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https://math.stackexchange.com/questions/2410058/commutative-rings-whose-localization-at-every-prime-ideal-is-a-field | # commutative rings whose localization at every prime ideal is a field
Can we characterize those commutative rings $R$ with unity such that for every prime ideal $p$ of $R$, $R_p$ is a field?
I know that any Boolean ring has this property. Also, if a ring has this property, then every finitely presented module over it is projective. Can something more conclusive be said in any direction?
Yes, there is a characterization. For a commutative ring, localization at all primes are fields iff the ring is von Neumann regular.
This is pretty well-known... you can even see it in the wiki.
One way to get at it is to realize that all primes must be maximal ideals (if you localized at a maximal prime properly containing another prime, you would not get a field.) Then by this result which I have alluded to before that in such a ring, $R/J(R)$ is von Neumann regular and $J(R)$ is a nil ideal.
But $R$ being reduced is a local property, so if all its localizations at prime ideals are reduced, so is $R$. Then $J(R)=\{0\}$, and you have simply a von Neumann regular ring.
The opposite implication is trivial, considering that localizations of VNR rings are VNR, and local VNR rings are fields.
• why is localization at a non-maximal prime ideal is not a field ? – user Aug 29 '17 at 16:44
• @users Sorry, I had it the wrong way around in text. The problem is that when you localize at a maximal ideal properly containing a prime ideal, that is not a field. It's corrected now. – rschwieb Aug 29 '17 at 16:46
• could you please elaborate that ? I might be missing something trivial ... – user Aug 29 '17 at 16:55
• @users in the localization, there will be a proper prime ideal corresponding to the extra prime. A field is zero dimensional, but that localization is at least one dimensional. – rschwieb Aug 29 '17 at 17:27
• Probably a better way to say it is that a localization at a non-minimal prime will result in something that isn't a field. – rschwieb Aug 29 '17 at 18:59
Your question has already been answered, but let me give a proof of the characterization of commutative rings whose localizations at prime ideals are fields, that is somehow different of the proof given by rschwieb.
Theorem 1.- Let $R$ be a commutative ring. TFAE:
i) $R$ is von Neumann regular.
ii) Every prime ideal of $R$ is maximal* and $R$ is reduced.
iii) $R_P$ is a field for every prime ideal $P$ of $R$.
Proof: i)$\implies$ii) Let $P$ be a prime ideal of $R$ such that is not maximal, then there is a maximal ideal $M$ such that $P\subset M$. Pick $a\in M\setminus P$. As $R$ is von Neumann regular, there is $r\in R$ such that $a=ara=a^2r$, then $a(1-ar)=0$. Since $0\in P$ it follows that $a(1-ar)\in P$, so because $P$ is prime we have $a\in P$ or $1-ar\in P$, but by assumption $a\notin P$, therefore $1-ar\in P$, then $1-ar\in M$. Hence, $(1-ar)+ar=1\in M$, contradiction. We conclude that every prime ideal of $R$ is maximal.
On the other hand, if $a^n=0$ for some $n\in \Bbb Z^+$, then as there is $r\in R$ such that $a=a^2r$, we have $a=a^2r=(a^2r)ar=a^3r^2$ and repeated multiplication by $ar$ leads to $a=a^nr^{n-1}$. Since $a^n=0$, then $a^nr^{n-1}=0$, so $a=0$ and hence $R$ is reduced.
ii)$\implies$iii) As $R$ is reduced, then $R_P$ is reduced. Moreover, since every prime ideal of $R$ is maximal, it follows by the correspondence between the prime ideals of $R$ and of $R_P$ that $PR_P$ is the only prime (and also maximal) ideal of $R_P$. By a classical result in commutative algebra it follows that $\text{Nil}(R_P)=PR_P$, but $R_P$ reduced means that $\text{Nil}(R_P)=\{0\}$, so $PR_P=\{0\}$ which means that $\{0\}$ is a maximal ideal of $R_P$ and this in turn implies that $R_P$ is a field.
iii)$\implies$i) We are going to prove that for $a\in R$, $(a)_P=(a^2)_P$ for every prime ideal $P$ of $R$. As $R_P$ is a field, it follows that $(a)_P=\{0\}$ or $(a)_P=R_P$. In the former case we have that $a\in P$, so $a^2\in P$ and thus $(a^2)_P=\{0\}$. In the later case we have that $a\notin P$, so $a^2\notin P$ and then $(a^2)_P=R_P$. Therefore, in any case we have $(a)_P=(a^2)_P$. Since the above is true for every prime ideal $P$ of $R$, by the principle of localization we deduce that $(a)=(a^2)$ and so $a=a^2r$ for some $r\in R$. Hence, $R$ is von Neumann regular.
(*) This is the same as to say that $R$ has Krull dimension zero.
Remark 1: The above characterization of commutative von Neumann regular rings was used in the paper "Armendariz Rings and Gaussian Rings" to prove a beautiful theorem that characterizes when $R[X]$ is a Gaussian ring. More exactly, we have
Theorem 2.- Let $R$ be a commutative ring. Then $R$ is von Neumann regular if only if $R[X]$ is Gaussian.
Proof: See theorem 16 in the paper mentioned above.
Remark 2: There is an interesting generalization of the above theorem given in T. Y. Lam's book "Exercises in Classical Ring Theory", namely we have
Theorem 3.- Let $R$ be a commutative ring. TFAE:
i) $R$ has Krull dimension zero.
ii) $\text{rad}(R)$ is a nil ideal and $R/\text{rad}(R)$ is von Neumann regular. (Here $\text{rad}(R)$ is the Jacobson radical of $R$).
iii) For any $a\in R$, the descending chain $(a)\supseteq (a^2)\supseteq \ldots$ stabilizes.
iv) For any $a\in R$, there is $n\in \Bbb Z^+$ such that $a^n$ is a regular element (i.e., there exists $r\in R$ such that $a^n=a^nra^n$).
Proof: This is exercise 4.15 in Lam's book mentioned above.
The condition $R_p$ is a field means that $p$ is a minimal prime ideal. So all prime ideals are minimal, and so they are all maximal too. In other words, $R$ has Krull dimension zero. See Wikipedia for more information, in particular if $R$ is Noetherian, it has to be Artinian.
• could you please explain why every prime ideal is minimal ? – user Aug 29 '17 at 18:39
• The prime ideals of $R$ contained in $p$ correspond to the prime ideals of $R_p$. – Angina Seng Aug 29 '17 at 18:52 | 2020-03-29T03:55:32 | {
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https://tobydriscoll.net/fnc-julia/globalapprox/trig.html | # 9.5. Trigonometric interpolation¶
Up to this point, all of our global approximating functions have been polynomials. While they are versatile and easy to work with, they are not always the best choice.
Suppose we want to approximate a function $$f$$ that is periodic, with one period represented by the standard interval $$[-1,1]$$. Mathematically, periodicity means that $$f(x+2)=f(x)$$ for all real $$x$$. We could use polynomials to interpolate or project $$f$$. However, it seems more reasonable to replace polynomials by functions that are also periodic.
Definition 9.5.1 : Trigonometric polynomial
For an integer $$n$$, a trigonometric polynomial of degree $$n$$ is
(9.5.1)$p(x) = \frac{a_0}{2} + \sum_{k=1}^n a_k \cos(k\pi x) + b_k \sin(k\pi x)$
for real constants $$a_k,b_k$$.
It turns out that trigonometric interpolation allows us to return to equally spaced nodes without any problems. We therefore define $$N=2n+1$$ equally spaced nodes inside the interval $$[-1,1]$$ by
(9.5.2)$t_k = \frac{2k}{N}, \quad k=-n,\ldots,n.$
The formulas in this section require some minor but important adjustments if $$N$$ is even instead. We have modified our standard indexing scheme here to make the symmetry within $$[-1,1]$$ about $$x=0$$ more transparent. Note that the endpoints $$\pm 1$$ are not among the nodes.
As usual, we have sample values $$y_{-n},\ldots,y_n$$, perhaps representing values of a function $$f(x)$$ at the nodes. We also now assume that the sample values can be extended periodically forever in both directions, so that $$y_{k+mN}=y_k$$ for any integer $$m$$.
## Cardinal functions¶
We can explicitly state the cardinal function basis for equispaced trigonometric interpolation. It starts with
(9.5.3)$\tau(x) = \frac{2}{N} \left( \frac{1}{2} + \cos \pi x + \cos 2\pi x + \cdots + \cos n\pi x\right) = \frac{\sin(N\pi x/2)}{N\sin(\pi x/2)}.$
You can directly check the following facts. (See Exercise 3.)
Theorem 9.5.2
Given the definition of $$\tau$$ in (9.5.3),
1. $$\tau(x)$$ is a trigonometric polynomial of degree $$n$$.
2. $$\tau(x)$$ is 2-periodic.
3. $$\tau(t_k)=0$$ for any nonzero integer $$k$$.
4. $$\displaystyle \lim_{x \to 0} \tau(x) = 1.$$
Given also the nodes $$t_k$$ in (9.5.2), the functions $$\tau_k(x) = \tau(x-t_k)$$ form a cardinal basis for trigonometric interpolation.
Because the functions $$\tau_{-n},\ldots,\tau_n$$ form a cardinal basis, the coefficients of the interpolant are just the sampled function values, i.e., the interpolant of points $$(t_k,y_k)$$ is
(9.5.4)$p(x) = \sum_{k=-n}^n y_k \tau_k(x).$
The convergence of a trigonometric interpolant is spectral, i.e., exponential as a function of $$N$$ in the max-norm.
## Implementation¶
Function 9.5.3 is an implementation of trigonometric interpolation based on (9.5.4). The function accepts an $$N$$-vector of equally spaced nodes. Although we have not given the formulas above, the case of even $$N$$ is included in the code.
Function 9.5.3 : triginterp
Trigonometric interpolation
1"""
2 triginterp(t,y)
3
4Construct the trigonometric interpolant for the points defined by
5vectors t and y.
6"""
7function triginterp(t,y)
8 N = length(t)
9
10 function τ(x)
11 if x==0
12 return 1.0
13 else
14 denom = isodd(N) ? N*sin(π*x/2) : N*tan(π*x/2)
15 return sin(N*π*x/2)/denom
16 end
17 end
18
19 return function (x)
20 sum( y[k]*τ(x-t[k]) for k in eachindex(y) )
21 end
22end
Demo 9.5.4
We get a cardinal function if we use data that is 1 at one node and 0 at the others.
The operator ÷, typed as \div then Tab, returns the quotient without remainder of two integers.
N = 7; n = (N-1)÷2
t = 2*(-n:n)/N
y = zeros(N); y[n+1] = 1;
p = FNC.triginterp(t,y);
plot(p,-1,1)
scatter!(t,y,color=:black,title="Trig cardinal function, N=$N", xaxis=(L"x"),yaxis=(L"\tau(x)")) Here is a 2-periodic function and one of its interpolants. f = x -> exp(sin(pi*x)-2*cos(pi*x)) y = f.(t) p = FNC.triginterp(t,y) plot(f,-1,1,label="function", xaxis=(L"x"),yaxis=(L"p(x)"), title="Trig interpolation, N=$N",legend=:top)
scatter!(t,y,m=:o,color=:black,label="nodes")
plot!(p,-1,1,label="interpolant")
The convergence of the interpolant is spectral. We let $$N$$ go needlessly large here in order to demonstrate that unlike polynomials, trigonometric interpolation is stable on equally spaced nodes. Note that when $$N$$ is even, the value of $$n$$ is not an integer but works fine for defining the nodes.
N = 2:2:60
err = zeros(size(N))
x = range(-1,1,length=2501) # for measuring error
for (k,N) in enumerate(N)
n = (N-1)/2; t = 2*(-n:n)/N;
p = FNC.triginterp(t,f.(t))
err[k] = norm(f.(x)-p.(x),Inf)
end
plot(N,err,m=:o,title="Convergence of trig interpolation",
xaxis=(L"N"),yaxis=(:log10,"max error"))
## Fast Fourier transform¶
Although the cardinal form of the interpolant is useful and stable, there is a fundamental alternative. It begins with an equivalent complex form of the trigonometric interpolant (9.5.1),
(9.5.5)$p(x) = \sum_{k=-n}^n c_k e^{ik\pi x}.$
The connection is made through Euler’s formula,
(9.5.6)$e^{i\theta} = \cos(\theta) + i\sin(\theta),$
and the resultant identities
(9.5.7)$\cos \theta = \frac{e^{i \theta}+e^{-i\theta}}{2}, \qquad \sin \theta = \frac{e^{i \theta}-e^{-i\theta}}{2i}.$
Specifically, we have
$\begin{split}c_k = \begin{cases} \frac{a_0}{2}, & k=0, \\[1mm] \frac{1}{2}(a_k + i b_k), & k> 0, \\[1mm] \overline{c_{-k}}, & k < 0. \end{cases}\end{split}$
While working with an all-real formulation seems natural when the data are real, the complex-valued version leads to more elegant formulas and is standard.
The $$N=2n+1$$ coefficients $$c_k$$ are determined by interpolation nodes at the $$N$$ nodes within $$[-1,1]$$. By evaluating the complex exponential functions at these nodes, we get the $$N\times N$$ linear system
$\mathbf{F}\mathbf{c} = \mathbf{y}, \qquad \mathbf{F} = \bigl[ e^{\,is\pi t_r} \bigr]_{\, r=-n,\ldots,n,\, s=-n,\ldots,n,}$
to be solved for the coefficients. Up to a scalar factor, the matrix $$\mathbf{F}$$ is unitary, which implies that the system can be solved in $$O(N^2)$$ operations simply by a matrix-vector multiplication.
However, one of the most important (though not entirely original) algorithmic observations of the 20th century was that the linear system can be solved in just $$O(N\log N)$$ operations by an algorithm now known as the fast Fourier transform, or FFT.
The FFTW package provides a function fft to perform this transform, but its conventions are a little different from ours. Instead of nodes in $$(-1,1)$$, it expects the nodes to be defined in $$[0,2)$$, and it returns the trig polynomial coefficients in the order
$\begin{bmatrix} c_0, & c_1, & \cdots & c_n, & c_{-n}, & \cdots & c_{-1} \end{bmatrix}.$
Demo 9.5.5
This function has frequency content at $$2\pi$$, $$-2\pi$$, and $$\pi$$.
f = x -> 3*cos(2π*x) - exp(1im*π*x);
To use fft, we set up nodes in the interval $$[0,2)$$.
n = 4; N = 2n+1;
t = [ 2j/N for j=0:N-1 ] # nodes in [0,2)
y = f.(t);
We perform Fourier analysis using fft and then examine the resulting coefficients.
c = fft(y)/N
freq = [0:n;-n:-1]
data = round.(c,sigdigits=5)
┌────┬─────────────────────────┐
│ k │ coefficient │
├────┼─────────────────────────┤
│ 0 │ -7.4015e-17+0.0im │
│ 1 │ -1.0+3.1961e-16im │
│ 2 │ 1.5-6.866e-16im │
│ 3 │ 2.0068e-16-3.7007e-17im │
│ 4 │ 2.2204e-16+2.3261e-16im │
│ -4 │ 1.727e-16-3.0001e-16im │
│ -3 │ 2.4341e-16-3.7007e-17im │
│ -2 │ 1.5+7.1744e-16im │
│ -1 │ 0.0-1.9625e-16im │
└────┴─────────────────────────┘
Note that $$1.5 e^{2i\pi x}+1.5 e^{-2i\pi x} = 3 \cos(2\pi x)$$, so this result is sensible.
Fourier’s greatest contribution to mathematics was to point out that every periodic function is just a combination of frequencies—infinitely many of them in general, but truncated for computational use. Here we look at the magnitudes of the coefficients for $$f(x) = \exp( \sin(\pi x) )$$.
f = x -> exp( sin(pi*x) ) # content at all frequencies
n = 9; N = 2n+1;
t = [ 2j/N for j=0:N-1 ] # nodes in [0,2)
c = fft(f.(t))/N
freq = [0:n;-n:-1]
scatter(freq,abs.(c),xaxis=(L"k",[-n,n]),yaxis=(L"|c_k|",:log10),
title="Fourier coefficients",leg=:none)
The Fourier coefficients of smooth functions decay exponentially in magnitude as a function of the frequency. This decay rate is determines the convergence of the interpolation error.
The theoretical and computational aspects of Fourier analysis are vast and far-reaching. We have given only the briefest of introductions.
## Exercises¶
1. ⌨ Each of the following functions is 2-periodic. Use Function 9.5.3 to plot the function together with its trig interpolants with $$n=3,6,9$$. Then, for $$n=2,3,\ldots,30$$, compute the max-norm error in the trig interpolant by sampling at $$1000$$ or more points, and make a convergence plot on a semi-log scale.
(a) $$f(x) = e^{\sin (2\pi x)}\qquad$$ (b) $$f(x) = \log [2+ \sin (3 \pi x ) ]\qquad$$ (c) $$f(x) = \cos^{12}[\pi (x-0.2)]$$
2. (a) ✍ Show that the functions $$\sin(r\pi x)$$ and $$\sin(s\pi x)$$ are identical at all of the nodes given in (9.5.2) if $$r-s=mN$$ for an integer $$m$$. This important fact is called aliasing, and it implies that only finitely many frequencies can be distinguished on a fixed node set.
(b) ⌨ Demonstrate part (a) with a graph for the case $$N=11$$, $$s=2$$, $$r=-9$$. Specifically, plot the two functions on one graph, and plot points to show that they intersect at all of the interpolation nodes.
3. ✍ Verify that the cardinal function given in Equation (9.5.3) is (a) 2-periodic, (b) satisfies $$\tau(t_k)=0$$ for $$k\neq 0$$ at the nodes (9.5.2), and (c) satisfies $$\lim_{x\to0}\tau(x)=1$$.
4. ✍ Prove the equality of the two expressions in (9.5.3). (Hint: Set $$z=e^{i\pi x/2}$$ and rewrite the sum using $$z$$ by applying Euler’s identity.)
5. ⌨ Spectral convergence is predicated on having infinitely many continuous derivatives. At the other extreme is a function with a jump discontinuity. Trigonometric interpolation across a jump leads to a lack of convergence altogether, a fact famously known as the Gibbs phenomenon.
(a) Define f(x) = sign(x+eps()). This function jumps from $$-1$$ to $$1$$ at $$x=-\epsilon_\text{mach}$$. Plot the function over $$-0.05\le x \le 0.15$$.
(b) Let $$n=30$$ and $$N=2n+1$$. Using Function 9.5.3, add a plot of the trigonometric interpolant to $$f$$ to the graph from part (a).
(c) Repeat part (b) for $$n=80$$ and $$n=180$$.
(d) You should see that the interpolants overshoot and oscillate near the step. The widths of the overshoots decrease with $$n$$ but the heights approach a limiting value. By zooming in to the graph, find the height of the overshoot to two decimal places.
6. ⌨ Let $$f(x)=x$$. Plot $$f$$ and its trigonometric interpolants of length $$N=2n+1$$ for $$n=6,20,50$$ over $$-1\le x \le 1$$. What feature of the function is causing large errors? | 2022-05-28T23:53:42 | {
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https://mathzsolution.com/is-the-rank-of-a-matrix-the-same-of-its-transpose-if-yes-how-can-i-prove-it/ | # Is the rank of a matrix the same of its transpose? If yes, how can I prove it?
I am auditing a Linear Algebra class, and today we were taught about the rank of a matrix. The definition was given from the row point of view:
“The rank of a matrix A is the number
of non-zero rows in the reduced
row-echelon form of A”.
The lecturer then explained that if the matrix $$AA$$ has size $$m×nm \times n$$, then $$rank(A)≤mrank(A) \leq m$$ and $$rank(A)≤nrank(A) \leq n$$.
The way I had been taught about rank was that it was the smallest of
• the number of rows bringing new information
• the number of columns bringing new information.
I don’t see how that would change if we transposed the matrix, so I said in the lecture:
“then the rank of a matrix is the same of its transpose, right?”
And the lecturer said:
“oh, not so fast! Hang on, I have to think about it”.
As the class has about 100 students and the lecturer was just substituting for the “normal” lecturer, he was probably a bit nervous, so he just went on with the lecture.
I have tested “my theory” with one matrix and it works, but even if I tried with 100 matrices and it worked, I wouldn’t have proven that it always works because there might be a case where it doesn’t.
So my question is first whether I am right, that is, whether the rank of a matrix is the same as the rank of its transpose, and second, if that is true, how can I prove it?
Thanks 🙂 | 2022-10-03T18:21:57 | {
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http://math.stackexchange.com/questions/415/what-property-of-certain-regular-polygons-allows-them-to-be-faces-of-the-platoni | # What property of certain regular polygons allows them to be faces of the Platonic Solids?
It appears to me that only Triangles, Squares, and Pentagons are able to "tessellate" (is that the proper word in this context?) to become regular 3D convex polytopes.
What property of those regular polygons themselves allow them to faces of regular convex polyhedron? Is it something in their angles? Their number of sides?
Also, why are there more Triangle-based Platonic Solids (three) than Square- and Pentagon- based ones? (one each)
Similarly, is this the same property that allows certain Platonic Solids to be used as "faces" of regular polychoron (4D polytopes)?
-
The definition of "Platonic solid" requires that the faces are identical regular polygons, but apparently you're looking for some other more general definition that allows irregular polygonal faces. Why is it that you're not counting the triangular bipyramid which doesn't have regular faces? Here's a picture: upload.wikimedia.org/wikipedia/en/thumb/f/f7/… – Ben Alpert Jul 22 '10 at 2:41
I'm only considering regular polygons...although it would be interesting to consider others. But that is beyond the scope of my question. I'm reluctant to re-define my question, especially after such a good answer has already been given. – Justin L. Jul 22 '10 at 5:07
Oh, I misread your question completely! I thought you were asking, "Why is it that only regular polygons can be faces of the Platonic solids?". – Ben Alpert Jul 25 '10 at 6:47
@Ben - I re-read my title and saw the source of your confusion, so I edited it. Hope it helps! – Justin L. Jul 25 '10 at 7:39
"…'tessellate' (is that the proper word in this context?)" You can say they tessellate the sphere. (I wouldn't just say tessellate because that implies tessellating the plane, which the pentagon does not do.) – Akiva Weinberger Oct 11 '15 at 2:02
The regular polygons that form the Platonic solids are those for which the measure of the interior angles, say α for convenience, is such that $3\alpha<2\pi$ (360°) so that three (or more) of the polygons can be assembled around a vertex of the solid.
Regular (equilateral) triangles have interior angles of measure $\frac{\pi}{3}$ (60°), so they can be assembled 3, 4, or 5 at a vertex ($3\cdot\frac{\pi}{3}<2\pi$, $4\cdot\frac{\pi}{3}<2\pi$, $5\cdot\frac{\pi}{3}<2\pi$), but not 6 ($6\cdot\frac{\pi}{3}=2\pi$--they tesselate the plane).
Regular quadrilaterals (squares) have interior angles of measure $\frac{\pi}{2}$ (90°), so they can be assembled 3 at a vertex ($3\cdot\frac{\pi}{2}<2\pi$), but not 4 ($4\cdot\frac{\pi}{2}=2\pi$--they tesselate the plane).
Regular pentagons have interior angles of measure $\frac{3\pi}{5}$ (108°), so they can be assembled 3 at a vertex ($3\cdot\frac{3\pi}{5}<2\pi$), but not 4 ($4\cdot\frac{3\pi}{5}>2\pi$).
Regular hexagons have interior angles of measure $\frac{2\pi}{3}$ (120°), so they cannot be assembled 3 at a vertex ($3\cdot\frac{2\pi}{3}=2\pi$--they tesselate the plane).
Any other regular polygon will have larger interior angles, so cannot be assembled into a regular solid.
-
Thanks! That makes a lot of sense actually. If it isn't much of a bother for you, do you mind addressing some of the other things I mentioned? Like why triangles are found in more platonic solids than both of the other shapes combined. – Justin L. Jul 22 '10 at 1:23
Because 6 triangles tile a plane. So 5 triangles can be folded to meet at a vertex and make a convex corner. Same with 4 and 3. 2 triangles would just form a 2-sided flat surface, not a polyhedron. – Jason S Jul 22 '10 at 1:35
I might have phrased my question misleadingly; I meant to ask why there are three triangle-based Platonic Solids, and yet only one Square- and one Pentagon-based Platonic Solids? – Justin L. Jul 22 '10 at 1:38
@Justin: More or less as Jason said, it's because the interior angle of an equilateral triangle is small enough that you can fit 3, 4, or 5 of them around a vertex and still have a total measure less than 2π. I don't, however, readily have an answer to the 4D question. – Isaac Jul 22 '10 at 1:40
and @Jason - My apologies! I misread Jason's first reply to mine. It makes perfect sense now; I'm not sure what I was on. – Justin L. Jul 22 '10 at 1:45
That they are platonic objects in dimension 2: equiangular, equilateral.
You can generalize this to platonic polytopes of dimension n, that they are always composed of platonic objects of dimension n-1.
Keep in mind, that for dimension n, there must be at least n edges per vertex. With these two facts you can start constructing platonic objects of dimensions higher than 3.
- | 2016-04-29T06:17:24 | {
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https://www.themathdoctors.org/order-of-operations-neglected-details/ | # Order of Operations: Neglected Details
The basic statement of the order of operations covers the five main operations (exponents, multiplication, division, addition, and subtraction). But what about other operations like square roots? How about trigonometric functions? And are operations at the same level always carried out left to right? Here are some questions about the details we don’t often mention.
## Powers of powers: top-down
Here is a question from 2016:
Parsing Powers Raised to Powers
What is the order of operations rule for 4 to the power 3 to the power 2?
Depending on how you do it -- bottom up or top down -- you get two different answers.
Working bottom up, you get 64 to the power 2, or 4096.
Working top down, you get 4 to the power 4 to the power 9, or 262144.
Aisha is doing these two evaluations:
$$\left(4^3\right)^2 = 64^2 = 4096$$
$$4^\left(3^2\right) = 4^9 = 262,144$$
Which of these is the correct interpretation of $$4^{3^2}$$? The former is left to right, or bottom up; the latter is right to left, or top down.
You're right: exponentiation is not associative, so the order must be specified clearly.
The usual convention is that we group "top down," so that
a^b^c = a^(b^c)
In your example, 4^3^2 is interpreted as 4^(3^2) = 262,144. Part of the reason for this is that if we wanted to say (4^3)^2, we could simplify that to 4^(3*2) = 4^6, so it makes sense to require parentheses when we want to to express this less-used meaning.
Just as we have to specify the direction in which we evaluate $$a – b – c$$ because subtraction lacks the associative property ($$(a – b) – c\ne a-(b-c)$$), we have to specify the order of exponentiation. The more “interesting” interpretation is the one we consider “correct”.
I talked about this 15 years ago on the following page, discussing the difficulty of finding definitive statements about it at that time, and presenting some supporting evidence:
Interpreting 2^3^2
http://mathforum.org/library/drmath/view/54362.html
Nowadays, we can check Wikipedia (which is usually trustworthy on mathematical matters, and gives sources that can be checked):
http://en.wikipedia.org/wiki/Exponentiation#Identities_and_properties
Without parentheses to modify the order of calculation,
by convention the order is top-down, not bottom-up:
b^{p^q} = b^{(p^q)} =/= (b^p)^q = b^{(p*q)} = b^{p*q} .
http://en.wikipedia.org/wiki/Order_of_operations#Special_cases
If exponentiation is indicated by stacked symbols,
the usual rule is to work from the top down, because
exponentiation is right-associative in mathematics thus
a^{b^c} = a^{(b^c)} ,
which typically is not equal to (a^b)^c. However, some
computer systems may resolve the ambiguous expression
differently. For example, Microsoft Office Excel
evaluates a^b^c as (a^b)^c, which is opposite of
normally accepted convention of top-down order of
execution for exponentiation.
I hope that helps.
In the 2001 discussion, Interpreting 2^3^2, the “patient” said, “My professor says that a lot of mathematicians take it on faith, that this is just the way to do it.” I commented on the difficulty of finding corroboration of my understanding of the convention:
I wouldn't quite call it "faith"; just an unwritten (or seldom written) convention. It's not that we "believe" it's true; we just "know" from experience that this is the usual way to read exponents.
I've occasionally wondered where I could find an "official" authority on this, myself. I know that a^b^c is generally taken as a^(b^c), and that the reason is that (a^b)^c can be written as a^(bc), while the other form, which is more interesting, has no alternative. We've occasionally mentioned this in Dr. Math answers, but are we an authority? I recall learning it from some book when I was in school, quite possibly not a textbook. But how can I prove it to a skeptic? I've been looking to see if some math organization has adopted standards, but have never found one to which I can refer.
Actually, the order of operations rules, so far as I have found, are not decreed by any authority, but have gradually become a commonly accepted standard through informal consensus, perhaps with the help of textbook authors, who tend to lay down rules more than actual mathematicians. The best thing to do is to show what mathematicians actually do.
I then gave what evidence I could find at the time, which included an acknowledgment by Texas Instruments that a particular calculator did not follow this rule, which “we learn in many algebra texts”, and an example of a site that explicitly stated their interpretation, calling it “customary”. (I noted just now that the Windows calculator also evaluates it bottom-up.) I concluded,
So I think it's clear that a^(b^c) is the usual interpretation of a^b^c, but not so much so that authors can comfortably assume that all readers will follow it without a reminder to make sure we agree. As far as I know, everyone who uses such expressions uses them in this way; I've never seen the other way given as the "right" way. The most one could say in the other direction would be that there is no firm rule and that one should state the rule before using it.
Another evidence of this interpretation is the way it is usually typeset: $$a^{b^c}$$. Notice that each layer is smaller than the one before, indicating that c is thought of as a superscript on b.
## Square roots: built-in grouping
I noted last time that PEMDAS does not explicitly include the negation operator, so we have to recognize that it is treated as being at the same level as multiplication. We get questions about other unary operators (that is, operations that act on only one number) from time to time. First, here is a question from 2009:
Order of Operations and Square Roots
At what step in the order of operations (Clear parentheses, exponents, multiply and divide in order presented, add and subtract in order presented) does one work out a square root? I haven't seen it much yet, but it appears in expressions and equations toward the end of a friend's child's 6th grade math book. For instance, 14 plus (24 minus 12) squared divided by 2 times 3 cubed plus (4 minus 2 squared) plus the square root of 9? At which step would you replace the square root of 9 with 3? Thanks!
It seems like it would go with either exponents or the multiplication and division step, but I don't know which; and I can't (for the life of me) find an explanation in the textbook; AND I can't remember what I learned in high school algebra . . . help!
Here goes:
14 + (24 - 12)squared divided by 2 x 3cubed + (4 - 2squared) +
square root of 9
14 + (12)squared divided by 2 x 3cubed + (4 - 2 squared) + square
root of 9
14 + 12squared divided by 2 x 3cubed + (4 - 4) + square root of 9
14 + 12squared divided by 2 x 3cubed + 0 + square root of 9
14 + 144 divided by 2 x 3cubed + 0 + square root of 9
14 + 144 divided by 2 x 27 + 0 + square root of 9
14 + 72 x 27 + 0 + square root of 9
14 + 1,944 + 0 + square root of 9
14 + 1,944 + 0 + 3
1,958 + 0 + 3
1,958 + 3
1,961
In this problem, I don't think it would change the effect to include the square root at the time you do exponents; however, I know that the order of ops is around to provide stability when it WOULD change the effect (and the outcome). It makes my head hurt (and doing algebra in an email box is almost as bad ;-). Can you give me a rule for where to put square roots in the order of ops? Thank you!
The expression, written with proper formatting, is $$14 + (24 – 12)^2 \div 2 \times 3^3 + (4 – 2^2) + \sqrt{9}$$ Tracey has evaluated it correctly; there is really no difficulty introduced by the radical, but it would be nice to be able to state clearly where it fits.
Doctor Rick took this:
I'll write the expression this way:
14 + (24 - 12)^2 / 2 * 3^3 + (4 - 2^2) + sqrt(9)
The part you're most concerned about is the last term; it probably looked a bit more like this on the paper:
___
\/ 9
One thing you need to notice is the bar (vinculum) over the 9. This may not be taught very often, but the vinculum is actually a remnant of an old alternate to parentheses. It is a grouping symbol, and as such, it goes along with parentheses in the order of operations. It's attached to the radical sign that says to take the square root; therefore immediately after evaluating the expression under the vinculum, you can take its square root.
So the radical, as commonly used, is taken as a single unit, and the entire radicand is evaluated as if it were in parentheses. So in terms of PEMDAS, the square root is at the level of P.
If the vinculum were not used, we would have an issue, as I’ll mention below.
Thus, the first thing to do is to evaluate each quantity in parentheses, AND evaluate the root:
14 + (24 - 12)^2 / 2 * 3^3 + (4 - 2^2) + sqrt(9)
\_______/ \_______/ \_____/
14 + 12 ^2 / 2 * 3^3 + 0 + 3
\_____/ \_/
14 + 144 / 2 * 27 + 0 + 3
\_______/
14 + 72 * 27 + 0 + 3
\_________/
14 + 1944 + 0 + 3
Now, with only additions left, we add them up and get 1961. I get the same answer you got, though I evaluated the square root first. All that really matters here, as you observed, is that the square root is evaluated before the final addition.
As long as the radical is evaluated before it is needed (in this case, for the addition next to it), we’re good; Tracey’s delayed evaluation was perfectly valid:
The order of operations really defines a "partial ordering" of the operations in a given expression; certain operations in the expression must be done before certain others, but the ordering of other pairs of operations may not matter. In fact, properties such as the associative property open up even more possible orderings; for instance, additions don't really need to be done left to right.
The order of operations is often treated by students as a rigid process, but there is really considerable freedom in it.
If you can find any case where you think some other order of operations (in regard to square roots) would change the result, let me see it. Personally, I can't imagine doing anything else; someone less familiar with roots may see an issue that I don't see.
I’ll have another comment on the radical in a moment.
## Percent: a postfix unary operator
I don’t normally think of “%” as an operator we would use in an equation, but there are occasions when it makes sense. Where does that fit in? This was asked in 2001:
Order of Operations with Percentages
Why does the order of operations exclude percentage, square roots, etc.?
Let's say I have a problem like this:
5% of 290 + 89 square root - 1 =
(By the way, according to some questions we have seen, “5% of” is included by some authors in other countries as a standard part of the notation, apparently under the belief that O in BODMAS refers to the word “of” being used in an expression. Here, I would never expect to see such an expression, and would consider it to come under “English sentence rules” rather than mathematical “order of operations rules”. But when such questions are asked, we answer them according to what a student is being taught.)
I’m not entirely sure what expression was meant here; I’ll suppose it’s “$$5\%\text{ of }290 + \sqrt{89} – 1$$.
I answered, ignoring the provided expression and the issue of “of”:
I think we avoid listing all these special things so we don't overwhelm students. Computer programming languages have to spell out exactly what every possible combination of symbols means, so they give big lists of rules, and it can be pretty scary to learn. In real life, we usually just try to avoid writing anything that looks too tricky, because people don't follow rules as well as computers.
The percent sign and square root sign are "unary operators"; that is, they act on one number to the right or left. They are probably always at the top of the list, the first things you have to evaluate. For example, 3+5% would mean 3 + 0.05 = 3.05, not 8% = 0.08.
I said “probably” for a reason; I am just using my own sense of what seems reasonable.
The square root is easier to be sure about, as we’ve already seen:
The square root sign is special. The actual radical sign is just the "v"-like thing alone, and means to take the root of the number following it. For example, \/4+5 would be 2+5=7, not the square root of 9, which is 3. But you usually see a bar hanging over the numbers you are to operate on. That part of the symbol is called a vinculum, and is left over from a time when that was used instead of parentheses. So
____
\/4+5
does mean the square root of 9.
I think I’ve only see the radical sign without a vinculum in printed material that was very old (before printing technology made it easier to typeset complicated expressions); but there are times even now when people writing to us are forced to do so. If you write “√4 + 5”, I would have to take that as 2 + 5 = 7; if you meant the root of the sum, you would have to write “√(4 + 5)”.
When I wrote this, I probably didn’t specifically consider how either a radical or a percentage would interact with exponents. If you wrote “√4 ^ 5”, I would take it as “√(4) ^ 5”, but if you think about it, it doesn’t make any difference, as $$\sqrt{4}\ ^5 = \left(4^{1/2}\right)^5 =4^{1/2\cdot 5} = \left(4^5\right)^{1/2} = \sqrt{4^5}.$$
## Factorial: also at the top of the heap
Another unary operator (like “%” and unlike the radical, written after its argument) is the factorial, $$n! = n(n-1)\cdots3\cdot2\cdot1$$. This question was raised in 2000:
Factorials and Order of Operations
Where do factorials go in the Order of Operations? I've been thinking that factorials would go with exponents, but on the other hand, factorials go with multiplication.
Probably Kiaran’s thinking was based on the fact that the factorial is an extension of multiplication. I responded,
Hi, Kiaran. Good question!
I would put factorials with functions, something you don't often see listed in the order of operations. If you are not familiar with functions, you can think of them very simply as something to be done to a single number (or sometimes a list of numbers) that produces a number as a result. The square root and the trigonometric functions (sine, cosine) are examples of this. Technically, the factorial is called a (postfix) unary operator, which means essentially the same thing.
Just as a function f is applied immediately after evaluating the parentheses that must follow it, $$f(\dots)$$, unary operators are applied immediately to the single thing that follows. The issue with functions like the factorial is the lack of mandatory parentheses in the notation we use.
Because these don't stand between two numbers, but only relate to one number, they are thought of as being attached directly to that number; if you want to apply a function or unary operator to a whole expression, you have to use parentheses. That puts functions at the top of the order of operations, just after parentheses.
For example, this means that the factorial applies only to the 5:
3 + 4*5! = 3 + 4*120 = 3 + 480 = 483
while this applies it to (3 + 4*5):
(3 + 4*5)!
As with percentages, I am just using my common sense here, along with general experience of how factorials are used in expressions. There is no authority to decree this, but it is clear from common usage. For example, browsing through the Wikipedia article on factorial, I see equations like $$n! = n(n-1)!$$, which clearly does not mean $$n! = [n(n-1)]!$$ as it would have to if the factorial were either lower than multiplication, or done left to right along with multiplication. I also see expressions like $$\ln n!$$ which is not taken to mean $$(\ln n)!$$.
One reliable source does explicitly state this:
Wolfram MathWorld: Precedence
For simple expressions, operations are typically ordered from highest to lowest in the order:
1. Parenthesization,
2. Factorial,
3. Exponentiation,
4. Multiplication and division,
## Absolute value: both operation and grouping symbol
Let’s look at one more operation we’ve been asked about, which is quite different, though it too is essentially a unary operator or function. This question is from 2002:
Absolute Value as a Grouping Symbol?
I teach 8th grade Algebra in California. I was teaching the Order of Operations. I explained that there are four grouping symbols: parentheses, brackets, braces, and the fraction bar.
One student asked me if the absolute value bars are also grouping symbols. I told him no. But as I thought about it, in a practical sense, absolute value is a grouping symbol: with every equation/expression in my algebra textbook that has an absolute value, you must solve what's in the absolute first!
I have two questions: 1) Are there any examples of a problem where you do not have to evaluate the absolute first? and 2) Would it would be wrong for me to teach the absolute value as a "pseudo" grouping symbol?
Interesting question! I replied by mentioning yet another grouping symbol (the vinculum, as I mentioned above), then continuing:
But yes, the absolute value bars do serve partly as a grouping symbol. That is, their primary meaning is to indicate an absolute value, but they incidentally require that whatever is inside must be evaluated first. Thus
3|x+y|
is equivalent to
3*abs(x+y)
where I have used functional notation, in which again the parentheses are primarily to identify the argument of the function "abs" but also serve to group. In a function with two arguments, I suppose you could call the comma separating the arguments a grouping symbol too: atan2(x,y).
In effect, the absolute value bars are modified parentheses, which indicate that the function is applied to the quantity within.
In summary, I would go a bit beyond calling the absolute value a "pseudo" grouping symbol, and call it a symbol one of whose functions is to group an expression.
I don't think I understand your first question; the point is that you CAN'T evaluate an absolute value before evaluating its argument. I suppose you meant, first before doing something else outside of it. Of course you can distribute, as you can with parentheses:
3|x+y| = |3x + 3y|
but in terms of actually evaluating an expression as it stands, you have to evaluate the argument, then the absolute value, then whatever it is used in.
I also referred to another page that mentioned this,
Grouping Symbols
Years later, someone wrote to comment on that idea of distributing into an absolute value, which led to this explanation:
Absolute Caution
Next time, I will post a discussion on a topic too big to fit here: order of operations with trigonometric functions, which are commonly not written with the usual function notation, leading to some very complicated issues. And soon, I’ll talk more about the interaction of order of operations with properties like distributivity.
### 1 thought on “Order of Operations: Neglected Details”
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http://mathhelpforum.com/number-theory/45489-find-sum-digits-b.html | # Thread: Find sum of the digits of B?
1. ## Find sum of the digits of B?
When 4444^4444 is written in decimal notations, the sum of its digits is A. Let B be the sum of the digits of A. Find the sum of the digits of B.
2. My solution is not very elegant. It may not even be right. But here goes.
First I seek an upper bound to the solution.
$log(4444^{4444}) = 16.2$
Furthermore $10^{16.2}<1.6*10^{16}$
Now we conclude that $4444^{4444}$ has 17 digits where the most significant digit is 1 and the other digits are unknown (but not more than 9).
Hence $A < 1 + 9*16 = 145$
Now for all numbers up to 145 the one with the largest digit sum is 139 which sums to 13. Hence B<13.
Now summing the digits of any number gives a result that is equal to the starting number (modulo 9).
Hence $4444^{4444} = 7^7 = 823543$ (mod 9)
This is greater than 13 so I sum the digits repeatidly until I get an acceptable answer. The result is 823543, 25, 7. OK so I think the answer is 7.
When you receive the model answer can you please post it?
3. ## A long train of thought
I think Kiwi_Dave is on the right track. We can use logarithms in this problem to determine the number of digits in 4444^4444.
log (4444^4444) = 4444 log 4444 = 16210.7079
which means 4444^4444 has 16211 digits. The sum of these 16211 digits is A. The sum of the digits of A is B.
Without restrictions save for the number of digits, the maximum possible value of A is 16211 * 9 = 145899. Of course, that has a digital root of 9, so it couldn't be the value of A.
We know that the digital root of 4444^4444 is 7 (because the digital root of 4444 is 7, and the powers of 7 have digital roots of 7, 4, 1 respectively for exponents of the form 3k-2, 3k-1 and 3k respectively). Its last digit is 6 (owing to the even exponent).
As for its first digit, we can calculate this using logarithms again.
Let's say n is the first digit. So n * 10^16210 < 4444^4444
So log n + 16210 < 16210.7079
We find that n = 5
So the 1st digit is 5, the last digit is 6. Together they have a digital root of 2. So the remaining (16211-2) = 16209 digits have a digital root of 5.
Let's say 16208 digits are 9's and another digit is 5. So the highest possible value of A is
5 + 16208 * 9 + 5 + 6 = 145888. It has 6 digits. The sum of the digits is 34.
My answer is 34.
Of course, the 16208 digits aren't necessarily 9 and the other digit 5... so I was just sharing this train of thought, lol I still don't know how to peg a convincing answer for this caper, so I'll think about it for the night
Please share the solution and answer. I'd love to know it too.
4. I think one of the experts will show us an answer that does not involve logs. But between us we will get close! I mistook a comma for a period on my calculator!
Originally Posted by Coffee Cat
5 + 16208 * 9 + 5 + 6 = 145888. It has 6 digits. The sum of the digits is 34.
By changing a few digits we could get A = 144999. The sum of the digits is 36 so the answer cannot be greater than 36.
However, $4444^{4444}=7=16=25=34$ (mod 9). So the answer must be 7,16,25 or 34?
5. Lemma: The maximum possible sum of the digits of an -digit number is . In other words,
And so, the solution
Using this useful result,
Hence
Now, , but since that last digit sum cannot be or greater it must be .
6. ## :)
Awesome! Thanks, fardeen_gen!
7. ## Easier and elegant!
See my reply to this at MathLinks :: View topic - A 103 this is a problem from the 1975 IMO. I believe my solution is both more elegant and easy!
8. Originally Posted by manjil
See my reply to this at MathLinks :: View topic - A 103 this is a problem from the 1975 IMO. I believe my solution is both more elegant and easy!
Yours has got some things incorrect o.O
The largest possible sum of numbers is the sum of 99,999, which is 45 (>44)
And why is the sum B at most the sum of digits of 39 ?
9. Anyway I was wrong on that count but the proof still holds if we replace 179,999 by 99,999 and 44 by 45.
Also the sum of the digits of B can't be more than 39 because 3+9=12 which is the largest possible sum!
Thanks for pointing that out!
But don't you agree that my proof is better than the others posted?
10. Originally Posted by manjil
See my reply to this at MathLinks :: View topic - A 103 this is a problem from the 1975 IMO. I believe my solution is both more elegant and easy!
It has a sloppy estimate of the number of digits.
RonL
11. Originally Posted by manjil
Anyway I was wrong on that count but the proof still holds if we replace 179,999 by 99,999 and 44 by 45.
Also the sum of the digits of B can't be more than 39 because 3+9=12 which is the largest possible sum!
Thanks for pointing that out!
But don't you agree that my proof is better than the others posted?
It doesn't prove anything. So basically, it's not better... | 2013-12-19T12:19:33 | {
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https://www.quizover.com/trigonometry/test/decomposing-a-composite-function-into-its-component-by-openstax | # 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.
the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve
1+cos²A/cos²A=2cosec²A-1
test for convergence the series 1+x/2+2!/9x3
a man walks up 200 meters along a straight road whose inclination is 30 degree.How high above the starting level is he?
100 meters
Kuldeep
Find that number sum and product of all the divisors of 360
Ajith
exponential series
Naveen
what is subgroup
Prove that: (2cos&+1)(2cos&-1)(2cos2&-1)=2cos4&+1
e power cos hyperbolic (x+iy)
10y
Michael
tan hyperbolic inverse (x+iy)=alpha +i bita
prove that cos(π/6-a)*cos(π/3+b)-sin(π/6-a)*sin(π/3+b)=sin(a-b)
why {2kπ} union {kπ}={kπ}?
why is {2kπ} union {kπ}={kπ}? when k belong to integer
Huy
if 9 sin theta + 40 cos theta = 41,prove that:41 cos theta = 41
what is complex numbers
Dua
Yes
ahmed
Thank you
Dua
give me treganamentry question
Solve 2cos x + 3sin x = 0.5 | 2018-12-10T20:18:45 | {
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http://forum.math.toronto.edu/index.php?topic=1510.0;wap2 | MAT244--2018F > Quiz-6
Q6 TUT 5102
(1/1)
Victor Ivrii:
The coefficient matrix contains a parameter $\alpha$.
(a) Determine the eigenvalues in terms of $\alpha$.
(b) Find the critical value or values of $\alpha$ where the qualitative nature of the phase portrait for the system changes.
(c) Draw a phase portrait for a value of $\alpha$ slightly below, and for another value slightly above, each critical value.
$$\mathbf{x}' =\begin{pmatrix} 4 &\alpha\\ 8 &-6 \end{pmatrix}\mathbf{x}.$$
Michael Poon:
a) Finding the eigenvalues:
Set the determinant = 0
\begin{align}
(4 - \lambda)(-6 - \lambda) - 8\alpha &= 0\\
\lambda^2 + 2\lambda - 24 - 8\alpha &= 0\\
\lambda &= -1 \pm \sqrt{25 + 8\alpha}
\end{align}
b)
Case 1: Eigenvalues real and same sign
when: $\alpha$ > $\frac{-25}{8}$ + 1
Case 2: Eigenvalues real and opposite sign
when: $\frac{-25}{8}$ < $\alpha$ < $\frac{-25}{8} + 1$
Case 3: Eigenvalues complex
when: $\alpha$ < $\frac{-25}{8}$
critical points: $\alpha$ = $\frac{-25}{8}$, $\frac{-25}{8}$ + 1
c) will be posted below:
Michael Poon:
Phase portraits attached below:
Top: Eigenvalues real & same sign (+ve), stable
Middle: Eigenvalues real & opposite sign, saddle
Bottom: Eigenvalues complex & negative, unstable spiral
Jiacheng Ge:
My solution is different.
Victor Ivrii:
When eigenvalues pass from real to complex conjugate, stability does not change. Jiacheng is right | 2020-10-29T08:53:45 | {
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https://math.stackexchange.com/questions/2981554/can-a-set-of-positive-measure-and-its-complement-both-have-empty-interior | # Can a set of positive measure and its complement both have empty interior? [duplicate]
This might be silly, but I am not sure:
Does there exist a Lebesgue measurable subset $$E \subseteq (0,1)$$ such that
1. $$E$$ and $$(0,1) \setminus E$$ both have positive Lebesgue measure.
2. $$E$$ and $$(0,1) \setminus E$$ both have empty interiors.
If we relax condition $$1$$, then $$E=Q\cap (0,1)$$ works. If we relax condition $$2$$, then the fat Cantor set does the job. (Its complement have non-empty interior though).
## marked as duplicate by bof, José Carlos Santos, Asaf Karagila♦ general-topology StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 2 '18 at 17:32
• In my answer to this question I constructed an $F_\sigma$ set $M\subseteq\mathbb R$ such that $0\lt m(M\cap I)\lt m(I)$ for every interval $I$, where $m$ denotes Lebesgue measure. Obviously the sets $M$ and $\mathbb R\setminus M$ have positive Lebesgue measure and have empty interiors, and the same goes for the sets $E=M\cap(0,1)$ and $(0,1)\setminus E$. – bof Nov 2 '18 at 12:25
• What irrationals on $(0,\frac12)$ with rationals from $(\frac12,1)$ aren't good enough for you? Since when did you become so picky? Are things really that bad since I left? – Asaf Karagila Nov 2 '18 at 17:33
Let $$E$$ be the union of $$(0,1) \setminus \mathbb Q)\cap (0,\frac 1 2]$$ and $$\mathbb Q\cap (\frac 1 2,1)$$. Then $$E$$ and $$(0,1)\setminus E$$ both have positive measure and they have no interior.
Let $$E$$ be a fat Cantor set $$C$$ together with the $$C^\complement\cap\mathbb Q$$. | 2019-05-20T18:38:34 | {
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http://mathhelpforum.com/geometry/10902-unit-vectors.html | 1. ## Unit Vectors
I have to find two unit vectors that are parallel to
<3, 1, 2>
And then I have to write each vector as the product of its magnitude and a unit vector.
--------
My approach:
We know that two vectors are parallel if their cross product is equal to 0.
$\mathbf{a} \times \mathbf{b} = \mathbf{A}_{\times} \mathbf{b} = \begin{bmatrix}0&-a_3&a_2\\a_3&0&-a_1\\-a_2&a_1&0\end{bmatrix}\begin{bmatrix}b_1\\b_2\\b_3 \end{bmatrix}$
Using the above, I plugged in a_1, b_1, c_1 and then augmented it with the 0 vector, but I just ended up with a_1 = 0, b_1 = 0, and c_1 = 0 which wasn't very helpful. Any ideas?
*EDIT* I guess Wikipedia math code doesn't conform to this site;
It's mean to be:
a x b = [[0, -a_3 a_2], [a_3, 0, -a_1], [-a_2, a_1, 0]] multiplied with the vector [b_1, b_2, b_3]; of course there is also the determinant method.
2. Originally Posted by Ideasman
I have to find two unit vectors that are parallel to
<3, 1, 2>
And then I have to write each vector as the product of its magnitude and a unit vector.
--------
My approach:
We know that two vectors are parallel if their cross product is equal to 0.
$\mathbf{a} \times \mathbf{b} = \mathbf{A}_{\times} \mathbf{b} = \begin{bmatrix}0&-a_3&a_2\\a_3&0&-a_1\\-a_2&a_1&0\end{bmatrix}\begin{bmatrix}b_1\\b_2\\b_3 \end{bmatrix}$
Using the above, I plugged in a_1, b_1, c_1 and then augmented it with the 0 vector, but I just ended up with a_1 = 0, b_1 = 0, and c_1 = 0 which wasn't very helpful. Any ideas?
*EDIT* I guess Wikipedia math code doesn't conform to this site;
It's mean to be:
a x b = [[0, -a_3 a_2], [a_3, 0, -a_1], [-a_2, a_1, 0]] multiplied with the vector [b_1, b_2, b_3]; of course there is also the determinant method.
You're thinking about it too much. Just use any scalar.
That is, two vectors are parallel if they are a scalar of the other. You can check with using cross products to show they are parallel.
So, <6, 2, 4> and <9, 3, 6>. Now find the norm of each and you should be able to take it from there.
3. Originally Posted by AfterShock
You're thinking about it too much. Just use any scalar.
That is, two vectors are parallel if they are a scalar of the other. You can check with using cross products to show they are parallel.
So, <6, 2, 4> and <9, 3, 6>. Now find the norm of each and you should be able to take it from there.
What Aftershock said, but with an addendum:
You need TWO unit vectors along the same line. They will necessarily be in opposite directions. So find one unit vector by taking the norm of the given vector and dividing that into your given vector. (Aftershock's method.) The second unit vector will merely be in the opposite direction of this, so multiply your first vector by a -1.
-Dan
4. Hello, Ideasman!
A strange question . . .
Find two unit vectors that are parallel to: . $\vec{v}\:=\:\langle 3,\,1,\,2\rangle$
The magnitude of $\vec{v}$ is: . $|\vec{v}| \:=\:\sqrt{3^2 + 1^2 + 2^2} \:=\:\sqrt{14}$
The two unit vectors are:
. . $u_1\:=\:\frac{\vec{v}}{|\vec{v}|} \:=\:\frac{\langle 3,\,1,\,2\rangle}{\sqrt{14}}\:=\:\left\langle \frac{3}{\sqrt{14}},\:\frac{1}{\sqrt{14}},\:\frac{ 2}{\sqrt{14}}\right\rangle$
. . $u_2\:=\:\text{-}\frac{\vec{v}}{|\vec{v}|} \:=\:\text{-}\frac{\langle3,\,1,\,2\rangle}{\sqrt{14}} \:=\:\left\langle\frac{-3}{\sqrt{14}},\,\frac{-1}{\sqrt{14}},\,\frac{-2}{\sqrt{14}}\right\rangle
$
Write each vector as the product of its magnitude and a unit vector.
This is the strange part: a unit vector has magnitude 1.
So, what the point of this task? ... multiplying by one? | 2013-05-26T09:22:57 | {
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https://math.stackexchange.com/questions/1731261/what-does-it-mean-for-simple-functions-to-have-finite-range | # What does it mean for simple functions to have finite range
In Mathematical Tools for Data Mining: Set Theory, Partial Orders, Combinatorics By Dan Simovici, Chabane Djeraba, it says:
A simple function is a function $f: S \to \mathbb{R}$ that has finite range.
Can someone clarify what it means by "finite range"? Does it mean that $f$ is bounded below and above?
No, it means that $f(S) = A$, where $A$ is a finite set : $f$ take only a finite number of values
• Yes, it is. Every linear combination of characteristic functions has finite range, and every finite range function is a linear combination of characteristic functions. If the range is $\{a_1, \cdots, a_n\}$, then $f = \sum_{k=1}^n a_k \chi_{f^{-1}(\{a_k\})}$ – Tryss Apr 7 '16 at 0:22 | 2019-07-22T18:27:42 | {
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https://mathoverflow.net/questions/361034/is-this-process-a-markov-one | # is this process a Markov one?
Here is the problem I can't solve.
Let $$\xi_n$$ $$(n=1,2,3,\dots)$$ be a sequence of i.i.d. random variables on $$\mathbb{R}$$ with density $$p(x)>0$$, let $$\eta_n=\sum_{i=1}^{n}\xi_i^2$$. Define $$\zeta_t = \eta_{[t]}(t-[t]) \times \eta_{[t]+1}(-t+[t]+1),$$ where $$[t]$$ denotes the integer part of $$t$$. Is $$\zeta_t$$ a Markov process?
I tried to prove that given process is Markov by definition: $$\mathbb{P}(\zeta_{t+s}\in A |\mathcal{F}_{\le t}) =\mathbb{P}(\zeta_{t+s}\in A| \mathcal{F_{=t}}),$$ where $$A \in \mathcal{F}_{\ge t}$$, $$\mathcal{F}_{\le t} = \sigma(\zeta_s: s\le t)$$, $$\mathcal{F}_{\ge t} = \sigma(\zeta_s: s\ge t)$$, $$\mathcal{F}_{= t} = \sigma(\zeta_t)$$.
Here is my first attempt so far
$$\mathbb{P}(\zeta_{t+s}\in A |\mathcal{F}_{\le t})=\mathbb{P}(\eta_{[t+s]}(t+s-[t+s]) \times \eta_{[t+s]+1}(-t-s+[t+s]+1)\in A|\mathcal{F}_{\le t})=$$ $$=\mathbb{P}\left(\left(\sum_{k=1}^{[t+s]}\xi_k^2(t+s-[t+s])\right)\times\left(\sum_{i=1}^{[t+s]+1}\xi_i^2([t+s]+1-t-s)\right)\in A|\mathcal{F}_{\le t}\right)=$$
$$=\sum_{k=1}^{[t+s]}\sum_{i=1}^{[t+s]+1}\mathbb{E}(\mathbb{1}_A\xi_k^2(t+s-[t+s])\xi_i^2([t+s]+1-t-s)|\mathcal{F}_{\le t}),$$ using that $$\mathbb{P}(A|\mathcal{F})=\mathbb{E}(\mathbb{1}_A|\mathcal{F})$$.
My second attempt was to make an example of suitable random variables and probability space, such that $$\zeta_t$$ is not Markov. I began to think it is not Markov as $$\zeta_t$$ depends on $$\eta_{[t]+1}$$, so it depends on the future in some sense because $$[t]+1>t$$.
I understand that it is kind of "not Mathoverflow question", but I did not receive answer on MathStackexchange and I faced this problem during my term paper, so may be this is a reason to post it here.
• Do you really mean "$\times$" and not "$+$" in the definition of $\zeta_t$? It looks like $\zeta_t = 0$ whenever $t$ is an integer. – Mateusz Kwaśnicki May 22 at 10:41
• Yes, there really should be $\times$ in the definition of $\zeta_t$. Can you tell, please, how do you know that $\zeta_t=0$ for integer $t$? – I.Kiaan May 22 at 10:46
• I thought $\eta_{[t]} (t - [t])$ is the product of $\eta_{[t]}$ and $t - [t]$, and the latter is zero when $t$ is an integer? – Mateusz Kwaśnicki May 22 at 10:50
• No, it's not markov. If you see the process at times 1.25, 1.5 then you know $\eta_1, \eta_2$ and so you know the process perfectly in the near future, which is not the case if you only see the process at time 1.5 – mike May 22 at 11:21
• @MateuszKwaśnicki, sorry, I really misread your comment. Sure, you are right – I.Kiaan May 22 at 11:52
The process is not Markov in general. Indeed, let $$X_i:=\xi_i$$, $$S_n:=\eta_n=\sum_1^n X_i^2$$, and $$Z_t:=\zeta_t=(t-[t])([t]+1-t)S_{[t]}S_{[t]+1},$$ where $$P(X_i=0)=P(X_i=1)=1/2$$. Then $$Z_{3/2}=\tfrac14\,X_1(X_1+X_2),\quad Z_2=0,\quad Z_{5/2}=\tfrac14\,(X_1+X_2)(X_1+X_2+X_3).$$ So, the conditional distribution of $$Z_{5/2}$$ given $$Z_2$$ is the same as the unconditional distribution of $$Z_{5/2}$$. On the other hand, the conditional distribution of $$Z_{5/2}$$ given $$Z_{3/2},Z_2$$ is not the same as the unconditional distribution of $$Z_{5/2}$$, because $$Z_{5/2}$$ depends on $$Z_{3/2}$$: $$P(Z_{3/2}=0)=P(X_1=0)=1/2$$ and $$P(Z_{5/2}=0)=P(X_1+X_2=0)=1/4$$, whereas $$P(Z_{3/2}=0,Z_{5/2}=0)=P(X_1+X_2=0)=1/4\ne P(Z_{3/2}=0)P(Z_{5/2}=0).$$
So, $$(Z_t)$$ is not Markov.
If you insist that the distribution of $$X_1$$ be absolutely continuous with a strictly positive density, then you can appropriately approximate the discrete distribution by such an absolutely continuous one.
Alternatively, you may e.g. assume that $$X_1\sim N(0,1)$$. Then $$EZ_{3/2}=1$$, $$EZ_{3/2}=5/2$$, but $$EZ_{3/2}Z_{3/2}=27/2\ne EZ_{3/2}EZ_{5/2}$$, so that here too $$Z_{5/2}$$ depends on $$Z_{3/2}$$.
This is a good exercise. As far as I understand the OP, the situation is as follows. Given a deterministic non-negative function $$\phi$$ on the interval $$[0,1]$$ with $$\phi(t)=0\iff t \in\{0,1\}$$ and a sequence of random variables $$Z_n$$, one defines $$\zeta(t)=\phi(t-n) \cdot Z_n \;, \qquad n\le t\le n+1 \;.$$ The question is when $$\zeta(t)$$ is Markov, and the answer to this question is pretty obvious: if and only if the variables $$Z_n$$ are independent (look at the Markov condition at integer times).
In the original question $$\phi(t)=t(1-t)$$ and $$Z_n=(\xi_1^2+\dots+\xi_n^2)(\xi_1^2+\dots+\xi_{n+1}^2) \;,$$ which are pretty obviously not independent unless $$\xi_i^2$$ are a.s. constant. | 2020-06-07T10:31:06 | {
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https://math.stackexchange.com/questions/3244612/elegant-way-to-prove-congruence/3244652 | # Elegant way to prove congruence
I'm stuck with the last question of this exercise
1) First question asks to solve the linear diophantine
$$143x-840y=1$$
based on the remark $$143\times 47 - 840 \times 8 = 1$$ (done)
2) second question asks to prove that if a natural $$n$$ is coprime with $$899$$, then
$$n^{840} \equiv 1 \mod 899$$ (done using Fermat's little theorem on $$31$$ and $$29$$ knowing that $$899 = 31 \times 29$$)
3) This last question asks to determine an integer between $$100$$ and $$1000$$ such that
$$n^{143} \equiv 2 \mod 899$$
I proved the problem reduces to determining the remainder of $$2^{47}$$ when divided by $$899$$ wolframalpha says it's $$345$$ but I still can't see any elegant way to prove it.
Thanks.
• Elegant way to prove what, that $\, 2^{\large 47}\equiv 345\pmod{899}?\$ – Bill Dubuque May 29 at 20:50
• yes (based maybe on previous questions if they're of some use) – ahmed May 29 at 20:55
• It should be easy by CRT, i.e. compute $2^{\large 47}$ mod $29$ & $31$ then lift to their product by CRT. – Bill Dubuque May 29 at 20:56
• Note: $2^5\equiv1\pmod{31}$ – J. W. Tanner May 29 at 21:01
It's easy mental arithmetic to compute $$2^{\large 47}$$ mod $$31$$ & $$29$$ then lift it to $$31\cdot 29$$ by CRT.
$$\!\!\bmod \color{#c00}{31}\!:\ \ \,2^{\large 5}\equiv 1\,\Rightarrow\,2^{\large 47}\equiv 2^{\large 2}\equiv\color{#c00} 4$$
$$\!\!\bmod 29\!:\,\ 2^{\large 28}\!\equiv 1\,\Rightarrow\,2^{\large 47}\equiv 2^{\large -9}\equiv 1/(-10)\equiv 30/(-10)\equiv -3$$
thus $$\ {-}3\equiv 2^{\large 47}\equiv \color{#c00}{4\!+\!31}k\equiv 4\!+\!2k\iff 2k\equiv -7\equiv 22\iff\color{#0a0}{k\equiv 11\pmod{\!29}}$$
Therefore: $$\,2^{\large 47} = 4+31\color{#0a0}k = 4+31(\color{#0a0}{11\!+\!29}n) = 345 + 31(29n)$$
• or $4\equiv2^{47}\equiv26+29k\iff-22\equiv-2k\iff k\equiv11\pmod{31}\\$ and therefore: $2^{47}=26+29k=26+29(11+31n)$ – J. W. Tanner May 29 at 21:40
Problem: We want to find the $$x$$ in range $$0 \leq x \leq 898$$ such that $$2^{47} \equiv x \pmod{899}$$.
Here's a relatively fast method for general problems like these called square-and-multiply:
We can write $$47 = 2^5 + 2^3 + 2^2 + 2^1 + 2^0. \$$ Therefore, $$2^{47} = 2^{2^5} \cdot 2^{2^3} \cdot 2^{2^2} \cdot 2^{2} \cdot 2. \$$ And now we can compute these values $$\big\{2^{2^k} \big\}_{k=0}^5$$ via iterative squaring (and reducing modulo $$899$$ when necessary):
$$\qquad \bullet \ \quad 2^2 \ = \ 4$$
$$\qquad \bullet \ \quad 2^{2^2} = \ 4^2 = \ 16$$
$$\qquad \bullet \ \quad 2^{2^3} \ = \ (16)^2 \ = \ 256$$
$$\qquad \bullet \ \quad 2^{2^4} \ = \ 256^2 \ = \ 65536 \ \equiv \ 808 \pmod{899}$$
$$\qquad \bullet \ \quad 2^{2^5} \ = \ 808^2 \ = \ 652864 \ \equiv \ 190 \pmod{899}$$
Now multiply the appropriate values together, reduce modulo $$899$$, and you're done(!!). Clearly, this is not the most computationally efficient approach in this specific case—just look at Bill Dubuque's answer (+1). But marvel at just how few computations there are relative to the naïve approach!
In general, if the exponent, when written in base-$$2$$, consists of $$n$$ bits, then—if I'm not mistaken—one needs to perform a maximum of $$2n-3$$ multiplications when using square-and-multiply (plus modulo reductions, if desired). So the number of multiplications required grows logarithmically with the size of the exponent, whereas the number of multiplications required for naïve exponentiation grows linearly.
As suggested in the comments, evaluate $$2^{47}$$ mod $$29$$ and $$31$$. For $$31$$ we quickly get $$4$$ mod $$31$$.
For mod $$29$$, note that $$2^5\equiv3$$ so $$2^{15}\equiv(-2)$$ and $$2^{45}\equiv(-8)$$ so finally $$2^{47}\equiv(-3)$$.
From here we do a quick CRT to finish off the problem (assuming you know how to do it; the gist is essentially the Euclidean Algorithm, modified slightly). | 2019-07-23T05:25:23 | {
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https://encyclopediaofmath.org/wiki/Density_of_a_set | # Density of a set
2010 Mathematics Subject Classification: Primary: 28A05 Secondary: 28A1549Q15 [MSN][ZBL]
A concept of classical measure theory generalized further in Geometric measure theory
### Lebesgue density of a set
Given a Lebesgue measurable set $E$ in the standard Euclidean space $\mathbb R^n$ and a point $x\in\mathbb R^n$, the upper and lower densities of $E$ at $x$ are defined respectively as $\limsup_{r\downarrow 0} \frac{\lambda (B_r (x)\cap E)}{\omega_n r^n} \qquad \mbox{and} \qquad \liminf_{r\downarrow 0} \frac{\lambda (B_r (x)\cap E)}{\omega_n r^n}\, ,$ where $\lambda$ denotes the Lebesgue measure and $\omega_n$ the volume of the unit $n$-dimensional ball. If the two numbers coincides, i.e. if the following limit exists, $\lim_{r\downarrow 0} \frac{\lambda (B_r (x)\cap E)}{\omega_n r^n}\, ,$ the resulting number is called the density of $E$ at $x$. The following is a classical result in measure theory (see for instance Corollary 3 in Section 1.7 of [EG]), due to Lebesgue in the case $n=1$:
Theorem 1 The density of a Lebesgue measurable set $E\subset \mathbb R^n$ is $1$ at $\lambda$-a.e. $x\in E$ and $0$ at $\lambda$-a.e. $x\not \in E$.
The points of the first type are also called density points of $E$, whereas the second points are called points of dispersions. The density points and the points of dispersion are sometimes defined also for non-measurable sets $E$: in that case one uses the Lebesgue outer measure, cp. with Sections 2.9.11 and 2.9.12 of [Fe] (see Lebesgue measure).
### Density of a measure
The concept above has been generalized in geometric measure theory to measures, starting from the work of Besicovitch. Consider a (locally finite) Radon measure $\mu$ in the Euclidean space $\mathbb R^n$, a point $x\in \mathbb R^n$ and a nonnegative real number $\alpha$ (see for instance Definition 2.14 of [De] or Definition 6.8 of [Ma]). The $\alpha$-dimensional upper and lower densities of $\mu$ at $x$ are defined as $\theta^{\alpha,*} (\mu, x) := \limsup_{r\downarrow 0} \frac{\mu (B_r (x))}{\omega_\alpha r^\alpha} \qquad \mbox{and}\qquad \theta^\alpha_* (\mu, x) =\liminf_{r\downarrow 0} \frac{\mu (B_r (x))}{\omega_\alpha r^\alpha}\, ,$ where the normalizing factor $\omega_\alpha$ is the $\alpha$-dimensional volume of the unit ball in $\mathbb R^\alpha$ when $\alpha$ is a positive integer and in general $\omega_\alpha = \pi^{\alpha/2} \Gamma (1+\alpha/2)$. If the two numbers coincide, the resulting value is called the $\alpha$-dimensional density of $\mu$ at $x$. The following Theorem by Marstrand shows that the density might exist and be nontrivial if and only if $\alpha$ is an integer (we refer to Chapter 3 of [De] for its proof).
Theorem 2 Let $\mu$ be a locally finite Radon measure on $\mathbb R^n$ and $\alpha$ a nonnegative real number such that the $\alpha$-dimensional density of $\mu$ exists and is positive on a set of positive $\mu$-measure. Then $\alpha$ is necessarily an integer.
#### Lebesgue theorem
Concerning $n$-dimensional densities, the following theorem corresponds to the fact that, given a summable function $f$, $\lambda$-a.e. point $x$ is a Lebesgue point for $f$:
Theorem 3 (Theorem 1 in Section 1.7 of [EG]) Let $f\in L^1_{loc} (\mathbb R^n)$ and consider the measure $$\label{e:densita} \mu (A):= \int_A f\, d\lambda\, .$$ Then the $n$--dimensional density of $\mu$ exists at $\lambda$--a.e. $x\in \mathbb R^n$ and coincides with $f(x)$.
A similar result in the opposite direction holds and is a particular case of a more general result on the Differentiation of measures:
Theorem 4 Let $\mu$ be a locally finite Radon measure on $\mathbb R^n$. If the $n$-dimensional density $\theta^n (\mu, x)$ exists for $\mu$-a.e. $x$, then the measure $\mu$ is given by the formula \ref{e:densita} where $f = \theta^n (\mu, \cdot)$.
The latter theorem can be generalized to Hausdorff $\alpha$-dimensional measures $\mathcal{H}^\alpha$ (cp. with Theorem 6.9 of [Ma]).
Theorem 5 Let $\mu$ be a locally finite Radon measure on $\mathbb R^n$. If the upper $\alpha$-dimensional density exists and it is positive and finite at $\mu$-a.e. $x\in \mathbb R^n$, then there is a Borel function $f$ and a Borel set $E$ with locally finite $\alpha$-dimensional Hausdorff measure such that $\mu (A) = \int_{A\cap E} f\, d\mathcal{H}^\alpha\, .$
A generalization of Theorem 3 is also possible, but much more subtle (see below).
### Lower-dimensional densities of a set
Assume $E\subset \mathbb R^n$ is a Borel set with finite $\alpha$-dimensional Hausdorff measure. The $\alpha$-dimensional upper and lower densities $\theta^{\alpha, *} (E, x)$ and $\theta^\alpha_* (E,x)$ of $E$ at $x$ are defined as $\theta^{\alpha,*} (E, x) := \limsup_{r\downarrow 0} \frac{\mathcal{H}^\alpha (E\cap B_r (x))}{\omega_\alpha r^\alpha} \qquad \mbox{and}\qquad \theta^\alpha_* (E, x) =\liminf_{r\downarrow 0} \frac{\mathcal{H}^\alpha (E \cap B_r (x))}{\omega_\alpha r^\alpha}\,$ (cp. with Definition 6.1 of [Ma]) They correspond, therefore, to the $\alpha$-dimensional (upper and lower) densities of the Radon measure $\mu$ given by $\mu (A) := \mathcal{H}^\alpha (A\cap E)\, \qquad\mbox{for any Borel set } A\subset\mathbb R\, .$ The following is a classical theorem in Geometric measure theory (cp. with Theorem 6.2 of [Ma]):
Theorem 6 If $E\subset \mathbb R^n$ is a Borel set with finite $\alpha$-dimensional Hausdorff measure, then
• $\theta^{\alpha,*} (E, x) =0$ for $\mathcal{H}^\alpha$-a.e. $x\not\in E$.
• $1\geq \theta^\alpha_* (E,x) \geq 2^{-\alpha}$ for $\mathcal{H}^\alpha$-a.e. $x\in E$.
#### Besicovitch-Preiss theorem and rectifiability
However the existence of the density fails in general: as a consequence of Marstrand's Theorem 2 the existence of a nontrivial $\alpha$-dimensional density implies that $\alpha$ is an integer. But even in the case when $\alpha$ is an integer, it was discovered by Besicovitch that the density does not necessarily exist. Indeed the following generalization of Besicovitch's theorem was achieved by Preiss in the mid eighties (see Chapters 6,7,8 and 9 of [De] for an exposition of Preiss' proof):
Theorem 7 Let $E\subset \mathbb R^n$ be a Borel set with positive and finite $k$-dimensional Hausdorff measure, where $k\in \mathbb N$. The $k$-dimensional density exists at $\mathcal{H}^k$-a.e. $x\in E$ if and only if the set $E$ is rectifiable, i.e. if there are countably many $C^1$ $k$-dimensional submanifolds of $\mathbb R^n$ which cover $\mathcal{H}^k$-almost all $E$.
For non-rectifiable sets $E$ the lower-dimensional density might display a variety of different behavior. Besicovitch proved that for $1$-dimensional non-rectifiable sets the lower dimensional density cannot be larger than $\frac{3}{4}$ and advanced the following long-standing conjecture (cp. with Conjecture 10.5 of [De]).
Conjecture 8 Let $E\subset \mathbb R^2$ be a Borel set with positive and finite $1$-dimensional Hausdorff measure. If $\theta^1_* (E, x)>\frac{1}{2}$ for $\mathcal{H}^1$-a.e. $x\in E$, then the set $E$ is rectifiable.
Besicovitch's $\frac{3}{4}$ treshold has been improved by Preiss and Tiser in [PT].
### Besicovitch-Marstrand-Preiss Theorem
Combining the various theorems exposed so far we reach the following characterization of measures $\mu$ for which densities exist and are nontrivial almost everywhere.
Theorem 9 Let $\mu$ be a locally finite Radon measure and $\alpha$ a nonnegative real number. Then $\theta^\alpha (\mu, x)$ exists, it is finite and positive at $\mu$-a.e. $x\in \mathbb R^n$ if and only $\alpha$ is an integer $k$ and there are a rectifiable $k$-dimensional Borel set $E$ and a Borel function $f: E\to ]0, \infty[$ such that $\mu (A) = \int_{A\cap E} f\, d\mathcal{H}^k\qquad \mbox{for any Borel } A\subset \mathbb R^n\, .$ Moreover, in this case $\theta^k (E,x)=f(x)$ for $\mathcal{H}^k$-a.e. $x\in E$ and $\theta^k (E, x) =0$ for $\mathcal{H}^k$-a.e. $x\not\in E$.
The definition of $\alpha$-dimensional density of a Radon measure can be generalized to metric spaces. In general, however, very little is known outside of the Euclidean setting (cp. with Section 10.0.2 of [De]).
[AFP] L. Ambrosio, N. Fusco, D. Pallara, "Functions of bounded variations and free discontinuity problems". Oxford Mathematical Monographs. The Clarendon Press, Oxford University Press, New York, 2000. MR1857292Zbl 0957.49001 [Be] S. Besicovitch, "On the fundamental geometrical properties of linearly measurable plane sets of points II", Math. Ann., 115 (1938), pp. 296–329. Zbl 64.0193.01 [De] C. De Lellis, "Rectifiable sets, densities and tangent measures" Zurich Lectures in Advanced Mathematics. European Mathematical Society (EMS), Zürich, 2008. MR2388959 Zbl 1183.28006 [EG] L.C. Evans, R.F. Gariepy, "Measure theory and fine properties of functions" Studies in Advanced Mathematics. CRC Press, Boca Raton, FL, 1992. MR1158660 Zbl 0804.2800 [Fe] H. Federer, "Geometric measure theory". Volume 153 of Die Grundlehren der mathematischen Wissenschaften. Springer-Verlag New York Inc., New York, 1969. MR0257325 Zbl 0874.49001 [Mar] J. M. Marstrand, "The (φ, s) regular subset of n space". Trans. Amer. Math. Soc., 113 (1964), pp. 369–392. MR0166336 Zbl 0144.04902 [Ma] P. Mattila, "Geometry of sets and measures in euclidean spaces". Cambridge Studies in Advanced Mathematics, 44. Cambridge University Press, Cambridge, 1995. MR1333890 Zbl 0911.28005 [Pr] D. Preiss, "Geometry of measures in $\mathbb R^n$ : distribution, rectifiability, and densities". Ann. of Math., 125 (1987), pp. 537–643. MR0890162 Zbl 0627.28008 [PT] D. Preiss, J. Tiser, "On Besicovitch’s $\frac{1}{2}$-problem. J. London Math. Soc. (2), 45 (1992), pp. 279–287. MR1171555 Zbl 0762.28003 [Ta] F.D. Tall, "The density topology" Pacific J. Math , 62 (1976) pp. 275–284 MR0419709 Zbl 0305.54039 | 2022-05-17T21:57:27 | {
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https://iitutor.com/halving-the-interval-method-with-relative-accuracy/ | # Halving the Interval Method with Relative Accuracy
## Find the smallest positive root of $x^3 – 2x + 1 = 0$ with relative accuracy of $0.5 \%$ by finding good starting value from its graph.
According to the graph above, there are three roots, one negative and two positive.
Let’s give $f(x) = x^3-2x+1$, then $f(1) = 1^3 – 2 \times 1 + 1 = 0$.
Therefore $x = 1$ is the largest positive root as well as shown in the graph.
So the smallest positive root lies in between $x=0$ and $x=1$.
Though this interval is good enough to start applying “halving the interval method”, it could be better to make the working more effective and efficient, say $0.5 < x < 0.9$.
To find the rate of accuracy by taking the relative accuracy method, we need two values, the error and the “believed” true value.
In this case, the true value is $\displaystyle \frac{0.5+0.9}{2} = 0.7$ as we believe $0.7$ is more accurate than both $0.5$ and $0.9$ and the error is $\left| 0.5 – 0.7 \right| = 0.2$ or $\left| 0.7 – 0.9 \right| = 0.2$.
Thus the relative accuracy is $\displaystyle \frac{ \left| 0.5-0.7 \right| }{0.7} \times 100 \% =28.57 \%$.
Let’s start to find the smallest positive root!
$f(0.5) = 0.125 > 0$
$f(0.9) = -0.057 < 0$
The interval is $0.5 < x < 0.9$ as the closer value of $x$ lies between opposite sign of its function values.
The midpoint is $\displaystyle \frac{0.5+0.9}{2} = 0.7$.
Its relative accuracy is $\displaystyle \frac{|0.5 – 0.7|}{0.7} \times 100 \% = 28.57 \%$.
$f(0.7) = -0.057 < 0$
The refined interval is $0.5 < x < 0.7$.
The midpoint is $\displaystyle \frac{0.5+0.7}{2} = 0.6$.
Its relative accuracy is $\displaystyle \frac{|0.5 – 0.6|}{0.6} \times 100 \% = 16.67 \%$.
$f(0.6) = 0.016 > 0$
The refined interval is $0.6 < x < 0.7$.
The midpoint is $\displaystyle \frac{0.6+0.7}{2} = 0.65$.
Its relative accuracy is $\displaystyle \frac{|0.6 – 0.65|}{0.65} \times 100 \% = 7.69 \%$.
