# Toronto Math Forum

## MAT334-2018F => MAT334--Tests => Term Test 2 => Topic started by: Victor Ivrii on November 24, 2018, 04:54:06 AM

Title: TT2A Problem 1
Post by: Victor Ivrii on November 24, 2018, 04:54:06 AM
Using Cauchy's integral formula calculate
$$\int_\Gamma \frac{z\,dz}{z^2-4z+5},$$
where $\Gamma$ is a counter-clockwise oriented simple contour, not passing through eiter
of $2\pm i$ in the following cases

(a) The point $2+i$ is inside  $\Gamma$ and $2-i$ is outside  it;

(b) The point $2-i$ is inside  $\Gamma$ and $2+i$ is outside it;

(c) Both points $2\pm i$ are inside  $\Gamma$.
Title: Re: TT2A Problem 1
Post by: ZhenDi Pan on November 24, 2018, 05:30:25 AM
We have

\int_\Gamma \frac{zdz}{z^2-4z+5}

Let

f(z) = \frac{z}{z^2-4z+5} = \frac{z}{(z-(2-i))(z-(2+i))}

Question a:
The point $2-i$ is outside of the contour $\Gamma$ and the point $2+i$ is inside of the contour $\Gamma$. Then let

g(z) =\frac{z}{z-2+i} \\
g(2+i) = \frac{2+i}{2i} \\
\int_\Gamma f(z)dz = \int_\Gamma \frac{g(z)}{(z-(2+i))}dz = 2\pi i g(z_0) = 2\pi i g(2+i) = 2\pi i \cdot \frac{2+i}{2i}= \pi(2+i)

Question b:
The point $2+i$ is outside of the contour $\Gamma$ and the point $2-i$ is inside of the contour $\Gamma$. Then let

g(z) =\frac{z}{z-2-i} \\
g(2-i) = -\frac{2-i}{2i} \\
\int_\Gamma f(z)dz = \int_\Gamma \frac{g(z)}{(z-(2-i))}dz = 2\pi i g(z_0) = 2\pi i g(2-i) = 2\pi i \cdot -\frac{2-i}{2i}= -\pi(2-i)

Question c:
Both points $2+i$ and $2-i$ are inside of the coutour $\Gamma$. Then we have

z_0 = 2+i \\
z_1 = 2-i \\
\left.Res(f;2+i) = \frac{z}{z-2+i} \right\vert_{z=2+i} = \frac{2+i}{2i} \\
\left.Res(f;2-i) = \frac{z}{z-2-i} \right\vert_{z=2-i} = - \frac{2-i}{2i}

So the Residue Theorem gives us

\int_\Gamma f(z)dz = 2\pi i(\frac{2+i}{2i}-\frac{2-i}{2i}) = 2\pi i \cdot 1= 2\pi i
Title: Re: TT2A Problem 1
Post by: Yifei Wang on November 24, 2018, 05:34:25 AM
We can rewrite the fraction as:

$let f(z) =\frac{z}{z^2-4z+5}$

as

$\frac{z}{(z-(2+i))(z-(2-i))}$

a. When $2+i$ is inside
$f(z) = \int \frac{\frac{z}{z-2+i}}{z-2-i}~dz$

By Cauchy's thm

$f(z) = \int \frac{\frac{z}{z-2+i}}{z-2-i}~dz= 2i\pi *f(2+i) = \pi * (2+i)$

b. When $2-i$ is inside

$f(z) = \int \frac{\frac{z}{z-2-i}}{z-2+i}~dz$

By Cauchy's thm

$f(z) = \int \frac{\frac{z}{z-2-i}}{z-2+i}~dz= 2i\pi *f(2-i) = -\pi * (2-i)$

c. When both points are inside

$f(z) = \int \frac{z}{(z-(2+i))(z-(2-i))}~dz = 2i\pi (Res(f, 2+i) + Res(f, 2-i)) = 2\pi i$

Title: Re: TT2A Problem 1
Post by: Yifei Wang on November 24, 2018, 05:36:15 AM

ZhenDi Pan

I think you are missing the $z$ on the numerator.
Title: Re: TT2A Problem 1
Post by: ZhenDi Pan on November 24, 2018, 06:23:00 AM
Yes thank you I corrected it. Still our answers are different, I don't know where went wrong though.
Title: Re: TT2A Problem 1
Post by: Zhuoer Sun on November 24, 2018, 12:50:28 PM
Yifei Wang

I think you did the last part wrong. For part (c), you don't need to multiply by 2ipi again. The answer should be the sum of what you got from part (a) and (b), as you've already included 2ipi in previous parts. The final answer should just be pi∗(2+i)−pi∗(2−i).

Title: Re: TT2A Problem 1
Post by: Yifei Wang on November 25, 2018, 02:38:21 PM
Thank you for the correction!
Title: Re: TT2A Problem 1
Post by: Victor Ivrii on November 29, 2018, 07:57:21 AM
Remark: Since integrand is $\frac{1}{z} +O(\frac{1}{z^2}$ the residue at $\infty$ is $-1$ and answer to (c) is $2\pi i$ (independently from(b),(c)). Yet another solution to (c)


# Finding a closed formula for $1\cdot2\cdot3\cdots k +\dots + n(n+1)(n+2)\cdots(k+n-1)$

Considering the following formulae:

(i) $1+2+3+..+n = n(n+1)/2$

(ii) $1\cdot2+2\cdot3+3\cdot4+...+n(n+1) = n(n+1)(n+2)/3$

(iii) $1\cdot2\cdot3+2\cdot3\cdot4+...+n(n+1)(n+2) = n(n+1)(n+2)(n+3)/4$

Find and prove a 'closed formula' for the sum

$1\cdot2\cdot3\cdot...\cdot k + 2\cdot3\cdot4\cdot...\cdot(k+1) + ... + n(n+1)(n+2)\cdot...\cdot (k+n-1)$

generalizing the formulae above.

I have attempted to 'put' the first 3 formulae together but i am getting know where and wondered where to even start to finding a closed formula.

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Do you know how to prove by induction? –  Clive Newstead Oct 23 '12 at 22:27
Yes, using the basic step n=1 and then induction step n+1 showing it follows. But don't i need to 'find' the formula before proving it using induction? –  Matt Oct 23 '12 at 22:29

The pattern looks pretty clear: you have

\begin{align*} &\sum_{i=1}^ni=\frac12n(n+1)\\ &\sum_{i=1}^ni(i+1)=\frac13n(n+1)(n+2)\\ &\sum_{i=1}^ni(i+1)(i+2)=\frac14n(n+1)(n+2)(n+3)\;, \end{align*}\tag{1}

where the righthand sides are closed formulas for the lefthand sides. Now you want

$$\sum_{i=1}^ni(i+1)(i+2)\dots(i+k-1)\;;$$

what’s the obvious extension of the pattern of $(1)$? Once you write it down, the proof will be by induction on $n$.

Added: The general result, of which the three in $(1)$ are special cases, is $$\sum_{i=1}^ni(i+1)(i+2)\dots(i+k-1)=\frac1{k+1}n(n+1)(n+2)\dots(n+k)\;.\tag{2}$$ For $n=1$ this is $$k!=\frac1{k+1}(k+1)!\;,$$ which is certainly true. Now suppose that $(2)$ holds. Then

\begin{align*}\sum_{i=1}^{n+1}i(i+1)&(i+2)\dots(i+k-1)\\ &\overset{(1)}=(n+1)(n+2)\dots(n+k)+\sum_{i=1}^ni(i+1)(i+2)\dots(i+k-1)\\ &\overset{(2)}=(n+1)(n+2)\dots(n+k)+\frac1{k+1}n(n+1)(n+2)\dots(n+k)\\ &\overset{(3)}=\left(1+\frac{n}{k+1}\right)(n+1)(n+2)\dots(n+k)\\ &=\frac{n+k+1}{k+1}(n+1)(n+2)\dots(n+k)\\ &=\frac1{k+1}(n+1)(n+2)\dots(n+k)(n+k+1)\;, \end{align*}

exactly what we wanted, giving us the induction step. Here $(1)$ is just separating the last term of the summation from the first $n$, $(2)$ is applying the induction hypothesis, $(3)$ is pulling out the common factor of $(n+1)(n+2)\dots(n+k)$, and the rest is just algebra.

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For each extra bracket (n+1) then (n+2) you're adding +1 on the bottom and then an addition bracket. im thinking the closed formula for $\sum_{k=1}^nn(n+1)(n+2)\dots(n+k-1)\;;$ would be $\frac1{k+1}(n+k-1)$ im not too sure –  Matt Oct 23 '12 at 22:40
@Matt: Look at the closed forms in $(1)$ again: your $\frac1{k+1}$ is fine, but you should have $k$ more factors, not just one. –  Brian M. Scott Oct 23 '12 at 22:47
Im thinking something $\frac1{k+1}n(n+k)$ but not sure how to 'repeat' brackets for more terms of $k$. so for $k=1$ it works as it gives $\frac12n(n+1)$ but then for $k=2$ all i get it $\frac13n(n+2)$ rather than $\frac13n(n+1)(n+2)$ –  Matt Oct 23 '12 at 22:52
@Matt: When $k=1$ the RHS has two factors involving $n$; when $k=2$ it has three factors involving $n$; and when $k=3$ it has four factors involving $n$. In each case the smallest is $n$ and the largest is $n+k$. In the general case, then you should expect to have $k+1$ factors involving $n$, running from $n$ up through $n+k$. –  Brian M. Scott Oct 23 '12 at 22:57
So it is simply $\frac1{k+1}(n+k)$ –  Matt Oct 23 '12 at 23:01

If you divide both sides by $k!$ you will get binomial coefficients and you are in fact trying to prove $$\binom kk + \binom{k+1}k + \dots + \binom{k+n-1}k = \binom{k+n}{k+1}.$$ This is precisely the identity from this question.

The same argument for $k=3$ was used here.

Or you can look at your problem the other way round: If you prove this result about finite sums $$\sum_{j=1}^n j(j+1)\dots(j+k-1)= \frac{n(n+1)\dots{n+k-1}}{k+1},$$ you also get a proof of the identity about binomial coefficients.

