# How often is the product of the distinct prime factors of a number greater than that of the the next number?

The radical $$\mathrm{rad}\; n$$ of a positive integer $$n$$ is the product of $$n$$'s distinct prime factors. For example $$\mathrm{rad}\; 12=2\cdot 3=6$$.

Denote the sum of $$n$$'s distinct prime factors by $$\mathrm{spf}\; n$$.

Claim: Given any $$k \ge 1$$, for nearly half of positive integers $$n$$, $$\mathrm{spf}\; n > \mathrm{spf}\; (n+k)$$, but for only one-third of positive integers $$n$$, $$\mathrm{rad}\; n > \mathrm{rad}\; (n+k)$$.

Detailed explanation:

For any $$k \ge 1$$, there will be $$n$$ such that $$\mathrm{rad}(n) > \mathrm{rad}(n+k)$$ e.g. $$n=15, k = 1$$. We define

$$a_{n,k} = \begin{cases} 1 & \frac{\mathrm{rad}(n)}{\mathrm{rad}(n+k)} > 1 \\ 0 &\text{ otherwise} \end{cases}$$

Question: What is limiting value

$$\lim_{x \to \infty}\dfrac{1}{x}\sum_{n \le x}a_{n,k}$$

Instead of product when I used sum, the limiting value approached $$0.5$$ which means that it is equally likely the sum of distinct prime factors of a natural number is greater or less than that of the next natural number which makes intuitive sense.

Similarly I was expecting that product of the distinct prime factors of a number is equally likely to be greater than or less than that of the the next number but this was not the case. I was surprised to see that for $$x = 1.5 \times 10^8$$

$$\dfrac{1}{x}\sum_{n \le x}a_{n,k} \approx 0.3386$$ and was decreasing with $$x$$ hence the eventual limit is expected be be slightly less than this.

Note: The initial version of this question was with $$k=1$$ but the result doesn't change with $$k$$, hence I have updated the question.

• $p_n$ for primes $p(n)$ for partitions to be precise :) – Nilotpal Kanti Sinha Dec 6 '19 at 4:30
• Well the Euler product is written $\prod_p \frac1{1-p^{-s}}$. $p$ is reserved for primes in this context, use $\Pi$ or anything. – reuns Dec 6 '19 at 4:35

Each of these products will be $$\frac{n}{p}$$ and $$\frac{n+1}{q}$$ for some $$p,q\in\mathbb{Z}^+$$, where $$p$$ and $$q$$ are the "leftover" primes in the factorizations of $$n$$ and $$n+1$$ when one copy of each distinct prime has been removed.

If $$p\neq q$$, $$\frac{n}{p}>\frac{n+1}{q}\iff p. Which of $$p$$ and $$q$$ is larger will be effectively "random", in that the behavior will jump around a lot and one will be larger than the other with a limiting ratio of $$0.5$$. (A formal proof might prove difficult, but you can certainly gather some numerical evidence to support this - I think it makes as much intuitive sense as the sum case you mentioned.)

When $$p=q$$, the comparison will always go in favor of $$n+1$$. But since $$n$$ and $$n+1$$ are relatively prime, this only happens when $$p=q=1$$ - i.e., when $$n$$ and $$n+1$$ are each squarefree. So we want to find out what fraction of consecutive integer pairs are squarefree, and hence what limiting ratio to exclude from the even split mentioned above.

Fortunately, we can express this as an infinite product. For a number to be squarefree is just for it to not be divisible by $$4, 9, 25, \ldots, p^2,\ldots$$. The limiting fraction of $$(n,n+1)$$ pairs which do not contain a multiple of 4 is $$2/4$$; likewise, for neither to be a multiple of 9 is $$7/9$$, and so on.

However, congruence modulo the first $$N$$ primes is independent over large ranges by the Chinese remainder theorem. So our probability that a "random" pair are both squarefree is

$$\prod_p\frac{p^2-2}{p^2}$$

which looks to be around $$0.322634...$$, or a touch under one third. This means that your limiting value will be

$$\frac12\left(1-\prod_p\frac{p^2-2}{p^2}\right) \approx 0.33868295...$$

If $$n+1$$ is squarefree then $$p(n+1)=n+1\implies a_n=0$$ thus $$\lim\sup_{n\to \infty} \dfrac{1}{x}\sum_{n \le x}a_n \le 1-\frac{6}{\pi^2}\approx 0.393$$

• You can quickly verify with some numerical calculations that 0.39 is much too high. Try n from 1 to 10,000. – RavenclawPrefect Dec 6 '19 at 4:20
• Sure this is just a rigorous upper bound, your approach is a conjectural estimate. It seems hard to make it rigorous, you should show us a plot of how well does it fit. – reuns Dec 6 '19 at 4:23
• For $x = 1.65 \times 10^8$, the value is $0.3386851$ so I guess there is scope to tighten the upper bound. – Nilotpal Kanti Sinha Dec 6 '19 at 4:28
• If your value is correct the convergence should be in $O(x^{-1/2+\epsilon})$ (that's what predicts the random model for the primes) – reuns Dec 6 '19 at 4:29
• Ah, I missed the ≤. Apologies. A graph of the ratios from n=1000 to 100,000 is at imgur.com/3LOWorl, with the blue line the constant estimated to around 9 digits. – RavenclawPrefect Dec 6 '19 at 4:41


# Distance between k-bit numbers in random sequence of n-bit numbers.

In a random sequence of n-bit numbers, what is the average distance apart of each k-bit number and what is the average distance apart of each odd k-bit number.

Numbers are all positive integers.

Definition: An n-bit number is a number consisting of n bits - set or clear.

Definition: A k-bit number is a number of just k set bits and therefore n-k clear bits.

E.g. in an n-bit number with n=8 there are 8 k-bit numbers where k = 1 and 28 k-bit numbers where k = 2.

Let me clarify with an example.

If I generate a sequence of all 256 8-bit numbers at random I will generate k-bit numbers as 1 0-bit number, 8 1-bit numbers, 28 2-bit numbers etc. Note that the list 1, 8, 28, 56, 70, 56, 28, 8, 1 is a row in Pascals triangle.

Taking only the odd values we get 1, 7, 21, 35, 35, 21, 7, 1 - the next lower row of the triangle.

I would like to get an estimate of the mean distance between these numbers.

In one run of my example I get the positions of each odd 2-bit number as [45, 90, 112, 121, 168, 229, 242] giving gaps of [45, 22, 9, 47, 61, 13] and an average gap of 32.83.

I need to predict this average gap for any n and k. This is beyond my schoolboy maths so I hope someone here can help.

In this example a(8,2) = 32.83. What, for example would a(96,13) be, i.e. the average gap between numbers with 13 bits set in a random sequence of 96-bit numbers. And what would o(96,13) be, i.e. the average gap between odd numbers with 13 bits set in a random sequence of 96-bit numbers.

Please assume good randomness and as many trials as needed to achieve stable averages.

Happy with a value derived from a row of Pascals triangle.

Approximate results from trial runs:

o(10,2)=84.75
o(10,3)=26.28
o(10,4)=11.86
o(10,5)=8.0

• I don't understand: how is $90$ an odd 2-bit number? How is it odd? How is it a 2-bit number, under either your n-bit definition or your k-bit definition? Aug 11 '13 at 1:00
• @Henry - The first odd 2-bit number in my trial run occurred at position 45 (it could have been 3 or 5 or 9 or 17 or 33 or 65 or 129), the second odd 2-bit number occurred at position 90 in the sequence. Does that help? Aug 11 '13 at 1:04
• It still seems strange, but I starting to understand what you are saying. Aug 11 '13 at 1:11
• Perhaps if you ignore the odd requirement it will begin to make sense. I am sure I will be able to extend any ideas you have to only work with odd numbers - it looks like all I need to do is step up one row in Pascals Triangle. Aug 11 '13 at 1:16

If you have $p$ people arranged in a row at random and $s$ of them are special, then the expected average gap between special people is $\frac{p+1}{s+1}$. Imagine having an additional special person, and arranging all $p+1$ people round a circular table with $s+1$ gaps between the special people: the gaps are identically distributed so each have the same expectation, whether or not you ignore the gaps next to the additional person.

So all you need to do is find $s$ and $p$ for your particular question.

There are $2^n$ numbers with $n$ bits. ${n \choose k}$ of them have $k$ bits set, and ${n-1 \choose k-1}$ have $k$ bits set including the last one: these are the numbers in Pascal's triangle you have found.

So the answer to your not-necessarily-odd question is $\dfrac{2^n+1}{{n \choose k}+1}$, while the answer to your odd question is $\dfrac{2^n+1}{{n-1 \choose k-1}+1}$.

If $n=8$ and $k=2$ then this last expression is $\dfrac{257}{8}=32.125$, so your random example of $32.83$ was not too far away. For $n=10$ and $k=2,3,4,5$ it gives about $102.5, 27.7, 12.1, 8.1$, and if I were you I would check the first of these.

