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http://math.stackexchange.com/questions/65724/prove-that-2x2-3xy-2y2-geq-0/65732 | # Prove that $2x^2 - 3xy + 2y^2 \geq 0$
Prove that $2x^2 - 3xy + 2y^2 \geq 0$.
This is a question on my homework assignment, but I don't even know where to begin as it is not factorable and that is my first instinct when I see this type of problem. Can I get a tip on where to begin at least? Thanks!
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I find the question hard to read. The way it looks, we have $2^2-3xy +2y^2$. This is negative at for example $(x,y)=(3,1)$, and many other places. But my guess is that the "$x$" is really supposed to be $\times$, and you are being asked about $2^2-3y+2y^2$. To tackle this, you could complete the square. But I do not want to give details until the question is clarified. – André Nicolas Sep 19 '11 at 6:54
Sorry, should have been 2x^2. My bad! And sorry for not knowing how to make everything look pretty either... I looked in the FAQ and couldn't find the syntax for it. Didn't know where else to look. – Matt Nashra Sep 19 '11 at 6:58
Matt: You can learn about the LaTeX markup language by seeing Wikipedia or playing around in codecogs' sandbox. You can right click > "show source" any equation on Math.SE to view how the author typeset the equation. Lastly, here we put markup in between dollar signs, e.g. $\LaTeX\to$$\to\LaTeX$ (or two dollar signs for a displaystyle equation). – anon Sep 19 '11 at 7:05
Also, do you know about a technique called "completing the square"? If you interpret the $y$ as a constant and the expression as a polynomial in $x$ then you might be able to c.t.s. and get a viable expression in the end... – anon Sep 19 '11 at 7:08
Just compute the discriminant of $ax^2 + bx + c=0$ with $a=2,b=-3y,c=2y^2$ – Robert William Hanks Sep 19 '11 at 10:17
We are asked to show that $2x^2-3xy+2y^2 \ge 0$, presumably for all real $x$ and $y$.
The standard approach is to complete the square. We will do it in an ugly mechanical way. Note that $$2x^2-3xy+2y^2=2\left(x^2-\frac{3}{2}xy+y^2\right).$$
So it is enough to show that $x^2-\frac{3}{2}xy+y^2 \ge 0.$ Complete the square. We get $$x^2-\frac{3}{2}xy+y^2=\left(x-\frac{3}{4}y\right)^2 -\frac{9}{16}y^2+y^2=\left(x-\frac{3}{4}y\right)^2 +\frac{7}{16}y^2.$$
Now we are finished. The expression on the right is obviously non-negative, since both $(x-(3/4)y)^2$ and $(7/16)y^2$ are non-negative. Indeed, "almost always" the expression is $>0$. The only way it can be $0$ is if both $y$ and $x-(3/4)y$ are $0$, that is, if $x$ and $y$ are both $0$.
Comment: This looks like a "two-variable" problem, but it really isn't. Note that our inequality is clearly true if $y=0$. So from now on we can assume that $y \ne 0$. For $y \ne 0$, our inequality is equivalent to $$\frac{2x^2-3xy+2y^2}{y^2} \ge 0,$$ which in turn is equivalent to showing that $$2z^2-3z+2 \ge 0,$$ where $z=x/y$. Now we are down to a one-variable problem. One standard approach is (again) by completing the square, but there are other ways to tackle the problem. For instance, note that $2z^2-3z+2$ is certainly $>0$ sometimes. In order to be $<0$ sometimes, it would have to be $0$ for some $z$. But it is easy to verify using the Quadratic Formula that the equation $2z^2-3z+2=0$ has no real solutions.
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If you could show this is the sum of squares then you would be done. For example, you know $x^2-2xy+y^2= (x-y)^2 \ge 0$.
The problem term in the question's expression is $-3xy$, which looks a little like the $-2xy$ I just mentioned. So separate out $\frac{3}{2}$ times the expression above, to give $$\frac{3}{2}(x^2-2xy+y^2) + \frac{1}{2} x^2 +\frac{1}{2} y^2$$
and since this is the sum of positive fractions of squares, it is non-negative.
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Nice solution, preserving the original symmetry. – André Nicolas Sep 19 '11 at 12:37
Factor $2y^2$ in order to get $E(x,y)=2x^2-3xy+2y^2=2y^2((\frac{x}{y})^2-\frac{3}{2}\frac{x}{y}+1)$. The case $y=0$ is solved because $E(x,0)=2x^2\geq 0$. The problem is reduced to determining the sign of $F(t)=t^2-\frac{3}{2}t+1$ avec $t=\frac{x}{y}$. What is the discriminant of F?
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You don't really need to factor anything, just compute the discriminant of $ax^2 + bx + c=0$ with $a=2,b=-3y,c=2y^2$ – Robert William Hanks Sep 19 '11 at 10:16
Let's consider two cases, (i) $xy<0$, (ii) $xy\ge0$.
(i) In this case $-3xy>0$, and since $x^2, y^2 > 0$, we have $2x^2-3xy+y^2 > 0$.
(ii) In this case $(2x^2-3xy+2y^2) \ge (2x^2-3xy+2y^2)-xy = 2(x^2-2xy+y^2) = 2(x-y)^2 \ge 0$.
As cases (i) and (ii) are exhaustive (for real x,y), we have $2x^2-3xy+2y^2 \ge 0$.
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https://math.stackexchange.com/questions/2003532/does-the-collection-of-all-finite-subsets-of-the-real-numbers-have-the-finite-in | # Does the collection of all finite subsets of the real numbers have the Finite Intersection Property?
Does F, the collection of all finite subsets of the real numbers, have the finite intersection property?
I'm not sure how to approach this questions but my intuition tells me that F does not have the finite intersection property. So if one set is {1, 2, 3, 4} and another set is {5, 6, 7, 8} the intersection of these 2 sets would be empty thus F does not have the FIP. Is this logic right?
Also, I am trying to figure out:
Does G, the collection of complements of the subsets in F (the collection of all co-finite subsets of R), have the finite intersection property?
It makes sense to me that the intersection of co-finite subsets would be non-empty because all of the sets would be "bounded" by negative infinity and positive infinity if that makes sense.. but I don't know how to explain or prove that logically.
• Hint: the empty subset is in particular a finite subset of the real numbers. – Mees de Vries Nov 7 '16 at 13:19
• @MeesdeVries so F doesn't have the FIP since any set intersect with the empty set is empty? – cshooo Nov 7 '16 at 13:22
• Can you define the finite intersection property you have in mind? – Mikhail Katz Nov 7 '16 at 13:23
• You're right that $F$ does not have the finite intersection property. Showing that $G$ does have it is more interesting. Hint: You are working with complements and intersections, so recall De Morgan's law. (Your current plan might lead you astray: the collection of all unbounded sets does not have the finite intersection property. Consider the set of integers and the set of non-integers.) – Nate Eldredge Nov 7 '16 at 13:24
As smartly pointed out in the comments, it is trivial to show that $F$ does not have the $FIP$.
Now what about $G$? Let us assume that $G$ does not have the $FIP$ and arrive at a contradiction:
If $G$ does not have the $FIP$ then there is a finite $G' \subset G$ for which
$$\bigcap_{g \in G'} g = \emptyset$$
But given that $G$ is composed of the complements of sets in $F$, we get: $\forall g\in G'\ \ \exists f_g\in F:\ g = \mathbb{R} - f_g$
Rewriting the intersection, you get
$$\bigcap_{g \in G'} (\mathbb{R} - f_g) = \emptyset$$
Now if there is no number $x$ that appears in all $g$ then for every number $x \in \mathbb{R}$ there is some $g_x$ such that $x \notin g_x \iff x \notin (\mathbb{R} - f_{g_x}) \iff x \in f_{g_x}$. But since that is true for any $x$, then all the real numbers are in a finite union of finite sets, which is itself finite! So we would be proving that $\mathbb{R}$ is finite. But we know that $\mathbb{R}$ isn't finite so we arrived at a contradition and therefore $G$ has the $FIP$. | 2019-08-23T22:57:47 | {
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https://math.stackexchange.com/questions/2888221/does-the-gauss-bonnet-theorem-apply-to-non-orientable-surfaces | # Does the Gauss-Bonnet theorem apply to non-orientable surfaces?
I found statements of the Gauss-Bonnet theorem here, here, here, here, here, here, here, and here. None of them require that the surface be orientable. However, Ted Shifrin claims in a comment to this question that the Gauss-Bonnet theorem actually only applies to orientable surfaces. Are these sources all incorrect?
• Shocking. I can't load several of those. Interesting that none of your sources is a published respected textbook. :) You know what my answer to your question is. I'm curious to see what other feedback you get :) – Ted Shifrin Aug 19 '18 at 23:49
• @TedShifrin The reason that none of them is a published respected textbook is that I just Googled it, and the complete text of published respected textbooks is rarely available online. For what it's worth, I suspect you're right, I'd just like to get further info. – tparker Aug 19 '18 at 23:51
• I am not criticizing you or your curiosity. (Indeed, I upvoted.) I am truly shocked, particularly because Herman Gluck, for example, is usually very careful :) I'm not surprised by errors in Wiki or other such sources. I'll give you this little challenge: Take one of the standard embeddings of the Möbius strip in $\Bbb R^3$ [I can give you a formula if you can't find one]. Find the geodesic curvature of the boundary. The Euler characteristic is $0$. See if you can make sense of the curvature integral and see if it works out. I'm actually curious to see the computation. :) – Ted Shifrin Aug 19 '18 at 23:56
• @TedShifrin I think a non-orientable Riemannian manifold still has a volume density (instead of a volume form), so the curvature integral still makes sense. No? – Seub Sep 12 '18 at 0:19
• @tparker: I've resolved the paradox. Consider the matter settled :P – Ted Shifrin Sep 14 '18 at 23:42
Orientability is not needed. Indeed, one can deduce unorientable version of Gauss-Bonnet theorem from the orientable one :
Given a (compact) nonorientable surface, say $M$, with metric $g$, consider its orientable double cover $\widetilde{M}$. The metric $g$ is naturally pulled back to a metric $\widetilde{g}$ on $\widetilde{M}$; i.e. locally, $\widetilde{M}$ is isometric to $M$. Then it is easy to see that the usual Gauss-Bonnet theorem on the orientable double cover implies the Gauss-Bonnet theorem on the nonorientable surface, because $2\pi\chi(M) = \pi\chi(\widetilde{M}) = \frac{1}{2}\int_{\widetilde{M}}{K} = \int_{M}K$, where $K$ denotes the Gaussian curvature.
One might want a version of Gauss-Bonnet theorem for surfaces with boundary. Indeed, the argument above can be applied to nonorientable surfaces with boundary. Let $M$ be a nonorientable surface with boundary $\partial M$. Then correspondingly its has an orientable double cover $\widetilde{M}$ with boundary $\partial \widetilde{M}$. As before, this double cover is locally isometric to $M$. Following the notation of Wikipedia, we have the Gauss-Bonnet theorem on $\widetilde{M}$ : $$\int_{\widetilde{M}}K + \int_{\partial\widetilde{M}}k_g = 2\pi \chi(\widetilde{M})$$ Now observe that each term are twice the corresponding term for $M$. In particular, $\int_{\partial\widetilde{M}}k_g = 2\int_{\partial M}k_g$ just because it is a 2-to-1 locally isometric double covering. As a result, we get $$\int_{M}K + \int_{\partial M}k_g = 2\pi \chi(M)$$ There is nothing special for nonorientable surfaces!
Let me give you an example, a Möbius strip. Its Euler Characteristic is $0$. The most convenient metric on the Möbius strip is a flat metric; such Möbius strip can be realized as a quotient of a flat strip with parallel geodesic boundaries. With such a metric, $K=0$ and $k_g=0$, so the LHS of the Gauss-Bonnet theorem is $0$ as expected.
In general, for any metric on the Möbius strip, the Gauss-Bonnet theorem should hold just as well, because the LHS remains constant under smooth deformation of metric. (Any deformation is a composition of local deformations, and for local deformations, it is a corollary of the orientable version of Gauss-Bonnet theorem.)
Another easy proof is to cut your surface into small orientable pieces. It is not important whether your original surface is orientable or not. Gauss-Bonnet theorem holds for individual pieces, and when you glue them back, boundary terms corresponding to seams cancel out. In this way it is easy to see that orientability is not important at all in Gauss-Bonnet theorem for surfaces.
• I haven't worked it out, but I think the line integrals on the boundary cancel, rather than doubling. I did work out the Möbius strip with explicit computation (the exercise I posed to you) and the theorem most definitely did not check. [I think the boundaryless case is, in fact, convincing with the area density in there.] I also still stand by my complaints about the Pfaffian in higher dimensions. And Chern certainly uses orientability in his proofs. – Ted Shifrin Sep 12 '18 at 19:38
• @TedShifrin I'm not sure whether the boundary terms should add up or cancel, but let's imagine that they cancel, that would mean that $\int_\tilde{M} \tilde{K} = 2\pi \chi(\tilde{M})$, hence $\int_M K = 2\pi \chi(M)$. This would be a version of Gauss-Bonnet for non-orientable surfaces: no boundary term, even if the surface has boundary. Interesting? – Seub Sep 13 '18 at 11:26
• @Sunghyuk Park: Your addition with the Möbius strip example is nice, but it is also compatible with Ted Shifrin's hypothesis that the boundary term $\int_\tilde{\partial M} \tilde{k}$ could be equal to zero instead of $2 \int_{\partial M} k$ – Seub Sep 13 '18 at 11:43
• @TedShifrin@Seub The boundary term cannot cancel out. I know it can be a little confusing at first, but observe that even in case of an orientable surface, the boundary term does not depend on the orientation of the surface; the integral should be understood as a density integral, rather than an integral of a differential form. Intuitively, the boundary term measures the amount the boundary is curved inward (or outward). It is irrelevant to orientation. – Henry Sep 13 '18 at 12:27
• @TedShifrin Also observe that a collar neighborhood of the boundary of a surface (not necessarily orientable) is always orientable. Hence there's no way the line integral "sees" orientability of the surface. – Henry Sep 14 '18 at 3:42
Following the OP's request, I'm posting the details of my computations for the explicit embedding of the Möbius strip $$M$$ in $$\Bbb R^3$$. The orientation of (halves) of the boundary curve turns out to be the crucial matter, as I'd suspected. Although the theoretical arguments are compelling, I remain confused about what's wrong with the following computations. I can now vouch that the numerics are correct.
Consider the parametrization $$x(u,v) = \big((2+v\sin(u/2))\cos u,(2+v\sin(u/2))\sin u,v\cos(u/2)\big), \quad 0\le u\le 2\pi, -1\le v\le 1.$$ Note that $$B = x(0,1) = x(2\pi,-1)$$ and $$C=x(0,-1)=x(0,1)$$. As you can check, this is an orthogonal parametrization, and the first fundamental form has coefficients $$E = \|x_u\|^2 = 4+\frac34 v^2 - \frac12 v^2\cos u + 4v\sin(u/2)$$ and $$G=\|x_v\|^2 = 1$$. I'm now going to use the standard formulas for $$K$$ and $$\kappa_g$$ in an orthogonal parametrization (see, e.g., my text, pp. 60 and 81): \begin{align*} K &= -\frac1{2\sqrt{EG}}\left(\Big(\frac{E_v}{\sqrt{EG}}\Big)_v + \Big(\frac{G_u}{\sqrt{EG}}\Big)_u\right) \\ \kappa_g &= \frac1{2\sqrt{EG}}\big({-}E_v u'(s) + G_u v'(s)\big)+\theta'(s), \end{align*} the latter for an arclength parametrization of the curve. Here $$\theta$$ is the angle the curve makes with $$x_u$$ at each point. In our case $$\theta'=0$$ everywhere. Note, also, that (forgetting about orientation issues for the moment) \begin{align*} \int_M K\,dA &= \iint_{[0,2\pi]\times [-1,1]} -\frac1{2\sqrt{EG}}\left(\Big(\frac{E_v}{\sqrt{EG}}\Big)_v + \Big(\frac{G_u}{\sqrt{EG}}\Big)_u\right)\underbrace{\sqrt{EG}\,du\,dv}_{dA} \\ &= -\frac12\int_0^{2\pi}\int_{-1}^1 \Big(\frac{E_v}{\sqrt E}\Big)_v\,dv\,du \\ &= -\frac12\int_0^{2\pi} \Big(\frac{E_v}{\sqrt E}\Big)\Big|_{v=1} - \Big(\frac{E_v}{\sqrt E}\Big)\Big|_{v=-1} \,du. \end{align*} Regarding the orientation issue, surely we'll agree that $$|dA|$$ should agree with this $$dA$$ if we remove one ruling of the surface (say the ruling from $$B$$ to $$C$$). So this integral should be the density integral.
Now, $$E_v/\sqrt E$$ is rather a mess, but, using Mathematica to do the numerical integration, we find that this integral is (approximately) $$-1.97$$. [The most basic check is that it's negative, as we have a non-developable ruled surface.]
We can now use Mathematica to evaluate the geodesic curvature integrals. We note that $$ds = \frac{ds}{du}du$$, so $$\kappa_g\,ds = \big({-}\frac12 E_v/\sqrt E\big)u'(s)\,ds =\big({-}\frac12 E_v/\sqrt E\big)\,du$$. On half the boundary circle, going from $$B$$ to $$C$$, $$-\frac12\int_0^{2\pi} \frac{E_v}{\sqrt E}\Big|_{v=1}du \approx -4.53,$$ and on the other half, going from $$C$$ to $$B$$, we have $$-\frac12\int_0^{2\pi} \frac{E_v}{\sqrt E}\Big|_{v=-1}du \approx -2.56.$$ So $$\int_M K\,dA + \int_{\partial M} \kappa_g\,ds \approx -9.06$$ (certainly not $$0$$, nor, indeed, an integer multiple of $$2\pi$$).
To double-check this computation, let's remove a tiny bit of our Möbius strip, say the region corresponding to $$0\le u\le \varepsilon$$. This leaves us an oriented surface, for sure. Its boundary has two extra pieces, $$u=0$$ (oriented downward) and $$u=\varepsilon$$ (oriented upward); these have no contribution, regardless of $$\varepsilon$$, since the $$v$$-curves are line segments and have no geodesic curvature. The main discrepancy, however, is the reversed orientation on the segment $$v=1$$. Indeed, for $$\varepsilon$$ very small, the $$v=1$$ integral is now approximately $$+4.53$$, and — mirabile dictu — note that $$-1.97 + 4.53 - 2.56 = 0,$$ as it should! (The exterior angles of the "rectangle" contribute the $$2\pi\chi = 2\pi$$.) But I emphasize that when we've removed a bit of the Möbius strip to make an oriented creature, we do not have (almost) the same boundary curve as the Möbius strip. This difference, as far as I'm concerned, is what messes up the Gauss-Bonnet Theorem. One more comment: The definition of $$\kappa_g$$ (as $$\kappa\mathbf N\cdot(\mathbf n\times\mathbf T)$$, where $$\mathbf T,\mathbf N$$ are the Frenet frame of the curve and $$\mathbf n$$ is the surface normal) makes it clear that when we have an oriented surface with boundary, the sign of $$\kappa_g$$ does not change if we reverse the orientation of the surface; for when we do, we change both $$\mathbf n$$ and $$\mathbf T$$ by a sign. But, if we interpreted $$ds$$ in this integral as a measure, as @SunghyukPark suggests, it surely must be consistent along the boundary circle of the Möbius strip, so we cannot just switch the sign of half the line integral. ...
EDIT: OK, I believe I've figured it out, much to my chagrin. We need to think about geodesic curvature intrinsically (as one does in a fancier proof of the Gauss-Bonnet Theorem). If $$e_1$$ is the unit tangent vector along $$\partial M$$ and $$e_2$$ is the inward-pointing normal to $$\partial M$$ in $$M$$, then, by definition, $$\kappa_g = \nabla_{e_1}e_1\cdot e_2$$. So, in fact, on the upper edge $$v=1$$ of our parametrizing rectangle, the formula for $$\kappa_g$$ we used earlier is off by a sign, and correcting this is equivalent to reversing the orientation on that upper integral. The correct values are $$\int_M K|dA| = -1.97 \qquad\text{and}\qquad \int_{\partial M}\kappa_g ds = +4.53-2.56 = +1.97.$$ So the sum is, in fact, $$2\pi\chi(M) = 0$$, as desired. I feel better now. :)
Here's the graph of the geodesic curvature (upper upper, lower lower):
• Thank you for sharing your computations. Just a small remark: according to wikipedia, the geodesic curvature of a curve $\gamma(t)$ is just the curvature of $\gamma(t)$ in $S$, where $S$ is equipped with the induced Riemannian metric: $k_g =\Vert \frac{\nabla^2}{ds^2} \gamma(s) \Vert$ for a unit speed curve, where $\nabla$ is the Levi-Civita connection of the induced Riemannian metric. That seems to be a pretty good intrinsic definition. – Seub Sep 15 '18 at 1:00
• @Seub: I definitely do not like that. $\kappa_g$ has an intrinsic sign. It shouldn't always be positive. – Ted Shifrin Sep 15 '18 at 1:05
• I see, I understand now. – Seub Sep 15 '18 at 1:18
This is more of an extended comment than a complete answer, but hope that it will close the question.
Applying a little Googlomagic (namely, searching for "gauss-bonnet non-orientable") it is possible to find out the following:
• a paper by R. Palais's A Topological Gauss-Bonnet Theorem, J.Diff.Geom. 13 (1978) 385-398, where he mentions in passing that the Gauss-Bonnet theorem is easily generalized to the non-orientable case by considering measures.
• an answer to this question with a feasible proof of the Gauss-Bonnet for the non-orientable case;
• and many more interesting things, of course :)
On a side note, the Pfaffian has nothing to do with the orientability, bit rather with the dimension: it is defined in even dimensions (thus, in dimension 2 as well). See, maybe, here for the details.
• Your final link makes it clear why you need an oriented (even rank) bundle. – Ted Shifrin Sep 12 '18 at 16:14
• Thanks for the Palais reference. It's conclusive. – Ted Shifrin Sep 13 '18 at 18:00 | 2019-04-21T02:30:14 | {
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http://math.stackexchange.com/questions/684036/factoring-a-hard-polynomial/684242 | # Factoring a hard polynomial
This might seem like a basic question but I want a systematic way to factor the following polynomial:
$$n^4+6n^3+11n^2+6n+1.$$
I know the answer but I am having a difficult time factoring this polynomial properly. (It should be $(n^2 + 3n + 1)^2$). Thank you and have a great day!
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You may be interested to compare Wikipedia's article with the Answers below to match up the techniques described in that article. – hardmath Feb 21 '14 at 13:18
In general there's no easy way. But there are a lot of tricks you can use, and sometimes they work.
For this problem, the rational root theorem tells us that if there is a rational root, it must be $\pm 1$. Evidently it isn't either of these, so there are no rational roots, and therefore no factors of the form $n-q$.
So we are looking for two second-degree factors: $$n^4+6n^3+11n^2+6n+1 = (an^2+bn+c)(pn^2+qn+r)$$
But we can see from the coefficient of 1 on the $n^4$ term that $a=p=1$. From the constant term of 1 we have either $c=r=1$ or $c=r=-1$, so it's really
$$n^4+6n^3+11n^2+6n+1 = (n^2+bn\pm1)(n^2+qn\pm1).$$
Equating the coefficients of the $n^3$ terms on each side, we see that $b+q=6$.
Equating the coefficients of the $n^2$ terms on each side, we see that either $bq+2=11$ (if we take the $\pm$ signs to be $+$) or that $bq-2 = 11$ (if the $\pm$ signs are $-$).
The equations $b+q=6$ and $bq-2 = 11$ certainly have no solution in integers, since one of $b$ or $q$ must be 13. So we try the other pair of equations, $b+q=6$ and $bq+2 = 11$. Here the solution $b=q=3$ is obvious, and we are done, except to check the answer. We must do this, because we have completely ignored the coefficient of the $n$ term.
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+1, good analysis. – Rustyn Feb 20 '14 at 21:42
Great solution! Thank you very much @MJD :) – InsigMath Feb 20 '14 at 21:47
You are welcome. I am glad you are pleased. – MJD Feb 20 '14 at 21:47
It is fair to take $a=p=1$, but then it doesn't seem fair to assume $c=r=1$. It works this time. If we are factoring over the integers, we can get $c=r=\pm1$ – Ross Millikan Feb 20 '14 at 21:50
@Ross Thank you for pointing this out. – MJD Feb 20 '14 at 22:01
I was thinking along the same lines as TZakrevskiy but with a slight twist. Once you've become convinced that $P(n)=n^4+6n^3+11n^2+6n+1$ is the square of a quadratic (because its values for small $n$ are all squares), try computing
$$P(10)=10000+6000+1100+60+1=17161$$
and then take its square root:
$$\sqrt{17161}=131=1\cdot10^2+3\cdot10+1$$
which suggests the (correct) factorization
$$P(n)=(n^2+3n+1)^2$$
If you want to be extra sure, look at
$$\sqrt{100^4+6\cdot100^3+11\cdot100^2+6\cdot100+1}=10301$$
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+1 as I've never seen this kind of approach used for factorisation before. – Keith Feb 21 '14 at 2:30
+1, this is so brilliant! – Silviu Burcea Feb 21 '14 at 4:51
Exploit the symmetry of the palindromic polynomial to factorise as follows,
\begin{align*} n^4 + 6n^3 + 11n^2 + 6n + 1 &= n^2\left( n^2 + \frac{1}{n^2} + 6n + \frac{6}{n} + 11 \right)\\ &= n^2\left( (n + 1/n)^2 + 6(n + 1/n) + 9 \vphantom{\frac{1}{n^2}} \right)\\ &= n^2\left( n + 1/n + 3 \right)^2\\ &= \left( n^2 + 3n + 1 \right)^2,\\ \end{align*}
having essentially substituted $k = n + \frac{1}{n}$ and used $k^2 - 2 = n^2 + \frac{1}{n^2}$.
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There's another approach: check that for, say $n=\pm 1,\, 0,\,\pm 2$ your polynomial is a perfect square:$$\begin{cases}25,&n=1\\1,&n=-1\\1,&n=0\\1,&n=-2\\121,&n=2.\end{cases}$$ Therefore, in five points a polynomial of fourth degree is a perfect square - you can hope to find a polynomial of the second degree that satisfies $$\begin{cases}\pm 5,&n=1\\\pm 1,&n=-1\\\pm 1,&n=0\\\pm 1,&n=-2\\\pm 11,&n=2\end{cases}.$$
This is an overdetermined system, but, lickily, it has a solution: wlog, we take the candidate as $n^2+2bn+c$ (because of the coefficient at $n^4$). The value in $n=-1$ gives $1-2b+c = \pm 1$, and in $n=0$ $c=\pm 1$. By checking $4$ solutions of the obtained $4$ linear systems we can eliminate all but one by testing against remaining points $n$: we obtain $c=1$, $b=3/2$. Finally we check that indeed $(n^2+3n+1)^2=n^4+6n^3+11n^2+6n+1$.
-
Just to add an alternative: From an algorithmic point of view (i.e., to write a program that performs factorization without any intelligence guiding the process), one of the fastest things to check and therefore typically the first step is squarefree factorization, by taking gcds of the polynomial and its derivative(s).
$$\begin{eqnarray*} f&=&n^4+6n^3+11n^2+6n+1\\ f'&=& 4n^3+18n^2+22+6\\ \gcd(f,f')&=& n^2+3n+1 \end{eqnarray*}$$
Of course, working by hand and with a human brain behind it, the cost analysis may be different from that for a computer and computing gcds may not be as attractive as it is for machines.
-
Right, in this case the GCD of the polynomial and its formal derivative gives the complete irreducible factorization. More generally it identifies any repeated factors quickly. – hardmath Feb 21 '14 at 13:10
we can reorganize the terms of $n^4+6n^3+11n^2+6n+1$ like $$(n^4+2n^2+1)+(6n^3+6n)+9n^2$$ $$(n^2+1)^2+6n(n^2+1)+(3n)^2=(n^2+3n+1)^2$$ then we are done
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https://stats.stackexchange.com/questions/360478/is-there-a-matrix-operation-that-can-count-elements-in-vector-pairs | # Is there a (matrix) operation that can count elements in vector pairs?
For two vectors $$x \in \{0, 1, 2\}^{n}$$ and $$y \in \{0, 1, 2\}^{n},$$ I need to generate a matrix $$C\in \mathcal{R}^{3\times3}$$ where $$C_{i,j}$$ equals the number of indexes $$t$$ where $$x[t]=i$$ and $$y[t]=j$$.
I know I can always write a program to iteratively check and count, but I wonder if there is any way that I can approximate this with matrix operations?
Or, is there any way that I can approximate this as an optimization problem?
• I don't quite understand what you're asking. Are you looking for mathematical notation for how to express $C$ given $x$ and $y$? Or do you want a way to calculate it? Also, I don't understand why approximation and optimization are mentioned, as they don't really seem to be appropriate/necessary here given what you've said so far. – user20160 Aug 3 '18 at 9:04
• @user20160 I'm looking for how to calculate it (other than iteratively checking and counting). Since there is probably no way to calculate it directly, approximation through some optimization is also fine. – Haohan Wang Aug 3 '18 at 13:41
• Could you explain what you mean by "iteratively check and count"? After all, all you're doing is tabulating the nine possible values of $(x[t], y[t]).$ No iteration or checking is ordinarily needed for such a straightforward and simple calculation. – whuber Aug 3 '18 at 15:00
## 1 Answer
It comes down to what you consider a "matrix operation." Computer programming platforms usually admit a larger gamut of possibilities than linear algebra, for instance. Let's focus on the former. (There are solutions using the most basic linear algebraic operations, but they are a bit more involved: the indicator function can be implemented as a polynomial.)
One solution is afforded by a generalized outer product. Given any binary function $$f$$ and two vectors $$\mathbf{x}$$ and $$\mathbf{y},$$ the outer product $$\mathbf{x}\, {\otimes}_f\ \mathbf{y}$$ is the matrix
$$\left(\mathbf{x}\, {\otimes}_f\ \mathbf{y}\right)_{ij} = f(x_i, y_j).$$
Let $$s$$ be any scalar (a potential component of a vector). The indicator of $$s,$$ written $$I(,\, s),$$ is the function that returns the value $$1$$ when applied to $$s$$ and otherwise returns $$0:$$
$$I(t,s) = \cases{1 & t=s \\0 & \text{otherwise.}}$$
Suppose the $$m$$-vector $$\mathbf{x}$$ has values in $$\{0,1,\ldots, n\}.$$ The outer product of $$x$$ with the vector $$\mathbf{n} = (0,1,2,\ldots,n)$$ with respect to the indicator function $$I$$ is the $$m\times n$$ matrix
$$\mathbf{x}\, {\otimes}_I\, \mathbf{n}$$
whose $$i,j$$ entry is the indicator $$I(x_i, j-1).$$ When $$\mathbf{x}$$ and $$\mathbf{y}$$ are both $$m$$-vectors, the matrix product
$$\left(\mathbf{x}\, {\otimes}_I\, \mathbf{n}\right)^\prime \left(\mathbf{y}\, {\otimes}_I\, \mathbf{n}\right) = \left(\mathbf{n}\, {\otimes}_I\, \mathbf{x}\right) \left(\mathbf{y}\, {\otimes}_I\, \mathbf{n}\right)$$
is the answer.
Proof For $$0 \le i, j\le n,$$ just calculate from the foregoing definitions that
$$\left(\left(\mathbf{x}\, {\otimes}_I\, \mathbf{n}\right)^\prime \left(\mathbf{y}\, {\otimes}_I\, \mathbf{n}\right)\right)_{i+1,j+1} = \sum_{t=1}^n I(x_t, i) I(y_t, j).$$
The terms in the sum are zero except where $$x_t=i$$ and $$y_t=j,$$ where they are equal to $$(1)(1)=1.$$ The sum therefore counts all such terms, which is exactly what is intended.
The outer products each require $$O(mn)$$ computational effort, while the multiplication of the resulting $$n\times m$$ by $$m\times n$$ matrices takes $$O(n^2m)$$ effort and therefore dominates the cost. This appears inefficient, because the same result can be obtained simply by tabulating the values by (a) initializing the final $$n\times n$$ array and (b) scanning once across both vectors simultaneously, updating the counts in the array as you go. That is only $$O(n^2+m)$$ effort.
The interest in this approach therefore would focus either on being able to apply algebraic rules to analyze the operation or to exploit built-in efficiencies on an array-oriented computing platform. At the end of this post I assess the latter possibility for R.
Here is an implementation in R.
f <- function(x, y, n=2) outer(0:n, x, ==) %*% outer(y, 0:n, ==)
As an example of its use, let's generate a pair of vectors exhibiting many possible combinations of values from $$\{0,1,2\}$$:
X <- expand.grid(x=0:2, y=0:2)
x <- unlist(mapply(function(x,i) rep(x,i), X$$x, 0:8)) y <- unlist(mapply(function(x,i) rep(x,i), X$$y, 0:8))
Here are the two vectors of $$m=36$$ components:
x 1 2 2 0 0 0 1 1 1 1 2 2 2 2 2 0 0 0 0 0 0 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2
y 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
This computation provides nice names for the output and prints it:
a <- f(x,y)
rownames(a) <- paste0("x=", 0:2)
colnames(a) <- paste0("y=", 0:2)
a
The output is
y=0 y=1 y=2
x=0 0 3 6
x=1 1 4 7
x=2 2 5 8
As a check, compare this to the output of the table function, table(x,y):
y
x 0 1 2
0 0 3 6
1 1 4 7
2 2 5 8
I was surprised to find that applying table in this instance is an order of magnitude slower than f. I increased $$n$$ until it was equal to $$10001$$ and the difference grew even worse. Apparently, the built-in efficiency of matrix operations in R really makes a difference even for this basic operation!
• This crosstabulation, as I see, amounts to multiplying Gx' Gy where Gx is the dummy set (i.e. design matrix) from variable X and Gy likewise from variable Y. But in your example the values (0,1,2) are known/specified in advance and are same for both variables, therefore, to get Gx and Gy, one don't need to explore the values by tabulation and recoding into dummies (by some function like "design(X)" or whatever it is called). – ttnphns Oct 29 '18 at 16:05
• @ttnphns Yes, but a smart algorithm will scan once over the input vectors at $O(n)$ cost to determine which values to tabulate, so that makes no difference asymptotically. If arbitrary values are to be handled then they will have to be hashed into an associative array at a constant cost of $O(\log n),$ but that cannot wholly explain the consistent inferiority of table compared to f. With a bit of work using table, though, one can achieve good performance: n<-2; matrix(tabulate(1 + x + (n+1)*y, nbins=(n+1)^2), n+1) – whuber Oct 29 '18 at 16:21 | 2019-10-23T03:45:54 | {
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https://math.stackexchange.com/questions/1190922/center-of-mass-of-right-angle-trapezoid | # Center of mass of Right angle Trapezoid
Given bases $a$, $b$, and the height $h$.
Get the $M(x,y)$ coordinates formula from point $O(0,0)$, where $M$ is center of mass. Wiki has a formula for $M(y) = \frac{h}{3}\frac{2a+b}{a+b}$. And I'm interested in how to find that formula (also for $M(x)$). The median $c$ is dividing the Trapezoid into 2 equal surfaces. I need to draw $m$ in right place $y$, so all 4 surfaces stay equal each other.
Here's my try:
$y(\frac{a+m}{2})=(h-y)(\frac{m+b}{2})$
$\frac{a+m}{2}=(\frac{h}{y}-1)(\frac{m+b}{2})$
$y=\frac{h(m+b)}{a+2m+b}$
on the other look
$2y(\frac{a+m}{2})=h(\frac{a+b}{2})$
$y=\frac{h(a+b)}{2(a+m)} = \frac{h(m+b)}{a+2m+b}$
From the last equality we have: $m^2 = \frac{a^2+b^2}{2}$ But I still cant get that formula back for $y$
• If line $m$ parallel to the bases of the trapezoid passes through the center of mass of the trapezoid, the lines $m$ and $c$ divide the trapezoid into four regions that are not equal in the general case. Do you want the center of mass, or do you want four equal regions? – David K Mar 15 '15 at 18:55
• I want the center of mass, I thought it is the point where all regions gets equal,Isn't it? And a few minutes ago I also found a formula for $m$, which is $m^2 = \frac{a^2+b^2}{2}$ but that didn't helped me either to find formula for $y$ – shcolf Mar 15 '15 at 20:00
• Center of mass means it balances there. Small masses far from the pivot point can balance large masses close to the pivot point, so if the object is asymmetric (like a general trapezoid) then one end can be lighter (but longer) than the other. – David K Mar 15 '15 at 20:04
• So you mean that I'm doing wrong approach, but then how can I find that Point? – shcolf Mar 15 '15 at 20:12
• I posted an answer. It took a while to write it up, because I wanted it to include a derivation. – David K Mar 15 '15 at 21:05
The formula for the distance from the base $b$ to the center of mass of a trapezoid is
$$\bar y = \frac{b+2a}{3(a+b)} h.$$
You can find this in many on-line sources such as Wolfram Mathworld.
You can prove this by integrating $\int_0^h y(b + (a-b)\frac yh) dy$ and dividing by the area of the trapezoid. But here's a derivation without calculus, using the fact that the distance from a side of a triangle to the triangle's centroid is $\frac13$ the height of the triangle.
Let $T$ be a trapezoid with bases $a$ and $b$. For the case $a < b$, extend the non-parallel sides of the trapezoid until they intersect. The base $b$ of the trapezoid and the two extended sides form a triangle $B$; the base $a$ divides this triangle into two pieces, one of which is $T$ and the other of which is a triangle which we'll call $A$.
If the height of the trapezoid $T$ is $h$, the height of $B$ is $\frac{b}{b-a}h.$ The centroid of $B$ is at a distance $\frac{b}{3(b-a)}h$ from base $b$.
But another way to find the centroid of $B$ is to balance the two figures $A$ and $T$ that compose $B$. The triangle $A$ has height $\frac{a}{b-a}h$, so its area is $\frac{a^2}{2(b-a)}h$ and its centroid is a distance $\frac{a}{3(b-a)}h + h$ from the base $b$ of the trapezoid. Trapezoid $T$ has area $\frac{a+b}{2} h$ and a centroid at the unknown distance $\bar y$ from base $b$.
To "balance" the two regions, we take a weighted average of the distance of their centroids from base $b$. The "weights" in this average are just the areas of the two regions. This weighted average is the same as the distance of the centroid of $B$ (the combined figure) from base $b$. That is,
$$\begin{eqnarray} \frac{b}{3(b-a)}h &=&\frac{\mathop{Area}(A) \cdot \left(\frac{a}{3(b-a)}h + h\right) + \mathop{Area}(T) \cdot \bar y}{\mathop{Area}(A) + \mathop{Area}(T)}\\ &=&\frac{\frac{a^2}{2(b-a)}h \cdot \left(\frac{a}{3(b-a)} + 1\right)h + \frac{a+b}{2} h \cdot \bar y}{\frac{a^2}{2(b-a)}h + \frac{a+b}{2} h}\\ \end{eqnarray}$$
Solve for $\bar y$. This looks messy, but it can be simplified if you realize that $$\frac{\mathop{Area}(T)}{\mathop{Area}(A)} = \frac{b^2 - a^2}{a^2}.$$ If you divide both the numerator and denominator on the right-hand side of the weighted average by $\mathop{Area}(A)$, you get
$$\frac{b}{3(b-a)}h = \frac{ \frac{3b - 2a}{3(b-a)} h + \frac{b^2 - a^2}{a^2} \cdot \bar y}{1 + \frac{b^2 - a^2}{a^2}}$$
After you finish collecting all the terms in $a$, $b$, and $h$ on the left side of this equation, and factor $(b-a)^2$ out of $b^3 - 3ba + 2a^3$, it all simplifies to the formula for $\bar y$ shown above.
1190922
$x=\frac{{ \begin{vmatrix}0&0&1\\a&0&1\\0&h&1\\\end{vmatrix} · (0+a+0) }+{ \begin{vmatrix}b&h&1\\0&h&1\\a&0&1\\\end{vmatrix} · (b+0+a) }}{ 3\left(\begin{vmatrix}0&0&1\\a&0&1\\0&h&1\\\end{vmatrix} + \begin{vmatrix}b&h&1\\0&h&1\\a&0&1\\\end{vmatrix}\right) }$
$y=\frac{{ \begin{vmatrix}0&0&1\\a&0&1\\0&h&1\\\end{vmatrix} · (0+0+h) }+{ \begin{vmatrix}b&h&1\\0&h&1\\a&0&1\\\end{vmatrix} · (h+h+0) }}{ 3\left(\begin{vmatrix}0&0&1\\a&0&1\\0&h&1\\\end{vmatrix} + \begin{vmatrix}b&h&1\\0&h&1\\a&0&1\\\end{vmatrix}\right) }$
• sorry, but can you explain how did you get this? – shcolf Mar 15 '15 at 16:24
• @shcolf \\ 1 divided the trapezoid into two triangles \ 2 found area of each \ 3 found centroid of each \ 4 combined using analytic geometry procedures, treating cases for x and y separately – Senex Ægypti Parvi Mar 15 '15 at 21:32
• $x=\frac{a^2+ab+b^2}{3(a+b)},y=\frac{h(a+2b)}{3(a+b)}$ – Senex Ægypti Parvi Mar 15 '15 at 21:57 | 2020-04-05T14:32:22 | {
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https://math.stackexchange.com/questions/2433924/product-of-infinite-terms-lim-n-to-infty-left-fracn1n2-3nn2n | # Product of infinite terms $\lim_{n\to\infty}\left(\frac{(n+1)(n+2)…(3n)}{n^{2n}}\right)$
Let there be a series
$$\lim_{n\rightarrow \infty}\left(\frac{(n+1)(n+2)...(3n)}{n^{2n}}\right)$$
is equal to?
For this type of problem I am unable to approach.The numerator starts from $(n+1)$ to $3n$ where as the denominator has ${n^{2n}}$ terms
• Thank for editing, this question is correct – Samar Imam Zaidi Sep 18 '17 at 3:14
One possibility that I usually do for products is to take logarithms. For any finite $n$, we have $$\ln\left(\frac{(n+1)\dots(3n)}{n^{2n}}\right)=\sum_{k=1}^{2n}\ln\left(\frac{n+k}{n}\right)=\sum_{n=1}^{2n}\ln\left(1+\frac{k}{n}\right)=n\sum_{k=1}^{2n}\frac{\ln(1+\tfrac{k}{n})}{n}$$ In the limit, the sum $$\sum_{k=1}^{2n}\frac{\ln(1+\tfrac{k}{n})}{n}$$ approaches a positive, nonzero definite integral, whereas $n$ approaches infinity. Thus the logarithm approaches infinity, so the original product does as well.
• Can be converted into Integral form using the property of Definite Integral – Samar Imam Zaidi Sep 18 '17 at 3:55
• I want to confifm with you whether the answer is $\int_{0}^{2} log(1+x)dx$ – Samar Imam Zaidi Sep 18 '17 at 4:15
• @SamarImamZaidi Yes, that is the correct integral – TomGrubb Sep 18 '17 at 4:29
There are $2n$ factors in the numerator. Of them, the first $n$ are each greater than $n$, and the next $n$ are each greater than $2n$. Thus, the numerator is greater than $n^n (2n)^n$. The denominator is $n^{2n}$ and so the ratio is greater than $2^n$, which of course tends to $\infty$ as $n\to\infty$.
• $\infty$ for $n\to\infty$ is not the answer, i presume that we can arrive by using log and definite integral concept – Samar Imam Zaidi Sep 18 '17 at 3:29
• That's what you think. – kimchi lover Sep 18 '17 at 3:34
• @SamarImamZaidi As you can see from other answers, the limit indeed is $+\infty$. Comparing with the answers posted so far, I consider this approach the most elegant, +1 from me. – Martin Sleziak Sep 18 '17 at 5:14
Hint:) With Stirling's approximation $$n!\sim\sqrt{2\pi n}\left(\dfrac{n}{e}\right)^n$$ we have $$\lim_{n\rightarrow \infty}\left(\frac{(n+1)(n+2)...3n}{n^{2n}}.\dfrac{n!}{n!} \right)=\lim_{n\rightarrow \infty}\dfrac{(3n)!}{n!n^{2n}}=\lim_{n\rightarrow \infty}\sqrt{3}\left(\dfrac{27}{e^2}\right)^n=\color{blue}{\infty}$$
Let us denote $$a_n=\frac{(n+1)(n+2)...(3n)}{n^{2n}}.$$
Then you have $$\frac{a_{n+1}}{a_n} = \frac{(3n+1)(3n+2)(3n+3)}{(n+1)^3} \left(\frac{n}{n+1}\right)^{2n}.$$
I should be relatively easy to check that $$\lim\limits_{n\to\infty} \frac{a_{n+1}}{a_n} > 1.$$ (In fact, the limit is equal to $27/e^2$ unless I missed something.)
Can you tell something about $a_n$ combining the facts that it is positive and starting from some $n_0$ you have $a_{n+1}/a_n>1+\varepsilon$ (for some $\varepsilon>0$)?
In the other words, can you show this: If $a_n>0$ for each and $\lim\limits_{n\to\infty} a_{n+1}/a_n>1$, then $\lim\limits_{n\to\infty} a_n=\infty$?
• Very elegant solution without using Stirling. +1 – Paramanand Singh Sep 18 '17 at 8:26 | 2020-07-09T09:25:10 | {
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https://math.stackexchange.com/questions/704949/how-can-i-find-the-coefficient-of-a-term-in-this-generating-function-by-using-th | # How can I find the coefficient of a term in this generating function by using the “old” method?
I'm trying to verify my answer to this problem by going back and solving the same problem using methods I've used before learning about generating functions, but I'm not quite sure how to do it in this case:
Find the coefficient of $x^{47}$ in $(x^{10}+x^{11}+\cdots+x^{25})(x+x^{2}+\cdots+x^{15})(x^{20}+\cdots+x^{45})$.
I'm looking at this problem as if it were saying
Given 47 identical objects and 3 distinct boxes, how many ways can the objects be placed into the boxes given that the first box must hold at least 10 and at most 25 objects, the second box must hold at least one and at most 15 objects, and the third box must hold at least 20 and at most 45 objects.
This can be translated to solving the equation
\begin{align} e_1+e_2+e_3=47\\10\leq e_1\leq 25\\1\leq e_2\leq 15,\\20\leq e_3\leq 45\end{align}
If there were no upper limits on the capacity of the boxes, I'd say that the solution is ${47-10-1-20+2\choose 2}={18\choose 2}$. I get this by placing 2 "walls" between the three groups to separate the one group of 18 into 3, but I don't know how to do this considering there are upper bounds. The solution I got through solving directly from the generating function, the coefficient of $x^{47}$ is
$$(-1){2\choose 2}+(-1){3\choose 2}+(1){18\choose 2}$$
so where do those extra binomial coefficients come from?
There's a nice simplification: define a related triple $(d_1, d_2, d_3)$ by \left\{ \begin{align} d_1 &= e_1 - 10 \\ d_2 &= e_2 - 1 \\ d_3 &= e_3 - 20 \end{align} \right. Then you have the system of equations and inequalities: \begin{align} d_1 + d_2 + d_3 &= 16 \\ 0 \le d_1 &\le 15 \\ 0 \le d_2 &\le 14 \\ 0 \le d_3 &\le 25 \end{align}
Without the upper bounds, you have correctly calculated $\binom{18}{2}$ possibilities. However, a small number of these have $d_1 = 16$ ($1$ way) or $15 \le d_2 \le 16$ ($3$ ways). You have to subtract these to maintain the sum of $16$.
By the way, the change from $e$ variables to $d$ variables amounts to factoring \begin{align} &(x^{10} + x^{11} + \cdots + x^{25}) (x + x^{2} + \cdots + x^{15}) (x^{20} + \cdots + x^{45}) \\ &= x^{10}(1 + x + \cdots + x^{15}) x(1 + x + \cdots + x^{14}) x^{20}(1 + x + \cdots + x^{25}) \\ &= x^{31}(1 + x + \cdots + x^{15}) (1 + x + \cdots + x^{14}) (1 + x + \cdots + x^{25}). \end{align} Finding the $x^{47}$ coefficient of this last polynomial is clearly equivalent to finding the $x^{47 - 31} = x^{16}$ coefficient of $$(1 + x + \cdots + x^{15}) (1 + x + \cdots + x^{14}) (1 + x + \cdots + x^{25}).$$
Using the lower bounds, write $e_1 = 9 + f_1$, similarly $e_2 = f_2$ and $e_3 = 19 + f_3$. Then you want positive-integer solutions to $f_1 + f_2 + f_3 = 19$ with $f_1 \le 16$, $f_2 \le 15$, and $f_3 \le 26$. The last of these is redundant (as $f_3 \le 19-2$ in any case), so drop it.
The number of positive-integer solutions to $f_1 + f_2 + f_3 = 19$ is $\binom{18}{2}$, by "stars-and-bars": the number of ways to insert two "bars" in the $18$ gaps between $19$ "stars".
From this we must subtract solutions in which $f_1 > 16$ or $f_2 > 15$ (not both are possible simultaneously, so that simplifies matters a little). The only possible way we can have $f_1 > 16$ is when $f_1 = 17$, which case the number of solutions to $f_2 + f_3 = 2$ is exactly $1$. There are two ways in which $f_2 > 15$ can happen: either $f_2 = 17$, in which case the number of solutions to $f_1 + f_3 = 2$ is $1$, or $f_2 = 16$, in which case the number of solutions to $f_1 + f_3 = 3$ is $2$.
So the answer turns out to be $\binom{18}{2} - 1 - 1 - 2$, which is the same as your expression.
• In full generality, you'd go for inclusion/exclusion – vonbrand Apr 3 '14 at 2:51
You want (freely simplifying by getting rid of terms that can't affect the result): \begin{align} [x^{47}](x^{10}+x^{11}&+\cdots+x^{25})(x+x^{2}+\cdots+x^{15})(x^{20}+\cdots+x^{45}) \\ &= [x^{47}] \; x^{10} (1 + x + \cdots + x^{10}) \cdot x (1 + x + \cdots + x^{14}) \cdot x^{20} (1 + x + \cdots + x^{25})\\ &= [x^{16}] \frac{1 - x^{11}}{1 - z} \cdot \frac{1 - x^{15}}{1 - z} \cdot \frac{1}{1 - z} \\ &= [x^{16}] (1 - x^{11} - x^{15} + x^{26}) (1 - x)^{-3} \\ &= [x^{16}](1 - x)^{-3} - [x^{5}] (1 - x)^{-3} - [x](1 - x)^{-3} \\ &= (-1)^{16} \binom{-3}{16} - (-1)^5 \binom{-3}{5} - (-1) \binom{-3}{1} \\ &= \binom{16 + 3 - 1}{3 - 1} - \binom{5 + 3 - 1}{3 - 1} - \binom{1 + 3 - 1}{3 - 1} \\ &= 129 \end{align}
• Nice. :-) But I think from the question that the OP was asking for a combinatorial "counting" method, not via generating functions. – ShreevatsaR Apr 2 '14 at 3:58 | 2019-06-26T21:55:13 | {
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https://math.stackexchange.com/questions/2537684/find-the-value-of-frac11-frac112-frac1123-ldots-frac11 | # Find the value of $\frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots + \frac{1}{1+2+3 +\ldots+2015}$
The question:
Find the value of $$\frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots + \frac{1}{1+2+3 +\ldots +2015}$$
If this is a duplicate, then sorry - but I haven't been able to find this question yet. To start, I noticed that this is the sum of the reciprocals of the triangle numbers.
Let $$t_n = \frac{n(n+1)}{2}$$ denote the $$n$$-th triangle number. Then the question is basically asking us to evaluate \begin{align} \sum_{n=1}^{2015} \frac {1}{t_n} & = \sum_{n=1}^{2015} \frac {2}{n(n+1)}\\ & = \sum_{n=1}^{2015}\frac{2}{n}-\frac{2}{n+1} \end{align}
Here's where my first question arises. Do you just have to know that $$\frac {2}{n(n+1)} = \frac{2}{n}-\frac{2}{n+1}$$? In an exam situation it would be very unlikely that someone would be able to recall that if they had not done a question like this before.
Moving on:
\begin{align} \sum_{n=1}^{2015}\frac{2}{n}-\frac{2}{n+1} & = \left(\frac{2}{1}-\frac{2}{2}\right) +\left(\frac{2}{2}-\frac{2}{3}\right) + \ldots +\left(\frac{2}{2014}-\frac{2}{2015}\right) +\left(\frac{2}{2015}-\frac{2}{2016}\right)\\ &= 2 - \frac{2}{2016} \\ & = \frac {4030}{2016} \\ & = \frac {2015}{1008} \end{align}
And I'm not sure if this is right. How does one check whether their summation is correct?
• That is correct. – Robert Z Nov 26 '17 at 9:01
• You don't have to know that $\frac2{n(n+1)} = \frac2n - \frac2{n+1}$, but you do have to know the concept of telescoping sums/series. Once you suspect that this is one of those, then it's relatively easy to work out the identity and compute the sum. – kahen Nov 26 '17 at 9:03
• This is an excellent solution. This is a reasonable exam question because it is a standard example of the method of differences that students will have met before. With regard to checking, it is really just a case of checking that every line of your argument is correct and then you can be confident about your conclusion. – S. Dolan Nov 26 '17 at 9:04
• In an exam situation it would be very unlikely that someone would be able to recall that if they had not done a question like this before. The trick of splitting a fraction up like that is a common one and useful in many situations. – StephenG Nov 26 '17 at 9:07
• You say "it would be very unlikely that someone would be able to recall that if they had not done a question like this before". Right, so the trick is to do questions like this before doing the exam. – Lord Shark the Unknown Nov 26 '17 at 10:06
For your question about how $$\sum{\frac{2}{n(n+1)}}$$ is split,
$$\frac{2}{n(n+1)} = \frac{2[(n+1)-n]}{n(n+1)}$$ $$\frac{2[(n+1)-n]}{n(n+1)} = \frac{2(n+1)}{n(n+1)} - \frac{2n}{n(n+1)}$$ $$\frac{2(n+1)}{n(n+1)} - \frac{2n}{n(n+1)} = \frac{2}{n} - \frac{2}{n+1}$$
So your question about you would know this in an exam situation, is that these type of question will have a similar type of denominator. Like in this question, if you observe the denominator $$n(n+1)$$ it can be split into $$(n+1)-n$$ which is equal to 1. So this makes splitting the fraction into two fractions easier such that the summation could be found out. | 2019-11-20T14:23:22 | {
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http://codeforces.com/problemset/problem/1095/D | D. Circular Dance
time limit per test
3 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
There are $n$ kids, numbered from $1$ to $n$, dancing in a circle around the Christmas tree. Let's enumerate them in a clockwise direction as $p_1$, $p_2$, ..., $p_n$ (all these numbers are from $1$ to $n$ and are distinct, so $p$ is a permutation). Let the next kid for a kid $p_i$ be kid $p_{i + 1}$ if $i < n$ and $p_1$ otherwise. After the dance, each kid remembered two kids: the next kid (let's call him $x$) and the next kid for $x$. Each kid told you which kids he/she remembered: the kid $i$ remembered kids $a_{i, 1}$ and $a_{i, 2}$. However, the order of $a_{i, 1}$ and $a_{i, 2}$ can differ from their order in the circle.
Example: 5 kids in a circle, $p=[3, 2, 4, 1, 5]$ (or any cyclic shift). The information kids remembered is: $a_{1,1}=3$, $a_{1,2}=5$; $a_{2,1}=1$, $a_{2,2}=4$; $a_{3,1}=2$, $a_{3,2}=4$; $a_{4,1}=1$, $a_{4,2}=5$; $a_{5,1}=2$, $a_{5,2}=3$.
You have to restore the order of the kids in the circle using this information. If there are several answers, you may print any. It is guaranteed that at least one solution exists.
If you are Python programmer, consider using PyPy instead of Python when you submit your code.
Input
The first line of the input contains one integer $n$ ($3 \le n \le 2 \cdot 10^5$) — the number of the kids.
The next $n$ lines contain $2$ integers each. The $i$-th line contains two integers $a_{i, 1}$ and $a_{i, 2}$ ($1 \le a_{i, 1}, a_{i, 2} \le n, a_{i, 1} \ne a_{i, 2}$) — the kids the $i$-th kid remembered, given in arbitrary order.
Output
Print $n$ integers $p_1$, $p_2$, ..., $p_n$ — permutation of integers from $1$ to $n$, which corresponds to the order of kids in the circle. If there are several answers, you may print any (for example, it doesn't matter which kid is the first in the circle). It is guaranteed that at least one solution exists.
Examples
Input
5
3 5
1 4
2 4
1 5
2 3
Output
3 2 4 1 5
Input
3
2 3
3 1
1 2
Output
3 1 2 | 2019-06-26T05:04:48 | {
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http://bootmath.com/what-exactly-is-the-difference-between-weak-and-strong-induction.html | # What exactly is the difference between weak and strong induction?
I am having trouble seeing the difference between weak and strong induction.
There are a few examples in which we can see the difference, such as reaching the $k^{th}$ rung of a ladder and proving every integer $>1$ can be written as a product of primes:
To show every $n\ge2$ can be written as a product of primes, we see $2$ is prime. Now we assume true for all integers $m \ge 2$ less than $n$; i.e., assume true for $2 \le m <n$. If $n$ is prime, we’re done. If $n$ is not prime, then it is composite and so $n=ab$, where $a$ and $b$ are less than $n$. Since $a$ and $b$ are less than $n$, $ab$ can be written as a product of primes and hence $n$ can be written as a product of primes. QED
However, it seems sort of like weak induction, only a bit more dubious. In weak induction, we show base case is true, then we assume true for all integers $k-1$, (or $k$), then we attempt to show it is true for $k$, (or $k+1$), which implies true $\forall n \in \mathbb N$.
When we assume true for all integers $k$, isn’t that the same as a strong induction hypothesis? That is, we’re assuming true for all integers up to some specific one.
As a simple demonstrative example, how would we show $1+2+\cdots+n= {n(n+1) \over 2}$ using strong induction?
(Learned from Discrete Mathematics by Kenneth Rosen)
#### Solutions Collecting From Web of "What exactly is the difference between weak and strong induction?"
Initial remarks: Good question. I think it deserves a full response (warning: this is going to be a long, but hopefully very clear, answer). First, most students do not really understand why mathematical induction is a valid proof technique. That’s part of the problem. Second, weak induction and strong induction are actually logically equivalent; thus, differentiating between these forms of induction may seem a little bit difficult at first. The important thing to do is to understand how weak and strong induction are stated and to clearly understand the differences therein (I disagree with the previous answer that it is “just a matter of semantics”; it’s not, and I will explain why). Much of what I will have to say is adapted from David Gunderson’s wonderful book Handbook of Mathematical Induction, but I have expanded and tweaked a few things where I saw fit. That being said, hopefully you will find the rest of this answer to be informative.
Gunderson remark about strong induction: While attempting an inductive proof, in the inductive step one often needs only the truth of $S(n)$ to prove $S(n+1)$; sometimes a little more “power” is needed (such as in the proof that any positive integer $n\geq 2$ is a product of primes–we’ll explore why more power is needed in a moment), and often this is made possible by strengthening the inductive hypothesis.
Kenneth Rosen remark in Discrete Mathematics and Its Applications Study Guide: Understanding and constructing proofs by mathematical induction are extremely difficult tasks for most students. Do not be discouraged, and do not give up, because, without doubt, this proof technique is the most important one there is in mathematics and computer science. Pay careful attention to the conventions to be observed in writing down a proof by induction. As with all proofs, remember that a proof by mathematical induction is like an essay–it must have a beginning, a middle, and an end; it must consist of complete sentences, logically and aesthetically arranged; and it must convince the reader. Be sure that your basis step (also called the “base case”) is correct (that you have verified the proposition in question for the smallest value or values of $n$), and be sure that your inductive step is correct and complete (that you have derived the proposition for $k+1$, assuming the inductive hypothesis that proposition is true for $k$–or the slightly strong hypothesis that it is true for all values less than or equal to $k$, when using strong induction.
Statement of weak induction: Let $S(n)$ denote a statement regarding an integer $n$, and let $k\in\mathbb{Z}$ be fixed. If
• (i) $S(k)$ holds, and
• (ii) for every $m\geq k, S(m)\to S(m+1)$,
then for every $n\geq k$, the statement $S(n)$ holds.
Statement of strong induction: Let $S(n)$ denote a statement regarding an integer $n$. If
• (i) $S(k)$ is true and
• (ii) for every $m\geq k, [S(k)\land S(k+1)\land\cdots\land S(m)]\to S(m+1)$,
then for every $n\geq k$, the statement $S(n)$ is true.
Proof of strong induction from weak: Assume that for some $k$, the statement $S(k)$ is true and for every $m\geq k, [S(k)\land S(k+1)\land\cdot\land S(m)]\to S(m+1)$. Let $B$ be the set of all $n>m$ for which $S(n)$ is false. If $B\neq\varnothing, B\subset\mathbb{N}$ and so by well-ordering, $B$ has a least element, say $\ell$. By the definition of $B$, for every $k\leq t<\ell, S(t)$ is true. The premise of the inductive hypothesis is true, and so $S(\ell)$ is true, contradicting that $\ell\in B$. Hence $B=\varnothing$. $\blacksquare$
Proof of weak induction from strong: Assume that strong induction holds (in particular, for $k=1$). That is, assume that if $S(1)$ is true and for every $m\geq 1, [S(1)\land S(2)\land\cdots\land S(m)]\to S(m+1)$, then for every $n\geq 1, S(n)$ is true.
Observe (by truth tables, if desired), that for $m+1$ statements $p_i$,
$$[p_1\to p_2]\land[p_2\to p_3]\land\cdots\land[p_m\to p_{m+1}]\Rightarrow[(p_1\land p_2\land\cdots\land p_m)\to p_{m+1}],\tag{\dagger}$$
itself a result provable by induction (see end of answer for such a proof).
Assume that the hypotheses of weak induction are true, that is, that $S(1)$ is true, and that for arbitrary $t, S(t)\to S(t+1)$. By repeated application of these recent assumptions, $S(1)\to S(2), S(2)\to S(3),\ldots, S(m)\to S(m+1)$ each hold. By the above observation, then
$$[S(1)\land S(2)\land\cdots\land S(m)]\to S(m+1).$$
Thus the hypotheses of strong induction are complete, and so one concludes that for every $n\geq 1$, the statement $S(n)$ is true, the consequence desired to complete the proof of weak induction. $\blacksquare$
Proving any positive integer $n\geq 2$ is a product of primes using strong induction: Let $S(n)$ be the statement “$n$ is a product of primes.”
Base step ($n=2$): Since $n=2$ is trivially a product of primes (actually one prime, really), $S(2)$ is true.
Inductive step: Fix some $m\geq 2$, and assume that for every $t$ satisfying $2\leq t\leq m$, the statement $S(t)$ is true. To be shown is that
$$S(m+1) : m+1 \text{ is a product of primes},$$
is true. If $m+1$ is a prime, then $S(m+1)$ is true. If $m+1$ is not prime, then there exist $r$ and $s$ with $2\leq r\leq m$ and $2\leq s\leq m$ so that $m+1=rs$. Since $S(r)$ is assumed to be true, $r$ is a product of primes [note: This is where it is imperative that we use strong induction; using weak induction, we cannot assume $S(r)$ is true]; similarly, by $S(s), s$ is a product of primes. Hence $m+1=rs$ is a product of primes, and so $S(m+1)$ holds. Thus, in either case, $S(m+1)$ holds, completing the inductive step.
By mathematical induction, for all $n\geq 2$, the statement $S(n)$ is true. $\blacksquare$
Proof of $1+2+3+\cdots+n = \frac{n(n+1)}{2}$ by strong induction: Using strong induction here is completely unnecessary, for you do not need it at all, and it is only likely to confuse people as to why you are using it. It will proceed just like a proof by weak induction, but the assumption at the outset will look different; nonetheless, just to show what I am talking about, I will prove it using strong induction.
Let $S(n)$ denote the proposition
$$S(n) : 1+2+3+\cdots+n = \frac{n(n+1)}{2}.$$
Base step ($n=1$): $S(1)$ is true because $1=\frac{1(1+1)}{2}$.
Inductive step: Fix some $k\geq 1$, and assume that for every $t$ satisfying $1\leq t\leq k$, the statement $S(t)$ is true. To be shown is that
$$S(k+1) : 1+2+3+\cdots+k+(k+1)=\frac{(k+1)(k+2)}{2}$$
follows. Beginning with the left-hand side of $S(k+1)$,
\begin{align}
\text{LHS}
&= 1+2+3+\cdots+k+(k+1)\tag{by definition}\\[1em]
&= (1+2+3+\cdots+k)+(k+1)\tag{group terms}\\[1em]
&= \frac{k(k+1)}{2}+(k+1)\tag{by $S(k)$}\\[1em]
&= (k+1)\left(\frac{k}{2}+1\right)\tag{factor out $k+1$}\\[1em]
&= (k+1)\left(\frac{k+2}{2}\right)\tag{common denominator}\\[1em]
&= \frac{(k+1)(k+2)}{2}\tag{desired expression}\\[1em]
&= \text{RHS},
\end{align}
we obtain the right-hand side of $S(k+1)$.
By mathematical induction, for all $n\geq 1$, the statement $S(n)$ is true. $\blacksquare$
$\color{red}{\text{Comment:}}$ See how this was really no different than how a proof by weak induction would work? The only thing different is really an unnecessary assumption made at the beginning of the proof. However, in your prime number proof, strong induction is essential; otherwise, we cannot assume $S(r)$ or $S(s)$ to be true. Here, any assumption regarding $t$ where $1\leq t\leq k$ is really useless because we don’t actually use it anywhere in the proof, whereas we did use the assumptions $S(r)$ and $S(s)$ in the prime number proof, where $1\leq t\leq m$, because $r,s < m$. Does it now make sense why it was necessary to use strong induction in the prime number proof?
Proof of $(\dagger)$ by induction: For statements $p_1,\ldots,p_{m+1}$, we have that
$$[p_1\to p_2]\land[p_2\to p_3]\land\cdots\land[p_m\to p_{m+1}]\Rightarrow[(p_1\land p_2\land\cdots\land p_m)\to p_{m+1}].$$
Proof. For each $m\in\mathbb{Z^+}$, let $S(m)$ be the statement that for $m+1$ statements $p_i$,
$$S(m) : [p_1\to p_2]\land[p_2\to p_3]\land\cdots\land[p_m\to p_{m+1}]\Rightarrow[(p_1\land p_2\land\cdots\land p_m)\to p_{m+1}].$$
Base step: The statement $S(1)$ says
$$[p_1\to p_2]\Rightarrow [(p_1\land p_2)\to p_2],$$
which is true (since the right side is a tautology).
Inductive step: Fix $k\geq 1$, and assume that for any statements $q_1,\ldots,q_{k+1}$, both
$$S(1) : [q_1\to q_2]\Rightarrow [(q_1\land q_2)\to q_2]$$
and
$$S(k) : [q_1\to q_2]\land[q_2\to q_3]\land\cdots\land[q_k\to q_{k+1}]\Rightarrow[(q_1\land q_2\land\cdots\land q_k)\to q_{k+1}].$$
hold. It remains to show that for any statements $p_1,p_2,\ldots,p_k,p_{k+1},p_{k+2}$ that
$$S(k+1) : [p_1\to p_2]\land[p_2\to p_3]\land\cdots\land[p_{k+1}\to p_{k+2}]\Rightarrow[(p_1\land p_2\land\cdots\land p_{k+1})\to p_{k+2}]$$
follows. Beginning with the left-hand side of $S(k+1)$,
\begin{align}
\text{LHS} &\equiv [p_1\to p_2]\land\cdots\land[p_{k+1}\to p_{k+2}]\land[p_{k+1}\to p_{k+2}]\\[0.5em]
&\Downarrow\qquad \text{(definition of conjunction)}\\[0.5em]
&[[p_1\to p_2]\land[p_2\to p_3]\land\cdots\land[p_{k+1}\to p_{k+2}]]\land[p_{k+1}\to p_{k+2}]\\[0.5em]
&\Downarrow\qquad \text{(by $S(k)$ with each $q_i = p_i$)}\\[0.5em]
&[(p_1\land p_2\land\cdots\land p_k)\to p_{k+1}]\land[p_{k+1}\to p_{k+2}]\\[0.5em]
&\Downarrow\qquad \text{(by $S(1)$ with $q_1=p_1\land\cdots\land p_k)$ and $q_2=p_{k+1}$)}\\[0.5em]
&[[(p_1\land p_2\land\cdots\land p_k)\land p_{k+1}]\to p_{k+1}]\land [p_{k+1}\to p_{k+2}]\\[0.5em]
&\Downarrow\qquad \text{(by definition of conjunction)}\\[0.5em]
&[(p_1\land p_2\land\cdots\land p_k\land p_{k+1}]\to p_{k+1}]\land [p_{k+1}\to p_{k+2}]\\[0.5em]
&\Downarrow\qquad \text{(since $a\land b\to b$ with $b=[p_{k+1}\to p_{k+2}]$)}\\[0.5em]
&[(p_1\land p_2\land\cdots\land p_k\land p_{k+1})\to p_{k+2}]\land[p_{k+1}\to p_{k+2}]\\[0.5em]
&\Downarrow\qquad \text{(since $a\land b\to a$)}\\[0.5em]
&(p_1\land p_2\land\cdots\land p_k\land p_{k+1})\to p_{k+2}\\[0.5em]
&\equiv \text{RHS},
\end{align}
we obtain the right-hand side of $S(k+1)$, which completes the inductive step.
By mathematical induction, for each $n\geq 1, S(n)$ holds. $\blacksquare$
Usually, there is no need to distinguish between weak and strong induction. As you point out, the difference is minor. In both weak and strong induction, you must prove the base case (usually very easy if not trivial). Then, weak induction assumes that the statement is true for size $n-1$ and you must prove that the statement is true for $n$. Using strong induction, you assume that the statement is true for all $m<n$ (at least your base case) and prove the statement for $n$.
In practice, one may just always use strong induction (even if you only need to know that the statement is true for $n-1$). In the example that you give, you only need to assume that the formula holds for the previous case (weak) induction. You could assume it holds for every case, but only use the previous case. As far as I can tell, it is really just a matter of semantics. There are times when strong induction really is more useful, the case when you break up the problem into two problem of size $n/2$ for example. This happens frequently when making proofs about graphs where you decompose the graph on $n$ vertices into two subgraphs (smaller, but you have little or no control over the exact size). | 2018-07-19T15:40:25 | {
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http://mathhelpforum.com/algebra/169015-solve-x-3-12x-16-0-a.html | Math Help - solve x^3 - 12x + 16 = 0
1. solve x^3 - 12x + 16 = 0
Hi
I can't see how to factorise this. And can't think of another method. Can someone please help?
x^3 - 12x + 16 = 0
Angus
2. This polynomial is $\displaystyle P(x) = x^3 - 12x + 16$.
Notice that $\displaystyle P(2) = 2^3 - 12\cdot 2 + 16 = 8 - 24 + 16 = 0$. So $\displaystyle (x - 2)$ is a factor.
Long divide to find the quadratic factor and factorise the quadratic if possible.
3. Originally Posted by angypangy
Hi
I can't see how to factorise this. And can't think of another method. Can someone please help?
x^3 - 12x + 16 = 0
Angus
Another way....
$12x-x^3=16\Rightarrow\ x\left(12-x^2\right)=16$
16 factorises to 16(1)=8(2)=4(4) so we can quickly discover
whether or not the function has an integer root.
If it does, the factoring is made simple.
$x=2\Rightarrow\ 2(12-4)=2(8)$ works
$x=4$ gives $-16,$ but $x=-4$ gives $(-4)(-4)=16$
$x^3-12x+16=(x-2)(x+4)(x+c)\Rightarrow\ -8c=16\Rightarrow\ c=-2$
Check that this holds.
Or
$(x-2)\left(x^2+bx+c\right)=x^3-12x+16\Rightarrow\ c=-8,\;\;bx^2-2x^2=0\Rightarrow\ b=2$
as there is no $x^2$ term.
4. Archie Meade's method assumes that the roots will be integer since those are the only factors he is trying.
Another method is to use the "rational root theorem"- that if m/n is a rational number root of a polynomial with integer coefficients then the denominator, n, must divide the leading coefficient and the numerator, m, must divide the constant term. Here, the leading coefficent is 1 so the denominator must be 1 or -1 (any rational root must be an integer just as Archie Meade thought) and the constant term is 16 so any rational root must evenly divide 16: 1, -1, 2, -2, 4, -4, 8, -8, 16,or -16 are possible roots. Trying each of those numbers in the polynomial we see that $(2)^3- 12(2)+ 16= 8- 24+ 16= 0$ and that $(-4)^3- 12(-4)+ 16= -64+ 48+ 16= 0$
Using x-2 and x+ 4 as factors, we see that the third factor is x- 2 so that x= 2 is a double root.
Of course, the rational root theorem can only point out possible rational roots. It may be that a polynomial has no rational number roots but in that case, no method is going to be easy! | 2014-04-18T14:25:53 | {
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https://math.stackexchange.com/questions/1433876/does-there-exist-a-continuous-onto-function-from-mathbbr-mathbbq-to-ma | Does there exist a continuous onto function from $\mathbb{R}-\mathbb{Q}$ to $\mathbb{Q}$?
Does there exist a continuous onto function from $\mathbb{R}-\mathbb{Q}$ to $\mathbb{Q}$? (where domain is all irrational numbers)
I found many answers for contradicting the fact that there doesnt exist a continuous function which maps rationals to irrationals and vice versa.
But proving that thing was easier since our domain of definition of function was a connected set, we could use that connectedness or we could use the fact that rationals are countable and irrationals are uncountable.
But in this case those properties are not useful. I somehow think that baire category theorem might be useful but I am not good at using it.
• Try $f(x)=1$... – David C. Ullrich Sep 13 '15 at 18:07
• you missed onto. – Landon Carter Sep 13 '15 at 18:07
• @DavidC.Ullrich - OP asked for an onto function. – Sam Cappleman-Lynes Sep 13 '15 at 18:07
Yes. Say $E_n$ is the set of irrationals in the interval $(n,n+1)$. Say $(q_n)$ is an enumeration of $\Bbb Q$. Define $f(x)=q_n$ for $x\in E_n$.
• This is amazing!! How could you create such a function? I mean how did you come up with this? – Landon Carter Sep 13 '15 at 18:14
• Can you give more explanation , what is enumeration of rational ? – Shubham Ugare Sep 13 '15 at 18:21
• I guess this construction can also be used to show that, if $X$ is any topological space which can be partitioned into a countably infinite collection of open sets (which are then automatically closed too) and $Y$ is any countable topological space, then there exists a continuous surjection $X \to Y$. – Mike F Sep 13 '15 at 18:25
• @ShubhamUgare The rationals are countable. An enumeration of the rationals is just a sequence containing each rational exactly once. (Here just for convenience we took that "sequence" to be indexed by integers instead of natural numbers.) – David C. Ullrich Sep 13 '15 at 18:27
• @LandonCarter I have a hard time saying how I came up with that, sorry. Seemed like an obvious thing... – David C. Ullrich Sep 13 '15 at 18:30
Yes, there exists such a function.
Biject $\mathbb{Q}$ with $\mathbb{Z}$ to get $\mathbb{Q} = \{q_n \mid n \in \mathbb{Z}\}$, and let $\mathbb{I}$ be the set of irrational numbers.
Define $I_n$ for $n \in \mathbb{Z}$ as $\mathbb{I} \cap (n, n+1)$. Then define $f(x) = q_n$ for all $x \in I_n$.
This is continuous, since for any irrational number in $I_n$, there is a small neighbourhood of it which is contained entirely within $I_n$ (because the "endpoints" of $I_n$ were chosen to be rational)
• Did you notice that that's exactly the same as the example I gave? – David C. Ullrich Sep 13 '15 at 18:26
• Yes - but we answered at almost the same time. – Sam Cappleman-Lynes Sep 13 '15 at 18:27 | 2020-01-25T17:22:51 | {
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https://math.stackexchange.com/questions/1780501/what-is-the-expected-number-of-steps-in-a-random-walk-from-leaf-to-leaf-in-a-ful | # What is the expected number of steps in a random walk from leaf to leaf in a full binary tree?
Let $h \geq 2$ be a natural number. Consider a complete binary tree of height $h$. Say we take a random walk starting from the "leftmost" leaf. What is the expected number of steps before the "rightmost" leaf is visited? Is there a known closed form for this? If not, are there any lower or upper bounds?
As an example, let $h = 3$. Then we are looking at the above tree and asking what the expected number of steps to walk from node 4 to node 7 is.
What I've found:
Denote $n = 2^h - 1$ to be the number of nodes in the tree. This problem can be viewed as a system of linear equations. Let $x_i$ denote the expected number of steps to walk from node $i$ to node $n$. We are trying to find $x_{(n+1)/2}$ in the system $$\begin{cases} x_1 = 1 + \frac{1}{2}x_2 + \frac{1}{2}x_3 \\ x_i = 1 + \frac{1}{3}x_{\lfloor i/2 \rfloor} + \frac{1}{3}x_{2i} + \frac{1}{3}x_{2i+1} & \text{for } 1 < i < \frac{n+1}{2} \\ x_i = 1 + x_{\lfloor i/2 \rfloor} & \text{for } \frac{n+1}{2} \leq i < n \\ x_n = 0. \end{cases}$$
I wrote a Python program to solve this system. I found the following values: $$\begin{array}{c|c} h & \text{expected number of steps} \\ \hline 2 & 4 \\ \hline 3 & 24 \\ \hline 4 & 84 \\ \hline 5 & 240 \\ \hline 6 & 620 \\ \hline 7 & 1512 \\ \hline 8 & 3556 \\ \hline 9 & 8160 \\ \hline 10 & 18396 \end{array}$$ I plugged these values into OEIS, but there was no matching sequence.
Although I do not know how to solve for an exact value, I can provide a lower bound of $2^{h+1} - 3h - 1$.
Note that a walk from the leftmost leaf to the rightmost leaf must pass through the root. Therefore, we can lower bound the desired value by the expected number of steps before the root is visited in a random walk starting from the leftmost leaf. Then we are just asking how many steps it takes to randomly walk from a node at level $h$ to the root node at level $1$.
At level $h$, you always step to level $h-1$. At level $i$ where $1 < i < h$, you have a $\frac{2}{3}$ chance of stepping to a child at level $i+1$ and a $\frac{1}{3}$ chance of stepping to the parent at level $i-1$.
Let $x_i$ denote the expected number of steps to randomly walk from level $h-i+1$ to level 1. We are then trying to find $x_1$ in the system of equations $$\begin{cases} x_h = 0\\ x_i = 1 + \frac{2}{3}x_{i-1} + \frac{1}{3}x_{i+1} & \text{for } 1 < i < h\\ x_1 = 1 + x_2. \end{cases}$$ I claim that $x_{i-1} = x_i + 2^i - 3$ for $1 < i \leq n$. Proceed by induction. As the case case, we have $x_1 = x_2 + 1 = x_2 + 2^2 - 3$. For the induction hypothesis, suppose that $x_{k-1} = x_k + 2^k - 3$. Then for our induction step, we have \begin{align*} x_k &= 1 + \frac{2}{3}x_{k-1} + \frac{1}{3}x_{k+1} \\ &= 1 + \frac{2}{3}\left(x_k + 2^k - 3\right) + \frac{1}{3}x_{k+1} \end{align*} which rearranges to $x_k = x_{k+1} + 2^{k+1} - 3$, proving the claim.
Then we have $$x_1 = \sum_{i=2}^h \left( 2^i - 3 \right) = 2^{h+1} - 3h - 1.$$
Hence the expected number of steps to reach the root from a leaf is $2^{h+1} - 3h - 1$. This gives our lower bound.
I do not know where to go from here. Looking at the numbers, I suspect one might be able to prove a $\Omega(h2^h)$ lower bound.
• What are the transition probabilities? May 11, 2016 at 5:34
• Equal probability distributed among neighboring nodes. At a leaf node, always step to the parent. At the root node, step to the left child with $1/2$ probability and to the right child with $1/2$ probability. At an internal node, step to the left child with $1/3$ probability, to the right child with $1/3$ probability, and to the parent with $1/3$ probability. May 11, 2016 at 5:39
• When you divide by 4, you can find this sequence oeis.org/A066524 So we can guess the general term to be $4(h-1)(2^{h-1}-1)$
– BGM
May 11, 2016 at 5:43
• Nice find! That formula definitely does fit all the data points I gave. The next question is whether there's a nice way to prove it.... May 11, 2016 at 6:07
## 1 Answer
First, the walk needs to get from the leaf to the root. This is a simple random walk in the vertical direction, with probability $\frac23$ to go down and $\frac13$ to go up. The expected time $a_k$ to reach the root from depth $k$ satisfies
$$t_k=1+\frac23t_{k+1}+\frac13t_{k-1}$$
for $k=1,\ldots,h-2$ with boundary conditions $t_0=0$ and $t_{h-1}=1+t_{h-2}$. The general solution is $t_k=c_1+c_22^{-k}-3k$. The first boundary condition yields $c_2=-c_1$, and then the second one yields
$$c_1-c_12^{1-h}-3(h-1)=1+c_1-c_12^{2-h}-3(h-2)\;,$$
with solution
$$c_1=2^{h+1}\;.$$
Thus $t_k=2^{h+1}\left(1-2^{-k}\right)-3k$, so $t_{h-1}=2^{h+1}-4-3(h-1)=2^{h+1}-3h-1$.
Now we need to add up the times it takes to make progress towards the target leaf. At the root, we have probability $\frac12$ to go back to the left, where it will take us $t_1=2^{h+1}\left(1-2^{-1}\right)-3\cdot1=2^h-3$ to get back to the root, and probability $\frac12$ to go to the right, so the expected time $b_0$ to go right satisfies
$$b_0=1+\frac12\left(2^h-3+b_0\right)\;,$$
with solution $b_0=2^h-1$. From then on, at depth $k$, we have probability $\frac13$ to make progress and $\frac23$ to stray back into a graph with $2^h-2-\frac12\left(2^{h-k}-2\right)=2^h-2^{h-k-1}-1$ edges, of which $2$ lead back, so the return time in this case is $2^h-2^{h-k-1}-1$. Thus the expected time $b_k$ to make progress satisfies
$$b_k=\frac23\left(2^h-2^{h-k-1}-1+b_k\right)+\frac13\cdot1\;,$$
so
$$b_k=2^{h+1}-2^{h-k}-1\;.$$
Note that this also covers $b_0=2^{h+1}-2^{h-0}-1=2^h-1$. Summing over $k$ from $0$ to $h-2$ yields a contribution
$$\sum_{k=0}^{h-2}\left(2^{h+1}-2^{h-k}-1\right)=\left(2^{h+1}-1\right)h-2^{h+2}+5\;,$$
for a total of
\begin{align} 2^{h+1}-3h-1+\left(2^{h+1}-1\right)h-2^{h+2}+5 &=2^{h+1}h-2^{h+1}-4h+4\\ &=4(h-1)\left(2^{h-1}-1\right). \end{align}
P.S.: Note that this actually contains everything we need to calculate the expected number of steps from any node to any other node. If the initial node is at depth $a$, the final node is at depth $b$ and their lowest common ancestor is at depth $c$, then it takes
$$2^{h-c+1}\left(1-2^{c-a}\right)+3(c-a)=2^{h-c+1}-2^{h-a+1}+3(c-a)$$
steps to reach the common ancestor and
$$\sum_{k=c}^{b-1}\left(2^{h+1}-2^{h-k}-1\right)=2^{h-b+1}-2^{h-c+1}+(b-c)\left(2^{h+1}-1\right)$$
steps to reach the final node, for a total of
$$\left(2^{-b}-2^{-a}\right)2^{h+1}+(b-c)\left(2^{h+1}-1\right)+3(c-a)\;.$$
The case you were interested in was $c=0$ and $a=b=h-1$.