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From (i), (ii) and (iii) it is reasonable to guess that your sum will be $$n(n+1)\cdot...\cdot(n+k)/(k+1)$$ Try to prove this by induction.

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For a fixed non-negative $k$, let $$f(i)=\frac{1}{k+1}i(i+1)\ldots(i+k).$$ Then $$f(i)-f(i-1)=i(i+1)\ldots(i+k-1).$$ By telescoping,
$$\sum_{i=1}^ni(i+1)(i+2)\dots(i+k-1)=\sum_{i=1}^n\left(f(i)-f(i-1)\right)=f(n)-f(0)=f(n)$$


Nth Term Test For Divergence Calculator

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Then use your descrip- tion to write a formula for the nth term of each sequence. # Optionally ignore the kl divergence term. 5n² = 5,20,45,80,125. The most common way to find the gcd is the Euclidean algorithm. Note: This test only determines the divergence of a series. How Does a Calculator Work? ~ A series that has no last. However, the equation above can be used to calculate the number of genotypes for a locus with any number alleles. Let the sequence $\sequence {a_n}$ be such that the limit superior $\displaystyle \limsup_{n \mathop \to \infty} \size {a_n}^{1/n} = l$. If ($\displaystyle \lim_{n \rightarrow \infty} a_n eq 0$), then the series ($\displaystyle \sum_{n=0}^\infty a_n$) diverges. Helps you plan out daily resin use by letting you know exactly what you need! Having trouble deciding between Artifacts? Use our trusty damage calculator to see which artifact has the edge!. ) (b ax Cos y 8. The limit of the nth term, [n+2]/ [5n+17], is lim n→∞ n+2 5n+17 =lim n→∞ 1+ 2 n 5+ 17 n = 1 5. The n th-term test for divergence is a very important test, as it enables you to identify lots of series as divergent. This calculator allows test solutions to calculus exercises. Nth term test for testing the series whether it is convergent or divergent Msc mathematics Mca entrance exam du And many more. Solving slope, simplify exponential expressions, multiply and divide monomials free worksheets, complex numbers solver, linear algebra for dummies, fractions math. 64 with lots of other numbers after the decimal point. The harmonic series is as follows: H 1 = 1 H 2 = H 1 + 1/2 H 3 = H 2 + 1/3 H 4 = H 3 + 1/4 H n = H n-1 + 1/n. Helps you plan out daily resin use by letting you know exactly what you need! Having trouble deciding between Artifacts? Use our trusty damage calculator to see which artifact has the edge!. 1 Nth Term Test for Divergence. Nth term test for divergence in hindi. Note that this is only a test for divergence. Calculator of Hardy-Weinberg equilibrium. Using the formula above you can quickly calculate any nth term. As part of our spreadcheats, today we will learn how to calculate moving average using excel formulas. Recall: Divergence Test If #lim_{n to infty}a_n ne 0#, then #sum_{n=1}^{infty}a_n# diverges. The Art of Convergence Tests. If that is the case, you may conclude that the series diverges by Divergence (Nth Term) Test. By following the above equation to reach an outcome where you are effectively showing that the terms of the target series will become less than. By using this website, you agree to our Cookie Policy. geometric series, nth term test (Test for Divergence). If desired, the precise definition of limit can be carefully explained; and students may even be made to memorize it, but it should not be emphasized. Please choose one option for each question then click Test Result to obtain your result and level (50 questions). Finding The nth Term Of A Geometric Sequence. Why is my calculator giving me a huge number for sin(3. An activity to calculate the nth term of a linear sequence. The task is to find the Nth Harmonic Number. com Test for Divergence This test, according to Wikipedia, is one of the easiest tests to apply; hence it is the first “test” we check when trying to determine whether a series converges or diverges. The formula for the nth term of the arithmetic sequence (also called the general term) is: A n = a + (n -1)d. We have ve tests for convergence: 1) the Divergence Test, 2) the Alternating Series Test, 3) the Ratio Test, 4) the Integral (comparison) Test, and 5) the Comparison Test. If the series is an alternating series, state whether it converges absolutely or conditionally. A summary of the Nth term test for divergence. This simpliﬁes ﬁnding say the 42nd term. 3n+5/(n*2^n) As. An explanation of how to use and how not to use the nth term test for divergence. Geometric series test. How do you use the Nth term test on the infinite series ∞ ∑ n=1 ln(2n + 1 n + 1) ? By the nth term test (Divergence Test), we can conclude that the posted series diverges. Then use your descrip- tion to write a formula for the nth term of each sequence. One test calculator to answer all your pre- and post-test analysis questions. Free series convergence calculator - test infinite series for convergence step-by-step This website uses cookies to ensure you get the best experience. Answer Save. Even if the terms are approaching zero the series could still diverge, though, hence the Harmonic Series, so this test isn't super-useful on its own. In mathematics, the nth-term test for divergence[1] is a simple test for the divergence of an infinite series When testing if a series converges or diverges, this test is often checked first due to its ease of use. Then its sum is. Determining Convergence or Divergence of an Infinite Series. The sum is. The nth term test for divergence The first, and usually simplest, test for divergence of a series is the nth term test. the true conversion rates (the numbers you would get if you ran the test forever) for Variation A and Variation B fall somewhere within their respective bell curves above. The next test for convergence or divergence of series works especially well for series involving powers. The task is to find the Nth Harmonic Number. Note: The Nth Term Test for Divergence is the special name given to the contrapositive of Theorem 2. com/yt/4758819/?ref=ytd. This tends to zero, showing that even if the nth term tends to zero, the series may still be divergent. This test cannot be used for convergence. Thank you for your questionnaire. Free series convergence calculator - test infinite series for convergence step-by-step This website uses cookies to ensure you get the best experience. ) (b ax Log y e 4. First Term in the Series. Let $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ be a series of real numbers $\R$ or complex numbers $\C$. The series might converge. 2) fails to provide. Example: the sequence {3, 5, 7, 9, } starts at 3 and jumps 2 every time: As a FormulaSaying "starts at 3 and jumps 2 every time" is fine, but it doesnt help us calculate the: 10th term, 100th term, or nth term, where n could be any term number we want. Displaying the steps of calculation is a bit more involved, because the Derivative Calculator can't completely depend on Maxima for this task. The most common way to find the gcd is the Euclidean algorithm. They also crop up frequently in real analysis. ifthe series converges absolutely. We offer free personalized SAT test prep in partnership with the test developer, the College Board. If the calculator did not compute something or you have identified an error, please write it in comments below. com Test for Divergence This test, according to Wikipedia, is one of the easiest tests to apply; hence it is the first “test” we check when trying to determine whether a series converges or diverges. It states that if the limit of a_n is not zero, or. There are plenty of tests to knwo the conv/div of a series. Integral Test and p-Series. Is the nth-term test not applicable in this case? Can you clarify?. The Integral Test. Sending completion. For example, Abel’s Test allows you to define convergence or divergence by the types of functions contained in the series. 2b) Diverges (nth Term Test) 2c) Diverges (Integral Test) 2d) Diverges by the nth Term Test n n lim 1 n 1 → = + 2e) Converges, Geometric 2 r 1 3 = 2f) Diverges (nth Term Test) 2g) Diverges (nth Term Test) 2h) Converges (Integral Test) 3a) Converges to 3 by Telescoping 3b) Diverges by Integral Test 3c) Converges to 25 23 by Geometric Series. App Downloads. The sum of the rst n terms of an A. 1A6: In addition to examining the limit of the sequence of partial sums of the series, methods for determining whether a series of numbers converges or diverges are the nth term test, the comparison test, the limit comparison test, the integral test, the ratio test, and the alternating series test. If ($\displaystyle \lim_{n \rightarrow \infty} a_n eq 0$), then the series ($\displaystyle \sum_{n=0}^\infty a_n$) diverges. If the in nite series X1 k=1 a k converges, then a k!0 as k !1. However, the equation above can be used to calculate the number of genotypes for a locus with any number alleles. Displaying the steps of calculation is a bit more involved, because the Derivative Calculator can't completely depend on Maxima for this task. This leaves us with an expression with nothing in the denominator. Step 2: The infinite geometric series is. Root Test: Suppose that the terms of the sequence in question are non-negative, and that there exists r such. Converges to 4 63 by Geo Series Test 2b. nth-Term Test for Divergence The contra-positive of Theorem 8 provides a useful test for divergence. The theorem then says ∂P (4) P k · n dS = dV. nth term test. so tuis series fails the nth term test. Online Integral Calculator » Solve integrals with Wolfram|Alpha. Limit Comparison Test Calculator. In mathematics, the nth-term test for divergence[1] is a simple test for the divergence of an infinite series When testing if a series converges or diverges, this test is often checked first due to its ease of use. Step-by-step Solutions » Walk through homework problems step-by-step from beginning to end. Proof of the divergence theorem. I hope that this was helpful. (-1)^(n+1)*(n/n^2+4) 8. One of the newer topics covered in the Maths GCSE as well as other qualifications is how to find the nth term of a quadratic sequence, but did you know there is a method of finding the nth term on the Casio Classwiz. The first term of the series is. com/yt/4758819/?ref=ytd. Free online beam calculator that calculates the reactions, deflection and draws bending moment and shear force diagrams for cantilever or simply supported beams. A summary of the Nth term test for divergence. The leading term in the inner expansion for the normal gravitational field gives the Bouguer formula. For example, the sum of the series n={1,1,1,1. Question 1 : Find the nth term of the following sequences. This is the special symbol that means "nth root", it is the "radical" symbol (used for square roots) with a little n to mean nth root. Practice using the nth term test to determine sequence divergence. Khan Academy has been translated into dozens of languages, and 100 million people use our platform worldwide. Example Cont. The Integral Test. The calculator does not go beyond 5 alleles and 15 possible genotypes. After reading through the example question below check out the worksheets and practice questions. So the 5-th term of a sequence starting with 1 and with a difference (step) of 2, will be: 1 + 2 x (5 - 1) = 1 + 2 x 4 = 9. Share yours for free!. The infinite series converges (i. The divergence of the harmonic series is proved by direct comparison with a series whose nth partial sum telescopes to the natural logarithm of n. In this situation the nth term test for divergence (Theorem 14. The integral test for convergence is only valid for series that are 1) Positive : all of the terms in the series are positive, 2) Decreasing : every term is less than the one before it, a_(n-1)> a_n, and 3) Continuous : the series is defined everywhere in its domain. There are divergent series whose nth terms approach 0. T-Test Calculator for 2 Independent Means. Testing convergence/divergence of a series using the nth term test for divergence, the geometric series test, and the telescoping series test. Thus, f_x(x,y) = Cx for some constant C = log y and f_y(x,y) = Clogy for some constant C = x. In this video we discuss the nth Term Test for Divergence, the conditions the test has to meet, how to perform it, and its limitations. nth term test, divergence test, zero test (KristaKingMath). Diverges because. • Determining convergence or divergence of sequences. Section 4-11 : Root Test. In other words, we do not have a definite conclusion. In mathematics, the nth-term test for divergence is a simple test for the divergence of an infinite series: If or if the limit does not exist, then diverges. Your budget deserves straightforward royalty-free pricing that lets you use an image just about anywhere, as often as you want. The following rule is a corollary of the comparison test: if. Step 4: Now, take these values (5n²) from the numbers in the original number sequence and work out the nth term of these numbers that form a linear sequence. The nth term is. It is difficult to explicitly calculate the sum of most alternating series, so. Get an answer for 'sum_(n=2)^oo n/ln(n) Determine the convergence or divergence of the series. Preliminary Question True of False? Given the series > an, if lim an = 0 then, the series must converge. S D ∂z The closed surface S projects into a region R in the xy-plane. Although many documents discuss excess returns caused by medium-term momentum and long-term reversal in the US market, we empirically test this conclusion using the method of Jegadeesh and Titman. In mathematics, the nth-term test for divergence is a simple test for the divergence of an infinite series Unlike stronger convergence tests, the term test cannot prove by itself that a series converges. 9: Use Divergence Test to determine whether r 1 ln. Example: the sequence {3, 5, 7, 9, } starts at 3 and jumps 2 every time: As a FormulaSaying "starts at 3 and jumps 2 every time" is fine, but it doesnt help us calculate the: 10th term, 100th term, or nth term, where n could be any term number we want. App Downloads. Simplifying terms. If the in nite series X1 k=1 a k converges, then a k!0 as k !1. Integral Test Let X1 n=1 a nbe a series with positive terms, and let f(x) be the function that results when n is replaced by xin the formula for a n. To determine if the series is convergent or divergent, apply the nth-Term Test for Divergence. A prerequisite of convergence for an infinite series is that its terms approach 0. You will not be able to see the correct answers to the questions. # Every Nth epoch, save weights: # Loop across test chunks # Calculate number of batches:. Justify your answer. This test helps determine if a series diverges. (1 point) The nth term test for Divergence of Series. The nth term test for divergence The first, and usually simplest, test for divergence of a series is the nth term test. ( , , ) Leave empty, if you don't need the divergence at a specific point. nth term = dn + (a - d). We Want To Check For Divergence Of The Series N > --- And We Have Decided To Try The N=2 Nth Term Test For Divergence. This video provides two examples of how to apply the nth term divergent test to determine if a infinite series is divergent. 