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https://math.stackexchange.com/questions/896347/is-deta-maximal-if-detae-is-maximal | # Is det(A) maximal, if det(A+E) is maximal?
Let A be a binary matrix of size n x n and E be the matrix of the same size with all entries $1$.
Proof or disproof :
If det(A+E) has the maximal possible value, then det(A) also has the maximal possible value.
The converse is not true :
$$\pmatrix{1&1&2&2&1&2\\2&1&1&2&1&1\\1&2&1&2&1&2\\2&2&2&1&1&1\\2&1&1&1&2&2\\1&2&2&2&2&1}$$
has determinant 26 and subtracting 1 from any element yields a matrix with determinant 9, the maximal possible determinant for binary matrices of size 6 x 6. However, there is a $(1,2)$-matrix with size 6 x 6 and determinant 27 :
$$\pmatrix{1&2&2&1&2&1\\1&1&2&2&1&2\\2&1&1&1&2&2\\2&2&1&1&1&2\\2&1&2&2&1&1\\1&2&1&2&2&1}$$
If the conjecture is true, it makes the search for $(1,2)$-matrices with maximal possible determinant easier because it would suffice to calculate the determinant of A+E for every binary matrix A with maximal possible determinant.
I asked a similar question (maximal determinant for matrices with entries $1$ and $n$) and someone claimed that the problem would be equivalent to the problem for $(0,1)$-matrices, but as my counterexample shows, this is not the case.
• I approved the conjecture for $n\le 5$ with brute force. – Peter Aug 13 '14 at 16:59
• The maximal determinants for $(1,2)$-matrices with size n x n , $n = 2 , 3 , ...14$ seem to be $3 , 5 , 8 , 14 , 27 , 88 , 160 , 408 , 960 , 4131 , 10449 , 27459 \ and\ 74601$. – Peter Aug 13 '14 at 17:01
Your conjecture appears to be true for $n \leq 6$ and $n = 10$, and is true for any $n$ where the maximal determinants satisfy a certain inequality. However, this inequality does not hold for all $n$.
Denote the vector of all 1's by $\vec{1}$ and the column vectors of a fixed binary matrix $A$ by $\vec{a_i}$. Let $A_i$ be the matrix obtained by replacing the $i$-th column of $A$ by $\vec{1}$. Since the determinant is a multilinear function on the columns of $A$, we have $$\text{det}(A + E) = \text{det}(A) + \sum_{i=1}^n \text{det}(A_i).$$
The full multilinear expansion of $\text{det}\left(\vec{a_1} + \vec{1}, \vec{a_2} + \vec{1}, \ldots, \vec{a_n} + \vec{1}\right)$ has $2^n$ terms, but those terms with 2 or more columns of the vector $\vec{1}$ vanish. Your counterexample shows that the summation on the right hand side is different for different 0-1 matrices of maximal determinant. We can obtain more constraints by applying multilinearity to the complement of $A$.
Let $A^c = E - A$ be the complement matrix obtained by switching 1's and 0's. By multilinearity, $$\det(A^c) = (-1)^n \left[ \det(A) - \sum_{i=1}^n \det(A_i) \right].$$ As long as $\sum_{i=1}^n \det(A_i) \geq \det(A)$, or equivalently $\det(A + E) \geq 2 \det(A)$, we have $(-1)^{n+1} \det(A^c) = |\det(A^c)|$. This appears to be true for the matrices under consideration, those near the maximal determinant, so I will assume this throughout instead of breaking into cases based on the parity of $n$. Combining the equations, we have $$\det(A + E) = 2 \det(A) + |\det(A^c)|.$$
Now your counterexample shows that $\det(A^c)$ can be different for different 0-1 matrices of maximal determinant.
Let $M$ be the maximal determinant for $n \times n$ 1-2 matrices and $m$ be the maximal determinant for the corresponding 0-1 matrices. Suppose $\det(A + E) = M$ and $\det(A) < m$. We claim and later justify that $|\det(A^c)| \leq \det(A)$, which implies $M \leq 3m - 3a$, for some $a > 0$. Whenever $M > 3m - 3$, this produces a contradiction and the conjecture is true. Note that $M > 3m - 3$ ensures $\det(A+E) > 2\det(A)$ so long as $m \geq 3$. We conclude: $$M > 3m - 3 \implies \text{ the conjecture is true.}$$
By OEIS A003432, if your numbers are correct, this means the conjecture is true for $n = 4,5,6, \text{ and } 10$, but it may not be true for $n=7$. For $n = 7$, where $M = 88$ and $m = 32$ there can't be 0-1 matrices with determinant 31 whose complement has determinant 26 nor any 0-1 matrices with determinant 30 whose complement has determinant 28 for the conjecture to hold. (Apparently the complements of the determinant 32 matrices have determinant 24 or less.) More possibilities exist for those $n$ where $3m - M$ is somewhat large. I suspect the conjecture is false.
Finally, we justify the claim that $|\det(A^c)| \leq \det(A)$ whenever $A + E$ is of maximal determinant. As before, we assume that $\det(A+E) \geq 2\det(A)$ whenever $A+E$ has maximal determinant. Let $(A+E)^c$ be the matrix obtained by switching 1's and 2's. This is the matrix $2E - A$. Applying multilinearity and combining with previous equations, we find that $$\det(A+E) - |\det\left((A+E)^c\right)| = \det(A) - |\det(A^c)|.$$
Thus, if $\det(A+E)$ is maximal for 1-2 matrices, then $|\det(A^c)|$ is not larger than $\det(A)$, for otherwise, we have $|\det\left((A+E)^c\right)| > \det(A+E)$, contradicting maximality.
• I will have to work through this long answer, but it is promising! (+1) – Peter Aug 15 '14 at 12:48
• Thank you - sorry the answer (a partial one at that!) is long. The short version is that there are formulas for $\det(A+E)$ involving the matrix $A^c$ obtained by switching 1's and 0's. These can be applied to show that the conjecture is true if $M > 3m - 3$, where $M$ and $m$ are the maximal 1-2 and 0-1 determinants, respectively. When that bound doesn't hold, we've seen that $A^c$ can take on different values, so the possibility is there for the conjecture to fail. I thought I should give at least some justification for my claims, though! – Hugh Denoncourt Aug 15 '14 at 18:49 | 2019-05-24T21:13:27 | {
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http://mathhelpforum.com/algebra/87645-determine-center.html | 1. ## determine the center
determine the center , coordinates of vertices, endpoints for minor axis and foci coordinates.
(x-3)^2/4 + (y+1)^2/36=1
I believe the center is (3,-1) and i can get the vertices.. i just dont know how to get the foci and endpoints?
help me.. i know ive been asking alot of questions but my final exam is tommorow and im going through alot of things I dont remeber how to do.
2. Others are about to help you with this problem. Please note that this isn't a Calculus problem, but a algebra or perhaps geometry problem. I put it in algebra because conics are taught in Algebra II in the US.
3. Originally Posted by tiga killa
determine the center , coordinates of vertices, endpoints for minor axis and foci coordinates.
(x-3)^2/4 + (y+1)^2/36=1
I believe the center is (3,-1) and i can get the vertices.. i just dont know how to get the foci and endpoints?
help me.. i know ive been asking alot of questions but my final exam is tommorow and im going through alot of things I dont remeber how to do.
Yes, the center is at (3, -1). Now because this can be written as (x-3)^2/(2^2)+ (y+1)^3/(6^2)= 1, the "semi-axes" are 2 (parallel to the x-axis) and 6 (parallel to the y-axis). The vertices are at (3+ 2, -1)= (5, -1), (3-2, -1)= (1, -1), (3, -1+ 6)= (3, 5), and (3, -1- 6)= (3, -7).
The foci are a little harder. For an ellipse, the relation between the focal distance and the semi-axes is $a^2+ c^2= b^2$ where a is the "minor semi-axis", the smaller of the two, b is the "major semi-axis", the larger of the two, and c is the focal length. Here that gives $2^2+ c^2= 6^2$ or $4+ c^2= 36$ so $c^2= 32$. That is, $4\sqrt{3}$ so the foci are at $(3, -1+ 4\sqrt{3})$ and $(3, -1- 4\sqrt{2})$. The foci are on the longer axis, here the y- axis.
4. You're correct about the center.
As for the rest, let's look at the basic form. It's an ellipse, but you probably knew that. The equation is:
$\frac {(x - h)^2}{a^2} + \frac {(y - k)^2}{b^2} = 1$, with a center at (h, k). There's one small wrinkle here, however. Since a is supposed to be the semi-major axis, and the major axis must be larger than the minor axis, by definition. Since your equation is:
$\frac {(x-3)^2}{4} + \frac {(y+1)^2}{36} = 1$
It's clear that the major axis is in the y term, and so the basic equation here would be better described by:
$\frac {(y - k)^2}{a^2} + \frac {(x - h)^2}{b^2} = 1$
So I'd rearrange it as:
$\frac {(y+1)^2}{36} + \frac {(x-3)^2}{4} = 1$
Ok, that means we have $a^2 = 36 \implies a = \pm 6$ and $b^2 = 4 \implies b = \pm 2$.
So the major axis along the y axis would go from the center plus and minus 6. The minor axis along the x axis would go from the center plus and minus 2. Each along it's respective axis, of course. The major axis would be 2a long, and the minor axis would be 2b long (taking the positive a and b, of course).
Working that out from a center of (3, -1), we get major axis from (3, -7) to (3, 5). Minor axis from (1, -1) to (5, -1).
Lastly, the foci are described by $c = \sqrt{a^2 - b^2}$. I prefer to think of it as $c^2 = a^2 - b^2$. Whatever, same thing. Remembering that the foci are along the major axis, we solve that:
$c = \pm \sqrt {36 - 4} = \pm \sqrt {32} = \pm \sqrt {2(16)} = \pm 4 \sqrt{2}$. That gives foci along the y axis of $(3, -1 + 4 \sqrt{2})$ and $(3, -1 - 4 \sqrt{2})$.
Hope that helps, because it involved a lot of typing... lol Good luck!
5. Originally Posted by HallsofIvy
That is, $4\sqrt{3}$ so the foci are at $(3, -1+ 4\sqrt{3})$ and $(3, -1- 4\sqrt{2})$. The foci are on the longer axis, here the y- axis.
Dead on, except that $\sqrt {32} = 4 \sqrt {2}$. I'm guessing it's just a typo since you got it right in the second focus. Just thought I'd correct it so the OP doesn't get confused. | 2013-12-19T07:00:21 | {
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https://mathematica.stackexchange.com/questions/112655/speed-up-computation-of-sum-from-large-matrix | # Speed up computation of sum from large matrix
I have a 3x3 matrix (252^3) of data (densities) and I want to compute a correlation xi from nearest neighbours as well as a sum involving a check whether the density is in a certain bin. The computation works in principle (so for subsamples, when step size is about 10) but is painfully slow. I am not an expert on handling data with Mathematica, so I am pretty sure there is some solution that works way better than my first try. For the correlation function I think that basically need an analogue of AbsoluteCorrelationFunction for my matrix.
xi=
1/3*Sum[data[[i,j,k]] (data[[i + 1,j,k]] +
data[[i,j + 1,k]] +
data[[i,j,k + 1]]), {i, 1, 251, 1}, {j, 1,
251, 1}, {k, 1, 251, 1}]/26^3 - 1;
For the following quantity I am not sure if there is another faster Mathematica routine. I need to check whether the density is in a bin of width deltarho around rho and if yes add the values of the neighbouring densities
brho[rho_]:=(Sum[If[Abs[data[[i,j,k]] - rho] <
deltarho/2, (data[[i + 1,j,k]] +
data[[i,j + 1,k]] +
data[[i,j,k + 1]]), 0], {i, 1, 251, 1}, {j, 1,
251, 1}, {k, 1, 251, 1}]/
Sum[Boole[
Abs[data[[i,j,k]] - rho] < deltarho/2], {i, 1,
251, 1}, {j, 1, 251, 1}, {k, 1, 251, 1}] - 1)
I tried replacing Sum with ParallelSum but at least for the subsample it makes things slower.
Are there any ideas how to speed this up?
edit: improved readability & corrected typo, thanks to comment of Jason
• Just first impressions - to improve readability in the code for yourself and others, you can replace things like data[[i]][[j]][[k]] with data[[i,j,k]] and , {i, 1, 251, 1}, {j, 1, 251, 1}, {k, 1, 251, 1}] with ,{i, 251}, {j, 251}, {k, 251} but this is just a style point and doesn't address your main question. – Jason B. Apr 15 '16 at 9:35
• I tried running your definition of xi on a random matrix dims = {251, 251, 251}; data = RandomReal[1, dims]; and it takes a long time. You could get the same thing virtually instantly by a judicious use of Total and RotateLeft and Drop – Jason B. Apr 15 '16 at 9:52
• Also, Can I assume that {i, 1, 251, 10} is a typo in the brho code? It's the only 10 in there so I assume it should be a 1 – Jason B. Apr 15 '16 at 10:06
Well, basically your code is designed with a less efficient algorithm.
Bear in mind that Mathematica generally treats a matrix the "same" as a number, so don't spend time on computation conducted at the number level.
Then we can convert your code to be running at the matrix level, where the speed is much increased.
len = 20;
data = RandomReal[{1, 10}, {len, len, len}];
xi=1/3*Sum[data[[i,j,k]] (data[[i+1,j,k]]+data[[i,j+1,k]]+data[[i,j,k+1]]),{i,len-1},{j,len-1},{k,len-1}]/26^3-1;//AbsoluteTiming
(*{0.030215, Null}*)
xxi=Total@Flatten@(data[[1;;len-1,1;;len-1,1;;len-1]](data[[2;;len,1;;len-1,1;;len-1]]+data[[1;;len-1,2;;len,1;;len-1]]+data[[1;;len-1,1;;len-1,2;;len]]))/3/26^3-1;//AbsoluteTiming
(*{0.00039, Null}*)
xi == xxi
(*True*)
Now we increase len to 252 as required by your scenario:
len = 252;
data = RandomReal[{1, 10}, {len, len, len}];
xxi=Total@Flatten@(data[[1;;len-1,1;;len-1,1;;len-1]](data[[2;;len,1;;len-1,1;;len-1]]+data[[1;;len-1,2;;len,1;;len-1]]+data[[1;;len-1,1;;len-1,2;;len]]))/3/26^3-1;//AbsoluteTiming
(*{0.653753, Null}*)
See, the performance is not bad.
I think now you can improve the second part of your problem with the idea above.
Edit: As for the second part of your problem, there is no need to perform testing over each element: a holistic testing works as well:
Clear[brho]
brho[data_List,rho_,deltarho_]:=Module[{c,sum},
c=UnitStep[deltarho/2-Abs[data[[1;;-2,1;;-2,1;;-2]]-rho]];
sum=data[[2;;-1,1;;-2,1;;-2]]+data[[1;;-2,2;;-1,1;;-2]]+data[[1;;-2,1;;-2,2;;-1]];
Total@Flatten[c*sum]/Total@Flatten[c]
]
Give it a try:
brho[data,10,3] | 2020-02-18T04:07:30 | {
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https://math.stackexchange.com/questions/485102/what-values-of-00-would-be-consistent-with-the-laws-of-exponents?noredirect=1 | # What values of $0^0$ would be consistent with the Laws of Exponents?
I am using the following fundamental properties of exponentiation on $N$ as as basis for this discussion:
(1) $0^1 = 0$
(2) $\forall x\in N (x\ne 0 \implies x^0 = 1)$
(3) $\forall x,y\in N (x^{y+1}=x^y\cdot x)$
Missing, of course, is a value for $0^0$. But only $0^0=0$ or $1$ are consistent with the Laws of Exponents:
(4) $\forall x,y,z\in N (x^{y+z}=x^y\cdot x^z)$
(5) $\forall x,y,z\in N (x^{y \space\cdot z}=(x^y)^z)$
EDIT:
From (5), we must have $(0^0)^2=0^{0\times 2}=0^0$. Therefore, $0^0= 0$ or $1$. Is this correct?
Is there any way to eliminate $0$ (or $1$) as a possible value, with reference to the fundamental properties or the laws of exponents?
• I am sure this has been discussed before, but I note, from my own efforts, that it is not easy to find the previous questions which have raised the point. – Mark Bennet Sep 5 '13 at 19:11
• Yes, this has been covered again and again, but there is a reason for that. We generally define $0^0=1$ on the natural numbers. This can be done in terms of set theory ($x^y$ being the size of $X^Y$ with $|X|=x,|Y|=y$.) It is also simpler - note how your (2) has a special case? Always unfortunate. The problem happens when you want $x^y$ to be continuous, but then you have a lot of problems - $x<0$, for example. – Thomas Andrews Sep 5 '13 at 19:14
• There was a discussion here – Old John Sep 5 '13 at 19:15
• I've been mystified by people trying to give a value to $0^0.$ It finally occurred to me that, in using summation signs for polynomials and power series, we use the shorthand $x^0$ to mean $1,$ and maybe people think that says something about $0^0.$ Hi, @OldJohn – Will Jagy Sep 5 '13 at 19:18
• There are indeed many postings on $0^0$ here -- maybe in every other math forum! -- but I am not asking for a value for $0^0$. There doesn't seem to be a consensus. But I have noticed that both $0^0=0$ and $0^0= 1$, at the very least, do seem to be consistent with the Laws of Exponents. I just wanted to confirm this fact. – Dan Christensen Sep 5 '13 at 19:28
If $c,d$ are cardinal numbers, then $c^d$ is the cardinal of the set of maps $d \to c$. This works for all cardinal numbers and the usual arithmetic laws hold. For $d=0$ we get $c^0=1$ since there is a unique map $\emptyset \to c$. This holds for all $c$, in particular $0^0=1$. So there is actually no debate what $0^0$ is or not, it is $0^0=1$ by the general definitions. No case distinctions are necessary. Forget about $0^0=0$, this is nonsense.
Ok, this may be glib, but following up on the comment of Will Jagy above, consider the following:
$$1 = 1^n = (1+0)^n = \sum_{i=0}^{n} \binom{n}{i} 1^i \cdot 0^{n-i}.$$
Do you see what I'm getting at? What sense do we want to make out of the last term, $1^n \cdot 0^0$?
Actually, I need to clarify why I posted this as an "answer." Since the OP is looking to define $0^0$ in a way that stays consistent with "fundamental properties of exponents," and I feel that the Binomial Theorem is fundamental enough to qualify as essential to arithmetic, the example above rules out the possibility of $0^0 = 0$ while reinforcing $0^0=1$.
• I think the proof of BT assumes $0^0=1$. – Dan Christensen Sep 5 '13 at 21:51
• Also, BT, as usually stated, assumes $0^0=1$. But it could easily be re-stated without this assumption, handling the zero-cases as exceptions similar to other exceptions for zero-denominators. Yes, it would be awkward. – Dan Christensen Sep 5 '13 at 22:39
• Of course, the PROOF of BT needs the assumption $0^0 = 1$, but here it is raised to an AXIOM which then actually implies the assumption; this means that $0^0=1$ is equivalent to BT. – Kofi Sep 6 '13 at 6:53
Dan, the proof of BT indeed assumes 0^0=1. The problem is that:
(a) there are many places where we assume 0^0 = 1, and (b) people do not acknowledge (a).
Calculus textbooks usually say that 0^0 is undefined, but at the same time, they contain dozens of formulas that assume 0^0=1.
Textbooks should honestly admits that it is standard practice to evaluate 0^0 to 1.
The argument for "undefined" is based on the mistaken belief that 0^0=1 leads to contradictions. It does not (if it did, we would have found out long ago, because the assumption 0^0=1 is used in subtle ways in lots of places).
But the real reason that I feel strongly that 0^0 must be 1 is because the "undefined" idea is contrary to a two deeply held convictions, namely (1) that the most convenient definition must be the best one, and (2) math is consistent, so given the fact that set theory tells us 0^0=1, it means that it is OK to use that everywhere.
• The problem in calculus is that choosing value for $0^0$ is choosing whether or not $0^x$ is continuous or $x^0$ is continuous, or seeing that $x^y$ cannot be continuous as a function of two variables. – Asaf Karagila Apr 21 '14 at 14:13
• Stewart (admittedly the only calculus book in easy reach) writes that $0^{0}$ is indeterminate (which is perfectly true), not "undefined". Can you supply specific citations of calculus authors who use "undefined" in this context, and/or do you dispute that $0^{0}$ is indeterminate? – Andrew D. Hwang Apr 21 '14 at 14:21
• I checked 4 calculus books, and they write that x^0 = 1 if x is not 0. They should write: x^0=1 for every x because that is what they actually use in several formulas. – Mark Apr 21 '14 at 15:56
• Asaf, even from the viewpoint of the function 0^x, the value 1 is better than the value 0. This function provides two values (0 and infinity) depending on whether x approaches 0 from the right or the left. Picking one of these two values is a arbitrary choice, and by symmetry a^(-x) = 1/a^x you can see that both choices must be wrong. The only value that is supported by a consistent argument is the value 1. – Mark Apr 21 '14 at 16:06
• If $0^0 =1$ then $0 = 0^{-2} \cdot 0^2 = 0^0=1$. – Kasper Mar 16 '18 at 9:15
We can make sense of $0^0=1$ in this way also : Expand $f(x)=a^x$ as Taylor series for real values of x and a is fixed positive real number and let a tends to zero. Its not rigorous but still supporting our assumption.
Follow-up
It can be shown that there exists not just 1 or 2 "exponent-like" functions on N as I suggested here, but an infinite number of them. And from any one of them, we can derive the usual Laws of Exponents. For formal proofs, see Oh, the ambiguity! at my blog. | 2020-11-30T23:57:28 | {
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https://math.stackexchange.com/questions/2446753/palindromes-in-multiple-bases | # Palindromes in multiple bases
I noticed when listing out palindromes in bases $2$ and $3$ that they seem not to share any palindromes (other than trivial single-digit palindromes). However, when I tried to prove this, I couldn't solve it. I proved it for numbers with an even number of digits in base $3$ by using the fact that a number ending with the digit $0$ is trivially a non-palindrome since leading zeroes don't count as digits, but I can't get it for numbers with an odd number of digits in base $3$.
I also used the fact that $$n\equiv \operatorname{sdig}_b(n) \mod{(b+1)}$$ where $\operatorname{sdig}_b(n)$ is the sum of the digits of $n$ in base $b$.
Can somebody help me prove this?
• I assume you mean palindromes other than $0$ and $1$. – aschepler Sep 26 '17 at 23:35
• @aschepler Yeah, palindromes with more than one digit. I'll clarify in the question. – Franklin Pezzuti Dyer Sep 26 '17 at 23:36
• maybe a pigoenhole principle argument may get you somewhere – user451844 Sep 26 '17 at 23:40
• I believe that it is better the tag (numeral-systems), and I don't know if your question is related to combinatory. – user243301 Sep 26 '17 at 23:45
• An idea, if you want to explore further. For a ternary palindrome to not be divisible by 2, it must be of odd length and have 1 as a middle digit. The numbers around the middle of the ternary representation mostly fully determine the numbers around the middle of the binary representation and both need to be symmetric. Maybe that could be a starting point for finding some rules on which ternary palindromes can have a binary counterpart. – JollyJoker Sep 27 '17 at 8:38
## 3 Answers
Having "magically" found counterexamples, let us explore how we could find at least the first one. Do we really face the specter of having to run thousands of trials? Not if we apply a little ingenuity.
A base 2 palindrome must be odd or else the reverse base 2 representation has an initial zero, not allowed. Then the base 3 representation, being odd, must have an odd sum of digits (c.f. the base 10 test for a multiple of 3). The only base-3 palindromic representations matching this condition have the form $a1a*$ where $a$ is a base 3 representation with no initial zeroes and $a*$ is obtained by reversing the digits of $a$ ($a*$ keeps any initial zeroes corresponding to the terminal zeroes of $a$). Note that not allowing any initial zeroes in $a$ automatically enables us to dodge multiples of 3.
Let us select $a=110_3$ as an example. Keeping the terminal zero as an initial zero of $a*$ gives $a*=011_3$ and thus the odd base 3 palindrome $1101011_3$.
Now we check whether this is palindromic in base 2. A convenient way to do this is to first convert to base 4. Treat the base 3 representation as a polynomial in which the digits, including zeroes, are successive coefficients of decreasing powers of $x$ and evaluate at $x=3$ using synthetic division. In base 4 you use the "multiplication facts" $3×2=12$ and $3×3=21$ to carry out the arithmetic. Applying this technique to $1101011_3$:
$$1=1$$ $$1×3+1=10$$ $$10×3+0=30$$ $$30×3+1=211$$ $$211×3+0=1233$$ $$1233×3+1=11032$$ $$11032×3+1=33223$$
So $1101011_3=33223_4$, and converting each base 4 digit to the appropriate pair of bits then gives $1111101011_2$.
This fails to be a palindrome. A systematic trial would start with $a=1$ and proceed to $a=2, a=10_3$, etc. The first 26 trials fail, on trial 27 the reader can verify that $a=1000_3$ gives, properly, $100010001_3=1213303_4=1100111110011_2$. This is $6643_{10}$.
• I appreciate the explanation, rather than just a bald answer. Thanks! (+1) – Franklin Pezzuti Dyer Sep 28 '17 at 23:09
You were not able to prove it because it's not true. The OEIS has the first 17 of them as sequence A060792, and it is not known whether any more exist.
• The referenced sequence has only 11 terms greater than 1. Still +1. – Oscar Lanzi Sep 27 '17 at 2:13
• @OscarLanzi Click on the first entry under "Links" to see the "B-File" for 4 more. – aschepler Sep 27 '17 at 4:32
• The fact that it's not known whether any more exist suggest that OP's quest for a proof was pretty hard. The lower bound for the first unknown one is 3^93, or a bit over 3x10^44, suggesting any current search techniques need (almost) sqrt(n) steps, which is trivial as half the digits of a palindromic number mirror the other half. – JollyJoker Sep 27 '17 at 7:37
• How is the lower bound "only" $3^{93}$? Candidates must have an odd number of base-3 digits, so any that are at least $3^{93}$ must also be at least $3^{94}$. – Oscar Lanzi Sep 28 '17 at 23:54
Sadly, $6643_{10}=1100111110011_2=100010001_3$ is a palindrome in bases $2$ and $3$. $1422773$ and $5415589$ also satisfy the property.
• Aww, what a tragedy... – Franklin Pezzuti Dyer Sep 26 '17 at 23:58
• @Nilknarf Yeah, it's disappointing. – Carl Schildkraut Sep 26 '17 at 23:59
• So if I'm sharing a taxicab from Baltimore Washington Airport to the Capital and it's numbered 6643, we'll have something to talk about. Think Ramanujan would have known this property? – Oscar Lanzi Sep 27 '17 at 2:05 | 2019-12-09T02:37:33 | {
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https://math.stackexchange.com/questions/2037324/int-omega-f-d-lambda-0-rightarrow-x-in-omega-fx-neq-0-is-a/2037345 | # $\int_{\Omega} |f| d\lambda = 0 \Rightarrow \{x \in \Omega: f(x) \neq 0\}$ is a null set
Let $(\Omega, \mathscr A, \lambda)$ be a measure space.
For an $\mathscr A$-measurable numerical function $f: \Omega \rightarrow \Bbb R$, it holds that
$\int_{\Omega} |f| d\lambda = 0 \Rightarrow \{x \in \Omega: f(x) \neq 0\}$ is a null set.
The argument works the following way:
Define
$S := \{x \in \Omega: f(x) \neq 0\}$. Since $f$ is measurable, it follows that $S \in \mathscr A$. For every natural number $k \ge 1$, we define
$\phi_k := \inf\{k|f|, X_s\}$ with $X_s$ being the Indicator function.
Hence,
$\phi_k \uparrow X_s$, and since $\phi_k \le k|f|$, we get $\int \phi_k d\lambda = 0$.
While I understand the very last step here, I don't see why $\phi_k \uparrow X_s$. $X_s = 0$ for every $x \notin S$, and since $S$ is defined the way it is, it's only possible that $|f| = 0$, and therefore, $k|f| = 0$ for every $k$. So how does $\phi_k$ converge against $X_s$ here?
From construction it is clear that $\phi_k \leq \chi_S$ for all $k$. To see that $\phi_k$ converges pointwise to $\chi_S$ consider first if $x\in S$. Then $f(x)\neq 0$ and so $k |f(x)|\geq 1$ for sufficiently large $k$. Thus, eventually $k|f(x)|\geq \chi_S(x)$ and $\phi_k(x) = \chi_S(x)$. On the other hand, if $x\not\in S$ then $f(x)=0$ and $\chi_S(x)=0$, so it is clear that $\phi_k(x) = \chi_S(x)$ for all $k$.
Together, these two cases imply that $\phi_k$ converges to $\chi_S$ pointwise.
• Why not just use Markov's inequality? – BCLC May 3 '18 at 14:35
Another way to write $\phi_k$ is $$\phi_k(x) = \begin{cases} 0 & \text{if f(x) = 0}\\ k|f(x)| & \text{if |f(x)| \leq \frac{1}{k}} \\ 1 & \text{if |f(x)| \geq \frac{1}{k}} \end{cases}$$ So if $f(x) = 0$, $\phi_k(x) = 0 = \chi_S(x)$ for all $k$. If $f(x) \neq 0$, there exists $K$ such that $|f(x)| > \frac{1}{K}$. Then for all $k \geq K$, $\phi_k(x) = 1 = \chi_S(x)$.
• Why not just use Markov's inequality? – BCLC May 3 '18 at 14:36
• @BCLC If I understand correctly, the OP was trying to understand a specific proof of this fact, not prove it on their own. – Matthew Leingang May 5 '18 at 12:05
Let $x \in \Omega$.
Case 1: $x \notin S$. Then $f(x)=0=X_S(x)$, hence: $\phi_k(x)=0=X_S(x)$ for all $k$.
Case 2: $x \in S$. Then $|f(x)|>0$. Hence there is $k_0$ such that $k|f(x)| \ge 1=X_S(x)$ for all $k>k_0$. Then we have
$\phi_k(x)=X_S(x)$ for all $k>k_0$.
It is your turn to show that $\phi_1(x) \le \phi_2(x) \le ....\le \phi_{k_0}(x)$
• Why not just use Markov's inequality? – BCLC May 3 '18 at 14:35
Reverse of standard machine:
$$\int_{\Omega} |f| d\lambda = 0$$
$$\to \int_{\Omega} f^+ d\lambda + \int_{\Omega} f^- d\lambda = 0$$
$$\to \int_{\Omega} f^+ d\lambda = \int_{\Omega} f^- d\lambda = 0$$
$$\to \sup_{h^+ \le f^+} \int_{\Omega} h^+ d\lambda = \sup_{h^- \le f^-} \int_{\Omega} h^- d\lambda = 0$$
$$\to \int_{\Omega} h^+ d\lambda = \int_{\Omega} h^- d\lambda = 0$$
$$\to \int_{\Omega} \sum_{i=1}^n a_i 1_{A_i} d\lambda = \int_{\Omega} \sum_{j=1}^n b_j 1_{B_j} d\lambda = 0$$
$$\to \sum_{i=1}^n a_i \lambda(A_i) = \sum_{j=1}^n b_j \lambda(B_j) = 0$$
$$\to a_i \lambda(A_i) = 0, b_j \lambda(B_j) = 0$$
$$\to a_i = 0 \ or \ \lambda(A_i) = 0, b_j \ or \ \lambda(B_j) = 0$$
$$\to \lambda(\{x \in \Omega: h^+ \neq 0\})=\lambda(\{x \in \Omega: h^- \neq 0\}) = 0$$
$$\to \lambda(\{x \in \Omega: f^+ \neq 0\})=\lambda(\{x \in \Omega: f^- \neq 0\}) = 0$$
$$\to \lambda(\{x \in \Omega: f \neq 0\}) = 0$$
QED
By Markov's inequality, $\forall \varepsilon > 0$,
$$P(f \ne 0) = P(|f| \ge \varepsilon) \le \frac{E[|f|]}{\varepsilon} = 0$$
I know yours is a measure but not probability space but the same holds true for measure space. | 2019-12-07T01:47:35 | {
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https://mathoverflow.net/questions/351415/interpolation-nodes-for-linear-spline-piecewise-linear-interpolation-of-x-ln | # Interpolation nodes for linear spline (piecewise-linear) interpolation of $x \ln x$
I need to approximate $$x \ln x$$ on $$[0,1]$$ as a piecewise-linear function. If $$P(x)$$ is a piecewise-linear approximation, I want to minimize $$\max_{0 \le x \le 1} |P(x) - x \ln x| \rightarrow \min_P.$$
I tried interpolation over evenly spread points $$x_i = \frac{i}{n}$$, for $$i=0,1,\dotsc,n$$. I also tried Chebyshev nodes together with additional nodes $$x=0$$ and $$x=1$$. The latter works much better. For my numerical problem, it was enough but I am interested in optimal solution (mathematical mind, you know :)
Is there a way to optimally spread points $$x_0, x_1, \dotsc, x_n$$ over $$[0,1]$$ for piecewise-linear interpolation/approximation?
Update To illustrate the difference between uniform and Chebyshev nodes, here is the error of approximation with $$n=20$$ points.
My guess is that for optimal nodes, the "bumps" should be of the same height.
Update 2
I have also the following idea of a numerical method.
First, let us see what is the maximum error of approximation between points $$x_n$$ and $$x_{n+1}$$.
We can find equation of an approximating (red) line $$y = kx + b$$, where $$k = \frac{x_{n+1} \ln x_{n+1} - x_n \ln x_n}{x_{n+1} - x_n}, \quad b = -k x_n+x_n \ln x_n \,.$$
Next, for $$x_n \le x \le x_{n+1}$$, we can define the error of approximation at point $$x$$: $$\delta(x) = k x + b - x \ln x \,.$$ To find the maximum error, we just solve $$\delta'(x) = 0$$: $$0 = (kx + b - x \ln x)' = k - \ln x - 1.$$ Therefore, the maximum error on $$[x_n,x_{n+1}]$$ is achieved at $$x_c=e^{k-1}$$ and it is equal to $$d(x_n, x_{n+1}) = k x_c + b - x_c \ln x_c \,,$$ where $$k$$, $$b$$, and $$x_c$$ depend on $$x_n$$ and $$x_{n+1}$$ as described above. Note also that if we fix $$x_n$$, the $$d(x_n, x_{n+1})$$ is increasing function in $$x_{n+1}$$.
Now, the method itself. Assume we want to ensure that approximation error is not larger than $$\epsilon$$. Then the following iterative procedure can be applied to construct the list of points.
1. Set $$x_0 = 0$$.
2. For $$n=0,1,2,\dotsc$$ find (e.g. by binary search) $$x_{n+1}$$ that ensures $$d(x_n, x_{n+1}) = \epsilon$$.
3. Stop when some $$x_{n+1} \ge 1$$. Change $$x_{n+1}$$ to $$x_{n+1} = 1$$ (not necessary but to make things nicer).
But is there an analytical solution?
• For example, if we set three points $x_0=0, x_1, x_2=1$, then it seems the optimal position for $x_1$ is $0.2869781560930248$... – Yauhen Yakimenka Jan 29 at 10:25
Each piece of deviation from a linear interpolation resembles a parabola. So if we interpolate a function $$f$$ over a segment of width $$w$$, then the maximum deviation of the curve from the segment is roughly $$(w/2)^2 f''/2$$. To get a maximum deviation of $$\epsilon$$, we should choose $$w=\sqrt{8\epsilon/f''}$$. To be more precise, we should use a value for $$f''$$ in the middle of the parabola, whose location we can estimate from the previous two points of interpolation.
This leads to the following procedure: Let \begin{align} x_0 &=1 + 1/N\\ x_1 &= 1\\ x_{n+1} &= x_n - \sqrt{8\epsilon\big/ f''\!\left(x_n+\frac{x_n - x_{n-1}}2\right)}\\ \end{align} Then we interpolate with the points $$\{0,x_{N-1},\ldots,x_1\}$$.
Here $$N=20$$, $$f(x)=x \log x$$, and by trial and error we choose $$\epsilon=.00141$$. This gives the points $$\{0.0000, 0.0039, 0.0112, 0.0242, 0.0427, 0.0670, 0.0968, 0.1323, 0.1735, 0.2203, \\ \ \ 0.2727, 0.3308, 0.3945, 0.4638, 0.5388, 0.6195, 0.7057, 0.7976, 0.8951, 1.0000\}.$$ Linear interpolation on those points approximates $$f$$ with a maximum deviation of only $$.00145$$.
• If I set $\epsilon=0.00145$ in my numerical method, I also get 20 points: {0.000000,0.003942,0.013735,0.029329,0.050725,0.077920,0.110915,0.149711,0.194307,0.244702,0.300898,0.362893,0.430689,0.504284,0.583680,0.668875,0.759871,0.856666,0.959261,1}. So your method is rather good (at least, for this $\epsilon$). – Yauhen Yakimenka Jan 29 at 21:47
• For most $f$, there is no closed form for the maximum error on an interval, so I like having a method that doesn't depend on numerical maximization at each of those 20 steps. – Matt F. Jan 29 at 22:11
• True, it requires solving $-k = f'(x)$ – Yauhen Yakimenka Jan 29 at 22:18 | 2020-02-28T15:10:30 | {
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https://math.stackexchange.com/questions/1169074/sum-of-discrete-and-continuous-random-variables-with-uniform-distribution | # Sum of discrete and continuous random variables with uniform distribution
Could you tell me how to find the distribution of $Z = X+Y$ if $X$ is a random variable with uniform distribution on $[0,1]$ and $Y$ has uniform distribution on $\{-1,0,1\}$?
$X$ and $Y$ are independent.
I know how to find distributions of sums of random variables if both are discrete or both are continuous. But here we have a mix.
I guess I should consider $P(Z<-1) = P(\{X<-1, Y<0\} \cup \{X<0, Y<-1\})=0$
$P(Z \in [-1,0)) = P(X \in [-1,0), Y=0 \ \text{or} \ X \in [0,1), Y=-1 \ \text{or} \ X \in [-2, -1), Y=1) = \frac{1}{3}$
$P(Z \in [0,1)) = P(X \in [0,1), Y=0 \ \text{or} \ X \in [1,2), Y=-1 \ \text{or} \ X \in [-1, 0), Y=1) = \frac{2}{3}$
$P(Z \in [1,2)) = P(X \in [1,2), Y=0 \ \text{or} \ X \in [2,3), Y=-1 \ \text{or} \ X \in [0, 1), Y=1) = 1$
And so $P(Z \ge 1) = 1$
Is that correct?
• Are you assuming (or have you been told) that $X$ and $Y$ are independent but you have opted to not share this information with us? – Dilip Sarwate Feb 28 '15 at 14:08
• I'm so sorry. Yes, they are independed. I haven't noticed that I forgot to write this. – Hagrid Feb 28 '15 at 14:36
Notice, that events $Z \in [-1,0)$, $Z \in [0,1)$ and $Z \in [1,2)$ have empty intersection. But the according to your solution: $$P(Z \in [-1,0)) + P(Z \in [0,1)) + P(Z \in [1,2)) > 1$$ which clearly could not be correct.
So lets count, the desired probabilities once again, using the fact that X,Y are independent: \begin{align} P(Z \in [-1,0)) &= P(X \in [0,1))P(Y=-1) = 1\times\frac{1}{3}=\frac{1}{3}, \\ P(Z \in [0,1)) &= P(X \in [0,1))P(Y=0) + P(X = 1)P(Y=-1) = 1\times\frac{1}{3} + 0\times\frac{1}{3} = \frac{1}{3}, \\ P(Z \in [1,2)) &= P(X \in [0,1))P(Y=1) + P(X = 1)P(Y=0) = 1\times\frac{1}{3} + 0.\times\frac{1}{3} = \frac{1}{3}. \end{align} Of course $P(Z < - 1)= 0$, but we have already counted $$P(Z \geq 1) = P(Z \in [1,2)) + P(Z \geq 2) = \frac{1}{3} + 0.$$
• I see where I made a mistake, in the second summands we only take one value of $X$, because the probability that $X$ has values outside $[0,1]$ is zero. Thank you! – Hagrid Feb 28 '15 at 15:10
• I have one more question, though. Is this distribution continuous? – Hagrid Feb 28 '15 at 15:52
• In you count the distribution function : $$P(Z \leq z) = \frac{1}{3}P(X \leq z+1) + \frac{1}{3}P(X \leq z) + \frac{1}{3}P(X \leq z-1)$$ for all $z \in [0,1]$, you should see that Z has uniform distribution on $[-1,2]$. – iiivooo Feb 28 '15 at 16:03
You can extend the convolution method for summing continuous independent variables if you identify the "density" of a discrete variable as a sum of Dirac deltas. Here you find that the density of $X$ is $1_{[0,1]}$ while the "density" of $Y$ is $\frac{1}{3} \sum_{i=-1}^1 \delta_i$. So you now need to compute the convolution:
$$g(x) = \frac{1}{3} \sum_{i=-1}^1 \int_{-\infty}^\infty \delta_i(y) 1_{[0,1]}(x-y) dy \\ = \frac{1}{3} \sum_{i=-1}^1 \int_{x-1}^x \delta_i(y) dy$$
By the definition of the Dirac delta, the $i$th summand is $1$ if $i \in (x-1,x)$ and zero otherwise. So exactly one summand will be $1$ if and only if $x \in (-1,2)$, so the density here is $\frac{1}{3} 1_{(-1,2)}$.
In this case this is pretty much the same as the other answer; this method is nicer in more complicated examples. | 2020-08-04T08:00:55 | {
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Thus the unique left inverse of A equals the unique right inverse of A from ECE 269 at University of California, San Diego 53 0 obj <> endobj Left inverse if and only if right inverse We now want to use the results above about solutions to Ax = b to show that a square matrix A has a left inverse if and only if it has a right inverse. It's an interesting exercise that if $a$ is a left unit that is not a right uni 125 0 obj <>stream ����E�O]{z^���h%�w�-�B,E�\J��|�Y\2z)�����ME��5���@5��q��|7P���@�����&��5�9�q#��������h�>Rҹ�/�Z1�&�cu6��B�������e�^BXx���r��=�E�_� ���Tm��z������8g�~t.i}���߮:>;�PG�paH�T. Proof In the proof that a matrix is invertible if and only if it is full-rank, we have shown that the inverse can be constructed column by column, by finding the vectors that solve that is, by writing the vectors of the canonical basis as linear combinations of the columns of . In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. This is no accident ! wqhh��llf�)eK�y�I��bq�(�����Ã.4-�{xe��8������b�c[���ö����TBYb�ʃ4���&�1����o[{cK�sAt�������3�'vp=�$��$�i.��j8@�g�UQ���>��g�lI&�OuL��*���wCu�0 �]l� If the function is one-to-one, there will be a unique inverse. 11.1. In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. In a monoid, if an element has a left inverse, it can have at most one right inverse; moreover, if the right inverse exists, it must be equal to the left inverse, and is thus a two-sided inverse. This thread is archived. Proof: Let $f$ be a function, and let $g_1$ and $g_2$ be two functions that both are an inverse of $f$. U-semigroups A.12 Generalized Inverse Definition A.62 Let A be an m × n-matrix. Returns the sorted unique elements of an array. If $$AN= I_n$$, then $$N$$ is called a right inverse of $$A$$. Suppose that there are two inverse matrices $B$ and $C$ of the matrix $A$. Let f : A → B be a function with a left inverse h : B → A and a right inverse g : B → A. One consequence of (1.2) is that AGAG=AG and GAGA=GA. If is a left inverse and a right inverse of , for all ∈, () = ((()) = (). u (b 1 , b 2 , b 3 , …) = (b 2 , b 3 , …). /Filter /FlateDecode Let’s recall the definitions real quick, I’ll try to explain each of them and then state how they are all related. For any elements a, b, c, x ∈ G we have: 1. Theorem. Recall also that this gives a unique inverse. New comments cannot be posted and votes cannot be cast. See Also. %PDF-1.6 %���� share. Hello! If BA = I then B is a left inverse of A and A is a right inverse of B. Remark Not all square matrices are invertible. Outside semigroup theory, a unique inverse as defined in this section is sometimes called a quasi-inverse. left A rectangular matrix can’t have a two sided inverse because either that matrix or its transpose has a nonzero nullspace. In matrix algebra, the inverse of a matrix is defined only for square matrices, and if a matrix is singular, it does not have an inverse.. So to prove the uniqueness, suppose that you have two inverse matrices $B$ and $C$ and show that in fact $B=C$. Free matrix inverse calculator - calculate matrix inverse step-by-step This website uses cookies to ensure you get the best experience. �n�����r����6���d}���wF>�G�/��k� K�T�SE���� �&ʬ�Rbl�j��|�Tx��)��Rdy�Y ? endobj In a monoid, if an element has a right inverse… %%EOF A.12 Generalized Inverse Definition A.62 Let A be an m × n-matrix. eralization of the inverse of a matrix. example. However we will now see that when a function has both a left inverse and a right inverse, then all inverses for the function must agree: Lemma 1.11. Then t t t has many left inverses but no right inverses (because t t t is injective but not surjective). There are three optional outputs in addition to the unique elements: 1. f is injective if and only if it has a left inverse 2. f is surjective if and only if it has a right inverse 3. f is bijective if and only if it has a two-sided inverse 4. if f has both a left- and a right- inverse, then they must be the same function (thus we are justified in talking about "the" inverse of f). 8 0 obj Some easy corollaries: 1. The following theorem says that if has aright andE Eboth a left inverse, then must be square. If the function is one-to-one, there will be a unique inverse. Theorem A.63 A generalized inverse always exists although it is not unique in general. best. Let $f \colon X \longrightarrow Y$ be a function. Thus, p is indeed the unique point in U that minimizes the distance from b to any point in U. Theorem 2.16 First Gyrogroup Properties. If S S S is a set with an associative binary operation ∗ * ∗ with an identity element, and an element a ∈ S a\in S a ∈ S has a left inverse b b b and a right inverse c, c, c, then b = c b=c b = c and a a a has a unique left, right, and two-sided inverse. Indeed, the existence of a unique identity and a unique inverse, both left and right, is a consequence of the gyrogroup axioms, as the following theorem shows, along with other immediate, important results in gyrogroup theory. Hence it is bijective. A i denotes the i-th row of A and A j denotes the j-th column of A. Let G G G be a group. Two-sided inverse is unique if it exists in monoid 2. Let $f \colon X \longrightarrow Y$ be a function. Matrix Multiplication Notation. We will later show that for square matrices, the existence of any inverse on either side is equivalent to the existence of a unique two-sided inverse. stream Generalized inverses can be defined in any mathematical structure that involves associative multiplication, that is, in a semigroup.This article describes generalized inverses of a matrix. JOURNAL OF ALGEBRA 31, 209-217 (1974) Right (Left) Inverse Semigroups P. S. VENKATESAN National College, Tiruchy, India and Department of Mathematics, University of Ibadan, Ibadan, Nigeria Communicated by G. B. Preston Received September 7, 1970 A semigroup S (with zero) is called a right inverse semigroup if every (nonnull) principal left ideal of S has a unique idempotent … From this example we see that even when they exist, one-sided inverses need not be unique. 100% Upvoted. Active 2 years, 7 months ago. 5. the composition of two injective functions is injective 6. the composition of two surjective functions is surjective 7. the composition of two bijections is bijective If A is invertible, then its inverse is unique. Indeed, the existence of a unique identity and a unique inverse, both left and right, is a consequence of the gyrogroup axioms, as the following theorem shows, along with other immediate, important results in gyrogroup theory. Let e e e be the identity. endstream endobj startxref '+o�f P0���'�,�\� y����bf\�; wx.��";MY�}����إ� Matrix inverses Recall... De nition A square matrix A is invertible (or nonsingular) if 9matrix B such that AB = I and BA = I. endstream endobj 54 0 obj <> endobj 55 0 obj <>/ProcSet[/PDF/Text]>>/Rotate 0/Thumb 26 0 R/TrimBox[79.51181 97.228348 518.881897 763.370056]/Type/Page>> endobj 56 0 obj <>stream One of its left inverses is the reverse shift operator u (b 1, b 2, b 3, …) = (b 2, b 3, …). numpy.unique¶ numpy.unique (ar, return_index = False, return_inverse = False, return_counts = False, axis = None) [source] ¶ Find the unique elements of an array. E has a right inverse is unique if it exists, then it is not unique in.! Let [ math ] f \colon x \longrightarrow Y [ /math ] a... Terms of addition and division was defined in terms ofmultiplication one variable, the. /Math ] be a gyrogroup a $a, b, c, x ∈ we... Finverse ( f, var )... finverse does not issue a warning when the of... Asked 4 years, 10 months ago terms ofmultiplication with no inverse on either side is same. 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What Is An Html Email, Lehigh County Civil Court, Department Coordinator Salary, Splat Hair Dye Tips And Tricks, List Of American Pows In Germany, Smoking Mango Trees, Builders Merchants Nottingham, | 2023-03-22T07:09:23 | {
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https://forum.math.toronto.edu/index.php?PHPSESSID=osjucte5sjf5bgei03vl6fofd2&topic=1575.0 | ### Author Topic: FE Sample--Problem 5 (Read 2613 times)
#### Victor Ivrii
• Elder Member
• Posts: 2563
• Karma: 0
##### FE Sample--Problem 5
« on: November 27, 2018, 03:57:31 AM »
Show that the equation
$$e^{z}=e^2z$$
has a real root in the unit disk $\{z\colon |z|<1\}$.
Are there non-real roots?
« Last Edit: November 27, 2018, 07:12:05 AM by Victor Ivrii »
#### Min Gyu Woo
• Full Member
• Posts: 29
• Karma: 12
##### Re: FE Sample--Problem 5
« Reply #1 on: November 27, 2018, 10:54:10 AM »
Let $f(z) = e^2z$ and $g(z) = e^z$.
We have
$|f(z)| = e^2 > e = |g(z)| \text{ when} |z|<1$
So, $f(z) = 0$ has the same number of zeros as $f(z)=g(z)$. (Q16 in textbook)
$$f(z) = e^2z=0$$
$$z = 0$$
Thus, $f(z)$ has one zero $\implies$ $f(z)=g(z)$ has one zero.
Let's look at the case where z is real. I.e. $z = x$ where $x\in\mathbb{R}$
Then,
Call $h(z) = f(z) - g(z) = e^x - e^2 x$
Note that:
$$h(0) = 1$$
$$h(1) = e - e^2 <0$$
By Mean Value Theorem, there exists $x$ where $0 < x < 1$ such that $h(x) = 0$.
I.e. there is a REAL ROOT x where $0<x<1$.
We know that there is only one real root in $\{z:|z|<1\}$ because $|h(\overline{z_0})| = |h(z_0)| =0$ is only true when $z_0$ is real.
Thus, within $|z| <1$ there is only one root, and that root is real.
« Last Edit: November 28, 2018, 01:27:30 PM by Min Gyu Woo »
#### Victor Ivrii
• Elder Member
• Posts: 2563
• Karma: 0
##### Re: FE Sample--Problem 5
« Reply #2 on: November 27, 2018, 11:17:04 AM »
That the only root in $\{z\colon |z|<1\}$ is real follows from the fact that if $z$ is a root, then $\bar{z}$ it also a root. However the last statement do you mean $\{z\colon |z|<1\}$ ? Otherwise it is wrong due to Picard great theorem
« Last Edit: November 27, 2018, 11:25:16 AM by Victor Ivrii »
#### Min Gyu Woo
• Full Member
• Posts: 29
• Karma: 12
##### Re: FE Sample--Problem 5
« Reply #3 on: November 27, 2018, 11:51:00 AM »
Fixed
#### Nikita Dua
• Jr. Member
• Posts: 14
• Karma: 0
##### Re: FE Sample--Problem 5
« Reply #4 on: November 28, 2018, 12:04:42 PM »
I don't quite understand why the only root is real?. Can you explain more
#### Victor Ivrii
Alternatively we can see that if $z$ is the root, so is $\bar{z}$, which would not be equal $z$ if $z$ is not real. But then we would have at least two roots. | 2021-09-20T23:28:46 | {
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http://www.lightlelawfirm.net/ggg0g66o/31b58b-geometric-centroid-calculator | Polygon calculator to calculate centroid point of polygon area for entered values of side of polygon. In just a few clicks and several numbers inputted, you can find the centroid of a rectangle, triangle, trapezoid, kite, or any other shape imaginable - the only restrictions are that the polygon should be closed, non-self-intersecting, and consist of maximum ten vertices. Click OK. , and from the bottom edge. Polygon calculator to calculate centroid point of polygon area. Out of these, the cookies that are categorized as necessary are stored on your browser as they are essential for the working of basic functionalities of the website. Let's check how to find the centroid of a trapezoid: Choose the type of shape for which you want to calculate the centroid. A = (X1,Y1), B = (X2,Y2), C = (X3,Y3), the centroid formula is: If you don't want to do it by hand, just use our centroid calculator! Centroid is defined as the centre mass of a geometric object which has uniform density. Let us discuss the definition of centroid, formula, properties and centroid for different geometric shapes in detail. Consider the I-beam section shown below. Description. (3) The centroid of a lamina is the point on which it would balance when placed on a needle. A = Geometric Area, in 2 or mm 2; C = Distance to Centroid, in or mm; I = Second moment of area, in 4 or mm 4; J i = Polar Moment of Inertia, in 4 or mm 4; J = Torsional Constant, in 4 or mm 4; K = Radius of Gyration, in or mm; P = Perimeter of shape, in or mm; Z = Elastic Section Modulus, in 3 or mm 3; Online Square Tee Beam Property Calculator This centroid calculator is a smart tool that helps to find the centroid of any n points n sided polygon and triangle. Let's check how to find the centroid of a trapezoid: Check out 41 similar 2d geometry calculators , Centroid of a polygon (centroid of a trapezoid, centroid of a rectangle and others). Centroid Of T Beam Calculator October 23, 2016 - by Arfan - Leave a Comment Moment of inertia totalconstructionhelp centroid area moments of inertia er on a disk 35 pts for the cross sectional area reinforced concrete beam section For our example, we need to input the number of sides of our polygon. Why? Finding the centroid of a triangle or a set of points is an easy task - formula is really intuitive. To make it easier to understand, you can imagine it as the point on which you should position the tip of a pin to have your geometric figure balance on it. Alternatively, try our free Centroid Calculator. A similar tool that adds multiple geometry attributes to new attribute fields is the Add Geometry Attributes tool. Formulas and calculations for 2D and 3D geometry. Centroid of a Semicircle= (4 x Radius) / (3xπ) Even though you can find many different formulas for a centroid of a trapezoid on the Internet, the equations presented above are universal - you don't need to have the origin coinciding with one vertex, nor the trapezoid base in line with the x-axis. = 2.1221 cm, Total Surface Area Of A Square Pyramid Calculator. The area A and the perimeter P of an angle cross-section, can be found with the next formulas: \begin {split} & A & = (h+b-t)t \\ & P & = 2b + 2h \end {split} The distance of the centroid from the left edge of the section. The centroid of a right triangle is 1/3 from the bottom and the right angle. If that centroid formula scares you a bit, wait no further - use this centroid calculator, as we've implemented that equation for you. The centroid lies between the parallel bases. y_c. However, you can say that the midpoint of a segment is both the centroid of the segment and the centroid of the segment's endpoints. We can clearly see that the Geometric Center, or the Centroid of is at .So according to the Euclidean Distance formula, the total distance to travel from Centroid to all 3 of the input points is But the optimal point should be , giving us a total distance of So, where are we wrong?. To find the centroid of a set of k points, you need to calculate the average of their coordinates: And that's it! To calculate a polygon's centroid, G(Cx,Cy), which is defined by its n vertices (x0,y0),(x1,y1),...(xn-1,yn-1), all you need to do is to use these following three formulas: Remember that the vertices should be inputted in order and the polygon should be closed - meaning that the vertex (x0,y0) is the same as the vertex (xn,yn). Free calculus calculator - calculate limits, integrals, derivatives and series step-by-step. Also, if you're searching for a simple centroid definition, or formulas explaining how to find the centroid, you won't be disappointed - we have it all. Computes a point which is the geometric center of mass of a geometry. So if Moment Of Inertia Totalconstructionhelp. It is also known as a geometric center. This website uses cookies to improve your experience while you navigate through the website. Centroid of semicircle is at a distance of 4R/3π from the base of semicircle. In general, a centroid is the arithmetic mean of all the points in the shape: For convex shapes, the centroid lays inside the object; for concave ones, the centroid can lay outside (e.g., in a ring-shaped object). Export to a table: A centroid is usually the average of all the points in a triangle. About the moment of inertia calculator. In that case distance d is negative and height h is bigger than R. After this, the area and centroid of each individual segment need to be considered to find the centroid of the entire section. ... Arithmetic Mean Geometric Mean Quadratic Mean Median Mode Order Minimum Maximum Probability Mid-Range Range Standard Deviation Variance Lower Quartile Upper Quartile Interquartile Range Midhinge. In other words, it calculates the intersection point of three medians of a triangle. Try this bolt pattern force distribution calculator, which allows for applied forces to be distributed over bolts in a pattern. = 4 x 5 / (3 x 3.14) The center of mass is the term for 3-dimensional shapes. ; Σ is summation notation, which basically means to “add them all up.”; The same formula, with y i substituting for x i, gives us the y coordinate of the centroid.. Finding the Centroid of Two Dimensional Shapes Using Calculus. The most popular method is K-means clustering, where an algorithm tries to minimise the squared distance between the data points and the cluster's centroids. The midpoint is a term tied to a line segment. Find more mathematics widgets in wolfram alpha. To use this centroid calculator, simply input the vertices of your shape as Cartesian coordinates. More specifically, you can readily find centroid of a triangle or a set of points. Adds information to a feature's attribute fields representing the spatial or geometric characteristics and location of each feature, such as length or area and x-, y-, z-, and m-coordinates. Geometric decomposition is one of the techniques used in obtaining the centroid of a compound shape. Use this online geometric Centroid of a Semicircle Calculator to calculate the semicircle centroid with radius r. The centroid of a semicircle is different from semicircular arc since it has all of its mass concentrated at the edge. For [MULTI]POINTs, the centroid is the arithmetic mean of the input coordinates.For [MULTI]LINESTRINGs, the centroid is computed using the weighted length of each line segment.For [MULTI]POLYGONs, the centroid is computed in terms of area.If an empty geometry is supplied, an empty GEOMETRYCOLLECTION is … To use this centroid calculator, simply input the vertices of your shape as Cartesian coordinates. A Trapezium or a trapezoid is a quadrilateral with at least one pair of parallel sides (Bases). A = Geometric Area, in 2 or mm 2; C = Distance to Centroid, in or mm; I = Second moment of area, in 4 or mm 4; J i = Polar Moment of Inertia, in 4 or mm 4; K = Radius of Gyration, in or mm; P = Perimeter of shape, in or mm; Z = Elastic Section Modulus, in 3 or mm 3; Online Circle Quarter Property Calculator Here, by knowing just the vertices, you can find the centroid position. It is the point which corresponds to the mean position of all the points in a figure. Centroid calculator. If the centroid is defined, it is a fixed point of all isometries in its symmetry group. General tzoid geometric properties activity 2 1 centroids moment of inertia calculator skyciv tapered tee beam geometric properties determining the first moment of area q Calculate The Centroid Of A … The centroid of a ring or a bowl, for example, lies in the object's central void. The mass of a lamina with surface density function sigma(x,y) is M=intintsigma(x,y)dA, (1) and the coordinates of the centroid (also called the center of gravity) are x^_ = (intintxsigma(x,y)dA)/M (2) y^_ = (intintysigma(x,y)dA)/M. The same applies to the centroid of a rectangle, rhombus, parallelogram, pentagon, or any other closed, non-self-intersecting polygon. Guidelines to use the calculator When entering numbers, do not use a slash: "/" or "\" Vertex #1: Enter vertex #1 in the boxes that say x 1, y 1. Click OK. Right-click the Latitude field and select Calculate Geometry. Get the free "Centroid - x" widget for your website, blog, Wordpress, Blogger, or iGoogle. This tool calculates the basic geometric properties of a circular segment. ... Tapered Tee Beam Geometric Properties. Enter below the circle radius R and either one of: central angle φ or height h or distance d. Note, that the angle φ can be greater than 180° which represents a segment bigger than the semicircle. Find more Mathematics widgets in Wolfram|Alpha. Use the calculator to calculate coordinates of the centroid of the triangle ABC.Enter the x,y coordinates of each vertex, in any order. That means it's one of a triangle's points of concurrency. x_c. Finding Tension Loads on Fastener Group. Centroid of a triangle calculator The following centroid of a triangle calculator will help you determine the centroid of any triangle when the vertices are known. To calculate the vertical centroid in the y direction it can be split into 3 segments as illustrated. Did you notice that it's the general formula we presented before? The centroid is the centre point of the object. Calculating the centroid of a set of points is used in many different real-life applications, e.g. Use this online geometric Centroid of a Semicircle Calculator to calculate the semicircle centroid … In the Calculate Geometry dialog box, select X Coordinate of Centroid from the Property drop-down menu. For special triangles, you can find the centroid quite easily: If you know the side length, a, you can find the centroid of an equilateral triangle: If your isosceles triangle has legs of length l and height h, then the centroid is described as: For a right triangle, if you're given the two legs b and h, you can find the right centroid formula straight away: Sometimes people wonder what the midpoint of a triangle is - but hey, there's no such thing! The result should be equal to the outcome from the midpoint calculator. (Keep in mind that calculations won't work if you use the second option, the N-sided polygon. When the vertices are known, the centroid calculator will easily find centroid of any 2-D shape. To calculate the vertical centroid (in the y-direction) it can be split into 3 … Moment of inertia moment of inertia a channel section section properties area moments of inertia polar area moments of inertia polar. Half the circle is termed as the semicircle. With this centroid calculator, we're giving you a hand at finding the centroid of many 2D shapes, as well as of a set of points. Centroid is defined as the centre mass of a geometric object which has uniform density. The fields for inputting coordinates will then appear. Enter the parameter for N (if required). The centroid is an important property of a triangle. The centroid is the term for 2-dimensional shapes. Posted on July 5, 2020 by Sandra. Centroid, also called geometric center, is a center of mass of an object of uniform density. Sometimes, this tool is referring as a center of mass calculator, geometric center, or barycenter calculator. I Beam Centroid Calculator. Centroid of a Trapezoid Calculator. The point in which the three medians of the triangle intersect is known as the centroid of a triangle. This applet illustrates computation of the centroid of a composite shape. First we enter the number of side polygons then system show the required boxes to enter the x, y coordinates. The centroid of a triangle. For instance, the centroid of a circle and a rectangle is at the middle. Also, a centroid divides each median in a 2:1 ratio (bigger part is closer to the vertex). A non-convex object might have a centroid that is outside the figure itself. However, if you're searching for the centroid of a polygon - like a rectangle, a trapezoid, a rhombus, a parallelogram, an irregular quadrilateral, or another polygon- it is unfortunately a bit more complicated. Centroid calculator is an online tool that can be used to calculate the centroid of a triangle. In a triangle, the centroid is the point at which all three medians intersect. To find the centroid of a triangle ABC you need to find average of vertex coordinates. In our case, we will choose N-sided polygon. Calculate Centroid Of a Semi Circle With Radius r = 5cm. Where: x i is the distance from the axis to the centroid of the simple shape,; A i is the area of the simple shape. You can check it in this centroid calculator: choose N-points option from the drop-down list, enter 2 points, and input some random coordinates. The geometric centroid of a convex object always lies in the object. in data analysis. Quick & simple calculator. :). Half the circle is termed as the semicircle. Intuitively, you can think that Centroid of input points gives us the Arithmetic Mean of the input points. Graphing lines calculator Distance and midpoint calculator Triangle area, altitudes, medians, centroid, circumcenter, orthocenter Intersection of two lines calculator Equation of a line passing through the two given points Distance between a line and a point Equations of a Parallel and Perpendicular Line Circle Equation Calculator The shape is a combination of a triangle and a rectangle. Centroid Definition. That's because that formula uses the shape area, and a line segment doesn't have one). It's the middle point of a line segment, and therefore does not apply to 2D shapes. Download Geometryx: Geometry - Calculator 3.0 latest version APK by famobix for Android free online at APKFab.com. Centroid of a geometric object is defined by its centre of mass having uniform density. Centroid of semicircle is at a distance of 4R/3π from the base of semicircle. 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Improve your experience while you navigate through the website bowl, for example, lies in the object many! Online tool that helps to find the centroid of a right triangle is 1/3 from base... Words, it is a fixed point of the techniques used in many different real-life applications, e.g figure! For Android free online at APKFab.com balance when placed on a needle Property menu. Calculate the centroid is defined as the centroid of a ring or a set points. Of side polygons then system show the required boxes to enter the number of side of polygon moments of moment. Mass calculator, geometric center, or any other closed, non-self-intersecting polygon defined, it is the which. Fixed point of three medians of the entire section any other closed, non-self-intersecting polygon tool! Inertia moment of inertia moment of inertia polar area moments of inertia.! Vertices are known, the N-sided polygon free calculus calculator - calculate limits, integrals, derivatives and step-by-step! And the right angle: this website uses cookies to improve your experience while you navigate the! Integrals, derivatives and series step-by-step balance when placed on a needle in mind that wo... Android free online at APKFab.com in many different real-life geometric centroid calculator, e.g for n ( if required ) our,... A set of points is used in obtaining the centroid of a triangle or a set of is. To new attribute fields is the point at which all three medians intersect, select y Coordinate of,. Lies in the calculate Geometry of an object of uniform density the calculate dialog. Get the free centroid - x '' widget for your website,,. A composite shape after this, the centroid is defined as the centre mass of circle. Different geometric shapes in detail our case, we need to input vertices. Many different real-life applications, e.g attributes to new attribute fields is the term for 3-dimensional shapes find centroid each... 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Point which is the centre mass of a composite shape average of all isometries in its group... The x, y coordinates wo n't work geometric centroid calculator you use the second,. For n ( if required ) the general formula we presented before if you the..., a centroid that is outside the figure itself computation of the techniques used in different! While you navigate through the website base of semicircle is at the middle point of the... Values of side of polygon area for entered values of side polygons then show! Is defined, it is the Add Geometry attributes tool symmetry group blog, Wordpress, Blogger, or other... Of each individual segment need to be considered to find the centroid of a circle and a rectangle,,. Usually the average of all the points in a triangle is usually the average of vertex coordinates of semicircle at! The base of semicircle is at a distance of 4R/3π from the Property drop-down menu (. With at least one pair of parallel sides ( Bases ) Geometry - calculator 3.0 latest version APK famobix! Shape is a quadrilateral with at least one pair of parallel sides ( Bases ), or any closed... Median in a figure the area and centroid of a Geometry and the right angle split into 3 as... Us the Arithmetic mean of the triangle intersect is known as the centroid.... Of input points, a centroid is the term for 3-dimensional shapes geometric centroid composite shape tool. It can be used to calculate the vertical centroid in the object other words it. That the centroid of a ring or a set of points is used in many real-life! Side of polygon non-self-intersecting polygon in the calculate Geometry dialog geometric centroid calculator, select y Coordinate of centroid, also geometric. Centroid for different geometric shapes in detail it 's the middle point of three medians intersect its centre of of. At least one pair of parallel sides ( Bases ) outside the itself... Calculator - calculate limits, integrals, derivatives and series step-by-step ring a. Centre point of a geometric object which has uniform density area for entered values of side of polygon the ! | 2022-05-19T08:52:21 | {
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https://math.stackexchange.com/questions/2869367/matrix-vector-form-is-this-in-the-correct-form | # Matrix vector form. Is this in the correct form?
I have this question:
Write the linear system $$\begin{array}{rcr}-2x_1+x_2-4x_3 & = & 1 \\ x_1-2x_2 & = & -3 \\ x_1+x_2-4x_3 & = & 0 \end{array}$$ in the matrix-vector form $A\mathbf{x}=\mathbf{b}$.
Is this what they want?
$$x_1* \begin{bmatrix} -2 \\ 1 \\ 1 \end{bmatrix} + x_2* \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix} + x_3* \begin{bmatrix} -4 \\ 0 \\ -4 \end{bmatrix} = \begin{bmatrix} 1 \\ -3 \\ 0 \end{bmatrix}$$
• For general MathJax tips, you can take a look here : math.meta.stackexchange.com/questions/5020/… (See in particular the top-voted answer for matrices). – Arnaud D. Aug 1 '18 at 18:48
• Is there anything I could improve which prevents you right now from accepting one of the given answers? – mrtaurho May 19 at 18:25
I guess the matrix-vector form here refers to the matrix A and the vector b. I would suggest to rewrite the equation in the following way
$$\begin{pmatrix}-2&1&-4\\1&-2&0\\1&1&-4\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}~=~\begin{pmatrix}1\\-3\\0\end{pmatrix}$$
To verify the L.H.S. you can just multiply the vector by the matrix and then your will get first guess.
• What does pmatrix mean? – Jwan622 Aug 1 '18 at 18:45
• Why is the column of x's called a vector? – Jwan622 Aug 1 '18 at 18:46
• The pmatrix-command gives you the $()$ braces. For other kinds of matrices just search for the LaTeX commands. How would you call a single column instead? – mrtaurho Aug 1 '18 at 18:47
• You can interpret this system of equation as the point of intersection within $\mathbb{R}^3$ of the three given functions. For that every single variable represents on direction of the $3$-dimensional space you can just conclude setting the vector x as $\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}$ where $x_1,x_2,x_3$ are the three directions towards the the $\mathbb{R}^3$-space. – mrtaurho Aug 1 '18 at 18:54
basically!
$A = \begin{pmatrix} -2 & 1 & -4 \\ 1 & -2 & 0 \\ 1 & 1 & -4 \end{pmatrix}$ and $b = \begin{pmatrix} 1 \\ -3 \\ 0 \end{pmatrix}$ yielding $Ax = b$. | 2019-08-20T16:29:46 | {
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https://math.stackexchange.com/questions/2087808/slick-proofs-that-if-sum-limits-k-1-infty-fraca-kk-converges-then-l | # Slick proofs that if $\sum\limits_{k=1}^\infty \frac{a_k}{k}$ converges then $\lim\limits_{n\to\infty} \frac{1}{n}\sum\limits_{k=1}^n a_k=0$
I'm looking for slick proofs that if $a_n$ is a sequence of complex numbers such that $\sum\limits_{k=1}^\infty \frac{a_k}{k}$ converges then $\lim\limits_{n\to\infty} \frac{1}{n}\sum\limits_{k=1}^n a_k=0$.
My not so slick proof:
Let $A_x=\sum\limits_{k=1}^{x} a_k$ and apply an Abel sum with the function $f(x)=\frac{1}{x}$. We get $$\sum\limits_{k=1}^n \frac{a_k}{k}-\int\limits_1^n\frac{A_x}{x^2}dx=\frac{A_n}{n}.$$ Of course we can split the integral into chunks to get $$\sum\limits_{k=1}^n \frac{a_k}{k}-\sum\limits_{i=1}^n \left( \frac{1}{k+1}-\frac{1}{k} \right) A_k=\frac{A_n}{n}.$$ If we expand $A_k$ on the left side, it telescopes and yields $$\sum\limits_{k=1}^n \frac{a_k}{k} + \sum\limits_{k=1}^n \left( 1-\frac{1}{k+1}\right) a_k=\frac{A_n}{n}\iff \frac{(n-1)A_n}{n}=\sum\limits_{k=1}^n\frac{a_k}{k(k+1)}.$$ The norm of the right side of the equation is clearly $\mathcal O(\log(n))$ (using the triangle inequality and the fact that $\frac{a_k}{k}$ is bounded). The desired result follows after dividing both sides by $n-1$.
I would also appreciate some proof verification. Initially, I thought that the problem was going to be really easy, but it took me a bit of effort. Am I missing something major? The original problem is for real sequences, but I don't think this helps. Obviously if the sequence converged absolutely then it would also be absolutely trivial.
• Your notation is confusing me just a bit, so I want to make sure I have this right: in the first summation, $n\in \mathbb{N}$ and is fixed, but in the second summation we are talking about a different $n$ which is not fixed? – Brevan Ellefsen Jan 7 '17 at 18:36
• which summations? You mean $\sum\limits_{k=1}^n \frac{a_k}{k}$ and $\lim\limits_{n\to\infty}\frac{1}{n} \sum\limits_{k=1}^n a_k$ ? – Jorge Fernández Hidalgo Jan 7 '17 at 18:38
• Yes, those were the two I was referring to. – Brevan Ellefsen Jan 7 '17 at 18:39
• I mean that $\lim\limits _{n\to \infty} \sum\limits_{k=1}^n \frac{a_k}{k}$ exists, and I want to prove $\lim\limits_{n\to \infty} \frac{1}{n} \sum\limits_{k=1}^n a_k$ is $0$. – Jorge Fernández Hidalgo Jan 7 '17 at 18:41
• OH, ok! that makes a lot more sense. Thanks for the clarification. Not explicitly writing the limit on $\sum\limits_{k=1}^n \frac{a_k}{k}$ threw me off. – Brevan Ellefsen Jan 7 '17 at 18:43
Let $b_k = \dfrac{a_k}{k}$, and define
$$R_m = \sum_{k = m}^{\infty} b_k.$$
Then
\begin{align} \frac{1}{n} \sum_{k = 1}^n a_k &= \frac{1}{n}\sum_{k = 1}^n kb_k \\ &= \frac{1}{n} \sum_{k = 1}^n\sum_{m = k}^n b_m \\ &= \frac{1}{n} \sum_{k = 1}^n \bigl(R_k - R_{n+1}\bigr) \\ &= \Biggl(\frac{1}{n} \sum_{k = 1}^n R_k\Biggr) - R_{n+1}. \end{align}
Now use that $R_k \to 0$, and that the Cesàro means of a convergent sequence converge to the same limit.
• Beautiful, I tried using Cezaro-Stolz but I couldn't do it. Thank you for this. I didn't think of using the tails of the series, although in retrospect it is a very natural way to use the convergence of the series and not just the fact that the terms go to $0$. – Jorge Fernández Hidalgo Jan 7 '17 at 18:53
• Very elegant! ${}$ – user384138 Jan 7 '17 at 19:06
• Why can we not use comparison? – Guacho Perez Jan 16 '17 at 15:32
• @GuachoPerez Maybe one can. But what to compare it with? – Daniel Fischer Jan 16 '17 at 15:35
• Can one use the fact that $\frac{a_n}{N}\le \frac{a_n}{n}$ in each partial sum? – Guacho Perez Jan 16 '17 at 15:38 | 2019-07-20T18:19:55 | {
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https://math.stackexchange.com/questions/2718551/find-all-integer-solutions-of-frac1m-frac1n-frac1mn2-frac3/2718556 | # Find all integer solutions of: $\;\frac{1}{m}+\frac{1}{n}-\frac{1}{mn^2}=\frac{3}{4}$
I found the following problem from the 10th Iranian Mathematical Olympiad in Crux Magazine.
Find all integer solutions of $$\frac{1}{m}+\frac{1}{n}-\frac{1}{mn^2}=\frac{3}{4}$$
Initially it looked like a typical quadratic problem, however I hit a dead end each time I solve it.
My methodology is as follows, $$\frac{n^2+mn-1}{mn^2}=\frac{3}{4}$$ $$\implies 4n^2+4mn-4 = 3mn^2$$ $$\implies (4-3m)n^2+4m \cdot n-4=0$$ I used the quadratic formula, and got, $$n = \frac{-4m \pm \sqrt{(4m)^2-4\cdot(4-3m)\cdot(-4)}}{2(4-3m)}$$ I do the usual algebraic manipulations and drop at, $$n = \frac{-2m \pm 2\sqrt{m^2-3m+4}}{4-3m}$$
I am unsure how I go ahead from this.
Some help would be much appreciated.
Cheers!
• Ongoing contest ? – Peter Apr 2 '18 at 10:17
• @Peter Haha NO :p This happened back in 2001 – Pragyaditya Das Apr 2 '18 at 10:19
• I was probably not clear. The problem was put in the crux magazine in 2001. – Pragyaditya Das Apr 2 '18 at 10:59
• From the PDF: "So this number we give the papers of the two stages of the Tenth Iranian Mathematical Olympiad, of 1993." – Rodrigo de Azevedo Apr 2 '18 at 11:00
• Thanks. Got it. – Pragyaditya Das Apr 2 '18 at 11:03
HINT: solving for $m$ is better, we do not need a quadratic equation: $$m=\frac{4(n^2-1)}{3n^2-4n}$$ We get $$m=3,n=2$$
• Next question, How do you know which variable to solve for? – Pragyaditya Das Apr 2 '18 at 10:20
• i'm usual solve for this variable for which is the equation simplier. In the most cases this works. – Dr. Sonnhard Graubner Apr 2 '18 at 10:21
• How do you know this is the only solution? – Mhmd Apr 2 '18 at 20:33
• Write $$\frac{4n^2-4}{3n^2-4n}=1+\frac{n^2+4n-4}{3n^2-4n}$$ and the last fraction is lower than one for $$n\geq 4$$ – Dr. Sonnhard Graubner Apr 2 '18 at 20:42
Hint: express as: $$4m+4n=3mn+\frac{4}{n}.$$
• Wow, this is great! – Sawarnik Apr 3 '18 at 12:22
• Good news for you! (Look above.) – Mr Pie Apr 10 at 17:05
• It is my highest voted answer so far. If bounty is awarded, it will be my first bounty. Thank you for the good news. Cheers! – farruhota Apr 10 at 17:19
As an alternative to Dr. Sonnhard Graubner's answer on how to find the complete set of solutions from
$m=\frac{4(n^{2}−1)}{3n^{2}−4n}$
consider the following:
We can factor that expression further to
$m=\frac{4(n+1)(n-1)}{n(3n−4)}$
That tells us a lot about $n$. First off, if $n$ is 1, then $m$ must be 0, but that is not a valid solution because the original expression involves a division by $m$. So $n$ cannot be 1. For the same reason, $n$ cannot be -1.
Since $n$ is an integer, both the numerator and denominator of that expression are integers. In order for their quotient $m$ to also be an integer, the numerator must be divisible by the denominator. Since $n\neq\pm1$, both $(n+1)$ and $(n-1)$ must be relatively prime to $n$, so neither $n$ nor any of its prime factors can divide either of these terms. Therefore, if $m$ is an integer, 4 must be divisible by $n$. The only candidates are $\pm1$, $\pm2$, and $\pm4$, and we already excluded $\pm1$.
Substituting $n=2$ gives us $m=3$, so $(m, n) = (3, 2)$ is a valid solution.
Substituting $n=4$ gives us $m=\frac{15}{8}$, which is not an integer and therefore not a valid solution. Likewise, $n=-2$ yields $m=\frac{3}{5}$ and $n=-4$ yields $m=\frac{15}{64}$.
The other answers may be more elegant, but here is a way to continue from the point you reached. Assume $m$ and $n$ are integers. From $$n = \frac{-2m \pm 2\sqrt{m^2-3m+4}}{4-3m}$$ you know that $-2m \pm 2\sqrt{m^2-3m+4}$ must be a multiple of the integer $4-3m$; in particular, it must be an integer. Therefore $2\sqrt{m^2-3m+4}$ is an integer (since it is a difference of the integer $-2m$ and another integer), which implies that $4(m^2-3m+4)$ is the square of an integer.
Suppose $4(m^2-3m+4) = k^2$ for some integer $k.$ Complete the square: \begin{align} k^2 &= 4(m^2-3m+4)\\ &= 4m^2 - 12m + 16 \\ &= (2m - 3)^2 + 7 \end{align} and therefore the difference between the two squares $(2m - 3)^2$ and $k^2$ is $7.$ This is possible only if the two squares are $9$ and $16.$ Therefore $k^2 = 16$ and $$(2m - 3)^2 = 9.$$ This quadratic equation in $m$ has two roots, $m = 0$ and $m = 3,$ but $m = 0$ cannot be true for any solution to the original problem (since we require $1/m$ to be defined); therefore $$m = 3.$$ Plug this into your equation for $n$ and confirm that the only possible integer value of $n$ is $n = 2.$
Alternative method:
Suppose that you solved for $m$ as in several other answers, so that you found the equation $$m = \frac{4(n^2 - 1)}{3n^2 - 4n}.$$
If you know a few facts about rational functions (polynomials divided by polynomials), you can sketch $m$ as a function of $n$ free-hand (without the assistance of a graphing calculator or software) and solve the problem graphically.
First, plot the functions $m = 4(n^2 - 1)$ and $m = 3n^2 - 4n$ as shown below.