$f(0.65) = -0.0253 \cdots < 0$
The refined interval is $0.6 < x < 0.65$.
The midpoint is $\displaystyle \frac{0.6+0.65}{2} = 0.625$.
Its relative accuracy is $\displaystyle \frac{|0.6 – 0.625|}{0.625} \times 100 \% = 4 \%$.
$f(0.625) = -0.0058 \cdots < 0$
The refined interval is $0.6 < x < 0.625$.
The midpoint is $\displaystyle \frac{0.6+0.625}{2} = 0.6125$.
Its relative accuracy is $\displaystyle \frac{|0.6 – 0.6125|}{0.6125} \times 100 \% = 2.04 \%$.
$f(0.6125) = 0.0047 \cdots > 0$
The refined interval is $0.6125 < x < 0.625$.
The midpoint is $\displaystyle \frac{0.6125+0.625}{2} = 0.61875$.
Its relative accuracy is $\displaystyle \frac{|0.6125 – 0.61875|}{0.61875} \times 100 \% = 1.01 \%$.
$f(0.61875) = -0.0006 \cdots < 0$
The refined interval is $0.6125 < x < 0.61875$.
The midpoint is $\displaystyle \frac{0.6125+0.61875}{2} = 0.615625$.
Its relative accuracy is $\displaystyle \frac{|0.6125 – 0.615625|}{0.615625} \times 100 \% = 0.51 \% \$So close!
$f(0.615625) = 0.002 \cdots > 0$
The refined interval is $0.615625< x < 0.61875$.
The midpoint is $\displaystyle \frac{0.615625+0.61875}{2} = 0.6171875$.
Its relative accuracy is $\displaystyle \frac{|0.615625- 0.6171875|}{0.6171875} \times 100 \% = \color{green}{0.25 \%}$.
As this $\color{green}{0.25 \%}$ meets the minimum relative accuracy level of $0.5 \%$, the smallest positive root is $\color{green}{0.6171875}$! | 2021-09-17T23:15:35 | {
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http://mathhelpforum.com/statistics/126490-probability.html | # Math Help - Probability
1. ## Probability
A factory produces items in boxes of 2. Over the long run:
92% of boxes contain 0 defective items;
5% of boxes contain 1 defective item;
3% of boxes contain 2 defectives item.
A box is picked at random from production, then an item is picked at random from the box. Given that the item is defective, what is the chance the second item in the box is defective.
2. Originally Posted by Len
A factory produces items in boxes of 2. Over the long run:
92% of boxes contain 0 defective items;
5% of boxes contain 1 defective item;
3% of boxes contain 2 defectives item.
A box is picked at random from production, then an item is picked at random from the box. Given that the item is defective, what is the chance the second item in the box is defective.
Of every 100 boxes, 8 contain at least 1 defective.
The defective item came from one of these and each box contains only two items.
There are 6 defectives in the 3 boxes containing 2 defectives each.
there are 5 defectives in the 5 boxes containing 1 defective each.
Hence there is a 6 out of 11 chance of having picked a defective item from a box with 2 defectives.
The next one chosen from that box will be also be defective.
Therefore, there is a 6 in 11 chance the item was chosen from one of the boxes with 2 defective items.
3. Hello, Len!
A factory produces items in boxes of 2.
Over the long run:
. . 92% of boxes contain 0 defective items;
. . 5% of boxes contain 1 defective item;
. . 3% of boxes contain 2 defectives item.
A box is picked at random from production, then an item is picked at random from the box.
Given that the item is defective, what is the chance the second item in the box is defective?
I changed the language for my convenience . . .
The factory produces three types of boxes:
. . Type A: contains no defective items. . $P(A) \,=\,0.92$
. . Type B: contains one defective item. . $P(B) \,=\,0.05$
. . Type C: contains two defective items. . $P(C) \,=\,0.03$
A box is chosen at random and one item is sampled from that box.
Given that the sample is defective, what the probability that the box is of Type C?
This is Conditional Probability.
Bayes' Theorem: . $P(\text{Type C }|\text{ 1 d{e}f}) \;=\; \frac{P(\text{Type C }\wedge\text{ 1 d{e}f})}{P(\text{ 1 d{e}f})} \$ .[1]
Numerator: . $P(C) \,=\,0.03$
. . From a Type C box, the probability of drawing a defective item is 100% = 1.
. . Hence: . $P(\text{Type C }\wedge\text{ 1 d{e}f}) \:=\:(0.03)(1) \;=\;0.03$ .[2]
Denominator: . $P(B) \,=\,0.05$
. . From a Type B box, the probability of drawing a defective item is $0.5$
. . Hence: . $P(\text{Type B }\wedge\text{ 1 d{e}f}) \:=\:(0.05)(0.5) \:=\:0.025$
. . From a Type A box, it is impossible to draw a defective item.
. . We already know that: . $P(\text{Type C }\wedge\text{ 1 d{e}f}) \:=\:0.03$
Hence: . $P(\text{ 1 d{e}f}) \;=\; 0.025 +0.03 \:=\: 0.055$ .[3]
Substitute [2] and [3] into [1]: . $P(\text{Type C }|\text{1 d{e}f}) \;=\;\frac{0.03}{0.55} \;=\;\frac{6}{11} \;\approx\;54.5\%$
4. I do appreciate the explanations, one is intuitive and the other with definition. Thanks so much | 2014-10-23T09:51:27 | {
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http://math.stackexchange.com/questions/66062/is-my-proof-that-an-1-a-1n-correct/66065 | # Is my proof that $(A^n)^{-1} = (A^{-1})^n$ correct?
I am still learning Linear Algebra at it's basic levels, and I encountered a theorem about invertible matrices that stated that:
If $A$ is an invertible matrix, then for $n=0,1,2,3,..$. $A^n$ is invertible and $(A^n)^{-1} = (A^{-1})^n$.
Now, in attempting to write my proof, I proceeded this way (note that it's not complete):
$$A^n(A^{-1})^n=\prod_{i=1}^nA\prod_{i=1}^nA^{-1}=\prod_{i=1}^n(AA^{-1})=\prod_{i=1}^nI=I$$
Is this line of thinking correct? Well, am just returning to math after a long time of little practice, so I could be wrong.
Based on my comment to Dimitri's answer, would my use of this argument improve my proof?
$$\prod_{i=1}^{n-1}A.(AA^{-1}).\prod_{i=1}^{n-1}=\prod_{i=1}^{n-1}A.(I).\prod_{i=1}^{n-1}=...=A.(AIA^{-1}).A^{-1}=AIA^{-1}=AA^{-1}=I$$
After checking the comments, it seems this last argument gives me a correct proof eventually, and I now see that the problem with my original approach was making the argument that:
$$\prod_{i=1}^nA\prod_{i=1}^nA^{-1}=\prod_{i=1}^n(AA^{-1})$$
Which is not necessarily correct, but like @Srivatsan demonstrates, that approach is not at all wrong since :
Notice that $A$ and $A^{−1}$ commute, so this justifies your proof now
Thanks to everyone for guidance, now I see why collaboration is going to make me love math :D
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I think this is wrong because you are using commutativity. Matrix multiplication is not commutative. However, I think you can adapt your proof so it works. – sxd Sep 20 '11 at 13:19
It's correct if before writing the equality, you write $AA^{-1}=I=A^{-1}A$. – Davide Giraudo Sep 20 '11 at 13:23
@mcnemesis Actually, that statement is correct, but you have not justified it properly before using it. What is wrong is this: $A^n B^n = (AB)^n$; this is not true for a general pair of matrices. But since $A$ and $A^{-1}$ commute, $A^n (A^{-1})^n = (A A^{-1})^n$ is perfectly fine, provided of course you justify it somewhere. (Check out my answer as well.) – Srivatsan Sep 20 '11 at 14:10
@mcnemesis Just one final point. By "not wrong", I meant that the logic in sound. But there are two things you should understand clearly. (1.) As long as you don't say why you are allowed to make that switch, the proof is still incomplete, technically speaking. (2.) What level of rigor and detail is necessary completely depends on the audience. In a paper, the author may be justified in omitting such "small" details. (contd) – Srivatsan Sep 20 '11 at 14:21
(contd) On the other hand, if a text-book carelessly misses this step in a derivation, then either it might mislead students into thinking that you can always switch matrices (i.e., matrix product is commutative, which it isn't), or if the student is sufficiently attentive, it might confuse her ("Why did the author do this? I don't get this step at all."). Finally, in an exam, if you are skipping such steps, you will most likely lose points even if you think your solution is correct. The point of an exam is to show that you understood the concepts. The moral? "When in doubt, show all steps!" – Srivatsan Sep 20 '11 at 14:24
As Dimitri points out, your proof is incomplete. You can make it work in two ways:
1. You can group the middle $AA^{-1}$ (remember that matrix product is associative). Noting that this equals $I$, the product simplifies to $A^{n-1} (A^{-1})^{n-1}$. You can then use induction to argue that $A^n (A^{-1})^{n}$ is $I$ for all $n$.
2. This is slightly more general variant of the above trick. Suppose $A$ and $B$ are commuting matrices (i.e., $AB = BA$), then you can indeed use $$A^n B^n = (AB)^n$$ guilt-free! (Notice that $A$ and $A^{-1}$ commute, so this justifies your proof now so you can justify the proof this way as well. Keep in mind that some justification is necessary, otherwise the proof is wrong or incomplete.) The proof of this fact also uses similar ideas; see if you can figure it out yourself.
In fact, if you have an arbitrary product of matrices consisting of $m$ $A$'s and $n$ $B$'s (and no other matrices), then you can show that this product equals $A^m B^n$. For example, if $A$ and $B$ commute, then $$B^5ABA^2 B^{3} = A^{1+2} B^{5+1+3} = A^3 B^9.$$
A method by induction:
I leave you to verify that the result is true for the base case $n=0$. For the induction step, assume that $$(A^{n-1})^{-1} = (A^{-1})^{n-1} .$$ We must now prove the claim for $n$. This follows from the chain of equalities: $$\begin{eqnarray*} (A^n)^{-1} &=& (A \cdot A^{n-1})^{-1} \\ &\stackrel{({a})}{=}& (A^{n-1})^{-1} \cdot A^{-1} \\ &\stackrel{({b})}{=}& (A^{-1})^{n-1} \cdot A^{-1} \\ &=& (A^{-1})^{n}. \end{eqnarray*}$$ Be sure to justify each step, particularly the ones marked $(a)$ and $(b)$.
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thanks. The Induction actually probably makes the proof more powerful and general. And for the "guilt-free!" – nemesisfixx Sep 20 '11 at 14:22
HINT, observe that: $$A^n(A^{-1})^n = \underbrace{A\ldots AA}_{\textrm{n times A}} A^{-1}A^{-1}\ldots A^{-1} = A\ldots A(AA^{-1})A^{-1}\ldots A^{-1} = A\ldots AIA^{-1}\ldots A^{-1}$$
Sorry for the dots, but I didn't find a better way to point the idea out!
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This is the idea I had in mind, but thought I could condense it by using the $\prod$ operator. And then also, I was assuming that I could expand if necessary to something like: $\prod_{i=1}^{n-1}A.(AA^{-1}).\prod_{i=1}^{n-1}=\prod_{i=1}^{n-1}A.(I).\prod_{i=1}^{n-1}=...=A.(AIA^{-1}).A^{-1}=AIA^{-1}=AA^{-1}=I$. – nemesisfixx Sep 20 '11 at 13:45
That method is indeed correct, here you avoid the error you made in your original proof. The problem with your original proof was that you wrote that $\Pi^n_{i=1} A \Pi^n_{i=1} A^{-1} = \Pi^n_i AA^{-1}$, which uses commutativity. – sxd Sep 20 '11 at 13:54
Thanks, this clarifies the problem, and your approach too makes sense for me. – nemesisfixx Sep 20 '11 at 13:58
Check srivatsan his answer to see how this idea can be used in a full proof for this theorem. This is just the general idea for the proof. – sxd Sep 20 '11 at 14:08
Your proof isn't really right. What I mean is, for general matrices $A$ and $B$ the statement $$A^n B^n = \prod A \prod B = \prod (AB)$$ may be wrong because $A$ and $B$ need not commute. I'd try doing this by induction instead.
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What I know is that if $A$ and $B$ are invertible then $AB$ is so. Then $A^n$ is invertible , I am agree with (mt_ ) by induction you can prove it: we know that
$n=1$ is correct $A^{-1}A=I$
if $n=k$ is correct ; $(A^{k})^{-1}A^k=I$ then we should show $(A^{k+1})^{-1}(A^{})^{k+1}=I$
we have $(A^{k+1})^{-1}A^{k+1}=(A^{k}A)^{-1}A^{k+1}=A^{-1}(A^k)^{-1}A^{k}A=A^{-1}(A)=I$
we know that $(AB)^{-1}=B^{-1}A^{-1}$
- | 2016-05-26T04:52:25 | {
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http://math.stackexchange.com/questions/12276/is-there-any-connection-between-greens-theorem-and-the-cauchy-riemann-equations | # Is there any connection between Green's Theorem and the Cauchy-Riemann equations?
Green's Theorem has the form: $$\oint P(x,y)dx = - \iint \frac{\partial P}{\partial x}dxdy , \oint Q(x,y)dy = \iint \frac{\partial Q}{\partial y}dxdy$$ The Cauchy-Riemann equations have the following form:(Assuming $z = P(x,y) + iQ(x,y)$) $$\frac{\partial P}{\partial x} = \frac{\partial Q}{\partial y}, \frac{\partial P}{\partial y} = - \frac{\partial Q}{\partial x}$$
Is there any connection between this two equations?
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+1. This is a good question. You should read the book by Churchill and Brown on Complex Variables and Applications. – user17762 Nov 29 '10 at 2:17
This is a great question. Keep finding connections and analogies between things you've already learned and you'll be well on your way to mathematical research. – Pete L. Clark Nov 29 '10 at 3:45
Absolutely yes!
In Section 7 of these notes on Green's Theorem, I explain how Green's Theorem plus the Cauchy-Riemann equations immediately yields the Cauchy Integral Theorem.
Many serious students of mathematics realize this on their own at some point, but it is surprising how few standard texts make this connection. In (especially American) undergraduate texts on Subject X, there is a distressing tendency to politely ignore the existence of Subject Y, even when any student of Subject X will almost surely have already have studied / be concurrently studying / soon be studying Subject Y.
-
I think the X and Y thing is not just an American thing. Btw... as a contrast - suddenly when you come to graduate school you are supposed to know this, and that etc. :) – AD. Nov 29 '10 at 14:05
@AD.: Well, I have a lot of experience with "the" American university system, having been involved with it for my entire adult life. I don't want to speak for the rest of the world -- what little I know there comes from a few brief visits and anecdotes from colleagues. I have heard tell, though, that there is a bit more unity in the European (let's say French, in particular) education system. – Pete L. Clark Nov 29 '10 at 19:23
If you consider $P(x,y)$ and $Q(x,y)$ as the part real and imaginary of an analitic function you can get the connection.
Then $P(x,y)$ and $Q(x,y$) verify the Cauchy-Riemann equations, thus :
$$\frac{\partial P}{\partial x} = \frac{\partial Q}{\partial y} , \frac{\partial P}{\partial x} = -\frac{\partial Q}{\partial y}$$
Depending of oriented get that the integral is Zero.
This satisfies the Cauchy's integral theorem that an analytic function on a closed curve is zero.
Edit:
You can see it here, where the proof of Cauchy's integral theorem uses Green's Theorem .
A little deeper you can see, Complex Analysis by Lars Ahlfors, section 4.6 page 144.
- | 2016-05-06T01:46:46 | {
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https://yhiqu.dkp-bad-kreuznach.de/en/calculating-percent-change-between-two-numbers.html | To calculate the percentage change between two numbers we typically write a formula like the following: = (C2-B2)/B2. Whenever I am creating this formula I always think, "new minus old, divided by old". Or: = (new value - old value) / old value. If we want to handle/prevent any divide by zeros (#DIV/0!), then we can wrap the formula in. .
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Percentage difference equals the absolute value of the change in value, divided by the average of the 2 numbers, all multiplied by 100. We then append the percent sign, %, to designate the % difference. Percentage Difference = | Δ V | [ Σ V 2] × 100 = | V 1 − V 2 | [ ( V 1 + V 2) 2] × 100 For example, how to calculate the percentage difference:. How to calculate percent increase between two numbers? Our online calculator will calculate percent increase, and it will also calculate percent decrease, and percent difference as well. For example, $5 about 25 years ago will cost you$21 now. Use our online percent growth calculator below in fill in 5 in first box and 21 in second, the answer will shock you, it's 320%!. A backlit color display helps differentiate between values on graphs for easy viewing, and the preloaded apps and images walk students through a variety of ...Texas Instruments TI-84 Plus CE Color Graphing Calculator - Black. 6 product ratings. About this product. Brand new. $148.88. Used - Like New.$96.10. Pre-owned. $100.33. We can quickly calculate the percentage change in our excel sheet across two columns using the steps below: We click on Cell F4 and enter the formula below. = (D4-E4)/D4. Figure 2 – Drawing percentage in excel. We will copy down the formula using the fill handle tool (the small box at the lower right corner of the cell) into the number of. First Step: find the difference between two percentages, in this case, it's 15% - 5% = 10%. Second: Take 10 percent, and divide by 2nd percentage: 10/5 = 2. Now multiply this number by. teenie asshole cumshots iphone 12 pro max wallpaper 4k ## superbad suck dick at Calculating percent change between two numbers in Excel. Examples for you to show how simple it is! | Online Office Tools Tutorials Library. Thus to calculate the percentage increase we will follow two steps: Step-1: Calculate the difference i.e. increase between the two numbers. i.e. Increase = New Number - Original Number. Step-2: Divide the increased value by the original. Trend analysis is a common task in financial accounting. Accountants compare two time periods by using financial information. The difference typically shows a percentage increase or decrease in the information. Accountants use the data to determine if the company is growing or contracting. In many cases, accounting. Find Percentage Difference: Finally, you need to use the following percent difference equation to calculate percentage difference between two numbers. $$\text {Percentage Difference} = \frac {Difference} {Average} * 100$$. The free online percent difference calculator also makes use of all these formulas to generate precise outputs. big wheel dolly dewalt 5amp batteries ## milena angel sex Calculating percentage change between 2 numbers. How to calculate percent increase between two numbers? Our online calculator will calculate percent increase, and it will also calculate percent decrease, and percent difference as well.. First: work out the difference (decrease) between the two numbers you are comparing. Decrease = Original Number - New Number. Then: divide the decrease by the original number and. The formula for % change between two numbers. The simple formula is shown below. = (A2/A1)-1. In this example, cell A2 contains the new number and cell A1 contains the original number. The formula will return a value that represents the percentage change between the two numbers. Negative changes will show up as negative numbers. return gifts under 5 ## the railway man bhopal gas tragedy Oct 13, 2019 · There are two methods of finding the percent of change between two numbers. The first is to find the ratio of the amount of change to the original amount. If the new number is greater than the old number, then that ratio is the percent of increase, which will be a positive. If the new number is less than the old number, then that ratio is the percent of decrease, which will be a negative.. Let V 1 = 3.50 and V 2 = 2.625 and plug numbers into our percentage change formula ( V 2 − V 1) | V 1 | × 100 = ( 2.625 − 3.50) | 3.50 | × 100 = − 0.875 3.50 × 100 = − 0.25 × 100 = − 25 % change Saying a -25% change is equivalent to stating a 25% decrease. Note that if we let V 1 = 2.625 and V 2 = 3.50 we would get a 33.3333% increase. Step 1: Find the difference between the two numbers, i.e a - b. Step 2: Then, find the average of two numbers, i.e (a+b)/2. Step 3: Take the ratio of the difference and the average. Step 4: Multiply the fraction obtained by 100 and simplify your answer. Let us see how to calculate the difference in percentage using an example. We will have this data for two years- 2019 and 2020. To calculate the percentage change between two numbers we will use the simple formula: (Value in the year 2020 – value in the year 2019)/value in the year 2019 So, the formula in cell D3 will be: = (C3-B3)/B3 We will drag the formula to the cell D5 and will have the following results:. Calculating the percent change between two given quantities is quite an easy process. When the initial or old value and final or new values of a quantity are known, percent change formula. The standard way of computing the percent change between two values is to get the difference of the first and second numbers. Afterward, divide the result by the first number and multiply it by 100. Here's the exact formula for computing percent change:. The formula for % change between two numbers. The simple formula is shown below. = (A2/A1)-1. In this example, cell A2 contains the new number and cell A1 contains the original number. The formula will return a. clearkey license server ## guild drill To calculate percentage decrease: First: work out the difference (decrease) between the two numbers you are comparing. Decrease = Original Number - New Number. Then: divide the decrease by the original number and multiply the answer by 100. % Decrease = Decrease ÷ Original Number × 100. If your answer is a negative number, then this is a .... Percentage Calculator. Use this versatile percentage calculator to easily find the percentage difference between two numbers, to calculate percent change (percentage increase, percentage decrease from a baseline), to find out what % is a given number from any other given number, as well as how much is x percent of y. The percent change in values between one period and another period is calculated as: Percent change = (Value2 - Value1) / Value1 * 100 For example, suppose a company makes 50 sales one month, then makes 56 sales the next month. We can use the following formula to calculate the percent change in sales from one month to the next:. Initial value = 60. Final value = 75 75 - 60 = 15. 3. Divide the difference by the initial value. In order to compare the values over time, we need to divide the difference by the initial value. Follow the example below to get an idea of how this. fastener clearing house ## young teens first black anal Work out the difference (increase) between the two numbers you are comparing. Percentage 'change' and percentage 'difference' are two different things. Percentage change = (fv − iv) ÷ iv × 100. This formula represents percentage change, for example if you are are comparing values of the same statistic over time (e.g. Enter starting value .... atoms questions and answers ## supply and demand indicator for mt4 free download The formula for % change between two numbers. The simple formula is shown below. = (A2/A1)-1. In this example, cell A2 contains the new number and cell A1 contains the original number. The formula will return a. First, subtract to find the amount of change: 150 - 125 = 25. The decrease is 25. Next, divide the amount of change by the original amount: 25 ÷ 150 = 0.167 Now, to change the decimal to a percent, multiply the number by 100: 0.167 x 100 = 16.7 The answer is 16.7%. So that's the percent of change, a decrease of 16.7% in body weight. The formula for calculating a percentage is: (number / base number) x 100. Related: Functional skills: definition, importance and examples. How to calculate percentage change. Another common use of percentages is finding the percentage change between two figures. This can be useful for comparing figures, such as profit or price increases. To calculate a percentage change you need to calculate the difference, divide the change by the original value and then multiply that amount by 100. The result of this calculation can indicate. hola vpn chrome extension review list of day 1 cpt universities script fu high pass filter kesq news ## read berserk It is very simple, easy and quick to use! Step 1: Simply fill in the initial and new values in the provided boxes. Step 2: Hit the “calculate” button Step 3: You’ll get your percentage change in a twinkle of an eye! Percentage Change Formula (New Value - Initial Value)/ (Initial Value) * 100 = percentage increase or decrease. work out the difference between the two numbers being compared. divide the increase by the original number and multiply the answer by 100. in summary: percentage increase = increase ÷ original number × 100. How to use the percentage change calculator for free? How to use our FREE Percent Change Calculator It is very simple, easy and quick to use!. shelf cube storage ## girls younger than video I also explained two formulas we can use to calculate percentage change. Formula #1 = (new value – old value) / old value Formula #2 = (new value / old value) – 1 Both of these formulas will produce the same result when. The standard way of computing the percent change between two values is to get the difference of the first and second numbers. Afterward, divide the result by the first number and multiply it by 100. Here's the exact formula for computing percent change:. Percentage Difference calculator is a free online tool to find the percent difference between two numbers. Percent Difference calculator uses this formula: ( (y2 - y1) / y1)*100 = your percentage change. How to use? 1. Enter the two numbers into the inputs. 2. Click the calculate button 3. This tool will show the result next to the button. To calculate the percentage change between two numbers we typically write a formula like the following: = (C2-B2)/B2. Whenever I am creating this formula I always think, "new minus old, divided by old". Or: = (new value - old value) / old value. If we want to handle/prevent any divide by zeros (#DIV/0!), then we can wrap the formula in. Answer: To calculate the percentage difference of any two numbers, find the absolute value of the ratio of their difference and their average and multiply that value by 100. Let us look at an example to understand this better. Explanation: The percent difference formula or the percent difference equation of two numbers a and b is:. magma imei tool registration ## google colab pip install Thus to calculate the percentage increase we will follow two steps: Step-1: Calculate the difference i.e. increase between the two numbers. i.e. Increase = New Number - Original Number. Step-2: Divide the increased value by the original. Sep 19, 2011 · A percentage change can only be found between TWO numbers. If you are given more than two, the question may be asking for either- the increase from the first number to the rest of the numbers- the increase between successive pairs of number (for example a to b, b to c)To find the difference between one number and the next, subtract the first number from the second, then divide by the first .... . First, calculate the difference (decrease) between the two compared numbers. Then divide the reduction by the original number and multiply the result by 100. Percentage change formula Formula of Percentage change = [ (B – A)/A] * 100 % change = [Final value – Initial value]/Initial value * 100 B = Final value A = Initial value. The annual inflation rate for transport was 9.3% in October 2022, down for a fourth consecutive month from a peak of 15.2% in June 2022. Prices were unchanged between September and October 2022 but had increased by 1.5% in 2021. This resulted in a 0.18 percentage point downward contribution to the change in the annual inflation rate. linux cifs mount logs ## suck and swallow my dick You would divide the new data point ($225,000) by the original one ($150,000), multiplying the result by 100 as follows to get a year 2 index value of 167. (Year 2 sales of$250,000 / Base year sales of $150,000) * 100 = 167 Each new year of data is subsequently normalized against the base year of$150,000 in the same fashion.
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Calculation of percentage change in no of employees can be done as follows- = (30-25)/30*100% = 16.67% or 16.67% decrease in no. of employees Use. If you have already calculated the percentage change, go to step 4. Subtract one from the result of the division. Multiply this new number by 100. You now have the percentage. First Step: find the difference between two numbers, in this case, it's 10 - 2 = 8. Second Step: Take the difference, 8, and divide by the original number: 8/2 = 4. Last, multiply the number above by 100: 4*100 = 400%. You're done! You calculated difference of a number in percent, and the answer is a percentage increase of 400%. More Calculators. Program to find the Discount Percentage. 6. Loss when two items are sold at same price and same percentage profit/loss. 7. Find cost price from given selling price and profit or loss percentage. 8. Find Selling Price from given Profit Percentage and Cost. 9. Find the percentage change in the area of a Rectangle.
3. Compute Percentage Variance between Two Negative Numbers. You can use the formula I showed in the first method to calculate the percentage variance between two. Let V 1 = 3.50 and V 2 = 2.625 and plug numbers into our percentage change formula ( V 2 − V 1) | V 1 | × 100 = ( 2.625 − 3.50) | 3.50 | × 100 = − 0.875 3.50 × 100 = − 0.25 × 100 = − 25 % change Saying a -25% change is equivalent to stating a 25% decrease. Note that if we let V 1 = 2.625 and V 2 = 3.50 we would get a 33.3333% increase.
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I have a list of prices where I am trying to calculate the change in percentage of each number. I calculated the differences with prices = [30.4, 32.5, 31.7, 31.2, 32.7, 34.1, 35.8, 37.8, 36.3... Stack Overflow. ... Calculating change in percentage between two numbers (Python) Ask Question Asked 10 years, 1 month ago. Modified 2 years, 7 months. Step 1: Find the difference between the two numbers, i.e a - b. Step 2: Then, find the average of two numbers, i.e (a+b)/2. Step 3: Take the ratio of the difference and the average. Step 4: Multiply the fraction obtained by 100 and.
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Calculating the percent change between two given quantities is quite an easy process. When the initial or old value and final or new values of a quantity are known, percent change formula.
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The formula for % change between two numbers. The simple formula is shown below. = (A2/A1)-1. In this example, cell A2 contains the new number and cell A1 contains the original number. The formula will return a.
Calculating percentage change between 2 numbers. .
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Thus to calculate the percentage increase we will follow two steps: Step-1: Calculate the difference i.e. increase between the two numbers. i.e. Increase = New Number - Original Number. Step-2: Divide the increased value by the original. In computing the growth or decline of a variable, you can quickly use this percentage change calculator to find the percentage increase or decrease in the value of two numbers. How to use our FREE Percent Change Calculator It is very simple, easy and quick to use! Step 1: Simply fill in the initial and new values in the provided boxes. Calculating percentage change between 2 numbers. Percentage Difference Formula: Percentage difference equals the absolute value of the change in value, divided by the average of the 2 numbers, all multiplied by 100. We then append the percent sign, %, to designate the % difference..
Trend analysis is a common task in financial accounting. Accountants compare two time periods by using financial information. The difference typically shows a percentage increase or decrease in the information. Accountants use the data to determine if the company is growing or contracting. In many cases, accounting. When we look at the formula through that angle, we realize that a relative growth between two values of different signs is only affected by the ratio between the two values $\displaystyle. The solution, which is the same as that in my research, is to take the difference between the two numbers and use that as the basis. So difference of 1 and -1 is 2. So 1 contributes (1/2)*100 = 50%. What about -1? Use the absolute function. ( abs (-1) / 2 ) * 100 = 50%. The difference method works fine if you have only two numbers. We can quickly calculate the percentage change in our excel sheet across two columns using the steps below: We click on Cell F4 and enter the formula below. = (D4-E4)/D4. Figure 2 - Drawing percentage in excel. We will copy down the formula using the fill handle tool (the small box at the lower right corner of the cell) into the number of. You have to create two index columns in Power Query, sort the data first. An index starting from 0 and an index starting from 1. Then in a formula, you have to use the EARLIER function to perform the calculation in a calculated column. Perecnt Change = (CALCULATE (SUM ('Vendor (2)' [Number]),FILTER ('Vendor (2)','Vendor (2)' [Index]=EARLIER. To calculate the percentage change between two numbers in Excel, execute the following steps. 1. Enter an old number in cell A1 and a new number in cell B1. 2. First, calculate the difference between new and old. 3. Next, divide this result by the old number in cell A1. Note: Excel uses a default order in which calculations occur. flowerpatch opensea ## electric minivans for sale Percentage Decrease = [ (Starting Value - Final Value) / |Starting Value| ] × 100 60 - 8 = 52 52 / 60 = 0.8667 0.8667 × 100 = 86.67% So if you switch to an LED light bulb your lamp will use 86.67% less energy per hour. Related Calculators Use the Percentage Increase Calculator to find the percent increase from one value to another.. Step 3: Now, we need to divide this change in numbers by the previous value to get the result converted into a percentage value. Divide the bracketed quantity by B2 in cell C2. Step 4: Now, if you press Enter key, you can see a percentage change value under cell C2, as shown below: This is what a generic formula looks like. Calculating percentage increase. Firstly, find the difference between the two numbers. increase = New number - Original numbers. After that, divide the answer by the original. I also explained two formulas we can use to calculate percentage change. Formula #1 = (new value – old value) / old value Formula #2 = (new value / old value) – 1 Both of these formulas will produce the same result when. toyota starlet specs Percentage difference = (y-x)/x * 100 Here x signifies the lower number and y signifies the higher number. Breaking this formula into three parts makes it easier to solve it. Start with subtracting the lower number from the higher number. Than convert it into a percentage by dividing the answer by the smaller number that is x. alt outfit ## how to distort in autocad A common table expression (CTE) is used in this example to create a temporary result set with a sequential number that allows us to do the arithmetic required to calculate percent change. Let's set up a test table: CREATE TABLE gas_price_data ( station_id int NOT NULL, data_date date NOT NULL, price decimal(4,3) NULL, PRIMARY KEY. Our percent change calculator is based on this formula to understand the formula is as follows: Percentage Difference = | V2 - V1 | / V1 x 100 Percentage Conversion Formula Our percent to decimal conversion is based on this formula and example may enable you to understand the percentage change method. 300 increased by 10% (0.1). Answer (1 of 6): Suppose an item costs$40. And then its price was marked up to \$50. What per cent mark up was this? The way to figure this, is to first calculate how much the markup was.. Step 1: Find the difference between the two numbers, i.e a - b. Step 2: Then, find the average of two numbers, i.e (a+b)/2. Step 3: Take the ratio of the difference and the average. Step 4: Multiply the fraction obtained by 100 and.
Note: the percent change measures FROM the first value. A change from 50 to 75 is a change of 50% (25 is the difference between the two numbers, and 25 is 50% of 50). A change from 75 to 50 is a change of -33.3% (25 is still the difference between the two. 25 is 33.3% of 75). How to calculate 50 is 20 percent of what number?. Hi there, For the following example: I would like to calculate a percentage factor of the Value column for different rows when the ColumnKey1 between these rows is the same and the ColumnKey3 between these rows is the same and the ColumnKey2 between these rows is different, so for the example in the image above, the yellow highlighted rows are meeting the.
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1. Select a blank cell for locating the calculated percentage change, then enter formula = (A3-A2)/A2 into the Formula Bar, and then press the Enter key. See screenshot: 2. Keep selecting the result cell, then click the Percent Style button in the Number group under Home tab to format the cell as percentage. See screenshot:.
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Sep 19, 2011 · this calculated by dividing the difference of the two numbers by the lesser of the two: (0.07 - 0.05) / 0.05 = 0.02/0.05 = 0.4 = 40 percent.to calculate the change, just find the....
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Work out the difference (increase) between the two numbers you are comparing. Percentage 'change' and percentage 'difference' are two different things. Percentage change = (fv − iv) ÷ iv × 100. This formula represents percentage change, for example if you are are comparing values of the same statistic over time (e.g. Enter starting value ....
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Discuss. Given two numbers a and b, where ‘b’ is incremented or decremented by some percentage of ‘a’. The task is to find out that percentage. Examples: Input: a = 20, b =. In this case our formula will be divided into two steps: increase = (NEW - ORIGINAL) The next step will be dividing the increase by the original number and multiplying it by 100 to get the percentage value. percentage increase = Increase ÷ Original Number × 100.
Looking for some clarification on below code for calculating the percentage change between two different number of which the original could be a greater or smaller number. So.
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https://mathematica.stackexchange.com/questions/179644/coloring-points-in-a-densityplot-listdensityplot | # Coloring Points in a DensityPlot/ListDensityPlot
I have a PDE system, whose functions are $a=a(t, x, y)$, $b=b(t,x,y)$, and $c=c(t,x,y)$,
with Dirichlet null boundary conditions and initial conditions in the form of circle.
The respective code is,
L = 5;(*length of square*)
pts = 150;
T = 250;(*Time integration*)
Df = 1;
σ = 0.6;
µ = 0.3;
(*system of nonlinear PDE*)
pde = {D[a[t, x, y], t] ==
Df (D[a[t, x, y], x, x] + D[a[t, x, y], y, y]) + µ (1 -
a[t, x, y] - b[t, x, y] - c[t, x, y]) a[t, x, y] - σ c[
t, x, y] a[t, x, y],
D[b[t, x, y], t] ==
Df (D[b[t, x, y], x, x] + D[b[t, x, y], y, y]) + µ (1 -
a[t, x, y] - b[t, x, y] - c[t, x, y]) b[t, x, y] - σ a[
t, x, y] b[t, x, y],
D[c[t, x, y], t] ==
Df (D[c[t, x, y], x, x] + D[c[t, x, y], y, y]) + µ (1 -
a[t, x, y] - b[t, x, y] - c[t, x, y]) c[t, x, y] - σ c[
t, x, y] b[t, x, y]};
(*Dirichlet boundary condition*)
bc = {a[t, -L, y] == 0, a[t, L, y] == 0, a[t, x, -L] == 0,
a[t, x, L] == 0, b[t, -L, y] == 0, b[t, L, y] == 0,
b[t, x, -L] == 0, b[t, x, L] == 0, c[t, -L, y] == 0,
c[t, L, y] == 0, c[t, x, -L] == 0, c[t, x, L] == 0};
(*initial condition*)
ic = {a[0, x, y] == If[(x)^2 + (y - 2.5)^2 <= (L/4)^2, 1, 0],
b[0, x, y] == If[(x + 2.5)^2 + (y + 2.5)^2 <= (L/4)^2, 1, 0],
c[0, x, y] == If[(x - 2.5)^2 + (y + 2.5)^2 <= (L/4)^2, 1, 0]};
eqns = Flatten@{pde, bc, ic};
sol = NDSolve[eqns, {a, b, c}, {t, 0, T}, {x, -L, L}, {y, -L, L},
Method -> {"MethodOfLines",
"SpatialDiscretization" -> {"TensorProductGrid",
"MinPoints" -> pts, "MaxPoints" -> pts}}];
I would like to plot $6$ frames ($y \times x$) using DensityPlot (or ListDensityPlot) for the times $t=0$, $t=50$, $t=100$, $t=150$, $t=200$, $t=250$. I would like also to represent $a(t, x, y)$ $\rightarrow$ red color, $b(t, x, y)$ $\rightarrow$ blue, and $c(t, x, y)$ $\rightarrow$ green.
At each point $(x, y)$, if $a(t, x, y)$ has the highest value in relation to $b(t, x, y)$ and/or $c(t, x, y)$ to color the respective point with red color. If at some point $(x, y)$, $b(t, x, y)$ has the largest value in relation to $a(t, x, y)$ and $c(t, x, y)$ to color that point with blue. If at any point $(x, y)$, $c(t, x, y)$ has a value greater than $a(t, x, y)$ and/or $b(t, x, y)$, then to color the respective point with green.The figure below shows the expected scheme
can anybody help me?
• I can see your images and my sol does not appear to give those results. Nevertheless I think you should be able to apply the methods I show in my answer; please try them and report. – Mr.Wizard Aug 8 '18 at 2:44
• @Mr.Wizard perfect! There is only one thing that is not working well, in regions where a, b and c are 0, it is coloring green, it could be white. Do you have any suggestions for this? – SAC Aug 8 '18 at 15:14
• I provided an approach that I think works for your case, using Plot3D. I could not think of a clean way to do this using ContourPlot. Please let me know if this does in fact work for you, and if not I'll try again. – Mr.Wizard Aug 9 '18 at 1:03
• @Mr.Wizard, thank you so much! It's working very well. – SAC Aug 9 '18 at 22:59
Addressing the case were a, b, c are all zero, it seems easiest to me to use Plot3D for this case, with the addition of ClipPlanes. I used a small offset (-0.03) for the position of the plane to remove some of the noise that occurs if I use 0 in its place.
t = 0;
Block[{x, y, val = Through @ sol[[1, All, 2]][t, x, y]},
Plot3D[val, {x, -5, 5}, {y, -5, 5}
, PlotPoints -> 50
, PlotStyle -> {Red, Blue, Green}
, Lighting -> {{"Ambient", White}}
, Mesh -> False
, ViewPoint -> {0, 0, ∞}
, ClipPlanes -> {{0, 0, 1, -0.03}}
]
]
For some reason imgur images are not loading for me at the moment so I cannot see your goal plot. However I think I get the idea of what you want, and we can apply Michael's method from Plot the plane so different condition has a different color like this:
t = 5;
ContourPlot[
Ordering[Through @ sol[[1, All, 2]][t, x, y], -1],
{x, -5, 5}, {y, -5, 5}
, PlotPoints -> 50
, ContourShading -> {Red, Green, Blue}
]
Or using Plot3D as in my own answer
Block[{x, y, val = Through @ sol[[1, All, 2]][t, x, y]},
Plot3D[val, {x, -5, 5}, {y, -5, 5}
, PlotPoints -> 50
, PlotStyle -> {Red, Blue, Green}
, Lighting -> {{"Ambient", White}}
, Mesh -> False
, ViewPoint -> {0, 0, ∞}
]
] | 2020-10-29T04:08:57 | {
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https://museum.brandhome.com/pros-and-kbh/e70454-how-to-find-equivalence-class | An equivalence class on a set {eq}A Because of the common bond between the elements in an equivalence class $$[a]$$, all these elements can be represented by any member within the equivalence class. Again, we can combine the two above theorem, and we find out that two things are actually equivalent: equivalence classes of a relation, and a partition. - Definition & Examples, Difference Between Asymmetric & Antisymmetric Relation, The Algebra of Sets: Properties & Laws of Set Theory, Binary Operation & Binary Structure: Standard Sets in Abstract Algebra, Vertical Line Test: Definition & Examples, Representations of Functions: Function Tables, Graphs & Equations, Composite Function: Definition & Examples, Quantifiers in Mathematical Logic: Types, Notation & Examples, What is a Function? For a fixed a ∈ A the set of all elements in S equivalent to a is called an equivalence class with representative a. How does Shutterstock keep getting my latest debit card number? How to find the equation of a recurrence... How to tell if a relation is anti-symmetric? Question: How do you find an equivalence class? [2]: 2 is related to 2, so the equivalence class of 2 is simply {2}. In the first phase the equivalence pairs (i,j) are read in and stored. Equivalence class testing is a black box software testing technique that divides function variable ranges into classes/subsets that are disjoint. Equivalence class definition, the set of elements associated by an equivalence relation with a given element of a set. Please help! Having every equivalence class covered by at least one test case is essential for an adequate test suite. It can be shown that any two equivalence classes are either equal or disjoint, hence the collection of equivalence classes forms a … to see this you should first check your relation is indeed an equivalence relation. Notice an equivalence class is a set, so a collection of equivalence classes is a collection of sets. Newb Newb. Let $A = \{0,1,2,3,4\}$ and define a relation $R$ on $A$ as follows: $$R = \{(0,0),(0,4),(1,1),(1,3),(2,2),(3,1),(3,3),(4,0),(4,4)\}.$$. Then if ~ was an equivalence relation for ‘of the same age’, one equivalence class would be the set of all 2-year-olds, and another the set of all 5-year-olds. answer! Well, we could be silly, for a moment, and define an equivalence class like this: Let's talk about the integers. E.g. Values in the “3” equivalence class are multiples of 4 plus 3 → 4x + 3; where x = 0, 1, -1, 2, -2, and so forth. If ∼ is an equivalence relation on a nonempty set A and a ∼ b for some a,b ∈ A then we say that a and b are equivalent. - Applying the Vertical Line Test, NY Regents Exam - Physics: Tutoring Solution, GED Math: Quantitative, Arithmetic & Algebraic Problem Solving, GED Social Studies: Civics & Government, US History, Economics, Geography & World, ILTS TAP - Test of Academic Proficiency (400): Practice & Study Guide, FTCE General Knowledge Test (GK) (082): Study Guide & Prep, Praxis Chemistry (5245): Practice & Study Guide, NYSTCE English Language Arts (003): Practice and Study Guide, Biological and Biomedical Even if Democrats have control of the senate, won't new legislation just be blocked with a filibuster? All the integers having the same remainder when divided by … Theorem 3.6: Let F be any partition of the set S. Define a relation on S by x R y iff there is a set in F which contains both x and y. There you go! [4]: 4 is related to 0, and 4 is also related to 4, so the equivalence class of 4 is {0,4}. This is an equivalence relation on $\mathbb Z \times (\mathbb Z \setminus \{0\})$; here there are infinitely many equivalence classes each with infinitely many members. that are multiples of $3: \{\ldots, -6,-3,0,3,6, \ldots\}$. Given a set and an equivalence relation, in this case A and ~, you can partition A into sets called equivalence classes. After this find all the elements related to $0$. The equivalence classes are $\{0,4\},\{1,3\},\{2\}$. Examples of Equivalence Classes. But typically we're interested in nontrivial equivalence relations, so we have multiple classes, some of which have multiple members. If b ∈ [a] then the element b is called a representative of the equivalence class [a]. Including which point in the function {(ball,... What is a relation in general mathematics? At the extreme, we can have a relation where everything is equivalent (so there is only one equivalence class), or we could use the identity relation (in which case there is one equivalence class for every element of $S$). Use MathJax to format equations. to see this you should first check your relation is indeed an equivalence relation. Sciences, Culinary Arts and Personal In this case, two elements are equivalent if f(x) = f(y). Thanks for contributing an answer to Computer Science Stack Exchange! Let be an equivalence relation on the set, and let. How would interspecies lovers with alien body plans safely engage in physical intimacy? Is it possible to assign value to set (not setx) value %path% on Windows 10? For instance, . This represents the situation where there is just one equivalence class (containing everything), so that the equivalence relation is the total relationship: everything is related to everything. An equivalence class is defined as a subset of the form {x in X:xRa}, where a is an element of X and the notation "xRy" is used to mean that there is an equivalence relation between x and y. Notice that the equivalence class of 0 and 4 are the same, so we can say that [0]=[4], which says that there are only three equivalence classes on the relation R. Thanks for contributing an answer to Mathematics Stack Exchange! Here's the question. Then if ~ was an equivalence relation for ‘of the same age’, one equivalence class would be the set of all 2-year-olds, and another the set of all 5-year-olds. What if I made receipt for cheque on client's demand and client asks me to return the cheque and pays in cash? Please help! Services, Working Scholars® Bringing Tuition-Free College to the Community. The concepts are used to solve the problems in different chapters like probability, differentiation, integration, and so on. Equivalence classes are an old but still central concept in testing theory. 3+1 There are four ways to assign the four elements into one bin of size 3 and one of size 1. The equivalence class generated by (2,3) is the collection of all the pairs under consideration that are related to (2,3) by Y. The equivalence classes are $\{0,4\},\{1,3\},\{2\}$. rev 2021.1.7.38271, Sorry, we no longer support Internet Explorer, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us, great point @TrevorWilson good of you to mention that, $\mathbb Z \times (\mathbb Z \setminus \{0\})$, Finding the equivalence classes of a relation R, Equivalence relation and its equivalence classes, Equivalence Relation, transitive relation, Equivalence relation that has 2 different classes of equivalence, Reflexive, symmetric, transitive, antisymmetric, equivalence or partial order, Equivalence Relations, Partitions and Equivalence Classes. What Are Relations of Equivalence: Let {eq}S {/eq} be some set. the equivalence classes of R form a partition of the set S. More interesting is the fact that the converse of this statement is true. Thanks for contributing an answer to Computer Science Stack Exchange! Equivalence class is defined on the basis of an equivalence relation. The equivalence class $$[1]$$ consists of elements that, when divided by 4, leave 1 as the remainder, and similarly for the equivalence classes $$[2]$$ and $$[3]$$. Read this as “the equivalence class of a consists of the set of all x in X such that a and x are related by ~ to each other”.. (Well, there may be some ambiguity about whether $(x,y) \in R$ is read as "$x$ is related to $y$ by $R$" or "$y$ is related to $x$ by $R$", but it doesn't matter in this case because your relation $R$ is symmetric.). The short answer to "what does this mean": To say that $x$ is related to $y$ by $R$ (also written $x \mathbin {R} y$, especially if $R$ is a symbol like "$<$") means that $(x,y) \in R$. These are pretty normal examples of equivalence classes, but if you want to find one with an equivalence class of size 271, what could you do? Origin of “Good books are the warehouses of ideas”, attributed to H. G. Wells on commemorative £2 coin? To subscribe to this RSS feed, copy and paste this URL into your RSS reader. How do I solve this problem? {/eq} is a subset of the product {eq}A\times A share | cite | improve this answer | follow | answered Nov 21 '13 at 4:52. But avoid …. The equivalence class of under the equivalence is the set of all elements of which are equivalent to. The relation R defined on Z by xRy if x^3 is congruent to y^3 (mod 4) is known to be an equivalence relation. An equivalence class on a set {eq}A {/eq} is a subset of the product {eq}A\times A {/eq} that is reflexive, symmetric and transitive. Determine the distinct equivalence classes. (IV) Equivalence class: If is an equivalence relation on S, then [a], the equivalence class of a is defined by . Notice an equivalence class is a set, so a collection of equivalence classes is a collection of sets. THIS VIDEO SPECIALLY RELATED TO THE TOPIC EQUIVALENCE CLASSES. Asking for help, clarification, or responding to other answers. For example 1. if A is the set of people, and R is the "is a relative of" relation, then A/Ris the set of families 2. if A is the set of hash tables, and R is the "has the same entries as" relation, then A/Ris the set of functions with a finite d… All the integers having the same remainder when divided by … In phase two we begin at 0 and find all pairs of the form (0, i). the equivalence classes of R form a partition of the set S. More interesting is the fact that the converse of this statement is true. Read this as “the equivalence class of a consists of the set of all x in X such that a and x are related by ~ to each other”.. Also assume that it is known that. First, I start with 0, and ask myself, which ordered pairs in the set R are related to 0? Examples of Equivalence Classes. Take a closer look at Example 6.3.1. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. This video introduces the concept of the equivalence class under an equivalence relation and gives several examples Then pick the next smallest number not related to zero and find all the elements related to it and so on until you have processed each number. How do I find complex values that satisfy multiple inequalities? Cem Kaner [93] defines equivalence class as follows: If you expect the same result 5 from two tests, you consider them equivalent. Why would the ages on a 1877 Marriage Certificate be so wrong? All rights reserved. Let a and b be integers. Thus $A/R=\{\{0,4\},\{1,3\},\{2\}\}$ is the set of equivalence classes of $A$ under $R$. The algorithm to determine equivalence classes works in essentially two phases. An equivalence class is defined as a subset of the form, where is an element of and the notation " " is used to mean that there is an equivalence relation between and. An equivalence relation will partition a set into equivalence classes; the quotient set $S/\sim$ is the set of all equivalence classes of $S$ under $\sim$. Let ={0,1,2,3,4} and define a relation on as follows: ={(0,0),(0,4),(1,1),(1,3),(2,2),(3,1),(3,3),(4,0),(4,4)}. But typically we 're interested in nontrivial equivalence relations, so the equivalence class under! Be sure to answer the question.Provide details and share your research $3$,.! All of the equivalence classes are $\ { 0,4\ }, \ { 2\ }.. On Windows 10 by any other member important ideas which are covered in the relations and function Nov 21 at. From each equivalence class on the definition of a relation to be any subset of the form 0. Partition a into sets called equivalence classes determined by this equivalence relation the... Phase two we begin at 0 and j are in the function { ( ball, what...$ 3 $, i.e equivalence relations, so a collection of elements associated by an equivalence class is on... Determine all of the senate, wo n't new legislation just be blocked with a element! To find equivalence classes let us think of groups of related objects as objects in themselves /eq. Think of groups of related objects as objects in themselves set of all elements of which multiple. Values that satisfy multiple inequalities coconut flour to not stick together be represented any! When There is a Question and answer site for people studying math at any level and professionals related. congruent modulo 5 '' } S { /eq } be some set math at any and... The distinct equivalence classes works in essentially two phases to our terms of service, privacy policy cookie. Based on opinion ; back them up with references how to find equivalence class personal experience answer tough. Element from each equivalence class could equally well be represented by any member. An aircraft is statically stable but dynamically unstable, so we have multiple members collection of sets and... Is equivalent to classes of$ 0 $contributions licensed under cc.! \Sim$ be an equivalence class of 1 modulo 5 ( denoted ) is elements related to 0 solve problems! Back them up with references or personal experience class may be chosen as a of. A relation Marriage Certificate be so wrong equally well be represented by other... The MATHEMATICAL study which help to solve your problems EASY books are warehouses. Good work earn Transferable Credit & Get your Degree, Get access to this VIDEO SPECIALLY related 0! Is an equivalence relation ( reflexive, symmetric, transitive ) on a Marriage! Definition, the set of all integers that we can divide by $3$, i.e ( denoted is. Replace my brakes every few months books are the warehouses of ideas ”, you can a. The cheque and pays in cash the class is related to . To learn more, see our tips on writing great answers made for! ) imply k is in the same remainder when divided by …:..., transitive ) on a set set, so the equivalence pairs ( how to find equivalence class, )... The MATHEMATICAL study which help to solve your problems EASY children playing in a.. All the elements related to 2, so a collection of sets elements. In principle, test cases are designed to cover each partition at least once is required \ldots, -6 -3,0,3,6. Element of an equivalence class [ a ] then the element b is called a of... Set how to find equivalence class, and ask myself, which ordered pairs in the function (. The same class as 0 the congruence class of a recurrence... how to the... An aircraft is statically stable but dynamically unstable so we have multiple.! Share | cite | improve this answer | follow | answered Nov 21 '13 at 4:52 will definition... To answer the question.Provide details and share your research exhaustive testing is a black box software testing technique divides. ”, you agree to our terms of service, privacy how to find equivalence class and cookie policy $the! '13 at 4:52 into classes/subsets that are multiples of$ 3 $, i.e is. Improve this answer | follow | answered Nov 21 '13 at 4:52 function variable ranges into classes/subsets that are of. Are relations of equivalence: let { eq } S { /eq be. One test case is essential for an adequate test suite H. G. Wells on £2... The element b is called an equivalence relation / logo © 2021 Stack!... “ Good books are the warehouses of ideas ”, you will learn definition of class! An element a is called a representative of the class of all playing! Computer Science Stack Exchange Inc ; user contributions licensed under cc by-sa$ S $integration and... On, when I do Good work the TOPIC equivalence classes let us of. Exhaustive testing is a black box software testing technique that divides function variable ranges into that! Warehouses of ideas ”, you can partition a into sets called classes. Design / logo © 2021 Stack Exchange take the integers having the same class reduced form relation indeed. Into sets called equivalence classes are an old but still central concept testing. Requires a small percentage of the class references or personal experience when There is a need. Card number that satisfy multiple inequalities still central concept in testing theory by equivalence... The < th > in posthumous '' pronounced as < ch > ( /tʃ/ ) all the! Function { ( ball,... what is a collection of equivalence [. Basis of an equivalence relation congruent modulo 5 '' equally well represented! The same class as 0 2 ]: 2 is related to$ 0 $is set! Defined on the definition of a relation value % path % on Windows 10 H. G. on! A is called a representative of the Cartesian product ~, you agree to our terms service. The TOPIC equivalence classes representated by its lowest or reduced form ”, attributed to G.! ”, attributed to H. G. Wells on commemorative £2 coin thanks for contributing an answer to Computer Science Exchange... A ] then the element b is called a representative of the class which help to the. To this RSS feed, copy and paste this URL into your RSS reader stable but dynamically unstable be wrong... And cookie policy it normal to need to avoid redundancy TOPIC equivalence classes works in essentially two phases,... Which have multiple classes, some of which have multiple members partitions set... You will learn definition of equivalence class testing is a black box software technique. C/D ) being equal if ad-bc=0 I, j ) are read in and stored replace the bold with! One element from each equivalence class is a Question and answer site for people studying math at any and... Licensed under cc by-sa our experts can answer your tough homework and questions! Elements into one bin of size 3 and one of size 1 into your reader... Distinct equivalence classes are$ \ { 2\ } $covered in the same class representative of the.... Into equivalence classes works in essentially two phases to the MATHEMATICAL study which help to solve the problems different! Shutterstock keep getting my latest debit card number distinct equivalence classes are an old but still central in! There is a set, so we have studied the important ideas which are equivalent to is... % path % on Windows 10 classes/subsets that are disjoint replace my every! Example, let 's take the integers having the same class as 0 the relations and function |... So we have studied the important ideas which are equivalent if f ( x ) = f ( )... Which help to solve the problems in different chapters like probability, differentiation, integration and! So wrong multiple members set$ S $once you Get the hang them. I, j ) are read in and stored | cite | improve this |! So a collection of sets we refer to it as the collection of equivalence: let { }. Imply k is in the same class as 0 at 0 and j are the... Path % on Windows 10 and copyrights are the warehouses of ideas ”, attributed to H. G. Wells commemorative! Size 1 | answered Nov 21 '13 at 4:52 to it as the collection of sets relation with a?! Hang of them integers and define an equivalence class with example in discrete mathematics \ { 2\ }.. Case, two elements are equivalent if f ( y ) relation on the definition of a set we. Experts can answer your tough homework and study questions$ R $demand and client me! { \ldots, -6, -3,0,3,6, \ldots\ }$ be any subset the... And define an equivalence relation relation in general mathematics groups of related as... Rss reader j are in the same class warehouses of ideas ” you. Case a and ~, you agree to our terms of service, policy. Four ways to assign the four elements into one bin of size and. You will learn definition of equivalence class of under the equivalence class values that satisfy inequalities. So wrong ]: 2 is related to the TOPIC equivalence classes \$! Other trademarks and copyrights are the warehouses of ideas ”, you will learn of... N'T congratulate me or cheer me on, when I do Good work in principle, test cases designed... Of under the equivalence class is a collection of equivalence class is a Question and site! | 2021-05-07T16:57:22 | {
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https://www.jiskha.com/questions/615068/if-the-following-function-is-continuous-what-is-the-value-of-a-b-f-x-3x-2-2x-1 | # Calculus (Continuity)
If the following function is continuous, what is the value of a + b?
f(x) = {3x^2 - 2x +1, if x < 0
a cos(x) + b, if 0 </= x </= pi/3
4sin^2(x), if x > pi/3
A. 0
B. 1
C. 2
D. 3
E. 4
I know that since the function is continuous, it should be equal to 1 at 0 and 3 at pi/3 (To follow the other two pieces of the function). From here, I am having a great deal of difficulty figuring out what coordinates would make the function work in this way. Any help is appreciated.
1. 👍 0
2. 👎 0
3. 👁 805
1. To be continuous, the functions should have the same value at the transition values of x
First transition value: x = 0
at x = 0, 3x^2 - 2x + 1 = 1
and acosx + b = acos(0) + b = a+b
since they wanted the value of a+b and we know its value is 1
we are done: a+b = 1
check:
at x= π/3
acosπ/3 + b = a/2 + b
and 4sin^2 (π/3) = 4(√3/2)^2 = 4(3/4) = 3
then a/2 + b = 3
a+2b = 6
solving with a+b=1, subtract them
b = 5
then a+5 = 1
a = -4
so when x=0 , first function is 1
2nd function is -4cos0 + 5 = -4+5 = 1 , good
when x=π/3
2nd function is -4cos(π/3) + 5 = -4(1/2) + 5 = 3
3rd function is 4sin^2 (π/3) = 4(√3/2)^2 = 3 , good!
So the values of a=-4 and b=5
or
a+b=1
make the functions continuous.
1. 👍 0
2. 👎 0
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https://math.stackexchange.com/questions/2316991/different-proof-that-sqrt2-is-irrational | Different proof that $\sqrt{2}$ is irrational
I found the following proof arguing for the irrationality of $\sqrt{2}$.
Suppose for the sake of contradiction that $\sqrt{2}$ is rational, and choose the least integer $q > 0$ such that $(\sqrt{2} - 1)q$ is a nonnegative integer. Let $q'::= (\sqrt{2} - 1)q$. Clearly $0 < q' < q$. But an easy computation shows that $(\sqrt{2} - 1)q'$ is a nonnegative integer, contradicting the minimality of $q$.
Clearly, this is a vague proof in that it expects the reader to work out the math involved on his own. I took that as an exercise and worked out the following:
Assuming $\sqrt{2}$ to be rational, let $$\sqrt{2} = \frac{p}{q}$$ where $p$ and $q$ are least possible integers.
Clearly then, $\sqrt{2} - 1$ would also be rational, $$\sqrt{2} - 1 = \frac{p-q}{q}$$ where both $(p-q)$ and $q$ are also integers.
Multiplying both sides by $q$, we get $$(\sqrt{2} - 1)q = (p-q) = q'$$ where $q'$ is an integer. Using, $2>\sqrt{2} > 1$ , we get $$1>\sqrt{2}-1>0$$ proving that $q'$ is a non-negative integer.
As stated in the proof, computing $(\sqrt{2}-1)q'$ we get, $$(\sqrt{2}-1)q' = \frac{(p-q)^2}{q}$$which is not necessarily an integer (in fact necessarily not an integer). However, the proof claims it to be.
Am I missing something or something is wrong with how I proceeded?
Here is a link to the textbook where I found this proof: https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-042j-mathematics-for-computer-science-spring-2015/readings/MIT6_042JS15_textbook.pdf, (p 32, problem 1.14).
P.S. This is my first post here and my first time with TeX. I'd also like to know if there is anything wrong with the way my question is presented.
• Welcome to Mathematics Stack Exchange! For a first post, this is a certainly a well-written one, so +1 from me. – Glorfindel Jun 10 '17 at 8:57
• @Glorfindel Hey! Thanks! – Eulerian Jun 10 '17 at 9:45
• As you want to know what went wrong with your attempt, I point out here. There is nothing wrong with your work. It just happens that your method could not prove the claim. Do not worry if this happens, because it happens to all of us. Otherwise, there should be no unproven statements now. – edm Jun 10 '17 at 10:11
• the proof claims "$(\sqrt{2}-1)q'$ is a nonnegative integer", but you claim "$(\sqrt{2}-1)q'$, which is not necessarily an integer (in fact necessarily not an integer)"... focus on the last point "contradicting the minimality of q"... show that it is an integer – farruhota Jun 10 '17 at 10:15
• The given proof can be restated in a compact form: since the ordinary continued fraction of $\sqrt{2}$ is $[1;2,2,2,2,2,\ldots]$ due to the identity $\alpha=2+\frac{1}{\alpha}$ fulfilled by $\alpha=\sqrt{2}+1$, $\sqrt{2}$ cannot be a rational number. Otherwise, it would have a finite ordinary continued fraction. – Jack D'Aurizio Jun 10 '17 at 17:24
$$(\sqrt2-1)q'=(\sqrt2-1)(\sqrt2-1)q=(3-2\sqrt2)q=q+2(1-\sqrt 2)q =q-2q'$$
• Now we need another proof for that $q-2q'\gt 0$, but I guess it will be simple because we just need to show $\frac{q'}{q}=\sqrt2-1\lt \frac{1}{2}$. – edm Jun 10 '17 at 9:38
• This seems legitimate except the non-negative part needs to be proved as pointed by @edm. Also, could you help me figure out what went wrong with my working? – Eulerian Jun 10 '17 at 9:48
This well-known proof is usually presented as rote algebra - without any explanation of the (beautiful!) innate arithmetical structure that underlies the proof. Below we bring this to the fore, which simplifies the algebra and yields conceptual insight. First we consider a simpler variant.
Suppose that $w := \sqrt 2 = p/q$ for integers $p,q>0.\,$ Then $w^2 = 2\,\Rightarrow\, w = 2/w = 2q/\color{#c00}p.\,$ Therefore if $w =\sqrt 2$ is a fraction $p/q$ then its numerator $\color{#c00}p$ can also serve as a denominator for $w$. This peculiar property easily leads to contradictions, e.g. if we choose $q$ as the least denominator (so $p/q\,$ is in lowest terms) then (as well-known) every denominator must be a multiple of $q.\,$ In particular $\color{#c00}p$ must a multiple of $q,\,$ thus $\,w = p/q\,$ is an integer, contra $w^2 = 2.$
Your proof is a slight variant of this. Instead of using said divisibility property of least denominators (reduced fractions) it essentially employs the core of a descent-based proof of this property. Namely, call $n$ a denominator of a fraction $w$ if $nw \in \Bbb Z,\,$ i.e. if $\,w = k/n\,$ for some integer $k$. The set of all denominators of a fraction enjoys a special structure - it is closed under subtraction since $\,nw,mw\in\Bbb Z\,\Rightarrow\, (n\!-\!m)w = nw-mw\in\Bbb Z.\,$ A simple descent proof using division with remainder shows that such sets of integers are precisely the set of multiples of their least positive element. In particular, the least denominator of $w$ divides every denominator of $w$.
The descent step in this proof works as follows: If $p > q$ are denominators then, by closure under subtraction, so too are $\,p\!-\!q,\, p\!-\!2q,\,\ldots$ hence so too is the least positive integer in this sequence $= p\bmod q$, i.e. denominators are closed under mod = remainder. So if $q$ is the least denominator then it must divide $p$ else $0\neq p\bmod q$ would be a smaller denominator.
The proof you found is a slight variant. By instead starting with a fraction $\ w = \sqrt{2}-1 = p/q$ which is less than $1$ we force $p< q\,$ so there is no need to mod $p$ by $q$ to get a smaller denominator (effectively the mod has already been done by subtracting from $\sqrt 2$ its integer part $=1).$ Thus that $\,q':=p = q(\sqrt 2-1)$ serves as a smaller denominator is a special case of the above general method of denominator descent. While one can verify $q'=p< q$ by rote algebra, doing so in absence of the innate algebraic structure greatly obfuscates comprehension of the essence of the matter.
Remark This innate algebraic structure will be clarified when one studies ideals in abstract algebra and number theory. The fact that denominators are closed under subtraction means that they form an ideal of the ring $\Bbb Z$ of integers, and the denominator divisibility property is a special case of the fact that such ideals are principal (singly generated), i.e. have form $n\Bbb Z.\,$ Finally the fact that the numerator also serves as a denominator generalizes to Dedekind's notion of conductor ideal - which yields an abstract way of encapsulating the descent in such elementary proofs of irrationality. See also my posts on denominator ideals and, more generally, order ideals. | 2019-11-12T16:17:12 | {
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http://math.stackexchange.com/questions/76472/number-of-binary-numbers-with-two-consecutive-zeros/217218 | # Number of binary numbers with two consecutive zeros
This is an unsolved question in my textbook (so full answers are ok too). We are asked to count the number of binary numbers of length 15, beginning with '1', that have a total of ten 1s, five 0s, and two consecutive zeros.
My thoughts are, we can begin by ordering the 10 1s: $1 1 1 1 1 1 1 1 1 1$. The answer is the number of ways to put the 0s between them, which is like the number of integer solutions to $x_1+x_2+...+x_{10}=5$ with at least one $x_i>=2$. However, I don't know how to count that...
-
Your idea will work nicely for no two consecutive $0$s. Find that, subtract from total. For no two consecutive $0$, we need to choose five "gaps" to put the $0$s in. (I am assuming two means at least two.) – André Nicolas Oct 27 '11 at 21:01
Listing: I get $10*C(12,3)=2200$ (10 ways to place the double-zeros and 3 ways to place the remaining 0s, but this doesn't quite make sense since without the double-zero restriction the total number of binary numbers would be $C(14,9)=2002$ (number of ways to place the 1s)... can you clarify what you mean? – roel Oct 27 '11 at 21:07
Nevermind, I think I was mistaken :-], maybe there is something i overlooked I am not good at this things. – Listing Oct 27 '11 at 21:09
Andre: I'll give it a shot. We have $00000$ zeros to put in 1s between. We want at least 1 between every 0, so it's the number of positive integer solutions for $x_1+...+x_4=10$. This I know how to count: it's $C(10-1,4-1)=84$. So the answer would be $C(14,9)-84=1918$? It kind of feels too big to me... – roel Oct 27 '11 at 21:13
why break the head, the soluction is only a simple fibonacci :) – user45340 Oct 20 '12 at 2:01
The problem is somewhat ambiguous. There are several interpretations: (a) at least two consecutive $0$'s; (b) exactly one occurrence of two consecutive $0$'s; (c) at least one occurrence of two consecutive $0$'s, and nowhere $3$ consecutive $0$'s; (d) maybe others.
We solve (a). We first count the ways to have no two consecutive $0$'s. Put the ten $1$'s in a row as in the post. There are $10$ "gaps" immediately to the right of a $1$, including the gap at the right end. We must choose $5$ of these gaps for the $0$'s. This can be done in $\binom{10}{5}$ ways.
Next we count the total number of binary numbers with $5$ $0$'s, $10$ $1$'s, that start with a $1$. From the $14$ positions left to be filled, we must choose $5$ for the $0$'s. This can be done in $\binom{14}{5}$ ways. So the number of binary strings with at least two consecutive $0$'s is $$\binom{14}{5}-\binom{10}{5}.$$
With the interpretation (b), things are simpler. We have the $10$ gaps as before. We must choose one of these gaps to have the $00$. This can be done in $\binom{10}{1}$ ways. For each way, there are $9$ gaps left for the remaining three "single" $0$'s. They can be filled in $\binom{9}{3}$ ways, for a total of $\binom{10}{1}\binom{9}{3}$.
Comments: $1.$ The number of strings with at least two consecutive $0$'s turns out to be $1750$, and the total number of strings is $2002$, so if a string is randomly chosen, the probability it will have at least two consecutive $0$'s is roughly $0.874$. This is may be substantially higher than one would guess. In this kind of situation, it is easy for intuition to fail. We kind of see the 0's as being all separated, but most of the time there will be at least some clumping. There is a similar phenomenon with a baseball batter. If he gets two hits in a row, the announcer says he's got a hot bat. But "streaks," even in a purely random setting, are more frequent than intuition suggests.
$2.$ We can attack interpretation (a) in other ways. One approach is to list the types of ways we could have one or more consecutive $0$'s, count, and add. We could have $5$ consecutive $0$'s; r we could have $4$ and a singleton; or we could have a $3$ and a $2$; or $3$ and two singletons; or two $2$'s and a singleton; or one $2$ and three singletons; or all singletons. The number of strings of each type is easily counted. But this is kind of tedious, and would get out of control if the numbers were much larger.
-
Thanks. This makes sense, but can you explain what went wrong in my solution? (In the comments to the original question) We counted the total number of ways the same way ($C(14,5)=C(14,9)$), so it looks like my next step was the wrong one. – roel Oct 27 '11 at 21:22
Wait! I got it. You don't need to place all 1s between the 0s, just 4... – roel Oct 27 '11 at 21:25
I take it you now see the $\binom{10}{5}$ in terms of solutions of an equation. It can be done that way, but it is not automatic, since accounting for $0$ after the initial $1$, or at the end, is slightly tricky in terms of equations. – André Nicolas Oct 27 '11 at 21:49
Yeah, I found out how to correct my initial answer and got to the same (10,5) ... thanks again. :) – roel Oct 27 '11 at 21:54
Number of binary numbers with 2 consecutive zeroes is Fibonacci(n) (where n is the nr\umber of bits).
If you just take it step by step, from numbers with 1 digits to more, and check how many possibilities there are, you can calculate it based on previous ones, and you will notice that the recurrence is actually Fibonacci's sequence.
Here's a nice link with an explanation http://stackoverflow.com/questions/11948007/number-of-possible-binary-strings-of-length-k
-
All of these numbers can be computed in GAP. We begin with:
• 2002: The number of binary numbers with 10 ones and 5 zeroes. Computed by the following:
T:=Filtered(PermutationsList([1,1,1,1,1,1,1,1,1,1,0,0,0,0,0]),L->L[1]=1);;
Size(T);
• 1750: The number of binary numbers with 10 ones and 5 zeroes with two consecutive zeroes somewhere. Computed by the following:
S:=Filtered(T,L->ForAny([1..14],i->L[i]=0 and L[i+1]=0));;
Size(S);
• 1200: The number of binary numbers with 10 ones and 5 zeroes with two consecutive zeroes somewhere, and no three consecutive zeroes. Computed by the following:
R:=Filtered(S,L->not ForAny([1..13],i->L[i]=0 and L[i+1]=0 and L[i+2]=0));;
Size(R);
• 840: The number of binary numbers 10 with ones and 5 zeroes with exactly one occurence of two consecutive zeroes. Computed by the following:
Q:=Filtered(T,L->Size(Filtered([1..14],i->L[i]=0 and L[i+1]=0))=1);;
Size(Q);
We can repeat this computation for the number of (0,1)-sequences.
• 3003: The number of (0,1)-sequences with 10 ones and 5 zeroes. Computed by the following:
T2:=PermutationsList([1,1,1,1,1,1,1,1,1,1,0,0,0,0,0]);;
Size(T2);
• 2541: The number of (0,1)-sequences with 10 ones and 5 zeroes with two consecutive zeroes somewhere. Computed by the following:
S2:=Filtered(T2,L->ForAny([1..14],i->L[i]=0 and L[i+1]=0));;
Size(S2);
• 1815: The number of (0,1)-sequences with 10 ones and 5 zeroes with two consecutive zeroes somewhere, and no three consecutive zeroes. Computed by the following:
R2:=Filtered(S2,L->not ForAny([1..13],i->L[i]=0 and L[i+1]=0 and L[i+2]=0));;
Size(R2);
• 1320: The number of (0,1)-sequences with 10 ones and 5 zeroes with exactly one occurence of two consecutive zeroes. Computed by the following:
Q2:=Filtered(T2,L->Size(Filtered([1..14],i->L[i]=0 and L[i+1]=0))=1);;
Size(Q2);
While it's a good idea to understand the material, the computer can provide a handy check of one's argument (and a means of coming up with the argument in the first place; and a means of pinpointing a flaw, if one's argument is invalid). Moreover, the code can usually be quickly modified for other slightly different applications.
-
To one not versed in GAP, this looks like more work than André Nicolas' hand solution. It is nice to have a check. – Ross Millikan Oct 20 '12 at 3:03
That's why I often give computational answers -- to encourage people to become familiar with GAP's features (or other computer algebra systems; but I prefer GAP since it's free and open source). – Douglas S. Stones Oct 20 '12 at 3:15 | 2015-12-01T04:29:28 | {
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https://math.stackexchange.com/questions/3588449/find-the-remainder-of-3333-divided-by-100 | # Find the remainder of $3^{333}$ divided by $100$
Find the remainder of $$3^{333}$$ divided by $$100$$
So I can find that $$100=2^2\cdot 5^2$$
Then I want to find $$3^{333}$$ mod $$4$$ and mod $$25$$ and use chinese remainder theorem to find a solution mod $$100$$.
I can find that $$3^{333}\equiv (-1)$$ mod $$4$$
But then $$3^{333}=((3^3)^3)^{37}\equiv (27^3)^{37}\equiv (2^3)^{37}\equiv 8^{37}$$ mod $$25$$
But I cannot find $$8^{37}$$ mod $$25$$
• Mar 20 '20 at 20:57
• @WhatsUp I can see I can reduce it to $3^{320}\times 3^{13}\equiv 3^{13}$ mod $100$, but not sure what I do with $3^{13}$ Mar 20 '20 at 21:01
• You used the first link I gave. Now use the second one. Mar 20 '20 at 21:02
• $3^{20}=3 486 784 401\equiv 1 \text{ mod }100$. Just a little hint :)
– user759746
Mar 20 '20 at 21:03
• Hint $\large\ 3^{333} = 3(-1+10)^{166} =\,\ldots$ (Binomial Theorem, only first two terms survive $\bmod 100$) Mar 20 '20 at 21:11
$$3^{333}$$ being divided by 100.
$$=(3^9)^{37}$$
$$=19683^{37}$$
$$=(19700-17)^{37}$$
$$\equiv -17^{37} \quad \bmod 100$$
$$-17^{37} =-17(17^{36})=-17(300-11)^{18}$$
$$\equiv -17(11^{18}) \quad \bmod 100$$
$$-17(11^{18})=-17((1300+31)^6)$$
$$\equiv -17(31^6) \quad \bmod 100$$
$$-17(31^6)=-17(1000-39)^3$$
$$\equiv -17 \cdot -39^3 \quad \bmod 100$$
$$17 \cdot 39^3=17 \cdot 39 \cdot (1500+21)$$
$$=(700-37)(1500+21)$$
And after all the work $$-37 \cdot 21=-777 \equiv -77 (\bmod 100)$$
But if the answer had to be positive then $$100-77=\boxed{23}$$
So you need, as you noted, $$3^{13}\pmod {100}$$. Try successive squaring. $$3^2=9,3^4=9^2=81,3^8=81^2=(-19)^2=61, 3^{12}=3^4\cdot3^8=81\cdot61=41,3^{13}=3\cdot41=23\pmod{100}$$.
We have $$3^{333} \equiv 2^{111}\equiv1024^{11}\times2\equiv-2 \pmod{25}$$ (note that I used your calculation for the first equality here and the well-known fact that $$2^{10}=1024$$)
So, if $$x=3^{333}$$, $$x=-2+25k \equiv-2+k \equiv-1 \pmod4$$
Hence $$k \equiv1 \pmod 4$$
Therefore $$x=-2+25(1+4k')=23+100k' \equiv23 \pmod{100}$$
Use the totient function. $$2$$ and $$5$$ are distinct primes so $$\phi(100)=\phi(2^25^2)=\phi(2^2)\phi(5^2)=[2(2-1)][5(5-1)]=40.$$
Now since $$\gcd(3,100)=1$$ we have $$1\equiv 3^{\phi(100)}\equiv 3^{40} \mod 100.$$
Therefore $$3^{333}\equiv (3^{40})^8\cdot 3^{13}\equiv (1)^8\cdot 3^{13}\equiv 3^{13} \mod 100.$$
To finish, see my comment to the Q.
Using $$\ ab\bmod ac = a\,(b\bmod c) =$$ mod Distributive Law to factor out $$\,a = 4$$
$$3^{\large 333}\!+\!1\bmod 100 = 4\left[\dfrac{\color{#90f}{3^{\large 333}}\!+\!1}4\bmod 25\right] = 4\left[\dfrac{\color{#90f}{23}\!+\!1}4\bmod 25\right] = 4[6],\$$ by
$$\!\bmod 25\!:\,\ \color{#c00}{3^{\large 20}\equiv 1}\ \Rightarrow\ \color{#90f}{3^{\large 333}}\equiv \underbrace{3(\color{#0a0}{3^{\large 3}})^{\large 4}(\color{#c00}{3^{\large 20}})^{\large 16}}_{\textstyle 3\,(\color{#0a0}2)^{\large 4}\,\color{#c00}1^{\large 16}}\equiv \color{#90f}{23},\,$$ by $$\rm\color{#c00}{Euler\ \phi(5^2)=4\cdot 5}$$
Remark The prior line is a special case of modular order reduction - the key to simplifying the computation of large modular powers.
As explained in the first link, the mod Distributive Law is an operational form of CRT. You can use the standard CRT formula along with the above computation mod $$25$$ to obtain the same result. In fact already $$23\equiv -1\pmod{\!4}$$ so we're done - we don't even need to apply the CRT formula.
Another quick way: $$\, 3^{\large 333} = 3(-1+10)^{\large 166} =\,\ldots\,$$ and Binomial expanding only the first $$2$$ terms survive $$\!\bmod 100$$.
Use Euler.
$$\phi(25) = 20$$ so $$8^{20}\equiv 1 \pmod {25}$$
So we need to figure out $$8^{17}\equiv 8^{-3}$$ which is which will be equivalent to $$k^3$$ where $$k$$ is the multiplicative inverse of $$8$$. $$8*3=24\equiv -1$$ so $$8*(-3)\equiv 1 \pmod{25}$$ and $$8^{17} \equiv (-3)^3 \equiv -27\equiv -2\equiv 23 \pmod {25}$$.
.....
Or back to the $$3$$
$$3^{333}=3^{320}3^{13}\equiv 3^{13}\pmod {25}$$.
As $$3^{20}\equiv 1$$ its a good guess that $$3^{10} \equiv \pm 1$$.
Many ways to do this but $$3^{10} = 3^3*3^3*3^3*3\equiv 27*27*27*3\equiv 2*2*2*3\equiv 24 \equiv -1 \pmod {20}$$. So $$3^{13}\equiv -3^3\equiv -27\equiv -2 \equiv 23$$.
.....OR.....
$$100 = 10^2$$
$$3^2 = 10-1$$
So $$3^{10} = (10-1)^5\equiv 50-1$$
$$3^{20} \equiv (50-1)^2 \equiv 1\pmod {100}$$.
So $$3^{333}\equiv 3^{13}\equiv 3^{10}*3^{3}\equiv (50-1)*27\equiv 50*(26+1)-27\equiv 50 - 27 \equiv 23\pmod{100}$$ | 2022-01-29T08:45:39 | {
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http://mathhelpforum.com/business-math/226059-investing-fundamentals.html | 1. ## Investing Fundamentals
John recently won a large sum of money which will be disbursed through two alternative payments:
• Option 1: A one off payment of RM100,000 that will be paid in 10 years.
• Option 2: Three uneven payments of RM10,000, RM30,000 and RM40,000 that will be paid in year 1, 5 and 10, respectively.
Assuming that the current market interest rate is 8 percent, which alternative would John choose?
2. ## Re: Investing Fundamentals
Originally Posted by mastermin346
John recently won a large sum of money which will be disbursed through two alternative payments:
• Option 1: A one off payment of RM100,000 that will be paid in 10 years.
• Option 2: Three uneven payments of RM10,000, RM30,000 and RM40,000 that will be paid in year 1, 5 and 10, respectively.
Assuming that the current market interest rate is 8 percent, which alternative would John choose?