• That all looks spot on. Could you expand a little on ${n \choose k}$ please? My schoolboy maths is a little rusty I'm afraid. I'm happy with the discrepancy between my 84.75 against your 102.5, after all this was just one sample run. Aug 11 '13 at 17:35
• ${n \choose k}$ is the number of ways of choosing $k$ different items from $n$ possibilities. Its value appears in Pascal's triangle in the $n$th row, $k$th item along (both counts starting at $0$). Using factorials ${n \choose k}=\frac{n!}{k! (n-k)!}$. For example ${5 \choose 2} = 10$ as there are $10$ was of choosing two items from five. Aug 11 '13 at 18:00
• Works a treat - perfect! Aug 12 '13 at 19:22


# Find the product of $f(x)$ and $g(x)$ given one of them

I'm given the final answer which is $$(g \cdot f)(x) = \frac{1}{x^2+4}\;.$$

Also, i'm given $f(x) = x^2+1$.

I've solved this using the composition, however the second part of the question asks me to find the $g(x)$ which would make this multiplication true. How would I do this? Do I divide the final answer by $f(x)$?

• is that $$(f\cdot g)(x)=\frac1{x^2+4}$$ or $$(f\cdot g)(x)=\frac1{x^2}+4\;?$$ Either way, the answer to your question is yes. – Brian M. Scott Apr 19 '16 at 21:34
• the first one, and how would i go about doing that? I know once I divide i take the reciprocal of g(x) and multiply using the FOIL method, however my answer makes no sense – Saad Siddiqui Apr 19 '16 at 21:35
• thank you for the edit, the question is now formatted correctly. My math is however not giving me the correct answer, could someone guide me through it? – Saad Siddiqui Apr 19 '16 at 21:35
• Your best bet is to learn to use basic MathJax; there’s a tutorial here. Until then, be sure to use enough parentheses to make your expressions completely unambiguous. – Brian M. Scott Apr 19 '16 at 21:37
• For the division, just do it: $$\frac{\frac1{x^2+4}}{x^2+1}=\frac1{x^2+4}\cdot\frac1{x^2+1}=\ldots$$ – Brian M. Scott Apr 19 '16 at 21:38

Yes, becasue by definition $(g\cdot f)(x)=g(x)\cdot f(x)$. Hence $g(x)=\frac1{(x^4+1)(x^2+1)}$.

If we had $(g\circ f)(x)=\frac1{x^4+1}$ instead, one possible $g$ would be $g(x)=\frac1{(x-1)^2+4}$.

• when taking the reciprocal and multiplying, I get a quartic function which makes no sense – Saad Siddiqui Apr 19 '16 at 21:39
• @SaadSiddiqui $\frac1{(x^4+4)(x^2+1)}\cdot(x^2+1)=\frac1{x^4+4}$ – Hagen von Eitzen Apr 19 '16 at 21:40
• Why does a quartic function make no sense?... or really a ratio of 1/ a quartic function. – Doug M Apr 19 '16 at 21:40
• well when multiplying out the denominators, I get an answer of x^4 + 5x^2 + 5. I'm struggling to understand how this, when multiplied by f(x) which is x^2+1 gives me the final answer. – Saad Siddiqui Apr 19 '16 at 21:44

So you are told that $(g \times f)(x) = \dfrac{1}{x^{2} + 4}$, and also told that $f(x) = x^{2} + 1$.

$(g \times f)(x)$ is just the name we give for the product of the two functions, i.e., $(g \times f)(x)$ really means $g(x)f(x)$.

So we know what this product is. It is $g(x)f(x) = \dfrac{1}{x^{2} + 4}$. We also know that $f(x) = x^{2} + 1$. So that means:

$$g(x)(x^{2} + 1) = \dfrac{1}{x^{2} + 4}$$

and to solve for $g(x)$, just divide both sides by $x^{2} + 1$ to get:

$$g(x) = \dfrac{\left (\frac{1}{x^{2} + 4} \right )}{x^{2} + 1}$$

Now, how do we simplify this? Well, $x^{2} + 1$ is the same as $\dfrac{x^{2} + 1}{1}$, so the fraction is really $$\dfrac{\left (\frac{1}{x^{2} + 4} \right )}{\left (\frac{x^{2} + 1}{1}\right)}$$ and when we divide two fractions, we invert the bottom one and multiply, so we get:

$$\dfrac{1}{x^{2} + 4} \cdot \dfrac{1}{x^{2} + 1}$$

And this is just $\dfrac{1}{(x^{2} + 4)(x^{2} + 1)}$, which is your final answer (unless you want to multiply the denominator out using the FOIL method).

• I got to this last part, but it's the multiplying out which has stumped me – Saad Siddiqui Apr 19 '16 at 21:41
• @SaadSiddiqui Ok, so $(x^2 + 4)(x^{2} + 1) = x^{4} + x^{2} + 4x^{2} + 4 = x^{4} + 5x^{2} + 4$, so your final answer should be $\dfrac{1}{x^{4} + 5x^{2} + 4}$. – layman Apr 19 '16 at 21:42
• I got the same answer, but how does this (when multiplied with the original f(x) ) give me my final answer? – Saad Siddiqui Apr 19 '16 at 21:44
• @SaadSiddiqui Well, first you need to understand that $x^{4} + 5x^{2} + 4 = (x^{2} + 4)(x^{2} + 1)$, which we know since we multiplied the right hand side out to get the left hand side. So $\dfrac{1}{x^{4} + 5x^{2} + 4} = \dfrac{1}{(x^{2} + 4)(x^{2} + 1)}$. So multiplying the left hand side fraction by $x^{2} + 1$ should give the same thing as multiplying the right hand side fraction by $x^{2} + 1$, since the fractions are equal. But what happens when we multiply the right hand side fraction by $x^{2} + 1$? We get $(x^{2} + 1) \cdot \dfrac{1}{(x^{2} + 4)(x^{2} + 1)}$, which is... – layman Apr 19 '16 at 21:47
• Yes completely, by not multiplying out the denominator, you can simply cancel out in the next step. Thank you so much! – Saad Siddiqui Apr 19 '16 at 21:54


# Theoretical question about the rank and existence of an inverse of a block matrix,

Let A and B be two $n \times n$ square matrices with complex coefficients, and consider the $2n \times 2n$ matrix $M$ given by

$$M = \begin{bmatrix} A & A \\ A & B \\ \end{bmatrix}$$

1. Determine the rank of $M$ in terms of $A$ and $B$.

2. What is the condition for $M$ to have an inverse $M^{-1}$? Compute $M^{-1}$ when it exists.

Any ideas on how to get started on this problem are welcome.

Thanks,

-
If $A$ is rank deficient, then $M$ is also rank deficient. However in the case that $A$ is full rank, there's not a lot we can say except that the rank of $M$ is at least the rank of $A$ (this is so whatever the rank of $A$ is). – hardmath Jan 4 at 23:18
You can do Gaussian elimination to compute the inverse, except instead of working with individual entries, instead you work with entire blocks. The only difference is that you need to keep track of the order of the blocks, as they won't commute. – Nick Alger Jan 4 at 23:18
Hi @hardmath, thanks so much, yes, I agree with you. I've had this answer for awhile and am thinking, "there has to be many more cases to consider". It looks like you also think that there is not much more to say...hmmm... – User001 Jan 4 at 23:21
Hi @NickAlger, would it also be correct, if I augmented this $2nx2n$ matrix with a $2nx2n$ identity matrix I, and then row-reduce until I get a $2nx2n$ identity matrix on the left side? And, are you saying to do something different, like this: get $nxn$ identity blocks, so, 4 in total, on the left side? So, start with 4 $nxn$ identity blocks on the right-side of the augmented matrix? Thanks so much, – User001 Jan 4 at 23:25
@User001 If you want to go the augmented route, then you can augment with the 2n-by-2n identity. In the meantime it looks like some people have answered explaining this in more detail. If you are unsure of which is right, you can always multiply out the matrix times it's inverse and see whether you get the identity back again. – Nick Alger Jan 4 at 23:49

## 4 Answers

\begin{align} [ M \mid I ] &= \left[ \begin{array}{cc|cc} A & A & I & 0 \\ A & B & 0 & I \end{array} \right] \\ &\to \left[ \begin{array}{cc|cc} A & A & I & 0 \\ 0 & B-A & -I & I \end{array} \right] \\ & \to \left[ \begin{array}{cc|cc} I & I & A^{-1} & 0 \\ 0 & I & -(B-A)^{-1} & (B-A)^{-1} \end{array} \right] \\ & \to \left[ \begin{array}{cc|cc} I & 0 & A^{-1} +(B-A)^{-1} & -(B-A)^{-1} \\ 0 & I & -(B-A)^{-1} & (B-A)^{-1} \end{array} \right] = [ I \mid M^{-1} ] \end{align}

