text stringlengths 1 2.12k | source dict |
|---|---|
Proof. Take $M=\begin{pmatrix}i\pi&1\\0&-i\pi\end{pmatrix}$.
• That is very sleek! Do you need Prop1 for Prop2? Do you get Prop1 also from Prop2? – Sebastian Schlecht Oct 29 '15 at 11:59
• And do you have any suggestion what to name $V$? I understand that there might be no agreed name. Unfortunately, I cannot access right now the Ikramov paper to have a look. – Sebastian Schlecht Oct 29 '15 at 12:01
• I would like to see the details for your Proposition 2. I don't believe that it is as easy as you seem to think. I thought of exactly this line, right up to your 'that is equivalent to there is a real matrix A...". I didn't see how to prove that directly, and I would be curious to see what method you use to prove it. Also, the smoothness of $V$ follows, as I wrote, easily from the fact that it is the fixed point set of an anti-holomorphic involution. I don't see why it needs anything from Kummer, unless you are quoting the surjectivity of the Cartan embedding, a nontrivial fact. – Robert Bryant Oct 29 '15 at 14:11
• @ Sebastian SchlechtClearly , Prop. 1 and 2 can be independently proved. About a name for $V$, I don't know. In general, the authors consider elements of $V$ and not the whole set. cf also Horn and Johnson; Matrix Analysis, Section 4.6 – loup blanc Oct 29 '15 at 14:59
• @loupblanc: I suspected as much. So it's somewhat misleading to claim that your proof is 'more elementary'; it just leaves out many details and even essential ideas. In fact, while I mention Cartan's results, I don't use them (or any result of Kummer) in the surjectivity argument, which, in fact, uses nothing more than than the semi-simple/nilpotent decomposition (i.e., Jordan normal form), and even the result over $\mathbb{C}$ needs this. – Robert Bryant Oct 29 '15 at 15:52 | {
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I totally lied about this not being a natural thing to ask! As loup blanc alludes to, in fact $n \times n$ matrices such that $M^{-1} = \overline{M}$ can be interpreted as Galois descent data for descending the complex vector space $\mathbb{C}^n$ to a real vector space, or equivalently as a $1$-cocycle in
$$Z^1(B \text{Gal}(\mathbb{C}/\mathbb{R}), GL_n(\mathbb{C})).$$
The statement that any such matrix must have the form $U \overline{U}^{-1}$ says that any such $1$-cocycle is cohomologous to zero, or equivalently that the Galois cohomology set
$$H^1(B \text{Gal}(\mathbb{C}/\mathbb{R}), GL_n(\mathbb{C}))$$
is trivial. This just says that there is only one real form of $\mathbb{C}^n$ up to isomorphism, namely $\mathbb{R}^n$. The space $GL_n(\mathbb{C})/GL_n(\mathbb{R})$ appearing in Robert Bryant's answer can then be interpreted as the space of real forms of $\mathbb{C}^n$. | {
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# Divisibility Challenge
#### anemone
##### MHB POTW Director
Staff member
Given unequal integers $a, b, c$ prove that $(a-b)^5+(b-c)^5+(c-a)^5$ is divisible by $5(a-b)(b-c)(c-a)$.
##### Well-known member
Let us put $f(x) = (x-b)^5 + (b-c)^5 + (c-x)^5$
Putting x = b we get f(b) = 0
so (x-b) is a factor
so a-b is a factor of $(a-b)^5 + (b-c)^5 + (c-a)^5$
similarly we have (b-c) and (c-a) are factors
now $(a-b)^5 = a^5 – 5a b^4 + 10a^2b^3 – 10 a^3 b^2 +5 a^4 b – b^5 = a^5 – b^5 + 5m$ where $m = - a b^4 + 2a^2b^3 – 2 a^3 b^2 + a^4 b b^4$
similarly $(b-c)^5 = b^5 – c^5 + 5n$
$(c-a)^5 = c^5 – a^5 + 5k$
Adding we get $(a-b)^5 + (b-c)^5 + (c-a)^5 = 5 (m+n+k)$
So $(a-b)^5 + (b-c)^5 + (c-a)^5$ is divisible by $5(a-b)(b-c)(c-a)$
#### anemone
##### MHB POTW Director
Staff member
Let us put $f(x) = (x-b)^5 + (b-c)^5 + (c-x)^5$
Putting x = b we get f(b) = 0
so (x-b) is a factor
so a-b is a factor of $(a-b)^5 + (b-c)^5 + (c-a)^5$
similarly we have (b-c) and (c-a) are factors
now $(a-b)^5 = a^5 – 5a b^4 + 10a^2b^3 – 10 a^3 b^2 +5 a^4 b – b^5 = a^5 – b^5 + 5m$ where $m = - a b^4 + 2a^2b^3 – 2 a^3 b^2 + a^4 b b^4$
similarly $(b-c)^5 = b^5 – c^5 + 5n$
$(c-a)^5 = c^5 – a^5 + 5k$
Adding we get $(a-b)^5 + (b-c)^5 + (c-a)^5 = 5 (m+n+k)$
So $(a-b)^5 + (b-c)^5 + (c-a)^5$ is divisible by $5(a-b)(b-c)(c-a)$
I like your solution because of the way you introduced a polynomial function for the problem and well done!
#### caffeinemachine
##### Well-known member
MHB Math Scholar
Given unequal integers $a, b, c$ prove that $(a-b)^5+(b-c)^5+(c-a)^5$ is divisible by $5(a-b)(b-c)(c-a)$.
Let $S_5=(a-b)^5+(b-c)^5+(c-a)^5$, $S^3=(a-b)^3+(b-c)^3+(c-a)^3$ and $S_2=(a-b)^2+(b-c)^2+(c-a)^2$.
From this thread http://mathhelpboards.com/challenge...^5-5=-^3-b^3-c^3-3-*-^2-b^2-c^2-2-a-8276.html we know that
$$\frac{S_5}{5}=\frac{S_3}{3}\frac{S_2}{2}$$.
Note that since $(a-b)+(b-c)+(c-a)=0$, we have $S_3=3(a-b)(b-c)(c-a)$.
Thus $$S_5=5(a-b)(b-c)(c-a)\frac{S_2}{2}$$. | {
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Thus $$S_5=5(a-b)(b-c)(c-a)\frac{S_2}{2}$$.
Clearly $2$ divides $S_2$ and thus we are done.
#### anemone
##### MHB POTW Director
Staff member
Let $S_5=(a-b)^5+(b-c)^5+(c-a)^5$, $S^3=(a-b)^3+(b-c)^3+(c-a)^3$ and $S_2=(a-b)^2+(b-c)^2+(c-a)^2$.
From this thread http://mathhelpboards.com/challenge...^5-5=-^3-b^3-c^3-3-*-^2-b^2-c^2-2-a-8276.html we know that
$$\frac{S_5}{5}=\frac{S_3}{3}\frac{S_2}{2}$$.
Note that since $(a-b)+(b-c)+(c-a)=0$, we have $S_3=3(a-b)(b-c)(c-a)$.
Thus $$S_5=5(a-b)(b-c)(c-a)\frac{S_2}{2}$$.
Clearly $2$ divides $S_2$ and thus we are done.
Hey caffeinemachine,
Thanks for participating and that's another trick for me to learn today! | {
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# Set of functions from empty set to $\{0,1\}$
How does the set of all functions $\{f \,|\, f: \emptyset \to \{0,1\}\}$ look like? Is it empty or does it contain infinitely many functions? Does the definition $f: \emptyset \to \{0,1\}$ make sense at all?
I was wondering because we know that the two sets $\{0,1\}^X$ and $\mathcal{P}(X)$ have the same cardinality. But this is only true if $X$ is non-empty, right?
• What have you tried so far? Have you tried considering the definition of "function"? – skyking Dec 11 '17 at 10:03
Yes, it makes sense. There is one and only one function from $\emptyset$ into $\{0,1\}$, which is the empty function. Think about the definition of function as a set of ordered pairs to see why.
• @philmcole If $X=\emptyset$ then $\mathcal P(X)$ has only one element. $\emptyset \in\mathcal P(X)$ but $\{\emptyset\}\notin\mathcal P(\emptyset)$ because $\{\emptyset\}\not\subseteq\emptyset.$ – bof Dec 11 '17 at 10:20
• @bof Oh, you're totally right. So the theorem holds also if $X=\emptyset$ since we have $\{0,1\}^X = \{f \,|\, f: \emptyset \to \{0,1\}\} = \{\text{the empty function}\}\sim \{\emptyset\} = \mathcal{P}(X)$ – philmcole Dec 11 '17 at 10:36
It is known as the empty function.
For any set $A$, there is exactly one function from the empty set to $A$, namely the empty function:
$$f_A: \emptyset \to A.$$
The graph of an empty function is a subset of the Cartesian product $\emptyset × A$. Since the product is empty the only such subset is the empty set $\emptyset$. | {
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• I have a weird (maybe naif) argument: since '$x \in \emptyset$ implies that any proposition about $x$ is true', can we also say that $x \in f^{-1}(1)$? If we do this for all $x \in \emptyset$ can we conclude that $f$ is constant? And analogously, if we pick $y \in \emptyset$ such that $y \in f^{-1}(0)$ whereas for any other $x \in \emptyset, x \in f^{-1}(1)$, can we conclude $f$ is not constant? This seems to be a contradiction, does not it? – Gibbs Dec 11 '17 at 10:12
• related – Siong Thye Goh Dec 11 '17 at 10:19
• I read that $f$ is a constant function, but I see no mention of the contradiction in my previous comment... – Gibbs Dec 11 '17 at 10:25
A function $f:A\to B$ is a set of ordered pairs $(a,b)$ with $a\in A,b\in B$. In this case there is no $a\in A$ so therefore $f$ is an 'empty function' | {
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# Numbering of each equation within cases with left alignment (both on the page and within the equations)
I am trying to create a set of equations with curly braces (like cases), each with their own set of numbers derived from the subsection (such as can be done with equations using \numberwithin{equation}{subsection}). I get a good result where I want one equation number representing all cases of the equation (code below).
$$\mathbb(E)(X) = \int_{-\infty}^{\infty} xdF(x) = \left\{ \begin{array}{lll} \sum_{x \in \mathcal{X}} xf_{X}(x) & =\sum_{x \in \mathcal{X}}x \mathbb{P}(X=x) &\text{ if } X \text{ is discrete}\\ \\ \int_{-\infty}^{\infty} xf_{X}(x)dx & &\text{ if } X \text{ is continuous} \end{array} \right.$$
What can I do to to have each of the cases numbered individually?
I have tried align and numcases, but each of them has some problem - either the equation or the text aligns right, or the spacing goes awry.
Would appreciate some help for the specific example above, with
1. Equation aligned to left of page
2. All text within cells left aligned
3. Equation numbers on extreme right
4. An empty cell in the 2nd case and
5. A blank line in between the two cases.
• this should be helpful: Separate labels in cases. rather than a full blank line, an optional dimension value for the gap, e.g. [.6\baselineskip] would seem preferable. – barbara beeton Aug 16 '16 at 19:37
• Off-topic: It should be \mathbb{E}(X), not \mathbb(E)(X). – Mico Aug 16 '16 at 19:48
• Thanks Mico :). I actually use macros for those and made a typo while changin it to regular (non-macro) code for the forum. – Py_Dream Aug 17 '16 at 2:13
Another solution (and numbering) with the empheq package (which loads mathtools, which loads amsmath):
\documentclass{article}
\usepackage[a4paper,margin=2.5cm]{geometry} % set page parameters
\usepackage{amsfonts, % for \mathbb and \mathcal macros
empheq}%
\numberwithin{equation}{subsection}
\setcounter{section}{1}
\setcounter{subsection}{1} | {
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\numberwithin{equation}{subsection}
\setcounter{section}{1}
\setcounter{subsection}{1}
\begin{document}
\begin{subequations}
\begin{empheq}[left={\mathbb{E}(X)=\displaystyle\int_{-\infty}^{\infty} x\,dF(x)=\empheqlbrace}]{alignat = 2}
& \sum_{x \in \mathcal{X}} xf_{X}(x) =\sum_{x \in \mathcal{X}} x\mathbb{P}(X=x)
&\qquad & \text{if $X$ is discrete}, \\
& \int_{-\infty}^{\infty} xf_{X}(x)\,dx
& &\text{if $X$ is continuous}. \end{empheq}
\end{subequations}
\begin{subequations}
\begin{empheq}[left={\mathbb{E}(X)=\displaystyle∫_{-∞}^{∞} x\,dF(x)=\empheqlbrace}]{flalign}
& ∑_{x ∈ \mathcal{X}} xf_{X}(x) =∑_{x ∈ \mathcal{X}} x\mathbb{P}(X=x)
& &\text{if $X$ is discrete},&\hspace{5em} & \\
& ∫_{-∞}^{∞} xf_{X}(x)\,dx
&& \text{if $X$ is continuous}. \end{empheq}
\end{subequations}
\end{document}
• Sorry. It's once more a problem with my editor. i'LL FIX IT. – Bernard Aug 16 '16 at 21:04
• Hi, many thanks for this. Can this be done without the global [fleqn] option? I don't want all equations left aligned. – Py_Dream Aug 17 '16 at 2:02
• Yes, absolutely. Actually, what formatting was not very clear, as you can see from some comments. So, to sum it up, you want equation numbers on the right, and centred alignments? – Bernard Aug 17 '16 at 8:01
• Hi Bernard, as mentioned in the original question, I am looking for this particular set of equations to be left aligned and the numbers to be right aligned. However, I do not want equations to be left-aligned globally, i.e., I do not want the global [fleqn] option. – Py_Dream Aug 18 '16 at 11:34
• @Py_Dream: You must use the flalign environment then. Please see my updated answer. – Bernard Aug 18 '16 at 11:56
You didn't specify the type of problem(s) you encounter with the numcases environment. At any rate, I don't seem to encounter any in the following example. (The fleqn option is set so that the entire equation is set flush-left instead of centered. If that's not needed, just drop that option.) | {
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\documentclass[fleqn]{article}
\usepackage[a4paper,margin=2.5cm]{geometry} % set page parameters
\usepackage{amsfonts} % for \mathbb and \mathcal macros
\usepackage{amsmath} % for \text and \numberwithin macros
\usepackage{cases} % for numcases environment
%% And, just for this example:
\numberwithin{equation}{subsection}
\setcounter{section}{1}
\setcounter{subsection}{1}
\begin{document}
\begin{numcases}{\mathbb{E}(X) = \int_{-\infty}^{\infty} x\,dF(x)=}
\sum_{x \in \mathcal{X}} xf_{X}(x) =
\sum_{x \in \mathcal{X}} x\mathbb{P}(X=x)
& \text{if $X$ is discrete} \\[1\baselineskip]
\int_{-\infty}^{\infty} xf_{X}(x)\,dx
& \text{if $X$ is continuous}
\end{numcases}
\end{document} | {
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• Many thanks. I do not want to use the global [fleqn] option. So I was loading cases using \usepackage[fleqn]{cases} as was suggested somewhere. When I do this with your code (as on my earlier attempts), I get multiple errors for both the begin{numcases} line and the end{numcases} line. E.g. Undefined control sequence. ...}(X) = \int_{-\infty}^{\infty} x\,dF(x)=} Missing number, treated as zero. ...}(X) = \int_{-\infty}^{\infty} x\,dF(x)=} Illegal unit of measure (pt inserted). ...}(X) = \int_{-\infty}^{\infty} x\,dF(x)=} Is there any way to do this without the global [fleqn] option? – Py_Dream Aug 17 '16 at 2:05
• @Py_Dream - Under no circumstance should you attempt to pass the option fleqn to the cases package -- what you say "was suggested somewhere" is faulty. If you don't want to set the fleqn option at the \documentclass stage, you should set it as an option when loading the amsmath package: \usepackage[fleqn]{amsmath}. – Mico Aug 17 '16 at 5:33
• @Py_Dream - If you don't want all displayed equations to be left-aligned, certainly don't use the option fleqn option in the first place. If the non-centering should apply to just one displayed equation, you'll have to fine-tune its position "by hand", e.g., by using one or more \qquad instructions at the very end of the equation's right-hand most portion. – Mico Aug 17 '16 at 5:44
• Hi Mico, do you happen to have any further information (or links) on why fleqn should never be passed to the cases package and why the suggestion was faulty? It is not very helpful having cardinal truths stated without any reasons ascribed. – Py_Dream Aug 18 '16 at 11:37
Equation aligned to left of page: \documentclass[12pt,leqno]{book}
• I think you meant to specify fleqn, not leqno. – Mico Aug 16 '16 at 20:00
• @Mico: I had misunderstood, I guess. I'll change my code. Thanks! – Bernard Aug 16 '16 at 20:52 | {
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# Why Gauss-Legendre Quadrature should keep the number of integral points less than about 50?
I wanted to use Gauss-Legendre Quadrature to calculate an integral as follows：
When n=10 and some other number(except odd numbers),the numerical result is the same as theoretical result.
Clear["Global*"];
n = 10;
L[x_] := D[(x^2 - 1)^n, {x, n}]/(n!*2^n);
A[x_] := 2/((1 - x^2) (D[L[x], x])^2);
f[x_] := E^(-x)/x;
sol = NSolve[L[x] == 0., x];
nodes = (x /. sol);
coef = Table[A[x] /. {x -> nodes[[i]]}, {i, 1, Length@nodes}];
Sum[coef[[i]] f[nodes[[i]]], {i, 1, Length@nodes}]
Integrate[E^(-x)/x, {x, -1, 1.}, PrincipalValue -> True]
-2.11450175075
-2.11450175075
But,when n>=50,the code will give wrong numerical result.
Clear["Global*"];
n = 50;
L[x_] := D[(x^2 - 1)^n, {x, n}]/(n!*2^n);
A[x_] := 2/((1 - x^2) (D[L[x], x])^2);
f[x_] := E^(-x)/x;
sol = NSolve[L[x] == 0., x];
nodes = (x /. sol);
coef = Table[A[x] /. {x -> nodes[[i]]}, {i, 1, Length@nodes}];
Sum[coef[[i]] f[nodes[[i]]], {i, 1, Length@nodes}]
Integrate[E^(-x)/x, {x, -1, 1.}, PrincipalValue -> True]
-43.8873164858
-2.11450175075
That is very puzzling.
NSolve has severe issues to solve for the roots (which is not surprising since finding the roots of a polynomial of degree 50 is a nontrivial task):
Max[Abs[L[x] /. sol]]
164.177
That seems to be a precision problem (the polynomial hase huge coefficients but the powers of x tend to be very small so that this easily lead to catastrophic cancellation. You will get better results with higher WorkingPrecision:
sol = NSolve[L[x] == 0, x, WorkingPrecision -> 100];
Max[Abs[L[x] /. sol]]
0.*10^-92
So the real art here is to determine the roots in a numerically stable way. Actually, Gauß-Legendre quadrature rule is already built into the system. For example, see | {
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n = 50;
prec = 200;
{nodes, coef, bla} = NIntegrateGaussRuleData[n, prec];
(* compute all quadrature points by convex combinations*)
nodes = (-1) * (1 - nodes) + nodes * 1;
coef = 2 coef;
a = coef.f[nodes];
b = Integrate[f[x], {x, -1, 1}, PrincipalValue -> True];
a - b
2.6225212*10^-190
PS.: I was quite astonished that Gauß quadrature works so well with this singular integral. Then it came back to my mind that it is constructed such that it neglects contributions of all odd functions (because they would integrate to 0 anyways)...
• thank you very much! I thought it was easy to solve for the roots of polynomial of large degree before. – Quere Sep 14 '18 at 12:25
• @Quere In the meantime, I found out that using higher WorkingPrecision helps to make your method more precise. – Henrik Schumacher Sep 14 '18 at 12:33
• Yes!I noticed it.The result was really amazing. – Quere Sep 14 '18 at 12:37
• Not really worth writing as a separate answer, so: the trick of using even-order Gauss-Legendre quadrature to evaluate principal value integrals with a singularity at the center was published by Piessens, but I seem to recall this being known as folklore. In case the pole is not exactly in the middle of the integration interval, one can either split the integral so that there is a regular part and a part symmetric about the pole, or one can use a substitution that centers the pole, allowing the use of Gaussian quadrature. – J. M.'s discontentment Sep 24 '18 at 19:45
If you add WorkingPrecision -> 60 in the NSolve function you get your solution.
n = 50;
A[x_] := 2/((1 - x^2) (D[LegendreP[n, x], x])^2);
f[x_] := E^(-x)/x;
sol = NSolve[LegendreP[n, x] == 0, x, WorkingPrecision -> 60];
nodes = (x /. sol);
coef = Table[A[x] /. {x -> nodes[[i]]}, {i, 1, Length@nodes}];
Sum[coef[[i]] f[nodes[[i]]], {i, 1, Length@nodes}]
Integrate[E^(-x)/x, {x, -1, 1.}, PrincipalValue -> True]
-2.11450175075145702914368470979175591804810787513962
-2.1145 | {
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-2.11450175075145702914368470979175591804810787513962
-2.1145
Or you can use the built in functions i mma
<< NumericalDifferentialEquationAnalysis
n = 50;
{pts, w} = Transpose[GaussianQuadratureWeights[n, -1, 1]];
Integrate[E^(-x)/x, {x, -1, 1.}, PrincipalValue -> True]
Sum[w[[i]] f[pts[[i]]], {i, 1, Length@pts}]
-2.1145
-2.1145
A more stable way of computing the roots of a polynomial family that satisfies a three-term recurrence is the method of Golub and Welsch (1969), which computes the eigenvalues of a matrix based on the recurrence, sometimes called the "comrade matrix."
