13}}u + 24 = 0$$ though $F(u)$ is no longer rationally factorable. Like the previous, by including the permutations of the $u_i$, there are again $n_r=24$ roots, but the product of the degrees is different now as $n_d=1\times2\times5\times4 = 40$ so $n_r\neq n_d$.
• Hi @Tito, Thank you for the answer. I think your first answer is what the the author of the recreational math problem has in his mind. – miracle173 Dec 13 '16 at 15:35
• @miracle173: You're welcome. Incidentally, If you recall, it is also the case that $\sum_{n=1}^k n^3 = \big(\sum_{n=1}^k n\big)^2$ which explains his value $(1+2+3+4)^2 = 10^2 = 100$. – Tito Piezas III Dec 14 '16 at 1:33 | 2019-08-24T02:26:49 | {
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http://math.stackexchange.com/questions/33743/fix-k-geq-2-convergence-of-sum-frac1nk/33751 | # Fix $k \geq 2$: convergence of $\sum \frac{1}{n^k}$?
I have a lot of sum questions right now ... could someone give me the convergence of, and/or formula for, $\sum_{n=2}^{\infty} \frac{1}{n^k}$ when $k$ is a fixed integer greater than or equal to 2? Thanks!!
P.S. If there's a good way to google or look up answers to these kinds of simple questions ... I'd love to know it...
Edit: Can I solve it by integrating $\frac{1}{x^k}$ ? I can show it converges, but to find the formula? Is my question just the Riemann Zeta function?
(edit)
Thanks guys! This got me the following result:
$\sum_{p} \frac{1}{p} \log \frac{p}{p-1} ~ ~ \leq ~ \zeta(2)$
summing over all primes $p$. (And RHS is Riemann zeta function.)
First, sum over all integers $p$ instead of primes. Then transform the log into $\sum_{m=1}^{\infty} \frac{1}{m p^{m}}$ (reference: wikipedia. I know.). Now we have (with rearranging):
$\leq ~ \sum_{m=1}^{\infty} \frac{1}{m} \sum_{p=2}^{\infty} \frac{1}{p^{m+1}}$
By the result of this question (Arturo's answer), this inner sum, which is $\zeta(m+1)$ is at most $\frac{1}{m+1-1} = \frac{1}{m}$. So we have
$\leq ~ ~ \sum_{m=1}^{\infty} \frac{1}{m} \frac{1}{m} = \zeta(2)$
I think this is a very pretty little proof. Thanks again, hope you math people enjoyed reading this....
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Try an integral test. You might also be interested in the Basel problem (don't miss Robin Chapman's notes linked to right at the bottom of the page) and the Riemann zeta function. – t.b. Apr 19 '11 at 1:34
If these are homework questions (as it seems to me, if you have a lot of them), you'll want to add the 'homework' tag. – Michael Chen Apr 19 '11 at 1:35
Thanks Michael, Theo. I'm actually doing research for a class project. I'm not getting any points for solving them or not (actually it's all just to try to prove a slightly tighter bound), so I didn't tag, hope that's ok. – bo1024 Apr 19 '11 at 1:43
joriki's answer is useful. – J. M. Apr 19 '11 at 1:57
Yes, this sum converges, and yes, you can prove this using the integral test. For $k$ even the exact value (plus $1$) turns out to be a rational multiple of $\pi^k$ (see, for example, Wikipedia), but for odd $k$ the exact values are much more mysterious.
The best way to find answers to these simple questions is to learn the basic techniques for solving them. In the context of discussing the convergence of series this means learning the basic convergence tests, as well as the knowledge of several specific examples such as might be covered in a typical calculus course. Beyond that, questions like this are hard to google for, so your best option is to ask them here!
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Thank you. I'm in the unfortunate position of having forgotten most of these techniques just when I need them again, so this is helping me pick it back up! Thanks for the answer! – bo1024 Apr 19 '11 at 2:42
The fact that these series converge follows from the integral test; their values are somewhat more complicated, as indicated by the comments: they correspond to evaluating Riemann's zeta function at $k$, and this is highly nontrivial, even for positive integral $k$ (called the zeta constants). There are formulas for even integer values of $k$, though in terms of Bernoulli numbers. Note that your series is actually one less than the zeta constants, since $$\zeta(k) = \sum_{n=1}^{\infty}\frac{1}{n^k}$$ and your sum starts with $n=2$, not $n=1$.
As mentioned, the Integral Test shows these series converge. But to do it explicitly:
Consider the sequence of partial sums $$s_n = \sum_{r=2}^n \frac{1}{r^k}.$$ Since all summands are positive, this is an increasing sequence: $$s_2\leq s_3\leq s_4\leq\cdots$$ so the convergence of the sequence of partial sums is equivalent to showing that the sequence is bounded.
To show the sequence is bounded, consider the function $\frac{1}{x^k}$. This function is decreasing, so if you approximate $$\int_1^b\frac{1}{x^k}\,dx$$ using a right hand sum, you will get an underestimate for the integral.
Consider $$\int_1^{n}\frac{1}{x^k}\,dx,$$ and approximate it using a right hand sum with the interval $[1,n]$ divided into $n-1$ equal parts (so we break up the integral into $[1,2]$, $[2,3],\ldots,[n-1,n]$). The right hand sum with that partition is given by: $$\mathrm{Right Sum} = \frac{1}{2^k} + \frac{1}{3^k} + \cdots + \frac{1}{n^k} = s_{n}.$$ Since right hand sums are underestimates, then for every positive integer $n$ we have that: $$s_n \leq \int_1^n\frac{1}{x^k}\,dx = \frac{1-n^{1-k}}{k-1}.$$ Since the bounds get larger for larger $n$, we have: $$s_n\leq \frac{1-n^{1-k}}{k-1} \leq \lim_{n\to\infty}\frac{1 - n^{1-k}}{k-1} = \frac{1}{k-1}.$$ So the sequence of $s_n$ is bounded above. Since it is increasing, the sequence of partial sums converges.
Since the sequence of partial sums converges, the series converges as well.
(This is the essence of the Integral Test: if you have a series $$\sum a_n$$ such that there is a positive, continuous, decreasing function $f(x)$ such that $f(n)=a_n$ for all $n$, then the convergence of the series is equivalent to the convergence of the improper integral $$\int_1^{\infty}f(x)\,dx$$ in the sense that if the integral converges, then the series converges; and if the integral diverges, then the series diverges. Of course, you can change the lower limit of the integral if necessary.)
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Wow, thanks for the thought-out answer. To clarify, this also proves that the sequence converges to something less than $\frac{1}{k-1}$, right? Thanks again! – bo1024 Apr 19 '11 at 2:04 @b01024: The argument shows the limit of the series is at most $\frac{1}{k-1}$, but not necessarily something strictly less than $\frac{1}{k-1}$. In fact, the limit is striclty smaller, but the argument above does not show it. – Arturo Magidin Apr 19 '11 at 2:09 Thanks, and thank you for the precision in answering (I was sloppy). – bo1024 Apr 19 '11 at 2:11 If you are interested, this allowed me to show that $\sum_{p} \frac{1}{p} \log \frac{p}{p-1}$ , I'll edit my question to show the steps! – bo1024 Apr 19 '11 at 2:15
You are looking for the Riemann zeta function $\zeta(k)$ (or close to it: the sum usually starts at $n=1$).
Since you are supposed to be doing research for a class project, perhaps you should search for it.
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Gotcha. That's really interesting because my problem is related to counting the primes.... Thanks! – bo1024 Apr 19 '11 at 1:52
It's also easy to show that this series converges through more elementary means (though without getting a good estimate of the value), by a version of the same elementary technique often used to show that the harmonic series diverges. Start by fixing $k=2$; since the terms for any other $k$ are smaller than this, then clearly if this series converges the others will. Next, break it into chunks of length $2^n$: $S = {1\over 1^2} + ({1\over 2^2}+{1\over 3^2}) + ({1\over 4^2}+{1\over 5^2}+{1\over 6^2}+{1\over 7^2}) + \ldots$ Now, replace each of the terms in the sets of parentheses by the first term there; since we know that e.g. ${1\over 7^2} \lt {1\over 4^2}$, our sum will be bounded by the result; $S\lt S'$, where $S' = {1\over 1^2} + ({1\over 2^2}+{1\over 2^2}) + ({1\over 4^2}+{1\over 4^2}+{1\over 4^2}+{1\over 4^2}) + \ldots$ - and here it's easy to see that, for instance, the $4$ terms of $1\over 4^2$ will add up to $1\over 4$, the (unshown) $8$ terms of $1\over 8^2$ will add up to $1\over 8$, etc; so $S'$ is just $1+{1\over 2}+{1\over 4}+{1\over 8}+\ldots = 2$, and so we know that the original series $S$ converges (and in fact that its value is less than $2$).
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The easiest way to prove that $1+\frac{1}{2^2}+...+\frac{1}{n^2}$ is convergent, is proving by induction that
$$1+\frac{1}{2^2}+...+\frac{1}{n^2} \leq 2 -\frac{1}{n} \,.$$
If $k >2$, convergence follows immediately from $\frac{1}{j^k} \leq \frac{1}{j^2}$.
I always find fascinating the fact that the above inequality is a trivial induction problem, while the easier inequality: $1+\frac{1}{2^2}+...+\frac{1}{n^2} \leq 2 \,.$ cannot be proven by induction....
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This is one interesting example where proving more is easier. But I prefer the less mysterious argument, where we upper bound $\frac{1}{n^2}$ by $\frac{1}{n(n-1)}$ and sum the latter series easily by telescoping. Is this what you mean by "proving by induction"? – Srivatsan Jul 26 '11 at 11:22 | 2013-05-24T07:48:02 | {
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https://math.stackexchange.com/questions/2530121/substitution-in-congruence-relations | # Substitution in congruence relations
I'm a noob to discrete math, and have a basic question for a sanity check. From studying basic modular arithmetic, the following are properties of congruences:
1) $a \equiv a \pmod{n}$
2) $a \equiv b \pmod{n} \implies b \equiv a \pmod{n}$
3) $a \equiv b \pmod{n}$ and $b \equiv c \pmod{n} \implies a \equiv c \pmod{n}$
4) $a \equiv b \pmod{n} \implies a+c \equiv b+c \pmod{n}$
5) $a \equiv b \pmod{n} \implies ac \equiv bc \pmod{n}$
6) $a \equiv b \pmod{n}$ and $c \equiv d \pmod{n} \implies a+c \implies b+d \pmod{n}$
7) $a \equiv b \pmod{n}$ and $c \equiv d \pmod{n} \implies ac \equiv bd \pmod{n}$
Powers of integers are just multiplications. More formally, property 5 plus induction can be used to prove:
8) $a \equiv b \pmod{n} \implies a^k \equiv b^k \pmod{n}$ for any nonnegative integer $k$
In addition, there is a "cancellation" lemma that states that: if $p$ is prime $k$ is not a multiple of $p$ by Fermat's little theorem, then
9) $ak \equiv bk \pmod{p} \implies a \equiv b \pmod{p}$
Here's my question, looking at the proof of https://en.wikipedia.org/wiki/Euler%27s_criterion
If there exists integer $x$ such that $a \equiv x^2 \pmod{p}$, and $a$ is coprime to $p$, then
$a^{(p-1)/2} \equiv (x^2)^{(p-1)/2} \equiv x^{p-1} \equiv 1$
My question is that I am confused about the substitution of $a$ with $x^2$. If $a=x^2$, then it's obvious we can make substitutions like this. But since in this case $a \equiv x^2$, I'm confused about when it's allowed to make substitutions in congruence relations? None of the properties of congruences 1-8 applies to "substitutions."
The closest I can come to "substitution" in congruence relations is: Suppose that:
A) $ab \equiv c \pmod{p}$, where $p$ is a prime
B) $d \equiv b \pmod{p}$, where $p$ is a prime
Then multiply A) and B) as in property 7, I get:
$adb \equiv bc \pmod{p}$
Now using the cancellation property for modulo a prime (property 9), I can cancel $b$ to get:
$ad \equiv c \pmod{p}$
which is as if I simply substituted $b$ from B) into A).
So, because property 9 only applies to modulo a prime, does my apparent substitution only work since A) and B) are modulo a prime?
Does it also mean the substitution of $a$ with $x^2$ in the Euler's criterion only valid since Euler's criterion is modulo a prime? Or, am I confused about "substitutions" somehow? I'm actually unsure if the substitution of $a$ with $x^2$ in Euler's criterion proof is actually a real substitution since we're substituting into an expression $a^{(p-1)/2}$ rather an congruence relation where $a^{(p-1)/2}$ is already congruent to some other expression.
• The possibility of substitutions is not specific to congruences, it is a general property of mathematical logic. As to cancellation, you can cancel $b$ even if the modulus is not prime. The condition is that $b$ be coprime to the modulus, because in this case, $b$ has a modular inverse, by Bézout's theorem. Nov 21 '17 at 2:04
There is nothing wrong here: you are using property $8$ with $b=x^2$ and $k=\frac{p-1}{2}$, which is an integer since $p$ is an odd number.
If $a\equiv b \pmod c$ then for every $n\in \Bbb N$ we have $a^n\equiv b^n \pmod c.$ There are several ways to prove this.
Method 1. Use the Binomial Theoerem. If $a\equiv b\pmod c$ then $b=a+kc$ for some $k\in \Bbb Z$. So for $2\leq n\in \Bbb N$ we have $$-a^n+b^n=-a^n+(a+kc)^n=-a^n+\sum_{j=0}^na^{n-j}(kc)^j\binom {n}{j}=$$ $$=-a^n+a^n+c\sum_{j=1}^na^{n-j}k^jc^{j-1}\binom {n}{j}=c\sum_{j=1}^na^{n-j}k^jc^{j-1}\binom {n}{j}$$ which is an integer multiple of $c.$ So $\;-a^n+b^n\equiv 0\pmod p.$
Method 2. Use induction on $n\geq 2.$ First, by your Rule # (7) with $c=a$ and $d=b$ we have $a\equiv b\pmod c\implies a^2\equiv b^2\pmod c.$
Second, if $2\leq n\in \Bbb N$ and $a\equiv b\pmod c$ and $a^n\equiv b^n \pmod c$ then by Rule #(7) with $c=a^n$ and $d=b^n$ we have $$a^{n+1}\equiv ac\equiv bd\equiv b^{n+1} \pmod c.$$ | 2021-09-24T16:07:58 | {
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https://math.stackexchange.com/questions/2918512/simplifying-log-e2e4aae4a/2918533 | # Simplifying $\log_{e^2}(e^{4a}+ae^{4a})$
## Problem
Simplify logaritm:
$$\log_{e^2}(e^{4a}+ae^{4a})$$
preferably in a way that end result contains only natural logarithm.
## Attempt to solve
I know few computational rules about logarithms:
$$\log_a(xy) = \log_a(x)+\log_a(y)$$
$$\log_a(\frac{x}{y})=\log_a{x}-\log_a(y)$$
$$\log_a(x^n)=n\log_a(x)$$
And formula for change of basis :
$$\log_a(x)=\frac{\log_b(x)}{\log_b(a)}$$
There is sum inside this logarithm and it appears we don't have formula for this. On wikipedia i found two formulas about summation / substraction inside logarithm:
$$\log_b(a+c)=\log_b a+ \log_b (1+\frac{c}{a})$$
$$\log_b(a-c)=\log_b a+ \log_b (1-\frac{c}{a})$$
I have no former experience / knowledge of these formulas but for now the assumption is these are correct.
I would try to first try change of basis to $\log_{e^2}()\rightarrow \log_e() = \ln()$
$$\log_{e^2}(e^{4a}+ae^{4a})=\frac{\log_e(e^{4a}+ae^{4a})}{\log_e(e ^2)}$$
$$\log_{e}(e^{4a}+ae^{4a})=\log_{e^2}(e^{4a}+ae^{4a})\cdot \log_e({e^2})$$
$$\log_{e}(e^{4a}+ae^{4a})=2 \cdot \log_{e^2}(e^{4a}+ae^{4a})$$
We get the original logarithm with base $e$
$$\log_{e^2}(e^{4a}+ae^{4a})=\frac{1}{2}\ln(e^{4a}+ae^{4a})$$
Now the sum:
$$\frac{1}{2}\ln((a+1)e^{4a})$$
$$4a \frac{1}{2} \ln ((a+1)e)$$ $$4a \cdot \frac{1}{2} \ln (a+1) + \ln(e)$$
$$\frac{4a}{2} \ln {(a+1)} + 1$$
$$2a \ln (a+1) + 1$$
If there is an error let me know.
Your answer is good, except for a slip in the second equation after your words “Now the sum”, because the exponent $4a$ does not apply to the factor $(1+a)$. But as usual when we’re beginners, there’s often a shorter, clearer way.
There are two questions here, the first of which is how to handle the unorthodox $\log_{e^2}$, while the second is how to handle the sum inside the parentheses.
For the first question, I recommend not depending on formulas that may seem to come from on high, but rather going back to the definitions: \begin{align} \log_b(x)=L\qquad&\text{means}\qquad b^L=x\\ \log_{e^2}(x)=Y\qquad&\text{means}\qquad (e^2)^Y=x&\text{so}\qquad e^{2Y}=x\\ \log_e(x)=2Y\qquad&\text{means}\qquad e^{2Y}=x&\text{so}\qquad Y={\scriptstyle\frac12}\log_e(x)\,. \end{align} That solves the first question: $\log_{e^2}(e^{4a}+ae^{4a})=\frac12\ln(e^{4a}+ae^{4a})$.
The second question now becomes easy: again, don’t rely on that funny rule from Wikipedia, but just factor the $e^{4a}$ from the two terms: $\frac12\ln(e^{4a}+ae^{4a})=\frac12\ln\bigl(e^{4a}(1+a)\bigr) =\frac12\bigl(\ln(e^{4a})+\ln(1+a)\bigr)$. Now remember that $\ln(e^{4a})=4a$ and get your answer $2a+\frac12\ln(1+a)$.
Factorizing the numerator and using change of basis :
$$\frac{\ln ((a+1)e^{4a})}{\ln e^2} =\frac 12 (\ln{(a+1)}+\ln e^{4a})$$ $$= \frac 12\ln{(a+1)} +2a$$
Using $$\log_{a}(x) = \frac{\ln(x)}{\ln(a)},$$ where $\ln(x) = \log_{e}(x)$, then \begin{align} \log_{e^{2}}(e^{4 a} + a e^{4 a}) &= \frac{\ln(e^{4 a}(1+a))}{\ln(e^{2})} \\ &= \frac{4 a + \ln(1 + a)}{2 \, \ln(e)} \\ &= 2 a + \frac{\ln(1+a)}{2}. \end{align}
Since $$\frac{\ln(1 +a)}{2} = \frac{\ln(1+a)}{2 \, \ln(e)} = \frac{\ln(1+a)}{\ln(e^{2})} = \log_{e^{2}}(1+a),$$ then $$\ln_{e^{2}}(e^{4 a} + a e^{4 a}) = 2a + \log_{e^{2}}(1+a).$$
You were doing fine down to $$\frac 12\ln((a+1)e^{4a})$$ but then you cannot pull out the $4a$ because it is not an exponent on $a+1$. Instead you should do $$\frac 12\ln((a+1)e^{4a})=\frac 12\left(\ln (a+1)+\ln(e^{4a})\right)\\=\frac 12\ln(a+1)+2a$$ | 2019-09-23T13:02:32 | {
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https://math.stackexchange.com/questions/155676/p-sylow-subgroups-of-a-group-of-order-53-cdot-292 | # $p$-Sylow subgroups of a group of order $5^3\cdot 29^2$
I need to calculate the $p$-Sylow subgroups of a Galois group with order $5^3 \cdot 29^2$, i.e. $|\mathrm{Gal}(K/F)|=5^3 \cdot 29^2$.
I've already established that there is only one 29-Sylow-subgroup (with $|G_{29}|=29^2$) by the following conditions: $n_p \equiv 1 \pmod p$ and $n_p \mid m$ where $|G|=m\cdot p^r$ and $p\nmid m$.
(Indeed $5,25,125 \not{\!\equiv}\; 1 \pmod {29}$ and they are the only $\ne1$ divisors of $5^3$, so $n_{29} = 1$.)
I want to apply it to find 5-Sylow subgroups. $n_5 \equiv 1\pmod 5$ and $n_5\mid 29^2$. We have that $29 \not{\!\equiv}\; 1 \pmod 5$ but $481 = 29^2 \equiv 1 \pmod 5$ so $n_5 = 1$ and $n_5 = 29^2$ are both valid options.
I want to show that $n_5 = 1$.
Edit:
I will explain the rational behind my question: we have $F \subset K$ a Galois extension of degree $5^3 \cdot 29^2 = | \mathrm{Gal}(K/F)|$ and I want to find two subfields $F \subset K_1 , K_2 \subset K$ which are Galois extensions and $K = K_1 K_2$.
My idea was to find the corresponding p-Sylow groups of $\mathrm{Gal}(K/F)$ and use the fundamental theorem of Galois theory and deduce the extensions. That's why I looked for a normal subgroup of order $5^3 = 125$.