6, 11, 16, 21, For this sequence d = 5, a = 6. you could try to find the general formula for the Nth Opt-in alpha test for a. You can operate the calculator directly from your keyboard, as well as. In mathematics, the nth-term test for divergence is a simple test for the divergence of an infinite series: If → ∞ ≠ or if the limit does not exist, then ∑ = ∞ diverges. Also, it can identify if the sequence is arithmetic or geometric. Find the nth term of a geometric sequence Contact Us If you are in need of technical support, have a question about advertising opportunities, or have a general question, please contact us by phone or submit a message through the form below. An activity to calculate the nth term of a linear sequence. Telescoping Series Pdf. Integral konvergence divergence. for Series TEST nth-term Geometric series p-series SERIES ∑ an ∞ ∑ ar (i) Converges with sum S = n −1 n =1 ∞ Useful for the comparison tests if the The comparison series ∑ bn is often a geometric series of a pseries. By following the above equation to reach an outcome where you are effectively showing that the terms of the target series will become less than. In mathematics, the nth-term test for divergence is a simple test for the divergence of an infinite series:. , the nth term is less than 0. and decreasing to show convergence, find a larger series. Sending completion. So in this case we would have X10 k=1 2k +1 = 3+5+7++21, and in this case the sum of the series is equal to 120. Helps you plan out daily resin use by letting you know exactly what you need! Having trouble deciding between Artifacts? Use our trusty damage calculator to see which artifact has the edge!. Diverges by Geo Series Test or nth Term Test 2g. Test Review 25 multiple choice formula sheet provided Calculator active *be able to do explicit rule/nth term for both arithmetic and geometric sequences *use and create a recursive formula *find finite and infinite sums *find a specific term *use sigma notation *divergence and convergence. This utility helps solve equations with respect to given variables. with the rst term a and the common difference d is n called an arithmetic series. How Does a Calculator Work? ~ A series that has no last. S D ∂z The closed surface S projects into a region R in the xy-plane. Only the variables i and j may occur in the sum term. ( , , ) Leave empty, if you don't need the divergence at a specific point. An activity to calculate the nth term of a linear sequence. Guidelines: Calculate cube root of 27 For example, use the square root calculator below to find the square root of 7. This is an excellent test to start with because the limit is often easy to calculate. diverges if (a) is not 0 know nothing if (a) equals 0. Many authors do not name this test or give it a shorter name. A function that calls itself is known as a recursive function. = , where ai = Then, = Therefore, by the Nth Term Test for Divergence, Exercise 6: Discuss the convergence or divergence of the series. Exercise 5: Discuss the convergence or divergence of the series. If the limit of sequence { an } doesn’t equal 0, then the series ∑ an is divergent. MIT License. If , then the series is convergent or; 2. Complete Solution. Thus, the harmonic series is a demonstration that the nth term test is a test for divergence only and cannot be used to show a series converges. fibonaciexp2. The n th-term test for divergence is a very important test, as it enables you to identify lots of series as divergent. The list of online calculators for sequences and series. an bn cancels the dominant terms in. This test helps determine if a series diverges. The divergence of the harmonic series is proved by direct comparison with a series whose nth partial sum telescopes to the natural logarithm of n. 12 3 45 The height of each bounce is three-fourths the height of the preceding bounce. In this section, we face the problem of deciding which method to use to test a series for convergence or divergence. The infinite series often contain an infinite number of terms and its nth term represents the nth term of a sequence. It can be used to determine that a series diverges, but it doesn't tell you when a series converges. If r = 1, the ratio test is inconclusive, and the series may converge or diverge. If that Show bx e y ax, sin 0) (2 2 1 2 y b a ay y 10. n"' term test for Divergence Geometric Series Test p-Series Test Telesco in Series Test Inte ral Test Direct Comparison Test Limit Com rison Test Alternating Series Test Ratio Test Root Test 10. 140625)? 12 answers. For instance, the terms of. The nth-term test for divergence is a very important test, as it enables you to identify lots of series as divergent. 3 Newton’s Newton Method Nature and Nature’s laws lay hid in night:. Test for Convergence v1. Using the fx-991EX and the Simul. Welcome to Clip from. Nth Term Test for Divergence - (3 Helpful Examples!) Nth term, georgia algebra 1 eoct online prep test, calculate square root fraction. Just because a(n) of a series approaches 0 does not mean it converges. See full list on gauravtiwari. In this radical calculator Y is a positive real number and x is a Nth root or radical power. Diverges because. This means our calculations are accurate and up-to-date to the practice materials shared from the test maker. Excluding the initial 1, this series is geometric with constant ratio r = 4/9. To use this calculator, enter the values in the given input boxes. Quick question: Doesn't the nth-term test for divergence state that if the limit of a function DOES NOT EXIST or is any number other than zero then it diverges? 1 and DNE both fit the criteria for divergence based from the nth-term test. Definition of Convergence and Divergence in Series. It states that if the limit of a_n is not zero, or. Then use your descrip- tion to write a formula for the nth term of each sequence. Unlike the nth term test for divergence from the last module, this module gives several tests which, if successfully applied, give a definitive answer of whether a series converges or not. terms tend to zero. We are Cambridge Assessment English. MEMORY METER. Simple Present, Present Progressive, Present Perfect, Simple Past, If-Satz Type I Level: lower intermediate. Is each term of the series positive? If so, a comparison test (page 417) may be used. M) of a series containing n observations is the nth root of the product of the values. The Secant Method, when it is working well, which is most of the time, is fast. 3n+5/(n*2^n) As. educreations. The main purpose of this calculator is to find expression for the n th term of a given sequence. You should start with a rm knowledge of each test and the ability to recall quickly the details of each test. If you're ever in doubt and would like to We made this SAT® score calculator because we saw that everyone else simply replicated the tables when creating what they called a "calculator". This online BMI Calculator is delivered to you at no charge and will allow you to obtain your Body Mass Index (BMI) to see whether you are underweight, of normal weight, overweight, or obese according to the classification employed by the World Health Organization (WHO). Is the first test test tests that be will lookingover. In exercise 22-28, test for convergence or divergence using each test at least once. use the ratio test to determine which of the following series converges and which diverges X∞ n=1 2n +5 3n, ∞ n=1 (2n)! (n!)2, X∞ n=1 n! nn;. Many authors do not name this test or give it a shorter name. = , where ai = Then, = Therefore, by the Nth Term Test for Divergence, Exercise 6: Discuss the convergence or divergence of the series. So I decided to knock this together so people can play around and test their nth's out!. If P b n diverges, and a n b n for all n beyond a certain value, then P a n also diverges. Since , the series is converges. Where: a n is the nth term of the sequence, a is the first term, d is the common difference. Here you can calculate a determinant of a matrix with complex numbers online for free with a very detailed solution. Basically if r = 1, then the ratio test fails and would require a different test to determine the convergence or divergence of the series. Converges to 4 63 by Geo Series Test 2b. The n th-term test for divergence is a very important test, as it enables you to identify lots of series as divergent. If the value of when n goes to inifinity the summation of the series will just continue to add up values, and. Get the free "Convergence Test" widget for your website, blog, Wordpress, Blogger, or iGoogle. It is possible to optimize our trainable weights. So the first term of the nth term is 5n². If the limit of sequence {a n} doesn’t equal 0, then the series ∑ a n is divergent. Geometric series test. I have seen so many tweets of late with people getting their heads frazzled with nth-child and nth-of-type. Since the series diverges. The divergence of the harmonic series is proved by direct comparison with a series whose nth partial sum telescopes to the natural logarithm of n. Consider a series S a n such that a n > 0 and a n > a n+1 We can plot the points (n,a n) on a graph and construct rectangles whose bases are of length 1 and whose heights are of length a n. The last two tests that we looked at for series convergence have required that all the terms in the series be positive. to show that this diverges should I use the $n^{th}$ term test? So far I have substituted infinity for $n$. The key idea is to apply the classical inequality x>=log(1+x) (valid for x>-1) with x=1/k and sum over k, 1<=k<=n-1. The series may or may not. Nth Root (Radical) calculator is a simple tool used to calculate the N th Root of the given real number which has the N th radical value. Test for Convergence v1. fibonaciexp2. 4 Series Tests for Convergence and Divergence. Converges by p-series Test 2f. Determining convergence or divergence of the sequence with the nth term. - Nth Term Test for Divergence - Geometric Series: Examining the ratio to determine Numerically calculate the value of the derivative of a function at a. One of the newer topics covered in the Maths GCSE as well as other qualifications is how to find the nth term of a quadratic sequence, but did you know there is a method of finding the nth term on the Casio Classwiz. Use the Standard Deviation Calculator to calculate your sample's standard deviation and mean. For example, sum_(n=1)^(infty)(-1)^n does not converge by the limit test. ﬁnd a formula for the nth term directly. Sending completion. Abstract: Based on the probability generating function of stuttering Poisson distribution (SPD), this paper considers some equivalent propositions of SPD. this proves the P-series test. n"' term test for Divergence Geometric Series Test p-Series Test Telesco in Series Test Inte ral Test Direct Comparison Test Limit Com rison Test Alternating Series Test Ratio Test Root Test 10. Looking at this function closely we see that f(x) presents an improper behavior at 0 and only. you could try to find the general formula for the Nth Opt-in alpha test for a. If the terms of a rather conditionally convergent series are suitably arranged, the series may be made to converge to any desirable value or even to diverge according to the. Incorrect Test Choice ITC Student chooses an incorrect test, such as an nth term test, or a geometric test. Suppose that there exists r such that. Fortunately, it’s also very easy to use. If the limit of sequence {a n} doesn’t equal 0, then the series ∑ a n is divergent. Test infinite series for convergence step-by-step. One of the newer topics covered in the Maths GCSE as well as other qualifications is how to find the nth term of a quadratic sequence, but did you know there is a method of finding the nth term on the Casio Classwiz. The calculator will approximate the definite integral using the Riemann sum and sample points of your choice: left endpoints, right endpoints, midpoints, and trapezoids. According to divergent test rules, we can't conclude anything about it. In mathematics, the nth-term test for divergence is a simple test for the divergence of an infinite series Unlike stronger convergence tests, the term test cannot prove by itself that a series converges. (1 point) Assume we are trying to determine the. To perform the test, first you need to calculate a measure known as U for each sample. Many authors do not name this test or give it a shorter name. Use the Standard Deviation Calculator to calculate your sample's standard deviation and mean. nth-Term Test for Divergence. The calculation. Step 4: Now, take these values (5n²) from the numbers in the original number sequence and work out the nth term of these numbers that form a linear sequence. Consider, if x 1, x 2 …. A series contain terms whose order matters a lot. The Nth Term Test is one of the most simple tests for convergence. 3 The nth Term Test for Divergence. Khan Academy has been translated into dozens of languages, and 100 million people use our platform worldwide. 60 – 75 min 75 min 75 min 75 min 60 – 75 min 75 min • grid paper • scientific calculator • graphing calculator (optional) Dynamic Activity; Animation • scientific calculator • graphing calculator (optional) Animations. Always check the Nth Term Test for Divergence. If → ∞ ≠ or if the limit does not exist, then ∑ = ∞ diverges. Although many documents discuss excess returns caused by medium-term momentum and long-term reversal in the US market, we empirically test this conclusion using the method of Jegadeesh and Titman. If you have a table of values, see Riemann sum calculator for a table. The limit test is inconclusive when the limit is zero. com online calculator provides basic and advanced mathematical functions useful for school or college. Share yours for free!. The nth term test is only a test for divergence. Determinant is calculated by reducing a matrix to row echelon form and multiplying its main diagonal elements. It's quick, free and gives an instant score. Always check first. (2 votes). or if the limit does not exist, then. Let s n(p)be the nth partial sum of the p-series ∞ k=1 1/k p. This means our calculations are accurate and up-to-date to the practice materials shared from the test maker. Alternating Series Test - Proof. Using the nth-Term Test for Divergence 613 The series in Example STUDY TIP 5(c) will play an important role in this chapter. so Etfe converges. The test that we are going to look into in this section will be a test for alternating series. This video explains how to apply the nth term divergence test to an infinite series. Find more Mathematics widgets in Wolfram|Alpha. Also, it can identify if the sequence is arithmetic or geometric. If a series converges, its nth term approaches 0. com/yt/4758819/?ref=ytd. (n^10 + 10)/n! 7. ParticipantsConversions. Equations and terms. 4 Comparison tests: the comparison test, limit comparison test. Suppose that the in nite series X1 k=1 a k converges. A function that calls itself is known as a recursive function. The Integral Test. Test for Divergence for Series, Two Examples. If , then the series is divergent or; 3. Nth Term Test. Learners analyze geometric series in detail. The ratio test is especially useful, but the integral test is one i dread to use. By the nth term test (Divergence Test), we can conclude that the posted series diverges. an Dr an1 or an an1 Dr: Again, in this case it is relatively easy to ﬁnd a formula for the nth term: an Da0rn. 1 Equation 2 Use 3 Explanation 4 Video Explanation Calculate the limit as stated above. The n and the 2 cancel from the numerator and the denominator. Example Cont. For example, students can be. n=1 n 2 + n. To each series we associate the sequence of nth terms of the series and also the sequence of partial sums of the series. • The nth-term test for divergence • The integral test and its relationship to improper integrals and areas of rectangles • Use of the integral test to introduce the test for p-series • Comparisons of series • Alternating series and the alternating series remainder • The ratio and root tests • Taylor polynomials and approximations. Nth term test for testing the series whether it is convergent or divergent Msc mathematics Mca entrance exam du And many more. First Term in the Series. If the th Term Divergence Test is inconclusive, the next step is to examine the corresponding series. This is used by this calculator. # Every Nth epoch, save weights: # Loop across test chunks # Calculate number of batches:. Function Grapher and Calculator Description:: All Functions. Use the NEB Tm Calculator to estimate an appropriate annealing temperature when using NEB PCR products. terms tend to zero. We use the product rule for f and the quotient rule for V as. Usually we need about 45 percent more iterations than with the Newton Method to get the same accuracy, but each iteration is cheaper. pdf Due December 2: Page 515: 3,4,24, 26, 49,50 Due December 4: Page 528: 7-12;14-18 (use LCT, DCT, nth term lor geometric series). Identify the test used. The calculation. 2b) Diverges (nth Term Test) 2c) Diverges (Integral Test) 2d) Diverges by the nth Term Test n n lim 1 n 1 → = + 2e) Converges, Geometric 2 r 1 3 = 2f) Diverges (nth Term Test) 2g) Diverges (nth Term Test) 2h) Converges (Integral Test) 3a) Converges to 3 by Telescoping 3b) Diverges by Integral Test 3c) Converges to 25 23 by Geometric Series. Nth Term Test For Convergence This test basically tells you that if your terms aren't approaching zero, there's no way the series converges.