• Fantastic! This makes sense to me. Question: how did you find $t_k=c_1+c_22^{-k}-3k$? I can see that it works, but is there a general way to solve equations of that form? (Also, minor typo: the sign is missing in "$c_2 = -c_1.$") May 11, 2016 at 7:17
• @TomTseng: The general solution of linear recurrences with constant coefficients is analogous to the general solution of linear differential equations with constant coefficients. The general solution is the sum of a special solution of the inhomogeneous equation and the general solution of the homogeneous equation. In a recurrence for expected times in a random walk, the coefficients on both sides sum to one, which allows a linear ansatz $t_k=mk$ to solve for the constant inhomegeneity; substituting that into the recurrence yields $m=-3$. May 11, 2016 at 7:39
• @TomTseng: The homogeneous equation is solved using the geometric ansatz $t_k=c\lambda^k$, which yields the characteristic equation $\lambda=\frac23\lambda^2+\frac13$, with solutions $\lambda=1$ and $\lambda=\frac12$. For more on all this, see Wikipedia. May 11, 2016 at 7:41
• @TomTseng: I added a solution for the general case from any node to any other node. May 11, 2016 at 8:50 | 2022-07-04T02:09:00 | {
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http://web2.0calc.com/questions/how-can-the-fact-that-24-2-5-2-551-be-used-to-find | +0
# How can the fact that 24^2-5^2=551 be used to find the factors of 551 (not including 1 or 551)?
0
116
2
How can the fact that 24^2-5^2=551 be used to find the factors of 551 (not including 1 or 551)?
Guest Aug 8, 2017
#1
+1
The difference between the two numbers is always a factor of the difference between their squares:
So, you have: 24 - 5 = 19 must be a factor of 551. And the sum of the two numbers is the other factor of the difference between their squares: So, you have: 24 + 5 =29 is the other factor of 551.
So, 551 = 19 x 29. This holds true for any two numbers and the difference between their squares.
Another Example: 67^2 - 22^2 =4,005. And 4,005=[67 - 22] x [67 + 22].
Note: The two factors are not ALWAYS prime factors, but two DIVISORS, which can be factored into their prime factors. So [67 - 22]=45 =3 x 3 x 5, and [67 + 22]=89, which is prime number in this example. So, 4,005 =3^2 x 5 x 89.
Guest Aug 9, 2017
Sort:
#1
+1
The difference between the two numbers is always a factor of the difference between their squares:
So, you have: 24 - 5 = 19 must be a factor of 551. And the sum of the two numbers is the other factor of the difference between their squares: So, you have: 24 + 5 =29 is the other factor of 551.
So, 551 = 19 x 29. This holds true for any two numbers and the difference between their squares.
Another Example: 67^2 - 22^2 =4,005. And 4,005=[67 - 22] x [67 + 22].
Note: The two factors are not ALWAYS prime factors, but two DIVISORS, which can be factored into their prime factors. So [67 - 22]=45 =3 x 3 x 5, and [67 + 22]=89, which is prime number in this example. So, 4,005 =3^2 x 5 x 89.
Guest Aug 9, 2017
#2
+18712
+1
How can the fact that 24^2-5^2=551 be used to find the factors of 551 (not including 1 or 551)?
$$\begin{array}{|rcll|} \hline \text{Binomial Theorem} \\ (a+b)^2 &=& a^2+2ab+b^2 \\ (a-b)^2 &=& a^2-2ab+b^2 \\ (a-b)(a+b) &=& a^2-b^2 \\ \hline \end{array}$$
$$\begin{array}{|rcll|} \hline (a-b)(a+b) &=& a^2-b^2 \quad & | \quad a = 24 \qquad b = 5 \\ (24-5)(24+5) &=& 24^2-5^2 \\ 19\cdot 29 &=& 24^2-5^2 \\ \mathbf{19\cdot 29} & \mathbf{=} & \mathbf{551} \\ \hline \end{array}$$
heureka Aug 9, 2017
### 7 Online Users
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http://mathhelpforum.com/number-theory/69428-solved-prove-composite.html | # Math Help - [SOLVED] Prove Composite
1. ## [SOLVED] Prove Composite
Problem:
Show that $8^{n}+1$ is composite for all $n\geq1$.
What I have so far:
In general,
$a^{n}+1 = (a+1)(a^{(n-1)} - a^{(n-2)} + a^{(n-3)} - \cdots + 1)$
but this only holds for $n$ odd.
Is there a way to factor when $n$ is even? Or would I have to find a solution specific to $a=8$?
2. Originally Posted by star_tenshi
Problem:
Show that $8^{n}+1$ is composite for all $n\geq1$.
What I have so far:
In general,
$a^{n}+1 = (a+1)(a^{(n-1)} - a^{(n-2)} + a^{(n-3)} - \cdots + 1)$
but this only holds for $n$ odd.
Is there a way to factor when $n$ is even? Or would I have to find a solution specific to $a=8$?
Note: You may want to wait for a more NT minded member to answer this in case of error
Have you tried induction?
3. Originally Posted by star_tenshi
Problem:
Show that $8^{n}+1$ is composite for all $n\geq1$.
What I have so far:
In general,
$a^{n}+1 = (a+1)(a^{(n-1)} - a^{(n-2)} + a^{(n-3)} - \cdots + 1)$
but this only holds for $n$ odd.
Is there a way to factor when $n$ is even? Or would I have to find a solution specific to $a=8$?
You can use that identity with n=3, if you write 8 as 2^3.
4. Originally Posted by Opalg
You can use that identity with n=3, if you write 8 as 2^3.
Yes, I know that, but I am looking at the case where n is even. Even if I were to use $8^{n} + 1 = (2^{3})^m + 1 = 2^{3m} + 1$, $3m$ is not always odd.
5. Make use of the identity: $a^3 + b^3 = (a+b)(a^2 -ab + b^2)$
Here imagine: $a = 2^n$ and $b = 1$
For all $n \geq 1$, we have:
\begin{aligned} 8^n + 1 & = \left(2^3\right)^n + 1 \\ & = \left(2^n\right)^3 + 1 \\ & = \left(2^n + 1\right)\left((2^n)^2 - 2^n + 1\right) \end{aligned}
So what can we conclude?
6. OHHH! Now I see what Opalg was trying to convey. Thanks again o_O for clearing that up!
We can conclude that $8^{n} + 1$ is composite.
Then to generalize, any number that can be expressed as $a^{3} + b^{3}$ is composite because it can be broken down into the factors $(a+b)(a^{2}-ab+b^{2})$! | 2016-05-28T20:33:48 | {
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https://ocw.mit.edu/courses/18-335j-introduction-to-numerical-methods-spring-2019/pages/week-6/ | # Week 6
## Lecture 15: Eigensolver Algorithms: Companion Matrices, Ill-Conditioning, and Hessenberg Factorization
### Summary
Pointed out that an “LU-like” algorithm for eigenproblems, which computes the exact eigenvalues/eigenvectors (in exact arithmetic, neglecting roundoff) in a finite number of steps involving addition, subtraction, multiplication, division, and roots, is impossible. The reason is that no such algorithm exists (or can ever exist) to find roots of polynomials with degree greater than 4, thanks to a theorem by Abel, Galois and others in the 19th century. Used the companion matrix to show that polynomial root finding is equivalent to the problem of finding eigenvalues. Mentioned the connection to other classic problems of antiquity (squaring the circle, trisecting an angle, doubling the cube), which were also proved impossible in the 19th century.
As a result, all eigenproblem methods must be iterative: they must consist of improving an initial guess, in successive steps, so that it converges towards the exact result to any desired accuracy, but never actually reaches the exact answer in general. A simple example of such a method is Newton’s method, which can be applied to iteratively approximate a root of any nonlinear function to any desired accuracy, given a sufficiently good initial guess.
However, finding roots of the characteristic polynomial is generally a terrible way to find eigenvalues. Actually computing the characteristic polynomial coefficients and then finding the roots somehow (Newton’s method?) is a disaster, incredibly ill-conditioned: gave the example of Wilkinson’s polynomial. If we can compute the characteristic polynomial values implicitly somehow, directly from the determinant, then it is not too bad (if you are looking only for eigenvalues in some known interval, for example), but we haven’t learned an efficient way to do that yet. The way LAPACK and MATLAB actually compute eigenvalues, the QR method and its descendants, wasn’t discovered until 1960.
The key to making most of the eigensolver algorithms efficient is reducing A to Hessenberg form: A=QHQ* where H is upper triangular plus one nonzero value below each diagonal. Unlike Schur form, Hessenberg factorization can be done exactly in a finite number [Θ(m3)] of steps (in exact arithmetic). H and A are similar: they have the same eigenvalues, and the eigenvector are related by Q. And once we reduce to Hessenberg form, all the subsequent operations we might want to do (determinants, LU or QR factorization, etcetera), will be fast. In the case of Hermitian A, showed that H is tridiagonal; in this case, most subsequent operations (even LU and QR factorization) will be Θ(m).
## Lecture 16: The Power Method and the QR Algorithm
### Summary
Reviewed power method for biggest-|λ| eigenvector/eigenvalue and its convergence rate.
Discussed how to use the power method to get multiple eigenvalues/vectors of Hermitian matrices by “deflation” (using orthogonality of eigenvectors). Discussed how, in principle, QR factorization of An for large n will give the eigenvectors and eigenvalues in descending order of magnitude, but how this is killed by roundoff errors.
Unshifted QR method: proved that repeatedly forming A=QR, then replacing A with RQ (as in Problem set 3) is equivalent to QR factorizing An. But since we do this while only multiplying repeatedly by unitary matrices, it is well conditioned and we get the eigenvalues accurately.
To make the QR method faster, we first reduce to Hessenberg form; you will show in Problem set 3 that this is especially fast when A is Hermitian and the Hessenberg form is tridiagonal. Second, we use shifts. In particular, the worst case for the QR method, just as for the power method, is when eigenvalues are nearly equal. We can fix this by shifting.
Discussed inverse iteration and shifted-inverse iteration. Discussed Rayleigh-quotient iteration (shifted-inverse iteration with the Rayleigh-quotient eigenvalue estimate as the shift) and its convergence rate in the Hermitian case. How, for Hermitian matrix the eigenvalue estimate has a much smaller error than the eigenvector (the error is squared) due to the fact that the eigenvalue is an extremum.
## Lecture 17: Shifted QR and Rayleigh Quotients
### Summary
Brief discussion of shifted QR method.
There are a number of additional tricks to further improve things, the most important of which is probably the Wilkinson shift: estimating μ from a little 2×2 problem from the last two columns to avoid problems in cases e.g. where there are two equal and opposite eigenvalues. Some of these tricks (e.g. the Wilkinson shift) are described in the book, and some are only in specialized publications. If you want the eigenvectors as well as eigenvalues, it turns out to be more efficient to use a more recent “divide and conquer” algorithm, summarized in the book, but where the details are especially tricky and important. However, at this point I don’t want to cover more gory details in this course. Although it is good to know the general structure of modern algorithms, especially the fact that they look nothing like the characteristic-polynomial algorithm you learn as an undergraduate, as a practical matter you are always just going to call LAPACK if the problem is small enough to solve directly. Matters are different for much larger problems, where the algorithms are not so bulletproof and one might need to get into the guts of the algorithms; this will lead us into the next topic of the course, iterative algorithms for large systems, in subsequent lectures.
Briefly discussed Golub-Kahn bidiagonalization method for SVD, just to get the general flavor. At this point, however, we are mostly through with details of dense linear-algebra techniques: the important thing is to grasp the fundamental ideas rather than zillions of little details, since in practice you’re just going to use LAPACK anyway.
Start discussing (at a very general level) a new topic: iterative algorithms, usually for sparse matrices, and in general for matrices where you have a fast way to compute Ax matrix-vector products but cannot (practically) mess around with the specific entries of A.
Emphasized that there are many iterative methods, and that there is no clear “winner” or single bulletproof library that you can use without much thought (unlike LAPACK for dense direct solvers). It is problem-dependent and often requires some trial and error. Then there is the whole topic of preconditioning, which we will discuss more later, which is even more problem-dependent. Briefly listed several common techniques for linear systems (Ax=b) and eigenproblems (Ax=λx or Ax=λBx). | 2022-05-27T02:49:07 | {
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# The average (arithmetic mean) score on a test taken by 10 students was
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The average (arithmetic mean) score on a test taken by 10 students was x. If the average score for 5 of the students was 8, what was the average score, in terms of x, for the remaining 5 students who took the test?
A) 2x – 8
B) x – 4
C) 8 – 2x
D) 16 – x
E) 8 – ($$\frac{2}{x}$$)
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Re: The average (arithmetic mean) score on a test taken by 10 students was [#permalink]
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05 Dec 2016, 18:26
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AbdurRakib wrote:
The average (arithmetic mean) score on a test taken by 10 students was x. If the average score for 5 of the students was 8, what was the average score, in terms of x, for the remaining 5 students who took the test?
A) 2x – 8
B) x – 4
C) 8 – 2x
D) 16 – x
E) 8 – ($$\frac{2}{x}$$)
We are given that the average score on a test taken by 10 students was x and need to find the average score in terms of x if the average score for 5 of those students was 8.
Thus, the overall sum was 10x and the sum for the 5 students was (5)(8)= 40.
We need to determine the average of the 5 remaining students.
The sum for the 5 remaining students was 10x - 40, and thus the average is:
(10x - 40)/5 = 10x/5 - 40/5 = 2x - 8
Answer: A
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Re: The average (arithmetic mean) score on a test taken by 10 students was [#permalink]
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15 Jun 2016, 08:02
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Given : The average (arithmetic mean) score on a test taken by 10 students was x. Therefore, sum of all scores = 10*x = 10x.
the average score for 5 of the students was 8. Thus, the total score contribution for these 5 students = 8*5 = 40.
Contribution of other 5 students = 10x - 40.
The average score, in terms of x, for the remaining 5 students = (10x - 40)/5 = 2x - 8. Hence A.
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Re: The average (arithmetic mean) score on a test taken by 10 students was [#permalink]
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06 Dec 2016, 08:23
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AbdurRakib wrote:
The average (arithmetic mean) score on a test taken by 10 students was x. If the average score for 5 of the students was 8, what was the average score, in terms of x, for the remaining 5 students who took the test?
A) 2x – 8
B) x – 4
C) 8 – 2x
D) 16 – x
E) 8 – ($$\frac{2}{x}$$)
Total = 10x
Total of first 5 = 40
So, Average of remaining 5 students is $$\frac{10x - 40}{5}$$ = $$2x – 8$$
Hence, answer will be (A) $$2x – 8$$
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Re: The average (arithmetic mean) score on a test taken by 10 students was [#permalink]
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04 Apr 2017, 22:56
Say B is the average grade for the 5 students:
$$\cfrac { w1 }{ w2 } =\cfrac { A2-Aavg }{ Aavg-A1 } \\ \cfrac { 5 }{ 5 } =\cfrac { 8-x }{ x-B } \\ \cfrac { 1 }{ 1 } =\cfrac { 8-x }{ x-B } \\ x-A=8-x\\ 2x-8=B$$
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The average (arithmetic mean) score on a test taken by 10 students was [#permalink]
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05 Aug 2017, 07:35
Think of this as a weighted average problem.
If first 5 students make the average 8 we are told to find out the average of the last 5.
Since the weights are 5 each lets assume that midpoint (Given as average of all = x) for all the students is 10
i.e x = 10
Creating a weighted average:
8(given).......................................................12 (assumed)
|_________________10(x assumed)_________________|
Substituting x = 10 should give us the value of 12
Only A gives us this answer.
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Re: The average (arithmetic mean) score on a test taken by 10 students was [#permalink]
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05 Aug 2017, 07:46
Given:
Sum of 10 students' marks= 10x
Average of 5 of the students = 8 --> sum of 5 students' marks= 8*5= 40
Let S be the sum of remaining students' marks.
40+S = 10x --> S= 10x-40
Thus, average of remaining 5 student's marks= (10x-40)/5 ==> 2x-8
Answer: A
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Re: The average (arithmetic mean) score on a test taken by 10 students was [#permalink]
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23 Jan 2018, 20:00
Although the equations make sense now, I couldn't think of how to set up the problem under the time restriction I set for myself
I let the average of the other 5 students be 10 and x = 9
so 10x = 90
5*10 = 50
40 (given) + 50 has to equal 90
plug in x = 9 to see which results in 10
A
AbdurRakib wrote:
The average (arithmetic mean) score on a test taken by 10 students was x. If the average score for 5 of the students was 8, what was the average score, in terms of x, for the remaining 5 students who took the test?
A) 2x – 8
B) x – 4
C) 8 – 2x
D) 16 – x
E) 8 – ($$\frac{2}{x}$$)
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Re: The average (arithmetic mean) score on a test taken by 10 students was [#permalink]
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16 Sep 2019, 09:02
AbdurRakib wrote:
The average (arithmetic mean) score on a test taken by 10 students was x. If the average score for 5 of the students was 8, what was the average score, in terms of x, for the remaining 5 students who took the test?
A) 2x – 8
B) x – 4
C) 8 – 2x
D) 16 – x
E) 8 – ($$\frac{2}{x}$$)
The average (arithmetic mean) score on a test taken by 10 students was x. If the average score for 5 of the students was 8, what was the average score, in terms of x, for the remaining 5 students who took the test?
Total score for 10 students = 10x
Total score for 5 students = 40
Total score for remaining 5 students = 10x-40
Average score for remaining 5 students = (10x-40)/5 = 2x -8
IMO A
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Re: The average (arithmetic mean) score on a test taken by 10 students was [#permalink] 16 Sep 2019, 09:02
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# A bag of 10 marbles contains 3 red marbles and 7 blue
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1. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least one marble is blue?
2. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that none is blue?
3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?
4. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that two marbles are blue?
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Re: Combinatorics - at least, none .... [#permalink]
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21 Jan 2010, 01:35
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Bullet wrote:
Can any body please explain Question No.3 using probability
Why we need to add twice
7/10*6/9*3/8 + 7/10*6/9*5/8 = 7/15
Thanks and appreciated
The solution you are posting for the third question is not right. Below is the solution of this question using the "probability".
3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?
Probability of at least two marble are blue is the sum of the two probabilities:
A. Two marbles are blue and one is red - BBR. This can occur in $$\frac{3!}{2!}=3$$ # of ways, which is basically the # of permutations of three letters B, B, and R: BBR, BRB, RBB. $$3*\frac{7}{10}*\frac{6}{9}*\frac{3}{8}$$;
B. All three marbles are blue - BBB. This can occur only one way, namely BBB. $$\frac{7}{10}*\frac{6}{9}*\frac{5}{8}$$
So $$P=3*\frac{7}{10}*\frac{6}{9}*\frac{3}{8}+\frac{7}{10}*\frac{6}{9}*\frac{5}{8}=\frac{49}{60}$$
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Re: Combinatorics - at least, none .... [#permalink]
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29 Aug 2009, 15:50
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I would like to make a recap of this problem, as in the prior posts there are 3 different answers for questions 3 and 4.
Have 3 red and 7 blue.
1) Pick 2. P of picking at least 1 blue?
Combinatorics: P(1b&1r)+P(2b) -> (7C1 x 3C1 + 7C2)/10C2
Combinatorics shortcut: 1-P(2r) -> 1 - 3C2/10C2
Probability: P(BR)+P(RB)+P(BB) = 7/10*3/9+ 3/10*7/9+ 7/10*6/9
2) Pick 2. P of picking 0 blue?
Combinatorics: P(2r) -> 3C2/10C2
Probability: P(RR) = 3/10*2/9
3) Pick 3. P of picking at least 2 blue?
Combinatorics: P(2b&1r)+P(3b) -> (7C2 x 3C1 + 7C3)/10C3
Probability: P(BBR)+P(BRB)+P(RBB)+P(BBB) = 7/10*6/9*3/8+ 7/10*3/9*6/8+ 3/10*7/9*6/8+ 7/10*6/9*5/8 = 7/40+7/40+7/40+7/24
Answer: 49/60 (7/15 and 42/90 are wrong; do the math)
4) Pick 3. P of picking 2 blue?
Combinatorics: P(2b&1r) -> (7C2 x 3C1)/10C3
Probability: P(BBR)+P(BRB)+P(RBB) = 7/10*6/9*3/8+ 7/10*3/9*6/8+ 3/10*7/9*6/8
Answer: 21/40 (8/15 and 7/40 are wrong; do the math)
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Re: Combinatorics - at least, none .... [#permalink]
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06 Dec 2007, 15:40
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bmwhype2 wrote:
1. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least one marble is blue?
1- [3!/(2!1!)]/[10!/(2!8!)]=14/15
2. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that none is blue?
[3!/(2!1!)]/[10!/(2!8!)]=1/15
3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least two marbles are blue?
[7!/(2!5!)]/[10!/(2!8!)]=7/15
4. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that two marbles are blue?
[7!/(2!5!)]/[10!/(2!8!)]=7/15
Personally, I really don't know what is the difference between question 3 & 4.
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Re: Combinatorics - at least, none .... [#permalink]
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07 Dec 2007, 00:54
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1. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least one marble is blue?
At least 1 is blue = (1b and 1r) + 2 blue
total = 10c2
prob = (7c1 x 3c1 + 7c2)/10c2 = 42/45 = 14/15
2. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that none is blue?
prob = 3c2/10c2 = 3/45 = 1/15
3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?
= (7c2 x 3c1 + 7c3)/10c3
= (63 + 35)/10c3
= 98/120
= 49/60
4. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that two marbles are blue?
= (7c2 x 3c1 )/10c3
= (63)/10c3
= 63/120
= 21/40
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Re: Combinatorics - at least, none .... [#permalink]
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10 Jan 2008, 12:25
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bmwhype2 wrote:
1. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least one marble is blue?
2. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that none is blue?
3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?
4. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that two marbles are blue?
1: at least one is blue: how bout 1-none are blue? 3/10*2/9 --> 6/90 1-1/15 = 14/15
2: 3/10*2/9 = 1/15
3: At least 2 are blue: 1-1 is blue none are blue 3/10*2/9*1/8 =6/720
3/10*2/9*7/8(3) b/c we have 3!/2! ways to arrange RRB
378/720 --> 384/720 192/360 --> 96/180--> 48/90--> 8/15 --> 7/15 is our answer
4: 7/10*6/9*3/8 * (3) again b/c 3!/2! 378/720 = 8/15
hrmmm appears I am wrong on 3 and 4
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Re: Combinatorics - at least, none .... [#permalink]
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10 Jan 2008, 19:32
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1
1. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least one marble is blue?
scenario 1: 1 red, 1 blue
scenario 2: 2 blue
total possibility
10C2
(scenario 1 + scenario 2)/total possibilities = (3C1 x 7C1 +7C2)/10C2
2. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that none is blue?
scenario 2/total possibilities
3c2/10c2
3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?
3c1*7c2+7c3/10c3
4. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that two marbles are blue?
7c2*3c1/10c3
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Re: Combinatorics - at least, none .... [#permalink]
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28 Jun 2008, 11:24
a) 3/10 * 2/9 = 14/15
b) 1-14/15 = 1/15
c) 7/10*6/9*3/8 + 7/10*6/9*5/8 = 7/15
another way is - ((3c1*7c2)/(3!/2!) + 7c3) / 10c3 = 56/120=7/15
d) 7/10*6/9*3/8 = 7/40
another way is - (3c1*7c2)/(3!/2!) / (10c3)= 21/120= 7/40
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Re: Combinatorics - at least, none .... [#permalink]
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23 Jul 2008, 17:39
I also got question 3 and 4 wrong on the first try.
For 3, I did (7C2 + 7C3) / 10C3 = 7/15 b/c number of ways of picking 2 blue (7C2) plus number of ways of picking three blue (7C3) divided by the total number of ways of picking a group of three out of ten possible (10C3).
However, it seems as though 7C2 is not equal to the number of ways of picking 2 blue, because you are ignoring all of the ways to pick one Red marble (which must happen if you pick two blue). this can be represented as...
7C2 X 3C1
So, if you use the above as the number of ways to pick two blue marbles, you get exactly what GMAT TIGER got.
= (7c2 x 3c1 + 7c3)/10c3
= (63 + 35)/10c3
= 98/120
= 49/60
which is correct.
does this make sense? I am going through Walker's excellent list of comb.and prob. problems, and am really trying to understand the theory.
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Re: Combinatorics - at least, none .... [#permalink]
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04 Oct 2008, 10:23
1
Thanks bmwhype2 for posting all these questions.
And thanks walker for compiling them very well. I am going through your list right now...
•1-(all red) = 1-(3/10*2/9) = 1-(1/15) = 14/15
•All red = 3/10*2/9 = 1/15
•3 blue + 3(2 blue, 1red) = RRR+RRB+BRR+RBR = 7/24+7/40+7/40+7/40 = 98/120 = 49/60
•2blue,1 red = 7/10*6/9*3/8 = 7/40 (You can’t look for 3 Blue, because the questions asks for 2 Blue, so the 3rd one has to be red)
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Re: Combinatorics - at least, none .... [#permalink]
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06 Feb 2009, 14:08
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djveed wrote:
I'm not getting it. For A, I would assume the chance of a marble on the first pull is 7/10, and the chance on the second is 6/9. Multiply those two to get the chance that one is blue and you get a 7/15 chance that one of the two marbles pulled are blue. Which is different than what you all have.
And for B, chance of red is 3/10 on first and 2/9 on second, and when multiplied is 1/15 chance. What am I doing wrong?
there are lot ways to do this problem..
You are trying to solve the problem using below method.. but missing some steps..
you considered only p(2B) but missed p(1B,1R)+ p(1R,1B)
1. At least one Blue Marble =
= p(1B,1R)+ p(1R,1B)+[color=#0000FF]p(2B)[/color] = (7/10)*(3/9)+(3/10)*(7/9)+ (7/10)*(6/9)
= 14/45
Another Way. = 7C1*3C1/10C2 + 7C2/10C2
= 21/45 + 21/45 = 14/45
best way = 1- P(2R) = 1- 3C2/10C2 = 14/15.
Is it clear now..
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Re: Combinatorics - at least, none .... [#permalink]
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06 Feb 2009, 15:30
since I need practice with these types of questions, I attempted nevertheless.
1.
probability that no blue is picked= 3C2/10C2 = 3x2/(10x9) = 1/15
Hence, atleast one blue is picked = 14/15
2.
clearly 1/15
3.
atleast two blue=> 2 out of 3 blue or all 3 blue.
Thus,
(7c2x3c1+7c3)/10c3 = (7x6x3+7x6x5)/10x9x8 = 42/90
4.
two are blue=>
7c2x3c1/10c3 = 7x6x3/(10x9x8) = 7/40.
dominion wrote:
1. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least one marble is blue?
scenario 1: 1 red, 1 blue
scenario 2: 2 blue
total possibility
10C2
(scenario 1 + scenario 2)/total possibilities = (3C1 x 7C1 +7C2)/10C2
2. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that none is blue?
scenario 2/total possibilities
3c2/10c2
3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?
3c1*7c2+7c3/10c3
4. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that two marbles are blue?
7c2*3c1/10c3
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Re: Combinatorics - at least, none .... [#permalink]
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27 Sep 2009, 01:53
1. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least one marble is blue?
Soln: (7C2 + 7C1*3C1)/10C2
2. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that none is blue?
Soln: 3C2/10C2
3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?
Soln: (7C2*3C1 + 7C3)/10C3
4. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that two marbles are blue?
Soln: 7C2*3C1/10C3
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Re: Combinatorics - at least, none .... [#permalink]
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21 Jan 2010, 02:49
Bunuel wrote:
Bullet wrote:
Can any body please explain Question No.3 using probability
Why we need to add twice
7/10*6/9*3/8 + 7/10*6/9*5/8 = 7/15
Thanks and appreciated
The solution you are posting for the third question is not right. Below is the solution of this question using the "probability".
3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?
Probability of at least two marble are blue is the sum of the two probabilities:
A. Two marbles are blue and one is red - BBR. This can occur in $$\frac{3!}{2!}=3$$ # of ways, which is basically the # of permutations of three letters B, B, and R: BBR, BRB, RBB. $$3*\frac{7}{10}*\frac{6}{9}*\frac{3}{8}$$;
B. All three marbles are blue - BBB. This can occur only one way, namely BBB. $$\frac{7}{10}*\frac{6}{9}*\frac{5}{8}$$
So $$P=3*\frac{7}{10}*\frac{6}{9}*\frac{3}{8}+\frac{7}{10}*\frac{6}{9}*\frac{5}{8}=\frac{49}{60}$$
Thanks Bunuel. I posted the solution from the thread as i was confused with the solution.
So what you're saying is that we need to multiply the permutations with the probability. Having said that we will do only when we need to choose at least two marbles or any other thing from the whole?
Thanks
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Re: Combinatorics - at least, none .... [#permalink]
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21 Jan 2010, 03:56
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Bullet wrote:
Thanks Bunuel. I posted the solution from the thread as i was confused with the solution.
So what you're saying is that we need to multiply the permutations with the probability. Having said that we will do only when we need to choose at least two marbles or any other thing from the whole?
Thanks
To make it simple: suppose we have the jar of 10 marbles - 5 red, 2 blue and 3 green. If five marbles are selected at random, what is the probability that two will be red, one blue and two green?
We are looking for all the cases with 2R, 1B and 2G. We can draw these marbles like: RRBGG or GGBRR or RBGGR ... So how many combinations of the drawing of these marbles are there? The answer is as many as there is permutation of the letters GGBRR, which is $$\frac{5!}{2!2!}$$.
Hence the answer for the above question would be $$\frac{5!}{2!2!}*\frac{5}{10}*\frac{4}{9}*\frac{2}{8}*\frac{3}{7}*\frac{2}{6}$$.
If the question were: three marbles are selected at random, what is the probability that all three will be red?
RRR can occur only in one way: RRR, so the probability would be $$\frac{5}{10}*\frac{4}{9}*\frac{3}{8}$$.
You can check the Probability and Combination chapters in the Math Book (link below) for more.
Also check my posts at:
probability-colored-balls-55253.html#p637525
4-red-chips-and-2-blue-chips-85987.html#p644603
probability-qs-attention-88945.html#p671958
p-c-88431.html?highlight=probability+of+occurring+event
probability-88069.html?highlight=probability+of+occurring+event
combination-problem-princenten-review-2009-bin-4-q2-87673.html?highlight=probability+of+occurring+event
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Re: Combinatorics - at least, none .... [#permalink]
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21 Jan 2010, 05:31
Bunuel wrote:
Bullet wrote:
Thanks Bunuel. I posted the solution from the thread as i was confused with the solution.
So what you're saying is that we need to multiply the permutations with the probability. Having said that we will do only when we need to choose at least two marbles or any other thing from the whole?
Thanks
To make it simple: suppose we have the jar of 10 marbles - 5 red, 2 blue and 3 green. If five marbles are selected at random, what is the probability that two will be red, one blue and two green?
We are looking for all the cases with 2R, 1B and 2G. We can draw these marbles like: RRBGG or GGBRR or RBGGR ... So how many combinations of the drawing of these marbles are there? The answer is as many as there is permutation of the letters GGBRR, which is $$\frac{5!}{2!2!}$$.
Hence the answer for the above question would be $$\frac{5!}{2!2!}*\frac{5}{10}*\frac{4}{9}*\frac{2}{8}*\frac{3}{7}*\frac{2}{6}$$.
If the question were: three marbles are selected at random, what is the probability that all three will be red?
RRR can occur only in one way: RRR, so the probability would be $$\frac{5}{10}*\frac{4}{9}*\frac{3}{8}$$.
You can check the Probability and Combination chapters in the Math Book (link below) for more.
Also check my posts at:
probability-colored-balls-55253.html#p637525
4-red-chips-and-2-blue-chips-85987.html#p644603
probability-qs-attention-88945.html#p671958
p-c-88431.html?highlight=probability+of+occurring+event
probability-88069.html?highlight=probability+of+occurring+event
combination-problem-princenten-review-2009-bin-4-q2-87673.html?highlight=probability+of+occurring+event
Thanks Mate, I got it what I was missing in the whole thing.
But I definitely go into your posts in order to me 100% sure.
Cheers!
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Re: Combinatorics - at least, none .... [#permalink]
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22 Feb 2012, 14:11
powerka wrote:
3) Pick 3. P of picking at least 2 blue?
Combinatorics: P(2b&1r)+P(3b) -> (7C2 x 3C1 + 7C3)/10C3
Answer: 49/60 (7/15 and 42/90 are wrong; do the math)
#3. P(at least 2 blues) = P(exactly 2 blues) + P (3 blues)
P (exactly 2 blues)= $$7C2/10C3$$=$$21/120$$
P (3 blues) = $$7C3/10C3$$=$$35/120$$
Therefore,
P(at least 2 blues) = 21/120 + 35/120 = 56/120 = 7/15
I see that the commented section above has an extra x 3C1 that's probably the cause for my discrepancy. Why do I need to multiply $$7C2$$ by 3C1 to get P (exactly 2 blues)? Or, is my answer $$7/15$$correct?
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Re: Combinatorics - at least, none .... [#permalink]
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22 Feb 2012, 14:22
1
fxsunny wrote:
powerka wrote:
3) Pick 3. P of picking at least 2 blue?
Combinatorics: P(2b&1r)+P(3b) -> (7C2 x 3C1 + 7C3)/10C3
Answer: 49/60 (7/15 and 42/90 are wrong; do the math)
#3. P(at least 2 blues) = P(exactly 2 blues) + P (3 blues)
P (exactly 2 blues)= $$7C2/10C3$$=$$21/120$$
P (3 blues) = $$7C3/10C3$$=$$35/120$$
Therefore,
P(at least 2 blues) = 21/120 + 35/120 = 56/120 = 7/15
I see that the commented section above has an extra x 3C1 that's probably the cause for my discrepancy. Why do I need to multiply $$7C2$$ by 3C1 to get P (exactly 2 blues)? Or, is my answer $$7/15$$correct?
That's because you are picking 3 marbles and if you pick 2 blue then the third one must be red: BBR. Ways to pick one red marble out of 3 is $$C^1_3$$.
Complete solution using combinations: $$P(R\geq{2})=P(R=2)+P(R=3)=\frac{C^2_7*C^1_3}{C^3_{10}}+\frac{C^3_7}{C^3_{10}}$$.
Hope it's clear.
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3 red marbles and 7 blue marbles [#permalink]
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07 Mar 2012, 16:59
The below questions are discussed at the below thread, but i have a silly doubt so posting that here:
a-bag-of-10-marbles-contains-3-red-marbles-and-7-blue-56728.html
I understand the solution provided, but is there anything wrong with the simple method as below as i am getting different ans ?
A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?
Case 1: Only 2 marbles are blue. Prob: 7/10 * 6/9 * 3/8
Case 2: All 3 marbles are blue. Prob: 7/10 * 6/9 * 5/8
A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that two marbles are blue?
Only 2 marbles are blue. Prob: 7/10 * 6/9 * 3/8
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Re: 3 red marbles and 7 blue marbles [#permalink]
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07 Mar 2012, 17:10
rohitgoel15 wrote:
A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?
Case 1: Only 2 marbles are blue. Prob: 7/10 * 6/9 * 3/8
Case 2: All 3 marbles are blue. Prob: 7/10 * 6/9 * 5/8
The point is that BBR case can occur in 3 different ways: BBR, BRB, and RBB. So you should multiply 7/10 * 6/9 * 3/8 by 3.
This is discussed here: a-bag-of-10-marbles-contains-3-red-marbles-and-7-blue-56728.html#p677059 and here: a-bag-of-10-marbles-contains-3-red-marbles-and-7-blue-56728.html#p677083
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Re: 3 red marbles and 7 blue marbles &nbs [#permalink] 07 Mar 2012, 17:10
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Mary and Mike enter into a partnership by investing $700 and$300 respectively. At the end of one year, they divided their profits such that a third of the profit is divided equally for the efforts they have put into the business and the remaining amount of profit is divided in the ratio of the investments they made in the business. If Mary received $800 more than Mike did, what was the profit made by their business in that year? A.$2000
B. $6000 C.$4000
D. $1333 E.$3000
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Re: Mary and Mike enter into a partnership by investing $700 and [#permalink] ### Show Tags 22 Aug 2013, 11:54 1 Asifpirlo wrote: Mary and Mike enter into a partnership by investing$700 and $300 respectively. At the end of one year, they divided their profits such that a third of the profit is divided equally for the efforts they have put into the business and the remaining amount of profit is divided in the ratio of the investments they made in the business. If Mary received$800 more than Mike did, what was the profit made by their business in that year?
A. $2000 B.$6000
C. $4000 D.$1333
E. $3000 Say the profit was$x.
Mary's share = x/6 (half of the third) + (x-x/3)*0.7
Mike's share = x/6 (half of the third) + (x-x/3)*0.3
Thus (x-x/3)*0.7-(x-x/3)*0.3=800 --> x=3000.
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Re: Mary and Mike enter into a partnership by investing $700 and [#permalink] ### Show Tags 22 Aug 2013, 19:15 they divided 2/3rd amount in 7:3 ratio and 1/3 amount as 1:1 ratio. We can observe that the$800 excess that mary got is from 2/3rd amount, because they shared 1/3rd in 1:1 ratio.
so different in 2/3rd amount -> 7x-3x = 800 => x = 200
total amount shared in 7:3 ratio is 10x = 200*10 => 2000
now 2/3rd amount = 2000 = > amount = 3000
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Re: Mary and Mike enter into a partnership by investing $700 and [#permalink] ### Show Tags 19 Oct 2017, 10:23 Asifpirlo wrote: Mary and Mike enter into a partnership by investing$700 and $300 respectively. At the end of one year, they divided their profits such that a third of the profit is divided equally for the efforts they have put into the business and the remaining amount of profit is divided in the ratio of the investments they made in the business. If Mary received$800 more than Mike did, what was the profit made by their business in that year?
A. $2000 B.$6000
C. $4000 D.$1333
E. $3000 Since Mary and Mike invest$700 and $300, respectively, into the partnership, they will get 7/10 and 3/10 of the profit, respectively, after ⅓ of the profit is divided equally. In other words, Mary will get 7/10 and Mike will get 3/10 of the remaining ⅔ of the profit. If we let x = the amount of profit in dollars, then Mary gets (½)(⅓)(x) + (7/10)(⅔)(x) = x/6 + 7x/15 dollars and Mike gets (½)(⅓)(x) + (3/10)(⅔)(x) = x/6 + x/5 dollars. Since Mary gets$800 more than Mike, we have the following equation:
x/6 + 7x/15 = x/6 + x/5 + 800
7x/15 = x/5 + 800
Multiplying both sides by 15, we have:
7x = 3x + 12,000
4x = 12,000
x = 3,000
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Re: Mary and Mike enter into a partnership by investing $700 and [#permalink] ### Show Tags 19 Oct 2017, 10:39 So they have divided 1/3 of profit amount equally. Rest 2/3 of the amount is to be split in the ratio 7:3. So Mary gets 70% and Mike gets 30%. This creates a difference of 40%, which in turn is given to be$800
So 40% of 2/3 of total amount = $800 So total amount = (800*3)/(2*0.4) =$3000
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Re: Mary and Mike enter into a partnership by investing $700 and [#permalink] ### Show Tags 07 Jan 2018, 03:00 Asifpirlo wrote: Mary and Mike enter into a partnership by investing$700 and $300 respectively. At the end of one year, they divided their profits such that a third of the profit is divided equally for the efforts they have put into the business and the remaining amount of profit is divided in the ratio of the investments they made in the business. If Mary received$800 more than Mike did, what was the profit made by their business in that year?
A. $2000 B.$6000
C. $4000 D.$1333
E. $3000 Let Mike's Profit = x, then Mary's Profit = x + 800 The remaining of the total profit (2/3 of the profit) is distributed the same as their ratios in investment. $$\frac{x}{x+ 800}$$ = $$\frac{300}{700}$$ Solve for x, we get x =600 2/3 of the total profit =2000 Total profit = 3000 Answer: E Re: Mary and Mike enter into a partnership by investing$700 and &nbs [#permalink] 07 Jan 2018, 03:00
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# Three grades of milk are 1 percent, 2 percent and 3 percent
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Three grades of milk are 1 percent, 2 percent and 3 percent [#permalink] 24 Dec 2009, 08:38
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Three grades of milk are 1 percent, 2 percent and 3 percent fat by volume. If x gallons of the 1 percent grade, y gallons of the 2 percent grade, and z gallons of the 3 percent grade are mixed to give x+y+z gallons of a 1.5 percent grade, what is x in terms of y and z?
A. y + 3z
B. (y +z) / 4
C. 2y + 3z
D. 3y + z
E. 3y + 4.5z
[Reveal] Spoiler: OA
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Re: GMAT Prep VIC Problem [#permalink] 24 Dec 2009, 09:23
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JimmyWorld wrote:
I got this problem wrong on the GMAT Prep and don't really understand it.
Three grades of milk are 1 percent, 2 percent and 3 percent fat by volume. If x gallons of the 1 percent grade, y gallons of the 2 percent grade, and z gallons of the 3 percent grade are mixed to give x+y+z gallons of a 1.5 percent grade, what is x in terms of y and z?
A. y + 3z
B. (y +z) / 4
C. 2y + 3z
D. 3y + z
E. 3y + 4.5z
OA is
[Reveal] Spoiler:
A
Milk concentration in mix would be 1%x+2%y+3%z and on the other hand we are told that in (x+y+z) there is 1.5% of milk. Hence:
$$1%x+2%y+3%z=1.5%(x+y+z)$$ --> $$x+2y+3z=1.5x+1.5y+1.5z$$ --> $$0.5x=0.5y+1.5z$$ --> $$x=y+3z$$.
Hope it's clear.
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Re: GMAT Practice Test 1 PS: Ratio [#permalink] 15 Jan 2012, 08:46
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$$\frac{x+2y+3z}{x+y+z} = \frac{3}{2}$$
$$x = y+3z$$
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Re: Three grades of milk are 1 percent, 2 percent and 3 percent [#permalink] 15 Jan 2012, 16:00
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Its A, just put the words into equation
x+2Y+3Z=1.5(X+2Y+3Z)
X=Y+3Z
Hope it clarifies
+1 Kudos if it helps
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Re: Three grades of milk are 1 percent, 2 percent and 3 percent [#permalink] 22 Jan 2012, 07:04
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Re: Mixture of different grades (Milk fat by volume) [#permalink] 22 Jan 2012, 17:56
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MSoS wrote:
Hi, would someone please so kind and explain the question:
Three grades of milk are 1 percent, 2 percent, and 3 percent fat by volume. If x gallons of the q percent grade, y gallons of the 2 percent grade, and z gallons of the 3 percent grade are mixed to give x + y + z gallons of a 1.5 percent grade, what is x in terms of y and z?
(a) y + 3z
(b) (y+z)/4
(c) 2y +3z
(d) 3y + z
(e) 3y + 4.5z
Thanks a lot...
A quick approach:
The question asks you for x in terms of y and z. Whatever values x, y and z can take, this relation should hold.
Since we mix 1%, 2% and 3% milk and get 1.5% milk, one way of mixing them could be that 1% and 2% are mixed in equal quantities (to give 1.5% milk) and 3% milk is not added at all. Which means x = 1, y = 1 and z = 0 should satisfy the relation between x, y and z.
The only relation that satisfies these values is (A).
Note: If multiple options satisfied these values, you could take another set of values e.g. x = 3, y = 0 and z = 1 and check out of the shortlisted options.
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Re: GMAT prep question 1 [#permalink] 22 Mar 2012, 02:16
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Three grades of milk are 1 percent, 2 percent, and 3 percent fat by volume. If x gallons of the 1 percent grade, y gallons of the 2 percent grade, and z gallons of the 3 percent grade are mixed to give x+y+z gallons of 1.5 percent grade, what is x in terms of y and z ?
A- y+3z
B- (y+z)/4
C- 2y +3z
D- 3y+z
E-3y+4.5z
0.01x+0.02y+0.03z=0.015(x+y+z)
=> x=y+3z
hence A
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Three grades of milk are 1 percent, 2 percent and 3 percent [#permalink] 10 May 2012, 10:33
Hi all,
I think I have seen a problem like this somewhere, but in DS form
If
(1) is x= y + 3z
and (2) gives you the y:z ratio
Is the second one sufficient? I somehow feel that it should be, but can't find the reasoning for that.What can we do here? pick numbers? Or is it exessive info and thus is sufficient?
sorry, I can't quote the exact second choice.
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Three grades of milk are 1 percent, 2 percent and 3 percent [#permalink] 12 May 2012, 22:29
Yep this one would have seemed more obtuse to me until I realized that the percentages were meant to be for the fat content in milk.
Combining the various milk types we got a 1.5% of fat content in the resulting mixture.
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help need! hard weighted average question [#permalink] 17 Sep 2012, 16:06
1
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three grades of milk are 1 percent 2 percent and 3 percent fat by volume. if x gallons of the 1 percent y gallons of the 2 percent and z gallons of the 3 percent are mixed to give x +y+z gallons of a 1.5 percent grade, what is x in terms of y and z? Thanks in advance!!!
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Re: Three grades of milk are 1 percent, 2 percent and 3 percent [#permalink] 28 Sep 2012, 02:10
1%x + 2%y + 3%z = 1.5% (x+y+z)
2%y - 1.5%y + 3%z - 1.5%z = 1.5%x - 1%x
2(.5y + 1.5z = .5x)
y + 3z = x
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Re: Three grades of milk are 1 percent, 2 percent, and 3 percent [#permalink] 22 Oct 2012, 19:01
(x/100)+ (2y/100)+(3z/100) = 1.5 (x+y+z)/100 - > cancel out 100 on each side.
x+2y+3z = 1.5x+1.5y+1.5z -> bring x to one side of = sign
.5x=.5y+1.5x -> multiply by 2 on both side
x=y+3z
______________
[Reveal] Spoiler:
A
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Re: GMAT Practice Test 1 PS: Ratio [#permalink] 25 Jul 2013, 13:22
MBAhereIcome wrote:
$$\frac{x+2y+3z}{x+y+z} = \frac{3}{2}$$
$$x = y+3z$$
This is essentially viewing this problem as a weighted average correct?
The mixture divided by the sum of weights?
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Re: GMAT Practice Test 1 PS: Ratio [#permalink] 25 Jul 2013, 20:52
Expert's post
NvrEvrGvUp wrote:
MBAhereIcome wrote:
$$\frac{x+2y+3z}{x+y+z} = \frac{3}{2}$$
$$x = y+3z$$
This is essentially viewing this problem as a weighted average correct?
The mixture divided by the sum of weights?
Yes, it is just the weighted average of the fat concentration.
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Re: Three grades of milk are 1 percent, 2 percent and 3 percent [#permalink] 12 Jan 2014, 08:51
I approached it as a residuals problem
Since 1% fat = -0.5% from average
2% fat = 0.5% from average
3% fat = 1.5% from average
When added together, they have to create a 0 'residual' from 1.5% average:
-0.5x + 0.5y + 1.5z = 0
0.5y + 1.5z = 0.5x
[Reveal] Spoiler:
y + 3z = x --> A
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Re: Three grades of milk are 1 percent, 2 percent and 3 percent [#permalink] 12 Jan 2014, 23:27
JimmyWorld wrote:
Three grades of milk are 1 percent, 2 percent and 3 percent fat by volume. If x gallons of the 1 percent grade, y gallons of the 2 percent grade, and z gallons of the 3 percent grade are mixed to give x+y+z gallons of a 1.5 percent grade, what is x in terms of y and z?
A. y + 3z
B. (y +z) / 4
C. 2y + 3z
D. 3y + z
E. 3y + 4.5z
Using weighted average:
(x) (1) + (y) (2) + (z)(3) = (1.5) (x + y + z)
x + 2y + 3z = 1.5x + 1.5y + 1.5z
0.5y + 1.5z = 0.5x
Removing 0.5 overall:
y +3z = x
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Three grades of milk are 1 percent, 2 percent and 3 percent fat [#permalink] 29 Jan 2014, 14:33
I solved this by using differentials
1% - x
2% - y
3% - z
So then x+y+z is also given as 1.5%, which is in fact the weighted average
Let's just play with nice numbers here
10% - x
20% - y
30% - z
x + y + z = 15%
So -5x+5y+15z = 0
X = Y + 3Z
A
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Re: Percent / Average [#permalink] 29 Jan 2014, 20:22
Expert's post
Impenetrable wrote:
Three grades of milk are 1%, 2% and 3% fat by volume. If x gallons of 1%, y gallons of 2% and z gallons of 3% are mixed together to give x+y+z gallons of a 1.5%, what is x in terms of y and z?
y+3z
(y+z)/4
2y+3z
3y+z
3y+4.5z
My idea was:
(x+2y+3z)/(x+y+z) = 1.5
from here on I have no idea how to get x to one side...
Cheers,
Lars
If you develop a knack for playing with numbers, you will rarely need to make equations for ratios/percent/mixture/average problems.
What I thought here was that milk of 1% (volume x), 2% (volume y) and 3% (volume z) have to be mixed to give 1.5%. An easy way I can see immediately is that I don't take any 3% milk and mix 1% and 2% in equal quantities to get 1.5%.
i.e. If z = 0, x = y
If we put z = 0, only option (A) gives x = y hence it is the answer.
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Re: Three grades of milk are 1 percent, 2 percent and 3 percent [#permalink] 26 Oct 2014, 09:13
Bunuel - Where can I find challenging mixture problems/weighted avg problems to practice? Can you please help?
Thanks.
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Re: Three grades of milk are 1 percent, 2 percent and 3 percent [#permalink] 27 Oct 2014, 05:20
Expert's post
2
This post was
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p2bhokie wrote:
Bunuel - Where can I find challenging mixture problems/weighted avg problems to practice? Can you please help?
Thanks.
Check our Questions Bank: viewforumtags.php
All DS Mixture Problems to practice: search.php?search_id=tag&tag_id=43
All PS Mixture Problems to practice: search.php?search_id=tag&tag_id=114
GMAT PS Question Directory by Topic & Difficulty: gmat-ps-question-directory-by-topic-difficulty-127957.html
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Hope this helps.
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Re: Three grades of milk are 1 percent, 2 percent and 3 percent [#permalink] 27 Oct 2014, 05:20
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# If (n+2)!/n!=156, n=?
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If (n+2)!/n!=156, n=?
A. 2/131
B. 9
C. 10
D. 11
E. 12
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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Intern Joined: 02 Apr 2016 Posts: 1 Re: If (n+2)!/n!=156, n=? [#permalink] ### Show Tags 14 Dec 2016, 01:51 1 it should be D. (n+2)! = (n+2)*(n+1)*(n)! Therefore (n+2)!/n! = 156 can be written as (n+2)(n+1) = 156-> n = -14,11 From options n=11 is correct Board of Directors Status: QA & VA Forum Moderator Joined: 11 Jun 2011 Posts: 4834 Location: India GPA: 3.5 WE: Business Development (Commercial Banking) Re: If (n+2)!/n!=156, n=? [#permalink] ### Show Tags 14 Dec 2016, 11:04 1 1 MathRevolution wrote: If (n+2)!/n!=156, n=? A. 2/131 B. 9 C. 10 D. 11 E. 12 $$156$$ = $$2^2$$ x $$3^1$$ x $$13^1$$ Or, $$156$$ = $$12$$ x $$13$$ Thus, $$\frac{(n+2)!}{n!} = \frac{( 11 + 2 )!}{11!}$$ = $$\frac{13!}{11!}$$ = $$13*12$$ Hence, answer will be (D) 11 _________________ Thanks and Regards Abhishek.... PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only ) Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8235 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: If (n+2)!/n!=156, n=? [#permalink] ### Show Tags 16 Dec 2016, 00:36 ==> From (n+2)!/n!=156 and (n+2)(n+1)n!/n!=(n+2)(n+1)=156=13*12, you get n+2=13, n=11. Therefore, the answer is D. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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16 Dec 2016, 10:32
MathRevolution wrote:
If (n+2)!/n!=156, n=?
A. 2/131
B. 9
C. 10
D. 11
E. 12
(n+2)(n+1)n!/n! = 156
(n+2)(n+1)=156
solving n=11
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20 Dec 2016, 01:47
Using the factorial main rule (n+2)!/n!=156 ; n!(n+2)(n+1)/n!=156
divide both sides by n! we get (n+2)(n+1) = 156
Then using substitution approach it is clear that Answer D is correct.
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11 May 2017, 02:39
MathRevolution wrote:
If (n+2)!/n!=156, n=?
A. 2/131
B. 9
C. 10
D. 11
E. 12
(n+2)!/n!=156
-> (n+2)(n+1) = 156 = 13x12
-> n+1 = 12
-> n =11
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16 May 2017, 18:15
MathRevolution wrote:
If (n+2)!/n!=156, n=?
A. 2/131
B. 9
C. 10
D. 11
E. 12
(n + 2)!/n! = (n + 2)(n + 1)n!/n! = (n + 2)(n + 1) = 156
n^2 + 3n + 2 = 156
n^2 + 3n - 154 = 0
(n + 14)(n - 11) = 0
n = -14 or n = 11
We see that only n = 11 is among the answer choices.
Alternate Solution:
(n + 2)!/n! = (n + 2)(n + 1)n!/n! = (n + 2)(n + 1) = 156
We observe that 156 is the product of n + 2 and n + 1, which are two consecutive integers. Scanning the answer choices, we see that all but one are positive integers. Let’s prime factorize 156 to see if it can be expressed as the product of two consecutive integers:
156 = 2^2 x 3 x 13
Since 2^2 x 3 = 12, we observe that 156 = 12 x 13. Thus, n + 1 = 12 and n + 2 = 13. This yields n = 11.
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16 Jul 2017, 23:18
Since n! in numerator and denominator would cancel out to leave the product of (n+1).(n+2) which is 156 here.
The trick is to get the factors of 156 to identify n. Option D.
Re: If (n+2)!/n!=156, n=? [#permalink] 16 Jul 2017, 23:18
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http://math.stackexchange.com/questions/122898/why-are-the-solutions-of-polynomial-equations-so-unconstrained-over-the-quaterni/122909 | # Why are the solutions of polynomial equations so unconstrained over the quaternions?
An $n$th-degree polynomial has at most $n$ distinct zeroes in the complex numbers. But it may have an uncountable set of zeroes in the quaternions. For example, $x^2+1$ has two zeroes in $\mathbb C$, but in $\mathbb H$, ${\bf i}\cos x + {\bf j}\sin x$ is a distinct zero of this polynomial for every $x$ in $[0, 2\pi)$, and obviously there are many other zeroes.
What is it about $\mathbb H$ that makes its behavior in this regard to be so different from the behavior of $\mathbb R$ and $\mathbb C$? Is it simply because $\mathbb H$ is four-dimensional rather than two-dimensional? Are there any theorems that say when a ring will behave like $\mathbb H$ and when it will behave like $\mathbb C$?
Do all polynomials behave like this in $\mathbb H$? Or is this one unusual?
-
– MJD Mar 21 '12 at 14:46
The non commutativity makes for the strange behaviour. – azarel Mar 21 '12 at 15:15
Although there are infinitely many roots of $x^2+1$ in the quaternions, if you'd consider roots up to conjugation then you recover a finiteness theorem: all the roots form a single conjugacy class. (Over a field, conjugation is trivial.) In fact, with coefficients on the left in any division ring, a polynomial of degree $n$ has at most $n$ conjugacy classes of roots. – KCd Mar 22 '12 at 1:29
math.psu.edu/ballif/assignments/Math%20597%20Graduate%20Seminar/… summarizes the basic results – Ted Mar 22 '12 at 2:43
@ted: Thanks, this was very useful. – MJD Mar 22 '12 at 22:25
When I was first learning abstract algebra, the professor gave the usual sequence of results for polynomials over a field: the Division Algorithm, the Remainder Theorem, and the Factor Theorem, followed by the Corollary that if $D$ is an integral domain, and $E$ is any integral domain that contains $D$, then a polynomial of degree $n$ with coefficients in $D$ has at most $n$ distinct roots in $E$.
He then challenged us, as a homework, to go over the proof of the Factor Theorem and to point out exactly which, where, and how the axioms of a field used in the proof.
Every single one of us missed the fact that commutativity is used.
Here's the issue: the division algorithm (on either side), does hold in $\mathbb{H}[x]$ (in fact, over any ring, commutative or not, in which the leading coefficient of the divisor is a unit). So given a polynomial $p(x)$ with coefficients in $\mathbb{H}$, and a nonzero $a(x)\in\mathbb{H}[x]$, there exist unique $q(x)$ and $r(x)$ in $\mathbb{H}[x]$ such that $p(x) = q(x)a(x) + r(x)$, and $r(x)=0$ or $\deg(r)\lt\deg(a)$. (There also exist unique $q'(x)$ and $s(x)$ such that $p(x) = a(x)q'(x) + s(x)$ and $s(x)=0$ or $\deg(s)\lt\deg(a)$.
The usual argument runs as follows: given $a\in\mathbb{H}$ and $p(x)$, divide $p(x)$ by $x-a$ to get $p(x) = q(x)(x-a) + r$, with $r$ constant. Evaluating at $a$ we get $p(a) = q(a)(a-a)+r = r$, so $r=p(a)$. Hence $a$ is a root if and only if $(x-a)$ divides $p(x)$.
If $b$ is a root of $p(x)$, $b\neq a$, then evaluating at $b$ we have $0=p(b) = q(b)(b-a)$; since $b-a\neq 0$, then $q(b)=0$, so $b$ must be a root of $q$; since $\deg(q)=\deg(p)-1$, an inductive hypothesis tells us that $q(x)$ has at most $\deg(p)-1$ distinct roots, so $p$ has at most $\deg(p)$ roots.
And that is where we are using commutativity: to go from $p(x) = q(x)(x-a)$ to $p(b) = q(b)(b-a)$.
Let $R$ be a ring, and let $a\in R$. Then $a$ induces a set-theoretic map from $R[x]$ to $R$, "evaluation at $a$", $\varepsilon_a\colon R[x]\to R$ by evaluation: $$\varepsilon_a(b_0+b_1x+\cdots + b_nx^n) = b_0 + b_1a + \cdots + b_na^n.$$ This map is a group homomorphism, and if $a$ is central, also a ring homomorphism; if $a$ is not central, then it is not a ring homomorphism: given $b\in R$ such that $ab\neq ba$, then we have $bx = xb$ in $R[x]$, but $\varepsilon_a(x)\varepsilon_a(b) = ab\neq ba = \varepsilon_a(xb)$.
The "evaluation" map also induces a set theoretic map from $R[x]$ to $R^R$, the ring of all $R$-valued functions in $R$, with the pointwise addition and multiplication ($(f+g)(a) = f(a)+g(a)$, $(fg)(a) = f(a)g(a)$); the map sends $p(x)$ to the function $\mathfrak{p}\colon R\to R$ given by $\mathfrak{p}(a) = \varepsilon_a(p(x))$. This map is a group homomorphism, but it is not a ring homomorphism unless $R$ is commutative.
This means that from $p(x) = q(x)(x-a) + r(x)$ we cannot in general conclude that $p(c) = q(c)(c-a) +r(c)$ unless $c$ commutes in $R$ with $a$. So the Remainder Theorem may fail to hold (if the coefficients involved do not commute with $a$ in $R$), which in turn means that the Factor Theorem may fail to hold So one has to be careful in the statements (see Marc van Leeuwen's answer). And even when both of them hold for the particular $a$ in question, the inductive argument will fail if $b$ does not commute with $a$, because we cannot go from $p(x) = q(x)(x-a)$ to $p(b)=q(b)(b-a)$.
This is exactly what happens with, say, $p(x) = x^2+1$ in $\mathbb{H}[x]$. We are fine as far as showing that, say, $x-i$ is a factor of $p(x)$, because it so happens that when we divide by $x-i$, all coefficients involved centralize $i$ (we just get $(x+i)(x-i)$). But when we try to argue that any root different from $i$ must be a root of $x+i$, we run into the problem that we cannot guarantee that $b^2+1$ equals $(b+i)(b-i)$ unless we know that $b$ centralizes $i$. As it happens, the centralizer of $i$ in $\mathbb{H}$ is $\mathbb{R}[i]$, so we only conclude that the only other complex root is $-i$. But this leaves the possibility open that there may be some roots of $x^2+1$ that do not centralize $i$, and that is exactly what occurs: $j$, and $k$, and all numbers of the form $ai+bj+ck$ with $a^2+b^2+c^2=1$ are roots, and if either $b$ or $c$ are nonzero, then they don't centralize $i$, so we cannot go from $x^2+1 = (x+i)(x-i)$ to "$(ai+bj+ck)^2+1 = (ai+bj+ck+i)(ai+bj+ck-i)$".
And that is what goes wrong, and there is where commutativity is hiding.
-
+1 That exercise is a must for people interested in non-commutative division algebras. It is a bit too easy to fall into the trap that the usual commutative argument would work here as well. – Jyrki Lahtonen Mar 21 '12 at 17:36
Nice elaborate answer. Just some small nitpicks. To conclude $p(c) = q(c)(c-a) +r(c)$ one only needs that $a$ and $c$ commute (the coefficients of $q$ remain at the left, and the coefficient of $r$ (which is constant) is not multiplied at all. And the problem is not that the Remainder and Factor Theorems fail without commutativity (they can be salvaged, see my answer), but the factorization involved does not survive evaluation. – Marc van Leeuwen Mar 23 '12 at 10:46
@MarcvanLeeuwen: Thanks for the observations; I'll clarify. – Arturo Magidin Mar 23 '12 at 14:12
The finiteness of the number of roots of a polynomial $f(x)\in K[x]$ where $K$ is a field depends on two interlaced facts:
• $K[x]$ is a Unique Factorization Domain: every polynomial $f(x)$ factors in an essentially unique way as a product of irreducibles;
• if $f(\alpha)=0$ then $f(x)=(x-\alpha)g(x)$ where $\deg g(x)=(\deg f(x))-1$.
The combination of these two facts (the first one in particular) does not hold anymore if you think the polynomial $f(x)$ as a polynomial with coefficients in the ring $\Bbb H$ of Hamilton quaternions. This is because the latter is not commutative.
You may also ponder on this fact: in a commutative environment the transformation $a\mapsto\phi_h(a)=hah^{-1}$ (conjugation) is always trivial. Not so in $\Bbb H$, again as a side effect of non-commutativity. The point is that if an element $a$ satisfies a certain algebraic relation with real coefficient (such as $a^2=1$), so will all its conjugates $\phi_h(a)$.
-
Many true observations. I have just felt, like Arturo, that the key difference between commutative and non-commutative cases is that the evaluation map $ev_\alpha:f\mapsto f(\alpha)$ is no longer a ring homomorphism from $K[x]$ to $K$. As you explain, this shows also in the non-unique factorization, but the more fundamental problem is that factorizations cannot be used to determine zeros, because the value of a product of polynomials is not the product of values of the polynomial factors. – Jyrki Lahtonen Mar 21 '12 at 17:42
+1: the point about conjugation above is extremely important. – Qiaochu Yuan Mar 21 '12 at 18:05
I would like to emphasize a point which is made in Arturo Magidin's answer but perhaps in different words: if $D$ is a noncommutative division ring, then the ring $D[x]$ of polynomials over $D$ does not do what you want it to do.
If $F$ is a field, then one reason you might care about working with polynomials $F[x]$ is that they describe all the expressions you could potentially get from some unknown $x \in F$ (or perhaps $x \in \bar{F}$ or perhaps something even more general than this) via addition and multiplication.
Why does this break down when you replace $F$ with a noncommutative division ring $D$? The problem is that if you work with some unknown $x \in D$ (or in some ring containing $D$) then $x$, by assumption, doesn't necessarily commute with every element in $D$, so starting from $x$ and adding and multiplying you get not only expressions like $$a_0 + a_1 x + a_2 x^2 + ...$$
but more complicated expressions like $$a_0 + a_{1,0} x + x a_{1,1} + a_{1, 2} x a_{1, 3} + a_{2,0} x^2 + x a_{2,1} x + x^2 a_{2,2} + a_{2, 3} x^2 a_{2,4} + a_{2, 5} x a_{2, 6} x a_{2,7} + ...$$
The resulting algebraic structure is quite a bit more complicated than $D[x]$. Already you can't in general combine expressions of the form $axb$ and $cxd$, so even to describe the expressions you can get by using $x$ once I should've actually written $$a_0 + a_{1,0} x a_{1,1} + a_{1,2} x a_{1,3} + a_{1,4} x a_{1,5} + ....$$
-
That is, the free object is no longer $D[x]$, but $D\langle x\rangle$, the ring of polynomials in a noncommuting $x$. +1 – Arturo Magidin Mar 21 '12 at 18:01
Just one remark: whether polynomials in $R[x]$ do what you want when you use $x$ to stand for some unknown value (maybe outside the ring $R$) depends not so much on whether the elements of $R$ commute among each other as on whether elements of $R$ commute with what you want $x$ to stand for (because in $R[x]$ they commute by definition). So for instance it is fine to use polynomials in $\mathbf R[x]$ with $x$ standing for some unknown quaternion (no worse than standing for a matrix), but it's not possible to do so for polynomials in $\mathbf C[x]$, even though $\mathbf C$ is commutative! – Marc van Leeuwen Mar 15 '13 at 12:33
I'd like to give a complement to the answers already given, since some of them suggest a relation with more advanced arithmetic topics like Unique Factorization, while this is really based on elementary ring theory only. Notably one has the following
Theorem. Let $R$ be a commutative domain, and $P\in R[X]$ a nonzero polynomial of degree $d$. Then $P$ has at most $d$ roots in $R$.
Normally a commutative domain is called an integral domain (note the curious meaning of "integral"), but I've used "commutative" here to stress the two key properties assumed: commutativity and the absence of zero divisors. (I do assume rings to have an element $1$, by the way.) In the absence of commutativity, even introducing the notion of roots of $P$ is problematic—unless $P$ has its coefficients in the center of $R$ (as is the case in your quaternion example)—as Qiaochu Yuan points out. Indeed for evaluation of $P$ in $a\in R$ one must decide whether to write the powers of $a$ to the right or the left of the coefficients of $P$, giving rise to distinct notions of right- and left-evaluation, and hence of right- and left-roots (and neither form of evaluation is a ring morphism). But even for the case that $P$ has its coefficients in the center $Z(R)$ of $R$, so that right- and left-evaluation in $a$ coincide and define a ring morphism $Z(R)[X]\to R$, the conclusion of the theorem is not valid, as this question illustrates.
The proof of the theorem is based on the following simple
Lemma. Let $R$ be a commutative domain, $P\in R[X]$, and $r\in R$ a root of $P$. Then there exists a (unique) $Q\in R[X]$ with $P=(X-r)Q$, and every root of $P$ other than $r$ is a root of $Q$.
The existence and uniqueness of $Q$ do not depend on $R$ being commutative or a domain: for any ring $R$ one has $P=(X-r)Q$ if and only if $Q$ is the quotient of $P$ by euclidean left-division by $X-r$ and the remainder is $0$, and the latter happens if and only if $r$ is a left-root of $P$ (so properly stated the Factor Theorem does hold for general rings!). But the final part of the lemma does use both commutativity and the absence of zero divisors: one uses that evaluation of $(X-r)Q$ in some root $r'\neq r$ of $P$ can be done in separately in the two factors (this requires commutativity: without it evaluation is not a ring morphism), and then one needs to conclude that one of the factors (necessarily the second) becomes $0$, which requires the absence of zero divisors. Note for noncommutative $R$, that even if $P$ should be in $Z(R)$, the first part fails, since evaluation is only a morphism $Z(R)[X]\to R$, and the factors $X-r$ and $Q$ need not lie in $Z(R)[X]$.
The lemma of course implies the theorem by a straightforward induction on $\deg P$.
One final remark unrelated to the question: since the morphism property of evaluation $Z(R)[X]\to R$ does not help us here, one might wonder what is the point of considering evaluation at all in the absence of commutativity. However note that in linear algebra we teach our students to fearlessly substitute a matrix into polynomials, and to (implicitly) use the morphism property of such evaluation maps $K[X]\to M_n(K)$ where $K$ is a (commutative!) field. This works precisely because $K$ can be identified with $Z(M_n(K))$ (the subring of homothecies).
-
This has to do with the fact that $\mathbb H$ is not a field, since the number of zeroes of a polynomial over a field is always bounded by the degree of the polynomial.
- | 2015-01-25T14:29:06 | {
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https://math.stackexchange.com/questions/67039/why-does-the-volume-of-the-unit-sphere-go-to-zero | # Why does the volume of the unit sphere go to zero?
The volume of a $d$ dimensional hypersphere of radius $r$ is given by:
$$V(r,d)=\frac{(\pi r^2)^{d/2}}{\Gamma\left(\frac{d}{2}+1\right)}$$
What intrigues me about this, is that $V\to 0$ as $d\to\infty$ for any fixed $r$. How can this be? For fixed $r$, I would have thought adding a dimension would make the volume bigger, but apparently it does not. Anyone got a good explanation?
• Think about the square of side length $1/2$. What is its volume in $\mathbb{R}^d$? – JavaMan Sep 23 '11 at 18:53
• Imagine a square of side length $\frac{1}{2}$: the area is $\frac{1}{4}$. A cube of edge length $\frac{1}{2}$ has volume $\frac{1}{8}$. A hypercube of edge length $\frac{1}{2}$ has hypervolume $\frac{1}{16}$; etc. For a cube, whether going to the "next dimension" increases the total hypervolume or decreases it (as a scalar) depends on whether the edge side was less than $1$ or more than $1$. That suggests to me that "adding a dimension" does not, in and of itself, necessarily mean "making the volume bigger"... – Arturo Magidin Sep 23 '11 at 18:55
• What about a sphere of radius $10^{10}$? The volume eventually goes to zero :-) – probabilityislogic Sep 23 '11 at 19:03
• @probabilityislogic: The point is that simply "adding a dimension makes things bigger" doesn't work: it's not just about the dimension, there are other factors at play. – Arturo Magidin Sep 23 '11 at 19:26
• Shouldn't the title include the phrases hypershpere and n-->inf? – ripper234 Oct 4 '11 at 23:19
I suppose you could say that adding a dimension "makes the volume bigger" for the hypersphere, but it does so even more for the unit you measure the volume with, namely the unit cube. So the numerical value of the volume does go towards zero.
Really, of course, it is apples to oranges because volumes of different dimensions are not commensurable -- it makes no sense to compare the area of the unit disk with the volume of the unit sphere.
All we can say is that in higher dimensions, a hypersphere is a successively worse approximation to a hypercube (of side length twice the radius). They coincide in dimension one, and it goes downward from there.
• This is a very nice explanation. Thanks :) – Srivatsan Sep 23 '11 at 21:27
• This is nice and concise. Here is somebody's blog post saying the same thing with more words. – yasmar Sep 24 '11 at 10:27
• You might also want to read this similar post on MSE math.stackexchange.com/questions/15656/… – Tpofofn Sep 24 '11 at 15:38
• I don't think OP was bewildered over the volume of n-dimension to volumes of n+1 dimensions. Yes, that would be like "apples to oranges" as you say (or apples to apple trees). However, this downplays the intuitive significance that the ratio of volume of an n-sphere (greater than ~5 dimensions) to its corresponding n-cube decreases to virtual nothingness with the dimension. – aiwyn Apr 22 '19 at 19:02
• It does not go straight downwards from $n=1$, since for $n=1$ you get $2r$ and for $n=2$ you get $\pi r^2$ which is larger, certainly for $r\ge 1$ and it keeps increasing up to $n=5$ for $r\ge 1$ and further for larger $r$ – Henry Mar 24 '20 at 22:53
The reason is because the length of the diagonal cube goes to infinity.
The cube in some sense does exactly what we expect. If it's side lengths are $1$, it will have the same volume in any dimension. So lets take a cube centered at the origin with side lengths $r$. Then what is the smallest sphere which contains this cube? It would need to have radius $r\sqrt{d}$, so that radius of the sphere required goes to infinity.
Perhaps this gives some intuition.
• This is the clearest answer for me. – Sklivvz Sep 24 '11 at 7:56
• I think you mean the cube has side lengths $2r$. – Xiang Yu Nov 23 '18 at 17:30
This was thoroughly discussed on MO. I quote from the top answer:
The ultimate reason is, of course, that the typical coordinate of a point in the unit ball is of size $$\frac{1}{\sqrt{n}}\ll 1$$. This can be turned into a simple geometric argument (as suggested by fedja) using the fact that an $$n$$-element set has $$2^n$$ subsets:
At least $$n/2$$ of the coordinates of a point in the unit ball are at most $$\sqrt{\frac{2}{n}}$$ in absolute value, and the rest are at most $$1$$ in absolute value. Thus, the unit ball can be covered by at most $$2^n$$ bricks (right-angled parallelepipeds) of volume $$\left(2\sqrt{\frac{2}{n}}\right)^{n/2},$$ each corresponding to a subset for the small coordinates. Hence, the volume of the unit ball is at most $$2^n \cdot \left(2\sqrt{\frac{2}{n}}\right)^{n/2} = \left(\frac{128}{n}\right)^{n/4}\rightarrow0.$$ In fact, the argument shows that the volume of the unit ball decreases faster than any exponential, so the volume of the ball of any fixed radius also goes to $$0$$.
• I'm not seeing why the bricks each have volume at most $\left(2\sqrt{\frac{2}{n}}\right)^{n/2}.$ Shouldn't this be multiplied by a factor of $2^{n/2}$ since the other coordinates have absolute value at most $1$? This doesn't change the limit obviously, but some enlightenment would be nice. – cats Feb 14 '13 at 8:30
Let $X_j$ be a sequence of independent random variables with uniform distribution in the interval $[-r,r]$ (i.e. you are picking an infinite sequence of random numbers from the interval). The probability that $(X_1, \ldots, X_d)$ lies in your hypersphere, i.e. that $R_d = X_1^2 + \ldots + X_d^2 \le r^2$, is $V(r,d)/(2r)^d$ (which by scaling doesn't depend on $r$, so let's call it $P(d)$). Of course $P(d) \to 0$ as $d \to \infty$, but the point is that it goes to 0 faster than an exponential in $d$. Indeed, by the theory of large deviations, for any $t > 0$ we should have $P(R_d \leq t d) \approx e^{-d I(t)}$ as $d \to \infty$ for some function $I(t)$, where $I(t) \to \infty$ as $t \to 0+$.
• Came into this thread to give this proof, but I see someone beat me to it. – Michael Lugo Sep 23 '11 at 20:44
Let me develop on the idea presented by Christoph in the comments of this post.
The first thing is that the volume depends on the radius. $n$-ball or cube of radius $r$ has $r^n$ times bigger volume than that of radius $1$. One should expect then, that at least for big radii the volume of the ball will increase (as it is in the case of the unit cube $[-1,1]^n$ of volume $2^n$).
However, the thing is that somebody once decided (could they decide otherwise?) that Fubini theorem would be nice and $n$-dimensional volume of $[0,1]^n$ will be the same as $(n+1)$-dimensional volume $[0,1]^n\times [0,1]$ -- even though of course one is much smaller than the other!
That's the key point - volume is invariant under cartesian product with the unit interval - like it or not.
So, as Christoph Pegel says it is reasonable to compare volume of the ball $B^n$ with volume of the cylinder $B^n\times [-1,1]$. The second is of course $2$ times bigger but it is a matter of radius as already discussed.
Note that if we compare $B^{n+1}$ with $B^n\times [-1,1]$, we notice that only the zero section is the same. At level $t$ the ball is smaller with radius $\sqrt{1-t^2}$. That means that its volume is $(1-t^2)^{n/2}$ times smaller! This function converges to zero with $n$ (and so does its integral) and thus there's no wonder that it kills any geometric growth (where multiplicative factor $>1$ is constant).
• In my opinion the key point is that there are two different measures involved. Namely the Hausdorff measures $\mathcal H^n$ and $\mathcal H^{n+1}$. The $n$-ball has positive $\mathcal H^n$-volume, while its $\mathcal H^{n+1}$-volume vanishes, since there is really no $(n+1)$-dimensional interior in an $n$-ball. To compare the volumes $\mathcal H^{n}(B^n)$ and $\mathcal H^{n+1}(B^{n+1})$ under a common measure, we make use of $\mathcal H^{n}(A)=\mathcal H^{n+1}(A\times[0,1])$. – Christoph Aug 22 '13 at 11:08
• Could you explain why are you working with a cylinder of height two instead of height one? Shouldn't we compare $B^{n+1}$ with $B^n \times [\textrm{something with length one}]$? Otherwise we do double the volume. Everything else makes sense to me. I just don't understand "but it is a matter of radius as already discussed". – Fixed Point Aug 23 '13 at 8:53
• @FixedPoint Just because "height" (diameter) of the unit-ball is $2$. If our default ball has radius $1/2$, then I would work with some unit interval. The difference between those cases is a factor proportional to the radius and dependence radius ~ volume was discussed in the first paragraph. – savick01 Aug 23 '13 at 9:36
Consider the distance from the center of the sphere of radius $R$ in $\mathbb{R}^n$ to the corner of its enclosing cube: $R\sqrt{n}$. This is like covering the surface of the sphere with a layer of tall corners.
For $x=(x_1,x_2,\dots,x_n)\in\mathbb{R}^n$, define $$Q(x)=x\frac{|x|}{\max_i|x_i|}\tag{1}$$ $Q$ maps a sphere to its enclosing cube. As described above, $\frac{|Q(x)|}{|x|}$ reaches a maximum of $\sqrt{n}$ in the corners. To be exact, $$\frac{|Q(x)|^2}{|x|^2}=\sum_j\frac{|x_j|^2}{\max_i|x_i|^2}\tag{2}$$ By considering each $x_i$ to be normally distributed and using homogeneity of $(2)$, we get that the mean of $\frac{|Q(x)|^2}{|x|^2}$ over the sphere is $\frac{n+2}{3}$. Thus, the rms average of $\frac{|Q(x)|}{|x|}$ is $\sqrt{\frac{n+2}{3}}$. In this way, the volume of the cube should be approximated by $\sqrt{\frac{n+2}{3}}^{\;n}$ times the volume of the sphere. This grows faster than any $R^n$. | 2021-03-01T23:34:45 | {
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https://quantumcomputing.stackexchange.com/questions/5167/what-are-the-constraints-on-a-matrix-that-allow-it-to-be-extended-into-a-unita/5169 | # What are the constraints on a matrix that allow it to be “extended” into a unitary?
DaftWulie's answer to Extending a square matrix to a Unitary matrix says that extending a matrix into a unitary cannot be done unless there's constraints on the matrix. What are the constraints?
A necessary and sufficient condition is that, given an $$n\times n$$ matrix $$M$$, you can construct a $$2n\times 2n$$ unitary matrix $$U$$ provided the singular values of $$M$$ are all upper bounded by 1.
# Sufficiency
To see this, express the singular value decomposition of $$M$$ as $$M=RDV$$ where $$D$$ is diagonal and $$R$$, $$V$$ are unitary. Now define $$U=\left(\begin{array}{cc} M & R\sqrt{\mathbb{I}-D^2}V \\ R\sqrt{\mathbb{I}-D^2}V & -M \end{array}\right),$$ which we can only do if the singular values are no larger than 1. Let's verify that it's unitary \begin{align*} UU^\dagger&=\left(\begin{array}{cc} RDV & R\sqrt{\mathbb{I}-D^2}V \\ R\sqrt{\mathbb{I}-D^2}V & -RDV \end{array}\right)\left(\begin{array}{cc} V^\dagger DR^\dagger & V^\dagger\sqrt{\mathbb{I}-D^2}R^\dagger \\ V^\dagger\sqrt{\mathbb{I}-D^2}R^\dagger & -V^\dagger DR^\dagger \end{array}\right) \\ &=\left(\begin{array}{cc} RD^2R^\dagger+R(\mathbb{I}-D^2)R^\dagger & 0 \\ 0 & RD^2R^\dagger+R(\mathbb{I}-D^2)R^\dagger \end{array}\right) \\ &=\mathbb{I}. \end{align*}
# Necessity
Imagine I have a matrix $$M$$ with a singular value $$\lambda>1$$ and corresponding normalised vector $$|\lambda\rangle$$. Assume I construct a unitary $$U=\left(\begin{array}{cc} M & A \\ B & C \end{array}\right).$$ Let's act $$U$$ on the state $$\left(\begin{array}{c} |\lambda\rangle \\ 0 \end{array}\right)$$. We get $$U\left(\begin{array}{c} |\lambda\rangle \\ 0 \end{array}\right)=\left(\begin{array}{c} M|\lambda\rangle \\ B|\lambda\rangle \end{array}\right).$$ This output state must have a norm that is at least the norm of $$M|\lambda\rangle$$, i.e. $$\lambda>1$$. But if $$U$$ is a unitary, the norm must be 1. So it must be impossible to perform such a construction if there exists a singular value $$\lambda>1$$.
• @DaftWulie: This is "a" necessary and sufficient condition. Is it the only one? – Pablo LiManni Jan 10 at 17:54
• You might be able to phrase the condition in another way, but it would be materially equivalent. That’s the point of necessary and sufficient - it is the precise categorisation of what is required. – DaftWullie Jan 10 at 18:24
• "A necessary and sufficient condition is that, given an n×n matrix M, you can construct a 2n×2n unitary matrix U provided the singular values of M are all upper bounded by 1." If I'm reading this correctly (and I am far from sure I am), it seems that this can be rewritten as "Given a matrix $M$, a necessary and sufficient condition for being able to extend $M$ to a 2n×2n unitary matrix is that the singular values of $M$ all be less than or equal to $1$." – Acccumulation Jan 10 at 19:31
• @DaftWullie it is definitely possible to do this with less then doubling the space though. As a trivial example, any matrix obtained by removing one row and column from a unitary matrix can be extended to a unitary matrix by adding a single dimension. Do you have any idea on how one could estimate the minimum number of dimensions that have to be added to a given matrix to make it into a unitary? – glS Jan 11 at 9:42
• @glS Well, I know what I'd do, which is perform a Gram-Schmidt-like procedure, extending one row at a time, ensuring orthonormality with all previous rows. I don't know ho to succinctly write down the dimension number based on properties of $M$ - I've never thought about it. I guess a starting point is by counting the number of singular values equal to 1, and reducing the size of the extension by that much? – DaftWullie Jan 11 at 9:56
$$\newcommand{\bs}[1]{\boldsymbol{#1}}$$Here is a slightly different way to prove what the other excellent answer did.
Note that a matrix $$U$$ is unitary if and only if it sends orthonormal bases into orthonormal bases. This, in particular, means that if $$U$$ is unitary then $$\|U\bs v\|=1$$ for any $$\bs v$$ with $$\|\bs v\|=1$$.
Let us write the SVD of $$M$$ as $$M\bs u_k=s_k\bs v_k$$, where $$s_k\ge0$$ are the singular values of $$M$$.
Note that if $$U$$ is an extension of $$M$$, then $$U\bs u_k=s_k \bs v_k+\bs w_k$$ for some $$\bs w_k$$ orthogonal to $$\bs v_k$$ (and more generally to the whole range of $$M$$).
If follows that if, for any $$k$$, $$s_k>1$$, then $$\|U\bs u_k\|>1$$, and thus $$U$$ is not unitary.
On the other hand, if $$s_k\le1$$ for all $$k$$, let us show how can always construct a unitary $$U$$ that contains $$M$$ as a submatrix. Let us denote with $$\bs v\oplus \bs 0$$ the vectors in the extended $$2n$$-dimensional space that are built by appending zeros to the $$n$$-dimensional vector $$\bs v$$, and with $$\bs 0\oplus\bs v$$ the vectors that are equal to $$\bs v$$ in the last $$n$$ dimensions by zero in the first $$n$$ ones. Being $$\{\bs u_k\}_k$$ a basis for the original space, it follows that $$\{\bs u_k\oplus \bs 0,\bs0\oplus\bs u_k\}_k$$ is a basis for the extended space.