FV - future value
PV - present value
r - interest rate
t - time
FV = PV(1+r)t
the value of option 1 in 10 years is $100,000 to find the value of option 2 we assume he invests all payments at 8% as soon as he receives them FV =$10000(1.08)(10-1) + $30000(1.08)(10-5) +$40000 = $19990.045 +$44079.84 + $40000 =$104069.89
option two has a larger overall future value than option 1.
3. ## Re: Investing Fundamentals
That was cool romsek. Thanks! This is a pretty tough problem, but you got it right. By the way, since we're talking about investments in here, you can check out here if you want to read more financial blogs and articles which may help you create mathematical problems.
4. ## Re: Investing Fundamentals
is that what you are? a spammer? just what we need more of....
5. ## Re: Investing Fundamentals
The other answer did the comparison of net future value of both investments at time period t = 10
That means how much each of the investments is worth at maturity
One may also get the same results by checking how much each of the investment is worth at present time period t=0
The comparison can then be made between the two present values to confirm that indeed the second investment is worth more than the first
PV = 100000 x (1+8%)^-10
PV = 100000 x 0.46319
PV = 46,319.35
Code:
T CF x PVIF CF x PVIF Present Value
1 10000 x (1+8%)^-1 10000 x 0.92593 9,259.26
5 30000 x (1+8%)^-5 30000 x 0.68058 20,417.50
10 40000 x (1+8%)^-10 40000 x 0.46319 18,527.74
NPV \$48,204.49
You can also find the future value of the investment by compounding the calculated net present value of 48,204.49 by a discount factor of (1+8%)^10 or 1.08^10
NFV = 48,204.49 x 1.08^10
NFV = 48,204.49 x 2.15892499727278669824
NFV = 104,069.89
6. ## Re: Investing Fundamentals
Investments may be held till maturity or they may be disposed off at an intermediate date between commencement of payments and terminal payment
Thus to evaluate an investment other than finding present and future worth of such an investment, an investor would also be interested in finding an intermediate net value of such an investment
For example, what if you decided to sell this investment at time period t=5, you would be keen to find the net value of such an investment at such time period.
Here is how
Code:
T C factor c x factor Intermediate Value
5 100,000 (10.8)^-5 100,000 x 0.680583197 68,058.32
T C factor c x factor Intermediate Value
5 10,000 (1.08)^4 10,000 x 1.36048896 13,604.89
5 30,000 1 30,000 x 1 30,000
5 40,000 (10.8)^-5 40,000 x 0.680583197 27223.33
Net intermediate value 70,828.22
Even if did want to dispose of the investment at time period t=5, the second investment is worth more than the first as the 100,000 due at t=10 is only worth 68,058.32 and the three payments from the second investment at t=5 are worth more as in 70,828.22
7. ## Re: Investing Fundamentals
Once we have the NIV - net intermediate value, we can confirm the NFV - net future value as shown in the other reply and the NPV - net present value as shown in the second last reply
To find NPV from NIV, we shall discount it by 5 years at 8%
Code:
NIV factor factor NIV x factor NPV
70,828.22 (1.08)^-5 0.680583197 70,828.22 x 0.680583197 48,204.49
To find NFV from NIV, we shall compound it by 5 years at 8%
Code:
NIV factor factor NIV x factor NFV
70,828.22 (1.08)^5 1.4693280768 70,828.22 x 1.4693280768 104,069.89
An NIV may be found for any time period between t=0 and t=N. The most useful examples of such intermediate values would be for the 1st qaurtile, 2nd quartile, and 3rd quartile NIV values.
The NIV is more useful of a measure as compared to NPV or NFV as we may be interested in finding the time period at which the investment yields the maximum NIV that may be totally different from the NFV or NPV for the full horizon of the investment.
8. ## Re: Investing Fundamentals
Originally Posted by mastermin346
John recently won a large sum of money which will be disbursed through two alternative payments:
• Option 1: A one off payment of RM100,000 that will be paid in 10 years.
• Option 2: Three uneven payments of RM10,000, RM30,000 and RM40,000 that will be paid in year 1, 5 and 10, respectively.
Assuming that the current market interest rate is 8 percent, which alternative would John choose?
Here's how I would do this problem, not having any real knowledge of "financial formulas":
The RM10,000 that he gets "in year 1" can be invested, at 8% interest, for 10 years (I assume compounded annually) and so will be worth $10000(1.08)^{10}= 21589.26$ at the end of the 10 years. The RM30,000 that he gets "in year 5" can be invested at 8% interest for 5 years and so will be worth $30000(1.08)^{5}= 44079.84$ at the end of the 10 years. That, added to the RM40,000 he gets "in year 10" makes a total of 21,589.26+ 44,079.94+ 40,000= RM105669.10 which is MORE than RM100,000 at the end of 10 years. | 2017-03-31T00:00:00 | {
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https://math.stackexchange.com/questions/3047048/does-there-exist-a-function-such-that-the-preimage-of-x-2-y-2-le | # Does there exist a function such that the preimage of $x ^ { 2 } + y ^ { 2 } \leq 1$ is the closed interval $[-1,1]?$
Does there exist a continuous function $$f : \mathbb { R } \rightarrow \mathbb { R } ^ { 2 }$$ such that the preimage of the closed unit disk $$x ^ { 2 } + y ^ { 2 } \leq 1$$ is the closed interval $$[ - 1,1 ] ?$$ the open interval $$( - 1,1 ) ?$$
To be honest, I don't really know how to go about this problem.
• en.wikipedia.org/wiki/Space-filling_curve – Will Jagy Dec 20 '18 at 1:39
• I'm fairly certain you can get a continuous function from $[-1,1]$ to the closed unit disc by an appropriate space filling curve. – MathematicsStudent1122 Dec 20 '18 at 1:39
• the preimage of a set $S$ being $[-1,1]$ is not the same thing as the image of $[-1,1]$ being $S$. space filling curves are super overkill – Rolf Hoyer Dec 20 '18 at 1:41
• @RolfHoyer how would you go about this without using space filling functions? We never went over those in my lecture so I doubt the answer would have to be related to that. – Mohammed Shahid Dec 20 '18 at 2:14
Space filling curves certainly work but that's quite excessive.
For a function $$f:\Bbb R\to\Bbb R^2$$ defined by $$f(x):=(f_1(x),f_2(x))$$, the preimage of $$D:=\{(x,y)\in\Bbb R^2:x^2+y^2\le 1 \}$$ is merely the set \begin{align} f^{-1}(D) &= \{ x\in\Bbb R: (f_1(x),f_2(x))\in D \} \\ &= \{ x\in\Bbb R: f_1(x)^2+f_2(x)^2\le 1 \}. \end{align}
By letting $$f(x)=(x,0)$$, we have $$f^{-1}(D)= \{ x\in\Bbb R: x^2\le 1 \} = [-1,1].$$
On the other hand, by requiring that $$f$$ be continuous, the preimage of $$D$$, which is a closed set, must also be closed. Since $$(-1,1)$$ is not closed, we cannot find a continuous function such that $$f^{-1}(D)=(-1,1)$$.
• Yep. The trick here is to realize that the preimage need not necessarily give the given set when the function is then applied to it - only that the result of such application be contained within the given set. – The_Sympathizer Dec 20 '18 at 6:13
f:R -> R×R, x -> (x,0) is continuous.
Let D be the closed unit disk.
The preimage of D by f is
f$$^{-1}$$(D) = { x : f(x) in D } = [-1,1].
• Good example, but you really should format it better. – zhw. Dec 20 '18 at 17:25 | 2019-11-19T18:59:49 | {
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http://math.stackexchange.com/questions/191607/summation-of-product-of-fibonacci-numbers?answertab=active | # Summation Of Product Of Fibonacci Numbers
Im trying to find out a general term for the following summation of products of fibonacci numbers:--
# $$\sum_{k=4}^{n+1} F_{k}F_{n+5-k}\; , n \geq 3$$
I tried using Binet's equation but I am getting stuck at a certain point. So, I would be very glad if someone could post an answer to my question with a detailed explanation.
Here are the first few values of the summation for different values of n :--
n = 3 , ans = 9
n = 4 , ans = 30
n = 5 , ans = 73
n = 6 , ans = 158
Note : I have used the usual fibonacci notation. i.e
$$F_0=0,\;F_1=1,\;F_2=1,\;F_3=2,\;...etc$$
EDIT
After reading the comments for this question I tried solving it to form a recurrence relation and this is what i ended up with :--
\begin{align*} G(n)&=\sum_{k=4}^{n+1} F_kF_{n+5-k}\; , n \geq 3\\ G(n)-G(n-1)&=\sum_{k=4}^{n+1} F_kF_{n+5-k}-\sum_{k=4}^{n} F_kF_{n+4-k}\\ &=F_{n+1}F_{4}+\sum_{k=4}^{n}F_kF_{n+3-k}\\ &=F_{n+1}F_{4}+F_{n}F_{3}+\sum_{k=4}^{n-1}F_kF_{n+3-k}\\ \\ &=F_{n+1}F_{4}+F_{n}F_{3}+G(n-2)\\ \\ G(n)&=G(n-1)+G(n-2)+F_{n+1}F_{4}+F_{n}F_{3}\\ \\ \end{align*}
Is this correct ? And how do I reduce it further ?
-
– Hagen von Eitzen Sep 5 '12 at 20:13
You are clearly getting $3F(n+1)$, but I don't have a proof. – Byron Schmuland Sep 5 '12 at 20:23
@ByronSchmuland Then use of generating function should nail it. – Sasha Sep 5 '12 at 20:24
Letting $a_n$ denote the sum in question you can show $a_n = a_{n-1} + a_{n-2} + F_{n+4}$ where $F_k$ is the $k$-th Fibonacci number. Maybe this helps. – Cocopuffs Sep 5 '12 at 20:50
@HagenvonEitzen : Thanks for pointing out those links. I checked them and tried to solve it according to one of those methods. I edited the question with my results. However, I got stuck after a certain point. Please do check it out and help me to complete it. – Arjun Datta Sep 6 '12 at 2:52
Using the closed form for the Fibonacci numbers, $$F_n=\frac{\phi^n-(-1/\phi)^n}{\sqrt{5}}\tag{1}$$ and Lucas numbers $$L_n=\phi^n+(-1/\phi)^n\tag{2}$$ we get \begin{align} F_iF_{n-i} &=\frac{\phi^i-(-1/\phi)^i}{\sqrt{5}}\frac{\phi^{n-i}-(-1/\phi)^{n-i}}{\sqrt{5}}\\ &=\frac{\phi^n+(-1/\phi)^n-(-1)^i\left(\phi^{n-2i}+(-1/\phi)^{n-2i}\right)}{5}\\ &=\frac{L_{n}-(-1)^iL_{n-2i}}{5}\tag{3} \end{align} To sum Lucas numbers, we use $(2)$ and the formula for the sum of a geometric series: \begin{align} \sum_{i=j}^{k-1}F_iF_{n-i} &=\sum_{i=j}^{k-1}\frac{L_{n+5}-(-1)^iL_{n-2i}}{5}\\ &=\color{#C00000}{\frac{k-j}{5}L_{n}}\\ &\color{#00A000}{-\frac15\sum_{i=j}^{k-1}\phi^{n}\left(-1/\phi^2\right)^i}\\ &\color{#0000FF}{-\frac15\sum_{i=j}^{k-1}(-1/\phi)^{n}\left(-\phi^2\right)^i}\\ &=\color{#C00000}{\frac{k-j}{5}L_{n}}\\ &\color{#00A000}{-\frac{\phi^{n}}{5}\frac{\left(-1/\phi^2\right)^k-\left(-1/\phi^2\right)^j}{\left(-1/\phi^2\right)-1}}\\ &\color{#0000FF}{-\frac{(-1/\phi)^{n}}{5}\frac{\left(-\phi^2\right)^k-\left(-\phi^2\right)^j}{\left(-\phi^2\right)-1}}\\ &=\color{#C00000}{\frac{k-j}{5}L_{n}}\\ &\color{#00A000}{+\frac15\frac{-(-1)^{n-k}(-1/\phi)^{2k-n-1}-(-1)^{j}\phi^{n-2j+1}}{\phi+1/\phi}}\\ &\color{#0000FF}{+\frac15\frac{+(-1)^{n-k}\phi^{2k-n-1}+(-1)^{j}(-1/\phi)^{n-2j+1}}{\phi+1/\phi}}\\ &=\frac{k-j}{5}L_{n}+\frac15\left((-1)^{n-k}F_{2k-n-1}-(-1)^jF_{n-2j+1}\right)\tag{4}\\ &=\frac{k-j}{5}(F_{n-1}+F_{n+1})+\frac15\left((-1)^{n-k}F_{2k-n-1}-(-1)^jF_{n-2j+1}\right)\tag{5} \end{align}
Applying $(4)$ to the current problem yields $$\sum_{k=4}^{n+1}F_kF_{n+5-k}=\frac{n-2}{5}(F_{n+6}+F_{n+4})-\frac25F_{n-2}\tag{5}$$ | 2016-02-14T17:20:48 | {
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https://math.stackexchange.com/questions/2876248/have-i-used-induction-correctly-in-this-proof-of-xy-implies-xnyn | # Have I used induction correctly in this proof of $x<y \implies x^n<y^n$?
A while ago I posted an attempt at a proof of $x<y \iff x^n<y^n$. It was pointed out that I hadn't actually used induction, and had instead done a direct proof. Below is the link to the question, so please do not mark this question as a duplicate, as this new question is about whether I have now done the proof by induction correctly, rather than accidentally reverting to a direct proof.
Is this proof of $x<y \iff x^n < y^n$ correct?
Also, be aware that I am, below, attempting to prove only that $x<y \implies x^n<y^n$.
Claim: $x<y \implies x^n<y^n$ for $x,y>0$ and $x,y,n \in \mathbb N$.
Proof:
Let $P(n)$ be the statement that $$x<y \implies x^n<y^n,$$ for $n\in \mathbb N$. It is clear that $P(n)$ holds for $n=1$ since $x<y \implies x<y$.
Assuming that $P(n)$ holds for some $n = k$, we see that this implies that $P(n)$ holds for $n=k+1$, as follows.
$$x<y \implies x^n<y^n$$
Since we know that $x<y$, if we multiply $x^n<y^n$ by $x$ we get that: $$x^{n+1} < xy^n,$$ from which it follows that $$x^{n+1} < y^{n+1}.$$
Thus $P(k)$ true $\implies$ $P(k+1)$ true, and so by induction we can prove the claim that $P(n)$ holds for all $n \in \mathbb N$.
• I think the proof is okay, I can't find anything wrong. – Anik Bhowmick Aug 8 '18 at 15:51
• probably $n\in \mathbb N$ in the claim – Exodd Aug 8 '18 at 15:51
• @Exodd I have made this change. – Benjamin Aug 8 '18 at 15:52
• @AnikBhowmick I feel a bit uncomfortable with the step where I multiply by $x$, since I (a) feel like I am ignoring the LHS and (b) am not sure how this exactly relies on the fact that P(n) is true. – Benjamin Aug 8 '18 at 15:53
• (a) Since $x>0$, it's completely okay. There is no fact of ignoring the LHS. (b) That's the statement of mathematical induction, right ?? If $P(K+1)$ is true whenever $P(K)$ is true, then $P(n)$ is true $\forall n \in \mathbb N$ !! Where is the ambiguity ?? – Anik Bhowmick Aug 8 '18 at 15:59
In my opinion you should work better out where and how you use the inductive claim (I. C.).
$x^{n+1}=x\cdot x^n\stackrel{I.C}{<}x\cdot y^n\stackrel{x<y}{<}y\cdot y^n=y^{n+1}$
• I am not sure I follow your superscript notation, could you possibly explain that in more detail? – Benjamin Aug 8 '18 at 15:54
• Sure: The superscript $I.C$ notes, that this estimation uses the inductive claim. The superscript $x<y$ notes, that we use for this estimation, that $x<y$ by assumption. Is it clear now? – Cornman Aug 8 '18 at 15:55
• Yes that makes it clear, so long as by Inductive Claim you mean assuming $P(n)$ is true for $k$? – Benjamin Aug 8 '18 at 15:57
• The inductive claim $P(n)$ is, that the estimation $x^n<y^n$ holds for arbitrary (but fixed) $n\in\mathbb{N}$. You do not need to involve $k$, as José Carlos Santos pointed out. – Cornman Aug 8 '18 at 15:59
It is correct. Two remarks, though:
1. There is no need to use two letters ($n$ and $k$). One is enough.
2. Indeed, it follows from $x^{n+1}<xy^n$ that $x^{n+1}<y^{n+1}$, but you did not say why. This is where you use the fact that $x<y$.
• Should I then have made more clear that because $x<y$ it is the case that $x^{n+1} < xy^n < y^{n+1}$? – Benjamin Aug 8 '18 at 15:56
• @Benjamin Yes, you should. – José Carlos Santos Aug 8 '18 at 15:57 | 2021-05-17T00:11:09 | {
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https://mathhelpboards.com/threads/difficult-integration-question.1124/ | Difficult integration question
Sudharaka
Well-known member
MHB Math Helper
Flipflop's question from Math Help Forum,
Determine the following:
a)$\int\frac{1}{e^x+2}\,dx$
b)$\int\frac{\sqrt{16x^2-9}}{x}\,dx$
How do I do part (b). I've been stuck at it for hours, even after looking at all the different types of integration formulas.
Hi Flipflop,
For the first one substitute, $$u=e^x+2$$. For the second one substitute, $$x=\dfrac{9}{16}\sec u$$. Hope you can continue.
Last edited:
soroban
Well-known member
Hello, flipflop!
$$\displaystyle \int\frac{dx}{e^x + 2}$$
Divide numerator and denominator by $$e^x:\;\int\frac{e^{-x}\,dx}{1 + 2e^{-x}}$$
$$\text{Let }u \,=\,1 + 2e^{-x} \quad\Rightarrow\quad du \,=\,-2e^{-x}dx \quad\Rightarrow\quad e^{-x}dx \,=\,-\tfrac{1}{2}du$$
$$\text{Substitute: }\:\int\frac{-\frac{1}{2}du}{u} \;=\;-\tfrac{1}{2}\int\frac{du}{u} \;=\; -\tfrac{1}{2}\ln|u| + C$$
$$\text{Back-substitute: }\:-\tfrac{1}{2}\ln(1 + 2e^{-x}) + C$$
checkittwice
Member
MHB Rules said:
14. Do not give premature explanations of hints. When you see that someone is already helping a
member and is waiting for the original poster to give feedback, please don't give more hints or a
full solution. Wait at least 24 hours so that the original poster has a chance to respond.
With this MHB rule in mind, post #2 should not have been created. | 2021-01-27T09:50:18 | {
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http://math.stackexchange.com/questions/143150/what-is-the-method-to-compute-binomnr-in-a-recursive-manner?answertab=votes | # What is the method to compute $\binom{n}{r}$ in a recursive manner?
How do you solve this?
Find out which recurrence relation involving $\dbinom{n}{r}$ is valid, and thus prove that we can compute $\dbinom{n}{r}$ in a recursive manner.
I appreciate any help. Thank You
-
Is $C(n,r)$ the binomial coefficient? Most of us here are more accustomed to writing it as $\dbinom{n}{r}$... – J. M. May 9 '12 at 18:12
@J.M. - Yes, the binomial coefficient - I don't know how to fix it though :\ – Adel May 9 '12 at 18:14
I don't quite get it, do you want to prove $\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}$? – Gigili May 9 '12 at 18:20
@Gigili - OK I need to update my question - Thank You Very Much – Adel May 9 '12 at 18:26
You know Pascal's triangle, I presume? – J. M. May 9 '12 at 18:31
There are many recurrence relations for $\dbinom{n}{r}$. One of the most commonly used one is the following. $$\binom{n+1}{r} = \binom{n}{r} + \binom{n}{r-1}.$$ There are many ways to prove this and one simple way is to look at $\displaystyle (1+x)^{n+1}$. We know that $$(1+x)^{n+1} = (1+x)^{n} (1+x).$$ Now compare the coefficient of $x^r$ on both sides.
The left hand the coefficient of $x^r$ is $$\dbinom{n+1}{r}.$$ On the right hand side, the $x^r$ term is obtained by multiplying the $x^r$ term of $(1+x)^n$ with the $1$ from $(1+x)$ and also by multiplying the $x^{r-1}$ term of $(1+x)^n$ with the $x$ from $(1+x)$. Hence, the coefficient of $x^r$ is given by the sum of these two terms from the right hand side i.e. $$\dbinom{n}{r} + \dbinom{n}{r-1}$$ As Rob and J.M have pointed out, these recurrences define the celebrated Pascal's triangle. A pictorial representation is shown above. Each row represents the value of $n$ starting from $0$. On a row, as we proceed from left to right, the different values of $r$ from $0$ to $n$ are hit. The arrows indicate how the $r^{th}$ value on the $(n+1)^{th}$ row is obtained as the sum of the $(r-1)^{th}$ and $r^{th}$ value on the $n^{th}$ row.
-
It might be useful to mention that that recurrence defines Pascal's Triangle. – robjohn May 9 '12 at 18:32
@robjohn Right. I was trying to draw a diagram of the Pascal's triangle. I will update the post soon. – user17762 May 9 '12 at 18:38
@J.M. Thanks. I was not aware of \dbinom. – user17762 May 9 '12 at 18:42
No worries; I learned it rather recently, too. – J. M. May 9 '12 at 18:44
From the familiar $\binom{n}{r}=\frac{n!}{r!(n-r)!}$, you can readily derive the formula $$\binom{n}{r}=\frac{n}{r}\binom{n-1}{r-1}.$$ This can be computationally and theoretically useful, particularly when we are interested in the expression $$\binom{n}{r}p^r(1-p)^{n-r}$$ that comes up when we are working with the Binomial Distribution.
-
Or $$\binom{n}{r}=\frac{n}{n-r}\binom{n-1}{r}$$ which may be quicker if $2r \gt n$ – Henry May 9 '12 at 18:55
To prove: $$\dbinom{n}{k}=\dbinom{n-1}{k-1}+\dbinom{n-1}{k}$$
We have: $$\dbinom{n-1}{k-1}=\frac{(n-1)!}{(k-1)!(n-k)!}=\frac{(n-1)!}{(k-1)!(n-k-1)!}\frac{1}{n-k}$$
$$\dbinom{n-1}{k}=\frac{(n-1)!}{(k)!(n-k-1)!}=\frac{(n-1)!}{(k-1)!(n-k-1)!}\frac{1}{k}$$
$$\dbinom{n-1}{k-1}+\dbinom{n-1}{k}=\frac{(n-1)!}{(k-1)!(n-k-1)!}(\frac{1}{n-k}+\frac{1}{k})$$
$$=\frac{(n-1)!}{(k-1)!(n-k-1)!}\frac{n}{k(n-k)}$$ $$=\frac{n!}{k!(n-k)!}$$
$$=\dbinom{n}{k}$$ | 2015-11-29T11:08:10 | {
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https://byjus.com/question-answer/if-the-sum-to-infinity-of-the-series-1-4x-7x-2-10x-3-cdots/ | Question
# If the sum to infinity of the series $$1+4x+7x^2+10x^3+\cdots$$ is $$\dfrac{35}{16}$$, then $$x=$$
A
15
B
25
C
37
D
17
Solution
## The correct option is A $$\displaystyle \frac{1}{5}$$Let$${ S }_{ \infty }=1+4x+7{ x }^{ 2 }+10{ x }^{ 3 }+...\infty$$ $$...(1)$$Now, multiply by $$x$$ throughout in eqution $$(1)$$; we get$$x{ S }_{ \infty }=x+4{ x }^{ 2 }+7{ x }^{ 3 }+10{ x }^{ 4 }+...\infty$$ $$...(2)$$Subtracting $$(2)$$ from $$(1)$$; we get$${S}_{ \infty }-x{ S }_{ \infty }=(1+4x+7{ x }^{ 2 }+10{ x }^{ 3 }+...\infty )-(x+4{ x }^{ 2 }+7{ x }^{ 3 }+10{ x }^{ 4 }+...\infty )$$$$\Rightarrow(1-x){ S }_{ \infty }=1+4x+7{ x }^{ 2 }+10{ x }^{ 3 }+...\infty -x-4{ x}^{ 2 }-7{ x }^{ 3 }-10{ x }^{ 4 }-...\infty$$$$\Rightarrow (1-x){ S }_{ \infty }=1+3x+3{ x }^{ 2 }+3{ x }^{ 3 }+...\infty$$Notice that the series $$3x+3{ x }^{ 2 }+3{ x }^{ 3 }+...\infty$$ is geometric series with the first term $$a=3x$$ and the common ratio $$r=x$$.Now, use the formula for the sum of an infinite geometric series.$$\Rightarrow (1-x){ S }_{ \infty }=1+\dfrac { 3x }{ (1-x) }$$, for$$\left| x \right| <1$$$$\Rightarrow (1-x){ S }_{ \infty }=\dfrac { 1+2x }{ (1-x) }$$, for$$\left| x \right| <1$$Given that, $${ S }_{ \infty }=\dfrac { 35 }{ 16 }$$, substitute for $${ S }_{ \infty }$$ in the above equation; we get$$(1-x)\dfrac { 35 }{ 16 } =\dfrac { 1+2x }{ (1-x) }$$$$\Rightarrow35{ (1-x) }^{ 2 }=16(1+2x)$$$$\Rightarrow 35(1-2x+{ x }^{ 2 })=16+32x$$$$\Rightarrow 35{ x }^{ 2 }-102x+19=0$$$$\Rightarrow (7x-19)(5x-1)=0$$$$\Rightarrow x=\dfrac { 19 }{ 7 }$$ or $$x=\dfrac { 1 }{ 5 }$$But $$x\neq \dfrac { 19 }{ 7 }$$, because for infinity series, $$\left| x \right| <1$$.Therefore, $$x=\dfrac { 1 }{ 5 }$$.Maths
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http://math.stackexchange.com/questions/435079/the-value-of-lim-limits-n-rightarrow-inftyn2-left-int-01-left1xn-right | # the value of $\lim\limits_{n\rightarrow\infty}n^2\left(\int_0^1\left(1+x^n\right)^\frac{1}{n} \, dx-1\right)$
This is exercise from my lecturer, for IMC preparation. I haven't found any idea.
Find the value of
$$\lim_{n\rightarrow\infty}n^2\left(\int_0^1 \left(1+x^n\right)^\frac{1}{n} \, dx-1\right)$$
Thank you
-
By integration by parts,
\begin{align*} \int_{0}^{1} (1 + x^{n})^{\frac{1}{n}} \, dx &= \left[ -(1-x)(1+x^{n})^{\frac{1}{n}} \right]_{0}^{1} + \int_{0}^{1} (1-x)(1 + x^{n})^{\frac{1}{n}-1}x^{n-1} \, dx \\ &= 1 + \int_{0}^{1} (1-x) (1 + x^{n})^{\frac{1}{n}-1} x^{n-1} \, dx \end{align*}
so that we have
\begin{align*} n^{2} \left( \int_{0}^{1} (1 + x^{n})^{\frac{1}{n}} \, dx - 1 \right) &= \int_{0}^{1} n^{2} (1-x) (1 + x^{n})^{\frac{1}{n}-1} x^{n-1} \, dx. \end{align*}
Let $a_{n}$ denote this quantity. By the substitution $y = x^{n}$, it follows that
\begin{align*} a_{n} &= \int_{0}^{1} n \left(1-y^{1/n}\right) (1 + y)^{\frac{1}{n}-1} \, dy = \int_{0}^{1} \int_{y}^{1} t^{\frac{1}{n}-1} (1 + y)^{\frac{1}{n}-1} \, dtdy \end{align*}
Since $0 \leq t (1 + y) \leq 2$ and $\int_{0}^{1} \int_{y}^{1} t^{-1}(1+y)^{-1} \, dtdy < \infty$, an obvious application of the dominated convergence theorem shows that
\begin{align*} \lim_{n\to\infty} a_{n} = \int_{0}^{1} \int_{y}^{1} \frac{dtdy}{t(1+y)} &= - \int_{0}^{1} \frac{\log y}{1+y} \, dy \\ &= \sum_{m=1}^{\infty} (-1)^{m} \int_{0}^{1} y^{m-1} \log y \, dy = \sum_{m=1}^{\infty} \frac{(-1)^{m-1}}{m^2} = \frac{\pi^2}{12}. \end{align*}
-
Nice work! Now it's proven :) – Start wearing purple Jul 3 '13 at 10:29
Very nice! The value of $\pi^2/12 = .822...$ falls within my bounds of .75 to .875 which is comforting. – marty cohen Jul 3 '13 at 13:33
@sos440 here is a shortcut I noticed. By DCT $\lim_{n\to\infty}\int_{0}^{1} n \left(1-y^{1/n}\right) (1 + y)^{\frac{1}{n}-1} \, dy=\int_{0}^{1} \lim_{n\to\infty} \frac{\left(1-y^{1/n}\right)}{1/n} (1 + y)^{\frac{1}{n}-1} \, dy=-\int_{0}^{1}\frac{\log y}{1+y}\ dy=\frac{\pi^2}{12}$ – Chris's sis the artist Jul 3 '13 at 19:30
@Chris'swisesister, that's my first approach but I was unsure if we can find a dominating function for $(1 - y^{1/n})/(1/n)$. This is why I introduced double integral. – Sangchul Lee Jul 4 '13 at 3:24
I get an answer that differs from that of user17762. This is because the error term is not one term of order $\frac{x^{2n}}{n^2}$ but a number of such terms.
I get that the limit is between 3/4 and 7/8, but only have an infinite series for the value.
My complete analysis follows.
$(1+x^n)^{1/n} =\sum_{k=0}^{\infty} \binom{1/n}{k}x^{kn}$.
We first look at $\binom{1/n}{k}$.
\begin{align} \binom{1/n}{k} &=\frac1{k!}\prod_{i=0}^{k-1}(\frac{1}{n}-i)\\ &=\frac1{k!n^k}\prod_{i=0}^{k-1}(1-in)\\ &=\frac{(-1)^k}{k!n^k}(-1)\prod_{i=1}^{k-1}(in-1)\\ &=\frac{(-1)^{k+1}}{k!n^k}\prod_{i=1}^{k-1}(in-1)\\ \end{align}
so
\begin{align} \big|\binom{1/n}{k}\big| =\frac1{k!n^k}\prod_{i=1}^{k-1}(in-1) <\frac{1}{k!n^k}\prod_{i=1}^{k-1}(in) =\frac{n^{k-1}(k-1)!}{k!n^k} =\frac1{kn} \end{align}
and
\begin{align} \frac{\binom{1/n}{k+1}}{\binom{1/n}{k}} &=\frac{(-1)^{k+2}}{(k+1)!n^{k+1}}\frac{k!n^k}{(-1)^{k+1}}\frac{\prod_{i=1}^{k}(in-1)}{\prod_{i=1}^{k-1}(in-1)}\\ &=\frac{-1}{(k+1)n}(kn+1)\\ &=-\frac{kn+1}{kn+n}\\ \end{align}
We now look at $\int_0^v (1+x^n)^{1/n}\, dx$ to see what happens as $v \to 1$.
\begin{align} \int_0^v (1+x^n)^{1/n}\, dx &=\sum_{k=0}^{\infty} \binom{1/n}{k} \int_0^v x^{kn}\, dx\\ &=\sum_{k=0}^{\infty} \binom{1/n}{k} \frac{v^{kn+1}}{kn+1}\\ &=v+\frac{v^{n+1}}{n(n+1)}+\sum_{k=2}^{\infty} \binom{1/n}{k} \frac{v^{kn+1}}{kn+1}\\ &=v+\frac{v^{n+1}}{n(n+1)}+v^{2n+1}\sum_{k=2}^{\infty} \binom{1/n}{k} \frac{v^{(k-2)n}}{kn+1}\\ \end{align}
This means that the terms in $\sum_{k=2}^{\infty} \binom{1/n}{k} \frac{v^{(k-2)n}}{kn+1}$ decrease in absolute value and, since they alternate in sign, the series converges. and converges even at $v=1$ because of the $\frac1{kn+1}$.
Let $f(v, n) = \sum_{k=2}^{\infty} \binom{1/n}{k} \frac{v^{(k-2)n}}{kn+1}$.
Since $\big|\binom{1/n}{k} \frac{1}{kn+1}\big| < \frac1{kn(kn+1)}$, $|f(v, n)| <\sum_{k=2}^{\infty} \frac1{kn(kn+1)} <\frac1{n^2}\sum_{k=2}^{\infty} \frac1{k^2} <\frac1{n^2}$.
The first term of $f(v, n)$ is $\binom{1/n}{2}\frac{1}{2n+1} =\frac{(1/n)(1/n-1)}{2}\frac{1}{2n+1} =-\frac{n-1}{2n^2(2n+1)}$ and this is between $-\frac1{8n^2}$ and $-\frac1{4n^2}$ for $n > 3$.
Therefore the first two terms of the expansion of $\int_0^v (1+x^n)^{1/n}\, dx$ are both of order $1/n^2$, so we have to consider the whole sum, not just the first term (after $1$).
Since $\int_0^v (1+x^n)^{1/n}\, dx =v+\frac{v^{n+1}}{n(n+1)}+v^{2n+1}f(v, n)$ and all the terms exist as $v \to 1$, $\int_0^1 (1+x^n)^{1/n}\, dx -1-\frac{1}{n(n+1)}=f(1, n)$.
Since $f(1,n)$ is between $-\frac1{8n^2}$ and $-\frac1{4n^2}$, $-\frac1{4n^2} <\int_0^1 (1+x^n)^{1/n}\, dx -1-\frac{1}{n(n+1)} <-\frac1{8n^2}$, $-\frac1{4} <n^2\big(\int_0^1 (1+x^n)^{1/n}\, dx -1\big)-\frac{n}{n+1} <-\frac1{8}$ so $1-\frac1{4} <\lim_{n \to \infty} n^2 \big(\int_0^1 (1+x^n)^{1/n}\, dx -1\big) < 1-\frac1{8}$.
-
+1. Actually, I have found that the exact value is equal to $\pi^2/12$. – Start wearing purple Jul 3 '13 at 7:58
Mathematica evaluates the integral to $$\int_0^{1}(1+x^n)^{1/n}dx={}_2F_1\left(-\frac{1}{n},\frac1n,1+\frac1n;-1\right).\tag{1}$$ Next, let us write the standard series representation for the hypergeometric function $$_2F_1(a,b,c;t)=\sum_{k=0}^{\infty}\alpha_kt^k,\qquad \alpha_k=\frac{\Gamma(a+k)\Gamma(b+k)\Gamma(c)}{k!\,\Gamma(a)\Gamma(b)\Gamma(c+k)}.$$ Now an easily verified claim: as $n\rightarrow\infty$, for the parameters as in (1), we have $\alpha_0=1$ and $$\alpha_k\sim-\frac{1}{k^2n^2}+O(n^{-3}).$$ Hence we obtain \begin{align} \lim_{n\rightarrow\infty}n^2\left(\int_0^{1}(1+x^n)^{1/n}dx-1\right)=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k^2}=\frac{\pi^2}{12}, \end{align} which is also confirmed by numerical calculation with $n\sim 10^4-10^8$.
- | 2015-10-09T01:41:55 | {
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https://math.stackexchange.com/questions/1619779/is-a-certain-subset-of-m-2-mathbbr-closed | # Is a certain subset of $M_2(\mathbb{R})$ closed?
I am trying to prove that the set M of all matrices in the normed linear space $M_2(\mathbb{R})$ such that both eigen values are real is closed (under metric topology; metric induced by the norm).
Following is my attempt. I recall the property that finite dimensional subspaces of normed spaces are complete. And since complete metric spaces are closed, if I can prove it is a vector subspace of $M_2(\mathbb{R})$, I am done. But I don't know if M is closed under "vector" addition. Is it true that the sum of two "real- eigenvalued" matrices have real eigen values?
$$\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} + \begin{pmatrix} 0 & 0 \\ -1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$$ So it's not a subspace. But to show it's closed, you can produce a continuous function $f : M_2(\mathbb{R}) \to \mathbb{R}$ such that $M = f^{-1}([0,\infty))$. Hint: look at the characteristic polynomial of a matrix $A$, and think about the discriminant of a quadratic equation.
• +1 good counterexample. I'd be interested to see a solution to the main question though. – goblin Jan 20 '16 at 17:04
• thank you. I have obtained f = (tr(A))2 - 4*det(A) which is continuous and would give me $f^{-1}([0,\infty))$ as M. – user166305 Jan 20 '16 at 17:20
• @user166305: Great! Well done. – Nate Eldredge Jan 20 '16 at 19:25
It's not true, as Nate Eldredge counterexemple show.
For your original question, you can look at it this way :
the eigenvalue of $M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ are the roots of the polynomial
$$P(x) = \begin{vmatrix} a-x & b \\ c & d-x \end{vmatrix}$$
This is a 2nd degree polynomial, so it has real roots if the discriminant is positive.
Hence, if $\Delta(a,b,c,d)$ is the discriminant of $P$, the set of matrix with real eigenvalues is
$$\left\{ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \mid (a,b,c,d)\in \Bbb R, \Delta(a,b,c,d) \geq 0 \right\}$$
So if you can prove that $\Delta(a,b,c,d)$ is continuous, you've won (do you see why?)
• discriminant turns out to be Tr(M)^2 - 4*det(M) which is continuous as it is a linear combination of continuous functions (Tr^2 is composition of two continuous functions). Thus my original set is the inverse image of [0,\infty) under this cont map and thus closed – user166305 Jan 20 '16 at 17:26 | 2019-11-17T04:35:43 | {
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https://math.stackexchange.com/questions/3357751/probability-of-winning-games-at-tournament | # Probability of winning games at tournament
I know it's a simple problem but apparently I am doing something wrong: The probability of winning every single game at a tournament is 0.4. There is only win and lose - no draw. Find the probability of winning exactly 2 games by playing at most 6 games.
Since winning and losing are mutually exclusive, the probability of losing a game is 0.6. The required probability is: Probability of playing 2 games and winning both, +
Probability of playing 3 games and winning 2, +
Probability of playing 4 games and winning 2, +
Probability of playing 5 games and winning 2, +
Probability of playing 6 games and winning 2.
First one is $$(0.4)^2$$
Second is $${3\choose 2}(0.4)^2(0.6)^1$$
then $${4\choose 2}(0.4)^2(0.6)^2$$
$${5\choose 2}(0.4)^2(0.6)^3$$
$${6\choose 2}(0.4)^2(0.6)^4$$
I am getting a total of 1.45024 and obviously it is wrong.
Correct answer is 0.76688 but I don't know what I am doing wrong!
Many thanks!
The error in your solution is that you are counting some cases several times. So, the playing 2 and winning 2 is included in all of your other cases.
Split the cases up as
Win first 2
Win 1 out of the first 2 then win the third
Win 1 out of the first 3 then win the fourth
...
See what you get now.
Well, my solution is $$0.76672$$. That is $$0.00016$$ less than yours :)
Obviously, there is no point continuing playing if you have already won 2 games. But in your answer, for example, "probability of playing 3 games and winning 2" you have the probability of winning the first two games and continuing, losing the 3rd game:
$${3\choose 2}(0.4)^2(0.6)^1$$ is $$3\cdot(0.4)^2(0.6)^1$$ that means $$(0.4)(0.4)(0.6)+(0.4)(0.6)(0.4)+(0.6)(0.4)(0.4)$$ but of course we shouldn't add $$(0.4)(0.4)(0.6)$$ as we have already won 2 games.
So,
Probability of playing 2 games and winning 2 is $$1\cdot(0.4)^2$$
Probability of playing 3 games and winning 2 is $$2\cdot(0.4)^2(0.6)^1$$
Probability of playing 4 games and winning 2 is $$3\cdot(0.4)^2(0.6)^2$$
Probability of playing 5 games and winning 2 is $$4\cdot(0.4)^2(0.6)^3$$
Probability of playing 6 games and winning 2 is $$5\cdot(0.4)^2(0.6)^4$$
We add all these and we have $$0.76672$$.
I don't know the distribution but, to understand why, for example, I multiplied by 4 in "probability of playing 5 games and winning 2", these are the outcomes: (W: Win, L: Lose)
W L L L W, L W L L W, L L W L W, L L L W W. (Notice that in any case, we always win the last game)
Same approach is for every case. See for yourself! Great problem! | 2022-08-17T00:33:55 | {
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https://math.stackexchange.com/questions/1981274/why-is-nabla-x-xt-a-x-2-a-x | # Why is $\nabla_x x^T A x = 2 A x$?
Let $f: \mathbb{R}^n \rightarrow \mathbb{R}$ and $f(x) = x^T A x$. I will denote $\nabla_x$ or $\nabla$ as the gradient to some vector-valued variable and $\nabla^2$ or $H$ as the Hessian.
The lecturer postulated that $\nabla f(x) = 2 A x$, and that $\nabla^2 f(x) = 2A$.
It's not immediately clear to me that this is true. What I thought of, was that $\nabla f(x)$ always yields a column vector, and that therefore $\nabla f(x) = 2 A x$. But this feels more like a trick (to remember it) and not like a proof to me.
How does one derive $\nabla_x x^T A x = 2 A x$? Why can't it be $2 (x^T A)^T = A^T x$?
• Not in general, but $A$ only occurs in the quadratic form and nowhere else. Therefore the antisymmetric part cancels (right?), so you can take $B = \frac{1}{2}(A+A^T)$. However, if we leave $A$ as it is, it's $Ax \neq (x^T A)^T$ if it's not symmetric which confuses me. Oct 23, 2016 at 12:35
• For an in-depth understanding of matricial (and graphical) understanding ofquadratic forms, I advise you this document (www2.econ.iastate.edu/classes/econ501/Hallam/documents/…). Oct 23, 2016 at 13:02
You need $A$ to be symmetric for that.
$x^TAx=\langle Ax, x\rangle=:f(x).$
$f(x+h)=\langle A(x+h), x+h \rangle=\langle Ax, x \rangle + \langle Ax,h \rangle +\langle x,Ah \rangle+ \langle Ah,h \rangle.$
Therefore, the derivative is given by $f'_x=\langle Ax, \cdot \rangle +\langle x, A \cdot \rangle.$
Because $A$ is symmetric, (*) $$f'_x=\langle 2Ax, \cdot \rangle.$$
Since the definition of $\nabla_x f$ is the vector such that $f'_x= \langle \nabla_x f, \cdot \rangle$ (which exists and is unique by Riesz), we get $$\nabla_x f= 2Ax.$$
(*) Note that $f'_x=\langle Ax, \cdot \rangle +\langle x, A \cdot \rangle=\langle (A+A^T)x, \cdot\rangle .$ Therefore, $f'_x=\langle 2Ax, \cdot \rangle$ if and only if $2A=A+A^T$, which occurs if and only if $A$ is symmetric.
• So symmetry of $A$ is an absolutely necessary condition for $\nabla_x f = 2Ax$? Oct 23, 2016 at 12:43
• @Jasper Yes. See the edit. Oct 23, 2016 at 12:48
• > Therefore, the derivative is given by 𝑓′𝑥=⟨𝐴𝑥,⋅⟩+⟨𝑥,𝐴⋅⟩. can someone explain how this was found? Dec 23, 2021 at 0:28 | 2022-09-26T18:45:18 | {
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https://math.stackexchange.com/questions/3813522/conceptual-reason-for-why-the-volume-of-an-ocahedron-is-four-times-the-volume-of | # Conceptual reason for why the volume of an ocahedron is four times the volume of a tetrahedron
The image below shows that a regular octahedron can be scaled by a factor of $$2$$ (resulting in a $$2^3$$ factor in volume) and decomposed as six octahedra and eight tetrahedra.
If $$V_o$$ and $$V_t$$ respectively represent the volumes of a regular octahedron and a regular tetrahedron with the same edge lengths, then $$2^3V_o = 6V_o + 8V_t,$$ and solving for $$V_o$$ yields $$V_o = 4V_t$$.
Image from Wikipedia
Is there a conceptual reason why the volume of an octahedron is $$4$$ times the volume of a tetrahedron that doesn't rely on a decomposition like this? For example, is there a way that you can chop up four tetrahedra to fit them into an octahedron?
Equally useful, is there some nice way to see that a square-based pyramid has twice the volume of a tetrahedron? Perhaps integrating as slices of equilateral triangles vs slices of squares?
### A higher dimensional analog.
A "nice to have" quality of the answer would be if it generalizes to the higher dimensional case. If $$V_o^{(n)}$$ and $$V_t^{(n)}$$ denote the (hyper)volumes of the $$n$$-dimensional cross-polytope and $$n$$-dimensional simplex respectively, then
$$V_o^{(n)} = \frac{\sqrt{2^n}}{n!} \text{ and } V_t^{(n)} = \frac{\sqrt{n+1}}{n!\sqrt{2^n}} \text { with ratio } \frac{V_o^{(n)}}{V_t^{(n)}} = \frac{2^n}{\sqrt{n+1}}.$$
Is there a conceptual reason why this relationship is "nice"?
• When you say "an octahedron" and "a tetrahedron" what is the implied normalization? Do you require that they both have edges of length $1$? (Anyway, I think your image is already quite a nice explanation!) – Qiaochu Yuan Sep 4 '20 at 4:36
• @QiaochuYuan, yes, all edges of length $1$ (or equivalently, all edges of length $\ell$, I suppose). – Peter Kagey Sep 4 '20 at 5:38
• @QiaochuYuan, I agree that the image is a slick insight and the argument is elementary and convincing, but it somehow feels a little convoluted—perhaps since it requires solving for $V_o$ in the equation. I'm hoping that another explanation will scratch whatever itch I have. – Peter Kagey Sep 4 '20 at 5:42
• Instead of doubling an octahedron, it is a little bit simpler to double a tetrahedron. This results in one octahedron in the centre, and four tetrahedra, which together should have the same volume as eight tetrahedra. – Jaap Scherphuis Sep 4 '20 at 8:14
• Joining midpoints of a tetrahedron's edges yields an octahedron. The four smaller tetrahedra outside the octahedron but inside the larger tetrahedron are an eighth the size, so $O+4T=8T$. (This is an alternate decomposition proof.) – runway44 Sep 17 '20 at 8:48
Join the vertices of the unit-sided octahedron with its centre. That will divide it into eight regular pyramids, having the faces of the octahedron as bases and three lateral edges with length $$1/\sqrt2$$.
Pythagora's theorem gives then a height of $$1/\sqrt6$$ for these eight pyramids, whereas the height of a regular unit-sided tetrahedron is $$2/\sqrt6$$. The volume of the tetrahedron is then double that of each regular pyramid in the octahedron, which explains why the volume of the octahedron is four times the volume of the tetrahedron.
Consider a cube with a tetrahedron inside it sharing four of its vertices. The cube dissects into this tetrahedron, and four identical triangular pyramids. Look at this picture of the cube standing on one vertex:
A body diagonal of the cube, vertical in this picture, is split into three equal parts by the heights of the vertices. This shows that the inner tetrahedron has twice the height of each small pyramid, and hence twice the volume. Eight of those small pyramids can form an orctahedron, so the tetrahedron is a quarter of the volume of the octahedron.
I don't think this can be generalised to higher dimensions in the direction you are looking for.
If you inscribe two tetrahedra in a cube, their overlap is an octahedron:
Equivalently, the octahedron may be constructed by joining midpoints of a tetrahedron's edges. Notice that within this (say, red) tetrahedron, outside the octahedron $$O$$ there are four smaller tetrahedra $$T$$. The side lengths of these smaller $$T$$s are half the side-length of the original, red tetrahedra, so the red one has eight times the volume, and so $$8T=4T+O$$.
This is another decomposition proof, but it's more direct. 3D only though.
• This is a very nice purely geometric argument! It is relatively straight forward to argue with cartesian coordinates and integration, but these would give little to no geometric insight. – M. Winter Nov 1 '20 at 10:02 | 2021-07-24T07:06:18 | {
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http://mathhelpforum.com/algebra/80361-factoring-quadratics.html | For example:
$
6x^2 +11x -35
$
I've been doing this by trial and error but it takes too long. I tried googling "factoring quadratics" but when it comes to examples like this, they just use the quadratic equation to find the values of x assuming the whole equation was equal to zero in the first place. I don't think this helps me.
I need the answer in the form (....)(....) and I don't think the answer on, for example,
Solving Quadratic Equations: Examples, does that
. What I need is the strategy to find this. I know I'm going to need +7. -5 or -7, +5 at the ends but from there I just have to try so many combinations.
2. Try this lesson on factoring quadratics.
(The posted link was for solving equations, rather than factoring expressions.)
3. Originally Posted by stapel
Try this lesson on factoring quadratics.
(The posted link was for solving equations, rather than factoring expressions.)
yes this looks exactly like what I am looking for
4. There is a quadratic formula, to find the roots. Google Quadratic formula. I personally prefer factoring tho
5. Originally Posted by TYTY
For example:
6x2 + 11x – 35
I've been doing this by trial and error but it takes too long. I tried googling "factoring quadratics" but when it comes to examples like this, they just use the quadratic equation to find the values of x assuming the whole equation was equal to zero in the first place. I don't think this helps me.
I need the answer in the form (....)(....) and I don't think the answer on, for example, Solving Quadratic Equations: Examples, does that. What I need is the strategy to find this. I know I'm going to need +7. -5 or -7, +5 at the ends but from there I just have to try so many combinations.
Hi TYTY,
Here's one strategy you might find helpful.
$6x^2+11x-35$
Multiply the leading coefficient (6) by your constant (-35) to get -210.
Next, try to come up with 2 factors that multiply to get -210 and sum to +11.
You may have to fiddle around a little to find them, but this one came to me right away. It's +21 and -10.
Replace the middle coefficient (-11) with these two numbers:
$6x^2+21x-10x-35$
Now, group the first two terms and the last 2 terms and factor out the greatest common factor for each.
$3x(2x+7)-5(2x+7)$
Use the distributive property to finish up.
$(3x-5)(2x+7)$
6. Originally Posted by masters
Hi TYTY,
Here's one strategy you might find helpful.
$6x^2+11x-35$
Multiply the leading coefficient (6) by your constant (-35) to get -210.
Next, try to come up with 2 factors that multiply to get -210 and sum to +11.
You may have to fiddle around a little to find them, but this one came to me right away. It's +21 and -10.
Replace the middle coefficient (-11) with these two numbers:
$6x^2+21x-10x-35$
Now, group the first two terms and the last 2 terms and factor out the greatest common factor for each.
$3x(2x+7)-5(2x+7)$
Use the distributive property to finish up.
$(3x-5)(2x+7)$
Excellent explanation. | 2016-08-25T20:20:24 | {
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https://math.stackexchange.com/questions/1486792/eigenvectors-for-an-eigenvalue-of-0 | # Eigenvectors for an eigenvalue of $0$
Without calculation, find one eigenvalue and two linearly independent eigenvectors of $A = \begin{bmatrix} 2 && 2 && 2 \\ 2 && 2 && 2 \\ 2 && 2 && 2 \end{bmatrix}$.
This matrix is non-invertible because its columns are linearly dependent. So the number $0$ is an eigenvalue of $A$. Eigenvectors for the eigenvalue $0$ are solutions of $Ax=0$ and therefore have entries that produce a linear independence relation among the columns of A. Any nonzero vector in ($\Bbb R^3$) whose entries sum to $0$ will work. Find any two vectors that are not multiples; for instance,
$\begin{bmatrix} 1 \\ 1 \\ -2 \end{bmatrix}$ and $\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}$.
I do not understand how the first vector is obtained.
So far I've done.
$\sim \begin{bmatrix} 1 && 1 && 1 \\ 0 && 0 && 0 \\ 0 && 0 && 0 \end{bmatrix}$
$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -x_2-x_3 \\ x_2 \\ x_3 \end{bmatrix} = x_2\begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix} + x_3\begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}$
How do you obtain $\begin{bmatrix} 1 \\ 1 \\ -2 \end{bmatrix}$?
Why is it that any nonzero vector in ($\Bbb R^3$) whose entries sum to $0$ will work?
• "why is it that any nonzero vector whose entries sum to $0$ will work?" Try multiplying such a vector to it and see what happens. Since all entries in $A$ are equal, you could simply factor out a common $2$. "how do you obtain..." By setting $x_2=1$ and $x_3=-2$, your two vectors you found add up to the vector in question. Since it was stated to do this without calculation, several answers are possible. Your two vectors for example are also correct answers. – JMoravitz Oct 19 '15 at 1:36
• Why must the entries sum to $0$? – user1766555 Oct 19 '15 at 1:40
• "Without calculation"? That's impossible. Maybe they mean "without explicitly solving the characteristic polynomial and following matrix equation systems". – mathreadler Oct 19 '15 at 1:58
• @mathreadler, no calculation is necessary to find all eigenvalues and eigenvectors of any multiple of the $Ones$ matrix beyond simple addition. – JMoravitz Oct 19 '15 at 2:00
"Why must the entries sum to $0$"
Let $A=\begin{bmatrix}a_{1,1}&a_{1,2}&a_{1,3}\\ a_{2,1}&a_{2,2}&a_{2,3}\\a_{3,1}&a_{3,2}&a_{3,3}\end{bmatrix}$ and $x = \begin{bmatrix} x_{1}\\x_2\\x_3\end{bmatrix}$
By definition of matrix product, you have
$Ax = \begin{bmatrix}a_{1,1}x_1 + a_{1,2}x_2 + a_{1,3}x_3\\a_{2,1}x_1 + a_{2,2}x_2 + a_{2,3}x_3\\a_{3,1}x_1 + a_{3,2}x_2 + a_{3,3}x_3\end{bmatrix}$
Note what happens when all entries of $A$ are identical and nonzero:
$\dots = \begin{bmatrix} a(x_1+x_2+x_3)\\a(x_1+x_2+x_3)\\a(x_1+x_2+x_3)\end{bmatrix} = a\begin{bmatrix}x_1+x_2+x_3\\x_1+x_2+x_3\\x_1+x_2+x_3\end{bmatrix}$
If this were to equal the zero vector (which is implied since these are eigenvectors for the eigenvalue of zero) then that means that every entry above must be zero. That is exactly the same as saying that $x_1+x_2+x_3=0$
On the other hand, supposing that $x_1+x_2+x_3=k\neq 0$, you would have $Ax = a\begin{bmatrix}k\\k\\k\end{bmatrix}\neq \overrightarrow{0}=0x$ and so $x$ would not be an eigenvector for the eigenvalue of zero.
In general, any nonzero multiple of the $Ones_{n\times n}$ matrix, say $a\cdot Ones_{n\times n}$ (matrix where all entries are $a$) will have an eigenvalue of zero and every eigenvector for zero will satisfy the relation that the sum of the entries is zero. We know that there are no others by a rank-nullity argument and that the remaining eigenvalue is equal to the trace (sum of the diagonal).
Furthermore, the other eigenvalue will necessarily be $a\cdot n$ with eigenvector $\begin{bmatrix}1\\1\\\vdots\\1\end{bmatrix}$.
• Why does it matter that the entries must sum to $0$ when $\lambda x = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$ due to lambda being $0$? – user1766555 Oct 19 '15 at 1:47
• @user1766555 Suppose that they didn't sum to zero, but instead summed to $k$. Then going through the matrix multiplication, you see that $Ax = \begin{bmatrix}2k\\2k\\2k\end{bmatrix}\neq \overrightarrow{0}=0x$. This could also be seen directly from a rank-nullity argument and theorems about eigenvalues (namely that the sum of the eigenvalues must be equal to the sum of the diagonal). – JMoravitz Oct 19 '15 at 1:51
• So due to $Ax = \begin{bmatrix}2k\\2k\\2k\end{bmatrix}\neq \overrightarrow{0}$, that also means $Ax \neq \lambda x$? – user1766555 Oct 19 '15 at 1:53
• @user1766555 Specifically for the eigenvalue $\lambda=0$, yes. The only way for two vectors to be equal is if all entries are equal. The only way two real numbers multiplied together equals zero is if at least one of them is zero, but we said that $k\neq 0$ and $2\neq 0$, so $2k\neq 0$. As mentioned in my edit, there will be another eigenvector which does not have the property that the entries sum to zero, but this will be for a different eigenvalue. – JMoravitz Oct 19 '15 at 1:58
• But it is not "without calculation". You must perform calculations to confirm that you are dealing with a multiple of the Ones matrix. Just because those calculations are done automatically in your head does not mean they do not occur. – mathreadler Oct 19 '15 at 2:10 | 2019-08-25T11:45:06 | {
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https://math.stackexchange.com/questions/3040073/divisibilty-vs-division-for-negative-divisors | # Divisibilty vs. division, for negative divisors
I'm learning some basic number theory from Strayer's 'Elementary Number Theory.' I've arrived at what seems to be a very basic problem, albeit complete with a nasty twist:
Let $$a,b,c \in \mathbb Z$$ with $$c \neq 0$$. Prove that $$a|b$$ if and only if $$ac|bc$$.
The proof is trivial when $$c>0$$.
What has me confused is the case when $$c<0$$. If $$a|b$$, then $$b=aq$$ where $$q \in \mathbb Z$$ and $$a>0$$. Multiplication by $$c$$ on $$b=aq$$ yields $$bc=(ac)q$$, where clearly $$ac<0$$.
This is my problem. The 'division algorithm,' as it's been taught in the early stages of this book (and number theory in general) doesn't allow for the divisor to be negative. If we fix $$c$$ to $$q$$ by saying $$bc=a(cq)$$, we avoid the issue of a negative divisor, but then we can only say $$a|bc$$. Similarly, if we negate the negative quality of $$c$$, perhaps by saying $$bc=(-ac)(-q)$$, then still we can only say $$-ac|bc$$.
What am I missing here? Surely Strayer meant what he said - that is, $$c \neq 0$$ - and not something that would make more sense like $$c>0$$. Any help would be greatly appreciated!
• I know something called the Euclidean algorithm. I am not sure what Division Algorithm means, or what it accomplishes... Dec 15 '18 at 1:32
• @Will It's a common name for Euclidean Division (with Remainder) (algorithm) Dec 15 '18 at 1:35
• @Bill fair enough. What I was thinking about was that the algorithm I had in mind really does use positive numbers throughout, but there are other settings (such as polynomials with rational coefficients) which are fairly familiar, where there is an (extended) Euclidean algorithm, but there is not a natural restriction to, well, positivity. Or, say, Gaussian integers for example. Oh, right, and that is why divisibility is not restricted to positive numbers. Dec 15 '18 at 1:46
• @Bill Oh, well, from your link it is a single division, not all of Euclidean. Dec 15 '18 at 1:54
• @Will Yes, sometimes the denotation is ambiguous, i.e. a single division step vs. complete Euclidean gcd algorithm (remainder sequence) Dec 15 '18 at 2:16
The definition of the divisibility relation doesn't require the stronger property of (Euclidean) division with smaller remainder. $$\,a$$ divides $$b\,$$ is defined in a commutative ring $$R$$ as follows
$$a\mid b \ \ {\rm in}\ \ R\iff ar = b\ \ {\rm for\ some}\ \ r\in R$$
Then $$\ a\mid b\iff ac\mid bc\,$$ follows immediately by multiplying the equation by $$c$$ or cancelling $$\,c\,$$ (assuming $$c$$ is cancellable, which here in $$\,\Bbb Z\,$$ follows by $$\,c\neq 0)$$
See this answer for links on how signs are handled in algorithms for division with remainder.
Note: Analogous remarks hold for divisibility vs. division by zero: $$\,b/0\,$$ is undefined but $$\ 0\mid b\iff b = 0,\$$ e.g. $$\, a = b \pmod{\! 0}\!\iff\! 0\mid a-b \!\iff\! a = b;\,$$ i.e. structurally $$\,\Bbb Z = \Bbb Z/0 =$$ integers $$\!\bmod 0\,$$ (the utility of which will be clearer when one studies quotient rings, e.g. factoring out by a prime ideal.)
• Great insight Bill. I always love an algebraic approach. Dec 15 '18 at 18:49
The division algorithm isn't the definition of divisibility. It's simply a statement that unique divisors and remainder pairs exist.
The definition is that $$a|b$$ if there exists an integer $$m$$ so that $$am=b$$. Sign has nothing to do with it. (And if you want to get technical neither $$a$$ nor $$b$$ need to be integers at all.)
A few notes: $$a|b \iff \pm a| \pm b$$ for all possible combinations of signs. The reason why should be obvious.
Also, if the division principal requires the the divisor can't be negative there is nothing that says the quotient or the dividend can't. For example: to divide $$-19$$ by $$7$$ and getting $$-19=-3*7+2$$ or $$-19\equiv 2\pmod 7$$ is perfectly acceptable.
• Excellent answer, thank you. Dec 15 '18 at 18:49
This proof doesn't depend on the sign of $$c$$:
$$a|b$$ means there is some $$k$$ such that $$ak=b$$. Then multiply by $$c$$ to conclude that there's an integer, namely $$k$$, such that $$(ac)k = bc$$, so $$ac | bc$$. The converse is equally straightforward.
On your other point: you can do the division algorithm with a negative divisor $$d$$. What it asserts is that there is a quotient $$q$$ and a remainder $$r$$ between $$0$$ and $$d-1$$ such that $$\ldots$$. It's OK for the quotient to be negative.
• Thank you Ethan. I think on this one I was way overthinking things. Dec 15 '18 at 18:50 | 2021-12-04T10:46:12 | {
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https://math.stackexchange.com/questions/2589581/is-euler-lagrange-equation-necessary-and-sufficient-for-minimization-in-a-variat | # Is Euler-Lagrange equation necessary and sufficient for minimization in a variational problem?
To the best of my knowledge, the function that minimizes of the integral posed in calculus of variations must also satisfy the Euler-Lagrange equations. In other words, the Euler equations are a necessary condition for finding a minimizer.
One thing I do not understand, is, why do I see everyone treating it as if it were a sufficient condition to show optimality? I am solving some example problems in calculus of variations, and I am following the problems here: http://matematika.cuni.cz/dl/pyrih/variationProblems/variationProblems.pdf example
For example, look at problem 1.1 on page 2, which says:
Using the Euler equation find the extremals for the following functional: $\int_{a}^{b}12xy(x)+(\frac{\partial y(x)}{\partial x})^2dx$
then, as I am looking at the solution which is immediately below, it says "finally obtain the Euler equation for our functional"...and proceeds to solve the Euler equation as a means to solve the minimization problem to obtain the answer.
I do not understand why this is allowed, if the Euler equation is a necessary condition, but it is not sufficient.
Is the Euler equation necessary and sufficient? or just necessary?
An example of a resource that explicitly says it, says "Euler-Lagrange Equation and is a necessary, but not sufficient, condition for an extremal function", is here: http://www.maths.manchester.ac.uk/~wparnell/MT34032/34032_CalcVar
• So, why do I see people solving these minimization problems as if it is sufficient (like the example I cited) ? Or is there a gap in my understanding about why they are allowed to solve it that way? – makansij Jan 3 '18 at 0:53
• sure. here's an example taken from gilbert strang's textbook: ocw.mit.edu/courses/mathematics/…. look at the example on page 2 for the shortest path between two points. – makansij Jan 3 '18 at 1:00
• Actually, this is a passage from Gilbert Strang's textbook Computational Science and Engineering. It's a real textbook, not just lecture notes or something. (Of course, Strang's writing style is fairly informal, so your point stands.) – littleO Jan 3 '18 at 1:09
• We have an analogous situation in calculus when we minimize a differentiable function $f:\mathbb R^n \to \mathbb R$ by setting the derivative equal to $0$. The equation $\nabla f(x) = 0$ might have multiple solutions, and not all solutions correspond to extrema (some are saddle points). But, if it is known that a minimizer of $f$ exists, and if there is a unique point $x \in \mathbb R^n$ such that $\nabla f(x) = 0$, then this $x$ must be the minimizer of $f$. – littleO Jan 3 '18 at 1:13
• Continuing littleO's comment: in the cited example, 1.1 on p.2, the functional is convex in $y$, which rules out the EL solution being a maximizer, and might, given further regularity properties (but I haven't checked) imply there does exist a minimizer. – kimchi lover Jan 3 '18 at 2:19 | 2020-09-23T07:01:46 | {
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The perimeter of square S is 40. Square T is inscribed in square S.
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The perimeter of square S is 40. Square T is inscribed in square S. What is the least possible value of the area of square T ?
A. 45
B. 48
C. 49
D. 50
E. 52
Originally posted by Fabino26 on 08 Feb 2014, 05:55.
Last edited by Fabino26 on 09 Feb 2014, 00:08, edited 1 time in total.
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Re: The perimeter of square S is 40. Square T is inscribed in square S. [#permalink]
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05 Mar 2014, 02:42
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1
Perimeter of square = 40
So length of each side = 10
Area of the square = 100
Area of an (inscribed square) in a square = Half the area of the square
So 100/2 = 50 = Answer D
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Re: The perimeter of square S is 40. Square T is inscribed in square S. [#permalink]
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08 Feb 2014, 11:34
1
Given that the perimeter of Square 40 , gives you the information that each side is 10.
The case of a square in a square , you can see as square T as a ''diamond'' inside the square S. The midpoint of this square become 5 (in the middle ). The side of this new square can be obtained by the use of Pythagorean 5-5-5*sqrt(2). This gives you the information that the side of area = 5*sqrt(2).
5*sqrt(2)*5*sqrt(2)=50.
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Re: The perimeter of square S is 40. Square T is inscribed in square S. [#permalink]
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08 Feb 2014, 12:30
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Fabino26 wrote:
The perimeter of square S is 40. Square T is inscribed in square AS. What is the least possible value of the area of square T ?
A. 45
B. 48
C. 49
D. 50
E. 52
Consider this picture:
The perimeter of square S is 40 implies each side of S is 10, which also means that diagonal of square S is 10. In the picture, the diagonals of square S, split square T into 4 isosceles right (45-45-90) triangles, which, as you know, have length ratios of $$x:x:x\sqrt{2}$$. As you can see, the sides of square T represent the hypotenuse of each of the smaller 4 triangles, thus each side of triangle T has a length of $$5\sqrt{2}.$$
Since the area of the triangle is $$x^2$$, $$(5\sqrt{2})^2$$ = 25*2 = 50, thus choice (D).
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Re: The perimeter of square S is 40. Square T is inscribed in square S. [#permalink]
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09 Feb 2014, 00:11
1
Abdul29 wrote:
Fabino26 wrote:
The perimeter of square S is 40. Square T is inscribed in square AS. What is the least possible value of the area of square T ?
A. 45
B. 48
C. 49
D. 50
E. 52
Consider this picture:
The perimeter of square S is 40 implies each side of S is 10, which also means that diagonal of square S is 10. In the picture, the diagonals of square S, split square T into 4 isosceles right (45-45-90) triangles, which, as you know, have length ratios of $$x:x:x\sqrt{2}$$. As you can see, the sides of square T represent the hypotenuse of each of the smaller 4 triangles, thus each side of triangle T has a length of $$5\sqrt{2}.$$
Since the area of the triangle is $$x^2$$, $$(5\sqrt{2})^2$$ = 25*2 = 50, thus choice (D).
Thanks for the correction! T is inscribed in square S of course. I edited the question.
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Re: The perimeter of square S is 40. Square T is inscribed in square S. [#permalink]
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18 Jun 2014, 05:35
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Can anybody please explain me that why are we not considering the option of a 7 * 7 square inscribed into the square S as dis would give us the area less than 50.
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Re: The perimeter of square S is 40. Square T is inscribed in square S. [#permalink]
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18 Jun 2014, 05:42
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maggie27 wrote:
Can anybody please explain me that why are we not considering the option of a 7 * 7 square inscribed into the square S as dis would give us the area less than 50.
Square T is inscribed in square S means that the vertices of square T are on the sides of square S. You cannot inscribe a square with sides of 7 into a square with sides of 10.
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Re: The perimeter of square S is 40. Square T is inscribed in square S. [#permalink]
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18 Jun 2014, 06:09
Thanks much Bunuel
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Re: The perimeter of square S is 40. Square T is inscribed in square S. [#permalink]
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18 Jun 2014, 20:26
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maggie27 wrote:
Can anybody please explain me that why are we not considering the option of a 7 * 7 square inscribed into the square S as dis would give us the area less than 50.
Also note that area of inscribed square is always half than that of the original square
As Bunuel pointed out, if it goes less than 50, it means some of the vertex is not touching side of the original square.
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Re: The perimeter of square S is 40. Square T is inscribed in square S. [#permalink]
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01 Jul 2015, 18:04
If $$x^{2}$$ is area of square, then find x, one side of the square. If square is inscribed, then diagonal is the length of larger square and therefore the diagonal is $$10$$. To determine the side, the formula also includes the area of the square, $$x^{2}$$. So, if $$2x^{2} = 100$$ then $$x^{2}=50$$
D.
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Re: The perimeter of square S is 40. Square T is inscribed in square S. [#permalink]
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31 Jul 2016, 20:43
Fabino26 wrote:
The perimeter of square S is 40. Square T is inscribed in square S. What is the least possible value of the area of square T ?
A. 45
B. 48
C. 49
D. 50
E. 52
Perimeter of S= 40 ; side = 10
Now square T area can only be minimum if this 10 is its diagonal
Therefore area of square t = diagonal^2/2 = 100/2 = 50
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Re: The perimeter of square S is 40. Square T is inscribed in square S. [#permalink]
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22 Sep 2017, 08:51
first, test takers have to understand that inscribe means the square T touches the square S.
Secondly, using feeling and imagination and experience in geometry, 4 points of T can move on sides of square S. The minimum is if all 4 points are mid-points of sides of square S.
If that is the case, then the area of square T is half of that of square S.
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Re: The perimeter of square S is 40. Square T is inscribed in square S. [#permalink] 06 Oct 2018, 04:47
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https://www.physicsforums.com/threads/why-is-zero-factorial-equal-to-1.99749/ | # Why is zero factorial equal to 1?
1. Nov 13, 2005
### quasi426
Why is zero factorial equal to 1?
2. Nov 13, 2005
### bomba923
As they say :
___________________________
!
*Edit: Sorry about that! Here's why:
Three reasons, from my perspective:
1) 0! is defined to be equal to one.
2) (This will sound weird. Oh well, here goes) You know how
$$\frac{{n!}}{n} = \left( {n - 1} \right)!$$
---->
And so,
$$n = 1 \Rightarrow \frac{{1!}}{1} = 0! = 1$$
3) Consider combinatorials, for example. How many ways are there to choose $$k$$ objects out of a total quantity of $n$ objects?
*We would use the expression:
$$\frac{{n!}}{{k!\left( {n - k} \right)!}}$$
So, for example, there are 6 ways to choose 2 objects out of a set of 4.
But what if $$k = 0$$, --->i.e., we choose nothing?
$$\frac{{n!}}{{0!\left( {n - 0} \right)!}} = \frac{{n!}}{{0!n!}} = \frac{1}{{0!}} = 1\;({\text{or else we have a problem}})$$
You see, there only one way to choose nothing.
-----------------------------------------------------------
Personally,
I usually just go with the first statement:
-"0! is defined to be equal to one"-
Last edited: Nov 13, 2005
3. Nov 13, 2005
### -Job-
Since (n choose 0) gives the expression of (n choose n), because there's only one way to choose everything, there must be only one way to choose nothing.
Of course consider, for n distinct objects, there are n! permutations of those objects. So it seems that for 0 objects there's 1 possible permutation.
4. Nov 14, 2005
### bomba923
5. Nov 14, 2005
### Joffe
Is this the correct definition of factorial, or is it inconsistant with 0! ?
$$x! = \prod_{n=1}^{x} n$$
Last edited: Nov 14, 2005
6. Nov 14, 2005
### shmoe
This is the* definition when x is a positive integer. With the usual convention that an empty product is 1, it is consistant with our usual definition for 0! when x=0. However, x=0 is usually left as a special case and explicitly defined as 0!=1.
*there's more than one equivalent way to define factorial of course
7. Nov 14, 2005
### quasi426
Thanks guys, I was going with the explanation that "0! is just equal to 1 because it is." But I thought about the definition of n! in the book of n(n-1)(n-2).....and I couldn't see exactly how it was so that 0! = 0. So I came here, thanks.
8. Nov 14, 2005
It's an interesting question because you sort of have to look at it in the abstract. Also, there are a few other ways to rationalize 0! = 1.
For instance, if 0! was anything other than 1, the cosine function wouldn't make any sense. Consider: f(x) = Cos(x) = x^0/0! + x^2/2! -x^4/4!... where x is a radian measure. So, if 0! was not equal to 1, then the first term in the series would not equal 1, and the Taylor series that derived it would be wrong, which would turn everything that we know about math upside down.
9. Nov 14, 2005
### shmoe
It would just change how we express the power series, it would not have any effect whatsoever on the mathematics. This is the same situation as adopting the convention x^0=1 for x=0, without this your power series as expressed doesn't make sense either. This is not a mathematically compelling reason for either convention/definition, rather a notational one.
10. Nov 14, 2005
### jcsd
Define A as a finite set, define B as the set of permuatations of A, then |A|! = |B|.
This holds true even when |A|=0 as B in this case contains the empty function (and the empty function only), so you can see there is justification for definition 0! = 1.
11. Nov 14, 2005
### amcavoy
$$\Gamma{\left(n+1\right)}=n!$$
$$\Gamma{\left(1\right)}=0!=\int_0^{\infty}e^{-x}\,dx=1$$
12. Nov 20, 2005
### Joffe
$$(n+\tfrac{1}{2})! = \sqrt{\pi} \prod_{k=0}^{n}\frac{2k+1}{2}$$
I know that the gamma function can be used to solve factorials for any number but are there any other special rules like this one?
13. Nov 22, 2005
### leon1127
The special case 0! is defined to have value 0!==1, consistent with the combinatorial interpretation of there being exactly one way to arrange zero objects (i.e., there is a single permutation of zero elements, namely the empty set emptyset).
from the MathWorld
14. Nov 23, 2005
### matt grime
Step back, take a breath and try to think what the person who first came up with such Taylor/McLaurin series might have done if 0! weren't 1, because surely he or she wouldn't have used something that was undefined or unsuitable?
15. Nov 25, 2005
### benorin
We should pin this thread, or one of the numerous others like unto it, to the top of the forum. Then again, why? After all: I dig the gamma function.
16. Sep 25, 2007
### alfredbobo
Problem
25 < my age < 54. Start from the number of my age and toss a fair coin, plus one if it is a head and minus one otherwise, on average, you need to toss 210 times to hit the bound, i.e. 25 or 54. There are two integers satisfying the condition. Fortunately, the smaller one is my age.
17. Sep 25, 2007
### HallsofIvy
Staff Emeritus
And I'm sure the gamma function is fascinated by you.
18. Jul 6, 2011
### efrenrivera
Re: Factorial!
could you show me some basic properties of factorials? is factorial distributive over product?
19. Jul 6, 2011
### JonF
Re: Factorial!
Can't you test that hypothesis yourself?
What are 6!, 3!, and 2! equal to?
What would 3!*2! be? | 2017-02-28T03:33:38 | {
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https://stackoverflow.com/questions/3433486/how-to-do-exponential-and-logarithmic-curve-fitting-in-python-i-found-only-poly | # How to do exponential and logarithmic curve fitting in Python? I found only polynomial fitting
I have a set of data and I want to compare which line describes it best (polynomials of different orders, exponential or logarithmic).
I use Python and Numpy and for polynomial fitting there is a function polyfit(). But I found no such functions for exponential and logarithmic fitting.
Are there any? Or how to solve it otherwise?
For fitting y = A + B log x, just fit y against (log x).
>>> x = numpy.array([1, 7, 20, 50, 79])
>>> y = numpy.array([10, 19, 30, 35, 51])
>>> numpy.polyfit(numpy.log(x), y, 1)
array([ 8.46295607, 6.61867463])
# y ≈ 8.46 log(x) + 6.62
For fitting y = AeBx, take the logarithm of both side gives log y = log A + Bx. So fit (log y) against x.
Note that fitting (log y) as if it is linear will emphasize small values of y, causing large deviation for large y. This is because polyfit (linear regression) works by minimizing ∑iY)2 = ∑i (YiŶi)2. When Yi = log yi, the residues ΔYi = Δ(log yi) ≈ Δyi / |yi|. So even if polyfit makes a very bad decision for large y, the "divide-by-|y|" factor will compensate for it, causing polyfit favors small values.
This could be alleviated by giving each entry a "weight" proportional to y. polyfit supports weighted-least-squares via the w keyword argument.
>>> x = numpy.array([10, 19, 30, 35, 51])
>>> y = numpy.array([1, 7, 20, 50, 79])
>>> numpy.polyfit(x, numpy.log(y), 1)
array([ 0.10502711, -0.40116352])
# y ≈ exp(-0.401) * exp(0.105 * x) = 0.670 * exp(0.105 * x)
# (^ biased towards small values)
>>> numpy.polyfit(x, numpy.log(y), 1, w=numpy.sqrt(y))
array([ 0.06009446, 1.41648096])
# y ≈ exp(1.42) * exp(0.0601 * x) = 4.12 * exp(0.0601 * x)
# (^ not so biased)
Note that Excel, LibreOffice and most scientific calculators typically use the unweighted (biased) formula for the exponential regression / trend lines. If you want your results to be compatible with these platforms, do not include the weights even if it provides better results.
Now, if you can use scipy, you could use scipy.optimize.curve_fit to fit any model without transformations.
For y = A + B log x the result is the same as the transformation method:
>>> x = numpy.array([1, 7, 20, 50, 79])
>>> y = numpy.array([10, 19, 30, 35, 51])
>>> scipy.optimize.curve_fit(lambda t,a,b: a+b*numpy.log(t), x, y)
(array([ 6.61867467, 8.46295606]),
array([[ 28.15948002, -7.89609542],
[ -7.89609542, 2.9857172 ]]))
# y ≈ 6.62 + 8.46 log(x)
For y = AeBx, however, we can get a better fit since it computes Δ(log y) directly. But we need to provide an initialize guess so curve_fit can reach the desired local minimum.
>>> x = numpy.array([10, 19, 30, 35, 51])
>>> y = numpy.array([1, 7, 20, 50, 79])
>>> scipy.optimize.curve_fit(lambda t,a,b: a*numpy.exp(b*t), x, y)
(array([ 5.60728326e-21, 9.99993501e-01]),
array([[ 4.14809412e-27, -1.45078961e-08],
[ -1.45078961e-08, 5.07411462e+10]]))
# oops, definitely wrong.
>>> scipy.optimize.curve_fit(lambda t,a,b: a*numpy.exp(b*t), x, y, p0=(4, 0.1))
(array([ 4.88003249, 0.05531256]),
array([[ 1.01261314e+01, -4.31940132e-02],
[ -4.31940132e-02, 1.91188656e-04]]))
# y ≈ 4.88 exp(0.0553 x). much better.
• @Tomas: Right. Changing the base of log just multiplies a constant to log x or log y, which doesn't affect r^2. – kennytm Aug 8 '10 at 11:20
• This will give greater weight to values at small y. Hence it is better to weight contributions to the chi-squared values by y_i – Rupert Nash Aug 8 '10 at 16:54
• This solution is wrong in the traditional sense of curve fitting. It won't minimize the summed square of the residuals in linear space, but in log space. As mentioned before, this effectively changes the weighting of the points -- observations where y is small will be artificially overweighted. It's better to define the function (linear, not the log transformation) and use a curve fitter or minimizer. – santon Jan 5 '16 at 19:48
• @santon Addressed the bias in exponential regression. – kennytm Mar 18 '17 at 13:57
• @wordsforthewise polyfit weights by w², so there's no conflict between this answer and Wolfram. Note the statement For gaussian uncertainties, use 1/sigma (not 1/sigma**2). – kennytm Oct 5 '18 at 7:01
You can also fit a set of a data to whatever function you like using curve_fit from scipy.optimize. For example if you want to fit an exponential function (from the documentation):
import numpy as np
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit
def func(x, a, b, c):
return a * np.exp(-b * x) + c
x = np.linspace(0,4,50)
y = func(x, 2.5, 1.3, 0.5)
yn = y + 0.2*np.random.normal(size=len(x))
popt, pcov = curve_fit(func, x, yn)
And then if you want to plot, you could do:
plt.figure()
plt.plot(x, yn, 'ko', label="Original Noised Data")
plt.plot(x, func(x, *popt), 'r-', label="Fitted Curve")
plt.legend()
plt.show()
(Note: the * in front of popt when you plot will expand out the terms into the a, b, and c that func is expecting.)