What is $\dfrac{i}{4-π} \int_{|z|=4} \dfrac{dz}{z \cos z}$?

How to evaluate the integral $$\dfrac{i}{4-π} \int_{|z|=4} \dfrac{dz}{z \cos z}?$$

Here $$f(z)=\dfrac{1}{z \cos z}$$ has poles at $$0$$ and $$\frac{\pm π}{2}$$ .

Residue at $$z=0$$ is 1 and residues at remaining poles add up to give 0. So the integral using Cauchy integral formula is $$2π(4-π)$$.

I think I am wrong. How to get the integral?

It's not true that residues add up to $$0$$. We have $$\cos z = \sin(\frac{\pi}{2}-z) = \sin(\frac{\pi}{2}+z)$$ so $${\rm Res}_{z=0} \frac{1}{z\cos z} = \lim_{z\rightarrow 0}\frac{1}{\cos z} = 1$$ $${\rm Res}_{z=\frac{\pi}{2}} \frac{1}{z\cos z} = \lim_{z\rightarrow \frac{\pi}{2}}\frac{(z-\frac{\pi}{2})}{z \sin(\frac{\pi}{2}-z)} = -\frac2\pi$$ $${\rm Res}_{z=-\frac{\pi}{2}} \frac{1}{z\cos z} = \lim_{z\rightarrow -\frac{\pi}{2}}\frac{(z+\frac{\pi}{2})}{z \sin(\frac{\pi}{2}+z)} = -\frac2\pi$$ and $$\sum {\rm Res} = 1 - \frac{4}{\pi} = \frac{\pi-4}{\pi}$$ so the final result is $$\frac{i}{4-\pi}\cdot 2\pi i\frac{\pi-4}{\pi} = 2$$

You made a mistake in your residues. The residues of $$1/(z\cos z)$$ at $$z=\pm\frac\pi2$$ are both $$-\frac2\pi$$.

• ohh...yes....thank you Jun 13 '19 at 16:20

The function $$f(z)=\frac{1}{z\cos(z)}$$ has simple poles at $$z=0$$ and $$z= (2n-1)\pi/2$$ for $$n\in \mathbb{Z}$$. The poles that are inside the circle $$|z|=4$$ are at $$z=0$$ and $$z=\pm \pi/2$$.