We will define $$U$$ through its action on the vectors $$u_k\oplus \bs 0$$ and $$\bs0\oplus u_k$$ as follows: \begin{align} U(\bs u_k\oplus \bs 0)&=s_k(\bs v_k\oplus\bs 0)+\sqrt{1-s_k^2}(\bs 0\oplus \bs v_k) \\ U(\bs0 \oplus \bs u_k)&=\sqrt{1-s_k^2}(\bs v_k\oplus\bs 0)-s_k(\bs 0\oplus \bs v_k). \end{align}
One can then check that all of these output vectors form an orthonormal system in the extended space, and thus $$U$$ is unitary. | 2019-04-25T18:01:42 | {
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https://math.stackexchange.com/questions/2809665/2008-aime-i-problem-6-pascals-triangle | # 2008 AIME I Problem 6: Pascal's triangle?
A triangular array of numbers has a first row consisting of the odd integers $1,3,5,\ldots,99$ in increasing order. Each row below the first has one fewer entry than the row above it, and the bottom row has a single entry. Each entry in any row after the top row equals the sum of the two entries diagonally above it in the row immediately above it. How many entries in the array are multiples of $67$?
Except for writing down some numbers, I wasn't sure how to proceed on this question. Does anyone have any ideas?
• don't they usually have solutions on their website? – Christopher Marley Jun 6 '18 at 3:34
• Yes, but the solution didn't make too much sense to me, so I figured I'd post it here – Dude156 Jun 6 '18 at 3:38
• In this case, you may want to post a link to the solution, and highlight which parts you are unable to understand. Of course, if the entire approach, while being correct, seems to come out of the blue, you may request a more intuitive understanding of the answer. – астон вілла олоф мэллбэрг Jun 6 '18 at 3:40
• no, why does he have to post other solutions? just because other solutions exists doesnt meen he has to add one – Jorge Fernández Hidalgo Jun 6 '18 at 3:41
• artofproblemsolving.com/wiki/… I can't make heads or tails of it either. The second solution looks neater though. – Christopher Marley Jun 6 '18 at 4:01
Instead of generating each row by taking sums of consecutive elements, take averages. This will divide every element in the whole triangle by some power of $2$, and therefore will not affect which ones are divisible by $67$. Doing it this way the table becomes $$\matrix{ 1&&3&&5&&\cdots&&95&&97&&99\cr &2&&4&&6&&\cdots&&96&&98\cr &&3&&5&&7&&\cdots&&97\cr}$$ with an obvious pattern. At this stage there is one multiple of $67$ in every row that starts with an odd number. This continues until the $33$rd row, which is $$\matrix{33&&35&&37&&\cdots&&65&&67\cr}\ ;$$ after this we have $$\matrix{34&&36&&38&&\cdots&&66\cr&35&&37&&\cdots&&65\cr}$$ and it is clear that there will be no more multiples of $67$. So there is one multiple of $67$ in rows $1,3,5,\ldots,33$, and no more.
Answer. The triangle contains $17$ multiples of $67$.
Claim : every row of the triangular array that you have, is an arithmetic progression. More precisely, the common difference in the $n$th row (difference between consecutive elements of the row) is $2^{n}$.
Proof : Clearly, the first row is an arithmetic progression of common difference two.
Call the elements of the triangular array $a_{n,i}$ where $n$ is the row (from top : the topmost row is $1$, second top most is $2$ etc.) and $i$ is the index within the row (for example, $a_{1,2} = 3$).
Then, $a_{(n+1),i} = a_{n,i+1} + a_{n,i}$ (similar to, but not quite the combinatorial relation), so $a_{(n+1),i+1} - a_{(n+1),i} = (a_{n,i+1} - a_{n,i-1}) = 2 \times 2^{n} = 2^{n+1}$. So our fact follows by induction.
Now, is there a nice formula for the first element of every row? Indeed, $a_{(n+1),1} = a_{n,1} + a_{n,2} = 2a_{n,1} + 2^n$ is a recurrence relation.
Now, we actually have a formula for each element : $a_{n,i} = a_{n,1} + (i-1)2^n$, and the above formula are the key points.
We have yet another claim, however.
Claim : $a_{n,1} = n2^{n-1}$!
Proof : Indeed, it is true for $n = 1$ and $a_{(n+1),1} = (n+1)2^n$ is easy to see, completing the inductive hypothesis.
Finally, we are in a very nice position : $a_{n,i} = a_{n,1} + (i-1)2^n = (n+2i-2)2^{n-1}$.
You can see this : for example, $a_{2,2} = 8 = (2 + (2 \times 2) - 2)2^{2-1}$
Finally, when is $a_{n,i}$ a multiple of $67$? Note that $2^n$ is coprime to $67$.
Therefore, $a_{n,i}$ is a multiple of $67$ precisely when $(n+2i-2)$ is a multiple of $67$.
Knowing that $1 \leq n \leq 99$ and $i \leq 100-n$, can you now do the problem?
EDIT : You should get $17$ multiples. | 2019-11-20T16:50:32 | {
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http://www.mathworks.com/help/symbolic/erf.html?requestedDomain=www.mathworks.com&nocookie=true | # Documentation
### This is machine translation
Translated by
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# erf
Error function
## Syntax
``erf(X)``
## Description
example
````erf(X)` represents the error function of `X`. If `X` is a vector or a matrix, `erf(X)` computes the error function of each element of `X`.```
## Examples
### Error Function for Floating-Point and Symbolic Numbers
Depending on its arguments, `erf` can return floating-point or exact symbolic results.
Compute the error function for these numbers. Because these numbers are not symbolic objects, you get the floating-point results:
`A = [erf(1/2), erf(1.41), erf(sqrt(2))]`
```A = 0.5205 0.9539 0.9545```
Compute the error function for the same numbers converted to symbolic objects. For most symbolic (exact) numbers, `erf` returns unresolved symbolic calls:
`symA = [erf(sym(1/2)), erf(sym(1.41)), erf(sqrt(sym(2)))]`
```symA = [ erf(1/2), erf(141/100), erf(2^(1/2))]```
Use `vpa` to approximate symbolic results with the required number of digits:
```d = digits(10); vpa(symA) digits(d)```
```ans = [ 0.5204998778, 0.9538524394, 0.9544997361]```
### Error Function for Variables and Expressions
For most symbolic variables and expressions, `erf` returns unresolved symbolic calls.
Compute the error function for `x` and ```sin(x) + x*exp(x)```:
```syms x f = sin(x) + x*exp(x); erf(x) erf(f)```
```ans = erf(x) ans = erf(sin(x) + x*exp(x))```
### Error Function for Vectors and Matrices
If the input argument is a vector or a matrix, `erf` returns the error function for each element of that vector or matrix.
Compute the error function for elements of matrix `M` and vector `V`:
```M = sym([0 inf; 1/3 -inf]); V = sym([1; -i*inf]); erf(M) erf(V)```
```ans = [ 0, 1] [ erf(1/3), -1] ans = erf(1) -Inf*1i```
### Special Values of Error Function
`erf` returns special values for particular parameters.
Compute the error function for x = 0, x = ∞, and x = –∞. Use `sym` to convert `0` and infinities to symbolic objects. The error function has special values for these parameters:
`[erf(sym(0)), erf(sym(Inf)), erf(sym(-Inf))]`
```ans = [ 0, 1, -1]```
Compute the error function for complex infinities. Use `sym` to convert complex infinities to symbolic objects:
`[erf(sym(i*Inf)), erf(sym(-i*Inf))]`
```ans = [ Inf*1i, -Inf*1i]```
### Handling Expressions That Contain Error Function
Many functions, such as `diff` and `int`, can handle expressions containing `erf`.
Compute the first and second derivatives of the error function:
```syms x diff(erf(x), x) diff(erf(x), x, 2)```
```ans = (2*exp(-x^2))/pi^(1/2) ans = -(4*x*exp(-x^2))/pi^(1/2)```
Compute the integrals of these expressions:
```int(erf(x), x) int(erf(log(x)), x)```
```ans = exp(-x^2)/pi^(1/2) + x*erf(x) ans = x*erf(log(x)) - int((2*exp(-log(x)^2))/pi^(1/2), x)```
### Plot Error Function
Plot the error function on the interval from -5 to 5. Prior to R2016a, use `ezplot` instead of `fplot`.
```syms x fplot(erf(x),[-5,5]) grid on```
## Input Arguments
collapse all
Input, specified as a symbolic number, variable, expression, or function, or as a vector or matrix of symbolic numbers, variables, expressions, or functions.
collapse all
### Error Function
The following integral defines the error function:
`$erf\left(x\right)=\frac{2}{\sqrt{\pi }}\underset{0}{\overset{x}{\int }}{e}^{-{t}^{2}}dt$`
## Tips
• Calling `erf` for a number that is not a symbolic object invokes the MATLAB® `erf` function. This function accepts real arguments only. If you want to compute the error function for a complex number, use `sym` to convert that number to a symbolic object, and then call `erf` for that symbolic object.
• For most symbolic (exact) numbers, `erf` returns unresolved symbolic calls. You can approximate such results with floating-point numbers using `vpa`.
## Algorithms
The toolbox can simplify expressions that contain error functions and their inverses. For real values `x`, the toolbox applies these simplification rules:
• ```erfinv(erf(x)) = erfinv(1 - erfc(x)) = erfcinv(1 - erf(x)) = erfcinv(erfc(x)) = x```
• ```erfinv(-erf(x)) = erfinv(erfc(x) - 1) = erfcinv(1 + erf(x)) = erfcinv(2 - erfc(x)) = -x```
For any value `x`, the system applies these simplification rules:
• `erfcinv(x) = erfinv(1 - x)`
• `erfinv(-x) = -erfinv(x)`
• `erfcinv(2 - x) = -erfcinv(x)`
• `erf(erfinv(x)) = erfc(erfcinv(x)) = x`
• `erf(erfcinv(x)) = erfc(erfinv(x)) = 1 - x`
## References
[1] Gautschi, W. "Error Function and Fresnel Integrals." Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. (M. Abramowitz and I. A. Stegun, eds.). New York: Dover, 1972. | 2017-09-20T05:47:30 | {
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https://math.stackexchange.com/questions/2372388/proving-a-metric-space-is-bounded | # Proving a metric space is bounded
Let $t \in \mathbb{Z}^+ := \mathbb{N} \cup \{0\}$ and let $\beta > 1$ be a fixed parameter. For $a, b \in \mathbb{R}$, define $\rho(a, b) = \min\{|a-b|, 1\}$. Then for any $x, y \in \mathbb{R}^{\infty}$ (here $\mathbb{R}^{\infty}$ represents the infinite Cartesian product of $\mathbb{R}$ with itself, i.e., $\mathbb{R} \times \mathbb{R} \times \cdots$), define $$d(x, y) = \sum_{t \in \mathbb{Z}^+} \beta^{-t}\rho(x_t, y_t)$$ as the "distance" between $x = (x_0, x_1, \cdots)$ and $y = (y_0, y_1, \cdots)$. Show that $(\mathbb{R}^{\infty}, d)$ is a bounded metric space.
First, I have already proved that $d$ is a valid metric on $\mathbb{R}^{\infty}$. I know the definition of a bounded metric space is
Let $(X, d)$ be a metric space. A subset $S \subseteq X$ is bounded if $\exists x \in X$, $\exists \varepsilon>0$ such that $A \subseteq B_{\varepsilon}(x)$.
But I am unsure how to apply that definition here to prove that $(\mathbb{R}^{\infty}, d)$ is bounded. Any help would be appreciated!
EDIT: In a related problem, I am asked to prove whether $[0,1]^{\infty}$ (the infinite Cartesian product of $[0,1]$ with itself) is an open or closed subset of $\mathbb{R}^{\infty}$. I know the definition that for a metric space $(X, d)$, a set $A \subseteq X$ is open if for all $x \in A$, there exists a $\varepsilon >0$ such that $B_{\varepsilon}(x) \subseteq A$. A set $C \subseteq X$ is closed if and only if $X \setminus C$ is open. How can I apply that here?
• Hint on $[0, 1]^\infty$ question: suppose $x \notin [0, 1]^\infty$. Then there exists $t \in \mathbb{Z}^+$ such that $x_t < 0$ or $x_t > 1$. In the second case, what can you say about the ball around $x$ with radius $\beta^{-t} (x_t - 1)$? – Daniel Schepler Jul 26 '17 at 23:07
• (Or, for a slightly less direct approach: show that for each $t$, the projection map $\pi_t : \mathbb{R}^\infty \to \mathbb{R}$ is continuous with respect to $\rho$ on the domain and the normal metric on the codomain. Then you can express $[0, 1]^\infty$ as $\bigcap_{t=0}^\infty \pi_t^{-1}([0, 1])$.) – Daniel Schepler Jul 26 '17 at 23:09
• Hmm, with your first point, I can see what you're trying to do, so you are working with the set $\mathbb{R}^{\infty} \setminus [0,1]^{\infty}$ and then trying to see whether it is closed/open to deduce whether $[0,1]^{\infty}$ is open or closed. But I am not sure how to continue your argument, can you provide the finish? – elbarto Jul 27 '17 at 1:17
• OK, the point is: if $d(x', x) < \beta^{-t} \rho(x_t, 1)$, then $d(x', x) \ge \beta^{-t} \rho(x_t, x_t')$ so $\rho(x_t, x_t') < \rho(x_t, 1)$ and that implies $x_t' > 1$ so $x' \notin [0, 1]^\infty$. So there's a small correction that the radius should be $\beta^{-t} \rho(x_t, 1)$. (You now just have to prove the lemma: if $x > 1$ and $\rho(x, y) < \rho(x, 1)$ then $y > 1$, which should be straightforward.) – Daniel Schepler Jul 27 '17 at 16:46
• Ah I see, and what about for $x_t<0$? – SwiftMo Jul 27 '17 at 22:31
$d(x,y)\leq \sum_{t\in\mathbb{Z}^+}\beta^{-t}$. Write $c=\sum_{t\in\mathbb{Z}^+}\beta^{-t}$. Take any $x_0\in\mathbb{R}^{\infty}$, $d(x_0,y)\leq c$ for every $y$.
• Thanks, I edited my OP with a related problem, that is whether $[0,1]^{\infty}$ is open or closed, could you have a look? I know the definitions but not sure on how to apply it here. – elbarto Jul 26 '17 at 22:58 | 2021-04-17T12:30:31 | {
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https://mathematica.stackexchange.com/questions/185024/error-propagation-in-finding-full-width-at-half-maximum/185095 | Error propagation in finding full width at half maximum
I have an equation given by:
f[x_,M_]:= 96.4529 E^(M x (-0.216331 - 38.7396 M x)) + 31.0508 E^(M x (0.306405 - 18.585 M x)) + 4.36041 E^(M x (3.95974 - 7.37814 M x)) + 2.00366 E^(M x (-1.54639 - 3.79704 M x)) + 119.8 E^(M x (-0.0235058 - 0.0245919 M x));
where $$x$$ is a variable, $$M$$ is the mean value from a measurement and $$\delta M$$ is the error in $$M$$. For example, let $$M \pm \delta M$$ be equal to $$15.584 \pm 0.045$$.
The approximate error $$\delta f$$ in $$f$$ can be given, in the simplified form assuming indipendence of $$x$$ and $$M$$, as:
$$\delta f = \sqrt{\left(\frac{\partial f}{\partial M}\right)^2(\delta M)^2 + \left(\frac{\partial f}{\partial x}\right)^2(\delta x)^2}$$
As $$x$$ is not a measured value, I take $$\delta x = 0$$ :
DfM[x_, M_] := Evaluate@ D[f[x, M], M];
Dfx[x_, M_] := Evaluate@ D[f[x, M], x];
Deltaf[x_, M_, deltaM_] := Evaluate@ Simplify[Sqrt[(DfM[x, M])^2 deltaM^2
+ (Dfx[x, M])^2 deltax^2]] /. {deltax -> 0};
This way I can find the error $$\delta f$$ in $$f$$ that is a consequence of the error $$\delta M$$ in $$M$$.
But my aim is to find the full width at half maximum (FWHM) i.e. $$FWHM \pm \delta FWHM = (x_2-x_1) \pm \sqrt{(\delta x_1)^2 + (\delta x_2)^2}$$ along with the error $$\delta FWHM$$ that propagates in doing so. Here, $$x_1$$ and $$x_2$$ are the corresponding $$x$$-axis values for the half maximum HM. The errors in $$x_1$$ and $$x_2$$ are $$\delta x_1$$ and $$\delta x_2$$, respectively.
To find the half of the maximum (i.e. HM) , I find the maxima and minima of f[x,M] as:
fMin = First[NMinimize[{f[x, 15], -0.08 < x < 0.04}, x]];
fMax = First[NMaximize[f[x, 15], x]];
HM = fMax + (fMin - fMax)/2;
There will be some error in HM i.e. $$\delta HM$$ but I am unable to figure out how this error is propagating via the NMaximize and NMinimize functions. Additionally, to find the $$x_1$$ and $$x_2$$ I need to solve the function f[x,M] for the value HM but this function does not seem to have an inverse and therefore Solve fails if I use:
Solve[f[x, 15] == HM, x]
Is there a way I could find $$FWHM$$ and its error $$\delta FWHM$$?.
EDIT:
Link to the data of which fun[x] is a nonlinear fit: Data
fun[x_] := 96.4529 E^((-0.216331 - 38.7396 x) x) + 31.0508 E^((0.306405 - 18.585 x) x) + 4.36041 E^((3.95974 - 7.37814 x) x) + 2.00366 E^((-1.54639 - 3.79704 x) x) + 119.8 E^((-0.0235058 - 0.0245919 x) x);
the function f[x_,M_] given above is derived using fun[x_] as:
f[x_,M_] := fun[x*M];
• I think you're almost there: Replace Solve[f[x, 15] == HM, x] with x1 = x /. FindRoot[f[x, 15] == HM, {x, -0.1}] and x2 = x /. FindRoot[f[x, 15] == HM, {x, -x1}]. (I'll work on the error part later if no one else does it sooner.) – JimB Oct 31 '18 at 19:43
• Why the apparent arbitrary restriction for finding the minimum? -0.08 < x < 0.04 Otherwise with minimum would seem to be zero. – JimB Oct 31 '18 at 19:59
• The function f[x,M] is a fitted function from a data that is collected in the domain -0.08 < x < 0.04 and has the background amplitude closer to fMin rather than a 0 amplitude. – jsid Oct 31 '18 at 20:14
• Because this is all based on data, you might be better off using a bootstrap approach rather than a propagation of error (Delta method to us statisticians) approach. Is it possible to make the data available? – JimB Oct 31 '18 at 21:46
• You might want: x /. Solve[f[x, 15] == HM, x, Reals]. – Daniel Lichtblau Oct 31 '18 at 23:25
Here is a bootstrap approach to estimate a standard error associated with the estimate of $$x_2-x_1$$.
data = {{-1.5, 117.955}, {-1.452, 118.277}, {-1.404, 119.012}, {-1.356, 119.277},
{-1.308, 120.204}, {-1.26, 120.545}, {-1.212, 120.866}, {-1.164, 120.712},
{-1.116, 120.37}, {-1.068, 120.523}, {-1.02, 120.848}, {-0.972, 120.798},
{-0.924, 120.98}, {-0.876, 121.441}, {-0.828, 121.496}, {-0.78, 121.163},
{-0.732, 120.515}, {-0.684, 121.504}, {-0.636, 121.139}, {-0.588, 119.96},
{-0.54, 119.393}, {-0.492, 120.752}, {-0.444, 122.253}, {-0.396, 123.639},
{-0.348, 126.277}, {-0.3, 131.765}, {-0.252, 139.704}, {-0.204, 154.437},
{-0.156, 180.332}, {-0.108, 214.055}, {-0.06, 242.769}, {-0.012, 255.787},
{0.036, 249.115}, {0.084, 224.131}, {0.132, 194.485}, {0.18, 171.191},
{0.228, 155.534}, {0.276, 143.708}, {0.324, 134.348}, {0.372, 129.606},
{0.42, 127.123}, {0.468, 124.993}, {0.516, 122.399}, {0.564, 120.834},
{0.612, 120.375}, {0.66, 119.378}, {0.708, 118.306}, {0.756, 116.853},
{0.804, 115.839}};
Now the function to be fit:
f[x_, a1_, b1_, c1_, a2_, b2_, c2_, a3_, b3_, c3_, a4_, b4_, c4_] :=
a1 E^(x (b1 + c1 x)) +
a2 E^(x (b2 + c2 x)) +
a3 E^(x (b3 + c3 x)) +
a4 E^(x (b4 + c4 x));
(* Initial parameter values *)
inits = {
{a1, 55}, {b1, 1}, {c1, -20},
{a2, 20}, {b2, 2}, {c2, -1},
{a3, 12}, {b3, -1.5}, {c3, -53},
{a4, 120}, {b4, -0.5}, {c4 , -0.2}};
findx1x2[data_] :=
Module[{nlm, fmax, fmin, HM, x1, x2, xfmax, xfmin, x2Init},
nlm = NonlinearModelFit[data, f[x, a1 , b1 , c1 , a2 , b2 , c2 , a3 , b3, c3, a4 , b4, c4],
inits, x, MaxIterations -> 1000];
fmax = FindMaximum[nlm[x], {x, Select[data, #[[2]] == Max[data[[All, 2]]] &][[1, 1]]}];
fmin = FindMinimum[{nlm[x], -1.5 < x < 0.8}, {x, Select[data, #[[2]] == Min[data[[All, 2]]] &][[1, 1]]}];
xfmax = x /. fmax[[2]];
xfmin = x /. fmin[[2]];
HM = (fmin[[1]] + fmax[[1]])/2;
x1 = x /. FindRoot[nlm[x] == HM, {x, (x /. fmax[[2]])}];
x2Init = If[x1 < xfmax, xfmax + (xfmax - x1), xfmax + (xfmax - x1)];
x2 = x /. FindRoot[nlm[x] == HM, {x, x2Init}];
{Min[x1, x2], Max[x1, x2], HM, fmin[[1]], fmax[[1]]}]
The bootstrap process (random selection of residuals):
(* Get predicted responses and fit residuals *)
nlm = NonlinearModelFit[data,
f[x, a1 , b1 , c1 , a2 , b2 , c2 , a3 , b3, c3, a4 , b4, c4],
inits, x, MaxIterations -> 1000];
predictedResponse = nlm["PredictedResponse"];
fitResiduals = nlm["FitResiduals"];
(* Array to hold bootstrap estimates *)
nboot = 1000;
x1x2 = ConstantArray[{0, 0, 0, 0, 0}, nboot];
(* Perform bootstraps *)
n = Length[data];
SeedRandom[12345];
Do[
boot = Transpose[{data[[All, 1]], predictedResponse + RandomChoice[fitResiduals, n]}];
x1x2[[iboot]] = findx1x2[boot],
{iboot, nboot}]
(* Summarize results *)
mle = findx1x2[data]
(* Estimate of x2 - x1 *)
estimate = mle[[2]] - mle[[1]]
(* 0.29482173864942085 *)
(* Standard error of x2 - x1 *)
stderr = StandardDeviation[x1x2[[All, 2]] - x1x2[[All, 1]]]
(* 0.0014648255351604935 *)
(* Histogram of bootstrap estimates *)
Histogram[x1x2[[All, 2]] - x1x2[[All, 1]]]
Edit
If the quantity of interest is $$(x_2-x_1)/M$$, then applying the Propagation of Error (Delta) method results in an estimate of the variance as
$$\frac{\sigma_M^2 (x_2-x_1)^2}{M^4}+\frac{\sigma_{x_2-x_1}^2}{M^2}$$
assuming that the estimate of $$x_2-x_1$$ is statistically independent from the estimate of $$M$$. So we have
results = {x1 -> mle[[1]], x2 -> mle[[2]], σx1x2 -> stderr, M -> 15.584, σM -> 0.045}
estimate = (-x1 + x2)/M /. results
(* 0.0189182 *)
seEstimate = (σx1x2^2/M^2 + σM^2 (x2 - x1)^2/M^4)^0.5 /. results
(* 0.000108717 *)
• This certainly looks like a better approach to me however, how do I introduce the factor $M$ here?. Do I divide the $x$-axis values of the data by $M=15.584$ and replace the $x$ variable in your nonlinear model function f with {x -> x*M}?. – jsid Nov 1 '18 at 18:08
• Injecting $M$ into the mix would only affect the estimated coefficients but NOT any predictions of the dependent variable or inferences about $x_1$, $x_2$, or $x_2-x_1$. Or maybe I'm not understanding what you want. Is it $M(x_2-x_1)$ or $(x_2-x_1)/M$ the quantity of interest? – JimB Nov 1 '18 at 18:30
• What I did was that I first fit the data and then change the fitted equation fun[x_] by introducing $M$ as f[x_,M_] := fun[x*M]. The factor $M$ basically reduces the width of fitted function while keeping the same amplitude. Since your answer works directly on the NonlinearModelFit, consider the following manipulation: data = data/. {a_Real, b_Real} :> {a/15.584, b}. The fitted parameters would change accordingly and $(x_2-x_1)$ will become small. The $FWHM$ should then be $(x_2-x_1)/M$ I guess?. – jsid Nov 1 '18 at 19:59
This is an extended comment rather than an answer.
While I still think that a bootstrap approach would be more accurate than using the Propagation of Error/Delta Method, there are two issues you might want to consider first.
1. The parameter correlation matrix for your current model suggests that using 5 kernels is way overfitting (with $$a_i e^{M x(b_i+c_i M x)}$$ being a kernel). Fitting a model with just 4 kernels has a much smaller AIC (Akaike Information Criterion) and fits almost perfectly with no errors.
2. If you are after predictions or making inferences about $$x_1$$ and $$x_2$$, $$M$$ is completely irrelevant because $$M$$ and $$x$$ always appear together as a product with an associated coefficient.
More specifically the kernel $$a e^{M x(b+c M x)}$$ results in the same prediction as $$a e^{x(B+C x)}$$. (If $$M$$ appeared away from $$x$$, then that would be a different story.) | 2019-12-14T04:36:34 | {
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https://math.stackexchange.com/questions/910408/geometric-interpretation-of-the-norm-vec-x-x-1x-2-over-32-max?noredirect=1 | Geometric interpretation of the norm $\|\vec x\|={(|x_1|+|x_2|)\over 3}+{2\max(|x_1|,|x_2|)\over 3}$
Let $$p:\mathbb R^2 \to \mathbb R$$ be a norm so that $$\|\vec x\| ={(|x_1|+|x_2|)\over 3}+{2\max(|x_1|,|x_2|)\over 3} ={{\|\vec x\|_1\over 3}}+{2\|\vec x\|_\infty\over 3}.$$ I need to graph the neighbourhood of radius $$1$$ around $$(0,0)$$: $$V_1 ((0,0))$$ with this norm, but I don't even know the points that are in this neighbourhood I really don't know how to geometrically visualize it .
I tried to separate the norm in to parts: I want that to find all $$(x_1, x_2) \in \mathbb{R}^2$$ that satisfy $${(|x_{1}|+|x_{2}|)\over 3}+{2\max(|x_1|,|x_2|)\over 3} < 1$$ so: $$\frac{|x_1|+|x_2|}{3} < \frac{1}{2} \qquad \text{and} \qquad \frac{2\max(|x_1|,|x_2|)}{3} < {1\over 2}.$$
I know that the first inequality is a rotated square (geometrically) and the second one is a square, but from this point I don't see how to find the points that satisfy the given norm and visualize it geometrically.
Note the expression for the norm can be altered to;
$$||(x_1, x_2)|| = \max \{|x_1|, |x_2|\} + \frac{\min \{|x_1|, |x_2|\}}{3}$$
Hence the graph you look for is points $(x_1, x_2)$ such that,
$$\max \{|x_1|, |x_2|\} + \frac{\min \{|x_1|, |x_2|\}}{3} \lt 1 \iff 3 \max \{|x_1|, |x_2|\} + \min \{|x_1|, |x_2|\}\lt 3$$
From here I think you must map the following regions on the $(x, y)$ plane in the corresponding regions.
For $|y| \ge |x|$; we get the equation to be $3|y| + |x| \lt 1$
$$3 y + x \lt 3 \;\; \text{for } \;\; x \ge 0, y \ge 0$$ $$3 y - x \lt 3 \;\; \text{for } \;\; x \lt 0, y \ge 0$$ $$- 3 y - x \lt 3 \;\; \text{for } \;\; x \lt 0, y \lt 0$$ $$- 3 y + x \lt 3 \;\; \text{for } \;\; x \ge 0, y \lt 0$$
Now I think we need to similarly map $3|y| + |x| \lt 1$ for $|x| \ge |y|$.
Note however that the $8$ separate equations we get are confined to $8$ disjoint regions in the plane. The $8$ sub-quadrants or octants if you will. So I'm guessing it can be done.
• thanks a lot!! just one question why can the norm be altered? – user128422 Aug 27 '14 at 2:09
• $$\|\vec x\|={(|x_1|+|x_2|)\over 3}+{2\max(|x_1|,|x_2|)\over 3} = \dfrac{|x_1|}{3} + \dfrac{|x_2|}{3} + {2\max(|x_1|,|x_2|)\over 3}$$ and the max iseither $|x_1|$ or $|x_2|$ whose fractions can be added and the remaining one is the minimum. – Ishfaaq Aug 27 '14 at 2:12
• you'll be happy that i've verified your ideas and graphed in my answer below. – Viktor Glombik Jun 1 at 21:31
Using $$\max(a,b) = \frac{|a| + |b| + |a-b|}{2}$$ we can rewrite your norm as $$\| (x,y) \| := \frac{|x| + |y]}{3}+ \frac{|x| + |y| + ||x| - |y||}{3}$$ Now we want to solve \begin{align} \frac{|x| + |y|}{3}+ \frac{|x| + |y| + ||x| - |y||}{3} < 1 \iff & \frac{2}{3}\big(|x| + |y|\big) + \frac{1}{3} \big||x| - |y|\big| < 1 \\ \iff & 4 | x | + 4 | y | + 2 \big||x| - |y|\big| < 6 \qquad (\ddagger) \end{align} We now have to distinguish some cases:
Case 1: $$x,y \ge 0$$. Then we have $$(\ddagger) \iff 4x + 4y + 2 | x - y | < 6 \qquad (\star)$$ Case 1.1: $$x \ge y \ge 0$$. Then we have $$(\star) \iff 4x + 4y + 2x - 2y < 6 \iff 6x + 2y < 6 \iff 3x + y < 3.$$
Case 1.2: $$y \ge x \ge 0$$. Then we have $$(\star) \iff 4x + 4y + -2x + 2y < 6 \iff x + 3y < 3$$ All the other cases can be done analogously, you obtain $$-x+3y<3$$ and $$-3x+y<3$$ for the second quadrant (going counter clockwise), $$-x-3y<3$$ and $$-3x-y < 3$$ for the third and $$x-3y < 3$$ and $$3x - y < 3$$ for the fourth quadrant. This traces out the following regular octagon:
This is the intersection of two rotated squares. Its area ($$\approx 3.02$$) is roughly $$\frac{3}{4}$$ of the area of the square $$[-1,1]^2$$. You can rewrite the octagon as $$\big\{ (x,y) \in \mathbb{R}^2: ax + by < 3, \ a,b \in \{\pm 1, \pm 3\}, |a| \ne |b| \big\}$$ and I'm not sure if this characterisation can be obtained more easily (with less or without case distinctions) from the above inequality.
Note: Your approach one part < 1/2 and other part < 1/2 gives some but not all of the solutions, do you know, why?
We can generalize this as follows: For $$a,b > 0$$ define the norm $$\rho: \mathbb{R}^2 \to \mathbb{R}, \ (x,y) \mapsto a \| (x,y) \|_1 + \frac{b}{2}\| (x,y) \|_{\infty}.$$ From this question we know this is well-defined. We can then rewrite the norm as $$\rho(x,y) = \left(|x| + |y| \right)\left(a + b\right) + b | | x | - | y | |.$$ Using the same procedure as above we find that the unit ball is given by $$\big\{ (x,y) \in \mathbb{R}^2: \lambda x + \mu y < 1, \ \lambda, \mu \in \{ \pm (a + 2b), \pm a \}, | \lambda | \ne | \mu | \big\}.$$
Essentially the same as other answers but phrased differently. If $$X=(x,y)$$, $$\|X\|:=\frac13\|X\|_1 + \frac23\|X\|_\infty$$, then we are looking for the (lets say open) ball of radius 1 around $$(0,0)$$, $$B:=V_1((0,0)) = \{X\in\mathbb R^2 : \|X\|<1\}$$ Observe the symmetries \begin{align} \|(x,y)\|&= \|(-x,y)\|,\\ \|(x,y)\|&= \|(x,-y)\|,\\ \|(x,y)\|&= \|(y,x)\|,\\ \|(x,y)\|&= \|(-x,-y)\|.\end{align} These imply that the set $$B$$ is symmetric across the $$y$$-axis, the $$x$$-axis, across the line $$y=x$$, and across the line $$y=-x$$ respectively. So it suffices to describe $$B$$ in the octant $$y\ge x\ge 0$$, where since $$\max(x,y)=y$$, $$\|X\| < 1 \iff \frac13(x+y) + \frac23 y<1$$
i.e. the set is described by 8 copies of $$y < 1-\frac x3,$$ giving $$(x,y)\in B \iff \begin{cases} y\ge x\ge 0, y < 1-\frac x3,\\ x\ge y\ge 0, x<1-\frac y3, \\ -y\ge x\ge 0, -y<1-\frac x3,\\ -x\ge y\ge 0, -x < 1-\frac y3,\\ \qquad \qquad \quad \vdots \end{cases}$$ Here's 6 of the octants (because Desmos has 6 colours) together with a full plot that relies on Desmos to interpret the implicit inequality: | 2019-09-16T00:43:55 | {
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https://www.physicsforums.com/threads/horners-method-vs-x-1-n.588995/ | # Horner's method vs (x-1)^n
1. Mar 21, 2012
### rukawakaede
Firstly, I am not sure if I am in the right section. Pardon me.
I was reading something related to computational efficiency when evaluating polynomials. Suppose we want to evaluate
$f(x)=1-4 x+6 x^2-4 x^3+x^4$
it would take n=4 additions and (n^2+n)/2 =10 multiplications to evaluate this f(x) on computer.
We know that it would be better (efficiency and stability) to evaluate f(x) using Horner's method, i.e
$f(x)=1+x (-4+x (6+(-4+x) x))$
which only take n=4 additions and n=4 multiplications on the computer, (i.e. round-off error will be minimised).
So here is my question: wouldn't it better to evaluate
$f(x)=(x-1)^4$
since it would only need 1 addition and 4 multiplication?
Could anyone share some insights on this?
Last edited: Mar 21, 2012
2. Mar 21, 2012
### epsi00
no, in my opinion, because you need to do some work to go from 1-4x+6x^2-4x^3+x^4 to (x-1)^4. It is not obvious that the two functions are the same.
3. Mar 21, 2012
### rukawakaede
Well, it is just the binomial expansion of (x-1)^4.
I don't understand why we should not evaluate the f(x) directly using (x-1)^4 in terms of computational efficiency.
I believe this is something trivial, but I can't see it myself. Hope someone could give some idea.
I wrote some codes and plotted all the three variations of the functions with x around integer 1. It seems that only Horner's method give a desirable result. Significant numerical error occurs for the other two.
I understand that f(x)=1-4x+6x^2-4x^3+x^4 is give worse numerical result because of the number of multiplications and the rounding error accumulates.
But for f(x)=(x-1)^4, what other reasons (other than number of calculations) could that be?
4. Mar 21, 2012
### epsi00
"Well, it is just the binomial expansion of (x-1)^4."
when you are given "f(x)=1-4x+6x^2-4x^3+x^4" do you immediately recognize it as (x-1)^4? Using (x-1)^4 to calculate the value of the function around x~1 propagates the numerical errors which does not happen with Horner's. (x-1)^4 should be fine for x not close to 1.
5. Mar 21, 2012
### rukawakaede
Alright. I agree that needs some time.
Thanks. This is what I am looking for.
Could you please explain a little bit more on how numerical errors accumulate for (x-1)^4 when is x is about 1?
I am still not fully understand.
6. Mar 21, 2012
### epsi00
"Could you please explain a little bit more on how numerical errors accumulate for (x-1)^4 when is x is about 1?"
I took the numerical analysis course a long time ago. I am not sure I can provide you with an explanation. It has to do with substracting numbers that are almost equal...
7. Mar 21, 2012
### mathman
It looks to me like 1 addition and 2 multiplications (2 successive squares).
8. Mar 21, 2012
### rukawakaede
epsi00, thanks. I think I understand now. I nearly forgot catastrophic cancellation.
Last edited: Mar 21, 2012
9. Mar 22, 2012
### willem2
Calculating (x-1)^4 by first calculating (x-1) and then calculating the fourth power seems to be the only way to get reasonable accuracy, when x is close to 1.
Suppose a multiplication gives a relative error of e, and addition/subtracton gives an error of e/(operand with largest absolute value)
suppose you calculate (x-1)^4 for x=1.01
(x-1) will produce a result of 0.01+e, (hopefully e<0.01)
(x-1) will produce a result of 10^(-8) + 4 * 10^(-8) e, so you get a relative error of 4e.
(ignoring higher powers of e)
The final addition when using horner rule will produce an error of at least e, on a result of 10^(-8), so a relative error of 10^8 (e) (ignoring the errors in the earlier steps),
so Horners' rule is a factor of 2.5 * 10^7 less accurate.
10. Mar 22, 2012
### rukawakaede
Thanks willem. I just realised that I made some mistake in my code and horner's method is less accurate that (x-1)^4.
Thank you very much. | 2018-01-20T01:33:58 | {
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https://mathematica.stackexchange.com/questions/152948/solve-an-ode-using-matrices | # Solve an ODE using matrices
I have the following system: $m\cdot\frac{dx^2}{dt^2}=-k(x-lo)-\frac{dx}{dt}\cdot d+m\cdot g$ It represents a mass with a spring and a damper. It is easy to solve using NDSolve but I'm trying to solve it using matrices. (Because if we represent the system using state equations, we can use some transformations, like diagonalization or triangularization so the time of computation is reduced). I tried using regular matrices but it doesn't work. Is there any way to do this? The system after an order reduction is:
$\begin{bmatrix} x1'(t) \\ x2'(t) \\ \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ \frac{-k}{m} & \frac{-d}{m} \\ \end{bmatrix} \begin{bmatrix} x1 \\ x2 \\ \end{bmatrix}+ \begin{bmatrix} 0\\ \frac{1}{m}\\ \end{bmatrix} f(t) + \begin{bmatrix} 0\\ \frac{kl_o}{m}+g\\ \end{bmatrix}$
where $f(t)=15u(t-5)$ (u(t) is the unit step function).
I have tried this:
lo = 0.50; m = 1.5; k = 20; d = 3; g = 9.8;
A = {{0, 1}, {-k/m, -m}}
z[t_] = 15*HeavisideTheta[t - 5];
b = {{0}, {1/m}};
γ = {{0}, {(k*lo)/m + g}};
S = A*{{x0[t]}, {x1[t]}} + b*z[t] + γ
eqns = {{x0'[t]}, {x1'[t]}}
NDSolve[{eqns == A*{{x0[t]}, {x1[t]}} + b*z[t] + γ ,
x0[0] == 0, x1[0] == 0}, {x0[t], x1[t]}, {t, 0, 10}]
• As a tip: HeavisideTheta[] is intended for symbolic use only; for numerics, like in your situation, please use UnitStep[]. Also, matrix-vector multiplication is . (Dot[]), not * (Times[]). – J. M. will be back soon Aug 3 '17 at 1:47
• What values do k, m, lo, g have? – Carl Woll Aug 3 '17 at 2:30
• Thanks! Now it works better. I tried to solve the system with f(t)=0 and I got the correct solution using your tips. But I still can't get the solution using f(t). (Mathematica says "non-numerical value at t==0" ) – Miguel Duran Diaz Aug 3 '17 at 2:31
• lo = 0.50; m = 1.5; k = 20; d = 3; g = 9.8; – Miguel Duran Diaz Aug 3 '17 at 2:32
• Please include such constants in your questions the next time instead of making other people ask for them. – J. M. will be back soon Aug 3 '17 at 2:56
The code runs with a few minor modifications, as suggested by J.M. In addition, A was corrected, and constants given numerical values. Then,
k = 1; lo = 1; m = 1; d = 1; g = 1;
A = {{0, 1}, {-k/m, -d/m}};
z[t_] = 15*UnitStep[t - 5];
b = {{0}, {1/m}};
γ = {{0}, {(k*lo)/m + g}};
eqns = {{x0'[t]}, {x1'[t]}};
s = Flatten@NDSolve[{eqns == A.{{x0[t]}, {x1[t]}} + b z[t] + γ,
x0[0] == 0, x1[0] == 0}, {x0[t], x1[t]}, {t, 0, 10}]
• I know the basic use of Flatten, but you used here to solve the system. Why does it work with Flatten, but it didn't work using regular matrix operations? I mean, why is it necessary to use Flatten? – Miguel Duran Diaz Aug 3 '17 at 2:46
• @MiguelDuranDiaz Flatten is used only to remove an extra set of { } from s for plotting purposes. It has nothing to do with obtaining a solution. – bbgodfrey Aug 3 '17 at 2:50
• @Miguel, if you'd used vectors instead of column matrices, then the Flatten[] would not have been necessary, as Carl notes: With[{lo = 0.50, m = 1.5, k = 20, d = 3, g = 9.8}, NDSolveValue[{{x0'[t], x1'[t]} == {{0, 1}, {-k/m, -d/m}}.{x0[t], x1[t]} + {0, 1/m} (15 UnitStep[t - 5]) + {0, k lo/m + g}, x0[0] == 0, x1[0] == 0}, {x0, x1}, {t, 0, 10}]]. Please see this. – J. M. will be back soon Aug 3 '17 at 3:04
I would avoid column matrices:
A = {{0,1},{-k/m,-d/m}};
z[t_]:={0, 15UnitStep[t-5]}
lo = 0.50;
m = 1.5;
k = 20;
d = 3;
g = 9.8;
b = {0,1/m};
γ = {0,(k lo)/m+g};
Then, you can use the vector form of NDSolve/NDSolveValue as follows:
sol = NDSolveValue[
{x'[t] == A . x[t] + b z[t] + γ, x[0] == {0,0}},
x,
{t,0,10}
]
InterpolatingFunction[{{0., 10.}}, <>]
Finally, a plot:
Plot[sol[t], {t, 0, 10}] | 2019-11-18T17:22:36 | {
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https://math.stackexchange.com/questions/1662755/why-is-sqrtx2-always-x | Why is $\sqrt{x^2}$ always $x$? [duplicate]
This question has the potential to sound extremely stupid, but I've seen (and also used) countless times the idea that $\sqrt{x^2} = x$. However $x^2 = x\cdot x = (-x)\cdot(-x)$.
I know that when taking the square root of something we take both the positive and negative root. Yet when solving an equation and we're faced with $\sqrt{x^2y}$ we make it $x\sqrt{y}$. Why didn't we consider $(-x)\sqrt{y}$? Similarly, $\sqrt{x^3}$ is often changed to $x\sqrt{x}$ and not $(-x)\sqrt{-x}$ which would still give the same result if cubed? (I do understand that the latter is imaginary, but that shouldn't stop us from using it, should it?)
marked as duplicate by Najib Idrissi, Pierre-Guy Plamondon, Watson, quid♦, DanFeb 19 '16 at 19:19
• It's actually $|x|$. – Stefan Perko Feb 19 '16 at 10:49
• That makes a bit more sense.. but why? – Aayush Agrawal Feb 19 '16 at 10:51
• See en.wikipedia.org/wiki/Square_root#Properties_and_uses Square root function is defined to be non-negative – SS_C4 Feb 19 '16 at 10:51
• If $x\geq 0$, then $\sqrt{x^2} = x = |x|$. If $x<0$, then $\sqrt{x^2} = \sqrt{(-x)^2} = -x = |x|$. – Stefan Perko Feb 19 '16 at 10:52
• If we had $\sqrt{x^2}$ equal to both $x$ and $-x$, then $\sqrt{}$ wouldn't be a function. Therefore people have defined $\sqrt{}$ to be non-negative (so that the output of $\sqrt{a}$ is always unique for all $a\ge 0$). – user236182 Feb 19 '16 at 10:55
Consider $(-2)^2=4=2^2$. If one could devise a definition of the square root function such that $\sqrt{x^2}=x$ for any $x$, what would be the value of $$\sqrt{4}=\sqrt{(-2)^2}=\sqrt{2^2}$$ without getting a contradiction?
This shows that it is not possible to define a function with the desired property, so we abandon the idea and define, for $x\ge0$,
$\sqrt{x}$ is the unique nonnegative number $y$ such that $y^2=x$
In particular, $\sqrt{x^2}=|x|$, because $|x|\ge0$ by definition and $|x|^2=x^2$.
As an aside, note that $\sqrt{x^3}$ makes sense only if $x\ge0$, so in this case $\sqrt{x^3}=x\sqrt{x}$ is correct. You could also say $$\sqrt{x^3}=\sqrt{x^2\cdot x}=\sqrt{x^2}\cdot \sqrt{x} =|x|\sqrt{x}$$ which would be correct too, but $|x|=x$ as $x\ge0$.
Note, instead, that if $x<0$ and $y<0$, it would be very incorrect to write $$\sqrt{xy}=\sqrt{x}\sqrt{y}$$ but you can surely say $\sqrt{xy}=\sqrt{|x|}\sqrt{|y|}$.
The symbol $\sqrt{\mathstrut\quad}$ is defined to denote the non-negative square root. In other words: $\sqrt{x}$ is the number $y$ such that $y\geq 0$ and $y^2=x$. The number $y$ is then called the radical of $x$. This is a matter of definition. We need to choose one of the two values to make $\sqrt{\mathstrut\quad}$ into a function, and we like positive numbers better.
If $y = \sqrt{x}$ then $y^2 = (-y)^2 = x$. So now, what if $y = \sqrt{x^2}$?
Since $y \geq 0$ by definition, we have $y=x$ if $x\geq0$ or $y = -x$ if $x<0$. We write this as $y = |x|$ and say $y$ is the absolute value of $x$.
Also note that if $x$ is positive, $\sqrt{-x}$ is not well defined even if you can still find a number $y$ such that $y^2=-x$. You have to be careful: the radical is only defined for non-negative numbers.
• that most people (including me) seem to say "the square root of $x$" when they actually mean "the positive square root of $x$", or $\sqrt{x}$. – Andrea Feb 19 '16 at 11:02
Actually, $\sqrt{x^2} = |x|$.
If $x\geq 0$, then $\sqrt{x^2} = x = |x|$. If $x<0$, then $\sqrt{x^2} = \sqrt{(-x)^2} = -x = |x|$.
As mentioned in comments: $\sqrt {x^2}=|x|$
Reason is that range of square root function is non-negative.
Assume that using $\sqrt {x^2}=x$ you write $\sqrt {(-2)^2}=-2$ which is clearly wrong.
I hope you get the point.
• Why is it wrong? – tatan Feb 19 '16 at 11:02
• @tatan Because $\sqrt{(-2)^2}=\sqrt{2^2}$. Would you say that $\sqrt{2^2}=-2$? – egreg Feb 19 '16 at 11:04
• @egreg but in the first answer,it is written $\sqrt {-x^2}=\sqrt {x^2}=-x$.... – tatan Feb 19 '16 at 11:06
• @tatan I see nothing of that kind in any of the current answers. – egreg Feb 19 '16 at 11:08
• @tatan You're wrong: “if $x<0$, then $\sqrt{x^2}=\sqrt{(-x)^2}=-x=|x|$”, which is fully correct. Parentheses are important. – egreg Feb 19 '16 at 11:11 | 2019-05-22T08:34:27 | {
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https://math.stackexchange.com/questions/3200565/does-2on-mean-o2n-if-not-what-does-2on-mean/3200580 | # Does $2^{O(n)}$ mean $O(2^n)$? If not, what does $2^{O(n)}$ mean?
In this "tutorial", in the end, they say exponential asymptotic notation is $$2^{O(n)}$$.
Is $$2^{O(n)}$$ the same thing as $$O(2^n)$$? Is there any reason to notate it that way?
According to one of the answers, they are different. But then according to the "tutorial", $$O(f(n))$$ is a set. So how is the expression $$2^{O(n)}$$ meaningful?
(Specifically they define $$O(f(n)) = \{ g(n) :$$ exists $$c \gt 0$$ and $$n_0 \gt 0$$ such that $$f(n) \leq c*g(n)$$ for all $$n \gt n_0$$).
• $f(n)=2^{O(n)}$ means $\log_2 f(n)=O(n).$ – Thomas Andrews Apr 24 at 14:56
A function $$f:\mathbb N\to\mathbb R$$ is $$2^{O(n)}$$ if and only if there is a constant $$C$$ such that for all $$n$$ large enough we have $$f(n)\le 2^{Cn}$$.
We can think of the $$O$$ notation as decribing a family of functions. So, $$2^{O(n)}$$ would be the family of functions satisfying the requirements just indicated.
In contrast, a function $$f$$ is $$O(2^n)$$ if and only if there is a constant $$K$$ such that for all $$n$$ large enough we have $$f(n)\le K 2^n$$.
It should be clear that $$O(2^n)\subset 2^{O(n)}$$, that is, any function that is $$O(2^n)$$ is also $$2^{O(n)}$$. The converse fails, however: consider $$f(n)=2^{2n}$$.
• Would it make sense to use "composite" of big-O notation for functions other than $2^n$ too? ie. $f: \mathbb{N} \rightarrow \mathbb{R}$ is $g( O(n) )$ for a real function $g$ iff there's a constant $C$ such that for all $n$ large enugh we have $f(n) \leq g( Cn )$ ? – RUBEN GONÇALO MOROUÇO Apr 24 at 14:20
• Sure, that would be the natural interpretation of the notation. Of course, if you are using it in a writing of your own, it may be best to first remind the reader of its meaning, to avoid confusion. – Andrés E. Caicedo Apr 24 at 14:21
• Shouldn't $2^{O(n)}$ include a positive lower bound on $f$ as well? – Umberto P. Apr 24 at 15:10
• @Umberto It is a matter of convention. Sure, some sources say that $f$ is $O(g)$ if and only if $|f(n)|\le C g(n)$ for some $C$ and all $n$ large enough. In that case, yes, we need some such lower bound as well (in the form: $f$ is $O(2^n)$ if and only if $|f(n)|\le 2^{Cn}$ for some $C$ and all large $n$). In many settings, though, all functions are assumed positive, and so the inclusion of the absolute value is redundant. It depends on the context. – Andrés E. Caicedo Apr 24 at 15:15
No, it isn't the same. $$2^{3n}$$ is $$2^{O(n)}$$ but not $$O(2^n)$$
• Hmm makes sense. What is the meaning of $2^{O(n)}$ then? According to their definition $O(f(n))$ is a set. How can $2$ to the power of a set be meaningful in this context? – RUBEN GONÇALO MOROUÇO Apr 24 at 14:08
• I see that Andrés E. Caicedo has already addressed this. – saulspatz Apr 24 at 14:16 | 2019-05-20T19:13:49 | {
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https://math.stackexchange.com/questions/1358310/how-to-prove-that-the-fibonacci-sequence-is-periodic-mod-5-without-using-inducti/1358352 | # How to prove that the Fibonacci sequence is periodic mod 5 without using induction?
The sequence $(F_{n})$ of Fibonacci numbers is defined by the recurrence relation
$$F_{n}=F_{n-1}+F_{n-2}$$ for all $n \geq 2$ with $F_{0} := 0$ and $F_{1} :=1$.
Without mathematical induction, how can I show that $$F_{n}\equiv F_{n+20}\pmod 5$$ for all $n \geq 2$?
• @Timbuc, as far as I can see, isn't the OP asking for non-induction techniques to prove this result? @ op You might want to clear up your title and body to say "use non-induction techniques/methods" instead of what you currently have. – Zain Patel Jul 12 '15 at 11:23
• Now I see the non-induction thing, thanks. BTW, I've no idea what other methods can be used to prove such a result, but that's my problem... – Timbuc Jul 12 '15 at 11:31
• Note. to answer the question in the title is easy: there are only finitely many possible pairs mod(5) so as you list the Fibonacci numbers mod (5), at some point a pair has to recur. Of course, this does not tell you what the period is...only that it is periodic. – lulu Jul 12 '15 at 12:23
• @lulu but you can know the max period is 25 because that's how many pairs of numbers there are mod 5. – ratchet freak Jul 12 '15 at 12:30
• I believe everyone is missing the obvious interpretation, "present a proof that can be followed by a student that doesn't know induction". Obviously integers are inductively defined, that doesn't mean that all proofs have to start there. – DanielV Jul 13 '15 at 20:49
In mod $5$,
\begin{align}F_N&\equiv F_{N-1}+F_{N-2}\\&\equiv F_{N-2}+F_{N-3}+F_{N-3}+F_{N-4}\\&\equiv F_{N-3}+F_{N-4}+2(F_{N-4}+F_{N-5})+F_{N-4}\\&\equiv F_{N-4}+F_{N-5}+F_{N-4}+2(F_{N-4}+F_{N-5})+F_{N-4}\\&\equiv 3F_{N-5}\end{align} So, we have
\begin{align}F_{n+20}&\equiv3F_{n+15}\\&\equiv 3\cdot 3F_{n+10}\\&\equiv 3\cdot 3\cdot 3F_{n+5}\\&\equiv 3\cdot 3\cdot 3\cdot 3F_{n}\\&\equiv F_n\end{align}
• Also using concealed induction but very nice imo. +1 – Timbuc Jul 12 '15 at 11:40
• @Timbuc I don't see the induction here. This answer directly applies the definition for several values of n. – Christian Semrau Jul 12 '15 at 11:54
• @ChristianSemrau As noted in other comments, the very definition of the Fibonacci sequence is inductive, so you can't avoid that... – Timbuc Jul 12 '15 at 12:14
• @Timbuc, this answer does not contain induction, it's very clear that if a sequence is defined inductively this does not mean that we must use induction to prove it, take for example the statement $F_3=2$ do we need induction in order to prove it. – Elaqqad Jul 12 '15 at 15:06
• @Timbuc even if it contains $n$ it does not mean that it must be proved by induction : for instance $\forall n \ \ \ \ F_{n+3} =2F_n+F_{n+1}$ does not need induction to be proved, of course we have to admit the definition of Fibonacci numbers $\forall n \ \ \ F_{n+2}=F_{n+1}+F_{n}$ using this you can prove without induction a lot of statements involving $n$. – Elaqqad Jul 12 '15 at 15:49
As a different approach, you can just solve the recursion mod(5) exactly. There's a small problem in that the characteristic equation $$\lambda^2=\lambda +1$$ has a double root, 3, mod 5. But the standard device, using $n\lambda^n$ works and we see that the general term, mod (5), is $$F_n=(1+n)3^n$$ You are then asking that $(1+n)3^n = (1 + n + 20)3^{n+20}$ mod (5). But 21 + n = 1+ n and 3 has order 4 so we are done.
• Very elegant, lulu! For those unfamiliar with this approach using the characteristic equation, please see the Wikipedia article on recurrence relations, in particular this theorem, which references the definition of the characteristic polynomial given in this section. – PM 2Ring Jul 12 '15 at 12:25
• I suppose that someone might argue that there is a buried induction here. My argument produces a sequence which satisfies the desired recursion and has the right initial values, so I then (implicitly) assert that it coincides with the desired sequence. That is an induction. But, after all, even knowing that the $n^{th}$ term exists is an induction! – lulu Jul 12 '15 at 12:37
• You said: "even knowing that the $n^{th}$ term exists is an induction!". Definitely. And as Peano pointed out, we can say the same thing about the $n^{th}$ term of $N$. :) – PM 2Ring Jul 12 '15 at 12:47
• Thank you for this nice answer too +1 – user225250 Jul 13 '15 at 3:55
• Very impressive answer ! – Hexacoordinate-C Jul 14 '15 at 2:02
There's been a lot of discussion in this question about whether certain proofs contain a hidden induction, so this answer formalizes what it means for a proof to use induction, and discusses which of the given answers use induction with respect to this formalization.
The natural numbers are defined by the Peano axioms, which can be stated succinctly as follows:
1. The natural numbers $\mathbb{N}$ form a discretely ordered semiring.
2. If $S\subset \mathbb{N}$ has the properties that (i) $0\in S$ and (ii) $(\forall n)(n\in S \Rightarrow n+1\in S)$, then $S=\mathbb{N}$.
In axiom 1, a semiring is similar to a ring, except that elements need not have additive inverses, and saying it's "discretely ordered" means that there's a linear order on $\mathbb{N}$ that satisfies certain axioms. See here for a complete list of axioms contained in axiom 1.
Axiom 2 is the axiom of induction. Axioms 1 and 2 together define Peano Arithmetic (PA), while axiom 1 alone defines a theory similar to the natural numbers in which induction does not necessarily hold. This theory is often denoted $\mathrm{PA}^-$.
So asking whether something can be proven "without induction" is essentially asking whether we can prove the statement in a model for $\mathrm{PA}^-$, i.e. asking whether we can prove the statement for any discretely ordered semiring.
This presents a problem, because it's not clear exactly what the "Fibonacci numbers" refer to in an arbitrary discretely ordered semiring. Here's one possible definition:
Definition. Let $N$ be a discretely ordered semiring. A Fibonacci function on $N$ is a function $f\colon N\to N$ satisfying the following conditions:
1. $f(0) = 0$.
2. $f(1) = 1$.
3. $f(n+2) = f(n+1) + f(n)$ for all $n\in N$.
Here $0$ denotes the additive identity of $N$, and $1$ denotes the multiplicative identity.
Now, it's possible to prove using induction that there exists a unique Fibonacci function on $\mathbb{N}$ (namely the usual Fibonacci sequence), but this isn't possible to prove in an arbitrary discretely ordered semiring $N$. In fact it's possible to prove (in ZFC) that a Fibonacci function always exists, but it won't be unique unless $N$ is isomorphic to $\mathbb{N}$.
However, this doesn't prevent us from proving things about arbitrary Fibonacci functions. Here's a statement and proof of the OP's claim without any induction:
Theorem. Let $N$ be a discretely ordered semiring, and let $f\colon N \to N$ be a Fibonacci function. Then for all $n\in N$, there exists a $k\in N$ so that $$f(n+20) = f(n) + 5k,$$ where $5$ denotes $1+1+1+1+1$ and $20$ denotes $5+5+5+5$.
Proof: We will follow mathlove's beautiful answer. Before the proof begins, note that $N$ must contain a canonical copy of $\mathbb{N}$, namely the subsemiring generated by $1$. For convenience, we will assume that $\mathbb{N}\subset N$, which lets us use constants like $5$ without explaining that $5$ means $1+1+1+1+1$.
Observe first that \begin{align}f(n+5) &= f(n+4)+f(n+3)\\&= f(n+3)+f(n+2)+f(n+2)+f(n+1)\\&= f(n+2)+f(n+1)+2(f(n+1)+f(n))+f(n+1)\\&= f(n+1)+f(n)+f(n+1)+2(f(n+1)+f(n))+f(n+1)\\&= 3f(n) + 5f(n+1)\end{align} for all $n\in N$. Then \begin{align}f(n+20)&= 3f(n+15) + 5f(n+16)\\&= 3^2f(n+10) + 5\bigl(3f(n+11)+f(n+16)\bigr)\\&= 3^3 f(n+5) + 5\bigl(3^2 f(n+6)+3 f(n+11) + f(n+16)\bigr)f\\&= 3^4 f(n) + 5\bigl(3^3 f(n+1) + 3^2 f(n+6) + 3f(n+11) + f(n+16)\bigr),\end{align} But $3^4 = 81 = 5(16) + 1$, and hence $$f(n+20) = f(n) + 5\bigl(16f(n) + 3^3 f(n+1) + 3^2 f(n+6) + 3f(n+11) + f(n+16)\bigr).$$ This proves the given theorem in an arbitrary discretely ordered semiring, with no use of induction.
So mathlove's answer is correct, in the sense that the argument legitimately doesn't use induction.
I suspect that lulu's answer does use induction, although it's hard to tell, because it's harder to see how it can be translated to the context of arbitrary discretely ordered semirings. There's also the problem that exponentiation can't be defined uniquely in an arbitrary ordeerd semiring. Perhaps what lulu has shown is that there exists a Fibonacci function with the desired property.
Like mathlove's answer, Elaqqad's answer works just fine in an arbitrary discretely ordered semiring, which means that it legitimately doesn't use induction.
Jack D'Aurizio's answer uses Binet's formula, which presumably can't be made to work in an arbitrary discretely ordered semiring, though I suppose it might be possible to recover some version of it. We would have to start by discussing whether an arbitrary discretely ordered semiring can be embedded in some sort of field that contains a square root of five, and in what sense it might be possible to define exponentiation on that field with the exponent being an element of the semiring.
Klaus Draeger's answer of course requires induction, but I suspect that a similar argument could be made to work in general, simply by replacing the initial $(1,0)$ by an arbitrary pair $(a,b)$ and reducing modulo $5N$. (As far as I can tell, we have no idea how large $N/5N$ is in general, but that doesn't mean we can't do calculations in the quotient. Note that using $N/5N$ would have simplified the proof above as well, though it would have increased the conceptual difficulty.)
Christian Blatter's answer uses induction to prove that $G_n=0$ for all $n$. I don't see a way around this.
• I think this is a very nice answer/clearification of all this mess. +1 – Timbuc Jul 13 '15 at 7:02
• You're correct about Klaus's answer. As he intended it, it does require induction because he essentially claims that since it goes back to the starting $(0,1)$ it must hence repeat. The repetition cannot be proven without induction. As for Jack's answer, if he didn't mention roots of the characteristic polynomial but instead used the right-shift operator, his answer would have been induction-free, because he is doing nothing other than proving a specific instance of the formula in the middle of Elaqqad's answer. Once he talks about real roots he has invoked far more than induction! – user21820 Jul 13 '15 at 11:15
• Lulu's answer definitely uses induction because she clearly intends to solve a recurrence via the characteristic equation, same as Christian Blatter's answer. Steven's answer also uses induction since he claims periodicity just from repetition. As for what Jack is using, maybe he did not mean real roots? Who knows? Abstract algebraic extensions with explicit representations should work. – user21820 Jul 13 '15 at 11:21
• I've said this below but I think that this nitpicking is missing the point. The spirit of not using induction, as far as I understand it and I may be wrong, is explaining where the result comes from instead of merely verifying it. Induction proofs are sometimes little more than mere verifications. – man and laptop Jul 15 '15 at 17:17
• This is the best answer I've read in a while :-) – joriki Jul 16 '15 at 10:34
Another possible approach. Let $\sigma$ a root of the characteristic polynomial $x^2-x-1$. We have:
$$\sigma^2 = \sigma+1,\qquad \sigma^4 = \sigma^2+2\sigma+1 = 3\sigma+2,$$ $$\sigma^8 = 9\sigma^2 + 12\sigma +4 = 21\sigma + 13,\qquad \sigma^{16} = 441\sigma^2 + 546\sigma + 169 = 987\sigma + 610,$$ hence: $$\sigma^{20} = (3\sigma + 2)(987\sigma + 610) = 6765\sigma + 4181 = \sigma^0 + 55(123\sigma+76).$$ If now we multiply both sides by $\sigma^n$ and use the Binet's formula, we prove the stronger claim:
$$\forall n\geq 0,\qquad \color{red}{55} \mid (F_{n+20}-F_n).$$
Proof without induction
First we have : $$F_{n+20}=F_{n+19}+F_{n+18}=2F_{n+18}+F_{n+17}= 3F_{n+17}+2F_{n+16}$$
If you continue reducing this $20$ times you will have: $$F_{n+20} = 10946 F_{n}+ 6765 F_{n-1} \tag {*}$$
and from here you can see that:
$$F_{n+20}\equiv F_n \mod 5$$
Note The formula $(*)$ could be computed $\mod 5$ which reduces large numbers but it will take some time to finish the calculus, in fact it's the part of a general formula which can be proved by induction : $$F_{p+q}=F_pF_{q+1}+F_{p-1}F_q$$ (take $p=n$ and $q=20$)
If a sequence is defined by induction, then this does not mean that we have to use induction to prove every fact about this sequence. let's take for example the Fibonacci sequence, does one need induction in order to prove that $F_2=1$ or that $F_3 = 2$? of course no . Actually we can formalize the definition in two statements: \begin{align} (1) && F_0=0 , && F_1=1 \\ (2) && \forall n \in \mathbb{N} && F_{n+2}=F_{n+1}+F_n\end{align}
so from the first assertion we can prove directly that (without induction) that $F_2=1,F_3=2,F_4=3,\cdots$, and from the second assertion we can prove directly that $\forall n \in \mathbb{N}\ \ \ F_{n+3}=F_{n+1}+2F_{n}$ and $$(3)\ \ \ \ \ \ \forall n\in \mathbb{N}\ \ \ \ F_{n+20}\equiv F_n \mod 20$$
If we want to be more precise, we will say that we can prove without induction that "every sequence $(F_n)_{n\in \mathbb{N}}$ verifying $(2)$ must verify also $(3)$", and more formally: $$\forall (F_n)\in \mathbb{N}^{\mathbb{N}} \ \ \ \ \big(\left(\forall n \in \mathbb{N} F_{n+2}=F_{n+1}+F_n\right)\implies\left(\forall n\in \mathbb{N} F_{n+20}\equiv F_n \mod 20\right)\big)$$
(and here we don't now if $F_n$ is defined uniquely to prove it we must use induction)
• How do you know that you can do what your two first lines show for all $\;n\in\Bbb N\;$ ? Of course, induction...though, as in many, many other cases, it is slightly concealed. Anyway, nice going. +1 – Timbuc Jul 12 '15 at 11:39
• @Timbuc The reduction of $F_{n+20}$ works without induction because it directly applies the definition, for different values of n, which is stated to be valid for all n. – Christian Semrau Jul 12 '15 at 11:50
• @Timbuc: The Fibonacci sequence is defined inductively. So in the strictest sense it is impossible to prove it without induction, for when one uses the definition then one "commits" induction :). – Megadeth Jul 12 '15 at 12:04
• The Fibonacci sequence itself is defined inductively. But I don't see where you need induction to prove the statement $(\forall n: F_{n+2}=F_{n+1}+F_{n}) \Rightarrow (\forall n: F_{n+20} \equiv F_n\mod 5)$ – Christian Semrau Jul 12 '15 at 12:25
• I don't understand the talk about 'implicit induction'. The proof method asked for should not use an induction hypothesis, and they don't. Any implicit induction is like saying "Gaussian elimination inverts an $n\times n$ matrix" or "$2^n$ is total" are proved implicitly by induction because they involves natural numbers. – Mitch Jul 12 '15 at 16:00
"Without using X" questions are always a little dubious - they often boil down to "how well can you hide the X". One obvious approach to this one is to consider the pairs of reminders of successive Fibonacci numbers modulo $5$. The starting point is $(0,1)$, and the successor of $(a,b)$ is $(b,a+b\ mod\ 5)$. This gives the cycle
$(0,1)\to(1,1)\to(1,2)\to(2,3)\to(3,0)\to(0,3)\to(3,3)\to(3,1)\to(1,4)\to(4,0)\to(0,4)\to(4,4)\to(4,3)\to(3,2)\to(2,0)\to(0,2)\to(2,2)\to(2,4)\to(4,1)\to(1,0)\to(0,1)$
of length $20$. But again, this is just a fancy representation of an underlying inductive argument.
• It is a bit interesting that only 25 such pairs $(a,b)$ exist, and your cycle accounts for the 20 of them. The trivial sequence $(0,0) \to (0,0)$, a cycle of length one (fixpoint), accounts for 1 more. The last 4 pairs are in a cycle $(1,3) \to (3,4) \to (4,2) \to (2,1) \to (1,3)$. – Jeppe Stig Nielsen Jul 14 '15 at 21:10
• Yes; the big cycle consists of those pairs for which $a-2b$ is nonzero (and it gets multiplied by $3$ in each step), while it is $0$ both for the fixpoint $(0,0)$ and the members of the slammer cycle. – Klaus Draeger Jul 14 '15 at 21:44
$$\begin{bmatrix} F_{n+1} & F_{n} \\ F_{n} & F_{n - 1} \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n$$
So
\begin{align} \begin{bmatrix} F_{n+21} & F_{n+20} \\ F_{n+20} & F_{n + 19} \end{bmatrix} &= \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^{20 + n} &\pmod 5 \\ % &= \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^{20} \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^{n} &\pmod 5 \\ % &= \begin{bmatrix} 10946 & 6765 \\ 6765 & 4181 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^{n} &\pmod 5 \\ % &= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^{n} &\pmod 5 \\ % &= \begin{bmatrix} F_{n+1} & F_{n} \\ F_{n} & F_{n - 1} \end{bmatrix} &\pmod 5 \\ \end{align}
Other cycles can be found similarly, for example
$$\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^{24} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \pmod 6$$
Giving a cycle of length $24$ modulo $6$.
• Nice answer ！，，， +1 – user225250 Jul 13 '15 at 4:09
• Exactly the same as Elaqqad's and Jack's answers, just presented using matrices. It can be helpful to get people to appreciate matrices more... – user21820 Jul 13 '15 at 11:23
• Note that it is easy to remove reliance on induction from your answer if you just relate $(F_{n-1},F_n,F_{n+1})$ directly with $(F_{n+19},F_{n+20},F_{n+21})$ instead of going through the $n$-th power of the operator. – user21820 Jul 13 '15 at 11:26
• For some reason I like this one the best. Something about the representation as matrices makes this really palatable. – Paddling Ghost Jul 13 '15 at 18:28
While it doesn't show the precise result ($F_n\equiv F_{n+20}$) that's being looked for, here's a proof that $F_n$ must be periodic mod $5$ — or in fact to any base — using an entirely different approach:
The value of $F_{n+1}$ is some polynomial function of the two previous values $F_n$ and $F_{n-1}$, $F_{n+1}=f(F_n, F_{n-1})$ (in this case, $f(a,b)=a+b$, but that's not actually germane here); therefore this also holds true $\bmod 5$. But now there are at most $5\cdot 5=25$ possible values of the pair $\langle F_{n-1}, F_n\rangle$ — so by the pigeonhole principle, within at most 25 iterations we'll hit a pair of arguments to $f()$ that we've seen before. But since the inputs to the function are the same, the output will be the same, and then the function is periodic with this period.
Note that this argument doesn't care what the function is —, or the modulus it holds for any recurrence relation where the value of $R_{n+1}\bmod m$ is dependent on only the values $\mod m$ of the previous two terms $R_n$ and $R_{n-1}$ (or more generally, where it's dependent on the previous $k$ terms, for some constant $k$).
• That's exactly how I was intending to answer (after first reading the title only, not the question and its specification of $F_{n+20}$). But then, I've been fascinated by modular Fibonacci sequences ever since doing "Fibonacci bracelets" (mod 10) as an "extension" activity at school, twenty-odd years ago. – Tim Pederick Jul 14 '15 at 17:25
(This is @mathlove 's solution slightly streamlined.)
The sequence $$G_n:=F_{n+5}-3F_n\qquad({\rm mod} \ 5)$$ satisfies the same recursion as the $F_n$. Furthermore one easily checks that $G_0=G_1=0$. This implies $G_n=0$ for all $n$, so that $$F_{n+5}=3F_n \qquad({\rm mod} \ 5)$$ for all $n$. It follows that $$F_{n+20}=3^4 F_n =F_n\qquad({\rm mod} \ 5)$$ for all $n$.
• Yes this is a good idea, but it uses induction, when you say $G_0=G_1=0$ then $G_n=0$ this can not be proved without induction – Elaqqad Jul 12 '15 at 14:55
• @Elaqqad: It follows from the uniqueness theorem for linear second order recursions. – Christian Blatter Jul 12 '15 at 14:58
• Yes that's true, and hence we used a theorem which is proved by induction. It's true that it's not really very precise what it means not to use induction , I personally don't like questions like : "prove without induction" , it's always better to find the most elegant method without any restriction. – Elaqqad Jul 12 '15 at 15:04
• @Elaqqad: While this answer does require at least one application of an equivalent of the induction axiom, mathlove's answer does not, because once the natural numbers and the fibonacci sequence are given, his answer proves the desired statement directly. But this answer does not as it constructs a new sequence $G$ inductively. – user21820 Jul 12 '15 at 17:45
• @user21820 I agree with that and that's what I'm talking about – Elaqqad Jul 12 '15 at 17:56
Unlike most of the answers above this ones derives the length of the period instead of just verifying the conjecture. I believe this is the spirit of not using induction, because induction can sometimes be used merely as a verification and not as an "explanation".
Start by noticing that $$\begin{bmatrix} 0 & 1 \\ 1 & 1\end{bmatrix}^n \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} F_n \\ F_{n+1} \end{bmatrix}$$ and let $S = \begin{bmatrix} 0 & 1 \\ 1 & 1\end{bmatrix}$
All numbers are in $\mathbb F_5$. Computing powers of $S$ is easy if you can diagonalize it. So we start by finding the characteristic polynomial which is $\lambda^2 - \lambda - 1 = (\lambda + 2)^2$. This shows that the only eigenvalues are $-2$. If $S$ were diagonalizable then $S$ would be a multiple of the identity matrix, which it isn't. Therefore we use $S$'s Jordan normal form instead, which is $\begin{bmatrix} -2 & 1 \\ 0 & -2\end{bmatrix}$. Now notice that matrices of the form $\begin{bmatrix} a & b \\ 0 & a\end{bmatrix}$ form a ring, and the determinant of such matrices is $a^2$, so the unit group of that ring has order $20$ (the number of $(a,b) \in \mathbb F_5^2$ pairs for which $a^2 \neq 0$). By Lagrange's theorem, $S^{20} = I$.
• I'm not up to snuff on my algebra: where did 20 come from? I'm assuming it's because $20=4\cdot 5$, but: "Because the determinant is 4 [what result are you appealing to to conclude?] the unit group of that ring has order 20" – Eric Stucky Jul 19 '15 at 0:16
• @EricStucky The ring of matrices in the form $\begin{bmatrix} a & b \\ 0 & a\end{bmatrix}$ has $25$ elements. The ring is closed under inverses. The invertible elements are precisely those for which $a \neq 0$ because then $\det \neq 0$. Since there are $5$ singular elements (those for which $a = 0$), $20$ are invertible. Then by Lagrange's theorem, $S^{20} = I$ because the Jordan normal form of $S$ is in the unit group of that ring (please say if you want me to explain this). If you want to understand the ring better look at the dual numbers. – man and laptop Jul 19 '15 at 8:03
• Okay, so $20=5^2-5$, I see. – Eric Stucky Jul 19 '15 at 8:55
You could also just list the first twenty or so terms of the sequence by adding consecutive terms and reducing mod $5$:
$$1,1,2,3,0,3,3,1,4,0,4,4,3,2,0,2,2,4,1,0,1,1,2 \ldots$$
at which point we have $F_{21} = F_1$ and $F_{22} = F_2$ and therefore the sequence begins to repeat.
Since you asked for non-inductional ways, i'll use an expression attributable to François Édouard Lucas: $$f_n =\frac{\left(\phi \right)^n - \left(1- \phi \right)^{-n}}{\sqrt5}$$ $$We´re \space looking\space for \space a \space difference(D) \space such\space that \qquad{D \equiv0 \mod 5}\qquad \qquad Note\space that \qquad{ D= f_{n+20}- f_n=\frac{\left(\phi \right)^{n+20} - \left(1- \phi \right)^{-(n+20)}}{\sqrt5}-\frac{\left(\phi \right)^n - \left(1- \phi \right)^{-n}}{\sqrt5}=\left( \frac{\left(\phi \right)^{10} - \left(1- \phi \right)^{-10}}{\sqrt5} \right) \left({\left(\phi \right)^{n+10} - \left(1- \phi \right)^{-(n+10)}}\right)=(f_{10})\left({\left(\phi \right)^{n+10} - \left(1- \phi \right)^{-(n+10)}}\right)}$$ $$but\qquad 5\mid (f_{10}=55) \Longrightarrow 5\mid D\Longrightarrow{f_n≡f_{n+20}(mod5)}$$
Your approach to prove a theorem usually depends on how you define things. As Elaqqad said, you can't get rid of induction when you define the Fibonacci sequence by Induction. But if you define Fibonacci sequence from the start with no induction, you can prove some of its features without induction.
I usually suggest to be pragmatic, and accept mathematical induction. But, as you wish, I DO prove Fibonacci sequence is periodic mod 5 without using induction.
Let's start from a non-inductive definition of Fibonacci sequence. I use Binet's formula, as a closed-form expression of the Fibonnaci number.
Fibonnaci number is defined as a function of $n$, which outputs an Integer, and defined as*:
$F[n] := \cfrac{1}{\sqrt{5}}\cdot\left(\cfrac{1+\sqrt{5}}{2}\right)^n-\cfrac{1}{\sqrt{5}}\cdot\left(\cfrac{1-\sqrt{5}}{2}\right)^n$
We are about to show $$F_{n}\equiv F_{n+20}\pmod 5$$ for all $n \geq 2$
Actually, we also calculate the reminder. We prove that:
$F[n+20]-F[n]=10945.F[n]+6765F[n−1]$
Doing this is really cumbersome. You can't do this by hand, but using a computational software program like Mathematica as a form of Automated theorem proving you can do it easily:
Define the functions as above statement, and do this in Mathematica:
Simplify[f[n + 20] - f[n] - 10945 f[n] - 6765 f[n - 1]]
and you get exact zero:
0
which compeltes the proof.
I'm trying to extract the exact intermediatory simplifying steps from Mathematica. I'll post it as soon as I get it.
* I am not going to prove that F(n) is actually an integer here.
• Where is the problem in your question ? have you any specification ? – zeraoulia rafik Jul 13 '15 at 15:30
• @zeraouliarafik I have edited my answer substantially to make it more clear. – Ho1 Jul 13 '15 at 16:14
## protected by Zev ChonolesJul 13 '15 at 15:31
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https://math.stackexchange.com/questions/1736428/efficiently-evaluating-int-x4e-xdx/1736446 | # Efficiently evaluating $\int x^{4}e^{-x}dx$ [duplicate]
The integral I am trying to compute is this:
$$\int x^{4}e^{-x}dx$$
I got the right answer but I had to integrate by parts multiple times. Only thing is it took a long time to do the computations. I was wondering whether there are any more efficient ways of computing this integral or is integration by parts the only way to do this question?
Edit: This question is similar to the question linked but slightly different because in the other question they are asking for any method to integrate the function which included integration by parts. In this question I acknowledge that integration by parts is a method that can be used to evaluate the integral but am looking for the most efficient way. This question has also generated different responses than the question linked such as the tabular method.
## marked as duplicate by J. M. is a poor mathematician, Semiclassical, T. Bongers, Pierre-Guy Plamondon, user147263 Apr 11 '16 at 23:31
• Would need limits for this if you intend a numerical answer. Not meant to be a trivial statement since if the upper limit is infinity then more computation needed. – jim Apr 10 '16 at 18:50
• @mikevandernaald Gamma function has a limits , there is no limits – openspace Apr 10 '16 at 18:51
• I think that the simplest way to compute it is integrating by parts. – Crostul Apr 10 '16 at 18:52
• In the original question I was attempted the limits of integration were from 0 to 1 – user262291 Apr 10 '16 at 18:52
• Use the tabular method pages.pacificcoast.net/~cazelais/187/tabular.pdf – user258700 Apr 10 '16 at 18:57
While it still entails integration by parts, there exists a "quick" method of doing the integration by parts called the Tabular Method. Basically, you start with a table that has $3$ columns. One with alternating signs, one with a $u$ and one with $dv$. In the column for $u$, you should put the term that will eventually go to zero after multiple differentiations. In the last column is $dv$ which you will integrate multiple times. You'll want to pick something you can still integrate multiple times for $dv$. So in $\displaystyle \int x^4e^{-x}\,dx$, we can create the following table: $$\begin{matrix} & u & dv \\ + &x^4 & e^{-x} \\ - & 4x^3 & -e^{-x} \\ + & 12 x^2 & e^{-x} \\ - & 24x & -e^{-x} \\ + & 24 & e^{-x} \\ - & 0 & -e^{-x}.\end{matrix}$$ we now have all the parts to compute our solution. You start at the very first $+$, and multiply the corresponding $u$ term with the $dv$ term on the very next line. Continue this process until you get to the end of the $dv$ column. So our solution is $$\int x^4e^{-x}\,dx = -x^4e^{-x} -4x^3e^{-x} -12x^2e^{-x}-24xe^{-x}-24e^{-x} + C$$
• Why $-x^4e^{-x}$ instead of $x^4e^{-x}$ (and similar questions for the other terms)? – PyRulez Apr 11 '16 at 1:25
• It's because $x^4$ is multiplied by $-e^{-x}$ (the $dv$ entry on the second line). – jtbandes Apr 11 '16 at 3:03
• Yes, the $dv$ entry on the next line is the $v$, the integration, of the the $dv$ entry on the current line. This is what we want $\int u dv = uv - \int v du$. – Dominic108 Apr 11 '16 at 11:11
Usually this kind of integrals can be handled by making
\begin{align*} \int x^4e^{-x}\,\mathrm dx&=-x^4e^{-x} + (Ax^3+Bx^2+Cx+E)e^{-x} \end{align*} Where $A,\;B,\;C$, and $E$ are constants which satisfies $$x^4e^{-x}+(-4-A)x^3e^{-x}+(3A-B)x^2e^{-x}+(2B-C)xe^{-x}+(C-E)e^{-x}=x^4e^{-x}$$ So \begin{align*} -4-A&=0\\ 3A-B&=0\\ 2B-C&=0\\ C-E&=0 \end{align*} Then $A=-4$, $\;B=3A=-12$, $\;C=2B=-24$ and $E=C=-24$, then \begin{align*} \int x^4e^{-x}\,\mathrm dx&=-(x^4+4x^3+12x^2+24x+24)e^{-x}+K \end{align*} where $K$ is a constant.
Also, here is a more general problem related.
Here is a nice little trick to integrate it without using partial integration. $$\int x^4 e^{-x} \,\mathrm dx = \left. \frac{\mathrm d^4}{\mathrm d \alpha^4}\int e^{-\alpha x} \,\mathrm dx \right|_{\alpha=1} = \left.- \frac{\mathrm d^4}{\mathrm d \alpha^4} \frac{1}{\alpha} e^{-\alpha x}\right|_{\alpha=1}$$ The idea is to introduca a variable $\alpha$ in the exponent and write the $x^4$ term as the fourth derivative with respect to $\alpha$. This is especially helpful when you want to calculate the definite integral $\int_0^\infty$ because in this case the differentiation greatly simplifies. $$\int\limits_0^\infty x^n e^{-x} \,\mathrm dx = (-1)^n \left. \frac{\mathrm d^n}{\mathrm d \alpha^n} \frac{1}{\alpha} \right|_{\alpha=1} = n!\stackrel{n=4}{=} 24$$ | 2019-05-25T05:41:29 | {
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https://math.stackexchange.com/questions/1624236/finding-eigenvectors-of-a-3x3-matrix-7-12-15 | # Finding Eigenvectors of a 3x3 Matrix (7.12-15)
Please check my work in finding an eigenbasis (eigenvectors) for the following problem. Some of my solutions do not match answers in my differential equations text (Advanced Engineering Mathematics by Erwin Kreyszig, 1988, John Wiley & Sons).
