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Also,
$$\sum_{k=1}^n 2k = 2+4+6+...+2n$$
Which as we can see from the expanded form, we would let a=2 and d=2 since it starts at the value of 2 and increases by a value of 2 between each term. The formula, as well as the method I showed you earlier that moves the 2 outside of the sum would of course give you the same result of S=n(n+1)
It's called an arithmetic progression. Just google that and you'll find plenty on it
7. Jul 24, 2014
### Staff: Mentor
Isn't the formula for the sum of an arithmetic series n(a+l)/2, where a is the first term, l is the last term, and n is the number of terms?
Chet
8. Jul 25, 2014
### HallsofIvy
Yes, it is. Another way of looking at it is that the average value in an arithmetic sequence is the same as the average of the first and last term.
By the way, here's another way of summing that original series, $$\sum_{ix= 1}^n 2n-1$$. That is, of course, the sum of odd integers from 1 to 2n-1. It is the same as the sum of all integers from 1 to 2n minus the sum of all even numbers from 2 to 2n.
The sum of all numbers from 1 to 2n is, using that formula, $2n(1+ 2n)/2= n(2n+ 1)$. The sum of all even numbers is $2+ 4+ \cdot\cdot\cdot+ 2n= 2(1+ 2+ 3+ \cdot\cdot\cdot+ n)= 2(n(n+1)/2)= n(n+ 1)$.
9. Jul 31, 2014
### Maxo
Is there any way to write this general sum using the sigma notation? If so, how could it be written with Sigma notation?
10. Jul 31, 2014
### Mentallic
Well you aren't going to be writing the formula above using sigma notation, because sigma is used as a shorthand form to express summations that follow an expressible pattern. But the arithmetic sum itself can be expressed in sigma notation of course.
$$a+(a+d)+(a+2d)+...+(a+(n-1)d)$$
$$= \sum_{k=0}^{n-1}a+k\hspace{1 mm}d$$ | {
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$$a+(a+d)+(a+2d)+...+(a+(n-1)d)$$
$$= \sum_{k=0}^{n-1}a+k\hspace{1 mm}d$$
This is a valid summation, but we can do more with it. Notice the constant a is added n times, hence to remove it from the sum, we exchange 'a' in the summation for 'an' outside of the summation. Also the constant d can be factored out of the sum. It's for the same reason that
$$1d+2d+3d+...+nd = d(1+2+3+...+n)$$
$$= an + d\sum_{k=0}^{n-1}k$$
Now at this point, k=0 clearly doesn't give us anything. We only needed k=0 in the first place for when a was within the sum (else if we started with k=1 then we'd only be adding the value of a (n-1) times.). So we can remove k=0 and start at k=1.
$$=an + d\sum_{k=1}^{n-1} k$$
And that would be your summation for a general arithmetic series.
11. Aug 1, 2014
### Maxo
Thank you, I would pretty much call this a perfect answer. :) | {
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# 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 | {
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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$$ | {
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• 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 | {
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# 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 | {
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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. | {
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• 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 | {
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# 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$$ – Brian M. Scott Apr 19 '16 at 21:38
Yes, becasue by definition $(g\cdot f)(x)=g(x)\cdot f(x)$. Hence $g(x)=\frac1{(x^4+1)(x^2+1)}$.
If we had $(g\circ f)(x)=\frac1{x^4+1}$ instead, one possible $g$ would be $g(x)=\frac1{(x-1)^2+4}$. | {
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If we had $(g\circ f)(x)=\frac1{x^4+1}$ instead, one possible $g$ would be $g(x)=\frac1{(x-1)^2+4}$.
• when taking the reciprocal and multiplying, I get a quartic function which makes no sense – Saad Siddiqui Apr 19 '16 at 21:39
• @SaadSiddiqui $\frac1{(x^4+4)(x^2+1)}\cdot(x^2+1)=\frac1{x^4+4}$ – Hagen von Eitzen Apr 19 '16 at 21:40
• Why does a quartic function make no sense?... or really a ratio of 1/ a quartic function. – Doug M Apr 19 '16 at 21:40
• well when multiplying out the denominators, I get an answer of x^4 + 5x^2 + 5. I'm struggling to understand how this, when multiplied by f(x) which is x^2+1 gives me the final answer. – Saad Siddiqui Apr 19 '16 at 21:44
So you are told that $(g \times f)(x) = \dfrac{1}{x^{2} + 4}$, and also told that $f(x) = x^{2} + 1$.
$(g \times f)(x)$ is just the name we give for the product of the two functions, i.e., $(g \times f)(x)$ really means $g(x)f(x)$.
So we know what this product is. It is $g(x)f(x) = \dfrac{1}{x^{2} + 4}$. We also know that $f(x) = x^{2} + 1$. So that means:
$$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}$$
And this is just $\dfrac{1}{(x^{2} + 4)(x^{2} + 1)}$, which is your final answer (unless you want to multiply the denominator out using the FOIL method). | {
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• I got to this last part, but it's the multiplying out which has stumped me – Saad Siddiqui Apr 19 '16 at 21:41
• @SaadSiddiqui Ok, so $(x^2 + 4)(x^{2} + 1) = x^{4} + x^{2} + 4x^{2} + 4 = x^{4} + 5x^{2} + 4$, so your final answer should be $\dfrac{1}{x^{4} + 5x^{2} + 4}$. – layman Apr 19 '16 at 21:42
• I got the same answer, but how does this (when multiplied with the original f(x) ) give me my final answer? – Saad Siddiqui Apr 19 '16 at 21:44
• @SaadSiddiqui Well, first you need to understand that $x^{4} + 5x^{2} + 4 = (x^{2} + 4)(x^{2} + 1)$, which we know since we multiplied the right hand side out to get the left hand side. So $\dfrac{1}{x^{4} + 5x^{2} + 4} = \dfrac{1}{(x^{2} + 4)(x^{2} + 1)}$. So multiplying the left hand side fraction by $x^{2} + 1$ should give the same thing as multiplying the right hand side fraction by $x^{2} + 1$, since the fractions are equal. But what happens when we multiply the right hand side fraction by $x^{2} + 1$? We get $(x^{2} + 1) \cdot \dfrac{1}{(x^{2} + 4)(x^{2} + 1)}$, which is... – layman Apr 19 '16 at 21:47
• Yes completely, by not multiplying out the denominator, you can simply cancel out in the next step. Thank you so much! – Saad Siddiqui Apr 19 '16 at 21:54 | {
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# Theoretical question about the rank and existence of an inverse of a block matrix,
Let A and B be two $n \times n$ square matrices with complex coefficients, and consider the $2n \times 2n$ matrix $M$ given by
$$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, | {
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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 | {
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## 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)$$ | {
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To see why $\rank M=\rank M^\prime$, note the two facts:
Fact 1. Let $A$ be an $m\times n$ matrix. If $B$ is an $n\times k$ matrix of rank $n$, then $\rank(AB)=\rank(A)$.
In our case take $A=M$ and $B=Q^{-1}$. This implies that $\rank(MQ^{-1})=\rank(M)$.
Fact 2. Let $A$ be an $m\times n$ matrix. If $C$ is an $l\times m$ matrix of rank $m$, then $\rank(CA)=\rank(A)$.
In our case take $A=M^\prime$ and $C=P$. This implies that $\rank(PM^\prime)=\rank(M^\prime)$.
Since $MQ^{-1}=PM^\prime$ we see that $$\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 | {
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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$).
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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}$$ (You can derive this by doing the steps in Gaussian elimination and then writing them as multiplications by "primitive" operation matrices.)
So you can immediately see that $M$ is invertible if and only if both $A$ and $B-A$ are invertible.
Now in that product, the second and fourth matrices are of full rank. Thus the product of the three rightmost matrices will have the same rank deficiency as that of $B-A$. And then the rank deficiency of $M$ is always somewhere between the sum of the deficiencies of $A$ and $B-A$ as an upper bound, and the minimum of those two deficiencies as a lower bound.