The residues of $$f$$ at the implicated poles are

\begin{align} \text{Res}\left(\frac{1}{z\cos(z)}, z=0\right)&=1\\\\ \text{Res}\left(\frac{1}{z\cos(z)}, z=\pi/2\right)&=-\frac2\pi\\\\ \text{Res}\left(\frac{1}{z\cos(z)}, z=-\pi/2\right)&=-\frac2\pi \end{align}

Therefore we have

$$\frac{i}{4-\pi}\oint_{|z|=4}f(z)\,dz=\frac{i}{4-\pi}\times 2\pi i\times\left(1-\frac{4}{\pi}\right)=2$$


# Proving a binomial sum identity

Mathematica tells me that

$$\sum _{k=0}^n { n \choose k} \frac{(-1)^k}{2k+1} = \frac{(2n)!!}{(2n+1)!!}.$$

Although I have not been able to come up with a proof.

Proofs, hints, or references are all welcome.

-

You could consider the integral $$\int_{0}^{1} (1-x^2)^n dx .$$

-
Aha this is the integral I was searching for! Using the integral $\int_0^1 (1-t)^n \, dt$ I was able to prove $\sum _{k=0}^n {n \choose k}\frac{(-1)^k}{k+1} = \frac{1}{n+1}$ I thought there should be a corresponding integral to my problem and after seeing your answer the integral I was searching for is obvious now! –  yjj Sep 8 '10 at 11:02

Sums of the form $$\sum_{k=0}^n(-1)^k{n\choose k}f(k)$$ can often be attacked via the Calculus of finite differences.

-

$$S_n=\sum\limits_{k=0}^n \dfrac{(-1)^k \binom{n}{k}}{2k+1}$$

We have:

\begin{align}S_n&=\dfrac{(-1)^n}{2n+1}+\sum\limits_{k=0}^{n-1} \dfrac{(-1)^k \binom{n}{k}}{2k+1}\\ &=\dfrac{(-1)^n}{2n+1}+\sum\limits_{k=0}^{n-1}\left[\dfrac{n}{n-k}.\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}\right]\\ &=\dfrac{(-1)^n}{2n+1}+\dfrac{1}{2n+1}\sum\limits_{k=0}^{n-1}\left[\dfrac{n(2n+1)}{n-k}.\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}\right]\\ &=\dfrac{(-1)^n}{2n+1}+\dfrac{1}{2n+1}\sum\limits_{k=0}^{n-1}\left[\dfrac{2n^2-2nk+2nk+n}{n-k}.\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}\right]\\ &=\dfrac{(-1)^n}{2n+1}+\dfrac{1}{2n+1}\sum\limits_{k=0}^{n-1}\left[\dfrac{2n^2-2nk}{n-k}.\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}\right]\\&+\dfrac{1}{2n+1}\sum\limits_{k=0}^{n-1}\left[\dfrac{2nk+n}{n-k}.\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}\right]\\ &=\dfrac{(-1)^n}{2n+1}+\dfrac{2n}{2n+1}\sum\limits_{k=0}^{n-1}\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}+\dfrac{1}{2n+1}\sum\limits_{k=0}^{n-1} \dfrac{n(-1)^k \binom{n-1}{k}}{n-k}\\ &=\dfrac{2n}{2n+1}\sum\limits_{k=0}^{n-1}\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}+\dfrac{1}{2n+1}\sum\limits_{k=0}^{n-1} \left[(-1)^k \binom{n}{k}\right]+\dfrac{(-1)^n}{2n+1}\\ &=\dfrac{2n}{2n+1}\sum\limits_{k=0}^{n-1}\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}+\dfrac{1}{2n+1}\sum\limits_{k=0}^n \left[(-1)^k \binom{n}{k}\right]\end{align} Therefore, $$S_n=\dfrac{2n}{2n+1}S_{n-1}+0 \Rightarrow S_{n-1}=\dfrac{2n-2}{2n-1}S_{n-2} ... \Rightarrow S_1=\dfrac{2}{3}S_0$$ and $S_0=1$

Hence, $$S_n=\dfrac{2n(2n-2)...2}{(2n+1)(2n-1)...3.1}=\dfrac{(2n)!!}{(2n+1)!!}$$

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As stated, you divide by zero since you sum to $k=n$ and $n-k$ is in the denominator. –  Julian Kuelshammer Nov 29 '12 at 7:36
Wah! Thanks for pointing out errors in my solution. –  hxthanh Dec 1 '12 at 12:12
Fix this problem as follows sum for $k=0$ to $n-1$ and $k=n$ is calculated separately –  hxthanh Dec 1 '12 at 12:34
You can edit your post accordingly. –  Julian Kuelshammer Dec 1 '12 at 12:37

Your identity is a special case of the Chu-Vandermonde identity.

${}_2 F_1(-n,b;c;1)=\frac{(c-b)_n}{(c)_n}$

with $b=\frac12$ and $c=\frac32$. More info on it is in A=B, as mentioned already by John.

-
It's on page 181. –  Ｊ. Ｍ. Sep 8 '10 at 11:05

I can't offer specific help, but I'd recommend thumbing through Concrete Mathematics looking for techniques. It has many sums that look similar, though of course the difficulty is in the details.

There's also the book A=B, but Concrete Mathematics gives an introduction to the content of A=B and in my opinion is easier to read, so I'd start with Concrete Mathematics.

-

There are a powerful algorithms generalizing telescopy (Gosper, Zeilberger et al.) that easily tackle this case of the Chu-Vandermonde identity and much more complicated sums. For example, see this paper which gives as an application a very interesting q-analogy - namely that L. J. Rogers' classical finite version of Euler's pentagonal number theorem is simply the dual of a special case of a q-Chu-Vandermonde.

-


# How many permutations of $\{1,2,3,…,n\}$ there are with no 2 consecutive numbers?

How many permutations of $$\{1,2,3,...,n\}$$ are there with no 2 consecutive numbers?

For example:
$$n=4$$, $$2143$$, $$3214$$, $$1324$$ are the permutations we look for and $$1234$$, $$1243$$, $$2134$$ are what we DON'T look for.

My solution:

we will sub from all the permutations the bads. All permutations= $$n!$$

bads:

Lets define $$A_i$$ to be a group of all permutations containing $$i,i+1$$ in it. ($$1\leq i\leq n-1$$)

My problem is to count to group of :$$|A_i \cap A_j|$$ where $$1 \leq i < j \leq n-1$$.

Because there is a big difference between $$A_i \cap A_{i+1}$$ and $$A_i \cap A_{i+5}$$ (for instance), we can't say that both are the same. So we need to divide $$|A_i \cap A_j|$$ to 2 options.

1. when $$i+1 = j$$ then $$|A_i \cap A_j|=(n-2)!$$
2. when $$i+1 < j$$ then $$|A_i \cap A_j|=(n-2)!$$

So $$|A_i \cap A_j|=(n-2)!+(n-2)!$$

This is a big mistake but I have no clue why.

(the rest of the solution is the same).

Any help would be welcome.

Stav

• It makes no sense to add your two figures of $(n-2)!$ together: the first applies only when $i+1=j$, and the second applies only when $i+1<j$, so they never both to the same pair of $i$ and $j$. – Brian M. Scott Jun 11 '16 at 19:07
• @BrianM.Scott Studying the subject and wasn't sure if the following statement was true.$N(A_iA_j)$ was same for all $i\neq j$(even if $i+1=j$) by calculation through inclusion principle. thus there was no contradiction? – user416486 Apr 23 '18 at 15:49

## 3 Answers

If, as appears to be the case, you’re planning to use an inclusion-exclusion argument, you’ll need $\left|\bigcap_{k\in I}A_k\right|$ for every non-empty $I\subseteq[n-1]$. The trick is to realize that no matter how the integers in $I$ are spaced, the cardinality is going to be the same. When $|I|=2$, the case that you’re discussing in the question, there are two possibilities: the two members of $I$ are consecutive, or they are not. But in both cases it turns out that the cardinality of the intersection is $(n-2)!$: there are different calculations for the two cases, but they lead to the same result. Note that there is no reason at all to add these calculations: no $I$ belongs to both cases.

The set $I$ can be divided into blocks of consecutive integers. For example, $I=\{2,3,5,6,7,9\}$ has $3$ blocks: $\{2,3\},\{5,6,7\}$, and $\{9\}$. Suppose that $I$ has a block $\{k,k+1,\ldots,k+r\}$. Then every permutation in $\bigcap_{k\in I}A_k$ must contain the subsequence $\langle k,k+1,\ldots,k+r,k+r+1\rangle$; call this an extended block. If $I$ has $b$ blocks altogether, each permutation of $[n]$ that belongs to $\bigcap_{k\in I}A_k$ must contain all $b$ extended blocks as subsequences, but it can contain them in any order. Those extended blocks contain altogether $|I|+b$ integers: the blocks themselves contain $|I|$ integers, and each extended block has one extra integer on the righthand end. That leaves $n-|I|-b$ members of $[n]$ that can be permuted arbitrarily, because they aren’t in any extended block of $I$. Thus, the permutations in $\bigcap_{k\in I}A_k$ are really permutations of

$$b+(n-|I|-b)=n-|I|$$

things: $b$ extended blocks, and the $n-|I|-b$ single elements that are not part of any extended block. It follows that

$$\left|\bigcap_{k\in I}A_k\right|=(n-|I|)!$$

whenever $\varnothing\ne I\subseteq[n-1]$.

Now the inclusion-exclusion principle tells you that

\begin{align*} \left|\bigcup_{k\in[n-1]}A_k\right|&=\sum_{\varnothing\ne I\subseteq[n-1]}(-1)^{|I|-1}\left|\bigcap_{k\in I}A_k\right|\\ &=\sum_{\varnothing\ne I\subseteq[n-1]}(-1)^{|I|-1}(n-|I|)!\\ &=\sum_{k=1}^{n-1}\binom{n-1}k(-1)^{k-1}(n-k)!\;, \end{align*}

and I’ll leave the rest for you to finish off.