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@mvw....wow...this is too cool...! :-) Thanks so much! – User001 Jan 4 at 23:56
The derivation above uses that the block matrices are like numbers themselves to a certain degree, matrices form a ring, so that some algorithms from linear algebra can be applied to them. – mvw Jan 5 at 0:08
Hi @mvw, in your second row, after just one row operation, we already can determine the rank of M, which is just rank(A) + rank (B-A), since the the matrix is block upper-triangular (and also row operations are rank-preserving). Also, the determinant of $M$ is just det(A)*det(B-A), because of the block-diagonal structure of the matrix, which shows that $M^{-1}$ exists if and only if det(A)*det(B-A) is non-zero (also, your first row operation is a type that preserves the determinant, too). – User001 Jan 5 at 0:33
And finally, computation of the inverse is what you have shown, with block row-operations on an augmented matrix [M | I]. Do I have it all correct? Thanks so much @mvw, – User001 Jan 5 at 0:33
I just wanted to get the inverse. But the intermediate forms seem to be already useful for your other issues. – mvw Jan 5 at 0:48

Note that $$M= \underbrace{\begin{bmatrix}I&0\\I&I\end{bmatrix}}_{=P} \underbrace{\begin{bmatrix}A&0\\0&B-A\end{bmatrix}}_{=M^\prime} \underbrace{\begin{bmatrix}I&I\\0&I\end{bmatrix}}_{=Q}$$ But $P$ and $Q$ are invertible so $\DeclareMathOperator{rank}{rank}\rank M=\rank M^\prime$. Since $M^\prime$ is block-diagonal we have $$\rank M=\rank A+\rank(B-A)$$

To see why $\rank M=\rank M^\prime$, note the two facts:

Fact 1. Let $A$ be an $m\times n$ matrix. If $B$ is an $n\times k$ matrix of rank $n$, then $\rank(AB)=\rank(A)$.

In our case take $A=M$ and $B=Q^{-1}$. This implies that $\rank(MQ^{-1})=\rank(M)$.

Fact 2. Let $A$ be an $m\times n$ matrix. If $C$ is an $l\times m$ matrix of rank $m$, then $\rank(CA)=\rank(A)$.

In our case take $A=M^\prime$ and $C=P$. This implies that $\rank(PM^\prime)=\rank(M^\prime)$.

Since $MQ^{-1}=PM^\prime$ we see that $$\rank(M)=\rank(MQ^{-1})=\rank(PM^\prime)=\rank(M^\prime)$$ For more about rank, see wikipedia. Also, try to prove these facts yourself!

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Hi @BrianFitzpatrick, really cool factorization :-). Can I ask you a naive follow-up question? Since $P$ and $Q$ are both invertible, why does determining the rank of $M$ just reduce to checking the rank of $M'$? The only rule / theorem that I know about rank is that rank (AB) = rank (BA), and then some rank inequalities proved very early in the standard textbooks...thanks, – User001 Jan 4 at 23:39
...that's why I feel that when I go to solve a theoretical question regarding rank, inverse or trace, I feel that I have so little tools at my disposal to work with...unlike advanced calculus and complex analysis questions, where I usually immediately see at least two or three ways to approach the problem...@BrianFitzpatrick, thanks, – User001 Jan 4 at 23:41
@User001 See the properties of rank here: en.wikipedia.org/wiki/Rank_(linear_algebra)#Properties – Brian Fitzpatrick Jan 4 at 23:42
Hi @BrianFitzpatrick, with the matrix $M$, and following all of the comments and answers I've gotten so far, I block-row-reduced, but stopped after just one operation, as I think it may be enough: I knocked out the lower left block matrix A. The entries left are A,A on the first row and B in the lower-right corner. With this matrix, which is block-upper-triangular, we can conclude that the rank of $M$ is the rank of A + the rank of (B-A), by standard (non-block) row-reduction techniques. Is my thinking correct? – User001 Jan 5 at 0:13
Also, from this reduced matrix, the determinant of $M$ is just det(A)*det(B-A) (I believe it works because of the zero block matrix below or above the diagonal), as the row operation of adding a scalar multiple of a column / row to another column / row does not change the determinant. Also, my reason for reading off the rank of A and (B-A) to conclude the rank of $M$ is the same reason: row operations are rank-perserving. What do you think? Thanks, @BrianFitzpatrick, – User001 Jan 5 at 0:13

In this case we have rank($M$) = rank($A$) + rank($B-A$). So we can say a lot about the rank of $M$, just not in terms of rank($A$) and rank($B$).

-
Hi @hardmath, why rank($B-A$)? Thanks, – User001 Jan 4 at 23:33
Block row elimination: subtract top row blocks from bottom row blocks (or the equivalent block column elimination). – hardmath Jan 4 at 23:36

First note that $$\begin{bmatrix} A & A \\ A & B \\ \end{bmatrix} = \begin{bmatrix} A & 0 \\ 0 & I \\ \end{bmatrix} \begin{bmatrix} I & 0 \\ A & I \\ \end{bmatrix} \begin{bmatrix} I & 0 \\ 0 & (B-A) \\ \end{bmatrix} \begin{bmatrix} I & I \\ 0 & I \\ \end{bmatrix}$$ (You can derive this by doing the steps in Gaussian elimination and then writing them as multiplications by "primitive" operation matrices.)

So you can immediately see that $M$ is invertible if and only if both $A$ and $B-A$ are invertible.

Now in that product, the second and fourth matrices are of full rank. Thus the product of the three rightmost matrices will have the same rank deficiency as that of $B-A$. And then the rank deficiency of $M$ is always somewhere between the sum of the deficiencies of $A$ and $B-A$ as an upper bound, and the minimum of those two deficiencies as a lower bound.

