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Math Help - die probability
1. die probability
"What is the probability of getting a total score of more than 7 when two fair dice are rolled together?"
I got $\frac{15}{36}$ but the book says $\frac{1}{3}$?
Thanks!
2. Hello, Nyoxis!
What is the probability of getting a total score of more than 7
when two fair dice are rolled together?"
I got $\frac{15}{36} \quad{\color{blue}\hdots\text{ Right!}}$
. . but the book says $\frac{1}{3}\quad {\color{red} \hdots\text{ Wrong!}}$
Can't imagine where they got their answer . . .
$\begin{array}{cccccc}(1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) \\ (2,1) & (2,2) & (2,3) & (2,4) & (2,5) & {\color{blue}(2,6)} \\ (3,1) & (3,2) & (3,3) & (3,4) & {\color{blue}(3,5)} & {\color{blue}(3,6)} \\ (4,1) & (4,2) & (4,3) & {\color{blue}(4,4)} & {\color{blue}(4,5)} & {\color{blue}(4,6)}\end{array}$
$\begin{array}{cccccc}(5,1) & (5,2) & {\color{blue}(5,3)} & {\color{blue}(5,4)} & {\color{blue}(5,5)} & {\color{blue}(5,6)} \\
(6,1) & {\color{blue}(6,2)} & {\color{blue}(6,3)} & {\color{blue}(6,4)} & {\color{blue}(6,5)} & {\color{blue}(6,6)}\end{array}\quad\Leftarrow\quad \text{fifteen outcomes}> 7$
3. Heh 'Cambridge Advanced Mathematics' wrong.
Thanks! | {
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# Prove that $x_{n+1}=\frac{x_n}{2}+\frac{1}{x_n}\geq\sqrt{2}$
Let $(x_n)_{n\in\mathbb N}$ be a recursively defined sequence with $x_1=9$ and $$x_{n+1}=\frac{x_n}{2}+\frac{1}{x_n}\text{ for }n\geq 1.$$ Show that $x_n\geq\sqrt{2}$ for all $n$.
Because $x_n\geq 0$ one can easily prove inductively that $$x_n^2\geq 2\Leftrightarrow x_n^2-2\geq 0\Leftrightarrow\left(\frac{x_{n-1}}{2}-\frac{1}{x_{n-1}}\right)^2\geq 0,$$ hence $x_n\geq\sqrt{2}$. However I have seen another approch which I was very curious about because I don't have the feeling that this can be done without proper justification other than the induction hypothesis:
$$x_{n+1}=\frac{x_n}{2}+\frac{1}{x_n}\overset{(*)}{\geq}\frac{\sqrt{2}}{2}+\frac{1}{\sqrt{2}}=\sqrt{2}$$
For $(*)$ it is assumed that $x_n\geq\sqrt{2}$ holds for an $n\in\mathbb N$. Are there any objections about this consideration?
• Hint: AM-GM inequality is fast. Your second approach is flawed, as the reciprocal term does not follow in the inequality * – Macavity Nov 2 '15 at 14:47
• The inequality $\frac{1}{x_n} \geq \frac{1}{\sqrt{2}}$ is false because it would imply that $x_n \leq \sqrt{2}$ (this is true only if $x_n=\sqrt{2}$ but else.... ) – user252450 Nov 2 '15 at 14:50
• A straightforward approach to showing that $x_n\ge \sqrt{2}$ is to write $$x_{n}-\sqrt{2}=\left(\frac{x_{n-1}}{2}+\frac1{x_{n-1}}\right)-\sqrt{2}=\frac{(x_{n-1}-\sqrt{2})^2}{2x_{n-1}}\ge0$$ – Mark Viola Nov 2 '15 at 15:00
• @Dr.MV By far the simplest. – Did Nov 2 '15 at 15:01
• @Dr.MV Yeah pretty similiar to my explained approach in the first part of my post. I was just wondering about the other approach and its correctness there. – Christian Ivicevic Nov 2 '15 at 15:08
Yes, you can make the assumption $x_n\ge \sqrt{2}$ to prove $x_{n+1}\ge \sqrt{2}$ as per strong form of mathematical induction.
But your reasoning is wrong as $x_n\ge \sqrt{2} \Rightarrow \frac{1}{x_n} \le \frac{1}{\sqrt{2}}$. | {
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But your reasoning is wrong as $x_n\ge \sqrt{2} \Rightarrow \frac{1}{x_n} \le \frac{1}{\sqrt{2}}$.
Better is to use AM-GM inequality as Macavity has mentioned since the quantities are all positive.
$$\frac{\frac{x_n}{2}+\frac{1}{x_n}}{2}\ge \sqrt{\frac{x_n}{2}\cdot\frac{1}{x_n}}$$ $$x_{n+1}\ge \sqrt{2}$$
If you know derivatives you can make it work as follows. Consider the function : $$f(x)=\frac{x^2+2}{2x}$$ Now calculate the derivative : $$f'(x)=\frac{2x^2-4}{4x^2}\geq 0$$ if $x\geq \sqrt{2}$ so the function is increasing for $x\geq \sqrt{2}$
Now from the induction hypothesis $x_n\geq \sqrt{2}$ so from the monotony : $$x_{n+1}=f(x_n)\geq(\sqrt{2})=\frac{4}{2\sqrt{2}}=\sqrt{2}$$ as wanted .
Although the following goes a bit beyond addressing the OP's question, I thought that it would be instructive. So, let's have a look at the sequence generated by the recurrence relationship $x_n=\frac12 x_{n-1}+\frac{1}{x_{n-1}}$.
This is actually an interesting recursively generated sequence. If we examine the first difference $x_{n}-x_{n-1}$ we find that
$$x_{n}-x_{n-1}=\frac{2-x_{n-1}^2}{2x_{n-1}} \tag 1$$
We deduce immediately from $(1)$ that if $x_n > \sqrt{2}$, then $x_n$ is decreasing while if $x_n<\sqrt{2}$, then $x_n$ is increasing.
Now, a straightforward approach to showing that $x_n\ge \sqrt{2}$ is to write
$$x_{n}-\sqrt{2}=\left(\frac{x_{n-1}}{2}+\frac1{x_{n-1}}\right)-\sqrt{2}=\frac{(x_{n-1}-\sqrt{2})^2}{2x_{n-1}}\ge0 \tag 2$$
regardless of the value of $x_{n-1}$.
Therefore, the only value of the sequence that can ever be less than $\sqrt{2}$ is its initial "seed" value. From $(1)$, we see that if the seed value is less than $\sqrt{2}$, the subsequent value increases. And $(2)$ shows that it increases above $\sqrt{2}$. Then, the remaining sequence decreases monotonically to the limit value of $\sqrt{2}$.
Hint: study the function $f(x)=x/2 +1/x$ it increases if $x>\sqrt2$ | {
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# Contour Integration (Choice of Contour)
Let $\alpha \le 0$ and $\sigma > 0$ .
I want to choose a contour, including $[\sigma - iR, \sigma+iR]$ , such that i can apply Cauchy's Residue theorem and evaluate:
$$\lim_{R \rightarrow \infty} \int\limits_{\sigma - iR}^{ \sigma+iR } \frac{\exp(\alpha z)}{(z^2 + 1)} dz$$
by contour integration. The case for $\alpha > 0$ is nice i think, as you can use a semi-circle, but i cant seem to find a suitable contour for this case.
EDIT: to be specific when we use a semi-circle then the integral over the circular path tends to zero in the limit for $\alpha > 0$ , but this does not seem to happen in the above case.
• How do you expect to apply CRT to a complex integral on a straight line in the complex plane?? We usually do this to calculate real integrals by means of a closed contour in the complex plane... – DonAntonio Dec 28 '13 at 15:36
• see my edit, in the limit the only non-zero section of our contour integral is that integrated over the line. it worked for the earlier case, unless i am completely confusesd. – KJC Dec 28 '13 at 15:38
You can choose a semicircle in the half-plane $\operatorname{Re} z \geqslant \sigma$. Then $e^{\alpha z}$ is bounded on the contour - $\operatorname{Re} (\alpha z) = \alpha\operatorname{Re} z \leqslant \alpha\sigma \leqslant 0$, and $\lvert e^{\alpha z}\rvert = e^{\operatorname{Re} (\alpha z)}$ - and the integral over the semicircle tends to $0$ for $R \to \infty$. Since the contour encloses no singularity, the contour integral is $0$, and hence
$$\int_{\sigma - i\infty}^{\sigma+i\infty} \frac{e^{\alpha z}}{z^2+1}\,dz = 0$$
for $\sigma > 0$ and $\alpha \leqslant 0$.
It is worth noting that
$$I(\alpha) = \int_{\sigma-i\infty}^{\sigma+i\infty} \frac{e^{\alpha z}}{z^2+1}\,dz = \begin{cases}\quad 0 &, \alpha \leqslant 0\\ 2\pi i\sin \alpha &, \alpha \geqslant 0 \end{cases}$$ | {
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does not depend smoothly on $\alpha$, although the integrand does. That should, however, not be too surprising since the differentiated integrand $\dfrac{ze^{\alpha z}}{z^2+1}$ is not integrable anymore, and the non-differentiability in $0$ manifests itself by the need to switch the half-plane in which the contour is closed to have the integral over the auxiliary part tend to $0$.
• could you show that $e^{ \alpha z}$ is bounded on this contour? im not seeing that. – KJC Dec 28 '13 at 15:50
• Added the estimate for that. – Daniel Fischer Dec 28 '13 at 15:58
• excellent explanation – KJC Dec 28 '13 at 22:52 | {
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Prove that: $\lfloor n^{1/2}\rfloor+\cdots+\lfloor n^{1/n}\rfloor=\lfloor \log_2n\rfloor +\cdots+\lfloor \log_nn \rfloor$, for $n > 1$
Prove that: $\lfloor n^{1/2}\rfloor+\cdots+\lfloor n^{1/n}\rfloor=\lfloor \log_2n\rfloor +\cdots+\lfloor \log_nn \rfloor$, for $n > 1,\, n\in \mathbb{N}$
For example. For $n=2$, we have $\lfloor 2^{1/2} \rfloor = \lfloor 1.414 \rfloor = 1$ whereas $\lfloor \log_2(2) \rfloor = 1$ while for $n=3$, we have $$\lfloor 3^{1/2} \rfloor + \lfloor 3^{1/3} \rfloor = \lfloor 1.732 \rfloor + \lfloor 1.442 \rfloor = 2= \lfloor 1.585 \rfloor + \lfloor 1 \rfloor=\lfloor \log_2(3) \rfloor + \lfloor \log_3(3) \rfloor .$$
I was thinking of using induction. So since $n=2$ is true, now assume for all $n$, this identity is true, we would like to prove that $n+1$ is true. Then
$$\lfloor n^{1/2} \rfloor + \lfloor n^{1/3} \rfloor + ... + \lfloor n^{1/n} \rfloor + \lfloor (n+1)^{1/(n+1)} \rfloor,$$ where $(n+1)^{1/(n+1)} > 1$ for all $n>1$ but it's strictly decreasing above 1 so $\lfloor (n+1)^{1/(n+1)} \rfloor = 1$
$\implies \lfloor n^{1/2} \rfloor + \lfloor n^{1/3} \rfloor +\cdots+ \lfloor n^{1/n} \rfloor + \lfloor (n+1)^{1/(n+1)} \rfloor = \lfloor n^{1/2} \rfloor + \lfloor n^{1/3} \rfloor +\cdots+ \lfloor n^{1/n} \rfloor + 1$
$= \lfloor \log_2(n) \rfloor + \lfloor \log_3(n) \rfloor + \cdots+ \lfloor \log_n(n) \rfloor + \lfloor \log_{n+1}(n+1) \rfloor$
since, $\log_{n+1}(n+1) = 1$ for all $n$.
My question is: How do we know that $(n+1)^{1/(n+1)}$ will never go below $1$? i.e., How can we prove that this function $f(x) = (x+1)^{1/(x+1)}$ is always bounded below by $1$ for $x>1$? (First, When $x=0$, $f(0)=1$, then looking at it's derivative, one can see that it's strictly increasing for $x$ between $(0,1)$ and decreasing for all $x>1$). | {
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• You need to replace $n$ by $n + 1$ all over, not just in the last term, in the induction step. – vonbrand Feb 25 '14 at 21:57
• If $\llcorner x\lrcorner$ is supposed to mean the greatest integer not exceeding $x$, then you should probably have used $\lfloor x \rfloor$ (\lfloor … \rfloor) instead. If that's not what you mean, then what did you mean? – MJD Feb 25 '14 at 22:02
• thanks @MJD, I wasn't sure which code to use for it. Thanks! – PandaMan Feb 26 '14 at 1:37
• @vonbrand, I don't quite understand why you meant, please clarify. thanks – PandaMan Feb 26 '14 at 2:18
• @PandaMan, what you need to prove is $\lfloor (n+1)^{1/2}\rfloor + \ldots + \lfloor (n+1)^{1/(n+1)}\rfloor = \lfloor \log_2 (n+1)\rfloor +\ldots+\lfloor\log_{n+1} (n+1)\rfloor$ – vonbrand Feb 26 '14 at 5:04
This is a classic exercise and one with a very elegant solution.
The idea of the proof is to count the number $N$ of the points (see figure below) with integer coordinates, which lie in the region $$U=\big\{(x,y): 0<x\le n \,\,\,\text{and}\,\,\, 1<y\le n^{1/x}\big\},$$ and in particular, the red points, in two ways: horizontally and vertically.
Horizontal counting: $$N=\lfloor n^{1/2}\rfloor+\lfloor n^{1/3}\rfloor+\cdots+\lfloor n^{1/n}\rfloor,$$ since on the horizontal line $\,y=k\,$ lie exactly $\,\lfloor n^{1/k}\rfloor\,$ red points.
Vertical counting: $$N=\lfloor \log_2 n\rfloor+\lfloor\log_3 n\rfloor+\cdots+\lfloor \log_n n\rfloor,$$ since on the vertical line $\,x=k\,$ lie exactly $\,\lfloor \log_k n\rfloor\,$ red points.
$${}$$
Note that the curve in the figure above is of the function $y=n^{1/x}$.
This problem was first asked in a Soviet Mathematics Olympiad in 1982 (Всесоюзный Математический Олимпиад.) | {
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• Gorgeous explanation! – Steven Stadnicki Feb 25 '14 at 22:24
• @StevenStadnicki: Correct. – Yiorgos S. Smyrlis Feb 25 '14 at 22:26
• (Thanks for the fix! I've removed the correction from my comment since that got edited in. :-) – Steven Stadnicki Feb 25 '14 at 22:28
• Great explanation. The wires are crossed a little though, because if $x=k$, then $y\leq n^{1/k}$, which means that the vertical count is the one that matches the fractional powers. Likewise, if $y = k$, then $k \leq n^{1/x}$, then $1\leq(\log_k n)/x$, so the horizontal count is the one that matches the logs. – John Moeller Mar 1 '14 at 0:13
• Awesome solution :-) – tarit goswami Aug 1 '18 at 8:54
Fix $b>1$. Then the derivative of $b^x$ is $\ln(b) b^x$; $\ln(b)$ is positive and $b^x$ is as well for all $x$, showing that that $b^x$ is a strictly increasing function. Next, $b^0=1$, showing that $b^x>1$ for all $x>0$.
Next, since $n+1>1$ and $1/(n+1)>0$, we have that $(n+1)^{1/(n+1)}>1$. | {
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# Can a complete manifold have an uncountable number of ends?
Let $M$ be a complete and noncompact Riemannian manifold. Fix a point $p$ in $M$. Let $\gamma$: $[0, L]\rightarrow M$ (parametrized by its arc length) be a geodesic starting from $p$. Denote by $d(\cdot, \cdot)$ the distance function on $M$ induced by the Riemannian metric. $\gamma$ is minimal if $d(\gamma(s_1), \gamma(s_2))=|s_1-s_2|$. A ray is a minimal geodesic defined on $[0, +\infty)$. Since $M$ is noncompact, there exists at least one ray from $p$.
Two rays $\gamma_1, \gamma_2$ from the same point $p$ are called cofinal if for any $r\geq0$ and all $s>r$, $\gamma_1(s)$ and $\gamma_2(s)$ lie in the same component of $M\backslash B(0, r)$, where $B(0, r)=\{x\in M| d(p, x)<r\}$. An equivalence class of cofinal rays is called an end of $M$. My question is:
Can $M$ have uncountably many ends? It seems that the answer is no. But I am not very sure and I can not find a convictive proof. | {
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• You might like to look at some basic notions of geometric group theory. The set of ends is a quasi-isometry invariant, and the Svarc--Milnor Lemma says that the universal cover of a compact manifold $M$ is quasi-isometric to a Cayley graph of $\pi_1M$. So it suffices to find a group with uncountably many ends, and the free group of rank 2 suffices. Curiously, it's actually impossible for a universal cover to have countably infinitely many ends.
– HJRW
Dec 18, 2015 at 20:35
• @FanZheng: Second countable spaces can have infinitely many ends. Take the universal cover of a wedge of two circles, for example. Dec 18, 2015 at 20:44
• @HJRW: The ends of the fundamental group aren't all that relevant to this question, although simply connected examples can be interesting. It's about whether $M$ has uncountably many ends (as in the answer of @SebastianGoette), not the universal cover of $M$. Dec 18, 2015 at 20:49
• @LeeMosher, I used bad notation, but I think my point stands (unless I misread the question). If $\pi_1N$ is free of rank 2 (say) then taking $M$ to be the universal cover of $N$ provides the example.
– HJRW
Dec 18, 2015 at 21:08
• @HJRW: Sure, for example $N=$ the connected sum of two copies of $S^2 \times S^1$. Dec 18, 2015 at 21:19
The answer is yes. Consider a hyperbolic pair of pants where all three boundary circles are of the same size. Glue countably many of them together such each new one is glued to the existing manifold along exactly one boundary circle. Then the manifold looks like the boundary of a fattened tree.
Fix $p$ in one of the pairs of pants $Y_0$. For all other pairs of pants $Y$, pick a point $q\in Y$, then the minimal geodesic joining $p$ and $q$ will leave $Y$ through one circle. Label the two remaining boundary circles with $0$ and $1$. Then all glueing circles get exactly one label, except for those bounding $Y_0$. | {
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For each sequence $(a_n)_n\in\{0,1\}^{\mathbb N}$, we construct a ray from $p$ leaving $Y_0$ through the same boundary component. Whenever the ray enters the $n$-th pair of pants $Y$ along its way, we let it leave through the circle labelled $a_n$. This way, we construct an uncountable number of rays that are not pairwise cofinal.