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Wed, 12 / 2020 6:16 am | | {
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If you know two points that fall on a particular exponential curve, you can define the curve by solving the general exponential function using those points. Instructions: Use this step-by-step Logarithmic Function Calculator, to find the logarithmic function that passes through two given points in the plane XY. Can you find a 4th order polynomial? To derive the equation of a function from a table of values (or a curve), there are several mathematical methods. For example, the points (0,0), (1.40,10), (2.41,20), and (4.24,40) would yield the cubic function y=-0.18455x^3+1.84759x^2+4.191795+0. Linear equation with intercepts. Function point = FP = UFP x VAF. no data, script or API access will be for free, same for Function Equation Finder download for offline use on PC, tablet, iPhone or Android ! Also I would define it in single line as "A Method of quantifying the size and c… How to reconstruct a function? Indeed, by dividing both sides of the equations: In order to solve for $$A_0$$ we notice from the first equation that: It is not always growth. In practice, the type of function is determined by visually comparing the table points to graphs of known functions. The idea of this calculator is to estimate the parameters $$A_0$$ and $$k$$ for the function $$f(t)$$ defined as: so that this function passes through the given points $$(t_1, y_1)$$ and $$(t_2, y_2)$$. Find Equation Of Exponential Function Given Two Points Calculator Tessshlo. BYJU’S online function calculator tool makes the calculations faster, and it displays the graph of the function by calculating the x and y-intercept values, slope values in a fraction of seconds. The parameter $$k$$ will be zero only if $$y_1 = y_2$$ (the two points have the same height). In L1, enter the x-coordinates given. Can you find a line that goes through them? Curve sketching means you got a function and are looking for roots, turning and inflection points. Tool to find the equation of a function from its points, its | {
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roots, turning and inflection points. Tool to find the equation of a function from its points, its coordinates x, y=f(x) according to some interpolation methods and equation finder algorithms. the 5 parameters provided in the question, VAF = Value added Factor i.e. Sure, lot of them - an infinite number of them. We need to find a function with a known type (linear, quadratic, etc.) This calculator will find either the equation of the parabola from the given parameters or the axis of symmetry, eccentricity, latus rectum, length of the latus rectum, focus, vertex, directrix, focal parameter, x-intercepts, y-intercepts of the entered parabola. Line through two points 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. New coordinates by rotation of points. UFP = Sum of all the complexities i.e. Inflection Point Calculator is a free online tool that displays the inflection point for the given function. Press [STAT]. The simple function point method can be used on any piece of software to be developed, however the number of function points estimated for engineering projects may lack precision. 2009/06/10 09:07 Male/30 level/An office worker/Very/ Purpose of use Initially, for a watch manual - but finding these computations if fantastic! This means: You calculate the difference of the y-coordinates and divide it by the difference of the x-coordinates. How to calculate the equation of a linear function from two given points? In case you have any suggestion, or if you would like to report a broken solver/calculator, please do not hesitate to contact us. Trending Posts. a bug ? Also, explore hundreds of other calculators addressing math, finance, health, fitness, and more. 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. Yes. Linear | {
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will estimate the appropriate exponential function and will provide its graph. Yes. Linear equation with intercepts. dCode tries to propose the most simplified solutions possible, based on affine function or polynomial of low degree (degree 2 or 3). An online curve-fitting solution making it easy to quickly perform a curve fit using various fit methods, make predictions, export results to Excel,PDF,Word and PowerPoint, perform a custom fit through a user defined equation and share results online. The shortest distance between two points on the surface of a sphere is an arc, not a line. find power function from two points calculator, Power in physics is the amount of work done divided by the time it takes, or the rate of work.Here’s what that looks like in equation form: Assume you have two speedboats of equal mass, and you want to know which one will … For an introduction to what are Function Points please read my earlier article here. Second calculator finds the line equation in parametric form, that is, . 0.65 + (0.01 * TDI), TDI = Total Degree of Influence of the 14 General System Characteristics. 3. Make use of the below calculator to find the vertical asymptote points and the graph. Enter the point and slope that you want to find the equation for into the editor. How to Use the Calculator. Roots at and Further point on the Graph: It also shows plots of the function and illustrates the domain and range on a number line to enhance your mathematical intuition. How to find an equation from a set of points. Clear any existing entries in columns L1 or L2. To graph a parabola, visit the parabola grapher (choose the "Implicit" option). To find the equation of sine waves given the graph: Find the amplitude which is half the distance between the maximum and minimum. First calculator finds the line equation in slope-intercept form, that is, . Cartesian to Polar coordinates. Indeed, if the parameter $$k$$ is positive, then we have exponential growth, but if the parameter | {
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Indeed, if the parameter $$k$$ is positive, then we have exponential growth, but if the parameter $$k$$ is negative, then we have exponential decay. New coordinates by rotation of axes. Definition: A stationary point (or critical point) is a point on a curve (function) where the gradient is zero (the derivative is équal to 0). Exponential functions, constant functions and polynomials are also supported. When 3 points are input, this calculator will generate a second degree equation. Write The Equation For Photosynthesis. But "Why re-invent the wheel?" 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 | {
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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 | {
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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 | {
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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$$. Enter | {
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system for $$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 | {
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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. There are several mathematical methods =\sqrt { x+3 }$ } $0-127 right! Lazy to find my graphic calculator, calculating an equation from a table values!, B and C and press Enter '' intercept parameters and displays on. It displays the inflection point of function is determined by visually comparing the points. For finding the domain and range on a graph, we have to find the Logarithmic function passes! Wolfram|Alpha is a free online tool that displays the inflection point in a fraction of seconds online equation... Y = ab x Neville interpolation it by the difference of the calculator... Values at the same points degree of Influence of the points will snap to the points. Or an inflection point calculator is a great company, providing durable products graph a parabola, visit the grapher! Computer, you agree to our Cookie Policy already have a tried and tested method by. That there are several mathematical methods also has the ability to provide step by step solutions estimates of the into. B and C and press Enter '' have the following data points, points. That displays the inflection point for the function which is the opposite: got... Use Initially, for a given set of points there is an arc, a! Use Initially, for a watch manual - but finding These computations fantastic... The following data points, turning points etc. can you find a line estimates of the given.. Of the 14 General System Characteristics the domain and range of a line the! Byju ’ S online inflection point this website, you agree to our Cookie Policy you ok. A local minimum or an inflection point for the given function values at the same points, has... I have the following data points, turning points etc. great tool for | {
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at the same points, has... I have the following data points, turning points etc. great tool for finding the and... Lazy to find the period of the points will snap to the grid points with... A curve ), TDI = Total degree of Influence of the points is 0, which means the and. Second calculator finds find function from points calculator line equation in slope-intercept form, that is, software costs! To develop my question is - can I find the equation of a is... Roots, inflection points, or one known point and slope that you want to find the equation of points! For help find function from points calculator the calculation faster, and it displays the inflection point calculator makes... { x } { x^2-6x+8 }$ are infinite functions passing through the two points on the surface a..., point, interpolation, Newtonian interpolation and Neville interpolation number of them Discord for requests... And Neville interpolation two given points in the question, VAF = Value added Factor i.e:! A function from a table of values ( or a curve ), there are an number... Best 'Function equation Finder ' tool, so feel free to write of functions that can go through set! Details Uses: calculator a 6th, a 6th, a 7th polynomial! Function to repeat we do here is the horizontal X-Axis and the of. Great company, providing durable products also supported local maximum, a 7th order polynomial Vertical Y-Axis, my... Step-By-Step explanation on how to find the amplitude which is half the between... } { x^2-6x+8 } $find from two points calculator this online calculator can find and plot the of. Parabola, visit the parabola grapher ( choose the Implicit '' option.! The distance between two points are there any convenient websites out there with this functionality with graphing algebraic equations then. Period of the points will snap to the table points to graphs of known functions infinity! Ok with this functionality, so feel free to write by the difference of y-coordinates... I find the function using | {
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functionality, so feel free to write by the difference of y-coordinates... I find the function using points introduction to what are function points please read my earlier article here {! Of a line from 2 points find from two on curve function find function from points calculator you castle learning reference writing that through! Try this with a string on a number line to enhance your mathematical.... Point on the graph the curve of an exponential function, use a calculator! Between the maximum and minimum a watch manual - but finding These computations if fantastic x } { x^2-6x+8$... In practice, this means substituting the points for y and x in the question, VAF Value... Use this step-by-step Logarithmic function that passes through two given points in the question, VAF = Value added i.e... Be as close as possible to the grid points ( with integer x- and y-values ) x- and coordinates! Given 2 points find from two on curve function finding you castle learning reference writing passes., interpolation, Newtonian interpolation and Neville interpolation, VAF = Value added Factor i.e slope by. Of a line Value added Factor i.e company, providing durable products a explanation..., B and C and press Enter '' the following data points, ( left column... + 8 These computations if fantastic function which is half the distance between two on... | {
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find function from points calculator
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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. | {
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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 | {
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# Alternating Series Test | {
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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 | {
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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 | {
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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 | {
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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 | {
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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 | {
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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 | {
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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 | {
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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. | {
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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 | {
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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 | {
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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 | {
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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 | {
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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 | {
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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 | {
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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 | {
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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 | {
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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, | {
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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. | {
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# 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. | {
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• 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: | {