• It's not true. It is true there is a subgroup of index $5$ which is normal, though. – user641 Jun 8 '12 at 16:56
• In particular, let $H=C_{29}\times C_{29}$, let $K=C_5\times C_5$, and let $\alpha\in GL(2,29)$ be an element of order $5$. Then $HK\rtimes\langle\alpha\rangle$ has more than one Sylow 5-group. – user641 Jun 8 '12 at 16:57
• $K \times (H \rtimes \langle \alpha \rangle)$ even. – Jack Schmidt Jun 8 '12 at 17:00
• It might help if I add that I do this calculation to find a normal subgroup of order $5^3$ and index $29^2$ in order to prove a claim about (at least 2) intermediate Galois extensions between $F$ and $K$. $G_{29}$ gives one Galois extension $[ K_{29} : F ] = 5^3$. – Zachi Evenor Jun 8 '12 at 17:46
• $29^2=841{}{}{}$ – Chris Eagle Jun 8 '12 at 23:53
Here is a survey of the groups of order 105125.
There are basically two kinds of groups of order $5^3 \cdot 29^2$: the nilpotent and the non-nilpotent.
Nilpotent: These are the groups with both $n_5=1$ and $n_{29}=1$. Such a group is the direct product of its Sylow 5-subgroup and its Sylow 29-subgroup. Since there are 5 isomorphism classes of groups of order $5^3$ and 2 different isomorphism classes of groups of order $29^2$, there are (5)(2) = 10 isomorphism classes of nilpotent groups of order $5^3 29^2$. They are: 125×481, 5×25×481, 5×5×5×481, (5⋉(5×5))×481, (5⋉25)×481, 125×29×29, 5×25×29×29, 5×5×5×29×29, (5⋉(5×5))×29×29, and (5⋉25)×29×29.
Every such group has normal subgroups of indices $5^i 29^j$ for 0 ≤ i ≤ 3, 0 ≤ j ≤ 2. In particular, any field with such a Galois group has a tower of Galois extensions of length 5 (but not 6).
Every such group has a direct product factorization $G_5 \times G_{29}$ as the product of two proper normal subgroups (in fact Sylow p-subgroups).
Non-nilpotent: These are the groups with $n_5=29^2$ and $n_{29}=1$. Such a group is a semi-direct product of a normal Sylow 29-subgroup of structure 29×29 with a Sylow 5-subgroup acting with a kernel of size $25$. There are 5 such Sylow 5-subgroups, and most of them have only a single action (up to iso): 125, 5×5×5, and 5⋉(5×5) all have a single action, so we get 3 groups 125⋉(29×29), 5×5×5⋉(29×29), and (5⋉(5×5))⋉(29×29). There are two actions for 25×5 (one where the kernel is 5×5 and one where it is 25×1), and 3 actions for 5⋉25. This gives a grand total of 8 non-nilpotent groups.
Every such group has normal subgroups of indices 1 and $5^i 29^j$ for 1 ≤ i ≤ 3 and j ∈ { 0, 2 }. In particular, any field with such a Galois group has a tower of Galois extensions of length 4 (but not 5).
No such group has a factorization $G=MN$ where M and N are normal, proper subgroups since every normal proper subgroup has index divisible by 5.
• Thank you very much, if $\mathrm{Gal}(K/F)$ is indeed nilpotent then the direct product of its Sylow subgroups (which are normal) yields two Galois extension as desired. It remains to show that this Galois group is nilpotent (which is stronger property than solvability). – Zachi Evenor Jun 8 '12 at 21:23
• Zachi: exactly, and need not be true given what you've said so far. – Jack Schmidt Jun 8 '12 at 21:24
• The edit to the question says the OP is looking for two intermediate normal extension such that their compositum is all of $K$. So he is looking for two normal subgroups whose intersection is trivial. These always exist. – user641 Jun 9 '12 at 14:31
It is possible to have $n_5 = 29^2$ as in the following group:
Let $F$ be a finite field of order $29^2$, and let $\zeta \in F$ have multiplicative order $5$. In other words, let $F$ be the splitting field of $\zeta^2 + 6\zeta + 1$ over $\mathbb{Z}/29\mathbb{Z}$ and let $\zeta$ be a root in $F$.
Then consider the matrix group:
$\displaystyle G = \left\{ ~~\begin{bmatrix} \zeta^i & 0 & 0 & \beta \\ 0 & \zeta^j & 0 & 0 \\ 0 & 0 & \zeta^k & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} : 0 ≤ i,j,k < 5, \beta \in F \right\} \leq \operatorname{GL}(4,29^2)$
which is a group of order $5^3 29^2$ with the following $481$ Sylow 5-subgroups indexed by the elements $\gamma$ of $F$:
$\displaystyle P_\gamma = \left\{ ~~\begin{bmatrix} \zeta^i & 0 & 0 & \frac{1-\zeta^i}{1-\zeta} \gamma \\ 0 & \zeta^j & 0 & 0 \\ 0 & 0 & \zeta^k & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} : 0 ≤ i,j,k < 5 \right\} \qquad (\beta \text{ has a special form })$
$\displaystyle H = \left\{ ~~\begin{bmatrix} \zeta^0 & 0 & 0 & \beta \\ 0 & \zeta^0 & 0 & 0 \\ 0 & 0 & \zeta^0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} : \beta \in F \right\} \qquad (i=j=k=0)$
$\displaystyle K = \left\{ ~~\begin{bmatrix} \zeta^0 & 0 & 0 & 0 \\ 0 & \zeta^j & 0 & 0 \\ 0 & 0 & \zeta^k & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} : 0 ≤ j,k < 5 \right\} \qquad (i=0, \beta = 0)$
$\displaystyle \alpha =~~~~~ \begin{bmatrix} \zeta & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \qquad (i=1, j=k=0, \beta=0)$
In terms of my favorite groups, $G \leq \operatorname{AGL}(1,29^2) \times K$, and in fact is a Hall {5,29}-subgroup of this group.
• Thank you for the example. So my claim is not true in general, but is the fact that the "mother" group of order $5^3 \cdot 29^2$ is Galois can give additional information that may imply that $n_5=1$? – Zachi Evenor Jun 8 '12 at 17:42
• @Zachi: I don't think so. I believe this group is in fact the galois group of a polynomial over the rationals, but it is a little large for me to give a specific polynomial (I think such a polynomial has degree 481). – Jack Schmidt Jun 8 '12 at 18:31
• It is a conjecture that any finite group can be obtained as the Galois group of some polynomial over the rationals, so being a Galois group does not really give you any extra information, unless you know something about the polynomial. – Tobias Kildetoft Jun 8 '12 at 18:55
• In this specific case, we only know that $[ K : F ] = 5^3 \cdot 29^2$. We are not even told to assume that $\mathrm{char}(F)=0$. So I don't think using a specific polynomial helps. – Zachi Evenor Jun 8 '12 at 19:12
• @ZachiEvenor: To add to what Tobias said, it is a theorem (of Shafarevich) that every finite solvable group is the Galois group of a polynomial over the rationals, and it is is a theorem of Burnside that every finite group whose order is divisible by only two primes is solvable. So you definitely gain no new information from just knowing your group is a Galois group. – Arturo Magidin Jun 8 '12 at 19:45 | 2020-02-20T22:02:19 | {
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https://math.stackexchange.com/questions/3239446/how-to-prove-that-left-frac1n2-right-is-cauchy-sequence | # How to prove that $\left\{\frac{1}{n^{2}}\right\}$ is Cauchy sequence
How can I prove that $$\left\{\frac{1}{n^{2}}\right\}$$ is a Cauchy sequence?
A sequence of real numbers $$\left\{x_{n}\right\}$$ is said to be Cauchy, if for every $$\varepsilon>0$$, there exists a positive integer $$N(\varepsilon)$$ such that $$\mid x_{n+p}-x_{n}\mid <\varepsilon$$ for all $$n\geq N$$ and $$p= 1, 2, 3,...$$
So I approached like this...
$$\mid \frac{1}{(n+p)^{2}}-\frac{1}{n^{2}}\mid = \frac{p(2n+p)}{n^{2}(n+p)^{2}}<\frac{p(2n+p)}{n^{2}} <\varepsilon$$
From here, I have to show that $$n>$$ some expression involving $$\varepsilon$$, because that expression will be the value of $$N$$. But I am getting stuck here.
Please anyone help me solve it. Thanks in advance.
• If a sequence is convergent, then it is Cauchy. (The converse is not necessarily true in non-complete spaces) – Julian Mejia May 25 at 16:27
• One of the simplest estimates may be $$\left|\frac1{(n+p)^2}-\frac1{n^2}\right|<\frac1{n^2}.$$ – Jyrki Lahtonen May 25 at 16:30
• @JulianMejia +1, Nothing more to say than this. – Michael Hoppe May 25 at 16:56
Since you already know what the limit is, this is not hard.
Let $$\epsilon > 0$$ be given. Choose $$N$$ such that $$\frac{1}{N^2} < \frac{\epsilon}{2}$$. Now assume $$n \ge N$$ and $$p \ge 1$$. Then $$|\frac{1}{n^2} - \frac{1}{(n+p)^2}| \le \frac{1}{n^2} + \frac{1}{(n+p)^2} \le \frac{2}{n^2} \le \frac{2}{N^2} < \epsilon \, .$$
• Ohh Thanks, Sir. Triangle inequality makes it so easy! – user587389 May 25 at 16:38
$$\frac{p(2n+p)}{n^2(n+p)^2}=\frac{p(2n+p)}{n^2(n^2+(2n+p)p)}\leq\frac{p(2n+p)}{n^2p(2n+p)}=\frac{1}{n^2}.$$
The triangular inequality is not needed, it is only application of $$0\le a\le b\implies 0\le b-a\le b$$
Let's have $$m\ge n$$ then apply to $$a=\frac 1{m^2}$$ and $$b=\frac 1{n^2}$$.
You get $$0\le \frac 1{n^2}-\frac 1{m^2}\le \frac 1{n^2}$$ | 2019-10-17T05:10:39 | {
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https://math.stackexchange.com/questions/1227769/can-this-be-shown-sqrt3a-sqrt3a-sqrt3a-dots-sqrt-a/1227775 | # Can this be shown: $\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\dots}}} = \sqrt a$?
$$\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\cdots}}}}}}}}}=\sqrt{a}$$
Just for fun. I would like to read the proof of this if it exists. Any references would be appreciated.
• As a first step, try to define the sequence rigorously, i.e., give the explicit recursive definition. Then you should be able to show the sequence is monotonic, bounded, and then figure out which equation does the limit satisfy. Apr 9 '15 at 23:41
• So this is equivalent to $$\prod_{i=1}^\infty a^{\frac 1{3^i}}$$ Apr 9 '15 at 23:42
• @abiessu hahah oh wow I did not see that. It's easy then. Thank you!
– Ivan
Apr 9 '15 at 23:53
• Please, try to make the title of your question more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. Apr 10 '15 at 11:09
• This is almost the same question as math.stackexchange.com/questions/815418/… Jun 3 '15 at 1:56
Let $x_n = \sqrt[k]{a\sqrt[k]{a\sqrt[k]{a\sqrt[k]{a\sqrt[k]{a\sqrt[k]{a\sqrt[k]{a\sqrt[k]{a\sqrt[k]{a\cdots}}}}}}}}}$, where the $\sqrt[k]{\cdot}$ appears $n$ times. Prove that this sequence is monotone and bounded. Now Setting $\lim_{n \to \infty} x_n = x$, we obtain $$x = \sqrt[k]{ax} \implies x^k = ax \implies x^{k-1} = a \implies x = \sqrt[k-1]{a}$$ In your case, $k=3$.
• $$x_{n+1}=\sqrt[k]{ax_n},$$ $$x=\lim_{n\to \infty} x_n= \lim_{n\to \infty} x_{n+1} =\lim_{n\to \infty} \sqrt[k]{ax_n}=\sqrt[k]{a \lim_{n\to \infty} x_n}=\sqrt[k]{ax}$$ Apr 10 '15 at 0:00
Let $\displaystyle b=\sqrt[3]{a\cdot\ \underbrace{\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\cdots}}}}}}}}}_{\text{This is$b$, which is allegedly$\sqrt a$.}}}$.
Then $\displaystyle b = \sqrt[3]{ab}$, so $b^3 = ab$, and then $b^2 = a$. Thus $b = \sqrt a$.
If we put $\sqrt{a}$ in place of the expression said to be $b=\sqrt{a}$, we get $\sqrt[3]{a\sqrt{a}}$. And it is easy to see that that is indeed $\sqrt a$.
As to convergence: let $g(x) = (ax)^{1/3}$. The question is the behavior of the sequence $$a,\ g(a),\ g(g(a)),\ g(g(g(a))),\ \ldots\ .$$ For $x$ between $\sqrt{a}$ and $a$, we have $0<g'(x)<1/3$, so $g$ is a contraction and thus has a unique attractive fixed point.
• Not sure why this got down votes... Apr 10 '15 at 1:20
• I think people misread "allegedly"
– k_g
Apr 10 '15 at 2:20
• From $b^3=ab$, how do you rule out the possibility $b=0$? Apr 10 '15 at 11:37
• @BarryCipra : Interesting question. I'll see if I can show (maybe by induction?) that $\sqrt a$ is a lower bound, at least if $a>1$. And $a$ itself if $a<1$. ${}\qquad{}$ Apr 10 '15 at 15:56
• @BarryCipra : OK, how about this argument: Iteration of a contraction moves you toward the attractive fixed point, so the sequence will go to $\sqrt a$, not to $0$. ${}\qquad{}$ Apr 10 '15 at 16:09
Taking the "redefinition" as $\prod_{i=1}^\infty a^{\frac 1{3^i}}$, we immediately have $\sum_{i=1}^\infty \frac 1{3^i}=\frac 12$, which is the effect of summing the exponents as would occur with the product described. Therefore, we have $$\prod_{i=1}^\infty a^{\frac 1{3^i}}=\sqrt a$$
Another way to go is to show that $x_n=a^{b_n}$ where $b_n$ satisfies
$$b_1=\frac13\quad\text{and}\\ b_{n+1}=(1+b_n)\frac13$$
then show $\lim b_n=\frac12$ in a manner similar to user17762's answer.
What we have is the $lim_{n\to\infty}$ $a^{\frac{1}{3}}a^{\frac{1}{9}}a^{\frac{1}{27}}..a^{\frac{1}{3^n}}$.
Using the property $(a^b)(a^c)=a^(b+c)$
We get:
$a^{\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...\frac{1}{3^n}}$
The exponent is in fact is geometric series which has a value of $(1/(1-(1/3))-1=\frac{1}{2}$ so:
We get your expression to be $a^{\frac{1}{2}}=\sqrt{a}$
• You should write your answer in $\rm\LaTeX$. I think you might have the right idea more or less and is quite similar in nature to @abiessu's answer. However it's hard to tell when it's written like this. Jun 3 '15 at 0:30
• May I ask you to join a chat room of mine? The chat room is at the bottom of my profile description. Feb 2 '17 at 1:27
this might not be entirely rigorous, but here is a simple intuitive way to think of it: \begin{align*}x &= \sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\cdots}}}}}}}}}\\ x^3&=a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\cdots}}}}}}}}}\\ \frac{x^3}{a}&=\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\cdots}}}}}}}}}\\ \frac{x^3}{a} &= x\\ x^2&=a\\ x&=\sqrt{a}\\ \sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\cdots}}}}}}}}}&=\sqrt{a} \end{align*}
• This assumes that the sequence converges.
– robjohn
Jun 3 '15 at 2:03
• Yes, so like I said, not entirely rigorous, just a solid intuition Jun 3 '15 at 2:05 | 2022-01-20T15:50:12 | {
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https://math.stackexchange.com/questions/1920907/difference-between-ab-22-and-ab | # Difference between $((a+b)/2)^2$ and $ab$?
I was thinking about the difference between the area of a rectangle that is not a square, and a square with sides whose lengths are at the midpoint between the lengths of $a$ and $b$. I did some algebraic manipulation and it seems that the difference between the area of the square, $((a+b)/2)^2$, and the area of the rectangle, $ab$, is $(a^2+b^2)/4 - (ab)/2$.
Now if you had a right triangle with sides $a$ and $b$, it's hypotenuse would be the square root of $a^2 + b^2$.
So what I'm wondering is why the difference between the area of the square and the rectangle is the same as the difference between one fourth the square of the hypotenuse of the right triangle with sides $a$ and $b$, and $1/2$ the rectangle $ab$? If you were trying to find this difference purely with geometry, what steps could you take to reach this conclusion, starting from the original square and rectangle?
• This is a wonderful question. – Juan Sebastian Lozano Sep 9 '16 at 22:10
• I have 2 points of confusion. The square whose lengths are at the mid-point of $a$ and $b$? And $a, b$ are the sides of a rectangle. The lenght of the segment between these two points is half the diagonal of the rectangle. And the square of that is $\frac 14 (a^2 + b^2)$ not $\big(\frac {(a+b)}{2}\big)^2$. – Doug M Sep 9 '16 at 22:19
• Actually it's more than half the diagonal of the rectangle. ((a+b)/2)^2 is 1/4(a^2+2ab+b^2) . I combined ab/2 and -ab to get -ab/2, aka half the area of the rectangle ab. – Hockeyfan19 Sep 9 '16 at 22:24
• Also ty @Juan Sebastian Lozano – Hockeyfan19 Sep 9 '16 at 22:25
• second point. This question seems to be asking a circular question about the proof of the Pythagorean theorem. i.e. $4$ right triangles with legs $( a, b)$ can be arranged such that the form a square with side $(a+b)$ and the 4 hypoteni forming a square. The $4$ triangles have combinded area $2ab.$ And the big square has area $(a+b)^2$, meaning that the square formed by the hypoteni has area $a^2 + b^2$ – Doug M Sep 9 '16 at 22:26
Here is my proposition building heavily on a well known proof of the Pythagorean Theorem:
The red rectangle is $ab$. You see how the four blue triangles with legs that are half of $a$ and $b$ respectively fit nicely in the left half of the rectangle. The white square is $1/4$ of the square on the hypotenuse.
So by subtracting the rectangle from the large square, we are essentially subtracting the right half of rectangle from the white square.
• This is the sort of thing I'm looking for, but I'm not seeing the geometric representation of the square with sides (a+b)/2. I'm looking for the difference of it and the rectangle ab, and it looks like you have the rectangle ab in your top left (times 4?). Maybe I'm just not understanding your proposition. If you multiplied the difference by 8, you would have 2 of those open squares surrounded by purple, minus that big red rectangle (4ab) and that would give 8 times the difference I'm looking for. – Hockeyfan19 Sep 9 '16 at 22:55
• @Hockeyfan19: the entire red rectangle is $ab$. The blue triangles have half of $a$ and $b$ as sides. – String Sep 9 '16 at 23:03
• The square of $(a+b)/2$ is enclosed by the legs of the blue triangles ... – String Sep 9 '16 at 23:10
• The big blue square is (a+b)/2 squared. – fleablood Sep 9 '16 at 23:10
• That makes a lot of sense String, I just wasn't reading it right. Then when you get rid of the blue squares and one half the red rectangle we have the difference I was looking for. Perfect! – Hockeyfan19 Sep 9 '16 at 23:14
One more note:
$\frac {(a^2 + b^2)}{4} - \frac {ab}2 = \big(\frac {(a-b)}{2}\big)^2$
Maybe a picture will help:
First figure: the larger square has dimensions $(A+B)/2 \times (A+B)/2$ Which has can be cut into smaller rectangles and squares of size $A/2 \times B/2,A/2 \times (B-A)/2, A/2 \times A/2$ leaving the green square that is $(B-A)/2 \times (B-A)/2$
The white rectangles can be re-arranged into an $A\times B$ rectangle.
Part 2. $C$ is the hypotenuse of the $A\times B$ rectangle.
$(C/2)^2 - (AB)/2 = ((B-A)/2)^2$
• This is a great answer, thank you for the information! You demonstrated the difference between the rectangle and square very well geometrically. – Hockeyfan19 Sep 9 '16 at 23:35 | 2021-06-25T00:33:50 | {
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https://www.albert.io/learn/abstract-algebra/question/group-action-on-itself-multiply-leftright-adjointconjugation | Limited access
Let $G$ be a group with neutral element $e$ and composition law $\ast$.
Let $G^{\rm \,op}$, called the opposite to $(G, \ast)$, be the group whose set of elements equals the set $G$, but whose group composition law $\bar{\ast}$ is $h\bar{\ast}g:= g\ast h$.
If $G$ is abelian, then $(G,\ast)=(G^{\rm \,op},\bar{\ast})$, so we assume in this question that $G$ is not necessarily abelian.
Let ${\rm Aut}(G)$ be the group of automorphisms of $G$, with composition law $\circ$ given by composition of maps.
An action of a group $(G, \ast)$ on a set $X$ is a map $g\mapsto \alpha_g$ from $G$ to the set of bijective maps from $X$ to $X$, such that $\alpha_e(x)=x$, for all $x\in X$, and $\alpha_{h\ast g}=\alpha_h\circ\alpha_g$, all $h,g\in G$, where $\circ$ is composition of maps.
For $g\in G$, let $L_g:G\rightarrow G$ be the map $L_g(h)=g\ast h$, let $R_g:G\rightarrow G$ be the map $R_g(h)=h\ast g$, and let ${\rm Ad}_g$ be the map ${\rm Ad}_g(h)=g\ast h\ast g^{-1}$, for all $h\in G$.
We call the $L_g$ left multiplications, the $R_g$ right multiplications, and the ${\rm Ad}_g$ adjoint maps, a.k.a. conjugations.
1) The map between sets $g\mapsto {\rm Ad}_g$, for $g\in G$,
Select Option is a group homomorphism from $G$ to ${\rm Aut}(G)$is a group homomorphism from $G^{\rm \,op}$ to ${\rm Aut}(G^{\rm \,op})$is a group homomorphism from $G$ to ${\rm Aut}(G^{\rm \,op})$is a group homomorphism from $G^{\rm \,op}$ to ${\rm Aut}(G)$
.