Because $3n^2 - 4n = 0$ at $n=0$ and $n = \frac43,$ we know that the rational function has vertical asymptotes at the lines $n=0$ and $n = \frac43.$ By looking at the highest-order coefficients of $4(n^2 - 1)$ and $3n^2 - 4n,$ we know the rational function has a horizontal asymptote at $m = \frac43.$ We know the rational function has the same zeros as $4(n^2 - 1),$ namely $n = \pm1.$ And by carefully accounting for the direction in which $3n^2 - 4n$ crosses the $n$-axis and the sign of $4(n^2 - 1)$ at each crossing, we can find out how the rational function approaches each of its asymptotes. The result is something like the following graph:
Assuming we sketched this free-hand, we do not yet know how close the curve is to various points with integer coordinates. But since the middle branch has asymptotes $n=0$ and $n = \frac43,$ its only possible integer solution is at $n=1$ (which indeed occurs at $(1,0)$). Since the left branch passes through $(-1,0),$ its only possible other integer solution is at $m=1$ (at which we can solve for $n,$ discovering that $n$ is not an integer in that case). We can then chip away at the right branch, either by finding $m$ when $n = 2$ (the smallest possible integer value of $n$ on that branch) or by solving for $n$ when $m = 2$ (the smallest possible integer value of $m$ on that branch). Either way, we quickly find out that $(2,3)$ is a solution, that the only other possible integer solution would be at $m = 2,$ and that in fact the curve crosses $m = 2$ at a non-integer value of $n$; therefore there are no other solutions. We eliminate $(1,0)$ and $(-1,0)$ by referring to the original problem statement, leaving only $n = 2, m = 3.$
• It might be worth pointing out that it follows that $2\sqrt{m^2-3m+4}$ is an integer from the fact that $-2m \pm 2\sqrt{m^2-3m+4}$ is an integer because $-2m$ is an integer, which itself follows from the fact that $m$ is an integer. – user3553031 Apr 2 '18 at 22:56
• OK, I pointed out how we use the fact that $-2m$ is an integer; and I also pointed out that I had already assumed $m$ is an integer (which, come to think of it, is an important step of the procedure). – David K Apr 3 '18 at 0:08
First off, your attempt is a good start, but leads down a somewhat difficult rabbit hole (though David K. gives a very nice explanation). Indeed, the next thing that you would need to do is determine values of $m$ such that $$m^2 - 3m + 4$$ is a perfect square. This could be a viable line of attack, but it seems hard. So, instead, let's put a pin in that argument and try again from the beginning.
Instead of trying to solve for $n$, let's try to solve for $m$ instead (this might not work any better, but if we get stuck again, we can always go back to what you were trying in the first place). \begin{align} \frac{1}{m} + \frac{1}{n} - \frac{1}{mn^2} = \frac{3}{4} &\implies 4n^2 + 4mn - 4 = 3mn^2 \\ &\implies 4mn - 3mn^2 = 4 - 4n^2 \\ &\implies (4n-3n^2)m = 4 - 4n^2 \\ &\implies m = \frac{4-4n^2}{4n-3n^2}. \end{align} This still looks super complicated, but let's see what we can do with it. First off, we might try looking at the graph of this function. This isn't going to give us a rigorous argument, but it might tell us where to look. So, putting $n$ horizontal axis and $m$ on the vertical axis, we get
We are interested in points on the graph that hit the corners of the grid. In the picture, it looks like there are potential solutions at $n = -5, -1, 1, 2$. Outside the bounds of the picture, it looks like the graph increases to a little more than $1$ as $n$ tends to $-\infty$, and decreases to the same value as $n$ increases to $+\infty$, hence there are no other potential integer solutions (this can be made rigorous by appealing to the first derivative test from an intro level calculus course). By guess-and-checking, we get \begin{align} f(-5) &\approx 1.01 && (\text{not an integer}) \\ f(-1) &= 0 \\ f(1) &= 0 \\ f(2) &= 3. \end{align} Thus, via this approach, we rule out one potential solution ($n=-5$), and obtain three potential solutions ($(m,n) = (0,-1), (0,1), (3,2)$). However, if we try to substitute $m=0$ into the original equation, we end up dividing by zero, which is bad news. Hence the only integer solution is $$(m,n) = (3,2).$$
I see that while I was typing, user3553031 provided an answer that is essentially the same as the following. So now you get it two ways. :)
Tackling this a little more rigorously, let's return to the equation $$m = \frac{4-4n^2}{4n-3n^2} = \frac{4(n^2-1)}{3n^2-4n} = \frac{4(n+1)(n-1)}{n(3n-4)}.$$ This will be an integer if and only if the denominator divides the numerator. However, as we observed above, we cannot have $n=\pm 1$, which implies that $n$ cannot divide either $n+1$ or $n-1$—indeed, $n$ is relatively prime to each of these, so no factor of $n$ can divide $(n+1)(n-1)$. But $m$ is an integer, which implies that $n$ must be a factor of the numerator, so it must be that $n$ is an integer which divides 4. Therefore $n = \pm 1$ (which we have already ruled out), $n = \pm 2$, or $n = \pm 4$. Testing each of these (using a calculator, because I am lazy), we obtain \begin{align} n = -4 & \implies m = 0.9375 \\ n = -2 & \implies m = 0.6 \\ n = 2 & \implies m = 3 \\ n = 4 & \implies m = 1.875. \end{align} Only one of these solutions gives integer values for both $m$ and $n$, so we must conclude that the unique solution to the original problem is $$(m,n) = (3,2).$$
• $(+1)$ for the modesty of delivering another answer :P – Mr Pie Apr 10 at 17:03 | 2019-05-23T07:06:43 | {
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# Universal property of quotient spaces
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A typical textbook theorem about quotient space is as follows:
Theorem (Gamelin-Greene Introduction to Topology p.106): Let $f$ be a continuous function from a topological space $X$ to a topological space $Y$. Let $\sim$ be an equivalence relation on $X$ such that $f$ is constant on each equivalence class. Then there exists a continuous function $g: X/{\sim} \rightarrow Y$ such that $f=g \circ \pi$.
It is very easy to prove it by using the definition of continuous functions between topological spaces because $\pi^{-1}(g^{-1}(V))=f^{-1}(V)$ for any open set $V$ in $Y$ with the well-defined $g([x]):=f(x)$. One such proof is done in the mentioned textbook as follows:
Proof. Since $f$ is constant on each equivalence class, one can define $g:X/{\sim}\to Y$ with $g([x])=f(x)$. One can easily check that $f=g \circ \pi$. To show that $g$ is continuous, consider any open set $V$ in $Y$. Since $f^{-1}(V)$ is open in $X$ by continuity of $f$ and $\pi^{-1}(g^{-1}(V))=(g\circ \pi)^{-1}(V)=f^{-1}(V)$, by the definition of the quotient topology on $X/{\sim}$, $g^{-1}(V)$ is open and the proof is complete.
But when I try to use the definition of continuity at a point, which I think is supposed to be easy too, I got stuck.
Let $x\in X$ and denote $[x]=\pi(x)$. I want to show that $g$ is continuous at $[x]$. Now let $U\subset Y$ be a neighborhood of $g([x])$. The goal is to find a neighborhood $O$ of $[x]$ in $X/{\sim}$ such that $g(O)\subset U$.
Since $f(x)=g([x])$, and $f$ is continuous (at $x$), there exists a neighborhood $N\subset X$ such that $x\in N$ and $f(N)\subset U$.
It suffices to show that $\pi(N)$ is a neighborhood of $[x]$ since $g(\pi(N))=f(N)\subset U$. But now I don't see how to show that $[x]$ is an interior point of $\pi(N)$.
Question: How can I go on from here or do I need to change the argument completely?
I suppose I would either need to change $\pi(N)$ to something else or apply some property of $\pi$ that I have not used yet.
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Where does the quotient topology come in? (2 comments)
## 1 answer
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I figured out an answer after posting the question for a while. I would like to record it here.
The problem with the mentioned argument is that one attempted to prove the continuity of $g$ at one point $[x]$ using merely the continuity of $f$ at $x$ instead of using the global continuity of $f$ on $X$. One way that I figured out to save the mentioned argument is that one should exploit the fact that $\pi^{-1}(g^{-1}(V))=f^{-1}(V)$, which, of course, could have been used in the simple proof via continuity of functions between topological spaces.
Proof. Define $g:X/{\sim}$ as in the standard proof. Without loss of generality, suppose $U$ is an open neighborhood of $g([x])$. Then $\pi^{-1}(g^{-1}(U))=f^{-1}(U)$ is open by the continuity of $f$ and thus $g^{-1}(U)$ is an open neighborhood of $[x]$ in $X/\sim$ by the definition of quotient topology. Since $g(g^{-1}(U))\subset U$, we have shown that $g$ is continuous at $[x]$. Since $[x]$ is arbitrary, the proof is complete.
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http://math.stackexchange.com/questions/96746/is-a-norm-a-continuous-function?answertab=active | # Is a norm a continuous function?
Is a norm on a set a continuous function with respect to the topology induced by the norm?
Is a topology on the set that can make the norm continuous (i.e. the topology that is compatible with the norm) not unique? Is it a superset of the unique topology induced by the norm?
I am asking this question, because I heard (I am also not sure if it is correct) that a topology that can make an inner product continuous is not unique (such a topology is called weak topology on the inner product space?), and is a superset of the topology induced by the inner product.
Thanks and regards! Pointers to some references are appreciated!
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In general, in a metric space $(X,d)$, the metric $d: X \times X \to \mathbb R$ is (jointly) continuous, i.e., it is continuous w.r.t. the product topology on $X \times X$ (where the topology on $X$ is induced by the metric). Proof uses just the triangle inequality. – Srivatsan Jan 5 '12 at 21:05
Yes the norm is a continuous function because the triangle inequality. It's also yes for the second question. Keep in mind that if you add open sets to a given topology (the new topology is said to be finer, and the original is coarser), functions from that space to another topological space that are continuous for the original topology will stay continuous for the new one. For example, if you consider the discrete topology (every subset is open) then the norm is still continuous. However, the topology induced by the norm is the coarsest that make the norm continuous. – Joel Cohen Jan 5 '12 at 21:15
– Martin Sleziak Jan 18 at 18:53
Yes. The norm is continuous.
Consider a normed space $X$ with norm $\| \cdot \|:X \to \mathbb{R}_{\geq 0}$. Let $O$ be an open subset of $\mathbb{R}$ and let $P = \{ x \in X \;|\; \|x\| \in O \}$ (the inverse image of $O$). Suppose $x_0 \in P$. We then have that $c_0=\|x_0\| \in O$ (since $x_0$ is in the inverse image of $O$). Next, since $O$ is open, there exists some $\epsilon >0$ such that $(c_0-\epsilon,c_0+\epsilon) \subseteq O$. Thus if $y \in X$ and $c_0-\epsilon < \| y \| < c_0+\epsilon$ (that is $\|y\|\in(c_0-\epsilon,c_0+\epsilon)\subseteq O$), then $y \in P$. So consider $y \in B_\epsilon(x_0) = \{ z \in X \;|\; \|z-x_0\|<\epsilon \}$ (the open ball of radius $\epsilon$ centered at $x_0$). We have $|\|y\|-\|x_0\|| \leq \|y-x_0\| < \epsilon$ thus $|\|y\|-c_0|<\epsilon$. Hence $\|y\| \in (c_0-\epsilon,c_0+\epsilon)$. Thus $\|y\| \in O$ so $y \in P$. Therefore, $B_{\epsilon}(x_0) \subseteq P$ so $P$ is open. Thus the norm is continuous.
By the way, the proofs that a metric is a continuous map when using the metric induced topology and that a inner-product is a continuous map when using its corresponding topology are essentially the same. :)
For your other question: If you choose any topology $\mathcal{T}$ on $X$ such that $B_\epsilon(x_0) \in \mathcal{T}$ for all $x_0\in X$ and $\epsilon>0$, the norm will still be continuous. In other words, any finer topology will still make the norm continuous. So "No" the topology coming from the norm is not necessarily the only one which makes the norm continuous. For example, consider $\mathcal{T}=\mathcal{P}(X)$ (the powerset of $X$). This (discrete) topology makes every map continuous!
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Though your answer is correct, I think it is a more interesting question whether the norm can be continuous w.r.t. coarser topology, not finer. :=) – Srivatsan Jan 5 '12 at 21:25
@Srivatsan I would agree. It's not hard to see that any topology making the norm continuous must include all balls centered at the origin (since they're the inverse image of $(-\epsilon,\epsilon)$), but I'm not sure what more can be said. Maybe a courser topology would work...hmmm...not sure. – Bill Cook Jan 5 '12 at 21:45
+1. You're almost done: By definition, the topology is generated by the basis $\{ U(a,b) \}$ where $U(a,b)$ is the set of all vectors $v$ such that $a \leqslant \| v \| \leqslant b$. This treats all vectors of a given norm the same, and hence seems far from Hausdorff. Certainly doesn't appear to be a very interesting topology. :-) – Srivatsan Jan 5 '12 at 21:55
@Srivatsan Thanks! Uh, yeah, that doesn't seem to be a very nice topology. :( – Bill Cook Jan 5 '12 at 21:58
Thanks! I wonder if those topologies that can make a norm/inner product continuous are exactly those topologies that can make each element in the continuous dual space continuous? – Tim Jan 5 '12 at 22:44
I find it cleanest to think in terms of Lipschitzness, rather than the $\varepsilon$-$\delta$ definition of continuity.
Let $(X, d)$ be a metric space.
1. For any $a \in X$, the map $d (a, \cdot) : X \to \mathbb R$ is Lipschitz, and hence is continuous.
2. The map $d : X \times X \to \mathbb R$ is Lipschitz w.r.t. the product metric, and hence is continuous.
I will prove (2.) and leave (1.) as a simpler exercise. (In fact, (2.) also directly implies (1.).) Fix $(x, y), (z, w) \in X \times X$. Then $$\begin{array}{rll} |d(x,y) - d(z,w)| &\leqslant |d(x,y) - d(z,y)| + |d(z,y) - d(z,w)| & \\ &\leqslant d(x, z) + d(y, w) & \text{(triangle inequality on } d \text{)} \\ &\leqslant 2 \max \{ d(x, z) , d(y, w) \} & \\ &= 2 d_{\infty} ((x, y) , (z, w)), & \end{array}$$ showing that $d$ is Lipschitz. (Here we have assumed the “max” metric on the product; certainly other choices are possible but they turn out to be equivalent.)
Coming to the OP's question, to prove that a norm on a linear space is continuous w.r.t. itself, notice that the linear space is also a metric space with $d(x,y) = \| x - y \|$. In other words, $\| x \|$ is just the distance of $x$ from the origin; hence applying item (1.) above (with $a = 0$) shows the continuity of the norm.
A point to ponder: Equivalent norms. The above discussion might suggest that continuity of a norm is not an interesting thing to study. However this question can be modified a little to give rise to a quite fruitful concept:
Given two norms $\| \cdot \|_1$ and $\| \cdot \|_2$ on a linear space $X$, are they continuous with respect to each other? I.e., when $\| \cdot \|_1$ is viewed as a function, is it continuous w.r.t. the norm $\| \cdot \|_2$ and vice versa?
It can be shown that this is true if and only if there exist numbers $0 < \alpha \leqslant \beta < \infty$ such that $$\alpha \| v \|_1 \leqslant \| v \|_2 \leqslant \beta \| v \|_1$$ for all vectors $v$. In this case, we say that the two norms are equivalent.
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+1 Really nice for pointing out more insights. – Tim Jan 5 '12 at 22:39
@Tim Not just the norm, but also the addition map $+ : X \times X \to X$ and scalar multiplication map $\bullet : \mathbb R \times X$ are also continuous in the respective topologies. These statements also require only the elementary properties of norms. – Srivatsan Jan 5 '12 at 23:13
Thanks! Do you mean that any topology that can make a norm continuous will also make addition and scalar product continuous? – Tim Jan 5 '12 at 23:18
@Tim No, that is not true. In fact, as t.b. pointed out in chat, the weird topology described by me and Bill under Bill's answer, that should not have this property (although I haven't checked the details myself). What I mean is that if you take the respective "standard" topologies, then the addition and scalar multiplication maps are continuous. In fact, you should think of this as saying that the norm is defined in a way to be "compatible" with these operations. – Srivatsan Jan 5 '12 at 23:21 | 2015-09-05T06:09:59 | {
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https://math.stackexchange.com/questions/1652107/convergence-and-value-of-infinite-product-prod-infty-n-1-n-sin-left-f/1905738 | # Convergence and value of infinite product $\prod^{\infty}_{n=1} n \sin \left( \frac{1}{n} \right)$?
Since the limit $\frac{\sin(x)}{x}=1$ for $x \rightarrow 0$, I wondered about the infinite product:
$$\prod^{\infty}_{n=1} n \sin \left( \frac{1}{n} \right)=\sin(1) \cdot 2 \sin\left( \frac{1}{2} \right) \cdot 3 \sin\left( \frac{1}{3} \right) \dots$$
By numerical experiment in Mathematica it seems to converge, even if very slowly (I mean to non-zero value):
$$P(14997)= 0.755371783$$
$$P(14998)= 0.755371782$$
$$P(14999)= 0.755371782$$
$$P(15000)= 0.755371781$$
I can prove the convergence by integral test for the series:
$$\sum^{\infty}_{n=1} \ln\left( n \sin \left( \frac{1}{n} \right) \right)$$
$$\int^{\infty}_{1} \ln\left( x \sin \left( \frac{1}{x} \right) \right) dx=\int^{1}_{0} \frac{1}{y^2} \ln \left( \frac{\sin (y)}{y} \right) dy=-0.168593$$
I think the integral test can work with negative function as long as it's monotone, otherwise I can just put the minus sign before the infinite sum.
By the way, this is a related question about the convergence of the sum above.
But I'm more interested in the infinite product itself.
I'm not sure if the value of this infinite product can be found and how to go about it. Is it zero or not? Any thoughts would be appreciated
• since $\sin(x) = \sum_k x^{2k+1}\frac{(-1)^{k}}{(2k+1)!}$ you get $n\sin(1/n) = 1-n^2/6+\mathcal{O}(1/n^4)$ which proves the convergence of the infinite product. and yes your integral test works too – reuns Feb 12 '16 at 15:48
• I assume its an ugly number since all of the products are transcendentals, but you never know. – Matthew Levy Feb 12 '16 at 15:53
• above I meant $n \sin(1/n) = 1-\frac{1}{6 n^2} + \mathcal{O}(1/n^4)$. and an infinite product which $=0$ is said divergent (the associated $\ln$ series diverges) here it is not so the result cannot be $0$ – reuns Feb 12 '16 at 15:56
• @user1952009, thank you, I didn't think about it – Yuriy S Feb 12 '16 at 15:57
• The observation from @ can be used to obtain lower bound on the value of the product: $n\sin(1/n) \geq 1-\frac{1}{6n^2}$ implies the infinite product cannot be smaller than $1-\sum_n \frac{1}{6n^2}=1-\frac{\pi^2}{36}\approx 0.7258$. – Peter Košinár Feb 18 '16 at 0:16
Thanks for all the comments. The closed form is unlikely, but some very good estimates can be provided, using Taylor series for $\sin$:
$$\prod_{n=1}^{\infty} n \sin \frac{1}{n}<\prod_{n=1}^{\infty} 1=1$$
$$\prod_{n=1}^{\infty} n \sin \frac{1}{n}>\prod_{n=1}^{\infty} \left(1-\frac{1}{6n^2} \right)=\frac{\sqrt{6}}{\pi} \sin \frac{\pi}{\sqrt{6}}=0.747529$$
$$\prod_{n=1}^{\infty} n \sin \frac{1}{n}<\prod_{n=1}^{\infty} \left(1-\frac{1}{6n^2}+\frac{1}{120n^4} \right)< \exp \left(-\frac{\pi^2}{36}+\frac{\pi^4}{10800} \right)=0.767101$$
Actually, for the last product Mathematica gives the closed form:
$$\prod_{n=1}^{\infty} \left(1-\frac{1}{6n^2}+\frac{1}{120n^4} \right)=\frac{5\sqrt{5}}{\pi^2} \sin \left(\frac{\pi \sqrt{1-\frac{i}{\sqrt{5}}}}{2 \sqrt{3}}\right) \sin \left(\frac{\pi \sqrt{1+\frac{i}{\sqrt{5}}}}{2 \sqrt{3}}\right)=$$
$$=\frac{\sqrt{30}}{\pi^2} \left(\cosh \left( \pi \sqrt{\frac{1}{\sqrt{30}}-\frac{1}{6}} \right)-\cos \left( \pi \sqrt{\frac{1}{\sqrt{30}}+\frac{1}{6}} \right) \right)=0.755542$$
We get the estimation:
$$\prod_{n=1}^{\infty} n \sin \frac{1}{n}<0.755542$$
Which gives at least three (maybe four) correct digits for the numerical value.
Alright, this is another answer, much better one.
We can evaluate this product numerically with excellent precision, if we get it into a better form.
$$P=\prod^{\infty}_{n=1} n \sin \left( \frac{1}{n} \right)=\prod^{\infty}_{n=1} \prod^{\infty}_{k=1} \left(1- \frac{1}{\pi^2 n^2 k^2} \right)$$
Now we take logarithm of the product:
$$\ln P=\sum^{\infty}_{n=1} \sum^{\infty}_{k=1} \ln \left(1- \frac{1}{\pi^2 n^2 k^2} \right)=-\sum^{\infty}_{n=1} \sum^{\infty}_{k=1}\sum^{\infty}_{l=1}\frac{1}{l~\pi^{2l} n^{2l} k^{2l}}=-\sum^{\infty}_{l=1}\frac{\zeta (2l)^2}{l~\pi^{2l}}$$
This last single sum Mathematica computes with great precision, so we can write:
$$\ln P=-0.280556336229155079602039680939198362173$$
And the product is:
$$P=\exp \left(-\sum^{\infty}_{l=1}\frac{\zeta (2l)^2}{l~\pi^{2l}} \right)=0.75536338851857321406336498617047655360$$
By the same logic we also have:
$$P_1=\prod^{\infty}_{n=1} n \sinh \left( \frac{1}{n} \right)=\exp \left(-\sum^{\infty}_{l=1}\frac{(-1)^l \zeta (2l)^2}{l~\pi^{2l}} \right)=1.307970936664283649012104476$$ | 2019-06-17T21:10:40 | {
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http://ttwj.alterauto.fr/unfair-coin-probability.html | # Unfair Coin Probability
p is the probability of. A probability of one means that the event is certain. are within the probability for a fair coin. Suppose I have an unfair coin and I want to turn it into a fair coin using the following way, Probability of generating head is equal for unfair coin; Flip unfair coin and only accept head; When a head is appearing, treat it as 1 (head for virtual fair coin), when another head is appearing, treat it as 0 (tail for virtual fair. If the coin is flipped 14 times, what is the probability that;a) it comes up tails exactly 6 times?;b) it comes up heads more than 11 times?. Coin A has a 90% chance of heads, coin B has a 5% chance heads. Simulating fair coins with unfair coins (and vice versa) → Flipping HHT before HTT? Not what you think. khanacademy. But again, it was possible that the coin had been fair. In This Part: Fair and Unfair Allocations The average for a set of data corresponds to the equal-shares allocation or fair allocation of the data. What is the probability that B got more heads than A? A fair coin is tossed until a head comes up for first time. In probability theory and statistics, a sequence of independent Bernoulli trials with probability 1/2 of success on each trial is metaphorically called a fair coin. The theory of probability has always been associated with gambling and many most accessible examples still come from that activity. , n Bernoulli trials). If the coin is tossed 3 times, what is the probability that at least one of the tosses will turn up tails?. Students often respond initially that the game is unfair because the player who wins when the coins match has two chances to win while the other player only has one chance. Now that in and of itself is not proof, so. And the probability that no one wins: Pr(no one wins) = 0. Case 2: One head. WEAR Abstract. If we see a coin tossed twice and we see 2 heads, we'd like to know if the coin is fair, or at least to be able to determine the probability that the coin is fair. In the preface, Feller wrote about his treatment of uctuation in coin tossing: \The results are so amazing and so at variance with common intuition that even sophisticated colleagues doubted that coins actually misbehave as theory predicts. If you know how to manage time then you will surely do great in your. In this paper, we explore some properties of the cumulative probability distribution of this random variable. Read about me, or email me. coin toss probability calculator,monte carlo coin toss trials. An example of such a vector for three possible outcomes is c(0. The standard (maybe overused) example is flipping a fair coin. When I flip the coin and get tails, I lose a dollar. 24 to estimate a probability of 0. After all, the probability that you will roll HHT right off the bat is , and the same for HTT and HHH, so there should be symmetry. Probability (unfair coin) Hello everyone, I need a little help with this problem I have. The remaining coins have heads on both sides. Two people take turns to ip a coin, Your opponent starts rst, you second. A common topic in introductory probability is solving problems involving coin flips. Not convinced? Suppose I give you a dollar for every head, and you give me a dollar for every tail, and we ip the coin 10 times (my coin!) and we get 0 heads and 10 tails. 9%: an unfair coin lands heads 65% of the time. Determining Dependent and Independent Events (HSS-CP. Read more about setting a seed below. The appeal of the coin toss that it is a simple, seemingly unbiased, method of deciding between 2 options. (b) What is the chance that the coin is flipped exactly $$i$$ times? (c) What is the chance that the coin is flipped more than twice? (d) Repeat the previous three questions for a unfair coin which has probability $$p$$ of getting Tails. In the case of coins, heads and tails each have the same probability of 1/2. Inequalities with Absolute Values. If you toss this coin twice, what is the pro… Get the answers you need, now!. In this case A is flipping 10 heads in a row and B is picking the two-headed coin. The remaining coins have heads on both sides. a) How to make an event with 50% probability? b) Expected number of flips until a realization occurs? c) Can you create a strategy to reduce the number of flips necessary? d) Can you create a strategy to reduce the number of flips necessary for an unfair coin with any bias?. With H0 = "coin is unfair", H1= "coin is fair", and S = "test succesful with p-value < 0. ) the probability that a coin flip will result in heads (set to a default of 0. Students often respond initially that the game is unfair because the player who wins when the coins match has two chances to win while the other player only has one chance. Otherwise, a student from a different class containing 12 boys and 9 girls is selected. Coin A has a 90% chance of coming up heads, coin B has a 5% chance of coming up heads. This is a common misconception that is best addressed through data collection and analyzing that data rather than through telling. What is the probability of getting 4 tails in 4 tosses of an unfair coin where probability of tails is 7?. But, could not arrive at a solution. 7 probability of landing 'heads'. Active 2 years, 7 months ago. Consider an unfair coin. 6 and tails with probability 0. But diving straight into Huszár (2017) or Chen et al (2017) can be a challenge, especially if you're not familiar with the basic concepts and underlying math. PROBABILITY SPACE A probability space has three components: (1) A sample space Ω that is the set of all possible outcomes of the random process modeled; (2) A family of sets F representing the allowable events , where each set in F is a subset of the sample space Ω; (3) A probability function. e a coin with equal probability of landing heads or tails) but would like to construct an outcome of biased probability , how would you do it? I remember this question. If you toss the coin 40 times, how many heads do you expect to see? A. An unfair coin is tossed; if a head appears uppermost, then a marble is selected from bag (X); otherwise, a marble is selected from bag (Y). An unfair coin has probability 0. Question 2 An insurance company writes policies for a large number of newly-licensed drivers each year. Find the probability of exactly one Heads. 4 y Expected Value Some Good Advice Pay careful attention to what notation tells you to do in performing a calculation. Build and represent graphically the probability distribution and the cumulative distribution of the function with random variable "X = number of time result is head". Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture. We express probability as a number between 0 and 1. 4 is tossed three times. Increasing the repetitions, you can compare the paths taken in repea. What is the probability categorized under Math and Probability. A random variableXis: PSYCH 2000-Consider and unfair coin. How do these numerical results compare to actual outcomes? Under the current overtime rules, there have been 51 overtime games. Mathematically, we find probability by comparing the number of favorable outcomes to the number of possible outcomes. , 𝑝 ℎ𝑒𝑎𝑑𝑠 = 0. The formula of probability of an event G: P (G) = Discussion: 1) If the coin is biased or unfair, what happens to the probability of getting “Heads”?. With dice rolling, your sample space is going to be every possible dice roll. Design a game between Alice and Bob so that Alice's winning probability is exactly α. Each biased coin has a probability of a head 4/5. Let a and b be the results of 2 tosses of the unfair coin. Instead, you only find a biased coin with probability 1/3 of getting heads. Competitive acquisition plan for low probability of detection data link networks. Make a fair coin from a biased coin You are given a function foo() that represents a biased coin. The probability is 0. In the above experiment, we used a fair coin. Can someone explain to me how can you get a fair (equal probability) outcome using only an unfair coin (where unfair means that it will land head with probability p and tails 1-p where p !=. The game of course has to end in a finite number of tosses with probability 1. Lec -7 Frequency Probability and Unfair Coins. P(result of a coin toss is heads). At first glance this might seem like an overcomplicated way of solving this. In calcu-lating expected value, you are told to first multiply the probability of each outcome by its. Here we will learn how to find the probability of tossing two coins. Hypothesis Testing 1 Hypothesis Testing Much of classical statistics is concerned with the idea of hypothesis testing. After you choose your first coin and flip it, you can base your decision of which coin to flip second on your results of the first flip. 7 is the probability of each choice we want, call it p. Luetkemeyer, Mr. If , discard observations, goto step 1. 1% probability that it will come up heads all ten times. The order does not matter as long as there are two head and two tails in the flip. This is a common misconception that is best addressed through data collection and analyzing that data rather than through telling. Example-Binomial #Suppose you have a biased coin that has a probability of 0. There is a die and a coin. For example, suppose we have three coins. Subscribe for more updates. Which answers, if any. PROBABILITY AND STATISTICS FOR SCIENTISTS AND ENGINEERS. Alice and Bob play a game as follows. The first two happen to Jackie with the same chance, but the third happens of the time, since the unfair coin is heads instead of tails. For example, suppose we have three coins. 2) - We look for all possible outcomes. Persi Diaconis has spent much of his life turning scams inside out. More generally, there are situations in which the coin is biased, so that heads and tails have different probabilities. , 𝑝 ℎ𝑒𝑎𝑑𝑠 = 0. If the coin is flipped 14 times, what is the probability that;a) it comes up tails exactly 6 times?;b) it comes up heads more than 11 times?. Output: Now, suppose we want to simulate 100 flips of an unfair two-sided coin. A random variableXis: PSYCH 2000-Consider and unfair coin. The Binomial Likelihood Function Note the similarity between the probability function and the likelihood function; the right hand sides are the same. Image Transcriptionclose. Probability Coin Flip Also, I've always had a problem with the oversimplified coin flip 50-50 thing. coin=randi([0:1], [100,1]) It should more or less give you 50 0's and 50 1's. Note that the. Dice Roll Probability for 6 Sided Dice: Sample Spaces. 9%: an unfair coin lands heads 65% of the time. If a fair coin (one with probability of heads equal to 1/2) is flipped a large number of times, the proportion of heads will tend to get closer to 1/2 as the number of tosses increases. Using Python 2. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. We're thinking about how the probability of an event can be dependent on another event occuring in this example problem. With dice rolling, your sample space is going to be every possible dice roll. 6 of landing heads. 25 " = 25% = 1/4 Probabilities are usually given as fractions. The coin is tossed four times. 7870 and the probability of getting three or more heads in a row or three or more tails in a row is 0. 2 is flipped. What about probabilities when we don't have equally likely events? Say, we have unfair coins? If you're seeing this message, it means we're having trouble loading external resources on our website. (Remember, to calculate probability when the question includes the word “and”, you multiply. The coin is flipped 50 times. What is the probability that you must flip the coin four or more times. Example The same unfair coin as in the previous example is flipped three times. Note that the. Here you can assume that if a child is a girl, her name will be Lilia with probability $\alpha \ll 1$ independently from other children's names. Click to left of y-axis to for a new run, to right of y-axis to pause. This theorem will justify mathematically both our frequency concept of probability and the interpretation of expected value as the average value to be expected in a large number of experiments. (a) What is the probability that the flipped coin will come up heads? We'll assume there is an equal chance (1/3) of picking any of the three coins. Suppose I have an unfair coin, and the probability of flip a head (H) is p, probability of flip a tail (T) is (1-p). - 11296804. Drivers spend on it is unfair that something like that To speak to a pet-related charity In group quarters - homes for the book value (i Rosario marchese: sorry; we’re going the profit and costing for healthcare software architect jobs Mail wasn't being forwarded correctly Infographics, insurance rate - get discounts up to $15,000 per person. What is the sample space for a fair coin ip? For a sequence of three coin ips? For a sequence of ve coin ips in which at least four ips turn out to be heads? Suppose you are told now that the coin is unfair, with the probability of a toss resulting in heads being 0. Suppose an unfair coin (P(head)=. If you know how to manage time then you will surely do great in your. Let’s start with our original problem: using a fair coin to simulate a coin with P(H) = 1/3, or P(T) = 2/3. The probability of having the unfair coin is$\frac13$. The pencils come in packages of 6. I was a mathematician, and now work in finance (systematic trading). With H0 = "coin is unfair", H1= "coin is fair", and S = "test succesful with p-value < 0. Probability of compound events Learn how to calculate the probability of at least 2 simple events. For example, suppose that each of 9 people has several dollars and altogether they have$45. Coin Toss Probability Calculator is a free online tool that displays the probability of getting the head or a tail when the coin is tossed. Frequency Stability Example: Range and mid-range This original Khan Academy video was translated into isiZulu by Wazi Kunene. Would you modify your approach to the the way you test the fairness of coins?. As above we can use R to simulate an experiment of rolling a die a number of times and compare our results with the theoretical probability. The coin is flipped 50 times. Competitive exams are all about time. Consider an unfair coin. The multinomial law gives the probability of exactly ki occurrances of the i-th face in n tosses of a die. If the coin shows tails, we draw a marble from urn T with 4 red and 2 blue marbles. Although it is. A box contains 5 fair coins and 5 biased coins. 3 Discrete Distributions A discrete distribution assigns a probability to every atom in the sample space of a random variable. Coin flipping game: how to make a fair toss from an unfair coin. Repeat task 4, but for a biased die with a probability of 0. A coin is tossed and comes up tails ten times: is this just random chance, or is an unfair coin being used? Learn when to reject the null hypothesis: if the probability (P) is less than the chosen significance level, the null must be. A bag contains 5 coins. We can explore this problem with a simple function in python. Binomial Distribution based on an Unfair Coin. Can you design a game where you and your opponent have an equal chance of winning? Show Answer. Find the probability of getting 4 heads. A random variableXis: PSYCH 2000-Consider and unfair coin. Please try solving this problem before jumping on the solution Click to learn. A classic example of a probabilistic experiment is a fair coin toss, in which the two possible outcomes are heads or tails. What about probabilities when we don't have equally likely events? Say, we have unfair coins? If you're seeing this message, it means we're having trouble loading external resources on our website. The probability of having the unfair coin is $\frac13$. The randomness comes from atmospheric noise, which for many purposes is better than the pseudo-random number algorithms typically used in computer programs. Question: Consider Two Coins, One Fair And One Unfair. Frequency Stability Example: Range and mid-range This original Khan Academy video was translated into isiZulu by Wazi Kunene. If the coin is tossed 3 times, what is the probability that at least 1 of the tosses will turn up tails? A. Students often respond initially that the game is unfair because the player who wins when the coins match has two chances to win while the other player only has one chance. If we have a biased coin (i. 12, 2012 Title 16 Commercial Practices Parts 0 to 999 Revised as of January 1, 2013 Containing a codification of documents of general applicability and future effect As of January 1, 2013. If we see a coin tossed twice and we see 2 heads, we'd like to know if the coin is fair, or at least to be able to determine the probability that the coin is fair. Determining Dependent and Independent Events (HSS-CP. ” “If that happens, then we’ll incorrectly decide that an unfair coin is really fair --called a ‘miss’. After all, the probability that you will roll HHT right off the bat is , and the same for HTT and HHH, so there should be symmetry. "Count line" can be moved by mouse. Well, that's the same as the probability that the first coin didn't come up heads when both coins are different, which is 1 - 1/2 = 1/2. Last time we talked about independence of a pair of outcomes, but we can easily go on and talk about independence of a longer sequence of outcomes. Pr[HHH] = Pr[TTT] = 1=2. ” “If that happens, then we’ll incorrectly decide that an unfair coin is really fair --called a ‘miss’. Download All Slides. What is the probability that a fair coin lands Heads 4 times out of 5 flips? Ans: C(5,4)/25 = 5/32. The coin is weighted so that the head {H} is 3 times more likely to occur than tails. When a coin is tossed, there lie two possible outcomes i. Consider the unfair coin with P(H) = 1/3 and P(T) = 2/3. So the probability of getting two heads is: 1 " in " 4 = 0. sim_fair_coin table (sim_fair_coin) Since there are only two elements in outcomes, the probability that we "flip" a coin and it lands heads is 0. The simplest case when you're learning to calculate dice probability is the chance of getting a specific number with one die. Taking many of the concepts he has covered in the last few videos, including probability, combinations, and conditional probability, Sal uses the example of fair and unfair coins in a bag to show various probability problems. Can you design a game where you and your opponent have an equal chance of winning? Show Answer. Design a game between Alice and Bob so that Alice's winning probability is exactly α. This is a formal framework that we can use to pose questions about a variety of topics in a consistent form that lets us apply statistical techniques to make statements about how results that we've gathered relate to questions that we're interested in. In fact, the probability for most other values virtually disappeared — including the probability of the coin being fair (Bias = 0. In theoretical studies, the assumption that a coin is fair is often made by referring to an ideal coin. For this problem we can use the normal approximation to the binomial distribution. Create a function like flip_heads(), but for a fair die instead, and have it focus on the probability of rolling a 4. The order does not matter as long as there are two head and two tails in the flip. For example, an unfair coin could have p(X=h) = 0. In your simulation of flipping the unfair coin 100 times, how many flips came up heads? Include the code for sampling the unfair coin in your response. Probability : We have a weighted coin which shows a Head with probability p, (0. Let's write a function that takes in two arguments: 1. Quiz CHAPTER 16 NAME:_____ UNDERSTANDING PROBABILITY AND LONG-TERM EXPECTATIONS 1. These games work with random events, so they are a useful way to learn how to use probabilities to predict events. Then, the probability of heads is not 0. In fact Bayes theorem gives us a way to compute this value, but we need more information: priors. a) What is the probability that the coin comes up tails more than 25 times? b) What is the probability that the coin comes up tails more than 30 times?. Construct the probability distribution for x , and calculate its mean. For example, consider a fair coin. What is the probability that it lands heads at least once? Log On. HOWEVER, the question does noy say it is a fair coin!!!! The fact that it has landed 61 times on heads in 100 tosses could be because it is an "UNFAIR" coin and is weighted to favour heads compared to tails. ) please help!. Tossing an unfair coin multiple times. If you keep rolling a fair coin, what is the probability that you will get HHT before HTT (H = heads, T = tails)? How about HHH before HTT? HHT before TTT?. Suppose Tori has an unfair coin which lands on Tails with probability 0. 51), then we would expect that the results would yield 25. For an unfair or weighted coin, the two outcomes are not equally likely. WEAR Abstract. My initial idea is that we need to choose appropriate. 2 What is the. Since there are only two elements in outcomes, the probability that we "flip" a coin and it lands heads is 0. Probability: Tossing an Unfair Coin. Click here for how to write a probability. 1 Coin Tossing. 2 Prediction (EMG52) Games of chance (EMG53). In probability theory and statistics, a sequence of independent Bernoulli trials with probability 1/2 of success on each trial is metaphorically called a fair coin. In some situations, such as in flipping an unfair coin, we cannot calculate the theoretical probability. You can load a die but you can't bias a coin Andrew Gelman∗ and Deborah Nolan† April 26, 2002 Abstract Dice can be loaded—that is, one can easily alter a die so that the probabilities of landing on the six sides are dramatically unequal. I'm a beginner with R and I am trying to design a coin flip simulation. Suppose you have an extremely unfair die: The probability of a 6 is 3/8, and the probability of each other number is 1/8. Knowing a little bit about the laws of probability, I quickly knew the fraction "2/6" for two dice and "3/6" for three dice was incorrect and spent a brief moment computing and then explaining the true percentages. Construct a probability model for this experiment. 9%: an unfair coin lands heads 65% of the time. If you toss a coin, it will come up a head or a tail. Please try solving this problem before jumping on the solution Click to learn. equals the probability that the expectation of a coin flip. B)If The Coin Is Flipped 5 Times, What Is The Probability Of Getting Exactly 2 Tails?. ) the number of games to be played, and 2. If he flips the coin three times, what is the probability that he flips more Heads than Tails? Express your answer as a common fraction. For a single toss of a balanced coin, let x = 1 for a head and x = 0 for a tail. So, after 500 flips most of the probability gets distributed around the value 0. Variational inference is all the rage these days, with new interesting papers coming out almost daily. As such, we will build a quick app to demonstrate an unfair coin. Question: Consider Two Coins, One Fair And One Unfair. But, could not arrive at a solution. This already is a pretty good estimate of the real bias! But you might want an even better estimate. The game is just like rolling a dice but with coins and tally marks. If we flip this coin three times, the sample space S is the following set of ordered triples: S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}. a) Calculate the theoretical probability of getting exactly 5 heads in 10 tosses when the probability of a head is 0. What is the probability that a fair coin lands Heads 6 times in a row? Ans: 1/26. Suppose that you're given a fair coin and you would like to simulate the probability distribution of repeatedly flipping a fair (six-sided) die. A fair coin gives you Heads. 322 Math Masters, p. The randomness comes from atmospheric noise, which for many purposes is better than the pseudo-random number algorithms typically used in computer programs. In fact, the probability for most other values virtually disappeared — including the probability of the coin being fair (Bias = 0. The coin is flipped 50 times. But, could not arrive at a solution. Example question: What is the probability of rolling a 4 or 7 for two 6. A box contains 5 fair coins and 5 biased coins. Conditional Probability & the Rules of Probability; Intersection and Union of Sets (HSS-CP. However, it is not possible to bias a coin flip—that. My initial idea is that we need to choose appropriate. In the fair coin experiment, there were 46 heads and 54 tails. But since there are 6 ways to get 2 heads, in four flips the probability of two heads is greater than that of any other result. For this problem we can use the normal approximation to the binomial distribution. An experiment in which a single action, such as flipping a coin, is repeated identically over and over. We can easily simulate an unfair coin by changing the probability p. In other words, the probability of getting 108 heads out of 200 coin tosses with a fair coin is 27%. Simulating a Biased Coin with a Fair Coin. share | cite | improve this question. (Or probability value) is a number between 0 and 1 inclusive associated with the likelihood of occurrence of a given event. The probability of having the unfair coin is $\frac13$. 5 Assuming an unfair coin (i. Probability is an estimate of the chance of winning divided by the total number of chances available. Now we must define the prior probability of seeing each of those values. In theoretical studies, the assumption that a coin is fair is often made by referring to an ideal coin. In some situations, such as in flipping an unfair coin, we cannot calculate the theoretical probability. We can have either HTT, THT, or TTH. Image Transcriptionclose. the probability of success, p, is constant. After all, the probability that you will roll HHT right off the bat is , and the same for HTT and HHH, so there should be symmetry. When asked the question, what is the probability of a coin toss coming up heads, most people answer without hesitation that it is 50%, 1/2, or 0. The Probability Of Getting Heads On A Given Flip Of The Unfair Coin Is 0. You toss each coin 10 times (100 tosses in total) and observe results. Bernoulli Trials. The Binomial Likelihood Function Note the similarity between the probability function and the likelihood function; the right hand sides are the same. We flip a fair coin. Challenge the students to make an argument not based on the data as whether the game is Fair or Unfair and why. (Or probability value) is a number between 0 and 1 inclusive associated with the likelihood of occurrence of a given event. Conditional Probability & the Rules of Probability; Intersection and Union of Sets (HSS-CP. asked 23 hours ago. What is the probability that both children are girls? In other words, we want to find the probability that both children are girls, given that the family has at least one daughter named Lilia. Let R_p(r,n) be the probability that a run of r or more consecutive heads appears in n independent tosses of a coin (i. So, after 500 flips most of the probability gets distributed around the value 0. suppose the outcome heads occurs with probability 0. Increasing the repetitions, you can compare the paths taken in repea. help_outline. How many different ways can a reader choose 3 books out of 4, ignoring the order of selection? a. What is the probability of choosing two books of different colors? a. The probability of getting AT MOST 2 Heads in 3 coin tosses is an example of a cumulative probability. In particular, it provides a "strategy" at the end to generate an unfair toss (which requires, e. You can change the weight or distribution of the coin by dragging the true probability bars (on the right in blue) up or down. Example question: What is the probability of rolling a 4 or 7 for two 6. The basic rule for probability is that you calculate it by looking at the number of possible outcomes in comparison to the outcome you’re interested in. I want it to start by having a dollar amount of x. Algebra -> Probability-and-statistics-> SOLUTION: An unfair coin has a probability 0. Up until now, we've looked at probabilities surrounding only equally likely events. 1 $\begingroup$ I am stuck on this question. e head or tail. Probability of switching coins = 0. Flipping coins is a classical way of thinking about probability. Additional figures show the probability distributions for n = 2,3,4,5,10. Coin toss probability Coin toss probability is explored here with simulation. Then a second coin is drawn at random from the box (without replacing the first one. If the coin is tossed 3 times, what is the probability that at least one of the tosses will turn up tails?. randomizing device is an unfair coin, with probability p ∈ (0,1) of heads. 6 and (P(tail)=. In this case, we can naturally assume that if the machines are unfair, the house will assign a lower probability to the expensive prizes, and higher probability to cheap ones. 6 that an "unfair" coin will turn up tails on any given toss. An unfair coin with 0. TOSSING A COIN M. Tutorials for Question - An unfair coin has a probability of coming up heads of 0. We express probability as a number between 0 and 1. An unfair coin has a probability of coming up heads of 0. Introduction The result of ntosses of a two-headed coin can be represented by. In simple terms, you have to figure out every possibility for what might happen. The probability of getting 3 lemons is 1/10 X 1/10 X 1/10, or 1/1000. Stating that it is "well known" that the probability of a coin landing on its side is 1/6000 is incorrect. It is not known whether a coin is fair or unfair. 2 is flipped. This is when the χ 2 test is important as it delineates whether 26:25 or 30:21 etc. The fact of the matter is, the human, not the coin (mostly, there is a slight weight bias that might be shown after approximately 10,000 flips), introduces the probability that the coin may land. An unfair coin is tossed; if a head appears uppermost, then a marble is selected from bag (X); otherwise, a marble is selected from bag (Y). probability of any continuous interval is given by p(a ≤ X ≤ b) = ∫f(x) dx =Area under f(X) from a to b b a That is, the probability of an interval is the same as the area cut off by that interval under the curve for the probability densities, when the random variable is continuous and the total area is equal to 1. Let t be the expected number of times you flip. Flipping the coin once is a Bernoulli trial, since there are exactly two complementary outcomes (flipping a head and flipping a tail), and they are both 1 2 \frac{1}{2} 2 1 no matter how many times the coin is flipped. | 2020-04-06T07:29:28 | {
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http://math.stackexchange.com/questions/271141/number-of-roots-of-a-complex-equation-rouches-theorem | # Number of roots of a complex equation/ Rouche's theorem
For $n\geq2$ consider the equation $z^n+z+n=0$ for $z\in \mathbb C$. Show that if $k$ is an integer with $1\leq k \leq n$ then inside the sector $$S_k=\left\{z\in \mathbb C: 0< Arg(z) < \dfrac{2\pi k}{n} \right\}$$ There are exactly $k$ roots of the above equation. $Arg (z)$ is the principal argument of $z$. (Hint: Prove that $x^n+n>x$ for real $x$)
The only thing I can think of is Rouche's theorem but then the region needs to be bounded to be able to use that. Can anybody give some pointers as to how I should proceed here. Thanks.
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If you make the substitution $z = n^{1/n} \zeta$, you get the equation $\zeta^n + \frac{n^{1/n}}{n}\zeta + 1 = 0$. Does this give you an idea of where to look for your roots? – Antonio Vargas Jan 5 '13 at 21:41
@AntonioVargas: I still don't see it. when $n$ is large the cofficient of $\zeta$ goes to zero. But that is not relevant, I think. We need to show this for all $n$ – Jack Dawkins Jan 5 '13 at 23:28
Exactly, when $n$ is large the equation is very similar to $\zeta^n+1=0$. So you should be looking for the roots of $\zeta^n + \frac{n^{1/n}}{n}\zeta + 1 = 0$ near the roots of $\zeta^n+1=0$. This should give you an idea of what region to use in Rouché's theorem. – Antonio Vargas Jan 5 '13 at 23:48
@AntonioVargas Could you please explain your answer in detail? – Arpan Dutta Aug 11 '13 at 5:28
@ArpanDutta sure, I've just posted an answer. – Antonio Vargas Aug 12 '13 at 21:01
Let
$$C_n(a) = \left\{z\in\mathbb C\,\colon |z^n+n| = a\right\}.$$
There are a few ways to see that $C_n(a)$ consists of exactly $n$ simple closed loops, one in each sector
$$\frac{2\pi k}{n} < \arg z < \frac{2\pi (k+1)}{n}, \qquad k=1,2,\ldots,n,$$
when $0 < a < n$. The simplest (and least general) is to note that the circle $|w + n| = a$ in the $w$-plane does not intersect the non-negative real axis, so under the conformal mapping $w=z^n$ it has exactly $n$ smooth preimages, one strictly inside each such sector.
The point of $|w + n| = a$ with largest modulus lies at $w=-a-n$, so the point of $C_n(a)$ with largest modulus lies on the circle $|z| = (a+n)^{1/n}$. Thus if $0 \leq a < n$ then
$$z \in C_n(a) \quad \Longrightarrow \quad |z| < (2n)^{1/n}.$$
Therefore if $n \geq 3$ and $z \in C_n(6^{1/3})$ we have
$$|z| < (2n)^{1/n} \leq 6^{1/3} = |z^n + n|$$
since $6^{1/3} < n$ and the function
$$f(x) = (2x)^{1/x}$$
is decreasing for $x \geq 3$.
By Rouché's theorem we may conclude that $z^n + z + n$ has precisely as many zeros inside each component of $C_n(6^{1/3})$ as does $z^n+n$. Since $6^{1/3} < n$ each component surrounds exactly one zero of $z^n+n$ and lies strictly inside one of the sectors
$$\frac{2\pi k}{n} < \arg z < \frac{2\pi (k+1)}{n}, \qquad k=1,2,\ldots,n.$$
It follows that $z^n + z + n$ likewise has exactly one zero in each of these sectors.
We assumed above that $n \geq 3$. We can address the case when $n=2$ by observing that the discriminant of $z^2 + z + 2$ is $-7$ and so its zeros are complex conjugates, one lying in the upper half plane and one in the lower.
Aside: This argument can be adapted to show that, for $n \geq 3$, the polynomial $z^n+z+n$ has a zero in each disk $$\left|z-\zeta_n^k n^{1/n}\right| \leq n^{1/n} - \left(n-(2n)^{1/n}\right)^{1/n} = O\left(\frac{1}{n^2}\right),$$ $k = 1,2,\ldots,n$, where $\zeta_n$ is the principal $n^\text{th}$ root of $-1$.
- | 2016-06-29T21:53:11 | {
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http://math.stackexchange.com/questions/161341/singular-points-of-a-set-of-matrices | # Singular points of a set of matrices
Let $k$ be an algebraically closed field. We identify the space $M_{2}(k)$ of $2 \times 2$ with $\mathbb{A}^{4}(k)$ with coordinates a,b,c,d. Let $X$ be the set of all matrices $A$ in $M_{2}(k)$ such that $A^{2}=0$.
Show that $X$ is isomorphic to the affine cone of the projective variety $W=V(x^{2}-yz) \subseteq \mathbb{P}^{2}$ and find the singular points of $X$. Finally compute the Zariski tangent space at such point.
My work:
Doing the computations shows that $X=V(ad-bc,a+d)$. Now note that the affine cone of $W$ is $V(x^{2}-yz) \subseteq \mathbb{A}^{3}$.
So let $f: X \rightarrow V(x^{2}-yz) \subseteq \mathbb{A}^{3}$ be given by $(x,y,z,w) \mapsto (x,-y,z)$ and let $g: V(x^{2}-yz) \subseteq \mathbb{A}^{3} \rightarrow X$ be given by $g(x,y,z)=(x,-y,z,-x)$.
(Unless I did something wrong) $g$ is the inverse of $f$ so $f$ is an isomorphism.
Now observe that $V(x^{2}-yz) \subseteq \mathbb{A}^{3}$ has only one singular point, namely the origin $(0,0,0)$ thus via the above isomorphism we see that $(0,0,0,0)$ is the only singular point of $X$.
The part I'm not sure is how to compute the Zariski tangent space we first need to find a set of generators for $I(X)$ no? or is there an easier way?
EDIT: Using software one can check that $I(X)=(xyz-x^{2}w+yzw-xw^{2})$ so the tangent space is $\mathbb{A}^{4}$ because all partials of $xyz-x^{2}w+yzw-xw^{2}$ vanish at $(0,0,0,0)$. However is it possible to solve this question without calculating $I(X)$?
-
a) The proof that $X$ is isomorphic to $V(x^2-yz)\subset \mathbb A^3$ is perfect: congratulations!
Let me call $V$ that variety $V(x^2-yz)\subset \mathbb A^3$
b) You haven't given a proof, but it is true that $V$ has $O=(0,0,0)$ as its only singularity.
A possible proof proof is to use the Jacobian criterion:
Letting $F(x,y,z)=x^2-yz$ we have $Jac(F)=(2x,-z,-y)$ and the singularities of the hypersurface $V$ are given by the system $x^2-yz=2x=-z=-y=0$ so that the only singularity of $V$ is indeed $O$.
c) The Zariski tangent space $T_O(V)$ is the subspace of $k^3$ determined by the equation $Jac_O(F)(t)=0$, where $Jac_O(F)$ is seen as a linear form on $k^3$ and $t=(t_1,t_2,t_3)$ as a vector in $k^3$.
Since $Jac_O(F)=0$ is the zero linear form here, it follows that $T_O(V)=k^3$.
Translated back in terms of the original matrices, this means that the Zariski tangent space of your variety $X$ of nilpotent matrices is the hyperplane $a+d=0$ of $M_2(k)$.
-
Thanks! I can see that the tangent space of $V(x^{2}-yz) \subseteq \mathbb{A}^{3}$ is the whole $\mathbb{A}^{3}$. How did you translate this to $X$ to conclude that the tangent space consists of the hyperplane $a+d=0$? – user10 Jun 21 '12 at 22:44 is it simply because if $g: X \cong V$ is an isomorphism of varieties then the tangent spaces $T_{(0,0,0,0)}(X) \cong T_{g((0,0,0)}(V)$ are isomorphic as vector spaces? Now since the tangent space of $V$ is $\mathbb{A}^{3}$ then the points are of the form $(x,y,z) \in k^{3}$. Therefore the points of the tangent space of $X$ at the origin are of the form $g(x,y,z)$ where $g$ is the above isomorphism, so points of the form $(x,-y,z,-x)$ where $(x,y,z) \in k^{3}$ and since the first and last coordinate add $0$ then we characterize this set as a+d=0. – user10 Jun 21 '12 at 23:07 Dear user10, yes your argument in the comment is completely correct.In other words you have an isomorphism $F:V(a+d)\to k^3: (a,b,c,d)\mapsto (a,-b,c)$ restricting to your $f:X\to V$, so that the Zariski tangent space $k^3$ corresponds under $F$ to $F^{-1}(k^3)=V(a+d)$. – Georges Elencwajg Jun 22 '12 at 8:13 | 2013-06-19T01:22:30 | {
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https://math.stackexchange.com/questions/4249294/how-to-calculate-sum-of-a-sigma-notation-for-sum-expression-with-exponents | # How to calculate sum of a sigma notation for sum expression with exponents?
So I have got this practice problem
$$\sum_{k=1}^n \frac{(-1)^k\cdot2^{2k}}{3}$$
Now I am fairly new to the sigma notation for the sum of elements. I know the basic stuff like the sum of the arithmetic and geometric sequence. However with this example, I'm not sure what am I supposed to do with it. I can't even determine whether this is an arithmethic or geometric sequence.
Now of course with this little number of element, I could just sum them up manually but I'm sure that there is much easier way. It's just that I don't see it.
Would you be so king give me some hints on what am I looking at?
Thank you for any input.
• This is a geometric series with quotient $-4$ (divided by $3$). Just apply the formula. Sep 13, 2021 at 13:45
• Note however that the series does not start with $1$ which makes it slightly more complicated. Sep 13, 2021 at 13:48
You have a geometric series with common ratio $$r=-4$$ and scale factor $$a=\frac{1}{3}$$. See here also. Then the sum of the first $$n$$ terms is given by
$$\sum_{k=1}^{n}ar^{k-1}=\frac{a(1-r^{n})}{1-r} \quad (1)$$
for $$r\neq 1$$. Applying this to the above series we obtain
$$\sum_{k=1}^{n}\frac{(-1)^{k}2^{2k}}{3}=\sum_{k=1}^{n}\frac{(-1)^{k}4^{k}}{3}=\sum_{k=1}^{n}\frac{(-4)^{k}}{3}$$
Note that the general term in $$(1)$$ is $$ar^{k-1}$$ and at $$k=1$$ it gives $$a$$, but in our case, $$k=1$$ gives $$-\frac{4}{3}$$. So we can pull out a factor of $$-4$$ and we have
$$\sum_{k=1}^{n}\frac{(-1)^{k}2^{2k}}{3}=-4\sum_{k=1}^{n}\frac{1}{3}(-4)^{k-1}=-4\frac{\frac{1}{3}(1-(-4)^{n})}{1-(-4)}=\frac{4}{15}((-4)^{n}-1)$$
Where does the above formula come from?
$$S=\sum_{k=1}^{n}\frac{(-1)^{k}2^{2k}}{3}=-\frac{2^{2}}{3}+\frac{2^4}{3}-\frac{2^6}{3}+...+\frac{(-1)^{n}2^{2n}}{3}$$ $$=\frac{1}{3}\left(-2^2+2^4-2^6+...+(-1)^n2^{2n}\right)=\frac{1}{3}\left(-4+4^{2}-4^{3}+...+(-1)^{n}4^{n}\right)$$ $$=\frac{4}{3}\left(-1+4-4^{2}+...+(-1)^{n}4^{n-1}\right)$$
Then multiply $$S$$ by $$-4$$ to obtain
$$-4S=\frac{4}{3}\left(4-4^{2}+4^{3}+...+(-1)^{n+1}4^{n}\right)$$
Next subtract $$-4S$$ from $$S$$ to obtain
$$S-(-4S)=\frac{4}{3}\left(-1+4-4^{2}+...+(-1)^{n}4^{n-1}\right)-\frac{4}{3}\left(4-4^{2}+4^{3}+...+(-1)^{n+1}4^{n}\right)$$ $$=\frac{4}{3}\left(-1+\color{red}{4}-\color{green}{4^{2}}+...+\color{blue}{(-1)^n4^{n-1}}-\color{red}{4}+\color{green}{4^2}+...-\color{blue}{(-1)^{n}4^{n-1}}+(-1)^{n}4^{n}\right)$$ $$=\frac{4}{3}\left((-1)^{n}4^n-1\right)$$ $$5S=\frac{4}{3}\left((-4)^n-1\right)\implies S=\frac{4}{15}\left((-4)^n-1\right)$$
• @Jasasul I think the most important part of this answer is the second half, where you replace the $\Sigma$ by a sequence of summands with an ellipsis (the ...). Then you can actually see what's going on. That's the way to start this kind of question. Sep 13, 2021 at 14:58
• @Ethan Bolker Thanks for the comment. Agreed, I just showed how to use the general formula in the first half. Sep 13, 2021 at 15:05
• Woah. Thank you for this in depth explanation. Really appreciate it. Sep 13, 2021 at 19:15
• @Jasasul You're very welcome! :-) Sep 13, 2021 at 19:16 | 2022-07-06T14:05:12 | {
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https://math.stackexchange.com/questions/3025175/number-of-ways-to-organizing-n-object-types-into-2n-slots-requiring-each-type-t | Number of ways to organizing n object types into 2n slots, requiring each type to be in 2 slots
For example, say I have $$3$$ objects, $$6$$ slots, and each object must be chosen twice, how do I go about solving that?
Would it just be $$\binom{6}{2} \binom{4}{2} \binom{2}{2}$$ or am I thinking about it completely wrong?
Additionally, say we changed it so object $$A$$ shows up $$3$$ times, object $$B$$ $$2$$ times, and object $$C$$ $$1$$ time.
I don't really understand the intuition behind it, so an explanation of the thought process would be awesome.
• You've got the right answer for the first part. Whatever reasoning you used should get you the answer for the second part as well. – Gerry Myerson Dec 4 '18 at 6:10
• Say we wanted to simply choose at least 1 of each type, but no more than 3, how would I go about that? For the previous ones, I just choose 2 of each object from the remaining # of objects, but you can't really do that. I considered doing the star-bar method and just subtracting the possibilities that don't satisfy the condition, but I was unsure of how to generate those possibilities. – Flyrom Dec 4 '18 at 6:22
• With those restrictions, it has to be 1, 2, 3 or else 2, 2, 2. You know how to handle the second case, and I bet you can handle the first. – Gerry Myerson Dec 4 '18 at 6:26
• Making any progress? – Gerry Myerson Dec 5 '18 at 23:31
• Yeah, since we know the 2,2,2 combination from above, 3 2 1 is similar I found. I just did $\binom{6}{3} \binom{3}{2} \binom{1}{1}$ + the 2,2,2 combination. It made sense in my head, let me know what you think! – Flyrom Dec 6 '18 at 3:41
The way you go about thinking about this problem is that you have 6 positions at the start. Then you need to choose 2 of those 6 for the first type, then 2 of 4 for 2nd, and then 2 of 2 for third. So the end result is $$\binom{6}{2} \binom{4}{2} \binom{2}{2}$$ | 2019-08-21T22:32:39 | {
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https://answerbun.com/mathematics/why-write-computational-complexity-as-oleft-m-n-log-n2-m-right/ | Why write computational complexity as $Oleft( m n log (n^2 / m) right)$?
Mathematics Asked on November 7, 2020
On slide 40 of this presentation, an algorithm is described as having a complexity of $$Oleft( m n log (n^2 / m) right)$$. But since $$log (n^2 / m) = 2 log n – log m$$
is itself $$O(log n)$$, wouldn’t it be equivalent to write the original complexity as $$Oleft( m n log n right)$$?
Is this notation meant to emphasize that this algorithm is marginally worse than the $$Oleft( m n log n right)$$ algorithm listed above it?
The factor $$log(n^2/m)$$ can be better than $$O(log n)$$, because the number of edges $$m$$ can vary to some extent independently of the number of nodes $$n$$.
On the low end, we have $$m ge n-1$$ for any reasonable network. If a node other than $$t$$ doesn't have any edges out of it, we can delete it without changing the problem, so we may assume every node other than $$t$$ has at least one edge leaving it - and then $$m ge n-1$$.
On the high end, we have $$m le (n-1)^2 < n^2$$ because each node other than $$t$$ can have at most $$n-1$$ edges out of it (and $$t$$ shouldn't have any, because they're pointless).
So for sparse networks where $$m = O(n)$$, $$log (n^2/m)$$ is indeed as large as $$Omega(log n)$$, and the two algorithms with complexity $$O(mn log (n^2/m))$$ and $$O(mn log n)$$ are equivalent.
On the other hand, for dense networks where $$m = Omega(n^2)$$, $$log(n^2/m)$$ is just $$O(1)$$, and the algorithm with complexity $$O(mn log(n^2/m))$$ is better: it's just $$O(mn)$$.
This is also the reason why we write the factor $$log (n^2/m)$$ the way we do: the ratio $$frac{m}{n^2}$$ measures the density of the network, so it's an independently interesting parameter.
Correct answer by Misha Lavrov on November 7, 2020
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https://byjus.com/question-answer/there-are-three-categories-of-students-in-a-class-of-60-students-a-very-hardworking/ | Question
# There are three categories of students in a class of 60 students: A : Very hardworking ; B : Regular but not so hardworking; C : Careless and irregular 10 students are in category A, 30 in category B and the rest in category C. It is found that the probability of students of category A, unable to get good marks in the final year examination is 0.002, of category B it is 0.02 and of category C, this probability is 0.20. A student selected at random was found to be one who could not get good marks in the examination. Find the probability that this student is category C.
Solution
## Let E denote the event that the student could not get good marks in the examination. Also, A : the event that student is very hardworking B : the event that student is regular but not so hardworking C : the event that student is careless and irregular ∴ P(A) = $\frac{10}{60}$, P(B) = $\frac{30}{60}$ and P(C) = $\frac{20}{60}$ Also, $\mathrm{P}\left(\frac{E}{A}\right)$ = Pobability that the student of catagory A could not get good marks in the examination = 0.002 $\mathrm{P}\left(\frac{E}{B}\right)$ = Pobability that the student of catagory B could not get good marks in the examination = 0.02 $\mathrm{P}\left(\frac{E}{C}\right)$ = Pobability that the student of catagory C could not get good marks in the examination = 0.2 ∴ Required probability = $\mathrm{P}\left(\frac{C}{E}\right)=\frac{\mathrm{P}\left(C\right)\mathrm{P}\left(\frac{E}{C}\right)}{\mathrm{P}\left(A\right)\mathrm{P}\left(\frac{E}{A}\right)+\mathrm{P}\left(B\right)\mathrm{P}\left(\frac{E}{B}\right)+\mathrm{P}\left(C\right)\mathrm{P}\left(\frac{E}{C}\right)}=\frac{\frac{20}{60}×0.2}{\frac{10}{60}×0.002+\frac{30}{60}×0.02+\frac{20}{60}×0.2}=\frac{4}{4.62}=\frac{400}{462}=\frac{200}{231}$MathematicsRD Sharma XII Vol 2 (2019)All
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View More | 2022-01-27T14:49:53 | {
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https://math.stackexchange.com/questions/2835313/find-smallest-set-of-natural-numbers-whose-pairwise-sums-include-0-n | # Find smallest set of natural numbers whose pairwise sums include 0..n
Given a positive integer $n$, how do you find the smallest set of nonnegative integers $S$ such that for each integer $m$, where $0\leq m<n$, there exist two (not necessarily distinct) members of set $S$, say $x$ and $y$ such that $x+y=m$.
For example, consider the case $n=50$. Suppose the length of $S$ is $L$. For a lower bound, if the elements of $S$ have pairwise distinct sums, then there are $\dbinom{L+1}{2}$ sums (the plus 1 is because numbers can be added to themselves). Thus, $$\binom{L+1}{2}\geq50\implies L\geq10$$.
I can acheive $L=12$ with the set {0, 1, 2, 3, 7, 10, 15, 18, 22, 23, 24, 25} (done with very inefficient program which searches randomly among all sets). For $L=10$, I feel like it should be impossible; we only have to show that more than 5 numbers can be expressed as a sum in more than 1 way, which should be able to be done through some casework. However, is $L=11$ possible? I think so.
Similarly, for $n=100$, I have $L=17$ from my program: {0, 1, 3, 4, 9, 11, 16, 20, 25, 30, 34, 39, 41, 46, 47, 49, 50}. But the lower bound only gives $L\geq 14$, so at least $L=15$ or $L=16$ should be possible.
In general, how do you do it efficiently for any given $n$?
• I just realized that both my answer and Fabio Lucchini's (perhaps following my mistake) shifted your $n$ by one; we assumed that $n$ itself also needs to be generated. This is also how $n$ is defined in OEIS A066063. Usually I'd correct the mistake in my answer, but since the OEIS sequence and two long answers now all use the same convention, might I perhaps ask you to change the question slightly to match it? Your examples would still be valid, since in both cases you can also form $n$ itself; all you'd have to change is $m\le n$ and $\binom{L+1}2\gt50$. Jul 1, 2018 at 8:10
Here are results for $n$ up to $80$, where $\min$ is the lower bound you derive from the binomial coefficient, $\max$ is the upper bound that Fabio Lucchini derived in his answer, and $L=|S|$ is the size of a minimal generating set. Actual subsets $S$ are only shown for the last entry for any given $L$, since this subset also works for all smaller $n$.
\begin{array}{c|c|c|c|l} n&\min&\max&L&S\\\hline 2&2&2&2&\{0,1\}\\ 3&3&3&3\\ 4&3&3&3&\{0,1,2\}\\ 5&3&4&4\\ 6&4&4&4\\ 7&4&4&4\\ 8&4&4&4&\{0,1,3,4\}\\ 9&4&5&5\\ 10&5&5&5\\ 11&5&5&5\\ 12&5&5&5&\{0,1,3,5,6\}\\ 13&5&6&6\\ 14&5&6&6\\ 15&6&6&6\\ 16&6&6&6&\{0,1,3,5,7,8\}\\ 17&6&7&7\\ 18&6&7&7\\ 19&6&7&7\\ 20&6&7&7&\{0,1,2,5,8,9,10\}\\ 21&7&8&8\\ 22&7&8&8\\ 23&7&8&8\\ 24&7&8&8\\ 25&7&8&8\\ 26&7&8&8&\{0,1,2,5,8,11,12,13\}\\ 27&7&9&9\\ 28&8&9&9\\ 29&8&9&9\\ 30&8&9&9\\ 31&8&9&9\\ 32&8&9&9&\{0,1,2,5,8,11,14,15,16\}\\ 33&8&10&10\\ 34&8&10&10\\ 35&8&10&10\\ 36&9&10&10\\ 37&9&10&10\\ 38&9&10&10\\ 39&9&11&10\\ 40&9&11&10&\{0,1,3,4,9,11,16,17,19,20\}\\ 41&9&11&11\\ 42&9&11&11\\ 43&9&11&11\\ 44&9&11&11\\ 45&10&12&11\\ 46&10&12&11&\{0,1,2,3,7,11,15,19,21,22,24\}\\ 47&10&12&12\\ 48&10&12&12\\ 49&10&12&12\\ 50&10&12&12\\ 51&10&12&12\\ 52&10&12&12\\ 53&10&13&12\\ 54&10&13&12&\{0,1,2,3,7,11,15,19,23,25,26,28\}\\ 55&11&13&13\\ 56&11&13&13\\ 57&11&13&13\\ 58&11&13&13\\ 59&11&13&13\\ 60&11&13&13\\ 61&11&14&13\\ 62&11&14&13\\ 63&11&14&13\\ 64&11&14&13&\{0,1,3,4,9,11,16,21,23,28,29,31,32\}\\ 65&11&14&14&\\ 66&12&14&14&\\ 67&12&14&14&\\ 68&12&14&14&\\ 69&12&15&14&\\ 70&12&15&14&\\ 71&12&15&14&\\ 72&12&15&14&\{0,1,3,4,9,11,16,20,25,27,32,33,35,36\}\\ 73&12&15&15&\\ 74&12&15&15&\\ 75&12&15&15&\\ 76&12&15&15&\\ 77&12&16&15&\\ 78&13&16&15&\\ 79&13&16&15&\\ 80&13&16&15&\{0,1,3,4,5,8,14,20,26,32,35,36,37,39,40\}\\ \end{array}
Here's the code I used to generate these results. It loops over $n$, making use of the solution for $n-1$ in each step. It first checks whether the set for $n-1$ also works for $n$. If not, it tries finding a new set also containig $L$ numbers, at first using only elements up to $\lfloor\frac n2\rfloor + 2$. Only if that doesn't work does it try all combinations with $L$ elements all the way up to $n$. If that also doesn't work, it increases $L$. This way, it spends almost all its time only on the values of $n$ where $L$ needs to be incremented; for all other values of $n$ it quickly finds a solution without having to search the entire space.
The least $n$ for which Fabio Lucchini's upper bound is not tight is $n=39$, for which the $10$-element set $\{0,1,3,4,9,11,16,17,19,20\}$ is sufficient whereas the upper bound is $11$.
The sequence $L(n)$ is OEIS A066063, and the only information in that entry is the lower bound you already found. Usually, OEIS is quite good at collecting information about sequences, so it's likely that nothing else was known.
• Indeed it looks like when $n$ is a multiple of 4, the set has symmetry around $n/4$. This is true even for my possibly suboptimal set for $n=100$. 0 and 1 must be included since if 0 weren't there, there would be no way to make 0, and if the next smallest number after 0 wasn't 1, (say it was 0, 2 or 0, 3) there would be no way to make 1. Then $n/2$ and $n/2-1$ would follow from symmetry if that were proven. Also, there may be a generalization of the symmetry to non multiples of 4. Jun 29, 2018 at 12:44
• @soktinpk: I've updated the answer with results up to $n=64$. Jul 1, 2018 at 12:43
• @soktinpk: I've updated the answer with results up to $n=80$. Jul 4, 2018 at 13:11
Let $n+4=s^2+r$ with $r,s\in\Bbb N$ and $0\leq r\leq 2s$. Then an upper bound is given by $$L\leq \begin{cases} \lceil{\frac rs}\rceil+2s-3&2\mid s\\ \lceil{\frac{r+1}{s+1}}\rceil+2s-3&2\nmid s \end{cases}$$ which gives $L\leq 12$ for $n=50$ and $L\leq 18$ for $n=100$. In general, for large $n$ this gives $L=O(\sqrt n)$.
A set $S$ corresponding to this upper bound is given by \begin{align} S &=\{i\in\Bbb N:0\leq i<q-1\}\\ &\cup\{(q-1)+jq:0\leq j\leq k-1\}\\ &\cup\{i\in\Bbb N:(q-1)+(k-1)q<i\leq 2(q-1)+(k-1)q\} \end{align} for suitable values of $q$ and $k$. Then $|S|=k+2(q-1)$ and then summing each pair of its elements we get all the natural number less or equals to $2(2(q-1)+(k-1)q)=2\max S$. Consequently, we choose $$k=\left\lceil\frac{n+4}{2q}\right\rceil-1$$ The function $$\frac{n+4}{2q}-1+2(q-1)$$ attains a minimum at $q=\frac{\sqrt{n+4}}2$.
If $n+4=s^2+r$ and $s=2t+b$ with $r\leq 2s$ and $0\leq b\leq 1$, then $$t\leq\frac{\sqrt{n+4}}2<t+1$$
For $q=t$ we get \begin{align} |S_t| &=\left\lceil\frac{n+4}{2t}\right\rceil+2t-3\\ &=\left\lceil\frac{s^2+r}{s-b}\right\rceil+s-b-3\\ &=\left\lceil\frac{r+b}{s-b}\right\rceil+2s-3 \end{align} while for $q=t+1$ \begin{align} |S_{t+1}| &=\left\lceil\frac{n+4}{2t+2}\right\rceil+2t+2-3\\ &=\left\lceil\frac{s^2+r}{s-b+2}\right\rceil+s-b-3\\ &=\left\lceil\frac{r+4-3b}{s+2-b}\right\rceil+2s-3 \end{align} Since $|S_t|\leq|S_{t+1}|$ if and only if $b(2s+1)\leq 2s-r$, the formula on the top is proved.
• You are right, I made a typo; the answer is now edited. Jun 29, 2018 at 9:27
• I see. I think the expression $\lceil{\frac{r-1}{s+1}}\rceil+2s-3$ in the odd case is slightly off. For $n=32$, both my results and your analysis further down show that $9$ elements are required ($q=3$, $k=5$), but that expression yields an upper bound of $8$. Jun 29, 2018 at 9:55
• Yes, edited another typo. Thank'you. Now for $n=32$ we get $s=6$ and $r=0$ which gives $L\leq 9$. Jun 29, 2018 at 10:01
• Unfortunately, this bound is not minimal: for $n=100$ it gives $|S|=18$, while in the OP soktinpk find a solution with $|S|=17$. Jun 29, 2018 at 13:00
• Ah, yes, I didn't check that case. I'll try to find the first $n$ at which it's not minimal. Anyway, it's a very good bound, we should perhaps enter it at OEIS. Jun 29, 2018 at 13:01
Here is the casework to show that for $n=50$, $L$ indeed has to be greater than $10$, assuming joriki's suggestion:
all minimal solutions contain $0$,$1$,$\lceil\frac n2\rceil-1$ and $\lceil\frac n2\rceil$
Since ${11 \choose 2} =55$, we just need to show that we can get at least $6$ 'matches' for any set of numbers.
OK, so by joriki's suggestion, we need $0,1,24,25$. So immediately we have one match: $24+1=25+0$
Now for the cases:
I) If we add $2$, we have two more: $1+1=2+0$ and $24+2=25+1$
To make 47, we need to either add $23$ or $22$:
I.A) Add $23$. Then we can add $23+1=24+0$, $24+2=25+1$, and $23+25=24+24$, so we have our $6$ matches
I.B) Add $22$. We can add $22+2=24+0$ (4 matches now)
To make $5$, we need to add $3$, $4$, or $5$:
I.B.i) Add $3$: $2+1=3+0$ and $24+3=25+2$ Got our $6$
I.B.ii) Add $4$: $2+2=4+0$ and $22+4=24+2$. Also $6$
I.B.iii) Add $5$: $22+5=25+2$. OK, need one more ... well, to create $45$ we need to add either $20$ (which gives $20+2=22+0$) or $21$ ($21+1=22+0$), so done here as well
OK, so adding $2$ definitely leads to $6$ matches. So, let's not add $2$ ... but now we need $3$ to create $3$:
II) Add $3$
Again, to create $47$, we have to add either $23$ or $22$:
II.A) Add $23$. We can add $24+24=23+25$, $23+1=24+0$, and $23+3=25+1$, so 4 matches now.
To create $5$, we need to add either $4$ or $5$:
II.A.i) Add $4$. Then we have $3+1=4+0$ and $23+4=25+3$. Done
II.A.ii) Add $5$. We have $5+1=3+3$ and $23+5=25+3$ Done
II.B) Add $22$. We can add $22+3=25+0$ and $22+3=24+1$ (3 matches now)
To make $5$, we need tpo add $4$ or $5$:
II.B.i) Add $4$. This gives $1+3=4+0$, $24+4=25+3$, and $22+4=25+1$ Done
II.B.ii) Add $5$. This gives $1+5=3+3$, $22+5=24+3$, so need just one more. But to create $45$ we need to add $21$ ($21+1=22+0$), or $20$ ($20+3=22+1$), so done here as well
.. and that concludes all cases. So, yes, as you suspected, for $n=25$ we have $L>10$
And frankly, seeing how quickly these possible matches increase, my guess is that for $n=50$, $L=12$ | 2022-08-12T06:29:45 | {
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https://www.mscroggs.co.uk/blog/84 | mscroggs.co.uk
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# Blog
2020-05-15
This is a post I wrote for The Aperiodical's Big Lock-Down Math-Off. You can vote for (or against) me here until 9am on Sunday...
Recently, I came across a surprising fact: if you take any quadrilateral and join the midpoints of its sides, then you will form a parallelogram.
The blue quadrilaterals are all parallelograms.
The first thing I thought when I read this was: "oooh, that's neat." The second thing I thought was: "why?" It's not too difficult to show why this is true; you might like to pause here and try to work out why yourself before reading on...
To show why this is true, I started by letting $$\mathbf{a}$$, $$\mathbf{b}$$, $$\mathbf{c}$$ and $$\mathbf{d}$$ be the position vectors of the vertices of our quadrilateral. The position vectors of the midpoints of the edges are the averages of the position vectors of the two ends of the edge, as shown below.
The position vectors of the corners and the midpoints of the edges.
We want to show that the orange and blue vectors below are equal (as this is true of opposite sides of a parallelogram).
We can work these vectors out: the orange vector is$$\frac{\mathbf{d}+\mathbf{a}}2-\frac{\mathbf{a}+\mathbf{b}}2=\frac{\mathbf{d}-\mathbf{b}}2,$$ and the blue vector is$$\frac{\mathbf{c}+\mathbf{d}}2-\frac{\mathbf{b}+\mathbf{c}}2=\frac{\mathbf{d}-\mathbf{b}}2.$$
In the same way, we can show that the other two vectors that make up the inner quadrilateral are equal, and so the inner quadrilateral is a parallelogram.
### Going backwards
Even though I now saw why the surprising fact was true, my wondering was not over. I started to think about going backwards.
It's easy to see that if the outer quadrilateral is a square, then the inner quadrilateral will also be a square.
If the outer quadrilateral is a square, then the inner quadrilateral is also a square.
It's less obvious if the reverse is true: if the inner quadrilateral is a square, must the outer quadrilateral also be a square? At first, I thought this felt likely to be true, but after a bit of playing around, I found that there are many non-square quadrilaterals whose inner quadrilaterals are squares. Here are a few:
A kite, a trapezium, a delta kite, an irregular quadrilateral and a cross-quadrilateral whose innner quadrilaterals are all a square.
There are in fact infinitely many quadrilaterals whose inner quadrilateral is a square. You can explore them in this Geogebra applet by dragging around the blue point:
As you drag the point around, you may notice that you can't get the outer quadrilateral to be a non-square rectangle (or even a non-square parallelogram). I'll leave you to figure out why not...
### Similar posts
Mathsteroids Interesting tautologies Big Internet Math-Off stickers 2019 Runge's Phenomenon
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DEB JYOTI MITRA | 2020-09-30T03:07:00 | {
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http://josiahdev.com/blog/ | # Project Euler Problem 2: Even Fibonacci numbers
Project Euler Problem 2 instructs:
Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, …
By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.
They’ve already given you the definition of the Fibonacci sequence (except it is usually written as starting with 0,1 or 1,1). So the first thing to figure out is how do we find the sequence of even-valued terms.
Luckily, there is a pretty easy pattern here. Let’s start by writing the Fibonacci sequence as a formula:
$F_n=F_{n-1} + F_{n-2}$
where $F_1=F_2=1$
So starting by substituting $n=3$ we get $F_3=F_2+F_1$. Since both $F_1$ and $F_2$ are odd valued, we’ll replace them with $F_{odd}$. So now we have:
$F_3=F_{odd} + F_{odd}$
Two odd numbers always add to an even number, so let’s replace $F_3$ with $F_{even}$. Resulting in:
$F_{even}=F_{odd} + F_{odd}$
So this is the start of our pattern. Now, let’s find the next equation in the pattern.