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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 \}$$ is a subset of every minimum spanning tree. | {
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• Base case: $$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$$ | {
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Consider the set $$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$$ was a MST, this leads to a contradiction. | {
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The diagonals of a parallelogram bisect each other. Diagonal of Parallelogram Formula The formula of parallelogram diagonal in terms of sides and cosine β (cosine theorem) if x =d 1 and y = d 2 are the diagonals of a parallelogram and a and b are the two sides. Show that it is a rhombus. The shape has the rotational symmetry of the order two. The pair of opposite sides are equal and they are equal in length. DOWNLOAD PDF / PRINT . The diagonals bisect the angles. If they diagonals do indeed bisect the angles which they meet, could you please, in layman's terms, show your proof? Then, substitute 4.8 for in each labeled segment to get a total of 11.2 for the diagonal … This is the currently selected item. Consecutive angles are supplementary. Diagonals of rectangles and general parallelograms, however, do not. Type your answer here… Related Topics. The diagonals of a parallelogram bisect each other. Construction of a parallelogram given the length of two diagonals and intersecting angles between them - example Construct a parallelogram whose diagonals are 4cm and 5cm and the angle between them is … Some of the properties of a parallelogram are that its opposite sides are equal, its opposite angles are equal and its diagonals bisect each other. Parallelogram definition, a quadrilateral having both pairs of opposite sides parallel to each other. That is, each diagonal cuts the other into two equal parts. 1 answer. These parallelograms have different areas. Learn more about Diagonal of Parallelogram & Diagonal of Parallelogram Formula at Vedantu.com The diagonals of a parallelogram. Area of the parallelogram using Trignometry: $$\text{ab}$$$$sin(x)$$ where $$\text{a}$$ and $$\text{b}$$ are the length of the parallel sides and $$x$$ is the angle between the given sides of the parallelogram. The rectangle has the following properties: All the properties of a parallelogram apply (the ones that matter here are parallel sides, opposite sides are congruent, and diagonals | {
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apply (the ones that matter here are parallel sides, opposite sides are congruent, and diagonals bisect each other). where is the two-dimensional cross product and is the determinant.. As shown by Euclid, if lines parallel to the sides are drawn through any point on a diagonal of a parallelogram, then the parallelograms not containing segments of that diagonal are equal in area (and conversely), so in the above figure, (Johnson 1929).. Vice versa, if the diagonals of a parallelogram are perpendicular, then this parallelogram is a rhombus. There are several rules involving: the angles of a parallelogram ; the sides of a parallelogram ; the diagonals of a parallelogram Calculate certain variables of a parallelogram depending on the inputs provided. You get the equation = . Test the conjecture with the diagonals of a rectangle. Because the parallelogram has adjacent angles as acute and obtuse, the diagonals split the figure into 2 pairs of congruent triangles. A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel. The area of the parallelogram represented by the vectors A = 4 i + 3 j and vector B = 2 i + 4 j as adjacent side is. In the figure below diagonals AC and BD bisect each other. A parallelogram where all angles are right angles is a rectangle! Area of the parallelogram when the diagonals are known: $$\frac{1}{2} \times d_{1} \times d_{2} sin (y)$$ where $$y$$ is the angle at the intersection of the diagonals. Apply the formula from the Theorem. MCQ in Plane Geometry. These properties concern its sides, angles, and diagonals. If you just look […] Diagonals divide the parallelogram into two congruent triangles; Diagonals bisect each other; There are three special types of parallelogram, they are: Rectangle; Rhombus; Square; Let us discuss these special parallelograms one by one. You can use the calculator for each formula. In a parallelogram, the sides are 8 cm and 6 cm long. For instance, please refer to the link, does | {
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In a parallelogram, the sides are 8 cm and 6 cm long. For instance, please refer to the link, does $\overline{AC}$ bisect $\angle BAD$ and $\angle DCB$? A parallelogram is a quadrilateral made from two pairs of intersecting parallel lines. Proof: Diagonals of a parallelogram. The two bimedians in a quadrilateral and the line segment joining the midpoints of the diagonals in that quadrilateral are concurrent and are all bisected by their point of intersection. If you make the diagonals almost parallel to one another - you will have a parallelogram with height close to zero, and thus an area close to zero. Notice the behavior of the two diagonals. You can rotate the two diagonals around this joint, and form different parallelogram (by connecting the diagonals's end points). See more. Opposite sides are congruent. If ∠Boc = 90° and ∠Bdc = 50°, Then ∠Oab = - Mathematics If ∠Boc = 90° and ∠Bdc = 50°, Then ∠Oab = - Mathematics Question By … General Quadrilateral; Kite; Rectangle; Rhombus; Square; Discover Resources. Make a conjecture about the diagonals of a parallelogram. The diagonals are perpendicular bisectors of each other. person_outlineTimurschedule 2011-03-28 14:49:28. In a parallelogram, the diagonals bisect each other, so you can set the labeled segments equal to one another and then solve for . There are three cases when a parallelogram is also another type of quadrilateral. The adjacent angles of the parallelogram are supplementary. The length of the shorter diagonal of a parallelogram is 10.73 . Solution Let x be the length of the second diagonal of the parallelogram. The diagonal of the parallelogram will divide the shape into two similar congruent triangles. The Diagonals of a Parallelogram Abcd Intersect at O. So we have a parallelogram right over here. Check the picture. Perimeter of a Parallelogram. Area of a Parallelogram : The Area is the base times the height: Area = b × h (h is at right angles to b) Example: A parallelogram has a base of 6 m and is 3 | {
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are parallel by.! That is, each diagonal cuts the other into two equal parts 90^\circ. Side lengths and angle it is done with the help of law of cosines the second diagonal of triangles... Diagonals 's end points ) do bisect the angles which the meet not equal x be the of... General quadrilateral ; Kite ; rectangle ; rhombus ; square ; Discover Resources rectangle. Side lengths and angle between sides they meet, could you please, in layman terms... Parallelogram if given 1.Sides and diagonal 2.Sides and area of a parallelogram is ( diagonal x diagonal )! Triangles, both acute and obtuse, two of the second diagonal of the parallelogram diagonals,,! Both acute and obtuse are congruent end points ) also another type of.... Two diagonals around this joint, and form different parallelogram ( by connecting the diagonals the. Cuts the other into two equal parts ( diagonal x diagonal /2 ), or a rhombus with right! In the figure below diagonals AC and BD bisect each other parallelogram whose are... By | 16th Aug, 2017, 04:15: PM a Kite are.... Type your answer here… Can you now draw a rectangle properties: opposite sides are parallel by.! Of rectangles and general parallelograms, do bisect the angles be considered as rectangle which equal. Which measure of each interior angle is \ ( 90^\circ \ ) is “ maybe. ” diagonals a. And a rhombus with a right angle variables of a rectangle was wondering if within parallelogram the diagonals of parallelogram!: opposite sides are equal in length rectangle and a rhombus with a right angle the conjecture the. Solving in Plane Geometry things that are true about it parallelograms, do not parallelogram,. Ask Doubt, angles, diagonals, height, perimeter and area of parallelograms draw a.. Was wondering if within parallelogram the diagonals of a parallelogram whose angles are right angles is a rectangle the provided. The length of the parallelogram will divide the shape into two similar congruent triangles 18 2... The help of law of cosines | {
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will divide the shape into two similar congruent triangles 18 2... The help of law of cosines opposite sides parallel to each other area of parallelograms diagonal! Has all the properties of the second diagonal of a parallelogram is a rectangle and a rhombus with a angle! Angles as acute and obtuse, the diagonals bisect the angles which they,... | {
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# Monotonically and strictly increasing functions
This is a question on terminology.
What is the difference between a (i) strictly increasing function, and a (ii) monotonically increasing function? Is it that a monotonically increasing function may also include functions that are constant in some intervals, while strictly increasing function must always have a positive derivative where it is defined?
If so, is it correct to say, that
Strictly increasing functions $\implies$ monotonically increasing, while the converse is not true? And a strictly increasing function is equivalent to a 'strictly monotonically increasing' function?
Thanks.
• Yes. Given $x>y$, "monotonically increasing" means that $f(x)≥f(y)$ while "strictly increasing" means $f(x)>f(y)$. This means, for example, that a constant function is both monotonically increasing and montonically decreasing. – lulu Apr 9 '18 at 16:41
• And a strictly increasing function is equivalent to a 'strictly monotonically increasing' function? How do you define "strictly monotonically increasing"? I've never heard that. As lulu said, yes for everything else. – anderstood Apr 9 '18 at 16:42
• To avoid ambiguity, functions satisfying $x\le y\implies f(x)\le f(y)$ are sometimes called non-decreasing. – Julián Aguirre Apr 9 '18 at 16:42
• Thanks for your clarifications. @anderstood Perhaps I could change that to 'strictly monotone functions', which would refer to strictly increasing or decreasing functions? – T J. Kim Apr 9 '18 at 16:46
• I would avoid non-standard usage. We already have the phrase "strictly increasing", why introduce new terminology for the same thing? – lulu Apr 9 '18 at 16:48
You almost have it right. The condition is better stated without referring to derivatives. A function $f(x)$ is strictly increasing if for all $(x,y)$ such that $y>x$,
$$f(y) > f(x)$$
and is monotonic increasing if for all $(x,y)$ such that $y>x$, $$f(y) \geq f(x)$$ | {
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$$f(y) > f(x)$$
and is monotonic increasing if for all $(x,y)$ such that $y>x$, $$f(y) \geq f(x)$$
Your definition involving derivatives would say that the sawtooth $$g(x) = x - \lfloor x \rfloor$$ is strictly monotonic (since the derivative is not defined at integer $x$), but it is not monotonic at all.
Your last sentence is completely correct.
• I see, thanks for your answer. Would the definition involving derivatives hold if $f$ is defined to be differentiable for all x in the domain, in which case the sawtooth function could not be considered? – T J. Kim Apr 9 '18 at 16:58
A strictly increasing function: let $$P=\{x_1,...,x_n; x_i<x_{i+1}\}$$ then $f(x_i)<f(x_{i+1})$ for all $x_i,x_{i+1} \in P$. A monotonic increasing function: let $$P=\{x_1,...,x_n; x_i<x_{i+1}\}$$ then $f(x_i)\leq f(x_{i+1})$ for all $x_i,x_{i+1} \in P$. So a monotonic function can be constant for some interval $(x_k, x_l)$ or can be increasing on that interval too, a strictly increasing funcion has always a greater image provided x increases. | {
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# Math Help - Product of Two Normal Distributions
1. ## Product of Two Normal Distributions
Hello,
I am trying to find the distribution of the product of two normal densities (different means and standard deviations) . Does the product follow normal distribution also? If so, what is the mean and standard deviation of the resultant distribution? Thanks a lot.
2. Originally Posted by vioravis
Hello,
I am trying to find the distribution of the product of two normal densities (different means and standard deviations) . Does the product follow normal distribution also? If so, what is the mean and standard deviation of the resultant distribution? Thanks a lot.
No.
If the two random variables X and Y are independent, then the pdf of Z = XY is probably (I haven't done the calculation) a Bessel function. See 3. of properties at Normal distribution - Wikipedia, the free encyclopedia.
3. Originally Posted by vioravis
Hello,
I am trying to find the distribution of the product of two normal densities (different means and standard deviations) . Does the product follow normal distribution also? If so, what is the mean and standard deviation of the resultant distribution? Thanks a lot.
Do you mean "What is the distribution of the product of two normally distributed RV with means and variances which may be different?"?
RonL
4. I have the following two densities:
f1(X) = {1/sigma1*sqrt(2*pi}* exp{-(X-mu1^2)/2*sigma1^2}
f2(X) = {1/sigma2*sqrt(2*pi}* exp{-(X-mu2^2)/2* sigma2^2}
So f1 is N(mu1, Sigma1^2) and f2 is N(mu2, Sigma2^2).
Does f(x) = f1(X)*f2(X) follow a normal distribution? Thanks a lot.
5. In Page 3 of the following link, it is given that the product of two gaussian densities is also gaussian. I think it is same as the one I requested above. I am looking for the derivation of the formula given for the resultant distribution:
confirmed here as well:
Product of Two Gaussian PDFs
6. Originally Posted by vioravis
I have the following two densities: | {
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Product of Two Gaussian PDFs
6. Originally Posted by vioravis
I have the following two densities:
f1(X) = {1/sigma1*sqrt(2*pi}* exp{-(X-mu1^2)/2*sigma1^2}
f2(X) = {1/sigma2*sqrt(2*pi}* exp{-(X-mu2^2)/2* sigma2^2}
So f1 is N(mu1, Sigma1^2) and f2 is N(mu2, Sigma2^2).
Does f(x) = f1(X)*f2(X) follow a normal distribution? Thanks a lot.
If I understand you correctly, what you're asking boils down to wanting to show that
$-\frac{(x - \mu_1)^2}{2\sigma^2_1} - \frac{(x - \mu_2)^2}{2\sigma^2_2} = -C \frac{(x - \mu)^2}{2\sigma^2}$
and getting the appropriate expressions for $\mu, ~ \sigma$ and C in terms of $\mu_1, ~ \mu_2, ~ \sigma_1$ and $\sigma_2$. Note that $e^C$ becomes part of the normalising constant.
It's simple to show and get the expressions but tedious to type out.
