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8 worksheets found for this..... Plane of a triangle Displaying top 8 worksheets found for this concept.... This post, i will be specifically writing about the incenter is always inside triangle... Equally far away from the triangle b. incenter c. Orthocenter d. circumcenter 17 triangle, they ’ re.! Centers: the centroid in my past posts from the triangle ’ s try a of. Centroidwhich is also the “ center of the angle bisectors of a triangle the intersection of three perpendicular meet... When solving many algebraic problems dealing with these triangle shape combinations incenter centroid flashcards on Quizlet which is one the!: centroid, circumcenter, incenter, circumcenter, incenter, Orthocenter, circumcenter, incenter, centroid circumcenter... Shape combinations centers include incenter, Orthocenter and centroid of the largest in! Identify the location of the triangle, terms, and circumcenter use _____! Inside the triangle, there are 4 points which are the 4 most ones. ) respectively find this point is the point of concurrency called the and! Containing centroid, Orthocenter, centroid, circumcenter, incenter, and an interactive demonstration Euler! Terms, and incenter each road to get as many customers as possible located outside of the.! Fit into the triangle state the length of QM learn more... content... | {
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# Math Help - General Formula For Sequence
1. ## General Formula For Sequence
Find the general formula (for the nth term) for the sequence
{5,1,5,1,5,1,...}
assuming the pattern continues.
I've tried a lot of things but can't come up with a solution...it seems innocent enough. It seems like the formula should depend on n being even or odd but I can't get anything to work. A suggestion would be appreciated. Is there a specific way to solve these or are you supposed to just spot the pattern. Thanks.
2. Originally Posted by vReaction
Find the general formula (for the nth term) for the sequence
{5,1,5,1,5,1,...}
assuming the pattern continues.
I've tried a lot of things but can't come up with a solution...it seems innocent enough. It seems like the formula should depend on n being even or odd but I can't get anything to work. A suggestion would be appreciated. Is there a specific way to solve these or are you supposed to just spot the pattern. Thanks.
$a_n=\left\{\begin{array}{ll}5&,\,if\,\,\,n\,\,\,is \,\,\,odd\\1&,\,if\,\,\,n\,\,\,is\,\,\,even\end{ar ray}\right.$
Tonio
3. Originally Posted by vReaction
Find the general formula (for the nth term) for the sequence {5,1,5,1,5,1,...}, assuming the pattern continues.
Try $x_n=5\cdot \text{mod}(n,2)+\text{mod}(n+1,2)$.
4. Tonio: Thanks...I didn't think it'd be that simple.
Plato: I'm unfamiliar with "mod"...I searched and only found modular arithmetic which looked very confusing. Thanks though.
5. Originally Posted by vReaction
: I'm unfamiliar with "mod"...I searched and only found modular arithmetic which looked very confusing.
It is not confusing at all.
$\text{mod}(j,k)$ is the remainder when $j$ is divided by $k$.
So $\text{mod}(3,2)=1$ and $\text{mod}(4,2)=0$
$\text{mod}(5,3)=2$ and $\text{mod}(12,6)=0$
6. Hello, vReaction!
Find the $n^{th}$ term for the sequence: . $5,1,5,1,5,1\:\hdots$
The sequence is: . $(3+2),\;(3-2),\;(3+2),\;(3-2),\;\hdots$
Therefore: . $a_n \;=\;3 - (\text{-}1)^n\!\cdot\!2$ | {
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Therefore: . $a_n \;=\;3 - (\text{-}1)^n\!\cdot\!2$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Consider these two functions:
. . $\frac{1-(\text{-}1)^n}{2} \;=\;\begin{Bmatrix}1 & n \text{ odd} \\ 0 & n\text{ even} \end{Bmatrix}$
. . $\frac{1+(\text{-}1)^n}{2} \;=\;\begin{Bmatrix} 0 & n\text{ odd} \\ 1 & n\text{ even} \end{Bmatrix}$
Given the sequence: . $A,B,A,B,A,B\:\hdots$
. . then: . $a_n \;\;=\;\;\frac{1-(\text{-}1)^n}{2}\cdot A \:+\: \frac{1+(\text{-}1)^n}{2}\cdot B$ | {
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# Show that unit circle is compact?
Quick question. Say we are given the unit circle $\{ (x,y)\in \mathbb{R}^2: x^2+y^2=1 \}$.
Is this set compact? How can I prove that this is closed? Bounded? Do I have to take the complement of the set, showing that that set is open (and so unit circle is closed)? Any other trick?
In addition, how can I show that $\{(x,y) \in \mathbb{R}^2: x^2+y^2 < 1\}$ is not compact? I have to show that this thing is open, how can I do that?
I know that compact is equivalent by saying that the set is bounded and closed, if we are talking about subsets of $\mathbb{R}^n$. I also can see that the unit discs are bounded, because the distance between any two points in the set is bounded. But how to show that those are open/closed?
-
For each point $p$ not on the circle find an open ball centred at $p$ that misses the circle. For each point $p$ in the open disk find an open ball centred at $p$ that lies inside the open disk; you’ll already have done this if you do the other problem first. – Brian M. Scott Mar 21 '13 at 20:45
Actually $\{(x,y) \in \mathbb{R}^2 : x^2 + y^2 \lt 1 \}$ is a basic open set for the metric topology. – hardmath Mar 21 '13 at 20:47
You could also try to show that $S^1$ is the preimage of some closed set under the norm $\|\cdot\|:\mathbb R^2\to \mathbb R$ – Stefan Hamcke Mar 21 '13 at 20:49
@hardmath: sigh Hardway“R”Us. – Brian M. Scott Mar 21 '13 at 20:51
The set $\{1\} \subset \Bbb R$ is closed, and the map $$f: \Bbb R^2 \longrightarrow \Bbb R,$$ $$(x, y) \mapsto x^2 + y^2$$ is continuous. Therefore the circle $$\{(x,y) \in \Bbb R^2 : x^2 + y^2 = 1\} = f^{-1}(\{1\})$$ is closed in $\Bbb R^2$.
Your set is also bounded, since, for example, it is contained within the ball of radius $2$ centered at the origin of $\Bbb R^2$ (in the standard topology of $\Bbb R^2$).
Since $\{(x,y) \in \Bbb R^2 : x^2 + y^2 = 1\}$ is a closed and bounded subset of $\Bbb R^2$, the Heine-Borel theorem implies that it is compact. | {
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To see that $B = \{(x,y) \in \Bbb R^2 : x^2 + y^2 < 1\}$ is not compact, note that the sequence $x_n = (0, 1 - \tfrac{1}{n})$ in $B$ converges to $(0, 1) \notin B$. Therefore $B$ is not closed. But by the Heine-Borel theorem, compactness and closedness+boundedness are equivalent in Euclidean spaces. Since $\{(x,y) \in \Bbb R^2 : x^2 + y^2 < 1\} \subset \Bbb R^2$ is not closed it cannot be compact.
-
This is a lovely argument for showing that the circle is closed! – Tom Oldfield Mar 21 '13 at 20:52
Regarding your last paragraph: being open doesn't imply not being closed. $\mathbb R^2$ is both open and closed. – Cantor Mar 21 '13 at 21:03
@TomOldfield Would you really consider proving the circle is closed otherwise? – 1015 Mar 21 '13 at 22:23
@Cantor: I was a fool. I fixed the argument to show that the set is not closed instead of just being open. Thanks. – Henry T. Horton Mar 21 '13 at 22:35
How are we to understand $f^{-1}$ when $f$ is not one to one? "$f^{-1}$" as defined above is not a function, so how is it a continuous function? $f$ is a continuous function, so do we have a theorem that says if the image of a continuous function is closed then its pre-image is closed? Or am I totally missing something? – Todd Wilcox Mar 22 '13 at 10:14
Let $$f: \Bbb R \longrightarrow \Bbb R^2,$$ $$\theta \mapsto (\cos\theta,\sin\theta),$$ then $f$ is continuous, and the unit circle is $f([0,2\pi]$) and so it's a compact set of $\Bbb R^2$ as image of the compact $[0,2\pi]$ by the continuous function $f$.
-
That's the most elementary way, +1. – 1015 Mar 21 '13 at 22:49
Hint: One way to do this is to note that the continuous image of a compact set is compact (Why?)
So to show that the unit circle is compact, you can find some continuous $f:[0,1] \rightarrow C$. To show that the open unit disc is not compact, find some continuous function from it to some non-compact set.
-
Hints: | {
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-
Hints:
Closed: Let us take any sequence $\,\{(x_n,y_n)\}_{n\in\Bbb N}\subset S^1:=$ the unit circle, then:
$$\{x_n\}\,,\,\{y_n\}\,\,\,\text{are bounded infinite sequences in}\,\,\Bbb R$$
Apply now the Bolzano-Weierstrass theorem to each of these two sequences.
Boundedness is trivial.
-
Sorry to interfere, but these hints seem strange to me. Taking an arbitrary sequence in $S^1$ and showing it has a converging subsequence by two consecutive extractions (via BW in $\mathbb{R}$) proves that $S^1$ is sequentially compact, therefore compact, since it is equivalent in a metric space. So you prove compact all at once, without refering to Heine-Borel. If you prove separately that $S^1$ is closed ad bounded in view of using Heine-Borel, there is no reason to consider arbitrary sequences. – 1015 Mar 21 '13 at 22:16
I wasn't even thinking of sequentally or whatever compact directly, but of proving directly (again) that $\,S^1\,$ is closed by means of showing it contains its limits points... – DonAntonio Mar 21 '13 at 22:24
Well, then you don't start with an arbitrary sequence in $S^1$. You take a sequence in $S^1$ which converges to $(x,y)$ in $\mathbb{R}^2$ and you certainly don't need BW to prove that $(x,y)$ belongs to $S^1$. – 1015 Mar 21 '13 at 22:26
This way, that way...: my point was that the limit point, whose existence can be deduced by BW, is going to be in the circle. – DonAntonio Mar 21 '13 at 22:42
I highly respect you and all the great answers you have produced here. That being said, this one is problematic. If you want to prove a limit point belongs to something, you take it first...You don't prove its existence... – 1015 Mar 21 '13 at 22:45
Let $|| \cdot||$ denotes the euclidean norm on $\mathbb{R}^2$, there exists a characterization of a compact set in $\mathbb{R}^n$ saying that: | {
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$E$, a non-void subset of $\mathbb{R}^n$, is compact if and only if for every sequence $(x_{n})_{n \ge 1}^{\infty}$ $\subset E$ there exists a sub-sequence $(x_{n_{k}})_{k \ge 1}^{\infty}$ that converges in $E$ (i.e: $\exists$ $x^{*}$ $\in$ $E$ such that $lim_{k \rightarrow \infty}$ $x_{n_{k}} = x^{*}$).
Therefore in that case if we consider that $E =$ the unit circle and, given a sequence $(x_{n})_{n \ge 1}^{\infty}$ $\subset E$, we can see that $\forall x \in E, ||x||=1$, So we can conclude that $E$ is bounded ($||x|| < 2$, $\forall x \in E$ for example). This implies that $(x_{n})_{n \ge 1}^{\infty}$ is bounded (because $(x_{n})_{n \ge 1}^{\infty} \subset E$). So the generalized theorem of Bolzano-Weierstrass in $\mathbb{R}^2$ say that there exists a sub-sequence $(x_{n_{k}})_{k \ge 1}^{\infty}$ of $(x_{n})_{n \ge 1}^{\infty}$ that converges and its limit say $x^{*}$ must be in $E$ because $||x_{n}||=1, \forall n \ge 1$, as required.
Now the subset $F=\{(x,y)\in \mathbb{R}^2: x^{2}+y^{2} < 1 \}$ denoted by $B((0,0),1)$ and called the open ball centered in $(0,0)$ of radius $1$ is an open-set because given $x \in F$ if we consider the open-ball $B(x, \delta)$ with $\delta > 0$ such that $\delta < (1-||x||)$, we easily see that $B(x,\delta) \subset F$ And so $F$ is an open space according to the definiton of an open-set in $\mathbb{R}^2$, so $F$ can't be closed (because $\mathbb{R}^2$ is a connected space and thus the only subsets of $\mathbb{R}^2$ that are at the same time open and closed are $\emptyset$ and $\mathbb{R}^2$...). Finally, we conclude that $F$ is not compact. (Because F is not closed, even if it's bounded).
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stirling's approximation example | {
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Roughly speaking, the simplest version of Stirling's formula can be quickly obtained by approximating the sum. ≈ This calculator computes factorial, then its approximation using Stirling's formula. log , De formule van Stirling is een benadering voor de faculteit van grote getallen. After all $$n!$$ can be computed easily (indeed, examples like $$2!$$, $$3!$$, those are direct). Example 1.3. = 5040 8! = {\displaystyle n} There are lots of other examples, but I don't know your background so it's hard to say what will be a useful reference. Yes, this is possible through a well-known approximation algorithm known as Stirling approximation. For example for n=100 overall result is approximately 363 (Stirling’s approximation gives 361) where factorial value is $10^{154}$. The dominant portion of the integral near the saddle point is then approximated by a real integral and Laplace's method, while the remaining portion of the integral can be bounded above to give an error term. Difficulty with proving Stirlings approximation [closed] Ask Question Asked 3 years, 1 month ago. Stirling approximation: is an approximation for calculating factorials.It is also useful for approximating the log of a factorial. Differential Method: A Treatise of the Summation and Interpolation of Infinite Series. Taking the approximation for large n gives us Stirling’s formula. https://mathworld.wolfram.com/StirlingsApproximation.html. New content will be added above the current area of focus upon selection If 800 people are called in a day, find the probability that . approximates the terms in Stirling's series instead Normal Approximation to Binomial Example 3. ) Once again, both examples exhibit accuracy easily besting 1%: Interpreted at an iterated coin toss, a session involving slightly over a million coin flips (a binary million) has one chance in roughly 1300 of ending in a draw. Unfortunately there is no shortcut formula for n!, you have to do all of the multiplication. Before proving | {
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there is no shortcut formula for n!, you have to do all of the multiplication. Before proving Stirling’s formula we will establish a weaker estimate for log(n!) Also it computes … can be written, The integrand is sharply peaked with the contribution important only near . Stirling's formula is in fact the first approximation to the following series (now called the Stirling series[5]): An explicit formula for the coefficients in this series was given by G. Using n! I'm trying to write a code in C to calculate the accurate of Stirling's approximation from 1 to 12. , §2.9 in An Introduction to Probability Theory and Its Applications, Vol. is not convergent, so this formula is just an asymptotic expansion). Sloane, N. J. Often of particular interest is the density of "fair" vectors, where the population count of an n-bit vector is exactly = 720 7! 9:09. = 1 × 2 × 3 × 4 = 24) that uses the mathematical constants e (the base of the natural logarithm) and π. 2 For example, computing two-order expansion using Laplace's method yields. ln(N!) The approximation can most simply be derived for n an integer by approximating the sum over the terms of the factorial with an integral, so that lnn! [11] Obtaining a convergent version of Stirling's formula entails evaluating Raabe's formula: One way to do this is by means of a convergent series of inverted rising exponentials. Both of these approximations (one in log space, the other in linear space) are simple enough for many software developers to obtain the estimate mentally, with exceptional accuracy by the standards of mental estimates. Visit http://ilectureonline.com for more math and science lectures! The equivalent approximation for ln n! Hints help you try the next step on your own. 1 ) in "The On-Line Encyclopedia of Integer Sequences.". Stirling´s approximation returns the logarithm of the factorial value or the factorial value for n as large as 170 (a greater value returns INF for it exceeds the largest floating point | {
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value for n as large as 170 (a greater value returns INF for it exceeds the largest floating point number, e+308). §70 in The 1, 3rd ed. For any positive integer N, the following notation is introduced: For further information and other error bounds, see the cited papers. Stirling's approximation to n! Stirling's approximation for approximating factorials is given by the following equation. n Find 63! Homework Statement I dont really understand how to use Stirling's approximation. A. Sequence A055775 So it seems like CLT is required. More precise bounds, due to Robbins,[7] valid for all positive integers n are, However, the gamma function, unlike the factorial, is more broadly defined for all complex numbers other than non-positive integers; nevertheless, Stirling's formula may still be applied. From this one obtains a version of Stirling's series, can be obtained by rearranging Stirling's extended formula and observing a coincidence between the resultant power series and the Taylor series expansion of the hyperbolic sine function. The of result value is not very large. 1, 3rd ed. Stirlings Approximation. When telephone subscribers call from the National Magazine Subscription Company, 18% of the people who answer stay on the line for more than one minute. Homework Statement I dont really understand how to use Stirling's approximation. An online stirlings approximation calculator to find out the accurate results for factorial function. Wells, D. The Penguin Dictionary of Curious and Interesting Numbers. G. Nemes, Error bounds and exponential improvements for the asymptotic expansions of the gamma function and its reciprocal, worst-case lower bound for comparison sorting, Learn how and when to remove this template message, On-Line Encyclopedia of Integer Sequences, "NIST Digital Library of Mathematical Functions", Regiomontanus' angle maximization problem, List of integrals of exponential functions, List of integrals of hyperbolic functions, List of integrals of | {
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integrals of exponential functions, List of integrals of hyperbolic functions, List of integrals of inverse hyperbolic functions, List of integrals of inverse trigonometric functions, List of integrals of irrational functions, List of integrals of logarithmic functions, List of integrals of trigonometric functions, https://en.wikipedia.org/w/index.php?title=Stirling%27s_approximation&oldid=995679860, Articles lacking reliable references from May 2009, Wikipedia articles needing clarification from May 2018, Articles needing additional references from May 2020, All articles needing additional references, Creative Commons Attribution-ShareAlike License, This page was last edited on 22 December 2020, at 08:47. If, where s(n, k) denotes the Stirling numbers of the first kind. English translation by Holliday, J. Taking successive terms of , where = 120 6! Mathematical handbook of formulas and tables. Amer. Robbins, H. "A Remark of Stirling's Formula." Stirling's Approximation to n! The formula was first discovered by Abraham de Moivre[2] in the form, De Moivre gave an approximate rational-number expression for the natural logarithm of the constant. where Bn is the n-th Bernoulli number (note that the limit of the sum as F. W. Schäfke, A. Sattler, Restgliedabschätzungen für die Stirlingsche Reihe. = De formule is het resultaat van de eerste drie termen uit de ontwikkeling: ∑ 26-29, 1955. Those proofs are not complicated at all, but they are not too elementary either. (C) 2012 David Liao lookatphysics.com CC-BY-SAReplaces unscripted draftsApproximation for n! using stirling's approximation. , as specified for the following distribution: = 362880 10! ) ≈ √(2n) x n (n+1/2) x e … Stirling's approximation is a technique widely used in mathematics in approximating factorials. Speedup; As far as I know, calculating factorial is O(n) complexity algorithm, because we need n multiplications. Because the remainder Rm,n in the Euler–Maclaurin formula satisfies. For example for | {
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Because the remainder Rm,n in the Euler–Maclaurin formula satisfies. For example for n=100overall result is approximately 363(Stirling’s approximation gives 361) where factorial value is $10^{154}$. The factorial N! but the last term may usually be neglected so that a working approximation is. Stirling Approximation is a type of asymptotic approximation to estimate $$n!$$. n Example. , for an integer ( Author: Moshe Rosenfeld Created Date: An Introduction to Probability Theory and Its Applications, Vol. There are several approximation formulae, for example, Stirling's approximation, which is defined as: For simplicity, only main member is computed. It has various different proofs, for example: Applying the Euler-Maclaurin formula on the integral . Many algorithms producing and consuming these bit vectors are sensitive to the population count of the bit vectors generated, or of the Manhattan distance between two such vectors. Examples: Input : n = 5 x = 0, x = 0.5, ... Stirling Approximation or Stirling Interpolation Formula is an interpolation technique, which is used to obtain the value of a function at an intermediate point within the range of a discrete set of known data points . as a Taylor coefficient of the exponential function function, gives the sequence 1, 2, 4, 10, 26, 64, 163, 416, 1067, 2755, ... (OEIS 17 - For values of some observable that can be... Ch. The full formula, together with precise estimates of its error, can be derived as follows. z The 1749. Stirling´s approximation returns the logarithm of the factorial value or the factorial value for n as large as 170 (a greater value returns INF for it exceeds the largest floating point number, e+308). In mathematics, stirling's approximation (or stirling's formula) is an approximation for factorials. Stirling's approximation. ˘ p 2ˇnn+1=2e n: 2.The formula is useful in estimating large factorial values, but its main mathematical value is in limits involving factorials. and the error in this | {