2) The map between sets $g\mapsto {\rm Ad}_{g^{-1}}$, for $g\in G$,
Select Option is a group homomorphism from $G$ to ${\rm Aut}(G)$is a group homomorphism from $G^{\rm \,op}$ to ${\rm Aut}(G^{\rm \,op})$is a group homomorphism from $G$ to ${\rm Aut}(G^{\rm \,op})$is a group homomorphism from $G^{\rm \,op}$ to ${\rm Aut}(G)$
.
3) The map $g\mapsto L_g$, for $g\in G$,
Select Option is an action of $G$ on the set $G$ is an action of $G^{\rm\, op}$ on the set $G$
.
4) The map $g\mapsto R_g$, for $g\in G$,
Select Option is an action of $G$ on the set $G$is an action of $G^{\rm\, op}$ on the set $G$
.
Select an assignment template | 2017-03-25T09:49:59 | {
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https://www.molympiad.net/2017/08/canadian-national-mathematical-olympiad-2014-solutions.html | 1. Let $a_1,a_2,\dots,a_n$ be positive real numbers whose product is $1$. Show that the sum $\frac{a_1}{1+a_1}+\frac{a_2}{(1+a_1)(1+a_2)}+\cdots+\frac{a_n}{(1+a_1)(1+a_2)\cdots(1+a_n)}$ is greater than or equal to $\dfrac{2^n-1}{2^n}$.
2. Let $m$ and $n$ be odd positive integers. Each square of an $m$ by $n$ board is coloured red or blue. A row is said to be red-dominated if there are more red squares than blue squares in the row. A column is said to be blue-dominated if there are more blue squares than red squares in the column. Determine the maximum possible value of the number of red-dominated rows plus the number of blue-dominated columns. Express your answer in terms of $m$ and $n$.
3. Let $p$ be a fixed odd prime. A $p$-tuple $(a_1,a_2,a_3,\ldots,a_p)$ of integers is said to be good if
• $0\le a_i\le p-1$ for all $i$, and
• $a_1+a_2+a_3+\cdots+a_p$ is not divisible by $p$, and
• $a_1a_2+a_2a_3+a_3a_4+\cdots+a_pa_1$ is divisible by $p$.
Determine the number of good $p$-tuples.
4. The quadrilateral $ABCD$ is inscribed in a circle. The point $P$ lies in the interior of $ABCD$, and $\angle PAB = \angle PBC = \angle PCD = \angle PDA$. The lines $AD$ and $BC$ meet at $Q$, and the lines $AB$ and $CD$ meet at $R$. Prove that the lines $P Q$ and $P R$ form the same angle as the diagonals of $ABCD$.
5. Fix positive integers $n$ and $k\ge 2$. A list of $n$ integers is written in a row on a blackboard. You can choose a contiguous block of integers, and I will either add $1$ to all of them or subtract $1$ from all of them. You can repeat this step as often as you like, possibly adapting your selections based on what I do. Prove that after a finite number of steps, you can reach a state where at least $n-k+2$ of the numbers on the blackboard are all simultaneously divisible by $k$.
MOlympiad.NET là dự án thu thập và phát hành các đề thi tuyển sinh và học sinh giỏi toán. Quý bạn đọc muốn giúp chúng tôi chỉnh sửa đề thi này, xin hãy để lại bình luận facebook (có thể đính kèm hình ảnh) hoặc google (có thể sử dụng $\LaTeX$) bên dưới. BBT rất mong bạn đọc ủng hộ UPLOAD đề thi và đáp án mới hoặc liên hệ[email protected]úng tôi nhận tất cả các định dạng của tài liệu: $\TeX$, PDF, WORD, IMG,...
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http://mathhelpforum.com/number-theory/185421-sum-numbers-product-numbers-problem.html | Thread: "sum of numbers = product of that numbers" problem
1. "sum of numbers = product of that numbers" problem
Hi guys,
find all solutions of equation a + b + c = a . b . c, where $\displaystyle a \leq b \leq c$ are positive integers.
Well, by trial and error method I found out that 1+2+3 = 1.2.3
Probably there is no other solution, but how can I proove that? I tried case analysis, but unsuccessfully. Any idea? Thanks!
2. Re: "sum of numbers = product of that numbers" problem
Originally Posted by jozou
Hi guys,
find all solutions of equation a + b + c = a . b . c, where $\displaystyle a \leq b \leq c$ are positive integers.
Well, by trial and error method I found out that 1+2+3 = 1.2.3
Probably there is no other solution, but how can I proove that? I tried case analysis, but unsuccessfully. Any idea? Thanks!
Hint(I believe):
$\displaystyle \frac{a+b+c}{3}\geq \sqrt[3]{abc}$
$\displaystyle a+b+c\geq 3\sqrt[3]{abc} <abc$
The last inequality holds for: $\displaystyle abc\geq 6$
3. Re: "sum of numbers = product of that numbers" problem
Originally Posted by jozou
Hi guys,
find all solutions of equation a + b + c = a . b . c, where $\displaystyle a \leq b \leq c$ are positive integers.
Well, by trial and error method I found out that 1+2+3 = 1.2.3
Probably there is no other solution, but how can I proove that? I tried case analysis, but unsuccessfully. Any idea? Thanks!
I try a straighforward approach.
The integers $\displaystyle a,b,c$ cannot be equal; otherwise $\displaystyle 3a=a+b+c=abc=a^3$ and then $\displaystyle a^2=3$, but $\displaystyle 3$ is not a perfect square.
So the following inequality holds: $\displaystyle a+b+c<c+c+c=3c$. Now if $\displaystyle ab\geq3$, then $\displaystyle 3c\leq abc$ implying $\displaystyle a+b+c<abc$.
Therefore, $\displaystyle ab<3$, i.e. $\displaystyle a=1$ and $\displaystyle b=2$. The equation $\displaystyle a+b+c=abc$ gives $\displaystyle 3+c=2c$ and hence $\displaystyle c=3$.
4. Re: "sum of numbers = product of that numbers" problem
Originally Posted by melese
I try a straighforward approach.
The integers $\displaystyle a,b,c$ cannot be equal; otherwise $\displaystyle 3a=a+b+c=abc=a^3$ and then $\displaystyle a^2=3$, but $\displaystyle 3$ is not a perfect square.
So the following inequality holds: $\displaystyle a+b+c<c+c+c=3c$. Now if $\displaystyle ab\geq3$, then $\displaystyle 3c\leq abc$ implying $\displaystyle a+b+c<abc$.
Therefore, $\displaystyle ab<3$, i.e. $\displaystyle a=1$ and $\displaystyle b=2$. The equation $\displaystyle a+b+c=abc$ gives $\displaystyle 3+c=2c$ and hence $\displaystyle c=3$.
Thank you. This proof is great!
5. Re: "sum of numbers = product of that numbers" problem
another nice proof (I'm not an author):
Let a ≤ b ≤ c be positive integers, s.t. equation a+b+c = a.b.c holds.
If a > 1, then 4c ≤ a.b.c = a+b+c. So 3c ≤ a+b, what contradicts a ≤ b ≤ c.
So a= 1. Now we have to solve 1+b+c = b.c. Then b.(c-1) = 1+b+c - b = 1 + c, (b-1).(c-1) = 1 + c - (c-1) = 2.
So (b-1).(c-1) = 2. Hence the only solution satistying assumptions is b = 2, c = 3. | 2018-04-22T13:36:29 | {
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http://math.stackexchange.com/questions/781218/numbers-whose-self-and-reciprocal-are-finitely-decimally-expressable-that-are-cl | # Numbers whose self and reciprocal are finitely decimally expressable that are close to one?
How would I go about finding numbers x such that x and 1/x are finitely decimally reciprocal and are also close to 1?
I'm not entirely certain how to phrase this question, but take for example 2. 2 and 1/2 can be represented with a finite number of decimal points.
The closest pair to 1 that I have found is 4/5 and 5/4. Are there closer pairs and how would I look for them?
-
Note that you can make them as close to $1$ as you like. This is because the irrational number $\ln 2/\ln 5$ can be approximated by rational fractions to any desired degree of accuracy. – NotNotLogical May 4 '14 at 19:34
The only rational numbers that have finite decimal representations are those whose denominator has only $2$ and $5$ as its prime factors. (because if $z$ has a finite decimal expansion, then $y=10^k z$ is an integer for some positive integer $k$, and thus $z = y/10^k$)
Thus, the pairs you are looking for are all of the form $2^m / 5^n$ and $5^n / 2^m$.
To see when they're close to 1, it's easier to turn it into an additive problem by taking logarithms: you want
$$m \ln 2 - n \ln 5 \sim 0$$
Rearranging, we want
$$\frac{m}{n} \sim \frac{\ln 5}{\ln 2}$$
so the problem is to find very good rational approximations to $\ln 5 / \ln 2$.
The first few approximations given by continued fractions is:
• $m/n = 2$, and thus $x = 4/5 = 0.8$
• $m/n = 7/3$, and thus $x = 2^7 / 5^3 = 1.024$ and $1/x = 0.9765625$
• $m/n = 65/28$, and thus $x = 2^{65}/5^{28} = 0.99035\ldots$
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Can you justify your first sentence? Also, is a number like $2^{m_1}5^{m_2}/2^{n_1}5^{n_2}$ also possible? – user103828 May 5 '14 at 9:03
Yes, but you can cancel things out. e.g. $8/10$ works, but it's the same number as $4/5$. – Hurkyl May 5 '14 at 10:18
The number could also be of the form $2^m 5^n$, $5^m / 2^m$, or $1 / (2^m 5^m)$. The first and the third options are not close to $1$. But I think the second option should be included. – Goos May 5 '14 at 18:48
@Goos: Remember we're looking for a pair ($x$, $1/x$); if $x=5^m/2^n$, then $1/x$ is of the form my post is looking for. However, I've been wondering if continued fractions would find a different set of approximations for $\ln 2 / \ln 5$, but not enough to grind through the algebra to see. – Hurkyl May 5 '14 at 20:27
@Hurkyl If $r > 1$, then to find the continued fraction for $r$ you first write $r = \frac{1}{1/r}$, and then you find a continued fraction for $1/r$. So the approximations to $\ln 2 / \ln 5$ and $\ln 5 / \ln 2$ are reciprocals of each other. There is also a definition of continued fraction approximation that makes this symmetry clear--good approximations $m/n$ are ordered pairs $(m,n)$ in the the plane that are close to the angle of the real number $r$ in some sense, and the plane is symmetric about $y = x$. – Goos May 6 '14 at 1:03
What you need is a power of $2$ and a power of $5$ which are close together. Your example has $4=2^2$ and $5=5^1$.
Since powers of $2$ are a multiple of $2$ apart, you can always get between $\frac 12$ and $2$.
To get the best results, though, we want $5^n=k2^m$ where $k$ is close to $1$.
Taking logs we have $n\log 5 = \log k + m\log 2$ and since $k$ is near to $1$ we have $\log k$ close to zero, and we see that $$\frac nm\approx\frac{\log 2}{\log 5}$$ and the best way of getting close is to use the continued fraction expansion of $\frac{\log 2}{\log 5}$.
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As an "extended" comment, adding to the above posts, we have $$\frac{\ln 2}{\ln 5}\approx 0.43067655807339306$$
The first few "best approximations" for that decimal are (research the Stern-Brocot Tree) $$[(2, 5, 0.4), (3, 7, 0.42857142857142855), (34, 79, 0.43037974683544306), (31, 72, 0.4305555555555556), (205, 476, 0.43067226890756305), (497, 1154, 0.4306759098786828), (643, 1493, 0.4306764902880107), (4647, 10790, 0.4306765523632993), (21306, 49471, 0.43067655798346505), (21306, 49471, 0.43067655798346505), (97879, 227268, 0.4306765580724079), (1740516, 4041353, 0.4306765580734967), (2034153, 4723157, 0.4306765580733395), (1936274, 4495889, 0.4306765580733866), (15392313, 35739844, 0.43067655807339283), (59632978, 138463487, 0.430676558073393)]$$ where the pairs are $(m,n,\text{decimal})$ for $5^m,2^n$.
And I broke my computer trying to compute the last decimal, so I'll be back with the outputs of those numbers...
Some results are $${5^{205}\over 2^{476}}=0.9967194951\dots$$ and $${5^{59632978}\over 2^{138463487}}=0.9999999850988389 \dots$$
- | 2015-01-30T17:02:17 | {
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http://mathhelpforum.com/calculus/142395-double-integration.html | 1. ## Double Integration
Any help with the following problem is appreciated (I have no idea how to approach it):
Consider the integral $I= \int^{\infty}_{-\infty} e^{-x^2} dx$. Express $I^2$ as a double integral involving $x$ and $y$.
2. Originally Posted by demode
Any help with the following problem is appreciated:
Consider the integral $I= \int^{\infty}_{-\infty} e^{-x^2} dx$. Express $I^2$ as a double integral involving $x$ and $y$.
It's well known that
$\int_{-\infty}^{\infty}e^{-x^2}\,dx = \sqrt{\pi}$.
So if $I = \sqrt{\pi}$, surely $I^2 = \pi$...
3. Originally Posted by Prove It
It's well known that
$\int_{-\infty}^{\infty}e^{-x^2}\,dx = \sqrt{\pi}$.
So if $I = \sqrt{\pi}$, surely $I^2 = \pi$...
But the problem says "express $I^2$ as a double integral". How do we need to express it like that?
4. $I= \int_{-\infty}^\infty e^{-x^2}dx$
$I= \int_{-\infty}^\infty e^{-y^2}dy$ since that is just the same integral with "dummy variable" y rather than x.
Then $I^2= \left(\int_{-\infty}^\infty e^{-x^2}dx\right)\left(\int_{-\infty}^\infty e^{-y^2}dy\right)$
by Fubini's theorem, that is the same as
$I^2= \int_{x= -\infty}^\infty \int_{y= -\infty}^\infty e^{-x^2}e^{-y^2} dy dx$
$I^2= \int_{x= -\infty}^\infty \int_{y= -\infty}^\infty e^{-(x^2+y^2)} dy dx$
5. Originally Posted by HallsofIvy
$I= \int_{-\infty}^\infty e^{-x^2}dx$
$I= \int_{-\infty}^\infty e^{-y^2}dy$ since that is just the same integral with "dummy variable" y rather than x.
Then $I^2= \left(\int_{-\infty}^\infty e^{-x^2}dx\right)\left(\int_{-\infty}^\infty e^{-y^2}dy\right)$
by Fubini's theorem, that is the same as
$I^2= \int_{x= -\infty}^\infty \int_{y= -\infty}^\infty e^{-x^2}e^{-y^2} dy dx$
$I^2= \int_{x= -\infty}^\infty \int_{y= -\infty}^\infty e^{-(x^2+y^2)} dy dx$
It should also be pointed out that this double integral
$I^2= \int_{-\infty}^\infty \int_{-\infty}^\infty e^{-(x^2+y^2)} \,dy\, dx$
can be evaluated to the value of $\pi$ by converting to polars...
6. Originally Posted by Prove It
It should also be pointed out that this double integral
$I^2= \int_{-\infty}^\infty \int_{-\infty}^\infty e^{-(x^2+y^2)} \,dy\, dx$
can be evaluated to the value of $\pi$ by converting to polars...
How did you evaluate it to the value of $\pi$?
Here's my attempt:
$\int_{0}^\infty [e^{-(r^2cos^2 \theta+r^2sin^2 \theta)}]r \,dr$
$= \lim_{k \to \infty}\int_{0}^k re^{-r^2} \,dr$
$= \frac{-1}{2} \lim_{k \to \infty} e^{-r^2} \,dr$
But how does this equal to $\pi$?
7. You know that
$= I^2$.
You can also say
Therefore
$I^2 = \pi$ which means $I = \sqrt{\pi}$.
Another handy application is using the Gaussian Integral to compute $\Gamma\left(\frac{1}{2}\right)$.
Since $e^{-x^2}$ is an even function, that means
$\int_{-\infty}^{\infty}{e^{-x^2}\,dx} = 2\int_0^{\infty}{e^{-x^2}\,dx}$.
Using the substitution $x =t^{\frac{1}{2}}$ yields
$-x^2 = -t$ and $dx = \frac{1}{2}t^{-\frac{1}{2}}$
So that the integral $2\int_0^{\infty}{e^{-x^2}\,dx}$ becomes
$2\int_0^{\infty}{\frac{1}{2}\,e^{-t}\,t^{-\frac{1}{2}}\,dt}$
$= \int_0^{\infty}{e^{-t}\,t^{-\frac{1}{2}}\,dt}$
$= \Gamma\left(\frac{1}{2}\right)$.
So that means $\Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}$.
8. Originally Posted by demode
How did you evaluate it to the value of $\pi$?
Here's my attempt:
$\int_{0}^\infty [e^{-(r^2cos^2 \theta+r^2sin^2 \theta)}]r \,dr$
$= \lim_{k \to \infty}\int_{0}^k re^{-r^2} \,dr$
$= \frac{-1}{2} \lim_{k \to \infty} e^{-r^2} \,dr$
But how does this equal to $\pi$?
It was a double integral, remember?
$I^2= \int_{\theta= 0}^{ 2\pi} \int_{r= 0}^\infty r e^{-r^2}drd\theta$ $= 2\pi \int_{r= 0}^\infty r e^{-r^2}dr$
Let $u= r^2$ so that $du= \frac{1}{2}r dr$ and $2du= rdr$. The integral becomes
$4\pi\int_0^\infty e^{-u}du$.
9. Originally Posted by HallsofIvy
It was a double integral, remember?
$I^2= \int_{\theta= 0}^{ 2\pi} \int_{r= 0}^\infty r e^{-r^2}drd\theta$ $= 2\pi \int_{r= 0}^\infty r e^{-r^2}dr$
Let $u= r^2$ so that $du= \frac{1}{2}r dr$ and $2du= rdr$. The integral becomes
$4\pi\int_0^\infty e^{-u}du$.
Actually, I believe you'll find that
$du = 2r\,dr$.
So the integral becomes
$2\pi \int_{r= 0}^\infty {r\,e^{-r^2}\,d} = \pi \int_{r = 0}^{\infty}{e^{-r^2}\,2r\,dr}$
$= \pi \int_{r = 0}^{\infty}{e^{-u}\,du}$
10. Prove It,
We know that $I=\int^{\infty}_{-\infty}e^{-x^2}= \sqrt{\pi}$
If I want to integrate the density function of the normal distribution:
$\int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}}e^{\frac{-x^2}{2}}dx$
using the value of I, what would be a suitable change of variable in this case?
11. Originally Posted by demode
Prove It,
We know that $I=\int^{\infty}_{-\infty}e^{-x^2}= \sqrt{\pi}$
If I want to integrate the density function of the normal distribution:
$\int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}}e^{\frac{-x^2}{2}}dx$
using the value of I, what would be a suitable change of variable in this case?
You can use the exact same change of variable as given to integrate the Gaussian function.
$I^2 = \left(\int_{-\infty}^{\infty}{e^{-\frac{1}{2}x^2}\,dx}\right)^2 = \left(\int_{-\infty}^{\infty}{e^{-\frac{1}{2}x^2}\,dx}\right)\left(\int_{-\infty}^{\infty}{e^{-\frac{1}{2}y^2}\,dy}\right)$
$= \int_{-\infty}^{\infty}{\int_{-\infty}^{\infty}{e^{-\frac{1}{2}(x^2 + y^2)}\,dy}\,dx}$
$= \int_0^{2\pi}{\int_0^{\infty}{e^{-\frac{1}{2}r^2}\,r\,dr}\,d\theta}$
$= 2\pi \int_0^{\infty}{r\,e^{-\frac{1}{2}r^2}\,dr}$
$= 2\pi \int_{-\infty}^0{e^u\,du}$
$= 2\pi \left[e^0 - e^{-\infty}\right]$
$= 2\pi \left[1 - 0\right]$
$= 2\pi$.
Since $I^2 = 2\pi$ that means $I = \sqrt{2\pi}$.
We have shown that
$\int_{-\infty}^{\infty}{e^{-\frac{1}{2}x^2}\,dx} = \sqrt{2\pi}$
So that means
$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{e^{-\frac{1}{2}x^2}\,dx} = \frac{1}{\sqrt{2\pi}}\sqrt{2\pi}$.
Therefore
$\int_{-\infty}^{\infty}{\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}\,dx} = 1$
which is what we require for any probability density function. | 2017-06-26T10:42:04 | {
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http://math.stackexchange.com/questions/368220/a-question-on-a-proof-of-yy1-le-x12-implies-yy-1-le-x2/368224 | A question on a proof of $y(y+1) \le (x+1)^2 \implies y(y-1) \le x^2$
I was reading the question posted here:Does $y(y+1) \leq (x+1)^2$ imply $y(y-1) \leq x^2$?. The solution posted was:
"Given $y^2+y≤x^2+2x+1$, if possible, let $x2<y(y−1)$. Clearly $y>1.$
Then $x^2+(2x+1)<y^2−y+(2x+1)$ So $y^2+y<y^2−y+2x+1$, which resolves to $y<x+1/2.$
Hence we also have $y−1<x−1/2.$ As $y>1$, the LHS is positive, and we can multiply the last two to get $y(y−1)<x^2−1/4⟹y(y−1)<x^2$, a contradiction." However, I don't think this is right because he had $y<x+1/2$, and in the last step he multiplied $y−1$ by $y$ and $x−1/2 by x+1/2$. However, this inequality is not in the same proportion anymore, because y does not equal x+1/2. It seems like he increased the value of the right side compared to the left, and then concluding that the right side is bigger/ It's like if I try to prove that $a \lt b$, and to solve this I say $a \lt b+5$. While this is true, this changes the inequality.