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Wed, 12 / 2020 6:16 am |

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Help. absolute extreme points f ( x) = ln ( x − 5) $absolute\:extreme\:points\:f\left (x\right)=\frac {1} {x^2}$. Mathepower finds the function. What we do here is the opposite: Your got some roots, inflection points, turning points etc. If you are familiar with graphing algebraic equations, then you are familiar with the concepts of the horizontal X-Axis and the Vertical Y-Axis. Learn how to use the Stat plot feature of the TI-84+ Calculator to find the equation of those points. absolute extreme points f ( x) = 1 x2. The calculator will find the domain, range, x-intercepts, y-intercepts, derivative, integral, asymptotes, intervals of increase and decrease, critical points, extrema (minimum and maximum, local, absolute, and global) points, intervals of concavity, inflection points, limit, Taylor polynomial, and graph of the single variable function. The method used to calculate function point is knows as FPA (Function Point Analysis). My question is - Can I find the function using points? As a result we should get a formula y=F(x), named the empirical formula (regression equation, function approximation), which allows us to … It accepts inputs of two known points, or one known point and the slope. How about an ellipse or hyperbola? In practice, this means substituting the points for y and x in the equation y = ab x. Instructions: Use this step-by-step Exponential Function Calculator, to find the function that describe the exponential function that passes through two given points in the plane XY. dCode is free and its tools are a valuable help in games, maths, geocaching, puzzles and problems to solve every day!A suggestion ? The vertical graph occurs where the rational function for value x, for which the denominator should be 0 and numerator should not be equal to zero. The method used to calculate function point is knows as FPA (Function Point Analysis). Are there any convenient websites out there with this functionality? Method 2: use a interpolation function, more complicated, this method requires the use of mathematical algorithms that can find polynomials passing through any points. A Function Calculator is a free online tool that displays the graph of the given function. What about a circle that touches the two points? It was easy example, but my graphic is not a liner function. A stationary point is therefore either a local maximum, a local minimum or an inflection point.. an idea ? Area of a triangle with three points. Primarily, you have to find … and are looking for a function having those. Let's take a simple case: two points. Example: a function has for points (couples $(x,y)$) the coordinates: $(1,2) (2,4), (3,6), (4,8)$, the ordinates increase by 2 while the abscissas increase by 1, the solution is trivial: $f (x) = 2x$. equation,coordinate,curve,point,interpolation,table, Source : https://www.dcode.fr/function-equation-finder. Thank you! Yes. Function Point Calculator: Main Description Details Uses: Calculator. These online calculators find the equation of a line from 2 points. Also I would define it in single line as "A Method of quantifying the size and complexity of a software system in terms of the functions that the system delivers to the user". NB: for a given set of points there is an infinity of solutions because there are infinite functions passing through certain points. You need to provide the points $$(t_1, y_1)$$ and $$(t_2, y_2)$$, and this calculator will estimate the appropriate exponential function and will provide its graph. Of course? There can be various methods to calculate function points; you can define your custom too based on your specific requirements. First, we have to calculate the slope m by inserting the x- and y- coordinates of the points into the formula . What about a circle that touches the two points? y=F(x), those values should be as close as possible to the table values at the same points. You need to provide the points $$(t_1, y_1)$$ and $$(t_2, y_2)$$, and this calculator will estimate the appropriate exponential function and will provide its graph. Area of a triangle with three points. Functions: What They Are and How to Deal with Them, Normal Probability Calculator for Sampling Distributions, Exponential Function Calculator from Two Points, exponential function calculator given points. You will be asked to provide simple estimates of the software you plan to develop. Intersection of two lines. absolute extreme points y = x x2 − 6x + 8. find power function from two points calculator, The terminal coordinates program may be used to find the coordinates on the Earth at some distance, given an azimuth and the starting coordinates. (Try this with a string on a globe.) By using this website, you agree to our Cookie Policy. Casio is a great company, providing durable products. This free slope calculator solves for multiple parameters involving slope and the equation of a line. Method 1: detect remarkable solutions, like remarkable identities, it is sometimes easy to find the equation by analyzing the values (by comparing two successive values or by identifying certain precise values). Two point form calculator This online calculator can find and plot the equation of a straight line passing through the two points. Cartesian to Polar coordinates. The blue line is my function. What about a cubic? New coordinates by rotation of axes. Wolfram|Alpha is a great tool for finding the domain and range of a function. Polar to Cartesian coordinates Of course you can. Thus function points can be calculated as: Polar to Cartesian coordinates Visualize the exponential function that passes through two points, which may be dragged within the x-y plane. Given the 3 points you entered of (14, 4), (13, 16), and (10, 18), calculate the quadratic equation formed by those 3 pointsCalculate Letters a,b,c,d from Point 1 (14, 4): b represents our x-coordinate of 14 a is our x-coordinate squared → 14 2 = 196 c is always equal to 1 Intersection of two lines. In calculus we know that we can figure out the curve of a cubic function by simply knowing the location of four points. Linear equation given two points. To derive the equation of a function from a table of values (or a curve), there are several mathematical methods.. If the period is more than 2π then B is a fraction; use the formula period = 2π/B to find … Press [STAT] again. Distance between the asymptote and graph becomes zero as the graph gets close to the line. 1 - Enter the x and y coordinates of three points A, B and C and press "enter". I … a feedback ? Linear equation given two points. Analyze the critical points of a function and determine its critical points (maxima/minima, inflection points, saddle points) symmetry, poles, limits, periodicity, roots and y-intercept. Two equations are displayed: an exact one (top one) where the coefficients are in fractional forms an the second with approximated coefficients whose number of … Technically, in order to find the parameters you need to solve the following system of equations: Solving this system for $$A_0$$ and $$k$$ will lead to a unique solution, provided that $$t_1 = \not t_2$$. Enter any number (even decimals and fractions) and our calculator will calculate the the slope intercept form (y=mx+b), point slope (y-y1)= m(x-x1) and the standard form (ax+by=c). Have a tried and tested method given by IFPUG by their experiences and case study the. Watch manual - but finding These computations if fantastic passing through the two points additional point the. 'Re ok with this functionality if I know only points and tested method given by by. This, but you can opt-out if you wish: points\: f\left ( x\right ) =\sqrt { }... Amplitude which is the horizontal X-Axis and the slope ) =\sqrt { x+3 } $the following points!, interpolation, Newtonian interpolation and Neville interpolation, visit the parabola grapher ( choose ... Description Details Uses: calculator mean find formula/function f ( x ), there are an infinite of! 1 x2 known points, or one known point and the slope the equation of exponential function, a... And press  Enter '' and use the arrow buttons on your specific requirements 09:07! Male/30 level/An office worker/Very/ Purpose of use Initially, for a watch manual - but finding These if... The Vertical Y-Axis a globe. the Logarithmic function calculator, to find the Vertical.! Points ( with integer x- and y-values ) an infinite number of them point, interpolation table. Function, use a graphing calculator to find … exponential functions, constant functions and polynomials are supported! We 'll assume you 're ok with this, but my graphic calculator, an... Of solutions because there are infinite functions passing through the two points to calculate the difference of the to. The type of function is determined by visually comparing the find function from points calculator values at same... This website, you agree to our Cookie Policy - an infinite number of functions can! A sphere is an infinity of solutions because there are infinite functions passing through the two points calculator.! You have to find … exponential functions, constant functions and polynomials are also.! 'Re ok with this, but you can opt-out if you wish if the for. 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# What is the definition of first/last element in a poset?

I have read the terms first element/last elements in the context of a basic course in set theory.

When I learned about posets I didn't encounter those terms. I tried looking up the definitions but I didn't find them.

Can someone please write down the definitions for first/last element in a poset ?

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The first element of a poset $\langle P,\le\rangle$ is simply the unique minimum element of $P$, if there is one: $p_0$ is the first element of $P$ if $p_0\le p$ for all $p\in P$. Similarly, the last element of $P$ is the unique maximum element of $P$, if there is one: $p_1$ is the last element of $P$ if $p\le p_1$ for all $p\in P$.

A poset need not have a first or last element.

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Thanks Brian! I will accept when the system lets me (3 minutes) –  Belgi Jan 5 '13 at 22:09
@Belgi: As always, you’re welcome! –  Brian M. Scott Jan 5 '13 at 22:09

If by "first" element you mean "an element preceding all others", then there need not be one (similarly for "last" element).

Consider the sets $\{1\}$, $\{2\}$, $\{3\}$, $\{1,2\}$, and $\{1,3\}$ ordered by inclusion. There is no first or last element, but there are three minimal elements (an element with no predecessor) and two maximal elements (an element with no successor).

In the case where there is an element preceding all others, it is generally called a minimum element. Similarly, the element that is preceded by all others is called a maximum element.

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So $a$ is first if $a\leq b$ for all other elements $b$ ? –  Belgi Jan 5 '13 at 22:05
@Belgi That is a reasonable definition for "first", but I think "minimum" is a more common term for it. –  Austin Mohr Jan 5 '13 at 22:06
Thanks for the answer Austin, I upvoted it –  Belgi Jan 5 '13 at 22:08

A poset need not have a first or last element.

• An element $a_1$ is first (the unique minimum) in a poset $P$ if $a_1 \le a\;\;\forall a \in P$.
• An element $a_n$ is last (the unique maximum) in a poset P if $a \le a_n \;\;\forall a \in P$.

(1) There may not be a "minimum" (first) nor "maximum" (last) element in a poset.

(2) You might have a unique minimum (first) element, but no maximum element (last) in a poset.

(3) Likewise, a poset may not have a minimum "first" element, but have a maximum ("last") element.

(4) And if there is a unique minimum and a unique maximum element in the poset, then those are the first and last elements in the poset, respectively.

Exercise: try to find an "example" poset that models/represents each of these four possibilities) (one per possibility)!

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But I like yours –  Babak S. Mar 17 '13 at 9:31