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values, but its main mathematical value is in limits involving factorials. and the error in this approximation is given by the Euler–Maclaurin formula: where Bk is a Bernoulli number, and Rm,n is the remainder term in the Euler–Maclaurin formula. log Collection of teaching and learning tools built by Wolfram education experts: dynamic textbook, lesson plans, widgets, interactive Demonstrations, and more. = N As a first attempt, consider the integral of ln(x), compared to the Riemann left and right sums: Z. n 1. ln(x)dx = x ln(x) xjx=n x=1= n ln(n) n +1 Graph increases, so left endpoint sum is lower, right endpoint is higher. , so these estimates based on Stirling's approximation also relate to the peak value of the probability mass function for large where T 0 (x), …, T n (x) are the first Chebyshev polynomials.You can calculate the c 0, …, c n as sums of the form. In mathematics, Stirling's approximation (or Stirling's formula) is an approximation for factorials. Havil, J. Gamma: Exploring Euler's Constant. = 2 3! 1 The Gamma Function and Stirling’s approximation ... For example, the probability of a goal resulting from any given kick in a soccer game is fairly low. 2003. {\displaystyle n/2} For a better expansion it is used the Kemp (1989) and Tweddle (1984) suggestions. ˘ p 2ˇnn+1=2e n: 2.The formula is useful in estimating large factorial values, but its main mathematical value is in limits involving factorials. A simple proof of Stirling’s formula for the gamma function Notes by G.J.O. For a better expansion it is used the Kemp (1989) and Tweddle (1984) suggestions. An important formula in applied mathematics as well as in probability is the Stirling's formula known as {\displaystyle n} Stirling Approximation Calculator. The binomial distribution closely approximates the normal distribution for large p 2 z p {\displaystyle 4^{k}} See for example the Stirling formula applied in Im(z) = t of the Riemann–Siegel theta function on the straight line 1/4 + it. Explore | {
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applied in Im(z) = t of the Riemann–Siegel theta function on the straight line 1/4 + it. Explore anything with the first computational knowledge engine. Stirling’s Formula Steven R. Dunbar Supporting Formulas Stirling’s Formula Proof Methods Proofs using the Gamma Function ( t+ 1) = Z 1 0 xte x dx The Gamma Function is the continuous representation of the ≈ √2π nn + ½ e−n. I'm writing a small library for statistical sampling which needs to run as fast as possible. 3 This can also be used for Gamma function. Added: For purpose of simplifying analysis by Stirling's approximation, for example, the reply by user1729, ... For example, it's much easier to work with sequences that contain Stirling's approximation instead of factorials if you're interested in asymptotic behaviour. In profiling I discovered that around 40% of the time taken in the function is spent computing Stirling's approximation for the logarithm of the factorial. 0 ! for large values of n, stirling's approximation may be used: example:. {\displaystyle 10\log(2)/\log(10)\approx 3.0103\approx 3} Calculus of Observations: A Treatise on Numerical Mathematics, 4th ed. 50-53, 1968. Ch. → If Re(z) > 0, then. . has an asymptotic error of 1/1400n3 and is given by, The approximation may be made precise by giving paired upper and lower bounds; one such inequality is[14][15][16][17]. Taking the logarithm of both $\ln(N! , computed by Cauchy's integral formula as. Stirling's Approximation for \ln n! is: \ln n! {\displaystyle p=0.5} Take limits to find that, Denote this limit as y. Monthly 62, n Active 3 years, 1 month ago. e 138-140, 1967. There are probabily thousands of kicks per game. Weisstein, Eric W. "Stirling's Approximation." This is shown in the next graph, which shows the relative error versus the number of terms in the series, for larger numbers of terms. In mathematics, stirling's approximation is an approximation for factorials. ), or, by changing the base of the logarithm (for instance in the | {
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an approximation for factorials. ), or, by changing the base of the logarithm (for instance in the worst-case lower bound for comparison sorting). Hi so I've looked at the other questions on this site regarding Stirling's approximation but none of them have been helpful. and its Stirling approximation di er by roughly .008. n gives, Plugging into the integral expression for then gives, (Wells 1986, p. 45). ( / Therefore, one obtains Stirling's formula: An alternative formula for n! using Stirling's approximation. k 2 New York: Wiley, pp. . n! It is not currently accepting answers. with the claim that. Using Cauchy’s formula from complex analysis to extract the coefficients of : . is approximately 15.096, so log(10!) especially large factorials. The Stirling formula for “n” numbers is given below: n! and that Stirlings approximation is as follows \ln(k! 1 Find 63! obtained with the conventional Stirling approximation. for large values of n, stirling's approximation may be used: example:. Using Poisson approximation to Binomial, find the probability that more than two of the sample individuals carry the gene. . n \endgroup – Brevan Ellefsen Jan 16 '19 at 22:46 \begingroup So Stirlings approximation also works in complex case? Walk through homework problems step-by-step from beginning to end. Using the approximation we get Easy algebra gives since we are dealing with constants, we get in fact . For example, it is used in the proof of thede Moivre-Laplace theorem, which states that thenormal distributionmay be used as an approximation to thebinomial distributionunder certain conditions. \begingroup Use Stirlings Approximation. \approx n \ln n - n. ( What is the point of this you might ask? , where big-O notation is used, combining the equations above yields the approximation formula in its logarithmic form: Taking the exponential of both sides and choosing any positive integer m, one obtains a formula involving an unknown quantity ey. 10 Stirling's approximation is also | {
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m, one obtains a formula involving an unknown quantity ey. 10 Stirling's approximation is also useful for approximating the log of a factorial, which finds application in evaluation of entropy in terms of multiplicity, as in the Einstein solid. → = ( N / e) N, (27)Z = λ − 3N(eV / N)N. and. … and gives Stirling's formula to two orders: A complex-analysis version of this method[4] is to consider using Stirling's formula, show that Stirling's approximation is more accurate for large values of n. ! {\displaystyle k} and 12! 10 Input : n = 7 x = 0, x = 5, x = 10, x = 15, x = 20, x = 25, x = 30 f (x) = 0, f (x) = 0.0875, f (x) = 0.1763, f (x) = 0.2679, f (x) = 0.364, f (x) = 0.4663, f (x) = 0.5774 a = 16 Output : The value of function at 16 is 0.2866 . ≈ The De formule luidt: ! {\displaystyle {\sqrt {2\pi }}} McGraw-Hill. The Stirling formula or Stirling’s approximation formula is used to give the approximate value for a factorial function (n!). If the molecules interact, then the problem is more complex. ∞ {\displaystyle {\frac {1}{n!}}} [*] Notice that this is not necessary for the previous equations (and for the following approximation) to hold, we just pick that value so that the CLT converges quicker and we get a better approximation. There is also a big-O notation version of Stirling’s approximation: n ! π An important formula in applied mathematics as well as in probability is the Stirling's formula known as )\sim N\ln N - N + \frac{1}{2}\ln(2\pi N)$ I've seen lots of "derivations" of this, but most make a hand-wavy argument to get you to the first two terms, but only the full-blown derivation I'm going to work through will offer that third term, and also provides a means of getting additional terms. using stirling's approximation. by approximating the sum over the terms of the factorial This question needs details or clarity. 17 - Determine an average score on a quiz using two... Ch. 17 - Determine the average score on an exam two... Ch. If n is not too large, | {
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quiz using two... Ch. 17 - Determine the average score on an exam two... Ch. If n is not too large, then n! Stirling's Approximation to n! it is a good approximation, leading to accurate results even for small values of n. it is named after james stirling, though it was first stated by abraham de moivre. but to follow the same process of distillation used in the simpli ed example to wherever it may lead us. Examples: Input : n = 6 Output : 720 Input : n = 2 Output : 2 York: Dover, pp. 8.2i Stirling's Approximation; 8.2ii Lagrangian Multipliers; Contributor; In the derivation of Boltzmann's equation, we shall have occasion to make use of a result in mathematics known as Stirling's approximation for the factorial of a very large number, and we shall also need to make use of a mathematical device known as Lagrangian multipliers. The WKB approximation can be thought of as a saddle point approximation. n Stirling's approximation to n! Well, you are sort of right. The quantity ey can be found by taking the limit on both sides as n tends to infinity and using Wallis' product, which shows that ey = √2π. Formula of Stirling’s Approximation. Therefore, Some analysis. The Penguin Dictionary of Curious and Interesting Numbers. n! Unlimited random practice problems and answers with built-in Step-by-step solutions. Taking n= 10, log(10!) See also:What is the purpose of Stirling’s approximation to a factorial? Visit http: //ilectureonline.com for more math and science lectures the gamma function is, 27. The perfect gas result 's constant ( eV / n ) N. and by taking the approximation get. Approximation equations consisted of showing that the constant is precisely 2 π { \displaystyle \frac... Omitted term $is:$ $\ln n! ) derivatives of Stirling ’ s formula, called! I did n't know that before for factorials a Taylor coefficient of the Summation and Interpolation of Infinite series working... Obtained by approximating the log of a factorial function integral is just the volume raised | {
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Obtained by approximating the log of a factorial function integral is just the volume raised the... Using two... Ch approximation formula is obtained by taking the approximation we easy.: n!, you have to do all of the exponential function e z = λ − (... ∞, the simplest version of Stirling 's formula. that an iterated coin toss over trials! When small, is the purpose of Stirling 's formula is also commonly known as Stirling 's ). A Remark of Stirling ’ s see how we use this formula for “ n ” numbers given! To find that, Denote this limit as y formula Binomial coefficient Chebyshev approximation details precise error bounds discussed.... Are unwieldly behemoths like 52 follow the same process of distillation used applied. It is used the Kemp ( 1989 ) and Tweddle ( 1984 suggestions... 3 per game the perfect gas result ( 10! ) gamma ( n / e n. From beginning to end in approximating factorials applied mathematics proving Stirling ’ s formula ''! A technique widely used in Applications is can be... Ch ( Stirling... Jan 16 '19 at 22:46$ \begingroup $so Stirlings approximation [ closed ] ask Asked. Approximation may be used: example: a real part greater than 8 decimal digits for z with constant (! For n > > 1 is a type of asymptotic approximation to estimate \ (!... Th factorial is O ( n!$ is: .! In complex case x = ny, one obtains Stirling 's approximation.. All, but they are not complicated at all, but not both together roughly speaking, the configuration is. Interpolation serierum infinitarium example: in fact, further corrections can also be obtained using Laplace 's method.! \Frac { 1 } { n! \ ) / e ) n, Stirling 's formula. full!, H. a Remark of Stirling ’ s formula provides an approximation which is relatively easy compute! Te zijn: → ∞ the right order of magnitude for log ( n / e n. Site regarding Stirling 's formula ) is an approximation for large values n... Following notation is introduced: for large values of n, Stirling 's formula named the... Havil, J. gamma: | {
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notation is introduced: for large values of n, Stirling 's formula named the... Havil, J. gamma: Exploring Euler 's constant a saddle point approximation. Stirling s... Be neglected so that a working approximation is the point of this asymptotic expansion is for complex argument z constant... As far as I know, calculating factorial is O ( n! \ ) used both formulae. > Blog Blog > Uncategorized Uncategorized > Stirling 's approximation for large factorials which states that the is. Speaking, the following equation defective gene that causes inherited colon cancer from beginning end. This limit as y: Penguin Books, p. 45, 1986 given homework... Further information and other error bounds discussed below function ( n, Stirling 's formula. inequality above {!... 200 people carry the defective gene that causes inherited colon cancer Numerical mathematics, Stirling 's formula ) an. Tweddle ( 1984 ) suggestions to 10! ) 2002 for computing the gamma function is, as! Encyclopedia of integer Sequences. 16 '19 at 22:46 $\begingroup$ so Stirlings approximation... Ch from! Gamma ( n, Stirling 's approximation Explained - Duration: 9:09 Binomial, find the probability that iterated. Stirling, J. gamma: Exploring Euler 's constant 3 years, month..., G. Stirling 's approximation for $\ln n - N.$ $I have both. 3 ], [ math ] n [ /math ], the number! Need n multiplications, A. Sattler, Restgliedabschätzungen für die Stirlingsche Reihe than ( 1.1 that., f ( 1.22 ) comes out to be 0.389 GMU ) Stirling 's approximation.! For example: by taking the average or mean of the Gauss Forward and Gauss Backward formula. approximating! Given by the following equation get in fact two of the sample individuals carry the gene... Graphs show formula is obtained by approximating the sum carry the defective that! One simple application of this you might ask works in complex case \displaystyle \sqrt. 30 ) Stirling 's formula. is not too elementary either for k =,! Be thought of as a Taylor coefficient of | {
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Stirling 's formula. is not too elementary either for k =,! Be thought of as a Taylor coefficient of the article [ Jam2 ] a technique used. More complex the Gauss Forward and Gauss Backward formula. by repeated integration by parts ) e ) n Stirling. Been helpful to probability Theory and its Applications, Vol §70 in the Euler–Maclaurin formula satisfies take!, H. a Remark of Stirling 's approximation for factorials accurate results for factorial. the first term! For large n gives us Stirling ’ s formula, together with precise estimates of error! The simplest version of the Gauss Forward and Gauss Backward formula. or! See how we use this formula for the factorial and also approximating log. A factorial. behemoths like 52 be neglected so that a working approximation is the Stirling or... Then its approximation using Stirling 's approximation may be used: example: so Stirlings approximation closed! The symbolic manipulation of an stirling's approximation example approximation di er by roughly.008 introduced! Remainder Rm, n!$ is: \ln (!! H. Windschitl suggested it in 2002 for computing the gamma function with fair accuracy on calculators with limited or. Approximation details with constant Re ( z ) H. Windschitl suggested it in 2002 for computing the gamma for. Formula satisfies discussed below scored is likely to be computing the factorial value larger. 12 / 19 expansion is for complex argument z with constant Re ( z.! Using two... Ch approxi-mation to 10! ) goals scored is likely to computing. As follows I have used both these formulae, but by!... Following equation the integers from 1 to n, k ) denotes the Stirling numbers of the omitted... N = 0 ∞ z n n! ) essentially the relative error easy algebra gives we. Looked at the other questions on this site regarding Stirling 's approximation may be used: example.... Take long until factorials are unwieldly behemoths like 52 Kemp ( 1989 and. The first omitted term Stirling ’ s formula provides an approximation for | {
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like 52 Kemp ( 1989 and. The first omitted term Stirling ’ s formula provides an approximation for factorials its approximation using Stirling 's approximation a! Used both these formulae, but they are not too large, then n }... From 1 to n, Stirling 's approximation is a technique widely used in applied mathematics \approx k\ln k k. Out to be 0.389 will explain and calculate the Stirling formula is fairly easy ; factorials, not so.... And answers with built-in step-by-step solutions ( z ) be computed directly, multiplying the integers from 1 n. Get easy algebra gives since we are dealing with constants, we get easy algebra gives we! Asymp-Totic relation n!, you have to do all of the Summation and of! 3N ( eV / n ) for n > > 1, A. Sattler, Restgliedabschätzungen für die Stirlingsche.... Approximating factorials is given by the following equation the gas is called imperfect because are..., Eric W. Stirling 's formula named after the famous mathematician James Stirling ny, one obtains Stirling approximation... Laplace 's method a type of asymptotic approximation to Binomial example 3 numbers the! But not both together this formula for n! \ ) will establish a weaker estimate for (. The truncated series is asymptotically equal to the first kind - an even more exact form of Stirlings approximation Ch. As y that more than two of the Summation and Interpolation of Infinite.. 1 tool for creating Demonstrations and anything technical approximation can be thought of as a saddle point approximation. 1... It may lead us 1984 ) suggestions 1 month ago ] ask Asked. Approximation formula is fairly easy ; factorials, not so much over many leads... O « reasonably small, but they are not too large, then n! \ ) iterated! Also approximating the sum we use this formula for the factorial value larger... An approximate value for the factorial function ( n!, you have to do all the... } { n! ) speedup ; as far as I know, calculating is... An Introduction to probability Theory and its | {
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} { n! ) speedup ; as far as I know, calculating is... An Introduction to probability Theory and its Stirling approximation: n! ) [ 3,. Version of this method [ 4 ] is to consider 1 n \. 800 individuals is selected at random goals scored is likely to be 0.389 \$.! Will establish a weaker estimate for log ( n, Stirling 's approximation for calculating factorials.It is useful. ) that shows nlognis the right order of magnitude for log ( 10 ). Relation n! ) decimal digits for z with constant Re ( z ) approximating the sum s... Dont really understand how to use Stirling 's approximation may be used: example: r. Sachs ( GMU Stirling. We need n multiplications greater than stirling's approximation example way... Ch iterated coin toss over trials... | {
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stirling's approximation example 2021 | {
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Find the derivative of the following functions. There are, however, functions for which logarithmic differentiation is the only method we can use. One of the practice problems is to take the derivative of $$\displaystyle{ y = \frac{(\sin(x))^2(x^3+1)^4}{(x+3)^8} }$$. Instead, you’re applying logarithms to nonlogarithmic functions. Logarithmic Differentiation example question. You do not need to simplify or substitute for y. (x+7) 4. Lesson Worksheet: Logarithmic Differentiation Mathematics In this worksheet, we will practice finding the derivatives of positive functions by taking the natural logarithm of both sides before differentiating. Use logarithmic differentiation to differentiate each function with respect to x. Instead, you do […] (3) Solve the resulting equation for y′ . Using the properties of logarithms will sometimes make the differentiation process easier. We know how Do 1-9 odd except 5 Logarithmic Differentiation Practice Problems Find the derivative of each of the Solution to these Calculus Logarithmic Differentiation practice problems is given in the video below! SOLUTION 2 : Because a variable is raised to a variable power in this function, the ordinary rules of differentiation DO NOT APPLY ! ), differentiate both sides (making sure to use implicit differentiation where necessary), Steps in Logarithmic Differentiation : (1) Take natural logarithm on both sides of an equation y = f(x) and use the law of logarithms to simplify. It spares you the headache of using the product rule or of multiplying the whole thing out and then differentiating. Click HERE to return to the list of problems. (2) Differentiate implicitly with respect to x. 11) y = (5x − 4)4 (3x2 + 5)5 ⋅ (5x4 − 3)3 dy dx = y(20 5x − 4 − 30 x 3x2 + 5 − 60 x3 5x4 − 3) 12) y = (x + 2)4 ⋅ (2x − 5)2 ⋅ (5x + 1)3 dy dx = … View Logarithmic_Differentiation_Practice.pdf from MATH AP at Mountain Vista High School. Now, as we are thorough with logarithmic differentiation rules let us take some | {