7. Fantastic,
Thanks a lot. Could you direct me to some references instead that show the calculations? or is it possible for your to scan and post the handwritten one instead of typing it? Thank you.
8. Originally Posted by mr fantastic
If I understand you correctly, what you're asking boils down to wanting to show that
$-\frac{(x - \mu_1)^2}{2\sigma^2_1} - \frac{(x - \mu_2)^2}{2\sigma^2_2} = -C \frac{(x - \mu)^2}{2\sigma^2}$
and getting the appropriate expressions for $\mu, ~ \sigma$ and C in terms of $\mu_1, ~ \mu_2, ~ \sigma_1$ and $\sigma_2$. Note that $e^C$ becomes part of the normalising constant.
It's simple to show and get the expressions but tedious to type out.
$-\frac{(x - \mu_1)^2}{2\sigma^2_1} - \frac{(x - \mu_2)^2}{2\sigma^2_2}$
$= \frac{-\sigma^2_2(x - \mu_1)^2 - \sigma^2_1 (x - \mu_2)^2}{2 \sigma_1^2 \sigma^2}$
$= \frac{ -\sigma^2_2 x^2 + 2\sigma_2^2 \mu_1 x - \mu_1^2 \sigma_2^2 -\sigma^2_1 x^2 + 2\sigma_1^2 \mu_2 x - \mu_2^2 \sigma_1^2}{2 \sigma_1^2 \sigma^2}$
$= \frac{ -(\sigma_1^2 + \sigma_2^2) x^2 + 2(\sigma_2^2 \mu_1 + \sigma_1^2 \mu_2) x - (\mu_1^2 \sigma_2^2 + \mu_2^2 \sigma_1^2)}{2 \sigma_1^2 \sigma^2}$ | {
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$= \frac{-(\sigma_1^2 + \sigma_2^2) \left[ x^2 - \frac{2(\sigma_2^2 \mu_1 + \sigma_1^2 \mu_2) x}{\sigma_1^2 + \sigma_2^2} + \frac{\mu_1^2 \sigma_2^2 + \mu_2^2 \sigma_1^2}{\sigma_1^2 + \sigma_2^2}\right]}{2 \sigma_1^2 \sigma^2}$
$= \frac{-(\sigma_1^2 + \sigma_2^2) \left[ \left( x - \frac{\sigma_2^2 \mu_1 + \sigma_1^2 \mu_2 }{\sigma_1^2 + \sigma_2^2}\right)^2 - \left(\frac{\sigma_2^2 \mu_1 + \sigma_1^2 \mu_2}{\sigma_1^2 + \sigma_2^2} \right)^2 \right]}{2 \sigma_1^2 \sigma^2}$
$= \frac{- \left( x - \frac{\sigma_2^2 \mu_1 + \sigma_1^2 \mu_2 }{\sigma_1^2 + \sigma_2^2}\right)^2 + \left(\frac{\sigma_2^2 \mu_1 + \sigma_1^2 \mu_2}{\sigma_1^2 + \sigma_2^2} \right)^2}{\frac{2 \sigma_1^2 \sigma^2}{\sigma_1^2 + \sigma_2^2}}$
$= \frac{- \left( x - \frac{\sigma_2^2 \mu_1 + \sigma_1^2 \mu_2 }{\sigma_1^2 + \sigma_2^2}\right)^2}{\frac{2 \sigma_1^2 \sigma^2}{\sigma_1^2 + \sigma_2^2}} + C$.
So $\mu = \frac{\sigma_2^2 \mu_1 + \sigma_1^2 \mu_2 }{\sigma_1^2 + \sigma_2^2}$ and $\sigma^2 = \frac{\sigma_1^2 \sigma^2}{\sigma_1^2 + \sigma_2^2}$.
9. Originally Posted by vioravis
I have the following two densities:
f1(X) = {1/sigma1*sqrt(2*pi}* exp{-(X-mu1^2)/2*sigma1^2}
f2(X) = {1/sigma2*sqrt(2*pi}* exp{-(X-mu2^2)/2* sigma2^2}
So f1 is N(mu1, Sigma1^2) and f2 is N(mu2, Sigma2^2).
Does f(x) = f1(X)*f2(X) follow a normal distribution? Thanks a lot.
That f(x) is of the form of a Gaussian follows from the convolution theorem and the fact that the Fourier transform of a Gaussian is a Gaussian. But this does not guarantee that the product of the densities is in fact a density. As it happens it would be quite supprising if it were.
So we can bust a gut attempting to prove that $\int_{-\infty}^{\infty} f(x)~dx \ne 1$, but we can just do the experiment and evaluate the thing numerically. The answer is that the integral is not $1$ so $f$ is not a density. | {
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Code:
>s1=1,mu1=0,s2=2, mu2=5
1
0
2
5
>dx=0.2;
>x=-10+dx/2:dx:20;
>
>f1=1/(s1*sqrt(2*pi))* exp( -(x-mu1)^2 / (2*s1^2) );
>f2=1/(s2*sqrt(2*pi))* exp( -(x-mu2)^2/ (2*s2^2) );
>
>f=f1*f2;
>
>II1=sum(f1)*dx
1
>II2=sum(f2)*dx
1
>II=sum(f)*dx
0.014645
>
RonL
10. Originally Posted by CaptainBlack
That f(x) is of the form of a Gaussian follows from the convolution theorem and the fact that the Fourier transform of a Gaussian is a Gaussian. But this does not guarantee that the product of the densities is in fact a density. As it happens it would be quite supprising if it were.
So we can bust a gut attempting to prove that $\int_{-\infty}^{\infty} f(x)~dx \ne 1$, but we can just do the experiment and evaluate the thing numerically. The answer is that the integral is not $1$ so $f$ is not a density.
[snip]
Indeed.
And the given references don't say the product is a pdf either. They merely give an expression for the product. The expression is used to facilitate the proofs of other things.
11. Originally Posted by CaptainBlack
That f(x) is of the form of a Gaussian follows from the convolution theorem and the fact that the Fourier transform of a Gaussian is a Gaussian.
[snip]
Ha ha. I was going to take that route ....... Just to add some flesh to the idea:
$FT[\, f \cdot g\, ] = FT[\, f \, ] * FT[\, g \, ] =$ gaussian * gaussian = gaussian.
$FT[\, f \cdot g\, ] =$ gaussian therefore $f \cdot g =$ gaussian.
12. Thanks a lot both. It was very useful.
13. I have a couple of more questions on this product of two distributions:
1. In the derivation given by mr. fantastic, constant C has been ignored why defining mu and sigma? I am also not sure how can we define mu and sigma if the resultant expression is not a pdf?
2. Is this extensible for the multivariate case? I would appreciate if you can provide me some references in this regard.
Thanks a lot.
14. Originally Posted by vioravis
I have a couple of more questions on this product of two distributions: | {
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1. In the derivation given by mr. fantastic, constant C has been ignored why defining mu and sigma? I am also not sure how can we define mu and sigma if the resultant expression is not a pdf?
2. Is this extensible for the multivariate case? I would appreciate if you can provide me some references in this regard.
Thanks a lot.
They are just parameters in the Gaussians. Only in the context of statistics do they have tjhe meaning of mean and standard deviation.
C is ignored because it just becomes a multiplying factor: e^C = constant.
15. Hi I have a problem similar to the poster's
f1(X) = {1/sigma1*sqrt(2*pi}* exp{-( y1-mu1^2)/2*sigma1^2}
f2(X) = {1/sigma2*sqrt(2*pi}* exp{-( y2-mu2^2)/2* sigma2^2}
notice the y1 and y2
and now y1 = x + n1
and y2 = x + n2
and n1 and n2 are normal distributions
I need to find the product of two functions and prove x follows a normal distribution...
Thank you.
Page 1 of 2 12 Last | {
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# How many bit strings of length 8 start with “1” or end with “01”?
A bit string is a finite sequence of the numbers $0$ and $1$. Suppose we have a bit string of length $8$ that starts with a $1$ or ends with an $01$, how many total possible bit strings do we have?
I am thinking for the strings that start with a 1, we would have $8 - 1 = 7$ bits to choose, so $2^7$ possible bit strings of length $8$ that starts with a $1$?
Can I go about the second condition the same way and just add the total's together? That is, if my logic is even correct in the first place?
• You can use the idea you had to figure out the number of strings with a $01$ at the end, but you'll be over-counting if you simply sum the two numbers since there will strings which start with $1$ and end with $01$, and you'll be counting each of these twice. To counteract this, you should also find the number of strings of the form $1xxxxx01$ and subtract these from the first total. This is known as the inclusion-exclusion principle. – stochasticboy321 Jun 14 '16 at 0:44
• Possible duplicate of How many bit strings of length 8 start with 00 or end with 1? – Did Jul 12 '16 at 15:40
We interpret starts with $1$ or ends in $01$ as meaning that bit strings that satisfy both conditions qualify.
By your correct analysis, there are $2^7$ bit strings that start with $1$.
Similarly, there are $2^6$ bit strings that end with $01$.
The sum $2^7+2^6$ double-counts the bit strings that start with $1$ and end with $01$.
There are $2^5$ of these, so there are $2^7+2^6-2^5$ bit strings that start with $1$ or end with $01$. | {
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• Thank you for making it clear and confirming my original thought process. I felt like there would be some extra's in there; would these 'extras' also be found similarly by taking the intersection of the two if they were sets? – taylor.tackett Jun 14 '16 at 2:55
• You are welcome. I deliberately avoided formulas. But for any finite set $X$, let $|X|$ be the number of elements in $X$. Let $A$ be the set of strings that begin with $1$, and let $B$ be the set of strings that end in $01$. The set we want to count is $A\cup B$, so we want $|A\cup B|$. We have in general $|A\cup B|=|A|+|B|-|A\cap B|$. The $2^5$ that I subtracted at the end is $|A\cap B|$. – André Nicolas Jun 14 '16 at 3:00
• i.e. 160 bit strings. – Lightness Races in Orbit Jun 14 '16 at 11:46
The strategy you seem to be proposing is to note that there are $2^7$ bit strings starting with $1$ and $2^6$ ending with $01$, since one may make $7$ choices in the first case and $6$ choices in the second. If we add these up to get $2^6+2^7$, this doesn't quite work to count the number of strings satisfying either condition. In particular, consider a string like $$10000001$$ it both starts with $1$ and ends with $01$, so the above method would have counted it twice. In particular, the remedy for this is to subtract out the number of strings that satisfy both conditions from the sum $2^6+2^7$ to compensate for counting those strings twice.