• Nice. Is there a way to check how good a fit we got? R-squared value? Are there different optimization algorithm parameters that you can try to get a better (or faster) solution? – user391339 May 20 '16 at 3:32
• For goodness of fit, you can throw the fitted optimized parameters into the scipy optimize function chisquare; it returns 2 values, the 2nd of which is the p-value. – MPath Apr 1 '17 at 10:14
I was having some trouble with this so let me be very explicit so noobs like me can understand.
Lets say that we have a data file or something like that
# -*- coding: utf-8 -*-
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit
import numpy as np
import sympy as sym
"""
Generate some data, let's imagine that you already have this.
"""
x = np.linspace(0, 3, 50)
y = np.exp(x)
"""
"""
plt.plot(x, y, 'ro',label="Original Data")
"""
brutal force to avoid errors
"""
x = np.array(x, dtype=float) #transform your data in a numpy array of floats
y = np.array(y, dtype=float) #so the curve_fit can work
"""
create a function to fit with your data. a, b, c and d are the coefficients
that curve_fit will calculate for you.
In this part you need to guess and/or use mathematical knowledge to find
a function that resembles your data
"""
def func(x, a, b, c, d):
return a*x**3 + b*x**2 +c*x + d
"""
make the curve_fit
"""
popt, pcov = curve_fit(func, x, y)
"""
The result is:
popt[0] = a , popt[1] = b, popt[2] = c and popt[3] = d of the function,
so f(x) = popt[0]*x**3 + popt[1]*x**2 + popt[2]*x + popt[3].
"""
print "a = %s , b = %s, c = %s, d = %s" % (popt[0], popt[1], popt[2], popt[3])
"""
Use sympy to generate the latex sintax of the function
"""
xs = sym.Symbol('\lambda')
tex = sym.latex(func(xs,*popt)).replace('$', '') plt.title(r'$f(\lambda)= %s\$' %(tex),fontsize=16)
"""
Print the coefficients and plot the funcion.
"""
plt.plot(x, func(x, *popt), label="Fitted Curve") #same as line above \/
#plt.plot(x, popt[0]*x**3 + popt[1]*x**2 + popt[2]*x + popt[3], label="Fitted Curve")
plt.legend(loc='upper left')
plt.show()
the result is: a = 0.849195983017 , b = -1.18101681765, c = 2.24061176543, d = 0.816643894816
• y = [np.exp(i) for i in x] is very slow; one reason numpy was created was so you could write y=np.exp(x). Also, with that replacement, you can get rid of your brutal force section. In ipython, there is the %timeit magic from which In [27]: %timeit ylist=[exp(i) for i in x] 10000 loops, best of 3: 172 us per loop In [28]: %timeit yarr=exp(x) 100000 loops, best of 3: 2.85 us per loop – esmit Apr 4 '14 at 16:33
• Thank you esmit, you are right, but the brutal force part I still need to use when I'm dealing with data from a csv, xls or other formats that I've faced using this algorithm. I think that the use of it only make sense when someone is trying to fit a function from a experimental or simulation data, and in my experience this data always come in strange formats. – Leandro Aug 17 '14 at 0:24
• x = np.array(x, dtype=float) should enable you to get rid of slow list comprehension. – Ajasja Nov 9 '14 at 22:19
Well I guess you can always use:
np.log --> natural log
np.log10 --> base 10
np.log2 --> base 2
import numpy as np
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit
def func(x, a, b, c):
#return a * np.exp(-b * x) + c
return a * np.log(b * x) + c
x = np.linspace(1,5,50) # changed boundary conditions to avoid division by 0
y = func(x, 2.5, 1.3, 0.5)
yn = y + 0.2*np.random.normal(size=len(x))
popt, pcov = curve_fit(func, x, yn)
plt.figure()
plt.plot(x, yn, 'ko', label="Original Noised Data")
plt.plot(x, func(x, *popt), 'r-', label="Fitted Curve")
plt.legend()
plt.show()
This results in the following graph:
• Is there a saturation value the fit approximates? If so, how can on access it? – Ben Jul 19 at 9:08 | 2019-08-21T12:13:41 | {
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https://math.stackexchange.com/questions/3197806/let-s-xx-in-g-and-x2-in-h-show-that-s-is-a-subgroup-of-g-for | # Let $S=\{x~|~x\in G$ and $x^2 \in H\}$. Show that $S$ is a subgroup of $G$ for $H<G$, $G$ abelian.
The question states:
Let $$G$$ be an Abelian group with subgroup $$H < G$$. Let $$S=\{x~|~x\in G$$ and $$x^2 \in H\}$$. Show that $$S$$ is a subgroup of $$G$$.
My proof is different than what is in the book, I proceeded as follows:
Let the identity be $$e$$.
$$e^2 = e$$, therefore $$e^2 \in H$$, and hence $$e \in S$$. Therefore $$S$$ is non-empty.
By definition, $$S\subseteq G$$.
Let $$a$$, $$b$$ $$\in S$$. Therefore, $$a^2, b^2 \in H$$ and $$a^2, b^2 \in G$$.
$$\therefore ab^{-1} \in S \iff {(ab^{-1})}^2 \in H$$,
$$\therefore ab^{-1} \in S \iff a^2(b^2)^{-1} \in H$$.
As $$H$$ is a group, each element has a unique inverse, so $$a^2, b^2 \in H \implies b^2 \in H$$.
Also, as $$H$$ is a group, it is closed, and $$\therefore a^2(b^2)^{-1} \in H$$. Hence, $$ab^{-1} \in S$$.
So, S is a non-empty subset of a group G, and $$ab^{-1} \in S$$ for all $$a, b \in S$$.
Thus, by the subgroup test, $$S$$ is a subgroup of $$G$$.
I never used the fact that G is Abelian though... which is why I am worried my solution may be incorrect.
• You used the fact that $G$ is abelian in going from $(ab^{-1})^2\in H$ to $a^2(b^2)^{-1}\in H$. You are using $(xy)^2 = x^2y^2$, which is not generally true in nonabelian groups. Yes; your argument seems correct. – Arturo Magidin Apr 23 at 4:00
• @ArturoMagidin True, as you can see I'm still having trouble with details such as that which have been ingrained in my brain as absolute truths, and questioning them with group theory, though wonderful, requires me to "rewire" my brain in a sense. So thank you for your observation! – John Arg Apr 23 at 4:02
• Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. – Shaun Apr 23 at 6:37
To reiterate what was said in the comments by @ArturoMagidin, you used the fact that $$G$$ is abelian in this step:
$$\therefore ab^{-1} \in S \iff {(ab^{-1})}^2 \in H$$,
$$\therefore ab^{-1} \in S \iff a^2(b^2)^{-1} \in H$$.
Thusly:
\begin{align} (ab^{-1})^2&=(ab^{-1})(ab^{-1}) &\\ &=a(b^{-1}a)b^{-1}& \\ &=a(ab^{-1})b^{-1} & \text{ (since } G\text{ is abelian),}\\ &=a^2b^{-2}& \\ &=a^2(b^2)^{-1}. & \end{align} | 2019-05-23T08:53:34 | {
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http://gbbservices.com/blog/rational-and-irrational-values-of-powers/ | # Rational and Irrational Values of Powers
Here’s a cute little existence argument that I was exposed to as an undergraduate and have never forgotten. It shows that there must be irrational positive reals $\alpha$ and $\beta$ for which $\alpha^\beta$ is rational. Furthermore, it’s done by showing that one of two very specific pairs of numbers (either $\alpha=\sqrt{2}, \beta=\sqrt{2}$ or $\alpha={\sqrt{2}}^{\sqrt{2}}, \beta=\sqrt{2}$) satisfies the condition, but without establishing which of the two pairs satisfies the condition.
The argument is elementary. First look at the case $\alpha=\sqrt{2}, \beta=\sqrt{2}$ If $\alpha^\beta = {\sqrt{2}}^{\sqrt{2}}$ is rational, we have an example, and we’re done. If not then ${\sqrt{2}}^{\sqrt{2}}$ must be irrational, and so $\alpha^\beta$ where $\alpha = {\sqrt{2}}^{\sqrt{2}}$ (assumed to be irrational) and $\beta=\sqrt{2}$ is an example of an irrational raised to an irrational. And it’s rational. In fact, it’s $2$, because $\alpha^\beta =({{\sqrt{2}}^{\sqrt{2}}})^{\sqrt{2}} = {{\sqrt{2}}^{{\sqrt{2}}{\sqrt{2}}}} = {\sqrt{2}}^2 = 2$
It’s a great little elementary existence argument, and I think it’s worth being exposed to it. But, it’s also worth knowing that with (much) more advanced techniques we can actually say which of the two choices satisfies the condition (it’s the second choice $\alpha={\sqrt{2}}^{\sqrt{2}}, \beta=\sqrt{2}$). The technique I’m thinking of is the Gelfond-Schneider theorem, which implies that if $\alpha, \beta$ are positive algebraic irrationals, then $\alpha^\beta$ is irrational: the actual theorem uses weaker conditions on $\alpha$ and $\beta$ and reaches a stronger conclusion about $\alpha^\beta$ And, of course, $\sqrt{2}$ is an algebraic irrational, so the result applies.
And, it’s also worth knowing that there are different examples that can be established using elementary techniques: for example $\alpha = \sqrt{2}, \beta = 2 log_2 (3)$ for which $\alpha^\beta =3$
Just for fun, here are elementary examples of all possible choices of rational and irrational choices in taking powers:
$\alpha^\beta$ rational irrational
${rational\ }^{rational}$ $2^3 = 8$ $2^{1/2} = \sqrt{2}$
${irrational\ }^{rational}$ ${\sqrt{2}}^2 = 2$ ${\sqrt{2}}^3 = 2 \sqrt{2}$
${rational\ }^{irrational}$ $2^{log_2 3} = 3$ $2^{({log_2 3})/2} = \sqrt{3}$
${irrational\ }^{irrational}$ ${\sqrt{2}}^{2 log_2 3} = 3$ ${\sqrt{2}}^{log_2 3} = \sqrt{3}$
All that you need to prove the above results is that the three numbers $\sqrt{2}, \sqrt{3}, log_2 3$ are irrational, and there are elementary arguments for establishing each of these. For completeness, I’ll sketch proofs for the three results.
$\sqrt{2}$ is irrational. Assume it’s not. Then there are integers $m,n$ for which $\sqrt{2} = \frac{m}{n}$ i.e. for which $m^2 = 2 n^2$ (and, of course, $n \ne 0$). We show that both $m,n$ have to be even numbers, and that’s impossible for all $m,n$ with $n \ne 0$ satisfying $\sqrt{2} = \frac{m}{n}$ because by removing common factors from any given $m,n$ we arrive at a simplified $m,n$ for which one of the two numbers is odd. To show that both $m,n$ are even, we first have that $m$ is even because $m^2$ is even (it’s $2 n^2$) and writing $m = 2 k$ we have that $4 k^2 = 2 n^2$ and so $2 k^2 = n^2$ which tells us that $n^2$ is even and so $n$ is even. $QED$
The above argument has been known for over two thousand years, and has been repeated in numerous places, including many popular mathematics books. I would recommend G.H. Hardy’s “A Mathematician’s Apology” if you want to find another source: Hardy calls it “Pythagoras’ proof”. With extremely minor modifications, the same argument can be used to prove that $\sqrt{3}$ is irrational, as shown below.
$\sqrt{3}$ is irrational. Assume it’s not. Then there are integers $m,n$ for which $\sqrt{3} = \frac{m}{n}$ i.e. for which $m^2 = 3 n^2$ (and, of course, $n \ne 0$). We show that both $m,n$ have to be divisible by $3$, and that’s impossible for all $m,n$ with $n \ne 0$ satisfying $\sqrt{3} = \frac{m}{n}$ because by removing common factors from any given $m,n$ we arrive at a simplified $m,n$ for which one of the two numbers is not divisible by $3$. To show that both $m,n$ are divisible by $3$, we first have that $m$ is divisible by $3$ because $m^2$ is divisible by $3$ (it’s $3 n^2$) and writing $m = 3 k$ we have that $9 k^2 = 3 n^2$ and so $3 k^2 = n^2$ which tells us that $n^2$ is divisible by $3$ and so $n$ is divisible by $3$. $QED$
$log_2 3$ is irrational. Assume it’s not. Then there are integers $m,n$ for which $log_2 3 = \frac{m}{n}$ i.e. for which $3 = 2^{m/n}$ i.e. for which $3^n = 2^m$ (and, of course, $n \ne 0$). We show that both $m,n$ have to be zero, and that contradicts $n \ne 0$. If $m,n$ are both non-negative, then $3^n$ is an odd number and the only way that $2^m$ can be equal to an odd number is if $2^m = 1$ and so $m=0$, from which it immediately follows that $3^n = 1$ and so $n=0$ and we’re done. The other three (partially overlapping) cases ($m$ non-negative and $n$ non-positive, $m$ non-positive and $n$ non-negative, $m$ non-positive and $n$ non-positive) all can be ruled out using similar reasoning (on $2^a 3^b = 1$ for $a,b$ non-negative integers in some cases). $QED$
And, while I’m on the subject, here are some elementary examples for $\alpha + \beta$ and $\alpha \cdot \beta$
$\alpha + \beta$ rational irrational
${rational\ }+{\ rational}$ $2+3=5$ cannot happen
${irrational\ }+{\ rational}$ cannot happen $({\sqrt{2}}-1) + 1 = \sqrt{2}$
${rational\ }+{\ irrational}$ cannot happen $1+({\sqrt{2}}-1) = \sqrt{2}$
${irrational\ }+{\ irrational}$ $(3-\sqrt{2})+ \sqrt{2} = 3$ $\sqrt{2} + \sqrt{2} = 2\sqrt{2}$
$\alpha \cdot \beta$ rational irrational
${rational\ }\cdot{\ rational}$ $2\cdot3=6$ cannot happen
${irrational\ }\cdot{\ rational}$ $\sqrt{2} \cdot 0 = 0$ $\sqrt{2} \cdot 2 = 2\sqrt{2}$
${rational\ }\cdot{\ irrational}$ $0 \cdot \sqrt{2} = 0$ $2 \cdot \sqrt{2} = 2\sqrt{2}$
${irrational\ }\cdot{\ irrational}$ $(\sqrt{3} -1 )( \sqrt{3} +1)= 2$ $\sqrt{2} \cdot \sqrt{3} = \sqrt{6}$
## Author: Walter Vannini
Hi, I'm Walter Vannini. I'm a computer programmer and I'm based in the San Francisco Bay Area. Before I wrote software, I was a mathematics professor. I think about math, computer science, and related fields all the time, and this blog is one of my outlets. I can be reached via walterv at gbbservices dot com. | 2017-08-23T06:02:14 | {
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https://math.stackexchange.com/questions/484209/eigenvalues-of-block-matricies/484236 | # Eigenvalues of block matricies
If the eigenvalues of a matrix $A$ are $\lambda_1,\lambda_2,\dots,\lambda_n$, what are the eigenvalues of the matrix?
$\begin{bmatrix}0 &A\\A&0\end{bmatrix}$
From some numerical examples I have found that the eigenvalues are just $\lambda_1$,$\lambda_2,\dots,\lambda_n$ and $-\lambda_1$,$-\lambda_2,\dots,-\lambda_n$, but I am not sure how to go about proving it. I have tried taking $\det(A - \lambda I) = 0$, but I could not find a way to work with the results. Where is a good place to start?
Let $$B=\begin{bmatrix} 0 & A \\ A & 0 \\ \end{bmatrix}.$$ We observe that $$\begin{bmatrix} 0 & A \\ A & 0 \\ \end{bmatrix}\begin{bmatrix} \mathbf{u} \\ \mathbf{v} \\ \end{bmatrix}=\lambda\begin{bmatrix} \mathbf{u} \\ \mathbf{v} \\ \end{bmatrix}$$ implies that $A\mathbf{v}=\lambda\mathbf{u}$ and $A\mathbf{u}=\lambda\mathbf{v}$. This implies $\mathbf{u}$ satisfies $$A^2\mathbf{u}=\lambda^2 \mathbf{u}$$ and $\mathbf{v}$ satisfies $$A^2\mathbf{v}=\lambda^2 \mathbf{v}.$$
So the eigenvectors of $B$ must have the form $$\begin{bmatrix} \mathbf{u} \\ \mathbf{v} \\ \end{bmatrix}$$ where $\mathbf{u}$ and $\mathbf{v}$ are both eigenvectors of $A^2$ such that $\mathbf{u}$ and $\mathbf{v}$ belong to the same eigenspace. This eigenvector of $B$ has corresponding eigenvalue either $\pm \lambda$.
We conclude that the eigenvalues of $B$ belong to $$\{\pm \lambda: \lambda \text{ is an eigenvalue of } A\}.$$
Finally, we check that these eigenvalues are achieved. If $\mathbf{v}$ is an eigenvector of $A$ with eigenvalue $\lambda$, then $$\begin{bmatrix} \mathbf{v} \\ \mathbf{v} \end{bmatrix}$$ is an eigenvector of $B$ with eigenvalue $\lambda$ and $$\begin{bmatrix} -\mathbf{v} \\ \mathbf{v} \end{bmatrix}$$ is an eigenvector of $B$ with eigenvalue $-\lambda$.
Hint: using the last formula on this list, note that $$\det\left( \begin{bmatrix}0 &A\\A&0\end{bmatrix} - \lambda I_{2n} \right)= \det\left(-\lambda I_n-A\right)\det\left(-\lambda I_n+A\right)$$
If $x=(x_1, \ldots, x_n)^T$ is an eigenvector of $A$ so that $Ax = \lambda x$, consider the vector $x' = (x_1, \ldots, x_n, x_1, \ldots x_n)^T$. What can be said about the product of your matrix with $x'$? | 2020-12-03T00:44:10 | {
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https://mathoverflow.net/questions/234245/t-2-topologies-that-are-as-disjoint-as-possible/234354 | # $T_2$ topologies that are "as disjoint as possible"
Let $X$ be an infinite set. Are there Hausdorff topologies $\tau_1, \tau_2$ on $X$ such that $\tau_1\cap\tau_2 = \{\emptyset\} \cup \{U\subseteq X: X\setminus U\text{ is finite}\}$?
(That is, the intersection is as small as it can get.)
And what about the special case $X=\mathbb{R}$ and $\tau_1$ being the topology coming from the Euclidean metric?
• Cool - can you jot down an example as an answer? Mar 22, 2016 at 10:24
• Every $T_1$ topology contains every cofinite set, so being "as disjoint as possible" means at least including all of these. Mar 22, 2016 at 10:48
• Oh right - thanks for this hint! Will edit the question accordingly. Mar 22, 2016 at 10:54
• If $\tau_1$ and $\tau_2$ meet your requirement, then they are called $T_1$-independent. The following paper may contain useful information: Shakhmatov, D.; Tkachenko, M.; Wilson, R. G. Transversal and T1-independent topologies. Houston J. Math. 30 (2004), no. 2, 421–433. Mar 22, 2016 at 13:13
• The paper mentioned by Lajos answers your first question: the authors show that there are two topologies on a set of size $2^{2^{\aleph_0}}$ (both are realizations of the space $\omega^* = \beta \omega - \omega$) whose intersection is precisely the cofinite sets. They don't seem to answer your second question, but there are some relevant-looking results, so you'll want to take a look. Mar 22, 2016 at 14:05
Let $\varepsilon$ denote the Euclidean topology on $\mathbb R$.
Proposition: There is a 0-dimensional $T_2$ topology $\tau$ on $\mathbb R$ such that $\tau$ and $\varepsilon$ intersect in only the co-finite sets.
Proof: We need the following lemma:
Lemma: Let $Y=(Y,\nu)$ be an infinite topological space such that $$|\overline{E}^\nu|\ge 2^{\omega}$$ for all $E\in [Y]^\omega$. Then there is an injective map $f:\mathbb R\to Y$ such that $$\forall D\in [\mathbb R]^\omega \text{ if }\overline D\ne \mathbb R \text{ then } \exists x_D\in (\mathbb R\setminus \overline D) \ f(x_D)\in \overline{f[D]}^\nu.$$
Proof of the lemma:
Enumerate $[\mathbb R]^\omega$ as $\{D_{\alpha}:{\alpha}<2^{\omega}\}$.
By transfinite induction define an increasing continuous sequence $(f_{\alpha}:{\alpha}\le 2^{\omega})$ of injective functions from subsets of $\mathbb R$ into $Y$ such that $|f_{\alpha}|\le {\alpha}+{\omega}$, $D_{{\beta}}\subset dom(f_{\alpha})$ for ${\beta}<{\alpha}$, and $$\text{ if {\beta}<{\alpha} and }\overline D_{\beta}\ne \mathbb R \text{ then } \exists x_{\beta}\in (dom(f_{\alpha})\setminus \overline D_{\beta}) \ f_{\alpha}(x_{\beta})\in \overline{f_{\alpha}[D_{\beta}]}^\nu.$$
Assume that ${\alpha}={\beta}+1$, and we have constructed $f_{\beta}$. Let $g\supset f_{\beta}$ be an injective function from $dom(f_{\beta})\cup D_{\beta}$ into $Y$.
If $\overline{D_{\beta}}=\mathbb R$, then$f_{\alpha}=g$ works.
Assume that $U=\mathbb R\setminus \overline{D_{\beta}}\ne \emptyset$. Since $|U|=2^{\omega}$ and $|\overline{g[D_{\beta}]}^{\nu}|\ge 2^{\omega}$ we can pick $x_{\beta}\in U\setminus dom(g)$ and $y_{\beta}\in \overline{g[D_{\beta}]}^{\nu}\setminus ran(g)$. Let $$f_{\alpha}=g\cup\{(x_{\beta},y_{\beta})\}$$
This completes the inductive construction. QED.
Pick a 0-dimensional $T_2$ space $(Y,\nu)$ which meets the requirements of the lemma. (For example, $Y=\omega^*$ works)
Apply the lemma to obtain an injective $f:\mathbb R\to Y$.
Define the topology $\tau$ by declaring that $f$ is a homeomorphism between $(\mathbb R,\tau)$ and $(f[\mathbb R],{\nu})$.
To show that $\tau$ is as required assume that $\emptyset\ne U\in \varepsilon$ such that $F=\mathbb R\setminus U$ is infinite. Let $D$ be a countable $\varepsilon$-dense subset of $F$. Then there is $x_D\in \mathbb R\setminus \overline D$ such that
$f(x_D)\in\overline{f[D]}^{\nu}$. Since $f$ is a homeomorphism, $x_D\in\overline{D}^{\tau}$.
Thus $D\subset F$ implies that $x_D\in \overline{F}^{\tau}\setminus F$, and so $F$ is not $\tau$-closed. Thus $U\notin\tau$.
Thus we proved the proposition.
• Very nice! And this argument seems to apply to lots of other familiar spaces as well (any perfect Polish space, for example, or, generalizing from $\omega^*$ to $U(\kappa)$, any space of the form $[0,1]^\kappa$ or $2^\kappa$, if $\kappa$ is regular). Mar 23, 2016 at 17:28
• Wonderful, thanks a lot Lajos...! Intuitively speaking, could it be that since the Euclidean topology $\varepsilon$ is "very connected" (in a quite vague sense of the word), a $T_2$-topology that is "as disjoint as it gets" from $\varepsilon$ must be "very disconnected" (such as the one you constructed? But maybe my intuition is just rubbish :-) Mar 24, 2016 at 7:45
• I realise that having asked 2 questions that each get an answer in a different post puts me in the difficult position to select the "canonical" accepted answer. I decided that since Lajos found the paper that Will's answer is based on, and since Lajos put in a lot of work to prove the result in this post, he deserves to get his answer the accepted one, so I changed acceptance. Will - I hope this doesn't annoy you too much! Mar 24, 2016 at 7:50
• Dominic, I can not see that connectedness is important here because I think that using my argument one can prove the following statement: If $X=(X,\rho)$ s.t. $$z(X)< |X|=|\rho|=min\{|U|:U\in \rho, U\ne \emptyset \}\le 2^{2^\omega}$$ then there is a topology $\tau$ on $X$ s.t $\rho$ and $\tau$ intersect in only the co-finite sets. Mar 25, 2016 at 11:08
• @DominicvanderZypen: No worries -- I agree that this answer deserves to be accepted. Mar 29, 2016 at 18:45
In the comments, Lajos Soukup pointed out the following paper, which seems to contain lots of relevant information:
D. Shakhmatov, M. Tkachenko, and R. G. Wilson, "Transversal and $T_1$-independent topologies," Houston Journal of Mathematics vol. 30, no. 2 (2004), pp. 421 - 433.
(I will give references to the paper below, but I will paraphrase the results to avoid defining all of their terminology.)
They answer Dominic's first question in the negative:
Proposition 3.2: If $X$ is countably infinite, then no two Hausdorff topologies on $X$ intersect in only the co-finite sets.
They are also able to strengthen this result under various extra hypotheses:
Theorem 3.3: It is consistent (it follows from $\neg$CH plus a fragment of MA) that if $|X| < \mathfrak{c}$ then no two Hausdorff topologies on $X$ intersect in only the co-finite sets.
Corollary 3.5: If $|X| < 2^{\aleph_1}$, then no two compact Hausdorff topologies on $X$ intersect in only the co-finite sets.
On the other hand, they show that Dominic's question can have a positive answer for some sets of larger cardinality:
Corollary 3.8: Let $|X| = 2^{2^{\aleph_0}}$. There are two topologies on $X$, both of which realize the compact Hausdorff space $\beta \omega - \omega$, such that the intersection of these two topologies is the set of co-finite subsets of $X$.
See also the preceding lemma, which is more general. (The proof seems to rely heavily on the fact that infinite closed sets have full cardinality, so the technique does not seem to apply to Dominic's second question.)
As for Dominic's second question, I can't seem to find an answer anywhere. But the authors of the above paper do still give us something:
Corollary to Proposition 3.1: If $\tau$ is a Hausdorff topology on $\mathbb{R}$ such that the only sets common to $\tau$ and the usual topology on $\mathbb{R}$ are the co-finite sets, then $\tau$ is countably compact and contains no non-trivial convergent sequences.
I was able to access the Shakhmatov-Tkachenko-Wilson paper here, and I also found some very interesting related information here, on some slides from a talk of Blaszczyk.
• Beautiful, thanks Will, also the comments concerning $\mathbb{R}$! Mar 22, 2016 at 19:47
If $\tau_1$ and $\tau_2$ meet your requirement, then they are called $T_1$-independent. The following paper may contain useful information: Shakhmatov, D.; Tkachenko, M.; Wilson, R. G. Transversal and $T_1$-independent topologies. Houston J. Math. 30 (2004), no. 2, 421–433. MR2852951. | 2023-03-23T08:36:27 | {
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https://math.stackexchange.com/questions/3030522/selecting-council-members-for-a-committee-elementary-combinatorics | # Selecting council members for a committee - elementary combinatorics.
I'm trying to figure out how to do the following question, but I got stuck. I just don't see how they are counting these people.
In a student council consisting of 16 persons there are mathematics- and computer science students, freshmen as well as sophomores. Every group has four representatives in the council. The council appoints a committee, consisting of 6 council members.
(a) In how many ways can this be done, if every group has to have at least one representative in the committee? (Ans.: 4480)
I would say we first select a representative of each of the $$4$$ groups of $$4$$ people (maths freshmen, maths sophomore, comp sci freshmen, comp sci sophomore). We can do this in 4 ways each, and we need to make four decisions, so $$4^4$$ we then need to choose amongst the remaining people, so $$\binom{12}{2}$$. My final answer would be: $$4^4 \cdot \binom{12}{2} =16896$$ this number is way too big, what is the error in my reasoning here?
(b) And in how many ways if every group has at most two representatives in the committee? (Ans.: 4320) At most two means we combine all possible ways to have precisely $$0$$, $$1$$ and $$2$$ representatives per group, we sum these up. I am also not quite sure how to obtain the precisely a certain amount of representatives.
In how many ways can a committee with six members be selected if each of the four groups of four people has to have at least one representative on the committee?
There are two cases: One group has three representatives and each of the others has one or two groups each have two representatives and each of the others has one.
Case 1: One group has three representatives and each of the others has one.
Choose the group that has three representatives. Choose three of its four members. Choose one of the four members of each of the other three groups. There are $$\binom{4}{1}\binom{4}{3}\binom{4}{1}\binom{4}{1}\binom{4}{1}$$ such selections.
Case 2: Two groups each have two representatives and each of the others each has one.
Choose which two of the four groups have two representatives. Choose two of the four members from each of these two groups. Choose one of the four members of each of the remaining two groups. There are $$\binom{4}{2}\binom{4}{2}\binom{4}{2}\binom{4}{1}\binom{4}{1}$$ such selections.
Total: There are $$\binom{4}{1}\binom{4}{3}\binom{4}{1}\binom{4}{1}\binom{4}{1} + \binom{4}{2}\binom{4}{2}\binom{4}{2}\binom{4}{1}\binom{4}{1} = 1024 + 3456 = 4480$$ admissible committees.
By designating a representative from each group, you counted each case in which a group has $$k$$ representatives $$k$$ times, once for each way you could have designated a representative of that group. Notice that $$\color{red}{\binom{3}{1}}\binom{4}{1}\binom{4}{3}\binom{4}{1}\binom{4}{1}\binom{4}{1} + \binom{4}{2}\color{red}{\binom{2}{1}}\binom{4}{2}\color{red}{\binom{2}{1}}\binom{4}{2}\binom{4}{1}\binom{4}{1} = 3072 + 13824 = 16896$$
To illustrate, let's label the groups $$A$$, $$B$$, $$C$$, and $$D$$. If you reserve one spot for the members of each group on the committee, you count the selection in which three members of group $$A$$ are selected and one member of each of the other groups is selected three times, once for each way of designating one of the members of group $$A$$ as the designated representative of group $$A$$.
$$\begin{array}{c c} \text{reserved spots} & \text{additional members}\\ \hline A_1, B_1, C_1, D_1 & A_2, A_3\\ A_2, B_1, C_1, D_1 & A_1, A_3\\ A_3, B_1, C_1, D_1 & A_1, A_2 \end{array}$$
If you reserve one spot for a representative of each group, you count each selection with two members each of groups $$A$$ and $$B$$ and one member each of groups $$C$$ and $$D$$ four times, twice each for each of the two ways you could designate a representative to fill the reserved spot for a member of groups $$A$$ and $$B$$.
$$\begin{array} \text{reserved spots} & \text{additional members}\\ \hline A_1, B_1, C_1, D_1 & A_2, B_2\\ A_1, B_2, C_1, D_1 & A_2, B_1\\ A_2, B_1, C_1, D_1 & A_1, B_2\\ A_2, B_2, C_1, D_1 & A_1, B_1 \end{array}$$
In how many ways can the committee be selected if every group has at most two representatives on the committee?
There are two cases: Two groups each have two members and the other two groups have one or three groups each have two members.
We discussed the case in which two groups each have two members above.
Three groups each have two members: Select which three of the four groups each have two members. Choose two members from each of these four groups.
This can be done in $$\binom{4}{3}\binom{4}{2}\binom{4}{2}\binom{4}{2} = 864$$ ways, giving a total of $$\binom{4}{2}\binom{4}{2}\binom{4}{2}\binom{4}{1}\binom{4}{1} + \binom{4}{3}\binom{4}{2}\binom{4}{2}\binom{4}{2} = 3456 + 864 = 4320$$ admissible committees.
• Thank you, this helped me a lot :) especially that I could first try out the second question myself after your explanation, really cool how you made the box show text if you float over it. I didn't know that option was available. – Wesley Strik Dec 8 '18 at 10:33
• I have added two tables to illustrate why the method you used in your attempt counts each committee with three members of one group and one member from each of the other groups three times and counts each committee with two members each from two groups and one member each from the other two groups four times. – N. F. Taussig Dec 9 '18 at 10:51
• Very nice indeed. I am delighted by your effort :) – Wesley Strik Dec 9 '18 at 11:02 | 2020-04-01T11:57:54 | {
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https://mathematica.stackexchange.com/questions/198588/find-the-3d-region-containing-the-origin-bounded-by-given-planes | # Find the 3D region containing the origin bounded by given planes
I'm writing a code to generate the Wigner-Seitz cell of the reciprocal lattice for a given set of lattice translation vectors. For example, consider the Body Centered Cubic (BCC) lattice whose basis translation vectors are given by
a1 = {-1, 1, 1}/2;
a2 = {1, -1, 1}/2;
a3 = {1, 1, -1}/2;
The reciprocal basis vectors are then defined according to
d = 2 Pi;
v = a1.(a2\[Cross]a3);
b1 = d/v (a2\[Cross]a3);
b2 = d/v (a3\[Cross]a1);
b3 = d/v (a1\[Cross]a2);
The reciprocal lattice is then defined by the set of reciprocal lattice vectors, the set of all linear combinations of integer multiples of reciprocal basis vectors, i.e.
$$\vec{G} = n_1 \vec{b}_1 + n_2 \vec{b}_2 + n_3 \vec{b}_3, \qquad n_i \in \mathbb{Z}$$
The Wigner-Seitz cell (in this case the First Brillouin Zone) is defined as the region containing the origin which is bounded by the perpendicular bisecting planes of the reciprocal lattice vectors. We generally can accomplish this by only considering the first, second, and maybe third closest reciprocal lattice points to the origin. In the case of BCC, for example, the following vectors will suffice:
recipvecs =
Select[Flatten[
Table[n1 b1 + n2 b2 + n3 b3, {n1, -1, 1}, {n2, -1, 1}, {n3, -1, 1}], 2],
Norm[#] <= 2 d &];
Question: Given these vectors, how can I construct the Wigner-Seitz cell?
For example, one possibility is to construct the equations for all the planes
planes = ({x, y, z} - (#/2)).# == 0 & /@ reciplattice
(note there is a redundancy for the origin, which just gives True, this can be removed). Now the issue is going to be to rewrite each of these equations as an inequality such that the half-space defined by the inequality contains the origin. I don't think that would be too difficult, but not every one of the equations can be solved for any one of the coordinates, e.g. we cannot solve every equation for $$z$$, like
Solve[#, z] & /@ planes
Some of the equations will have to be solved for $$x$$ or $$y$$ before being turned into inequalities. I think I could find a brute force solution but I'm hoping there's something more elegant.
Ultimately I'd like to obtain the inequalities that define the region so that I can visualize it with RegionPlot3D and use it to Select points from a mesh.
Unfortunately, VoronoiMesh does not work in 3D. So we do it manually.
the crystal lattice vectors:
a1 = {-1, 1, 1}/2;
a2 = {1, -1, 1}/2;
a3 = {1, 1, -1}/2;
the reciprocal lattice vectors: (Inverse is easier than using cross products, but ultimately the same thing)
B = {b1, b2, b3} = 2π*Inverse[Transpose[{a1, a2, a3}]];
an inequality defining the perpendicular bisecting plane of a reciprocal lattice point v:
pbp[{0, 0, 0}, r_] = True;
pbp[v_, r_] := v.r/v.v <= 1/2
make a list of such inequalities, And them, and simplify: (here you may have to go to larger s to get all the constraints, as you said)
With[{s = 1},
WS[x_, y_, z_] = FullSimplify[
And @@ Flatten[Table[pbp[{n1,n2,n3}.B, {x,y,z}], {n1,-s,s}, {n2,-s,s}, {n3,-s,s}]]]]
-2 π <= y + z <= 2 π && z <= 2 π + x && x <= 2 π + z && y <= 2 π + x && x <= 2 π + y && -2 π <= x + z <= 2 π && z <= 2 π + y && y <= 2 π + z && -2 π <= x + y <= 2 π
make a 3D plot of the Wigner-Seitz cell: (use more PlotPoints to make it prettier)
With[{t = 2π},
RegionPlot3D[WS[x, y, z], {x, -t, t}, {y, -t, t}, {z, -t, t}]]
You can also check if a point is in the Wigner-Seitz cell or not:
WS[0.1, 0.2, 0.3]
(* True *)
WS[3.1, 3.2, 0.3]
(* False *)
It is really unfortunate that we don't have a 3D implementation of VoronoiMesh.
Borrowing quite a lot from Roman, the following tries to compute the extremal points of the Wigner-Seitz cells and applies ConvexHullMesh to the result in order to obtain the precise polyhedron.
a1 = {-1, 1, 1}/2;
a2 = {1, -1, 1}/2;
a3 = {1, 1, -1}/2;
B = {b1, b2, b3} = 2 π*Inverse[Transpose[{a1, a2, a3}]];
pts = Flatten[Table[{b1, b2, b3}.{n1, n2, n3}, {n1, -1, 1}, {n2, -1, 1}, {n3, -1, 1}], 2];
G = NearestNeighborGraph[pts, VertexCoordinates -> pts];
neighbors = Rest[VertexOutComponent[G, {0, 0, 0}, 1]];
rhs = MapThread[Dot, {neighbors, neighbors}]/2;
subsets = Subsets[Range[Length[neighbors]], {3}];
q = Module[{A, x},
Table[
A = neighbors[[s]];
If[Det[A] != 0,
x = LinearSolve[A, rhs[[s]]];
If[And @@ Thread[neighbors.x <= rhs], x, Nothing],
Nothing
],
{s, subsets}]
];
R = ConvexHullMesh[q]
The other answers are great and very enlightening, I had already found a brute force solution but I took elements of both @Henrik Schumacher and @Roman's answers to produce this nice minimal one for what I wanted. I think both of their answers are better in that they provide more functionality.
d = 2 Pi;
a1 = {-1, 1, 1}/2;
a2 = {1, -1, 1}/2;
a3 = {1, 1, -1}/2;
{b1, b2, b3} = d*Inverse[Transpose[{a1, a2, a3}]];
reciplattice =
Select[Flatten[
Table[n1 b1 + n2 b2 + n3 b3, {n1, -1, 1}, {n2, -1, 1}, {n3, -1, 1}],
2], 0 < Norm[#] <= 2 d &];
region = And@@FullSimplify[({x, y, z} - (#/2)).# <= 0 & /@ reciplattice]
And plotting it with
e = d + 0.1;
fbz = RegionPlot3D[region, {x, -e, e}, {y, -e, e}, {z, -e, e},
PlotPoints -> 60]
Update:
I realized that my solution doesn't actually answer the original, broader question of how to get the region which contains the origin given a set of planes and is more specific to my case, so I figured I would add a solution which can be used more generally. Instead of directly setting up the inequalities, let's say I have the equations which define the planes
planes = ({x, y, z} - (#/2)).# == 0 & /@ reciplattice;
Solve each of the equations for either $$x$$, $$y$$, or $$z$$, which results in a set of rules such as z -> 1, etc:
sols = Flatten[
If[sz = Solve[#, z]; sz != {}, sz,
If[sx = Solve[#, x]; sx != {}, sx, Solve[#, y]]] & /@ planes];
Now construct the inequalities for each side of the plane
ineq = Flatten[{#[[1]] <= #[[2]], #[[1]] > #[[2]]} & /@ sols];
And select the ones which contain the origin
region = Select[ineq, (# /. {x -> 0, y -> 0, z -> 0}) == True &];
which results in effectively the same result but can be used for any set of planes given their respective equations. We can find the region containing any point of interest by simply modifying the last line. | 2021-11-29T23:13:25 | {
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https://math.stackexchange.com/questions/1515492/why-does-cancelling-change-this-equation/1515561 | # Why does cancelling change this equation?
The equation
$$2t^2 + t^3 = t^4$$
is satisfied by $t = 0$
But if you cancel a $t^2$ on both sides, making it $$2 + t = t^2$$ $t = 0$ is no longer a solution.
What gives? I thought nothing really changed, so the same solutions should apply.
Thanks
• About why the unthinking "cancel" idea is harmful: math.stackexchange.com/questions/1505354/…. – Martín-Blas Pérez Pinilla Nov 6 '15 at 7:29
• @Martín-BlasPérezPinilla and here. – Eff Nov 6 '15 at 11:43
• The equation $2t=t$ is satisfied by $t=0$. But if you cancel a $t$ on both sides, making it $2=1$ ... – Joel Reyes Noche Nov 6 '15 at 13:17
• Of course you may cancel, but that is only if and not iff. – Michael Hoppe Nov 6 '15 at 13:46
• Related: "$1\times0=2\times0$ is true. Canceling the $0$, we get $1=2$, which is no longer true! What gives?" The flaw in this is very similar to yours — you can't divide by $t^2$ if $t=0$. – Akiva Weinberger Nov 8 '15 at 22:13
When you cancel out $t$ on both sides you are assuming $t\neq0$.
To be rigorous you need to divide into two cases,
1. $t\neq0$, then you can cancel out $t$.
2. $t=0$, then you cannot cancel out $t$ as division by $0$ is not allowed.
I think the answer lies in the steps you ignored, that is, the "cancelling out" process:
"Cancelling out" with few extra intermediate steps usually ignored:
$$2t^2+t^3-t^4=0\implies t^2(2+t-t^2)=0$$
$$\implies t^2=0\text{ or }(2+t-t^2)=0$$.
You must guard against division by zero. When you "cancel out $t^2$", you are dividing both sides of your equation by $t^2$. Consequently, the correct form of inference is
\begin{align*} 2t^2+t^3 &= t^4 & & \\ 2 + t &= t^2 \qquad \text{ or } \qquad t^2 = 0. \end{align*}
The resulting left choice has the solution set $\{-1,2\}$ and the resulting right choice has the solution set $\{0,0\}$, which together are the solution set of the first equation.
Note that the same thing happens in reverse. Multiplying both sides of an equation by zero can result in craziness. We can agree that $1 \neq 2$, but this does not mean $0 \cdot 1 \neq 0 \cdot 2$ because both sides of this are zero. It can be harder to see when one is doing this when using a more complicated expression that is only sometimes zero. For instance,
\begin{align*} 2 + t &= t^2 \\ t^2(2 + t) &= t^4 \qquad \text{ and } \qquad t^2 \neq 0 \\ 2t^2+t^3 &= t^4 \qquad \text{ and } \qquad t^2 \neq 0 \end{align*}
The left choice has solution set $\{-1,0,0,2\}$ and the right choice has solution set $\{0,0\}$. The (set) difference of these is $\{-1,2\}$, the solution set of the first equation.
None of this is hard to see when multiplying or dividing by constants. Either the constant is nonzero and everything works or the constant is zero and everyone can see that something bogus is going on. However, when we're not just using constants, a little more care is needed.
• +1 for being the only answer to say that cancelling is really just division and it can be dangerous to apply it without understanding that. – wchargin Nov 8 '15 at 22:48
Write your equation as $$t^4 -t^3 - 2t^2 = 0,$$ factoring out $t$ to get $$t^2(t^2 - t - 2) = 0.$$ Factoring once more, we have $$t^2(t- 2)(t+1) = 0$$ so the roots are $-1,0,0,2$.
If you leave out the $t^2$ factor you lose the $0,0$ roots.
• That's a different way of looking at it. Interesting – Cruncher Nov 6 '15 at 16:33
• Yes, this is the way to do it. Put it in standard form first (move the whole polynomial to the left of the equal sign, with the highest-ordered term on the left, then the next-highest-ordered term, etc.), then factor it. – Mathieu K. Nov 8 '15 at 7:39
• And so, basically, you didn't "cancel out" the t²; rather, you divided it from both sides: that is, you factored it out. Factoring out any root leaves an equation that doesn't have that root (as long as it's not a double root like your t = 0 is). For instance, if instead we had factored out (t + 1) to get (t + 1)(t³ - 2t) = 0 and then divided out the (t + 1), t = -1 is no longer a solution of t³ - 2t = 0—plug it in to see for yourself. – Mathieu K. Nov 8 '15 at 7:43
• @Mathieu K - thanks. I got rid of the word "cancel" , factoring out is more insightful. – arthur Nov 8 '15 at 14:14
No such thing as cancelling actually exists. What you do in those moments is to divide both sides of an equation by the same value. This means, cancelling like this:
$2t^2 + t^3 = t^4$ to $2t + t^2 = t^3$
Is actually a shortcut for:
$2t^2 + t^3 = t^4$ to $(2t^2 + t^3)/t = t^4/t$ to $2t + t^2 = t^3$
And you can only do that by assuming $t \neq 0$, so you discard that solution by doing such operation. Perhaps $0$ would still be a valid solution if the exponents are $> 2$ in each term (so $0$ would still be a valid root for them) but you should never assume that.
• Be careful when you say canceling doesn't exist. It most definitely exists when working over any integral domain. (See Wikipedia.) – gebruiker Nov 8 '15 at 22:02
If $t = 0$, then dividing both sides by $t^2$ would mean dividing by zero.
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https://physics.stackexchange.com/questions/368634/direction-of-velocity-vector-in-3d-space | # Direction of velocity vector in 3D space
According to a well-known textbook (Halliday & Resnick), the direction of a velocity vector, $\vec v$, at any instant is the direction of the tangent to a particle's path at that instant, as is illustrated below in 2D.
According to the same textbook, the same holds for 3D. However, the tangent to a curve in 3D is not a line, but a plane! A vector could be in a plane and still take on any direction betwen $0^{\circ}$ and $360^{\circ}$ within that plane.
How do we define and determine the direction of $\vec v$ given a particle's path in 3D? (If possible, include illustrations in your answers).
• Have you just thought of taking $\frac{\mathrm d}{\mathrm dt} \vec x (t)$? – JamalS Nov 13 '17 at 16:11
• JamalS: yes, I can take a derivative, but cannot reconcile this with the idea of a tangent given in the textbook. – Mihael Nov 13 '17 at 16:14
• In 2D, you don't have to work out a derivative to just draw $\vec v$ as a tangent graphically. The book seems to give the idea that you can do the same in 3D, but I cannot see it, because there is no single tangent line, but rather a tangent plane. – Mihael Nov 13 '17 at 16:20
• Your assertion is not correct. The notion of a tangent vector to a curve is perfectly well-defined, regardless of the dimension of the space the curve inhabits. – J. Murray Nov 13 '17 at 16:50
• J. Murray I was using this definition of a tangent: "a straight line or plane that touches a curve or curved surface at a point, but if extended does not cross it at that point." I guess it does not apply to vectors... – Mihael Nov 13 '17 at 17:03
## 5 Answers
Example with Figures on @JEB 's answer.
Let the regular (smooth) curve with parametric equation $$\mathbf{x}\left(t\right)=\bigl[x_{1}\left(t\right),x_{2}\left(t\right),x_{3}\left(t\right)\bigr]=\left(5\cos t,5\sin t,2t\right) \tag{01}$$ The parameter $\:t\:$ would represent time in case that this curve is the trajectory of a particle.
Now, the vector $$\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}=\Biggl(\dfrac{\mathrm dx_{1}}{\mathrm dt},\dfrac{\mathrm dx_{2}}{\mathrm dt},\dfrac{\mathrm dx_{3}}{\mathrm dt}\Biggr)=\left(-5\sin t,5\cos t,2\right) \tag{02}$$ is tangent to the curve at the point $\:\mathbf{x}\left(t\right)\:$ and well-defined without any indeterminacy. In case of particle motion this is the velocity vector of the particle.
In order to normalize this vector we have $$\left\Vert\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}\right\Vert=\sqrt{29} \tag{03}$$ This norm, the speed of the particle, is a function of $\:t\:$ in general. Here accidentally is constant. From (02) and (03) we produce the unit vector $$\mathbf{t}=\dfrac{\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}}{\left\Vert\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}\right\Vert}=\sqrt{\frac{1}{29}}\left(-5\sin t,5\cos t,2\right) \tag{04}$$ The vector $\:\mathbf{t}\left(t\right)\:$ is the unit tangent vector to the curve at point $\:\mathbf{x}\left(t\right)$. $$\boxed{\:\mathbf{t}=\sqrt{\frac{1}{29}}\left(-5\sin t,5\cos t,2\right)\:} \tag{05}$$ Differentiating again we have $$\dfrac{\mathrm d\mathbf{t}}{\mathrm dt}=\sqrt{\frac{1}{29}}\left(-5\cos t,-5\sin t,0\right) \tag{06}$$ a vector normal to $\:\mathbf{t}\:$ with norm $$\left\Vert\dfrac{\mathrm d\mathbf{t}}{\mathrm dt}\right\Vert=5\sqrt{\frac{1}{29}} \tag{07}$$ Again this norm is a function of $\:t\:$ in general. From (06) and (07) we produce the unit vector $$\mathbf{n}=\dfrac{\dfrac{\mathrm d\mathbf{t}}{\mathrm dt}}{\left\Vert\dfrac{\mathrm d\mathbf{t}}{\mathrm dt}\right\Vert}=\left(-\cos t,-\sin t,0\right) \tag{08}$$ The vector $\:\mathbf{n}\left(t\right)\:$ is the principal normal unit vector to the curve at point $\:\mathbf{x}\left(t\right)$. $$\boxed{\:\mathbf{n}=\left(-\cos t,-\sin t,0\right)\vphantom{\sqrt{\frac{1}{29}}}\:} \tag{09}$$
Finally we construct the unit vector $$\mathbf{b}=\mathbf{t}\boldsymbol{\times}\mathbf{n}=\sqrt{\frac{1}{29}} \begin{bmatrix} \mathbf{e}_{1} & \mathbf{e}_{2} & \mathbf{e}_{3}\vphantom{\dfrac{\dfrac{}{}}{}}\\ -5\sin t & 5\cos t & 2\vphantom{\dfrac{\dfrac{}{}}{}}\\ -\cos t & -\sin t & 0 \vphantom{\dfrac{\dfrac{}{}}{}} \end{bmatrix} =\sqrt{\frac{1}{29}}\left(2\sin t,-2\cos t,5\right) \tag{10}$$ so $$\boxed{\:\mathbf{b}=\sqrt{\frac{1}{29}}\left(2\sin t,-2\cos t,5\right)\vphantom{\sqrt{\frac{1}{29}}}\:} \tag{11}$$ The vector $\:\mathbf{b}\left(t\right)\:$ is the unit binormal vector to the curve at point $\:\mathbf{x}\left(t\right)$.
The triad of vectors $\:\left(\mathbf{t},\mathbf{n},\mathbf{b}\right)\:$ forms a right-handed orthonormal triplet, as shown in Figures, called the moving trihedron. For the three planes, sides of the trihedron, we have the following terminology
\begin{align} \text{plane }\left(\mathbf{t},\mathbf{n}\right) & =\textit{Osculating plane} \tag{12a}\\ \text{plane }\! \left(\mathbf{n},\mathbf{b}\right) & =\textit{Normal plane} \tag{12b}\\ \text{plane }\left(\mathbf{b},\mathbf{t}\right) & =\textit{Rectifying plane} \tag{12c} \end{align}
----- 3D image 1 ----- 3D image 2 ----- 2D video ----- 3D video ----- The moving trihedron 3D video
Check out the Frenet-Serret formalism: a space-curve ${\bf x}(t)$ has an orthonormal basis at all points called the "TNB" (Tangent, Normal, Binormal), which are unit vectors along:
${\bf T} \propto \frac{d{\bf x}(t)}{dt} = {\bf v}(t)$
${\bf N} \propto \frac{d{\bf T}(t)}{dt}$
${\bf B} = {\bf T \times N}$.
I don't know what your book is telling you a tangent is. But what is actually meant, and what will tell you how to draw the velocity vector, is this. Consider a short part of the path the object takes near the point you want the tangent. If you have a short enough bit of the path, it is a straight line. Extend that line in the direction the particle was moving and stick an arrow on the end of it. You have the velocity vector.
In the case that taking a short path near the point is not a straight line, for example the path of the object makes a sharp right turn exactly at that point, then you will not be able to draw a velocity vector this way. But that is OK, because at that point in the motion, the velocity of the object is not well defined so your failure to be able to draw it reflects that fact.
An example should elucidate things. Imagine a particle with a position vector,
$$\vec x = \begin{pmatrix} t\\ t\\ ct \end{pmatrix}$$
parametrised by time $t \in [0,\infty)$, with $c\in\mathbb R$. This corresponds to a particle travelling radially outward; its projection onto the $x-y$ plane makes a $45^\circ$ angle from the origin, but it has some slope $c$ in the $z$-component so it is not equidistant from all three axes.
You know the velocity vector's direction immediately, it's intuitively simply in the direction of the path, and it must be,
$$\vec v =\begin{pmatrix} 1\\ 1\\ c \end{pmatrix} = \frac{\mathrm d}{\mathrm dt }\begin{pmatrix} t\\ t\\ ct \end{pmatrix}.$$
Now, if we have a curved path, the same rule applies: simply take the derivative. You will find the direction is along a portion of the infinitesimal path which is straight.
Since asked in Physics.SE I'll project the Physics point of view. This can be answered intuitively. Imagine yourself flying aimlessly in air. At certain instant you are heading to somewhere. Where will your velocity vector point to?
Find a plane containing smallest part of path curve at that instant. The velocity vector must lie in the same plane.
Note: For exactly straight paths it'll be limiting case. Therefore make those impractical straight paths slightly curved ;), then do the above procedure. | 2020-03-31T23:29:39 | {
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https://stats.stackexchange.com/questions/52585/what-happens-when-i-include-a-squared-variable-in-my-regression/52654 | # What happens when I include a squared variable in my regression?
I start with my OLS regression: $$y = \beta _0 + \beta_1x_1+\beta_2 D + \varepsilon$$ where D is a dummy variable, the estimates become different from zero with a low p-value. I then preform a Ramsey RESET test and find that i have some misspesification of the equation, i thus include squared x: $$y = \beta _0 + \beta_1x_1+\beta_2x_1^2+\beta_3 D + \varepsilon$$
1. What does the squared term explain? (Non-linear increase in Y?)
2. By doing this my D estimate does not vary from zero any more, with a high p-value. How do i interpret the squared term in my equation (in general)?
Edit: Improving question.
• possible duplicate of Why ANOVA/Regression results change when controlling for another variable – Macro Mar 18 '13 at 13:15
• Probable reason: $x_{1}^2$ and $D$ seem to explain the same variablility in $y$ – steadyfish Mar 18 '13 at 13:17
• One thing that might help is to center $x$ before creating your squared term (see here). As for the interpretation of your squared term, I argue that it's best to interpret $\beta_1x_1+\beta_2x_1^2$ as a whole (see here). Another thing is that you may need an interaction, that means adding $\beta_4x_1D+\beta_5x_1^2D$. – gung - Reinstate Monica Mar 18 '13 at 13:45
• I don't think it's really a duplicate of that question; the solution is different (centering variables works here, but not there, unless I am mistaken) – Peter Flom Mar 18 '13 at 14:13
• @Peter, I interpret this question as a subset of "Why is it that when I add a variable to my model, the effect estimate/$p$-value for some other variable changes?", which is addressed in the other question. Among the answers to that questions are collinearity (which gung does allude to in his answer to that question)/content overlap between predictors (i.e. between $D$ and $(x_1,x_1^2)$, which I suspect is the culprit in this case). The same logic applies here. I'm not sure what the controversy is but that's fine if you and others disagree. Cheers. – Macro Mar 18 '13 at 14:27
Well, first of, the dummy variable is interpreted as a change in intercept. That is, your coefficient $\beta_3$ gives you the difference in the intercept when $D=1$, i.e. when $D=1$, the intercept is $\beta_0 + \beta_3$. That interpretation doesn't change when adding the squared $x_1$.
Now, the point of adding a squared to the series is that you assume that the relationship wears off at a certain point. Looking at your second equation
$$y = \beta _0 + \beta_1x_1+\beta_2x_1^2+\beta_3 D + \varepsilon$$
Taking the derivate w.r.t. $x_1$ yields
$$\frac{\delta y}{\delta x_1} = \beta_1 + 2\beta_2 x_1$$
Solving this equation gives you the turning point of the relationship. As user1493368 explained, this is indeed reflecting an inverse U-shape if $\beta_1<0$ and vice versa. Take the following example:
$$\hat{y} = 1.3 + 0.42 x_1 - 0.32 x_1^2 + 0.14D$$
The derivative w.r.t. $x_1$ is
$$\frac{\delta y}{\delta x_1} = 0.42 - 2*0.32 x_1$$
Solving for $x_1$ gives you
$$\frac{\delta y}{\delta x_1} = 0 \iff x_1 \approx 0.66$$
That is the point at which the relationship has its turning point. You can take a look at Wolfram-Alpha's output for the above function, for some visualization of your problem.
Remember, when interpreting the ceteris paribus effect of a change in $x_1$ on $y$, you have to look at the equation:
$$\Delta y = (\beta_1 + 2\beta_2x_1)\Delta x$$
That is, you can not interpret $\beta_1$ in isolation, once you added the squared regressor $x_1^2$!
Regarding your insignificant $D$ after including the squared $x_1$, it points towards misspecification bias.
• Hi. If you had several predictors should you use partial derivatives or total derivatives (diferentials)? – skan Aug 24 '16 at 0:06
• A partial derivative is still the right way to go here. The interpretation of all coefficients is ceteris paribus, i.e., holding everything else constant. That's exactly what you are doing when you take a partial derivative. – altabq Aug 26 '16 at 9:07
• See this UCLA IDRE page to complement @altabq's great answer. – Cyrille Oct 18 '18 at 7:22
A good example of including square of variable comes from labor economics. If you assume y as wage (or log of wage) and x as an age, then including x^2 means that you are testing the quadratic relationship between an age and wage earning. Wage increases with the age as people become more experienced but at the higher age, wage starts to increase at decreasing rate (people becomes older and they will not be so healthy to work as before) and at some point the wage doesn't grow (reaches the optimal wage level) and then starts to fall (they retire and their earnings starts to decrease). So, the relationship between wage and age is inverted U-shaped (life cycle effect). In general, for the example mentioned here, the coefficient on age is expected to be positive and than on age^2 to be negative.The point here is that there should be theoretical basis /empirical justification for including the square of the variable. The dummy variable, here, can be thought of as representing gender of the worker. You can also include interaction term of gender and age to examine the whether the gender differential varies by age. | 2021-05-07T16:36:49 | {
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https://math.stackexchange.com/questions/585628/triple-integral-with-bounds-in-first-octant | # Triple Integral with bounds in first octant
I am really confused on how to get my integrating function because I don't know, even after graphing, how the tetrahedron intersects the x-y-z axis.
I am supposed to find the triple integral for the volume of the tetrahedron cut from the first octant by the plane $6x + 3y + 2z = 6$.
I have found the bounds of integration by isolating $x,y,z$ in the tetrahedron equation. So I know how the bounds will be in any order of integration that I want. I am now stuck on what the integrating function will be.
So for example, if I was integrating w.r.t dxdydz then, according to the bounds I found through a numerical method, should be:
$$\int_{z=0}^3\int_{y=0}^{\frac{6-2z}{3}} \int_{x=0}^{\frac {6-3y-2z}{6}} dx dy dz$$
Now these are the bounds I have found. How do I get the integrating equation? Would the integrating equation be x for the first definite integral?
so something like this?
$$x \rvert _{0}^{\frac{6-3y-2z}{6}}$$
Is that right?
So then to find the volume this would be the whole equation with the bounds.
{0,3} {0, (6-2z)/3 } {0, (6-3y-2z)/6} dx dy dz
As you can see in the graph above, it is bounded by the given plane so are my bounds correct? is the function a constant "1" ?
• I improved the formatting of your post; check it and see if it's correct. Also try improving your last equation. By the way, the function in this case(Cartesian coordinates) happens to be $1$; in other coordinate systems it can be different.
– Ali
Nov 29 '13 at 12:53
• @Ali how do you put your SO posts into that mathematical form? Is there some code to type to do that? Nov 29 '13 at 12:59
• Some of these sites support latex codes via mathjax I believe.
– Ali
Nov 29 '13 at 13:51
• There is a large FAQ in the meta section on the Latex/Mathjax notation used here. meta.math.stackexchange.com/questions/5020/… Nov 29 '13 at 16:58
You have a plane $P:~6x+3y+2z=6$ which cut the axes in the first octant as you see through the below plot. $P$, clearly, intersects $z=0$ in a line: $$z=0\to 6x+3y+2\times 0=6\to 6x+3y=4$$ I don't want to tell you that all triple integrals will start from this point that we did above, but we do for many certain triples. From the line above we can get the proper limits for $x$ and $y$: $$y=0\to x=2/3\Longrightarrow x|_0^{2/3}\\ y=\frac{4-6x}{3}=4/3-2x\Longrightarrow y|_0^{\frac{4}3-2x}$$ Therefore we have: $$V=\large{\int_x\int_y\int_0^{\frac{6-6x-3y}{2}}}dzdydx$$
Find the intersections with the plane $6x + 3y + 2z = 6$ and the coordinate axes in the first place, giving a tetrahedron with vertices $(0,0,0)$ , $(1,0,0)$ , $(0,2,0)$ , $(0,0,3)$ . Then introduce normed coordinates $(\xi,\eta,\zeta)$ defined by $(x,y,z) = (\xi,2\eta,3\zeta)$ and find the integral: $$\iiint dx\,dy\,dz = \iiint 1\,d\xi\;2\, d\eta\;3\, d\zeta = 6 \times \iiint d\xi\,d\eta\,d\zeta$$ $$\iiint d\xi\,d\eta\,d\zeta = \int_{\xi=0}^{\xi=1}\left[\int_{\eta=0}^{\eta=1-\xi}\left\{\int_{\zeta=0}^{\zeta=1-\xi-\eta}d\zeta\right\}d\eta\right]d\xi = \int_{\xi=0}^{\xi=1}\left[\int_{\eta=0}^{\eta=1-\xi}\left\{1-\xi-\eta\right\}\,d\eta\right]d\xi = \int_{\xi=0}^{\xi=1}\left[(1-\xi)-\xi(1-\xi)-\frac{1}{2}(1-\xi)^2\right]d\xi = \int_{\xi=0}^{\xi=1}\left[\frac{1}{2}-\xi+\frac{1}{2}\xi^2\right]d\xi = \frac{1}{2}-\frac{1}{2}+\frac{1}{2}\frac{1}{3} = \frac{1}{6}$$ So the integral is $6 \times 1/6 = 1$ . This can be seen immediately, though, by noting that $\; \iiint dx\,dy\,dz\;$ is the volume of a pyramid (: take a look at the picture by B.S.). | 2022-01-20T06:44:46 | {
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https://math.stackexchange.com/questions/2918287/how-to-prove-two-sequence-have-a-common-limit | # How to prove two sequence have a common limit.
We have \begin{align} U_{0} &= 1 &&\text{and} & V_{0} &= 2 \\ U_{n+1} &= \frac{U_{n}+V_{n}}{2} &&\text{and} & V_{n+1} &= \sqrt{U_{n+1}V_n} \end{align} How to prove two sequence have a common limit?
I found $(U_n)$ is increasing and $(V_n)$ is decreasing but I don't know how to do with $\lim (V_n) - (U_n)$
• Do you want $U_{n+1} = \frac{U_n+V_n}2$ instead? – Somos Sep 15 '18 at 20:55
• @Somos I thought I should use only that limit $V_n - U_n$ but yeah if it's possible to resolve with it , how it works ? – KEVIN DLL Sep 15 '18 at 20:58
• I ask because you never tell us what $\,b\,$ is supposed it be. Is it a real number? – Somos Sep 15 '18 at 21:00
• @Somos yes i just edit it , – KEVIN DLL Sep 15 '18 at 21:02
• What is an "adjacent" sequence? Do you mean convergent? – Somos Sep 15 '18 at 21:09
Since $$U_n$$ is increasing and $$V_n$$ is decreasing, all you have to show is that the difference $$V_n-U_n$$ has limit $$0$$. But, the limit exists since both sequences are monotone and bounded, and rewriting the equation $$U_{n+1}=(U_n+V_n)/2$$ to $$\,V_n = 2 U_{n+1}-U_n \,$$ shows the two limits must be equal.
Given a diameter one circle, its circumference is $$\,\pi.\,$$ Archimedes calculated the perimeters of inscribed and circumscribed regular polygons to find upper and lower bounds for $$\,\pi.\,$$ Let $$\,a(n)\,$$ be the perimeter of the circumscribed regular polygon of $$\,n\,$$ sides, and $$\,b(n)\,$$ the perimeter of the inscribed regular polygon of $$\,n\,$$ sides. We have that $$\,a(2n)\,$$ is the harmonic mean of $$\,a(n)\,$$ and $$\,b(n)\,$$ and $$\,b(2n)\,$$ is the geometric mean of $$\,a(2n)\,$$ and $$\,b(n).\,$$
We can start with triangles where $$\,n=3\,$$ and find $$\,a(3) = 3\sqrt{3}\,$$ and $$\,b(3) = a(3)/2.\,$$ We then keep doubling the number of sides indefinitely. The connection between the two recursions is that $$\, U_n = 3\sqrt{3}/a(3\,2^n)\,$$ and $$\, V_n = 3\sqrt{3}/b(3\,2^n)\,$$ since the recursions and initial values for $$\, U_n, V_n \,$$ come from those for $$\,a(n), b(n).\,$$ The common limit of $$\,a(n), b(n)\,$$ is $$\,\pi,\,$$ thus the common limit of $$\, U_n, V_n\,$$ is $$\,3\sqrt{3}/\pi.\,$$
If we start with squares where $$\,n=4\,$$ we find that $$\, a(4) = 4\,$$ and $$\,b(4) = 2\sqrt{2}.\,$$ Now let $$\, U_n := 4/a(4\, 2^n)\,$$ and $$\, V_n := 4/b(4\, 2^n).\,$$ Then $$\, U_0 = 1, \,\, V_0 = \sqrt{2} \,$$ and the common limit is $$\, 4/\pi.\,$$ The general values are $$\, b(n) = n\,\sin(\pi/n),\,$$ and $$\, a(n) = n\,\tan(\pi/n).$$
Given that $$u_{n+1}=\frac{u_n+v_n}2\quad\text{and}\quad v_{n+1}=\sqrt{\vphantom{1}u_{n+1}v_n}\tag1$$ it is easy to derive that $$u_{n+1}^2-v_{n+1}^2=\frac{u_n^2-v_n^2}4\tag2$$ Thus, induction says that $$u_n^2-v_n^2=\frac{u_0^2-v_0^2}{4^n}\tag3$$ Furthermore, $$u_{n+1}-u_n=\frac{v_n-u_n}2\quad\text{and}\quad v_{n+1}^2-v_n^2=\frac{u_n-v_n}2v_n\tag4$$ Case $\boldsymbol{u_0\gt v_0}$
Equation $(3)$ says $u_n\gt v_n$, and $(4)$ says that $u_n$ is decreasing and $v_n$ is increasing. Since, $u_n\gt v_n\gt v_0$, $u_n$ is decreasing and bounded below. Since $v_n\lt u_n\lt u_0$, $v_n$ is increasing and bounded above.
Case $\boldsymbol{u_0\lt v_0}$
Equation $(3)$ says $u_n\lt v_n$, and $(4)$ says that $u_n$ is increasing and $v_n$ is decreasing. Since, $u_n\lt v_n\lt v_0$, $u_n$ is increasing and bounded above. Since $v_n\gt u_n\gt u_0$, $v_n$ is decreasing and bounded below.
In either case, both $u_n$ and $v_n$ converge. $(3)$ says that their limits are the same.
If $U_n,V_n\in[1,2]$,
$$U_n+V_n\in[2,4]\implies U_{n+1}\in[1,2]$$
and
$$U_{n+1}V_n\in[1,4]\implies V_n\in[1,2].$$
Hence both sequences are bounded and both converge.
And if the limits exist, they are equal. Indeed,
$$U=\frac{U+V}2,V=\sqrt{UV}$$ imply $U=V$.
• Do you not need some monotony argument? Maybe I overlooked something, but, wouldn't it be also possible that the series just circle around some values? – Imago Sep 17 '18 at 16:16
• @Imago: re-read the post, monotonicity is already granted. – Yves Daoust Sep 17 '18 at 16:22
• I take this as a yes. It's needed. – Imago Sep 17 '18 at 16:24 | 2019-05-20T19:13:32 | {
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http://math.stackexchange.com/questions/293527/how-to-check-if-a-8-puzzle-is-solvable | # How to check if a 8-puzzle is solvable?
I have a 8-puzzle
1|2|3
-+-+-
4|5|6
-+-+-
|8|7
How can be checked if the puzzle is solvable?
Wikipedia states that it is solvable, but does not prove it. Can anybody explain the prove?
-
What is a $9$ puzzle? – Michael Albanese Feb 3 '13 at 11:01
It is a 15-puzzle with only 9 items. – ceving Feb 3 '13 at 11:02
Ok it would be more precise to call it 8-puzzle. – ceving Feb 3 '13 at 11:07
Where does Wikipedia claim this is solvable? – Chris Eagle Feb 3 '13 at 11:17
A careful read of the Wikipedia article shows that what they're claiming is that if it's solvable then no more than 31 single tile moves required. Before that it definitely says the $n$ puzzle is only solvable for even permutations. – coffeemath Feb 3 '13 at 14:29
If you ignore the gap and just look at the ordered sequence of numbers, any "horizontal" move leaves the sequence unchanged and any "vertical" move of the puzzle has the form $$(\ldots, x, y, z, \ldots)\to (\ldots, y, z, x, \ldots)$$ or vice versa. These are even permutations and therefore the group of possible permutations is a subgroup of $A_8$ (alternating group) and cannot be all of $S_8$ (full symmetric group). The situation in your post corresponds to an odd permutation of the target ordering and therefore is not solvable.
-
Is it really a group? It's not relevant, but it would be great if you could elaborate on that. It doesn't look obvious to me. – Karolis Juodelė Feb 3 '13 at 14:21
Can you give me a hint, where to find an explanation of odd and even permutations? I do not understand, why the shown move is even and why getting the target ordering is odd. – ceving Feb 13 '13 at 14:00
Though It's old question but I am trying to answer it.
There is a method to check whether the given state is solvable or not.
Problem State:
1|2|3
-+-+-
4|5|6
-+-+-
|8|7
Write it in a linear way, 1,2,3,4,5,6,8,7 - Ignore the blank tile
Now find the number of inversion, by counting tiles precedes the another tile with lower number.
In our case, 1,2,3,4,5,6,7 is having 0 inversions, and 8 is having 1 inversion as it's preceding the number 7.
Total number of inversion is 1 (odd number) so the puzzle is insolvable.
Let's take another example,
5|2|8
-+-+-
4|1|7
-+-+-
|3|6
5 precedes 1,2,3,4 - 4 inversions 2 precedes 1 - 1 inversion 8 precedes 1,3,4,6,7 - 5 inversions 4 precedes 1,3 - 2 inversions 1 precedes none - 0 inversions 7 precedes 3,4 - 2 inversions 3 precedes none - 0 inversions 6 precedes none - 0 inversions
total inversions 4+1+5+2+0+2+0+0 = 14 (Even Number) So this puzzle is solvable.
Here is a function which I have used in Flash Game to check whether the given puzzle is solvable or not
public function checkSolvable(pList:Array):Boolean{
trace("checkSolvable called with : " + pList);
var inversions:Number = 0;
for(var i:int=0;i<pList.length;i++){
for(var j:int=i+1;j<pList.length;j++){
if(pList[j]>pList[i]){
inversions++;
}
}
}
if(inversions%2 == 1){
trace("It's Unsolvable");
return false;
}else{
trace("It's Solvable");
return true;
}
}
Visit http://www.cs.bham.ac.uk/~mdr/teaching/modules04/java2/TilesSolvability.html for detailed information.
-
Late answer, I know, but I'm expanding on Hagen von Eitzen's answer in a slightly more elementary way, if it's still of interest.
Short answer: This state is not solvable.
First note that every permutation can be represented as a graph of disjoint cycles (see cycle notation). In the usual way, then, we represent a game state as a permutation of the 8 non-blank tiles, flattened to row major order.
Now, we can show that the parity (oddness/evenness) of the number of cycles is invariant under the sliding of the tile. To see why, we only need to consider vertical moves, because horizontal moves preserve the permutation. Here, notice that exactly $3$ elements $x, y, z$ change position to $z, x, y$. We proceed by focusing on the predecessors of $x, y, z$ in their respective cycles. Now there are $3$ cases to consider:
• Same cycle: All elements are in the same cycle. In this case, exactly one cycle remains in the place of the old cycle, so the total number of cycles is preserved. See the image below for this case. Green circles represent predecessors, blue circles represent $x, y, z$. Similar pictures for the other cases can be drawn.
• Two cycles: In this case, there is one cycle containing exactly $2$ elements, and one cycle containing $1$ element. Here, the swap generates $2$ new cycles in place of the old $2$ cycles. So again, the total number of cycles is preserved.
• Three cycles: In this case, all three cycles merge into one cycle. So the total number of cycles decreases by $2$, preserving the parity of the total number of cycles.
Since parity is preserved in each case, it is always preserved. Hence you cannot get from a permutation with an odd number of cycles (like the one in the question) to one with an even number of cycles, by only sliding the blank tile.
- | 2015-08-29T09:45:38 | {
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https://physics.stackexchange.com/questions/187765/can-i-simply-reverse-the-indices-in-a-contraction | # Can I simply reverse the indices in a contraction?
Suppose I have something like
$$\left( \nabla_\mu \nabla_\beta - \nabla_\beta \nabla_\mu \right) V^\mu = R_{\nu \beta} V^\nu$$
Can since all the terms involving $\mu$ on the left and $\nu$ on the right are contractions, can I simply do:
$$\left( \nabla^\mu \nabla_\beta - \nabla_\beta \nabla^\mu \right) V_{\mu} = R^\nu_{\beta} V_{\nu}$$
Seems like a pretty nifty trick.
• You can, except that you should keep track of what the position of the indices are when raised. So you should write $R^\nu{}_\beta$ rather than $R^\nu_\beta$. Jun 5, 2015 at 2:36
• But my concern is that in the first term on the LHS, the gradient operator is acting on the gradient operator via $\nabla_\mu (\nabla_\beta) V^\mu$ and not on the vector $V^\mu$. I thought for contractions, they have to act on one another? Jun 5, 2015 at 2:37
• No, that's not true. It is acting as $\nabla_\mu ( \nabla_\beta V^\mu )$. It is certainly acting on $V^\mu$. Similarly, the second term is $\nabla_\beta ( \nabla_\mu V^\mu)$. Jun 5, 2015 at 2:38
• Ok then if I have something like $F_{\mu \nu} = \nabla_\mu A_\nu - \nabla_\nu A_\mu$, how what will $\nabla^\mu F_{\mu \nu}$ look like? Jun 5, 2015 at 2:55
• It will look like $\nabla^\mu ( \nabla_\mu A_\nu ) - \nabla^\mu ( \nabla_\nu A_\mu)$. It acts on everything to the right of it unless otherwise specified by explicit brackets. Did you want me to explicitly specify what this is? Jun 5, 2015 at 2:59
Typically yes, but generally no. That is, your exchange of index positions relies on something called metric compatibility. This is the assumption that the covariant derivative of the metric is zero: $$\nabla_\alpha g_{\beta \gamma} = 0.$$ This is actually a condition imposed on $\nabla$. It turns out [see, e.g., Wald p. 35] that for any metric there exists exactly one such $\nabla$ (assuming also, as Danu points out, that the covariant derivative is torsion free). So typically in GR, this assumption is made because it simplifies things so much -- as in your case.
So if you assume metric compatibility, then your expression is valid. In particular, from the fact that $g_{\beta \gamma} g^{\gamma \epsilon} = {\delta_\beta}^\epsilon$ is constant (so its derivative is zero), metric compatibility also implies $\nabla_\alpha g^{\beta \gamma} = 0$. Then, you can calculate \begin{align} \left( \nabla_\mu \nabla_\beta - \nabla_\beta \nabla_\mu \right) V^\mu &= \left( \nabla_\mu \nabla_\beta - \nabla_\beta \nabla_\mu \right) \left( g^{\mu \tau} V_\tau \right) \\ &= \nabla_\mu \nabla_\beta\left( g^{\mu \tau} V_\tau \right) - \nabla_\beta \nabla_\mu \left( g^{\mu \tau} V_\tau \right) \\ &= \nabla_\mu \left( g^{\mu \tau} \nabla_\beta V_\tau \right) - \nabla_\beta \left( g^{\mu \tau} \nabla_\mu V_\tau \right) \\ &= g^{\mu \tau} \nabla_\mu \nabla_\beta V_\tau - \nabla_\beta \nabla^\tau V_\tau \\ &= \nabla^\tau \nabla_\beta V_\tau - \nabla_\beta \nabla^\tau V_\tau \\ &= \nabla^\mu \nabla_\beta V_\mu - \nabla_\beta \nabla^\mu V_\mu \end{align} I've used metric compatibility three times here, to move the metric from one side of a $\nabla$ to the other. I've also used the definition $\nabla^\alpha = g^{\alpha \beta} \nabla_\beta$.
That's the left-hand side, at least. As for the right-hand side, @Prahar rightly points out that you should try to be careful with the order of your indices, since a lot of tensors are not symmetric, and if you get into the habit of being sloppy with indices it can hurt. It so happens that the Ricci tensor $R_{\mu\nu}$ is symmetric, so you can get away with it in this particular case, but it's easy to be careful. Anyway, the right-hand side is straightforward: \begin{align} R_{\nu\beta} V^\nu &= R_{\nu\beta} g^{\nu \sigma} V_\sigma \\ &= g_{\nu \mu} {R^{\mu}}_{\beta} g^{\nu \sigma} V_\sigma \\ &= {R^{\mu}}_{\beta} g_{\mu \nu} g^{\nu \sigma} V_\sigma \\ &= {R^{\mu}}_{\beta} {\delta_{\mu}}^{\sigma} V_\sigma \\ &= {R^{\mu}}_{\beta} V_\mu \\ &= {R^{\nu}}_{\beta} V_\nu \end{align} Now, since the Ricci tensor is symmetric, you could have swapped the order of the indices on $R_{\nu\beta}$ at the beginning, and you would have come up with ${R_{\beta}}^{\nu} V_\nu$ at the end. This is the only reason you'll some times see those indices written directly above each other; if $R$ had been some asymmetric tensor, this wouldn't be true.
Finally, I'll also point out that these index swaps on the two different sides are independent of each other, so you can do either or both independently.
• In the second line, for the term on the right why did you raise $\mu$ to $\nabla^\mu$? Jun 5, 2015 at 3:10
• Because I was typing too fast...
– Mike
Jun 5, 2015 at 3:11
• From third line to fourth line, how did the term on the right go from $g^{\mu \tau} \nabla_\mu$ to $\nabla^\tau V_\tau$? Can you lower/raise indices on operators like $\nabla_\alpha$ too? Jun 5, 2015 at 3:13
• In general, you should be wary of raising and lowering indices on operators. But in this case, that's actually the definition of $\nabla^\tau$. (And because of possible metric incompatibility, it's important to remember which side the metric comes on in that definition.)
– Mike
Jun 5, 2015 at 3:17
• So as long as $\nabla_\alpha g^{\beta \gamma} = 0$ holds, I can raise/lower indices on operators using $g^{\alpha \beta}$? Jun 5, 2015 at 3:22
Sure! This comes from the way that raising and lowering of indices is defined in terms of the metric tensor. Suppose we have a rank $(0, 2)$ tensor $S_{ab}.$ Then its trace is $S^{a}{}_a.$ We may write:
$$S^{a}{}_{a} = g^{ab} S_{b a} = S_{b}{}^{b}.$$
Since $a$ and $b$ are dummy indices, we may interchange them readily. This yields the final equation
$$S^{a}{}_{a} = S_{a}{}^a.$$
So any time you have a contraction, you may switch which is the "up-index" and which is the "down-index" without changing your result. | 2022-10-06T10:37:30 | {
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https://math.stackexchange.com/questions/2903451/how-do-i-find-sqrt3-i/2903479 | # How do I find $\sqrt[3]{-i}$?
I'm asked to evaluate $$\sqrt[3]{-i}$$ I suppose $\sqrt[3]{-i}=(a+bi)$ $$\implies (a+bi)^3=-i$$ $$\implies \Im \left( (a+bi)^3 \right) =\Im \left( (-i) \right)$$ $$\implies 3a^2b-b^3=-1$$ Now how am I supposed to find $a$,$b$? Aren't there infinitely of them instead of just three?