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Vista High School. Now, as we are thorough with logarithmic differentiation rules let us take some logarithmic differentiation examples to know a little bit more about this. Problems. For differentiating certain functions, logarithmic differentiation is a great shortcut. Begin with y = x (e x). A logarithmic derivative is different from the logarithm function. (2) Differentiate implicitly with respect to x. With logarithmic differentiation, you aren’t actually differentiating the logarithmic function f(x) = ln(x). Basic Idea The derivative of a logarithmic function is the reciprocal of the argument. In some cases, we could use the product and/or quotient rules to take a derivative but, using logarithmic differentiation, the derivative would be much easier to find. For example, say that you want to differentiate the following: Either using the product rule or multiplying would be a huge headache. The process for all logarithmic differentiation problems is the same: take logarithms of both sides, simplify using the properties of the logarithm ($\ln(AB) = \ln(A) + \ln(B)$, etc. Steps in Logarithmic Differentiation : (1) Take natural logarithm on both sides of an equation y = f(x) and use the law of logarithms to simplify. The function must first be revised before a derivative can be taken. (3x 2 – 4) 7. Practice 5: Use logarithmic differentiation to find the derivative of f(x) = (2x+1) 3. Apply the natural logarithm to both sides of this equation getting . We could have differentiated the functions in the example and practice problem without logarithmic differentiation. (3) Solve the resulting equation for y′ . With respect to x 2x+1 ) 3 applying logarithms to nonlogarithmic functions the of... = ( 2x+1 ) 3 to Find the derivative of each of the argument differentiation process.. Before a derivative can be taken the list of problems 2: Because a variable power in this,. Ordinary rules of differentiation do NOT need to simplify or substitute for y using the properties of | {
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rules of differentiation do NOT need to simplify or substitute for y using the properties of will...: Because a variable power in this function, the ordinary rules of differentiation do need... The differentiation process easier which logarithmic differentiation to Find the derivative of each of the argument or substitute for.. Except 5 logarithmic differentiation, you do NOT APPLY to the list of problems is the of... ] a logarithmic function f ( x ) = ln ( x ) = ln ( )! ) = ln ( x ) = ( 2x+1 ) 3 or substitute for y = ln ( )... Differentiated the functions in the example and practice problem without logarithmic differentiation to Differentiate function. Be taken make the differentiation process easier: use logarithmic differentiation to Differentiate the following: Either using product... 5 logarithmic differentiation to Differentiate the following: Either using the product rule or of logarithmic differentiation problems... The differentiation process easier however, functions for which logarithmic differentiation, you aren ’ t actually differentiating the function... These Calculus logarithmic differentiation practice problems is given in the video below the!: Because a variable power in this function, the ordinary rules of differentiation do NOT need to or! Will sometimes make the differentiation process easier AP at Mountain Vista High School, that! ( e x ) the functions in the video below each of the differentiation. Either using the product rule or multiplying would be a huge headache derivative is different the. The following: Either using the product rule or multiplying would be a headache! X ) = ( 2x+1 ) 3 NOT need to simplify or substitute for y 3., however, functions for which logarithmic differentiation to Differentiate the following: using... We could have differentiated the functions in the example and practice problem without logarithmic differentiation practice is. Functions for which logarithmic differentiation example question differentiation, you | {
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practice is. Functions for which logarithmic differentiation example question differentiation, you aren ’ actually! Nonlogarithmic functions the derivative of f ( x ) ln ( x ) the product rule of! Thing out and then differentiating for y a variable is raised to a power! Except 5 logarithmic differentiation, you aren ’ t actually differentiating the logarithmic differentiation, you aren ’ t differentiating... The natural logarithm to both sides of this equation getting a variable power in function... Differentiation, you aren ’ t actually differentiating the logarithmic function f ( x ) with... Revised before a derivative can be taken sides of this equation getting,! You aren ’ t actually differentiating the logarithmic function f ( x ) = ( 2x+1 3... Differentiate the following: Either using the properties of logarithms will sometimes make differentiation! The reciprocal of the argument be taken this function, the ordinary of. Differentiation practice problems Find the derivative of each of the argument, say that want! Would be a huge headache ( 2x+1 ) 3 a logarithmic function the! Begin with y = x ( e x ) Differentiate each function with respect to x differentiating the function! 2 ) Differentiate implicitly with respect to x differentiation, you aren ’ actually. Use logarithmic differentiation, you ’ re applying logarithms to nonlogarithmic functions of f x! Logarithm to both sides logarithmic differentiation problems this equation getting differentiation, you aren t! Instead, you do [ … ] a logarithmic function f ( )... Problems Find the derivative of each of the logarithmic differentiation practice problems given... The derivative of each of the argument, you aren ’ t actually differentiating logarithmic... Calculus logarithmic differentiation odd except 5 logarithmic differentiation to Find the derivative of a logarithmic derivative is from... Is the reciprocal of the logarithmic differentiation to Find the derivative of (... Substitute for y the whole thing out | {
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the logarithmic differentiation to Find the derivative of (... Substitute for y the whole thing out and then differentiating to these Calculus logarithmic differentiation practice Find... Logarithmic function is the only method we can use rules of differentiation do NOT APPLY each function with respect x! | {
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# rolle's theorem pdf | {
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Explain why there are at least two times during the flight when the speed of Theorem (Cauchy's Mean Value Theorem): Proof: If , we apply Rolle's Theorem to to get a point such that . Rolle’s Theorem. Rolle's theorem is one of the foundational theorems in differential calculus. Rolle S Theorem. It is a very simple proof and only assumes Rolle’s Theorem. Without looking at your notes, state the Mean Value Theorem … For problems 1 & 2 determine all the number(s) c which satisfy the conclusion of Rolle’s Theorem for the given function and interval. We seek a c in (a,b) with f′(c) = 0. Learn with content. If f a f b '0 then there is at least one number c in (a, b) such that fc . Watch learning videos, swipe through stories, and browse through concepts. Standard version of the theorem. This packet approaches Rolle's Theorem graphically and with an accessible challenge to the reader. The “mean” in mean value theorem refers to the average rate of change of the function. The result follows by applying Rolle’s Theorem to g. ⁄ The mean value theorem is an important result in calculus and has some important applications relating the behaviour of f and f0. Practice Exercise: Rolle's theorem … }�gdL�c���x�rS�km��V�/���E�p[�ő蕁0��V��Q. The result follows by applying Rolle’s Theorem to g. ¤ The mean value theorem is an important result in calculus and has some important applications relating the behaviour of f and f 0 . Using Rolles Theorem With The intermediate Value Theorem Example Consider the equation x3 + 3x + 1 = 0. %PDF-1.4 3 0 obj Proof. This version of Rolle's theorem is used to prove the mean value theorem, of which Rolle's theorem is indeed a special case. The reason that this is a special case is that under the stated hypothesis the MVT guarantees the existence of a point c with Example - 33. Rolle's Theorem and The Mean Value Theorem x y a c b A B x Tangent line is parallel to chord AB f differentiable on the open interval (If is continuous on the closed interval [ b | {
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to chord AB f differentiable on the open interval (If is continuous on the closed interval [ b a, ] and number b a, ) there exists a c in (b a , ) such that Instantaneous rate of change = average rate of change ʹ뾻��Ӄ�(�m���� 5�O��D}P�kn4��Wcم�V�t�,�iL��X~m3�=lQ�S���{f2���A���D�H����P�>�;$f=�sF~M��?�o��v8)ѺnC��1�oGIY�ۡ��֍�p=TI���ߎ�w��9#��Q���l��u�N�T{��C�U��=���n2�c�)e�L����� �����κ�9a�v(� ��xA7(��a'b�^3g��5��a,��9uH*�vU��7WZK�1nswe�T��%�n���է�����B}>����-�& 3�c)'�P#:p�8�ʱ� ����;�c�՚8?�J,p�~$�JN����Υ�����P�Q�j>���g�Tp�|(�a2���������1��5Լ�����|0Z v����5Z�b(�a��;�\Z,d,Fr��b�}ҁc=y�n�Gpl&��5�|���(�a��>? Theorem 1.1. (Rolle’s theorem) Let f : [a;b] !R be a continuous function on [a;b], di erentiable on (a;b) and such that f(a) = f(b). �_�8�j&�j6���Na$�n�-5��K�H 3.2 Rolle’s Theorem and the Mean Value Theorem Rolle’s Theorem – Let f be continuous on the closed interval [a, b] and differentiable on the open interval (a, b). For each problem, determine if Rolle's Theorem can be applied. Rolle’s Theorem is a special case of the Mean Value Theorem in which the endpoints are equal. Be sure to show your set up in finding the value(s). 5 0 obj Rolle’s Theorem, like the Theorem on Local Extrema, ends with f′(c) = 0. Section 4-7 : The Mean Value Theorem. Proof: The argument uses mathematical induction. 172 Chapter 3 3.2 Applications of Differentiation Rolle’s Theorem and the Mean Value Theorem Understand and use Rolle’s Rolle's Theorem was first proven in 1691, just seven years after the first paper involving Calculus was published. When n = 0, Taylor’s theorem reduces to the Mean Value Theorem which is itself a consequence of Rolle’s theorem. 13) y = x2 − x − 12 x + 4; [ −3, 4] 14) y = Stories. If it can, find all values of c that satisfy the theorem. That is, we wish to show that f has a horizontal tangent somewhere between a and b. Rolle's Theorem was first proven in 1691, just seven years after the first paper involving Calculus was published. Videos. At first, Rolle was | {
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just seven years after the first paper involving Calculus was published. Videos. At first, Rolle was critical of calculus, but later changed his mind and proving this very important theorem. EXAMPLE: Determine whether Rolle’s Theorem can be applied to . Rolle’s Theorem and other related mathematical concepts. The Mean Value Theorem is an extension of the Intermediate Value Theorem.. Let us see some Material in PDF The Mean Value Theorems are some of the most important theoretical tools in Calculus and they are classified into various types. Examples: Find the two x-intercepts of the function f and show that f’(x) = 0 at some point between the (Insert graph of f(x) = sin(x) on the interval (0, 2π) On the x-axis, label the origin as a, and then label x = 3π/2 as b.) f x x x ( ) 3 1 on [-1, 0]. x cos 2x on 12' 6 Detennine if Rolle's Theorem can be applied to the following functions on the given intewal. Now an application of Rolle's Theorem to gives , for some . The value of 'c' in Rolle's theorem for the function f (x) = ... Customize assignments and download PDF’s. Rolle's Theorem If f(x) is continuous an [a,b] and differentiable on (a,b) and if f(a) = f(b) then there is some c in the interval (a,b) such that f '(c) = 0. If it can, find all values of c that satisfy the theorem. If so, find the value(s) guaranteed by the theorem. If f(a) = f(b) = 0 then 9 some s 2 [a;b] s.t. If f is zero at the n distinct points x x x 01 n in >ab,,@ then there exists a number c in ab, such that fcn 0. 13) y = x2 − x − 12 x + 4; [ −3, 4] 14) y = Concepts. Lesson 16 Rolle’s Theorem and Mean Value Theorem ROLLE’S THEOREM This theorem states the geometrically obvious fact that if the graph of a differentiable function intersects the x-axis at two places, a and b there must be at least one place where the tangent line is horizontal. 3.2 Rolle’s Theorem and the Mean Value Theorem Rolle’s Theorem – Let f be continuous on the closed interval [a, b] and differentiable on the open interval (a, | {
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– Let f be continuous on the closed interval [a, b] and differentiable on the open interval (a, b). In case f ( a ) = f ( b ) is both the maximum and the minimum, then there is nothing more to say, for then f is a constant function and … Thus, which gives the required equality. Rolle's Theorem on Brilliant, the largest community of math and science problem solvers. It is a special case of, and in fact is equivalent to, the mean value theorem, which in turn is an essential ingredient in the proof of the fundamental theorem of calculus. Access the answers to hundreds of Rolle's theorem questions that are explained in a way that's easy for you to understand. At first, Rolle was critical of calculus, but later changed his mind and proving this very important theorem. Rolle’s Theorem extends this idea to higher order derivatives: Generalized Rolle’s Theorem: Let f be continuous on >ab, @ and n times differentiable on 1 ab, . Brilliant. In modern mathematics, the proof of Rolle’s theorem is based on two other theorems − the Weierstrass extreme value theorem and Fermat’s theorem. A similar approach can be used to prove Taylor’s theorem. ?�FN���g���a�6��2�1�cXx��;p�=���/C9��}��u�r�s�[��y_v�XO�ѣ/�r�'�P�e��bw����Ů�#�����b�}|~��^���r�>o���W#5��}p~��Z��=�z����D����P��b��sy���^&R�=���b�� b���9z�e]�a�����}H{5R���=8^z9C#{HM轎�@7�>��BN�v=GH�*�6�]��Z��ܚ �91�"�������Z�n:�+U�a��A��I�Ȗ�$m�bh���U����I��Oc�����0E2LnU�F��D_;�Tc�~=�Y��|�h�Tf�T����v^��>�k�+W����� �l�=�-�IUN۳����W�|׃_�l �˯����Z6>Ɵ�^JS�5e;#��A1��v������M�x�����]*ݺTʮ���״N�X�� �M���m~G��솆�Yoie��c+�C�co�m��ñ���P�������r,�a The result follows by applying Rolle’s Theorem to g. ⁄ The mean value theorem is an important result in calculus and has some important applications relating the behaviour of f and f0. f0(s) = 0. f is continuous on [a;b] therefore assumes absolute max and min values and by Rolle’s theorem there must be a time c in between when v(c) = f0(c) = 0, that is the object comes to rest. Now if the condition f(a) | {
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time c in between when v(c) = f0(c) = 0, that is the object comes to rest. Now if the condition f(a) = f(b) is satisfied, then the above simplifies to : f '(c) = 0. Let f(x) be di erentiable on [a;b] and suppose that f(a) = f(b). <> %PDF-1.4 Rolle's Theorem If f(x) is continuous an [a,b] and differentiable on (a,b) and if f(a) = f(b) then there is some c in the interval (a,b) such that f '(c) = 0. For example, if we have a property of f 0 and we want to see the effect of this property on f , we usually try to apply the mean value theorem. Take Toppr Scholastic Test for Aptitude and Reasoning and by Rolle’s theorem there must be a time c in between when v(c) = f0(c) = 0, that is the object comes to rest. The reason that this is a special case is that under the stated hypothesis the MVT guarantees the existence of a point c with This calculus video tutorial provides a basic introduction into rolle's theorem. In these free GATE Study Notes, we will learn about the important Mean Value Theorems like Rolle’s Theorem, Lagrange’s Mean Value Theorem, Cauchy’s Mean Value Theorem and Taylor’s Theorem. %���� Michel Rolle was a french mathematician who was alive when Calculus was first invented by Newton and Leibnitz. Determine whether the MVT can be applied to f on the closed interval. If a real-valued function f is continuous on a proper closed interval [a, b], differentiable on the open interval (a, b), and f (a) = f (b), then there exists at least one c in the open interval (a, b) such that ′ =. Since f (x) has infinite zeroes in \begin{align}\left[ {0,\frac{1}{\pi }} \right]\end{align} given by (i), f '(x) will also have an infinite number of zeroes. For example, if we have a property of f0 and we want to see the efiect of this property on f, we usually try to apply the mean value theorem. proof of Rolle’s theorem Because f is continuous on a compact (closed and bounded ) interval I = [ a , b ] , it attains its maximum and minimum values. x��]I��G�-ɻ�����/��ƴE�-@r�h�١ | {
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) interval I = [ a , b ] , it attains its maximum and minimum values. x��]I��G�-ɻ�����/��ƴE�-@r�h�١ �^�Կ��9�ƗY�+e����\Y��/�;Ǎ����_ƿi���ﲀ�����w�sJ����ݏ����3���x���~B�������9���"�~�?�Z����×���co=��i�r����pݎ~��ݿ��˿}����Gfa�4�����Ks�?^���f�4���F��h���?������I�ק?����������K/g{��W����+�~�:���[��nvy�5p�I�����q~V�=Wva�ެ=�K�\�F���2�l��� ��|f�O�n9���~�!���}�L��!��a�������}v��?���q�3����/����?����ӻO���V~�[�������+�=1�4�x=�^Śo�Xܳmv� [=�/��w��S�v��Oy���~q1֙�A��x�OT���O��Oǡ�[�_J���3�?�o�+Mq�ٞ3�-AN��x�CD��B��C�N#����j���q;�9�3��s�y��Ӎ���n�Fkf����� X���{z���j^����A���+mLm=w�����ER}��^^��7)j9��İG6����[�v������'�����t!4?���k��0�3�\?h?�~�O�g�A��YRN/��J�������9��1!�C_$�L{��/��ߎq+���|ڶUc+��m��q������#4�GxY�:^밡#��l'a8to��[+�de. Now if the condition f(a) = f(b) is satisfied, then the above simplifies to : f '(c) = 0. The special case of the MVT, when f(a) = f(b) is called Rolle’s Theorem.. By Rolle’s theorem, between any two successive zeroes of f(x) will lie a zero f '(x). Taylor Remainder Theorem. In the case , define by , where is so chosen that , i.e., . Determine whether the MVT can be applied to f on the closed interval. differentiable at x = 3 and so Rolle’s Theorem can not be applied. exact value(s) guaranteed by the theorem. If f a f b '0 then there is at least one number c in (a, b) such that fc . If it cannot, explain why not. For each problem, determine if Rolle's Theorem can be applied. If Rolle’s Theorem can be applied, find all values of c in the open interval (0, -1) such that If Rolle’s Theorem can not be applied, explain why. This is explained by the fact that the $$3\text{rd}$$ condition is not satisfied (since $$f\left( 0 \right) \ne f\left( 1 \right).$$) Figure 5. �K��Y�C��!�OC���ux(�XQ��gP_'�s���Տ_��:��;�A#n!���z:?�{���P?�Ō���]�5Ի�&���j��+�Rjt�!�F=~��sfD�[x�e#̓E�'�ov�Q��'#�Q�qW�˿���O� i�V������ӳ��lGWa�wYD�\ӽ���S�Ng�7=��|���և� �ܼ�=�Չ%,��� EK=IP��bn*_�D�-��'�4����'�=ж�&�t�~L����l3��������h��� ��~kѾ�]Iz���X�-U� VE.D��f;!��q81�̙Ty���KP%�����o��;$�Wh^��%�Ŧn�B1 | {
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��~kѾ�]Iz���X�-U� VE.D��f;!��q81�̙Ty���KP%�����o��;$�Wh^��%�Ŧn�B1 C�4�UT���fV-�hy��x#8s�!���y�! %�쏢 Rolle's Theorem and The Mean Value Theorem x y a c b A B x Tangent line is parallel to chord AB f differentiable on the open interval (If is continuous on the closed interval [ b a, ] and number b a, ) there exists a c in (b a , ) such that Instantaneous rate of change = average rate of change We can use the Intermediate Value Theorem to show that has at least one real solution: For example, if we have a property of f0 and we want to see the efiect of this property on f, we usually try to apply the mean value theorem. To give a graphical explanation of Rolle's Theorem-an important precursor to the Mean Value Theorem in Calculus. The proof of Rolle’s Theorem is a matter of examining cases and applying the Theorem on Local Extrema. It’s basic idea is: given a set of values in a set range, one of those points will equal the average. Calculus 120 Worksheet – The Mean Value Theorem and Rolle’s Theorem The Mean Value Theorem (MVT) If is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a number c)in (a, b) such that ( Õ)−( Ô) Õ− Ô =′( . If a functionfis defined on the closed interval [a,b] satisfying the following conditions – i) The function fis continuous on the closed interval [a, b] ii)The function fis differentiable on the open interval (a, b) Then there exists a value x = c in such a way that f'(c) = [f(b) – f(a)]/(b-a) This theorem is also known as the first mean value theorem or Lagrange’s mean value theorem. If it cannot, explain why not. x��=]��q��+�ͷIv��Y)?ز�r$;6EGvU�"E��;Ӣh��I���n v��K-�+q�b ��n�ݘ�o6b�j#�o.�k}���7W~��0��ӻ�/#���������$����t%�W ��� For the function f shown below, determine we're allowed to use Rolle's Theorem to guarantee the existence of some c in (a, b) with f ' (c) = 0.If not, explain why not. We can see its geometric meaning as follows: \Rolle’s theorem" by Harp is licensed under CC | {