(*
* n-point Gauss quadrature (Golub-Welsch 1969)
*)
(* p[n][x] == (a[n]x+b[n])p[n-1][x] - c[n]p[n-2][x] *)
comradeMatrix[{a_, b_, c_}, {n_, n0_Integer}, prec_: Infinity] :=
Block[{n = Range@n0},
With[{beta = Sqrt[Rest@c/(Most@a*Rest@a)]},
N[SparseArray[{Band[{1, 1}] -> -b/a, Band[{2, 1}] -> beta,
Band[{1, 2}] -> beta}, {n0, n0}], prec]
]];
Module[{n}, comradeMatrix[{2 n - 1, 0, n - 1}/n, {n, n0}, prec]];
(*
{-0.998866, -0.994032, -0.985354, -0.972864, -0.956611, -0.936657, \
-0.913079, -0.885968, -0.85543, -0.821582, -0.784556, -0.744494, \
-0.701552, -0.655896, -0.607703, -0.557158, -0.504458, -0.449806, \
-0.393414, -0.3355, -0.276288, -0.216007, -0.154891, -0.0931747, \
-0.0310983, 0.0310983, 0.0931747, 0.154891, 0.216007, 0.276288, \
0.3355, 0.393414, 0.449806, 0.504458, 0.557158, 0.607703, 0.655896, \
0.701552, 0.744494, 0.784556, 0.821582, 0.85543, 0.885968, 0.913079, \
0.936657, 0.956611, 0.972864, 0.985354, 0.994032, 0.998866}
*)
The integration weights can be computed from these nodes as in the OP. | {
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The integration weights can be computed from these nodes as in the OP.
• As a terminological note: the tridiagonal matrix with the orthogonal polynomial's recurrence coefficients is called a Jacobi matrix. (The comrade matrix, as you might recall, is a Jacobi matrix for the Chebyshev polynomials, with a correction term representing the Chebyshev series coefficients.) I talked about this here. – J. M.'s discontentment Sep 24 '18 at 11:27
• @J.M. Barnett (1975) seems to introduce the name "comrade matrix" of a polynomial with respect to an arbitrary family of polynomials that satisfies a three-term recurrence and presents it as a generalization of the Chebyshev colleague matrix of Good (1961). But, you know, I'm not that confident about current terminology in this area. -- Yes, I found the link to that Q&A shortly after posting an answer and put it under the question above. – Michael E2 Sep 25 '18 at 17:34
• Argh, I switched up "comrade" and "colleague" again... I keep switching the general and specific cases. :D – J. M.'s discontentment Sep 25 '18 at 17:40 | {
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# Finding resonant frequency or damping ratio from Bode Plot
I am working on a question where I have to estimate a transfer function from its bode plot.
I plotted the asymptotes of this bode diagram, and was able to find out that this is 3rd order system with a pole at s = 0 and two complex poles.
$$\dfrac{w_n^2}{s(s^2 + 2\zeta w_n s + w_n^2)}$$ where I found out that $$w_n = 6.7 rad/s$$ from the asymptotic bode plot.
I was also able to find the system gain $$\K\$$ from the magnitude plot.
I am unable to find $$\\zeta\$$ here. I tried finding the resonant frequency $$\w_r\$$ from the bode plot so that I can calculate $$\\zeta\$$ using: $$w_r = w_n\sqrt{1 - 2\zeta^2}$$ but I was unable to.
I know that the frequency at which the phase plot crosses zero is the resonant frequency but the phase plot here doesn't cross zero.
I tried approximating $$\\zeta\$$ using the fact that maximally flat response is obtained for $$\\zeta = 0.707 \$$, so that for the given plot, $$\\zeta < 0.707 \$$. But I wasn't able to exactly find a value.
Is there any other way to find $$\\zeta\$$ or will I be just able to approximate it?
The solution says that the value of $$\\zeta\$$ is $$\0.447\$$.
• Curiously, it changed May 18, 2021 at 13:01
• @TonyStewartEE75 Sorry I added the wrong bode plot. May 18, 2021 at 13:02
• Compute d phi /dt max in the same units(rads) May 18, 2021 at 13:06
• Or rather $d\phi /d\omega$ then determine the ratio constant May 18, 2021 at 13:18
• If the plot is not on a paper (i.e. you have access to the datapoints), subtract the 1/s from it and you'll be left with a more manageable 2nd order magnitude/phase. May 18, 2021 at 15:09 | {
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The following graphical solution relies on the idea that it's possible to separate the contribute of the zero in w=0 by referring the magnitude measurements to a line slanted with a -20dB/decade slope. Thanks to the properties of logarithms, division becomes translation on the magnitude Bode plot. We also have to take into account the -90 degrees contribution in the phase - it's basically a constant -90° addition since, being the pole in the origin, it has already 'run its course'.
1. the first step is to find wn. In a second order system with no zeros, the phase resonance happens exactly at wn, the undamped natural frequency (a frequency that is in general different from wpeak, the peak frequency of the magnitude, and also from the damped natural frequency wd). Since we need to separate the phase contribution of the pole in the origin, instead of finding the frequency where the phase is -90° we need to find the frequency where the phase is -180° By eyeballing the scale on the tiny plot I have I believe I can locate it at 6.7 rad/s. You might end up with a better estimate.
Now I want to find the 3dB corner frequency the system would have without the pole in the origin . I therefore...
1. ...trace a slanted line, translated 3dB under the asymptotic behavior at low frequencies and
2. ...and then I look for the intersection of said line with the magnitude of the transfer function to find w3dB. Eyeballing again I find w3dB = 8.7 rad/sec
3. We are now in the position to computed the ratio w3dB/wn = 1.298 = 1.3
We can now either solve the expression for w3dB as a function of zeta
or, if we have a graph like this, 5) use it to find the value of zeta corresponding to w3dB/wn = 1.3.
again by eyeballing I get a zeta value of around 0.48, a value not dissimilar from that found by solving the equation
for zeta, which gives zeta = 0.477 | {
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for zeta, which gives zeta = 0.477
And this value is reasonably close, considering the amount of lazy eyeballing employed, to the correct answer of 0.447. Try your estimates on a bigger graph, counting the pixels and report back. Did it work?
Caveat emptor: it is imperative that the second order function be without additional zeroes (apart for the one we have been able to separate). The relevant frequencies have different expressions form system with one or more zeroes.
• Thank you, yeah it did work. I got the w3dB to be around 8.9 rad/s from the bode plot, which gave me a ratio of 1.328. Got zeta = 0.4482. Thanks a lot! Jun 13, 2021 at 5:39 | {
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# plot graph of function with different constants
I need to plot the graph of :
$$v(t) = \frac{-mg}{b} (e^{-\frac{b}{m}t}+1)$$ , where $$g = 9.8 m/s^2$$ and $$b$$ and $$m$$ are positive constants.
Is there a way to plot this without having to define random values for $$b$$ and $$m$$? I mean a way in which I can see how the graph behaves for different values of $$b$$ and $$m$$ ?
• Unfortunately no, I am new to Mathematica, I only know the basics – Physmath Apr 22 at 2:17
• I will take a look, thank you ! – Physmath Apr 22 at 2:22
• To simplify little bit, you can introduce a new variable, say $\lambda=\frac{b}{m}$. Then you can Plot a sequence of curves for a list of $\lambda$ values. – yarchik Apr 22 at 3:03
v[s_, t_] := -g/s (Exp[-s t] + 1)
g = 9.8;
slist = {0.1, 0.2, 0.4, 0.8};
Plot[Evaluate[v[#, t] & /@ slist], {t, 0, 10},
PlotTheme -> {"BoldColors", "Frame", "Grid"},
FrameLabel -> {Automatic, v[t]},
FrameStyle -> Directive[14, Black],
PlotLabels -> slist
]
• Plot[Evaluate[v[slist, t] , {t, 0, 10},...] works too. – Ulrich Neumann Apr 22 at 8:02
• @UlrichNeumann That is good suggestion, it makes the syntax more transparent. – yarchik Apr 22 at 8:12 | {
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# Showing that if $3x$ is even then $3x+5$ is odd
I'm learning the absolute basics of how to do proofs, and am really struggling.
If 3x is even then 3x+5 is odd.
This is the solution:
I get that even numbers are 2n and odd numbers are 2n+1. For the life of me, I CANNOT get it into that form shown below. I feel so dumb. I tried looking up other answers before posting, but nothing I found is this basic.
Work: -Assumptions- 3x = 2n 3x+5 = 2k+1
-Trying to make sense of 3x- 3x+5 = 2k+1 3x = 2k-4
-Plugging in 2k-4 for 3x- 2k-4 = 2n 2k = 2n+4 k = n+2
-Plugging in n+2 for k- 3x+5 = 2(n+2)+1
...This is where I gave up. I don't know where I'm going with this anymore.
• You should calm down, everyone can have difficulties. Explain us where you get lost. If you don't know where you get lost I can try to explain you the proof step by step. – Eureka Dec 5 '20 at 17:42
• I'm good, just frustrated. I'd like step by step. I'll edit in my work above. Warning, there is no rhyme or reason to it. – visualbread Dec 5 '20 at 17:44
• If $y$ is even, can you prove that $y+5$ is odd? Try this first, then try your problem. There’s no substantial difference. – Benjamin Wang Dec 5 '20 at 17:46
• @visualbread your resolution may be a bit convoluted but it's correct. You just proved that $3x+5=2(n+2)+1$. And $2(n+2)+1$ is odd ,because it is of the form $2h+1$ with $h=n+2$ – Eureka Dec 5 '20 at 17:52
• @BenjaminWang I am unable to do that one either. From y = 2n and y+5 = 2k+1, I can get that all the way to n = k -2 (or k = n+2), but I can't see what that tells me. The final solution has everything nicely equal to one another, and I can't seem to set it up that way. – visualbread Dec 5 '20 at 17:54
I think you're getting confused by the $$3x$$ part. The $$3x$$ plays no role in the problem.
Suppose you're given any number that is even. I'll call it y. Now we want to show $$y+5$$ is odd.
Therefore by definition
$$y=2k$$ for some integer $$k$$
Now
$$y+5 = 2k+5$$ | {
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Therefore by definition
$$y=2k$$ for some integer $$k$$
Now
$$y+5 = 2k+5$$
Now we just need to show that $$2k+5$$ is 2 times an integer plus 1
$$2k+5 = 2(k+2)+1$$
So $$2k+5$$ is odd because it can be written in the form 2*integer +1 where the integer here is $$k+2$$. So $$y+5$$ is odd since $$y+5 = 2k+5$$
So if any number is even. Then that number plus 5 is odd. It doesn't matter if the original number is 3x or 8z or 3x^2-5x+x^3 etc... | {
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• I might be there, but I need just a little bit more help. How did you write 2k+5 in the form 2(k+2) + 1? I can see that those are the same, but how could I do it without trial and error? – visualbread Dec 5 '20 at 18:16
• So the first step is to look at 5. We see it is an odd integer. So I decide it is more convenient to write it as $4+1$. So I write $2k+5$ as $2k+4+1$. Now I can factor $2k+4 = 2(k+2)$. So putting it all together $2k+5 = 2k+4+1 = 2(k+2)+1$. Hope this makes sense. let me know. Suppose I had $2k+6$ instead. since 6 is even, I know I can factor immediately $2k+6 = 2(k+3)$. – Ameet Sharma Dec 5 '20 at 18:19
• Thanks! This shows me all the steps I need to follow my book's logic. I don't know what happened in my math education, but I never did learn some of these little things that make some of these proofs actually understandable. At least there's websites like this one to fill in the gaps. – visualbread Dec 5 '20 at 18:25
• No problem. I highly recommend doing lots of exercises with factoring, simplification and algebra so that these things become second nature. It takes a lot of practice but it is worth it. As you get into more advanced math, these things really need to be 2nd nature where you don't need to think about them anymore... otherwise you'll get bogged down at every step. – Ameet Sharma Dec 5 '20 at 18:31
• I'm sure there are lots of online websites. Here's one I just found as part of an online algebra textbook: openstax.org/books/college-algebra/pages/1-review-exercises. Each chapter has review exercises and a practice test. I recommend doing all these, maybe multiple times till you can do them in your sleep. In my opinion... if you want online resources, look for online algebra "textbooks"... and do all the exercises in each chapter. If you have doubts you can review the chapter. – Ameet Sharma Dec 5 '20 at 18:51
Your problem is that $$3x + 5 = 2k +1$$ is not an assumption. It is the conclusion you need to prove. | {
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Your one and only assumption is that $$3x = 2n$$ for some integer $$n$$.
$$3x = 2n$$.
.... then you do a bunch of steps ....
.... steps .....
.... and get in the end ........
Conclusion: $$3x + 5 = 2(????????) + 1$$ where $$??????$$ is some integer you come up with in you steps.
Let's see what happens when we try. Let's take it nice and slow:
=======
$$3x = 2n$$.
$$3x +5 = 2n + 5$$
....hmmm, we want $$2(??????) + \color{red}1$$ in the end so let's pull out the $$+\color{red}1$$ first.....
$$3x + 5 = 2n+5 = 2n + (4 + \color{red}1)=(2n+4) +\color{red}1$$
.... hmmm, okay that's the $$+1$$ now we want $$2(\color{red}{??????}) + 1$$. To get the So we need to factor then $$2$$ out of $$2n+4$$ and see what we have left.... that will bee the $$\color{red}{??????}$$
$$3x + 5 = (\color{red}{2n+4}) + 1$$
$$3x + 5= 2(\color{red}{n + 2}) + 1$$
.... and that's it......
Conclusion: $$3x+5 = 2(\color{red}{n + 2})+1$$.
The $$??????$$ we wanted turns out to be $$\color{red}{n+2}$$ an we have
$$3x + 5 = (3x+4) + 1 = (2n+4) + 1 = 2(\color{red}{n+2}) + 1$$.
And because $$\color{red}{n+2}$$ is an integer if we let $$k = n+2$$ be that integer $$3x+5 = 2k + 1$$ and so... $$3x + 5$$ is odd.
=======
Although if you want to work backwards
Conclusion: $$3x + 5 = 2k +1$$ ..... and we want to solve for $$k$$ to show it is possible...
$$3x + 5 -1 =2k + 1-1$$
$$3x +4 = 2k$$
$$k = \frac {3x + 4}2 = \frac {3x}2 + 2$$.
.... but is $$\frac {3x}2 + 2$$ an integer?????
Well, $$3x$$ is even. So there is an integer $$n$$ so that $$3x = 2n$$ so
$$k = \frac {3x}2 +2 = \frac {2n}2 + 2 = n+2$$.
So $$k=n+2$$ is the integer we want to conclude $$3x+5 =2k +1$$.
If we did it this way our proof would go:
$$3x$$ is even so there is an integer $$n$$ so that $$3x = 2n$$. Let $$k = n+2$$; that is an integer.
$$2k + 1 = 2(n+2)+1 = 2n + 5 = 3x + 5$$.
So $$3x+5 = 2k +1$$ and that is odd. | {
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$$2k + 1 = 2(n+2)+1 = 2n + 5 = 3x + 5$$.
So $$3x+5 = 2k +1$$ and that is odd.
Ok so we know that $$3x$$ is even, that means we can write $$3x=2n$$ for a suitable $$n$$, since even means that the number is divisible by two without remainder. But then we have $$3x+5=2n+5=2n+(4+1)=2n+2\cdot 2+1=2(n+2)+1$$ which clearly is odd.
• Thanks for the reply. I'm still not there yet. I can't figure out where 2n+5 came from or why it can be set equal to 2x+5. I need this broken down even further if that's even possible. – visualbread Dec 5 '20 at 17:55
• We set it equal to $3x+5$ NOT $2x+5$. I will try to break it down further: so – Mo145 Dec 5 '20 at 18:04
• 1)a natural number $N$ is called even if it is a multiple of $2$. So per definition we can write every even number $N$ as $N=2M$. Note that we can write any natural number as $N=2\frac{N}{2}$ but in general $\frac{N}{2}$ will not be a natural number, for example take $N=3$, then clearly $\frac{3}{2}$ is not a natural number. In fact this will only be the case if $N$ is even. – Mo145 Dec 5 '20 at 18:08
• 2)Since we assume $3x$ to be even we can, by 1) write it as $3x=2n$ where $n=\frac{3x}{2}$ so we have the following equality $3x=2n$ adding a number on both sides preserves the equality hence we have $3x+5=2n+5$ – Mo145 Dec 5 '20 at 18:11
• I hope this clarifies your concerns :) – Mo145 Dec 5 '20 at 18:11
We know that $$3x=2n$$. So: $$\color{blue}{3x}+5=\color{blue}{2n}+5$$ Because the "blue quantities" are equal. Now: $$3x+5=2n+5=2n+4+1$$ In this step I just wrote $$5$$ as $$4+1$$:
$$3x+5=2n+5=2n+4+1=2n+2\cdot 2+1$$ In this step I just wrote $$4$$ as $$2 \cdot 2$$.
$$3x+5=2n+5=2n+4+1=\color{green}{2n+2\cdot 2}+1=\color{green}{2(n+2)}+1$$ The last step is valid because the green quantities are equal. In the end: $$3x+5=2(n+2)+1$$ This means that $$3x+5$$ is odd because is of the form $$2h+1$$ with $$h$$ integer(in particular $$h=(n+2)$$) | {
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Remember here that $$n$$ represents ANY natural number. You got to the answer but you didn’t even realize it. That’s probably because you are thinking syntactically rather than semantically. What I mean is the literal string of symbols $$2(n+2)$$ didn’t register to you as even because it is not the same as the string $$2n$$. But $$n+2$$ is a natural number just like $$n$$ is. So the the strings $$2(n+2)$$ and $$2n$$ both represent even numbers, and so $$2(n+2) + 1$$ is odd just as you have shown in your last line.
You are dealing with a problem where you are given too much information. Here is another way of doing it (we'll use $$m$$ for an integer).
If $$3x$$ is even then $$x$$ must be even, so we can put $$x=2m$$. [This is a consequence of the fact that $$2$$ is a prime, or can be proved in various ways]
Then $$3x+5=6m+5=6m+4+1=2(3m+2)+1$$.
Now we can put $$3m=n$$, an integer, to get $$2(n+2)+1$$.
For an alternative proof, we could put $$3m+2=n$$ and then we get $$2n+1$$.
Note that the last two sentences are alternatives to one another. They both use $$n$$, but $$n$$ is defined differently in the two cases.
What I'm trying to do here is to unpack how the different expressions for the same thing relate to each other. If you get your head round that you will be flying.
• I find this answer to be pretty overwhelming for the OP, since you are talking about "alternative proofs" and "consequences of the fact that 2 is prime that can be proved in various ways". I'm not underestimating the OP, but I think that when you learn fron basic you should do things calmly step by step. – Eureka Dec 5 '20 at 18:23
• @Eureka What I was trying to do was to show what happens when you keep the factor $3$ for an extra step. If it is "overwhelming" it might help OP to understand why the factor $3$ is best dropped ("too much information"). Either it helps or it doesn't. – Mark Bennet Dec 5 '20 at 18:35 | {
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# Evaluate $\int_0^\frac{\pi}{4} \cos^{-1}({\sin x}) \,dx$
I came across this integral in a math competition and couldn't solve it $$\int_0^\frac{\pi}{4} \cos^{-1}({\sin x})\, dx$$
I tried a $u$-substitution, with $u=\sin x$ and ended up with the integral $$\int_0^\frac{\pi}{4} \cos^{-1}(\text{u}) \cdot \frac{1}{\sqrt{1-u^2}}\,du$$ which is not much simpler and I cannot figure out how to solve this. Any hints/solutions for this problems?
I also tried drawing a triangle for the problem but it didn't really help with the solution.