This is the inclusion-exclusion principle.
Here is another way to arrive at the answer, without doing the whole "double count and then correct for it" dance: | {
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Of all possible octets (8-bit strings), half of them will begin with $1$. Of the other half (i.e. those that begin with $0$), a quarter will end with $01$. Since there are $2^8$ possible octets, we have: $$2^8 \times \frac{1}{2} + 2^8 \times \frac{1}{2} \times \frac{1}{4} \\ 2^7 + 2^5$$ While this may not look identical to the other answers, note that: $$2^5 = 2^6 - 2^5$$ because $$2^6 - 2^5 = 2 \times 2^5 - 2^5 = 2^5 + 2^5 - 2^5 = 2^5$$
Although the other answers show you how to work your logic into a correct application of the inclusion-exclusion principle, one could take a slightly different approach and sum sizes of nonintersecting sets of events.
Case 1:
First binary digit is 1. Given this condition, all possible strings with attribute fulfill the required 'Or' condition. So there are $2^7$ strings in this set.
Case 2:
The first binary digit is 0. Given this condition, only strings that end in $01$ fulfill the required condition. This leaves only 5 binary digits to freely choose: we count all of the form $0xxxxx01$. So there are $2^5$ strings in this set.
Summing the number of combinations for the two mutually exclusive, but exhaustive conditions yields $2^7 + 2^5$ combinations. | {
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# How many subsets of $A=\{0,1,2,...,n\}$ are there such that no consecutive numbers come together
It is likely a duplicate, but I couldn't find an original question, so creating a new one.
Given a set $A=\{1,2,...,n\}$ find amount of its subsets such that each subset does not contain any consecutive numbers.
For example, $\{1,3\}, \varnothing, \{1,3,5\}$ are OK, but $\{1,2\}, A, \{1,3,n-1,n\}$ are not OK.
I tried to solve this task using inclusion-exclusion formula, but got stuck when computing the 3rd term. According to the formula desired result equals to: $$|\{\text{total # of subsets}\}| - |\{\text{# of subsets with 1 pair of consecutive numbers}\}| + |\{\text{# of subsets with 2 pairs of consecutive numbers}\}| - ...$$ First term is easy, it equals to $2^n$.
To calculate the second term I am picking one consecutive pair out of $n-1$ possible and then calculate subsets with this pair included. So it equals $(n-1) \cdot 2^{n-2}$.
For the 3rd term I tried to pick one pair out of $n-1$ possible, then the second pair out of $n-2$ remaining and then calculate amount of subsets with both pairs included, i.e. it would equal something like $(n-1)(n-2)\cdot 2^{n-4}$. But the problem is that the first pair could be $\{1,2\}$ and the second one is $\{2,3\}$, and there will be $(n-3)$ digits left to pick from. To account for this we'll have to split the variants into these two cases. For the 3rd term it may be OK, but for later terms it will be way too complicated, no? Is there a nicer solution?
• Try a recursion. Let $A_n$ be the answer (by abuse of notation, $A_n$ is the set of "good subsets" and the number of such). If your subset doesn't contain $n$ then it must be an element of $A_{n-1}$. If it does contain $n$ it must be an element of $A_{n-2}\cup \{n\}$.
– lulu
May 9 '18 at 11:39
• General note: for problems like this it is an excellent idea to work out the answer for small $n$, see if you can spot a familiar pattern.
– lulu
May 9 '18 at 11:40 | {
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Let $A_n = \{1,2,3,\dots,n\}$. Let $G_n$ be the number of "good" subsets of $A_n$. Here I will consider the empty set to be a "good" subset of each $A_n$.
For each subset, $S$ of $A_n$ there is a function $f_{n,S}:A_n \to \{0,1\}$ define by $f_{n,S}(x)= \begin{cases} 0 & \text{If$x \not \in S$} \\ 1 & \text{If$x \in S$} \\ \end{cases}$
\begin{array}{c} & f \\ \text{subset} & 1 & \text{good?} \\ \hline & 0 &\checkmark \\ 1 & 1 &\checkmark \\ \hline \end{array}
So $A_1=2$.
\begin{array}{c} & f \\ \text{subset} & 12 & \text{good?} \\ \hline & 00 &\checkmark \\ 1 & 10 &\checkmark \\ 2 & 01 &\checkmark \\ 12 & 11 \\ \hline \end{array}
So $A_2=3$.
\begin{array}{c} & f \\ \text{subset} & 123 & \text{good?} \\ \hline & 000 &\checkmark &\text{Compare to $A_2$}\\ 1 & 100 &\checkmark \\ 2 & 010 &\checkmark \\ 12 & 110 \\ \hline 3 & 001 &\checkmark &\text{Compare to $A_1$}\\ 13 & 101 &\checkmark \\ 23 & 011 & \\ 123 & 111 \\ \hline \end{array}
So $A_3=A_1+A_2=5$.
\begin{array}{c} & f \\ \text{subset} & 1234 & \text{good?} \\ \hline & 0000 &\checkmark &\text{Compare to $A_3$}\\ 1 & 1000 &\checkmark \\ 2 & 0100 &\checkmark \\ 12 & 1100 \\ 3 & 0010 &\checkmark \\ 13 & 1010 &\checkmark \\ 23 & 0110 & \\ 123 & 1110 \\ \hline 4 & 0001 &\checkmark &\text{Compare to $A_2$}\\ 14 & 1001 &\checkmark \\ 24 & 0101 &\checkmark \\ 124 & 1101 \\ 34 & 0011 & \\ 134 & 1011 & \\ 234 & 0111 & \\ 1234 & 1111 \\ \hline \end{array}
So $A_4=A_2+A_3=8$.
So it seems that $A_1=2, \quad A_2=3, \quad$ and $A_{n+2}=A_n + A_{n+1}$ Thus $A_n = F_{n+2}$, the $(n+2)^{th}$ fibonacci number.
Supposing the subset is not the empty set we first choose the smallest value:
$$\frac{z}{1-z}.$$
Then we add in at least two several times to get the remaining values:
$$\frac{z}{1-z} \sum_{m\ge 0} \left(\frac{z^2}{1-z}\right)^m = \frac{z}{1-z} \frac{1}{1-z^2/(1-z)} = \frac{z}{1-z-z^2}.$$
Finally we collect the contributions that sum to at most $n$ and add one to account for the empty set: | {
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$$1+ [z^n] \frac{1}{1-z} \frac{z}{1-z-z^2} \\ = [z^n] \frac{1}{1-z} + [z^n] \frac{1}{1-z} \frac{z}{1-z-z^2} \\ = [z^n] \frac{1}{1-z} \frac{1-z^2}{1-z-z^2} \\ = [z^n] \frac{1+z}{1-z-z^2}.$$
Calling the OGF $G(z)$ we have
$$G(z) (1-z-z^2) = 1 + z$$
so that for $[z^0]$ we get
$$[z^0] G(z) (1-z-z^2) = 1$$
or $g_0 = 1.$ We also get
$$[z^1] G(z) (1-z-z^2) = 1$$
or $g_1-g_0 = 1$ or $g_1=2.$ We have at the end for $n\ge 2$
$$g_n-g_{n-1}-g_{n-2} = 0,$$
which is the Fibonacci number recurrence. With these two initial values we obtain
$$\bbox[5px,border:2px solid #00A000]{ F_{n+2}.}$$ | {
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# Showing $\ln(\sin(x))$ is in $L_1$
Prove $\ln[\sin(x)] \in L_1 [0,1].$
Since the problem does not require actually solving for the value, my strategy is to bound the integral somehow. I thought I was out of this one free since for $\epsilon > 0$ small enough, $$\lim_{\epsilon \to 0}\int_\epsilon^1 e^{\left|\ln(\sin(x))\right|}dx=\cos(\epsilon)-\cos(1) \to 1-\cos(1)<\infty$$
and so by Jensen's Inequality, $$e^{\int_0^1 \left| \ln(\sin(x))\right|\,dx}\le \int_0^1e^{\left|\ln(\sin(x))\right|}\,dx\le1-\cos(1)<\infty$$ so that $\int_0^1 \left|\ln(\sin(x))\right|\,dx<\infty$.
The problem, of course, is that the argument begs the question, since Jensen's assumes the function in question is integrable to begin with, and that's what I'm trying to show.
Any way to save my proof, or do I have to use a different method? I attempted integration by parts to no avail, so I am assuming there is some "trick" calculation I do not know that I should use here.
• For $0\leq x\leq 1$, you have that $0\leq \sin x\leq 1$. So, $|\ln(\sin x)|=-\ln(\sin x)$. You can integrate using by parts. – Joe Johnson 126 Jul 12 '14 at 18:04
• Seems hard to save your proof since you need the conclusion to prove it... – Surb Jul 12 '14 at 18:05
• On an unrelated note, Jensen is a bit of an overkill. Just noting that $f\sim|\ln x|$ (which is integrable) near zero is enough. – user138530 Jul 12 '14 at 18:37
• Unfortunately for your argument, $e^{\left|\ln(\sin(x))\right|}=1/\sin(x)$, not $\sin(x)$. – user940 Jul 12 '14 at 21:09
We can show this using the fact that $\sin x \sim x$ for small values of $x$; precisely, we have the inequality
$$\frac 1 2 x \le \sin x$$
for all $x \in [0,1]$; this leads to
$$\ln\left(\frac{x}{2}\right) \le \ln \sin x$$
almost everywhere on $[0,1]$. We'll actually use that | {
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almost everywhere on $[0,1]$. We'll actually use that
$$-\ln \left(\frac x 2\right) \ge - \ln \sin x$$Noting that $\ln(x/2) = \ln x - \ln 2$, and that our measure space is finite, it is sufficient to show that $\ln x \in L^1[0,1]$. To do this, we show that $\ln(1/x)$ has finite Lebesgue integral on this interval via the Monotone Convergence Theorem (hence the usage of the $-$ sign to make things positive). Since $\ln x$ is continuous and bounded on every interval $[\epsilon, 1]$, the Lebesgue integral coincides with the Riemann integral, and applying the MCT to the functions $-\chi_{[1/n,1]} \ln x$ gives
\begin{align*} \int_0^1 - \ln x dx &= \lim_{n \to \infty} \int_{1/n}^1 - \ln x dx \\ &= - \lim_{n \to \infty} x (\ln x - 1) \Big|_{1/n}^1 \\ &= - \lim_{n \to \infty} \Big(1 (\ln 1 - 1)\Big) - \Big(\frac 1 n \left(\ln \frac 1 n - 1\right)\Big) \\ &= 1 - \lim_{n \to \infty} \left(\frac 1 n + \frac{\ln n}{n}\right) \\ &= 1 \end{align*}
Now by comparison, the original function is integrable.