• First of all thank you very much for your time. Secondly Im having hard time to undestand how I={2,3,5,6,7,9}I={2,3,5,6,7,9} has 3 blocks: {1,2},{5,6,7}{1,2},{5,6,7}, and {9}. how and why did you choose those blcoks specifically? what does a block mean? – Stav Alfi Jun 11 '16 at 19:53
• @Stav: By block I simply mean a maximal set of consecutive integers. In the set $\{2,3,5,6,7,9\}$ the sets $\{2,3\},\{5,6,7\}$, and $\{9\}$ are the longest strings of consecutive integers, so they’re the blocks. The blocks of $\{1,5,6,7,8,12,13,14,20\}$ are $\{1\},\{5,6,7,8\},\{12,13,14\}$, and $\{20\}$: in each block the numbers are consecutive, and the blocks are as long as possible. Note that in the case $|I|=2$, your two cases correspond to one block, when $i+1=j$, and two blocks, when $i+1<j$. – Brian M. Scott Jun 11 '16 at 19:58
• No, Thank you! I could not get a better answer! The second paragraph provided the answer to my question in the best way. Combining all was perfectlly clear. Thank you agian and good night from isreal! :) – Stav Alfi Jun 11 '16 at 20:24
• @Stav: You’re welcome! (And have a good night.) – Brian M. Scott Jun 11 '16 at 20:26
• @MarkoRiedel I think Brian's expression counts the permutations that do have 2 consecutive numbers. If you take $n!$ minus his sum, it reproduces your sequence. – David Zhang Jun 11 '16 at 22:05

Consider the following recurrence: the desired count $Q_n$ is $1$ for $n=1$ and $1$ for $n=2.$ For $n\gt 2$ we obtain an admissible permutation either by placing the value $n$ anywhere at $n-1$ possible positions of an admissible permutation from $Q_{n-1}$ (this is not $n$ because we may not place $n$ next and to the right of $n-1$) or we construct a permutation having exactly one pair of consecutive numbers and place the value $n$ between these two. This can be done by taking a permutation from $Q_{n-2}$ and replacing one of the $n-2$ values by a fused pair containing the value and its successor and incrementing the values that are larger than the first element of the fused pair.

We get the recurrence

$$Q_n = (n-1) Q_{n-1} + (n-2) Q_{n-2}$$

and $Q_1= Q_2= 1.$ This yields the sequence

$$1, 3, 11, 53, 309, 2119, 16687, 148329, 1468457, 16019531, 190899411, \\ 2467007773, 34361893981, 513137616783,\ldots$$

which is OEIS A000255, where a detailed entry may be found.

The Maple code for this was as follows.

with(combinat);

C :=
proc(n)
option remember;
local perm, pos, res;

res := 0;
perm := firstperm(n);

while type(perm, list) do
for pos to n-1 do
if perm[pos] + 1 = perm[pos+1] then
break;
fi;
od;

if pos = n then
res := res + 1;
fi;

perm := nextperm(perm);
od;

res;
end;

Q :=
proc(n)
option remember;

if n = 1 or n = 2 then return 1 end if;

(n - 1)*Q(n - 1) + (n - 2)*Q(n - 2)
end;

Addendum. Here is my perspective on the inclusion-exclusion approach. We take as the underlying partially ordered set the set $P$ of subsets (these are the nodes of the poset) of $\{1,2,\ldots,n-1\}$ where a subset $S\in P$ represents permutations where the elements of $S$ are next to their successors, plus possibly some other elements also next to their successors. The partially ordered set is ordered by set inclusion. To compute the cardinality of the permutations corresponding to $S$ suppose that the elements of $S$ listed in order form $m$ blocks. We first remove these from $[n].$ We must also remove the elements that are consecutive with the rightmost element of each block, so we have now removed $|S|+m$ elements. We then put the augmented and fused blocks back into the permutation and permute them. We have added in $m$ blocks, therefore the net change is $-|S|-m +m = -|S|.$ Hence by inclusion-exclusion we compute the quantity

$$\sum_{S\in P, S\ne\emptyset} (-1)^{|S|} (n-|S|)!.$$

Now this depends only on the number $q$ of elements in $S$ so we get

$$\sum_{q=1}^{n-1} {n-1\choose q} (-1)^q (n-q)!.$$

We must now ask about the weight assigned to a permutation with $p$ elements (call this set $T$) next to their successor. This permutation is included in or rather represented by all sets $S$ that are subsets of the set $T$, which is the poset spanned by the singletons and $T$ being the topmost node. We obtain

$$\sum_{q=1}^p (-1)^q {p\choose q} = -1 + (1-1)^p = -1.$$

The count assigns the weight minus one to the permutation. It follows that when we add $n!$ exactly those permutations remain that do not contain consecutive adjacent values, for a result of

$$n! + \sum_{q=1}^{n-1} {n-1\choose q} (-1)^q (n-q)!.$$

We may simplify this to

$$\sum_{q=0}^{n-1} {n-1\choose q} (-1)^q (n-q)!.$$

In both cases, there are $(n-2)!$ permutations.
In the first case, there is the triple $(i,i+1,i+2)$ and $n-3$ other numbers.
In the second case, there are two pairs, and $n-4$ other numbers.

• You say that becouse in both cases we will permutate n-2 numbers , the answer should be (n-2)!. but my quastion is why you neglect the big deffrence between those 2 cases? – Stav Alfi Jun 11 '16 at 14:25
• They are different; but they both have the same number of permutations. – Empy2 Jun 11 '16 at 14:27
• Then how do we know when to sum 2 options instead of taking only one as an answer like now? – Stav Alfi Jun 11 '16 at 14:28
• @Stav: For each fixed pair of $i$ and $j$ with $1\le i<j\le n-1$ there are $(n-2)!$ permutations in $A_i\cap A_j$; the reasoning explaining this is different when $j=i+1$ from when $j>i+1$, but in both cases the result of the reasoning is $(n-2)!$. – Brian M. Scott Jun 11 '16 at 19:06
• Thank you for your comment. Just to clarify the rest of the prove, I see that |Ai∩Aj| for all cases is (n-2)! but very hard to guess for the |Ai1∩Ai2∩....Aik|=(n-k)!, how do you know that with out proving it? – Stav Alfi Jun 11 '16 at 19:21


# Converting “n is an odd negative integer” Into Formal Logic

I am learning how to think about mathematical truth at the level of formal logic and I am tasked with converting the following basic statement into something formal: "n is a negative integer that is odd"

My first attempt at this was the following: Let $$n \in \mathbb{Z}$$ such that $$2n + 1 < 0$$

From my point of view this statement is saying that "n is an integer and it's values are restricted by the inequality $$2n + 1 < 0$$". But how this actually translated to a mathematician was "$$n$$ is an integer satisfying the inequality $$2n + 1<0$$" and after some thought, I concluded my initial statement was indeed wrong.

So I went back to the drawing board and came up with this:

$$\exists k \in \mathbb{N} \quad n = -2k + 1$$

But I feel that this formulation is still missing something, and I just can't put my finger on it. Should something be said about $$n$$ as well? Another way I thought of writing this was:

$$\exists k \in \mathbb{N} \quad (n = -2k + 1) \Rightarrow n \in \mathbb{Z}$$

$$\exists n \in \mathbb{Z} \quad \exists k \in \mathbb{N} \quad n = -2k + 1$$ Any help thinking through this would be appreciated.

• “n is an odd negative integer” is a predicate; thus, the formula must be of form $P(n)$ with $n$ free. – Mauro ALLEGRANZA Jan 19 at 14:41
• @MauroALLEGRANZA Could you perhaps provide an example demonstrating what you mean? It doesn't need to be related precisely to my question, but just to give a bit more insight. I understand that $P(n)$ is a statement about $n$, but I am missing what you mean by "$n$ free". – GrayLiterature Jan 19 at 14:48
• $\text {OddNegInt}(n) \leftrightarrow [n \in \mathbb Z \land (n < 0) \land \exists k \in \mathbb N (n=2k+1)]$ – Mauro ALLEGRANZA Jan 19 at 14:56
• @MauroALLEGRANZA I see now. "N is an odd negative integer" can be translated as "n is an integer AND n is odd AND n is negative" – GrayLiterature Jan 19 at 15:02
• @MauroALLEGRANZA Although, I think that the last part in your formulation for $k$ would require that $k \in \mathbb{Z}$. Since if $n = 2k + 1$ then n is strictly positive for any $k \in \mathbb{N}$. Would you agree with this observation? – GrayLiterature Jan 19 at 15:15

We can write "$$n$$ is a negative integer" as $$n\in\mathbb{Z}^-$$ And "$$n$$ is odd" as $$\exists k\in\mathbb{Z}.\ n=2k-1$$ Combine both: $$\exists k\in\mathbb{Z}.\ n=2k-1 \wedge n\in\mathbb{Z}^-$$ Or alternatively, $$\exists k\in\mathbb{Z}.\ \mathbb{Z}^-\ni n=2k-1$$

Hope this helps. :)

• No, $\mathbb{Z}$ also contains all positive integers. "$n$ is a negative integer" is writen $n \in \mathbb{Z} \wedge n < 0$. – Olivier Roche Jan 19 at 15:29
• @OlivierRoche: As far as the popular convention says, $\mathbb {Z}^+$ is the set of positive integers, and $\mathbb {Z}^-$ is the set of negative integers. – ultralegend5385 Jan 19 at 23:40
• Oh sorry I didn't see the superscript! – Olivier Roche Jan 20 at 0:50

This answer follows the comment provided by Mauro ALLEGRANZA in the comments:

The statement "$$n$$ is an odd negative integer" can be translated to "$$n$$ is an integer AND $$n$$ is odd AND $$n$$ is negative". Converting this to formal logic we can write:

• "$$n$$ is an integer" == $$n \in \mathbb{Z}$$
• "$$n$$ is negative" == $$n < 0$$
• "$$n$$ is odd" == $$\exists k \in \mathbb{Z} \quad (n = 2k +1)$$ (edit : if one takes $$k \in \mathbb{N}$$, then $$n > 0$$)

and combining these statements about $$n$$:

• "n is an odd negative integer" == ($$n \in \mathbb{Z}) \quad \land \quad (n<0) \quad \land \quad \exists k \in \mathbb{Z} \thinspace (n = 2k +1)$$

edit : Since $$\exists k \in \mathbb{Z} \thinspace (n = 2k +1)$$ already implies that $$n \in \mathbb{Z}$$, one can omit it and state :

$$(n<0) \quad \land \quad \exists k \in \mathbb{Z} \thinspace (n = 2k +1)$$

• Thank you for the edit. – GrayLiterature Jan 19 at 16:01


## The Element’s-Eye View

There are three things to keep in mind when it comes to set-theory proofs (which most of our proofs hereafter will be):

1. Sets are defined by their membership. Because of this, we nearly always take an element’s-eye view of any proof involving sets. Notice that in the proofs we presented above, the first thing we did to get going was say something like “Let $x\in$. . .”.
2. Generally, the way to go to prove two sets are equal is to show (separately) that each is a subset of the other.
3. Keep in mind what your goal is. That is, pay attention to the consequent. Your goal is to translate from the language of the consequent into the language of the antecedent.