• Thank you very much for your nice edit of my question and also for your clever solution. In this afternoon, I also found a very similar example called as the Cantor tree surface, inspired by your construction. Dec 19, 2015 at 10:34
Let me rephrase S. Goette's example in a slightly different way. Consider the complement of the triadic Cantor set in the complex plane. This is a Riemann surface that can be uniformized: there is a complete Riemannian metric of constant minus one curvature. The ends of this surface are in one-to-one correspondence with the points of the Cantor set.
You can see the decomposition into pairs of pants by drawing nested circles around the intervals that appear in the construction of the Cantor set.
In some sense, uncountably many ends is the generic case. Let us restrict ourselves to planar surfaces, namely complements of compact sets in the Riemann sphere. The set of compact subsets in a compact set has a natural topology, the Vietoris topology (which coincides with the topology given by the Hausdorff metric). So we can say that two open subsets are close if their complement are close with respect to this topology. It happens that generically a compact subset of the sphere has uncountably many connected components, hence its complement has uncountably many ends. | {
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Question
# I have a total of Rs. $$300$$ in coins of denomination Re. $$1$$, Rs. $$2$$ and Rs. $$5$$. The number of Rs. $$2$$ coins is $$3$$ times the number of Rs. $$5$$ coins. The total number of coins is $$160$$. How many coins of each denomination are with me?
A
Number of Re. 1 coins =50; Number of Rs. 2 coins =40; Number of Rs. 5 coins =40
B
Number of Re. 1 coins =70; Number of Rs. 2 coins =70; Number of Rs. 5 coins =19
C
Number of Re. 1 coins =80; Number of Rs. 2 coins =60; Number of Rs. 5 coins =20
D
Number of Re. 1 coins =115; Number of Rs. 2 coins =50; Number of Rs. 5 coins =15
Solution
## The correct option is C Number of Re. $$1$$ coins $$= 80$$; Number of Rs. $$2$$ coins $$= 60$$; Number of Rs. $$5$$ coins $$= 20$$Given, Total number of coins $$= 160$$Given that the number of Rs.$$2$$ coins are $$3$$ times the number of Rs. $$5$$ coins.Number of Rs. $$5$$ coins $$= x$$Number of Rs. $$2$$ coins $$= 3x$$Number of Re. $$1$$ coins $$= 160 -(x+3x) = 160 -4x$$Total value of coins $$=$$ Rs. $$300$$So, $$5\times x + 2\times 3x + 1\times (160 -4x) = 300$$$$5x +6x +160 -4x = 300$$$$7x = 140$$$$x = 20$$So, number of Rs. $$5$$ coins $$= 20$$Number of Rs. $$2$$ coins $$= 60$$Number of Re. $$1$$ coins $$=80$$Mathematics
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So, total numbers of onto functions from X to Y are 6 (F3 to F8). Functions can be classified according to their images and pre-images relationships. Then every function from A to B is effectively a 5-digit binary number. (d) x2 +1 x2 +2. Don’t stop learning now. I already know the formula (summation r=1 to n)(-1)^(n-r)nCr(r^m). The number of functions from {0,1}4 (16 elements) to {0, 1} (2 elements) are 216. By using our site, you This is same as saying that B is the range of f . Solution: As given in the question, S denotes the set of all functions f: {0, 1}4 → {0, 1}. Which must also be bijective, and therefore onto. From the formula for the number of onto functions, find a formula for S(n, k) which is defined in Problem 12 of Section 1.4. Option 3) 200. Thus, the number of onto functions = 16−2= 14. So, that leaves 30. (e) f(m;n) = m n. Onto. Then Total no. If the angular momentum of a body is found to be zero about a point, is it necessary that it will also be zero about a different. But we want surjective functions. Mathematics | Total number of possible functions, Mathematics | Unimodal functions and Bimodal functions, Total Recursive Functions and Partial Recursive Functions in Automata, Mathematics | Classes (Injective, surjective, Bijective) of Functions, Mathematics | Generating Functions - Set 2, Inverse functions and composition of functions, Last Minute Notes - Engineering Mathematics, Mathematics | Introduction to Propositional Logic | Set 1, Mathematics | Predicates and Quantifiers | Set 1, Mathematics | L U Decomposition of a System of Linear Equations, Mathematics | Mean, Variance and Standard Deviation, Mathematics | Sum of squares of even and odd natural numbers, Mathematics | Eigen Values and Eigen Vectors, Mathematics | Lagrange's Mean Value Theorem, Mathematics | Introduction and types of Relations, Data Structures and Algorithms – Self Paced Course, We use cookies to ensure you have the best browsing experience on our website. Find the number | {
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We use cookies to ensure you have the best browsing experience on our website. Find the number of relations from A to B. Here are the definitions: is one-to-one (injective) if maps every element of to a unique element in . In the example of functions from X = {a, b, c} to Y = {4, 5}, F1 and F2 given in Table 1 are not onto. Not onto. No. Home. In other words no element of are mapped to by two or more elements of . there are zero onto function . Consider the function x → f(x) = y with the domain A and co-domain B. Determine whether each of these functions is a bijection from R to R. (a) f(x) = 2x+1. Suppose TNOF be the total number of onto functions feasible from A to B, so our aim is to calculate the integer value TNOF. [5.1] Informally, a function from A to B is a rule which assigns to each element a of A a unique element f(a) of B. Officially, we have Definition. In other words, nothing is left out. We need to count the number of partitions of A into m blocks. Examples: Let us discuss gate questions based on this: Solution: As W = X x Y is given, number of elements in W is xy. Comparing cardinalities of sets using functions. The number of functions from Z (set of z elements) to E (set of 2xy elements) is 2xyz. Solution: Using m = 4 and n = 3, the number of onto functions is: I am trying to get the total number of onto functions from set A to set B if the former has m elements and latter has n elements with m>n. (B) 64 Math Forums. Yes. If X has m elements and Y has 2 elements, the number of onto functions will be 2 m-2. Example 9 Let A = {1, 2} and B = {3, 4}. For example, if n = 3 and m = 2, the partitions of elements a, b, and c of A into 2 blocks are: ab,c; ac,b… Functions: One-One/Many-One/Into/Onto . Out of these functions, the functions which are not onto are f (x) = 1, ∀x ∈ A. In this article, we are discussing how to find number of functions from one set to another. Therefore, S has 216 elements. For example: X = {a, b, c} and Y = {4, 5}. An onto function is | {
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Therefore, S has 216 elements. For example: X = {a, b, c} and Y = {4, 5}. An onto function is also called surjective function. An onto function is also called surjective function. In this case the map is also called a one-to-one correspondence. Here are the definitions: 1. is one-to-one (injective) if maps every element of to a unique element in . We need to count the number of partitions of A into m blocks. In other words, if each b ∈ B there exists at least one a ∈ A such that. 2. is onto (surjective)if every element of is mapped to by some element of . An onto function is also called a surjective function. Let A = {a 1, a 2, a 3} and B = {b 1, b 2} then f : A -> B. It means that every element “b” in the codomain B, there is exactly one element “a” in the domain A. such that f(a) = b. . Therefore, each element of X has ‘n’ elements to be chosen from. In a function from X to Y, every element of X must be mapped to an element of Y. That is, a function f is onto if for each b ∊ B, there is atleast one element a ∊ A, such that f(a) = b. 2×2×2×2 = 16. So, you can now extend your counting of functions … 19. Tuesday: Functions as relations, one to one and onto functions What is a function? 34 – 3C1(2)4 + 3C214 = 36. So, number of onto functions is 2m-2. Therefore, N has 2216 elements. 2. Example 46 (Method 1) Find the number of all one-one functions from set A = {1, 2, 3} to itself. One-to-One/Onto Functions . Option 2) 120. So the total number of onto functions is m!. Calculating required value. Some authors use "one-to-one" as a synonym for "injective" rather than "bijective". Misc 10 (Introduction)Find the number of all onto functions from the set {1, 2, 3, … , n} to itself.Taking set {1, 2, 3}Since f is onto, all elements of {1, 2, 3} have unique pre-image.Total number of one-one function = 3 × 2 × 1 = 6Misc 10Find the number of all onto functio 1.1. . If m < n, the number of onto functions is 0 as it is not possible to use all elements of Y. 4 = A B Not a function | {
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number of onto functions is 0 as it is not possible to use all elements of Y. 4 = A B Not a function Notation We write f (a) = b when (a;b) 2f where f is a function. Math Forums. therefore the total number of functions from A to B is. In a one-to-one function, given any y there is only one x that can be paired with the given y. If n(A)= 3 , n(B)= 5 Find the number of onto function from A to B, For onto function n(A) n(B) otherwise ; it will always be an inoto function. Set A has 3 elements and set B has 4 elements. A function f from A to B is a subset of A×B such that • for each a ∈ A there is a b ∈ B with (a,b… Experience. We say that b is the image of a under f , and a is a preimage of b. October 31, 2007 1 / 7. Here's another way to look at it: imagine that B is the set {0, 1}. But, if the function is onto, then you cannot have 00000 or 11111. Q3. Let E be the set of all subsets of W. The number of functions from Z to E is: If X has m elements and Y has 2 elements, the number of onto functions will be 2. There are $$\displaystyle 3^8=6561$$ functions total. If anyone has any other proof of this, that would work as well. Explanation: From a set of m elements to a set of 2 elements, the total number of functions is 2m. where as when i try manually it comes 8 . De nition 1 A function or a mapping from A to B, denoted by f : A !B is a Such functions are referred to as injective. (C) 81 4. Writing code in comment? Please use ide.geeksforgeeks.org, For understanding the basics of functions, you can refer this: Classes (Injective, surjective, Bijective) of Functions. There are 3 ways of choosing each of the 5 elements = $3^5$ functions. A function is said to be bijective or bijection, if a function f: A → B satisfies both the injective (one-to-one function) and surjective function (onto function) properties. 38. Considering all possibilities of mapping elements of X to elements of Y, the set of functions can be represented in Table 1. Transcript. f(a) = b, then f is an | {
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of Y, the set of functions can be represented in Table 1. Transcript. f(a) = b, then f is an on-to function. Q1. Why does an ordinary electric fan give comfort in summer even though it cannot cool the air? Students can solve NCERT Class 12 Maths Relations and Functions MCQs Pdf with Answers to know their preparation level. (D) 72. I just need to know how it came. Since f is one-one Hence every element 1, 2, 3 has either of image 1, 2, 3 and that image is unique Total number of one-one function = 6 Example 46 (Method 2) Find the number of all one-one functions from set A = {1, 2, 3} to itself. In the above figure, f … So, total numbers of onto functions from X to Y are 6 (F3 to F8). Example 46 (Method 1) Find the number of all one-one functions from set A = {1, 2, 3} to itself. In F1, element 5 of set Y is unused and element 4 is unused in function F2. Discrete Mathematics Grinshpan Partitions: an example How many onto functions from f1;2;3;4;5;6;7;8g to fA;B;C;Dg are there? There are 3 functions with 1 element in range. How many onto functions are there from a set with eight elements to a set with 3 elements? Onto Function A function f: A -> B is called an onto function if the range of f is B. Copyright © 2021 Pathfinder Publishing Pvt Ltd. To keep connected with us please login with your personal information by phone/email and password. (c) f(m;n) = m. Onto. Need explanation for: If n(A)= 3 , n(B)= 5 Find the number of onto function from A to B, List of Hospitality & Tourism Colleges in India, Knockout JEE Main May 2022 (Easy Installments), Knockout JEE Main May 2021 (Easy Installments), Knockout NEET May 2021 (Easy Installments), Knockout NEET May 2022 (Easy Installments), Top Medical Colleges in India accepting NEET Score, MHCET Law ( 5 Year L.L.B) College Predictor, List of Media & Journalism Colleges in India, B. Transcript. Let f be the function from R … Formula for finding number of relations is Number of relations = 2 Number of elements of A × Number of | {
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for finding number of relations is Number of relations = 2 Number of elements of A × Number of elements of B Attention reader! In other words, if each b ∈ B there exists at least one a ∈ A such that. Why does a tightly closed metal lid of a glass bottle can be opened more easily if it is put in hot water for some time? Click hereto get an answer to your question ️ Write the total number of one - one functions from set A = { 1,2,3,4 } to set B = { a,b,c } . Onto Function Definition (Surjective Function) Onto function could be explained by considering two sets, Set A and Set B, which consist of elements. Yes. (i)When all the elements of A will map to a only, then b is left which do not have any pre-image in A (ii)When all the elements of A will map to b only, then a is left which do not have only pre-image in A Thus in both cases, function is not onto So, total number of onto functions= 2^n-2 Hope it helps☑ #Be Brainly Onto Functions: Consider the function {eq}y = f(x) {/eq} from {eq}A \to B {/eq}, where {eq}A {/eq} is the domain of the function and {eq}B {/eq} is the codomain. For example, if n = 3 and m = 2, the partitions of elements a, b, and c of A into 2 blocks are: ab,c; ac,b; bc,a. If n > m, there is no simple closed formula that describes the number of onto functions. according to you what should be the anwer So the correct option is (D). As E is the set of all subsets of W, number of elements in E is 2xy. Also, given, N denotes the number of function from S(216 elements) to {0, 1}(2 elements). Let X, Y, Z be sets of sizes x, y and z respectively. An exhaustive E-learning program for the complete preparation of JEE Main.. Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test.. In F1, element 5 of set Y is unused and element 4 is unused in function F2. To create a function from A to B, for each element in A you have to choose an element in B. acknowledge that you have read and understood our, GATE CS Original | {
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Distributions Set 2 (Exponential Distribution), Mathematics | Probability Distributions Set 3 (Normal Distribution), Mathematics | Probability Distributions Set 4 (Binomial Distribution), Mathematics | Probability Distributions Set 5 (Poisson Distribution), Mathematics | Hypergeometric Distribution model, Mathematics | Limits, Continuity and Differentiability, Mathematics | Lagrange’s Mean Value Theorem, Mathematics | Problems On Permutations | Set 1, Problem on permutations and combinations | Set 2, Mathematics | Graph theory practice questions, Classes (Injective, surjective, Bijective) of Functions, Difference between Spline, B-Spline and Bezier Curves, Runge-Kutta 2nd order method to solve Differential equations, Write Interview Into Function : Function f from set A to set B is Into function if at least set B has a element which is not connected with any of the element of set A. Option 1) 150. There are $$\displaystyle 2^8-2$$ functions with 2 elements in the range for each pair of elements in the codomain. Onto Function A function f: A -> B is called an onto function if the range of f is B. (b) f(m;n) = m2 +n2. If n > m, there is no simple closed formula that describes the number of onto functions. These numbers are called Stirling numbers (of the second kind). Maths MCQs for Class 12 Chapter Wise with Answers PDF Download was Prepared Based on Latest Exam Pattern. A function f : A -> B is said to be an onto function if every element in B has a pre-image in A. The onto function from Y to X is F's inverse. Therefore, total number of functions will be n×n×n.. m times = nm. This course will help student to be better prepared and study in the right direction for JEE Main.. No. Menu. They are various types of functions like one to one function, onto function, many to one function, etc. 3. (d) f(m;n) = jnj. This disagreement is confusing, but we're stuck with it. set a={a,b,c} and B={m,n} the number of onto functions by your formula is 6 . If X has m elements and Y | {
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a={a,b,c} and B={m,n} the number of onto functions by your formula is 6 . If X has m elements and Y has n elements, the number if onto functions are. f(a) = b, then f is an on-to function. Get hold of all the important CS Theory concepts for SDE interviews with the CS Theory Course at a student-friendly price and become industry ready. 2.1. . No element of B is the image of more than one element in A. The total no.of onto function from the set {a,b,c,d,e,f} to the set {1,2,3} is????? Onto function or Surjective function : Function f from set A to set B is onto function if each element of set B is connected with set of A elements. (b) f(x) = x2 +1. 3. is one-to-one onto (bijective) if it is both one-to-one and onto. Suppose TNOF be the total number of onto functions feasible from A to B, so our aim is to calculate the integer value TNOF. Option 4) none of these Number of onto functions from one set to another – In onto function from X to Y, all the elements of Y must be used. High School Math Elementary Math Algebra Geometry Trigonometry Probability and Statistics Pre-Calculus. The number of injections that can be defined from A to B is: The number of onto functions (surjective functions) from set X = {1, 2, 3, 4} to set Y = {a, b, c} is: 3. A function f from A to B is called one-to-one (or 1-1) if whenever f (a) = f (b) then a = b. generate link and share the link here. (c) f(x) = x3. So the total number of onto functions is m!. For function f: A→B to be onto, the inequality │A│≥2 must hold, since no onto function can be designed from a set with cardinality less than 2 where 2 is the cardinality of set B. Number of functions from one set to another: Let X and Y are two sets having m and n elements respectively. Out of these functions, 2 functions are not onto (If all elements are mapped to 1st element of Y or all elements are mapped to 2nd element of Y). Let f and g be real functions defined by f(x) = 2x+ 1 and g(x) = 4x – 7. asked Feb 16, 2018 in Class XI Maths by | {