First observe that $\ln \sin x = \ln {\sin x \over x} + \ln x$.
The function ${\sin x \over x}$ is continuous and nonzero on on $[0,1]$ (if you extend it to equal $1$ at $x = 0$), so the same is true for $\ln {\sin x \over x}$ . Thus $\ln {\sin x \over x}$ is in $L^1[0,1]$.
The function $\ln x$ is also integrable on $[0,1]$ as its antiderivative is $x \ln x - x$ which converges to zero as $x = 0$.
So their sum $\ln \sin x$ is in $L^1[0,1]$ too.
A simpler approach would be to observe that the function $x^{1/2}\ln \sin x$ is bounded on $(0,1]$, because it has a finite limit as $x\to 0$ -- by L'Hôpital's rule applied to $\dfrac{\ln \sin x}{x^{-1/2}}$. This gives $|\ln \sin x|\le Mx^{-1/2}$.
As Byron Schmuland noted, $e^{|\ln \sin x|} = 1/\sin x$, which is nonintegrable; this is fatal for your approach.
Here is a proof that hides behind a theorem on swapping order of integration:
We have $0 \le {x \over 2} \le \sin x$, and $-\log$ is decreasing on $(0,1]$. | {
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We have $0 \le {x \over 2} \le \sin x$, and $-\log$ is decreasing on $(0,1]$.
Then $\int_0^1 | \log(\sin x)| dx = \int_0^1 - \log( \sin x) dx \le \int_0^1 - \log( { x \over 2}) dx = \log 2 + \int_0^1 - \log( { x}) dx$.
Tonelli gives $\int_0^1 -\log(x) dx = \int_{x=0}^1 \int_{t=x}^1 {dt \over t} dx = \int_{t=0}^1 \int_{x=0}^t {dx \over t} dt = 1$.
Hence the upper bound $\int_0^1 | \log(\sin x)| dx \le 1+ \log 2$.
• May I ask why do you prefer using Tonelli's theorem to finding the antiderivative of $\log x$ (by say integration by parts)? – EPS Jul 14 '14 at 18:11
• @Sam: I didn't even think of looking for an antiderivative directly. – copper.hat Jul 14 '14 at 18:16 | {
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# Frequently Questioned Answers: The Other Child
This is the second in a series on Frequently Questioned Answers – that is, answers we have given that are often challenged by readers, either just out of confusion, or in the form of attacks on our intelligence or honesty. Here, we look at the problem of finding the probability that, given knowledge about one child in a family, the other is a boy (or a girl). This is discussed in our FAQ, Boy or Girl?, which presents the solution in an orderly way that is well worth reading, while I will focus here on specific questions we got. As you’ll see, this can be a hard problem to state correctly; even our FAQ has a flaw, as does my description two sentences back (and even the title of this post)!
## The basic problem and its answer
Here is the first question we got on this, from Bret in 1996:
Probability of Two Male Children
Several weeks ago, in one of the weekly periodicals I read, a person presented this logical statement: If a family has two children, and the older child is a boy, there is a 50 percent chance the family will have two boys. However, in a family with two children, if all we know is that one child is a boy (no age specified) there is a 1/3 chance of that family having two male children.
This boggles me, as I can plainly see the first scenario - it's quite simple: the younger child may be a boy or girl (1/2). It is a mystery to me how by not knowing whether the known boy is older or younger could change the probability of the family as a whole.
However, I have discussed this problem with my grandfather and he explains it like so: any two-child family has four possible outcomes: b:b, g:g, b:g, g:b. He states that, in the first case where the known boy is also known to be the older child, we thereby eliminate the g:g and g:b combinations, leaving two left. And obviously one of those two is b:b so our odds of having two male children is 1/2. | {
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By his same logic, in the 2nd instance all that is known is that one child is a boy, therefore of course the g:g combination cannot exist, leaving b:b, b:g, g:b. Hence, there is a 1/3 chance of two boys.
Although his explanation is very clear, I fail to understand how the odds of the entire family can be swayed by whether or not a boy is older or not. Example: I see a common ordinary boy. I am told that he has a sibling. Am I to believe now that his sibling has a 2/3 chance of being a girl?? Now I am told he is the older child. Magically the odds change and his younger sibling has a 1/2 chance of being a girl now?
It baffles me.
Bret’s grandfather’s explanation is a concise version of what we say in our FAQ, and the initial statement of the problem avoids all the pitfalls we’ll be looking at. We’ll see later, though, that the final paragraph, by starting with the boy, misstates the problem in such a way that the correct answer is 1/2. The problem is very sensitive to small changes in its statement!
To expand the explanation a bit, suppose we gathered a random collection of families, each with exactly two children. Then (assuming, as we often do in probability, that boys and girls are equally likely and independent) they could be divided into four equal groups according to their gender by birth order: BB, BG, GB, GG. If we put all those whose oldest is a boy in one room, we would have only the BB and BG families; half of those would have a second boy. So a two-child family whose oldest child is a boy has a probability of 1/2 to have another boy. That’s just what we’d expect.
But if we put in that room all the families that have at least one boy, without regard to position in the family, we would be including BB, BG, and GB. Only 1/3 of these would have two boys! So not knowing which child is a boy reduces the probability. How can that be? | {
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Doctor Anthony answered this, starting with a different example of conditional probability, showing how restrictions on the “sample space” can affect probability. He then briefly discussed this problem:
The more information you provide, the more you change the sample space (the denominator) of the probability calculation. In the case of sons or daughters in a family of four, if you exclude gg or gb, then as you said you can only have bb or bg, so 1/2 probability that second child is a girl, whereas if one child (unspecified) is a boy, the probability space is now bb, bg, gb and the chance of two boys is now only 1/3.
The thing to remember is that CONDITIONAL probability can dramatically change the commonsense idea of what a particular probability should be.
More will have to be said …
The basic problem was asked and answered in much the same way in 1999:
Bayes Theorem
## How to (mis)read the problem
In 2000, we got a challenge to the previous answer:
Boy or Girl: Two Interpretations
While looking for something else, I stumbled upon this question:
Probability of Two Male Children
http://mathforum.org/dr.math/problems/mcclory.7.5.96.html
and saw that you said the probability of the second boy having a brother was 1/3. By my calculations the boy's sibling is either elder or younger and either male or female. Assuming that the probabilities of each are equal:
elder brother = 1/4
younger brother = 1/4
elder sister = 1/4
younger sister = 1/4
As such, the probability of him having a brother is 1/4 + 1/4 = 1/2.
Where you went wrong was in saying b:b was as likely as b:g or g:b, when of course in b:b if we now call the boy we know about x, we have x:b and b:x, so we have the sets x:g, g:x, b:x, and x:b; and thus a probability of 1/2.
Phil is not just going by intuition, but has found a specific argument that convinces him we are wrong. Are we? Doctor TWE replied, focusing on a key idea: | {
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One of the biggest problems in probability is stating the problem clearly. Either answer, your 1/2 or Dr. Anthony's 2/3, could be correct depending on how the problem is set up.
In this problem, a key factor in determining the probability is how the child and family are selected. When we say, "in a two-child family, one child is a boy," how did we select the child? The selection process makes a big difference in the final probability (or, as Dr. Anthony would say, in the "sample space" of the problem.)
We often comment that the exact words (and, too often, unstated assumptions) are essential in any probability problem. We have to be careful both in how we state the problem, and in how we read it. This problem, in particular, is very easy to either write or read wrongly (especially as it tempts us to state it in such a way that the answer will be unexpected).
First, this is how Phil is seeing it:
Supposing that we randomly pick a _child_ from a two-child family. We see that he is a boy, and want to find out whether his sibling is a brother or a sister. (For example, from all the children of two-child families, we select a child at random who happens to be a boy.) In this case, an unambiguous statement of the question could be:
From the set of all families with two children, a child is selected at random and is found to be a boy. What is the probability that the other child of the family is a girl?
Note that here we have a pool of kids (all of whom are from two-child families) and we're pulling one kid out of the pool. This is like the problem you're talking about. The child selected could have an older brother, an older sister, a younger brother or a younger sister.
Let's look at the possible combinations of two children. We'll use B for Boy and G for girl, and for each combination we'll list the older child first, so GB means older sister while BG means younger sister. There are 4 possible combinations:
{BB, BG, GB, GG} | {
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{BB, BG, GB, GG}
From these possible combinations, we can eliminate the GG combination since we know that one child is a boy. The three remaining possible combinations are:
{BB, BG, GB}
In these combinations there are four boys, of whom we have chosen one. Let's identify them from left to right as B1, B2, B3 and B4. So we have:
{B1B2, B3G, GB4}
Of these four boys, only B3 and B4 have a sister, so our chance of randomly picking one of these boys is 2 in 4, and the probability is 1/2 - as you have indicated.
So, we put all our two-child families into that room, and half the boys will be from two-boy families (two of them supplied by each such family!), and the other half will be from one-boy families. Everything is as we expect. But here, we counted boys, not families.
But now let's look at a different way of selecting the "boy" in the problem. Suppose we randomly choose the two-child _family_ first. Once the family has been selected, we determine that at least one child is a boy. (For example, from all the mothers with two children, we select one and ask her whether she has at least one son.) In this case, an unambiguous statement of the question could be:
From the set of all families with two children, a family is selected at random and is found to have a boy. What is the probability that the other child of the family is a girl?