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Yes, it is right and a nice one, moreover.
He had $y \lt x + \frac{1}{2}$
which implies $y-1 \lt x - \frac{1}{2}$
He then multiplied them. He was even careful to mention that the terms were positive (which was correct, based on the assumption he started out with: $y(y-1) \gt x^2$ and combined with the assumption in the problem statement, $y \gt 0$).
I don't understand what you mean by: "not even in the same proportion anymore".
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Well yea but if you wanna solve the original inequality aren't you limited to just multiplying both sides by the same thing? That way the RHS is increased in proportion to the LHS. However, what he multiplied the RHS by is larger than what he multiplied the LHS by. – Ovi Apr 21 '13 at 14:43
@Ovi: There is no "solving" original inequality. One wants to show a statement about $x,y$ implies another statement. The proof is by contradiction. btw, Why does it have to be exact? Say, I want to prove that $2^{10} \lt 1600$. I can always prove it this way: $2^{5} \lt 40$, squaring gives $2^{10} \lt 1600$. There is nothing wrong there. – Aryabhata Apr 21 '13 at 14:45
O oh I see. Thanks – Ovi Apr 21 '13 at 14:49
As mentioned, $$0<y-1<x-\frac12,$$ and $$0<y<x+\frac12.$$ Hence, $$y(y-1)<y\left(x-\frac12\right)<\left(x+\frac12\right)\left(x-\frac12\right)=x^2-\frac14<x^2.$$
More generally, if $a<b$ and $c<d$, and we happen to know that $a,c>0,$ then we can always conclude that $ac<bd$.
-
Yea I understand that but if you didn't know that a<b, could you use as an argument that ac<bd to prove that a<b? – Ovi Apr 21 '13 at 14:46
Not in general, but that's not remotely what was done, here. It was simply a chain of inequalities, using ordered field properties of the reals, to induce a contradiction. That's a fairly standard technique of analysis. – Cameron Buie Apr 21 '13 at 14:51
Oh ok I see it now haha I can't believe how I thought this was wrong. Thanks – Ovi Apr 21 '13 at 14:56
To reword the proof given there: We start with real numbers $x,y$ such that $$\tag1 y\ge0$$ $$\tag2 y^2+y\le x^2+2x+1$$ $$\tag3 x^2<y(y-1)$$ If we had $y\le1$, then the right hand side in $(3)$ would be the product of a nonnegative (according to $(1)$) and a nonpositve (by assumtion) factor, hence nonnegative, contradicting $x^2\ge0$. Therefore $$\tag4 y>1.$$ Adding $2x+1$ to $(3)$ and combining with $(2)$ we find $y^2+y\le x^2+2x+1<y^2-y+2x+1$, i.e. $$\tag5 y\le x+\frac12.$$ By subtracting $1$, this becomes $$\tag6 y-1\le x-\frac12.$$ Because of $(4)$ and $(6)$, the number $x-\frac12$ is positive, hence we are allowed to multiply $(5)$ with $x-\frac12$. Also, we can multiply $(6)$ with the (again by $(4)$) positive number $y$ and combine this to find $$x^2\stackrel{(3)}<y(y-1)\stackrel{y\cdot(6)}\le y\left(x-\frac12\right)\stackrel{(x-\frac12)\cdot(5)}\le \left(x+\frac12\right)\left(x-\frac12\right)=x^2-\frac14<x^2,$$ contradiction. | 2014-07-13T17:29:07 | {
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https://math.stackexchange.com/questions/1085808/number-of-ways-to-arrange-n-items-in-m-positions-having-exactly-k-items-ad | # Number of ways to arrange $n$ items in $m$ positions having exactly $k$ items adjacent to each other
It was over 20 years since I studied maths and I am stuck. I'd really appreciate some help understanding this (probably quite simple) problem.
I have $n$ items that I can place on $m$ positions. $m$ ≥ $n$. The total number of combinations is the old and trusty ${m \choose n}$, but I also need to partition the combinations on k - the number of items directly adjacent to each other
What I am looking for is best shown with a couple of examples.
Five positions and four items can be combined in two ways when $k$ is four, two ways when $k$ is three and one way when $k$ is two
.
Five positions and three items can be combined in three ways when $k$ is three, six ways when $k$ is two and one way when $k$ is one.
Five positions and one item can be combined in five ways when $k$ is one.
What I am trying to find is a function $f(m, n, k)$ that give me the number of combinations for some value $k$. For example:
$f(5, 3, 3)$ => $3$
$f(5, 3, 2)$ => $6$
$f(5, 3, 1)$ => $1$
Of course, adding up the number of combinations for all valid $k$ is ${m \choose n}$.
Just by toying with pen and paper I have observed some edge cases
$f(x, x, x)$ => $1$
$f(x, 1, 1)$ => $x$
$f(x, y, y)$ => $x-y+1$
• great diagrams, how did you created them? – RE60K Dec 30 '14 at 17:12
• Nothing fancy. I have solved some of the smaller problems by hand using excel. The pictures above are just screenshots from my worksheet. – Frustrated Jan 7 '15 at 19:00
• You seem to allow "leftover" items when $k$ does not divide $n$. However, it's not clear from your examples what your general rule is. One possibility is that all groups must have size $k$ except that when $k$ doesn't divide $n$, you're allowed one additional group of size $n$ mod $k$. Is this what you intend? – Tad Jan 20 '15 at 5:18
• I should have ended the title with "in the largest group". Any combination of groups smaller than $k$ is valid as a solution as long as there is one or more groups of size $k$. – Frustrated Jan 20 '15 at 7:36
It suffices to consider the function $g(m,n,k)$, which counts the number of ways to select $n$ positions from $m$ such that at most $k$ consecutive positions are chosen; note that $f(m,n,k)=g(m,n,k)-g(m,n,k-1)$. The function $g$ satisfies a simpler recurrence than $f$, namely $$g(m,n,k)=\begin{cases} {m\choose n},&n\le k\\ \sum_{r=1}^{k+1} g(m-r,n-r+1,k),&n>k \end{cases}$$ where the sum arises from breaking into cases according to the location of the last vacant cell.
By induction, we can show that, for fixed $n$ and $k$, $g(m,n,k)$ is a polynomial in $m$ of degree $n$, for $m>k$. This is clear for $n\le k$ by the initial conditions. For $n>k$ we see that $g(m,n,k)-g(m-1,n,k)$ is a sum of polynomials of degree $n-1, n-2, \ldots, n-k$, i.e. $g(m,n,k)-g(m-1,n,k)$ has degree $n-1$. So finite difference calculus tells us $g(m,n,k)$ is a degree $n$ polynomial; indeed, it tells us how to write down this polynomial explicitly, having used the recurrence to generate $n+1$ terms. That may suffice for your needs.
Alternatively, you can get sort of a closed form using generating functions. Denote empty cells by $0$ and filled cells by $1$. For each $r$ with $0\le r\le k$, define an $r$-block to be $r$ $1$'s followed by a $0$. If we temporarily append an extra vacant cell to the end, each valid configuration can be uniquely described by laying down a sequence of $r$-blocks, for various $r$, so that the total number of $1$'s is $n$ and the total number of cells is $m+1$. Note that in order to get density $n$, we need to use $m+1-n$ blocks. Standard exponential generatingfunctionology implies that $g(m,n,k)$ is the coefficient of $x^{m+1}\,z^{m+1-n}/(m+1-n)!$ in $$G_k(x,z) := {\rm exp}\left( z(x + x^2 + \cdots + x^{k+1})\right).$$
(Basically, $z$ tags the total number of blocks, so the g.f. is exponential in $z$, and $x$ tags the lengths of the blocks. The $m$ turns into $m+1$ because of the extra empty cell we added.) Bringing things full circle, this says
$$f(m,n,k) = \left[\frac{x^{m+1}\,z^{m+1-n}}{(m+1-n)!} \right]\left( G_k(x,z) - G_{k-1}(x,z)\right).$$
There is no simple formula, but you can use a simple recursive property to efficiently compute $f$.
Border conditions :
1. if $k>n$ or $n>m$ then $f(m,n,k)=0$
2. if $k=m=n$ then $f(m,n,k)=1$
Then, any valid (m,n,k) configurations either :
1. begins with $i$ occupied cells followed by an empty cell and a $(m-i-1,n-i,k)$ configuration (for any $i\le k$)
2. begins with $k$ occupied cells followed by an empty cell and a $(m-k-1,n-k,i)$ configuration (for any $i<k$)
$$f(m,n,k)=\sum_{i=0}^kf(m-i-1,n-i,k)+\sum_{i=0}^{k-1}f(m-k-1,n-k,i)$$
Using that, you can very efficiently compute $f$ with some dynamic programming. An example in python :
_db={}
def f(m,n,k) :
if k>n or n>m: return 0
if k==n and k==m : return 1
if (m,n,k) in _db : return _db[(m,n,k)]
_db[(m,n,k)]=sum(f(m-i-1,n-i,k) for i in range(k+1))
_db[(m,n,k)]+=sum(f(m-k-1,n-k,i) for i in range(k))
return _db[(m,n,k)]
And then compute $f(300,60,8) = 3093843836691461760481283005507531400353771346949728892427692$ in a few seconds. | 2021-06-17T17:32:39 | {
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https://irbisgs.com/is/bali/236059723e5b5d9cd344ad557de10-spread-statistics-example | Cancel the common factor of 10+74+78+102 10 + 74 + 78 + 102 and 4 4. Examining the raw data is an essential first step before proceeding to statistical analysis. It is mostly used in the variable array where there is more than 1 values are expected. The average is the addition of all the numbers in the data set and then having those numbers divided by the number of numbers within that set. Xlsx. RAND () Returns a pseudorandom number, equal to or greater than zero and less than one.
You can calculate set-up costs, profit and loss forecast, breakeven forecast and balance sample sheet forecast by this template. Or download one of the many sample data files in Excel format. ; The central tendency concerns the averages of the values. With this shape, the odds of anything happening are equal. There are 3 main types of descriptive statistics:The distribution concerns the frequency of each value.The central tendency concerns the averages of the values.The variability or dispersion concerns how spread out the values are. Types of descriptive statistics. An important characteristic of any set of data is the variation in the data. Statistics. A measure of spread gives us an idea of how well the mean, for example, represents the data. However, as we are often presented with data from a sample only, we can estimate the population standard deviation from a sample standard deviation. Inferential statistics have two main uses: making estimates about populations (for example, the mean SAT score of all 11th graders in the US). The original article indicated that kurtosis was a measure of the flatness of the distribution or peakedness. Range defines the spread, or variability, of a sample. The range of the data is given as the difference between the maximum and the minimum values of the observations in the data. Measures of the Spread of the Data. Finally all pictures we've been displayed in this site will inspire you all. The smaller the Standard Deviation, the closely grouped the data point are. A sample statistic is a characteristic or measure obtained by using data values from a sample.
Thank you for visiting. The easiest way to describe the spread of data is to calculate the range. Appsloveworld allows developers to download a sample Excel file with a large dummy data for testing purposes. When you have collected data from a sample, you can use inferential statistics to understand the larger population from which the sample is taken. A person who has sex with an infected partner can become infected with the virus. value is the column where values will fill in under the new variables created from key. When data are skewed, the majority of the data are located on the high or low side of the graph. Bond yield is the internal rate of return of the bond cash flows. This is technically not correct (see below).
Historically, the average credit spread between 2-year BBB-rated corporate bonds and 2-year U.S. Treasuries is 2%. Sample excel data for analysis. For example, SD = +/- 0.5 cm. On the menu bar, click Excel Templates > Create Excel Template. The range tells us that the spread from the tallest to the shortest is 0.33m (33cm). The hepatitis B virus can be found in the blood, semen, and other body fluids of an infected person. Lets look at the following data set.
An array value is also defined. An important characteristic of any set of data is the variation in the data. Dispersion Statistics. Its the most common way to measure how spread out data values are. Excel is a popular tool for data analysis, especially among non-statisticians. Yes. Table 3.7 shows the numbers that can be used to summarize measurement data. The sample standard deviation s is equal to the square root of the sample variance: s = 0.5125 = 0.715891, The range covered by the data is the most intuitive measure of spread and is exactly the distance between the smallest data point (min) and the largest one (Max). Yield spread is the difference between the yield to maturity on different debt instruments. We can use the range and the interquartile range to measure the spread of a sample. Another measure of spread is the inter-quartile range (IQR), which is the range covered by the middle 50% of the data. Inferential statistics, by contrast, allow scientists to take findings from a sample group and generalize them to a larger population. Can hepatitis B be spread through food? Basically, it is the square-root of the Variance (the mean of the differences between the data points and the average). 2. Real world data is often normally distributed. Find the value that is two standard deviations below the mean. 10, 14, 8, 10, 15, 4, 7. Measures of spread summarise the data in a way that shows how scattered the values are and how much they differ from the mean value. The following numbers would be 27, 54, 13, 81, and 6. The sample variance, s2, is equal to the sum of the last column (9.7375) divided by the total number of data values minus one (20 1): s2 = 9.7375 20 1 = 0.5125. The two types of statistics have some important differences. Size: 100.9KB. The most common measure of variation, or spread, is the standard deviation. The sample standard deviation s is equal to the square root of the sample variance: s But since there is an outlier of \$110,000 in this sample, the standard deviation is inflated such that average distance is about \$17,936. It is intuitively obvious why we define range in statistics this way - range should suggest how diversely spread out the values are, and by computing the difference between the maximum and minimum values, we can get an estimate of the spread of the data. In datasets with a What is a It allows us the privilege to obtain a list of parameters from an array. Give any two examples of collecting data from day-to-day life. The smaller the Standard Deviation, the closely grouped the data point are. The current yield on a 2-year BBB-rated corporate bond is 5%, while the current yield on a 2-year U.S. Treasury is 2%. For example, finding the median is simply discovering what number falls in the middle of a set. Go to Sales > Opportunities > My Open Opportunities. The final part of descriptive statistics that you will learn about is finding the mean or the average. The Range (Statistics) The Range is the difference between the lowest and highest values. Xls. In some data sets, the values are concentrated closely, while in others the are more spread out. Development on spread () is complete, and for new code we recommend switching to pivot_wider (), which is easier to use, more featureful, and still under active development. #ofSTDEV does not need to be an integer. Click Upload. Excel has different types of formats like Xls and Xlsx. Explore a vast collection of premium Excel templates made available when you subscribe to Microsoft 365, or check out an expansive selction of free Excel templates.
Statistics is a form of mathematical analysis that uses quantified models, representations and synopses for a given set of experimental data or real-life studies. Measures of spread together with measures of location (or central tendency) are important for identifying key features of a sample to better understand the population from which the sample comes from. Covering popular subjects like HTML, CSS, JavaScript, Python, Available as Google Sheets . Basically, it is the square-root of the Variance (the mean of the differences between the data points and the average). The Interquartile Range (IQR), also called the mid-spread, is a measure of statistical dispersion, being equal to the difference between 75th and 25th percentiles, or between upper and lower quartiles, IQR = Q3 Q1. For example, the mean score for the group of 100 students we used earlier was 58.75 out of 100. In descriptive statistics, the mean may be confused with the median, mode or mid-range, as any of these may be called an "average" (more formally, a measure of central tendency).The mean of a set of observations is the arithmetic average of the values; however, for skewed distributions, the mean is not necessarily the same as the middle value (median), or the most likely value (mode). The range is the difference between the highest and lowest values from a sample . The below is one of the most common descriptive statistics examples. You can modify any time and update as per your requirements and uses. 1. [su_note note_color=#d8ebd6] The For example, if your minimum value is $10 and the maximum value is$100 then the range is $90 ($100 $10). The more spread out a data distribution is, the greater its standard deviation. Limitations of Range . Data Analysis & Quality Control. Standard Deviation is the measure of how far a typical value in the set is from the average. Variability for our Descriptive Statistics Example. In this instance, the IQR is the preferred measure of spread because the sample has an outlier. For instance, when the variance of data in a set is large, the data is widely scattered. Examine the shape of your data to determine whether your data appear to be skewed. Central tendency. If the spread of values in the data set is large, the mean is not as representative of the data as if the spread of data is small. Spreadsheets can be stored in a digital format with any amount of data. For example, the mean of our data is 29.0 years. File Format. Standard Deviation is the measure of how far a typical value in the set is from the average. With the use of tidyverse package is become easy to manage and create new datasets. This Excel spreadsheet example can be useful in creating a financial plan for your business. Spread operator allows an iterable to expand in places where 0+ arguments are expected. Usually, we are interested in the standard deviation of a population. The range represents the difference between the largest and smallest value. Example: if our 5 dogs are just a sample of a bigger population of dogs, we divide by 4 instead of 5 like this: Sample Variance = 108,520 / 4 = 27,130. Here is a histogram of the age of Nobel Prize winners when they won the prize: The normal distribution drawn on top of the histogram is based on the population mean ($$\mu$$) and standard deviation ($$\sigma$$) of the real data. Related Topics: Common Core (Statistics & Probability) Common Core for Mathematics Examples, solutions, videos, and lessons to help High School students learn how to interpret differences in shape, center, and spread in the context of the data sets, accounting for possible effects of extreme data points (outliers). Point estimate in statistics is calculated from sample data and used to estimate an unknown population parameter. The minimum and maximum are the smallest and largest values.Q2 is the median of the dataset.The 1st and 2nd quartiles are the medians on both sides of Q2.25% of our data falls before Q1 it represents the 25th percentile.75% of our data falls before Q3 it represents the 75th percentile.More items Divide the sum by the total number of data. However, pivot_wider is a more flexible alternative to spread. You can draw similar conclusions from the weight data. The sample variance, s2, is equal to the sum of the last column (9.7375) divided by the total number of data values minus one (20 1): s 2 = 9.7375 20 1 = .5125. s 2 = 9.7375 20 1 = .5125. Some rough measures of spread we have already seen are the range and IQR. Questions on Statistics with Answers. For example, two data sets can have very different spreads but still have the same range. Find ( + 1s). Sample Excel Sheet With Large Data And Microsoft Sample Excel Spreadsheets can be beneficial inspiration for people who seek an image according specific topic, you can find it in this site. Spread. It is described by the minimum and maximum values of the variables. The range gives the spread between the lowest and highest values. 3, 3, 4, 6, 7, 9, 10, 11, 11, 12 A measure of spread gives us an idea of how well the mean, for example, represents the data. The mean is the most common measure of central tendency used by researchers and people in all kinds of professions. One has the liberty of accessing a decade long data on a spreadsheet from any corner of the world by storing it on a drive. To describe spread, a number of statistics are available, including the range, quartiles and standard deviation. A normal distribution, sometimes called the bell curve (or De Moivre distribution [1]), is a distribution that occurs naturally in many situations.For example, the bell curve is seen in tests like the SAT and GRE. The standard deviation is a measure of the spread of scores within a set of data. W3Schools offers free online tutorials, references and exercises in all the major languages of the web. One is two standard deviations less than the mean of five because: 1 = 5 + ( 2) ( 2) 1 = 5 + (2) (2) 1 = 5+(2)(2) . The parameters and statistics with which we first concern ourselves attempt to quantify the "center" (i.e., location) and "spread" (i.e., variability) of a data set. Download. spread.Rd. This test examines the hypothesis about the median 0 of a population. A Real Data Example of Normally Distributed Data. But since there is an outlier of \$110,000 in this sample, the standard deviation is inflated such that average distance is about \\$17,936. gather and spread with dplyr easy example. Maximum Value in the data = 15. Spread, in the context of statistics, describes the variation of your data and how dispersed or spread out it is. Sign test. The other type of descriptive statistics is known as the measures of spread. Example 3: Lets say you have a sample of 5 girls and 6 boys. Step-by-Step Examples.
Example. There are 3 main types of descriptive statistics: The distribution concerns the frequency of each value. The range is a descriptive statistic that gives a very crude indication of how spread out a set of data is by subtracting the minimum from maximum values. A Sample: divide by N-1 when calculating Variance. Median. ; You can apply these to assess only one variable at a time, in univariate analysis, or to compare two or more, in
football trends and facts | 2023-01-31T03:53:46 | {
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https://math.stackexchange.com/questions/3938624/derivative-of-sqrtxtax?noredirect=1 | # Derivative of $\sqrt{x^TAx}$
Let $$x\in \mathbb{R}^n$$ and $$A\in\mathbb{R}^{n\times n}$$ be a positive semi-definite matrix.