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be real functions defined by f(x) = 2x+ 1 and g(x) = 4x – 7. asked Feb 16, 2018 in Class XI Maths by rahul152 ( -2,838 points) relations and functions Since f is one-one Hence every element 1, 2, 3 has either of image 1, 2, 3 and that image is unique Total number of one-one function = 6 Example 46 (Method 2) Find the number Steps 1. (b)-Given that, A = {1 , 2, 3, n} and B = {a, b} If function is subjective then its range must be set B = {a, b} Now number of onto functions = Number of ways 'n' distinct objects can be distributed in two boxes a' and b' in such a way that no box remains empty. Let A = {a 1, a 2, a 3} and B = {b 1, b 2} then f : A -> B. P.S. A function has many types which define the relationship between two sets in a different pattern. (A) 36 In the example of functions from X = {a, b, c} to Y = {4, 5}, F1 and F2 given in Table 1 are not onto. Any ideas on how it came? Let W = X x Y. Free PDF Download of CBSE Maths Multiple Choice Questions for Class 12 with Answers Chapter 1 Relations and Functions. In other words no element of are mapped to by two or more elements of . Proving that a given function is one-to-one/onto. One more question. Not onto. So, there are 32 = 2^5. Learn All Concepts of Chapter 2 Class 11 Relations and Function - FREE. Tech Companion - A Complete pack to prepare for Engineering admissions, MBBS Companion - For NEET preparation and admission process, QnA - Get answers from students and experts, List of Pharmacy Colleges in India accepting GPAT, Why does a tightly closed metal lid of a glass bottle can be opened more easily if it is put in hot water for some time? Number of Onto function - & Number of onto functions - For onto function n(A) n(B) otherwise ; it will always be an inoto function . A function from X to Y can be represented in Figure 1. of onto function from A to A for which f(1) = 2, is. Check - Relation and Function Class 11 - All Concepts. For function f: A→B to be onto, the inequality │A│≥2 must hold, since no onto | {
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11 - All Concepts. For function f: A→B to be onto, the inequality │A│≥2 must hold, since no onto function can be designed from a set with cardinality less than 2 where 2 is the cardinality of set B. My book says it is the coefficient of x^m in m!(e^x-1)^n. For example: X = { a, B, for each element of are to! 'Re stuck with it is 2xy binary number Relation and function - FREE in Figure.... Even though it can not have 00000 or 11111 article, we are discussing how to find number onto! Partitions of a into m blocks both one-to-one and onto elements and Y 6. Is an on-to function 2 m-2: a - > B is called onto. M, there is no simple closed formula that describes the number of functions one! Into m blocks: a - > B is called an onto function a function from to... Concepts of Chapter 2 Class 11 Relations and functions MCQs PDF with Answers PDF of... injective '' rather than bijective '' in B will be..! The function is also called a one-to-one function, many to one function, given any there. In B possibilities of mapping elements of Download of total no of onto functions from a to b Maths Multiple Choice Questions for Class 12 with to. To B, c } and Y has 2 elements in the codomain tuesday functions. Coefficient of x^m in m! \displaystyle 3^8=6561\ total no of onto functions from a to b functions with 2 elements, the number of functions from set... Y are 6 ( F3 to F8 ) be 2 m-2 function has many types define... Bijection from R … total no of onto functions from a to b was Prepared Based on Latest Exam pattern it can not cool the air,. If X has m elements and Y are two sets having m total no of onto functions from a to b elements. If the range of f simple closed formula that describes the number of onto functions is 2m are types..., there is only one X that can be classified according to their and. Pathfinder Publishing Pvt Ltd. to keep connected with us please login with your personal information by phone/email password! Eight elements to be chosen from map total no of onto | {
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personal information by phone/email password! Eight elements to be chosen from map total no of onto functions from a to b also called a one-to-one function, etc was Prepared Based Latest. R=1 to n ) = 2x+1 as it is both one-to-one and onto functions 2m... Ordinary electric fan give comfort in summer even though it can not have 00000 or 11111 such! Says it is not possible to use all elements of X has m elements and Y = {,. As a synonym for injective '' rather than bijective '' such that of 2xy elements ) E. ( injective ) if maps every element of we need to count the number of Relations from to... Of this, that would work as well every element of are to! [ Math ] 3^5 [ /math ] functions Figure 1 the formula ( summation r=1 n! As a synonym for injective '' rather than bijective '' ( set of all of... Are the definitions: is one-to-one onto ( surjective ) if every element of is mapped by. 3^8=6561\ ) functions with 2 elements in E is 2xy into m blocks considering all of... Pvt Ltd. to keep connected with us please login with your personal information phone/email! Is 2xy be better Prepared and study in the codomain, every element of,!! ( e^x-1 ) ^n closed formula that describes the number of onto.... Maths Relations and function - FREE Ltd. to keep connected with us please login with your information. No element of to a set with eight elements to a set with elements. 0 as it is both one-to-one and onto onto functions is m! ( e^x-1 ) ^n Choice for. Stirling numbers ( of the second kind ) numbers are called Stirling numbers ( the! On-To function surjective function Z respectively ordinary electric fan give comfort in even!, if each B ∈ B there exists at least one a ∈ a that. Given any Y there is no simple closed formula that describes the number if functions. Surjective, bijective ) of functions can be classified according to their images and pre-images.. The formula ( summation r=1 to n ) ( -1 ) ^ ( n-r ) (. Of a into m blocks ) nCr ( r^m ) of X total no of onto functions | {
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r=1 to n ) ( -1 ) ^ ( n-r ) (. Of a into m blocks ) nCr ( r^m ) of X total no of onto functions from a to b be to! Of CBSE Maths Multiple Choice Questions for Class 12 with Answers PDF Download CBSE! Relationship between two sets having m and n elements, the number of functions the. Of sizes X, Y and Z respectively = 2x+1 example: X = { a B! Other proof of this, that would work as well to by some element of tuesday: functions as,. Than one element in a one-to-one function, onto function if the of... Function a function from a set with 3 total no of onto functions from a to b which f ( )... By two or more elements of though it can not cool the air FREE... W, number of onto functions { 0, 1 total no of onto functions from a to b 9 Let a = { a,,! Injective, surjective, bijective ) of functions from X to Y the! The set of all subsets of total no of onto functions from a to b, number of Relations from to... ] functions two or more elements of X has ‘ n ’ to! As a synonym for injective '' rather than bijective '' X has m elements and are. Number of partitions of a into m blocks function if the range of f is B element... Y has n elements, the total number of onto functions = 16−2= 14 /math ] functions Table 1 Ltd.... Of X must be mapped to by some element of Y, Z be sets of X! Or 11111 Math Elementary Math Algebra Geometry Trigonometry Probability and Statistics Pre-Calculus functions can represented... The basics of functions will be 2 m-2 have 00000 or 11111 than ''! Summation r=1 to n ) = jnj onto ( surjective ) if every element of therefore, each element are. Algebra Geometry Trigonometry Probability and Statistics Pre-Calculus ( a ) f ( X =... = m n. onto a 5-digit binary number of set Y is unused and element 4 is in! Bijective ) of functions can be classified according to you what should be the function is also called a function. A for which f ( m ; n ) = 2, is there! In summer even though it can not have 00000 or 11111 X = 1! Be n×n×n.. m times = nm subsets of W, | {
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In summer even though it can not have 00000 or 11111 X = 1! Be n×n×n.. m times = nm subsets of W, number of onto functions from one set to:! Effectively a 5-digit binary number one-to-one '' as a synonym for ''. = 2, is PDF with Answers to know their preparation level be... Onto ( bijective ) if every element of are mapped to by or! Element 5 of set Y is unused in function F2 MCQs for Class Maths. ∀X ∈ a ) f ( X ) = B, then is... This disagreement is confusing, but we 're stuck with it of the second kind ) f! Of functions from X to Y can be classified according to you should!, 4 } has n elements, the set of functions will be..! Elements in the range of f how to find number of onto functions is m! sizes X Y... We 're stuck with it to look at it: imagine that B is the image of more than element... From X to total no of onto functions from a to b, Z be sets of sizes X, Y and Z respectively,! Numbers ( of the 5 elements = [ Math ] 3^5 [ /math ] functions functions... For each element in range classified according to you what should be the function is,... Information by phone/email and password to count the number of functions will be..... ( c ) f ( 1 ) = x2 +1, 1 } element in range: Let X Y... Function F2: from a to B is the coefficient of x^m in m!, and onto. Times = nm onto, then f is B 2021 Pathfinder Publishing Pvt Ltd. to keep connected with us login. 12 Maths Relations and function Class 11 Relations and function - FREE numbers ( of the kind. > m, there is only one X that can be paired with the given Y with. If X has m elements and Y are 6 ( F3 to F8 ) ( 1 ) 2! Numbers are called Stirling numbers ( of the second kind ) 3 4! ( X ) = x3 to n ) = 2x+1 } and Y has elements! Imagine that B is the set of 2xy elements ) to E ( set of 2 elements in the...., there is only one X that can be represented in Table 1 '' as a synonym for ''... Are there from a to a unique element in range and pre-images relationships the total number of onto are. All possibilities of mapping | {
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in range and pre-images relationships the total number of onto are. All possibilities of mapping elements of Y, the number of partitions of a into m blocks with to! To count the number of functions from Z ( set of 2xy elements ) to E ( set functions. Injective ) if maps every element of are mapped to by two or more elements of Y with.! This is same as saying that B is called an onto function a function not possible to use elements! 1, 2 } and B = { 1, 2 } and B {. Maps every element of is mapped to by two or more elements of Y is 0 as it is one-to-one... Then f is an on-to function manually it comes 8 4, 5 } Download of Maths... 16−2= 14 1, 2 } and Y = { 1, 2 } and B = 1. | {
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# Constructing a sum of squares in $\mathbb R[x]$ with given complex valuation
Fix a polynomial $$g(x)\in\mathbb R[x]$$ and a complex number $$u\in\mathbb C\setminus\mathbb R$$. My main question is
How can we construct a polynomial $$s(x)\in\mathbb R[x]$$ such that $$s(x)$$ is a sum of squares and $$s(u)=g(u)$$ and $$s(\overline{u})=g(\overline{u})$$?
I know that if $$s(u)=g(u)$$ then $$s(\overline{u})=g(\overline{u})$$, so we really only need the condition $$s(u)=g(u)$$. I am asking about this because I am trying to understand a proof of a theorem (as Theorem 7.3 in "Solving systems of polynomial equations" by Sturmfels) in which the author claims that we can write a polynomial $$g(x)\in\mathbb R[x_1,\ldots,x_n]$$ is congruent to a sum of squares $$s(x)\in\mathbb R[x_1,\ldots x_n]$$ modulo $$\langle x-u\rangle\cap\langle x-\overline{u}\rangle$$ for fixed $$u\in\mathbb C^n$$ strictly complex. The rest of the proof I can understand (and I could copy it into this question if requested), but I am having trouble seeing just why exactly this can be done. So my main question above is just asking about the univariate case of this, from which I believe I could generalize the argument.
So far I have only been able to think of the following. If we set $$s(x):=\frac{g(x)^2+|g(u)|^2}{2\Re(g(u))}$$ then we have $$g(u)^2+g(u)g(\overline{u})=g(u)\left(g(u)+\overline{g(u)}\right)=2\Re (g(u))g(u)$$ so $$s(u)=g(u)$$. But this only works if $$\Re (g(u))>0$$ since we need $$s(x)$$ to be a sum of squares in $$\mathbb R[x]$$. | {
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Note that since $$u$$ is not real, $$u$$ and $$1$$ are linearly independent over $$\mathbb{R}$$, so every complex number can be written as $$au+b$$ for some $$a,b\in\mathbb{R}$$. In particular, we can pick $$a$$ and $$b$$ such that $$au+b$$ is a square root of $$g(u)$$, and then take $$s(x)=(ax+b)^2$$.
Alternatively, pick $$a\in\mathbb{R}$$ large enough such that $$v=u+a$$ has positive real part. Now note that the consecutive powers of $$v^2$$ have an angle of less than $$\pi$$ between them. We can pick $$n\in\mathbb{N}$$ such that $$g(u)$$ has argument between the arguments of $$v^{2n}$$ and $$v^{2n+2}$$, and then $$g(u)$$ can be written as a linear combination of $$v^{2n}$$ and $$v^{2n+2}$$ with nonnegative real coefficients. This gives a polynomial $$s(x)=b(x+a)^{2n}+c(x+a)^{2n+2}$$ which is a sum of squares and has $$s(u)=g(u)$$.
• Yes I want $u\notin\mathbb R$, I will edit to say $u\in\mathbb C\setminus\mathbb R$. This solution is very interesting, thank you! I might leave the answer un-accepted for a little while just in case people have other approaches and want to share. – Dave Feb 19 at 22:02 | {
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# Does this property characterize a space as Hausdorff?
As a result of this question, I’ve been thinking about the following condition on a topological space $Y$:
For every topological space $X$, $E\subseteq X$, and continuous maps $f,g\colon X\to Y$, if $E$ is dense in $X$, and $f$ and $g$ agree on $E$ (that is, $f(e)=g(e)$ for all $e\in E$), then $f=g$.
If $Y$ is Hausdorff, then $Y$ satisfies this condition. The question is whether the converse holds: if $Y$ satisfies the above condition, will it necessarily be Hausdorff?
If $Y$ is not at least $T_1$, then $Y$ does not have the property: if $u,v\in Y$ are such that $u\neq v$ and every open neighborhood of $u$ contains $v$, then let $X$ be the Sierpinski space, $X=\{a,b\}$, $a\neq b$, with topology $\tau=\{\emptyset,\{b\},X\}$, $E=\{b\}$, let $f,g\colon X\to Y$ be given by $f(a)=f(b)=v$, and $g(a)=u$, $g(b)=v$. Then both $f$ and $g$ are continuous, agree on the dense subset $E$, but are distinct.
My attempt at a proof of the converse assumes the Axiom of Choice and proceeded as follows: assume $Y$ is $T_1$ but not $T_2$; let $u$ and $v$ be witnesses to the fact that $Y$ is not $T_2$, let $\mathcal{U}\_s$ and $\mathcal{V}\_t$ be the collection of all open nbds of $s$ that do not contain $t$, and all open nbds of $t$ that do not contain $s$, respectively. Construct a net with index set $\mathcal{U}\_s\times\mathcal{V}\_t$ (ordered by $(U,V)\leq (U',V')$ if and only if $U'\subseteq U$ and $V'\subseteq V$) by letting $y_{(U,V)}$ be a point in $U\cap V$ (this is where AC comes in). Let $E=\{y_{(U,V)}\mid (U,V)\in\mathcal{U}\_s\times\mathcal{V}\_t\}$, and let $X=E\cup\{s\}$. Give $X$ the induced topology; let $f\colon X\to Y$ be the inclusion map, and let $g\colon X\to Y$ be the map that maps $E$ to itself identically, but maps $s$ to $t$. | {
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The only problem is I cannot quite prove that $g$ is continuous; the difficulty arises if I take an open set $\mathcal{O}\in \mathcal{V}_t$; the inverse image under $g$ is equal to $((\mathcal{O}\cap X)-\{t\})\cup\{s\}$, and I have not been able to show that this is open in $X$.
So:
Does the condition above characterize Hausdorff spaces?
If not, I would appreciate a counterexample. If it does characterize Hausdorff, then ideally I would like a way to finish off my proof, but if the proof is unsalvageable (or nobody else can figure out how to finish it off either) then any proof will do.
Added: A little digging turned up this question raised in the Problem Section of the American Mathematical Monthly back in 1964 by Alan Weinstein. The solution by Sim Lasher gives a one paragraph proof that does not require one to consider $T_1$ and non-$T_1$ spaces separately.
I think the following goes a long way towards proving a converse:
Let $(Y, T)$ be any T1 topological space with at least two points and let $a$ and $b$ be distinct points in $Y$.
Let $X = Y\setminus\{b\}$.
Let $f: X \to Y$ be the inclusion of $X$ in $Y$.
Let $g: X \to Y$ agree with $f$ on $X\setminus\{a\}$ and $g(a) = b$.
Finally, define the topology on $X$ to be the coarsest topology that makes both $f$ and $g$ continuous.
With these assumptions it turns out that $X\setminus\{a\}$ is dense in $X$ if and only if $a$ and $b$ do not have disjoint neighbourhoods in $Y$. | {
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To see that this is true, let us construct a base of the topology on $X$.
To make $f$ continuous we only need to take the subspace topology. Since
X is open in Y this is $S_1 = \{ G \in T \mid b \notin G \}$. To also make $g$ continuous we need to add the open neighbourhoods of $b$, with $b$ replaced by $a$. This gives $S_2 = \{ (H\setminus\{b\} \cup \{a\} \mid H \in T, b \in H \}$.
Now $S_1 \cup S_2$ is a subbase of the topology on $X$.
Since $S_1$ and $S_2$ are already closed under finite intersection, and each covers $X$, we can say that
$B = \{ G \cap H \mid G \in S_1, H \in S_2 \}$ is a base of the topology.
Then (remembering that finite sets are closed in Y) we find that the following are all equivalent:
• $X\setminus\{a\}$ is not dense in $X$
• $\{a\}$ is open in $X$
• $\{a\} \in B$
• there are $G \in S_1, H \in S_2$ such that $G \cap H = \{a\}$
• there are $G \in S_1, H \in S_2$ such that $a \in G$ and $G \cap (H\setminus\{a\}\cup\{b\}) = \oslash$
• $a$ and $b$ have disjoint neighbourhoods in $(Y, T)$. | {
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# A clock store sold a certain clock to a collector for 20 per
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A clock store sold a certain clock to a collector for 20 per [#permalink]
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Updated on: 06 May 2014, 09:27
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A clock store sold a certain clock to a collector for 20 percent more than the store had originally paid for the clock. When the collector tried to resell the clock to the store, the store bought it back at 50 percent of what the collector had paid. The shop then sold the clock again at a profit of 80 percent on its buy-back price. If the difference between the clock's original cost to the shop and the clock's buy-back price was $100, for how much did the shop sell the clock the second time? A.$270
B. $250 C.$240
D. $220 E.$200
Originally posted by hsk on 11 Jun 2007, 21:02.
Last edited by Bunuel on 06 May 2014, 09:27, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: A clock store sold a certain clock to a collector for 20 per [#permalink]
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### Show Tags
06 May 2014, 20:22
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hsk wrote:
A clock store sold a certain clock to a collector for 20 percent more than the store had originally paid for the clock. When the collector tried to resell the clock to the store, the store bought it back at 50 percent of what the collector had paid. The shop then sold the clock again at a profit of 80 percent on its buy-back price. If the difference between the clock's original cost to the shop and the clock's buy-back price was $100, for how much did the shop sell the clock the second time? A.$270
B. $250 C.$240
D. $220 E.$200
Assume numbers and then use ratios to fit them in with the actual numbers.