You can use the trigonometric formula : $$\forall x \in\mathbb{R},\cos^{-1}(x)+\sin^{-1}(x)=\frac{\pi}{2}$$
Thus $\forall x\in\left[0,\frac{\pi}{4}\right]$ you have $\sin^{-1}(\sin(x))=x$ so : $$\cos^{-1}(\sin(x))+\sin^{-1}(\sin(x))=\frac{\pi}{2}\Rightarrow \cos^{-1}({\sin x})=\frac{\pi}{2}-x$$
Finally : $$\int_0^\frac{\pi}{4} \cos^{-1}({\sin x}) dx=\int_0^\frac{\pi}{4} \frac{\pi}{2}-xdx$$ Can you finish ?
• What is $\sin^{-1}(\sin x)=?$ – lab bhattacharjee Apr 11 '17 at 15:48
• it is $x$ since $x$ is in $[0,\pi/4]$ – Jennifer Apr 11 '17 at 15:54
HINT:
If $\cos^{-1}(\sin x)=y,0\le y\le\pi\ \ \ \ (1)$
and $\cos y=\sin x=\cos\left(\dfrac\pi2-x\right)$
$y=2m\pi\pm\left(\dfrac\pi2-x\right)$ where $m$ is an integer such that $(1)$ is satisfied
For $0\le2m\pi+\dfrac\pi2-x\le\pi\iff 2m\pi-\dfrac\pi2\le x\le2m\pi+\dfrac\pi2$
Here $m=0$
Here's a barely different route to take, continuing from the direction you've taken.
First, note that substituting $u=\sin x$ would actually give
$$\int_{x=0}^{x=\pi/4}\cos^{-1}(\sin x)\,\mathrm dx=\int_{\color{red}{u=0}}^{\color{red}{u=1/\sqrt2}}\frac{\cos^{-1}u}{\sqrt{1-u^2}}\,\mathrm du$$
Now, recall that $\dfrac{\mathrm d}{\mathrm du}\cos^{-1}u=-\dfrac1{\sqrt{1-u^2}}$, which means you can make another intermediate substitution of, say, $v=\cos^{-1}u$. Then you have
$$-\int_{v=\cos^{-1}0=\pi/2}^{v=\cos^{-1}(1/\sqrt2)=\pi/4}v\,\mathrm dv=\int_{\pi/4}^{\pi/2}v\,\mathrm dv=\dfrac{3\pi^2}{32}$$ | {
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which agrees with the other answers above.
$$\int \cos^{-1}(\sin x) dx$$ $$= \int 1 \cdot \cos^{-1}(\sin x) dx = x \cdot \cos^{-1}(\sin x) - \int x \cdot -\frac{\cos x}{\sqrt{1-\sin^{2}x}} dx + C$$ (integration by parts)
$$= x \cdot \cos^{-1}(\sin x) + \int x \cdot \frac{\cos x}{\sqrt{1-\sin^{2}x}} dx + C$$ $$= x \cdot \cos^{-1}(\sin x) + \int x \cdot \frac{\cos x}{\sqrt{\cos^2 x}} dx + C$$
$$= x \cdot \cos^{-1}(\sin x) \pm \int x \cdot \frac{\cos x}{\cos x} dx + C$$ $$= x \cdot \cos^{-1}(\sin x) \pm \frac{x^2}{2} + C$$
Now do it for the definite integral.
1)
$(u^2)^{\prime}=2\times u\times u^{\prime}$
therefore $\dfrac{1}{2}u^2$ is an antiderivative of $u\times u^{\prime}$
2) derivative of $\text{cos}^{-1}(x)=-\dfrac{1}{\sqrt{1-x^2}}$
$\displaystyle J= \int_0^{\tfrac{\pi}{4}} \text{cos}^{-1}(\sin x)dx$
Perform the change of variable $y=\sin x$,
$\displaystyle J=\int_0^{\tfrac{\sqrt{2}}{2}} \dfrac{\text{cos}^{-1}(x)}{\sqrt{1-x^2}}dx$
Using $(1)$,
\begin{align}J&=-\dfrac{1}{2}\Big[\text{cos}^{-1} (x)^2\Big]_0^{\tfrac{\sqrt{2}}{2}}\\ &=-\dfrac{1}{2}\left[\dfrac{\pi^2}{16}-\dfrac{\pi^2}{4}\right]\\ &=\boxed{\dfrac{3}{32}\pi^2} \end{align} | {
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This section details the validation for the Beam calculator. Several examples are worked through to determine expected results such as deflections, forces, and stresses. These expected results are then compared to the actual output of the calculator.
The standard beam equations for a cantilever beam loaded at the free end are given below:
Max Deflection: $$\delta_{max} = {P L^3 \over 3EI}$$ @ x = L Max Slope: $$\theta_{max} = {P L^2 \over 2EI}$$ @ x = L Shear: $$V = +F$$ constant Moment: $$M_{max} = -FL$$ @ x = 0
For the validation case, an Aluminum 6061-T6 beam with a 1 inch diameter circular cross-section and a length of 10 inches is loaded at 1000 lbf. The inputs are:
L = 10 in
F = 1000 lbf
E = 9.9e6 psi (Al 6061-T6)
$$A = {\pi d^2 \over 4} = 0.7854 ~in^4$$
$$I = {\pi d^4 \over 64} = 0.04909 ~in^4$$
For the given inputs, the expected results are:
Max Deflection: $$\delta_{max} = 0.6859 \text{ in}$$ @ x = L Max Slope: $$\theta_{max} = 0.1029 \text{ rad}$$ @ x = L Shear: $$V = +1000 \text{ lbf}$$ constant Moment: $$M_{max} = -10,000 \text{ in-lbf}$$ @ x = 0
### Comparison With Calculator
#### Free Body Diagram
The Free Body Diagram (FBD) is shown below. From the FBD it can be seen that the forces balance and the beam is in static equilibrium.
#### Deflection & Slope
A screenshot of the results table giving the maximum deflection and slope is shown below. It can be seen that these values match the expected values above:
The deflection plots are shown below. From these plots it can be seen that a maximum deflection of 0.6859 inches occurs at the free end of the beam as well as a maximum slope of 0.1029 radians.
#### Shear-Moment Diagram
The shear-moment diagram for the beam is shown below. This diagram is what would be expected for the current case. A constant shear force of 1000 lbf exists along the length of the beam, and the moment increases linearly from 0 in-lbf at the free end of the beam to the full value of -10,000 in-lbf at the fixed end. | {
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#### Stresses
There are two points of interest for validating stresses. Before calculating stress, the forces at these points need to be determined.
Axial (lbf) Shear (lbf) Moment (in-lbf)
Forces @ x = 0: $$F_{ax} = 0$$ $$F_{sh} = 1000$$ $$M = 10000$$
Forces @ x = L: $$F_{ax} = 0$$ $$F_{sh} = 1000$$ $$M = 0$$
The stresses are calculated using the following equations:
Axial Stress Shear Stress Bending Stress Von Mises Stress
$$\sigma_{ax} = {F_{ax} \over A}$$ $$\tau_{sh} = {F_{sh} \over A}$$ $$\sigma_{b} = {Mc \over I}$$ $$\sigma_{vm} = \sqrt{ (\sigma_{ax} + \sigma_{b})^2 + 3\tau_{sh}^2 }$$
Based on the forces at each point and the equations above, the expected stresses at the points of interest are:
Tensile (psi) Shear (psi) Bending (psi) Von Mises (psi)
Stress @ x = 0: $$\sigma_{ax} = 0$$ $$\tau_{sh} = 1273$$ $$\sigma_{b} = 101859$$ $$\sigma_{vm} = 101883$$
Stress @ x = L: $$\sigma_{ax} = 0$$ $$\tau_{sh} = 1273$$ $$\sigma_{b} = 0$$ $$\sigma_{vm} = 2205$$
The stress plots are shown below. It can be seen that these plots agree with the stresses calculated above.
There is one discrepancy that can be noted, which is that the values for bending stress and von Mises stress at x = L are shown to be higher in the stress plots than predicted. The reason for this is that the stress plots are showing the maximum nodal values for each element, and the maximum bending stress for the element at x = L will occur at the left-most node in that element. Therefore, the values of bending stress and von Mises stress reported in the plots will be somewhat higher than what was calculated above. However, the stresses for each node within any element can be interrogated in the result tables. This table will be discussed following the stress plots.
##### Stress At End of Beam (x = L) | {
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##### Stress At End of Beam (x = L)
A screenshot of one of the result tables for this case is shown below with the element of interest highlighted. The node at the very end of the beam (i.e. at x = L) is "Node 2" for the highlighted element. It can be seen that the bending stress at this node is 0 psi and the von Mises stress at this node is 2205 psi. This matches what was predicted. | {
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# Find the remainder when $528528528…$up to $528$ digits is divided by $27$?
Find the remainder when $528528528...$up to $528$ digits is divided by $27$?
Here's what I have done: The number can be written as $528\cdot 10^{525}+528\cdot 10^{522}+...+528$ which has $176$ terms and each term is $\equiv15 \mod 27$ thus the number should be $176*15 \mod 27$ hence $21$ should be the remainder. But book says it is $6$. I don't understand the flaw in my logic. Please correct me.
• you have $21+6=27$ perhaps you are off by a sign? – gt6989b Oct 16 '17 at 15:26
• I think that your answer is correct. – alexp9 Oct 16 '17 at 15:29
• Brute force in python gives 21. – Gribouillis Oct 16 '17 at 15:31
• Are you sure the book says $6$ and not $-6$? After all, $21\equiv -6 \pmod{27},$ so then both answers would agree. – David K Oct 16 '17 at 15:36
• @fleablood Also a good point about negative remainders. It looks like there's probably an error in the book. (Unless the transcription of the problem is very confused, it's only $176$ "copies" of the group of digits $528,$ which makes $528$ digits altogether since each group has three digits. Also, we don't need $10^k$ to be congruent to $1$; we only need $10^{3k}\equiv 1 \pmod{27},$ which is true.) – David K Oct 16 '17 at 17:00
Here is a python3 session
>>> s = '528' * 176
>>> len(s)
528
>>> int(s) % 27
21
• Is it possible to multiply a string by a number? My god...python is awesome. – Integral Oct 16 '17 at 15:35
• @Integral yes strings, lists and tuples can be repeated by multiplying them by a number. – Gribouillis Oct 16 '17 at 15:36
You can see that $6$ cannot be correct by casting out $9$'s: Since $5+2+8=5+5+5$, we have
$$528528\ldots528\equiv5+5+5+\cdots+5+5+5=5\cdot528\equiv5(5+2+8)\equiv5\cdot6\equiv3\mod 9$$
so the remainder mod $27$ must be either $3$, $12$, or $21$. Your approach gave the correct answer, $21$. | {
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• Anyway it is also $-6$ so there is a typo in the book or Anuran has not have seen the minus sign. – Piquito Oct 16 '17 at 15:54
• @Piquito, I agree, a negative sign would fix things (as David K noted in comments). But remainders are usually understood to be nonnegative, so I'm inclined to think it's a typo. – Barry Cipra Oct 16 '17 at 15:57
Since $$3\mid111$$, we know that $$27\mid999$$, Therefore, $$1000\equiv1\pmod{27}$$ Thus, \begin{align} \sum_{k=0}^{175}528\cdot1000^k &\equiv528\cdot176\pmod{27}\\ &\equiv3\cdot176^2\pmod{27}\\ &\equiv3\cdot14^2\pmod{27}\\ &\equiv3\cdot7\pmod{27}\\ &\equiv21\pmod{27} \end{align}
You are incorrect in assuming $10^{k} \equiv 1 \mod 27$. As $10^k \not \equiv 1$ we do not have $528*10^k \equiv 15 \mod 27$.
What you need instead is $528528... = 528(1001001001......)$
And $1001001..... =\sum_{k=0}^{175} 10^{3k}$
$10^3 \equiv 1 \mod 27$... Oh... we do have that and you were not wronng after all.... so $\sum 10^{3k}\equiv 176 \equiv 14 \mod 27$.
So $528528....... \equiv 15*14 \equiv 21 \mod 27$.
And... the book is wrong.
Had it been 527 iterations of 528 the answer would be $6$.
Note that $$27 \times 37 = 999$$.
To find the remainder you get when you divide $$528528\cdots 528$$ by $$999$$, you can "cast out" $$999$$s.
$$\underbrace{528 + 528 + \cdots 528}_{\text{176 times} } \to 176 \times 528 \to 92928 \to 92+928 \to 1020 \to 21$$
So the remainder is $$21$$.
Note. If the remainder was bigger than 26, you would have had to divide it by 27 to get the correct remainder. | {
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# Why do we study Cantor Set?
For finding counter examples. That does not sound convincing enough, at least not always. Why as a object in its own right the study of Cantor Set has merit ?
• Why are you not convinced that counterexamples are an important thing to study? (Re: "at least not always", the beauty of a counterexample is that only needs to be used once, in a particular key spot, to change our understanding forever.) – Daniel R. Collins Sep 30 '18 at 1:07
1. in beginning real analysis: to counter the naive notion that a "closed set" is a union of closed intervals, plus a few single points.
2. In beginning Lebesgue measure: the easiest example of an uncountable set of measure zero
3. In general topology: sets homeomorphic to the Cantor set are useful in proofs
4. Fractal geometry: many fractals are homeomorphic to the Cantor set. Mandelbrot calls such a thing a Cantor dust to suggest its appearance.
• Can one claim and substantiate that the study of cantor set is something fundamental and not just a source of ready counterexample ? – Vagabond Sep 29 '18 at 15:21
• Point (4) seems to address this. The Cantor set is one of the simplest examples of a fractal, and is therefore studied in detail by students who are first learning fractal geometry or analysis (to the point that a colleague of mine, after writing a masters thesis on Cantor-like sets, started her PhD by stating that she never wanted to see the Cantor set again). – Xander Henderson Sep 30 '18 at 14:51
The Cantor set is quite useful in its own right. My preferred way to think of the Cantor set is as "the most general compact metrizable space." That it is the most general such space means that it is often good for counterexamples because it lacks the particulars. At the same time, one can construct things by specialization. | {
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The formal expression of the Cantor set being the most general compact metrizable space is that every compact metrizable space is the image of the Cantor set under a continuous function. One can use this as a method of proof. For example, to show that a continuous surjection from $$[0,1]$$ onto $$[0,1]^2$$ exists, one can use the fact that $$[0,1]^2$$ is the image of the Cantor set under a continuous function and then extend this continuous function by linear interpolation (the complement of the Cantor set consists of open intervals) to all of $$[0,1]$$.
A lot more can be proven this way, including a number of very surprising results. A wonderful source for such arguments is the following paper, which (deservedly) won an award for mathematical exposition:
Benyamini, Yoav. "Applications of the universal surjectivity of the Cantor set." The American mathematical monthly 105.9 (1998): 832-839.
Among the results proven there is the entirely classical result of Banach and Mazur that every separable Banach space is linearly isometric to a subspace of $$C[0,1]$$; a result as far from being a counterexample as can be.
I think that you may be selling short the value of a counterexample! They are quite useful for making sure that you have not proven too much. i.e. When you have a plausible but only semi-formal argument, how do you tell if it is worth the effort in making it rigorous? Checking against counterexamples is often a useful step.
Still, there are relations to unbounded paths in an infinite binary tree without leaves. That is, suppose that you start at the root and write $$0.$$, thought of as a ternary number. As we travel to the left or right, write a $$0$$ or $$2$$, respectively, for the following digit. Continuing on we get a ternary expansion defining a real number in the Cantor set. Correspondingly, the ternary expansion of an element of the Cantor set gives rise to a path from the root in the tree. | {
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Spaces that are homeomorpic to the Cantor set arise naturally in many mathematical settings, particularly in dynamical systems.
For one dynamical example, the Cantor set is homeomorphic to the phase space of any infinite Bernoulli process.
For another, the "nonescaping set" of many simple dynamical systems in the real line (or the complex plane) is homeomorphic to the Cantor set (this is a "Cantor dust" example as in the answer of @GeraldEdgar). Consider for example the dynamical system $$z_n = (z_{n-1})^2 + 10$$ (You can replace $$10$$ by any real or complex number of magnitude $$>2$$). One can prove that there is a subset $$C \subset \mathbb C$$ homeomorphic to the Cantor set such that if $$z_0 \in C$$ then the sequence $$(z_n)$$ is bounded (in fact it stays in $$C$$), whereas if $$z_0 \not\in C$$ then $$\lim_{n \to \infty} |z_n| =\infty$$. In short, points not in $$C$$ escape to infinity, points in $$C$$ do not.
Also, there are important theoretical descriptions/properties of the Cantor set, for example:
• A topological space is homeomorphic to the Cantor set if and only if it is compact, metrizable, has no isolated points, and every component is a point.
• Any compact zero-dimensional metrizable topological space is homeomorphic to a subspace of the Cantor set.
Cantor sets even occur naturally in number theory! The $$p$$-adic integers $$\mathbb Z_p$$ are homeomorphic to the Cantor set.
• Do you mean some subset in $\mathbb{C}$? I don't get the example; for any $z_{-1}$ real one has $z_n \geq 10^{2^n} \to +\infty$. – Vandermonde Oct 7 '18 at 16:01
• Oops, you're right, that was careless of me. Fixed. – Lee Mosher Oct 7 '18 at 21:26 | {
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# Sequence queue - converges or diverges?
I have the next sequence queue: $$\sum_{n=1}^{\infty}{\frac{n^2-n-1}{n^4+n^2+1}}$$. Does the queue converges or diverge?
My attempt: I have tried to show that $$\frac{1}{n^2}>|\frac{n^2-n-1}{n^4+n^2+1}|$$ for any $$n>0$$, and then by the comparation test we get that since $$\sum_{n=1}^{\infty}{\frac{1}{n^2}}$$ converges, we have that $$\sum_{n=1}^{\infty}{|\frac{n^2-n-1}{n^4+n^2+1}|}$$ converges too, and by a theorem we have that since $$\sum_{n=1}^{\infty}{|\frac{n^2-n-1}{n^4+n^2+1}|}$$ $$\implies$$ $$\sum_{n=1}^{\infty}{\frac{n^2-n-1}{n^4+n^2+1}}$$ converges. Is that right? | {
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• Yeah, a direct comparison test works for this. You were right to compare the following: $$\frac{1}{n^2}>\left|\frac{n^2-n-1}{n^4+n^2+1}\right|$$ Now because the $$\sum_{n=1}^{\infty} \frac{1}{n^2}$$ converges by the p-test, you can conclude that the original series also converges. Mar 16, 2021 at 22:22
• @ObsessiveInteger So my attempt is right? but can you show me how I satisfy that $\frac{1}{n^2}>|\frac{n^2-n-1}{n^4+n^2+1}|$. for $\frac{1}{n^2}>\frac{n^2-n-1}{n^4+n^2+1}$ it's easy since we get by algebra that $n^3+1>0$ for $n>0$. However, for the negative it's not that easy. Mar 16, 2021 at 22:31
• You could make an argument by saying that the numerator of $\frac{n^2-n-1}{n^4+n^2+1}$ grows smaller faster compared to that of $\frac{1}{n^2}$ and similarly, the denominator of $\frac{n^2-n-1}{n^4+n^2+1}$ grows bigger much faster compared to that of $\frac{1}{n^2}$. This means that $\frac{n^2-n-1}{n^4+n^2+1}$ is reaching $0$ faster compared to $\frac{1}{n^2}$, hence, the inequality is true. Of course you can see that graphically as well. Plug in both functions into desmos and you will see. Mar 16, 2021 at 22:41
• @ObsessiveInteger seeing that throw desmos is good, but can I actually verify that by using just inequality algebra? and thank you for the answer. Mar 16, 2021 at 22:44
• Why don't you do the limit comparison test instead? That way you do not have to worry about proving the inequality. Mar 16, 2021 at 22:45
I continue from where you stopped,
Easy to show that :
$$\frac{n^2-n-1}{n^4+n^2+1} < \frac{n^2-1}{n^4+1} < \frac{n^2}{n^4} < \frac{1}{n^2}$$
When we know that $$\sum_{n=1}^{\infty}{\frac{1}{n^2}}$$ converges.
Therefore, by comparison test :
$$\underset{converges}{\underbrace{\frac{n^2-n-1}{n^4+n^2+1}}} < \underset{converges}{\underbrace{\frac{1}{n^2}}}$$ | {
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• However, what stops us from doing so for just the expression without ||, is that when $n$ has small values then the whole expression is negative, and then you cannot use the comparison test, so we have to show it with ||. Mar 16, 2021 at 22:48
• @aasc232 By the limit test we get the equation aims to zero so can see that only the first element in the sum has negative value do not forget is the sum of this equation .