Is there a way to express in closed form the following derivative?
$$\frac{\partial}{\partial x} \sqrt{x^TAx}$$
• In this context, are positive semidefinite matrices necessarily symmetric? Dec 7, 2020 at 15:47
• In any case, the answer will be $\frac{1}{2\sqrt{x^TAx}}(A + A^T)x$. Dec 7, 2020 at 15:48
• Thanks, Ben. If you post this as an answer I can close this thread and confirm the answer. It would be nice also if you can add some more information on the derivation. Dec 7, 2020 at 15:51
First of all, using the chain rule, we have $$\frac{\partial }{\partial x} \sqrt{x^TAx} = \frac{1}{2 \sqrt{x^TAx}}\cdot \frac{\partial }{\partial x} x^TAx.$$ One approach to this partial derivative is to write the expression $$f(x + h)$$ in the form $$f(x) + g(x)^Th + o(h)$$; by definition, the $$g(x)$$ for which this holds is equal to $$\frac{\partial f}{\partial x}$$. With that in mind, $$(x+h)^TA(x + h) = x^TAx + x^TAh + h^TAx + h^TAh\\ = x^TAx + x^TAh + [h^TAx]^T + o(h)\\ = x^TAx + x^TAh + x^TA^Th + o(h)\\ = x^TAx + [(A + A^T)x]^Th + o(h).$$ With that, the derivative of $$x \mapsto x^TAx$$ is $$(A + A^Tx)$$, and we have $$\frac{\partial }{\partial x} \sqrt{x^TAx} = \frac{1}{2 \sqrt{x^TAx}}\cdot (A + A^T)x.$$
$$\def\o{{\tt1}}\def\p{\partial} \def\L{\left}\def\R{\right}\def\LR#1{\L(#1\R)} \def\trace#1{\operatorname{Tr}\LR{#1}} \def\grad#1#2{\frac{\p #1}{\p #2}}$$The simplest approach is to square the function and then use implicit differentiation \eqalign{ f^2 &= x^TAx \\ 2f\;df &= \LR{dx^TAx+x^TA\,dx} = \LR{Ax+A^Tx}^Tdx \\ \grad{f}{x} &= \frac{\LR{A+A^T}x}{2f} = \frac{\LR{A+A^T}x}{2\sqrt{x^TAx}} \\ }
If $$A$$ is symmetric, you can rewrite it as $$A = B^T B$$ for some matrix $$B$$ (see more information on how to find $$B$$ here).
Hence we have $$x^TAx = x^TB^TBx = (Bx)^TBx = \|Bx\|^2$$
Where naturally $$\|Bx\|^2$$ is a scalar. | 2022-07-07T11:41:45 | {
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http://math.stackexchange.com/questions/457103/word-problem-in-probability | # word problem in probability
i would like to exactly what is asked in this problem:
The probability that A can solve the problem is $1/4$ and that B can solve it is $1/3$. If both of them try, what is the probability that problem will be solved.
$A. 1/4$
$B. 7/12$
$C. 1/3$
$D. 1/2$
$E. 1/12$
i will say what is a point of my confusion,when we are trying to find probability of two independent events,we are multiplying probability of each other to get probability that both event occur .now first it is what i have tried and got $1/12$,but it shows me that it is not correct,now i have this question and please could help me,how can i translate if both of them will try into probability language?
$P(A or B)$ is not correct because it means that one or second will solve,so it means that $P(A and B)$,but they are independent are this would not be product of their probabilities?let us denote probability that $A$ will solve by $P(A)$, and probability that $B$ will solve by $P(B)$
$P(A)=1/4$
$P(B)=1/3$
now what would be $P(A and B)$?would not it be $1/4*1/3$?
or i am calculating wrongly and it would be $P(A or B)$
-
It's 1 minus probability of "not A and not B". – Gerry Myerson Aug 1 '13 at 9:29
but in terms of $A$ and $B$ itself? – dato datuashvili Aug 1 '13 at 9:30
probability of not-A is 1 minus probability of A. – Gerry Myerson Aug 1 '13 at 9:32
@GerryMyerson this one i know,i am asking different thing – dato datuashvili Aug 1 '13 at 9:45
OK: what thing is it that you are asking? – Gerry Myerson Aug 1 '13 at 9:47
You're trying to find the probability that either A or B solves the question. An easier way to think about it is this - the only way that the problem does not get solved is if both cannot solve it. We can find the probability that A and B cannot solve it, and subtract this from one.
As you mentioned, the probability of A not solving AND B not solving is simply the product of these. Hence, we find the probability that A cannot solve, which is $1 - \frac 14 = \frac 34$. Similarly, we find the probability that B cannot solv, which is $1 - \frac13 = \frac 23$. Multiplying, we have $1- \frac34 \times \frac 23$. This gives us an answer of $1- \frac12 = \frac 12$.
Hope that helps. Cheers!
-
i see,i understood thanks very much – dato datuashvili Aug 1 '13 at 9:37
only one thing is why i can't use formula for $P(A and B)$ – dato datuashvili Aug 1 '13 at 9:39
for such problem,we are saying that it is equal to 1- neither A nor B,but why it could not be simply product? – dato datuashvili Aug 1 '13 at 9:44
Because the product obviously gives the wrong answer. The product is smaller than either of the individual probabilities, whereas the answer must be bigger than either of the individual probabilities. – Gerry Myerson Aug 1 '13 at 9:48
@dato: The question asks for the probability that the problem gets solved. This can happen in any of three ways: A solves it and B does not, B solves it and A does not, or both of them solve it. The formula for P(A and B) only gives you the probability of the third "way". – HJ32 Aug 1 '13 at 10:03
If you want to do it directly (though the proposed way by Gerry is both easier to understand and easier to carry on, imo), you can argue as follows:
The event whose probability we want is "$A$ solves the problem or $B$ solves the problem", and this in numbers is given by "prob. that $A$ solves + prob. that $B$ solves minus the prob. that they both solve
$$\frac14+\frac13-\frac1{12}=\frac6{12}=\frac12$$
Can you see why we have to substract the probability that they both solve the problem?
-
but last one probability that they both solve it is asked exactly,i am confused – dato datuashvili Aug 1 '13 at 9:54
@dato, the fact that both $\,A,B\,$ solve the problem already is contained or observed in the "either $A$ solves the problem or $B$ solves the problem" part, since "or" in mathematics is inclusive, not exclusive. – DonAntonio Aug 1 '13 at 9:57
i see ,so in other word,probability that if they both will try,problem would be solved is the same as either A or B will solve? – dato datuashvili Aug 1 '13 at 10:00
consider that it has answer $1/12$ for confusion – dato datuashvili Aug 1 '13 at 10:01
Yes to your first question, as long as we're clear that saying "either A solves the problem or B solves the problem" contains within it the even "Both A and B solve the problem". – DonAntonio Aug 1 '13 at 10:13 | 2015-04-22T04:01:43 | {
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https://math.stackexchange.com/questions/426191/concavity-and-convexity | # Concavity and Convexity
A set $X \subseteq \mathbb{R}^n$ is said to be convex if $tx + (1-t)y \in X$ for all $x,y \in X$ and $t \in (0,1)$. Given a convex set $X \subseteq \mathbb{R}^n$, a function $f: X \to \mathbb R$ is said to be concave if $f(tx + (1-t)y) \ge tf(x) + (1-t)f(y)$ for all $x,y \in X$ and $t \in (0,1)$.
1) Show that $f: \mathbb{R}^n \to \mathbb R$ is concave iff $\sum_{i=1}^r t_if(x_i) \le f\left(\sum_{i=1}^r t_ix_i\right)$
for every positive integer r, for all $x_1, \dots, x_r \in \mathbb R^n$, and all $t_1,\dots,t_r \in (0,1)$ with $\sum_{i=1}^r t_i = 1$
2) Use (1) to show that $\prod_{i=1}^r x_i^{t_i} \le \sum_{i=1}^r t_ix_i$ for all non negative $x_1, \dots, x_r \in \mathbb R$ and all $t_1,\dots,t_r \in (0,1)$ with $\sum_{i=1}^r t_i = 1$
3) Show that the function $f: \mathbb{R}^n \to \mathbb{R}$ is convex iff the set $\{(x,r) \in \mathbb{R}^n \times \mathbb{R} \mid f(x) \le r \}$ is convex.
• why isnt my text getting formatted? :O – Mathy Jun 21 '13 at 14:12
• You have to put your LaTeX commands between \$\$ – math Jun 21 '13 at 14:13
• I followed the MathJax basic tutorial. Not even a single one is working. or do we need to add something extra to activate these formats? check \ge – Mathy Jun 21 '13 at 14:16
• please yous \to for $\to$ and \mathbb{R} for $\mathbb{R}$ – math Jun 21 '13 at 14:20
• Please, see my edits to your code for (1) and (2). – Nick Peterson Jun 21 '13 at 14:21
For part 1), try induction on $r$. For 2), think about log. For 3) think about the pairs $(x, f(x)), (y, f(y))$.
I'm being somewhat mysterious because I think these are healthy exercises to solve for oneself.
EDIT:
Hey sorry for the delay. For the first part, the second condition implies concavity so we only need to prove the first condition implies the second. For the base case, this is easy. Suppose it holds for k and pick $x_1, ..., x_k, x_{k+1} \in \mathbb{R}^n$ and $t_1, ..., t_{k+1} \in (0, 1)$ such that $\sum\limits_{i=1}^{k+1} t_i=1$. Now, let $x'_k=\frac{t_kx_k+t_{k+1}x_{k+1}}{t_k+t_{k+1}}$ and $t'_k= t_k+t_{k+1}$. Then $\sum\limits_{i=1}^{k-1} t_i + t'_k=1$ and by inductive assumption we therefore have $t'_kf(x'_k)+\sum\limits_{i=1}^{k-1} t_if(x_i)$ $\leq f( t'_kx'_k+\sum\limits_{i=1}^{k-1} t_ix_i)=f(\sum\limits_{i=1}^{k+1} t_ix_i)$. To finish up, we need to show that $t_kf(x_k)+t_{k+1}f(x_{k+1}) \leq t'_kf(x'_k)$. Now, $0<t_k+t_{k+1}<1$ so there is some $r \in \mathbb{R}$ so that $rt_k+rt_{k+1}=1$. Then by the base case, $rt_kf(x_k)+rt_{k+1}f(x_{k+1}) \leq f(rt_kx_k+rt_{k+1}x_{k+1})=f(rt'_kx'_k)=f(x'_k)$. But by definition, $r= \frac{1}{t_k+t_{k+1}}$ and so dividing by $r$, this gives what we want.
• thanx for your answer. I am trying it right now. As far as you being mysterious is concerned: I can give you the link, from where these questions are taken. This question was asked in an exam in the year 2004. – Mathy Jun 21 '13 at 15:20
• Cool. Also, if anything I wrote is too vague, I can definitely clarify. – Alexander Jun 21 '13 at 15:24
• First part: I am not able to prove it via induction. Base case: it is obvious because r=1 so $t_1f(x_1) \le f(t_1x_1)$ and $t_1$ = 1. but I got stuck when proving it for r=k+1 assuming the inequality to hold good for r=k;i.e. $t_1f(x_1) + ... +t_kf(x_k)$ + $t_{k+1}x_{k+1}$ $\le f(t_1x_1 + ... t_kx_k$ + $t_{k+1}x_{k+1}$) . how to prove this?? – Mathy Jun 21 '13 at 16:21
• Alexander I am still waiting for you to clarify more. thanx – Mathy Jun 22 '13 at 12:15
• Sorry, I was away from my computer for a day. I have updated my answer to give a more thorough solution to part 1. – Alexander Jun 22 '13 at 14:44 | 2019-08-22T20:04:59 | {
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https://math.stackexchange.com/questions/2273950/find-the-arc-length-of-a-path | # Find the arc length of a path
I'm trying to solve the arc length for the following. $c(t) = (2, 6t^2, 4t^3)$ from $0\le t\le 1.$
I've checked on WolframAlpha and I get the answer $8\sqrt2 - 4$ but when I work it out I get $16\sqrt2$ where am I going wrong?
\begin{align*} L&= \int_0^1 \sqrt{(12t)^2 + (12t^2)^2}\, dt\\ &= \int_0^1 \sqrt{144t^2 + 144t^4}\, dt\\ &= \int_0^1 12t\sqrt{1 + t^2}\,dt \end{align*}
Then I do $$8t\sqrt{(1+t^2)^3} = 16\sqrt2.$$
• How did you get to $16\sqrt{2}$? May 10 '17 at 0:22
From $$\int_{0}^{1}12t\sqrt{1+t^2}dt$$ Use $u=1+t^2\implies \frac{1}{2}du=tdt$ to get $$\int_{0}^{1}12t\sqrt{1+t^2}dt=\int12\sqrt{u}\cdot\frac{1}{2}du=6u^{3/2}\cdot\frac{2}{3}=4(1+t^2)^{3/2}\bigg\vert_{0}^{1}=4\cdot2^{3/2}-4\cdot1$$ $$=4\cdot2\cdot2^{1/2}-4\cdot1=8\sqrt{2}-4$$
From $$\int_0^1 12t\sqrt{1 + t^2}\,dt,$$ let \begin{align*} u &= 1+t^{2}\\ du &= 2t\,dt. \end{align*}
This yields $$\int_{1}^{2}6 u^{1/2}\,du =4u^{3/2}\bigg|_{1}^{2}=4\sqrt{8}-4 = 8\sqrt{2}-4.$$
$$z=6t^2+i4t^3\\ \dot z=12t+i12t^2\\ |\dot z|=12t\sqrt{1+t^2}\\ s=\int_0^1|\dot z|dt=6\int_0^12t\sqrt{1+t^2}dt=6\int_0^1\sqrt{1+u}\ du,\quad u=t^2, du=2tdt\\ s=4(1+u)^{3/2}\big|_0^1=8\sqrt{2}-4$$ | 2022-01-23T16:18:22 | {
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https://math.stackexchange.com/questions/595483/finding-indicated-trigonometric-value-in-specified-quadrant | # Finding indicated trigonometric value in specified quadrant
If I have csc $\theta$ = - $\dfrac{10}{3}$ and have to find tan $\theta$ in quadrant III, would I use 1 + $\cot^2\theta$ = $\csc^2\theta$ then find reciprocal which would be tan $\theta$?
If so, I get $\dfrac{3\sqrt{91}}{91}$ as tan, but that doesn't seem right as if I wanted to get cot $\theta$ from tan now, it would be different. I hope that made sense. Any help would be appreciated!
$$\csc \theta = \frac 1{\sin\theta} = -\frac {10}{3} \implies \sin\theta = -\frac 3{10} = \frac{\text{opposite}}{\text{hypotenuse}}$$
$$\tan \theta = \dfrac{\sin\theta}{\cos\theta} = \frac{\text{opposite}}{\text{adjacent}}$$
All you need is to find $\cos\theta$ in the third quadrant to compute tangent. Use the right angle that $\theta$ forms with the x-axis and the Pythagorean Theorem: $$3^2 + \text{adjacent}^2 = 10^2 \implies \text{adjacent} = \sqrt{91}$$ or else use the identity:
$$\sin^2\theta + \cos^2\theta = 1$$
knowing that $\cos \theta < 0$ in the third quadrant.
However, you are also correct having used your method: $$\tan\theta = \dfrac 3{\sqrt {91}} = \dfrac{3\sqrt{91}}{91}$$
• Oh wow, I was actually correct. I tried using the theorem you gave me, but I kept getting a positive answer. I get sqrt(91)/10. When I move 9/100 over to other side I subtracted it by 100/100 then took square root. Am I doing basic math wrong? D:
– o.o
Dec 6, 2013 at 13:30
• $\cos \theta$, in Quadrant III, is negative, (remember, the identity above is in $\cos^2 \theta$, so $\cos\theta = \pm \sqrt {\cos^2\theta}$. So $\cos\theta = -\dfrac{\sqrt{91}}{10}$, giving us $$\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{-3/10}{-\sqrt{91}/10} = \dfrac 3{\sqrt{91}} \cdots$$ Dec 6, 2013 at 13:34
• Oh I see. Thank you so much for your help! Makes a lot more sense now.
– o.o
Dec 6, 2013 at 13:36
• And no, you're not doing basic math wrong, it's just a matter of knowing which quadrant you're working in, and the corresponding signs of the trig functions. You're welcome! Dec 6, 2013 at 13:36
• @amWhy: Needs a TU w/G! +1 Dec 7, 2013 at 0:14 | 2023-03-21T10:11:15 | {
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https://math.stackexchange.com/questions/4061071/in-definite-integration-do-i-have-to-watch-out-for-the-behaviour-of-a-function/4065306 | # In definite integration, do I have to watch out for the behaviour of a function between the limits?
Calculate - $$\int^{1}_{-1}\frac{e^\frac{-1}{x}dx}{x^2(1+e^\frac{-2}{x})}$$
This is my approach-
Substitute $$t=e^\frac{-1}{x}$$
$$\therefore \int^{\frac{1}{e}}_{e}\frac{dt}{1+t^2}$$
which gives us $$\frac{\pi}{2}- 2\arctan{e}$$, but the answer given is $$\pi- 2\arctan{e}$$. I'm guessing that my wrong answer has something to do with the exponential term.
• For what it's worth, the "exact" answer from WA is wrong too in the same way: wolframalpha.com/input/… (but the decimal approximation is correct). Mar 14 at 4:46
• Correction: WA is not wrong in the same way. I neglected the fact that $\tan(1/e) = \frac\pi2 - \tan(e).$ Mar 14 at 5:59
• You mean arctan, of course, @DavidK . Mar 14 at 6:46
• @TedShifrin Yes, I mean $\arctan(1/e) = \frac\pi2 - \arctan(e).$ Mar 14 at 15:03
You're pretty much correct. The usual theorem for integration by substitution says that we can use a substitution $$t=\phi(x)$$ if $$\phi$$ is $$C^1$$ (continuously differentiable) in the interval of integration. (This is just a sufficient condition of course).
But your substitution $$t=\phi(x)=e^{-\frac 1x}$$ is not even continuous at $$x=0$$. In fact $$\lim_{x\to0^-} \phi(x) = \infty$$ $$\lim_{x\to0^+} \phi(x) = 0$$ So we have to split the integral up like this: $$\int_{-1}^0 \frac{e^{-\frac{1}{x}}}{x^2(1+e^{-\frac{2}{x}})} dx = \int_e^\infty \frac{1}{1+t^2} dt = \arctan{\infty} - \arctan e = \frac{\pi}{2} - \arctan e$$ and $$\int_0^1 \frac{e^{-\frac{1}{x}}}{x^2(1+e^{-\frac{2}{x}})} dx = \int_0^{e^{-1}} \frac{1}{1+t^2} dt = \arctan{e^{-1}} - \arctan 0 = \frac{\pi}{2} - \arctan e$$ so $$\int_{-1}^1 \frac{e^{-\frac{1}{x}}}{x^2(1+e^{-\frac{2}{x}})} dx = \pi - 2\arctan e$$
Bottom line: If $$\phi$$ is not continuous (or $$C^1$$, but that would be rarer) on the interval, then split the interval up.
BONUS: In this case we could also have taken a common short cut. We can notice that $$f(x) = \frac{e^{-\frac{1}{x}}}{x^2(1+e^{-\frac{2}{x}})}$$ is an even function (i.e. $$f(x)= f(-x)$$). Therefore we can also write $$\int_{-1}^1f(x)dx = 2\int_{0}^1 f(x)dx$$ and then the substitution is legal again.
The answer to the question in the header is yes, and this is a really interesting example.
When you substitute a variable, the substitution applies to the whole interval of integration. If you say $$t=e^{\frac{-1}{x}}$$, you mean that all the values that $$x$$ takes on are related to the values that $$t$$ takes by that relation. Really you're applying the function $$e^{\frac{-1}{x}}$$ to the whole interval of the $$x$$'s, i.e. $$[-1,1]$$. Normally, when we do substitution the function you use maps an interval to an interval, so it suffices to just look at the endpoints. Not so with this function! What is the image of $$[-1,1]$$ under $$e^{-\frac1{x}}$$? It's $$[0,\frac1e]\cup [e,\infty]$$, which you can see in a number of ways, e.g. the image of $$[-1,1]$$ under $$1/x$$ is $$[-\infty,-1]\cup [1,\infty]$$, then map it through $$e^{-x}$$ to get the right interval. Thus the correct substitution ought to be $$\begin{eqnarray} &&\int_{-1}^1\frac{e^{-\frac1x}}{x^2(1+e^{-\frac2x})}dx = \int_{[0,\frac1e]\cup [e,\infty]} \frac1{1+t^2} dt = \int_0^{\frac1e} \frac1{1+t^2} + \int_e^\infty \frac1{1+t^2} dt\\ &=& \arctan{\frac1e} - \arctan0 + \arctan{\infty} - \arctan{e} = \arctan\frac1e + \frac\pi2 - \arctan(e) \\&=& \pi-2\arctan(e) \end{eqnarray}$$
• +1 I think this shows the heart of the problem better than mine. Mar 17 at 16:26
You can start from the substitution $$\;y=-\dfrac1x.\;$$ Then $$I=\int\limits_{-\infty}^{-1}\dfrac{e^y\,\text dy}{e^{2y}+1} +\int\limits_1^\infty\dfrac{e^y\,\text dy}{e^{2y}+1} = \int\limits_1^\infty\dfrac{e^{-y}\,\text dy}{e^{-2y}+1} + \int\limits_1^\infty\dfrac{e^y\,\text dy}{e^{2y}+1} = 2\int\limits_1^\infty\dfrac{e^y\,\text dy}{e^{2y}+1}$$ $$= 2\int\limits_e^\infty\dfrac{\text de^y}{1+e^{2y}} =2\arctan e^y\bigg|_1^\infty =\pi-2\arctan e,$$ without doubts in the result.
• This doesn't change much. A "blind" Riemann-style substitution would get us $\int_{1}^{-1}dy$, analogous to OP's attempt. Mar 18 at 15:15
• @Milten Thank you for adice! Done. Mar 18 at 15:38 | 2021-10-25T10:12:03 | {
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# A nickel, a dime, and 2 identical quarters are arranged
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A nickel, a dime, and 2 identical quarters are arranged along a side of a table. If the quarters and the dime have to face heads up, while the nickel can face either heads up or tails up, how many different arrangements of coins are possible?
A. 12
B. 24
C. 48
D. 72
E. 96
[Reveal] Spoiler: OA
Last edited by Bunuel on 05 Feb 2012, 19:44, edited 1 time in total.