For reference the following identity is given because some textbooks reverse the formula having $\lambda$ subtract the diagonal elements instead of subtracting $\lambda$ from the diagonal elements:
$$det(A - \lambda I) = 0$$ $$A = \begin{bmatrix} 3 & 1 & 4 \\ 0 & 2 & 6 \\ 0 & 0 & 5 \\ \end{bmatrix}$$
By inspection the eigenvalues are the entries along the diagonal for this upper triangular matrix. \begin{align*} \lambda_1 = 3 \qquad \lambda_2 = 2 \qquad \lambda_3 = 5 \end{align*} When $\lambda_1 = 3$ we have: $$A - 3I = \begin{bmatrix} 3-3 & 1 & 4 \\ 0 & 2-3 & 6 \\ 0 & 0 & 5-3 \\ \end{bmatrix} = \begin{bmatrix} 0 & 1 & 4 \\ 0 & -1 & 6 \\ 0 & 0 & 2 \\ \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}$$ \begin{align*} x_1 = 1 \: (free \: variable) \qquad x_2 = 0 \qquad x_3 = 0 \\ \end{align*} $$v_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \\ \end{bmatrix} \qquad (matches \: answer \: in \: text)$$ When $\lambda_2 = 2$ we have: $$A - 2I = \begin{bmatrix} 3-2 & 1 & 4 \\ 0 & 2-2 & 6 \\ 0 & 0 & 5-2 \\ \end{bmatrix} = \begin{bmatrix} 1 & 1 & 4 \\ 0 & 0 & 6 \\ 0 & 0 & 3 \\ \end{bmatrix} = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}$$ \begin{align*} x_1 = -x_2 \qquad x_2 = 1 \: (free \: variable) \qquad x_3 = 0 \\ \end{align*} $$v_2 = \begin{bmatrix} -1 \\ 1 \\ 0 \\ \end{bmatrix} \qquad but \: answer \: in \: text \: is \qquad \begin{bmatrix} 1 \\ -1 \\ 0 \\ \end{bmatrix}$$ What happened? Is it from a disagreement in what we should consider arbitrary or am I doing something fundamentally wrong?
When $\lambda_3 = 5$ we have: $$A - 5I = \begin{bmatrix} 3-5 & 1 & 4 \\ 0 & 2-5 & 6 \\ 0 & 0 & 5-5 \\ \end{bmatrix} = \begin{bmatrix} 2 & 1 & 4 \\ 0 & -3 & 6 \\ 0 & 0 & 0 \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 & -3 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \\ \end{bmatrix}$$ \begin{align*} x_1 = 3x_3 \qquad x_2 = 2x_3 \qquad x_3 = 1 \: (free \: variable) \\ \end{align*} $$v_3 = \begin{bmatrix} 3 \\ 2 \\ 1 \\ \end{bmatrix} \qquad (matches \: answer \: in \: text)$$
• Both your and the text book's answer work. If $(\lambda ,v)$ is an eigenpair of a matrix $M$, then so is $(\lambda, \mu v)$, for all scalars $\mu$. – Git Gud Jan 23 '16 at 23:15
Eigenvectors are never unique. In particular, for the eigenvalue $2$ you can take, for example, $x_2=-1$ which gives you the answer in the book.
• Precisely, and in fact you could take for $x_2$ any nonzero real number. – John B Jan 23 '16 at 23:26 | 2019-07-21T02:10:13 | {
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https://genenetto.com/china-cabinet-nrl/pythagorean-theorem-distance-formula-hw-answers-80f63d | Sketch a right triangle with the segment as the hypotenuse. Remember that this formula only applies to right triangles. The Distance Formula Date_____ Period____ Find the distance between each pair of points. Apply the formula and you should come up with the answer: You need to throw the ball 127.3 feet to get it from first base to third base. Distance Formula and Pythagorean Theorem. VIDEO. WORD ANSWER KEY. Students practice using the Pythagorean Theorem and Distance Formula. Simplifying Radicals. I can use the Pythagorean Theorem to find distance in a coordinate plane. The distance formula is a variant of the Pythagorean theorem. DISTANCE FORMULA AND PYTHAGOREAN THEOREM.. Slide 18 / 31 16 A designer wants to create an apartment in the ... answer to the nearest tenth. The distance between any two points. (1, -4) (5,6) (-2,3) Please explain to me how you do it. Share practice link. 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Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem contains 8th Standard Go Math solutions which will make students understand the concepts easily help the students to score well in the exams.This Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem. C. The Pythagorean theorem can only be used with right triangles. Unformatted text preview: Due Thursday 4/23 Name_____ ©r B2h0r2i0I KKuwtGai ASSorf[tIwAa]rYeH DLxLGCC.J x ^Alhli lrmiaggh_tKsc orTeHsje\rLveeWdo.The Distance Formula, Pythagorean Theorem, and Equations Find the distance between each pair of points. Some other items to compare: distance versus time relationships, population growth, growth functions for diseases, growth of minimum wage, auto insurance tables, etc. 1. Email your homework to your ... (6, 2). Pythagorean Theorem/Distance Formula DRAFT. Solo Practice. ... determine what percent of the students typically spent 3hours or more doing homework ... practice B for we don't give out the answers to homework or practice. you get C. hope this helped Question 7 (request help) b. Identify the legs and the hypotenuse of … HW5-3_sec1h_pythagorean_theorem.pdf: File Size: 48 kb: File Type: pdf: Download File. Homework Answers. MAC 1105 Pre-Class Assignment: Pythagorean Theorem and Distance formula Read section 2.8 Distance and Midpoint Formulas; Circles' and 4.5 "Exponential Growth and Decay, Modeling Dato' to prepare for class In this week's pre-requisite module, we covered the topics completing the square, evaluating radicals and percent increase. Each answer is listed on all of the MathO cards, ensuring students can check their work while solving the problems. (hypotenuse)2 5 (leg)2 1 (leg)2(AB)2 5 32 1 42 Substitute. 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Many proofs of the missing side length vertices of a right triangle to the nearest tenth, if necessary,! Side length of the MathO cards, ensuring students can check their work! to! We are finding the distance between two points - Displaying top 8 found... World Link: 1. right triangle with the segment as the hypotenuse [ longest side of legs! The hypotenuse of 7 will reach up a 25 foot tall house when placed 10 away. ( leg ) 2 5 ( leg ) 2 1 5 3 form a triangle! Triangles as right, Obtuse, or the Pythagorean Theorem Assignment Pythagorean Theorem quiz answers to their. Between two points in a coordinate context or the Pythagorean distance formula and hypotenuse... With one side 4 and the hypotenuse of … the Pythagorean Theorem See if the measurements below can form right! | 2021-04-16T03:20:19 | {
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http://math.stackexchange.com/questions/677928/factorization-of-a-degree-three-polynomial | # Factorization of a degree three polynomial
So I was doing some Vector Calculus homework and was working with Lagrange Multipliers, but then I came across a polynomial that I either forgot how to factor or never learned. I plugged it into Wolfram and was able to get the real solution, and finished the problem, but I'd like to know if there is a simple way to factor this out and if so what is it?
$3y^3 +y -4 = 0$
I know the real solution to this is $y=1$, but could someone give me a step by step process to receive this answer in the simplest way they know?
Edit: I'm sorry if I didn't make this clear. I Actually didn't know before hand $y=1$ this came after I looked for the solution on wolfram alpha. What I am looking for is actually how to factor that polynomial with no given information of the roots.
Thank You,
Valentino
-
Well, since you are in a class (i.e., this is not a polynomial from the wild), if you are expected to find the root of a polynomial of degree 3 or more, it is quite likely that you are expected to start by trying small integer solutions. Zero is clearly out, so the next thing to try is $y=1$. If that hadn't worked, I'd try $y=-1$, and probably $y=2$ and $y=-2$, and then I'd start thinking that I made an error in getting the polynomial. – Matthew Conroy Feb 16 at 1:07
Do you want to now know how to find the root $\,y=1,\,$ or how to continue once that root is know to find the other roots, or do you want to know how to do both? – Bill Dubuque Feb 16 at 1:07
I believe the rational root theorem is what you're looking for. It states that the only possible rational roots can be written as a fraction whose numerator is a factor of -4 and whose denominator is a factor of 3. Therefore, you can search for a root among $\pm1,\pm2,\pm4$ (and then if those fail, $\pm4/3$, $\pm2/3$, $\pm1/3$). If you find one root, then as Andrew said, you can divide and be left with a quadratic.
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Yes I believe you are correct. Man I can't believe I forgot about that haha. Thanks though man. I would rate the answer, but I have no reputation – Valentino Feb 16 at 3:35
If you know that $y = 1$ is a solution, then you can use polynomial division to divide $3y^3+y-4$ by $y-1$. This yields $3y^3+y-4 = (y-1)(3y^2+3y+4)$. From here you can use the quadratic formula to get the remaining complex roots.
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I did not know that y=1 until I checked Wolfram Alpha. So I would not have been able to do this method. – Valentino Feb 16 at 2:41
@Valentino Read Mark S. link, this is a really common method – chubakueno Feb 16 at 3:15
Possible roots: $$\pm[1, 2, 4]$$ Plug in each possible root to see if it is an actual root. It is a root if the equation evaluates to 0.
For $y=1$: $$3(1)^3+1-4=0$$ $$3+1-4=0$$ $$0=0$$ $y=1$ is a root of $3y^3+y-4=0$. This means that $(y-1)$ is a factor. Using synthetic division (hopefully you know how to do this; if you don't look it up), we know that the other factor is $(3y^2+3y+4)$. It takes about three seconds of thinking to realize that $(3y^2+3y+4)$ does not have real roots, because its discriminant is less than $0$. Let's use the quadratic formula to find the two complex roots. $$3y^2 + 3y + 4$$ $$y=\dfrac{-3\pm\sqrt{3^2-4(3)(4)}}{2(3)}$$ $$y=\dfrac{-3\pm\sqrt{9-48}}{6}$$ $$y=\dfrac{-3\pm\sqrt{-39}}{6}$$ $$y=\dfrac{-3\pm i\sqrt{39}}{6}$$ The full factoring is: $$(y-1)\left(y+\dfrac{3+i\sqrt{39}}{6}\right)\left(y-\dfrac{i\sqrt{39}-3}{6}\right)$$ I hope I did the factoring right D:
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I like Matthew Conroy's comment above; reminding oneself of the pragmatic context can be either reassuring or maddening (when the answer doesn't come readily enough).
In any case, there's a general solution to a cubic equation, due to Cardano, I believe: Take any monic polynomial of degree three, $x^3 + a_2 x^2 + a_1 x + a_0$, and let $x = z - a_2 / 3$. (Monic just means the leading coefficient is $1$.) Substitute this into your original polynomial; it will simplify to something of the form $z^3 + bz + c$, which is called a "depressed" cubic equation. Then there exists a root of the form
$$z = \sqrt[3]{-\frac{c}{2}+\sqrt{\frac{c^2}{4}+\frac{b^3}{27}}}+\sqrt[3]{-\frac{c}{2}-\sqrt{\frac{c^2}{4}+\frac{b^3}{27}}}$$
Needless to say, this will come out messy most of the time; therefore, you should probably start with the rational roots, as an earlier answer recommended. But this will give you an answer.
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When the coefficients add to zero +1 will be a root. If the sum of the coefficients of the even power terms minus the sum of the coefficients of the odd power terms then -1 will be a root.
- | 2014-07-28T16:59:19 | {
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https://math.stackexchange.com/questions/3492372/show-that-the-number-of-1s-in-all-the-partitions-equals-the-sum-of-number-of-d | # Show that the number of $1$s in all the partitions equals the sum of number of distinct elements in each partition
Consider the number $$n$$ a partition $$P$$ of $$n$$. Denote by $$f_n(P)$$ the number of $$1$$s in $$P$$ and by $$g_n(P)$$ the number of distinct elements in $$P$$. Show that $$\displaystyle\sum_{P} f_n(P) = \displaystyle\sum_{P} g_n(P)$$. Note that a partition is a non-decreasing sequence of integers that add upto $$n$$.
Here is my take on the problem :
Denote by $$p(n)$$ the number of partitions of $$n$$. Now any partition of $$n+1$$ such that it has a $$1$$ is basically $$1$$ $$+$$ some partition of $$n$$, which gives -
$$\displaystyle\sum_P f_{n+1}(P) = p(n) + \displaystyle\sum_P f_{n}(P)$$
Similarly just add $$1$$ to the largest element of every partition of $$n$$ to get $$n+1$$. Consider the partitions of $$n+1$$ into two categories - one having the largest integer appear only once in the partition and another having the largest integer repeat. In the second category, reducing the last number (by $$1$$) either keeps the number of distinct elements same or decreases by $$1$$, which isn't desired, while in the first category, reducing the largest element by $$1$$ yields a partition of $$n$$ that has exactly one less number of distinct elements. We basically generate
$$\displaystyle\sum_P g_{n+1} (P) = p(n) + \displaystyle\sum_P g_{n} (P)$$, as desired.
Is there a way to solve this without induction? Elegant methods requiring generating functions are fine too, though I'd appreciate any solution that relies on construction and not recurrences/GFs.
• I'm unable to follow the second part of your proof. It seems flawed to me, but perhaps I just don't understand it. – joriki Dec 30 '19 at 18:36
• if $[a_1, a_2, \cdots, a_k]$ is a partition of $n$ (with $a_i \leq a_j$ for $i\leq j$), just make it $[a_1, a_2, \cdots, a_{k-1}, a_k +1]$ to get a partition of $n+1$ having one more distinct element. Do this for every partition of $n$. I'm sorry, perhaps the explanation isn't very clear. – charlesh Dec 30 '19 at 20:27
• Why does it have one more distinct element? $a_k$ could already have been distinct? – joriki Dec 30 '19 at 21:05
Let $$a_n$$ be the number of 1s occurring in all partitions of $$n$$, let $$b_n$$ be the number of distinct parts summed over all partitions of $$n$$, and let $$p_n$$ be the number of partitions of $$n$$. We show that $$a_n = \sum_{k=0}^{n-1} p_k = b_n$$.
First, we have \begin{align*} a_n &= \sum_{p \vdash n} \sum_{1 \in p} 1 = \sum_{\substack{p \vdash n \\ p \ni 1}} \sum_{1 \in p} 1 = \sum_{\substack{p \vdash n \\ p \ni 1}} \left(1 + \sum_{1 \in p-1} 1 \right) \\ &= \sum_{q \vdash n-1} \left(1 + \sum_{1 \in q} 1 \right) = \sum_{q \vdash n-1} 1 + \sum_{q \vdash n-1} \sum_{1 \in q} 1 = p_{n-1} + a_{n-1}, \end{align*} which immediately implies that $$a_n = \sum_{k=0}^{n-1} p_k$$. Now $$b_n = \sum_{p \vdash n} \sum_{\substack{\text{distinct} \\k \in p}} 1 %= \sum_{p \vdash n} \sum_{\substack{k \ge 1 \\ k \in p}} 1 = \sum_{k=1}^n \sum_{\substack{p \vdash n \\ p \ni k}} 1 = \sum_{k=1}^n p_{n-k} = \sum_{k=0}^{n-1} p_k,$$ as desired.
A constructive proof
We will consider two different one-to-many mappings between the partitions of a number $$N$$ and the partitions of numbers smaller than $$N$$. It is convenient to adopt the convention that $$P(0)=1.$$
FIRST METHOD
Let the given partition of $$N$$ contain $$k$$ copies of $$1$$. Map this partition onto the $$k$$ partitions obtained by deleting one $$1$$, two $$1$$s, ...
E.G. $$9=1+1+2+5$$ maps to $$1+2+5$$ and $$2+5$$.
SECOND METHOD
Let the given partition of $$N$$ contain $$l$$ distinct numbers. Map this partition onto the $$l$$ partitions obtained by deleting one of each distinct number in turn.
E.G. $$9=1+1+2+5$$ maps to $$1+2+5$$,$$1+1+5$$ and $$1+1+2$$.
Each partition of a number smaller than $$N$$ (including $$0$$) is mapped onto by precisely one partition of $$N$$ for each of these methods and so $$\sum l=\sum k$$.
The generating function $$f(x,y)=\sum_{n,k}f_{nk}x^ny^k$$ that counts the partitions of $$n$$ with $$k$$ parts $$1$$ is
$$\begin{eqnarray} f(x,y) &=&\sum_{j=0}^\infty(xy)^j\prod_{m=2}^\infty\sum_{j=0}^\infty x^{jm} \\ &=& \frac1{1-xy}\prod_{m=2}^\infty\frac1{1-x^m}\;. \end{eqnarray}$$
The generating function $$g(x,y)=\sum_{n,k}g_{nk}x^ny^k$$ that counts the partitions of $$n$$ with $$k$$ distinct parts is
$$\begin{eqnarray} g(x,y) &=& \prod_{m=1}^\infty\left(1+y\sum_{j=1}^\infty x^{jm}\right) \\ &=& \prod_{m=1}^\infty\frac{1-x^m(1-y)}{1-x^m}\;. \end{eqnarray}$$
We can extract the desired sums from these generating functions:
$$\begin{eqnarray} \sum_Pf_n(P) &=& \left[x^n\right]\left.\frac\partial{\partial y}f(x,y)\right|_{y=1} \\ &=& \left[x^n\right]\left.\frac x{(1-xy)^2}\prod_{m=2}^\infty\frac1{1-x^m}\right|_{y=1} \\ &=& \left[x^n\right]\frac x{1-x}\prod_{m=1}^\infty\frac1{1-x^m} \end{eqnarray}$$
and likewise
$$\begin{eqnarray} \sum_Pg_n(P) &=& \left[x^n\right]\left.\frac\partial{\partial y}g(x,y)\right|_{y=1} \\ &=& \left[x^n\right]\left.\sum_{m=1}^\infty\frac{x^m}{1-x^m}\prod_{m'\ne m}\frac{1-x^m(1-y)}{1-x^m}\right|_{y=1} \\ &=& \left[x^n\right]\left(\sum_{m=1}^\infty x^m\right)\left(\prod_{m=1}^\infty\frac1{1-x^m}\right) \\ &=& \left[x^n\right]\frac x{1-x}\prod_{m=1}^\infty\frac1{1-x^m}\;. \end{eqnarray}$$
Sorry about the generating functions ;-)
• Could you please tell me what you did by $[x^n]$? – charlesh Dec 30 '19 at 20:32
• @charlesh: $\left[x^n\right]$ is meant to denote the extraction of the coefficient of $x^n$. Since $y=1$ is substituted, the resulting function is a function of $x$ only, and $\left[x^n\right]$ extracts the coefficient of $x^n$ in that function. – joriki Dec 30 '19 at 20:37 | 2020-09-28T10:21:47 | {
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https://math.stackexchange.com/questions/392309/solve-the-roots-of-a-cubic-polynomial | # Solve the roots of a cubic polynomial?
I have had trouble with this question - mainly due to the fact that I do not fully understand what a 'geometric progression' is:
"Solve the equation $x^3 - 14x^2 + 56x - 64 = 0$" if the roots are in geometric progression.
Any help would be appreciated.
It means the roots are of the form $a, ar, ar^2$.
Here it is not too difficult to see that $2, 4, 8$ are ok, just look at the constant term, which is $- (a \cdot ar \cdot ar^2) = - (a r)^3$, and check that $2, 4, 8$ fit with the other coefficients $$14 = 2 + 4 + 8, \qquad 56 = 2 \cdot 4 + 2 \cdot 8 + 4 \cdot 8.$$
• Thanks, but are there different types of geometric progessions? – missiledragon May 15 '13 at 10:21
• @missiledragon, not that I am aware of, it means a sequence of the form $a_{n} = a r^{n}$ for some $a, r$, see the Wikipedia link. – Andreas Caranti May 15 '13 at 10:23
• Thanks, if you don't mind, could you briefly explain what an arithmetic progression is? I had a similar question with it. – missiledragon May 15 '13 at 10:25
• @missiledragon, the Wikipedia is your friend en.wikipedia.org/wiki/Arithmetic_progression it means a sequence of the form $a_{n} = a + n r$, for fixed $a, r$, and $n \ge 0$. – Andreas Caranti May 15 '13 at 10:27
• Alright, thanks for your help – missiledragon May 15 '13 at 10:32
Let the roots be $a,a\cdot r,a\cdot r^2$
Using Vieta's formula $a+a\cdot r+a\cdot r^2=14\implies a(1+r+r^2)=14$
and $a(a\cdot r)+a\cdot r(a\cdot r^2)+a(a\cdot r^2)=56\implies a^2\cdot r(1+r+r^2)=56$
On division, $ar=4$ as $a\cdot r\ne0$
$$\implies a=\frac 4r\implies \frac{4(1+r+r^2)}r=14\implies 2r^2-5r+2=0\implies r=2\text{ or }\frac12$$
Alternatively, using Vieta's formula $a\cdot(a\cdot r) \cdot (a\cdot r^2)=64\implies (ar)^3=64$
So, $a\cdot r$ can be one of $4,4w,4w^2$ where $w$ is the cube root of $1$
Using Polynomial Remainder Theorem, observe that $4$ is a root of the given equation
$\implies a\cdot r=4\iff a=\frac4r$
Again, using Vieta's formula $a+a\cdot r+a\cdot r^2=14\implies 2r^2-5r+2=0$ | 2021-01-27T23:24:07 | {
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https://mathematica.stackexchange.com/questions/271511/minimizing-the-combined-neighbourhood-of-a-path-in-a-grid-graph | # Minimizing the combined neighbourhood of a path in a grid graph
Given two vertices in a simple grid graph, how to find the path connecting them such that the combined neighbourhood of the vertices in the path is the smallest?
Consider this path between the vertices $$1$$ and $$18$$ in a $$5\times5$$ graph,
The neighbourhood graph of all the vertices in the path would give,
The combined neighbourhood has a total of $$13$$ vertices.
For another path,
,
The combined neighbourhood would be,,
with a total of $$14$$ vertices.
I want to find a path that minimizes the number of vertices in the combined neighbourhood.
• +1 for the clear question. We can always assume the starting point has coordinates $(0,0)$. Up to reflection symmetries and so on, we can also assume that the target point has coordinates $(m,n)$ for some $0 \leq n \leq m$. (Your first picture would be $m=3$ and $n=2$.) Suppose I first move to $(n,n)$ using right-up-right-...-up, and then move right-right-...-right to $(m,n)$. Do you think that is optimal? If yes, and you can prove it, then done. If no, can you give an example where that is not optimal? Jul 31 at 9:10
• Counter Example; Path from 1 to 22 in the 5x5 grid. Path you defined would be (1,6,7,12,17,22) with combined neighbourhood 14. A more optimal path would be (1,6,11,16.21,22) with combined neighbourhood 11. Jul 31 at 9:28
• Ok good, so boundary plays a role. Is your graph always a square or rectangle? Is the starting point always a corner point, or can start and stop be anywhere? Jul 31 at 9:32
• It can be square or rectangle; It can start and stop anywhere Jul 31 at 9:45
I think this question of OP might be more appropriate for a site with more graph theory people. Anyhow, the following seems to be a reasonable strategy: Given the graph G, try to construct an auxiliary graph auxG, with directed edges and weights, so that the available function FindShortestPath on auxG solves the problem that OP is asking on G. This will be useful if auxG can be constructed reasonably quickly and is of a similar size as G.
To see what I mean: If one moves right-right-right-right-... then one extends the combined neighborhood by 3 with each hop. If one moves right-up-right-up-... then on average one extends the not by 3+3 for each right-up-combination, but only by 5. So we could give weight 3 to each existing edge, and add diagonal edges with weight 5. In this case, auxG would have the same vertices as G, but more edges.
Because of the boundary, which is important in this problem, we have to do something similar near the boundary. I will not describe this in words and just drop the code:
(* original grid graph *)
n=10;
G=GridGraph[{n,n},VertexLabels->Automatic];
coords=Round[VertexCoordinates/.AbsoluteOptions[G,VertexCoordinates]];
(* neighborhood computations *)
nbh0[vertex_,dist_:1]:=VertexList[NeighborhoodGraph[G,vertex,dist]];
nbh0[vertex_,dist_,distmax_]:=Select[nbh0[vertex,dist],(Norm[coords[[vertex]]-coords[[#]],Infinity]<=distmax)&];
nbh[vertex_,x___]:=DeleteCases[nbh0[vertex,x],vertex];
sizenbh[vertices_]:=Length[DeleteDuplicates[Flatten[Map[nbh0,vertices]]]];
(* construct auxG *)
aux[from_,to_]:=sizenbh[FindShortestPath[G,from,to]]-sizenbh[{from}];
aux[from_]:=Map[{DirectedEdge[from,#],aux[from,#]}&,nbh[from,2,1]];
{auxedges,auxweights}=Transpose[Flatten[Map[aux,VertexList[G]],1]];
auxG=Graph[auxedges,EdgeWeight->auxweights,VertexLabels->Automatic];
(* find path *)
findPath[from_,to_]:=Flatten[Map[FindShortestPath[G,#[[1]],#[[2]]]&,
Partition[FindShortestPath[auxG,from,to],2,1]]]//.{a___,b_,b_,c___}:>{a,b,c};
I stress that the construction of auxG is currently ad-hoc, I have not verified that it really generates the paths that OP wants.
Here is G:
Here is auxG. It has directed edges with weights. Away from the boundary, the weights are $$3$$ for horizontal or vertical hop, and $$5$$ for diagonal hop. Near the boundary it is a bit more complicated:
Examples: Corner to opposite corner:
p1 = findPath[1,100]
(* {1,2,3,4,5,6,7,8,9,10,20,30,40,50,60,70,80,90,100} *)
Let us compare with the diagonal path:
p2 = {1,11,12,22,23,33,34,44,45,55,56,66,67,77,78,88,89,99,100};
They have equal neighborhood size:
Map[sizenbh,{p1,p2}]
(* {36,36} *)
The first example of OP is essentially
p3 = findPath[1,33]
(* {1,11,12,22,23,33} *)
which is the path OP has given. Finally, consider two paths from the left boundary to the right boundary:
p4=findPath[3,93]
(* {3,2,1,11,21,31,41,51,61,71,81,91,92,93} *)
p5=findPath[4,94]
(* {4,14,24,34,44,54,64,74,84,94} *)
Note that p4 walks along the boundary, p5 does not.
Map[sizenbh,{p4,p5}]
(* {26,30} *)
This shows that p4 is indeed better than walking horizontally, which would have neighborhood size 30. And p5 is as good as walking along the boundary, which also has neighborhood size 30.
Again, I do not claim that this always produces the paths that OP wants, but it seems to go in the right direction. It is based on calling FindShortestPath once (on auxG) so should be reasonably efficient.
• This is very interesting and works really fast for large graphs. I've been looking for some counter examples but haven't been able to find any. Jul 31 at 15:51
• There are plenty counterexamples. For 5x5 grid and {start,stop}=={2, 24} it returns {2, 3, 4, 5, 10, 15, 20, 19, 24} which has total neighborhood 17, but there are other six paths with total neighborhood 16, for example: {2, 3, 8, 9, 14, 15, 20, 25, 24}. Jul 31 at 17:46
• @azerbajdzan Please Quit[] and try again. For n=5, with findPath[2,24] I get {2,3,4,5,10,15,20,25,24} with size $16$. But perhaps you find another counterexample (I am quite sure there is one). Jul 31 at 18:03
• And this is definitely counterexample: For n=8 findPath[1, 47] returns {1, 2, 3, 4, 5, 6, 7, 8, 16, 24, 32, 40, 39, 47} which has total neighborhood 27, but there are two paths with 25: {{1,2,10,11,19,20,28,29,37,38,46,47}, {1,2,3,11,12,20,21,29,30,38,39,47}}. Jul 31 at 18:23
• The last example I gave does not work even if I use fresh kernel. Jul 31 at 18:26
Minimal combined neighborhood is given by cn, in this case it is 15 for three different paths.
g = GridGraph[{5, 5}, VertexLabels -> Automatic];
{start, stop} = {7, 20};
paths = FindPath[g, start, stop, All, All];
len = Length /@ (AdjacencyList[g, #] & /@ paths);
cn = Min[len]
bestpaths = paths[[Flatten[Position[len, cn]]]];
HighlightGraph[g, PathGraph[#]] & /@ bestpaths
Clear[g, paths, len, bestpaths, start, stop, cn]
(* 15 *)
For {start,stop}={2,22}:
(* 12 *)
• Take {start,stop}={2,22}. Your code produces {2,7,12,17,22}, but {2,1,6,11,16,21,22} is better. Jul 31 at 10:56
• This code only considers paths with length equal to that of the shortest path. In general it wont hold Jul 31 at 11:02
• @user293787: I modified my code, now it works correctly, but it is more or less a brute-force. Jul 31 at 11:03
• Hmmm...Its not even showing an output for 6x6 grid Jul 31 at 11:10
• Limiting the length of the paths at 2n where n is the grid length gives good results Jul 31 at 11:26 | 2022-10-05T22:56:23 | {
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https://math.stackexchange.com/questions/3760642/small-angle-approximation-of-frac-sin2-xx2-sqrt1-frac-sin2-x3 | # Small-angle approximation of $\frac{\sin^2 x}{x^2 \sqrt{1-\frac{\sin^2 x}{3}}}$
I need to show the following:
$$\frac{\sin^2 x}{x^2 \sqrt{1-\frac{\sin^2 x}{3}}} \approx 1-\frac{x^2}{6}$$ when $$x$$ is small.
I think this problem is trickier than most other questions like it because in the original source there is comment saying "if you got $$1+\frac{x^2}{6}$$ [what I got] then think again!". My attempt was:
When $$x$$ is small, $$\sin x \approx x$$ so
$$\frac{\sin^2{x}}{x^2 \sqrt{1-\frac{\sin^2 x}{3}}} = \frac{x^2}{x^2 \sqrt{1-\frac{x^2}{3}}} = \left ( 1 - \frac{x^2}{3} \right )^{-\frac{1}{2}}$$
Then using the binomial series approximation,
$$\left ( 1 - \frac{x^2}{3} \right )^{-\frac{1}{2}} \approx 1 - \frac{1}{2}\left ( -\frac{x^2}{3} \right ) + ... = 1 + \frac{x^2}{6}$$
...and so it looks like I've fallen into whatever trap the question set.
Where is my error?
$${\sin x\over x}\approx1-{1\over6}x^2$$
so
$${\sin^2x\over x^2}\approx\left(1-{1\over6}x^2\right)^2\approx1-{1\over3}x^2$$
not just $$1$$. We get
$${\sin^2x\over x^2\sqrt{1-{\sin^2x\over3}}}\approx\left(1-{1\over3}x^2\right)\left(1+{1\over6}x^2\right)\approx1-{1\over6}x^2$$
• Is there any way to know when I should use more than one term to approximate sin(x) (like you did here)? – Nick_2440 Jul 17 at 21:04
• @Nick_2440, it might be a good idea to carry asymptotics along explicitly in the form of $\sin x=x-{1\over6}x^3+O(x^5)$, etc. That way if you try $\sin x=x+O(x^3)$ you quickly see that ${\sin x\over x}=1+O(x^2)$ isn't good enough to give you an answer of the form $1-{1\over6}x^2+O(x^3)$. In fact it might be a good exercise to write out your own answer here doing that. – Barry Cipra Jul 17 at 21:21
Suppose for simplicity you had two polynomials $$P(z) = 1+ z + z^2 + 4z^4 + 7z^5$$ and $$Q(z) = 1 + 2z + 3z^2 + 4z^3$$, and I asked you to calculate the product $$P(z)Q(z)$$... but not the entire thing. Suppose I only want the terms up to quadratic term; i.e if we write \begin{align} P(z)Q(z) &= a_0 + a_1z + a_2z^2 + a_3z^3 + a_4z^4 + a_5z^5 +a_6z^6 + a_7z^7 +a_8z^8 \end{align} then I'm asking you to find the coefficients $$a_0,a_1,a_2$$ (but for now, let's just say for some reason I'm interested in what happens when $$|z|$$ is very small up to an accuracy of quadratic order, so I don't really care about the rest of the terms ). Well, we just multiply everything out: \begin{align} P(z)Q(z) &= (1+ z + z^2 + 4z^4 + 7z^5)(1 + 2z + 3z^2 + 4z^3) \\ &= 1 + (1\cdot 2z + z \cdot 1) + (1\cdot 3z^2 + z\cdot 2z + z^2 \cdot 1) \\ &+ \text{(terms involving z^3 or higher, which I don't care about for now)} \\ &= 1 + 3z + 6z^2 + O(z^3) \end{align} In other words, because in my final product, I'm only interested in calculating up to the quadratic term, I can simply truncate the polynomials $$P$$ and $$Q$$ to quadratic order, and then multiply them out (and then again only keep terms up to quadratic order): \begin{align} P(z)Q(z) &= (1 + z + z^2 + \cdots)(1 + 2z + 3z^2 + \cdots) \\ &= 1 + 3z + 6z^2 + O(z^3) \end{align}
Again, because I'm only interested up to quadratic order, there's no need for me to keep any terms beyond that for $$P(z)$$ and $$Q(z)$$, because if I approximate $$P(z) \approx 1+ z + z^2 + \color{red}{4z^4}$$ (i.e I keep the $$4^{th}$$ order term) and I multiply with $$Q(z) = 1+2z+3z^2 + 4z^3$$, then the red term multiplied with anything in $$Q(z)$$ will yield terms which are $$4^{th}$$ order or higher (which I don't care about).
But what you should not do is truncate $$P(z)$$ and $$Q(z)$$ up to linear order, and say that \begin{align} P(z)Q(z) & \approx (1+z)(1+2z) = 1 + 3z + 2z^2 \end{align} Because in this way, you're missing out other second order contributions (by multiplying constant term of $$P$$ with quadratic term of $$Q$$ and vice-versa).
This is how you know how many terms you need to use in your approximation. In your case, you want to approximate \begin{align} f(x) &= \dfrac{\sin^2x}{x^2\sqrt{1 - \frac{\sin^2x}{3}}} \end{align} up to $$2^{nd}$$ order. So, what do we do? We write things as a product first: \begin{align} f(x) &= \left(\dfrac{\sin x}{x}\right)\cdot\left(\dfrac{\sin x}{x}\right) \cdot \left(\dfrac{1}{\sqrt{1- \frac{\sin^2x}{3}}}\right)\tag{1} \end{align} Now, we have to expand each bracketed term up to atleast second order in $$x$$ and then multiply the result together. First: \begin{align} \dfrac{\sin x}{x} &= \dfrac{x - \dfrac{x^3}{6} + O(x^4)}{x} = 1 - \dfrac{x^2}{6} + O(x^3) \tag{2} \end{align} Next, we recall that \begin{align} \dfrac{1}{\sqrt{1-z}} &= 1+ \dfrac{z}{2} + \dfrac{3z^2}{8} + O(z^3) \end{align} Now, plug in $$z= \frac{\sin^2x}{3} = x + O(x^3)$$, to get \begin{align} \dfrac{1}{\sqrt{1-\frac{\sin^2x}{3}}} &= 1+ \dfrac{1}{2}\left(\dfrac{\sin^2x}{3}\right) + \dfrac{3}{8}\left(\frac{\sin^2x}{3}\right)^2 + O((\sin^2 x)^3) \\ &= 1 + \dfrac{1}{2}\left(\dfrac{x^2 + O(x^4)}{3}\right) + O(x^4) + O(x^6) \\ &= 1 + \dfrac{1}{6}x^2 + O(x^4)\tag{3}, \end{align} where in the second line, hopefully it's clear how I got the various terms: for example $$\sin x = x + O(x^3)$$, so $$\left(\frac{\sin^2x}{3}\right)^2 = \frac{1}{9}\sin^4x = \frac{1}{9} (x + O(x^3))^4 = O(x^4)$$. Therefore, the final answer is obtained by plugging $$(2)$$ and $$(3)$$ into $$(1)$$ : \begin{align} f(x) &= \left(1 - \dfrac{x^2}{6} + O(x^3)\right)^2 \cdot \left(1 + \dfrac{1}{6}x^2 + O(x^4)\right) \\ &= 1 - \dfrac{x^2}{6} + O(x^4) \end{align}
Long story short, if your end goal is to calculate up to second order, then at each stage of your algebra make sure you're keeping terms atleast up to $$x^2$$.
• +1 for a fantastic answer. While experienced users may not really want to read the long story, it's a boon for beginners. – Paramanand Singh Jul 18 at 0:49
• I wish I could choose 2 answers as accepted...this was very well explained, thanks :) – Nick_2440 Jul 18 at 11:19
You can also note the the singularity at $$x=0$$ is removable and that $$f$$ is in fact at least 4 times differentiable. Taylor's expansion gives you the answer...
$$f(x)=f(0)+f'(0)x + \frac 12 f''(0) x^2 + O(x^3)$$
where
$$f(0)=\lim_{x\to 0}f(0) = 1, \quad f'(0) = \lim_{x\to 0}f'(x)=0, \quad f''(0)=\lim_{x\to 0}f''(x)= -\frac 13, \quad f'''(0)=0$$
yielding
$$f(x)=1-\frac 16 x^2 + O(x^4)\approx 1-\frac 16 x^2 (\textrm{for small } x ).$$
• Typo in last line. – spalein Jul 17 at 21:47
• @spalein thanks. – PierreCarre Jul 18 at 10:35 | 2020-11-30T23:52:06 | {
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Fundamentals 17 2. Smith A copy that has been read, but remains in clean condition. The new edition includes revised and greatly expanded sections on stability based on the Lax-Richtmeyer definition, the application of Pade approximants to. Finite Difference Method for Heat Equation Simple method to derive and implement Hardest part for implicit schemes is solution of resulting linear system of equations Explicit schemes typically have stability restrictions or can always be unstable Convergence rates tend not to be great – to get an. (1979) Explicit Solutions of Fisher's Equation for a Special Wave Speed. 20, Corporate Author : MARYLAND UNIV COLLEGE PARK DEPT OF MATHEMATICS. Discretization methods, including finite difference & finite-volume schemes, spectral. Numerical Methods for Partial Differential Equations I. This package performs automation of the process of numerically solving partial differential equations systems (PDES) by means of computer algebra. important application of finite differences is in numerical analysis, especially i n numerical differential equations, which aim at the numerical solution of ordinary and partial differential equations respectively. Introduction 10 1. The aim is to obtain a numerical solution within a prescribed tolerance using a minimal amount of work. Start your review of Numerical Solution of Partial Differential Equations: Finite Difference Methods Write a review Oct 15, 2015 Chand added it. pdf), Text File (. The solution of PDEs can be very challenging, depending on the type of equation, the number of independent variables, the boundary, and initial. Solution of linear systems by iterative methods and preconditioning. "Finite volume" refers to the small volume surrounding each node point on a mesh. The concepts of stability and convergence. 1 BACKGROUND OF STUDY. the accuracy of the numerical approximations depends on the truncation errors in the formulas used to convert the partial differential equation into a difference equation. Methods Partial Differ. Indo-German Winter Academy, 2009 20. The focuses are the stability and convergence theory. Numerical Partial Differential Equations: Finite Difference Methods. (1979) Explicit Solutions of Fisher's Equation for a Special Wave Speed. It can be used as a reference book for the PDElPROTRAN user* who wishes to know more about the methods employed by PDE/PROTRAN Edition 1 (or its predecessor, TWODEPEP) in solving two-dimensional partial differential equations. Introduction to Finite Differences. 35—dc22 2007061732. Learn to write programs to solve ordinary and partial differential equations. The course will concentrate on the key ideas underlying the derivation of numerical schemes and a study of their stability and accuracy. The aim is to obtain a numerical solution within a prescribed tolerance using a minimal amount of work. Morton and D. It is not the only option, alternatives include the finite volume and finite element methods, and also various mesh-free approaches. There are many forms of model hyperbolic partial differential equations that are used in analysing various finite difference methods. Includes bibliographical references and index. for the numerical solution of partial differential equations with mixed initial and boundary conditions specified. & Joulia, X. The results obtained for these numerical examples validate the ef-ficiency, expected order and accuracy of the method. These range from simple one-dependent variable first-order partial differential equations. Low Order Approximations; Difference Calculus; High Order Methods. Finite difference method (FDM) is the most practical method that is used in solving partial differential equations. * Inverse Problems. Numerical methods are needed to solve partial differential equations (PDEs). Let's set the hype and anti-hype of machine learning aside and discuss the opportunities it can provide to the field of metal casting. 1 The Finite Difference Method The heat equation can be solved using separation of variables. Laplace Equation in 2D. Numerical Methods for Partial Differential Equations: Finite Difference and Finite Volume Methods focuses on two popular deterministic methods for solving partial differential equations (PDEs), namely finite difference and finite volume methods. It is not the only option, alternatives include the finite volume and finite element methods, and also various mesh-free approaches. Published: (2005) Methods for the numerical solution of partial differential equations, by: Von Rosenberg, Dale U. LeVeque, SIAM, 2007. Thus, it has become customary to test new approaches in computational fluid dynamics by implementing novel and new approaches to Burger equation yielding in various finite-differences, finite volume, finite-element and boundary element methods etc. ! Show the implementation of numerical algorithms into actual computer codes. We present qualitative and numerical results on a partial differential equation (PDE) system which models a certain fluid-structure dynamics. Chapter 2 Introduction to Finite Differences Numerical Partial Differential Equations. Numerical Solution of Partial Differential Equations : Finite Difference Methods by Gordon D. The finite element method became a very widely used method in practice. We use finite differences with fixed-step discretization in space and time and show the relevance of the Courant-Friedrichs-Lewy stability criterion for some of these discretizations. Introductory Finite Difference Methods for PDEs Contents Contents Preface 9 1. FINITE ELEMENT METHODS FOR THE NUMERICAL SOLUTION OF PARTIAL DIFFERENTIAL EQUATIONS Vassilios A. Numerical Methods for Partial Differential Equations supports. The solution of PDEs can be very challenging, depending on the type of equation, the number of independent variables, the boundary, and initial conditions, and other factors. , Le Lann, J. Accuracy, temporal performance and stability comparisons of discretization methods for the numerical solution of Partial Differential Equations (PDEs) in. of numerical analysis, the numerical solution of partial differential equations, as it developed in Italy during the crucial incubation period immediately preceding the diffusion of electronic computers. * Numerical Partial Differential Equations. Numerical Methods for Partial Differential Equations is an international journal that aims to cover research into the development and analysis of new methods for the numerical solution of partial differential equations. ISBN: 0198596251 019859626X 9780198596257 9780198596264: OCLC Number: 3915448: Description: xii, 304 pages : illustrations ; 22 cm: Contents: 1. Frequently exact solutions to differential equations are unavailable and numerical methods become. The Finite Difference Method provides a numerical solution to this equation via the discretisation of its derivatives. Integrate initial conditions forward through time. The goal of this course is to provide numerical analysis background for finite difference methods for solving partial differential equations. This is an introduction to solution techniques for partial differential equations that has been developed from an introductory course on the subject for junior and senior-level mathematics or physical science majors with an undergraduate background in calculus, introductory linear algebra, and ordinary differential equations. ference schemes, and an overview of partial differential equations (PDEs). The heat equation is a simple test case for using numerical methods. A numerical is uniquely defined by three parameters: 1. Survey of PDEs; Hyperbolic Systems; Finite Difference Approximations. Solution of the Diffusion Equation by Finite Differences The basic idea of the finite differences method of solving PDEs is to replace spatial and time derivatives by suitable approximations, then to numerically solve the resulting difference equations. • Finite Element (FE) Method (C&C Ch. Here, the techniques of functional analysis and partial differential equations are applied to the classical problem of numerical integration, to establish many important and deep analytical properties of cubature formulas. Numerical approximations of autonomous SPDEs are thoroughly investigated in the literature, while the non-autonomous case is not yet understood. Finite Difference Method for Heat Equation Simple method to derive and implement Hardest part for implicit schemes is solution of resulting linear system of equations Explicit schemes typically have stability restrictions or can always be unstable Convergence rates tend not to be great – to get an. We present qualitative and numerical results on a partial differential equation (PDE) system which models a certain fluid-structure dynamics. •• Introduction to Finite Differences. Replace continuous problem domain by finite difference mesh or grid u(x,y) replaced by u i, j = u(x,y) u i+1, j+1 = u(x+h,y+k) Methods of obtaining Finite Difference Equations - Taylor. Finally the numerical solutions obtained by FDM, FEM and MCM are compared with exact solution to check the accuracy of the developed scheme Keywords - Dirichlet Conditions, Finite difference Method, Finite Element Method, Laplace Equation, Markov chain Method. Numerical Solution of Partial Differential Equations An Introduction K. Numerical Solution of Partial Differential Equations: Finite Difference Methods Paperback - April 30 1999 by G. Wellposedness is established by constructing for it a nonstandard semigroup generator representation; this representation is accomplished by an appropriate elimination of the pressure. In this work, we studied the matter of heat transfer by the natural convection for a dissipatable fluid which flows in a tube, its walls composed from porous material, and a mathe. Lecture 3 Numerical Methods - Free download as Powerpoint Presentation (. 4 Systems of ordinary differential equations 156. Numerical Solution of Partial Differential Equations - Finite Element Methods proficient in basic numerical methods, linear algebra, mathematically rigorous. In numerical analysis, finite-difference methods (FDM) are discretizations used for solving differential equations by approximating them with difference equations that finite differences approximate the derivatives. It is a comprehensive presentation of modern shock-capturing methods, including both finite volume and finite element methods, covering the theory of hyperbolic. The goal of this course is to introduce theoretical analysis of finite difference methods for solving partial differential equations. Differentiate arrays of any number of dimensions along any axis with any desired accuracy order; Accurate treatment of grid boundary; Includes standard operators from vector calculus like gradient, divergence. This is because many mathematical models of physical phenomena result in one or more. LeVeque, R. The aim is to obtain a numerical solution within a prescribed tolerance using a minimal amount of work. The study on numerical methods for solving partial differential equation will be of immense benefit to the entire mathematics department and other researchers that desire to carry out similar research on the above topic because the study will provide an explicit solution to partial differential equations using numerical methods. Personal Author(s) : Babuska,I ; Liu,T -P ; Osborn,J. So the first goal of this lecture note is to provide students a convenient textbook that addresses both physical and mathematical aspects of numerical methods for partial differential equations (PDEs). The convergence and stability analysis of the solution methods is also included. Parabolic equations ; 3. For example, if the derivatives are with respect to several different coordinates, they are called Partial Differential Equations (PDE), and if you do not know everything about the system at one point, but instead partial information about the solution at several different points they are called. This history is inextricably intertwined with that of modern mathematical analysis (in particular, functional analysis and the calculus. See all 8 formats and editions Hide other formats and editions. Smith Substantially revised, this authoritative study covers the standard finite difference methods of parabolic, hyperbolic, and elliptic equations, and includes the concomitant theoretical work on consistency, stability, and convergence. Numerical Solution of Partial Differential Equations: An Numerical Solution of Partial Differential Equations by the Finite. Buy Numerical Solution of Differential Equations : Introduction to Finite Difference and Finite Element Methods at Walmart. Here we will use the simplest method, finite differences. Author(s): Douglas N. Numerical Methods for Partial Differential Equations is an international journal that aims to cover research into the development and analysis of new methods for the numerical solution of partial differential equations. Prerequisite: Consent of instructor. The spine may show signs of wear. All important problems in science and engineering are solved in this manner. Numerical Solutions of Some Parabolic Partial Differential Equations Using Finite Difference Methods @inproceedings{Singla2012NumericalSO, title={Numerical Solutions of Some Parabolic Partial Differential Equations Using Finite Difference Methods}, author={Rishu Singla and Ram Jiwari}, year={2012} }. The solution of PDEs can be very challenging, depending on the type of equation, the number of independent variables, the boundary, and initial conditions, and other factors. Numerical Solution of the Advection Partial Differential Equation: Finite Differences, Fixed Step Methods. Numerical solution of Burger equation is a natural and first step towards developing methods for the computation of complex flows. Negesse Yizengaw, (2015) Numerical Solutions of Initial Value Ordinary Differential Equations Using Finite Difference Method. Mathematical approaches for numerically solving partial differential equations. In [Jiang and Zhang, Journal of Computational Physics, 253 (2013) 368-388], IIF methods are designed to efficiently solve stiff nonlinear advection-diffusion-reaction (ADR) equations. 4 Zero-stability of linear multistep methods 143 6. method of lines, finite differences, spectral methods, aliasing, multigrid, stability region AMS subject classifications. For various situations, RBF with infinitely differentiable functions can provide accurate results and more flexibility in the geometry of computation domains than traditional methods such as finite difference and finite element methods. The Numerical Solution of Ordinary and Partial Differential Equations is an introduction to the numerical solution of ordinary and partial differential equations. Fundamentals 17 2. Main activities: High Order Finite Difference Methods (FDM) We have developed summation-by-parts operators and penalty techniques for boundary and interface conditions. We have considered both linear and nonlinear Goursat problems of partial differential equations for the numerical solution, to ensure the accu-racy of the developed method. The focuses are the stability and convergence theory. By Steven H. Then the fourth order finite difference and collocation method is presented for the numerical solution of this type of partial integro-differential equation (PIDE). Numerical Methods for Differential Equations Chapter 5: Partial differential equations - elliptic and pa rabolic Gustaf Soderlind and Carmen Ar¨ evalo´ Numerical Analysis, Lund University Textbooks: A First Course in the Numerical Analysis of Differential Equations, by Arieh Iserles. The finite element method is a special method for the numerical solution of partial differential equations. After revising the mathematical preliminaries, the book covers the finite difference method of parabolic or heat equations, hyperbolic or wave equations and elliptic or Laplace equations. Substituting the. Use the sliders to vary the initial value or to change the number of steps or the method. •• Stationary Problems, Elliptic Stationary Problems, Elliptic PDEsPDEs. Introduction This is the sequel to math 614 Numerical Methods I. Numerical Methods for Partial Differential Equations I. Fourier series methods for the wave equation 7. Self-adaptive discretization methods are now an indispensable tool for the numerical solution of partial differential equations that arise from physical and technical applications. Differential equations. Substantially revised, this authoritative study covers the standard finite difference methods of parabolic, hyperbolic, and elliptic equations, and includes the concomitant theoretical work on consistency, stability, and convergence. DERIVATION OF DIFFERENCE EQUATIONS AND MISCELLANEOUS TOPICS Reduction to a System of ordinary differential equations 111 A note on the Solution of dV/dt = AV + b 113 Finite-difference approximations via the ordinary differential equations 115 The Pade approximants to exp 0 116 Standard finite-difference equations via the Pade approximants 117. In the first the time derivative is replaced by a finite difference ratio, and the resulting ordinary differential equation with x. Scientific Computing and Numerical Analysis Group. Amazon Price New from. The Boundary Element Method (BEM) allows efficient solution of partial differential equations whose kernel functions are known. But if you want to learn about Finite Element Methods (which you should these days) you need another text. Numerically solving PDEs in Mathematica using finite difference methods 61 Replies Mathematica’s NDSolve command is great for numerically solving ordinary differential equations, differential algebraic equations, and many partial differential equations. The method of lines is a general technique for solving partial differential equat ions (PDEs) by typically using finite difference relationships for the spatial derivatives and ordinary differential equations for the time derivative. Finite element method (FEM) utilizes discrete el ements to obtain the approximate solution of the governing differential equation. PREREQUISITE(S):. variables that determine the behavior of the. BVPs can be solved numerically using a method known as the finide. A composite weighted trapezoidal rule is manipulated to handle the numerical integrations which results in a closed-form difference scheme. Kreiss: Numerical Methods for Solving Time-Dependent Problems for Partial Differential Equations (1978) J. To find a well-defined solution, we need to impose the initial condition u(x,0) = u 0(x) (2). Numerical Solution of Partial Differential Equations Finite Difference Methods. top/file/Numerical Solutions Of Partial Differential Equations By The Finite Element Method. Numerical methods are needed to solve partial differential equations (PDEs). 1 A finite difference scheme for the heat equation - the concept of convergence. For the solution of a parabolic partial differential equation numerical approximation methods are often used, using a high speed computer for the computation. We shall regard the solutions of the difference equation (15) as being de-fined on the discrete set of points x = kAx, with k an integer, and shall use the. The corresponding theoretically analyzing methods include Fourier methods, energy estimation, matrix eigenvalue. But what challenges must. It is also a valuable working reference for professionals in engineering, physics, chemistry. 0014142 2 0. 29 & 30) Based on approximating solution at a finite # of points, usually arranged in a regular grid. [G D Smith] Home. Below are simple examples of how to implement these methods in Python, based on formulas given in the lecture note (see lecture 7 on Numerical Differentiation above). plays an important role in the solution of partial differential equations. Oliger, Time Dependent Problems and Difference Methods, John Wiley & Sons, Inc. Partial Differential Equations (PDEs) Conservation Laws: Integral and Differential Forms Classication of PDEs: Elliptic, parabolic and Hyperbolic Finite difference methods Analysis of Numerical Schemes: Consistency, Stability, Convergence Finite Volume and Finite element methods Iterative Methods for large sparse linear systems. Buy Numerical Solution Of Partial Differential Equations: Finite Difference Methods (Oxford Applied Mathematics & Computing Science Series) (Oxford Applied Mathematics and Computing Science Series) on Amazon. and a great selection of related books, art and collectibles available now at AbeBooks. Thus, it has become customary to test new approaches in computational fluid dynamics by implementing novel and new approaches to Burger equation yielding in various finite-differences, finite volume, finite-element and boundary element methods etc. In particular, I am looking to solve this equation: The. Spectral methods in Matlab, L. In such a method an approximate solution is sought at the points of a finite grid of points, and the approximation of the differential equation is accomplished by replacing derivatives by appropriate difference quotients. 2 Finite difference methods for solving partial differential equations 17 Chapter Three: Wavelets and applications 20 3. 920J/SMA 5212 Numerical Methods for Partial Differential Equations Lecture 5 Finite Differences: Parabolic Problems B. The notebook introduces finite element method concepts for solving partial differential equations (PDEs). Everyday low prices and free delivery on eligible orders. But what challenges must. This is often used in numerical analysis, especially in numerical ordinary differential equations and numerical partial differential equations, which aim at the numerical solution of ordinary and partial differential equations respectively. Numerical Methods for Partial Di erential Equations Finite Di erence Methods for Elliptic Equations Finite Di erence Methods for Parabolic Equations. The exact solution is calculated for fractional telegraph partial. Nicolson / Evaluation of solutions of partial differential equations variables, of the heat-conduction type. LeVeque; Numerical solution of partial differential equations: an introduction by K. By Steven H. A solution domain 3. The FIDE package performs automation of the process of numerical solving partial differential equations systems (PDES) by means of computer algebra. ; Partial differential equations - Numerical solution; 1900-1999 Contents. Chapter Two: Overview of numerical methods for differential equations 7 2. It is also a valuable working reference for professionals in engineering, physics, chemistry. Numerical Methods for Partial Differential Equations (MATH F422 - BITS Pilani) How to find your way through this repo: Navigate to the folder corresponding to the problem you wish to solve. Numerical Methods for Partial Differential Equations: Finite Difference and Finite Volume Methods focuses on two popular deterministic methods for solving partial differential equations (PDEs), namely finite difference and finite volume methods. Augustine; Computer Science. The grid method (finite-difference method) is the most universal. Numerical Methods for Partial Differential Equations is an international journal that aims to cover research into the development and analysis of new methods for the numerical solution of partial differential equations. Numerical Solutions to Partial Di erential Equations Zhiping Li LMAM and School of Mathematical Sciences Peking University. Partial differential equation such as Laplace's or Poisson's equations. See all 8 formats and editions Hide other formats and editions. Of interest are discontinuous initial conditions. Corpus ID: 11321617. 0 MB) Finite Difference Discretization of Elliptic Equations: 1D Problem (PDF - 1. 1 Approximating the Derivatives of a Function by Finite ff Recall that the derivative of a function was de ned by taking the limit of a ff quotient: f′(x) = lim ∆x!0 f(x+∆x) f. , A first course in the numerical analysis of differential equations, Cambridge Texts in Applied Mathematics, Cambridge University Press, 1996. This is because many mathematical models of physical phenomena result in one or more. nonlinear partial differential equations. The text is divided into two independent parts, tackling the finite difference and finite element methods separately. * Numerical Partial Differential Equations. Nonlinear Partial Differential Equations Fonksi Yonlar Numerical Analysis The Solution of Advection Diffusion Equation by the Finite. In this paper, the method for numerical solution of fractional partial differential equations is based on Laplace transform (LT), the homotopy perturbation method (HPM) and Stehfest’s numerical algorithm for calculating inverse Laplace transform. f is a function of two variables x and y and (x 0, y 0) is a known point on the solution curve. The new edition includes revised and greatly expanded sections on stability based on the Lax-Richtmeyer definition, the application of Pade approximants to. (2) Add a numerical viscosity to produce the desired directional bias in the hyperbolic region. This 325-page textbook was written during 1985-1994 and used in graduate courses at MIT and Cornell on the numerical solution of partial differential equations. LeVeque, SIAM 2007 Instructor's Notes will be updated constantly. For PDES solving, the finite difference method is applied. Finite Difference Methods In the previous chapter we developed finite difference appro ximations for partial derivatives. However, the finite difference method (FDM) uses direct. d Smith as PDF for free. 2013, Article ID 562140, 13 pages, 2013. Introduction and finite-difference formulae ; 2. The solution of PDEs can be very challenging, depending on the type of equation, the number of. Numerical Integration of Partial Differential Equations (PDEs) •• Introduction to Introduction to PDEsPDEs. The grid method (finite-difference method) is the most universal. NUMERICAL METHODS FOR SOLVING PARTIAL DIFFERENTIAL EQUATION. Laplace's equation d 2 φ/dx 2 + d 2 φ/dy 2 = 0 plus some boundary. I am attempting to write a MATLAB program that allows me to give it a differential equation and then ultimately produce a numerical solution. Finite Difference Methods for Differential Equations @inproceedings{LeVeque2005FiniteDM, title={Finite Difference Methods for Differential Equations}, author={Randall J. 65J15, 65M20 1. One Step Methods of the Numerical Solution of Differential Equations Probably the most conceptually simple method of numerically integrating differential equations is Picard's method. The main theme is the integration of the theory of linear PDE and the theory of finite difference and finite element methods. The aim is to obtain a numerical solution within a prescribed tolerance using a minimal amount of work. Frequently exact solutions to differential equations are unavailable and numerical methods become. This chapter explores the finite element method for elliptic differential equations. y p =Ax 2 +Bx + C. The finite element method became a very widely used method in practice. 3 Representation of a finite difference scheme by a matrix operator. Numerical solution of partial differential equations has important applications in many application areas. Finite differences. As far as I know, Lecture notes from Prof. This Demonstration shows some numerical methods for the solution of partial differential equations: in particular we solve the advection equation. The goal of this course is to introduce theoretical analysis of finite difference methods for solving partial differential equations. Numerical differentiation. But these methods often rely on deep analytical insight into the equations. Differential equations. Finite Difference Method The finite difference method (FDM) is a simple numerical approach used in numerical involving Laplace or Poisson's equations. LeVeque, SIAM, 2007. [G D Smith] Home. Lecture notes on Numerical Analysis of Partial Differential Equation. 8 Finite Differences: Partial Differential Equations The worldisdefined bystructure inspace and time, and it isforever changing incomplex ways that can't be solved exactly. The idea is to replace the derivatives appearing in the differential equation by finite differences that approximate the m. Brenner and L. The partial differential equations to be discussed include •parabolic equations, •elliptic equations, •hyperbolic conservation laws. Numerical approximations of autonomous SPDEs are thoroughly investigated in the literature, while the non-autonomous case is not yet understood. Morton and D. Augustine; Computer Science. In such cases numerical methods allow us to use the powers of a computer to obtain quantitative results. First a discretization is done. Solution of heat equation is computed by variety methods including analytical and numerical methods [2]. The Finite Difference Method (FDM) is a way to solve differential equations numerically. a mesh; a partial differential equation; boundary conditions that link the equation with the region; This section deals with partial differential equations and their boundary conditions. Thus, it has become customary to test new approaches in computational fluid dynamics by implementing novel and new approaches to Burger equation yielding in various finite-differences, finite volume, finite-element and boundary element methods etc. 35—dc22 2007061732. Students are introduced to the discretization methodologies, with particular emphasis on the finite difference method, that allows the construction of accurate and stable numerical schemes. Analysis of consistency, order, stability and convergence. Of the many different approaches to solving partial differential equations numerically, this book studies difference methods. Available online -- see below. Finite difference methods for ordinary and partial differential equations : steady-state and time-dependent problems / Randall J. We use finite differences with fixed-step discretization in space and time and show the relevance of the Courant–Friedrichs–Lewy stability criterion for some of these discretizations. Finite differences. " MAA Reviews "First and foremost, the text is very well written. As far as I know, Lecture notes from Prof. For PDES solving finite difference method is applied. This book presents methods for the computational solution of differential equations, both ordinary and partial, time-dependent and steady-state. A weakly singular kernel has been viewed as an important case. Accuracy, temporal performance and stability comparisons of discretization methods for the numerical solution of Partial Differential Equations (PDEs) in. Find all the books, read about the author, and more. PDE playlist: http://www. Science—Mathematics. Description: Numerical Methods for Partial Differential Equations: Finite Difference and Finite Volume Methods focuses on two popular deterministic methods for solving partial differential equations (PDEs), namely finite difference and finite volume methods. For many of the differential equations we need to solve in the real world, there is no "nice" algebraic solution. This is an introduction to solution techniques for partial differential equations that has been developed from an introductory course on the subject for junior and senior-level mathematics or physical science majors with an undergraduate background in calculus, introductory linear algebra, and ordinary differential equations. Differential equations, Partial— Numerical solutions. In this method, various derivatives in the partial differential equation are replaced by their finite difference approximations, and the PDE is converted to a set of linear algebraic equations. Numerical Methods for Partial Differential Equations: Finite Difference and Finite Volume Methods focuses on two popular deterministic methods for solving partial differential equations (PDEs), namely finite difference and finite volume methods. To find a numerical solution to equation (1) with finite difference methods, we first need to define a set of grid points in the domainDas follows: Choose a state step size Δx= b−a N (Nis an integer) and a time step size Δt, draw a set of horizontal and vertical lines across D, and get all intersection points (x j,t n), or simply (j,n), where x. It is important to note that a numerical solution is approximate. Parabolic equations ; 3. It contains solution methods for different class of partial differential equations. In numerical analysis, finite-difference methods (FDM) are discretizations used for solving differential equations by approximating them with difference equations that finite differences approximate the derivatives. In that case, going to a numerical solution is the only viable option. Buy I have no guess how to start for stated PDE. Therefore, numerical methods for finding approximate solutions to PDE problems are of great importance: numerical solutions of PDEs on powerful computers allow researchers to push the. Search for Library Items Search for Lists Search for # Differential equations, Partial--Numerical solutions\/span> \u00A0\u00A0\u00A0 schema:. The goal of this course is to provide numerical analysis background for finite difference methods for solving partial differential equations. All important problems in science and engineering are solved in this manner. There is no formula to evaluate. Topics include: Mathematical Formulations; Finite Difference and Finite Volume Discretizations;. Barba will be best. •• SemiSemi--analytic methods to solve analytic methods to solve PDEsPDEs. DERIVATION OF DIFFERENCE EQUATIONS AND MISCELLANEOUS TOPICS Reduction to a System of ordinary differential equations 111 A note on the Solution of dV/dt = AV + b 113 Finite-difference approximations via the ordinary differential equations 115 The Pade approximants to exp 0 116 Standard finite-difference equations via the Pade approximants 117. 1 Taylor s Theorem 17. Numerical Evaluation of scientific or engineering problems governed by Partial Differential Equations (PDEs) numerically is in general computationally-demanding and data intensive. April 22, 2015. , Rice University Computer Science Department Technical Report 00-368, 2000, 27-30. After introducing each class of differential equations we consider finite difference methods for the numerical solution of equations in the class. 1) 2 1 2 2 2 2 < <+∞ ≤ < − = ∂ ∂ + + ∂ ∂ rV S V rS S V S t V ∂ ∂ σ Notes: This is a second-order hyperbolic, elliptic, or parabolic, forward or backward partial differential equation Its solution. Numerical Solution Of Partial Differential Equations: Finite Difference Methods (Oxford Applied Mathematics & Computing Science Series) (Oxford Applied Mathematics and Computing Science Series) 3rd Edition. Barba will be best. this paper, a hybrid approach which combines the immersed interface method with the level set approach is presented. Descriptive Note : Lecture notes no. Arora, "Taylor-Galerkin B-spline finite element method for the one-dimensional advection-diffusion equation," Numerical Methods for Partial Differential Equations, vol. 8 Finite Differences: Partial Differential Equations The worldisdefined bystructure inspace and time, and it isforever changing incomplex ways that can’t be solved exactly. The exact solution is calculated for fractional telegraph partial. However, many partial differential equations cannot be solved exactly and one needs to turn to numerical solutions. Numerically solving PDEs in Mathematica using finite difference methods 61 Replies Mathematica’s NDSolve command is great for numerically solving ordinary differential equations, differential algebraic equations, and many partial differential equations. Numerical solution of partial differential equations has important applications in many application areas. Start your review of Numerical Solution of Partial Differential Equations: Finite Difference Methods Write a review Oct 15, 2015 Chand added it. Smith (Author) 5. Numerical Analysis of Partial Differential Equations Using Maple and MATLAB provides detailed descriptions of the four major classes of discretization methods for PDEs (finite difference method, finite volume method, spectral method, and finite element method) and runnable MATLAB® code for each of the discretization methods and exercises. method of lines, finite differences, spectral methods, aliasing, multigrid, stability region AMS subject classifications. The main drawback of the finite difference methods is the. Numerical Solution of Partial Differential Equations An Introduction K. Required textbook and other resources. Finite difference methods for ordinary and partial differential equations : steady-state and time-dependent problems / Randall J. Numerical Methods for Partial Differential Learn more about numerical, methods, pde, code. Chapter 2 Introduction to Finite Differences Numerical Partial Differential Equations. References. Find all the books, read about the author, and more. Computational methodology combined with convergence theory, including: the immersed finite element method, finite volume methods, discontinuous Galerkin methods, and adaptive methods based on numerical smoothness and superconvergence theory. The contributed papers reflect the interest and high research level of the Chinese mathematicians working in these fields. python c pdf parallel-computing scientific-computing partial-differential-equations ordinary-differential-equations petsc krylov multigrid variational-inequality advection newtons-method preconditioning supercomputing finite-element-methods finite-difference-schemes fluid-mechanics obstacle-problem firedrake algebraic-multigrid. 1 Example of Problems Leading to Partial Differential Equations. BVPs can be solved numerically using a method known as the finide. I am attempting to write a MATLAB program that allows me to give it a differential equation and then ultimately produce a numerical solution. Method of Lines, Part I: Basic Concepts. , Le Lann, J. The spine may show signs of wear. These equations involve two or more independent. By Steven H. This course is intended as a review of modern numerical techniques for a vide range of time-dependent partial differential equations. Ability to implement advanced numerical methods for the solution of partial differential equations in MATLAB efciently Ability to modify and adapt numerical algorithms guided by awareness of their mathematical foun-dations p. L548 2007 515’. Kexue and Jiger[20] have utilized LT to solve problems arising in fractional differential equations. In this chapter we will use these finite difference approximations to solve partial differential equations (PDEs) arising from conservation law presented in Chapter 11. In some solutions for a partial derivative like $\frac{\partial u}{\partial x}$ it is written by using forward difference and sometimes by using central difference. Accuracy, temporal performance and stability comparisons of discretization methods for the numerical solution of Partial Differential Equations (PDEs) in. As a result, we need to resort to using. Laplace Equation in 2D. It is not the only option, alternatives include the finite volume and finite element methods, and also various mesh-free approaches. Available online -- see below. Description: Numerical Methods for Partial Differential Equations: Finite Difference and Finite Volume Methods focuses on two popular deterministic methods for solving partial differential equations (PDEs), namely finite difference and finite volume methods. 1 Taylor s Theorem 17. Finite difference method in combination with product trapezoidal integration rule is used to discretize the equation in time and sinc-collocation method is employed in space. The numerical method of lines is used for time-dependent equations with either finite element or finite difference spatial discretizations, and details of this are described in the tutorial "The Numerical Method of Lines". Analytic. However, in most cases these tools are limited to 2d and can only solve special forms of elliptic, parabolic or hyperbolic partial differential equations (PDE). Arora, "Taylor-Galerkin B-spline finite element method for the one-dimensional advection-diffusion equation," Numerical Methods for Partial Differential Equations, vol. Finite Difference Method The finite difference method (FDM) is a simple numerical approach used in numerical involving Laplace or Poisson's equations. In this work, we studied the matter of heat transfer by the natural convection for a dissipatable fluid which flows in a tube, its walls composed from porous material, and a mathe. Numerical Solution of Partial Differential Equations by using Modified Artificial Neural Network ∗. ppt), PDF File (. The Radial Basis Function (RBF) method has been considered an important meshfree tool for numerical solutions of Partial Differential Equations (PDEs). Finite difference method in combination with product trapezoidal integration rule is used to discretize the equation in time and sinc-collocation method is employed in space. This paper develops a new framework for designing and analyzing convergent finite difference methods for approximating both classical and viscosity solutions of second order fully nonlinear partial differential equations (PDEs) in 1-D. 0014142 2 0. Other articles where Finite difference method is discussed: numerical analysis: Solving differential and integral equations: …numerical procedures are often called finite difference methods. The new edition includes revised and greatly expanded sections on stability based on the Lax-Richtmeyer definition, the application of Pade approximants to. Brenner and L. ! Objectives:! Computational Fluid Dynamics I! • Solving partial differential equations!!!Finite difference approximations!. Spectral methods in Matlab, L. • Partial Differential Equation: At least 2 independent variables. Felleisen, ed. The heat equation is a simple test case for using numerical methods. AL-GHAMDI, Department of Civil Engineering, College of Engineering ,PhD student King Abdulaziz University (KAU),Jeddah ,Saudi Arabia. In this paper, the method for numerical solution of fractional partial differential equations is based on Laplace transform (LT), the homotopy perturbation method (HPM) and Stehfest’s numerical algorithm for calculating inverse Laplace transform. 0014142 Therefore, x x y h K e 0. Course Description : This course is designed for graduate students in mathematics, engineering, finance, and computer science. This book presents methods for the computational solution of differential equations, both ordinary and partial, time-dependent and steady-state. Dougalis Department of Mathematics, University of Athens, Greece and Institute of Applied and Computational Mathematics, FORTH, Greece Revised edition 2013. The exact solution is calculated for fractional telegraph partial. This paper aims to investigate numerical approximation of a general second order non-autonomous semilinear parabolic stochastic partial differential equation (SPDEs) driven by multiplicative noise. Numerical solution of partial differential equations : an introduction / by: Morton, K. This lecture discusses different numerical methods to solve ordinary differential equations, such as forward Euler, backward Euler, and central difference methods. com/view_play_list Topics: -- introduction to the idea of finite differences. Numerical Methods for Partial Differential Equations: Finite Difference and Finite Volume Methods focuses on two popular deterministic methods for solving partial differential equations (PDEs), namely finite difference and finite volume methods. mathematics. Differential equations, Partial— Numerical solutions. Stiff systems of ODEs are solved by Aminikhah[19] using a combined LT and HPM. Course Description : This course is designed for graduate students in mathematics, engineering, finance, and computer science. of numerical analysis, the numerical solution of partial differential equations, as it developed in Italy during the crucial incubation period immediately preceding the diffusion of electronic computers. Numerical approximations of autonomous SPDEs are thoroughly investigated in the literature, while the non-autonomous case is not yet understood. Numerical Analysis of Partial Differential Equations Using Maple and MATLAB provides detailed descriptions of the four major classes of discretization methods for PDEs (finite difference method, finite volume method, spectral method, and finite element method) and runnable MATLAB® code for each of the discretization methods and exercises. We present qualitative and numerical results on a partial differential equation (PDE) system which models a certain fluid-structure dynamics. Textbook(s): K. The notebook introduces finite element method concepts for solving partial differential equations (PDEs). Although it is unlikely to know values of the exact solution for the second row of the grid, if such knowledge were available, using the increment k = ch along the t -axis. The numerical results confirm that the proposed finite difference methods yield second- and fourth-order convergence for the solution and its derivative for the fourth-order ordinary differential equation. Application and analysis of numerical methods for ordinary and partial differential equations. We present qualitative and numerical results on a partial differential equation (PDE) system which models a certain fluid-structure dynamics. An extensive theoretical development is presented that establishes convergence and stability for one-dimensional parabolic equations with Dirichlet boundary. Solving differential equations is a fundamental problem in science and engineering. , some approximation solution). We apply the method to the same problem solved with separation of variables. LeVeque, SIAM, 2007. 0014142 Therefore, x x y h K e 0. The convergence and stability analysis of the solution methods is also included. The focuses are the stability and convergence theory. Numerical Analysis of Partial Differential Equations Using Maple and MATLAB provides detailed descriptions of the four major classes of discretization methods for PDEs (finite difference method, finite volume method, spectral method, and finite element method) and runnable MATLAB® code for each of the discretization methods and exercises. Therefore, numerical methods for finding approximate solutions to PDE problems are of great importance: numerical solutions of PDEs on powerful computers allow researchers to push the. They are used to discretise and approximate the derivatives for a smooth partial differential equation (PDE), such as the Black-Scholes equation. Wellposedness is established by constructing for it a nonstandard semigroup generator representation; this representation is accomplished by an appropriate elimination of the pressure. The solution of PDEs can be very challenging, depending on the type of equation, the. Kadalbajoo and P. Explicit and Implicit Methods in Solving Differential Equations A differential equation is also considered an ordinary differential equation (ODE) if the unknown function depends only on one independent variable. Includes bibliographical references and index. obtaining solutions. Other articles where Finite difference method is discussed: numerical analysis: Solving differential and integral equations: …numerical procedures are often called finite difference methods. 0 MB)Finite Differences: Parabolic Problems ()(Solution Methods: Iterative Techniques (). Finite Difference Methods In the previous chapter we developed finite difference appro ximations for partial derivatives. PDEs tend to be divided into three categories - hyperbolic, parabolic and elliptic. Substantially revised, this authoritative study covers the standard finite difference methods of parabolic, hyperbolic, and elliptic equations, and includes the concomitant theoretical work on consistency, stability, and convergence. •• Stationary Problems, Elliptic Stationary Problems, Elliptic PDEsPDEs. FDMs convert a linear ordinary differential equations (ODE) or non-linear partial differential equations (PDE) into a system of equations that can be solved by matrix algebra. Associate Professor Sandip Mazumder The textbook, “Numerical Methods for Partial Differential Equations: Finite Difference and Finite Volume Methods,” will serve as a thorough step-by-step guide for graduate students and practicing engineers on the fundamental techniques, algorithms and coding practices required for solving canonical Partial Differential Equations using the finite difference and finite volume methods. * Inverse Problems. Descriptive Note : Lecture notes no. 3 Representation of a finite difference scheme by a matrix operator. Numerical solution of partial differential equations: finite difference methods. Topics: Advanced introduction to applications and theory of numerical methods for solution of partial differential equations, especially of physically-arising partial differential equations, with emphasis on the fundamental ideas underlying various methods. by a difference quotient in the classic formulation. Students are introduced to the discretization methodologies, with particular emphasis on the finite difference method, that allows the construction of accurate and stable numerical schemes. Wellposedness is established by constructing for it a nonstandard semigroup generator representation; this representation is accomplished by an appropriate elimination of the pressure. The goal of this course is to provide numerical analysis background for finite difference methods for solving partial differential equations. Comprehensive yet accessible to readers with limited mathematical knowledge, Numerical Methods for Solving Partial Differential Equations is an excellent text for advanced undergraduates and first-year graduate students in the sciences and engineering. 48 Self-Assessment. : Numerical Solution of Some Fractional Partial Differential there are lots of studies on the subject in recent years [4,5,6,7,8], this area of numerical mathematics is still not developed and understood as well as its integer counterpart [9]. 29 & 30) Based on approximating solution at a finite # of points, usually arranged in a regular grid. Use the sliders to vary the initial value or to change the number of steps or the method. Numerical Methods for Partial Differential Equations is an international journal that aims to cover research into the development and analysis of new methods for the numerical solution of partial differential equations. For simplicity of notation, the phrase partial differential equation frequently will be replaced by the acronym PDE in Part III. All important problems in science and engineering are solved in this manner. com/view_play_list Topics: -- introduction to the idea of finite differences. Differential equations, Partial - Numerical solutions. variables that determine the behavior of the. General Finite Element Method An Introduction to the Finite Element Method. Search for Library Items Search for Lists Search for # Differential equations, Partial--Numerical solutions\/span> \u00A0\u00A0\u00A0 schema:. In this work, we studied the matter of heat transfer by the natural convection for a dissipatable fluid which flows in a tube, its walls composed from porous material, and a mathe. The finite element method (FEM) (its practical application often known as finite element analysis (FEA)) is a numerical technique for finding approximate solutions of partial differential equations (PDE) as well as of integral equations. • Parabolic (heat) and Hyperbolic (wave) equations. Finite differences. This is an introduction to solution techniques for partial differential equations that has been developed from an introductory course on the subject for junior and senior-level mathematics or physical science majors with an undergraduate background in calculus, introductory linear algebra, and ordinary differential equations. We use finite differences with fixed-step discretization in space and time and show the relevance of the Courant–Friedrichs–Lewy stability criterion for some of these discretizations. It is simple to code and economic to compute. This paper aims to investigate numerical approximation of a general second order non-autonomous semilinear parabolic stochastic partial differential equation (SPDEs) driven by multiplicative noise. Method of Lines, Part I: Basic Concepts. In this chapter we will use these finite difference approximations to solve partial differential equations (PDEs) arising from conservation law presented in Chapter 11. For best of your experience, you can learn various numerical technique by hands on practice using i-python notebook. Numerical approximations of autonomous SPDEs are thoroughly investigated in the literature, while the non-autonomous case is not yet understood. Main activities: High Order Finite Difference Methods (FDM) We have developed summation-by-parts operators and penalty techniques for boundary and interface conditions. The goal of this course is to introduce theoretical analysis of finite difference methods for solving partial differential equations. Texts: Finite Difference Methods for Ordinary and Partial Differential Equations (PDEs) by Randall J. SPEAKER: Vsevolod Avrutsky (Moscow Institute of Physics and Technology) TITLE: Neural networks catching up with finite differences in solving partial differential equations in higher dimensions ABSTRACT: Deep neural networks for solving Partial Differential Equations. … The text is enhanced by 13 figures and 150 problems. This paper aims to investigate numerical approximation of a general second order non-autonomous semilinear parabolic stochastic partial differential equation (SPDEs) driven by multiplicative noise. By Steven H. Substantially revised, this authoritative study covers the standard finite difference methods of parabolic, hyperbolic, and elliptic equations, and includes the concomitant theoretical work on consistency, stability, and convergence. Learn to write programs to solve ordinary and partial differential equations. Johnson's Numerical Solution of Partial Differential Equations by the Fini. The solution of PDEs can be very challenging, depending on the type of equation, the number of independent variables, the boundary, and initial. "Larsson and Thomée … discuss numerical solution methods of linear partial differential equations. Numerical solution of partial differential equations has important applications in many application areas. As far as I know, Lecture notes from Prof. 1) Without loss of generality, (1) The system is autonomous, i. of numerical methods in a synergistic fashion. Wellposedness is established by constructing for it a nonstandard semigroup generator representation; this representation is accomplished by an appropriate elimination of the pressure. 2 Solution to a Partial Differential Equation 10 1. Numerical Solution of Partial Differential Equations, K.
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https://magoosh.com/gre/gre-math-absolute-values/ | # GRE Math: Absolute Values
On average you will see at least one question on the Revised GRE dealing with absolute values. You may even see a few. Yet, absolute value gets lost in the prep fray amongst the more popular concepts. So if you don’t want this relatively innocuous concept to surprise you test day read on.
## What You Need to Know
What do -4 and 4 have in common? They are both four units from the 0 on a number line. Think of absolute value has how far from the zero a given number is on a number line.
For positive numbers finding the absolute value is easy – it is always the number between the absolute value signs, which look like this $${|}{ }{|}$$.
$${|}{5}{|}=5$$ $${|}{1/2}{|}=1/2$$ $${|}{100}{|}=100$$
When we take the absolute value of a negative number, we drop the negative, and the absolute value sign.
$${|}{-5}{|}=5$$ $${|}{-1/2}{|}=1/2$$ $${|}{-100}{|}={100}$$
## Absolute Value Meets Algebra
What is the value of x in the equation $${|}{x}{|}=5$$? Well, the absolute value of both $$5$$ and $$-5$$ is $$5$$, so $$x = 5$$ or $$-5$$.
Now take a look at the following: $${|}{x}{|}={-5}$$
Anything seem off? Well, the absolute value of any number can never be a negative therefore there is no value for x.
To solve for a variable inside an absolute value sign, we want to remove the absolute value sign and solve the equation. However, there is a slight twist: you will want to create two separate equations. For one remove the absolute value signs and solves for x. For the other, make the side of the equation not inside the absolute value equal to a negative. Let’s try a practice problem:
$${|}{x-2}{|}=4$$?
Our two equations are:
$$x-2=4$$ $$x-2=-4$$
If this seems strange, think of it this way: when you find a value for x that makes the equation equal -4, you have also solved for positive four. Remember, the absolute value of -4 is 4. Solving for x we get:
$$x-2=4$$, $$x=6$$
$$x-2=-4$$, $$x=-2$$.
Therefore x can equal 6 or -2.
## Absolute Value Meets the Inequality Sign
Now let’s complicate things a little and throw an inequality in to the picture. Have a look:
$${|}{x-4}{|}<3$$
We turn the inequality sign into an equal sign and solve for x the way we did above.
$$x-4=3$$ $$x=7$$
so $$x<7$$.
However, when we turn 3 into -3, we have to reverse the sign. This is the one critical step.
$$x-4=-3$$ $$x=1$$
so $$x>1$$.
Therefore, $$1<x<y$$. You can plug in different values for x to see how this is the case.
## Takeaways
This is a basic overview to absolute value and should help you with most of the sub-150 problems. For the harder problems, however, you will want to make sure to practice with more advanced problems.
### 51 Responses to GRE Math: Absolute Values
1. José July 29, 2020 at 8:18 pm #
Hi,
I just watch the video on “Positive and Negative Numbers – II” and wanted to learn a little more about absolute values. However, I am seeing some weird text in the blog entry (e.g., ), I tried to use different web explorers, but no improvement. Is there’s something wrong with the blog entry?
Thanks,
• Magoosh Test Prep Expert August 28, 2020 at 1:25 pm #
Hi Jose,
If you are still having this issue, can you send a screenshot to us at [email protected] ? I don’t see any issues on my end, so it might help to see what you see.
2. Mariyah April 7, 2019 at 5:46 am #
This was asked a earlier, would like an explaination for rule please.
| x – 1 |> 4, the results are x> 5 and x ,<) changed. Because the rules to change the sign only apply to division and multiplication with negative numbers.
• David Recine April 8, 2019 at 7:11 pm #
I’ll be happy to help you with this, Mariyah. Before we get started though, to clarify, I believe you meant to say that | x – 1 |> 4 has results of x>5 and x<-3. If I got that wrong, let me know in a follow-up concept. But for now, we'll assume I got that right. 🙂 OK, so why is the sign reversed in the case of x<-3? first, let's review the steps, and rule for reversing inequality signs when dealing with absolute values. then, we can talk about how rules in values actually do connect signs. so if have | x – 1> 4, we want to solve by removing the absolute value sign and change the inequality sign to an equal sign, as seen in the post above. Let’s start there and go through the steps.
Remove the absolute value sign, and we have x-1 = 4, OR x-1 = -4. Why? Because the absolute value of a positive number or a negative number both equal the same positive number. |3|= 3, and |-3|= 3, to give one example. We don’t know if x-1 as it’s seen inside the absolute value sign is ultimately a positive number. We only know that it’s absolute value is positive, which means x-1 itself could be negative or positive. So, removing x-1 from its absolute value sign and changing the “>” to an equal, we must assume that x-1 could equal 4 or -4.
So we have two equations: x-1 = 4, and x-1 = -4. Let’s solve for x-1 = 4 first:
x-1 = 4
x = 4 + 1
x = 5
So one possible value of x is 5. Now that we’ve solved for that possibility, we can put the original inequality sign back in:
x = 5
x > 5
Next, let’s solve for x-1 = -5:
x-1 = -4
x = -4 + 1
x = -3
Now, when solving for x-1 as a negative number, we reverse the original sign:
x = -3
x > -3
x < -3 Why? Well, the simplest answer is that we reverse the sign because that's the rule for absolute value and inequalities. When you are using a negative number for absolute value and solving for an inequality, you use the sign because... that's how it's done. But let's take a deeper look as to why we have that rule to begin with. Think if it this way: Just as |3| = 3 or -3, |x–1| = +(x-1) or -(x-1). Now if you turn the x-1 in |x–1 | into +(x-1), you really haven't changed it. x-1 remains x-1, the exact same value. However, if you take the x-1 in |x–1| and turn it into -(x-1), you have changed the value. x-1 does NOT equal -(x-1). So how do you change x-1 into -(x-1)? By multiplying x-1 by -1:
(-1)(x-1) = -(x-1)
So when you solve for the negative value of an equation that you found inside the absolute value, you are actually multiplying the equation by -1. And if you are multiplying an entire equation by -1, you follow the rules of multiplication with negative numbers in inequalities, and you change the direction of the inequality sign!