These three principles are not only good advice for writing a proof, but they are in fact good advice for {\em figuring out} a proof as well. The following examples will illustrate principles 1 and 3.

Theorem. If $A$ and $B$ are sets with $2^A\subseteq 2^B$, then $A\subseteq B$.

Proof. Let $A$ and $B$ be set with $2^A\subseteq 2^B$. Since we have to show $A\subseteq B$, let $x\in A$. Then $\{x\}\subseteq A$, so $\{x\}\in 2^A$. Since $2^A\subseteq 2^B$, we have $\{x\}\in 2^B$. But this means $\{x\}\subseteq B$. Since $x\in\{x\}$, we see that $x\in B$.

So we’ve shown every $x\in A$ also has $x\in B$, i.e. $A\subseteq B$. $\Box$

Theorem. If $A$ and $B$ are sets with $A\subseteq B$, then $2^A\subseteq 2^B$.

Proof. Let $A$ and $B$ be sets with $A\subseteq B$. Since we have to show $2^A\subseteq 2^B$, let $x\in 2^A$. Then $x\subseteq A$. Since $A\subseteq B$, and set inclusion is transitive, we have $x\subseteq B$. But this is exactly the statement that $x\in 2^B$.

So we’ve show that every $x\in 2^A$ also has $x\in 2^B$, i.e. $2^A\subseteq 2^B$. $\Box$

Notice that in both cases, we were concerned with the {\em consequent} of the conditional we’d be asked to show (principle III), and that we proceeded by following elements (principle I). In the first proof, we translated from the language of elements of $A$ and $B$ to the language of subsets of $A$ and $B$ to the language of elements of $2^A$ and $2^B$, applied our assumption, and translated back. In the second proof, we translated from the language of elements of $2^A$ and $2^B$ to the language of subsets of $A$ and $B$, applied our assumptions, and translated back.

Activity Consider the following couplet:

Heffalumps and woozles are very confuzle.
A heffalump or woozle’s very sly.

Explain why both lines are talking about the same thing as follows. First, give a mathematical name to what they describe. Then explain why each line takes a different perspective on that mathematical object, and why those perspectives are not contradictory, the difference between “and” and “or” notwithstanding.

## Meta Data

Title: The Element’s-Eye View
Date Posted: October 10, 2018
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Category: sets
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Largest odd factors summing to a square

I just found this awesome puzzle from the Tournament of the Towns (though I'm sure it's appeared other places too). The connection between odd factors and square is surprising, and the proof has a lovely 'aha' moment. Enjoy!

For any natural number $x$, prove that $$\sum_{y=x+1}^{2x}(\text{largest odd factor of y})=x^2.$$

• I'm never around when the good math puzzles are posted :( – Quintec Feb 14 '18 at 1:00
• dude same :( aiyaaa – NL628 Feb 14 '18 at 4:42

Let $H(x) = (\text{largest odd factor of x})$. We can make two observations about this function:

1. $H(x) = H(\frac{x}{2})$, if x is even
2. $H(x) = x$, if x is odd

Let $$F(x) = \sum_{y=x+1}^{2x}H(x).$$

$F(1) = 1 = 1^2$. By induction, if $F(x) = x^2$, then

\begin{align*} F(x+1) &= F(x) + H(2x+2) + H(2x+1) - H(x+1) \\&= F(x) + H(x+1) + (2x+1) - H(x+1) \\&= F(x) + 2x+1 \\&= x^2 + 2x + 1 \\&= (x+1)^2 \end{align*}

• Whoa! Nice proof! What I had in mind was more like Gareth's proof, but this is really neat and fast. – Rand al'Thor Feb 14 '18 at 12:11

ffao posted his answer while I was writing this up. I like my solution better, though in some sense it's equivalent to ffao's, so I'm posting mine too. I will of course be entirely unoffended if Rand gives the mighty green checkmark to ffao, who after all got there first.

It is well known that

the sum of all odd numbers below $2x$ is equal to $x^2$. And there are $x$ of these, from $2\cdot1-1$ to $2x-1$.

So presumably the sum we have here,

which has the right number of terms and consists entirely of odd numbers, will simply be a permutation of this.

Well,

take any odd number $t$ between $1$ and $2x$ inclusive. Keep doubling it until the next doubling would go beyond $2x$. You will end up with a number from $x+1$ to $2x$ inclusive, and clearly each such number appears just once. And the largest odd divisor of this number will be exactly $t$. We're done.

Or, to put it a bit more formally (possibly harder to follow quickly but easier to convince yourself it's definitely correct):

let $A=\{\,y\,:\,x+1\leq y\leq2x\,\}$ and $B=\{\,z\,:1\leq z\leq2x\,\&\,z \textrm{ odd}\,\}$;
define $f:A\rightarrow B$ and $g:B\rightarrow A$ by
$f(y)=\textrm{largest odd divisor of$y$}$ and
$g(z)=\textrm{largest$2^kz$that's$\leq2x$}$;
then $f,g$ are inverses and therefore our sum, which is $\sum_{y\in A}f(y)$,
equals $\sum_{z\in B}z$, which famously equals $x^2$.

• Yes, in some sense this is actually equivalent to what I ended up doing. The good side to my solution is that there is no actual thinking required to reach it, but it ends up being less clean. – ffao Feb 14 '18 at 0:10
• You can also word it this way: Let: $x+1 = O_{x+1}*2^{E_{x+1}}$ ... $2x = O_{2x}*2^{E_{2x}}$ None of the numbers in the range can be a multiple of another, but if the $O$s of two numbers are identical, one of the numbers will be the multiple of other, so all $O$s must be different, which are the odd numbers up to $2x-1$. The sum will be $x^2$. – Nautilus Feb 14 '18 at 9:30
• "clearly each such number appears just once" - I think this could do with just a little more explanation, as it's really the crux of this proof. – Rand al'Thor Feb 14 '18 at 12:13
• This is the proof I had in mind when I posted the question (the 'aha' moment being what's in your first two spoilertags). But ffao's proof is so unexpectedly short and slick! I'm not sure where to put the checkmark now, but knowing me, it'll probably be a while until I award it anyway ;-) – Rand al'Thor Feb 14 '18 at 12:14
• One of the reasons for the more-formal spoilered paragraph was to be more explicit about that "clearly". (I claim it really is obvious that those functions f and g are inverses, and that's all you need.) – Gareth McCaughan Feb 14 '18 at 13:40


Math Help - dependent probability question?

1. dependent probability question?

Hello, I had a question about how to calculate dependent probability. Calculating independent probability is easy, simply multiply the probability of each event: P1 * P2 * P3.

Therefore, if 3 different resources are working on 3 independent and unrelated tasks, and each resource has an 80% probability of completing his task, then the total probability of all resources completing all tasks, successfully, would be .80 * .80 * .80 = 51.2%

However, I need to figure out the probability formula, for a more advanced scenario. Consider a scenario, where 1 resource is working on 3 tasks in sequence, and the probability of the resource completing each individual task is 80%.

In this scenario, the first task would be independent, but the 2nd task would be dependent on completion of the 1st task, and the 3rd task would be dependent on completion of the 1st and 2nd task. So can you tell me the formula I would use to calculate total probability in this scenario?

2. Re: dependent probability question?

Let's look at failure rather than success. Then the overall probability of success is 1 - probability of failure.

You can fail at step 1, step 2, or step 3.

Pr[fail at step 1] = 1 - .8 = .2

Pr[fail at step 2] = Pr[fail at step 2 | success at step 1]

Now assuming failure at step 2 is independent of success at step 1 we get

Pr[fail at step 2] = Pr[fail at step 2] Pr[success at step 1] = .2 * .8 = .16

Pr[fail at step 3] = Pr[fail at step 3 | success at 1 and 2] = .2 * .8 * .8 = .128

The total probabily of failure is then the sum of these (.2 + .16 + .128) = .488

So Pr[Success] = 1 - .488 = .512

This is identical to the non sequenced case. Why? Because in order to succeed all 3 steps have to be succeeded at in either case.
Where this methodology becomes very useful is when there are multiple paths to success as well as failure.

3. Re: dependent probability question?

My probability abilities are not advanced but i can get a surprising number of problems correct by using probability trees and other basic probability techniques.
When I learn more advanced probability theories they are more likely to make sense because I can 'see' that the formula is logical.

Learn to use trees effectively and the answer to these problems becomes much easier to discover. (You don't always need to draw the whole tree)
That's what I think anyway.

4. Re: dependent probability question?

Originally Posted by Melody2
My probability abilities are not advanced but i can get a surprising number of problems correct by using probability trees and other basic probability techniques.
When I learn more advanced probability theories they are more likely to make sense because I can 'see' that the formula is logical.

Learn to use trees effectively and the answer to these problems becomes much easier to discover. (You don't always need to draw the whole tree)
That's what I think anyway.
In industry probability trees are used EXTENSIVELY.

5. Re: dependent probability question?

romsec - thanks for your feedback. However, I think the total probability should be lower, if 1 resource works on 3 tasks in sequence. Here's why: If the resource fails at task 1, then he will never get to task 2 or task 3. If the resource fails at task 2, then he will never get to task 3.

On the other hand, if 3 different resources work on these 3 different tasks independently, then this approach will completely sidestep the dependencies/risks that I described with the single resource above, so it seems like the realistic probability of success should be greater.

It seems like probability theory should have a formula that factors in the dependencies I described above. Can you provide an updated probability formula based on the additional context I provided above?

6. Re: dependent probability question?

I tell you what. You look at my derivation and tell me specifically what bits you think are incorrect. I took into account all the dependencies you put in by requiring serial operation.