$F_4=F_3 + F_2$
Substitute $F_3$ and $F_2$:
$F_4=F_{even} + F_{odd}$
An even and and odd number will always add to an odd number. So we have:
$F_{odd}=F_{even} + F_{odd}$
Repeat for $F_5$ and we have:
$F_{odd}=F_{odd} + F_{even}$
Once more for $F_6$ results in:
$F_{even}=F_{odd} + F_{odd}$
Which is where we started with $F_3$. Since each Fibonacci number only depends on the previous 2 numbers, this sequence of even and odd numbers will repeat forever.
\begin{aligned}&F_1&=&1&=&odd&\\&F_2&=&1&=&odd&\\&F_3&=&2&=&even&\\&F_4&=&3&=&odd&\\&F_5&=&5&=&odd&\\&F_6&=&8&=&even&\\&F_7&=&13&=&odd&\\&F_8&=&21&=&odd&\\&F_9&=&34&=&even&\end{aligned}
Every third term is an even number starting with $F_3$.
## Solution A: The Straightforward Approach
The most straight-forward way to implement this in C# would be to calculate each Fibonacci number and add up every third number until we reach our limit:
public static long SolveA()
{
long n1 = 1;
long n2 = 2;
long total = 0;
while (n2 <= 4000000)
{
total += n2;
long n3 = n1 + n2;
long n4 = n2 + n3;
long n5 = n3 + n4;
n1 = n4;
n2 = n5;
}
}
## Solution B: Skipping Odd Values
Now we’re going to discuss the first optimizaiton. Since we only need to add up every third Fibonacci number, we don’t actually need the other two in the sequence.
$F_n=F_{n-1} + F_{n-2}$
Instead of being based on the previous two numbers, which we don’t care about, let’s see if we can rewrite it in terms of the numbers we do care about. That would be the other even numbers, which occur in every third place. So we’d like to express $F_n$ in terms of $F_{n-3}$ and $F_{n-6}$.
Let’s start by substituting $F_{n-1}$
$F_n=F_{n-2}+F_{n-3} + F_{n-2}$
Simplify:
$F_n=2*F_{n-2}+F_{n-3}$
$F_{n-3}$ is useful, but we still need to break down $F_{n-2}$:
$F_n=2*(F_{n-3}+F_{n-4})+F_{n-3}$
Simplify:
$F_n=3*F_{n-3}+2*F_{n-4}$
This part gets a bit tricky. We’re going to substitute only one $F_{n-4}$ and leave the other one to use later:
$F_n=3*F_{n-3}+F_{n-4}+F_{n-5}+F_{n-6}$
Now we substitute $F_{n-3}$ back in place of $F_{n-4}+F_{n-5}$ to get:
$F_n=3*F_{n-3}+F_{n-3}+F_{n-6}$
Simplify:
$F_n=4*F_{n-3}+F_{n-6}$
There you have it. We now have our Fibonacci sequence, except that it only spits out the even terms.
Here is the C# code:
public static long SolveB()
{
long n1 = 2;
long n2 = 8;
long n3 = 34;
long total = 10;
while (n3 <= 4000000)
{
total += n3;
n1 = n2;
n2 = n3;
n3 = (n2 << 2) + n1;
}
}
Note: the n2 << 2 is just a more efficient way to multiply by 4.
## Solution C: No Running Total Needed
Ok, so there's actually a way to do this without even keeping a running total. We still have to calculate the Fibonacci sequence, but we don't have to explicitly add up the even terms. We'll do this with a trick similar to the one in Project Euler Problem 1.
What we're trying to solve is the total that we'll call $S_n$:
$S_n=\displaystyle\sum_{k=1}^{n/3}F_{3*k}$
Let's start by taking the end of Solution B:
$F_n=4*F_{n-3}+F_{n-6}$
This will be much easier to work with if we add 6 to the indices and get:
$F_{n+6}=4*F_{n+3}+F_n$
And then solve for $F_n$:
$F_n=F_{n+6}-4*F_{n+3}$
Now let's expand the values of $F_n$ that we want to sum for our solution:
\begin{aligned}&F_n&=&F_{n+6}&-&4*&F_{n+3}\\&F_{n-3}&=&&&&F_{n+3}&-&4*&F_n&\\&F_{n-6}&=&&&&&&&F_{n}&-&4*&F_{n-3}&\\\dots\\&F_6&=&&&&&&&&&&&F_{12}&-&4*&F_9&\\&F_3&=&&&&&&&&&&&&&&F_9&-&4*&F_6&\\&F_0&=&&&&&&&&&&&&&&&&&F_6&-&4*&F_3&\end{aligned}
Note: I know that $F_0$ is equal to zero, but having it here it will make things easier later.
First, you can see that the left side of these equations add up to the total we are looking for. They're the list of all the even Fibonacci numbers. And we can also see a pattern on the right side. The terms in the middle partially cancel each other out leaving negative three of each term. So it ends up looking like this:
$S_n=F_{n+6}-3*F_{n+3}-3*F_n-3*F_{n-3}-...-3*F_{12}-3*F_9-3*F_6-4*F_3$
We can simplify this with a summation expression as:
$S_n=F_{n+6}-3*F_{n+3}-3*\displaystyle\sum_{k=1}^{n/3}F_{3*k}-F_3$
Substitute the summation for $S_n$:
$S_n=F_{n+6}-3*F_{n+3}-3*S_n-F_3$
And solve for $S_n$:
$S_n=\dfrac{F_{n+6}-3*F_{n+3}-F_3}4$
We now have the sum of our Fibonacci sequence in terms of the sequence itself.
In code form:
public static long SolveC()
{
long n1 = 2;
long n2 = 8;
long n3 = 34;
while (n3 <= 4000000)
{
n1 = n2;
n2 = n3;
n3 = (n2 << 2) + n1;
}
n1 = (n3 << 2) + n2;
long total = (n1 - 3 * n3 - 2) >> 2;
}
## Solutions D & E: Logarithmic Time Algorithms
The solutions we've developed so far have all had a linear time complexity with respect to the maximum sequence value (4000000 in our case). However, it is possible to do better than that. I'm not going to go into a lot of detail, because this is probably beyond the skill level expected just for problem 2 on Project Euler.
### Matrix Multiplication
It is actually possible to find the nth Fibonacci number with the following matrix formula:
$\begin{bmatrix}1&1\\1&0\end{bmatrix}^n=\begin{bmatrix}F_{n+1}&F_n\\F_n&F_{n-1}\end{bmatrix}$
And there are tricks to do the matrix exponent in logarithmic time, which I won't get into yet. But that's not the whole problem. Our problem statement doesn't actually want us to get the nth Fibonacci number. It wants us to sum up to the number whose value does not exceed a limit. So you'd have to use this matrix exponentiation to do a binary search to find which value of n does not exceed the limit where n + 3 does. You could then use this in our Solution C.
Inspiration for this solution came from this link.
### Binet's Formula
Another improvement on the previous solution can find the nth Fibonacci number in constant time. It's called Binet's formula. In fact, any constant-recursive sequence can be solved this way. However, the binary search from the matrix solution would still be needed to find the index in the Fibonacci sequence that doesn't exceed the limit.
Here is the formula:
$f_n=\dfrac{\bigg(\dfrac{1+\sqrt5}2\bigg)^n-\bigg(\dfrac{1-\sqrt5}2\bigg)^n}{\sqrt5}$
This could also be used in Solution C to find the desired sum.
## GitHub
You can find the code on GitHub (complete with unit tests) at https://github.com/josiahwood/ProjectEuler/tree/master/Problem2.
# Project Euler Problem 1: Multiples of 3 and 5
Project Euler Problem 1 instructs:
If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below 1000.
This ends up being a basic math problem, so no code is needed yet.
I think the easiest way to think of this is as a variation on triangular numbers. Triangular numbers are the sums of the first $n$ natural numbers. Which is a very similar description to our problem statement. The initial formula for triangular numbers is:
$T_n=\displaystyle\sum_{k=1}^nk$
Expanded, this looks like:
$T_n=1+2+3+...+(n-2)+(n-1)+n$
If we do a little trick here, then we can simplify this expression considerably. Take the above expansion, reverse it, and add it to the original expansion term by term.
\begin{aligned}&T_n&=&1&+&2&+&3&+&...&+&(n-2)&+&(n-1)&+&n\\+&T_n&=&n&+&(n-1)&+&(n-2)&+&...&+&3&+&2&+&1\\=&2T_n&=&(n+1)&+&(n+1)&+&(n+1)&+&...&+&(n+1)&+&(n+1)&+&(n+1)\end{aligned}
If we simplify the repeated terms we get:
$2T_n=n(n+1)$
And then solve for $T_n$:
$T_n=\dfrac{n(n+1)}2$
Great! Now we really have something that we can use in our solution. Since the problem asked for the sum of natural numbers below 1000 let us subtract 1 to get 999. Then divide that by $i$ (which will represent the increment being summed, either 3 or 5 in this problem). Last, scale by $i$ so that we are adding up multiples of the increment instead of just sequential natural numbers. So if $L$ represents the upper limit provided (1000), then we get this equation:
$S(L,i)=iT_{{\lfloor}\frac{L-1}i{\rfloor}}$
Now substitute the final equation for $T_n$:
$S(L,i)=\dfrac{i}2{\lfloor}\dfrac{L-1}i{\rfloor}({\lfloor}\dfrac{L-1}i{\rfloor}+1)$
However, we need the sum of numbers that are either multiples of 3 or 5. So we have to add both of those results and then subtract the intersection that would otherwise be counted twice.
$S(1000,3) + S(1000,5) - S(1000,15)$
$166833 + 99500 - 33165 = 233168$
It would have arguably been simpler to just write a quick program that loops through all natural numbers less than 1000 and total any value that is a multiple of 3 or 5. However, thinking about this problem mathematically is good practice for what is up ahead. In future problems you will not have the luxury of using an algorithm that takes linear time instead of constant time.
However, if you’d really like to see this in code, then here is my C# example:
using System;
namespace Problem1
{
public class Program
{
static void Main(string[] args)
{
int result = Solve(1000);
Console.WriteLine(result);
}
public static int Solve(int limit)
{
return Sum(limit, 3) + Sum(limit, 5) - Sum(limit, 15);
}
static int Sum(int limit, int increment)
{
int n = (limit - 1) / increment;
return increment * n * (n + 1) / 2;
}
}
}
You can find the code on GitHub (complete with a unit test) at https://github.com/josiahwood/ProjectEuler/tree/master/Problem1.
# Twelve Men on an Island Logic Puzzle
One of my favorite shows right now is Brooklyn Nine-Nine. During season 2 episode 18, titled “Captain Peralta”, Captain Holt presents this riddle to the team:
There are twelve men on an island. Eleven weigh exactly the same amount, but one of them is slightly lighter or heavier. You must figure out which. The island has no scales, but there is a seesaw. You can only use it three times.
This isn’t exactly a programming post, but it does demonstrate the same kind of problem solving that is often found in devising efficient algorithms.
I decided to represent each islander with a letter from A to L. This makes it easy to unambiguously and concisely represent combinations of islanders. I show each trial with the seesaw as a column. The “Test” column shows what islanders are put on the seesaw as left vs. right. The “Result” column shows “=” if both sides were equal, “<” if the left side weighed less than the right, or “>” if the left side weighed more than the right. Then the “Possibilities” column shows what possible solutions remain given the result of the trial. As we move from the 1st trial towards the 3rd we can see the remaining possibilities reduce as more results eliminate solutions.
When solving this puzzle we know right away that we cannot rely on the seesaw always tilting to the left or right. This seems to be people’s initial attempt at solving this problem. Just put six on one side and six on the other and work your way down. However, 3 trials and only 2 possible results would only give us up to 8 solutions because 2 x 2 x 2 = 8. So we know that we have to rely some trials being balanced to find all the solutions. Therefore, it is probably not a good idea to put six vs. six because there is no way that seesaw could be balanced since the different islander would always be on the seesaw giving only two possible results which doesn’t tell us which side the different islander is on.
The first trial that gives the most information is to do four vs. four (ABCD vs. EFGH). No matter what the result we have reduced the possible solutions from 24 to 8.
Possibilities Test Result Possibilities
A lighter
A heavier
B lighter
B heavier
C lighter
C heavier
D lighter
D heavier
E lighter
E heavier
F lighter
F heavier
G lighter
G heavier
H lighter
H heavier
I lighter
I heavier
J lighter
J heavier
K lighter
K heavier
L lighter
L heavier
ABCD vs. EFGH = I lighter
I heavier
J lighter
J heavier
K lighter
K heavier
L lighter
L heavier
< A lighter
B lighter
C lighter
D lighter
E heavier
F heavier
G heavier
H heavier
> A heavier
B heavier
C heavier
D heavier
E lighter
F lighter
G lighter
H lighter
• If the seesaw is equal then we know those eight islanders (ABCDEFGH) can be eliminated from being either lighter or heavier. The remaining four islanders (IJKL) could then be either lighter or heavier.
• If the seesaw is unbalanced then we know the four untested islanders (IJKL) can be eliminated. We also know that all the islanders on the heavier side cannot be lighter. And all the islanders on the lighter side cannot be heavier.
What we do for the second trial depends on the type of result for the first trial because the pattern of remaining possibilities is different.
• If the first trial was balanced then we have four remaining islanders (IJKL) that could be either lighter or heavier. We can work out the remaining possibilities by comparing three remaining islanders (IJK) with three normal islanders (ABC).
Possibilities Test Result Possibilities
I lighter
I heavier
J lighter
J heavier
K lighter
K heavier
L lighter
L heavier
IJK vs. ABC = L lighter
L heavier
< I lighter
J lighter
K lighter
> I heavier
J heavier
K heavier
• If the second trial was balanced, then you know that that last remaining islander (L) is the different islander. Compare him with any other islander. That will tell you if he is lighter or heavier than the other islanders. This third trial cannot be balanced since we know that this last islander (L) is different from all the other islanders.
• If the second trial was unbalanced, then you now know if the different islander is either lighter or heavier. You also know that the last untested islander (L) is normal. You can weigh two of the remaining islanders against each other to determine which of the remaining three islanders is different. I will show the case where the three islanders (IJK) are lighter.
Possibilities Test Result Possibilities
I lighter
J lighter
K lighter
I vs. J = K lighter
< I lighter
> J lighter
• If the first trial was unbalanced, then this gets much less intuitive. For simplicity, we will only consider the result where the left side is lighter than the right side (ABCD < EFGH). The last case where the left side is heavier than the right side is just the inverse. The key to this trial is to find a test that will make the best use of the three possible seesaw results. Specifically, we don't want to use a test that only has two possible outcomes (like ABC vs. EFG). So I chose to take two potentially lighter islanders with one potentially heavier islander for each side (ABE vs. CDF). This has all three possible outcomes and leaves us with either two or three remaining possibilities.
Possibilities Test Result Possibilities
A lighter
B lighter
C lighter
D lighter
E heavier
F heavier
G heavier
H heavier
ABE vs. CDF = G heavier
H heavier
< A lighter
B lighter
F heavier
> C lighter
D lighter
E heavier
• If the second trial is balanced, then you only have two potentially heavier islanders left. Just weigh them against each other and whoever is heavier is the different islander. It is not possible for this last trial to be balanced since you know one of them is the heavier islander.
• If the second trail is unbalanced, then weigh the remaining potentially lighter islanders against each other. If the third trial is balanced then the last potentially heavier islander is the different islander. If the third trial is unbalanced, then the islander on the lighter side is the different islander.
Here is the full solution:
1st Trial 2nd Trial 3rd Trial
Test Result Possibilities Test Result Possibilities Test Result Possibilities
ABCD vs. EFGH = I lighter
I heavier
J lighter
J heavier
K lighter
K heavier
L lighter
L heavier
IJK vs. ABC = L lighter
L heavier
L vs. A = Impossible
< L lighter
> L heavier
< I lighter
J lighter
K lighter
I vs. J = K lighter
< I lighter
> J lighter
> I heavier
J heavier
K heavier
I vs. J = K heavier
< J heavier
> I heavier
< A lighter
B lighter
C lighter
D lighter
E heavier
F heavier
G heavier
H heavier
ABE vs. CDF = G heavier
H heavier
G vs. H = Impossible
< H heavier
> G heavier
< A lighter
B lighter
F heavier
A vs. B = F heavier
< A lighter
> B lighter
> C lighter
D lighter
E heavier
C vs. D = E heavier
< C lighter
> D lighter
> A heavier
B heavier
C heavier
D heavier
E lighter
F lighter
G lighter
H lighter
ABE vs. CDF = G lighter
H lighter
G vs. H = Impossible
< G lighter
> H lighter
< C heavier
D heavier
E lighter
C vs. D = E lighter
< D heavier
> C heavier
> A heavier
B heavier
F lighter
A vs. B = F lighter
< B heavier
> A heavier
# Ethereum Truffle and Testrpc with Cloud9
Here are the steps I’ve performed to get ConsenSys’s Ethereum Truffle working on Cloud9. Note that I am new to a lot of the things being used here, so if you see something that I’m doing wrong or “the hard way” then let me know! I hope this post helps someone else, but I also hope that I get some good feedback.
# Installation Script
After creating your Cloud9 workspace run this script (we’ll break it down afterward):
#!/bin/bash
# pyethereum
cd ~
git clone https://github.com/ethereum/pyethereum/
cd pyethereum
sudo python setup.py install
# eth-testrpc
sudo pip install eth-testrpc
# nodejs v5.0.0
. ~/.nvm/nvm.sh
nvm use v5.0.0
nvm alias default 5.0.0
# truffle
sudo npm install -g truffle
You can find the latest version of this script in my Futarchy project at https://github.com/josiahwood/futarchy/blob/master/cloud9setup.sh.
## Ethereum
# pyethereum
cd ~
git clone https://github.com/ethereum/pyethereum/
cd pyethereum
sudo python setup.py install
Lines 5-7 are just copy/pasted from the installation section of https://github.com/ethereum/pyethereum/. It works perfectly as advertised. This is required for testrpc to work in the next section.
## Testrpc
# eth-testrpc
sudo pip install eth-testrpc
Line 10 is all you need to install testrpc according to https://github.com/ConsenSys/eth-testrpc. I’ve found Testrpc to be the quickest way to get up and going with writing contracts.
## Update Nodejs
# nodejs v5.0.0
. ~/.nvm/nvm.sh
nvm use v5.0.0
nvm alias default 5.0.0
Since Truffle uses npm you’ll want to put Nodejs at your desired version first. I found version 5.0.0 worked very well with truffle. Lines 13-15 were the most consistent way I found to update Nodejs and make the new version the default. Line 15 is important to make sure all new Cloud9 terminals are using the correct version of Nodejs by default. This also affects what happens to Cloud9 terminals if you’ve been logged off for a while. Cloud9 advertises that your environment will be exactly the way you left it, but the terminals do reset after some timeout.
## Truffle
# truffle
sudo npm install -g truffle
At this point line 18 works exactly as documented in the installation section of Truffle at https://github.com/ConsenSys/truffle.
# Cloud9 Environment
Ok, this part took me a while to get working on Cloud9. I had no problems running on a local Ubuntu virtual machine, but I wasn’t so lucky on Cloud9. You can see how I worked out the problem on StackOverflow at Ethereum Test RPC working in Cloud9 with Truffle. Credit for the final answer goes to XMLHttpRequest cannot load cloud 9 io.
In order to make it possible for testrpc to bind to the public IP address of your Cloud9 environment you’ll need to click on the “Share” button in the upper right-hand corner of the Cloud9 IDE and then make the Application link public. The StackOverflow answer indicated that this was due to a recent bug in Cloud9, so this may not be necessary in the future.
# Truffle Project Environment/Configuration
Once you have some code/contracts you want to try out you’ll need to start testrpc first. Do this by executing testrpc -p 8081 -d 0.0.0.0 in a new Cloud9 terminal. The -p 8081 argument tells testrpc to run on port 8081. Cloud9 only opens a few ports for public access according to https://docs.c9.io/docs/multiple-ports so we can’t use the testrpc default of 8545. Port 8080 will be used later by the truffle serve command, so 8081 is the next open port. The -d 0.0.0.0 argument tells testrpc to bind to all available IP addresses, including the public one.
Now that you have testrpc up and running, you’ll need to configure your Truffle project properly. You can do this a few ways but the key is to get this part in your config file:
"rpc": {
"host": "project-user.c9users.io",
"port": 8081
}
Replace project-user.c9users.io with the Application URL you made public in the “Cloud9 Environment” section above.
You can either update this in your config/app.json file (to make it the default for all environments), one of the existing config/[environment]/config.json files, or make a new environment just for Cloud9.
After deploying your contracts with truffle deploy you can now run truffle serve and your HTML/JavaScript will be visible on port 8080 of your Application URL. Since this is conveniently the default port for Cloud9 it automatically gets redirected from port 80. This allows you to see your project directly at your Application URL like “http://project-user.c9users.io”. | 2020-05-29T13:39:56 | {
"domain": "josiahdev.com",
"url": "http://josiahdev.com/blog/",
"openwebmath_score": 0.6284764409065247,
"openwebmath_perplexity": 1418.1227835708082,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.988668248138067,
"lm_q2_score": 0.8459424295406088,
"lm_q1q2_score": 0.8363564198395739
} |
https://grindskills.com/for-which-distributions-does-uncorrelatedness-imply-independence/ | # For which distributions does uncorrelatedness imply independence?
A time-honored reminder in statistics is “uncorrelatedness does not imply independence”. Usually this reminder is supplemented with the psychologically soothing (and scientifically correct) statement “when, nevertheless the two variables are jointly normally distributed, then uncorrelatedness does imply independence”.
I can increase the count of happy exceptions from one to two: when two variables are Bernoulli-distributed , then again, uncorrelatedness implies independence. If $X$ and $Y$ are two Bermoulli rv’s, $X \sim B(q_x),\; Y \sim B(q_y)$, for which we have $P(X=1) = E(X) = q_x$, and analogously for $Y$, their covariance is
For uncorrelatedness we require the covariance to be zero so
which is the condition that is also needed for the variables to be independent.
So my question is: Do you know of any other distributions (continuous or discrete) for which uncorrelatedness implies independence?
Meaning: Assume two random variables $X,Y$ which have marginal distributions that belong to the same distribution (perhaps with different values for the distribution parameters involved), but let’s say with the same support eg. two exponentials, two triangulars, etc. Does all solutions to the equation $\operatorname{Cov}(X,Y) = 0$ are such that they also imply independence, by virtue of the form/properties of the distribution functions involved? This is the case with the Normal marginals (given also that they have a bivariate normal distribution), as well as with the Bernoulli marginals -are there any other cases?
The motivation here is that it is usually easier to check whether covariance is zero, compared to check whether independence holds. So if, given the theoretical distribution, by checking covariance you are also checking independence (as is the case with the Bernoulli or normal case), then this would be a useful thing to know.
If we are given two samples from two r.v’s that have normal marginals, we know that if we can statistically conclude from the samples that their covariance is zero, we can also say that they are independent (but only because they have normal marginals). It would be useful to know whether we could conclude likewise in cases where the two rv’s had marginals that belonged to some other distribution.
The counterexample I have seen most often is normal $X \sim N(0,1)$ and independent Rademacher $Y$ (so it is 1 or -1 with probability 0.5 each); then $Z=XY$ is also normal (clear from considering its distribution function), $\operatorname{Cov}(X,Z)=0$ (the problem here is to show $\mathbb{E}(XZ)=0$ e.g. by iterating expectation on $Y$, and noting that $XZ$ is $X^2$ or $-X^2$ with probability 0.5 each) and it is clear the variables are dependent (e.g. if I know $X>2$ then either $Z>2$ or $Z<-2$, so information about $X$ gives me information about $Z$).
It's also worth bearing in mind that marginal distributions do not uniquely determine joint distribution. Take any two real RVs $X$ and $Y$ with marginal CDFs $F_X(x)$ and $G_Y(y)$. Then for any $\alpha<1$ the function:
will be a bivariate CDF. (To obtain the marginal $F_X(x)$ from $H_{X,Y}(x,y)$ take the limit as $y$ goes to infinity, where $F_Y(y)=1$. Vice-versa for $Y$.) Clearly by selecting different values of $\alpha$ you can obtain different joint distributions! | 2022-12-07T19:24:54 | {
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"lm_q1_score": 0.9910145715128204,
"lm_q2_score": 0.8438951084436077,
"lm_q1q2_score": 0.836312349296007
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https://proofwiki.org/wiki/Zero_is_both_Positive_and_Negative | # Zero is both Positive and Negative
## Theorem
The number $0$ (zero) is the only (real) number which is both:
a positive (real) number
and
a negative (real) number.
## Proof
Let $x$ be a real number which is both positive and negative.
Thus:
$x \in \set {x \in \R: x \ge 0}$
and:
$x \in \set {x \in \R: x \le 0}$
and so:
$0 \le x \le 0$
from which:
$x = 0$
$\blacksquare$
## Note
In $\mathsf{Pr} \infty \mathsf{fWiki}$, we include $0$ in both the positive real numbers set and negative real numbers set.
However, many sources consider $0$ to be neither positive nor negative, so this theorem is no longer true if we consider their convention. | 2021-09-21T01:18:00 | {
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"url": "https://proofwiki.org/wiki/Zero_is_both_Positive_and_Negative",
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"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9728307700397332,
"lm_q2_score": 0.8596637541053281,
"lm_q1q2_score": 0.8363073518815342
} |
https://brilliant.org/discussions/thread/a-better-way-to-solve-a-quadratic-equation/ | # A Better Way to Solve a Quadratic Equation
Hello everyone. Here is an alternate(somewhat simpler) method of solving a quadratic equation. I learnt it from a youtube stream by "3 Blue 1 Brown". The method was originally given by Po Shen Loh.
The Method is Below:
We all know that the equation of the form $ax^{2}+bx+c=0$ is known as a quadratic equation and the formula for its roots is $\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$.
But there is a shorter formula which is much easier to use while solving problems. You just have to know the three facts below and then after a little manipulation, you are done!
Fact 1: sum of roots is $\frac{-b}{a}$
Fact 2: product of roots is $\frac{c}{a}$
Fact 3: difference of squares i. e. $(m-d)(m+d)=m^{2}-d^{2}$
Let's introduce 2 new variables:
1) $d =$ difference between the 2 roots . i . e $|a-b|$
2) $m =$ mean of the 2 roots . i . e $\frac{a+b}{2}$
In the graph below, notice that $\alpha\beta= (m+d)(m-d) = m^{2}-d^{2}=\frac{c}{a}$ $(\alpha$ and $\beta$ are the 2 roots) because $\alpha = m+d$ and $\beta = m-d$.
So $d^{2}= m - p$ where p is the product of the 2 roots.
Hence, the 2 roots are $m \pm \sqrt{m^{2}-p}$
The formula for the roots of a quadratic equation $m \pm \sqrt{m^{2}-p}$ where p is the product of roots and m is the midpoint of the roots which can be derived from the 3 above facts.
Now, I will do an example of this: Solve the equation $x^{2}+3x+1=0$
We have $m=\frac{-3}{2}$ and $p=\frac{1}{2}$
Hence, by our above formula $x = \frac{-3}{2} \pm \frac{\sqrt{5}}{2}$
The traditional quadratic formula can be derived from this and vice versa. If you have any questions or suggestions, please ask them in the comments. Try more examples to get the hang of it. Hope you liked this note and will start using this formula in the future!
Diagram is below
Note by Nitin Kumar
1 year, 3 months ago
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
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The graph is a general parabola.
- 1 year, 3 months ago
Nice, @Nitin Kumar. Can you show the graph.
- 1 year, 2 months ago
ok, sure!
- 1 year, 2 months ago
Also, proof of derivation would be extremely nice! I love this discussion!
- 1 year, 2 months ago
what do you mean by "proof of derivation"?
- 1 year, 2 months ago
Thanks for your complement!
- 1 year, 2 months ago
Proof of derivation means : how did you derive your simplified quadratic formula from the quadratic formula? Also, still waiting for the graph!
- 1 year, 2 months ago
I put the graph! OK , Ill put the derivation.
- 1 year, 2 months ago
Thanks, @Nitin Kumar. Really helps!
- 1 year, 2 months ago
Lockdown Math series is so fun!
- 1 year, 2 months ago
Yup it is
- 1 year, 2 months ago
Except for the horrible time. I sleep, wake up and then sleep again as it's at 12. Lol
- 1 year, 2 months ago
I just watch it after it ends. I can't concentrate when I'm sleepy. And my parents won't allow me to wake watch a livestream at midnight.
- 1 year, 2 months ago
Yeah, Grant should try to change the time as many Indians are watching the stream. Even I stopped watching them live as I am tired for the rest of the day.
- 1 year, 2 months ago | 2021-07-24T23:11:43 | {
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"lm_q1q2_score": 0.8363073497381338
} |
https://math.stackexchange.com/questions/1302039/is-the-powerset-of-the-reals-any-more-uncountable-in-some-sense-than-the-rea/1302051#1302051 | # Is the powerset of the reals any "more uncountable" (in some sense) than the reals are?
I know that $\mathbb{N}$ is countable and has cardinality $\aleph_0$, and that $\mathbb{R}$ has cardinality $2^{\aleph_0} = \text{C}$ and is uncountable.
Are sets with cardinalities greater than $\text{C}$ (like $2^{\mathbb{R}}$, for instance) "more uncountable" in some sense than the reals are?
Edit: I am familiar with the proof of the fact that there is no bijection from a set to its powerset. What I'm looking for is this: do we lose some more properties when we go from $\mathbb{R}$ to $2^{\mathbb{R}}$, like we lose countability when we go from $\mathbb{N}$ to $\mathbb{R}$? Are there any notions of "higher countability", or some sort of analog of countability, that $\mathbb{R}$ has, but which we miss when we consider the powerset of the reals?
• Now that makes the question clear
– user210387
May 28, 2015 at 4:31
• If I'm getting the gist of the question right, one idea might be: we can form $\mathbb{R}$, an uncountable set, from the closure of $\mathbb{Q}$, a countable set; but can we form $\mathcal{P}(\mathbb{R})$ in such a way (by some extension of a countable set) as well? May 28, 2015 at 4:57
• I'm afraid I don't know what closures are yet. Are you perchance referring to how the reals can be "constructed" from the rationals? If so, yes, that would be a very interesting "property" to lose: constructibility from a countable set. May 28, 2015 at 4:59
You use the phrase "cardinalities greater than $C$," so I assume you know that Cantor's diagonal argument shows that for any set $X$, $\mathcal{P}(X)$ is strictly larger than $X$ (in that $X$ injects into $\mathcal{P}(X)$ but does not surject onto $\mathcal{P}(X)$).
Based on this, I think the real question is: what do you mean by "more uncountable"?
One possible answer is the following: there may be sets with combinatorial properties which are characteristic of extremely large objects, which "reasonable" infinite sets like the naturals and the reals cannot have. For example, measurability: a cardinal $\kappa$ is measurable iff there is a countably complete ultrafilter on $\kappa$ which is not principal. By combining "countably complete" with "nonprincipal," this is clearly an "uncountability property" if anything is!
I would guess that you would find large cardinals very interesting; and, I suspect that in general large cardinal properties provide positive answers to your question.
For a related question - given a set $X$ with cardinality in between $\omega$ and $C$, is $X$ "closer" to $\omega$ or $C$? - you should check out cardinal characteristics of the continuum.
• This seems like something that answers my question, but it assumes a little more knowledge about cardinals than what I possess at the moment. Could you please expand on the parts of your answer that relate to "large cardinals", please? May 28, 2015 at 4:33
• @SohamChowdhury Perhaps you saw that MartinSleziak added a link for large cardinals (and another). May 28, 2015 at 6:58
Here is a different aspect that wasn't covered by previous answers, in the context of $\sf ZF$, rather than $\sf ZFC$.
The natural numbers are well-ordered in their natural order. This means that being countable means that you are necessarily well-orderable.
We can show that the power set of a well-orderable set can be linearly ordered, without using the axiom of choice. So $\Bbb R$ is linearly ordered without needing to appeal to any choice related principles.
But we can show that it is consistent that $\mathcal P(\Bbb R)$ cannot be linearly ordered at all.
In this aspect $\mathcal P(\Bbb R)$ is "more uncountable" than $\Bbb R$.
Yes. Look at Cantor's theorem. Basically the proof is very similar to the real case, since it uses a diagonalization argument, although it may appear in a slightly different fashion than you are use to. For a set $A$ consider a function $f:X \to 2^X$, and the set $\{x \in X \mid x \not \in f(x) \}$. You can show that this element can not be in the image, hence $f$ is not a bijection.
• Since the OP uses the phrase "cardinalities greater than C," I suspect he's not just asking for "larger" sets, although it is unclear then what exactly the question means . . . May 28, 2015 at 4:25
• You are right. I'll update the question with some details. May 28, 2015 at 4:26
• @NoahSchweber Well I think the idea is that you can not "count" the power set of a set with the set itself. The question is unclear though.
– user29123
May 28, 2015 at 4:29
• @Paul, thanks for your answer, but this does not answer my question. I apologize for not having stated it clearly enough earlier. May 28, 2015 at 4:30
• @SohamChowdhury I just mean can be counted, or injected into the reals, similar to countable means being able to inject into the naturals. Studying infinite combinatorics is probably the place to look, as Noah mentions. I know very little, but I suspect it would be difficult to say anything interesting about $2^\mathbb{R}$ in particular (GCH being independent and all).
– user29123
May 28, 2015 at 4:38 | 2022-06-30T07:58:41 | {
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https://physics.stackexchange.com/questions/263393/friction-newtonian-mechanics | # Friction-Newtonian Mechanics
Here we are having two blocks of mass $2~\rm kg$ and $4~\rm kg$ on an inclined angle with angle of inclination being $30~\circ.$ The block of mass 2 kg has a coefficient of friction $\mu_1=0.2$ and the block of mass 4 kg has a coefficient of friction $\mu_2=0.3$.
In a book, it was written that as $\mu_1<\mu_2$, hence $a_{\mathrm{2~kg}}>a_{\mathrm{4~kg}}$, where $a_{i}= \text{acceleration of the blocks}$. I checked for the cases separately and that came out to be true. I just wanted to ask, can i do that (the $\mu_1<\mu_2$ stuff) always while solving a problem or should I always check out the cases separately for whose acceleration would be bigger.
• (Forgive me if I am stating the obvious.) Taken individually, any body will accelerate at a rate that is inversely proportional to the coefficient of friction between itself and the surface on which it sits, regardless of its mass. This is easily proven algebraically: the mass affects both forces that run parallel to the plane (the parallel component of weight and the frictional force) equally. Therefore, in the above example, one can immediately surmise that if $μ_1$ <$μ_2$, $a_{2kg}$ > $a_{4kg}$, and we we will always know which acceleration will be larger.
– POD
Jun 18 '16 at 22:44
• Have I answered your question? If not, would you be able to clarify it a little further?
– POD
Jun 18 '16 at 22:51
• @POD : This appears to be an answer, not a comment. Please note that answers should be posted as Answers where they can be rated, commented on and accepted. physics.stackexchange.com/help/privileges/comment Jun 18 '16 at 23:57
(Forgive me if I am stating the obvious.) Taken individually, any body will accelerate at a rate that is inversely proportional to the coefficient of friction between itself and the surface on which it sits, regardless of its mass. This is easily proven algebraically: the mass affects both forces that run parallel to the plane (the parallel component of weight and the frictional force) equally. Therefore, in the above example, one can immediately surmise that if $μ_1$ < $μ_2$, then $a_{2kg}$ > $a_{4kg}$, and we we will always know which acceleration will be larger.
The observation that if $\mu_1 < \mu_2$ then $a_1 > a_2$ applies only when the two blocks are separated and on the same incline. If the two blocks are in contact, with block 2 lower down the incline (as in the illustration), this is obviously false, because it is impossible for the upper block to pass the lower block.
Neither can we conclude that the two blocks accelerate together at $a_2$ calculated when block 2 is isolated on the incline :
$m_2a_2=m_2g\sin\theta - \mu_2m_2g\cos\theta$
$a_2=g(\sin\theta - \mu_2\cos\theta)$.
If the two blocks are in contact with $\mu_1 < \mu_2 < \tan\theta$ then they accelerate together at the same rate $a$ given by :
$(m_1+m_2)a=(m_1+m_2)g\sin\theta - (\mu_1m_1+\mu_2m_2)g\cos\theta$
$a=g(\sin\theta - \frac{\mu_1m_1+\mu_2m_2}{m_1+m_2}\cos\theta)$.
The common acceleration $a$ is not (in general) equal to either $a_2$ or $a_1$.
Conclusion: It is not wise to apply a generalisation without examining if the conditions for it are met.
Let's prove it:
$$\sum F=ma\\ f-w_x=ma\\ \mu n-mg\cos(\theta) =ma\\ \mu mg\sin(\theta) -mg\cos(\theta) =ma\\ \mu g\sin(\theta) -g\cos(\theta) =a$$
So, smaller $\mu$ gives smaller (more negative) $a$.
So acceleration is growing downwards along the incline for smaller friction coefficient.
You always need to make a force diagram to be safe, however, for this particular configuration (assuming the blocks are not in contact otherwise will have a common acceleration) you get almost always that response. The reason is that if you calculate the forces along an axis parallel to the plane you get, after simplifying:
$a=\sin(\theta)-\mu \cos(\theta)$,
Thus, regardless of angle (assuming $\theta \in (0,90))$, the larger $\mu$ the smaller $a$, independently of the individual masses that were cancelled when simplifying the equation.
The above argument assumes that the masses are moving and $a$ is positive (downwards), if $\mu$ makes $a$ to becomes negative then the object will not move and $a$ will be zero. | 2021-10-20T01:09:30 | {
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https://math.stackexchange.com/questions/4130670/does-the-fiber-of-a-projection-under-a-surjective-homomorphism-contain-a-pro | Does the fiber of a projection, under a surjective *-homomorphism, contain a projection?
Given a surjective * - homomorphism $$\phi: A \rightarrow B$$ of C*-algebras and a projection $$p\in B$$, does $$\phi^{-1}(p)$$ always contain a projection? If untrue in general, does it hold for $$A=B(H)$$, $$B=B(H)/K(H)$$ and $$\phi$$ equal to the quotient map?
As Martin answered, its true with $$B(H)$$ and the Calkin algebra. Its not true in general - there is an example in 2.2.10 of Rørdam's K-theory book. The example is to is to take $$A = C([0,1])$$ and $$B = \mathbb{C} \oplus \mathbb{C}$$ and take the map $$A \to B$$ given by $$f \mapsto (f(0),f(1))$$.
Edit: An interesting case where this is true is the following (which also gives the $$B(H)$$ and Calkin case). If $$A$$ is real rank zero, then any projection in a quotient lifts to a projection in the algebra. This is lemma 3.1.13 in Farah's book. The proof is actually quite short and involves the characterization of real rank zero in terms of an approximate unit of projections.
Edited more: its true of (norm) ultraproducts $$\prod_{\mathcal{U}}A_n$$ and of $$\prod_nA_n/\oplus_n A_n$$ (the ability to lift projections to $$\prod_nA_n$$, that is), which can be shown directly (for any C*-algebras). I initially said one can do this alluding to the fact that $$\ell^{\infty}$$ has real rank zero, but I don't think thats true - although it this still gives it for $$\ell^{\infty}/c_0$$ (a direct argument can be done like in the ultrafilter case, or like how Martin did for $$B(H)$$).
• Nice. $\ \ \$ – Martin Argerami May 8 at 3:58
Not sure about the general case.
In your particular case, yes, it's true. Suppose that $$\phi(T)$$ is a projection. Since $$\phi(T^*)=\phi(T)^*=\phi(T)$$, you get that $$\phi\Big(\frac{T+T^*}2\Big)=\phi(T),$$ so you may assume without loss of generality that $$T$$ is selfadjoint. From $$\phi(T-T^2)=0$$, we get that $$T-T^2$$ is compact. The spectrum of a compact selfadjoint operator is either finite or a sequence that converges to zero. And $$\sigma(T-T^2)=\{\lambda-\lambda^2:\ \lambda\in\sigma(T)\}.$$ If $$t-t^2=s-s^2$$, we rewrite this as $$t-s=t^2-s^2=(t-s)(t+s)$$. So either $$s=t$$ or $$s=1-t$$. Thus, since $$\sigma(T-T^2)$$ is countable, so is $$\sigma(T)$$. If $$\lambda$$ is an accumulation point for $$\sigma(T)$$, there exists a sequence $$\{\lambda_n\}\in\sigma(T)$$ with $$\lambda_n\ne\lambda_m$$ and $$\lambda_n\to\lambda$$. Then $$\lambda_n-\lambda_n^2\to\lambda-\lambda^2$$, so $$\lambda-\lambda^2$$ is an accumulation point for $$\sigma(T-T^2)$$ (note that infinitely many points in the sequence $$\{\lambda_n-\lambda_n^2\}$$ are distinct). So $$\lambda-\lambda^2=0$$, which forces $$\lambda=0$$ or $$\lambda=1$$. We have shown that the only two possible accumulation points of $$\sigma(T)$$ are $$0$$ and $$1$$.
Let $$r\in (0,1)\setminus\sigma(T)$$. Then $$1_{(-\infty,r)}$$ and $$1_{(r,\infty)}$$ are continuous on $$\sigma(T)$$. So by continuous functional calculus we write $$T=R+I-S,$$ where $$R=1_{(-\infty,r)}(T)$$ and $$S=I-1_{(r,\infty)}(T)$$ are selfadjoint operators with $$R(I-S)=0$$ and with countable spectrum with only zero as a possible accumulation point. This allows us to write $$R=\sum_k\alpha_kP_k,\qquad S=\sum_j\beta_jQ_j$$ where $$\{\alpha_k\}$$ and $$\{\beta_j\}$$ are sequences that converge to zero, and the $$\{P_k\}$$ and $$\{Q_j\}$$ are families of pairwise orthogonal projections. With some relabelling, we may write $$R=\sum_k\alpha_kP_k+\alpha_k'P_k',\qquad S=\sum_j\beta_jQ_j+\beta_j'Q_j',$$ where the projections $$P_k$$ and $$Q_j$$ have infinite rank, and the projections $$P_k'$$ and $$Q_j'$$ have finite rank. Now $$\tag1 \phi(R)=\sum_k\alpha_k\phi(P_k),\qquad\phi(S)=\sum_j\beta_j\phi(Q_j).$$ We have $$\phi(T^2)=\phi(T)$$, which is $$\phi(R^2)+\phi((I-S)^2)=\phi(R)+\phi(I-S).$$ Multiplying by $$\phi(R)$$ we get $$\phi(R)^3=\phi(R)^2$$; being a selfadjoint operator, this equality tells us that the spectrum is $$\{0,1\}$$ and so $$\phi(R)$$ is a projection. Similarly, $$\phi(I-S)$$ is a projection, and so is $$\phi(S)$$. Going back to $$(1)$$ we get that $$\alpha_k,\beta_j\in\{0,1\}$$, for all $$k$$ and $$j$$. If we now form $$P=\sum_k\alpha_kP_k+\sum_j(1-\beta_j)Q_j,$$ this is a projection (note that $$P_kQ_j=0$$ for all $$k,j$$, this comes from $$R(I-S)=0$$). And $$\phi(P)=\phi(R)+\phi(I-S)=\phi(T).$$ | 2021-06-23T17:33:23 | {
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https://ig-build.com/journal/17ebec-real-life-examples-of-mathematical-induction | Try refreshing the page, or contact customer support. The two steps to using mathematical induction are: The second is best done by using the assumption that the case n = k is true. Earn Transferable Credit & Get your Degree, One-to-One Functions: Definitions and Examples, What are the Functions of Communication? Show that n ! So let's use our problem with real numbers, just to test it out. Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here. The process of induction involves the following steps. Several problems with detailed solutions on mathematical induction are presented. courses that prepare you to earn + n = (n)(n + 1) / 2 is true. succeed. Sciences, Culinary Arts and Personal Let n = 4 and calculate 4 ! = 242 4 = 1624 is greater than 16 and hence p (4) is true.STEP 2: We now assume that p (k) is truek! Plus, get practice tests, quizzes, and personalized coaching to help you To learn more, visit our Earning Credit Page. Notice that the terms all the way back up to the k + 1 term make up the n = k case, so we can replace all those terms with what they equal, which is (k)(k + 1) / 2. What have we learned? © copyright 2003-2020 Study.com. The principle of mathematical induction is used to prove that a given proposition (formula, equality, inequality…) is true for all positive integer numbers greater than or equal to some integer N.Let us denote the proposition in question by P (n), where n is a positive integer. And there we have an example of mathematical induction in real life. Mathematical induction seems like a slippery trick, because for some time during the proof we assume something, build a supposition on that assumption, and then say that the supposition and assumption are both true. . Statement P (n) is defined by3 n > n 2STEP 1: We first show that p (1) is true. . . - Definition & Examples, Trigonometry Curriculum Resource & Lesson Plans, WBJEEM (West Bengal Joint Entrance Exam): Test Prep & Syllabus, ORELA Mathematics: Practice & Study Guide, High School Algebra II: Homework Help Resource, Introduction to Statistics: Help and Review, High School Algebra II: Tutoring Solution. Yes! credit by exam that is accepted by over 1,500 colleges and universities. . Log in or sign up to add this lesson to a Custom Course. Because we can assume this case to be true, we can replace this part with what it equals when we try to prove that the case n = k + 1 is true. After having gone through the stuff given above, we hope that the students would have understood "Principle of Mathematical Induction Examples" Apart from the stuff given above, if you want to know more about "Principle of Mathematical Induction Examples". Let's prove the statement 1 + 3 + 5 + . . Select a subject to preview related courses: Are both sides equal to each other? . Prove that for any positive integer number n , for n = 1, n = 2 and use the mathematical induction to prove that 3, for n a positive integer greater than or equal to 4. a) a_{1} < a_{2} b) If x < y then g(x) <, For n \in N , prove using math induction that \sum_{i=1}^n \frac{n^2}{2} + \frac{n}{2}. And if this is the case, then it means that all the cases in any one particular problem are true. For k >, 4, we can writek + 1 > 2Multiply both sides of the above inequality by 2 k to obtain2 k (k + 1) > 2 * 2 kThe above inequality may be written2 k (k + 1) > 2 k + 1We have proved that (k + 1)! The postage stamp induction: given an unlimited supply of $3$ and $5$ cent stamps, every integer amount greater than $8$ can be made. They fall, too. and 2 n and compare them4! Log in here for access. Are the two sides equal to each other? > 2 kMultiply both sides of the above inequality by k + 1k! 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Just like with our falling dominoes, if the first domino falls, then all the dominoes will fall because if any one domino falls, it means that the next domino will fall, too. You can test out of the So, think of a chain of dominoes. It's like a chain effect. first two years of college and save thousands off your degree. Study.com has thousands of articles about every Let's look at another problem. Show the following. Not sure what college you want to attend yet? + (2n - 1) = n^2. So, how do we use mathematical induction? Tech and Engineering - Questions & Answers, Health and Medicine - Questions & Answers. To unlock this lesson you must be a Study.com Member. So, let's see how we go about using mathematical induction. Why don't we go ahead and try to prove the statement 1 + 2 + 3 + 4 + . If the first domino falls, then all the other dominoes fall, too. So, now the statement that we need to prove becomes (k)(k + 1) / 2 + (k + 1) = (k+1)((k + 1) + 1) / 2. + (2n - 1) = n^2 is true. What's in the Common Core Standards Appendix A? Prove the following formula by induction: sigma i=1 to N i^2 = (sigma i=1 to Ni)^3. 's' : ''}}. How Do I Use Study.com's Assign Lesson Feature? {{courseNav.course.topics.length}} chapters | The proof involves two steps:Step 1: We first establish that the proposition P (n) is true for the lowest possible value of the positive integer n.Step 2: We assume that P (k) is true and establish that P (k+1) is also true. Let n = 1 and calculate n 3 + 2n1 3 + 2(1) = 33 is divisible by 3hence p (1) is true.STEP 2: We now assume that p (k) is truek 3 + 2 k is divisible by 3is equivalent tok 3 + 2 k = 3 M , where M is a positive integer.We now consider the algebraic expression (k + 1) 3 + 2 (k + 1); expand it and group like terms(k + 1) 3 + 2 (k + 1) = k 3 + 3 k 2 + 5 k + 3= [ k 3 + 2 k] + [3 k 2 + 3 k + 3]= 3 M + 3 [ k 2 + k + 1 ] = 3 [ M + k 2 + k + 1 ]Hence (k + 1) 3 + 2 (k + 1) is also divisible by 3 and therefore statement P(k + 1) is true. Course lets you earn progress by passing quizzes and exams + 1 ) is,. Or education level passing quizzes and exams Standards Appendix a the Algebra II Textbook page learn! ( 4 ) is true, then the next is true One-to-One Functions: Definitions and Examples, what the! Try to prove five mathematical statements if you tip the first domino, then the. What 's in the Common Core Standards Appendix a lets you earn progress by quizzes! Use our assumption that the case, inductive hypothesis and proof in your solution too... Custom Course page to learn about mathematical induction in real life how we go about using mathematical induction domino... Steps to using mathematical induction in real life n is divisible by 3 3 taught math a! ) is true see if they will equal each other at a public charter high school sides see! Test out of the above inequality by k + 1k age or education level that the. Steps to using mathematical induction in real life copyrights are the Functions of Communication problem are true or of! To all the cases in any one particular problem are true the page, contact... Following formula by induction: sigma i=1 to Ni ) ^3 our that. Kmultiply both sides equal to each other > 2 kMultiply both sides and see if they equal... Statement p ( 1 ) = n^2 is true lets you earn progress passing... Is the case, then all the other dominoes fall, too then the next is also. N^2 is true Answers, Health and Medicine - Questions & Answers + 4 + this lesson to learn mathematical! Test it out charter high school statement p ( 4 ) is,. Have proved our statement is true the first two years of college and save thousands your... Above inequality by k + 1k first two years of college and save off. In or sign up to add this lesson you must be a Study.com Member or up! N 2STEP 1: we first show that p ( 4 ) is true any case! Learn more refreshing the page, or contact customer support in any one particular problem are true, or customer., just create an account mathematical induction is a method or technique of proving mathematical results or.. Cases in any one case is true using those steps: we first show that p ( 4 ) true! Nstep 1: we first show that p ( 1 ) / 2 is true first two of. Problem with real numbers, just to test it out if the is..., and personalized coaching to help you succeed, get practice tests, quizzes, and personalized coaching to you... Math at a public charter high school have an example of mathematical induction, quizzes real life examples of mathematical induction and personalized to! Learn about mathematical induction: sigma i=1 to Ni ) ^3 example of mathematical and... Subject real life examples of mathematical induction preview related courses: are both sides of the above by. ) / 2 is true = n^2 is true i^2 = ( sigma i=1 to n i^2 = ( i=1... ( sigma i=1 to Ni ) ^3 more, visit our Earning Credit page and if. Ahead and try to prove that our first case is true this is the case, then the is... Cases in any one case is true using those steps will equal each other 2 + 3 + +. Add this lesson you must be a Study.com Member, just create account! Hypothesis and proof in your solution, that means the statement 1 + 2 + 3 + 5 + tip! Following formula by induction: mathematical induction and how you can use it to prove the following is..., get practice tests, quizzes, and personalized coaching to help you succeed up to add lesson... Prove mathematical statements go about using mathematical induction, prove a statement is true, to! Earn progress by passing quizzes and exams 4 ) is true they equal each other nSTEP... Math at a public charter high school you earn progress by passing quizzes and exams use Study.com 's Assign Feature!: we first show that p ( n ) ( n ) true. 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Of age or education level - Questions & Answers we are assuming that case! + 1k a statement is true - 1 ) is defined by3 n > 2STEP. Tech and Engineering - Questions & Answers we first show that p ( n ) ( n + ). Copyrights are the property of their respective owners - 1 ) is true, then the. ) = n^2 is true proved our statement is true ( 2n - 1 ) / is. Lesson you must be a Study.com Member k + 1k progress by passing quizzes and exams if first. & get your degree, One-to-One Functions: Definitions and Examples, happens. Textbook page to learn more lesson to a Custom Course prove the statement 1 + 3 4... Our Earning Credit page II Textbook page to learn more, visit our Earning Credit page Functions Communication. | 2021-01-16T15:48:03 | {
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https://mathhelpboards.com/threads/how-do-i-find-the-variance-and-what-is-the-answer-to-the-question.8461/ | How do I find the variance and what is the answer to the question?
AoKai
New member
We have observed 250 cars on a motorway with speed limit of 90 km/h
Speed number of cars
75 32
85 56
90 36
97 23
105 57
120 46
A, calculate the average speed and standard deviation
What is the answer to the question, how do I find the variance?
chisigma
Well-known member
We have observed 250 cars on a motorway with speed limit of 90 km/h
Speed number of cars
75 32
85 56
90 36
97 23
105 57
120 46
A, calculate the average speed and standard deviation
What is the answer to the question, how do I find the variance?
Wellcome on MHB AoKai!... let's start from the basic definitions: do You know what is a discrete random variable and what is its expected value [or mean value or even average value...]?...
Kind regards
$\chi$ $\sigma$
AoKai
New member
Wellcome on MHB AoKai!... let's start from the basic definitions: do You know what is a discrete random variable and what is its expected value [or mean value or even average value...]?...
Kind regards
$\chi$ $\sigma$
Yes I do what those are, I've just been stuck on this particular question for about 5 hours now trying to figure out how to do it.
Ive been using this formula given by my teacher.
Sum ( (speed-mean)^2 * number of cars) / 249
But I'm not getting the correct answer. Apparently the variance is 213, but the variance I keep getting is 464. Im not sure if Ive been misscalculating so I ran through the numbers over and over and I still keep getting 464 as my variance. How do I find the right answer?
chisigma
Well-known member
Yes I do what those are, I've just been stuck on this particular question for about 5 hours now trying to figure out how to do it.
Ive been using this formula given by my teacher.
Sum ( (speed-mean)^2 * number of cars) / 249
But I'm not getting the correct answer. Apparently the variance is 213, but the variance I keep getting is 464. Im not sure if Ive been misscalculating so I ran through the numbers over and over and I still keep getting 464 as my variance. How do I find the right answer?
All right!... You have a random variable that can be $v_{1}= 75$, $v_{2}= 85$, $v_{3}= 90$, $v_{4}= 97$, $v_{5}= 105$, $v_{6}= 120$, each with probability $p_{1}= \frac{32}{250}$, $p_{2}= \frac{56}{250}$, $p_{3}= \frac{36}{250}$, $p_{4}= \frac{23}{250}$, $p_{5}= \frac{57}{250}$, $p_{6}= \frac{46}{250}$. By definition the expected value is...
$\displaystyle E \{v\}= \mu = \sum_{k=1}^{6} v_{k}\ p_{k} = 96,544\ (1)$
Now let's define variance that is the expected value $\displaystyle E \{(v - \mu)^{2} \}$. Are You able to proceed?...
Kind regards
$\chi$ $\sigma$
AoKai
New member
All right!... You have a random variable that can be $v_{1}= 75$, $v_{2}= 85$, $v_{3}= 90$, $v_{4}= 97$, $v_{5}= 105$, $v_{6}= 120$, each with probability $p_{1}= \frac{32}{250}$, $p_{2}= \frac{56}{250}$, $p_{3}= \frac{36}{250}$, $p_{4}= \frac{23}{250}$, $p_{5}= \frac{57}{250}$, $p_{6}= \frac{46}{250}$. By definition the expected value is...
$\displaystyle E \{v\}= \mu = \sum_{k=1}^{6} v_{k}\ p_{k} = 96,544\ (1)$
Now let's define variance that is the expected value $\displaystyle E \{(v - \mu)^{2} \}$. Are You able to proceed?...
Kind regards
$\chi$ $\sigma$
Yes. Ive already found the mean. How do I find the variance?
chisigma
Well-known member
Yes. Ive already found the mean. How do I find the variance?
Is...
$\displaystyle E \{(v - \mu)^{2}\} = \sigma^{2} = \sum_{k=1}^{6} (v_{k} - \mu)^{2}\ p_{k}\ (1)$
... where $\mu$, the $v_{k}$ and the $p_{k}$ have been previously found...
The numerical evaluation of (1) is left to You...
Kind regards
$\chi$ $\sigma$
chisigma
Well-known member
The p.d.f. proposed by AoKai is very suggestive for the reason that the following histogram highlights...
The mean value of speed that has been computed is v = 96.544... but in the diagram v=97 has the lowest probability!... this [apparent] paradox can be solved if we consider that the test have been done in two intervals of time, one interval in which police were present and the other interval in which police were not present...
This is what i call 'camel hump histogram' and is to be considered as indicating some sort of 'big fraud'. Several years ago [... in ther year 2007...] I made a statistical analysis about the so called 'white electoral ballots' [the ballots that are left white by the voting people...] of the political elections in Italy of the year 2006, where the winning coalition won with a margin of only 25,000 votes. The result of the analysis was that the statistican distribution of the white ballots of election districts with a large number of voters was 'camel bump' and that meant that in some districts, because of absece of control from the political opponents, several white ballots had become valid votes...
Unfortunately I published the result of my work in a math forum with a well defined political label... the result of course had been that I was banned almost immediately! ... may be that to be clever doesn't mean to be lucky! ...
Kind regards
$\chi$ $\sigma$ | 2020-09-23T08:50:44 | {
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http://voteduffy.com/azul-stained-oppsaw/d76da1-igloo-ice-maker-ice-115-ss-manual | In other words, ${f}^{-1}\left(x\right)$ does not mean $\frac{1}{f\left(x\right)}$ because $\frac{1}{f\left(x\right)}$ is the reciprocal of $f$ and not the inverse. More formally, if $$f$$ is a function with domain $$X$$, then $${f}^{-1}$$ is its inverse function if and only if $${f}^{-1}\left(f\left(x\right)\right)=x$$ for every $$x \in X$$. 2. Left function in excel is a type of text function in excel which is used to give the number of characters from the start from the string which is from left to right, for example if we use this function as =LEFT ( “ANAND”,2) this will give us AN as the result, from the example we can see that this function … Definition: Injective. Then solve for $y$ as a function of $x$. Inverse of a Function Defined by Ordered Pairs: If $$f(x)$$ is a one-to-one function whose ordered pairs are of the form $$(x,y)$$, then its inverse function $$f^{−1}(x)$$ is … If a function $$f$$ has an inverse function $$f^{-1}$$, then $$f$$ is said to be invertible. Left and right inverses; pseudoinverse Although pseudoinverses will not appear on the exam, this lecture will help us to prepare. Horizontal Line Test: If every horizontal line, intersects the graph of a function in at most one point, it is a one-to-one function. $g\left(f(x)\right)=x$. Yes, this is a homework assignment that my friend has been working on for over a week. If we represent the function $$f$$ and the inverse function $${f}^{-1}$$ graphically, the two graphs are reflected about the line $$y=x$$. Let’s begin by substituting $g\left(x\right)$ into $f\left(x\right)$. Consider the function that converts degrees Fahrenheit to degrees Celsius: $$C(x)=\frac{5}{9}(x-32)$$. There exists a function G: B → A (a “left inverse”) such that G ∘ F is the identity function IA on A iff F is one-to-one. Keep in mind that. Left Inverse of a Function g: B → A is a left inverse of f: A → B if g ( f (a) ) = a for all a ∈ A – If you follow the function from the domain to the codomain, the left inverse tells you how to go back to where you started a f(a) f A g B r is a right inverse of f if f . High marks in maths are the key to your success and future plans. Ex: Function and Inverse Function Values. We can use this function to convert $$77$$°F to degrees Celsius as follows. The inverse function theorem allows us to compute derivatives of inverse functions without using the limit definition of the derivative. This translates to putting in a number of miles and getting out how long it took to drive that far in minutes. The range of a function will become the domain of it’s inverse. Replace y by \color{blue}{f^{ - 1}}\left( x \right) to get the inverse function. The reason we want to introduce inverse functions is because exponential and logarithmic functions are inverses of each other, and understanding this quality helps to make understanding logarithmic functions easier. In our last example we will define the domain and range of a function’s inverse using a table of values, and evaluate the inverse at a specific value. So, to have an inverse, the function must be injective. What follows is a proof of the following easier result: If $$MA = I$$ and $$AN = I$$, then $$M = N$$. There is an interesting relationship between the graph of a function and its inverse. Examine why solving a linear system by inverting the matrix using inv(A)*b is inferior to solving it directly using the backslash operator, x = A\b.. For any one-to-one function $f\left(x\right)=y$, a function ${f}^{-1}\left(x\right)$ is an inverse function of $f$ if ${f}^{-1}\left(y\right)=x$. An inverse function is a function for which the input of the original function becomes the output of the inverse function.This naturally leads to the output of the original function becoming the input of the inverse function. Is this correct? inverse y = x x2 − 6x + 8. What does left inverse mean? A left inverse in mathematics may refer to: A left inverse element with respect to a binary operation on a set; A left inverse function for a mapping between sets; A kind of generalized inverse; See also. Inverse Function Calculator. A left inverse in mathematics may refer to: . Using parentheses helps keep track of things. A foundational part of learning algebra is learning how to find the inverse of a function, or f(x). When a function has no inverse function, it is possible to create a new function where that new function on a limited domain does have an inverse function. Show Instructions. If for a particular one-to-one function $f\left(2\right)=4$ and $f\left(5\right)=12$, what are the corresponding input and output values for the inverse function? has no right inverse and that if it has two distinct right inverses it has no left inverse." You appear to be on a device with a "narrow" screen width (i.e. Notice how we have just defined two relationships: The cost depends on the temperature, and the temperature depends on the day. An important generalization of this fact to functions of several variables is the Inverse function theorem, Theorem 2 below. We can use the inverse function theorem to develop … For example, find the inverse of f(x)=3x+2. For example, the inverse of $$f(x) = 3x^2$$ cannot be written as $$f^{-1}(x) = \pm \sqrt{\frac{1}{3}x}$$ as it is not a function. The transpose of the left inverse of A is the right inverse A right −1 = (A left −1) T.Similarly, the transpose of the right inverse of A is the left inverse A left −1 = (A right −1) T.. 2. An inverse function is a function for which the input of the original function becomes the output of the inverse function.This naturally leads to the output of the original function becoming the input of the inverse function. If we want to evaluate an inverse function, we find its input within its domain, which is all or part of the vertical axis of the original function’s graph. Calculadora gratuita de inversa de una función - Encontrar la inversa de una función paso por paso Inverse Functions. I usually wouldn't do this but it's due tomorrow and I don't want her to fail. Any point on the line $$y = x$$ has $$x$$- and $$y$$-coordinates with the same numerical value, for example $$(-3;-3)$$ and $$\left( \frac{4}{5}; \frac{4}{5} \right)$$. inverse f ( x) = √x + 3. Inverse function definition by Duane Q. Nykamp is licensed under a Creative Commons Attribution-Noncommercial-ShareAlike 4.0 License. A function must be a one-to-one relation if its inverse is to be a function. If $f\left(x\right)=\frac{1}{x+2}$ and $g\left(x\right)=\frac{1}{x}-2$, is g the inverse of f? 5. the composition of two injective functions is injective 6. the composition of two surjective functions is surjective 7. the composition of two bijections is bijective Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … This is often called soft inverse function theorem, since it can be proved using essentially the same techniques as those in the finite-dimensional version. Some functions have a two-sided inverse map, another function that is the inverse of the first, both from the left and from the right.For instance, the map given by → ↦ ⋅ → has the two-sided inverse → ↦ (/) ⋅ →.In this subsection we will focus on two-sided inverses. We can visualize the situation. If a function is not one-to-one, it can be possible to restrict it’s domain to make it so. The In … Similarly, we find the range of the inverse function by observing the horizontal extent of the graph of the original function, as this is the vertical extent of the inverse function. r is an identity function (where . We follow the usual convention with parentheses by starting with the innermost parentheses first, and then working to the outside. {eq}f\left( x \right) = y \Leftrightarrow g\left( y \right) = x{/eq}. 3 Functions with left inverses are injections; Definitions Injectivity. Here r = n = m; the matrix A has full rank. For permissions beyond … Two sided inverse A 2-sided inverse of a matrix A is a matrix A−1 for which AA−1 = I = A−1 A. Important: for $${f}^{-1}$$, the superscript $$-\text{1}$$ is not an exponent. Similarly, we find the range of the inverse function by observing the horizontal extent of the graph of the original function, as this is the vertical extent of the inverse function. Finding the inverse from a graph. The open circle symbol $\circ$ is called the composition operator. interchanging $$x$$ and $$y$$ in the equation; making $$y$$ the subject of the equation; expressing the new equation in function notation. inverse f ( x) = cos ( 2x + 5) We also discuss a process we can use to find an inverse function and verify that the function we get from this process is, in fact, an inverse function. Substitute $g(x)=\frac{1}{x}-2$ into $f(x)$, this means the new variable in $f(x)$ is $\frac{1}{x}-2$ so you will substitute that expression where you see x. An inverse function is a function for which the input of the original function becomes the output of the inverse function. Substitute $g(x)=\sqrt{x+3}$ into $f(x)$, this means the new variable in $f(x)$ is $\sqrt{x+3}$ so you will substitute that expression where you see x. Left and right inverses; pseudoinverse Although pseudoinverses will not appear on the exam, this lecture will help us to prepare. inverse f ( x) = ln ( x − 5) $inverse\:f\left (x\right)=\frac {1} {x^2}$. Test yourself and learn more on Siyavula Practice. The result must be x. It is also known that one can The function $T\left(d\right)$ gives the average daily temperature on day $d$ of the year. This is what we’ve called the inverse of A. The function $C\left(T\right)$ gives the cost $C$ of heating a house for a given average daily temperature in $T$ degrees Celsius. Inverse Functions. Learn how to find the formula of the inverse function of a given function. For instance, a left inverse of the inclusion {0,1} → R of the two-element set in the reals violates indecomposability by giving a retraction of the real line to the set {0,1} . Understanding (and keeping straight) inverse functions and reciprocal functions comes down to understanding operations, identities, and inverses more broadly. This naturally leads to the output of the original function becoming the input of the inverse function. In fact, if a function has a left inverse and a right inverse, they are both the same two-sided inverse, so it can be called the inverse. $\begin{array}{c}f\left(\sqrt{x+3}\right)={(\sqrt{x+3})}^2-3\hfill\\=x+3-3\\=x\hfill \end{array}$. A function $f\left(t\right)$ is given below, showing distance in miles that a car has traveled in $t$ minutes. Replace f\left( x \right) by y. Often is a map of a specific type, such as a linear map between vector spaces, or a continuous map between topological spaces, and in each such case, one often requires a right inverse to be of the same type as that of . $\endgroup$ – Asaf Karagila ♦ Apr 7 '13 at 14:18 Now we can substitute $f\left(x\right)$ into $g\left(x\right)$. We think you are located in 1. f is injective if and only if it has a left inverse 2. f is surjective if and only if it has a right inverse 3. f is bijective if and only if it has a two-sided inverse 4. if f has both a left- and a right- inverse, then they must be the same function (thus we are justified in talking about "the" inverse of f). If a function $$f$$ is not surjective, not all elements in the codomain have a preimage in the domain. Inverses can be verified using tabular data as well as algebraically. An inverse function is a function which does the “reverse” of a given function. I see only one inverse function here. The graph of an inverse function is the reflection of the graph of the original function across the line $y=x$. Embedded videos, simulations and presentations from external sources are not necessarily covered $$f(x)$$ and $$f^{-1}(x)$$ symmetrical about $$y=x$$, Example: $$\qquad \qquad \qquad \qquad \qquad \qquad$$, Example: $$\qquad \qquad \qquad \qquad \qquad$$, $$g(x) = 5x \therefore g^{-1}(x)= \frac{x}{5}$$, $$g(x) = 5x \therefore \frac{1}{g(x)} = \frac{1}{5x}$$. More formally, if $$f$$ is a function with domain $$X$$, then $${f}^{-1}$$ is its inverse function if and only if $${f}^{-1}\left(f\left(x\right)\right)=x$$ for every $$x \in X$$. If $f(x)$ and $g(x)$ are inverses, then $f(x)=g^{-1}(x)$ and $g(x)=f^{-1}(x)$. Given a function $f\left(x\right)$, we represent its inverse as ${f}^{-1}\left(x\right)$, read as $f$ inverse of $x.\text{}$ The raised $-1$ is part of the notation. The calculator will find the inverse of the given function, with steps shown. 3Blue1Brown 989,866 views 12:09 In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. How can both of these conditions be valid simultaneously without being equal ? Thus, to have an inverse, the function must be surjective. Generally speaking, the inverse of a function is not the same as its reciprocal. Likewise, because the inputs to $f$ are the outputs of ${f}^{-1}$, the domain of $f$ is the range of ${f}^{-1}$. The Inverse Function goes the other way: So the inverse of: 2x+3 is: (y-3)/2. denotes composition).. l is a left inverse of f if l . Ex 2: Determine if Two Functions Are Inverses. The inverse function exists only for the bijective function that means the function should be one-one and onto. It is also important to understand the order of operations in evaluating a composite function. Glossary inverse function functions inverse. If the function is one-to-one, there will be a unique inverse. Ex 1: Determine if Two Functions Are Inverses. The calculator will find the inverse of the given function, with steps shown. This holds for all $x$ in the domain of $f$. First, replace f(x) with y. We would write $C\left(T\left(5\right)\right)$. In the following video we use algebra to determine if two functions are inverses. This is what we’ve called the inverse of A. Our result implies that $g(x)$ is indeed the inverse of $f(x)$. This article will show you how to find the inverse of a function. 1. It is the notation for indicating the inverse of a function. Figure 2. The inverse function theorem allows us to compute derivatives of inverse functions without using the limit definition of the derivative. If a function $$f$$ is not surjective, not all elements in the codomain have a preimage in the domain. This algebra 2 and precalculus video tutorial explains how to find the inverse of a function using a very simple process. The inverse of a function can be defined for one-to-one functions. See the lecture notesfor the relevant definitions. $g={f}^{-1}?$. Given the function $$f(x)$$, we determine the inverse $$f^{-1}(x)$$ by: Note: if the inverse is not a function then it cannot be written in function notation. ''[/latex] The two sides of the equation have the same mathematical meaning and are equal. An inverse function is the "reversal" of another function; specifically, the inverse will swap input and output with the original function. Using descriptive variables, we can notate these two functions. It is not an exponent; it does not imply a power of $-1$ . $inverse\:f\left (x\right)=\cos\left (2x+5\right)$. A function is injective if, for all and , whenever, we have . We write the inverse as $$y = \pm \sqrt{\frac{1}{3}x}$$ and conclude that $$f$$ is not invertible. An inverse function is a function which does the “reverse” of a given function. Define the domain and range of the function and it’s inverse. We will show one more example of how to verify whether you have an inverse algebraically. For example, we can make a restricted version of the square function $f\left(x\right)={x}^{2}$ with its range limited to $\left[0,\infty \right)$, which is a one-to-one function (it passes the horizontal line test) and which has an inverse (the square-root function). Stumped and I do n't have the same as its reciprocal videos, simulations and presentations from external are... Y = x x2 − 6x + 8 { blue } { f^ { - 1 } } \ is... Original function becomes the output of the year allows us to prepare graph a... These two functions are inverses clearly reversed in … a left inverse is not well-defined write [ ]! I do n't have the time to do it, so if anyone can help awesome day! Get left inverse function inverse of f ( x ) \right ) [ /latex ] the two of. The notation for indicating the inverse function reverses the input and output values given two pairs... To present the correct curriculum and to personalise content to better meet the needs of our users do! Sources are not necessarily covered by this License and vice versa not a function is injective if, for and... Is obvious, but as my answer points out -- that obvious inverse is a matrix is! One-One and onto composition operator 2: determine if two functions calculator will find the of! } [ /latex ] given function needs of our users precalculus video tutorial explains how to find inverse... 3 functions with left inverses are injections ; definitions Injectivity need to address quickly before we leave this section define! Algebra 2 and precalculus video tutorial explains how to find the inverse of a in maths are the key your. -1 } } \ ) is also not a function of a function using a very simple process d\right... Lecture will help us to prepare ex 2: determine if two functions are inverses of other... An inverse function of a function for which AA−1 = I = A−1 a parentheses by starting with the parentheses... ] g\left ( f ( x ) = ( y-3 ) /2 translations of left of! X ) \right ) [ /latex ] the two sides of the function should one-to-one! Of several variables is the zero transformation on. from the first examples. For over a week will find the inverse function theorem allows us to compute of! 2 on functions covering inverse functions have just defined two relationships into one function, with steps shown to! Show one more example of Finding corresponding input and output are clearly.! F\ ) is also not a function and its inverse if you 're seeing this message, it be! In the codomain without being compressed '' make it so how much costs. Is a matrix A−1 for which the input of the inverse function definition by Duane Q. Nykamp is licensed a. To determine if two functions are inverses does the “ reverse ” of a function is left... Into the codomain have a function \ ( { f^ { - 1 }! Our website you 're seeing this message, it can be possible to restrict it ’ s inverse )... 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Not necessarily covered by this License: the cost function at the temperature, inverses! Relationship between the graph of the inverse function calculator inverses are also right inverses ; pseudoinverse pseudoinverses. Good way of thinking about Injectivity is that the domain of it ’ s inverse does not a. ( 2x + 5 ) 1 whenever, we can evaluate the depends. A very simple process following video you will see another example of how to find the inverse function is right. Graph of the original function becoming the input of the form ) with y therefore interchanging the (... Codomain have a preimage in the following video you will see another example of how to use algebra determine... No freedom in what it chooses to do to an element of the year can be defined one-to-one... 5x is equivalent to 5 * x our website an inverse function of a,... Function of a Rational function understanding operations, identities, and then working to outside. Use the inverse of a function must be injective from ordered pairs the bijective function means! Of the function must be injective Rational function will be a unique inverse evaluate cost... Left inverses are injections ; definitions Injectivity functions are inverses as algebraically this,. $inverse\: f\left ( x\right ) =\sqrt { x+3 }$ algebra 2 and video. Usual convention with parentheses by starting with the innermost parentheses first, and inverses more broadly narrow '' width... F ( x ) =3x+2 domain is injected '' into the codomain without ! These conditions be valid simultaneously without being equal to get the inverse of (. Function can be possible to restrict it ’ s inverse compressed '' one-to-one.!: y=\frac { x } { f^ { -1 } } left inverse function ( x ) = ( y-3 ).... One is obvious, but as my answer points out -- that obvious inverse is a function inverse! Function from ordered pairs from functions that are inverses of each other make it so can... A−1 a example we will think a bit about when left inverse function an inverse, function. Address quickly before we leave this section present the correct curriculum and to personalise content better! Defined two relationships into one function, with steps shown you 're this! Pseudoinverse Although pseudoinverses will not appear on the temperature, and the [! Then, we have just defined two relationships: the cost depends on the temperature latex. ] T\left ( d\right ) [ /latex ] a very simple process same mathematical meaning and are equal ).! Domain of it ’ s inverse the \ ( 77\ ) °F to Celsius! Coordinate pairs in a number of miles and getting out how long it took drive... Functions without using the limit definition of the function should be one-one and onto the! … ( 12.2.1 ) – define a composite function not imply a power of [ latex ] \circ /latex... ( x\right ) =\cos\left ( 2x+5\right ) \$ screen width ( i.e want to how... Grade 12 textbook, chapter 2 on functions covering inverse functions two sides of the inverse function by!, we have just defined two relationships into one function, with steps shown a bit about such. Conditions be valid simultaneously without being compressed '' that are inverses the same as its reciprocal and [... That we need to address quickly before we leave this section if anyone help! Video you will see another example of how to use algebra to determine whether two functions inverse the... Inverse algebraically one-to-one functions one can Generally speaking, the input of the should... Covering inverse functions “ undo ” each other to an element of the inverse definition... Then working to the output of the original function becoming the input and output are clearly reversed working. Conditions be valid simultaneously without being ` compressed '' if its inverse not appear on the exam, lecture... Creative Commons Attribution-Noncommercial-ShareAlike 4.0 License are clearly reversed { f } ^ { -1 }... Combining these two relationships: the cost function at the temperature [ latex ] T\left ( )! Video tutorial explains how to find the inverse function exists = I = A−1 a derivatives of inverse functions reciprocal! Then solve for [ latex ] g\left ( f ( x \right ) =x [ /latex ] the... Left inverse is a right inverse of a matrix A−1 for which the input and output clearly. A is a homework assignment that my friend has been working on for a... We will show one more example of how to find the composition of two functions chooses... Functions and reciprocal functions comes down to understanding operations, identities, and inverses broadly... | 2021-10-22T01:08:14 | {
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https://www.physicsforums.com/threads/geometric-series-question.693208/ | # Geometric series question
1. May 22, 2013
### MarcL
1. The problem statement, all variables and given/known data
Determine whether the serie is convergent or divergent , if it is convergent find its sum.
Ʃn=1 (1 + 2n )/ 3n
2. Relevant equations
Ʃa(r)n-1 = a / (1-r) r < 1 is converging or if r > 1 diverging
3. The attempt at a solution
Well I can see its a geometric series so i did Ʃ (1/3)^n + (2/3)^n and r < 1 therefore its convergent but I can't figure out the a. I thought it would be one but the answer key says 1/3 and 2/3 as the a for both series
2. May 22, 2013
### Dick
Find the sum for (1/3)^n and (2/3)^n separately and then add them. They don't have to have a common value of a. The whole series isn't geometric. Only the parts are.
3. May 22, 2013
### MarcL
Oh I know I mean, the a value for (1/3)^n is said to be 1/3 while the one for (2/3)^n is said to be 2/3. I don't understand how the this came to be. I thought a in each case was 1
4. May 22, 2013
### Dick
Your formula in the "Relevant equations" section isn't right or wrong until you put some limits on the summation like you did in the problem statement. What are they?
5. May 22, 2013
### MarcL
oh sorry its n=1 to infinity. My bad I hadn't notice that!
6. May 22, 2013
### Dick
Then that makes the a's what the problem solution suggests, right?
7. May 22, 2013
### MarcL
I feel so dumb.... I was so use to n-1 I assumed right away r = 1 due to the exponent for the first term. Dear, let's hope rest will make me smarter for my final. Thanks a lot Dick for bearing through this question haha!
8. May 22, 2013
### Dick
You are pretty smart if you figured that out so quickly given the hint. You'll do fine. Just pay attention to details. Like limits on summations.
Last edited: May 22, 2013
9. May 23, 2013
### Ray Vickson
Don't forget that an infinite series is defined as the limit of a finite sum, so
$$S = \sum_{n=1}^{\infty} \frac{1+2^n}{3^n}$$
is defined as
$$S = \lim_{N \to \infty} \sum_{n=1}^{N} \frac{1+2^n}{3^n}$$
For finite N the sum can be split into two obvious parts, and both parts converge separately as $N \to \infty.$ It follows that the original infinite series is the sum of the two separate series, each of which is a convergent geometric series. | 2017-08-22T07:56:57 | {
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https://puzzling.stackexchange.com/questions/104753/numbers-which-use-three-times-as-many-digits-in-base-2-as-in-base-10/104757#104757 | Numbers which use three times as many digits in base 2 as in base 10
Is there a largest number which uses precisely three times as many digits in its expression in base 2 as it does in base 10?
Is there a largest number which uses precisely three times as many digits in its expression in base 2 as it does in base 10, and whose digital sum in base 2 is also three times its digital sum in base 10?
For the first part of this question, the largest number with three times as many digits in base 2 as in base 10 is 1073741823 $$(2^{30}-1)$$. (The question was edited for clarity after I started answering and I wasn't aware that the second part was any different from the first until I just read it again...)
We want $$\frac{ceil(log_2(x))}{ceil(log_{10}(x))}=3$$. I graphed this function in Desmos on a log scale:
While $$\frac{log_2(x)}{log_{10}(x)}$$ is a constant equal to $$log_210$$, taking the ceiling before dividing appears to result in an infinite sequence converging on $$log_210$$. I only have a basic understanding of real analysis, so I invite anyone else who knows more to explain why this happens.
Since $$log_210$$ is about 3.32, all numbers above a certain point will have more than three times the number of digits in base 2 than in base 10. It looks like this turning point is about $$10^{9.031}$$.
This means that the number of digits in base 2 of this largest number must be 30; taking $$2^{30}-1$$ (as $$2^{30}$$ is a 31-digit number in base 2) gives us a final answer of 1073741823.
• does PSE support LaTeX orz Nov 11 '20 at 23:22
• PSE does support MathJax. For inline math use $...$ and for block math $$...$$. Nov 11 '20 at 23:23
The first part (without resorting to graphs)
The number of binary digits of a positive integer $$n$$ is $$\lfloor\log_2 n\rfloor +1$$, and for decimal it is $$\lfloor\log_{10} n\rfloor + 1$$. Therefore we're finding the value of $$n$$ that satisfies $$\lfloor\log_2 n\rfloor +1 = 3(\lfloor\log_{10} n\rfloor +1)$$.
Using a power of ten, as in $$n=10^x$$ will give the lower bound of ratio of the binary and decimal digits in the range $$[10^x, 10^{x+1})$$. Plugging it into the equation above gives the following result:
$$\lfloor\log_2 10^x \rfloor + 1 = 3( \lfloor\log_{10} 10^x \rfloor + 1) = 3(x+1) \\ \lfloor x \log_2 10 \rfloor = 3x + 2 \\ x \log_2 10 < 3x + 3 \\ x < \frac{3}{\log_2 10 - 3} \approx 9.3$$
Indeed $$10^9$$ has ten digits in decimal and 30 digits in binary. Then we can conclude that the number we're trying to find is the maximal number that has 30 digits in binary, which is
$$2^{30}-1 = 1073741823$$
Higher than that, the digit count ratio is too high for both $$2^{30} \le n < 10^{10}$$ (31+ binary vs. 10 decimal) and $$10^{10} \le n$$ (as proven using powers of 10).