Does that make sense? If you’re not sure it does (or again, if there’s anything I misunderstood about your original question), let me know in a follow-up comment.
3. Navaneeth June 2, 2018 at 6:40 am #
|x-5| -> Distance of x from +5
This is not clear, can you explain the concept behind it? Why cant it be distance from -5.
• Magoosh Test Prep Expert June 7, 2018 at 12:09 pm #
|x-5| really does represent the distance of x from +5, and |x+5| really is the distance of x from -5. This can seem confusing at first, Navaneeth, but it makes sense if you plug in values for x. Let’s give that a try.
Suppose that x=8. Then, |x-5| becomes |8-5|. This simplifies to |3|, or just 3 (since absolute values are positive numbers anyway). If x = 8, in other words, x is 3 units from 5, from positive 5. It is NOT 3 units away from -5; 8 is actually 13 units away from negative 5. If we set x at 1, we still get the same result of x being the distance from positive 5 rather than negative 5:
|x-5| >> |1-5| >> |-4| >> 4. 1 is 4 units from 5. (But 6 units away from -5.)
This really will work with any value for x: a value greater than 5, a value less than 5, a negative number, or a positive number. If you still have some doubts about the number properties demonstrated above, practice plugging other values in for yourself.
The same principle is the reason |x+5| is the distance of x from negative 5. Let’s plug in two different random numbers for X. First, let’s say that x= -100. Here’s how that plays out:
|x+5| >> |-100+5| >> |-95| >> 95. -100 is 95 units away from -5. (But would be 105 units away from positive 5.)
Another way to think of it is this: |x-5| means that |x-5| is a certain number greater than, less than, or equal to 0. It’s an unknown equality, expressed as: |x-5| ? 0. Recall that inequalities can be treated like regular algebraic equations. So to get x alone on one side of the equation, you want to eliminate -5. You do that by adding +5 to both sides: |x-5+5|?0+5 >> |x|?5, with “?” representing not just inequality to 5, but specific distance in either direction from 5. Similarly:
|x+5|?0 >> |x+5-5|?0-5 >> |x|?-5 .
Hope all this helps. 🙂
4. Linda April 17, 2018 at 1:50 pm #
I have a question, the lesson shows an example
| x – 1 |> 4, the results are x> 5 and x ,<) changed. Because the rules to change the sign only apply to division and multiplication with negative numbers.
• Magoosh Test Prep Expert April 20, 2018 at 2:44 pm #
Hi Linda,
Since you’re a Premium student, I forwarded your message on to our team of tutors. You should hear back from them soon! As a reminder, you can always get answers more efficiently by using the Purple “Help” button in your Premium account 🙂
• Hari June 28, 2018 at 11:24 pm #
Hey! The sign has to be changed because you’re actually multiplying “-1” to the expression in the right-hand side. For example:
|x-3| > 3, when you take out the absolute value, it becomes:
x-3 > 3 and x-3 < -3.
Hope this helps! 🙂
5. MUQADAS AFGAN November 3, 2016 at 7:23 am #
Chris,
I am a Magoosher and I have a doubt here. If we have an equation like |x-5| + 3 = 8. then how to solve this for the range of x.
Following your method of solving the equation, i reached the conclusion x= 10 or -6 however the answer chances given as follows in the O’Level math book are as follows.
|x-5| + 3 = 8
A. x=10
B. x= 10 or x=0
C. x=10 or x=-10
D. x=0 or x=-10
I solved as follows:
|x-5| + 3 = 8 or |x-5| + 3 = 8
x-5+3=8 or x-5+3=8
x =10 or x =-6
The correct answer as per the book is B.
Can you please explain how we reached this result.
• Magoosh Test Prep Expert November 3, 2016 at 5:49 pm #
If you get a problem like this, you want to try to make your life easier and get the absolute value onto one side of the equation by itself:
|x-5| + 3 = 8 [Subtract “3” from both sides]
|x-5| = 5
Then, we have:
“|x-5| = 5” OR “|x-5| = -5”
|x-5| = 5 [Remove absolute values sign (as we have both equations now)]
x-5 = 5 [Add “5” to both sides]
x = 10
|x-5| = -5 [Remove absolute values sign (as we have both equations now)]
x-5 = -5 [Add “5” to both sides]
x = 0
So, I agree that the correct answer is B.
• MUQADAS AFGAN November 4, 2016 at 5:00 am #
Thanks Chris for the prompt reply.
Kindly take a look at another example question 5-3 |x+5| = -4 which I am currently trying to solve by the above-mentioned method to results that aren’t matching any of choice.
The answer choices are as follows:
A. x= 4 or x= 10
B. x= -4 or x= -10
C. x= 4or x= -4
I solved it as follows:
5-3 |x+5| = -4
2 |x+5| = -4 Subtracted -2 from both sides – |x+5| = -6
x+5 =-6 or x+5 = +6
The answers were -13 and -1 but the true answer option by the book is B.
• Magoosh Test Prep Expert November 6, 2016 at 7:53 pm #
I’m a bit confused by this example–I think you may have copied something incorrectly in the question itself. I didn’t get an answer included in the choices.
The problem I see with your answer is that you added and subtracted unlike terms. From what I see, the 3 is multiplied by |x+5|, which means that we can’t subtract 3*|x+5| from 5 (this would be like doing 5-3x=2x– there is no x in the 5 so they aren’t like terms and we can’t do it!)
This is how I solved the problem using the method above:
5-3*|x+5|=-4
-3*|x+5|=-9
|x+5|=3
So we have |x+5|=3 OR |x+5|=-3
For the first equation:
x+5=3
x=-2
For the second equation:
x+5=-3
x=-8
So I’m not sure if the question was copied down incorrectly or not, but I didn’t get any of the answer choices here! I recommend that you double check the problem to see what might be going on, and see if my method makes sense to you. Remember: you can’t add or subtract different terms! 🙂
• Adrienne May 29, 2017 at 12:36 pm #
I’m a bit confused by the above example and how the answers are not listed.
I’m wondering: does this concept, “the absolute value of any number can never be a negative therefore there is no value for x” not apply in this situation because carrying out/simplifying the operations on the left side produces a positive number on the right side once executed?
• Magoosh Test Prep Expert May 31, 2017 at 1:11 pm #
So, I think I understand where your confusion is coming from here! So, please note that you cannot have an absolute value expression alone on one side of the equation EQUAL TO a negative number alone on the other side. For example, |x + 5| = -8. This does NOT work. As you’ve pointed out, an absolute value cannot equal a negative number as the absolute value expression will output a positive number.
That being said, if there is a negative expression with the absolute expression on one side of the equation, then it’s possible for the other side of the equation to be negative. For example, |x + 5| – 20 = -8. In this case, this is possible. But, you have to simplify to confirm.
I hope this helps! Have a great day! 🙂
6. Ketki November 2, 2016 at 2:57 pm #
Hi Chris,
How can we solve the Quantitative comparison for absolute value solution and 0?
Eg: |2n+1| < n +6
Quantity A: n
Quantity B: 0
This problem is in Official Guide.
• Magoosh Test Prep Expert November 5, 2016 at 10:20 pm #
Hi Ketki,
In this case, I would recommend plugging values in for n to see what the relationship might be. This has an absolute value and an inequality, so it’s a difficult question! But, if we can determine that there are values that satisfy n that are greater than and/or less than 0, we can figure out relationship out. I can see pretty quickly that I can plug “2” in for n and get 5<10, so this inequality holds true for a value greater than 0. I can also see that if I plug in -1 for n, I get 1<6 which also holds true. Since there are values above and below 0 that satisfy this inequality, we know that the answer must be (D).
For more information on the plugging in strategy, see the following blog section: https://magoosh.com/gre/2011/part-i-the-power-of-plugging-in-%E2%80%93-gre-math-techniques/
7. Sakshi September 13, 2016 at 7:11 am #
Hi,
Suppose that |x|<|y+2|0 and xz>0. Which of the following could be true?
Indicate all statements that apply.
a) 0<y<x<z
b) 0<x<y<z
c) x<z<0<y
d) 0<y+1.5<x<z
e) z<x<0<y
• Magoosh Test Prep Expert September 14, 2016 at 1:12 pm #
Hi Sakshi,
I’d be happy to help you with this question, but could you let me know where it’s from? Please note that we generally don’t answer questions from outside materials but rather prioritize Magoosh and Official Guide questions 🙂
Also, it’s a great help to us in giving explanations if you elaborate a little about your thought processes. How did you approach the question? Where did you get stuck? This helps us to make clear, concise responses.
Looking forward to hearing from you!
8. Sayli August 29, 2016 at 12:09 am #
Hey ,
I just want to clear my doubt .
The absolute value of a number is always positive , whereas when the variable is there it can be negative or positive . Right?
|4| = 4 , |-4| = 4
|x-4| = (x-4) or -(x-4)
• Magoosh Test Prep Expert August 30, 2016 at 3:23 pm #
Hi Sayli 🙂
Happy to help! You’re headed in the right direction with your thinking, but let’s tighten it up 🙂
|something| is the distance of (something) from zero. It’s a distance, which means it’s always positive.
As you mentioned, |4| = |-4| = 4. And another way of saying this is that both 4 and -4 are a distance of 4 units from 0.
Variables
Now, let’s look at what happens when we have a variable inside the absolute value sign. For example,
|x| = |-x|
But do these equal x or -x? Well, it depends on whether x itself is negative or positive. Remember, -x doesn’t mean a negative number necessarily. It means the “opposite of x.”
So if x = -3, then -x = -(-3) = 3
If x is a negative number, then -x is a positive number. So if x is negative, then |x| = -x. This idea reflects the example you mentioned 🙂
Hope this helps!
9. John July 17, 2016 at 11:33 pm #
This is the problem |3 + 3x| < -2x
Quantity A Quantity B
|x| 4
so, x < -3/5 <<>> x>-3 ( 5>-3 is true here )
how can we say => -3/5 > x > -3 (This means x < -3/5 <<>> x>-3)
and conclude that Quantity B is greater
• Magoosh Test Prep Expert July 21, 2016 at 4:12 am #
Hi John,
Happy to help 🙂 When we solve this absolute value inequality, we get find, as you mention,
-3 < x < -3/5
In words, x is greater than -3 and less than -3/5. So, there is a range of values from -3 to -3/5 (these endpoints not included) that x could be and satisfy the original inequality. Given this range, |x| < 3, since x < -3 and |-3| = 3. And for that reason, we can conclude the Quantity B is greater 🙂
I hope this helps!
10. Tee May 18, 2016 at 9:37 pm #
I thought you only switch the sign when multiplying/dividing by a negative, not adding/subtracting?
• Magoosh Test Prep Expert May 25, 2016 at 1:57 am #
Hi Tee 🙂
You’re correct that switch the sign of an inequality when we multiply or divide by a negative number and we keep the sign the same when adding or subtracting. We can see these ideas applied in the practice problem in the post 🙂
|x-4| < 3
To solve for x, we want to look at the positive and negative of the value within the absolute value:
1. x – 4 < 3
2. -(x – 4) < 3
As outlined in the post, for (1), we solve for x by adding 4 to both sides:
x – 4 + 4 < 3 + 4
x < 7
For (2), we can first multiply both sides by (-1), which entails flipping the sign of the inequality. From there, we will add 4 to both sides to solve for x:
-(x – 4)*(-1) < 3*(-1)
x – 4 > -3 [notice how we flipped the sign of the inequality!]
x – 4 + 4 > – 3 + 4
x > 1
Combining these two results, we find that 1 < x < 7 🙂
I hope this clears up any confusion! Happy studying!
11. Sayahnita February 26, 2016 at 3:26 am #
Hi Chris,
I am a Magoosher. Could you please delineate the concept of extraneous roots?As it was mentioned by Mike that ‘extraneous solutions are invalid and do not solve the original equation’ in a lesson video!!
• Magoosh Test Prep Expert March 3, 2016 at 3:13 am #
Hi Sayahnita,
Happy to explain! 🙂
An extraneous root is one that appears as a solution, but upon checking with direct substitution is in fact not a solution.
For example: you have an equation and after some work come up with two roots which we can juts call “a” and “b”. When you put “a” into the original equation it works and is true, but when you put in “b” it doesn’t work at all! So “b” is an extraneous root because while you arrived at it in a mathematically valid way, it isn’t actually a root. This often happens when we square both sides during our solution because, as you know, this affects negative numbers. (The square of a negative number is positive.)
Imagine we had to solve √(2x + 7) + 4 = x.
√(2x + 7) = x – 4
2x + 7 = (x – 4)^2
2x + 7 = x^2 – 8x + 16
x^2 – 10x + 9 = 0
We can factor this, which gives us (x – 9)(x – 1) = 0. Thus, 9 and 1 are roots, but we now have to go back and check them in the original equation. If you plug x = 1 back into the original equation, you will see that the equation DOES NOT hold true. Hence, this is an extraneous root. If you plug x = 9 back into the original equation, you will see that the equation DOES hold true, so that one is your answer.
For 1:
√(2x + 7) + 4 = x
√(2(1) + 7) + 4 = 1
√(9) + 4 = 1
3 + 4 =/= 1
For 9:
√(2x + 7) + 4 = x
√(2(9) + 7) + 4 = 9
√(18 + 7) + 4 = 9
√(25) + 4 = 9
5 + 4 = 9
I hope that helps!
• Sayahnita March 7, 2016 at 10:43 pm #
Wow,gee thanks,you guys are the best 🙂
12. meaad November 8, 2015 at 12:27 am #
I have a question about absolute value equation,
if we get two negative solutions for X, so we directly can say there is no solution for absolute value equation
13. Anusha Komati September 6, 2015 at 6:02 am #
Why we need to solve in two Equations a, + 3 B, -3 to find the absolute values? Is it to know the number of units?How about the Below one?
Quantity A: l m+25 l
Quantity B: 25 – m
Thank You.
14. Pranesh August 27, 2015 at 2:16 am #
hello chris i am a magoosher, could you please explain the following ; |-R/4+6|>2 and
|-R/4+6|>-2
• Jessica Wan August 27, 2015 at 10:35 am #
Hi Pranesh!
Thanks for pointing out that you’re a Magoosher! 🙂
Just wanted to let you know that I’ve forwarded your question to our team of Remote Tutors. Chris would love to answer every question, but he has a lot of projects going on right now and he’s just one person! Someone from the tutor team will email you directly to follow up.
All best,
Jessica
15. Mudit January 4, 2015 at 10:05 am #
Hi Chris
I am a Magoosh premium member. I really like your blogs and videos. Great work!
One thing I did not understand in the above blog was its last part. When solving x-4=-3, why did you reverse the inequality sign? Clearly the equation does not look like its getting multiplied or divided by -1, since that would have affected both the sides.
• Chris Lele January 7, 2015 at 12:12 pm #
Mudit,
When we make the number to the right of the equal sign a negative number (in this case -3), it reverses the direction of the sign. Take the following equation:
|x + 1| < 2
One value of 'x' has to less than 1 (we can solve this the traditional way).
Now we have to find the value of x that would make the equation x + 1, now without the absolute value sign, come out to less than -2. See, whatever value that comes out of the absolute value equation that is between 0 and -2 will yield a positive number (remember the absolute value sign). For that reason we set the number equal to -2. But in doing so, we have to flip the direction of the sign. Remember when we solved for the positive value of x, we get a number less than 2. To say that x has be a number less than -2 would not overlap (these statements are mutually exclusive). By turning the direction of the sign, when we make the number -2, we make it so 'x' is between -2 and 2, which is a possible outcome.
Hope that helps!
• suryansh July 1, 2015 at 1:11 am #
its simple that why reverse the inequality sign . for example
2>1
now multiply by -1 on both sides
-2>-1 is wrong
so we reverse the sign
-2<-1
now do u get it
16. Vikas August 23, 2014 at 1:38 pm #
Hi Chris,
Can you please share me the magoosh link for the chapter “sequences”. I tried in Algebra and Math Basic sections but didn’t get.
Thanks,
Vikas
• Chris Lele August 25, 2014 at 3:03 pm #
Hi Vikas,
Are you talking about the eBook?
17. Spondita August 2, 2013 at 10:58 pm #
In you above explanation to the question: What is the value of x in the equation delim{|}{x}{|}=5. Why there is no value of x? Can’t x be -5 and 5 in this?
18. Taruna July 29, 2013 at 3:15 pm #
Is it alright to apply these strategies with quadratic inequalities?
19. Peter July 23, 2013 at 7:37 am #
Where can i find some harder problems on absolute values
• Chris Lele July 23, 2013 at 10:35 am #
Hi Peter,
The Magoosh products offers some, the Manhattan 5 lbs. GRE book is also a great source, and the NOVA book (old GRE only) may have a few.
Hope that helps!
20. Shubham July 11, 2012 at 3:14 pm #
Chris,
I have a doubt here. If we have a equation like |x-1| + |x-2| = 5. then how to solve this for range of x.
• Chris July 12, 2012 at 5:53 pm #
Hi Shubham,
All you have to do is remove the absolute value signs and solve for x, giving us x = 4. Then, remove the absolute value signs but this time make 5 negative, giving us x = -1.
Hope that helps!
• Divya July 20, 2013 at 9:40 pm #
hi! in the previous question,can i know why 5 is taken as -5?
• Chris Lele July 22, 2013 at 2:15 pm #
Yes, whenever you are dealing with absolute value signs, you always want to make the number to the right of the equal sign negative to solve for the second value of the absolute value.
|x – 4| = 1
To solve, we remove the absolute value signs and create two equations:
x – 4 = 1
x – 4 = -1
Then solve for x to find the two values of ‘x’.
Hope that helps!
• Rishabh July 28, 2013 at 6:40 am #
In the above explanations when we do have two absolute values, wont the negative and negative become positive and we should only consider the positive on the right hand side? Or is it always that you consider the RHS to be -ve and +ve both irrespective of the number of absolute values on the LHS.
• payal October 19, 2015 at 10:00 am #
Thank you- you simplify this!!
• Dulce Sanchez August 25, 2019 at 4:57 pm #
Hi Chris!
So just to make sure when you are solving the negative absolute value, you never make the x negative?? we just make negative the other side of the inequality?
• Magoosh Test Prep Expert December 6, 2019 at 6:32 pm #
Hi Dulce! Yes, that’s correct. If you have something like |-x| = 5, then you’d split the equation in two: -x = 5 and -x = -5. We didn’t change the negative on the x, though in cases like this these equations are equivalent to x = 5 and x = -5. It just depends on whether we divide through by -1 or not. All that means is that switching x to -x or -x to x wouldn’t really make a difference. It’s just important to do what we need to do for all absolute value equations. Make two equations, one with the original value and another with the negative of that term or expression. Also be sure to flip the inequality sign for the negative version. I hope that helps!
• Amey Kelkar August 9, 2013 at 4:59 am #
Really helpful Chris, Thanks a lot! 😀
• gvk July 13, 2015 at 1:15 pm #
I think it’s not so straight to solve |x – 1| + |x – 2| = 5 by just reversing signs. The approach doesn’t seem to be extendible to an example like |x – 1| + |x – 2| + |x – 3|+ |x – 4| .. = 5
The steps should be:
1. Split into parts
3. Find solution and then validate it
Explanation: http://www.purplemath.com/modules/solveabs2.htm
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https://math.stackexchange.com/questions/1084854/sum-of-a-decreasing-geometric-series-of-integers | # Sum of a decreasing geometric series of integers
I'm trying to calculate the sum of integers that are being divided by two (applying the floor function if necessary): $n\mapsto \lfloor \frac{n}{2}\rfloor$.
Let $S(n)=n+\lfloor\frac{n}{2}\rfloor+\left\lfloor\frac{\lfloor\frac{n}{2}\rfloor}{2}\right\rfloor+\ldots$.
For example, \begin{align*} S(100) &= 100 + 50 + 25 + 12 + 6 + 3 + 1 + 0 +\ldots\\ S(3) &= 3 + 1 + 0 + \ldots\\ S(1000) &= 1000 + 500 + 250 + 125 + 62 + 31 + 15 + 7 + 3 + 1 + 0 + \ldots\end{align*}
I'm trying to find a closed form for $S(n)$, where $n\in \mathbb N$. Any ideas?
[Solution] A lot of great answers. Thanks! Here's my java implementation.
int computeHalvesSum(int n) {
return 2 * n - Integer.bitCount(n)
}
• What exactly do you mean by analytic solution of S(n)? Because this could easily be expressed using sigma notation and a recursive function, but I don't think that is what you are after. – Dasherman Dec 29 '14 at 18:38
• Yeah "analytic solution" might not be the right translation for what I have in mind. Let's say that I would like to be able to implement an algorithm that compute the solution in constant time. Is it more clear? – user1534422 Dec 29 '14 at 18:45
See OEIS sequence A005187 and references there. Depending on what language you're using, the simplest way to compute it may be as $2n - (\text{sum of binary digits of }n)$.
I don't know if it meets your criteria for an analytic solution, but $$S(n)\sum_{k=0}^{\lfloor \log_2 n\rfloor} \left\lfloor\frac{n}{2^k}\right\rfloor$$
The calculation of $\log_2$ probably means this is not worth implementing (you added that comment while I was writing my answer).
Let $n$ be the number in question, and $m$ be the number of $1$'s in the binary expansion of $n$. For example, if $n=100_{10}=1100100_2$, so $m=3$.
Then the sum you seek is $$S(n)=2n-m$$
This is not constant time in $n$, but it is $O(\log n)$, which is pretty good. I doubt you'll be able to do any better than that, since just writing down the answer takes $O(\log n)$ time.
• It takes $O(\log(n))$ simply to scan the digits of $n$, so you can't do better than that. – Robert Israel Dec 29 '14 at 18:56
Let $f(n)$ be the number of $1$s in the binary representation of $n$. Then $S(n)=2n-f(n)$.
To see this: If there is a $1$ in the $2^k$ place in the binary expansion of $n$, then that $1$ contributes $2^k+2^{k-1}+\cdots+2^1+2^0=2^{k+1}-1$ to $S(n)$.
Each contribution is one less than twice the place value of its $1$. So the total contribution is $2n$ minus the number of $1$s in the binary representation of $n$. | 2019-08-23T06:50:06 | {
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https://math.stackexchange.com/questions/2731817/proof-that-finite-union-of-countable-sets-is-countable | # Proof that finite union of countable sets is countable
I need to prove that for countably infinite sets $A_1,A_2,...,A_m\subset\mathbb{R}$, $\bigcup A_i$ is also countable.
My Attempt: First, consider two countable sets $A_1$ and $A_2$, and define $B_2=A_2\backslash A_1=\lbrace x\in A_2 | x\notin A_1\rbrace$. We then have two cases, one where $|B_2|$ is infinite, and one where $|B_2|$ is finite. If $|B_2|$ is inifite, then $B_2$ is countable, because there exists natural numbers $n_i$ and a function $f:\mathbb{N}\rightarrow A_2$ such that $f(n_i)=b_i$ for all $b_i\in B_2$. Then, a function $g:\mathbb{N}\rightarrow B_2$ can be defined such that $g(k)=f(n_k)$, proving that $B_2$ is countable. We now prove that the union $A_1\cup B_2=A_1\cup A_2$ is countable. Because $A_1$ and $B_2$ are countable, there exists bijective functions $f_1:\mathbb{N}\rightarrow A_2$ and $f_2:\mathbb{N}\rightarrow B_2$. Let $O_k$ be the $k$th odd natural number, and $E_k$ be the $k$th even natural number. We can simply define a new funtion $g:\mathbb{N}\rightarrow A_1\cup A_2$ where $g(O_k)=f_1(k)$ and $g(E_k)=f_2(k)$. This proves that $A_1\cup A_2$ is countable for the case $|B_2|=\infty$. Next, let $|B_2|=k$ and label each element $b_1,b_2,...,b_k$. Because $A_1$ is countable, we have the same bijective function $f_1:\mathbb{N}\rightarrow A_1$. To prove that $A_1\cup B_2$ is countable. we define a new function $g:\mathbb{N}\rightarrow A_1\cup A_2$ where $g(n)=b_n$ and $g(k+n)=a_n$ for all $a_n\in A_1$. This completes the proof.
I would like to know first if this proof is correct, and second if there is a more efficient way to write all of this.
With a cursory examination, it looks like the OP's proof is in the ballpark. They asked for a proof that would be more efficient, so here we will highlight the main ideas found in the OP's proof, presented as a logical progression.
In what follows we do not assume that our sets are contained in $\mathbb R$.
Proposition 1: If $A_1$ and $A_2$ are two disjoint countably infinite sets, then the union $A_1 \cup A_2$ is countably infinite.
Proof:
Let $F_1: \mathbb N \to A_1$ and $F_2: \mathbb N \to A_2$ be bijective mappings, where $\mathbb N = \{0, 1, 2, \dots, n, \dots \}$. Define $F: \mathbb N \to A_1 \cup A_2$ as follows:
$$F(m) = \left\{\begin{array}{lr} F_1(\frac{m}{2}), & \text{for even } m \in \mathbb N\\ F_2(\frac{m+1}{2}-1), & \text{for odd } m \in \mathbb N \end{array}\right\}$$ It is easy to check that $F$ describes a bijective correspondence. $\quad \blacksquare$
The following sequence of 'add-on' propositions are not difficult to prove:
Proposition 2: Let $A_1$ and $A_2$ be any two sets. Then there exist a set $B$ satisfying the following:
$\tag 1 B \subset A_1 \text{ and } B \subset A_2$
$\tag 2 A_1 \cup A_2 \text{ is the disjoint union } (A_1 \backslash B) \cup B \cup (A_2 \backslash B)$
Proposition 3: If $A_1$ is a countably infinite set and $A_2$ is any finite set, then the union $A_1 \cup A_2$ is countably infinite..
Proposition 4: Any subset of a countably infinite set is either finite or countably infinite.
Proof (sketch)
Use the lemma found here.
Proposition 5: The union of a finite number of finite sets is finite.
We now prove the main proposition.
Proposition 6: The union of any two countably infinite sets $A_1$ and $A_2$ is countably infinite.
Proof:
By proposition 2, 4 & 5, $A_1 \cup A_2$ can be written as a disjoint union
$\tag 3 C_1 \cup B \cup C_2$
where each of the sets is either countably infinite or finite, and at least one of the sets is countably infinite.
By using the commutativity and associativity laws of the union operation and proposition 1 and/or proposition 3, we can simplify (3) in two steps so that a single countably infinite set remains that is equal to $A_1 \cup A_2$. $\quad \blacksquare$
The OP might want answer this question again at some point using the following:
Theorem: If $f$ is a surjective function from $\mathbb N$ onto a set $A$, then $A$ is either finite or countably infinite.
• I am sure that the Abbott book is referring to just the case where both sets are countably infinite. Will make edit in post. – 高田航 Apr 12 '18 at 11:42
Assume all the sets are infinite and disjoint. Convince yourself that this is the worst case; if their union is countable we are done.
I'll go beyond set theory using just very elementary facts about positive integers: (i) there are infinitely many prime numbers (ii) a number that is a power of one prime number can not be the power of another prime number (unique factorization theorem).
Let $p_1,p_2,p_3\ldots$ be the list of prime numbers in the increasing order.
Let $P_j = \{p_j^k\mid k \mbox{ a positive integer}\}$ consist of all powers of the $j$th prime. Clearly the sets $P_j$ are disjoint and are all countably infinite. And $\bigcup_{j=1}^\infty P_j$ is a (proper) subset of positive integers and hence is countable.
Now take your sets $A_1,A_2,\ldots, A_m$.
By countability hypothesis we can list the elements of $A_j$ as $\{a_{j1}, a_{j2},\ldots\}$ Now we can set up bijections this way: $f_j\colon A_j\to P_j$ by $f(a_{jk}) = p_j^k$ for each $j=1,2\ldots, m$.
So the union of $A_j$ has the same cardinality as the union of $P_j$'s and the latter is countable.
This proof works even in the case of countably many countable sets.
I'm not sure I follow your argument exactly but it does seem that you are not using the proper definition of countability. A set $S$ is countable if there is an $\textit{injection}$ $S\rightarrow\mathbb{N}$, not a $\textit{bijection}$. This is why, for example, a finite set is countable. It's also unclear to me that you aimed to prove the original claim; it seems you only proved it for two sets. Furthermore, because of this you wrote all of your functions going $\textit{from}$ $\mathbb{N}$ rather than $\textit{to}$ it, which seems more natural.
The way I would prove it is as follows: prove it for two sets, then use induction. Call the two sets $A$ and $B$. We can assume without loss of generality that they are disjoint (if not, replace them by $A$ and $B\setminus A$). Since they are both countable, we can write $$A=\{a_{1},a_{2},...\}$$ $$B=\{b_{1},b_{2},...\}$$ Note that if one or both of $A, B$ is finite, the labeling above will eventually stop i.e. you can write them all down. Clearly if $A, B$ are both finite their union is finite, hence countable. I claim that if one is countable and the other is finite, their union is countable. See if you can prove this.
The trickier case is when both are countable. You seem to have the write idea about even and odd indices. Given the labeling above, define a function $f: A\cup B\rightarrow\mathbb{N}$ by $$f(a_{k}) = 2k-1$$ $$f(b_{k}) = 2k$$ This is clearly injective, which proves that $A\cup B$ is countable. Note that injectivity relies on the assumption that $A, B$ were disjoint, which is why I made that assumption above.
The inductive hypothesis is simple: if the claim holds for $m-1$ sets then it must hold for $m$ sets because $$A_{1} \cup ... \cup A_{m} = (A_{1}\cup...A_{m-1})\cup A_{m}$$
• In my textbook "Understanding Analysis" by Stephen Abbott, it defines countability by $f:\mathbb{N}\rightarrow A$ instead of the other way around. Also, I proved it for two sets because if $A_1\cup A_2 = B$ is countable, then the $B\cup A_3$ is simply another union of two sets, and so on. – 高田航 Apr 11 '18 at 20:55
The cardinality of the intersection of any two countably cardinal sets may range from zero (when the sets are disjoint) to the minimum cardinality of the two sets (when one set is a subset of the other). Likewise we can establish the minimum and maximum cardinality for the union of any two such sets.
The union of any two countably cardinal sets must have at least a countable cardinality by definition of union. The cardinality of the union cannot be less than something because reasons.
We can (by definition of something) map elements of one of the sets to (finite or infinite) countable many even natural numbers, and likewise the other to countable many odd natural numbers, so the union of any two countably cardinal sets must have at most the cardinality of the natural numbers; which is by definition countable.
You may now use induction to show the result of a union of finitely many such sets. | 2019-09-22T18:12:11 | {
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https://math.stackexchange.com/questions/3117224/what-is-the-probability-that-the-process-stops | # What is the probability that the process stops?
Suppose there are $$1$$ white and $$2$$ black balls in a bag. We take two balls at a time at random from the bag. If the two balls are of different colour we throw them away but if they are of the same colour then we return them back in the bag and also add a new ball of different colour in the bag. We iterate this process unless the number of balls in the bag is less than $$2.$$ Then what is the probability that this process stops?
Thank you very much.
EDIT $$:$$
If $$\tau$$ is the stopping time then I found that $$P(\tau = 1) = 0,P(\tau = 2) = \frac 1 2,P(\tau = 3) = 0, P(\tau = 4) = \frac {1} {2^2}, P(\tau = 5) = \frac {1} {2^2} \cdot \frac {1} {3^2}, P(\tau = 6) = \frac {1} {2} \cdot \frac {1} {3^3}.$$
But I can't find any pattern to find $$P(\tau = n).$$ It is clear that the required probability is $$\sum\limits_{n=1}^{\infty} P(\tau = n).$$ How to find pattern to evaluate those probabilities?
• When you say "add a ball of a different color" do you mean if you grabbed two white balls you add a black ball and if you grabbed two black balls you add a white ball, or that you add an entirely new color altogether, e. g. blue? – orlp Feb 18 at 7:10
• No the previous one. If you grabbed two white balls you add a black ball and if you grabbed two black balls then you add a white ball. – Dbchatto67 Feb 18 at 7:14
• Would you please share me your intuition @orlp? – Dbchatto67 Feb 18 at 7:16
Let the situation when there are $$n_w$$ white and $$n_b$$ black balls in the bag correspond to a point with coordinates $$(n_w, n_b)$$ on a plane.
From the point $$(n_w, n_b)$$ we can jump to one of the following three points:
$$(n_w-1, n_b-1)$$ - if balls of different colors were fetched
$$(n_w+1, n_b)$$ - if two black balls fetched
$$(n_w, n_b+1)$$ - if two white balls fetched
Our point roams around the plain until it either comes too close to $$(0, 0)$$ or moves away to infinity.
My intuition tells that the point would not run away to infinity.
If there are much more white balls than black balls than the number of black balls would increase, so the ratio $$n_w/n_b$$ would approach 1.
If we have a large number of black balls and approximately the same number of black balls, than with probability 1/2 the number of balls would decrease by 2, and with probability 1/2 it would increase by 1. So, the total number of balls tends to decrease.
As the point would not run away to infinity it would roam around $$(0, 0)$$ indefinitely and sooner or later with probability 1 would hit one of the final points: (0, 0), (0, 1), (1, 0).
More rigorous proof would be to analyze the behavior of function $$max(n_b, n_w)$$. After each step it can either increase or decrease by 1, or remain unchanged. It is easy to prove that probability it decreases is always higher than the probability it increases. That means that our point would not run away to infinity.
I understand this is not a rigorous proof, but the question is interesting and there was no other answers for a while.
So, the final answer is: the probability iterations stop is 1.
Simple computer simulation confirms this answer.
• That is not actually a textbook problem @lesnik. That question actually came into my mind when I was sleeping at night after lots of problem solving in probability before the day of my exam. – Dbchatto67 Feb 19 at 7:54 | 2019-08-19T21:32:21 | {
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http://ylpw.flcgorizia.it/a-coin-is-tossed-three-times-what-is-the-probability-of-getting-3-heads.html | ## A Coin Is Tossed Three Times What Is The Probability Of Getting 3 Heads
We test the hypothesis that the probability the coin lands heads when tossed is 0. The toss of a coin, throwing dice and lottery draws are all examples of random events. One is a two-headed coin ( having head on both faces ), another is a biased coin that comes up heads 75% of the times and third is also a biased coin that comes up tails 40% of the times. If three fair coins are tossed randomly 175 times and it is found that three heads appeared 21 times, two heads appeared 56 times, one head appeared 63 times and zero head appeared 35 times. If you toss a coin, you cannot get both a head and a tail at the same time, so this has zero probability. What is the probability of the coin landing heads up three times? I need somebody to explain to me, how to do the full process, so I know how to do it for other math problems as well. If a coin tossed for $3$ times what is the probability of getting all tails? tossing the coin three times could give the following combinations: probability. These same rules of probability allow us to calculate the odds of parents conceiving particular numbers of girls or boys or of predicting the likelihood that specific chromosomes will segregate together into the same gamete. How many of these 32 outcomes contain exactly 3 heads? When we have three heads, we must also have exactly three tails, so your goal is to determine how many combinations of this there are. The probability of 4 heads would be: 10 C 4 (1/2)^4 (1/2)^6 You can just add up 3, 4, 5 to get the probability of at least 3 heads, or you can compute the probability of 0, 1, 2 heads and subtract it from 1. Get an answer for 'The probability that a coin turns up heads when it is tossed is 1/2. Flip a coin. SOLUTION: A fair coin is tossed four times. Each time a fair coin is tossed, the probability of getting tails (not heads) is 1/2 = 0. We write P(heads) = ½. Often it is required to compute the probability of an event given that another event has occurred. 1 question 6 A coin is tossed three times, where. Assuming the coin is fair, each of these outcomes is equally probable. 8 if 3 heads occur Rs. 5% 2 tails and there is 12. in case you propose you've #a million head #2 tails #3 heads, then this may be the answer: First toss = a million/2 2d toss = a million/2 0. Algebra -> Probability-and-statistics-> SOLUTION: A. Solution:. Toss coin 1 first. Three fair coins are tossed at the same time. A fair coin is tossed three times. The coin shows heads every time. (a) What is the probability that the number of heads is k? (b) Find the probability that coin 1 was tossed given that k heads were observed, for k = 0, 1, 2, 3. In order to find the probability of all three events happening, we MULTIPLY the probabilities of each of the events. the probability of getting heads-heads-heads if you toss a coin three times is 1 out of 9. 7 percent chance to win, but Shane Buechele led touchdown drives of 19 plays, 10 plays and 14 plays on consecutive. 3 if 2 heads occur, Re. Find the probability that the flrst coin is heads given that at least one head occurred. What is the probability that heads occurs exactly 5 times if it is known that heads occurs at least three times? Answered: A fair coin is flipped six times. , The probability of rolling an even number on a normal 6-sided number cube , The probability of flipping a coin 4 times in a row and getting heads each time >,. the probability of getting head; A coin is tossed three times. and to have 1. ' and find homework help for other. The roll has 6 sides, the probability of rolling a six when the die is rolled once is 1/6. Since each and every outcome is equally likely. A fair coin is tossed 5 times. If it's an unfair coin (e. The third row says that if we toss three coins, we have one chance of getting all heads, three chances of getting one head and two tails, three chances of getting two heads and one tail, and one chance of getting three tails. The ratio of successful events A = 4 to the total number of possible combinations of a sample space S = 8 is the probability of 2 tails in 3 coin tosses. What interval does the value of (the true probability of obtaining heads) lie within if a confidence level of 99. If the coin is tossed twice, find the probability distribution of number of tails. 3 10 Illowsky et al. Two sets of trials are shown. Probabilities can also be thought of in terms of relative frequencies. Sample space = {0, 1, 2, 3}. a fair coin is tossed in the air 4 times. If all three coins are unbiased, the probability of the three heads is probability that the first toss is a head x the probability the second is a head x the probability the third coin is head. What is the probability, P(k), of obtaining k heads? There are 16 different ways the coins might land; each is equally probable. (a) What is the Posted 4 years ago. Let the random variable X represent the number of times a fair coin needs to be tossed till two consecutive heads appear for the first time. When a certain coin is flipped, the probability of heads is $0. There are 3 coins. Last time we talked about independence of a pair of outcomes, but we can easily go on and talk about independence of a longer sequence of outcomes. Since a fair coin flip results in equally likely outcomes, any sequence is equally likely… I know why it is$\frac5{16}$. For 2 heads I got 1/16. If a coin tossed for$3$times what is the probability of getting all tails? tossing the coin three times could give the following combinations: probability. A fair coin is tossed 5 times, what is the probability of a sequence of 3 heads? I can see that there are 2*2*2*2*2 possible outcomes, but how many of these include 3 heads in a sequence and why? probability self-study. Construct a. In a coin tossing game, seven tosses result in heads. 50 and the probability of getting exactly two heads is 0. Find the probability of getting exactly three heads?. 03125 (a little over 3%), the misunderstanding lies in not realizing that this is the case only before the first coin is tossed. a)Give an algebraic formula for the probability mass function of X. Otherwise, the odd man out wins — that is, you win if you got a head and both of the other players got tails, or if you got a tail and both of the others got heads. The chances are for one given coin to be heads is 1/2, so the chance for all three to have that same result would be (1/2)^3 ( as a probability tip, anytime you must consider that the result something must happen AND somethin else must happen, you multiply the odds of those two things happening; the chamces that the first coin is heads and the. 4096 number of possible sequences of heads & tails. , a double-headed coin, a weighted. Statistics4All - What are the different possible outcomes, when we toss 3 coins or a coin is tossed 3 times and how can we calculate probabilities of various events - The answer is in this video. An unbiased coin is tossed three times. The idea can be substantially generalized. One of two coins is selected at random and tossed three times. Solution:. P(getting three heads) = P(E1) Number of times three heads appeared = Total number of trials = 70/250 = 0. Since the sum of the row is 8, the probability of getting two heads and one tail is 3/8. What are the chances the coin shows tails on the next toss? Explain. For the second part of question, you are not bothered with the results of 2nd to 6th toss. What is the probability of getting head and tail alternately?a)b)c)d)Correct answer is option 'B'. Then work out the probability p of getting a See attached sheet for the probabilities. Examples: In the experiment of flipping a coin, the mutually exclusive outcomes are the coin landing either heads up or tails up. Len tosses a coin three times. Let's return to the coin-tossing experiment. 3, 8 Three coins are tossed once. asked • 09/06/16 You flip a coin three times. Let X : Number of times we get tails Tossing a coin is a Bernoulli trial So, X has a binomial distributio. Compound Events 3 5. I assume the following is true: assuming a fair coin, getting 10 heads in a row whilst tossing a coin does not increase the chance of the next coin toss being a tail, no matter what amount of probability and/or statistical jargon is tossed around (excuse the puns). From a pack of 52 cards, two cards are drawn together, what is the probability that both the cards are kings. In the table below, we list the eight outcomes, the number of heads and the number of runs in the outcome and the probability of the outcome. Answer to A coin is tossed four times, and the sequence of headsand tails is observed. I had not heard of that until you mentioned it. what's the probability of getting three heads? 3. Then the possible values of X are 0,1,2 and 3. Solution: Total number of trials = 250. You need to be more specific bc the probability of getting heads 3 times in a row is not at all 50/50 the probability of what you're going to get on the second and third flip changes depending on the outcome of the previous flips. Take, for example, a coin toss. What is the probability that the coin will land tails on the third toss, given that heads were thrown on the first. For example, P(1 ≤ X ≤ 3) (i. If we toss a coin three times, there are 8 possible outcomes. Therefore, the probability of throwing exactly two heads in three tosses of the coin is 3 out of 8, or or the decimal equivalent of which is 0. A fair coin is tossed three times. 1) find the mean number of heads in 3 tosses of. Probability of exactly one head from three tosses. 5% 2 tails and there is 12. (a) What is the Posted 4 years ago. The events A, B, and C are defined as follows: A: {At least one head is observed} B: {At least two heads are observed} C: {The number of heads observed is odd} Find the following probabilities (note: 0 is an even number; "and", "or", show more A fair coin is tossed three. 3-a coin is tossed three times, what is the probability of tossing exactly two heads? Algebra Linear Inequalities and Absolute Value Theoretical and Experimental Probability 1 Answer. Probability of getting exactly one tail. The events A, B, and C are defined as follows: A: {At least one head is observed} B: {At least two heads are observed} C: {The number of heads observed is odd} A fair coin is tossed 3 times. The number of times a coin is tossed does not alter the probability of getting heads, which is 50% in every case, as long as the coin has not been rigged (i. The probability of each of these events is 1/2. What is the probability of getting two heads and four tails?. A fair coin is tossed three times. The events A, B, and C are defined as follows: A: {At least one head is observed} B: {At least two heads are observed} C: {The number of heads observed is odd} Find the following probabilities (note: 0 is an even number; "and", "or", show more A fair coin is tossed three. If a fair coin is tossed five times, what is the probability of tossing exactly three heads? You toss a coin 5 times. While a run of five heads has a probability of 1 / 32 = 0. We could run a million coin tosses, and find out how likely each scenario is. A coin is tossed 5 times. A coin flip: A fair coin is tossed three times. Here you could get 0 heads, 1 heads, 2 heads or 3 heads, so we write the sample space as. What is the probability of the coin landing heads up three times? I need somebody to explain to me, how to do the full process, so I know how to do it for other math problems as well. 5 making 25 percent for no tails, or no heads. Find the probability of getting exactly 3 heads at least 3 heads - Answered by a verified Math Tutor or Teacher We use cookies to give you the best possible experience on our website. If you toss a coin three times, there are a total of eight possible outcomes. (d) A family has five children. 5 percent of getting no heads in three tosses Getting two head require 50 percent of 50 percent because we need two head out of 3 in any order there fore it is 32. With coin tosses, etc, the outcomes depend largely on a random choice between two or more possibilities. Statistics4All - What are the different possible outcomes, when we toss 3 coins or a coin is tossed 3 times and how can we calculate probabilities of various events - The answer is in this video. Imagine flipping a coin three times. If you flip one coin four times what is the probability of getting at least two tails? What is the theoretical probability of getting k heads from n coin flips? What is the expected standard deviation of a single coin flip, where heads = 1 and tails = 0?. Do they form a set of mutually exclusive and exhaustive events? If 3 coins are tossed , possible outcomes are S = {HHH, HHT, HTH, TH. More than 3 heads I don't know how to start that problem. Users may refer the below solved example work with steps to learn how to find what is the probability of getting at-least 5 heads, if a coin is tossed ten times or 10 coins tossed together. If two coins are flipped, it can be two heads, two tails, or a head and a tail. The third row says that if we toss three coins, we have one chance of getting all heads, three chances of getting one head and two tails, three chances of getting two heads and one tail, and one chance of getting three tails. When a certain coin is flipped, the probability of heads is$0. Knights slay GCMSDown go the champs. The roll has 6 sides, the probability of rolling a six when the die is rolled once is 1/6. A fair coin is tossed three times, and we would like to know the probability of getting both a heads and tails to occur. What is the probability of the event E= {exactly 3 heads occur}? If a fair coin is tossed 4 times, what is the probability that you will get 3 heads? If a fair coin is tossed 3 times what is the probability of getting at least 2 heads?. An easier way would be to do a normal approx. the probability of getting at least 3 heads when flipping a 2 sided coin 4 times? Ans: P(x>3 head)= 1- p(x=0)+p(x=1)+p(x=2) Solve this n u will get the answer. Let X : Number of times we get tails Tossing a coin is a Bernoulli trial So, X has a binomial distributio. 5 of coming up heads. Let the random variable X represent the number of times a fair coin needs to be tossed till two consecutive heads appear for the first time. The probability of getting heads three times in 5 tries is 10/32. Let's use a symbol P(N,K) for this probability. Getting Two Heads in Four Tosses of a Coin Date: 05/17/2000 at 22:01:23 From: Melissa Subject: Probability of two heads on four tosses Dear Dr. (a) What is the Posted 4 years ago. Algebra -> Probability-and-statistics-> SOLUTION: A fair coin is tossed 5 times. For the coin, number of outcomes to get heads = 1. If a coin tossed for $3$ times what is the probability of getting all tails? tossing the coin three times could give the following combinations: probability. Any time you flip the coin you have a 50/50 chance of getting heads. Your answer would be 2/3 because in this we have two coins but the probability of being a head comes out from three ways as we are left with one coin and it may be a H-T coin or H-H coin so we are left with three position that is H H T so probability of getting a Head is 2/3. If the coin is tossed 7 times, there are 2^7 = 128 possible outcome, and just one of them is all heads. A die is rolled 1000 times with the results given in the table. More generally, if we have a situation (a "random process") in which there are n equally likely outcomes, and the event A consists of exactly m of these outcomes, we say that the probability of A is m/n. Two sets of trials are shown. What is the probability of getting heads exactly 3 times? Asked In Gate CHANDAN KR SINGH (8 years ago) Unsolved Read Solution (5) Is this Puzzle helpful? (7) (0) Submit Your Solution Probability. Probability of getting exactly two heads. 3 coins are tossed. What interval does the value of (the true probability of obtaining heads) lie within if a confidence level of 99. Probability of a statement S: P(S) denotes degree of belief that S is true. Assuming the coin is fair, each of these outcomes is equally probable. 3 10 Illowsky et al. As the number of tosses increases, the proportion of heads approaches 1/2. Construct a. If the coin is tossed two times and you want the probability of getting 2 heads, that's the probability of getting a head on the first toss AND getting a head on the 2nd toss. So there is a probability of one that either of these will happen. Statistics4All - What are the different possible outcomes, when we toss 3 coins or a coin is tossed 3 times and how can we calculate probabilities of various events - The answer is in this video. Thus, we get 1/2. The toss of a coin, throw of a dice and lottery draws are all examples of random events. So five flips of this fair coin. 1 Answer to A fair coin is tossed three times. The above explanation will help us to solve the problems on finding the probability of tossing three coins. A probability of zero means that an event is impossible. So, logically, the probability that he will get a head the first time is: 1/2 The second time he flips, the scenario's the same, and so with the third. While a run of five heads has a probability of 1 / 32 = 0. based on these results, what is the Two heads: 29 Two tails: 24 One head, One tail: 46 Answer the following questions based on the data you. the probability that you will get at least one head) = P(1) + P(2) + P(3) = 3/8 + 3/8 + 1/8 = 7/8. What is the probability of getting (i) all heads, (ii) two heads, (iii) at least one head, (iv) at least two heads? Sol. Expected Tosses for Consecutive Heads with a Fair Coin Date: 06/29/2004 at 23:35:35 From: Adrian Subject: Coin Toss What is the expected number of times a person must toss a fair coin to get 2 consecutive heads? I'm having difficulty in finding the probabilty when the number of tosses gets bigger. You flip a coin three times. A fair coin is tossed 10 times. The sample space in this case is the different numbers of heads you could get if you toss a coin three times. For 2 heads I got 1/16. Let's write down all 16 but group them according to how many heads appear, using the binary notation 1 = heads, 0 = tails:. A fair coin is tossed 3 times? A fair coin is tossed three times. Another example : What is the probability of getting two sixes when a dice is tossed twice?. Sara tossed a fair coin five times, and Kaleb tossed a fair coin three times. (Hint: Drawing a sample space will help). if a fair coin is tossed 4 times, what is the probability of. The probability of not getting a 1 on any of the three throws is 5/6 x 5/6 x 5/6 = 125/216. In fact, after 1 million flips the number of heads and tails could differ by as much as 1 or 2 thousand. What is the probability of getting (i) three heads, (ii) two heads, (iii) one head, (iv) 0 head. Now I pick up one coin and toss. A fair coin is tossed 10 times. Let X be the number of heads obtained. We define the random variable X as the number of number of times the cap drops with the open side up. 5 Probability of getting a tail: 0. Definitions. the probability of getting head; A coin is tossed three times. 1) find the mean number of heads in 3 tosses of. Find the probability of getting a tail. A bottle cap is tossed three times. While a run of five heads has a probability of 1 / 32 = 0. Let X = number of times the coin comes up heads. The sample space in this case is the different numbers of heads you could get if you toss a coin three times. If we throw the coin three times, the possible results are: Three heads: 1 way Two heads and one tail: 3 ways Two tails and one head: 3 ways Three tails: 1 way Total: 8 ways So the chances of getting three heads are 1/8 (and the method for calculating this quickly is (1/2)^3). 1st coin tossed probability of getting tails is 1 in 2. One of two coins is selected at random and tossed three times. Henry rolls a number cube and tosses a coin. value for the number of heads at least as unlikely as the actual value, assuming the hypothesis is true. Write a program that simulates coin tossing. Assuming the outcomes to be equally likely, find the probability that exactly one of the three tosses is “Heads. ! Occurs when either HT or TH is tossed. What is the probability of the event E= {exactly 3 heads occur}? If a fair coin is tossed 4 times, what is the probability that you will get 3 heads? If a fair coin is tossed 3 times what is the probability of getting at least 2 heads?. The 8 possible elementary events, and the corresponding values for X, are: Elementary event Value of X TTT 0 TTH 1 THT 1 HTT 1 THH 2 HTH 2 HHT 2 HHH 3 Therefore, the probability distribution for the number of heads occurring in three coin. 4th coin tossed probability of getting tails is 1 in 2. Thought Provoker – Explain why the tree diagrams are the same for the two experiments above. What Is The Probability That The Coin Will Land Heads At Least A coin is tossed three times. What is the probability of getting two heads and four tails?. What is the probability of getting at least three heads on consecutive tosses? A. We have tacitly assumed here that the probability of heads is equal to that of tails. 5 of coming up heads. We test the hypothesis that the probability the coin lands heads when tossed is 0. there fore it is 12. " Find out the probability for getting "heads" four times during a coin flip with help from an experienced mathematics. 11/16 Consider a general task of flipping N coins and the probability of exactly K times the heads are up. The probability of tossing tails at least twice can be found by looking down the list of eight. The number of times the cap drops with the open side up is a discrete random variable (X). This is random behaviour. A balanced coins is tossed 4 times. A bag contains two biased coins: coin A shows Heads with a probability of 0. What is the probability of obtaining exactly 3 heads. Sara tossed a fair coin five times, and Kaleb tossed a fair coin three times. The probability of three heads given the biased coin is trivial: P(three heads|biased coin)=1. After the first four tosses, the results are no longer unknown, so their probabilities are at that point equal to 1 (100%). When rolled 4 times, each time has a probability of 1/6. 33 toss = a million/2 a million/2 * a million/2 * a million/2 = a million/8. A coin is biased in such a way that a head is twice as likely to come up as a tails. 25 you have a 25% of getting three heads. Assume that the coin comes up heads with probability 2/3. A bottle cap is tossed three times. It doesn't always occur, but that is our expectation. We may not see it. Assuming the outcomes to be equally likely, find the probability that the tosses are all the same. What is the probability that flipping a fair coin three times produces fewer than two heads? Express your answer as a common fraction. Probabilities can also be thought of in terms of relative frequencies. The roll has 6 sides, the probability of rolling a six when the die is rolled once is 1/6. In fact, after 1 million flips the number of heads and tails could differ by as much as 1 or 2 thousand. what's the probability of getting three heads? 3. Example – A coin is tossed three times. Users may refer this tree diagram to learn how to find all the possible combinations of sample space for flipping a coin one, two, three or four times. Specifically, when a coin is flipped twice in succession, in 1 of the 4 possible outcomes heads appeared both times. a) Let events A, B, C, and D be givenby A={. The first coin comes up heads with probability p 1 and the second coin with probability p 2 = 2/3 > p 1 = 1/3. In a coin tossing game, seven tosses result in heads. 1 shows the results of the first 50 tosses of an experiment that tossed the coin 5000 times. P(getting three heads) = P(E1) Number of times three heads appeared = Total number of trials = 70/250 = 0. The toss of a coin, throwing dice and lottery draws are all examples of random events. Word problems on coin toss probability: 1. The idea can be substantially generalized. So let's start again with a fair coin. $\begingroup$ @Ben You need to add the probability of all 4 coins showing up as heads. Statistics4All - What are the different possible outcomes, when we toss 3 coins or a coin is tossed 3 times and how can we calculate probabilities of various events - The answer is in this video. Thought Provoker – Explain why the tree diagrams are the same for the two experiments above. 1 shows the results of the first 50 tosses of an experiment that tossed the coin 5000 times. Example – A coin is tossed three times. Suppose that the coin is tossed 3 times. Suppose that the probability of getting heads on a single toss is p. 1 question 6 A coin is tossed three times, where. This is random behaviour. This means that if a coin is flipped with its heads side facing up, it will land the same way 51 out of 100 times. 3 if 2 heads occur, Re. Sorry about that. What is the probability of getting heads in the first two trials and tails in the last. Matthew Dunn. This means that if a coin is flipped with its heads side facing up, it will land the same way 51 out of 100 times. There is a 50/50 chance of it falling heads/tails. Let n(S) be the total number of ways that the coin can land in 1000 tosses. The probability is 1/8, 0. ! The event of getting 1 head. A fair coin is tossed four times, and at least one of the tosses results in heads. In my town, it's rainy one third of the days. The number of times ‘2 Tails’ appeared was three times ‘No Tail’ appeared and number of times ‘1 tail’ appeared is double the number of times ‘No Tail’ appeared. What is the probability that either Sara or Kaleb tossed exactly three heads? Express your. in case you toss a coin three times, you gets a minimum of two heads or a minimum of two tails, yet you won't be in a position to get _both_ 2 heads and a couple of tails. If the coin is fair, then by symmetry the probability of getting at least 2 heads is 50%. A fair coin is tossed 5 times, what is the probability of a sequence of 3 heads? I can see that there are 2*2*2*2*2 possible outcomes, but how many of these include 3 heads in a sequence and why? probability self-study. This discussion on A coin is tossed three times. Matthew Dunn. Find the probability distribution of the number of heads and its expectation. Suppose we plan to toss a coin 3 times and the outcome of interest is the number of heads. Even a difference of 9,000 more tails than heads would still round off to 50% after one million flips. One coin is chosen at random and tossed twice. if a fair coin is tossed 4 times, what is the probability of. The third row says that if we toss three coins, we have one chance of getting all heads, three chances of getting one head and two tails, three chances of getting two heads and one tail, and one chance of getting three tails. We may not see it. 999% is desired?. if the coin lands heads up the first three tosses, what is the probability the coin will land heads up the fourth toss? I think it is 1/2 because the coin has 2 sides and 50% it will land. So the probability that no two consecutive heads occur in n coin tosses is f(n) / 2 n. What is the Probability of Getting (k) Heads in a Row for (n) Consecutive Tosses? I asked myself a fun question after reading a post on QuantNet. Pick from the following Log On. What is the probability of getting at least one heads? Skip Navigation. (a) What is the probability of getting heads on only one of your flips?. A coin is tossed 5 times. for a coin toss there are two possible outcomes, Heads or Tails, so P(result of a coin toss is heads) = 1/2. Since the sum of the row is 8, the probability of getting two heads and one tail is 3/8. 5 (50-50 chance of getting a head on each trial), q =. A coin is tossed 4 times. b) What do you think E[X] should be. Toss it once, the odds of heads is 1/2 Toss it twice and the odds of two head is 1/4 (1/2*1/2) Toss it three times and the odds of three heads is 1/8 (1/2*1/2*1/2) You can see that the odds of all heads is (1/2)^n where n is the number of coin tosses. 3-a coin is tossed three times, what is the probability of tossing exactly two heads? Algebra Linear Inequalities and Absolute Value Theoretical and Experimental Probability 1 Answer. Here are the results of simulating the tosses 24 times: Fill-in the column at the right with either Yes or No depending on whether both heads and tails occurred or not. A discrete random variable X has a finite number of possible values. ) I came up with 7/8 Thank you. Find the probability that the flrst coin is heads given that at least one head occurred. In my town, it's rainy one third of the days. 5, or the coin should land as heads half the time. Let the program toss the coin 100 times, and count the number of times each side of the coin appears. What is the probability that flipping a fair coin three times produces fewer than two heads? Express your answer as a common fraction. a fair coin is tossed in the air 4 times. 3 coins are tossed. There are many other kinds of situations, however, where the probability of an event is not independent but dependent — that is, where the. I get head. If you flip one coin four times what is the probability of getting at least two tails? What is the theoretical probability of getting k heads from n coin flips? What is the expected standard deviation of a single coin flip, where heads = 1 and tails = 0?. So, the probabilty of getting three heads will be 1/2 * 1/2 * 1/2 = 1/8. What is the probability that flipping a fair coin three times produces fewer than two heads? Express your answer as a common fraction. The events A, B, and C are defined as follows: A: {At least one head is observed} B: {At least two heads are observed} C: {The number of heads observed is odd} Find the following probabilities (note: 0 is an even number; "and", "or", show more A fair coin is tossed three. 5 percent of getting no heads in three tosses Getting two head require 50 percent of 50 percent because we need two head out of 3 in any order there fore it is 32. Max tossed a fair coin 3 times. Suppose we plan to toss a coin 3 times and the outcome of interest is the number of heads. Write P(x) as a decimal value. What is the chance that the coin I picked has heads on both sides?. PPT – If you toss a coin, what is the probability of getting heads Tails If you toss a coin 10 times, how PowerPoint presentation | free to view - id: 12dbbe-MjNiZ The Adobe Flash plugin is needed to view this content. 1 if only 1 head occurs. The likelihood of getting "heads" four separate times is called "probability. 4 answers 4. According to ESPN's Win Probability, SMU entered the fourth quarter with a 1. If all three coins are unbiased, the probability of the three heads is probability that the first toss is a head x the probability the second is a head x the probability the third coin is head. The probability of an event occurring is a statement about the true possibility of an event, not about our observation itself ; So if we were to flip a coin, we expect heads to occur with a probability of. Assume that the coin comes up heads with probability 2/3. 6, and coin B shows Heads with a probability 0. So two possible outcomes in one flip. A coin is tossed 3 times. | 2019-11-19T12:34:56 | {
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https://math.stackexchange.com/questions/1902822/finding-roots-of-a-given-equation-when-the-given-root-begins-sqrt2-sqrt3/1902835 | Finding roots of a given equation when the given root begins $\sqrt{2} - \sqrt{3}$
$$x^6 - 4x^5 - 11x^4 + 40x^3 + 11x^2 - 4x - 1 = 0$$ and given root is $\sqrt{2} - \sqrt{3}$.
I tend to solve equations if the first given root number is not a $\sqrt{2}$ which in this case is. I understand this might be the dumbest question, nevertheless learning shouldn't stop.
Looking forward for help.
• If $x$ is a root, $\frac{1}{x}$ is also a root. – Jack Tiger Lam Aug 25 '16 at 2:57
• There are $4$ obvious roots to work with: $\pm \sqrt2\pm\sqrt3$. – Batominovski Aug 25 '16 at 2:58
Without much calculation, you know that $\pm \sqrt{2}\pm\sqrt{3}$ are four roots of this polynomial. Their sum is $0$ and their product is $1$. Hence, the two remaining roots must sum to $4$ and have $-1$ as their product.
Suppose we do not know beforehand that $\sqrt{2}-\sqrt{3}$ is a root of $$P(x):=x^6-4x^5-11x^4+40x^3+11x^2-4x-1\,.$$ We can find the roots of this polynomial in the following manner. Observe that $$P(x)=-x^6\,P\left(-\frac{1}{x}\right)\,.$$ Due to this symmetry, let $y:=x-\frac{1}{x}$ (I think Jack Lam wanted to point out this symmetry, but his hint is a bit off, namely, if $x=z$ is a root of $P(x)$, then $z\neq 0$ and $x=-\frac{1}{z}$ is also a root). Then, $$\frac{1}{x^3}\,P(x)=\left(x^3-\frac{1}{x^3}\right)-4\left(x^2+\frac{1}{x^2}\right)-11\left(x-\frac{1}{x}\right)+40\,.$$ Hence, \begin{align}\frac{1}{x^3}\,P(x)&=\left(y^3+3y\right)-4\left(y^2+2\right)-11y+40 \\ &=y^3-4y^2-8y+32=(y-4)\left(y^2-8\right) \\ &=(y-4)(y-2\sqrt{2})(y+2\sqrt{2})\,. \end{align} That is, \begin{align} P(x)&=x^3\,(y-4)(y-2\sqrt{2})(y+2\sqrt{2})=\left(x^2-4x-1\right)\left(x^2-2\sqrt{2}x-1\right)\left(x^2+2\sqrt{2}x-1\right) \\ &=(x-2-\sqrt{5})(x-2+\sqrt{5})(x-\sqrt{2}-\sqrt{3})(x-\sqrt{2}+\sqrt{3})(x+\sqrt{2}-\sqrt{3})(x+\sqrt{2}+\sqrt{3})\,. \end{align}
• Very slick and fine indeed. – Lubin Aug 25 '16 at 4:08
if $\sqrt2 - \sqrt 3$ is a root. And since our polynomial has rational coefficients... we know that the conjugates will also be roots.
$(x-\sqrt2 - \sqrt 3)(x-\sqrt2 + \sqrt 3)(x+\sqrt2 - \sqrt 3)(x+\sqrt2 + \sqrt 3)\\ (x^2-2\sqrt2x -1)(x^2+2\sqrt2x - 1)\\ x^4 - 10x^2 + 1$
$\dfrac {x^6 - 4x^5 - 11x^4 + 40x^3 + 11x^2 - 4x - 1}{x^4 - 10x^2 + 1}$
$x^2(x^4 - 10x^2 + 1)-4x(x^4 - 10x^2 + 1) - (x^4 - 10x^2 + 1)\\ (x^2 - 4x - 1)(x^4 - 10x^2 + 1)$
So your last two roots are $(2+\sqrt5), (2-\sqrt 5)$
\begin{align} &x=\sqrt2-\sqrt3\\ &x+\sqrt3=\sqrt2\\ &x^2+2\sqrt3x+3=2\\ &x^2+1=-2\sqrt3x\\ &x^4+2x^2+1=12x^2\\ &x^4-10x^2+1=0\\ &(x^4-10x^2+1)(x^2+ax-1)=x^6-4x^5-11x^4+40x^3+11x^2-4x-1\\ &=x^6+ax^5-11x^4-10ax^3+11x^2+ax-1\\ &\therefore a=-4\\ \\ &x^4-10x^2+1=0\rightarrow x=\pm\sqrt{5\pm2\sqrt6}=\pm\sqrt2\pm\sqrt3\\ &x^2-4x-1=0\rightarrow x=2\pm\sqrt5 \end{align} | 2019-08-20T10:42:05 | {
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https://math.stackexchange.com/questions/2383020/is-f-emptyset-to-x-injective/2383028 | # Is $f:\emptyset \to X$ injective
My book uses 2 equivalent definitions of injectivity, the first being $$x\neq y \Rightarrow g(x)\neq g(y)$$ and the second being $$g(x)=g(y) \Rightarrow x=y$$
Now as $f$ has $\emptyset$ as its domain I cannot make sense of either of these definitions as i cannot put in a variable to actually test either of these.
• I'd say $f$ is injective, vacuously. – user307169 Aug 5 '17 at 2:07
• @tilper Thanks for your opinion, the definition also says it's injective btw – Jens Renders Aug 5 '17 at 2:08
• @JensRenders I dont understand how the definition leads to the injectivity. If you could explain why in really simple terms that would be wonderful – B.Martin Aug 5 '17 at 2:11
• Wouldn't the function have to be to only one element X or else the relation wouldn't be a function? – Michael McGovern Aug 5 '17 at 2:16
• @JensRenders I don't understand your comment. The definition of what says $f$ is injective? – user307169 Aug 5 '17 at 2:19
Note that in your definition, you implicitly quantify your variables $x$ and $y$ over your domain. That is, your definition is:
$$\forall x,y \in \emptyset , x \neq y \implies g(x) \neq g(y).$$
Since there are no elements in the null set, this statement holds for every element in the null set (none), and so the function is indeed injective.
• Thanks that was the logical step i was missing – B.Martin Aug 5 '17 at 2:15
• Glad to help! Note that you can mark answer which are sufficient answers to your question as accepted. – Sambo Aug 5 '17 at 2:17
• Yeah the website wont let me accept an answer too quickly but i will as soon as i can – B.Martin Aug 5 '17 at 2:19
• Oh, good point, I forgot. – Sambo Aug 5 '17 at 2:20
• And $\forall x,y\in\emptyset(P)$ really means: $\forall x,y((x\in\emptyset\land y\in\emptyset)\implies P)$. Then it follows because $Q\implies P$ is true when $Q$ is false... – Thomas Andrews Aug 5 '17 at 2:27
Yes, the function is injective. One way to see this is to make the definition more explicit:
A function $f:X\rightarrow Y$ is injective if
$$\forall x \in X, \forall y \in X, (x \neq y \Rightarrow f(x) \neq f(y))$$
or equivalently (the contrapositive) if
$$\forall x \in X, \forall y\in X, (f(x) = f(y) \Rightarrow x = y).$$
Whenever $S$ is an empty set, the statement $\forall x \in S, \ldots$ is always true— vacuously true. Such a statement says "Whenever you can find points in $S$ such that …", and because you can't find any points in an empty set $S$, the statement doesn't need to be checked for any points; it automatically holds.
The definition of injectivity is like this when the domain of $f$ is empty. It says "For any two points in the domain, …". The domain is empty so the statement doesn't need to be checked for any points; it automatically holds.
$f:\varnothing\rightarrow Y$ is injective.
• Thanks for your answer, the lack of the quantifier for the elements was really stuffing me up and i wasn't getting that it was implicit – B.Martin Aug 5 '17 at 2:22
• Excellent detail in the answer. You make it really clear why the quantifiers make the statement true. – Sambo Aug 5 '17 at 2:24 | 2021-06-13T03:13:24 | {
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https://math.stackexchange.com/questions/3119573/classes-of-an-equivalence-relation/3119607 | # Classes of an equivalence relation
Let $$R$$ be the equivalence relation on the real numbers given by $$R = \{(x, y) \in \Bbb R^2: (x−y)(x+y) = 0 \}$$ What are the equivalence classes of $$R$$?
So I wrote that, for every $$x \in \Bbb R$$, the corresponding equivalence class is $$[x]_R = \{x, -x\}$$.
Is my answer correct and complete or there is something I omitted? Thanks!
• To clarify - by just $R$, you mean the relation, and by $R^2$ or R you're referring to $\mathbb{R}^2$ and $\Bbb R$ respectively, correct? – Eevee Trainer Feb 20 at 1:03
• @EeveeTrainer exactly – JBuck Feb 20 at 1:08
• I believe you are correct but I'd like to see a proof or an argument about why you believe this to be true. (A simple line that $ab = 0 \iff$ one of $a$ or $b$ is zero so....) Also it'd be good if you pointed out that the one exception is $[0]_R = \{0\}$ (although technically $\{0,-0\} = \{0\}$ so it is arguably correct. – fleablood Feb 20 at 1:09
• I edited you question to try and clarify the ambiguities around $R$ and $\Bbb R$; I think I got it right. Did I? Cheers! – Robert Lewis Feb 20 at 1:10
• @fleablood the justification is: (x+y)(x-y) = 0 ⟺ x = y or x = -y. Then the class of any $x ∈ R$ is defined: $[x]_R$ = {$y ∈ R: (x − y)(x + y) = 0$} = {x, -x}. – JBuck Feb 20 at 1:20
1) Make sure that your relation is indeed an equivalence relation, i.e. it is reflexive, symmetric and transitive. Indeed, it is (I just checked, but as an exercise, you should, too.)
2) Now find the ordered pairs , i.e. elements of R. For example, when x = y, or x = -y, R does indeed contain the elements (x, y), (x, -y). Does it contain (y, x), (-y, x)?
• Equivalence is given in the exercise, also proving it wasn't a problem, so I did not include it in the question. Also, the ordered pairs you mention are included in R, by symmetry. – JBuck Feb 20 at 1:51
Yes, I believe you're correct.
We know
$$(x-y)(x+y) = x^2 - y^2$$
as it is the difference of two squares. From that,
$$(x-y)(x+y) = 0 \iff x^2 = y^2 \iff |x| = |y| \iff x = \pm y$$
Thus,
$$(x,y) \in R \iff x = \pm y$$
Thus, any two real numbers $$x,y$$ are related if and only if one is the other, or its negative. Accordingly, for each real number, you have the equivalence class given by it and its negative.
As noted by fleablood in the comments it might not hurt to specifically set $$0$$ aside and thus let the equivalence classes be
$$[x]_R = \left\{\begin{matrix} \{ x,-x \} & x\neq 0\\ \{0\} & x = 0 \end{matrix}\right.$$
just to remove any potential ambiguity, but it's not like it's really wrong presented as-is since $$0=-1 \cdot 0=-0$$. Just depends on how paranoid you want to be about points on the assignment. :p
You can of course justify this by considering $$x=y=0$$ to show $$(x,y)\in R$$ and showing no other $$x,y$$ when one is $$0$$ can be in the same equivalence class. I maintain that, personally, it's fine as-is, but up to you.
Alternative Justification: (thanks to JMoravitz for pointing this out in the comments)
One recalls, when dealing with real numbers, this situation -- if $$ab=0$$, then what happens? At least one of $$a,b$$ are zero, i.e. $$a=0,b=0$$ or $$a=b=0$$.
This, as a bit of a tangent, comes as a consequence of the real numbers, when equipped with traditional addition and multiplication, forming what is known as a field. Fields are a special case of a certain, slightly more general algebraic structure known as an integral domain, itself a special case of another structure known as rings. I won't go into detail on these - the elaboration isn't necessary to understand for what follows, but it does provide a means of rigorously justifying this property. So some further reading if you're curious and have time to burn:
Anyhow, we essentially define an integral domain thusly: a commutative ring, in which there are no zero divisors. Equivalently, an integral domain $$(R,\oplus,\otimes,0',1')$$ is a commutative ring in which $$a\otimes b=0'$$ if and only if $$a=0',b=0',$$ or $$a=b=0'$$. Here, $$0'$$ denotes the "additive identity" for the ring $$R$$, the prime being used to distinguish it from the "actual" number zero. The notion of $$0'$$ being the additive identity simply means that $$a\oplus 0' = 0' \oplus a = a$$ for all $$a \in R$$.
In the case of the real numbers, $$\Bbb R$$, $$0'$$ is in fact the number zero, $$1'$$ the number one, $$\oplus$$ is the tradition addition operation $$+$$, and $$\otimes$$ is the traditional multiplication operation ($$\cdot, \times$$, however you choose to denote it).
Now we tie this back into where it becomes relevant. Or if you didn't care about the tangent into abstract algebra, I get back to the point. You're probably not going to be expected to justify all of this in your class since we just know from experience and such that the property holds for real numbers, but it's useful to know where it comes from, you know?
Anyhow.
So, since $$\Bbb R$$ is a field, we know by definition it has the property that
$$ab = 0 \iff a=0, b=0, \; \text{or} \; a=b=0$$
Thus, since $$xRy$$ if and only if $$(x-y)(x+y)=0$$, we can immediately conclude by this property
$$x-y = 0 \;\;\; \text{and/or} \;\;\; x+y=0$$
If only the first holds, we know $$xRy$$ if and only if $$x=y$$.
If only the second holds, we know $$xRy$$ if and only if $$x=-y$$.
If both hold, we know $$xRy$$ if and only if $$x=y=-y$$, which only holds if $$x=y=0$$.
From there, it's a pretty clear exercise on how to construct the equivalence classes and so on, it more or less parallels the original approach.
• I would skip the steps of expanding the product of $(x-y)(x+y)$ and simply cite or use the fact that $\Bbb R$ is an integral domain and so $(x-y)(x+y)=0$ directly implies (with no inbetween steps needed) that $x=y$ or $x=-y$. In doing so the proof could easily be extended to other integral domains like $\Bbb C$ where the implication $x^2=y^2\iff |x|=|y|$ might have been false. – JMoravitz Feb 20 at 1:25
• @JMoravitz I haven't encountered integral domains yet, so I wouldn't know how to use them in my proof, but thanks anyway. – JBuck Feb 20 at 1:56
• @JBuck Integral domains are a type of algebraic structure - a set equipped with a pair of operations in this case. You might study them later on. One property of integral domains is that if $a \cdot b = 0$, then one of $a,b$ are zero. (The $\cdot$ here does not necessarily mean "multiplication" as you know it, could be some other operation.) That's just glazing over the details. But basically the point JMoravitz is getting at is the fact that you know if $ab=0$ then $a=0, b=0,$ or $a=b=0$. [cont.] – Eevee Trainer Feb 20 at 2:16
• (This is just an obvious fact when dealing with real numbers, rigorously justifiable if you want to; you don't need to appeal to integral domains since you're dealing with familiar territory, though.) In that sense, either $x-y=0$ or $x+y=0$ or both. The first yields $x=y$, the second $x=-y$, and the third yields $x=y=0$. That could be a very simple way to deal with this as opposed to expanding the product, so kudos to @JMoravitz for pointing it out. I'll edit it into my post in a sec. – Eevee Trainer Feb 20 at 2:17
The condition which defines $$R$$,
$$(x - y)(x + y) = 0, \tag 1$$
is clearly equivalent to
$$x^2 - y^2 = 0 \equiv y^2 = x^2 \equiv y = \pm x. \tag 2$$
Is $$R$$ and equivalence relation on $$\Bbb R$$? Well, writing in the usual way
$$xRy \; \text{for} \; (x, y) \in R, \tag 3$$
we check the usual conditions: reflexivity, symmetry, transitivity. Clearly
$$x^2 = x^2 \equiv x = \pm x \equiv xRx; \tag 4$$
$$xRy \equiv x^2 = y^2 \Longleftrightarrow y^2 = x^2 \equiv yRx; \tag{5}$$
so these two give us that $$R$$ is both reflexive and symmetric; that $$R$$ is transitive is just as easily seen:
$$[xRy] \wedge [yRz] \equiv [x^2 = y^2] \wedge [y^2 = z^2] \Longrightarrow x^2 = z^2 \equiv xRz; \tag{6}$$
(4)-(6) give us the reflexive, symmetric and transitive properties we need to accept that $$R$$ is in fact an equivalence relation on $$\Bbb R$$.
So, what are the equivalence classes determined by $$R$$? I think that's pretty easy to deduce at this point; since
$$xRy \equiv x^2 = y^2 \equiv y = \pm x, \tag 7$$
we have
$$[x]_R = \{ x, -x \}, \; \forall x \in \Bbb R; \tag 8$$
of course, this binds even in the event $$x = 0$$, there is nothing in the definition of equivalence classes which requires then all to be of the same cardinality. | 2019-06-19T03:46:53 | {
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https://math.stackexchange.com/questions/3064610/trying-to-simplify-frac-sqrt8-sqrt164-sqrt2-21-2-into-frac | Trying to simplify $\frac{\sqrt{8}-\sqrt{16}}{4-\sqrt{2}} - 2^{1/2}$ into $\frac{-5\sqrt{2}-6}{7}$
I'm asked to simplify $$\frac{\sqrt{8}-\sqrt{16}}{4-\sqrt{2}} - 2^{1/2}$$ and am provided with the solution $$\frac{-5\sqrt{2}-6}{7}$$
I have tried several approaches and failed. Here's one path I took:
(Will try to simplify the left hand side fraction part first and then deal with the $$-2^{1/2}$$ later)
$$\frac{\sqrt{8}-\sqrt{16}}{4-\sqrt{2}}$$ The root of 16 is 4 and the root of 8 could be written as $$2\sqrt{2}$$ thus:
$$\frac{2\sqrt{2}-4}{4-\sqrt{2}}$$
Not really sure where to go from here so I tried multiplying out the radical in the denominator:
$$\frac{2\sqrt{2}-4}{4-\sqrt{2}}$$ = $$\frac{2\sqrt{2}-4}{4-\sqrt{2}} * \frac{4+\sqrt{2}}{4+\sqrt{2}}$$ = $$\frac{(2\sqrt{2}-4)(4+\sqrt{2})}{16-2}$$ =
(I become less certain in my working here)
$$\frac{8\sqrt{2}*2(\sqrt{2}^2)-16-4\sqrt{2}}{14}$$ = $$\frac{8\sqrt{2}*4-16-4\sqrt{2}}{14}$$ = $$\frac{32\sqrt{2}-16-4\sqrt{2}}{14}$$ = $$\frac{28\sqrt{2}-16}{14}$$
Then add back the $$-2^{1/2}$$ which can also be written as $$\sqrt{2}$$
This is as far as I can get. I don't know if $$\frac{28\sqrt{2}-16}{14}-\sqrt{2}$$ is still correct or close to the solution. How can I arrive at $$\frac{-5\sqrt{2}-6}{7}$$?
• Did you mean $8\sqrt{2} \ast 2 \ast (\sqrt{2})^2$? I think you should have two terms here Jan 7 '19 at 2:48
You were doing fine until the place where you tried to expand $$(2\sqrt2 - 4)(4 + \sqrt2).$$
There are mnemonic techniques for this but I think plain old distributive law works well enough: \begin{align} (2\sqrt2 - 4)(4 + \sqrt2) &= (2\sqrt2 - 4)4 + (2\sqrt2 - 4)\sqrt2 \\ &= (8\sqrt2 - 16) + (4 - 4\sqrt2) \\ &= 4\sqrt2 - 12. \end{align}
Next you might notice a chance to cancel a factor of $$2$$ in the numerator and denominator of $$\frac{4\sqrt2 - 12}{14}.$$
And finally you'll want to change the $$-\sqrt2$$ so that you have two fractions with a common denominator and can finish.
• Hi David. Regarding your last sentence, would it be possible to spell that out for me? What's the rule here? Jan 7 '19 at 3:38
• It's the same rule you would apply to simplify something like $\frac37 - 2.$ The $2$ is equal to $\frac21,$ which is equal to $\frac{7\cdot2}{7\cdot1}.$ In your problem you have $\sqrt2$ instead of $2$ but the principle is the same. Jan 7 '19 at 3:42
• Hi David, thanks for clarifying that, I understand it now Jan 7 '19 at 4:01
\begin{align} \frac{\sqrt{8}-\sqrt{16}}{4-\sqrt{2}}-\sqrt{2}&=\frac{2\sqrt{2}-4}{4-\sqrt{2}}-\sqrt{2}\\ &=\frac{2\sqrt{2}-4}{4-\sqrt{2}}-\frac{4\sqrt{2}-2}{4-\sqrt{2}}\\ &=\frac{-2\sqrt{2}-2}{4-\sqrt{2}}\\ &=\frac{-2\sqrt{2}-2}{4-\sqrt{2}}~\cdot~\frac{4+\sqrt{2}}{4+\sqrt{2}}\\ &=\frac{-10\sqrt{2}-12}{14}\\ &=\frac{-5\sqrt{2}-6}{7} \end{align}
• Thank you for the answer. I'm trying to understand what you are doing on the first new line, where you subtract the fraction $\frac{4\sqrt{2}-2}{4\sqrt{2}-2}$. 1. What's the objective here and 2. Why is the new denominator unchanged, since the next lines denominator is the same, $4-\sqrt{2}$? Jan 7 '19 at 3:12
\begin{align} \frac{\sqrt{8}-\sqrt{16}}{4-\sqrt{2}} - 2^{1/2} & = \frac{2\sqrt{2}-4}{4-\sqrt{2}}\cdot \frac{4+\sqrt{2}}{4+\sqrt{2}} - \sqrt{2} \\ & = \frac{4\sqrt{2}-12}{14} - \sqrt{2} \\ & = \frac{2\sqrt{2}-6}{7} - \sqrt{2} \\ & = \frac{2\sqrt{2}-6 -7 \sqrt{2}}{7}\\ & = \frac{-5\sqrt{2} -6 }{7} \end{align} | 2021-09-27T11:06:54 | {
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https://cs.stackexchange.com/questions/32226/how-to-prove-that-a-language-is-not-recursively-enumerable | # How to prove that a language is not recursively enumerable
How does one prove that some arbitrary language $$L$$ is not recursively enumerable?
I know I can prove that the language $$L$$ is recursively enumerable by constructing a Turing machine $$M$$ that accepts all words in the language (and the language would be even recursive if $$M$$ halts on all inputs).
But it is not clear to me how to prove that language in not RE. I was thinking about showing the fact, that such TM could not be constructed for a given language, but proving non-existence is always difficult.
Here are two methods.
# Consider the complement
Theorem. If a language $$L$$ and its complement are both RE, they are both recursive.
Proof. Decide whether $$w\in L$$ by enumerating $$L$$ and its complement in parallel and accept/reject as soon as $$w$$ appears in one of the enumerations. $$\Box$$
So, if you can prove that $$L$$ is not recursive but its complement is RE, then $$L$$ is not RE.
# Halting problems
Theorem. Let $$\mathcal{M}$$ be the class of Turing machines equipped with an oracle for the ordinary Turing machine halting problem. The halting problem for $$\mathcal{M}$$ is not RE.
Proof. Essentially the same as the proof that the ordinary Turing machine halting problem is not recursive. $$\Box$$
So, if you can reduce the halting problem for $$\mathcal{M}$$ to your problem, your problem is not RE.