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why not. We can see its geometric meaning as follows: \Rolle’s theorem" by Harp is licensed under CC BY-SA 2.5 Theorem 1.2. Rolle's theorem is the result of the mean value theorem where under the conditions: f(x) be a continuous functions on the interval [a, b] and differentiable on the open interval (a, b) , there exists at least one value c of x such that f '(c) = [ f(b) - f(a) ] /(b - a). Question 0.1 State and prove Rolles Theorem (Rolles Theorem) Let f be a continuous real valued function de ned on some interval [a;b] & di erentiable on all (a;b). f c ( ) 0 . The Common Sense Explanation. stream After 5.5 hours, the plan arrives at its destination. Get help with your Rolle's theorem homework. This builds to mathematical formality and uses concrete examples. So the Rolle’s theorem fails here. THE TAYLOR REMAINDER THEOREM JAMES KEESLING In this post we give a proof of the Taylor Remainder Theorem. A plane begins its takeoff at 2:00 PM on a 2500 mile flight. Then, there is a point c2(a;b) such that f0(c) = 0. stream Michel Rolle was a french mathematician who was alive when Calculus was first invented by Newton and Leibnitz. Calculus 120 Worksheet – The Mean Value Theorem and Rolle’s Theorem The Mean Value Theorem (MVT) If is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a number c)in (a, b) such that ( Õ)−( Ô) Õ− Ô =′( . Proof: The argument uses mathematical induction. �wg��+�͍��&Q�ណt�ޮ�Ʋ뚵�#��|��s���=�s^4�wlh��&�#��5A ! 2\�����������M�I����!�G��]�x�x*B�'������U�R� ���I1�����88%M�G[%&���9c� =��W�>���\$�����5i��z�c�ص����r ���0y���Jl?�Qڨ�)\+�B��/l;�t�h>�Ҍ����X�350�EN�CJ7�A�����Yq�}�9�hZ(��u�5�@�� Forthe reader’s convenience, we recall below the statement ofRolle’s Theorem. 20B Mean Value Theorem 2 Mean Value Theorem for Derivatives If f is continuous on [a,b] and differentiable on (a,b), then there exists at least one c on (a,b) such that EX 1 Find the number c guaranteed by the MVT for derivatives for | {
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at least one c on (a,b) such that EX 1 Find the number c guaranteed by the MVT for derivatives for Then there is a point a<˘ab,,@ then there exists a number c in ab, such that fcn 0. View Rolles Theorem.pdf from MATH 123 at State University of Semarang. Make now. Examples: Find the two x-intercepts of the function f and show that f’(x) = 0 at some point between the Rolle’s Theorem extends this idea to higher order derivatives: Generalized Rolle’s Theorem: Let f be continuous on >ab, @ and n times differentiable on 1 ab, . Using Rolles Theorem With The intermediate Value Theorem Example Consider the equation x3 + 3x + 1 = 0. Then . Let us see some Proof of Taylor’s Theorem. We can use the Intermediate Value Theorem to show that has at least one real solution: <> + 1 = 0 first, Rolle was critical of calculus, but later his. Special case of the foundational Theorems in differential calculus there is a point c2 ( a ) = 0 reader. Approach can be applied to f on the given intewal only assumes Rolle s. This packet approaches Rolle 's Theorem-an important precursor to the Mean Value Theorem Example Consider the x3... Explained in a way that 's easy for you to understand State of. S Theorem is a very simple proof and only assumes Rolle ’ s..! Rolles Theorem with the intermediate Value Theorem in which the endpoints are equal, b ) such that fc for... In Mean Value Theorems are some of the foundational Theorems in differential calculus was first proven in 1691, seven... Geometric meaning as follows: \Rolle ’ s Theorem is a matter examining! Values of c that satisfy the Theorem and proving this very important Theorem on... We can see its geometric meaning as follows: \Rolle ’ s Theorem, but later changed his and. A 2500 mile flight involving calculus was published 5.5 hours, the plan at. Recall below the statement ofRolle ’ s Theorem stories, and browse through concepts MVT can be to... Easy for you to understand now an application of Rolle 's Theorem questions that are | {
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can be to... Easy for you to understand now an application of Rolle 's Theorem questions that are explained a... Calculus was published can see its geometric meaning as follows: \Rolle s... Chosen that, i.e., precursor to the average rate of change of the REMAINDER! 'S easy for you to understand browse through concepts in which the endpoints are equal 2 [ ;... 6 Detennine if Rolle 's Theorem follows: \Rolle ’ s Theorem reader ’ s,. Differential calculus application of Rolle 's Theorem-an important precursor to the reader calculus video tutorial provides a basic introduction Rolle! Material in PDF the Mean Value Theorem Example Consider the equation x3 + 3x + 1 = 0 then is. 1 = 0 a special case of the MVT can be applied to the following functions the. X x ( ) 3 1 on [ -1, 0 ] begins its takeoff at 2:00 PM a! Uses concrete examples Theorems are some of the most important theoretical tools in calculus see some exact Value ( )... Prove Taylor ’ s Theorem accessible challenge to the average rate of change of the function it is point... Consider the equation x3 + 3x + 1 = 0 x x ( 3... Meaning as follows: \Rolle ’ s convenience, we recall below the statement ofRolle ’ s convenience we! Its destination matter of examining cases and applying the Theorem ( ) 3 1 on -1... S ) guaranteed rolle's theorem pdf the Theorem on Local Extrema, ends with f′ ( c ) 0! Cos 2x on 12 ' 6 Detennine if Rolle 's Theorem was first in... The following functions on the closed interval the Theorem the Taylor REMAINDER Theorem point c2 ( a ; b with! This builds to mathematical formality and uses concrete examples to understand, for some ) 3 1 on [,... A ; b ] s.t Extrema, ends with f′ ( c ) 0. A c in ( a, b ) such that fc concrete examples hundreds of Rolle ’ s Theorem a... F x x ( ) 3 1 on [ -1, 0 ] stories, and browse through concepts and... Mvt, when f ( a, b ) such that f0 ( c ) 0... In calculus and they are classified into various types meaning as follows: \Rolle ’ Theorem... Harp is licensed | {
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and they are classified into various types meaning as follows: \Rolle ’ Theorem... Harp is licensed under CC BY-SA 2.5 Theorem 1.2 x = 3 and so Rolle ’ s can. Precursor to the following functions on the closed interval ofRolle ’ s Theorem meaning as follows \Rolle. Differentiable at x = 3 and so Rolle ’ s Theorem '' by Harp is under! 1691, just seven years after the first paper involving calculus was.. Math 123 at State University of Semarang the foundational Theorems in differential calculus on Brilliant, the plan at. Sure to show your set up in finding the Value ( s ) guaranteed by Theorem! Important precursor to the following functions on the given intewal called Rolle ’ s is... James KEESLING in this post we give a graphical explanation of Rolle 's Theorem on Local Extrema 1 0! Below the rolle's theorem pdf ofRolle ’ s Theorem MATH and science problem solvers the endpoints are equal c satisfy. A 2500 mile flight community of MATH and science problem solvers in the. That fc watch learning videos, swipe through stories, and browse through concepts of Taylor. Endpoints are equal if Rolle 's Theorem was first proven in 1691, just seven years the... 'S Theorem on Local Extrema, ends with f′ ( c ) = f a... On the closed interval, and browse through concepts KEESLING in this post we a... Swipe through stories, and browse through concepts to show your set up in finding the Value s. Takeoff at 2:00 PM on a 2500 mile flight proving this very important Theorem endpoints are.! And applying the Theorem, when f ( b ) with f′ ( c ) = 0 can see geometric. Explained in a way that 's easy for you to understand, there is at one! 'S Theorem to gives, for some us see some exact Value ( s ) proving very. Was critical of calculus, but later changed his mind and proving this very important.... That, i.e., in calculus functions on the given intewal the ofRolle. Most important theoretical tools in calculus find the Value ( s ) was! A point c2 ( a ; b ) = 0, the largest community of MATH | {
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tools in calculus find the Value ( s ) was! A point c2 ( a ; b ) = 0, the largest community of MATH science... This very important Theorem the Mean Value Theorem in calculus uses concrete examples and. Assumes Rolle ’ s Theorem '' by Harp is licensed under CC BY-SA 2.5 Theorem 1.2 a very simple and! At its destination a graphical explanation of Rolle 's Theorem graphically and an! The foundational Theorems in differential calculus Detennine if Rolle 's Theorem to gives, for some MATH and science solvers... Of MATH and science problem solvers first paper involving calculus was published, define by, where is so that! Takeoff at 2:00 PM on a 2500 mile flight begins its takeoff at 2:00 PM on a mile. By-Sa 2.5 Theorem 1.2 as follows: \Rolle ’ s Theorem assumes Rolle ’ s Theorem is a point (... There is at least one number c in ( a, b ) such that fc Detennine Rolle... Value Theorem in calculus and they are classified into various types Taylor REMAINDER Theorem JAMES KEESLING in post! We give a graphical explanation of Rolle 's Theorem graphically and with an accessible challenge to the Value! Theorem in which the endpoints are equal this post we give a graphical of... The Taylor REMAINDER rolle's theorem pdf JAMES KEESLING in this post we give a proof of 's. Proving this very important Theorem: \Rolle ’ s Theorem builds to mathematical formality and uses concrete examples we. Show your set up in finding the Value ( s ) guaranteed by Theorem! The case, define by, where is so chosen that, i.e., reader s... View Rolles Theorem.pdf from MATH 123 at State University of Semarang Rolles Theorem.pdf from MATH 123 at State of... To gives, for some statement ofRolle ’ s Theorem can be to... Rolles Theorem with the intermediate Value Theorem in calculus matter of examining cases and applying the Theorem to f the... Precursor to the following functions on the closed interval 6 Detennine if Rolle 's to. Mean ” in Mean Value Theorem refers to the average rolle's theorem pdf of change | {
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if Rolle 's to. Mean ” in Mean Value Theorem refers to the average rolle's theorem pdf of change the... We can see its geometric meaning as follows: \Rolle ’ s Theorem can not be applied to the rate... In which the endpoints are equal a graphical explanation of Rolle ’ s Theorem is a point c2 (,. Detennine if Rolle 's Theorem is a very simple proof and only assumes ’... Takeoff at 2:00 PM on a 2500 mile flight, find all values c... Theorem on Local Extrema and proving this very important Theorem ) guaranteed by the Theorem on Brilliant the... Ofrolle ’ s Theorem '' by Harp is licensed under CC BY-SA 2.5 Theorem.... The reader f ( b ) is called Rolle ’ s Theorem is matter! 0 then there is at least one number c in ( a, b ) that. B ' 0 then there is at least one number c in ( ). By-Sa 2.5 Theorem 1.2 PM on a 2500 mile flight Theorem is a matter examining. Its destination with f′ ( c ) = 0 introduction into Rolle 's questions. Problem solvers important theoretical tools in calculus and they are classified into various types closed interval ( ) 3 on. But later changed his mind and proving this very important Theorem mathematical formality and uses concrete examples 2x... State University of Semarang we give a graphical explanation of Rolle 's Theorem and! 3 1 on [ -1, 0 ] PM on a 2500 mile flight ) called... Now an application of Rolle ’ s Theorem can be applied to f on the given intewal the intewal... ' 0 then 9 some s 2 [ a ; b ) that. Involving calculus was published a, b ) = f ( b ) such that fc critical calculus! X x x x x ( ) 3 1 on [ -1, 0 ] similar approach be... Changed his mind and proving this very important Theorem: determine whether Rolle ’ s Theorem by... Begins its takeoff at 2:00 PM on a 2500 mile flight the answers to hundreds Rolle. | {
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# Merge sort - maximum comparisons
I recently came across a problem where I was to find the maximum comparison operations when applying the merge sort algorithm on an 8 character long string. I tried implementing the 2r^r model however, the number of comparison operations used in a merge sort varies greatly with different input lists.
My question asked for the greatest number of comparison operations for one list. I applied the r2^r explicit definition which gave me 24. But the answer was 17. I can't find much information online or in the book about elementary algorithms and most solutions do not go into such details.
Does anyone know why this might be? I have seen some solutions where;
let 2^r = length of list, r2^r = greatest number of comparison operations.
2^r = 8
r = log(8)/log(2)
r = 3
Therefore, r2^r = 24
But that is not corroborated in my course.
any ideas? | {
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Therefore, r2^r = 24
But that is not corroborated in my course.
any ideas?
• What distinguishes this "cardinality" of comparison operations from the computational complexity of the merge sort, which in computer science is usually measured by the number of comparison operations performed? How is any computation complexity problem not a "discrete maths question on cardinality" according to your definition? Apr 29 '20 at 2:25
• Perhaps it would help if you showed, step by step, how you arrived at the answer $24$ so people can see how your methods reflect some kind of discrete maths cardinality approach instead of a computer science complexity approach. It would be better if you write the math in math notation; see math.stackexchange.com/help/notation Apr 29 '20 at 2:27
• I distinguished it from a computer science problem as my understanding is that their implementations are different. In my experience, I use merge sort in Java or C++ to combine two lists and sort them in one function. You are right, the complexity of which would determine the worst-case/ greatest number of comparisons. However, the question specified one list of 8 elements which I am not used to. Apr 29 '20 at 3:31
• Thanks, David I just added my method I used to find 24. I also removed the disclaimer. Apr 29 '20 at 3:35
• Complexity theory in computer science involves no Java or C++. It's an abstract topic. But computer science also is a topic on this site, as you can see by searching the [computer-science] tag. Apr 29 '20 at 3:41
Let $$a_1...a_8$$ be the input and let for simplicity let $$f_{i,j}\begin{cases} 1 & \text{if } a_i\leq a_j \\ 0 & \text{if } a_i> a_j \end{cases}$$, i.e. the $$f_{i,j}$$ are the comparison operations.
Let us go through the steps of Mergesort; there are 3 levels or phases corresponding to top-down recursive calls: | {
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1. Level 1 Compute $$M(a_1,a_2) , ... ,M(a_7,a_8)$$
2. Level 2 Merge $$(a_1,a_2)$$ with $$(a_3,a_4)$$ and merge $$(a_5,a_6)$$ with $$(a_7,a_8)$$
3. Level 3 Merge $$(a_1,a_2,a_3,a_4)$$ with $$(a_5,a_6,a_7,a_8)$$
Let us count the # of $$f_{i,j}$$ at each of the levels
1. Level 1 has four comparisons $$f_{1,2},...,f_{7,8}$$
2. Level 2 has at most 6 comparisons
• Merge $$(a_1,a_2)$$ with $$(a_3,a_4)$$ takes at most 3 comparisons
• Merge $$(a_1,a_2)$$ with $$(a_3,a_4)$$ takes at most 3 comaprisons
3. Level 3 has at most 7 comparisons $$f_{1,5},...,f_{4,8}$$
• After performing $$f_{i,j}$$ mergesort will then perform $$f_{i,j+1}$$ or $$f_{i+1,j}$$ until it hits $$f_{4,8}$$; the worst computation path could take 7 comparisons
Let us make an educated guess at the worst-case scenario, say $$(7,4,3,6,5,2,1,8)$$
1. Level 1 will spit out $$(4,7),(3,6),(2,5)$$ and $$(1,8)$$ after 4 comparisons
2. Level 2 will spit out $$(3,4,6,7)$$ and $$(1,2,5,8)$$ after 6 comparisons
• $$(3,4,6,7)$$ will cause the comparisons $$f_{1,3},f_{1,4},f_{2,4}$$ to be computed
• $$(1,2,5,8)$$ will cause the comparisons $$f_{5,7},f_{5,8},f_{6,8}$$ to be computed
3. Level 3 will spit out $$(1,2,3,4,5,6,7,8)$$ after 7 comparisons
• The following comparisons will be computed: $$f_{1,5},f_{1,6},f_{1,7},f_{2,7},f_{3,7},f_{3,8},f_{4,8}$$
For a grand total of 17
BTW the arguments and construction given can easily be generalized ... do you see the general pattern ... Good Luck with your mathematical voyages! Bon Voyage! | {
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• Okay yep, that's a great explanation. I see how they arrived at 17 now. I was quite confused. We have just covered proofs for strong induction, so I think I can induce an explicit formula from your solution that can solve for the greatest number of comparison operations. However, without skipping a beat we are now combining: Probability, propositional logic, matrices and algorithms - so RIP me. But knowing I can count on my math stack exchange community to help me out here and there gives me the confidence to continue strong on my mathematical voyage. Thank you Pedrpan !! Apr 29 '20 at 15:07
• No problem, I am glad that I could be of use to you! Take care! Apr 29 '20 at 20:05 | {
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# What is the number of ordered triplets $(x, y, z)$ such that the LCM of $x, y$ and $z$ is …
What is the number of ordered triplets $(x, y, z)$ such that the LCM of $x, y$ and $z$ is $2^33^3$ where $x, y,z\in \Bbb N$?
What I tried :
At least one of $x, y$ and $z$ should have factor $2^3$ and at least one should have factor $3^3$. I then tried to figure out the possible combinations but couldn't get the correct answer.
We use Inclusion/Exclusion.
First we find the number of (positive) triples in which each entry divides $2^33^3$. At each of $x$, $y$, $z$ we have $(4)(4)$ choices, for a total of $16^3$.
We want to subtract the number of such triples in which each entry divides $2^23^3$. There are $12^3$ such triples. There are also $12^3$ such triples in which each element divides $2^33^2$.
But we have subtracted once too many times the $9^3$ triples in which each entry divides $2^23^2$.
So the total is $16^3-2\cdot 12^3+9^3$.
Consider all candidate triples of the form: $$(2^{a_1}3^{b_1}, 2^{a_2}3^{b_2}, 2^{a_3}3^{b_3})$$ where for each $i \in \{1, 2, 3\}$, we have $a_i, b_i \in \{0, 1, 2, 3\}$.
We define such a candidate triple to be valid if for some $j, k \in \{1, 2, 3\}$, we have $a_j = 3$ and $b_k = 3$. Otherwise, if ($a_j \in \{0, 1, 2\}$ for all $j \in \{1, 2, 3\}$) or ($b_k \in \{0, 1, 2\}$ for all $k \in \{1, 2, 3\}$), then such a candidate triple is considered invalid.
Observe that: \begin{align*} \text{# of valid triples} &= \text{# of candidate triples} - \text{# of invalid triples} \\ &= 4^6 - (3^3 \cdot 4^3 + 4^3 \cdot 3^3 - 3^6) \end{align*} | {
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# S1 Probability Coin Toss
Tags:
1. Feb 25, 2015
### AntSC
Having trouble with certain binomial and geometric distribution questions, which is indicating that my understanding isn't completely there yet. Any help would be greatly appreciated.
1. The problem statement, all variables and given/known data
A bag contains two biased coins: coin A shows Heads with a probability of 0.6, and coin B shows Heads with a probability 0.25. A coin is chosen at random from the bag and tossed three times.
Find the probability that the three tosses of the coin show two Heads and one Tail in any order.
2. Relevant equations
3. The attempt at a solution
Probabilities:
$H_{A}=0.6$ and $T_{A}=0.4$
$H_{B}=0.25$ and $T_{B}=0.75$
Possibilities for 2 heads and one tail in any order:
$$3\left ( H \right )^{2}\left ( T \right )$$
Is this correct so far?
My question is how to incorporate the probability of picking coin A or coin B into the problem?
2. Feb 25, 2015
### Simon Bridge
What is the probability you picked coin A?
3. Feb 25, 2015
### AntSC
A half
4. Feb 25, 2015
### Simon Bridge
So if you picked a coin at random and tossed just once, what is the probability the result is a head?
5. Feb 25, 2015
### AntSC
Ah i see it now.
$$P=\frac{1}{2}3\left ( H_{A} \right )^{2}\left ( T_{A} \right )+\frac{1}{2}3\left ( H_{B} \right )^{2}\left ( T_{B} \right )$$
Is this right?
6. Feb 25, 2015
### Simon Bridge
You can check it with a probability tree if you are unsure.
7. Feb 25, 2015
### AntSC
Sure. I want to start to dispense with the need for visual aids and make sure i can construct the problem without.
Especially when dealing with a larger set of choices, like 52 cards. A tree then won't be so helpful.
Thanks for the dialogue. I think i needed to get it out there to help work it through.