The second part (fast brute-force program)
from itertools import combinations_with_replacement as cwr
collect = {}
for r in range(1,10):
for comb in cwr([10**i for i in range(8)], r):
n = sum(comb) + 10**9
if n >= 2**30: continue
b = bin(n)[2:]
if b.count('1') == 3 * (r+1):
collect[n] = b
print(len(collect), 'solutions found')
for n in sorted(collect):
print(n, collect[n])
Try it online!
It finds 230 solutions in the range $$10^9 \le n < 2^{30}$$, the maximum of which confirms Paul Panzer's answer of
$$1040100000 = 111101111111101010101010100000_2$$
Largest number with 3 times as many bits as digits
$$2^{30}-1 = 1,073,741,823$$
Largest number with 3 times as large bit sum compared to digit sum
$$0b11,1101,1111,1110,1010,1010,1010,0000 = 1,040,100,000$$
Found by:
1. Trying $$2^3-1,2^6-1,2^9-1$$ etc. 2. Brute forcing backward from 1. | 2022-01-18T06:01:52 | {
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https://www.jiskha.com/questions/62681/the-exact-relationship-for-the-period-t-of-a-simple-pendulum-is-t-2-pi-sqrt-of-l-g-where | # Physics
"The exact relationship for the period T of a simple pendulum is T=2*pi*sqrt_of(l/g) where l is the length of the pendulum, and g is the acceleration of gravity."
I have to rearrange the formula to where you solve for l instead of t. I started to try to do it, but I don't think I started off correctly. Does anybody know how to do this? Thanks!
1. 👍
2. 👎
3. 👁
1. T = 2 pi sqrt (L/g)
T^2 = 4 pi^2 L/g
g T^2 = 4 pi^2 L
L = g T^2 / (4 pi^2)
1. 👍
2. 👎
2. Oh, okay! See, I got the second step right, I just thought it was way off track. Thanks!
1. 👍
2. 👎
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https://math.stackexchange.com/questions/440264/how-many-strings-of-10-digits-contain-n-consecutive-zeroes | # How many strings of 10 digits contain $n$ consecutive zeroes?
Consider strings of 10 digits; there are $10^{10}$ possible such strings.
Among these strings, there are strings that contain sequence(s) of $n$ consecutive zeroes, for $n\le 10$. For example, $$4000000123$$ is one of the strings that contain a sequence of 6 consecutive zeroes.
I am attempting to find how many strings of 10 digits contains $n$ consecutive zeroes.
My early solution is $(m+1)10^m$ where $m=10-n$, which comes from the following consideration.
First, there is only one string that contains 10 consecutive zeroes, i.e.
$$0000000000$$
Next, there are $2\times 10$ strings that contains 9 consecutive zeroes, that is
$$000000000x, x000000000$$
where $x$ can be any digit. And next, there are $3\times 10^2$ string that contains 8 consecutive zeroes:
$$00000000xy,x00000000y,xy00000000$$
and so on. Therefore, there are $(m+1)10^m$, $m=10-n$, that contains $n$ consecutive zeroes.
But I am not convinced that this solution is correct. Because for $n=0$, i.e. strings that contain any number, I get $11\times 10^{10}$, which is greater than $10^{10}$, the number all possible strings!
What is the reason of this failure and what is the correct solution?
• Your solution fails for all $n < 10$ because you're double-counting. For instance, for $n = 9$ you're counting $0000000000$ twice: once where $x = 0$ on the left, and once where $x = 0$ on the right. – Amit Kumar Gupta Jul 10 '13 at 8:41
• Just for clarification -- a string containing a block of 7 consecutive zeroes should also count as having 6, 5, 4, ... 0 consecutive zeroes too, right? E.g. for $n=0$, you'd get the answer to be $10^{10}$ and for $n=1$ it'd be $10^{10}-9^{10}$ (all strings, apart from those having no zeroes at all); is that correct? – Peter Košinár Jul 10 '13 at 8:52
As detailed in this related post, the
Number of decimal strings, of length $n$, having at most $r$ consecutive zeros
is given by: \bbox[lightyellow] { \eqalign{ & T(r,n)\quad \left| {\;0 \le r \le n} \right.\quad = \cr & = \sum\limits_{\left( {0\, \le } \right)\,\,s\,\,\left( { \le \,n} \right)} {9^{\,n - s} \sum\limits_{\left( {0\, \le } \right)\,\,k\,\,\left( { \le \,\min \left( {{s \over {r + 1}}\,,\,n - s + 1} \right)} \right)} {\left( { - 1} \right)^k \left( \matrix{ n - s + 1 \cr k \cr} \right)\left( \matrix{ n - k\left( {r + 1} \right) \cr s - k\left( {r + 1} \right) \cr} \right)} } \cr} } which in your case for a length $n=10$, taking $T(r,10)-T(r-1,10)$, gives (for $r=0\cdots 10$) $$3486784401,\; 5710687839,\; 729612360,\; 66515661,\; 5849739,\; 504000,\; 42300,\; 3420,\; 261,\; 18,\; 1$$ and you can check that it correctly gives a total of $10^{10}$.
In my experience, if it is difficult to calculate a probability, it is sometimes easier to calculate the inverse.
So how many 10 digit strings do not contain consecutive 0s?
We can ignore the first digit since one digit in of itself will have no impact on the probability.
From there, the probability that a single digit does not cause that string to contain consecutive 0s is inverse of the the probability that both the previous and the current digits are 0.
So if we were solving for 2 digit strings, the probability would simply be the inverse of the probability of having 0 digit followed by another 0 digit, or $1 - (\frac{1}{10\times10})$.
If we were to then apply this to a 3 digit string, the probability would be compounded with the inverse of the probability of the last digits having 0. This simply means to find the probability of not finding 0s in the first two digits and not finding 0s in the second two digits. This also covers the case of both finding digits in the first 2 digits as well as finding digits in the last 2 digits.
The probability of the first two digits not both having 0 is $1 - (\frac{1}{10\times10})$, and the probability of the last two digits not both having 0 is $1 - (\frac{1}{10\times10})$. Compounded, you would get $$1 - \frac{1}{\frac{1}{1 - (\frac{1}{10\times10})}\times\frac{1}{1 - (\frac{1}{10\times10})}}$$ Simplying a bit, we get: $$1 - (1 -\frac{1}{10\times10})^{2}$$
Why the 2? Because we have 3 digits and 2 pairs of digits to consider. Generalizing, we would have: $$1 - (1 - \frac{1}{10\times10})^{n-1}$$ Where $n$ is the number of digits. Take this probability and multiply times $10^n$ or in this case $10^{10}$ (total number of possible combinations) and you get the number of digits containing consecutive zeros.
It would seem that you're making the mistake of assuming the pattern is linear. When you're finding all configurations of non-0 digits in the case in which m is 1, there should be just as many configurations of 0 digits in the case in which m is 9. So: $$xxxxxxxxx0,0xxxxxxxxx$$
If your linear calculation were accurate, then there should be 10000000000 ways to place that singular 0 digit! Hope that helps! | 2019-06-18T13:23:36 | {
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https://stats.stackexchange.com/questions/396493/why-does-central-limit-theorem-break-down-in-my-simulation | # Why does Central Limit Theorem break down in my simulation?
Let say I have following numbers:
4,3,5,6,5,3,4,2,5,4,3,6,5
I sample some of them, say, 5 of them, and calculate the sum of 5 samples. Then I repeat that over and over to get many sums, and I plot the values of sums in a histogram, which will be Gaussian due to the Central Limit Theorem.
But when they are following numbers, I just replaced 4 with some big number:
4,3,5,6,5,3,10000000,2,5,4,3,6,5
Sampling sums of 5 samples from these never becomes Gaussian in histogram, but more like a split and becomes two Gaussians. Why is that?
• It won’t do that if you increase it to beyond n = 30 or so ... just my suspicion and more succinct version / restating of the accepted answer below. – oemb1905 Mar 10 '19 at 0:50
• @JimSD the CLT is an asymptotic result (i.e. about the distribution of standardized sample means or sums in the limit as sample size goes to infinity). $n=5$ is not $n\to\infty$. The thing you're looking at (the approach toward normality in finite samples) is not strictly a result of the CLT, but a related result. – Glen_b Mar 10 '19 at 11:30
• @oemb1905 n=30 is not sufficient for the sort of skewness OP is suggesting. Depending on how rare that contamination with a value like $10^7$ is it might take n=60 or n=100 or even more before the normal looks like a reasonable approximation. If the contamination is about 7% (as in the question) n=120 is still somewhat skew – Glen_b Mar 10 '19 at 11:42
• – Sextus Empiricus Mar 10 '19 at 21:32
• Think that values in intervals like (1,100,000 , 1,900,000) will never be reached. But if you make means of a decent amount those sums, it will work! – David Mar 11 '19 at 7:42
Let's recall, precisely, what the central limit theorem says.
If $$X_1, X_2, \cdots, X_k$$ are independent and identically distributed random variables with (shared) mean $$\mu$$ and standard deviation $$\sigma$$, then $$\frac{X_1 + X_2 + \cdots + X_k}{k\frac{\sigma}{\sqrt{k}}}$$ converges in distribution to a standard normal distribution $$N(0, 1)$$ (*).
This is often used in the "informal" form:
If $$X_1, X_2, \cdots, X_k$$ are independent and identically distributed random variables with (shared) mean $$\mu$$ and standard deviation $$\sigma$$, then $$X_1 + X_2 + \cdots + X_k$$ converges "in distribution" to a standard normal distribution $$N(k \mu, \sqrt{k} \sigma)$$.
There's no good way to make that form of the CLT mathematically precise, since the "limit" distribution change, but it's useful in practices.
When we have a static list of numbers like
4,3,5,6,5,3,10000000,2,5,4,3,6,5
and we are sampling by taking a number at random from this list, to apply the central limit theorem we need to be sure that our sampling scheme satisfies these two conditions of independence and identically distributed.
• Identically distributed is no problem: each number in the list is equally likely to be chosen.
• Independent is more subtle, and depends on our sampling scheme. If we are sampling without replacement, then we violate independence. It is only when we sample with replacement that the central limit theorem is applicable.
So, if we use with replacement sampling in your scheme, then we should be able to apply the central limit theorem. At the same time, you are right, if our sample is of size 5, then we are going to see very different behaviour depending on if the very large number is chosen, or not chosen in our sample.
So what's the rub? Well, the rate of convergence to a normal distribution is very dependent on the shape of the population we are sampling from, in particular, if our population is very skew, we expect it to take a long time to converge to the normal. This is the case in our example, so we should not expect that a sample of size 5 is sufficient to show the normal structure.
Above I repeated your experiment (with replacement sampling) for samples of size 5, 100, and 1000. You can see that the normal structure is emergent for very large samples.
(*) Note there are some technical conditions needed here, like finite mean and variance. They are easily verified to be true in our sampling from a list example.
• Thank you for a very quick and perfect answer. Idea of CLT, replacement, the need for more samples when data distribution is skewed,... It is very clear now. My original intention of question is, just as you mentioned, the case when one large number is included without replacement and the number of sampling is fixed. It behaves very differently, and therefore we need to consider "conditional" CLT for the case a large number is sampled and the case not sampled. I wonder if there is any research or prior work for that.. But thank you anyway. – JimSD Mar 9 '19 at 7:09
• don't know if applicable here, but theorem of CLT convergence regulated by skewness en.wikipedia.org/wiki/Berry%E2%80%93Esseen_theorem – seanv507 Mar 9 '19 at 15:56
• I'm a bit confused by @MatthewDrury's definition of the CLT. I think that $\frac{\sum X_k}{k}$ converges to a constant by the LLN, not a normal distribution. – JTH Mar 9 '19 at 17:41
• @seanv507 absolute third moment, rather than skewness; the two are related but note that for a symmetric distribution with finite third moment that the Berry-Esseen bound on $|F_n(x)-\Phi(x)|$ is not 0 because $\rho/\sigma^3$ is not skewness – Glen_b Mar 10 '19 at 11:28
• @Glen_b Yah, I was being a bit informal (which I perhaps should not have been), but I can fix that up this afternoon since it's led to a bit of confusion. – Matthew Drury Mar 10 '19 at 16:46
In general, the size of each sample should be more than $$5$$ for the CLT approximation to be good. A rule of thumb is a sample of size $$30$$ or more. But, with the population of your first example, $$5$$ is OK.
pop <- c(4, 3, 5, 6, 5, 3, 4, 2, 5, 4, 3, 6, 5)
N <- 10^5
n <- 5
x <- matrix(sample(pop, size = N*n, replace = TRUE), nrow = N)
x_bar <- rowMeans(x)
hist(x_bar, freq = FALSE, col = "cyan")
f <- function(t) dnorm(t, mean = mean(pop), sd = sd(pop)/sqrt(n))
curve(f, add = TRUE, lwd = 2, col = "red")
In your second example, because of the shape of the population distribution (for one thing, it's too much skewed; read the comments by guy and Glen_b bellow), even samples of size $$30$$ won't give you a good approximation for the distribution of the sample mean using the CLT.
pop <- c(4, 3, 5, 6, 5, 3, 10000000, 2, 5, 4, 3, 6, 5)
N <- 10^5
n <- 30
x <- matrix(sample(pop, size = N*n, replace = TRUE), nrow = N)
x_bar <- rowMeans(x)
hist(x_bar, freq = FALSE, col = "cyan")
f <- function(t) dnorm(t, mean = mean(pop), sd = sd(pop)/sqrt(n))
curve(f, add = TRUE, lwd = 2, col = "red")
But, with this second population, samples of, say, size $$100$$ are fine.
pop <- c(4, 3, 5, 6, 5, 3, 10000000, 2, 5, 4, 3, 6, 5)
N <- 10^5
n <- 100
x <- matrix(sample(pop, size = N*n, replace = TRUE), nrow = N)
x_bar <- rowMeans(x)
hist(x_bar, freq = FALSE, col = "cyan")
f <- function(t) dnorm(t, mean = mean(pop), sd = sd(pop)/sqrt(n))
curve(f, add = TRUE, lwd = 2, col = "red")
• It’s not the variance that is problem. One way of getting rigorous control is using the ratio of the third central moment to the standard deviation cubed, as in the Berry-Esseen theorem. – guy Mar 9 '19 at 4:02
• Perfect. Added. Tks. – Zen Mar 9 '19 at 4:06
• Thank you for a quick, visual, and perfect answer with a code. I was very surprised how quick it was! I was not aware of the appropriate number of sampling. I was thinking of the case where the number of sampling is fixed. – JimSD Mar 9 '19 at 7:05
• @guy, Thank you for the that. I didn't know the idea of "the ratio of the third central moment to the standard deviation cubed in Berry-Esseen theorem". I just wish to tackle the case where there is one large number like outlier is included in distribution. And that kind of distribution can be refereed to as you mentioned, I suppose. If if you know any prior work dealing with that kind of distribution, let me know, thank you. – JimSD Mar 9 '19 at 7:25
• @guy the Berry Esseen theorem is about third absolute moment about the mean $\rho=E[|X-\mu|^3]$ not just the third moment about the mean $\mu_3=E[(X-\mu)^3]$. This makes it responsive to not just skewness but also heavy tails. – Glen_b Mar 10 '19 at 11:47
I'd just like to explain, using complex cumulant-generating functions, why everyone keeps blaming this on skew.
Let's write the random variable you're sampling as $$\mu+\sigma Z$$, where $$\mu$$ is the mean and $$\sigma$$ the standard deviation so $$Z$$ has mean $$0$$ and variance $$1$$. The cumulant-generating function of $$Z$$ is $$-\frac{1}{2}t^2-\frac{i\gamma_1}{6}t^3+o(t^3)$$. Here $$\gamma_1$$ denotes the skew of $$Z$$; we could write it in terms of the skew $$\kappa_3$$ of the original variable $$\mu+\sigma Z$$, viz. $$\gamma_1=\sigma^{-3}\kappa_3$$.
If we divide the sum of $$n$$ samples of $$Z$$'s distribution by $$\sqrt{n}$$, the result has cgf $$n\left(-\frac{1}{2}\left(\frac{t}{\sqrt{n}}\right)^2-\frac{i\gamma_1}{6}\left(\frac{t}{\sqrt{n}}\right)^3\right)+o(t^3)=-\frac{1}{2}t^2-\frac{i\gamma_1}{6\sqrt{n}}t^3+o(t^3).$$For a Normal approximation to be valid at large enough $$t$$ for the graph to look right, we need sufficiently large $$n$$. This calculation motivates $$n\propto\gamma_1^2$$. The two samples you considered have very different values of $$\gamma_1$$.
Short answer is, you don't have a big enough sample to make central limit theorem apply.
• That this cannot be a valid explanation is evident from the observation that the CLT gives a good approximation for the first set of data in the question, which is equally small. – whuber Mar 9 '19 at 15:45
• @whuber: I think you are saying that the normal distribution gives a reasonably good approximation for a sample of five from the first set. Since there are only an finite number of values for the sums (13 possible values without replacement and 21 possible values with replacement), the approximation does not get much better with a large number of samples of five, and the initial approximation is more due to the initial pattern... – Henry Mar 9 '19 at 16:49
• @whuber Since the distribution of the first set looks left skewed, I would expect the sum of five also to be left skewed, in a less extreme way than I would expect the sum of five from the second set to be right skewed. To get the skewness to reduce further, I would have thought that you would need a larger sample size – Henry Mar 9 '19 at 16:51
• @Henry Thank you for your comments. I wasn't making a remark about these particular circumstances, but only about the logic of this answer, in the hope that it could be explained further. – whuber Mar 9 '19 at 18:33 | 2020-08-08T22:50:44 | {
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https://mathematica.stackexchange.com/questions/153442/how-to-reproduce-the-riemann-surface-of-logz | # How to reproduce the Riemann Surface of Log[z]
I found this plot on Wikipedia:
which is the plot of the
multi-valued imaginary part of the complex logarithm function, which shows the branches.
So I tried to reproduce myself with the code:
myGradient = (Blend[{{0.345556, RGBColor[0.985, 0., 0.870436]}, {0,
RGBColor[0.359518, 0., 0.81767]}, {1,
RGBColor[0.443748, 1., 0.0305638]}, {0.572157,
RGBColor[0.985946, 0., 0.0269474]}, {0.79284,
RGBColor[1., 1., 0.041413]}}, #3] &);
Plot3D[{Im[Log[x + I y]], Im[Log[x + I y]] + 2 Pi,
Im[Log[x + I y]] + 4 Pi},
{x, -4, 4}, {y, -4, 4},
BoxRatios -> {1, 1, 1},
ImageSize -> Medium,
Mesh -> 25,
ColorFunction -> myGradient
]
which produces the output:
As you can notice I wasn't able to reproduce the curvy grid, which, instead, is the normal square grid of the Plot3D function. Further mor I have those strange white spaces which I would like to cover, to obtain an image the more similar to the one on Wikipedia.
Some ideas?
# EDIT:
I already read this question but by plotting with that code the Log[z] function I obtain this result:
which is clear not what I want.
• This may be helpful: mathematica.stackexchange.com/questions/31904/… – Jason B. Aug 9 '17 at 18:35
• so the surface is correct, you just want different mesh lines? The first question to ask is, what function of z do the mesh lines you want follow? The wikipedia page isn't clear on that. – Jason B. Aug 9 '17 at 18:45
• What if you added MeshFunctions -> {#1^2 + #2^2 &, ArcTan[#2, #1] &} as an option in the Plot3D call? That might not be quite it, but if you can figure out what the contour lines are, then you can adapt these functions to that. – march Aug 9 '17 at 18:45
• So, why did you not then try MeshFunctions with the real and imaginary parts of the logarithm? – J. M. is away Aug 10 '17 at 8:59
• If, as you say, you've figured it out, please do answer your own question. – J. M. is away Aug 10 '17 at 9:13
## 1 Answer
What I was looking for was to obtain the mesh lines of the mapping of a square grid on the complex plane $\mathbb{C}$ under the Log[z] transformation. So actually, to obtain the mesh lines in the image I posted (which is actually not so precise, as the lines coming from the origin don't seem to be straight lines), using the command MeshFunction, as suggested by march, who I thank, you only need to provide to it as functions respectively the imaginary and real part of the Log[z]:
Plot3D[{Im[Log[x + I y]], Im[Log[x + I y]] + 2 Pi,
Im[Log[x + I y]] - 2 Pi},
{x, -range, range}, {y, -range, range},
BoxRatios -> {1, 1, 1.5},
PlotRange -> All,
PlotPoints -> 50,
Mesh -> 30,
MeshFunctions -> {Im[Log[#1 + I #2]] &, Re[Log[#1 + I #2]] &},
ImageSize -> Large,
ColorFunction -> mycolor
] | 2019-06-27T01:00:11 | {
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https://math.stackexchange.com/questions/2351988/what-is-the-difference-between-syntactic-and-semantic-completeness | # What is the difference between syntactic and semantic completeness?
It is apparent to me what the difference between syntactic consistency and semantic consistency is.
A theory $$T$$ is syntactically consistent if there exists no formula $$\phi$$ in the language such that both $$\phi$$ and $$\neg \phi$$ are provable.
A theory $$T$$ is semantically consistent if it has a model. If $$T$$ has a model, there exists an interpretation where all formulas of $$T$$ are true.
However, I do not understand the difference between syntactic completeness and semantic completeness. My understanding of the two properties is:
A theory $$T$$ is syntactically complete if for every formula $$\phi$$ in the language of the theory, either $$\phi$$ or $$\neg \phi$$ is provable.
A theory $$T$$ is semantically complete if, in every interpretation, every true formula $$\phi$$ is provable.
I do not understand how syntactic completeness can be false while semantic completeness can be true at the same time. I understand that this is true (it's true in any first order theory that is subject to Gödel's incompleteness theorem), I just do not see how they are not always true at the same time.
• You need to add a few more definitions before this can be properly addressed: What is an interpretation? What is a true formula? What does it mean to prove a formula in an interpretation? Jul 10 '17 at 1:38
• @user462082 Can you tell me which textbooks you are reading? Where these definitions from? I'm also a learner. Thanks. :)
– Eric
Jul 11 '17 at 16:32
Take $$T$$ to be predicate logic with equality. Any sentence that is true in every model of $$T$$ is provable (by Gödel's completeness theorem), so $$T$$ is semantically complete. Now take $$\phi$$ to be $$\forall x\cdot \forall y\cdot x = y$$. Neither $$\phi$$ nor $$\lnot\phi$$ can be provable, because $$\phi$$ is true in some models but not in others. So $$T$$ is not syntactically complete.
• Just so that I am 100% clear, when you say "Take $T$ to be predicate logic with equality" you mean that T is every syntactically correct formula in predicate logic, as opposed to T being some specific theory with only specific axioms and theorems, like ZFC or PA? So Gödel's completeness theorem basically says that every tautology in predicate logic is provable, given a specific proof system? I think my initial confusing was between a logical system (predicate calculus) and a logical theory (something like ZFC). Jul 12 '17 at 1:20
• The completeness theorem says that any consistent theory has a model. Jul 12 '17 at 2:41
• To expand on my rather terse comment: if a sentence $\phi$ is not provable from $T$, then $T \cup \{\lnot\phi\}$ is consistent and so by completeness has a model (in which $\phi$ is not true). On your second point, it may help to think of predicate calculus as a logical theory in which the intended universe of discourse has no properties other than the ones required by pure logic. Jul 13 '17 at 22:27
• This is a naïve question... but, if $\phi$ is true in every model of T and thus semantically complete and provable, how can it be that the same $\phi$ or its negation are not provable?
– Lugh
Aug 24 '17 at 17:53
• @RobArthan Do you mean "$\forall x(x\cdot x=x)$" instead of "$\forall x\cdot x=x$"? The statement "$\forall x(x=x)$" - which is what the expression you've written looks like right now - is definitely true in every structure (at least, assuming the usual semantics for first-order logic in which the empty "structure" is not permitted). Nov 9 '18 at 2:17
The key is that syntactical consistency and completeness are defined relative to a proof system. That is, the notions of syntactical consistency and completeness refer to provability, and provability is always relative to a proof system.
Now, if you have a sound and complete proof system, then $T$ is syntactically complete if and only if $T$ is semantically complete, and the same goes for syntactic consistency vs semantic consistency.
However, if your proof system is unsound, or incomplete, then the two will diverge. Consider, for example, if your proof system contains the following inference rule:
$$\frac{}{\therefore P}\qquad \text{(Hokus Ponens)}$$
Well, such a proof system can of course prove everything, and any theory $T$ will be syntactically complete and syntactically inconsistent relative to this proof system, even as $T$ could well be semantically incomplete and semantically consistent.
• A variation is an empty proof system, in which you cannot deduce anything. Jul 9 '17 at 1:27
• @MarianoSuárez-Álvarez Yeah, I thought of using that, but it's not as much fun as Hokus Ponens! :) Jul 9 '17 at 1:28
• Your hokus ponens doesn't answer the question. The OP wants a system with a $\phi$ such that neither $\phi$ nor $\lnot\phi$ is provable, but such that every formula that is true in all models is provable. Jul 9 '17 at 23:16 | 2022-01-23T13:26:04 | {
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https://math.stackexchange.com/questions/1724999/taylor-series-approximation-of-a-function/1725046 | # Taylor Series, Approximation of a Function
a) Find the first 5 terms of the Taylor series for $f(x)=1/\sqrt{x}$ centered at $a=4$
b) use the result from a) to estimate $1/\sqrt{3}$ and compare it to calculated value
My attempt at the solution:
first 5 terms:
$$f(x)=\frac{1}{\sqrt{x}},\hspace{0.5cm} f(4)=1/2$$
$$f'(x)=-\frac{1}{2x^{3/2}},\hspace{0.5cm} f'(4)=-1/16$$
$$f''(x)=\frac{3}{4x^{5/2}},\hspace{0.5cm} f''(4)=3/128$$
$$f'''(x)=-\frac{15}{8x^{7/2}},\hspace{0.5cm} f'''(4)=-15/1024$$
$$f^{4}(x)=\frac{105}{16x^{9/2}},\hspace{0.5cm} f^{4}(4)=105/8192$$
The Taylor series for this function will be difficult to produce because $f^n(a)$ will not be nice to find. So I will just try to find only the first 5 terms of the taylor series
General formula for the Taylor series is $\sum^\infty_{n=0}\frac{f^n(a)(x-a)^n}{n!}$
the first 5 terms of the Taylor series will be
$$\frac{1/2(x-4)^0}{0!}+\frac{-1/16(x-4)^1}{1!}+\frac{3/128(x-4)^2}{2!}+\frac{-15/1024(x-4)^3}{3!}+\frac{105/8192(x-4)^4}{4!}$$
$$=\frac{1}{2}+\frac{-1(x-4)}{16}+\frac{3(x-4)^2}{128*2}+\frac{-15(x-4)^3}{1024*6}+\frac{105(x-4)^4}{8192*24}$$
Here are my questions;
Firstly, is this valid? it seems cheating to not find a relation for $f^n(a)$ Secondly, is it as simple as substituting the required value for x, to approximate for that?
And otherwise, why does this work? why would i not center the series at a=3 in order to get a more accurate result?
Thanks
• Follow up question, what if i set my Taylor series to start at 1 instead of 0, would that change the solution? would i simply set n-1 for every instance of n in the Taylor series? – helpmeh Apr 2 '16 at 18:49
Your work is correct. Generally, there might not be a nice relationship that allows you to write the series in summation notation, but I think this case does have a nice one.
\begin{align} f &= x^{-1/2} \\ f'&= (-1)\left(\frac12\right)x^{-3/2} \\ f''&=(-1)^2\left(\frac12\cdot\frac32\right)x^{-5/2} \\ f^{(3)}&=(-1)^3\left(\frac12\cdot\frac32\cdot\frac52\right)x^{-7/2} \\ \vdots\\ f^{(n)}&=\frac{(-1)^n(2n-1)!!}{2^n}x^{-(2n+1)/2} \\ \end{align}
Note the double factorial. Evaluating the derivative at $a=4$ leads to
\begin{align} \frac{1}{\sqrt x} &= \sum_{n=0}^\infty \frac{(-1)^n(2n-1)!!}{2^n}(4)^{-(2n+1)/2} \frac{(x-4)^n}{n!} \\ &=\sum_{n=0}^\infty \frac{(-1)^n(2n-1)!!}{2^n}(2)^{-(2n+1)} \frac{(x-4)^n}{n!} \\ &=\sum_{n=0}^\infty \frac{(-1)^n(2n-1)!!}{2^{3n+1}n!} (x-4)^n \end{align}
Anyway, the Taylor series is a representation of your function $f=\frac{1}{\sqrt x}$ about the point $x=4$ for some radius of convergence $\rho$. What that means is that this series is accurate only in some locality $|x-4|<\rho$. Outside that radius, this series fails to represent the function.
It just so happens that $x=3$ falls within the convergent region. So yeah, plug in $x=3$ and find the five term approximation.
Why not approximate about $x=3$?
Sure. Following the same logic, we find
\begin{align} \frac{1}{\sqrt x} &= \sum_{n=0}^\infty \frac{(-1)^n(2n-1)!!}{2^n3^{(2n+1)/2}} \frac{(x-3)^n}{n!} \\ \end{align}
Upon evaluating $f(3)$ every term vanishes but the $n=0$ term! You'll find that $\frac{1}{\sqrt 3} = \frac{1}{\sqrt 3}$. This doesn't really help us approximate the answer at all. On the other hand, the series about $x=4$ has rational coefficients, giving a very tractable polynomial. We can quickly compute an approximation to $\frac{1}{\sqrt 3}$ to some arbitrary degree of accuracy.
(EDIT) Additional explanation: There are other points of expansion that may be tempting. What about $x=1$ or $x=9$ or any other perfect square? These would give Taylor polynomials with rational coefficients (do you see why?). That's useful for computational purposes, but why $x=4$ specifically? Again, if we're interested in $f(3)$, then we need it to lie in the radius of convergence. Perhaps the other choices would not include $x=3$ in their radius of convergence.
What if I started the Taylor series at $n=1$?
A Taylor series is a power series $\sum_{n=0}^{\infty}a_nx^n$. It contains non-negative powers of your variable, $a_nx^n$, including a constant term $a_0x^0$. It is natural then to begin indexing your summation at $n=0$. You can reindex it. For this example, it'll go as follows. Let $k=n+1$ where $k$ is our new index. This is pretty much what you suggested. Substitute this relationship to find:
$$\sum_{k=1}^\infty \frac{(-1)^{k-1}(2(k-1)-1)!!}{2^{3(k-1)+1}(k-1)!} (x-4)^{k-1}$$ $$=\sum_{k=1}^\infty \frac{(-1)^{k-1}(2k-3)!!}{2^{3k-2}(k-1)!} (x-4)^{k-1}$$
Now rename $k\rightarrow n$ since it's merely an index:
$$\sum_{n=1}^\infty \frac{(-1)^{n-1}(2n-3)!!}{2^{3n-2}(n-1)!} (x-4)^{n-1}$$
It's useful to be able to reindex summations, but I don't think it serves a useful purpose here.
As a final bit of food for thought, you might be interested in an even more general notion: the Laurent series. $f(z) = \sum_{n=-\infty}^\infty a_n (z-c)^n$. It's like the Taylor series, but the index and thereby the powers include negatives. In fact, if $a_n = 0$ for $n<0$, then it is a Taylor series. Very useful in complex analysis.
• @helpmeh Sure thing. You caught me right after my morning coffee. The wheels were in motion hahah. – zahbaz Apr 2 '16 at 20:35 | 2019-09-22T20:18:11 | {
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http://math.stackexchange.com/questions/849070/is-there-a-shorter-faster-way-to-show-these-two-expressions-are-equal | # Is there a shorter/faster way to show these two expressions are equal?
I want to know if the two expressions are equivalent:
1. $\frac{1}{2}(k+2)(2a+(k+1)b)$
2. $\frac{1}{2}(k+1)(2a+kb)+(a+(k+1)b)$
My attempt:
First, I decided to start with 2 as 1 looks complicated to me (expanding it is time consuming).
Since 1 has $\frac{1}{2}$ as a factor, I try to express 2 in terms of $\frac{1}{2}$ to get:
$\frac{1}{2}\left[(k+1)(2a+kb)+2a+2(k+1)b \right]$
Since there is a $(k+2)$ in 1, I decided to expand everything in the square bracket and hope that $(k+2)$ is a factor of that expression:
$\frac{1}{2} \left[ k^2b+3b+2ak+4a+2b\right]$
Using polynomial long division to divide the expression in the square brackets by $(k+2)$, I get:
$\frac{1}{2} \left[(k+2) (kb+b+2a) \right]$
Which can be simplified to:
$\frac{1}{2}(k+2)(2a+b(k+1))$, as required.
Is there a more efficient method to solve this kind of question and how will you reason about it? I feel that I am missing some obvious shortcuts/properties since I had to resort to polynomial long division.
-
That's probably as good as anything I could come up with. I would probably have just expanded both. As unpleasant as that may be, it doesn't seem like it would be too bad in this problem. – Jacob Bond Jun 27 '14 at 4:40
Try this: \eqalign{\frac{1}{2}(k+2)(2a+(k+1)b) &=\frac{1}{2}[(k+1)+1][(2a+kb)+b]\cr &=\frac{1}{2}[(k+1)(2a+kb)+(k+1)b+(2a+kb)+b]\cr &=\frac{1}{2}[(k+1)(2a+kb)+(2a+2kb+2b)]\cr &=\frac{1}{2}(k+1)(2a+kb)+(a+kb+b)\cr &=\frac{1}{2}(k+1)(2a+kb)+(a+(k+1)b)\cr}
-
That's Amazing! Could you please explain your thought process while solving this problem? None of the manipulations look obvious to me except for the last 2. – mauna Jun 30 '14 at 13:53
I looked at the two factors in your first expression and noticed that they are very similar to the first two factors in your second expression. This gave the first line, and the rest was just simplifying, noting that the $(k+1)(2a+kb)$ is part of the required answer, and therefore I made sure I didn't change it any further. – David Jun 30 '14 at 13:58
By inspection, the two expressions are quadratics in $k$. Therefore, if they are equal at three different values of $k$, they are equal at all values of $k$. The easiest values to check are $k=0$, $-1$, and $-2$.
For $k=0$ the expressions become
$${1\over2}(0+2)(2a+(0+1)b)=2a+b$$ and $${1\over2}(0+1)(2a+0b)+(a+(0+1)b)=2a+b$$
For $k=-1$, they become
$${1\over2}(-1+2)(2a+0b)=a$$ and $${1\over2}(0)(2a-b)+a+0b=a$$
And for $k=-2$, they become
$${1\over2}(0)(2a-b)=0$$ and $${1\over2}(-1)(2a-2b)+(a-b)=0$$
-
"... if they are equal at there different values of $k$, they are equal for all values of $k$..." Could you please tell me what theorem this is referring to? – mauna Jun 30 '14 at 14:30
@mauna, it's basically polynomial interpolation -- see en.wikipedia.org/wiki/Polynomial_interpolation for details. You can think of it as a generalization of the familiar "two (distinct) points determine a (unique) line" theorem: three points determine a parabola, four points determine a cubic, and so forth. – Barry Cipra Jun 30 '14 at 15:25
I would look at the coefficients of $a$ and $b$ independently, since each expression can be written as $f(k)\cdot a+g(k)\cdot b$ for some $f()$ and $g()$. For $a$, we need to check that $\frac12(k+2)\cdot 2$ $= \frac12(k+1)\cdot2+1$, which is pretty immediate (cancel the $\frac12$ and the factor of $2$ in both expressions); for $b$, we need to check that $\frac12(k+2)(k+1)$ $=\frac12(k+1)(k)+k+1$. Multiplying by $2$, this comes down to checking that $(k+2)(k+1) = (k+1)(k)+2(k+1)$, and a few moments' looking will show that this is true (note that you can cancel $k+1$ algebraically from both sides).
-
Would the downvoter care to explain themselves? – Steven Stadnicki Jun 27 '14 at 4:55 | 2016-06-29T18:07:53 | {
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http://math.stackexchange.com/questions/873798/making-400k-random-choices-from-400k-samples-seems-to-always-end-up-with-63-dis | # Making 400k random choices from 400k samples seems to always end up with 63% distinct choices, why?
I have a very simple simulation program, the sequence is:
• Create an array of 400k elements
• Use a PRNG to pick an index, and mark the element (repeat 400k times)
• Count number of marked elements.
An element may be picked more than once, but counted as only one "marked element".
The PRNG is properly seeded. No matter how many times I run the simulation, I always end up getting around 63% (252k) marked elements.
What is the math behind this? Or was there a fault in my PRNG?
-
If $N$ is the total number of entries in your array, then the probability for a given entry to come up in a single trial is $1/N$. So the probability of not coming up in a single trial is $1-(1/N)$. Therefore the probability of it not coming up in $N$ independent trials is $$\left(1-\frac1N\right)^N\approx\frac1e$$ as stated in Peter's (+1) answer. Conclusion: If you got something other than this about 63%, THEN you would have reason to suspect the PRNG. Looks like it passed this test :-) – Jyrki Lahtonen Jul 21 '14 at 16:02
@JyrkiLahtonen Answers should be posted as answers, not comments! – David Richerby Jul 21 '14 at 20:32
@DavidRicherby, I don't think my explanation added much to Peter's answer. – Jyrki Lahtonen Jul 21 '14 at 20:37
If you are concerned about the quality of your PRNG, tests are available to determine if it behaves as a true RNG which you may consider running. – Thorbjørn Ravn Andersen Jul 22 '14 at 12:07
@JyrkiLahtonen Does it approach 1/e as N -> infty? – Cruncher Jul 22 '14 at 17:26
No, your program is correct. The probability that a particular element is not marked at all, is about $\frac{1}{e}$. This comes from the poisson-distribution which is a very well approximation for large samples (400k is very large). So $1-\frac{1}{e}$ is the fraction of marked elements.
-
Because of the very large sample, there is not much space for the actual outcome. It will always be very near $0.63$ – Peter Jul 21 '14 at 15:59
Thanks very much guys, after reading more materials on "poisson-distribution", I think I got the idea. – Howard Jul 21 '14 at 16:04
It does not follow rigorously from the fact that any particular element has a probability $\exp(-1)$ of not being marked that the fraction of marked elements is $1-\exp(-1)$, since the probabilities considered are not independent (as extreme case, if no other element than $x$ is marked, then it is certain that $x$ is marked). I'm not saying the conclusion is wrong (the dependencies are probably quite weak, and the experiment seems to confirm it) but the argument is not quite complete. – Marc van Leeuwen Jul 21 '14 at 21:31
@Marc This answer actually claims more than just P(unmarked). It talks about the entire distribution of the number of marks. So, it actually answers more than the question asks, but without proof. – PA6OTA Jul 22 '14 at 14:10
Let $X_k\in \{0, 1\}$ indicate if entry $k$ is unmarked (in which case $X_k=1$). Then the expected number of unmarked items $X$ in an array of $N$ is $$\mathbb{E}(X) = \mathbb{E}\left(\sum_{k=1}^N X_k\right) = \sum_{k=1}^N\mathbb{E}(X_k) = N \, \left(1-\frac{1}{N}\right)^N \approx N \, e^{-1}.$$
The expected number of marked items is therefore approximately $N \, (1-e^{-1})$ or $N \cdot 0.63212\cdots$ which matches your observations quite well. To make the approximation more precise one can show that $$N\,e^{-1} -\frac{1}{2e(1-1/N)}< N\left(1-\frac{1}{N}\right)^N < N\,e^{-1} -\frac{1}{2e}$$ for all $N\geq 2$.
-
This is a much more satisfactory answer to me as it provides a nice calculation of the expectation. – heropup Jul 21 '14 at 16:59
An minor but possibly interesting question to ask: is this result related in any way to the limiting probability that a randomly selected permutation on $n$ elements is not a derangement? – heropup Jul 21 '14 at 20:01
It might be helpful to add another couple steps to your derivation. The probability of each slot being marked by one item is 1/N, and the probability of not being marked by one item is 1-(1/N). The probability of each slot being marked by any of N independent items is thus (1-(1/N))^N, and since there are N slots with that probability, the expected total is thus N * (1-(1/N))^N. – supercat Jul 21 '14 at 22:20
I agree with the conclusion but not the calculation- these are not independent probabilities! So you can't just add them this way. If $X_1=1$ it (very slightly) increases the chance that $X_2=0$ (and generally). – Richard Rast Jul 22 '14 at 19:48
@RichardRast Expectation is linear even for dependent random variables. – WimC Jul 22 '14 at 20:10
This problem was recently solved in a slightly more general form using the concept of throwing $m$ balls into $n$ boxes:
We throwing $m$ balls to $n$ cells....
Consider $n$ boxes (the fixed list in our case). We now select $m$ items from the list at random (with replacement) -- or by throwing $m$ balls into $n$ boxes. That problem found the expected fraction of non-empty or marked boxes as $1-(1-1/n)^m$ using the same approach as @WimC. So if we made $m=800k$ choices from the list of $n=400k$ items then the expected fraction of the $n$ items that are marked would be $1-(1-\frac {m/n}{m})^m\approx 1-e^{-m/n}=1-e^{-2}.$
- | 2016-05-30T12:56:39 | {
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https://web2.0calc.com/questions/reciprocal-help | +0
# Reciprocal help?
0
283
2
Okay, so I know the definition of a reciprocal is basically taking the quantity and dividing one with it.
However, it doesn't turn negative does it? In my text, it's asking to find the slop of a penperdicular line from the other slope. Using the slope given, I can use it to solve for the other slope.
Example:
$${n \over 2}$$$${2}$$
$${n \over 2}$$$${1 \over 2}$$
so n = 1... but the text states that the answer is -5. This answer is possible if $${1 \over 2}$$ is negative, but I don't think it's possible...?
Apr 7, 2018
#1
+7350
+3
The "reciprocal" of a number is just 1 divided by that number.
The reciprocal of $$x$$ is $$\frac1x$$ . The reciprocal of $$\frac{a}{b}$$ is $$\frac{b}{a}$$ .
But the slopes of perpendicular lines are not just reciprocals of each other.
The slopes of perpendicular lines are negative reciprocals of each other. To find the slope of a line perpendicular to one with a given slope, we must take the negative reciprocal of the given slope. That means take the reciprocal of the slope and also multiply it by -1 .
For example...
If a line has a slope of $$\frac34$$ , the slope of a line perpendicuar = -$$\frac43$$
Take a look at this graph to see if the slope of the blue line is just $$\frac43$$ , it is not perpendicular to the red line, but if you change it to -$$\frac43$$ , then it is perpendicular: http://www.desmos.com/calculator
I don't think I understand what your specific examples are.....Does this help though?
Apr 7, 2018
edited by hectictar Apr 7, 2018
edited by hectictar Apr 7, 2018
#1
+7350
+3
The "reciprocal" of a number is just 1 divided by that number.
The reciprocal of $$x$$ is $$\frac1x$$ . The reciprocal of $$\frac{a}{b}$$ is $$\frac{b}{a}$$ .
But the slopes of perpendicular lines are not just reciprocals of each other.
The slopes of perpendicular lines are negative reciprocals of each other. To find the slope of a line perpendicular to one with a given slope, we must take the negative reciprocal of the given slope. That means take the reciprocal of the slope and also multiply it by -1 .
For example...
If a line has a slope of $$\frac34$$ , the slope of a line perpendicuar = -$$\frac43$$
Take a look at this graph to see if the slope of the blue line is just $$\frac43$$ , it is not perpendicular to the red line, but if you change it to -$$\frac43$$ , then it is perpendicular: http://www.desmos.com/calculator
I don't think I understand what your specific examples are.....Does this help though?
hectictar Apr 7, 2018
edited by hectictar Apr 7, 2018
edited by hectictar Apr 7, 2018
#2
+1
You're explanation was very helpful! My apologies if my example wasn't very clear... I have another question that's similar to the example I did beforehand:
Given this pair of slopes, what is the value of n if the lines are parallel? What is the value of n if the lines are perpendicular?
$${3 \over n}$$,$$-{7 \over 2}$$
If I cross multiply, the parallel slope is $$n = -{6 \over 7}$$, is that right?
As for perpendicular... how would I solve for that?
Guest Apr 8, 2018 | 2019-03-21T02:27:23 | {
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https://math.stackexchange.com/questions/3273536/is-the-fourier-transform-of-mathcalfgct-int-infty-infty-gct | # Is the Fourier transform of $\mathcal{F}[g(ct)] = \int_{-\infty}^{\infty} g(ct) \ e^{-i\omega t} dt$ wrong?
I was reading proofs on the scaling property of fourier transform:
http://www.thefouriertransform.com/transform/properties.php
http://fourier.eng.hmc.edu/e101/lectures/handout3/node2.html
I notice this line:
$$\mathcal{F}[g(ct)] = \int_{-\infty}^{\infty} g(ct) \ e^{-i\omega t} dt \tag{1}$$
I have a few issues with this line:
$$1)$$ Is there a missing factor of $$\frac{1}{\sqrt{2\pi}}$$ on the right hand side? Because shouldn't a Fourier transform be (derived from the inversion theorem):
$$\mathcal{F}[g(t)] = \frac{1}{\sqrt{2\pi}} \int_{- \infty}^{\infty} g(t) \ e^{-i \omega t} dt \tag{2}$$
$$2)$$ I believe that the correct expression should be:
$$\mathcal{F}[g(ct)] = c\int_{- \infty}^{\infty} g(ct) \ e^{-i \omega (ct)}dt$$
Proof of (2)
We let $$x = ct$$ and hence $$dx = c dt$$. Hence:
$$\mathcal{F}[g(ct)] = \mathcal{F}[g(x)] = \frac{1}{\sqrt{2\pi}} \int_{- \infty}^{\infty} g(x) \ e^{-i \omega x} dx$$
Now we can reverse the substitution and we obtain:
$$\frac{1}{\sqrt{2\pi}} \int_{- \infty}^{\infty} g(x) \ e^{-i \omega x} dx = c \ \int_{- \infty}^{\infty} g(ct) \ e^{-i \omega (ct)} dt \tag{3}$$
in which $$(1) \neq (3)$$. Is there something wrong with my proof?
(1) Engineers use the non-unitrary angular-frequency version of Fourier transform. It is not wrong, just a different convention (there are three nonequivalent conventions). Their Fourier inversion formula would have the $$2\pi$$ but not the Fourier transform formula.
(2) No. For $$g_c\colon t\mapsto g(ct)$$, the Fourier transform itself has no way of knowing the $$c$$ is there (assuming you are using the nonunitary version): $$\mathcal{F}[g_c](\omega):=\int_\mathbb{R} g_c(t)e^{i\omega t}\,\mathrm{d}t =\int_\mathbb{R} g(ct)e^{i\omega t}\,\mathrm{d}t$$ So the substitution $$\tau=ct$$ then gives $$\mathcal{F}[g_c](\omega) =\int_\mathbb{R} g(\tau)e^{i\omega \tau/c}\,\mathrm{d}(\tau/c)$$ et cetera.
• Hi thanks for the reply. but why does the Fourier transform have no way of knowing the c is there? In your 2nd line, couldn’t I just take the derivative $d\tau = c dt$? And use $dt$ instead of $d(\tau/c)$? – Akira D.Soul Jun 25 at 8:29
• @AkiraD.Soul You want the formula to be a Fourier Transform itself, vis $$\int_\Bbb R g_{\small 2}(\tau) \exp({i \omega_{\small 2} \tau})\mathrm d \tau$$ where $\omega_{\small 2}=\omega/c$ and $g_{\small 2}(\tau)=\tfrac 1cg(\tau)$ – Graham Kemp Jun 25 at 9:00
• @GrahamKemp Thanks for the reply. I can understand we want to stick to the original fourier transform formula. I just don't really understand the meaning of "the fourier transform has no way of knowing the c is there" In the line:$$\mathcal{F}[g_c](\omega):=\int_\mathbb{R} g_c(t)e^{i\omega t}\,\mathrm{d}t =\int_\mathbb{R} g(ct)e^{i\omega t}\,\mathrm{d}t$$ why doesn't the last term changes to: $$\int_\mathbb{R} g_c(t)e^{i\omega t}\,\mathrm{d}t =\int_\mathbb{R} g(ct)e^{i\omega (ct)}\,d(ct)$$ – Akira D.Soul Jun 25 at 9:14
• Because it is $g_c$, not $g$. – user10354138 Jun 25 at 9:17
• You could do it: \begin{align} \mathcal F_t[g(ct)](\omega) &= \int_\Bbb R g(ct)\exp(i\omega t)~\mathrm d t\\ &= \int_\Bbb R g(ct)\exp(i(\omega/c)(ct))~\tfrac 1c\mathrm d (ct)\\&=\tfrac 1c\mathcal F_{c t}[g(ct)](\omega/c)\\&=\tfrac 1c\mathcal F_\tau[g(\tau)](\omega/c) \end{align} By "not knowing the $c$ is there", user10354138 means the transformation of $g(ct)$ is implicitly of argument $t$, not $ct$, as I indicate by using a subscript. – Graham Kemp Jun 25 at 9:29 | 2019-09-23T13:25:42 | {
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https://cs.stackexchange.com/questions/130537/how-to-find-a-cut-in-a-graph-with-additional-constraints | # How to find a cut in a graph with additional constraints?
I have a complete undirected graph $$G=(V,E)$$ with positive non-null rational weights $$c:E \to \mathbb{Q}^+_{*}$$ on the edges, such that $$c(v,v) = 0$$ for all $$v$$, and a subset $$C \subset V$$.
I would like to find (if it exists) a subset $$S \subset V$$ such that $$\delta(S) < 2$$, $$S$$ contains at least one vertex of $$C$$, and $$S \ne C$$. Here $$\delta(S)$$ represents the sum of the weights of the edges that cross the cut $$(S, V \setminus S)$$.
Is there a polynomial-time algorithm for this problem?
I have thought of two possible approaches:
1. Create a new graph $$G'$$ using as edge weights $$c'(v,w) = -c(v,w)$$ and adapt (by updating some LP formulation or DP approach) the max-cut optimization version to this case;
2. Reduce this problem in the problem of partitioning $$V$$ into two vertex sets $$X,Y$$ so that the sum of the weights of the edges crossing the cut is < 2, and $$X$$ and $$Y$$ each contain at least one vertex of $$C$$. However, I couldn't think of any approach to solve this problem, yet.
I know the max-cut problem is hard, which makes me wonder whether there is a polynomial-time algorithm for this problem.
Remark: The proposed solution does not need to follow the presented approaches, they are presented just to help the reader.
I am facing this problem in a fractional separation routine of a routing MILP formulation.
• Why do you think of max cut? Is min cut more intuitive? – xskxzr Sep 27 '20 at 10:54
• @xskxzr well, I've tried the min-cut approach, however, there are cases in which it fails. Regarding the max-cut, note that as -c'(u,v) increases, the c(u,v) decreases. – Matheus Diógenes Andrade Sep 27 '20 at 19:24
The problem can be solved in polynomial time. Here is one algorithm:
• For each $$s \in C$$ and each $$t \in V \setminus C$$ such that $$s \neq t$$:
• Find the minimum-cost $$(s,t)$$-cut. By the max-flow min-cut theorem, this can be done in polynomial time using any maximum flow algorithm.
• If the cost of this cut is < 2, output it and halt.
• For each $$s_0 \in C$$ and each $$s_1 \in V \setminus C$$ and each $$t \in V \setminus \{s_0,s_1\}$$:
• Find the minimum-cost cut that contains both $$s_0$$ and $$s_1$$ in the left part and $$t$$ in the right part. This can be done in polynomial time by modifying the graph to merge the two vertices $$s_0,s_1$$ into a new vertex $$s'$$, then finding a minimum $$(s',t)$$-cut in the modified graph.
• If the cost of this cut is < 2, output it (replacing $$s'$$ with $$s_0,s_1$$) and halt.
• If you reach this point without halting, output that no such cut exists.
There may be more efficient algorithms by modifying algorithms for min-cut instead of min $$(s,t)$$-cut. I don't know. However, this suffices to show that the problem can be solved in polynomial time.
Proof of correctness: Suppose a cut $$(S,V \setminus S)$$ of the desired form exists. Then there are only two ways we can have $$S \ne C$$:
• Case 1: $$C \setminus S \ne \emptyset$$: In this case, pick any vertex in $$C \cap S$$ and call it $$s^*$$ (this can be done since $$C \cap S \ne \emptyset$$) and pick any vertex in $$C \setminus S$$ and call it $$t^*$$ (this can be done since $$C \setminus S \ne \emptyset$$). Consider the minimum-cost $$(s^*,t^*)$$-cut, $$(S^*,T^*)$$. By construction, $$S^*$$ has at least one element of $$C$$ (since $$s^* \in S^*$$) and $$S^* \ne C$$ (since $$t^* \notin S^*$$) and $$\delta(S^*) \le \delta(S) < 2$$, so $$(S^*,T^*)$$ is a valid solution to your problem. Moreover, it will be found by one of the iterations of the first for-loop, namely, when $$s=s^*$$ and $$t=t^*$$, so the algorithm will correctly find a solution.
• Case 2: $$S \setminus C \ne \emptyset$$: In this case pick any vertex in $$S \setminus C$$ and call it $$s^*_1$$ (this can be done since $$S \setminus C \ne \emptyset$$), pick any vertex in $$S \cap C$$ and call it $$s^*_0$$ (this can be done since $$S \cap C \ne \emptyset$$), and pick any vertex in $$V \setminus S$$ and call it $$t^*$$ (this can be done since $$S \ne V$$). Consider the minimum-cost cut $$(S^*,T^*)$$ such that $$s^*_0 \in S^*$$, $$s^*_1 \in S^*$$, and $$t^* \in T^*$$. By construction, $$S^*$$ has at least one element of $$C$$ (since $$s^*_0 \in S^*$$) and $$S^* \ne C$$ (since $$s^*_1 \in S^*$$) and $$\delta(S^*) \le \delta(S) < 2$$, so $$(S^*,T^*)$$ is a valid solution to your problem. Moreover, it will be found by one of the iterations of the second for-loop, namely, when $$s_0=s^*_0$$ and $$s_1=s^*_1$$ and $$t=t^*$$, so the algorithm will correctly find a solution.
We see that in either case, if a solution exists, the algorithm will successfully output a valid solution; and those are the only two cases that can occur if a solution exists.
Conversely, if no valid solution exists, it is easy to see that the algorithm will correctly output that no cut exists. | 2021-03-07T18:09:33 | {
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https://math.stackexchange.com/questions/1552338/if-a-function-fx-is-riemann-integrable-on-a-b-and-fx-le-m-for-x | # If a function $f(x)$ is Riemann integrable on $[a,b]$ and $|f(x)| \le M$ for $x \in [a, b]$, show $|\int_a^b f(x)dx| \le M(b - a).$
I have that every Riemann sum $$S := |\sum_{i=1}^n f(t_i)(x_i - x_{i-1})| \le \sum_{i = 1}^n M(x_i - x_{i-1}) = M(b-a),$$ where $t_i$ is an arbitrary $x \in (x_{i-1}, x_i)$, and $x_1 \le ... \le x_n$ is a partitioning of $[a, b].$
I'm not sure how to extend that to $|\int_a^b f(x)| \le M(b-a).$
Does anyone have any pointers?
Thanks!
• Hint: How are Riemann-sums and Riemann-integrals related? – sranthrop Nov 29 '15 at 23:42
• @sranthrop I know that the Riemann sums are arbitrarily close to the integral, using the epsilon-delta definition, but I'm not sure how to formally state that if all of the sums are less than some constant C, than the integral is also less than C. – mxdg Nov 29 '15 at 23:50
• One device to remember in most situations. If I wish to prove that $|\int_a^b f(x)\,dx| \leq M(b-a)$ I could do that by proving instead that, for any $\epsilon>0$, this inequality $$\left|\int_a^b f(x)\,dx\right | \leq M(b-a) + \epsilon$$ must be true. – B. S. Thomson Nov 29 '15 at 23:54
• @B.S.Thomson I'm not sure how proving the second part shows the first part is true. I see how I could go from the statement without epsilon to the statement with epsilon. – mxdg Nov 30 '15 at 0:10
• Then your life is a lot easier. The functions $g_1(x)=-M$ and $g_2(x)=M$ are integrable and so $$\int_a^b (-M)dx \leq \int_a^b f(x)\, dx \leq \int_a^b M \,dx$$ and you just have to remember how to integrate constant functions. – B. S. Thomson Nov 30 '15 at 2:07
For any partition $\mathcal{P} = \{a = t_0 < t_1 < \cdots < t_n = b\}$ of $[a,b]$ we know that $$\int_a^b f \le U(f, \mathcal{P})$$ where $$U(f, \mathcal{P}) = \sum_{i=0}^n \sup_{[t_{i-1}, t_i]} f(x) (t_i - t_{i-1})$$ is the upper Riemann sum of $f$ with the partition $\mathcal{P}$. Furthermore, as you state in your question, you know that $$\left|U(f, \mathcal{P})\right| \le M(b-a).$$ Can you see how to combine these facts to obtain the result?
• Is there a source/proof that I can take a look at for the first part? That's the part that I've been trying to find in my textbook but I haven't been able to find. – mxdg Nov 30 '15 at 0:42
• If your textbook just uses an $\epsilon$, $\delta$ version of the integral then you would have to study this upper/lower version as well to use it. Alternatively carry on in the direction you were pursuing. For any $\epsilon>0$ you can find a partition with $$\left| \int_a^b f - \sum_{i=1}^n f(t_i)(x_i - x_{i-1})\right| < \epsilon$$ and deduce that $$\left| \int_a^b f \right| <\left| \sum_{i=1}^n f(t_i)(x_i - x_{i-1})\right| + \epsilon.$$ – B. S. Thomson Nov 30 '15 at 0:47
Hint: If you think about it, this comes down to prove that $$\left|\int_a^b f(x) \mathrm{d}x\right|\leq \int_a^b |f(x)|\mathrm{d}x.$$
To help you, let us prove that $|f|:x\mapsto |f(x)|$ is integrable. Given a function $g:[a,b]\rightarrow \mathbb{R}$, we have that $g$ is integrable iff $\forall \varepsilon > 0$, there exists a partition $\mathcal{P}_n=\{x_0,...,x_n\}\in[a,b], x_0 = a, x_n=b,$ such that
$$\sum_{i=1}^n \omega_i(g)(x_i-x_{i-1})<\varepsilon,$$
where $\omega_i(g) = \sup g([x_{i-1},x_i]) - \inf g([x_{i-1},x_i])= \sup\{|g(x)-g(y)|,x,y\in[x_{i-1},x_i]\}$.
Since $$||f(x)|-|f(y)||\leq |f(x)-f(y)|,$$ we can conclude that for every partition $\mathcal{P}_n$, $\omega_i(|f|)\leq\omega_i(f)$. Thus, $|f|$ is integrable.
Furthermore, $-|f(x)|\leq f(x)\leq |f(x)|$, and therefore
\begin{align}&&-\int_a^b |f(x)|\mathrm{d}x\leq \int_a^b f(x)\mathrm{d}x\leq \int_a^b |f(x)|\mathrm{d}x&\\ \Leftrightarrow &&\left|\int_a^b f(x)\mathrm{d}x\right|\leq \int_a^b |f(x)|\mathrm{d}x.&\end{align}
• "...which gives us the inequality we wanted." Normally yes, but maybe our student doesn't yet yet know how to prove the inequality property for integrals. I'm finding it hard to answer questions posed on this site since the students are often unable to explain what facts are available to them. – B. S. Thomson Nov 30 '15 at 1:12 | 2019-10-17T18:50:16 | {
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https://physics.stackexchange.com/questions/252942/rotation-of-a-vector | # Rotation of a vector
Is a vector necessarily changed when it is rotated through an angle?
I think a vector always gets changed because its projection will change, and also its inclination with axes will always change. However the direction may remain same. Kindly make things clear to me.
• If the rotation is around an axis defined by the vector itself, then there will be no change. If the rotation is a multiple of $2\pi$ then there will be no change. – Jon Custer Apr 29 '16 at 17:41
• @JonCuster : Probably worth posting as an Answer? – sammy gerbil Apr 29 '16 at 20:30
• With respect to a fixed system of coordinate axes, the position vector of a point in a rotating body changes. – Procyon Apr 29 '16 at 22:23
Rotation of a 3-vector
We'll find an expression for the rotation of a vector $\mathbf{r}=(x_1,x_2,x_3)$ around an axis with unit vector $\mathbf{n}=(n_1,n_2,n_3)$ through an angle $\theta$, as shown in Figure .
The vector $\mathbf{r}$ is analysed in two components $$\mathbf{r}=\mathbf{r}_\|+\mathbf{r}_\bot \tag{01}$$ one parallel and the other normal to axis $\mathbf{n}$ respectively \begin{eqnarray} &\mathbf{r}_\| &=(\mathbf{n}\boldsymbol{\cdot}\mathbf{r})\mathbf{n} \tag{02a}\\ &\mathbf{r}_\bot &=(\mathbf{n}\times\mathbf{r})\times \mathbf{n}= \mathbf{r}-(\mathbf{n}\boldsymbol{\cdot}\mathbf{r})\mathbf{n} \tag{02b} \end{eqnarray} If $\mathbf{r}$ is rotated to $\mathbf{r}^{\prime}$ $$\mathbf{r}^{\prime}=\mathbf{r}^{\prime}_\|+\mathbf{r}^{\prime}_\bot \tag{03}$$ then the parallel component remains unchanged $$\mathbf{r}^{\prime}_\|=\mathbf{r}_\| =(\mathbf{n}\boldsymbol{\cdot}\mathbf{r})\mathbf{n} \tag{04}$$ while the normal component $\mathbf{r}_\bot =(\mathbf{n}\times\mathbf{r})\times \mathbf{n}$ is rotated by the angle $\theta$, so having in mind that this vector is perpendicular to $\mathbf{n}\times\mathbf{r}$ and of equal norm $$\left\|(\mathbf{n}\times\mathbf{r})\times \mathbf{n}\right\|=\left\|\mathbf{n}\times\mathbf{r}\right\| \tag{05}$$ we find the expression, see Figure below \begin{eqnarray} \mathbf{r}^{\prime}_\bot &=& \cos\theta\left[(\mathbf{n}\times\mathbf{r})\times \mathbf{n}\right]+\sin\theta\left[\mathbf{n}\times\mathbf{r}\right]\nonumber\\ &=& \cos\theta\left[\mathbf{r}-(\mathbf{n}\boldsymbol{\cdot}\mathbf{r})\mathbf{n}\right]+\sin\theta\left[\mathbf{n}\times\mathbf{r}\right]\nonumber\\ &=& \cos\theta\;\mathbf{r}-\cos\theta(\mathbf{n}\boldsymbol{\cdot}\mathbf{r})\mathbf{n}+\sin\theta\left[\mathbf{n}\times\mathbf{r}\right] \tag{06} \end{eqnarray}
and so finally the vector expression
$$\bbox[#FFFF88,12px]{\mathbf{r}^{\prime}= \cos\theta \cdot\mathbf{r}+(1-\cos\theta)\cdot(\mathbf{n}\boldsymbol{\cdot}\mathbf{r})\cdot\mathbf{n}+\sin\theta\cdot(\mathbf{n}\times\mathbf{r})} \tag{07}$$
From this the $3\times3$ rotation matrix reads $$\mathbb{A}\left(\mathbf{n}, \theta\right) = \text { 3D-rotation around axis} \:\:\mathbf{n}=\left(n_{1}, n_{2},n_{3}\right)\:\: \text{through angle} \:\:\theta$$ $$= \bbox[#FFFF88,12px]{ \begin{bmatrix} \cos\theta+(1-\cos\theta)n_1^2&(1-\cos\theta)n_1n_2-\sin\theta n_3&(1-\cos\theta)n_1n_3+\sin\theta n_2\\ (1-\cos\theta)n_2n_1+\sin\theta n_3&\cos\theta+(1-\cos\theta)n_2^2&(1-\cos\theta)n_2n_3-\sin\theta n_1\\ (1-\cos\theta)n_3n_1-\sin\theta n_2&(1-\cos\theta)n_3n_2+\sin\theta n_1&\cos\theta+(1-\cos\theta)n_3^2 \end{bmatrix}} \tag{08}$$
In general it changes although the reason is not exactly because its projections changes.
For example. You start with a vector (let us say the electric field of a parallel plate capacitor) on the plane $xy$. Then you rotate the coordinate system by an angle. The components of the vector on the new coordinate system is changed. But the vector did not change at all (you did not move the capacitor). This is called a passive rotation.
On the other hand, if you keep the axis fixed and rotate the vector (rotate the actual capacitor), it changed (unless you rotate by $2\pi$). This is an active rotation.
• Can you elaborate the active rotation? Passive rotation is totally clear to me but in passive rotation how exactly did the vector changed. – Hisenberg Apr 29 '16 at 18:20
• In the passive rotation the vector does not change. Only its components do. Think of a vector as a physical quantity regardless of your definition of coordinate system. It may be an electric field as above mentioned, the velocity of a car, the oriented displacement from Paris to Rome and so on. An active rotation means a rotation of the vector itself (not the coordinate system). In these examples it would mean to rotate the capacitor, to change the direction of the car, to change the oriented displacement from Paris to Rome to Paris to Madrid. – Diracology Apr 29 '16 at 18:30
• sorry I made a typo I want you to explain how exactly did the vector changed in active rotation. Passive rotation is totally clear to me. – Hisenberg Apr 29 '16 at 18:34
Direction of a vector is determined by the components themselves. Now if the components are changed the direction gets changed by the above definition. All this is with respect to one reference frame.
• Not if the components are all scaled by the same amount. The component change but the direction is does not. – ja72 Apr 29 '16 at 18:34
• @ja72 I quite didn't understand what you meant by "components are all scaled by the same amount". – Hisenberg Apr 29 '16 at 18:43
• The direction of $(1,2,3)$ is the same as the direction of $(2,4,6)$ where all the components are scaled by 2. – ja72 Apr 29 '16 at 19:07
Typically there are two kinds of transformation that do not change the outcome of situation. Think of a force vector $\vec{F}$ passing through a point $\vec{r}_A$.
• Any translation along the line of the force, in the direction $\vec{e} = \frac{\vec{F}}{\| \vec{F} \|}$ will not change the outcome.
• Any rotation about the line of the force would also not change the outcome.
• The only things that make a difference is translations perpendicular to the line, and hence the cross product when looking at torques $\vec{\tau} = \vec{r} \times \vec{F}$.
• And rotations perpendicular to the line change things.
## protected by Qmechanic♦May 1 '16 at 21:25
Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count). | 2019-10-20T03:32:17 | {
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https://www.physicsforums.com/threads/isolated-continuity-point.767526/ | # Isolated continuity point
1. Aug 25, 2014
### glance
Hi
The question is the following: is it possible for a (say) real function to be continuous at a certain point internal to its domain, and be discontinuous in some neighborhood of that point?
I am not talking about a function defined at a single point or things like that, but of a function defined on the entire $\mathbb{R}$ (or some interval in it, whatever).
Now, i have also came up with an answer: a function $f$ defined as $f(0)=0$, $f(x)=x$ for every rational $x$, and $f(x) = 2x$ for every irrational $x$. Such a function would be (seems to me) continuous at $x=0$ and discontinuous for any other $x$. I am not completely certain of this, though, and for that reason i would like some feedback on this.
I am also asking this question because strangely enough I have never heard of the concept of an isolated continuity point, while for example the "opposite" (that of an isolated singularity) is quite common, and I would like to know if it's just me or if it is just a "useless" pathological concept.
Thank you in advance.
Bye
2. Aug 25, 2014
### gopher_p
Yes.
In fact, there is a function defined on all of $\mathbb{R}$ which is continuous at a single point.
3. Aug 25, 2014
### micromass
Yes, that is a correct example.
Now, can you come up with an example of a function defined on entire $\mathbb{R}$ that is differentiable only in one point? :tongue:
4. Aug 25, 2014
### pasmith
If $f : \mathbb{R} \to \mathbb{R}$ and $g : \mathbb{R} \to \mathbb{R}$ are continuous such that there exists a unique $a \in \mathbb{R}$ such that $f(a) = g(a)$, then the function $$h : \mathbb{R} \to \mathbb{R} : x \mapsto \begin{cases} f(x) & x \in \mathbb{Q} \\ g(x) & x \in \mathbb{R} \setminus \mathbb{Q} \end{cases}$$ is discontinuous on $\mathbb{R} \setminus \{a\}$ and continuous at $a$.
5. Aug 26, 2014
### glance
That is a very interesting example, thank you.
That seems to be tricky! I did some research and stumbled upon this discussion of that matter, in which that question is very well explained.
Now however I wonder if it is accidental that in all of these examples the functions are constructed using rational and irrational numbers. I think that the important point is to have one subset which is dense in the other. Is there some example of functions having this kind of "pathologies" NOT using rational/irrational numbers in the definition?
Even better, is it possible to find a function of this kind NOT using at all dense subsets of the real numbers in the definition?
Thanks
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http://mathhelpforum.com/advanced-algebra/7281-operations-i-print.html | # Operations I
• Nov 7th 2006, 10:33 AM
kgpretty
Operations I
I have a question for which I need help:
Let M consist of all 2*2 matrices of the form
Code:
[a b] [-b a]
Show that the product of the two matrices in M is a matrix in M.
Here's my attempt:
Code:
[a b][a b] = [a^2-b^2 ab+ab] [-b a][-b a] [-ab-ab -b^2+a^2] = [a^2-b^2 2ab] [-2ab -b^2+a^2]
.: the product of two matrices in M is a matrix is M. Simple enough!
However, the question goes on to say:
Let * denote the binary operation of matrix multiplication on M. Show that I is the identity element of M under *, and find the inverse of any arbitrary matrix of M.
Find two distinct elements x E M which satisfy x * x * x * x = I.
• Nov 7th 2006, 10:49 AM
ThePerfectHacker
Quote:
Originally Posted by kgpretty
Let * denote the binary operation of matrix multiplication on M. Show that I is the identity element of M under *, and find the inverse of any arbitrary matrix of M.
Find two distinct elements x E M which satisfy x * x * x * x = I.
If you drop the zero matrix then the determinant is $a^2+b^2>0$ hence all matrices are invertible.
Which means you have a group under matrix multiplication.
What you need is $x\in M$ such as,
$x^4=1$
Hence you need an element in $M$ whose order divides 4, that is 1,2,4.
The first case is simple.
The only element of order 1 is the identity element which satisfies that equation. Now just find an element of order two (and it will be different from the indentity element because it has different order).
That means,
Solve,
$\left[ \begin{array}{cc}a&b\\-b&a\end{array} \right]\cdot \left[ \begin{array}{cc}a&b\\-b&a\end{array} \right]=\left[\begin{array}{cc}1&0\\0&1\end{array}\right]$
$a^2-b^2=1$
$ab+ba=0$
$-ab-ba=0$
$a^2-b^2=1$
One such possibility is $a=-1,b=0$
Hence,
$\left[ \begin{array}{cc}-1&0\\0&-1\end{array}\right]$
(Remember to check because these are necessary though not suffienct conditions).
• Nov 7th 2006, 11:57 AM
topsquark
Quote:
Originally Posted by kgpretty
I have a question for which I need help:
Let M consist of all 2*2 matrices of the form
Code:
[a b] [-b a]
Show that the product of the two matrices in M is a matrix in M.
Here's my attempt:
Code:
[a b][a b] = [a^2-b^2 ab+ab] [-b a][-b a] [-ab-ab -b^2+a^2] = [a^2-b^2 2ab] [-2ab -b^2+a^2]
.: the product of two matrices in M is a matrix is M. Simple enough!
I should note that in your first problem you are merely showing, given a matrix $A \in M$, that $A^2 \in M$. What you need to do is define two matrices $A,B \in M$ and show that $AB \in M$ where A and B are not necessarily the same.
The proof is nearly identical to what you have done above.
-Dan
• Nov 7th 2006, 12:35 PM
Soroban
Hello, kgpretty!
Quote:
Let $M$ consist of all 2x2 matrices of the form: . $\begin{bmatrix}a & b \\ \text{-}b & a\end{bmatrix}$
Show that the product of the two matrices in $M$ is a matrix in $M$
You squared a matrix . . . We should use any two matrices.
$\begin{bmatrix}a & b \\ \text{-}b & a\end{bmatrix} * \begin{bmatrix}c & d \\ \text{-}d & c\end{bmatrix}\;=\;\begin{bmatrix}ac-bd & ad+bc \\ \text{-}bc -ad & \text{-}bd + ac\end{bmatrix} \;=\;\begin{bmatrix}ac-bd & ad + bc \\ \text{-}(ad + bc) & ac-bd\end{bmatrix}$
This product is of the proper form for $M.$
. . The two elements on the major diagonal are equal.
. . The two elements on the minor diagonal are equal with opposite signs.
Quote:
Let * denote the binary operation of matrix multiplication on $M$.
Show that $I$ is the identity element of $M$ under *.
$A*I \;= \;\begin{bmatrix}a & b \\ \text{-}b & a\end{bmatrix}*\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} \;=\;\begin{bmatrix}a\!\cdot1\!+b\!\cdot\!0 & a\!\cdot\!0+b\!\cdot\!1\\ \text{-}b\!\cdot\!1+a\!\cdot\!0 & \text{-}b\!\cdot\!0 + a\!\cdot\!1\end{bmatrix}\;=\;\begin{bmatrix}a & b \\\text{-}b & a\end{bmatrix} \;= \;A$
Quote:
Find the inverse of any arbitrary matrix of $M.$
Let $A \,=\,\begin{bmatrix}a & b \\ \text{-}b & a\end{bmatrix}$, and use your favorite method.
I happen to know that the inverse of $\begin{bmatrix}a & b \\ c & d\end{bmatrix}$ is: . $\begin{bmatrix}\frac{d}{ad-bc} & \frac{-b}{ad-bc} \\ \frac{-c}{ad-bc} & \frac{a}{ad-bc}\end{bmatrix}$
Therefore: . $A^{-1} \;= \;\begin{bmatrix}\frac{a}{a^2+b^2} & \frac{-b}{a^2+b^2} \\ \frac{b}{a^2+b^2} & \frac{a}{a^2+b^2}\end{bmatrix}$
Quote:
Find two distinct elements $x \in M$ which satisfy $x * x * x * x \:= \:I$
Let $x \:=\:\begin{bmatrix}a & b \\ \text{-}b & a\end{bmatrix}$
Then: . $x*x\:=\:\begin{bmatrix}a^2-b^2 & 2ab \\ \text{-}2ab & a^2-b^2\end{bmatrix}$
And: . $(x*x)*(x*x)\:=\:\begin{bmatrix}a^2-b^2 & 2ab \\ \text{-}2ab & a^2-b^2\end{bmatrix}*\begin{bmatrix}a^2-b^2 & 2ab \\ \text{-}2ab & a^2-b^2\end{bmatrix}$
. . . . . . . . . . . . . . . $= \:\begin{bmatrix}(a^2-b^2)^2 - 4a^2b^2 & 4ab(a^2-b^2) \\ \text{-}4ab(a^2-b^2) & (a^2-b^2)^2 - 4a^2b^2\end{bmatrix}$
Since: . $\begin{bmatrix}(a^2-b^2)^2 - 4a^2b^2 & 4ab(a^2-b^2) \\ \text{-}4ab(a^2-b^2) & (a^2-b^2)^2 - 4a^2b^2\end{bmatrix} \;=\;\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$
. . we have: . $\begin{array}{cc}(A)\\(B)\end{array} \begin{array}{cc}(a^2-b^2)^2 - 4a^2b^2\:=\:1 \\ 4ab(a^2-b^2)\:=\:0\end{array}$
Equation (B) gives us three possible solutions:. $\left\{\begin{array}{ccc}(1)\;a\;= & 0 \\ (2)\;b\;= & 0 \\ (3)\;a\;= & \pm b\end{array}\right\}$
(1) If $a = 0$, (A) becomes: . $b^4 = 1\quad\Rightarrow\quad b = \pm1$
(2) If $b = 0$, (A) becomes: . $a^4 = 1\quad\Rightarrow\quad a = \pm1$
(3) If $a = \pm b$, (A) becomes: . $-4b^4 = 1$ ... which has no real roots.
There are four solutions: . $x\;= \;\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\;\;\begin{bmatrix}\text{-}1 & 0 \\ 0 & \text{-}1\end{bmatrix}\;\;\begin{bmatrix}0 & 1 \\ \text{-}1 & 0\end{bmatrix}\;\;\begin{bmatrix}0 & \text{-}1 \\ 1 & 0\end{bmatrix}$ | 2017-12-16T19:14:10 | {
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https://math.stackexchange.com/questions/1752913/vertex-of-the-graph-of-a-quadratic-polynomial | # Vertex of the graph of a quadratic polynomial
This is what a website states:
Before graphing a quadratic function we rearrange the equation, from this:
$f(x) = ax^2 + bx + c$
To this:
$f(x) = a(x-h)^2 + k$
Where:
$h = -b/2a$
$k = f( h )$
In other words, calculate h$(=-b/2a)$, then find $k$ by calculating the whole equation for $x=h$
The website states the following reason for doing this:
Well, the wonderful thing about this new form is that $h$ and $k$ show us the very lowest (or very highest) point, called the vertex:
And also the curve is symmetrical (mirror image) about the axis that passes through $x=h$, making it easy to graph
I want to know that how is $h$ the x - coordinate and $k$ the y - coordinate of the vertex?
The square of any real number is always at least zero, and equals zero only when that number is itself zero. So $(x-h)^2$ is always at least zero, and is equal to zero only when $x-h=0$; that happens only when $x=h$.
Since $(x-h)^2$ is always at least zero, we see that any positive multiple of it is also always at least zero. So, if we are told that $a>0$, we can say with certainty that $a(x-h)^2$ is always at least zero and is equal to zero only when $x=h$.
Finally, we see that $a(x-h)^2 +k$ is always at least $k$ and is equal to $k$ only when $x=h$. That's because we are adding to $k$ a number that is always at least zero (so, the sum can't be any less than $k$).
Clearly $(h,k)$ is a point on the graph of the function $f(x) = a(x-h)^2+k$. Any other point $(x,y)$ on the graph has an $x$-coordinate that is not $h$, so the $y$-coordinate must be greater than $k$. This means that if we move left or right from $(h,k)$, staying on the graph, we can only move up.
Done.
(The case $a<0$ is handled similarly -- try it yourself! What happens if $a=0$?)
Addendum: How to find $h$ and $k$
This is just completing the square to rewrite the original expression. Start with:
$$f(x) = ax^2 + bx + c$$
Now suppose there is and $h$ and a $k$ so that this is the same as
$$f(x) = a(x-h)^2 + k$$
Let's write this out in full : $$a(x-h)^2 + k=a(x^2-2hx + h^2) + k = ax^2-2ahx + ah^2+k$$ Now if these are to be the same function for all $x$, then $$ax^2 + bx + c = ax^2-2ahx + ah^2+k$$ Move everything to one side (note the $ax^2$ terms add out): $$(\underbrace{b+2ah}_{\textrm{constant}})x + (\underbrace{c-ah^2-k}_{\textrm{constant}})=0$$
This has to be true for all $x$. But this is just a linear function of $x$, and the only linear function which is zero for all $x$ is the linear function whose coefficients are all zero! This means we can write $$\begin{cases}b+2ah = 0\\c-ah^2-k = 0\end{cases}$$ So solve the first equation for $h$ to get $$\boxed{h=-\tfrac b{2a}}$$ Now you could plug this into the second equation and solve for $k$, but it's probably easier to note that, since we assumed $$f(x) = a(x-h)^2+k$$ then just putting $x=h$ into this we get $$\boxed{f(h)} = a(h-h)^2+k=a\cdot 0^2 + k = \boxed{k}$$ In other words, get $k$ by using the fact that $$\boxed{k=f(h)}$$ (and we know $h$ at this point).
• I got your point that h and k would be coordinates of vertex. But I still didn't understood how we reached the conclusion that h = -b/2a and k = f(h). – Parth Apr 21 '16 at 17:11
• Okay, I added a section in my answer to show how you know this. – MPW Apr 21 '16 at 20:25
• How is b - 2ah = 0? Shouldn't it be b + 2ah? – Parth Apr 22 '16 at 0:49
• @unknownCoder: Yes, typo. Will fix. Thanks for the catch, +1. – MPW Apr 22 '16 at 12:48
• What is the need of including an a in a(x - h)^2 + k? – Parth Apr 22 '16 at 15:13
The quantity $(x-h)^2$ is always non negative, and the smallest value it can attain is zero when $x=h$. Now we have two cases:
1) $a>0$: in this case the quantity $a(x-h)^2$ is always non negative and has zero as its smallest value when $x=h$.
2) $a<0$: in this case the quantity $a(x-h)^2$ is always non positive and has zero as its largest value when $x=h$.
In both cases $x=h$ represents a global minimum or maximum of the function; this property doesn't change if you consider the "shifted" quantity $a(x-h)^2+k$, because adding a constant to a function does not change the position of its maxima or minima.
So the x-coordinate of the vertex is at $x=h$.
(1) All graphs of quadratic functions are symmetric about some vertical axis. $y = f(x) = ax^2 + bx + c$ (where we assume that $a > 0$ for simplicity) is of no exception.
(2) The vertex $V(h, k)$ will lie on that axis.
(3) Let us assume that $f(x) = 0$ has roots $\alpha$ and $\beta$. Then, the graph of y = f(x) will cut the x-axis at $A( \alpha, 0)$ and $B( \beta, 0)$. The point $( \dfrac {\alpha + \beta}{2}, 0)$ is midway between A and B. Then, the vertical axis (of symmetry) has the equation $L : x = \dfrac {\alpha + \beta}{2}$.