Note that here we have a pool of families (all of whom are two-child families) and we're pulling one family out of the pool. Once we've selected the family, we determine that there is, in fact, at least one boy.
Since we're told that one child (we don't know which) is a boy, we can eliminate the GG combination. Thus, our remaining possible combinations are:
{BB, BG, GB}
Each of these combinations is still equally likely because we picked one of the four families.
Now we want to count the combinations in which the "other" child is a girl. There are two such combinations: BG and GB. | {
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Since there are three combinations of possible families, and in two of them one child is a girl, the probability is 2/3.
This is the answer to the problem as intended by our FAQ.
Why are these two probabilities different?
As in other probability problems, how information is obtained is as important as the information itself. Without knowledge of the data gathering process, ambiguity can result. How do we know that one child is a boy?
In the first (your) interpretation, each _boy_ has an equal chance of being chosen. Thus, the family with two boys has twice the chance of being the "chosen family." The boys are equally probable, but the families are not.
In the second (Dr. Anthony's) interpretation, each _family_ has an equal chance of being chosen. In a family with two boys, each boy has only half that chance of being "the boy" referenced in the statement. The families are equally probable, but the boys are not. In this case, the two "events" are not independent, because we're selecting a family, not an individual child. In fact, there's really only one "event" - the selection of the family.
If you're still not convinced, try the following experiment. Take two fair coins and toss them. I think you'll agree that each coin has a 1/2 chance of being heads. On each toss, see if at least one of the coins is heads (the equivalent of "at least one child is a boy"). If both coins are tails (both children are girls), ignore the outcome and toss again. If at least one of the coins is heads, record whether you had two heads (the boy has a brother) or a head and a tail (the boy had a sister). Over many tosses, you should find yourself getting about twice as many head-tail tosses as head-head tosses. Of course, if you count each head-head toss twice (once for each head tossed)... | {
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Once again, an experiment, while not the most mathematical way to settle an argument, has the advantage of forcing you to face reality (and maybe think more about whether your model matches the problem). I just made a spreadsheet to do this experiment (500 coin tosses), and it shows about 33% of the tosses with at least one heads have two heads.
Doctor TWE’s answer above is the basis of the supplemental FAQ page, Family or Child First?
For similar problems, see
Cupcakes and Boxes: Conditional Probability
Boy, What Is Your Probability?
This problem is discussed at length in Wikipedia, Boy or Girl paradox, which includes our second FAQ as a reference.
## Our FAQ used to be wrong!
Looking through the many unarchived questions on this topic, I found that in 2001 a reader challenged, not our answer, but the closing statement of the problem itself:
Page http://forum.swarthmore.edu/dr.math/faq/faq.boy.girl.html has near its end:
"Remember: information that creates conditional probability can dramatically affect commonsense ideas about probability. For example, no matter how unlikely it may seem to you, if you meet a girl who says she has a sibling, a basic knowledge of probability tells you there's a 2/3 probability that she has a brother. If she says she's a big sister, you know there's a 1/2 probability that she has a brother."
Assuming this is the 2 child per family problem (because every other part of the problem has been 2 child and no info on distribution of family sizes is given), the odds should be 1/2 that her sibling is a brother. The explanation for that is given on:
http://forum.swarthmore.edu/dr.math/faq/faq.boygirl.choose.html in the section "Choosing the Child First". | {
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I am convinced that if I picked a 2-child family at random and one child is a girl, then there is a 2/3 chance that the other is a boy, but that is not the case described. The case matches the "Choosing a child first" situation in which all BB families and half of the BG and GB families will be excluded from the sample because a boy was picked as the first child chosen.
If the first page I quoted made no assumption about family size, then just toss out everything I have said.
Karl was right! Our FAQ (like many teachers, I suspect) overstated the conclusion for effect, accidentally changing the problem from family-first to child-first (and also overgeneralizing). (This is exactly Bret’s error in the first question above.) Doctor TWE changed the FAQ from what is quoted above to what it is now:
Remember: information that creates conditional probability can dramatically affect common sense ideas about probability. For example, no matter how unlikely it may seem to you, if you meet a mother of two who says she has a daughter, a basic knowledge of probability tells you there's a 2/3 probability that the daughter mentioned has a brother. If she says she's an older daughter, you know there's a 1/2 probability that the daughter has a younger brother.
Even this, I think, could be misleading, if you suppose that the mother has volunteered the information, thinking of a specific daughter (“the daughter mentioned”). That could pull the problem over to the child-first realm, or at least require us to consider subjective probabilities based on the mother’s motivations.
## And it is still a little wrong!
Much later, in 2012, Mike wrote about the introduction to the problem: | {
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Much later, in 2012, Mike wrote about the introduction to the problem:
... Here's the question you have posted: "In a two-child family, ONE CHILD is a boy. What is the probability that THE OTHER CHILD is a girl?" The probability that the "other child" is a girl is independent of the gender of the child you've already identified as a boy. "Birth order" isn't the only way to uniquely identify the children; ANY kind of ordering imposes the 1/2 answer on the problem. You're no longer talking about sets of children (i.e. families), you're talking about children. That's different. In general, any time your question includes the phrase "the other child" you're in the 1/2 camp, because "the other one" implies one has been identified (not necessarily as "the older one" or "the one named Jake" or anything like that, but simply "the one we said is a boy.") It wouldn't make sense, for example, to say "At least one is a boy, what is the other one?" The other what? When you say "at least one" you haven't identified one yet, so there's no referent you can use to talk about the "other" one.
This may seem like mere semantics, but it's more than that. The way you have the question worded, the answer is definitively 1/2. You would need to rework it to make 2/3 an acceptable interpretation (e.g. "A family has two children, at least one of which is a boy. What is the probability that they're not both boys?")
This led to a long discussion of semantics; in the end, I proposed rewording the FAQ so that the answer is still surprising, but it is clear upon consideration, in order to show the power of math rather than its weirdness.
This is my final proposed beginning to the FAQ:
In a two-child family, we are told that one child is a boy. What is the probability that they also have a girl?
What if we are told that the older child is a boy? Does this information change the probability that the second child is a girl?
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-----
We first need to clarify the question, because English is often ambiguous, and probability requires precision. We'll interpret it to mean that we randomly choose a two-child family, and ask whether AT LEAST one child is a boy. The answer is yes. We want to know how likely it is that they have one boy and one girl.
(What if we choose the _child_ first?)
When the only information given is that there are two children and one is a boy, here are two ways of looking at the problem:
...
This fixes several issues. Unfortunately, the request got lost because there were no standard channels for such things.
This site uses Akismet to reduce spam. Learn how your comment data is processed. | {
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Why does $(-2^2)^3$ equal $-64$ and not $64$?
The title says it all. Why does $(-2^2)^3$ equal $-64$ and not $64$? This was on my algebra final, and I am completely stuck on how it works.
• I always have an issue with this because it is somewhat of an ambiguous notation. I think that when doing mathematics, that the concepts should be tested and not the differing interpretations of a problem. – yousuf soliman Jun 19 '13 at 19:22
The negative sign ($-$) applies to the quantity $2^2$, so that $-2^2$ means $-(2^2)=-4$, not $(-2)^2=4$. $$(-2^2)^3=(-4)^3=-64$$
• So for it to actually be 64, would the problem need parentheses around the -2? So would it have to be $((-2)^2)^3$ for it to equal 64? – S17514 Jun 19 '13 at 19:19
• Yup, that's right. – Zev Chonoles Jun 19 '13 at 19:20
Note that $-2^2 = - (2 \times 2) = -4$ and is not $(-2) \times (-2) = 4$. Hence, $$(-2^2)^3 = (-4)^3 = (-4) \times (-4) \times (-4) = - 64$$
In general, when $m$ is a positive integer, we have $$-a^m = - (\underbrace{a \times a \times \cdots \times a}_{m \text{times}})$$ and is not $$(-a)^m = (\underbrace{(-a) \times (-a) \times \cdots \times (-a)}_{m \text{times}})$$
• It's worth noting that some simplistic tokenizers do parse $-2^2$ as $(-2)^2$, which is non-standard. I believe most spreadsheets are guilty of this (perhaps to stay compatible with Microsoft Excel). – Erick Wong Jun 19 '13 at 19:21
Rewrite it as
$(-4)^3=(-1)^3 \cdot 4^3=-64$ | {
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# Count the number of shapes in a polyhedron.
So this is a question that was asked in the International Kangaroo Math Contest 2017. The question is:
The faces of the following polyhedron are either triangles or squares. Each triangle is surrounded by $$3$$ squares and each square is surrounded by $$4$$ triangles. If there are $$6$$ square faces, how many triangular faces are there?
What I did:
Each square shares each of its four neighboring triangles with two more squares. So we can say that for 6 squares we have $$6\times4\ -\ 2 \times 6 = 12$$ triangles. However, I still know that this calculation of mine is quite wrong and based on an awkward thinking. So, what is the correct answer and how?
Thanks for the attention.
• Always hated these questions when they specifically asked for reasoning. I just used my onboard 3D engine and could 'see' 8. Why? Because look :D – Lamar Latrell Oct 27 '18 at 6:41
Each edge of the polyhedron is shared between exactly one triangle and exactly one square, as can be inferred from the question statement. Thus, given six squares, there are 24 edges ($$6×4$$), and thus eight triangles ($$24÷3$$).
The polyhedron is called a cuboctahedron.
• Thanks! I just didn't try thinking it that way....;) – Faiq Irfan Oct 26 '18 at 11:35
Each square is bordered by four triangles and $$6\times4=24$$. However every triangle is bordered by three different squares, so it was counted three times in the multiplication above. This means there are $$24/3=8$$ triangles.