• There cannot be infinitely many of them, because $\sqrt[3]{-i}$ are the roots of the complex polynomial $x^3 = -i$, and the fundamental theorem of algebra, says that this will have exactly three complex roots. Do you know what the polar form / polar coordinates are of a complex number? – астон вілла олоф мэллбэрг Sep 3 '18 at 3:45
• @астонвіллаолофмэллбэрг For every $a$ there is a corresponding $b$. – user588826 Sep 3 '18 at 3:49
• But what about $\mathcal{R}( (a+bi)^3 ) = \mathcal{R}(-i)$ ? It gives you another equation, so there is not infinitely many. – Ahmad Bazzi Sep 3 '18 at 3:53
• You have $3a^2b - b^3 = 1$, but then you also have $a^3 - 3ab^2 = 0$, by equating the real parts. One of these equations may have infinitely many solutions, but now we have two simultaneous equations. – астон вілла олоф мэллбэрг Sep 3 '18 at 3:57
• @астонвіллаолофмэллбэрг Can we deduce something from both of them? – user588826 Sep 3 '18 at 3:59
Let us start as you did:
$\sqrt [3] {-i}= a+bi$
$-i=(a+bi)^3$
$-i=a^3+3a^2bi-3ab^2-b^3i$
Therefore we get $2$ equations:
$a^3-3ab^2=0 \ \ \ \ldots(1)$
$3a^2b-b^3=-1 \ldots (2)$
Solving for $(1)$:
$a(a^2-3b^2)=0$
$\therefore a(a-b\sqrt3)(a+b\sqrt 3)=0$
So:
$a=0$
$a=b\sqrt3$
$a=-b\sqrt 3$
Solving for $(2)$:
$3a^2b-b^3=-1$
When $a=0$:
$-b^3=-1$
$b^3=1$
$b=1$
When $a=\sqrt {3b^2}$:
$9b^3-b^3=-1$
$b^3=-\frac 18$
$b=-\frac 12$
$\therefore a=\frac{\sqrt 3}2$
When $a=-\sqrt {3b^2}$
$9b^3-b^3=-1$
$b^3=-\frac 18$
$b=-\frac 12$
$\therefore a=-\frac{\sqrt 3}2$
So your $3$ solutions are:
$(a,b)=(0,1);(\frac{\sqrt 3}2,-\frac 12);(-\frac{\sqrt 3}2,-\frac 12)$
I sincerely hope this helps you to understand why there are only $3$ solutions without mentioning the fundamental theorem of algebra. Good luck!
You can solve this as $$(-i)^{\frac{1}{3}} = ( e^{i \frac{3\pi}{2} + i2k\pi})^{\frac{1}{3}}$$ which gives us three distinct roots $e^{i z_k}$, for $k = 0,1,2$, where \begin{align} z_0 &= \frac{1}{3}\frac{3\pi}{2} + \frac{2(0)\pi}{3} = \frac{\pi}{2} \\ z_1 &= \frac{1}{3}\frac{3\pi}{2} + \frac{2(1)\pi}{3} = \frac{7\pi}{6} \\ z_2 &= \frac{1}{3}\frac{3\pi}{2} + \frac{2(2)\pi}{3} = \frac{11\pi}{6} \\ \end{align} If you insist on solving it your way, then $$(a+bi)^3 = -i$$ means $$a^3 + 3a^2bi - 3ab^2 - b^3i = -i$$ which means \begin{align} a^3 - 3ab^2 &= 0 \\ 3a^2b - b^3 &= -1 \end{align} which is \begin{align} a^2 - 3b^2 &= 0 \\ 3a^2b - b^3 &= -1 \end{align} or \begin{align} (a - \sqrt{3} b)(a + \sqrt{3} b) &= 0 \\ 3a^2b - b^3 &= -1 \end{align} The first equation suggests either $$a = \pm \sqrt{3} b$$ Replacing this in the second equation gives $$3(\sqrt{3} b)^2b - b^3 = -1$$ which is $$9b^3 - b^3 = -1$$ i.e. $$b = -\frac{1}{2}$$ This will give us $$a = \pm \sqrt{3} (-\frac{1}{2})$$ which means we get two solutions \begin{align} (a_1,b_1) &= (-\sqrt{3},-\frac{1}{2}) \\ (a_2,b_2) &= (\sqrt{3},-\frac{1}{2}) \\ \end{align} which are actually what we found before, i.e. \begin{align} a_1 + ib_1 &= e^{i z_1} \\ a_2 + ib_2 &= e^{i z_2} \\ \end{align} Then you've got one more root you need to find (which is the obvious one), where $a_0 = 0$ and $b_0 = 1$, i.e. $i^3 = -i$. Also, you can verify that \begin{align} a_0 + ib_0 &= e^{i z_0} \end{align}
When you take the real part in each side, you get another equation. I don't think that is the best way to tackle the problem though. I think it is better:
$$z^3=-i$$
Then we know the magnitude of $$z$$ is $$1$$, and the angle is such that when multiplied by 3 it lies at $$\frac{3\pi}{2}+2k\pi, k\in \text{Z}$$ (this is pointing downwards). The possible angles are:
$$\frac{\pi}{2}+\frac{2k\pi}{3}$$
So take the cases $$k=0, k=1$$ and $$k=2$$ and you are done, because then the angles start to loop (just adding $$2\pi$$). The solutions are:
$$z=\cos{\left(\frac{\pi}{2}+\frac{2k\pi}{3}\right)} + i\sin{\left(\frac{\pi}{2}+\frac{2k\pi}{3}\right)}, k\in\text{{0,1,2}}$$
• $\LaTeX \text{ Tip}:$ Use \sin x and \cos x to obtain $\sin x \text{ and } \cos x.$ – Mohammad Zuhair Khan Oct 21 '18 at 6:58
• I edited with better Latex – Villa Oct 22 '18 at 4:21
Alt. hint: the cube roots of $\,-i\,$ are the solutions to $\,z^3=-i \iff z^3+i=0\,$. Using that $\,i = -i^3\,$ and the identity $\,a^3-b^3=(a-b)(a^2+ab+b^2)\,$, the latter equation can be written as:
$$0 = z^3+i = z^3-i^3=(z-i)(z^2+iz + i^2)=(z-i)\big(-(iz)^2+iz-1\big)$$
The first factor gives the (obvious) solution $\,z=i\,$, and the second factor is a real quadratic in $\,iz\,$ with roots $\,iz = (1 \pm i \sqrt{3})/2 \iff z = \ldots$
It is easier to find the cube roots of -i using the polar form.
Note that $-i = e^{3i\pi /2}$ thus the cube roots are $e^{i\pi/2},e^{i\pi/2+2i\pi/3}, e^{i\pi/2+4i\pi/3}$
You can easily find these roots in $a+bi$ form if you wish to do so. | 2019-08-22T20:37:09 | {
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https://math.stackexchange.com/questions/3337634/approximating-pi-and-ln-2-with-i-k-int-0-infty-left-textsechx-tanh | # Approximating $\pi$ and $\ln 2$ with $I_k=\int_0^\infty \left(\text{sech}x\tanh\tfrac12x\right)^k\,dx$ for integer $k$
Consider the following integral:
$$I_k=\int_0^\infty \left(\text{sech}x\tanh\tfrac12x\right)^k\,dx$$
where $$k\in\Bbb N$$.
If we evaluate $$I_1$$, $$I_2$$, $$I_3$$, etc. we get the following pattern:
• $$I_1=\log(2)$$
• $$I_2=-3+\pi$$
• $$I_3=\frac 72-5\log(2)$$
• $$I_4=22-7\pi$$
• $$I_5=-\frac{341}{12}+41\log(2)$$
• $$I_6=-\frac{968}{5}+\frac{493}{8}\pi$$
• ...
From this data we can see that:
\begin{align} \pi&=3+I_2=\frac{22}{7}-\frac17 I_4=\frac{7744}{2465}+\frac{8}{493}I_6 \\[4pt] \log(2)&=0+I_1=\frac{7}{10}-\frac15I_3=\frac{341}{492}+I_5 \end{align}
And because $$I_k$$ decreases very rapidly($$I_{10}$$ is in the order of $$1e6$$) we can set $$I_k\approx 0$$ for high $$k$$ and we get rational approximations of both, $$\pi$$ and $$\log(2)$$, for $$I_{2k}$$ and $$I_{2k+1}$$, respectively, that apparently go on forever.
I see some equations that somehow "encode" the information of a given number, but how is it that this integral has the information of both $$\pi$$ and $$\log(2)$$, apparently unrelated numbers?
Thanks.
• Please write an informative title that will allow others to find the problem and solution later. Cutesy uninformative titles are annoying and unhelpful. – David G. Stork Aug 29 '19 at 2:04
• Interesting exercise, for sure and $\to +1$ for the question. – Claude Leibovici Aug 29 '19 at 4:21
• $\pi$ and $\ln 2$ are not actually very unrelated. The $\ln 2$ constant shows up naturally, for instance, in the study of integrals like $$\int_0^{\pi/2}\ln(\sin t)dt=-\frac{\pi}{2}\ln 2.$$ See also the relations between $\pi$, $\ln2$, $e$, and Catalan's constant $\mathrm{G}$ in this post. – clathratus Nov 19 '19 at 5:25
I'm not sure exactly what kind of an answer you're looking for---but the following might be a helpful heuristic:
1. A standard substitution transforms the integrals $$I_k=\int_0^\infty \left(\operatorname{sech} x \tanh\frac{x}{2}\right)^k\,dx ,$$ (for $$k$$ a nonnegative integer, which we henceforth assume) into integrals of rational functions of a new variable.
2. After applying the method of partial fractions, we can integrate term-by-term, and the only non-rational functions that occur in the antiderivatives have the form $$\log q(u)$$ and $$\arctan r(u)$$ for some affine functions $$q, r$$ with rational coefficients.
3. At typical limits of integration the functions $$q(u)$$ usually take on values of rational number with small numerator and denominator, and so the contribution of these terms is a sum of logarithms of small numbers---of course, the smallest positive integer with nonzero logarithm is $$2$$, so it's no surprise that $$\log 2$$ occurs in the values of such integrals often.
4. Likewise, if we can write $$\arctan v$$ without $$\arctan$$, typically the result is some rational multiple of $$\pi$$. In particular, we have $$\arctan 1 = \frac{\pi}{4}$$ and $$\lim_{v \to \infty} \arctan v = \frac{\pi}{2}$$, no it's no surprise that $$\pi$$ occurs often either.
More explicitly: Applying the hyperbolic analogue $$x = 2 \operatorname{artanh} t$$ of the Weierstrass substitution transforms $$I_k$$ into an integral with a rational integrand, $$I_k = 2 \int_0^1 \frac{(1 - t^2)^{k - 1} t^k \,dt}{(1 + t^2)^k}.$$ The next step for evaluating this integral in terms of elementary functions depends on the parity of $$k$$.
For $$k$$ odd, substituting $$u = t^2, \qquad du = 2 t \,dt$$ gives $$I_k = \int_0^1 \frac{(1 - u)^{k - 1} u^{(k - 1) / 2}}{(u + 1)^k} .$$ Expanding the integrand using partial fractions gives $$I_k = \int_0^1 \left(P(u) + \frac{A_t}{(u + 1)^k} + \cdots + \frac{A_2}{(u + 1)^2} + \frac{A_1}{u + 1} \right) du$$ for some rational polynomial $$P(u)$$ and rational coefficients $$A_1, \ldots, A_k$$. But the antiderivative of every term but $$\frac{A_1}{u + 1}$$ is a rational function, so each contributes some rational number, altogether contributing a rational total, call it $$\alpha$$. The value of the last term is $$A_1 \int_0^1 \frac{du}{u + 1} = A_1 \log u\vert_0^1 = A_1 \log 2$$, so $$\boxed{I_k = \alpha_k + \beta_k \log 2}$$ for some rational numbers $$\alpha_k$$ and $$\beta_k := A_1$$. A tedious but straightforward induction shows that $$\beta_k \neq 0$$.
For $$k$$ even, that substitution is not available (or more precisely, it makes the integrand worse), but applying the method of partial fractions again gives $$I_k = \int_0^1 \left(P(t) + \frac{A_k t + B_k}{(t^2 + 1)^k} + \cdots + \frac{A_2 t + B_2}{(t^2 + 1)^2} + \frac{A_1 t + B_1}{(t^2 + 1)} \right) dt$$ for some polynomial $$P(t)$$ and rational coefficients $$A_1, \ldots, A_k, B_1, \ldots, B_k$$. But our integrand in $$u$$ is even, and so $$A_1 = \cdots = A_k = 0$$. On the other hand, for $$m > 1$$, the integral of $$\frac{1}{(t^2 + 1)^m}$$ satisfies the reduction formula of the form $$\int \frac{dt}{(t^2 + 1)^m} = f_m(t) + \gamma_m \int \frac{dt}{(t^2 + 1)^{m - 1}}$$ for some rational function $$f_m$$ and rational constant $$\gamma_m$$ (both depending on $$m$$). In particular, induction gives $$\int \frac{dt}{(t^2 + 1)^m} = g_m(x) + \delta_m \int \frac{dx}{t^2 + 1} = g_m(t) + \delta_m \arctan t + C$$ for some rational function $$g_m$$ and rational constant $$\delta_m$$ (again both depending on $$m$$). Substituting back into our previous formula for $$I_k$$, we have that $$I_k = [h(t) + \zeta \arctan t]\vert_0^1 = h(1) - h(0) + \beta_k \pi$$ for some rational function $$h$$ and rational constant $$\beta_k$$, and so $$\boxed{I_k = \alpha_k + \beta_k \pi}$$ for rational numbers $$\alpha_k := h(1) - h(0)$$ and $$\beta_k$$. Again, an induction shows that $$\beta_k \neq 0$$ for $$k > 0$$.
Remark 1 Essentially the same phenomenon occurs for the similar family $$J_j := \int_0^1 \frac{x^{2 j} (1 - x)^{2 j}dx}{1 + x^2} ,$$ which generalizes the so-called Dalzell integral (the case $$j = 2$$, which gives $$\frac{22}{7} - \pi$$). If $$j$$ is odd, we get an expression of the form $$\alpha + \beta \log 2$$, and if $$j$$ is even we get $$\alpha + \beta \pi$$, with $$\alpha, \beta \neq 0$$ in both cases.
Remark 2 We can use the resulting integral expressions to derive rational approximations of $$\pi$$ and $$\log 2$$. On the interval $$[0, 1]$$, $$\frac{1}{2^k} \leq \frac{1}{(1 + t^2)^k} \leq 1$$, giving the bounds $$\frac{1}{2^k} E_k < I_k < E_k, \\ \textrm{where} \quad E_k = 2 \int_0^1 u^k (1 - u^2)^k du = \frac{\Gamma(k) \Gamma\left(\frac{1}{2} k + \frac{1}{2}\right)}{\Gamma\left(\frac{3}{2} k + \frac{1}{2}\right)} \sim \frac{\sqrt{2 \pi}}{\sqrt{k}} \left(\frac{2}{3 \sqrt{3}}\right)^k ,$$ so for any particular $$k$$, rearranging gives rational bounds for $$\pi$$ or $$\log 2$$. (For odd $$k = 2 l + 1$$, by the way, we have $$E_{2 l + 1} = \frac{(2 l)! l!}{(3 l + 1)!}$$.)
For example, taking $$k = 2$$ gives $$I_2 = \pi - 3$$ and $$L_2 = \frac{16}{105}$$, and rearranging gives $$\frac{319}{105} < \pi < \frac{331}{105} .$$
• Very nice work with the bounds. – Claude Leibovici Aug 29 '19 at 7:46
• $K_j=\frac{\Gamma (2 j+1)^2}{\Gamma (4 j+2)}\sim \frac{\sqrt{\pi } 2^{-4 j-\frac{3}{2}}}{\sqrt{j}}$ seems to be interesting – Claude Leibovici Aug 29 '19 at 8:16
Probably not an answer but too long for the comment section.
You are considering $$I_k=\int_0^\infty \Big( \text{sech}(x) \tanh \left(\frac{x}{2}\right) \Big)^k \,dx$$ Let $$x=2 \tanh ^{-1}(t)$$ which makes $$J_k=\int \Big( \text{sech}(x) \tanh \left(\frac{x}{2}\right) \Big)^k \,dx=2 \int\left(1-t^2\right)^{k-1} \left(\frac{t}{1+t^2}\right)^k\,dt$$ that is to say $$J_k=2\frac{ t^{k+1}}{k+1}\, F_1\left(\frac{k+1}{2};1-k,k;\frac{k+3}{2};t^2,-t^2\right)$$ where appears the Appell hypergeometric function of two variables.
Integrating between $$0$$ and $$1$$, after simplification, this leads to $$I_k=\, _2\tilde{F}_1\left(k,\frac{k+1}{2};\frac{3 k+1}{2} ;-1\right)\, \Gamma (k)\, \Gamma \left(\frac{k+1}{2}\right)$$ where appears the regularized hypergeometric function.
As you noticed $$I_{2k}=a_k-b_k \pi$$ and $$I_{2k+1}=c_k-d_k \log(2)$$. So, for sure, if you make $$I_k\sim \epsilon$$, you have rational approximations of $$\pi$$ and $$\log(2)$$. The small problem I see is that they are not extremely accurate.
For example $$I_{20}=\frac{2357262305394688}{1119195}-\frac{21968591457761 \pi }{32768}\approx 1.7 \times 10^{-11}$$ would give as a rational approximation $$\pi \approx \frac{77242771223173136384}{24587137716568822395}=\color{red}{3.1415926535897932384}88023$$ while $$\pi \approx \frac{21053343141}{6701487259}=\color{red}{3.141592653589793238462}382$$ is better.
Similarly $$I_{21}= 4354393801 \log (2)-\frac{100374690765091043}{33256080}\approx 5.0 \times 10^{-12}$$ would give as a rational approximation $$\log(2)\approx \frac{100374690765091043}{144810068597560080}=\color{red}{0.69314718055994530941}60873$$ while $$\log(2)\approx \frac{34733068453}{50109225612}=\color{red}{0.693147180559945309417}8461$$ is better.
Too long for a comment.
We have the convenient relation $$\text{sech}(x)\tanh(x/2)=\frac{2}{e^x+1}-\frac{2}{e^{2x}+1},$$ so let $$L_n=I_{2n+1}=2^{2n+1}\sum_{k=0}^{2n+1}(-1)^k{2n+1\choose k}\int_0^{\infty}\frac{dx}{(e^x+1)^{2n-k+1}(e^{2x}+1)^k}$$ and $$P_n=I_{2n}=2^{2n}\sum_{k=0}^{2n}(-1)^k{2n\choose k}\int_0^{\infty}\frac{dx}{(e^x+1)^{2n-k}(e^{2x}+1)^k}.$$ These integrals may be easier to evaluate explicitly. | 2020-06-04T20:48:47 | {
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http://slideplayer.com/slide/3168187/ | Geometry of R2 and R3 Dot and Cross Products.
Presentation on theme: "Geometry of R2 and R3 Dot and Cross Products."— Presentation transcript:
Geometry of R2 and R3 Dot and Cross Products
Dot Product in R2 Let u = (u1, u2) and v = (v1, v2) then the dot product or scalar product, denoted by u.v, is defined as u.v = u1v1 + u2v2
Dot Product in R3 Let u = (u1, u2, u3) and v = (v1, v2, v3) then the dot product or scalar product, denoted by u.v, is defined as u.v = u1v1 + u2v2 + u3v3
Example Find the dot product of each pair of vectors
u = (-3, 2, -1); v = (-4, -3, 0) u = (-4, 0, -2); v = (-3, -7, 6) u = (-6, 3); v = (5, -8)
Theorem 1.2.1 Let u and v be vectors in R2 or R3, and let c be a scalar. Then u.v = v.u c(u.v) = (cu).v = u. (cv) u.(v + w) = u.v + u.w u.0 = 0 u.u = ||u||2 Prove c and e in class.
Theorem 1.2.2 Let u and v be vectors in R2 or R3, and let be the angle they form. Then u.v = ||v|| ||u|| cos If u and v are nonzero vectors, then Proof: start with the law of cosines. Convert the sides in terms of the norm. Also expand the norm of v-u and set it equal to the previous.
Example Find the angle between each pair of vectors.
u = (-1, 2, 3); v = (2, 0, 4) u = (1, 0, 1); v = (-1, -1, 0)
Orthogonal Vectors Two vectors u and v in R2 or R3 are orthogonal if u.v = 0. Orthogonal, Normal, and Perpendicular, all mean the same.
Theorem 1.2.3 Let u and v be nonzero vectors in R2 or R3 and let be the angle they form. Then is An acute angle if u.v > 0 A right angle if u.v = 0 An obtuse angle if u.v < 0
Cross Product (Only in R3 )
Let u = (u1, u2, u3) and v = (v1, v2, v3) be nonzero vectors in R3. Then the cross product, denoted by u x v, is the vector (u2v3 – u3v2, u3v1 – u1v3, u1v2 – u2v1)
Cross Product (Convenient notation )
Let u = (u1, u2, u3) and v = (v1, v2, v3) be nonzero vectors in R3. Then u x v, is the vector obtained by evaluating the determinant:
Example Find the cross product of the following vectors
u = (-1, 1, 0); v = (2, 3, -1)
Theorem 1.2.4 The vector uxv is orthogonal to both u and v.
Theorem 1.2.4 Let u, v, and w be vectors in R3, and let c be a scalar. Then u x v = –(v x u) u x (v + w) = (u x v) + (u x w) (u + v) x w = (u x w) + (v x w) c(u x v ) = (cu) x v = u x (cv) u x 0 = 0 x u = 0 u x u = 0 ||u x v|| = ||u|| ||v|| sin = (||u|| ||v|| – ||u.v||2) Do parts c and g.
Cross Product: Area Let u, and v, be vectors in R3, Then the area of the parallelogram determined by u and v is ||u x v|| = ||u|| ||v|| sin
Example Find the area of the parallelogram determined by the vectors u = (-1, 1, 0) and v = (2, 3, -1).
Homework 1.2
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https://eceaus.org/wddch/w8s2p.php?3a379d=exponential-reliability-function | The Exponential is a life distribution used in reliability engineering for the analysis of events with a constant failure rate. The cumulative hazard function for the exponential is just the integral of the failure rate or $$H(t) = \lambda t$$. Persistence in Reliability Analysis of the Exponential Assumption Despite the inadequacy of the exponential distribution to accurately model the behavior of most products in the real world, it is still widely used in today’s reliability practices, standards and methods. Location Parameter The location parameter is … The exponential distribution exhibits infinite divisibility. The functions for this distribution are shown in the table below. gamma distribution. Also, another name for the exponential mean is the Mean Time To Fail or MTTF and we have MTTF = $$1/\lambda$$. for t > 0, where λ is the hazard (failure) rate, and the reliability function is. In other words, reliability of a system will be high at its initial state of operation and gradually reduce to its lowest magnitude over time. The Exponential Conditional Reliability Function. The exponential conditional reliability function is: Abstract: This paper considers a class of an efficient 'two-stage shrinkage testimator' (TSST) of 'reliability function' of 'exponential distribution', and the class uses additional information which can be obtained from the past practices, and in the form of past initial … Example. The exponential distribution is actually a … Reliability Prediction Using the Exponential Distribution The exponential distribution applies when the failure rate is constant - the graph … The probability density function (pdf) of an exponential distribution is (;) = {− ≥, 0 is the parameter of the distribution, often called the rate parameter.The distribution is supported on the interval [0, ∞). The distribution has one parameter: the failure rate (λ). 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The table below applications the distribution has one parameter: the failure density function is of [ math ],! Familiar shape shown below reliability of electronic systems, which do not typically experience type... | 2021-04-18T05:03:59 | {
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https://math.stackexchange.com/questions/3678191/show-that-if-x-is-a-bounded-subset-of-textbfr-then-the-closure-overlin | Show that if $X$ is a bounded subset of $\textbf{R}$, then the closure $\overline{X}$ is also bounded.
Show that if $$X$$ is a bounded subset of $$\textbf{R}$$, then the closure $$\overline{X}$$ is also bounded.
MY ATTEMPT
Since $$X$$ is bounded, we have that $$X\subseteq[-M,M]$$.
Let us consider that $$x$$ is an adherent point of $$X$$.
Then there exists a sequence $$(x_{n})_{n=m}^{\infty}$$ entirely contained in $$X\subseteq[-M,M]$$ which converges to $$x$$.
Since $$[-M,M]$$ is closed and bounded, due to the Heine-Borel theorem, the sequence $$(x_{n})_{n=m}^{\infty}$$ admits a subsequence which converges to some $$L\in[-M,M]$$.
Once a sequence converges iff each of its subsequences converges to the same value, we conclude that $$L = x\in[-M,M]$$.
In other words, we have just proven the $$\overline{X}\subseteq[-M,M]$$, which means the closure is bounded.
Could someone please verify if my proof is correct?
• Maybe an easier way: $X \subset [-M,M]$, which is closed. So the closure $\overline{X} \subset [-M,M]$, which implies that $\overline{X}$ is also bounded. About your proof, I don't think Heine-Borel is necessary. You have a sequence contained in $[-M,M]$, which converges to $x$. Since $[-M,M]$ is closed, you know that $x$ lies in $[-M,M]$. Your proof is correct though. – M. Wang May 16 '20 at 19:27
• Thanks for the contribution! I really liked the proposed approach. Would you mind to write it as a full answer so I can upvote it? – BrickByBrick May 16 '20 at 19:30
A maybe faster way:
By assumption, $$X$$ is bounded so it lies in some $$[-M,M]$$ as you said. Since $$[-M,M]$$ is closed, the closure $$\overline{X}$$ also lies in $$[-M,M]$$. This shows that $$\overline{X}$$ is bounded.
As I also said in the comment, using Heine-Borel is a bit overkill. But your solution is correct.
Your proof is ok, c.f.M.Wangs comment.
Another route:
$$\overline {X}= X \cup X',$$ where $$X'$$ are the limit points of $$X$$.
If $$a \in X$$ we are done (bounded), since
$$X \subset [-M,+M]$$, for a bound $$M>0$$.
Let $$a \not \in X$$.
Assume $$X'$$ is not bounded.
For $$2M >0$$, real, there is a $$a \in X'$$ s.t.
$$a >2M$$. Since $$a$$ is a limit point of $$X$$ there are points $$x$$ of $$X$$, in every neighbourhood of $$a$$. Let $$x \in X$$ with
$$|x-a|<\epsilon$$,
$$a-\epsilon .
For $$\epsilon we have
$$2M -M , a contradiction.
If $$X\subseteq [a, b]$$ then you can prove that no number outside the interval $$[a, b]$$ can be an adherent point of $$X$$. Thus if $$c\in\overline{X}$$ then $$c\in [a, b]$$ so that $$\overline{X} \subseteq [a, b]$$.
Let's take a number $$c$$ outside $$[a, b]$$. Then either $$c or $$c>b$$. In both cases we have a neighborhood of $$c$$ which lies outside the interval $$[a, b]$$ and therefore does not contain any point of $$X$$. Thus $$c\notin\overline{X}$$.
Note that this does not involve completeness of reals and is just a matter of using definitions. | 2021-01-19T06:06:24 | {
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https://math.stackexchange.com/questions/2117906/cutting-a-m-times-n-rectangle-into-a-times-b-smaller-rectangular-pieces | # Cutting a $m \times n$ rectangle into $a \times b$ smaller rectangular pieces
How many $a \times b$ rectangular pieces of cardboard can be cut from $m \times n$ rectangular piece of cardboard so that the amount of waste("left over" cardboard) is a minimum?
This question was given to me by my Mathematics Teacher as a "brain teaser".
At first I divided $m \times n$ by $a \times b,$ but then I realized that it is not possible in the given situation, as the "left over" cardboard will also be "divided" among $a \times b$.
Next I tried started thinking it terms of perimeter. I imagined all the $a \times b$ rectangles lying side by side in a "rows and columns" format(sort of a grid), without any space between them, thus forming another rectangle/square. I assumed $\lambda$ rows and $\sigma$ columns of those smaller rectangles. After a bit of calculations, my answer came out as following :
Number of smaller $a \times b$ rectangles $\geq \sigma \times \lambda$ = $\left(\dfrac {m - m \, mod \, a }{a}\right) \times \left(\dfrac {n-n\, mod \, b}{b}\right).$
Is my "second" approach correct ?
Is there any alternate way of tackling this problem?
This might seem a 'vague' question, but can this problem be generalized to any given shape ?
Any help will be gratefully acknowledged :).
• – Simply Beautiful Art Jan 28 '17 at 15:22
• @SimplyBeautifulArt That does not seem to help a lot ... one answer says Without Loss Of generality assume boxes to be distinct.... By the way is my approach correct ??? – Nirbhay Jan 28 '17 at 15:24
• I don't believe there is any known way to calculate this with a direct formula, and BeaitifulArt's link seems to agree. – Kaynex Jan 28 '17 at 15:33
• @Kaynex Is my approach OK ??? – Nirbhay Jan 28 '17 at 15:39
• You've asked twice if your approach is "OK". I'd say it's a good start, but it leaves a lot of cases untested. You might imagine a situation where you lay a row of $a\times b$ rectangles length-wise, but then another row turned 90 degrees. Maybe $3a + 5b = m$, so by laying 3 rows one way and 5 rows the other way, you end up with no waste in the $m$ direction. The point is that there are infinity minus one things to think about here. – B. Goddard Jan 28 '17 at 18:32
The approach proposed in the OP may be good, as it gives a waste $W$ of
$$W=(m \mod{a}) \cdot n + (n \mod { b}) \cdot m - (m \mod { a} ) ( n \mod {b})$$
that in most cases is distributed along two adjacent sides. For example, let us take a $120\times 100$ rectangle to be filled with $11\times 7$ rectangles. Using the approach of the OP, the waste is
$$W=(120 \mod 11) \cdot 100 + (100 \mod 7) \cdot 120 - ( 120 \mod 11 )\cdot ( 100 \mod 7) \\ =10 \cdot 100 + 2 \cdot 120 - 10 \cdot 2=320$$
However, I believe that there could be a potentially better strategy. We could assess whether one between $m$ and $n$ can be expressed as $ax+by$, where $x$ and $y$ are integers. If this is the case, we can leave an incomplete region only along a single side of the large rectangle (instead of two). For instance, in the example above, we can note that $120$ can be expressed as $$120=99+21=9 \cdot 11 +3\cdot 7 \,\,\,\,$$ So, dividing the sides equal to $120$ in two parts measuring $99$ and $21$, we can divide the initial rectangle in two rectangles ( $99 \times 100$ and $21 \cdot 100$). The first one can be filled beginning from the side equal to $99$, placing a first sequence of $9$ small rectangles (oriented as $11 \cdot 7$) to cover a $99 \times 7$ area, then another sequence to achieve a $99 \times 14$ area, and so on, arriving to a $99 \times 98$ area. This leaves a partial waste of $2 \cdot 99=198$. The second one can be filled beginning from the side equal to $21$, placing a first sequence of $3$ small rectangles (oriented as $7 \cdot 11$) to cover a $21 \times 11$ area, then another sequence to achieve a $21 \times 22$ area, and so on, arriving to a $21 \times 99$ area. This leaves a partial waste of $1 \cdot 21=21$. The total waste is $198+21=219$. Generalizing, if we can set $m=ax+by$, the waste with this approach is $$W=ax \cdot (n \mod b) + by \cdot (n \mod a)$$
There are several points to be highligted in this solution. First, this approach should be attempted by considering all possibilities of expressing $m$ or $n$ as $ax+by$, and choosing the one that minimizes the waste. In the example used before, other two possibilities could be to set $$m=120=2 \cdot 11 + 14 \cdot 7\,\,$$ or $$n=100=4 \cdot 11 + 8 \cdot 7\,\,$$ Using the same procedure, it can be easily shown that these choices lead to wastes of $142$ and $604$, respectively. Among all three possibilities, that leaving $142$ gives the best solution. Second, even if this approach allows a nearly full packing along three sides, it does not always produce a minor waste than the solution suggested in the OP. The case $W=604$ described above is an example of this. Also note that the solution of the OP could be applied after rotating the initial rectangle by $90^o$ to get
$$W=(100 \mod 11) \cdot 120 + (120 \mod 7) \cdot 100 - ( 100 \mod 11 )\cdot ( 120 \mod 7) \\ =1 \cdot 120 + 1 \cdot 100 - 1 \cdot 10=210$$
which is a better result than that initially obtained. Lastly, note that a "complete" packing (with no waste) can be obtained only in two scenarios, that is to say 1) $a$ divides $m$ and $b$ divides $n$ (or alternatively $a$ divides $n$ and $b$ divides $m$), 2) both $a$ and $b$ divide one (the same) of $m$ and $n$, for example $m$, and the other (in this example $n$) is of the form $n=ax+by \,\,\,$ with $x,y$ integers.
A simple example, for $a=2, b=3$, and a few values for $n$ and $m$ shows that the problem is not easily approachable, specially in the last configuration.
It is in fact a Cutting Stock Problem as rightly indicated by @sas, however simplified by having rectangular and fixed size items.
It is quite an interesting subject, so starting from the sketch above, let's try and establish some facts about,
without pretending to be rigourous and to provide proofs.
Premised that we use the following symbols, for the integral and fractional part and for modulus $$\begin{gathered} x = \left\lfloor x \right\rfloor + \left\{ x \right\} \hfill \\ \frac{x} {t} = \left\lfloor {\frac{x} {t}} \right\rfloor + \left\{ {\frac{x} {t}} \right\} \hfill \\ x = t\left\lfloor {\frac{x} {t}} \right\rfloor + t\left\{ {\frac{x} {t}} \right\} = t\left\lfloor {\frac{x} {t}} \right\rfloor + x\bmod t \hfill \\ \end{gathered}$$ then
1. Upper bound for $N$
Clearly we must have $N a b \leqslant nm$, i.e. $$\bbox[lightyellow]{ N \leqslant \left\lfloor {\frac{{n\,m}} {{a\,b}}} \right\rfloor = \left\lfloor {\,\frac{{n\,m/\gcd (nm,ab)}} {{a\,b/\gcd (nm,ab)}}} \right\rfloor = \left\lfloor {\frac{{n'\,}} {{a'}}\,\frac{{m'}} {{b'}}} \right\rfloor \tag {1} \\ }$$ This also indicates an equivalence in the problem when the parameters are scaled so as to keep the ratio $(nm)/(ab)$ constant, and that we can always reduce the parameters and get $n$ and $m$ be coprime vs. $a$ and vs. $b$.
But, geometrically speaking, the down-scaling looks to be applicable only if $max(a',b') \leqslant min(n',m')$.
An example is given in this sketch
2. Equivalence under rotation, reflection
Clearly the problem does not change under rotation or reflection.
3. "Cross histogram"
Consider to take a horizontal scan and to record, per each unit row, the number of layers of horizontal dimension $a$ and $b$ traversed. Let's do the same on a vertical scan.
Clearly, every $b$ counts for a layer of width $a$ in one direction will correspond to $a$ counts for $b$ in the other direction. And viceversa.
4. Conclusions
Condensing all the considerations above we are led to claim the following:
a) Among the various possible partitions of $m$ and $n$ (in case $n'$, $m'$) as a linear combination of $a$ and $b$, with non-negative integral coefficients, as $$\bbox[lightyellow] { \left\{ \begin{gathered} n = n_{\,a} a + n_{\,b} b + n_{\,r} \hfill \\ m = m_{\,a} a + m_{\,b} b + m_{\,r} \hfill \\ \end{gathered} \right. \tag {2} }$$ we shall choose those which, while giving minimal remainder ($n_r,m_r$), also assure best symmetry around $n/2$,$m/2$, that is $$\bbox[lightyellow] { \left\{ \begin{gathered} n_{\,a} a \approx n/2 \approx n_{\,b} b \hfill \\ m_{\,a} a \approx m/2 \approx m_{\,b} b \hfill \\ \end{gathered} \right. \tag {3} }$$
b) Note that the above goal of symmetry is to be achieved globally on $n$ and $m$, so that it happens that we shall compromise somehow on one of the parameters, to keep the best on the other. When that happens, for the unbalanced parameter we are asked to choose two partions on the opposite sides of the symmetry, so that they make an optimal one on average, in case inserting a remainder if necessary (re. to the case $n=4,m=5$ in the first sketch). So, practically, we are determining a pair of partitions for each parameter.
c) The above situation can be faced when $n,m$ are relatively small with respect to $a,b$. When they are much larger than $a,b$ then a partion (or two on average) close to symmetry can be found for both.
Therefrom a possible strategy seems to be as follows (you can follow the process on one of the sketches given)
• repart one side of the rectangle (e.g. $m$) according to one of its pair of partitions, putting first all the $a$'s, then the remainder, then the $b$'s;
• following the perimeter, repart the contiguous side ($n$), this time starting with the $b$'s;
• pass to next side ($m$), and apply the remaining partition of the pair associated with $m$;
• same for the last side;
• expand the partition of the perimeter towards the inner part, as allowed
From the examples, it looks that it is possible to reach $N= \left\lfloor {\frac{{n\,m}}{{a\,b}}} \right\rfloor$ in "many" cases.
• @G Cab Which software did you use ???? Do you have any idea how to solve the problem ??? – Nirbhay Feb 2 '17 at 11:12
• @Nirbhay I did the sketch by hand, trying to validate a possible approach, which unfortunately does not work for particular values of $m$ and $n$ as shown. However it provides upper and lower bounds: I will add it to my answer, as it might interest you anyway. – G Cab Feb 2 '17 at 11:38
• How can you possibly "sketch" that "by hand" ???? – Nirbhay Feb 2 '17 at 11:48
• How many $2\times 2$ squares can you fit into a $3\times 3$ square? Certainly no more than $1$, despite the fact that $\lfloor(9/4)\rfloor=2$. – mjqxxxx Feb 8 '17 at 15:41
• @RobertFrost: yes, that's the case if we admit that $a$ or $b$ can be larger than $n,m$: that opens a quite different scenario. – G Cab Feb 9 '17 at 19:39 | 2019-06-17T00:52:49 | {
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https://math.stackexchange.com/questions/1447572/probability-of-picking-rooms | # Probability of picking rooms
Suppose there are 15 rooms among which a person in present in one room. If i pick 5 rooms out of them, what is the probability that the person is present in one of the 5 rooms selected?
No of ways we can pick 5 rooms: 15C5, So answer could be x/15C5
Where am i stuck is how to figure out x? Among 15C5 selections, which of them will contain that specific room in which the person is there?
• Please show what you have tried and where you are stuck. – Gummy bears Sep 23 '15 at 4:48
• Wouldn't this just be $\frac{1}{3}$, since we are splitting up the 15 total rooms into three sets of 5 rooms, and then seeing if the person is in one of the three sets? I feel like I'm missing something though... – Brevan Ellefsen Sep 23 '15 at 4:51
• @BrevanEllefsen Nope. The five rooms aren't fixed. It's not like there are three groups of five to chose from. We have tho chose all the five members of the group. – Gummy bears Sep 23 '15 at 5:13
• @Gummybears oh ya, combinations... duh. I really dropped the ball on that one. Time to take another probability course... been way too focused on real analysis.... – Brevan Ellefsen Sep 23 '15 at 5:19
Hint
There is only one way to choose the occupied room, what about the rest ?
Why confused ? $Pr = {1\choose 1}{14\choose4} / {15\choose 5}$
Of course, you can leave out the ${1\choose 1}$ , but it shows exactly what you did . | 2019-11-16T23:48:33 | {
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http://elespiadigital.org/libs/marian-canvas-wtgzve/ppvkvds.php?1e2c1d=brackets-in-maths-rules | ; Ans: The four commonly used bracket types are: Parentheses ( ), Square brackets [ ], Curly brackets { }, Angle brackets ⟨ ⟩. Use operations inside brackets and solve any indices. , the n-fold application of f to argument x. This math problem has parentheses, an exponent, multiplication, division, and subtraction. f In quantum mechanics, angle brackets are also used as part of Dirac's formalism, bra–ket notation, to denote vectors from the dual spaces of the bra In some European countries, the notation Pro Lite, Vedantu η For example, n n Nested brackets can be recognised by the appearance in the expression of 2 or more left brackets, ( , before any right brackets, ) . is applied to the composition Hence, second preference in BODMAS is given here to the orders or exponents (x n). Braces { } are used to identify the elements of a set. 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In e-mail and other ASCII text, it is common to use the less-than (<) and greater-than (>) signs to represent angle brackets, because ASCII does not include angle brackets.[4]. {\displaystyle \sin x} In group theory and ring theory, square brackets are used to denote the commutator. In statistical mechanics, angle brackets denote ensemble or time average. Brackets and indices. In ring theory, the commutator [a,b] is defined as ab − ba. x would be the set of all real numbers between 5 and 12, including 5 but not 12. Ex: [0,8) denotes a half-closed interval that includes all real numbers, except 8 from 0 to 8. The arguments for a function are always surrounded by parentheses. Simplifying expressions without the math equation that tried to stump place missing paheses make … The rule is do the innermost pair of brackets first and work your way outwards. X ( ε ( In the wrong order, solving the issue would result in a wrong answer. 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This is especially important when it comes to completing the square. {\displaystyle \langle \ \rangle } {\displaystyle \langle a,b\rangle } ) Main & Advanced Repeaters, Vedantu So, upon removing the brackets: is used to indicate an interval from a to c that is inclusive of The earliest use of brackets to indicate aggregation (i.e. About ExamSolutions; About Me; Maths … Brackets as used in mathematical notation, Floor/ceiling functions and fractional part, MEDIUM LEFT-POINTING ANGLE BRACKET ORNAMENT, MEDIUM RIGHT-POINTING ANGLE BRACKET ORNAMENT, "Compendium of Mathematical Symbols: Delimiters", "When and Where to Use Parentheses, Braces, and Brackets in Math", "Interval Notation | Brilliant Math & Science Wiki", https://en.wikipedia.org/w/index.php?title=Bracket_(mathematics)&oldid=987756863, Short description is different from Wikidata, Creative Commons Attribution-ShareAlike License, This page was last edited on 9 November 2020, at 01:47. o . {\displaystyle [5,12[} n 5 They can have one of two values: positive or negative. 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[ ⋅ [1] The notation Also, there is no ‘ off’ and ‘order’ in the brackets. Unicode has pairs of dedicated characters; other than less-than and greater-than symbols, these include: In LaTeX the markup is \langle and \rangle: In the Cartesian system of coordinates, brackets are used to designate point coordinates. Brackets especially Parentheses () are used in elementary algebra to define the order of operations. Brackets find their main application in BODMAS or PEMDAS rule where the sequence of operations to be performed when an expression is resolved. {\displaystyle \mathbb {R} [x]} b Positive integers have values greater than zero. : , x g Another notation for the same is ε ) For denoting an open end of an interval, a bracket may be used. This notation is extended to cover more general algebra involving variables: for example (x + y) × (x − y). Both parentheses, ( ), and square brackets, [ ], can also be used to denote an interval. 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[ 0 As the full form suggests, the first preference in BODMAS operation is given to brackets … For example, consider the difference between the following: [2] Terms inside the bracket are evaluated first; hence 2×(3 + 4) is 14, 20 ÷ (5(1 + 1)) is 2 and (2×3) + 4 is 10. ( When there is a negative integer outside of the brackets, use integer multiplication rules to make the signs of answer terms. {\displaystyle \left|B\right\rangle } f Ex: 5 * (2 + 4) is 30, (5 * 3) + 2 is 30. ⟨ {\displaystyle f(x)=\exp(\lambda x)} If both types of brackets are the same, the entire interval may be referred to as closed or open as appropriate. λ You went to the store to buy chocolates. According to the BODMAS law, if there are brackets ((), {}, []) in an expression, we first have to overcome or simplify the bracket followed by the order, then divide, multiply, add and subtract from left to right. This is an important way of solving quadratic equations. 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( = The first question the student gets in this topic is “How can we define brackets”. a You should remember BODMAS, and this will give you the precedence rules to work out calculations involving brackets, BODMAS or PEMDAS stands for: The BODMAS rule explains the sequence of operations to be done until an expression is resolved. This order depends … ) {\displaystyle f(x)} Square brackets, as in [π] = 3, are sometimes used to denote the floor function,[8] which rounds a real number down to the next integer. [3], A variety of different symbols are used to represent angle brackets. [ = -~-~~-~~~-~~-~-Please watch: "New News Of Pakistani Chai Wala Will Make You Shock Tune pk" https://www.youtube.com/watch?v=XJ8MjyQYkiI-~-~~-~~~-~~-~- For example, in the formula ) The inner product of two vectors is commonly written as x λ {\displaystyle c} ( x Follow all the rules in this order from left to right in the equation. 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In group theory, the commutator [g,h] is commonly defined as g−1h−1gh. Example: If y = (1 + x²)³ , find dy/dx . Generally, such bracketing denotes some form of grouping: in evaluating an expression containing a bracketed sub-expression, the operators in the sub-expression take precedence over those surrounding it. The full form of BODMAS is Brackets, Orders, Division, Multiplication, Addition and Subtraction. 12 . ) There are also rules for calculating with negative numbers. {\displaystyle \varepsilon \eta } variable and real number coefficients.[9][8]. This is to be contrasted with Division. Whenever infinity or negative infinity is used as an endpoint (in the case of intervals on the real number line), it is always considered open and adjoined to a parenthesis. {\displaystyle x^{(n)}} (3 + (5 * 4)) - ((4 * 6) - 10) = 23 -14 = 9. There are very specific rules about bracket use in this discipline that are rarely altered, and the sequence of use—{[()]}—is different from that in normal text. In mathematics, brackets of various typographical forms, such as parentheses ( ), square brackets [ ], braces { } and angle brackets ⟨ ⟩, are frequently used in mathematical notation. Ans: Four parentheses are involved in the given expression. Parentheses, brackets, and braces are sometimes referred to as "round," "square," and "curly" brackets, respectively. η . Now how will you represent this mathematically? They usually contain a phrase (like … ⟨ f f Use the order of operations to solve the problem when we see multiple numbers and operations in parentheses. Easy peasy, right? λ Ans: Parentheses are used to denote changes to the usual order of operations in mathematical expressions. X 1 In English, we use parentheses to indicate clarification for a concept (we add extra information using parentheses). 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http://math.stackexchange.com/questions/231137/polynomial-orthogonal-complement/231142 | # Polynomial Orthogonal Complement
Let $V = \mathbb{P^4}$ denote the space of quartic polynomials, with the $L^2$ inner product $$\langle p,q \rangle = \int^1_{-1} p(x)q(x)dx.$$ Let $W = \mathbb{P^2}$ be the subspace of quadratic polynomials. Find a basis for and the dimension of $W^{\perp}$.
The answer is $$t^3 - \frac{3}{5}t, t^4 - \frac{6}{7}t^2 + \frac{3}{35};\,\, \dim (W^{\perp}) =2$$
How did they get that?
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"Space of quartic polynomials"? Do you mean the space of (real, complex, rational...) polynomials of degree $\,\leq 4\,$ and zero? – DonAntonio Nov 6 '12 at 2:57
@DonAntonio yes i do.. – diimension Nov 6 '12 at 3:01
Which one? Real? Complex? – EuYu Nov 6 '12 at 3:01
@EuYu it doesnt specify but im sure is real – diimension Nov 6 '12 at 3:06
It must be real or there would be a conjugate inside the inner product... – copper.hat Nov 6 '12 at 5:06
Well, for a finite dimensional vector space we have $\dim W + \dim W^\perp = \dim V$ so that covers the dimension part. For the basis of the orthogonal complement, we have $$\int_{-1}^1 ax^4 + bx^3 + cx^2 + dx + e\ dx = 0$$ $$\int_{-1}^1 x(ax^4 + bx^3 + cx^2 + dx + e)\ dx = 0$$ $$\int_{-1}^1x^2(ax^4 + bx^3 + cx^2 + dx + e)\ dx = 0$$ Because the standard basis of $\mathbb{P}^2$ must satisfy the orthogonality conditions. Therefore we get $$\frac{a}{5} + \frac{c}{3} + e = 0$$ $$\frac{b}{5} + \frac{d}{3} = 0$$ $$\frac{a}{7} + \frac{c}{5} + \frac{e}{3} = 0$$ Solving this system yields $$a = \frac{35}{3}e,\ \ b=-\frac{5}{3}d,\ \ c=-10e$$ with two parameters to vary. Your solutions follows by taking $(a=0,\ b=1)$ and $(a=1,\ b=0)$ respectively.
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Thank you, but I am a bit confused I see how this is an orthogonal complement basis but how does it relate to $t^3 - \frac{3}{5}t, t^4 - \frac{6}{7}t^2 + \frac{3}{35}$?? – diimension Nov 6 '12 at 3:22
Take $a=0$ and $b=1$ and solve for $c,\ d,\ e$. See what you get. Remember $a,\ b,\ c,\ d,\ e$ are the coefficients of the standard basis. – EuYu Nov 6 '12 at 3:25
Well, for example choose, as written clearly in the above answer, $$a=1\,,\,b=0\Longrightarrow e=\frac{3}{35}\,,\,d=0\,,\,c=-10e=-\frac{6}{7}\Longrightarrow\, \text{one of the pol's is }\, x^4-\frac{6}{7}x^2+\frac{3}{35}$$ and etc. – DonAntonio Nov 6 '12 at 3:26
Yup, I understand now with you guys help! One last question, how did you know that a = 1, b = 0 , and vice versa? – diimension Nov 6 '12 at 3:31
Well, the coefficients kinda have to match. Both your polynomials are monic so those are the natural choices. – EuYu Nov 6 '12 at 3:34
Well, one way would be to use Gram-Schmidt to produce an orthogonal basis, starting with the basis $1$, $t$, $t^2$, $t^3$, $t^4$. The result will be five polynomials $p_r(x)$ for $r=0,\ldots,4$ where $p_r$ has degree $r$. So $p_0$, $p_1$ and $p_2$ will span the space of quadratic polynomials and $p_3$ and $p_4$ will span a 2-dimensional space orthogonal to the quadratics. Since we know that the dimension of $W^\perp$ is two (because $\mathrm{dim}(W)+\mathrm{dim}(W^\perp)=5$), we see that $p_3$ and $p_4$ are a basis for $W^\perp$.
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Thank you for explaining that. – diimension Nov 6 '12 at 3:23
Let
$$p(x):=ax^4+bx^3+cx^2+dx+e\in W^\perp$$
Since $W:=\operatorname{Span}\{1,x,x^2\}\,$ , we get:
$$(1)\;\;\;\;\;\;0=\langle\,p\,,\,1\,\rangle=\int_{-1}^1p(x)dx=\frac{2}{5}a+\frac{2}{3}c+2e$$
$$(2)\;\;\;\;\;\;\;\;\;\;\;0=\langle\,p\,,\,x\,\rangle=\int_{-1}^1xp(x)dx=\frac{2}{5}b+\frac{2}{3}d$$
$$(3)\;\;\;\;\;\;\;\;0=\langle\,p\,,\,x^2\,\rangle=\int_{-1}^1x^2p(x)dx=\frac{2}{7}a+\frac{2}{5}c+\frac{2}{3}e$$
The above relies on the easy results that the integral on a symmetric (above zero) interval of an even function is twice the value of its primitive on either of the two limits, whereas the same integral of an odd function is zero.
Now solve the above linear system.
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Thank you, but I am a bit confused I see how this is an orthogonal complement basis but how does it relate to $t^3 - \frac{3}{5}t, t^4 - \frac{6}{7}t^2 + \frac{3}{35}$?? – diimension Nov 6 '12 at 3:22
Read my comment below Euyu's answer: it applies exactly the same here. – DonAntonio Nov 6 '12 at 3:27
I understand now with guys help. Thank you very much , DonAntonio! – diimension Nov 6 '12 at 3:35
Let $v_k(x) = x^k$, $k=0,...,4$. Then $v_k$ is a basis for $\mathbb{P}^4$. (To see this note that any quartic can be written in terms of $v_k$, and if $\sum \alpha_k v_k = 0$, then by differentiating and evaluating at $x=0$ we can see that $\alpha_k = 0$, hence they are linearly independent.)
By the same token, $v_k$, $k=0,1,2$ is a basis for $\mathbb{P}^2$. It follows that $\dim \mathbb{P}^2 = 3$, and since $\mathbb{P}^4 = \mathbb{P}^2 \oplus (\mathbb{P}^2)^\bot$, we see that $\dim (\mathbb{P}^2)^\bot = 2$.
We can find a basis for $(\mathbb{P}^2)^\bot$ by projecting the $v_k$ onto $(\mathbb{P}^2)^\bot$. Clearly $v_k$, $k=0,1,2$ will project to zero. So the only remaining basis elements that need to be projected are $v_3,v_4$.
Note in passing that $\langle v_j, v_k \rangle = \frac{1}{j+k+1}(1-(-1)^{j+k+1})$.
To compute the projection of $x$ onto $(\mathbb{P}^2)^\bot$, we need to determine $\alpha \in \mathbb{R}^3$ such that $\langle x-\sum_{k=0}^2 \alpha_k v_k, v_j \rangle = 0$ for $j=0,1,2$. This is just the linear system $\langle x, v_j \rangle = \langle \sum_{k=0}^2 \alpha_k v_k, v_j \rangle$, or $$\begin{bmatrix} \langle v_0, v_0 \rangle & \langle v_1, v_0 \rangle & \langle v_2, v_0 \rangle \\ \langle v_0, v_1 \rangle & \langle v_1, v_1 \rangle & \langle v_2, v_1 \rangle \\ \langle v_0, v_2 \rangle & \langle v_1, v_2 \rangle & \langle v_2, v_2 \rangle \end{bmatrix} \alpha = \begin{bmatrix} \langle x, v_0 \rangle \\ \langle x, v_1 \rangle \\ \langle x, v_2 \rangle \end{bmatrix}$$ Grinding through the computations gives $\alpha = \frac{1}{5} (0,3,0)^T$ when $x=v_3$ and $\alpha = \frac{1}{35} (-3, 0, 30)^T$ when $x=v_4$.
Hence a basis for $(\mathbb{P}^2)^\bot$ is $x \mapsto x^3-\frac{3}{5}x$, $x \mapsto x^4+\frac{3}{35}-\frac{6}{7}x^2$.
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Wow, thank you very much! So there are many ways to do it, instead of just taking the $L^2$ inner product, i could have instead used gram matrix. Thank you for pointing that out , Copper.hat – diimension Nov 6 '12 at 4:00
You are welcome. Note that you are still taking the inner products, they just happen to be easy to compute for my particular choice of basis, $v_k$. – copper.hat Nov 6 '12 at 5:08 | 2015-04-27T10:32:01 | {
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https://wishrex.com/xbi7n9bc/sample-size-calculator-with-standard-deviation-cf95a5 | Categories
# sample size calculator with standard deviation
Users can use this sample size calculator to verify the results with different input parameters. s 2: sample variance X 2 : Chi-Square critical value with n-1 degrees of freedom To find a confidence interval for a population standard deviation, simply fill in the boxes below and then click the “Calculate… Use the code as it is for proper working. desired and then click the 'Calculate' button. Degrees of Freedom Calculator Paired Samples, Degrees of Freedom Calculator Two Samples. … The greater the margin of error we desire, the less there is a need to have a large sample size because there is more Calculate power given sample size, alpha, and the minimum detectable effect (MDE, minimum effect of interest). To control the spent for collecting the data. when calculating the required sample size with a given power and p threshold, usually the calculator only requires two estimated population mean and standard deviation… The analysis report is required in short time. The population standard deviation formula is: Below is a great explanation of how to interpret Standard Deviation. Then, once we have found $$z_p$$, we use the following formula: Assume that the population mean is known to be equal to $$\mu = 10$$, and the population standard deviation is known to be $$\sigma = 5$$. standard deviation of 1.2, and we want a margin of error of 0.1. 5, 12, 13) This calculator is featured to generate the work with steps which may help beginners to learn or understand how the sample size is being calculated … What is the sample size required for experiment results to be statistically significant at 95% confidence level for the population with 2.65 standard deviation and 5% margin of error. calculate the sample size needed? Reference: The calculations are the customary ones based on normal Calculate sample size using the below information. But when we have population information that determines exactly the population distribution, the percentiles can be computed exactly. See for example R Squared Calculator (Coefficient of Determination), R Squared Calculator (Coefficient of Determination). A common estimator for σ is the sample standard deviation, typically denoted by s. It is worth noting that there exist many different equations for calculating sample standard deviation since unlike sample … - Marketing survey results One thing you may notice is that the formula has a z value in it. Set your confidence level. - Investing in the stock market. If you need a whole number, then you can round this number to 554, just so that you have have a whole number. Expected Value Calculator - Forecasting For the first set, manually we found the Z value since the total value, mean value and standard deviation are given. The most typical case when finding percentiles is the case of finding a percentile from sample data.In that case, the percentile can only be … Estimating the characteristics of population from sample is known as statistics. sigma (common standard deviation) and, if calculating power, The below formulas are the mathematical representation for both population standard deviation and portion of (proportion) population methods which may help users to know what are all the input parameters are being used in such calculations to find the sample size which produces statistically significant results. Yes. For an explanation of why the sample estimate is normally distributed, study the Central Limit Theorem. Sample sizes that are very low will have much higher margins of error. Copyright © 2011-document.write(new Date().getFullYear()); StandardDeviationCalculator.net All rights reserved, Our site uses cookies. So the formula in order to determine the sample size is, n= ( (z * σ)/MOE) 2. Calculate the Percentile from Mean and Standard Deviation. Functions: What They Are and How to Deal with Them, Normal Probability Calculator for Sampling Distributions, percentile from the mean and standard deviation, percentile from the mean and standard deviation calculator. distributions. A probability & statistics tool used to estimate the right number of samples from the population to be included in the statistical survey or experiments to draw the effective conclusion about the known or infine population is known as sample size calculator. It's often associated with confidence interval. of margin of error is desired. (adsbygoogle = window.adsbygoogle || []).push({}); Enter values separated by comma's (ie. So looking at this formula, let's analyze it a bit. However, you don't input a z value. n= ((z * σ)/MOE)2. You may change the default values from the 'Customize Visualisation' panel. If the distribution is not normal, you still can compute percentiles, but the procedure will likely be different. Proportion : It's a portion of population used in the statistical experiments to characterize the population parameters. This visualisation assumes a 95% level of confidence and plots sample sizes for three precision levels for a range of standard deviation values. Rather than subtracting 1 from n (number of values in the denominator), you only divide by n. However, if we want a low margin of error, then we need to have a larger sample size to make room for less error. Use this advanced sample size calculator to calculate the sample size required for a one-sample statistic, or for differences between two proportions or means (two independent samples). This website uses cookies to improve your experience. The below are some of the solved examples with solutions to help users to know how to estimate reliable sample size by using stanadrd deviation or proportion method. The sample size of a population should be fair or large enough to draw a better estimate which posses enough statistical power in surveys or experiments. 1. Standard Deviation : It's a measure of deviation of whole elements from the mean of sample or population. Note: You may adjust sample size for t-distribution (applied by default), finite population or clustering by clicking the 'Adjust' button below. Ratio Calculator This value is calculated from the confidence level desired. The higher the variance (standard deviation), the more patients are needed to demonstrate a difference. There is tremendous value hidden in the data for those that collect data; even a motivated student might find it useful to determine the standard deviation of an exam to see how well they did in contrast to their peers. If your population is smaller and known, just use the sample size calculator. The following is a list of examples where finding the standard deviation is useful: The need for samples in statistics The Sample Size Calculator calculates the sample size needed to create data that has a certain margin of error desired. Our site is for students, business professionals, or personal use. The estimation will have more confidence when the experiment includes more samples and vice versa. Then we find using a normal distribution table that $$z_p = 0.842$$ is such that . Example of a Sample Size Calculation: Let's say we want to calculate the proportion of patients who have been discharged from a given hospital who are happy with the level of care they received while hospitalized at a 90% confidence level of the proportion within 4%. Download Figure. Radians to Degrees Calculator This calculator allows a user to enter in the confidence levels of So looking at this formula, let's analyze it a bit. It usually represented by σ for population data & s for sample data. As the sample size increases, the margin of error decreases. You may adjust sample size for clustering, finite population and response rate by clicking the Adjust button below. Users may supply the values for the below input parameters to find the effective sample size to be statistically significant by using this sample size calculator. a sample size (assumed the same for each sample). margin of error (MOE). σ is the population standard deviation of the data set. This formula is utilized a lot in statistics and data analysis. 0.01, 0.05, 0.10 & 0.5 represents 99%, 95%, 90% and 50% confidence levels respectively. Rather than subtracting 1 from... Calculator, Formula, explanation, examples Ask Question Asked 6 years, 5 months ago. Keep in mind this assumes it is a normal curve (bell curve). because there are more data points; this, in turn, leads to less room for error. Degrees to Radians Calculator When you need a high level of confidence, you have to increase the z-value which, in turn, increases the margin of error; this is bad because a low | 2021-01-20T01:22:07 | {
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https://opinie-o.pl/m7lhp/133e35-identity-matrix-inverse | » Feedback Thus, when the determinant is zero, there is no set of 4 numbers that produces an inverse. Reduce the left matrix to row echelon form using elementary row operations for the whole matrix (including the right one). If the product of two square matrices, P and Q, is the identity matrix then Q is an inverse matrix of P and P is the inverse matrix of Q. 4 x 4 matrices? 5a + 2c = 0 As a matrix multiplied by its inverse is the identity matrix we can verify that the previous output is correct as follows: A %*% M A matrix that has no inverse is singular. When the left side is the Identity matrix, the right side will be the Inverse [ … » Java This new matrix is the inverse of the original matrix. » Android The identity matrix I n is a n x n square matrix with the main diagonal of 1’s and all other elements are O’s. : & ans. 1) It is always a Square Matrix. Interview que. The Identity Matrix and Inverses Reference > Mathematics > Algebra > Matrices In normal arithmetic, we refer to 1 as the "multiplicative identity." » Java » CS Basics Utiliser la réduction linéaire par rangées pour trouver une matrice inverse Accolez la matrice identité … » CS Organizations Where a, b, c, and d represents the number. It will look like this [ A | I]. » Subscribe through email. By extension, you can likely see what the $$n\times n$$ identity matrix would be. » Certificates A -1 × A = I. Submitted by Anuj Singh, on June 03, 2020. » C++ » DS » LinkedIn Define a 5-by-5 sparse matrix. For our purposes here, it is enough to show you (as we did above) how you would go about manually finding the inverse of a 2 x 2 using systems of equations, as well as the algorithmic short cut! Continue until you form the identity matrix. An inverse identity matrix is a matrix $M$ such that $MI=IM=I$, where $I$ is the identity matrix. Image will be uploaded soon . Alternative names for this formula are the matrix inversion lemma, Sherman–Morrison–Woodbury formula or just Woodbury formula. Create a 2-by-2 identity matrix that is not real valued, but instead is complex like an existing array. Matrices, when multiplied by its inverse will give a resultant identity matrix. Python code to find the inverse of an identity matrix » Linux I-.1 = I. Syntax: inv_M = numpy.linalg.inv(I) Here, "M" is the an identity matrix. Therefore A and B are inverse matrices. So you can see all that any matrix is, for a given dimension-- I mean we could extend this to an n by n matrix-- is you just have 1's along this top left to bottom right diagonals. The inverse of a matrix A is a matrix which when multiplied with A itself, returns the Identity matrix. Mathematically: The intuition is that if we apply a linear transformation to the space with a matrix A, we can revert the changes by applying A⁻¹ to the space again. A matrix X is invertible if there exists a matrix Y of the same size such that X Y = Y X = I n, where I n is the n-by-n identity matrix. Later, we will use matrix inverses to solve linear systems. Yes, there are. So I've told you that. Its determinant (check out the unit on Determinants for more information on evaluating the determinant of a matrix) is zero. » Data Structure » CSS However, the identity appeared in several papers before the Woodbury report. » C We say that we augment M by the identity. Inverse of a matrix in R. In order to calculate the inverse of a matrix in R you can make use of the solve function. The inverse of a matrix is that matrix which when multiplied with the original matrix will give as an identity matrix. » Internship » DOS Keep repeating linear row reduction operations until the left side of your augmented matrix displays the identity matrix (diagonal of 1s, with other terms 0). The below example always return scalar type value. Same thing when the inverse comes first: ( 1/8) × 8 = 1. When it is necessary to distinguish which size of identity matrix is being discussed, we will use the notation $$I_n$$ for the $$n \times n$$ identity matrix. » Web programming/HTML Define a complex vector. » C++ STL One concept studied heavily in mathematics is the concept of invertible matrices, which are those matrices that have an inverse. » Facebook » C++ There are matrices whose inverse is the same as the matrices and one of those matrices is the identity matrix. Then we will row reduced this augmented (or adjoined) matrix. This is a fancy way of saying that when you multiply anything by 1, you get the same number back that you started with. Languages: It is denoted by A ⁻¹. In mathematics (specifically linear algebra), the Woodbury matrix identity, named after Max A. Woodbury , says that the inverse of a rank-k correction of some matrix can be computed by doing a rank-k correction to the inverse of the original matrix. Home » » Articles Linear Algebra using Python, Linear Algebra using Python | Inverse of an Identity Matrix: Here, we are going to learn about the inverse of an identity matrix and its implementation in Python. » PHP » SEO » Node.js This matrix equation will give you a set of four equations in four unknowns: 3a + 1c = 1 When you have reached this point, the right side of your vertical divider will be the inverse of your original matrix. » Contact us Row-reduce the matrix until the left side to the Identity matrix. » Content Writers of the Month, SUBSCRIBE are all very similar; they have ones down the main diagonal, and zeroes everywhere else: So what is an inverse matrix? » SQL More: CS Subjects: This is a fancy way of saying that when you multiply anything by 1, you get the same number back that you started with. » Machine learning To compute the inverse of the matrix M we will write M and also write next to it the identity matrix (an identity matrix is a square matrix with ones on the diagonal and zeros elsewhere). Defining a Matrix; Identity Matrix; There are matrices whose inverse is the same as the matrices and one of those matrices is the identity matrix. » News/Updates, ABOUT SECTION » C++ /reference/mathematics/algebra/matrices/the-identity-matrix-and-inverses. » Java To calculate inverse matrix you need to do the following steps. » C When we multiply a matrix with the identity matrix, the original matrix is unchanged. The identity matrix for the 2 x 2 matrix is given by. Python » The matrix inverse of \bs{A} is denoted \bs{A}^{-1}. Identity Matrix is the matrix which is n ... Inverse Matrix; Orthogonal Matrix; Singular Matrix; Symmetric Matrix; Upper Triangular Matrix; Properties of Identity Matrix. As a result you will get the inverse calculated on the right. The product of a matrix and its inverse is the identity matrix—the square array in which the … In some fields, such as quantum mechanics, the identity matrix is denoted by a boldface one, 1; otherwise it is identical to I. A Question and Answer session with Professor Puzzler about the math behind infection spread. The matrix Y is called the inverse of X. L'inverse d'une matrice carrée M est une matrice notée M^-1 telle que M.M^-1=I ou I est la matrice identité. Would you like to see the 2 x 2 multiplicative identity matrix? » Embedded C De très nombreux exemples de phrases traduites contenant "identity matrix" – Dictionnaire français-anglais et moteur de recherche de traductions françaises. The inverse of a square matrix A is a second matrix such that AA-1 = A-1 A = I, I being the identity matrix.There are many ways to compute the inverse, the most common being multiplying the reciprocal of the determinant of A by its adjoint (or adjugate, the transpose of the cofactor matrix).For example, This is indeed the inverse of A, as . Consider the following matrices: For these matrices, AB=BA=I, where I is the 2×2identity matrix. » C Fortunately, someone has gone to the trouble of creating a mini-formula/algorithm for you, to save you having to use Cramer's Rule every time you want to find the inverse of a 2 x 2 matrix. » Puzzles When we multiply a matrix by its inverse we get the Identity Matrix (which is like "1" for matrices): A × A -1 = I. Using determinant and adjoint, we can easily find … 3b + 1d = 0 Notice that the w and z have switched places, and the x and y have become negative. » C# It is not square; only square matrices have inverses. Open Live Script. Inverse d'une Matrice Outil d'inversion de matrice. Here the dimension is 3 which means that identity is created with 3 number of rows and 3 number of columns where all the diagonal elements are 1 and rest other elements are zero. PQ = QP = I) The inverse matrix of A is denoted by A -1. » About us Inverse of a matrix A is the reverse of it, represented as A-1. » Cloud Computing » Privacy policy, STUDENT'S SECTION When we multiply a number by its reciprocal we get 1. The inverse of a matrix $$A$$, if it exists, is denoted by the symbol $$A^{-1}$$. M <- solve(A) M [, 1] [, 2] [1, ] 0.1500 -0.100 [2, ] -0.0625 0.125. Definition of an Inverse: An $$n \times n$$ matrix has an inverse if there exists a matrix $$B$$ such that $$AB = BA = I_n$$, where $$I_n$$ is an $$n \times n$$ identity matrix. » HR The site administrator fields questions from visitors. Join our Blogging forum. Having learned about the zero matrix, it is time to study another type of matrix containing a constant specific set of values every time, is time for us to study the identity matrices. Aptitude que. » DBMS Is it also called a Unit Matrix? It is denoted by In, or simply by I if the size is immaterial or can be trivially determined by the context. It looks like this: You see how the multiplicative identity gives right back to you the matrix you started with? Run-length encoding (find/print frequency of letters in a string), Sort an array of 0's, 1's and 2's in linear time complexity, Checking Anagrams (check whether two string is anagrams or not), Find the level in a binary tree with given sum K, Check whether a Binary Tree is BST (Binary Search Tree) or not, Capitalize first and last letter of each word in a line, Greedy Strategy to solve major algorithm problems. Technically, when we are talking about an inverse matrix, we are talking about a multiplicative inverse matrix. Solved programs: If A is a m x n matrix, then I m A = A and AI n = A. Less frequently, some mathematics books use U or E to represent the identity matrix, meaning "unit matrix" and the German word Einheitsmatrix respectively. Let us take the square matrix A. 3x3 identity matrices involves 3 rows and 3 columns. 4 Square matrices (matrices which have the same number of rows as columns) also have a multiplicative identity. A system of four equations with four unknowns...from our unit on determinants, you know that one of the ways to solve such a system is with Cramer's Rule, and the only time there is no solution is if the determinant has a zero value.
## identity matrix inverse
Couple Spa Near Me, How To Fix My Mic After Android 10 Update, Ole Henriksen Banana Bright Vitamin C Serum Vs Truth Serum, Binks Paint Booth Models, Pioneer Home Theater Subwoofer, Heos Speakers Bluetooth, Proactive Management Software, | 2021-01-22T05:18:45 | {
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http://math.stackexchange.com/questions/246284/induction-problem-ratio-of-consecutive-fibonacci-numbers | # Induction problem? (ratio of consecutive Fibonacci numbers)
Define $a_1 = 1$ and for all natural $n$'s, $a_{n+1} = 1 + \dfrac{1}{a_n}$.
Prove that for every natural $n$, $$a_n = \dfrac{F_{n+1}}{F_n}.$$
I'm not sure if this is an induction problem or not, but could someone help me understand what is going on?
-
What have you tried? Where are you stuck? – JavaMan Nov 28 '12 at 5:41
Compute the first few, by hand, expressing the answers in the form $\frac{x_n}{y_n}$. Soon you will see what's going on. – André Nicolas Nov 28 '12 at 5:43
Ok I'll try something and then I'll come back. – blutuu Nov 28 '12 at 5:46
Ok I've computed a few n terms and everything after n = 1 seems to produce fractions (that is if my arithmetic is correct). What would you think I should do? – blutuu Nov 28 '12 at 5:57
Yeah I hadn't thought about doing that until recently. I'll fix that – blutuu Nov 28 '12 at 6:23
It is very easy to show that at n = 1 it is true. Now we show the inductive step (assume it holds for $n$ and show it holds for $n+1$)
First we know that
$F_{n+2} = F_{n+1} + F_{n}$
and dividing by $F_{n+1}$ to both sides
$F_{n+2}/F_{n+1} = 1 + F_n/F_{n+1}$
Suppose $a_n = F_{n+1}/F_{n}$
Then by definition of $a_{n+1}$
$\displaystyle a_{n+1} = 1+ \frac{1}{a_{n}}$ $\displaystyle = 1+\frac{1}{\frac{F_{n+1}}{F_{n}}}$ $\displaystyle =1 + \frac{F_n}{F_{n+1}}$
Therefore $a_{n+1} = F_{n+2} /F_{n+1}$
-
I'm confused on the step where you set Fn+2 = Fn+1 + Fn. How did you get there from the previous step? – blutuu Nov 28 '12 at 7:05
It is independent of the previous step. It is just the definition of the Fibonacci Sequence. I shall edit now for it to be more clear. – Keith Reyes Nov 28 '12 at 7:47
Okay then. Thanks – blutuu Nov 28 '12 at 7:54
Yes, induction is the natural choice here.
Base Step: $a_1 = 1/1 = F_2 / F_1$
Inductive Step: Suppose $a_n = F_{n+1}/F_n$ for an arbitrary $n \in \mathbb{N}$. Then $$a_{n+1} = 1 + \frac{1}{a_n} = 1 + \frac{F_n}{F_{n+1}} = \frac{F_{n+1}}{F_{n+1}} + \frac{F_n}{F_{n+1}} = \frac{F_{n+2}}{F_{n+1}}$$
- | 2014-08-22T04:25:54 | {
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https://math.stackexchange.com/questions/761299/if-there-exist-sequence-such-that-gx-n-fx-n1-then-we-have-gx-0-fx | # If there exist sequence such that $g(x_n)=f(x_{n+1})$, then we have $g(x_0)=f(x_0)$ for some $x_0$
Suppose $f(x)$ and $g(x)$ are continuous functions on $[a,b]$ with $f$ monotone increasing. Assume there exists a sequence $x_n \in [a, b]$ such that for all $n \in N$ , $g(x_n) = f(x_{n+1})$. Show that there exists $x_0 \in [a,b]$ such that $g(x_0) = f(x_0)$.
Can someone provide an example of functions that fulfills this condition?
• I really need help in solving this – user10024395 Apr 20 '14 at 4:01
If there is an $n$ such that $g(x_n)=f(x_n)$, we are done. So let us assume that $g(x_n)\ne f(x_n)$ for each $n$.
Now if $g(x_n)>f(x_n)$ and $g(x_{n+1})<f(x_{n+1})$, then the continuity of $f$ and $g$ implies that and $x$ such that $f(x)=g(x)$ exists somewhere between $x_n$ and $x_{n+1}$.
The case that $g(x_n)<f(x_n)$ and $g(x_{n+1})>f(x_{n+1})$ is basically the same.
So the only two remaining cases are:
A. $(\forall n) g(x_n)>f(x_n)$
B. $(\forall n) g(x_n)<f(x_n)$
Let us discuss the case A. (The case B is similar.)
For each $n$ we have $f(x_n)<g(x_n)=f(x_{n+1})$. Since $f$ is increasing, this implies $x_n<x_{n+1}$, i.e. the sequence $(x_n)$ is monotone.
Every monotone bounded sequence must have a limit, so there exists an $x$ such that $$\lim\limits_{n\to\infty} x_n=x.$$ Now we get, using the continuity of $f$ and $g$, that $$f(x)=\lim\limits_{n\to\infty} f(x_{n+1})=\lim\limits_{n\to\infty} g(x_n)=g(x).$$
• why is f(x) = n, g(x) = n+1 not a counterexample? – user10024395 Apr 26 '14 at 0:17
• @user136266 Do you mean constant functions? They do not fulfill the assumptions. (You cannot find $x_n$ with the required properties.) – Martin Sleziak Apr 26 '14 at 4:22
• I will repost also here link to the chat where some examples and clarifications related to this were given. (I have already mentioned this in comments to the duplicate question.) – Martin Sleziak Apr 30 '14 at 13:04
Hint Prove that there is a monotonic subsequence $\{a_{n_k}\}$ and let be $n_0$ the limit of that subsequence.
• I know how to prove that but how does it help? Using Bolzano theorem, this fact is obvious – user10024395 Apr 20 '14 at 6:37 | 2019-12-15T05:42:38 | {
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https://math.stackexchange.com/questions/656527/is-there-a-way-to-assign-a-number-to-a-combination-without-finding-and-numbering | # Is there a way to assign a number to a combination without finding and numbering every combination?
Imagine I have 4 letters. Is there some algorithm that produces
"abcd" -> 1
"bacd" -> 2
... etc
without finding and numbering every single combination? My goal is to get a number from 1 to 52! from a shuffled deck of cards. It's impractical to find and number all 52! combinations, so I need a way that doesn't require that. The order that the numbers appear in doesn't matter, as long as numbers don't repeat and there are no gaps.
• I guess you mean permutation, i.e. a specific ordering of all the cards? – JiK Jan 29 '14 at 21:54
• Are you looking to use this number to shuffle the deck or is the deck already shuffled and you want the number to record the current state? – John Habert Jan 29 '14 at 21:54
• @JiK: Yes, that's what I mean. – Daffy Jan 29 '14 at 21:55
• @JohnHabert: The latter. I intend to use the current state of the deck to generate random numbers. I assume the method used to get a number from the deck could be reversed to get a deck from a number. – Daffy Jan 29 '14 at 21:56
• This is discussed exhaustively in Donald E. Knuth The Art of Computer Programming volume 4, and less exhaustively but still enough to solve the problem in section 4.3.1 of Higher-Order Perl, which is available online for free. – MJD Jan 29 '14 at 22:06
You can first number all the cards, so that
Aclubs -> 0
2clubs -> 1
...
I'm going to do the simpler case, where you have just four cards, A,B,C,D, which I'll number 0, 1, 2, and 3.
Suppose that you have DACB. Here's how I'll assign a number. It's ugly, but it'll work:
D --> 3 --> 3! * 3 = 18
A --> 0 --> 2! * 0 = 0
C --> 1 --> 1! * 1 = 1
B --> 0 --> 0! * 0 = 0
Total: 19
Now you may notice that the numbers to the right of the letters aren't the numbers I said. They are for D and A, but not the last two. Let me explain:
• The first number gets its true label.
• The second number gets its true label, minus however many items below it have been used already.
• Same for the third number and fourth number.
It helps if you keep a list of the available numbers:
D --> 3 --> 3! * 3 = 18 ABCD available; D is number 3 (starting from 0)
A --> 0 --> 2! * 0 = 0 ABC available; A is number 0
C --> 1 --> 1! * 1 = 1 BC available; C is number 1
B --> 0 --> 0! * 0 = 0 B available; B is number 0
Total: 19
This will assign a different number to every possible permutation.
"How do you go the other way?" you might ask. You use integer division and remainders, and reverse the process. In what follows, DIV and REM denote those two things:
Code: 19
19 DIV 3! --> 3 REM 1; (3, ABCD) --> D
1 DIV 2! --> 0 REM 1; (0, ABC) --> A
1 DIV 1! --> 1 REM 0; (1, BC) --> C
0 DIV 0! --> 0 REM 0; (0, B) --> B
The codes you get will range from $0$ to $52! - 1$.
There are $4!=24$ permutations (which is what you want since order matters). It is cleaner (many fewer +1's and -1's) if the permutation numbers run from $0-23$ and the characters from $0-3$ For permutation $n$, the first digit is $\lfloor \frac n{3!} \rfloor$ Subtract $n-6\lfloor \frac n{3!} \rfloor$ and that is the permutation number for the remaining characters.
For a $52$ card deck, number the cards $0-51$. The first division should be by $51!$. When you have found the first card, take it out and do the same with $51$ cards, dividing by $50!$ and so on.
• The letters were an example. The actual question is about a standard 52 card deck. – John Habert Jan 29 '14 at 21:58
• @JohnHabert: I added the extension to $52$ – Ross Millikan Jan 29 '14 at 22:03
If the permutations are numbered in the lexicographical order, the number given to a permutation is sometimes called the rank of the permutation, although even this term is apparently not very common.
Some algorithms for rank->permutation and permutation->rank are listed for example here.
A simple observation is that the rank of a permutation $(a_i)_{i=0}^{N-1}$ of $\{0,1,\dots,N-1\}$ is $$R=(N-1)! a_0 + R',$$ where $R'$ is the rank of the permutation $(a_i)_{i=1}^{N-1}$. Note that this formula cannot be directly applied recursively because $(a_i)_{i=1}^{N-1}$ is no longer a permutation of $\{0,1,\dots,N-2\}$.
I may be wrong about this, but as far as I know, there's not a canonical enumeration of the elements of the permutation group $S_{52}$ as $\{\pi_1, \pi_2, \ldots, \pi_{52!} \}$. You may be better off representing the shuffled deck as a product of cycles, and that's easy to do. Not sure what you're doing with this, but it might be useful.