– ATB
Mar 16, 2021 at 22:54
• Can you remind me the limit test? Mar 16, 2021 at 22:56
• And the comparison test is when you have $0\le a_k\le b_k$ for all $k>0$, with the first elements of $a_k$ included, then if $b_k$ converges we get that $a_k$ converges too. So since we have that $a_k$ isn't positive for all $k>0$ we cannot use it. That is what I tried telling you. Isn't right? Mar 16, 2021 at 23:01
• @aasc232 Do the $n=1$ term separately $$|\frac{1^2-1-1}{1^{4}+1^2+1}|<\frac{1}{1}$$ and then ATB's answer applies for all $n>1$ as $\frac{n^{2}-n-1}{n^{4}+n^{2}+1}$ is positive for $n>1$. Of course, a finite number of terms does not determine the convergence and can be disregarded in general when checking convergence
– user649348
Mar 16, 2021 at 23:03
We are trying to prove that the series, $$\sum_{n=1}^{\infty}{\frac{n^2-n-1}{n^4+n^2+1}}$$ converges.
Let us use the Limit Comparison Test. That is, if $$a_n>0$$ and $$b_n>0$$, then if $$\lim_{n\to \infty}\frac{a_n}{b_n}=L$$ for some real number $$L$$. Then, if $$b_n$$ converges, by the limit comparison test, $$a_n$$ must also converge.
Let $$a_n=\frac{n^2-n-1}{n^4+n^2+1}$$ and we say $$b_n=\frac{1}{n^2}$$. Then, $$\lim_{n\to \infty} \frac{\frac{n^2-n-1}{n^4+n^2+1}}{\frac{1}{n^2}}\\ \lim_{n\to \infty}\left(\frac{(n^2-n-1)(n^2)}{n^4+n^2+1}\right)\\ \lim_{n\to \infty}\frac{n^4-n^3-n^2}{n^4+n^2+1}\cdot \frac{\frac{1}{n^4}}{\frac{1}{n^4}}\\ \lim_{n\to \infty} \left(\frac{{1-\frac{1}{n}}-\frac{1}{n^2}}{1+\frac{1}{n^2}+\frac{1}{n^4}}\right)\\ \lim_{n\to \infty}\frac{1-0-0}{1+0+0}=\frac{1}{1}=1$$ | {
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Hence, we have shown that the $$\lim_{n\to \infty}\frac{a_n}{b_n}=L$$ converges because it equals some constant; in this case, $$1$$.
Now, because $$\sum_{n=1}^{\infty}\frac{1}{n^2}$$ converges by the p-test, (since $$p=2>1$$), we can conclude that the series, $$\sum_{n=1}^{\infty}{\frac{n^2-n-1}{n^4+n^2+1}}$$ must also converge by the limit comparison test.
• Ohhh I didn't try using this, in this way. Thank you! Mar 16, 2021 at 23:03
• One last thing, by limit comparison test we have to show that the sequence queue $\frac{n^2−n−1}{n^4+n^2+1}$ is positive. How do we show that? or you saying we disregard the first negative elements of the sum?' Mar 16, 2021 at 23:12
• Technically you do not have to show that, note that the sum $\sum_{n=1}^{\infty}$ starts at $n=1$, so we are only considering values for $n$ that are positive. Notice that for values greater than $1$, $\frac{n^2-n-1}{n^4+n^2+1}$ can never be negative. Mar 16, 2021 at 23:19
• So you are saying that becuase the only negative fraction of the sum is $-\frac{1}{3}$ then there exist elements that if we sum them we get a positive number bigger then $\frac{1}{3}$? Mar 16, 2021 at 23:24
• Thank you a lot! Mar 16, 2021 at 23:29 | {
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Normal subgroup of automorphisms of a free group
Let $F_2=\langle X,Y\rangle$ be the free group of rank $2$ and consider $A,B,C\in Aut(F_2)$ given by: $$A(X,Y)\mapsto(YX^{-1}Y^{-1},Y^{-1})$$ $$B(X,Y)\mapsto(X^{-1},Y^{-1})$$ $$C(X,Y)\mapsto(X^{-1},XY^{-1}X^{-1})$$ The abelianisation of $F_2$ is isomorphic to $\mathbb{Z}^2$ whose automorphism group is $GL(2,\mathbb{Z})$. Thus we get an induced surjective map $m\colon Aut(F_2)\to GL(2,\mathbb{Z})$ which maps an element in $Aut(F_2)$ to the matrix associated to the corresponding element in $Aut(\mathbb{Z}^2)$. I would like to show that the subgroup $\langle A,B,C\rangle$ is normal in $Aut(F_2)$ by proving that $m(\langle A,B,C\rangle)$ is the centre of $GL(2,Z)$. So far I wasn't able to have any good idea... could you help me? Thanks, bye.
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Just calculate the matrices: m(A) = m(B) = m(C) = -1 – Jack Schmidt Nov 22 '11 at 18:22
You may want to use \langle and \rangle instead of < and >; the spacing is correct for the former. – Arturo Magidin Nov 22 '11 at 18:27
how did you calculate explicitly the matrices? $A$,$B$ and $C$ do not belong to $Aut(\mathbb{Z}^2)$ so I should compute their image in the quotient, first. Do I have to write how, say $A$, acts on the cosets $F_2'X$ and $F_2'Y$ in $F_2/F_2'$? P.S. I did use \langle and \rangle! :) – fatoddsun Nov 22 '11 at 19:22
@fatoddsun: Since $[F_2,F_2]$ is fully invariant, any homomorphism $f\colon F_2\to F_2$ induces a homomorphism $f_{\rm ab}\colon \mathbb{Z}^2\to\mathbb{Z}^2$ by $f_{\rm ab}(\overline{X}) = \overline{f(X)}$ and $f_{\rm ab}(\overline{Y}) =\overline{f(Y)}$. So just "abelianize" the image. – Arturo Magidin Nov 22 '11 at 20:24
@fatoddsun: Note, however, that having the image be normal is not sufficient; that will only show that the preimage under the abelianization map $\mathrm{Aut}(F_2)\to\mathrm{GL}(2,\mathbb{Z})$ is normal, and it's not immediately clear to me that this preimage is just $\langle A,B,C\rangle$. – Arturo Magidin Nov 22 '11 at 20:57 | {
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Arturo Magidin's excellent answer explains how to show that $m(\langle A,B,C\rangle)=\{\pm 1\}$, and also why this does not in general imply that your subgroup is normal. However, I want to explain why in this case it is true that $\langle A,B,C\rangle$ is the full preimage of $\{\pm 1\}$, and thus is normal.
The key is a theorem of Nielsen (1) on what is now called $IA_n$: this is by definition the kernel of the natural surjection $Aut(F_n)\to GL_n\mathbb{Z}$, so it fits into a short exact sequence
$$1\to IA_n\to Aut(F_n)\to GL_n\mathbb{Z}\to 1$$
This group of automorphisms which are the Identity on the Abelianization has been well-studied, starting with Magnus (2), who proved that $IA_n$ is finitely generated by giving an explicit set of generators: type (A) which for given $i$ and $j$ conjugates $x_i$ by $x_j$ and fixes all other generators, and type(B) which for given $i$, $j$, and $k$ multiplies $x_i$ by $x_jx_kx_j^{-1}x_k^{-1}$ and fixes all other generators.
But anyway, we are interested here in $IA_2$, and Nielsen proves in (1) that $IA_2$ is just $Inn(F_2)$, the group of automorphisms obtained by conjugation. This is isomorphic to $F_2$ and generated by $conj_X\colon X\mapsto X, Y\mapsto XYX^{-1}$ and by $conj_Y\colon X\mapsto YXY^{-1}, Y\mapsto Y$.
We know that your group $\langle A,B,C\rangle$ subjects to $\{\pm 1\}$, so to show it is the full preimage of $\{\pm 1\}$ it suffices to show that $\langle A,B,C\rangle$ contains the full preimage of $\{1\}$, which is $IA_2$. But we can compute by hand that $AB=conj_Y$ and $CB=conj_X$. So by Nielsen's theorem your group contains $IA_2$, and thus fits into a short exact sequence
$$1\to IA_2\to \langle A,B,C\rangle\to \{\pm 1\}\to 1$$
In particular, your subgroup is normal in $Aut(F_2)$. | {
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In particular, your subgroup is normal in $Aut(F_2)$.
Finally, here is another method of proving that $\langle A,B,C\rangle$ is normal which does not require quoting any papers written in German. We know that the elementary Nielsen transformations provide a generating set for $Aut(F_n)$. This generating set becomes particularly simple in the case when $n=2$: we have three generators $R\colon X\mapsto X^{-1}, Y\mapsto Y$, $S\colon X\mapsto Y, Y\mapsto X$, and $T\colon X\mapsto XY,Y\mapsto Y$. To show that $\langle A,B,C\rangle$ is normal in $Aut(F_2)$ means that it is invariant under conjugation under any element of $Aut(F_2)$. But since $R,S,T$ generate $Aut(F_2)$, it is enough just to show that $\langle A,B,C\rangle$ is invariant under conjugation by $R$, by $S$, and by $T$.
To do this for $R$, for example, you just need to compute the conjugates $RAR^{-1}$, $RBR^{-1}$, and $RCR^{-1}$ and show that each can be written as a combination of $A$, $B$, and $C$. In this case this is quite easy: we have $RAR^{-1}=A$, $RBR^{-1}=B$, and $RCR^{-1}=C^{-1}$. Now all you have to do is do the same for $SAS^{-1}$, $SBS^{-1}$, $SCS^{-1}$, $TAT^{-1}$, $TBT^{-1}$, and $TCT^{-1}$, and you've proved that $\langle A,B,C\rangle$ is normal!
(1) Nielsen, Die Isomorphismen der allgemeinen unendlichen Gruppe mit zwei Erzeugenden, Math. Ann. 78 (1964), 385-397
(2) Magnus, Über n-dimensinale Gittertransformationen, Acta Math. 64 (1935), 353-367.
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Nice; thanks for filling the gap! – Arturo Magidin Nov 22 '11 at 21:39
If $G$ is a group, $N$ is a normal subgroup of $G$, and $f\colon G\to G$ is a homomorphism with $f(N)\subseteq N$, then $f$ induces a homomorphism $\overline{f}\colon G/N\to G/N$ by $\overline{f}(gN) = f(g)N$. To see this, note that $\widehat{f}\colon G\to G/N$ given by $\widehat{f}(g) = gN$ is a homomorphism, and since $f(N)\subseteq N$, then $N$ is contained in the kernel of $\widehat{f}$, so $\widehat{f}$ factors through $G/N$. | {
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Since $[F_2,F_2]$ is fully invariant, this argument will work for $A$, $B$, and $C$.
For $A$, $x$ is mapped to $yx^{-1}y^{-1}$, which gets mapped to $\overline{x}^{-1}$ in $F_2^{\rm ab}$; $y$ is mapped to $\overline{y}^{-1}$. So the induced map $A_{\rm ab}\colon F_2^{\rm ab}\to F_{2}^{\rm ab}$ sends $(a,b)$ to $(-a,-b)$.
This is the same as what $B_{\rm ab}$ does; again the same thing as $C_{\rm ab}$ does. So the image of the subgroup generated by $A$, $B$, and $C$ in $\mathrm{GL}(2,\mathbb{Z})$ is the subgroup generated by their images, which is just the subgroup generated by $-I$.
Note, however, that the fact that $m(\langle A,B,C\rangle) = \{I,-I\}$ does not show, by itself, that $\langle A,B,C\rangle$ is normal in $\mathrm{Aut}(F_2)$; it shows that the pre-image is normal. The preimage is the subgroup of $\mathrm{Aut}(F_2)$ generated by $A$, $B$, $C$, and all automorphisms that induce the identity on $F_2^{\rm ab}$. It is was not clear to me whether this is just $\langle A,B,C\rangle$. For this, see Tom Church's great answer.
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Thank you very much guys! – fatoddsun Nov 23 '11 at 18:23 | {
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# Deriving Equations of a Parabola
1. May 29, 2008
### PFStudent
Hey,
1. The problem statement, all variables and given/known data
How do you derive the equations of the parabola from the general equation of a Conic Section?
2. Relevant equations
General Equation of a Conic Section,
$${{{A}{{x}^{2}}}+{{B}{x}{y}}+{{C}{{y}^{2}}}+{{D}{x}}+{{E}{y}}+{F}} = {0}$$
Where (for a parabola),
$${{\{}A,B,C,D,E,F{\}}}{{\,}{\,}{\,}}{\in}{{\,}{\,}{\,}}{\mathbb{R}}$$
$${{\{}A, C{\}}}{{\,}{\,}{\,}}{\neq}{{\,}{\,}{\,}}{0}$$
$${{{B}^{2}}-{{4}{A}{C}}} = {0}$$
3. The attempt at a solution
From the general equation of a Conic Section,
$${{{A}{{x}^{2}}}+{{B}{x}{y}}+{{C}{{y}^{2}}}+{{D}{x}}+{{E}{y}}+{F}} = {0}$$
How do I derive the following formulas for a parabola:
General Form for a Parabola,
$${{{{\left(}{{{H}{x}}+{{I}{y}}}{\right)}}^{2}}+{{J}{x}}+{{K}{y}}+{L}} = {0}$$
Where,
$${{{I}^{2}}-{{4}{H}{J}}} = {0}$$
Analytic Geometry Equations,
Vertical Axis of Symmetry
$${{{\left(}x-h{\right)}}^{2}} = {{{4}{p}}{{{\left(}y-k}{\right)}}}$$
$${y} = {{{a}{{x}^{2}}}+{{b}{x}}+{c}}$$
Horiztonal Axis of Symmetry
$${{{\left(}y-k{\right)}}^{2}} = {{{4}{p}}{{{\left(}x-h{\right)}}}}$$
$${x} = {{{d}{{y}^{2}}}+{{e}{y}}+{f}}$$
I read that every parabola is a combination of transformations of the parabola, $${{y}={{x}^{2}}}$$; but I'm not quite sure how that helps.
Thanks,
-PFStudent
Last edited: May 29, 2008
2. May 29, 2008
### Dick
I'm not precisely sure of what exactly you are supposed to do, but the only way that the quadratic terms Ax^2+Bxy+Cy^2 can be factored as (Hx+Iy)^2 is that B^2-4AC=0. Just multiply (Hx+Iy)^2 out and identify the terms. Does that help?
3. May 30, 2008
### HallsofIvy
First you need to find the "principal axes" (axis of symmetry and perpenpendicular to it through the vertex) and rotate the coordinate system so that new x' and y' axes are the principal axes. | {
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There are two ways to do that, basically using the quadratic terms $Ax^2+ Bxy+ Cx^2$ (By the way, you don't have to put braces, { }, around every term. They are only necessary when you want an entire expression in a particular place. For example, [i t e x]e^{xy+ b}[/i t e x] gives $e^{xy+ b}$ while [i t e x]e^xy+ b[/i t e x] gives $e^xy+ b$.)
Rotating the axes by angle $\theta$, so that the new x'y'- axes are at angle $\theta$ to the old xy- axex, x and y are given, in terms of the new x', y' variables, by $x= x'cos(\theta)+ y'sin(\theta)$ and $y= -x'sin(\theta)+ y'cos(\theta). Replace x and y by those in $Ax^2+ Bxy+ Cx^2$ and choose$\theta[/itex] so that the coefficient of x'y' is 0. Use those formulas with the correct value of $\theta$ to replace x and y in the entire formula. If it really is a parabola, both the x'y' and y'2 terms should vanish.
The other way is to write it as a matrix problem:
$$Ax^2+ Bxy+ Cy^2= \left[\begin{array}{cc}x & y \end{array}\right]\left[\begin{array}{cc}A & \frac{B}{2} \\ \frac{B}{2} & C\end{array}\right]\left[\begin{array}{c}x \\ y\end{array}\right]$$
You can "diagonalize" that matrix (so writing the quadratic without the xy term) by finding the eigenvalues and eigenvectors of the coefficient matrix. The eigenvectors will point along the principal axes. Since this is a parabola, one of the eigenvalues should be 0.
4. May 31, 2008
### PFStudent
Hey,
Thanks for the replies: Dick and HallsofIvy.
Yea, to be honest I had not seen that general form for a parabola before until I saw it on wikipedia, maybe it is wrong? - here is the link, (Scroll down a bit to the section "General Parabola").
http://en.wikipedia.org/wiki/Parabola
Let me know what you think. | {
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http://en.wikipedia.org/wiki/Parabola
Let me know what you think.
Hmm, ok that makes sense, I will try that. However, I thought since a parabola - in its' most general form - could be expressed as a conic section with the constraint that $${{B^2} = {4AC}}$$. Then shouldn't I be able to derive the more specific forms of a parabola from the general equation of a conic section with the constraint that $${{B^2} = {4AC}}$$?
Where the general equation of a Conic Section is,
$${{{A}{{x}^{2}}}+{{B}{x}{y}}+{{C}{{y}^{2}}}+{{D}{x}}+{{E}{y}}+{F}} = {0}$$
Also, yea I tend to use a bit too many braces ({ }), just a habit from first learning $LaTeX$ way back.
Thanks,
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Updated on: 12 Jan 2019, 03:25
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How many nails did Rudy buy if he purchased them at a price of $0.25 per four nails, sold them at$0.22 per three nails, and made a profit of $2.60? A. 300 B. 240 C. 180 D. 160 E. 120 Originally posted by Nums99 on 12 Jan 2019, 01:42. Last edited by Bunuel on 12 Jan 2019, 03:25, edited 1 time in total. Renamed the topic, edited the question, moved to PS and added the OA. Senior Manager Joined: 15 Jan 2018 Posts: 371 Concentration: General Management, Finance GMAT 1: 720 Q50 V37 Re: How many nails did Rudy buy if he purchased them at a price of$0.25 [#permalink]
### Show Tags
12 Jan 2019, 02:12
1
Hi there,
Here's the solution.
Let the number of nails Rudy purchased be x.
Rudy purchased nails at a price of $0.25 per four nails. So, Cost Price of 4 nails =$0.25
Cost Price of 1 nail = = $0.25/4 Cost price of x nails = ($0.25/4)x
Also given that Rudy sold nails at $0.22 per three nails. So, Selling Price of 3 nails =$0.22
Selling Price of 1 nail = = $0.22/3 Selling price of x nails = ($0.22/3)x | {
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Also, Rudy made a profit of $2.60. Profit = Selling Price - Cost Price$2.60 = ($0.22/3)x - ($0.25/4)x = x(0.88 - 0.75)/12
$2.60 = x(0.13)/12 x = 240 Hence, the the number of nails Rudy purchased is 240. Therefore, the Correct Answer is Option B. 240 _________________ New to GMAT Club or overwhelmed with so many resources? Follow the GMAT Club Study Plan! Not happy with your GMAT score? Retaking GMAT Strategies! Game of Timers - Join the Competition to Win Prizes Senior Manager Status: Manager Joined: 27 Oct 2018 Posts: 365 Location: Egypt Concentration: Strategy, International Business GPA: 3.67 WE: Pharmaceuticals (Health Care) Re: How many nails did Rudy buy if he purchased them at a price of$0.25 [#permalink]
### Show Tags
16 Jan 2019, 02:57
Nums99 wrote:
How many nails did Rudy buy if he purchased them at a price of $0.25 per four nails, sold them at$0.22 per three nails, and made a profit of $2.60? A. 300 B. 240 C. 180 D. 160 E. 120 Profit = Sales - Cost $$2.6 = \frac{0.22}{3}n - \frac{0.25}{4}n = \frac{0.88n}{12} - \frac{0.75n}{12}$$ $$\frac{13}{5} = \frac{0.13n}{12} = \frac{13n}{1200}$$ $$n = \frac{13*1200}{13*5} = 240$$ B _________________ RC Moderator Joined: 24 Aug 2016 Posts: 802 Location: Canada Concentration: Entrepreneurship, Operations GMAT 1: 630 Q48 V28 GMAT 2: 540 Q49 V16 Re: How many nails did Rudy buy if he purchased them at a price of$0.25 [#permalink]
### Show Tags
16 Jan 2019, 08:15
As we are dealing with two numbers 3 & 4 .... lets take the LCM of them i.e.,12 .