Edited the question and added the OA
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Re: still permutations with repetitions [#permalink]
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05 Feb 2012, 17:14
can somone explain this please? 4!/2! to get to diff ways the coins can be placed taking into acct the repeat in quarters but i don't get the *2! part. Thanks!
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T740qc wrote:
can somone explain this please? 4!/2! to get to diff ways the coins can be placed taking into acct the repeat in quarters but i don't get the *2! part. Thanks!
Welcome to Gmat Club.
THEORY.
Permutations of $$n$$ things of which $$P_1$$ are alike of one kind, $$P_2$$ are alike of second kind, $$P_3$$ are alike of third kind ... $$P_r$$ are alike of $$r_{th}$$ kind such that: $$P_1+P_2+P_3+..+P_r=n$$ is:
$$\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}$$.
For example number of permutation of the letters of the word "gmatclub" is $$8!$$ as there are 8 DISTINCT letters in this word.
Number of permutation of the letters of the word "google" is $$\frac{6!}{2!2!}$$, as there are 6 letters out of which "g" and "o" are represented twice.
Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be $$\frac{9!}{4!3!2!}$$.
BACK TO THE ORIGINAL QUESTION:
A nickel, a dime, and 2 identical quarters are arranged along a side of a table. If the quarters and the dime have to face heads up, while the nickel can face either heads up or tails up, how many different arrangements of coins are possible?
A. 12
B. 24
C. 48
D. 72
E. 96
# of arrangements of a nickel, a dime, and 2 quarters, or arrangements of 4 letters NDQQ out of which 2 Q's appear twice will be $$\frac{4!}{2!}$$. Next, as nickel can face either heads up or tails up then we should multiple this number by 2, so finally we'll get: $$\frac{4!}{2!}*2=24$$.
Hope it's clear.
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Re: still permutations with repetitions [#permalink]
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13 Dec 2012, 09:09
Bunuel wrote:
T740qc wrote:
can somone explain this please? 4!/2! to get to diff ways the coins can be placed taking into acct the repeat in quarters but i don't get the *2! part. Thanks!
Welcome to Gmat Club.
THEORY.
Permutations of $$n$$ things of which $$P_1$$ are alike of one kind, $$P_2$$ are alike of second kind, $$P_3$$ are alike of third kind ... $$P_r$$ are alike of $$r_{th}$$ kind such that: $$P_1+P_2+P_3+..+P_r=n$$ is:
$$\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}$$.
For example number of permutation of the letters of the word "gmatclub" is $$8!$$ as there are 8 DISTINCT letters in this word.
Number of permutation of the letters of the word "google" is $$\frac{6!}{2!2!}$$, as there are 6 letters out of which "g" and "o" are represented twice.
Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be $$\frac{9!}{4!3!2!}$$.
BACK TO THE ORIGINAL QUESTION:
A nickel, a dime, and 2 identical quarters are arranged along a side of a table. If the quarters and the dime have to face heads up, while the nickel can face either heads up or tails up, how many different arrangements of coins are possible?
A. 12
B. 24
C. 48
D. 72
E. 96
# of arrangements of a nickel, a dime, and 2 quarters, or arrangements of 4 letters NDQQ out of which 2 Q's appear twice will be $$\frac{4!}{2!}$$. Next, as nickel can face either heads up or tails up then we should multiple this number by 2, so finally we'll get: $$\frac{4!}{2!}*2=24$$.
Hope it's clear.
I completely understood the concept. But i am little doubtful about one more condition given " If the quarters and dime have to face heads up"
Do we assume that they are facing up or we should subtract the possibility of having tails up??
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Re: still permutations with repetitions [#permalink]
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13 Dec 2012, 09:17
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Bunuel wrote:
T740qc wrote:
can somone explain this please? 4!/2! to get to diff ways the coins can be placed taking into acct the repeat in quarters but i don't get the *2! part. Thanks!
Welcome to Gmat Club.
THEORY.
Permutations of $$n$$ things of which $$P_1$$ are alike of one kind, $$P_2$$ are alike of second kind, $$P_3$$ are alike of third kind ... $$P_r$$ are alike of $$r_{th}$$ kind such that: $$P_1+P_2+P_3+..+P_r=n$$ is:
$$\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}$$.
For example number of permutation of the letters of the word "gmatclub" is $$8!$$ as there are 8 DISTINCT letters in this word.
Number of permutation of the letters of the word "google" is $$\frac{6!}{2!2!}$$, as there are 6 letters out of which "g" and "o" are represented twice.
Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be $$\frac{9!}{4!3!2!}$$.
BACK TO THE ORIGINAL QUESTION:
A nickel, a dime, and 2 identical quarters are arranged along a side of a table. If the quarters and the dime have to face heads up, while the nickel can face either heads up or tails up, how many different arrangements of coins are possible?
A. 12
B. 24
C. 48
D. 72
E. 96
# of arrangements of a nickel, a dime, and 2 quarters, or arrangements of 4 letters NDQQ out of which 2 Q's appear twice will be $$\frac{4!}{2!}$$. Next, as nickel can face either heads up or tails up then we should multiple this number by 2, so finally we'll get: $$\frac{4!}{2!}*2=24$$.
Hope it's clear.
I completely understood the concept. But i am little doubtful about one more condition given " If the quarters and dime have to face heads up"
Do we assume that they are facing up or we should subtract the possibility of having tails up??
The nickel can face either heads up or tails up, thus we multiply the total # of ways in which we can arrange NDQQ by 2. The quarters and the dime have to face heads up, so only 1 choice for both of them, thus we don't need to multiply further.
Hope it's clear.
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Re: A nickel, a dime, and 2 identical quarters are arranged [#permalink]
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28 Dec 2012, 07:29
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marcodonzelli wrote:
A nickel, a dime, and 2 identical quarters are arranged along a side of a table. If the quarters and the dime have to face heads up, while the nickel can face either heads up or tails up, how many different arrangements of coins are possible?
A. 12
B. 24
C. 48
D. 72
E. 96
How many ways to arrange {N}{D}{Q}{Q} ? $$=\frac{4!}{2!} = 12$$
$$=12*1*1*1*(2)$$ Since there are two ways to arrange the nickel...
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Re: A nickel, a dime, and 2 identical quarters are arranged [#permalink]
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Re: A nickel, a dime, and 2 identical quarters are arranged [#permalink]
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06 Dec 2015, 11:51
wrote 4!/2! * 2
and rewritten it as 4*3/1*2 and forgot to put on top 2!
because of this, I got 12
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Re: A nickel, a dime, and 2 identical quarters are arranged [#permalink]
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07 May 2016, 01:42
Hi, shouldn't the answer be 48.
For the Nickel and the Dime, we can select 2 places out of 4 in 4C2 ways followed by the arrangement in 2! ways. The two quarters then can have 4 combinations (HH, HT, TH, TT).
Hence the answer = 4C2 * 2! * 4 = 48. Please suggest Bunuel.
Please ignore any typos as I am new to this forum.
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Re: A nickel, a dime, and 2 identical quarters are arranged [#permalink]
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07 May 2016, 04:31
[quote="marcodonzelli"]A nickel, a dime, and 2 identical quarters are arranged along a side of a table. If the quarters and the dime have to face heads up, while the nickel can face either heads up or tails up, how many different arrangements of coins are possible?
A. 12
B. 24
C. 48
D. 72
E. 96
No of ways when nickel face head=4!/2!=12
no of ways when nickel face tail= 4!/2!=12
total no of ways =24
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Re: A nickel, a dime, and 2 identical quarters are arranged [#permalink]
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07 May 2016, 04:39
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Hi, shouldn't the answer be 48.
For the Nickel and the Dime, we can select 2 places out of 4 in 4C2 ways followed by the arrangement in 2! ways. The two quarters then can have 4 combinations (HH, HT, TH, TT).
Hence the answer = 4C2 * 2! * 4 = 48. Please suggest Bunuel.
Please ignore any typos as I am new to this forum.
Hi,
It says that quarters and dime have to be faced heads up..
ONLY Nickel can be either head or tail..
so Nickel can be placed in two ways..
ans 4C2*2!*2= 24..
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Re: A nickel, a dime, and 2 identical quarters are arranged [#permalink]
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12 May 2016, 10:08
chetan2u - yes, misread the question. thanks.
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Re: A nickel, a dime, and 2 identical quarters are arranged [#permalink]
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11 Sep 2016, 03:57
This clears it up!! Thanks a lot!!!
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Re: A nickel, a dime, and 2 identical quarters are arranged [#permalink]
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14 Jun 2017, 12:03
Can anyone give a complete explanation for this
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Re: A nickel, a dime, and 2 identical quarters are arranged [#permalink]
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14 Jun 2017, 12:06
ashisplb wrote:
Can anyone give a complete explanation for this
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There are several solutions given above. You should be more specific when asking a question.
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Re: A nickel, a dime, and 2 identical quarters are arranged [#permalink]
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14 Jun 2017, 12:17
+ B = 24
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Re: A nickel, a dime, and 2 identical quarters are arranged [#permalink]
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14 Jun 2017, 12:23
Please let me know whether my approach is correct or not
For 2 identical quarters =4C2
For dime =2 heads up are available
Nickel =either heads up or tails up =2
Total=4C2*2*2=24
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Re: A nickel, a dime, and 2 identical quarters are arranged [#permalink] 14 Jun 2017, 12:23
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Display posts from previous: Sort by | 2017-06-26T00:34:35 | {
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https://brilliant.org/discussions/thread/some-general-questions/ | ×
# Some General Questions
1. The first term of a sequence is 2014. Each succeeding term is the sum of the cubes of the digits of the previous term. What is the 2014th term of the sequence?
2. In a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is 17. What is the greatest possible perimeter of the triangle?
3. If real numbers a, b, c, d, e satisfy a + 1 = b + 2 = c + 3 = d + 4 = e + 5 = a + b + c + d + e + 3, what is the value of a^2 + b^2 + c^2 + d^2 + e^2
4. Let ABCD be a convex quadrilateral with perpendicular diagonals. If AB = 20, BC = 70 and CD = 90, then what is the value of DA?
5. A natural number k is such that k^2 < 2014 < (k + 1)^2. What is the largest prime factor of k?
6. Let S be a set of real numbers with mean M. If the means of the sets S∪{15} and S∪{15, 1} are M + 2 and M + 1, respectively, then how many elements does S have?
7. What is the smallest possible natural number n for which the equation x^2 − nx + 2014 = 0 has integer roots?
8. In a triangle ABC, X and Y are points on the segments AB and AC, respectively, such that AX : XB = 1 : 2 and AY : Y C = 2 : 1. If the area of triangle AXY is 10 then what is the area of triangle ABC?
9. Let XOY be a triangle with angleXOY = 90◦ . Let M and N be the midpoints of legs OX and OY , respectively. Suppose that XN = 19 and Y M = 22. What is XY ?
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## Comments
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All of these are from the Pre - RMO in Mumbai region, right?
- 2 years, 10 months ago
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Yes !!
- 2 years, 10 months ago
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If u have some more question of pre emo send to [email protected]. ..... k?
- 2 years, 10 months ago
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sure ... actually I too am hunting for these papers !! :)
- 2 years, 10 months ago
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This Set has all of the Pre - RMO questions of 2014. You can also find them on the HBCSE's site I think.
- 2 years, 10 months ago
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thanks ... If you have any more sets or papers of this kind then kindly send me the links or stuff... also HBCSE site copied ... thanks again ..!!!
- 2 years, 10 months ago
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waiting fr ur email dude.......
- 2 years, 10 months ago
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I said I have this pre rmo paper only ...you can check the websites suggested by Siddhartha Srivastava ..... kindly have a look. PS: I've got few more papers regarding class 10 and 11 course ..but i'll mail them to you after my exams are over .. so plz wait till 24th ..
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I wanted the papers so am asking since you have fr 10 th and 11 th pls do mail them earliest....pls
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https://math.stackexchange.com/questions/1939107/series-summation-of-polynomial-on-x-multiplied-by-x-power-summation/1939849 | # Series summation of polynomial on x multiplied by x power summation
An unbiased coin is tossed repeatedly and outcomes are recorded. What is the expected no of toss to get HT ( one head and one tail consecutively) ?
I tried to solve above question using the following probability tree diagram.
Probability in each branch is = $0.5$. I double circled the satisfying toss events. While observing the diagram I noticed that, from 2nd toss onward our required event starts showing up. Additionally,
1. in the $\text{2nd}$ toss (or the 3rd level) we have one satisfying case.
2. in the $\text{3rd}$ toss (or the 4th level) we have two satisfying case.
3. in the $\text{4th}$ toss (or the 5th level) we have three satisfying case.
4. in the $\text{5th}$ toss (or the 6th level) we have four satisfying case.
5. etc.
i.e. in the $\text{kth}$ toss we would have $(k-1)$ satisfying case. So,
$\\ E = \sum_{k=2}^{\infty } k.P(k)\ \\ E = \sum_{k=2}^{\infty } k.\left \{ (k-1)*(0.5)^k \right \}\\ \\ E = \sum_{k=2}^{\infty } \left \{ (k^2-k)*(0.5)^k \right \}\\$
I understand the series summation would converge because we have the $0.5^{k}$ term which eventually becomes zero.
But could not solve this polynomial and power series summation.
Please suggest how to solve this and or any other method.
We can describe all valid tosses as follows:
The valid tosses are
• zero or more tosses of T followed by
• zero or more tossed of H followed by
• HT
This corresponds to the tree diagram where we can see following valid paths starting from the root
Note the entries in a diagonal have the same length, starting with the first node $HT$ of length $2$. The entries in the diagonal with length $5$ are marked in blue.
From this scheme we can derive the expectation value. We obtain following the same arrangement
We can add up these values along the diagonals. Note, the diagonal marked in blue has $4$ entries
$$\{HHHHT,THHHT,TTHHT,TTTHT\}$$ of length $5$.
We obtain \begin{align*} E[X]&=\sum_{n=1}^\infty n(n+1)\left(\frac{1}{2}\right)^{n+1}\\ &=\left.\sum_{n=1}^\infty n(n+1)x^{n+1}\right|_{x=\frac{1}{2}}\\ &=x^2\left.\sum_{n=1}^\infty n(n+1)x^{n-1}\right|_{x=\frac{1}{2}}\\ &=x^2\left.\sum_{n=2}^\infty (n-1)nx^{n-2}\right|_{x=\frac{1}{2}}\\ &=\left.\left(x^2D_x^2\sum_{n=0}^\infty x^n\right)\right|_{x=\frac{1}{2}}\\ &=\left.\left(x^2D_x^2\frac{1}{1-x}\right)\right|_{x=\frac{1}{2}}\\ &=\left.\frac{2x^2}{(1-x)^3}\right|_{x=\frac{1}{2}}\\ &=4 \end{align*}
Here we use the differential operator $D_x:=\frac{d}{dx}$ and apply it to the geometric series.
• The wiki link directly points me the exact required formula. Thanks. – Debashish Sep 26 '16 at 5:39
• @Debashish: You're welcome! Good to see the answer is helpful! – Markus Scheuer Sep 26 '16 at 6:32
You can treat this as a Markov Chain with the following states:
End, Head, Tail (not end) (or equivalently worded, HT has occured, heads is most recent result, and no heads have occured respectively)
This has the transition matrix $\begin{bmatrix}1&0.5&0\\0&0.5&0.5\\0&0&0.5\end{bmatrix}$, which is in the standard form for an absorbing chain: $\left[\begin{array}{c|c}I&S\\\hline0&R\end{array}\right]$
We look to the fundamental matrix: $(I-R)^{-1}$ which in this case is $\begin{bmatrix}0.5&-0.5\\0&0.5\end{bmatrix}^{-1} = \frac{1}{1/4}\begin{bmatrix}0.5&0.5\\0&0.5\end{bmatrix}=\begin{bmatrix}2&2\\0&2\end{bmatrix}$
The fundamental matrix holds the information about how many steps are expected until we reach an absorbing state. Since our initial state can be thought of as being a tail (or more correctly worded, as being that no head has yet occurred), we look to the corresponding column and add the entries to find the expected time.
In this case, we see it will take an expected number of $2+2=4$ flips to have flipped a tails after a heads.
• Thanks ! please suggest or give some good reference regarding this Markov Chain Method. – Debashish Sep 24 '16 at 6:38
• The wikipedia links I already included have a good bit of information in them. Otherwise, PatrickJMT covers the basics rather well, though his notation differs from mine and textbooks from which I've taught and studied from in the past in that he uses row-stochastic matrices instead of column-stochastic matrices, so his state vectors will appear on the left and have rows summing to one, whereas I would write it with column vectors on the right of matrix with columns summing to one. He begins discussing absorbing chains in part 7 of his videos. – JMoravitz Sep 24 '16 at 6:46 | 2019-10-14T23:39:24 | {
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http://stonebrookacademy.com/how-many-wnd/prove-that-the-diagonals-of-a-parallelogram-bisect-each-other-ef38e9 | Theorem 8.6 The diagonals of a parallelogram bisect each other Given : ABCD is a Parallelogram with AC and BD diagonals & O is the point of intersection of AC and BD To Prove : OA = OC & OB = OD Proof : Since, opposite sides of Parallelogram are parallel. The Equation 2 gives. Verify your number to create your account, Sign up with different email address/mobile number, NEWSLETTER : Get latest updates in your inbox, Need assistance? Hence diagonals of a parallelogram bisect each other [Proved]. Thus the two diagonals meet at their midpoints. This is exactly what we did in the general case, and it's the simplest way to show that two line segments are equal. Thus the two diagonals meet at their midpoints. google_ad_client = "pub-9360736568487010"; . In AOD and BOC OAD = OCB AD = CB ODA = OBC AOD BOC So, OA = OC & OB = OD Hence Proved. Prove that the diagonals of a parallelogram bisect each other. That is, each diagonal cuts the other into two equal parts. Then the two diagonals are c = a + b (Eq 1) d = b - a (Eq 2) Now, they intersect at point 'Q'. We show that these two midpoints are equal. Start studying Geometry. In a quadrangle, the line connecting two opposite corners is called a diagonal. The position vectors of the midpoints of the diagonals AC and BD are (bar"a" + bar"c")/2 and (bar"b" + bar"d")/2. If possible I would just like a push in the right direction. The angles of a quadrilateral are in the ratio 3: 5: 9: 13. In this video, we learn that the diagonals of a parallelogram bisect each other. To prove that AC and BD bisect each other, you have to prove that AE = EC = BE = ED. In any parallelogram, the diagonals (lines linking opposite corners) bisect each other. Google Classroom Facebook Twitter Angles EDC and EAB are equal in measure for the same reason. 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D.K. Metcalf Womens Jersey | 2021-05-09T14:20:13 | {
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https://www.physicsforums.com/threads/every-cauchy-sequence-of-real-numbers-converges.370206/ | # Homework Help: Every Cauchy sequence of real numbers converges
1. Jan 17, 2010
### kingwinner
1. The problem statement, all variables and given/known data
I understand everything except the last two lines. I am really confused about the last two lines of the proof. (actually I was never able to fully understand it since my first year calculus)
I agree that if ALL three of the conditions n≥N, k≥K, and nk≥N are satisfied, then the last line is true. But why does the condition n≥N alone imply that the last line is true? Does n≥N guarantee that k≥K, and nk≥N?
Why is it true that n≥N => |an-an_k| < ε/2 ?
And why is it true that n≥N => |an_k-L| < ε/2 ?
Can somebody kindly explain the last two lines of the proof in more detail?
2. Relevant equations
N/A
3. The attempt at a solution
N/A
Any help is much appreciated!
Last edited: Jan 17, 2010
2. Jan 17, 2010
### HallsofIvy
Knowing that $n\ge N$ doesn't tell you anything about k. That was why the proof says "Pick any $k\ge K$" k is chosen to be larger than K. That has nothing to do with n. What the proof is really saying is "K is a fixed number so certainly there exist k> K. Also N is a fixed number and nk is a sequence that increases without bound so there exist numbers in the subsequence such that nk> N. Of all the nk> N, choose one for which k is also > K.
3. Jan 17, 2010
### kingwinner
But at the end, we are supposed to prove that:
an->L
i.e. for all ε>0, there exists N such that n≥N => |an-L| < ε ?
(note that here only n appears, there is nothing about k)
Looking at the definition of an->L above, all we have to do is to construct an N that works. Just like every ε-limit proof, we have to find an N that works. And if we can find such an N, then the only restriction should be n≥N and nothing else, but in this proof we also have other restrictions k≥K and nk≥N which does not even appear in |an-L| < ε . How come? Please help...I really don't understand :(
Do we need all three conditions (n≥N, k≥K, and nk≥N) to be simultaneously satisfied in order for |an-L| < ε to hold??
4. Jan 17, 2010
### vela
Staff Emeritus
All you have to show is that given an $\epsilon$, an N exists. It doesn't really matter how you found it; you just have to show it exists.
5. Jan 17, 2010
### kingwinner
But what is that "N" in this case?
6. Jan 17, 2010
### vela
Staff Emeritus
It says what N is in the fifth sentence of the proof.
7. Jan 17, 2010
### kingwinner
So the exact same N there would work at the end?
Last edited: Jan 17, 2010
8. Jan 17, 2010
### vela
Staff Emeritus
Yes. That's the point of the rest of the proof.
9. Jan 18, 2010
### kingwinner
I've seen another proof of this theorem:
Given ε>0.
There exists N s.t. m,n≥N => |an-am|<ε/2
There exists M s.t. k≥M => |ank-L|<ε/2 and also M≥N.
Take N'=max{N,nM}
Then if n>N',
|an-L|≤|an-anM|+|anM-L|<ε/2+ε/2=ε
=====================
Is this proof right or wrong? If it's right, I have the following questions:
1) Why is M≥N?