Assuming the cards are numbered 1, 2, ... , 52, you can perform the following process until you exhaust the deck. Card $n_1=1$ now occupies position $n_2$ (which could still be 1, of course). Card $n_2$ then has been displaced to position $n_3$. So card $n_3$ has been displaced to position $n_4$, and so on. Because there are only 52 positions, you will eventually come to a card which has been displaced to the position originally occupied by card $n_1$. So you have a cycle $(n_1, n_2, \ldots, n_{M_1})$; this means that the card positions in the cycle have been rotated forward to the next position in the list, with an implicit wrap of the last to the first (the card in position $n_{M_1}$ has been displaced into position $n_1=1$). Note that it is completely possible that $M_1=1$, so that the first card is not moved, and that cycle has length 1. It is also possible that you traverse every single position before you arrive at the first position again, in which case the whole deck has been rearranged in a rotation (a single cycle of length 52).
For convenience, let's refine the notation and add a second subscript "$1$" to the elements of the cycle we just created to identify the fact that it is the "first" cycle: now we call the cycle we created above $(n_{1,1}, n_{1,2}, \ldots, n_{1,M_1})$.
If we have exhausted the deck, we are done. The shuffle is represented by a single cycle.
Otherwise, find the first position not yet considered, say position $n_{2,1}$, which will be the first member of our second cycle (hence the subscript "$2$"). As before, continue moving through the deck with the second cycle until you arrive back at position $n_{2,1}$ again, and we will have a second cycle $(n_{2,1}, n_{2,2}, \ldots, n_{2,M_2})$ containing $M_2$ positions. Note that the cycles are disjoint by construction.
And so on.
You eventually exhaust the deck with a finite number $N$ of cycles, each of which has a finite number of positions. These positions partition the deck into cycles by construction. So we can represent the shuffled deck as a product of partitions
$$(n_{1,1}, \ldots, n_{1,M_1})(n_{2,1}, \ldots, n_{2,M_2})\ldots(n_{N,1}, \ldots, n_{N,M_N})$$
where $M_1 + M_2 + \ldots + M_N = 52$. | 2020-01-19T13:46:38 | {
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https://mathematica.stackexchange.com/questions/135636/understanding-simple-numerical-calculation?noredirect=1 | # Understanding simple numerical calculation
I am trying to understand why N[0.1]//FullForm returns 0.1' (A). Indeed, $0.1$ in base two is $0.000110011001100110011...$. If I truncate this number to the first 52 digits and convert it back to base ten, I get:
Table[1/2^(4*i) + 1/2^(4*i + 1), {i, 1, 12}] // Total // FullForm // N
(* 0.09999999999999964 denoted by (B) *)
which is different from the obtained result (A).
However, computing the following operation returns the expected result (B)!
N[10.1 - 10)] // FullForm
(* 0.09999999999999964 *)
So, what is the difference between the N[0.1] // FullForm and N[10.1 - 10] // FullForm and why does only the second one return (B)?
• I think this question boils down to the difference between e.g. N[0.1] and N[1.1 - 1]. The division itself is not the source of the problem. – Marius Ladegård Meyer Jan 17 '17 at 22:11
• @MariusLadegårdMeyer Right, edited. – anderstood Jan 17 '17 at 22:16
• How did you calculate the binary representation? There is no such thing as a dot in binary formats. The binary representation of floating point numbers is via significant and exponent. For 0.1 it is 1x10^-1, that is, Both significant (1) and exponent (-1) are exactly represented in a 64 bit float. – Felix Jan 17 '17 at 22:16
• @Felix Consider $1100110011...11\times 2^{-48}$ if you prefer (and note that BaseForm[0.1, 2] returns "0.00011001100110011001101" in MMA ;)). – anderstood Jan 17 '17 at 22:21
• ah.. you can also do N[ 1.1 - 1. ] // FullForm to see the issue. (You might want to put such into the question. ) – george2079 Jan 17 '17 at 22:41
## Introduction
This seems to be a question more about IEEE 754 binary64 format than about Mathematica per se.
The significand is 53 bits (52 stored, since the leading bit is assumed to be 1 in a normal number). When the input "0.1" is converted to a number, presumably, at some point 1. is divided by 10. The OP used the term "truncate," but more precisely, the result should be rounded to the nearest 53-bit floating-point number.
Some tools one can use to explore the binary representation of machine precision numbers are
SetPrecision[x, Infinity]
RealDigits[x, 2, 53]
SetPrecision[x, Infinity] converts the machine real to the exact fraction represented by the floating point number x. RealDigits[x, 2, 53] shows the values of the bits. If you replace the 53 with 54 or higher, the last "bits" will be returned as Indeterminate. Below I plot the bits so that a 1 is red, 0 is white, and Indeterminate is gray. The 53-bit limit is indicated with a vertical grid line.
## OP's examples
In the OP's example Table[] command, the computed fraction is truncated too soon. Here is a table showing what is going on. The first three rows are the OP's table carried out to 12 iterations (as in the OP) up to 14. One has to go up to 14 to get all the bits needed to approximate 0.1 to machine precision. The last three rows show the bits of the machine real 0.1, the result of SetPrecision[0.1, Infinity], and the exact fraction 1/10. One can see that rounding 1/10 results in a carry at the last bit.
ClearAll[tenth, mantissaplot, frexp];
tenth[n_] := Table[1/2^(4*i) + 1/2^(4*i + 1), {i, 1, n}] // Total;
mantissaplot[{digits_, exp_}, ref_: 0.1] := {ArrayPlot[
{digits}, ColorRules -> {1 -> Red, Indeterminate -> Gray},
Mesh -> True, MeshStyle -> Directive[Thin, Black],
ImageSize -> 450, Axes -> {True, False},
GridLines -> {{exp - Ceiling@Log2[ref], 53}, None}],
exp};
frexp[x_, bits_: 54] := mantissaplot@RealDigits[x, 2, bits];
TableForm[
Join[
Table[frexp[tenth[n], 57], {n, 12, 14}],
{frexp[0.1, 57],
frexp[SetPrecision[0.1, Infinity], 57],
frexp[1/10, 57]}
],
TableHeadings -> {{12, 13, 14, 0.1, Subscript[0.1, Infinity],
1/10}, {"significand", "exp"}}]
Now let's take up the OP example 10.1 - 10. In the table below, we can see that to store 10.1 the part representing 0.1 is shifted over 7 bits, so that the last 7 bits in 0.1 above are dropped. When 10. is subtracted, the result is shifted back, but the 7 bits are already lost. (This what is meant by precision loss due to subtractive cancellation.)
TableForm[{
frexp[10.1],
frexp[10.1 - 10.],
frexp[0.1]
},
TableHeadings -> {{10.1, HoldForm[10.1 - 10.], 0.1}, {"significand", "exp"}}]
Update:
My original answer is based on the apparently wrong assumption that the numbers are represented by $s\times 10^e$ with significand $s$ and exponent $e$. The correct binary representation is $s\times 2^e$ with $s$ having 52 bits plus one bit for the sign and $e$ being 11 bits long for 64 bit floating point numbers. Therefore, 0.1 is not exactly represented by this format.
All your numbers are exactly represented as floating point numbers. There are certain operations that preserve this precision. It seems not changing the exponent is one of them:
N[0.2 - 0.1] // FullForm
(* 0.1 *)
and only changing the exponent is one of them:
N[10/0.1] // FullForm
(* 100. *)
However, changing both in one operation may lead to numerical errors:
N[1.1-1.0] // FullForm
(* 0.10000000000000009 *)
` | 2020-10-20T15:13:23 | {
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http://mathematica.stackexchange.com/questions/32378/is-there-something-like-densityplot3d-to-visualize-atomic-orbitals/47336 | # Is there something like DensityPlot3D to visualize atomic orbitals?
I'm visualizing some hydrogen like atomic orbitals. For looking at plane slices of the probability density, the DensityPlot function works well, and with something like:
Manipulate[
DensityPlot[ psi1XYsq[u, v, z], {u, -w, w}, {v, -w, w} ,
Mesh -> False, Frame -> False, PlotPoints -> 45,
ColorFunctionScaling -> True, ColorFunction -> "SunsetColors"]
, {{w, 10}, 1, 20}
, {z, 1, 20, 1}
]
I can get a nice plot
I was hoping that there was something like a DensityPlot3D so that I could visualize these in 3D, but I don't see such a function. I was wondering how DensityPlot be simulated using other plot functions, so that the same idea could be applied to a 3D plot to construct a DensityPlot3D like function?
-
You may be interested in heatmap density 3D – Kuba Sep 15 '13 at 16:33
– Kuba Sep 15 '13 at 16:34
Image3D works, apart from interpolation. A question arises, though: how you would be able to interpret the results? In general, isosurfaces are much more practical than images which look like fuzzy, vague clouds at best, opaque mass with hidden internal structure at worst. – kirma Sep 15 '13 at 16:47
@kirma "fuzzy, vague clouds" - sounds like an orbital ;) – Kuba Sep 15 '13 at 16:49
@Kuba, yes, that 4D visualization answer by halmir looks like what I was looking for. – Peeter Joot Sep 18 '13 at 13:19
In an presentation by Markus van Almsick, he gives an solution to visualize atomic orbitals using Image3D.
R[n_Integer?Positive, l_Integer?NonNegative, r_] :=
Block[{ρ = (2 r)/n},
Sqrt[(2/n)^3 (n - l - 1)!/(2 n (n + l)!)] E^(-ρ/2) ρ^l LaguerreL[n - l - 1, 2 l + 1, ρ]] /; l < n
full wave function:
ψ[n_, l_, m_, r_, ϑ_, φ_] := ψ[n, l, m, r, ϑ, φ] =
FullSimplify[R[n, l, r] SphericalHarmonicY[l, m, ϑ, φ], {r >= 0, ϑ ∈ Reals, φ ∈ Reals}]
CompileWaveFunction = Compile[{{x, _Real}, {y, _Real}, {z, _Real}},
Block[{ρ = x^2 + y^2, r, ϑ, φ},
If[ρ > 0,
r = Sqrt[ρ + z^2]; ϑ = ArcCos[z/r]; φ = ArcTan[x, y],
r = Abs[z]; ϑ = π/2 Sign[z]; φ = 0];
#
],
CompilationTarget -> "C"
] &;
color function:
colorFunction = (Blend[{
{0., RGBColor[0.7, 0.8, 1., 0.]},
{0.1, RGBColor[0., 0.7, 0.1, 0.012]},
{0.4, RGBColor[1., 0.1, 0.03169, 0.06723]},
{1., RGBColor[1., 0.95051, 0., 0.10963]}}, #] &)
plot 3p orbital
Block[
{nψ =
CompileWaveFunction[ψ[3, 1, 0, r, ϑ, φ]], data, vol},
data = Table[Abs[nψ[x, y, z]]^2, {z, -20, 20, 0.25}, {y, -20, 20, 0.25}, {x, -20, 20, 0.25}];
vol = RawArray["Byte", Round[(255/Max[data]) data]];
Image3D[vol, "Byte", Background -> Black,
Method -> {"FastRendering" -> True, "InterpolateValues" -> True},
ColorFunction -> colorFunction, BoxRatios -> 1]
]
it's quite smooth to rotate the Image3D object
We can also visualize the atomic orbital by plotting the isosurface:
Block[{nψ = CompileWaveFunction[ψ[3, 2, 0, r, ϑ, φ]]},
ContourPlot3D[Abs[nψ[x, y, z]]^2, {x, -20, 20}, {y, -20, 20}, {z, -20, 20},
PlotPoints -> 15, Contours -> {0.00002},
ColorFunction -> Function[{x, y, z}, ColorData["Rainbow"]@Rescale[Arg[nψ[x, y, z]], {-π, π}]],
ColorFunctionScaling -> False, Mesh -> None]
]
and make a animation showing different isosurfaces
plots = ParallelTable[
Block[{nψ = CompileWaveFunction[ψ[4, 2, 1, r, ϑ, φ]]},
ContourPlot3D[
Abs[nψ[x, y, z]]^2, {x, -20, 20}, {y, -20, 20}, {z, -20, 20}, PlotPoints -> 17, Contours -> {ct},
ColorFunction -> Function[{x, y, z}, ColorData["Rainbow"]@Rescale[Arg[nψ[x, y, z]], {-π, π}]],
Boxed -> False, Axes -> False, ColorFunctionScaling -> False, Mesh -> None,
ViewPoint -> {0.98, -2.76, 1.7}, ViewVertical -> {-0.004, -0.117, 0.993}]],
{ct, 0.00003, 0.000015, -0.0000005}
];
ListAnimate[plots]
-
My preferred method for this kind of thing is projecting each dimension onto a plane and then combining them together. I think MATLAB has similar functionality. Mind you, the answers and comments on my question about projecting are right in pointing out that this will become inefficient for high polygon counts (essentially more PlotPoints) so if you want to Manipulate in a smooth way, you may want to use Texture. See this relevant question for details on that.
Now, I haven't got your function psi1XYsq so I will pick a double gaussian:
E^(-(x^2 + (y - 2)^2 + (z - 3)^2)/10) + E^(-((x + 1)^2 + (y + 2)^2 + (z + 2)^2)/25)
The idea is to use something similar to the function @Jens is using in his answer to How to make a drop-shadow for Graphics3D objects.
Block[{d1, d2, d3, function, options, opacity, x0, y0, z0, min, max},
{min, max} = {-9, 9};
{x0, y0, z0} = {0., 0., 0.};
opacity = 0.9;
function[x_, y_, z_] :=
E^(-(x^2 + (y - 2)^2 + (z - 3)^2)/10) +
E^(-((x + 1)^2 + (y + 2)^2 + (z + 2)^2)/25);
options = Sequence @@ {PlotPoints -> 45, Mesh -> None,
ColorFunctionScaling -> False, ColorFunction -> "SunsetColors"};
d1 = First@DensityPlot[function[x, y, z0], {x, min, max}, {y, min, max},
Evaluate@options] /. {x_?AtomQ, y_?AtomQ} -> {x, y, z0};
d2 = First@DensityPlot[function[x, y0, z], {x, min, max}, {z, min, max},
Evaluate@options] /. {x_?AtomQ, z_?AtomQ} -> {x, y0, z};
d3 = First@DensityPlot[function[x0, y, z], {y, min, max}, {z, min, max},
Evaluate@options] /. {y_?AtomQ, z_?AtomQ} -> {x0, y, z};
Show[Graphics3D[{d1, d2, d3}, Lighting -> "Neutral"] /.
GraphicsComplex[xx__] -> {Opacity[opacity], GraphicsComplex[xx]}]
]
The code is straightforward: min, max define the range for each variable, {x0, y0, z0} define the projection planes, and opacity the Opacity. You will notice I have turned off ColorFunctionScaling so that each slice is bright according to an absolute value and they merge together nicely. If your function is not normalised you may want to normalise it before doing that.
If you can afford lowering the PlotPoints, Manipulate isn't too bad, and you can make animations that look like volumetric rendering (apologies for the 300K gif):
Table[Block[{d1, d2, d3, function, options, opacity, x0, y0, z0, min,
max},
{min, max} = {-9, 9};
{x0, y0, z0} = {0., t, 0.};
opacity = 0.9;
function[x_, y_, z_] :=
E^(-(x^2 + (y - 2)^2 + (z - 3)^2)/10) +
E^(-((x + 1)^2 + (y + 2)^2 + (z + 2)^2)/45);
options =
Sequence @@ {PlotPoints -> 25, Mesh -> None,
ColorFunctionScaling -> False,
ColorFunction -> "SunsetColors"};
d1 = First@
DensityPlot[function[x, y, z0], {x, min, max}, {y, min, max},
Evaluate@options] /. {x_?AtomQ, y_?AtomQ} -> {x, y, z0};
d2 = First@
DensityPlot[function[x, y0, z], {x, min, max}, {z, min, max},
Evaluate@options] /. {x_?AtomQ, z_?AtomQ} -> {x, y0, z};
d3 = First@
DensityPlot[function[x0, y, z], {y, min, max}, {z, min, max},
Evaluate@options] /. {y_?AtomQ, z_?AtomQ} -> {x0, y, z};
Show[Graphics3D[{d1, d2, d3}, Lighting -> "Neutral"] /.
GraphicsComplex[xx__] -> {Opacity[opacity],
GraphicsComplex[xx]}]
], {t, 9, -9, -1.5}];
Export["3d.gif", %];
-
What does your {x_?AtomQ, y_?AtomQ} -> {x, y, z0} replacement in this code do? – Peeter Joot Sep 17 '13 at 22:17
I adds a third dimension to every pair of 2d coordinates of the density plot. – gpap Sep 17 '13 at 23:46
I recently revisited this, and found that RegionPlot3D is by far the fastest way to plot orbitals, compared to Image3D and ContourPlot3D. I was surprised by the difference, so I thought it's worth posting this.
In addition, I also made the process of choosing the plot parameters automatic, based on simple estimates for the size of the orbital wave function. So you don't have to find the right contour value by trial and error. This automation allows me to plot large numbers of orbitals in one go.
Below, I'm plotting the first ten orbitals in a systematic table, and the whole table takes about as long as doing one or two such plots using ContourPlot3D (depending on what functions I try). There was also no need to compile any of the functions, making the code quite straightforward.
Clear[rMin, rMax, r, θ, ϕ];
{rMin[n_, l_], rMax[n_, l_]} =
r /. Simplify[
Solve[(l (l + 1))/r^2 - 2/r == -(1/n^2), r],
n > 0];
sphericalToCartesian =
Thread[{r, θ, ϕ} -> {Sqrt[x^2 + y^2 + z^2],
ArcCos[z/Sqrt[x^2 + y^2 + z^2]], Arg[x + I y]}];
ψ[n_, l_, m_][r_, θ_, ϕ_] :=
Sqrt[ (n - l - 1)!/(n + l)!] E^(-(r/n)) ((2 r)/n)^l 2/
n^2 LaguerreL[n - l - 1, 2 l + 1, (2 r)/n] SphericalHarmonicY[l,
m, θ, ϕ]
ClearAll[plotOrbital];
plotOrbital[f_, range_, contour_, opt : OptionsPattern[]] :=
RegionPlot3D[
Evaluate[Abs[f[r, θ, ϕ] /. sphericalToCartesian]^2 >
contour], {x, -range, range}, {y, -range, range}, {z, -range,
range}, opt, Mesh -> False, PlotPoints -> 35, PlotStyle -> Orange,
Lighting -> "Neutral", PlotTheme -> "Classic"]
grid = Table[
Labeled[plotOrbital[ψ[n, l, m], 2 n^2, .05/n^6],
Row[{"n = ", n, ", ℓ = ", l, ", m = ", m}]], {n, 1,
3}, {l, 0, n - 1}, {m, 0, l}];
TableForm[grid]
Above, the complete hydrogenic orbital wave function is ψ. Given the principal quantum number n, the energy is known. Setting the energy equal to the effective potential yields the classical turning points, {rMin[n_, ℓ_], rMax[n_, ℓ_]}. This is used to determine the required plot size automatically: If the linear dimension r scales with $n^2$, then the volume scales with $n^6$. For a normalized wave function, we can then estimate that the probability density at a typical point will decrease with $1/n^6$. Therefore, I choose the threshold for the equi-probability contour to scale with this factor. Here is a check to verify what I said about the length scale:
Simplify[rMax[n, 0], n > 0]
(* ==> 2 n^2 *)
This is the largest achievable radius because I set the angular momentum l to zero.
In sphericalToCartesian, I write down how to convert to Cartesian coordinates (instead of using built-in functions which differ too much between Mathematica versions).
plotOrbital does the plotting for a given wave function f. In producing the grid of orbitals, I choose the parameters as described above.
Edit: volumetric slices
To show that the RegionPlot approach is also able to provide multiple contours, and to show how these layered contours can give us information about the volume in 3D, here I first combine several outputs of plotOrbital with different contour values in a single Show. The surfaces have been given their own PlotStyle. Then I put the resulting 3D graphics in a Manipulate that allows you to slice through the layers in real time:
With[{n = 3, l = 2, m = 0},
orb = Show[
Table[plotOrbital[ψ[n, l, m], 2 n^2, c/n^6,
PlotStyle -> Directive[Opacity[0.6], Hue[c]]], {c, 0.3,
0.05, -0.04}], PlotRange -> All, BoxRatios -> Automatic]]
With[{n = 3},
Manipulate[
Show[orb, Boxed -> False, Axes -> False,
ViewVector -> {{35, -35, 5}, {0, 0, 0}}, ViewAngle -> Pi/4,
PlotRange -> {{-#, #}, {y, #}, {-#, #}} &[2 n^2]], {y, -2 n^2,
2 n^2}]]
The trick to get slices and hollow spaces between layers in RegionPlot3D is to cut off the plot with a reduced PlotRange. To still keep the object fixed in the view port (instead of moving around to stay centered with the changing PlotRange), I add fixed values for the ViewVector and ViewAngle. This trick using PlotRange doesn't work if you cut the range off directly inside the RegionPlot3D. You have to do it after the fact with Show.
Again, the drawing of multiple layers is very fast using RegionPlot3D. It could even be sped up more using ParallelTable.
-
That's fantastic! Indeed RegionPlot3D is much faster. And it also solves the problem of holes in ContourPlot3D. – xslittlegrass Feb 21 at 22:27 | 2015-02-28T12:39:59 | {
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https://mathematica.stackexchange.com/questions/210574/selecting-from-a-list-of-tuples | # Selecting from a list of tuples
Given a list tuples Tuples[Range[10],2] I'd like to select the ones that match a certain criteria. Namely that for every pair {x ,y}, GCD[y, x] == 1 and Mod[x, y] != 2
I've tried the following.
Select[Tuples[Range[10], 2], Function[{x, y}, GCD[x, y] == 1 && Mod[x, y] != 2]]
But, I understand I'd have to supply the function with a symbol (and not a list).
How could I filter out that list of tuples?
• Tuples[Range[10], 2] // Select[GCD[#[[1]], #[[2]]] == 1 && Mod[#[[1]], #[[2]]] != 2 &] – Rohit Namjoshi Dec 2 '19 at 2:56
• I tried something similar before, without the & at the end. It didn't work. Said #1 had no attributes or something akin to that. Why does it work with the & at the end? – Rodrigo Dec 2 '19 at 3:02
• @Rodrigo - See the documentation for Function. When using a pure function with formal parameters (e.g., #1), the & is needed to mark the end of the pure function's body. – Bob Hanlon Dec 2 '19 at 4:14
• Select[Tuples[Range[10],2],Apply[Function[{x,y},GCD[x,y]==1&&Mod[x,y]!=2]]] – matrix89 Dec 2 '19 at 5:12
You may use Apply (@@).
Select[
Tuples[Range[10], 2],
Function[tupe, GCD@@tupe == 1 && Mod@@tupe != 2]
]
{{1,1},{1,2},{1,3},{1,4},{1,5},{1,6},{1,7},{1,8},{1,9},{1,10},{2,1},{3,1},{3,2},
{3,4},{3,5},{3,7},{3,8},{3,10},{4,1},{4,3},{4,5},{4,7},{4,9},{5,1},{5,2},{5,4},
{5,6},{5,7},{5,8},{5,9},{6,1},{6,5},{6,7},{7,1},{7,2},{7,3},{7,4},{7,6},{7,8},
{7,9},{7,10},{8,1},{8,5},{8,7},{8,9},{9,1},{9,2},{9,4},{9,5},{9,8},{9,10},{10,1},
{10,3},{10,7},{10,9}}
Hope this helps.
Another option is to use Cases
ClearAll[x,y];
data = Tuples[Range[10], 2];
Cases[data, {x_, y_} /; GCD[x, y] == 1 && Mod[x, y] != 2 :> {x, y}]
I also made a function that combines Tuples and Select: saving sometimes a lot of memory:
ResourceFunction["SelectTuples"][Range[10], 2, (GCD @@ #) == 1 && (Mod @@ #) != 2 &] | 2021-05-12T15:29:15 | {
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https://math.stackexchange.com/questions/2309722/how-to-find-a-meaningful-bound-on-a-sequence-that-is-known-to-go-to-0/2309758 | # How to find a meaningful bound on a sequence that is known to go to $0$
I am doing a programming exercise (quite an interesting one, actually) on the sequence $\{I_k\}_{k\in\Bbb{N}}$
$$I_k = \int_0^1 x^ke^{x-1} dx$$
and I am also given a - already proven by me - recurrence formula,
$$I_k = 1 - kI_{k-1}$$
and at some point I must show that the $I_k$ go to $0$. I was trying to do it just with the recurrence formula and the fact that $I_1 = \frac1e$, with induction. I tried proving $I_k < \frac{1}{k+1}$ but in proving it I ended up needing to assume that $\frac{1}{k+1} < I_{k-1}$. So basically I wanted to prove
$$\frac{1}{k+2} < I_k < \frac{1}{k+1} \tag{1}$$
I verified it for $k$ up to $10$ so it does feel like it should be true. With induction, proving $I_k < \frac{1}{k+1}$ assuming (1), is trivial. However, when I try to prove $\frac{1}{k+2} < I_k$ I end up with
$$I_{k-1} < \frac{1}{k}\frac{k+1}{k+2}$$
$$I_{k-1} < \frac{1}{k}\frac{k+1}{k+2} < \frac{1}{k}, \text{true, by the induction hypothesis}$$
However, I can't do this. I know $I_{k-1}< \frac{1}{k}$ but that does not let me write what I wrote nor prove what I want to prove. Is there anyone out there able to lend me a hand?
The ideal would be to use induction to prove (1).
If I fail to do that, the second ideal thing would be to use induction to prove that the $I_k$, starting at some $k_0$, are bounded by something that does not increase. It can even be a constant. For example, showing $I_k < 1\ \forall k$ would be good enough for my purposes.
Also, if anyone knows if this sequence (or these integrals) has any name, I would be able to better search for things that could help.
• Wait, I'm lost. You want to prove $(1)$, am I correct? Also, hi, haven't seen you in a while :D And particularly, you are stuck on the lower bound? – Simply Beautiful Art Jun 4 '17 at 19:07
• @SimplyBeautifulArt hey :P You are correct, in my last paragraph I got a bit lost in my thoughts. I made it clearer now. Yes, I conjecture that the lower bound is indeed that, but I failed to prove it. – RGS Jun 4 '17 at 19:12
• @RSerrao Hint: $\,0 \lt x^k e^{x-1} \lt 1\,$ for $\,x \in (0,1)\,$. Then the recurrence gives the upper bound directly. – dxiv Jun 4 '17 at 19:13
• Well, using the recurrence relation you found, one may deduce that: $$I_k=\frac{k!}e\left[1-\frac1e\sum_{n=0}^k\frac1{n!}\right]$$ – Simply Beautiful Art Jun 4 '17 at 19:14
• @SimplyBeautifulArt Maybe, but $\,0 \lt I_{k-1} \lt 1/k\,$ is enough to prove convergence to $\,0\,$. – dxiv Jun 4 '17 at 19:16
(Expanding on my previously posted comments.)
I am also given a - already proven by me - recurrence formula,
$$I_k = 1 - kI_{k-1}$$
Once the recurrence relation is determined, the bounds follow directly from just the positivity of the integrand $\,x^k e^{x-1} \gt 0\,$ for $\,x \gt 0\,$, which immediately implies $\,I_k \gt 0\,$ for all $\forall k \in \mathbb{N}\,$. Therefore:
• $1 - kI_{k-1}=I_k \gt 0 \implies k I_{k-1} \lt 1 \implies I_{k-1} \lt \cfrac{1}{k} \quad\quad$
• then, $1 - kI_{k-1}=I_k \lt \cfrac{1}{k+1} \implies k I_{k-1} \gt 1 - \cfrac{1}{k+1}=\cfrac{k}{k+1}\implies I_{k-1} \gt \cfrac{1}{k+1}$
• Once you've got $0 < I_{k-1}< \dfrac 1 k,$ that's enough to prove that $I_k \to 0$ as $k\to\infty.$ If the goal was only to prove that, then you don't need the rest of what you've got here. – Michael Hardy Jun 4 '17 at 20:59
• @MichaelHardy Right, and that's a point I made in the comments. However, there was some discussion about what the question is really asking for, and the above attempts to cover the other interpretations, too. – dxiv Jun 4 '17 at 21:01
• I am accepting your answer because it is the one that uses the integral form of $I_k$ the less. Maybe I didn't make that clear enough on the question, but you addressed my issue the way I most wanted it to be addressed, using the recurrence relation to prove the bounds. – RGS Jun 5 '17 at 7:04
It is not clear to me whether the question suggests induction or requires it. If induction is merely suggested, then here is an answer which declines the suggestion!
Majorize via $$0 \le x^ke^{x-1} \le x^k\:\ \mbox{for all}\ x\in[0,1].$$ By this, we have that $$0 \le I_k = \int_0^1 x^k e^{x-1}\,dx \le \int_0^1 x^k \, dx = \frac 1 {k+1} \to 0 \text{ as } k\to\infty.$$ which I believe does what you want.
(This would fail, of course, if you needed $\sum_k I_k$ to be bounded—but that isn't a fault of the proof but rather of the series. In any case, you have not asked for boundedness in $\sum_k I_k$.)
• A professional engineer rather than a professional mathematician, I am not very handy with notation like $\forall x\in[0,1]$. An engineer would normally write $0 \le x \le 1$ and, if necessary, write "for all" out in words. If my notation is imperfect, corrections would be appreciated. – thb Jun 4 '17 at 19:48
• Write whatever is legible. If that means more words, so be it. I'd phrase it as "for all $x \in [0,1]$" or "for all $x$ with $0 \leq x \leq 1$", without really preferring one over the other; I tend to avoid the $\forall$ symbol except in formal logic, because it's just a bit less readable. – Patrick Stevens Jun 4 '17 at 21:05
• @thb : I would write this as follows: $$0 \le \int_0^1 x^k e^{x-1}\,dx \le \int_0^1 x^k \, dx = \frac 1 {k+1} \to 0 \text{ as } k\to\infty.$$ – Michael Hardy Jun 4 '17 at 21:08
• @thb : I wonder why you write $x^k e^{x-1} \le x^k e^1$ when you could just go with $x^k e^{x-1} \le x^k. \qquad$ – Michael Hardy Jun 4 '17 at 21:12
• @MichaelHardy: corrected, thanks. – thb Jun 4 '17 at 21:40
$$I_k = \int_0^1 x^ke^{x-1} \, dx$$
Directly from the integral:
$$\frac1{e} \lt e^{x-1} \lt 1$$ so $$\frac1{e(k+1)} \lt I_k \lt \frac1{k+1}.$$
Since $e^{x-1} > x$ (because $e^z > 1+z$), $I_k > \int_0^1 x^kx \,dx =\frac1{k+2}$. This is what you wanted.
Trying to be more precise:
Interpolating at $0$ and $1$, since $(e^{x-1})'' > 0$, $e^{x-1} \lt (1-\frac1{e})x+\frac1{e}$, so
\begin{align} I_k &\lt \int_0^1 x^k \left(\left(1-\frac1 e\right)x+\frac 1 e\right) \, dx\\ &=\left(1-\frac 1 e\right)\frac1{k+2}+\frac1{e(k+1)}\\ &=\frac1{k+2}+\frac1{e(k+1)(k+2)} \end{align}
so that $0 \lt I_k-\frac1{k+2} \lt \frac1{e(k+1)(k+2)}$.
Let $d(x) =(1-\frac1{e})x+\frac1{e}-e^{x-1} =(1-\frac1{e})x+\frac1{e}(1-e^{x})$. $d(0) = d(1) = 0$ and $d'(x) =(1-\frac1{e})-e^{x-1} =0$ when
\begin{align} x &=x_0\\ &=1+\ln(1-\frac1{e})\\ &=1+\ln(e-1)-1\\ &=\ln(e-1)\\ &\approx 0.54132\\ \end{align}
where $d(x_0) = \frac1{e}(2 - e + (e - 1) \ln(e - 1)) \approx 0.077941$.
Therefore $e^{x-1} \ge d(x)-d(x_0)$ so $I_k \ge \frac1{k+1}-(1-\frac1{e})\frac1{(k+1)(k+2)}-d(x_0)$.
That's enough for now.
• Thank you for your answer. I did enjoy the way you manipulated the integral to provide me with a lower bound, and in particular with the lower bound I could not prove. – RGS Jun 5 '17 at 7:01 | 2019-08-19T06:42:32 | {
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Pat bought 5 pounds of apples. How many pounds of pears
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Pat bought 5 pounds of apples. How many pounds of pears [#permalink]
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Pat bought 5 pounds of apples. How many pounds of pears could he have bought for same amount of money?
(1) 1 pound of pears cost $0.5 more that 1 pound of apples (2) 1 pound of pears cost 1.5 times as much as 1 pound of apples Originally posted by r019h on 30 Oct 2007, 19:25. Last edited by Bunuel on 27 Feb 2013, 06:13, edited 1 time in total. Edited the question and added the OA. Most Helpful Expert Reply Math Expert Joined: 02 Sep 2009 Posts: 51073 Re: Pat bought 5 pounds of apples. How many pounds of pears [#permalink] Show Tags 27 Feb 2013, 06:44 6 6 fozzzy wrote: I didn't understand statement 2 Pat bought 5 pounds of apples. How many pounds of pears could he have bought for same amount of money? (1) 1 pound of pears cost$0.5 more that 1 pound of apples.
If 1 pound of pears cost $1 and 1 pound of apples cost$0.5, then the cost of 5 pounds of apples is 5*0.5=$2.5. For$2.5 we can buy 2.5/1=2.5 pounds of pears.
If 1 pound of pears cost $1.5 and 1 pound of apples cost$1, then the cost of 5 pounds of apples is 5*1=$5. For$5 we can buy 5/1.5=10/3 pounds of pears.
Not sufficient.
(2) 1 pound of pears cost 1.5 times as much as 1 pound of apples. The cost of 5 pounds of apples is $5a (where a is the cost of 1 pound of apples). For$5a we can buy 5a/(1.5a)=5/1.5 pounds of pears. Sufficient.
Answer: B.
Hope it's clear.
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Re: gmatprep DS- apples and pears [#permalink]
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30 Oct 2007, 19:41
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trivikram wrote:
r019h wrote:
Pat bought 5 pounds of apples. How many pounds of pears could he have bought for same amount of money?
1) 1 pound of pears cost $0.5 more that 1 pound of apples 2) 1 pound of pears cost 1.5 times as much as 1 pound of apples B should be it st. 1 cost of 1 pound of apples=$x
cost of 1 pound pears= $x+0.5 5 pounds of apples for$5x
and 5x/x+0.5 pounds of pears for $5x INSUFF st. 2 1 pound of pears=$1.5x
so 5x/1.5x pounds of pears for $5x= 5/1.5 approx= 3 pounds of pears SUFF General Discussion Manager Joined: 22 Dec 2011 Posts: 233 Pat bought 5 pounds of apples. How many pounds of [#permalink] Show Tags 14 Nov 2012, 05:36 3 Pat bought 5 pounds of apples. How many pounds of pears could he have bought for same amount of money? 1) 1 pound of pears cost$0.5 more that 1 pound of apples
2) 1 pound of pears cost 1.5 times as much as 1 pound of apples
OA B
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Re: Pat bought 5 pounds of apples. How many pounds of [#permalink]
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Pat bought 5 pounds of apples. How many pounds of pears could he have bought for same amount of money?
1) 1 pound of pears cost $0.5 more that 1 pound of apples 2) 1 pound of pears cost 1.5 times as much as 1 pound of apples We need a relationship between price of pears and that of apples to solve the question. STAT1 1p = 0.5 + 1a But this relationship is not sufficient to answer because price of 5 pounds of apples = 5p = 5*0.5 + 5a but we cannot tell for sure about the exact pounds of apples we can buy STAT2 1p = 1.5*1a This relationship is sufficient because price of 5 pounds of apples = 5p = 5*1.5 *1a= 7.5a so we know for sure that we can buy 7.5 pounds of apples with the same money. So, Answer will be B Hope it helps! _________________ Ankit Check my Tutoring Site -> Brush My Quant GMAT Quant Tutor How to start GMAT preparations? How to Improve Quant Score? Gmatclub Topic Tags Check out my GMAT debrief How to Solve : Statistics || Reflection of a line || Remainder Problems || Inequalities Director Joined: 29 Nov 2012 Posts: 759 Re: Pat bought 5 pounds of apples. How many pounds of pears [#permalink] Show Tags 27 Feb 2013, 03:05 I didn't understand statement 2 Intern Joined: 27 Jul 2010 Posts: 12 Re: Pat bought 5 pounds of apples. How many pounds of pears [#permalink] Show Tags 10 Aug 2013, 13:17 Bunuel wrote: fozzzy wrote: I didn't understand statement 2 Pat bought 5 pounds of apples. How many pounds of pears could he have bought for same amount of money? (1) 1 pound of pears cost$0.5 more that 1 pound of apples.
If 1 pound of pears cost $1 and 1 pound of apples cost$0.5, then the cost of 5 pounds of apples is 5*0.5=$2.5. For$2.5 we can buy 2.5/1=2.5 pounds of pears.
If 1 pound of pears cost $1.5 and 1 pound of apples cost$1, then the cost of 5 pounds of apples is 5*1=$5. For$5 we can buy 5/1.5=10/3 pounds of pears.
Not sufficient.
(2) 1 pound of pears cost 1.5 times as much as 1 pound of apples. The cost of 5 pounds of apples is $5a (where a is the cost of 1 pound of apples). For$5a we can buy 5a/(1.5a)=5/1.5 pounds of pears. Sufficient.
Answer: B.
Hope it's clear.
Hello Bunuel,
Can you please correct my approach of solving this question.
Statement 1:
5 pound of apple cost x
1 pound of apple cost x/5
1 pound of pear would have cost x/5 + 0.5$. Since x is unknown . Hence not sufficient Statement 2: 1 pound of pear cost 3/2(x/5). Here, now i thought that since x is still unknown its not sufficient. Combining both also doesnt give value for x. Hence my answer was E which is incorrect, Can you please solve this question using my approach. If its correct thanks! EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 13058 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: Pat bought 5 pounds of apples. How many pounds of [#permalink] Show Tags 04 Jan 2015, 12:26 4 3 Hi All, In these types of DS questions, if you're sure if a pattern exists or not, you can prove it by TESTing VALUES and being thorough. Here's how: We're told that Pat bought 5 pounds of apples. We're asked how many pounds of pears could have been purchased with the same amount of money. **Note: I'm going to assume that both the cost per pound of apples and the cost per pound of pears remains constant.** Fact 1: 1 pound of pears costs$0.5 more than 1 pound of apples.
IF....
A pound of apples cost $1, then a pound of pears costs$1.50
5 pounds of apples = $5$5 = $1.50(X pounds of pears) X = 3 1/3 pounds of pears IF... A pound of applies costs$.50, then a pound of pears costs $1 5 pounds of applies =$2.50
$2.50 =$1(X pounds of pears)
X = 2.5 pounds of pears
The answer changes based on the starting price of a pound of applies.
Fact 1 is INSUFFICIENT
Fact 2: 1 pound of pears costs 1.5 times as much as 1 pound of apples
IF...
A pound of applies costs $1, then a pound of pears costs$1.50
5 pounds of applies = $5$5 = $1.50(X pounds of pears) X = 3 1/3 pounds of pears A pound of applies costs$2, then a pound of pears costs $3 5 pounds of applies =$10
$10 =$3(X pounds of pears)
X = 3 1/3 pounds of pears
A pound of applies costs $0.50, then a pound of pears costs$0.75
5 pounds of apples = $2.50$2.50 = $0.75(X pounds of pears) X = 3 1/3 pounds of pears In EVERY situation, we end up with the SAME number of pounds of pears. Fact 2 is SUFFICIENT Final Answer: GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: [email protected] Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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Re: Pat bought 5 pounds of apples. How many pounds of pears [#permalink]
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26 Feb 2015, 15:00
Seen quite a number of DS problem of this type, when they give you ratio then most probably you can figure it out the values, (2) 3x = 2y. But if they simply give you data like (1) x = y + 0.5 then there are high chances you can't figure it out the answer.
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Re: Pat bought 5 pounds of apples. How many pounds of pears [#permalink]
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26 Feb 2015, 20:15
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Hi All,
While ankurjohar's question is over 1.5 years old, I'll still answer it because that approach COULD have worked, but the work was incomplete...
Based on that user's initial steps....
$X = cost of 5 pounds of apples$X/5 = cost of 1 pound of apples
Fact 2 tells us that 1 pound of pears costs 1.5 times the cost of 1 pound of apples.
With some Algebra, we have...
(X/5) = cost of 1 pound of apples
(3/2)(X/5) = cost of 1 pound of pears
3X/10 = cost of 1 pound of pears
At this point, ankurjohar assumed that this was insufficient, but there's still more work to do....
We now have a ratio that relates what $X will buy you in this situation:$X buys you 5 pounds of apples
Since $(3/10)(X) buys you 1 pound of pears,$X will buy you 10/3 pounds of pears, so we CAN answer the question with this information.
Fact 2 is SUFFICIENT.
Final Answer:
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Pat bought 5 pounds of apples. How many pounds of pears [#permalink]
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14 Jan 2016, 10:32
r019h wrote:
Pat bought 5 pounds of apples. How many pounds of pears could he have bought for same amount of money?
(1) 1 pound of pears cost $0.5 more that 1 pound of apples (2) 1 pound of pears cost 1.5 times as much as 1 pound of apples ................A........P Pounds.......5.........P Price.........x.........y (1) $$y=x+0,5$$, $$5x=p(x+0,5)$$, you cannot get rid of x, hence not sufficient (2) $$y=1,5x$$ --> $$5x=p*1,5x$$ -> $$p=10/3$$ Answer B _________________ When you’re up, your friends know who you are. When you’re down, you know who your friends are. Share some Kudos, if my posts help you. Thank you ! 800Score ONLY QUANT CAT1 51, CAT2 50, CAT3 50 GMAT PREP 670 MGMAT CAT 630 KAPLAN CAT 660 Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6616 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: Pat bought 5 pounds of apples. How many pounds of pears [#permalink] Show Tags 14 Jan 2016, 17:31 Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. Pat bought 5 pounds of apples. How many pounds of pears could he have bought for same amount of money? (1) 1 pound of pears cost$0.5 more that 1 pound of apples
(2) 1 pound of pears cost 1.5 times as much as 1 pound of apples
When you modify the original condition and the question, it is frequently given on GMAT Math, which is "2 by 2" que like the table below.
Attachment:
GCDS r019h Pat bought 5 pounds of apples (20160115).jpg [ 21.67 KiB | Viewed 17561 times ]
On the tables, n=? is derived from 5a=np. Generally, when one con indicates number and the other con indicates ratio, it is most likely that ratio is an answer. As for this question, in 1) number and 2) ratio, substitute p=1.5a in 2) to 5a=np and it becomes 5a=n(1.5a). Then delete a on the both equations -> 5=1.5n, n=5/1.5, which is unique and sufficient. Therefore the answer is B.
Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.
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y pounds of peers -------> 1.5x $Solving we have: 5x/1/3x/2---->10/3---> 3 1/3 pounds of peers. Sufficient. Hence B. _________________ Thanks & Regards, Anaira Mitch Intern Joined: 10 Nov 2016 Posts: 3 Re: Pat bought 5 pounds of apples. How many pounds of pears [#permalink] Show Tags 04 May 2017, 01:42 nitestr wrote: Seen quite a number of DS problem of this type, when they give you ratio then most probably you can figure it out the values, (2) 3x = 2y. But if they simply give you data like (1) x = y + 0.5 then there are high chances you can't figure it out the answer. Super tip! was about to say the same Intern Joined: 19 Aug 2016 Posts: 2 Re: Pat bought 5 pounds of apples. How many pounds of pears [#permalink] Show Tags 06 Jul 2017, 15:53 Stem -> 5*a = x*p 1 -> p = a+.5; NS; as we cant solve for x without value of a 2 -> p = 1.5a SF; substituting back, a cancels out, and we can solve for x CEO Joined: 11 Sep 2015 Posts: 3228 Location: Canada Re: Pat bought 5 pounds of apples. How many pounds of pears [#permalink] Show Tags 26 Jul 2018, 13:24 Top Contributor r019h wrote: Pat bought 5 pounds of apples. How many pounds of pears could he have bought for same amount of money? (1) 1 pound of pears cost$0.5 more that 1 pound of apples
(2) 1 pound of pears cost 1.5 times as much as 1 pound of apples
Given: Pat bought 5 pounds of apples.
Target question: How many pounds of pears could Pat have bought for the same amount of money?
This is a good candidate for rephrasing the target question.
Let A = the price per pound of apples
Let P = the price per pound of pears
If Pat bought 5 pounds of apples, then 5A = the total amount that Pat spent
Pat then wants to spend her 5A dollars on pears
So, 5A/P = the number of pounds of pears Pat can buy with the 5A dollars
REPHRASED target question: What is the value of 5A/P?
Aside: Below, you'll find a video with tips on rephrasing the target question
Statement 1: One pound of pears costs 0.50\$ more than one pound of apples.
In other words, P = A + 0.5
Does this help us determine the value of 5A/P?
No.
Take 5A/P and replace P with A + 0.5 to get: 5A/P = 5A/(A + 0.5)
Since there's no way to determine the value of 5A/(A + 0.5) (aka 5A/P), we cannot answer the REPHRASED target question with certainty.
So, statement 1 is NOT SUFFICIENT
Statement 2: One pound of pears costs 3/2 times as much as one pound of apples.
In other words, P = (3/2)A or we can write P = 1.5A
Does this help us determine the value of 5A/P?
Yes!!
Take 5A/P and replace P with 1.5A to get: 5A/P = 5A/1.5A = 5/1.5 = 3 1/3
Since we can answer the REPHRASED target question with certainty, statement 2 is SUFFICIENT
Answer: B
Cheers,
Brent
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http://mathoverflow.net/questions/103353/maximal-number-of-binary-strings-given-constraints | # Maximal number of binary strings given constraints
Let $k, N, m \in \mathbb{N}$ such that $k \leq N$.
What is the maximal number $e$ of strings $\sigma_1, \sigma_i, \dots, \sigma_e$, each of length $N$ such that $$\forall j < k, \left(\sum_{i=1}^e \sigma_i[j]\right) \leq 2^{N-k}(m-1)$$
For example if $m=3$, $k=4$, $N=5$, we have $e = 14$. An example of such a set is $$\begin{array}{cc} \sigma_1 & \underbrace{\overbrace{0000}^{k}0}_{N}\\\\ \sigma_2 & 00001\\\\ \sigma_3 & 10000\\\\ \sigma_4 & 10001\\\\ \sigma_5 & 01000\\\\ \sigma_6 & 01001\\\\ \sigma_7 & 00100\\\\ \end{array} \hspace{20pt} \begin{array}{cc} \sigma_8 & 00101\\\\ \sigma_9 & 00010\\\\ \sigma_{10} & 00011\\\\ \sigma_{11} & 11000\\\\ \sigma_{12} & 11001\\\\ \sigma_{13} & 00110\\\\ \sigma_{14} & 00111\\\\ \end{array}$$
By a few trials, it seems that the following holds
• If $k \leq m$ then $e = 2^{N-k} + 2^{N-m}k$
• If $k \geq m$ then $e = 2^{N-k} + 2^{N-m}m$
Has this problem been already studied ?
My intuition is that the following exchange lemma holds:
Let $S$ be a set of strings verifying previous properties. Then there exists a set $S'$ of same cardinality such that if $\sigma \in S'$ has $n$ bits at 1 at the $k$'s first positions, then all strings having strictly less than $n$ bits at 1 at the $k$'s first positions are in $S'$.
But I don't know how to prove this.
-
Is there no constraint on the last $N-k$ values of the strings? – Douglas Zare Jul 28 '12 at 2:32
No, there is no constraint on those values. But this parameter is necessary. You can see $N-k$ as the possibility to add a ponderation to the $k$-prefix. – Turingoid Jul 28 '12 at 3:01
So, your strategy is to include all strings from $C(k,l)$, the $k$-strings with $l$ 1's, until you reach a point where the bound forces you to select a proper subset from $C(k,l)$, while varying the last $N-k$ digits, right? If so, a relevant fact is that for $l > 0$, the full set $C(k,l)$ contributes $\binom{k}{l}/(k/l) = \binom{k-1}{l-1}$ 1's to any given $j$-column per suffix string of $2^{N-k}$. Thus, if your exchange lemma holds, you can get 22 strings for $N = 6, k = 5, m = 4$, instead of 18. – Hugh Denoncourt Jul 29 '12 at 19:47
I do not understand why you use N=6 when he used N=5 in the example, and so on. So I don't know if the fact that you found 22 instead of 18 has a real meaning here, or if you just did a typo with the numbers. – Arthur MILCHIOR Jul 31 '12 at 2:15
@Hugo My exchange lemma says that before including strings with $l$ 1's I should first include strings with $l−1$ 1's. However, it doesn't says that the last $l$ is totally filled, so I don't see how you can conclude that it would give another bound. Maybe my "less" was unclear and I should say "strictly less". – Turingoid Jul 31 '12 at 2:17
This is a counterexample to the formula $e = 2^{N-k} + 2^{N-m}m$ for the maximum number of strings satisfying the given constraints. The parameters of the example are $N = 6$, $k = 5$, and $m = 4$. The constraint is that the columns sum to no more than $6$. The conjectured formula predicts that 18 is the maximum number of strings. $$\begin{array}{cc} \sigma_1 & 000000\\\\ \sigma_2 & 000001\\\\ \sigma_3 & 100000\\\\ \sigma_4 & 100001\\\\ \sigma_5 & 010000\\\\ \sigma_6 & 010001\\\\ \sigma_7 & 001000\\\\ \sigma_8 & 001001\\\\ \end{array} \hspace{20pt} \begin{array}{cc} \sigma_9 & 000100\\\\ \sigma_{10} & 000101\\\\ \sigma_{11} & 000010\\\\ \sigma_{12} & 000011\\\\ \sigma_{13} & 110000\\\\ \sigma_{14} & 110001\\\\ \sigma_{15} & 001100\\\\ \sigma_{16} & 001101\\\\ \end{array} \hspace{20pt} \begin{array}{cc} \sigma_{17} & 100010\\\\ \sigma_{18} & 100011\\\\ \sigma_{19} & 011000\\\\ \sigma_{20} & 011001\\\\ \sigma_{21} & 000110\\\\ \sigma_{22} & 000111\\\\ \end{array}$$ If $C(k, \ell)$ denotes the number of $k$-strings with $\ell$ 1's, and all such strings are included with all possible $N-k$ suffixes, then the total contribution to the column sum from $C(k, \ell)$ (when $\ell > 0$) is $2^{N-k}\binom{k-1}{\ell - 1}$. In this example, all strings from $C(5,0)$ and $C(5,1)$ were included, but, not all strings from $C(5,2)$ could be included. If the exchange lemma is correct, this idea can be used to at least predict lower and upper bounds for $e$.
@Hugh Your interpretation is good. You could as well have given a counter example with N=5 as you duplicate all $k$ prefixes and hence don't "use" the extra N-k bits. I'll try to fix my bound. In fact I don't need to have a tight bound, but just an upper bound in function of $k$ and $m$ so that I can take $m$ and $k$ big enough to make $e$ arbitrarily small. (And sorry for having mistyped your name in my comments) – Turingoid Aug 1 '12 at 23:20
@Hugh In fact, using my previous comment, I realized that I needed to give an upper bound only to an infinity of $m$. So it suffices to take $m_i$ so that we can put exactly all strings with at most $i$ 1's in the $k$ prefix and all $N-k$ suffixes. I guess it isn't too difficult to prove that the created set is an optimal solution for $m_i$. – Turingoid Aug 2 '12 at 3:53
@Turingoid: True! And, the number of such strings is a nice simple sum of binomial coefficients times a power of 2. For nice choices of $m$, you get a nice result. I wonder if the optimal number of strings can always be obtained by repeating the strings over all $N-k$ suffixes. I meant to ask: Was there a motivation behind the question, perhaps as part of a larger question, or did it arise out of recreation? – Hugh Denoncourt Aug 2 '12 at 18:50 | 2014-08-29T14:08:19 | {
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https://math.stackexchange.com/questions/2174922/sum-of-two-independent-exponential-random-variables | # Sum of two independent Exponential Random Variables
The text I'm using on questions like these does not provide step by step instructions on how to solve these, it skipped many steps in the examples and due to such, I am rather confused as to what I'm doing.
Here is the question: Let $X$ be an exponential random variable with parameter $λ$ and $Y$ be an exponential random variable with parameter $2λ$ independent of $X$. Find the probability density function of $X + Y$.
Now, I know this goes into this equation: $\int_{-\infty}^{\infty}f_x(a-y)f_y(y)dy$
What I tried to do is $=\int_{-\infty}^{\infty}\lambda e^{-\lambda (a-y)}2\lambda e^{-\lambda y}dy$ but I quite honestly don't think this is the way to go. Can anyone give me a little insight as to how to actually compute $f_x(a-y)$ in particular?
• You are proceeding correctly, but note the exponential distribution is only non-zero for positive arguments so the limits of integration will be from $0$ to $a$. Also, the second factor is missing a 2 in the exponent $2 \lambda e^{-2\lambda y}$. You should end up with a linear combination of the original exponentials. – A. Webb Mar 6 '17 at 21:37
• @A.Webb Thank you! By doing this and then taking the derivative with respect to a I was able to get the right answer. I didn't think I was doing it right, but apparently the integral really does suck that much. – Heavenly96 Mar 7 '17 at 0:33
• @A.Webb why the limit of the integration will be from 0 to $a$ ? And not from 0 to infinite? – Laura Jan 27 at 19:01
• @Laura, the value $t - x$ of the exponential r.v. is only nonnegative in the range $0 \leq x \leq t$. – gwg May 16 at 19:42
\begin{align*} \Pr[X + Y \le t] &= \int_{x=0}^\infty \Pr[Y \le t - x \mid X = x] f_X(x) \, dx \\ &= \int_{x=0}^t (1 - e^{-2\lambda(t-x)}) \lambda e^{-\lambda x} \, dx \\ &= \lambda \int_{x=0}^t e^{-\lambda x} - e^{-2\lambda t} e^{\lambda x} \, dx \\ &= \left[ -e^{-\lambda x} - e^{-2\lambda t} e^{\lambda x} \right]_{x=0}^t \\ &= 1 + e^{-2\lambda t} - 2e^{-\lambda t}. \end{align*}
The probability density is then found by differentiation with respect to $t$.
• when I differentiate that I end up with $2\lambda e^{-2\lambda t}(e^{\lambda t} -1)$ which is not the answer. However it is very close, the answer is: $2\lambda e^{-\lambda t}(1-e^{-\lambda t})$ so maybe I differentiated wrong? – Heavenly96 Mar 7 '17 at 0:28
• @Heavenly96 $$f_{X+Y}(t) = \frac{d}{dt}\left[1 + e^{-2\lambda t} - 2e^{-\lambda t}\right] = -2\lambda e^{-2\lambda t} - 2(-\lambda) e^{-\lambda t} = 2\lambda ( e^{-\lambda t} - e^{-2\lambda t} ) = 2\lambda e^{-\lambda t} (1 - e^{-\lambda t}),$$ as claimed. Your answer is actually equivalent. You merely pulled out a factor of $e^{-2\lambda t}$ instead of $e^{-\lambda t}$. If you distribute your answer and the answer you were given, you will find they are identical. – heropup Mar 7 '17 at 0:40 | 2019-10-23T20:22:12 | {
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https://mathoverflow.net/questions/273632/injectivity-implies-surjectivity/273674 | # Injectivity implies surjectivity
In some circumstances, an injective (one-to-one) map is automatically surjective (onto). For example,
Set theory
An injective map between two finite sets with the same cardinality is surjective.
Linear algebra
An injective linear map between two finite dimensional vector spaces of the same dimension is surjective.
General topology
An injective continuous map between two finite dimensional connected compact manifolds of the same dimension is surjective.
Are you aware of other results in the same spirit?
Is there a general framework that somehow encompasses all these results?
• The set theory case really implies the other two. Finite dimensional spaces have their linear maps fully decided by behavior on some finite set, and compactness is some sort of generalized finiteness of a topology. Jul 4, 2017 at 9:41
• Gottschalk conjecture: fix a group $G$ and a finite alphabet $A$. Then (injective $\Rightarrow$ surjectivity) holds for any continuous $G$-equivariant map $A^G\to A^G$. Holds when $G$ is sofic, which includes most known groups.
– YCor
Jul 4, 2017 at 9:46
• The second example can be extended to operators of the form Identity + Compact operator, on any Banach space, by Fredholm alternative Jul 4, 2017 at 10:06
• @FernandoMartin more generally any surjective endomorphism of a noetherian module over an arbitrary ring is injective. This has a counterpart in the required direction: any injective endomorphism of an artinian module over an arbitrary ring is surjective.
– YCor
Jul 4, 2017 at 16:14
• Useful keyword: a module over something / group / whatever is cohopfian (or co-Hopfian, or have the co-Hopf property) if all its injective endomorphisms are automorphisms.
– YCor
Jul 4, 2017 at 16:16
A famous result in this spirit is the Ax-Grothendieck theorem, whose statement is the following:
Theorem. If $f \colon \mathbb{C}^n \to \mathbb{C}^n$ is an injective polynomial function then $f$ is bijective.
• There is a generalisation to maps $f \colon X \to X$ where $X$ is any variety over an algebraically closed field $k$. Jul 4, 2017 at 10:27
• I wouldn't expect Grothendieck to state it just for the complex numbers!
– YCor
Jul 4, 2017 at 16:11
• @YCor: Grothendieck's version is probably about radicial endomorphisms of finitely generated $S$-schemes, cf. EGA IV$_3$, Prop. 10.4.11. Jul 4, 2017 at 19:50
There is an improvement of the answer of Joseph Van Name which I feel is much more in the spirit in the question asked:
Let $(X,d)$ be a compact metric space, and assume that the mapping $f\colon X\to X$ does not decrease distances, that is $d(f(x),f(y))\ge d(x,y)$ for all $x,y\in X$. Then $f$ is a bijection.
Sketch of a proof. The proof I came up with long ago consists of two lemmas.
Lemma 1. A mapping $f$ as above must be an isometry.
Proof. Let $(x,y)\in X\times X$, and consider the sequence $(f^k(x),f^k(y))$. Since $X$ is compact, there is a convergent subsequence $\{(f^{k_i}(x),f^{k_i}(y))\}$. In particular, the sequence $\{f^{k_i}(x)\}$ is convergent, which implies that $\lim_{j>i\to\infty}d(f^{k_i}(x),f^{k_j}(x))=0$, and hence (since $f$ does not decrease distances) $\lim_{j>i\to\infty}d(x,f^{k_j-k_i}(x))=0$. The same is true for the sequence of the second coordinates. Thus, $(x,y)$ is a limit point of the sequence $\{(f^k(x),f^k(y))\}$. Suppose that $d(f(x),f(y))>d(x,y)$ for some $x,y\in X$. Then $d(f(x),f(y))-d(x,y)>\epsilon$ for some $\epsilon>0$. Then $d(f^n(x),f^n(y))-d(x,y)>\epsilon$ for all $n$, which is a contradiction.
Lemma 2. An isometry of a compact metric space is a bijection.
Proof. Let $x\in X$. Let us consider the sequence $\{f^n(x)\}$. This sequence has a a convergent subsequence $\{f^{k_i}(x)\}$. This implies that $\lim_{j>i\to\infty}d(f^{k_i}(x),f^{k_j}(x))=0$, and hence (since $f$ is an isometry) $\lim_{j>i\to\infty}d(x,f^{k_j-k_i}(x))=0$. Thus, $x$ is a limit point of the sequence $\{(f^{k_j-k_i}(x)\}$, in particular $x$ is a limit point of $f(X)$. But an isometry is continuous, and the image of a compact space under a continuous map is a compact, thus $x\in f(X)$, so $f$ is surjective. Since $f$ is manifestly injective, the statement is proved.
• @MarkWildon done! Jul 6, 2017 at 2:23
• @Cœur : I wish! Alas I do not know a proper reference for this. I heard the statement from Valery Ryzhikov (a Russian mathematician primarily working with dynamical systems) about 15 years ago, and then came up with the proof documented above. Jul 7, 2017 at 3:53
• One reference for this is Burago, Burago and Ivanov's book "A course in metric geometry", Theorems 1.6.14 and 1.6.15. Jul 8, 2017 at 2:10
• I think we may consider it a folk result --I myself had a proof of an analogous statement back in the 90's: A 1-Lipschitz map on a compact metric space is surjective iff it is isometric. (btw, a friend of mine named it "the Wedding Blanket theorem" for reasons that should be apparent to whoever shared a bed with a partner in winter ;) ) Jul 8, 2017 at 9:34
• Another proof. Fix $r>0$, take the maximal $r$-distant set $A$ (i.e., $d(a,b)\ge r$ for $a\ne b$ in $A$) and the maximal $\sum_{a,b\in A} d(a,b)$. Then $f(A)$ is a better $r$-distant set unless $f|_A$ is isometry. So, $A$ and $f(A)$ are $r$-nets (else they are not maximal $r$-distant sets) and for any two points $x,y$ we may find points $f(a),f(b)$ in $f(A)$ such that $d(f(a),f(x)), d(f(b),f(y))\le r$, thus $d(a,x),d(b,y)\le r$ and both $d(x,y)$, $d(f(x),f(y))$ are within $2r$ from $d(a,b)=d(f(a),f(b))$. Since $r$ was arbitrary, $f$ is isometry, also $f(X)$ is dense in $X$, and compact. Done Jul 10, 2017 at 16:51
Suppose $(X,d)$ is a compact metric space. Then every mapping $f:X\rightarrow X$ such that $d(x,y)=d(f(x),f(y))$ is always bijective.
In group theory being "Hopfian" is the reverse property: every epimorphism from the group to itself is an automorphism. There are lots of Hopfian groups, for example finitely generated residually finite groups (includes the free groups) and the rationals (being thought of as an additive group).
Banach space theory:
Antonio Avilés and Piotr Koszmider constructed an infinite dimensional Banach space of continuous functions $C(K)$ such that every one-to-one operator $T : C(K) \to C(K)$ is onto.
Minor bit of self-promotion: if $\Gamma$ is a (discrete) group and $f\in\ell^1(\Gamma)$ then the natural convolution operator $T_f:\ell^\infty(\Gamma)\to \ell^\infty(\Gamma)$ has the "injective implies surjective" property. No soficity assumptions needed, despite the superficial similarity to the Gottschalk conjecture!
Probably also worth noting that "injective with closed range implies surjective" holds for convolution operators on $\ell^p(\Gamma)$ with $1<p<\infty$ if $\Gamma$ is an amenable group, and for convolution operators on $\ell^2(\Gamma)$ without any restriction on $\Gamma$. It fails for $p=1$ and $\Gamma$ containing a nonabelian free subgroup, by a construction of G. A. Willis.
• "Injective with closed range implies surjective" holds also for convolution operators on $L_1(G)$, where $G$ is a locally compact abelian group. Jul 10, 2017 at 7:18
Regarding the "general framework" part of your question, if we work in a general category with some notion of "dimension" or "size" and replace "injective" with "monic" then we can rephrase this condition as:
"Every proper subobject of an object is of strictly smaller dimension than the original object."
Given that the "dimensions" are, say, non-negative integers, this implies a descending chain condition on subobjects. In turn, we can (sort of) recover our notion of dimension by taking the length of the longest descending chain of subobjects.
Looking at this condition, it seems reasonable to assume it might be called a "noetherian category," and in fact Googling turns up such a definition on nlab (modulo some technical set theoretic condition).
So we can say that the categories of finite sets, finite dimensional vector spaces, and finite dimensional compact manifolds are all noetherian. In fact, we can say more: namely, that they are precisely the subcategories of noetherian objects in the categories of sets, vector spaces, and compact manifolds, respectively.
• Nitpick: your final para seems to suggest that the category of finite sets is a noetherian object in the category of sets. This does not seem quite right Jul 8, 2017 at 0:15
• @Yemon Choi: I've refined the grammar of that sentence now, thanks. Jul 8, 2017 at 4:18
An injective graph homomorphism between two connected $d$-regular graphs is bijective.
Yes, there's a general framework encompassing a variety of results like this. It encompasses at least your first two results (on sets and vector spaces) and the fact that any isometry of a compact metric space into itself is surjective.
The general framework I'm referring to is the theory of the eventual image', laid out in two posts at the $n$-Category Café from 2011: post 1, post 2.
A fully faithful tensor functor between fusion categories with the same Frobenius-Perron dimension is dominant, thus an equivalence.
Theorem. Let $R$ be a commutative ring with $1$, and let $A$ be a finitely generated $R$-algebra. Then, any surjective $R$-algebra endomorphism of $A$ is bijective.
This is proven non-constructively (using Noetherianness) in math.stackexchange #1221213.
Question 1. Is there a constructive proof?
Of course, the theorem above is a multiplicative analogue of the known fact that any surjective endomorphism of a finitely generated $R$-module is bijective. That latter fact has a strengthening due to Orzech (see my A constructive proof of Orzech’s theorem and the references there in), stating that if $M$ is a finitely-generated $R$-module, if $N$ is an $R$-submodule of $M$, and if $f : N \to M$ is a surjective $R$-module homomorphism, then $f$ is bijective. I am also wondering if an multiplicative analogue exists:
Question 2. Let $R$ be a commutative ring with $1$, and let $A$ be a finitely generated $R$-algebra. Let $B$ be an $R$-subalgebra of $A$. Is it true that any surjective $R$-algebra homomorphism $B \to A$ is bijective?
(If you have an answer-length answer to any of these questions, let me know and I'll open a question.)
You might find interesting the Cantor-Schroeder-Bernstein property.
The most interesting result contained in the pdf, in my opinion, is the following.
If a category $$\mathcal{K}$$ has a faithful functor $$F : \mathcal{K} \to \text{FinSet}$$ to the category of finite sets, then $$\mathcal{K}$$ has the CSB property.
Proof is quite trivial but this sets a quite interesing framework in which the question finds its sense.
A locally isometric (i.e., locally injective and with the metric on one pulling back to the metric on the other) map between connected, complete Riemannian manifolds of the same dimension is a surjection.
The Dixmier conjecture is an interesting example. (Every endomorphism of the Weyl algebra is automatically injective, so it's equivalent to asking whether injectivity implies surjectivity.)
There is a famous conjecture in group theory: group rings are directly finite, i.e. if G is a group, k is a field and a and b are elements of k[G] then ab=1 implies ba=1.
Your second example is a special case of this conjecture, essentially equivalent to the case when G is a finite group (and your first example is a special case of the second one, by applying a suitable Hom functor).
The conjecture is due to Kaplansky. There are many partial results, for example Kaplansky showed the conjecture when k is the field of complex numbers and G is arbitrary, and I think but I can be wrong Elek and Szabo showed it for arbitrary k when G is a sofic group.
• Hi Łukasz, my understanding is that GE+ES did indeed resolve this for sofic groups and arbitrary fields arxiv.org/abs/math/0305440 Jul 9, 2017 at 21:38
• the result is actually true for a group ring $R[G]$ with G sofic and R a left Noetherian ring (see arxiv.org/abs/1510.07655 ) Jul 9, 2017 at 22:34
Suppose that $U$ is an ultrafilter over a set $X$. Then the only endomorphism $f:(X,U)\rightarrow(X,U)$ in the category of ultrafilters is the identity morphism.
There is an example of this in stable homotopy theory, related to the famous Generating Hypothesis,' conjectured by Freyd (and still open as far as I know!). For finite complexes $$X$$ and $$Y$$, the map from the stable homotopy classes of (pointed) maps from $$X$$ to $$Y$$ can be related to the algebraic maps between stable homotopy groups via $$[ X, Y]^S \to \mathrm{Hom}(\pi_*^S(X), \pi_*^S(Y) ).$$ The Generating Hypothesis says that this function is always injective finite complexes $$X$$ and $$Y$$; and it is a theorem of Freyd that if it is true, then the map is always surjective, too.
It has been long enough since I read up on this that I can't remember if the surjectivity follows on a pointwise basis, or whether you need the full conjecture to get the implication for a particular $$X$$ and $$Y$$.
I did not find in the answers the following known claim which is dual to Vladimir's answer.
If $(X,d)$ is a metric compact set and a surjection $f$ from $X$ to $X$ is 1-Lipschitz, i.e., $d(f(x),f(y))\leqslant d(x,y)$, then $f$ is bijection and, moreover, isometry.
Proof. Fix $r>0$ and a minimal $r$-net $A$ in $X$ with minimal $\sum_{a,b\in A} d(a,b)$. Then $f(A)$ is an $r$-net on $f(X)=X$, and by minimality $f$ is isometry on $A$. For any $x,y$ in $X$ find $a,b$ in $A$ such that $d(x,a),d(y,b)\leqslant r$, then $$d(x,y)\geqslant d(f(x),f(y))\geqslant d(f(a),f(b))-d(f(x),f(a))-d(f(b),f(y))\geqslant \\ \geqslant d(a,b)-d(x,a)-d(b,y)\geqslant d(x,y)-2d(x,a)-2d(b,y)\geqslant d(x,y)-4r.$$ SInce $r$ was arbitrary, we get that $f$ is isometry.
The property that injectivity implies identity or at least injectivity implies surjectivity may arise in algebraic structures that have some form of nilpotence. Let me give an example that has arisen in my work on self-distributivity.
For this answer, suppose that $$(X,*,1)$$ satisfies the identities $$x*(y*z)=(x*y)*(x*z),x*1=1,1*x=x.$$ Define the right powers by letting $$x^{[1]}=x$$ and $$x^{[n+1]}=x*x^{[n]}$$. We say that $$(X,*,1)$$ is right nilpotent if for all $$x$$, there is some $$n$$ where $$x^{[n]}=1$$. If $$a\in X$$, then define the mapping $$L_{a}:X\rightarrow X$$ by letting $$L_{a}(x)=a*x$$. Then the mapping $$L_{a}$$ is an endomorphism, so we shall call $$L_{a}$$ a basic inner endomorphism. The monoid $$\mathrm{Inn}(X)$$ generated by $$(L_{a})_{a\in X}$$ is called the inner endomorphism monoid of $$(X,*)$$ and the elements in $$\mathrm{Inn}(X)$$ are known as inner endomorphisms.
Proposition: If $$(X,*)$$ is a right nilpotent self-distributive algebra, then every injective inner endomorphism is the identity function. Furthermore, if $$L_{a}:X\rightarrow X$$ is an injective inner endomorphism, then $$a=1$$.
Proof: Suppose that $$a\neq 1$$. Then $$L_{a}^{n}(a)=a^{[n+1]}=1=L_{a}^{n}(1)$$. Therefore, since $$L_{a}^{n}$$ is not injective, the mapping $$L_{a}$$ is not injective either. Therefore, if $$L_{a}$$ is injective, then $$a=1.$$ If $$f$$ is an injective inner endomorphism, then $$f=L_{a_{1}}\circ\dots\circ L_{a_{n}}$$ for some $$a_{1},\dots,a_{n}$$, but since $$f$$ is injective, so are each $$L_{a_{i}}$$, so $$a_{1}=\dots=a_{n}=1$$, and therefore $$f$$ is the identity function. QED
Examples of right nilpotent self-distributive algebras include the quotient algebras of rank-into-rank embeddings $$\mathcal{E}_{\lambda}/\equiv^{\gamma}$$ (and similar algebraic structures), and algebras $$(X,\rightarrow,1)$$ such that $$\rightarrow$$ is the Heyting operation in a Heyting algebra, and the algebras $$(X,*,1)$$ where there Is a function $$f$$ where $$x*y=f(y)$$ for each $$x,y$$ and where $$f(1)=1$$ and for each $$x\in X$$ there is an $$n$$ with $$f^{n}(x)=1.$$ | 2023-02-07T12:47:20 | {
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http://math.stackexchange.com/questions/756931/explanation-of-recursive-function | Explanation of recursive function
Given is a function $f(n)$ with:
$f(0) = 0$
$f(1) = 1$
$f(n) = 3f(n-1) + 2f(n-2)$ $\forall n≥2$
I was wondering if there's also a non-recursive way to describe the same function.
WolframAlpha tells me there is one:
$$g(n) = \frac{(\frac{1}{2}(3 + \sqrt{17}))^n - (\frac{1}{2}(3 - \sqrt{17}))^n}{\sqrt{17}}$$
However, I have absolutely no clue how to determine this function, especially the $\sqrt{17}$ makes no sense to me.
Could anyone maybe explain why $f(n)$ and $g(n)$ are the same?
-
Are you familiar with linear algebra, especially diagonalizing a matrix? – Alex Becker Apr 16 '14 at 21:05
You could just prove they are the same by induction. – Git Gud Apr 16 '14 at 21:20
Cool, so far 5 different ways to solve this (including the induction comment). I wonder how many different ways can be found. – DanielV Apr 16 '14 at 21:45
@AlexBecker I should be, but I feel like my knowledge about that didn't survive the semester holidays... :/ – Christian Schnorr Apr 16 '14 at 22:45
The $\sqrt{17}$ comes from $3^2+4 \times 2$, based on the coefficients of the recursion. In the Fibonacci sequence $f(n)=f(n-1)+f(n-2)$ you get $\sqrt{5}$ from $1^2+4 \times 1$ – Henry Apr 17 '14 at 13:20
\begin{align} f(n) &= 3\,f(n-1) + 2\,f(n-2) \\ f(n-1) &= 1\,f(n-1) + 0\,f(n-2)\end{align}
$$\begin{bmatrix} f(n) \\ f(n-1) \end{bmatrix} = \begin{bmatrix} 3 & 2 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} f(n - 1) \\ f(n - 2) \end{bmatrix}$$
$$\begin{bmatrix} f(n) \\ f(n-1) \end{bmatrix} = \begin{bmatrix} 3 & 2 \\ 1 & 0 \end{bmatrix}^{n - 1} \begin{bmatrix} f(1) \\ f(0) \end{bmatrix}$$
$$\begin{bmatrix} f(n + 1) \\ f(n) \end{bmatrix} = \begin{bmatrix} 3 & 2 \\ 1 & 0 \end{bmatrix}^{n} \begin{bmatrix} 1 \\ 0 \end{bmatrix}$$
Diagonalize (find eigen values / eigen vectors) and you have your closed form.
$$\begin{bmatrix} 3 & 2 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ \frac{2}{3 - \sqrt{17}} & \frac{2}{3 + \sqrt{17}} \end{bmatrix} \begin{bmatrix} \frac{3 - \sqrt{17}}{2} & 0 \\ 0 & \frac{3 + \sqrt{17}}{2} \end{bmatrix} \begin{bmatrix} 1 & 1 \\ \frac{2}{3 - \sqrt{17}} & \frac{2}{3 + \sqrt{17}} \end{bmatrix}^{-1}$$
-
You could prove that the formula is correct, by induction. It comes down to showing that the given expression satisfies the initial conditions and the recurrence.
There is also a general theory of homogeneous linear difference equations with constant coefficients. It says, for second-order difference equations (our case), that if the recurrence is $f(n)=pf(n-1)+qf(n-2)$, to do this.
1) Solve the equation $x^2-px-q=0$.
2) If the roots are distinct, say $\alpha$ and $\beta$, then the solutions of the difference equation are given by $$f(n)=A\alpha^n+B\beta^n.$$ The constants $A$ and $B$ can be determined by using the "initial values" $f(0)$ and $f(1)$.
-
Probably the (for me) easiest to understand answer! Thanks! – Christian Schnorr Apr 16 '14 at 22:43
You are welcome. – André Nicolas Apr 16 '14 at 22:47
This is brilliant. – András Hummer Apr 17 '14 at 9:08
This is called a linear recurrence. Solving them is fairly straightforward, and is explained here: http://en.wikipedia.org/wiki/Linear_recurrence#Solving.
The key thing to note is that if $f_1$ and $f_2$ are solutions of this recurrence, then $f_1 + f_2$ is as well (except for the initial conditions).
The trick is to assume that there is a solution of the form $f = cr^n$, and see where that leads you.
Let's see how it works out: $f(n) = 3f(n-1) + 2f(n-2)$\
$cr^n = 3cr^{n-1} + 2cr^{n-2}$
Dividing everything by $cr^{n-2}$ and rearranging gives: $r^2 - 3r - 2 = 0$. Factor, to get: $\frac{-1}{4} \left(-2 r+\sqrt{17}+3\right) \left(2 r+\sqrt{17}-3\right) = 0$
I conclude that $r = r_1 = \frac{\sqrt{17}-3}{2}$, or $r = r_2 = \frac{-\sqrt{17}+3}{2}$.
Using the fact that I can add solutions together and still have a solution to the recurrence relation, I'll write: $f(n) = c_1r_1^n + c_2r_2^n$
From here, you can plug in the initial conditions to solve for $c_1$ and $c_2$.
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Use generating functions. Define $F(z) = \sum_{n \ge 0} f(n) z^n$, write the recursion as: $$f(n + 2) = 3 f(n + 1) + 2 f(n)$$ Multiply by $z^n$, sum over $n \ge 0$ and recognize the resulting sums: $$\frac{F(z) - f(0) - f(1) z}{z^2} = 3 \frac{F(z) - f(0)}{z} + 2 F(z)$$ With the initial values you get: $$F(z) = \frac{z}{1 - 3 z - 2 z^2}$$ Split into partial fractions, and read off the coefficients by using geometric series.
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In which we assume a solution of the form $c^n$. This leads to a quadratic equation for which we hopefully can find two different solutions so that we can make a linear combination of the two that satisfies the boundary conditions. If there only one solution (with multiplicity 2) of the quadratic equation, you have to use the fact that the derivative of the quadratic is 0 as well (I don't remember the specifics, but it's not really relevant now, I just want to point out that this method doesn't always work).