Cost price for 12 nails = $0.25*12/4=$0.75 & Selling price for 12 nails = $0.22*12/3=$0.88
Profit made per 12 nails = $0.88 -$0.75 = $0.13 So total '12 units' sold =$2.60/$0.13 =20 Total nails sold = 20*12 = 240 .... Thus Ans would be option B. _________________ Please let me know if I am going in wrong direction. Thanks in appreciation. Director Joined: 12 Feb 2015 Posts: 863 Re: How many nails did Rudy buy if he purchased them at a price of$0.25 [#permalink]
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### Show Tags
25 Feb 2019, 09:26
Nums99 wrote:
How many nails did Rudy buy if he purchased them at a price of $0.25 per four nails, sold them at$0.22 per three nails, and made a profit of $2.60? A. 300 B. 240 C. 180 D. 160 E. 120 LCM approach is good in such questions:- Rudy purchased nails at a price of$0.25 per four nails, or $0.75 per 12 nails sold them at$0.22 per three nails, or $0.88 per 12 nails and made a profit of$2.60 (total) or $(0.88-0.75) =$ 0.13 per 12 nails
(2.60 * 12)/0.13 = 240 nails (Ans)
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How many nails did Rudy buy if he purchased them at a price of $0.25 [#permalink] ### Show Tags 28 Feb 2019, 11:09 Nums99 wrote: How many nails did Rudy buy if he purchased them at a price of$0.25 per four nails, sold them at $0.22 per three nails, and made a profit of$2.60?
A. 300
B. 240
C. 180
D. 160
E. 120
Cost Price of $$4$$ nails $$= 0.25$$
Cost Price of $$3$$ nails $$= \frac{0.25*3}{4} = 0.1875$$
Selling Price of $$3$$ nails $$= 0.22$$
Profit on $$3$$ nails $$= 0.22 - 0.1875 = 0.0325$$
Let $$"x"$$ be total number of nails to make profit of $$2.60$$.
$$\frac{0.0325}{3} = \frac{2.60}{x}$$
$$0.0325*x =3*2.60$$
$$x = \frac{3*2.60}{0.0325} = 240$$
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Re: How many nails did Rudy buy if he purchased them at a price of $0.25 [#permalink] ### Show Tags 02 Mar 2019, 10:08 Nums99 wrote: How many nails did Rudy buy if he purchased them at a price of$0.25 per four nails, sold them at $0.22 per three nails, and made a profit of$2.60?
A. 300
B. 240
C. 180
D. 160
E. 120
We can let n = the number of nails and create the equation: | {
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A. 300
B. 240
C. 180
D. 160
E. 120
We can let n = the number of nails and create the equation:
(0.22/3)n - (0.25/4)n = 2.6
Multiplying by 12, we have:
0.22 x 4n - 0.25 x 3n = 31.2
0.88n - 0.75n = 31.2
0.13n = 31.2
n = 240
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Re: How many nails did Rudy buy if he purchased them at a price of $0.25 [#permalink] 02 Mar 2019, 10:08 Display posts from previous: Sort by # How many nails did Rudy buy if he purchased them at a price of$0.25
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# Aternating sum of an increasing sequence of positive integers
Suppose $$A = (a_n) = (a_1, a_2, a_3, . . .)$$ is an positive, increasing sequence of integers.
Define an $$A$$- expressible number $$c$$ if $$c$$ is the alternating sum of a finite subsequence of $$A.$$ To form such a sum, choose a finite subset of the sequence $$A,$$ list those numbers in increasing order (no repetitions allowed), and combine them with alternating plus and minus signs. We allow the trivial case of one-element subsequences, so that each an is $$A-$$expressible.
Definition. Sequence $$A = (a_n)$$ is an “alt-basis” if every positive integer is uniquely $$A-$$ expressible. That is, for every integer $$m > 0,$$ there is exactly one way to express $$m$$ as an alternating sum of a finite subsequence of $$A.$$
Examples. Sequence $$B = (2^{n−1}) = (1, 2, 4, 8, 16, . . .)$$ is not an alt-basis because some numbers are B-expressible in more than one way. For instance $$3 = −1 + 4 = 1 − 2 + 4.$$
Sequence $$C = (3^{n−1}) = (1, 3, 9, 27, 81, . . .)$$ is not an alt-basis because some numbers (like 4 and 5) are not C-expressible.
An example of an alt-basis is $$\{2^n-1\}=\{1,3,7,15,31,\ldots\}$$
Is there a fairly simple test to determine whether a given sequence is an alt basis?
I have attempted to solve this from a limited knowledge in sequences and have found out various kinds of sequences do not work but fail to see what it is that could make it work. | {
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• Seems slightly (at least) related to this: encyclopediaofmath.org/index.php/Additive_basis But I think you know already about additive bases. – Masterphile Mar 25 at 0:37
• See math.stackexchange.com/questions/3579462. (Not an exact duplicate because it doesn't ask for a general test to determine whether a sequence is an alt-basis.) What is the source of this question? – joriki Mar 25 at 6:47
• Do you know of any alt-bases that are not of the form $\{b_1,b_2,\dots,b_k,a_{k+1},a_{k+2},\dots\}$ where $A=\{a_1,a_2,\dots\}=\{2^n-1\}$ and $\{b_1,\dots,b_k\}$ is some finite increasing set of integers? – Vepir Mar 25 at 21:54
• For example I believe that $\{5\cdot2^{n - 2} + (-1)^{n + 1} 2^{n - 2} - 1\}$ is an alt-basis. – Vepir Mar 25 at 22:50
I can’t answer the question, but I can at least give you a systematic large family of alt-bases.
If $$A$$ is a finite set of positive integers, let $$S(A)$$ be the set of $$A$$-expressible integers, and let $$S^+(A)$$ be the set of $$A$$-expressible positive integers. Then
$$S(A)=S^+(A)\cup\{-a:a\in S^+(A)\},$$
and if $$b>\max A$$, then
$$S^+\left(A\cup\{b\}\right)=S^+(A)\cup\{b-s:s\in S^+(A)\}\cup\{b\}.$$
Thus, if $$|A|=n$$, the maximum number of $$A$$-expressible positive integers is $$2^n-1$$, and $$\max S(A)=\max A$$.
Now suppose that $$A=\{a_n:n\in\Bbb Z^+\}$$, where $$a_n for each $$n\in\Bbb Z^+$$. For $$n\in\Bbb Z^+$$ let $$A_n=\{a_k\in A:1\le k\le n\}$$. Then each $$m\in S(A)$$ is uniquely $$A$$-expressible iff $$|S^+(A_n)|=2^n-1$$ for each $$n\in\Bbb Z^+$$. Moreover, $$S^+(A)=\Bbb Z^+$$ iff for each $$k\in S^+(A)$$ there is a minimal $$n(k)\in\Bbb Z^+$$ such that $$k\in S^+(A_{n(k)})$$. Note that either $$n(k)=1$$, or $$k\in S^+(A_{n(k)})\setminus S^+(A_{n(k)-1})=\{a_{n(k)}-s:s\in S^+(A_{n(k)-1})\}$$.
For $$n\in\Bbb Z^+$$ let
$$a_n=2^n-1=\underbrace{1\ldots 1}_n\text{ in binary},$$
and let $$A=\{a_n:n\in\Bbb Z^+\}$$. It’s not hard to see that
$$S^+(A_n)=\{1,\ldots,2^n-1\}$$ | {
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and let $$A=\{a_n:n\in\Bbb Z^+\}$$. It’s not hard to see that
$$S^+(A_n)=\{1,\ldots,2^n-1\}$$
for each $$n\in\Bbb Z^+$$, so $$A$$ is, as you already observed, an alt-basis. For instance, working in binary, we see that
\begin{align*} 22&=10110_{\text{two}}\\ &=11111_{\text{two}}-1111_{\text{two}}+111_{\text{two}}-1_{\text{two}}\\ &=31-15+7-1\\ &=a_5-a_4+a_3-a_1. \end{align*}
Now let $$\ell,m\in\Bbb Z^+$$. For $$n=1,\ldots,\ell$$ let
$$\color{red}{a_n^{(\ell,m)}}=2^ma_n=\underbrace{1\ldots 1}_n\underbrace{0\ldots 0}_m\text{ in binary}.$$
For $$n=\ell+k$$, where $$k=1,\ldots,m$$, let
$$\color{blue}{a_n^{(\ell,m)}}=2^{m-k}a_n=\underbrace{1\ldots 1}_n\underbrace{0\ldots 0}_{m-k}\text{ in binary}.$$
Finally, for $$n>\ell+m$$ let $$a_n^{(\ell,m)}=a_n$$, and let $$A_{(\ell,m)}=\left\{a_n^{(\ell,m)}:n\in\Bbb Z^+\right\}$$; then $$A_{(\ell,m)}$$ is an alt-basis.
For example,
\begin{align*} A_{(4,2)}&=\{\color{red}{4},\color{red}{12},\color{red}{28},\color{red}{60},\color{blue}{62},\color{blue}{63},127,\ldots\}\\ &=\{\color{red}{100},\color{red}{1100},\color{red}{11100},\color{red}{111100},\color{blue}{111110},\color{blue}{111111},1111111,\ldots\}\text{ in binary}. \end{align*}
To verify this it suffices to show that $$S^+\left(\left\{a_n^{(\ell,m)}:1\le n\le \ell+m\right\}\right)=S^+(A_{\ell+m})$$. The argument is a bit messy to write out, but the idea is straightforward; I’ll illustrate it with $$A_{(4,2)}$$. First, it’s clear from the discussion of $$A$$ that
\begin{align*} S^+\left(\{4,12,28,60\}\right)&=S^+\left(4\{1,3,7,15\}\right)\\ &=4S^+\left(\{1,3,7,15\}\right)\\ &=4\{1,2,\ldots,15\}\\ &=\{4,8,12,\ldots,60\}\\ &=4S^+(A_4). \end{align*}
Then
\begin{align*} S^+(\{4,&12,28,60,62\})=\\ &4S^+(A_4)\cup\left\{|62-s|:s\in S^+(\{4,12,28,60\})\right\}\cup\{62\}=\\ &4S^+(A_4)\cup\left\{|62-s|:s\in\{4,8,12,\ldots,60\}\right\}\cup\{62\}=\\ &4S^+(A_4)\cup\{2,6,10,\ldots,58,62\}=\\ &\{2,4,6,8,\ldots,60,62\}=\\ &2S^+(A_5), \end{align*} | {
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and a similar calculation shows that $$S^+(\{4,12,28,60,62,63\})=S^+(A_6)$$.
I did not collect the set of all alt-bases, but I did find some useful observations, including:
Alt-basis must contain an infinite number of terms of form $$a_k=2^{k}-1,k\in N\subseteq\mathbb N$$.
The converse does not hold. At the end, I give examples of alt-bases and not-alt-bases in this context.
Do correct me If I missed anything.
Let $$A=\{a_1,a_2,\dots\}$$ such that $$a_1\lt a_2 \lt \dots$$ are positive integers.
Definition. For $$A$$ to be an "alt-basis", we need to have both the "uniqueness" and "completeness". In other words, every number is expressible in exactly one way via alternating summation of subsets of $$A$$, which are summed in increasing order.
Definition. A finite (sub)sequence $$A|_n=\{a_1,\dots,a_n\}$$ is an "alt-prefix" if every integer in $$[1,2^{n}-1]$$ is uniquely expressible via alternating summation of subsets of $$A|_n$$ when summed in increasing order. The element $$a_n$$ is called an "anchor element".
Definition. "Anchor sequence" is a set $$\mathcal A(A):=\{a_{n_1},a_{n_2},\dots\}$$ of all "anchor elements" $$a_{n_1},a_{n_2},\dots$$
Notice that a set has $$2^n$$ subsets minus the empty set and that every subset can be rearranged in an increasing order. We want to assign a distinct value to each of those subsets via the alternating summation, to have an alt-basis. The alt-prefix is defined to cover exactly those $$2^n-1$$ subsets. It follows that:
Lemma. $$A$$ is an alt-basis $$\iff$$ $$A$$ is a union of alt-prefixes $$A=A|_{n_1}\cup A|_{n_2}\cup \dots$$
That is, $$A$$ is an alt-basis if and only if there exists a corresponding infinite anchor sequence $$\mathcal A(A)$$.
We add two more definitions to write all of this more easily: | {
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We add two more definitions to write all of this more easily:
Definition. Let $$s(\{b_1,\dots,b_n\})$$ be the result of the alternating summation of $$b_1\lt b_2\lt \dots\le b_n$$. Let $$s_+$$ and $$s_-$$ always start the alternating summation with $$+,-$$ respectively. Then $$s_+=-s_-$$. If $$n$$ is odd then $$s=s_+$$ and if $$n$$ is even then $$s=s_-$$. This guarantees $$s\gt 0$$ because the largest element $$b_n$$ will have a positive sign.
Definition. Define "$$n$$-th partial subset-sum set" of a positive increasing integer sequence $$A$$ as:
$$\mathcal S_n(A):=\{s(A_i):A_i\in\mathcal P(A|_n)\}$$
Where $$\mathcal P(A|_n)$$ is the set of all subsets of $$A|_n=\{a_1,a_2,\dots,a_n\}$$.
The set of all "anchor elements" $$\mathcal A(A)=\{a_{n_1},a_{n_2},\dots\}\subseteq A$$ satisfies $$S_{n_i}=[1,2^{n_i-1}-1]$$ for all $$n_i$$.
Corollary. $$A$$ is an alt-basis if and only if it "is covered by the anchor sequence": $$\max \mathcal A(A)\to \infty$$.
Notice that $$\max S_n = a_n$$. If $$a_n$$ is an anchor element, then $$\max S_n = 2^n-1$$. This gives:
Proposition. If $$a_n$$ is an anchor element, then $$a_n=2^n-1$$.
The converse does not hold. For example, in $$\{1,4,7\}$$ the $$a_3=7=2^3-1$$ but $$a_3$$ is not an anchor element, because $$S_3=\{1,3,4,6,7\}\ne[1,7]$$.
Example $$1$$. It is not hard to see that $$\mathcal A(\{2^n-1\})=\{2^n-1\}$$. This is because:
• $$S_1=\{(+1)\}$$ $$\implies$$ $$a_1$$ is an anchor element.
• $$S_2=\{(+1),(-1+3),(3)\}$$ $$\implies$$ $$a_2$$ is an anchor element.
• $$S_3=\{(+1),(-1+3),(3),(-3+7),(+1-3+7),(-1+7),(7)\}$$ $$\implies$$ $$a_3$$ is an anchor element.
• $$\dots$$ proceed via induction to show every $$a_n$$ is an anchor element.
Since $$\mathcal A(\{2^n-1\})$$ exists and covers the entire $$\{2^n-1\}$$, the $$\{2^n-1\}$$ is an alt-basis.
Example $$2$$. The $$\mathcal A(\{n\})=\{1\}$$ does not cover the entire $$\{n\}$$, hence $$\{n\}$$ is not an alt-basis. | {
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It is not hard to see that $$\max S_n = n \lt 2^n-1\implies a_n$$ is not an anchor element, for every $$n\gt 1$$.
Example $$3.$$ We construct an alt-basis where every $$2$$nd element is an anchor element.
$$A=\begin{cases} 2^n-1, & n\text{ is even} \\ 2^n+2^{n-1}-1, & n\text{ is odd} \end{cases}$$
Use an inductive argument. Assume $$n=2k$$, $$a_{n}=2^{n}-1$$ is an anchor element, which means we have uniquely constructed all $$I_0=[1,2^n-1]=S_{n}$$ elements. Now we can subtract numbers in this interval from $$a_{n+1}$$ to see that:
• $$a_{n+1}=2^{n+1}+2^{n}-1$$ will cover $$I_1=[a_{n+1}-a_{n}, a_{n+1}]=[2^{n+1},2^{n+1}+2^{n}-1]$$
Here we see that $$I_0\cup I_1 \ne [1,2^{n+1}-1]$$ $$\implies$$ $$a_{n+1}$$ is not an anchor elemetn.
To see that $$a_{n+2}=2^{n+2}-1$$ is an anchor element, lets see what will we cover with it:
• $$a_{n+2}$$ combined with $$I_0$$ will cover $$I_2=[a_{n+2}-a_{n}, a_{n+2}]=[2^{n+1}+2^{n},2^{n+2}-1]$$
• $$a_{n+2}$$ combined with $$I_1$$ will cover $$I_3=[a_{n+2}-a_{n+1},a_{n+2}-2^{n+1}]=[2^n,2^{n+1}-1]$$
Now observe $$I=I_0\cup I_3\cup I_1\cup I_2$$ is equal to:
$$I=[1,2^n-1]\cup[2^n,2^{n+1}-1]\cup[2^{n+1},2^{n+1}+2^{n}-1]\cup[2^{n+1}+2^{n},2^{n+2}-1]=[1,2^{n+2}-1]$$
Implying $$a_{n+2}$$ covers $$I=[1,2^{n+2}-1]=S_{n+2}$$, $$\implies$$ $$a_{n+2}$$ is an anchor.
It is not hard to check base cases $$n=1,2$$, and we are done. We have:
$$\mathcal A\left(\left.\begin{cases} 2^n-1, & n\text{ is even} \\ 2^n+2^{n-1}-1, & n\text{ is odd} \end{cases}\right\}\right)=\{2^{2n}-1\}$$
So we have an alt-basis $$A$$.
Example $$4$$. It is not hard to show that:
$$A=\{2^k,2^k+1,2^k+3,2^k+7,\dots,2^k+2^{k}-1,2^{k+2}-1,2^{k+3}-1,\dots\}$$
Is an alt-basis for every $$k=0,1,2,\dots$$, whose anchors are all elements $$a_n,n\gt k$$.
Example $$5$$. The sequence of natural, triangular, tetrahedral,... numbers, or in general, any diagonal of the pascals triangle, is not an alt basis. | {
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This is because for every fixed $$d$$, there exists $$n_0$$, such that for all $$n\ge n_0$$, we have $$\binom{n+d-1}{d}<2^n$$ implying that $$\max S_n\lt 2^n-1$$ for all $$n\ge n_0$$. This implies the sequence of anchors has at most $$n_0$$ elements, implying $$\max\mathcal A(A)\lt \infty$$, hence we do not have an alt-basis becuase of inevetable duplicates.
• I think => part of the initial lemme requires proof. – balcinus Mar 31 at 10:07 | {
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What is the sum of the $81$ products in the $9 \times 9$ multiplication grid?
What is an easy way to solve this problem? I believe that the value in each box is the product of $x$ and $y$.
Suppose the 9 × 9 multiplication grid, shown here, were filled in completely. What would be the sum of the 81 products?
• I asked a question that used this result (and has a nice, clean answer!) several years ago: See MSE 226983 for more! – Benjamin Dickman Jun 12 '16 at 3:29
• (1+2+...+8+9)^2 – Evorlor Jun 12 '16 at 18:45
The sum of the products in the top row would just be $(1+2+3+4+5+6+7+8+9)=45$
Then the next row would be $(2+4+6+8+10+12+14+16+18) = 2\times45 = 90$
So the top two rows sum to $(1+2)\times 45 = 135$
Then it becomes obvious that the full sum of the products is the product of the sums, ie. $45\times45 = 2025$
The sum is essentially $$\sum_{a=1}^9 \sum_{b=1}^9 ab =\sum_{a=1}^9 a \sum_{b=1} ^9 b=\sum_{a=1}^9 a \frac{9\cdot 10}{2}=\left(\frac{9\cdot 10}{2}\right)^2$$
We want to find the following: $$\sum_{i=1}^9 \sum_{j=1}^9 ij$$ Factor out the $i$ from the first summation: $$\sum_{i=1}^9 i\left(\sum_{j=1}^9 j\right)$$ Note that $\sum_{j=1}^9 j=45$. $$\sum_{i=1}^9 i\cdot 45$$ Factor out the $45$: $$45\left(\sum_{i=1}^9 i\right)$$ Note that $\sum_{i=1}^9 i=45$. $$45\cdot 45=2025$$
Use the distributive principle. The sum of the entries in the $2$ column is $2$ times the sum of the numbers $1$ through $9$, so the sum of all the entries is the sum of the numbers $1$ through $9$ times (what?)