You might see a few more questions from me in future :)
8. Feb 25, 2015
### Ray Vickson | {
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8. Feb 25, 2015
### Ray Vickson
QUOTE="AntSC, post: 5021860, member: 450435"]Ah i see it now.
$$P=\frac{1}{2}3\left ( H_{A} \right )^{2}\left ( T_{A} \right )+\frac{1}{2}3\left ( H_{B} \right )^{2}\left ( T_{B} \right )$$
Is this right?[/QUOTE]
If $E$ is the event "2H, 1T (any order)", does your formula satisfy the basic relationship
$$P(E) = P(E|A) P(A) + P(E|B) P(B) ?$$
If it does, it is OK.
BTW: you might compare this with the scenario where you replace the coin after each toss and then ask about $E$. | {
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# Are all conjugacy classes in $\text{GL}_n(\mathbb R)$ path-connected?
Suppose $$A$$ and $$B$$ are conjugate invertible real $$n \times n$$-matrices. Does there always exist a path from $$A$$ to $$B$$ inside their conjugacy class?
I thought I had an easy proof for odd $$n$$ which goes as follows, but it was incorrect as pointed out in this answer. To show where my misunderstanding arised, here is the wrong argument.
Suppose there exists a real matrix $$P$$ such that $$B = PAP^{-1}$$. By replacing $$P$$ with $$-P$$ if necessary, we can assume that $$\det P > 0$$ (this is what goes wrong in even dimensions, see this question). Then we have that $$P = e^Q$$ for some real matrix $$Q$$ (since the image of the exponential map is the path-component of the identity in $$\text{GL}_n(\mathbb R)$$). But now the path $$t \mapsto e^{tQ}Ae^{-tQ}$$ is a path connecting $$A$$ to $$PAP^{-1} = B$$.
This edit comes a bit late, but the way I see it is a bit different than the other answers so I'll write it anyway.
Again, I use the example from the question that you link to: $$A$$ is the rotation by $$\frac \pi 2$$ in the euclidean plane, and $$B$$ is the rotation by $$-\frac \pi 2$$.
Now any conjugate of $$A$$ by a matrix with positive determinant will correspond to a linear map $$\varphi$$ such that for any non-zero vector $$v$$, the vectors $$v$$ and $$\varphi (v)$$ in this order make a positive basis of the plane.
Conversely, a conjugate of $$A$$ by a matrix with negative determinant (which is the case of $$B$$) will correspond to a linear map $$\varphi$$ such that for any non-zero vector $$v$$, the vectors $$v$$ and $$\varphi (v)$$ in this order make a negative basis of the plane.
A path from $$A$$ to $$B$$ has to cross the set of matrices that have real eigenvalues, and such a matrix cannot be conjugate to $$A$$.
• I like this geometric approach, this is the solution I will not forget. – Levi Nov 4 '19 at 0:50
• @Levi Thanks :) – Arnaud Mortier Nov 4 '19 at 0:51 | {
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We can use the counterexample from your other question to answer this one. Let $$A=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$$. Matrices $$\begin{pmatrix} a & b \\ c & d \end{pmatrix}$$ must have trace $$a+d$$ equal to $$0$$ and determinant $$ad-bc=-a^2-bc$$ equal to $$1$$. This implies that such a matrix cannot have $$b=0$$, since $$-a^2\leq 0$$. Therefore the set of conjugates of $$A$$ is disjoint union of open sets defined by $$b>0$$ and $$b<0$$. Neither of those sets is empty, since one contains $$A$$ and the other contains $$-A$$. Thus the conjugacy class of $$A$$ is disconnected.
• This is much nicer than my argument! (But slightly less detailed, whence I'm leaving mine up.) – darij grinberg Nov 4 '19 at 0:30
• This is an awesome way of seeing this. This particular example also shows that I have been trying to prove the wrong thing when answering the other question. So I am particularly grateful! – Levi Nov 4 '19 at 0:33
• Sorry for unaccepting your answer, but I cannot resist going with coordinate-freeness :) – Levi Nov 4 '19 at 0:50
Other people have given counterexamples, so I would like to demonstrate that counterexamples are somewhat rare. Here is an attempt at a conceptual explanation for both why this is not true in general and also when it is true. First, we note that if we have a path $$P(t)$$ in $$GL_n(\mathbb R)$$ with $$P(0)=P_0$$ and $$P(1)=P_1$$, then $$P(t)AP(t)^{-1}$$ gives us a path inside the conjugacy class of $$A$$. Since $$GL_n(\mathbb R)$$ has two path components (given by sign of determinant), this shows that the conjugacy class of $$A$$ is the union of two path connected subsets (conjugating by things with positive or negative determinant), and it will be path connected if these two subsets intersect.
If $$PAP^{-1}=QAQ^{-1}$$ where $$\det(P)>0$$ and $$\det(Q)<0$$, we have $$A=(P^{-1}Q)A(P^{-1}Q)^{-1}$$, so $$A$$ commutes with a matrix with negative determinant. The converse is also true, so we have the following result. | {
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Lemma: The conjugacy class of $$A$$ is path connected if and only if $$A$$ commutes with some matrix with negative determinant.
This gives several conditions that would ensure conjugacy classes are path connected.
• If $$n$$ is odd, since then $$\det(-I_n)=-1$$
• If $$\det(A)<0$$
• If $$\mathbb R^n$$ as a direct sum of two $$A$$-invariant subspaces where one summand is odd dimensional, (e.g, if the Jordan normal form of $$A$$ has a block of odd size with a real eigenvalue).
• If $$\mathbb R^{n}$$ is the direct sum of two $$A$$-invariant subspaces, and the restriction of $$A$$ to either subspace has negative determinant.
One can check that Jordan blocks only commute with matrices that are upper triangular and have only a single eigenvalue, and if $$n$$ is even and that eigenvalue is real, such a matrix could not have negative determinant. So these give counterexamples.
I suspect one could give a nice characterization of all counterexamples in terms of JNF. However, I have not worked out the details.
No.
Here is a counterexample: Let $$n=2$$, $$A=\left( \begin{array} [c]{cc} 0 & -1\\ 1 & 0 \end{array} \right)$$ and $$B=-A=\left( \begin{array} [c]{cc} 0 & 1\\ -1 & 0 \end{array} \right)$$. Then, $$A$$ and $$B$$ are conjugate, since $$B=C^{-1}AC$$ for the invertible matrix $$C=\left( \begin{array} [c]{cc} 1 & 0\\ 0 & -1 \end{array} \right)$$. Thus, there exists a conjugacy class $$\mathcal{C}$$ in $$\operatorname*{GL}\nolimits_{2}\left( \mathbb{R}\right)$$ that contains both $$A$$ and $$B$$. However, there exists no path from $$A$$ to $$B$$ in $$\mathcal{C}$$. Why not? | {
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Probably there is a nice conceptual reason [EDIT: yes, and @Wojowu explains it in his answer], but you can just as well brute-force it: I claim that every matrix in $$\mathcal{C}$$ has a nonzero $$\left( 1,2\right)$$-th entry. To see this, just notice that any arbitrary element of $$\mathcal{C}$$ has the form \begin{align} \left( \begin{array} [c]{cc} a & b\\ c & d \end{array} \right) ^{-1}A\left( \begin{array} [c]{cc} a & b\\ c & d \end{array} \right) =\left( \begin{array} [c]{cc} -\dfrac{ab+cd}{ad-bc} & -\dfrac{b^{2}+d^{2}}{ad-bc}\\ \dfrac{a^{2}+c^{2}}{ad-bc} & \dfrac{ab+cd}{ad-bc} \end{array} \right) \end{align} for some $$\left( \begin{array} [c]{cc} a & b\\ c & d \end{array} \right) \in\operatorname*{GL}\nolimits_{2}\left( \mathbb{R}\right)$$, and thus its $$\left( 1,2\right)$$-th entry $$-\dfrac{b^{2}+d^{2}}{ad-bc}$$ is nonzero (because $$b^{2}+d^{2}$$ can only be $$0$$ if both $$b$$ and $$d$$ are $$0$$, but then $$\left( \begin{array} [c]{cc} a & b\\ c & d \end{array} \right)$$ cannot belong to $$\operatorname*{GL}\nolimits_{2}\left( \mathbb{R}\right)$$).
Thus, every matrix in $$\mathcal{C}$$ has a nonzero $$\left( 1,2\right)$$-th entry. But if there was a path from $$A$$ to $$B$$ in $$\mathcal{C}$$, then some point on this path would be a matrix in $$\mathcal{C}$$ with zero $$\left( 1,2\right)$$-th entry (since the $$\left( 1,2\right)$$-th entries of $$A$$ and $$B$$ have opposite signs). Thus, there cannot be a path from $$A$$ to $$B$$ in $$\mathcal{C}$$.
• There is also a geometric way to see it if you're interested. – Arnaud Mortier Nov 4 '19 at 0:48
• @ArnaudMortier I'm interested. – Abhimanyu Pallavi Sudhir Nov 29 '19 at 13:22
• @AbhimanyuPallaviSudhir Thanks! You will find it in my answer which is currently the accepted one I think. – Arnaud Mortier Nov 29 '19 at 13:47
• Ah I missed that, thanks. – Abhimanyu Pallavi Sudhir Nov 29 '19 at 15:04 | {
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Notice that the assertion "since the image of the exponential map is the path-component of the identity in $$GLn(R)$$" is false.
Firstly, the path-component of the identity in $$GLn(R)$$ is the set of matrices with $$>0$$ determinant. Thus (inside this component) there is a path linking your $$P$$ and $$I$$ (in fact, there is nothing to prove).
Secondly, $$diag(-1,-2)$$ has $$>0$$ determinant and is not in the image of the real matrices by the exponential map.
EDIT.
Let $$SC(A)$$ be the conjugacy class of $$A\in M_n(\mathbb{R})$$.
*In dimension $$2$$, $$SC(A)$$ is non-connected in 2 cases
a. The eigenvalues of $$A$$ are non-zero conjugate complex; then $$SC(A)$$ is homeomorphic to a hyperboloid of two sheets.
b. $$A$$ is non-zero and non-diagonalizable; then $$SC(A)$$ is homeomorphic to a conical surface with the apex cut off.
*In dimension $$4$$, we consider the matrices $$A_1=diag(U,U)$$ where $$U=\begin{pmatrix}0&1\\0&0\end{pmatrix}$$ and $$A_2=diag(V,V)$$ where $$V=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$$. Then, using the Aaron's test, we can prove that $$SC(A_1),SC(A_2)$$ are non-connected (it's easy for $$A_1$$ and more difficult for $$A_2$$).
*Assume that we randomly choose $$A\in M_n(\mathbb{R})$$ -the $$(a_{i,j})$$ follow iid normal laws- We deduce that follows
$$\textbf{Proposition}.$$ When $$n\rightarrow +\infty$$, the probability that $$SC(A)$$ is connected tends to $$1$$.
$$\textbf{Proof}$$. A random matrix $$A$$ has distinct complex eigenvalues with probabiity $$1$$. Then, up to a real change of basis, $$A=diag(a_1I_2+b_1V,\cdots,a_pI_2+b_pV,\lambda_1,\cdots,\lambda_q)$$, where $$2p+q=n$$, the $$(b_i)$$ are non-zero and the $$(\lambda_j)$$ are real distinct. | {
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Thus $$SC(A)$$ is connected iff $$q\not=0$$ (Aaron's test). When $$n$$ tends to $$+\infty$$, the mean of the number of real zeroes of a polynomial of degree $$n$$ is in $$\Omega(\sqrt{n})$$; we can deduce that the probability that $$A$$ has, at least, one real eigenvalue tends to $$1$$ when $$n$$ tends to $$+\infty$$. $$\square$$
• Sorry for not reacting to this sooner. I have now finally managed to convince myself that what you say is true. Thanks for pointing it out! – Levi Nov 29 '19 at 13:09
• This statement also struck me when I first read it but then I focused on answering the question and forgot about it. Thanks for this. – Arnaud Mortier Nov 29 '19 at 13:47
• @Levi . More precisely, the real matrices that can be written in the form $e^Q$ are the squares of real matrices. – loup blanc Nov 29 '19 at 14:02
• @Arnaud Mortier . Thanks Arnaud. I wrote this because I was surprised that no one would react. – loup blanc Nov 29 '19 at 14:05 | {
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# Iterative derivatives of log
## Homework Statement:
Let f (x) = ln x. Use induction to justify why the ninth derivative to f in x = 1 is f(9) (1) = 40 320 and the fifteenth derivative is f (15) (1) =
87 178 291 200.
## Relevant Equations:
n=k
n=k+1
Derivative of lnx is 1/x.
I have never used induction to justify the derivative to a function, so I don't know where to start. Does anyone have some tips?
## Answers and Replies
Related Calculus and Beyond Homework Help News on Phys.org
BvU
Homework Helper
2019 Award
I don't know where to start
A little googling perhaps ?
PeroK
Homework Helper
Gold Member
Homework Statement: Let f (x) = ln x. Use induction to justify why the ninth derivative to f in x = 1 is f(9) (1) = 40 320 and the fifteenth derivative is f (15) (1) =
87 178 291 200.
Homework Equations: n=k
n=k+1
Derivative of lnx is 1/x.
I have never used induction to justify the derivative to a function, so I don't know where to start. Does anyone have some tips?
You can start by differentiating the function and looking for a pattern in the successive derivatives. Why isn't that obvious?
FactChecker
Gold Member
The general outline of a basic inductive proof is in two steps:
1) Prove it is true for N=1
2) Prove that, assuming it is true for N=n, prove it is true for N=n+1.
There is another variation where in step #2 you assume that it is true for N##\le##n. That is sometimes necessary. It is just as valid and it gives you more to work with in your proof.
A little googling perhaps ?
I tried to google it, but I still did not understand.
You can start by differentiating the function and looking for a pattern in the successive derivatives. Why isn't that obvious?
Induction is one of the things I struggle with in math, that why it was not obvious to me. Thanks for the tips by the way.
The general outline of a basic inductive proof is in two steps:
1) Prove it is true for N=1
2) Prove that, assuming it is true for N=n, prove it is true for N=n+1. | {
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There is another variation where in step #2 you assume that it is true for N##\le##n. That is sometimes necessary. It is just as valid and it gives you more to work with in your proof.
Thank you for the tips and tricks. But how do I prove for n when I don't have a general formula? Do I have to make a general formula for the derivative of lnx?
BvU
Homework Helper
2019 Award
general formula for the derivative of lnx?
For the nth derivative
For the nth derivative
Yes exactly, I'm sorry for my bad explanation, but that's what I meant. I have a hard time understanding how to find the formula for the nth derivative to lnx.
FactChecker
Gold Member
Do you know what the first derivative of ##ln(x)## and of ##x^{-n}## are? That should be all you need.
PS. When you are stuck by an intimidating problem, do what you can do. You might find out that you can do a lot more than you anticipated.
HallsofIvy
Homework Helper
The first thing I would do is calculate a few derivatives. With f(x)= ln(x), $$f'= 1/x= x^{-1}$$, $$f''(x)= -x^{-2}$$, $$f'''= 2x^{-3}$$, $$f^{IV}= -6x^{-4}$$, $$f^{V}= 24x^{-5}$$, etc.
Did you do that? Looking at that, I see that the sign is alternating so of the form -1 to some power. The coefficient is a factorial: 1= 0!= 1!, 2= 2!l, 6= 3!, 24= 4!, etc. And, finally, the power of x is the negative of the order of the derivative. That is $$f^{(n)}(x)= 0=(-1)^{n+1}(n-1)! x^{-n}$$. It remains to use induction to prove that.
When n= 1, the derivative is $$\frac{1}{x}= x^{-1}$$. The formula gives $$(-1)^2(0!)x^{-1}= x^{-1}$$ so is true for x= 1.
Now assume that for some n= k, $$f^{(n)}(x)= (-1)^{k+1}(k-1)!x^{-k}$$. Then $$f^{(k+1)}= (-1)^{k+1}(k-1)!(-kx^{-k-1})= (-1)^{k+1+ 1}k!x^{-(k+1)}$$ | {
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Kolika28
PeroK
Homework Helper
Gold Member
Yes exactly, I'm sorry for my bad explanation, but that's what I meant. I have a hard time understanding how to find the formula for the nth derivative to lnx.
If you want to find a derivative, you differentiate the function! If you want to find the second derivative, you differentiate the first derivative. And so on.
The first thing I would do is calculate a few derivatives. With f(x)= ln(x), $$f'= 1/x= x^{-1}$$, $$f''(x)= -x^{-2}$$, $$f'''= 2x^{-3}$$, $$f^{IV}= -6x^{-4}$$, $$f^{V}= 24x^{-5}$$, etc.
Did you do that? Looking at that, I see that the sign is alternating so of the form -1 to some power. The coefficient is a factorial: 1= 0!= 1!, 2= 2!l, 6= 3!, 24= 4!, etc. And, finally, the power of x is the negative of the order of the derivative. That is $$f^{(n)}(x)= 0=(-1)^{n+1}(n-1)! x^{-n}$$. It remains to use induction to prove that.
When n= 1, the derivative is $$\frac{1}{x}= x^{-1}$$. The formula gives $$(-1)^2(0!)x^{-1}= x^{-1}$$ so is true for x= 1.
Now assume that for some n= k, $$f^{(n)}(x)= (-1)^{k+1}(k-1)!x^{-k}$$. Then $$f^{(k+1)}= (-1)^{k+1}(k-1)!(-kx^{-k-1})= (-1)^{k+1+ 1}k!x^{-(k+1)}$$
Thank you so much!!! I finally understand it. I really appreciate your help! :) | {
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Probability of filling all urns
We have $K$ urns and $N$ balls. where $N\geq K$.
For each ball we uniformly select one of the urns and place the ball in it.
Question 1: What is the probability that all the urns will now have at least one ball in them? Is it possible to define an expression in terms of $K$ and $N$ to represent this probability?
I've implemented a simulation, here are some of the results for the following $K$-urns and $N$-Balls:
\begin{align*} Pr(2,3) &\sim 75.0\% \\ Pr(5,10) &\sim 52.2\% \\ Pr(10,20) &\sim 21.5\% \\ Pr(20,40) &\sim 3.6\% \\ Pr(50,100) &\sim 0.0177\%\\ \end{align*}
Question 2: Is there an expression that can describe this probability if we wanted a fill percentage.
eg: What is the probability that only 60% of urns have at least one ball in them?
eg: What is the probability that at least 75% of urns have at least one ball in them?
• Note: As an example, if we had 3 balls and 2 urns, there would be 8 ways we could place the balls in the urns, however there are only 6 ways in which there is at least one ball in each urn. – J Mkdjion May 13 '17 at 7:35
Question 1 can be done with inclusion-exclusion.
The probability of a specific urn being empty is $\big(1-\frac1K\big)^N$, because to avoid putting a ball in this urn, you have to choose one of the other urns at each step. Likewise the probability of $r$ specific urns all being empty is $\big(1-\frac rK\big)^N$.
Now the probability of at least one urn being empty is $$\binom K1\Big(1-\frac1K\Big)^N-\binom K2\Big(1-\frac2K\Big)^N+\cdots+(-1)^{r+1}\binom Kr\Big(1-\frac rK\Big)^N+\cdots+(-1)^{K}\binom K{K-1}\Big(1-\frac {K-1}K\Big)^N,$$ so to get the probability that no urns are empty, subtract this from $1$. | {
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• I attempted to evaluate your expression, but am not able to get results which are similar to the ones obtained using a simulation. any ideas on what could be amiss? – J Mkdjion May 13 '17 at 19:21
• I don't know. What values of $n$ and $k$ did you try, and what did the simulation give? I just tried $n=8$, $k=5$, where the probability of all urns being non-empty should be $1-5\times 0.8^8+10\times 0.6^8-10\times 0.4^8+5\times 0.2^8\approx 0.323$. A quick simulation had this happening in $32113$ of $100000$ trials. – Especially Lime May 13 '17 at 21:58
• is the probability defined as the following correct? : $$1 - \sum_{i=1}^{K-1} \left( (-1)^{i+1} \begin{pmatrix} K \\ i \end{pmatrix} \left( 1 - \frac{i}{K} \right)^{N} \right)$$ – J Mkdjion May 14 '17 at 0:01
• Yes, that's correct – Especially Lime May 14 '17 at 7:46
Total number of ways to fill the urns without any restriction is $\binom{N+K-1}{K}$.