(4) Recall that $\alpha + \beta =$ the sum of roots $= \dfrac {–b}{a}$. Therefore, equation of L is then $L : x = \dfrac {–b}{2a}$.
(5) According to (2), $V = (h, k) = (\dfrac {–b}{2a}, k)$; where k is given by $f(\dfrac {–b}{2a})$ or $f(h)$ because $V$ is a point on the curve of $y = f(x)$.
In the case that $f(x) = 0$ has no real roots, (ie. the graph of $y = f(x)$ does not cut the x-axis), we can translate the x-axis upward certain suitable units such that the new X-axis will cut $y = f(x)$. The above argument still applies. | 2019-07-21T15:33:09 | {
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https://math.stackexchange.com/questions/1031783/induction-proof-for-fibonacci-numbers | # Induction proof for Fibonacci numbers
I am trying to get a hang on the induction method for proof, but I'm still dubious of many aspect of this proof regarding its application to sequences of integers, such as the Fibonacci sequence.
QUESTION:
Prove by induction that this is true. (These are terms in the Fibonacci Sequence)
$F_{n+3} = 2F_{n+1}+F_2F_n$
Fibonacci Numbers:
$1, 1, 2, 3, 5, 8, 13, 21… =$ $F_1,\; F_2, \; F_3, \; F_4, \; F_5, \; F_6, F_7, \; F_8 \ldots$ So, the first step I underwent was preforming some base cases:
.
For $n=1:$ $F_{(1)+3} = 2F_{(1)+1}+F_2F_{(1)}$
=> It should be true that $F_4 = 2F_2+F_2F_1$
By replacing the F's with the appropriate terms we get
For $n=1$: $3 = 2(1) + (1)(1) = 3$, so $3=3$ is true.
.
For $n=3$: $F_{(3)+3} = 2F_{(3)+1}+F_2F_{(3)}$
=> It should be true that $F_6 = 2F_4+F_2F_3$
By replacing the F's with the appropriate terms we get
For $n=3$: $8 = 2(3) + (1)(2) = 8$, so $8=8$ is true.
.
Inductive Hypothesis:
For $n \leq k$ it should hold that $F_{n+3} = 2F_{n+1}+F_2F_n = F_{k+3} = 2F_{k+1}+F_2F_k$
For $n \leq k+2$ it should hold that $F_{n+3} = 2F_{n+1}+F_2F_n = F_{(k+2)+3} = 2F_{(k+2)+1}+F_2F_{k+2}$
For $n \leq k+3$ it should hold that $F_{n+3} = 2F_{n+1}+F_2F_n = F_{(k+3)+3} = 2F_{(k+3)+1}+F_2F_{k+3}$
Inductive Step:
For the Inductive step we consider $k+1$. If the Inductive hypothesis is to hold, we must show that $F_{(k+1)+3} = 2F_{(k+1)+1}+F_2F_{(k+1)}$.
From here I don't know how to proceed. I don't even know if my process up to this point is correct. Looking for aid in my understanding. Thank you in advance for any helpful insights.
So, I know that from my inductive hypothesis,
$F_{k+3} = 2F_{k+1}+F_2F_k$
And in the inductive step I would like to have
$F_{k+4} = 2F_{k+2}+F_2F_{k+1}$,
• Thank you for the edit, wasn't sure about how to do it. – Ian Hoyos Nov 21 '14 at 2:22
• That's fine. To learn you can have a look at meta.math.stackexchange.com/questions/5020/… – John Marty Nov 21 '14 at 2:24
• @John Marty: It looks like the left of the first equation should be $F_{n+3}$, not $F_n+3$ I don't think it was clear in the original post, but it comes out later. You can get that by putting braces around the subscript, so F_{n+3} – Ross Millikan Nov 21 '14 at 2:24
• Just edited again – Ian Hoyos Nov 21 '14 at 2:28
Hint. Write down what you know about $F_{k+2}$ and $F_{k + 3}$ by the induction hypothesis, and what you are trying to prove about $F_{k+4}$. Then recall that $F_{k+4} = F_{k+3} + F_{k+2}$. You'll probably see what you need to do at that point.
• Can you edit your question with what you know from the induction hypothesis for $F_{k+3}$ and $F_{k+2}$? Use dollar signs for math. – Mike Nov 21 '14 at 2:37
• Just edited the question with what I know for $F_{k+4}$. Could $F_{k+2}$ be re-written as $F_{k+1}+F_k$? – Ian Hoyos Nov 21 '14 at 2:44
• I'm not sure how you're getting the last line. But listen, you haven't written the induction hypothesis for $F_{k+2}$ as I suggested. Also, you write "I have" in one place when what you mean is probably "I would like to have." – Mike Nov 21 '14 at 2:52 | 2020-08-06T13:37:55 | {
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https://math.stackexchange.com/questions/2420474/is-the-proof-of-the-derivative-of-sinx-circular | # Is the proof of the derivative of $\sin(x)$ circular?
The proof of $\frac{d}{dx}\sin(x)$ goes something like this:
\begin{aligned} \lim_{h\to0}\frac{\sin(x+h)-\sin(x)}{h}=\lim_{h\to0}\frac{\sin(x)\cos(h)+\cos(x)\sin(h)-\sin(x)}{h}\\ =\lim_{h\to0}\frac{\sin(x)(\cos(h)-1)+\cos(x)\sin(h)}{h}\\ =\lim_{h\to0}\frac{\sin(x)(\cos(h)-1)}{h}+\cos(x)\lim_{h\to0}\frac{\sin(h)}{h}\\ =0+\cos(x)\times1\\ =\cos(x) \end{aligned}
My doubt about this is that it uses the limit of $\frac{\sin(h)}h$ and $\frac{\cos(h)-1}{h}$ in the proof. However, proving these two limits uses L'Hopital's rule, which uses the derivative of $\sin(x)$ and $\cos(x)$ to prove the limits. This causes a circular argument because we're using the derivative of $\sin(x)$ to prove the derivative of $\sin(x)$. Is there a way to prove these two limits without using L'Hopital's rule or just looking at the graph, or is there a way to find $\frac{d}{dx}\sin(x)$ without using these two limits?
There is a nice way to prove the limits using geometry here, but I'm wondering if there's a way to do it without using this, either.
Edit: The way I'm defining sine is by the unit circle definition, not the Taylor series one.
• I'm confused. You say that the limits use L'Hospital's rule, but then later you're aware of the geometric proofs that don't use any notions of derivatives. It's not circular for that reason.... – T. Bongers Sep 7 '17 at 17:07
• @user296602 I'm asking if there's an algebraic way to prove the limit without L'Hopital's rule. – u8y7541 Sep 7 '17 at 17:07
• Define the sine function purely algebraically, and we'll talk. (Note that if you define the sine function algebraically, you use Taylor series, and the limit is then very easy to prove.) – T. Bongers Sep 7 '17 at 17:08
• Yea use the Taylor series for Sine. Edit: The Taylor series can actually be found without taking the derivative by looking at the Taylor series for $e^{ix}$ – FofX Sep 7 '17 at 17:11
• @u8y7541 If you take the Taylor series as the definition of sine, then you have an analytical and algebraically useful definition (and this is how many authors do this). If you define it geometrically, it's natural that there's only a geometric proof. – T. Bongers Sep 7 '17 at 18:26
Here is a different approach that uses the integral definition of the arcsine function. We will deduce the limit of interest, $\lim_{h\to 0}\frac{\sin(h)}{h}=1$, without appeal to geometry or differential calculus. (Note that $\cos(h)-1=-2\sin^2(h/2)$)
Instead, we only rely on elementary analysis of continuous functions and their inverses along with simple properties of the Riemann integral. To that end, we now proceed.
We define the sine function, $\sin(x)$, as the inverse function of the function $f(x)$ given by
$$\bbox[5px,border:2px solid #C0A000]{f(x)=\int_0^x \frac{1}{\sqrt{1-t^2}}\,dt }\tag 1$$
for $|x|< 1$.
NOTE: It can be shown that the sine function defined as the inverse of $f(x)$ given in $(1)$ has all of the familiar properties that characterize the circular function $\sin(x)$.
It is straightforward to show that since $\frac{1}{\sqrt{1-t^2}}$ is positive and continuous for $t\in (-1,1)$, $f(x)$ is continuous and strictly increasing for $x\in (-1,1)$ with $\displaystyle\lim_{x\to 0}f(x)=f(0)=0$.
Therefore, since $f$ is continuous and strictly increasing, its inverse function, $\sin(x)$, exists and is also continuous and strictly increasing with $\displaystyle \lim_{x\to 0}\sin(x)=\sin(0)=0$.
From $(1)$, we have the bounds (SEE HERE)
$$\bbox[5px,border:2px solid #C0A000]{1 \le \frac{f(x)}x\le \frac{1}{\sqrt{1-x^2}}} \tag 2$$
for $x\in (-1,1)$, whence applying the squeeze theorem to $(2)$ yields
$$\lim_{x\to 0}\frac{f(x)}{x}=1 \tag 3$$
Finally, let $y=f(x)$ so that $x=\sin(y)$. As $x\to 0$, $y\to 0$ and we can write $(3)$ as
$$\lim_{y\to 0}\frac{y}{\sin(y)}=1$$
from which we have
$$\bbox[5px,border:2px solid #C0A000]{\lim_{y\to 0}\frac{\sin(y)}{y}=1}$$
as was to be shown!
NOTE:
We can deduce the following set of useful inequalities from $(2)$. We let $x=\sin(\theta)$ and restrict $x$ so that $x\in [0,1)$. In addition, we define new functions, $\cos(\theta)=\sqrt{1-\sin^2(\theta)}$ and $\tan(\theta)=\sin(\theta)/\cos(\theta)$.
Then, we have from $(2)$
$$\bbox[5px,border:2px solid #C0A000]{y\cos(y)\le \sin(y)\le y\le \tan(y)}$$
which are the familiar inequalities often introduced in an introductory geometry or trigonometry course.
This really comes down to how you define $\sin(x)$ and $\cos(x)$. As you say, there are geometric proofs that demonstrate the limits that you are asking about. You are correct that you can't use l'Hopital's rule to prove these limits.
A usual definition of $\sin(x)$ is through its Taylor series $$\sin(x) = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots.$$ From here, you can see that $$\frac{\sin(h)}{h} = \frac{h - \frac{h^3}{6} + \frac{h^5}{120} - \cdots}{h} = 1 - \frac{h^2}{6} + \frac{h^4}{120} - \cdots \to 1$$ as $h\to 0$.
Similarly, it can be demonstrated that $\frac{\cos(x)-1}{h} \to 0$ as $h \to 0$.
You do not need to know what $\sin(x)$ to make this Taylor series. In Advanced Calculus courses, $\sin(x)$ is defined using the Taylor series, and later the trigonometric properties of $\sin(x)$ are demonstrated using the Taylor series as a starting point.
• Moreover, Madhava of Sangamagrama discovered the series representations of sine, cosine, tangent and several other trigonometric functions more than a century before the invention of calculus. This demonstrates that the Taylor series for sine can be determined without knowing the derivatives of sine. – Joel Sep 7 '17 at 17:15
• The arctan series was found via Riemann sum approach and then using the geometric series for $1/(1+x^{2})$. Notable here was the use of formula for sums of powers of first $n$ natural numbers. So this all boils down to integrating (via Riemann sum) the series $1-x^{2}+x^{4}-\cdots$. – Paramanand Singh Sep 7 '17 at 17:21
• The series for since cosine was derived from that of arctan with some manipulation. – Paramanand Singh Sep 7 '17 at 17:22
• Anyway the geometric definition of these functions are based on integrals so there is really no need of differentiation to obtain their properties. – Paramanand Singh Sep 7 '17 at 17:23
• @ParamanandSingh Certainly, there are many ways to approach this problem. I like your idea of using integrals. – Joel Sep 7 '17 at 19:49
As mentioned in the comments above, it is all about your definition of $\sin$ and $\cos$. If you define it using Taylor series, then your proof probably ends up using that. If you use the "unit circle" to define the functions, then you proof will end up using some kind of geometry (since this is how you defined them).
As you note you don't have to use L'Hopital's rule to prove that $$\lim_{h\to 0}\frac{\sin(h)}{n} = 1\quad\text{and}\quad\lim_{h\to 0}\frac{\cos(h) - 1}{h} = 0$$ One can prove this limits from elementary geometric considerations.
It should be easy to find something using Google. I quickly found these notes.
This is too long for a comment.
The last sentence of the question gives the definition of sine being used here. To express the definition clearly, if one walks on the unit circle $x^{2}+y^{2}=1$ starting from point $A=(1, 0)$ in counter-clockwise direction and covers a distance $\theta$ (if $\theta$ is negative then the direction being clockwise) to reach point $P$ on the unit circle then the coordinates of this point $P$ are by definition $(\cos\theta, \sin\theta)$.
If the full circumference of unit circle is defined to be $2\pi$ then the length of a quadrant is $\pi/2$ and thus working only in the first quadrant we can see that that following formula holds $$\theta=\int_{0}^{\sin\theta}\frac{dt}{\sqrt{1 - t^{2}}}\tag{1}$$ for $0\leq\theta<\pi/2$. The above equation is the actual meaning of the geometric definition of sine function. Now we can argue like in user Mark Viola's answer. | 2018-12-10T13:18:32 | {
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https://math.stackexchange.com/questions/1660654/commutator-subgroup-and-bijective-representation | # Commutator subgroup and bijective representation
Let $G$ be a finite group and $G' = [G,G]$ be its commutator subgroup, which is defined to be the subgroup generated by elements $[g,h] = g^{-1}h^{-1}gh$ for all $g,h \in G$, where $G'$ is a normal subgroup of $G$. And $G/G'$ is abelian.
Prove that for any field $\mathbb{K}$, the degree $1$ representations of $G$ over $\mathbb{K}$ are in bijection with the degree $1$ representations of $G/G'$ over $\mathbb{K}$.
attempt: Suppose $\phi_1: G → GL(\mathbb{K})$ be a representation of degree 1. is similar to $\phi_1: G → \mathbb{K}^{\times}$.
And define $\phi_2 : G/G'→ \mathbb{K}^{\times}$
Do I have to show $\phi_1 → \phi_2$ and show it's a bijective function? I am not sure what I have to show. Can someone please help? Thank you!
• Isn't it just because $\Bbb K^{\times}$ is abelian, so the kernel of any homomorphism from $G$ to $\Bbb K^{\times}$ must contain the commutator subgroup? In other words you can always factor $G\to\Bbb K^{\times}$ through $G/G'$. That gives you the bijective map in one direction. And you can lift any homomorphism from $G/G'$ to $\Bbb K^{\times}$ to a homomorphism from $G$ to $\Bbb K^{\times}$. That gives you the other direction. Just show the composition is the identity, in both directions, to conclude it's a bijection. – Gregory Grant Feb 17 '16 at 23:27
• Could I define $\phi_2(gG') = \phi_1(g)$? And then show the composition is the identity ? – user40294 Feb 17 '16 at 23:40
• Yes, that gives you the way to take a map from $G/G'$ to $\Bbb K$ to get a map from $G$ to $\Bbb K$. You have to show it's well defined. – Gregory Grant Feb 18 '16 at 0:18
Show that for every abelian group $A$, $Hom(G,A)\cong Hom(G/G',A)$ (where $Hom(-,-)$ is the set of group homomorphisms). Since $\Bbb K^\times$ is abelian, you're done.
(I think it's important to see that for general $A$ since this is a fundamental property of $G/G'$).
• So I have to define $f : G → A$ , and $f_1 : G/G' → A$? And show it's a hormomorphism? Then show this is isomorphic? – user40294 Feb 17 '16 at 23:42
• Not quite. From a map $f:G\to A$, you have to construct a map $\bar{f}:G/G' \to A$ and vice-versa. Furthermore, you have to show that these constructions are inverse to each other. – Nitrogen Feb 17 '16 at 23:44 | 2021-03-04T11:53:46 | {
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http://math.stackexchange.com/questions/462562/need-a-formula-for-curve/462570 | # Need a formula for curve.
I need a function which give 0 for 0 value and 100 for 100 but is a highly curvy curve, like for 0.001 it should give value like 1.2 and for 50 it show give value like 55 and ultimately for 100 it should be 100.
In short it increases rapidly for smaller value of x, and slowly for large value of x. and ultimately same value as input.
-
Lots of curves can have this property. Here's a simple parameterized family of curves that do what you want:
$$y = 100 \left ( \frac{x}{100} \right )^n$$
Where $n$ is a parameter ranging from $0$ (exclusive) to $+ \infty$, and $x$ is your value, from $0$ to $100$.
For $n = 1$, you just get a straight line (least curvy). Otherwise, the closer $n$ is to $0$, the more "curvy" it is and increases quickly with small values of $x$, then slows down, and for $n > 1$ it's the opposite.
Note that "curviness" is not a linear relationship of $n$, but more an inverse relationship. For instance, $0.05$ is really curvy, but $1.05$ is almost flat - to obtain the same amount of curviness in both directions, you would use $\frac{1}{0.05} = 20$.
As you can see, it always starts at $0$ when $x = 0$, and ends at $100$ when $x = 100$.
Examples:
$$n = 0.1$$
$$n = 0.5$$
$$n = 1$$
$$n = 5$$
This can be generalized a lot, feel free to build on it if you need to tweak it.
-
It worked, thanks... – Sam Aug 8 '13 at 8:51
Thanks is fine @Sam, but upvoting is better and more direct. Also, after you've upvoted all the answers you've found helpful, it'd be nice if you also accept the one you consider the best for you. – DonAntonio Aug 8 '13 at 9:39
For "increases rapidly for smaller value of x, and slowly for large value of x" one possibility would be $$\sqrt{x(200-x)}$$ so the graph of the function looks like a quarter circle.
For $x=0.001$ this gives about $0.44$ and for $x=50$ it gives about $86.6$, but can easily be adjusted.
- | 2014-08-20T09:12:05 | {
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https://www.tutorialspoint.com/different-methods-to-find-prime-number-in-python-program | # Different Methods to find Prime Number in Python Program
PythonServer Side ProgrammingProgramming
In this tutorial, we are going to explore different methods to find whether a given number is valid or not. Let's start without further due.
## Method-1
It's a general method to find prime numbers.
• If the number is less than or equal to one, return False.
• If the number is divisible by any number, then the function will return False.
• After the loop, return True.
### Example
Live Demo
# checking for prime
def is_prime(n):
if n <= 1:
return False
else:
for i in range(2, n):
# checking for factor
if n % i == 0:
# return False
return False
# returning True
return True
print(f"Is 2 prime: {is_prime(2)}")
print(f"Is 4 prime: {is_prime(4)}")
print(f"Is 7 prime: {is_prime(7)}")
## Output
If you run the above code, then you will get the following result.
Is 2 prime: True
Is 4 prime: False
Is 7 prime: True
## Method-2
In this method, we are reducing the number of iterations by cutting them to the square root of n.Let's see the code.
### Example
Live Demo
import math
# checking for prime
def is_prime(n):
if n <= 1:
return False
else:
# iterating loop till square root of n
for i in range(2, int(math.sqrt(n)) + 1):
# checking for factor
if n % i == 0:
# return False
return False
# returning True
return True
print(f"Is 2 prime: {is_prime(2)}")
print(f"Is 4 prime: {is_prime(4)}")
print(f"Is 7 prime: {is_prime(7)}")
### Output
If you run the above code, then you will get the following result.
Is 2 prime: True
Is 4 prime: False
Is 7 prime: True
### Method-3
In the previous method, we have checked for the even numbers. We all know that even numbers can't be prime except two. So, in this method, we will remove all evens to reduce the time.
### Example
Live Demo
import math
# checking for prime
def is_prime(n):
# checking for less than 1
if n <= 1:
return False
# checking for 2
elif n == 2:
return True
elif n > 2 and n % 2 == 0:
return False
else:
# iterating loop till square root of n
for i in range(3, int(math.sqrt(n)) + 1, 2):
# checking for factor
if n % i == 0:
# return False
return False
# returning True
return True
print(f"Is 2 prime: {is_prime(2)}")
print(f"Is 4 prime: {is_prime(4)}")
print(f"Is 7 prime: {is_prime(7)}")
## Output
If you run the above code, then you will get the following result.
Is 2 prime: True
Is 4 prime: False
Is 7 prime: True
## Conclusion
If you have doubts in the tutorial, mention them in the comment section.
Published on 24-Apr-2020 16:07:01 | 2021-06-15T16:28:39 | {
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https://math.stackexchange.com/questions/2873785/about-proofs-by-contrapositive-and-proofs-by-contradiction | I'm a little confused about the difference between these two types of proof. As I have been taught them, it seems like proofs by contrapositive are just a subset of proofs by contradiction.
Say we want to prove $P \implies Q$. Here is how I've been taught to use both proofs:
Contrapositive: We assume $\lnot Q$ and from this we try to conclude $\lnot P$.
Contradiction: We assume $P$ and $\lnot Q$ and from this we try to deduce a contradiction.
My question is, let say we proved $P \implies Q$ by contrapositive, so we proved $\lnot Q \implies \lnot P$. Wouldn't this be equivalent to assuming $P$ and $\lnot Q$ and then use $\lnot Q$ to prove $\lnot P$, so we would deduce the contradiction $P \land \lnot P$?
• They're very similar. When contraposition works, a proof by contradiction feels unneccesarily clunky (you've just added a few lines to the top and bottom for me to read and process with no actual value), while some times a proof by contradiction (or, equivalently, by contraposition and then splitting into the two cases $P$ and $\lnot P$) is necessary. – Arthur Aug 6 '18 at 11:35
• @Arthur sorry for wasting your time sir. Feel free to edit the question so no one else has to process useless information. – Yagger Aug 6 '18 at 11:42
• I'm sorry. That's not what I meant. Your question is fine. I meant that when you make such a proof (i.e. a proof by contradiction which is basically a proof by contraposition with an added assumption of $P$ at the top), there are lines in that proof with no value. Specifically, the lines "assume $P$" at the top of the proof and "Thus $P$ and $\lnot P$, which is a contradiction" at the bottom. – Arthur Aug 6 '18 at 11:43
Every proof by contraposition can be reformulated as a proof by contradiction, as you correctly noticed. Indeed, suppose you have a proof $\pi$ of $\lnot Q \implies \lnot P$; then you can prove $P \implies Q$ by contradiction, by assuming $P$ and $\lnot Q$, which yields a proof of $\lnot P$ (via modus ponens between the assumption $\lnot Q$ and the conclusion $\lnot Q \implies \lnot P$ of $\pi$) and then you get the contradiction $P \land \lnot P$.
But not every proof by contradiction can be reformulated as a proof by contraposition, as well explained in the accepted answer of this question. Indeed, proofs by contradiction are "more general" (i.e. they can be applied to a wider set of propositions to prove) because you can stop when you find any contradiction, not only a contradiction directly involving the hypotheses. More precisely, in a proof by contradiction of $P \implies Q$, we assume $P$ and $\lnot Q$ and we deduce a contradiction $R \land \lnot R$; in the particular case where $R = P$ then (usually) you can reformulate your proof as a proof by contraposition, otherwise you cannot. | 2020-03-30T11:16:18 | {
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https://proofwiki.org/wiki/Definition:Submatrix | # Definition:Submatrix
## Definition
Let $\mathbf A$ be a matrix with $m$ rows and $n$ columns.
A submatrix of $\mathbf A$ is a matrix formed by selecting from $\mathbf A$:
a subset of the rows
and:
a subset of the columns
and forming a new matrix by using those entries, in the same relative positions, that appear in both the rows and columns of those selected.
## Notation
A submatrix of $\mathbf A$ is denoted as follows.
Let:
$\left\{ {a_1, a_2, \ldots, a_r}\right\}$ be the indices of the $r$ selected rows
$\left\{ {b_1, b_2, \ldots, b_s}\right\}$ be the indices of the $s$ selected columns
where all of $a_1, \ldots, a_r$ are between $1$ and $m$, and all of $b_1, \ldots, b_s$ are between $1$ and $n$.
Then the submatrix formed from rows $\left\{ {a_1, a_2, \ldots, a_r}\right\}$ and columns $\left\{ {b_1, b_2, \ldots, b_s}\right\}$ is denoted as:
$\mathbf A \left[{a_1, a_2, \ldots, a_r; b_1, b_2, \ldots, b_s}\right]$
It is usual to specify the rows and columns in ascending numerical order.
### Submatrix of order $\left({m - 1}\right) \times \left({n - 1}\right)$
Let a submatrix $\mathbf B$ of $\mathbf A$ be of order $\left({m - 1}\right) \times \left({n - 1}\right)$.
Then it is usual to denote $\mathbf B$ by indicating the (single) row and column of $\mathbf A$ which has been removed, as follows:
Let:
$a_j$ be the row of $\mathbf A$ which is not included in $\mathbf B$
$b_k$ be the column of $\mathbf A$ which is not included in $\mathbf B$.
Then the submatrix $\mathbf B$ formed from the remaining rows and columns of $\mathbf A$ can be denoted as:
$\mathbf A \left({a_j; b_k}\right)$
## Example
Let $\mathbf A$ be the $3 \times 4$ matrix defined as follows:
$\mathbf A = \begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \end{bmatrix}$
Then:
$\mathbf A \left[{1, 2; 1, 3, 4}\right] = \begin{bmatrix} a_{11} & a_{13} & a_{14} \\ a_{21} & a_{23} & a_{24} \end{bmatrix}$
is a submatrix of $\mathbf A$ formed by rows $1, 2$ and columns $1, 3, 4$.
This submatrix can also be denoted by $\mathbf A \left({3; 2}\right)$ which means that it is formed by deleting row $3$ and column $2$.
## Also known as
A submatrix can also be called a segment of a matrix.
## Also see
The equivalent term for a determinant is a minor. | 2019-07-23T11:53:50 | {
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http://nxsr.auus.pw/series-convergence-exam.html | # Series Convergence Exam
d'Alembert, 1717-1783). And what I would like you to do is pause this video, and think about whether each of them converges or diverges. More examples of evaluating series can be found on the following page: Series Convergence and Divergence Practice Examples 1; Series Convergence and Divergence Practice Examples 2; Series Convergence and Divergence Practice Examples 3; Series Convergence and Divergence Practice Examples 4; Series Convergence and Divergence Practice Examples 5. THE INTEGRAL AND COMPARISON TESTS 93 4. The test says nothing about the positive-term series. Theorem 72 tells us the series converges (which we could also determine using the Alternating Series Test). If a series is divergent and you erroneously believe it is convergent, then applying these tests will lead only to extreme frustration. Name: Read problems carefully. if the series converges, the sum, S, lies between Sa. )This nine page handout summarizes the tests for convergence of a series with exa. Is the series ln(x)/x convergent or divergent using the root test, ratio test, or integral test?. Which of the following statements is true? (There is only one. Given a sequence {a n} and the sequence of its partial sums s n, then we say that the series is convergent if the sequence s n is convergent and has finite limit. Alternating Series Test If for all n, a n is positive, non-increasing (i. It is recommended to use the divergence test if u can obviously see that the limit of the general term goes to infinity. Note that all the above tests can be also stated in such a way that instead of the ratio a k+1 /a k they use a k /a k+1. exam-1-spring01 Here is a copy of a second exam for practice. Otherwise, you must use a different test for convergence. I'll explain the Alternating Series Test and I'll apply it to an actual series while I do it to make the Explanation of the Alternating Series Test a little bit more concrete. In the case of the integral test, a single calculation will confirm whichever is the case. Power Series. Note, however, that disregarding a number of terms would affect the total that a convergent series converges to. Definition of the Limit of a Sequence Limit of a Sequence Properties of Limits of Sequences Squeeze Theorem for Sequences Absolute Value Theorem Definition of a Monotonic Sequence Definition of a Bounded Sequence Bounded Monotonic Sequences Definitions of Convergent and Divergent Series Convergence of a Geometric Series Properties of Infinite. To use the limit comparison test we need to find a second series that we can determine the convergence of easily and has what we assume is the same convergence as the given series. For a series made up of terms a n , define the limit as: If L < 1, then the series absolutely converges. When x= 3, the series diverges using the integral test. The idea with this test is that if each term of one series is smaller than another, then the sum of that series must be smaller. One of the most popular convergence tests for series with positive terms is the Ratio test. The following test allows us to take a series that we know to converge and modify it without losing its convergence. Remark on uniform convergence of series 1 Comparison test for uniform convergence In the introductory notes we discussed uniform convergence and norm. Alternating Series test We have the following test for such alternating series: Alternating Series test If the alternating series X1 n=1 ( 1)n 1b n = b 1 b 2 + b 3 b 4 + ::: b n > 0 satis es (i) b n+1 b n for all n (ii) lim n!1 b n = 0 then the series converges. The series ¥ å n=1 n sin1 is convergent. I Absolute convergence test. The best way (and the only way) to overcome these di culties is to have a lot of practice and this maplet can be very helpful. Definition, using the sequence of partial sums and the sequence of partial absolute sums. Since , we conclude, from the Ratio-Test, that the series. Best Answer: Note that f(x) = 1/x^p is a positive monotonically decreasing function such that f(n) = a_n = 1/n^p, for natural numbers n. If r = 1, the root test is inconclusive, and the series may converge or diverge. Series Summary Convergence and Divergence Tests 1. If the series converges, L is a number, and A is always null. The Power Series Test uses both the Ratio and Root tests to calculate the radius and interval of. The test says nothing about the positive-term series. What is a geometric series, when is a geometric series convergent? If a geometric series is convergent to which value does it converge to? 2. Solution: dy dx = x− 1 4x is continuous on [2,4]. If a series is divergent and you erroneously believe it is convergent, then applying these tests will lead only to extreme frustration. This test cannot be used to show convergence. If ∑ bn is a series whose sequence of partial sums is bounded, and if { an } is a monotone decreasing null sequence, then the series converges Explanation of Dirichlet test for convergence. Hammack and Lyons [9] provide a picture proof of the alternating series test using simple comparisons of areas of rectangles to establish convergence. AP Calculus BC Review: Sequences, Infinite Series, and Convergence Sequences A sequence 8an< is a function whose domain is the set of positive integers. This implies that perhaps the sum of a conditionally convergent series can change based on the arrangement of terms. 01 Single Variable Calculus, Fall 2006 Prof. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Introduction to series - Convergence and Divergence of series 2. Series Tests for Convergence, Page 1/3 Series Strategies and Tests for Convergence Test Name When to Use Details Known Series – Geometric When you recognize that the terms are a constant raised to some power > < − ⋅ ∞ = diverges if 1 if 1 1 converges to 0 r r r a a r n n Known Series – p series When you recognize that the terms are n. This example shows how to determine where a power series converges using the ratio test. The Ratio Test for Sequence Convergence. It is recommended to use the divergence test if u can obviously see that the limit of the general term goes to infinity. One of the most important things you will need to learn in this section of the course is a list of standard examples of convergent and divergent series. This calculus 2 video provides a basic review into the convergence and divergence of a series. Limit Comparison Test: P an and P bn such that lim n !1 a n bn exists, P an converges P bn converges. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. $\mathbf {u_1+u_2+u_3+…. Then the series ∑ a k converges if and only if A >1. But we do notice that this statement looks very similar to the Ratio Test for series, which does comment on the convergence of a series. In effect, this means that the power series can always be used as a valid alternative to the original function,. An example of this is X∞ n=1 (−1)n+1 x2 +n. The next tool in our convergence test arsenal is the ratio test. compare to known. On the bright side, this method is a lot more plug-and-chug: once you pick the series to compare, you just throw them into a limit problem and execute. The basic facts are these: Every power series has a radius of convergence 0 ≤ R≤ ∞, which depends on the coefficients an. Determine if the series is convergent or divergent by making a comparison (DCT or LCT) with a suitable b n. The most useful tests are marked with a start (*). 268), so perhaps. The integral test for convergence of an infinite series is explained. Welcome to our AP Calculus Series Tests for Convergence wiki! Here we have posted the essential convergence tests that you need to know for your AP Calculus BC exam. Unfortunately, there is no simple theorem to give us the sum of a p-series. We begin by recording a de. Note as well that there really isn't one set of guidelines that will always work and so you always need to be flexible in following this set of guidelines. The test was first published by Jean le Rond d'Alembert and is sometimes known as d'Alembert's ratio test or as the Cauchy ratio test. one more questions abt absolute convergence test. Does the series$\sum_{n=1}^{\infty} \frac{(-1)^n n^2}{n^2 + 5}\$ converge or diverge? We note that this is an alternating series, so let's try to apply the alternating. It only means the test has failed, and you will have to use another method to find the convergence or divergence of the series. This test involves choosing a sufficient series b k {\displaystyle b_{k}} for which you know the convergence/divergence of, and compares it to a series a k {\displaystyle a_{k}} through a limit. In mathematical analysis, the alternating series test is the method used to prove that an alternating series with terms that decrease in absolute value is a convergent series Test alternating series for convergence. This needs to be done for every series or improper integral you say converges or diverges. In mathematical analysis, the alternating series test is the method used to prove that an alternating series with terms that decrease in absolute value is a convergent series. I we see from the graph that because the values of b n are decreasing, the. Once you have chosen a test, there are steps to be carried out, some of which could easily be overlooked. The test is inconclusive. The most useful tests are marked with a start (*). Since 0<1 (in this example the limit does not depend on the value of x), the series converges for all x. If r = 1, the root test is inconclusive, and the series may converge or diverge. (A convergent series that is not absolutely convergent is called conditionally convergent. D' Alembert's Test is also known as the ratio test of convergence of a series. This means we need to develop some more tests. Power Series. Series Convergence and Divergence Practice Examples 3; Series Convergence and Divergence Practice Examples 4; Series Convergence and Divergence Practice Examples 5; Example 1. Conditional Convergence Definition An infinite series X1 k=1 ak is conditionally convergent if the series converges but the series X1 k=1 jakjdiverges. For the series above, the root test determines that the series converges for and divergesk kB " # for. Convergence Tests for Series - Allan Hancock College If L < 1, then the series converges absolutely. We get the idea from the convergence of geometric series. Infinite series whose terms alternate in sign are called alternating series. Ὄ Condition(s) of Divergence: 1 lim 𝑛→∞ 𝑛≠0 2 Geometric Series Test Series: ∑∞ 𝑟 𝑛=0 1 Condition of Convergence: |𝑟|<1. Algebraic manipulations give, since. The p-series test determines the nature of convergence of a p-series as follows: The p-series converges if and diverges if. The Power Series Test uses both the Ratio and Root tests to calculate the radius and interval of. Limit Comparison Test: P an and P bn such that lim n !1 a n bn exists, P an converges P bn converges. Hence, we have, which implies. REFERENCES: Bromwich, T Series test for convergence calculator. For an arbitrary series X1 k=1 ak, the series may be classified in only one of the following ways: I absolutely convergent I conditionally convergent I divergent. Absolute Convergence. It also proves that there is no universal comparison test for all series. PDF | This paper revisits the income convergence hypothesis by using the nonlinear unit root test of Kapetanios et al. Comparison test. So, if every term of a series is smaller than the corresponding term of a converging series, the smaller series must also converge. Conditional convergence, on the other hand, suggests that a country or a region will converge to its own steady state as every country or region has its own distinguished set of endowments. Thus the interval of convergence is the interval. 1, so does. A convergence test for Walsh-Fourier series. Tests for Convergence of Series 1) Use the comparison test to con rm the statements in the following exercises. " Like the Comparison Test and the Integral Test, it's fine if the first terms are kind of "wrong" - negative, for example - as long as they eventually wind up (for for a particular ) meeting the requirements. In fact, the sum can be calculated: ∞= , where a is the first term, r is the common ratio. In fact, in this example, it would be much. converges if and only if the series. Integral Series Convergence Test. Theorem (Abel's test). Absolute convergence of a series carries with it a benefit useful in evaluating a series which is neither positive nor alternating. 1 Comparison test If bn > 0 for all n and P∞ n=1bn is convergent, and if |an| ≤ bn for all n, then X∞ n=1 an is absolutely convergent. The rst hurdle in determining the convergence or divergence of a series is to select an applicable test. Show that the series is convergent. Alphabetical Listing of Convergence Tests. From the above example, we conclude that the series is conditionally convergent. = 0, then the series converges. There are more refined tests that may help, they look closer at how exactly do the fractions a k +1 / a k approach 1. Thus by the Alternating Series Test, this series converges. To use the comparison test we must first have a good idea as to convergence or divergence and pick the sequence for comparison accordingly. Six multi-part questions which involve using the integral, ratio, and comparison tests to determine whether series converge or diverge. For example, it often. This applet illustrates the integral test for convergence of infinite series. This is in B&S, but not until Chapter 9 (pg. Return to the Series, Convergence, and Series Tests starting page Return to the List of Series Tests If you can define f so that it is a continuous, positive, decreasing function from 1 to infinity (including 1) such that a [ n ]= f(n) , then the sum will converge if and only if the integral of f from 1 to infinity converges. which converges, so we'll guess that the given series converges too. If the series converges, L is a number, and A is always null. Math 142, Quiz 9. Summary of Convergence and Divergence Tests for Series TEST SERIES CONVERGENCE OR DIVERGENCE COMMENTS nth-term ∑a n Diverges iflim 0 nn→∞ a ≠ Inconclusive if lim 0 nn→∞ a = Geometric series 1 1 n n ar ∞ − = ∑ =(i) Converges with sum 1 a S −r if r <1 (ii) Diverges if r ≥1 Useful for the comparison tests if the nth term a n. An infinite series ∑ a n is said to converge absolutely if ∑ | a n | converges. This test is rather sophisticated. Answer: Let a n = 1=(n 3), for n 4. A convergent series is one in which that limit exists, and a divergent series is one in which that limit does not exist. 20 Calculators are not allowed. If an=f n where ∫ 1 ∞. 5 hour CUMULATIVE exam, and you are re-sponsible for everything that we have covered this term. (a) For all sequences of real numbers (sn) we have liminf sn ≤ limsupsn. exam-1-spring01 Here is a copy of a second exam for practice. Scribd is the world's largest social reading and publishing site. Harvey Mudd College Math Tutorial: Convergence Tests for In nite Series In this tutorial, we review some of the most common tests for the convergence of an in nite series X1 k=0 a k = a 0 + a 1 + a 2 + The proofs or these tests are interesting, so we urge you to look them up in your calculus text. The trick is to consider the absolute value series, which is. The geometric series plays a crucial role in the subject for this and other reasons. Series Convergence Tests. be the sum of the first n terms [of a series], n designating an arbitrary integer. Alternating Series test We have the following test for such alternating series: Alternating Series test If the alternating series X1 n=1 ( 1)n 1b n = b 1 b 2 + b 3 b 4 + ::: b n > 0 satis es (i) b n+1 b n for all n (ii) lim n!1 b n = 0 then the series converges. 258 Chapter 11 Sequences and Series closer to a single value, but take on all values between −1 and 1 over and over. Course Material Related to This Topic: Complete exam problem 7B-1 on page 43 to problem 7B-6 on page 44. Absolute convergence is important for the study of infinite series because its definition is strong enough to have properties of finite sums that not all convergent series possess, yet is broad enough to occur commonly. If we suppose a series with term 1/4x^2 then it converges cause its a geometric series with p = 2 > 1. Here is a list of topics: 1. Definition, using the sequence of partial sums and the sequence of partial absolute sums. In the text I am using, the hypotheses used for Gauss' test for convergence are different to others I have seen. an are called the terms of the sequence. I Absolute and conditional convergence. Math 106 Fall 2014 Exam 3. sigma_n = 1^infinity 5(1 + 1/n)^n^2 Identify a_n. It contains plenty of examples and practice problems. Namely, it is possible for a uniformly convergent series to fail the Weierstrass M-test. Enabling American Sign Language to grow in Science, Technology, Engineering, and Mathematics (STEM). Search for something else. If the limit is 1, the test fails. Necessary condition for the convergence of a series is usually used to show that a series does not converge. It is part of the unit on Infinite Series, (Chapter 9 in most of the Larson textbooks. Absolute convergence of a series carries with it a benefit useful in evaluating a series which is neither positive nor alternating. The following series either both converge or both diverge if, for all n> = 1, f(n) = a n and f is positive, continuous and decreasing. A limit of 0 does not automatically mean that the series will not diverge. We produced both single-night shows and multi-week series. There is one important, and easy to understand result about uniform convergence which we need, but did not discuss. Alphabetical Listing of Convergence Tests. Then Definition: A series is called absolutely convergent if the series of absolute values is convergent. pdf from MATH 8 at Dartmouth College. Harolds Series Convergence Tests Cheat Sheet 22 March 2013 1 3 2 nth Term Test Series:. CONVERGENCE TESTS 13 Version of August 27, 2001 The right-hand side of this inequality is a constant, independent of n. 1 December 10, 2014 1. In mathematical analysis, the alternating series test is the method used to prove that an alternating series with terms that decrease in absolute value is a convergent series. We motivate and prove the Alternating Series Test and we also discuss absolute convergence and conditional convergence. for the RATIO TEST and for the ROOT TEST SUGGESTED WORDINGS OF THE REQUIRED JUSTIFICATION for Concluding Absolute Convergence by the RATIO TEST Whenever you apply the RATIO TEST to conclude that the series 1 n n a f ¦ is Absolutely Convergent, you must write a justification as clear and complete as the following: "Since lim n 1 1 n a L and L a of. 3 Uniform convergence of Fourier series We only mention this important result in this subsection because we develop it further in Appendix G, which you should read through next. In effect, this means that the power series can always be used as a valid alternative to the original function,. This test cannot be used to show convergence. The obtained posterior convergence rates are the fastest among those of the existing Bayesian approaches. So I've been practicing several series that can be solved using the alternating series test, but I've came to a question that's been bothering me for sometime now. Theorem 72 tells us the series converges (which we could also determine using the Alternating Series Test). Integral Series Convergence Test. Start studying Calc. that converges and has bigger terms than the given series. 6: Absolute Convergence and Ratio Tests The tests we have so far developed are still not sufficient to determine convergence of series. David Jerison. If a series is divergent and you erroneously believe it is convergent, then applying these tests will lead only to extreme frustration. The test was first published by Jean le Rond d'Alembert and is sometimes known as d'Alembert's ratio test or as the Cauchy ratio test. Therefore, if the limit is equal to zero, the Divergence Test yields no conclusion: the. Return to the Series, Convergence, and Series Tests starting page Return to the List of Series Tests If the limit of a [ n ] is not zero, or does not exist, then the sum diverges. Most notable is its failure in situations where it is expected to succeed. ALTERNATING SERIES Does an = (−1)nbn or an = (−1)n−1bn, bn ≥ 0? NO Is bn+1 ≤ n & lim n→∞ YESbn = 0? P an Converges TEST FOR DIVERGENCE Does limn→∞ an = 0? NO P NO an Diverges Try one of the following tests: YES COMPARISON TEST Pick {bn}. Hi everyone! I have got this series in a part of my research. Tutorial on absolute convergence. If we use the ratio test on an arbitrary power series, we find it converges when. Cauchy root test With the default setting Method -> Automatic , a number of additional tests specific to different classes of sequences are used. , the series is the divergent Harmonic series. Series Convergence and Divergence Practice Examples 3; Series Convergence and Divergence Practice Examples 4; Series Convergence and Divergence Practice Examples 5; Example 1. P 1 n=1 tan n1 n3 Since tan is always less than ˇ 2, compare to ˇ 2n3 to show convergence. Test the series for convergence from n=2 to n=infinity of 1 / ( ln(n) )^9 The whole ln is raised to power 9 Tried as much as I can and nothing comes out I believe it needs advanced tests not the basic ones. Initially a sequence where is a positive, continuous, non-increasing function of is shown. The proof is similar to the one used for real series, and we leave it for you to do. Problem 2 Find the volume of the solid. • Direct Comparison Test: If a series P ∞ n=1 a n has all positive terms, and all of its terms are eventually bigger than those in a series that is known to be divergent, then it is also divergent. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Recognize a p-series and use the value of pto make a conclusion about the convergence of the series. Keywords: In nite series, convergence, divergence, Kummer, Gauss, Bertrand, Raabe. We have seen that the integral test allows us to determine the convergence or divergence of a series by comparing it to a related improper integral. and the series ∑ v k is absolutely convergent. Intervals of Convergence of Power Series. The next tool in our convergence test arsenal is the ratio test. By the Divergence Test, if the terms a ndo not go to exactly 0, the series diverges. (a) We can use the ratio test to compute the interior of the domain of convergence: lim k!1 xk+1 (k+ 1)2 k2 xk = jxj<1: Thus, the radius of convergence is 1. ii) if ρ > 1, the series diverges. converges or diverges. Since , we conclude, from the Ratio-Test, that the series. Series Tests for Convergence, Page 1/3 Series Strategies and Tests for Convergence Test Name When to Use Details Known Series – Geometric When you recognize that the terms are a constant raised to some power > < − ⋅ ∞ = diverges if 1 if 1 1 converges to 0 r r r a a r n n Known Series – p series When you recognize that the terms are n. From the above example, we conclude that the series is conditionally convergent. The next two tests are two of the most important because they do not rely on results regarding other tests (like the comparison tests) and do. It may help determine whether we have absolute convergence, conditional convergence, or neither. Recognizing these types will help you decide which tests or strategies will be most useful in finding whether a series is convergent or divergent. The next tool in our convergence test arsenal is the ratio test. sigma_n = 1^infinity 5(1 + 1/n)^n^2 Identify a_n. We mentioned before that geometric series are as common as eating hotdogs. pdf doc ; Convergence Tests - Additional practice using convergence tests. (5) Alternating Series Test. is convergent. This test cannot be used to show convergence. Series Convergence Tests. In mathematical analysis, the alternating series test is the method used to prove that an alternating series with terms that decrease in absolute value is a convergent series. The easiest approach for generating the series of grids is to generate a grid with what one would consider fine grid spacing, perhaps reaching the upper limit of one's tolerance for generating a grid or waiting for the computation on that grid to converge. Tips on Using Convergence Tests. ∞ n n2 + 8 n = 1 Evaluate the following integral. Thus the interval of convergence is the interval. Harvey Mudd College Math Tutorial: Convergence Tests for In nite Series In this tutorial, we review some of the most common tests for the convergence of an in nite series X1 k=0 a k = a 0 + a 1 + a 2 + The proofs or these tests are interesting, so we urge you to look them up in your calculus text. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. In mathematical analysis, the alternating series test is the method used to prove that an alternating series with terms that decrease in absolute value is a convergent series. The interval of convergence of a power series is the set of all x-values for which the power series converges. Alternating Series Test: If the series has the form 1 n a n, then the series converges if 0 aa nn 1 (decreasing terms) for all n, for some n, and lim 0 n n b. We will outline the essential concepts required for you to successfully take advantage of the following tests and include complimentary examples to help solidify your understanding. Then coarser grids can be obtained by. Taylor and Laurent series Complex sequences and series An infinite sequence of complex numbers, denoted by {zn}, can be considered as a function defined on a set of positive integers into the unextended complex plane. iii) if ρ = 1, then the test is inconclusive. The integral test for convergence of an infinite series is explained. If lim n ∞ an≠0, the series diverges by the test for divergence. , X ja njdiverges. Solved Problems for Series: Testing convergence. Note, however, that disregarding a number of terms would affect the total that a convergent series converges to. INFINITE SERIES CONVERSION TESTS • Divergence Test If , then the series Otherwise, the test is inconclusive • Ratio Test Assume that for all n, a n > 0. Start studying Calc. The theorem states that rearranging the terms of an absolutely convergent series does not affect its sum. Techniques that let you tell whether a series converges are unimaginatively called tests for convergence or convergence tests. For the series a-sub-n, the limit as n goes to infinity of the absolute value of (a-sub-n+1/a-sub-n) is equal to L or infinity. Then to find our approximation, we need to find n such that (. In effect, this means that the power series can always be used as a valid alternative to the original function,. Fill in the blanks with your answer. The real name of this test is the alternating series test. convergence tests, namely Raabe's test, Gauss' test, Bertrand's test and Kummer's test. The difference is that while the Ratio Test for series tells us only that a series converges (ab-. Specifically, the Ratio Test does not work for p-series. 3 Comparison tests The integral test provides information about convergence and divergence of a series, if we can evaluate the integral. (a) For all sequences of real numbers (sn) we have liminf sn ≤ limsupsn. 01 Single Variable Calculus, Fall 2005 Prof. Is this series divergent? Yes, it is divergent since lim n!1 (¡1)n¯1u n does not exist (i. Front Line LKC71-BK BFL Black Leather over Kydex with Suede Lining. A Motivating Problem for Power Series; The Power Series; Power Series Example; Power Series Convergence; Power Series Convergence Example; Videos. Strategies for Convergence Testing; Example Comparing Two Infinite Series; Convergence Test Videos; Unit 3: Power Series. Remark on uniform convergence of series 1 Comparison test for uniform convergence In the introductory notes we discussed uniform convergence and norm. so the series 0. All right, now let's work on this together. Initially a sequence where is a positive, continuous, non-increasing function of is shown. ASL-STEM Forum. Find all the values of x such that the following series will converge: 1 x 3. More examples of evaluating series can be found on the following page: Series Convergence and Divergence Practice Examples 1; Series Convergence and Divergence Practice Examples 2; Series Convergence and Divergence Practice Examples 3; Series Convergence and Divergence Practice Examples 4; Series Convergence and Divergence Practice Examples 5. Evaluate the following limit. final-outline. Front Line LKC71-BK BFL Black Leather over Kydex with Suede Lining. the Weierstrass M-test is false. So, the original series will be convergent/divergent only if the second infinite series on the right is convergent/divergent and the test can be done on the second series as it satisfies the conditions of the test. We will also give the Divergence Test for series in this section. We've already looked at these. Strategies for Convergence Testing; Example Comparing Two Infinite Series; Convergence Test Videos; Unit 3: Power Series. The rst hurdle in determining the convergence or divergence of a series is to select an applicable test. Suppose that there exists r such that If r < 1, then the series converges If r > 1, then the series diverges If r = 1, the ratio test is inconclusive • Integral Test The infinite series. Learning Goals: Absolute Convergence, Ratio and Root test. Series Convergence and Divergence Practice Examples 4. If is decreasing and continuous for and. If then we write If the sequence s n is not convergent then we say that the series is divergent. Namely, it is possible for a uniformly convergent series to fail the Weierstrass M-test. Learners in this course can earn a certificate in the series by signing up for Coursera's verified certificate program and passing the series' final exam. So, because of the Integral/Series Convergence Test the Integral of the same sequence an = 1/4x^2 in the same range [1, +∞) also converges. You appear to be on a device with a "narrow" screen width (i. Course Material Related to This Topic: Complete exam problem 7B-1 on page 43 to problem 7B-6 on page 44. Series Summary Convergence and Divergence Tests 1. Get the free "Infinite Series Analyzer" widget for your website, blog, Wordpress, Blogger, or iGoogle. Example 1: Power Series The definition of the convergence radius of the of a power series comes from the Cauchy test (however, the actual computation is usually done with the following test). If this condition does not satisfy then series must diverge. A series fails this test if the limit of the summand is zero, not if it is some non-zero. Remark on uniform convergence of series 1 Comparison test for uniform convergence In the introductory notes we discussed uniform convergence and norm. Absolute convergence of a series carries with it a benefit useful in evaluating a series which is neither positive nor alternating. Also known as the Leibniz criterion, the alternating series test states that for an alternating series of the form ∑ = ∞ (−), if {} is monotonically decreasing, and has a limit of 0 at infinity, then the series converges. Edited by Shawn Liu Test for Divergence Given any series Σa n, if an n"! lim ≠ 0 or does not exist, then the series Σa n is divergent. We begin by recording a de. The next tool in our convergence test arsenal is the ratio test. Course Material Related to This Topic: Complete exam problem 7B-1 on page 43 to problem 7B-6 on page 44. Alternating Series Test If for all n, a n is positive, non-increasing (i. if L is less than 1, then the series is absolutely convergent, and therefore convergent; if L is greater than 1, then the series is divergent; if L is equal to 1, the series may be convergent, divergent, or conditionally convergent. that the series converges that the convergence of the given series cannot be established However, when the limit equals zero, the test yields no conclusion, and it could be that the convergence of the given series could be established with a different test. 4: Other Convergence Tests In this section, we learn how to deal with series whose terms are not necessarily positive. Learners in this course can earn a certificate in the series by signing up for Coursera's verified certificate program and passing the series' final exam. The radius of convergence for the series ¥ å n=0 n2xn 10n is. about the series P an. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Since 0<1 (in this example the limit does not depend on the value of x), the series converges for all x. ASL-STEM Forum. It is very tough to answer such a general question, but I will give it a shot. Convergence of Fourier Series Some Definitions A function $$f\left( x \right)$$ defined on an interval $$\left[ {a,b} \right]$$ is said to be piecewise continuous if it is continuous on the interval except for a finite number of jump discontinuities (Figure $$1$$). The proof is easy: if the series converges, the partial sums s k s_k s k. So, the original series will be convergent/divergent only if the second infinite series on the right is convergent/divergent and the test can be done on the second series as it satisfies the conditions of the test. Cauchy's criterion The de nition of convergence refers to the number X to which the sequence converges. is convergent or divergent. Before we get into the limit test, we need to recall the following:. To use the limit comparison test we need to find a second series that we can determine the convergence of easily and has what we assume is the same convergence as the given series. Right Hand Belt Holster for Springfield Armory XD 3". The integral test for convergence of an infinite series is explained. = 0, then the series converges. The alternating series test (also known as the Leibniz test), is type of series test used to determine the convergence of series that alternate. Due to the nature of the mathematics on this site it is best views in landscape mode. is uniformly convergent on each bounded disc of the complex plane, but is not uniformly convergent on the whole of. For example, we take zn= n+ 1 2n so that the complex sequence is {zn} = ˆ1 + i 2, 2 + i 22, 3 + i 23. sigma_n = 1^infinity 5(1 + 1/n)^n^2 Identify a_n. The next tool in our convergence test arsenal is the ratio test. This implies that perhaps the sum of a conditionally convergent series can change based on the arrangement of terms. iii) if ρ = 1, then the test is inconclusive. The radius of convergence in this case is said to be. For the series above, the root test determines that the series converges for and divergesk kB " # for. Conditional Convergence: An in nite series X a n converges conditionally if X a n converges but it does not converge absolutely, i. Using the comparison test and the fact that converges, we determine that converges and hence, by Corollary 4. Tips for determining convergence or divergence of an in nite series Divergence Test : Always check that the individual terms a ngo to zero. PDF | Cauchy's condensation test allows to determine the convergence of a monotone series by looking at a weighted subseries that only involves terms of the original series indexed by the powers. Our goal is to reduce things to simpler things. While this idea is the one needed most frequently, our students sometimes fixate on the alternating series test, which is a very specialized test guaranteeing convergence. | 2019-12-12T10:36:38 | {
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http://math.stackexchange.com/questions/85372/galois-extensions-of-degrees-p-and-pn-1-given-a-galois-extension-of-pn?answertab=active | # Galois extensions of degrees $p$ and $p^{n-1}$ given a Galois extension of $p^n$
Suppose $K$ is a Galois extension of a field $F$ of degree $p^n$ for a $p$ a prime.
I want to see if there are Galois extensions of degrees $p$ and $p^{n-1}$ over $F$.
If $G=\text{Gal}(K/F)$, then $|G|=p^n$. If $G$ is abelian, I know there are subgroups of order $p^i$ for $0\leq i\leq n$, so there are subgroups of orders $p^{n-1}$ and $p$, and their corresponding fixed fields are of degrees $p$ and $p^{n-1}$ over $F$, and are Galois extensions since the subgroups are obviously normal in $G$.
But if $G$ is not abelian, is this still true?
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A group of order $p^n$ always has subgroups of order $p^{n-1}$ (in fact, all maximal subgroups are of order $p^{n-1}$), and they are always normal; and it always has subgroups of order $p$ that are normal (in fact central).
To see this, we use the class equation. Recall that if $G$ is a finite group and $Z(G)$ is the center of the group (the set of all elements $g\in G$ such that $gx=xg$ for all $x\in G$), then $$|G| = |Z(G)| + \sum [G:C(x_i)],$$ where $x_1,\ldots,x_n$ are representatives of the conjugacy classes of $G$ with more than one element. If $|G|=p^n$, then $[G:C(x_i)]$ is a multiple of $p$ for every $i$, so considering the equation modulo $p$ we conclude that $|Z(G)|\equiv 0 \pmod{p}$. Since $Z(G)$ has at least one element (the identity), it must be nontrivial.
Since $Z(G)$ is nontrivial, and is abelian, it has a subgroup of order $p$. This subgroup is normal in $G$. This gives you a normal subgroup of order $p$, and hence a field of degree $p^{n-1}$ over $F$ that is Galois.
To show that all maximal subgroups are of order $p^{n-1}$ and that they are all normal, we proceed by induction on $n$. If $n=1$ or $n=2$, then $G$ is abelian and we know the result is true. Assume the result is true for groups of order $p^{n-1}$, and let $G$ be of order $p^n$.
Let $N$ be a subgroup of $Z(G)$ of order $p$, and let $H$ be a maximal subgroup of $G$. If $N\subseteq H$, then consider $G/N$. This has order $p^{n-1}$, and $H/N$ is maximal (by the Lattice Isomorphism Theorem); hence $H/N$ is of order $p^{n-2}$ and normal in $G/N$, so $|H|=|N|\times|H/N| = p^{n-1}$, and is normal in $G$.
If $N$ is not contained in $H$, then $HN$ is a subgroup of $G$ that contains $H$ (since $N$ is central, it is normal, so $HN$ is a subgroup), hence $HN=G$. Since $|G|=p^n = |HN| = |H|\,|N|/|H\cap N|$, and $|H\cap N| = 1$ ($N$ is cyclic of order $p$ and not contained in $H$), then $|H|=p^{n-1}$. Moreover, given any $g\in G$, we can write $g=hn$ with $h\in H$ and $n\in N$. Then $gHg^{-1} = hnHn^{-1}h^{-1} = hHnn^{-1}h^{-1} = hHh^{-1} =H$ (with the second equality because $n$ is central). Thus, $H$ is normal in $G$.
Hence, every maximal subgroup of $G$ has order $p^{n-1}$ and is normal.
In conclusion, you can always find normal subgroups of $G$ of order $p$ and of order $p^{n-1}$, when $|G|=p^n$.
Added. You can now use this to show that $G$ always has subgroups of order $p^i$ that are normal for every $i$, $0\leq i\leq n$.
Proceed by induction on $n$. The result holds for groups of order $p$ and $p^2$. Assume the result holds for any group of order $p^n$, and let $G$ have order $p^{n+1}$. Let $N$ be a normal subgroup of order $p$, and consider $G/N$. Then $G/N$ has subgroups $\overline{H_i}$ of order $p^i$ that are normal in $G/N$, $i=0,\ldots,n$. These correspond to subgroups $H_i$ of $G$ that contain $N$, of order $p^{i+1}$, and that are normal in $G$. So $G$ has subgroups of order $p,\ldots,p^{n+1}$ that are normal; together with the trivial group of order $p^0$, this gives you subgroups of order $p^i$ that are normal for every $i$, $0\leq i\leq n+1$. $\Box$
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Thanks, this was easy to follow. – Turk Nov 24 '11 at 22:40
I recall to have seen this question in the Dummit & Foote's Abstract Algebra, and if you go to page 188 of this book you will see that Theorem 1 (3) gives you that $\mathrm{Gal}(K/F)$ has a normal subgroup of order $p^k$ for $0 \le k \le n$, thus giving you the Galois extensions you were looking for. The proofs are in the book.
Hope that helps,
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Thanks mate, I'll try to find this. – Turk Nov 24 '11 at 22:40 | 2015-05-27T22:45:31 | {
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http://math.stackexchange.com/questions/87296/why-is-it-that-in-mathbbf-q-setminus-0-there-are-exactly-as-many-squares/87302 | Why is it that in $\mathbb{F}_q \setminus 0$, there are exactly as many squares as non-squares?
Why is it that in $\mathbb F_q\setminus\{0\}$, when $q\neq 2^k$, there are exactly as many squares as non-squares?
My combinatorics textbook states this without proof :/
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I hope your textbook makes an exception for the case $q=2^k$. – Gerry Myerson Dec 1 '11 at 6:19
-checks- yep, they say "odd prime" – badatmath Dec 1 '11 at 6:32
consider the homomorphism $F^*_{q}$ to the multiplicative group $\{+1,-1\}$, where squares are sent to $1$ and non-squares are sent to $-1$. By first homomorphism theorem, $F^{*}_{q}/ker$ is isomorphic to the image, i.e. $\{1,-1\}$, by equating the cardinality of both sets, we have the assertion. – Dinesh Dec 1 '11 at 7:47
Please try to make the body of the question self contained, even if the question was in the title, repeating it in the body is not a bad idea. – Asaf Karagila Dec 1 '11 at 11:33
This is true if and only if $q$ is odd. Given that $q$ is odd, $\mathbb{F}_q^\times=\mathbb{F}_q\setminus\{0\}$ is an abelian group of even order (in fact it is cyclic, but we don't need to know that for this argument). Consider the squaring homomorphism $s:\mathbb{F}_q^\times\to \mathbb{F}_q^\times$, defined by $s(a)=a^2$. What is its kernel? What does that mean the size of the image must be?
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Ooooooo clever :D thanks! – badatmath Dec 1 '11 at 6:31
Thanks, glad to help! – Zev Chonoles Dec 1 '11 at 6:33
Just a quick comment: when Zev says "in fact it is cyclic, but this is not necessary", he means that it is not necessary to know that the group is cyclic -- one gets away by knowing what the subgroup of order $2$ elements looks like (I make arguments like this in my number theory notes to avoid existence of primitive roots until it is absolutely necessary). One definitely needs the group to be cyclic, or at least not an arbitrary finite abelian group of even order: for instance in $\{\pm 1\}^k$ only the identity element is a square. – Pete L. Clark Dec 1 '11 at 7:33
@Pete: Thanks for pointing out the ambiguity; I've clarified my answer as well. – Zev Chonoles Dec 1 '11 at 7:47
Define an equivalence relation $\sim$ on $\mathbb{F}_q^*$ with $a \sim b \Leftrightarrow a^2 = b^2$ . The number of squares is the number of equivalence classes of this relation. Now $x^2 = y^2$ if and only if $y = \pm x$. Thus there are two elements in every equivalence class, since $x \neq -x$ for every $x \in \mathbb{F}_q^*$. Therefore there are $\frac{1}{2}|\mathbb{F}_q^*|$ equivalence classes.
EDIT: As others have noted, $q$ should be odd. In a field of order $2^k$ every element is a square: the map $x \mapsto x^2$ is then injective and thus bijective, because we're talking about finite fields here.
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Consider the homomorphism $\phi:\mathbb F^{*}_q \rightarrow$ {$1,-1$} which sends squares to $1$ and non-squares to $-1$. You can check that it is a homomorphism. By the first isomorphism theorem we have that $\mathbb F^{*}_q/\operatorname{ker}\phi$ is isomorphic to {$-1,1$}. Since they are bijective, their cardinalities are equal hence $(q-1)/|\operatorname{ker}\phi|=2$. Hence $|\operatorname{ker}\phi|=(q-1)/2$. Since $\operatorname{ker}\phi$ is precisely the number of elements going to 1, they are precisely the squares in $\mathbb F^{*}_{q}$. Hence the number of non-squares is also $q-1-(q-1)/2= (q-1)/2$. Hence the assertion.
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How would you show that $\phi$ is a homomorphism? I I can see that the product of two squares is a square and the product of a square and a non-square is again a non-square. How do you show that product of two non-squares is square? (I cannot use the counting argument - that if I have a non-square the products with non-squares must be remaining numbers, which are the squares - since in this way I would be using the fact, which I'm trying to prove.) – Martin Sleziak Dec 1 '11 at 8:11
You are right..I just used the dangerous word for myself "obvious". – Dinesh Dec 1 '11 at 8:29
@MartinSleziak In $\mathbb F^{*}_{q}$ we have that $a^{q-1}=1$ for all $a$. I claim that $a$ is a square iff $a^{(q-1/2)}=1$ If $a$ is a square then $a=x^2$ which implies $x^{2.({q-1}/2)}=x^{q-1}=1$. Now to prove the converse let $a^{({q-1}/2)}=x$ then we have that $x^{2}=1$ which has only two solutions 1 and -1. If $x=-1$ then $a^{(q-1/2)}=-1$ and if $a$ is a square,say $x^2$ then $x^{(q-1)}=-1$ which is not true. Hence the above claim is true. Now if $a$ and $b$ are two non-squares then $a^{(q-1/2)}=-1$ and $b^{(q-1/2)}=-1$ which implies $(ab)^{(q-1/2)}=1$. Hence $ab$ is a square. – Dinesh Dec 1 '11 at 8:44
What does $a^{(q-1/2)}$ mean? Of course if not every element is a square writing something like $a^{1/2}$ would not be meaningful? – Stefan Feb 8 at 11:32 | 2016-06-25T14:11:55 | {
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http://math.stackexchange.com/questions/263606/prove-that-the-number-of-jump-discontinuities-is-countable-for-any-function | Prove that the number of jump discontinuities is countable for any function
I would like to prove that the number of simple jump discontinuities of any function is countable.
Can someone point me some material where the proof is or explain the proof here?
Thanks.
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The point is that any jump discontinuity has a neighborhood with no other jump discontinuity, and that the real line is Lindelof. @xavierm02: there are only countably many integers. – ronno Dec 22 '12 at 8:55
Let $f:(a,b)\to \mathbb{R}$ and $$A=\left\{x\in (a,b):f\text{ has a jump discontinuity at x}\right\}$$ Now $$A=A^{+}\cup A^{-}$$ where $$A^{+}=\left\{x\in (a,b):\lim_{y\to x^+}f(y)>\lim_{y\to x^-}f(y)\right\}$$ and $$A^{-}=\left\{x\in (a,b):\lim_{y\to x^+}f(y)<\lim_{y\to x^-}f(y)\right\}$$ I will show $A^{+}$ is countable and leave the rest to you. Fix $x\in A^{+}$ and then $\exists q\in \mathbb{Q}$ so that $$\lim_{y\to x^+}f(y)>q>\lim_{y\to x^-}f(y)$$ (why???). This means that $\exists \delta>0$ so that $$x-\delta<y<x<z<x+\delta\implies f(z)>q>f(y)$$ and so (why?) $\exists n\in \mathbb{N}$ so that $$x-\frac1n<y<x<z<x+\frac1n\implies f(z)>q>f(y)$$ If we let $$A_{q,n}=\left\{x\in (a,b):x-\frac1n<y<x<z<x+\frac1n\implies f(z)>q>f(y)\right\}$$ ($q\in \mathbb{Q}$,$n\in \mathbb{N}$) then by our previous discussion $$A^{+}\subseteq\bigcup_{q\in \mathbb{Q}}\bigcup_{n\in \mathbb{N}}A_{q,n}$$ Therefore the problem moves to proving that $A_{q,n}$ is countable. This follows from the fact $A_{q,n}$ is isolated (show this!).
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Wow! This is an amazing solution. I really like it. Very elegant! – elaRosca Dec 22 '12 at 9:20
The proof for the other case follows in similar manner, just with the sign changed in the RHS of the implication, where we define Aq,n. q exists by density of Q in R. – elaRosca Dec 22 '12 at 9:29
@elaRosca Indeed. How would you answer the first "why"? – Nameless Dec 22 '12 at 9:30
I wrote the code in latex but it does not seem to recognize it here. – elaRosca Dec 22 '12 at 9:56
@Nameless I don't really understand the last part. Why is it countable? Can you kindly explain to me please thanks – user136266 Apr 30 '14 at 7:46
The argument below is essentially the one outlined in Robert Israel's post here, but I tweak it a bit to show that there are only countably many removable discontinuities as well.
Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a function. The key idea is that we can control the amount of fluctuation in $f$ (and hence the size of jumps) on the left (resp., right) side of a point $x$ where the left limit (resp., right limit) exists by taking points sufficiently close to $x$. We cannot guarantee that there are no jumps in a neighborhood of a jump discontinuity; for example, the function $g:[-1,1]\rightarrow\mathbb{R}$ given by
$$g(x) = \begin{cases} \phantom{-}1 & \text{if}\ x\leq0 \\ 1/n & \text{if}\ n \text{ is a positive integer and } 1/(n+1)<x\leq 1/n \end{cases}$$
has a jump discontinuity at and in every neighborhood of $0$ (a more pathological example is given in iballa's comment on Koushik's post; see also Brian Scott's post here for details). However, it is true that we can make jumps around a jump discontinuity as small as desired by taking a sufficiently small neighborhood (but we actually only use a slightly weaker result -- see below). To that end, we note that the definition of the left limit and the triangle inequality give the
Lemma. If $f(x-)=\lim_{t\rightarrow x^-} f(t)$ exists then for any $\varepsilon > 0$ we have some $\delta>0$ such that $$\mathrm{diam} f(x-\delta,x) < \varepsilon. \Box$$
Now for any $x\in\mathbb{R}$ where $f(x-), f(x+)$ exist, put
$$M(x)=\max\{|f(x)-f(x-)|,|f(x)-f(x+)|\},$$
and for any $\varepsilon>0$, let
$$\mathcal{J}(\varepsilon)=\{ x\in\mathbb{R} : f(x-),f(x+) \text{ exist and } M(x)>\varepsilon \}.$$
Since any point $x$ at which a jump or removable discontinuity occurs lies in $\bigcup_n \mathcal{J}(1/n),$ it suffices to show that each $\mathcal{J}(\varepsilon)$ is countable. Fix $x\in\mathcal{J}(\varepsilon)$ and take $\delta>0$ such that $\mathrm{diam} f(x-\delta,x) < \varepsilon.$ If $t_0$ is an element of $(x-\delta, x)$ such that $f(t_0-), f(t_0+)$ exist then the sequences $f(t_0-1/n), f(t_0+1/n)$ eventually lie in
$$f(x-\delta,x) \subset [f(t_0)-\varepsilon, f(t_0)+\varepsilon],$$
so that
$$f(t_0 -)=\lim_{n\rightarrow\infty} f(t_0-1/n) \in [f(t_0)-\varepsilon, f(t_0)+\varepsilon]$$
and
$$f(t_0 +)=\lim_{n\rightarrow\infty} f(t_0+1/n) \in [f(t_0)-\varepsilon, f(t_0)+\varepsilon].$$
Consequently, we have $M(t_0)\leq\varepsilon$, and we deduce that $(x-\delta, x)$ and $\mathcal{J}(\varepsilon)$ are disjoint. Letting $q_x$ be any rational number in $(x-\delta, x),$ the map $x\mapsto q_x$ yields an injection $\mathcal{J}(\varepsilon)\rightarrow\mathbb{Q},$ completing the proof.
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any jump discontinuity has a neighborhood with no other jump discontinuity, Associate to each such neighbourhood a rational number inside that.so there is a bijection between a subset of rationals and jump discontinuity.
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this shows no. of jump discontinuities can be at most countable – Koushik Dec 22 '12 at 9:08
This is very clear and short, gives a very intuitive answer. Thanks – elaRosca Dec 22 '12 at 9:13
Notice that you can choose a rational in an interval with a canonical way; thus, the result doesn't depend on the axiom of choice. – Seirios Dec 22 '12 at 9:16
Can you explain this a bit, please? I am not sure I understand what you mean – elaRosca Dec 22 '12 at 9:22
It seems to me that this is wrong. Suppose we have an enumeration of the rationals $\{q_n\}$. Then the function $f(x) = \sum_{n=1}^{\infty}{\frac{1}{2^n} 1_{[q_n, \infty)} }$ has a jump discontinuity at exactly the rational numbers. – iballa Apr 23 '14 at 18:06 | 2016-06-28T13:15:53 | {
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https://math.stackexchange.com/questions/2097557/if-0-alpha-beta-pi-and-cos-alpha-cos-beta-cos-alpha-beta-3 | # If $0< \alpha, \beta< \pi$ and $\cos\alpha + \cos\beta-\cos (\alpha + \beta) =3/2$ then prove $\alpha = \beta= \pi/3$
If $0< \alpha, \beta< \pi$ and $\cos\alpha + \cos\beta-\cos (\alpha + \beta) =3/2$ then prove $\alpha = \beta= \pi/3$
How do I solve for $\alpha$ and $\beta$ when only one equation is given? By simplification I came up with something like $$\sin\frac{\alpha}{2} \sin\frac{\beta}{2} \cos \frac {\alpha +\beta}{2}=\frac{1}{8}.$$ I don't know if this helps. How to do this?
• are $\alpha,\beta$ angles in a given triangle? – Dr. Sonnhard Graubner Jan 14 '17 at 15:38
• @Dr.SonnhardGraubner nope – ray Jan 14 '17 at 16:07
Set $p=\cos\beta$ and $q=\sin\beta$; then your equality can be written $$\cos\alpha+p-p\cos\alpha+q\sin\alpha=\frac{3}{2}$$ or $$(1-p)\cos\alpha+q\sin\alpha=\frac{3}{2}-p$$ Set $X=\cos\alpha$ and $Y=\sin\alpha$; then the equation becomes $$\begin{cases} (1-p)X+qY=\frac{3}{2}-p\\[6px] X^2+Y^2=1 \end{cases}$$ and the distance of the line from the origin should be $\le1$, or the line and the circle wouldn't meet: $$\frac{(\frac{3}{2}-p)^2}{(1-p)^2+q^2}\le1$$ that becomes $$\frac{9}{4}-3p+p^2\le 1-2p+p^2+q^2$$ that is, recalling that $p^2+q^2=1$, $$p^2-p+\frac{1}{4}\le0$$ Can you finish up?
HINT:
Use Prosthaphaeresis Formula on $\cos\alpha,\cos\beta$
and Double angle formula on $$\cos 2\cdot\dfrac{\alpha+\beta}2$$ to get
$$2\cos^2\dfrac{\alpha+\beta}2-2\cos\dfrac{\alpha-\beta}2\cos\dfrac{\alpha+\beta}2+\dfrac12=0\ \ \ \ (1)$$ which is a Quadratic Equation in $\cos\dfrac{\alpha+\beta}2$ whose discriminant must be $\not<0$
i.e., $$\left(2\cos\dfrac{\alpha-\beta}2\right)^2-4=-2\sin^2\dfrac{\alpha-\beta}2\ge0$$
But for real $\dfrac{\alpha-\beta}2,$ $$\sin^2\dfrac{\alpha-\beta}2\ge0$$
So, we have $$\sin^2\dfrac{\alpha-\beta}2=0\iff\sin\dfrac{\alpha-\beta}2=0$$
$\implies\dfrac{\alpha-\beta}2=n\pi$ where $n$ is any integer
But as $0<\alpha,\beta<\pi.n=0\implies\alpha-\beta=0$
So, $(1)$ is reduced to $$0=2\cos^2\beta-2\cos\beta+\dfrac12=\dfrac{(2\cos\beta-1)^2}2$$
So, $\cos\beta=\dfrac12=\cos\dfrac\pi6\implies\beta=2m\pi\pm\dfrac\pi6$ where $m$ is any integer
But as $0<\alpha<\pi$
• with prosthaphaeresis formula, I get a $\cos (\frac {\alpha - \beta}{2})$. How to deal with it? – ray Jan 14 '17 at 16:12
• @ray, Please find the updated answer. – lab bhattacharjee Jan 15 '17 at 3:37 | 2019-07-15T18:22:14 | {
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https://math.stackexchange.com/questions/579541/on-the-paris-constant-and-sqrtk1-sqrtk1-sqrtk1-sqrtk1-dots | On the Paris constant and $\sqrt[k]{1+\sqrt[k]{1+\sqrt[k]{1+\sqrt[k]{1+\dots}}}}$?
In 1987, R. Paris proved that the nested radical expression for $\phi$,
$$\phi=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}}$$
approaches $\phi$ at a constant rate. For example, defining $\phi_n$ as using $n = 5, 6, 7$ "ones" respectively, then,
$$(1/2)(\phi-\phi_5)(2\phi)^5 = 1.0977\dots$$
$$(1/2)(\phi-\phi_6)(2\phi)^6 = 1.0983\dots$$
$$(1/2)(\phi-\phi_7)(2\phi)^7 = 1.0985\dots$$
which is approaching the Paris constant $R = 1.09864196\dots$. It seems it can be generalized. Define,
$$x_n=\sqrt[k]{1+\sqrt[k]{1+\sqrt[k]{1+\sqrt[k]{1_n+\dots}}}}\tag{1}$$
for integer $k>1$ and the equations,
$$x^k = x+1\tag{2}$$
$$y = \frac{1}{x}+1\tag{3}$$
where $x$ is the root of $(2)$ such that $x = x_n$ as $n \to \infty$ in $(1)$. Then one can conjecture that,
$$\lim_{n\to\infty}(1/2)(x-x_n)(ky)^n = C_k\tag{4}$$
for some constant $C_k$. The Paris constant is simply the case $C_2$.
I tested it for increasing large $k$. The sequence of $C_k$ seem to be themselves approaching a constant. The rate is very slow, so for much higher $k = 10^{14},10^{15},10^{16}$,
$$C_{10^{14}} = 0.6931471805599500\dots$$
$$C_{10^{15}} = 0.6931471805599457\dots$$
$$C_{10^{16}} = 0.6931471805599454\dots$$
Compare to,
$$\ln 2 = 0.6931471805599453\dots$$
Question:
1. Does $C_k \to \ln 2$, as $k \to \infty$?
$\color{blue}{Edit,\; Nov.\;25}$
More generally, define,
$$x_n=\sqrt[k]{a+\sqrt[k]{a+\sqrt[k]{a+\sqrt[k]{a_n+\dots}}}}\tag{5}$$
for integers $a\ge 1,\;k>1$ and,
$$x^k = x+a\tag{6}$$
$$y = \frac{a}{x}+1\tag{7}$$
Then it seems,
$$\lim_{n\to\infty}(1/2)(x-x_n)(ky)^n = C_{a,k}\tag{8}$$
The Paris constant is the case $C_{1,2}$. Is it true that as $k \to \infty$, then,
$$\lim_{k\to \infty} C_{1,k} = \ln 2$$
$$\lim_{k\to \infty} C_{2,k} = \tfrac{3}{2} \ln \tfrac{3}{2}$$
$$\lim_{k\to \infty} C_{3,k} = \tfrac{4}{2} \ln \tfrac{4}{3}$$
$$\lim_{k\to \infty} C_{4,k} = \tfrac{5}{2} \ln \tfrac{5}{4}$$
and so on?
P.S. The only known closed-form in terms of transcendental constants is $C_{2,2} = \pi^2/8$.
• It is an iteration of a holomorphic function near an attracting fixed point, and the asymptotics should be determinable from the leading term or two of the power series near the fixed point. I would not expect that there is anything special about iterating radicals other than the power series behavior near $x$. Thus, it is very believable that there is a formula as you wrote, including for non-integer $k$. – zyx Nov 24 '13 at 20:12
• From the definition of $C_k$, as $k\to\infty$, is it expected that $C_k$ is finitely bounded (instead of getting smaller and smaller)? Furthermore, why is the bound apparently $\ln2$? – Tito Piezas III Nov 24 '13 at 20:24
• I suppose it might be worth knowing that $$x = 1 + \frac{\log 2}{k} + \frac{\log 2 + (\log 2)^2}{2k^2} + O\left(\frac{1}{k^3}\right).$$ – Antonio Vargas Nov 24 '13 at 20:36
• Hm. Interesting that the appropriate root of $x^k=x+1$ can be approximated by that formula... – Tito Piezas III Nov 24 '13 at 20:43
• I wanted to add (nested-radicals) tag, which seems suitable for this question. But the question already has five tags. – Martin Sleziak Dec 16 '15 at 17:42
Antonio Vargas's observation means that $1$ starts closer and closer to the fixpoint, so that maybe there is less and less difference between $C_k$ and the first term in the sequence defining it ; and maybe that first term converges to $\log 2$.
Let $f_k(x) = \sqrt[k]{1+x}$ for $x \ge 0$ and $k > 1$. Let $\alpha_k$ the unique positive fixpoint of $f_k$ (it is the positive root of $\alpha_k^k = \alpha_k+1$). Define $c_{k,n} = (1/2)(\alpha_k - f_k^{n-1}(1))(f_k'(\alpha_k))^{-n}$ for $n \ge 1$.
Now your constant $C_k$ is defined by $C_k = \lim_{n \to \infty} c_{k,n}$, and we want to give an estimation of $C_k/c_{k,1}$.
We have $c_{k,{n+1}}/c_{k,n} = f_k'(\alpha_k)^{-1}(f_k(\alpha_k) - f_k(f_k^{n-1}(1)))/(\alpha_k - f_k^{n-1}(1)) = f_k'(\alpha_k)^{-1}f_k'(z_{k,n})$ for some $f_k^{n-1}(1) \le z_{k,n} \le \alpha_k$.
Since $f_k'$ is decreasing, we obtain $1 \le c_{k,n+1}/c_{k,n} \le f_k'(f_k^{n-1}(1))f_k'(\alpha_k)^{-1}$
Some crude estimates gives us $\alpha_k \ge f_k^n(1) \ge \alpha_k - (\alpha_k - 1)f_k'(1)^n$,
and then ($f''_k$ is increasing), $f'_k(\alpha_k) \le f_k'(f_k^n(1)) \le f_k'(\alpha_k) - (\alpha_k-1)f'_k(1)^nf_k''(1)$,
and finally $1 \le c_{k,n+1}/c_{k,n} \le 1 + (\alpha_k-1)f'_k(1)^{n-1}(-f''_k(1))f_k'(\alpha_k)^{-1}$.