Hint $\quad \begin{eqnarray} &\color{#c00}{1+2+3}\\ +\ &\color{#0a0}{2+4+6}\\ +\ &\color{blue}{3+6+9}\end{eqnarray}$ $\quad =\quad \begin{eqnarray} &\color{#c00}1\,(1+2+3)\\ +\ &\color{#0a0}2\,(1+2+3)\\ +\ &\color{blue}3\,(1+2+3)\end{eqnarray}$ $\quad =\quad (\color{#c00}1 + \color{#0a0}2 + \color{blue} 3)(1+2+3)\ \ =\ \ 6\times 6$
Let's prove by induction that the sum of an ${n}\times{n}$ grid is $\frac{n^4+2n^3+n^2}{4}$:
First, show that this is true for $n=1$: | {
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First, show that this is true for $n=1$:
$\sum\limits_{x=1}^{1}\sum\limits_{y=1}^{1}xy=\frac{1^4+2\cdot1^3+1^2}{4}$
Second, assume that this is true for $n$:
$\sum\limits_{x=1}^{n}\sum\limits_{y=1}^{n}xy=\frac{n^4+2n^3+n^2}{4}$
Third, prove that this is true for $n+1$:
$\sum\limits_{x=1}^{n+1}\sum\limits_{y=1}^{n+1}xy=$
$\color\red{\sum\limits_{x=1}^{n}\sum\limits_{y=1}^{n}xy}+\left(\sum\limits_{x=1}^{n}x(n+1)\right)+\left(\sum\limits_{y=1}^{n}y(n+1)\right)+(n+1)(n+1)=$
$\color\red{\frac{n^4+2n^3+n^2}{4}}+\left(\sum\limits_{x=1}^{n}x(n+1)\right)+\left(\sum\limits_{y=1}^{n}y(n+1)\right)+(n+1)(n+1)=$
$\frac{n^4+2n^3+n^2}{4}+(n+1)\left(\sum\limits_{x=1}^{n}x\right)+(n+1)\left(\sum\limits_{y=1}^{n}y\right)+(n+1)(n+1)=$
$\frac{n^4+2n^3+n^2}{4}+(n+1)\left(\frac{n^2+n}{2}\right)+(n+1)\left(\frac{n^2+n}{2}\right)+(n+1)(n+1)=$
$\frac{n^4+2n^3+n^2}{4}+\frac{n^3+2n^2+n}{2}+\frac{n^3+2n^2+n}{2}+n^2+2n+1=$
$\frac{n^4+2n^3+n^2}{4}+n^3+2n^2+n+n^2+2n+1=$
$\frac{n^4+2n^3+n^2}{4}+n^3+3n^2+3n+1=$
$\frac{n^4+6n^3+13n^2+12n+4}{4}=$
$\frac{(n+1)^4+2(n+1)^3+(n+1)^2}{4}$
Please note that the assumption is used only in the part marked red.
Therefore, the sum of a ${9}\times{9}$ grid is $\frac{9^4+2\cdot9^3+9^2}{4}=2025$.
I will focus on an intuition for finding the result. Once you have the formula, other answers are perfect in providing the means for proving it technically.
Let us start with a small size version. If you take the top $3\times 3$ table only with numbers $1$, $2$, $3$ (to save some electrons), the sum of products is:
$$s= 1\times 1 +1\times 2 + 1\times 3 + 2\times 1 +2\times 2 + 2\times 3 + 3\times 1 +3\times 2 + 3\times 3 \,.$$ You see that you can rewrite this sum of nine products as: $$s = (1+2+3)\times (1+2+3)\,,$$ using the distribution of multiplication over additions.
This form is more interesting, because you can see a pattern. The sum of the first integers, multiplied by itself. | {
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It can be reassuring to verify it works: you get $36$, which you can check by hand ($6+12+18$). You can test it with the $2\times2$ or $4\times4$ matrix, to verify it is not a coincidence.
The hint is apparent in the squared shape of the table. The same works for bigger tables too. Each product of $i$ and $j$ in this order is contained once and only once in the product $(1,\ldots,i,\ldots,n)\times(1,\ldots,j,\ldots,n)$.
Now you have to find the sum of integers. If you forget the generic expression for the sum of the $n$ first integer, have this figure in mind:
You pack together two triangles which give you a $n\times (n+1)$ rectangle, of area $n(n+1)$, the double of the area of each triangle.
So finally the answer is $\left(n(n+1)/2\right)^2$. With $n=9$, you get $2025$.
Once you have the method, which is relatively simple, you can easily turn it into a more formal proof, via induction for instance, as given in other answers.
• The link to "I like visual proofs" is dead. Besides that, I think the name "visual proof" is not suitable here. Perception may be betraying, so visual remarks should be backed up by some other argument. This makes it not related to "visual"! – Jasper Jun 12 '16 at 16:23
• @Jasper I have modified the answer. The visual aspect was just a reminder for a formula given in other answers – Laurent Duval Jun 12 '16 at 16:49 | {
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# Absolute Value
## Definition
For a real number, the absolute value of $$x$$, denoted $$\lvert x \rvert$$, is defined as
$\left|x \right| = \begin{cases} x, & \text{if } x \geq 0 \\ -x, & \text{if } x < 0. \end{cases}$
## Technique
### Evaluate $\left| 3\left( 17-55 \right) \right|$.
\begin{aligned} \left| 3\left( 17-55 \right) \right| &= \left| 3\left( -38 \right) \right| \\ &= \left| -114 \right| \\ &= 114 _\square \end{aligned}
### If $x=\frac{5+\sqrt{11}i}{2}$ and $y=\frac{5 -\sqrt{11}i}{2}$, what is the value of $\left| x + 3y \right|^2$?
\begin{aligned} \left| x + 3y \right|^2 &= \left| \frac{5+\sqrt{11}i}{2} + 3\left( \frac{5 -\sqrt{11}i}{2} \right) \right|^2 \\ &= \left| \frac{5+\sqrt{11}i + 15 -3\sqrt{11}i}{2} \right|^2 \\ &= \left| \frac{20-2\sqrt{11}i}{2} \right|^2 \\ &= \left| 10 - \sqrt{11}i \right|^2 \\ &= \left( \sqrt{10^2 + \sqrt{11}^2} \right)^2 \\ &= \sqrt{111}^2 \\ &= 111 _\square \end{aligned}
### What is the absolute value of $2 - 10a$?
The answer depends on the value of $a$. From the definition above, we have
$\left|2-10a \right| = \begin{cases} 2-10a, & \text{if } 2-10a \geq 0 \\ -(2-10a), & \text{if } 2-10a < 0 \end{cases}$
Solving the inequality on the right, we obtain
\begin{aligned} 2-10a &\geq 0 \\ 2 & \geq 10a \\ \frac{1}{5} &\geq a. \end{aligned}
It follows that $2-10a \geq 0 \Longleftrightarrow a \leq \frac{1}{5}.$
Therefore,
$\left|2-10a \right| = \begin{cases} 2-10a, & \text{if } a \leq \frac{1}{5} \\ 10a - 2, & \text{if } a > \frac{1}{5} \end{cases} _\square$
Note by Arron Kau
7 years ago
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thank you very much :-)
- 6 years, 11 months ago
I like this format of note, I think I'll Start Share So !
Explanation More Exercises Solved !
- 7 years ago
I didn't quite understand the problem having $i$. At one stage the $i$ disappears. Can you please explain?
- 7 years ago
It is because the absolute value of a complex number ($a + bi$) is defined as $\sqrt{a^2 + b^2}$ , i.e., $|a + bi|$ = $\sqrt{a^2 + b^2}$. In the above question, $|10 - \sqrt11|$ = $\sqrt{10^2 + \sqrt{11}^2}$
- 7 years ago
Didn't know that, thanks!!
- 7 years ago | {
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- 7 years ago
Didn't know that, thanks!!
- 7 years ago
I am not sure but as far as my mind think I can say this is because when the sign of absolute value was removed.Then sign was changed to positive & i(iota) has nothing to do when there is is postive root.So,It would have been removed...
- 7 years ago
$i$ is defined as $\sqrt{-1}$ i.e. imaginary unit. So, those were complex numbers (real number + imaginary numbers).
Excellent...
- 7 years ago
In ending step ..note that to take modulus of imag eniry number is different than simple modulas
- 6 years, 5 months ago | {
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# A randomized definition of the natural logarithm constant
The number $e$ is the base of the natural logarithm. It is an important constant in mathematics, which is approximately 2.718281828. This post discusses a charming little problem involving the the number $e$. This number can be represented in many ways. In a calculus course, the number $e$ may be defined as the upper limit of the following integral:
$\displaystyle \int_1^e \frac{1}{t} \ dt=1$
Another representation is that it is the sum $\displaystyle e=\sum_{n=1}^\infty \frac{1}{n!}$. Still another is that it is the limit $\displaystyle e=\lim_{n \rightarrow \infty} \biggl( 1+\frac{1}{n} \biggr)^n$. According to the authors of a brief article from the Mathematics Magazine [1], these textbook definitions do not give immediate insight about the number $e$ (article can be found here). As a result, students come away from the course without a solid understanding of the number $e$ and may have to resort to rote memorization on what the number $e$ is. Instead, the article gives six probability oriented ways in which the number $e$ can occur. These occurrences of $e$ are more interesting and, according to the authors of [1], can potentially increase students’ appreciation of the number $e$. In two of these six examples (Example 2 and Example 5), the number $e$ is defined by drawing random numbers from the interval $(0,1)$. This post discusses Example 2.
________________________________________________________________________
Random Experiment
Here’s the description of the random experiment that generates the number $e$. Randomly and successively select numbers from the interval $(0, 1)$. The experiment terminates when the sum of the random numbers exceeds 1. What is the average number of selections in this experiment? In other words, we are interested in the average length of the experiment. According to the article [1], the average length is the number $e$. The goal here is to give a proof of this result. | {
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As illustration, the experiment is carried out 1 million times using random numbers that are generated in Excel using the Rand() function. The following table summarizes the results.
$\left[\begin{array}{rrrrrr} \text{Length of} & \text{ } & \text{ } & \text{ } & \text{Relative} \\ \text{Experiment} & \text{ } & \text{Frequency} & \text{ } & \text{Frequency} \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\ 2 & \text{ } & 500777 & \text{ } & 0.500777 \\ 3 & \text{ } & 332736 & \text{ } & 0.332736 \\ 4 & \text{ } & 124875 & \text{ } & 0.124875 \\ 5 & \text{ } & 33465 & \text{ } & 0.033465 \\ 6 & \text{ } & 6827 & \text{ } & 0.006827 \\ 7 & \text{ } & 1130 & \text{ } & 0.001130 \\ 8 & \text{ } & 172 & \text{ } & 0.000172 \\ 9 & \text{ } & 14 & \text{ } & 0.000014 \\ 10 & \text{ } & 4 & \text{ } & 0.000004 \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\ \text{Total} & \text{ } & 1000000 & \text{ } & \text{ } \end{array}\right]$
The average length of experiment in these 1 million experiments is 2.717001. Even though the rate of convergence to the number $e$ is fairly slow, the simulated data demonstrates that on average it takes approximately $e$ number of simulations of numbers in the unit interval to get a sum that exceeds 1.
________________________________________________________________________
A proof
Let $U_1,U_2,U_3,\cdots$ be a sequence of independent and identically distributed random variables such that the common distribution is a uniform distribution on the unit interval $(0,1)$. Let $N$ be defined as follows:
$\displaystyle N=\text{min}\left\{n: U_1+U_2+\cdots+U_n>1 \right\} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$ | {
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The objective is to determine $E(N)$. On the surface, it seems that we need to describe the distribution of the independent sum $X_n=U_1+U_2+\cdots+U_n$ for all possible $n$. Doing this may be possible but the result would be messy. It turns out that we do not need to do so. We need to evaluate the probability $P(X_n \le 1)$ for all $n$. We show that
$\displaystyle F_n(x)=P(X_n \le x)=\frac{x^n}{n!} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$
for $0 \le x \le 1$ and for $n=1,2,3,\cdots$. This is accomplished by an induction proof. This is true for $n=1$ since $X_1=U_1$ is a uniform distribution. Suppose (2) holds for the integer $n-1$. This means that $\displaystyle F_{n-1}(x)=P(X_{n-1} \le x)=\frac{x^{n-1}}{(n-1)!}$ for $0 \le x \le 1$. Note that $X_n=X_{n-1}+U_n$, which is an independent sum. Let’s write out the convolution formula for this independent sum:
\displaystyle \begin{aligned} F_n(x)&=\int_{0}^x F_{n-1}(x-y) \cdot f_{U_n}(y) \ dy \\&=\int_{0}^x \frac{(x-y)^{n-1}}{(n-1)!} \cdot 1 \ dy \\&=\frac{1}{(n-1)!} \int_0^x (x-y)^{n-1} \ dy \\&=\frac{x^n}{n!} \end{aligned}
The above derivation completes the proof of the claim. We now come back to the problem of evaluating the mean of the random variable $N$ defined in (1). First, note that $N>n$ if and only if $X_n=U_1+U_2+\cdots+U_n \le 1$. So we have the probability statement $\displaystyle P(N>n)=F_n(1)=\frac{1}{n!}$.
As a result, the following is the probability function of the random variable $N$.
\displaystyle \begin{aligned} P(N=n)&=P(N>n-1)-P(N>n) \\&=F_{n-1}(1)-F_n(1) \\&=\frac{1}{(n-1)!}-\frac{1}{n!} \\&=\frac{n-1}{n!} \end{aligned}
Now evaluate the mean.
\displaystyle \begin{aligned} E(N)&=\sum \limits_{n=2}^\infty \frac{n(n-1)}{n!} \\&=\sum \limits_{n=2}^\infty \frac{1}{(n-2)!} \\&=\sum \limits_{m=0}^\infty \frac{1}{m!} \\&=e \end{aligned} | {
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With the above derivation, the proof that $e=$ 2.718281828… is the average number of random numbers to select in order to obtain a sum that exceeds 1 is completed.
________________________________________________________________________
Reference
1. Shultz H. S., Leonard B., Unexpected Occurrences of the Number e,Mathematics Magazine, October 1989, Volume 62, Number 4, pp. 269–271.
________________________________________________________________________
$\copyright \ \text{2015 by Dan Ma}$ | {
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# Can the product of infinitely many elements from $\mathbb Q$ be irrational?
I know there are infinite sums of rational values, which are irrational (for example the Basel Problem). But I was wondering, whether the product of infinitely many rational numbers can be irrational. Thank you for your answers.
• Do you know Wallis's product for $\pi$? Mar 28, 2018 at 13:23
• $e=\lim_{n\to \infty}{(1+1/n)^n}$ Mar 28, 2018 at 13:29
• @Vasya Correct limit of rationals, but not an infinite product. Mar 28, 2018 at 13:29
• @DzamoNorton Not a limit of partial products. The number of factors increases each time, but the factors change: $(1 + 1/2)(1+1/2)$, $(1+1/3)(1 + 1/3)(1+ 1/3)$ and so on. Mar 28, 2018 at 19:51
• @toliveira: An "infinite product" is not multiplication. It's a limit. Fundamentally, multiplication has two operands; you can inductively extend that to any finite number of operands. But "any finite number" does not include "infinitely many". Mar 29, 2018 at 15:10
Yes, it can.
Consider any sequence $(a_n)$ of non-zero rational numbers which converges to an irrational number. Then define the sequence $b_n$ by $b_1 = a_1$ and $$b_n = \frac{a_n}{a_{n-1}}$$ for $n > 1$.
We then have that $$b_1 b_2 \cdots b_n = a_1 \frac{a_2}{a_1} \frac{a_3}{a_2} \cdots \frac{a_n}{a_{n-1}} = a_n.$$
We thus see that every term of $(b_n)$ is rational, and that the product of the terms of $(b_n)$ is the same as the limit of $a_n$, which is irrational. | {
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• Great answer. Very simple. Mar 29, 2018 at 9:57
• But it does look like a "cheat" since everything is divided out by itself. So, is there an infinite product which does not contain a collection of $\frac{a_n}{a_n}$ ? both your answer and mohammad Riazi's have this "cheat", while Kumar's does not. Mar 29, 2018 at 18:56
• @Carl This isn't a cheat. In fact you can rewrite any infinite product in this way. Mar 29, 2018 at 19:23
• +1 But I think it needs at least a short explanation (or link to one, maybe the question is already out here too?) that such a sequence "of non-zero rational numbers which converges to an irrational number" exists. It is easier too see than the initial question (I think) but I guess some of those who will fins the initial question interesting may not see this as obvious Mar 29, 2018 at 20:08
Yes, every irrational number is an infinite product of rationals.
We can write an infinite sum of rationals as an infinite product of rationals.
\begin{align} a&=a,\\ a+b&=a\times\frac {a+b}{a}\\ a+b+c &= a \times \frac {a+b}{a}\times\frac {a+b+c}{a+b}\\.\\.\\.\\.\end{align}
For example, $$\sqrt 2 =1.414213....=1+.4+.01+.004+.....=$$
$$1\times \frac {1.4}{1}\times \frac {1.41}{1.4}\times\frac {1.414}{1.41}\times .....$$
• Your answer is chosen for a review audit of mine, which I upvoted (+1) because it is simple, generalized and easy to understand. Mar 29, 2018 at 5:10
• @TrầnThúcMinhTrí Thanks for your attention. Mar 29, 2018 at 11:19
• I took the liberty to align equalities at '=' sign. Feel free to rollback if you don't like it. Mar 30, 2018 at 7:29
• Thanks, I like it Mar 30, 2018 at 11:12
Yes!
$\cfrac{\pi}{2} = \cfrac{2}1 \cfrac 23 \cfrac 43 \cfrac 45 \cfrac 65 \cfrac 67 \cdots$ | {
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Yes!
$\cfrac{\pi}{2} = \cfrac{2}1 \cfrac 23 \cfrac 43 \cfrac 45 \cfrac 65 \cfrac 67 \cdots$
• This looks cool, but what is the geometric picture to go along with it? I an almost see polygons in there... Mar 28, 2018 at 16:40
• @MikeWise I din't understand which polygon you are talking about. Mar 28, 2018 at 16:42
• General term is $\cfrac{2n}{(2n – 1)}\cfrac{2n}{(2n + 1)}$ Mar 28, 2018 at 16:43
• @MikeWise: I was curious about that as well. Here's a short note on how when you draw rectangles of those areas, they add up to a quarter circle in the limit: math.chalmers.se/~wastlund/monthly.pdf Mar 28, 2018 at 17:39
• Cool, thanks for that, will print it out and read it on my flight in the morning. :) Mar 28, 2018 at 17:47
Too big to be a comment: it should be noted that the order is more crucial in infinite products than in infinite sums, which is strikingly seen on the example cited many times already:
\begin{align*}\cfrac{\pi}{2}&=\cfrac{2}1 \cfrac 23 \cdot \cfrac 43 \cfrac 45 \cdot\cfrac 65 \cfrac 67\cdot \ldots\\ &= \cfrac{2^2}{2^2-1}\cdot \cfrac{4^2}{4^2-1}\cdot \cfrac{6^2}{6^2-1}\ldots\\ \end{align*} is an infinite product with partials starting at $\frac43$ and increasing towards $\frac\pi 2$ (every factor is greater than $1$), whereas the seemingly identical
\begin{align*}0&=\cfrac{2}3 \cfrac 23 \cdot \cfrac 45\cfrac 45\cdot\cfrac 67 \cfrac 67 \cdot\ldots\end{align*}
starts below $1$ and decreases, towards $0$. All that happened was a shift of denominators one step to the left. | {
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• Products are isomorphic to sum of logs, so any order effect that shows up in one can be generated in the other. By allowing the denominators to be shifted independently of the numerators, you are treating the fraction (a/b) as ab^-1, so when you take the log, it's log(a)-log(b), which is an alternating series, and of course order is more important in alternating series than monotonic ones. It's the alternation generated by treating the numerator and denominator separately, not the product, that makes order important. Mar 28, 2018 at 15:00
• Allowing the denominators to be shifted independently of the numerators can also be interpreted as replacing each term a/b by two terms, a and 1/b. Thus when you take logs, you get two terms, log a and - log b. Mar 28, 2018 at 22:10
• These explanations are of course correct, but the point is that it is not obvious to realise that it can matter, easy to make a mistake here. Mar 29, 2018 at 8:32
• I think the claim is not that one shouldn't point out that order matters in an infinite product, but rather that one shouldn't claim that it matters more in infinite products than in infinite sums, since (as @Acccumulation points out) for products of positive real numbers and sums of real numbers it is literally the same phenomenon. Mar 29, 2018 at 22:38
• @LSpice when I said that it matters more, what I meant is that one should beware of the intuition that a permutation of terms acting only locally does not affect the final result (you need a violent perturbation of terms to give $\sum{(-1)^n\over n}$ a different limit). Of course here what is happening is that the terms are not only reordered, but broken to pieces and the pieces reordered. *This natural way to think of a rational factor as being made of two pieces does not happen with series. * Mar 29, 2018 at 23:44 | {
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Consider the Riemann-Zeta Function: $$\sum_{n=1}^{\infty}\frac{1}{n^{s}}=\prod_{p\text{ prime}}\frac{1}{1-p^{-s}}.$$ For $s=2$, the infinite sum on the left is $\pi^{2}/6$, which is irrational. Thus, $\pi^{2}/6$ is an infinite product of rationals.