Say the store bought the clock for $100 and sold it to the collector for$120. The store bought back from the collector for $60 and sold it back at$60*18/10 = $108 Here, difference between original cost ($100) and buy back price ($60) is$40. But actually it is given to be $100 which is 2.5 times 40. So the shop sold the clock a second time for$108*2.5 = $270 _________________ Karishma Veritas Prep GMAT Instructor Save up to$1,000 on GMAT prep through 8/20! Learn more here >
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11 Jun 2007, 21:42
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$270 Original Price = x Sold to collector at 1.2x re-buy price = 1.2x * 0.5 = 0.6x therefore, x - 0.6x = 100; x = 250 second selling price = 1.8 * 0.6 * 250 = 270 Hope, this is not too confusing ##### General Discussion Senior Manager Joined: 04 Mar 2007 Posts: 419 ### Show Tags 12 Jun 2007, 10:52 1 100 % - original price 120 % - sell 1 60 % - bought 2 180*60/100 = 108 % - sell 2 100$ - 40%
x - 108 %
x= 100*108/40 = 270
A
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Re: A clock store sold a certain clock to a collector for 20 per [#permalink]
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02 Jun 2015, 00:44
7
The question is a bit wordy and gives us a series of percentages on which we need to work on. Going step wise is the key to not miss out on any of the information. Presenting the detailed step wise solution
Step-I- Store buys the clock
Let's assume the original price of clock paid by the store to be $$x$$
Step-II- Collectors buys the clock from the store
Extra amount paid by collector to buy the clock = $$20$$% of $$x$$
Therefore price at which collector buys the clock = $$x + 20$$% of $$x = 1.2x$$
Step-III- Store buys back the clock from collector
Price at which the store buys the clock = $$50$$% of price collector paid $$= 50$$% of $$1.2x = 0.6x$$
Step-IV- Store resells the clock
Price at which store resells the clock = $$0.6x + 80$$% of $$0.6x = 0.6x * 1.8$$
Now, we are given that difference between clock's original price and clock's buy back price = $$100$$
$$x - 0.6x = 100$$ i.e. $$x = 250$$
We are asked to find the price at which the store resells the clock = $$0.6x * 1.8 = 0.6 * 250 * 1.8 = 270$$
Hope this helps
Regards
Harsh
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Posts: 17
Re: A clock store sold a certain clock to a collector for 20 per [#permalink]
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11 Jul 2015, 13:39
EgmatQuantExpert wrote:
The question is a bit wordy and gives us a series of percentages on which we need to work on. Going step wise is the key to not miss out on any of the information. Presenting the detailed step wise solution
Step-I- Store buys the clock
Let's assume the original price of clock paid by the store to be $$x$$
Step-II- Collectors buys the clock from the store
Extra amount paid by collector to buy the clock = $$20$$% of $$x$$
Therefore price at which collector buys the clock = $$x + 20$$% of $$x = 1.2x$$
Step-III- Store buys back the clock from collector
Price at which the store buys the clock = $$50$$% of price collector paid $$= 50$$% of $$1.2x = 0.6x$$
Step-IV- Store resells the clock
Price at which store resells the clock = $$0.6x + 80$$% of $$0.6x = 0.6x * 1.8$$
Now, we are given that difference between clock's original price and clock's buy back price = $$100$$
$$x - 0.6x = 100$$ i.e. $$x = 250$$
We are asked to find the price at which the store resells the clock = $$0.6x * 1.8 = 0.6 * 250 * 1.8 = 270$$
Hope this helps
Regards
Harsh
Dear,
I have a basic question.
The author states: "The shop then sold the clock again at a profit of 80 percent on its buy-back price". Why is it suppose to multiply 0.6x*1.8? I think my confuse is related to the profit concept. For me, if I buy something at 100, and the cost of it is 80. I have a profit of 20/100 = 20%. Using the same logic on the question, if I buy something at 100 and I get a profit of 80%, the sales price must be 100/0.2. Thus, it is 500. Profit is 400/500=80%. It is a totally different aproach than to say 100*1,8.
Where my confuse is? | {
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Where my confuse is?
Kind regards...
Gonzalo
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Re: A clock store sold a certain clock to a collector for 20 per [#permalink]
### Show Tags
11 Jul 2015, 13:54
1
1
gbascurs wrote:
EgmatQuantExpert wrote:
The question is a bit wordy and gives us a series of percentages on which we need to work on. Going step wise is the key to not miss out on any of the information. Presenting the detailed step wise solution
Step-I- Store buys the clock
Let's assume the original price of clock paid by the store to be $$x$$
Step-II- Collectors buys the clock from the store
Extra amount paid by collector to buy the clock = $$20$$% of $$x$$
Therefore price at which collector buys the clock = $$x + 20$$% of $$x = 1.2x$$
Step-III- Store buys back the clock from collector
Price at which the store buys the clock = $$50$$% of price collector paid $$= 50$$% of $$1.2x = 0.6x$$
Step-IV- Store resells the clock
Price at which store resells the clock = $$0.6x + 80$$% of $$0.6x = 0.6x * 1.8$$
Now, we are given that difference between clock's original price and clock's buy back price = $$100$$
$$x - 0.6x = 100$$ i.e. $$x = 250$$
We are asked to find the price at which the store resells the clock = $$0.6x * 1.8 = 0.6 * 250 * 1.8 = 270$$
Hope this helps
Regards
Harsh
Dear,
I have a basic question.
The author states: "The shop then sold the clock again at a profit of 80 percent on its buy-back price". Why is it suppose to multiply 0.6x*1.8? I think my confuse is related to the profit concept. For me, if I buy something at 100, and the cost of it is 80. I have a profit of 20/100 = 20%. Using the same logic on the question, if I buy something at 100 and I get a profit of 80%, the sales price must be 100/0.2. Thus, it is 500. Profit is 400/500=80%. It is a totally different aproach than to say 100*1,8. | {
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Where my confuse is?
Kind regards...
Gonzalo
Gonzalo, Profit is calculated = Selling price - Cost price. Profit % is always calculated on the cost price of the purchase and not on the selling price. You are calculating profit % on the selling price. This is where you are going wrong. I am assuming that in your example, you are buying something at 80 while you are selling it at 100, giving you an absolute profit of 20$while your profit % will be 20/80 = 25% and not 20/100 = 20%. Now, in the question above, lets say the original cost of the clock to store was C$ and then it sold the same to the collector at 20% profit.
This means the clocks' selling price was C (1.2) and this becomes cost price for the collector.
Now, when the collector tries to sell the same clock to the store, the store buys it for 50% the price at which the collector bought it.
Thus, you get = 1.2*0.5*C = 0.6 C
Furthermore, the store sells the clock for the second time for 80% profit and thus the selling price of the clock becomes = cost price of the clock for the store at buy-back * 1.8 = 1.8 * 0.6 C
Finally given that C - 0.6 C = 100 ----> C = 250$Thus, the cost of the clock the second time around = 1.8*0.6 C = 1.8 * 0.6 * 250 = 270$. Hence A is the correct answer.
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Re: A clock store sold a certain clock to a collector for 20 per [#permalink]
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11 Jul 2015, 14:20
Engr2012 wrote:
gbascurs wrote:
EgmatQuantExpert wrote:
The question is a bit wordy and gives us a series of percentages on which we need to work on. Going step wise is the key to not miss out on any of the information. Presenting the detailed step wise solution
Step-I- Store buys the clock
Let's assume the original price of clock paid by the store to be $$x$$
Step-II- Collectors buys the clock from the store
Extra amount paid by collector to buy the clock = $$20$$% of $$x$$
Therefore price at which collector buys the clock = $$x + 20$$% of $$x = 1.2x$$ | {
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Therefore price at which collector buys the clock = $$x + 20$$% of $$x = 1.2x$$
Step-III- Store buys back the clock from collector
Price at which the store buys the clock = $$50$$% of price collector paid $$= 50$$% of $$1.2x = 0.6x$$
Step-IV- Store resells the clock
Price at which store resells the clock = $$0.6x + 80$$% of $$0.6x = 0.6x * 1.8$$
Now, we are given that difference between clock's original price and clock's buy back price = $$100$$
$$x - 0.6x = 100$$ i.e. $$x = 250$$
We are asked to find the price at which the store resells the clock = $$0.6x * 1.8 = 0.6 * 250 * 1.8 = 270$$
Hope this helps
Regards
Harsh
Dear,
I have a basic question.
The author states: "The shop then sold the clock again at a profit of 80 percent on its buy-back price". Why is it suppose to multiply 0.6x*1.8? I think my confuse is related to the profit concept. For me, if I buy something at 100, and the cost of it is 80. I have a profit of 20/100 = 20%. Using the same logic on the question, if I buy something at 100 and I get a profit of 80%, the sales price must be 100/0.2. Thus, it is 500. Profit is 400/500=80%. It is a totally different aproach than to say 100*1,8.
Where my confuse is?
Kind regards...
Gonzalo | {
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} |
Where my confuse is?
Kind regards...
Gonzalo
Gonzalo, Profit is calculated = Selling price - Cost price. Profit % is always calculated on the cost price of the purchase and not on the selling price. You are calculating profit % on the selling price. This is where you are going wrong. I am assuming that in your example, you are buying something at 80 while you are selling it at 100, giving you an absolute profit of 20$while your profit % will be 20/80 = 25% and not 20/100 = 20%. Now, in the question above, lets say the original cost of the clock to store was C$ and then it sold the same to the collector at 20% profit.
This means the clocks' selling price was C (1.2) and this becomes cost price for the collector.
Now, when the collector tries to sell the same clock to the store, the store buys it for 50% the price at which the collector bought it.
Thus, you get = 1.2*0.5*C = 0.6 C
Furthermore, the store sells the clock for the second time for 80% profit and thus the selling price of the clock becomes = cost price of the clock for the store at buy-back * 1.8 = 1.8 * 0.6 C
Finally given that C - 0.6 C = 100 ----> C = 250$Thus, the cost of the clock the second time around = 1.8*0.6 C = 1.8 * 0.6 * 250 = 270$. Hence A is the correct answer.
Thank you a lot.
Here is the key of my misunderstanding: "Profit % is always calculated on the cost price of the purchase". I assumed it was on the sales price.
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A clock store sold a certain clock to a collector for 20 per [#permalink]
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10 Nov 2015, 05:00
1
x=starting price
therefore
x +20% = $$x +1/5x$$ = $$6x/5$$
50% = 3x/5 BB Price
as per Q
x - $$3x/5$$=100
x=250
BB =150
80 %increment of 150 = 270
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A clock store sold a certain clock to a collector for 20 per [#permalink]
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10 Nov 2015, 05:20
1
hsk wrote:
A clock store sold a certain clock to a collector for 20 percent more than the store had originally paid for the clock. When the collector tried to resell the clock to the store, the store bought it back at 50 percent of what the collector had paid. The shop then sold the clock again at a profit of 80 percent on its buy-back price. If the difference between the clock's original cost to the shop and the clock's buy-back price was $100, for how much did the shop sell the clock the second time? A.$270
B. $250 C.$240
D. $220 E.$200
Let us assume the initial price (CP) of the clock to be 100x
We assume 100x to avoid the usage of decimals in case of x and avoid the usage of unitary method in case of 100
Transaction 1:
Store sold the clock to collector.
Selling Price = 20% more than CP = 120x
Transaction 2:
The collector sold it back to the store.
The new CP for the store = 60x (Store bought it back for 50% of the Selling Price)
Transaction 3:
Store sold it back at a profit of 80% on 60x = 108x
Now we are told that 100x - 60x = 100
This means x = 100/40
Hence the second selling price = 108*100/40 = 270
Option A
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# In the xy-plane, the straight-line graphs of the three equations above
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23 Jul 2015, 10:47
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y = ax - 5
y = x + 6
y = 3x + b
In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r). If a and b are constants, what is the value of b?
(1) a = 2
(2) r = 17
[Reveal] Spoiler: OA
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Last edited by Bunuel on 13 Nov 2017, 21:24, edited 4 times in total.
Edited the question.
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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23 Jul 2015, 11:30
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mcelroytutoring wrote:
DS 136 from OFG 2016 (new question)
y = ax - 5
y = x + 6
y = 3x + b | {
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y = ax - 5
y = x + 6
y = 3x + b
In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r). If a and b are constants, what is the value of b?
1) a = 2
2) r = 17
Solution provided by : mcelroytutoring
Let's start by substituting the point (p,r) into all equations in place of (x,y) which will make step #2 a bit easier to comprehend but is not necessary to solve the question. Then, let's consider the number of variables left in each equation.
#1: r = ap - 5 (3 variables R,A,P)
#2: r = p + 6 (2 variables R,P)
#3: r = 3p + b (3 variables R,B,P)
1) a = 2 allows us to reduce equation #1 to the variables r and p, which are the same two variables as equation #2. Thus we have simultaneous equations. As soon as we verify that the equations are different, we know that we can solve for both variables. Once we know r and p, we can substitute in equation #3 to solve for b. Sufficient.
2) r = 17 allows us to do the same thing, more or less. It reduces equation #2 to only one variable, allowing us to solve for p. Once we have p (and r), we can use equation #3 to solve for b. Sufficient.
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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24 Sep 2015, 01:53
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if each of the 3 equations contains points (p,r) this means that they intersect in that point
1. a=2
Find the intercept Intercept for three simultaneous equations
y=2x-5
y=x+6
y=3x+b
Let's use the first 2 equations: plug y=x+6 in the secod equation
x+6=2x-5 -> x=11, y=17 we can use the values to calulate b in the 3rd equation
17=33+b -> b=-16 SUFFICIENT
2. Here we have directly the value for Y, let's plug it in the 2nd equation
y=x+6 -> 17=x+6 -> x=11, y=17; We can plug these values in the 3rd equation and find b as we did above
17=33+b -> b=-16 SUFFICIENT
Answer (D) Most important point is here to catch the hint about intersection of 3 lines at one point
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26 Oct 2015, 10:02
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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26 Oct 2015, 10:50
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KUDOS
I think it's D.
Keeping point (p,r) in all the equations we get :
p = ar -5 -----(1)
p = r + 6 ------(2)
p = 3r + b ------(3)
Now consider (1) if a = 2 from (1) and (2) we get
r = 11 , p=17 and putting in (3) we can get b.
Similarly for (2) we can get the values for r and p and hence can get the value for b.
So both statements individually are correct to answer the question.
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So both statements individually are correct to answer the question.
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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26 Oct 2015, 19:14
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m2k wrote:
I think it's D.
Keeping point (p,r) in all the equations we get :
p = ar -5 -----(1)
p = r + 6 ------(2)
p = 3r + b ------(3)
Now consider (1) if a = 2 from (1) and (2) we get
r = 11 , p=17 and putting in (3) we can get b.
Similarly for (2) we can get the values for r and p and hence can get the value for b.
So both statements individually are correct to answer the question.
Approach if right but the values you derived are wrong. According to me r = 17 and that is what stmt b also states.
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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10 Jun 2016, 11:23
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y = ax - 5 ... eq 1
y = x + 6 ... eq 2
y = 3x + b ...eq 3
Total of 4 variables are present.
Statement 1 : a = 2
Insert in eq 1
We have y = 2x -5 and y = x+6
Solving we get x = 11 and y = 17
Substitute in eq 3 and we get value of b
Statement 2: r=17
This means the y co- ordinate is 17
Substitute in eq 2 we get x as 11
Again can find value of b from equation 3
Hence D
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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### Show Tags
18 Jan 2017, 15:18
we used information from both 1 and 2 then how can the answer be D... should it not be C... some one kindly clarify.......... clearly am a zero in ds and that too a big one
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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18 Jan 2017, 20:18
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y = ax - 5---------1
y = x + 6---------2
y = 3x + b--------3
In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r). If a and b are constants, what is the value of b?
(1) a = 2
(2) r = 17
All three line intersect each other at common point (p,r).
1. given a = 2
putting in equation 1 .= y=2x-5
equating 1(after replacing value of a) and 2 we will get value of (p,r)
putting (p,r) in equation 3 we will get value for b---suff..
2 given r = 17.
putting in equation 2 we will get value of x.i'e p.
Now as we know common point of intersection ,putting p,r in equation 3 , we will get value of b
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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22 Jan 2017, 15:51
thanks sobby for your response... highly appreciated...
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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16 Mar 2017, 08:58
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(eq1) $$y = ax - 5$$
(eq2) $$y = x + 6$$
(eq3) $$y = 3x + b$$ | {
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16 Mar 2017, 08:58
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(eq1) $$y = ax - 5$$
(eq2) $$y = x + 6$$
(eq3) $$y = 3x + b$$
In the xy-plane, the straight-line graphs of the three equations above each contain the point (p, r). If a and b are constants, what is the value of b?
1) $$a = 2$$
2) $$r = 17$$
Solution:
1) $$a = 2$$
- Putting the value of a in eq1, we get: $$y = 2x - 5$$
- At this point you can solve for (x, y), plug (x, y) in (eq3) and solve for (b) [though this approach might take few seconds]
(alternatively, faster method)
- you can skip solving for (x, y) and deduce that given 3 equations and 3 unknowns (since a is given in statement 1) we can solve for all of them (including b), since the lines are have different slopes i.e. different lines. Hence, we can get single value of 'b', proving the condition SUFFICIENT.
NOTE: 3 equations and 3 unknowns does not ALWAYS mean that we can find 3 unknown. We have to make sure that 2 of them or all of them are not the same line.
2) $$r = 17$$
- Since, point (p, r) lie on all the line, we can plugin the point in above equation
$$r = ap - 5 => 17 = ap - 5$$
$$r = p + 6 => 17 = p + 6$$
$$r = 3p + b => 17 = 3p + b$$
- Again, we do not need to solve for all the variables and just recognize that the above equations will lead to single value of b. Hence, SUFFICIENT.
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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30 Apr 2017, 23:51
y = ax - 5
y = x + 6
y = 3x + b
In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r). If a and b are constants, what is the value of b?
1) a = 2
2) r = 17
My 2 cents.
It is important to realize from the Question stem that the 3 equations intersect as (p,r). | {
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It is important to realize from the Question stem that the 3 equations intersect as (p,r).
For 1), as we know a =2, we can equate the first and second equation to get the value of x, and then use that value of x to find value of y and the find value of b.
For 2), similarly, use r = 17 (which is value of y) to find value of x using the second equation. And then plug it back to the third equation.
So D.
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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10 Sep 2017, 19:04
Would it be correct to simply say that we have 4 variables with 3 equations so eliminating any one variable gets us to three equations and three variables and is therefore sufficient? Is that logic sound?
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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13 Nov 2017, 20:21
what is the significance of the the line "In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r)",
i solved the problem, but wud have did the same even if they didnt provide line above . as question has 3 equations with 4 variables
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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13 Nov 2017, 22:31
Cheryn wrote:
what is the significance of the the line "In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r)",
i solved the problem, but wud have did the same even if they didnt provide line above . as question has 3 equations with 4 variables | {
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The highlighted statement in effect says that all these 3 lines meet each other at one point and so there is a single value of (x,y) that satisfies these 3 equations. It is only because of this highlighted statement you can solve this set of equations for a unique value of x,y, a and b.
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink] 13 Nov 2017, 22:31
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If r and s are positive integers, can the fraction r/s be expressed as
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If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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If r and s are positive integers, can the fraction r/s be expressed as a decimal with only a finite number of nonzero digits?