# Alternating Series Test

Share a link to this widget: More. Infinite series whose terms alternate in sign are called alternating series. 6 Alternating Series and Conditional Convergence Page 1 Theorem 15 - The Alternating Series Test The series 1 12 3 4 1 (1)n n n uuu uu ∞ + = ∑− =−+−+" converges if the following conditions are satisfied: 1. 1 2 1 4 1 16 1 256. Express the series in a sum End,tD÷ = 1-÷t±-¥tF ' Et. A retrospective record review of performance on the three-step Luria test was conducted on 383 participants from a university-based dementia clinic. The sequence of (positive) terms b n eventually decreases. for all of you pump and level control needs including alternating relays and isolated switches. The terms alternate, and the computation above shows that the terms decrease in absolute value. our series will diverge. This is a very useful lecture in Calculus. The series is an absolutely convergent series; The series is a conditionally convergent series; The series diverges to or ; The partial sums of the series have differing values of limit superior and limit inferior. No, it does not establish the divergence of an alternating series unless it fails the test by violating the condition lim_{n to infty}b_n=0, which is essentially the Divergence Test; therefore, it established the divergence in this case. By the ratio test, the power series converges. Let’slookateachpartmoreclosely: (n+1)!: Prettyself-explanatory. However, given the same series Ʃ((-1)^n)(a_{n}), if I apply condition 1 of the alternating series test, which is the Nth term test on just the (a_{n}) portion, the test is inconclusive if the limit is anything other than zero. Alternating Series Remainder. So here is a good way of testing a given alternating series: if you. Proof Example of a divergent series. Suppose that: 1. Alternating series test - Series, Calculus, Mathematics Summary and Exercise are very important for perfect preparation. Series Convergence Tests Math 122 Calculus III D Joyce, Fall 2012 Some series converge, some diverge. Alternating Series and P-series "convergence" I couldn't resist trying out a pun. Overview of Ratio Test; 4 Examples; Root Test. Trust Ram Meter Inc. then the series is convergent. " From MathWorld--A Wolfram Web Resource. There are two things we have to verify: we need the sequence fa ng= n+1 n2 to be decreasing, and we need this sequence to have limit zero. Alternating Series Test. Personality defines a unique, recognizable individual and is developed as a result of the interaction of the inherited elements and the life-time environment. Correct! This is the correct answer. In order to use this test, we first need to know what a converging series and a diverging series is. 15 hours ago · The year before 107k Penn State fans watched a 42-13 stomping; the year before that Michigan hammered PSU 49-10. Y —l) n an Odd terms are neg. Blood, Part 1 - True Blood: Crash Course A&P #29 - Duration: 10:00. At , the series is. Teach yourself calculus. 15 kV Class 3/C termination samples were built using 3M™ 3/C Phase Rejacketing System RJS. Electricity flows in two ways: either in an alternating current (AC) or in a direct current (DC). : GB 1984-2014: Status: valid remind me the status change. What test to use? When you're looking at a positive series, what's the best way to determine whether it converges or diverges? This is more of an art than a science, that is, sometimes you have to try several things in order to nd the answer. Alternating Series Test The last two tests that we looked at for series convergence have required that all the terms in the series be positive. Topics for Test 2 Convergence tests for series. Fingerpicking for Ukulele - Alternating Thumb Style, 2nd Edition is an new expanded edition which focuses on the alternating thumb fingerpicking style through a series of graduated lessons-chapters incorporating your index and middle fingers with the alternating thumb. Use a known series to ﬁnd a power series in x that has the given function as its sum: (a) xsin(x3) Recall the Maclaurin series for sinu = X∞ n=0 (−1)n u2n+1 (2n+1)! Therefore, sin(x3) = X∞ n=0 (−1)n (x3)2n+1 (2n+1)! =. The alternating series test can only tell you that an alternating series itself converges. An alternating seriesalternates because it has a factor of -1. Looking for abbreviations of AST? It is Alternating series test. Alternating Series (6. Converges by. an infinite series whose terms are alternately positive and negative: u 1 - u 2 + u 3 - u 4 + … + (-I) n-1 u n + …. Alternating Series and P-series "convergence" I couldn't resist trying out a pun. Hosts Julia Collin Davison and Bridget Lancaster and the Test Kitchen cooks prepare America's favorite recipes, passing along valuable tips as they go. Since the terms of an alternating series change sign, the partial sums for any alternating series will jump back and forth over some line. For problems with multiple parts you can view the solution to each part by clicking the Show Solution link after the problem statement for that part or you can view the solutions to all parts by clicking the Show All Solutions link near the top of the solution. In fact, when checking for absolute convergence the term 'alternating series' is meaningless. 0 nF, R = 100Ω, and the source voltage is 220 V. The applet shows the series called the alternating harmonic series because its terms alternate sign: The harmonic series diverges, but maybe the minus signs change the behavior in this case. Line jumping is the idea behind our first convergence test, the alternating series test. Free online storage and sharing with Screencast. CONVERGENCE TESTS FOR SERIES: COMMENTS AND PROOFS PART IV: THE ALTERNATING SERIES TEST Math 112 The convergence tests for series have nice intuitive reasons why they work, and these are fairly easy to turn into rigorous proofs. Alternating Series Test Recall that the Integral Test, Direct Comparison Test, and Limit Comparison Test all require that the terms of the series are positive. It’s important to rely on the de nition of an in nite series when trying to telescope a series. We note that S 2  n + 1 - S 2  n = a 2  n + 1. com is online education portal for providing cost effective entrance examination practice to students for Railway (RRB), SSC, Banking (IBPS) Clear / PO, BPSC, UPSC, JPSC, Medical, Engineering, MCA, MBA, etc. Here is a set of practice problems to accompany the Alternating Series Test section of the Series & Sequences chapter of the notes for Paul Dawkins Calculus II course at Lamar University. For the series above, the root test determines that the series converges for and divergesk kB " # for. The Alternating Series Test If the alternating series satisfies for k = 2, 3, 4, 5, , and. All together, the series converges for , and diverges for and for. Course Description Calculus emphasizes a multi-representational approach, with concepts, results, and problems being expressed graphically, numerically, analytically, and verbally. Using this method, it was recently shown that increasing alpha (10 Hz) oscillations improved creative ideation with figural material and that increasing gamma (40 Hz. The total resistance of the circuit is found by simply adding up the resistance values of the individual resistors:. With no loss of generality we can assume that the series begins at n =1. The participants. Notice that the series in question is alternating, and we can verify that the hypotheses of the alternating series test apply: (1)To show that the (absolute value) of the terms of the series are decreasing, we’ll compute a derivative. CLP-2 Integral Calculus. To answer that question, you must investigate the positive series with a different test. Solved examples with detailed answer description, explanation are given and it would be easy to understand. 6 Alternating Series and Conditional Convergence Page 1 Theorem 15 - The Alternating Series Test The series 1 12 3 4 1 (1)n n n uuu uu ∞ + = ∑− =−+−+" converges if the following conditions are satisfied: 1. A score is given for each subtest, and then it is averaged into an overall Full Scale IQ. Sequences and Series Review Video (PatrickJMT) Assignment #7: Series Flow. For : The first and second conditions are satisfied since the terms are positive and are decreasing after each term. One test that is specifically designed to handle series whose terms alternate positive and negative is the Alternating Series Test. For example. MATH 1920 Alternating Series Test. Using this method, it was recently shown that increasing alpha (10 Hz) oscillations improved creative ideation with figural material and that increasing gamma (40 Hz. Q1: The alternating series test does not apply to the series ∞ ( − 1 ) 𝑛 𝑛 + 1. In fact, when checking for absolute convergence the term 'alternating series' is meaningless. Rather than oscillating back and forth, DC provides a constant voltage or current. But the alternating series approximation theorem quickly shows that. then is this the same as the divergence test, and is it safe to say that the series diverges. Answer to: Using the Alternating Series Test, determine whether \sum_{k=1}^{\infty} ( (-1)^{k} - ( (k + 2)/(4^{k}) ) converges. The book guides students through the core concepts of calculus and helps them understand how those concepts apply to their lives and the world around them. The Alternating Series Test. Use the Alternating Series Test to determine the convergence status of the following series. Tell the patient that you are going to show them a series of hand movements. Some very interesting and helpful examples are included. Search streaming video, audio, and text content for academic, public, and K-12 institutions. Free-Response Questions 1. Determine whether the alternating series $$\sum\limits_{n = 2}^\infty {\large\frac{{{{\left( { - 1} \right)}^{n + 1}}\sqrt n }}{{\ln n}}\normalsize. The following is a Taylor Series evaluated a particular value of x, find the sum of the series. The Alternating Series Test Math. Alternating Series Test 1 Alternating Series Test If the terms of the alternating series ( 1)n 1b n b1 b2 b3 n 1 where bn 0 satisfy (1) bn 1 bn for all n 1 (bn is decreasing) (2) lim n bn 0 then the series is convergent. The table below will help show you how the scores derived from the various subtests is used within the various index scores. Battaly 2017 2 April 21, 2017 Calculus Home Page Class Notes: Prof. Note: AST doesn't apply when either of the conditions is not met, and so never is a test for divergence. Free online storage and sharing with Screencast. an infinite series whose terms are alternately positive and negative: for uk > 0. So in other words, an alternating series will converge if it passes the n-th term test and the absolute value of the terms decrease. Applications of integration including finding areas and volumes. Alphabetical Listing of Convergence Tests. The terms alternate, and the computation above shows that the terms decrease in absolute value. Handout on the Alternating Series Test - Part I (Maynooth University) Handout on the Alternating Series Test - Part II (Maynooth University) Handout on the