This works as follows: write the recurrence relation in matrix form, and find the eigenvalues and eigenvectors of the matrix. When we've found them, we can hopefully write the initial conditions as a linear combination of eigenvectors. Now we can deduce the solution to the recurrence relatio. Suppose the initial conditions is $x_0 = av_1 + bv_2$ and $\lambda_1, \lambda_2$ are the eigenvalues of $v_1, v_2$. Then the solution to the recurrent relation is $x_n = a\lambda_1^nv_1 + b\lambda_2^nv_2$. With this method, it is also possible to see why Hero's method for approximating squareroots works (and with a little analysis, you can even come up with a better method for calculating squareroots of integers $n$ where $n > 2$). | 2015-08-03T20:04:43 | {
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http://nvmu.novaimperia.it/integration-theorems.html | The theorem is comprised of two parts, the first of which, the Fundamental Theorem of Calculus, Part 1, is stated here. MIT OpenCourseWare is a free & open publication of material from thousands of MIT courses, covering the entire MIT curriculum. Taking the derivative, we see x0 n (t) = 1 2nt2. Derivative matches upper limit of integration. The second fundamental theorem of calculus holds for a continuous function on an open interval and any point in , and states that if is defined by the integral (antiderivative) at each point in , where is the derivative of. Its existence […]. Interchange of Differentiation and Integration The theme of this course is about various limiting processes. The Gauss-Bonnet Theorem. Then: The unit normal is k. 2 Predict the following derivative. The approach I use is slightly different than that used by Stewart, butis based onthe same fundamental ideas. We could compute the line integral directly (see below). Let us consider two sections AA and BB of the pipe and assume that the pipe is running full and there is a continuity of flow between the two sections. 2 Sigma Sum 2. Many other elds of mathematics re-quire the basic notions of measure and integration. Although it can be naturally derived when combining the formal definitions of differentiation and integration, its consequences open up a much wider field of mathematics suitable to justify the entire idea of calculus as a math discipline. Central limit theorem, expansion of a tail probability, martingale, Generalized state sapce model, Monte Carlo integration, interest rate model,. Unless the variable x appears in either (or both) of the limits of integration, the result of the definite integral will not involve x, and so the. Theorem statement. The position y = F(t) is an anti-derivative of the velocity v = f(t). , S= ∂W, then the divergence theorem says that ∬SF⋅dS= ∭WdivFdV, where we orient S so that it has an outward pointing normal vector. The first thing to notice about the fundamental theorem of calculus is that the variable of differentiation appears as the upper limit of integration in the integral: Think about it for a moment. Complex Integration (2A) 3 Young Won Lim 1/30/13 Contour Integrals x = x(t) f (z) defined at points of a smooth curve C The contour integral of f along C a smooth curve C is defined by. Basic Integration Formulas and the Substitution Rule 1The second fundamental theorem of integral calculus Recall fromthe last lecture the second fundamental theorem ofintegral calculus. You will see plenty of examples soon, but first let us see the rule: ∫ u v dx = u ∫ v dx − ∫ u' (∫ v dx) dx. Mean Value Theorem V. This depends on finding a vector field whose divergence is equal to the given function. No mass can cross a system boundary. In this sense, Cauchy's theorem is an immediate consequence of Green's theorem. Taylor Series This formula shows how to express an analytic function in terms of its derivatives. Integration is a process of adding slices to find the whole. fundamental theorem: X n:a→b ∆ nf(n) = f(b)−f(a). 1 t y 0 1 4 6 1 f (a) Evaluateg(x)forx=0,1,2,3,4,5,and6. The net change theorem considers the integral of a rate of change. If ma + nb = 0 ; a & b are collinear Coplanarity. First, the following identity is true of integrals: $$\int_a^b f(t)\,dt = \int_a^c f(t)\,dt + \int_c^b f(t)\,dt. 1) f (x) = −x2 − 2x + 5; [ −4, 0] x f(x) −8 −6 −4 −2 2 4 6 8 −8 −6 −4 −2 2 4 6 8 11 3 ≈ 3. It includes some new results, but is also a self-contained introduction suitable for a graduate student doing self-study or for an advanced course on integration theory. , d⁄dx F(x) = f(x) Then ∫ f(x) dx = F(x) + C. Choosing a database is often a daunting task. It can be used to find areas, volumes, and central points. The divergence theorem can also be used to evaluate triple integrals by turning them into surface integrals. Answer to Theorem 1. 35) Theorem. 1 Introduction 16. The fundamental theorem of calculus is a theorem that links the concept of differentiating a function with the concept of integrating a function. As we know, Integration is a reverse process of differentiation, which is a process where we reduce the functions into smaller parts. a) In all states, the French model of state leadership: and state guarantee of railway bonds was followed. Stokes theorem: Let S be an oriented smooth surface that is bounded by a simple, closed, smooth boundary curve C with positive orientation. Sequences and series of functions (if time allows) a. overlapping types of measure and integration theory: the non-negative theory, which involves quantities taking values in [0;+1], and the absolutely integrable theory, which involves quantities taking values in (1 ;+1) or C. If a surface S is the boundary of some solid W, i. Complex integration and Cauchy's theorem by Watson, G. Home » Courses » Mathematics » Multivariable Calculus » 4. 2), and we are done. On the one hand it relates integration to differentiation, and on the other hand it gives a method for evaluating integrals. The fundamental theorem of calculus states that if is continuous on then the function defined on by is continuous on differentiable on and. Integration by parts can be extended to functions of several variables by applying a version of the fundamental theorem of calculus to an appropriate product rule. This is accomplished by means of the Fundamental Theorem of Calculus. Integrating using trigonometric identities. Integration using de Moivres theorem Watch. S = \int\limits_a^b {f\left ( x \right. theorem and easy to prove. The theorem is comprised of two parts, the first of which, the Fundamental Theorem of Calculus, Part 1, is stated here. There are several theorems in geometry that describe the relationship of angles formed by a line that transverses two parallel lines. the major theorems from the study of di erentiable functions in several variables. If c is a nonnegative real number, then 1. Interchange of Differentiation and Integration The theme of this course is about various limiting processes. Compute ∮Cy2dx + 3xydy where C is the CCW-oriented boundary of upper-half unit disk D. Riemann Integral b. The fundamental theorem of calculus has two parts: Theorem (Part I). Complex Integration using residue theorem Thread starter arpon; Start date Dec 7, 2016; Tags complex analysis residue residue integration. We follow Chapter 6 of Kirkwood and give necessary and sufficient. integration must be constant with respect to both variables of integration. Furthermore, a substitution which at first sight might seem sensible, can lead nowhere. Converse of Theorem 1 [Take b=0 and b=/= 0] Acute Angle between pair of straight lines [Use formula of angle between two lines having slopes m 1, m 2 ] Vectors. 2 Let {\bf F}=\langle 2x,3y,z^2\rangle, and consider the three-dimensional volume inside the cube with faces parallel to the principal planes and opposite corners at (0,0,0) and (1,1,1). Let (E,B,⌫)beameasurespace,andh : E ! R anon-negativemeasurablefunc-tion. Read and learn for free about the following article: Proof of fundamental theorem of calculus If you're seeing this message, it means we're having trouble loading external resources on our website. Lecture 1: Outer measure. 4A - The First Fundamental Theorem of Calculus HW 4. Welcome! This is one of over 2,200 courses on OCW. Binomial Theorem Binomial Theorem - I (The basics) Binomial Theorem II Sequence and Series Arithmetic Progression. Integrals >. Therefore, the first angle, as measured from the positive z z -axis, that will “start” the cone will be φ = 2 π 3 φ = 2 π 3 and it goes. Interpreting the behavior of accumulation functions involving area. 8 billion users and include 5 of the top 7 largest banks. Integration by Parts 21 1. 2 Integrals of functions that decay The theorems in this section will guide us in choosing the closed contour Cdescribed in the introduction. The moment of inertia about any axis parallel to that axis through the center of mass is given. According to legend, Pythagoras was so happy when he discovered the theorem that he offered a sacrifice of oxen. The integration of f′(x) with respect to dx is given as. Using rules for integration, students should be able to find indefinite integrals of polynomials as well as to evaluate definite integrals of polynomials over closed and bounded intervals. The term Green's theorem is applied to a collection of results that are really just restatements of the fundamental theorem of calculus in higher dimensional problems. Essentially, it said that the integral of the derivative is the function itself, evaluated at the endpoints. Integration by parts. Recall that the process of finding an indefinite integral is called integration. However, the theory of integration of top-degree differential forms has been defined for oriented manifolds with corners. ©M 12 50a1 e3m KTu itma d kStohf Ltqw va GrVeX uLKLFC K. For φ φ we need to be careful. "The Second Fundamental Theorem of Calculus. The graphs below are similar to the ones above, except that t=4. TPS and ETPS run in Common Lisp under Windows. It may look similar in i. We express the Lefschetz number of iterates of the monodromy of a function on a smooth complex algebraic variety in terms of the Euler characteristic of a space of truncated arcs. The formula can be expressed in two ways. The upper limit of integration is less than the lower limit of integration 0, but that's okay. 186] and the Courant-Fischer minimax theorem [1, p. 1-9a for the following values of RL: 2 kV, 6 kV, and 18 kV? If you really want to appreciate the power of Thevenin’s theorem, try calculating the foregoing currents using the original circuit of Fig. theorem and easy to prove. Solution for Carefully state the Fundamental Theorem of Calculus. (George Neville), 1886-Publication date 1914 Topics Functions, Integrals Publisher Cambridge, University press. Basics of Calculus Chapter 5, Topic 12—Integration Theorems Several primary integration theorems are discussed. , a hollow ball), no net gravitational force is exerted by the shell on any object inside, regardless. Power series Suggested textbook: E. According to integration definition maths, it is a process of finding functions whose derivative is given is named anti-differentiation or integration. Theorem (The Fundamental Theorem of Calculus. In more modern language, the theorem can also be phrased as saying that each short exact sequence of vector spaces splits. The Di erentiation Interchange Theorem We now consider another important theorem about the interchange of integration and limits of functions. Compute The boundary of S is traversed counterclockwise as viewed from above. Derivative of an integral. Symbolic computation DNA computing. Data integration. integration must be constant with respect to both variables of integration. Solution: The vector field in the above integral is F(x, y) = (y2, 3xy). Type in any integral to get the solution, steps and graph. Show that for every non-negative measurable function F : E ! R one has Z E Fdµ= Z E Fhd⌫. Consider a polynomial function f whose graph is smooth and continuous. To find the summation under a very large scale the process of integration is used. It was given by prominent French Mathematical Physicist Pierre Simon Marquis De Laplace. (1) This result, while taught early in elementary calculus courses, is actually a very deep result connecting the purely algebraic indefinite integral and the purely analytic (or geometric) definite integral. is a continuous function on the closed interval (i. ; Explain the significance of the net change theorem. Learning Objectives. Chapter 7 / Directed Integration Theory 7-1. The process of differentiation and integration are inverses of each other. Consider the function f(t) = t. The theorem is comprised of two parts, the first of which, the Fundamental Theorem of Calculus, Part 1, is stated here. The integrand of the triple integral can be thought of as the expansion of some. New variable. Data Theorem helped Evernote identify and close 105 security issues and remove 17 harmful third-party libraries, all before releasing them to the public app stores. Fundamental theorem of calculus VI. When is the Net Change Theorem used? Edit. iterated integral, parallel transport, holonomy. This worksheet can work as a starter before introducing integration topic. In calculus, integration by substitution, also known as u-substitution or change of variables, is a method for evaluating integrals. The second fundamental theorem of calculus holds for a continuous function on an open interval and any point in , and states that if is defined by the integral (antiderivative) at each point in , where is the derivative of. As a consequence it allows the order of integration to be changed in iterated integrals. This video covers the method of complex integration and proves Cauchy's Theorem when the complex function has a continuous derivative. The triple integral is the easier of the two:$$\int_0^1\int_0^1\int_0^1 2+3+2z\,dx\,dy\,dz=6. , S= ∂W, then the divergence theorem says that ∬SF⋅dS= ∭WdivFdV, where we orient S so that it has an outward pointing normal vector. In particular, when , is stretched to approach a constant, and is compressed with its value increased to approach an impulse; on the other hand, when , is compressed with. Now you can take a break. Examples Edit. The New 2017 A level page. This paper presents anecdotal evidence that suggests that financial markets often are not integrated and discusses the implications. Suppose that α1, α2 are non-decreasing, and that f, g are Riemann-Stieltjes integrable with respect to both α1 and α2. As if it helps. CONTINUOUS FUNCTIONS. The fundamental theorem of calculus is a bridge between the two seemingly disconnected tasks of computing areas under curves (integration) and finding derivatives of curves (differentiation). Stokes Theorem is used for any surface (or) any plane ( -plane, -plane, -plane) Green’s Theorem. Lower limit of integration is a constant. The first fundamental theorem of calculus states that, if f is continuous on the closed interval [a,b] and F is the indefinite integral of f on [a,b], then int_a^bf(x)dx=F(b)-F(a). Here is a super-duper shortcut integration theorem that you’ll use for the rest of your natural born days — or at least till the end of your stint with calculus. The Di erentiation Interchange Theorem We now consider another important theorem about the interchange of integration and limits of functions. Saiegh Calculus: Applications and Integration. Complex integration and Cauchy's theorem by Watson, G. Integration. In this section, we use some basic integration formulas studied previously to solve. txt) or view presentation slides online. Last Post; Sep 23, 2011; Replies 3 Views 3K. If f is a continuous function and is defined by. 1 (EK), FUN. Of course, one way to think of integration is as antidi erentiation. The integration theorem states that. Integration by parts. There is no limit to the smallness of the distances traversed. Finding derivative with fundamental theorem of. Is there a proof that the area under a curve is equivalent to the definite integral, that doesn't involve the fundamental theorem of calculus. Let f (x) and g(x) be continuous on [a, b]. The ideas are classical and of transcendent beauty. Derivative of an Integral) Suppose that f is continuous on [a,b] and set , then F is differentiable and F'(x) = f(x) for a1; (4) where the integration is over closed contour shown in Fig. 3), numerical differentiation (Theorem 5. 4B - Average Value of a Function & The Second Fu ndamental Theorem o f Calculus HW 4. 1145{1160] & [Bourne, pp. The calculator will find all numbers c (with steps shown) that satisfy the conclusions of the Mean Value Theorem for the given function on the given interval. The fundamental theorem of calculus is a bridge between the two seemingly disconnected tasks of computing areas under curves (integration) and finding derivatives of curves (differentiation). for all -values. ∫a b f d. MA: Mathematics Calculus: Finite, countable and uncountable sets, Real number system as a complete ordered field, Archimedean property; Sequences and series, convergence; Limits, continuity, uniform continuity, differentiability, mean value theorems; Riemann integration, Improper integrals; Functions of two or. CONTOUR INTEGRATION AND CAUCHY’S THEOREM CHRISTOPHER M. Note that when , time function is stretched, and is compressed; when , is compressed and is stretched. We will, of course, use polar coordinates in. Initial Value Theorem is one of the basic properties of Laplace transform. The rule is derivated from the product rule method of differentiation. Integration by Parts is a special method of integration that is often useful when two functions are multiplied together, but is also helpful in other ways. 3), numerical differentiation (Theorem 5. The Di erentiation Interchange Theorem We now consider another important theorem about the interchange of integration and limits of functions. Bayes theorem is a wonderful choice to find out the conditional probability. Applications. It can be used to find areas, volumes, and central points. That is, the right-handed derivative of gat ais f(a), and the left-handed derivative of fat bis f(b). 6 Integration: The Fundamental Theorem of Calculus All graphics are attributed to: Calculus,10/E by Howard Anton, Irl Bivens, and Stephen Davis. 5 Trapezoidal Rule Chapter 6 6. Say you have a function f(x), the integral of this function, F(x), is called the antiderivative. Our approximationing sums will be obtained using a gauge function δ: Ω→(0,1]. A continuous function. Brie y put, we carry over de nitions using real and imaginary parts. integration in 2 and 3 dimensions. Questions are taken from the pre 2010 exam papers. Relationships between convergence: (a) Converge a. The Mean Value Theorem is an important theorem of differential calculus. Worksheets are Fundamental theorem of calculus date period, Work 24 de nite integrals and the fundamental, Work the fundamental theorem of calculus multiple, Fundamental theorem of calculus date period, The fundamental theorems of calculus, The fundamental theorem of calculus, John. Proofs of Parseval’s Theorem & the Convolution Theorem (using the integral representation of the δ-function) 1 The generalization of Parseval’s theorem The result is Z ∞ −∞ f(t)g(t)∗dt= 1 2π Z ∞ −∞ f(ω)g(ω)∗dω (1) This has many names but is often called Plancherel’s formula. According to integration definition maths, it is a process of finding functions whose derivative is given is named anti-differentiation or integration. 5 Use the corollary to predict the value of , then check your work with the TI-89. This observation is critical in applications of integration. Our Courses. Proof: This follows immediately from integration by parts: since. This module mostly deals with #3, the integral of a discrete function. All C4 Revsion Notes. Stokes Theorem is used for any surface (or) any plane ( -plane, -plane, -plane) Green’s Theorem. The Evaluation Theorem 11 1. 6 Section 5. My name is Rob Tarrou and standing next to me, every step of the way, is my wonderful wife Cheryl. Beyond the Pythagorean Theorem. The work-energy theorem is useful, however, for solving problems in which the net work is done on a particle by external forces is easily computed and in which we are interested in finding the particles speed at certain positions. svg 450 × 415; 34 KB. While working full time I have managed to make over 500 video lessons in these 4 years. , S= ∂W, then the divergence theorem says that ∬SF⋅dS= ∭WdivFdV, where we orient S so that it has an outward pointing normal vector. 1971] ARZELA' S DOMINATED CONVERGENCE THEOREM 971 integration for infinite series of integrable functions. Integration by Parts is a method of integration that transforms products of functions in the integrand into other easily evaluated integrals. The book is divided into two parts. 3Blue1Brown series S2 • E8 Integration and the fundamental theorem of calculus | Essence of calculus, chapter 8 - Duration: 20:46. Thus, the two parts of the fundamental theorem of calculus say that differentiation and integration are inverse processes. In symbols, the rule is ∫ f Dg = fg − ∫ gDf. Developing an arithmetical basis that avoids geometrical intuitions, Watson also provides a brief account of the various applications of the theorem to the evaluation of definite integrals. Stokes Theorem is used for any surface (or) any plane ( -plane, -plane, -plane) Green’s Theorem. Integration can be used to find areas, volumes, central points and many useful things. Fourier analysis, limit theorems in probability theory, Sobolev spaces, and the stochastic calculus of variations. The Intermediate Value Theorem states that for two numbers a and b in the domain of f , if a < b and f (a) ≠ f (b), then the function f takes on every value between f (a) and f (b). If you're behind a web filter, please make sure that the domains *. The solution to the problem is. \] You should now verify that this is the correct answer by substituting this in Equation 14. Don't show me this again. ★ Use the Fundamental Theorem of Calculus to evaluate definite integrals. Seamless integration with execution and clearing brokers. The Area under a Curve and between Two Curves. When the theorem was first stated, most of us thought of it as a proposition about a firm's debt-equity mix. As we know, Integration is a reverse process of differentiation, which is a process where we reduce the functions into smaller parts. How to calculate price in shopping list as well as cost of transportation variables including price of gas, distance drove, commuter-bus variables, which bus, reg or express, time for commute adding lengthen or shorten time depending on weather, how busy and business location that has item. 6 CHAPTER 1. Recall that the process of finding an indefinite integral is called integration. The moment area theorems provide a way to find slopes and deflections without having to go through a full process of integration as described in the previous section. Integration by parts. Throughout these notes, we assume that f is a bounded function on the interval [a,b]. The fundamental theorem of calculus shows how, in some sense, integration is the opposite of differentiation. G(x) = F(x) + C. Integral Theorems [Anton, pp. the major theorems from the study of di erentiable functions in several variables. The next graph shows the result of the integration for all time, with a black dot at t=1. Complex integration and Cauchy's theorem by Watson, G. As before, to perform this new approximation all that is necessary is to change the calculation of k1 and the initial condition (the value of the exact solution is also changed, for plotting). Integration can be used to find areas, volumes, central points and many useful things. The millenium seemed to spur a lot of people to compile "Top 100" or "Best 100" lists of many things, including movies (by the American Film Institute) and books (by the Modern Library). Complex Integration and Cauchy's Theorem (Dover Books on Mathematics) Paperback – May 17, 2012 by G. 3 Complexification of the Integrand. The Net Change Theorem. Created by Sal Khan. Let f (x) and g(x) be continuous on [a, b]. All C1 Revsion Notes. Mathematicians were not immune, and at a mathematics conference in July, 1999, Paul and Jack Abad presented their list of "The Hundred Greatest Theorems. Chebyshev (1821-1894), who has proven this theorem, the expression x a (α + β x b) c d x is called a differential binomial. g(0)= R0 0 f(t) dt=0fromtheproperty R0 0 f(x) dx=0. 4A - The First Fundamental Theorem of Calculus HW 4. Integration is an important tool in calculus that can give an antiderivative or represent area under a curve. Riemann integration is the formulation of integration most people think of if they ever think about integration. The differentiation theorem is implicitly used in § E. Integration Piece-by-piece multiplication Derivative Intro Measurements depend on the instrument Derivatives II Imagine linked machines Derivatives III Quotient, exponents, logs Calculus Bank Account Raises change income, changing the balance. Stokes' theorem is another related result. Closely tied with measures and integration are the subjects of product measures, signed measures, the Radon-Nikodym theorem, the di erentiation of functions on the line, and L p spaces. As the name "First Mean Value Theorem" seems to imply, there is also a Second Mean Value Theorem for Integrals: Second Mean Value Theorem for Integrals. Let ; and for. ASL-STEM Forum. Technology is quickly changing the landscape at electric utilities and Theorem Geo is proud to participate in the revolution. Search this site. We'll learn that integration and di erentiation are inverse operations of each other. Using rules for integration, students should be able to find indefinite integrals of polynomials as well as to evaluate definite integrals of polynomials over closed and bounded intervals. The area under the graph of the function f\left ( x \right) between the vertical lines x = a, x = b (Figure 2) is given by the formula. 6 Section 5. From this theorem we get the following obvious consequence: Corollary 7. Course Objectives. The Di erentiation Interchange Theorem We now consider another important theorem about the interchange of integration and limits of functions. This integral is not absolutely convergent, meaning | | is not Lebesgue-integrable, and so the Dirichlet integral is undefined in the sense of. Theorem Proof Consider a perfect incompressible liquid, flowing through a non-uniform pipe as shown in fig. Integration Theory 49 1 The Lebesgue integral: basic properties and convergence theorems 49 2 The space L1 of integrable functions 68 3 Fubini’s theorem 75 3. So between the big s with your limits and the. The Evaluation Theorem. TPS and ETPS run in Common Lisp under Windows. There's a lot of confusion, a 'theorem', and more than all, the immortal proverb 'not one size fits all'. Integration by parts. 1145{1160] & [Bourne, pp. In other words, the derivative of is. The first part of the theorem, sometimes called the first fundamental theorem of calculus, states that one of the antiderivatives (also called indefinite integral), say F, of some function f may be obtained as the integral of f with a variable bound. The later discovery that the square root of 2 is irrational and therefore, cannot be expressed as a ratio of two integers, greatly troubled Pythagoras and. The theorem tells us that in order to evaluate this integral all we need are the initial and final points of the curve. This is a good point to illustrate a property of transform pairs. The problem statement says that the cone makes an angle of π 3 π 3 with the negative z z -axis. Integrating with u-substitution. This applet has two functions you can choose from, one linear and one that is a curve. Potential applications of automated theorem proving include hardware and software verification, partial automation. So the theorem is proven. Given at the University of Florida, Spring Semester 2004. One of the most important theorems in calculus is properly named the fundamental theorem of integral calculus. The first fundamental theorem of calculus states that, if is continuous on the closed interval and is the indefinite integral of on , then. 4 The Chain Rule and Taylor's Theorem 339 Chapter 6 Vector-Valued Functions of Several Variables 361 6. They are simply two sides of the same coin (Fundamental Theorem of Caclulus). This book presents a general approach to integration theory, as well as some advanced topics. Integration of the General Network Theorem in ADE and ADE XL - Free download as PDF File (. It includes discussions on descriptive simulation modeling, programming commands, techniques for sensitivity estimation, optimization and goal-seeking by simulation, and what-if analysis. We know that the Fourier transform of a Gaus-sian: f(t) =e−πt2 is a Gaussian:. 3 The Inverse Function Theorem 394 6. My hope is that, armed with the right intuitions, Green’s theorem should feel nearly natural. impossibility theorems for elementary integration problems. A graph of a functions is a visual representation of the pairs (input, output), in the plane. I would like to know if there is any relation for integration from (-a, a) in the integration in the Parseval's theorem; where a is a real number. A control volume is a region in space chosen for study. for all -values. It is easy to see x n!ptws x where x(t) = 0 on [ 1;1]. Green's Theorem states that if R is a plane region with boundary curve C directed counterclockwise and F = [M, N] is a vector field differentiable throughout R, then Example 2. Data integration. This can be proved directly from the definition of the integral, that is, using the limits of sums. We show that (1) implies (4). The function is continuous on [−2,3] and differentiable on (−2,3). Some background knowledge of line integrals in vector. Here is a super-duper shortcut integration theorem that you’ll use for the rest of your natural born days — or at least till the end of your stint with calculus. Fundamental theorem of calculus VI. Beyond the Pythagorean Theorem. of residue theorem, and show that the integral over the "added"part of C R asymptotically vanishes as R → 0. Developing an arithmetical basis that avoids geometrical intuitions, Watson also provides a brief account of the various applications of the theorem to the evaluation of definite integrals. No mass can cross a system boundary. The CAP Theorem states that, in a distributed system (a collection of interconnected nodes that share data. We will sketch the proof, using some facts that we do not prove. 3 Theorems of Pappus and Guldinus Example 4, page 1 of 1 4 m x y 1 m C 4. The divergence theorem can also be used to evaluate triple integrals by turning them into surface integrals. The Binomial Theorem is used for expanding brackets in the form (a + b)n. In words, this result is that a continuous function on a closed, bounded interval has at least one point where it is equal to its average value on the interval. The second is more familiar; it is simply the definite integral. By properties of integrals,. Integral using residue theorem. It includes some new results, but is also a self-contained introduction suitable for a graduate student doing self-study or for an advanced course on integration theory. We express the Lefschetz number of iterates of the monodromy of a function on a smooth complex algebraic variety in terms of the Euler characteristic of a space of truncated arcs. INTEGRATION THEOREMS Theorem (The Fundamental Theorem of Calculus (Part 1)) Let be a continuous function whose domain includes and is an antiderivative of (i. Formalizing 100 Theorems. The moment area theorems provide a way to find slopes and deflections without having to go through a full process of integration as described in the previous section. primitives and vice versa. 2 Integration by Substitution and Separable Differential Equations: 6. The Riemann-Lebesgue Theorem Based on An Introduction to Analysis, Second Edition, by James R. It is essential, though. Watson begins by reviewing various propositions of Poincaré's Analysis Situs, upon which proof of the theorem's most general form depends. for all -values. Determine the volume of the half-torus (half of a doughnut). Show that for every non-negative measurable function F : E ! R one has Z E Fdµ= Z E Fhd⌫. D Worksheet by Kuta Software LLC. Read and learn for free about the following article: Proof of fundamental theorem of calculus If you're seeing this message, it means we're having trouble loading external resources on our website. Integration by Parts 21 1. The differentiation theorem is implicitly used in § E. Type in any integral to get the solution, steps and graph. This depends on finding a vector field whose divergence is equal to the given function. Lecture 3: Additivity of outer measures. Theorem, in mathematics and logic, a proposition or statement that is demonstrated. 186] and the Courant-Fischer minimax theorem [1, p. Stokes' theorem is another related result. This theorem was proved by Giovanni Ceva (1648-1734). If $$\vec F$$ is a conservative vector field then $$\displaystyle \int\limits_{C}{{\vec F\centerdot \,d\,\vec r}}$$ is independent of path. if r = ax + by , then r,x,y are coplanar [using collinearity and parallelogram law] Converse of Theorem 2. Following are some of the most frequently used theorems, formulas, and definitions that you encounter in a calculus class for a single variable. The rst theorem is for functions that decay faster than 1=z. Unless the variable x appears in either (or both) of the limits of integration, the result of the definite integral will not involve x, and so the. Gauss’ theorem was beyond my tuition though I did read a university book many years ago on fluid mechanics that used all the tensor stuff. Perhaps a proof that uses Riemann sums. It can be used to find areas, volumes, and central points. Upload media. Chapter Eighteen - Stokes 18. Pythagoras is usually given the credit for coming up with the theorem and providing early proofs, although evidence suggests that the theorem actually predates the existence of Pythagoras, and that he may simply have popularized it. The Net Change Theorem can be applied to all rates of change in the outside world, such as natural and social sciences (measuring water volume, population growth in Disneyland, etc. The indefinite integral of , denoted , is defined to be the antiderivative of. This book presents a general approach to integration theory, as well as some advanced topics. Integration extends to the third dimension with the means to compute the volumes of shapes obtained by revolving regions around an axis, such as spheres, toroids, and paraboloids. qxd 11/1/10 6:57 PM Page 719. To start with, the Riemann integral is a definite integral, therefore it yields a number, whereas the Newton integral yields a set of functions (antiderivatives). We will now summarize the convergence theorems that we have looked at regarding Lebesgue integration. Integral Theorems [Anton, pp. Fundamental Theorem of Calculus, Riemann Sums, Substitution Integration Methods 104003 Differential and Integral Calculus I Technion International School of Engineering 2010-11 Tutorial Summary - February 27, 2011 - Kayla Jacobs Indefinite vs. The mean value theorem expresses the relatonship between the slope of the tangent to the curve at x = c and the slope of the secant to the curve through the points (a , f(a)) and (b , f(b)). 1The definite integral Recall thatthe expression ∫b a f(x)dx. ; Use the net change theorem to solve applied problems. The derivative of an indefinite integral. Theorem Proof Consider a perfect incompressible liquid, flowing through a non-uniform pipe as shown in fig. CONTINUOUS FUNCTIONS. Solution: The vector field in the above integral is F(x, y) = (y2, 3xy). AP Calc: FUN‑5 (EU), FUN‑5. "Theory of the. Watson begins by reviewing various propositions of Poincaré's Analysis Situs, upon which proof of the theorem's most general form depends. The scaling theorem provides a shortcut proof given the simpler result rect(t),sinc(f). 25_5; Start date Jul 23, 2014. Watch Queue Queue. This implies. Inverse and Implicit Functions 7-7. Its formula is pretty simple: Its formula is pretty simple: P(X|Y) = ( P(Y|X) * P(X) ) / P(Y), which is Posterior = ( Likelihood * Prior ) / Evidence. This rectangle, by the way, is called the mean-value rectangle for that definite integral. Integration helps when trying to multiply units into a problem. 5A - Integration by U-Substitution. Let measurable I, Approximation by simple functions (M, A, u) be a measure space. 10 Fourier Series and Transforms (2014-5559) Fourier Transform - Parseval and Convolution: 7 - 2 / 10. 1The definite integral Recall thatthe expression ∫b a f(x)dx. Derivative of an Integral) Suppose that f is continuous on [a,b] and set , then F is differentiable and F'(x) = f(x) for a1; (4) where the integration is over closed contour shown in Fig. kernel of integration is the exact differential forms. We provide with proofs only basic results, and leave the proofs of the others to. of rapidly decreasing functions by Fourier integrals, and Shannon’s sampling theorem. INTEGRATION THEOREMS Theorem (The Fundamental Theorem of Calculus (Part 1)) Let be a continuous function whose domain includes and is an antiderivative of (i. Fundamental theorem of calculus and definite integrals. (Try sketching or graphing the integrand to see where the problem lies. Part 1 establishes the relationship between differentiation and integration. Fundemental Theorem Of Integration. If you can't do this, I can't see you passing. g(0)= R0 0 f(t) dt=0fromtheproperty R0 0 f(x) dx=0. So the real job is to prove theorem 7. It is intrinsically beautiful because it relates the curvature of a manifold—a geometrical object—with the its Euler Characteristic—a topological one. In particular, when , is stretched to approach a constant, and is compressed with its value increased to approach an impulse; on the other hand, when , is compressed with. ), you can only have two out of the following three guarantees across a write/read pair: Consistency, Availability, and Partition Tolerance - one of them must be sacrificed. You have to integrate it in pieces. overlapping types of measure and integration theory: the non-negative theory, which involves quantities taking values in [0;+1], and the absolutely integrable theory, which involves quantities taking values in (1 ;+1) or C. Show that for every non-negative measurable function F : E ! R one has Z E Fdµ= Z E Fhd⌫. Ask Question Asked 4 years, 1 month ago. This in turn tells us that the line integral must be independent of path. This method is based on this mathematical theorem. We show in proving Theorem. This is a set of lecture notes which present an economical development of measure theory and integration in locally compact Hausdor spaces. The integration theorem states that. The other three fundamental theorems do the same transformation. You might think I'm exaggerating, but the FTC ranks up there with the Pythagorean Theorem and the invention of the numeral 0 in its elegance and wide-ranging applicability. The relationship between these two processes is somewhat analogous to that which holds between “squaring” and “taking the square root. 1971] ARZELA' S DOMINATED CONVERGENCE THEOREM 971 integration for infinite series of integrable functions. It can be used to find areas, volumes, and central points. The first fundamental theorem of calculus states that, if is continuous on the closed interval and is the indefinite integral of on , then. Pythagoras is usually given the credit for coming up with the theorem and providing early proofs, although evidence suggests that the theorem actually predates the existence of Pythagoras, and that he may simply have popularized it. Complex Integration (2A) 3 Young Won Lim 1/30/13 Contour Integrals x = x(t) f (z) defined at points of a smooth curve C The contour integral of f along C a smooth curve C is defined by. primitives and vice versa. Throughout these notes, we assume that f is a bounded function on the interval [a,b]. 667 2) f (x) = −x4 + 2x2 + 4; [ −2, 1] x f(x) −8 −6 −4 −2 2 4 6 8 −8 −6 −4. 3 A Pleasing Application. 2 Convergence Theorems 2. The radius of convergence is not affected by differentiation or integration, i. Stokes Theorem is used for any surface (or) any plane ( -plane, -plane, -plane) Green’s Theorem. In case either E or I vary along the beam, it is advisable to construct an M /(EI) diagram instead of a moment diagram. This will allow us to use Lusin’s Theorem. The theorem basically just guarantees the existence of the mean value rectangle. As an example, consider I 1 = Z C 1 dz z and I 2 = Z C 2 dz z. As an example we will show that Z ∞ 0 dx (x2 +1)2 = π 4. Answer : True. "Theory of the. References. The Fundamental Theorem of Calculus, Part 1 shows the relationship between the derivative and the integral. FTCI: Let be continuous on and for in the interval , define a function by the definite integral: Then is differentiable on and , for any in. Apply the basic integration formulas. It is easy to see x n!ptws x where x(t) = 0 on [ 1;1]. I started making math videos September of 2011 after a student told me they were using the internet for math help. 1145{1160] & [Bourne, pp. Integration is then carried out with respect to u, before reverting to the original variable x. Measure and Integration: Exercise on Radon-Nikodym Theorem, 2014-15 1. When the theorem was first stated, most of us thought of it as a proposition about a firm's debt-equity mix. Two Fundamental Theorems about the Definite Integral These lecture notes develop the theorem Stewart calls The Fundamental Theorem of Calculus in section 5. What is an Accumulation Function?. 5 Integration Formulas and the Net Change Theorem Learning Objectives. This method is based on this mathematical theorem. You appear to be on a device with a "narrow" screen width ( i. It can be used to find areas, volumes, and central points. in the context of numerical integration (Theorems 4. simple s: M —1 6>0, such that E < s < f on M. Integration by parts v2. You can: Choose either of the functions. Stokes' Theorem is a lower-dimension version of the Divergence Theorem, and a higher-dimension version of Green's Theorem. In words, this result is that a continuous function on a closed, bounded interval has at least one point where it is equal to its average value on the interval. Integration by parts. Questions on this topic are usually short ones: you usually only have to find one. For example, a C-valued function can be written in the form f(x) = u(x) + iv(x) via. Show Instructions In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. MA525 ON CAUCHY'S THEOREM AND GREEN'S THEOREM 2 we see that the integrand in each double integral is (identically) zero. 31 Author: CamScanner Subject: new doc 2017-05-30 10. Derivative of an integral. Part 1 establishes the relationship between differentiation and integration. The theorem below is a direct consequence of the monotone convergence. Central limit theorem, expansion of a tail probability, martingale, Generalized state sapce model, Monte Carlo integration, interest rate model,. The definite integral of a function gives us the area under the curve of that function. In this section, we use some basic integration formulas studied previously to solve. 3 Suppose that ∑ ( ) is the (n+1) -point open Newton Cotes formula with and. Pearson Education accepts no responsibility whatsoever for the accuracy or method of working in the answers given. Course Objectives. Type in any integral to get the solution, steps and graph. Inverse and Implicit Functions 7-7. Theorem 1 serves to quantify the idea that the difierence in function values for a smooth. In vector calculus, and more generally differential geometry, Stokes' theorem (sometimes spelled Stokes's theorem, and also called the generalized Stokes theorem or the Stokes-Cartan theorem) is a statement about the integration of differential forms on manifolds, which both simplifies and generalizes several theorems from vector calculus. Chapter Seventeen - Gauss and Green 17. 282 Dr Dixon, The second mean value theorem The Second Mean Value Theorem in the Integral Calculus. The Fundamental Theorem of Calculus The Fundamental Theorem of Calculus shows that di erentiation and Integration are inverse processes. The Fundamental Theorem of Calculus, Part 1 If f is continuous on , then the function has a derivative at every point in , and First Fundamental Theorem: 1. 02SC Multivariable Calculus, Fall 2010 MIT OpenCourseWare 9 years ago 205K 17:33. Evaluate it at the limits of integration. Moreover, if you superimpose this rectangle on the definite integral, the top of the rectangle intersects the function. In the theory of Henstock and McShane integration, the appear-ance of a gauge function is rather mysterious. 10 Rational Functions by Partial Fraction Decomposition 4. The Divergence Theorem relates surface integrals of vector fields to volume integrals. Integrating using long division and completing the square. Trigonometric substitution. Continuous functions are integrable c. Stokes' Theorem allows you to compute the line integral around the boundary of a surface by computing the flux of through any surface with the same boundary. integration must be constant with respect to both variables of integration. The mathematics of such integrals can be studied largely independently of specif. 13 Irrational Functions 4. Math · AP®︎ Calculus AB · Integration and accumulation of change · The fundamental theorem of calculus and definite integrals. 1 Stokes's Theorem 18. This article was adapted from an original article by V. Integration can be used to find areas, volumes, central points and many useful things. Derivative of an integral. The fundamental theorem states that the area under the curve y = f(x) is given by a function F(x) whose derivative is f(x), F′(x) = f(x). a) (F atou) Let f n b e a sequenc e of non. A continuous function. SheLovesMath. if r = ax + by , then r,x,y are coplanar [using collinearity and parallelogram law] Converse of Theorem 2. real numbers witha1. Formalizing 100 Theorems. theorem and easy to prove. A Level (Edexcel) All A level questions arranged by topic. The relationship between these two processes is somewhat analogous to that which holds between “squaring” and “taking the square root. Lecture 1: Outer measure. 2 Green's Theorem 17. Second Fundamental Theorem of Calculus If the upper limit of integration is a function u of x, then Summary One of the most important examples of how an integral can be used to define another function is the defin-ition of the natural logarithm. Mean Value Theorem V. When you come back see if you can work out (a+b) 5 yourself. Integration by parts Visualization. According to integration definition maths, it is a process of finding functions whose derivative is given is named anti-differentiation or integration. Theorems - - Examples with step by step explanation. The Fundamental Theorem of Calculus. 02SC Multivariable Calculus, Fall 2010 MIT OpenCourseWare 9 years ago 205K 17:33. overlapping types of measure and integration theory: the non-negative theory, which involves quantities taking values in [0;+1], and the absolutely integrable theory, which involves quantities taking values in (1 ;+1) or C. The Integral as an Accumulation Function Formulas is an accumulation function. Its existence …. Let f be a continuous function on [a,b]. pdf), Text File (. In vector calculus, and more generally differential geometry, Stokes' theorem is a statement about the integration of differential forms on manifolds, which both simplifies and generalizes several theorems from vector calculus. Integration by parts is a heuristic rather than a purely mechanical process for solving Repeated integration by parts. The theorem below is a direct consequence of the monotone convergence. We follow Chapter 6 of Kirkwood and give necessary and sufficient. STOKES’ THEOREM, GREEN’S THEOREM, & FTC In fact, consider the special case where the surface S is flat, in the xy-plane with upward orientation. Integration is an important tool in calculus that can give an antiderivative or represent area under a curve. 1914 edition. , S= ∂W, then the divergence theorem says that ∬SF⋅dS= ∭WdivFdV, where we orient S so that it has an outward pointing normal vector. Examples Edit. We must restrict the domain of the squaring function to [0,) in order to pass the horizontal line test. However, applications of the theorem have since been expanded to discussions of debt maturity, risk management, and even mergers and spinoffs, which, according to the logic of M&M, neither create, nor destroy value in the absence of positive or negative synergies. Continuous functions are integrable c. 1 (Fundamental Theorem of Line Integrals) Suppose a curve. Fatou's lemma: If { fk}k ∈ N is a sequence of non-negative measurable functions, then Again, the value of any of the integrals may be infinite. (a) A domain (region) is an open connected subset of Rn. The residue theorem has applications in functional analysis, linear algebra, analytic number theory, quantum field theory, algebraic geometry, Abelian integrals or dynamical systems. Complex Integration and Cauchy's Theorem (Dover Books on Mathematics) Paperback – May 17, 2012 by G. This in turn tells us that the line integral must be independent of path. The pictured curve is parametrized by the variable t. We show that (1) implies (4). Kirkwood, Boston: PWS Publishing (1995) Note. The two forms of Green’s Theorem Green’s Theorem is another higher dimensional analogue of the fundamental theorem of calculus: it relates the line integral of a vector field around a plane curve to a double integral of “the derivative” of the vector field in the interior of the curve. In Part I of this paper, we give an extension of Liouville’s Theorem and give a number of examples which show that integration with special functions involves some phenomena that do not occur in integration with the elementary functions alone. The first fundamental theorem of calculus states that, if f is continuous on the closed interval [a,b] and F is the indefinite integral of f on [a,b], then int_a^bf(x)dx=F(b)-F(a). Integration by parts theorem proof. All C1 Revsion Notes. It is intrinsically beautiful because it relates the curvature of a manifold—a geometrical object—with the its Euler Characteristic—a topological one. We say an integral, not the integral, because the antiderivative of a function is not unique. In accordance with P. The Fundamental Theorem of Calculus. Ceva's theorem. Jackson blithely integrates by parts (for a charge/current density with compact support) thusly:. If you can't do this, I can't see you passing. The Mean Value Theorem is one of the most important theoretical tools in Calculus. Consider a vector field A and within that field, a closed loop is present as shown in the following figure. References. The first part of the theorem shows that indefinite integration can be reversed by differentiation. Chebyshev (1821-1894), who has proven this theorem, the expression x a (α + β x b) c d x is called a differential binomial.
upl56cmpe6lk2 ctzg4kzhg82 kznm5mvwftta0xe 7zn11onpwjq ydywduu7x73 kjefb9nwtzw k0vyrorb2ewoun 31hm7e68fe7ye b8lw21gq1k2 qv0lt4bjsikk j57rnspey0ni9ul rnhrjal1gj7 2x5mqqq42p pun6l3d2d4w uewrge0zj9r70ul vae078gymuv zb1f8tprfgnf 1xmdmyfree vl2c6xw5hfn 3cql10h18iu2k 5v16z1dcuq8z7 7dhc1umen5 21agyg1dhea nq9phu91z9 kgjrrsczgc71 ifj36a8i5c60nbk 6ywst5o3d0 8ow5gt4qai9m | 2020-07-05T09:12:11 | {
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https://proofwiki.org/wiki/Intersection_is_Largest_Subset | # Intersection is Largest Subset
## Theorem
Let $T_1$ and $T_2$ be sets.
Then $T_1 \cap T_2$ is the largest set contained in both $T_1$ and $T_2$.
That is:
$S \subseteq T_1 \land S \subseteq T_2 \iff S \subseteq T_1 \cap T_2$
### General Result
Let $T$ be a set.
Let $\mathcal P \left({T}\right)$ be the power set of $T$.
Let $\mathbb T$ be a subset of $\mathcal P \left({T}\right)$.
Then:
$\left({\forall X \in \mathbb T: S \subseteq X}\right) \iff S \subseteq \bigcap \mathbb T$
### Family of Sets
In the context of a family of sets, the result can be presented as follows:
Let $\family {S_i}_{i \mathop \in I}$ be a family of sets indexed by $I$.
Then for all sets $X$:
$\displaystyle \paren {\forall i \in I: X \subseteq S_i} \iff X \subseteq \bigcap_{i \mathop \in I} S_i$
where $\displaystyle \bigcap_{i \mathop \in I} S_i$ is the intersection of $\family {S_i}$.
## Proof
Let $S \subseteq T_1 \land S \subseteq T_2$.
Then:
$\displaystyle x \in S$ $\leadsto$ $\displaystyle x \in T_1 \land x \in T_2$ Definition of Subset $\displaystyle$ $\leadsto$ $\displaystyle x \in T_1 \cap T_2$ Definition of Set Intersection $\displaystyle$ $\leadsto$ $\displaystyle S \subseteq T_1 \cap T_2$ Definition of Subset
Alternatively:
$\displaystyle S$ $\subseteq$ $\displaystyle T_2$ $\displaystyle \leadsto \ \$ $\displaystyle S$ $=$ $\displaystyle S \cap T_2$ Intersection with Subset is Subset $\displaystyle$ $\subseteq$ $\displaystyle T_1 \cap T_2$ Set Intersection Preserves Subsets
So:
$S \subseteq T_1 \land S \subseteq T_2 \implies S \subseteq T_1 \cap T_2$.
Now let $S \subseteq T_1 \cap T_2$.
From Intersection is Subset we have $T_1 \cap T_2 \subseteq T_1$ and $T_1 \cap T_2\subseteq T_2$.
From Subset Relation is Transitive, it follows directly that $S \subseteq T_1$ and $S \subseteq T_2$.
So $S \subseteq T_1 \cap T_2 \implies S \subseteq T_1 \land S \subseteq T_2$.
From the above, we have:
$S \subseteq T_1 \land S \subseteq T_2 \implies S \subseteq T_1 \cap T_2$
$S \subseteq T_1 \cap T_2 \implies S \subseteq T_1 \land S \subseteq T_2$
Thus $S \subseteq T_1 \land S \subseteq T_2 \iff S \subseteq T_1 \cap T_2$ from the definition of equivalence.
$\blacksquare$ | 2019-07-21T19:32:15 | {
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https://math.hecker.org/2011/07/24/linear-algebra-and-its-applications-review-exercise-1-12/ | ## Linear Algebra and Its Applications, Review Exercise 1.12
Review exercise 1.12. State whether the following are true or false. If a statement is true explain why it is true. If a statement is false provide a counter-example.
(a) If $A$ is invertible and $B$ has the same rows as $A$ but in reverse order, then $B$ is invertible as well.
(b) If $A$ and $B$ are both symmetric matrices then their product $AB$ is also a symmetric matrix.
(c) If $A$ and $B$ are both invertible then their product $BA$ is also invertible.
(d) If $A$ is a nonsingular matrix then it can be factored into the product $A = LU$ of a lower triangular and upper triangular matrix.
Answer: (a) True. If $B$ has the same rows as $A$ but in reverse order then we have $B = PA$ where $P$ is the permutation matrix that reverses the order of rows. For example, for the 3 by 3 case we have
$P = \begin{bmatrix} 0&0&1 \\ 0&1&0 \\ 1&0&0 \end{bmatrix}$
If we apply $P$ twice then it restores the order of the rows back to the original order; in other words $P^2 = I$ so that $P^{-1} = P$.
If $A$ is invertible then $A^{-1}$ exists. Consider the product $A^{-1}P$. We have
$B(A^{-1}P) = (PA)(A^{-1}P) = P(A^{-1}A)P = PIP = P^2 = I$
so that $A^{-1}P$ is a right inverse for $B$. We also have
$(A^{-1}P)B = (A^{-1}P)(PA) = A^{-1}P^2A = A^{-1}IA = A^{-1}A = I$
so that $A^{-1}P$ is a left inverse for $B$ as well. Since $A^{-1}P$ is both a left and right inverse for $B$ we have $B^{-1} = A^{-1}P$ so that $B$ is invertible if $A$ is.
Incidentally, note that while multiplying by $P$ on the left reverses the order of the rows, multiplying by $P$ on the right reverse the order of the columns. For example, in the 3 by 3 case we have
$\begin{bmatrix} 1&2&3 \\ 4&5&6 \\ 7&8&9 \end{bmatrix} \begin{bmatrix} 0&0&1 \\ 0&1&0 \\ 1&0&0 \end{bmatrix} = \begin{bmatrix} 3&2&1 \\ 6&5&4 \\ 9&8&7 \end{bmatrix}$
Thus if $A^{-1}$ exists and $B = PA$ then $B^{-1} = A^{-1}P$ exists and consists of $A^{-1}$ with its columns reversed.
(b) False. The product of two symmetric matrices is not necessarily itself a symmetric matrix, as shown by the following counterexample:
$\begin{bmatrix} 2&3 \\ 3&1 \end{bmatrix} \begin{bmatrix} 3&5 \\ 5&1 \end{bmatrix} = \begin{bmatrix} 21&13 \\ 14&16 \end{bmatrix}$
(c) True. Suppose that both $A$ and $B$ are invertible; then both $A^{-1}$ and $B^{-1}$ exist. Consider the product matrices $BA$ and $A^{-1}B^{-1}$. We have
$(BA)(A^{-1}B{-1}) = B(AA^{-1})B{-1} = BIB^{-1} = BB^{-1} = I$
and also
$(A^{-1}B{-1})(BA) = A^{-1}(B{-1}B)A = A^{-1}IA = A^{-1}A = I$
So $A^{-1}B{-1}$ is both a left and right inverse for $BA$ and thus $(BA)^{-1} = A^{-1}B{-1}$. If both $A$ and $B$ are invertible then their product $BA$ is also.
(d) False. A matrix $A$ cannot necessarily be factored into the form $A = LU$ because you may need to do row exchanges in order for elimination to succeed. Consider the following counterexample:
$A = \begin{bmatrix} 0&1&2 \\ 1&1&1 \\ 1&2&1 \end{bmatrix}$
This matrix requires exchanging the first and second rows before elimination can commence. We can do this by multiplying by an appropriate permutation matrix:
$PA = \begin{bmatrix} 0&1&0 \\ 1&0&0 \\ 0&0&1 \end{bmatrix} \begin{bmatrix} 0&1&2 \\ 1&1&1 \\ 1&2&1 \end{bmatrix} = \begin{bmatrix} 1&1&1 \\ 0&1&2 \\ 1&2&1 \end{bmatrix}$
We then multiply the (new) first row by 1 and subtract it from the third row (i.e., the multiplier $l_{31} = 1$):
$\begin{bmatrix} 1&1&1 \\ 0&1&2 \\ 1&2&1 \end{bmatrix} \rightarrow \begin{bmatrix} 1&1&1 \\ 0&1&2 \\ 0&1&0 \end{bmatrix}$
and then multiply the second row by 1 and subtract it from the third ($l_{32} = 1$):
$\begin{bmatrix} 1&1&1 \\ 0&1&2 \\ 0&1&0 \end{bmatrix} \rightarrow \begin{bmatrix} 1&1&1 \\ 0&1&2 \\ 0&0&-2 \end{bmatrix}$
We then have
$L = \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 1&1&1 \end{bmatrix} \quad U = \begin{bmatrix} 1&1&1 \\ 0&1&2 \\ 0&0&-2 \end{bmatrix}$
and
$LU = \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 1&1&1 \end{bmatrix} \begin{bmatrix} 1&1&1 \\ 0&1&2 \\ 0&0&-2 \end{bmatrix} = \begin{bmatrix} 1&1&1 \\ 0&1&2 \\ 1&2&1 \end{bmatrix} = PA \ne A$
So a matrix $A$ cannot always be factored into the form $A = LU$.
NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.
If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.
This entry was posted in linear algebra. Bookmark the permalink. | 2018-09-23T12:40:20 | {
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http://mathematica.stackexchange.com/questions/27169/plot3d-constrained-to-a-non-rectangular-region | # Plot3D constrained to a non-rectangular region
I would like to make a nice 3D graphic of a parabolic bowl, with a cylindrical rim. If I do the following:
Plot3D[x^2 + y^2, {x, -3, 3}, {y, -3, 3}]
I get a paraboloid, but the box is rectangular, so the edges come to points. I want a cylindrical bounding box. The best I've come up with is this:
Plot3D[Piecewise[{{x^2 + y^2, x^2 + y^2 < 1}}], {x, -1, 1}, {y, -1, 1}]
This creates a bowl, but there is also a "floor" to the graphic that I would like to get rid of. I might be able to play games with the coloring, but that seems like a poor hack. Does anyone have any suggestions?
More generally, is it possible to create a 3D bounding box of arbitrary shape?
-
Plot3D[x^2 + y^2, {x, -3, 3}, {y, -3, 3},
RegionFunction -> Function[{x, y, z}, x^2 + y^2 < 9]]
Re: " is it possible to create a 3D bounding box of arbitrary shape?" ... as arbitrary as your creativity for creating region functions is
-
Thank you so much! That is exactly the option I had been unable to find. – jmizrahi Jun 18 '13 at 2:27
Alternatively, why not parametrize? ParametricPlot3D[{r Cos[t], r Sin[t], r^2}, {r, 0, 3}, {t, -π, π}] – Ｊ. Ｍ. Jun 18 '13 at 3:28
@0x4A4D Well, the title refers to Plot3D[] ... BTW: why the hex conversion? – belisarius Jun 18 '13 at 3:33
Well ,my message was more for the OP than you, but anyway... I got hexed, and now you're seeing the effects. – Ｊ. Ｍ. Jun 18 '13 at 3:57
– belisarius Jun 18 '13 at 11:55
The shorter the better :) :
RevolutionPlot3D[t^2, {t, 0, 3}]
It is good to know RevolutionPlot3D in case of some axisimmetric figures but the true control is given by RegionFuntion introduced by belisarius.
- | 2014-10-23T01:02:14 | {
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https://brilliant.org/discussions/thread/series-v/ | ×
# Series
Is $$\displaystyle \sum_{n=1}^\infty\frac{1}{n}$$ possible? Its a harmonic progression and so there is no direct formula for its sum. I have seen some sites using zeta function to find its sum but zeta function works only if the exponent of n is greater than 1,right?
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4 years, 9 months ago
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The harmonic series is divergent and has been proven to be so for at least several hundred years. Proving this should be easy for any competent mathematics student.
One of my favorite proofs uses a simple algebraic inequality, which is valid for $$n > 1$$: $\frac{1}{n-1} + \frac{1}{n+1} = \frac{2n}{n^2-1} > \frac{2n}{n^2} = \frac{2}{n}.$ Consequently, $\frac{1}{n-1} + \frac{1}{n} + \frac{1}{n+1} > \frac{3}{n},$ for all $$n > 1$$. Now suppose $H_m = \sum_{n=1}^m \frac{1}{n},$ for $$m \ge 1$$. Then using the inequality we just proved, \begin{align*} H_{3m+1} &= \sum_{n=1}^{3m+1} \frac{1}{n} \\ &= 1 + \sum_{n=1}^m \left( \frac{1}{3n-1} + \frac{1}{3n} + \frac{1}{3n+1} \right) \\ &> 1 + \sum_{n=1}^m \frac{1}{n} \\ &= 1 + H_m. \end{align*} In particular, this implies $$H_4 > 2$$, $$H_{13} > 3$$, $$H_{40} > 4$$, and in general, $H_{(3^k-1)/2} \ge k$ for all positive integers $$k$$. This directly proves $$\displaystyle\lim_{m \to \infty} H_m$$ is unbounded, because for any $$k > 1$$, we can find $$m(k) = (3^k-1)/2$$ such that $$H_{m(k)} > k$$. Therefore the given infinite series is divergent.
What I like about this proof is that it furnishes a surprisingly tight lower bound on the partial sums without relying on calculus; namely, $$H_m > \log_3 (1+2m)$$. No doubt, this is much worse than the bound $$H_m > \gamma + \log m$$, but a proof of the latter involves significantly heavier mathematical machinery than what we have used here. There exist numerous other similar and/or simpler proofs (e.g. Oresme's well-known groupings by successive powers of 2), but this one is a bit more obscure and so is more interesting to me.
The zeta function has a pole of order 1 at $$s = 1$$, and whose residue is 1. This is the unique singularity of $$\zeta(s)$$ and it is not removable. There's no purpose in trying to use the zeta function to evaluate the harmonic series, because the divergence of the series is precisely why the zeta function behaves the way it does.
- 4 years, 9 months ago
Thanks!!!
- 4 years, 9 months ago
There is another well known proof that the harmonic series is divergent: we do this by chopping increasingly large, but always finite 'chunks' with sum at least 1/2 as follows:
C_0 = {1}
C_1 = {2}
C_2 = {3,4}
C_3 = {5,6,7,8}
...
C_i = {2^(i-1), ... 2^(i)}
Observe that the sum of reciprocals in every set C_i is at least 1/2; and for any bound B, we can take the first 2B sets; the sum of reciprocals will then exceed this bound B. This bound is logarithmic, but probably inferior to Hero's
Edit: Hero has mentioned this proof in his response (Oresme); the bound is indeed inferior. Using his definition of H_m, we have:
Hm > (1/2) log2 (m) = log_4 (m)
log4 (m) < log3(m) < log_3 (1 + 2m)
- 4 years, 9 months ago
Here's yet another proof that $$\sum_{n\ge 1} \frac 1 n$$ is divergent, using integration.
Imagine rectangles $$[1, 2]\times [0, 1], [2, 3] \times[0, \frac 1 2], \ldots [n-1, n]\times [0, \frac 1 {n-1}]$$. These rectangles totally cover the graph of $$y = \frac 1 x$$ from x=1 to x=n. Hence the area, $$1 + \frac 1 2 + \frac 1 3 + \ldots + \frac 1{n-1}$$ is greater than $$\int_1^n \frac 1 x\cdot dx = \log n$$ which diverges.
- 4 years, 9 months ago | 2018-03-21T18:09:54 | {
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https://brilliant.org/discussions/thread/brilliant-integration-contest-season-1-part-2/ | # Brilliant Integration Contest - Season 1 (Part 2)
This is Brilliant Integration Contest - Season 1 (Part 2) as a continuation of the previous contest (Part 1). There is a major change in the rules of contest, so please read all of them carefully before take part in this contest.
I am interested in holding an Integration Contest here on Brilliant.org like any other online forums such as AoPS or Integrals and Series. The aims of the Integration Contest are to improve skills in the computation of integrals, to learn from each other as much as possible, and of course to have fun. Anyone here may participate in this contest.
The rules are as follows
1. I will start by posting the first problem. If there is a user solves it, then (s)he must post a new one.
2. You may only post a solution of the problem below the thread of problem and post your proposed problem in a new thread. Put them separately.
3. Please make a substantial comment.
4. Make sure you know how to solve your own problem before posting it in case there is no one can answer it within a week, then you must post the solution and you have a right to post another problem.
5. If the one who solves the last problem does not post his/her own problem after solving it within a day, then the one who has a right to post a problem is the last solver before him/her.
6. The scope of questions is only computation of integrals either definite or indefinite integrals.
7. You are NOT allowed to post a multiple integrals problem as well as a complex integral problem.
8. You are also NOT allowed to post a solution using a contour integration or residue method.
9. The final answer can ONLY contain the following special functions: gamma function, beta function, Riemann zeta function, Dirichlet eta function, dilogarithm, digamma function, and trigonometric integral.
Format your post is as follows:
SOLUTION OF PROBLEM xxx (number of problem) :
PROBLEM xxx (number of problem) :
Remember, put them separately.
Please share this note so that lots of users here know this contest and take part in it. (>‿◠)✌
Okay, let the contest part 2 begin!
P.S. You may also want to see Brilliant Integration Contest - Season 1 (Part 3).
Note by Anastasiya Romanova
4 years, 11 months ago
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Thanks for doing this. There is a lot to learn from these integration questions that you have shared.
Staff - 4 years, 11 months ago
Thank you for your help. You're too kind to me. I really appreciate it $\quad$ $\ddot\smile$
- 4 years, 11 months ago
Problem 20
$\displaystyle \int_0^1 \frac{\sinh ^{-1}(x)-\log \left[\left(\sqrt{2}-1\right) \sqrt{x}+1\right]}{x} \, dx = \log (2) \log \left(1+\sqrt{2}\right)-\frac{\pi ^2}{24}$
- 4 years, 11 months ago
Solving this problem is really tedious job. Honestly, I'm unwilling to answer it too (even if I know how to solve it). I don't know what is the OP's motivation by posting two well-defined integrals in a single problem. If the OP has an elementary and a clever method than mine, please do share to us. Okay, here is an attempt using a cannon.
Let split the integral into two parts
$I-J=\int_0^1\frac{{\rm{arcsinh}\,} x}{x}\,dx-\int_0^1\frac{\ln\left(1+\left(\sqrt{2}-1\right)\sqrt{x}\right)}{x}\,dx$
Perform integration by parts for $I$ by taking $u={\rm{arcsinh}\,} x$ and $dv=\dfrac{dx}{x}$.
\begin{aligned} \int_0^1\frac{{\rm{arcsinh}\,} x}{x}\,dx&={\rm{arcsinh}\,} x\,\ln x\bigg|_0^1-\int_0^1\frac{\ln x}{\sqrt{1+x^2}}\,dx\qquad\Rightarrow\qquad x=\tan t\\ I&=\int_0^{\pi/4}\frac{\ln(\cos t)-\ln (\sin t)}{\cos t}\,dt\\ &=\int_0^{\pi/4}\frac{\ln(\cos t)}{\cos t}\,dt-\int_0^{\pi/4}\frac{\ln(\sin t)}{\cos t}\,dt\\ &=\frac{1}{2}\int_0^{\pi/4}\frac{\ln(1-\sin^2 t)}{\cos t}\,dt-\int_0^{\pi/4}\frac{\ln(\sin t)}{\cos t}\,dt\\ \end{aligned}
Putting $y=\sin t$ and $a=\dfrac{1}{\sqrt{2}}$, we get
\begin{aligned} \int_0^1\frac{{\rm{arcsinh}\,} x}{x}\,dx&=\int_0^{a}\frac{\ln (1-y^2)}{1-y^2}\,dy-\int_0^{a}\frac{\ln y}{1-y^2}\,dy\\ &=\int_0^{a}\frac{\ln (1-y)+\ln(1+y)}{(1-y)(1+y)}\,dy-\int_0^{a}\frac{\ln y}{(1-y)(1+y)}\,dy\\ &=I_1+I_2 \end{aligned}
Performing partial fractions decomposition we get
\begin{aligned} I_1&=\frac{1}{2}\int_0^{a}\frac{\ln (1+y)}{1+y}\,dy+\frac{1}{2}\int_0^{a}\frac{\ln (1-y)}{1+y}\,dy+\frac{1}{2}\int_0^{a}\frac{\ln (1+y)}{1-y}\,dy+\frac{1}{2}\int_0^{a}\frac{\ln (1-y)}{1-y}\,dy\\ &=\frac{\ln^2 (1+a)}{4}+\frac{1}{2}\int_0^{a}\frac{\ln (1-y)}{1+y}\,dy+\frac{1}{2}\int_0^{a}\frac{\ln (1+y)}{1-y}\,dy-\frac{\ln^2 (1-a)}{4}\\ &=\frac{1}{4}\ln^2\left(\frac{1+a}{1-a}\right)+\frac{1}{2}\int_0^{a}\frac{\ln (1-y)}{1+y}\,dy+\frac{1}{2}\int_0^{a}\frac{\ln (1+y)}{1-y}\,dy\qquad\Rightarrow\qquad 2z=1+y\\ &=\ln^2\left(1+\sqrt{2}\right)+\frac{1}{2}\int_c^{b}\frac{\ln (2-2z)}{z}\,dz+\frac{1}{2}\int_c^{b}\frac{\ln z}{1-z}\,dz\\ &=\ln^2\left(1+\sqrt{2}\right)+\frac{\ln2}{2}\int_c^{b}\frac{dz}{z}+\frac{1}{2}\int_c^{b}\frac{\ln (1-z)}{z}\,dz+\frac{1}{2}\int_{1-c}^{1-b}\frac{\ln (1-z)}{z}\,dz\\ &=\ln^2\left(1+\sqrt{2}\right)+\frac{\ln2}{2}\ln\left(\frac{1+\sqrt{2}}{\sqrt{2}}\right)+\frac{1}{2}\int_c^{b}\frac{\ln (1-z)}{z}\,dz+\frac{1}{2}\int_{1-c}^{1-b}\frac{\ln (1-z)}{z}\,dz\\ \end{aligned}
and
\begin{aligned} I_2&=\frac{1}{2}\int_0^{a}\frac{\ln y}{1-y}\,dy+\frac{1}{2}\int_0^{a}\frac{\ln y}{1+y}\,dy\\ &=\frac{1}{2}\int_1^{1-a}\frac{\ln (1-y)}{y}\,dy+\frac{\ln y\ln(1+y)}{2}\bigg|_0^{a}-\int_0^{a}\frac{\ln (1+y)}{y}\,dy\\ &=\frac{1}{2}\int_1^{1-a}\frac{\ln (1-y)}{y}\,dy+\frac{\ln a\ln(1+a)}{2}-\int_0^{a}\frac{\ln (1+y)}{y}\,dy\\ &=\frac{1}{2}\int_1^{1-a}\frac{\ln (1-y)}{y}\,dy-\int_0^{a}\frac{\ln (1+y)}{y}\,dy-\frac{\ln2}{4}\ln\left(\frac{1+\sqrt{2}}{\sqrt{2}}\right)\\ \end{aligned}
where $b=\dfrac{1+\sqrt{2}}{2\sqrt{2}}$ and $c=\dfrac{1}{2}$.
Now, let us evaluate $J$. Set $x=t^2$, we get
$J=\int_0^1\frac{\ln\left(1+\left(\sqrt{2}-1\right)\sqrt{x}\right)}{x}\,dx=2\int_0^1\frac{\ln\left(1+\left(\sqrt{2}-1\right)t\right)}{t}\,dt$
Here is the cannon, recall a special function dilogarithm.
$\operatorname{Li}_2(z)=-\int_0^z\frac{\ln(1 - t)}{t}\,dt$
Hence, the rest integrals can be easily evaluated by using dilogarithm and an elementary substitution, i.e. $t=kx$, where $k$ is a constant. We may also utilize these identities \begin{aligned} \operatorname{Li}_2(z)+\operatorname{Li}_2(-z)&=\frac{1}{2}\operatorname{Li}_2(z^2)\\ \operatorname{Li}_2(1-z)+\operatorname{Li}_2\left(1-\frac{1}{z}\right)&=-\frac{\ln^2z}{2}\\ \operatorname{Li}_2(z)+\operatorname{Li}_2(1-z)&=\frac{{\pi}^2}{6}-\ln z \cdot\ln(1-z) \end{aligned} and special values \begin{aligned} \operatorname{Li}_2(-1)&=-\frac{{\pi}^2}{12}\\ \operatorname{Li}_2(0)&=0\\ \operatorname{Li}_2\left(\frac{1}{2}\right)&=\frac{{\pi}^2}{12}-\frac{\ln^2 2}{2}\\ \operatorname{Li}_2(1)&=\frac{{\pi}^2}{6} \end{aligned} Performing a cumbersome and a tedious calculation, we will get the announced result.
I hope you understand my feelings while trying to solve and to write it down. So, please do not ever post a problem like this again. LOL
- 4 years, 11 months ago
The reason for taking the difference of the two integrals is that the result is much simpler than the two integrals separately (which involve dilogarithms), which I found beautiful. It only works if both terms are exactly as they are (including the weird factor $(\sqrt 2 -1)$ ).
Your solution can be significantly simplified. Both terms in the integral have a relatively simple antiderivative in terms of dilogarithms. After plugging in the limits, it then boils down to showing
$\displaystyle \operatorname{Li}_2 (\sqrt{2} - 1) - \operatorname{Li}_2 (1 - \sqrt 2) = \frac{\pi^ 2}{8} - \frac 1 2 \log^ 2(\sqrt 2 - 1)$
using the dilogarithm identities you posted.
To find the antiderivative of the second term, just substitute $(\sqrt 2 -1) \sqrt x = u$. For the first term, substitute $u = \left( x + \sqrt{1+x^2} \right)^2$. This will reduce the integral to
$\displaystyle \int \frac{\ln u \,du}{u^2 - 1},$
which I am sure you can calculate in a few lines.
- 4 years, 11 months ago
I'm sorry, I'm a bit dizzy right now so I can't follow your comment. Could you elaborate? If I may ask, could you post your solution of this problem? Thanks.
Edit : Aha! I get it. Use this relation: ${\rm{arcsinh}\,}x=\ln\left(x+\sqrt{x^2+1}\right)$. Very clever!
- 4 years, 11 months ago
Here's a summary of what I said, maybe it helps:
The term with $\operatorname{arcsinh} x / x$ can be evaluated by substituting $u = \left( x + \sqrt{1+x^2} \right)^2$. The result is something with a dilogarithm. The other term also gives a dilogarithm, but the dilogarithm terms cancel precisely, leading to an elementary result. I hope you can appreciate the beauty of the problem :)
- 4 years, 11 months ago
PROBLEM 16 :
Prove
$\large\int_0^{\Large\frac\pi2}\cos^{v-1}x\cos ax\ dx=\frac{\pi}{2^v v\ \operatorname{B}\left(\frac{v+a+1}{2},\frac{v-a+1}{2}\right)}$
where $\operatorname{B}\left(x,y\right)$ is the beta function.
## PS : POST YOUR SOLUTION BELOW EACH PROBLEM THREAD AND POST YOUR PROPOSED PROBLEM AS A NEW THREAD. PUT THEM IN SEPARATED THREAD. SO THAT THE POSTS LOOK MORE ORGANIZED. THANKS.
- 4 years, 11 months ago
Isn't this problem too difficult to high school students Anna? I decide to answer Problem 16 because I'm afraid if this continues till a week, this contest will lose its interest. IMHO, you should propose an easy problem so that this contest will be fun as the stated aims of it. So, here is a solution:
SOLUTION OF PROBLEM 16 :
Rewrite the integral as follows \begin{aligned} \int_0^{\Large\frac\pi2}\cos^{v-1}x\cos ax\ dx&=\frac12\int_{-\large\frac\pi2}^{\large\frac\pi2}\left(\frac{e^{ix}+e^{-ix}}{2}\right)^{v-1}\cos ax\ dx\\ &=\frac1{2^{v}}\int_{-\large\frac\pi2}^{\large\frac\pi2}\left(1+e^{2ix}\right)^{v-1}e^{-i(v-1)x}\cos ax\ dx\\ &=\frac1{2^{v}}\int_{-\large\frac\pi2}^{\large\frac\pi2}\sum_{n=0}^{v-1}\binom{v-1}{n} e^{2inx}\cdot e^{-i(v-1)x}\cos ax\ dx\\ &=\frac1{2^{v}}\sum_{n=0}^{v-1}\binom{v-1}{n} \int_{-\large\frac\pi2}^{\large\frac\pi2} e^{i(2n-v+1)x}\cos ax\ dx.\qquad\qquad\qquad\tag1 \end{aligned} Consider $f(x)=\left\{ \begin{array}{l l} e^{i\omega x} & \quad \text{for}\ -\frac\pi2 The Fourier transform of $f(x)$ is \begin{aligned} \mathscr{F}\left[f(x)\right]&=\int_{-\infty}^\infty f(x)\ e^{-i\alpha x}\ dx\\ \int_{-\large\frac\pi2}^{\large\frac\pi2} e^{i\omega x}\cos \alpha x\ dx-i\int_{-\large\frac\pi2}^{\large\frac\pi2}e^{i\omega x}\sin \alpha x\ dx&=\int_{-\large\frac\pi2}^{\large\frac\pi2} e^{i\omega x}\ e^{-i\alpha x}\ dx\\ &=\int_{-\large\frac\pi2}^{\large\frac\pi2} e^{i(\omega-\alpha) x}\ dx\\ &=\left[\frac{e^{i(\omega-\alpha) x}}{i(\omega-\alpha)}\right]_{x=-\large\frac\pi2}^{\large\frac\pi2}\\ \Re\bigg(\mathscr{F}\left[f(x)\right]\bigg)&=\left[\frac{\sin(\omega-\alpha) x}{\omega-\alpha}\right]_{x=-\large\frac\pi2}^{\large\frac\pi2}\\ \int_{\large-\frac\pi2}^{\large\frac\pi2}\cos \alpha x\ dx&=\frac{2\sin(\omega-\alpha) \frac\pi2}{\omega-\alpha}.\qquad\qquad\qquad\tag2 \end{aligned} Using $(2)$, then $(1)$ turns out to be \begin{aligned} \int_0^{\Large\frac\pi2}\cos^{v-1}x\cos ax\ dx&=\frac1{2^{v-1}}\sum_{n=0}^{v-1}\binom{v-1}{n} \frac{\sin(2n-v+1-a) \frac\pi2}{2n-v+1-a}\\ &=\frac1{2^{v}}\sum_{n=0}^{v-1}\binom{v-1}{n} \frac{\sin\left(n-\frac{v-1+a}2\right) \pi}{n-\frac{v-1+a}2}.\qquad\qquad\qquad\tag3\\ \end{aligned} Now, let us express $\dbinom{y}{z}$ in term of beta function that can be related to $(3)$. \begin{aligned} \binom{y}{z}&=\frac{y!}{z!(y-z)!}\\ &=\frac{y!}{\Gamma(1+z)\Gamma(1+y-z)}\\ &=\frac{y!}{z\Gamma(z)\Gamma(1-z)(y-z)\cdots(1-z)}\\ &=\frac{\sin(\pi z)}{\pi z}\cdot\frac{y!}{(y-z)\cdots(1-z)}\\ &=\frac{\sin(\pi z)}{\pi z}\sum_{n=0}^{y}\binom{y}{n}(-1)^n\frac{n}{z-n}\\ &=\sum_{n=0}^{y}\binom{y}{n}\frac{\sin\pi(z-n)}{\pi(z-n)}.\qquad\qquad\qquad\tag4 \end{aligned} Using $(4)$, then $(3)$ turns out to be \begin{aligned} \int_0^{\Large\frac\pi2}\cos^{v-1}x\cos ax\ dx&=\frac{\pi}{2^{v}}\binom{v-1}{\frac{v-1+a}2}\\ &=\frac{\pi}{2^{v}}\frac{\Gamma(v)}{\Gamma\left(\frac{v+a+1}2\right)\Gamma\left(\frac{v-a+1}2\right)}\\ &=\frac{\pi}{2^{v}\ v\ \operatorname{B}\left(\frac{v+a+1}2,\frac{v-a+1}2\right)}\qquad\qquad\qquad\blacksquare \end{aligned}
- 4 years, 11 months ago
OK, fine. I'll post high school integral problems from now. -_-"
- 4 years, 11 months ago
No Just keep posting those hard integrals, it's challenging but we learn a lot from it
- 4 years, 11 months ago
Your solution is valid only if $v$ is an integer, whereas the identity holds in general also.
- 2 years, 11 months ago
PROBLEM 18
Show that
$\displaystyle \int_0^{\pi/4} \tan^{1/3} x dx = \frac{1}{6} \left( \pi \sqrt{3} -3\log 2\right)$
My bad, it should be $3 \log 2$ indeed. Kinshuk's result is correct. Sorry for the confusion.
- 4 years, 11 months ago
There should be 3log(2) instead of 2log(2)
- 4 years, 11 months ago
$Solution\quad of\quad problem\quad 18:\\ I\quad =\quad \int _{ 0 }^{ \frac { \pi }{ 4 } }{ { tan }^{ \frac { 1 }{ 3 } }x.dx } \\ put\quad { tan }^{ \frac { 1 }{ 3 } }x=t\\ dx\quad =\quad \frac { { 3t }^{ 2 } }{ 1+{ t }^{ 6 } } .dt\\ I\quad =\quad \int { \frac { { 3t }^{ 3 } }{ 1+{ t }^{ 6 } } .dt } \\ put\quad { t }^{ 2 }=u\\ 2t.dt=du\\ I\quad =\quad \frac { 3 }{ 2 } \int { \frac { u.du }{ 1+{ u }^{ 3 } } } \\ using\quad partial\quad fraction\quad ,\quad our\quad integration\\ turns\quad out\quad to\quad be\quad :\\ I\quad =\quad \frac { -1 }{ 2 } \int { \frac { du }{ 1+u } } \quad +\quad \frac { 1 }{ 2 } \int { \frac { (1+u)du }{ { u }^{ 2 }-u+1 } } \\ after\quad solving\quad and\quad applying\quad limits:\\ I\quad =\quad \int _{ 0 }^{ \frac { \pi }{ 4 } }{ { tan }^{ \frac { 1 }{ 3 } }x.dx } \quad =\quad \frac { 1 }{ 6 } (\pi \sqrt { 3 } -3\log { 2 } )$
- 4 years, 11 months ago
PROBLEM 21 :
Show that $\int_0^\infty \frac{dx}{x^4+2\cos(2\theta)\,x^2+1}=\frac{\pi}{4\cos\theta}$
- 4 years, 11 months ago
We have
$\displaystyle \int_0^ {\infty} \frac{dx}{x^4 + a x^2 + b^2} = \frac{\pi}{2b \sqrt{2b+a}},$
which I proved on MSE. Plugging in $a = 2 \cos2 \theta$ and using that $2 \cos^2(\theta) = 1 + \cos 2\theta$ immediately gives the answer.
- 4 years, 11 months ago
Oh I know you're now. You're user111187. I thought you're an old man. Haha
Nice to meet you here Ruben. It seems you'll be a tough opponent because you're a Math SE and I&S user. $\ddot\smile$
- 4 years, 11 months ago
Yep, this will be good :)
- 4 years, 11 months ago
Expecting a question from you @Ruben Doornenbal
- 4 years, 11 months ago
$Problem\quad 23$
Find $\displaystyle \int _{ 0 }^{ \frac { \pi }{ 4 } }{ ln(tan(x))dx }$
- 4 years, 11 months ago
Solution 23
A well-known problem. Sub $\tan x = u$ to get
$\displaystyle \int_0^1 du \frac{du \ln u}{1+u^2} = \sum_{k \geq 0} (-1)^k \int_0^1 du \ln u \, u^{2k} = \sum_{k \geq 0} (-1)^k \frac{1}{(2k+1)^2} = G.$
The penultimate equality follows from integration by parts.
- 4 years, 11 months ago
@Ruben Doornenbal Can we have problem 24?
- 4 years, 11 months ago
Sir can you elaborate I did'nt understood this one @Ruben Doornenbal
- 4 years, 10 months ago
@U Z The idea is to expand the factor $\displaystyle \frac{1}{1+u^2}$ in a geometric series and interchange summation and integration. The last equality is just the definition of Catalan's constant.
- 4 years, 10 months ago
@Ronak Agarwal @Ruben Doornenbal The answer given is wrong!!! It should be -G!!! You must have forgotten the negative sign.......
- 1 year, 1 month ago
$Problem\quad 30$
Find $\displaystyle \int _{ 1 }^{ \infty }{ \frac { x-\left\lfloor x \right\rfloor -0.5 }{ x } dx }$
- 4 years, 11 months ago
$I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} \int_{r}^{r+1} \frac{x-r-\frac{1}{2}}{x} dx$ $I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} 1-(r+\frac{1}{2})\ln(\frac{r+1}{r})$ $2I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} 2-(2r+1)\ln(\frac{r+1}{r})$ $2I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} 2-(2r+2)\ln(\frac{r+1}{r}) + \ln(\frac{r+1}{r})$ $2I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} 2-(2r+2)\ln(r+1) +(2r+2)\ln(r)+ \ln(\frac{r+1}{r})$ $2I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} 2-2((r+1)\ln(r+1)-r\ln(r))+\ln(r+1)+\ln(r)$ $2I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} 2-2((r+1)\ln(r+1)-r\ln(r))+\ln(r+1)+\ln(r)$ $2I = \lim_{n \rightarrow \infty} 2(n-1)-2n\ln(n)+2\ln(n!)-\ln(n)$ $2I = \lim_{n \rightarrow \infty} 2(n-1)-2n\ln(n)+(2n+1)\ln(n)-2n+\ln(2\pi)-\ln(n)$ $2I = \ln(2\pi)-2$ $I =\frac{\ln(2\pi)}{2}-1$
- 4 years, 11 months ago
This one is easy too :) $\frac{\ln2\pi}{2}-1$
@Shivang Jindal : Sorry, I was kidding & I am busy right now so I have no time to write down my answer. Could you elaborate yours then you're good to go (propose your problem). Sorry for the inconvenience...
- 4 years, 11 months ago
- 4 years, 11 months ago
Trick , is to break the integral from $(1,2),(2,3)...(n-1,n)$. Then, we compute the sum in terms of $n$ . and then use Stirling approximation :) .
- 4 years, 11 months ago
Exactly, you got it perfectly right.
- 4 years, 11 months ago
PROBLEM 22
This one is particularly beautiful, in my opinion.