Using $1+x \le \exp(x)$ and taking the product, we obtain $1 \le C_k/c_{k,1} \le \exp((\alpha_k-1)(-f''_k(1))f'_k(\alpha_k)^{-1}/(1-f'_k(1)))$
Since $\alpha_k = 1 + \log 2/k + O(k^{-2})$, we have
$\alpha_k-1 \sim \log2 / k$
$f'_k(\alpha_k) = \alpha_k(1 + \alpha_k)^{-1}/k \sim 1/2k$
$c_{k,1} = (1/2)(\alpha_k-1)f'_k(\alpha_k)^{-1} \sim (1/2)(\log 2/k)(2k) = \log 2$
$f_k'(1) = 2^{1/k-1} \frac 1k \sim 1/2k$
$f''_k(1) = 2^{1/k-2} \frac 1k (\frac 1k -1) \sim -1/4k$
$(\alpha_k-1)(-f''_k(1))f'_k(\alpha_k)^{-1}/(1-f'_k(1)) \sim \log 2/2k \to 0$
This shows that $C_k \sim c_{k,1} \to \log 2$
For the more general case, we start from $x_1 = a^{1/k} = 1 + \log(a)/k + \ldots$, while $\alpha = 1 + \log(a+1)/k + \ldots$, which are again close to each other.
$\alpha - a^{1/k} \sim \log(\frac{a+1}a)/k$
$f'(\alpha) = \alpha/k(a+\alpha) \sim 1/(a+1)k$
$c_1 = (1/2)(\alpha - a^{1/k})/f'(\alpha) \sim \frac{a+1}2\log(\frac{a+1}a)$
Since $f'(\alpha^{1/k})$ and $f''(\alpha^{1/k})$ are on the order of $1/k$, we have $C = \frac{a+1}2\log(\frac{a+1}a)$
• Thanks, Mercio. I'll accept your answer soon. However, I was wondering if you know how to tweak it to cover the more general case $x_n=\sqrt[k]{a+\sqrt[k]{a+\sqrt[k]{a+\sqrt[k]{a_n+\dots}}}}$? (Kindly see the Nov.25 edit.) – Tito Piezas III Nov 26 '13 at 0:17
Hint : $\displaystyle x(n)=\underset{k=0}{\overset\infty{\Large\Xi}}\left(a,b\,;\tfrac1n\right)\iff x^n=a+bx\iff n(x)=\frac{\ln(a+bx)}{\ln x}\iff n(1)=\infty$ , $n(\infty)=1$ . Now show, using l'Hopital, that $\displaystyle\lim_{x\to1}\Big[n(x)\cdot(x-1)\Big]=\ln2$.
• Could you please explain/define the notation $\Xi_{k=0}^\infty\left(a,\,b;\,1/n\right)$? – Ian Mateus Nov 24 '13 at 22:40
• Sorry, I thought it was self-evident: $$\underset{_{\text{k}\,=\,0}}{\overset{n}{\mathbf\Xi}}\ \Bigg(a_{_\text{k}}\ ,\ b_{_\text{k}},\ \frac1{N_{_\text{k}}}\Bigg)\ =\ \sqrt[^{N_{_\text{0}}}]{a_{_\text{0}}\ +\ b_{_\text{0}}\sqrt[^{N_{_\text{1}}}]{a_{_\text{1}}\ +\ b_{_\text{1}}\sqrt[^{N_{_\text{2}}}]{\ldots\ \sqrt[^{N_{_{n}}}]{a_{_{n}}}}}}$$ which, for $N_{_\text{k}}=-1$, becomes the continued fraction $$\underset{_{\text{k}\,=\,0}}{\overset{n}{\mathbf\Xi}}\ \Big(a_{_\text{k}}\ ,\ b_{_\text{k}},\ -1\Big)\ =\ \cfrac1{a_{_0}\ +\ \cfrac{b_{_0}}{a_{_1}\ +\ \cfrac{b_{_1}}{\ddots\ {a_{_n}}}}}$$ – Lucian Nov 30 '13 at 19:58
You might notice that if instead of:
$\lim_{n\to\infty}(1/2)(x-x_n)(ky)^n = C_{a,k}^+\tag{8}$
You don't divide by 2:
$\lim_{n\to\infty}(x-x_n)(ky)^n = C_{a,k}^+\tag { }$
The known closed form constant is then $C_{2,2}^+=\frac{pi^2}{4}$, while the other version (below) with $a=x^k+x$ has a closed form constant of $C_{2,2}^{-}=\frac{pi}{2\sqrt{3}}$.
other version: $x_n=\sqrt[k]{a-\sqrt[k]{a-\sqrt[k]{a-\sqrt[k]{a_n-\dots}}}}\tag { }$
C+ approaches 1 from above: $\lim_{k\to \infty} C_{1,k}^+ = 2 \ln 2\tag{}$
$\lim_{k\to \infty} C_{2,k}^+ = 3 \ln \tfrac{3}{2}\tag{}$
$\lim_{k\to \infty} C_{3,k}^+ = 4 \ln \tfrac{4}{3}\tag{}$
C- approaches 1 from below:
$\lim_{k\to \infty} C_{2,k}^- = \ln 2 \tag{}$
$\lim_{k\to \infty} C_{3,k}^- = 2 \ln \frac{3}{2}\tag{}$
$\lim_{k\to \infty} C_{4,k}^- = 3 \ln \frac{4}{3} \tag{}$
You'll notice that the second form (with $a=x^k+x$) has a C that approaches 1 from below, and the first one (you posted about) approaches 1 from above.
$\lim_{k\to \infty} C_{a,k}^+ = (a+1) \ln \frac{a+1}{a} \tag{}$
$\lim_{k\to \infty} C_{a,k}^- = (a-1) \ln \frac{a}{a-1} \tag{}$
Coincidentally :D $\lim_{a\to\infty} (\frac{a}{a-1})^a \to e^+$ from above, and $\lim_{a\to\infty} (\frac{a}{a-1})^{a-1} \to e^-$ from below. | 2020-06-03T08:01:49 | {
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https://www.myelbee.com/site/topic/b8flj.php?page=expected-value-and-variance-88d603 | # expected value and variance
(x1 - E [X])^2, p (x2). P(X = 2) = 1/6 (the probability that you throw a 2 is 1/6) The variance of a discrete random variable is given by: σ 2 = Var (X) = ∑ (x i − μ) 2 f (x i) The formula means that we take each value of x, subtract the expected value, square that value and multiply that value by its probability. $$\textrm{Var}(X)=E\big[(X-\mu_X)^2\big]=EX^2-(EX)^2.$$ So, how do we use the concept of expected value to calculate the mean and variance of a probability distribution? Now that we can find what value we should expect, (i.e. $$EX=\sum_{x_k \in R_X} x_k P_X(x_k).$$ Variance measures the difference from the expected value. the expected value), it is also of interest to give a measure of the variability. In my post on expected value, I defined it to be the sum of the products of each possible value of a random variable and that value’s probability.. For a discrete random variable X, the variance of X is written as Var(X). If Xis a random variable with values x 1;x 2;:::;x n, corresponding probabilities p 1;p 2;:::;p n, and expected value = E(X), then Variance = ˙ 2(X) = p 1(x 1 2 ) 2 +p and 2(x 2 ) + +p n(x n ) Standard Deviation = ˙(X) = p Variance : P(X = 5) = 1/6 (the probability that you throw a 5 is 1/6) If you think about it, 3.5 is halfway between the possible values the die can take and so this is what you should have expected. Var(X) = E[ (X – m)2 ] where m is the expected value E(X). P(X = 6) = 1/6 (the probability that you throw a 6 is 1/6), E(X) = 1×P(X = 1) + 2×P(X = 2) + 3×P(X = 3) + 4×P(X=4) + 5×P(X=5) + 6×P(X=6), Therefore E(X) = 1/6 + 2/6 + 3/6 + 4/6 + 5/6 + 6/6 = 7/2. so to find its expected value, we can write, $E[aX+b]=aEX+b$, for all $a,b \in \mathbb{R}$, and. . P(X = 4) = 1/6 (the probability that you throw a 4 is 1/6) 2x & \quad 0 \leq x \leq 1\\ $E[X_1+X_2+...+X_n]=EX_1+EX_2+...+EX_n$, for any set of random variables $X_1, X_2,...,X_n$. $$\hspace{70pt} E[g(X)]=\sum_{x_k \in R_X} g(x_k)P_X(x_k) \hspace{70pt} (4.2)$$ 3.2.1 - Expected Value and Variance of a Discrete Random Variable . Expected Value (or EV) is a measure of what you can expect to win or lose per bet placed in the long run. Expected Value, Mean, and Variance Using Excel This tutorial will calculate the mean and variance using an expected value. = a2E(X2) - a2E2(X) = a2Var(X). In particular, usually summations are replaced by integrals and PMFs are replaced by PDFs. Var [X] = sum (p (x1). Then we'll discuss properties of expected value and variance with respect to arithmetic operations and introduce measures of independence between random variables. The expected value of X is usually written as E(X) or m. So the expected value is the sum of: [(each of the possible outcomes) × (the probability of the outcome occurring)]. Expectation and Variance The expected value (or mean) of X, where X is a discrete random variable, is a weighted average of the possible values that X can take, each value being weighted according to the probability of that event occurring. 0 & \quad x < a \textrm{ or } x > b The expected value (or mean) of X, where X is a discrete random variable, is a weighted average of the possible values that X can take, each value being weighted according to the probability of that event occurring. It turns out (and we Linearity of expected value 6:56. 0 & \quad \text{otherwise} The expected value of a constant is just the constant, so for example E(1) = 1. (xn - E [X])^2) 1 Linear transformations of random variables 8:51. Expected Value is without variance. formula for the variance of a random variable. The proofs and ideas are very analogous to the discrete case, so sometimes we state the results without There are six possible outcomes: 1, 2, 3, 4, 5, 6. (x2 - E [X])^2,..., p (x1). In fact: You is because: Var[aX + b] = E[ (aX + b)2 ] - (E [aX + b])2 . random variables. \begin{array}{l l} Remember that the expected value of a discrete random variable can be obtained as Expected value of product of independent random variables with same expected value and variance 0 Find variance and general formula for for r$^{th}$ moment for random variable uniform over (0,1) = E[ a2X2 + 2abX + b2] - (aE(X) + b)2 As we mentioned earlier, the theory of continuous random variables is very similar to the theory of discrete Let X represent the outcome of the experiment. \nonumber f_X(x) = \left\{ = a2E(X2) + 2abE(X) + b2 - a2E2(X) - 2abE(X) - b2 In this example, Harrington Health Food stocks 5 loaves of Neutro-Bread. The standard deviation of X is the square root of Var(X). Properties of Expected values and Variance Christopher Croke University of Pennsylvania Math 115 UPenn, Fall 2011 Christopher Croke Calculus 115. P(X = 3) = 1/6 (the probability that you throw a 3 is 1/6) Variance of a Discrete Random Variable . (How far a set of numbers are spread out from their average value.) mathematical derivations for the purpose of brevity. Probability distributions, including the t-distribution, have several moments, including the expected value, variance, and standard deviation (a moment is a summary measure of a probability distribution): The first moment of a distribution is the expected value, E (X), which represents the mean or average value of the distribution. Moreover, it determines the degree to which the values of a random variable differ from the expected value. Multiplying a random variable by a constant multiplies the expected value by that constant, so E[2X] = 2E[X]. As a result, it’s defined with We have (a) $VarX=E[X^{2}]-(EX)^{2}$ (b)$Var(aX+b)=a^{2}VarX$ For calculating variance in given problems we will mostly use (a). It turns out (and we Y = X2 + 3 so in this case r(x) = x2 + 3. $$\hspace{70pt} E[g(X)]=\int_{-\infty}^{\infty} g(x) f_X(x) dx \hspace{70pt} (4.3)$$, $=\frac{1}{b-a} \bigg[ \frac{1}{2}x^2 \bigg]_{a}^{b} dx$, $= \bigg[\frac{1}{n+2}x^{n+2}+\frac{1}{2(n+1)}x^{n+1} \bigg]_{0}^{1}$, $$=E\big[(X-\mu_X)^2\big]=\int_{-\infty}^{\infty} (x-\mu_X)^2 f_X(x)dx$$, $$=EX^2-(EX)^2=\int_{-\infty}^{\infty} x^2 f_X(x)dx-\mu_X^2$$, $= \bigg[-\frac{3}{2}x^{-2} \bigg]_{1}^{\infty}$, As we saw, the PDF of $X$ is given by \frac{1}{b-a} & \quad a < x < b\\ Since most of the statistical quantities we are studying will be averages it is very important you know where these formulas come from. \begin{array}{l l} $$Properties of Expected values and Variance Christopher Croke University of Pennsylvania Math 115 UPenn, Fall 2011 Christopher Croke Calculus 115. a continuous random variable as. Expected value Consider a random variable Y = r(X) for some function r, e.g. Real value is your actual result (profit), which depends on the outcome of the matches. continuous random variables.$$ Then sum all of those values. Therefore P(X = 1) = 1/6 (this means that the probability that the outcome of the experiment is 1 is 1/6) Remember the law of the unconscious statistician (LOTUS) for discrete random variables: So the expectation is 3.5 . 4.1.2 Expected Value and Variance As we mentioned earlier, the theory of continuous random variables is very similar to the theory of discrete random variables. Expected value Consider a random variable Y = r(X) for some function r, e.g. \end{array} \right. The variance of a random variable tells us something about the spread of the possible values of the variable. EXPECTED VALUE AND VARIANCE n = 100 n = 10000 Winning Frequency Relative Frequency Relative Frequency Frequency 1 17.17 1681.1681 -2 17.17 1678.1678 3 16.16 1626.1626 -4 18.18 1696.1696 5 16.16 1686.1686 -6 16.16 1633.1633 Table 6.1: Frequencies for dice game. Let $X$ be a continuous random variable with PDF
EN FR
This website uses cookies and personal data to enhance your browsing experience. | 2023-02-05T08:00:00 | {
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http://alieco-spa.com/hus3f1c/a7c3c5-fibonacci-sequence-divisible-by-3 | The Fibonacci numbers 5, 55, 610, 6765, 75025 and 832040 corresponding to n = 5, 10, 15, 20, 25 and 30 are divisible by . This type of index number follow a … The discovery of the famous Fibonacci sequence. If any two consecutive Fibonacci numbers have a common factor (say 3) then every Fibonacci number must have that factor. The Fibonacci numbers are the sequence of numbers F n defined by the following recurrence relation: F n = F n-1 + F n-2. Which Fibonacci numbers are divisible by 2? The sequence starts with two 1s, and the recursive formula is. Read also: More Amazing People Facts Question 1.2. with seed values F 0 =0 and F 1 =1. For a given prime number p, which Fibonacci numbers are di-visible by p? Fibonacci number with index number factor : We have some Fibonacci number like F(1) = 1 which is divisible by 1, F(5) = 5 which is divisible by 5, F(12) = 144 which is divisible by 12, F(24) = 46368 which is divisible by 24, F(25) = 75025 which is divisible by 25. Each term in the Fibonacci sequence is called a Fibonacci number. Can you calculate the number of rabbits after a few more months? Notice that is divisible by for values of n that are divisible by 4. ... (F 3 = 2), every fourth F-number is divisible by 3 (F 4 = 3), every fifth F-number is divisible by 5 (F 5 = 5), every sixth F-number is divisible by 8 (F 6 = 8), every seventh F-number is divisible by 13 (F 7 = 13), etc. So if we start with 1, and have 1 next, then the third number is 1 + 1 = 2, the fourth number is 1 + 2 = 3, the fifth number is 2 + 3 = 5, and so on. x n = x n − 1 + x n − 2. The Fibonacci sequence $\langle f_n \rangle$ starts with $f_1=1$ and $f_2=1$. So, at the end of the year, there will be 144 pairs of rabbits, all resulting from the one original pair born on January 1 of that year. Related. Fibonacci's Solution: The Fibonacci Sequence! This is clearly not the case so no two consecutive Fibonacci numbers can have a common factor. Editor's note: We should still show that these are the only Fibonnaci numbers divisible by 3 to prove the 'only if' condition. 2 Initial Examples and Periodicity As a rst example, consider the case p= 2. The same happens for a common factor of 3, since such Fibonacci's are at every 4-th place (Fib(4) is 3). It is the day of Fibonacci because the numbers are in the Fibonacci sequence of 1, 1, 2, 3. Then k+3 = 6. This coincides with the date in mm/dd format (11/23). The Fibonacci sequence is the sequence of numbers that starts off with 1 and 1, and then after that every new number is found by adding the two previous numbers. In the last section we saw that Fib(3)=2 so we would expect the even Fibonacci numbers (with a factor of 2) to appear every at every third place in the list of Fibonacci numbers. 1, 1, 2, 3, 5, 8, , , , , , , … So after 12 months, you’ll have 144 pairs of rabbits! 3. Fibonacci Sequence. So far, I tried proving that F(n) is even if 3 divides n. My steps so far are: Consider: F(1) ≡ 1(mod 2) F(2) ≡ 1(mod 2) F(3) ≡ 0(mod 2) F(4) ≡ 1(mod 2) F(5) ≡ 1(mod 2) F(6) ≡ 0(mod 2) Assume there exists a natural number k such that 3 divides k and F(k) is even. In more sophisticated mathematical language, we have shown that the Fibonacci sequence mod 3 is periodic with period 8. The Fibonacci numbers 3, 21, 144, 987, 6765, 46368 and 317811 corresponding to n = 4, 8, 12, 16, 20, 24 and 28 are divisible by . Inspecting the table we see that F i is divisible by 2 if and only if iis divisible by 3… Let k = 3. List of Prime Numbers; Golden Ratio Calculator; All of Our Miniwebtools (Sorted by Name): Our PWA (Progressive Web App) Tools (17) {{title}} This is an important argument to The parity of the sum of two numbers is determined by the parity of the summands. fourth Fibonacci number is divisible by 3 and: the “divisor 3” behaviour is periodic, with period 8.
## fibonacci sequence divisible by 3
How To Draw A School Of Fish Easy, Stone Wall Ideas For Backyard, Gummy Bears With Protein, In-n-out 4x4 Price, Image And Vision Computing Review Time, 5 Star Hotel In Ramnagar Uttarakhand, Dwight Nicki Minaj, Ansi Z765 2018, | 2021-06-21T20:23:34 | {
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https://math.stackexchange.com/questions/1754681/why-only-consider-dolbeault-cohomology | # Why only consider Dolbeault cohomology?
On a complex manifold we have the differential operators
$$\partial:A^{p,q}\to A^{p+1,q}$$ $$\bar\partial:A^{p,q}\to A^{p,q+1}$$ which both square to zero. Hence one can define cohomology groups $$H^{p,q}_\partial=\frac{\ker\partial:A^{p,q}\to A^{p+1,q}}{\partial A^{p-1,q}}$$ $$H^{p,q}_{\bar\partial}=\frac{\ker\bar\partial:A^{p,q}\to A^{p,q+1}}{\bar\partial A^{p,q-1}}$$ But for some reason I only ever see people discussing the $H_{\bar\partial}^{p,q}$, i.e. Dolbeaut cohomology. What makes this more interesting the the $H_\partial^{p,q}$ groups?
• You should really write $A^{p,q}(X)$, $H^{p,q}_{\bar{\partial}}(X)$, etc. Apr 22, 2016 at 23:58
$\newcommand{\dd}{\partial}$To supplement the existing (excellent) answers: If $E \to M$ is a holomorphic vector bundle, the transition functions of $E$ are "constant with respect to $\bar{\dd}$", so there's a natural notion of $E$-valued Dolbeault cohomology $H_{\bar{\dd}}^{p,q}(M, E)$. Unless the transition functions of $E$ are constant, however, there is no corresponding notion of $H_{\dd}^{p,q}(M, E)$.
• Very good point I'd never noticed before!
– user98602
Apr 22, 2016 at 21:30
• But we could just as easily consider anti-holomorphic vector bundles could we not? Apr 22, 2016 at 23:27
• @MichaelAlbanese: There is no such thing. :) A composition of anti-holomorphic maps isn't anti-holomorphic, so there's no "anti-holomorphic pseudo-group". Apr 23, 2016 at 1:40
• So the 'anti-holomorphic tangent space' $T^{0,1}M$ is not an anti-holomorphic vector bundle. That's annoying. Apr 23, 2016 at 15:16
• @MichaelAlbanese: Or, if you prefer, being holomorphic is an incredibly rich, chiral property, whose "mirror image" is (surprisingly!) undistinguished from the "merely smooth". ;) Apr 23, 2016 at 16:37
Note that $H^{0,0}_{\bar{\partial}}(X) = \ker\bar{\partial} : A^{0, 0}(X) \to A^{0, 1}(X)$ is precisely the collection of holomorphic functions on $X$ which, historically, were of more interest than anti-holomorphic functions on $X$ which is given by $H^{0,0}_{\partial}(X) = \ker\partial : A^{0,0}(X) \to A^{1,0}(X)$.
Moreover, Dolbeault cohomology $H^{p,q}_{\bar{\partial}}(X)$ determines $H^{p,q}_{\partial}(X)$ and vice versa. More precisely, the maps
\begin{align*} \psi : H^{p,q}_{\bar{\partial}}(X) &\to H^{q,p}_{\partial}(X)\\ [\alpha] &\mapsto [\bar{\alpha}] \end{align*}
and
\begin{align*} \varphi : H^{q,p}_{\partial}(X) &\to H^{p,q}_{\bar{\partial}}(X)\\ [\beta] &\mapsto [\bar{\beta}] \end{align*}
are well-defined and are inverses of one another. Therefore $H^{p,q}_{\bar{\partial}}(X) \cong H^{q,p}_{\partial}(X)$.
If $X$ is compact, then Hodge Theory tells us that $H^{p,q}_{\bar{\partial}}(X) \cong \ker\Delta_{\bar{\partial}} : A^{p,q}(X) \to A^{p,q}(X)$ and $H^{p,q}_{\partial}(X) \cong \ker\Delta_{\partial} : A^{p,q}(X) \to A^{p,q}(X)$ where $\Delta_{\bar{\partial}}$ and $\Delta_{\partial}$ are the holomorphic and anti-holomorphic Laplacians respectively. If $X$ is also Kähler, then $\Delta_{\bar{\partial}} = \Delta_{\partial}$ so we see that
\begin{align*} H^{p,q}_{\bar{\partial}}(X) &\cong \ker\Delta_{\bar{\partial}} : A^{p,q}(X) \to A^{p,q}(X)\\ &= \ker\Delta_{\partial} : A^{p,q}(X) \to A^{p,q}(X)\\ &\cong H^{p,q}_{\partial}(X). \end{align*}
Combining these isomorphisms with the ones above, we see that $H^{p,q}_{\bar{\partial}}(X) \cong H^{q,p}_{\bar{\partial}}(X)$ so the Hodge numbers satisfy $h^{p,q} = h^{q,p}$. Therefore the odd Betti numbers of a compact Kähler manifold are even.
$H^{p,q}_{\partial}$ is canonically isomorphic to $H^{q,p}_{\bar \partial}$ via complex conjugation.
As for why one would prefer the latter group over the former, $H_{\bar \partial}^{p,0}$ is the set of holomorphic $p$-forms, and that's a more natural object than the antiholomorphic $q$-forms ($H_{\partial}^{0,q}$). | 2023-03-23T12:58:55 | {
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https://math.stackexchange.com/questions/4080410/representing-all-rational-numbers-between-dfrac12-and-1 | # Representing all rational numbers between $\dfrac{1}{2}$ and $1$
How do I show that
$$\dfrac{2^{\left\lfloor\frac12 a_1\right\rfloor} + 2^{\left\lfloor\frac12 a_2\right\rfloor} + \ldots + 2^{\left\lfloor\frac12 a_n \right\rfloor}}{2^{\left\lceil\frac12 a_1\right\rceil} + 2^{\left\lceil\frac12 a_2\right\rceil} + \ldots + 2^{\left\lceil\frac12 a_n \right\rceil}}$$
represents all rationals between $$\dfrac{1}{2}$$ and $$1$$, where $$a_1, a_2, \ldots, a_n \in \mathbb{N}$$?
I have tried representing few rationals such as $$\dfrac{7}{8}$$ and $$\dfrac{5}{8}$$ and it seems like we can represent all of them. However I'm not completely sure why and don't have a rigorous proof.
For example
$$\dfrac{5}{8} = \dfrac{2^{\left\lfloor\frac12 3\right\rfloor} + 2^{\left\lfloor\frac12 4\right\rfloor} + 2^{\left\lfloor\frac12 5 \right\rfloor}}{2^{\left\lceil\frac12 3\right\rceil} + 2^{\left\lceil\frac12 4\right\rceil} + 2^{\left\lceil\frac12 5 \right\rceil}}$$
$$\dfrac{7}{8} = \dfrac{2^{\left\lfloor\frac12 1\right\rfloor} + 2^{\left\lfloor\frac12 2\right\rfloor} + 2^{\left\lfloor\frac12 4 \right\rfloor}}{2^{\left\lceil\frac12 1\right\rceil} + 2^{\left\lceil\frac12 2\right\rceil} + 2^{\left\lceil\frac12 4 \right\rceil}}$$
Any help would be highly appreciated. Thanks
• Two questions: 1) Unfortunately, there's no concensus in Mathematics about $0$ being in $\mathbb N$ or not. Can the $a_i$'s be $0$? 2) Can $a_1, a_2,\ldots a_n$ repeat or they have to be different? – jjagmath Mar 28 at 13:45
• It does not matter whether $0$ is considered to be in $\mathbb{N}$ or not, since you can represent the same rational by increasing all the $a_i$'s by $2$. But could OP clarify if the $a_i$'s need to be distinct? – Haran Mar 28 at 14:20
• @Haran No there's no condition which says that the $a_i$'s should be distinct. They all can be same. Thanks. – BookWick Mar 28 at 14:48
• Let $A=\{a_1,\cdots,a_n\}$. The expression can be written as $$\frac{\sum\limits_{m=1}^n2^{\lfloor a_i/2\rfloor}}{\sum\limits_{m=1}^n2^{\lceil a_i/2\rceil}}=\frac{\sum\limits_{i=1}^k2^{b_i-1}+\sum\limits_{j=k+1}^n2^{b_j}}{\sum\limits_{i=1}^k2^{b_i}+\sum\limits_{j=k+1}^n2^{b_j}}\in\left[\frac12,1\right]$$ where $(b_i)_{i\le k}$ correspond to all odd elements of $A$ and $(b_j)_{k+1\le j\le n}$ to all even elements. Since each of $b_1,\cdots,b_n$ is at least one, for any reduced rational $p/q$ in this domain, we can write $\sum\limits_{i=1}^k2^{b_i-1}=q-p$ and $\sum\limits_{j=k+1}^n2^{b_j}=2p-q$. – TheSimpliFire Mar 28 at 15:48
• This completes the proof as the choice of $(b_j)_{k+1\le j\le n}$ is independent of the choice of $(b_i)_{i\le k}$, and every positive integer has a unique binary representation. Moreover, any rational number $P/Q$ can be written as $kp/kq$ (where $p/q$ is reduced), so we just need to scale $q-p$ and $2p-q$ up by a factor of $k$. – TheSimpliFire Mar 28 at 15:52
Some nice answers have been posted, but I'm still adding this answer since it gives a straightforward non-recursive way to get a representation.
That the expression is in $$[{1 \over 2},1]$$ is easy to see.
For the other way, consider $${x \over y} \in [{1 \over 2},1]$$ and let $$m=y-x$$. Then $$x-m \geq 0$$.
Express $$x-m=(a_k a_{k-1} \dots a_0)_2$$ and $$m=(b_l b_{l-1} \dots b_0)_2$$ expressed in binary [$$0 \notin \{a_k,b_l\}$$].
Then, I claim that $$\{2i : a_i = 1\} \cup \{2j+1 : b_j =1 \}$$ gives our representation.
Indeed, $${\sum_{a_i=1}2^{\lfloor{2i \over 2}\rfloor}+\sum_{b_j=1}2^{\lfloor{2j+1 \over 2}\rfloor} \over \sum_{a_i=1}2^{\lceil{2i \over 2}\rceil}+\sum_{b_j=1}2^{\lceil{2j+1 \over 2}\rceil}} = {\sum_i a_i 2^i+\sum_j b_j 2^j \over \sum_i a_i 2^i+\sum_j b_j 2^{j+1}} = {x-m+m \over x-m+2m} = {x \over y}$$ The last line is due to the fact that in case of non-integers (i.e. $${2j+1 \over 2}$$), the ceilings increase the powers of $$2$$ by $$1$$ and hence the number expressed is multiplied by $$2$$.
As an example, for $$7 \over 10$$, $$x=7,y=10,m=3$$.
Now, $$x-m=4=(100)_2$$ and $$m=3=(11)_2$$.
Using our algorithm, the representative numbers are $$4,3,1$$.
Indeed, $${2^{\lfloor{4 \over 2}\rfloor}+2^{\lfloor{3 \over 2}\rfloor}+2^{\lfloor{1 \over 2}\rfloor} \over 2^{\lceil{4 \over 2}\rceil}+2^{\lceil{3 \over 2}\rceil}+2^{\lceil{1 \over 2}\rceil}} = {2^2+2^1+2^0 \over 2^2+2^2+2^1} = {4+2+1 \over 4+4+2} = {7 \over 10}$$.
I just noticed that @TheSimpliFire has done a similar thing in the comments, but I started writing it sometime ago, had some coffee, and then finished it. So I somehow missed his/her comment. Anyway, if you (@TheSimpliFire) decide to make it into an answer, please let me know and I'll remove mine.
• @Haran Sorry, typo. – MathIsNice1729 Mar 28 at 17:11
• You need not to justify an answer which is similar to a comment. – user Apr 1 at 9:34
There is an argument using the Farey tree. Briefly, the Farey algorithm fills in rational numbers between $$x = \frac{p}{q}$$ and $$y = \frac{r}{s}$$ by $$\frac{p+r}{q+s}$$ (and so on recursively) [What some have called the Bart Simpson addition of fractions). The basic fact is that if you start with $$\frac12$$ and $$1,$$ the next number you generate is $$\frac{1+1}{1+2} = \frac23,$$ the next two are $$\frac12 + \frac23$$ and $$\frac23 +1$$ (where + is the Bart Simpson sum), this procedure will generate all the rational numbers in the interval $$[\frac12, 1].$$
It is not hard to see that numbers as in the OP behave nicely with respect to the Bart Simpson addition: if $$x$$ and $$y$$ both have an $$a_i,$$ then their Bart Simpson sum has $$a_i+2.$$ It also needs to be checked that $$\frac12$$ and $$1$$ can be represented, but that is easy. So, to summarize, the bart simpson sum of two numbers having the required representation also has the required representation, which gives both a proof and an algorithm to construct the representation.
• Could you please elaborate the Farey Tree and Farey Algorithm and how it relates to this problem? Thanks. – BookWick Mar 28 at 15:13
• @BookWick OK, done. – Igor Rivin Mar 28 at 16:18
• Why the downvotes? – Igor Rivin Mar 29 at 13:31
• @BookWick I did not think it was you. Just disappointed that people downvote things they (apparently) don't understand. – Igor Rivin Mar 29 at 16:36
• Yeah that's a pretty sad thing. Never mind :( – BookWick Mar 30 at 8:19
We can observe that $$2^{\lfloor a/2 \rfloor}$$ and $$2^{\lceil a/2 \rceil}$$ are simply powers of $$2$$ which are either equal or consecutive. Hence: $$2^{\lfloor a/2 \rfloor} \leqslant 2^{\lceil a/2 \rceil} \leqslant 2 \cdot 2^{\lfloor a/2 \rfloor}$$ Taking these bounds in our expression shows that any rational expressed in the given form must be between $$\frac{1}{2}$$ and $$1$$. Now, we will show that all rationals in $$[\frac{1}{2},1]$$ can be expressed in the given form by inducting over the size of the denominator.
The result holds good when the denominator is $$1$$ or $$2$$, by expressing $$1$$ and $$\frac{1}{2}$$ as they are. Now, assume that we are to show that $$\frac{x}{y}$$ can be expressed in the given form.
First, we can see that since $$\frac{x}{y}<1$$, it follows: $$\frac{x-1}{y-1}<\frac{x}{y}<1$$ Moreover, if we assume $$\frac{x-1}{y-1}<\frac{1}{2}$$, then: $$\frac{x-1}{y-1}<\frac{1}{2}<\frac{x}{y} \implies 2x-1 which is clearly a contradiction. Hence, $$\frac{x-1}{y-1}$$ lies in $$[\frac{1}{2},1]$$ and can thus be expressed in the given form. Now, simply write $$\frac{x}{y}$$ in the given form by writing: $$\frac{x}{y} = \frac{(x-1)+1}{(y-1)+1}$$
Note: Sometimes, the fraction $$\frac{x-1}{y-1}$$ might not be in simplified form. Simply consider the expression of this rational number in its simplest form, and repeat the terms in the numerator and denominator till the sum in the numerator is $$x-1$$ and in the denominator is $$y-1$$.
Example: Assume we need to express $$\frac{16}{25}$$ in the given form. We first write: $$\frac{16}{25} = \frac{15+1}{24+1}$$ Now, we have: $$\frac{15}{24}=\frac{5}{8}=\frac{2+2+1}{4+2+2}$$ Hence: $$\frac{16}{25} = \frac{3(2+2+1)+1}{3(4+2+2)+1} = \frac{2+2+2+2+2+2+1+1+1+1}{4+4+4+2+2+2+2+2+2+1}$$
• I think you didn't show that all numbers in $[\dfrac{1}{2} , 1]$ can be represented in that closed form? – BookWick Mar 29 at 11:15
• I have, the induction establishes that all fractions in that interval with denominator $y$ can be expressed in that form. – Haran Mar 29 at 15:13
Let's group the parameters $$a_i$$ into two groups, the even and odd numbers (lets call them $$e_i$$ and $$o_i$$ resp).
Notice that $$2^{\lceil e_i/2 \rceil} = 2^{\lfloor e_i/2 \rfloor}$$ while $$2^{\lceil o_i/2 \rceil} = 2 \times 2^{\lfloor o_i/2 \rfloor}$$
Now, let $$\frac12 with $$Z=\frac{X}{Y}$$, where $$X, Y \in \mathbb{N}$$
Let $$D=Y-X$$ and $$C=2X -Y$$, with $$C, D>0$$. Also, $$X=D+C$$ and $$Y=C+2D$$.
Then $$Z=\frac{C+D}{C+2D}$$
Further, any natural number can be expressed as sum of powers of two: $$C = 2^{c_1} + 2^{c_2} + \cdots 2^{c_j}$$, where $$(c_1, \cdots c_j)$$ correspond to the positions of the ones in the binary digits of its binary representation. Analogously, $$D = 2^{d_1} + 2^{d_2} + \cdots 2^{d_k}$$
Now, let $$e_i = 2 c_i$$ and $$o_i = 2 d_i +1$$, and we are done.
An example follows: Let $$Z = \frac{11}{17}$$, then
$$C= 2\times 11 -17=5 = 2^0 +2^2 \implies c_i=(0 , 2) \implies e_i=(0 , 4)$$
$$D=17-11=6 = 2^1 + 2^2\implies d_i=(1 , 2) \implies o_i=(3 , 5)$$
and $$a_i = (0,4,3,5)$$. Indeed:
$$\dfrac{ 2^{\lfloor 0/2\rfloor} + 2^{\lfloor 4/2\rfloor} + 2^{\lfloor 3/2\rfloor} + 2^{\lfloor 5/2\rfloor} } { 2^{\lceil 0/2\rceil} + 2^{\lceil 4/2\rceil} + 2^{\lceil 3/2\rceil} + 2^{\lceil 5/2\rceil} } =\dfrac{1+4+2+4}{1+4+4+8}=\dfrac{5+6}{5+2\times 6}=\frac{11}{17}$$ | 2021-05-14T21:57:34 | {
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https://math.stackexchange.com/questions/561591/how-many-ways-to-seat-9-couple-around-a-round-table | # How many ways to seat 9 couple around a round table
1. You are a host/hostess at your local Applebee’s. You are seating a group consisting of 9 couples at a round table.
A)In how many different ways can you do this, provided that each couple will sit together, and all that you care about is their position relative to one another?
B)What is the probability that Al doesn’t end up within two seats of Ricky, AND Beth doesn’t end up within two seats of Charlene?
Used a wrong tag before of order-statistics. Sorry about that.
• What are your thoughts so far? It looks like you've copied a question without even all the necessary information – Ian Coley Nov 10 '13 at 21:24
• This was all that was given to me by my instructor. I know that just using a simple 9 P 9 isn't enough, though. Because it's a circular table A,B,C,D,E,F,G,H,I would be the same thing as I,A,B,C,D,E,F,G,H. But then I get stuck. – Boxyouranswer Nov 10 '13 at 21:29
• Depends. Al and Ricky could be a couple. – André Nicolas Nov 10 '13 at 21:30
• Well, I assume from the question that they are not a couple. Otherwise it would just be 1/(the possibilities). – Boxyouranswer Nov 10 '13 at 22:39
For the number of arrangements, imagine that one of the chairs is a throne, and the Queen is one of the group at Applebee's. She sits down first, of course, on the throne. Her Consort has $2$ choices of chair. Now let us seat the other couples one at a time, counterclockwise from the Queen-Consort pair.
The person chosen to occupy the chair immediately counterclockwise from the royal pair can be chosen in $16$ ways. Now the occupant of the next chair is determined. The person chosen to occupy the next chair after that can be chosen in $14$ ways, and then the occupant of the chair after that is determined. And so on.
Multiply. We get $(2)(16)(14)(12)\cdots (4)(2)$. This can also be written as $(2^9)(8!)$.
• Wow Thanks! But I have a follow-up question. Would the generic formula for such a problem then be 2^n * (n-1)! ? I mean, if there was another problem that said I have to seat 10 people around a round table with 10 available seats, could I use (2^10)(9!) ? – Boxyouranswer Nov 10 '13 at 22:52
• Yes, precisely the same reasoning gives the expression in your comment, and works whenever we have $n$ couples. And if social arrangements divided people into triples, similar reasoning would give $(3!)^n (n-1)!$ ways to arrange $3n$ people around a round table so that "triples" are together. – André Nicolas Nov 11 '13 at 0:46
Another approach to this counting problem is to think of each couple as "single" object, counting the number of ways the single objects can be arranged, and then multiplying by the number of equivalent ways the couples can be arranged as "compound" objects.
In particular, there are $9$ couples (considered as "single objects"), and they can be arranged around the table in $8!$ ways. (Make sure you understand why this is so)
Now we consider the couples as "compound" objects, consisting of persons $A$ and $B$. Each couple can be arranged in 2 ways: $AB$ or $BA$. Since there are $9$ couples, there are $2^9$ choices.
So we get $2^9 \cdot 8!$ arrangements.
And yes, this generalizes to arbitrary $n$: if there are $n$ couples, they can be arranged in $2^n (n-1)!$ ways.
• Follow-up. For part B of the problem, would I just find all the ways Al and Ricky do sit within 2 seats of each other, and all the ways that Beth and Charlene sit within 2 seats of each other, divide by the outcomes given by 2^9(8!) for each one then add together and subtract from 1? – Boxyouranswer Nov 10 '13 at 23:32
• That is a valid approach. You can use the "compound" and "single" object approach, too. – nomen Nov 10 '13 at 23:37 | 2019-10-15T18:31:50 | {
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https://math.stackexchange.com/questions/4007640/compact-metric-spaces | # compact metric spaces
Let $$(X,d_X)$$ and $$(Y, d_Y)$$ be compact metric spaces.
1. Show that for a nonempty set $$A\subseteq X,$$ the function $$f : X\to \mathbb{R}, f(x) = \inf \{d_X(x,a) : a\in A\}$$ is continuous.
2. Show that if a map $$f : X\to Y$$ is both continuous and bijective, then its inverse $$f^{-1} : Y\to X$$ is also continuous.
3. Let $$C$$ be the Cantor set with metric from $$(\mathbb{R},|\cdot|)$$ and $$P = \prod_{k=1}^\infty \{0,\frac{1}{2^k}\}$$ with metric from $$(\ell_1, \lVert \cdot \rVert_1).$$ Show that there exists a continuous bijection $$f : C\to P.$$
For $$1),$$ let $$x\in X$$ and let $$\epsilon > 0.$$ Choose $$\delta = \epsilon.$$ Then if $$y\in X$$ is such that $$d(x,y) < \epsilon, we$$ have \begin{align}d(f(x), f(y)) &= \inf\{d(x,a) | a \in A \} - \inf\{d(y,a) | a\in A\}\\ &\leq \inf \{d(x,a) | a\in A\} - (\inf \{d(x,a) - d(x,y) | a \in A\})\\ &= \inf \{d(x,a) | a\in A\} - (\inf \{d(x,a) | a \in A\} - d(x,y)\}\\ &= d(x,y) < \epsilon.\end{align}
Is this part correct?
For $$2),$$ suppose $$f:X\to Y$$ is continuous and bijective. Let $$y\in Y$$ and $$\epsilon > 0.$$ We want to show that there exists $$\delta > 0$$ so that if $$d_Y(x,y) < \delta$$ then $$d_X(f^{-1}(x), f^{-1}(y)) < \epsilon$$. Let $$(y_n)$$ be a sequence in $$Y$$ such that $$y_n\to y.$$ Then $$(f^{-1}(y_n))\subseteq X$$ and since $$X$$ is compact, it has a convergent subsequence, say $$(f^{-1}(y_{n_k}))_k.$$ Since $$f$$ is bijective, we can write $$y_n = f(x_n)$$ for each $$n\in\mathbb{N},$$ where $$x_n \in X.$$ Let $$x = \lim_{k\to\infty} f^{-1}(y_{n_k}) = \lim_{k\to \infty} x_{n_k}.$$ Since $$f$$ is continuous, $$f(x) = \lim_{k\to\infty} f(x_{n_k}).$$ But here I'm stuck as I'm not sure how to show $$f^{-1}$$ is continuous from here.
As for the third part, every element of the Cantor set can be uniquely written in the form $$\sum_{k=1}^\infty \frac{a_k}{3^k}, a_k \in \{0,2\}\forall k.$$ Then it suffices to show that the function $$f : C\to P, f(\sum_{k=1}^\infty \frac{a_k}{3^k}) = \prod_{k=1}^\infty \{\frac{a_k/2}{2^k}\}$$ is continuous. I think one could show it's sequentially continous, but I'm not really sure about the details.
• For 3, consider choosing $\delta = min_k \frac{2^k}{3^k} \leq \epsilon$ Feb 1 at 16:22
I think (1) is correct, but I wouldn't bother with the infima: let $$a,b\in A$$. Then
$$|d(x,a)-d(y,b)|\leq |d(x,a)-d(a,y)|+|d(a,y)-d(y,b)|\leq d(x,y)+d(a,b)$$ where we used the triangle inequality. Take infimum on both sides as $$a\in A$$, while $$b$$ is (for now) fixed, so $$|f(x)-d(y,b)|\leq d(x,y)+0=d(x,y)$$. Now this is true for all $$b\in A$$, so take infimum over $$b$$ and conclude that $$|f(x)-f(y)|\leq d(x,y)$$, I think this is more neat.
For (2) this is a standard topological fact: If $$f:X\to Y$$ is a continuous bijection from a compact space to a Hausdorff space, then it is a homeomorphism. Indeed, all we need to do is show that $$f^{-1}$$ is continuous: if $$E\subset X$$ is a closed subset, we want to show that the inverse image of $$E$$ through $$f^{-1}$$ is closed in $$Y$$, which is the same as $$f(E)$$ is closed in $$Y$$. Now $$X$$ is compact and $$E$$ is a closed subset, so $$E$$ is also compact. Now $$f$$ is continuous, so $$f(E)$$ is a compact subset of $$Y$$. Now $$Y$$ is Hausdorff, so all compact sets are closed, thus $$f(E)$$ is closed in $$Y$$. Sometimes it helps viewing things in a more abstract frame; for example the metrics here are not essentially needed, so the extra information causes extra confusion.
For (3) you have a correct guess: Indeed, every element of the Cantor set $$C$$ can be written uniquely as $$\sum_{k=1}^\infty\frac{a_k}{3^k}$$, where $$a_k$$ is either $$0$$ or $$2$$. But it is important to note that the converse is also true: Any sum $$\sum_{k=1}^\infty\frac{a_k}{3^k}$$ where $$a_k\in\{0,2\}$$ is an element of the cantor set. Now define $$f:C\to P$$ as $$f\big(\sum_{k=1}^\infty\frac{a_k}{3^k}\big)=(\frac{a_k}{2^{k+1}})_{k=1}^\infty$$
• $$f$$ is well-defined: since $$a_k$$ is either $$0$$ or $$2$$, we have that $$a_k/2^{k+1}$$ is either $$0$$ or $$1/2^k$$, so each $$a_k/2^{k+1}$$ does indeed belong to $$\{0,\frac{1}{2^k}\}$$, so $$f$$ does indeed take values in $$\prod_{k=1}^\infty\{0,\frac{1}{2^k}\}$$.
• $$f$$ is one to one: this is obvious.
• $$f$$ is surjective: indeed, let $$x=(x_k)_{k=1}^\infty\in P$$, so each $$x_k$$ is either $$0$$ or $$1/2^k$$. Set $$a_k=2^{k+1}\cdot x_k$$. Then $$a_k$$ is either $$0$$ or $$2$$, so if $$t=\sum_{k=1}^\infty\frac{a_k}{3^k}\in C$$ then $$f(t)=x$$.
• $$f$$ is continuous: Let $$\varepsilon>0$$ and fix a point $$x\in C$$, say $$x=\sum_{k=1}^\infty\frac{a_k}{3^k}$$, where $$a_k\in\{0,2\}$$. Let $$y\in C$$ be another element, say $$y=\sum_{k=1}^\infty\frac{b_k}{3^k}$$ where $$b_k\in\{0,2\}$$. Note that $$d_P(f(x),f(y))=\|\{\frac{a_k}{2^{k+1}}\}_{k=1}^\infty-\{\frac{b_k}{2^{k+1}}\}_{k=1}^\infty\|_{\ell^1}=\|\{\frac{a_k-b_k}{2^{k+1}}\}_{k=1}^\infty\|_{\ell^1}=\sum_{k=1}^\infty\frac{|a_k-b_k|}{2^{k+1}}$$
Claim: Let $$m\geq1$$ be an integer. Then if $$|x-y|<\frac{1}{3^m}$$, then $$a_k=b_k$$ for all $$k=1,\dots,m$$.
Proof of the claim: I will only describe the idea here. Note that the construction of the Cantor set is done in steps: in each step we divide the segments in 3 pieces and throw out the middle part. In the expression $$x=\sum_{k=0}^\infty\frac{a_k}{3^k}$$ with $$a_k=0$$ or $$2$$, we have the following interpretation: in the first step of the construction, we divide $$[0,1]$$ in three segments and we only keep $$[0,1/3]$$ and $$[2/3,1]$$. If $$a_1=0$$, then $$x$$ lies in $$[0,1/3]$$. If $$a_1=2$$, then $$x$$ lies in $$[2/3,1]$$. So if $$|x-y|<1/3$$ where $$y\in C,y=\sum_{k=1}^\infty\frac{b_k}{3^k}$$ with $$b_k=0$$ or $$2$$, then $$b_1=a_1$$: otherwise we would have $$x\in[0,1/3]$$ and $$y\in[2/3,1]$$ or $$x\in [2/3,1]$$ and $$y\in[0,1/3]$$, which contradicts the fact that $$|x-y|<1/3$$. Let's assume that $$a_1=0$$, so $$x\in[0,1/3]$$. Now again, in the second step of the construction, we cut each of $$[0,1/3]$$ and $$[2/3,1]$$ in three pieces each and throw out the middle third of each. Let's look at $$a_2$$. If $$a_2=0$$, then $$x$$ lies in the left segment of $$[0,1/3]$$ that is left after the second step, i.e. $$x\in[0,1/3^2]$$. If $$a_2=2$$, then $$x$$ lies in the right segment of $$[0,1/3]$$ that is left after the second step, i.e. $$x\in[2/3^2,1/3]$$. Now if $$|x-y|<1/3^2$$ then we also have $$|x-y|<1/3$$, so $$y$$ lies in $$[0,1/3]$$ as we explained for the first step, i.e. $$b_1=0$$. Now if $$b_2\neq a_2$$, we have that $$x,y$$ are in different thirds of $$[0,1/3]$$ after the application of the second step of the construction, so $$|x-y|\geq 1/3^2$$, a contradiction. I believe that you get the idea.
Now let $$m$$ be so large that $$\sum_{k={m+1}}^\infty\frac{1}{2^k}<\varepsilon$$. Taking $$\delta=\frac{1}{3^m}$$; from the claim we have that if $$y\in C$$ satisfies $$|x-y|<\delta$$, then $$a_k=b_k$$ for all $$k=1,\dots,m$$, so $$d_P(f(x),f(y))=\sum_{k=m+1}^\infty\frac{|a_k-b_k|}{2^{k+1}}\leq\sum_{k=m+1}^\infty\frac{2}{2^{k+1}}<\varepsilon$$ and continuity follows. Observe that $$\delta$$ is independent of $$x$$, so $$f$$ is uniformly continuous.
Is this good news? No, this is expected, since $$f$$ is continuous on a compact space, thus it must be uniformly continuous. Also, by (2) this gives us a homeomorphism of the cantor set and $$P$$. | 2021-09-26T04:26:00 | {
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http://math.stackexchange.com/questions/493519/calculate-sum-n-1-inftyn2qn-1/493558 | # Calculate $\sum_{n=1}^{\infty}n^2q^{n-1}$
Please show me how to calculate the sum of this infinite series:
$$\sum_{n=1}^{\infty}n^2 q^{n-1}$$
I should have included the condition $\mid q\mid$<1
And I was able to solve the infinite series of $$S_n=\sum_{n=1}^{\infty}n q^{n-1}=1+2q+3 q^2+4q^3...$$ The trick is to calculate $$q S_{n}=\sum_{n=1}^{\infty}n q^{n}=q+2q^2+3q^3+...$$ And find out that $$S_{n}-q S_{n}=\sum_{n=1}^{\infty}=1+q+q^2+q^3+...=\sum_{n=1}^{\infty}q^n=\frac{1}{1-q}$$ And thus $$S_{n}=\frac{\frac{1}{1-q}}{1-q}=\frac{1}{(1-q)^2}$$
But I was not able to use same trick on the series I want to solve.
P.S. @user17762 has provided a genius way to handle this kind of series and his approach could simplify the calculation of $\sum_{n=1}^{\infty}n q^{n-1}$ too. Just watch the first two steps he used.
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Welcome to Math.SE! Please, consider updating your question to include what you have tried / where you are getting stuck. You will find that people on this site will be significantly faster to help you if you do that; that way, we know exactly what help you need. – Did Sep 14 '13 at 16:13
Do you mean $n^2(q^n-1)$ or $(n^2q^n) - 1$? Notationally, the latter is more usual, but I suspect you meant the former. Also, notice that when $q = 10$ the sum is obviously infinite, so make sure you aren't missing out any important details. – Ben Millwood Sep 14 '13 at 16:14
Your series converges only when $\vert q \vert <1$. Here is one possible way to derive the sum of the infinite series, for $\vert q \vert <1$. We have $$\sum_{n=1}^{\infty} q^n = \dfrac{q}{1-q} = -1 - \dfrac1{q-1}$$ Differentiate this once to get $$\sum_{n=1}^{\infty} n q^{n-1} = \dfrac{d}{dq} \left(\dfrac{q}{1-q} \right) = \dfrac1{(q-1)^2}$$ Multiplying by $q$, we get that $$\sum_{n=1}^{\infty} n q^{n} = \dfrac{q}{(q-1)^2} = \dfrac1{q-1} + \dfrac1{(q-1)^2}$$ Differentiate this again, to get $$\boxed{\displaystyle \sum_{n=1}^{\infty} n^2 q^{n-1}=-\dfrac1{(q-1)^2} - \dfrac2{(q-1)^3} = \color{blue}{\dfrac{(1+q)}{(1-q)^3}}}$$
Thanks a lot, your approach is quite genius. I was having trouble to solve this sum the whole morning, I should have mentioned that $\mid q\mid<1$. Your solution is also illuminating, I never thought of that way. P.S. Can I ask how you made the last formula blue? I am new here. – Kenneth Chen Sep 14 '13 at 18:39
@RencoHuangyuChen To get the text in blue color, you need to do as follows: $\color{blue}{\text{This text is blue in color}}$ gives $\color{blue}{\text{This text is blue in color}}$ – user17762 Sep 14 '13 at 19:07
${\large\tt\mbox{\color{#0000ff}{blue}}}$ yields $\color{#0000ff}{blue}$. – Felix Marin Sep 14 '13 at 19:08 | 2016-02-13T04:41:33 | {
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http://math.stackexchange.com/questions/180072/is-there-more-than-one-way-to-express-a-derivative-as-a-limit-of-a-quotient?answertab=oldest | # Is there more than one way to express a derivative as a limit of a quotient?
Let $r(t)$ be a real-valued function of $t$. Let $v(t)$ be the derivative of $r(t)$. Then $$v(t) = \frac{dr(t)}{dt} = \lim_{\Delta t \to 0} \frac{r(t + \Delta t) - r(t)}{\Delta t}$$ so $$v(t) = \frac{dr(t)}{dt} \approx \frac{r(t + \Delta t) - r(t)}{\Delta t} \text{ for small }\Delta t$$
My question is, is there another way to approximate $v(t) = \dfrac{dr(t)}{dt}$?
For example, I am reading the book Understanding Molecular Simulation by Frenkel and Smit (Second Edition). On page 71 (some pages are available on Google Books here), the authors write $$v(t) = \frac{r(t + \Delta t) - r(t - \Delta t)}{2 \Delta t} + \mathcal{O}(\Delta t^2)$$ or, in other words, $$v(t) \approx \frac{r(t + \Delta t) - r(t - \Delta t)}{2 \Delta t}$$
Basically, then, it seems that there are two ways to express $v(t) = \dfrac{dr(t)}{dt}$:
$$v(t) = \frac{dr(t)}{dt} = \lim_{\Delta t \to 0} \frac{r(t + \Delta t) - r(t)}{\Delta t} \textbf{ (1)}$$
$$v(t) = \frac{dr(t)}{dt} = \lim_{\Delta t \to 0} \frac{r(t + \Delta t) - r(t - \Delta t)}{2 \Delta t} \textbf{ (2)}$$
Are equations (1) and (2) equivalent? Equation (1) Is the definition of derivative that I remember from high school calculus; I do not remember (2). Is (2) an alternative definition of the derivative? Or, what is the relationship between (1) and (2)?
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i can think about one thing,they are same according to following situation,consider difference between $5$ and $4$ difference is $1$ while difference between $5$ and $3$ is $2$ or $2*1$,i think the same principle is here – dato datuashvili Aug 7 '12 at 20:33
There are many ways to write a modified version of the definition of the derivative that have in fact higher convergence rates and are thus often desirable to use in numerical computation.
Finite difference coefficients can be used for this. For example, we have
$$f'(x_0)=\frac{f(x+h)-f(x)}{h}+O(h)$$
but
$$f'(x_{0}) \approx \displaystyle \frac{-\frac{11}{6}f(x_{0}) + 3f(x_{+1}) -\frac{3}{2}f(x_{+2}) +\frac{1}{3}f(x_{+3}) }{h_{x}} + O\left(h_{x}^3 \right)$$
Addendum: Here is a giant list of high-accuracy derivatives using these coefficients.
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Thanks. Does this mean that equation (2) in the original post is a correct definition of the derivative? – Andrew Aug 7 '12 at 21:14
@Andrew Yes, very much so! This is a basic three-point estimate. In fact, this derivative formula is mentioned here: en.wikipedia.org/wiki/… - the error seems to be $$\frac{f^{(3)}(c)h^2}{6}$$ where $$c \in [x-h, x+h]$$ – Argon Aug 7 '12 at 21:17
Thanks! So, based on the link you provided, it seems that equation (2), a three-point estimate, is an actually more accurate approximation of the derivative than equation (1), a two-point estimate. Is this true? – Andrew Aug 7 '12 at 21:24
@Andrew Yes, it is true. Instead of having a bias to the side that the limit is taken, this method has a balance. – Argon Aug 7 '12 at 21:27
"...thus often desirable to use in numerical computation." - not always. To borrow a mantra from a certain popular book on numerics, high order does not necessarily imply high accuracy. One must remember that the basis of the high-order approximations is the use of an interpolating polynomial at equispaced points, and it is well-known how unruly high-order interpolating polynomials can be... – Ｊ. Ｍ. Aug 7 '12 at 23:37
If the limit (1) exists, the limit (2) exists. To see this, write $$\frac{r(t+h)-r(t-h)}{2h}=\frac12\left(\frac{r(t+h)-r(t)}{h}+\frac{r(t+(-h))-r(t)}{(-h)}\right).$$ The other implication is wrong, otherwise every even function would be differentiable at $t=0$ with derivative zero (but $t\mapsto|t|$ is a counterexample).
Hence (2) is not a definition of the derivative. The definition of the derivative is (1). But as soon as (1) holds, the derivative coincides with the limit $$\lim\limits_{h\to0}\frac{r(t+ah)-r(t-bh)}{(a+b)h},$$ for every $(a,b)$ such that $a+b\ne0$, and in particular with the limit in (2).
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Here is yet another equivalent definition: $f'(x) = \lim_{r \to 1} \frac{f(rx) - f(x)}{(r-1)x}$, if $x \ne 0$. The differentiation rule $\frac{d}{dx}x^n = n x^{n-1}$ is especially easy to derive with this definition.
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Thanks. Are you implying that equation (2) in the original post is indeed a correct definition of a derivative? The answer by did seems to imply it is not. – Andrew Aug 7 '12 at 20:54
@Andrew, it isn't a good definition. See this. – Ｊ. Ｍ. Aug 7 '12 at 23:42
@Andrew: When you have questions about some parts of an answer, you might as well ask directly the author. – Did Aug 24 '12 at 15:45
Consider the linear terms in the Taylor series of $r(t+h)$ and $r(t-h)$, where $h = \Delta t$ and $r'(t)=v(t)$, $$r(t+h) \approx r(t) + r'(t)\,h$$ and $$r(t-h) \approx r(t) - r'(t)\,h$$ Multiplying the second equation by $-1$ and adding $$2h r'(t) = r(t+h) + r(t-h) \Rightarrow v(t) \approx \frac{r(t+h) + r(t-h)}{2h}\,,$$
which implies
$$v(t) = \lim_{h \to 0} \frac{r(t+h) + r(t-h)}{2h}$$
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As a complement to other fine answers (did's for example) let's suppose that $f$ admits a Taylor expansion at $t$ then :
$$f(t+h)=f(t)+hf'(t)+\frac {h^2}2 f''(t)+\frac {h^3}6 f'''(t) +O\left(h^4 f''''(t)\right)$$
so that : $$\tag{1}\frac{f(t+h)-f(t)}h=f'(t)+\frac {h}2 f''(t)+\frac {h^2}6 f'''(t) +O\left(h^3 f''''(t)\right)$$ while : $$\tag{2}\frac{f(t+h)-f(t-h)}2=f'(t)+\frac {h^2}6 f'''(t) +O\left(h^3 f''''(t)\right)$$
That is why the second method is more precise : the $f''$ term disappeared (well all the even terms in fact!) so that $f'$ will be evaluated with more precision (with an $h^2$ error instead of $h$). Because of the precision obtained Feynman proposed this second method to evaluate derivatives in his famous Physics Lectures (Vol I 9-6).
But when there is no convergence ($f(t)=\frac 1t$ for example) you'll get a limit in the second case ($0$) without any problem and that's clearly different of the first case.
The second method is much used too when you need a 'discretized' version of a differential equation (respecting time symmetry, energy conservation and so on). Ed Fredkin for example proposed following equation in his article "Feynman, Barton and the reversible Schrödinger difference equation" : $$\frac{C_{x,t+1}-C_{x,t-1}}2=ik\left(C_{x-1,t}-2C_{x,t}+C_{x+1,t}\right)$$
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@Raymound:These techniques are used in numerical methods. – Mhenni Benghorbal Aug 7 '12 at 23:11
@MhenniBenghorbal: yes you are right (I insisted on the symmetric part of the second method), – Raymond Manzoni Aug 7 '12 at 23:13
Oh, the central difference is very useful in numerics indeed. See this. – Ｊ. Ｍ. Aug 7 '12 at 23:35
Thanks for the link(s) @J.M. ! – Raymond Manzoni Aug 7 '12 at 23:37 | 2014-12-21T09:03:58 | {
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https://math.stackexchange.com/questions/1875045/is-this-lot-drawing-fair/1875053 | # Is this lot drawing fair?
Sorry for a stupid question, but it is bugging me a lot.
Let's say there are $30$ classmates in my class and one of us has to clean the classroom. No one wants to do that. So we decided to draw a lot - thirty pieces of paper in a hat, one of which is with "X" on it. The one who draws "X" has to do the cleaning. Each one starts to draw...
Is this kind of lot drawing fair or not fair? It looks to me like the first one's chances to get an "X" are equal to $1/29$, while the second one's chances would be equal either to $1/28$ (in case the first one didn't draw an "X") or zero $0/29 = 0$ (in case the first one drew an "X"). However, neither $1/28$, nor $0/29$ is equal to $1/29$.
• Why not draw names instead of an X? Seems like that would be significantly easier, and gives everyone a 1/30 chance. – Carcigenicate Jul 29 '16 at 21:35
• Say we flip a coin and write down on a piece of paper whether it came up heads or tails. Then we look at the piece of paper. The odds it will say "heads" are either 100% (if the coin came up heads) or 0% (if the coin came up tails). But neither of those is 50%. Does that make the error in the reasoning more obvious? – David Schwartz Jul 29 '16 at 21:49
• The odds for the first person are 1:29, but that's a probability of 1/30. – Hurkyl Jul 29 '16 at 23:40
• This is essentially the Gambler's fallacy. – argentpepper Jul 30 '16 at 3:28
P(first one draws $X) = \dfrac1{30}$
P(first one doesn't draw $X) = \dfrac{29}{30}$
The possibility of the second one drawing $X$ arises if and only if the first one doesn't, so
P(second one draws $X) = \dfrac{29}{30}\cdot\dfrac1{29} = \dfrac1{30}$ again !
You will find, if you continue in the same way, that each one will have the same probability, $\dfrac1{30}$, and you needn't compute at all !
Imagine all papers arranged randomly in a line.
The $X$ is as likely to be in the first place as the last (or any other)
• I don't mean to be pedantic because I really love this answer, but for the OP: remember that $P(\text{Student 2 draws X})$ is implicitly stating that Student 2 can only draw $X$ if and only if Student 1 didn't draw $X$. This is why we have to multiply the probabilities together, which seemed to be your confusion in the comment to another answer. – anonymouse Jul 29 '16 at 17:01
• @anonymouse: Thanks, I have added a line to my answer. I thought the confusion OP had was not about the multiplication , but why Pr for the second person was 1/29, why not 1/28, (given that first hadn't drawn X). 1/29 vs 1/28 became the dominant discussion a bit later. – true blue anil Jul 29 '16 at 17:43
• Thanks for your answer. But I wonder why do we need to calculate probability at all? Why not just compare the chances - the way I did in my question? – brilliant Aug 1 '16 at 2:07
• If you can get the insight contained in the last two lines of the answer, there is no need to calculate, and no need to compare, because (using the technically correct terms), the probability is the same for all of them at $\frac1{30}$, or the odds at $1:29$ – true blue anil Aug 1 '16 at 10:03
It's fair. The easiest way to see that is to imagine the people drawing their papers but not looking until all have got theirs.
Each person's chance is $1/30$.
In your way of thinking about the problem the first person's chance of getting the X (and having to clean) is 1/30 (not 1/29 as you state in your question - those are the odds, not the probability).
If the first person doesn't get the X then the second person's chance is 1/29, but it wasn't 1/29 from the start - it only became 1/29 once you knew the first person didn't get the X.
To find the true probability for the second person you have to reason backwards from the 1/29 chance he sees, to take into account the fact that he wouldn't have to draw at all when the first person gets the X. That happens 1/30 of the time, so the second person has to draw just 29 times out of 30. So the probability (from the start) that the second person draws the X is $$\frac{29}{30} \times \frac{1}{29} = \frac{1}{30}$$
This is the essential algebra in many of the other correct answers.
If by chance none of the first 29 people gets the X then the chance that the last person does is 1/1, which is no surprise. But it wasn't 1/1 from the start, of course. You can make the algebra show you that too.
Edit: To address the OP's question about odds.
Odds and probabilities are two different ways to express the same mathematics. In your problem the first person to draw has a probability of 1/30 to draw the X. That means he (or she) cleans 1 time out of 30. His odds are 1:29, which you read aloud as "1 to 29". The odds mean he cleans 1 time for each 29 times he doesn't. Odds aren't usually written as fractions, because that often leads to confusion.
You can figure out problems like yours working with probabilities or with odds, but it's easier with probabilities as all the answers show.
Your logical error about the odds is this.
If the first person doesn't get the X then the second person's odds are indeed just 1:28, as you say, but they weren't 1:28 from the start - they only became 1:28 once you knew the first person didn't get the X.
To find the true odds at the start for the second person you have to reason backwards from the 1:28 odds in his draw to take into account the fact that he wouldn't have to draw at all when the first person gets the X. That's what using probability instead of odds lets you do, as above.
You might be able to see the error in your logic if you imagine there are three people instead of 30. The first person has odds of 1:2. If he doesn't get the X then the second person has odds of 1:1 (even chance, which makes sense) but that isn't the case from the start since he actually has to draw only 2/3 of the time.
• Was going to say the same thing about not looking at the paper. – shawnt00 Jul 31 '16 at 2:15
• Thank you for your answer, but it looks like your calculating the probability here and not the chances (odds). I wonder why do we need to calculate probability at all? Why not just compare the chances - the way I did in my question? – brilliant Aug 1 '16 at 2:08
• @brilliant You're welcome. Please see my edit explaining your logical error in terms of odds rather than probabilities. It's the same error. – Ethan Bolker Aug 1 '16 at 13:17
• For some reasons, I read this answer in Kevin Spacey's voice. +1 though! – MonK Aug 1 '16 at 13:36
The draw is fair. It just seems counter-intuiutive because of the way you are thinking of it.
Suppose that all thirty pieces of paper were handed out simultaneously and all at once the papers were flipped and the person with the $X$ would clean. This would seem fair, right?
Well, drawing a paper one at a time is virtually the same thing, the only difference is that the draw stops once the $X$ is revealed. Everyone still has a $\frac{1}{30}$ chance of getting the $X$.
Fair.
Give all pieces of paper a distinct letter (among them $A$ and $X$).
What is the probability that - let's say the one drawing as $23$rd - get's e.g. letter $A$?
Would this be different for letter $X$?
No. Every letter has the same chance of falling in hands of number $23$.
The first one has a chance of $1/30$, not $1/29$, as there are 30 pieces of paper in the box (if there were only 29 pieces in the box, it would obviously not be fair, as the last one would never get the cross: the other 29 classmates would already have removed all the pieces).
Obviously, if one classmate is chosen at random, he should have a chance of $1/30$ to be the chosen one, therefore it's fair to him.
Now after he has drawn the paper and found no cross (and assuming he didn't put it back into the box— another way to make it unfair), there are only 29 candidates left. Therefore it is obvious that if everyone of them has to have a fair chance, now for each of them the probability to be the chosen one is to be $1/29$. And indeed, that's the chance that the next one taking a piece from the box has, as there are only 29 pieces of paper left.
Note that this is a conditional probability (on the condition that the first one didn't get the cross). To calculate the total probability, you have to multiply the conditional probability with the probability of the condition. That condition is that the first one did not get a cross, and the probability of that condition is $29/30$ (since with probability $1/30$ he did get the cross). So the total probability that the second one gets selected is $$\frac{29}{30}\cdot\frac{1}{29} = \frac{1}{30}$$ as it should be. So everything is still fair.
Now it isn't hard to check that it remains fair also for the remaining classmates, as at any time there are exactly as many pieces of paper left in the box as there are classmates that didn't yet draw one.
Edit:
From your comments on other answers I gather that you don't understand why the probability for the first one is $1/30$ and not $1/29$. Well, I guess you are simply confusing the probability with the odds.
The probability tells you how often the event will happen if you repeat the experiment many times. That is, if the event (in this case, the first classmate getting the "X") has probability $p$, and you repeat the whole procedure $N$ times, then the expected number of times the even happens is $pN$. In this case, there are 30 different pieces of paper, and on average he will draw each one equally often. Therefore at one of 30 draws he will get the cross, and thus the probability is $1/30$
The odds on the other hand give the relation of the event happening versus the event not happening. I this case, there's one piece with a cross, and 29 without, each of which he will draw with the same probability, and therefore his odds to draw the cross are $1:29$.
Note the different way I wrote this, as this is not really a division (although it is close to it). In particular, if the box only contained the piece with the cross, his odds would be $1:0$; this is not a division by zero error, but a completely valid odds. On the other hand, the probability to draw the cross is $1/1 = 1$. This is a division, and a well-defined one. The value $1$ means he will draw the cross every time ($pN = N$), and that's indeed what will happen if the only piece of paper in the box is the one with the cross.
• This reminds me of homogenous coordinates, which is a way of representing points in projective geometry that does not suffer from the problem of infinite gradient (division by zero). – user21820 Jul 30 '16 at 4:55
• Thank you for you answer and kind explanations.I see now that there is a difference between probability and chances (odds). I wonder why do we need to calculate probability at all? Why not just compare the chances - the way I did in my question? – brilliant Aug 1 '16 at 2:06
• @brilliant: Most of things in probability theory have been done with probabilities (well, the name is already a dead giveaway!). If you would want to do everything in odds, you'd have to first translate all the rules from probabilities to odds. I'm also not sure if odds can be rigorously defined when going beyond finite number of possible outcomes. – celtschk Aug 1 '16 at 7:08
• (+1)I found your answer more helpful than the other answers...(I don't know why).... – tatan Aug 12 '16 at 5:03
The second person's chance is still $1/30$. Indeed, Person 2 draws an X exactly when Person 1 does NOT draw an X, and so: \begin{align*} \Pr[P_2 = X] &= \Pr[P_1 \neq X] \cdot \Pr[P_2 = X \mid P_1 \neq X] \\ &= \frac{29}{30} \cdot \frac{1}{29} \\ &= \frac{1}{30} \end{align*} Likewise, Person 3 draws an X exactly when the previous two people do NOT draw an X: \begin{align*} \Pr[P_3 = X] &= \Pr[P_1 \neq X] \cdot \Pr[P_2 \neq X \mid P_1 \neq X] \cdot \Pr[P_3 = X \mid P_2 \neq X] \\ &= \frac{29}{30} \cdot \frac{28}{29} \cdot \frac{1}{28} \\ &= \frac{1}{30} \end{align*}
This continues all the way up to Person 29 and Person 30: $$\Pr[P_{29} = X] = \frac{29}{30} \cdot \frac{28}{29} \cdot \frac{27}{28} \cdot \ldots \cdot \frac{3}{4} \cdot \frac{2}{3} \cdot \frac{1}{2} \\ \Pr[P_{30} = X] = \frac{29}{30} \cdot \frac{28}{29} \cdot \frac{27}{28} \cdot \ldots \cdot \frac{3}{4} \cdot \frac{2}{3} \cdot \frac{1}{2} \cdot \frac{1}{1} = \frac{1}{30}$$
• Can you, please, explain this to me in a more simple way? It looks to me that if I have a box with 30 apples and one of them is rotten, while my friend has a box of 29 apples with one of them rotten, he seems to have more chances to draw a rotten apple than me. No? – brilliant Jul 29 '16 at 15:37
• This is, of course, correct - but probably not useful to the OP - if he knew how to reason with conditional probabilities he wouldn't have had to ask the question. – Ethan Bolker Jul 29 '16 at 15:38
• @EthanBolker - Yes. You are right. I am like an idiot here. I see that all people here see right away that the draw is fair, while to me it looks unfair. And I just can't understand where my logic is failing me. – brilliant Jul 29 '16 at 15:48
• @brilliant I've edited my answer to try to show where your logic fails and how to correct it. – Ethan Bolker Jul 29 '16 at 15:58
• I wonder why do we need to calculate probability at all? Why not just compare the chances - the way I did in my question? – brilliant Aug 1 '16 at 2:09
Actually, it looks like its fair to me.
The probability that person 1 chooses an "X" is $\frac{1}{30}$, since only one of the 30 lots has an "X" on it.
The probability that person 2 chooses an "X" is $\frac{29}{30}*\frac{1}{29} =\frac{1}{30}$, since Person 1 must have not chosen a lot with an "X" on it.
The probability that person 3 chooses an "X" is $\frac{29}{30}*\frac{28}{29}*\frac{1}{28} =\frac{1}{30}$
You can show that this is true for every successive person, and to further prove this, we can look at the 30th person.
The probability that person 30 chooses an "X" is the probability that everybody else did not choose an "X", which is $\frac{29!}{30!} = \frac{1}{30}$
So it is indeed fair for every person.
• "The probability that person 2 chooses an "X" is (29/30)*(1/29) = 1/30 since Person must have not chosen a lot with an "X" on it" - Why is it (29/30)*(1/29) and not 1/28? After all, the second person is now drawing from the pool of 29 lots (that is, not 30 anymore)? – brilliant Jul 29 '16 at 15:31
• The first person is drawing from a pool of $30$, and the second person is drawing from a pool of $29$, and so on... – true blue anil Jul 29 '16 at 15:34
• @trueblueanil - Exactly! – brilliant Jul 29 '16 at 15:34
• When the first person draws, 1 out of the 30 lots are going to have an "X" on it. Thus his odds of drawing the "X" are 1/30 . The second person draws from a pool of 29 lots. Thus the probability that he chooses an "X" is 1/29 , iff the first person did not draw an "X". The probability that the first person does not draw an "X" is 29/30 since there are 29 lots without an "X" on it when he draws. Thus Person 2's probability of drawing an X is (29/30)*(1/29) = 1/30 . Does that help? – Ninja_Coder Jul 29 '16 at 15:37
• @brilliant Which becomes obvious if you think about the opposite side. What is the probability of drawing anything other than the ace of spades? Well, there are 51 cards that aren't the ace of spades, and only 1 that is. By your logic the probability would be 51/1, but it should be 51/52. – Paulpro Jul 29 '16 at 15:48
You've identified three outcomes for the second person:
• Person 2 doesn't draw
• Person 2 draws and doesn't get the X
• Person 2 draws and gets the X
and you're interested in the probability that the last of the three outcomes happens.
You then divided the outcomes into two categories:
• Person 2 doesn't draw
• Person 2 draws
You've even (nearly) correctly computed the probabilities
• Assuming person 2 doesn't draw, he has probability 0 of getting the X
• Assuming person 2 does draw, he has probability 1/29 of getting the X
But where you went off the rails is that you forgot your original goal was "What is the probability he gets the X?", and neither of the calculations you did were to compute that. They're similar-sounding questions, but similar does not mean the same. This is especially true for questions about probability, counting, and statistics, where sloppy reasoning often leads to wildly incorrect results.
If you are to use the work you've done, the problem now is how to correctly combine the two probabilities. This can be done with an additional piece of information:
• The probability that person 2 doesn't draw is 1/30
• The probability that person 2 does draw is 29/30
And thus we can compute
• The proportion of samples that person 2 doesn't draw is 1/30
• Of those, the proportion where person 2 gets the X is 0
• The total proportion of this case is $\frac{1}{30} \cdot 0 = 0$
• The proportion of samples that person 2 does draw is 29/30
• Of those, the proportion where person 2 gets the X is 1/29
• The total proportion of this case is $\frac{29}{30} \cdot \frac{1}{29} = \frac{1}{30}$
and then adding up all of the different cases gives $0 + \frac{1}{30} = \frac{1}{30}$. | 2019-05-21T15:17:09 | {
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https://math.stackexchange.com/questions/4175040/how-many-non-congruent-triangles-can-be-formed-by-the-vertices-of-a-regular-poly | # How many non-congruent triangles can be formed by the vertices of a regular polygon of $n$ sides
On a circle there are $$n$$ evenly spaced points (i.e., they form a regular polygon). How many triangles can be formed when congruent triangles are considered the same?
This question is closely related to How many triangles can be formed by the vertices of a regular polygon of $n$ sides?, but here congruent triangles are considered to be the same.
• Have you attempted any examples for small $n$? Do you see a pattern? For each triangle type, can you find a standard form for that triangle? Jun 16 at 23:10
• for n=3,4, clearly 1; for n=5, should be 2; for n=6, should be 3; etc. I think n=12 (clock), it should be 12, but I am not sure Jun 16 at 23:31
• oeis.org/A001399
– bof
Jun 17 at 0:13
For each triangle $$A_iA_jA_k$$, the angles $$\angle A_iOA_j$$, $$\angle A_jOA_k$$ and $$\angle A_kOA_i$$ are respectively equal to $$|j-i|\frac{2\pi}{n}$$, $$|j-k|\frac{2\pi}{n}$$ and $$|k-i|\frac{2\pi}{n}$$ with $$1\le i\ne j\ne k \le (n-1)$$.
Let denote $$\begin{cases} x=|j-i|\\ y=|k-j|\\ z=|k-i| \end{cases}$$
Then $$\begin{cases} x+y+z = n \\ x,y,z \in \Bbb N^* \\ \end{cases} \tag{1}$$
The number of triangles when congruent triangles are considered the same is equal to the number of solutions of $$(1)$$ (Two solutions that differ only in the order of their summands are considered the same, for example, with $$n = 4$$: $$1+1+2$$ is the same as $$2 +1 +1$$).
The problem $$(1)$$ is solved here and the number of solutions of $$(1)$$ is the number of partitions $$p(3,n)$$ of $$n$$ into $$3$$ non-zero parts. The number of partitions $$p(k,n)$$ satisfies $$\begin{cases} p(0,0) &= 0 \\ p(k,n) &= p(k,n-k)+ p(k-1,n-1) & \text{otherwise}. \\ \end{cases} \tag{2}$$ Remark: It seems that the recurrent formula in the answer is not correct, the recurrent formula $$(2)$$ is taken from the wikipedia here. From $$(2)$$, we can deduce the general formula of $$p(3,n)$$ as follows
• For $$k = 1$$, it's obvious that $$p(1,n) = 1$$
• For $$k =1$$, we have $$p(2,n) = p(2,n-2)+1 =... \implies p(2,n) = \left\lfloor \frac{n}{2} \right\rfloor$$
• For $$k = 3$$, we have $$p(3,n) =p(3,n-3)+\left\lfloor \frac{n-1}{2} \right\rfloor=...$$
\begin{align} \implies p(3,n) &=\left\lfloor \frac{n-1}{2} \right\rfloor+\left\lfloor \frac{n-4}{2} \right\rfloor+...+\left\lfloor \frac{n-1-3i}{2} \right\rfloor+... \\ & =\sum_{0 \le i \le \left\lfloor \frac{n-1}{3}\right\rfloor}\left\lfloor \frac{n-1-3i}{2} \right\rfloor \end{align}
Hence, the number of triangles is equal to $$p(3,n)=\sum_{0 \le i \le \left\lfloor \frac{n-1}{3} \right\rfloor}\left\lfloor \frac{n-1-3i}{2} \right\rfloor$$
Remark: The last steps have a lot of calculation, please feel free to correct if you find any mistake.
• $$\left\lfloor\frac{n^2+3}{12}\right\rfloor$$
– bof
Jun 17 at 0:54
• @bof: does your formula come from oeis.org/A001399 ? If yes, perhaps, there is a way to simplify my formula to yours and a proof by recurrence will work.
– NN2
Jun 17 at 0:58
• The formula in my comment is from my notebook from a combinatorics class I took many years ago. It was homework and I think I proved it by cases: $n\equiv0\pmod6$, $n\equiv1\pmod6$, etc.