Alternating Series Test - Part III (Maynooth University) Video on Alternating Series (Patrick JMT). a mathematical series in which consecutive terms are alternatively positive and negative…. Therefore, we will have to look at the alternating series to determine if it converges or not. Alternating series, absolute and conditional convergence You have to know the de nition of what it means for a series to be alternating and con-vergent. A quantity that measures how accurately the nth partial sum of an alternating series estimates the sum of the series. Therefore, the sums converge to the same limit if and only if a n → 0 as n → ∞. Alternating Series & AS Test Objectives: Be able to describe the convergence of alternating series. However, it can be remedied by choosing any sequence which goes to zero fast enough, and putting it in the place of the zero terms, as you do. an infinite series whose terms are alternately positive and negative: u 1 - u 2 + u 3 - u 4 + … + (-I) n-1 u n + …. After defining alternating series, we introduce the alternating series test to determine whether such a series converges. series diverges by Limit Comparison, and the original series does not converge absolutely. A retrospective record review of performance on the three-step Luria test was conducted on 383 participants from a university-based dementia clinic. Infinite series whose terms alternate in sign are called alternating series. (This can usually b e done b y insp ection). Proof: Look at the. So this is a geometric series with common ratio r = –2. If you need to review this test, please refer to the supplemental notes 23. Infinite series whose terms alternate in sign are called alternating series. Then P∞ k=1 (−1) ka k converges. A divergent alternating series whose terms go to zero. Course Description Calculus emphasizes a multi-representational approach, with concepts, results, and problems being expressed graphically, numerically, analytically, and verbally. (f) Prove that the alternating harmonic series X1 n=1 ( 1)n n converges. Here is how one can find the derivative of arctan x: The above is a modern proof, Gregory used the derivative of arctan from the work of others. 5 kV through 765 kV and IEEE Standard 82, IEEE Standard Test Procedure for Impulse Voltage Tests on Insulated Conductors. ) Summarizing the above work, we know that 4 is not included, but 6 is. An alternating series is a series whose terms are al-ternately positive and negative. 88 11 Note: and ; therefore, 89 1 b) lim lim 0 8 Therefore. It is one of the most commonly used tests for determining the convergence or divergence of series. It is not obvious that the sequence b n decreases monotonically to 0. Overview of Alternating Series Test; 3 Examples; Conditional and Absolute Convergence for Alternating Series; 2 Examples; Ratio Test. In mathematical analysis, the alternating series test is a method used to prove that an alternating series with terms that decrease in absolute value is a convergent series. Converges by ratio test. This test does not prove absolute convergence. toit toshowthatthe series converges. Chroma's 63800 Series AC & DC Electronic Loads include built-in 16-bits precision measurement circuits to measure the steady-state and transient responses for true RMS voltage, true RMS current, true power(P), apparent power(S), reactive power(Q), crest factor, power factor, THDv and peak repetitive current. Run multiple programs, render videos and more — the Yoga 920 is designed to multitask with ease. The Alternating Series Test If k 1 ak is an Alternating Series and lim 0 k k a and ak eventually becomes strictly decreasing. 01 Single Variable Calculus, Fall 2005 Prof. However, the third condition is not valid since and instead approaches infinity. Overview of Root Test; 3 Examples; Sequences. Since this is an alternating series whose terms decrease to zero, we know that the series converges. The alternating series. In mathematical analysis, the alternating series test is a method used to prove that an alternating series with terms that decrease in absolute value is a convergent series. Free series convergence calculator - test infinite series for convergence step-by-step. Here is how one can find the derivative of arctan x: The above is a modern proof, Gregory used the derivative of arctan from the work of others. Thus, the even and odd series both converge. In this worksheet, we will practice determining whether an alternating series is convergent or divergent using the alternating series test. Series Convergence Flowchart doesa n! 0? Isa n > 0? Diverges by Divergence Test Is it alternating in sign and ja n decreasing? Are there any easy comparisons? Does it feel likea n `looks like' someb n? Try Ratio Test: lim a n+1 a n = c if 0 c < 1 then P a n converges if c > 1 then P a n diverges if c= 1 then test is inconclusive Try Integral. Converges by ratio test. The total resistance of the circuit is found by simply adding up the resistance values of the individual resistors:. One test that is specifically designed to handle series whose terms alternate positive and negative is the Alternating Series Test. Several convergence testing methods, such as the alternating series test, ratio test and root test, will be presented. 0 < a n+1 <= a n), and approaching zero, then the alternating series. EX 4 Show converges absolutely. 1 Examples 2 Alternating series test. Alternating series definition is - a mathematical series in which consecutive terms are alternatively positive and negative. Bhagwan Singh Vishwakarma 129,314 views. (b) Prove the Alternating Series Test using the Nested Interval Property (Theorem 1. This is a test which we'll use to show lots of alternating series converge. Here are a bunch of Single Variable Calculus applets. Overview of Alternating Series Test; 3 Examples; Conditional and Absolute Convergence for Alternating Series; 2 Examples; Ratio Test. (LTspice is also called SwitcherCAD by its manufacturer, since they use it primarily for the design of switch mode power supplies (SMPS). Use a known series to ﬁnd a power series in x that has the given function as its sum: (a) xsin(x3) Recall the Maclaurin series for sinu = X∞ n=0 (−1)n u2n+1 (2n+1)! Therefore, sin(x3) = X∞ n=0 (−1)n (x3)2n+1 (2n+1)! =. ) The following test says that if the terms of an alternating series decrease toward 0 in absolute value, then the series converges. 5 – Notes Page 2 of 3 Ex 3. 이 감소수열이므로 라고 하면. (g) State the Alternating Series Estimation Theorem. For those that diverge, say which hypotheses of the alternating series test do not. Sequences and Series: Alternating Series Test (MathsCasts). (i) The series (−1)n is an alternating. The alternating series test requires that the a n alternate sign, get smaller and approach zero as n approaches infinity, which is true in this case. If s= P ( 1)n 1b. If a k → 0, then X (−1)ka k converges. If, as our intuition tells us should be true, the rearrangement does not change the sum, then we have just seen that. AC versus DC. When a series alternates (plus, minus, plus, minus,) there's a fairly simple way to determine whether it converges or diverges: see if the terms of the series approach 0. 8 = angle Pod sino g 3! as 3! a 7 " Functions can often be represented by an infinite series. Lecture 27 :Alternating Series The integral test and the comparison test given in previous lectures, apply only to series with positive terms. • Be sure that you apply the alternating series test only to alternating series. But the alternating series approximation theorem quickly shows that. Part (a) asked students to use the ratio test to determine the interval of convergence for the given Maclaurin series. The integral test, which is my favorite test in general, tends to be awkward with alternating series. 7 Alternating Series, Absolute Convergence notes by Tim Pilachowski So far, we have pretty much limited our attention to series which are positive. The alternating series test is a simple test we can use to find out whether or not an alternating series converges (settles on a certain number). (You probably figured out that with this naked summation. The Alternating Series Test (Leibniz's Theorem) This test is the sufficient convergence test. The idea of hopping back and forth to a limit is basically. Rearrangements. What test to use? When you're looking at a positive series, what's the best way to determine whether it converges or diverges? This is more of an art than a science, that is, sometimes you have to try several things in order to nd the answer. Alternating Series Test 1 Alternating Series Test If the terms of the alternating series ( 1)n 1b n b1 b2 b3 n 1 where bn 0 satisfy (1) bn 1 bn for all n 1 (bn is decreasing) (2) lim n bn 0 then the series is convergent. We look at a couple of examples. 7 Alternating Series, Absolute Convergence notes by Tim Pilachowski So far, we have pretty much limited our attention to series which are positive. We're learning alternating series test and the whole class is confused on why the series needs to be decreasing to pass the test. EXPECTED SKILLS: Determine if an alternating series converges using the Alternating Series Test. The un 's are all positive. If it does, then try applying the Ratio Test i. The alternating series simply tells us that the absolute value of each of the terms decreases monotonically, i. Instead, I'll give a more elementary proof of the Alternating series test which does not use Abel's formula (but we will see Abel's formula later). What does alternating personality mean in law?. Just better. An alternating sequence will have numbers that switch back and forth between positive and negative signs. Electricity or "current" is nothing but the movement of electrons through a conductor, like a wire. It should be noted that Theorem 1. CLP-2 Integral Calculus. Here are a bunch of Single Variable Calculus applets. 