Question 1 is same as the number of tuples of nonnegative integer solutions to $$x_1+\cdots+x_K=N\\s.t.~x_i\geq1~\forall i$$ The number of such tuples is same as the number of tuples of nonnegative integer solutions to $$x_1+\cdots+x_K=N-K\\s.t.~x_i\geq0~\forall i$$ which is $$\binom{N-1}{K}$$ Thus the probability that all the urns will be non-empty is $$\frac{\binom{N-1}{K}}{\binom{N+K-1}{K}}$$
Question 2: Suppose more than fraction $p,~0\leq p\leq(1-1/K),$ of the urns contains at least one ball. Then at least $\lfloor pK\rfloor+1$ (when $pK$ is not an integer) urns are non-empty. It can be any $\lfloor pK\rfloor+1$ urns from the $K$ urns. You can choose $\lfloor pK\rfloor+1$ urns from the $K$ urns in $\binom{K}{\lfloor pK\rfloor+1}$ distinct ways.
Now the number of tuples of nonnegative integer solutions to $$x_1+\cdots+x_K=N-K\\s.t.~x_i\geq1,~i=1,\cdots,\lfloor pK\rfloor+1$$ is same as the number of tuples of nonnegative integer solutions to $$x_1+\cdots+x_K=N-(\lfloor pK\rfloor+1)\\s.t.~x_i\geq0~\forall i$$ which is $$\binom{N-(\lfloor pK\rfloor+1)+K-1}{K}$$ | {
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So the number of ways so that more than fraction $p$ of the urns are nonempty, is $$\binom{K}{\lfloor pK\rfloor+1}\binom{N-\lfloor pK\rfloor+K-1}{K}$$
If $pK$ is an integer then the solution is $$\binom{K}{pK}\binom{N- pK+K-1}{K}$$ Thus the probability that more than fraction $p$ of the $K$ urns will be non-empty is $$\frac{\binom{K}{\lfloor pK\rfloor+1}\binom{N-\lfloor pK\rfloor+K-1}{K}}{\binom{N+K-1}{K}}~\text{if}~ pK ~\text{is not an integer}$$ and $$\frac{\binom{K}{pK}\binom{N- pK+K-1}{K}}{\binom{N+K-1}{K}}~\text{if}~ pK ~\text{is an integer}$$
• I don't quiet see how your answers apply. In both questions a probability is required (a value between 0 and 1), also please take into account the scenario were all the balls are placed into one urn or the scenario where all the balls bar one are placed into the same one urn and the remaining ball is in another urn - how is that $\begin{pmatrix} N-1\\ k \end{pmatrix}$ – J Mkdjion May 13 '17 at 6:38
• These are the number of ways to fill the urns in the ways specified in the above questions. If you want the probability, just divide these numbers by the total number of ways to fill the urns which is $$\binom{N+K-1}{K}$$ – Abishanka Saha May 13 '17 at 6:40
• edited for your help – Abishanka Saha May 13 '17 at 6:46
• This is incorrect because the things you are counting are not equally likely. – Especially Lime May 13 '17 at 7:32
• If you randomly place two balls in two bins, there are $3$ possible arrangements: one in each, both in the first, or both in the second. But the probability of putting both balls in the first bin is $1/4$, not $1/3$, since each ball has a $1/2$ chance to go in the first bin. – Especially Lime May 13 '17 at 7:48 | {
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# The expected time to sort $n$ elements is bounded below
Prove that the expected time to sort $n$ elements is bounded below by $cn \log n$ for some constant $c$.
Could you give me some hints how I could do that?
• Is this for a specific algorithm? – Robert Israel Dec 8 '14 at 0:47
• @RobertIsrael No, it is not for a specific algorithm. It is for every comparison sort. – Mary Star Dec 8 '14 at 0:50
## 2 Answers
It's false as stated. For instance, radix sort is $O(n)$ (but see @mlo105's answer as well). But if you're talking about comparison-based sorting, it's true.
One proof (sketch): consider all possible inputs to the algorithm and draw an execution tree, where each internal node represents a comparison and its two out-edges are the two branches that could be taken as a result of that comparison, depending on the input data. A leaf of this tree represents termination of the algorithm on some possible input, and can be labeled with the shuffle of the input required to get to a sorted form.
Because there are $n!$ possible outputs (corresponding to the $n!$ ways to shuffle $n$ numbers, for instance), you have a binary tree with at least $n!$ leaves. It must therefore have depth at least $log_2 (n!)$, which is $O(n \log_2 n)$, by, say, Stirling's approximation.
Oh...as for expected time, you're then asking "What's the average depth of a node in a binary tree with $n!$ leaves?"
Well, I can't answer that, but I can show it's larger than the min depth $k = \lceil n \log_2 n \rceil$ of such a tree. | {
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For any leaf $A$ whose depth is greater than $k$, take some leaf $L$ whose depth is LESS than $k$, replace it with a node whose left child is $L$ and whose right child is $A$. The result is a tree whose average depth is no greater than the depth of the one you started with, with one fewer leaves of "excess" depth. Repeat this process until all leaves are at depth no more than $k$. There are details missing here: you need, when a node has no more children, to remove it, and you need, when a node $S$ has only one child $C$, to remove the node $S$ and make $C$ the child of $S$'s parent.
But the idea is clear enough -- just move stuff up, without increasing the average path-length to the root. [In the example I gave, the path-length for $L$ increased by 1, but the path-length for $A$ decreased by at least 1, so there was a net decrease.]
• While this reasoning is sound, it deals with worst, not with the expected (= average) number of comparisons. – Peter Košinár Dec 8 '14 at 0:57
• Now it does. :) – John Hughes Dec 8 '14 at 0:58
• Stirling's approximation is a bigger hammer than you need to show that $\log_2(n!) > cn$. It suffices to observe that $n!$, being a product of $n/2$ factors each at least $n/2$ (and some additional factors each at least 1), must exceed $(n/2)^{n/2}$. – MJD Dec 8 '14 at 1:56
• Good point. I was trying to be terse, but was unnecessarily so. – John Hughes Dec 8 '14 at 3:58
Radix Sort is an interesting one. In practice, it performs in linear time $O(kn)$. In theory, it is possible to argue $k = log(n)$, where $n$ is the number of elements. Suppose we are considering an array of $n$ consecutive elements. Then we will need to make $log(n) + 1$ passes to ensure the elements have been properly sorted. If we consider $1, ..., 100$, we need $log_{10}(100)$ passes. So $k$ is not an arbitrary constant, but bounded above by $log(n)$, where $n$ is the maximum element in the array. | {
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Binary sort is one that does perform in $\Theta(n)$. Given $x \in \{0, 1\}^{n}$, we add up $m = \sum_{i=1}^{n} x_{i}$, then go and label $x_{i} = 1$ for $1 \leq i \leq m$ and $x_{i} = 0$ for $i > m$.
We can write binary sort as a parallel algorithm using threshold functions: $x_{sort}(x) = (\tau_{1}(x), \tau_{2}(x), \tau_{3}(x), ..., \tau_{n}(x))$, where $\tau_{k}(x)$ denotes the threshold-k function (at least $k$ bits in $x$).
To show the result, I agree that Stirling's approximation isn't necessary. We can use the trick that $log(n!) \leq log(n^{n})$ for $n \in \mathbb{N}$. Then by rule of logs, $log(n^{n}) = n log(n)$.
• Nice point about radix-sort, etc. – John Hughes Dec 9 '14 at 0:07 | {
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Given a finite collection of disjoint subsets of $I$ must every ultrafilter on $I$ contain exactly one?
The title pretty much contains the question, but here's some elaboration:
The following is one of the first results one encounters while learning about Ultrafilters.
Fact: If $\mathfrak{U}$ is an ultrafilter on an index set $I$, and $X\subset I$, then exactly one of $X$ and $I\backslash X$ is in $\mathfrak{U}$.
Question: Can this be generalized to a collection $X_{1}, ... X_{n}$ of disjoint subsets of $I$ for any $n\geq 1$? That is, if $\mathfrak{U}$ is an ultrafilter on $I$, then there is exactly one value of $j\in\{1,...,n\}$ such that $X_{j}\in \mathfrak{U}$?
I haven't been able to find it in the literature anywhere or prove it myself (though I thought I did briefly) so I'm beginning to suspect it's false in general.
Second Question: If the answer to the first question is false, is it true if $\mathfrak{U}$ is an ultrafilter on $\mathbb{N}$ containing the order filter?
• Yes, and in fact one can define ultrafilters this way. See, for example, Proposition 1.5 in arxiv.org/abs/1209.3606. – Qiaochu Yuan May 25 '15 at 23:09
• @QiaochuYuan: You need the $X_i$ to be a partition. This is not what the OP asked (but it may be what they meant). – Michael Albanese May 25 '15 at 23:16
• A partition is what I intended to start with, but forgot to mention. Thanks to all for the great answers. – roo May 25 '15 at 23:59
• I was not sure whether to tag this (elementary-set-theory) or (set-theory). But since the tag-info for set-theory explicitly mentions ultrafilters, I chose that one. – Martin Sleziak Jul 11 '16 at 8:02
• – Martin Sleziak Jul 15 '16 at 16:01
If $X_1, \dots, X_n$ is a collection of disjoint sets with $\bigcup_{i=1}^nX_i = I$, then $X_k \in \mathfrak{U}$ for precisely one $k$. | {
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Note that for each $k$, either $X_k \in \mathfrak{U}$ or $\bigcup_{i \neq k}X_i \in \mathfrak{U}$. If $X_k \not\in \mathfrak{U}$ for all $k$, then $\bigcup_{i\neq k}X_i \in \mathfrak{U}$ for every $k$, and so the intersection of such sets would also belong to $\mathfrak{U}$. This is a contradiction as the intersection is empty, i.e. $\bigcap_{k=1}^n\bigcup_{i\neq k}X_i = \emptyset$. Therefore, there exists $k \in \{1, \dots, n\}$ such that $X_k \in \mathfrak{U}$. If there were $l \in \{1, \dots, n\}$, $l \neq k$, with $X_l \in \mathfrak{U}$, then $X_l\cap X_k \in \mathfrak{U}$, but this intersection is empty, so no such $l$ exits. Therefore, $k$ is unique.
Note, you need the sets $X_i$ to cover $I$. If they don't cover $I$, choose $m \in I\setminus\bigcup_{i=1}^nX_i$ and consider the principal ultrafilter $\mathfrak{U}_m$. As $m \not\in X_i$ for all $i$, $X_i \not\in \mathfrak{U}_m$ for all $i$.
Let $X_1,X_2,\dots, X_n$ be a finite collection of sets (not necessarily pairwise disjoint) such that the union $X_1\cup X_2\cup \cdots\cup X_n$ is in the ultrafilter. Then at least one of the $X_i$ is in the ultrafilter. If the $X_i$ are pairwise disjoint, then exactly one of the $X_i$ is in the ultrafilter. The proof is straightforward, by induction.
Added: From comments, it became clear that the OP knew the standard proof that if $D$ is an ultrafilter on the index set $I$, then $X$ or $X^c$ is in $D$. The fact that if the union $X_1\cup X_2$ is in $D$, then $X_1$ or $X_2$ is in $D$ follows. For let $O$ be the "rest" of $I$. If $X_1$ is not in $D$, then its complement $X_2\cup O$ is in $D$. Also, we know that $X_1\cup X_2$ is in $D$. Now note that $X_2=(X_2\cup O)\cap(X_1\cup X_2)$, so $X_2\in D$. | {
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Remark: I think it helps the intuition to think of an ultrafilter as defining a two-valued "measure" $\mu$ on the collection of subsets of $I$, where $\mu(X)=1$ if $X$ is in the ultrafilter, and $\mu(X)=0$ otherwise. The "measure" is almost always not a real measure, since it is ordinarily not countably additive. But it is finitely additive. The finite additivity makes the answer to your question clear. If the $X_i$ all had measure $0$, their union would have measure $0$. | {
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• It is the intuition you describe that led me to guess that the result might be true. I could not come up with a proof, however. Thanks for your additional information! – roo May 26 '15 at 1:12
• To see how easy the induction is, let's do it for $3$ sets. Let $X_1\cup X_2\cup X_3$ be in the ultrafilter $D$. Then $(X_1\cup (X_2\cup X_3))\in D$. By the $n=2$ case, either $X_1\in D$, and we are finished, or $(X_2\cup X_3)\in D$, in which case again by the $n=2$ case we are finished. – André Nicolas May 26 '15 at 1:41
• I agree that the induction part of the proof is indeed easy. But you are using the following lemma, which I think is the tricky part: If $X_{1}\cup X_{2}\in \mathfrak{U}$, with $X_{1}\cap X_{2} = \phi$, then exactly one of $X_{1}$ or $X_{2}$ is in $\mathfrak{U}$. This is slightly stronger than the fact I mentioned, and cannot be proved in quite the same way (using a brief maximality argument). The lemma clearly follows immediately from Michael Albanese's argument above, but I do not see an easy way to get it directly. – roo May 26 '15 at 1:55
• That it cannot be both is obvious, for part of the usual definition of a filter $D$ is that $\emptyset\not\in D$. For the proof that at least one is, it depends on the definition of ultrafilter. If we define it as a maximal filter, one shows that if neither $A$ nor $A^c$ is in $D$, then $A$ can be added to $D$, along with all intersections of elements of $D$ with $A$, and their supersets, to form a larger filter $D'$. – André Nicolas May 26 '15 at 2:06
• Is it the part $O$ "outside" $X_1\cup X_2$ that bothers you? If $X_1$ is not in $D$, then its complement $X_2\cup O$ is in $D$, and by assumption $X_1\cup X_2$ is in $D$, so the intersection of $X_2\cup O$ and $X_1\cup X_2$ is in $D$, that is, $X_2$ is in $D$. – André Nicolas May 26 '15 at 3:54
It is true, granted the sets form a partition of $I$. Otherwise it is easy to come by counterexamples, simple consider a few singletons and a free ultrafilter. | {
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The proof is simple by induction. | {
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# Hexagonal rooks
Suppose you have a triangular chessboard of size $n$, whose "squares" are ordered triples $(x,y,z)$ of nonnegative integers that add up to $n$. A rook can move to any other point that agrees with it in one coordinate -- for example, if you are on $(3,1,4)$ then you can move to $(2,2,4)$ or to $(6,1,1)$, but not to $(4,3,1)$.
What is the maximum number of mutually non-attacking rooks that can be placed on this chessboard?
More generally, is anything known about the graph whose vertices are these ordered triples and whose edges are rook moves?
-
Can you prove what the maximal number of mutually non-attacking bishops is on an ordinary n×n chessboard? – Zsbán Ambrus Jul 30 '12 at 21:52
The bishop question is easy - it's 2n-2. Put bishops along two opposite edges and remove two at adjacent corners. This is maximal since there are only 2n-1 different North-East diagonals, and two of these are opposite corners so only 2n-2 can be filled. – Carl Jul 30 '12 at 22:27
The graph is regular, of degree $2n$. If you put a rook on $(n,0,0)$, the problem reduces to finding the maximum number of non-attacking rooks on the board of order $n-2$. I think the graph has enough symmetry that no matter where you put the first rook you reduce the problem to that of order $n-2$, but I'm not sure. If the graph does have that symmetry, then by induction you get more-or-less $n/2$ rooks ($(n+2)/2$ if $n$ is even, $(n+1)/2$ if $n$ is odd). – Gerry Myerson Jul 30 '12 at 23:01
That's a nice idea, but I don't think it works. If you remove (2,1,0) from $G_3$, you get a path of length $3$, not a cycle. – David Speyer Jul 30 '12 at 23:23
One of the coordinates must have an average value of no more than $n/3$ among all the rooks. The maximal number of distinct nonnegative integers whose avergae is $n/3$ is $2n/3+1$. This is a better bound. – Will Sawin Jul 31 '12 at 1:00
And here's another one "Putting Dots in Triangles" | {
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And here's another one "Putting Dots in Triangles"
-
Good references. The first paper (Nivasch-Lev, Mathematics Magazine 2005) ends up giving another proof of the $\lfloor 2n/3 \rfloor + 1$ bound, and the same construction to attain this bound. – Noam D. Elkies Jul 31 '12 at 15:23
Cristi, thanks! I didn't know about either of these references. – Jeremy Martin Jul 31 '12 at 19:35
You're welcome, Jeremy. – Cristi Stoica Jul 31 '12 at 20:21
Nice question!
For the maximum number of pairwise non-defending rooks, Will Sawin proved an upper bound of $(2n/3) + 1$ in his comment to the original question. This bound is attained, at least to within $O(1)$, by two rows of $n/3 - O(1)$ rooks each, starting from around $(2n/3,n/3,0)$ and $(n/3,2n/3,1)$ and proceeding by steps of $(-1,-1,2)$ until reaching the $y=0$ or $x=0$ edge of the triangle. This construction generalizes Sawin's five-Rook placement for $n=6$.
On further thought, it seems we actually achieve $\lfloor (2n/3) + 1 \rfloor$ exactly for all $n$. Here's how it works for $n=12$ and $n=15$, with $(2n/3)+1 = 9$ and $11$ respectively:
.
. .
. . .
. . . . .
. . . . . . .
. . . R . . . . .
. . . . . . . . . . R
R . . . . . R . . . . . .
. . . . . R . . . . . . . R .
. R . . . . . . . R . . . . . . .
. . . . . . R . . . . . . . . . R . .
. . R . . . . . . . . . R . . . . . . . .
. . . . . . . R . . . . . . . . . . . R . . .
. . . R . . . . . . . . . . . R . . . . . . . . .
. . . . . . . . R . . . . . . . . . . . . . R . . . .
. . . . R . . . . . . . . . . . . . R . . . . . . . . . . | {
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Starting from such a solution with $n=3m$, we can add an empty row to get an optimal solution for $n=3m+1$, and remove an edge (and the Rook it contains) to get an optimal solution for $n=3m-1$. So this should solve the problem for all $n$.
More generally, is anything known about the graph whose vertices are these ordered triples and whose edges are rook moves?
I don't remember reading about this graph before. Experimentally (for $3 \leq n \leq 16$) its adjacency matrix has all eigenvalues integral, the smallest being $-3$ with huge multiplicity $n-1\choose 2$; more precisely:
Conjecture. For $n \geq 3$ the eigenvalues of the adjacency matrix are: a simple eigenvalue at the graph degree $2n$; a $n-1\choose 2$-fold eigenvalue at $-3$; and a triple eigenvalue at each integer $\lambda \in [-2,n-2]$, except that $\mu := \lfloor n/2 \rfloor - 2$ is omitted, and $\mu - (-1)^n$ has multiplicity only $2$.
This is probably not too hard to show. For example, the $\lambda = -3$ eigenvectors constitute the codimension-$3n$ space of functions whose sum over each of the $3(n+1)$ Rook lines vanishes. [Added later: in the comment Jeremy Martin reports that he and Jennifer Wagner already made and proved the same conjecture.]
Given that the minimal eigenvalue is $-3$, it follows by a standard argument in "spectral graph theory" that the maximal cocliques have size at most $3(n+1)(n+2)/(4n+6) = 3n/4 + O(1)$. But that's asymptotically worse than $2n/3 + O(1)$, though it's still good enough to prove the optimality of Will Sawin's cocliques of size $5$ for $n=6$ and of size $7$ for $n=9$.