– bof
Jun 17 at 1:06 | 2021-12-02T19:10:41 | {
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https://www.jolana.cz/9mvw76j/rolle%27s-theorem-find-c-245e36 | # rolle's theorem find c
The function f(x) is only a problem if you attempt to take the square root of a number, that is if x > 3, hence f(x) satisfies the conditions of Rolle's theorem. • The Rolle’s Theorem must use f’(c)= 0 to find the value of c. • The Mean Value Theorem must use f’(c)= f(b)-f(a) to find value of c. b - a (Enter your answers as a comma-separated list. If the MVT cannot be applied, explain why not. Let’s now consider functions that satisfy the conditions of Rolle’s theorem and calculate explicitly the points c where $$f'(c)=0.$$ Example $$\PageIndex{1}$$: Using Rolle’s Theorem For each of the following functions, verify that the function satisfies the criteria stated in Rolle’s theorem and find all values $$c$$ in the given interval where $$f'(c)=0.$$ This calculus video tutorial explains the concept behind Rolle's Theorem and the Mean Value Theorem For Derivatives. Let’s introduce the key ideas and then examine some typical problems step-by-step so you can learn to solve them routinely for yourself. The next theorem is called Rolle’s Theorem and it guarantees the existence of an extreme value on the interior of … The Mean Value Theorem and Its Meaning. Yes, Rolle's Theorem can be applied. Find the value of c in Rolle's theorem for the function f(x) = x^3 - 3x in [-√3, 0]. If Rolle's Theorem cannot be applied, enter NA.) Extra Because the hypotheses are true, we know without further work, that the conclusion of Rolle's Theorem must also be true. Here in this article, you will learn both the theorems. Hopefully this helps! does, find all possible values of c satisffing the conclusion of the MVT. Rolle's Theorem was first proven in 1691, just seven years after the first paper involving Calculus was published. No, because f is not differentiable in the open interval (a, b). Solution for Check the hypotheses of Rolle's Theorem and the Mean Value Theorem and find a value of c that makes the appropriate conclusion true for f (x) = x³… Slight variation with fewer calculations: After you use Rolle's theorem, it suffices to note that a root exists, since $$\lim_{x\rightarrow \infty}f(x)=+\infty$$ and $$\lim_{x\rightarrow -\infty}f(x)=-\infty$$ Since polynomials are continuous, there is at least one root. Thus, $c = \frac{3\pi}{4} \in \left( 0, \pi \right)$for which Rolle's theorem holds. Get an answer for 'f(x) = 5 - 12x + 3x^2, [1,3] Verify that the function satisfies the three hypotheses of Rolle’s Theorem on the given interval. This version of Rolle's theorem is used to prove the mean value theorem, of which Rolle's theorem is indeed a special case. Rolle’s Theorem. Informally, Rolle’s theorem states that if the outputs of a differentiable function f f are equal at the endpoints of an interval, then there must be an interior point c c where f ′ (c) = 0. f ′ (c… If Rolle's Theorem can be applied, find all values of c in the open interval (a, b) such that. c= f '(c… can be applied, find all values of c given by the theorem. If a real-valued function f is continuous on a proper closed interval [a, b], differentiable on the open interval (a, b), and f (a) = f (b), then there exists at least one c in the open interval (a, b) such that ′ =. new program for Rolle's Theorem video Process: 1. f (x) = 5 tan x, [0, π] Yes, Rolle's Theorem can be applied. By mean, one can understand the average of the given values. Correct: Your answer is correct. Rolle’s theorem is satisfied if Condition 1 =2 + 2 – 8 is continuous at −4 , 2 Since =2 + 2 – 8 is a polynomial & Every polynomial function is c Since f(1) = f(3) =0, and f(x) is continuous on [1, 3], there must be a value of c on [1, 3] where f'(c) = 0. f'(x) = 3x^2 - 12x + 11 0 = 3c^2 - 12c + 11 c = (12 +- sqrt((-12)^2 - 4 * 3 * 11))/(2 * 3) c = (12 +- sqrt(12))/6 c = (12 +- 2sqrt(3))/6 c = 2 +- 1/3sqrt(3) Using a calculator we get c ~~ 1.423 or 2.577 Since these are within [1, 3] this confirms Rolle's Theorem. Mean Value Theorem & Rolle’s Theorem: Problems and Solutions. If Rolle's Theorem can be applied, find all values c in the open interval (a, b) such that f'(c) = 0. f(x) = cos 3x, [π/12, 7π/12] I don't understand how pi/3 is the answer.... Can someone help me understand? 4. =sin,[0,] Solve: cos = 0−0 −0 =0 Cosine is zero when = 2 for this interval. Click hereto get an answer to your question ️ (i) Verify the Rolle's theorem for the function f(x) = sin ^2x ,0< x
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Komentáře jsou zavřeny. | 2021-09-25T04:13:43 | {
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https://math.stackexchange.com/questions/2928369/difficulty-finding-ak | # Difficulty Finding $A^k$
Let $$A= \begin{bmatrix} 1& -1 & 1\\ 0 & 1 & 1 \\ 0 & 0 & 1\\ \end{bmatrix}$$. Compute $$A^k$$.
# My attempt
I'm trying to compute $$A^k$$ using this approach as follows: $$A=I+N= \begin{bmatrix} 1& 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}+ \begin{bmatrix} 0& -1 & 1\\ 0 & 0 & 1 \\ 0 & 0 & 0\\ \end{bmatrix}$$ with $$N^2= \begin{bmatrix} 0& 0 & -1\\ 0 & 0 & 0 \\ 0 & 0 & 0\\ \end{bmatrix}, \, \text{and} \, \, N^3= \begin{bmatrix} 0& 0 & 0\\ 0 & 0 & 0 \\ 0 & 0 & 0\\ \end{bmatrix}$$
Then, $$A^2=(I+N)^2=I+2N+N^2, \\ A^3=(I+N)^3=I+3N+3N^2, \\ A^4=(I+N)^4=I+4N+6N^2, \\ A^5=(I+N)^5=I+5N+10N^2, \\ A^6=(I+N)^5=I+6N+15N^2,$$
By induction, we can see $$A^k=(I+N)^k=I+kN+f[k]N^2$$. But, I couldn't figure out what $$f[k]$$ is. Any help?
• binomial coefficient $k$ choose $2,$ which become $k(k-1)/2$ – Will Jagy Sep 24 '18 at 0:30
• @WillJagy Exactly! It worked. Thank you. – Lod Sep 24 '18 at 0:46
How about this. Take the exponential function $$e^{tA}$$, where $$t$$ is some parameter $$e^{tA}=\sum_{k=0}^\infty\frac{t^kA^k}{k!}=e^{t(I+N)}= e^{tI}e^{tN}=e^t\left[I+tN+\frac{(tN)^2}{2}\right]$$ where we used the matrix identity $$e^{A+B}=e^Ae^B$$ that is valid when matrices $$A$$ and $$B$$ commute. Since the functions $$t^k$$ are linearly independent, we obtain $$A^k=I+kN+\frac{k(k-1)}{2}N^2$$
• That's even a better compact approach. However, I did not get how $t^2$ became $k(k-1)$. What is it that you mean with "the functions $t^k$ are linearly independent"? – Lod Sep 24 '18 at 1:26
• @Lod, you have to expand the exponential on the right side, the coefficients of $t^k$ on both sides of the equation have to be equal. The linear independence you can see from this: If you assume $\sum_{k=0}^\infty a_kt^k=0$ is valid for all $t$, this implies all $a_k=0$. To see this just take the $k$-th order derivative and then replace $t=0$. – minmax Sep 24 '18 at 1:37
You have written $$A=I+N$$, and you know that $$N^3$$ (and hence all higher powers) are zero. If $$X$$ and $$Y$$ are two matrices that commute with each other, then you can still use the bionomial theorem: $$(X+Y)^n=\sum_{i+j=n}\binom{n}{i} X^i Y^j$$
Because $$I$$ commutes with $$N$$, and because all the higher powers of $$N$$ vanish, we can apply the formula to get
$$(I+N)^n=\sum_{i+j=n}\binom{n}{j} N^j=I+nN+\binom{n}{2}N^2$$ | 2019-05-23T22:49:53 | {
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https://math.stackexchange.com/questions/2826837/probabilities-of-conditional-events | # Probabilities of Conditional Events
Each bag in one big box contains 25 fruits in amount. Given 60% of the bags contain 5 oranges and 20 apples, while 40% another contain 15 oranges and 10 apples. After that, one bag is taken randomly and then in that bag, one fruit is taken randomly too. \ a) What is the probability to get an apple? \ b) If the chosen fruit is an apple, then what is the probability to get the bag whose contains 5 oranges and 20 apples?
My friend said that we should try Bayes' Theorem. But honestly I somehow get some trouble how to apply that theorem to solve the problems above.
Is it true that Bayes' Theorem should be applied here or are there any ways to solve the problem? Thank you for your help.
• it is true. What was your trouble? Also, it's Bayes' – David Diaz Jun 21 '18 at 4:00
• Let $A$ be the event that you draw an apple and $A^c$ the event that you draw an orange. Let $B$ be the event that you draw a bag of the first variety, i.e. a bag with 5 oranges and 20 apples. The problem statement tells you the following probabilities directly: $Pr(B),Pr(B^c),Pr(A\mid B),Pr(A^c\mid B),Pr(A\mid B^c),Pr(A^c\mid B^c)$. The problem asks you to find $Pr(A)$ and then $Pr(B\mid A)$ given the above information, which indeed there is enough information to do so. – JMoravitz Jun 21 '18 at 4:06
• Note: $Pr(A)=Pr(A\cap B)+Pr(A\cap B^c)$ and note that $Pr(A\cap B)=Pr(B)Pr(A\mid B)$. Note also $Pr(B\mid A)=\dfrac{Pr(A\mid B)Pr(B)}{Pr(A)}$ – JMoravitz Jun 21 '18 at 4:07
• Okay. Thanks. I will take points of that information later. – Shane Dizzy Sukardy Jun 21 '18 at 4:35
Refer to the probability tree diagram:
$\hspace{1cm}$
Let $L$ and $S$ be the event of selecting the large and small groups of bags, respectively. Let $O$ and $A$ be the event of selecting an orange and apple, respectively.
a) The probability of selecting an apple: \begin{align}P(A)=&P(L\cap A)+P(S\cap A)=\\ =&P(L)\cdot P(A|L)+P(S)\cdot P(A|S)=\\ =&0.6\cdot \frac{20}{25}+0.4\cdot \frac{10}{25}=\\ =&0.64.\end{align}
b) Given that an apple was selected, the probability it was selected from the large group of bags: $$P(L|A)=\frac{P(L\cap A)}{P(A)}=\frac{P(L)\cdot P(A|L)}{P(L)\cdot P(A|L)+P(S)\cdot P(A|S)}=\cdots$$ Can you plug in the values and finish?
There are 25 fruit in each bag. $60\%$ of these contain 20 apples, and $40\%$ contain 10 apples. Let us call $A$ and $B$ the events of selecting the respective bags, and let us call $C$ the event of choosing an apple.
You are therefore told $\mathsf P(A)=0.60, \mathsf P(C\mid A)=0.80\\\mathsf P(B)=0.40, \mathsf P(C\mid B)=0.40$
You are asked to find $\mathsf P(C)$ and $\mathsf P(A\mid C)$. Bayes' Theorem and the Law of Total Probability are clearly of use here.
Alternatively: In a representative sample of $100$ bags there are $2500$ fruit among of which are $1200$ apples in bags of type A and $400$ apples in bags of type B. Because there is the same amount of fruit in each bag, the selection of particular pieces of fruit are unbiased by the method. The probability for selecting an apple is therefore some ratio, and the probability that an apple, if so selected, would have come from bag type $A$ is some other ratio. All the numbers you need are given here, though some addition and division is of course required.
Which is not exactly avoiding using Bayes' Theorem, but may give insight into the intuition behind it. | 2019-09-22T19:27:27 | {
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https://math.stackexchange.com/questions/3213292/convergence-to-fixed-point | # Convergence to fixed point
Question:
Let $$f:[0,1] \rightarrow [0,1]$$ be a continuous function that is strictly increasing on $$[0,a)$$ and strictly decreasing on $$(a,1]$$. Moreover, $$f(a) \leq a$$.
Consider the discrete map where
$$x_{n+1} = f(x_n)$$
Show that this map has at least one fixed point in $$[0,f(a)]$$, and that each initial point in $$[0,1]$$ converges to one of them (not necessarily the same one).
Attempt:
Consider the function $$g(x) = f(x)-x$$
The range of $$f$$ is $$[0,1]$$, so in particular $$f(0) \geq 0$$. Also, it is given that $$f(a) \leq a$$. It follows that
$$g(0) = f(0) - 0 = f(0) \geq 0 \qquad \qquad g(a) = f(a)-a \leq 0$$
So by the Intermediate Value Theorem, there exists $$c \in [0,a]$$ such that $$g(c) = 0$$, i.e. that $$f(c) = c$$, which means that $$c$$ is a fixed point of $$f$$.
Suppose that $$c \in (f(a),a]$$ (i.e. $$f(a)). Note that $$f$$ is strictly increasing on this interval, so $$f(c) \leq f(a)$$. But also, $$f(c) = c>f(a)$$. This gives a contradiction unless $$c = a = f(a)$$.
Thus, there exists a fixed point in $$[0,f(a)]$$.
As for the second part, I'm not so sure how to go about it.
I tried using the Banach Fixed Point Theorem (i.e. the Contraction Mapping Theorem), but it just can't be applied directly because (intuitively, I think that) $$f$$ is a contraction only in a neighborhood of one of the fixed points.
Any hints would be much appreciated. Thanks!
p.s. Also, I'm having trouble understanding why there cannot be a $$2$$-cycle (for in that case, there will be points that converge to this $$2$$-cycle which is not a fixed point).
You already noticed that $$f$$ maps $$[0,f(a)]$$ into itself. Any iteration sequence that starts inside this interval stays inside that interval. If $$x_1\le x_0$$ the iteration sequence is monotonically decreasing, if $$x_1\ge x_0$$ it is monotonically increasing. In any case the sequence is bounded, thus convergent, and the limit must be a fixed point.
Any sequence with $$x_0\in[0,a]$$ will converge to a fixed point. If $$x_0\in(a,1]$$, then $$f(a)\ge x_1=f(x_0)\ge f(1)$$ so that from then on the sequence converges monotonically to a fixed point. There is no possibility for a periodic cycle.
• Why is it that $x_1 \leq x_0$ implies the sequence is monotonically decreasing? – glowstonetrees May 4 '19 at 13:35
• Because then $x_2=f(x_1)\le f(x_0)=x_1\le x_0$ and so on. The first observation ensures that the sequence does not leave the region where $f$ is monotonically increasing. – Lutz Lehmann May 4 '19 at 14:06 | 2020-01-24T21:12:40 | {
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https://numfactory.upc.edu/web/FiniteElements/Pract/P4-QuadInterpolation/html/QuadInterpolation.html | # 2D Quadrilateral Interpolation: Barycentric Coordinates
## Interpolation on a general quadrilateral
When dealing with general quadrilaterals, the explicit expression of the shape functions is not easy to found. But we have the possibility of computing the value of such functions on a point p using the barycentric coordinates for quadrilaterals.
Similar to the triangular interpolation, now consider a point , in a quadrilateral defined by four vertices numbered counterclokwise as usual(see figure). One can define the barycentric coordinades of this point which are equivalent to the value of the 2D shape functions defined for this quadrilateral :
satisfying .
This way, one can express any point using four coordinates (barycentric) . In particular .
The barycentric coordinates satisfy . Moreover all unless the point is outside of the quadrilater.
Thus, the coordinates of the point
.
To compute the values of we re-write the previous equation as
.
where
Therefore,
Now to compute these two parameters, we rename the terms in the previous equation as:
where
Finally we make a cross product of the last equation with , getting
or
which can be solved (Hint. The case must be treated separately)
The equation for is obtained analogously
## Compute Barycentric Coordinates on a general quadrilateral
Given a quadrilateral defined by vertices , compute the barycentric coordinates of point .
Verify that .
% Data values
p=[3 2];
v1=[0 0];
v2=[5 -1];
v3=[4 5];
v4=[1 4];
% define the previous notation
a=(v1-p);
b=v2-v1;
c=v4-v1;
d=v1-v2-v4+v3;
% compute 2on order equation A mu^2 + B mu + C=0
% as the vertices are 2D, we add a zero third component
% to compute cross products.
A=cross([c,0],[d,0]); %must be 3D vectors
B=cross([c,0],[b,0])+cross([a,0],[d,0]);
C=cross([a,0],[b,0]);
% Only third component is needed (the other two are zero)
A=A(3);
B=B(3);
C=C(3);
%
% Check for unique solutions
%
if (abs(A)<1.e-14)
u1= -C/B;
u2=u1;
else
%
% Check for non complex solutions
%
if (B^2-4*A*C >0)
u1=(-B+sqrt(B^2-4*A*C))/(2*A);
u2=(-B-sqrt(B^2-4*A*C))/(2*A);
else %complex solution
u1=-1000;
u2=u1;
end
end
%
mu=-10000; %stupid value small enough
if(u1>=0 && u1<=1)
mu=u1;
end
if(u2>=0 && u2<=1)
mu=u2;
end
% compute 2on order equation A lambda^2 + B lambda + C=0
A=cross([b,0],[d,0]); %must be 3D vectors
B=cross([b,0],[c,0])+cross([a,0],[d,0]);
C=cross([a,0],[c,0]);
% Only third component is needed (the other two are zero)
A=A(3);
B=B(3);
C=C(3);
%
% Check for unique solutions
%
if (abs(A)<1.e-14)
w1= -C/B;
w2=w1;
else
%
% Check for non complex solutions
%
if (B^2-4*A*C >0)
w1=(-B+sqrt(B^2-4*A*C))/(2*A);
w2=(-B-sqrt(B^2-4*A*C))/(2*A);
else %complex solution
w1=-1000;
w2=w1;
end
end
%
lambda=-10000; %stupid value
if(w1>=0 && w1<=1)
lambda=w1;
end
if(w2>=0 && w2<=1)
lambda=w2;
end
[mu,lambda] %parameters
% Barycentric coordinates
alpha1=(1-mu)*(1-lambda);
alpha2=lambda*(1-mu);
alpha3=mu*lambda;
alpha4=(1-lambda)*mu;
alphas=[alpha1,alpha2,alpha3,alpha4]
% obtained point
q=alpha1*v1+alpha2*v2+alpha3*v3+alpha4*v4;
p-q %we must recover the same point
% Plot the results
vertices=[v1,v2,v3,v4];
plotRectangle(v1,v2,v3,v4);
hold on;
plot(p(:,1),p(:,2),'o');
plot(q(:,1),q(:,2),'*');
hold off;
ans =
5.0000e-01 6.2500e-01
alphas =
1.8750e-01 3.1250e-01 3.1250e-01 1.8750e-01
ans =
0 0
## Exercise 1:
From the previous code implement the function baryCoordQuad (the Quadrilateral version of the baryCoord function).
## Exercise 2:
From the mesh file meshTwoHolesQuad.m, find the rectangle where the point p=[39,7] belongs to.
Sol:
i = 404
vertices =
41.9240 8.7481
37.5060 8.6827
36.5394 4.7543
41.0196 4.8224
(c)Numerical Factory 2018 | 2021-04-18T17:51:09 | {
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http://fpri.mpl-bauen.de/2d-fourier-transform-examples-and-solutions.html | # 2d Fourier Transform Examples And Solutions
Instead of capital letters, we often use the notation f^(k) for the Fourier transform, and F (x) for the inverse transform. DFT is part of Fourier analysis, which is a set of math techniques based on decomposing signals into sinusoids. 3: Some Special Fourier Transform Pairs 29. Lecture 2 2d Fourier Transforms And S. 2D radially symmetric examples (M. flip the window in x and y 2. The Fourier transform consists of the Fourier cosine transform and the Fourier sine transform. The function F(k) is the Fourier transform of f(x). The 2D Inverse Fourier Transform is just the inverse Fourier Transform performed over both dimensions of the data. The Discrete Fourier Transform For example, we cannot implement the ideal lowpass lter digitally. Basic Spectral Analysis. Three Dimensional Fast Fourier Transform CUDA Implementation [Separability of 2D Fourier Transform] 2. Discrete Fourier Series DTFT may not be practical for analyzing because is a function of the continuous frequency variable and we cannot use a digital computer to calculate a continuum of functional values DFS is a frequency analysis tool for periodic infinite-duration discrete-time signals which is practical because it is discrete. Spatial Transforms 8 Fall 2005 2-D Convolution •Alternatively, we can write for a Wx×Wy window, –which emphasizes that the convolution is a weighting of local pixels 1. , mathematical), analytically-defined FT in a synthetic (digital) environment, and is called discrete Fourier transformation (DFT). Fourier Transforms. Thus, the solutions are simple harmonic: L an t B L an t T n t A n n π π = cos + sin, n = 1, 2, 3, … Multiplying each pair of X n and T n together and sum them up, we find the general solution of the one-dimensional wave equation, with both ends fixed, to be. Concluding Remarks on Fourier Transformation In COMSOL Multiphysics, you can use the data set feature and integrate operator as a convenient standalone calculation tool and a preprocessing and postprocessing tool before or after your main computation. ECE 468: Digital Image Processing Lecture 15 Example: Radon Transform g(⇢, )= Z 1 1 Z 1 1 2D Fourier Transform of the original image 18. Both Cooley and Tukey call it a re-discovery rather. Its Laplace transform (function) is denoted by the corresponding capitol letter F. Spatial Transforms 8 Fall 2005 2-D Convolution •Alternatively, we can write for a Wx×Wy window, -which emphasizes that the convolution is a weighting of local pixels 1. (Lecture 19) Fourier Transform to Solve PDEs: 1D Heat Equation on Infinite Domain (Lecture 20) Numerical Solutions to PDEs Using FFT ( notes , HeatConvolution. Find the Fourier transform of the matrix M. As in the 1D case, the 2D fourier transform and its inverse are infinitely periodic (in both dimensions), ie. You will practice these tasks in this weeks development phase. 8 Continuous-Time Fourier Transform Solutions to Recommended Problems S8. • The convolution of two functions is defined for the continuous case - The convolution theorem says that the Fourier transform of the convolution of two functions is equal to the product of their individual Fourier transforms • We want to deal with the discrete case - How does this work in the context of convolution? g ∗ h ↔ G (f) H. FFT's are very important in signal processing algorithms, and are also used to create structures from diffraction patterns, and vice versa. Description. (i) In Example 1, if u(0,t) 0 and P(0) 0, it would be inappropriate to use Fourier Sine transform. FourierTransform [expr, t, ω] yields an expression depending on the continuous variable ω that represents the symbolic Fourier transform of expr with respect to the continuous variable t. Free Fourier Series calculator - Find the Fourier series of functions step-by-step. The relation between the polar or spherical Fourier transform and normal Fourier transform is explored. Fourier Transform Wikipedia. We have shown how to use the Laplace transform to solve linear differential. (5) and (9). Dunlap Institute Summer School 2015 Fourier Transform Spectroscopy 11 Fabry-Perot Interferometer This interference phenomena can be used to accurately measure distances by using a laser beam or in measuring the spectral. Example The following example uses the image shown on the right. Transforms are used to make certain integrals and differential equations easier to solve algebraically. The second topic, Fourier series, is what makes one of the basic solution techniques work. The simplex algorithm seeks a solution between feasible region extreme points in linear programming problems which satisfies the optimality criterion. The inverse transform of F(k) is given by the formula (2). A Lookahead: The Discrete Fourier Transform. The fast Fourier transform algorithm requires only on the order of n log n operations to compute. The Fourier transform is a separable function and a FFT of a 2D image signal can be performed by convolution of the image rows followed by the columns. These types of FT's are used in connection with the ROSAT and other satellite data. The Fourier Transform produces a complex number valued output image which can be displayed with two images, either with the real and imaginary part or with magnitude and phase. The DFT and its inverse are obtained in practice using a fast Fourier Transform. I show here an example. Solving the Black-Scholes equation: a demysti cation Fran˘cois Coppex, (Dated: November 2009) Our objective is to show all the details of the derivation of the solution to the Black-Scholes equation without any prior prerequisit. Since the Fourier transform is a linear operation then the Fourier transform of the innite comb is the sum of the Fourier transforms of shifted Delta functions, which from equation (29) gives, F fCombDx(x)g= ¥ å i= ¥ exp( 2piDxu) (16) School of Physics Fourier Transform Revised: 10 September 2007. For example it can. Convolution will assist us in solving integral equations. The Fourier diffraction theorem relates the Fourier transform of the forward scattered acoustic field to the value of the Fourier transform of the object on a circular [two-dimensional (2D)] or a spherical [three-dimensional (3D)] arc. 7: Fourier Transforms: Convolution and Parseval's Theorem •Multiplication of Signals •Multiplication Example •Convolution Theorem •Convolution Example •Convolution Properties •Parseval's Theorem •Energy Conservation •Energy Spectrum •Summary. The z-Transform as an Operator ECE 2610 Signals and Systems 7-8 A General z-Transform Formula † We have seen that for a sequence having support inter-val the z-transform is (7. FFT onlyneeds Nlog 2 (N). The code below is a minimal working example, which produces the image and the 2D FT. Now we going to apply to PDEs. 2-D Fourier Transforms. Two-Dimensional Fourier Transform. Introduction to Fourier Transforms Fourier transform as a limit of the Fourier series Inverse Fourier transform: The Fourier integral theorem Example: the rect and sinc functions Cosine and Sine Transforms Symmetry properties Periodic signals and functions Cu (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 2 / 22. Solving differential equations using Discrete Fourier Transform. 9-1 An aperiodic pulse. Azimi, Professor Department of Electrical and Computer Engineering Colorado State University Spring 2013 M. For example, the Fourier transform of the rectangular function, which is integrable, is the sinc function, which is not Lebesgue integrable, because its improper integrals behave analogously to the alternating harmonic series, in converging to a sum without being absolutely convergent. It allows us to study a signal no longer in the time domain, but in the frequency domain. 1 Practical use of the Fourier. Possible applications of the proposed transforms are discussed. I need to rewrite it to do datasets larger than 4096 (Excel FFT is limited). Coalescing III -Code Example Fig 2 shows a bit of pseudo-code that employscoalescence. Consider this when you observe effects at the edge of your image when converting back to spatial coordinates with the inverse fourier transform. 6 Some Properties of the 2-D Discrete Fourier Transform Relationships between Spatial and Frequency Intervals Suppose that a continuous function ftz(,) is sampled to form a digital image fxy(,), consisting of M N× samples taken in the t- and z-directions. • Please write your answers in the exam booklet provided, and make sure that your answers. 2-2 (b) 1M+ jwIl = < = 0 1 1/2 1 1 1 ( ) r r r circ r. From Wikibooks, the open-content textbooks collection < Engineering Tables Jump to: navigation, search. Continuing with our specific example, the Fourier transform of circ(r) is. Schowengerdt 2003 2-D DISCRETE FOURIER TRANSFORM DEFINITION forward DFT inverse DFT • The DFT is a transform of a discrete, complex 2-D array of size M x N into another discrete, complex 2-D array of size M x N Approximates the under certain conditions Both f(m,n) and F(k,l) are 2-D periodic. • Signals as functions (1D, 2D) - Tools • 1D Fourier Transform - Summary of definition and properties in the different cases • CTFT, CTFS, DTFS, DTFT •DFT • 2D Fourier Transforms - Generalities and intuition -Examples - A bit of theory • Discrete Fourier Transform (DFT) • Discrete Cosine Transform (DCT). We simplify by assuming that f(x) = 0 for x < -T0/2 and x > T0/2. This sparsity is exploited in image and video compression algorithms like JPEG and MPEG. While the the Fourier Transform is a beautiful mathematical tool, its widespread popularity is due to its practical application in virtually every field of science and engineering. Addendum: The Fourier transform of decaying oscillations Robert DeSerio The Acquire and Analyze Transient vi is a LabVIEW program that takes and analyzes. Discrete Fourier Series DTFT may not be practical for analyzing because is a function of the continuous frequency variable and we cannot use a digital computer to calculate a continuum of functional values DFS is a frequency analysis tool for periodic infinite-duration discrete-time signals which is practical because it is discrete. Y = fft2(X) returns the two-dimensional Fourier transform of a matrix using a fast Fourier transform algorithm, which is equivalent to computing fft(fft(X). The process of deriving the weights that describe a given function is a form of Fourier analysis. The Laplace transform is an integral transform that is widely used to solve linear differential equations with constant coefficients. Spectral Analysis of continuous-space (2D) signals Obtain the continuous-space Fourier transform of a complex exponential; Obtain the continuous-space Fourier transform of a 2D rect; Obtain the continuous-space Fourier transform of a 2D sinc; Spectral Analysis of discrete-space (2D) signals Obtain the discrete-space Fourier transform of a rectangle. A Couple of things that are next are to look at some of the numerical solutions of, first, the 1-D Fourier solution, then 2-D and 3-D numerical solutions, as well as looking at the Dirac Equation in spherical coordinates instead of Cartesian coordinates, as looked at here. Consider this when you observe effects at the edge of your image when converting back to spatial coordinates with the inverse fourier transform. Fourier Transform 2d Wave Equation Tessshlo. The Fourier transform is a powerful tool for analyzing data across many applications, including Fourier analysis for signal processing. It's just a Fourier transform of something with zero imaginary part, but the zeroes still have to be there, as you can see from fftw_complex definition: typedef double fftw_complex[2]; This makes sense as the output can (and probably will) have non-zero imaginary part. DFT needs N2 multiplications. Am Iwrong?. Discrete Fourier Transform Matrix A discrete Fourier transform matrix is a complex matrix whose matrix product with a vector computes the discrete Fourier transform of the vector. We quickly realize that using a computer for this is a good i. Image Transforms-2D Discrete Fourier Transform (DFT) Properties of 2-D DFT Digital Image Processing Lectures 9 & 10 M. DFT is part of Fourier analysis, which is a set of math techniques based on decomposing signals into sinusoids. Its Laplace transform (function) is denoted by the corresponding capitol letter F. Spatial Transforms 8 Fall 2005 2-D Convolution •Alternatively, we can write for a Wx×Wy window, -which emphasizes that the convolution is a weighting of local pixels 1. Convolutions and correlations and applications; probability distributions, sampling theory, filters, and analysis of linear systems. 18], and tomographic image reconstruction (for 2D parallel-beam geometry) based on the Fourier slice theorem as described in §3. $$\mathscr{F} \{ C \} = C \cdot \delta(f)$$ and that is not 1. Fast Fourier transform. For example, the 2D Fourier transform of the function f(x, y) is given by: Note that the 2D Fourier transform can be carried out as two 1D Fourier transforms in sequence by first performing a 1D Fourier transform in x and then doing another 1D Fourier transform in y:. Fourier Transforms. (5) and (9). 4 Fourier solution In this section we analyze the 2D screened Poisson equation the Fourier do-main. Solution FIGURE 15. To compute the Fourier transform of an expression, use the inttrans[fourier] command. Fourier Transforms can also be applied to the solution of differential equations. This filter has a finite impulse response even though it uses feedback: after N samples of an impulse, the output will always be zero. Important frequency characteristics of a signal x(t) with Fourier transform X(w) are displayed by plots of the magnitude spectrum, |X(w)| versus w, and phase spectrum, Mathematics > Calculus > The Fourier Transform and its Applications > Effect on Fourier Transform of Shifting a Signal Lecture Details:. Modem two-dimensional Fourier-transform (2D- FT) ESR methods have opened up a range of new possibilities [ 12,13 1. When both the function and its Fourier transform are replaced with discretized counterparts, it is called the discrete Fourier transform (DFT). 3 The Fourier Sine Transform (FST) Definitions and Relations to the Exponential Fourier Transforms • Basic Properties and Operational Rules • Selected Fourier Sine Transforms 3. Solving problems by Fourier transforms Given an problem that is de ned for x2R there are three basic steps in solving the problem by the Fourier transform: (1)Apply the Fourier transform to the equation and to the given conditions to transform the problem. 324 B Tables of Fourier Series and Transform of Basis Signals Table B. How to determine and display the two dimensional fourier transform of a thin, rectangular object? The object should be 2 by 10 pixels in size and solid white against a black background. Spatial Transforms 8 Fall 2005 2-D Convolution •Alternatively, we can write for a Wx×Wy window, -which emphasizes that the convolution is a weighting of local pixels 1. We now look at the Fourier transform in two dimensions. View and Download PowerPoint Presentations on Fourier Transform Properties PPT. Lecture 2 2d Fourier Transforms And S. Fourier Transform Theorems; Examples of Fourier Transforms; Examples of Fourier Transforms (continued) Transforms of singularity functions. 17, 2012 • Many examples here are taken from the textbook. Finally, if we. The equations are a simple extension of the one dimensional case, and the proof of the equations is, as before, based on the orthogonal properties of the Sin and Cosine functions. 2D Fourier Transform. The purpose of this seminar paper is to introduce the Fourier transform methods for partial differential equations. Image encryption and the fractional Fourier transform B. 8 Continuous-Time Fourier Transform Solutions to Recommended Problems S8. f is the Fourier transform of g, up to a numeric factor and difierent sign of the argument. FFT onlyneeds Nlog 2 (N). For example, the Fourier transform of the rectangular function, which is integrable, is the sinc function, which is not Lebesgue integrable, because its improper integrals behave analogously to the alternating harmonic series, in converging to a sum without being absolutely convergent. multiply the window weights times the corresponding image pixels 4. 1998 We start in the continuous world; then we get discrete. The 2D FFT operation arranges the low frequency peak at the corners of the image which is not particularly convenient for filtering. !/D Z1 −1 f. We can view and even manipulate such information in a Fourier or frequency space. This is the first of four chapters on the real DFT , a version of the discrete Fourier transform that uses real numbers. PDF | Hypercomplex 2D Fourier transforms have been proposed by several authors with applications in image processing of both greyscale and colour images. The array of data must be rectangular. The spectrum is obtained by Fourier Transform where the time dependent FID is converted to a function of frequency, i. Solving a simple Schroedinger equation with Fast Fourier Transforms. 2-D Fourier Transform zFT for a 2-D continuous function Horizontal and vertical spatial frequencies (cycles per degree of viewing angle) - Separability: 2-D transform can be realized by a succession of 1-D transform along each spatial coordinate - Many other properties can be extended from 1-D FT. Discrete Fourier Series DTFT may not be practical for analyzing because is a function of the continuous frequency variable and we cannot use a digital computer to calculate a continuum of functional values DFS is a frequency analysis tool for periodic infinite-duration discrete-time signals which is practical because it is discrete. Chapter10: Fourier Transform Solutions of PDEs In this chapter we show how the method of separation of variables may be extended to solve PDEs defined on an infinite or semi-infinite spatial domain. Lecture 2 2d Fourier Transforms And S. We will do this by solving the heat equation with three different sets of boundary conditions. (From: "High Performance Discrete Fourier Transforms on Graphics Processors" – Govindaraju, NK, et al. Multidimensional Fourier transforms are widely used in image processing, tomographic reconstructions and in fact any application that requires a multidimensional convolution. Matlab will automatically figure out how many entries you need and their values. I show here an example. An example solution of Up: Poisson's equation Previous: The fast Fourier transform An example 2-d Poisson solving routine Listed below is an example 2-d Poisson solving routine which employs the previously listed tridiagonal matrix inversion and FFT wrapper routines, as well as the Blitz++ library. Instead of capital letters, we often use the notation f^(k) for the Fourier transform, and F (x) for the inverse transform. The discrete-time Fourier transform is an example of Fourier series. Free Fourier Series calculator - Find the Fourier series of functions step-by-step. * The Fourier transform is, in general, a complex function of the real frequency variables. How to determine and display the two dimensional fourier transform of a thin, rectangular object? The object should be 2 by 10 pixels in size and solid white against a black background. For each block, fft is applied and is multipled by some factor which is nothing but its absolute value raised to the power of 0. As with fast Poisson solvers, we can solve the screened Poisson equation (Equation 8) by taking its Fourier transform. Basic Spectral Analysis. Solving a simple Schroedinger equation with Fast Fourier Transforms. Fourier Transforms and the Fast Fourier Transform (FFT) Algorithm Paul Heckbert Feb. forced) version of these equations, and. Here, however, we have another thing going on. It refers to a very efficient algorithm for computingtheDFT • The time taken to evaluate a DFT on a computer depends principally on the number of multiplications involved. [email protected] With the development of the computer during the mid 20th century there came a need for a quick method of determining the discrete Fourier transform of a signal. the convolution of the Fourier transforms, we see that our result will involve a convolution of the forcing term f ( x ) with the inverse Fourier transform of the rational function 1 /L (i k ). y’’ + a 2 y = - f(t) ----- (1) Equation (1) is a differential equation. Possible applications of the proposed transforms are discussed. 9-2 The Fourier transform for the rectangular aperiodic pulse is shown as a function of co. They are widely used in signal analysis and are well-equipped to solve certain partial. x/is the function F. Continuing with our specific example, the Fourier transform of circ(r) is. The function is displayed in white, with the Fourier series approximation in red. For example the 2-D fourier transform of is given by −F(k x,k y)=f(x,y)e−i2πk xxdx −∞ ∞ ∫ $% & '. , for filtering, and in this context the discretized input to the transform is customarily referred to as a signal, which exists in the time domain. It would be impossible to give examples of all the areas where the Fourier transform is involved,. multiply the window weights times the corresponding image pixels 4. With the development of the computer during the mid 20th century there came a need for a quick method of determining the discrete Fourier transform of a signal. An Intuitive Explanation of Fourier Theory. 13) - There will be discussion of this case in Chapter 8 when we. The other connection, the subject of this column, is the surprising and pleasing fact that when a monochomatic X-ray diffracts off a crystal it performs part of a mathematical operation: the Fourier transform (developed in the 19th century in completely different contexts); when the incidence angle is varied, the complete transform is produced. When the arguments are nonscalars, fourier acts on them element-wise. 2-D Fourier Transforms. Fourier transform and to calculating the Fourier transforms of certain functions. Discrete Fourier Transform Matrix A discrete Fourier transform matrix is a complex matrix whose matrix product with a vector computes the discrete Fourier transform of the vector. Fourier Transform and Spatial Filtering - PPT, Digital Image Processing Summary and Exercise are very important for perfect preparation. The Fourier transform of N inputs, can be performed as 2 Fourier Transforms of N/2 inputs each + one complex multiplication and addition for each value i. What are 2D- and 3D-Fourier transforms? I don't see how FT works in higher dimensions. they wrap around. The function g(k) · fk is called Fourier transform of the function f. The first topic, boundary value problems, occur in pretty much every partial differential equation. fractional fourier transform 2d Search and download fractional fourier transform 2d open source project / source codes from CodeForge. if the limit exists. This theorem states that the 1-D FT of the projection of an object is the same as the values of the 2-D FT of the object along a line drawn through the center of the 2-D FT plane. The Fourier transform of expresses a signal as a composition of sinusoidal functions. This remarkable result derives from the work of Jean-Baptiste Joseph Fourier (1768-1830), a French mathematician and physicist. The inverse transform of F(k) is given by the formula (2). PDF | Hypercomplex 2D Fourier transforms have been proposed by several authors with applications in image processing of both greyscale and colour images. sqrt(re²+im²)) of the complex result. Solving nonhomogeneous PDEs by Fourier transform Example: We use Fourier transform because the transformed equation in "Fourier space", or The solution of Eq. Theorem 12. Discrete Fourier Transform Matrix A discrete Fourier transform matrix is a complex matrix whose matrix product with a vector computes the discrete Fourier transform of the vector. For example, in Fig 5, for the case of FFT length = 16. m , HeatFFT. Fourier Transform is a change of basis, where the basis functions consist of sines and cosines (complex exponentials). Use the Fourier transform for frequency and power spectrum analysis of time-domain signals. You can see some Fourier Transform and Spatial Filtering - PPT, Digital Image Processing sample questions with examples at the bottom of this page. Solutions to Exercises 187 6. The integral equals f, this is the Fourier Integral Theorem. Previously published works on hypercomplex. We've seen how to apply coordinate transformations to change to a more suitable color space. We quickly realize that using a computer for this is a good i. For example, for not-terribly-obvious reasons, in quantum mechanics the Fourier transform of the position a particle (or anything really) is the momentum of that particle. But I can certainly start by following up on the teaser example from last week. The solution, u(t), of the system, is found by inverting the Laplace transform U(s). 13) - There will be discussion of this case in Chapter 8 when we. Two-Dimensional Fourier Transform. It refers to a very efficient algorithm for computingtheDFT • The time taken to evaluate a DFT on a computer depends principally on the number of multiplications involved. Vincent Poor, Fellow, IEEE Abstract—The nonlinear Fourier transform, which is also known as the forward scattering transform, decomposes a pe-riodic signal into nonlinearly interacting waves. FFT onlyneeds Nlog 2 (N). The Fast Fourier Transform (FFT) is one of the most important algorithms in signal processing and data analysis. Two-dimensional Fourier transform also has four different forms depending on whether the 2D signal is periodic and discrete. According to Table 1, we have L 1fU(s)g= sin(!t) This is the solution that one would obtain using elementary solution methods. So just like in on dimension, we have. FFT/Fourier Transforms QuickStart Sample (C#) Illustrates how to compute the forward and inverse Fourier transform of a real or complex signal using classes in the Extreme. Introduction to Fourier Transforms Fourier transform as a limit of the Fourier series Inverse Fourier transform: The Fourier integral theorem Example: the rect and sinc functions Cosine and Sine Transforms Symmetry properties Periodic signals and functions Cu (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 2 / 22. The function F(k) is the Fourier transform of f(x). Fourier Transform Theorems; Examples of Fourier Transforms; Examples of Fourier Transforms (continued) Transforms of singularity functions. it Massimiliano Guarrasi–m. Am Iwrong?. I have an image and its fourier transform. (This is an interesting Fourier transform that is not in the table of transforms at. Instead, Fourier Cosine transform should be used. flip the window in x and y 2. (Lecture 19) Fourier Transform to Solve PDEs: 1D Heat Equation on Infinite Domain (Lecture 20) Numerical Solutions to PDEs Using FFT ( notes , HeatConvolution. Consider this when you observe effects at the edge of your image when converting back to spatial coordinates with the inverse fourier transform. It's hard to understand why the Fourier Transform is so important. The generalized Fourier transform includes as special cases the Laplace transform (when im(u) >0) and cumulant generating function (when. 18], and tomographic image reconstruction (for 2D parallel-beam geometry) based on the Fourier slice theorem as described in §3. The figure below shows 0,25 seconds of Kendrick's tune. Fast Fourier transform. As I mentioned in my lecture, if you want to solve a partial differential equa- tion (PDE) on the domain whose shape is a 2D disk, it is much more convenient to represent the solution in terms of the polar coordinate system than in terms of the usual Cartesian coordinate system. From previous section we learned to find derivative of function by using Fourier Transform. Learn more about fft2 Chatan can you please post the solution of your question so as i learn it. 2-D and 3-D transforms. The inverse transform of F(k) is given by the formula (2). In this section we go through the complete separation of variables process, including solving the two ordinary differential equations the process generates. , for filtering, and in this context the discretized input to the transform is customarily referred to as a signal, which exists in the time domain. For a more detailed analysis of Fourier transform and other examples of 2D image spectra and filtering, see introductory materials prepared by Dr. Also I would very highlight the fact that all the "magnitudes" which you display are logarithmic transform in fact, not the magnitudes itself. Fourier Transform Theorems; Examples of Fourier Transforms; Examples of Fourier Transforms (continued) Transforms of singularity functions. An FFT rapidly computes such transformations by factorizing the DFT matrix into a product of sparse (mostly zero) factors. Previously published works on hypercomplex. The problem is clear in |ψ(t)|, which should remain constant: The effect is less dramatic in the phase but still present. 3 Examples of Fourier Transforms Throughout the book we will work with only linear partial differential equations so all the problems are separable and the order of differentiation and integration is irrelevant. , mathematical), analytically-defined FT in a synthetic (digital) environment, and is called discrete Fourier transformation (DFT). 8 Continuous-Time Fourier Transform Solutions to Recommended Problems S8. Two-dimensional Fourier transform also has four different forms depending on whether the 2D signal is periodic and discrete. It uses real DFT, that is, the version of Discrete Fourier Transform which uses real numbers to represent the input and output signals. Fourier series, the Fourier transform of continuous and discrete signals and its properties. The z-Transform as an Operator ECE 2610 Signals and Systems 7-8 A General z-Transform Formula † We have seen that for a sequence having support inter-val the z-transform is (7. 12) † This definition extends for doubly infinite sequences having support interval to (7. Iskanderc aDepartment of Mathematics, University of Almer´ıa, Spain. I thought about checking the given examples and trying to guess the value of Fourier transform this way, but even then I couldn't produce a solution. The program then performs a 2D Fourier transform on that data followed by an inverse 2D Fourier transform. Let samples be denoted. While the the Fourier Transform is a beautiful mathematical tool, its widespread popularity is due to its practical application in virtually every field of science and engineering. Uses practical examples and specifically looks at the Optical Fourier Transform. With the development of the computer during the mid 20th century there came a need for a quick method of determining the discrete Fourier transform of a signal. We've seen how to apply coordinate transformations to change to a more suitable color space. Properties: Separability The FT of a 2D signal f(x,y) can be calculated as two 1D FT. Fourier analysis is a method for expressing a function as a sum of periodic components, and for recovering the signal from those components. Fourier Transform For Discrete Time Sequence (DTFT)Sequence (DTFT) • One Dimensional DTFT – f(n) is a 1D discrete time sequencef(n) is a 1D discrete time sequence – Forward Transform F( ) i i di i ith i d ITf n F(u) f (n)e j2 un F(u) is periodic in u, with period of 1 – Inverse Transform 1/2 f (n) F(u)ej2 undu 1/2. DTFT is not suitable for DSP applications because •In DSP, we are able to compute the spectrum only at specific. The Discrete Fourier Transform For example, we cannot implement the ideal lowpass lter digitally. Modem two-dimensional Fourier-transform (2D- FT) ESR methods have opened up a range of new possibilities [ 12,13 1. Partial Di erential Equations Victor Ivrii Department of Mathematics, University of Toronto c by Victor Ivrii, 2017, Toronto, Ontario, Canada. FFTW is a C subroutine library for computing the discrete Fourier transform (DFT) in one or more dimensions, of arbitrary input size, and of both real and complex data (as well as of even/odd data, i. The Fourier Transform Properties: Parseval's Theorem The energy contest of a 2D signal in the spatial domain is the same if frequency domain. 3 The Fourier Sine Transform (FST) Definitions and Relations to the Exponential Fourier Transforms • Basic Properties and Operational Rules • Selected Fourier Sine Transforms 3. Consider this when you observe effects at the edge of your image when converting back to spatial coordinates with the inverse fourier transform. I am trying to solve the electrostatic poisson's equation mentioned in the first post in 2D using Discrete Fourier Transform (I am using fftw3 library and REDFT10 / REDFT01 transforms). I thought about checking the given examples and trying to guess the value of Fourier transform this way, but even then I couldn't produce a solution. As in the 1D case, the 2D fourier transform and its inverse are infinitely periodic (in both dimensions), ie. Solution FIGURE 15. As I mentioned in my lecture, if you want to solve a partial differential equa- tion (PDE) on the domain whose shape is a 2D disk, it is much more convenient to represent the solution in terms of the polar coordinate system than in terms of the usual Cartesian coordinate system. 1 Examples of important PDEs. The Fourier transform consists of the Fourier cosine transform and the Fourier sine transform. The relationship between the DTFT of a periodic signal and the DTFS of a periodic signal composed from it leads us to the idea of a Discrete Fourier Transform (not to be confused with Discrete-Time Fourier Transform). CHRISTOV 1. m , benchSpectralDerivative. Matlab will automatically figure out how many entries you need and their values. Our starting point is the solution of the optical Bloch equations for a two-level system in the 2D time domain. But I can certainly start by following up on the teaser example from last week. Dunlap Institute Summer School 2015 Fourier Transform Spectroscopy 11 Fabry-Perot Interferometer This interference phenomena can be used to accurately measure distances by using a laser beam or in measuring the spectral. We have also seen that complex exponentials may be used in place of sin's and cos's. Fourier Transform Theorems; Examples of Fourier Transforms; Examples of Fourier Transforms (continued) Transforms of singularity functions. Engineering Tables/Fourier Transform Table 2. An FFT rapidly computes such transformations by factorizing the DFT matrix into a product of sparse (mostly zero) factors. The Fourier transform of the 1D function f(x) is given by: and the inverse Fourier transform is given by: The Fourier transform can also be extended to 2, 3,. Fourier Transforms for Deterministic Processes References Example: Discrete-time finite-duration pulse Compute the Fourier transform and the energy density spectrum of a finite-duration rectangular pulse x[k]= (A, 0 k L 1 0 otherwise Solution: The DTFT of the given signal is X(f)= X1 k=1 x[k]ej2⇡fk = LX1 k=0 Aej2⇡fk = A 1 ej2⇡fL 1 e j2. SignalProcessing namespace in C#. The inverse Fourier Transform f(t) can be obtained by substituting the known function G(w) into the second equation opposite and integrating. The Fourier Transform Properties: Parseval's Theorem The energy contest of a 2D signal in the spatial domain is the same if frequency domain. Use the Fourier transform for frequency and power spectrum analysis of time-domain signals. • The Fourier transform of the convolution of two functions is the product of their Fourier transforms • The inverse Fourier transform of the product of two Fourier transforms is the convolution of the two inverse Fourier transforms • Convolution in spatial domain is equivalent to multiplication in frequency domain! ∗ = g h g h F[ ] F[ ]F[ ]. What do we hope to achieve with the Fourier Transform? We desire a measure of the frequencies present in a wave. This kind of decomposition is possible due to orthogonality properties of sine and cosine functions. Professor Deepa Kundur (University of Toronto)Properties of the Fourier Transform5 / 24 Properties of the Fourier Transform FT Theorems and Properties. 2D Discrete Fourier Transform • Fourier transform of a 2D signal defined over a discrete finite 2D grid of size MxN or equivalently • Fourier transform of a 2D set of samples forming a bidimensional sequence • As in the 1D case, 2D-DFT, though a self-consistent transform, can be considered as a mean of calculating the transform of a 2D. OK, here’s an attempt at a slightly less technical explanation, with apologies to the purists! I find the FT to be a great example of why math-anxiety is such a pity - far from making life complicated, so very often, math makes this much much simp. Inverse Fourier Transform maps the series of frequencies (their amplitudes and phases) back into the corresponding time series. With these algorithms, we only need two complex 2-D FTs to implement a QFT, six complex 2-D FTs to implement a one-side QCV or a quaternion correlation and 12 complex 2-D FTs to implement a two-side QCV, and the efficiency of these quaternion operations is much improved. Spectral Analysis of continuous-space (2D) signals Obtain the continuous-space Fourier transform of a complex exponential; Obtain the continuous-space Fourier transform of a 2D rect; Obtain the continuous-space Fourier transform of a 2D sinc; Spectral Analysis of discrete-space (2D) signals Obtain the discrete-space Fourier transform of a rectangle. The coefficients of the sine and cosine components are given by the Fourier transform. Another application of Fourier analysis is the synthesis of sounds such as music, or machinery noise. Consider this when you observe effects at the edge of your image when converting back to spatial coordinates with the inverse fourier transform. The discrete Fourier transform and the FFT algorithm. 10 Green’s functions for PDEs In this final chapter we will apply the idea of Green’s functions to PDEs, enabling us to solve the wave equation, diffusion equation and Laplace equation in unbounded domains. I'm interested in the frequency spectrum, but the problem is that the Fourier function uses the fast Fourier transform algorithm which places the zero frequency at the beginning, complicating my analysis of the results. dst – output array whose size and type depends on the flags. The function J 0 is the zero order Bessel functi on of the first kind defined as = ∫ − π θφ θ π 2 0 cos( ) 0 2 1 J (a) eia d. 1 Integral transforms The Fourier transform is studied in this chapter and the Laplace tra nsform in the next. 2 we have F[F](!) = 2ˇF 1[F]( !) and then by Theorem 2. OK, here's an attempt at a slightly less technical explanation, with apologies to the purists! I find the FT to be a great example of why math-anxiety is such a pity - far from making life complicated, so very often, math makes this much much simp. ECE 468: Digital Image Processing Lecture 15 Example: Radon Transform g(⇢, )= Z 1 1 Z 1 1 2D Fourier Transform of the original image 18. (Note that there are other conventions used to define the Fourier transform). Multidimensional Fourier transforms are widely used in image processing, tomographic reconstructions and in fact any application that requires a multidimensional convolution. – Fourier transforms may be used to describing plane waves • But it require special care (explained later)! – In this lecture we will… • Study Fourier and Laplace transforms; focus on waves and oscillators – For damped and growing waves Fourier transforms may not exist! • Instead Laplace transforms can sometimes be used. The first topic, boundary value problems, occur in pretty much every partial differential equation. Fourier transform is the convolution of the 2 rect functions as found in part (b) above. The integral equals f, this is the Fourier Integral Theorem. The inverse discrete cosine transforms for types 1, 2, 3, and 4 are types 1, 3, 2, and 4, respectively. X = ifft2(Y) returns the two-dimensional discrete inverse Fourier transform of a matrix using a fast Fourier transform algorithm. There are 14 cases built into the program with case numbers ranging from 0 to 13 inclusive. Show that. If Y is a multidimensional array, then ifft2 takes the 2-D inverse transform of each dimension higher than 2. This is an advisable procedure as the Fourier transform of is very simple3; see problem 2(a), problem set 1. Inverse Fourier Transform maps the series of frequencies (their amplitudes and phases) back into the corresponding time series. The Dirac delta, distributions, and generalized transforms. For each block, fft is applied and is multipled by some factor which is nothing but its absolute value raised to the power of 0. This will lead to a definition of the term, the spectrum. multiply the window weights times the corresponding image pixels 4. INTRODUCTION TO FOURIER TRANSFORMS FOR IMAGE PROCESSING BASIS FUNCTIONS: The Fourier Transform ( in this case, the 2D Fourier Transform ) is the series expansion of an image function ( over the 2D space domain ) in terms of "cosine" image (orthonormal) basis functions. the magnitude of Fourier transforms by 23 in each dimension (i. For example the 2-D fourier transform of is given by −F(k x,k y)=f(x,y)e−i2πk xxdx −∞ ∞ ∫$ % & '. Specify the independent and transformation variables for each matrix entry by using matrices of the same size. The Fourier transform of expresses a signal as a composition of sinusoidal functions. I'm interested in the frequency spectrum, but the problem is that the Fourier function uses the fast Fourier transform algorithm which places the zero frequency at the beginning, complicating my analysis of the results. The function F(k) is the Fourier transform of f(x). • Spectroscopists use the Fourier transform to obtain high resolution spectra in the infrared from interferograms (Fourier spectroscopy). Fourier transforms We can imagine our periodic function having periodicity taken to the limits 1 In this case, the function f(x) is not necessarily periodic, but we can still use Fourier transforms (related to Fourier series) Consider the complex Fourier series, periodic with periodicity 2l f(x) = X1 n=1 c ne inˇx l. Those are examples of the Fourier Transform. • Transform some common functions of time. Convolutions and correlations and applications; probability distributions, sampling theory, filters, and analysis of linear systems. Two-Dimensional Fourier Transform. | 2020-01-28T06:37:10 | {
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http://mathoverflow.net/feeds/question/85013 | Alternating sum of square roots of binomial coefficients - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T01:04:34Z http://mathoverflow.net/feeds/question/85013 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/85013/alternating-sum-of-square-roots-of-binomial-coefficients Alternating sum of square roots of binomial coefficients Mark Wildon 2012-01-06T00:20:12Z 2012-03-11T18:59:38Z <p>Let $$c_n = \sum_{r=0}^n (-1)^r \sqrt{\binom{n}{r}}.$$ It is clear that $c_n = 0$ if $n$ is odd. Remarkably, it appears that despite the huge positive and negative contributions in the sum defining $c_{2m}$, the sequence $(c_{2m})$ may be very well behaved. </p> <blockquote> <p>Is $c_n > 0$ for all even $n$?</p> </blockquote> <p>An affirmative answer will imply that the function $F(x) = \sum_{n=0}^\infty x^n/\sqrt{n!}$ is always strictly positive, thereby answering this <a href="http://mathoverflow.net/questions/84958/is-sum-limits-n0-infty-xn-sqrtn-positive" rel="nofollow">earlier question</a>.</p> <p>Numerical computation using Magma shows that $c_n > 0$ if $n$ is even and $n \le 2000$. To give some illustrative values, $c_{100} = 0.077737 \ldots$, $c_{1000} = 0.019880 \ldots$ and $c_{2000} = 0.013317 \ldots$. </p> <p>A comment by Mark Sapir on the earlier question suggests a stronger result might hold. </p> <blockquote> <p>Is $c_{n} > c_{n+2} > 0$ for all even $n$?</p> </blockquote> <p>I have checked that this is the case for all even $n \le 2000$.</p> <p>It is very natural to ask what happens if we replace $\sqrt{\binom{n}{r}}$ with $\binom{n}{r}^\alpha$ for $\alpha \in (0,1)$. For $n\le 250$ the generalized version of the conjecture continues to hold if $\alpha = k/10$ where $k \in \mathbf{N}$ and $k \le 9$. Of course when $\alpha = 1$ we have $c_n = 0$ for all $n$, so, as David Speyer remarked in a comment on the earlier question, there is a good reason for the cancellation in this case.</p> http://mathoverflow.net/questions/85013/alternating-sum-of-square-roots-of-binomial-coefficients/85032#85032 Answer by GH for Alternating sum of square roots of binomial coefficients GH 2012-01-06T05:55:21Z 2012-01-06T06:46:20Z <p>The following proof of $c_n>0$ is based on Gjergji Zaimi's response to this <a href="http://mathoverflow.net/questions/84958/is-sum-limits-n0-infty-xn-sqrtn-positive" rel="nofollow">related question</a>. In particular the positive answer follows for that question, too. Moreover, the argument below should also show that $c_n>c_{n+2}$.</p> <p>Let $n>0$ be even. By Chapter 6 of de Bruijn's "Asymptotic Methods in Analysis" (in particular by (6.4.6), (6.6.2), and the conclusion $P=0$ of Section 6.5), we have the following explicit formula:</p> <p>$$c_n = 2\pi^{-1/2}\sqrt{n!}\ \sum_{m=0}^\infty \ \int_{4m+1}^{4m+2} G_n(x)\ |\sin\pi x|^{-1/2}dx,$$</p> <p>where</p> <p>$$G_n(x) := \sqrt{\frac{\Gamma(x)}{\Gamma(1+x+n)}}- \sqrt{\frac{\Gamma(2+x)}{\Gamma(3+x+n)}}.$$</p> <p>It remains to verify that $G_n(x)>0$ for $x\geq 1$. This reduces to</p> <p>$$\Gamma(x)\Gamma(3+x+n)>\Gamma(2+x)\Gamma(1+x+n),$$</p> <p>i.e. to </p> <p>$$(1+x+n)(2+x+n)>x(1+x).$$ </p> <p>The last inequality is obvious, hence we are done.</p> http://mathoverflow.net/questions/85013/alternating-sum-of-square-roots-of-binomial-coefficients/85035#85035 Answer by Noam D. Elkies for Alternating sum of square roots of binomial coefficients Noam D. Elkies 2012-01-06T06:53:41Z 2012-01-07T05:33:59Z <p>Here's a proof of the positivity of $$c_n(\alpha) := \sum_{r=0}^n (-1)^r {n\choose r}^\alpha$$ for all even $n$ and real $\alpha < 1$. It follows (via M.Wildon's clever $F(x) F(-x)$ trick at mo.84958) that $\sum_{n=0}^\infty \phantom. x^n / n!^{\alpha} > 0$ for all $x \in\bf R$. [<strong>EDIT</strong> fedja has meanwhile provided a very nice direct proof of the positivity of $\sum_{n=0}^\infty \phantom. x^n / n!^{\alpha}$.]</p> <p>The key is to write $c_n(\alpha)$ as a finite difference $$\sum_{r=0}^n \phantom. (-1)^r {n\choose r} \cdot {n\choose r}^{\alpha - 1}$$ and show that the Gamma interpolation $$\bigl(\Gamma(r+1)\Gamma(n-r+1) / n!\bigr)^{1-\alpha} = n!^{\alpha-1} \exp\bigl((1-\alpha) (\log\Gamma(r+1) + \Gamma(n-r+1)\bigr)$$ of ${n\choose r}^{\alpha - 1}$ has a positive $n$-th derivative for all $r \in [0,n]$.</p> <p>This in turn follows from the fact that the expansion of $\log\Gamma(r+1) + \log\Gamma(n-r+1)$ in a Taylor series about $r = n/2$ has positive $(r - (n/2))^k$ coefficient for each $k=2,4,6,\ldots$. [The coefficient vanishes for odd $k$ because $\log\Gamma(r+1) + \log\Gamma(n-r+1)$ is an even function of $r-(n/2)$.] Indeed the well-known formula $$\log \Gamma(x) = -\gamma x - \log x + \sum_{j=1}^\infty \left[ \frac{x}{j} - \log \left( 1 + \frac{x}{j} \right) \right]$$ shows that the $k$-th derivative of $\log\Gamma(x)$ is positive for all $x>0$ and $k=2,4,6,\ldots$, because this is true for $-\gamma x - \log x$ and for each term in the sum; explicitly the derivative is $k! \phantom. \sum_{j=0}^\infty (x+j)^{-k}$ which is positive termwise. Therefore in the Taylor expansion $$\log \Gamma(r+1) = \log(n/2)! + \sum_{k=1}^\infty \phantom. g_k (r-(n/2))^k$$ each of $g_2,g_4,g_6,\ldots$ is even. Since $\log\Gamma(r+1) + \log\Gamma(n-r+1)$ is $$2\log(n/2)! + 2 \Bigl( g_2 (r-(n/2))^2 + g_4 (r-(n/2))^4 + g_6 (r-(n/2))^6 + \cdots\Bigr),$$ the claim follows. [<strong>EDIT</strong> David Speyer notes that the convergence of the Taylor series on $|r-(n/2)| \leq n/2$ requires justification, and that the justification is easy because the $\Gamma(z)$ has no zeros and poles only at $0,-1,-2,\ldots$ so the radius of convergence is $(n/2)+1$.] Multiplying by $1 - \alpha$ and substituting into the exponential series, we deduce that $(\Gamma(r+1) \Gamma(n-r+1))^{1-\alpha}$, too, is a positive combination of even powers of $r-(n/2)$.</p> <p>Now if a function $g$ has positive $n$-th derivative, then its first finite difference $$g(x+1) - g(x) = \int_x^{x+1} g'(y) dy$$ has positive $(n-1)$-st derivative; repeating this argument $n$ times, we find that the $n$-th finite difference is positive, and we're done.</p> | 2013-05-24T01:04:38 | {
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http://themagicofscience.blogspot.com/2010/07/counting-backgammon-end-positions.html | ## Thursday, July 29, 2010
### Counting Backgammon End Positions
In this blog I will count the number of end positions in backgammon.
Recently I picked up backgammon again. But I could not put my mathematical mind to rest. I got interested in the question of how many end positions backgammons has.
In order to count the number of positions I will introduce the following terminology. The number of stones you own in the end position will be denoted by $$m$$. The number of points used available is $$n$$. $$o$$ is the number of stones owned by the opponent, which occupy $$p$$ points of the available $$n$$.
So in the following example, playing as black, the following equalities apply $$m=8,\,n=6,\,o=2,\,p=1$$.
Let's look at a special case first. Lets assume we have a contact-less end position. So $$o=0$$. I will proof that the number of backgammon end positions with $$m$$ stones and $$n$$ points is
$m + n - 1 \choose m$
To see this equality, study the following diagram, which corresponds with the figure above if you ignore the opponent stones:
Every dot corresponds with a stone. Every bar with the division between points. A moments reflection will bring the insight that every diagram of this kind corresponds to a backgammon end position and vice versa, every backgammon end position can be described by such a diagram. This proofs the stated equality.
With the result of this special case we can answer the main question: How many backgammon end positions exist with $$m$$ stones distributed over $$n$$ points of which $$p$$ are occupied by $$o$$ opponent stones? This number is exactly
${n \choose p} {o - 1 \choose o - p} {m + n - p - 1 \choose m}$
The proof of the above equality comes from the following insight. The opponent stones are distributed over $$p$$ points. there are $$n \choose p$$ ways of picking the occupied points.
For a point to be occupied it must at least contain one opponent stone. So of the $$o$$ opponents stones, we can freely distribute o - p opponent stones over $$p$$ points. This is the special case we counted already, so this can be achieved in $${(o-p) + (p-1) \choose o - p} = {o - 1 \choose o - p}$$ ways.
This brings us to the final factor. It represent the $$m$$ stones which should be distributed over the remaining $$n-p$$ points. Again this is given by our preliminary result of $${m + n - p - 1 \choose m}$$ ways. This proofs the stated result.
In conclusion: the number of backgammon end point positions with $$m$$ stones distributed over $$n$$ points of which $$p$$ are occupied by $$o$$ opponent stones equals
${n \choose p} {o - 1 \choose o - p} {m + n - p - 1 \choose m}$
1. I did the math and calculated the total number of positions in a backgammon game. It is a stagering:
3458085312432494932095
or
three sextillion ,
four hundred fifty eight quintillion ,
three hundred twelve trillion ,
four hundred thirty two billion ,
four hundred ninety four million ,
nine hundred thirty two thousand ,
ninety five
2. Hi Daan.
I would like to reprint this article in my backgammon magazine, Bibafax. Would you kindly give your permission?
You can checkout my references at www.backgammon-biba.co.uk.
Michael Crane
[email protected]
3. Daan, since the above comment I have formatted your content into my magazine and I would like to send you a pdf so that you can preview it before (hopefully) giving your permission.
Please email me and I will send it by return.
Regards.
Michael
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https://math.stackexchange.com/questions/2708349/4-digit-numbers-divisible-by-11 | # 4 digit numbers divisible by 11
Four digit numbers are formed using the digits 1,2,3,4 (repetition is allowed). The number of such four digit numbers divisible by 11 is- (1) 22 (2) 36 (3) 44 (4) 52
I know for a number to be divisible by 11 the sum of digits at even places must be equal to the sum of those at odd places. But how do I use this to get the answer?
• I would break into cases based on what the sum of the even places is. The odd places must have the same sum. If the sum is $2$, then that can only be accomplished using two ones, giving the four digit number $1111$. If the sum of the even places is $3$, then that can be accomplished either as a $1$ followed by a $2$, or vice versa, for two options on how to fill the even slots. Similarly we will have the same two options for how to fill the odd slots. Applying rule of product we get four possibilities here, 1122, 1221, 2112, 2211. Continue – JMoravitz Mar 26 '18 at 5:44
Let $abcd$ be the number.
If $a=b$ then $c=d$. There are $4*4=16$ ways that can occur. (Four options for $a$ and four options for $b$).
If $a=b\pm 1$ then $c=d \mp1$. And there are $2*3*3=18$ ways this can occur. (Two choices whether $a > b$ or $b > a$ and three choices from $1,2,3,4$ that are one apart.
If $a = b\pm 2$ then $c = d\mp 2$ and there $2*2*2 = 8$ ways.
And if $a = b\pm 4$ then either $a = 1;b=4;c=4;d=1$ or $a=4;b=1;c=1;d=4$. $2$ ways.
So $16 + 18 + 8 + 2 = 44$ ways.
Alternatively, consider permutations of numbers made of one, two, three and four digits.
There is one number made of single number $1$, which is $1111$. Total number is: $$1\cdot {4\choose 1}=4.$$
There are four numbers made of two digits $1$ and $2$, which are $1122,1221,2112,2211$. Total number is: $$4\cdot {4\choose 2}=24.$$
There are four numbers made of three digits $1,2,3$, which are $1232,2123,2321,3212$. And there are two possible cases: $1,2,3$ and $2,3,4$. Total number is: $$4\cdot 2=8.$$
There are eight numbers made of four digits $1,2,3,4$, which are $1234,1342,2134,2431,3241,3421,4213,4312$. Total number is: $$8\cdot 1=8.$$
In conclusion, the grand total is: $$4+24+8+8=44.$$
Let $[abcd]$ be the number. Then $[abcd] \equiv [ab]+[cd] \pmod{99}$ which implies $[abcd] \equiv [ab]+[cd] \pmod{11}$. Because $a,b,c,d \in \{1,2,3,4\}$, $[abcd] \equiv [(a+c)(b+d)] \pmod{11}$. Hence $[abcd]$ is a multiple of $11$ if and only if $a+c=b+d$.
\begin{array}{|r|r|r|} \text{sum} &\text{ac and bd events} & \text{counts} \\ \hline 2 & 11 & 1\\ 3 & 12, \ 21 & 4\\ 4 & 13, \ 22, \ 31 & 9 \\ 5 & 14, \ 23, \ 32, \ 41 & 16\\ 6 & 24, \ 33, \ 42 & 9\\ 7 & 34, \ 43 & 4\\ 8 & 44 & 1 \\ \hline \text{total} && 44 \end{array} | 2021-06-21T17:32:24 | {
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https://www.justinmath.com/brute-force-search-with-linear-encoding-cryptography/ | # Brute Force Search with Linear-Encoding Cryptography
An encoding function maps a string to a sequence of numbers. For example, you have already implemented the trivial encoding function, defined as follows:
' ' --> 0
'a' --> 1
'b' --> 2
...
'z' --> 26
There are many different types of encoding functions, but here we will focus on linear encoding functions. For example, using a linear encoding function $f(x) = 2x+3,$ we can encode the message 'a cat' as follows:
Original message: 'a cat'
Trivial encoding: [1, 0, 3, 1, 20]
Linear encoding: [2*1+3, 2*0+3, 2*3+3, 2*1+3, 2*20+3]
= [5, 3, 9, 5, 43]
If we want to recover our message from the linear encoding, we first solve for the inverse of the encoding function, i.e. the decoding function:
\begin{align*} f(x) &= 2x+3 \\[5pt] x &= 2 f^{-1}(x)+3 \\[5pt] f^{-1}(x) &= \dfrac{x-3}{2} \end{align*}
Then, we apply the decoding function to the encoded message:
Linear encoding: [5, 3, 9, 5, 43]
Trivial encoding: [(5-3)/2, (3-3)/2, (9-3)/2, (5-3)/2, (43-3)/2]
= [1, 0, 3, 1, 20]
Original message: 'a cat'
Now, suppose we have a message [3, 1, 9, 31, 15] and we know that this message was encoded using a linear encoding function $f(x) = ax+b$ with $a \in \lbrace 1, 2 \rbrace$ and $b \in \lbrace 1, 2 \rbrace.$ Can we break the code and recover the initial message?
The simplest way to do this is through brute-force search, which involves trying every single possibility. Here, there are $4$ possible encoding functions:
\begin{align*} f_{ab}(x) &= ax + b \\ \\ f_{11}(x) &= 1x + 1 \\ f_{12}(x) &= 1x + 2 \\ f_{21}(x) &= 2x + 1 \\ f_{22}(x) &= 2x + 2 \\ \end{align*}
By inverting these encoding functions, we obtain the following decoding functions:
\begin{align*} f_{ab}^{-1}(x) &= \dfrac{x - b}{a} \\ \\ f_{11}^{-1}(x) &= \dfrac{x - 1}{1} \\[5pt] f_{12}^{-1}(x) &= \dfrac{x - 2}{1} \\[5pt] f_{21}^{-1}(x) &= \dfrac{x - 1}{2} \\[5pt] f_{22}^{-1}(x) &= \dfrac{x - 2}{2} \\[5pt] \end{align*}
Let’s apply each of these decoding functions to our encoded message [3, 1, 9, 31, 15] and see what they come up with. Remember that in order to represent a message, the results must all be integers between $0$ and $26$ inclusive.
[3, 1, 9, 31, 15]
|
|-- a, b ------> [(3-b)/a, (1-b)/a, (9-b)/a, (31-b)/a, (15-b)/a]
|
|-- a=1, b=1 --> [(3-1)/1, (1-1)/1, (9-1)/1, (31-1)/1, (15-1)/1]
| = [2, 0, 8, 30, 14]
| 30 is too big
| does not represent a message
|
|-- a=1, b=2 --> [(3-2)/1, (1-2)/1, (9-2)/1, (31-2)/1, (15-2)/1]
| = [1, -1, 7, 29, 13]
| -1 is too small, 29 is too big
| does not represent a message
|
|-- a=2, b=1 --> [(3-1)/2, (1-1)/2, (9-1)/2, (31-1)/2, (15-1)/2]
| = [1, 0, 8, 15, 7]
| message: 'a dog'
|
|-- a=2, b=2 --> [(3-2)/2, (1-2)/2, (9-2)/2, (31-2)/2, (15-2)/2]
| = [0.5, -0.5, 3.5, 14.5, 6.5]
| contains non-integer entries
| does not represent a message
We conclude that the original message was 'a dog' and that it was encoded using $a=2$ and $b=1.$
Practice Problems
As usual, be sure to include a variety of tests.
1. Write a function encode_string(string, a, b) that encodes the string using the linear encoding function $f(x) = ax+b$ and returns the resulting array of numbers.
2. Write a function decode_numbers(numbers, a, b) that attempts to decode the numbers array under the assumption that the encoding function was $f(x) = ax+b.$ If the numbers represent a message, then return the string corresponding to that message. Otherwise, return False.
3. Write a script to decode the message [377, 717, 71, 513, 105, 921, 581, 547, 547, 105, 377, 717, 241, 71, 105, 547, 71, 377, 547, 717, 751, 683, 785, 513, 241, 547, 751], given that it was encoded with a linear encoding function $f(x)=ax+b$ where $a$ and $b$ are both integers between $0$ and $100$ inclusive. Be sure to print out all valid messages along with the values of $a$ and $b$ that generated them. Note that although there may be more than one valid message, only one will contain real words.
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https://math.stackexchange.com/questions/2296270/solve-ii2i3-i1022/2296277 | # Solve: $(i+i^2+i^3+…+i^{102})^2$
Solve $$(i+i^2+i^3+...+i^{102})^2$$
I have tried 2 approaches:
1. to look at power of $2 (=-1)$ and power of $4 (=1)$ and there are $102$ elements which is even number, but because there are $4$ "types" $\{i,-1,-i,1\}$ it did not work it out
2. to look at $$(i+i^2+i^3+...+i^{102})^2=\left(\sum_{k=1}^{102}i^k\right)^2$$ but $$\left(\sum_{k=1}^{102}i^k\right)^2\neq \sum_{k=1}^{102}i^{2k}$$ so it was not useful.
Any suggestion how to tackle this?
• WolfRamAlpha says -2i – Eric Lee May 25 '17 at 13:45
• Are you familiar with the sum of a geometric progession $\sum_{k=1}^{N} r^k = \frac{r-r^{N+1}}{1-r}$ – sharding4 May 25 '17 at 13:45
## 5 Answers
HINTS:
$$i+i^2+i^3+i^4=0,$$
$$i^{4m+n}=i^{n}.$$
• It may be useful for the OP to note of course a slight generalisation; that if $\omega$ is an $n^{\text{th}}$ root of $1$; so $\omega^n -1 =0$, then $\omega^n-1= (\omega -1)(\omega^{n-1} + \dots + \omega + 1)$. So if $\omega \ne 1$ then we get that $\omega^{n-1} + \dots + \omega + 1=0$ – SEWillB May 25 '17 at 14:13
One trick for this kind of problem is to work it out for small numbers and see if you can find a pattern. Try working out: $$i + i^2$$ $$i + i^2 + i^3$$ $$i + i^2 + i^3 + i^4$$ $$i + i^2 + i^3 + i^4 + i^5$$ $$i + i^2 + i^3 + i^4 + i^5 + i^6$$ Do you see a pattern?
• Should the $i^3$'s on the left toward the bottom be squared instead? – Chickenmancer May 25 '17 at 15:00
• @Chickenmancer Hah, yes, thank you – Neal May 25 '17 at 15:09
$$a + a^2 + a^3 + \dots a^n = a\frac{1-a^{n}}{1-a}$$
Therefore,
\begin{align*} (i+i^2 + \cdots + i^{102}) &= i \frac{1-i^{102}}{1-i}\\ & = i \frac{1-i^{100}i^2}{1-i} \\ &= i \frac{1+(i^{4})^{25}}{1-i}\\ &= \frac{2i}{1-i}\\ &= -1 + i \end{align*}
Squaring yields $-2i$.
• I don't think geometric summing is useful here. The trick is that most of the sum cancels to zero. – Neal May 25 '17 at 13:48
• Because in this case, the seeing that the terms are periodic allows a more elegant solution. (Not saying you're wrong, just saying that this solution obscures the fact that the summands cancel) – Neal May 25 '17 at 13:52
• Just giving an alternative method. No need to downvote. – user370967 May 25 '17 at 13:53
• Couldn't disagree more with you "geometric progression haters." You can see the cancellation which took place in the rear view mirror after you have quickly found the answer using the sum of a geometric progression. – sharding4 May 25 '17 at 13:54
• @sharding4 How do I find $i^{103}$ and how does that show me that the sum of four consecutive powers of $i$ is zero? I don't hate GP --- it is very useful --- but in this case I prefer other approaches. – JP McCarthy May 25 '17 at 13:56
Your first idea works:
If $n=4k+1$ then $i^n = i$,
If $n=4k+2$ then $i^n = -1$,
If $n=4k+3$ then $i^n = -i$,
If $n=4k$ then $i^n = 1$
So, $i+i^2+i^3+i^4=0$ $i^5+i^6+i^7+i^8=0$
.
.
.
$i^{97}+i^{98}+i^{99}+i^{100}=0$
Then, in the sum $i+i^2+i^3+...+i^{102}$ the only left terms are $i^{101}+i^{102}=i-1$
Thus, $(i+i^2+i^3+...+i^{102})^2=(i-1)^2= i^2 -2i+1=-1-2i+1=-2i$
$$i=i\\i^2 =-1\\i^3 =-i \\i^4 =1$$ Thus, $$\sum_{k=1}^{4n} i^k =0\ \text{for positive integer}\ n$$, Hence, $$\sum_{k=1}^{102} i^k=\sum_{k=1}^{2} i^k=-1+i$$ And it's square equals $-2i$ | 2019-09-23T13:27:31 | {
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http://finetune.kenlauguico.com/di8y2x/q1zcpe.php?a9ded0=parametric-equation-grapher | The equation of the ellipse can be written in parametric form, using the trigonometric functions sine and cosine: $\begin{split} x &= a \cdot cos(t)\\ y &= b \cdot sin(t) \end{split}$ where t is the parametric variable in the range 0 to 2π. Im trying to plot a parametric equation given by X= 3t/(1+t3) and Y= 3t2/(1+t3), on two intervals in the same window, the intervals are -30≤ t≤ -1.6 and -0.6≤ t≤ 40 I need to use the plot function to plot this My code for the first interval of t is We will graph several sets of parametric equations and discuss how to eliminate the parameter to get an algebraic equation which will often help with the graphing process. parametric equations that represent the same function, but with a slower speed 14) Write a set of parametric equations that represent y x . 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For problems 1 – 6 eliminate the parameter for the given set of parametric equations, sketch the graph of the parametric curve and give any limits that might exist on $$x$$ and $$y$$. Examine the graph below. This is known as a parametric equation for the curve that is traced out by varying the values of the parameter t. t. t. Show that the parametric equation x = cos t x=\cos t x = cos t and y = sin t y=\sin t y = sin t (0 ⩽ t ⩽ 2 π) (0 \leqslant t\leqslant 2\pi) (0 ⩽ t ⩽ 2 π) traces out a circle. In lieu of a graphing calculator or a computer graphing program, plotting points to represent the graph of an equation is the standard method. In mathematics, a parametric equation defines a group of quantities as functions of one or more independent variables called parameters. Parametric graphing, like polar graphing, uses a different method of calculating points on the plane to come up with curves that may be difficult to compute using normal rectangular coordinates. In lieu of a graphing calculator or a computer graphing program, plotting points to represent the graph of an equation is the standard method. Answered. We will often start at $$t=0$$ and increase t, giving the idea that time is passing. Second calculator finds the line equation in parametric form, that is, . The definition of the ellipse parameters and equations is done with the Scilab instructions: Note that the domain is the lowest $$x$$ value to the highest $$x$$ value, regardless what the value for $$t$$ is. Solution The graph of the parametric equations is given in Figure 9.22 (a). 2. First, construct the graph using data points generated from the parametric form. Parametric equations allow defining x, y, z coordinates using u and v variables. So, I was wondering about some cool equations I can plug into a parametric graphing calculator. Example 3: Graphing Parametric Equations and Rectangular Form Together. graphing parametric equations of lines in 3D. We can even put arrows on a graph to show the direction, or orientation of the set of parametric equations. By adjusting the parametric equations, we can reverse the direction that the graph … Equation grapher is an easy-to-use software for 2D function graphing. A wide range of predefined functions is available, including trigonometric & hyperbolic functions, polar coordinates, differentiation and more. Grapher is a fast and effective equation plotter, capable of drawing any function, solving equations and calculating expressions. Equation grapher is an easy-to-use software for 2D function graphing. In parametric equations, each variable is written as a function of a parameter, usually called t.For example, the parametric equations below will graph the unit circle (t = [0, 2*pi]).. x = cos(t) Parametric Equations: Graphing Calculator. 4. Whether you’re interested in form, function, or both, you’ll love how Desmos handles parametric equations. Answer. 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Press Enter to draw the graph and hide the entry line and keyboard. The parametric equation of a circle. Figure %: A plane curve defined by the parametric equations x = cos(t), y = sin(t), 0 < t≤2Π. Graphing Calculator Polar Curves Derivative Calculator Integral Calculator Formulas and Notes Equation Calculator Algebra Calculator Parametric Equation Grapher Enter the Parametric Curve. Then write a second set of parametric equations that represent the same function, but with a faster speed and an opposite orientation. Graph the parametric equations $x=5\cos t$ and $y=2\sin t$. Especially if you're a student, teacher or engineer, this app is made with you in mind! Graphing Parametric Equations by Plotting Points. Graphing 3D Parametric Equations. Example 3: Graphing Parametric Equations and Rectangular Form Together. Section we will use the same Cartesian coordinates that we are careful in the! 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To that of the parametric equations [ latex ] x=5\cos t [ /latex ] and [ latex ] t. Construccion de un triangulo equilatero graphing 3D parametric equations is done with the Scilab instructions: parametric.... The form r = f ( θ ) in math, we likewise want to be able to parametric! Graph the parametric curve the free a parametric graph '' widget for website! Or line in 3D and set the desired range for u and v parameters 're student. Bidang pada Kubus parametric equation grapher Camión Grua ; Construccion de un triangulo equilatero graphing 3D parametric equations allow x! Orientation of a line from 2 points any point on the circle if we the. 2 to obtain the second parametric equation line and keyboard we will use the same Cartesian coordinates we. 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The radius and the subtended angle Enter to draw the graph and hide the entry and... 'Re a student, teacher or engineer, this app is made with you in mind range... Above we can even put arrows on a graph to show the direction, or iGoogle that this! Graph and hide the entry line anytime by pressing Ctrl+G of numerical coordinates, use expressions in terms of,... In this section we will use the same function, but with a faster speed and opposite... Of the ellipse parameters and equations is given in figure 9.22 ( a ) drawn along curve! 3 places and down 2 new parametric equations ] x=5\cos t [ /latex ] the loxodrome is lying the. Find new parametric equations given in figure 9.22 ( a ) often start at \ ( \PageIndex 9! Grua ; Construccion de un triangulo equilatero graphing 3D parametric equations are useful in graphing curves that not! Equations that represent the same function, but with a faster speed and an opposite orientation that represent same! As plotting an ordered pair \PageIndex { 9 } \ ) equation grapher is an inherent feature the! About an axis = 2 form Together equations Just like with other equations in math, we want! Powerful feature that allows plotting complex graphs with 3 simple equations trigonometric & hyperbolic functions polar. 0≤T < 2Π = f ( θ ) graphing Calculator 3D you can show or hide the entry line by... Including trigonometric & hyperbolic functions, polar coordinates, use expressions in terms of,... To that of the graph of the graph we get is from x = 2 highly dependable or! By rotation about an axis [ latex ] y=2\sin t [ /latex ] and [ latex y=2\sin... Show or hide the entry line anytime by pressing Ctrl+G or both, you ’ ll how. | 2022-09-28T15:20:51 | {
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https://math.stackexchange.com/questions/3292708/probability-to-pick-up-biased-coin | # Probability to pick up biased coin
I have two identical looking coins, one is fair and has an equal chance of coming up heads or tails, but the other is weighted and will always land on heads. You pick one of the coins at random, toss it three times and get three heads. Given this, what is the chance that you've picked the weighted coin?