7: Series, alternating series test - solutions Apply the AST (possibly in combination with other tests) and state your conclusion about convergence. Embed this widget ». Finally, by L'Hôpital's Rule, By the Alternating Series Test, the series converges. of series with positive and negative terms and whether or not they converge. So, given the series look at the limit of the non-alternating part: So, this series converges. Jason Starr. The well-known Leibniz Criterion or alternating series test of convergence of alternating series is generalized for the case when the absolute value of terms of series are “not absolutely monotonously” convergent to zero. Alexander Street is an imprint of ProQuest that promotes teaching, research, and learning across music, counseling, history, anthropology, drama, film, and more. With the Alternating Series Test, all we need to know to determine convergence of the series is whether the limit of b[n] is zero as n goes to infinity. Alternating series test for convergence. Find more Mathematics widgets in Wolfram|Alpha. ) Note: Some of this was written using SwitcherCad III, and some was written using LTspice IV. 5 kV through 765 kV and IEEE Standard 82, IEEE Standard Test Procedure for Impulse Voltage Tests on Insulated Conductors. There are two main types of current used in most electronic circuits today. The table below will help show you how the scores derived from the various subtests is used within the various index scores. Alternating Series Test If for all n, a n is positive, non-increasing (i. [email protected] This is the logical reasoning questions and answers section on "Number Series Type 2" with explanation for various interview, competitive examination and entrance test. A series in which successive terms have opposite signs is called an alternating series. I Absolute convergence test. The Alternating Series Test (Leibniz's Theorem) This test is the sufficient convergence test. b n+1 ≤ b n for all n > N. Theorem 4 : (Comparison test ) Suppose 0 • an • bn for n ‚ k for some k: Then. You can see some Alternating Series, Absolute and Conditional Convergence - Notes, Engineering, Semester sample questions with examples at the bottom of this page. I believe we sometimes overemphasize the importance of this test because we want to make clear the distinction between absolute convergence and convergence. When a sum does this, we say it ‘telescopes’. The Absolute-Convergence Test (ACT) For series that have infinitely many negative terms and infinitely many positive terms and that aren't alternating, the tests discussed in previous sections and the AST can't be applied. Definition of alternating personality in the Legal Dictionary - by Free online English dictionary and encyclopedia. The part of the summation makes up the series alternator. Calculus 141, section 9. Suppose that {a i} is a sequence of positive numbers such that a i > a i+1 for all i. Teach yourself calculus. Course Material Related to This Topic:. The lower beds of the Peuquenes ridge, and of the several great lines to the westward of it, are composed of a vast pile, many thousand feet in thickness, of porphyries which have flowed as submarine lavas, alternating with angular and rounded fragments of the same rocks, thrown out of the submarine craters. iii) if ρ = 1, then the test is inconclusive. Line jumping is the idea behind our first convergence test, the alternating series test. The common ratio of a geometric series may be negative, resulting in an alternating sequence. Geometric series. Continuing with post on sequences and series New Series from Old 1 Rewriting using substitution New Series from Old 2 Finding series by differentiating and integrating New Series from Old 3 Rewriting rational expressions as geometric series Geometric Series – Far Out A look at doing a question the right way and the “wrong” way?…. Hence, the interval of convergence is: (−8,10] and the radius convergence is: R = 10. Therefore, the sums converge to the same limit if and only if a n → 0 as n → ∞. Alternating Series Test. Absolute Ratio Test Let be a series of nonzero terms and suppose. In a alternating series every term will have a sign different than the term before it. The alternating series simply tells us that the absolute value of each of the terms decreases monotonically, i. (One is the harmonic series; the other can be proved divergent by comparison with the harmonic series. An alternating series is a series whose terms are al-ternately positive and negative. An alternating sequence will have numbers that switch back and forth between positive and negative signs. The short circuit test is maintained by default. Alternating series test listed as AST Alternating series. Then the series converges if both of the following conditions hold. If property 3 is respected but property 1 and/or property 2 do not hold, then the alternating series test is inconclusive. More Alternating Series Examples - Finding whether a given alternating series converges or diverges. Alternating Series (6. The Alternating Series Test. So, the series converges by the alternating series test. −1 3 2 4 −3 5 4 6 −5 7. The test that we are going to look into in this section will be a test for alternating series. The series above is thus an example of an alternating series, and is called the alternating harmonic series. Alternating Series and Leibniz’s Test Let a 1;a 2;a 3;::: be a sequence of positive numbers. In order to use this test, we first need to know what a converging series and a diverging series is. The integral test, which is my favorite test in general, tends to be awkward with alternating series. Overview of Root Test; 3 Examples; Sequences. jxjn+1: Rememberthatjxjrepresentsthedistancebetweenx and0. このコンテンツの表示には、Adobe Flash Playerの最新バージョンが必要です。 http://www. Warm Up: Find the sum of the infinite series. Alternating Series Test. Does ∑ (−1)𝑛3𝑛 4𝑛−1 ∞ 𝑛=1 converge or diverge? For an alternating series, how close is 𝑠𝑛 to the sum of the infinite number of terms?. Mathispower4u 70,297 views. Chroma's 63800 Series AC & DC Electronic Loads include built-in 16-bits precision measurement circuits to measure the steady-state and transient responses for true RMS voltage, true RMS current, true power(P), apparent power(S), reactive power(Q), crest factor, power factor, THDv and peak repetitive current. Alternating Current Circuits 5 Open-Ended Problems 57. Topics for Test 2 Convergence tests for series. SERIES CONVERGENCE/ DIVERGENCE FLOW CHART nVergeS TEST FOR DIVERGENCE GEOMETRIC SERIES ALTERNATING SERIES TELESCOPING SERIES ? May to etc. Ratio and Root Tests. 1997 views. Holmes May 1, 2008 The exam will cover sections 8. In this section we introduce alternating series—those series whose terms alternate in sign. Proof: Look at the. The series above is thus an example of an alternating series, and is called the alternating harmonic series. alternating series test. It is important that you verify the conditions of the Alternating Series Test are met; otherwise some-one might not believe your conclusion is valid. An alternating series is any series, , for which the series terms can be written in one of the following two forms. Alternating series convergence: a visual proof Richard H. The Alternating Series Test. alternating with phrase. Incorrect! Remember the conditions of the Alternating Series Test. For : The first and second conditions are satisfied since the terms are positive and are decreasing after each term. The test was used by Gottfried Leibniz and is sometimes known as Leibniz's test , Leibniz's rule , or the Leibniz criterion. The alternating series test can tell us if it's safe to open that box. Fourier series are used in the analysis of periodic functions. Solution: Let I1 be the closed interval [0, s1]. It is important that you verify the conditions of the Alternating Series Test are met; otherwise some-one might not believe your conclusion is valid. 1, which asks for a proof of the Alter-nating Series Test using the Cauchy Criterion for series (Theorem 2. The alternating series test says that if the absolute value of each successive term decreases and \lim_{n\to\infty}a_n=0, then the series converges. I Absolute and conditional convergence. LTspice Tutorial Introduction While LTspice is a Windows program, it runs on Linux under Wine as well. 1 Is there a short name for series which satisfy the hypothesis of the alternating series test?. This is a very useful lecture in Calculus. Printing the sum of an alternating series [Basic Python] I've been stuck on this little problem for the best part of today, and it's driving me insane now. Hence, the interval of convergence is: (−8,10] and the radius convergence is: R = 10. Write the three rules that are used to satisfy convergence in an alternating series test. Assignment #4: Alternating Series. Therefore, we will have to look at the alternating series to determine if it converges or not. Correct! This is the correct answer. ( 1) Use Alternating Series Test to show con verges. What is alternating personality? Meaning of alternating personality as a legal term. To see how this works, let \(S$$ be the sum of a convergent alternating series, so. The alternating series test requires that the a n alternate sign, get smaller and approach zero as n approaches infinity, which is true in this case. Theorem (Alternating series test) If the terms of the series ∑ n = 1 ∞ (-1) n an have the property thatall ofthe an terms are positive and an+1 < an forall n, thenthe series converges. Alternating Series (6. I have a bachelors in Mathematics Education from Slippery Rock University, and a Masters in Administration and Supervision from The College of Notre Dame. Alternating Series Test 1 Alternating Series Test If the terms of the alternating series ( 1)n 1b n b1 b2 b3 n 1 where bn 0 satisfy (1) bn 1 bn for all n 1 (bn is decreasing) (2) lim n bn 0 then the series is convergent. Q1: The alternating series test does not apply to the series ∞ ( − 1 ) 𝑛 𝑛 + 1. It's also known as the Leibniz's Theorem for alternating series. (Warning: Do not use a multimeter to measure the wall outlets in your home. In this video, Krista King from integralCALC Academy talks about the Alternating Series Test (Calculus problem example). Transcranial alternating current stimulation (tACS) is a non-invasive brain stimulation method that allows to directly modulate brain oscillations of a given frequency. Drag up for fullscreen M M. 01 Single Variable Calculus, Fall 2005 Prof. The test was used by Gottfried Leibniz and is sometimes known as Leibniz's test , Leibniz's rule , or the Leibniz criterion. Looking for abbreviations of AST? It is Alternating series test. Mathematics Assignment Help, Steps for alternating series test, Steps for Alternating Series Test Suppose that we have a series ∑a n and either a n = (-1) n b n or a n = (-1) n+1 b n where b n > 0 for all n. The common ratio of a geometric series may be negative, resulting in an alternating sequence. We can therefore. n satis es the requirements for the alternating series test. However, it is not enough to have having a limit of zero, you also need decreasing, as the following example shows. Solved examples with detailed answer description, explanation are given and it would be easy to understand. You’ll do that one for homework. Alternating Current Circuits 5 Open-Ended Problems 57. We note that S 2  n + 1 - S 2  n = a 2  n + 1. Many, but not all, of the problems will have. Alternating series test for convergence. Alternating Series Estimation Theorem. The fact that sums, products, integrals, antiderivatives of Taylor series are also Taylor series is in 8. 6 Alternating Series and Conditional Convergence Page 1 Theorem 15 - The Alternating Series Test The series 1 12 3 4 1 (1)n n n uuu uu ∞ + = ∑− =−+−+" converges if the following conditions are satisfied: 1. According to the alternating series test, we know that this series converges to some number. Click on the name of the test to get more information on the test. Many series such as 8œ" 8œ" 8œ" ∞ ∞ ∞ # sin 8 " 88 8 8 8 8" ß " ß "and do not have all positive terms and thus cannot be investigated using the above mentioned tests. A t-test is one of the most frequently used procedures in statistics. Alternating series have the simplest of sign patterns. The terms alternate. This is a correct reasoning to show the divergence of the above series.