2) Why should we take N'=max{N,nM}?
Does anyone have any idea?
(I increased the font size to make the subscripts legible) | 2018-12-16T17:41:29 | {
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https://mathematica.stackexchange.com/questions/146065/plotting-a-taylor-series-of-two-variable-trigonometric-function | # Plotting a Taylor Series of two-variable trigonometric function
I am pretty sure I have made an obvious mistake. I want to plot a Taylor polynomial of a degree 4 around the point (0, 0). So I have defined a function
f[x_, y_] := Cos[x + y]
Then I write a Taylor series up to 4th degree
Series[f[x, y], {x, 0, 4}, {y, 0, 4}]
I can plot the function f, but when I try to plot series I get errors. I believe it is because. when computing the series, along from the terms I will also get a remainder, which can not be plotted.
Any help would be much appreciated.
• Greetings! Make the most of Mma.SE and take the tour now. Help us to help you, write an excellent question. Edit if improvable, show due diligence, give brief context, include minimal working examples of code and data in formatted form. As you receive give back, vote and answer questions, keep the site useful, be kind, correct mistakes and share what you have learned. – rhermans May 15 '17 at 12:34
• There are things to do after your question is answered. It's a good idea to stay vigilant for some time, better approaches may come later improving over previous replies. Experienced users may point alternatives, caveats or limitations. New users should test answers before voting and wait 24 hours before accepting the best one. Participation is essential for the site, please come back to do your part tomorrow – rhermans May 15 '17 at 12:34
Use Normal and Evaluate,e.g.:
t[n_] := Plot3D[ | 2019-10-14T05:04:47 | {
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https://math.stackexchange.com/questions/2055207/how-to-find-int-04-sqrt16-x2-dx | # How to find $\int_{0}^{4} \sqrt{16-x^2} dx$?
What is the easiest way of solving this integral:
$$\int_{0}^{4} \sqrt{16-x^2} dx$$
My idea was to substitute $x$ with $4\sin u$ and to get under the square root $\cos^{2}{u}$ so i can get rid of it, but then i get again $\cos^{2}u$. I suppose that could be solved using formula
$$\cos^{2}{\frac{u}{2}} = \frac{1+\cos u}{2}$$
But then I got troubles with getting back substitution.
Am I making somewhere mistake and if not how should I precede, or is there some easier way of solving it?
• If you want the easiest way (without substitution) then you can interpret the integral as the quarter area under the circle $x^2+y^2=16$, so the answer will be $1/4(\pi)(4^2)$. – Anurag A Dec 12 '16 at 9:38
• General tip: when you do a substitution for a definite integral, you can change the limits of integration and then not have to substitute backwards. – Brian J Dec 12 '16 at 21:22
The easiest way to solve this integral is to notice that the curve $y=\sqrt{16-x^2}$ is one quarter of a circle, so the area under this curve will be one quarter the area of the circle. The circle has radius $4$, so area $\pi 4^2 = 16\pi$.
This means the integral must evaluate to $4\pi$.
• It seems like very simple answer but i have trouble with figuring out how is $y = \sqrt{16-x^{2}}$ one quarter of circle. – Zvnoky Brown Dec 12 '16 at 9:54
• @ZvnokyBrown If $y=\sqrt{16-x^2}$, then $y^2=16-x^2$ and $x^2+y^2=4^2$... – 5xum Dec 12 '16 at 9:55
• and $y$ is always non-negative,... – DJohnM Dec 12 '16 at 16:49
• How do you rigorously get the area of a circle though? I am a little afraid this reasoning might be a little... circular. – Brian Moths Dec 12 '16 at 18:41
With the suggested substitution,
$$\int_0^4\sqrt{16-x^2}dx=\int_0^{\frac\pi2}16\cos^2u\,du=8\int_0^{\pi/2}(\cos2u+1)\,du.$$
As $\cos2u$ runs from $1$ to $-1$ symmetrically, this contribution vanishes and $4\pi$ remain.
• I may be mistaken, but I think you lost your square root around $16\cos^2(u)$ – Brian J Dec 12 '16 at 21:23
• Nevermind, I definitely forgot the dx => du portion. – Brian J Dec 12 '16 at 21:27
Does the answer have to be algebraically derived? If not, just draw it, recognize that it's a quarter of a circle with radius 4, thus the answer is $$\frac{\pi 4^2}{4} = 4\pi.$$
If you require something more analytic, you can work backward through double integrals: $$\int_0^4 \sqrt{16-\cos^2 x} \operatorname{d} x = \int_0^4 \int_0^{\sqrt{16-\cos^2 x}} 1 \operatorname{d} y \operatorname{d} x.$$ From there, you again draw the integration region, and do a change of variables to polar coordinates: \begin{align} x &= r\cos\theta \\ y &= r\sin\theta \Rightarrow \\ I &= \int_0^{\pi/2} \int_0^4 r \operatorname{d} r \operatorname{d} \theta \\ & = \frac{\pi}{2} \left(\frac{4^2}{2}\right) = 4\pi \end{align}
First notice that $$\int_0^r\sqrt{r^2-x^2}dx=r^2\int_0^1\sqrt{1-x^2}dx.$$
Then by parts,
$$I=\int_0^1\sqrt{1-x^2}dx=\left.x\sqrt{1-x^2}\right|_0^1+\int_0^1\frac{x^2}{\sqrt{1-x^2}}dx.$$
But $x^2=1-(1-x^2)$ so that
$$I=\int_0^1\frac{1}{\sqrt{1-x^2}}dx-I.$$ The remaining integral is solved with an arc sine and yields $\pi/2$.
• Notes: 1) u-substitution: let $x=r u \rightarrow dx = r du$ – FundThmCalculus Dec 12 '16 at 17:15 | 2020-07-06T18:27:34 | {
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https://math.stackexchange.com/questions/1892568/given-series-of-complex-terms-converges-locally-uniformly | # Given series of complex terms converges locally uniformly
Prove that the series $$f(z)=\frac{z}{1-z^2}+\frac{z^2}{1-z^4}+\frac{z^4}{1-z^8}+ \cdots$$ conveges locally uniformly to $\frac{z}{1-z}$ when $z$ is in the unit disc $D$ and to $\frac{1}{1-z}$ when $z \notin \overline{D}.$ Are these two limiting functions analytic continuations of each other? If not, is $\partial D$ the natural boundary of analyticity for $f(z)$ ?
I came across this problem while studying for my complex preliminaries. I tried to factor out the term $\frac{z}{1-z}$ and consider the difference $|f(z)-\frac{z}{1-z}|$ and go by to show that it's less than $\epsilon$ for all $z.$ But it wasn't successful. And any help in this part and other two parts are much appreciated. I wouldn't ask if I was able to solve the problem. Thank you for your time.
Consider the partial sums $$f_n(z)=\sum_{k=0}^{n-1}\frac{z^{(2^k)}}{1-z^{(2^{k+1})}}.$$ By induction on $n$, we have $$f_n(z)=\frac1{1-z}-\frac1{1-z^{(2^n)}}.$$ Indeed $$f_1(z)=\frac{z}{1-z^2}=\frac{1+z}{1-z^2}-\frac{1}{1-z^2} =\frac1{1-z}-\frac{1}{1-z^2}.$$ Assuming the formula holds for $n$, we have $$\begin{eqnarray*} f_{n+1}(z) &=&\frac1{1-z}-\frac1{1-z^{(2^n)}}+\frac{z^{(2^n)}}{1-z^{(2^{n+1})}}\\ &=&\frac1{1-z}-\frac{1+z^{(2^n)}}{1-z^{(2^{n+1})}}+\frac{z^{(2^n)}}{1-z^{(2^{n+1})}}\\ &=&\frac1{1-z}-\frac{1}{1-z^{(2^{n+1})}}, \end{eqnarray*}$$ as required. If $|z|>r>1$ then $|z^{(2^n)}|\geq r^{(2^n)}$, so $f_n(z)\rightarrow\frac1{1-z}$ uniformly. On the other hand $$f(z)=\frac{z}{1-z}-\frac{1}{z^{(-2^n)}-1}.$$ If $|z|<r<1$ then $|z^{(-2^n)}|\geq r^{(-2^n)}$, so $f_n(z)\rightarrow\frac{z}{1-z}$ uniformly.
• Thank you. Appreciate it. That's very elaborate. It's all about coming up with two different general functions for $f_n(z)$. Then the convergence is clear enough. – user358174 Aug 15 '16 at 4:48
The local uniform convergence can be seen (without knowing the limit) pretty easily: Let's take the $|z|<1$ case. If $|z| \le r < 1,$ then
$$\left |\frac{z^{2^n}}{1-z^{2^n+1} }\right | \le \frac{|z|^{2^n}}{1-|z|^{2^n+1} }\le \frac{r^{2^n}}{1-r^2 }.$$
Since $\sum r^{2^n}/(1-r^2) < \infty,$ we get uniform convergence on $\{|z|\le r\}$ by Weierstrass M.
As for what the sum converges to, note the sum equals
$$\sum_{n=0}^{\infty}\sum_{m=0}^{\infty} z^{(2m+1)2^n}.$$
Because every positive integer power of $z$ occurs once and only once here, and $|z|<1$ gives us absolute convergence, the above is just the sum $\sum_{k=1}^{\infty}z^k = z/(1-z).$
I assume similar shenanigans will work for $|z|>1.$
The fact that $\{|z|=1\}$ is a natural boundary for $f$ follows from this, but actually there is a more direct way. Note that set $E=\{e^{j\pi i/2^k}: j,k\in \mathbb N\}$ is dense in the unit circle. If you fix any $\zeta \in E,$ you'll see that along the radius terminating at $\zeta,$ the terms of the series defining $f$ in the disc equal $r^{2^n}/(1-r^{2^{n+1}})$ for large $n.$ That shows $f$ blows up at each $\zeta\in E,$ which implies the unit circle is a natural boundary for $f.$
• Thank you. The first explanation with W-M test is pretty clear. May be for $|z|>1$ we can do the inequality the other way round: I'm not quite understand how you get the limit and the natural boundary. For the latter I haven't seen such an argument before. The two limiting functions being not analytic continuation is clear because one couldn't be obtained from the other. I was thinking of showing that the unit circle has a singularity at every point on it. Or else the unit circle is dense with the set of singularities. – user358174 Aug 15 '16 at 22:10 | 2020-07-04T05:32:19 | {
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https://math.stackexchange.com/questions/2781904/integrate-sin-1-frac2x1x2/2782785 | Integrate $\sin^{-1}\frac{2x}{1+x^2}$
Integrate $\sin^{-1}\frac{2x}{1+x^2}$
The solution is given in my reference as: $2x\tan^{-1}x-\log(1+x^2)+C$.
But, is it a complete solution ?
My Attempt $$\int 2\tan^{-1}x \, dx=\int \tan^{-1}x \cdot 2\,dx=\tan^{-1}x\int2\,dx-\int\frac{1}{1+x^2}\int2\,dx\cdot dx\\ =\tan^{-1}x \cdot2x-\int\frac{2x}{1+x^2}\,dx=2x\tan^{-1}x-\log(1+x^2)+C$$ $$2\tan^{-1}x=\begin{cases}\sin^{-1}\frac{2x}{1+x^2}\text{ if }|x|\leq{1}\\ \pi-\sin^{-1}\frac{2x}{1+x^2}\text{ if }|x|>{1}\text{ and }x>0\\ -\pi-\sin^{-1}\frac{2x}{1+x^2}\text{ if }|x|>{1}\text{ and }x<0 \end{cases}\\ \sin^{-1}\frac{2x}{1+x^2}=\begin{cases}2\tan^{-1}x\text{ if }|x|\leq{1}\\ \pi-2\tan^{-1}x\text{ if }|x|>{1}\text{ and }x>0\\ -\pi-2\tan^{-1}x\text{ if }|x|>{1}\text{ and }x<0 \end{cases}$$ $$\int\sin^{-1}\frac{2x}{1+x^2}\,dx=\begin{cases}\int2\tan^{-1}x\,dx&\text{ if } |x|\leq{1}\\\int\pi\, dx-\int2\tan^{-1}x\,dx&\text{ if }|x|>{1}&\text{ and } x>0\\-\int\pi \,dx-\int2\tan^{-1}x\,dx&\text{ if }|x|>{1}&\text{ and } x<0\end{cases}=\begin{cases}\color{red}{2x\tan^{-1}x-\log(1+x^2)+C\text{ if } |x|\leq{1}}\\\pi x-2x\tan^{-1}x+\log(1+x^2)+C&\text{ if }|x|>{1}&\text{ and } x>0\\-\pi x-2x\tan^{-1}x+\log(1+x^2)+C&\text{ if }|x|>{1}&\text{ and } x<0\end{cases}$$
So don't we have two more cases for our solution rather than that is given in my reference, right ?
• Are you asking why $\sin^{-1}\frac{2x}{1+x^2}=2\tan^{-1}x$? – A.Γ. May 15 '18 at 8:50
• @A.Γ. i mean my reference seems to take only that. But, what about the other two cases ? – ss1729 May 15 '18 at 8:55
You're right; but it depends on what the full question is; if one is asked to compute $$\int_{0}^{1}\arcsin\frac{2x}{1+x^2}\,dx$$ then just the antiderivative $2\arctan x$ is sufficient. Not if one wants to compute an integral involving points outside the interval $[-1,1]$.
Here's a shorter way to get at your result.
Consider $$f(x)=\arcsin\frac{2x}{1+x^2}$$ Then $$f'(x)=\frac{1}{\sqrt{1-\dfrac{4x^2}{(1+x^2)^2}}}\frac{2(1+x^2)-4x^2}{(1+x^2)^2} =\frac{1+x^2}{|1-x^2|}\frac{2(1-x^2)}{(1+x^2)^2}$$ hence $$f'(x)=\begin{cases} \dfrac{2}{1+x^2} & x\in(-1,1) \\[4px] -\dfrac{2}{1+x^2} & x\in(-\infty,-1)\cup(1,\infty) \end{cases}$$ Therefore, knowing that $f(0)=0$, $$f(x)=\begin{cases} -\pi-2\arctan x & x<-1 \\[4px] 2\arctan x & -1\le x\le 1 \\[4px] \pi-2\arctan x & x>1 \end{cases}$$
In order to find an antiderivative we can consider $$\int\arctan x\,dx=x\arctan x-\int\frac{x}{1+x^2}\,dx =x\arctan x-\frac{1}{2}\log(1+x^2)$$ Thus an antiderivative of $f$ has the form $$F(x)=\begin{cases} c_- -\pi x-2x\arctan x+\log(1+x^2) &\qquad x<-1 \\[4px] 2x\arctan x-\log(1+x^2) &\qquad -1\le x\le 1 \\[4px] c_+ +\pi x-2x\arctan x+\log(1+x^2) &\qquad x>1 \end{cases}$$ and you just need to determine $c_-$ and $c_+$ to ensure continuity at $-1$ and $1$.
The other antiderivatives differ from $F$ by a constant.
• thanx. I like ur observation about the function from the derivative. thats something i never thought of, though i have seen it many times. – ss1729 May 16 '18 at 4:43
The "identity"
$$\arcsin\left(2x\over1+x^2\right)=2\arctan x$$
does not hold for all $x$. You can see this by comparing the ranges of the two sides. The arcsine function has $[-\pi/2,\pi/2]$ as its range, but twice the arctangent function has $(-\pi,\pi)$ as its range. In particular, if $x\gt1$, then $2\arctan x\gt\pi/2$, and likewise if $x\lt-1$ then $2\arctan x\lt-\pi/2$ (since $\arctan1=\pi/4$).
The correct identity is
$$\arcsin\left(2x\over1+x^2\right)= \begin{cases} \pi-2\arctan x\qquad\text{if }x\ge1\\ 2\arctan x\qquad\quad\text{ }\text{ if }-1\le x\le1\\ -\pi-2\arctan x\quad\text{ if }x\le-1 \end{cases}$$
(Note, I've intentionally overlapped the intervals at $x=\pm1$ to stress the agreement of $\pi-2\arctan1=2\arctan1$ and $2\arctan(-1)=-\pi-2\arctan(-1)$.) The OP's observations regarding the integral follow.
According to the standard tangent half-angle substitution \begin{align} & x=\tan\frac\theta 2 \\[8pt] & 2\arctan x = \theta \\[8pt] & \frac {2\,dx}{1+x^2} = d\theta \\[8pt] & dx = \sec^2\frac\theta 2 \,\,\frac{d\theta} 2 \\[8pt] & \frac{2x}{1+x^2} = \sin\theta \\[8pt] & \frac{1-x^2}{1+x^2} = \cos\theta \end{align} Thus we have $\theta = \arcsin\dfrac{2x}{1+x^2}.$ So \begin{align} & \int \arcsin\frac{2x}{1+x^2} \, dx = \int \theta \left( \sec^2\frac\theta 2 \,\, \frac{d\theta} 2 \right) = \int\theta \, dv = \theta v - \int v\,d\theta \\[10pt] = {} & \theta\tan\frac\theta 2 - \int \tan\frac\theta 2 \, d\theta = \theta\tan\frac\theta 2 +2\log\left|\cos\frac\theta 2\right| + C \\[10pt] = {} & \theta \tan \frac \theta 2 + \log\left|\frac 1 2 + \frac 1 2 \cos\theta \right| + C = \theta\tan\frac \theta 2 + \log\left| \frac 1 2 + \frac{1-x^2}{2(1+x^2)} \right| + C \\[10pt] = {} & 2x\arctan x - \log(1+x^2) + C \end{align}
• It follows that $\arcsin\frac{2x}{1+x^2}=2\arctan x$, for all $x$, which is obviously false. – egreg May 16 '18 at 6:52
• @egreg : It's true if $-1\le x\le1,$ so I guess some piecewise stuff needs to be added$\,\ldots\qquad$ – Michael Hardy May 16 '18 at 16:30
We know that $\tan(\theta/2) = \frac{\sin(\theta)}{1+\cos(\theta)}$ and $\cos(\theta) = \sqrt{1-\sin^2(\theta)}$. So, letting $x = \sin(\theta)$ $(\implies \theta = \arcsin(x))$ we have:
$\tan(\theta/2) = \frac{x}{1+\sqrt{1-x^2}} \implies \frac{\theta}{2} = \arctan\left(\frac{x}{1+\sqrt{1-x^2}}\right) \implies \arcsin(x) = 2\arctan\left(\frac{x}{1+\sqrt{1-x^2}}\right)$.
With that in hands, we just have to calculate $\arcsin\left(\frac{2x}{1+x^2}\right) = 2\arctan\left(\frac{\frac{2x}{1+x^2}}{1+\sqrt{1-\frac{4x^2}{1+2x^2+x^4}}}\right)$.
Now observe that $\frac{\frac{2x}{1+x^2}}{1+\sqrt{1-\frac{4x^2}{1+2x^2+x^4}}} = \frac{\frac{2x}{1+x^2}}{1+\sqrt{\frac{1-2x^2+x^4}{1+2x^2+x^4}}} = \frac{\frac{2x}{1+x^2}}{1+\frac{1-x^2}{1+x^2}} = \frac{\frac{2x}{1+x^2}}{\frac{2}{1+x²}} = x$. I thik this solve the equality you were struggling with.
• It is true only for $x\in[-1\ 1]$ (otherwise $\arcsin x$ is not defined). Compare the graphs. – A.Γ. May 15 '18 at 11:13
• For all $x \in \mathbb{R}$ we have $2x/(1+x^2) \in [-1,1]$, otherwise $2x/(1+x²) < -1 \implies 2x < -1-x^2 \implies 0 < -(1+x)^2$. Contradiction. Analogously we can't have $1 < 2x/(1+x^2) \implies 1+x^2 < 2x \implies (1-x)^2 < 0$. – Xablau123 May 15 '18 at 11:36
• i'm sorry i dont get wht u r trying to say. my doubt wht was whether it has three cases for the solution ? – ss1729 May 15 '18 at 12:29
• When you take your solution and derivate it, you'll have $2\arctan(x)$. What I showed is that actually $2\arctan(x) = \arcsin(\frac{2x}{1+x^2})$ for every $x \in \mathbb{R}$ and, by my previously comment, it's "well-defined". – Xablau123 May 15 '18 at 12:42
• @Xablau123 ... and you will notice that it is not true if you look at their graphs in the link in my previous comment.Your mistake: when you take the square root you will get $|1-x^2|$. You forgot the absolute value there. – A.Γ. May 15 '18 at 13:36 | 2021-01-23T08:19:24 | {
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https://mathematica.stackexchange.com/questions/72801/floating-point-operations-division-vs-multiplication-how-do-they-affect-accur/72806 | # Floating point operations — division vs multiplication; how do they affect accuracy?
Given two functions:
f[x_] := x ((x + 1)^(1/2) - x^(1/2))
g[x_] := x/((x + 1)^(1/2) + x^(1/2))
Which one is more accurate?
Side note: If you could explain why, that would really help me out, I have a feeling it's f(x) as there isn't a denominator but I'm not 100% certain.
• I believe accuracy would be the same for both division and multiplication, because the main culprit for loss of precision is adding (or subtracting) a very large number to a very small number, as opposed to adding two numbers of comparable size. Division and multiplication have near the same effect on precision, so there is no expected difference in precision between your f and g. – Manuel --Moe-- G Jan 29 '15 at 18:37
• Welcome to Mathematica.SE! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Read the faq! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! – user9660 Jan 29 '15 at 18:46
• More accurate at doing what ? – Sektor Jan 29 '15 at 18:49
• @Sektor Just when x=500. I know that f(x) can be re-arranged to be g(x) but when computed with a calculator, there are rounding errors. I'm not quite sure which one is more "accurate" – Claire Blackman Jan 29 '15 at 19:10
• Are you sure this question is about Mathematica and not Mathematics, in general ? – Sektor Jan 29 '15 at 19:24
ybeltukov showed empirically that $g(x)$ is numerically more precise than $f(x)$, despite the two expressions being formally mathematically equal. Why is this the case?