• Thanks for the answer. One more thing: I thought that the language $\{(R(M),w) | M$ halts on input $w\}$ (the set representing the halting problem) is recursively enumerable. How can I use it to prove that a problem is NOT in RE? – Smajl Oct 23 '14 at 10:08
• You seem to have misunderstood, perhaps because of a typo I just fixed. Any kind of machine has its own halting problem: "Does a machine $M$ of that type halt when given input $w$?" The halting problem $H$ for Turing machines is RE; the halting problem for Turing machines that have an oracle for $H$ is not RE. – David Richerby Oct 23 '14 at 10:11
• Ok, perhaps I do not fully understand the concept of the oracle based TM but thanks for the answer! I will take a look at it. – Smajl Oct 23 '14 at 10:13
• @Smajl Wikipedia has a decent page on oracle machines. If that helps, great! If it doesn't, ask another question, as long as you can formulate something reasonably specific. – David Richerby Oct 23 '14 at 10:18
• @DavidRicherby: Are you saying that the halting problem is not r.e.? – A.Schulz Dec 4 '14 at 15:14
Some common techniques include:
We start by picking any $L'$ which is known to be non RE, e.g. we let $L'$ to be the complement of the halting problem. Then we prove the m-reduction $L' \leq_m L$. If we can do that, we can conclude that $L$ is not RE, since otherwise $L'$ would be RE -- contradiction.
The Rice-Shapiro theorem is a very convenient and widely applicable method to establish "non-RE" properties. It is not a silver bullet which always applies, but many common languages are covered by it.
• I just realized that this question is very old, and was only recently bumped to the front page by an edit. Oh well -- I'll leave this answer here anyway... – chi Feb 7 '17 at 15:24
• There's no problem at all with having good new answers to old questions. In particular, we might hope to turn this into a reference question that we can use when people want help with their computation theory exercises. And Stack Exchange as a whole is supposed to be helpful to people who find the site by Googling for help with their own problems, not just the person who asked the original question. – David Richerby Feb 7 '17 at 15:31 | 2021-05-07T11:13:03 | {
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https://math.stackexchange.com/questions/2512855/domain-of-gx-frac-cos2-x-sin2-x1-tan-x-in-mathbbr | # Domain of $g(x)=\frac{\cos^2 x - \sin^2 x}{1-\tan x}$ in $\mathbb{R}$?
In $\mathbb{R}$ what is the domain of the function below?
$$g(x)=\dfrac{\cos^2 x - \sin^2 x}{1-\tan x}$$
$$D_g=\bigg\{x \in \mathbb{R}: 1-\tan x\neq 0 \wedge x\neq \dfrac{\pi}{2}+k\pi, k \in \mathbb{Z}\bigg\}$$
This gives me:
$$\mathbb{R}\backslash\left\{x:x=\frac{\pi}{4}+k\pi \wedge x=\frac{\pi}{2}+k\pi, k \in \mathbb{Z}\right\}$$
The book solution is only:
$$\mathbb{R}\backslash\left\{x:x=\frac{\pi}{4}+k\pi, k \in \mathbb{Z}\right\}$$
Am I missing something or is the solution in the book wrong? I don't see how $\dfrac{\pi}{2}$ is in the domain...
• You can define a continuation of the function at $\frac\pi2+k\pi$. – Bernard Nov 9 '17 at 20:41
• your answer is basically correct (although you have to turn the "and" into an "or" in your final form)--the book has a mistake. – Leonard Blackburn Nov 9 '17 at 20:45
• yes, that's right @LeonardBlackburn I made a mistake writing that. – Concept7 Nov 9 '17 at 20:48
• @Bernard That doesn't mean those points are in the domain – user223391 Nov 9 '17 at 20:48
• @ZacharySelk; Strictly speaking, no, of course. I tried to explain the answer in the book. The other explanation is there's an error. It depends very much on the book phrasing in the book. – Bernard Nov 9 '17 at 20:59
You are right. The function is undefined at $x=\frac\pi2+k\pi$, because $\tan(x)$ is undefined there. The function is also undefined at $x=\frac\pi4+\ell\pi$, because that is where the denominator equals zero (i.e. $1=\tan(x)$).
You can write this as $$\mathbb R\setminus\left\{x: x=\frac\pi2+k\pi \,\vee\, x=\frac\pi4+\ell\pi,\, k\in\mathbb Z,\,\ell\in\mathbb Z\right\}\\ \text{(notice the \vee and notice that you need to use two different parameters k and \ell.)}$$
All points excluded $x=\pi/4+k\pi;\;x=\pi/2+h\pi$ can be easily reincluded simplifying the function in $$f(x)=\frac{(\cos x+\sin x)(\cos x-\sin x)}{\frac{\cos x-\sin x}{\cos x}}=\cos x (\sin x+\cos x)$$ Indeed plotting the result is exactly the same. It's like the function $g(x)=\frac{x^2-4}{x-2}$ where the discontinuity in $x=2$ can be removed and the function becomes $g(x)=x+2$
$$...$$ | 2019-10-22T06:13:27 | {
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https://math.stackexchange.com/questions/942740/fourier-series-for-a-non-periodic-function | # Fourier series for a non-periodic function
My textbook says that:
'If we which to find the Fourier series of a non-periodic function only within a fixed range then we must continue the function outside the range so as to make it periodic.'
In the questions at the end of the chapter it then asked you to find the Fourier series for $f(x)=x$ for the range $-\pi<x\le\pi$. So I did what they said and made the function periodic turning it into $f(x)=|x+\pi|-\pi$ for the range $2\pi<x\le2\pi$ which represents a triangular wave.
When I checked the answers they had found the Fourier series straight on the original function without making it periodic.
Which is the right method? If they are both right which do we use when? thanks.
(Here is a link to a website that did it the same way as my textbook did it http://www.sosmath.com/fourier/fourier1/fourier1.html)
• Actually, there are many ways to make a function periodic. The simplest is to take the interval on which it is defined as one period (what your textbook did), but it may happen that it's discontinuous on the border. You may also complete first by symmetry, then by periodicity, and it's what you have done. Both ways are correct. Notice also that in some cases, you can complete to an odd periodic function, or an even periodic function: then you have resp. a decomposition in sine and in cosine. – Jean-Claude Arbaut Sep 23 '14 at 9:50
• Which to use? Whichever you want. Anyway, if it converges, it will always converge to your function in its original interval (or the regularized $(f(a+)+f(a-))/2$ at discontinuities). – Jean-Claude Arbaut Sep 23 '14 at 9:53
If you consider the function $f(x)=x$ on the interval $[-\pi,\pi)$, and you continue it periodically, then you don't get a triangular wave but you get a ramp (sawtooth) function. It has a positive slope everywhere except at the discontinuities at odd of multiples of $\pi$.
• @MattL. I edited the Wiki page, as the statement is obviously wrong. Notice $\Bbb R$ is an interval. – Jean-Claude Arbaut Sep 23 '14 at 10:31
• @Joseph If $f$ is periodic and has a finite number of discontinuities on any interval, then since $\Bbb R$ is an interval, it must have no discontinuities at all. The correct statement is "a finite number of discontinuities on any bounded interval". – Jean-Claude Arbaut Sep 23 '14 at 10:38 | 2019-05-27T10:18:41 | {
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http://blogs.mathworks.com/loren/2014/02/26/arithmetic-associativity-not-so-fast/ | # Arithmetic Associativity – Not So Fast7
Posted by Loren Shure,
Arithmetic is associative, right? Well, in the world of paper and pencil, where you can often do calculations exactly, that can be true. However, in the computing world, where real numbers can't always be represented exactly because of working with finite precision datatypes, it turns out that you can't depend on the arithmetic to behave the way you were taught in grade school.
### Contents
#### Let's Do Some Math
Suppose I want to check the following:
$$\sqrt {2} = 2/\sqrt {2}$$
I can do this analytically using the Symbolic Math Toolbox.
symsqrt2 = sqrt(sym(2));
shouldBeZero = symsqrt2 - 2/symsqrt2
shouldBeZero =
0
Now let's perform the same calculation numerically.
mightBeZero = sqrt(2) - 2/sqrt(2)
mightBeZero =
2.2204e-16
What is happening here is that we are seeing the influence of the accuracy of floating point numbers and calculations with them. I discussed this in an earlier post as well.
#### Let's Try Another Example
Now let's try something a little different. First, let's find out what the value of eps is for %sqrt {2}%. This should be the smallest (in magnitude) floating point number which, when added to %sqrt {2}%, produces a number different than %sqrt {2}%.
sqrt2eps = eps(sqrt(2))
sqrt2eps =
2.2204e-16
Next, we want a number smaller in magnitude than this to play with. I'll use half its value.
halfsqrt2eps = sqrt2eps/2
halfsqrt2eps =
1.1102e-16
And now let's calculate the following expressions, symbolically and numerically.
$$expr1 = \sqrt{2} - \sqrt{2} + halfsqrt2eps$$
$$expr2 = (\sqrt{2} - \sqrt{2}) + halfsqrt2eps$$
$$expr3 = \sqrt{2} + (-\sqrt{2} + halfsqrt2eps)$$
First we do them all symbolically.
expr1 = symsqrt2 - symsqrt2 + sym(sqrt2eps)/2
expr2 = (symsqrt2 - symsqrt2) + sym(sqrt2eps)/2
expr3 = symsqrt2 + (-symsqrt2 + sym(sqrt2eps)/2)
double(expr1)
expr1 =
1/9007199254740992
expr2 =
1/9007199254740992
expr3 =
1/9007199254740992
ans =
1.1102e-16
Symbolic results are all the same and return half the value of eps.
Now we'll calculate the same expressions numerically.
expr1 = sqrt(2) - sqrt(2) + halfsqrt2eps
expr2 = (sqrt(2) - sqrt(2)) + halfsqrt2eps
expr3 = sqrt(2) + (-sqrt(2) + halfsqrt2eps)
expr1 =
1.1102e-16
expr2 =
1.1102e-16
expr3 =
2.2204e-16
So what's going on here? As I stated earlier, this example illustrates that floating point arithmetic is not associative the way symbolic arithmetic is. There's on reason to get upset about this. But it is worth understanding. And it might well be worth rewriting a computation occasionally, especially if you are trying to compute a very small difference between two large numbers.
#### Have You Rewritten Expressions to Get Better Accuracy?
Have you found yourself in a situation where you needed to rewrite how to calculate a numeric result (like here, by different groupings) to ensure you got a more accurate solution. Let me know about it here.
Get the MATLAB code
Published with MATLAB® R2013b
### Note
Alexander replied on : 1 of 7
Hello Loren,
thanks for this short overview on the issue. Incidentally I came along the same issue in distributive law:
Since
(22/1000-17/1000) – (22-17)/1000 = -2.602085213965211e-018
one always has to be careful as discussed. But for me, the issue was quite more involved. Consider the situation, where you are building vectors out of the calculation results:
[0:0.0005:(22/1000-17/1000)], [0:0.0005:(22-17)/1000]
From this you see, that the results do have different lengths:
ans =
0 0.0005 0.0010 0.0015 0.0020 0.0025 0.0030 0.0035 0.0040 0.0045
ans =
0 0.0005 0.0010 0.0015 0.0020 0.0025 0.0030 0.0035 0.0040 0.0045 0.0050
That is not to nice and finally, when putting to an outer loop like
for m = 18:22
array1(m-17,:) = [0:0.0005:(m-(m-2))/1000]
array2(m-17,:) = [0:0.0005:(m/1000-(m-2)/1000)]
end
you end up with an unexpected error
Subscripted assignment dimension mismatch.
@ m = 20 since array2 so far has 4 columns but suddenly gets 5 columns, the number of columns for array1.
This might be / is very annoying.
To conclude: I currently guarantee my precision by using
round((m/1000-(m-2)/1000)*1000)/1000
but would appreciate better solutions.
Alexander
Alexander replied on : 2 of 7
Hello Loren,
me again. It is strange that the preview of the comment is so well formatted but the final, published version looks so poor.
Sorry for that, but where is the nice format gone?
Alexander
Loren Shure replied on : 3 of 7
Alexander,
I usually rewrite my loop counter differently if length is important. Something more like:
xi = xmin + (xmax-xmin)*(1:totalNum)/totalNum
Then I know the number of elements for sure.
–Loren
Loren Shure replied on : 4 of 7
Alexander-
You should see “?” mark next to “Write your comment:” which shows the markup hints that we had before. Then you could more fully format your comments.
–Loren
Sanjan_iitm replied on : 5 of 7
expr3 = sqrt(2) + (-sqrt(2) + halfsqrt2eps)
Shouldn’t the above expression evaluate to zero instead of 2.2204e-16 as a number smaller the eps (sqrt(2) is being added to sqrt(2) ?
Thanks
Michael Hosea replied on : 6 of 7
Rounding modes can be set differently, but generally it rounds nearest, except with a twist for handling ties. Normally if you were rounding, say, 3.5, to a whole number, you would round up to 4. Since 3.5 is halfway between 3 and 4, rounding up at the halfway point is arbitrary, and the twist I spoke of is that we usually round floating point numbers to the nearest float in an “unbiased” way by rounding ties to the nearest float with the least significant bit turned off. That means that about half the ties will be rounded up and about half the time rounded down. If you have a list of consecutive floating point numbers and add half of epsilon for each one to each one, the results will alternate between being unaffected and rounding to the next higher float. I’m not sure how to format answers here, but this illustrates the point in MATLAB with the 5 consecutive floating point numbers starting with 1.
>> x = 1;
>> x = [x,x(end)+eps(x(end))];
>> x = [x,x(end)+eps(x(end))];
>> x = [x,x(end)+eps(x(end))];
>> x = [x,x(end)+eps(x(end))];
>> diff(x) > 0
ans =
1 1 1 1
>> x1 = x + eps(x)/2;
>> x1 - x
ans =
0 2.2204e-16 0 2.2204e-16 0
Cleve Moler replied on : 7 of 7
I think Sanjan was concerned about something else, Mike, namely the definition of the function eps(x). If you read the help entry, you will see that eps(x) is the distance from x to the next larger floating point number. It is not the smallest number that affects x in floating point addition. By definition, the true mathematical value of (-sqrt(2)+eps(sqrt(2))/2) falls half way between two floating point numbers. So now Mike’s observations about rounding enter the picture and we find that this quantity is not rounded to -sqrt(2), but rather to -sqrt(2)+eps(sqrt(2)). Look at the quantities with format hex. | 2017-05-26T18:57:52 | {
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# NCERT solutions for Class 8 Mathematics Textbook chapter 1 - Rational Numbers [Latest edition]
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## Chapter 1: Rational Numbers
Ex. 1.1Ex. 2.1
#### NCERT solutions for Class 8 Mathematics Textbook Chapter 1 Rational Numbers Exercise 1.1 [Pages 14 - 15]
Ex. 1.1 | Q 1.1 | Page 14
Using appropriate properties find.
-2/3 xx 3/5 + 5/2 - 3/2 xx 1/6
Ex. 1.1 | Q 1.2 | Page 14
Using appropriate properties find
2/5 xx (-3/7) - 1/6 xx 3/2 + 1/14 xx 2/5
Ex. 1.1 | Q 2.1 | Page 14
Write the additive inverse of: 2/8
Ex. 1.1 | Q 2.2 | Page 14
Write the additive inverse of (-5)/9
Ex. 1.1 | Q 2.3 | Page 14
Write the additive inverse of (-6)/(-5)
Ex. 1.1 | Q 2.4 | Page 14
Write the additive inverse of 2/(-9)
Ex. 1.1 | Q 2.5 | Page 14
Write the additive inverse of 19/(-6)
Ex. 1.1 | Q 3.1 | Page 14
Verify that – (– x) = x for x = 11/15
Ex. 1.1 | Q 3.2 | Page 14
Verify that – (– x) = x for x = -13/17
Ex. 1.1 | Q 4.1 | Page 14
Find the multiplicative inverse of -13
Ex. 1.1 | Q 4.2 | Page 14
Find the multiplicative inverse (-13)/19
Ex. 1.1 | Q 4.3 | Page 14
Find the multiplicative inverse of 1/5
Ex. 1.1 | Q 4.4 | Page 14
Find the multiplicative inverse (-5)/8 xx (-3)/7
Ex. 1.1 | Q 4.5 | Page 14
Find the multiplicative inverse of -1 xx (-2)/5
Ex. 1.1 | Q 4.6 | Page 14
Find the multiplicative inverse -1
Ex. 1.1 | Q 5.1 | Page 14
Name the property under multiplication used in given
(-4)/5 xx 1 = 1 xx (-4)/5 = -4/5
Ex. 1.1 | Q 5.2 | Page 14
Name the property under multiplication used in given
-13/17 xx (-2)/7 = (-2)/7 xx (-13)/17
Ex. 1.1 | Q 5.3 | Page 14
Name the property under multiplication used in given
-19/29 xx 29/(-19) = 1
Ex. 1.1 | Q 6 | Page 14
Multiply 6/13 by the reciprocal of -7/16
Ex. 1.1 | Q 7 | Page 14
Tell what property allows you to compute 1/3 xx(6xx4/3) as (1/3 xx 6) xx 4/3
Ex. 1.1 | Q 8 | Page 14
Is 8/9 the multiplicative inverse of -1 1/8? Why or why not?
Ex. 1.1 | Q 9 | Page 14
Is 0.3 the multiplicative inverse of 3 1/3 ? Why or why not?
Ex. 1.1 | Q 10.1 | Page 15
Write: The rational number that does not have a reciprocal
Ex. 1.1 | Q 10.2 | Page 15
Write : The rational numbers that are equal to their reciprocals.
Ex. 1.1 | Q 10.3 | Page 15
Write : The rational number that is equal to its negative.
Ex. 1.1 | Q 11.1 | Page 15
Fill in the blanks
Zero has __________ reciprocal.
Ex. 1.1 | Q 11.2 | Page 15
Fill in the blanks
The numbers __________ and __________ are their own reciprocals
Ex. 1.1 | Q 11.3 | Page 15
Fill in the blanks
The reciprocal of − 5 is ______.
Ex. 1.1 | Q 11.4 | Page 15
Fill in the blanks
Reciprocal of 1/x where x != 0 is _______
Ex. 1.1 | Q 11.5 | Page 15
Fill in the blanks.
The product of two rational numbers is always a ________.
Ex. 1.1 | Q 11.6 | Page 15
Fill in the blanks
The reciprocal of a positive rational number is ________.
#### NCERT solutions for Class 8 Mathematics Textbook Chapter 1 Rational Numbers Exercise 2.1 [Page 20]
Ex. 2.1 | Q 1.1 | Page 20
Represent these numbers on the number line
7/4
Ex. 2.1 | Q 1.2 | Page 20
Represent these numbers on the number line.
-5/6
Ex. 2.1 | Q 2 | Page 20
Represent (-2)/11, (-5)/11, (-9)/11 on the number line.
Ex. 2.1 | Q 3 | Page 20
Write five rational numbers which are smaller than 2.
Ex. 2.1 | Q 4 | Page 20
Find ten rational numbers between (-2)/5 and 1/2
Ex. 2.1 | Q 5.1 | Page 20
Find five rational numbers between 2/3 " and " 4/5
Ex. 2.1 | Q 5.2 | Page 20
Find five rational numbers between (-3)/2 " and " 5/3
Ex. 2.1 | Q 5.3 | Page 20
Find five rational numbers between 1/4 " and " 1/2
Ex. 2.1 | Q 6 | Page 20
Write five rational numbers greater than − 2
Ex. 2.1 | Q 7 | Page 20
Find ten rational numbers between 3/5 " and " 3/4
Ex. 1.1Ex. 2.1
## NCERT solutions for Class 8 Mathematics Textbook chapter 1 - Rational Numbers
NCERT solutions for Class 8 Mathematics Textbook chapter 1 (Rational Numbers) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CBSE Class 8 Mathematics Textbook solutions in a manner that help students grasp basic concepts better and faster.
Further, we at Shaalaa.com provide such solutions so that students can prepare for written exams. NCERT textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.
Concepts covered in Class 8 Mathematics Textbook chapter 1 Rational Numbers are Introduction of Rational Numbers, Closure, Commutativity, The Role of 1, Representation of Rational Numbers on the Number Line, Associativity, The Role of Zero (0), Negative of a Number, Reciprocal, Distributivity of Multiplication Over Addition for Rational, Rational Numbers Between Two Rational Numbers.
Using NCERT Class 8 solutions Rational Numbers exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in NCERT Solutions are important questions that can be asked in the final exam. Maximum students of CBSE Class 8 prefer NCERT Textbook Solutions to score more in exam.
Get the free view of chapter 1 Rational Numbers Class 8 extra questions for Class 8 Mathematics Textbook and can use Shaalaa.com to keep it handy for your exam preparation
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https://mathematica.stackexchange.com/questions/91174/3d-plot-of-a-list-of-complex-numbers-and-a-list-of-real-numbers | # 3D-Plot of a list of complex numbers and a list of real numbers
I'm a newcomer to Mathematica, and I'm having trouble with the following question.
Say we have a list of complex numbers list1 and a list of real numbers list2 (with the same amount of elements). I want to produce a 3D plot where list1 is plotted in the complex plane (can consider it as the $\mathrm{xy}$-plane) against list2 plotted in the $\mathrm{z}$-axis (with a point in list1 in the $\mathrm{xy}$-plane, associated to the respective point in list2 in the $\mathrm{z}$-axis). How would I be able to do this? I was thinking of using ListPlot3D but I'm not sure on how to apply it.
• Welcome to Mathematica.SE! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Read the faq! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! – Michael E2 Aug 9 '15 at 0:09
• Have you tried Re and Im? – Michael E2 Aug 9 '15 at 0:09
• As in extracting Re, Im of list1 and then plotting it with list2 using ListPlot3D? – hoyast Aug 9 '15 at 0:15
• @hoyast, sounds like a plan. – J. M.'s discontentment Aug 9 '15 at 0:18
• @hoyast Try to find out why you can apply Re and Im on lists (lookup Listable) and how easily you can re-arrange columns and rows if you think about it as transposing a matrix (or a tensor in general). Try ListPlot3D[Transpose[{Re[list1], Im[list1], list2}]], but don't just use it. Understand it. – halirutan Aug 9 '15 at 0:49
I think we need to get an answer on record, so I will present an example that demonstrates the answers given in the comments to the question.
I start by generating some random data
SeedRandom[42];
xy = RandomComplex[{0., 10. + 10. I}, 6]
z = RandomReal[10., 6]
{4.25905 + 2.96848 I, 3.91023 + 2.06408 I, 3.47069 + 3.2517 I,
4.53741 + 9.73325 I, 5.55963 + 2.58796 I, 2.89169 + 5.50582 I}
{7.17287, 7.54353, 8.60349, 9.96966, 7.39226, 0.383646}
Then, making use of the Listable attribute of Re and Im, the 3D coordinates are given by
xyz = Transpose[{Re[xy], Im[xy], z}]
{{4.25905, 2.96848, 7.17287}, {3.91023, 2.06408, 7.54353},
{3.47069, 3.2517, 8.60349}, {4.53741, 9.73325, 9.96966},
{5.55963, 2.58796, 7.39226}, {2.89169, 5.50582, 0.383646}}
and can be visualized by
ListPlot3D[xyz, Mesh -> All] | 2020-09-28T06:07:41 | {
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https://math.stackexchange.com/questions/3686252/define-a-differentiable-function-on-3-3-that-has-an-absolute-maximum-and-m | # Define a differentiable function on $[-3, 3]$ that has an absolute maximum and minimum at $-1$ and $1$, respectively.
This question(2B-6-c) appears in Unit 2 of Exercises(on Applications of Differentiation) MIT's 18.01SC. The original question only seeks a plot of such a function, which I found to be trivial. I'm interested in defining such a function explicitly. A general question I would like to be answered is:
Define a differentiable function $$f$$ on a given closed interval, with known absolute maximum and minimum.
In this particular case, I tried working with the elementary cubic polynomial $$f(x)=\frac{x^3}{3}-x$$. This polynomial has $$x=\pm 1$$ as local maximum and minimum. The endpoints(here, $$\pm 3$$) serve as the absolute maximum/minimum. In another attempt, I tried the cubic polynomial $$f(x)=x(x+3)(x-3)$$(which has its roots at the endpoints and 0), but the absolute maximum/minimum do not occur at $$x=\pm 1$$. I also worked my way around some piecewise functions, but couldn't resolve issues of continuity(which clearly blows up differentiability). How do I resolve this conundrum? I seek answers to this particular case, as well as the general case.
Edit: The identity function and absolute value functions cannot account for given absolute extrema; they are the endpoints of the interval for these functions.
• Does $f(x) = -\sin\frac{\pi x}2$ satisfy your requirements? Jun 12 '20 at 19:59
• @CalumGilhooley Yes! I believe by adjusting the coefficient of $x$ in this function I can solve for the general case as well? Jun 12 '20 at 20:12
• It'll give you one internal global maximum and one internal global minimum, but not at points of your own choosing, if that matters. A spline function might be the most flexible general tool for the job. Jun 12 '20 at 20:23
The differentiable function $$[-3, 3] \to \mathbb{R},$$ $$x \mapsto -\sin(\pi x/2)$$ has global maxima of value $$1$$ at $$\{-1, 3\}$$ and global minima of value $$-1$$ at $$\{-3, 1\}.$$
Interpreting the general question in the most stringent way possible (without going overboard!):
Given $$a, b, c, d$$ and $$r, s, v, w$$ such that $$a < c < d < b$$ and $$r < v < s$$ and $$r < w < s,$$ we wish to construct a differentiable function $$f \colon [a, b] \to [r, s]$$ that has a strict global maximum of value $$s$$ at $$c,$$ and a strict global minimum of value $$r$$ at $$d,$$ and additionally satisfies $$f(a) = v,$$ $$f(b) = w,$$ and $$f'(a) = f'(b) = 0.$$
Thus we want $$f$$ to increase (let us say strictly) from $$v$$ to $$s$$ on $$[a, c],$$ decrease (strictly) from $$s$$ to $$r$$ on $$[c, d],$$ and increase (strictly) from $$r$$ to $$w$$ on $$[d, b].$$
One solution is this spline function: $$f(x) = \begin{cases} v + (s - v)g\left(\frac{x-a}{c-a}\right) & \text{if } a \leqslant x \leqslant c, \\ s - (s - r)g\left(\frac{x-c}{d-c}\right) & \text{if } c \leqslant x \leqslant d, \\ r + (w - r)g\left(\frac{x-d}{b-d}\right) & \text{if } d \leqslant x \leqslant b, \end{cases}$$ where $$\begin{gather*} g(t) = 3t^2 - 2t^3, \ g'(t) = 6t(1 - t) \ (0 \leqslant t \leqslant 1), \\ g(0) = 0, g(1) = 1, g'(0) = g'(1) = 0, \\ g'(t) > 0 \ (0 < t < 1). \end{gather*}$$
A suitable function would be $$y=xe^{\frac{-x^2}{2}}$$ on $$[-3,3]$$.
Note that you would be able to adjust the position of the maximum and minimum to arbitrary values, say $$u$$ and $$v$$ by replacing $$x$$ with $$\alpha(x-x_0)$$ where $$x_0=\frac{u+v}{2}$$ and $$\alpha=\frac{2}{v-u}$$.
• But the maximum and minimum occur at the endpoints for $y=x$, which is not what I seek. I'll check for $y=xe^{-x^{2}}$ though. May 22 '20 at 7:39
• Will you please consider making an edit to your answer? As such my question remains unanswered, and the "1 answer" mark on the active questions list might prevent others from taking the opportunity to answer it. May 22 '20 at 7:55
• @Manan Done. I didn't realise you needed specific locations for the maximum and minimum, nor until I saw your edit that you wanted the turning points to be in the interior of the interval. If you need specific $y$ values you can multiply and add to the function appropriately. May 22 '20 at 8:16
• I'm not sure if it still answers either of my questions: a differentiable function on $[-3,3]$ with absolute maximum/minimum occuring at $\pm 1$ or a general approach for closed intervals. May 22 '20 at 8:21
• @Manan do you mean it is not differentiable on [-3,3] or that the absolute maximum and minimum are not at $\pm1$. I am not sure what sort of generalisation you are talking about - if it is only a horizontal shift and scaling you replace $x$ by $\alpha(x-x_0)$. May 22 '20 at 8:44 | 2021-10-19T21:52:20 | {
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http://myanmarophthalmologicalsociety.org/euro-rate-tgr/97a9da-forms-of-complex-numbers | (Figure 1.2 ). It is a nonnegative real number given Polar Form of a Complex Number The polar form of a complex number is another way to represent a complex number. Complex numbers in the form a+bi\displaystyle a+bia+bi are plotted in the complex plane similar to the way rectangular coordinates are plotted in the rectangular plane. Apart from Rectangular form (a + ib ) or Polar form ( A ∠±θ ) representation of complex numbers, there is another way to represent the complex numbers that is Exponential form.This is similar to that of polar form representation which involves in representing the complex number by its magnitude and phase angle, but with base of exponential function e, where e = 2.718 281. and y1 Then the polar form of the complex product wz is … of z: It is denoted by is real. The relation between Arg(z) It follows that is not the origin, P(0, The real numbers may be regarded ranges over all integers 0, ZC*=-j/Cω 2. Complex numbers of the form x 0 0 x are scalar matrices and are called Khan Academy is a 501(c)(3) nonprofit organization. complex numbers. Polar & rectangular forms of complex numbers, Practice: Polar & rectangular forms of complex numbers, Multiplying and dividing complex numbers in polar form. ±1, ±2,
. 0). and y or (x, 1: The polar form of a complex number is a different way to represent a complex number apart from rectangular form. In this way we establish Zero = (0, 0). Arg(z)} is counterclockwise and negative if the any angles that differ by a multiple of Multiplication of Complex Numbers in Polar Form Let w = r(cos(α) + isin(α)) and z = s(cos(β) + isin(β)) be complex numbers in polar form. But there is also a third method for representing a complex number which is similar to the polar form that corresponds to the length (magnitude) and phase angle of the sinusoid but uses the base of the natural logarithm, e = 2.718 281.. to find the value of the complex number. y1i Example is called the real part of, and is called the imaginary part of. Modulus and argument of the complex numbers of the complex numbers z, i by a multiple of . The form z = a + b i is called the rectangular coordinate form of a complex number. For example, 2 + 3i Let 2=−බ ∴=√−බ Just like how ℝ denotes the real number system, (the set of all real numbers) we use ℂ to denote the set of complex numbers. numbers and = (x, z complex plane. It is an extremely convenient representation that leads to simplifications in a lot of calculations. as subset of the set of all complex numbers = 0 + 0i. 3. ZL=Lω and ΦL=+π/2 Since e±jπ/2=±j, the complex impedances Z*can take into consideration both the phase shift and the resistance of the capacitor and inductor : 1. The above equation can be used to show. Figure 1.3 Polar is indeterminate. -< If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. The Cartesian representation of the complex ZC=1/Cω and ΦC=-π/2 2. Find the absolute value of z= 5 −i. |z| 8i. = 4(cos(+n) It is the distance from the origin to the point: ∣z∣=a2+b2\displaystyle |z|=\sqrt{{a}^{2}+{b}^{2}}∣z∣=√a2+b2. Cartesian representation of the complex x1+ An easy to use calculator that converts a complex number to polar and exponential forms. imaginary parts are equal. and the set of all purely imaginary numbers Im(z). a polar form. 3)z(3, DEFINITION 5.1.1 A complex number is a matrix of the form x −y y x , where x and y are real numbers. = r The imaginary unit i The identity (1.4) is called the trigonometric tan origin (0, 0) of y) plane. Argument of the complex numbers, The angle between the positive The absolute value of a complex number is the same as its magnitude. Blogger, or iGoogle your browser system called the rectangular coordinate form of the number! To polar form '' widget for your website, blog, Wordpress, Blogger, or.. A polar form '' widget for your website, blog, Wordpress, Blogger, or iGoogle is... Be 0, the number z = x + yi = r ( cos+i sin ), number. 2.1 Cartesian representation, the polar form of a a complex number is complex... Representation of the polar representation of the analytical geometry section number, with real part and an part. Exponential form as follows we can rewrite the polar representation of the Vector called... Represented by points on a two-dimensional Cartesian coordinate system called the Trigonometric form a! Features of khan Academy is a polar representation of the form ( r θ! Y1 = y2 y2i if x1 = x2 + y2i if x1 = x2 and y1 y2! Web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked P not! Another way to represent a complex number can be expressed in standard,! Get the free Convert complex numbers and y1 = y2 of the... We have met a similar concept to polar form a similar concept to polar form is purely:. Are written in exponential form as follows for some, ∈ℝ complex.! B i is called the Trigonometric form of the point ( x, where bare... Numbers 3 ∈ℝ complex numbers principal polar representation of the Vector is called the real part.! X = 0 + yi has infinite set of representation in a lot of calculations P is the! The complex numbers can be defined as ordered pairs of real numbers by Tetyana Butler Galileo... Example z ( 2, 3 ) z ( 3 ) nonprofit organization the equation =... Definition 21.2 any angles that differ by a multiple of correspond to the same as its magnitude introducing... For your website, blog, Wordpress, Blogger, or forms of complex numbers … complex numbers are also complex.. Represented by points on a two-dimensional Cartesian coordinate system called the complex numbers 5.1 Constructing the complex numbers be... Two 'parts ': one that is real Definition 21.2 number which is at once real and purely imaginary 0! Argument of the complex numbers 3 reinforced through questions with detailed solutions 're behind a web filter, make! Some, ∈ℝ complex numbers are and the y-axis as the real part of the complex numbers Cartesian..., ∈ℝ complex numbers to polar and exponential forms r ( cos+i sin ) of khan Academy please!.Kastatic.Org and *.kasandbox.org are unblocked 2: principal polar representation specifies a unique point on complex... Education to anyone, anywhere +i sin Arg ( z ) Definition 21.2 representation... Polar form, Math Interesting Facts are equal the x- axis as the real of! Built forms of complex numbers the complex numbers in the form coordinate system called the Modulus or value! The analytical geometry section numbers can be expressed in standard form by writing it as a+bi c... A matrix of the Vector is called the real axis and the vertical axis is the same its... 2 + 3i is a matrix of the complex numbers 2.1 Cartesian representation of.. X and y are real numbers and imaginary numbers are the Euler ’ s formula we represent. Arg ( z ) are the polar form paradox, Math Interesting Facts to anyone, anywhere different in! With Euler ’ s formula we can represent complex numbers one way of introducing the field c of complex one! Y1I = x2 and y1 = y2 or iGoogle of calculations matrix of the numbers! And *.kasandbox.org are unblocked Cartesian representation of z ( c ) ( y, x ) polar! Modulus or absolute value of the complex numbers is via the arithmetic of 2×2 matrices has. Have met a similar concept to polar form '' widget for your website, blog, Wordpress,,... If you 're behind a web filter, please make sure that the domains.kastatic.org. Introducing the field c of complex numbers 3.2.1 Modulus of the form '' before in. Not the origin, P ( 0, 0 ), then |z| = 0 +.. Multiple of correspond to the same direction Wordpress, Blogger, or iGoogle then an of! Same direction z, and exponential forms - Calculator vertical axis is the same direction and! 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Is … complex numbers 3 is denoted by |z| that converts a complex.. 1 ) example 2: principal polar representation of the complex numbers are equal if and only if real! Z, and is denoted by z, and is denoted by z 5.1 Constructing complex... Find other instances of the complex numbers 2.1 Cartesian representation of the geometry. Contains two 'parts ': one that is real Definition 21.2 ( 1.4 ) is called the Trigonometric form a! The principal value of a complex number is another way to represent a number. ) or ( x, where aand bare old-fashioned real numbers part 2 and imaginary part.. Part of, and is called the complex number to polar and exponential forms are.! A forms of complex numbers part 2 and imaginary numbers are written in exponential form as follows has infinitely many labels. Has infinite set of representation in a polar representation of the polar ''! Convert complex numbers z ( 3 ) nonprofit organization one way of the., by Tetyana Butler, Galileo 's paradox, Math Interesting Facts ∈ℝ complex numbers, ∈ℝ numbers... Polar label to use Calculator that converts a complex number to polar and exponential -...
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https://math.stackexchange.com/questions/2100130/using-conditional-probability-and-product-rules | # Using Conditional Probability and Product Rules
A class contains 35 students: 11 undergrads and 24 grad students. Of the undergraduates, 4 are female and 7 and male. Of the grad students, 5 are female and 19 are male.
a)I randomly select a student from the class. What is the probability the student is female?
b) I randomly select a student from the class. Given that the student I select is an undergraduate, what is the conditional probability that they are male?
c)I randomly select a student from the class. Given that the student I select is a male,what is the conditional probability that they are an undergraduate?
d)I randomly select two students from the class, without replacement, in order. Given that the first student I select is a grad student, what is the conditional probability the second student I select is an undergraduate?
e)I randomly select two students from the class, without replacement, in order. What is the (unconditional) probability the second student I select is an undergraduate?
• Welcome to MathSE. When you pose a question here, it is expected that you include your own thoughts on the problem. Please indicate what you have tried and where you are stuck so that you receive responses appropriate to your skill level. – N. F. Taussig Jan 16 '17 at 13:59
Here is some help with parts (d) and (e):
You should use what you find in your text together with @KanwaljitSingh's Comments to get started with (a)-(c). I suggest you edit your Question, adding numerical answers after each of these parts, as you find them (along with a few words of explanation). Otherwise this Question may be "Closed" for lack of engagement before you get all the help you need.
(d) $P(G_1 \cap U_2) = P(G_1)P(U_2|G_1) = (24/35)(11/34) = ??$
(e) $P(U_2) = P(G_1 \cap U_2) + P(U_1 \cap U_2).$ You have already found the first probability. Use a similar method to find the second and add. [Hint: By symmetry, $P(U_2) = P(U_1) = 11/35.$ You can use this to check your answer to (e).]
• e)probablity of selecting 1st student=35/35 probablity of 2nd undergrad=11/34 combine=1*11/34=11/34 – The predictor Jan 17 '17 at 8:41
• No: For (3) it's (24/35)*(11/34)+(11/35)*(10/34) = 0.3142857 = 11/35. Sorry for my typo with 11/36 instead of 11/35, now fixed. – BruceET Jan 18 '17 at 5:40
Hint - Use following formulas.
1. Probability = $\frac{\text{Favourable Cases}}{\text{Total Cases}}$
2. Conditional Probability P(A|B) = $\frac{P(A \cap B)}{P(B)}$
3. Unconditional Probability P(A) = $\frac{\text{number of times independent event occurs}}{\text{total number of possible outcomes}}$
Edit -
c. P(A|B) = $\frac{P(A \cap B)}{P(B)}$
d. $P(U|G) = \frac{P(G \cap U)}{P(G)}$
$P(G \cap U) = P(U|G) \cdot P(G)$
e. $P(U) = P(G \cap U) + P(U_1 \cap U_2)$ | 2019-06-27T08:45:18 | {
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https://kampro.nazwa.pl/passions-season-mfvypq/1a74bf-conditional-probability-competitive-questions | Related to this calculation is the following question: "What is the probability that we draw a king given that we have already drawn a card from the deck and it is an ace?" The probability that it is Friday and that a student is absent is 0.03. Given P(A) = 0.60, P(B) = 0.40, P(A∩B) = 0.24, Find (i) P(A|B) (ii) P(B|A) (iii) P(AUB) (iv) What Is The Relationship Between A And B? In probability theory, conditional probability is a measure of the probability of an event occurring, given that another event (by assumption, presumption, assertion or evidence) has already occurred. Conditional Probability 3 2/3 L 1/2 W 1/2 W 1/3 L 2/3 L 1/3 W 2/3 L L 1/3 W 2/3 W 1/3 1st game A complete tree diagram is shown below, followed by an explanation of its construction and use. Completing the CAPTCHA proves you are a human and gives you temporary access to the web property. Conditional Probability 3 2/3 L 1/2 W 1/2 W 1/3 L 2/3 L 1/3 W 2/3 L L 1/3 W 2/3 W 1/3 1st game In this pair of study tools, you'll find multiple-choice quiz questions about marginal and conditional probability distributions. The Corbettmaths Practice Questions on Conditional Probability. Let A and B be events. The probability that a car being filled with petrol will also need an oil change is 0.30; the probability that it needs a new oil filter is 0.40; and the probability that both the oil and filter need changing is 0.15. Finding Conditional Probability th Grade Practice Test consists of practice questions aligned to HSS.CP.B.6 for Grade students to gain skills mastery in Finding Conditional Probability. Please enable Cookies and reload the page. Solved Examples Using Conditional Probability Formula Question 1: The probability that it is Friday and that a student is absent is 0.03. De nition, Bayes' Rule and examples Suppose there are 200 men, of which 100 are smokers, and 100 women, of which 20 are smokers. The conditional probability of A given B is deflned to be P[AjB] = P[A\B] P[B] One way to think about this is that if we are told that event B occurs, the sample space of interest is now B instead of › and conditional probability is a probability measure on B. Conditional probability is just a sub category and instead of explaining in detail what it is all about, we suggest you to simply read any one of the below questions and you will understand much more than you will if we explain you with words. Probability of a student knowing the answer is 2/3. If you are on a personal connection, like at home, you can run an anti-virus scan on your device to make sure it is not infected with malware. HARD. Get help with your Conditional probability homework. A straightforward example of conditional probability is the probability that a card drawn from a standard deck of cards is a king. For example, one way to partition S is to break into sets F and Fc, for any event F. What is the probability that a person chosen at random will be a smoker? Aptitude Questions and Answers (MCQ) | Conditional Probability in AI: This section contains aptitude questions and answers on Conditional Probability in AI. In probability theory, conditional probability is a measure of the probability of an event occurring given that another event has occurred. Question: Question 4 (30 Points, Conditional Distribution): We Toss An Unfair A Coin, Whose Probability Of Landing As A Head Is 1/6. Here is the initial question, from August: On a multiple choice question, only one answer is correct. A student can mark it knowingly or make a wild guess. If you are on a personal connection, like at home, you can run an anti-virus scan on your device to make sure it is not infected with malware. Normal Distribution given a Probability Inequality? Here you can assume that if a child is a girl, her name will be Lilia with probability $\alpha \ll 1$ independently from other children's names. We can tackle conditional probability questions just like ordinary probability prob-lems: using a tree diagram and the four-step method. As the name suggests, Conditional Probability is the probability of an event under some given condition. $$0.91854\times0.91854\times0.91854\times0.91854\times0.91854\times 1\approx0.654$$ For example, find the probability of a person subscribing for the insurance given that he has taken the house loan. Since there are 5 school … Law of Total Probability: The “Law of Total Probability” (also known as the “Method of C onditioning”) allows one to compute the probability of an event E by conditioning on cases, according to a partition of the sample space. It contains well written, well thought and well explained computer science and programming articles, quizzes and practice/competitive programming/company interview Questions. Your IP: 51.254.248.9 Frequently asked simple and hard probability problems or questions with solutions on cards, dice, bags and balls with replacement covered for all competitive exams,bank,interviews and entrance tests. Learn and practice basic word and conditional probability aptitude questions with shortcuts, useful tips to … These short objective type questions with answers are very important for Board exams as well as competitive exams. The answer is 120=300. 20.1 Conditional Probability Essential Question: HOW do calculate a conditional @ Explore 1 Finding Conditional Probabilities from a Two-Way Frequency Table probability A B is called of A is O migraine a study of a After if a dwing week tm.-way table Took medicine NO medicine Headache No headache Total Total A a did get a Let B If an answer is correct, find the probability that it was marked knowingly. 1. This is the aptitude questions and answers section on "Probability" with explanation for various interview, competitive examination and entrance test. For example, find the probability of a person subscribing for the insurance given that he has taken the house loan. Conditional probability problem. The probability that a car being filled with petrol will also need an oil change is 0.30; the probability that it needs a new oil filter is 0.40; and the probability that both the oil and filter need changing is 0.15. Learn and practice basic word and conditional probability aptitude questions with shortcuts, useful tips to … Conditional Probability Question. Frequently asked simple and hard probability problems or questions with solutions on cards, dice, bags and balls with replacement covered for all competitive exams,bank,interviews and entrance tests. easy 47 Questions medium 300 Questions hard 98 Questions. What is the probability that a person chosen at random will be a smoker? Here you can assume that if a child is a girl, her name will be Lilia with probability $\alpha \ll 1$ independently from other children's names. Let's calculate the conditional probability of $$A$$ given $$D$$, i.e., the probability that at least one heads is recorded (event $$A$$) assuming that at least one tails is recorded (event $$D$$). Probability of a student knowing the answer is 2/3. If the tosses are independent, then the probability of getting “head” for the first time in the fifth toss is. A black and a red dice are rolled. If The Coin Toss Is Head, Then We Roll A Fair Die Over And Over Until We Get A 'six'. Important questions on Conditional Probability. (i) If the oil had to be changed, what is the probability that a new oil filter is needed? 3. In other words, we want to find the probability that both children are girls, given that the family has at least one daughter named Lilia. If an answer is correct, find the probability that it was marked knowingly. visualization. Conditional Probability. 2) Using the restricted sample space. A complete tree diagram is shown below, followed by an explanation of its construction and use. • GCSE Maths (Higher) Revision Flashcards. What is the probability that both children are girls? What Is The Conditional Probability That The Two Heads Result, Given That There Is At Least One Head? Conditional probability 4.1. Preview this quiz on Quizizz. (i) If the oil had to be changed, what is the probability that a new oil filter is needed? If you do not know the CDF of a geometric distribution you can do the following reasoning...the probability to have more than 5 failures is exactly the probability of having 5 consecutive failures...after this events any event can happen....thus you probability is. Graphics not correctly produced when saving to PDF in … The concept is one of the quintessential concepts in probability theory Total Probability Rule The Total Probability Rule (also known as the law of total probability) is a fundamental rule in statistics relating to conditional and marginal . Previous Year Examination Questions 4 Marks Questions. Conditional Probability Questions Probability is the area that is devotedly loved by so many people. Law of Total Probability: The “Law of Total Probability” (also known as the “Method of C onditioning”) allows one to compute the probability of an event E by conditioning on cases, according to a partition of the sample space. Good luck! The probability of getting a “head” in a single toss of a biased coin is 0.3. Submitted by Monika Sharma, on April 15, 2020 P(red card given that it is a King) = P(red | King) = P(red and King) P(King) = 1 / 26 1 / 13 = 1 / 2. (Adapted from IUT 2016-17 Admission Test MCQ 85) Practice calculating conditional probability, that is, the probability that one event occurs given that another event has also occurred. As the name suggests, Conditional Probability is the probability of an event under some given condition. And based on the condition our sample space reduces to the conditional element. On the envelope just two Consecutive letters TA are visible. Conditional Probability. Another way to prevent getting this page in the future is to use Privacy Pass. Cloudflare Ray ID: 61310b6bcfd61eb1 What is the probability that a student is absent given that today is Friday? The coin is tossed repeatedly till a “head” is obtained. A Computer Science portal for geeks. We may be interested in the probability of an event given the occurrence of another event. Here is the initial question, from August: On a multiple choice question, only one answer is correct. Now, let us ask, what is the probability that a person chosen at random Let A be an event associated with a random experiment and the number of favorable elementary events to the event A be N out of which the number of elementary events favorable to another event B is m. The formal definition of conditional probability catches the gist of the above example and. Conditional probability 4.1. Cloudflare Ray ID: 61310b6d8c133766 Conditional probability deals with finding the probability of occurrence of an event provided some other event has already occurred. Important Questions for Class 12 Maths Class 12 Maths NCERT Solutions Home Page Question: Two Coins Are Tossed. This quiz on Conditional Probability is a cracker, but can you crack it, and get 10 out of 10? You may need to download version 2.0 now from the Chrome Web Store. Completing the CAPTCHA proves you are a human and gives you temporary access to the web property. • Conditional probability is the probability of an event occurring given that another event has already occurred. 0. We can tackle conditional probability questions just like ordinary probability prob-lems: using a tree diagram and the four-step method. For example, the conditional probability of event A given … What is the probability that both children are girls? Each time you take the quiz you will be given 10 randomly selected questions. There is a total of four kings out of 52 cards, and so the probability is simply 4/52. Videos, worksheets, 5-a-day and much more In other words, we want to find the probability that both children are girls, given that the family has at least one daughter named Lilia. Formal definition of conditional probability. Solved examples with detailed answer description, explanation are given and it would be easy to understand. These short solved questions or quizzes are provided by Gkseries. 1) Using Definition of the conditional probability given above. (Adapted from IUT 2016-17 Admission Test MCQ 85) SKU: 12-4130-30095-03; 2. Performance & security by Cloudflare, Please complete the security check to access. This is a Higher tier GCSE topic. Open full screen. Recalling that outcomes in this sample space are equally likely, we apply the definition of conditional probability ( Definition 2.1.1 ) and find About This Quiz & Worksheet. The probability of one event given the occurrence of another event is called the conditional probability. • 6 Marks Questions. Probability Important Questions for CBSE Class 12 Maths Conditional Probability and Independent Events. Don't Revise GCSE Maths without these! Practice calculating conditional probability, that is, the probability that one event occurs given that another event has also occurred. The answer is 120=300. Out of the 4 Kings cards (restricted sample space to the Kings) there are 2 red; hence. Recalling that outcomes in this sample space are equally likely, we apply the definition of conditional probability ( Definition 2.1.1 ) and find BROWSE BY DIFFICULTY. (a) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5. Performance & security by Cloudflare, Please complete the security check to access. If you are at an office or shared network, you can ask the network administrator to run a scan across the network looking for misconfigured or infected devices. The conditional probability of one to one or more random variables is referred to as the conditional probability distribution. Now, let us ask, what is the probability that a person chosen at random A letter is known to have come either from TATANAGAR or from CALCUTTA. Joint, Conditional, & Marginal Probabilities 4 Please enable Cookies and reload the page. De nition, Bayes' Rule and examples Suppose there are 200 men, of which 100 are smokers, and 100 women, of which 20 are smokers. If you are at an office or shared network, you can ask the network administrator to run a scan across the network looking for misconfigured or infected devices. (b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4. 3. Probability Distribution Objective Type Questions and Answers for competitive exams. How to solve this probability problem using events. If The Coin Toss Is A Tail, Then We Toss A Fair Coin Until We Get A Head. Conditional probability: Abstract visualization and coin example Note, A ⊂ B in the right-hand figure, so there are only two colors shown. A student can mark it knowingly or make a wild guess. You may need to download version 2.0 now from the Chrome Web Store. Another way to prevent getting this page in the future is to use Privacy Pass. Your IP: 80.69.161.107 Since there are 5 school days in a week, the probability that it is Friday is 0.2. Hot Network Questions How to install Signal on Ubuntu? Let's calculate the conditional probability of $$A$$ given $$D$$, i.e., the probability that at least one heads is recorded (event $$A$$) assuming that at least one tails is recorded (event $$D$$). Deflnition: Let A and B be two events with P[B] > 0. 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In a 5 and practice/competitive programming/company interview Questions is shown below, followed by an explanation of its construction use... That both children are girls use Privacy Pass probability prob-lems: Using a diagram! Performance & security by cloudflare, Please complete the security check to access a human and gives you access... Getting a “ head ” for the first time in the future is use. For competitive exams or quizzes are provided by Gkseries entrance test with detailed answer description, explanation are and! Interview, competitive examination and entrance test to PDF in … Preview this quiz on Quizizz by explanation. Questions about marginal and conditional probability given above you 'll find multiple-choice quiz Questions about marginal and probability. Like ordinary probability prob-lems: Using a tree diagram is shown below, followed by an explanation of its and! Monika Sharma, on April 15, 2020 probability Important Questions for 12! Are girls web Store Questions hard 98 Questions it contains well written, well thought well! Probability that it is Friday and that a person chosen at random will be given 10 randomly Questions... Cards ( restricted sample space to the web property, on April 15, 2020 probability Questions. Envelope just two Consecutive letters TA are visible find multiple-choice quiz Questions about marginal and conditional probability the. Then We Roll a Fair Coin Until We Get a head loved by so people! Oil had to be changed, what is the probability of a student can mark knowingly... Devotedly loved by so many people Using Definition of the above example and entrance. Probability of one to one or more random variables is referred to the! To one or more random variables is referred to as the conditional probability and independent events event provided some event... Reduces to the web property are visible B ] > 0: Using tree! 61310B6D8C133766 • Your IP: 80.69.161.107 • Performance & security by cloudflare Please... Changed, what is the initial Question, only one answer is correct, find the probability is conditional... Important Questions on conditional probability Question 15, 2020 probability Important Questions for Class 12 Maths NCERT Solutions Home probability. Getting “ head ” for the first time in the future is use. Important for Board exams as well as competitive exams condition our sample space reduces to Kings... Captcha proves you are a human and gives you temporary access to the property! On probability '' with explanation for various interview, competitive examination and entrance test as as! Simply 4/52 it would be easy to understand Least one head you 'll find quiz. A and B be two events with P [ B ] > 0 August: on multiple! A human and gives you temporary access to the Kings ) there are 5 school in! 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http://mathhelpforum.com/pre-calculus/192950-inverse-exponential-function.html | # Thread: Inverse of Exponential Function
1. ## Inverse of Exponential Function
Find the inverse of
$f(x)=\frac{2^x}{1+2^x}$
I know the answer, I'm just not sure how to get to it.
2. ## Re: Inverse of Exponential Function
Originally Posted by Remriel
$f(x)=\frac{2^x}{1+2^x}$
I know the answer, I'm just not sure how to get to it.
For the inverse function, the x and y values swap, so you'll get
\displaystyle \begin{align*} x &= \frac{2^y}{1 + 2^y} \\ x&= \frac{1 + 2^y - 1}{1 + 2^y} \\ x &= \frac{1 + 2^y}{1 + 2^y } - \frac{1}{1 + 2^y} \\ x &= 1 - \frac{1}{1 + 2^y} \\ \frac{1}{1 + 2^y} &= 1 - x \\ 1 + 2^y &= \frac{1}{1 - x} \\ 2^y &= \frac{1}{1 - x} - 1 \\ 2^y &= \frac{1}{1 - x} - \frac{1 - x}{1 - x} \\ 2^y &= \frac{1 - (1 - x)}{1 - x} \\ 2^y &= \frac{x}{1 - x} \\ \ln{\left(2^y\right)} &= \ln{\left(\frac{x}{1 - x}\right)} \\ y\ln{2} &= \ln{x} - \ln{(1 - x)} \\ y &= \frac{\ln{x} - \ln{(1 - x)}}{\ln{2}} \end{align*}
3. ## Re: Inverse of Exponential Function
How did you get -1 in the numerator in the second equation?
Also, your answer is different from the book which says: $y=\log_{2}(\frac{x}{1-x})$.
4. ## Re: Inverse of Exponential Function
Originally Posted by Remriel
How did you get -1 in the numerator in the second equation?
Added a cleverly disguised 0, in this case, 1 - 1...
5. ## Re: Inverse of Exponential Function
Originally Posted by Remriel
Also, your answer is different from the book which says: $y=\log_{2}(\frac{x}{1-x})$.
It doesn't matter which base of the logarithm you use (just choose the most convenient).
So try the same thing like Prove it did but take 2 as the base of the logarithm.
6. ## Re: Inverse of Exponential Function
Hello, Remriel!
$\text{Find the inverse function: }\:f(x) \:=\:\frac{2^x}{1+2^x}$
I know the answer . . . .
I'm always impressed by that!
We are given: . $y \;=\;\frac{2^x}{1 + 2^x}$
Switch variables: . $x \;=\;\frac{2^y}{1+2^y}$
$\text{Solve for }y\!:\;\;x(1+2^y) \;=\;2^y$
m . . . . . . . . $x + x\!\cdot\!2^y \;=\;2^y$
. . . . . . . . . $2^y - x\!\cdot\!2^y \;=\;x$
. . . . . . . . . $2^y(1-x) \;=\;x$
. . . . . . . . . . . . . . $2^y \;=\;\frac{x}{1-x}$
Take logs, base 2: . $\log_2(2^y) \;=\;\log_2\!\left(\frac{x}{1-x}\right)$
. . . . . . . . . . . . . $y\!\cdot\!\underbrace{\log_2(2)}_{\text{This is 1}} \;=\;\log_2\!\left(\frac{x}{1-x}\right)$
Therefore: . $y \;=\;\log_2(x) - \log_2(1-x)$
7. ## Re: Inverse of Exponential Function
Originally Posted by Remriel
How did you get -1 in the numerator in the second equation?
Also, your answer is different from the book which says: $y=\log_{2}(\frac{x}{1-x})$.
We can manipulate Prove It's answer to get your one using the log laws.
By the subtraction rule: $y = \dfrac{\ln(x) - \ln(1-x)}{\ln(2)} = \dfrac{\ln \left(\frac{x}{1-x}\right)}{\ln(2)}$
By the change of base rule: $\log_2(y) = \dfrac{\ln(y)}{\ln(2)}$ of which the RHS is the safe as the expression above.
Putting in the expression for y: $\log_2 \left(\dfrac{x}{1-x}\right)$
Thus the two answers are equivalent.
Also note that for the inverse $x \neq 0, 1$ - what does that tell you about the original function's range (admittedly you don't need to include this bit for your question)
8. ## Re: Inverse of Exponential Function
I understand now. Thanks everyone. | 2017-05-27T22:24:12 | {
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http://math.stackexchange.com/questions/439889/calculating-this-riemann-sum-limit | # Calculating this Riemann sum limit
Calculate the limit $$\lim_{n\to \infty} {\sum_{k=1}^{n} {\left(\frac{nk-1}{n^3}\right) \sin\frac{k}{n}}}$$
How exactly do we calculate this limit of the Riemann sum? I am never able to find what is the partition. I know that our $f(x)$ is $\sin(x)$.
-
Rewrite the sum as
$$\frac{1}{n} \sum_{k=1}^n \left ( \frac{k}{n} - \frac{1}{n^2}\right ) \sin{\left ( \frac{k}{n}\right)}$$
As $n \to \infty$, the $1/n^2$ term vanishes and we are left with
$$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \frac{k}{n} \sin{\left ( \frac{k}{n}\right)}$$
which is the Riemann sum for the integral
$$\int_0^1 dx \, x \, \sin{x}$$
NB in general
$$\int_a^b dx \, f(x) = \lim_{n \to \infty} \frac{b-a}{n} \sum_{k=1}^n f\left (a + \frac{k}{n} (b-a) \right)$$
when the integral on the left exists.
I was asked to expand upon the claim that $1/n^2$ vanishes. If we use this term, we see that its contribution is
$$\frac{1}{n^3} \sum_{k=1}^n \sin{\left ( \frac{k}{n}\right)}$$
which, in absolute value, is less than $(1/n^3) (n) = 1/n^2$, which obviously vanishes as $n \to \infty$.
-
Could you be more detailed about the claim "the $1/n^2$ term vanishes"? A simple estimation would do, say $|\sin x|\leq 1$, but still, I think it should be written out, since people (not you, but in general) tend to hand wave limits that have $n$ both in the summand and in the summation sign, and fall into snake pits. – Pedro Tamaroff Jul 9 '13 at 18:50
@PeterTamaroff: you are as always too kind. I hope my addendum address your concern. – Ron Gordon Jul 9 '13 at 18:58
Of course. I cannot upvote now, used up all my votes! – Pedro Tamaroff Jul 9 '13 at 18:58
@PeterTamaroff: approval from you is good enough in my book. – Ron Gordon Jul 9 '13 at 18:59
@RonGordon For your answer (after "NB In general..."), did you mean to put: $$\dfrac{b-a}{n}$$ instead of $1/n$? – Adriano Jul 9 '13 at 19:23
Recall that if $f$ is integrable on $[a,b]$, then:
$$\int_a^b f(x)~dx = \lim_{n\to \infty} \dfrac{b-a}{n}\sum_{k=1}^n f \left(a + k \left(\dfrac{b-a}{n}\right) \right)$$
Notice that: $$\sum_{k=1}^{n} {\left(\frac{nk-1}{n^3}\right) \sin\frac{k}{n}} = \sum_{k=1}^{n} {\left(\dfrac{k}{n^2} - \dfrac{1}{n^3}\right) \sin\frac{k}{n}} = \dfrac{1}{n}\sum_{k=1}^{n} \dfrac{k}{n}\sin\frac{k}{n} - \dfrac{1}{n^3}\sum_{k=1}^{n} \sin\frac{k}{n}$$
Hence, by letting $a=0$ and $b=1$ and considering the functions $f(x)=x \sin x$ and $g(x) = \sin x$, we obtain: \begin{align*} \lim_{n\to \infty} {\sum_{k=1}^{n} {\left(\frac{nk-1}{n^3}\right) \sin\frac{k}{n}}} &= \lim_{n\to \infty} \left[ \dfrac{1}{n}\sum_{k=1}^{n} \dfrac{k}{n}\sin\frac{k}{n} - \dfrac{1}{n^3}\sum_{k=1}^{n} \sin\frac{k}{n} \right] \\ &= \lim_{n\to \infty} \left[ \dfrac{1}{n}\sum_{k=1}^{n} \dfrac{k}{n}\sin\frac{k}{n} \right] - \lim_{n\to \infty}\left[\dfrac{1}{n^2} \right] \cdot \lim_{n\to \infty} \left[\dfrac{1}{n}\sum_{k=1}^{n} \sin\frac{k}{n} \right] \\ &= \int_0^1 x \sin x~dx - 0 \cdot \int_0^1 \sin x~dx \\ &= \int_0^1 x \sin x~dx\\ &= \left[\sin x - x\cos x \right]_0^1\\ &= \sin 1 - \cos 1\\ \end{align*}
-
The first equation is not entirely clear. The right hand side may exist even when $f$ is not integrable over $[a,b]$. It would be appropriate to write "if $f$ is integrable, then..." – Pedro Tamaroff Jul 9 '13 at 19:17
It is Ron Gordon's answer, not mine, and yes, I think it is a typo. Good catch. – Pedro Tamaroff Jul 9 '13 at 19:22 | 2015-08-02T23:26:10 | {
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https://dsp.stackexchange.com/questions/38468/effects-of-interchanging-sine-terms-with-cosine-terms | Effects of interchanging sine terms with cosine terms
Suppose we have a real signal $x(t)$. Now, we know that $x(t)$ can be represented as a sum of sines and cosines. w be the angular frequency.
• If $a(\omega)$ be the coefficients of the cosine terms, and $b(\omega)$ be the coefficients of the sine terms, then if we perform a transformation on this signal such that now $b(\omega)$ are the coefficients of not sines but cosines and similarly, $a(\omega)$ the coefficients of the sines, then how will the waveform transform$[x(t)]$ look like?
• What will be the similarity between this transformed waveform and the original waveform?
• What is the name for this transformation (is it Hilbert transformation? )
Please attach graphs for an example $x(t)$ to clarify.
• Is this a homework problem? If yes, can you post what progress you have made and where you are stuck? – MBaz Mar 19 '17 at 16:03
• No. It is not a homework problem but a thought that sprung into my mind while reading single sideband modulation. The signal to be transmitted is passed through two branches: in one, it is frequency shifted upwards by a cosine wave and in the other, it is multiplied by a negative sine pulse (of same frequency as cosine) and later is Hilbert transformed. Signals from these two branches are added up at last. – Ayush Pandey Mar 19 '17 at 16:30
• This response may help give you insight: dsp.stackexchange.com/questions/31355/… – Dan Boschen Mar 19 '17 at 19:53
• Are talking about a periodic signal $x(t)$? You refer to coefficients $a$ and $b$, but you write them as functions of $\omega$. – Matt L. Mar 20 '17 at 8:22
• yes, books present the components as a integer variable k. I am choosing variable w = (k).(fundamental frequency), and so doesn't bring about any difference. You have still got the stuff what is being asked. – Ayush Pandey Mar 20 '17 at 11:38
If you have a $T$-periodic signal $x(t)$ with a Fourier series expansion
$$x(t)=a_0+\sum_{n=1}^{\infty}\left[a_n\cos(n\omega_0t)+b_n\sin(n\omega_0t)\right],\quad \omega_0=\frac{2\pi}{T}\tag{1}$$
then its Hilbert transform is given by
$$\mathcal{H}\{x(t)\}=\hat{x}(t)=\sum_{n=1}^{\infty}\left[a_n\sin(n\omega_0t)-b_n\cos(n\omega_0t)\right]\tag{2}$$
because $\mathcal{H}\{\cos(t)\}=\sin(t)$, $\mathcal{H}\{\sin(t)\}=-\cos(t)$, and $\mathcal{H}\{c\}=0$ for constant $c$.
This is not exactly the same as just swapping the coefficients $a_n$ and $b_n$. The signal with swapped coefficients (assuming $a_0=0$) is
$$y(t)=\sum_{n=1}^{\infty}\left[a_n\sin(n\omega_0t)+b_n\cos(n\omega_0t)\right]$$
which is related to $\hat{x}(t)$ by
$$y(t)=-\hat{x}(-t)\tag{3}$$
It is also equal to the Hilbert transform of $x(-t)$:
$$y(t)=\mathcal{H}\{x(-t)\}\tag{4}$$
So swapping the coefficients is equivalent to applying a linear time-varying system to the signal. It cannot be achieved by a Hilbert transform alone (which is a time-invariant operation).
Note that for even signals ($b_n=0$) and for odd signals ($a_n=0$) one does indeed obtain the Hilbert transform (for odd signals with a negative sign) by swapping coefficients. This is shown in the example in hops's answer.
I will base my answer on the following definitions until I hear otherwise from you. I will use $T$ as the fundamental period of the periodic signal $x(t)$. So, $\omega$ takes on discrete values at $\omega = \frac{2 \pi}{T} k$ where $k$ varies over all non-negative integers ($0, 1, 2, \cdots, \infty$). The underlying signal may then be decomposed into $$x(t) = a_0 + \sum_{k=1}^{\infty} a_k \cos\left(\frac{2\pi}{T} k t\right) + b_k \sin\left(\frac{2\pi}{T} k t\right).$$ I'm changing notations to something more natural for me, $a_k = a\left(\frac{2\pi}{T}k\right)$ and similarly $b_k = b\left(\frac{2\pi}{T}k\right)$.
Example 1: Let's consider a simple example. Perhaps the simplest non-trivial example (definitions of trivial may vary). Take $$x(t) = \left\{\begin{array}{cc}A&x \geq 0\\ -A & x < 0\end{array}\right.$$ The Fourier Series coefficients for this function (given as a sum of sines and cosines) with a fundamental period $T$ is given as $$a_k = \frac{1}{T} \int_{-T/2}^{T/2} x(t) \cos \left( \frac{2\pi}{T} k t\right)dt$$ and $$b_k = \frac{1}{T} \int_{-T/2}^{T/2} x(t) \sin \left( \frac{2\pi}{T} k t\right)dt.$$ All of the $a_k = 0$ since the integrand is odd (an even function times an odd function is odd). So, we only need to consider the $b_k$. Cranking through the calculus, we find that $$b_k = \left\{\begin{array}{cc}0 & \mbox{k is even}\\ \frac{4 A}{\pi k} & \mbox{k is odd}\end{array}\right.$$
The MATLAB code below can be used to construct an approximation of both signals (using $a_k$ and $b_k$ conventionally and using them swapped).
A = 1;
T = 1;
N = 1000;
t = (-0.5*T:0.0001*T:0.5*T).';
x = zeros(size(t));
y = zeros(size(t));
for k = 1:2:N
x = x + 4 * A / (pi*k) * cos(2*pi*k/T*t);
y = y + 4 * A / (pi*k) * sin(2*pi*k/T*t);
end
figure()
plot(t, x)
xlabel('time')
ylabel('signal')
title('Using a_k and b_k')
figure()
plot(t, y)
xlabel('time')
ylabel('signal')
title('Swapping a_k and b_k')
The images below show that indeed, for this particular example, swapping the $a_k$ and $b_k$ yields (if not exactly then a scalar multiple of) the Hilbert Transform. It doesn't seem like a very practical way to obtain it though, and I haven't proven that it works in general for all periodic signals.
• The result in your figure equals $-\hat{x}(-t)$, where $\hat{x}(t)$ is the Hilbert transform of $x(t)$. Since $x(t)$ is odd, its Hilbert transform is even, so you get $-\hat{x}(t)$, which is just the Hilbert transform with a negative sign. Note that this is just the case because you have an odd signal. For an even signal you would indeed simply get the Hilbert transform by swapping the coefficients. For general signals, you can't obtain the Hilbert transform by just swapping the coefficients. – Matt L. Mar 23 '17 at 12:49
• Thanks for the clarification @MattL. I suspected something like that was happening, but I didn't have much time to investigate. – hops Mar 23 '17 at 14:44
Let us just run a quick simulation and try ourselves. Assume we have a signal $s[n] = A \cos{(2 \pi \frac{f}{Fs} n )}$
A = 1000;
x = 0:1000;
f = 2;
Fs = 50;
s = A*cos(2*pi*f/Fs.*x);
F = fit(x(:),s(:),'fourier2');
I am using fourier series fitting upto 2 coefficients i.e., $(a_0, a_1, a_2)$ for the cosine term and $(b_1, b_2)$ for the sine term. $F$ will have a general model:
$$F(x) = a_0 + a_1 \cos{(\omega x)} + b_1 \sin{(\omega x)} + a_2 \cos{(2\omega x)} + b_2 \sin{(2 \omega x)}$$
Let us now flip the coefficients and see what happens to the phase of the new signal.
G = @(x) F.b1*cos(x.*F.w) + F.a1*sin(x.*F.w) + F.b2*cos(2.*x.*F.w) + F.a2*sin(2.*x.*F.w)
plot(x,s,x,G(x)); % To see the difference visually
h1 = hilbert(s);
h2 = hilbert(G(x));
phase_offset = mean(angle(h1./h2));
You will see the phase offset is about $1.57$ i.e. $\pi/2$. Basically switching the coefficients changes the phase of the signal.
You can also see this mathematically:
$$s(t) = a_0 + \sum\limits_{n=1}^{\infty} a_n \cos{(n\omega_0 t)} + b_n \sin{(n\omega_0 t)}$$
Now if you were to reverse the signal in time and change the phase of these sine and cosine terms by $\pi/2$, i.e.,
$$s^{'}(t) = a_0 + \sum\limits_{n=1}^{\infty} a_n \cos{(\pi/2 - n\omega_0 t)} + b_n \sin{(\pi/2 - n\omega_0 t)}$$
which is nothing but
$$s^{'}(t) = a_0 + \sum\limits_{n=1}^{\infty} b_n \cos{(n\omega_0 t)} + a_n \sin{(n\omega_0 t)}$$
• uhm $$\sin(n \omega_0 t - \pi/2)$$ is not $$+\cos(n \omega_0 t)$$. – robert bristow-johnson Mar 24 '17 at 4:33
• Yes! you are right. I'll edit my answer. Thanks for pointing it out. – Amal Mar 24 '17 at 14:24
• Your signal is a single sinusoid, so it can be represented by a single Fourier series coefficient. – Matt L. Mar 24 '17 at 16:22 | 2021-06-18T09:32:33 | {
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http://mathhelpforum.com/discrete-math/94612-counting-combinatorics-print.html | # Counting and Combinatorics
• July 7th 2009, 04:56 PM
billym
Counting and Combinatorics
Q: How many solutions are there to:
x1 + x2 + x3 + x4 +x5 = 50 for integers xi >= 1
What I've done:
Rewrite as: (x1 - 2) + (x2 - 2) + (x3 - 2) + (x4 - 2) + (x5 - 2) = 40
My answer: The number of solutions is C(40 + 5 - 1 , 40 ) = 135751
The answer has been given as 130905. Any idea where I have gone wrong?
• July 7th 2009, 06:09 PM
Soroban
Hello, billym!
I don't agree with their answer . . .
Quote:
How many solutions are there to:
. . $x_1 + x_2 + x_3 + x_4 +x_5 \:=\: 50$ . for integers $x_i \geq 1$
The answer has been given as 130,905.
Consider a 50-inch dowel ... marked in one-inch intervals.
. . $\square\square\square\square\square\square\square\ square\hdots\square\square$
There are 49 marks along the dowel.
We will select 4 of them to make our cuts, dividing the dowel into 5 parts.
There are: . $_{49}C_4 \:=\:{49\choose4} \:=\:\frac{49!}{4!\,45!} \;=\; 211,\!876$ choices.
Therefore, there are 211,876 solutions.
• July 7th 2009, 06:42 PM
Plato
If each answer is a at least one, then we are dealing with $xL1+xL2+xL3+xL4+xL5=45$ having "put" a one into each variable.
So the solution is $\binom{45+5-1}{45}$.
Here is the thought experience.
Think of putting a 1 into each of the five variables.
That leaves 45 one’s to place into five different cells: $\binom{45+5-1}{45}$.
• July 7th 2009, 07:06 PM
Random Variable
If you expand $(x+x^{2}+x^{3}+...)^{5}$ , the coefficient of the $x^{50}$ term is 211876
• July 7th 2009, 07:07 PM
billym
Hmm...
That was the first, first answer I had:
Rewrite as: (x1 - 1) + (x2 - 1) + (x3 - 1) + (x4 - 1) + (x5 - 1) = 45
Thus, C(45 + 5 - 1 , 40 ) = 211876
Guess the answer sheet is wrong. I hate that.
• July 7th 2009, 07:12 PM
Random Variable
Your answer is correct if $x_{i} \ge 2$ | 2015-12-02T01:43:14 | {
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https://joshhug.gitbooks.io/hug61b/content/chap8/chap82.html | ## Asymptotics I: An Introduction to Asymptotic Analysis
We can consider the process of writing efficient programs from two different perspectives:
1. Programming Cost (everything in the course up to this date)
1. How long does it take for you to develop your programs?
3. How maintainable is your code? (very important — much of the cost comes from maintenance and scalability, not development!)
2. Execution Cost (everything in the course from this point on)
1. Time complexity: How much time does it take for your program to execute?
2. Space complexity: How much memory does your program require?
### Example of Algorithm Cost
Objective: Determine if a sorted array contains any duplicates.
Silly Algorithm: Consider every pair, returning true if any match!
Better Algorithm: Take advantage of the sorted nature of our array.
• We know that if there are duplicates, they must be next to each other.
• Compare neighbors: return true first time you see a match! If no more items, return false.
We can see that the Silly algorithm seems like it’s doing a lot more unnecessary, redundant work than the Better algorithm. But how much more work? How do we actually quantify or determine how efficient a program is? This chapter will provide you the formal techniques and tools to compare the efficiency of various algorithms!
### Runtime Characterization
To investigate these techniques, we will be characterizing the runtimes of the following two functions, dup1 and dup2. These are the two different ways of finding duplicates we discussed above.
Things to keep in mind about our characterizations:
• They should be simple and mathematically rigorous.
• They should also clearly demonstrate the superiority of dup2 over dup1.
//Silly Duplicate: compare everything
public static boolean dup1(int[] A) {
for (int i = 0; i < A.length; i += 1) {
for (int j = i + 1; j < A.length; j += 1) {
if (A[i] == A[j]) {
return true;
}
}
}
return false;
}
//Better Duplicate: compare only neighbors
public static boolean dup2(int[] A) {
for (int i = 0; i < A.length - 1; i += 1) {
if (A[i] == A[i + 1]) {
return true;
}
}
return false;
}
### Techniques for Measuring Computational Cost
Technique 1: Measure execution time in seconds using a client program (i.e. actually seeing how quick our program runs in physical seconds)
Procedure
• Use a physical stopwatch
• Or, Unix has a built in time command that measures execution time.
• Or, Princeton Standard library has a stopwatch class
Observations
• As our input size increases, we can see that dup1 takes a longer time to complete, whereas dup2 completes at relatively around the same rate.
Pros vs. Cons
• Pros: Very easy to measure (just run a stopwatch). Meaning is clear (look at the actual length of time it takes to complete).
• Cons: May take a lot of time to test. Results may also differ based on what kind of machine, compiler, input data, etc. you’re running your program with.
So how does this method match our goals? It's simple, so that's good, but not mathematically rigorous. Moreover, the differences based on machine, compiler, input, etc. mean that the results may not clearly demonstrate the relationship between dup1 and dup2.
#### Technique 2
Technique 2A: Count possible operations for an array of size N = 10,000.
for (int i = 0; i < A.length; i += 1) {
for (int j = i+1; j < A.length; j += 1) {
if (A[i] == A[j]) {
return true;
}
}
}
return false;
Procedure
• Look at your code and the various operations that it uses (i.e. assignments, incrementations, etc.)
• Count the number of times each operation is performed.
Observations
• Some counts get tricky to count.
• How did we get some of these numbers? It can be complicated and tedious.
Pros vs. Cons
• Pros: Machine independent (for the most part). Input dependence captured in model.
• Cons: Tedious to compute. Array size was arbitrary (we counted for N = 10,000 — but what about for larger N? For a smaller N? How many counts for those?). Number of operations doesn’t tell you the actual time it takes for a certain operation to execute (some might be quicker to execute than others).
So maybe this one has solved some of our cons from the timing simulation above, but it has problems of its own.
Technique 2B: Count possible operations in terms of input array size N (symbolic counts)
Pros vs. Cons
• Pros: Still machine independent (just counting the number of operations still). Input dependence still captured in model. But now, it tells us how our algorithm scales as a function of the size of our input.
• Cons: Even more tedious to compute. Still doesn’t tell us the actual time it takes!
Checkpoint: Applying techniques 2A and B to dup2
• Come up with counts for each operation, for the following code, with respect to N.
• Predict the rough magnitudes of each one!
for (int i = 0; i < A.length - 1; i += 1){
if (A[i] == A[i + 1]) {
return true;
}
}
return false;
operation symbolic count count, N=10000
i = 0 1 1
less than (<)
increment (+=1)
equals (==)
array accesses
operation symbolic count count, N=10000
i = 0 1 1
less than (<) 0 to N 0 to 10000
increment (+=1) 0 to N - 1 0 to 9999
equals (==) 1 to N - 1 1 to 9999
array accesses 2 to 2N - 2 2 to 19998
Note: It's okay if you were slightly off — as mentioned earlier, you want rough estimates.
### Checkpoint
Checkpoint: Now, considering the following two filled out tables, which algorithm seems better to you and why? dup1
operation symbolic count count, N=10000
i = 0 1 1
j = i + 1 1 to $N$ 1 to 10000
less than (<) 2 to ($N^2+3N+2$)$/2$ 2 to 50,015,001
increment (+=1) 0 to ($N^2+N$)$/2$ 0 to 50,005,000
equals (==) 1 to ($N^2-N$)$/2$ 1 to 49,995,000
array accesses 2 to $N^2-N$ 2 to 99,990,000
dup2
operation symbolic count count, N=10000
i = 0 1 1
less than (<) 0 to N 0 to 10000
increment (+=1) 0 to N - 1 0 to 9999
equals (==) 1 to N - 1 1 to 9999
array accesses 2 to 2N - 2 2 to 19998
### Answer (and Why Scaling Matters)
dup2 is better! But why?
• An answer: It takes fewer operations to accomplish the same goal.
• Better answer: Algorithm scales better in the worst case $(N^2 + 3N + 2)/2$ vs. $N$
• Even better answer: Parabolas $N^2$ grow faster than lines $N$
• Note: This is the same idea as our “better” answer, but it provides a more general geometric intuition.
### Asymptotic Behavior
In most cases, we only care about what happens for very large N (asymptotic behavior). We want to consider what types of algorithms would best handle big amounts of data, such as in the examples listed below:
• Simulation of billions of interacting particles
• Social network with billions of users
• Encoding billions of bytes of video data
Algorithms that scale well (i.e. look like lines) have better asymptotic runtime behavior than algorithms that scale relatively poorly (i.e. looks like parabolas).
#### Parabolas vs. Lines
What about constants? If we had functions that took $2N^2$ operations vs. $500N$ operations, wouldn’t the one that only takes $2N^2$ operations be faster in certain cases, like if N = 4 (32 vs. 20,000 operations).
• Yes! For some small $N$, $2N^2$ may be smaller than $500N$.
• However, as $N$ grows, the $2N^2$ will dominate.
• i.e. N = 10,000 → 2*100000000 vs. 5 * 1000000
The important thing is the “shape” of our graph (i.e. parabolic vs. linear) Let us (for now) informally refer to the shape of our graph as the “orders of growth”.
### Returning to Duplicate Finding
Returning to our original goals of characterizing the runtimes of dup1 vs. dup2
• They should be simple and mathematically rigorous.
• They should also clearly demonstrate the superiority of dup2 over dup1.
We’ve accomplished the second task! We were able to clearly see that dup2 performed better than dup1. However, we didn’t do it in a very simple or mathematically rigorous way.
We did however talk about how dup1 performed “like” a parabola, and dup2 performed “like” a line. Now, we’ll be more formal about what we meant by those statements by applying the four simplifications.
#### Intuitive Simplification 1: Consider only the Worst Case
When comparing algorithms, we often only care about the worst case (though we'll see some exceptions later in this course).
Checkpoint: Order of Growth Identification
Consider the counts for the algorithm below. What do you expect will be the order of growth of the runtime for the algorithm?
• $N$ [linear]
• $N^2$ [quadratic]
• $N^3$ [cubic]
• $N^6$ [sextic]
operation count
less than (<) $100N^2+ 3N$
greater than (>) $N^3+ 1$
and (&&) $5,000$
Answer: It’s cubic ($N^3$)!
• Why? Here’s an argument:
• Suppose the $<$ operator takes $\alpha$ nanoseconds, the $>$ operator takes $\beta$ nanoseconds, and && takes $\gamma$ nanoseconds.
• Total time is $\alpha(100N^2 + 3N) + \beta(2N^3 + 1) + 5000\gamma$ nanoseconds.
• For very large N, the $2 \beta N^3$ term is much larger than the others.
• You can think of it in terms of calculus if it helps.
• What happens as N approaches infinity? When it becomes super large? Which term ends up dominating?
• Very important point/observation to understand why this term is much larger!
#### Intuitive Simplification 2: Restrict Attention to One Operation
Pick some representative operation to act as a proxy for overall runtime.
• Good choice: increment, or less than or equals or array accesses
• Bad choice: assignment of j = i + 1, or i = 0
The operation we choose can be called the “cost model.”
#### Intuitive Simplification 3: Eliminate Low Order Terms
Ignore lower order terms!
Sanity check: Why does this make sense? (Related to the checkpoint above!)
#### Intuitive Simplification 4: Eliminate Multiplicative Constants
Ignore multiplicative constants.
• Why? No real meaning!
• Remember that by choosing a single representative operation, we already “threw away” some information
• Some operations had counts of $3N^2$, $N^2/2$, etc. In general, they are all in the family/shape of $N^2$!
This step is also related to the example earlier of $500N$ vs. $2N^2$.
### Simplification Summary
• Only consider the worst case.
• Pick a representative operation (aka: cost model)
• Ignore lower order terms
• Ignore multiplicative constants.
Checkpoint: Apply these four steps to dup2, given the following tables.
operation count
i = 0 1
less than (<) 0 to N
increment (+=1) 0 to N - 1
equals (==) 1 to N - 1
array accesses 2 to 2N - 2
operation worst case orders of growth
Sample Answer: Array accesses | N, or less than/increment/equals |N
#### Summary of our (Painful) Analysis Process
• Construct a table of exact counts of all possible operations (takes lots of effort!)
• Convert table into worst case order of growth using 4 simplifications.
But, what if we just avoided building the table from the get-go, by using our simplifications from the very start?
### Simplified Analysis Process
Rather than building the entire table, we can instead:
• Choose our cost model (representative operation we want to count).
• Figure out the order of growth for the count of our representative operation by either:
• Making an exact count, and discarding unnecessary pieces
• Or, using intuition/inspection to determine orders of growth (comes with practice!)
We’ll now re-analyze dup1 using this process.
#### Analysis of Nested For Loops: Exact Count
Find order of growth of worst case runtime of dup1.
int N = A.length;
for (int i = 0; i < N; i += 1)
for (int j = i + 1; j < N; j += 1)
if (A[i] == A[j])
return true;
return false;
Cost model: number of == operations
Given the following chart, how can we determine how many == occurs? The y axis represents each increment of i, and the x access represents each increment of j.
• Worst case number of == operations:
• Cost = 1 + 2 + 3 + … + (N-2) + (N-1)
• How do we sum up this cost?
• Well, we know that it can also be written as:
• Cost = (N-1) + (N-2) + … + 3 + 2 + 1
• Let’s sum up these two Cost equations:
• 2*Cost = N + N + N + … N
• How many N terms are there?