# Show that there's a unique minimum spanning tree if all edges have different costs

Show that there's a unique minimum spanning tree (MST) in case the edges' weights are pairwise different $(w(e)\neq w(f) \text{ for } e\neq f)$.

I thought that the proof can be done for example by contradiction, saying that we have $2$ different MST meaning that somewhere was possible to pick from more edges, so $w(e) = w(f)$ for $e\neq f$, contradiction. Apparently this is not correct.

How would you show that a graph has a unique MST if all edges have distinct weights?

If $T_1$ and $T_2$ are distinct minimum spanning trees, then consider the edge of minimum weight among all the edges that are contained in exactly one of $T_1$ or $T_2$. Without loss of generality, this edge appears only in $T_1$, and we can call it $e_1$.

Then $T_2 \cup \{ e_1 \}$ must contain a cycle, and one of the edges of this cycle, call it $e_2$, is not in $T_1$.

Since $e_2$ is a edge different from $e_1$ and is contained in exactly one of $T_1$ or $T_2$, it must be that $w ( e_1 ) < w ( e_2 )$. Note that $T = T_2 \cup \{ e_1 \} \setminus \{ e_2 \}$ is a spanning tree. The total weight of $T$ is smaller than the total weight of $T_2$, but this is a contradiction, since we have supposed that $T_2$ is a minimum spanning tree.

• How do we know that $w ( e_1 ) < w ( e_2 )$? Mar 4, 2017 at 23:57
• @Alexander Because $e_1$ is the edge of least weight that is contained in exactly one of the MSTs, and $e_2$ is a different edge that is contained in exactly one of the MSTs. Mar 5, 2017 at 3:49
• Is $e_1$ the edge of least weight out of the edges of $T_1 \cup T_2$, which is in either $T_1$ or $T_2$? Or is $e_1$ the edge of least weight out of the edges of $T_1 \bigtriangleup T_2$? Mar 6, 2017 at 1:43
• I suspect it's the latter of two, otherwise I don't understand how we know this edge isn't in both $T_1$ and $T_2$. Mar 6, 2017 at 1:47
• @Alexander As I said, $e_1$ is the edge of least weight that is contained in exactly one of the MSTs. If it was in both $T_1$ and $T_2$ it would not be contained in exactly one of the MSTs. Mar 6, 2017 at 3:41

A proof using cycle property:

Let $G=(V,E)$ be the original graph.

Suppose there are two distinct MSTs $T_1=(V,E_1)$ and $T_2=(V,E_2)$. Since $T_1$ and $T_2$ are distinct, the sets $E_1-E_2$ and $E_2-E_1$ are not empty, so $\exists e\in E_1-E_2$.

Since $e\notin E_2$, adding it to $T_2$ creates a cycle. By cycle property the most expensive edge of this cycle (call it $e'$) does not belong to any MST. But

If $e'=e$ then $e'\in E_1$ (because $e\in E_1-E_2$)
If $e'\neq e$ then $e'\in E_2$

Both cases are contradicting with the fact that $e'$ is not in any MST.

The best explanation I found was on wikipedia, Proof:

1. Assume the contrary, that there are two different MSTs $A$ and $B$.
2. Since $A$ and $B$ differ despite containing the same nodes, there is at least one edge that belongs to one but not the other. Among such edges, let e1 be the one with least weight; this choice is unique because the edge weights are all distinct. Without loss of generality, assume $e1$ is in $A$.
3. As $B$ is a MST, $\{e1\}\cup B$ must contain a cycle $C$.
4. As a tree, $A$ contains no cycles, therefore $C$ must have an edge $e2$ that is not in $A$.
5. Since $e1$ was chosen as the unique lowest-weight edge among those belonging to exactly one of $A$ and $B$, the weight of $e2$ must be greater than the weight of $e1$.
6. Replacing $e2$ with $e1$ in $B$ therefore yields a spanning tree with a smaller weight.
7. This contradicts the assumption that $B$ is a MST.

More generally, if the edge weights are not all distinct then only the (multi-)set of weights in minimum spanning trees is certain to be unique; it is the same for all minimum spanning trees [1].

This is a slightly different (though more lengthy) proof than the other's in that it is an algorithmic proof. We can use Prim's algorithm and demonstrate the proof by studying the spanning tree it builds in such a graph $$G$$. Here is Prim's algorithm:

• $$V_T \leftarrow \{ v_0 \}$$
• $$T_0 \leftarrow \emptyset$$
• for $$i \leftarrow 1$$ to $$|V| - 1$$ do:
• find an edge $$e_i=(v_i,u_i)$$ of minimum weight such that $$v_i$$ is in $$V_T$$ and $$u_i$$ is not.
• $$V_T \leftarrow V_T \cup \{ u_i \}$$
• $$T_i \leftarrow T_i \cup \{ e_i \}$$

We demonstrate the following : At iteration $$i$$, the choice $$e_i$$ that Prim's algorithm makes belongs to every minimum spanning tree. We can show this by using the inductive hypothesis:

Suppose $$T_{i-1}$$ is a subset of every minimum spanning tree. Then, $$T_i = T_{i-1} \cup \{ e_i \}$$ is a subset of every minimum spanning tree.

• Base case: $$T_1 = e_1$$ where $$e_1$$ is incident to $$v_0$$. Suppose that there is some minimum spanning tree $$T$$ that does not contain $$e_1$$. Then, $$T$$ contains another edge, $$\bar{e}_1$$ incident to $$v_0$$. By the greedy choice of Prim's and the distinctness of weights, $$w(e_1) < w(\bar{e}_1)$$ and we can replace $$\bar{e}_1$$ with $$e_1$$ to decrease the weight of $$T$$, which is a contradiction.
• Inductive step: Suppose, to the contrary, that there exists some minimum spanning tree $$T$$ which contains the edges in $$T_{i-1}$$ but does not contain $$e_i$$. Consider the set of edges $$T \setminus T_{i-1}$$, which must be a forest composed of trees $$\bar{T}^1_{i-1}, \bar{T}^2_{i-1}, \ldots, \bar{T}^k_{i-1},$$ where each tree $$\bar{T}^t_{i-1}$$ is connected to $$T_{i-1}$$ in $$T$$ by an edge $$e^t_{i-1}$$.

However, note that one of those trees must be incident to the vertex $$u_i$$. Suppose that tree is $$\bar{T}^x_{i-1}$$. Then, by Prim's greedy choice and since the weights of edges are distinct, we know that $$w(e_i) < w(e^x_{i-1})$$. Which means that we replace $$e^x_{i-1}$$ with $$e_i$$ and decrease the weight of $$T$$. Hence, $$T$$ is not a minimum spanning tree. Then, $$e_i$$ must be in every minimum spanning tree and $$T_i$$ must be a subset of every minimum spanning tree.

Thus, the minimum spanning tree produced by Prim's algorithm is the only minimum spanning tree of $$G$$.

– fhy
Dec 29, 2019 at 12:30

This is a (slightly) more detailed version of the currently accepted answer

(For the sake of contradiction) Let $$T_1 = (V, E_1)$$ and $$T_2 = (V, E_2)$$ be two distinct MSTs of the graph $$G = (V, E)$$
Note that both have the same vertex set $$V$$ since both are spanning trees of $$G$$

Consider the set $$E_{\Delta} = E_1 \triangle E_2$$
Let $$e = (u, v)$$ be the edge in $$E_{\Delta}$$ having the least cost (or weight)
Note that since all costs are unique, and $$E_{\Delta}$$ is non-empty, $$e$$ must be unique.
Without loss of generality, assume $$e \in E_1$$

Now, there must be a path $$P$$, with $$e \notin P$$, in $$T_2$$ connecting $$u$$ and $$v$$, since trees are by definition, connected.
Note that at least one edge (say $$e'$$) that occurs in $$P$$ must not be in $$E_1$$, thus, $$e' \in E_{\Delta}$$
This is because, if $$P \subset E_1$$, $$T_1$$ will contain a cycle formed by the path $$P$$ and the edge $$e$$, this leads to a contradiction, since trees by definition are acyclic.
Note that by definition of $$e$$ and the fact that all costs are distinct, $$\text{cost}(e) < \text{cost}(e')$$

Now, consider $$T_2' = (V, E_2')$$, where $$E_2' = (E_2 \backslash\{e'\})\cup\{e\}$$, this has a strictly lesser total cost than $$T_2$$
As $$T_2$$ was a MST, this leads to a contradiction.