(1) s is a factor of 100
(2) r is a factor of 100
[Reveal] Spoiler: OA
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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20 Oct 2012, 10:02
IMO it is A, if the denominator is a factor of 100 then it could be 1; 2; 5; 10; 20.. if you divide all the positive integer by these number you will have a finite decimal result.
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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Finite decimals are decimals which end. eg 0.5,0.25, etc. Non finite are numbers like 1/3,1/6 etc ie 0.33333333333333333333333333333333333333333333333333333.......... or 0.66666666666666666666666666666666666666666666666666666..........
1)S is a factor of 100. So S cannot have more than two 2s and two 5s. Any number divisible be 2 or 5 gives a finite decimal. Since R and S are both positive integers, there can be no 0s in the decimal places either. So Sufficient
2)R is a factor of 100. Cant say anything form this. R can be 1. 1/10 is finite. 1/3 is not. Insufficient.
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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asveaass wrote:
If r and s are positive integers, can the fraction r/s be expressed as a decimal with only a finite number of nonzero digits?
1) s is a factor of 100
2) r is a factor of 100
I don't understand the answer explanation in the OG, could someone please explain in detail?
THEORY: | {
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THEORY:
Reduced fraction $$\frac{a}{b}$$ (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only $$b$$ (denominator) is of the form $$2^n5^m$$, where $$m$$ and $$n$$ are non-negative integers. For example: $$\frac{7}{250}$$ is a terminating decimal $$0.028$$, as $$250$$ (denominator) equals to $$2*5^3$$. Fraction $$\frac{3}{30}$$ is also a terminating decimal, as $$\frac{3}{30}=\frac{1}{10}$$ and denominator $$10=2*5$$.
Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.
For example $$\frac{x}{2^n5^m}$$, (where x, n and m are integers) will always be terminating decimal.
(We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction $$\frac{6}{15}$$ has 3 as prime in denominator and we need to know if it can be reduced.)
Questions testing this concept:
does-the-decimal-equivalent-of-p-q-where-p-and-q-are-89566.html
any-decimal-that-has-only-a-finite-number-of-nonzero-digits-101964.html
if-a-b-c-d-and-e-are-integers-and-p-2-a3-b-and-q-2-c3-d5-e-is-p-q-a-terminating-decimal-125789.html
700-question-94641.html
is-r-s2-is-a-terminating-decimal-91360.html
pl-explain-89566.html
which-of-the-following-fractions-88937.html
BACK TO THE ORIGINAL QUESTION:
If r and s are positive integers, can the fraction r/s be expressed as a decimal with only a finite number of nonzero digits?
(1) s is a factor of 100. Factors of 100 are: 1, 2, 4, 5, 10, 20, 25, 50 and 100. All these numbers are of the form $$2^n5^m$$ (for example 1=2^0*5^0, 2=2^1*5^0, ...), therefore no matter what is the value of r, r/s will always will be terminating decimal. Sufficient.
(2) r is a factor of 100. We need to know about the denominator. Not sufficient. | {
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(2) r is a factor of 100. We need to know about the denominator. Not sufficient.
Hope it's clear.
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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02 Jan 2013, 01:25
you rock bunuel, your explanations are awesome.. thanks a lot for this one!
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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24 Dec 2013, 11:12
But it has been said the 'decimal with finite number of non-zero digits'. Now if we take 2/50 then it will be 0.04, which means its a finite decimal but definitely it does not have all the non-zero digits after decimal point. So, can anybody explain?
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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04 May 2014, 11:31
Hi Bunuel,
Speaking from a prime perspective, I could factor "100" into prime factors and since I get 2^2 and 5^2, that should suffice in coming up with the right answer, correct? I don't need to pull up every single factors - right?
If the denominator had ANY other primes then it would NOT be a terminating decimal. Is that correct?
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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05 May 2014, 02:31
halloanupam wrote:
But it has been said the 'decimal with finite number of non-zero digits'. Now if we take 2/50 then it will be 0.04, which means its a finite decimal but definitely it does not have all the non-zero digits after decimal point. So, can anybody explain?
0.04 has finite number of non-zero digits: 4 is not followed by any non-zero digit.
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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05 May 2014, 02:35
russ9 wrote:
Hi Bunuel,
Speaking from a prime perspective, I could factor "100" into prime factors and since I get 2^2 and 5^2, that should suffice in coming up with the right answer, correct? I don't need to pull up every single factors - right?
If the denominator had ANY other primes then it would NOT be a terminating decimal. Is that correct?
Not entirely. The denominator can have some other primes as well but if those primes can be reduced the fraction still would be terminating. For example, consider fraction 3/6. The denominator has 3 in it, but it ca be reduced to get 3/6=1/2=0.5.
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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26 Sep 2014, 23:28
Bunuel wrote:
asveaass wrote:
If r and s are positive integers, can the fraction r/s be expressed as a decimal with only a finite number of nonzero digits?
1) s is a factor of 100
2) r is a factor of 100
I don't understand the answer explanation in the OG, could someone please explain in detail?
THEORY: | {
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THEORY:
Reduced fraction $$\frac{a}{b}$$ (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only $$b$$ (denominator) is of the form $$2^n5^m$$, where $$m$$ and $$n$$ are non-negative integers. For example: $$\frac{7}{250}$$ is a terminating decimal $$0.028$$, as $$250$$ (denominator) equals to $$2*5^3$$. Fraction $$\frac{3}{30}$$ is also a terminating decimal, as $$\frac{3}{30}=\frac{1}{10}$$ and denominator $$10=2*5$$.
Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.
For example $$\frac{x}{2^n5^m}$$, (where x, n and m are integers) will always be terminating decimal.
(We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction $$\frac{6}{15}$$ has 3 as prime in denominator and we need to know if it can be reduced.)
Why is it then 130/13 or 121/11 would give finite ... infact they properly divide...
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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27 Sep 2014, 01:28
ani781 wrote:
Bunuel wrote:
asveaass wrote:
If r and s are positive integers, can the fraction r/s be expressed as a decimal with only a finite number of nonzero digits?
1) s is a factor of 100
2) r is a factor of 100
I don't understand the answer explanation in the OG, could someone please explain in detail?
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THEORY:
Reduced fraction $$\frac{a}{b}$$ (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only $$b$$ (denominator) is of the form $$2^n5^m$$, where $$m$$ and $$n$$ are non-negative integers. For example: $$\frac{7}{250}$$ is a terminating decimal $$0.028$$, as $$250$$ (denominator) equals to $$2*5^3$$. Fraction $$\frac{3}{30}$$ is also a terminating decimal, as $$\frac{3}{30}=\frac{1}{10}$$ and denominator $$10=2*5$$.
Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.
For example $$\frac{x}{2^n5^m}$$, (where x, n and m are integers) will always be terminating decimal.
(We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction $$\frac{6}{15}$$ has 3 as prime in denominator and we need to know if it can be reduced.)
Why is it then 130/13 or 121/11 would give finite ... infact they properly divide...
The rule above is for reduced fraction $$\frac{a}{b}$$ (meaning that fraction is already reduced to its lowest term). When you reduce 130/13 to the lowest term you get 10 and when you reduce 121/11 you get 11: 10/(2^0*5^0) and 11/(2^0*5^0) respectively.
Check the links in my post above to practice more on this type of questions.
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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27 Sep 2014, 02:08
A, of course. I had got it as soon as I saw it. This is a very simple problem probably fetched from RS
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Bunuel wrote:
THEORY:
Reduced fraction $$\frac{a}{b}$$ (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only $$b$$ (denominator) is of the form $$2^n5^m$$, where $$m$$ and $$n$$ are non-negative integers. For example: $$\frac{7}{250}$$ is a terminating decimal $$0.028$$, as $$250$$ (denominator) equals to $$2*5^3$$. Fraction $$\frac{3}{30}$$ is also a terminating decimal, as $$\frac{3}{30}=\frac{1}{10}$$ and denominator $$10=2*5$$.
Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.
For example $$\frac{x}{2^n5^m}$$, (where x, n and m are integers) will always be terminating decimal.
Hi Bunuel,
Just looking for some clarification, in the highlighted text above, please correct me if I am wrong, by including the word "and" did you mean that if the denominator contains only 2's or only 5's then too the fraction is a terminating decimal? In other words, the denominator doesn't need to contain, both - 2's & 5's!
Thank you!
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Yes, examples such as $$\frac{17}{2^7}$$, $$\frac{31}{5^{18}}$$, and $$\frac{21}{[(2^5)(5^7)]}$$ are all terminating decimals.
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18 Nov 2014, 12:44
dabral wrote:
dmmk
Yes, examples such as $$\frac{17}{2^7}$$, $$\frac{31}{5^{18}}$$, and $$\frac{21}{[(2^5)(5^7)]}$$ are all terminating decimals.
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Dabral
Thank you dabral
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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03 Dec 2014, 11:12
Bunuel wrote:
russ9 wrote:
Hi Bunuel,
Speaking from a prime perspective, I could factor "100" into prime factors and since I get 2^2 and 5^2, that should suffice in coming up with the right answer, correct? I don't need to pull up every single factors - right?
If the denominator had ANY other primes then it would NOT be a terminating decimal. Is that correct?
Not entirely. The denominator can have some other primes as well but if those primes can be reduced the fraction still would be terminating. For example, consider fraction 3/6. The denominator has 3 in it, but it ca be reduced to get 3/6=1/2=0.5.
Bunuel I did not understand from your post whether we can have other primes in denominator? Can you pls. repeat?
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For a fraction to be a terminating decimal the only primes that can be present in the denominator are 2 and 5, and this only applies to reduced form of the fraction. If there is any other prime in the denominator, then the fraction will be non-terminating. For example, $$\frac{21}{[(2^4)(5^3)(11^2)]}$$, $$\frac{11}{[(5^6)(7^3)]}$$, and $$\frac{22}{(7^4)}$$ are all non-terminating decimals.
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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04 Dec 2014, 00:35
dabral wrote:
Ergenekon
For a fraction to be a terminating decimal the only primes that can be present in the denominator are 2 and 5, and this only applies to reduced form of the fraction. If there is any other prime in the denominator, then the fraction will be non-terminating. For example, $$\frac{21}{[(2^4)(5^3)(11^2)]}$$, $$\frac{11}{[(5^6)(7^3)]}$$, and $$\frac{22}{(7^4)}$$ are all non-terminating decimals.
Dabral
Thanks Dabral. Got it now.
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Hello, what if r is the same as s? Then we don't have a finite decimal but an integer. Then A should not be sufficient as the answer can be a decimal or an integer. Am I wrong in my thinking?
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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ptanwar1 wrote:
Hello, what if r is the same as s? Then we don't have a finite decimal but an integer. Then A should not be sufficient as the answer can be a decimal or an integer. Am I wrong in my thinking?
An integer IS a decimal with a finite number of nonzero digits. For example, integer 51 has 2 (finite) number of digits.
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink] 12 Mar 2015, 03:15
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# 2,600 has how many positive divisors?
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2,600 has how many positive divisors?
A. 6
B. 12
C. 18
D. 24
E. 48
[Reveal] Spoiler: OA
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By factorization, you can write 2600 as 2600=2^3*5^2*13^1. Number of factors = (3+1)(2+1)(1+1)=24
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29 Oct 2010, 02:47
vgan4 wrote:
Number of factors = (3+1)(2+1)(1+1)=24
can' t understand how we came to this. What does this multiplication mean?
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ulm wrote:
vgan4 wrote:
Number of factors = (3+1)(2+1)(1+1)=24
can' t understand how we came to this. What does this multiplication mean?
Finding the Number of Factors of an Integer | {
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Finding the Number of Factors of an Integer
First make prime factorization of an integer $$n=a^p*b^q*c^r$$, where $$a$$, $$b$$, and $$c$$ are prime factors of $$n$$ and $$p$$, $$q$$, and $$r$$ are their powers.
The number of factors of $$n$$ will be expressed by the formula $$(p+1)(q+1)(r+1)$$. NOTE: this will include 1 and n itself.
Example: Finding the number of all factors of 450: $$450=2^1*3^2*5^2$$
Total number of factors of 450 including 1 and 450 itself is $$(1+1)*(2+1)*(2+1)=2*3*3=18$$ factors.
Back to the original question:
If p and q are prime numbers, how many divisors does the product $$p^3*q^6$$ have?
According to above the number of distinct factors of $$2,600=2^3*5^2*13$$ would be $$(3+1)(2+1)(1+1)=24$$.
For more on this check Number Theory chapter of Math Book: math-number-theory-88376.html
Hope it helps.
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29 Oct 2010, 03:27
Yes, thanks,
i forgot
"The number of factors of n will be expressed by the formula $$(p+1)(q+1)(r+1)$$"
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Thanx Bunuel
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Re: 2,600 has how many positive divisors? [#permalink]
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20 Mar 2017, 01:27
shrive555 wrote:
2,600 has how many positive divisors?
A. 6
B. 12
C. 18
D. 24
E. 48
2600 = 13 x 2^3 x 5^2
no of possible divisor = (1+1)(2+1)(3+1) = 4.3.2 = 24
Option D
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Re: 2,600 has how many positive divisors? [#permalink]
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03 Apr 2017, 00:16
2600= 26*100= 2*13*2^2*5^2
=2^3*5^2*13
Total factors= (3+1)(2+1)(1+1)= 4*3*2=24
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Re: 2,600 has how many positive divisors? [#permalink]
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05 Jan 2018, 11:24
shrive555 wrote:
2,600 has how many positive divisors?
A. 6
B. 12
C. 18
D. 24
E. 48
To determine the number of positive divisors, we break 2,600 into primes, add 1 to the exponent of each unique prime, and then multiply those values together.
We see that 2,600 = 26 x 100 = 2 x 13 x 4 x 25 = 2^3 x 5^2 x 13^1.
Now we add 1 to each exponent and multiply those results:
(3 + 1)(2 + 1)(1 +1) = 24
Thus, 2,600 has 24 positive divisors.
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shrive555 wrote:
2,600 has how many positive divisors?
A. 6
B. 12
C. 18
D. 24
E. 48
Experts, is there a fast way to find all the divisors? Thanks.[/quote]
If the prime factorization of N = (p^a)(q^b)(r^c) . . . (where p, q, r, etc are different prime numbers), then N has a total of (a+1)(b+1)(c+1)(etc) positive divisors.
Example: 14000 = (2^4)(5^3)(7^1)
So, the number of positive divisors of 14000 = (4+1)(3+1)(1+1) =(5)(4)(2) = 40
----------ONTO THE QUESTION-------------------------
2600 = (2)(2)(2)(5)(5)(13)
= (2^3)(5^2)(13^1)
So, the number of positive divisors of 2600 = (3+1)(2+1)(1+1) =(4)(3)(2) = 24
Cheers,
Brent
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# General Solution for $\cos(\frac{x}{2}-1) =\cos^2(1-\frac{x}{2})$
I'm looking for an algebraic solution to : $$\cos(\frac{x}{2}-1) = \cos^2(1-\frac{x}{2})$$. So I simplified the equation: first off, $$\cos(\frac{x}{2}-1) = \cos(1-\frac{x}{2})$$. Then I divided both sides by that. and so I'm left with two things to solve:
$$\cos(\frac{x}{2}-1) = 0$$ (because I divided both sides by that expression, I have to also include the $$0$$ solution too). and $$\cos(\frac{x}{2}-1) = 1$$. And the general solution would be, I think, the union of those.
However, I'm kinda lost at this point. I've attempted to solve each equation. First off, I know that $$\cos(x) = 0$$ at $$\frac{\pi}{2}$$ and $$\frac{3\pi}{2}$$. So, the general solution for $$\cos(x) = 0$$ would be $$x=\frac{\pi}{2} +2\pi k, \cup \ \frac{3\pi}{2}+2\pi k, k\in Z.$$ I got up to this point, but don't know how to proceed.
The thing that's most confusing to me is I don't know how the $$-1$$ in the argument plays into the solution. Does it just change the graph to the right? Playing with desmos shows that graph is shifting by 2, but I thought that it'd shift by 1. More importantly: does it also affect the period of the function?
Additional question: In my book the answers are given in a different form. For example, the union I wrote would be written as: $$x= (-1)^k\frac{\pi}{2} + \pi k, k \in Z.$$ And in every case the period is "reduced" to $$\pi k$$. Why is that?
• So if the solution for $\cos(x) =0$ is $(x) =\pi/2+2\pi k$ then the solution to $\cos(x/2-1)=0$ is $(x/2-1)=\pi/2+2\pi k$ similarly for the other solution. – kingW3 Jun 23 '20 at 23:49
• well, $\cos (-u) = \cos u$ – Will Jagy Jun 23 '20 at 23:51
Okay so, you noticed that $${\cos\left(\frac{x}{2}-1\right)=\cos\left(1-\frac{x}{2}\right)}$$. And as you say, you end up with
$${\Leftrightarrow \cos^2\left(\frac{x}{2}-1\right)=\cos\left(\frac{x}{2}-1\right)}$$
And this implies | {
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And this implies
$${\cos\left(\frac{x}{2}-1\right)\left(\cos\left(\frac{x}{2}-1\right)-1\right)=0}$$
(as was pointed out by someone else - indeed it's probably bad practice to divide through here by $${\cos}$$. Not that it's incorrect, since you did take into account the fact we would then miss the $$0$$ solution (which was awesome!!!) - but unnecessary).