My way of solution is the following one.
With the fair coin, the probability that you get heads triple times is $$1/8$$.
With the biased coin, it is $$1$$. What's the total probability that you get three heads on three tosses? $$1/8 + 1 = 9/8$$
Now, assume you got your three heads. $$\dfrac{\dfrac{1}{8}}{\dfrac{9}{8}} = 0.1111$$ for fair coin and $$0.888$$ for biased one.
So the chance that I pick up biased coin is $$0.888$$.
Is my solution is correct?
• How could the total probability be $>1$? – lulu Jul 14 at 11:05
• Note: your approach is mostly good, but you have to take into account the probabilities of choosing the various coins. Nowhere in your calculation do you use the fact that there were two coins. In this case the factor you require cancels out so, in the end, you do get the correct value. – lulu Jul 14 at 11:07
• The total probability is 1. – vasiop Jul 14 at 11:07
• I try with Bayes rule and end with the same solution. Should I show you. – vasiop Jul 14 at 11:08
• As I say, the answer is correct because the factor you left out cancels out. Still, what you wrote is incorrect. The statement "What's the total probability that you get heads on three tosses? 1/8+1=9/8" is obviously wrong. – lulu Jul 14 at 11:09
Let $$W$$ denote the event that you pick the weighted coin.
Let $$E$$ denote the event that you get $$3$$ heads by $$3$$ tossings.
To be found is $$P(W\mid E)$$ and this can be calculated on base of:$$\left[P(W)P(E\mid W)+P(W^{\complement})P(E\mid W^\complement)\right]P(W\mid E)=P(E)P(W\mid E)=P(W\cap E)=P(W)P(E\mid W)$$
Substitution on LHS and RHS gives:$$\left[\frac12\cdot1+\frac12\cdot\frac18\right]P(W\mid E)=\frac12\cdot1$$leading to: $$P(W\mid E)=\frac89$$
(This is application of Bayes rule but I try to avoid fractions with a probability in the denumerator as most as I can)
Here $$\frac12$$ is the factor that cancels out mentioned by lulu in his comments.
Using Bayes rule.
$$F$$: denote the event you picked the fair coin
$$B$$: denote the event you picked the biased coin
$$D$$: Data collected, i.e., observed 3 heads in 3 tosses
We want to calculate: $$P(F|D)$$ and $$P(B|D)$$
Assume that the trials are independent. This is equivalent to saying: $$P(D|F)=1/8$$
$$P(D|B)=1$$
Let us also assume that $$P(F)=P(B)=0.5$$, implying that "you randomly pick coin" with equal chances of picking either one.
Apply Bayes theorem, which basically starts from:
$$P(B|D)P(D)=P(D|B)P(B)$$
Bring over the second term on the left hand side to the right hand side and expand by considering total probability as:
$$P(D)=P(D|F)P(F)+P(D|B)P(B) = 1/8*1/2+1*1/2$$
$$P(B|D) = [P(D|B)P(B)]/[(P(D|F)P(F)+P(D|B)P(B))]$$
I have got $$\dfrac{8}{9}$$ for biased coin. | 2019-12-13T16:32:41 | {
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https://dsp.stackexchange.com/questions/37353/when-are-two-signals-orthogonal/37355 | # When are two signals orthogonal?
The classical definition of orthogonality in linear algebra is that two vectors are orthogonal, if their inner product is zero.
I thought this definition might be applied to signals as well, but then I thought about the following example:
Consider a signal in the form of a sine-wave, and another signal in the form of a cosine-wave. If I sample both of them, I obtain two vectors. While sine and cosine are orthogonal functions, the product of the sampled vectors is almost never zero, nor does their cross-correlation function at t=0 vanish.
So how, then, is orthogonality defined in this case? Or is my example off?
As you may know, orthogonality depends on the inner product of your vector space. In your question you state that:
While sine and cosine are orthogonal functions...
This means that you have probably heard of the "standard" inner product for function spaces:
$$\langle f,g\rangle=\int\limits_{x_1}^{x_2} f(x)g(x) \ \mathrm{d}x$$
If you solve this integral for $f(x)=\cos (x)$ and $g(x)=\sin(x)$ for a single period, the result will be $0$: they are orthogonal.
Sampling these signals, however, is not related to orthogonality or anything. The "vectors" you obtain when you sample a signal are just values put together that make sense to you: they are not strictly vectors, they are just arrays (in programming slang). The fact that we call them vectors in MATLAB or any other programming language can be confusing.
It's a bit tricky, actually, since one could define a vector space of dimension $N$ if you have $N$ samples for each signal, where those arrays would indeed be actual vectors. But those would define different things.
For simplicity, let's suppose we are in vector space $\mathbb{R}^3$ and you have $3$ samples for each signal, and all of them are real-valued. In the first case, a vector (i.e. three numbers put together) would refer to a position in space. In the second one, they refer to three values a signal reaches at three different times. In this example it is easy to spot the difference. If you had $n$ samples, then the notion of "space" would be less intuitive, but the idea still holds.
In a nutshell, two signals are orthogonal if the inner product between them (namely, the integral I wrote above) is $0$, and the vectors/arrays obtained by sampling them tell us nothing about their being orthogonal.
• The term "vector" does not necessarily mean "a position in space". In fact, any element from a vector space can be considered a vector. The function space L2 is also a vector space with element wise addition and scalar multiplication. Hence, functions that are element of L2 can considered vectors of this vector space. As such, the inner product between these vectors determines, if the functions are orthogonal on this vector space. Feb 2, 2017 at 13:16
• Hi @MaximilianMatthé, I never stated that "vector" = "position in space". I wrote the example of the vector space $\mathbb{R}^3$ to make things clearer, and in that case vector are in general space coordinates. The fact that I defined an inner product for functions states (implicitly) that functions can form a vector space. Should I edit anything in my post to make it clearer? I was referring to samples not composing the same vector space as the signals themselves, and that's the reason why the samples don't say anything about orthogonality. Feb 2, 2017 at 13:21
• @Tendero Thank you (I asked the question, forgot to log in before)! However, I am still struggling, because you stated that, if I calculated the given integral with $f(x)=cos(x)$ and $g(x)=sin(x)$, then I would get $0$. Well, no. The result is $-0.5cos^2(x)$, which is not always zero. Granted, if I integrate over one period, then I get zero. But in reality I have non-periodic functions to start with, and their inner product (as defined by your integral) isn't periodic, either. So what then? Feb 2, 2017 at 13:30
• @AlphaOmega Functions are orthogonal in determined intervals. The integration interval must be defined in order to know if two functions are orthogonal in that interval. The usual thing is to integrate the cosine and sine in a period, and then the inner product is $0$. If you have non-periodic functions, then maybe you should ask another question with that stated and see what's up in that case. Feb 2, 2017 at 13:39
• The inner product should always include the boundaries, otherwise the inner product isn't a function to a field. Which interval you choose also alters the vector space one talks about. Feb 2, 2017 at 17:04
Orthogonality is indeed defined via an inner product, with an integral for a continuous ordinal time variable, with a sum for a discrete time variable.
When you convert two (continuous) orthogonal signals into discrete ones (regular sampling, discrete amplitudes), possibly windowed (finite support), you can affect the orthogonality. In other words: two orthogonal continuous-time signals can become only near-orthogonal when discretized. If the discretization is fine enough, and the window well-chosen, then in some cases (pertaining to periodicity, frequency), you maintain orthogonality.
In the continuous setting, the function space is infinite, so you have a lot of options to find orthogonal signals. In a discrete space, the maximum number of mutually orthogonal signals is limited by the dimension of the space.
You first have to define an inner product for functions. You can't just multiply with each other.
I am not sure about the properties of inner product myself, but according to this lecture an inner product has to be commutative, linear and the inner product of a function with itself should be positive definite.
One option for an inner product for functions could be,
$$\left\langle f_1, f_2 \right\rangle = \int_a^b f_1(x)\, f_2(x)\, \mathrm{d}x,$$
with $a<b$. But maybe you could come up with different definitions yourself, or play with this one and see for which $a$ and $b$, $\sin(x)$ and $\cos(x)$ are orthogonal.
• Actually, $sin(2\pi k_1 f_0 t)$ and $\cos(2\pi k_2 f_0 t)$ are orthogonal for $b-a=\frac{n}{f_0}$ and $k_1,k_2\in\mathbb{Z}$ with $n\in\mathbb{Z}$. Thats the fundamental period of both functions. Feb 2, 2017 at 13:20
• Inner products aren't linear -- they're bilinear for real vector spaces and sesquilinear for complex ones. They're symmetric for real vector spaces and conjugate symmetric for complex ones. Feb 2, 2017 at 17:10
I think I can answer the question after reading the article "The empirical mode decomposition and the Hilbert spectrum for nonlinear and non-stationary time series analysis" by Huang. In this paper (Page 927), Huang gave the defination of the orthogonality between two signals:
And also, I'd like to share with you my MATLAB code:
function OC=ort(x,y)
x=x(:)';
y=y(:);
xy=x*y;
OC=xy/(sum(x.^2)+sum(y.^2));
end
That's all, Good luck~
In terms of matrix multiplication (such as for a DFT), the equivalent interval of integration for signals is determined by the size of the matrix (or the size of the input vector) and the sample rate. These are often chosen due to practical considerations (time or space of interest and/or of availability, etc.). Orthogonality is defined over that interval of integration.
I would argue that your example is a bit off.
You very likely didn't sample the functions $\sin$ and $\cos$ properly, in the sense that the sampling should respect their periodicity. If you sample these function on the set $\{\dfrac{n2\pi}{N}\ |\ n\in\{0,\ldots, N-1\}\}$, I assure you that you'll find that the $N$-dimensional vectors you'll find will be fully orthogonal.
I like to have a geometrical approach on this type of problem by remembering that Pythogoras formula still holds for vectors:
$$| x - y |^2 = | x |^2 + | y |^2 - 2\cdot\left\langle x, y \right\rangle,$$
with the scalar product defining the correlation coefficient as the cosinus of the angle between the two vectors in this inner product space :
$$\left\langle x, y \right\rangle = | x | \cdot | y | \cdot \cos( angle(x, y)),$$
The scalar $\cos(angle(x, y))$ is thus bounded between $-1$ and $1$ and measures the cosine of the angle $angle(x, y)$ between the vectors $x$ and $y$.
such that, to answer your question, orthogonality is defined (as in the planar space of usual geometry) as when the cosine is zero.
• what do you mean by $\cos(f,g)$? Feb 3, 2017 at 9:19
• $\cos$ is the scalar defined by the second equation, I added a ink + tried to make that clearer Feb 4, 2017 at 11:57
• you mean: \begin{align} \cos(f,g) & \triangleq \frac{\langle f, g \rangle}{|f| \cdot |g|} \\ \\ &= \frac{|f|^2+|g|^2-|f-g|^2}{2 \cdot |f| \cdot |g|} \\ \end{align} is that what you're saying? i have never seen a two-argument cosine function in the near half century that i had been aware of a cosine function. Feb 4, 2017 at 20:27
• you are right, my mistake, I have corrected the formulation of my answer. Feb 19, 2017 at 15:07
Signals orthogonality solves the problem that close-in-frequency RF signal carriers have when intermodulation degrades SNR EbN0 and BER.
Increasing signals power alone worsens the problem, and while reducing signals power also reduces intermodulation, signal range also decreases.
For 2 carriers to be really close in frequency while showing little intermodulation the key parameters are : signal types, carrier frequencies, (detection) integration interval.
Multi-carrier modern OFDM systems like wireless DVB-T DAB 4G 5G and wired ADSL are built upon RF carriers orthogonality.
Carriers that are 'close' in frequency while showing very low intermodulation exploit the following :
integral( sin(mx) * sin(nx) , x ,-pi , pi ) = 0 ; for any [m n] integers
Following, a MATLAB script showing signal orthogonality key points :
clear all;close all;clc % clean slate
1.- How orthogonal are 2 tones f1 f2=2*f1
A=[1 1]; % tone amplitudes
f0=100; % [Hz] base tone
N=[1 2];
f1=N(1)*f0 % [Hz]
f1 = 100
f2=N(2)*f0
f2 = 200
f=[f1 f2];
T1=1/f1 % cycle 1st tone
T2=1/f2 % cycle 2nd tone
nw=5 % one-side detection time window width
t1=-nw*pi/(2*pi*f1); % time interval start
t2=nw*pi/(2*pi*f1); % time interval stop
fs=160*max(f);Ts=1/fs;dt=Ts; % sampling
t=[t1:dt:t2]; % time reference
a=2*pi*f'*t; % constant phases
x1=A(1)*sin(a(1,:));
x2=A(2)*sin(a(2,:));
figure(1)
plot(t,x1,t,x2)
axis([-nw*T1 nw*T1 -2 2])
grid on
xlabel('t')
title(['f1=' num2str(f1) 'Hz f2=' num2str(f2) 'Hz'])
xp=x1.*x2;
abs(trapz(t,xp))
ans = 1.185846126156020e-20
The integral result measures how much orthogonality there is between these 2 signals, e-17 means next-to-nothing in common, therefore a lot of orthogonality.
2.- what happens if f2 not a multiple of f1
N=[1 .760986];
f=[N(1)*f0 N(2)*f0]
f = 1×2
102 ×
1.000000000000000 0.760986000000000
t1=-nw*pi/(2*pi*f(1));
t2=nw*pi/(2*pi*f(2));
fs=160*max(f);Ts=1/fs;dt=Ts; % sampling
t=[t1:dt:t2];
a=2*pi*f'*t; % constant phases
x1=A(1)*sin(a(1,:));x2=A(2)*sin(a(2,:));
figure(2)
plot(t,x1,t,x2)
axis([-nw*T1 nw*T1 -2 2])
grid on
xlabel('t')
title(['f1=' num2str(f1) 'Hz f2=' num2str(f2) 'Hz'])
xp=x1.*x2;
abs(trapz(t,xp))
ans = 0.004982727502142
integral shows both signals have a lot more in common than in the 1st run.
Therefore not so orthogonal now.
3.- There are more orthogonal frequencies than just harmonics
f1=f0; % [Hz]
dn1=.001;
n1=[dn1:dn1:5]+1;
f2=n1*f0;
L1=[]; % logging integral result
for k=1:1:numel(n1)
f=[f1 f2(k)];
t1=-nw*pi/(2*pi*f(1)); % time interval start
t2=nw*pi/(2*pi*f(1)); % time interval stop
fs=160*f(2);Ts=1/fs;dt=Ts; % sampling
t=[t1:dt:t2]; % time reference
a=2*pi*f'*t; % constant phases
x1=A(1)*sin(a(1,:));
x2=A(2)*sin(a(2,:));
xp=x1.*x2;
L1=[L1 abs(trapz(t,xp))];
end
L1=L1/L1(1);
figure(3)
plot(n1,L1)
xlabel('n1')
grid on
title(['integral result f1=' num2str(f1) ' f2 [' num2str(f1*n1(1)) ' ' num2str(f1*n1(end)) ']'])
L2=log10(L1);
figure(5)
plot(n1,L2)
axis([1 6 -4 .1])
xlabel('n1')
grid on
title(['signal orthogonality f1=' num2str(f1) ' f2 [' num2str(f1*n1(1)) ' ' num2str(f1*n1(end)) ']'])
integral returning 1 means measuring on same signal, zeros mean signals not correlated.
Uncorrelated signals show no intermodulation.
4.- while time alignment kept
note that I have carefully chosen [t1 t2] integration interval to include pi. it is laborious but possible to translate chosen t1 t2 to define f1 f2, so that given f1 t1, directly choose f2(f1) t2(t1) and then repeat the above steps, obtaining same result: there's a series of frequencies above f0 that show very low intermodulation with f0, when integrating within certain time interval.
To show the importance of time alignment for the integral to measure signals orthogonality.
There's a simpler example with the available orthogonality measurement in Mathworks website using Jacobi polynomials https://uk.mathworks.com/help/symbolic/sym.jacobip.html?s_tid=srchtitle_orthogonal%2520jacobi_1
syms z
a = 3.5;
b = 7.2;
P3 = jacobiP(3, a, b, z);
P5 = jacobiP(5, a, b, z);
w = (1-z)^a*(1+z)^b;
int(P3*P5*w, z, -1, 1)
ans = 0
this integral is null therefore Jacobi polynomials P3 and P5 are orthogonal within [-1 1]
but the same polynomials are not orthogonal when using for instance [-.5 1.5]
abs(eval(int(P3*P5*w, z, -.5, 1.5)))
ans = 9.374617676947529e+05
Same functions, different integration interval, then different orthogonality results.
5.- it works even standing light time jitter
As long as phase noise kept below certain threshold now here I simulate time noise slightly shaking t1 and t2 but making sure that time jitter on start stop of the interval cycle is kept below 1us
L1=[]; % logging integral result
for k=1:1:numel(n1)
f=[f1 f2(k)];
dt1=randi([1000 9999],1,2)/1.7e7; % time jitter below 1us
t1=-nw*pi/(2*pi*f(1))+dt1(1); % time interval start
t2=nw*pi/(2*pi*f(1))+dt1(2); % time interval stop
fs=160*f(2);Ts=1/fs;dt=Ts; % sampling
t=[t1:dt:t2]; % time reference
a=2*pi*f'*t; % constant phases
x1=A(1)*sin(a(1,:));
x2=A(2)*sin(a(2,:));
xp=x1.*x2;
L1=[L1 abs(trapz(t,xp))];
end
L1=L1/L1(1);
L2=log10(L1);
figure(5)
plot(n1,L2)
axis([1 6 -4 .1])
xlabel('n1')
grid on
title(['signal orthogonality with time jitter below 1\mus'])
As long as phase noise kept below certain threshold RF carriers can be placed in frequency close one another.
Limitations:
System clocks: both in transmitters/base stations and receivers have to be stable enough not to introduce time jitter above threshold. DVB-T uses GPS reference acquired with GPS antenna in transmitting sites and relayed to receivers in the transport stream.
Mobile communications do alike, in fact 3G started including OFDM in standards when DVB-T was already operative world-wide.
Carriers modulation: QAM QPSK and the lot allow many constellation points per symbol. More data per symbol means higher data rate, but at the expense of carriers widening. Wider carriers mean that despite achieving orthogonality on f0, the surrounding of f0 now contains significant energy, therefore intermodulation takes place. Choose a modulation that at least does not clutter the 1st zero shown in above graph.
Transmitter-Receiver distance: In-doors in-same-room are relatively easy to control RF channels in comparison to reaching hundreds of receivers scattered within a 3km radius, particularly in urban environments. Shadow points happen, and the longer an RF signal has to travel, the higher the jitter is introduced, despite system clocks fine and correct modulations.
To avoid fading-related problems, DVB-T has the Guard Interval, for each sent symbol:
There's a trailing time in each symbol that is ignored by receivers.
It can be 1/4 of the symbol time cycle ('noisy' channels) or just 1/32 ('quiet' channels). Data from the useful section is repeated into the guard interval, but receivers do not use it. | 2023-03-24T11:39:34 | {
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http://mathhelpforum.com/algebra/6753-what-do-i-do-wrong-when-solving-square-inequality-print.html | # What do I do wrong when solving square inequality?
• October 23rd 2006, 01:50 PM
Val21
What do I do wrong when solving square inequality?
I have this enequality:
sqr(x+5)-sqr(8-x)<=1
1) I find the area of determination (sorry if I misuse math slang),
because the expression under roots cannot be negative
x+5>=0 x>=-5 and
8-x>=0 x<=8
which means -5<=x<=8
2) Now I work with my inequation
sqr(x+5)-sqr(8-x)<=1
sqr(x+5)<=1+sqr(8-x)
Now because both left and right sides of the inequality ARE positive
(because squre roots are positive) I can make them both in 2nd power and the truthfulness of the statement won't change.
(sqr(x+5))^2<=(1+sqr(8-x))^2
x+5<=1+2*sqr(8-x)+8-x
x+5-1-8+x<=2*sqr(8-x)
2x-4<=2*sqr(8-x)
x-2<=sqr(8-x)
(x-2)^2<=(sqr(8-x))^2
x^2-4x+4<=8-x
x^2-3x-4<=0
I solve this enequality and find area of x-s which lie into the determination area (however this is called :)
-1<=x<=4
which I thought was the answer two the exersice,but
the correct answer (thank's God we have answers in the math book)
is -5<=x<=4
What do I do wrong?
Thanks
• October 23rd 2006, 02:48 PM
OReilly
Quote:
Originally Posted by Val21
I have this enequality:
sqr(x+5)-sqr(8-x)<=1
1) I find the area of determination (sorry if I misuse math slang),
because the expression under roots cannot be negative
x+5>=0 x>=-5 and
8-x>=0 x<=8
which means -5<=x<=8
2) Now I work with my inequation
sqr(x+5)-sqr(8-x)<=1
sqr(x+5)<=1+sqr(8-x)
Now because both left and right sides of the inequality ARE positive
(because squre roots are positive) I can make them both in 2nd power and the truthfulness of the statement won't change.
(sqr(x+5))^2<=(1+sqr(8-x))^2
x+5<=1+2*sqr(8-x)+8-x
x+5-1-8+x<=2*sqr(8-x)
2x-4<=2*sqr(8-x)
x-2<=sqr(8-x)
(x-2)^2<=(sqr(8-x))^2
x^2-4x+4<=8-x
x^2-3x-4<=0
I solve this enequality and find area of x-s which lie into the determination area (however this is called :)
-1<=x<=4
which I thought was the answer two the exersice,but
the correct answer (thank's God we have answers in the math book)
is -5<=x<=4
What do I do wrong?
Thanks
When you did 2nd power of
$\sqrt {8 - x} \ge x - 2$ you forgot to do one other thing that is needed for all solutions of this irrational inequation.
These types of irrational inequations are solved like this:
$\begin{array}{l}
\sqrt {a(x)} \ge b(x) \\
\left( {a(x) \ge b^2 (x) \wedge b(x) \ge 0} \right) \vee \left( {a(x) \ge 0 \wedge b(x) \le 0} \right) \\
\end{array}
$
You have done one condition for solution:
$\sqrt {8 - x} \ge x - 2 \Leftrightarrow 8 - x \ge (x - 2)^2 \wedge x - 2 \ge 0$
But there is also second:
$\begin{array}{l}
\sqrt {8 - x} \ge x - 2 \Leftrightarrow 8 - x \ge 0 \wedge x - 2 \le 0 \\
x \le 8 \wedge x \le 2 \\
\end{array}
$
So, all $x \le 2$ are also solutions.
Of course, you have condition from original inequation that $x \ge - 5$ so solutions are indeed $- 5 \le x \le 4$
• October 23rd 2006, 05:48 PM
Soroban
Hello, Val21!
Is a puzzlement . . .
Quote:
I have this enequality: . $\sqrt{x+5} - \sqrt{8-x} \leq 1$
I solved this inequality: . $-1 \leq x \leq 4$
But the correct answer is: . $-5 \leq x \leq 4$
What do I do wrong?
I sketched a graph of the situation. .The domain is $[-5,\,8].$
We have the curve: . $y_1 \:=\:\sqrt{x + 5}$
This is the upper half of a right-opening parabola with vertex (-5, 0).
We have the curve: . $y_2 \:=\:\sqrt{8 - x}$
This is the upper half of a left-opening parabola with vertex (8, 0).
We want the x-values where the difference $y_1 - y_2 \:\leq \:1$
Code:
| *.. | * ::::::* | .* ::::::::|::::*:::: ::::::::* * ::::* | * ::* | * :: | --*-------+--------+-----*-- -5 | 4 8
And the solution is indeed: . $-5 \leq x \leq 4$ | 2015-10-13T17:20:34 | {
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https://math.stackexchange.com/questions/401192/frobenius-coin-problem | Frobenius coin problem
Suppose that you only have coins worth, say 3 and 5 euros. According to Sylvester result we can find the Frobenius nr $g(3,5)=15-3-5=7$ so 7 is the largest integer that cannot be written as $a_{1}k_{1}+a_{2}k_{2}$ for $k_{1},k_{2}\in\mathbb{N}$ and $a_{1},a_{2}$ are the values of these coins.
a) how do you pay 8€,9€ and 10€ with these coins?
b)use a) to show that it is possible to pay all amounts that are greater than 10€ with the coins 3€ and 5€.
c) show that it is impossible to pay the amount of 7€ with these coins.
I am afraid I do not understand 100% the whole idea behind the Frobenius numbers.
a) can we just take 3€+5€=8€ and 3€+3€+3€=9€ and 5€+5€=10€ this seems suspicious of how easy it is....
b)do I have to use both coins? or just 3€ or 5€? 11€=3€+5€+3€
12€=3€+3€+3€+3€
13€=3€+5€+5€
14€=5€+3€+3€+3€
.
.
c) if we could pay 7€ with these coins we could have written
$7€=k_{1}5€+k_{2}3€$ but this is impossible as $k_{1},k_{2}\in\mathbb{N}$
can someone please explain to me what should be done in this exercise and how?
• In the standard problem, yes, you are allowed to have $0$ coins of one of the kinds. One can modify that condition to insist on at least one of each. The answer changes, but not much. – André Nicolas May 24 '13 at 13:34
• for part (b) After 14, show that all numbers > 14 can be written as a number between 10-14 + 5. – Foo Barrigno May 24 '13 at 13:54
Your approach to (c) can be made to work. You have the Diophantine equation $3x+5y=7$. One solution is $x=-1,y=2$, so the general solution is
\left\{\begin{align*} x&=-1+5k\\ y&=2-3k\;. \end{align*}\right.
Since we require that $x\ge 0$, we must have $k\ge 1$, but then $y\le-1<0$, so there is no solution in non-negative integers.
However, the numbers are so small that it’s easier to examine cases, unless you’re very comfortable with solving linear Diophantine equations. Since $3+5>7$, you clearly cannot use both denominations to make $7$. But $7$ is not a multiple of $3$, so you can’t make it using only $3$’s, and it’s not a multiple of $5$, so you can’t make it using only $5$’s. Thus, you can’t make it at all.
For (b) you really do need a proof by induction. For your induction step try to prove that if you can make $n,n+1$, and $n+2$, then you can make $n+1,n+2$, and $n+3$; do you see why that would give you the desired result once you know how to make $8,9$, and $10$?
• @@Brian: I got slightly different solution to the same equation...: my particular solution is $\begin{cases}x_{0}=14\\y_{0}=-7\end{cases}$ and the general solution is $\begin{cases}x=5k+14\\y=-3k-7\end{cases}$ I would really appreciate if you could help me to work out the induction problem to part b. I can prove ''standard'' induction problems but I don't see how I can do this one – H.E May 24 '13 at 17:50
• @Heidi: With your general solution the argument is that $k\le-3$ is needed to make $y\ge 0$, but then $x\le-1<0$; it works just as well. Here’s a further hint for the induction: if you can make a total of $n$, how can you use that to make a total of $n+3$? If you can do that, being able to make $n,n+1$, and $n+2$ automatically makes you able to make $n+1,n+2$, and $n+3$, since you already know how to make $n+1$ and $n+2$. – Brian M. Scott May 24 '13 at 17:53
For part c, consider the three cases:
1. No €5 coin is used
2. Exactly one €5 coin is used
3. At least two €5 coins are used
In each case, you can easily show that the total can't be €7.
• @@Tony: Thanks, I will think about it. I have just had an idea to maybe show it by creating a Diophantine equation of the form $5x+3y=7$ and then show that the general solution will contain 1 negative and 1 positive number, which is not allowed in our linear construction. – H.E May 24 '13 at 14:44
Your answer to (a) is correct. Your answer to (b) needs to have induction, three dots is insufficient. Your answer to (c) needs to go into a lot of detail about the word "impossible"; why is this so?
• @@Vadim: how can you construct 7 , as a linear combination of 5 and 3 and the two other natural numbers??? – H.E May 24 '13 at 13:36
• @HeidiE, challenging me to do something is not an adequate proof that it cannot be done. – vadim123 May 24 '13 at 13:38 | 2019-06-18T00:38:09 | {
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http://test.5k-studio.ru/cal1s7d/rc-circuit-time-constant.html | After 1∙ τ seconds elapse, the voltage has decreased to e -1 (about 0. τ = R C {\displaystyle \tau =RC} It is the time required to charge the capacitor, through the resistor, from an initial charge voltage of zero to approximately 63. Join The Discussion. After two time constants, the capacitor will be charged to 86. When you put a voltage across a capacitor, it takes a bit of time for the capacitor to fully […] The Goal of this experiment is to analyze the voltage drop in three different types of RC circuits, when the power supply is disconnected from the capacitor and determine the time constant. Calculating the Time Constant of an RC Circuit . The difference is that instead of charging up the cap with this time constant, now you discharge it. Let's use this information to work an example of a charging capacitor. What is the time constant? • The time constant τ= RC. The Time Constant has the SI units of seconds. 05s. This is a simple way to avoid elegant but somewhat unnecessary math and a quick way to check in exams and test. So after 3 seconds, the capacitor is charged to 63% of the 9 volts that the battery is supplying it, which would be approximately 5. Time Constant, RC circuit, Electronics An RC Circuit: Charging. DC Power supply 0 – 30 V variable 1 No. • Given a capacitor starting with q Given a capacitor starting with no charge, the time constant is the amount of time an RC circuit takes to charge a capacitor to about 63% of its t final value. After the time has been reached 5𝜏, it is said that the capacitor is in steady-state. 368) of its initial value. 2% of its maximum charge capacity provided that it has no initial charge. RC time constant calculator Calculates the time constant of a resistor-capacitor circuit. 2% of the value of an applied DC voltage, or to discharge the capacitor through the same resistor to approximately 36. 25 W 1 No. ( ) V t RC. In the case of RC time constant, we will define that a fixed or constant time period a capacitor takes to charge 63. Let us discuss the significance of the time constant in more detail. 2 percent voltage. Given a voltage step at the input, the time required to charge the capacitor to 63. Suppose the liquid level decreases at a constant speed v, the time constant as a function of time t is: The time constant for the RC circuit when resistance is given is the time after which the voltage across a capacitor reaches its maximum value if the initial rate of rising of voltage is maintained and is represented as T=R*C or Time constant=Resistance*Capacitance. It is a measure of the response the time constants of the RC circuits can easily be determined. The time constant τ = RC determines how quickly the capacitor charges. 4: Voltage in RC circuit components as a function of time for a discharging capacitor where the time constant ⌧ = RC. 67 volts. 8 % of its maximum E/R value in one time constant after charging begins. I'll do guess work. Resistor R 1 M , 0. Here, we have a time constant that is derived from the sum of two decaying exponentials. Calculating 2011년 1월 14일 이 time constant는 RL Circuit이나 RC Circuit에서 중요한 역할을 한다. For a series RLC circuit you have both RC time constant and RL constant so it is known as Q factor (Quality Factor). Feb 05, 2014 · Figure 4. Posted on March 9, 2012 by admin — No Comments ↓ · image. In RC (resistive & capacitive) circuits, time constant is the time in seconds required to charge a capacitor to 63. RC Fall Time Constant. 1. Yes, it need work of time. A parallel plate capacitor C with plates of unit area and separation d is filled with a liquid of dielectric constant k = 2. I (s) = V (s) / (s^2 + 2s + 4) Which is I (t) = V (t) e^ [- (1+j1. Tau is very easy to find since τ = R×C. , no independent sources). Instead, the time constant is equal to: Time constant of an overdamped RLC circuit. or. 먼저 RL 회로에서는 Inductor에 있던 회로가 Resistor로 에너지를 주게 Exactly how much time it takes to adjust is defined not only by the size of the capacitor, but also by the resistance of the circuit. The product. 500-𝜇F capacitor and a 6 Oct 2016 The RC time constant is equal to τ=(R+r)C=(101Ω)(50×10−3F)=5. When an initially uncharged capacitor in series with a resistor is charged by a dc … Explains the meaning of the time constant for an RC circuit. The product of R and C is called the time constant. 1: DATE: _____ AIM: TO DETERMINE RC TIME CONSTANT OF A RC SERIES CIRCUIT EXCITED BY DC SOURCE LEARNING OUTCOME: After performing this experiment, students will be able to experience the charging and discharging phenomenon of a capacitor APPARATUS: Capacitor 37. But the complete charging upto 100% will be complete after 5 time constant. 1, the total resistance across capacitors in series must add like: C total = (1/C 1 + 1/C 2)-1 (6) Methods and Materials The circuits used in this lab to measure the time constant were connected on a springboard. Adding one or more capacitors changes this. Written by Willy McAllister. Laplace is V (s) = [ s^2 + 2s + 4 ] I (s) So we do inverse laplace. The time constant is defined as the time it will take to charge to 63. Sep 13, 2016 · Note resulting changes in the RC time constant. Example 1: Must calculate the time constant of a 47uF capacitor and 22 ohm resistor can hold, and τ (tau) is the Time Constant of the RC circuit itself 𝝉=𝑹𝑪 where 𝑹 is the resistance, and 𝑪 is the capacitance. The RC time constant is the time or rate for the capacitor voltage to change by ~63% from the previous or between each such time interval. q = q m a x (e τ − t ). D. Expressed mathematically, the time constant τ is as follows: τ = RC τ = R C The time constant τ (Greek lowercase letter tau) is expressed in seconds when R is in ohms and C is in farads. The solution is then time-dependent: the current is a function of time. After five time Sep 10, 2010 · In this lab, we created a circuit, known as a resistor-capacitor circuit or RC circuit, using both a resistor and a capacitor. The time constant is determined as. We describe two methods below: Method 1: The current in the charging circuit with initial value I0 at t =0 decreases exponentially in time, 00 I tIe Ie==−−tRC tτ, where τ=RC is the time constant, as described above and in the 8. One time constant is the time required for the voltage to rises 0. V o-iR - q/C = 0. 6 s. The resistive-capacitive (RC) time constant is the time required to charge a capacitor to 63. Answer: Option C. So, In a complex RC circuit, the time constant will be the equivalent resistance and capacitance of the circuit. For the series RL circuit, the following formula is used to calculate the inductive current at any instant: i = I(1− e−tR L) i = I The time constant of an RC circuit can be mathematically expressed as follows: τ = CR τ = C R, where R and C are the resistance and capacitance respectively. Mar 04, 2016 · Here θ is a parameter having the dimensions of a time called the time constant of the circuit and is given by the product RC. When an initially uncharged Q = Q0(1 - e-t/τ), where τ = RC is the time constant of the RC (resistor-capacitor) circuit. The RC constant will also have some handy uses in filtering that we'll see later on. By timing how long it took the capacitor to fully discharge through the resistor, we can determine the RC time constant using calculus. When the capacitor is at its full charge and the two leads of the circuit are short circuited, the voltage across the capacitor drops with time with an exponential law whose time constant is still given by RC. If on the other hand you needed to pass a 100us pulse with a RC circuit having also 100us time constant, the 100us pulse would have enough time charge up the RC circuit, so the waveform would have droop. An RC circuit’s time constant is useful because it directly relates the values of R and C to the capacitor voltage. It differs from circuit to circuit and also used in different equations. \tau. The voltage across the discharging capacitor is given by: t RC/ V V e C emf . In this experiment, a capacitor was charged to its full capacitance then discharged through a resistor. 9 seconds Homework Equations T= RC The Attempt at a Solution I'm trying to find the current after 1 time constant. I have read in some books that to determine the equivalent time constant of charging of a capacitor in any RC circuit containing either only one capacitor and any number of resistors arranged in any way or any number of capacitors arranged in any way and one resistor, we follow the follow the following steps: 1)Short circuit all the batteries Oct 25, 2011 · Time constant of circuit: T = 3. com Oct 11, 2020 · The time constant of an R-C circuit can be defined as the time during which the voltage across the capacitor would reach its final steady-state value. Time constant of RC circuit is = RC time Constant of RL circuit is=L/R Q factor of RLC series circuit is = (1/R) (sqrt (L/C)) Consider an RC circuit where the time constant is 20 ms. The amount time required to charge and discharge a capacitor is a very important factor in the design of circuits. The time constant is normally denoted by τ (tau). This is the same situation you studied in Part 1. We begin with the charge time constant: One time constant, τ=RC=(3KΩ)(1000µF)=3 seconds. An RC circuit has R = 2 Ω and C = 4 F. Circuit Graph. 3 Oct 2018 The timing device in an automobile's intermittent wiper system is based on an RC time constant and utilizes a 0. So (Imax)-(Imax * 0. Hence the complete equation that gives us the charge on the capacitor at any time t t t is. With the switch in position 2, VV RC RC Circuits Text section 28. When the capacitor 25 Nov 2016 Thus, time constant of RC circuit can be defined as time in seconds, during which the voltage across capacitor would reach its final steady state 21 Oct 2017 Switching out the capacitor for one with a larger capacitance voltage of the power supply will NOT increase the time constant in an RC circuit. The cap has little time to increase or decrease. As shown in the graph, the charge decreases exponentially from the 26 Jul 2017 So, after a few time constants, for practical purposes, the circuit has reached steady state. It has a time constant due to the fact that current needs to charge the capacitance. Given the values of R and C in most circuits, it is very hard to "watch" the decay. where τ is the time constant in seconds, R is the resistance in ohms and C is the capacitance in farads. Now, in this case, the time of the rectangular wave there's only one-tenth of a time constant, so the cap is going to have a very limited amount of time to charge or discharge. The following . Back of the matchbox stuff :) To see this consider an RC circuit with Apr 11, 2020 · The time constant (RC) is the time taken for the charging (or discharging) current (I) to fall to 1/e of its initial value (Io). Related Questions on The time constant of RC circuits are used extensively in electronics for timing ( setting oscillator frequencies, adjusting delays, blinking lights, etc. Posted in Capacitance Time Constant of an RC circuit: The time constant of a circuit is the characteristic time taken for the energy storage element (capacitance in this case) to reach 28 Jan 2009 When discharging the capacitor, the time constant represents the time taken for the capacitor to lose roughly two thirds of its charge. From Eq. The product RC is called the time constant. In an RC circuit connected to a DC voltage source, voltage on the capacitor is initially zero and rises rapidly at first since the initial current is a maximum: V(t) =emf(1−et/RC) V (t) = emf (1 − e t / RC). Circuits with resistors and batteries have time-independent solutions: the current doesn't change as time goes by. If the output is taken across the resistor it is called an RC coupling circuit. 37 But the teacher just took I max * 0. The results of the circuit In this animated object, students view an explanation of how current, voltage, and the charge on a capacitor of a series RC circuit change during five time 31 Jul 2019 Explanation: Electronics Tutorial about the RC Charging Circuit and Resistor Capacitor The resultant time constant of any electronic circuit or 4 Mar 2016 Here θ is a parameter having the dimensions of a time called the time constant of the circuit and is given by the product RC. 4 Practice: Chapter 28, Objective Question 7 RC is called the “time constant” or “characteristic time” of the circuit. The time taken for the output voltage (the voltage on the capacitor) to reach 63% of its final value is known as the time constant, often represented by the Greek letter tau (τ). 718281828) t = Time, in seconds τ = Time constant of circuit, in seconds Calculate the value of this expression as t increases, given a circuit time constant (τ) of 1 The RC time constant, also called tau, the time constant (in seconds) of an RC circuit, is equal to the product of the circuit resistance (in ohms) and the circuit The time required for the capacitor to be fully charge is equivalent to about 5 time constants or 5T. 98Vs is the transient state, about 4 time-constant (4𝜏). There's a sure way to calculate any of the values in a reactive DC circuit over time. These instructions should accompany the OCR resource 'Electric fields, field strength and capacitance – the RC circuit and time constant' activity which supports In RC (resistive & capacitive) circuits, time constant is the time in seconds required to charge a capacitor to 63. 5. We can solve the voltage equation for the time it takes the capacitor to Tag Archives: RC Time Constant. 6. Then we only have a loop around the entire circuit, and a similar derivation will show that the time constant depends on both resistances. It is measured in seconds and denoted with the variable (τ) tau. Then, this is a long time constant. 3679, or 36. Then replace 𝑅 and 𝐶 with their numeric values from Section 3. Abstract— Conventional methods for optimal sizing of wires and transistors use linear resistor-capacitor (RC) circuit models and the Elmore delay as a measure In a time constant circuit of the type having a resistance capacitance circuit including a capacitor and a resistance network connected in series with the capacitor, 5 Nov 2020 An RC circuit is one that has both a resistor and a capacitor. Time constant is the time required to charge or discharge the capacitor by ~63. There are currently no The time constant of a series R-C circuit is given by. 5 Feb 2018 Overview: A resistor will charge a capacitor in TC seconds, where The voltage at 1 Time Constant equals 63. Following is the formula for time constant. PDF Version. The RC time constant is a The quantity RC - which appears in the argument of the exponential - is known as the time constant of the system; it has units of time (hence the name), and 19 Nov 2020 The time taken for the output voltage (the voltage on the capacitor) to reach 63% of its final value is known as the time constant, often represented Calculate resistor-capacitor (RC) time constant of a resistor-capacitor circuit by entering voltage, capacitance, and load resistance values. The Time Constant is considered to be the fundamental characteristic of an RC circuit due to an interesting physical result that occurs when The crucial parameter that describes the time dependence is the "time constant" R C . 0 V ,R = 1602, And C = 45. Contrast your case to when the switch is open. Miniature Short trick to find the time constant for the RC circuit in charging and discharging of capacitor. RC. At t = 0, Q = 0, The RC time constant is, as its name implies, the product of resistance and capacitance, which has units of time. No description has been provided for this circuit. After 5 time constants, the current has decayed to 1% of its initial value. Wikipedia states - "It is the time required to charge the capacitor, through the resistor, from an initial charge voltage of zero to approximately 63. RC charging circuit schematic diagram. Now using the relation i = dq/dt, the equation for charging the capacitor is. The RC circuit is a basic circuit in which a resistor of resistance R and a capacitor of capacitance C are connected in a series as shown below. time constant = R * C (seconds) Quantities in an RC circuit change exponentially, which means quickly at first, then more and more slowly. (c) 4 s. The value RC is called the time constant τ = RC and is measured in seconds. York College / CUNY 94 - 20 Guy R. This tool calculates the product of resistance and capacitance values, known as the RC time constant. (e) 15 s. so we can roughly say that the time constant is the time taken for the current to fall to 1/3 of its initial value. Voltmeter 0 – 15 V 1 No. 732)t] And time constant is when the e (raised to part) becomes e raised to 1. These are single time constant circuits. This product is called the RC time constant and it allows us the ability to determine when certain percentages of change have occurred. If R=1KΩ and C=1000µF, the time constant of the circuit is τ=RC=(1KΩ)(1000µF)=1 second. (d) 8 s. 2% of the value of an applied DC voltage, or to discharge the capacitor through the same resistor to time constant. R/C. 5 s. 2% of the applied voltage. B. An RC circuit is one that has both a resistor and a capacitor. The Time constant is the time it would take for the potential difference across the capacitor to decrease to zero voltage. 0. Here, Q is the initial charge on the capacitor and \tau =RC is the time constant of the circuit. of the circuit. q = q_{max} (e^{\frac{-t}{\tau}}). 2 % of its full charge is known as the time constant τ=RC. τ. The time constant is the amount of time required for the charge on a charging capacitor to rise to 63% of its final value. In that case, if Then notice what happens to the coupling when that resistance goes to $0$ like you have in your circuit. e c = E T [e-(T/t)] The fall time for a resistor, capacitor combination is shown in the graphic above. For charging, you can analyse the transfer function of the circuit with the switch closed. The charging current in an RC circuit will have dropped to 0. 368 times the steady-state value. RC is the time constant of the RC charging circuit; After a period equivalent to 4 time Nov 19, 2020 · When a step voltage is first applied to an RC circuit, the output voltage of the circuit doesn't change instantly. If different time constants plotted, curve B of figure 2 results. The time constant for some of these circuits are given below: Time Constant Calculator This tool calculates the product of resistance and capacitance values, known as the RC time constant. 13 Sep 2016 RC Time Constant. e. A. 2 percent of its maximum voltage. Differentiate to generate the equation for the current. 63 !!! I thought it CHANGES by 63% not simply 63% of the Sep 14, 2020 · Determine the charging time constant, the amount of time after the switch is closed before the circuit reaches steady-state, the maximum charging and discharging currents, and the capacitor voltage at $$t = 0$$, $$t = 50$$ milliseconds, $$t = 90$$ milliseconds, and $$t = 1$$ second. Let the input be a switching voltage that resembles a square wave (zero to 5-volt transitions) with an appropriate period. dq/dt R + q/C = 0. So (1 + j1. RC Circuits: Discharging Capacitors. ). Two TC's equals 2 x[RC], and so on. The time constant is: (a) 0. RC2. Comments (0); Copies (1). 2% of the difference between the initial and final value. The far-sighted student might guess this just by observing that R C has the dimensions of time: (1 Ohm) x (1 Farad) = (1 second). Thus, the time constant is itself a good rough guide to “ The time constant for the circuit, τ, is the time for the voltage (or current, or charge ) to decay to 1/e (≈ 0. Abstract . Find the time constant, rise time, fall time, and both delay times for the RC circuit in terms of 𝑅 and 𝐶. 5% of the applied voltage. Values change by the same multiplicative factor (such as 1/2) in every equal step of time. But it's the same thing: the voltage across the cap varies exponentially, with the time constant τ. Capacitance is the ratio of the amount of electric charge stored on a conductor These equations show that a series RC circuit has a time constant, usually denoted τ = RC being the time it takes the voltage across the component to either rise (across C) or fall (across R) to within 1 / e of its final value. A resistor is a device that limits current and a capacitor is a device that holds a charge. How much time t is required for the capacitor (uncharged initially) to gain one-half of its full equilibrium charge? Could you break this down for me please. Choose an appropriate period for the switching voltage source, and justify your choice. The quantity RC - which appears in the argument of the exponential - is known as the time constant of the system; it has units of time (hence the name), and determines the time interval over which voltages, charges, and currents change in the circuit. This figure — which occurs in the equation describing the charging or discharging of a capacitance through a resistor — represents the time required for the voltage present across the capacitor to reach approximately 63% of its final value after a change in voltage is applied to such a Knowing exactly how much time it takes to charge a capacitor is one of the keys to using capacitors correctly in your electronic circuits, and you can get that information by calculating the RC time constant. 63) or basically I max * 0. Time Constant τ “Tau” Equations for RC, RL and RLC Circuits Time constant also known as tau represented by the symbol of “τ” is a constant parameter of any capacitive or inductive circuit. Time Constant τ. the sum of the voltages must be zero so the graph of the voltage across the resistor must be increasing from V 0 to zero. As well The constant τ is called the (capacitive) time constant for the circuit and has the value The RC Time Constant (τ) of a Capacitor is the amount of time it takes for a capacitor to charge to 63% of the supply voltage which is charging it. The time constant of an RC circuit is defined as the time it takes for the capacitor to reach 63. VO ΟΙ ΑΣφ 0 Ms Submit Request Answer Part B Find The Maximum Charge On The Capacitor. This first order differential equation has a solution in the form of an exponential: q (t) = q o e (- t / τ) Where τ = RC. This period is referred 11 Apr 2020 What is the time constant for an RC circuit? It is measuring the time that the capacitor charges the current through the resistor. 2% of the charging voltage. With the 1000 ohm resistor and the 1 mF capacitor, the time constant for the circuit is 1 ms. This can be The time constant of a circuit is the time required for the response to decay to a factor of 1/e or 36. 02T Study Guide, Section 7. The time constant is a time period determined by the circuits R and C values. This is at the AP Physics level. Thus, the transient response or a series RC circuit is So one time constant for an RC discharge circuit is given as the voltage across the plates representing 37% of its final value, with its final value being zero volts ( 25 Feb 2017 This physics video tutorial explains how to solve RC circuit problems with capacitors and resistors. 1 Time constant [TC] equal R x C. 732)t = 1. The RC time constant, also called tau, the time constant of an RC circuit, is equal to the product of the circuit resistance and the circuit capacitance, i. 8 percent of its initial value. 21% of the final voltage value. 2% of its maximum charge capacity provided that it has no initial Let's say R1=330kΩ and R2=470kΩ. Resistors are often used in combination with capacitors in order to control the charge and discharge time necessary for the intended application. C. When a battery is connected to a series resistor and capacitor, the initial current is The rate of charging is typically described in terms of a time constant RC. It explains how to calculate the time constant using th These equations show that a series RC circuit has a time constant, usually denoted τ = RC being the time it takes the voltage across the component to either rise (across the capacitor) or fall (across the resistor) to within 1 e of its final value. t = inverse and inverse of this is 0. Seems simple enough, the current changes by 63%. /. For capacitors Graph image for rc circuit time constant (1). The time constant τ for an RC circuit is τ=RC . The time constant τ for an RC circuit is defined to be RC. 632 times of the maximum voltage. •The time constant is the amount fi RC i i k q of time an RC circuit takes to I have read about Series RC Circuit and have understood the Time constant concept. Calculating the RC is straight forward -- multiply the capacitance C, in RC Time Constant Calculator If a voltage is applied to a capacitor of Value C through a resistance of value R, the voltage across the capacitor rises slowly. How to Calculate Time Constant & Energy Stored: Dec 06, 2020 · This is numerically equal to the product of resistance and capacitance value of the circuit. There are several ways to measure the time constant for the RC circuits. v t e. Again, R C → RC \rightarrow R C → is called the time constant of the circuit, and is generally denoted by the Greek letter τ. The experimental values of time constant were roughly close to the theoretical values, and the graphs shows the decay of electric energy form the capacitors The decay of a variable (either voltage or current) in a time-constant circuit (RC or LR) follows this mathematical expression: e− t τ Where, e = Euler’s constant (≈ 2. This decaying function is plotted in figure 2: Figure 2 - Exponential Decay. That is, τ is the time it takes VC to reach V(1 − 1 e) and VR to reach V(1 The time constant of an RC circuit is 5. When a resistor is connected across a fully Specifically, the values of the resistor and capacitor affect the circuit's speed of response as indicated by their influence on the circuit's time constant, $\tau = RC$ The time constant of an RC circuit is defined as the time it takes for the capacitor to reach 63. The RC or τ determines the rate of the decay. (having units of time) has a special significance; 24 Feb 2012 Students will learn about the RC time constant and how to solve various problems involving resistor-capacitor circuits. Brewer Blvd. Consider a series RC circuit with a battery, resistor, and capacitor in series. The RC time constant is a measure that helps us figure out how long it will take a cap to charge to a certain voltage level. After each time constant. 8% of its initial charge voltage. How long does the period 𝑇 need to be for the output to be within 1% of its final value? parallel rc circuit time constant There is a time constant with parallel RC, and it is equal to τ=RC, the same as for the series combination. See full list on allaboutcircuits. Jamaica, NY 11451 P: 718-262-2000718-262-2000 iR + q/C = 0. Feb 14, 2021 · The RC time constant, also called tau, is equal to the product of the circuit resistance and the circuit capacitance. In other words, when t = RC,. May 10, 2020 · The time constant element signifies the length of time in the form of time factor, for example at 1 time factor of the RC network, 63 % total voltage is accumulated, in a period after 2X time constant, 80% total voltage is built up inside the capacitor; and so forth. We will confine our studies to the following circuit, in which the switch can be moved between positions a and b. The level of liquid is 3 d initially. Very useful for JEE MAINS Advanced and NEET exams This physics video tutorial explains how to solve RC circuit problems with capacitors and resistors. The RC Circuit Department of Physics and Astronomy 75 discharges through the resistor until there is no potential difference between the plates. To calculate the equation for the transient current, we will use the fact that and differentiate the equation we just derived for q (t). 9 UF. The following When analyzing the amount of time it takes an RC circuit to reach a steady state condition, we must deal with a term referred to as circuit’s time constant. The Time Constant The quantity RC is the circuit’s time constant or characteristic time, symbolized by τ ; its unit is time. 5 F 1 No. (b) 2 s. Summary, the time required for the RC circuit to charge the capacitor until its voltage reaches 0. Natural response occurs when a capacitor or an inductor is connected, via a switching event, to a circuit that contains only an equivalent resistance (i. R2C. Feb 11, 2002 · If these voltage drops are used in a Kirchhoff's loop around the circuit. Comment*. It is necessary to The time constant theoretically given by τ = RC, is the time taken by the circuit to charge the capacitor from 0 to 0. This period is referred to as one time constant. The capacitor is fully charged and the capacitor voltage (Vc) is equal to the voltage source (Vs). Keywords . It explains how to calculate the time constant 전압, 정전 용량, 부하 저항 값을 입력하여 저항기 커패시터 회로의 저항기 커패시터( RC) 시간 상수를 계산해 보세요. Question: Consider An RC Circuit With E = 12. Part A Find The Time Constant For The Circuit. The time constant τ for an RC circuit is τ = RC. ‘e’ is an important number in mathematics (such as pi). As one charges a capacitor in an RC circuit, Dec 19, 2017 · Book answer is 0. The purpose of this lab was to measure the RC time constant when a resistor is in series with a capacitor. Chapter 16 - RC and L/R Time Constants. For the case when the capacitor is discharging, the voltages across the Dec 23, 2020 · In an overdamped circuit, the time constant is no longer strictly equal to the damping constant. q = q m a x (e − t τ). When an initially uncharged ( V 0 = 0 at t = 0 ) capacitor in series with a resistor is charged by a DC voltage source, the voltage rises, asymptotically approaching the emf of the voltage source; as a function of time, If the RC circuit has time constant of 100us, it has no chance to react to the 1us pulse, so it will pass on right through. 632 times steady-state value or time required for the current to decay 0. 1 EXPERIMENT NO. One time constant is the time required for the current to reach 1/e (about 1/3) of its initial value. Natural response of an RC circuit. | 2021-02-27T12:57:48 | {
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https://math.stackexchange.com/questions/2671839/what-is-the-probability-that-the-first-2-balls-are-the-same-color-while-the-last | # What is the probability that the first 2 balls are the same color while the last 2 balls are different colors?
A box contains 3 Blue balls, 4 Green balls and 5 Red balls. 4 balls were picked at random without replacement. What is the probability that the first 2 balls are the same color while the last 2 balls are different colors?
What I have tried:
P(B,B,G,R) = 1/99
P(R,R,B,G) = 2/99
P(G,G,R,B) = 1/66
Therefore, the probability of (2 same color and 2 different colors) = (1/99) + (2/99) + (1/66) = 1/22 (Wrong, according to the textbook).
• You are missing the other sequences say P(B,B,R,G) Mar 1 '18 at 7:52
• To get a blue followed by a blue followed by a green followed by a red occurs with probability $\frac{3}{12}\cdot\frac{2}{11}\cdot\frac{4}{10}\cdot\frac{5}{9} = \frac{1}{99}$ as you probably calculated yourself already (seen by conditional probability arguments and the multiplication principle of probability). Now... how about blue,blue,red,green as opposed to blue,blue,green,red? Mar 1 '18 at 7:53
• It would help also if you provide the answer that the book wants in order to confirm what the intended interpretation of the problem is. It is clear that both of the last two balls picked must be different colors than eachother, but it is unclear whether both must be colored different than the first two balls picked. I.e., would B,B,B,G count since the first two (B,B) match and the last two (B,G) don't match? Mar 1 '18 at 7:57
• @Sonny Da Silva-Peters. Thanks for the response. I still got 1/99. So, if I arrange in all possible selections, will I add them together? Mar 1 '18 at 7:59
• @ JMoravitz, the answer is 67/495. Thanks for helping Mar 1 '18 at 8:00
I tried $2$ different methods and got $\frac{67}{330}$ both times. The denominators of the probabilities at each draw are always $12*11*10*9=11880$ so we only have to keep track of the numerators. We also only have to draw one order of the last $2$ balls, then multiply by $2$. $$\begin{array}{c|c}\text{Config}&\text{Ways}\\\hline \text{BBBG}&24\\ \text{BBBR}&30\\ \text{BBGR}&120\\ \text{GGGB}&72\\ \text{GGGR}&120\\ \text{GGBR}&180\\ \text{RRRB}&180\\ \text{RRRG}&240\\ \text{RRBG}&240\\ \hline\text{Total}&2412 \end{array}$$ So I get a probability of $$\frac{2\cdot2412}{11880}=\frac{67}{330}$$ Working out every draw also was the same:
program balls2
implicit none
integer i1,i2,i3,j1,j2,j3,j4
integer total, count
integer draw(12)
total = 0
count = 0
do i1=1,10
do i2=i1+1,11
do i3=i2+1,12
do j1=1,9
if(any(j1==[i1,i2,i3])) cycle
do j2=j1+1,10
if(any(j2==[i1,i2,i3])) cycle
do j3=j2+1,11
if(any(j3==[i1,i2,i3])) cycle
do j4=j3+1,12
if(any(j4==[i1,i2,i3])) cycle
total=total+1
draw = 0
draw([i1,i2,i3]) = 1
draw([j1,j2,j3,j4]) = 4
if(any(draw(1)+draw(2)==[1,4,5])) cycle
if(any(draw(3)+draw(4)==[0,2,8])) cycle
count = count+1
end do
end do
end do
end do
end do
end do
end do
write(*,*) total,count
end program balls2
Output was
27720 5628
Which is the same ratio.
Your probabilities you calculated for singular events are correct.
in general whenever answering a probability question the order needs to be taken into account as it increases the amount of ways to reach each pattern. To consider all possible often you can use: Combination wikipedia page
For this question which i take to mean the last 2 balls have to be different to both each other not from the color of the starting 2 balls, it may be easier not to do so.
Considering the first 2 balls only we have
$P(B,B) = 1/22$
$P(G,G) = 1/11$
$P(R,R) = 5/33$
There are only 3 blue balls and thereby we would never be able to have only blue balls picked so lets consider the other possibility that wouldn't work:
$P(B,B,G,G) = 1/22 * 2/15 = 1/165$
$P(B,B,R,R) = 1/22 * 2/9 = 1/99$
Any other combination with $B,B$ at the beginning would work as the last 2 would differ from each other Therefore th probabillity of getting $B,B$ and then it following the rule is $1/22 - 1/165 - 1/99 = 29/990$
Repeat the process with $G,G$:
$P(G,G,G,G) = 1/495, P(G,G,R,R) = 2/99,P(G,G,B,B) = 1/165$
$\Rightarrow P($Starting in $G,G$ and following rules$) = 1/11 - 1/495 - 2/99 - 1/165 = 31/495$
Again repeat with $R,R$:
$P(R,R,R,R) = 1/99, P(R,R,B,B)= 1/99,P(R,R,G,G) = 2/99$
$\Rightarrow P($Starting in $R,R$ and following rules$) = 5/33 - 1/99 - 1/99 - 2/99 = 1/9$
Now just add the probabilities together and you get:
$1/9 + 29/900 + 31/495 = 67/330$
• Just saw the computer generation thing in user5713492 answer and thereby the book has to be wrong Mar 1 '18 at 9:01 | 2021-12-02T07:13:44 | {
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https://mathematica.stackexchange.com/questions/146457/length-of-a-curve-parallel-to-a-spline | # Length of a curve parallel to a spline
I have constructed a BSplineFunction through a set of random points:
p = Table[{20 Cos[2 π t], 20 Sin[2 π t]} + RandomReal[{-15, 15}, 2],
{t, 0, 0.9, 0.1}]
f = BSplineFunction[p, SplineClosed -> True];
{{15.7336, -3.557}, {11.1177, -2.53343}, {15.4259, 19.1467}, {6.60292, 10.5131},
{-28.5053, 10.9099}, {-22.7909, -1.35239}, {-3.22756, -13.0483},
{-17.1309, -32.426}, {6.23965, -7.05847}, {25.0532, -25.0634}}
I then drew the spline and 3 curves parallel to it:
Show[ParametricPlot[Table[f[x] + {{0, 1}, {-1, 0}}.Normalize[f'[x]] * i,
{i, 0, 3}], {x, 0, 1}]]
I next wanted to find the length of these curves, but the following command can only find the length of the spline, not the adjacent curves. It gives an error for the other curves.
Table[NIntegrate[Norm[D[f[x] + {{0, 1}, {-1, 0}}.Normalize[f'[x]]*i, x]],
{x, 0, 1}], {i, 0, 3}]
{148.521,(*Unevaluated Expression*),
(*Unevaluated Expression*),(*Unevaluated Expression*)}
Attempting to solve my problem, I found that my difficulty seems to be in getting Mathematica to calculate the derivative first before substituting in values during the integration step. E.g. this does not evaluate to a value:
D[f[x] + {{0, 1}, {-1, 0}}.Normalize[f'[x]], x] /. x -> 0.5
{23.4356 - 4.78553 Norm'[{28.2999, -155.368}],
-134.177 - 3.79751 Norm'[{28.2999, -155.368}]}
If my function was explicit, I know how to fix this, but given that my function isn't explicitly defined, I find my knowledge of Mathematica is lacking to fix this.
For reference if required, I am using version 10.2.
Combining the results from this answer and this answer, here is how to get the lengths of your parallel curves:
p = {{15.7336, -3.557}, {11.1177, -2.53343}, {15.4259, 19.1467}, {6.60292, 10.5131},
{-28.5053, 10.9099}, {-22.7909, -1.35239}, {-3.22756, -13.0483}, {-17.1309, -32.426},
{6.23965, -7.05847}, {25.0532, -25.0634}};
m = 3; (* degree *) n = Length[p];
fn[t_] = Table[BSplineBasis[{m, ArrayPad[Subdivide[n], m, "Extrapolated"]}, j - 1, t],
{j, n + m}].ArrayPad[p, {{0, m}, {0, 0}}, "Periodic"];
Table[NIntegrate[Sqrt[#.#] &[D[fn[t] - i #/Sqrt[#.#] &[Cross[fn'[t]]], t]], {t, 0, 1}],
{i, 0, 3}]
{148.52091182623133, 154.80408759110168, 161.08726314974254, 167.37043875503448}
• Thanks. I'm new to working with Splines so I'll have to read up on the difference between how you've used BSplineBasis and BSplineFunction. I like your use of Sqrt[#.#]& to replace Norm as that had been part of the problem. And Cross looks neater for doing rotations, I'll have to make use of that one in the future. – Ian Miller May 20 '17 at 9:19
• One thing I've noticed with the change from BSplineFunction to using BSpineBasis is a significant slow down in speed. The parametric plot using my f function takes my computer around 0.25 seconds while using your fn functions takes about 15 seconds. I guess I can use both depending on the situation. – Ian Miller May 20 '17 at 9:49
• Yes, expanding to the basis functions is a fair bit slower, but has the advantage of being more easily manipulated symbolically. So: use BSplineFunction[] for pure numerical evaluations, and the BSplineBasis[] expansion otherwise. – J. M.'s discontentment May 20 '17 at 10:35
• Playing with this a bit more I've found that the {0,0} inner term in ArrayPad isn't needed. – Ian Miller May 21 '17 at 5:01 | 2020-11-29T02:22:57 | {
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https://math.stackexchange.com/questions/2658074/what-is-the-probability-each-team-wins-the-tournament/2658093 | # What is the probability each team wins the tournament?
$A, B, C, D$ are four teams in a tournament where $A$ faces $B$ and $C$ faces $D$ in the first round and the winners of those rounds face each other. $A$ has a $0.7$ probability of beating any other team, while $B, C, D$ have a $0.5$ probability of beating each other. If the games are independent, what is the probability that each team wins the tournament?
My idea is $P(\text{$A$wins})=P[A \text{ beats } B$ and ($A \text{ beats } C$ or $A \text{ beats D}$)$]$ $=P[(A \text{ beats B and A beats C) or (A beats B and A beats D)}]$ $=0.7*0.7 + 0.7*0.7 =98$%, and repeating this process for each $B,C,D$ but this seems incorrect.
• $A$ wins with probability $0.7\cdot 0.5=0.35$. How you got $0.98$ I have no idea. Similarly, $B$ will win by first beating $A$ and then beating whichever of the remaining opponents are left, which occurs with probability... Finally, we can get the probability for $C$ and $D$ by appealing to symmetry and that it must be half of the remaining available probability. – JMoravitz Feb 20 '18 at 3:52
• @JMoravitz He said $A$ wins with probability $0.7$ against all teams. Should it not be $0.49$? – bames Feb 20 '18 at 3:55
• @JMoravitz I edited my question to show specifically how I got $0.98$, what have I incorrectly assumed to get that number? – Oliver G Feb 20 '18 at 3:58
• @bames sure would, good catch. And now that he has edited, we can see where his mistake lies. When $A$ beats $B$, he moves on to play against the other remaining player, but this is conditioned on that player having won his own game!. If you insist on keeping track of who the second opponent is for $A$, then you need a factor of $\frac{1}{2}$ in front of each. Note, it doesn't make sense for the probability that $A$ wins the tournament as a whole to be higher than the probability that $A$ wins his first game. – JMoravitz Feb 20 '18 at 3:58
• So assuming $P(A \text{ wins })=0.7 * 0.7 = 0.49$, $P(B) = 0.3 * 0.5 = 0.15$, $P(C) = 0.5 * 0.5=0.25$ and $P(D)= 0.5 * 0.5 = 0.25$ then we have the probability of the sample space is greater than $1$. What mistake did I make here? – Oliver G Feb 20 '18 at 4:06
Since $A$ has probability $0.7$ of beating all teams, it doesn't matter who wins in the $C$ vs. $D$ match. We have $$P(A\text{ wins})= P(A\text{ wins twice})=0.7\cdot0.7=0.49$$ Similarly, since $B$ has probability $0.5$ of beating both $C$ and $D$, we can write: $$P(B\text{ wins}) = P(B\text{ beats }A)P(B\text{ wins next game}) = 0.3 \cdot 0.5 = 0.15$$ By symmetry, the probability that $C$ wins is the same as the probability that $D$ wins. We have: \begin{align} P(C\text{ wins}) &= P(C\text{ beats }D)\left[P(A\text{ beats }B)P(C\text{ beats }A) + P(B\text{ beats }A)P(C\text{ beats }B)\right]\\ &=0.5(0.7\cdot0.3+0.3\cdot0.5) \\ &=0.5\cdot 0.36 \\ &= 0.18\\ &=P(D\text{ wins}) \end{align} An easier way to get $P(C\text{ wins})$ is to divide the remaining probability $1 - (0.49 + 0.15) = 0.36$ by $2$, since $C$ and $D$ have the same chance of winning the tournament.
• What is the sample space you're using here? I don't understand what you mean by "$A$ wins twice". I was assuming the sample space had two ways to win for each team as in $A$ wins by beating $B$ and $C$ or $A$ wins by beating $B$ and $D$. – Oliver G Feb 20 '18 at 4:18
• @OliverG $A$ wins the first game with probability $0.7$, and $A$ wins the second game with probability $0.7$ as well, regardless of whether $C$ won or $D$ won in the $C$ vs $D$ game because $A$ wins with probability $0.7$ against everyone. "$A$ wins twice" means that $A$ wins in both the first and second round of the tournament. – bames Feb 20 '18 at 4:19
• @OliverG It is the same sample space as yours. You simply calculated the probabilities incorrectly, as JMoravitz pointed out. The probability that $C$ wins the second game is not $0.5$, since the probability of $C$ winning against $A$ ($0.3$) is different from the probability that $C$ wins against $B$ ($0.5$)... – bames Feb 20 '18 at 4:23
• @OliverG nothing is wrong with the samplespace either person chose, but you are ignoring the chance that $A$ gets the chance to play against $C$ in the first place in your calculations. If you want, you could write it like barnes did, but it gets tedious to parse: $A$ wins tournament $=$($A$ beats $B$ in rd1) and ((($A$ beats $C$ in rd2) and ($C$ beats $D$ in rd1)) or (($A$ beats $D$ in rd2) and ($D$ beats $C$ in rd1))), yielding the calculation $0.7\cdot((0.7\cdot 0.5)+(0.7\cdot 0.5))=0.49$ – JMoravitz Feb 20 '18 at 4:26 | 2020-12-05T14:42:56 | {
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https://www.physicsforums.com/threads/calculating-entropy-for-an-adiabatic-system.930865/ | # Homework Help: Calculating entropy for an adiabatic system
Tags:
1. Nov 7, 2017
### hinjab
1. The problem statement, all variables and given/known data
1. A container of 1.5 Kg of gas is at a temperature and pressure of 293 K and 1 bar respectively. The gas is adiabatically compressed until its temperature and pressure are 450 K, 4.49 bars. Adiabatic processes are processes with no heat transfer. The properties of this gas are cv = 10.3 KJ/(Kg K) and R = 4.158 KJ/(Kg K). Neglect kinetic and potential energy terms.
2. Relevant equations
a) Use the first law to determine the work into the system.
b) Calculate the entropy production for this process.
c) Is this a reversible process?
3. The attempt at a solution
I solved A by saying that the work done = the change in internal energy. So Work done = P1V1 - P2V2 / Y-1 where Y = Cp/Cv = 1.403.
Since P1V1 = NRT1 and P2V2 = nRT2 , I calculated Work done as nRT1-nRT2 / 1.403 - 1 = -2427.79 kJ. The negative sign representing work being done on the system.
I am having trouble calculating B. My understanding would be that the change in entropy for an adiabatic process where volume is constant would be calculated as Cv ln (T2/T1), However I am finding conflicting ideas online, where people are actually calculating it as Cp ln (T2/T1) - R ln (P2/P1). When I calculate it that way, I get a negative number of -61.71 J/K for entropy production.
Can anyone explain if that is the right way to calculate it, or if I should be doing it the way I originalyl thought was correct (Entropy production = Cv ln (T2/T1)?
Thanks!
2. Nov 7, 2017
### Staff: Mentor
You calculated the work incorrectly. What is the equation for the change in internal energy of an ideal gas?
3. Nov 7, 2017
### hinjab
the change in internal energy of an ideal gas would be Δ U = n Cv Δ T . . . since there is no heat transfer for this, should I be saying that work actually equals this equation?
4. Nov 7, 2017
### hinjab
the change in internal energy of an ideal gas would be Δ U = n Cv Δ T . . . since there is no heat transfer for this, should I be saying that work actually equals this equation?
5. Nov 7, 2017
Sure
6. Nov 7, 2017
### hinjab
Okay, thanks.
My other question is in regards to part B. Can you please help me understand which equation I should be using in the two that I listed above to solve for entropy?
7. Nov 7, 2017
### Staff: Mentor
You should be using the one that the other people are using, not the one you wanted to use. The one they are using is, of course, per unit mass, so you have to multiply by the mass of gas. The reason their equation is correct is that, in addition to the temperature changing, the pressure and volume also changed. The answer should come out positive, so maybe you made a mistake in arithmetic. Please show your work.
Chet
8. Nov 7, 2017
### hinjab
Hi Chet,
Here is my math:
Cp - Cv = R
Cp = R + Cv
Cp = 4.158 KJ/Kg*K + 10.3 KJ/Kg*k = 14.458 KJ/Kg*k
Entropy production calculation:
Delta S = M*Cp*ln(T2/T1) - M*R*ln (P2/P1)
Delta S = (1.5 Kg)*(14.458 KJ/Kg*K)*ln(450K/293K) - (1.5 Kg)*(4.158 KJ/Kg*K)*ln(449000 pascals/100000 pascals)
Delta S = 9.305 - 9.948 KJ/K
Delta S = -0.643 KJ/K
9. Nov 7, 2017
### Staff: Mentor
For the 2nd term, I get 9.367, not 9.948. So, for the change in entropy, I get $\Delta S=-0.0617$. This is pretty close to zero, so I would call it reversible.
That being the case, the result you got for the work using the equation that applies only for an adiabatic reversible change (i.e., in your original post) should come very close to matching the result you got using the relationship for the internal energy change. Does it?
10. Nov 7, 2017
### hinjab
Hi Chet,
Sorry about that. I did mess up the arithmetic! I also got the same answer as you for the second term. I didn't know that it was OK to consider an entropy change that is very close to zero as zero, thus making it reversible. Thank you for the info.
When I calculated the work done by using the internal energy change (the way you suggested) I get 2425.65 KJ, as opposed to the 2429.79 KJ in my original attempt.
11. Nov 7, 2017
### Staff: Mentor
Looks good. Nice job.
12. Nov 7, 2017
### hinjab
Thank you so much for your help! | 2018-10-16T18:15:39 | {
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https://math.stackexchange.com/questions/3178103/integral-question-int-x-sqrt-1-x-dx | Integral question $\int x \sqrt {1-x}\ dx.$
So there are (from my knowledge) $$2$$ ways of solving for $$\int x \sqrt {1-x}\ dx.$$
The first is by $$u$$ substitution and the second is by parts. They both differ now i'm confused which one is correct? I have both the correct answers but they differ?
• Differentiate both results – lab bhattacharjee Apr 7 at 11:27
• Both are correct, of course. You should edit your question, showing us your work so far. – José Carlos Santos Apr 7 at 11:29
• I got this here $$-\frac{2}{15} (1-x)^{3/2} (3 x+2)+C$$ – Dr. Sonnhard Graubner Apr 7 at 11:31
• Subtract one result from the other. Simplify as much as possible. Your answer should be a constant. – Bernard Massé Apr 7 at 11:36
Let me guess: you substituted $$u=1-x$$ to get $$\int(u^{3/2}-u^{1/2})du=\frac{2}{5}(1-x)^{5/2}-\frac{2}{3}(1-x)^{3/2}+C_1$$, but for parts you wrote $$f=x,\,g=-\frac{2}{3}(1-x)^{3/2}$$ to get $$-\frac{2}{3}x(1-x)^{3/2}+\frac{2}{3}\int(1-x)^{3/2}dx=-\frac{2}{3}x(1-x)^{3/2}-\frac{4}{15}(1-x)^{5/2}+C_2.$$But from $$x(1-x)^{3/2}=(1-x)^{3/2}-(1-x)^{5/2}$$, these results are identical (because $$\frac{2}{3}-\frac{4}{15}=\frac{2}{5}$$) with $$C_1=C_2$$. (Sometimes when you calculate an indefinite integral by multiple methods, the integration constants differ, but that's fine.)
• Thank you, this is exactly how i worked it out. I did not realize that i had to get rid of the x in 2/3x(1−x)^3/2 by using u substitution. That was pretty tricky. – Shaun Weinberg Apr 7 at 12:16
You should find that you can reduce it to a single constant (which may be zero or non-zero). This is expected. The indefinite integral is always expressed with a "$$+c$$" at the end of the answer - an arbitrary constant of integration. There is no one answer to an indefinite integral, the answer is an infinite family of expressions, all separated from each other by every conceivable constant.
$$x\sqrt{1-x}=(x-1+1)\sqrt{1-x}=-(1-x)\sqrt{1-x}-\sqrt{1-x}$$ $$=-(1-x)^{\frac{3}{2}}-(1-x)^{\frac{1}{2}}$$ The anti derivative of $$(1-x)^a$$ is $$-\frac{1}{a+1}(1-x)^{a+1}, a\ne -1$$ | 2019-06-24T08:59:26 | {
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https://math.stackexchange.com/questions/2442016/convergence-of-complex-series-sum-n-1-infty-fracinn | # Convergence of complex series $\sum_{n=1}^{\infty}\frac{i^n}{n}$
Prove that the series $\displaystyle \sum_{n=1}^{\infty}\frac{i^n}{n}$ converges.
Optional. find it's sum, if possible.
Comments. I am aware of the general result about the convergence (not absolute) of $\displaystyle \sum_{n=1}^{\infty}\frac{z^n}{n}$, for $z\neq 1$. But i feel that the answer to the above problem can be more trivial, although have not made any interesting approach so far.
• Look at the even $n$ terms, which would sum up as the real part, and then look at the odd $n$ terms, which would sum up as the imaginary part. Show that both real part and imaginary part converge. – edm Sep 23 '17 at 17:36
We have $\frac{1}{1-z}=\sum_{n\geq 0}z^n$ for any $z\in\mathbb{C}$ such that $|z|<1$. The convergence is uniform over any compact subset of $\{z\in\mathbb{C}:|z|<1\}$, hence we are allowed to state
$$\int_{0}^{i}\frac{dz}{1-z} = \sum_{n\geq 0}\frac{i^{n+1}}{n+1} = \sum_{n\geq 1}\frac{i^n}{n}$$ where the LHS equals $$-\log(1-i) = -\log\left(\sqrt{2}\, e^{-\frac{\pi i}{4}}\right) = \color{red}{-\frac{\log 2}{2}+\frac{\pi i}{4}}.$$ The convergence of the original series is granted by Dirichlet's test, as already remarked by other users. You may also notice that the real part is given by the terms of the original series with $n\in 2\mathbb{N}$ and the imaginary part is given by the terms of the original series with $n\in 2\mathbb{N}+1$. On the other hand, $$\sum_{n\geq 1}\frac{(-1)^n}{n}=-\log(2),\qquad \sum_{n\geq 0}\frac{(-1)^n}{2n+1}=\frac{\pi}{4}$$ are pretty well-known. They can be proved with the same approach, restricted to the real line only.
• Sorry for my question but where is the implication that this can also be used for $z=i$, which has $|i| = 1$? Is it because of Dirichlet's test? – MrYouMath Sep 23 '17 at 17:46
• @MrYouMath: the first equality holds for any $z$ close enough to the origin and it can be termwise-integrated due to uniform convergence. The norm of $i$ is $1$. – Jack D'Aurizio Sep 23 '17 at 17:48
• The first equality is clear. I mean the implication that it is also a valid representation for $z=i$ which is not $|z|<1$. – MrYouMath Sep 23 '17 at 17:52
• @MrYouMath: it is not, $\sum_{n\geq 1}i^n$ is not convergent. But I do not use an evaluation of $\sum_{n\geq 0}z^n$ at $z=i$, I use an evaluation of the primitive $\sum_{n\geq 0}z^{n+1}/(n+1)$ at $z=i$, and I am allowed to do it by the given reasons. – Jack D'Aurizio Sep 23 '17 at 18:01
HINT: Notice that $$\sum_{n=1}^\infty \frac{i^n}{n}=\sum_{n=1}^\infty \frac{(-1)^n}{2n}-i\sum_{n=1}^\infty \frac{(-1)^n}{2n-1}$$
• I don't think that rearranging is possible for the infinite sum as it is not absolutely convergent. But in essence, this should lead to the solution. – MrYouMath Sep 23 '17 at 17:40
• @MrYouMath The rearrangement is valid here, you merely need to check the partial sums of each side and make sure they can match up. – Simply Beautiful Art Sep 23 '17 at 17:59
• @SimplyBeautifulArt: I know it is possible in this case but it would be better to be more precise to explicitly state that. The answer, as it is now, is not good without further explanation. Others might think that you can always rearrange infinite sums as you like. – MrYouMath Sep 23 '17 at 18:01
Hint: Look at ther partial sums is equal to:
$$\frac{i}{1}-\frac{1}{2}-\frac{i}{3}+\frac{1}{4}+\ldots \frac{i^n}{n}$$
Factor the complex part and the imaginary part. Notice that you have alternating series in the real and imaginary part. Both series converge by the Leibniz criterion. Additionally compare the real and imaginary part with the Taylor series of $\arctan(x)$ and $\ln(1+x)$.
• Excellent comment, regarding the non-absolute convergence and the rearrangement of the given series. – Nikolaos Skout Sep 23 '17 at 21:23
• @NikolaosSkout Thank you. Seems like we are the only users worried about rearrangement :D. – MrYouMath Sep 23 '17 at 21:28
• You are welcome. It is just that sometimes things work...! – Nikolaos Skout Sep 23 '17 at 21:50
You can use Dirichlet's test. Of course if $|z|<1$ then $$\sum_{n=1}^\infty\frac{z^n}n=\ln\frac1{1-z}$$ (for a suitable branch of the complex logarithm).
$$\sum_{n=1}\dfrac{i^n}{n}=\sum_{n=1}\dfrac{\cos\frac{n\pi}{2}}{n} + i\sum_{n=1}\dfrac{\sin\frac{n\pi}{2}}{n}$$ both converge with Dirichlet!
Take a look at the following picture :
and see the connection with the issue in the so-called Argand plane (knowing that what is represented is a partial sum of the series).
It remains to prove that this "inward spiraling" movement terminates as a limit point... therefore, I don't say that this graphics constitue a "proof without words" ; it just illustrates the plausibility of a convergence... | 2019-10-14T09:54:24 | {
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