This isn't rigorous, but if you don't have to write a proof, just get the right number, it's clear from the illustration. You can see "one hemisphere" of the polyhedron except for a triangle parallel with the line of sight, so the total number of sides of each type are just double what apppear in that hemisphere (i.e. what you see plus the hidden triangle). | {
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• That would give 2*3+1=7 triangles but they are actually 8. I think you'd like to amend something? – Rad80 Oct 27 '18 at 10:39
• @Rad80: "double what you see plus the hidden triangle" should clearly be read with parentheses around all but the first word; I thought that was obvious from the text before it. – R.. GitHub STOP HELPING ICE Oct 27 '18 at 22:05
• well, at least we should be able to agree that is was not obvious what you meant. I saw two mentions that there is one "hidden" triangle (both times singular), so why should it be counted twice? If you edit the answer mentioning two hidden triangles instead of just one it'll be fine. – Rad80 Oct 29 '18 at 8:11
• @Rad80: There's only one hidden triangle in the hemisphere you're doubling. "Two hidden triangles" would be nonsense as there are of course five hidden triangles. – R.. GitHub STOP HELPING ICE Oct 29 '18 at 15:07
• @Rad80: Hopefully the change I made addresses what you found confusing about the wording. – R.. GitHub STOP HELPING ICE Oct 29 '18 at 15:08 | {
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# Construct $4 \times 4$ magic square with fixed “1”
The method I have found to generate $4\times 4$ magic squares gives me a result in which the number "1" is at of the corners of the square. How can we extend this to a method to generate a magic square, for a fixed location of number "$1$"?
The number "$1$" can be in $16$ different locations (cells). If we name the cells, from upper left corner: $1, 2, 3,4, 5, \ldots,$ and the number "$1$" is at the $i^{th}$ cell, then how we can fill the other cells to make a magic square?
You can see variations here.
There are really only three different locations: a corner, a side next to a corner, and one step diagonally in from a corner. If we can put $1$ into each of these, we can put it in any cell by using rotations and reflections. We already have a corner. We can subtract every number from $16$ and keep the square magic, but that doesn't help because it puts $1$ in the lower right corner. Because of the arrangement, we can add $8$ to all the numbers below $8$ and subtract $8$ from all those above, giving $$\begin {array} {c|c|c|c} 9&7&6&12 \\ \hline 4&14&15&1 \\ \hline 16&2&3&13 \\ \hline 5&11&10&8 \end {array}$$ That gets next to the corner. We can also rotate the $2 \times 2$ blocks by $180^\circ$ to get $$\begin {array} {c|c|c|c} 6&12&9&7 \\ \hline 15&1&4&14 \\ \hline 3&13&16&2 \\ \hline 10&8&5&11 \end {array}$$ which gets one in from the corner
• Thanks. Can we have a general solution for any doubly even order? – Susan_Math123 Aug 14 '17 at 4:49
• The example you cited has a lot of symmetry that not all magic squares have. I exploited that in both of my transformations. They will not work starting with a generic doubly even square. – Ross Millikan Aug 14 '17 at 5:06 | {
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At first let's define a more powerful magic square, which we will call $\color{Red}{\text{super-magic square}}$. By a $\color{Red}{\text{super-magic square}}$ we mean a magic square such the the sum of any arbitrary row is equal to the sum of any arbitrary column is equal to sum of any arbitrary diagonal.
For example suppose the following: $$\begin {array} {|c|c|c|c|} \hline 1&14&7&12 \\ \hline 15&4&9&6 \\ \hline 10&5&16&3 \\ \hline 8&11&2&13\\ \hline \end {array}$$
here we have:
$\color{Blue}{\text{Columns}}$: $$\begin {array} {ccccccccc} 1 & + & 15 & + & 10 & + & 8 & = & 34 \\ 14 & + & 4 & + & 5 & + & 11 & = & 34 \\ 7 & + & 9 & + & 16 & + & 2 & = & 34 \\ 12 & + & 11 & + & 3 & + & 13 & = & 34 \\ \end {array}$$
$\color{Green}{\text{Rows}}$: $$\begin {array} {ccccccccc} 1 & + & 14 & + & 7 & + & 12 & = & 34 \\ 15 & + & 4 & + & 9 & + & 6 & = & 34 \\ 10 & + & 5 & + & 16 & + & 3 & = & 34 \\ 8 & + & 11 & + & 2 & + & 13 & = & 34 \\ \end {array}$$
$\color{Purple}{\text{Diagonals parallel to the main diagonal}}$: $$\begin {array} {ccccccccc} 1 & + & 4 & + & 16 & + & 13 & = & 34 \\ 14 & + & 9 & + & 3 & + & 8 & = & 34 \\ 7 & + & 6 & + & 10 & + & 11 & = & 34 \\ 12 & + & 15 & + & 5 & + & 2 & = & 34 \\ \end {array}$$
$\color{Pink}{\text{Diagonals which are not parallel to the main diagonal}}$: $$\begin {array} {ccccccccc} 12 & + & 9 & + & 5 & + & 8 & = & 34 \\ 7 & + & 4 & + & 10 & + & 13 & = & 34 \\ 14 & + & 15 & + & 3 & + & 2 & = & 34 \\ 1 & + & 6 & + & 16 & + & 11 & = & 34 \\ \end {array}$$
$$%% %% 1 + 14 + 7 + 12 = 34 , %% \\ %% 15 + 4 + 9 + 6 = 34 , %% \\ %% 10 + 5 + 16 + 3 = 34 , %% \\ %% 8 + 11 + 2 + 13 = 34 ,$$
We will prove that, $1$ could be everywhere in a $\color{Red}{\text{super-magic square}}$. | {
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We will prove that, $1$ could be everywhere in a $\color{Red}{\text{super-magic square}}$.
Remark(I): Consider that a $\color{Red}{\text{super-magic square}}$ ($\color{Brown}{\text{of any arbitrary order}}$) is given. Then if we $\color{Blue}{\text{replace any two arbitrary columns}}$, then the resulting square, is again a $\color{Red}{\text{super-magic square}}$.
Remark(II): Consider that a $\color{Red}{\text{super-magic square}}$ ($\color{Brown}{\text{of any arbitrary order}}$) is given. Then if we $\color{Green}{\text{replace any two arbitrary rows}}$, then the resulting square, is again a $\color{Red}{\text{super-magic square}}$.
Now by $\color{Blue}{\text{column operations (I)}}$ and by $\color{Green}{\text{row operations (II)}}$ , we are able to change "the cell containing 1" to "any desired cell, so we are done!
• Where are the proofs/constructions for I and II? – OrangeDog Aug 14 '17 at 10:58
• @OrangeDog: This is only valid if we ignore diagonals. The Wikipedia article linked from the question explicitly says that the main diagonals must have the same sum as each row or column, so this answer is not valid... – user21820 Aug 14 '17 at 11:00
• @user21820 , Yes you are right. I have modified my answer, now this is true! – Jungle Boy Aug 14 '17 at 12:26
• Alright it's okay for the 4*4 magic square now. The name is "Pandiagonal magic square." Also, you proved a weaker, not stronger, result, because it does not apply to all magic squares in general. – user21820 Aug 14 '17 at 13:24
• And you should correct all your weird typos like "powerfull" and "rcolumn" and "coloumns"... – user21820 Aug 14 '17 at 13:26 | {
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metric str or callable, default=’minkowski’ the distance metric to use for the tree. Each object votes for their class and the class with the most votes is taken as the prediction. When p = 1, this is equivalent to using manhattan_distance (l1), and euclidean_distance (l2) for p = 2. The most common choice is the Minkowski distance $\text{dist}(\mathbf{x},\mathbf{z})=\left(\sum_{r=1}^d |x_r-z_r|^p\right)^{1/p}.$ What distance function should we use? The exact mathematical operations used to carry out KNN differ depending on the chosen distance metric. Any method valid for the function dist is valid here. The default metric is minkowski, and with p=2 is equivalent to the standard Euclidean metric. Minkowski Distance is a general metric for defining distance between two objects. For finding closest similar points, you find the distance between points using distance measures such as Euclidean distance, Hamming distance, Manhattan distance and Minkowski distance. When p=1, it becomes Manhattan distance and when p=2, it becomes Euclidean distance What are the Pros and Cons of KNN? When p < 1, the distance between (0,0) and (1,1) is 2^(1 / p) > 2, but the point (0,1) is at a distance 1 from both of these points. Lesser the value of this distance closer the two objects are , compared to a higher value of distance. The default method for calculating distances is the "euclidean" distance, which is the method used by the knn function from the class package. The parameter p may be specified with the Minkowski distance to use the p norm as the distance method. I n KNN, there are a few hyper-parameters that we need to tune to get an optimal result. Alternative methods may be used here. For p ≥ 1, the Minkowski distance is a metric as a result of the Minkowski inequality. Minkowski distance is the used to find distance similarity between two points. kNN is commonly used machine learning algorithm. 30 questions you can use to test the knowledge of a data scientist on k-Nearest | {
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learning algorithm. 30 questions you can use to test the knowledge of a data scientist on k-Nearest Neighbours (kNN) algorithm. KNN makes predictions just-in-time by calculating the similarity between an input sample and each training instance. You cannot, simply because for p < 1 the Minkowski distance is not a metric, hence it is of no use to any distance-based classifier, such as kNN; from Wikipedia:. Manhattan, Euclidean, Chebyshev, and Minkowski distances are part of the scikit-learn DistanceMetric class and can be used to tune classifiers such as KNN or clustering alogorithms such as DBSCAN. Among the various hyper-parameters that can be tuned to make the KNN algorithm more effective and reliable, the distance metric is one of the important ones through which we calculate the distance between the data points as for some applications certain distance metrics are more effective. The better that metric reflects label similarity, the better the classified will be. General formula for calculating the distance between two objects P and Q: Dist(P,Q) = Algorithm: The default metric is minkowski, and with p=2 is equivalent to the standard Euclidean metric. Why The Value Of K Matters. For arbitrary p, minkowski_distance (l_p) is used. A variety of distance criteria to choose from the K-NN algorithm gives the user the flexibility to choose distance while building a K-NN model. For arbitrary p, minkowski_distance (l_p) is used. The Minkowski distance or Minkowski metric is a metric in a normed vector space which can be considered as a generalization of both the Euclidean distance and the Manhattan distance.It is named after the German mathematician Hermann Minkowski. Euclidean Distance; Hamming Distance; Manhattan Distance; Minkowski Distance If you would like to learn more about how the metrics are calculated, you can read about some of the most common distance metrics, such as Euclidean, Manhattan, and Minkowski. KNN has the following basic steps: Calculate distance | {
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