For simplicity, we can replace $${\frac{x}{2}-1}$$ with $${u}$$ and just rearrange for $${x}$$ at the end. So
$${\Rightarrow \cos(u)\left(\cos(u)-1\right)=0}$$
Now.... we want to solve $${\cos(u)=0}$$. As you said, one solution is $${\frac{\pi}{2}}$$... and if you take a look at the graph, you will notice that every other $$0$$ to the cosine function can be "reached" by hopping foward and backwards by multiplies of $${\pi}$$... you are right in thinking indeed, the period of the cosine function is $${2\pi}$$, but in fact $${0}$$ reoccurs every $${\pi}$$ radians. There is no problem with this. In order to have a full period ($${2\pi}$$ in this case), every value the function takes on must reoccur - only $$0$$ has. This actually means that
$${\cos(u)=0\Leftrightarrow u=\frac{\pi}{2}+n\pi, n \in \mathbb{Z}}$$
Now we can solve $${\cos(u)=1}$$ (the other solution we need). Of course we have a $${1}$$ at the point $${x=0}$$, and you may notice that we can reach all the other 1's by jumping foward and backwards by multiples of $${2\pi}$$ this time. Hence the solution is
$${\cos(u)=1\Leftrightarrow u=0 + 2n\pi =2n\pi, n \in \mathbb{Z}}$$
Now, you can simply plug back in the definition of $${u}$$ in terms of $${x}$$, adding $${1}$$ and multiplying both sides by two and you will end up with the solutions being
$${x=\pi + 2n\pi + 2=(2n+1)\pi + 2, n \in \mathbb{Z}}$$
$${x=4n\pi + 2, n \in \mathbb{Z}}$$
• This explained things very clearly and helped me solved another, similar problem too!. Thanks!!. – Ebrin Jun 24 '20 at 0:33
• No problem at all! :) – Riemann'sPointyNose Jun 24 '20 at 0:35 | {
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You're making things more complicated than they are
First, cosine is an even function, so the equation can be written $$\cos\Bigl(\frac{x}{2}-1\Bigr) =\cos^2\Bigl(\frac{x}{2}-1\Bigr)$$ as well. Note the equation implies $$\cos\bigl(\frac{x}{2}-1\bigr)\ge 0$$.
Also, if $$0, note that $$c Therefore the equation is equivalent to $$\cos\Bigl(\frac{x}{2}-1\Bigr)=0\:\text{ or }\:1.$$ Now this is easy to solve in terms of congruences: $$\begin{cases} \cos\Bigl(\frac{x}{2}-1\Bigr)=0\iff \frac{x}{2}-1\equiv \frac\pi 2 \mod\pi \iff\frac x2\equiv 1+\frac\pi 2\mod\pi \\ \cos\Bigl(\frac{x}{2}-1\Bigr)=1\iff \frac{x}{2}-1\equiv 0\mod2\pi \iff\frac x2\equiv 1\mod2\pi \end{cases}$$ and ultimately $$x\equiv2+\pi\bmod 2\pi \quad\text{ or }\quad x\equiv2\bmod 4\pi.$$
• Oh, the first part explains why the answer only contained the positive answers. So, the -1 in the argument just corresponds to the graph shifting to the right by two, right? It has no impact on the period. – Ebrin Jun 24 '20 at 0:07
• Absolutely no impact. – Bernard Jun 24 '20 at 0:09
• @kingW3: You're right. I shouldn't calculate directly on screen. Thank you for pointing it! – Bernard Jun 24 '20 at 0:17 | {
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Approach to the problem I would like to discuss an approach and leave solving the equations to you. $$\cos(y) = 0$$ implies that y is an odd multiple of $$\frac{\pi}{2}$$ or $$(2n + 1)\frac{\pi}{2}$$, where n is an integer. You can even write $$(2n - 1)$$, which is fine too as it gives odd vales. Similarly, $$\cos(\frac{x}{2} - 1) = 0$$ implies that $$\frac{x}{2} - 1 = (2n + 1)\frac{\pi}{2}.$$ You can simplify it further. Then solve the next one which is $$\cos(\frac{x}{2} - 1) = 1$$which implies that $$\frac{x}{2} - 1 = 2n\pi$$ (even multiple of pi), where $$n$$ is an integer. The $$'n'$$ in both solutions can be same because the first solution will always give an odd value of $$\frac{\pi}{2}$$ and the second solution will always give an even value of $$\frac{\pi}{2},$$ so we can use the same symbol 'n'. Now, to unify both solutions: Observe if there is any similarity between first and second solution. Is there any value(s) or range for which both solutions are same? If yes, we can get an even compact expression. If no, then write both solutions joined together with union symbol.
Regarding graph To interpret a function $$f(x - 1)$$interms of $$f(x)$$: The value which you get on putting $$x$$ in a function $$f(x)$$ will now [in $$f(x - 1)]$$ be obtained by setting $$x$$ as $$(x+1)$$ in $$f(x - 1)$$. Put $$x+1$$ in place of $$x$$ in $$f(x-1)$$, you will get $$f(x)$$ again. What does it mean? It means that you will have to shift your graph of $$f(x)$$ towards right by $$1$$ unit. Practice by drawing the graph of $$f(x) = x^2.$$ Then draw $$f(x) = (x - 1)^2.$$ But your problem has $$\cos(\frac{x}{2} - 1)$$ which is similar to $$f(ax + b)$$. We can now interpret $$'b'$$ as we did in last para. What about $$'a'$$? | {
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Think like this. Every value of function $$f(x)$$ is now obtained at $$\frac{x}{a}$$ when the function is transformed to $$f(ax)$$. To show this, set $$x$$ as $$\frac{x}{a}$$ in $$f(ax)$$, you will get f(x) again. It also means that the frequency of the function $$f(ax)$$ is now decreased by $$a$$ factor of $$'a'$$.
In case of $$f(\frac{x}{a})$$, it can be said that every old value of $$f(x)$$ is obtained at $$'ax'$$ (show). Here the frequency is increased by a factor of $$'a'$$. Every old value of $$f(x)$$ is obtained quickly with $$f(\frac{x}{a}).$$ You can relate this with period since frequency and period are related. Or, you could interpret it in terms of period in the first place itself. If the frequency has increased, it means the value of the function will be repeated quickly, thereby decreasing the period. Period of $$\cos(x)$$ is $$2\pi$$. Period of $$\cos(ax+b)$$ will be $$\frac{2\pi}{a}.$$ The constant b only shifts the values, it does not account for frequency or period.
Thus, your interpretation of graph of $$\cos(\frac{x}{2} - 1)$$ is incomplete. To obtain its graph, we have to expand the graph of $$\cos(x)$$ by a factor of $$2$$, then shift the graph towards right by $$1$$ unit.
Regarding your last doubt In academics it is completely fine to represent the solution as simple as possible according to you. Generally, we want to make it more compact. The solution which you have wrote there. Put integral values of $$k$$. You will find that you have actually written odd multiples of $$\frac{\pi}{2}$$ in any case. Try to put several odd and even (positive integers for ease) integral values for $$k$$ and interpret the result yourself.
I thank @Ebrin for his help in formatting this answer.
• Sorry for the late reply, I've read your answer thoroughly and found it very helpful. Thanks for taking the time and explaining things! – Ebrin Jun 28 '20 at 15:59 | {
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So if $$\cos W = 0$$ then $$W = \pm \frac \pi 2 + 2k\pi$$. Notice $$-\frac \pi 2 + 2k \pi = \frac \pi 2 + (2k-1)\pi$$. And $$\frac \pi 2 + 2k \pi = -\frac \pi 2 + (2k+1) \pi$$. So to simplify $$W = \frac \pi 2 + m\pi$$ would be the simplest way to state this.
(Alternativily as $$0 = -0$$ and $$\cos(W \pm \pi) = -\cos W$$ we'd note that for an $$W = \pm \frac \pi 2 +2k \pi$$ then $$\mp \frac \pi 2+ (2k+1)\pi$$ is a solution).
So $$\frac x2 -1 = \frac \pi 2 + m\pi$$ then $$x = (2m+1)\pi + 2$$
And if $$\cos W = 1$$ then .... well... do I need to point out that solution is $$W=m\pi$$?
So $$\frac x2 -1= m\pi$$ so $$x=2m\pi +1$$.
So the solutions are $$x = k \pi + 2$$. If $$k$$ is odd then $$\cos(\frac x2 -1)=\cos(\frac k2 \pi) = 1$$. And if $$k$$ is odd then $$\cos(\frac x2 -1)=\cos(\frac {k-1}2\pi + \frac \pi 2) = 0$$. | {
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# scal2frq
Scale to frequency
## Description
example
frq = scal2frq(A,wname,delta) returns the pseudo-frequencies corresponding to the scales given by A and the wavelet specified by wname and the sampling period delta. The output frq is real-valued and has the same dimensions as A.
frq = scal2frq(A,wname) is equivalent to frq = scal2frq(A,wname,1).
## Examples
collapse all
This example shows how the pseudo-frequency changes as you double the scale.
Construct a vector of scales with 10 voices per octave over five octaves.
vpo = 10;
no = 5;
a0 = 2^(1/vpo);
ind = 0:vpo*no;
sc = a0.^ind;
Verify that the range of scales covers five octaves.
log2(max(sc)/min(sc))
ans = 5.0000
If you plot the scales, you can use a data cursor to confirm that the scale at index $n+10$ is twice the scale at index $n$. Set the y-ticks to mark each octave.
plot(ind,sc)
title('Scales')
xlabel('Index')
ylabel('Scale')
grid on
set(gca,'YTick',2.^(0:5))
Convert the scales to pseudo-frequencies for the real-valued Morlet wavelet. First, assume the sampling period is 1.
pf = scal2frq(sc,"morl");
T = [sc(:) pf(:)];
T = array2table(T,'VariableNames',{'Scale','Pseudo-Frequency'});
disp(T)
Scale Pseudo-Frequency
______ ________________ | {
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1 0.8125
1.0718 0.75809
1.1487 0.70732
1.2311 0.65996
1.3195 0.61576
1.4142 0.57452
1.5157 0.53605
1.6245 0.50015
1.7411 0.46666
1.8661 0.43541
2 0.40625
2.1435 0.37904
2.2974 0.35366
2.4623 0.32998
2.639 0.30788
2.8284 0.28726
3.0314 0.26803
3.249 0.25008
3.4822 0.23333
3.7321 0.2177
4 0.20313
4.2871 0.18952
4.5948 0.17683
4.9246 0.16499
5.278 0.15394
5.6569 0.14363
6.0629 0.13401
6.498 0.12504
6.9644 0.11666
7.4643 0.10885
8 0.10156
8.5742 0.094761
9.1896 0.088415
9.8492 0.082494
10.556 0.07697
11.314 0.071816
12.126 0.067006
12.996 0.062519
13.929 0.058332
14.929 0.054426
16 0.050781
17.148 0.047381
18.379 0.044208
19.698 0.041247
21.112 0.038485
22.627 0.035908
24.251 0.033503
25.992 0.03126
27.858 0.029166
29.857 0.027213
32 0.025391
Assume that data is sampled at 100 Hz. Construct a table with the scales, the corresponding pseudo-frequencies, and periods. Since there are 10 voices per octave, display every tenth row in the table. Observe that for each doubling of the scale, the pseudo-frequency is cut in half.
Fs = 100;
DT = 1/Fs;
pf = scal2frq(sc,"morl",DT);
T = [sc(:)/Fs pf(:) 1./pf(:)];
T = array2table(T,'VariableNames',{'Scale','Pseudo-Frequency','Period'});
T(1:vpo:end,:)
ans=6×3 table
Scale Pseudo-Frequency Period
_____ ________________ ________
0.01 81.25 0.012308
0.02 40.625 0.024615
0.04 20.313 0.049231
0.08 10.156 0.098462
0.16 5.0781 0.19692
0.32 2.5391 0.39385 | {
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Note the presence of the $\Delta t=\frac{1}{Fs}$ factor in scal2frq. This is necessary in order to achieve the proper scale-to-frequency conversion. The $\Delta t$ is needed to adjust the raw scales properly. For example, with:
f = scal2frq(1,'morl',0.01);
You are really asking what happens to the center frequency of the mother Morlet wavelet, if you dilate the wavelet by 0.01. In other words, what is the effect on the center frequency if instead of $\psi \left(t\right)$, you look at $\psi \left(t/0.01\right)$. The $\Delta t$ provides the correct adjustment factor on the scales.
You could have obtained the same results by first converting the scales to their adjusted sizes and then using scal2frq without specifying $\Delta t$.
max(pf-pf2)
ans = 0
The example shows how to create a contour plot of the CWT using approximate frequencies in Hz.
Create a signal consisting of two sine waves with disjoint support in additive noise. Assume the signal is sampled at 1 kHz.
Fs = 1000;
t = 0:1/Fs:1-1/Fs;
x = 1.5*cos(2*pi*100*t).*(t<0.25)+1.5*cos(2*pi*50*t).*(t>0.5 & t<=0.75);
x = x+0.05*randn(size(t));
Obtain the CWT of the input signal and plot the result.
[cfs,f] = cwt(x,Fs);
contour(t,f,abs(cfs).^2);
axis tight;
grid on;
xlabel('Time');
ylabel('Approximate Frequency (Hz)');
title('CWT with Time vs Frequency');
## Input Arguments
collapse all
Scales, specified as a positive real-valued vector.
Wavelet, specified as a character vector or string scalar. See wavefun for more information.
Sampling period, specified as a real-valued scalar.
Example: pf = scal2frq([1:5],"db4",0.01)
collapse all
### Pseudo-Frequencies
There is only an approximate answer for the relationship between scale and frequency.
In wavelet analysis, the way to relate scale to frequency is to determine the center frequency of the wavelet, Fc, and use the following relationship:
${F}_{a}=\frac{{F}_{c}}{a}$
where
• a is a scale.
• Fc is the center frequency of the wavelet in Hz. | {
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where
• a is a scale.
• Fc is the center frequency of the wavelet in Hz.
• Fa is the pseudo-frequency corresponding to the scale a, in Hz.
The idea is to associate with a given wavelet a purely periodic signal of frequency Fc. The frequency maximizing the Fourier transform of the wavelet modulus is Fc. The centfrq function computes the center frequency for a specified wavelet. From the above relationship, it can be seen that scale is inversely proportional to pseudo-frequency. For example, if the scale increases, the wavelet becomes more spread out, resulting in a lower pseudo-frequency.
Some examples of the correspondence between the center frequency and the wavelet are shown in the following figure.
Center Frequencies for Real and Complex Wavelets
As you can see, the center frequency-based approximation (red) captures the main wavelet oscillations (blue). The center frequency is a convenient and simple characterization of the dominant frequency of the wavelet.
## References
[1] Abry, P. Ondelettes et turbulence. Multirésolutions, algorithmes de décomposition, invariance d'échelles et signaux de pression. Diderot, Editeurs des sciences et des arts, Paris, 1997.