There is a simple way to obtain any irrational number as an infinite product:
• take any sequence $s_n$ of rational numbers converging to the targeted irrational one (say the approximations of $\pi$ to $n$ decimals);
• form the product of the numbers $f_n:=\dfrac{s_{n+1}}{s_n}$, with $f_0=1$.
$$\pi=\prod_{n=0}^\infty f_n=\frac{31}{10}\cdot\frac{314}{310}\cdot\frac{3141}{3140}\cdot\frac{31415}{31410}\cdot\frac{314159}{314150}\cdots$$
• Ooops, I just noticed that my answer is a duplicate.
– user65203
Mar 30, 2018 at 15:29 | {
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# A gravitation problem with two masses
1. Apr 11, 2014
### toothpaste666
1. The problem statement, all variables and given/known data
A uniform sphere has mass M and radius r. A spherical cavity (no mass) of radius r/2 is then carved within this sphere (the cavity's surface passes through the sphere's center and just touches the sphere's outer surface). The centers of the original sphere and the cavity lie on a straight line, which defines the x axis.
With what gravitational force will the hollowed-out sphere attract a point mass m which lies on the x axis a distance d from the sphere's center? [Hint: Subtract the effect of the "small" sphere (the cavity) from that of the larger entire sphere.]
2. Relevant equations
$F = G\frac{m_1m_2}{r^2}$
3. The attempt at a solution
Let mass $M_f$ = the mass of the sphere without the cavity, $M_c$ = the mass of the cavity, and $R_f$ = the distance between the two masses.
$M_f = M - M_c$
$R_f = d + R_2$
$F = G\frac{(M_f) m}{(R_f)^2}$
is this right?
2. Apr 11, 2014
### SammyS
Staff Emeritus
Did you intend for MC to be negative?
I interpret the Hint to be to calculate the gravitational force that would be exerted on the point mass if the sphere were not hollow, but had the same density. From that, subtract the gravitational force that would be exerted on the point mass by an isolated uniform sphere the size of the cavity, also having the same density.
That would give MF = M + MC .
When you say, "Rf = the distance between the two masses", what two masses are you referring to ?
3. Apr 12, 2014
### toothpaste666
Here is the picture. The two masses are the small particle mass on the right and the sphere on the left with the cavity
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4. Apr 12, 2014
### haruspex | {
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4. Apr 12, 2014
### haruspex
The hint does not say subtract the masses, it says subtract the gravitational effects.
What force would be exerted by the complete sphere?
Suppose we take the "complement" of the cavitated sphere, i.e. throw away the larger sphere and instead have just the smaller sphere (at the position of the cavity). What force would that exert?
5. Apr 13, 2014
### toothpaste666
so if M is the mass of the sphere before the cavity was taken out and Mc is the mass of the cavity then we have:
$G\frac{Mm}{d^2} - G\frac{M_c m}{(d-r_2)^2}$
6. Apr 13, 2014
### SammyS
Staff Emeritus
Yes, if M is the mass of the sphere before the cavity is taken out. That is as the problem is stated. (So, I believe I was incorrect in my previous post.)
You can determine Mc in terms of M.
7. Apr 13, 2014
### toothpaste666
If the mass is evenly distributed wouldn't subtracting the volume be the same as subtracting the mass?
8. Apr 13, 2014
### SammyS
Staff Emeritus
What are you subtracting from what?
What is the mass of a sphere the size of the cavity? How is that related to the mass of the large sphere before the cavity is formed in the sphere?
9. Apr 13, 2014
### toothpaste666
Subtracting the volume of the cavity from the volume of the original sphere i mean. Wouldn't the mass be the same as the volume if its evenly distributed?
10. Apr 13, 2014
### SammyS
Staff Emeritus
Mass is be proportional to volume in this case.
What fraction of the initial solid sphere's mass is removed to make the cavity ?
11. Apr 13, 2014
### toothpaste666
To find the volumes let Vc = Volume of cavity:
$r_2 = \frac{r}{2}$
$V_c = \frac{4pi(\frac{r}{2})^3}{3}$
$V_c = \frac{\frac{4pi(r^3)}{8}}{3}$
$V_c = \frac{\frac{pi(r^3)}{2}}{3}$
$V_c = \frac{pi(r^3)}{6}$
since $V_M =\frac{4pi(r^3)}{3}$
we have
$V_M = 8V_c$
does this get me closer to expressing Mc in terms of M?
12. Apr 14, 2014
### haruspex
That's it. | {
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does this get me closer to expressing Mc in terms of M?
12. Apr 14, 2014
### haruspex
That's it.
13. Apr 14, 2014
### toothpaste666
but how do i relate that back to mass?
14. Apr 14, 2014
### haruspex
Through density, which is the same for both.
15. Apr 14, 2014
### toothpaste666
$D = \frac{M}{V_M} = \frac{3M}{4(pi)r^3}$
$M_c = D V_c = D \frac{(pi)r^3}{6} = (\frac{3M}{4(pi)r^3})(\frac{(pi)r^3}{6}) = \frac{M}{8}$
which gives
$G\frac{Mm}{d^2} - G\frac{Mm}{8(d-r_2)^2}$
how do we know the density is equal?
16. Apr 14, 2014
### haruspex
See the OP:
17. Apr 14, 2014
Thank you. | {
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# Limits and substitution
As in @RobertZ's answer to this question, we often perform substitutions when evaluating limits. For instance, if you're asked to show that $$L = \lim_{t \to 0} \frac{\sin t^3}{t^3} = 1,$$ it's pretty common to say "Let $x = t^3$; then as $t \to 0$, we have $x \to 0$, so $$L = \lim_{x \to 0} \frac{\sin x}{x}$$ which we know is $1$, and we're done."
What's going on here in general is an application of the following "Theorem"
Theorem 1: If the function $g$ satisfies [fill in missing properties] and $$\lim_{t \to a} g(t) = b,$$ then $$\lim_{t \to a} f(g(t)) = \lim_{x \to b} f(x),$$ i.e., one limit exists if and only if the other does, and if they both exist, they're equal.
In the example above, $f(x) = \frac{\sin x}{x}$ and $g(t) = t^3$ and $a = b = 0$.
There's an alternative form, in which we're asked to show that $$L = \lim_{t \to 0} \frac{\sin \sqrt[3]{t}}{\sqrt[3]{t}} = 1,$$ it's pretty common to say "Let $t = x^3$; then as $t \to 0$, we have $x \to 0$, so $$L = \lim_{x \to 0} \frac{\sin x}{x}$$ which we know is $1$, and we're done."
In this case, the implicit theorem is very similar to the other, but with the role of $g$ reversed (i.e., we're substituting $t = x^3$ instead of $x = t^3$, so the natural form of the theorem puts $g$ on the other side):
Theorem 2: If the function $g$ satisfies [fill in missing properties] and $$\lim_{x \to b} g(x) = a,$$ then $$\lim_{t \to a} f(t) = \lim_{x \to b} f(g(x)).$$
In the second example above, we have $a = b = 0$, $f(t) = \frac{\sin \sqrt[3]{t}}{\sqrt[3]{t}}$, and $g(x) = x^3$.
The two theorems are obviously the same: if you swap $a$ and $b$, $x$ and $t$, and reverse the equality in the last line, they're identical. But each represents a different approach to working with limits, so I've stated both. | {
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In the second form, it's clearly important that $g$ be surjective near $a$ (i.e., for every small enough interval $I = (b-\epsilon, b + \epsilon)$, there's an interval $I' = (a-\delta, a + \delta)$ such that $I- \{b\} \subset g(I' - \{a\})$. (Hat-tip to MathematicsStudent1122 for the observation that I need to delete $a$ and $b$ from those intervals). Otherwise you could use things substitutions like $s = t^2$, which would turn a two-sided limit into a one-sided one (or vice versa), in which case one limit might exists and the other might not.
Addendum to clarify why this might matter, for @MathematicsStudent1122:
Consider $$f(x) = \begin{cases} 1 & x \ge 0 \\ 0 & x < 0 \end{cases}.$$
and look at $L = \lim_{x \to 0} f(x^2)$. It's clear that $L$ exists and is $1$. But if we substitute $t = x^2$, then we get $L = \lim_{t \to 0} f(t)$, which does not exist; hence this "substitution" is not valid: I've turned what amounts to a 1-sided limit (which exists) into a two-sided limit (which does not exist). The domains of $f$ and $g$ are both all of $\Bbb R$.
My question is this:
What is a reasonable set of missing properties for each of these theorems? (I can work out the exact properties easily enough by running through the definitions, but they don't seem to be very helpful/checkable.)
One answer might be "$g$ is locally a bijection", but that rules out things like $y = x + x\sin \frac{1}{x}$ near $x = 0$, so it seems too limited. (It also rules out things like $x \mapsto x + \sin x$ for limits as $x \to \infty$, which is a pity.)
I recognize that this is not a strictly mathematical question. But my goal is to come up with a "calculus student's theorem", one that says "if you're trying to work out a limit, which may or may not exist, then it's OK to do substitutions of this sort along the way," and which will cover the vast majority of the problems that they might encounter in a standard calculus book, or even in Spivak's book. | {
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This question gives two theorems, but both have assumptions about the existence of limits. This one comes a little closer, but still isn't entirely satisfactory.
I'd love any nice-enough condition to be broadly useful. In particular, I think it's completely reasonable to require, for instance, that the "substitution function" $g$ be continuous, and perhaps even differentiable (although I doubt that's of much use).
• – Jack D'Aurizio Aug 20 '17 at 15:03
• Great point, @JackD'Aurizio. That may be the first place where this particular stunt bothered me. :) – John Hughes Aug 20 '17 at 15:04
• So that's part of the quest for the Holy Grail of all lazy students, the Grand Unified Formula Of All Textbook Exercises? ;-) – Professor Vector Aug 20 '17 at 15:06
• Substitutions like $t=x\sin \frac{1}{x}$ simply don't always work. If we let $g(x) = x\sin \frac{1}{x}$ and $f(x) = 1$ for $x=0$ and $f(x) = 0$ for $x \neq 0$, then we can note that $$\lim_{x \to 0} f(g(x))$$ does not exist, but $$\lim_{y \to 0} f(y)$$ does exist. – MathematicsStudent1122 Aug 20 '17 at 15:56
• @ParamanandSingh: Well...I've mused about it on miscellaneous occasions for about 45 years, but I thank you for pointing out that I've bene wasting my time. :) Your answer in that linked case is the easy one --- the one that only goes one direction. I want to know that if I make a substitution and the origiinal limit does NOT exist, then the new limit is also guaranteed to not exist. – John Hughes Aug 20 '17 at 18:07
I'm afraid this isn't quite what you're after, but it seems to be a decent starting point. These conditions, while restrictive, appear to be necessary for the following proof.
Assume $\lim_{x\to a}g(x)=b$, where the function $g$ also satisfies: $$\text{there exists a neighborhood U of a such that b\not\in g(U\setminus\{a\})} \tag{1}$$ and for each $(y_n)$ converging to $b$ with $y_n\ne b$, $$\text{there exists (x_n) such that g(x_n)=y_n for large n and x_n\to a}. \tag{2}$$ | {
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Then we show that $$\lim_{x\to a}f(g(a)) = \lim_{y\to b}f(y).$$
Proof:
Let $L_1:=\lim_{x\to a}f(g(a))$ and $L_2:=\lim_{y\to b}f(y)$, either of which may or may not exist. We will use the sequential criterion.
First suppose $L_2$ exists and let $(x_n)$ be a sequence such that $x_n\to a$ and $x_n\ne a$ for all $n$. Since $\lim_{x\to a}g(x)=b$, we have $g(x_n)\to b$. Due to $(1)$, we have $g(x_n)\ne b$ for $n$ large enough. Then, because $L_2$ exists, we have $f(g(x_n))\to L_2$. This proves by the sequential criterion that $L_1$ exists and $L_1=L_2$.
Now assume $L_1$ exists and let $(y_n)$ be a sequence such that $y_n\to b$ and $y_n\ne b$ for all $n$. By $(2)$ there exists a sequence $(x_n)$ such that $g(x_n)=y_n$ for large $n$ and $x_n\to a$. Since $y_n\ne b$ for all $n$, we have $x_n\ne a$ for large $n$. Then $$\lim_{n\to\infty}f(y_n)=\lim_{n\to\infty}f(g(x_n))=L_1,$$ and so the sequential criterion implies $L_2$ exists and $L_2=L_1$.
• You're right --- that's not quite what I'm after. For one thing, it's not a condition that most calc students could test. :( But I agree that it's a decent starting point -- thanks. – John Hughes Aug 21 '17 at 19:58 | {
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Prove that if $A$ is normal, then eigenvectors corresponding to distinct eigenvalues are necessarily orthogonal (alternative proof)
The problem statement is as follows:
Prove that for a normal matrix $A$, eigenvectors corresponding to different eigenvalues are necessarily orthogonal.
I can certainly prove that this is the case, using the spectral theorem. The gist of my proof is presented below.
If possible, I would like to find a simpler proof. I was hoping that there might be some sort of manipulation along these lines, noting that $$\langle Av_1,A v_2\rangle = \langle v_1,A^*Av_2\rangle = \langle v_1,AA^*v_2\rangle = \langle A^* v_1,A^* v_2 \rangle$$
Any ideas here would be appreciated.
My proof:
Let $\{v_{\lambda,i}\}$ be an orthonormal basis of eigenvectors (as guaranteed by the spectral theorem) such that $$A v_{\lambda,i} = \lambda v_{\lambda,i}$$ Let $v_1,\lambda_1$ and $v_2,\lambda_2$ be eigenpairs with $\lambda_1 \neq \lambda_2$. We may write $v_1 = \sum_{i,\lambda}a_{i,\lambda}v_{i,\lambda} .$ We then have $$0 = Av_1 - \lambda_1 v_1 = \sum_{i,\lambda}(\lambda - \lambda_1)a_{i,\lambda}v_{i,\lambda}$$ So that $a_{i,\lambda} = 0$ when $\lambda \neq \lambda_1$. Similarly, we may write $v_2 = \sum_{i,\lambda}b_{i,\lambda}v_{i,\lambda}$, and note that $b_{i,\lambda} = 0$ when $\lambda \neq \lambda_2$. From there, we have $$\langle v_1,v_2 \rangle = \sum_{i,\lambda}a_{i,\lambda}b_{i,\lambda}$$ the above must be zero since for each pair $i,\lambda$, either $a_{i,\lambda}=0$ or $b_{i,\lambda} = 0$. | {
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• "Let $v_{\lambda,i}$ be an orthonormal basis of eigenvectors..." I'd guess that this might be something very close to what would need to be proved. – Algebraic Pavel May 3 '14 at 2:13
• @PavelJiranek I was worried that I had used circular logic at some point. However, the existence of such a basis (i.e. the spectral theorem) comes directly from the Schur triangularization theorem, which says nothing about normal matrices in particular. – Omnomnomnom May 3 '14 at 19:11
Assume $\;\lambda\neq \mu\;$ and
$$\begin{cases}Av=\lambda v\;\,\implies\; A^*v=\overline \lambda v\\{}\\Aw=\mu w\implies A^*w=\overline\mu w\end{cases}$$
From this we get:
$$\begin{cases}\langle v,Aw\rangle=\langle v,\mu w\rangle=\overline\mu\langle v,w\rangle\\{}\\ \langle v,Aw\rangle=\langle A^*v,w\rangle=\langle\overline\lambda v,w\rangle=\overline\lambda\langle v,w\rangle \end{cases}$$
and since $\;\overline\mu\neq\overline\lambda\;$ , we get $\;\langle v,w\rangle =0\;$
Question: Where did we use normality in the above?
• How de we know that $v$ is also an eigenvector for $A^*$? – Berci May 2 '14 at 22:55
• @Berci, read and think of the question at the end of the answer. – DonAntonio May 2 '14 at 22:56
• From the other two answers this follows, of course, using that $A$ is normal. But can you show it directly? – Berci May 2 '14 at 22:57
• Why directly? Use the definition and work it out. I usually don't give full answers and, after all, the OP is trying to get a more or less simpler proof to the claim than the one using the spectral theorem... – DonAntonio May 2 '14 at 23:00
• I like it! Showing that $Av=\lambda v\;\,\implies\; A^*v=\overline \lambda v$ seems to be the step in the spirit of what I was looking for. Thanks. – Omnomnomnom May 2 '14 at 23:14 | {
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Specializing your identity to $v_1=v_2=v$, we get $\|Av\|=\|A^*v\|$. Hence $\ker A=\ker A^*$. Recalling that $\ker A^* = (\operatorname{ran} A)^\perp$ for general $A$, we conclude that the kernel and range of a normal matrix are mutually orthogonal.
It remains to apply the above conclusion to $A-\lambda I$ where $\lambda$ is an eigenvalue of $A$.
• Neat! Although DonAntonio's proof is more along the lines I was thinking of, this is a nice perspective. – Omnomnomnom May 2 '14 at 23:16
I try to give another simple proof to $$T^*v=\bar{\lambda}v ~\text{ if }~ Tv=\lambda v$$ where $T$ is a normal operator on a Hilbert space $H$.
Suppose $V=\ker(T-\lambda I)$. Since $T^*$ communicate with $T$, $$T^*V\subset V.$$ Because $$\langle v,T^*v\rangle =\langle Tv,v\rangle =\langle \lambda v,v\rangle=\langle v,\bar{\lambda}v\rangle ~~\forall v \in V,$$ $\langle u,T^*v\rangle =\langle u, \bar{\lambda}v\rangle ~\forall u,v\in V$ by polarisation identity, and thus $T^*v=\bar{\lambda}v.$
REMARK: Let $\sigma:V\times V\to W$ be a sesquilinear form, where $V$ and $W$ are linear vector spaces over $\mathbb{C}$. The follwing formula is called Polarisation Identity : $$\sigma(u,v)=\sum_{k=0}^3 i^k\sigma(u+i^k v, u+i^kv).$$
• I don't really see how this answer addresses the question being asked. Also, the question being asked was answered 3 years ago. – Omnomnomnom Jun 12 '17 at 15:01
• @Omnomnomnom This is why $Av=\lambda v\Rightarrow A^*v=\bar{\lambda}v$ using no spectral theorem which is the point to the question I think. – C.Ding Jun 13 '17 at 2:37
• That is not the point to the question, actually. The point is to show that eigenvectors corresponding to different eigenvalues are necessarily orthogonal. – Omnomnomnom Jun 13 '17 at 2:58
• Oh, sorry to disturb you. As I give some additional remarks to the right answer you have accepted. – C.Ding Jun 13 '17 at 3:04
A normal matrix is unitarily similar to diagonal matrix.
$$A = UDU^{-1}$$ where $U$ is Unitary matrix. | {
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$$A = UDU^{-1}$$ where $U$ is Unitary matrix.
Eigen decompositions tells that $U$ is a matrix composed of columns which are eigenvectors of $A$. And matrix $D$ is Diagonal matrix with eigenvalues on diagonal.
Property: Columns of Unitary matrix are orthogonal.
So, columns of $U$ (which are eigenvectors of $A$) are orthogonal. | {
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# Within a rectangular courtyard of length 60 feet, a graveled path,
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Within a rectangular courtyard of length 60 feet, a graveled path, [#permalink]
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Within a rectangular courtyard of length 60 feet, a graveled path, 3 feet wide, is laid down along all the four sides. The cost of graveling the path is Rs 2 per sqft. If the path had been twice as wide, the gravek would have cost Rs 984 more. The width of the courtyard is :
A 24
B 30
C 40
D 45
E 54
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Within a rectangular courtyard of length 60 feet, a graveled path, [#permalink]
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24 Sep 2018, 11:12
Being x the width of the rectangle, we have the following equation:
$$2[2(60*3)+2*3(x-6)] + 984 = 2[2(60*6)+2*6(x-12)]$$ | {
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$$2[2(60*3)+2*3(x-6)] + 984 = 2[2(60*6)+2*6(x-12)]$$
where $$2(60*3)$$ is the area of the path next to the longer side (60 ft long and 3 ft wide), and $$2*3(x-6)$$ the area of the path next to the shorter side, avoiding to count the area from the corner twice. Then we need to repeat the same idea with the bigger path. Both areas are multiplied by 2 because we are measuring the cost, which is Rs 2 per sq ft. We have to consider that the path with double size is Rs 984 more expensive, so we add this into our equation. Then, we can solve the equation:
$$2[360+6x-36]+984=2(720+12x-144)$$
$$360+6x-36+492=720+12x-144$$
$$324+6x+492=576+12x$$
$$816-576=6x$$
$$240=6x$$
$$x=40$$
Thus, C is the correct answer.