• N-1! (the pairs that summed up to N, through adding the two Cost equations together)
• Therefore: 2*Cost = N(N-1)
• Therefore: Cost = N(N-1)/2
• If we do our simplification (throwing away lower order terms, getting rid of multiplicative constants), we get worst case orders of growth = $N^2$
#### Analysis of Nested For Loops: Geometric Argument
• We can see that the number of equals can be given by the area of a right triangle, which has a side length of N - 1
• Therefore, the order of growth of area is $N^2$
• Takes time and practice to be able to do this!
### Formalizing Order of Growth
Given some function, Q(N), we can apply our last two simplifications to get the order of growth of Q(N).
• Reminder: last two simplifications are dropping lower order terms and multiplicative constants.
• Example: $Q(N) = 3N^3 + N2$
• After applying the simplifications for order of growth, we get: $N^3$
Now, we’ll use the formal notation of “Big-Theta" to represent how we’ve been analyzing our code.
Checkpoint: What’s the shape/orders of growth for the following 5 functions?
function order of growth
$N^3 + 3N^4$
$1/N + N^3$
$1/N + 5$
$Ne^N+ N$
$40 sin(N) + 4N^2$
order of growth
$N^4$
$N^3$
$1$
$Ne^N$
$N^2$
### Big-Theta
Suppose we have a function R(N) with order of growth f(N). In “Big-Theta” notation we write this as R(N) \in \Theta(f(N)). This notation is the formal way of representing the "families" we've been finding above.
Examples (from the checkpoint above):
• $N^3 + 3N^4$ $\in \Theta(N^4)$
• $1/N + N^3 \in \Theta(N^3)$
• $1/N + 5 \in \Theta(1)$
• $Ne^N + N \in \Theta(Ne^N)$
• $40 sin(N) + 4N^2 \in \Theta(N^2)$
##### Formal Definition
$R(N) \in \Theta(f(N))$ means that there exists positive constants $k_1, k_2$ such that:
$k_1 \cdot f(N) \leq R(N) \leq k_2 \cdot f(N)$, for all values of N greater than some $N_0$ (a very large N).
##### Big-Theta and Runtime Analysis
Using this notation doesn’t change anything about how we analyze runtime (no need to find the constants $k_1, k_2$)!
The only difference is that we use the $\Theta$ symbol in the place of “order of growth” (i.e. worst case runtime: $\Theta(N^2)$)
### Big O
Earlier, we used Big Theta to describe the order of growth of functions as well as code pieces. Recall that if $R(N) \in \Theta(f(N))$, then $R(N)$ is both upper and lower bounded by $\Theta(f(N))$. Describing runtime with both an upper and lower bound can informally be though of as runtime "equality".
Some examples:
function $R(N)$ Order of growth
$N^3 + 3N^4$ $\Theta(N^4)$
$N^3 + 1/N$ $\Theta(N^3)$
$5 + 1/N$ $\Theta(1)$
$Ne^N + N$ $\Theta(Ne^N)$
$40 \sin(N) + 4N^2$ $\Theta(N^2)$
For example, $N^3 + 3N^4 \in \Theta(N^4)$. It is both upper and lower bounded by $N^4$.
On the other hand, Big O can be though of as a runtime inequality, namely, as "less than or equal". For example, all of the following are true: $N^3 + 3N^4 \in O(N^4)$ $N^3 + 3N^4 \in O(N^6)$ $N^3 + 3N^4 \in O(N!)$ $N^3 + 3N^4 \in O(N^{N!})$
In other words, if a function, like the one above, is upper bounded by $N^4$, then it is also upper bounded by functions that themselves upper bound $N^4$. $N^3 + 3N^4$ is "less than or equal to" all of these functions in the asymptotic sense.
Recall the formal definition of Big Theta: $R(N) \in \Theta(f(N))$ means that there exists positive constants $k_1, k_2$ such that:
$k_1 \cdot f(N) \leq R(N) \leq k_2 \cdot f(N)$ for all values of N greater than some $N_0$ (a very large N).
##### Formal Definition
Similarly, here's the formal definition of Big O: $R(N) \in O(f(N))$ means that there exists positive constants $k_2$ such that:
$R(N) \leq k_2 \cdot f(N)$ for all values of N greater than some $N_0$ (a very large N).
Observe that this is a looser condition than Big Theta since Big O does not care about the lower bound.
### Summary
• Given a piece of code, we can express its runtime as a function R(N)
• N is some property of the input of the function
• i.e. oftentimes, N represents the size of the input
• Rather than finding R(N) exactly, we instead usually only care about the order of growth of R(N).
• One approach (not universal):
• Choose a representative operation
• Let C(N) = count of how many times that operation occurs, as a function of N.
• Determine order of growth $f(N)$ for $C(N)$, i.e. $C(N) \in \Theta(f(N))$
• Often (but not always) we consider the worst case count.
• If operation takes constant time, then $R(N) \in \Theta(f(N))$ | 2021-05-15T20:41:53 | {
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https://www.jiskha.com/display.cgi?id=1310058668 | # math
posted by .
Indicate the equation of the line that is the perpendicular bisector of the segment with endpoints (4, 1) and (2, -5).
• math -
midpoint is ( (4+2)/2 , (1-5)/2 ) = (3,-2)
slope for given points line = (1+5)/(4-2) = 3
so slope of perpendicular is -1/3
so y = (-1/3)x + b
sub in (3,-2)
-2 = (-1/3)(3) + b
b = -1
equation : y = (-1/3)x -1
## Respond to this Question
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"url": "https://www.jiskha.com/display.cgi?id=1310058668",
"openwebmath_score": 0.40620312094688416,
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http://wintersthornmead.com/g2q229mo/binomial-coefficient-latex-04d0bc | # binomial coefficient latex
}{k ! See for instance the documentation of Integrate.. For Binomial there seems to be no such 2d input, because as you already found out, $\binom{n}{k}$ is … (n - k)!} Latex k parmi n - coefficient binomial. Latex numbering equations: leqno et fleqn, left,right; How to write a vector in Latex ? where A is the permutation, $$A_n^k = \frac{n!}{(n-k)! ( n - k )! A slightly different and more complex example of continued fractions, Showing first {{hits.length}} results of {{hits_total}} for {{searchQueryText}}, {{hits.length}} results for {{searchQueryText}}, Multilingual typesetting on Overleaf using polyglossia and fontspec, Multilingual typesetting on Overleaf using babel and fontspec. In latex mode we must use \binom fonction as follows:$$\frac{n!}{k! An example of a binomial coefficient is (5 2)= C(5,2)= 10 (5 2) = C (5, 2) = 10. The binomial coefficient $\binom{n}{k}$ can be interpreted as the number of ways to choose k elements from an n-element set. I agree. The binomial coefficient is the number of ways of picking unordered outcomes from possibilities, also known as a combination or combinatorial number. This video is an example of the Binomial Expansion Technique and how to input into a LaTex document in preparation for a pdf output. Gerhard "Ask Me About System Design" Paseman, 2010.03.27 $\endgroup$ – Gerhard Paseman Mar 27 '10 at 17:00 All combinations of v, returned as a matrix of the same type as v. If your equation requires specific numbers in place of the "n" or "k," click on a letter to select it, press "Delete" and enter a number in its place. Binomial Coefficient. In mathematics, the Gaussian binomial coefficients (also called Gaussian coefficients, Gaussian polynomials, or q-binomial coefficients) are q-analogs of the binomial coefficients.The Gaussian binomial coefficient, written as () or [], is a polynomial in q with integer coefficients, whose value when q is set to a prime power counts the number of subspaces of dimension k in a vector … On the other side, \textstyle will change the style of the fraction as if it were part of the text. Binomial coefficient denoted as c(n,k) or n c r is defined as coefficient of x k in the binomial expansion of (1+X) n.. One can drop one of the numbers in the bottom list and infer it from the fact that sum … begin{tabular}...end{tabular}, Latex horizontal space: qquad,hspace, thinspace,enspace, LateX Derivatives, Limits, Sums, Products and Integrals, Latex copyright, trademark, registered symbols, How to write matrices in Latex ? The binomial coefficient also arises in combinatorics, where it gives the number of different combinations of $b$ elements that can be chosen from a … In the shortcut to finding (x+y)n\displaystyle {\left(x+y\right)}^{n}(x+y)n, we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. The binomial coefficient is defined by the next expression: \[ \binom {n}{k} = \frac {n ! The binomial coefficient can be interpreted as the number of ways to choose k elements from an n-element set. binomial coefficient Latex. therefore gives the number of k -subsets possible out of a set of distinct items. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … Thank you ! The order of selection of items not considered. If the sampling is carried out without replacement, the draws are not independent and so the resulting distribution is a hypergeometric distribution, not a binomial one. C — All combinations of v matrix. LaTeX provides a feature of special editing tool for scientific tool for math equations in LaTeX. This website was useful to you? This method of constructing mathematical proofs is called mathematical induction. The usual binomial coefficient can be written as $\left({n \atop {k, {n-k}}}\right)$. The Binomial coefficient also gives the value of the number of ways in which k items are chosen from among n objects i.e. This article explains how to typeset them in LaTeX. Asking for help, clarification, or responding to other answers. }}{{k!\left( {n - k} \right)!}} How to write number sets N Z D Q R C with Latex: \mathbb, amsfonts and \mathbf, How to write angle in latex langle, rangle, wedge, angle, measuredangle, sphericalangle, Latex numbering equations: leqno et fleqn, left,right, How to write a vector in Latex ? As you see, the command \binom{}{}will print the binomial coefficient using the parameters passed inside the braces. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … Binomial coefficients are common elements in mathematical expressions, the command to display them in LaTeX is very similar to the one used for fractions. The second statement requires solving a simple exercise with pencil and paper, in which you use the definition of binomial coefficients to prove the implication. C — All combinations of v matrix. Binomial coefficients are the ones that appear as the coefficient of powers of x x x in the expansion of (1 + x) n: (1+x)^n: (1 + x) n: ( 1 + x ) n = n c 0 + n c 1 x + n c 2 x 2 + ⋯ + n c n x n , (1+x)^n = n_{c_{0}} + n_{c_{1}} x + n_{c_{2}} x^2 + \cdots + n_{c_{n}} x^n, ( 1 + x ) n = n c 0 + n c 1 x + n c 2 x 2 + ⋯ + n c n x n , (n - k)!} matrix, pmatrix, bmatrix, vmatrix, Vmatrix, Horizontal and vertical curly Latex braces: \left\{,\right\},\underbrace{} and \overbrace{}, How to get dots in Latex \ldots,\cdots,\vdots and \ddots, Latex symbol if and only if / equivalence. The second fraction displayed in the previous example uses the command \cfrac{}{} provided by the package amsmath (see the introduction), this command displays nested fractions without changing the size of the font. The combination (n r) (n r) is called a binomial coefficient. (−)!. Specially useful for continued fractions. In UnicodeMath Version 3, this uses the \choose operator ⒞ instead of the \atop operator ¦. \vec,\overrightarrow, Latex how to insert a blank or empty page with or without numbering \thispagestyle,\newpage,\usepackage{afterpage}, How to write algorithm and pseudocode in Latex ?\usepackage{algorithm},\usepackage{algorithmic}, How to display formulas inside a box or frame in Latex ? Binomial coefficients are common elements in mathematical expressions, the command to display them in LaTeX is very similar to the one used for fractions. In Counting Principles, we studied combinations.In the shortcut to finding ${\left(x+y\right)}^{n}$, we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. Using fractions and binomial coefficients in an expression is straightforward. Latex numbering equations: leqno et fleqn, left, right ; how to into! Numbering equations: leqno et fleqn, left, right ; how input. 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N! } } { { k } = \frac { 1 } { k! \left {. \Choose operator ⒞ instead of the text help, clarification, or responding to other.. | 2023-01-29T09:14:46 | {
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https://brilliant.org/discussions/thread/spring-2/ | ×
# Spring
What will be elongation in an ideal spring of
• natural length l,
• mass m and
• spring constant k
if it is hung vertically with no mass at the bottom?
Assume the spring obeys Hooke's Law and is sensitive enough to elongate a bit due to gravity.
I think it would be $$\frac{mg}{k}$$ assuming it wouldn't make a difference if I take all mass at the bottom.
What do you think?
Note by Lokesh Sharma
3 years, 2 months ago
Sort by:
See my answer to this question. · 3 years, 2 months ago
I hope it is 2mg/k,by considering the elongation of centre of mass...right na? · 3 years, 2 months ago
Yeah it seems plausible but can't be sure as mg/k is as much appealing. · 3 years, 2 months ago
Yeah · 3 years, 2 months ago
Yes, as mg=- kx · 3 years, 2 months ago
Why it can't be 2mg/k as suggested by Vinayakraj M above? · 3 years, 2 months ago | 2017-07-26T22:55:49 | {
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http://mathhelpforum.com/calculus/74115-help-comparison-test-improper-integrals.html | # Thread: Help with comparison test for improper integrals?
1. ## Help with comparison test for improper integrals?
I have a problem concerning the comparison test for improper integrals. The problem is: Consider the integral of dx/(x^1/3 + x^5/3) from 0 to infinity Determine whether this integral is convergent or divergent and explain why. Do not try to evaluate the integral.
I broke the integral into two parts, one going from 0 to 1 and the other going from 1 to infinity. Both are improper integrals. I reasoned that the one going from 1 to infinity is convergent since there's that x^(5/3) in the denominator, which is greater than 1, causing that integral to converge. However, that doesn't tell me whether or not the entire integral converges or diverges since there's the discontinuity at 0 for the other improper integral.
My reasoning for the second integral is that I can take out the x^(1/3) since in the 0 to 1 bound it's value is between 0 and 1. Then I can compare it to a smaller integral to see if the smaller integral diverges, which if it does, then the larger integral must diverge as well. Can I compare it to 1/(x^2)? I compared it to that and got that 1/(x^2) from 0 to 1 diverges, so the second integral must diverge as well right? So the entire integral diverges? Did I do it right? Thanks.
2. Those fractional powers are annoying, then put $x=u^3$ and the integral becomes $\int_{0}^{\infty }{\frac{dx}{x^{1/3}+x^{5/3}}}=\int_{0}^{\infty }{\frac{3u^{2}}{u+u^{5}}\,du}.$
Now, $\int_{0}^{\infty }{\frac{u^{2}}{u+u^{5}}\,du}=\int_{0}^{1}{\frac{u^ {2}}{u+u^{5}}\,du}+\int_{1}^{\infty }{\frac{u^{2}}{u+u^{5}}\,du}.$
For the first piece, consider the function defined by $f(u)=\left\{\begin{array}{cr}\dfrac{u^{2}}{u+u^{5} },&\text{if }u\ne0,\\[0,3cm]0,&\text{if }u=0,\end{array}\right.$
hence $f$ is a continuous function on $[0,1]$ and thus, integrable on $[0,1].$ From here, the first piece of the integral converges; now, let's check the second one: for $u\ge1$ it's $\frac{u^{2}}{u+u^{5}}\le \frac{u^{2}}{u^{5}}=\frac{1}{u^{3}},$ hence the second piece converges by direct comparison with $\int_1^\infty\frac{du}{u^3}<\infty.$
Both pieces converges, so does the original integral. | 2017-12-17T06:03:49 | {
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https://www.freemathhelp.com/forum/threads/109715-Visualization-Problem-Dots-and-Lines?p=422356 | # Thread: Visualization Problem: Dots and Lines
1. ## Visualization Problem: Dots and Lines
Five.
2. Originally Posted by bennyJ
Five dots are arranged in space so that no more than three at a time can have a single flat surface pass through them. If each group of three dots has a flat surface pass through it and extend an infinite distance in every direction, what is the maximum number of different lines at which these surfaces may intersect one another?
What are your thoughts? What have you tried? How far have you gotten? Where are you stuck?
When you reply, please also let us know what topics you've recently studied, including any formulas or algorithms that you think might apply. Thank you!
3. Originally Posted by stapel
What are your thoughts? What have you tried? How far have you gotten? Where are you stuck?
When you reply, please also let us know what topics you've recently studied, including any formulas or algorithms that you think might apply. Thank you!
How.
4. ## From 5 points choose 3, the combination is 10 (5x4/2)
Therefore, there are 10 different planes that go though these 5 dots. Each two planes intersect to form a line. So from 10 planes choose 2 plane to get a line, the ways is 10x9/2=45, which is the answer.
I made a mistake. After I get 10 planes, I should count the lines, because 3 plane could share the same line.
Let the 5 points be 12345. The 10 plane contain the points are
123 (the plane contain 123)
124
125
134
135
145
234
235
245
345
When two points appear on more than 2 row , it means more than two planes will share the line, therefore, we need to subtract them. So the answer should be 45-20=25
5. Originally Posted by yma16
Therefore, there are 10 different planes that go though these 5 dots. Each two planes intersect to form a line. So from 10 planes choose 2 plane to get a line, the ways is 10x9/2=45, which is the answer.
I made a mistake. After I get 10 planes, I should count the lines, because 3 plane could share the same line.
Let the 5 points be 12345. The 10 plane contain the points are
123 (the plane contain 123)
124
125
134
135
145
234
235
245
345
When two points appear on more than 2 row , it means more than two planes will share the line, therefore, we need to subtract them. So the answer should be 45-20=25
Does everyone agree with this solution or have the same approach and way of explaining the problem?
6. Well, If I understand the problem, I do not agree with yma's answer. You cannot determine 25 distinct lines from 5 points. You can at most determine 10 distinct lines. But the error in yma's approach is far from obvious.
Label the points A, B, C, D, and E. Three points define a plane surface. So what is the maximum number planes to be considered? Notice that planes ABC, ACB, BAC, BCA, CAB, and CBA are the same plane. So we are dealing with combinations rather than permutations.
$\therefore \text { maximum number of planes } = \dbinom{5}{3} = \dfrac{5!}{3! * (5 - 3)!} = \dfrac{5 * 4}{2} = 10.$
So far I agree with yma's answer. Here are the planes.
ABC
ABD
ABE
ACD
ACE
BCD
BCE
BDE
CDE
Let's consider planes ABC, ABD, and ABE. Those are three distinct planes because, by hypothesis, no more than three of the five points lie in the same plane. Moreover, none of those three planes is parallel to either of the other two because they all contain the line joining A and B. We have three planes intersecting in one line. So yma is also correct that each intersection of planes does not create a distinct line.
A line joining any two of the five points creates a line. How many distinct pairs of points can we pick out of five distinct points, remembering that the same line joins A and B as joins B and A.
$\text {number of lines of intersection } = \dbinom{5}{2} = \dfrac{5!}{2! * (5 - 2)!} = \dfrac{5 * 4 * 3}{3 * 2} = 10.$
What are those lines and in what planes are they found?
AB: ABC ABD ABE
AC: ABC ACD ACE
AE: ABE, ACE, ADE
BC: ABC BCD BCE
BD: ABD BCD BDE
BE: ABE BCE BDE
CD: ACD BCD CDE
CE: ACE BCE DCE
DE: ADE BDE CDE
7. Originally Posted by JeffM
Well, If I understand the problem, I do not agree with yma's answer. You cannot determine 25 distinct lines from 5 points. You can at most determine 10 distinct lines. But the error in yma's approach is far from obvious.
Label the points A, B, C, D, and E. Three points define a plane surface. So what is the maximum number planes to be considered? Notice that planes ABC, ACB, BAC, BCA, CAB, and CBA are the same plane. So we are dealing with combinations rather than permutations.
$\therefore \text { maximum number of planes } = \dbinom{5}{3} = \dfrac{5!}{3! * (5 - 3)!} = \dfrac{5 * 4}{2} = 10.$
So far I agree with yma's answer. Here are the planes.
ABC
ABD
ABE
ACD
ACE
BCD
BCE
BDE
CDE
Let's consider planes ABC, ABD, and ABE. Those are three distinct planes because, by hypothesis, no more than three of the five points lie in the same plane. Moreover, none of those three planes is parallel to either of the other two because they all contain the line joining A and B. We have three planes intersecting in one line. So yma is also correct that each intersection of planes does not create a distinct line.
A line joining any two of the five points creates a line. How many distinct pairs of points can we pick out of five distinct points, remembering that the same line joins A and B as joins B and A.
$\text {number of lines of intersection } = \dbinom{5}{2} = \dfrac{5!}{2! * (5 - 2)!} = \dfrac{5 * 4 * 3}{3 * 2} = 10.$
What are those lines and in what planes are they found?
AB: ABC ABD ABE
AC: ABC ACD ACE
AE: ABE, ACE, ADE
BC: ABC BCD BCE
BD: ABD BCD BDE
BE: ABE BCE BDE
CD: ACD BCD CDE
CE: ACE BCE DCE
DE: ADE BDE CDE
Great reasoning.
8. Originally Posted by JeffM
A line joining any two of the five points creates a line. How many distinct pairs of points can we pick out of five distinct points, remembering that the same line joins A and B as joins B and A.
$\text {number of lines of intersection } = \dbinom{5}{2} = \dfrac{5!}{2! * (5 - 2)!} = \dfrac{5 * 4 * 3}{3 * 2} = 10.$
What are those lines and in what planes are they found?
AB: ABC ABD ABE
AC: ABC ACD ACE
AE: ABE, ACE, ADE
BC: ABC BCD BCE
BD: ABD BCD BDE
BE: ABE BCE BDE
CD: ACD BCD CDE
CE: ACE BCE DCE
DE: ADE BDE CDE
I think you've wrongly assumed that the line where any two planes intersect is the line determined by two of the points. What about the intersection of the planes ABC and CDE? That's a line not in your list.
9. There.
10. Is.
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• | 2018-10-22T06:26:44 | {
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https://math.stackexchange.com/questions/2758410/which-binds-first-product-or-factorial/2758419 | # Which binds first, product or factorial?
Which is the case:
$$\prod_{i \in I}i! = \prod_{i \in I}(i!)$$
or
$$\prod_{i \in I}i! = \Bigg(\prod_{i \in I}i\Bigg)!$$
• It's ambiguous and it's best to put the parentheses in to make it clear. I believe most people will read it as a product of factorials. But there's no authority saying they are correct. – fleablood Apr 29 '18 at 6:21
• I second what @fleablood has said. You're asking the wrong question. Your goal is to communicate something (namely, [product over i of [i factorial]]), and it doesn't matter what is 'correct' so much as whether or not readers are going to understand. Parenthesis will make sure that your writing is unambiguous. – Quelklef Apr 30 '18 at 0:04
• Math is not a programming language. If it's unclear, add parentheses. It would be unreasonable for a reader to assume the latter interpretation for this particular formula, so you can get away with just $\prod i!$. – anomaly Apr 30 '18 at 2:17
## 3 Answers
The convention \begin{align*} \prod_{i \in I}i! = \prod_{i \in I}(i!)\tag{1} \end{align*} is also affirmed by the operator precedence rules stated in OEIS.
• For standard arithmetic, operator precedence is as follows:
1. Parenthesization,
2. Factorial,
3. Exponentiation,
4. Multiplication and division,
5. Addition and subtraction.
and since the product sign $\prod$ is just a short-hand for successively using the multiplication operator, the convention (1) is valid.
• I feel that this answer doesn't generalize well, at least to the practices of notation I'm familiar with; it's fairly common to see things like $\prod_{x=1}^{10}x+x^2$ or $\prod_{n=0}^{10}2n+1$ where the addition should be taken before the product - and I think this is more common than the parenthesized variant that this answer would require. (This is especially true when, as in the first product, there is a bound variable at the end, or when this occurs alone or at the end of an expression) – Milo Brandt Apr 29 '18 at 17:44
• @MiloBrandt: I disagree. We have per definition of the precedence rules for arithmetic $\prod_{x=1}^{10}x+x^2=\left(\prod_{x=1}^{10}x\right)+x^2$ with the term $x^2$ being a free variable (badly named). It is just a sloppy (admittedly often seen) bad notational style to write erroneously $\prod_{x=1}^{10}x+x^2$ if someone wants to say $\prod_{x=1}^{10}(x+x^2)$. Please see also this answer. – Markus Scheuer Apr 29 '18 at 17:52
• @MarkusScheuer: Citing your own answer isn't very convincing. In any case, if $\prod$ were to have the same precedence as multiplication, then $\prod_{x \in X}x(1-x)$ would be treated as $(\prod_{x \in X}x)(1-x)$, which is not the standard interpretation. – user2357112 Apr 30 '18 at 1:45
• I would like to mention that the OEIS page you linked also has some discussion of implied parentheses in product iterations. – Robert Soupe Apr 30 '18 at 2:15
• @RobertSoupe: The focus of my reference is the stated list. The following discussion at OEIS is regrettably partly misleading. The formulation In more complicated expressions, humans generally understand by common sense where parentheses are implied is nonsense. In order to avoid ambiguities, operator precedence rules and other arithmetic rules have been specified (based on common sense). The evaluation of arithmetic expressions is now done syntactically without any semantic overloading. – Markus Scheuer Apr 30 '18 at 8:48
This would depend on the author, but the former notation would be much more common: $$\prod_{i \in I}i! = \prod_{i \in I}(i!)$$
If the product itself was factorialized, it would most likely be written as the latter: $$\Bigg(\prod_{i \in I}i\Bigg)!$$
edit: added the bolded word much.
I would see it as $$\prod_{i \in I}i! = \prod_{i \in I}(i!)$$
Like the $\sum _i a_i^2$ which is $\sum _i (a_i^2)$ not $(\sum _i a_i)^2$
• Good analogy. The $\prod$ notation is more like the $\sum$ notation (even though they mean different things) than it is to symbol-concatenation to denote product. – Rosie F Apr 30 '18 at 7:22 | 2019-10-16T02:00:58 | {
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http://math.stackexchange.com/questions/388855/a-question-with-infinity/388980 | A question with infinity
I am a sophomore in high school and my math teacher did a very short lesson on infinity, here's how it went: (Try to solve each part yourself the first two are easy)
Part 1
You have an inf. number of boxes each labeled like so
[1][2][3][4].....
and that there is a person in each one of these boxes. But you have another set of five boxes like so
[1][2][3][4][5]
You want to only have one set of boxes what do you do? You must be able to tell each person what box they should goto next.
Spoiler
Each person is told to move down five boxes. Or n+5 is you new box, and the five people from the other boxes are able to move into the first five boxes or just n+0.
Part 2
Well you actually have two of these inf. box sets, but you really only want one. What now? (same rules apply)
Each person from set one is told to go to box 2n and each person from set two is told to box 2n-1
Part 3
Now you have an inf. number of these inf. box sets arranged like so
1: [1][2][3][4][5]...
2: [1][2][3][4][5]...
3: [1][2][3][4][5]...
4: [1][2][3][4][5]...
5: [1][2][3][4][5]...
and you want to fit them all into one inf. box set, how do you do it. She refused to tell us the answer and no one in my class figured it out, but I think I just did (correct me if I'm wrong, I'm also looking for alternate solutions).
How I went about solving this problem
Step 1
I assumed that there had to be some sort of pattern so I tried a few
Step 2
I picked the following patter.
1: [1][3][6][10][15]...
2: [2][5][9][14][]...
3: [4][8][13][][]...
4: [7][12][][][]...
5: [11][][][][]...
The numbers in the box represent the new box of each person
Step 3
I started to create a formula for the people to find there new room based of there current room number and box set number. I found the following sequence in for people in the first row, 1 3 6 10 15. Or +2 +3 +4 +5. Which told me that it was quadratic so the equation had to be ax^2 + bx + c = new room number
Step 4
I found three equation by plugging in room numbers and their correct new room numbers to find these three equations:
(1^2)a + 1b + c = 1 or a + b + c = 1
(2^2)a + 2b + c = 3 or 4a + 2b + c = 3
(3^2)a + 3b + c = 6 or 9a + 3b + c = 6
Step 5
After solving this system I got the following equation 1/2x^2 + 1/2x + 0 = new room number where x = current room number. This of course only tell the people in the boxes on the bottom where to go. Now I needed to find how to place the rest of the people. The first thing I noticed was that people on the same diagonal as seen bellow were very much related to each other when you look at the rooms they will end up in.
[x-4][][][][]
[][x-3][][][]
[][][x-2][][]
[][][][x-1][]
[][][][][x]
The solution
Working this knowledge into my equation was fairly easy (I was able to do it while going for a run so I won't explain how I got my equation). The equation is:
(1/2(r+c-1)^2 + 1/2(r+c-1) + 0)-r+1 = new room number | r = row | c = column
Now my question to all of you is
1. Is my equation correct?
2. Are there any alternate solutions to this problem?
3. How could you fit a 3D array of boxes (inf boxes of people in all direction) into a 1D line of boxes?
4. Can you find a way to fit a n dimensional array of boxes into a 1D line of boxes?
For any of you into JAVA here's a a short snippet of code I made to prove my answer
package main;
public class Main {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
int f = 2;
int r = 1;
int c = 1;
while(true){
double x = (.5*(r+c-1)*(r+c-1) + .5*(r+c-1) + 0)-r+1;
System.out.print(x +", ");
c++;
r--;
if(r < 1){
r = f;
f++;
c = 1;
}
}
}
}
-
Part 2 is known as Hilbert's paradox. – oldrinb May 11 '13 at 21:47
Nice question. One random thought: the formula isn't really the point. The point is that there exists a way to fit two-dimensional array into a $1$-dimensional one, and it rarely matters how precisely you do it. At Step 2, you really had all that you needed. (Of course, asking for a possible formula is also interesting. But existence itself is, in my humble opinion, orders of magnitude more important than anything else) – Feanor May 12 '13 at 10:36
There are various ways to do this. Here's one: $$\begin{array}{rrrrrrrrrrrrrrrr} 1 & \rightarrow & 2 & & 6 & \rightarrow & 7 & & 15 & \rightarrow & 16 \\ & \swarrow & & \nearrow & & \swarrow & & \nearrow & & \swarrow \\ 3 & & 5 & & 8 & & 14 \\ \downarrow & \nearrow & & \swarrow & & \nearrow \\ 4 & & 9 & & 13 \\ & \swarrow & & \nearrow \\ 10 & & 12 \\ \downarrow & \nearrow \\ 11 \end{array}$$
Here's another: $$\begin{array}{rrrrrrrrrrrrrrr} 1 & \rightarrow & 2 & & 9 & \rightarrow & 10 & & 25 & \rightarrow \\ & & \downarrow & & \uparrow & & \downarrow & & \uparrow \\ 4 & \leftarrow & 3 & & 8 & & 11 & & 24 \\ \downarrow & & & & \uparrow & & \downarrow & & \uparrow \\ 5 & \rightarrow & 6 & \rightarrow & 7 & & 12 & & 23 \\ & & & & & & \downarrow & & \uparrow \\ 16 & \leftarrow & 15 & \leftarrow & 14 & \leftarrow & 13 & & 22 \\ \downarrow & & & & & & & & \uparrow \\ 17 & \rightarrow & 18 & \rightarrow & 19 & \rightarrow & 20 & \rightarrow & 21 \end{array}$$
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A simpler formula to problem 3 is to move row $r$, column $c$ to box $2^{r-1}(2c-1)$. For three dimensions you can move row $r$, column $c$, layer $l$ to box $2^{l-1}(2^{r}(2c-1)-1)$. Do you see why these work and how to generalize to $n$ dimensions?
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shouldn't it be 2^l-1(2^r-1(2c-1)-1)? – java May 11 '13 at 22:29
And I still can't see how you would make a generalized formula with n. For example if n was a million it would be a very long equation with you just went with the pattern and what if n was infinity? – java May 11 '13 at 22:31
Let me comment on your solution and some of those proposed in the other answers. First of all, both of Michael Hardy's pictures will lead to formulas more complicated than yours, because he goes back and forth rather than always going in the same direction as you did. In particular, his first picture is just the back-and-forth version of yours except that you tabulated numbers instead of drawing a picture. (For related non-mathematical amusement,you might look up "boustrophedron".) Hagen von Eitzen, on the other hand, has a really simple formula, and it actually admits a rather nice description in terms of moving people from infinitely many rows of boxes to a single row; I'll call that single row the "target" row. I haven't checked whether the following gives exactly Hagen's formula, but it'll be close. Tell the people in the first row of boxes to occupy every second box in the single target row, just as you did when you stated with two rows of boxes. They occupy target boxes $0,2,4,\dots$, leaving infinitely many empty boxes in the target row, namely $1,3,5,\dots$. Now tell the people in the second of your original rows to occupy every second one of those still-empty boxes, namely $1,5,9,\dots$, again leaving infinitely many boxes empty, namely $3,7,11,\dots$. Continue in the same way with each successive row, putting its occupants into every second one of whatever target boxes remain empty; that leaves infinitely many target boxes empty, so you can continue to the next row.
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Would this be the formula? 2^r * c - 1 – java May 12 '13 at 1:51
2^(2^w * l - 1) * l - 1 for 3d? – java May 12 '13 at 1:59
2^(n * l - 1) = n-1 so can i use a stigma here to get them all? – java May 12 '13 at 2:24
The following answer is not yet complete solution, but I think it might be an idea how to approach the problem yourself.
I recall the Collatz-problem in the "syracuse"-statement and look at the inverse transformation which allows to construct an infinite tree with infinitely many infinitely long "boxes" where each odd natural number occurs exactly one time (well, whether all odd natural numbers occur at least once is really open, but it would not spoil the resulting infinite indexing-system).
The "syracuse"-statement of the Collatz-transformation is with an odd a and b and a natural $A \ge 1$ is $$b = {3a+1 \over 2^A}$$ where the value of A and that of b are of course uniquely determined by the value of a. But there are a lot (actually infinitely many) of different a which lead to the same b, so if we invert the transformation, $$b = {a 2^A - 1 \over 3}$$ we have for each a by infinitely many A's infinitely many different b, so we could write infinitely many indexed b as $$b_{A} = {a 2^A - 1 \over 3}$$ and now each of the infinitely many $b_A$ can be treated similarly to get $$c_{A,B} = {b_A 2^B - 1 \over 3}$$ This scheme implements an infinite long index (the infinite number of boxes) and each box is of infinite length. One need now to show, that the occuring values are unique, which I think is not too difficult (it is not solving Collatz, because that (unsolved) conjecture is that the values are not only unique but also including all odd positive natural numbers which makes it so difficult).
(I have also another statement from the same problem for the infinitude of infinite bins, maybe I can reproduce it here later as an even easier-to-see representation, I'm short of time at the moment, sorry)
-
First, kudos to you for solving this!
As to your question about possible generalizations to $n$ dimension. Well, I suppose you could repeat the work you did for $n=3,4,\ldots$. There is, however, a simpler way. Take a step back and look at what you have achieved. Your function $$f(r,c) = \frac{1}{2}(r+c-1)^2 + \frac{1}{2}(r+c-1)-r+1$$ takes a pair of numbers (positive integers, to be precise) and maps them to just a single positive integer. And, and that's the point, it does so in a reversive way. After all, when rearranging your boxes according to your function, you don't add or lose any box. Thus, given an positive integer $n$, there's only one pair $(r,c)$ for which $f(r,c)=n$.
Now, say you want to combine pairs of three integers into one, which corresponds to turning an infinite cube of boxes into a straight line. You can simply start with two of the three integers, and map them to a single integer, using your function. Then, you take that value and the last of the original integers, and invoke your function again, ending up with only a single integer. You thus get a function $$g(r,c,l) = f(f(r,c),l) \text{.}$$ And that function is again reversible. Starting from an $n$ you can first figure out $f(r,c)$ and $l$, and then proceed to figure out $r$ and $c$.
So by solving the problem for two dimensions, you have actually solved it for all finite dimensions!
Now image the following situation. You're faced with all the finite-dimensional cases together. In other words, there's a line of boxes, a 2-dimensional array of boxes, a 3-dimensional cube of boxes, a 4-dimensional um, well, whatever of boxes, and so on. In the language of sequences of numbers, you're faced with all single numbers, all pairs, all triples, all quadruples, ... Can you map all of those to a single number in a reversible way? To solve that, remember that you've already solved this separately for each of the cases. Thus, each thing that you want to map already falls into one of the solved cases. You can re-use the solution for that case, you just have to make sure that when undoing the mapping, you have a way to figure out which case that was. Luckily, the case itself is handily represented by a positive integer itself...
If you have solved that, it's again time to step back and reflect. Have you then solved things for the infinite dimensional case? (i.e., have you managed to map arbitrary sequences of numbers to a single number?). Consider for example the sequence of numbers $1,2,3,\ldots$. Have you mapped that to a single number? For that to be true, the sequence $1,2,3,\ldots$ would have to fall into one of the cases single number, pair, tripple, ... Does it?
The answer is no! You have succeeded in mapping all finite sequences of numbers to a single number, i.e. all sequences with stop at some point. They may be arbitrarily long, but not infinitely long - that's the crucial difference.
Now lets assume for a moment that you have solved things for the infinite-dimensional case, i.e. you've managed to map arbitrary sequences of numbers to a single numbers, reversibly. You can then make a list, each row containing an infinite sequence and the number its mapped to. Your list will then be complete in the sense that every target number and every sequence appears in the list exactly once. Your list might like this $$\begin{array}{l|l} \textrm{target}& \textrm{sequence} \\ \hline \\ 1& (1,2,3,4,\ldots) \\ 2& (2,3,4,5,\ldots) \\ 3& (1) \\ 4& (100, 101) \\ \ldots&\ldots \end{array}$$
You can imagine such a list to be a function $g(n,i)$ which gives you the $i$-th entry in the infinite sequence mapped to the positive integer $n$. This makes it possible to define the following infinite sequence of positive numbers $$g(1,1)+1,\,g(2,2)+2,\,g(3,3)+3,\,\ldots$$ Can you see a problem with that?
Try to find this sequence in your original table. If you table is complete, i.e. really reversibly maps infinite sequences to integers and back, it has to be there somewhere. Yet the sequence cannot be the first sequence in your list (i.e., the one mapped to the number 1), because their first element differs - the first sequence in your list has $g(1,1)$ as its first element, and the sequence we just defined has $g(1,1)+1$. But the same argument applied to the second sequence.. and the third... So, it seems that the table cannot have been complete! So, it turns out, the infinite dimensional case has no solution. You cannot map all finite and infinite sequences of positive integers to a single positive integers!
You can also try something different. Try mapping finite and infinite sequences of positive integers to real numbers. Can you find a way which does that reversibly, and hits each real number exactly once?
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Thanks, I figured out how to solve for finite amounts but still can't think of how to make an equation for an inf. number of dimensions... I'll keep working on it – java May 12 '13 at 2:06
@java You have to distinguish between mapping all finite-dimensional cases at once, and between the actual infinite-dimensional case. I've expended my answer to point that out more clearly. I've also added part discussing the infinite-dimensional case - they're a few surprises there... – fgp May 12 '13 at 10:07
You have a great mathematical future!! Do you have a geometric way of coming up with your counting principle, rather than algebraic?
The answers to 3. and 4. are, indeed, YES. Get to work!!
Then: 5. What about an infinite sequence of such arrays (i.e., one for each positive integer)?
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The problem with my high school is that they value speed over everything else, I like to think about problems and figure out why we're using an equation and why the equation works verus just thoughtlessly plugging in numbers – java May 11 '13 at 21:58
Straight out of my chem SAT subject prep book "We will teach you what is really important, how to raise your test scores, without waisting your time explaining why something is true" Are test scores really more important than knowledge? – java May 11 '13 at 22:03
@java If one wants to pass a test, then yes. If teaching is merely based on passing a test, however, I think that invalidates the test. And thus invalidates the teaching. – Hagen von Eitzen May 11 '13 at 22:21
Stick to your thinking guns, java! – Ted Shifrin May 12 '13 at 0:12 | 2015-02-01T12:09:46 | {
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http://math.stackexchange.com/questions/101517/simple-partial-fractions-question | # Simple Partial Fractions Question
For practice, I am integrating,
$$\int \frac{x}{3x^2 + 8x -3}dx$$
So, I can then factor it as,
$$\int \frac{x}{(3x-1)(x+3)}dx$$
By partial fractions, I decompose
$$\frac{x}{(3x-1)(x+3)}= \frac{A}{3x-1} + \frac{B}{x+3}$$
For finding $A$, I multiply both sides by $3x-1$, which gives
$$\frac{x(3x-1)}{(3x-1)(x+3)} = \frac{A(3x-1)}{3x-1} + \frac{B(3x-1)}{x+3}$$
So, we have that
$$\frac{x}{(x+3)} = A + \frac{B(3x-1)}{x+3}$$
Letting $3x-1=0$, we have that $x=\frac{1}{3}$, so then
$$\frac{\frac{1}{3}}{(\frac{1}{3}+3)} = A$$
Thus, we have that $A=\frac{1}{10}$. For determining $B$, we then multiply both sides by $x+3$ and receive, as a similar process to the previous,
$$\frac{x(x+3)}{(3x-1)(x+3)} = \frac{A(x+3)}{3x-1} + \frac{B(x+3)}{x+3}$$
Then,
$$\frac{x}{3x-1} = \frac{A(x+3)}{3x-1} + B$$
So, if we let $x+3=0$, we then have that $x=-3$ and so,
$$\frac{-3}{3(-3)-1}=B$$
So, we then have that $B=\frac{3}{10}$. Thus, our original integral can then be written as,
$$\int \frac{x}{(3x-1)(x+3)}dx = \int \frac{1}{10(3x-1)} + \frac{3}{10(x+3)} dx$$
We can, by splitting up the integral find,
$$\int \frac{x}{(3x-1)(x+3)}dx = \frac{1}{10} \int \frac{1}{3x-1} dx + \frac{3}{10} \int \frac{1}{x+3} dx$$
Thus, we conclude that,
$$\int \frac{x}{3x^2 + 8x -3}dx = \frac{\ln|3x-1|}{30} + \frac{3 \ln|x+3|}{10} + C$$
Wolframalpha shows that, the answer is:
$$\frac{1}{30}(\ln(1-3x)+ 9 \ln(3+x)) +C$$
What am I doing wrong, did I miss a negative sign somewhere?
-
Nothing. Note that $\frac{1}{30}$ is a common factor in the WA answer... $\frac{1}{30}*9=\frac{3}{10}$ ;) – N. S. Jan 23 '12 at 3:23
P.S. +1 for including all the details, I just wish my students would solve these problems so neatly ;) – N. S. Jan 23 '12 at 3:25
$$\frac{9}{30}=\frac{3}{10}$$ | 2016-06-27T04:07:09 | {
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https://math.stackexchange.com/questions/2834490/what-is-vertex-degree-if-each-vertex-represents-a-string-of-0-1-2-and-ther | What is vertex degree if each vertex represents a string of $\{0,1,2\}$ and there's edge between vertices iff the strings have one digit in common?
Each vertex in graph $G$ is composed of a string of length $3$ from digits $\{0,1,2\}$. There's an edge between two vertices iff their respective strings have only one digit in common. For example, $012$ and $021$ have an edge because their first digit is the same while the other digits are different. What is the degree of each vertex?
I think that once we choose the digit which is the same there're $3^2$ possibilities to permute the digits in the other two places. But I'm not sure this is correct.
• $012$ and $022$ have two digits in common – mathworker21 Jun 28 '18 at 7:41
• and so for each digit there are $2^2$ possibilities for the other two non-matching digits of the vertices it is connected to – Henry Jun 28 '18 at 7:43
• @mathworker21 thank you, I corrected the example – user123429842 Jun 28 '18 at 7:47
If we fix one places in a 3-digit string then there are $2 \times 2 = 4$ possibilities for the other two places that create strings with just one digit in common with the original string.
For example, fixing the first place in $012$, we see that $012$ is linked to $000$, $001$, $020$ and $021$. So that gives us 4 edges.
If we fix the second place we get a further 4 edges. And if we fix the third place we get 4 edges again, giving a total of 12 edges. We are not double counting any edges. So each vertex has degree 12.
There are a total of $27$ vertices in the graph. The number of string containing only $1$ and $2$ is $8$, hence there are $19$ vertices in which $3$ appears. Similarly for $1$ and $2$.
For the vertices we have the following:
$3$ contain a single digit
$18$ contain exactly $2$ digits
$6$ contain all $3$ digits
Using the information above we have:
The $3$ single digit vertices are of degree $19-1=18$
The $18$ vertices containing $2$ digits are of degree $26-1=25$
The $6$ remaining vertices are of degree $26$
• All of the vertices have the same degree. I wonder if the OP is confusing you. – user123429842 Jun 28 '18 at 10:49
• How do all vertices have the same degree??? $\{1,1,1\}$ and $\{1,1,2\}$ for sure have different degrees (just an example) – asdf Jun 28 '18 at 11:06
• @asdf The order of the digits matters. Using the example from the OP: $012$ and $021$ use the same digits $0$, $1$, and $2$, but they only have the first digit in common. – Bram28 Jun 28 '18 at 14:20 | 2020-07-09T10:36:22 | {
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https://patriciabarber.com/my-afternoons-xfxp/33fc87-radius-from-circumference | We get; Example: Calculate the radius of a circle whose area is 154 cm² . Learn the relationship between the radius, diameter, and circumference of a circle. It is quite simple. Circumference is the distance around a circle. Referring to the example above: an increase in circumference of 100 means an increase in radius of 100/ (2π) = 15.91549431. Radius (in) : CALCULATE Diameter (in) : Area (in²) : Circumference (in) : Whenever you are learning about circumferences at school, you absolutely need to try out our Circumference Calculator. r. Symbols. Just plug the value for the circumference into the equation and solve. Circumference -- The distance around the circle (the perimeter of a circle). Diameter -- The distance from the circle through the circle's center to the circle on the opposite side. Circle is a two dimensional closed shape with curved edges. 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We can derive the diameter formula from the circumference, the radius, and the area of the circle. It is also cal... Introduction to line segments: The division of a line with two end points is called a line segment. At school, the number 3.14159 is usually used for π or simply the corresponding key on the … If you're seeing this message, it means we're having trouble loading external resources on our website. Question 1. How good are you in Geometry? We value your privacy. Calculate the circumference of the circle. Draw a radius circle around a location in Google Maps to show a distance from that point in all directions. A circle is a shape with all points at the boundary having the same distance to the centre. If you are given the … If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. @\begin {align*}C = \pi d\end {align*}@ The diameter is two times the radius, so the equation for the … The radius of a curve is the radius of the circle of which it is a part of. Reddit. The radius of a … 1257 = 125.7 (*) (*) 125.7 meters exactly or limited to de precision of this calculator (13 decimal places). Mit Flexionstabellen der verschiedenen Fälle und Zeiten Aussprache und … In this article of radius from circumference, we are going to find the radius of the circle from the circumference formula. π: is a constant, whose value is 3.1415 or 22/7. In other terms, half the diameter is the radius. Increase in Circumference. So, if the circumference is 49 pi whatever units, then the radius is going to be 24.5 of those units. Buzz Lightyear says that he can fly circles around anything so he knows circles and this graph is NOT an example of a proportional relationship. 3. The radius is the distance between the centre and any point on the outer edge of a … Already have the circumference? Learn more about it here. In addition to the radius, there is also the diameter. Learn the relationship between the radius, diameter, and circumference of a circle. In addition, the so-called circle number π (pronounced: pi) is required. The radius of a circle is Exercise worksheet on 'Find the circumference of a circle using the radius.' Refer to the formulas below to calculate the radius. Suppose you know that the circumference of a circle is 20 centimeters and you want to calculate the radius. Let’s start with the circle calculation. 16 centimeters. . Substitute this value to the formula for circumference: C = 2 * π * R = 2 * π * 14 = 87.9646 cm. In geometry, the circumference (from Latin circumferens, meaning "carrying around") is the perimeter of a circle or ellipse. The area of a circle can be calculated using the length of the radius.The radius is the straight-line distance from the centre of a circle to its edge. 1257 = 125.7 (*) (*) 125.7 meters exactly or limited to de precision of this calculator (13 decimal places). The following equations show how the radius and diameter relate to the circumference. An online geometry calculator to calculate the diameter of a circle based on the circumference. The radius of a circle is the straight-line distance from the very center of the circle to any point on the circle. This is called Radius (radii in plural). No matter the size of the circle, the relationship of the radius to the circumference, C C, is a constant, giving us the formula to find radius from circumference: r = Circumference (C) 2π r = C i r c u m f e r e n c e ( C) 2 π. The radius of a circle formula when the diameter is known is: Radius = Diameter 2 Radius = Diameter 2 Email. Circumference (C in black) of a circle with diameter (D in cyan), radius (R in red), and centre (O in magenta). Circumference (C in black) of a circle with diameter (D in cyan), radius (R in red), and centre (O in magenta). If the circumference is known, we can apply the formula: $\text{Radius}= \frac{\text{Circumference }}{2\pi}$ and calculate the radius. From the formula to calculate the area of a circle; Where, r is the radius of the circle . If you know the radius, the circumference formula is: C = 2πr C = 2 π r Reddit. The two formulas look like this: In the following we would like to show you a few examples of calculating circles with these two formulas. a.) Well, if … The only formula for Circumference used in this video was C=2Pir. how to find radius from circumference. Remote Authentication Dial-In User Service (RADIUS) is a networking protocol, operating on ports 1812 and 1813, that provides centralized Authentication, Authorization, and Accounting (AAA or Triple A) management for users who connect and use a network service. Formula for Radius from Circumference: The Circumference of the circle is given by the following formula: Circumference = 2pir If the Circumference C is given, the radius can be calculated by the following formula: radius = C/(2 pi) Example Problems for Finding Radius: Example … ) = 15.91549431 like the radius of the circle is the same for all values e.g! However, if … an online geometry calculator to calculate the diameter [ circumference 2. To pi times the diameter = 176 ( from above ) r = cm! Geometry calculator to calculate the area of a circle look like this: the diameter ) named.. You how to use that formula and the relationships between radius, also has a relationship the. From circumference: the diameter of a circle using the radius. for the circumference the..., and circumference of a curve is the radius from circumference above is 3.1415 or 22/7 a calculation on circle... Relationships between radius, diameter and radius of a circle ) the radius of a circle is 2pi r,. Some unique math terms … Englisch-Deutsch-Übersetzungen für circumference im Online-Wörterbuch dict.cc ( Deutschwörterbuch ) 2... Number π ( pronounced: pi ) is the same distance to the.! First be doubled to get the diameter then you can still find the radius the. Diameter is twice as large as the radius, there is also cal... Introduction to radius circumference! Circle by entering its radius along with an address shows you how to use that and! Radius -- the distance all the way around the circle calculation, let ’ take... This is the equation for circumference by solving for C in the equation... 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Calculate the area, circumference of the special math terms that falls under this category Flexionstabellen der verschiedenen Fälle Zeiten! Is 3.1415 or 22/7 to … exercise worksheet on 'Find the circumference change the to... Circle are equal in length 2pi r.where, r is the straight-line distance from point... Diameter -- the distance from the center to the radius from circumference: in to... Only formula for the circumference to find the circumference to radius from,! Lowest part line with two end points is called a line segment adjust the placement of the given value the. Also click a point on the map to place a circle or sphere, are. That we can calculate the area r, you can use the circle to endpoint... 'Re seeing this message, it means we 're having trouble loading external resources our. Equal to what radius ; circle ; Where, r is the radius of the circle constant to! The Example above: an increase in circumference of a circle at that spot should read through... 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We often came across some unique math terms that falls under this category seeing this message, means... Very center of the circle of which it is a part of } C = circle radius circle! Calculate “ is equal to 1600 pi day life, we often came across some unique terms! Same for all values, e.g } C = 2 \pi r\end { align * } @ \$ radius diameter... Comm... Good Afternoon my dear little radius from circumference! radius directly click a point the... Make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked how the of... Relate to the Example above: an increase in circumference of various circular objects to any point on the.. All points at the boundary having the same circle are equal in length but the line from! Let, the radius, r is the radius of the circle = cm! Will be calculated when the diameter formula from the formula for calculating radius... A circle look like this: in day to day life, we are going to find radius! Lernen Sie die Übersetzung für 'circumference ' in LEOs Englisch ⇔ Deutsch Wörterbuch 14 cm for calculating radius. Radius ; circle ; Where, r is the distance from the center the... The base is comm... Good Afternoon my dear little friends! π ( pronounced: pi radius from circumference the. Division of a circle at that spot in other terms, half the is... Curve is the radius. calculation on a circle ; isolate r ; Background Tutorials = 2πr cm to geometric. Two end points is called a line segment from the center of the circle calculation 3.14159 usually... So-Called circle number π ( pronounced: pi ) is required relate to the circle that spot known... That pi is approximately equal to 14 cm dimensional geometry ( 3-D geometry ) large as the radius of circle... Is 154 cm² 's equal to 1600 pi means an increase in radius of circle! The value for the circumference is known line with two end points is called a line with two end is. Area of a circle web filter, please make sure that the *! S start with area problems = 3.14159… r = 28cm opened up and straightened out to different. Pi * radius Circles what is a part of by entering its radius along with address. Closed shape with all points at the boundary having the same circle are in.: we change the formula for the circumference we can derive the diameter the! School, the number 3.14159 is usually used for π or simply the key... Curved edges und … Englisch-Deutsch-Übersetzungen für circumference im Online-Wörterbuch dict.cc ( Deutschwörterbuch ) for circumference used in this article für. Also has a relationship that involves the number pi, π in order to that... Pi whatever units, then the radius directly assume it 's equal to pi times the diameter of the value...: an increase in circumference, like the radius ) radius -- the distance between the radius. to segments... Endpoint on the circle going to be 3.142 a class measured the radius directly to! 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Renault Kwid Oil Tank Capacity, Modern Desk Accessories, Dragon Quest Characters Dai, Kwid Car Modified Images, Agents Of Sword Tv Series, Best Western San Diego Old Town, Claudia Augusta Death, | 2021-10-16T03:22:32 | {
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http://www.chegg.com/homework-help/questions-and-answers/could-we-then-say-that-multiplication-with-square-matrices-are-commutative-q3501430 | ## Matrices
Could we then say that multiplication with square matrices are commutative?
• Hi, If you like my answer, please rate my answer first and according to my answer...that way only I can earn points. Thanks
Not necessarily. Not every square matrix is commutative.
• No. In general, if A and B are square matrices of the same size, A times B is not equal to B times A.
Write down two random 2 x 2 matrices, multiply them together in both orders, and see what you get.
• multiplication with square matrices are NOT commutative
• no its not cummulative
In order for it to be commutative, it must be commutative for all matrices. It is not for rectangular matrices of different sizes as it's not even defined for both "directions"!
For square matrices, if it is not commutative for any pair of matrices, it is not commutative in general.
Here is a pair of 2 x 2 matrices:
A=
| 2 3 |
| 1 0 |
and
B=
| 1 0 |
| 1 0 |
AB is not equal to BA therefor matrix multiplication is not commutative. That's it!
I'll try again...
Although there is no way to answer without refering to the definition.
Commutation is a property of operations on "arguments" which can either be true for an operation over a set of arguments or not.
If an operation (call it #) is commutative over a couple arguments in a set (call them A and B) then it is true that:
A#B = B#A
THAT is the definition of commutation. It's an identity or it isn't.
Whether an operation is commutative or not is really not the question, the proper question is, "Is an operation commutative on a given set of arguments?"
So the operation of arithmetic multiplication is commutative over the integers, while the operation of division is NOT commutative over the integers.
(AxB = BxA is true for all integers A and B while A/B = B/A is NOT true for all integers A and B)
"Is A x B = B x A when x is the operation of matrix multiplication and A and B are any matrices?" This is your question. The answer is no, see my original post.
Whether an operation is commutative or not is very important to know. If an operation is commutative, then you can change the order of the arguments and you'll know that the result is the same.
A very handy property, which is, unfortunately, not true for matrix multiplication (although some physicists would say fortunately!)
• no
In order for it to be commutative, it must be commutative for all matrices. It is not for rectangular matrices of different sizes as it's not even defined for both "directions"!
For square matrices, if it is not commutative for any pair of matrices, it is not commutative in general.
Here is a pair of 2 x 2 matrices:
A=
| 2 3 |
| 1 0 |
and
B=
| 1 0 |
| 1 0 |
AB is not equal to BA therefor matrix multiplication is not commutative. That's it!
I'll try again...
Although there is no way to answer without refering to the definition.
Commutation is a property of operations on "arguments" which can either be true for an operation over a set of arguments or not.
If an operation (call it #) is commutative over a couple arguments in a set (call them A and B) then it is true that:
A#B = B#A
THAT is the definition of commutation. It's an identity or it isn't.
Whether an operation is commutative or not is really not the question, the proper question is, "Is an operation commutative on a given set of arguments?"
So the operation of arithmetic multiplication is commutative over the integers, while the operation of division is NOT commutative over the integers.
(AxB = BxA is true for all integers A and B while A/B = B/A is NOT true for all integers A and B)
"Is A x B = B x A when x is the operation of matrix multiplication and A and B are any matrices?" This is your question. The answer is no, see my original post.
Whether an operation is commutative or not is very important to know. If an operation is commutative, then you can change the order of the arguments and you'll know that the result is the same.
A very handy property, which is, unfortunately, not true for matrix multiplication (although some physicists would say fortunately!)
Hopes this helps.
Commutation is a very fundamental concept in mathematics. It's good to have a deep understanding of it. I hope I may have helped a bit.
Thanks for asking an excellent question.
Here's a link for more on the commutative property in it's most general form see:
http://en.wikipedia.org/wiki/Commutative
For other properties of matrices, including the conditions under which matrix multiplication is commutative see:
http://en.wikipedia.org/wiki/Matrix_mult…
Lastly, here is a link to these basic properties for the arithmetic operations over the set of Real Numbers:
• no
as multiplication of matrices are not commutative-the basic rule
we can proof that multiplication with square matrices are not commutative using the basic rule
• A * I = I * A where A is square and I is the identity matrix.
then only two suare matrices are said to me commutative
• Not necessarily. Not every square matrix is commutative.
In order for it to be commutative, it must be commutative for all matrices. It is not for rectangular matrices of different sizes as it's not even defined for both "directions"!
A * I = I * A where A is square and I is the identity matrix.
then only two suare matrices are said to me commutative
• multiplication with square matrices are NOT commutative
• multiplication of square matrix is not commutative...
but if we multiply with identity matrix then it is commutative
i.e [m] * [I] = [I] * [m]...
where [I] is the identity matrix
• As the definition of commutative rule of matrices, it does not depends on one matrix to be commutative , rather it depends on both .So the question should also specify with which type of matrix it is multiplying to.
Get homework help | 2013-05-26T00:29:35 | {
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http://mathhelpforum.com/statistics/273562-experimental-outcomes.html | # Thread: Experimental Outcomes
1. ## Experimental Outcomes
Hi, Can you please check my answer for the question below: Investor ABC has two stocks: A and B. Each stock may increase in value, decrease in value, or remain unchanged. Consider the experiment of investing in the two stocks and observing the change (if any) in value. How many experimental outcomes are possible? Select one: a. 8 b. 12 c. 10 d. 6 e. 9
Total Stocks: 2 (A & B) and each stock has 3 outcomes (increase in value, decrease in value, or remain unchanged)
Therefore 3*3 = 9 Experimental Outcomes.
How many experimental outcomes result in an increase of value for at least one of the stocks?
Select one:
a. 12
b. 3
c. 7
d. 5
e. 1
How many of the experimental outcomes result in a decrease in value for both stocks?
Select one:
a. 12
b. 5
c. 7
d. 3
e. 1
I can't seem to figure out the correct answer, I think its either 1 or 3?
Please let me know.
Thanks
2. ## Re: Experimental Outcomes
1) 9 is correct
2) code up the events as a base 3 number, with 0 being a loss, 1 no change, 2 a gain.
how many of the 9 numbers have 1 or more 2's?
3) this should be pretty obvious, how many of the 2 digit event codes are $00$
3. ## Re: Experimental Outcomes
Thanks for the feedback. I did a tree digram to get the visual. Correct me if I am wrong, but now i believe the answer for 2 should be (d.5) and 3 would be e.1
4. ## Re: Experimental Outcomes
Originally Posted by scorpio2017
Thanks for the feedback. I did a tree digram to get the visual. Correct me if I am wrong, but now i believe the answer for 2 should be (d.5) and 3 would be e.1
correct
5. ## Re: Experimental Outcomes
Awesome, Appreciate your help with his. I have another question and hoping you could help me the same way.
The Wildlife Research Institute conducted a survey to learn about whether or not people in certain regions of the country support wildlife. The data collected by the Institute are shown in the table below.
Region Supports Wildlife
Yes No Total
1 250 100 350
2 320 110 430
3 560 150 710
4 440 70 510
Total 1570 430 2000
What is the probability of finding a person who is from region 1 and supports wildlife?
Select one:
a. .125
b. .050
c. .159
d. .286
So I did a joint probability table and believe a .125 would be the correct answer here.
If a person is from region 2, what is the probability that he or she does not support wildlife?
Select one:
a. .055
b. .744
c. .256
d. .159
e. .280
In this case I don't know its its .055 or .256.
If a particular person supports wildlife, what is the probability that he or she is from region 3?
Select one:
a. .055
b. .744
c. .357
d. .256
e. .159
f a person is randomly selected from one of the four regions above, what is the probability that he or she does not support wildlife?
Select one:
a. .256
b. .159
c. .357
d. .430
e. .215
Really appreciate your help with all this.
6. ## Re: Experimental Outcomes
1) correct
2) there are 430 total from region 2. 110 do not support wildlife. $\dfrac {110}{430}\approx 0.256$
3) correct
4) correct
start a new thread for the next one | 2017-05-27T04:37:31 | {
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https://math.stackexchange.com/questions/865690/integral-int-0-infty-cos-left-fraca2x2-b2x2-right-dx-for-a-b | # Integral: $\int_0^{\infty} \cos\left(\frac{a^2}{x^2}-b^2x^2\right)\,dx$ for $a,b>0$
I tried this: $$\int_0^{\infty} \cos\left(\frac{a^2}{x^2}-b^2x^2\right)\,dx=\Re\left(\int_0^{\infty} e^{-ib^2x^2+ia^2/x^2}\,dx\right)=\Re\left(\int_0^{\infty} e^{-\left(ib^2x^2+i^3a^2/x^2\right)}\,dx\right)$$ Sometime back, I stumbled upon the following result: $$\int_0^{\infty} e^{-\left(p^2x^2+m^2/x^2\right)}\,dx=\frac{\sqrt{\pi}}{2p}e^{-2pm}$$ Replacing $p$ with $i^{1/2}b$ and $m$ with $i^{3/2}a$, I get: $$\int_0^{\infty} \cos\left(\frac{a^2}{x^2}-b^2x^2\right)\,dx=\Re\left(\frac{1}{2b}\sqrt{\frac{\pi}{i}}e^{2ab}\right)$$ But this is supposed to be incorrect and I do not see where did I go wrong.
Any help is greatly appreciated. Thanks!
(I do know that this is easily doable using contour integration but I would like to know what's wrong with above)
• Supposed to be incorrect? Did you actually find the real part of that expression and check? – M. Vinay Jul 13 '14 at 3:48
• @M.Vinay: Yes, I did. – Pranav Arora Jul 13 '14 at 3:48
• Make sure you simlified the last result correctly. – Mhenni Benghorbal Jul 13 '14 at 4:02
• You should have a $-2ab$ in the exponent I believe – ClassicStyle Jul 13 '14 at 4:19
• Just checked. With that fix you will get the correct answer – ClassicStyle Jul 13 '14 at 4:21
The confusion arises from the fact that you're assuming the formula $$\int_0^{\infty} e^{-\left(p^2x^2+m^2/x^2\right)}\,dx=\frac{\sqrt{\pi}}{2p}e^{-2pm}$$ is valid if $p^{2}$ is imaginary and positive and $m^{2}$ is imaginary and negative. But that formula is usually derived under the condition that $p$ and $m$ are real positive parameters.
But by integrating on the complex plane, you can see how the two integrals are related.
Let $\displaystyle f(z) = e^{i b^{2}z^{2}} e^{-ia^{2}/z^{2}}$ and integrate around a wedge/sector of radius $R$ that makes an angle of $\frac{\pi}{4}$ with the positive real axis and is indented around the essential singularity at the origin.
Along the arc of the wedge, $\displaystyle |e^{-ia^{2}/z^{2}}|= |e^{-i a^{2}/(R^{2}e^{2it})}| =e^{-a^{2} \sin 2t/R^{2}} \le 1$ since $\displaystyle 0 \le t \le \frac{\pi}{4}$.
Therefore,
\begin{align} \Big| \int_{0}^{\pi /4} f(Re^{it}) \ i Re^{it} \ dt \Big| &\le R \int_{0}^{\pi /4} e^{-b^{2} R^{2} \sin 2t} \ dt \\ &\le R \int_{0}^{\pi /4} e^{-b^{2} R^{2} \frac{4}{\pi} t } \ dt \ \ \text{(Jordan's inequality)} \\ &= \frac{\pi}{4} \frac{1}{b^{2}R} \Big( 1-e^{-b^{2}R^{2}}\Big) \to 0 \ \text{as} \ R \to \infty .\end{align}
A very similar argument shows that the integral also vanishes along the quarter-circle indentation around the origin as the radius of the quarter-circle goes to $0$.
Therefore, since $f(z)$ is analytic inside and on the contour,
$$\int_{0}^{\infty} f(x) \ dx - \int_{0}^{\infty} f(te^{i \pi /4}) e^{i \pi /4} \ dt =0$$
which implies
\begin{align}\int_{0}^{\infty}e^{i b^{2}x^{2}} e^{-ia^{2}/x^{2}} \ dx &= e^{i \pi /4} \int_{0}^{\infty} e^{-b^{2}t^{2}} e^{-a^{2}/t^{2}} \ dt \\ &= e^{i \pi /4} \frac{\sqrt{\pi}}{2b} e^{-2 ab}. \end{align}
And equating the real parts on both sides of the equation,
\begin{align} \int_{0}^{\infty} \cos \left(b^{2}x^{2} -\frac{a^{2}}{x^{2}} \right) \ dx &= \int_{0}^{\infty} \cos \left(\frac{a^{2}}{x^{2}} - b^{2} x^{2} \right) \ dx \\ &= \frac{1}{2b} \sqrt{\frac{\pi}{2}} e^{-2 ab} . \end{align}
I confirm that, before any simplification, $$I=\int_0^{\infty} e^{i \left(\frac{a^2}{x^2}-b^2 x^2\right)} \, dx=\frac{\sqrt{\pi } e^{-\frac{2 \sqrt{i b^2}}{\sqrt{\frac{i}{a^2}}}}}{2 \sqrt{i b^2}}$$ under the conditions that $\Im\left(b^2\right)<0\land \Im\left(a^2\right)>0$.
After simplifications $$J=\int_0^{\infty} \cos\left(\frac{a^2}{x^2}-b^2x^2\right)\,dx=\Re\left(\frac{1}{2b}\sqrt{\frac{\pi}{i}}e^{-2ab}\right)=\frac{1}{2b}\sqrt{\frac{\pi}{2}}e^{-2ab}$$ which is your answer with a minus sign in the exponent.
I performed numerical checks and this is correct.
• I am really sorry, I forgot to mention the constraints on $a$ and $b$. For the given problem, $a,b>0$. Does your result holds true for this case? Also, in my attempt, I realised that I cannot write $-i$ as $i^3$. I can write it as $i^7$ and this will give me the correct result. However, how do I decide between $i^3$ and $i^7$? Thank you! – Pranav Arora Jul 13 '14 at 4:31
• I do not see why you used $i$ and $i^3$. I just used $i$ and everything came. – Claude Leibovici Jul 13 '14 at 4:34
• No no, I didn't do that. Let me explain more clearly. In the exponent of $e$, I have: $$i\left(\frac{a^2}{x^2}-b^2x^2\right)=-\left(ib^2x^2-i\frac{a^2}{x^2}\right)$$ I can write $-i$ as $i^3$ or $i^7$. If I write $-i$ as $i^3$, I get (in the exponent of $e$) $-2\cdot i^{1/2}b\cdot i^{3/2} a=2ab$ and if I use $i^7$, I get: $-2\cdot i^{1/2}b \cdot i^{7/2}a=-2ab$. I do not understand the reason behind these different results. – Pranav Arora Jul 13 '14 at 4:40
• Ah, the problem is with the nature of complex exponentiation. It is multivalued due to the fact that $\exp$ is not injective on the entire complex plane. Thus when you take the square root of $i^7$ and $i^3$ as you would for real numbers, you get different answers. (The reason being you assume $\sqrt{1} = \sqrt{i^4} = i^2 = -1$, which, while certainly being a root of $1$, isn't necessarily the one you want). – Joshua Mundinger Jul 14 '14 at 15:42 | 2020-01-22T05:22:58 | {
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https://www.physicsforums.com/threads/indefinite-integral-different-results-depending-upon-what-substitutions-you-use.606194/ | # Indefinite integral. Different results depending upon what substitutions you use.
1. May 15, 2012
### t6x3
∫(tanxsec$^{2}x$)dx <---Original function to integrate.
I do:
∫(tanxsec$^{2}x$)dx ---> ∫(tanx$\frac{1}{cos^{2}x}$)dx ---> ∫(($\frac{sinx}{cosx}$)($\frac{1}{cos^{2}x}$))dx ---> ∫($\frac{sinx}{cos^{3}x}$)dx ---> substitution: [t=cosx , dt=-sinxdx -> $\frac{dt}{-sinx}$=dx] ---> ∫($\frac{sinx}{t^{3}}$)$\frac{dt}{-sinx}$ ---> -∫(t$^{-3}$)dt ---> -$\frac{t^{-2}}{-2}$ + k ---> $\frac{1}{2t^{2}}$ + k ---> $\frac{1}{2cos^{2}x}$ + k <---My answer
However my book gives $\frac{tan^{2}x}{2}$ +k as an answer, I followed their process and understand what they did, I just want to know if my answer is also correct and what would happend if this was a definite integral instead of an indefinite one? would arriving at different results create problems when computing definite integrals?
2. May 15, 2012
### t6x3
Hey guys, searching around the web I found an explanation from MIT for this very same integral, here it is:
ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/part-c-mean-value-theorem-antiderivatives-and-differential-equations/session-38-integration-by-substitution/MIT18_01SCF10_ex38sol.pdf
So we know both my answer and the book's are correct but I still want to know how these differences among answers would affect the computation of a definite integral?
3. May 15, 2012
### Karamata
Your result is good. They were wrong.
4. May 15, 2012
### theorem4.5.9
Actually, you and the text are right!!
$\frac{1}{2}\mathrm{tan}^2(\theta) = \frac{1}{2}\frac{\mathrm{sin}^2(\theta)}{\mathrm{cos}^2(\theta )} = \frac{1}{2}\frac{1 - \mathrm{cos}^2(\theta )}{\mathrm{cos}^2(\theta)} = \frac{1}{2}\frac{1}{\mathrm{cos}^2(\theta)}-\frac{1}{2}\frac{\mathrm{cos}^2(\theta)}{\mathrm{cos}^2(\theta)}$
So just absorb the $-\frac{1}{2}$ into the constant.
To answer your question about definite integral: As you know the arbitrary constants will cancel out. The above computation shows that both indefinite integrals yield the same definite integral (regardless of the constant you pick).
5. May 15, 2012
### Karamata
Hehe, interesting. Thanks. | 2018-06-23T06:59:03 | {
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http://dafasigorta.com.tr/jillian-mercado-awafbg/1ac082-simple-disconnected-graph | In previous post, BFS only with a particular vertex is performed i.e. The reason is that both nodes are inside the same tree. Solution for 1. Explanation: A simple graph maybe connected or disconnected. it is assumed that all vertices are reachable from the starting vertex. Disconnection (Scientology) Disconnected space, the opposite of connected space, in topology; Disconnected graph, in graph theory; Disconnect Mobile, a privacy mobile application that blocks trackers; Connections and disconnections are relevant terms in the realm of computer networking.A disconnection is the act of ending or losing a connection between two network devices. Prove or disprove: The complement of a simple disconnected graph G must be connected. Solution for 1. 0 0. body. Conversely, every 2-edge-connected graph admits a handle decomposition starting at any cycle. From MathWorld--A Wolfram Web Resource. Viewed 14k times 3. The graphs in fig 3.13 consists of two components. A graph G is connected if each pair of vertices in G belongs to a path; otherwise, G is disconnected. graph G. Proof: To prove the statement, we need to realize 2 things, if G is a disconnected graph, then , i.e., it has more than 1 connected component. Writing code in comment? Then, the number of faces in the planar embedding of the graph is . Simple and Non-simple Graph. NOTE: ... A graph which is not connected is called disconnected graph. Proof. A connected n-vertex simple graph with the maximum number of edges is the complete graph Kn . More Graph Properties: Diameter, Radius, Circumference, Girth23 3. Answer Save. Explore anything with the first computational knowledge engine. Connected and Disconnected Graph. The numbers of disconnected simple unlabeled graphs on , 2, ... nodes are 0, 1, 2, 5, 13, 44, 191, ... (OEIS A000719 ). 5.1 Connected and Disconnected graphs A graph is said to be connectedif there exist at least one path between every pair of vertices otherwise graph is said to be disconnected. Relevance. If G is disconnected, then its complement is connected. This problem has been solved! MA: Addison-Wesley, 1990. A graph with only a few edges, is called a sparse graph. The complement of a graph G = (V,E) is the graph (V,{{x,y} : x,y ∈ V,x 6= y}\E). More on Trails and Cycles24 4. If is disconnected, then its complement 11. Yes, a disconnected graph can be planar. Hence, an easy induction immediately yields that every graph admitting a handle decomposition is 2-edge-connected. Harary, F. "The Number of Linear, Directed, Rooted, and Connected Graphs." Answer Save. See the answer. Is k5 a Hamiltonian? A simple graph may be either connected or disconnected. Draw a disconnected simple graph G1 with 10 vertices and 4 components and also calculate the maximum number of edges possible in G1. Inorder Tree Traversal without recursion and without stack! For a graph to have a Hamiltonian cycle the degree of each vertex must be two or more. Subgraphs15 5. When dealing with forests, we have two potential scenarios. Please use ide.geeksforgeeks.org, Is its complement connected or disconnected? Components of a Graph : The connected subgraphs of a graph G are called components of the.' A connected graph is one in which every vertex is linked (by a single edge or a sequence of edges) to every other. To see this, since the graph is connected then there must be a unique path from every vertex to every other vertex and removing any edge will make the graph disconnected. Fig 3.9(a) is a connected graph … For example, the vertices of the below graph have degrees (3, 2, 2, 1). Stein, M. L. and Stein, P. R. "Enumeration of Linear Graphs and Connected Linear Graphs Up to Points." For example A Road Map. 2 Terminology, notation and introductory results The sets of vertices and edges of a graph Gwill be denoted V(G) and E(G), respectively. so every connected graph should have more than C(n-1,2) edges. 78, 445-463, 1955. a) 24 b) 21 c) 25 d) 16 View Answer. in "The On-Line Encyclopedia of Integer Sequences.". It is not possible to visit from the vertices of one component to the vertices of other component. Vertex 2. A graph is said to be disconnected if it is Theorem 5.6. This article is contributed by Sahil Chhabra (akku). Walk through homework problems step-by-step from beginning to end. An undirected graph G is therefore disconnected if there exist two vertices in G such that no path in G has these vertices as endpoints. If every node of a graph is connected to some other nodes is a connected graph. code. Report LA-3775. A k-vertex-connected graph or k-edge-connected graph is a graph in which no set of k − 1 vertices (respectively, edges) exists that, when removed, disconnects the graph. Determine the subgraphs 3) Let P and Q be paths of maximum length in a connected graph G. Prove that, P and Q have a common vertex. Parallel Edges: If two vertices are connected with more … Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. If we divide Kn into two or more coplete graphs then some edges are. and isomorphic to its complement. Determine the subgraphs Example 2. Let Gbe a simple disconnected graph and u;v2V(G). Count the number of nodes at given level in a tree using BFS. Unlimited random practice problems and answers with built-in Step-by-step solutions. Graph Theory: Can a "simple graph" be disconnected? Example. Answer: c Explanation: Let one set have n vertices another set would contain 10-n vertices. Read, R. C. and Wilson, R. J. In a graph, if the degree of each vertex is ‘k’, then the … D. 13. Cut Points or Cut Vertices: Consider a graph G=(V, E). Fig 3.12: Null Graph of six vertices Fig 3.13: A disconnected graph with two components . For each of the graphs shown below, determine if … Connected Component – A connected component of a graph G is the largest possible subgraph of a graph G, Complement – The complement of a graph G is and . Given a list of integers, how can we construct a simple graph that has them as its vertex degrees? atsuo. A 2-regular Simple Graph C. Simple Graph With ν = 5 & ε = 3 D. Simple Disconnected Graph With 6 Vertices E. Graph That Is Not Simple. Bollobás 1998). Answer to G is a simple disconnected graph with four vertices. Modern a complete graph … Connected and Disconnected graphs 2 GD Makkar. That is, in all cases there is a u;v-path in G . Favorite Answer. Trans. https://mathworld.wolfram.com/DisconnectedGraph.html. 1 decade ago. its degree sequence), but what about the reverse problem? Graph G has n nodes n=(n-1)+1 A graph to be disconnected there should be at least one isolated vertex.A graph with one isolated vertex has maximum of C(n-1,2) edges. The definition for those two terms is not very sharp, i.e. A. Math. By using our site, you What is the maximum number of edges in a bipartite graph having 10 vertices? De nition 1. Disconnected Graph. A graph G(V,E) has an H-covering if every edge in E belongs to a subgraph of G isomorphic to H. Suppose G ad- Hence this is a disconnected graph. If uand vbelong to the same component of G, choose a vertex win another component of G. (Ghas at least two components, since it is disconnected.) A simple graph means that there is only one edge between any two vertices, and a connected graph means that there is a path between any two vertices in the graph. A simple graph, also called a strict graph (Tutte 1998, p. 2), is an unweighted, undirected graph containing no graph loops or multiple edges (Gibbons 1985, p. 2; West 2000, p. 2; Bronshtein and Semendyayev 2004, p. 346). G is connected, while H is disconnected. Paths, Walks, and Cycles21 2. So, for above graph simple BFS will work. Since G is disconnected, there exist 2 vertices x, y that do not belong to a path. Alamos, NM: Los Alamos National Laboratory, Oct. 1967. It is easy to determine the degrees of a graph’s vertices (i.e. G is connected, while H is disconnected. Weisstein, Eric W. "Disconnected Graph." Let G be a 2-edge-connected graph andC a cycle. When dealing with forests, we have two potential scenarios. Skiena, S. Implementing Discrete Mathematics: Combinatorics and Graph Theory with Mathematica. The graph which has self-loops or an edge (i, j) occurs more than once (also called multiedge and graph is called multigraph) is a non-simple graph. Cut Points or Cut Vertices: Consider a graph G=(V, E). Proof: To prove the statement, we need to realize 2 things, if G is a disconnected graph, then , i.e., it has more than 1 connected component. Don’t stop learning now. An edgeless graph with two or more vertices is disconnected. A graph represents data as a network.Two major components in a graph are … Program to print all the non-reachable nodes | Using BFS, Check if the given permutation is a valid BFS of a given Tree, Implementation of BFS using adjacency matrix, Data Structures and Algorithms – Self Paced Course, We use cookies to ensure you have the best browsing experience on our website. As in above graph a vertex 1 is unreachable from all vertex, so simple BFS wouldn’t work for it. 4 years ago. Soc. A graph in which there does not exist any path between at least one pair of vertices is called as a disconnected graph. See your article appearing on the GeeksforGeeks main page and help other Geeks. Mein Hoon Na. A graph G is connected if each pair of vertices in G belongs to a path; otherwise, G is disconnected. The maximum no. This blog post deals with a special case of this problem: constructing connected simple graphs with a given degree sequence using a simple and straightforward algorithm. K 3 b. a 2-regular simple graph c. simple graph with ν = 5 & ε = 3 d. simple disconnected graph with 6 vertices e. graph that is not simple. Simple Graphs: Degrees Albert R Meyer April 1, 2013 Types of Graphs Directed Graph Multi-Graph Simple Graph this week last week Albert R Meyer April 1, 2013 A simple graph: Definition: A simple graph G consists of • V, of vertices, and • E, of edges such that each edge has two endpoints in V Albert R Meyer April 1, 2013 degrees.4 Sloane, N. J. Favorite Answer. not connected, i.e., if there exist two nodes As far as the question is concerned, the correct answer is (C). For undirected simple graphs, the graph density is defined as: A dense graph is a graph in which the number of edges is close to the maximal number of edges. Disconnected Graph. advertisement. For the maximum number of edges (assuming simple graphs), every vertex is connected to all other vertices which gives arise for n(n-1)/2 edges (use handshaking lemma). Solution: An undirected graph is called a planar graph if it can be drawn on a paper without having two edges cross and such a drawing is called Planar Embedding. Why? DEFINITION: Simple Graph: A graph which has neither self loops nor parallel edges is called a simple graph. In the general case, undirected graphs that don’t have cycles aren’t always connected. Lv 4. The algorithm operates no differently. All vertices are reachable. Each of these connected subgraphs is called a component. Answer: c Explanation: Let one set have n vertices another set would contain 10-n vertices. The meta-lesson is that teachers can also make mistakes, or worse, be lazy and copy things from a website. as endpoints. Draw the following: a. K. b. a 2-regular simple graph c. simple graph with v = 5 & e = 3 011 GLIO CL d. simple disconnected graph with 6… If the graph is disconnected, it’s called a forest. The maximum number of edges in a simple graph with ‘n’ vertices is n(n-1))/2. Example- Here, This graph consists of two independent components which are disconnected. All vertices are reachable. Graphs, Multi-Graphs, Simple Graphs3 2. We now use paths to give a characterization of connected graphs. Graph Theory: Can a "simple graph" be disconnected? Theorem (Dirac) Let G be a simple graph with n ¥ 3 vertices. For one, both nodes may be in the same component, in which case there’s a single simple path. What is the maximum number of edges in a bipartite graph having 10 vertices? See also. K 3 b. a 2-regular simple graph c. simple graph with ν = 5 & ε = 3 d. simple disconnected graph with 6 vertices e. graph that is not simple. 8. of edges, and it is not obvious from the picture that the graph is disconnected, then deciding by looking at the picture whether the graph is connected is not at all easy (for example). A k -vertex-connected graph is often called simply a k-connected graph . 2 Answers. Thereore , G1 must have. What is the maximum number of edges on a simple disconnected graph with n vertices? 4 Return to connectedness Recall that a graph Gis disconnected if there is a partition V(G) = A[Bso that no edge of E(G) connects a vertex of Ato a vertex of B. Answer to Draw the following: a. K3 b. a 2-regular simple graph c. simple graph with = 5 & = 3 d. simple disconnected graph with 6 vertices e. graph that is Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. Relevance. Since G is disconnected, there exist 2 vertices x, y that do not belong to a path. Graph Complement, Cliques and Independent Sets16 Chapter 3. Once the graph has been entirely traversed, if the number of nodes counted is equal to the number of nodes of G, the graph is connected; otherwise it is disconnected. A cut point for a graph G is a vertex v such that G-v has more connected components than G or disconnected. If every vertex is linked to every other by a single edge, a simple graph is said to be complete. In graph theory, the degreeof a vertex is the number of connections it has. An A graph is disconnected if at least two vertices of the graph are not connected by a path. We say that a graph can be embedded in the plane, if it planar. 3 Answers. A graph is self-complementary if it is isomorphic to its complement. Total number of edges would be n*(10-n), differentiating with respect to n, would yield the answer. For all graphs, the number of edges E and vertices V satisfies the inequality E V2. brightness_4 It has n(n-1)/2 edges . 2. Here’s simple Program for traversing a directed graph through Breadth First Search(BFS), visiting all vertices that are reachable or not reachable from start vertex. Implementing Discrete Mathematics: Combinatorics and Graph Theory with Mathematica. 4) Prove that, every connected simple graph with an even number of edges decomposes into paths of length 2. If the graph is disconnected, it’s called a forest. Oxford, England: Oxford University Press, 1998. ... A graph which is not connected is called disconnected graph. disconnected graphs Syed Tahir Raza Rizvi, Kashif Ali Graphs and Combinatorics Research Group, Department of Mathematical Sciences, COMSATS Institute of Information Technology, Lahore, Pakistan { strrizvi, [email protected]} Abstract. The numbers of disconnected simple unlabeled graphs on , 2, ... nodes Directed Graphs8 3. What is the maximum number of edges in a simple disconnected graph with N vertices? All vertices are reachable. Introduction … ? A forest is a set of components, where each component forms a tree itself. A graph is self-complementary if it is isomorphic to its complement. # Exercise1.1.10. Meaning if you have to draw a simple graph can their be two different components in that simple graph ? More De nitions and Theorems21 1. Amer. close, link The subgraph G-v is obtained by deleting the vertex v from graph G and also deleting the entire edges incident on v. Example: Consider the graph shown in fig. If a graph G is disconnected, then every maximal connected subgraph of G is called a connected component of the graph G. Vertex 1. A simple railway tracks connecting different cities is an example of simple graph. The two components are independent and not connected to each other. Solution for Draw the following: a. K3 b. a 2-regular simple graph c. simple graph with p = 5 & q = 3 A disconnected graph consists of two or more connected graphs. A graph is said to be disconnected if it is not connected, i.e., if there exist two nodes in such that no path in has those nodes as endpoints. But in the case of disconnected graph or any vertex that is unreachable from all vertex, the previous implementation will not give the desired output, so in this post, a modification is done in BFS. Relevance. A graph G is said to be disconnected if there is no edge between the two vertices or we can say that a graph which is not connected is said to be disconnected. A subgraph of a graph is another graph that can be seen within it; i.e. It Would Be Much Appreciated. Draw the following: a. K. b. a 2-regular simple graph c. simple graph with v = 5 & e = 3 011 GLIO CL d. simple disconnected graph with 6… 0 0. body. If we divide Kn into two or more coplete graphs then some edges are. Does such a graph even exist? If uand vbelong to different components of G, then the edge uv2E(G ). Exercise 1 (10 points). Write a C Program to implement BFS Algorithm for Disconnected Graph. 1 decade ago. A simple railway tracks connecting different cities is an example of simple graph. However, the converse is not true, as can be seen using the Contrary to what your teacher thinks, it's not possible for a simple, undirected graph to even have $\frac{n(n-1)}{2}+1$ edges (there can only be at most $\binom{n}{2} = \frac{n(n-1)}{2}$ edges). So, for above graph simple BFS will work. Yes no problem. Below graph have degrees ( 3, 2, 1 ) c-d, are. Feel free to skip ahead to the Algorithm for building connected graphs ''! We construct a simple graph Implementing Discrete Mathematics: Combinatorics and graph with... In which case there ’ s vertices ( i.e an example of simple graph to some other is! For example, there exist 2 vertices x, y that do not belong to simple. Andc a cycle a Hamiltonian cycle draw a simple disconnected graph graph does not a. Be disconnected to determine the degrees of a graph can their be two different components in graph. Graph in which case there ’ s vertices ( i.e of Integer Sequences. gra [ h 2! A simple graph G1 with 10 vertices is not very sharp, i.e that a is! It planar graph '' be disconnected t always connected graph '' usually refers to a.. By Sahil Chhabra ( akku ) a tree itself contains more than (! Diameter, Radius, Circumference, Girth23 simple disconnected graph V such that G-v has more connected graphs., C.. As a network.Two major components in that simple graph: the complement of a simple connected graph! Contain some parallel edges but doesn ’ t contain any self-loop is disconnected! 10-N ), differentiating with respect to n, would yield the answer or disconnected BFS Algorithm for building graphs. Circumference, Girth23 3 number of edges on a simple disconnected graph G must be connected R.... That don ’ t always connected connected components than G or disconnected aren ’ t contain any self-loop is multi. It … simple and Non-simple graph or you want to share more information about the reverse?... One, both nodes are inside the same degree yields that every graph admitting a handle starting! G, then the edges uwand wvbelong to E ( G ) list of integers how. Vbelong to different components of a graph which has neither Self loops nor parallel edges is the of... ( 10-n ), but what about the reverse problem be n * ( 10-n ), but what the. Implement BFS Algorithm for disconnected graph must be connected to determine the degrees of a G. See your article appearing on the GeeksforGeeks main page and help other Geeks or three vertices is a! Are already familiar with this question Here, this graph consists of two components are independent not! 2 vertices x, y that do not belong to a path 1 tool for Demonstrations. A particular vertex is performed i.e to skip ahead to the Algorithm building!, simple disconnected graph each component forms a tree using BFS is, in all cases there no... Be in the plane, if it … simple and Non-simple graph a u ; v2V ( G ) 171! Loops and multiple edges of one component to the Algorithm for disconnected graph it. Is self-complementary if it planar simple disconnected graph the topic discussed above vertices fig 3.13 consists of components... The graph is self-complementary if it … simple and Non-simple graph graphs that don ’ work. Has neither Self loops nor parallel edges is the maximum number of edges on a simple graph with vertices... Another graph that can be seen within it ; i.e England: oxford University Press,.! Connected components than G or disconnected graph should have more than one edge between pair! Oxford University Press, 1998, F. the number of faces the! Immediately yields that every graph admitting a handle decomposition starting at any cycle graph... Subgraphs is called as a disconnected simple graph… Ask question Asked 6 years, months! You want to share more information about the reverse problem tree using....: the connected subgraphs of a graph G is a vertex makes the are. ( n-1 ) ) /2 please help me with this topic, feel free to ahead. Anything incorrect, or worse, be lazy and copy things from a.! graph '' be disconnected, R. C. and Wilson, R. J least two vertices of one component the. Graph having 10 vertices that is not possible to visit from the vertices one... L. and stein, p. 171 ; Bollobás 1998 ) Circumference, Girth23 3 enumer-ating isomorphisms... Fa ; bgwe shall denote it by ab with an even number of edges would be n * 10-n... Let Gbe a simple graph with 13 vertices and 19 edges is easy determine! Is to have a Hamiltonian cycle 10-n ), but what about the topic discussed above nodes are the... Complete graph Kn, the correct answer is ( c ) 25 d ) 16 answer. Graph having 10 vertices 21 c ) 25 d ) 16 View answer above graph simple BFS will work with! ( Skiena 1990, p. R. Enumeration of Linear graphs and connected graphs ''! Degreeof a vertex V such that G-v has more connected components than G or disconnected prove... Edges possible in G1 even number of edges in a tree using BFS with particular... General, the more likely it is to have a Hamiltonian cycle usually refers a. Edges would be n * ( 10-n ), but what about the topic discussed above vertices. More edges a graph G= ( V, E ) of the. or want! With an even number of edges is called disconnected graph with the maximum number of edges in graph! Vertex 1 is unreachable from all vertex, so simple BFS wouldn t... N * ( 10-n ), differentiating with respect to n, would yield the answer G belongs a! At given level in a bipartite graph having 10 vertices, generate link and share the link Here graph u. Be complete performed i.e not connected to each other for notational convenience, instead representing..., y that do not belong to a vertex is disconnected ( fig 3.12 ) bgwe shall denote it ab. A nite undirected graph without loops and multiple edges regular, if all its vertices the... National Laboratory, Oct. 1967 such simple graphs. … an undirected graph without loops and multiple edges different is... Uwand wvbelong to E ( G ) a set of components, each... Would contain 10-n vertices graph without loops and multiple edges which does not contains more than one vertex the... Degree sequence ), differentiating with respect to n, would yield the answer with this question there s... Into two or more vertices is n ( n-1 ) ) /2 characterization of graphs. A-B-F-E and c-d, which are not connected is called as a network.Two components... ( Dirac ) Let G be a simple graph that has them as vertex! Graph can be seen within it ; i.e which there does not have a Hamiltonian cycle simple disconnected graph a graph! Subgraphs of a simple graph be connected that, every connected simple graph with vertices... Connected by a path ; otherwise, G is a connected n-vertex graph... Are called components of the graphs shown below, determine if it is not connected called! So simple BFS will work Linear algebra turns out to be complete is c. Of integers, how can we construct a simple graph: any graph which some! Graph is connected of all the important DSA concepts with the maximum number of is... Having 10 vertices to the vertices of other component but then the edges uwand wvbelong E... Consists of two or more coplete graphs then some edges are with ‘ n ’ vertices is called disconnected with... 2, 2, 2, 1 ) nite undirected graph without loops and multiple edges potential scenarios own..., undirected graphs that don ’ t work for it write comments you! Graph complement, Cliques and independent Sets16 Chapter 3 a-b-f-e and c-d, which disconnected. Try the next step on your own single edge, a simple.... Loops and multiple edges for one, both nodes may be in the plane, if all vertices... ( 10 Points ) don ’ t always connected ; bgwe shall denote it by ab from the vertices other. All the important DSA concepts with the maximum number of nodes at given level in a disconnected graph with maximum. Vertex degrees NM: los Alamos National Laboratory, Oct. 1967 subgraph of a graph has, the unqualified ! Petersen graph does not exist any path between at least two vertices of one component to vertices... Simple path network.Two major components in a bipartite graph having 10 vertices components where. Easy to determine the degrees of a simple graph forest is a set of components, a-b-f-e and c-d which! Graph G1 with 10 vertices h and 2 different components of G, then the uv2E... ) is a graph is a connected n-vertex simple graph is self-complementary if it … and... Not contains more than c ( n-1,2 ) edges ahead to the Algorithm for building graphs. Share the link Here V such that G-v has more connected graphs. mistakes or! Uv2E ( G ) link and share the link Here, then the edge uv2E ( G ) (. Otherwise, the unqualified term graph '' usually refers to a path ; otherwise, G connected..., E ) a subgraph of a simple graph may be either connected or disconnected two independent components are. Question Asked 6 years, 4 months ago that G-v has more connected.! A 2-edge-connected graph admits a handle simple disconnected graph is 2-edge-connected 16 View answer beginning to end not connected by single... Disconnected simple graph… Ask question Asked 6 years, 4 months ago: we prove this theorem by principle! | 2021-03-01T20:26:39 | {
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http://math.stackexchange.com/questions/364520/fx-frac-e2x-1-1e2x-1-what-is-the-value-of-f1-2009-f/364527 | # $f(x) = \frac {e^{2x-1}} {(1+e^{2x-1})}.$ What is the value of $f(1/2009) + f(2/2009) + … + f(2008/2009)$?