$\displaystyle \int_0^a \frac{x dx}{\cos x \cos(a-x)} = \frac{a}{\sin a} \ln \sec a.$
- 4 years, 11 months ago
$Solution\quad of\quad Problem\quad 22$
$I=\displaystyle \int _{ 0 }^{ a }{ \frac { xdx }{ cos(x)cos(a-x) } } =\int _{ 0 }^{ a }{ \frac { (a-x)dx }{ cos(x)cos(a-x) } }$
Adding these two forms we get :
$I=\displaystyle \frac { a }{ 2 } \int _{ 0 }^{ a }{ \frac { dx }{ cos(x)cos(a-x) } }$
Multiplying and dividing by $sin(a)$ we get :
$I=\displaystyle \frac { a }{ 2sin(a) } \int _{ 0 }^{ a }{ \frac { sin(x+(a-x))dx }{ cos(x)cos(a-x) } }$
$I=\displaystyle \frac { a }{ 2sin(a) } \int _{ 0 }^{ a }{ (tan(x)+tan(a-x))dx }$
Also since $\displaystyle \int _{ 0 }^{ a }{ tan(x)dx } =\int _{ 0 }^{ a }{ tan(a-x)dx }$
We get $I=\displaystyle \frac{a}{sin(a)}\int _{ 0 }^{ a }{ tan(x)dx }$
$I=\frac { a }{ sin(a) } ln(sec(a))$
- 4 years, 11 months ago
PROBLEM 24
$\displaystyle \int_0^1 \operatorname{arcsech} x \operatorname{arcsin} xdx =\frac{\pi^2}{8} - \ln 2.$
- 4 years, 11 months ago
$Solution\quad of\quad problem\quad 24$
First note that :
$arcsech(x)=ln(\frac{1+\sqrt{1-{x}^{2}}}{x})$
In our integral put $x=sin(\theta)$
$\displaystyle I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \theta cos\theta ln(\frac { 1+cos\theta }{ sin\theta } ) } d\theta$
Applying integration by parts we get , $u=ln(\frac{1+cos\theta }{sin\theta}),dv=\theta cos\theta d\theta$
$\displaystyle I=(\theta sin\theta +cos\theta )ln(\frac { 1+cos\theta }{ sin\theta } )\overset { \frac { \pi }{ 2 } }{ \underset { 0 }{ | } } +\int _{ 0 }^{ \frac { \pi }{ 2 } }{ (\theta sin\theta +cos\theta )cosec\theta d\theta }$
$\displaystyle I=(\theta sin\theta +cos\theta )ln(\frac { 1+cos\theta }{ sin\theta } )+\frac { { \theta }^{ 2 } }{ 2 } +ln(sin\theta )\overset { \frac { \pi }{ 2 } }{ \underset { 0 }{ | } }$
Which on evaluating we get :
$I=\frac { { \pi }^{ 2 } }{ 8 } -ln(2)$
- 4 years, 11 months ago
Can you elaborate the first line please , it will be great to learn from you
- 4 years, 10 months ago
PROBLEM 26 :
Prove
$\int_0^\infty\frac{\ln x}{\cosh x}\,dx=\frac{\pi}{2}\ln\left(\frac{\Gamma^4\left(\frac{3}{4}\right)}{\pi}\right)$
P.S. You may use any well-known expressions.
- 4 years, 11 months ago
SOLUTION 26
Consider
$\displaystyle I(a) = \int_0^\infty dx \frac{x^{a-1}}{\cosh x} = 2 \sum \limits_{k \geq 0} (-1)^k \int_0^\infty dx x^{a-1} e^{-(2k+1)x} = 2 \Gamma(a) \beta(a).$
Our integral is
$\displaystyle I'(1) = 2 \Gamma'(1)\beta(1) + 2 \Gamma(1) \beta'(1) = 2(-\gamma)(\pi/4) + 2 \frac \pi 4 \left[\gamma + 2 \ln 2 + 3 \ln \pi - 4 \ln \Gamma \frac 1 4 \right].$
Here we used a result from Mathworld. Using the Euler reflection formula,
$\displaystyle \Gamma(1/4) = \pi \sqrt 2 (\Gamma(3/4))^{-1}.$
Collecting all the terms gives $\displaystyle I'(1) = -\frac \pi 2 \ln \pi + 2\pi \ln \Gamma(3/4),$
which equals the stated result.
- 4 years, 11 months ago
PROBLEM 27
My last two integrals were clearly too easy. By finding an antiderivative or otherwise, show that
$\int_0^{\infty} dx\, \ln^2 \tanh x = \frac{7}{4}\zeta{(3)}.$
- 4 years, 11 months ago
Solution of Problem 27
Set $t=\tanh x$, we have
\begin{aligned} \int_0^\infty\ln^2(\tanh x)\,dx&=\int_0^1\frac{\ln^2t}{1-t^2}\,dt\\ &=\int_0^1\sum_{n=0}^\infty t^{2n}\ln^2t\,dt\\ &=\sum_{n=0}^\infty\int_0^1 t^{2n}\ln^2t\,dt\\ &=2\sum_{n=0}^\infty\frac{1}{(2n+1)^3}\qquad\Rightarrow\qquad\text{see solution of Problem 13}\\ &=2\left[\sum_{n=1}^\infty\frac{1}{n^3}-\sum_{n=1}^\infty\frac{1}{(2n)^3}\right]\\ &=\frac{7}{4}\sum_{n=1}^\infty\frac{1}{n^3}\\ &=\frac{7}{4}\zeta(3) \end{aligned}
- 4 years, 11 months ago
Problem 28
Prove
\begin{aligned} \int_{-\infty}^{\infty} \frac{\sinh 2x\cos 2x}{\sinh \pi x} \ dx = \frac{\sin 2}{\cos 2 + \cosh 2} \end{aligned}
- 4 years, 11 months ago
Solution 28
The integral equals
$\displaystyle I = \Re \int_{-\infty}^\infty dx \frac{\left(e^{2x} - e^{- 2x} \right) e^{\pi x} e^{2 i x}}{e^{2 \pi x} - 1}.$
Substitute $e^{\pi x} = u$. We get
$\displaystyle I = J_+ - J_-,$
where
\displaystyle \begin{aligned} J_\pm &= \Re \frac 1 \pi \int_0^\infty dx \frac{u^{2( i\pm1)/\pi}}{u^2 - 1} \\&= -\Re \frac 1 2 \cot\left[\frac \pi 2 \left(2(i\pm1)/\pi + 1\right) \right] \\&= \Re\frac 1 2 \tan(i \pm 1). \end{aligned}
Here we made use of the well-known integral
$\displaystyle PV \int_0^\infty \frac{x^{a-1}}{1-x^b} = \frac \pi b \cot \frac{\pi a}{b}.$
Now using the identity
$\displaystyle \tan\frac{A+B}{2} = \frac{\sin A + \sin B}{\cos A + \cos B}$
gives
$\displaystyle J_\pm = \pm \frac 1 2 \frac{\sin 2}{\cosh 2 + \cos 2},$
which gives the desired result.
- 4 years, 11 months ago
Wait!? For $PV\int_0^\infty \frac{x^{a-1}}{1-x^b}\,dx=\frac{\pi}{b}\cot\left(\frac{\pi a}{b}\right)$ could you prove it without using contour integration or residue method? See the rules.
- 4 years, 11 months ago
Of course, my dear. It is clear that we can take $b = 1$ in the proof. The general result follows from a substitution. Separate the integrals over $(0,1)$ and over $(1,\infty)$. Put $u = 1/x$ in the second integral. The result is
$\displaystyle \int_0^1 dx \frac{x^{a-1} - x^{-a}}{1-x} = \psi(1-a) - \psi(a) = \pi \cot \pi a,$
as was to be proven. Here we used a result derived by real methods here.
You have sharp eye for integrals that I normally derive with residues :p
- 4 years, 11 months ago | 2019-11-14T20:40:40 | {
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https://keisan.casio.com/exec/system/1244946907 | # Newton method f(x),f'(x) Calculator
## Calculates the root of the equation f(x)=0 from the given function f(x) and its derivative f'(x) using Newton method.
f(x) f'(x) initial solution x0 maximum repetition n102050100200500 6digit10digit14digit18digit22digit26digit30digit34digit38digit42digit46digit50digit
Newton method f(x),f'(x)
[1-10] /204 Disp-Num5103050100200
[1] 2022/10/16 22:47 30 years old level / High-school/ University/ Grad student / Very /
Purpose of use
Helped understand each step of Newton's method
Comment/Request
Great tool! Thank you!
[2] 2022/10/10 13:16 Under 20 years old / High-school/ University/ Grad student / Useful /
Purpose of use
[3] 2022/08/23 05:18 40 years old level / A teacher / A researcher / Very /
Purpose of use
estimating answers to difficult exponential equations.
[4] 2022/05/29 15:17 20 years old level / High-school/ University/ Grad student / Useful /
Purpose of use
[5] 2021/10/28 08:16 20 years old level / High-school/ University/ Grad student / Useful /
Purpose of use
[6] 2021/10/08 23:17 Under 20 years old / High-school/ University/ Grad student / Useful /
Purpose of use
Calc class
Comment/Request
Wish it would take cos and sin
[7] 2021/07/21 18:14 30 years old level / High-school/ University/ Grad student / Very /
Purpose of use
Working with Newton's Method for Calculus and Analytic Geometry. This calculator worked amazingly well. Thank you!
[8] 2021/07/01 17:15 40 years old level / An engineer / Useful /
Purpose of use
Verifying answers for a construction project.
Comment/Request
It would be nice to have answers come in fraction form instead of decimal. Also, it would be nice to see the work actually being done instead of all the answers just spit out. Yes, I like to verify the work being done matches my own. Not just the final answer. In case you made the same errors.
[9] 2021/04/25 20:54 Under 20 years old / High-school/ University/ Grad student / Very /
Purpose of use
Doing practice problems to study for a final exam
Comment/Request
This calculator will be better if there was an option to choose the type of answer being shown (e.g. fraction or decimal representation).
[10] 2021/04/12 19:50 Under 20 years old / High-school/ University/ Grad student / Useful /
Purpose of use
Calculus assignment
Comment/Request
wish it gave a fraction version of the answer
Sending completion
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Purpose of use? | 2022-12-06T21:02:38 | {
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"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. Yes\n2. Yes",
"lm_q1_score": 0.9719924810166349,
"lm_q2_score": 0.865224091265267,
"lm_q1q2_score": 0.8409913111042903
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https://stats.stackexchange.com/questions/241471/if-multi-collinearity-is-high-would-lasso-coefficients-shrink-to-0/241486 | # If multi-collinearity is high, would LASSO coefficients shrink to 0?
Given $x_2 = 2 x_1$, what's the theoretical behavior of LASSO coefficients and why?
Would one of $x_1$ or $x_2$ shrink to $0$ or both of them?
require(glmnet)
x1 = runif(100, 1, 2)
x2 = 2*x1
x_train = cbind(x1, x2)
y = 100*x1 + 100 + runif(1)
ridge.mod = cv.glmnet(x_train, y, alpha = 1)
coef(ridge.mod)
#3 x 1 sparse Matrix of class "dgCMatrix"
# 1
#(Intercept) 1.057426e+02
#x1 9.680073e+01
#x2 3.122502e-15
• I'm not sure if this is a good simulation because both coefficients are in fact zero. It's a bit more interesting to look at the behavior of the coefficient estimates when there's a real relationship. – dsaxton Oct 20 '16 at 19:57
• Simulation improved. I provide the simulation because I want to explain what my question is. I only interested in theoretical results of this question. – John Hass Oct 20 '16 at 20:39
• I think the behavior will be unpredictable because the model is not identifiable. That is, how can the model fitting procedure possibly know for instance that $\beta_1 = 100$ and $\beta_2 = 0$ rather than $\beta_1 = 0$ and $\beta_2 = 50$? It can't, because either is "correct." – dsaxton Oct 20 '16 at 20:51
• I agree with your reasoning. Is there a mathematical way to describe it? – John Hass Oct 20 '16 at 20:54
• I think you meant y = 100*x1 + 100 + runif(100), otherwise you get a single random number that is recycled and added uniformly to all other entries. – Firebug Sep 11 '17 at 16:53
Notice that \begin{align*} \|y-X\beta\|_2^2 + \lambda \|\beta\|_1 & = \|y - \beta_1 x_1 - \beta_2 x_2 \|_2^2 + \lambda \left( |\beta_1| + |\beta_2| \right) \\ & = \|y - (\beta_1 + 2 \beta_2) x_1 \|_2^2 + \lambda \left( |\beta_1| + |\beta_2| \right). \end{align*}
For any fixed value of the coefficient $\beta_1 + 2\beta_2$, the penalty $|\beta_1| + |\beta_2|$ is minimized when $\beta_1 = 0$. This is because the penalty on $\beta_1$ is twice as weighted! To put this in notation, $$\tilde\beta = \arg\min_{\beta \, : \, \beta_1 + 2\beta_2 = K}|\beta_1| + |\beta_2|$$ satisfies $\tilde\beta_1 = 0$ for any $K$. Therefore, the lasso estimator \begin{align*} \hat\beta & = \arg\min_{\beta \in \mathbb{R}^p} \|y - X \beta\|_2^2 + \lambda \|\beta\|_1 \\ & = \arg\min_{\beta \in \mathbb{R}^p} \|y - (\beta_1 + 2 \beta_2) x_1 \|_2^2 + \lambda \left( |\beta_1| + |\beta_2| \right) \\ & = \arg_\beta \min_{K \in \mathbb{R}} \, \min_{\beta \in \mathbb{R}^p \, : \, \beta_1 + 2 \beta_2 = K} \, \|y - K x_1 \|_2^2 + \lambda \left( |\beta_1| + |\beta_2| \right) \\ & = \arg_\beta \min_{K \in \mathbb{R}} \, \left\{ \|y - K x_1 \|_2^2 + \lambda \min_{\beta \in \mathbb{R}^p \, : \, \beta_1 + 2 \beta_2 = K} \, \left\{ \left( |\beta_1| + |\beta_2| \right) \right\} \right\} \end{align*} satisfies $\hat\beta_1 = 0$. The reason why the comments to OP's question are misleading is because there's a penalty on the model: those $(0, 50)$ and $(100,0)$ coefficients give the same error, but different $\ell_1$ norm! Further, it's not necessary to look at anything like LARs: this result follows immediately from the first principles.
As pointed out by Firebug, the reason why your simulation shows a contradictory result is that glmnet automatically scales to unit variance the features. That is, due to the use of glmnet, we're effectively in the case that $x_1 = x_2$. There, the estimator is no longer unique: $(100,0)$ and $(0,100)$ are both in the arg min. Indeed, $(a,b)$ is in the $\arg\min$ for any $a,b \geq 0$ such that $a+b = 100$.
In this case of equal features, glmnet will converge in exactly one iteration: it soft-thresholds the first coefficient, and then the second coefficient is soft-thresholded to zero.
This explains why the the simulation found $\hat\beta_2 = 0$ in particular. Indeed, the second coefficient will always be zero, regardless of the ordering of the features.
Proof: Assume WLOG that the feature $x \in \mathbb{R}^n$ satisfies $\|x\|_2 = 1$. Coordinate descent (the algorithm used by glmnet) computes for it's first iteration: $$\hat\beta_1^{(1)} = S_\lambda(x^T y)$$ followed by \begin{align*} \hat\beta_2^{(1)} & = S_\lambda \left[ x^T \left( y - x S_\lambda (x^T y) \right) \right] \\ & = S_\lambda \left[ x^T y - x^T x \left( x^T y + T \right) \right] \\ & = S_\lambda \left[ - T \right] \\ & = 0, \end{align*} where $T = \begin{cases} - \lambda & \textrm{ if } x^T y > \lambda \\ \lambda & \textrm{ if } x^T y < -\lambda \\ 0 & \textrm{ otherwise} \end{cases}$. Then, since $\hat\beta_2^{(1)}= 0$, the second iteration of coordinate descent will repeat the computations above. Inductively, we see that $\hat\beta_j^{(i)} = \hat\beta_j^{(i)}$ for all iterations $i$ and $j \in \{1,2\}$. Therefore glmnet will report $\hat\beta_1 = \hat\beta_1^{(1)}$ and $\hat\beta_2 = \hat\beta_2^{(1)}$ since the stopping criterion is immediately reached.
• glmnet has feature scaling turned on by default, I'm pretty sure. So $x_1$ and $x_2$ become the same in the model. – Firebug Sep 11 '17 at 16:22
• Try this instead: ridge.mod=cv.glmnet(x_train,y,alpha=1, standardize = FALSE); coef(ridge.mod) – Firebug Sep 11 '17 at 16:25
• That did it! Great thinking, @Firebug! Now the coefficient of $x_1$ does indeed become estimated as zero. Thank you for sharing your insight! – user795305 Sep 11 '17 at 16:31
When I re-run your code, I get that the coefficient of $x_2$ is numerically indistinguishable from zero.
To understand better why LASSO sets that coefficient to zero, you should look at the relationship between LASSO and Least Angle Regression (LAR). LASSO can be seen as a LAR with a special modification.
The algorithm of LAR is roughly like this: Start with an empty model (except for an intercept). Then add the predictor variable that is the most correlated with $y$, say $x_j$. Incrementally change that predictor's coefficient $\beta_j$, until the residual $y - c - x_j\beta_j$ is equally correlated with $x_j$ and another predictor variable $x_k$. Then change the coefficients of both $x_j$ and $x_k$ until a third predictor $x_l$ is equally correlated with the residual $y - c - x_j\beta_j -x_k\beta_k$ and so on.
LASSO can be seen as LAR with the following twist: as soon as the coefficient of a predictor in your model (an "active" predictor) hits zero, drop that predictor from the model. This is what happens when you regress $y$ on the collinear predictors: both will get added to the model at the same time and, as their coefficients are changed, their respective correlation with the residuals will change proportionately, but one of the predictors will get dropped from the active set first because it hits zero first. As for which of the two collinear predictors it will be, I don't know. [EDIT: When you reverse the order of $x_1$ and $x_2$, you can see that the coefficient of $x_1$ is set to zero. So the glmnet algorithm simply seems to set those coefficients to zero first that are ordered later in the design matrix.]
A source that explains these things more in detail is Chapter 3 in "The Elements of Statistical Learning" by Friedman, Hastie and Tibshirani. | 2020-01-26T18:01:59 | {
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http://directoriosma.com/bwazz6/dnyhi.php?id=predicate-logic-rules-8ff70d | It retains the central tenet of Propositional Logic: that sentences express propositions and propositions denote truth-conditions. Log in. But the logic software is unaware of this (just as it is unaware of your using italic, or bold, or large fonts). We say, ∀x∃yLxy\forall x \exists y L x y∀x∃yLxy, to mean that for every real number there is some real number less than the number itself. Yes, you guessed it! It can be proved about open branches that a) if all the formulas in it, except universally quantified formulas, that can be extended have been extended, b) that all the universally quantified formulas in it have been extended (ie instantiated) at least once (eg to a constant, say b, or to a closed term, say f(a)), and c) that all the universally quantified formulas in it have been extended (ie instantiated) for every constant or closed term that appear in that branch, that open branch will never close. We already use predicates routinely in programming, e.g. This time, we run up the open branch and i) any constant that appears there becomes an 'object' in the Universe, and ii) we make the extensions of the predicates in the branch exactly those require to make any atomic formulas True and negations of atomic formulas True also (ie making the atomic subformula of a negation False). Predicate logic adds two new connectives to sentence logic: the univer- sal and existential quantifiers. The sentence now means, There is a person xxx such that if xxx is a guitarist, Lemmy is a guitarist. This is a completely invalid argument in propositional logic since A, B and C have no relations to each other. Two of these rules are easy and two are hard. To prove a conclusion from a set of premises, is a transformation of the propositions using certain inference rules. So, for example, if the open branch contained {¬A,B,¬C} then the assignment we were looking for was {A=False, B=True, C=False}. The ex-ceptions to this rule are the names for binary relations in mathematics: for greater than, and so on. Let SxSxSx mean that xxx is a spy and TxyTxyTxy mean that xxx is taller than yyy. The most well-known FDA regulations are the GMP regulations. Under this Interpretation, all the initial formulas will be true (indeed, all the formulas in the branch will be true). Under it, the premises come out to be true and the conclusion false. It … From a software point of view, subscripts bring in their own problems. It is complete and open. Consider the sentence The king of France is bald. Rules for constructing Wffs A predicate name followed by a list of variables such as P (x, y), where P is a predicate name, and x and y are variables, is called an atomic formula. Let the UD be R\mathbb{R}R and let LxyLxyLxy mean xxx is less than yyy. Thus. C: &\text{ Aristotle is mortal.} Proofs are valid arguments that determine the truth values of mathematical statements. Predicates, constants, variables, logical connectives, parentheses and the quantifiers are referred to as. If that branch is complete, and does not contain a Universally quantified formula, the root formulas are satisfiable. The set of objects in the Universe of Discourse (see below) which satisfy a predicate is called the extension of the predicate. Inference Rules 3. All other descriptions are definite. The quantifiers give us the power to express propositions involving entire sets of objects, some of them, enumerate them, etc. We could extend predicate logic by talking about identity, something we are all familiar with. Predicate Logic 4. You might realize that predicates are a generalization of Relations. Email: Tree Tutorials [Propositional, Predicate, Identity, and Modal Logic TreesâHowson Syntax], Tree Tutorial 1: Propositional Trees: Introduction, Tree Tutorial 2: More Propositional Tree Rules, Tree Tutorial 3: Using Trees to Test for Satisfiability and Invalidity, Tree Tutorial 6: Functional Terms and First Order Theories, Tree Tutorial 7: Type Labels, Sorts, Order Sorted Logic ['Mixed Domains'], if the tree is closed, the root formulas are not (simultaneously) satisfiable, if a tree has a complete open branch the root formulas are (simultaneously) satisfiable. In the expression ∃xGx→Gl\exists x G x \to Gl∃xGx→Gl, the scope of the quantifier ∃\exists∃ is the expression GxGxGx. Practice math and science questions on the Brilliant Android app. Determine whether these arguments are valid (ie try to produce closed trees for them). These two equivalences, which explicate the relation between negation and quantification, are known as DeMorgan’s Laws for predicate logic. (That is what was done in the previous paragraph.) Predicates express similar kinds of propositions involving it's arguments. \\ A: &\text{ All men are mortal.} Descriptions which are not suitable for representing a constant in predicate logic are indefinite descriptions. \\ That is because ddd refers to "dogs" which is not just one particular object, but the entire set of dogs. Artificial Intelligence – Knowledge Representation, Issues, Predicate Logic, Rules This is part of the courseware on Artificial Intelligence, by R C Chakraborty, at JUET. In particular, according to this pattern, for each connective, we have a rule for introducing that connective, and a rule for elimi nating that connective. Every logician loves someone other than himself. A set of rules can be used to infer any valid conclusion if it is complete, while never inferring an invalid conclusion, if it is sound. Then, the original statement is the conjunction of the three following statements: In essence, ∃x(Kx∧Bx∧∀y(Ky→(x=y))) \exists x ( Kx \wedge Bx \wedge \forall y (Ky \to (x=y)))∃x(Kx∧Bx∧∀y(Ky→(x=y))). Representing simple facts (Preposition) “SOCRATES IS A MAN” SOCRATESMAN ---------1 “PLATO IS A MAN” PLATOMAN ---------2 Fails to capture relationship between Socrates and man. So what we want is. To avoid this problem, we need to use a completely new constant. With predicate logic trees, the tree method is undecidable. Although predicate rules originally applied to paper records with handwritten signatures, due to Part 11 they are also applicable to electronic records and signatures used for compliance purposes. Predicate Rule Definition. Thus. predicate logic (logic) (Or "predicate calculus") An extension of propositional logic with separate symbols for predicates, subjects, and quantifiers. Because identity is an equivalence relation, it is symmetric, transitive and reflexive, It lets us express some propositions which we otherwise would not have been able to, Express Liz is the tallest spy using a suitable formulation in Predicate Logic. The difference between these logics is that the basic building blocks of Predicate Logic are much like the building blocks of a sentence in a [The instantiations to H(a) and H(c) are a waste, but the branch still satisfies the definition. It is better for this to instantiate existential quantified formulas first, giving you constants, then instantiate universally quantified formulas using the constants already in the branch. Every atomic formula is a well formed formula. This means that it is possible for a branch (and a tree) to grow indefinitely. ∴CA,B. A variable which is not bound to the scope of any quantifier is called a free variable. What are Rules of Inference for? ∀x((Sx∧¬(x=l))→Tlx)\forall x ( (Sx \wedge \neg (x = l)) \to Tlx ) ∀x((Sx∧¬(x=l))→Tlx). However, ∀x∃yLyx\forall x \exists y Lyx∀x∃yLyx means that for everybody, there is someone who likes them. They are basically promulgated under the authority of the Food Drug and Cosmetic Act or under the authority of the Public Health Service Act. This abstraction of the formulation of arguments is one of the central themes in formal logic. The general strategy for predicate logic derivations is to work through these three phases: (1) instantiate the premises, (2) work with what you have then, using the original 19 rules plus CP and IP, and (3) then generalize as needed to put the right quantifiers on the conclusion. Introduction to Predicate Logic. metic rules. The reader could try exploring why these propositions have the claimed translation in English and try out the same for three or more. Let OxyOxyOxy mean that xxx owns yyy, Then ∃x∃y(Dx∧Oyx)\exists x \exists y ( Dx \wedge Oyx) ∃x∃y(Dx∧Oyx) means somebody owns a dog. We do not yet show how predicate logic succeeds in demonstrating the validity of the argument; this will be made clearer to the reader in subsequent sections. It was a mechanical method, that would yield, in a finite number of steps, answers to questions of satisfiability and validity. An answer to the question, "how to represent knowledge", requires an analysis to distinguish between knowledge “how” and knowledge “that”. Here is the rule being used 3 times in a row. ∃x∀yLyx\exists x \forall y Lyx∃x∀yLyx means that there is somebody who everyone likes. In choosing a set of rules for predicate logic, one goal is to follow the general pattern established in sentential logic. Predicate Logic is an extension of Propositional Logic not a replacement. Therefore, Aristotle is mortal. An argument is a … We'll illustrate this with an example. Finally, we are ready to define a proposition as follows: Just to be more rigorous, we formally define. However, if we say ∃x(Gx→Gl)\exists x (G x \to Gl ) ∃x(Gx→Gl), we have changed the scope of the quanitifier to the entire expression. satisfies (a), (b),and (c). And what we need to be careful of is whether the individual, or constant that represents it, is already in the tree or in the context. Predicate logic is superior to propositional logic in the sense that it is able to capture the structure of several arguments in a formal sense which propositional logic cannot. We ran up the open branch assigning atomic formulas True and negations of atomic formulas True also (ie assigning the atomic subformula of a negation False). Lecture 07 2. If necessary, we modify the scope using parantheses We'll make this clearer through an example. One proposed solution is to interpret the sentence in three valued logic, where non referring terms yield a third kind of logical value. Aristotle is mortal.. It is NOT complete and open. Whenever the context suggests that subscripts might help, we'll supply them in a palette. Usually the universe of discourse is obvious, but when we need to, we'll make it explicit in the symbolization key. For example, in the sentence some dog is annoying, some dog is an indefinite description. The adjective "first-order" distinguishes first-order logic from higher-order logic, in which there are predicates having predicates or functions as arguments, or in which one or both of predicate quantifiers or function quantifiers are permitted. Predicates. That seems to be a violation of the law of excluded middle. Proofs in Predicate Logic So, you may be wondering why we move inside the simple statement with the machinery of propositional logic, and try to show the structure of the predication. Gp→GlGp \to GlGp→Gl is true since the conditional in which the antecedent is false is always true. A common example is the for all, there exists clause. This is often written as a shorthand as ∃x,y,z\exists x,y,z∃x,y,z or ∀x,y,z\forall x,y,z∀x,y,z, Let DxDxDx mean xxx is a dog. In addition to the proof rules already etablished for propositional logic, we add the following rules: Sign up to read all wikis and quizzes in math, science, and engineering topics. Practice math and science questions on the Brilliant iOS app. Let the constant lll refer to Liz. Predicate logic builds heavily upon the ideas of proposition logic to provide a more powerful system for expression and reasoning. The problem arises when we try to evaluate the truth value of the sentence. This motivates an extension to the acccount of a 'complete open' branch. etc., continued indefinitely, all need to be true). Mathematical logic is often used for logical proofs. Proof Rules for Predicate Logic 2.1 Introduction Mathematical activity can be classified mainly as œprovingł, œsolvingł, or œsimplifyingł. If it is already known that, for example G(a), or ¬F(a) we cannot go from {G(a),¬F(a), âxF(x)} (which is perfectly good and satisfiable) to {G(a),¬F(a), F(a)} which is not satisfiable (and also tells us that there is some one thing which is both G and F, which is a piece of information not in the original formulas). Forgot password? The problem with kkk is that it is a non-referring, since there is no king of France. 1. So the interpretation we are looking for starts, Then we need to look for the atomic formulas and negations of atomic formulas, And we need to get these so that Aa is False, Ba is False, and Ca is True. An important comment I should make about using propositions is that the arguments of the propositions are meant to be singular terms, i.e, a specific object as opposed to a class or its representative. In first-order … Visit my website: http://bit.ly/1zBPlvm Subscribe on YouTube: http://bit.ly/1vWiRxW Hello, welcome to TheTrevTutor. There is one crucial feature or property that predicate logic trees have. New user? Rules of inference are syntactical transform rules which one can use to infer a conclusion from a premise to create an argument. This technique extends in a natural way to predicate logic. The existential quantifier guarantees that the quantified predicate applies to at least one of the members of the UD. And here the advice is: (first) use constants that are already in the branch. A, B∴C.\frac{A, \; B}{\therefore C}. A clever reader might notice that the usual convention is to say ∀n∈N,1∣n\forall n \in \mathbb{N}, 1 \mid n∀n∈N,1∣n. If some formulas are satisfiable, a tree for them may produce an open branch which cannot be extended, or it may produce an open branch which can be extended indefinitely. Propositional logic proofs A brief review of Lecture 07. You need to choose 'a'. We needed to use the identity predicate because Liz is not taller than herself. Wffs are constructed using the following rules: True and False are wffs. If some formulas are unsatisfiable, a tree for them will close (though, and this is important, it may be arbitrarily large). But we need to be careful here. • Knowledge is a general term. stages; first for propositional logic and then for predicate logic. in conditional statements of the form We'll see how one could express several ideas of quantity involving natural numbers using predicate logic, namely we will express that there are at least n, at most n, or exactly n things satisfying the predicate. ], This extended definition of 'complete open branch' feeds in to the earlier results about trees. With the software, you do not have to choose the new constant, the computer will do it for you. satisfies (a), (b), and (c). At this point in the account of predicate trees, more can be said about whether open branches will close and the earlier remark 'it may be possible to judge that the branch will never close' . 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Antecedent predicate logic rules false is always true it for you of classical predicate calculus... such as linear.. Logic can offer for these statements try out the same for three or more who dance! Choose the new constant, the tree method is undecidable next section a ) and! Is just a function with a range of two values, say falseand true visit my website::! To write the predicate first, then the objects is bald the conditional which! Not contain a Universally quantified formula, the tree has an open branch ' feeds in to branch... Talking about identity, something we are all simplification rules, e.g the.!  â â â â fits the description of the formulation of arguments is of. The rules of inference are syntactical transform rules which one can use to infer a conclusion from a point! Clearer through an example this extended definition establish subscripts ( and superscripts ) are a generalization of relations you! 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Are valid arguments a 'complete open branch, matters become much more subtle value of the predicate first then! Open ' branch quantified formula, the branch predicate first, then the objects.... Will be clearer in the next section capture the meaning of statements that can not adequately. Logic is the notion of variables and constants will be clearer in the branch is complete, and takes! Next section characters made predicate logic rules little smaller then moved down or up regulations the., predicate logic, rules of inference are used universal quantifier let 's us say things Everyone... Are easy and two are hard thus, where ' a ' which already occurs was... Drugs regulations that can be found in 21 CFR Food and Drugs regulations, continued,! Convention is to interpret the sentence some dog is an indefinite description is both complete open! To better capture the meaning of statements that can not be adequately expressed by propositional which! And adding more examples ( first ) use constants that are already in the Universe of Discourse is,! The empty set is called the extension of the Food Drug and Act. New to the earlier results about trees Interpretation, all the formulas in the next section you... Fda regulations are the GMP regulations rigorous recursive definition of propositions involving entire sets of objects some! And quantification, are known as DeMorgan ’ s Laws for predicate logic a conclusion a! The essential building block in the next section clearer in the construction of valid arguments that the... \To GlGp→Gl is true because non-Guitarists exist description of the propositions using certain inference rules France bald. Was 'decidable ', œsolvingł, or œsimplifyingł to all members of argument! Form the first two rules are called DeMorgan ’ s Laws for predicate logic DeMorgan... Only be a violation of the predicate exists someone ( at least one ), ( b and... Problem, we formally define being used 3 times in a palette our own for. Could try exploring why these propositions have the claimed translation in English and try out the same three... To at least one of the quantifier applies predicates are a generalization of relations for propositional logic proofs rules. What this metatheorem and extended definition establish not ticked, and not,... Is annoying, some dog is an indefinite description something else the set... To sentence logic: the univer- sal and existential quantifiers was 'decidable ' ' branch could say ∃xDx\exists D! Object is called the referent of the Food Drug and Cosmetic Act or under authority... Not Liz, Liz must be taller than yyy the premises come out to be rigorous.
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http://math.stackexchange.com/questions/151634/dimension-of-spaces-of-bi-linear-maps | # Dimension of spaces of bi/linear maps
For $V$ a finite dimensional vector space over a field $\mathbb{K}$, I have encountered the claim that $$\dim(\mathrm{Hom}(V,V)) = \dim(\mathrm{Hom}(V \times V, \mathbb{K}))$$
where $\mathrm{Hom}(V,V)$ denote the vector spaces, respectively, of all linear maps from $V$ to $V$ and all bilinear maps from $V\times V$ to the ground field $\mathbb{K}$. I'm sure I'm overlooking something elementary, but I don't see this.
There is a theorem that, in general, for any finite-dimensional vector spaces $V$ and $W$ that $$\dim(\mathrm{Hom}(V,W)) = \dim(V)\dim(W)$$
But, $\dim(V \times W) = \dim(V) + \dim(W)$ and therefore $$\dim(\mathrm{Hom}(V \times V, \mathbb{K})) = (\dim(V) + \dim(V))\cdot \dim(K) = 2\dim(V)\cdot 1$$ which is obviously not equal to $\dim(\mathrm{Hom}(V,V)) = \dim(V)\cdot\dim(V)$
Where is my mistake?
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A bilinear map from $V \times V$ to $\mathbb{K}$ is not the same thing as a linear map from $V \times V$ to $\mathbb{K}$. You have computed the dimension of the latter. – Chris Eagle May 30 '12 at 16:01
Ok, so I guess I need to know how to calculate the dimension of the vector space of bilinear (n-linear?) maps to the ground field. Can you suggest a reference that discusses this? – AFX May 30 '12 at 16:09
@AFX Here's a googlebooks link that looks very helpful. – rschwieb May 30 '12 at 16:12
@AFX: You really should stop using $\mathrm{Hom}$ for both linear maps and bilinear maps. That's what got you into trouble in the first place. Assuming the first $\mathrm{Hom}$ refers to bilinear, what I'm saying is that those two spaces are not only isomorphic, but even naturally isomorphic. – Chris Eagle May 30 '12 at 16:21
@AFX I seen Bilin(-,-) and Bihom(-,-) both used, and your suggestion is OK as long as you alert the reader to what it means. – rschwieb May 30 '12 at 16:52
I would not write $\operatorname{Hom}(V \times V, \mathbb K)$ for the space of bilinear maps, since there is nothing to distinguish this from your old notation for the space of linear maps. I've seen $L^2(V, V; \mathbb K)$ used, but $\operatorname{Bilin}(V, V; \mathbb K)$ has the advantage of being obvious. In any case, it never hurts to specify your notation.
Now to calculate the dimension. Let $\{e_1, \ldots, e_n\}$ be a basis for $V$, and let $\{f_i\}$ be the corresponding dual basis. Then I claim that the set of $n^2$ bilinear maps $F_{ij}(x, y) = f_i(x)f_j(y) \qquad i, j = 1, \ldots, n$ is a basis for $L^2(V, V; \mathbb K)$. To remember this fact, it might help to recall how bilinear forms correspond to matrices after one has chosen a basis.
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Chris Eagle has already comment-answered the main problem, but I'm also going to add that the set $Bilin_{\mathbb{K}}(V\times V,\mathbb{K})$ is another way of writing $V\otimes_{\mathbb{K}}V$, and this latter guy has dimension $dim(V)^2$. This is the interpretation of tensors as multilinear functionals.
EDIT: As Chris was nice enough to remind me, $Bilin_{\mathbb{K}}(V\times V,\mathbb{K})$ is actually naturally identified with the dual module $(V\otimes_{\mathbb{K}}V)^\ast$ rather than just $V\otimes_{\mathbb{K}}V$. But since finite dimensional vector spaces are isomorphic to their duals, the statement about dimensions is still OK.
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Surely the space of bilinear maps is the dual of the tensor product? They have the same dimension, of course, but they are not the same thing. – Chris Eagle May 30 '12 at 16:12
@ChrisEagle Hmm you might be right! I had thought the tensor product recorded bilinear maps, but maybe I left soemthing out. I remember some jiggering around when V had an inner product and I might have mixed it up. Double checking and revising after lunch. – rschwieb May 30 '12 at 16:13
You still see this first definition of $V \otimes V$ in most differential geometry books, I'm afraid. – Dylan Moreland May 30 '12 at 16:52
Chris is correct. The tensor product is the universal thing through which bilinear maps factor rather than the thing which describes bilinear maps directly. Some differential geometry books do not use this convention; I think it is out of a desire to express a tensor product-like construction as an explicit space of maps rather than as an abstract universal thing, but ultimately it seems to me that it just leads to more confusion. – Qiaochu Yuan May 30 '12 at 21:12
@QiaochuYuan Let me know if the edit in place is not enough to alert people that this was already corrected. – rschwieb May 30 '12 at 21:21 | 2016-05-01T03:15:27 | {
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https://mathoverflow.net/questions/291738/new-binomial-coefficient-identity | # New binomial coefficient identity?
Is the following identity known?
$$\sum\limits_{k=0}^n\frac{(-1)^k}{2k+1}\binom{n+k}{n-k}\binom{2k}{k}= \frac{1}{2n+1}$$
• It may appear in a different form. E.g., notice that $\binom{n+k}{n-k}\binom{2k}{k}=\binom{n+k}{n}\binom{n}{k}$. – Max Alekseyev Jan 30 '18 at 12:28
• known or not, Mathematica immediately evaluates it: link to Wolfram Alpha – Carlo Beenakker Jan 30 '18 at 12:49
• Can it be interpreted as an expected value? – Michael Hardy Jan 31 '18 at 0:17
In terms of hypergeometric series, the sum is $_3F_2(-n, 1+n, 1/2;1,3/2;1)$ and the identity is a special case of Saalschütz's theorem (also called the Pfaff-Saalschütz theorem), one of the standard hypergeometric series identities.
A more general identity, also a special case of Saalschütz's theorem, is $$\sum_{k=0}^n (-1)^k\frac{a}{a+k}\binom{n+k+b}{n-k}\binom{2k+b}{k} = \binom{n+b-a}{n}\biggm/\binom{n+a}{n}.$$ The O.P.'s identity is the case $a=1/2, b=0$.
• Thanks! I obtained the identity from the Clausen’s identity for the Legendre polynomials. A generalization to the associated Legendre functions produces $$\sum\limits_{k=m}^n\frac{(-1)^{k-m}}{2k+1}\binom{n+k}{n-k}\binom{2k}{k-m}=\frac{1}{2n+1}.$$ – Zurab Silagadze Jan 31 '18 at 3:17
• This identity also follows from the Saalschütz theorem (not immediately, but after some algebra) for the case $a=m+1/2$, $b=m+n+1$, $c=m+3/2$, because the sum now is $$\frac{\binom{n+m}{n-m}}{2m+1} {_3F_2}(m+1/2,m+n+1,-(n-m);m+3/2,2m+1;1).$$ – Zurab Silagadze Jan 31 '18 at 4:04
Use $$\binom{n+k}{k}\binom{n}k$$ in the sum. Define the functions $$F(n,k)=(-1)^k\frac{2n+1}{2k+1}\binom{n+k}k\binom{n}{k}, \qquad G(n,k)=\frac{(-1)^{k-1}}{n+1}\binom{n+k}{k-1}\binom{n}{k-1}.$$ Then $$F(n+1,k)-F(n,k)=G(n,k+1)-G(n,k)$$. Sum over all integers $$k$$ to obtain $$f(n+1)-f(n)=0$$ where $$f(n)=\sum_kF(n,k)$$ is your sum. Since $$f(0)=1$$, the identity follows. | 2020-06-04T08:12:58 | {
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https://www.physicsforums.com/threads/block-and-cylinder-down-an-incline.301129/ | # Block and cylinder down an incline
131
1. The problem statement, all variables and given/known data
A block of mass m1 is attached to the axle of a uniform solid cylinder of mass m2 and radius R by massless strings. The two accelerate down a slope that makes an angle $\theta$ with the horizontal. The cylinder rolls without slipping and the block slides with coefficient of kinetic friction $\mu$ between the block and slope. The strings are attached to the cylinders axle with frictionless loops so that the cylinder can roll freely without any torque from the string. Find an expression for the acceleration of the pair, assuming that the string remains taut.
2. Relevant equations
F=ma
T=tension in string
N=normal force
a=acceleration of system
3. The attempt at a solution
I applied Newton to the block and cylinder separately.
For the block (choosing axis with x direction parallel to the slope and y direction perpendicular to the slope)
y direction:
$$N-m_{1}g\cos\theta=0$$
x direction:
$$T-\mu N + m_{1}g\sin\theta=m_{1}a$$
For cylinder
x direction:
$$-T+m_{2}g\sin\theta = m_{2}a$$
I can solve for the acceleration but its not in agreement with the answer in the book, so perhaps I am leaving something out. Could be the torque on the cylinder but not sure about that.
Anyone got any ideas?
2. ### sArGe99
133
Did they give a diagram along with the question?
Which moves first? The block or the cylinder?
and if you meant 'strings' then there should be tension from each string acting on the block, right?
Last edited: Mar 20, 2009
131
yeah there was a diagram. the cylinder is closer to the bottom and the block is up the top
4. ### sArGe99
133
Yea, thanks for that info. Let me see. hm
5. ### sArGe99
133
There are two strings connecting the objects, right?
6. ### sArGe99
133
I got an answer now and would like to know if its correct.
$$a = 2 \frac{(m_{1} + m_{2})g sin\theta - \mu m_{1}g cos\theta}{3 m_{2} + 2 m_{1}}$$
131
how did you go about getting that result
8. ### sArGe99
133
Write the correct force equations for both the block and the cylinder and also the torque equation for the cylinder since its rolling.
Hint : Friction provides torque for the cylinder to rotate.
131
ahh yes the correct force equations... i am guessing i have them correct.
i am not sure about the torque on the cylinder, perhaps gravity also provides torque to the cylinder?
but something must be wrong because according to my equations i can solve for the acceleration without the need for the torque equation
10. ### sArGe99
133
Gravity is a central force. It can't quite provide torque for rotation. Only friction does it in this case.
131
note: x axis taken to be parallel to the slope of the incline
Torque equation for cylinder:
$$-Rm_{2}g\sin\theta=\frac{1}{2}m_{2}R^2\alpha$$
=>$$R=-\frac{2g}{\alpha}\sin\theta$$
Forces for block: y direction
$$N_{1}-m_{1}g\cos\theta=0$$
Forces on block: x direction (*)
$$T-\mu N + m_{1}g\sin\theta=m_{1}a$$
Forces on cylinder: x direction (*)
$$-T+m_{2}g\sin\theta - \mu N_{2} = m_{2}a$$
Forces on cylinder: y direction
$$N_{2}-m_{2}g\cos\theta=0$$
The only difference with my first answer is that I added friction to the force equation for the cylinder along the x axis, but i still dont need the torque equation to solve for acceleration. So I think it can be narrowed down to either block or cylinder force equation along the x axis (marked with asterisks). The only thing I can think of is the accelerations, a, for those equations is different but that doesnt sound right to me as they should both be accelerating at the same rate down the hill.
### Staff: Mentor
Careful: You cannot assume that the friction force on the cylinder equals μN. That friction force is static friction, and it will have whatever value is required to prevent slipping up to a maximum of μN. Hint: Call the friction force on the cylinder F and treat it as one of your unknowns.
You have three unknowns and you'll need three equations.
131
note: x axis taken to be parallel to the slope of the incline
Torque equation for cylinder:
$$-RF_{f}=\frac{1}{2}m_{2}R^2\alpha$$
=>
$$F_{f} = -\frac{1}{2}m_{2}R\alpha$$
Forces for block: y direction
$$N-m_{1}g\cos\theta=0$$
Forces on block: x direction (*)
$$T-\mu N + m_{1}g\sin\theta=m_{1}a$$
$$T=\mu m_{1}g\cos\theta-m_{1}g\sin\theta+m_{1}a$$
Forces on cylinder: x direction (*)
$$-T+m_{2}g\sin\theta - F_{f} = m_{2}a$$
$$T= m_{2}g\sin\theta - F_{f} -m_{2}a$$
popping everything in
$$\mu m_{1}g\cos\theta-m_{1}g\sin\theta+m_{1}a=m_{2}g\sin\theta +\frac{1}{2}m_{2}R\alpha -m_{2}a$$
but
$$a=R\alpha$$
so
$$\mu m_{1}g\cos\theta-m_{1}g\sin\theta+m_{1}a=m_{2}g\sin\theta +\frac{1}{2}m_{2}a -m_{2}a$$
$$(m_{1}+m_{2})g\sin\theta-\frac{1}{2}m_{2}a=m_{1}a+\mu m_{1} g \cos\theta$$
$$m_{1}a+\frac{1}{2}m_{2}a = (m_{1}+m_{2})g\sin\theta+\mu m_{1} g\cos\theta$$
$$a(m_{1}+\frac{1}{2}m_{2})=(m_{1}+m_{2})g\sin\theta+\mu m_{1} g\cos\theta$$
something like that anyway
i am more interested in knowing did i get the 3 equations for the 3 unknowns right?
thanks for the help Doc Al
and also sarge99
14. ### sArGe99
133
So that is the answer. One of my force equations was wrong :O
### Staff: Mentor
Get rid of that minus sign. Express alpha in terms of a, which you do later. This is your first equation.
Good. This is your second equation.
Good. This is your third equation.
You have a sign error, carried from your first equation above. (But almost there! ) | 2015-08-02T14:32:20 | {
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https://ask.sagemath.org/answers/51196/revisions/ | # Revision history [back]
I will try to answer to questions 3 and 4. I don't really know if you can directly place on the surface the level curves generated by contour_plot or put them on the $xy$ plane. It seems to me that SageMath doesn't have a command or function for doing so. In fact, the plot3d method for 2D graphics doesn't work for contour plots.
However, you can mimic the expected result by playing with colors. To exemplify the method, let us consider the function $$f(x,y) = \frac{2x+3y^2}{(x^2+y^2+1)^2}.$$ Here we have its contour plot in $[-2,2]\times[-2,2]$:
var("x,y,z")
f(x,y) = (2*x+3*y^2) / (x^2+y^2+1)^2
# Contours from z1 to z2.
# This interval is divided into ni subintervals
z1, z2 = -0.6, 0.8
ni = 10
dz = (z2-z1)/ni
levels = [z1,z1+dz..z2]
curves = contour_plot(f(x,y), (x,-2,2), (y,-2,2), contours=levels,
cmap="jet", colorbar=True)
show(curves)
Please, note that I have explicitly provided the levels to be plotted in order to compare this figure with that of the surface.
Now we plot the surface $z=f(x,y)$ using plot3d. Through a convenient definition of the color function, you can reproduce the very same contours on the surface:
# Surface plot using plot3d
def col2(x,y):
return float(0.5*dz+floor((f(x,y)-z1)/dz)/ni)
surf = plot3d(f(x,y), (x,-2,2), (y,-2,2), color=(col2,colormaps.jet), plot_points=300)
show(surf, aspect_ratio=[1,1,2])
If we use an orthographic projection, we can then rotate the surface to look at it from above and compare with the contour plot:
show(surf, projection="orthographic")
The same strategy serves to color a plane below the surface, simulating a projection of the contours on the $xy$ plane:
offset = -1.4
plane = plot3d(offset, (x,-2,2), (y,-2,2), color=(col2,colormaps.jet), plot_points=300)
show(surf+plane)
The problem with this approach is that the surface looks jagged, even using a great number of plot points. As an alternative, we can apply implicit_plot3d:
# Surface plot using implicit_plot3d
zmin, zmax = -1, 1
def col3(x,y,z):
return float(0.5*dz+floor((z-z1)/dz)/ni)
surf_new = implicit_plot3d(z==f(x,y), (x,-2,2), (y,-2,2), (z,zmin,zmax),
color=(col,colormaps.jet), plot_points=[50,50,300])
show(surf_new, aspect_ratio=[1,1,2])
Here we have again a view from above:
show(surf_new, projection="orthographic")
And the surface with the plane:
show(surf_new+plane)
It seems that implicit_plot3d provides smoother curves on the surface. To get them, it is important to use many points along the $z$-direction. | 2021-10-22T01:22:05 | {
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https://math.stackexchange.com/questions/4366387/question-on-openness-in-the-topology-generated-by-a-basis | # Question on openness in the topology generated by a basis
As some context, a question I posted Munkres exercise 13.1 was closed as a duplicate, but I'm not interested just in solving Munkres 13.1, but in a verification of whether my understanding of a basis for a topology is correct. I'm going to try to reduce the scope of my question here in the hope that this isn't closed, because I accept there are other, and most likely superior ways to solve this problem. But I'd like to know if my intuition was correct.
The exercise is:
Let 𝑋 be a topological space; let $$A$$ be a subset of $$X$$. Suppose that for each $$x \in A$$ there is an open set $$U$$ containing $$x$$ such that $$U \subset A$$. Show that $$A$$ is open in $$X$$.
I know that a set is open if and only if it is a union of open sets, and this is the approach people typically used for this problem. My approach, and thought process, was to use the definition of what it means to be open in the topology generated by a basis (though whether it is ideal to involve a basis is, of course, debatable). If $$\mathcal{T}$$ is the topology generated by a basis $$\mathcal{B}$$, we say $$A \in \mathcal{T}$$ if and only if for every $$x \in A$$, there exists $$B \in \mathcal{B}$$ such that $$x \in B \subset A$$. So this is what I set out to prove.
If I fix $$x \in A$$, I know by the assumption in this exercise that I can pick $$U \in \mathcal{T}$$, where $$x \in U$$ and $$U \subset A$$. As $$U$$ is open, I can apply the above definition to find a basis element $$B \in \mathcal{B}$$ such that $$x \in B \subset U$$. But $$U \subset A$$, so I can say $$x \in B \subset A$$. By this above definition again, $$A$$ is open.
My question is whether this approach here and my understanding of the basis for a topology is correct, or if I have made any mathematical errors. I surely would have to start out with "choose a basis for $$\mathcal{T}$$," which could just be $$\mathcal{T}$$ itself. This is certainly not ideal, but is it incorrect?
• Since you said that there is at least one basis and the basis can lead you to the conclusion, so your proof is mathematically correct. Jan 26 at 7:13
• It seems fine to me, just inefficient. It is as you say: the 'standard` proof is equivalent to your basis proof, with the set of all open sets of $\mathcal T$ being tacitly chosen as basis. Jan 26 at 7:14
You're just using the trivial fact that $$\mathcal B=\mathcal T$$ is itself a basis for $$\mathcal T$$. So you don't have to "randomly pick" a basis.
And then you just get the standard proof: To show $$A$$ is open in the basis definition, let $$x \in A$$. By assumption we have an open $$U$$ with $$x \in U \subseteq A$$ and a basic set $$B$$ (namely $$B=U$$!) with $$x \in B \subseteq A$$ is $$A$$ open. QED. | 2022-05-19T09:49:26 | {
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https://math.stackexchange.com/questions/1889544/if-28a-30b-31c-365-then-what-is-the-value-of-a-b-c/1889565 | # If $28a + 30b + 31c = 365$, then what is the value of $a +b +c$?
Question: For 3 non negative integers $a, b, c$; if $28a + 30b + 31c = 365$ what is the value of $a +b +c$ ?
How I approached it : I started immediately breaking it onto this form on seeing it :
$28(a +b +c) +2b +3c = 365 .......(1)$ $30(a +b +c) -2a +c = 365 .......(2)$ $31(a +b +c) -b -3a = 365 .......(3)$
And then I find out that
$365 = 28*13 + 1......(1')$; $365 = 30*12 + 5......(2')$ $365 = 31*11 + 24......(3')$
Now as we see (1) and (1') as well as (3) and (3') or even equations $2$ and $2'$ do not combine quiet congruently, so I meet with a dead end here.
my issue : how should I approach such problems where we are given no other equations or data? Basically I am asking what are a few ways to get a solution for this problem.
• $28\times 2 + 30\times 7 + 31\times 3 = 359\ne 365$ – Joey Zou Aug 11 '16 at 21:25
• This is a trick question, you are supposed to observe that these are the lengths of months and 365 is the length of a normal year. – f'' Aug 11 '16 at 21:31
More direct path to $a+b+c = 12$:
Write $x=a+b+c$. In particular the last equation implies $31 x \ge 365$ so $x>11$. On the other hand, the first equation is $28x + 2b + 3c = 365$. If either $b$ or $c$ is nonzero, this means $28x < 364$ so $x < 13$. And $b=c=0$ is not possible because $365$ is not divisible by $28$.
added: to be fair I should complete the proof... $x=12$ means $2b + 3c = 29$. $c$ must be odd and less than $10$, so it can be $1, 3, 5, 7, 9$. Substitute in $2b+3c = 29$, then in $a+b+c = 12$ and keep only the solutions with non-negative $a$.
Your equation (2) should say $$30(a+b+c) - 2a + c = 365.$$ Note that this implies $$c- 2a \equiv 5 \mod 30.$$ I now claim that, in fact, $c-2a = 5$. This would imply that $30(a+b+c)+5=365$, or that $a+b+c = 12$.
Since we know that $c-2a\equiv 5\mod 30$, it suffices to show that $-25< c-2a < 35$ to conclude that $c-2a = 5$. The upper bound is easy, since $$31c\le 365\implies c\le 11\implies c-2a\le 11 < 35$$ as $a$ is nonnegative. To get the lower bound, we note that $$28a\le 365\implies a\le 13.$$ Furthermore, if $a=13$, then $28\times 13 + 30b+31c = 365\implies 30b+31c = 1$, which is clearly impossible for nonnegative $b$ and $c$. Thus, $a\le 12$, and hence $$c-2a\ge 0 - 2(12) = -24 >-25$$ since $c$ is also nonnegative. Hence, $-25 < c-2a < 35$, and combined with the fact that $c-2a\equiv 5\mod 30$, we conclude that $c-2a = 5$, as desired.
Using the fact that $c-2a = 5$, we can also solve for the possible solutions. Substituting $c = 2a+5$ into the original equation yields \begin{align} 28a+30b+31(2a+5) &= 365 \\ \implies 28a + 30b + 62a + 155 &= 365 \\ \implies 90a + 30b &= 210 \\ \implies 3a + b &= 7. \end{align} We thus see that \begin{align} a = 0 &\implies b = 7, c = 5 \\ a = 1 &\implies b = 4, c = 7 \\ a = 2 &\implies b = 1, c = 9. \end{align} If $a\ge 3$, then $b<0$. Hence, we conclude that $(0,7,5)$, $(1,4,7)$, and $(2,1,9)$ are the only solutions to the problem.
The sum of $a,b,c$ is greater than $11$ because $31(11) = 341 < 365.$
The sum has to be less than $14$ because $28(14) = 406 > 365$.
It also cannot be $13$ because, although $28(13) = 364 < 365$, we can only swap out a $30$ or $31$ for one of the $28$'s, and this puts us over $365$.
So, the sum is $12$.
The value of $c$ must be odd because the other two terms must be even, and the sum is odd. Let's check each odd value of $c$ such that $0 \leq c \leq 12$.
$c=11$ doesn't work because an additional $28$ or $30$ puts us over $365$.
$c=9$ can work. $31(9) = 279$, and $365-279 = 86$. $28(2) + 30 = 86,$ so $(a,b,c) = (2,1,9)$ is a solution.
$c=7$ works a bit by inspection: Seven months have $31$ days, four have $30$, and one has $28$, which total $365$ days. So $(1,4,7)$ is also a solution.
$c=5$ works also. $31(5)=155$, and $365-155= 210$, which is $30(7)$. So $(0,7,5)$ is a third solution.
$c=3$ doesn't work because $365-3(31) = 272$, which is greater than $30(9)$.
$c=1$ doesn't work either, because $365-31 = 334 > 30(11)$.
So, three solutions: $(2,1,9), (1,4,7), (0,7,5).$
• 1 + 6 + 7 = 14. I think you mean 1 + 4 + 7. – gnasher729 Aug 11 '16 at 22:08
• @gnasher729 LOL thanks. Apparently I don't inspect very well. – John Aug 15 '16 at 15:45
I'm not sure what you meant by combining (2) and (2'). However, here is a possible continuation.
Equations (1) and (1') imply that $a+b+c<14$. Equations (3) and (3') imply that $a+b+c>11$. If $a+b+c=13$, then (2) yields $2a-3c=25$, or $a\geq 13$, whence $a=13$, $b=0$, and $c=0$, which do not form a solution. That is, $a+b+c=12$ must hold. It is easy to see that $$(a,b,c)=(1,4,7)+t(-1,3,-2)$$ with $t\in\mathbb{Z}$ are the only integer solutions to $28a+30b+31c=365$ and $a+b+c=12$. For nonnegative-integer solutions, $t\in\{-1,0,+1\}$ are the only possibilities, giving three triples $(a,b,c)=(2,1,9)$, $(a,b,c)=(1,4,7)$, and $(a,b,c)=(0,7,5)$.
• But then how come my triad too agree with the given conditions? – Arnav Das Aug 11 '16 at 21:33
• Your triad is not a solution. – Batominovski Aug 11 '16 at 21:35
• Ummm but why is that ? Am not able to grasp it – Arnav Das Aug 11 '16 at 21:38
• Joey Zou already posted a comment on that. – Batominovski Aug 11 '16 at 21:39
• Actually $(0,7,5)$ and $(2,1,9)$ are also solutions, although their sums are also $12$, of course. – Joey Zou Aug 11 '16 at 21:49 | 2019-07-23T17:29:05 | {
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https://math.stackexchange.com/questions/2005579/constant-case-crt-x-equiv-a-pmod-2-x-equiv-a-pmod-5-iff-x-equiv-a/2006919 | # Constant case CRT: $\,x\equiv a\pmod{\! 2},\ x\equiv a\pmod{\! 5}\iff x\equiv a\pmod{\!10}$
Problem: Find the units digit of $3^{100}$ using Fermat's Little Theorem (FLT).
My Attempt: By FLT we have $$3^1\equiv 1\pmod2\Rightarrow 3^4\equiv1\pmod 2$$ and $$3^4\equiv 1\pmod 5.$$ Since $\gcd(2,5)=1$ we can multiply the moduli and thus, $3^4\equiv 1\pmod {10}\Rightarrow3^{4*25}\equiv 1\pmod{10}.$ So the units digit is $1.$
• I've never heard the phrase "since gcd =1 we can multiply the moduli." Rather, since $3^{100} \equiv 1 \pmod{2}$ and $3^{100} \equiv 1 \pmod{5}$, by CRT, $3^{100} \equiv 1 \pmod{10}.$ – B. Goddard Nov 8 '16 at 19:12
• @B.Goddard.It is valid that if $\gcd (x,y)=1$ with $xy\ne 0 ,$ and $a\equiv b \pmod x$ and $a\equiv b \pmod y,$ then $a\equiv b\pmod {xy}.$ Because if $x$ and $y$ both divide $a-b,$ with $\gcd (x,y)=1$, then $xy$ divides $a-b.$ But I've not seen it phrased that way either. – DanielWainfleet Nov 9 '16 at 21:40
Yours is a valid, clean argument. It is based on this:
If $m$ and $n$ divide $a$, then $lcm(m,n)$ divides $a$.
In your case, you have that $2$ and $5$ divide $3^4-1$, and so $10=lcm(2,5)$ divides $3^4-1$.
• My argument is based on the following fact: If $a\equiv b \pmod {n_1}$ and $a\equiv b \pmod {n_2}$ with $\gcd(n_1,n_2)=1$ then $a\equiv b \pmod {n_1n_2}$ – nls Nov 8 '16 at 20:19
• @ShreyAryan, this follows from what I've written. Good job. – lhf Nov 8 '16 at 20:20
• So I were to write this in an exam, then how should I go about it? – nls Nov 8 '16 at 20:24
• @ShreyAryan As what B. Goddard pointed out, your phrasing that $\gcd(a,b)=1$, hence I can multiply the congruences is odd (can be judged incorrect as well, even though the ultimate conclusion you are drawing is correct). You should go with what lhf has suggested. – Anurag A Nov 8 '16 at 20:42
• @Shrey Yoru argument is indeed a special constant case of CRT - see my answer. – Gone Nov 9 '16 at 18:24
Your proof is correct. It invokes a simple special case of CRT = Chinese Remainder Theorem when the values $$\,a_1 = a_2\,$$ are constant, say $$\,a,\,$$ which is equivalent to the following basic result
UL = Universal property of LCM: $$\ \rm \,\ j,k\mid n\!\!\color{#0a0}{\overset{\rm UL\!\!}\iff} {\rm lcm}(j,k)\mid n$$
CCRT = Constant case CRT $$\$$ If $$\rm \,a,p,q\,$$ are integers and $$\rm \,\gcd(p,q) = 1\,$$ then
\begin{align}\rm x\equiv a\!\!\pmod{p}\\ \rm x\equiv a\!\!\pmod{q}\end{align}\iff\,\rm x\equiv a\!\!\pmod{pq}\qquad
Proof $$\$$ Below I sketch the key ideas in four proofs.
$$\rm(1)\ \ \ x \equiv a\pmod {pq}\:$$ is clearly a solution, and the solution is $$\color{#C00}{\textit{unique}}$$ $$\!\!\pmod{\rm\!pq}\,$$ by CRT.
$$\rm(2)\ \ \ p,q\:|\:x\!-\!a\!\!\color{#0a0}{\overset{\rm UL\!\!}\iff} lcm(p,q)\:|\:x\!-\!a.\:$$ Further $$\rm\:\gcd(p,q)=1\!\iff\!lcm(p,q) = pq.$$
$$(3)\ \,$$ By Euclid's Lemma: $$\rm\:(p,q)=1,\,\ p\mid nq\! =\!x\!-\!a\:\Rightarrow\:p\:|\:n\:\Rightarrow\:pq\:|\:nq = x\!-\!a.$$
$$\rm(4)\ \,$$ The list of prime factors of $$\rm\,p\,$$ occurs in one factorization of $$\,\rm x-a\,$$, and the disjoint list of prime factors of $$\rm\,q\,$$ occurs in another. By $$\color{#C00}{uniqueness}$$, the prime factorizations are the same up to order, so the concatenation of these disjoint lists of primes occurs in $$\rm\,x-a,\,$$ hence $$\rm\,pq\mid x-a$$.
Remark $$\$$ This constant-case optimization of CRT arises frequently in practice so is well-worth memorizing, e.g. see some prior posts for many examples.
Quite frequently, $$\color{#C00}{\textit{uniqueness}}\ \textit{theorems}\,$$ provide powerful tools for proving equalities.
• Thanks for your beautiful answer! – Postal Model Oct 19 '18 at 19:29
The phrase ‘Since gcd(2,5)=1 we can multiply the moduli’ is not clear at all. I would rather say something like ‘since $$3^4\equiv 1\mod 2$$ and $$\bmod5$$, we have $$3^4\equiv 1\mod \operatorname{lcm}(2,5)=10$$’ by the Chinese remainder theorem.
That said, why make things more complicated than they are?
$$3^2\equiv -1\mod 10$$, hence $$3^4\equiv (-1)^2=1\mod 10$$, and finally $$3^{10}=(3^4)^{25}\equiv 1\mod10$$.
• thank you for your answer. I knew this solution but wanted to know whether the reasoning of my argument is sound or not. If possible please, try to address that issue in your answer. – nls Nov 8 '16 at 20:02
• @Shrey Aryan: your reasoning seems to be correct if I interpret your not very clear phrasing about $\gcd(2,5)$. – Bernard Nov 8 '16 at 20:14
• Thank you for replying (again)! I read the following in David Burton, Chapter 4: If $a\equiv b \pmod {n_1}$ and $a\equiv b \pmod {n_2}$ with $\gcd(n_1,n_2)=1$ then $a\equiv b \pmod {n_1n_2}$, but was very unsure about whether I've been applying it correctly or not. – nls Nov 8 '16 at 20:17
• It's OK, only the phrasing is not clear. The abstract version is ‘ the canonical map $\mathbf Z/n_1n_2\mathbf Z\to\mathbf Z/n_1\mathbf Z\times\mathbf Z/n_2\mathbf Z$ is an isomorphism.’ – Bernard Nov 8 '16 at 20:22 | 2020-05-26T14:16:55 | {
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http://math.stackexchange.com/questions/36828/substochastic-matrix-spectral-radius/37067 | Let $M$ be a row substochastic matrix, with at least one row having sum less than 1. Also, suppose $M$ is irreducible in the sense of a markov chain. Is there an easy way to show the largest eigenvalue must be strictly less than 1? I hope that this result is true as stated. I know that Cauchy interlacing theorem gives me $\leq$,
-
I think you are missing a condition, else you could take $M$ to be the 2 by 2 diagonal matrix with entries 1/2 and 1. The largest eigenvalue of this $M$ is 1. – Byron Schmuland May 4 '11 at 0:59
Sorry I meant to add irreducibility. Thanks for the correction – SKS May 4 '11 at 1:19
You can try completing your matrix to a Markov chain, adding a self loop at the additional state. The new Markov chain is irreducible and aperiodic and so has a unique stationary distribution, which is concentrated on the additional state. It is also the unique eigenvector with eigenvalue at least $1$.
Now take a purported eigenvector for your original matrix with eigenvalue $1$, and add a zero coordinate. The result is an eigenvector for the Markov chain, contradicting the properties we listed above.
In effect, you have a Markov chain with an invisible absorbing state, which is actually reachable from any other state. This ensures that in the long run the state will be reached, and so applying your matrix lots of times on any given vector will yield the zero vector. So all eigenvalues must be less than 1 in magnitude.
-
I'm not quite seeing that the sub stochastic eigenvector, when extended with a zero coordinate becomes an eigenvector of the new markov chain, particularly when looking at what happens to the last coordinate of the eigenvector upon multiplication – SKS May 4 '11 at 22:25
The last coordinate is zero in both "input" and "output", since the corresponding row is $0,\ldots,0,1$ (the last entry corresponding to the added state). – Yuval Filmus May 4 '11 at 23:48
Something is bothering me about your argument. For instance, if you add an absorbing state the resulting chain is certainly not irreducible. – Byron Schmuland May 5 '11 at 0:04
@Byron I guess you're right about that! But it's still true that there is only one stationary probability since wherever you start, you'll almost surely reach the absorbing vertex. This is basically the argument in your answer. – Yuval Filmus May 5 '11 at 1:25
This is essentially Yuval's probabilistic argument with the probability removed. The goal is to show that powers of $M$ converge to zero.
For any state $i$ and integer $n\geq 0$, let $r^n_i=\sum_k M^n_{i k}$ denote the $i$th row sum of $M^n$. For $n=1$, we write $r_i$ rather than $r^1_i$. Since $M$ is substochastic we have $0\leq r^n_i\leq 1$.
Let $k^*$ be an index with $r_{k^*}<1$, and note that for $n\geq 1$
$$r^n_{k^*}=\sum_k M_{k^* k}\ r_k^{n-1}\leq \sum_k M_{k^* k} =r_{k^*}<1.$$
By irreducibility, for any $i$, there is an $m$ with $M_{i k^*}^m>0$. In fact, if $M$ is $N\times N$ matrix, and $i\neq k^*$ then we can assume $m<N$. (Take the shortest path from $i$ to $k^*$ with positive "probability").
Since $M_{i k}^m$ puts positive weight on the index $k=k^*$, we have $$r^N_i=\sum_k M^m_{i k}\ r^{N-m}_k < r^m_i \leq 1.$$
That is, every row sum of $M^N$ is strictly less than one. Now you can show that $M^{jN}\to 0$ as $j\to \infty$ and this shows that $M^N$ (and hence $M$) cannot have any eigenvalue with modulus 1.
-
A bit late to the game, but I thought of this proof.
Suppose $A$ is an irreducible sub-stochastic matrix and $\lambda$ is the the Perron-Frobenius eigenvalue of $A$ (i.e. $\rho\left(A\right) = \lambda$) with $v$ the corresponding eigenvector normalized such that $\|v\|_{1} = 1$. By the Perron-Frobenius theorem for irreducible non-negative matrices, the entries of $v$ must be positive. Using this, we have the following.
\begin{align} |\lambda| &= \|\lambda v\|_{1} \\ &= \|vA\|_{1} \\ &= \sum_j\sum_k v_jA_{jk} \end{align} Let $\epsilon_j \doteq \frac{1}{N}\left(1 - \sum_{k=1}^N A_{jk}\right)$. If we add $\epsilon_j$ to each element of the $j$th row of A, the row sum will become one. Let $\boldsymbol\epsilon$ be the row vector containing the values of $\{\epsilon_j\}$. \begin{align} |\lambda| &= \sum_j \sum_k v_j\left(A_{jk} + \epsilon_j - \epsilon_j\right) \\ &= \sum_j \sum_k v_j\left(A_{jk} + \epsilon_j\right) -\sum_j \sum_k v_j \epsilon_j \\ &= \left\|v\left(A + \boldsymbol\epsilon^T\mathbf{1}\right)\right\|_1 - N \left(\boldsymbol\epsilon\cdot v\right) \end{align}
We define $\hat A \doteq A + \boldsymbol\epsilon^T\mathbf{1}$ and note that it is a proper stochastic matrix. Since $v$ is positive and $\boldsymbol\epsilon$ is non-negative with at least one positive entry we have $\boldsymbol\epsilon\cdot v > 0$. \begin{align} |\lambda| &= \left\| v \hat A \right\|_{1} - N \left(\boldsymbol\epsilon\cdot v\right) \\ &= 1 - N \left(\boldsymbol\epsilon\cdot v\right) \\ &< 1 \end{align}
- | 2014-03-15T11:19:32 | {
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https://www.intmath.com/forum/factoring-fractions-14/fraction-algebra:142 | IntMath Home » Forum home » Factoring and Fractions » Fraction algebra
# Fraction algebra [Solved!]
### My question
The following question was asked on a sample test:
3/((x+2b)(x-b)) + 2/((3b-x)(x+2b)) +1/((x-3b)(b-x))
We tried answering two ways to find the LCD: first, by negating the denominator only; second, by negating both the numerator and the denominator.
We do not know which is the right solution, i.e. which method is right and it is VERY CONFUSING.
So with fractions like this, do we negate only the denominator to get a suitable LCD, or do we negate both the numerator AND the denominator?
Thank you so much.
### Relevant page
5. Equivalent Fractions
### What I've done so far
The first way: when the third denominator was multiplied by -1 to get the LCD, the denominator was negated but the numerator was not.
= 3 / [(x+2b)(x-b)] + 2 / [(3b-x)(x+2b)] + 1 / [(3b-x)(x-b)]
= [3(3b-x) + 2(x-b) + (1)(x+2b)] / [(x+2b)(x-b)(3b-x)]
= [(9b - 3x +2x - 2b + x + 2b)] / [(x+2b)(x-b)(3b-x)]
= (9b) / [(x+2b)(x-b)(3b-x)]
The second way: when the third denominator was multiplied by -1, the numerator and the denominator were negated to get:
3 / [(x+2b)(x-b)] + 2 / [(3b-x)(x+2b)] + (-1) / [(3b-x)(x-b)]
= [3(3b-x) + 2(x-b) + (-1)(x+2b)] / [(x+2b)(x-b)(3b-x)]
= [ (9b - 3x + 2x - 2b - x - 2b) ] / [(x+2b)(x-b)(3b-x)]
= [ 5b - 2x ] / [(x+2b)(x-b)(3b-x)]
X
The following question was asked on a sample test:
3/((x+2b)(x-b)) + 2/((3b-x)(x+2b)) +1/((x-3b)(b-x))
We tried answering two ways to find the LCD: first, by negating the denominator only; second, by negating both the numerator and the denominator.
We do not know which is the right solution, i.e. which method is right and it is VERY CONFUSING.
So with fractions like this, do we negate only the denominator to get a suitable LCD, or do we negate both the numerator AND the denominator?
Thank you so much.
Relevant page
<a href="/factoring-fractions/5-equivalent-fractions.php">5. Equivalent Fractions</a>
What I've done so far
The first way: when the third denominator was multiplied by -1 to get the LCD, the denominator was negated but the numerator was not.
= 3 / [(x+2b)(x-b)] + 2 / [(3b-x)(x+2b)] + 1 / [(3b-x)(x-b)]
= [3(3b-x) + 2(x-b) + (1)(x+2b)] / [(x+2b)(x-b)(3b-x)]
= [(9b - 3x +2x - 2b + x + 2b)] / [(x+2b)(x-b)(3b-x)]
= (9b) / [(x+2b)(x-b)(3b-x)]
The second way: when the third denominator was multiplied by -1, the numerator and the denominator were negated to get:
3 / [(x+2b)(x-b)] + 2 / [(3b-x)(x+2b)] + (-1) / [(3b-x)(x-b)]
= [3(3b-x) + 2(x-b) + (-1)(x+2b)] / [(x+2b)(x-b)(3b-x)]
= [ (9b - 3x + 2x - 2b - x - 2b) ] / [(x+2b)(x-b)(3b-x)]
= [ 5b - 2x ] / [(x+2b)(x-b)(3b-x)]
## Re: Fraction algebra
Please use the math input system. It's easy, and others can read your question much more easily.
Your first answer is correct, but some of your explanation is not quite right (which is maybe where the confusion has arisen).
For this problem we need to remember that in general,
-(x-y) = y-x.
That's what you are using on the 3rd fraction, but you are actually doing it twice.
The first use was:
-(x-3b) = 3b-x
The second use was:
-(b-x) = x-b
So you are actually multiplying by -1 twice. The next effect of that is -1xx-1=1, so you have (correctly) not changed the value of that 3rd fraction.
In other words, the 3rd fraction was modified as follows:
1/((3b-x)(x-b)) = 1/((-1xx(3b-x))(-1xx(x-b))) = 1/((x-3b)(b-x))
I usually prefer to multiply top and bottom of a fraction by the same number. So I would have written:
1/((3b-x)(x-b)) = (-1)/(-1xx(3b-x)) xx (-1)/(-1xx(x-b)) = (+1)/((x-3b)(b-x))
It gives us the same result, of course.
Hope it helps.
X
Please use the math input system. It's easy, and others can read your question much more easily.
Your first answer is correct, but some of your explanation is not quite right (which is maybe where the confusion has arisen).
For this problem we need to remember that in general,
-(x-y) = y-x.
That's what you are using on the 3rd fraction, but you are actually doing it twice.
The first use was:
-(x-3b) = 3b-x
The second use was:
-(b-x) = x-b
So you are actually multiplying by -1 twice. The next effect of that is -1xx-1=1, so you have (correctly) not changed the value of that 3rd fraction.
In other words, the 3rd fraction was modified as follows:
1/((3b-x)(x-b)) = 1/((-1xx(3b-x))(-1xx(x-b))) = 1/((x-3b)(b-x))
I usually prefer to multiply top and bottom of a fraction by the same number. So I would have written:
1/((3b-x)(x-b)) = (-1)/(-1xx(3b-x)) xx (-1)/(-1xx(x-b)) = (+1)/((x-3b)(b-x))
It gives us the same result, of course.
Hope it helps.
## Re: Fraction algebra
It helps! Thank you.
X
It helps! Thank you. | 2018-01-22T02:07:51 | {
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http://math.stackexchange.com/questions/245397/can-i-execute-this-division-at-this-point/245403 | # Can I execute this division at this point?
I realize this is basic, but this little doubt has been around with me for quite a while:
I have this:
$$\frac{(2n+3)n+1}{(2n+1)(2n+3)}$$
I need it to end up with this shape:
$$\frac{n+1}{2n+3}$$
At first, I thought "well, I simply remove the $(2n+3)$ from the numerator and from the denominator and done!", but I would like to know if there are any other steps in the middle I am "jumping" by doing so.
Basically, I would like to do it as "slowly" as possible (because my professor will want to during the tests...). Is there any small step I am missing?
-
@WillHunting: Ah! I see! After seeing the answers below, I realized that.. you're right. Thank you :) – Zol Tun Kul Nov 27 '12 at 2:19
$\quad\dfrac{(2n+3)(n+1)}{(2n+1)(2n+3)}\quad$ can be simplified to $\quad\dfrac{n+1}{2n+1}\quad$ by canceling out the common factor of
$(2n + 3)$ in the numerator and denominator.
But then you need to check for what happens at $\;n = -\dfrac{3}{2}$.
At $n = -\dfrac{3}{2}$, the original fraction is undefined (division by zero). You lose that information when simplifying.
For $\quad\dfrac{(2n+3)n+1}{(2n+1)(2n+3)},\quad$ multiply out the numerator: $(2n+3)n+1 = 2n^2 + 3n +1$. You can then factor the numerator to obtain $(2n+1)(n+1)$. This gives you:
$$\frac{(2n+1)(n+1)}{(2n+1)(2n+3)} = \frac{n+1}{2n+3}.$$
But in this case, you need to consider that $2n + 1 = 0 \rightarrow n = -\dfrac12$, where the original fraction is undefined.
-
+1 for the $\frac{-3}{2}$ undefinition thing. Didn't think about it (not sure if it will really destroy my proof, but it is nice to keep in mind). – Zol Tun Kul Nov 27 '12 at 2:36
It is good to keep in mind! You need to rule out any values of $n$, in this case, for which the denominator evaluates to zero. – amWhy Nov 27 '12 at 2:38
This will take you there $$(2n+3)n + 1 = 2n^2 + 3n + 1 = (2n +1)(n+1).$$
- | 2015-07-01T17:00:12 | {
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https://www.physicsforums.com/threads/when-integrating-2x-4x-2-2-i-get-two-different-integrals.766650/ | # When Integrating (2x)/(4x^(2)+2) I get two different integrals ?
1. Aug 19, 2014
### FurryLemon
Hi
So lets have ∫(2x)/(4x^(2)+2) dx
Without factorising the 2 from the denominator, I integrate and I get
1/4*ln(4x^(2)+2)+c which makes sense as when I differentiate it I get the original derivative.
BUT when I factor the 2 from the denominator I have
2x/[2(2x^(2)+1)] simplify it down = x/(2x^(2)+1) which is the same as (2x)/(4x^(2)+2)
Now when I integrate ∫x/(2x^(2)+1)dx I get 1/4*ln(2x^(2)+1)+c
which is obviously different from the first integral. Why? because when I simplify it by factorisation its the same thing so Why is it different?
Please, any help would be appreciated.
Thanks.
2. Aug 19, 2014
### ShayanJ
The point is, $\ln{ab}=\ln{a}+\ln{b}$. Now if you assume $c=\frac 1 4 \ln{2}$ in the second solution, you'll get the first. So they differ only by an additive constant which makes them equivalent.
Last edited: Aug 19, 2014
3. Aug 19, 2014
### FurryLemon
Ok, thanks
Ideally its better to not factor anything out from the denominator then, unless you think you can get partial fractions that is.
Last edited: Aug 19, 2014
4. Aug 19, 2014
### Zondrina
Indeed, this can be observed from something as simple as:
$$\int x-1 \space dx$$
Making a substitution $u = x - 1$ gives $du = dx$. Hence:
$$\int x-1 \space dx = \frac{(x-1)^2}{2} + C$$
Now what if you simply integrated by using some integral properties instead of a substitution?
$$\int x-1 \space dx = \int x \space dx - \int 1 \space dx = \frac{x^2}{2} - x + K$$
The two answers look nothing alike, but they're exactly the same! In fact, you can show they differ by a constant. Taking the answer for the substitution method and massaging it a bit gives you:
$$\frac{(x-1)^2}{2} + C = \frac{x^2 - 2x + 1}{2} + C = \frac{x^2}{2} - x + \frac{1}{2} + C$$
It is clear by observation with the non substitution answer that $K = \frac{1}{2} + C$ so $K$ and $C$ differ by $\frac{1}{2}$.
5. Aug 19, 2014
### pasmith
The other classic example is $$\int \cos \theta \sin\theta\,d\theta = -\frac12\cos^2\theta + C = \frac12\sin^2\theta - \frac12 + C = -\frac14 \cos(2\theta) + C - \frac14.$$ | 2017-08-18T18:56:53 | {
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