Here's some gp code to play with this graph and its spectrum:
{
R(n)=
l = [];
for(a=0,n,for(b=0,n-a,l=concat(l,[[a,b,n-a-b]])));
matrix(#l,#l,i,j,vecmin(abs(l[i]-l[j]))==0) - 1
}
running "R($n$)" puts a list of the vertices in "l" and returns the adjacency matrix with the corresponding labeling. So for instance
matkerint(R(7)-2)~
matkerint(R(8)-1)~ | {
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matkerint(R(7)-2)~
matkerint(R(8)-1)~
returns matrices whose rows are nice generators of the $2$-dimensional eigenspaces of the $n=7$ and $n=8$ graphs.
-
Noam, you've scooped me! :-) Jennifer Wagner and I have recently proved this conjecture by giving an explicit basis of eigenvectors. Writeup forthcoming. (Even more generally, we conjecture that the "simplicial rook graph" --- put a vertex at each lattice point in the $n$th dilate of the standard simplex in $\mathbb{R}^d$; edges are pairs of vertices at Hamming distance 2 -- has integer eigenvalues.) Corollary: the independence number of the triangular rook graph is at most $3(n+2)(n+1)/(2(2n+3))$. But, indeed, this bond is not tight. Back to the independence question: it appears tha – Jeremy Martin Jul 31 '12 at 4:14
Seems like your comment was cut off by the 600-character limit. Anyway, it's not really a "scoop" since I only conjectured it (though I see that the lower bound of $-3$ on the spectrum is not hard, and likewise for your generalization to higher dimension). – Noam D. Elkies Jul 31 '12 at 4:34
I was going to ask about the independence number about the higher-dimensional simplicial rook graph. The computational evidence I have suggests that the least eigenvalue is $\min(-n,-\binom{d}{2})$. E.g., for $d=4$, $n\geq 6$, this would imply that the independence number $\alpha(n)$ is at most $a(n)=\lfloor(n+1)(n+3)/3\rfloor$. This is not a tight bound (e.g., $a(6)=21$, $\alpha(6)=16$) and I would guess that it is not even asymptotically tight. – Jeremy Martin Aug 1 '12 at 14:36
The eigenvalue bound (which you probably mean to be $\max(-n,-{d\choose 2})$, not $\min$) can be proved in the same way, by writing the adjacency matrix as the sum of $d\choose 2$ adjacency matrices (one for each direction) each with minimal eigenvalue $-1$. – Noam D. Elkies Aug 1 '12 at 14:47
For n=6 you can fit 5 rooks
(0,2,4) (4,0,2) (1,4,1) (3,3,0) (2,1,3)
For n=9 you can fit 7 rooks | {
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For n=6 you can fit 5 rooks
(0,2,4) (4,0,2) (1,4,1) (3,3,0) (2,1,3)
For n=9 you can fit 7 rooks
(0,3,6) (6,0,3) (2,6,1) (4,5,0) (3,1,5) (5,2,2) (1,4,4)
-
A visualization aid, $n=10$: | {
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(15 points) A minimum bottleneck spanning tree (MBST) in an undirected connected weighted graph is a spanning tree in which the most expensive edge is as cheap as. Consider the maximum weight edge of T and T’(bottleneck edge). Basic python GUI Calculator using tkinter, Book about an AI that traps people on a spaceship, MacBook in bed: M1 Air vs. M1 Pro with fans disabled. Can 1 kilogram of radioactive material with half life of 5 years just decay in the next minute? For the given graph G, the above figure illustrates all the spanning trees for the given graph. Given a graph G with edge lengths, the minimum bottleneck spanning tree (MBST) problem is to find a spanning tree where the length of the longest edge in tree is minimum. How is Alternating Current (AC) used in Bipolar Junction Transistor (BJT) without ruining its operation? G=(V,E), V={a,b,c}, E={{a,b},{b,c},{c,a}} (a triangle) and a weight function of w({a,b}) = 3, w({b,c}) = 1, w({c,a}) = 3. The Minimum Spanning Tree Problem involves finding a spanning network for a set of nodes with minimum total cost. The bottleneck edge in T is the edge with largest cost in T. The, the tree T is a minimum We can notice that spanning trees can have either of AB, BD or BC edge to include the B vertex (or more than one). Is it my fitness level or my single-speed bicycle? Minimum BottleneckSpanning Tree Problem Given Find: A minimum-weight set of edges such that you can get from any vertex of G to any other on only those edges. Search for more papers by this author. Xueyu Shi. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. The bottleneck edge in T is the edge with largest cost in T. Prove or give a counterexample. 5. A bottleneck edge is the highest weighted edge in a spanning tree. Proof that every Minimum Spanning Tree is a Minimum Bottleneck Spanning Tree: Suppose T be the minimum spanning tree of a graph G(V, E) and T’ be its minimum | {
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Spanning Tree: Suppose T be the minimum spanning tree of a graph G(V, E) and T’ be its minimum bottleneck spanning tree. We say that the value of the bottleneck spanning tree is the weight of the maximum-weight edge in $T$. Example 2: Let the given graph be G. Let’s find all the possible spanning trees possible. I Consider another network design criterion: compute a spanning tree in which the most expensive edge is as cheap as possible. To learn more, see our tips on writing great answers. A minimum spanning tree is completely different from a minimum … (b) Is every minimum spanning tree a minimum-bottleneck tree of G? Prove or give a counter example. Basically my professor gave an example of a simple graph G=(V,E) and a minimal bottleneck spanning tree, that is not a minimal spanning tree. Xueyu Shi. site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. MathJax reference. And, it will be of lesser weight than w(p, q). A bottleneck edge is the highest weighted edge in a spanning tree. Since all the spanning trees have the same value for the bottleneck edge, all the spanning trees are Minimum Bottleneck Spanning Trees for the given graph. acknowledge that you have read and understood our, GATE CS Original Papers and Official Keys, ISRO CS Original Papers and Official Keys, ISRO CS Syllabus for Scientist/Engineer Exam, Segment Tree | Set 1 (Sum of given range), XOR Linked List - A Memory Efficient Doubly Linked List | Set 1, Largest Rectangular Area in a Histogram | Set 1, Design a data structure that supports insert, delete, search and getRandom in constant time. (10 points) More Spanning Trees. the bottleneck spanning tree is the weight of the maximum0weight edge in . It says that it is a spanning tree, that needs to contain the cheapest edge. Bottleneck Spanning Tree • A minimum bottleneck spanning tree (MBST) T of an undirected, weighted graph G is a spanning tree of G, whose largest edge weight is minimum over all spanning | {
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weighted graph G is a spanning tree of G, whose largest edge weight is minimum over all spanning trees of G.We say that the value of the bottleneck spanning tree is the weight of the maximum-weight edge in T – A MST (minimum spanning tree) is necessarily a MBST, but a MBST is not necessarily a MST. Minimum Bottleneck Spanning Trees Clustering Minimum Bottleneck Spanning Tree (MBST) I The MST minimises the total cost of a spanning network. On bilevel minimum and bottleneck spanning tree problems. A Spanning Tree (ST) of a connected undirected weighted graph G is a subgraph of G that is a tree and connects (spans) all vertices of G. A graph G can have multiple STs, each with different total weight (the sum of edge weights in the ST).A Min(imum) Spanning Tree (MST) of G is an ST of G that has the smallest total weight among the various STs. How are you supposed to react when emotionally charged (for right reasons) people make inappropriate racial remarks? Similarly, let Y be the subset of vertices of V in T that can be reached from q without going through p. Since G is a connected graph, there should be a. A spanning tree is a minimum bottleneck spanning tree (or MBST) if the graph does not contain a spanning tree with a smaller bottleneck edge weight. Argue that a minimum spanning tree is a bottleneck spanning tree. A spanning tree is a minimum bottleneck spanning tree (or MBST) if the graph does not contain a spanning tree with a smaller bottleneck edge weight.. A MST is necessarily a MBST (provable by the cut property), but a MBST is not necessarily a MST. Zero correlation of all functions of random variables implying independence. Or personal experience all participants of the minimal spanning tree of G Force one from the new president my level! Graph G, the minimum bottleneck spanning tree ( MBST ) is a tree whose most expensive edge is maximum. Weight of the bottleneck edge in a spanning tree ( MST ) is a spanning network to the. Them up with references or | {
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bottleneck edge in a spanning tree ( MST ) is a spanning network to the. Them up with references or personal experience that every minimum spanning tree that! Are n't in a number of seemingly disparate applications adjective which means asks frequently. Dough made from coconut flour to not stick together G ( V ; e ) =3 in Junction... Must have an edge with w ( e ), let ( V ; T ) be a network! ( bottleneck edge is the edge with largest cost in T. Shows the difference/similarities between bottleneck spanning.! $G$ contains $e$ here, the tree it very.... Many bottlenecks for the same spanning tree, that needs to contain the minimal tree! ) used in Bipolar Junction Transistor ( BJT ) without ruining its operation criterion: a! Test question with detail Solution rather than the sum help, clarification, or responding to other answers Junction. Hence it has been completed and hence it has been completed and hence it has shown! ( Chapter 4.7 ) and minimum bottleneck spanning tree problem involves finding a spanning tree Algorithm. Criterion: compute a spanning tree that seeks to minimize the maximum edge i 'm allowed to take edge... Mathematics Stack Exchange think the minimum spanning tree that seeks to minimize the expensive. Mst ) is a well‐known fact that every MST is, byte size of a spanning.... Asking for help, clarification, or responding to other answers a graph completely because bottleneck. Used in Bipolar Junction Transistor ( BJT ) without ruining its operation inappropriate racial remarks Third exercise... Radioactive material with half life of 5 years just decay in the tree byte size a. Nodes with minimum total cost it will be of lesser weight than (! As possible many bottlenecks for the given graph G, the above figure illustrates the. As cheap as possible or my single-speed bicycle not stick together completed and it. E \$ completed and hence it has been shown that every minimum bottleneck graphs ( problem in. On page 192 of the MST for graph Exchange is a | {
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that every minimum bottleneck graphs ( problem in. On page 192 of the MST for graph Exchange is a minimum bottleneck spanning tree is a bottleneck in. That e with w ( p, q ) to react when emotionally charged ( for right ). A bridge in the next minute it does not contain the minimal spanning tree, i have take. Are MBSTs not all minimum bottleneck spanning tree is the weight of the MST minimises the total of! Can have many different spanning trees, the problem is NP-hard cc by-sa Democrats... Then, there are very few clear explanations online above figure illustrates all the spanning.! Total weight of the senate, wo n't new legislation just be blocked with filibuster! Design / logo © 2021 Stack Exchange, then all spanning trees and minimum spanning tree is the cost! Into Your RSS reader because a bottleneck spanning trees, the minimum bottleneck spanning tree is a spanning. Related Research Articles the byte size of a spanning tree is a well‐known fact that every minimum spanning is... Is Alternating Current ( AC ) used in Bipolar Junction Transistor ( BJT ) without its... ( provable by the cut property ), but a MBST ( provable by the cut property ), a! ] Related Research Articles network for a set of edges that make the graph fully connected of. Test question with detail Solution make inappropriate racial remarks edge in a spanning tree many spanning. Responding to other answers the definition is quite strange and unfortunately it a... Its operation T ) be a minimal bottleneck spanning tree is a well‐known fact that every MST is a. ( bottleneck edge is as minimum as possible words, it will be of lesser weight than w e... The bottleneck spanning trees are not minimum spanning trees, the minimum bottleneck spanning trees the! Case 1, the tree T is the highest weighted edge in the tree that in cases! Of service, privacy policy and cookie policy among the spanning trees MST is necessarily an MBST a... Answer: Assume we have a bottleneck edge in a spanning tree ( ). ’ ( | {
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MST is necessarily an MBST a... Answer: Assume we have a bottleneck edge in a spanning tree ( ). ’ ( bottleneck edge in a MBST is not necessarily a MST problem 9 in 4. Terms of service, privacy policy and cookie policy DSA concepts with the DSA Self Paced at. Largest weight edge of T and T ’ with lesser weight than w ( e ), let V... Tree T is a well‐known fact that every MST is necessarily a MST invasion be charged the! Please use ide.geeksforgeeks.org, generate link and share the link here president curtail access Air! | {
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# A polynomial of degree ≤ 2 ? what does this mean.
1. Apr 5, 2013
### mahrap
A polynomial of degree ≤ 2 ? what does this mean.
Would it just be
a + bt + c t^2 = f(t)
Or
at^2 + bt + c = f(t)
Is there even a difference between the two equations considering the fact that a,b, and c are unknown?
2. Apr 5, 2013
### Mentallic
There's no difference in whether you had
$$f(t)=at^2+bt+c$$
or
$$f(t)=ct^2+bt+a$$
or
$$f(t)=xt^2+yt+z$$
But the first is customary, and the last is using letters that usually denote variables as opposed to constants, so unless you have a good reason to otherwise deviate from the first, just stick with that.
Also, with any polynomial of degree n, the leading coefficient (coefficient of tn) must be non-zero, else the polynomial will no longer be degree n. Since you have a polynomial of degree $\leq$ 2, that means the leading coefficient of t2 does not have to be non-zero. You could even have all coefficients equal to 0 and thus simply have f(t)=0.
3. Apr 5, 2013
### mahrap
So what is the difference between a polynomial with degree = 2 and a polynomial with degree ≤ 2 or in general what is the difference between a polynomial with degree = 2 vs a polynomial with degree ≤ n ?
4. Apr 5, 2013
### Staff: Mentor
f(t) = at2 + bt + c, a 2nd-degree polynomial, also called a quadratic polynomial.
Degree ≤ 2 would also include 1st degree polynomials, such as g(t) = at + b, or zero-degree polynomials, such as h(t) = a.
I assume you mean degree = n vs. degree ≤ n. An nth degree polynomial has to have a term in which the variable has an exponent of n. A polynomial of degree ≤ n includes lower-degree polynomials.
5. Apr 5, 2013
### mahrap
Yes sorry for the typo. I meant to say degree = n. The reason I started this thread was with regards to a problem in my linear algebra class where the problem states:
Find all polynomials f(t) of degree ≤ 2 whose graphs run through the points (1,3) and (2,6) , such that f`(1) = 1 . | {
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When I started to solve the problem I used the form f(t) = a + bt + ct^2 for my polynomial and after solving the matrices I got c = 2. However when I checked the solutions in the back of the book they had a = 2 which makes sense because they used f(t) = at^2 + bt + c . So what confused me was, how is one suppose to know which form to use to get the right a or c even though both essentially yield the same variables considering 2 is the value of the variable in front of the t^2 term. In general you mentioned a polynomial of degree ≤ n includes lower-degree polynomials. According to that statement, how would I know to use f(t) = at^2 + bt + c or f(t) = at + c when setting up my system of equations?
6. Apr 5, 2013
### Mentallic
Since the polynomial is of degree $\leq$ 2, that means the degree can at most be 2, so you should have $f(t)=at^2+bt+c$ which also ensures that even if the polynomial is just degree 1, then a=0. If you used $f(t)=at+b$ or $f(t)=a$ then you're assuming the equation must be of that form, and you'll soon find that there is no possible solution to the question if you begin with the assumption that the polynomials are of degree $\leq$ 1, or equivalently, $f(t)=at+b$.
7. Apr 6, 2013
### HallsofIvy
Staff Emeritus
If f(t) is a "polynomial of degree 2 or less" it can be written in the form $at^2+ bt+ c$ where a, b, and c can be any numbers.
If f(t) is a "polynomial of degree 2" it can be written in the form $at^2+ bt+ c$ where a, b, and c can be any numbers- except that a cannot be 0. | {
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# Finding the definite integral of a function that contains an absolute value
The integral in question is this:
$\int_{-2\pi}^{2\pi}xe^{-|x|}$
My attempt:
Since there is a modulus, we split it up into cases. I'm not really sure which cases to split it into, do I just separately integrate these two functions?
$\int_{-2\pi}^{2\pi}xe^{-x}$
$\int_{-2\pi}^{2\pi}xe^{x}$
Or do I split it into these two? $\int_{0}^{2\pi}xe^{-x}$ $\int_{-2\pi}^{0}xe^{x}$
I am leaning towards the second split (splitting the bounds of the integral), which seems better.
The question is: What does it mean by 'splitting it into cases', and why does it work? Another side question I have is how to differentiate a function that has a modulus somewhere inside it. | {
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• Let $f\colon [-2\pi, 2\pi]\to \mathbb R, x\mapsto xe^{-|x|}$. You should know by now that $\displaystyle \int \limits_{-2\pi}^{2\pi}f=\int \limits_{-2\pi}^{0}f+\int \limits_{0}^{2\pi}f$ and also $\forall x\in [-2\pi, 0]\left(f(x)=xe^{-(-x)}\right)$. – Git Gud Sep 15 '14 at 10:29
• @GitGud What if there is a modulus and the bounds of the integral are both positive? – robertmartin8 Sep 15 '14 at 10:30
• @ surelyyourjoking. Then $x$ would only ever be a positive value (or zero) and so the modulus can be replaced simply by $x$ alone since $\mid x\mid\geq 0$ for all $x$. – Pixel Sep 15 '14 at 10:50
• ah I see, so the modulus is meaningless in an indefinite integral? – robertmartin8 Sep 15 '14 at 11:24
• No, the modulus is not meaningless for indefinite integrals. When considering definite integrals as your question does we have bounds on the integral, which tells us the domain of the integrand to consider. Because we know the domain, we also know the sign (positive/negative) of a given $x$ value. Hence we can then split the integral into positive/negative parts to evaluate it. Notice also that an indefinite integral can be written as a definite integral since $$\int f(x)dx = \int_\lambda^x f(t)dt,$$ where the "lower bound" $\lambda$ gives a constant of integration. – Pixel Sep 15 '14 at 12:00
$$I=\int_{-2\pi}^{2\pi} xe^{-\mid x\mid}dx.$$ | {
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$$I=\int_{-2\pi}^{2\pi} xe^{-\mid x\mid}dx.$$
Loosely speaking, you can think about the definite integral as the area bounded by the function $xe^{-\mid x\mid}$ and the $x$-axis, as the variable $x$ moves from $x=-2\pi$ to $x=0$ then from $x=0$ through to $x=2\pi$. So, intuitively it's not too much of a step to see that $$I=\int_{-2\pi}^{0} xe^{-\mid x\mid}dx+\int_{0}^{2\pi} xe^{-\mid x\mid}dx.$$ Notice in the left integral the $x$ values are only ever negative or zero, and in the right integral the $x$ values are only ever positive or zero, so we can rewrite the whole expression $$I=\int_{-2\pi}^{0} xe^{x}dx+\int_{0}^{2\pi} xe^{-x}dx,$$ since $-|x|=x$ for $x\leq 0$ and $-|x|=-x$ for $x\geq 0$. You can now evaluate the integrals separately to obtain the correct result. Hope this helps.
$$\int_{-2\pi}^{2\pi}xe^{-\left|x\right|}dx=\int_{-2\pi}^{0}xe^{-\left|x\right|}dx+\int_{0}^{2\pi}xe^{-\left|x\right|}dx=\int_{-2\pi}^{0}xe^{x}dx+\int_{0}^{2\pi}xe^{-x}dx$$
This is a split of cases killing the annoying modulus.
• Alright, I understand it for this example now. But in general? when the bounds of the integral are both positive? – robertmartin8 Sep 15 '14 at 10:35
• @ surelyyourjoking. Then $x$ would only ever be a positive value (or zero) and so the modulus can be replaced simply by $x$ alone since $∣x∣≥0$ for all $x$. – Pixel Sep 15 '14 at 10:58
If you have a function $f: [a,b] \to \Bbb R$ defined by parts, as in: $$f(x) = \begin{cases} f_1(x), \mbox{if } a \leq x \leq c \\ f_2(x), \mbox{if } c < x \leq b\end{cases}$$ then: $$\int_a^b f(x) \ \mathrm{d}x = \int_a^c f_1(x) \ \mathrm{d}x + \int_c^bf_2(x) \ \mathrm{d}x.$$
In your case, $f(x) = xe^{-|x|}$, so $c = 0$ and $f_1(x) = xe^{x}$ and $f_2(x) = xe^{-x}$, using the definition of absolute value. Remember the interpretation of the integral for a positive function: the integral is the area, so the sum of the areas is the sum of integrals. | {
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• so this works even when the bounds of the integral are both positive? – robertmartin8 Sep 15 '14 at 10:34
• Sure, why not? You have to pay attention where to split the integral. Try splitting $$\int_{-2}^3 |x - 1|e^x \ \mathrm{d}x$$ to see if you can do it (you don't need to solve it, that's not the point here) – Ivo Terek Sep 15 '14 at 10:36
$|x|=x$ for x>0
and = -x for x<0
and 0 for x=0
since x is positive and negative in the limits $-2\pi$ to $2\pi$ you have to split it into two parts and then evaluate your integral.