7. Re: dependent probability question?

@romsek - I'm only a beginner at statistics. Maybe there's some type of statistics concept that we're not accounting for here?

I was under the impression that the total probability for independent events was calculated differently than the total probability for dependent events, and the total probability for dependent events would be lower. Maybe it will help if I simplify my example?:

Scenario 1: 3 independent events, each event has a probability of 80%: .80 * .80 * .80 = 51.2%

Scenario 2: 3 events with sequential dependency, each distinct event has a probability of 80%: ???

Maybe part of the problem here, is that when I say each event has a probability of 80%, that implies that the dependency relationship has no impact, compared to scenario 1. I was assuming that the total probability formula for dependent events would have a built-in function to lower the total probability.

Does my simplified example change anything? Or do you still assert that the total probability for these scenarios should be identical?

8. Re: dependent probability question?

Originally Posted by random512
@romsek - I'm only a beginner at statistics. Maybe there's some type of statistics concept that we're not accounting for here?

I was under the impression that the total probability for independent events was calculated differently than the total probability for dependent events, and the total probability for dependent events would be lower. Maybe it will help if I simplify my example?:

Scenario 1: 3 independent events, each event has a probability of 80%: .80 * .80 * .80 = 51.2%

Scenario 2: 3 events with sequential dependency, each distinct event has a probability of 80%: ???

Maybe part of the problem here, is that when I say each event has a probability of 80%, that implies that the dependency relationship has no impact, compared to scenario 1. I was assuming that the total probability formula for dependent events would have a built-in function to lower the total probability.

Does my simplified example change anything? Or do you still assert that the total probability for these scenarios should be identical?
In both cases there is a single outcome that denotes success of the entire venture. All 3 tasks must be completed successfully. There is only a single way this can be achieved and that is that all 3 tasks are done successfully.

The probability that your 3 independent folks each succeeding is .8. The probability that all 3 independently succeed is (.8)^3 = .512

Task 1 succeeds with pr. .8. Only then does it proceed to Task 2. The probability that we get to Task 2 is thus .8. We don't care what happens if we fail at Task 1, if we do we fail.

We have arrived at Task 2 with pr .8. Now we have .8 pr of completing Task 2. There's no other dependency of our success on Task 1, it got here, that's all we need to know. So our pr of success of Task 2 is again .8 but also times the pr. that it got here at all, which we know to be .8. So the overall pr of completing Task 2 is (.8)^2 = .64

Now we make it to Task 3 with pr .64. Once again there is no dependency on how well things went at task 1 or 2, the thing got here and that's all we know and we have pr .8 of success at Task 3. So the overall pr of completing Task 3 is .8 times the pr it got here at all .64, or (.8)^3 = .512.

There are all sorts of dependencies you could put between your tasks that would show the sort of conditional probability effects you are thinking of. But you haven't done that here.

9. Re: dependent probability question?

@Romsek - Thanks for your knowledge and patience. I think I figured out what I was missing: total probability of overrun.

3 independent events can execute concurrently. Therefore, if each event has an overrun probability of 50% then the total probability of overrun can be kept to 50%. However, if the 3 events must execute in sequence, then the probability of overrun is:

1 - (.50 * .50 * .50) = 87.5%

So that's how the concept of total probability can be used to prove the cost/impact of dependent events, as opposed to independent events.

10. Re: dependent probability question?

If you want to set up an example demonstrating what you have in mind consider this.

Let's say we're making toy dolls since it's the holidays. We have to cut the parts, assemble the parts, paint the doll.
We have 3 elves and they have different abilities regarding making these dolls. For elf A his pr success at those tasks is say {0.85, .6, .4}
elf B could be {0.6, 0.9, 0.6}, and elf C could be {0.5, 0.5, 0.8}

You could have each elf build the entire doll at their station but it's pretty clear that there's a better strategy and that's the pair the elf's task with their best ability.

That's where your sequential operation comes in since you have to build the dolls in a sequence.

If you really want to get fancy you can have the probability of success at a station depend on the quality that the previous station produced. It's harder to assemble poorly cut parts. It's harder to paint a poorly assembled doll, and incorporate all that in to further show how important your strategy of matching worker and task is.

All this will certainly show the effect of sequential vs parallel operation.

11. Re: dependent probability question?

Originally Posted by random512
@Romsek - Thanks for your knowledge and patience. I think I figured out what I was missing: total probability of overrun.

3 independent events can execute concurrently. Therefore, if each event has an overrun probability of 50% then the total probability of overrun can be kept to 50%. However, if the 3 events must execute in sequence, then the probability of overrun is:

1 - (.50 * .50 * .50) = 87.5%

So that's how the concept of total probability can be used to prove the cost/impact of dependent events, as opposed to independent events.
ok I'm glad you figured it out.

12. Re: dependent probability question?

One more related question. What's the proper way to describe the probability multiplication factor? For example:

* the probability of a single risk event occurring is 50%
* the probability of at least 1/3 risk events occurring is:
1 - ( .50 * .50 * .50) = 87.5%

Is it valid to say that additional risk factors create an "exponential" probability that at least 1 risk event will occur? "Exponential" is a good word, because it's descriptive, and powerful. However, I want to make sure that I use the word appropriately. So can I describe this multiplication effect as "exponential" or should I choose a different word?


# Find Smallest contraction coefficient

I have been given the following function $$f:[-1,1]\to \mathbb{R}$$: $$f(x)=\ln(x+2)-x$$ And I have been asked whether it is a contraction or not, and if it is, I have to find the smallest contraction coefficient, such that $$0.

Attempt

Since $$0<|f'(x)|<1$$ for $$(-1,1)$$ we must have that the function is a contraction, and i would intuitively say that $$\left|-\frac{2}{3}\right|=\frac{2}{3}$$ is the smallest contraction coefficient.

Doubts

I have no concrete theorem or example to support my claim, and therefore I am skeptical.

Since $$f$$ is differentiable $$\forall x \in [-1,1], |f'(x)|= |\cfrac{1}{x+2} - 1| = |\cfrac{-x-1}{x+2}| = \cfrac{1+x}{2+x}$$ and $$\forall x \in [-1,1], f''(x) = \cfrac{1}{(x+2)^2} \ge 0$$ so $$f'$$ is growing. And we have $$| f'(1)| = \cfrac{2}{3}$$ So $$\forall x \in [-1,1], |f'(x)| \le \cfrac{2}{3}$$ Therefore $$\cfrac{2}{3}$$ is indeed the smallest contraction coefficient.
Hint: Because of the mean value inequality, you want to maximize $$|f'(x)|$$ for $$x \in [-1,1]$$.
I think that Rasmus' realized that $$L=\frac 23$$ is the maximum value for $$f'$$ and therefore will work as a contraction constant. This was probably not the question. The question, I think, was about the possibility of getting a contraction constant $$\tilde L < L$$. The answer is no, it is not possible. If you consider a smaller constant you immediately get an interval of the form $$[\xi, 1]$$ where the contraction inequality does not hold.


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 Eigen  3.3.9
Solving linear least squares systems

This page describes how to solve linear least squares systems using Eigen. An overdetermined system of equations, say Ax = b, has no solutions. In this case, it makes sense to search for the vector x which is closest to being a solution, in the sense that the difference Ax - b is as small as possible. This x is called the least square solution (if the Euclidean norm is used).

The three methods discussed on this page are the SVD decomposition, the QR decomposition and normal equations. Of these, the SVD decomposition is generally the most accurate but the slowest, normal equations is the fastest but least accurate, and the QR decomposition is in between.

# Using the SVD decomposition

The solve() method in the BDCSVD class can be directly used to solve linear squares systems. It is not enough to compute only the singular values (the default for this class); you also need the singular vectors but the thin SVD decomposition suffices for computing least squares solutions:

Example:Output:
#include <iostream>
#include <Eigen/Dense>
using namespace std;
using namespace Eigen;
int main()
{
cout << "Here is the matrix A:\n" << A << endl;
cout << "Here is the right hand side b:\n" << b << endl;
cout << "The least-squares solution is:\n"
<< A.bdcSvd(ComputeThinU | ComputeThinV).solve(b) << endl;
}
Here is the matrix A:
-1 -0.0827
-0.737  0.0655
0.511  -0.562
Here is the right hand side b:
-0.906
0.358
0.359
The least-squares solution is:
0.464
0.043

This is example from the page Linear algebra and decompositions .

# Using the QR decomposition

The solve() method in QR decomposition classes also computes the least squares solution. There are three QR decomposition classes: HouseholderQR (no pivoting, so fast but unstable), ColPivHouseholderQR (column pivoting, thus a bit slower but more accurate) and FullPivHouseholderQR (full pivoting, so slowest and most stable). Here is an example with column pivoting:

Example:Output:
MatrixXf A = MatrixXf::Random(3, 2);
VectorXf b = VectorXf::Random(3);
cout << "The solution using the QR decomposition is:\n"
<< A.colPivHouseholderQr().solve(b) << endl;
The solution using the QR decomposition is:
0.464
0.043

# Using normal equations

Finding the least squares solution of Ax = b is equivalent to solving the normal equation ATAx = ATb. This leads to the following code

Example:Output:
MatrixXf A = MatrixXf::Random(3, 2);
VectorXf b = VectorXf::Random(3);
cout << "The solution using normal equations is:\n"
<< (A.transpose() * A).ldlt().solve(A.transpose() * b) << endl;
The solution using normal equations is:
0.464
0.043

If the matrix A is ill-conditioned, then this is not a good method, because the condition number of ATA is the square of the condition number of A. This means that you lose twice as many digits using normal equation than if you use the other methods.

Eigen
Namespace containing all symbols from the Eigen library.
Definition: Core:309
Eigen::ComputeThinU
@ ComputeThinU
Definition: Constants.h:385
Eigen::DenseBase::Random
static const RandomReturnType Random()
Definition: Random.h:113
Eigen::MatrixBase::bdcSvd
BDCSVD< PlainObject > bdcSvd(unsigned int computationOptions=0) const
Definition: BDCSVD.h:1269
Eigen::ComputeThinV
@ ComputeThinV
Definition: Constants.h:389
Eigen::Matrix
The matrix class, also used for vectors and row-vectors.
Definition: Matrix.h:180