In your original question, you guessed that
If you could explain why, that would really help me out, I have a feeling it's f(x) as there isn't a denominator but I'm not 100% certain.
To explain why $g(x)$ is more accurate, recall Manuel --Moe-- G's comment:
...the main culprit for loss of precision is adding (or subtracting) a very large number to a very small number, as opposed to adding two numbers of comparable size. Division and multiplication have near the same effect
In the case of $f(x)$, for large values of $x$ we have $x$ (a large number), times $\sqrt{x+1}-\sqrt{x}$, which tends towards zero as $x\to\infty$. Thus, there is catastrophic loss of precision for large $x$, which is consistent with the first figure that ybeltukov's answer provided.
In contrast, $g(x)$ is the ratio of two large numbers whose sizes aren't that much different (or at least, become more different at a slower rate than in the case of $f(x)$).
### More formal explanation
First, examine $f(x)$. The first term $x$ asymptotically tends as $x$, whereas the second term $\sqrt{x+1}-\sqrt{x}$ asymptotically tends as $1/\sqrt{x}$:
Series[Sqrt[x + 1] - Sqrt[x], {x, Infinity, 2}]
(*Sqrt[1/x]/2 - 1/8 (1/x)^(3/2)*)
This means that the relative sizes of the terms diverge as $x^{3/2}$.
Now examine $g(x)$. The first term $x$ asymptotically tends as $x$, whereas the second term $\sqrt{x}+\sqrt{x+1}$ tends as $\sqrt{x}$:
Series[Sqrt[x + 1] + Sqrt[x], {x, Infinity, 2}]
(*2 Sqrt[x] + Sqrt[1/x]/2 - 1/8 (1/x)^(3/2)*)
This means that their relative sizes diverge as $x^{1/2}$, which is slower than in the case of $f(x)$. This means that $g(x)$ will be more accurate for large $x$.
You can compare the precision empirically:
prec[f_, x_] := Abs[SetPrecision[f[x], 30] - f[SetPrecision[x, 30]]]/
Abs@f[SetPrecision[x, 30]];
LogPlot[{prec[f, x], $MachineEpsilon}, {x, -10, 10}] LogPlot[{prec[g, x],$MachineEpsilon}, {x, -10, 10}]
So g has better numerical precision (around \$MachineEpsilon). It is because f contains the difference of potentially big numbers.
• I love your prec function, but can you elaborate a bit on how does it actually work? – plasmacel Jun 4 '18 at 19:49 | 2019-09-20T19:28:38 | {
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http://allenarebene.it/ti-84-statistics-functions-cheat-sheet.html | # Ti 84 Statistics Functions Cheat Sheet
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The key-by-key correspondence is relatively the same, but the 84 features some improved hardware. factoring polynomial cubed. Statistics There is a great amount of material here. michelleott28. Casio FX-260 Solar Scientific Calculator features functions, related to General Math, Trigonometry, Statistics, and Arithmetic. STATISTICS. Calculate F (b) - F (a). frame (date=c ('10/30/2021', '11/18/2021', '11/13/2021', '11/19/2021'), sales=c (3, 15, 14, 9)) df. It is related to the factorial by. Example 212 = 2 1 x2 enter The correct answer is 441. The (complete) gamma function is defined to be an extension of the factorial to complex and real number arguments. Learn Desmos: Graphing. If you haven't used the calculator before, you may want to get rid of everything that was there. This is a walkthrough of the statistics features of your TI-84 graphing calculator. Suppose you are provided with a bell-shaped, normal distribution that has a mean,$\mu$, of 50, and a standard deviation,$\sigma%$, of 5. " This provides a handy reference you can stuff in a book. You do this by pressing 2nd [MEM] (above the + sign), select [4:ClrAllLists] and press ENTER twice. Poster une nouvelle réponse: Répondre au sujet : Не могу создать новую тему в tedpublications. Create the worksheets you need with Infinite Pre-Algebra. ninth grade math symbols. There is also an updated version of TI Connect for the TI-84; if you have any calculator in the TI-84 family you can download TI Connect CE Software. Statfun83 provides each function as both a Basic program and as an App. In this post, I have compiled a list of 6 of the most important statistics. This is a cheat sheet that can be given to students on how to use the TI-84 Silver Plus Graphing Calculator. Argument name. Enter data separated by commas or spaces. How to calculate a definite integral. 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Is there any way to essentially convert or port games from the original TI-84 Plus to the TI-84 Plus CE? 8. Press F2 to put the cell in the edit mode, and then press Enter to accept the formula. 074318267 b. For example, the formula =IF(B5<0),"Not valid",B5*1. Vincennes University Jasper Academic Center for Excellence page 1 of 3 General Sites Vocabulary Study Geek Math Vocabulary-Find needed math terms quickly as well as math expressions, terms and formulas. The software will calculate the test statistic and the P-value for the test statistic. Press STAT, scroll over to CALC and select 1:1-Var Stats. Guide to TI-83/84 Statistics Functions Histograms Enter a list of data in L1 by pressing [STAT], then EDIT and [ENTER]. Video Transcript. So when you compare what you got t equal to and what our common since grid at the top of the page they are the same. For example, the X key contains the trigonometry functions sin and sin/ as well as the hyperbolic functions sinh and sinh/. 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Linear relationships are examined in practically every discipline of the natural and social sciences. Probability & Statistics. Steps 1-3 are the same as I described previously for the TI-83 calculator (except that you only need to clear L1 and L2). Ti-89 solves system by solving algebra equations with trigonometry functions, dividing hyperballs, algebra software II. It supports all of the existing models in this series (TI-73, TI-76. Solution: The command would be weighted. FREE Delivery Across Papua New Guinea. )TilEm features detailed emulation of all aspects of the calculator hardware, and includes a debugger for writing assembly programs. ( More info, Hours, & Instructions) The MSLC also has additional online resources listed below for your course. There is a $50 fee you must pay to use WeBWorK. Select 2: t Test, and indicate that you will be using Stats as the data input method. You do not need to know all of it. 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http://math.stackexchange.com/questions/167935/write-cscx-in-terms-of-secx/167942 | # Write $\csc(x)$ in terms of $\sec(x)$
I was working on trig homework and came across a question that I didn't understand how to even begin to approach. It asked us to use trigonometric identities to write $\csc(x)$ in terms of $\sec(x)$.
I'm not sure what I can do. I had though the cofunction identity $\sec \left(\frac{\pi}{2} - x \right) = \csc x$ was the answer, but I was told I was wrong, with no explanation. This identity was on a handout provided by the teacher.
How would I go about solving this problem and problems like it? Also, why was my initial answer wrong? Any help would be greatly appreciated.
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It seems like you should rather contact your teacher and ask him why your answer is wrong, instead of questioning here. – Listing Jul 7 '12 at 17:59
I am curious. What was the official answer? – André Nicolas Jul 7 '12 at 18:02
It's an online course. The teacher hasn't been good about responding to questions. – user18260 Jul 7 '12 at 18:35
There was no official answer from the teacher – user18260 Jul 7 '12 at 18:36
It's just that the answer should be a little different depending on the location of $x$, for example $\frac{\sec x}{\sqrt{\sec^2 x-1}}$ between $0$ and $\pi/2$, $-\frac{\sec x}{\sqrt{\sec^2 x-1}}$ between $\pi/2$ and $\pi$. – André Nicolas Jul 7 '12 at 20:52
It’s wrong because you didn’t write $\csc x$ in terms of $\sec x$: $\frac{\pi}2-x$ isn’t $x$. I suspect that you were intended to do something like this:
$$\csc x=\frac1{\sin x}\;,$$ and $$\sec x=\frac1{\cos x}\;,$$ so
\begin{align*}&\frac1{\csc^2 x}+\frac1{\sec^2x}=1\;,\\ &\frac1{\csc^2x}=1-\frac1{\sec^2x}=\frac{\sec^2x-1}{\sec^2x}\;,\\ &\csc^2x=\frac{\sec^2x}{\sec^2x-1}\;,\\ &\csc x=\pm\frac{\sec x}{\sqrt{\sec^2x-1}}\;. \end{align*}
For a complete answer you still have to sort out where $\csc x$ is positive and where it’s negative, which is just a bit messy.
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Thank you! I found you're explanation easiest to follow. – user18260 Jul 7 '12 at 20:00
Probably your teacher wanted a relation between $\csc(x)$ and $\sec(x)$ and not $\sec \left(\frac{\pi}2 - x \right) = \csc(x)$. If you want a relation between $\csc(x)$ and $\sec(x)$, then recall that $$\sin^2(x) + \cos^2(x) = 1$$ This gives us $$\dfrac1{\csc^2(x)} + \dfrac1{\sec^2(x)} = 1 \implies \csc(x) = \pm\sqrt{\dfrac{\sec^2(x)}{\sec^2(x)-1}}$$
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This looks like something that teacher could have wanted...+1 – DonAntonio Jul 7 '12 at 18:02 | 2014-10-02T05:04:59 | {
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https://www.physicsforums.com/threads/is-our-complementary-solution-guaranteed-to-be-linearly-independent.633640/ | # Is our complementary solution guaranteed to be linearly independent?
## Main Question or Discussion Point
I had kind of a general question. Say I have a second order, homogeneous ODE. Say I use one of the general techniques to generate a complementary solution for my ODE and I end up with something of the form y = C1(solution1) + C2(solution2)
Am I gauranteed that these two solutions will be linearly independent or do I need to verify with the Wronskian each time?
Also, can I think of linearly independent solutions as a "basis" like I do in linear algebra? It seems like it because its my understanding that any solution for the ODE is simply a linear combination of solutions in this "basis"
Also, can we tell the dimensions of this basis based on the order of the ODE? Like, a second order ODE has two solutions in its basis, a third order has 3, a first order has one, etc.
Thanks!
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Yes, you can think of the solutions as a basis in a vector space. In this case, they're basis functions in a space of functions, but most of the ideas from vector algebra still apply.
I think often the techniques chosen to generate solutions are meant to come up with solutions out of orthogonal functions--sines and cosines, for instance, are automatically linlearly independent as long as none of their frequencies are the same. Someone may know more on this matter, though.
HallsofIvy
Homework Helper
A complementary solution can't be "dependent" on the solution to the homogeneous equation. Suppose the equation is L(f)= g(x) where L is the linear differential operator, u(x) a linear combination of the independent solutions to the homogenelus equation, and v(x) a complementary solution. To be dependent means that there exist numbers, a and b, non-zero, such that au(x)+ bv(x)= 0. Applying L to both sides of that, L(au(x)+ bv(x))= aL(u)+ bL(v)= a(0)+ b(g)= 0 so that bg(x)= 0 for all x. Since g is not always 0, b= 0. That makes our original equation au(x)= 0 for all x and so a= 0.
The solutions to any linear homogeneus nth order differential equation form a vector space of dimension n. The solutions to a non-homogeneous linear equation do NOT form a vector space but do form a "linear manifold". If you think of a one-dimensional vector space as a line through the origin of a xy-coordinate system, or a two dimensional vector space as a plane through the origin of a xyz-coordinate system, then you can think of a one dimensional "linear manifold" as a line NOT through the origin or a two dimensional "linear manifold" NOT through the origin.
Last edited by a moderator:
If you think of a one-dimensional vector space as a line through the origin of a xy-coordinate system, or a two dimensional vector space as a plane through the origin of a xyz-coordinate system, then you can think of a one dimensional "linear manifold" as a line NOT through the origin or a two dimensional "linear manifold" NOT through the origin.
Thank you, this really does help me visualize this.
So I want to make sure I understand. So for example, when dealing with non homogeneous equations, does it mean we can't establish a "basis" or fundamental set of solutions so that every particular solution can be expressed as a linear combination of those solutions?
y'' - 5y' + 4y = 8ex
I can generate a basis for the vector space of solutions for the associated homogeneous equation and its B = {e4x,ex} (I got this by solving the auxilary equation and attaining distinct real roots x=4 and x=1)
Using variation of parameters, I can then generate a particular solution for this sytem:
yp=(-8/9)ex-(8/3)xex
To make sure it is indeed a particular solution I took y'p and y''p and plugged it in to the original ODE and it does check out. However, yp is not a linear combination of the functions in B.
oh wait I think I get it, the "basis" as I'm thinking of it for a nohomogeneous linear 2nd order ODE would be B={b1,b2,yp}
where b1 and b2 are the functions in the basis for the associated homogeneous ODE and yp is a particular solution satisfying the non homogeneous ODE.
EDIT: Oh waaaait... No what I just wrote is wrong. Any solution Y(x) of the non homogeneous system can be expressed as:
Y=C1b1+C2b2+yp
So I can see why {b1,b2,yp} is NOT a "basis" because we can't just have any linear combination of these to generate solutions, the solutions are linear combinations of b1 and b2 with yp simply added on (not any multiple of yp)! So this is why it's not a vector space but a linear manifold as you described it
Last edited: | 2020-01-23T23:02:20 | {
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https://math.stackexchange.com/questions/1840634/power-series-as-approximation | # Power series as approximation
I have to estimate the error when I approximate the function $$e^{\sin x}$$ to $$1+x+x^{2}+x^{3}$$ when $|x|<0.1$.
I really don't know how to do because my teacher didn't teach me. But what I did it was: I found the Taylor's polynom of $e^{\sin x}$ for $a=0$: $$T_{0}(x)=f(0)+f'(0)(x-0)+\frac{f''(0)}{2!}(x-0)^{2}+\frac{f'''(0)}{3!}(x-0)^{3}$$ then $$T_{0}(x)=1+x+\frac{1}{2}x^{2}$$ My question is: until here, is that the right way to solve it? If yes, what should I do now? How I use the fact that $|x|<0.1$? If it is wrong, where can I find some solved example to understand it? I didn't find anything good on google.
• Your education isn't restricted to what your teacher did or didn't teach. If you have a textbook, presumably that book discusses the remainder theorem which gives an expression for the error involved. Read that theorem. Understand it. Use it. – John Coleman Jun 26 '16 at 20:22
• "Estimate the error" is pretty broad, we must know what is the measure for that in your class. Usually, the error measure means the measure of the quadratic error. For a function $f$ approximated by $g$ within the interval $(a,b)$ the quadratic error is $$\int_a^b (f(x) - g(x))^2 \;\text{d}x$$ Since $|x| < 0.1$ we have $-0.1 < x < 0.1$ as the bounds, $f(x) = e^{\sin x}$ and $g(x) = 1+x+x^2+x^3$. Throwing this in WA gives numerical result $1.032\cdot10^{-6}$. But as as I said, we need more background here. – Maximilian Gerhardt Jun 26 '16 at 20:23
• @JohnColeman I know, actually, I've already read three textbooks, internet articles, tried to understand, did some other exercises, but I couldn't solve it, I'm not asking you to solve it for me, just some tips, or some article so I could understand it better. – mvfs314 Jun 26 '16 at 20:24
The Taylor expansion of $\exp\sin x$ around zero is $1+x+x^{2}/2+O(x^{4})$. Therefore, the error is \begin{align*} \left|\exp\sin x-(1+x+x^{2}+x^{3})\right| & =\left|-x^{2}/2+O(x^{3})\right|\\ & \leq|x^{2}|/2+|O(x^{3})|\\ & \approx|x^{2}|/2 & \text{for }|x|\text{ small}. \end{align*}
• To finish the solution is only this? – mvfs314 Jun 27 '16 at 2:44
• When $|x|\leq0.1$, $|x^{2}|/2\leq0.1^{2}/2=0.005$. To verify that your estimate is OK, note that $1+0.1+0.1^{2}+0.1^{3}=1.111$ and $$1.111-\exp\sin0.1\approx0.006\approx0.005.$$ – parsiad Jun 27 '16 at 14:36
• Also, here is a plot of the error, which you can see is always approximately $\leq 0.006$: wolframalpha.com/input/… – parsiad Jun 27 '16 at 14:39
As you noted the Taylor series for $f(x) = e^x$ is $f(x) \approx 1 + x + \frac{x^2}{2} + \frac{x^3}{6}$. Now plug in $\sin x$ and use the fact that for values close to $x$ we have $\sin x \approx x$. So we have:
$$e^{\sin x} \approx 1 + x + \frac{x^2}{2} + \frac{x^3}{6}$$
Now the error of the first approximation is around $\frac{x^2}{2} + \frac{5x^3}{6}$. Use the fact that $\mid x \mid \le 0.1$ and plug it into the estimation to get the numerical approximation of the error. | 2019-10-23T23:24:32 | {
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https://math.stackexchange.com/questions/2024173/question-about-a-proof-of-f-overline-a-subseteq-overlinefa-for-all-a-su/2025153 | # Question about a proof of $f(\overline A)\subseteq\overline{f(A)}$ for all $A\subset{\rm dom}(f)$ implies that $f$ is continuous
If $f(\overline A)\subseteq\overline{f(A)}$ for all $A\subset{\rm dom}(f)$ then $f$ is continuous.
I know there are others questions about this topic but I want to know what more I need to conclude my proof, if possible. The way Im trying to prove this is different to what I see in other answers about this topic. The context of this proof is
$$f:X\to Y$$
where $X$ and $Y$ are metric spaces.
Then to proof that if $f(\overline A)\subseteq\overline{f(A)}$ then $f$ is continuous I will reason with convergent sequences.
We have that for any sequence $(x_n)\to x$ with $x_n\in A$ for all $n\in\Bbb N$ then $x\in\overline A$.
Then observe that the statement $f(\overline A)\subseteq\overline{f(A)}$ implies that if $f(x_n)\in f(A)$ then $f(x)\in\overline{f(A)}$, i.e. the image of the limit of any convergent sequence in $A$ is defined in the closure of the image of $A$, in other words
$$f(\lim x_n)=f(x)\in\overline{f(A)}$$
But I dont know if I can conclude that
$$\lim f(x_n)=f(x)$$
what is the definition of continuity on metric spaces. In other words: what I need to conclude
$$\lim f(x_n)=f(\lim x_n)$$
in this context?
• You haven't made use of the fact that the subset condition holds for all $A \subset dom(f)$. There may be some mileage in considering a subset consisting just of the points$(x)_n$. – Tom Collinge Nov 21 '16 at 13:58
The proof in topological spaces:
Assume that $f$ is not continuous, that is there exists some open set $V\subset Y$ with $U:=f^{-1}(V)$ is not open. Let $A = X\setminus U$. Notice that for every $x\in X$ we have $$x \in A = X\setminus f^{-1}(V) \iff x\notin f^{-1}(V) \iff f(x) \notin V,$$ that is $f(A) \subseteq Y\setminus V$.
Now, as $U$ is not open, there exists a $x\in U \cap \bar A$ and we have
1. $x\in U = f^{-1}(V) \iff f(x) \in V$ and
2. $x\in \bar A \implies f(x) \in f(\bar A) \subseteq \overline{f(A)} \subseteq Y \setminus V$,
Proof only for metric spaces:
Assume the converse: Let $x_n\to x$ and $f(x_n)\not\to f(x)$. By going to a subsequence we may assume that there exists a $\epsilon > 0$ such that $|f(x_n) - f(x)| \ge \epsilon$ holds for all $n$. On other hand we have $$f(x) \in \overline{\{ f(x_n) \mid n\in \mathbb N \}}.$$ That is, there exists a sequence $f(x_{n_k}) \to f(x)$, which is a contradiction.
• Good answer: I added some notes. – Tom Collinge Nov 22 '16 at 14:20
• To me, there is a gap in your Old answer. I cannot see why you can assume that one can always find a subsequence such that for all $n$, $|f(x)-f(a)|>\epsilon>0$. Can you clarify that please? It is easy to imagine that for some $n$, $|f(x)-f(a)|>\epsilon$, but not for all $n$. – Drake Marquis Feb 6 '18 at 7:01
• @DrakeMarquis "by going to a subsequence" ... that subsequence must be infinite, otherwise, $f(x_n)$ converges. – user251257 Feb 6 '18 at 12:59
• Would you please explicitly construct such a subsequence? – Drake Marquis Feb 7 '18 at 5:37
• @DrakeMarquis you should ask a new question. Similar questions are probably answered before. – user251257 Feb 7 '18 at 9:07
This is too long for a comment, but hopefully a useful addition.
The topological proof given by user251257 is neat, but perhaps worth some expansion.
If $U$ is not open then $X \setminus U$ is not closed and therefore doesn't contain all its limit points. So there is a limit point $x$ of $X \setminus U$ in $X \setminus (X \setminus U) = U$.
$x$ and other limit points of $X \setminus U$ are in the closure of $X\setminus U$, so as stated there is $x \in U \cap \bar A$ and it follows (1) $f(x) \in V$.
For (2), $\overline{f(A)}$, the closure of $f(A)$ is a subset of every closed set that contains $f(A)$; since $V$ is open then $Y \setminus V$ is closed, and as shown initially $f(A) \subset Y \setminus V$: therefore $Y \setminus V$ is a closed set containing $f(A)$ and therefore contains $\overline{f(A)}$.
(Since it took me a while to work this out I thought I'd share it).
• Thx. It was late and I was lazy :D – user251257 Nov 22 '16 at 14:34 | 2019-12-13T21:59:03 | {
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