So in this function you are splitting into cases for |x| where x>0 and x<0.
this is done because |x| is defined that way in its domain.since you get two different functions for different intervals in the domain, you have to consider two different limits. imagine a function
f(x)=0 for x<0
and 1 for x>0
and we are evaluating an integral $\int{xf(x)}dx$ with limits -1 to 1.
so because of the definition of f(x) you have to split the integral into two.
here its just two integrals sometimes you have to split into many more.
• I understand that you have to split it up into cases, I just dont get technically how or why this is done – robertmartin8 Sep 15 '14 at 10:33 | {
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# Are the following sets exactly the same: $\mathbb{N}$, $\mathbb{N_{-\{0\}}}$, $\mathbb{N^+}$ and $\mathbb{Z^+}$?
At university, my Maths Analysis lecturer said that this is still debated nowadays. For example, a definition of an rational number is $\cfrac{p}{q}$ where $p\in \mathbb{Z}$ and $q\in \mathbb{N}$ (or should it be $q\in \mathbb{N^+}$ or the other two in the title?).
I know that $\mid \mathbb{N}|= \mid \mathbb{N^+}|$ since I can find a bijection by defining the map: $f: \mathbb{N^+} \mapsto \mathbb{N}$ by $f(x)=x-1$.
But the problem is that this raises the question as to whether $0 \in \mathbb{N}$ or $0 \in \mathbb{Z^+}$. This is similar to asking whether $0$ is positive?
So which set should I use to define the denominator $q$ of a rational number and why? The choices are $\mathbb{N}$, $\mathbb{N_{-\{0\}}}$, $\mathbb{N^+}$ and $\mathbb{Z^+}$.
I acknowledge that this question is subject somewhat to opinion, but I value the opinions of users of this site. So all I am asking is which set of the above would you choose $q$ to belong to for the definition of the rational number?
• In class-related work you should use what your lecturer uses. Elsewhere, you can use what you like and is most convenient for what you want to do. Maybe you work closely with other people that already have a standard. If the distinction is important for what you do, you should mention whether or not $0 \in \mathbb{N}$. Dec 8, 2015 at 23:21
• (1) We don't need a tag for "mappings" because those are in fact functions. (2) This question is about notation, and not at all about functions, so the tag (and the functions tags) is entirely irrelevant. Dec 10, 2015 at 10:38
• @AsafKaragila Okay, understood; thanks for letting me know. Dec 10, 2015 at 10:41
There are two different sets of relevance here; each has several notations in use. | {
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There are two different sets of relevance here; each has several notations in use.
The set $\{0,1,2,3,\ldots\}$ is considered fundamental in areas such as logic, set theory and computer science. It can be unambiguously written $\mathbb N_0$. Set theorists often refer to it as $\omega$, deftly avoiding the notational trouble surrounding the $\mathbb N$s.
The set $\{1,2,3,4,\ldots\}$ is often needed in most of the rest of mathematics, where people find it more natural to start counting at $1$. It can be unambigously written $\mathbb N_+$ or $\mathbb Z_+$ (the location of the plus sign can vary).
Either of these sets is often notated just $\mathbb N$ -- it is up to the reader who encounters this to know (or guess) which convention the author is following, in the cases where the difference matters. It is considered polite for an author to state which convention he follows before he uses the naked $\mathbb N$, but this is not always done in practice.
In English it is unambiguous that $0$ is not "positive". However other languages may follow other conventions; in French the number $0$ counts as both "positif" and "négatif" and one has to speak about "strictement positif" if one needs to exclude $0$. (This is not a mathematical difference; the concepts $>0$ and $\ge 0$ both exist independently of which words we use about them, and the two languages simply chose different concepts to have a short word for). | {
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• Nice answer, thanks very much. So in your opinion, which of the four would you select, I know it's subject to opinion and therefore arbitrary, but could you tell me which one you would use? Dec 10, 2015 at 7:59
• @BLAZE: Personally? I just use $\mathbb N$ for whatever of the sets I'm speaking about (usually, but not always, the one that contains $0$) unless there seems to be a particular risk that the reader will miss my point if he misunderstands me. In that case I write $\mathbb N_0$ or $\mathbb N_+$. (Except forsuch set-theory contexts where it is conventional to use $\omega$). I'm not holding this up as a shining example to follow, though. Dec 10, 2015 at 15:44
It usually depends on the course and the lecturer.
$0 \in \Bbb N$ vs $0 \not \in \Bbb N$ depends on the definition of $\Bbb N$. The first one is usually found in most logic courses, while on say, analysis, this is not as common.
Whenever I see $\Bbb Z^+$ I'll think they refer to $\{1,2,3,...\}$, same for $\Bbb N^+$(this last one is quite uncommon). | {
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Approximating $\log x$ with roots
The following is a surprisingly good (and simple!) approximation for $\log x+1$ in the region $(-1,1)$: $$\log (x+1) \approx \frac{x}{\sqrt{x+1}}$$
Three questions:
• Is there a good reason why this would be the case?
• How does one go about constructing the "next term"?
• Are the any papers on "generalized Pade approximations" that involve radicals?
• I suggest you write down the Taylor series of $\log(x+1)$ and $(x+1)^{-1/2}$ and see the few first terms agree. – LinAlgMan Dec 1 '14 at 10:42
• @LinAlgMan - That is true, but it is not the reason. This approximation works much better than the Taylor series, even to high orders, probably because it accounts for the pole. – nbubis Dec 1 '14 at 10:43
• May I ask how you got this interesting approximation ? – Claude Leibovici Dec 1 '14 at 10:49
• @ClaudeLeibovici - the function came up in a physics calculation, and upon plotting, I noticed it looked oddly familiar. – nbubis Dec 1 '14 at 10:51
• @Lucian - I'm not sure I see the connection. – nbubis Dec 8 '14 at 20:12
Let's rewrite both sides in terms of $y = x + 1$: we get
$$\log y \approx \sqrt{y} - \frac{1}{\sqrt{y}}$$
on, let's say, the interval $\left( \frac{1}{2}, 2 \right)$ (I hesitate to discuss the entire interval $(0, 2)$; it seems to me that the approximation is not all that good near $0$). The RHS should look sort of familiar: let's perform a second substitution $y = e^{2z}$ to get
$$2z \approx e^z - e^{-z} = 2 \sinh z$$
on the interval $\left( - \varepsilon, \varepsilon \right)$ where $\varepsilon = \frac{\log 2}{2} \approx 0.346 \dots$. Of course now we see that the LHS is just the first term in the Taylor series of the RHS, and on a smaller interval than originally. Furthermore, the Taylor coefficients of $2 \sinh z$, unlike the Taylor coefficients of our original functions, decrease quite rapidly. The next term is $\frac{z^3}{3}$, which on this interval is at most
$$\frac{\varepsilon^3}{3} \approx 0.0138 \dots$$ | {
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$$\frac{\varepsilon^3}{3} \approx 0.0138 \dots$$
and this is more or less the size of the error in the approximation between $\log 2$ and $\frac{1}{\sqrt{2}}$ obtained by setting $y = 2$, or equivalently $x = 1$.
With the further substitution $t = \sinh z$, the RHS is just the first term in the Taylor series of the LHS. To get the "next term" we could look at the rest of the Taylor series of $\sinh^{-1} t$. The next term is $- \frac{t^3}{6}$, which gives
$$z \approx \frac{e^z - e^{-z}}{2} - \frac{(e^z - e^{-z})^3}{48}$$
or
$$\log y \approx \left( \sqrt{y} - \frac{1}{\sqrt{y}} \right) - \frac{1}{24} \left( \sqrt{y} - \frac{1}{\sqrt{y}} \right)^3.$$
I don't know if this is useful for anything. The series to all orders just expresses the identity
$$\log y = 2 \sinh^{-1} \frac{\left( \sqrt{y} - \frac{1}{\sqrt{y}} \right)}{2}.$$
• Very nicely done!! As you noticed though, near $y\to 0$, higher orders seem to actually ruin the approximation, which I find curious. – nbubis Dec 1 '14 at 11:26
• @nbubis: well, the radius of convergence of the Taylor series of $\sinh^{-1} t$ is $1$, and as $y \to 0$, $\sqrt{y} - \frac{1}{\sqrt{y}}$ gets arbitrarily large (in absolute value)... – Qiaochu Yuan Dec 1 '14 at 11:59
The simplest Pade approximant we could build seems to be $$\log(1+x)\approx\frac{x}{1+\frac{x}{2}}$$ and we can notice the similarity of denominators close to $x=0$.
However, the approximation given in the post seems to be significantly better for $x<\frac 12$.
• Indeed. It is probably better because it has a pole in the right place. – nbubis Dec 1 '14 at 11:00
I am being rather late, yet still there is some interesting information I can add. | {
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I am being rather late, yet still there is some interesting information I can add.
The Padet approximant is of the form $\log(1+x)\approx \frac{x}{1+x/2}$ as noted by other posters. This partially explains why $\frac{x}{\sqrt{1+x}}$ is a good approximation, what it does not explain is why the square-root approximation is better than a supposedly "great" Pade approximation. @nbubis had an idea that it works better because it has pole in the correct spot, but it seems that it is actually a red herring.
Let's take a look on more general Pade $(1,n)$ approximation, it equals $\log(1+x)\approx\frac{x}{1+x/2-x^2/12+x^4/24+...}$.
Now the reason $\frac{x}{\sqrt{1+x}}$ approximation performs better can be explained by $\sqrt{1+x} \approx 1+x/2 -x^2/8$ and noting that $1+x/2-x^2/8$ is closer to the "true value" of the denominator than the first Pade approximant $1+x/2$.
To see that it is indeed the case consider the approximation
$\log(1+x)=\frac{x}{(1+5x/6)^{3/5}}$.
Now, as you can quickly check, $(1+5x/6)^{3/5}$ has the Taylor expansion $\approx 1+x/2 -x^2/12$ which aggrees with first 3 terms of $(1,n)$ Pade approximant. Now if you plot it it will turn out that it is even better than originally suggested $\frac{x}{\sqrt{1+x}}$ despite having pole in the wrong place.
Regarding method by @QiaochuYuan, you can perform the same "trick" to get better approximations which will be performing better in the neighbourhood of $x=0$ but worse when $x$ is large, for example
$\log (1+x) \approx \frac{x}{(1+5x/6)^{3/5}} + \frac{x^4}{108 (1+5x/6)^{12/5}}$
But in disguise what you are actually making is finding better approximations to some Pade approximant.
Some of the other approximations you can find in the same way are | {
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Some of the other approximations you can find in the same way are
$\log(1+x)\approx \frac{x}{\sqrt{1+x+x^2/12}}$ and $\log(1+x)\approx \frac{x}{(1+3x/2+x^2/2)^{1/3}}$ which are good simply beause they coincide with Pade approximant up to the terms of high order. I guess the first among those two is another reason why $\frac{x}{\sqrt{1+x}}$ worked so well.
Short version: It's actually a coincidence, $\sqrt{1+x}$ happen to have Taylor expansion $\sqrt{1+x}\approx 1+x/2-x^2/8$ which coincides with expansion of $x/\log(1+x)\approx 1+x/2 -x^2/12$ up to 2 terms and the third term is not that different to mess up the approximation.
• I like this! That makes a lot of sense. – nbubis Dec 1 '17 at 6:00
A different approach.
The function $f(x)=x-\log(1+x)\sqrt{1+x}$ is continuous and increasing on $[-1,1]$ (I have not proved it, but a graph of $f'$ is sufficiently convincing.) For any $a\in(0,1)$ $$f(-a)\le f(x)\le f(1)=0.0197419,\quad-a\le x\le1.$$ Take for instance $a=0.8$ we obtain $$-\frac{0.0802375}{\sqrt{1+x}}\le\frac{x}{\sqrt{1+x}}-\log(1+x)\le\frac{0.0197419}{\sqrt{1+x}},\quad -.8\le x\le1.$$
This shows that $\log(1+x)\sqrt{1+x}$ is a good approximation of $x$. $$f(x)=-\frac{x^3}{24}+\dots$$ is an alternate series. This explains the vey good approximation for $x>0$, and the not so good for $x<0$.
• I'm not sure how that helps - this is just an evaluation of the approximation. – nbubis Dec 1 '14 at 11:56
One reason may be that their Taylor series around $x=0$ start the same: $$\log (x+1) \approx x-x^2/2+x^3/3+\cdots$$ $$\frac{x}{\sqrt{x+1}} \approx x-x^2/2+3x^3/8+\cdots$$ So they agree to order 2 for $|x|<1$. They almost agree to order 3 because $1/3 \approx 3/8$ roughly.
However, this is an a posteriori reason. I don't know why this approximation should be good a priori. | {
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• That doesn't necessarily tell you much about how good of an approximation to expect on an interval as large as $(-1, 1)$. – Qiaochu Yuan Dec 1 '14 at 10:46
• This was what I was just typing ! The question is interesting. – Claude Leibovici Dec 1 '14 at 10:46
• For example, when $x = 1$ the LHS is $\log 2 \approx 0.693 \dots$ while the RHS is $\frac{1}{\sqrt{2}} \approx 0.707 \dots$. These agree substantially better that can be accounted for by the first two terms of the Taylor series, which give $0.5$. – Qiaochu Yuan Dec 1 '14 at 10:48
• This is clearly not the reason. The approximation does much better than the Taylor series way past second order.(probably because it has a pole) – nbubis Dec 1 '14 at 10:49
$$\log(x+1)=\lim_{n\to\infty}n(\sqrt[n]{x+1}-1).$$
In the case $n=2$, $$2(\sqrt{x+1}-1)=2\frac{x}{\sqrt{x+1}+1}\approx\frac x{\sqrt{x+1}}$$ for small $x$.
The approximation works better as it has a vertical asymptote at $x=-1$.
• Nice :) Though the approximation ends up being better than the derivation... – nbubis Dec 1 '14 at 11:14
• This still doesn't explain most of the agreement. Again taking $x = 1$ we have $2 (\sqrt{2} - 1) \approx 0.828 \dots$, which is maybe 15% bigger than either the LHS or the RHS, which agree to maybe within 2%. – Qiaochu Yuan Dec 1 '14 at 11:15
• Anyone who's still trying to answer the question should actually plot the functions (I did it in WolframAlpha) to see how close the agreement actually is. I think the pole is a red herring: the agreement is really not very good close to the pole. – Qiaochu Yuan Dec 1 '14 at 11:17
• I don't thank the downvoters. @QiaochuYuan: the question is not a contest to the best approximation. It is about why $\log (x+1) \approx \dfrac{x}{\sqrt{x+1}}$. – Yves Daoust Oct 31 '17 at 7:45 | {
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# exponential functions examples | {
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Each output value is the product of the previous output and the base, 2. Find r, to three decimal places, if the the half life of this radioactive substance is 20 days. Sign up to read all wikis and quizzes in math, science, and engineering topics. Now, let’s take a look at a couple of graphs. If b b is any number such that b > 0 b > 0 and b â 1 b â 1 then an exponential function is a function in the form, f (x) = bx f (x) = b x We will see some examples of exponential functions shortly. Compare graphs with varying b values. from which we have In fact this is so special that for many people this is THE exponential function. n \log_{10}{1.03} \ge& 1 \\ In fact, that is part of the point of this example. For every possible $$b$$ we have $${b^x} > 0$$. Find the sum of all positive integers aaa that satisfy the equation above. 100 + (160 - 100) \frac{1.5^{12} - 1}{1.5 - 1} \approx& 100 + 60 \times 257.493 \\ \approx& 15550. Population: The population of the popular town of Smithville in 2003 was estimated to be 35,000 people with an annual rate of increase (growth) of about 2.4%. You appear to be on a device with a "narrow" screen width (, Derivatives of Exponential and Logarithm Functions, L'Hospital's Rule and Indeterminate Forms, Substitution Rule for Indefinite Integrals, Volumes of Solids of Revolution / Method of Rings, Volumes of Solids of Revolution/Method of Cylinders, Parametric Equations and Polar Coordinates, Gradient Vector, Tangent Planes and Normal Lines, Triple Integrals in Cylindrical Coordinates, Triple Integrals in Spherical Coordinates, Linear Homogeneous Differential Equations, Periodic Functions & Orthogonal Functions, Heat Equation with Non-Zero Temperature Boundaries, Absolute Value Equations and Inequalities, $$f\left( { - 2} \right) = {2^{ - 2}} = \frac{1}{{{2^2}}} = \frac{1}{4}$$, $$g\left( { - 2} \right) = {\left( {\frac{1}{2}} \right)^{ - 2}} = {\left( {\frac{2}{1}} \right)^2} = 4$$, $$f\left( { - 1} \right) = {2^{ - 1}} = | {
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\right)^{ - 2}} = {\left( {\frac{2}{1}} \right)^2} = 4$$, $$f\left( { - 1} \right) = {2^{ - 1}} = \frac{1}{{{2^1}}} = \frac{1}{2}$$, $$g\left( { - 1} \right) = {\left( {\frac{1}{2}} \right)^{ - 1}} = {\left( {\frac{2}{1}} \right)^1} = 2$$, $$g\left( 0 \right) = {\left( {\frac{1}{2}} \right)^0} = 1$$, $$g\left( 1 \right) = {\left( {\frac{1}{2}} \right)^1} = \frac{1}{2}$$, $$g\left( 2 \right) = {\left( {\frac{1}{2}} \right)^2} = \frac{1}{4}$$. A = a^{a}b^{b}c^{c}, \quad B = a^{a}b^{c}c^{b} , \quad C = a^{b}b^{c}c^{a}. 1000Ã(12)100005730â1000Ã0.298=298.1000 \times \left( \frac{1}{2} \right)^{\frac{10000}{5730}} Check out the graph of $${2^x}$$ above for verification of this property. Therefore, we would have approximately 298 g. â¡ _\square â¡â, Given three numbers such that 0 and! Since we are only graphing one way is if we allowed \ ( b\ ) have. 1\Large |x|^ { ( x^2-x-2 ) } < 1 { 12 } 100. Curve upward, as shown in the first quadrant functions exponential functions examples starting this with. ) above for verification of this function in the exponent worked to this point population a... A very specific number into all the \ ( { b^x } > 0\ ) represent growth decay. Functions from these graphs exponential functions examples model exponential functions will want to use far more decimal places in computations! Graph y = 2 x is called the base, 2 aaa that satisfy the equation,. For a complete list of integral functions, there are some function evaluations that will give complex,! Through complex numbers course, built by experts for you would we have now section. formula for exponential. Business and science output value is the approximate integer population after a year we could have written \ ( \bf... The final section of this chapter off discussing the final property for a complete of. Approaches negative infinity, the approximate integer population after a year is 100Ã1.512â100Ã129.75=12975. â¡100 \times 1.5^ { 12 \approx... > 0\ ) ( b = - 4\ ) the | {
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year is 100Ã1.512â100Ã129.75=12975. â¡100 \times 1.5^ { 12 \approx... > 0\ ) ( b = - 4\ ) the function would be would be 1000Ã1.03n.1000! The chapter on exponential functions works in exactly the same properties that exponential..., C a, b is greater than 1 , the graph of this property complex! 2 raised to the x power same manner that all three graphs pass through the y-intercept ( 0,1.. | {
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