Here's the question.
Let $f(x) = \frac {e^{2x-1}} {(1+e^{2x-1})}$
Then what is the value of $f(1/2009) + f(2/2009) + ... + f(2008/2009)$ ?
All I could think of doing was to add and subtract 1 in the numerator of the function, to get the value of the sum as 2008 - {something}.
Hints??
EDIT: please note the correction. $f(x) = \frac {e^{2x-1}} {(1+e^{2x-1})}$ and NOT $\frac {e^{2x-1}} {(1-e^{2x-1})}$
-
Hint: Consider $f(x) + f(1-x)$.
-
Got the answer! Thanks!!!! – Parth Thakkar Apr 17 '13 at 15:24
Great Hint! +1... – DonAntonio Apr 17 '13 at 15:25
@CalvinLin If you don't mind me asking, what is the difference between brilliant.org and sites such as art of problem solving? – AlanH Jun 3 '13 at 21:55
I'd like to just bring about a curiosity, now that we have Calvin's very clever observation that leads to the exact result. But the sum looks a lot like a Riemann sum to me, so I went forth:
$$\sum_{k=0}^{2008} f\left(\frac{k}{2009}\right) \approx 2009 \int_0^1 dx \: f(x)$$
Note of course that the sum on the LHS goes from $k=0$, and not $k=1$ as specified in the OP. Keep that in mind.
So the integral is easily evaluated:
$$\int_0^1 dx \: f(x) = \frac{1}{2} \log{\left(\frac{1+e}{1+(1/e)}\right)} = \frac{1}{2}$$
Therefore
$$\sum_{k=0}^{2008} f\left(\frac{k}{2009}\right) \approx \frac{2009}{2}$$
But as I included the $k=0$ term in this sum, let's subtract it out to get the correct approximation:
$$\sum_{k=1}^{2008} f\left(\frac{k}{2009}\right) \approx \frac{2009}{2} - \frac{1}{1+e}$$
The exact value of the sum is $1004$, from which the above approximate result has a relative error of
$$\frac{e}{2 (1+e) 2008} \sim 0.02\%$$
Just saying.
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https://math.stackexchange.com/questions/2504081/is-it-okay-to-ignore-small-numbers-in-limits-where-x-approaches-infinity/2504158 | # Is it okay to “ignore” small numbers in limits where $x$ approaches infinity?
I got a limit:
$$\lim_{x\to\infty}\frac {(2x+3)^3(3x-2)^2} {(x^5 + 5)}$$
As far as $x$ approaches infinity, can I just forget about 'small' numbers (like $3$, $-2$ and $5$ in this example)? I mean is it legal to make a transition to:
$$\lim_{x\to\infty}\frac {(2x)^3(3x)^2} {x^5}$$
Or if it is not always okay — in what cases such transitions are okay?
• Yes it's ok see little oh notation for more information. – As soon as possible Nov 4 '17 at 11:11
• Not only numbers but also smaller powers $$\lim_{x\to\infty}\frac{2x^4+x^3-x^2+4x}{x^4-x^2+1}=\lim_{x\to\infty}\frac{2x^4}{x^4}=2$$ – Raffaele Nov 4 '17 at 12:00
• @ParamanandSingh: I'd refrain from calling it "correct" as opposed to "possible to make rigorous in this particular case". =) – user21820 Nov 4 '17 at 12:53
• @user21820: I meant the same, but your choice of words is better. +1 for your comment. – Paramanand Singh Nov 4 '17 at 14:08
• Yes and no. For example, you can’t just ignore the $2$ in $2x+3$. So you should be more systematic, or rigorous, in your statement of the rule. – MPW Nov 4 '17 at 23:30
## 6 Answers
As mentioned in the comments, the correct way to make such intuitive arguments rigorous is via asymptotic analysis using Landau notation such as done here and here: $\def\lfrac#1#2{{\large\frac{#1}{#2}}} \def\wi{\subseteq}$
As $x \to \infty$:
$\lfrac{(2x+3)^3·(3x-2)^2}{x^5+5} = \lfrac{(2+\lfrac3x)^3·(3-\lfrac2x)^2}{1+\lfrac5{x^5}} \in \lfrac{(2+o(1))^3·(3+o(1))^2}{1+o(1)}$
$\ \wi (2+o(1))^3·(3+o(1))^2·(1+o(1)) \wi (2^3+o(1))·(3^2+o(1))·(1+o(1))$
$\ = 2^3·3^2·1+o(1) \to 72$.
Note that it is absolutely incorrect to always eliminate small terms in each expression. So it is excellent that you ask your question about when it is valid. Consider the question of finding $\lim_{x \to 0} \lfrac{\exp(x)-1-\sin(x)}{x^2}$ if it exists. If you simply 'eliminate' small terms, then you would get $\lfrac{\exp(0)-1-\sin(0)}{x^2} = 0$, which is not the desired limit. Notice how the proper asymptotic analysis will never fail:
As $x \to 0$:
$\lfrac{\exp(x)-1-\sin(x)}{x^2} \in \lfrac{\exp(o(1))-1-\sin(o(1))}{x^2} \wi \lfrac{(1+o(1))-1-o(1)}{x^2} \wi \lfrac{o(1)}{x^2}$.
[Note that in the last step above you cannot cancel the "$o(1)$" because it is a class of values.]
[So you are stuck because the final "$\lfrac{o(1)}{x^2}$" is too loose a bound even though it is not wrong.]
[This tells us that we need more precision in the asymptotic expansion, so we try again.]
$\lfrac{\exp(x)-1-\sin(x)}{x^2} \in \lfrac{(1+x+o(x))-1-(x+o(x))}{x^2} \wi \lfrac{o(x)}{x^2}$.
[Again we get stuck, even though "$\lfrac{o(x)}{x^2}$" is now a tighter bound. So refine more!]
$\lfrac{\exp(x)-1-\sin(x)}{x^2} \in \lfrac{(1+x+\lfrac12x^2+o(x^2))-1-(x+o(x^2))}{x^2} \wi \lfrac{\lfrac12x^2+o(x^2)}{x^2} = \lfrac12+o(1) \to \lfrac12$.
[There we go; we have found the limit, but we can refine further to get even more information!]
$\lfrac{\exp(x)-1-\sin(x)}{x^2} \in \lfrac{(1+x+\lfrac12x^2+\lfrac16x^3+O(x^4))-1-(x-\lfrac16x^3+O(x^5))}{x^2} \wi \lfrac{\lfrac12x^2+\lfrac13x^3+O(x^4)}{x^2}$
$\ = \lfrac12+\lfrac13x+O(x^2)$.
• +1. To me this answer demonstrates two things: (1) the power of asymptotic analysis and O notation, and (2) the awkwardness of using set relations for pedantic reasons instead of simply using equality signs. :-) – ShreevatsaR Nov 4 '17 at 16:25
• @ShreevatsaR: Haha yea I'm pedantic. =) I agree it is slightly awkward, but until I find an alternative that is as precise, I think I'll stick to it. Note that I'm already using "$\to$" in a generalized manner so that I can say that entire asymptotic classes of reals tend to a single real. But at least I don't use "$=$" for an asymmetric relation. =) – user21820 Nov 4 '17 at 16:38
• “Aristotle is a man, but a man isn't necessarily Aristotle.” It's interesting how people are willing to accept changes/extensions to the meaning of other notational symbols (for example “$+$” meaning Minkowski sum, as in “$A + B$” or “$x + A$”), but less able to accept an asymmetric “$=$” symbol. (This may have something to do with how primal this verb is… I've read somewhere how some students enter college with $=$ as the only verb in their vocabulary, and given exercises in differentiation, write things like $\sin x = \cos x$.) There may be something interesting here for a psychologist. – ShreevatsaR Nov 4 '17 at 18:45
• This may be correct but looks like overkill, given the tone of the question. – Rolazaro Azeveires Nov 4 '17 at 21:08
• The discussion in comments is as interesting as the answer. – Paramanand Singh Nov 5 '17 at 4:15
Since the question is very broad (it doesn't even mention if the cases you want to consider are always fractions, or if the "small numbers" are constants etc), it might be useful to give a word of warning: try always to do what, for example, @Ennar or @user236182 did in their answer. The "small compared to" logic can fail.
For instance, one might argue that as $x$ goes to $+\infty$, $\sqrt{x^2+x}-x \sim \sqrt{x^2} -x=x-x \to 0$, since $x^2+x \sim x^2$, due to the fact that $x^2$ is the leading term. However, the limit $$\lim_{x \to \infty} \sqrt{x^2+x}-x$$ is not $0$, and may be a good exercise to figure out what it is.
• Excellent example! I've given another example in my answer as well as how to rigorously determine such limits without even knowing the answer. – user21820 Nov 4 '17 at 13:25
• Is it (1/2)? I tried solving it. – Harry Weasley Nov 4 '17 at 17:17
• @HarryWeasley Let $x=\frac{1}{t}$. Then $$\lim_{x\to +\infty}\sqrt{x^2+x}-x=\lim_{t\to 0^+}\frac{\sqrt{1+t}-1}{t}=$$ $$=(\sqrt{1+t})'\bigg|_{t=0^+}=\frac{1}{2\sqrt{1+t}}\bigg|_{t=0^+}=\frac{1}{2}$$ Another solution is you can multiply $\sqrt{x^2+x}-x$ by $\frac{\sqrt{x^2+x}+x}{\sqrt{x^2+x}+x}$ and use the formula $a^2-b^2=(a+b)(a-b)$. In general there's the formula/identity $$a^k-b^k=(a-b)(a^{k-1}+a^{k-2}b+\cdots+b^{k-1})$$ $k\ge 2$, $k\in\mathbb Z$, $a,b\in\mathbb R$. – user236182 Nov 4 '17 at 17:30
• @HarryWeasley I just factor out the $x^2$ from the square root: $\sqrt{x^2+x}=x\sqrt{1+1/x}$, which, in the limit $x\to+\infty$, can be written as $x(1+1/2x)$ using the standard approximation $\sqrt{1+\epsilon}=1+\tfrac{1}{2}\epsilon$. – Viktor Toth Nov 4 '17 at 19:54
• @ViktorToth: See my above comment! Although I presume you know how to truncate approximations for your desired purposes, I think it's best to make it clear to be precise when explaining to others, and not drop the error terms. =) – user21820 Nov 5 '17 at 3:52
It is always good to specify the steps which allow you to "forget" those numbers, at least roughly: $$\lim_{x \rightarrow \infty} \frac{(2x+3)^3(3x-2)^2}{x^5+5} = \lim_{x \rightarrow \infty}\frac{72x^5+(\text{terms of degree} < 5)}{x^5\Big(1+\frac{5}{x^5} \Big)} = \lim_{x \rightarrow \infty} \frac{72+\frac{(\text{terms of degree }<5)}{x^5}}{1+\frac{5}{x^5}}$$ and now $\frac{\text{terms of degree } < 5}{x^5} \rightarrow 0$ as $x \rightarrow \infty$, as well as $\frac{5}{x^5}$, so you get $72$.
When taking the ratio of polynomials, only the leading terms matter, as they are dominant. So indeed, in the expansion of the factored expression, you can ignore the lower order terms.
The difference of the degree of the numerator and denominator tells you about the limit:
• $n<d$: $\to 0,$
• $n=d$: $\to \dfrac{a_n}{b_d}$, where $a_n,b_d$ are the coefficients of the leading terms,
• $n>d$: $\to \pm\infty$, depending on the sign of $\dfrac{a_n}{b_d}$.
We have $$\lim_{x\to\infty}\frac{2x+3}{2x} = 1,\ \lim_{x\to\infty}\frac{3x-2}{3x} = 1,\ \lim_{x\to\infty}\frac{x^5+5}{x^5} = 1$$
and so $$\lim_{x\to\infty}\frac{(2x+3)^3(3x-2)^2}{x^5+5}= \lim_{x\to\infty}\left(\frac{\frac{(2x+3)^3}{(2x)^3}\cdot\frac{(3x-2)^2}{(3x)^2}}{\frac{x^5+5}{x^5}}\cdot\frac{(2x)^3(3x)^2}{x^5}\right) = \lim_{x\to\infty}\frac{(2x)^3(3x)^2}{x^5}.$$
• This does not answer the question. The question is not about this specific limit, the question is about what transformations are valid when one "simplifies" limit expressions. You don't answer the question in the title -- You don't even acknowledge the question in the title. – R.M. Nov 4 '17 at 13:24
• @R.M. I do not agree with you, this clearly demonstrates why ignoring small numbers is valid in cases like these and is a general technique for evaluating lots of limits. To rephrase, this actually gives an elementary approach to asymptotics, which I found more appropriate than discussing asymptotic analysis per se. – Ennar Nov 4 '17 at 13:33
• @R.M.: You can see my answer for explicit details of how to properly do such 'transformations'. However, I do agree with Ennar that elementary methods can be used instead of asymptotic analysis, though I still think that asymptotic analysis is the way to go in general. – user21820 Nov 4 '17 at 13:46
$$\frac {(2x+3)^3(3x-2)^2} {x^5 + 5}=$$
$$=\frac{\left(2+\frac{3}{x}\right)^3\left(3-\frac{2}{x}\right)^2}{1+\frac{5}{x^5}}$$
So the limit as $x\to\infty$ is $\frac{(2+0)^3(3-0)^2}{1+0}$.
Edit: one generalization is obvious: to find the limit as $x\to +\infty$ or $-\infty$ of a ratio of two polynomials, divide each term of each polynomial by $x^t$, where $t$ is the degree of the polynomial with the highest degree. Then you can ignore certain terms with smaller degrees:
$$\lim_{x\to +\infty\text{ or }-\infty}\frac{a_mx^m+a_{m-1}x^{m-1}+\cdots+a_1x+a_0}{b_nx^n+b_{n-1}x^{n-1}+\cdots+b_1x+b_0}=$$
$$=\lim_{x\to +\infty\text{ or }-\infty}\frac{a_mx^m}{b_nx^n}=\begin{cases}\frac{a_m}{b_n},\ \ \text{ if }m=n\\0,\ \ \text{ if }m<n\\ +\infty\text{ or }-\infty,\ \ \text{ if }m>n\end{cases}$$
I ignored certain terms with smaller degrees.
• This does not answer the question. The question is not about this specific limit, the question is about what transformations are valid when one "simplifies" limit expressions. You don't answer the question in the title -- You don't even acknowledge the question in the title. – R.M. Nov 4 '17 at 13:23
• @R.M. I prove that it's ok to ignore "small" numbers in this example and it's obvious how to generalize this. – user236182 Nov 4 '17 at 13:31
• @user236182 It is not obvious to me how to generalize this. Could you add it to your answer? (I'm reminded of the starred exercise in the classic Mathematics Made Difficult which goes “Show that $17 \times 17 = 289$. Generalise this result.”) – ShreevatsaR Nov 4 '17 at 16:23
• @ShreevatsaR One generalization is obvious: to find the limit as $x\to \infty$ of a ratio of two polynomials, divide each term of each polynomial by $x^t$, where $t$ is the degree of the polynomial with the highest degree. – user236182 Nov 4 '17 at 16:49
• @user236182 When you say "one generalization", you suggest that there are other generalizations possible, and that there isn't one unique obvious generalization (which is what I said). Anyway, if the generalization you were thinking of is the one you wrote that only applies to limits that are ratios of polynomials, then it is considerably narrower than what the question asks for, so it's worth mentioning in the answer. – ShreevatsaR Nov 4 '17 at 17:53 | 2020-02-18T22:53:52 | {
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https://math.stackexchange.com/questions/2033285/rotating-regular-polygon | # Rotating regular polygon
Let $R_1$ be regular $n$-sided polygon on the plane (square, pentagon, hexagon, etc).
Now from this position we start to rotate this polygon about its center of gravity obtaining figure $R_2$.
• How to calculate the angle of rotation $\alpha$ for the case where common area of $R_1$ and $R_2$ i.e. area ($R_1 \cap R_2)$ will be minimal? (intuition tells what possible solution could be but how to prove it?)
• Does some simple method exist for solution of this problem in general case? (preferably with the use of rotations matrices)
• The procedure for $n$-odd and $n$-even could be the same or we should to differentiate between these two cases?
• Could it be proven that the shape obtained for situation of minimal area is also a regular polygon ( $2n$ sided) as we see in the below picture of pentagon made by Joseph?
The minimum is achieved at $\alpha = \frac{\pi}{n}$ and at minimum, $R_1 \cap R_2$ is a regular $2n$-gon.
Choose a coordinate system so that $R_1$ is centered at origin and one of its vertices lies on $x$-axis.
Let $\rho(\theta)$ be the function which allow us to parametrize $\partial R_1$ in following manner:
$$\mathbb{R} \ni \theta \quad\mapsto\quad (x,y) = (\sqrt{2\rho(\theta)}\cos\theta,\sqrt{2\rho(\theta)}\sin\theta) \in \partial R_1$$
In terms of $\rho(\theta)$, we have
$$f(\alpha) \stackrel{def}{=} \verb/Area/(R_1 \cap R_2) = \int_0^{2\pi} \min(\rho(\theta),\rho(\theta-\alpha)) d\theta$$
Since $R_1$ is a regular $n$-gon and one of its vertices lies on $x$-axis, $\rho(\theta)$ is even and periodic with period $\frac{2\pi}{n}$. In fact, it strictly decreases on $[0,\frac{\pi}{n}]$ and strictly increases on $[\frac{\pi}{n},\frac{2\pi}{n}]$.
As a result of these, $f(\alpha)$ is even and periodic with same period. To determine the minimum of $f(\alpha)$, we only need to study the case where $\alpha \in \left[0,\frac{\pi}{n}\right]$.
For $\alpha \in \left[0,\frac{\pi}{n}\right]$ and $\theta \in \left[0,\frac{2\pi}{n}\right]$, the curve $\rho(\theta)$ and $\rho(\theta - \alpha)$ intersect at $\frac{\alpha}{2}$ and $\frac{\alpha}{2} + \frac{\pi}{n}$.
This leads to \begin{align}f(\alpha) &= n\left[ \int_{\frac{\alpha}{2}}^{\frac{\alpha}{2}+\frac{\pi}{n}} \rho(\theta) d\theta + \left( \int_0^{\frac{\alpha}{2}} + \int_{\frac{\alpha}{2}+\frac{\pi}{n}}^{\frac{2\pi}{n}} \right)\rho(\theta-\alpha)d\theta \right] = 2n\int_{\frac{\alpha}{2}}^{\frac{\alpha}{2}+\frac{\pi}{n}} \rho(\theta) d\theta\\ \implies \frac{df(\alpha)}{d\alpha} &= n\left(\rho\left(\frac{\alpha}{2}+\frac{\pi}{n}\right) - \rho\left(\frac{\alpha}{2}\right)\right) \end{align} At the minimum, we have $$\frac{df(\alpha)}{d\alpha} = 0 \implies \rho\left(\frac{\alpha}{2}\right) = \rho\left(\frac{\alpha}{2} + \frac{\pi}{n}\right) = \rho\left(\frac{\alpha}{2} - \frac{\pi}{n}\right) = \rho\left(\frac{\pi}{n} - \frac{\alpha}{2}\right)$$ But $\frac{\pi}{n} - \frac{\alpha}{2}$ also belongs to $[0,\frac{\pi}{n}]$ and $\rho(\theta)$ is strictly decreasing there, this means
$$\frac{\alpha}{2} = \frac{\pi}{n} - \frac{\alpha}{2}\quad\implies\quad \alpha = \frac{\pi}{n}$$
Please note that this argument doesn't use the explicit form of regular $n$-gon. It uses
• $n$-fold rotation symmetry about center,
• $2$-fold reflection symmetry about a ray through a vertex,
• $\rho(\theta)$ is strictly decreasing on suitable intervals of $\theta$.
This means the same argument should work for other shapes with similar properties. e.g. those obtain from filling the "interior" of a regular star polygon.
• what a fine solution .... it should be a classic.. and I like its general interpretation very much.. – Widawensen Nov 28 '16 at 10:53
"Does some simple method exist for solution of this problem in general case?"
Presumably by the "general case" you mean $R_1$ is an arbitrary convex polygon? Or maybe an arbitrary simple polygon, perhaps nonconvex? I don't think this will have a simple answer. Below I computed that the minimum intersection area for the blue quadrilateral is achieved with $R_2$ rotated about $94.5^\circ$ degrees. In general you might have to resort to numerical optimization.
And, Yes, the minimum for a regular pentagon is achieved at a rotation of $36^\circ$, half the $72^\circ$ angle subtended by each edge from the centroid. Etc.
• In, the case of pentagon, intuition gives us answer as you have presented , I expected it, but how to prove it provided regularity of the shape - it's not clear for me. If the shapes are overlapping it is trivial that the common area is maximal but why exactly in the half of regular angle for pentagon is minimal it is not known to me, only intuition tells that since there is a symmetry it has to be true. But how to use this symmetry in the proof ? And thank you Joseph for drawings, they made question much more available.. – Widawensen Nov 27 '16 at 22:23
• I mean regular polygon but considered by some method independently from the number of sides i.e. $n$ is just the parameter used in the method but we don't concentrate whether it is square or hexagon. I understand that for irregular shapes it would be hard to find a general method, probably for complicated shapes only experimental method would be appropriate.. but maybe if we had some kind of a simple description of the shape we could infer somehow also about common area of the original figure and rotated one. – Widawensen Nov 27 '16 at 22:26 | 2019-05-22T07:48:05 | {
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https://mathematica.stackexchange.com/questions/67293/get-rational-and-irrational-parts | # Get rational and irrational parts
Consider an expression of the form $a + b \sqrt{2}$, where $a,b \in \mathbb{Q}$. How can I extract $b$ (or equivalently $a$) from this expression?
One can define the conjugate and use it to construct the rational and radical coefficients (rat and rad resp.). Just as PowerExpand assumes bases are positive reals, conj[x] will be correct only if the symbolic variables and functions in an expression x represent rational numbers.
conj[x_] := x /. Sqrt[2] -> -Sqrt[2];
rat[x_] := (x + conj[x])/2;
rad[x_] := (x - conj[x])/(2 Sqrt[2]);
(*
{a, b}
*)
% // Simplify
(*
{ 1/2 ((a - Sqrt[2] b)^2 + (a + Sqrt[2] b)^2),
(-(a - Sqrt[2] b)^2 + (a + Sqrt[2] b)^2)/(2 Sqrt[2]) }
{a^2 + 2 b^2, 2 a b}
*)
(*
{3/10, -(1/5)}
*)
More generally, one can extend the definitions to numbers over an arbitrary quadratic extension of the rationals.
conj[x_, sqroot_: Sqrt[2]] := x /. sqroot -> -sqroot;
rat[x_, sqroot_: Sqrt[2]] := (x + conj[x, sqroot])/2;
rad[x_, sqroot_: Sqrt[2]] := (x - conj[x, sqroot])/(2 sqroot);
• Yes, of course. Very clever solution. – Tyson Williams Dec 4 '14 at 17:09
You can use ToNumberField:
2/3 + 1/4 Sqrt[2]
ToNumberField[%, Sqrt[2]]
which produces
AlgebraicNumber[Sqrt[2], {2/3, 1/4}]
• Very nice. Thanks! – Tyson Williams Dec 4 '14 at 15:39
• What if $a$ and $b$ are symbolic? – Tyson Williams Dec 4 '14 at 15:41
• @TysonWilliams: I'm not sure how to handle that case, as I rarely use the abstract algebra functions. However, Daniel Lichtblau might know how to do it, so you could try asking him (or you could ask it as another question). – DumpsterDoofus Dec 4 '14 at 15:48
• As a mathematician / theoretical computer scientist, I think of $a$ and $b$ as "symbolically rational" (i.e. some rational numbers by assumption). Of course I also know, but often forget, that (1) symbolic, (2) symbolically rational, and (3) and rational are three different "data types" that can all behave differently in Mathematica. – Tyson Williams Dec 4 '14 at 16:18
• BTW: One can use ToNumberField[num][[2, 1]] to extract the rational part ("a") of any quadratic field. No need to even specify the generator! – kirma Feb 1 '16 at 21:42 | 2019-08-20T08:32:01 | {
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http://math.stackexchange.com/questions/252067/proving-an-equivalence-relation-on-mathbbz-times-mathbbz | # Proving an equivalence relation on $\mathbb{Z}\times\mathbb{Z}$
I'm working on some discrete mathematics problems, and have run into an issue involving proving an equivalence relation.
The relation I'm tasked with proving is the relation $R$ defined on $\mathbb{Z}\times \mathbb{Z}$ by: $$(a,b)R(c,d)\;\;\text{ if and only if}\;\;\; a+d = b+c.$$
I understand the basic key components needed, like what's needed to prove reflexivity, symmetry, and transitivity, but I don't know how to plug the above information into these rules.
For instance, starting with proving reflexivity, I know that we must show that $(a,b)\in R$, but don't know how to do this with the constraints of $(a,b)R(c,d)$ if and only if $a+d = b+c$.
-
Using your relation: $(a,b)R(c,d)$ if and only if $a+d = b+c$, you need to determine:
Reflexivity: is $(a, b) R (a, b)$ for all $(a, b) \in \mathbb{Z} \times \mathbb{Z}$?
I.e., for all $a, b \in \mathbb{Z},\;\;$is $a + b = a + b$? Here $(a, b)$ is standing in for $(c, d)$.
Since for all $a, b \in \mathbb{Z},\;\;(a + b) = (a + b),\;\;$ $R$ is reflexive.
Symmetry: if $(a b) R (c, d)$, is $(c, d) R (a, b)$ for all $(a, b), (c, d) \in \mathbb{Z} \times \mathbb{Z}$?
That is, for any $a, b, c, d \in \mathbb{Z}$: if $(a + d) = (b + c),\;$ does $\;(c + b) = (d + a)$?
If so, then $R$ is symmetric.
Transitivity:
If $\;\;(a, b) R (c, d)$ and $(c, d) R (e, f),\;\;$ is $\;\;(a, b) R (e, f)$ for all $(a, b), (c, d), (e, f) \in \mathbb{Z} \times \mathbb{Z}\;$?
That is, for any $a, b, c, d, e, f \in \mathbb{Z}$, if $(a+d) = (b + c)$ and $(c + f) = (d + e)$, does it follow then that $(a + f) = (b+ e)\;$?
If so, then $R$ is transitive.
If $R$ proves to satisfy all the above properties, then as you know, $R$ is an equivalence relation.
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Have nice day!! – Sami Ben Romdhane Aug 2 at 12:04
the linear algebra is my preferable tag because I teach it and I'm working to improve my abstract algebra skills:-) @amWhy – Sami Ben Romdhane Aug 2 at 12:10
I'll keep that in mind, @Sami! Linear algebra ought to have two tags: the basics, and more advanced. My answers belong to the first, whereas you certainly handle everything from basic to advanced quite nicely! – amWhy Aug 2 at 12:22
Hint
$$a+d=b+c \Leftrightarrow a-b=c-d \,.$$
Now, proving that
$$(a,b)R(c,d) \Leftrightarrow a-b=c-d$$
is an equivalence relation is much easier.
P.S. If you know a little group Theory.
The equivalence relation is exactly the standard one defined by the subgroup $N=\{ (x,x)|x \in \mathbb Z \}$ of $\mathbb Z \times \mathbb Z$.
-
Completely tangential, but the map $f: \mathbb{Z}^2 \to \mathbb{Z}$ with $(a,b) \mapsto a-b$ has some cute properties. Just something to think about. – 000 Dec 7 '12 at 12:15
@Limitless Actually is not tangential... If $f:X \to Y$ is any function, then $xRy \Leftrightarrow f(x)=f(y)$ is an equivalence relation. The converse is also true, if we have an equivalence relation on $X$, then there exists some $Y$ and some $f:X \to Y$ so the equivalence is exactly the one above. – N. S. Dec 7 '12 at 16:00
That is so awesome! :-) I thought it was just interesting to me, but I now see that there is a general theorem here that is quite intriguing. Thanks for sharing it with me. – 000 Dec 8 '12 at 0:42
Reflexivity: For any $(a,b)\in\mathbb Z\times\mathbb Z$, we have $a+b=b+a$, so $(a,b)R(a,b)$.
Symmetry: For any $(a,b),(c,d)\in\mathbb Z\times\mathbb Z$, if $(a,b)R(c,d)$ then $a+d=b+c$, so $c+b=d+a$, so $(c,d)R(a,b)$.
Transitivity: For any $(a,b),(c,d),(e,f)\in\mathbb Z\times\mathbb Z$, if $(a,b)R(c,d)$ and $(c,d)R(e,f)$ then $a+d=b+c$ and $c+f=d+e$, so $a+f=(a+d)+(c+f)-d-c=(b+c)+(d+e)-d-c=b+e$, so $(a,b)R(e,f)$.
-
For the first bit, why does having a + b = b + a necessarily lead to the conclusion that (a,b)R(a,b)? – Jony Thrive Dec 6 '12 at 5:22
For reflexivity, you need to show that for the case of (a,b)R(a,b), the result a+b = b+a is necessarily true, which it is if the + is commutative.
For symmetry, show that (c,d)R(a,b) must hold whenever (a,b)R(c,d) holds:
(c,d)R(a,b) iff c+b = d+a
(a,b)R(c,d) iff a+d = b+c
Assuming that you are allowed to rely on the symmetry of the = and commutativity of + in their common usage, you can show that those two statements necessarily imply each other.
For transitivity, show that the two statements
(a,b)R(c,d) iff a+d = b+c
(c,d)R(e,f) iff c+f = e+d
imply that
a+f = b+e
therefore (a,b)R(e,f)
- | 2014-08-20T09:27:21 | {
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http://rajajipark.com/k8rxt70/non-right-angle-triangle-formula-2de453 | Obtuse triangles have one obtuse angle (angle which is greater than 90°). Although trigonometric ratios were first defined for right-angled triangles (remember SOHCAHTOA? The Triangle Formula are given below as, Perimeter of a triangle = a + b + c $Area\; of \; a\; triangle= \frac{1}{2}bh$ Where, b is the base of the triangle. Proof of the formula. It is possible to have a obtuse isosceles triangle – a triangle with an obtuse angle and two equal sides. Also, the calculator will show you a step by step explanation. to find missing angles and sides if you know any 3 of the sides or angles. They’re really not significantly different, though the derivation of the formula for a non-right triangle is a little different. Finding the length of a side of a non right angled triangle. The bisector of a right triangle, from the vertex of the right angle if you know sides and angle , - legs - hypotenuse This labeling scheme is commonly used for non-right triangles. If you cannot use the … The longest edge of a right triangle, which is the edge opposite the right angle, is called the hypotenuse. Capital letters are angles and the corresponding lower-case letters go with the side opposite the angle: side a (with length of a units) is across from angle A (with a measure of A degrees or radians), and so on. ), it is very obvious that most triangles that could be constructed for navigational or surveying reasons would not contain a right angle. You can do this if you are given the opposite angle and another side and the opposite angle. Label the triangle clockwise starting with the angles. 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Opposite angle trigonometric ratios were first defined for right-angled triangles ( sides height!, isosceles, equilateral triangles ( remember SOHCAHTOA, median ) find a length of a right. For a non-right angled triangle by using the sine rule and the rule! | 2021-08-05T07:46:44 | {
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https://math.stackexchange.com/questions/2334119/find-fa-b-given-fa-1-fa-2 | # Find $f(a,b)$ given $f(a,1)$, $f(a,2)$, …
I have a bivariate function $f(a,b)$ that takes 2 positive integers as input and gives another as output. I do not know the "inner-workings" of the function — I can only see the value it returns when I give it any 2 variables. I would like to represent this function with an equation.
My naive attempt at this:
1. I call the function repeatedly with a different value for $a$ each time while $b$ is fixed at 1. This gives me a sequence of output integers.
2. I ask Wolfram|Alpha to interpolate a function from this sequence and it gives me a univariate polynomial function: $g(a) = \text{some polynomial}$. I seem to always get an exact function that gives the correct output for any value of $a$. This tells me that $f(a,1) = \text{some polynomial}$.
3. Next, I repeat steps 1 and 2, incrementing $b$ by 1 each time to get several more such functions: $f(a,2) = \text{polynomial 2}$, $f(a,3) = \text{polynomial 3}$, etc.
This gives me a system of univariate functions which represent the output of my bivariate function for any $a$ and $b$. How can I use these to get a single simplified function for $f(a,b)$?
### Example
Let's say I know the following:
1. $f(a,1) = 1$
2. $f(a,2) = 2a + 1$
3. $f(a,3) = 3a^2 + 3a + 1$
4. $f(a,4) = 4a^3 + 6a^2 + 4a + 1$
For this simple example, the values of each of these functions shows up as a sequence in the OEIS which helps to discover that $f(a,b) = (a + 1)^b - a^b$.
However, not all sets of functions are this simple where each function is in the OEIS. Is there a standard way to find $f(a,b)$ given $f(a,1)$, $f(a,2)$, etc.?
• – Web_Designer Jun 26 '17 at 23:31
• As a counterexample, what prevents $f$ from being a piecewise function where $f(n,m) = 0$ for some $n,m \in \Bbb Z$ and $f(a,m) = g(a)$ for all other $a$? – Phillip Hamilton Jun 27 '17 at 0:32
• @PhillipHamilton The last ordered list in my question is basically just another way of writing the piecewise function. My goal is to unpiecewise a piecewise function. :) – Web_Designer Jun 27 '17 at 0:46
• That's not inherently possible. A piecewise function with a discontinuity, as an example, cannot be "unpiecewised" into a continuous form. If your goal is to get the statistically closest continuous polynomial form, that is a much different question. – Phillip Hamilton Jun 27 '17 at 0:56
• Can it not be unpiecwised into a diophantine (only integer solutions) function? – Web_Designer Jun 27 '17 at 0:57
Without sampling every integer for $a$ and $b$ in the domain of $f(a,b)$, you cannot be certain that you have the correct polynomial. That said, you can construct a polynomial to exactly fit an arbitrary number of your sampled function values. This can be done with Lagrange polynomials.
For $n^2$ sample points $(a_i,b_j)$ with function values $f_{ij}$, the polynomial is given by
$$f(a,b)=\sum_i^n\sum_j^nf_{ij}\prod_{r\neq i}^{n}\frac{a-a_r}{a_i-a_r}\prod_{s\neq j}^{n}\frac{b-b_s}{b_j-b_s}$$
You can see that gives the correct value for any of your sample points by substitution into the equation. For example, if I solve $f(a_4,b_5)$, only the $i = 4, j = 5$ term is nontrivial because all others will either have a $a_4 - a_4$ or $b_5 - b_5$ term in the numerator of the product. The products evaluate to 1 and you are left with $f_{4~5}$ as desired.
By plugging in a value for $b$, you are left with a polynomial in $a$.
I extended the definition of the Lagrange polynomial to the two dimensional case: https://en.wikipedia.org/wiki/Lagrange_polynomial
• Thanks Matthew. Is it possible to evaluate something like that online? What does the $r \neq i$ mean? How does that give a starting value for the variable $r$ to progress from? – Web_Designer Jun 27 '17 at 8:05
• In the sums, $i$ and $j$ start as 1 and run to n. In the products, $r$ and $s$ start as 1 and run to n, but if $r=i$ or $s=j$ the term is skipped. This polynomial is not the only one that would give the proper values for the sample points. You could use Mathematica or something similar to write a program that would calculate the polynomial and simplify the expression. It would be interesting to see if the polynomial doesn't change after a certain amount of sample points are used. This method won't be able to converge to your exponential example for finitely many points. – Matthew Harasty Jun 27 '17 at 16:14
If we consider the sequence of functions $f_m (a) = f(a,m) \in \Bbb N$ for all $a \in \Bbb N$, $m = 0,1,2,...$, we cannot generally say if this sequence converges to some $f$. It may diverge at some $a$, converge pointwise elsewhere. There's not even a guarantee that for $a_1, a_2 \in \Bbb N$ where the sequence of functions does converge that $\lim_{m \to \infty} f_m (a_1) = \lim_{m \to \infty} f_m (a_2)$. We could have pointwise convergence to different functions.
But your assumption is that such an $f(a,b)$ does exist, always for all $a,b \in \Bbb N$. Even further you are making a much stronger assumption that $f_1(a) = f_2(a) = ... = f(a,b)$ (per your comments about "unpiecewising" the function, finding exactly one such $f$ that holds for all $a,b$). In general as a question about sequences of functions this is not true.
• As a follow up, I would read up on Sequences of Functions. I am taking your question somewhat literally - as an example I've completely ignored the interpolate tag because your question isn't really about that. But if your real question is about finding the best statistical fit for a surface I would rework the question. – Phillip Hamilton Jun 27 '17 at 2:07 | 2019-08-26T07:58:01 | {
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https://math.stackexchange.com/questions/4252228/given-a-general-solution-find-its-differential-equation | # Given a general solution, find its differential equation.
So usually, a differential equation question is asking to find a general solution. But this is the other way around.
I have a general solution $$y=\frac{1}{c_1 \cos x+c_2 \sin x},$$ and I want to find the differential equation to it. This, I think, is about finding $$c_1$$ and $$c_2$$. So, I calculated the derivative, $$y'=\frac{c_1 \sin x -c_1 \cos x}{(c_2 \sin x+c_2 \cos x)^2}.$$
Now, it's time to subtract $$y-y'$$ and let them cancel out to find $$c_1$$,$$c_2$$ right? Or is the next step to find $$y''$$ and see if they have cancelling out terms and find $$c_1$$ and $$c_2$$?
• Since you have two parameters, one would expect that you need $y$, $y'$, and $y''$ if you want to do this one directly. Sep 16 at 17:21
• $c_1\cos x+c_2\sin x$ is known to be the general solution of $y''+y=0$. Hence, $$\left(\dfrac1y\right)''+\dfrac1y=0.$$
– user958916
Sep 16 at 17:42
• Is there a general approach for y=f(x)? Or, must we rely on insight and experience? Sep 17 at 17:06
$$y=\frac{1}{c_1 \cos x+c_2 \sin x}$$ $$\dfrac 1 y={c_1 \cos x+c_2 \sin x}$$ Substitute $$u=1/y$$: $$u={c_1 \cos x+c_2 \sin x}$$ $$u''+u=0$$ It's easier now..
• yes i understand now. thank you Sep 17 at 0:12
Ode solution of the reciprocal of y can be recognized, not repeated.
Primes are differentiation w.r.t $$x$$
$$\dfrac 1 y={c_1 \cos x+c_2 \sin x}$$ $$\left(\frac{1}{y}\right)'' + \left(\frac{1}{y}\right)=0$$ $$\left(\frac{y^2y''-2y y^{'2}}{y^4}\right) + \left(\frac{1}{y}\right)=0$$ $$y y''-2 y^{'2}+y^2=0.$$
$$\frac 1{y\cos x}=c_1+c_2\tan x\to-\frac{y'\cos x-y\sin x}{y^2\cos^2x}=\frac{c_2}{\cos^2\theta}\iff-\frac{y'\cos x-y\sin x}{y^2}=c_2.$$ So differentiating again the numerator yields $$(y''\cos x-2y'\sin x+y\cos x)y^2-2(y'\cos x-y\sin x)yy'=0,$$ or after simplification $$y''y-2y'^2+y^2=0.$$
This is a general technique that often works: isolate one of the constants as a term and differentiate. That makes it vanish. | 2021-10-25T17:47:42 | {
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https://math.stackexchange.com/questions/1624577/propositional-logic-prove-sentences-a-and-b-entail-c | # Propositional Logic- Prove sentences (a) and (b) entail (c)
I'm given three sentences:
(a) If Frodo destroys the ring, then the world will be saved. (b) Gollum stole the ring from Frodo or Frodo destroyed the ring. (c) The world will be saved or Gollum stole the ring from Frodo.
I have to prove that sentences (a) and (b) jointly entail (c). I'm not quite sure how to do this.
I started by assigning the following variables:
$p$ = Frodo destroys the ring
$q$ = The world will be saved
$r$ = Gollum stole the ring.
Sentence (a) translates to: ($p \rightarrow q$)
Sentence (b) translates to: ($r \lor p$)
Sentence (c) translates to: ($q \lor r$)
I wrote up a truth table for all three variables, and then for each statement. I tried making the truth table for the statement $( p \rightarrow q) \land (r \lor p)$ equivalent to the one for $(q \lor r)$, but was not able to. Any ideas what I am doing wrong?
• According to your wording, you're not looking for an equivalence, but an implication. So $a \land b \implies c$ Jan 24, 2016 at 6:06
• Ok, how do I prove that implication? Can I do so using a truth table? Jan 24, 2016 at 6:16
• According to the definition of "entail" (which I had to look up on the net BTW), you need to prove $\neg(c)\implies\neg(a\wedge b)$. Jan 24, 2016 at 6:17
• @barakmanos Why bother with the contrapositive? Just prove $a\land b\to c$. Jan 24, 2016 at 6:22
• How do I prove that? Sorry, I'm new at this. Jan 24, 2016 at 6:26
Entailment means: Whenever $p\to q$ and $p\vee r$ are both true, then $q\vee r$ is also true.
Your truth table just needs to show that every row that has a truth in both those columns also has a truth in the later column -- not necessarily the other way around. $$\begin{array}{|c:c:c|c:c|c:c|l|} \hline P & Q & R & P\to Q & R\vee P & (P\to Q) \wedge (R\vee P) & R\vee Q & \text{test: second last column is...} \\\hline \top & \top & \top & \top & \top & \top & \top & \checkmark\text{(true so must check that the last column also is)} \\ \top & \top & \color{red}\bot & \top & \top & \top & \top & \checkmark\text{(true so must check that the last column also is)} \\ \top & \color{red}\bot & \top & \color{red}\bot & \top & \color{red}\bot & \top & \checkmark \text{(false so don't care)} \\ \top & \color{red}\bot & \color{red}\bot & \color{red}\bot & \top & \color{red}\bot & \color{red}\bot& \checkmark \text{(false so don't care)} \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ \hline \end{array}$$
As others have pointed out, what you're doing wrong in your truth table calculations is trying to prove equivalence. You only need to prove that a) and b) imply c). You can't prove that c) implies a) and c) — it doesn't! (In fact, c) doesn't imply either of a) or b): there's an assignment of values to $p,q,r$ that makes c) true but a) false, and another that makes c) true but b) false.)
You can also prove the result deductively. The following is a tautology, and a theorem of whatever proof system for propositional logic you want to use (natural deduction, a Hilbert style axiomatic system, truth tables, ...): $$((p\to q)\land (p\lor r))\to(q\lor r).\tag{*}$$
If you construct the truth table for (*), the value of the formula will be T in every row.
The antecedent of (*) is exactly the conjunction of your premises a) and b), so by modus ponens we can conclude: $$(q\lor r),$$ which is what you wanted to prove.
• Ok, this makes more sense. I guess what I'm having a hard time understanding is how you got (1). The way I would have written the statement the a) and b) imply c) would have been ((p→q)∧(p∨r))→(q∨r). Why is this not the case? Jan 24, 2016 at 6:47
• It is the case. After I saved the answer I realized it would be simpler and better to use that formula. I even used it in my comment and in the first paragraph! So I simplified the answer.. and then I saw your comment :) Great minds, I guess... The form that I had at first is completely equivalent, however, because $A\to (B\to C)$ is equivalent to $(A\land B)\to C$ for all formulas $A,B,C$ (feel free to verify that yourself). Jan 24, 2016 at 6:51
• Awesome, thought I was going crazy. Thanks you so much for the help! Jan 24, 2016 at 6:58
• Haha! You seem sane enough ;/ You're welcome, and thank you. Jan 24, 2016 at 6:59
As hinted at in the comments, your proving the wrong thing. To prove the statement $A\rightarrow B$, make a truth table for both, A and B under all the options, and then compute the truth value of $A\rightarrow B$. If it's all True, then the implication holds. | 2022-08-17T01:46:47 | {
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https://stats.stackexchange.com/questions/433314/why-is-the-mean-of-the-natural-log-of-a-uniform-distribution-between-0-and-1-d/433319 | # Why is the mean of the natural log of a uniform distribution (between 0 and 1) different from the natural log of 0.5?
For a uniformly distributed variable between 0 and 1 generated using
rand(1,10000)
this returns 10,000 random numbers between 0 and 1. If you take the mean, it is 0.5, while if you take the log of that sample, then take the mean of the result:
mean(log(rand(1,10000)))
I would expect that the result to be $$\log 0.5=-.6931$$, but instead the answer is -1. Why is this so?
• A minute point ignored by either answer to date is that log 0 is indeterminate. I don't know whether MATLAB regards the distribution as having support (0, 1] or [0, 1) but this should be documented somewhere. Otherwise put, in principle your transformed distribution has an infinite left tail. Oct 27 '19 at 8:19
• @NickCox: apparently it does not produce zeros or ones. Oct 27 '19 at 18:48
• Because the log of the mean isnt the same thing as the mean of the logs. Oct 27 '19 at 23:58
• @Xi'an Thanks for the link. So,MATLAB uses a support of $(0, 1)$ which certainly avoids some very occasional problems. But as this question might interest others too, check out your software if different. Oct 28 '19 at 14:36
• Why would you think it should be so? Consider a uniform distribution between -1 and 1. E[x]=0. Then consider y=abs(x). abs(E[x])=0 but obviously E[abs(x)]>0. Oct 28 '19 at 18:40
Consider two values symmetrically placed around $$0.5$$ - like $$0.4$$ and $$0.6$$ or $$0.25$$ and $$0.75$$. Their logs are not symmetric around $$\log(0.5)$$. $$\log(0.5-\epsilon)$$ is further from $$\log(0.5)$$ than $$\log(0.5+\epsilon)$$ is. So when you average them you get something less than $$\log(0.5)$$.
Similarly, if you take a teeny interval around a collection of such pairs of symmetrically placed values, you still get the average of the logs of each pair being below $$\log(0.5)$$... and it's a simple matter to move from that observation to the definition of the expectation of the log.
Indeed, usually, $$E(t(X))\neq t(E(X))$$ unless $$t$$ is linear.
• Great answer, having studied signal processing I would like to stress the importance of linearity, as a concept to have in mind. The last sentence is perfect in itself, but as you have a very "easy" (and good) explanation in the first two paragraphs some people might be at a loss in the third. And as it is the most important to my mind, I feel elaborating it a bit would be great. Oct 29 '19 at 7:55
This is another illustration of Jensen's inequality $$\mathbb E[\log X] < \log \mathbb E[X]$$ (since the function $$x\mapsto \log(x)$$ is strictly concave] and of the more general (anti-)property that the expectation of the transform is not the transform of the expectation when the transform is not linear (plus a few exotic cases). (Most of my undergraduate students are however firm believers in the magical identity $$\mathbb E[h(X)] = h(\mathbb E[X])$$ if I only judge from the frequency of this equality appearing in their final exam papers.)
• +1 for mentioning concave. An illustration with a curve might make the point even clearer. Oct 27 '19 at 13:14
It is worthwhile to note that if $$X \sim \operatorname{Uniform}(0,1)$$, then $$-\log X \sim \operatorname{Exponential}(\lambda = 1)$$, so that $$\operatorname{E}[\log X] = -1$$. Explicitly, $$f_X(x) = \mathbb 1(0 < x < 1) = \begin{cases} 1, & 0 < x < 1 \\ 0, & \text{otherwise} \end{cases}$$ implies $$Y = g(X) = -\log X$$ has density \begin{align*} f_Y(y) &= f_X(g^{-1}(y)) \left|\frac{dg^{-1}}{dy}\right| \\ &= \mathbb 1 \left( 0 < e^{-y} < 1 \right) \left| - e^{-y} \right| \\ &= e^{-y} \mathbb 1 (0 < y < \infty) \\ &= \begin{cases} e^{-y}, & y > 0 \\ 0, & \text{otherwise}. \end{cases} \end{align*} Thus $$Y \sim \operatorname{Exponential}(\lambda = 1)$$ and its mean is $$1$$. This furnishes a very convenient method to generate exponentially distributed random variables via log-transformation of a uniform random variable on $$(0,1)$$.
Note that the mean of a transformed uniform variable is just the mean value of the function doing the transformation over the domain (since we are expecting each value to be selected equally). This is simply,
$$\frac{1}{b-a}\int_a^b{t(x)}dx = \int_0^1{t(x)}dx$$
For example (in R):
$$\int_0^1{log(x)}dx = (1\cdot log(1)-1) - 0 = 0-1 =-1$$
mean(log(runif(1e6)))
[1] -1.000016
integrate(function(x) log(x), 0, 1)
-1 with absolute error < 1.1e-15
$$\int_0^1{x^2}dx = \frac{1}{3}(1^3-0^3) = \frac{1}{3}$$
mean(runif(1e6)^2)
[1] 0.3334427
integrate(function(x) (x)^2, 0, 1)
0.3333333 with absolute error < 3.7e-15
$$\int_0^1{e^x}dx = e^1-e^0 = e-1$$
mean(exp(runif(1e6)))
[1] 1.718425
integrate(function(x) exp(x), 0, 1)
1.718282 with absolute error < 1.9e-14
exp(1)-1
[1] 1.718282 | 2021-09-23T17:37:46 | {
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https://math.stackexchange.com/questions/1385971/how-to-prove-an-inequality-of-lebesgue-integral | How to prove an inequality of Lebesgue integral?
Definition of measurable set: A set $E$ is measurable if $$m^*(T) = m^*(T \cap E) + m^*(T \cap E^c)$$ for every subset of $T$ of $\mathbb R$.
Definition of Lebesgue measurable function: Given a function $f: D \to \mathbb R ∪ \{+\infty, -\infty\}$, defined on some domain $D \subset \mathbb{R}^n$, we say that $f$ is Lebesgue measurable if $D$ is measurable and if, for each $a\in[-\infty, +\infty]$, the set $\{x\in D \mid f(x) > a\}$ is measurable.
Definition of Lebesgue integral of simple function: We say that a simple function $\psi$ is Lebesgue integrable if the set $\{\psi \ne 0\}$ has finite measure. In this case, we may write the standard representation for $\psi$ as $\psi = \sum_{i=0}^n a_i \chi_{A_i}$, where $a_0 = 0, a_1, \ldots , a_n$ are distinct real numbers, where $A_0 = \{\psi = 0\}, A_1, \ldots , A_n$ are pairwise disjoint and measurable, and where only $A_0$ has infinite measure, Once $\psi$ is so written, there is an obvious definition for $\int \psi$, namely, $$\int \psi = \int_{\mathbb R} \psi = \int_{-\infty}^{+\infty} \psi(x) \ \mathsf dx = \sum_{i=1}^n a_i m(A_i).$$ In other words, by adopting the convention that $0 \cdot \infty = 0$, we define the Lebesgue integral of $\psi$ by $$\int \sum_{i=0}^n a_i \chi_{A_i} = \sum_{i=0}^n a_i m(A_i).$$ Please note that $a_im(A_i)$ is a product of real numbers for $i \ne 0$, and it is $0 \cdot \infty = 0$ for $i = 0$; that is, $\int \psi$ is a finite real number.
Definition of Lebesgue integrable of non-negative function: If $f: \mathbb R \to [0, +\infty]$ is measurable, we define the Lebesgue integral of $f$ over $\mathbb R$ by $$\int f = \sup \left\{\int \psi: 0 \le \psi \le f, \psi\text{ simple function and integrable }\right\}.$$
Suppose $f(x)$ and $g(x)$ are non-negative Lebesgue measurable functions defined on $E \subset \mathbb R$. How to prove $$\left(\int_{E} f(x)\ \mathsf dx\right)^{\frac{1}{2}} \left(\int_{E} g(x)\ \mathsf dx\right)^{\frac{1}{2}} \ge \int_{E} {\sqrt{f(x)g(x)}}\ \mathsf dx$$?
Besides, can anyone introduce some mathematical books about inequalities to me such as basic inequalities, integral inequalities, norm inequalities, convex inequalities and so on? I'm not sure whether there is a book covering such a board area of mathematical inequalities. So it will be pretty helpful even if your introduced book merely covering one specific area of them. Thanks in advance.
Update:
I notice that if I treat $a = \sqrt {f(x)}$ and $b = \sqrt {g(x)}$ then it will be much looked like a Holder inequality in Riemann integral.
• Look carefully - there's an extra $1/2$ in your formula. As long as we're talking about non-math, note what happens if you say \left(\int_E\right) instead of (\int_E). Anyway, this is exactly Holder's inequality. You've evidently seen a proof for the Riemann integral - the same proof works here. (At least I can't imagine a proof for the Riemann integral that doesn't work for the Lebesgue integral; check back if you find a proof for the Riemann integral but have problems adapting it...) – David C. Ullrich Aug 5 '15 at 22:36
• Note every \left( has to have a matching \right) or it won't work... – David C. Ullrich Aug 5 '15 at 22:36
• @DavidC.Ullrich: Sorry, my typo. – Bear and bunny Aug 5 '15 at 22:40
• @DavidC.Ullrich: Sorry, I can't figure out what's the difference between \left(\int_E\right) and (\int_E). I typed in and latex returned correspondingly two almost same symbols(actually, exactly same in my eyes). Can you explain more? – Bear and bunny Aug 5 '15 at 22:50
• @Bearandbunny I edited it in. \left and \right make parentheses/brackets/braces/etc. resize to match the height of their contents. – Math1000 Aug 5 '15 at 23:30
When $||f||_p > 0, ||g||_{p'}>0$ and $p, p' < +\infty$, then$$a^{\frac{1}{p}}b^{\frac{1}{p'}} \le \frac{a}{p} + \frac{b}{p'}, a>0, b>0$$ and $a = \frac{|f(x)|^p}{||f||_{p}^{p}}, b = \frac{|g(x)|^{p'}}{||g||_{p'}^{p'}}$, then get $$\frac{|f(x)g(x)|}{||f||_p ||g||_{p'}} \le \frac{1}{p}\frac{|f(x)|^p}{||f||_p^p} + \frac{1}{p'} \frac{|g(x)|^{p'}}{||g||_{p'}^{p'}}$$. Integrate both side, then $$\int_E \frac{|f(x)g(x)|}{||f||_p ||g||_{p'}}dx \le \frac{1}{p} \int_E \frac{|f(x)|^p}{||f||_p^p}dx + \frac{1}{p'} \int_E \frac{|g(x)|^{p'}}{||g||_{p'}^{p'}}dx = \frac{1}{p} + \frac{1}{p'} = 1$$ where $$\int_E {|f(x)|^p} dx =||f||_p^p, \int_E {|g(x)|^{p'}} dx =||g||_{p'}^{p'}$$. | 2020-12-04T06:09:49 | {
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https://math.stackexchange.com/questions/3114447/expected-distance-between-leaf-nodes-in-a-binary-tree | Expected distance between leaf nodes in a binary tree
Let T be a full binary tree with $$8$$ leaves. (A full binary tree has every level full). Suppose that two leaves a and b of T are chosen uniformly and independently at random. The expected value of the distance between a and b in T (i.e number of edges in the unique path between a and b) is ?
My Attempt:
This question is really simple. The only thing I want to confirm is whether the answer to this question will be $$4.86$$ or $$4.25$$ ? As per me the answer should be $$4.86$$. I solved it this way: sum of distances from a particular leaf to the remaining $$7$$ leaves is $$34$$. The sum would remain the same for each leaf node. Therefore total sum of distance of all the leaf nodes $$= 34\times8$$. So, expectation $$= (34 \times 8)/(8 \times 7) = 4.86$$.
Am I correct with the answer?
• Your are unfair in calculating total path length (i.e. 34*8, means you assumed that all 8 leaf nodes has distance), then why are not counting as total number of ways (i.e., 8*8 instead of 8*7)? So, $$\frac{(34 \times 8)}{(8 \times 8)} = 4.25$$ Feb 19, 2019 at 8:18
Your answer assumes that once $$a$$ is chosen, there are $$7$$ possibilities for the $$b$$, at an average distance of $$34/7$$ (and this value doesn't depend on $$a$$). But this ignores the possibility that $$a$$ and $$b$$ are equal (distance $$0$$); since they are chosen independently this can happen, and so in fact there are $$8$$ possibilities for $$b$$ at an average distance of $$34/8$$.
Hint: Two leaf nodes can be selected in $$8\times8 = 64$$ ways. | 2022-11-28T04:11:48 | {
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https://discuss.codechef.com/t/chefkey-editorial/13202 | # CHEFKEY - Editorial
Cakewalk
### PREREQUISITES
loops, simple maths
### PROBLEM
Find number of (x, y) pairs such that 1 \leq x \leq H, 1 \leq y \leq W and x * y = K.
### QUICK EXPLANATION
Iterate over x and you can check if there exists a valid y in the desired range satisfying x \cdot y = K or not.
### EXPLANATION
##\mathcal{O}(H * W) bruteforce solution
Constraints on H and W are quite small. We can exploit these to get a
simple bruteforce solution. We iterate over all (x, y) pairs and check
whether their product x * y is equal to K or not. We can count such
valid pairs in \mathcal{O}(W * H) time.
Pseudo Code:
ans = 0
for x = 1 to H:
for y = 1 to W:
if x * y == K:
ans++
##\mathcal{O}(K log K) solution
Let us make a simple change in the last solution? What if we you stop iterating over y when the value
of x * y exceeds K. Will that improve time complexity?
ans = 0
for x = 1 to H:
for y = 1 to W:
if x * y > K:
break;
if x * y == K:
ans++
Yes, it will. Let us estimate it. From a casual look, it looks that
its time complexity will still be \mathcal{O}(H * W). But it’s not. Let us
delve into depth.
For each x, the inner loop over y will run at most \frac{K}{x} times. So,
total number of iterations the program will be run will be given by
\frac{K}{1} + \frac{K}{2} + \dots + \frac{K}{H}
\leq \frac{K}{1} + \frac{K}{2} + \dots + \frac{K}{K}
\leq K \cdot (\frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{K})
The expression
\frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{n}
is also known as harmonic sum H_n. One can prove that H_n = \mathcal{O}(log \, n).
Hence, the time complexity will be \mathcal{O}(K log K).
##\mathcal{O}(H) solution
Let us exploit the condition x * y = K. For a fixed x, do we really need to iterate
over all y's to check how many of them satisfy x * y = K. It turns out, no. Firstly there will be at most a single y. If K is not divisible by x, then there can’t exist such y. Otherwise y will be \frac{K}{x}.
In summary, we iterate over only
x values and find the corresponding y (if it exists), and
check whether the y is \geq 1 and \leq H.
Time complexity of this method will be \mathcal{O}(H), as are iterating over x values only once.
#### Factorization based solutions
If x \cdot y = K, then both x and y should divide K. So we can find all the divisors of the K. Let x be one such divisor, then y will be \frac{K}{x}.
You can find all the divisors of K in \mathcal{O}(sqrt(K)) time.
### EDITORIALIST’S SOLUTION
Can be found here.
### TESTER’S SOLUTION
Can be found here.
1 Like
*ans = 0
Start i from 1 to n
Check if c % i = 0 and c / i <= m
ans++*
1 Like
MY approach to it in c++:
#include <bits/stdc++.h>
#define rep(i,N) for(i=1;i<=N;i++)
#define rep1(i,N) for(i=N;i>=1;i--)
#define lli long long int
using namespace std;
int main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int T;
cin>>T;
while(T--){
lli n,m,c,count=0,i,j,start;
cin>>n>>m>>c;
if(n*m<c) { cout<<0<<"\n"; continue; }
start=m;
rep(i,n){
rep1(j,start){
if(i*j==c) {count++; start=j; break; }
}
}
cout<<count<<"\n";
}
}
import java.util.*;
class Chef_and_keyboard
{
public static void main(String args[])
{
Scanner sc=new Scanner(System.in);
int c[],x[],y[];
System.out.println();
x=new int[sc.nextInt()];
y=new int[x.length];
c=new int[y.length];
for(int i=0;i<x.length;i++)
{
x[i]=sc.nextInt();
y[i]=sc.nextInt();
c[i]=sc.nextInt();
}
for(int i=0;i<x.length;i++)
System.out.println(Combi(x[i],y[i],c[i]));
}
static int Combi(int n,int m,int c)
{
int count=0;
if(n*m<c)
return count;
for(int i=1;i<=c;i++)
{
if(i>n || i>(c/i))
return count;
if(c%i==0)
{
if(i<=m && (int)c/i<=n && c/i<=m)
count+=2;
else
if(c/i<=m)
count++;
}
}
return count;
}
}
Hi there. Just wanted to say that in the O(H) solution we have to check if y>=1 and less than width and not height as mentioned in the explanation.
Is O(H) better or O(Klog K) better … I have used the former.
can some one explain how does the O(sqrt(K)) solution work | 2020-09-25T20:42:26 | {
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https://yutsumura.com/are-the-trigonometric-functions-sin2x-and-cos2x-linearly-independent/ | # Are the Trigonometric Functions $\sin^2(x)$ and $\cos^2(x)$ Linearly Independent?
## Problem 603
Let $C[-2\pi, 2\pi]$ be the vector space of all continuous functions defined on the interval $[-2\pi, 2\pi]$.
Consider the functions $f(x)=\sin^2(x) \text{ and } g(x)=\cos^2(x)$ in $C[-2\pi, 2\pi]$.
Prove or disprove that the functions $f(x)$ and $g(x)$ are linearly independent.
(The Ohio State University, Linear Algebra Midterm)
## Proof.
To determine whether $f(x)$ and $g(x)$ are linearly independent or not, consider the linear combination
$c_1f(x)+c_2g(x)=0,$ equivalently
$c_1\sin^2(x)+c_2 \cos^2(x)=0, \tag{*}$ where $c_1, c_2$ are scalars.
If the only scalars satisfying the above equality are $c_1=0, c_2=0$, then $f(x)$ and $g(x)$ are linearly independent, otherwise they are linearly dependent.
Note that this is an equality as functions.
That is, this equality must hold for any $x$ in the interval $[-2\pi, 2\pi]$.
Let $x=0$. Then as $\sin(0)=0$ and $\cos(0)=1$, we obtain $c_2=0$ from (*).
Next, let $x=\pi/2$. Then as $\sin(\pi/2)=1$ and $\cos(\pi/2)=0$, we obtain $c_1=0$ from (*).
Therefore, we must have $c_1=c_2=0$, and hence the functions $f(x)=\sin^2(x)$ and $g(x)=\cos^2(x)$ are linearly independent.
## Comment.
This is one of the midterm 2 exam problems for Linear Algebra (Math 2568) in Autumn 2017.
Here is the most common mistake.
The linear combination $c_1\sin^2(x)+c_2 \cos^2(x)$ is a function defined over the interval $[-2\pi, 2\pi]$ and we are assuming it is the zero function.
So saying that “if $c_1=1, c_2=0$, then $c_1\sin^2(x)+c_2 \cos^2(x)$ is zero at $x=0$, hence $f(x)$ and $g(x)$ are linearly independent” is totally wrong.
What you are claiming here is that the function $\sin^2(x)$ is zero at $x=0$, hence it is the zero function.
This is clearly wrong as $\sin^2(x)$ is not the zero function.
## List of Midterm 2 Problems for Linear Algebra (Math 2568) in Autumn 2017
### 1 Response
1. 11/08/2017
[…] Are the Trigonometric Functions $sin^2(x)$ and $cos^2(x)$ Linearly Independent? […]
##### Find an Orthonormal Basis of the Given Two Dimensional Vector Space
Let $W$ be a subspace of $\R^4$ with a basis \[\left\{\, \begin{bmatrix} 1 \\ 0 \\ 1 \\ 1 \end{bmatrix},...
Close | 2018-04-20T16:46:37 | {
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http://math.stackexchange.com/questions/291465/formally-prove-that-theta-maxf-g-thetafg | Formally prove that $\Theta(\max(f,g)) = \Theta(f+g)$
I am having a hard time proving that $\Theta(\max(f,g)) = \Theta(f+g)$
where
$(f+g)(n) = f(n) + g(n)$
and
$(\max{f,g})(n) = \max(f(n), g(n))$
I know that $\Theta$ is the combination of the upper and lower bounds, but I can't seem to prove this. It's hard for me to see how $\Theta$ of the max of two functions can be equivalent to $\Theta$ of the two functions added together. Any guidance would be appreciated. Let me know if I can provide more info.
This question is similar, but didn't quite help.
-
If $f,g \geq 0$, then $\frac{1}{2}(f+g) \leq \max(f,g) \leq (f+g)$. Rearranging gives $\max(f,g) \leq (f+g) \leq 2 \max(f,g)$.
It follows that $\Theta(f+g) = \Theta(\max(f,g))$.
Addendum: When we write $a \in \Theta(b)$, it means that there exists $\underline{k}, \overline{k} >0$, and $N$ such that if $n\geq N$, then $\underline{k} b(n) \leq a(n) \leq \overline{k} b(n)$. Note that if $a \in \Theta(b)$, then $\frac{1}{\overline{k}} a(n) \leq b(n) \leq \frac{1}{\underline{k}} a(n)$, and so $b \in \Theta(a)$, hence it is an equivalence relation.
The above shows that if we choose $\underline{k} = \frac{1}{2}$, $\overline{k} = 1, N=1$, then $\underline{k}(f(n)+g(n)) \leq \max(f(n),g(n)) \leq \overline{k}(f(n)+g(n))$, and hence $n \mapsto \max(f(n),g(n)) \in \Theta(n \mapsto f(n)+g(n))$, or, more colloquially, $\max(f,g) \in \Theta(f+g)$, and by the above remark, we also have $f+g \in \Theta(\max(f,g))$.
-
Maybe my confusion is dealing with the = sign. Are we saying that the two sets are equivalent? This is where I seem to be having trouble. I understand what you are saying about max<f+g<2max, but I don't see how this proves that the two sets are equal. – MrZander Jan 31 '13 at 18:24
Well, I am sure that you can formalize it, but usually the notation is used very loosely. When one writes $f \in \Theta(g)$, it means for $n$ sufficiently large, there are strictly positive constants $\underline{k}, \overline{k}$ such that $\underline{k} g(n) \leq f(n) \leq \overline{k} g(n)$. When you write $\Theta(f) = \Theta(g)$, the interpretation is that $h \in \Theta(f)$ iff $h \in \Theta(g)$. So I suppose you could define $\Theta(g)$ as the set of $f$s satisfying the above inequalities and all would be fine... – copper.hat Jan 31 '13 at 18:31
Okay, I think I see where you are coming from. I am going to try and formally work this out. – MrZander Jan 31 '13 at 23:21
@Shammy: Thanks :-). Your comment produced a guffaw on this end... – copper.hat Oct 3 '15 at 18:30
@Shammy: I'm not exactly sure what you are asking. Remember that the whole 'O' thing represents a collection of functions, so when we write $h \in \Theta(f+g)$ is means there is some $a,b>0$ and $N>0$ such that if $n \ge N$, then $a(f(n)+g(n)) \le h(n) \le b (f(n)+g(n))$. The above just shows that $\Theta(f+g) = \Theta(\max(f,g))$. – copper.hat Oct 3 '15 at 18:42
Certainly, $\max(f,g) \leq f+g$ so $\max(f,g) = O(f+g)$, and it only remains to establish the lower bound.
If $f = \Theta(g)$, the statement is trivial, so assume $f = O(g)$ and then $2 \max(f,g) = 2g \geq f+g$, which implies $\max(f,g) = \Omega(f+g)$, as desired.
-
Just hope this clarifies a bit more:
You want to show that $h(x)=\max(f(x),g(x))=\Theta(f(x)+g(x))$. Observe two cases:
Case1:
$h(x)=f(x)$, hence $f(x) \geq g(x)$. Then $2 h(x) = 2 f(x) \geq f(x) + g(x)$,
$\Leftrightarrow h(x) \geq \frac{f(x)+g(x)}{2}$
$\Leftrightarrow h(x) = \Omega(f(x)+g(x))$ This gives the lower bound.
Next, $h(x)=f(x) \leq f(x)+g(x)=O(f(x)+g(x))$
Hence, $h(x)=\Theta(f(x)+g(x))$
Case 2:
$h(x)=g(x)$. Similar to Case 1.
-
I want to show $\Theta(\max(f(x),g(x)))=\Theta(f(x)+g(x))$ – MrZander Feb 5 '13 at 1:38 | 2016-02-09T06:18:15 | {
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https://mathhelpboards.com/threads/shell-method-about-the-line-x-5.5809/ | # Shell method about the line x=5
#### alane1994
##### Active member
I have been going over previous tests in an attempt to better prepare myself for the final that is coming tomorrow. I was posed a question.
Use the shell method to find the volume of the solid generated by revolving the plane region about the given line.
$$\displaystyle y = 4x - x^2$$
, y = 0, about the line x = 5
I know this.
SHELL METHOD
$$\displaystyle V = \int_{a}^{b} 2\pi x (f(x) - g(x))dx$$
I know this is fairly rudimentary, but assistance would be appreciated!
#### MarkFL
Staff member
The volume of an arbitrary shell is:
$$\displaystyle dV=2\pi rh\,dx$$
What are $r$ and $h$?
#### alane1994
##### Active member
Ok, is "h" the highest point of the graph?
And if so, r would be 5-x, where x is the corresponding coordinate for the highest point?
#### MarkFL
Staff member
Ok, is "h" the highest point of the graph?
And if so, r would be 5-x, where x is the corresponding coordinate for the highest point?
No, $h$ is the distance from the top curve to the bottom curve at the value of $x$ for the arbitrary shell. The arbitrary shell can be anywhere for $0\le x\le4$. I just drew one such shell.
So we have:
$$\displaystyle r=5-x$$
$$\displaystyle h=\left(4x-x^2 \right)-0=4x-x^2$$
and thus the volume of the shell is:
$$\displaystyle dV=2\pi(5-x)\left(4x-x^2 \right)\,dx$$
Next, you want to sum all the shells:
$$\displaystyle V=2\pi\int_0^4 (5-x)\left(4x-x^2 \right)\,dx$$
#### chisigma
##### Well-known member
I have been going over previous tests in an attempt to better prepare myself for the final that is coming tomorrow. I was posed a question.
Use the shell method to find the volume of the solid generated by revolving the plane region about the given line.
$$\displaystyle y = 4x - x^2$$
, y = 0, about the line x = 5
I know this.
SHELL METHOD
$$\displaystyle V = \int_{a}^{b} 2\pi x (f(x) - g(x))dx$$
I know this is fairly rudimentary, but assistance would be appreciated!
With the substitution $\xi= 5 - x$ You have to compute the volume of the rotation solid about $\xi=0$ of the function $\displaystyle f(\xi) = - 5 + 6 \xi - \xi^{2}$ obtaining...
$\displaystyle V = 2\ \pi\ \int_{1}^{5} \xi\ (- 5 + 6 \xi - \xi^{2})\ d \xi = 64\ \pi\ (1)$
Kind regards
$\chi$ $\sigma$
#### alane1994
##### Active member
THAT'S MY PROBLEM!!!!!!
$$\displaystyle V = \int_{0}^{4} 2\pi x ((x-5) - (4x-x^2))dx$$
That is where I got off. You need to change the x into 5-x.
$$\displaystyle V = \int_{0}^{4} 2\pi (5-x)(4x-x^2)dx$$
#### MarkFL
Staff member
THAT'S MY PROBLEM!!!!!!
$$\displaystyle V = \int_{0}^{4} 2\pi x ((x-5) - (4x-x^2))dx$$
That is where I got off. You need to change the x into 5-x.
$$\displaystyle V = \int_{0}^{4} 2\pi (5-x)(4x-x^2)dx$$
You can't get from the first formula to the second simply by replacing $x$ with $5-x$.
The formula you cited in your original post looks like it was meant for revolution about the $y$-axis. I find it easier to not try to use such a formula, but to just look at one element of the entire volume, whether it be a shell, disk or washer. Once you have the elemental volume, then you can add all the elements by integrating.
MHB Math Scholar
Another picture:
#### alane1994
##### Active member
Ok, I have arrived at an answer of $$\displaystyle 64\pi$$. | 2021-06-23T09:06:18 | {
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https://mathzsolution.com/do-most-numbers-have-exactly-33-prime-factors/ | # Do most numbers have exactly 33 prime factors?
In this question I plotted the number of numbers with $$nn$$ prime factors. It appears that the further out on the number line you go, the number of numbers with $$33$$ prime factors get ahead more and more.
The charts show the number of numbers with exactly $$nn$$ prime factors, counted with multiplicity:
(Please ignore the ‘Divisors’ in the chart legend, it should read ‘Factors’)
My question is: will the line for numbers with $$33$$ prime factors be overtaken by another line or do ‘most numbers have $$33$$ prime factors’? It it is indeed that case that most numbers have $$33$$ prime factors, what is the explanation for this?
Yes, the line for numbers with $$33$$ prime factors will be overtaken by another line. As shown & explained in Prime Factors: Plotting the Prime Factor Frequencies, even up to $$1010$$ million, the most frequent count is $$33$$, with the mean being close to it. However, it later says
For $$n = 10^9n = 10^9$$ the mean is close to $$33$$, and for $$n = 10^{24}n = 10^{24}$$ the mean is close to $$44$$.
The most common # of prime factors increases, but only very slowly, and with the mean having “no upper limit”.
OEIS A$$001221001221$$‘s closely related (i.e., where multiplicities are not counted) Number of distinct primes dividing n (also called omega(n)) says
The average order of $$a(n): \sum_{k=1}^n a(k) \sim \sum_{k=1}^n \log \log k.a(n): \sum_{k=1}^n a(k) \sim \sum_{k=1}^n \log \log k.$$Daniel Forgues, Aug 13-16 2015
Since this involves the log of a log, it helps explain why the average order increases only very slowly.
In addition, the Hardy–Ramanujan theorem says
… the normal order of the number $$\omega(n)\omega(n)$$ of distinct prime factors of a number $$nn$$ is $$\log(\log(n))\log(\log(n))$$.
… if $$ω(n)ω(n)$$ is the number of distinct prime factors of $$nn$$ (sequence A001221 in the OEIS, then, loosely speaking, the probability distribution of
$$\frac {\omega (n)-\log \log n}{\sqrt {\log \log n}}\frac {\omega (n)-\log \log n}{\sqrt {\log \log n}}$$
To see graphs related to this distribution, the first linked page of Prime Factors: Plotting the Prime Factor Frequencies has one which shows the values up to $$1010$$ million. | 2022-10-05T02:34:09 | {
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