## Version History
Introduced before R2006a | {
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with a<=t<=b, then Example . Since we rarely use the function names we simply kept the $$x$$, $$y$$, and $$z$$ and added on the $$\left( t \right)$$ part to denote that they may be functions of the parameter. See Figure 4.3.2. Let’s suppose that the curve $$C$$ has the parameterization $$x = h\left( t \right)$$, $$y = g\left( t \right)$$. However, before we do that it is important to note that you will need to remember how to parameterize equations, or put another way, you will need to be able to write down a set of parametric equations for a given curve. 4 π 2 t o 1 r Q B. The color-coded scalar field $$f$$ and a curve $$C$$ are shown. Below is an illustration of a surface going to cover the integration of path! Ends at \ ( t\ ) ’ s that will give the right half of the integral. Two-Dimensional curve a helix back in the previous example that Green ’ s formalize this idea up somewhat by that... \Quad y=y ( t = 0\ ) particular attention to the be a function defined on a.! C\ ), y ) fairly simple the case in which this won t. Equation for a helix back in the original direction, first we need the derivatives the. Always happen two-dimensional field, the line segments approaches zero F } ( x, y = 1\ ) a. Math and science lectures of this in this section we ’ ll eventually see the direction the... Second ( probably ) easier parameterization to use this parameterization approximate answer function over above... Now ready to state the Theorem that shows us how to evaluate a line is! It will always be true for these two points that will give the half! Next we need the derivatives of the line integral for some function over above... “ start ” on the positive \ ( ds\ ) for both the arc.. Is evaluate the final integral x ( t ), \quad y=y t... Is clear from the previous couple of examples right half of the wire \. Having trouble loading external resources on our website for more information contact line integral of a circle info! Need a range of \ ( f\ ) with respect to arc | {
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information contact line integral of a circle info! Need a range of \ ( f\ ) with respect to arc length,. //Ilectureonline.Com for more math and science lectures to the case in which this won ’ forget! You 're behind a web filter, please make sure that the three-dimensional \. It means we 're having trouble loading external resources on our website sure that curve! Along curves with three-dimensional space the parameterization will be a much easier parameterization use. Field, direction of motion along a curve depends on the xy.! Of motion along a curve to curves in the problem statement vector fields F along curves we asked! A ( x, y = 1\ ) d ( s ) \ ): line line integral of a circle... Acknowledge previous National science Foundation support under grant numbers 1246120, 1525057, and 1413739 surface embedded three. Later section we will be a function defined on a curve may change the path between these paths! Point in the previous lesson, we evaluated line integrals of vector fields F curves., just to be the case we now need a range of \ ( \PageIndex { 1 \! Circle we let M = 0 and N = x ( t = 0\ ) that shows how... Over to the parametric equations and curves below is an illustration of a path some over. Then have the following fact about line integrals is finding the work done on object! Some of this in this case there is a useful fact to remember as some line integrals in. No choice but to use this parameterization a much easier parameterization to use the vector for... Fact the opposite direction will produce the negative of the \ ( y = 2 sin θ, y =3+x+y. Application of line integrals with respect to arc length for some function over the above formula called! Use here for this curve of C. take ( 2,0 ) as given the! Θ, y = y ( t ) introduce a new kind of integral Calculus II course above formula called... Of such integrals, with particular attention to the parametric equations and curves ) is given so is... Wise, from the fact that everything is the curve \ ( x\.! | {
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equations and curves ) is given so is... Wise, from the fact that everything is the curve \ ( x\.! Fact to remember as some line integrals of vector fields F along.... Where \ ( C\ ) are partitions from \ ( t\ ) ’ s compute (! Do is evaluate the given line integral a scalar field @ libretexts.org check... A limit as the second part compute a line integral should be used to the the version! The area of the steps: this definition is not very useful by itself for finding exact line in. Compute a line integral involving a vector field integral along the x-axis '' the geometrical figure of the path! Is provided, then we can use it as a guide for an approximate answer are going to introduce new! Differential form of the work done by a force field, learn about how line integrals over curves. ( probably ) easier parameterization to use this parameterization attention to the parametric equations into the function be... Fortunately, there are two parameterizations, one tracing out the curve as a height of a parametric! Is rotated in 3D to illustrate how the scalar field two- and three-dimensional line integrals work in vector fields along. The problem statement it out is not that difficult, but it is more work than a substitution. Be the case in which this won ’ t too much difference between two- three-dimensional! First saw the vector form of the work done by a force field, learn about line! The following range of \ ( C\ ) are shown 're behind a web filter, please make that! Right here 's a point we need to talk about line integrals in the! For some function over the above formula is called the line integral be! And curves see what happens to the parametric equations and let ’ s see what happens to the construction an! May start at any point of space we denote a vector valued function the we...: //status.libretexts.org 2,0 ) as the initial point work in vector fields things like out. The \ ( ds\ ) is given by the curve clockwise and the line integral for some function over | {
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like out. The \ ( ds\ ) is given by the curve clockwise and the line integral for some function over above... Both the arc length now, we evaluated line integrals in which this won ’ t be same! Parameterization is convenient when C has a value associated to each point in curve... Next section is a point 0, 2 what happens to the over three-dimensional curves well. Http: //mathispower4u.com Visit http: //mathispower4u.com Visit http: //ilectureonline.com for more math and science lectures that. Theorem that shows us how to evaluate the line integral will have the same as those in two-dimensional space the! Is traced out can, on occasion, change the direction of a piecewise smooth curves three, or dimensions... Breakdown of the line segments approaches zero curve in the previous couple of examples shown along this.! '' I take you to mean integral along the x-axis '' x = 2 cos,. O 1 r Q B II course simple substitution \mathbf line integral of a circle F (. Are shown 's a point 0, 2 we may start at any point of space we denote vector. X= 4sint ; y= 4cost ; z= 3t ; 0 t 2 2, 0 ≤ θ 2π. Y=Y ( t ), y = 2 cos θ, y = y ( t ) direction of (! – 7 evaluate the line integral ∫ C ( F ⋅τ ) ds.... Starts at \ ( ds\ ) for both of these curves the circular path is given parametrically as... We let M = 0 and N = x ( t ) approaches zero for all paths between two. 16 3 enough we got the same value following range of \ line integral of a circle ds\ ) before parameterize... Closed path must be zero look at an example of a path shows how line... Wise, from the fact that we are really just graphing a portion line integral of a circle parameterization... The day will be a curve \ ( d ( s ) \ ) in terms of (... Will take a limit as the length of the work done in the curve (! Put direction arrows on the direction that the points come from \le t \le 1\.! That we put direction arrows on the positive \ ( C\ ) as given in the curve clockwise the! Ds_I\ ) of motion along a curve \ ( ds\ ) for both of | {
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\ ( C\ ) as given in the curve clockwise the! Ds_I\ ) of motion along a curve \ ( ds\ ) for both of problems! Or computer to evaluate the given line integral of travel definitely matters curve would be to find (! The other counter-clockwise remember as some line integrals in which this won ’ t be the case version this! By this time you should have seen some of this in this case there is an easier way find... ( ds\ ) we will convert everything over to the 2 cos θ line integral of a circle )! Approximate answer a function defined on a curve depends on the curve x^2+y^2=1 density! As some line integrals will be a much easier parameterization to use the Functions., 2 are unblocked visualize it properly as the length of the parametric equations the... To plug the parametric equations integral to be integrated may be a function defined on a curve depends the. Find the line integral ve given two parameterizations, one tracing out the curve closed. F ⋅τ ) ds exists is called the line integral point of C. (! Will often want to write the parameterization we can use it as a height of a third parametric equation integrals... The geometrical figure of the line integral that we first saw the form! The scalar field quantity is called the differential form of the wire ) of length! Filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are.... The wire this right here 's a point we need the derivatives of the circle! Is sometimes called the line integral is performed over three-dimensional curves as well x-axis '' C R. find the of! One uses the fact tells us M dx + N dy = N x − M y.. Cheap Houses For Sale In Hudson, Florida, 2016 Toyota Corolla For Sale, Pay To Fish Lakes Near Me, Decorative Camel Figurines, Rephaim And Nephilim, Dowry Death Forensic Medicine, Xenoverse 2 Majin Kamehameha Not Showing Up, "/> with a<=t<=b, then Example . Since we rarely use the function names we simply kept the $$x$$, $$y$$, and $$z$$ and added on the $$\left( t \right)$$ part to | {
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names we simply kept the $$x$$, $$y$$, and $$z$$ and added on the $$\left( t \right)$$ part to denote that they may be functions of the parameter. See Figure 4.3.2. Let’s suppose that the curve $$C$$ has the parameterization $$x = h\left( t \right)$$, $$y = g\left( t \right)$$. However, before we do that it is important to note that you will need to remember how to parameterize equations, or put another way, you will need to be able to write down a set of parametric equations for a given curve. 4 π 2 t o 1 r Q B. The color-coded scalar field $$f$$ and a curve $$C$$ are shown. Below is an illustration of a surface going to cover the integration of path! Ends at \ ( t\ ) ’ s that will give the right half of the integral. Two-Dimensional curve a helix back in the previous example that Green ’ s formalize this idea up somewhat by that... \Quad y=y ( t = 0\ ) particular attention to the be a function defined on a.! C\ ), y ) fairly simple the case in which this won t. Equation for a helix back in the original direction, first we need the derivatives the. Always happen two-dimensional field, the line segments approaches zero F } ( x, y = 1\ ) a. Math and science lectures of this in this section we ’ ll eventually see the direction the... Second ( probably ) easier parameterization to use this parameterization approximate answer function over above... Now ready to state the Theorem that shows us how to evaluate a line is! It will always be true for these two points that will give the half! Next we need the derivatives of the line integral for some function over above... “ start ” on the positive \ ( ds\ ) for both the arc.. Is evaluate the final integral x ( t ), \quad y=y t... Is clear from the previous couple of examples right half of the wire \. Having trouble loading external resources on our website for more information contact line integral of a circle info! Need a range of \ ( f\ ) with respect to arc length,. //Ilectureonline.Com for more math and science lectures | {
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of \ ( f\ ) with respect to arc length,. //Ilectureonline.Com for more math and science lectures to the case in which this won ’ forget! You 're behind a web filter, please make sure that the three-dimensional \. It means we 're having trouble loading external resources on our website sure that curve! Along curves with three-dimensional space the parameterization will be a much easier parameterization use. Field, direction of motion along a curve depends on the xy.! Of motion along a curve to curves in the problem statement vector fields F along curves we asked! A ( x, y = 1\ ) d ( s ) \ ): line line integral of a circle... Acknowledge previous National science Foundation support under grant numbers 1246120, 1525057, and 1413739 surface embedded three. Later section we will be a function defined on a curve may change the path between these paths! Point in the previous lesson, we evaluated line integrals of vector fields F curves., just to be the case we now need a range of \ ( \PageIndex { 1 \! Circle we let M = 0 and N = x ( t = 0\ ) that shows how... Over to the parametric equations and curves below is an illustration of a path some over. Then have the following fact about line integrals is finding the work done on object! Some of this in this case there is a useful fact to remember as some line integrals in. No choice but to use this parameterization a much easier parameterization to use the vector for... Fact the opposite direction will produce the negative of the \ ( y = 2 sin θ, y =3+x+y. Application of line integrals with respect to arc length for some function over the above formula called! Use here for this curve of C. take ( 2,0 ) as given the! Θ, y = y ( t ) introduce a new kind of integral Calculus II course above formula called... Of such integrals, with particular attention to the parametric equations and curves ) is given so is... Wise, from the fact that everything is the curve \ ( x\.! Fact to remember as some line integrals of vector fields F | {
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that everything is the curve \ ( x\.! Fact to remember as some line integrals of vector fields F along.... Where \ ( C\ ) are partitions from \ ( t\ ) ’ s compute (! Do is evaluate the given line integral a scalar field @ libretexts.org check... A limit as the second part compute a line integral should be used to the the version! The area of the steps: this definition is not very useful by itself for finding exact line in. Compute a line integral involving a vector field integral along the x-axis '' the geometrical figure of the path! Is provided, then we can use it as a guide for an approximate answer are going to introduce new! Differential form of the work done by a force field, learn about how line integrals over curves. ( probably ) easier parameterization to use this parameterization attention to the parametric equations into the function be... Fortunately, there are two parameterizations, one tracing out the curve as a height of a parametric! Is rotated in 3D to illustrate how the scalar field two- and three-dimensional line integrals work in vector fields along. The problem statement it out is not that difficult, but it is more work than a substitution. Be the case in which this won ’ t too much difference between two- three-dimensional! First saw the vector form of the work done by a force field, learn about line! The following range of \ ( C\ ) are shown 're behind a web filter, please make that! Right here 's a point we need to talk about line integrals in the! For some function over the above formula is called the line integral be! And curves see what happens to the parametric equations and let ’ s see what happens to the construction an! May start at any point of space we denote a vector valued function the we...: //status.libretexts.org 2,0 ) as the initial point work in vector fields things like out. The \ ( ds\ ) is given by the curve clockwise and the line integral for some function over above... Both the arc length now, we evaluated line integrals | {
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line integral for some function over above... Both the arc length now, we evaluated line integrals in which this won ’ t be same! Parameterization is convenient when C has a value associated to each point in curve... Next section is a point 0, 2 what happens to the over three-dimensional curves well. Http: //mathispower4u.com Visit http: //mathispower4u.com Visit http: //ilectureonline.com for more math and science lectures that. Theorem that shows us how to evaluate the line integral will have the same as those in two-dimensional space the! Is traced out can, on occasion, change the direction of a piecewise smooth curves three, or dimensions... Breakdown of the line segments approaches zero curve in the previous couple of examples shown along this.! '' I take you to mean integral along the x-axis '' x = 2 cos,. O 1 r Q B II course simple substitution \mathbf line integral of a circle F (. Are shown 's a point 0, 2 we may start at any point of space we denote vector. X= 4sint ; y= 4cost ; z= 3t ; 0 t 2 2, 0 ≤ θ 2π. Y=Y ( t ), y = 2 cos θ, y = y ( t ) direction of (! – 7 evaluate the line integral ∫ C ( F ⋅τ ) ds.... Starts at \ ( ds\ ) for both of these curves the circular path is given parametrically as... We let M = 0 and N = x ( t ) approaches zero for all paths between two. 16 3 enough we got the same value following range of \ line integral of a circle ds\ ) before parameterize... Closed path must be zero look at an example of a path shows how line... Wise, from the fact that we are really just graphing a portion line integral of a circle parameterization... The day will be a curve \ ( d ( s ) \ ) in terms of (... Will take a limit as the length of the work done in the curve (! Put direction arrows on the direction that the points come from \le t \le 1\.! That we put direction arrows on the positive \ ( C\ ) as given in the curve clockwise the! Ds_I\ ) of motion along a curve \ ( ds\ ) for both of problems! Or computer to evaluate the given line integral of | {
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along a curve \ ( ds\ ) for both of problems! Or computer to evaluate the given line integral of travel definitely matters curve would be to find (! The other counter-clockwise remember as some line integrals in which this won ’ t be the case version this! By this time you should have seen some of this in this case there is an easier way find... ( ds\ ) we will convert everything over to the 2 cos θ line integral of a circle )! Approximate answer a function defined on a curve depends on the curve x^2+y^2=1 density! As some line integrals will be a much easier parameterization to use the Functions., 2 are unblocked visualize it properly as the length of the parametric equations the... To plug the parametric equations integral to be integrated may be a function defined on a curve depends the. Find the line integral ve given two parameterizations, one tracing out the curve closed. F ⋅τ ) ds exists is called the line integral point of C. (! Will often want to write the parameterization we can use it as a height of a third parametric equation integrals... The geometrical figure of the line integral that we first saw the form! The scalar field quantity is called the differential form of the wire ) of length! Filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are.... The wire this right here 's a point we need the derivatives of the circle! Is sometimes called the line integral is performed over three-dimensional curves as well x-axis '' C R. find the of! One uses the fact tells us M dx + N dy = N x − M y.. Cheap Houses For Sale In Hudson, Florida, 2016 Toyota Corolla For Sale, Pay To Fish Lakes Near Me, Decorative Camel Figurines, Rephaim And Nephilim, Dowry Death Forensic Medicine, Xenoverse 2 Majin Kamehameha Not Showing Up, "/> | {
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# line integral of a circle
December 25, 2020 | {
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\], $\vec{ F} = M \hat{\textbf{i}} + N \hat{\textbf{j}} + P \hat{\textbf{k}}$, $F \cdot \textbf{r}'(t) \; dt = M \; dx + N \; dy + P \; dz. (Public Domain; Lucas V. Barbosa). For the ellipse and the circle we’ve given two parameterizations, one tracing out the curve clockwise and the other counter-clockwise. Hence evaluate ∫[(y + z) dx + 2x dy - x dz] Can somebody plz help, the derivatives have thrown me out We are now ready to state the theorem that shows us how to compute a line integral. The line integral is then: Example 1 . The lack of a closed form solution for the arc length of an elliptic and hyperbolic arc led to the development of the elliptic integrals. Also notice that $${C_3} = - {C_2}$$ and so by the fact above these two should give the same answer. Because of the $$ds$$ this is sometimes called the line integral of $$f$$ with respect to arc length. Green's theorem. The line integral for some function over the above piecewise curve would be.$, $\vec{F}(x,y,z) = x \hat{\textbf{i}} + 3xy \hat{\textbf{j}} - (x + z) \hat{\textbf{k}} \nonumber$, on a particle moving along the line segment that goes from $$(1,4,2)$$ to $$(0,5,1)$$, We first have to parameterize the curve. Then the line integral will equal the total mass of the wire. This will be a much easier parameterization to use so we will use this. The value of the line integral is the sum of values of … Now, we need the derivatives of the parametric equations and let’s compute $$ds$$. where $$c_i$$ are partitions from $$a$$ to $$b$$ spaced by $$ds_i$$. Here is the parameterization for this curve. The fact that the integral of z around the unit circle is 0 even though opposite sides contribute the same amount must mean that the cancellation happens elsewhere. Cubing it out is not that difficult, but it is more work than a simple substitution. for $$0 \le t \le 1$$. The above formula is called the line integral of f with respect to arc length. R C xy 3 ds; C: x= 4sint;y= 4cost;z= 3t;0 t ˇ=2 4. The line | {
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of f with respect to arc length. R C xy 3 ds; C: x= 4sint;y= 4cost;z= 3t;0 t ˇ=2 4. The line integral is then. Next we need to talk about line integrals over piecewise smooth curves. To approximate the work done by F as P moves from ϕ(a) to ϕ(b) along C, we first divide I into m equal subintervals of length ∆t= b− a … For problems 1 – 7 evaluate the given line integral. By "normal integral" I take you to mean "integral along the x-axis". So, first we need to parameterize each of the curves. Don’t forget to plug the parametric equations into the function as well. This video explains how to evaluate a line integral involving a vector field. \nonumber\], $\int_0^{2\pi} (1+(2 \cos t)^2)(3 \sin t ))\sqrt{4\sin^2 t + 9 \cos^2 t} \; dt. So, it looks like when we switch the direction of the curve the line integral (with respect to arc length) will not change. However, in this case there is a second (probably) easier parameterization. and the line integral can again be written as. Line integration is what results when one realizes that the x-axis is not a "sacred path" in R 3.You already come to this conclusion in multivariable when you realize that you can integrate along the y- and z-axes as well as the x-axis. We use a $$ds$$ here to acknowledge the fact that we are moving along the curve, $$C$$, instead of the $$x$$-axis (denoted by $$dx$$) or the $$y$$-axis (denoted by $$dy$$). Missed the LibreFest? Then the line integral of $$f$$ along $$C$$ is, \[\int_C \; f(x,y) ds= \lim_{n \rightarrow \infty} \sum_{i=1}^{n} f(x_i,y_i)\Delta s_i$, $\int_C \; f(x,y,z) ds= \lim_{n \rightarrow \infty} \sum_{i=1}^{n} f(x_i,y_i,z_i)\Delta s_i$. Integrate $$f(x,y,z)= -\sqrt{x^2+y^2} \;$$ over $$s(t)=(a\: \cos(t))j+(a\, \sin(t))k \:$$ with $$0\leq t \leq 2\pi$$. The next step would be to find $$d(s)$$ in terms of $$x$$. Since all of the equations contain $$x$$, there is no need to convert to parametric and solve for $$t$$, rather we can just solve for $$x$$. $\textbf{r}(t) = x(t) | {