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Re: Within a rectangular courtyard of length 60 feet, a graveled path, [#permalink]
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24 Sep 2018, 11:33
sultanatehere wrote:
Within a rectangular courtyard of length 60 feet, a graveled path, 3 feet wide, is laid down along all the four sides. The cost of graveling the path is Rs 2 per sqft. If the path had been twice as wide, the gravek would have cost Rs 984 more. The width of the courtyard is :
A 24
B 30
C 40
D 45
E 54
A = Area of outer rectangle = 60 x Width
Ax = Area of inner rectangle with width of 3 = 54 x (Width-6)
Ay = Area of inner rectangle with twice the width, i.e 6 = 48 x (Width - 12)
Area of gravel with normal width = A-Ax
Area of gravel with twice width = A-Ay
Given, cost of gravelling twice width - cost of gravelling normal width = 984.
Given, cost of gravelling = Rs2/sq ft.
2(A-Ay) - 2(A-Ax) = 984.
A-Ay-A+Ax=492.
Ax-Ay=492.
On substituting Ax and Ay from above and solving the above equation, we get width=40.
Hence C.
Cheers!
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Re: Within a rectangular courtyard of length 60 feet, a graveled path, [#permalink]
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24 Sep 2018, 13:48
Tried to solve using an Ans choices and got lucky with the 1st option
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Re: Within a rectangular courtyard of length 60 feet, a graveled path, [#permalink]
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27 Sep 2018, 17:20
sultanatehere wrote:
Within a rectangular courtyard of length 60 feet, a graveled path, 3 feet wide, is laid down along all the four sides. The cost of graveling the path is Rs 2 per sqft. If the path had been twice as wide, the gravek would have cost Rs 984 more. The width of the courtyard is :
A 24
B 30
C 40
D 45
E 54
We can let w = the width of the courtyard. Therefore, the courtyard, including the graveled path, has an area of 60w sq ft. Excluding the path, the area of the courtyard would be (60 - 2(3))(w - 2(3)) = 54(w - 6) = 54w - 324 sq ft. So the path alone has an area of 60w - (54w - 324) = 6w + 324 sq ft. Since the cost of graveling the path is Rs 2 per sq ft, the cost of graveling the path is Rs 12w + 648.
If the path had been twice as wide, then the courtyard, including the path, would still have an area of 60w sq ft. However, excluding the path, the area of the courtyard would be (60 - 2(6))(w - 2(6)) = 48(w - 12) = 48w - 576 sq ft. So the path alone would have an area of 60w - (48w - 576) = 12w + 576 sq ft, and the cost of the path would be Rs 24w + 1152. We are told that this would cost Rs 984 more, so we can set up the following equation to solve for w:
12w + 648 + 984 = 24w + 1152
12w + 1632 = 24w + 1152
480 = 12w
40 = w
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Re: Within a rectangular courtyard of length 60 feet, a graveled path, &nbs [#permalink] 27 Sep 2018, 17:20
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# 5 drill machines can drill 15 meters in 5 hours.
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5 drill machines can drill 15 meters in 5 hours. [#permalink]
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5 drill machines can drill 15 meters in 5 hours. Assuming the speed of each machine to be the same, how many machines would be needed to drill 60 meters in 2 hours?
A)8
B)16
C)23
D)50
E)60
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Re: 5 drill machines can drill 15 meters in 5 hours. [#permalink]
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06 Nov 2017, 07:27
Mac 5 meters 15 hours 5
Mac x meters 60 hours 2
Do it with proportion.
X/5= 60/15 * 5/2
X=50
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Re: 5 drill machines can drill 15 meters in 5 hours. [#permalink]
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06 Nov 2017, 07:32
ManishKM1 wrote:
5 drill machines can drill 15 meters in 5 hours. Assuming the speed of each machine to be the same, how many machines would be needed to drill 60 meters in 2 hours?
A)8
B)16
C)23
D)50
E)60 | {
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A)8
B)16
C)23
D)50
E)60
to drill 15 mtr in 5 hrs, 5 machines are required..
so to drill 15*4 mtr in 5 hrs, 5*4 machines are required..
to drill 60 mtr in 1 hrs, 20*5 machines are required..
finally to drill 60 mtr in 2 hrs, 100/2 or 50 machines are required..
D
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5 drill machines can drill 15 meters in 5 hours. [#permalink]
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06 Nov 2017, 09:17
1
ManishKM1 wrote:
5 drill machines can drill 15 meters in 5 hours. Assuming the speed of each machine to be the same, how many machines would be needed to drill 60 meters in 2 hours?
A)8
B)16
C)23
D)50
E)60
Find a machine's individual rate from the first information, and use it to solve the question.
W = (# of machines) * (indiv.) rate * time
W = 15, # of machines = 5, rate = ??, time = 5
15 = 5 * rate * 5
15 = 25 * rate
Individual rate = $$\frac{15}{25}=\frac{3}{5}$$ in meters/hour
At that rate, how many machines are needed to drill 60 meters in 2 hours?
W = (# of machines) * (indiv.) rate * time
W = 60, # of machines = ??, rate = $$\frac{3}{5}$$, time = 2
60 = (# of machines) * $$\frac{3}{5}$$ * 2
60 = (# of machines) * $$\frac{6}{5}$$
# of machines = $$\frac{60}{(\frac{6}{5})}= 60 * \frac{5}{6}=$$ 50
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Re: 5 drill machines can drill 15 meters in 5 hours. [#permalink]
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### Show Tags
12 Nov 2017, 08:47
1
1
ManishKM1 wrote:
5 drill machines can drill 15 meters in 5 hours. Assuming the speed of each machine to be the same, how many machines would be needed to drill 60 meters in 2 hours?
A)8
B)16
C)23
D)50
E)60
The rate for 5 drill machines is 15/5 = 3 meters per hour. We need to determine how many drill machines we need for a rate of 60/2 = 30 meters per hour. Since 30 is 10 times 3, we would need 5 x 10 = 50 drill machines.
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# What is the average (arithmetic mean) of eleven consecutive
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23 Apr 2010, 18:05
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What is the average (arithmetic mean) of eleven consecutive integers?
(1) The average of the first nine integers is 7
(2) The average of the last nine integers is 9
[Reveal] Spoiler: OA
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Re: Average ( arithmetic mean ) of eleven consecutive integers? [#permalink]
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23 Apr 2010, 18:11
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Some how i got E
1. (63+ X10 + X11) / 11 = ?? ..sitill missing two numbers? so insuff??
2. (X1+X2 +81)/ 11 = ?? Still missing two numbers so Insuff???
and combining 1 and 2 still missing some info ? .. Sorry friends, i need help here.
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Re: Average ( arithmetic mean ) of eleven consecutive integers? [#permalink]
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Re: Average ( arithmetic mean ) of eleven consecutive integers? [#permalink]
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24 Apr 2010, 06:10
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zz0vlb wrote:
What is the average ( arithmetic mean ) of eleven consecutive integers?
1.The avg of first nine integers is 7
2. The avg of the last nine integers is 9
[Reveal] Spoiler:
D
friends, please explain this to me.
Consecutive integers represent evenly spaced set. For every evenly spaced set mean=median, in our case $$mean=median=x_6$$.
(1) $$x_1+x_2+...+x_9=63$$ --> there can be only one set of 9 consecutive integers to total 63. Sufficient.
If you want to calculate: $$(x_6-5)+(x_6-4)+(x_6-3)+(x_6-2)+(x_6-1)+x_6+(x_6+1)+(x_6+2)+(x_6+3)=63$$ --> $$x_6=8$$.
OR: Mean(=median of first 9 terms=5th term)*# of terms=63 --> $$x_5*9=63$$ --> $$x_5=7$$ --> $$x_6=7+1=8$$
(2) $$x_3+x_4+...+x_{11}=81$$ --> there can be only one set of 9 consecutive integers to total 81. Sufficient.
If you want to calculate: $$(x_6-3)+(x_6-2)+(x_6-1)+x_6+(x_6+1)+(x_6+2)+(x_6+3)+(x_6+4)+(x_6+5)=81$$ --> $$x_6=8$$.
OR: Mean(=median of last 9 terms=7th term)*# of terms=81 --> $$x_7*9=81$$ --> $$x_7=9$$ --> $$x_6=9-1=8$$
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Re: Average ( arithmetic mean ) of eleven consecutive integers? [#permalink]
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24 Apr 2010, 07:11
Thanks for clarifying this Bunuel. +1Kudos to you.
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Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink]
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12 Aug 2013, 04:00
Bumping for review and further discussion.
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Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink]
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Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink]
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12 Aug 2013, 04:39
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zz0vlb wrote:
What is the average (arithmetic mean ) of eleven consecutive integers?
(1) The avg of first nine integers is 7
(2) The avg of the last nine integers is 9
Here is a neat little trick for such kind of problems:
Will be better illustrated using a numerical example: take the set {2,3,4,5,6}. Here the common difference (d)=1. The initial average = 4. Now, the averge of the set, after removing the last integer of the set(i.e. 6)will be reduced by exactly $$\frac{d}{2} units \to$$ The new Average = $$4-\frac{1}{2} = 3.5$$
Again, for the new set of {2,3,4,5} the average is 3.5 . Now, if the last integer is removed, the new average will again be = 3.5-0.5 = 3.
Similarly, for the same set {2,3,4,5,6}, if we remove the first integer from the given set, the average increases by 0.5 and so on and so forth.
Back to the problem:
From F.S 1, we know that the average of the first 9 integers is 7. Thus, the average with the original 11 integers must have been 7+0.5+0.5 = 8. Sufficient.
From F.S 2, we know that the average of the last 9 integers is 9, thus the average with the initial 11 integers must have been 9-0.5-0.5 = 8. Sufficient.
D.
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Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink]
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12 Aug 2013, 23:15
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zz0vlb wrote:
What is the average (arithmetic mean ) of eleven consecutive integers?
(1) The avg of first nine integers is 7
(2) The avg of the last nine integers is 9
As a general rule whenever there is a AP the average of the series is always the median of the series. Here it is a AP with difference 1 | {
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1. First 9 integers average is 7 . So the median that is the 5th digit is 7. Hence we can easily find the series and the average of the 11 consecutive digit series. Sufficient
2. Average of last 9 integers is 9 hence we know that for this subset of 9 integers the 5th integer would be 9 and we can find the series on the basis of this and the average. Sufficient
And is D
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Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
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Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
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09 Mar 2015, 19:15
I considered following approach
if the smallest number in set is x , then sum of 11 consecutive numbers = 11x+(1+2+...10)=11x+55--->A
if largest number in set is x ,then sum of 11 consecutive numbers=11x-(1+2+10)=11x-55
Now as per statement 1 , average of first 9 numbers is 7 i.e sum =63
sum of 11 numbers =63+x+9+x+10----->B
Equating A& B
11X+55=63+X+9+10 ,which can be solved to get x=3
statement I is sufficient
similar approach for Statement II
11X-55=8+2X-19 ,can be solved to get X=13
statement 2 is sufficient | {
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statement 2 is sufficient
OA=D
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Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
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09 Mar 2015, 19:26
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Hi All,
When you look at this question, if you find yourself unsure of where to "start", it might help to break down everything that you know into small pieces:
1st: We're told that we have 11 consecutive integers. That means the 11 numbers are whole numbers that are in a row. If we can figure out ANY of the numbers AND it's place "in line", then we can figure out ALL of the other numbers and answer the question that's asked (the average of all 11 = ?)
2nd: Fact 1 tells us that the average of the FIRST 9 integers is 7. For just a moment, ignore the fact that there are 9 consecutive integers and let's just focus on the average = 7.
What would have to happen for a group of consecutive integers to have an average of 7?
Here are some examples:
7
6, 7, 8
5, 6, 7, 8, 9
Notice how there are the SAME number of terms below 7 as above 7. THAT'S a pattern.
With 9 total terms, that means there has to be 4 above and 4 below:
3, 4, 5, 6,.......7.......8, 9, 10, 11
Now we have enough information to figure out the other 2 terms (12 and 13) and answer the question. So Fact 1 is SUFFICIENT
With this same approach, we can deal with Fact 2.
The key to tackling most GMAT questions is to be comfortable breaking the prompt into logical pieces. Don't try to do every step at once and don't try to do work in your head. Think about what the information means, take the proper notes and be prepared to "play around" with a question if you're immediately certain about how to handle it.
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D | {
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[Reveal] Spoiler:
D
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Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
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Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
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29 Jun 2016, 03:10
wow such complex explanations for such a simple problem?
given :
11 consec integers
let them be x,x+1,x+2,...,x+10
Q: what is their mean?
mean is (11x+55)/11 = x+5.
Q becomes what is x+5
1) mean first 9 is 7.
so (9x+36)/9 = x+4 = 7 , so x+5 =8 ,--> sufficient A or D
2) mean of last 9 is 9.
so (9x+54)/9 = x+6= ---> x+5=8, sufficient . so D
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Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
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15 Jul 2016, 00:10
zz0vlb wrote:
What is the average (arithmetic mean) of eleven consecutive integers?
(1) The average of the first nine integers is 7
(2) The average of the last nine integers is 9 | {
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(1) The average of the first nine integers is 7
(2) The average of the last nine integers is 9
For odd number of consecutive integer the mean and median both is the middle value. Use this property to solve th question
(1) The average of the first nine integers is 7
7 will be the middle value; there will be 4 consecutive integers to the left and also to the right of 7
we will have {3,4,5,6,7,8,9,10,11}
now we can add last two consecutive integer after 11, they will be 12,13
our new set will become = {3,4,5,6,7,8,9,10,11,12,13}
again since the number of total elements in the set is odd, Mean will simply be the middle value = 8
SUFFICIENT
(2) The average of the last nine integers is 9
Again number of element in the set are odd, 9 will be the middle value; 4 consecutive integers will lie to its left and right
Middle value will be
{5,6,7,8,9,10,11,12,13}
Add 3,4 at the start of the set
new set = {3,4,5,6,7,8,9,10,11,12,13}
Mean will be 8
Sufficient
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Re: What is the average (arithmetic mean) of eleven consecutive [#permalink] 15 Jul 2016, 00:10
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## Phase Portrait Nonlinear System | {
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1) Find all equilibrium points by solving the system 2) Let a standard software (e. for the analysis of nonlinear systems; to introduce controller design methods for nonlinear systems. Specific topics include maps and flows in one and two dimensions, phase portraits, bifurcations, chaos, and fractals. Phase Portraits and Time Plots for Cases A (pplane6) Saddle Ex. 1), which was diagnosed using a set of four features extracted from the phase plane trajectory of the system to characterize the nonlinear response in the periodic regime. Local Phase Portrait of Nonlinear Systems Near Equilibria. “Proof”: Consider trajectory sufficiently close to origin time reversal symmetry. Save the phase portraits to submit on Gradescope. Keywords: nonlinear dynamics, chaos, electrical circuits. Biological Models: Predator-prey models, Competition models, Survival of one species, Co-existence, Alligators, doomsday and extinction. • As much as possible, piece the phase portraits of the linearized systems together to get an approximate phase portrait of the full non-linear system. A: The origins and evolution of predator-prey theory, Ecology 73, 1530-1535 (1992). Consider the following phase portraits of two two-dimensional linear dynamic sys-tem What can you say about the real parts of the two eigenvalues for both systems? What creates the di erence between the two phase portraits? Is the equilibrium point in phase portrait (b) an attractor? Exercise 3 This is exercise 3. Often, mathematical models of real-world phenomena are formulated in terms of systems of nonlinear differential equations, which can be difficult to solve explicitly. This diagram clearly illustrates for what values of r, the system exhibits chaotic and non-chaotic behavior. See phase portrait below. 1 In each problemfind the critical points and the corresponding linear system. Albu-Schaffer. The author starts off with an introduction to nonlinear systems, then moves on to phase portraits for 2-D systems, before moving | {
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introduction to nonlinear systems, then moves on to phase portraits for 2-D systems, before moving on to advanced concepts of stability theory and feedback linearization. Stable and unstable manifolds of equilibrium points and periodic orbits are important objects in phase portraits. In this paper, based on the classic Chua's circuit, a charge-controlled memristor is introduced to design a novel four-dimensional chaotic system. Introduction to systems of differential equations 2. (reductor and multipliers). While nonlinear systems of-ten require complex idiosyncractic treatments, phase potraitshaveevolved as apowerfultool forglobal anal-ysis ofthem. The system lives in a state space or phase. Weak non-linear oscillators and. Existence, uniqueness, and strong topological consequences for two-dimensions. For a much more sophisticated phase plane plotter, see the MATLAB plotter written by John C. Simmons, Differential Equations with Applications and Historical Notes, New York: McGraw-Hill, 1991. We also observe that the two xed points are progressively pushed apart in the amplitude direction. For a general 2 × 2 matrix A, the phase portrait will be equivalent to one of the four cases above, obtained by a linear transformation of coordinates (similarity transformation). Fixed points are stagnation points of the flow. The three examples will all be predator-prey models. 1 Concepts of Phase Plane Analysis 18 2. Consider the following phase portraits of two two-dimensional linear dynamic sys-tem What can you say about the real parts of the two eigenvalues for both systems? What creates the di erence between the two phase portraits? Is the equilibrium point in phase portrait (b) an attractor? Exercise 3 This is exercise 3. Click on the button corresponding to your preferred computer algebra system (CAS). 3 Symmetry in Phase Plane Portraits 22 2. Thus, a system has a limit cycle if and only if it has an isolated, closed. Numerical Construction of Phase Portraits. Two integral | {
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if and only if it has an isolated, closed. Numerical Construction of Phase Portraits. Two integral constraints on the amplitude and phase variation of the oscillations of an autonomous multi-degree of freedom system were obtained. 2 Draw the phase portraits of the following systems, using isoclines (a) 8+8+0. Generally, the nonlinear time series is analyzed by its phase space portrait. The book is very readable even though it has a lot of jargon (read heavy mathematics). A phase portrait is a graphical tool to visualize long term. Phase portrait generator. 5 0xy (7) which itself is a dynamical equation, the phase portrait is a trajectory along the switching line σ = 0. By plotting phase portrait on the computer, show that the system undergoes a Hopf bifurcation at 휇 = 0. shown to produce sharp phase portraits in long-term simulations, see e. Note: If you want a more traditional treatment of phase portraits, I recommend exploring Nonlinear Dynamics and Chaos by Strogatz. On this page I explain how to use Matlab to draw phase portraits for the the two linear systems. This video deals with. Existence, uniqueness, and strong topological consequences for two-dimensions. Simmons, Differential Equations with Applications and Historical Notes, New York: McGraw-Hill, 1991. The y nullcline is given by 3 4 1 4 3 y 2 3 x y = 0 (12) which gives the lines y = 0 or y = 3 4. , non-linear) 2 × 2 autonomous system discussed at the beginning of this chapter, in sections 1 and 2: x = f (x, y); (1) y = g(x, y). The nonlinear system's phase portrait near the fixed point is topologically unchanged due to small perturbations, and its dynamics are structurally stable or robust. Phase plane analysis for linear systems. 50= 1 (c) 8+82+0. m: A demonstration that plots the linearized phase portraits and the full phase plane. The nonlinear gyroscope model, which is employed in aerospace engineering [24], generally exhibits chaotic behavior. of problems that are described by nonlinear differential | {
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[24], generally exhibits chaotic behavior. of problems that are described by nonlinear differential equations. The complex dynamics of the novel chaotic system such as equilibrium points, stability, dissipation, bifurcation diagrams, Lyapunov exponent spectra and phase portraits are investigated. EECS 222 Nonlinear Systems: Analysis, Stability and Control Shankar Sastry 299 Cory Hall Tu-Th 11-12:30 pm. The book is very readable even though it has a lot of jargon (read heavy mathematics). Phase portraits of nonlinear systems: predator-prey, van der Pol (MATLAB examples). Nonlinear Models and Nonlinear Phenomena. However, these behaviors are not properly depicted in phase portraits when dealing with sys-tems that could be described as rotating systems,. The department offers project courses where you may choose/propose a project on topics related to Nonlinear Dynamical Systems. ) Lecture, three hours; discussion, one hour. A two-state phase portrait approach has been used to analyse vehicle dynamics and provides an illustrative view of the state trajectories at constant speed. Analyze the stability and its margins. by graphing and the use of phase portraits; D. First, let us look at the phase space portraits for a range of phase advances from 0:2 2ˇto 0:5 2ˇ. The phase portraits is able to perfectly capture all of the nonlinear trajectories and display them in a way that would be otherwise difficult. The phase portrait behavior of a system of ODEs can be determined by the eigenvalues or the trace and determinant (trace = λ 1 + λ 2, determinant = λ 1 x λ 2) of the system. In previous work, it was shown that bang-bang trajectories with low values of the energy integral are optimal for arbitrarily large times. The equation governing the dynamics of the nonlinear gyro, enriched with linear and nonlinear smoothening terms [24], is given by x_1 = x2. The author starts off with an introduction to nonlinear systems, then moves on to phase portraits for 2-D systems, before | {
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