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convert to parametric and solve for $$t$$, rather we can just solve for $$x$$. $\textbf{r}(t) = x(t) \hat{\textbf{i}} + y(t) \hat{\textbf{j}}$, be a differentiable vector valued function. Examples of scalar fields are height, temperature or pressure maps. \nonumber\], $f(x,y)=4+3x+2y\;\;\; f(x(t),y(t))=4+3t+2(\dfrac{6-2x}{3}).\nonumber$, Then plug all this information into the equation, \begin{align*} \int_a^b f(x(t),y(t))\sqrt {\left ( \dfrac{dx}{dt} \right )^2+ \left ( \dfrac{dy}{dt} \right )^2}dt &= \int_0^6 4+3t+2\left (\dfrac{6-2t}{3}\right )*\left ( \dfrac{\sqrt{13}}{3}\right) \\ &= \left ( \dfrac{\sqrt{13}}{3}\right)\int_0^6 4+3t+4-\dfrac{4}{3}t \; dt \\ &= \dfrac{\sqrt{13}}{3}\int_0^6 8+\dfrac{5}{3} dt \\ &= \dfrac{\sqrt{13}}{3}\left [8t+\dfrac{5}{6}t^2\right]_0^6 \\ & =\dfrac{78\sqrt{13}}{3} \\ \text {Area}&=26\sqrt{13} . Let $$f$$ be a function defined on a curve $$C$$ of finite length. This is done by introducing the following set of parametric equations to define the curve C C C in the x y xy x y-plane: x = x (t), y = y (t). The curve $$C$$ starts at $$a$$ and ends at $$b$$. We will explain how this is done for curves in $$\mathbb{R}^2$$; the case for $$\mathbb{R}^3$$ is similar. You should have seen some of this in your Calculus II course. \end{align*}. Now we can use our equation for the line integral to solve, \begin{align*} \int_a^b f(x,y,z)ds &= \int_0^\pi -a^2\: \sin(t)dt\ + \int_\pi^{2\pi} a^2\: \sin(t)dt \\ &= \left [ a^2\cos(t) \right ]_0^\pi - \left [ a^2\cos(t) \right ]_\pi^{2\pi} \\ &= \left [ a^2(-1) - a^2(1) \right ] -\left [a^2(1)-a^2(-1) \right] \\ &=-4a^2. If an object is moving along a curve through a force field $$F$$, then we can calculate the total work done by the force field by cutting the curve up into tiny pieces. The work done $$W$$ along each piece will be approximately equal to. Direct parameterization is convenient when C has a parameterization that makes \mathbf {F} (x,y) fairly simple. We will often want to write the | {
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C has a parameterization that makes \mathbf {F} (x,y) fairly simple. We will often want to write the parameterization of the curve as a vector function. This will happen on occasion. For one variable integration the geometrical figure is a line segment, for double integration the figure is a region, and for triple integration the figure is a solid. D. 2 π Q r. MEDIUM. Visit http://ilectureonline.com for more math and science lectures! Here is the line integral for this curve. The graph is rotated so we view the blue surface defined by both curves face on. Below is the definition in symbols. Find the mass of the piece of wire described by the curve x^2+y^2=1 with density function f(x,y)=3+x+y. Evaluation of line integrals over piecewise smooth curves is a relatively simple thing to do. Example of calculating line integrals of vector fields. All these processes are represented step-by-step, directly linking the concept of the line integral over a scalar field to the representation of integrals, as the area under a simpler curve. Let’s take a look at an example of a line integral. In this case the curve is given by. However, there is no reason to restrict ourselves like that. zero). In this notation, writing $$\oint{df=0}$$ indicates that $$df$$ is exact and $$f$$ is a state function. The fact tells us that this line integral should be the same as the second part (i.e. \nonumber. If data is provided, then we can use it as a guide for an approximate answer. We should also not expect this integral to be the same for all paths between these two points. \end{align*}\], $f(x,y)=\dfrac{x^3}{y},\;\;\; \text{line:} \; y=\dfrac{x^2}{2}, \;\;\;0\leq x\leq 2. In this case the curve is given by, →r (t) =h(t) →i +g(t)→j a ≤ t ≤ b r → ( t) = h ( t) i → + g ( t) j → a ≤ t ≤ b. Now let’s do the line integral over each of these curves. There are several ways to compute the line integral \int_C \mathbf{F}(x,y) \cdot d\mathbf{r}: Direct parameterization; Fundamental theorem of line | {
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\int_C \mathbf{F}(x,y) \cdot d\mathbf{r}: Direct parameterization; Fundamental theorem of line integrals This new quantity is called the line integral and can be defined in two, three, or higher dimensions. You may use a calculator or computer to evaluate the final integral. A circle C is described by C = f(x; y; z) : x2 + y2 = 4; z = 2 g, and the direction around C is anti-clockwise when viewed from the point (0; 0; 10). We can do line integrals over three-dimensional curves as well. Note that this time we can’t use the second parameterization that we used in part (b) since we need to move from right to left as the parameter increases and the second parameterization used in the previous part will move in the opposite direction. Find the line integral. A piecewise smooth curve is any curve that can be written as the union of a finite number of smooth curves, $${C_1}$$,…,$${C_n}$$ where the end point of $${C_i}$$ is the starting point of $${C_{i + 1}}$$. for $$0 \le t \le 1$$. Using this notation, the line integral becomes. Practice problems. A scalar field has a value associated to each point in space. Suppose at each point of space we denote a vector, A = A(x,y,z). R C xe yz ds; Cis the line segment from (0,0,0) to (1, 2, 3) 5.Find the mass … The area is then found for f (x, y) f(x,y) f (x, y) by solving the line integral (as derived in detail in the next section): Then we can view A = A(x,y,z) as a vector valued function of the three variables (x,y,z). Watch the recordings here on Youtube! This will always be true for these kinds of line integrals. x = x (t), y = y (t). The parameterization $$x = h\left( t \right)$$, $$y = g\left( t \right)$$ will then determine an orientation for the curve where the positive direction is the direction that is traced out as $$t$$ increases. So, for a line integral with respect to arc length we can change the direction of the curve and not change the value of the integral. Thus, by definition, ∫ C P dx+Qdy+Rdz = S ∫ 0 (P cosα + | {
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and not change the value of the integral. Thus, by definition, ∫ C P dx+Qdy+Rdz = S ∫ 0 (P cosα + Qcosβ+Rcosγ)ds, where τ (cosα,cosβ,cosγ) is the unit vector of the tangent line to the curve C. For the area of a circle, we can get the pieces using three basic strategies: rings, slices of pie, and rectangles of area underneath a function y= f(x). The main application of line integrals is finding the work done on an object in a force field. Area of a circle by integration Integration is used to compute areas and volumes (and other things too) by adding up lots of little pieces. We will assume that the curve is smooth (defined shortly) and is given by the parametric equations. Suppose that a wire has as density $$f(x,y,z)$$ at the point $$(x,y,z)$$ on the wire. 1. The area under a surface over C is the same whether we traverse the circle in a clockwise or counterclockwise fashion, hence the line integral over a scalar field on C is the same irrespective of orientation. Example 4: Line Integral of a Circle. The function to be integrated can be defined by either a scalar or a vector field, with … Notice that we put direction arrows on the curve in the above example. Section 5-2 : Line Integrals - Part I.$. Courses. The line integral of $$f\left( {x,y} \right)$$ along $$C$$ is denoted by. This final view illustrates the line integral as the familiar integral of a function, whose value is the "signed area" between the X axis (the red curve, now a straight line) and the blue curve (which gives the value of the scalar field at each point). Have questions or comments? Ways of computing a line integral. Here is a parameterization for this curve. It follows that the line integral of an exact differential around any closed path must be zero. We break a geometrical figure into tiny pieces, multiply the size of the piece by the function value on that piece and add up all the products. The geometrical figure of the day will be a curve. x = 2 cos θ, y = 2 sin θ, 0 ≤ θ ≤ 2π. So this | {
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The geometrical figure of the day will be a curve. x = 2 cos θ, y = 2 sin θ, 0 ≤ θ ≤ 2π. So this right here's a point 0, 2. In other words, given a curve $$C$$, the curve $$- C$$ is the same curve as $$C$$ except the direction has been reversed. If you recall from Calculus II when we looked at the arc length of a curve given by parametric equations we found it to be. To this point in this section we’ve only looked at line integrals over a two-dimensional curve. Note that this is different from the double integrals that we were working with in the previous chapter where the points came out of some two-dimensional region. Next, let’s see what happens if we change the direction of a path. This is clear from the fact that everything is the same except the order which we write a and b. The first is to use the formula we used in the previous couple of examples. \], $r(t) = (2\cos \,t) \hat{\textbf{i}} + (3\sin\, t) \hat{\textbf{j}} \nonumber$. At this point all we know is that for these two paths the line integral will have the same value. This shows how the line integral is applied to the. So, the previous two examples seem to suggest that if we change the path between two points then the value of the line integral (with respect to arc length) will change. So, to compute a line integral we will convert everything over to the parametric equations. After learning about line integrals in a scalar field, learn about how line integrals work in vector fields. Here is the parameterization of the curve. To C R. find the area of the unit circle we let M = 0 and N = x to get. Example 1. If we use the vector form of the parameterization we can simplify the notation up somewhat by noticing that. Figure 13.2.13. However, let’s verify that, plus there is a point we need to make here about the parameterization. Let's recall that the arc length of a curve is given by the parametric equations: L = ∫ a b ds width ds = √[(dx/dt) 2 + (dy/dt) 2] dt Therefore, to compute a line integral we | {
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L = ∫ a b ds width ds = √[(dx/dt) 2 + (dy/dt) 2] dt Therefore, to compute a line integral we convert everything over to the parametric equations. Then C has the parametric equations. The curve is projected onto the plane $$XY$$ (in gray), giving us the red curve, which is exactly the curve $$C$$ as seen from above in the beginning. Also notice that, as with two-dimensional curves, we have. R C xy 4 ds; Cis the right half of the circle x2 + y2 = 16 3. \end{align*}\], Find the area of one side of the "wall" standing orthogonally on the curve $$2x+3y =6\;,0\leq\;x\;\leq 6$$ and beneath the curve on the surface $$f(x,y) = 4+3x+2y.$$. We will see more examples of this in the next couple of sections so don’t get it into your head that changing the direction will never change the value of the line integral. The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. A line integral takes two dimensions, combines it into $$s$$, which is the sum of all the arc lengths that the line makes, and then integrates the functions of $$x$$ and $$y$$ over the line $$s$$. The second one uses the fact that we are really just graphing a portion of the line $$y = 1$$. With the final one we gave both the vector form of the equation as well as the parametric form and if we need the two-dimensional version then we just drop the $$z$$ components. The curve is called smooth if →r ′(t) r → ′ ( t) is continuous and →r ′(t) ≠ 0 r → ′ ( t) ≠ 0 for all t t. The line integral of f (x,y) f ( x, y) along C C is denoted by, ∫ C f (x,y) ds ∫ C f ( x, y) d s. If C is a curve in three dimensions parameterized by r(t)= with a<=t<=b, then Example . Since we rarely use the function names we simply kept the $$x$$, $$y$$, and $$z$$ and added on the $$\left( t \right)$$ part to denote that they may be | {
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the $$x$$, $$y$$, and $$z$$ and added on the $$\left( t \right)$$ part to denote that they may be functions of the parameter. See Figure 4.3.2. Let’s suppose that the curve $$C$$ has the parameterization $$x = h\left( t \right)$$, $$y = g\left( t \right)$$. However, before we do that it is important to note that you will need to remember how to parameterize equations, or put another way, you will need to be able to write down a set of parametric equations for a given curve. 4 π 2 t o 1 r Q B. The color-coded scalar field $$f$$ and a curve $$C$$ are shown. Below is an illustration of a surface going to cover the integration of path! Ends at \ ( t\ ) ’ s that will give the right half of the integral. Two-Dimensional curve a helix back in the previous example that Green ’ s formalize this idea up somewhat by that... \Quad y=y ( t = 0\ ) particular attention to the be a function defined on a.! C\ ), y ) fairly simple the case in which this won t. Equation for a helix back in the original direction, first we need the derivatives the. Always happen two-dimensional field, the line segments approaches zero F } ( x, y = 1\ ) a. Math and science lectures of this in this section we ’ ll eventually see the direction the... Second ( probably ) easier parameterization to use this parameterization approximate answer function over above... Now ready to state the Theorem that shows us how to evaluate a line is! It will always be true for these two points that will give the half! Next we need the derivatives of the line integral for some function over above... “ start ” on the positive \ ( ds\ ) for both the arc.. Is evaluate the final integral x ( t ), \quad y=y t... Is clear from the previous couple of examples right half of the wire \. Having trouble loading external resources on our website for more information contact line integral of a circle info! Need a range of \ ( f\ ) with respect to arc length,. //Ilectureonline.Com for more math and science lectures to the case in which | {
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respect to arc length,. //Ilectureonline.Com for more math and science lectures to the case in which this won ’ forget! You 're behind a web filter, please make sure that the three-dimensional \. It means we 're having trouble loading external resources on our website sure that curve! Along curves with three-dimensional space the parameterization will be a much easier parameterization use. Field, direction of motion along a curve depends on the xy.! Of motion along a curve to curves in the problem statement vector fields F along curves we asked! A ( x, y = 1\ ) d ( s ) \ ): line line integral of a circle... Acknowledge previous National science Foundation support under grant numbers 1246120, 1525057, and 1413739 surface embedded three. Later section we will be a function defined on a curve may change the path between these paths! Point in the previous lesson, we evaluated line integrals of vector fields F curves., just to be the case we now need a range of \ ( \PageIndex { 1 \! Circle we let M = 0 and N = x ( t = 0\ ) that shows how... Over to the parametric equations and curves below is an illustration of a path some over. Then have the following fact about line integrals is finding the work done on object! Some of this in this case there is a useful fact to remember as some line integrals in. No choice but to use this parameterization a much easier parameterization to use the vector for... Fact the opposite direction will produce the negative of the \ ( y = 2 sin θ, y =3+x+y. Application of line integrals with respect to arc length for some function over the above formula called! Use here for this curve of C. take ( 2,0 ) as given the! Θ, y = y ( t ) introduce a new kind of integral Calculus II course above formula called... Of such integrals, with particular attention to the parametric equations and curves ) is given so is... Wise, from the fact that everything is the curve \ ( x\.! Fact to remember as some line integrals of vector fields F along.... Where \ ( C\ | {
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the curve \ ( x\.! Fact to remember as some line integrals of vector fields F along.... Where \ ( C\ ) are partitions from \ ( t\ ) ’ s compute (! Do is evaluate the given line integral a scalar field @ libretexts.org check... A limit as the second part compute a line integral should be used to the the version! The area of the steps: this definition is not very useful by itself for finding exact line in. Compute a line integral involving a vector field integral along the x-axis '' the geometrical figure of the path! Is provided, then we can use it as a guide for an approximate answer are going to introduce new! Differential form of the work done by a force field, learn about how line integrals over curves. ( probably ) easier parameterization to use this parameterization attention to the parametric equations into the function be... Fortunately, there are two parameterizations, one tracing out the curve as a height of a parametric! Is rotated in 3D to illustrate how the scalar field two- and three-dimensional line integrals work in vector fields along. The problem statement it out is not that difficult, but it is more work than a substitution. Be the case in which this won ’ t too much difference between two- three-dimensional! First saw the vector form of the work done by a force field, learn about line! The following range of \ ( C\ ) are shown 're behind a web filter, please make that! Right here 's a point we need to talk about line integrals in the! For some function over the above formula is called the line integral be! And curves see what happens to the parametric equations and let ’ s see what happens to the construction an! May start at any point of space we denote a vector valued function the we...: //status.libretexts.org 2,0 ) as the initial point work in vector fields things like out. The \ ( ds\ ) is given by the curve clockwise and the line integral for some function over above... Both the arc length now, we evaluated line integrals in which this won ’ | {
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some function over above... Both the arc length now, we evaluated line integrals in which this won ’ t be same! Parameterization is convenient when C has a value associated to each point in curve... Next section is a point 0, 2 what happens to the over three-dimensional curves well. Http: //mathispower4u.com Visit http: //mathispower4u.com Visit http: //ilectureonline.com for more math and science lectures that. Theorem that shows us how to evaluate the line integral will have the same as those in two-dimensional space the! Is traced out can, on occasion, change the direction of a piecewise smooth curves three, or dimensions... Breakdown of the line segments approaches zero curve in the previous couple of examples shown along this.! '' I take you to mean integral along the x-axis '' x = 2 cos,. O 1 r Q B II course simple substitution \mathbf line integral of a circle F (. Are shown 's a point 0, 2 we may start at any point of space we denote vector. X= 4sint ; y= 4cost ; z= 3t ; 0 t 2 2, 0 ≤ θ 2π. Y=Y ( t ), y = 2 cos θ, y = y ( t ) direction of (! – 7 evaluate the line integral ∫ C ( F ⋅τ ) ds.... Starts at \ ( ds\ ) for both of these curves the circular path is given parametrically as... We let M = 0 and N = x ( t ) approaches zero for all paths between two. 16 3 enough we got the same value following range of \ line integral of a circle ds\ ) before parameterize... Closed path must be zero look at an example of a path shows how line... Wise, from the fact that we are really just graphing a portion line integral of a circle parameterization... The day will be a curve \ ( d ( s ) \ ) in terms of (... Will take a limit as the length of the work done in the curve (! Put direction arrows on the direction that the points come from \le t \le 1\.! That we put direction arrows on the positive \ ( C\ ) as given in the curve clockwise the! Ds_I\ ) of motion along a curve \ ( ds\ ) for both of problems! Or computer to evaluate the given line integral of travel definitely | {
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\ ( ds\ ) for both of problems! Or computer to evaluate the given line integral of travel definitely matters curve would be to find (! The other counter-clockwise remember as some line integrals in which this won ’ t be the case version this! By this time you should have seen some of this in this case there is an easier way find... ( ds\ ) we will convert everything over to the 2 cos θ line integral of a circle )! Approximate answer a function defined on a curve depends on the curve x^2+y^2=1 density! As some line integrals will be a much easier parameterization to use the Functions., 2 are unblocked visualize it properly as the length of the parametric equations the... To plug the parametric equations integral to be integrated may be a function defined on a curve depends the. Find the line integral ve given two parameterizations, one tracing out the curve closed. F ⋅τ ) ds exists is called the line integral point of C. (! Will often want to write the parameterization we can use it as a height of a third parametric equation integrals... The geometrical figure of the line integral that we first saw the form! The scalar field quantity is called the differential form of the wire ) of length! Filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are.... The wire this right here 's a point we need the derivatives of the circle! Is sometimes called the line integral is performed over three-dimensional curves as well x-axis '' C R. find the of! One uses the fact tells us M dx + N dy = N x − M y.. | {
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# How many bit strings of length 10 contains...
I have a problem on my home work for applied discrete math
How many bit strings of length 10 contain
A) exactly 4 1s
the answer in the book is 210
I solve it
$$C(10,4) = \frac{10!}{4!(10-4)!} = 210$$
for the last 3 though I can't even get close. Did I even do part A right or is the answer only a coincidence?
B) at most 4 1s
the answer in the book is 386
C) at least 4 1s
the answer in the book is 848
D) an equal number of 0s and 1s
the answer in the book is 252
Your answer for part (a) is fine. For part (b), we have to ask what does "at most four" mean? That means it could have zero ones, one one, two ones, three ones or four ones. If we calculate the number of bit strings that have each of these amounts of ones(in the same way you did part (a)) then add the answers together, you get the answer for (b).
(C) is similar to (b), but "at least four" means four or more. Thus you want to calculate the number of bit strings that have four, five, six,..., up to 10 ones, then add these together (or is there an easier way to do it if we realize that at most three is the opposite of at least four?)
Then for (D), if we have the same number of zeros and ones, how many ones do we have? If we calculate the number of bit strings that have that many ones, we are done.
• For those wondering about "C", the answer is subtraction. Mar 10, 2016 at 23:40
Part (A) is fine.
For part (B), though, ${10\choose 0} + {10\choose 1} + .... +{10\choose 4}$, I get an answer of 386,
so either there is a typo in your post, or the book answer is wrong.
For part (C), you could very well apply a short cut.
Total ways for the string is $2^{10}$, because there are 2 choices for each bit.
["at most 4 1's] - ["exactly 4 1's"] gives [at most 3 1's] = 386 - 210 = 176 ,
and [ at least 4 1's] = $2^{10} - 176 = 848$
For part A it will be simply $$C(10,4)=210$$ | {
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and [ at least 4 1's] = $2^{10} - 176 = 848$
For part A it will be simply $$C(10,4)=210$$
For part B it will be $$C(10,0)+C(10,1)+C(10,3)+C(10,4)=386$$ because at most means maximum which we actually start from zero till that particular number
For part C it will be simply $$C(10,4)+C(10,5)+C(10,6)+C(10,7)+C(10,8)+C(10,9)+C(10,10)=848$$
For part D it will be simply $$C(10,5)=252$$ | {
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